Structural Mechanics NoRestriction

Structural Mechanics

Bao ShihuaGong Yaoqing

WUPTWuhan University of Technology Press

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附注

第二作者龚耀清教授对本书的贡献参见Preface第二页的下划线部分或书后语中的下划线部分。

本书简介

本书是根据中国高等学校土木工程专业的教学计划、结构力学课程的教学大纲和课程基本要求，

为中国高等学校土木工程专业结构力学课程‘双语教学’编写的英文结构力学教材。全书共分 14 章，包括：绪论，结构的几何组成分析，静定梁，静定刚架，三铰拱，静定桁架和组合结构，静定结构

总论，影响线，虚功原理和结构的位移，力法，位移法，渐进法和超静定结构的影响线，矩阵位移

法和超静定结构总论。除第一章外，每章均有提要、小结、思考题和习题，书后附有答案。本书与第一作者主编的中文结构力学（参考文献 1）在章节次序和内容上基本上是对应的，便

于中文与英文教学中交互使用。本书可作为土木工程专业，即‘大土木’的房建、路桥、水利等各类专门化方向的‘双语’教

材及参考用书，也可供工程技术人员学习专业外语之用。

Brief Introduction of the Book

The textbook, written in terms of the national education program, syllabus and requirement of the course—structural Mechanics—for the undergraduates majoring civil engineering in China, is intend to be a sort of bilingually teaching material. It includes 14 chapters, which are: introduction, geometric construction analysis of structures, statically determinate beams, plane statically determinate rigid frames, three hinged arches, plane statically determinate trusses and composite structures, general remarks on statically determinate structures, influence lines, principle of virtual work and displacement of structures, force method, displacement method, method of successive approximations, matrix displacement method, and general remarks on statically indeterminate structures. All chapters except for chapter 1 are arranged by abstract, text, summary, problems for reflecting, and problems for solution; the answers to selected problems are attached to the back as well.

It should be pointed out that the arrangement of the content and the order of the chapters for the book is almost the same as its Chinese counterpart (reference book 1), so as to give an alternant usage for Chinese readers and teachers.

The book can be not only used as a textbook and/or reference book for the students majoring civil engineering structures, but also employed as a specialized English book for the engineer and technicians interested in civil engineering and English.

Preface

Background

When time reaches to the 21st century, strong pressures developed from many ranks to write a textbook of Structural Mechanics in English for Chinese undergraduates majoring in civil engineering specialty. The naissance of the book is the balance of the pressures and the request of WUTP, the Wuhan University of Technology Press, a press in china.

In recent years, the manner of teaching technical courses in English or partially in English is strongly promoted by Chinese Ministry of Education. In the circumstance, the course of Structural Mechanics has been taught in English or partially in English in Civil Engineering Department of Tsinghua University since middle 80th of last century. Here, ‘taught in English’ means that English language is employed in all teaching procedure, such as in class, textbooks and the like; whereas ‘taught partially in English indicates that only Chinese language is used in class and English language are adopted in other teaching process, such as textbook, students’ homework and so on. The two teaching ways are strongly dependent on the English oral level of the teachers. Nowadays, undergraduates learning Structural Mechanics in Civil Engineering Department of Tsinghua University were grouped into two classes, an English class and a Chinese class. In the English class, class language is English, that is, the teacher must use English to prelect and the students must read English textbooks and use English do their home works; while in the Chinese class, class language is Chinese. The two teaching ways have been normalized into teaching plan of Tsinghua University.

The first author of the book, Professor Bao Shihua, has been taught Structural Mechanics partially in English in Tsinghua University since middle 80th of last century. The second author of the book, professor Gong Yaoqing, one of PHD students of the first author, has been taught Structural Mechanics partially in English in Ningxia University and Henan Polytechnic University for many years. The first draft of the book has also been used in Henan Polytechnic University. Apparently, the didactical experience has provided a bed for the appearance of the book.

In China, every specialty in a university has its own teaching plan and every course has its own teaching program and demand depend on different specialties. So does Structural Mechanics. The course has formed its own teaching system after nearly 50 years’ didactical experience. Nowadays, the teaching system is quite different from that of America or the Great British because the content and system of the textbooks of Structural Mechanics coming from the two countries do not have unified syllabus and requirement. So it is hard for the original English textbook of Structural Mechanics to meet Chinese

syllabus and requirement. In addition, in the universities of China the teaching plan and program of Structural Mechanics must be unchangeable no matter how the course is taught, in English or in Chinese. The tenet has been obeyed for many years by the teachers who teach Structural Mechanics in Tsinghua University. Based on the tenet, the book is written by means of teaching plan and program of Structural Mechanics used in civil engineering specialty of Chinese universities. In other words, the content and system of the Structural Mechanics written in English are identical to those of the Structural Mechanics written in Chinese. The original intention of the consistence is to facilitate Chinese readers and teachers.

Organization and approach

Since there are too many contents pertinent to Structural Mechanics, the contents are divided, by Chinese teaching program and demand, into two portions, fundamental portion that will be a required course for the common students majoring civil engineering specialty and advanced portion that is planed to be a selected course for the undergraduates or graduate students who have some advanced requirement for structural analysis. The contents of the fundamental portion are organized by the book, named Structural Mechanics including the analysis of statically determinate and indeterminate structures, matrix structural analysis and so on; whereas the contents of the advanced portion are composed by another book, known as Advanced Topics of Structural Mechanics or Advanced Structural Mechanics comprising the structural dynamics, stabilities of structures, plastic analysis and the like. It is actually the sister book of Structural Mechanics.

The compiling outline, content and requirement are designed by the first author, Professor Bao Shihua of Tsinghua University. The initial draft of the book including English composition, figure design and so on is completed by the second author, professor Gong Yaoqing of Henan Polytechnic University. The final manuscript of the book is also revised and checked by the first author.

Each chapter of the book beginning with an abstract introducing its objective, proceeding with text presenting its contents, ending with a summary outlining its salient features, and providing with problems for both reflecting and solution.

The nature of the book is consistent with that of the book written in Chinese.

Contents

This book consists of 14 chapters. Chapter 1 provides a brief introduction of the various types of structural forms and loads. The structurology (or geometric construction) of framed structures is discussed in chapter 2. The analysis of statically determinate structures is covered in the next 5 chapters. Chapter 3 through 6 discuss the analysis of statically determinate beams, rigid frames, three hinged arches and plane trusses and composite structures, respectively. Chapter 7 presents general remarks on statically determinate structures so as to enhance and deepen the comprehension about the types of structures and their analytical

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下划线

methods. In chapter 8 influence lines for beams, girders and trusses are discussed by static and mechanical methods, respectively. Chapter 9 covers the principle of virtual work and evaluation of the displacements of statically determinate structures.

In chapter 10 through 12, the analysis of statically indeterminate structures is discussed by the force method (chapter 10), displacement method (chapter 11) and successive approximation method (chapter 12), respectively.

Chapter 13 presents an analytical method by computer program, matrix displacement method. In the chapter, a computer program for the analysis of plane framed structures is also attached.

Chapter 14 makes a general discussion about the analytical methods, behavioral characteristics and computing models of statically indeterminate structures.

Acknowledgements

The development of this book was strongly influenced by the authors’ colleagues, their students and numerous books published here and abroad. We would like to hereby acknowledge all of their valuable suggestions, comments, help and English composition. In particular we would like to thank the second author’s students, Gong Yun and He Shufeng, who prepared the most of the figures on the book.

The authors will be greatly indebted to the readers who dig out the errors slipped the authors’ notice in the book because the book is written in a relatively short time, and not in the authors’ mother language.

Bao Shihua Tsinghua University

Gong Yaoqing ([email protected]) Henan Polytechnic University

June, 2005

作者简介

清华大学土木工程系教授，中国力学学会《工程力

学》编委，中国建筑学会高层建筑结构委员会委员。

1985~1986 年美国伊利诺大学（University of Illinois）土木工程系访问学者，1991~1993 年为香

港理工大学土木与结构系研究员。长期从事高层建筑结构、结构力学、弹性

力学、能量原理及有限元、板壳结构和薄壁杆件结构等领域的教学和研究工

作。

，包世华

出版教材和专著 25 本。教材有《高层建筑结构设计》、《结构力学》、《结

构力学教程》等，分别于 1987 年获建设部优秀教材二等奖，1988 年、1992 年获国家教委国家优秀

教材奖，1988 年度获教育部科学技术进步奖一等奖，1999 年度获国家科学技术进步奖二等奖，2002年获全国普通高等学校优秀教材一等奖。专著有《薄壁杆件结构力学》、《高层建筑结构计算》、《新

编高层建筑结构》等。

在国内外发表学术论文 110 多篇。其中，壳体研究成果被收入国家行业标准《钢筋混凝土薄壳

结构设计规程》。提出和创建高层建筑结构解析和半解析常微分方程求解器法系列。科研成果 1993年获北京市科委技术成果奖，1986 年、1992 年、1994 年分别获国家教委科学技术进步奖一、二、

三等奖。 Bao Shihua, professor in Civil Engineering Department of Tsinghua University, is one of consulting

editors of Engineering Mechanics, an academic periodical superintended by the China Science Association (CSA) and the Chinese Society of Theoretical and Applied Mechanics (CSTAM); one of commissioners in the Tall Building Committee of the Chinese Society of Architecture. He was a visiting Scholar in Civil Engineering Department of University of Illinois during 1985~1986, research fellow in Civil Engineering and Structure Department of Hong Kong Polytechnic University during 1991~1993. He is engaged in the education and academic research for the fields of tall building structures, structural mechanics, elastic theory, energy theorem and finite element method, plate and shell structures, thin walled member structures and so on.

He has been written totally 25 books involving textbooks and monographs. His textbooks written in Chinese are: Design of Tall Building Structures, Structural Mechanics and so on, which were born off the palms of second-class prize awarded by China Construction Ministry in 1987, advanced textbook prize awarded by China Education Committee in 1988 and 1992, first-class prize of technology advancement awarded by China Education Ministry in 1988, second-class prize of national technology advancement awarded by China Science And Technology Ministry in 1999, and first-class advanced textbook awarded

本书简介

本书是根据中国高等学校土木工程专业的教学计划、结构力学课程的教学大纲和课程基本要求，

为中国高等学校土木工程专业结构力学课程‘双语教学’编写的英文结构力学教材。全书共分 14 章，包括：绪论，结构的几何组成分析，静定梁，静定刚架，三铰拱，静定桁架和组合结构，静定结构

总论，影响线，虚功原理和结构的位移，力法，位移法，渐进法和超静定结构的影响线，矩阵位移

法和超静定结构总论。除第一章外，每章均有提要、小结、思考题和习题，书后附有答案。本书与第一作者主编的中文结构力学（参考文献 1）在章节次序和内容上基本上是对应的，便

于中文与英文教学中交互使用。本书可作为土木工程专业，即‘大土木’的房建、路桥、水利等各类专门化方向的‘双语’教

材及参考用书，也可供工程技术人员学习专业外语之用。

Brief Introduction of the Book

The textbook, written in terms of the national education program, syllabus and requirement of the course—structural Mechanics—for the undergraduates majoring civil engineering in China, is intend to be a sort of bilingually teaching material. It includes 14 chapters, which are: introduction, geometric construction analysis of structures, statically determinate beams, plane statically determinate rigid frames, three hinged arches, plane statically determinate trusses and composite structures, general remarks on statically determinate structures, influence lines, principle of virtual work and displacement of structures, force method, displacement method, method of successive approximations, matrix displacement method, and general remarks on statically indeterminate structures. All chapters except for chapter 1 are arranged by abstract, text, summary, problems for reflecting, and problems for solution; the answers to selected problems are attached to the back as well.

It should be pointed out that the arrangement of the content and the order of the chapters for the book is almost the same as its Chinese counterpart (reference book 1), so as to give an alternant usage for Chinese readers and teachers.

The book can be not only used as a textbook and/or reference book for the students majoring civil engineering structures, but also employed as a specialized English book for the engineer and technicians interested in civil engineering and English.

by China Education Ministry in 2002, respectively. His monographs written in Chinese are: Structural Mechanics for Thin Walled Member Structures, Analysis of Tall Building Structures, Newly Edited Tall Building Structures and so forth.

His published papers both in his mother language and foreign language exceed 110 pieces, of which papers pertinent to shell structures are adopted by national profession code—Design Code of Reinforce Concrete Thin Shell Structures. His pioneering work is analytical and semi-analytical methods by Ordinary Differential Equation Solver used for analyzing tall building structures. His research findings were brought down the persimmon, and awarded for technology advancement prize by Beijing government in 1993, for technology advancement fist-class prize, second-class prize and third-class prize by China Education Committee in 1986, 1992 and 1994, respectively.

作者简介

1999 年清华大学土木工程系毕业并获结构工程

专业工学博士学位，师从我国著名土木工程专家

龙驭球院士和我国著名高层建筑专家包世华教

授。现为河南理工大学特聘教授，研究生导师; 国家一级学会中国力学学会主办的中文核心期刊《工程力学》编委会委员; 国家自然科学基金委员会工程与材料科学部项目评审专家。1994 年被中国

力学学会评为优秀力学教师。长期从事《结构力学》的教学工作，近年来

主要从事超高层建筑结构与大型桥跨结构的分析与研究，发表学术论文 30多篇，其中部分已被 EI 收录；出版了首部用半解析方法分析超高层建筑结构的专著《弹性地基上高

层建筑结构及半解析法研究》；先后参加和主持国家级、省部级科研基金项 9 项，研究成果已得到国

内外同行的认可和关注。目前正在英国爱丁堡大学 (Edinburgh University) 作高级研究者。

，龚耀清

Gong Yaoqing, graduated and obtained doctor’s degree of structural engineering in Civil Engineering Department of Tsinghua University, is a student of Academican Long Yuqiu who a well known expert in civil engineering and Professor Bao Shihua who is a famous expert in tall building structures, a specially engaged professor and supervisor of postgraduate students in Henan Polytechnic University, one of editors of Engineering Mechanics, an academic periodical superintended by the China Science Association (CSA) and the Chinese Society of Theoretical and Applied Mechanics (CSTAM), one of experts examining and commenting proposal projects in Department of Engineering & Materials Science of National Natural Science Foundation of China. He got an honor of excellent teacher from Chinese Society of Theoretical and Applied Mechanics in 1994.

He has been engaged in the education and academic research for structural mechanics, super tall building and very long-span bridge structures for many years. His published papers have been more than 30 pieces, some of which are accepted by EI Index. In 2003, he published a monograph—Tall Building Structures on Elastic Subgrade and Research of Semi-Analytical Method (in Chinese), which is a first monograph pertinent to the semi-analytical analysis about tall and super tall building structures. He has finished 9 researching projects involving with those helped by National Natural Science Foundation of China and local government. His research findings have been paid attention to and interested in by pertinent fellows. He has been doing some research work in Edinburgh University of United Kingdom.

benq

附注

第二作者龚耀清教授对本书的贡献参见Preface第二页的下划线部分或书后语中的下划线部分。

Contents

Chapter 1 Introduction 1

1.1 Structures and Their Classification 1 1.2 Objective and Learning Method of Structural Mechanics 3 1.3 Analytical Models of Structures 5 1.4 Classification of Framed Structures 13 1.5 The Classification of Loads 17

Summary 18 Problems for Reflecting 18

Chapter 2 Geometric Construction Analysis of Structures 19

2.1 Purpose of Analyzing Geometric Construction of Structures, Stable and Unstable Structural Systems 19

2.2 The Concept of Degrees of Freedom and Restraints 20 2.3 Geometric Construction Rules of Planar Stable Framed Systems Without Redundant

Restraints 25 2.4 Illustration of Geometric Construction Analysis 31 2.5 The Relationship between Static Determinacy and Geometric Construction of Structures

32 Summary 33 Problems for Reflecting 34 Problems for Solution 34

Chapter 3 Statically Determinate Beams 38

3.1 The Analysis of Single Span Beams 38 3.2 Construction of Bending Moment Diagram By Principle of Superposition for Straight

Members 52 3.3 The Analysis of Simply Supported Inclined Beams 55 3.4 The Restraint Force Calculation and Geometric Construction of Statically Determinate

Multispan Beams 59

i

ii Contents

3.5 The Construction of Internal Force Diagrams of Statically Determinate Multispan Beams 62 Summary 66 Problems for Reflecting 67 Problems for Solution 68

Chapter 4 Plane Statically Determinate Rigid Frames 73

4.1 Geometric Construction and Characteristics of Plane Statically Determinate Rigid Frames 74

4.2 The Analyzing of Reactions for Statically Determinate Rigid Frames 76 4.3 Determination of Internal Forces of Member Ends by Using The Method of Sections

79 4.4 The Construction of Internal Force Diagrams of Statically Determinate Frames 84 4.5 Internal Force Diagrams of Statically Determinate Three-hinged Frames, Multispan and

Multistory Frames 87 Summary 92 Problems for Reflecting 94 Problems for Solution 95

Chapter 5 Three Hinged Arches 101

5.1 The Constitution and Type of Three Hinged Arches 101 5.2 The Reactions of Three Hinged Arches Under The Action of Vertical Loads 102 5.3 The Formula for Calculating Internal Forces of Three Hinged Arches Under Action of

Vertical Loads 105 5.4 Stressing Performance of Three Hinged Arches 110 5.5 Rational Axial Lines of Three Hinged Arches 111

Summary 116 Problems for Reflecting 117 Problems for Solution 118

Chapter 6 Plane Statically Determinate Trusses and Composite Structures 120

6.1 Characteristics and Classification of Trusses 120

Contents iii

6.2 The Method of Joints 126 6.3 The Method of Sections 132 6.4 The Combination of The Method of Joints and The Method of Sections 136 6.5 Form and Stressing Characteristics of Girder Trusses 138 6.6 Composite Structures 142


Chapter 7 General Remarks on Statically Determinate Structures 155

7.1 Analytical Methods For Statically Determinate Structures 155 7.2 General Property of Statically Determinate Structures 160 7.3 Stressing Characteristics of Various Types of Structures 164

Problems for Reflecting 167

Chapter 8 Influence Lines 169

8.1 Concept of influence lines 169 8.2 Influence Lines for Statically Determinate Single Span Beams by Static Method 172 8.3 Influence Lines for Girders with Floor Systems 180 8.4 Influence Lines for Trusses by Using Static Method 184 8.5 Virtual Displacement Method for Constructing Influence Lines for Statically Determinate

Beams 187 8.6 Application Of Influence Lines 197


Chapter 9 Principle of Virtual Work and Displacement of Structures 220

9.1 Introduction for Calculation of Structural Displacement 220 9.2 Virtual Work and Principle of Virtual Work 222 9.3 General Equation and Unit Load Method for Computing Displacements 235

iv Contents

9.4 Calculation of Displacements Caused by Loads 236 9.5 Graph-Multiplication Method 245 9.6 Calculation of Displacements Caused by Temperature Changes 260 9.7 Calculation of Displacements Caused by Support Movement 263 9.8 Reciprocal Laws for Linear Elastic System 266


Chapter 10 Force Method 280

10.1 Statically Indeterminate Structures and Degrees of Indeterminacy 280 10.2 Basic Concept of Force Method 283 10.3 Canonical Equations of Force Method 287 10.4 Statically Indeterminate Beams, Rigid Frames and Bent Frames 292 10.5 Statically Indeterminate Trusses and Composite Structures 304 10.6 Analysis of Symmetric Structures 310 10.7 Statically Indeterminate Arches 326 10.8 Internal Forces due to Support Settlements and Temperature Changes 340 10.9 Computation of Displacements of Statically Indeterminate Structures 347 10.10 Verification of Calculation of Statically Indeterminate Structures 352


Chapter 11 Displacement Method 370

11.1 Fundamental Concepts of Displacement Method 370 11.2 Member End Forces of Various Single-Span Indeterminate Prismatic Beams Due to Their

End Displacements and External Loadings 374 11.3 Primary Unknowns and Primary Systems in Displacement Method 382 11.4 Displacement-Method Equations 387 11.5 Analysis of Statically Indeterminate Beams and Rigid Frames with No Sidesway 393 11.6 Analysis of Statically Indeterminate Rigid and Bent Frames with Sidesway 401 11.7 Analysis of Statically Indeterminate Symmetric Structures 411

Contents v

11.8 Development of Displacement-Method Equations by Direct Stiffness Method 416 Summary 421 Problems for Reflecting 423 Problems for Solution 425

Chapter 12 Method of Successive Approximations and Influence Lines for Indeterminate structures 431

12.1 General Remarks for Method of Successive Approximations 431 12.2 Concepts and Terminology in Moment Distribution Method 432 12.3 Moment Distribution at A Single Joint—Moment-Distribution Process 439 12.4 Moment Distribution at Multi-Joints—Successive Approximations 446 12.5 No-Shear Moment Distribution Method 461 12.6 Shear Distribution Method 469 12.7 Influence Lines for Forces of Statically Indeterminate Structures 476 12.8 Most Unfavorable Arrangement of Live Loads and Envelops for Internal Forces of

Continuous Beams 483 Summary 488 Problems for Reflecting 489 Problems for Solution 490

Chapter 13 Matrix Displacement Method 495

13.1 Introduction 495 13.2 Member Stiffness Matrix in Local Coordinate System 500 13.3 Member Stiffness Matrix in Global Coordinate System 507 13.4 Global Stiffness Matrixes of Continuous Beams 513 13.5 Global Stiffness Matrixes of Rigid Frames 525 13.6 Equivalent Nodal Loads 531 13.7 Procedure for Analysis and Examples 538 13.8 Global Analysis of Rigid Frames without Considering Axial Deformation 549 13.9 Block Diagram and Computer Programs for Plane Framed Structures 556


vi Contents

Chapter 14 General Remarks on Statically Indeterminate Structures 588

14.1 Classification and Comparison of Fundamental Method used to Analyze Statically Indeterminate Structures 588

14.2 Combined Method—Simultaneously Applying Moment Distribution and Displacement Methods to Frames with Sidesway 591

14.3 Approximate Analysis of Statically Indeterminate Structures 596 14.4 Properties of Statically Indeterminate Structures 604 14.5 Complementary Discussion about Computing Models of Structures 611


Answers to selected problems Ⅰ

Bibliography

CHAPTER 1 INTRODUCTION

The abstract of the chapter The chapter will introduce four questions such as the objective and learning method of structural

mechanics, the analytical models of structures, the classification of framed structures and their loads. Among the four questions, the analytical models are the most significant since they will lay the foundations of the other chapters of the book.

1.1 Structures and Their Classification

In civil engineering project, the generic term an engineering structure or briefly named a structure is referred to as a frame or skeleton used to carry loads applied on it and composed by members of buildings or other constructions made of construction materials. The following figures show some photographs of engineering structures. Fig.1.1 is a tall building suspended structure, Fig.1.2 is a bridge structure, Fig.1.3 is

the structure of a hydraulic power station, Fig.1.4 is the structure of an industrial premises. Speak in detail, the roof panel, the roof truss, the beams, the columns, the foundation and their combination of the

Fig. 1.2 Bridge structure

Fig. 1.3 Hydraulic power station

Fig. 1.4 Industrial premisesFig.1.1 Tall building suspended structure

1

2 Chapter 1 Introduction

one-storey workshop of a plant shown in Figure 1.18 are all structures. Structures can be classified into three categories by their geometric characteristics. (1) Framed structure

A framed structure is comprised of members whose cross-sectional dimension (e.g. the width b and depth h of a rectangular cross section, the radius of a circular cross section, etc.) is much smaller than the length l as shown in Fig. 1.5. The most commonly used types of structures

tures which

b

h

lb

h

Fig.1.5 Bar

in the structural engineering are framed strucwill be the main attention the book focuses on.

(2) Plate, Slab and Shell (or thin-walled structure) When the thickness of the structure is very small in

comparison with its other two dimensions (length and width), the structure is referred to as a thin-walled structure. The geometric characteristic of a thin-walled

(3) Massive Structure ssive structure have the same order of

mag ple, the retaining wall shown in Fig. 1.9 and the dam shown in Fig. 1.10 are two of projects applying massive structures.

l

Fig.1.6 Plate

structure is that its thickness h is much smaller than its length l and width b. The plate shown in Fig. 1.6 is one of instances of thin-walled structures. The combination of finite number of plates would develop a floded plate shown as in Fig. 1.7 (a). Figure 1.7 (b) shows a building with a roof structure composed of floded plates. If a structure has a curved middle surface, it is called a shell, as shown in Fig. 1.8. In that context, a plate or a slab can be considered as a thin-walled structure with a plane middle surface.

The three dimensions (length l, width b and depth h) of a ma

(b) floded plate roof

Fig.1.7 Floded plate structure

(a) floded plate

nitude. Consider, for exam

1.2 Objective and Learning Method of Structural Mechanics 3

(a) shell (b) shell roof

Fig.1.8 Shell structure

Fig.1.9 Retaining wall Fig.1.10 Dam

1.2 Objective and Learning Method of Structural Mechanics

1.2.1 The relationship between structural mechanics and other curricula

In structural mechanics, the primary focus will be on the analysis of. In this context, you can name the course as Mechanics of Framed Structure. For simplicity, Structural Mechanics is adopted.

Structural Mechanics belongs to one of technically fundamental courses, and plays a very important role and a connecting link between the preceding courses and the following (or subsequent) courses learned by the undergraduates majoring in the specialty pertinent to civil engineering.

Structural Mechanics is the following course of Theoretical Mechanics and Strength of Materials. The objective of Theoretical Mechanics is the investigation of essential rules and analysis of mechanical motion (including static state and equilibrium) of rigid bodies. The attention paid by Strength of Materials is the strength, stiffness and stability of a single member (or a bar). While the contents treated of in Structural Mechanics are the strength, stiffness and stability of framed structures, which are composed of many members. Therefore, Theoretical Mechanics and Strength of Materials would provide primary principles and basement of mechanical analysis for the studying of Structural Mechanics.


Structural Mechanics is meanwhile the preceding course of Theory of Elasticity (focusing on the strength, stiffness and stability of plate, shell and massive structures), Reinforced Concrete Design, Masonry Structure, Steel Structure and other specialized curricula associated with building construction, structural engineering, highway engineering, bridge engineering, water conservancy engineering and underground engineering. By this token, Structural Mechanics will provide basic mechanical knowledge for the studying of the subsequent courses and play a very important role in the specialty pertinent to civil engineering.

1.2.2 Objective and learning method of Structural Mechanics

The objective of Structural Mechanics comprises following aspects: （1） Discuss principles for constructing structures, rational configurations of structures and

selections of computing models for analyzing structures; （2） Investigate the methods for analyzing internal forces and displacements of structures so as to

check their strength and stiffness; （3） Study structural stability and structural response induced by dynamic loading. The analyzing problems involved in Structural Mechanics can be classified into two categories: the

first category is statically determinate problems, which could be solved by means of force equilibrium conditions, the first fundamental condition listed in next paragraph; the second category is statically indeterminate problems, which could be figured out only by satisfying all of the following three types of fundamental conditions. The three types of fundamental conditions are:

（1） Force equilibrium conditions The entire structure or part of it must be balanced under the action of a system of forces.

（2） Compatibility conditions or geometrical conditions of displacements The continuity of a structure must be maintained after the structure has deformed under the action of the loadings applied on it. That is, there are no overlap and gap existing in the materials composing the structure, and meanwhile the deformation and displacement of the structure should satisfy the restraint conditions provided by the supports and the connecting joints.

（3） Physical conditions This is the physical performance condition linking stress and strain or forces and displacements yielding in a structure, i.e. physical or constitutive equation.

The all analyzing methods formulated in Structural Mechanics are permeated by aforementioned three types of fundamental conditions, i.e. every analyzing method discussed in the course has to utilize the three conditions in some kind of degree and order.

During studying the course, we should focus our attention on the relationship between Structural Mechanics and other curricula. A necessary review about the knowledge associated with Theoretical Mechanics and Strength of Materials should be taken and the knowledge also should be consolidated and advanced during the study.

1.3 Analytical Models of Structures 5

In the process of studying the course, we have to pay attention to the leaning method and clue of solving problems. All of the analyzing methods discussed in the book are the concrete application of above mentioned three types of fundamental conditions. How the three types of conditions are applied to the computing process of every analyzing method is the point we should focus on. During studying the course, the clues of solving problems must be mastered; especially the general method of analyzing problems should be holden. For instances, the method of how passing through unknown field to known field; the method of how partitioning the entire structure into elements then integrating them into the very beginning structure, the method of how contrasting pertinent problems between which some sort of relationship exist in some extent, etc.

Study should associate with practice. Doing exercises is a very important approach for learning Structural Mechanics. We will not predominate the primary concept, principle and method of Structural Mechanics without doing some amount of exercises. However, doing exercises should avoid silliness. The number one silliness is only doing exercises without reading and review the book. The number two silliness is doing exercises fast and more without understanding them completely. The number three silliness is doing exercises with referring to the answers but being not able to verify the exercises by self. The number four silliness is doing exercises without rectifying fault happened in doing exercises and extracting lesson from them.

1.3 Analytical Models of Structures

their real practical states. a tails, as analyzing models

priore is referred to

as it

xperience and knowledge of desig

intain principal factors and

1.3.1 The rules of developing analytical model

Real structures are usually too complex to perform a rational analysis in They h ve to be simplified or idealized, by throwing away some unimportant de

to computation. An analytical model is a simplified representation of sketch, or an ideal sketch, of a real structure for the purpose of analysis. The simplified representation of sketch of a structur

s analytical model. All the analysis of structures is performed in their analytical models. Therefore, the development of analytical model is the basement of structural analysis. Establishment of an analytical model is one of the most important steps of the analysis process; it requires e

n practices in addition to a thorough understanding of the behavior of structures. Remember that the structural response predicted from the analysis of the model is valid only to the extent that the model represents the actual structure.

Development of the analytical model should generally obey the following rules: （1） The analytical model should reflect, as accurately as practically possible, the main stressed and

strained characteristics of the structure of interest to the analyst; （2） In order to facilitate the analysis, the analytical model should ma


disca

tructure; while in technical design stage we can select a more precise analytical model for the models of structures should be used;

while for computer-oriented method, complex models of structures might be selected.

1.3.2

ergo the loadings likely acting on it in various sense. Fortunately, many actua e restraints, into plan

framed structure is comprised of mem

ar cross section, etc.) is much smaller than the length l. Also cross-sectional defor

etween the members is re

oach is only suitable for the members whose ratios between their

join members of structures:

rd much of the detail about the members, connections, and so on, that is expected to have little effect on the desired characteristics.

It should be point out that the development of analytical model should meet the site and time conditions under the above rules. It should have alternative ways. For instance, a more precise analytical model should be developed for an important structure; while a less precise analytical model should be used for an unimportant structure. Furthermore, in schematic design stage we can develop a rough analytical model for a ssame structure. For hand-oriented method, the simplest analytical

Simplifying points of analytical model

(1) The simplification of structural system Generally, the actual structures are space, or three-dimensional, structures, whose members are

connected as a space frame to undl three-dimensional structures can be subdivided, by discarding some subsidiary spac

e structures for simplifying analysis. If all the members of a structure as well as the applied loads lie in a single plane, the structure is called a plane structure. The book will mainly discuss the calculation problems of plane structures.

(2) The simplification of members The main attention the book focuses on is the framed structures. Abers (or bars), whose cross-sectional dimensions (e.g. the width b and depth h of a rectangular cross

section, the radius of a circulmation of the members satisfies plane-section assumption, and stress acting on the cross sections of

members can be determined by the stress resultants (i.e. internal forces, bending moment, shear force and axial force). The cross-sectional deformation can be also evaluated by the strain components of the centroidal axes of the members. In the circumstances, the analytical model of a two- or three-dimensional structure selected for analysis is represented by a line diagram. On this diagram, each member of the structure is represented by a line coinciding with its centroidal axis; each connection b

presented by a kind of joint; the length of each member is represented by the distance between joints to which the member be attached; the position of the loads acting upon members is also transmitted to their centroidal axes. However, the simplifying appr

longitudinal and lateral dimensions are great than 4. (3) The simplification of connections The connections between members of a structure are commonly simplified into joints. Two types of

joints are commonly used to


Flexible, or hinged, joint A hinged joint prevents only relative translations of member ends connected to it; that is, all member

ends connected to a hinged joint have the same translation but may have different rotations. Such joints are thus capable of transmitting forces but not moments between the connected members. Fig.1.11 (a) shows a roof structure, the relative translations of each member are restrained by the gussets, but slight rotation between members will exist. So the gussets or connections of the roof structure are idealized into hinges. Hinged joints are usually depicted by small circles at the intersections of members on the line diagram of the structure, as shown in Fig. 1.11(b).

Rigid joint

joints are, therefore, capable of transmitting forces as well as moments betw rete side column and beam

ete. So the connection is simplified into a rigid

A rigid joint prevents relative translations and rotations of the member ends connected to it, that is, all member ends connected to a rigid joint have the same translations and rotation. In other words, the original angles between the members intersecting at a rigid joint are maintained after the structure has deformed

under the action of loads. Such

(a)

Fig.1.12 Connection of reinforced concrete bars and its computing model

een the connected members. Fig.1.12 (a) shows a connection of a reinforced conc, the relative translations and rotation of the column and beam are restricted by the arrangement of the

reinforcements which are cast into the whole body by the concrjoint. Rigid joints are usually represented by monolithic point at the intersections of members on the

(b)

(a) (b)

ucture and its computing modelFig.1.11 Roof str


line diagram of the structure, as shown in Fig. 1.12 (b). (4) The simplification of connections between structure and its foundation

. he ne structures are grouped into four categories, depending

on th on the structures.

Supports are used to attach structures to their foundations or other bodies, thereby restricting their movements under the action of applied loads. The loads tend to move the structures; but supports prevent the movements by exerting opposing forces, or reactions, to neutralize the effects of loads, thusly keeping the structures in equilibrium. The type of reaction depends on the type of supporting device used and the type of movement it prevents. A support that prevents translation of the structure in a particular direction exerts a reaction force on the structure in that direction. Similarly, a support that prevents rotation of the structure about a particular axis exerts a reaction couple on the structure about that axis

T types of supports commonly used for plae number of reactions (1, 2, or 3) they exert

Roller support Fig.1.13 (a) shows a photo of the roller support of a bridge structure; Fig.1.13 (b) and (c) depict the

roller and rocker supports used in bridge structures. The supports only prevent translation perpendicular to the supporting surface. So the reaction force AR acts perpendicular to the supporting surface and may be direc t The magnitude of ted either into or away from the struc ure. AR is the unknown. The support is thusl

(c) the rocker support of a bridge

y idealized, in according with its behavioral characteristics, as a roller-liked symbol shown as in Fig.1.13 (d) or a link shown as in Fig.1.13 (e).

(a) Photo of the roller support of a bridge

A

A A

(b) the roller support of a bridge

A

AR

A

(d) computing models of roller support

(e) computing model of roller support

Fig.1.13 Roller support


Hinged support The conformations of this kind supports are depicted in Fig.1.14 (a) and (b), they are simply referred

to as hinge. The supports are able to prevent translations in any direction. So the reaction force AR may

act in any direction. It is usually convenient to represent AR by its rectangular components, AX and AY . The magnitudes of AX and AY are the two unknowns. The support is thusly idealized, in according with its behavioral characteristics, as a hinge-liked symbol shown as in Fig.1.14 (c) or two concurrent links

Fig.1.15 Practical hinged supports(a) precast reinforced concrete column embedded in cup-liked foundation;(b) arc-shaped sluice with hinged support

(a) (b)

asphaltum with jute fiber

vertical beam

main trusssupport arm

supportpanel

lattice beam

A

Fig.1.14 Hinged support(a) and (b) conformations of hinged support; (c), (d) and (e) computing models of hinged support

AX

AY AR

A A

(a) (b)

AA A

(d) (e)(c)


shown as in Fig.1.14 (d) and (e). Fig.1.15 (a) shows a precast reinforced concrete column embedded in cup-liked foundation and the

inter

of moment The

in Fig.1.16 (b). Fig.1.16 (c) shows a precast reinforced concrete column embedded in cup-liked foundation and the

interspace around the column bottom end is filled by crushed stone concrete. In practice, there is no movement between the column and the foundation when the depth of embedment reaches some kind of degree. So the attachment generally is simplified as a fixed support.

Directional support The directional support (or double-link support) restricts all relative movement between structure and

its fo ong its supporting surface. So the reactions consist of a force perpendicular to th

ig.

order to simplify the structural analysis, all materials

(a) (b)

Fig.1.16 Fixed support(a) conformation of fixed support; (b) computing model of fixed support; (c) precast reinforced concrete cup-liked foundation

space around the column bottom end is filled by asphaltum with jute fiber. In practice, there is slight rotation between the column and the foundation but no translations. So the attachment can be simplified as a hinge. Fig.1.15 (b) shows another example of hinged support, an arc-shaped sluice, there is a rotation about pin A when the sluice is being opened.

Fixed support The fixed support can prevent both relative translations and rotation between structure and its

foundation. So the reactions consist of two force components AX , AY and a couple AM . magnitude of AX , AY and AM are the three unknowns. The support is thusly idealized, in

according with its behavioral characteristics, as a symbol shown as

crushed stone concrete

AX

AY

AMA

(c)

AYundation but slides ale supporting surface and a couple of moment AM . The magnitudes of AY and AM are the two

unknowns. The support is thusly idealized, by its behavioral characteristics, as a symbol shown as in Fig.1.17 (b) or two parallel links shown as in F 1.17 (c).

(5) The simplification of property of material Generally, the structures constructed in civil engineering are made of materials such as steel, concrete,

bricks, stone, timber, and so on. However, in


com g sumed to be continuous, homogeneous, isotropic, perfectly elastic or plastic.

Above assumption is suitable for metal within some stressing extent, but for concrete, reinforced concrete, bricks, stone and the like, the assumption will have some degree of approximation. As far as timber, because the property along the timber grain is quite different from that cross the timber grain, the attention should be paid when applying the assumption.

(6) The simplification of load The loads applied on structures may be divided into two categories such as body force and surface

force. The body force indicates the gravities and inertial forces of structures and the like; the surface force means the action upon structures transmitted by other bodies attached to them, for instance, the compression of soil and vehicular wheels and so on. In framed structures, the members are represented by lines coinciding with their centroidal axes, so both body force and surface force can be simplified as the forces acting on the lines. The loads can be simplified into concentrated loads and distributed loads according to their distribution. The simplification and determination of loads are of complexity. The special discussion about them will be given in a latter section.

1.3.3 Illustration of analytical models of structures

Fig.1.18 (a) shows an illustrative diagram of a workshop of a factory. Now its analytical models will be discussed.

(1) The simplification of structural system of the workshop The structure of the workshop is, like clockwork along its longitudinal direction, collocated by a series

of planar elements, composed by roof structure, columns and foundation shown as dashed area or in Fig.1.18 (b), and then connected by roof panels and other longitudinal members as a space structure. The loads acting upon the workshop are usually uniformly distributed along its longitudinal sense. Therefore,

mn ticatur e t

(a) (b)

Fig.1.17 Directional support(a) conformation of directional support; (b) schematic symbol of directional sup

posin structures are as

the portion between adjacent centroidal axes of two colu s could be extracted as the analy l element from the space struc e, whil he loads acting upon the structure could be distributed to each analytical element by the longitudinal members. Because in each analytical element, its members as well as its loads

port;(c) computing model of directional support

(c)AYAM

A

A

A


lie in a single plane, the space structure would be thusly divided into plane structures shown as in Fig.1.18 (b).

cti ho ig.

by a line coinciding with its centroidal axis; dealized as a hinged joint;

For the plane structure shown in Fig.1.18 (b), the analytical models of its roof structure and columns will be discussed in the following.

roof structure

(a)

foundation

roof panelcolumn

(b)

(c) (d)

Fig.1.18 Analytical model of the structure of one-storey workshop(a) the structure of one-storey workshop; (b) planar analytical element;(c) analytical model of roof structure under vertical load;

l model of framed bent under horizontal load(d) analytica

(2) The analytical model of roof structure under vertical loads The analytical model of roof structure under the a on of vertical loads is s wn in F 1.18 (c), here

the following simplification has made: Each member of the roof structure is represented Each connection between the members is i The two ends of the roof structure welded together with the columns below it by steel gussets are

simplified as one roller and one hinged support.

1.4 Classification of Framed Structures 13

The loads are recognized as concentrated forces located at four corners of the roof slab and acting on the top chords of the roof truss.

(3) The analytical model of the columns under action of horizontal loads The analytical model of the columns of the workshop structure under the action of horizontal loads

(for example, the lateral wind load) is shown in Fig.1.18 (d), here the following simplification has made: Each column of the workshop structure is represented by a line coinciding with its centroidal

axis; The connections between two ends of the roof structure and the support heads of the columns are

simplified as hinged joints. Here the function of the roof structure is realized as a link which connect the two columns;

l factories. model is the most important step of the analysis process; it is complex

too.

two- or three-dimensional structure selected for anal

bers on the line diagrams. This is done, when necessary, for clarity of presentation; in such cases, the r depth.

1.4

Because the crushed stone concrete would be used to fill the interspace between the bottom ends of the columns and their foundations, the supports are simplified as fixed supports.

The structure shown as in Fig.1.18 (b) or (d) are referred as framed bent, it is one of common types of structures to be used in one-storey workshops in industria

Establishment of the analytical It requires experience and knowledge of design practices in addition to a thorough understanding of the

behavior of structures. For a new type of structures, a rational analytical model would be developed by many tests and practices sometimes. However, for commonly used types of structures, their analytical models will be straight developed according to the experience of the predecessors devoted to structural analysis. Regrettably it is nearly impossible to give the training of ability of developing analytical models of structures only in school except in a most indirect sense. A rational analytical model comes from intuition based on our experience, judgment, knowledge, and our mastery of information concerning the engineering structures and civil engineering projects.

It should be note that the analytical model of the ysis is represented by a line diagram. On this diagram, two lines ( ) are often used in this text to

represent memdistance between the lines does not represent the membe

Classification of Framed Structures

In practice, perhaps the most important decision made by a structural engineer in implementing an engineering project is the selection of the type of the structure used for supporting or transmitting the loadings. Commonly used structures can be classified according to their composition and stressing characteristics or their method being analyzed or the arrangement of their members as well as the locations of the applied loads acting on. Essentially, the classification of framed structures is the classification of


their analytical models. Commonly used structures can be classified into following categories, depending on their composition

and s

(1) Beams The centroidal axis of a beam is generally a straight line. For a horizontal beam, there is no horizontal

reaction under transverse loads, its internal forces are bending moment and shear force. There are single span beam [Fig.1.19 (a) and (b)] and multispan beam [Fig.1.20 (a) and (b)].

under the action of vertical loads. The horizontal thrusts would imp

t, shear and axial force in rigid frame members, of which

Fig.1.21 Arch(a) three-hinged arch; (b) hingeless arch

(a)

tressing characteristics.

(2) Arches The centroidal axis of an arch is a curved line. Horizontal thrusts H [Fig.1.21 (a) and (b)] will be

developed at the supports of the archrove, out and away, the stressing property of the arch. (3) Rigid frames A rigid frame is a system of several members, hinged or rigidly connected at their ends as shown in

Fig.1-22. Generally, there exist bending momen

H H

(b)H H

(a)

(b)

Fig.1.19 Single span beam(a) statically determinate beam; (b) statically indeterminate beam

(a)

(b)

Fig.1.20 Multispan beam(a) statically determinate multispan beam; (b) statically indeterminate multispan beam

1.4 Classification of Framed Structures 15

bend

d to the loads applied at its joints, it trans

Framed structures can be also classified into two categories, statically determinate structures and l zing method applied to them.

(1) Statically determinate structures

(b) statically indeterminate truss

ing moment is the dominated internal force.

(4) Trusses A truss consists of a group of members which are assumed to be connected at their joints with

frictionless hinges as shown in Fig. 1-23. When a truss is subjectefers the loads by developing axial forces in its members.

Fig.1.23 Truss(a) statically determinate truss;

(b)

(5) Composite structures A composite structure is a combination of beam-typed members (mainly resist bending) and links or

two-force members as shown in Fig.1.24.

statical y indeterminate structures, according to analy

A stable structure is considered to be statically determinate if all its support reactions and internal forces can be determined by solving the equations of equilibrium. The structures shown in figure (a) from Fig.1.19 to Fig.1.24 are all statically determinate structures.

(2) Statically indeterminate structure Conversely, a stable structure is considered to be statically indeterminate if all its support reactions

(a)

Fig.1.22 Rigid frame

(a) (b)

(a) statically determinate rigid frame; (b) statically indeterminate rigid frame


and internal forces can not be determined by solving the equations of equilibrium. The structures shown in figur

(1) Plane structures If all the members of a structure as well as the applied loads lie in a single plane, the structure is called

a plane structure. The foregoing mentioned structures are the primary structures concerned with in the book.

action of nonplanar load

zP

e (b) from Fig.1.19 (b) to Fig.1.24 (b) are all statically indeterminate structures.

Furthermore, framed structures can be classified into another two categories, plane structures and space structures, according to the arrangement of their members as well as the locations of the applied loads acting on.

(2) Space structures Some structures, such as latticed domes, aerospace structures, and transmission towers, cannot, due to

their shape, arrangement of members, or applied loadings, be subdivided into planar components. Such

Fig.1.25 Space structure(a) space structure; (b) plane structure under the

(a) (b)

Py

x

Fig.1.24 Composite structure(a) statically determinate composite structure; (b) statically indeterminate com

A

B(a)

(b)

posite structure

1.5 The Classification of Loads 17

structures, called space structures, are analyzed as three-dimensional bodies subjected to three-dimensional force systems.

Fig.1.25 (a) is a space structure whose members do not lie in a single plane; although the members of the structure shown in Fig.1.25 (b) lie in a single plane, the applied load lie in another plane, the structure belongs to space structure.

1.5 The Classification of Loads

The objective of a structural engineer is to design a structure that will be able to withstand all the loads to which it is subjected while serving its intended purpose throughout its intended life span. In designing a structure an engineer must, therefore, consider all the loads that can realistically be expected to act on the structure during its planned life span. Well then, what is load? In fact, the loads are the external active forces im ctures. The deadweights (or gravities) of structures, the crane loads acting on the struc

ccording to their duration and the manner of application.

(1) According to their duration, the loads can be divided as: Dead loads Dead loads are loads of constant magnitudes and fixed positions that act

permanently on the structure. For instance, the weights of the structural system itself and of all other materials and equipment permanently attached to the structural system and the like.

Live loads Live loads are loads of varying magnitudes and/or positions caused by the use of the structure. Sometimes, the term live loads are used to refer to all loads on the structure that are not dead loads, including environmental loads, such as crowd loads on floors, crane loads, snow loads or wind loads, etc. into following two categories according to the situation of their varying posi

re those which may be located at one position or another on a struc

oving loads, such as a serie t

d due to no effect of structural vibration. are loads of varying magnitudes and positions with time.

Dyn

posed on strutures of workshops, the loads due to vehicular traffic on bridges, hydraulic pressure and soil pressure

acting on the hydraulic structures are the examples of loads. The loads may be divided into the following categories a

Live loads can be subdivided tions: a. Movable loads Movable loads ature, such as furniture and goods on a building floor and snow on a roof, etc. b. Moving loads Live loads that move under their own power are said to be ms of rucks or railroad trains, etc. (2) According to their manner of application, the loads can be divided as:

Static loads The magnitudes and positions of static loads are not varying or varying very slowly with time, so that the inertial forces could be neglecte

Dynamic loads Dynamic loadsamic loads give rise to vibration of structures. So the inertial forces have to be taken into consideration.

18 CHAPTER 1 INTRODUCTION

Blast loads caused by explosion, impact loads, centrifugal forces and the like are examples of dynamic loads.

Some external effects such as support settlement, temperature change, manufacture errors, shrinkage and creep of materials may cause internal forces and deformations in structures. These factors may be called loads in generalized sense.

SUMMARY

In this chapter we have discussed four questions: (1) objective and learning method of structural mechanics, (2) the analytical models of structures, (3) classification of framed structures and (4) the classification of loads. They are all important questions throughout the book, but a primary understand will satisfy the requirement of the introduction. The profound comprehension needs our gradually leaning the other content of the book.

It is worthy of emphasize that the analytical models of the structures are the key point of the chapter and

ts and supports) ought to be paid to during the studying, so as to lay a solid foundation of know e internal forces and deformations of structures.

he analytical or computing model of a structure? What is the relationship and difference between the model and its realistic structure? Why we have to simplify actual structures into analytical models?

1.2 Which two kinds of joints are the connections of a plane structure commonly simplified? What are the conformation, the characteristics of restricting movement and restraint forces of the joints?

1.3 What kinds of supports for plane structures are usually simplified into? What are the conformation, the characteristics of restricting movement and restraint reactions of the supports?

1.4 How many categories are there of commonly used framed structures?

also the starting point of computing structures in the latter study. Some special attention about the principle of developing analytical models and the key points of simplification (especially, the simplification of connecting join

ledge for analyzing th

Problems for Reflecting

1.1 What is t

CHAPTER 2 GEOMETRIC CONSTRUCTION ANALYSIS

OF STRUCTURES

The abstract of the chapter The chapter is the essential requisition of structural analysis. Firstly, degrees of freedom are

introduced; then, geometric construction rules of plane stable structures with no redundant restraint are discussed in the chapter. For plane structures, there are three construction rules recognized as the key point of the chapter.

2.1 Purpose of Analyzing Geometric Construction of Structures, Stable and Unstable Structural Systems

In order to withstand and transmit loads the geometric shape of a structure must not be changed under loads. If the shape of a structural system is variable under loads the structural system cannot be used as a structure.

It should be realized that all physical bodies deform when subjected to loads; the deformations in most engineering structures under service conditions are so small that their effect on the geometric construction analysis of the structures can be neglected. This means that all members composing of the structures are recognized as rigid bodies. For plane structures, the rigid bodies are referred to as rigid sheets. In the following discussion of the chapter, the all members of structures are considered to be rigid bodies or rigid sheets; and for simplicity we call a structure as a system. For instance, framed structures are named framed systems.

BA

C D

(a)

A B

(b)

B

According to the shape and location of their members, framed systems can be classified into two categories: A

(c)

Fig.2.1 Stable system

C

(1) Geometrically stable system Under the action of the loads, the system still maintains its shape and remains its location if the small deformations of the members are neglected as shown in Fig.2.1 (a) and (b).

(2) Geometrically unstable system Under the action of the loads, the system will change its

19

20 Chapter 2 Geometric Construction Analysis of Structures

shape and its location if the small deformations of the members are neglected as shown in Fig.2.2 (a) and (b).

Corresponding to geometrically stable and unstable systems, there are internally stable and unstable systems as well. A structure is considered to be internally stable, or rigid, if it maintains its shape and remains a rigid body when detached from the supports. Conversely, a structure is termed internally unstable (or nonrigid) if it cannot maintain its shape and may undergo large displacements under small disturbances when not supported externally. Fig.2.1 (c) shows a triangular structure conformed by three members and detached from the supports. Because the triangle can maintain its shape and remains a rigid body although it can only move in its own plane as an integer, it is referred to as an internally stable system, simply the internally stable. However, the structure shown as in Fig2.2 (c), conformed to a parallelogram by four bars and connected by pins each other, cannot maintain its shape and may undergo large displacements under small disturbances when not supported externally. This kind of structures is thusly named as internally unstable system, simply the internally unstable.

Purpose of analyzing geometric construction of structures is as following: (1) To estimate whether or not a system is geometrically stable so as to determine whether or not the

system can be used as a structure; (2) To discuss geometric construction rules of stable systems. In addition above two points, analyzing geometric construction of structures are helpful to make their

static analysis due to the close relationship between their geometric construction analysis and mechanical analysis.

D

2.2 The Concept of Degrees of Freedom and Restraints

In the analyzing geometric construction of structures, it is very feasible to consider one part of the members (i.e., rigid bodies) or joints of a system as an object which possesses degrees of freedom, whereas other part of the members or joints of the system as restraints which restricts the movement of the object. The relationship of these tow parts are then analyzed and whether or not the system is stable or unstable will be determined. Accordingly, the concept of degrees of freedom and restraints of a system is discussed

B′(b)

A B

Fig.2.2 Unstable system

(a)A B

C

(c)A B

C D

2.2 The Concept of Degrees of Freedom and Restraints 21

first of all.

2.2.1 The degrees of freedom

The degrees of freedom of a system are the numbers of independent movements (or coordinates) which are required to locate the system fully. Obviously, a rigid body has three degrees of freedom in a planar coordinate system (six degrees of freedom in a three dimensional coordinate system), e. g., the position of member AB may be determined by three parameters Ax , and Ay Aθ [Fig. 2-3 (b)].

The degrees of freedom of a joint The movement of a point in a planar coordinate system could be decomposed to two translations in

any different directions, i.e., a point possesses two independent moving styles or two independent coordinates are needed to locate its position in a planar coordinate system. So a joint (or a point) has two degrees of freedom in a planar coordinate system (three degrees of freedom in a three dimensional coordinate system). In Fig. 2-3 (a) the parameters Ax , will locate joint A. Ay

The degrees of freedom of a rigid body The movement of a rigid body (or rigid sheet) in a planar coordinate system could be decomposed to

two translations in any different directions and a rotation about some point in the system, i.e., a rigid body possesses three independent moving styles or three independent coordinates are needed to locate its position in a planar coordinate system. Therefore, a rigid body has three degrees of freedom in a planar coordinate system (six degrees of freedom in a three dimensional coordinate system), e.g., the position of member AB may be determined by three parameters

yy

BA

Ax , and Ay Aθ [Fig. 2-3 (b)].

2.2.2 Restraints

The devices or connections which can reduce the degrees of freedom of a system are defined as restraints. The number of the degrees of freedom of a system reduced by the device or connection is named the number of its restraints. There are two kinds of restraints, support restraints and connecting restraints between rigid bodies.

1. Support restraints (1) The roller support [Fig.2.4 (a)] can restrict the translation of joint A in the direction perpendicular

Ay

Ax(a) (b)

Ay

Ax

x

A Aθ

x

Fig.2.3 The degree of freedom(a) the degree of freedom for a joint;(b) the degree of freedom for a rigid body


its moving surface but cannot prevent its translation along its moving surface and rotation about joint A, i.e., one roller support reduces one degree of freedom and is equivalent to one restraint.

(2) The hinged support [Fig.2.4 (b)] can restrict the translations of joint A in vertical and horizontal directions but cannot prevent the rotation about joint A, i.e., one hinged support reduces two degrees of freedom and is equivalent to two restraints.

(3) The fixed support [Fig.2.4 (c)] can restrict the translations of joint A in vertical and horizontal directions and the rotation about joint A, i.e., one fixed support reduces three degrees of freedom and is equivalent to three restraints.

AA A

(a) (b) (c)

Fig.2.4 Support restraint(a) roller support; (b) hinged support; (c) fixed support

2. Connecting restraints between rigid bodies We will pay more attention to connecting restraints between two rigid bodies. One rigid body has

three degrees of freedom and two independent rigid bodies have six degrees of freedom in a planar coordinate system. When connecting them together their degrees of freedom would be reduced. Now we will discuss the equivalent restraints of a few kinds of commonly used connections.

A A

(1) Link [Fig.2.5 (a)] The two independent rigid bodies AB and DC have three relative degrees of freedom with respect to each other, when they are connected by link EF which is a bar with hinges at its ends, the relative translation between the two rigid bodies in the direction of the link is restricted, i.e., one of the degrees of freedom between them is removed. Therefore, one link is equivalent to one restraint.

(2) Simple hinge (the hinge which connects two rigid bodies) [Fig.2.5 (b)] The two independent

(a) (b) (c)

C

D

A

E FB CB

B

C

Fig.2.5 Connecting restraint(a) restraint of a link; (b) restraints of a simple hinge; (c) restraints of a simple rigid joint

2.2 The Concept of Degrees of Freedom and Restraints 23

rigid bodies AB and AC have three relative degrees of freedom with respect to each other, when they are connected by hinge A only one relative degree of freedom between them is existed, i.e., the relative rotation about A, and two translation degrees of freedom between them are removed. Therefore, one simple hinge is equivalent to two restraints.

(3) Simple rigid joint (the joint which connects two rigid bodies rigidly) [Fig.2.5 (c)] The two independent rigid bodies AB and AC have three relative degrees of freedom with respect to each other, when they are connected by a rigid joint A there exist no relative degree of freedom between them in the plane, i.e., three of the degrees of freedom between them are removed. Therefore, one simple rigid joint is equivalent to three restraints.

By similar analysis, a hinge connecting three rigid bodies [Fig.2.6 (a)] makes the system remove four degrees of freedom; the hinge is thusly equivalent to two simple hinges. While a rigid joint connecting three rigid bodies [Fig.2.6 (b)] set the system remove six degrees of freedom; the rigid joint is then equivalent to two simple rigid joints. Therefore, if a hinge or a rigid joint is used to connect more than two rigid bodies, it is called a multiple hinge or multiple rigid joint. A multiple hinge connecting n bars is equivalent to n-1 simple hinges or restraints, while a multiple rigid joint connecting n bars is equivalent to n-1 simple rigid joints or restraints.

2( 1)n−3( 1)n −

2.2.3 Restraint substitution and virtual hinges

Foregoing discussion tells us that one hinge is equivalent to two restraints and two links are also equivalent to two restraints; whereas restraints can be substitute. In order to extend the concept of restraint substation the concept of virtual or instantaneous hinge are introduced.

Two noncollinear links which connects two rigid bodies are equivalent to one hinge. If two noncollinear links AB and AC intersect at point A which lies on the rigid bodyⅠ[Fig.2.7 (a)], the point A will be a actual simple hinge. If the intersectional point A does not lie on the two rigid bodies, i.e., the intersecting point lies outside of the two rigid bodies, the intersection is referred to as a virtual or instantaneous hinge, which has a variable position.

Two links AB and AC connecting two rigid bodiesⅠandⅡshown as in Fig.2.7 (b) intersect at point E which lies outside of the two rigid bodies. Because the two rigid bodies are connected by two links the two of their relative degrees of freedom are removed, but their relative rotation still exist. Where is their

AB

C

D

(a)(b)

Fig.2.6 Multiple joint(a) multiple hinge; (b) multiple rigid joint


instantaneous rotational center? Obviously it is the intersectional point E about which the two rigid bodies are able to rotate relatively and instantaneously and soon the point E will move to a new position E′ .

2.2.4 Necessary restraints and redundant restraints

In a framed system, the restraints which can restrict the degrees of freedom of the system are named as necessary restraints; whereas the restraints which are not able to restrict the degrees of freedom of the system are referred to as redundant restraints.

A free joint A shown as in Fig.2.8 (a) has two degrees of freedom; if it were attached to the ground with two noncollinear links 1 and 2 shown as in Fig.2.8 (b), its position would be determined and its two degrees of freedom would be restricted. Therefore, the link 1 and 2 are all necessary restraints in order to locate the position of the joint. If a link 3 were added to attach the joint with the ground shown as in Fig.2.8 (c) the link 3 would be redundant (you can also recognize any one of the three links as a redundant restraint).

A

A A

(a) (b) (c)

1 2 1 23

Fig.2.8 Connection of a joint(a) a free joint; (b) two links attached to a joint; (c) three links attached to a joint

Ⅰ

A

B C

(a)Ⅱ

Fig.2.7 Actual hinge and virtual hinge(a) actual hinge; (b) virtual hinge

E′E

A

B

C

DⅠ

Ⅱ(b)

2.3 Geometric Construction Rules of Planar Stable Framed Systems without Redundant Restrains 25

A free member AB shown as in Fig.2.9 (a) has three degrees of freedom; if it were attached to the ground with three noncollinear links 1 and 2 and 3 shown as in Fig.2.9 (b), its position would be

determined and its three degrees of freedom would be restricted. Actually, it is a simply supported stable beam. Therefore, the link 1, 2 and 3 are all necessary restraints in order to locate the position of the member AB. If a link 4 were added to attach the member with the ground as shown in Fig.2.9 (c) the link 4 would be redundant (you can recognize any one of the three vertical links as a redundant restraint as well). It should be noted that in Fig.2.9 (c) you cannot recognize the horizontal link 1 as a redundant restraint, and it is a real necessary restraint. If the three links connecting the member AB intersected to one point as shown in Fig.2.9 (d) only one link of link 1 and 3 would be necessary, the other one would be redundant, while the vertical link 2 would be necessary restraint. It seems unclear for a system which restraints are necessary or redundant restraints. In fact, the stability of a system depends on not only the number of its restraints but also the arrangement of its restraints.

12 34

12 3

12 3

A BBA

B A

(a)

(b)

(c)

(d)

Fig.2.9 Connection of a member(a) a free member; (b) three links attached to a member; (c) four links attached to a member; (d) three links attached to a member

BA

2.3 Geometric Construction Rules of Planar Stable Framed Systems without Redundant Restraints

There are three elementary geometric construction rules for planar stable framed systems without redundant restraints. Now we will discuss them through examples. It is worth of mentioning in anticipation that a hinged triangle [Fig.2.1 (c)] is the simplest stable form of framed system. It is the basic stable shape from which the general rules for stability may be developed.

2.3.1 The connection between a point and a rigid body

There are two degrees of freedom for a free point A [Fig.2.10 (a)] in planar coordinate system; when the point is connected with a rigid body by a single link AB only the motion along the direction of the link


is restricted. So the point still has a degree of freedom in planar coordinate system. If another link AC, which is noncollinear with link AB, is used to connect the point [Fig.2.10 (b)], all of the degrees of freedom

of the point is restricted; an internally stable system without redundant restraints is then composed by the point A, the two links AB and AC and the rigid body.

A AA

BC C

Thereby, the following rule will be obtained: Rule 1: A point and a rigid body connected by two noncollinear links form an internally stable system

without redundant restraint. The rule can be reworded by the construction of binary system: If a binary system is attached to a rigid body to form a new system the new system is still a stable

system without redundant restraints. Here, the binary system indicates the construction of a new joint (connection) which is formed by two

noncollinear links (or members) connected each other. The attachment of binary systems will not change the initial degrees of freedom of a system.

Lastly, it should be mentioned that in the geometric construction analysis the foundation of a structure or the ground can be also regarded as a rigid body [Fig.2.10 (c)], thus rule 1 will be a standard model to fix a point to the ground or foundation. And it is helpful to note that a hinged support is equivalent to two noncollinear links and a roller support is equivalent to one link.

Example 2-1 Analyze the geometric construction of the framed system shown as in Fig.2.11. Solution

B B C

(a) (b) (c)

Fig.2.10 Connection between a point and a rigid body(a) a link connecting a point and a rigid body; (b) two links connecting a point and a rigid body; (c) a point attached to the ground by two links

A EC

FDB

Fig.2.11 A framed system


According to rule 1, (1) starting from stationary points A and B, use two noncollinear links AC and BC to fix point C; (2) beginning at stationary points B and C, add a binary system to fix point D; reincrease a binary system to fix point E; again starting from stationary points D and E, reincrease a binary system to fix point F.

Therefore, the integral system is a stable system without redundant restraint.

2.3.2 The connection of two rigid bodies in the same plane

In planar coordinate system，if rigid bodies AB and BC are connected by one hinge the relative rotation between them will still exist [Fig.2.12 (a)]; If a link AC, which will not intersect with hinge B, is used to connect rigid bodies AB and BC, the degrees of freedom of the new system are restricted and the system will become an internally stable system without redundant restraints [Fig.2.12 (b)].

A AC

Thereby, the following rule will be achieved: Rule 2: Two rigid bodies, connected by one hinge and one link that do not cross the hinge, form an internally

stable system with no redundant restraint. By the concept that one hinge is equivalent to two links and the concept of virtual or instantaneous

hinge, rule 2 can be reworded in another way: Two rigid bodies connected by three links, which are nonparallel and nonconcurrent, will form an

internally stable system with no redundant restraint [Fig.2.12 (d)]. When the ground is regarded as a rigid body [Fig.2.12 (c)], rule 2 will be a standard model to fix a

rigid body to the ground or foundation. Example 2-2 Analyze the geometric construction of the system shown as in Fig.2.13. Solution

CB B C

(a) (b) (c) (d)

1 2 3

Ⅰ

Ⅱ

Fig.2.12 Connection between two rigid bodies(a) one hinge connecting two rigid bodies; (b) one hinge and one link connecting two rigid bodies; (c) and (d) three links connecting two rigid bodies

28 Chapter 2 Geometric Construction Analysis Of Structures

Firstly, assume rigid body AB to be a moveable object, it is connected with the foundation by a fixed support; so AB and the foundation are fixed together and they form a new rigid body with no redundant restraint.

A B C D

Fig.2.13 The system of example 2-2Secondly, regard rigid body BC as a moveable

object, AB and the foundation as a new rigid body, by rule 2, hinge B and roller C (a roller is equivalent to a link) are used to connect the two rigid bodies and the roller C does not cross hinge B. Therefore, the system is stable with no redundant restraint.

2.3.3 The connection of three rigid bodies in the same plane

In nature, a hinged triangle is the simplest internally stable system with no redundant restraints [Fig.2.14 (b)]. Based on this system rule 3 will be acquired [Fig.2.14 (a)]:

A A

CB B C

(a) (b)

Ⅰ Ⅱ

(Ⅰ,Ⅲ)

(Ⅰ,Ⅱ)

(Ⅱ,Ⅲ) (Ⅰ,Ⅲ)

(Ⅰ,Ⅱ)

(Ⅱ,Ⅲ)

Ⅲ

Ⅰ Ⅱ

Ⅲ

Fig.2.14 Connection of three rigid bodies(a) connection of three rigid bodies; (b) two rigid bodies attached to the foundation

Rule 3: Three rigid bodies (I, II and III) joined pairwise by hinges, provided that the three hinges (A, B and C)

do not lie on the same straight line, form an internally stable system with no redundant restraint. When the ground is regarded as a rigid body [Fig.2.14 (b)], rule 3 will be a standard model to fix two

rigid bodies to the ground or foundation. It is worthwhile to emphasize that all the three rules are only different explanations of the geometric

construction of the identical basic hinged triangle [Fig. 2-1 (c)] from which any one of the rules developed formerly may be derived.

Example 2-3 Analyze the geometric construction of the system shown as in Fig.2.15 (a), (b) and (c).


Solution Fig.2.15 (a): According to rule 3, rigid bodies ADC, BEC and the foundation which is regarded as the

third rigid body are connected pairwise by hinges A, B and C to form a stable system with no redundant restraint.

Fig.2.15 (b): AFD is a rigid body, then a binary system is added to AFD to fix point C and to form a larger rigid bodyⅠ. Similarly, BGEC can be composed to another larger rigid body II. The foundation is regarded as rigid body Ⅲ. Upon that, three rigid bodiesⅠ, II and Ⅲ are connected pairwise by hinges A, B and C to form a stable system with no redundant restraint.

Fig.2.15 (c): The system can be recognized that link FG is attached to the system shown in Fig.2.15 (b). Accordingly, the system shown as in Fig.2.15 (c) is a stable system with one redundant restraint.

2.3.4 Instantaneously unstable system

In the construction of framed system, although the number of restraints is sufficient or even more than sufficient, it is still unclear whether the system is stable or unstable, because the stability of a system depends on not only the number but also the arrangement of restraints. If the arrangement of restraints of a system does not satisfy the requirement of geometric construction rules as discussed in previous subsections, the system will not remain stable.

In Fig.2.16 (a), the ground and joint A are connected by two links lying on the same straight line. The system is unstable, because link AB and AC can not prevent the infinitesimal normal displacement of joint A. However, when the system undergoes an infinitesimal displacement, A′ is the new position of joint A, the two links will not lie on the same line any more. In this case, the system is termed an instantaneously unstable system.

A B

CD E

(Ⅰ,Ⅱ)

(Ⅱ,Ⅲ)(Ⅰ,Ⅲ)

Ⅰ Ⅱ

Ⅲ

(a)

A B

CD E

(Ⅰ,Ⅱ)

(Ⅱ,Ⅲ)(Ⅰ,Ⅲ)

Ⅰ Ⅱ

Ⅲ

(b)

G

ECD

FF

B

(c)

G

A

Fig.2.15 The systems of example 2-3


In Fig.2.16 (b), three links are intersected at point O, an infinitesimal rotation of rigid bodyⅠaround rigid bodyⅡwill exist instantaneously. After the infinitesimal rotation happened the three links will not intersect at one point and the system will not change again.

The system existing infinitesimal instantaneous displacement is referred to as an instantaneously unstable system. Although the displacement of an instantaneously unstable system is an infinitesimal value, the internal forces of its members are infinitely large under the action of the loads which have finite values. Therefore, instantaneously unstable system can not be used in practice.

Example 2-4 Analyze the geometric

construction of the system shown in Fig.2.17 (a), and (b).

Solution Fig.2.17 (a): Rigid body AB is

fixed with the foundation to form a larger rigid body; then the new rigid body is connected with rigid body CDE by a hinged support D and a link BC, but BC cross the hinge D. Thereby, the system does not meet rule 2. It is an instantaneously unstable one.

Fig.2.17 (b): Analyzing similarly as Fig.2.17 (a), the stable portion composed by AB and the foundation is connected with rigid body CDE by a hinged support D and a link BC, but BC does not cross the hinge D. According to rule 2 the system is stable one with no redundant restraint.

(b)(a)

AB C

′

Ⅱ

Ⅰ

O

Fig.2.16 Instantaneously unstable system

A

C DB EA

(a)

EC D

BA(b)

Fig.2.17 The systems of example 2-4

2.4 Illustration of Geometric Construction Analysis 31

2.4 Illustration of Geometric Construction Analysis

Example 2-5 Analyze the geometric construction of the systems shown as in Fig.2.18 (a), and (b). Solution

Fig.2.18 (a): The hinged triangle CGH is the basic stable portion (the hinged triangle GDE, or GHE, HEF can be also regarded as a basic stable portion). By using rule 1, adding binary system from triangle CGH, joints E, D, F, B and A are connected in the mentioned order. Consequently the portion becomes internally stable with no redundant restraint. Afterward, connect the stable portion with the ground with one hinged support and one roller support (equivalent one link, by rule 2) to form a stable system with no redundant restraint.

Fig.2.18 (b): The hinged triangle AFD can be regarded as a basic rigid body. By using rule 1, then add a binary system to AFD to form a larger rigid body AFCD. Similarly, another larger rigid body BGCE can be obtained. And then, connect the two rigid bodies AFCD and BGCE with one hinge and one link (rule 2) to form an internally stable system with no redundant restraint. Finally, attach the system to the ground with a hinged support and one roller support which does not cross the hinged support. Therefore, the integral system is stable with no redundant restraint.

Example 2-6 Analyze the geometric construction of the systems shown as in Fig.2.19 (a), and (b). Solution 1. Fig.2.19 (a): Rigid bodies AC, BC and the foundation which is regarded as the third rigid body

are connected pairwise by hinge C and virtual hinges D and E, and the three hinges do not lie on the same straight line. By rule 3, the system is a stable one with no redundant restraint.

2. Fig.2.19 (b): Although the rigid bodies AD and BE are angle bars, their ends are connected by hinges. So AD and BE can be recognized as two links. The two links and roller support C (the roller support

A

B

C

DCD E

G

H

(a)

FE

FA

(b)

G

Fig.2.18 Systems of example 2-5

B


is equivalent to one link) are used to connect rigid body CDE and the ground. However, the three links are concurrent. Thereupon, the system is an instantaneously unstable system.

2.5 The Relationship between Static Determinacy and Geometric Construction of Structures

Another important function of geometric construction analysis is that the analysis can be used to determinate the static determinacy or indeterminacy of a structure by analyzing whether or not a system has redundant restraints.

A stable structure is considered to be statically determinate if all its support reactions and internal forces can be determined by solving the equations of equilibrium. For example, the simply supported beam shown in Fig.2.20 (a) has one hinged support and one roller support, so it is a stable system with no redundant restraint; obviously, the two supports will yield three reactions to the beam. Because the action lines of the three reactions are not concurrent at one same point, the three equations of equilibrium

A BP

PAX

AY BY(a)

A BCP

PAX

AY BYCY

(b)

Fig.2. 20 Structures(a) statically determinate structure; (b) statically indeterminate structure

A BC

D E

O

Ⅰ

Ⅱ

2 3

(b)

C

D E

A BⅠ Ⅱ

12

3

Ⅲ

4

(a)

Fig.2.19 Systems of example 2-6

Summary 33

( , and ) of a coplanar force system can be used to solve the three reactions. Once the reactions are solved all internal forces of the beam will be determined.

0X =∑ 0Y =∑ 0M =∑

Conversely, if all support reactions and internal forces of a stable structure cannot be determined by solving the equations of equilibrium the structure is recognized to be statically indeterminate. For the structure shown in Fig.2.20 (b), the beam is attached to the foundation by one hinged support and two roller supports, and the supports will develop four reactions. It is clear that the number of reactions is in excess of those of equilibrium equations of a coplanar force system.

Thereby, statically determinate structures are stable with no redundant restraint systems whose mechanical characteristics are that all of their support reactions and internal forces can be uniquely determined by their static equilibrium equations; whereas statically indeterminate structures are stable systems but with redundant restraints whose mechanical characteristics are that all of their support reactions and internal forces cannot be uniquely determined by their static equilibrium equations.

SUMMARY

1. The main purpose of geometric construction analysis of a system is to determinate whether or not the system is stable or unstable so as to make a decision that the system whether or not can be utilized as a structure. Analyzing the geometric construction rules of a stable system with no redundant restraint will help us to select correct method and procedure of static analysis, the idea will be frequently referred to in the consequent chapters.

2. There are three geometric construction rules: (1) one point and one rigid body are connected with two noncollinear links; (2) two rigid bodies are connected with one hinge and one link which does not cross the hinge or with three nonparallel and nonconcurrent links; (3) three rigid bodies are connected pairwise with hinges.

The essential of the three rules is a triangle rule, i.e., when the three side length of a triangle is determined its geometric configuration is uniquely definite. Understanding the three rules is not difficult. The main attention of the chapter should be paid to the sound application of the rules to analyze a variety of framed structures. The difficulty of a novice is how to begin with analyzing a structure. Obviously, doing some amounts of exercise will overcome the difficulty.

While using the three rules to analyze a system, their strictness should be pay attention to: distinguish restricted objects and their restraints which play a role to restrict them; the number of the restraints and their arrangement whether or not meet the requirement of geometric construction rules. On the other hand, flexibility should be also paid when analyzing a system, such as the relation of substitution between restricted objects and their restraints, the concept of virtual or instantaneous hinge and the like, and not confused by the variation of configuration of the system.


3. Make a sound understanding of relationship between geometric construction and static characteristics:

A stable with no redundant restraint system means that it is a statically determinate structure. A stable with redundant restraints system means that it is a statically indeterminate structure. An unstable (include instantaneously unstable) system means that it cannot be used as a structure.


2.1 Whether the instability of an unstable system can be changed by adding or discarding binary system in turn from it? Contrariwise, whether an unstable system can be changed to a stable one by adding or discarding binary system in turn from it?

2.2 What kind of relationship is there among the three rules mentioned in the chapter? Why it can be said that their essential is the same?

2.3 What is the instantaneous unstable system? Why instantaneous unstable system cannot be used to engineering structure?

2.4 What kind of characteristic of geometric construction is there for a statically determinate structure? What kind of characteristic of static analysis is there for a statically determinate structure?

2.5 What is the difference between the geometric construction and characteristics of static analysis of a statically indeterminate structure and those of statically determinate one?

Problems for Solution

Analyze the geometric construction of following systems.

problem 2-1

problem 2-2

problem 2-3

Problems for Solution 35

problem 2-4 problem 2-6problem 2-5

problem 2-7 problem 2-8 problem 2-9

problem 2-10problem 2-11





problem 2-18 problem 2-19





CHAPTER 3 STATICALLY DETERMINATE BEAMS

The abstract of the chapter The objective of this chapter is to present the analysis of support reactions and internal forces

that may develop in beams under the action of coplanar systems of external forces and the construction of internal force diagrams. The analysis of single span beams is the foundation of the analysis of multispan beams; the content pertinent to single span beams should be soundly mastered. Pay attention to the characteristics of constructing bending moment diagrams of straight members, which are plotted by using the method of superposition segment by segment.

3.1 The Analysis of Single Span Beams

In engineering practice, single span beams have a variety of application. The analysis pertaining to single span beams is the foundation of statically determinate multispan beams, frames and so forth. The commonly used single span beams are the following three kinds: (1) simply supported beams or simple beams [Fig.3.1 (a)], (2) cantilever beams [Fig.3.1 (b)] and (3) overhanging beams [Fig.3.1 (c)].

(b)(a) (c)

Fig.3.1 Single span beams(a) simple beam; (b) cantilever beam; (c) overhanging beam

The chapter will review and discuss the calculation of internal forces and construction of their diagrams of statically determinate single span beams.

3.1.1 Internal forces and their sign convention

For a planar member, it is generally subjected to shear forces and bending moments as well as axial forces under the action of external loads.

The internal force component oriented in the direction of the centroidal axis of a beam at the section under consideration is referred to as the axial force. Positive axial forces make the beam in tensile state while negative in compressing state [Fig.3.2 (a)]

The internal force component oriented in the direction perpendicular to the centroidal axis of a beam

38


A

B

C0.6m

6m

10kN10kN

(f)

1k

N/m

q=

3m

(e)

P B

Paa

a

Problem 3-1

C

A

A BC D

qM

A BC

(a) (b)

diagramM diagramM

diagramQ diagramQ

BAC D

diagramM

(c)

diagramQ

BA

diagramQ

C

(d)

diagramM

problem 3-2 (contd)

70 Chapter 3 Statically Determinate Beams

diagramQ diagramM diagramQ diagramM

(e)

C

BB

AA

(f)

Problem 3-2

2

8ql

A C B

2

8ql

q

(a)

2

8ql

q

A C B

2

8ql

(b)

q2

8ql 2

8ql

A C B

(c)

4Pl

P

A C B

(d)

P

A C B

(e)

A C B

(f)

2l

2l

2l

2l

2l

2l

2l

2l

2l

2l

2l

2l

4Pl

problem 3-3 (contd)

4Pl

4Pl

4Pl

4Pl

P


AD C E

B

3kN/m

2m 2m 2m 2m

(g)

2m 2m 2m 2m

(h)

2kN/m

DA B

CE

10kN 3kN

2m 2m 2m

(j)

2m 4m 4m 2m

(i)Problem 3-3

DA

C B E

4kN2kN2kN/m

2kN/m

A D C B

90o

q

AD

C

B

3m 3m

(b)

q

B

D

C

A

1m 1m 2m

(a)

Problem 3-4

1m 1m 2m


20kNA B

(a)

20kN m⋅

C10kN

D EF

2kN/m

3m 3m 1.5m 2m 2m 4m

(b)

2m 2m 2m 2m 2m 2m 2m

EDCBA

5kN/m

Problem 3-5

2.5m

10kN10kN

B CP

D E F

l

A

Problem 3-6

2l

2l

2l

2l l

B

2a

2a

2a

2a

2a

C D E FP

a

A

Problem 3-7

3.1 The Analysis of Single Span Beams 39

at the section under consideration is called the shear force (or, simply, shears). As shown in Fig.3.2 (b), the shears are considered to be positive when they tend to make the portion of the member on the left of the section rotate clockwise and vice versa.

The internal couple which is the moment about (the centroid of the cross section of a beam at) the section under consideration of all the internal forces is termed the bending moment. The positive bending moment is shown in Fig.3.2 (c). The bending moments are considered to be positive when they tend to bend a horizontal beam concave upward, causing compression in the upper fibers and tension in the lower fibers of the beam at the section and vice versa.

3.1.2 The method of computing internal forces— method of sections

The method of sections is an elementary method to determinate the internal force components in a member. The method means to pass an imaginary section through the member and cut the member into two parts and then consider the free body of the either side of the section. We may obtain the three internal force components at the section by equilibrium equations of the free body.

Now we will discuss the method of sections by concrete examples.

Example 3-1

Fig.3.3 (a) shows a simple beam, determine the internal forces at sectionsC , and . 1D 2DSolution

1. Before determining the internal forces the reactions can be obtained by considering the equilibrium conditions of the free body of the entire beam [see Fig.3.3 (a)].

0X =∑ ，， 10 0AX − = 10kN( )AX = →

，， 0AM =∑ 2 2 1 10 0.2 8 0BY× × + × − × = 0.25kN( )BY = ↑

，， 0BM =∑ 2 2 7 10 0.2 8 0AY− × × − × + × = 3.75 kN( )AY = ↑

Check: 0Y =∑ ， 2 2 3.75 0.25 0× − − =

(c)

−M M

+M M

(b)

−Q Q

Q Q+

(a)

+

−

N N

N N

Fig.3.2 Sign convention of internal forces(a) sign convention of axial force; (b) sign convention of shear force; (c) sign convention of bending


2. The internal forces at section C An imaginary section is passed through section C, cutting the beam into two portions, AC and CB. The

portion AC shown in Fig.3.3 (c), which is to the left side of the section, is used here to compute the internal forces. By utilizing the three equilibrium conditions, the internal forces can be determined.

0X =∑ , 10 0CN+ = , 10kNCN = −

0Y =∑ , 3.75 2 2 0CQ− × − = , 0.25kNCQ = −

0CM =∑ , 3.75 2 2 2 1 0CM× − × × − = , 3.5kN mCM = ⋅ (tension in the lower fibers)

The negative and indicate that their real sense are opposite to those shown on the free CN CQ

AX C

AY BY

(b)(c)

3.75kN

10kN

CQ

CMCN

A

2kN/m0.2me =

10kNBA

C 1D2m

2D4m2m

1DQ

1DM1DN

1D

1DQ

1DM1DN

1D 2DB

10kN

0.25kN(d)

(e)

(f)

(g)2DQ 2DM

2DN1D 2D10kN

2DQ

2DM2DN

2D

0.25kN

B

C

A

3.75kN

10kN

C

A

3.75kN

10kN

2kN/m

10kN

2m4m2mC

0.2me = BA2kN/m

2kN/m

1D 2D

(a)

2kN/m

Fig.3.3 Diagrams of example 3-1


CMbody, i.e., axial force is a pressure. The shear is really negative. The sense of is positive means that its actual direction is the same as that shown on the free body, that is, CM causes tension in the lower fibers of the beam at section C.

3. The internal forces at section 1DAn imaginary section is passed through section , cutting the beam into two portions, 1D 1AD and . The portion 1D B 1AD shown in Fig.3.3 (d), which is to the left side of the section, is used here to

compute the internal forces. By utilizing the three equilibrium conditions, the internal forces can be determined.

0X =∑ 110 0DN+ = 1 10kNDN = −, ，

0Y =∑ , ， 13.75 2 2 0DQ− + × + = 1 0.25kNDQ = −

0DM =∑ , ， 13.75 6 2 2 5 0DM× − × × − = 1 2.5kN mDM = ⋅ ( tension in the lower fibers)

The meanings of the internal forces are the similar as the above. 4. The internal forces at section 2DAn imaginary section is passed through section , cutting the beam into two portions, 2D 2AD and . The portion 2D B 2AD shown in Fig.3.3 (e), which is to the left side of the section, is used here to

compute the internal forces. By utilizing the three equilibrium conditions, the internal forces can be determined.

0X =∑ 210 10 0DN− + = 2 0DN =，，

0Y =∑ , ， 23.75 2 2 0DQ− + × + = 2 0.25kNDQ = −

, ，2 0DM =∑ 23.75 6 2 2 5 0DM× − × × − = 2 0.5kN mDM = ⋅ ( tension in the lower fibers)

The meanings of the internal forces are similar as the foregoing. The selection of the free body can be the either side of the section under consideration when use the

method of sections. For instance, in above example we can select the left side of the section under consideration or right side of the section as well. Although either of the two sides of the beam can be employed for computing internal forces, we should select the side that will require the least amount of computational effort, such as the side that does not have any reactions acting on it or that has the least number of external loads and reactions applied to it. Based on this argument, select segment [Fig.3.3 (f)] and segment [Fig.3.3 (g)] can more easily find internal forces at sections and respectively. It is recommended for readers to check them.

1D B2D B lD 2D

The procedure for determining internal forces at a specified location on a beam can be summarized as follows:

(1) Pass an imaginary section perpendicular to the centroidal axis of the beam at the section where the internal forces are desired, thereby cutting the beam into two portions. Then select any one of the two


portions (commonly the simplest portion) as the free body. (2) Draw the free-body diagram upon which reactions and applied loads and desired internal forces

should be actually imposed. (3) Apply the equilibrium conditions of the free body and determinate the desired three internal

forces. Obviously, in order to establish the correct equilibrium equations so as to obtain correct internal forces,

the key step is presenting the free-body diagram correctly. Following consideration has to take when present free-body diagrams:

(1) The free body should be isolated completely with its restraints and all the restraints should be substituted by their corresponding constraint forces.

(2) The constraint forces should meet the characteristics of restraints. For example, while remove one roller support, one hinged support and one fixed support, one support reaction, two support reactions and three support reactions should be imposed on the free body respectively; while cut one link, one hinge and one rigid joint, a axial force, a axial force and a shear, and a axial force and a shear and a couple should be exerted on the free body respectively.

(3) All the forces acting on the free body should be indicated actually. Do not leave out the forces directly exerting on it and do not superimpose the forces not directly exerting on it. The forces exerting on a free body can be classed as two groups. One is the externally applied loads and the other is internal forces exerted on by the other portion corresponding to the free body.

(4) The presentation of a free body should be clear and accurate. Generally, given forces ought to indicate according to their actual locations and magnitudes and directions, whereas unknown forces should be indicated at their locations in their positive directions. Based on this regulation, if consequent unknowns are positive they are actually positive; if some of the unknowns are negative they are actually negative, their direction indication are just in the opposite direction. Therefore, you can avoid the ambiguity caused by directions of unknown forces.

Above mentioned procedure of method of sections is a normal progress to calculate internal forces. Its counterpart method, which does not need free-body diagram and may be developed by the definition of internal forces, can be stated as following:

Axial force: The internal axial force on any section of a beam is equal in magnitude but opposite in direction to the algebraic sum (resultant) of the components in the direction parallel to the axis of the beam of all the external loads and support reactions acting on either side of the section under consideration.

According to the sign convention adopted in the preceding paragraphs, if the portion of the beam to the left side of the section is being used for computing the axial force, then the external forces acting to the left are considered positive, whereas the external forces acting to the right are considered to be negative. If


the right portion is being used for analysis, then the external forces acting to the right are considered to be positive and vice versa.

Shear: The shear on any section of a beam is equal in magnitude but opposite in direction to the algebraic sum (resultant) of the components in the direction perpendicular to the axis of the beam of all the external loads and support reactions acting on either side of the section under consideration.

If the left portion of the beam is being used for analysis, then the external forces acting upward are considered positive, whereas the external forces acting downward are considered to be negative. If the right portion has been selected for analysis, then the downward external forces are considered positive and vice versa.

Bending moment: The bending moment on any section of a beam is equal in magnitude but opposite in direction to the algebraic sum of the moments about the centroid of the cross section under consideration of all the external loads and support reactions acting on either side of the section.

If the left portion is being used for analysis, then the clockwise moments are considered to be positive, and the counterclockwise moments are considered negative. If the right portion has been selected for analysis, then the counterclockwise moments are considered positive and vice versa.

Example 3-2

When the reactions of example 3-1 [Fig.3.3 (a)] are calculated try to determine the internal forces at sections , and again by employing the counterpart of method of sections. 1D 2DC

Solution

1. The internal forces at section C

The portion (from AAC to ), which is to the left side of the section, is used here to compute the internal forces.

C

Axial force: Considering the external forces acting to the left to be positive, we write 10kNCN = −

Shear: Considering the external forces acting upward to be positive, we write 3.75 2 2 0.25kNCQ = − × = −

Bending moment: Considering the clockwise moment of external forces about to be positive, we write

C

3.75 2 2 2 1 3.5kN mCM = × − × × = ⋅

2. The internal forces at section 1DA1ADThe portion (from to ), which is to the left side of the section, is used here to compute the

internal forces. 1D

Axial force: Considering the external forces acting to the left to be positive, we write


1 10kNDN = −Shear: Considering the external forces acting upward to be positive, we write

1 3.75 2 2 0.25kNDQ = − × = −


1D

1 3.75 6 2 2 5 2.5kN mDM = × − × × = ⋅

3. The internal forces at section 2DA2ADThe portion (from to ), which is to the left side of the section, is used here to compute

the internal forces. 2D

Axial force: Considering the external forces acting to the left to be positive, we write

2 10 10 0DN = − + =

Shear: Considering the external forces acting upward to be positive, we write

2 3.75 2 2 0.25kNDQ = − × = −


2D

2 3.75 6 2 2 5 10 0.2 0.5kN mDM = × − × × − × = ⋅

B1BDIt is clear that in the above computations if we select right segment (from to ) and segment 1D2BD B (from to ) to compute the internal forces at section and respectively, the analysis

will be more easily. It is recommended for readers to check them. 2D 1D 2D

3.1.3 Relationships between loads, shears and bending moments

The construction of internal force diagrams can be considerably expedited by using the basic relationships that exist between the loads, the shears, and the bending moments.

1. Differential relationships To derive these relationships, consider a beam subjected to an arbitrary loading, as shown in Fig.3.4

(a). All the external loads shown in this figure are assumed to be acting in their positive directions. As indicated in this figure, the external distributed and concentrated loads acting downward (in the positive y direction) are considered positive; the external couples acting clockwise are also considered to be positive and vice versa. Next, we consider the equilibrium of a differential element of length , isolated from the beam by passing imaginary sections at distances

dxx x dx+ and from the origin A, as shown in Fig.3.4

(a). The free-body diagram of the element is shown in Fig.3.4 (b), in which Q and M represent the shear and bending moment, respectively, acting on the left face of the element (that is at distance x from the origin A), and denote the changes in shear and bending moment, respectively, over the distance

. As the distance is infinitesimally small, the distributed load q acting on the element can be dQ dM

dx dx


considered to be uniform of magnitude . In order for the element to be in equilibrium, the forces and couples acting on it must satisfy the two equations of equilibrium,

)(xq

0 0M =∑0Y =∑ and . The third equilibrium equation, 0X =∑ , is automatically satisfied, since no horizontal forces are acting on the element. Applying the equilibrium equations the following formulae can be obtained.

(1) Distributed loads Considering the equilibrium of an arbitrary free body of a differential element of length isolated

from the segment subjected to distributed loads of the beam, we write dx

0

0, ( )

0, ( )

dQY qdx

dM

a

M Q bdx

⎫= = − ⎪⎪⎬⎪= =⎪⎭

∑

∑ (3-1)

Combine Eqs. (3-1) (a) and (b), we obtain 2

2

d M qdx

= − (3-2)

Above differential relationships represent the following geometric meanings: Eq. (3-1) (a) can be expressed as: Slope of shear diagram at a point equals to the intensity of

distributed load at that point but opposite sign. Eq. (3-1) (b) can be stated as: Slope of bending moment diagram at a point equals to the shear at that

point. Eq. (3-2) can be presented as: Curvature of bending moment diagram at a point equals to the intensity

of distributed load at that point but opposite sign.

q

M d M+

Q d Q+M

Q

dxy

x

(b)

ELM

ELQ

y

xE

P

ERM

ERQ

(c)

A BC D E F

mPq

(a)

FLM

FLQ

y

m M FR

xF FRQ

(d)



Based of above differential relationships, the internal force diagrams located in the segments imposed by distributed loads possess the following geometric characteristics:

At the segment imposed by uniformly distributed loads (q=constant) Q diagram is a linear function of x, the curve of which is an inclined straight line. If consider upward

direction as positive ordinate of Q and q acting downward in the segment, the Q curve will be straightly inclined downward from the left to the right of the segment.

While M diagram is a quadratic function of x, the moment curve is a parabolic curve, which is concave upward for a downward load q.

At no loading segment (q=0) Q is a constant and its diagram is a horizontal line. While M diagram is a linear function of x, the moment curve is an inclined straight line, the inclined

sense of which is the same as the sign of Q. (2) Concentrated loads The relationships between the loads and shears derived thus far [Eqs. (3-1) through (3-2)] are not valid

at the points of application of concentrated loads, because at such a point the shear changes abruptly by an amount equal to the magnitude of the concentrated load. To verify this relationship, we consider the equilibrium of a differential element that is isolated from the beam of Fig.3.4 (a) by passing imaginary sections at infinitesimal distances to the left and to the right of the point of application E of the concentrated load P. The free-body diagram of this element is shown in Fig.3.4(c). Applying the equilibrium equations we write

0, ( )

0, ( )ER EL

E EL ER

Y Q Q P a

M M M b

⎫= = − ⎪⎬

= = ⎪⎭

∑∑

(3-3)

Eq. (3-3) (a) can be expressed as: The magnitudes of shears at the two sides of the point of application of concentrated load are unequal. The shear difference at the two sides is equal to the magnitude of P. Therefore, the shear diagram will have abrupt change at the point of application of concentrated load P by an amount equal to the magnitude of the concentrated load P.

Eq. (3-1) (b) can be stated as: The bending moments at the two sides of the point of application of concentrated load are identical. Note that because of the abrupt change in the shear diagram at such a point, there will be an abrupt change in the slope of the bending moment diagram at that point. That is, there are cusps (angle points) on the moment diagram at the locations where concentrated loads are applied on. When concentrated loads are downward the cusps are downward as well.

(3) Couples or concentrated moments Although the relationships between the loads and shears derived thus far [Eq. (3-1) (a), Eq. (3-2) and

Eq. (3-3) (a)] are valid at the points of application of couples or concentrated moments, the relationships


between the shears and bending moments as given by Eqs. (3-1) (b) and (3-3) (b) are not valid at such points, because at such a point the bending moment changes abruptly by an amount equal to the magnitude of the moment or the couple. To derive this relationship, we consider the equilibrium of a differential element that is isolated from the beam of Fig.3.4 (a) by passing imaginary sections at infinitesimal distance to the left and to the right of the point F of application of the couple m. The free-body diagram of this element is shown in Fig.3.4 (d). Applying the equilibrium equations we write

0, ( )

0, ( )EL ER

F ER EL

Y Q Q a

M M M m b

⎫= = ⎪⎬

= = + ⎪⎭

∑∑

(3-4)

Eq. (3-4) (a) can be expressed as: The magnitudes of shears at the two sides of the point of application of concentrated moment m are equal; thusly the shear diagram will not change at the two sides.

Eq. (3-4) (b) can be stated as: The magnitudes of bending moments at the two sides of the point of application of concentrated moment m are unequal. Their difference at the two sides is equal to the magnitude of m. Thereby, the moment diagram will have abrupt change at the point of application of concentrated moment m by an amount equal to the magnitude of the concentrated moment m, whereas the slope of the moment diagram will not change since the constant value of shears at the two sides of the point, i.e., the slopes at the two sides remain parallel.

2. Integral relationships To determine the change in shear and bending moment between points A and B along the axis of the

member (see Fig.3.5), we integrate Eq. (3-1) from A to B to obtain the following integral relations between loads and internal forces [Eq. (3-5)].

( ) ( )

( )

B

B A AB

B A A

Q Q q x dx a

M Qdx b

⎫= − ⎪⎬⎪= + ⎭

∫∫

(3-5) M

These equations imply that: (1) The shear at end B is equal to the difference

between the shear at end A and the resultant of load q between A and B, or the change in shear between sections A and B is equal to the area of the distributed load diagram between sections A and B.

(2) The bending moment at end B is equal to the sum of that at end A and the area of the shear diagram between A and B, or the change in bending moment between sections A and B is equal to the area of the shear diagram between sections A and B.

If not only the distributed but also concentrated loads are exerted on segment AB, Eq. (3-5) may be written as

( )q xM B

BAM

AQ BQA

Fig.3.5 Segment AB of a member


( ) ( )

( )

B

B A AB

B A A

Q Q q x dx P a

M M Qdx m b

⎫= − − ⎪⎬⎪= + + ⎭

∑∫∑∫

(3-6)

Where represents the algebraic sum of the vertical concentrated loads between points A and B, and is the sum of the concentrated moments exerted on segment AB. The downward P and clockwise

m are assumed to be positive in the equations.

P∑m∑

3.1.4 The construction of internal force diagrams

An internal force diagram is the diagram on which the variation law along the location of member sections of the internal force is clearly represented. On such diagrams, the abscissas represent the axial lines of members, whereas the ordinates denote the internal force values of member sections (i.e., plot the internal force values in the direction perpendicular to the axial lines of members). The regulations of constructing internal force diagrams are as followings:

The ordinates of bending moment diagrams have to be plotted on the sides where the fibers of the members are under tensile action, and no sign indication is necessary on bending moment diagrams. Shear and axial force diagrams can be depicted on any sides of the members, but sign indication must be made (generally, for horizontal members positive shears and axial forces are plotted on the top sides of the members).

The basic differential relationships that exist between the loads and internal forces tell us that there is direct relationship between the shape of internal forces and load distributions. In the segments imposed by distributed loads or no load, the curves of internal forces are continuous; while at the points where loads vary abruptly, internal force curves vary abruptly as well. Utilizing these features, we could depict internal force diagrams very soon. The key points to construct them are stated as followings:

1. Select control sections The control sections (or points) indicate the sections at which loads vary discontinuously, such as the

starting and ending points of distributed loads, the locations of concentrated forces and couples. The control sections are the sections where internal force curves vary discontinuously as well.

2. Determinate the internal force values of control sections The primary method to determinate internal forces is method of sections. See subsection 3.1.2. 3. Construct internal force diagrams segment by segment Divide a member into several segments by control sections, and then consider the internal force values

of control sections as ordinates and lay off them based on corresponding abscissas, finally plot the internal force diagrams by means of their curve properties dominated by the basic differential relationships that exist between the loads and internal forces.


Now illustrate above statement through the following example.

Example 3-3

Depict the internal force diagrams for the simple beam shown in Fig.3.6 (a).

Solution

1. Find reactions Applying the equilibrium conditions of entire beam, we obtain

0X =∑ 0AX = ，

0AM =∑ 1 4 4 4 8 0BY× + × × − × =， 16 ， 10kN( )BY = ↑

0BM =∑ 8 16 7 4 4 4 0AY × − × − × × =，， 22kN( )AY = ↑

0Y =∑ 22 16 4 4 10 0− − × + =，

2. Construct shear diagram (1) Determinate the internal force values of control sections. The control sections are sections A, C,

D, E and B, at which loads vary discontinuously. The five sections divide the beam into four segments, AC, CD, DE and EB. As no load is applied to the segments AC, CD and EB, the shear curves in these segments are horizontal lines, which could be plotted just by one numerical value of their shears. As the uniformly distributed loads are applied in segment DE the shear curve in this segment is an inclined straight line, which needs two numerical values of the shears to be plotted.

22kNA CL AQ Q Y= = =

22 16 6kNCR D AQ Q Y P= = − = − =

10kNE B BQ Q Y= = − = −

(2) Construct Q diagram Take axial line AB as the abscissa axis; indicate the locations of control sections (A, C, D, E and B);

lay off the ordinates at points A and (just to the left of section C) by the magnitude of 22kN upwardly, we obtain points and ; do so at points (just to the right of section C) and D by the magnitude of 6kN upwardly, we get points and ; do so at points E and B by the magnitude of 10kN downwardly, we obtain points and . Then connect points

leftC1C rightC1A

2C 1D1A1E 1B and by a horizontal line; do

so between points and , 1C

2C 1D 1E 1Band respectively; and connect points and 1D 1E by a straight, sloping line (downward to the right). Finally, indicate positive sign on upward side and negative sign on downward side of the abscissa. The final shear diagram will be completed as shown in Fig.3.6 (b)

3. Construct moment diagram (1) Determinate the magnitude of bending moments of control sections. The control sections are

section A, B, C, D and E as well. The magnitude of the bending moments is:


0AM =

22 1 22kN mCM = × = ⋅ (Tension in the lower fibers)

22 2 16 1 28kN mDM = × − × = ⋅ (Tension in the lower fibers)

10 2 20kN mEM = × = ⋅ (Tension in the lower fibers)

0BM =

A B

C D E

1A

F

16kN 4kN/m

0AX =

2m 2m 2m1m 1m

22kNAY = 10kNBY =

22 221C

2C6

1D

AB

C DEG

1B1E

6

10 101.5m

(kN)Q

C DA BEG F

221C

1A

281D 32.5

1G32

1F

201E

1B (kN m)M ⋅

(b)

(c)

(a)

(d)

16kN 4kN/m

22kN 1m 1m

AG

x 0Q =

maxM



(2) Construct M diagram On downward side of the abscissa axis, lay off the values of the bending moments of each control

sections(A, C, D, E and B) by their corresponding magnitude (0, 22, 28, 20, 0), whose corresponding points are ( 1A , , ,1C 1D 1E 1Band ) shown as in Fig.3.6 (c).

1 1ACIn the segments to which no load is applied (i.e., segments AC, CD and EB), connecting , and

1 1C D1 1E B by straight sloping lines will finish the construction of bending moment diagram about these

segments of the beam. In the segment DE to which uniformly distributed loads is applied, the bending moment diagram

should be a quadratic parabola. Three points are needed to determinate the parabola. Since we only have two points ( and 1D 1E ), another additional point is necessary. We may take the value of bending moment at the middle point F of segment DE or the maximum value of the bending moments at the point G where the shear is equal to zero.

maxM

The two bending moment quantities are calculated as following: 22 4 16 3 4 2 1 32kN mFM = × − × − × × = ⋅ (Tension in the lower fibers)

Consider AG as a free body shown in Fig.3.6 (d); assume the maximum value of the bending moments occurs at section G at a distance x from the starting point of distributed loads, applying method of sections, we obtain

maxM

22 16 0GQ q= − − =x

22 16 1.5m4

x −= =

2

max1.522 3.5 16 2.5 4 32.5kN.m

2M = × − × − × =

Now we can lay off the ordinate at point F downwardly by the magnitude of , obtain point ; or do so at point G downwardly by the magnitude of

1F32kN m⋅

32.5kN m⋅ , obtain point .Then connecting the three points , and

1G1D 1F 1E or points , and 1D 1G 1E by a parabola will accomplish the construction of

the bending moment diagram of the beam as shown in Fig.3.6 (c). 4. The check of the shape characteristics of internal force diagrams By analyzing the load diagram, shear diagram and bending moment diagram shown in Fig.3.6 (a), (b)

and (c) respectively, shear and bending moment curves in segments AC, CD and EB, to which no load is applied, are horizontal and inclined lines respectively. Since a downward concentrated force of16 magnitude acts at section C, the shear diagram decreases abruptly from to 6kN at this section; while bending moment curve is not derivable at the location of this section, that is, there are two slopes or there is a downward cusp at the location of the section for the bending moment curve. In segment DE, the beam is subjected to a downward uniformly distributed load of . Because the load intensity is

22kNkN

4kN / m


constant and positive in segment DE, the shear diagram in this segment is a straight line with negative slope; whereas the bending moment diagram in this segment is a second-order upward-concave parabola with a minus constant curvature of . Note that the bending moment curve should be derivable at points

and 4kN / m

1D 1E , i.e., there is only one slope or tangent line at each of the two points for the curve.

3.2 Construction of Bending Moment Diagram by Principle of Superposition for Straight Members

The principle of superposition is extremely convenient in structural analysis. If a structure is subjected to a variety of loads, the displacements in the structure vary linearly with the applied loads, that is, any increment in displacement is proportional to the loads causing it, and if all deformations of a structure are small enough so that the resulting displacement of the structure does not significantly affect the geometry of the structure and hence do not change the initially acting property of the loads in the members. Under such conditions, reactions, internal forces and displacements due to the loads can be obtained by utilizing the principle of superposition.

The principle can be expressed as, if a structure is linearly elastic, the forces acting on the structure may be separated or divided into some expediently individual forms and then the structure may be separately analyzed for each individual forms. The final results can then be obtained by adding up the individual results.

Now we will discuss how the principle of superposition be used to facilitate the construction of bending moment diagrams for straight members.

3.2.1 Superposition method of bending moments for simple beams

Fig.3.7 (a) shows a simply supported beam subjected to not only the uniformly distributed loads in the span but also the end couples. By principle of superposition, divide the loads applied on the beam into two groups, one is the two end couples [Fig.3.7 (b)] and the other is uniformly distributed loads [Fig.3.7 (c)]. If the beam undergoes the end couples only, the moment diagram is simply a straight line as shown in Fig.3.7 (b), it is denoted by M diagram; and if the beam is subjected to the uniformly distributed loads in the span only, the bending moment diagram 0 ( )M x is a second-order parabola and may be plotted as shown in Fig.3 7 (c). Then the final bending moment diagram of the beam may be obtained by the principle of superposition. That is, depicting the 0 ( )M xM diagram first, then the ordinates of diagram being superimposed on the M diagram will complete the final bending moment diagram of the beam as shown in Fig.3.7 (d).

Note that it is the ordinates not the figures of the two diagrams which are superimposed together. The ordinate ( )M x at any arbitrary section shown in Fig.3.7 (d) should meet the following formula.

3.2 Construction of Bending Moment Diagram by Principle of Superposition for Straight Members 53

0( ) ( ) ( )M x M x M x= +

A B′ ′0 ( )M x are perpendicular to the beam axis AB not the dashed line Where the ordinates of .

3.2.2 Superposition method segment by segment

Now extend the superposition method discussed in above subsection to the construction of bending moment for any arbitrary segment of a straight member. Fig.3.8 (a) shows a simply supported beam subjected to the uniformly distributed loads only in segment AB. The bending moments AM at section A and BM at section B are determined by the method of sections. Next we will discuss how to construct the bending moment diagram of segment AB by utilizing principle of superposition.

q

(b)

(c)

(a)

(d)

MqM

Take segment AB of the beam shown in Fig.3.8 (a) as an example and compare the free body diagram of the segment [Fig.3.8 (b)] with the simply supported beam in Fig.3.8 (c). We may find that in these two cases both the load q and the end couples are the same. By employing the equilibrium equations, we also find 0

A AY Q= and . Since the stressed state of segment AB [Fig.3.8 (b)] is the same as that of the simple beam [Fig.3.8 (c)], the bending moment diagrams of these two cases must be identical. Thereby,

0BY Q= − B

A′0M

M

B′

BMAM

A B

BMAM

A B

BQAQ

0BY0

AY

M

A B

q

A B

BMAM q

BA

BA

MAM BA

B

M

q

BAM 0

A B(d)

A′0

M MM MAB

B′M

(b)

(c)

(a)

Fig.3.7 Construction of moment diagram by superposition method

Fig.3.8 Construction of moment diagram for arbitrary segment by superposition method


the superposition method used for constructing bending moment diagram of the simple beam AB [Fig.3.8 (b)] can be employed to construct the bending moment diagram of the segment AB shown in Fig.3.8 (a). Therefore, for a segment of a member, the moment diagram may be constructed as follows.

AM BM at section A and Firstly, determine the end moments at section B of the segment and take the end couples AM BM and as ordinates (indicate on tension side of the segment), draw a straight dashed line which represent the M diagram [Fig.3.8 (d)].

Secondly, take the dashed line as the base line, superimpose the ordinates (in the direction perpendicular to AB) of the M0M diagram of segment AB [Fig.3.8 (d)]. diagram, we accomplish the final

Example 3-4

Construct the bending moment diagram for the overhanging beam shown in Fig.3.9 (a).

Solution

It is no necessary for constructing bending moment diagram to find reactions. 1. Determination of bending moments of control sections The sections A, B and D are control sections at which loads vary discontinuously, which divide the

beam into two segments AB and BD. By the method of sections we obtain the bending moments 3kN mAM = − ⋅ (Tension in the upper fibers)

0DM =

1 2 1 2kN mBM = − × × = − ⋅ (Tension in the upper fibers)

2. Constructing bending moments diagram segment by segment Take axial line AD as the abscissa axis; indicate the locations of control sections (A, B, and D); lay off

the ordinates at points A, B, and D on tension side of the beam by the magnitudes of AM BM, and

DM 1A 1Brespectively, we obtain points , and [Fig.3.8 (b)]. 1D1A 1BConnect points , and with dashed lines, then take the dashed lines as base lines and

impose the ordinates of the corresponding bending moment diagram 1D

0 ( )M x . The final bending moment diagram is shown as in Fig.3.9 (b).

Note that since a concentrated load of 4kN is applied at the middle of the segment AB, the bending moment diagram 0 ( )M x in segment AB should be that of a simple beam with the same span of AB and subjected to the same external load of segment AB (i.e., a simple beam with a span of 5m and subjected to a concentrated downward force of 4kN at the middle of the beam). While the bending moment diagram

0 ( )M x in segment BD should be that of a simple beam with the same span of BD and subjected to the same uniformly distributed load of segment BD (i.e., a simple beam with a span of 2m and subjected to uniformly distributed load of 1kN/m in the whole span of the beam).

3.3 The Analysis of Simply Supported Inclined Beams 55

0MKnowing from strength of materials, the bending moment value at the middle of segment AB

will be4 5 5kN m

4 4Pl ×

= = ⋅ , the total value of the bending moment at section C will be

(3 2) 5 2.5kN m2CM +

= − + = ⋅ (Tension in the lower fibers)

221 1 2 0.5kN m

8 8ql

= × × = ⋅0M at the middle of segment BD will be The bending moment value ,

the total value of the bending moment at the middle section E will be (2 0) 0.5 0.5kN m

2CM += − + = − ⋅ (Tension in the upper fibers)

The bending moment diagram in segment BD can be plotted by considering the segment as a cantilever beam and directly drawing the moment diagram with no need of method of superposition as well.

B

3.3 The Analysis of Simply Supported Inclined Beams

In the civil engineering construction, inclined beams shown as in Fig.3.10 (b) and (c), are commonly used as the analyzing models of stairs [Fig.3.10 (a)].

The axial line of an inclined beam is straight slope line; its slope angle is denoted by θ . The loads imposed to the inclined beams can be classified into two groups: (1) vertically distributed loads along the axial lines of inclined beams, such as deadweight loads; (2) vertically distributed loads along the horizontal

(a)A C E D

2m2.5m2.5m

4kN3kN m⋅ 1kN/m

(b)

A BC

D

1C

1D

1B

1E

5

32 (0.5)

(0.5)

1A

2.5

diagram(KN m)M

⋅

Fig.3.9 Diagram of example


lines (the projections of axial lines of inclined beams), such as live loads. In order to facilitate the analysis, vertically distributed loads along the axial lines of inclined beams [Fig.3.10 (b)] are usually altered to

vertically distributed loads along the horizontal projections of the beams [Fig.3.10 (c)]. The final analyzing results under the action of vertically distributed loads along the axial lines of inclined beams are the same as those under the action of vertically distributed loads along the horizontal projections of the beams. If the distributed loads are uniformly distributed the alternative relationship of this two kinds loads will be

(b) (c)(a) q

'cos cos

l qql q l q qθ θ

′′ ′= = ∴ =

In which, is vertically distributed loads per unit length along the axial lines of inclined beams, while is vertically distributed loads per unit length along the horizontal projections of the beams.

q′ q

Next we will discuss the analytical nature, by an example, of statically determinate inclined beams subjected to vertically uniformly distributed loads along their horizontal projections.

Example 3-5

Construct the internal force diagrams for the simply supported inclined beam shown in Fig.3.11 (a).

Solution

1. By employing equilibrium conditions of entire beam, the reactions will be 0, 0AX X=∑ =

102A B ( )M Y ql= =∑ ↑

10, ( )2B AM Y ql= = ↑∑

θθθ

q′

l′

l l

Fig.3.10 Simple inclined beam

3.3 The Analysis of Simply Supported Inclined Beams 57

02 2ql qlql − − =0Y =∑Check: ,

(a)

2. Determinate the internal forces on an arbitrary section C by employing method of sections An imaginary arbitrary section at a horizontal distance x from the origin A is passed through section C,

cutting the beam into two portions, AC and CB. The portion AC shown in Fig.3.11 (b), which is to the left side of the section, is used here to compute the internal force components. Note that since the axial line of the beam is slope straight line the action directions of axial force and shear are all inclined. When utilizing the equilibrium conditions to evaluate internal force components, in order to avoid solving simultaneous equations in the computations two projective equations, one of which is along the direction s tangent to the axial line and the other of which is along the direction r normal to the axial line of the inclined beam, should be employed to directly obtain the axial force and shearing force . Therefore, both reaction cN cQ

Fig.3.11 Diagram of example 3-5

(e)

sin2ql θ

B

A

diagramN

sin2ql θ

(d) diagramQ

cos2ql θ

cos2ql θ

A

B

(c) diagramM

A

B

218

ql

12BY ql=

B

C

q

l

12AY ql= x

0AX = A θ

q

sinqx θ qx(b)

cosqx θs

CN

CQ

r

M C

C

A θ

θsin

2ql θ 1

2ql

xcos

2ql θ


AY and resultant of uniformly distributed load qx ought to be decomposed along the direction s and r. Thusly, we obtain

21 10, 0, ( )2 2 2C C C

xM qlx qx M M q lx x= − − = =∑ − (a)

1 10, cos cos 0, ( 2 )cos2 2C Cr ql qx Q Q q l xθ θ θ= − + + = = −∑ (b)

1 10, sin sin 0, ( 2 )sin2 2S C Cql qx N N q l xθ θ θ= − + = = − −∑ (c)

3. Construct M, Q and N diagrams (1) M diagram Since all the forces are vertical, only the horizontal distances are involved in the bending moment

calculation, hence the bending moment at any arbitrary section C, denoted by CM , of the inclined beam is the same as that of a simply supported horizontal beam with the same span and subjected to the same load. The bending moment diagram is depicted as shown in Fig.3.11 (c).

(2) Q diagram Recalling from equation (b), the shape of the shear of the inclined beam is a linear function with

respect to the horizontal distance x from the origin A, which will be plotted by two numerical values of its ordinates. The two numerical values are

1 cos2AQ ql θ=0x = ,

1 cos2BQ ql θ= −x l= ,

When comparing the two shear values with those of its corresponding horizontal beam (the same span subjected to the same load), their difference is only a constant cosθ , i.e., the magnitude of shears of inclined beam is the projection of that of its corresponding horizontal beam along the normal direction of the axial line of the inclined beam. The Q diagram is shown in Fig.3.11 (d).

(3) N diagram Knowing from equation (c), the shape of the axial force of the inclined beam is a linear function with

respect to the horizontal distance x from the origin A, which will be depicted by two numerical values of its ordinates. The two numerical values are

1 sin2AN ql θ= −0x = ,

1 sin2BN ql θ=x l= ,

When comparing the two axial force values with those of its corresponding horizontal beam, their

3.4 The Restraint Force Calculation and Geometric Construction of Statically Determinate Multispan Beams 59

sinθdifference is only a constant , i.e., the magnitude of axial forces of inclined beam is the projection of shears of its corresponding horizontal beam along the tangent direction of the axial line of the inclined beam. The N diagram is shown in Fig.3.11 (e).

The example shows that: (1) For simply supported inclined beam, the method to calculate its reactions and internal forces is

still the method of sections and application of equilibrium conditions of free bodies. (2) Under the action of vertical loads, the reactions of a simply supported inclined beam are the

same as those of a simply supported horizontal beam with the same span and subjected to the same load. (3) Under the action of uniformly distributed loads, the bending moment diagram of a simply

supported inclined beam are the same as those of a simply supported horizontal beam with the same span and subjected to the same load.

(4) Under the action of vertical loads, a simply supported inclined beam will be imposed by axial force; the magnitude of shears and axial forces of the inclined beam is the two projections of shears of its corresponding horizontal beam along the normal and tangent directions of the axial line of the inclined beam respectively.

Finally, the additional explanation is needed to mention, which indicate that although the axial line of an inclined member is a straight slope line, the superposition method can be applied to plot bending moment diagrams for all segments with straight axial lines of the member.

3.4 The Restraint Force Calculation and Geometric Construction of Statically Determinate Multispan Beams

A statically determinate multispan beam is constituted by several single span beams connected by hinged joints each other based on the geometric construction rules of a stable system with no redundant restraint. This type of structure is a commonly used structural form in road bridges [Fig.3.12 (a)] and purlines of buildings [Fig.3.13 (a)], whose corresponding computing model is shown in Fig.3.12 (b) and Fig.3.13 (b) separately.

Generally, the reactions of a statically determinate multispan beam are more than three. The statically determinate multispan beam shown in Fig.3.14 (a) have five reactions, which is greater than the number of equilibrium equations provided by the free body of the entire beam. Obviously, the three equilibrium equations are not sufficient to determine the five unknown reactions at the supports for this multispan beam. However, the presence of the internal hinges at C and E yields two additional equations that can be used with the three equilibrium equations to determine the five unknowns. The additional equations are based on the conditions that internal hinges cannot transmit moments; that is, the moments at the internal hinged joints must be equal to zero. Therefore, when an internal hinge is used to connect two portions of a


structure, the algebraic sum of the moments about the hinge due to loads and reactions acting on the structure from either side of the hinge must be zero. Such equation is commonly referred to as the equation

of condition or construction. For the multispan beam of Fig.3.14 (a), there are two internal hinges at C and E which yield two equations of condition. For simplicity, we use the symbols , express the equations. Therefore, all five unknown reactions for the beam of Fig. 3.14(a) can be determined by solving the three equations of equilibrium plus two equations of condition ( and ).

A B C D

1680cm1680cm640cm 490cm490cm 700cm

(a)

0CM = 0EM =

0CM = 0EM =Although the total number of equilibrium equations and condition equations of a statically determinate

multispan beam is sufficient to determine its all reactions, the difficulty of solving simultaneous equations will sometimes occur. In order to avoid solving simultaneous equations the best procedure for determining reactions of a multispan beam can be stated as: first, analyze the geometric construction of the beam, then, find reactions in the order which is the reverse of that of its geometric construction.

The constituent characteristics of statically determinate multispan beams are that the ends of some constituent members serve as supports for the other constituent members. Therefore, the former is considered as the main portion of a multispan beam, and the latter as a subsidiary one. For example, in the

A CB D(b)

GA CB D(a)

E F

Top chord of roof

P urlinH I J

CB D E F I(b) HGA J

Fig.3.12 Multispan road bridge

Fig.3.13 Multispan purlin

640cm

3.4 The Restraint Force Calculation and Geometric Construction of Statically Determinate Multispan Beams 61

statically determinate multispan beam shown in Fig.3.14 (a), the overhanging beam ABC (the portion is attached to the foundation by one hinged support and one roller support to form a stable system) is regarded as the main one with respect to the subsidiary member CDE which is supported at end C by beam ABC; the beam CDE is the main one with respect to the subsidiary member EFG at the same time. The geometric construction relationship between members of the statically determinate multispan beam is presented by the figure shown in Fig.3.14 (b).

The figures shown in Fig.3.14 (b) through (c) give a better idea about the transmitting behavior of the forces of the entire beam: The load applied on the main beam produces no reactions and internal forces in the subsidiary portions, whereas the load applied on the subsidiary portion always gives rise to reactions and internal forces in the main portion of the beam; a multispan beam remains stable if only a subsidiary portion is removed, but becomes unstable when a main portion is removed.

When analyzing the multispan beam shown in Fig.3.14 (a), take analyzing sequence as the order which is opposite to that of the geometric construction, that is, first, calculate subsidiary portion EFG, then, compute portion CDE, finally, analyze the main portion ABC [Fig.3.14 (c)]. We can find the restraint forces

EX , and reaction by applying three equilibrium equations of the free body EFG [Fig.3.14(c)], EY FY

(a) A CB D E

FG

q

G

(b)

q

C

B

D

EF

A

(c) E

B

C D

E

q

0EX =G

F

A

0EX =

FYEY

0CX =

0CXAY

CY DY

=

BY

AX

Fig.3.14 Statically determinate multispan

C


EXthen and are considered as the loads which are applied on member CDE as shown in Fig.3.14(c). Similarly, the restraint forces

EYCX and and reaction CY DY may be obtained through the free body

CDE. Finally, the reactions AX , and may be obtained by applying three equilibrium equations of the free body ABC where

AY BYCX and are regarded as the external loads applied on. CY

In conclusion, it may be worth of mentioning that dividing a multispan beam into main and subsidiary portions will facilitate its analysis. The analysis may be implemented in an order which is the reverse of that of its geometric construction. The loads exerted on a subsidiary portion would be transmitted to its main portion through the connecting device between them. By the analysis of the subsidiary portion we may determine the restraint forces provided by the main portion, and these restraint forces (in opposite direction) turn to the load acting on the main portion. Ultimately, all restraint forces and reactions may be obtained successfully without solving simultaneous equations.

3.5 The Construction of Internal Force Diagrams of Statically Determinate Multispan Beams

1. The sign convention of the internal forces and construction convention of the internal forces diagrams of statically determinate multispan beams are the same as those of single span beams

2. The internal force calculation of a statically determinate multispan beam is, in fact, the internal force computation of its each individual single span beam. Connecting internal force diagrams of all of its individual single span beams together will accomplish the construction of the internal force diagrams of the entire multispan beam.

3. The relationships between loads, shears and bending moments derived from straight members and characteristics of internal force diagrams occurring on straight members are still suitable for constructing internal force diagrams of statically determinate multispan beams, which can be employed to check the correctness of the internal force diagrams of multispan beams and to depict the diagrams quickly. It should be realized that at the locations of middle hinges of multispan beams bending moments must be equal to zero ( ). 0M =

Example 3-6

Construct the internal force diagrams for the statically determinate multispan beam shown in Fig.3.15 (a).

Solution

1. Geometric construction analysis, draw construction relationship diagrams of members The geometric construction relationship between members of the statically determinate multispan

beam is expressed by the figure shown in Fig.3.15 (b). The beam ABC is fixed to the foundation by a fixed

3.5 The Construction of Internal Force Diagrams of Statically Determinate Multispan Beams 63

support to form the main portion of the statically determinate multispan beam. At the right end of member CE, there is a hinge E which will provide horizontal restraint for member EFG, but under the action of

CB D E FA G(a)

6kN 6kN2kN 2kN/m

6m2m2m2m 1m 1m

(b)B C

DE

GAF

(c)B C

DE

GAF

B C

6kN2kN

2kN/m

D

6kN

E

(d)

A FG

5kN 3kN

5kN 3kN32kN m⋅

11kN10kN 5kN

B CD

EA F G

32(e)

10

3

6 (9)

6

diagram(kN m)M

⋅

(f )

diagram(kN)Q

11 11

3 3

3 3

7

5

5 5

B C DE

FG

Fig.3.15 Statically determinate multispan beam of example 3-6

A


vertical load no horizontal restraint force will exist at hinge E. Therefore, removing the horizontal restraint at hinge E to location G would not change the stressing state of the statically determinate multispan beam. So doing yields a construction relationship diagram shown in Fig.3.15 (c), in which cantilever ABC and overhanging EFG are the main portions of the statically determinate multispan beam and simple beam CDE is a subsidiary portion supported on the two main portions.

2. Compute restraint forces and reactions In order to avoid solving simultaneous equations we should find restraint forces and reactions in the

order which is the reverse of that of the geometric construction of the beam. On this point of view, the restraint forces of beam CDE is first calculated, which are considered as the loads applied on beam ABC and EFG at the positions C and E (in opposite direction) respectively; then reactions of ABC or EFG separately. The final results are shown in Fig.3.15 (d).

Check: applying the equilibrium condition in vertical direction Y of the free body of the entire beam, we obtain

0Y =∑ , 6 2 6 2 6 11 10 5 0+ + + × − − − =

3. Construction of internal force diagrams As soon as the restraint forces and reactions are determined, the bending moment and shear diagrams of

these individual members may be plotted by method of superposition segment by segment; put these diagrams together, we obtain the bending moment and shear diagrams of the multispan beam. They are shown in Fig.3.15 (e) and (f) separately.

4. characteristics of internal force diagrams In Fig.3.15 (e) and (f), it is indicated that at hinges C and E the ordinates of bending moment diagram

are equal to zero; in the segments where no load is applied to, such as segments AB, BC, CD, DE and EF, the shape of moment diagrams are straight slope lines while that of shear diagrams are horizontal lines. All of these features of the diagrams coincide with the relationship existing between loads and shears and bending moments of straight members.

Example 3-7

Construct the bending moment diagram for the statically determinate two-span beam shown in Fig.3.16 (a) and then compare it with that of two neighboring simply supported beams.

Solution

1. Geometric construction analysis The geometric construction relationship existing between members of the statically determinate

two-span beam is expressed by the figure shown in Fig.3.16 (b). The beam BCD is the main portion of the beam. Whereas the beam AB is a subsidiary portion of the two-span beam.

3.5 The Construction of Internal Force Diagrams of Statically Determinate Multispan Beams 65

2. Compute reactions In order to avoid solving simultaneous equations, the reactions of subsidiary beam AB is first

calculated, one (at position B) of which is considered as the load (in opposite direction) applied on beam BCD; then reactions of BCD can be found easily. The final results are shown in Fig.3.16 (c).

1kN/m

6m0.9m5.1m

(a)CB

DA

AC

BD

(b)

1kN/m

A B

2.55kN 2.55kN

1kN/m2.55kN

BC

D

2.55kN6.9kN

Check: applying the equilibrium condition in vertical direction Y to the free body of the entire beam, we obtain

0Y =∑ , 1 12 2.55 6.9 2.55 0× − − − =

3. Construction of bending moment diagram

(d)

diagram(kN m)M

⋅A

C

3.25

2.7

D

3.15

(e)

4.5

diagram(kN m)M

⋅

A C D

4.5

Fig.3.16 Statically determinate two-span beam of example 3-7

( )c


The bending moment diagrams of individual beam AB and BCD may be plotted separately; put these diagrams together, we obtain the bending moment diagram of the two-span beam. It is shown in Fig.3.16 (d).

4. Comparison of bending moment diagrams Compare the bending moment diagram for the statically determinate two-span beam shown in Fig.3.16

(d) with that for two neighboring simply supported beams shown in Fig. 3.16 (e), we can find the difference between the two cases. The maximum bending moment in each span of the two-span beam is smaller (in a percentage of 28%) than that in the simple beams. The bending moment distribution of the two-span beam is more reasonable than that of the simple beams, because the negative bending moment due to the load acting on the overhang BC decreases the positive moment distributed in the span. The maximum and minimum moments for the two-span beam can be controlled by adjusting the length of the overhang

SUMMARY

The primary content of the chapter is the calculation of reactions and construction of internal force diagrams for both statically determinate single span beams and multispan beams.

Although the analysis of statically determinate single span beams has been learned in the course of strength of materials, it is yet the foundation of analysis of statically determinate multispan beams and rigid frames; the requirement now is that internal force diagrams of statically determinate single span beams must be constructed as soon as correctly possible. It should be realized that the more practice would enhance your sound proficiency about the construction of internal force diagrams. The key points of the chapter are stated as followings:

1. Calculating step: firstly, find reactions (except cantilever beam); then, determinate internal forces; finally, construct internal force diagrams.

The attention that the order to find the reactions is the reverse of that of their geometric construction should be paid for multispan beams

2. The method of sections is the elementary method to determinate the internal force components in a member. But the other form of method of sections should be applied skillfully. That is: The internal axial force at any section of a member is equal in magnitude, but opposite in direction, to the algebraic sum of the components in the direction parallel to the axis of the member of all the external loads and reactions acting on either side of the section. We consider it to be positive when the external forces tend to produce tension. The shear at any section of a member is equal in magnitude, but opposite in direction, to the algebraic sum of the components in the direction perpendicular to the axis of the member of all the external loads and reactions acting on either side of the section. We consider it to be positive when the external forces tend to push the portion of the member on the left of the section upward with respect to the portion on the right of the section. The bending moment at any section of a member is equal in magnitude, but


opposite in direction, to the algebraic sum of the moments about the section of all the external loads and reactions acting on either side of the section. We consider it to be positive when the external forces and couples tend to bend the member concave upward, causing compression in the upper fibers and tension in the lower fibers at the section.

3. Shear, bending moment, and axial force diagrams depict the variations of these quantities along the length of the member. Such diagrams can be constructed by determining and plotting the equations expressing these stress resultants in terms of the distance of the section from an end of the member. The construction of shear and bending moment diagrams can be considerably expedited by applying the differential relationships that exist between the loads, shears, and bending moments.

4. The practical method for constructing bending moment diagram is the method of superposition. The steps are: (1) dividing a member into several segments by control sections; (2) determining bending moments of the control sections; (3) in segment where there is no load applied on, directly connecting the ordinates with a straight line between two adjacent sections; (4) in segments where there are loads applied on, connecting the ordinates with a straight dashed line between two adjacent sections and imposing the ordinates of the corresponding bending moment diagram of a corresponding simple beam on the dashed line, which is considered as a base line. Shear and axial force diagrams can be depicted in the similar way.

5. Depict the ordinates of internal forces in the direction perpendicular to the axial line of the member. The ordinates of bending moment diagrams have to be plotted on the sides where the fibers of the members are under tensile state, and no sign indication is necessary on bending moment diagrams. Shear and axial force diagrams can be depicted on any sides of the members, but sign indication must be made (for horizontal members, positive shears and axial forces are generally plotted on the top sides of the members).


3-1 Why the moment diagram for a certain straight segment of a member may be constructed by the method of superposition of a simple beam with the same span and subjected to the same loads?

3-2 Why the bending moment diagram of an inclined beam subjected to vertical loads is the same as that of a horizontal beam with the same span and subjected to the same loads?

3-3 Would internal forces yield on the subsidiary portion when loads were applied on the main portion for a statically determinate multispan beam? Why?

3-4 Why, generally the maximum bending moment in each span of a multispan beam is smaller than that in a series of simple beams with the same spans?



3-1 Construct the bending moment and shear diagrams of following statically determinate single span beams or columns (please follow the regulation required by the chapter).

3-2 Check and revise the following internal diagrams by the differential relationship existing between the loads and shears and bending moments.

3-3 Construct bending moment diagrams of the following beams by using the method of superposition segment by segment and calculate the bending moment at section C.

3-4 Construct internal force diagrams of the following inclined beams and calculate the internal forces at sections C and D.

3-5 Calculate the reactions of the following statically determinate multispan beams and depict their internal force diagrams.

3-6 A concentrated load P is applied on the multispan beam, please construct the bending moment diagrams in the following three cases and compare them with each other: (1) P is only applied at the middle of segment CD; (2) P is only applied at the middle of segment AB; (3) P is only applied at the middle of segment EF.

3-7 Depict the bending moment diagram straightforwardly according the stressing characteristics and the differential relationship existing between the loads and shears and bending moments of the multispan beam.

2

2ql

A

q

C B

2l

2l

(a) (b)

4kN3kN/m

2m 2m 2mC

A BD

Problem 3-1 (contd)

C DA B

6kN1kN/m 2kN/m 10kN mi 8kN

4m 2m 2mA C B D

2m 4m 2m

(c)

4kN m⋅

(d)

CHAPTER 4 PLANE STATICALLY DETERMINATE RIGID FRAMES

The abstract of the chapter The chapter will discuss the analysis of plane statically determinate rigid frames. Firstly, present

the analysis of the support reactions; then, the internal forces and the construction of their diagrams. When finding the reactions of three-hinged frames the condition of moment at middle hinge, i.e.,

, must be used; when calculating the reactions of multispan and multistory rigid frames the geometric construction analysis must be made in order to proceed the analysis in the order which is the reverse of that of their geometric constructions; while computing internal forces the attention should be paid on the equilibrium of rigid joints where several members are connected together; while constructing bending moment diagrams the method of superposition segment by segment is still powerful.

0M =

In practical engineering, there is a variety of employment for plane statically determinate rigid frames, such as awnings in train or bus stations, car sheds [Fig.4.1 (a) and (b)], one-storey industrial workshops, warehouses [Fig.4.2 (a) and (b)] and the like. The analyzing for statically determinate rigid frames is the foundation of calculation of displacements of rigid frames and the base of analysis of statically indeterminate rigid frames as well. Therefore, the content of this chapter must be really mastered.

(a) (b)

C20 crushed stone concrete

Fig.4.1 Y-shaped rigid frame

The procedure to analyze a plane statically determinate rigid frame has the following steps: (1) find reactions; (2) calculate internal forces of control sections by using the method of sections; (3) construct internal force diagrams by employing the method of superposition.

73

74 Chapter 4 Plane Statically Determinate Rigid Frames

4.1 Geometric Construction and Characteristics of Plane Statically Determinate Rigid Frames

Rigid frames, usually referred to simply as frames, are composed of straight members (columns and beams) connected either by rigid (moment-resisting) connections or by hinged connections to form stable configurations. The frame shown in Fig.4.3 (a) is a portal frame, whose joints C and D are rigid joints. If all of the joints of the frame shown in Fig.4.3 (a) were changed into hinged joints, the system would be unstable truss; the dashed lines shown in Fig.4.3 (b) represent its movement. In order to make the system [shown in Fig.4.3 (b)] be stable a restraint must be added, for instance, a slope member is attached to it as shown in Fig.4.3 (c) to form a stable truss.

Comparing the frame shown in Fig.4.3 (a) with the truss shown in Fig.4.3 (c), it can be seen that the members of the frame are connected by rigid joints which are able to transmit bending moments and make the system be stable; while the members of the truss are connected by hinges which are only able to transmit forces but not moments, and in order to prevent the relative rotations of the members connected to

Ridge joint

Asphaltum with jute fiberA B

CD E

(a)

Fig.4.2 Three hinged rigid frame

C

D E

(b)BA

090090 090

090

P

A B

C D

(a)

P

B

D

A

C

(c)

Fig.4.3 Frame and truss(a) a gate-shaped frame; (b) an unstable system; (c) a truss

P

B

C D

(b)A

4.1 Geometric Construction and Characteristics of Plane Statically Determinate Rigid Frames 75

hinges inclined member has to be employed. Therefore, the members of rigid frames are able to transmit not only forces but also bending moments,

whe

rigid joints, although hinged connections are som

the columns shown in Fig.4.4 (a) and their disab

pressures and with no bending deformation. However, because of the rigid joints between the beam and the

reas those of trusses are only able to transmit forces; a frame structure can develop larger internal space with fewer members than a truss with the same dimension.

The members of frames are usually connected by etimes used. A rigid joint prevents relative translations and rotation of the member ends connected to it,

so the joint is capable of transmitting two rectangular force components and a couple between the connected members. Under the action of external loads, the member ends connected to one rigid joint of a frame have the same displacements, that is, their ends have the same translations and rotate the same angle, as dashed lines shown in Fig.4.3 (a) (a rectangular angle still remains rectangular under the deformation of the frame); whereas a hinged joint only prevents relative translations but not the rotations of the member ends connected to it, so a hinged joint is only capable of transmitting two rectangular force components between the connected members. Under the action of external loads, the member ends connected to one hinged joint of a structure only have the same translations, that is, their ends rotate different angles or they have relative rotations, as dashed lines shown in Fig.4.3 (c).

Due to the hinged connections between the beam andility of transmitting moments, under the action of the load distributed on beam CD, only the beam CD

yields bending moments and bending deformation, while the columns AC and BD are only subjected axial

columns shown in Fig.4.4 (b) and their ability of transmitting moments, under the action of the load distributed on beam CD, not only the beam CD but also the columns AC and BD develop bending moments and bending deformation. Contrasting the two situations, the maximum bending moment and bending

q

C D

Δ

A B

2

8ql

l(a)

q

A B

C D

(b)

2

8ql

Fig.4.4 Frames(a) a frame with hinged joints; (b) a gate-shaped frame

l

Δ


deformation of beam CD shown in Fig.4.4 (b) have reduced a lot in comparison with those of beam CD shown in Fig.4.4 (a), whose dashed curves represent the deformation configurations.

The geometric construction rules for statically determinate rigid frames must obey the rules of stable syste

g.4.5 (a) and (b)];

ig.4.5 (d) and (e)].

4.2 The Analyzing of Reactions for Statically Determinate Rigid Frames

Generally, in the analysis of a statically determinate rigid frame the reactions may be determined first, and t

(d)

ms. Commonly utilized plane rigid frames have the following types: (1) simply supported rigid frames and overhanging rigid frames [Fi(2) three hinged rigid frames [Fig.4.5 (c)]; (3) multispan and multistory rigid frames [F

(a)

(b)

(c)

(e)

Fig.4.5 Frames(a) simply supported frame; (b) Y-shaped frame; (c) three hinged frame;(d) multispan frame; (e) multistory frame

hen the internal forces on control sections are calculated; finally the internal force diagrams may be depicted. Due to the simplicity of simply supported and overhanging rigid frames, the following discussion will only involve the analysis of reactions of statically determinate three hinged and multispan rigid frames.

It is worth of emphasizing that the order of calculating reactions of a frame should be the reverse of

4.2 The Analyzing of Reactions for Statically Determinate Rigid Frames 77

that

Example 4-1

Calculate the reactions of the three hinged frame shown in Fig.4.6 (a).

Solution

There are totally four unknown support reactions on the frame, i.e.,

of its geometric construction.

C

AX and AY , BX and From the viewpoint of geometric construction, three rigid bodies (AC, BC and the foundation) joined pairwise by hinges A, B and C, which do not lie on the same straight line, form an internally stable system with no redundant restraint. Thus the steps in the determination of reactions of the frame are:

(1) to evaluate the reactions

BY .

AY and by taking the entire frame as a free body and then

applying the equations of moment equilibrium t hinges B and A (BY

abou 0BM =∑ and 0AM =∑ ), i.e.,

0BM =∑ , 1 02 AP Y l× + × = , ( )

2APY = − ↓

0AM =∑ , 1 02 BP Y l× − × = , ( )

2BPY = ↑

check: 0Y =∑ , 02 2P P− =

(2) to determine the reaction AX or BX by considering either the right or the left portion as a free

body and then employing the equations of moment equilibrium about hinge C, if taking the right portion as a free body we write

0CM = ,1 02B BX l Y× − × = , ( )

2 4B

BY PX = = ←

P

A B

2l

2l

2l

P2l

(a)

A

C

(b)

AX

AY

B BX

BY

Fig.4.6 A three hinged frame of example 4-1


AX(3) to evaluate reaction by taking the entire frame as a free body and then applying the equations

of force equilibrium in horizontal direction, i.e.,

, , 3 ( )4A BX X P P= − = − ←0X =∑ 0A BP X X+ − =

Example 4-2

Calculate the reactions of the two-span frame shown in Fig.4.7 (a).

Solution

There are totally four unknown support reactions [Fig.4.7 (b)] on the frame, i.e.,

(a)2kN/m

AX , BY , CX anThe geometric construction order can be stated as: in the beginning of geometric construction, a rigid frame BCEFG is fixed on the foundation by a hinged support C and a roller support B to form a stable system with no redundant restraint, and then the left portion is constructed. Thus the steps in the determination of reactions of the frame are:

(1) To evaluate the reactions and restraint forces acting on the subsidiary portion ADE [Fig.4.7 (c)] by

d CY .

A B C

DE F G

2m 2m 4m

4m(b)2kN/m

A B C

DE F G

AX

BY CY

CX

(c)2kN/m EY

Fig.4.7 A two-span frame of example 4-2

4kN

1kN

EX

4m

ED

AX

2mA B C

E F G

BY CY

CX2m 4m

4m

(d)

4.3 Determination of Internal Forces of Member Ends by Using The Method of Sections 79

considering the left portion as a free body, i.e.，

0EM =∑ , 1kN( )AX = − ← 2 2 1 4 0AX× × + × = ,

0Y =∑ , 2 2 0EY× − = EY 4kN( )= ↑

0AM =∑ , , 1kN( )EX = → 2 2 1 4 2 4 0EX× × − × + × =

(2) To determinate reactions on the main portion BCEFG [Fig.4.7 (d)] by taking the portion as a free body and then applying the equations of equilibrium, i.e.,

0X =∑ , 1kN( )CX = − →1 0CX + = ,

, 0,BM =∑ 4 2 1 4 4 0CY− × − × − × = 3kN( )CY = − ↓

, 0,CM =∑ 4 6 1 4 4 0BY− × − × + × = 7kN( )BY = ↑

check:

The foregoing steps for calculating reactions are the commonly used procedure for determinate unknown forces, that is, the calculating order should be the reverse of that of its geometric construction of the frame. However, the reactions of the example can be calculated in another way, that is, (1) taking the subsidiary portion ADE as a free-body and utilizing the condition that the bending moment at hinge E must be eq

0Y =∑ , 4 3 7 0+ − =

ual to zero, 0EM = , to find reaction AX

0EM = , 2 2 1 4 0AX× × + × = , 1kN( )AX = − ←

(2) then considering the entire frame [Fig.4.7 (b)] as a free body and employing the equations of equi tio

)CYlibrium to obtain the other all reac ns

0BM =∑ , 2 2 3 4 0CY− × × − × = , ( 3kN= − ↓

,

0CM =∑ 2 2 7 4 0BY− × × + × = 7kN( )BY = ↑,

0X =∑ , 1kN( )CX = − →0A CX X− = ,

4.3 Determination of Internal Forces of Member Ends by Using The Method of Sections

es an ei sub ted to sh

he bending moments on the tension side of a member. The sign convention for shears and axial forces are the hose of beams discussed . That is, a positithe section rotate clockwise and positive axial forces make the section in tensile state.

1. Internal forces of rigid fram d th r sign convention For a rigid frame, it is generally jec ear forces and bending moments as well as axial forces

under the action of external loads. It is no need for bending moments to regulate sign convention if depicting the ordinate of t

same as t previously ve shear makes


2. The member-end sec ns and the representation of internal forces of member ends at a rigid joint A rigid frame is composed of straight m

tioembers (columns and beams) with different axial-line

directions connected by rigid joints. There are thusly different member-end sections at a rigid joint. For example, there are tw -end sections id joint C of the frame show(a). We will represent the internal forces of a memb ds by using two subscripts, of which the first repre of the nearend of the member whereas the second represents that of the farend of the mem and are represented by

o member and 2C at rig n in Fig.4.8 1C er en

sents the label

1C 2Cber. For instance, the bending moments of member-end sections

CAM CBM (C is the nearend of member CA while A is the farend of the member) and (C is the nearend of member CB while B is the farend of the member) respectively, and the sh d axial forces are indicated by and and separately as shown in Fig.4.8 (b).

ethod of sections. We state the method as following again.

CA

ears an

CAQ CBQ , CAN CBN

N

0

C

1C

2C B

3. The calculation of internal forces of member-end sections The member-end sections here indicate the cross sections of members which are concurrent at rigid

joints, and we take this kind of sections as control sections. Then, the internal forces of the control sections are computed by using the m

The bending moment on the end section of a member is equal in magnitude, but opposite in direction,

A

8kN

6kN

8kN

6kN4m

3m

(a) (b)

CCAQ

CBQ

CBN

CAN

CAM

CBM

(c)

CAQ CAM

C

A

3m

8kN

6kN

CBQ

CBN

CBMB

6kN4m

(d)

8kN6kN

C

24kN m

C

8kN 24kN m

⋅6kN

(e)

Fig.4.8 A frame and free bodies of its members

⋅


to the algebraic sum of the moments about the section of all the external loads and restraint forces acting on either side of the section. We co it to be positive when al and restraint fmake the member rotate in the op site sense as yo s ar on the end seequal in magnitude, but oppo n direction, to th ponents iperpendicular to axis of the er of all the external loads and int forces actin

ection of a member is equal in magnitude, but opposite in direction, to the algebraic s e compone arallel to the axis the external loads and restraint forces acting on either side of the section. We consider it to be positive when the external forces and re ces tend to produce tens

We can also use free-body ms of each members of a f ulate ths, and in ses the calc ng the

free-body of rigid jo

N =

we obtain

⋅ (Tension lower fiber)

N

Check: take rigid joint C as a free body [Fig.4.8 (b)] and we find that the three equilibrium it

nsider the extern orces and couples po u a sumed. The she ction of a member is

site i e algebraic sum of the com n the direction the memb restra g on either side of

the section. We consider it to be positive when the external forces and restraint forces tend to make the section rotate clockwise. The axial force on the end s

um of th nts in the direction p of the member of all

straint for ion. diagra rame to calc e internal forces of the

control end section some ca ulation will be accomplished efficiently by taki diagrams ints.

The control sections for the frame shown in Fig.4.8 (a) are sections 1C and 2C at rigid joint C. in order to find the internal forces acting on the two sections we use method of sections with no free-body diagram first. For section 1C (consider its underside), we obtain

8 3 24kN mCAM = × = ⋅ (Tension in the right fiber)

8kNCA Q =

6kNCA

For section 2C (consider its right side),

6 4= × = in the 24kN mCBM

CBQ 6kN= −

0B = C

cond ions, 0X =∑ , 0Y =∑ and 0M =∑ are satisfied. ou can useY each free-body diagram shown in Fig.4.8 (c) to (e) to check the correctness of the

calculation.

ns with no free-body

E e 4-3

Calculate member end internal forces at rigid joints of the frame shown in Fig.4.9 (a).

Solution

The control sections for the frame shown in Fig.4.9 (a) are sections 1C , 2C and 3C at rigid joint C. In order to find the internal forces acting on the three sections we use method of sectio

xampl


diag f

N = −

(Tension in the upper fiber)

For section (consider its right side), we obtain

(Tension in the upper fiber)

For finding and

ram irst. For section 1C (consider its left side), we obtain

C

D1C

A

B 2C3C

1kN / m2kN

3kN

4m

(a) (b) (c)

(d)

1kN / m2kN

C

2m 2m

B

2mCBQ

CBN

CBM

3kN1kN / m2kN C

DB3C CAQ

CANCAM

CDQ

CD D2C2m

(e)

C

CAQ

CA

M CD

NM CA

N

22

2 3 60

Fig.4.9 The frame and the free bodies of its members of example 4-3

3kN

2kNCB

1 2 2kNCBQ = − × = −

1 2 1 2kN mCBM = × × = ⋅

2C0CDN =

3kNCDQ =

3 2 6kN mCDM = − × = − ⋅

CAN , CAQ CAM , we take rigid joint C as a free body [Fig.4.9 (e)] and obtain

left fiber)

2 3 5kNCAN = − − = −

2kNCAQ =

2 6 4kN mCAM = − + = ⋅ (Tension in the

The readers can check the results by using free-body diagrams shown in Fig.4.9 (b) to (d).

Example 4-4

Find member end internal forces at rigid joints C and D of the frame shown in Fig.4.10 (a).


Solution

6m

2CC D

A B

1C

2D

1D

3kN

/m

12kN

4kN

4m

DCQ

DCNDCM

0DBQ =

4DBN = −

0DBM =

D

(b)4kN(a)

Fig.4.10 The frame and the free body of rigid joint D of example 4-4

1. Reactions considering entire frame as a free body we will find the reactions as shown in Fig.4.10 (a).

2. Find member-end ternal forces at rigid joint C inFor section (consider its underside), we ob

CAQ = − ×

⋅ (Tension in the right fiber)

For section (consider its left side), we obtai

CDN = × −

CDQ

1C tain

4kNCAN =

3 4 12 0+ =

12 4 3 4 2CAM = − × + × × = −24kN m

2C n

3 4 12 0=

4kN= −

12 4 3 4 2 24kN mCDM = − × + × × = − ⋅ (Tension in the lower fiber)

3. Fin r-en al fo d joint D d membe d intern rces at rigi

For section (consider its right side), we obtain

For section D (consider its underside), we obtain 1

4kNDBN = −

0DBQ =

0M = DB

2

0DCN =

D


4kNDC = − Q

We can also use the free-body diagram [Fig.4.10 (b)] of rigid joint D to find the end forces

0DCM =

DCN ,

DCQ DCM DBN , DBQ DBM and after the end forces and are obtained. The results are the same as the above calculation.

4.4 Construction of Internal Force Diagrams of Statically Determinate Frames

4.4.1 Illustration of construction of internal force diagrams of statically determinate rigid frames

Internal force diagrams of a statically determinate rigid frame are bending moment, shear and axial force diagrams. Each kind of internal force diagram is composed of the diagrams for all the members of the frame. In general, by following the steps: (1) evaluating the internal forces at the both ends of each member, (2) determining the shape of the internal force diagram by the differential relationships existing between the internal forces and external loads, (3) employing the principle of superposition when the member is subjected to external loads as well as the end forces, the diagram may be conveniently depicted. Nexample to illustrate the procedure for cointernal force diagrams.

5

Construct internal force shown in Fig.4.11.

Solution

1. Calculation of reactions [Fig.4.11]

， + ×

ow we utilize an nstructing

Example 4-

diagrams of the frame

0∑ Y× × − × =AM = 4 5 1 4 2 4 0B ， 7kN( )BY = ↑

0X =∑ , + × −4 1 4 0AX = ， 8kN( )AX = ←

0BM =∑ , × − × × + ×4 1 1 4 2 8 4 4 0AY+ × = , 7kN( )AY = − ↓

2. Construct M diagram (1) Compute tBy the method

he bending moments on the end sections of each member of sections, the bending moment at the end section of a member is equal in magnitude,

4kN

D

C

1mB

1kN/m BY

4m

A

AX

4mAY

Fig.4.11 The frame of example 4-5

4.4 Construction of Internal Force Diagrams of Statically Determinate Frames 85

but o direction, to the algebraic sum of the moments about the section of all the external loads and restraint forces acting on either side of the section. We consider it to be positive when the external and restraint forces and couples make the member rotate in the opposite sense you assumed.

Member CD, from C to D:

(Tension in the left fiber)

Member BD, from B to D:

(Tension in the lower fiber)

Member AD, from A to D:

M = × − × × = ⋅ (Tension in the right fiber)

M diagrams of all the members of the frame.

Afte luated, the shape of its M diagram can be de the dbe plotted by using the principle of supas the end moments.

There embers in the frame. We may drawordin l ber) of these members one by one as follows:

Member CD and BD: Since the two memtheir bending moment curves is a slope straighmoments of each member with a straig

Member AD: since the member carries uniformly distributequad c ollowings: firstly, connect the ordinates of the two

pposite in

0CDM =

4 1 4kN mDCM = × = ⋅

0BDM =

7 4 28kN mDBM = × = ⋅

0ADM =

8 4 1 4 2 24kN mDA

(2) Construct M diagram of each member of the frame For a rigid frame，the M diagram is composed of the

m ber are evar the bending oments at the both ends of each memtermined by ifferential relation existing between M and external loads, and then the diagram may

erposition when the member is subjected to an external load as well

are totally three m the bending moment diagrams (whose ates should be aid off on the tensile side of a mem

bers carry no external load but the end moments, each of t line. We can directly connect the ordinates of its two end

ht line. d loads, the bending moment curve is a

rati parabola. The parabola can be accomplished as fend moments ( ADM and DAM ) of the member with a straight dashed line; then, superpose the

ordinates of the corresponding bending same span and is subjected to the same loads)on the dashed line. The final M diagram shown in Fig.4.12 (a) and the final bending moment at th

moment diagram of a corresponding simple beam(which has the

e section E (which is located at the middle length of the member AD) is

( ) 214 1 4 14kN m1 0 2M2 8E = + + (Tension in the right fiber)

3. Construct Q diagram

× × = ⋅


(1) Compute the shears of each memMember CD, from C to D: Q

ber end 4kNDCQ= = CD

Member BD, from B to D: DB BDQ Q 7kN= = −

AD

Member , from A to D: ADQ 8kN=

8 1 4 4kNDAQ = − × =

C4

(2) Construct Q diagram of each memThe ordinates of a shear diagram of a m

ber ember can be plotted on either side of the member, but the sign

has to be indicated on iagram. For the frame show , directly conthe shears of each en e members of the f wing of the sheFig.4.12 (b).

4. Construct N m of each member of the frame es of each member end

the d n in Fig.4.11 necting the ordinates of d of th rame will finish the dra ar diagram as shown in

diagra(1) Evaluate the axial forcMember CD, from C to D: 0CD DCN N= = Member BD, from B to D: 0DB BDN N= = Member AD, from A to D: 7kNAD DAN N= =

E

C

4 D 24

2814

A

B

diagram (kN m)M

⋅

(a)A

BD

m

4

8

7 7

diagra (kN)

Q

(b)

A

C

(c)7

28kN mi

24kN mi

4kN7kN

4kN7kN

Fig.4.12 Internal force diagram of example 4-5

DD

diagram (kN)N

D 7 B

4kN mi(d) (e)

4.5 Internal Force Diagrams of Statically Determinate Three-Hinged Rigid Frames, Multispan and Multistory Rigid Frames 87

tly connecting the ordin h end of the members of the frame will complete the drawing of the axial force diagram as shown in Fig.4.12 (c).

5. Check Taking the rigid joint D of the frame as a free body, all of the equilibrium conditions are satisfied by

the f

(2) Depict N diagram of each member The ordinates of an axial force diagram of a member can be depicted on either side of the member, but

the sign has to be indicated on the diagram. For the frame shown in Fig.4.11, direcates of the axial forces of eac

ollowing equations:

DM 0=∑ , 4 24 28 0+ − =

0X =∑ 4 4 0− =,

0 7 7 0Y =∑ , − =

4.4.2 The key points of constructing the internal force diagramframes

method of constructing internal force diag

me.

mber. The

icated on the diagram. Magnitudes of the internal forces on som

ou an be

4.5.1 Construction of internal force diagrams of statically dmultispan and multistory rigid frames

ternal force diagrams for a statically determinate three-hinged rigid frame,

s for statically determinate rigid

From foregoing examples, we observe that the fundamentalrams of rigid frames is the method of decomposition and composition. It can be stated as following: (1) Calculate the internal forces on the member-end sections of each member of a rigid frame;

construct the internal force diagrams of each of the members respectively by employing the method of sections; compose each kind of internal force diagrams of all members together to form the internal force diagram of the entire fa

(2) The regulation of constructing internal force diagrams: The ordinates of internal force diagrams of each member of a rigid frame must be depicted in the sense perpendicular to the axial line of the me

ordinates of bending moment diagram of a member ought to be laid off on the tensile side of the member, whereas the ordinates of a shear and an axial force diagram of a member can be plotted on either side of the member, but the sign has to be ind

e special sections must be indicated on the diagrams. (3) Since the all members of a rigid frame are straight bars, the method of superposition segment by

segment presented previ sly c implemented for each straight member.

4.5 Internal Force Diagrams of Statically Determinate Three-Hinged Rigid Frames, Multispan and Multistory Rigid Frames

eterminate three-hinged rigid frames,

The steps of constructing in


or a t he same as those discussed previously. They are: (1) com g the internal forces of member ends for each member; (3) constructing inter be noted again th the order to determinate reactions ought to be the reve ction of a structure. The following examples explain the steps of cons ms for statically det inate three-hinged rigid frames, multispan and multistory rigid frames.

Example 4-6

Construct bending moment diagram of the three-hinged gable frame shown in Fig.4.13 (a) and compare it with that of the frame shown in Fig.4.13 (b).

Solution

1. Calculation of reactions [Fig.4.13 (a)] e actions on the frame, i.e.,

mul ispan or multistory rigid frame remain tputing reactions; (2) evaluatinnal force diagrams. It should at rse of that of geometric construtructing internal force diagra erm

1kN/m

AX

AY4m 4m

Ther are totally four unknown support re AX and AY , BX and From

BY . the viewpoint of geometric construction, three rigid bodies (ADC, BEC and the foundation) joined

pairwise by hinges A, B and C, which do not lie on the same straight line, form a stable system with no redundant restraint. Thus the steps in the determination of reactions of the frame are: (1) to evaluate the reactions AY and by taking the entire frame as a free body and then applying the equations of moment equil u

BYibri m about hinges B and A ( 0BM =∑ and AM 0=∑ ), i.e.,

0AM =∑ , 1 8 4 BY 8 0× × − × = ， 4kN( )BY = ↑

0BM =∑ , 8 1 4 8 0AY × − × × = ， 4kN( )AY = ↑

Check: 0Y =∑ , 4 1 8 4 0− + × − =

A

D

C

B

E

BY

BX

2m2m

(a)

1kN/m

AX

AY4m 4m

DC E

B BA

BY

X

4mFig.4.13 Frames of example 4-6

(b)


(2) t eo det rmine the reaction AX (or BX ) by 0CM = . using the construction condition of hinge C, i.e., We can employ the algebraic sum of the moments about hinge C of all the external loads and reactions from the either side (left or right) of C to establish it. If taking the right side of C, we write

0CM = , 1 4 2 4 4 4 0BX× × − × + × = , 2kN( )BX = ←

(3) to evaluate reaction AX by taking the entire frame as a free body and then applying the equations of force equilibrium in horizontal direction, i.e.,

0X =∑ , A BX X= , 2kN( )AX = →

2. Construct M diagram (1) Compute the bending moments at the end sections of each member Member AD: ， 0ADM = 2 2 4kN mDAM = × = (tension in the left fiber) ⋅

Member DC: ， 0CDM = 2 2 4kN mDCM = × = (tension in the upper fiber) ⋅

Member BE: 0BEM = ， 2 2 4kN mEBM = × = ⋅ (tension in the right fiber)

mber CEMe : ， 0CEM = 2 2 4kN mECM = × = ⋅ (tension in the upper fiber)

between M and external loads, h

to an e

e

(2) Construct M diagram of each member of the frame

Fig.4.14 Bending moment diagrams of exam

For the three-hinged gable rigid frame, the M diagram is composed of the M diagrams of all above listed members of the frame. After the bending moments at the both ends of each member are evaluated, the shape of its M diagram can be determined by the differential relation existingand t en the diagram may be plotted by using the principle of superposition when the member is subjected

xternal load as well as the end moments. There are totally four members in the frame. We may draw the bending moment diagrams (whose

ordinat s should be laid off on the tensile side of a member) of these members one by one as follows: Member AD and BE: Since the two members carry no external load but the end moments, each of

ple 4-6

444

(2) (2)

4

(a)

A B

C

D diagrM Eam (kN m)⋅

88 8

8

(2) (2)

CD E

diagramM

A B

(kN m)⋅

(b)


their bending moment curves is a slope straight line. We can directly connect the ordinates of its two end moments of each member with a straight line.

Member DC and CE: since the two members carry uniformly distributed loads, each of the two bending moment curves is a quadratic parabola based on its inclined axial line. The parabola can be accomplished as followings: firstly, connect the ordinates of the two end moments of the member with a straight dashed line; then, superpose the ordinates of the corresponding bending moment diagram of a corre le beam (which has the same span, the length of horizontal projection of member DC or CE, a

sponding simpnd is subjected the same loads) on the dashed line. The final M diagram shown in Fig.4.14 (a) and the

final bending moment at the middle section of the member DC or CE is

( ) 21 10 4 1 4 02 8

M = − + + × × =

3. Discussion Fig.4.14 (b) presents the bending moment diagram of the three-hinged rigid frame shown in Fig.4.13

(b), of which the span and height of middle hinge and acting loads are the same as the frame shown in Fig.4.13 (a). Comparing the two bending moment diagrams, we observe that the bending moment magnitude of beams and columns of the frame shown in Fig.4.13 (a) is smaller than that of beams and columns of the frame shown in Fig.4.13 (b)， because of the reason that the beams of the frame shown in Fig.4.13 (a) have inclined axes and the columns have smaller heights. Therefore, the stressing state of the frame shown in Fig.4.13 (a) is more rational than the frame shown in Fig.4.13 (b).

Example 4-7

Construct bending moment diagram of the two-span frame shown in Fig.4.15 (a) [the same as shown in Fig.4.7 (a)]

Solution

1. Calculation of reactions and restraint forces en determined in example 4-2, and now they are shown

again

ber

The reactions and restraint forces have be in Fig.4.15 (b). 2. Construct M diagram The followings are the calculation of bending moments of member ends of the frame, whose

subsidiary and main portion are ADE and CBEFG, respectively. Mem AD: 0ADM = ， 1 4 4kN mDAM = × = ⋅ (tension in the right fiber)

Member DE: 0M = ， 1 4 4kN mM = × (ten ED DE = ⋅ sion in the lower fiber)

Member EF: ， 0EFM = M 2 2 4kN mFE = × = ⋅ (tension in the upper f

Member FG: FGM = × = (tension in the upper fiber), GFM = ×iber)

4 2 8kN m⋅ 4kN m= ⋅ (tension 1 4


)

ers of the frame. After the bending moments th ends of each member are e ted, the shape of

ferenti

The fin diagram s ig.4.15(c) oment at hii.e., oment curve at location E is derivable, i.e., there is only one tangent line at E.

in the lower fiberMember CG: 0CGM = 1 4 4kN m (tension in the left fiber)

Member BF: 0BF FBM M= =

For the frame, the M diagram is composed of the M diagrams of all above listed memb

GCM = × = ⋅,

at the bo valua its M diagram can be determined by the dif al relationships between M and external loads, and then the diagram may be plotted by using the principle of superposition when the member is subjected to an external load as well as the end moments.

al M hown in F , on which the m nge E is equal to zero, 0E and the bending mM =

2kN (a)/m

A B C

E G

4m2m2m

4m

2kN/m4kN

1kN

4mD

F

4m

1kN2m

A

DE

(b)

1kN

4kN

2m 4m

7kN 3kN

1kNB C

Fig.4.15 A frame and its fr

E

F G

ee-body and bending moment diagrams of example 4-7

G4

(c)

A B C

D

diagram (kN m)M

⋅

4

84 4

E F


4.5.2

ch as a rigid joint, a member or the entire frame, must keep in equi

The check of internal force diagrams of rigid frames

1. The check of equilibrium conditions since any portion isolated from a rigid frame, su

librium, the equilibrium conditions that may not be used in calculation of internal forces ought to be employed as possible as you can when check the correctness of internal force diagrams.

It should be emphasized that the three equilibrium conditions ( 0M =∑ , 0X =∑ , 0Y =∑ )for each rigid joint of a frame must be employed. The three equilibrium conditions can be used to calculate the internal forces on member end sections as well.

It should be noted that the characteristics of the bending moment diagram at a rigid joint of a rigid frame must satisfy (a)

the followings: when a rigid joint is composed by two member ends, the magnitude of the two bending moments othe t m must be identical and the tensile fiber of th

own in Fig.4.9 (a), the algebrai hree bending moments are equal to zero, the clockwise is 6kN·m and the count 2 kN·m +4 kN·m, the bending moments keep in equilibrium.

bending moments hip existing between the loads, shears and bending moments for straight members is still

suita r the rigid frames on which middle hinges exist, the bend

f wo ember endse rigid joint must be yielded on the same side; while a

rigid joint is consisted of more than two member ends the bending moments of the member ends must keep in equilibrium. Such as the bending moments of member ends at the rigid joint D [see Fig.4.16 (a)] of the frame shown in Fig.4.15 (a), the two bending moments are equal in magnitude, but opposite in direction of rotation; they keep the joint in equilibrium and put the internal side of the joint in tension. And the bending moments of member ends at the rigid joint C [see Fig.4.16 (b)] of the frame sh

(b)

4

4

4

2 6D C

Fig.4.16 Free-body diagram of rigid joint

c sum of the terclockwise is 2. The check of characteristics of internal force diagrams and relationship existing between the

loads, shears andThe relationsble for the beams and columns of rigid frames. Foing moments at hinges must be equal to zero, i.e., 0M = must be satisfied at hinges.

SUMMARY

The chapter has discussed the analysis of plane statically determinate rigid frames and its content includes:

1. Steps to cal atically deculate st terminate rigid frames ions; (2) c ternal forces of control ) by

using f sections uct internal force diagram(1) Find react alculate in sections (the sections of member ends the method o ; (3) constr s by employing the method of

Summary 93

supe reactions

d the t f solving simultaneous the order of calcu s of a rigid frame hould be the reverse of that of its geometric construction. Therefore, for

on ought to be paid.

A rigides of the member ends at a rigid joint is still the method of sections. For emphasis, it is stated again as

the followings: The bending moment on the end sections of a member is equal in magnitude, but opposite in direction,

to the algebraic sum of the moments about the section of all the external loads and restraint forces acting on either side of the section. We consider it to be positive when the external and restraint forces and couples make the member rotate in the opposite sense as you assumed. The shear on end sections of a member is equal in magnitude, but opposite in direction, to the algebraic sum of the components in the direction perpendicular to the axis of the member of all the external loads and restraint forces acting on either side of the section. We consider it to be positive when the external forces and restraint forces tend to make the section rotate clockwise. The axial force at end sections of a member is equal in magnitude, but opposite in direction, to the algebraic sum of the components in the direction parallel to the axis of the member of all the external loads and restraint forces acting on either side of the section. We consider it to be positive when the external forces and restraint forces tend to produce tension.

5. Construction of internal force diagrams The ordinates of a shear or axial force diagram must be depicted in the direction perpendicular to the

axial line of the member and can be plotted on the either side of the member, but the sign has to be indicated on the diagrams.

The ordinate of a bending moment diagram must be depicted on the side where the fiber of the member is in tensile state. No sign indication is necessary for bending moment diagrams. The powerful way for constructing bending moment diagrams is the method of superposition member by member. That is, for members imposed by no external load but the end moments, directly connect the ordinates of its two end moments of each member with a straight line; for members imposed by loads, firstly, connect the ordinates of the two end moments of the member with a straight dashed line, then, superpose the ordinates

rposition. 2. Calculation ofIn order to avoi rouble o equations, it is preferable that lating reaction s

a rigid frame whose geometric construction is quite complex, we must analyze its geometric construction or stability first, and then calculate reactions in the order opposite to that of its geometric construction.

3. The internal forces of a rigid frame include bending moment, shear and axial force, the attention to their sign conventi

4. The method to determine the internal forces on the member-end sections joint may include several member ends. Generally, the method for calculating the internal

forc


of the rresponding bending moment diagram of a correspondingco simple beam (which has the same span and i b d line.

,

s su jected to the same loads) on the dashe6. The check of internal force diagrams The shapes of internal force diagrams must satisfy the relationships existing between the loads, shears

and bending moments. Each rigid joint of a frame must keep in equilibrium, i.e., M =∑ 0Y =∑0 and

∑ e hinge of a rigid frame, and

ting

on of rigid frames. hinged joint based on their features of

diagrams. statically determinate rigid frame. ntly determine its reactions when they are

eir differences? 4-8

onstructing internal force diagrams of statically determinate rigid

oin eck the bending t diagram of a rigid frame?

the characteristics of internal force diagrams at rigid joint C of the rigid frame shown in the

ber ends at rigid joint C must be satisfied?

0= . The bending moment curves must get across each middl the bending moment diagrams must be located in the same side of the rigid joints composed of two member ends.

X

Problems for Reflec

4-1 Simply reword the characteristics of geometric constructi4-2 Simply reword the difference between a rigid joint and a

transmitting forces and deformation as well as free-body4-3 Simply reword the features of calculating reactions for a4-4 For a statically determinate rigid frame, how convenie

more than three? 4-5 What is the sign convention of internal forces in the members of a rigid frame? Why we have to use

two subscripts to represent the internal forces of member end sections of a rigid frame? 4-6 How orient unknown bending moment directions when you isolate a free-body diagram from a rigid

frame? How determine the tensile side by the calculating result of bending moments? 4-7 Simply reword the calculating points of internal forces of statically determinate rigid frames.

Comparing the points with those of statically determinate beams, what are thWhat are the steps for constructing internal force diagrams of statically determinate rigid frames?

4-9 What are the regulations for cframes?

4-10 How use the equilibrium conditions to check the correctness of internal force diagrams? How employ the conditions that the bending moments at hinged j ts must be equal to zero to chmomen

4-11 What areproblem for reflecting [Fig.4.11 (a) and (b)]? Try to list the equations that the internal forces of mem


(a) (b)A A


4-1 Calculate the reactions of following plane statically determinate rigid frames.

m

C

D

a

a b

PC

D

B B

a

a b

Figs. of problem for reflecting 4.11

A B

C D

6m

4m

2kN/mq =

(a)

0.85

m

C

A

1.5kN

5kN/m

1m

(b)

Problem 4-1 (contd)

B


C D

4-2 Calculate the internal forces on the designated section C for the frame shown in the figure of problem 4-1(a).

4-3 Calculate the internal forces on the designated section C for the frame shown in the figure of

4m

q

a

a

(c)

10kN

C

A B

4m 4mD

(d)

2kN

/m

BA

a a

aa

A

B

C DE

E

(e) (f)5m 5m

Pa

D C

2kN/m

5m

2p

A B

6kN

a a

a

A B

D

C

EM

A B

CD

E

3m 3m 2m

6m

(h)Problem 4-1

(g)


problem 4-1(c). 4-4 Calculate the internal forces on the designated section C for the frame shown in the figure of

problem 4-1(d). 4-5 Calculate the internal forces on the designated sections B and C for the frame shown in the figure of

problem 4-1(e).

4-6 Calculate the internal forces on the designated section D for the frame shown in the figure of problem 4-6.

4-7 Calculate the internal forces on the designated section E for the frame shown in the figure of problem 4-7.

4-8 Construct the internal force diagrams for the following frames.

3kN

/m

4kN

6m 2m

A B

C D E

2kN 3m

1m

Problem 4-6

3kN

/m

BA

D E F

C

8kN 8kN

2m 2m 2m 2m

3m

Problem 4-7

acts at points , and respectively

P DE B

4m

4m

2kN/m

A

BC

(a)

A B

C D

E

P

l

(b)

2h2h

Problem 4-8


4-9 Construct the internal force diagrams for the frame shown in the figure of problem 4-1 (b). 4-10 Construct the internal force diagrams for the frame shown in the figure of problem 4-1 (d). 4-11 Construct the bending moment diagrams for the following frames.

(a) (b)

Pa

4-12 Construct the internal force diagrams for the three-hinged frame shown in the figure of problem 4-1 (f).

4-13 Construct the internal force diagrams for the three-hinged frame shown in the figure of problem 4-1 (g).

A B

C D

a

a

A B

C D

2kN/m

2m 3m 2m

4m

2kN

4kN(c)

(d)B

A

C

2m 2m 2m

3m 5m

2kN/m

4m 6m 4m

6m

B

A

C

Problem 4-11

4m4m


4-14 Construct the bending moment diagrams for the three-hinged frames shown in the figures of problem 4-14.

4-15 Construct the bending moment diagram for the two-span frame shown in the figure of problem 4-15. 4-16 Construct the bending moment diagram for the two-storey frame shown in the figure of problem

4-16. 4-17 Check the mistakes for the following bending moment diagrams. If there are mistakes in the

diagrams please revise them.

4kN/m

A B

CD E

5m 5m2.5m 2.5m

10m

(a)

P

BA

2a

2a

E

D

C

P

a a

aa

(b)

Problem 4-14

4kN/m 4kN8kN

2m 2m 2m 2m

4m

A B C

D E F G H

Problem 4-15

A B

CF

E D

3m 3m

8kN

Problem 4-16

3m 3m


(a) (b)

Problem 4-17

CHAPTER 5 THREE HINGED ARCHES

The abstract of the chapter The chapter will discuss the analysis of three hinged arches. There are two important

characteristics for a three hinged arch. They are: (1) curved members; (2) horizontal thrusts yielded under vertical loads. The two characteristics make the analysis and stressing performance of three hinged arches quite different from those of other type of structures. For instance, when finding the horizontal thrusts of a three hinged arch the condition of moment at middle hinge, i.e., , must be used; when constructing the internal force diagrams of a three hinged arch the values of the internal forces at a number of sections must be evaluated in order to finish the diagrams; the dominating internal force in a three hinged arch is axial compression, its bending moments and shears are much smaller in comparison with its axial compression; there exists a rational or optimal axial line for a three hinged arch.

0=M

5.1 The Constitution and Type of Three Hinged Arches

We may find reactions in the way as we do in the previous analysis. It is worth of mentioned that the analyzing order ought to be the reverse of that of its geometric construction of a three hinged arch to avoid solving simultaneous equations. Three hinged arches composed of curved members and hinges are

statically determinate arch-typed structures. They have a variety of application in bridge structure, water power engineering, underground engineering and building roof structures. Fig.5.1 (a) and (b) show the two

CP

CP

f f

AHBA B

H

l

(a) (b)AV

BA

BV AV BV

Fig.5.1 Two types of three hinged arches

101

102 Chapter 5 Three Hinged Arches

primary types of three hinged arches. The arch shown in Fig.5.1 (a) is a three hinged arch with no tie. Two curved members AC and BC are

connected each other by hinge C and then connected on their foundation by two hinges A and B, the three hinges do not lie on the same straight line, to form a stable system with no redundant restraint. Under the action of vertical loads, a horizontal thrust H will develop on two foot supports. The thrust impose very important influence on the internal forces in the arch. The arch shown in Fig.5.1 (b) is a three hinged arch with a tie. Hinged support B as shown in Fig.5.1 (a) is now replaced by a roller support. Under the action of vertical loads, the tensile force in the tie substitute the function of the horizontal thrust developed in the foot supports of the arch with no tie.

Some of terminology pertinent to three hinged arches will be explained by employing the arch shown in Fig.5.1 (a) as an example. The curve ACB is referred to as centre or axial line of the arch, the axial line of a commonly used arch is a parabola or circular arc. The hinge C is named top or central hinge, whose height, the distance between C and the line connecting the hinges of supports A and B, is called the rise of the arch and denoted by f. The horizontal distance between the hinges of supports A and B is referred to as the span and denoted by l. The value of /f l is termed as the ratio of rise to span. The ratio /f l is one

of the basic parameters of an arch, which varies in the extent of . 1 ~ 1/10

5.2 The Reactions of Three Hinged Arches under Action of Vertical Loads

There are four reaction components acting on the arch shown in Fig.5.2 (a), AV and AH ,

and . We must need four equilibrium

equations to determinate the four unknown reactions. Apparently, the equilibrium condition of the entire arch can contribute three equilibrium equations; in addition, another construction condition at hinge C can be obtained that the bending moment at hinge C must be equal to zero, i.e., . Since the number of unknown

reactions is just equal to that of the equilibrium and construction equations, it is evident that a three hinged arch is a statically determinate structure.

BV BH

0CM =0

AV0

BV

1a1b

2a 2b

1d1P 2P

AAH B BH

AVBV

CD

D f

l

y

1l 2l

(a)

(b)1P

2P

BACD

l1l 2l

Fig.5.2 A three hinged arch and its corresponding simple beam

5.2 The Reactions of Three Hinged Arches under Action of Vertical Loads 103

Usually, the two support hinges of a commonly used three hinged arch are mounted on the same level, the directions of action lines of its loads are vertical and there is a vertical axis of symmetry in the arch. The calculation of reactions for this kind of commonly utilized three hinged arches will be discussed as below.

In order to make a clear comprehension and compare the difference of the reactions between an arch and a beam, a corresponding simply supported beam covering the same span and carrying the same vertical loads is depicted as shown in Fig.5.2 (b). Under the action of vertical loads, the horizontal reaction of the simple beam is equal to zero, i.e., ; vertical reactions (0H = 0

AV and ) of the simple beam can be

determined by the equilibrium conditions of the entire beam,

0BV

0BM =∑ and 0AM =∑ , respectively.

01 1 2 2

1 ( )AV Pb P bl

= + ，0

1 1 2 21(BV Pa P al

= + ) (5-1)

Considering the entire arch as a free body and applying equilibrium conditions ( and ) to it, we obtain 0BM =∑ 0AM =∑

1 1 2 21( )AV Pb P bl

= + ， 1 1 2 21(BV Pa P al

= + ) (5-2)

Comparing equation (5-1) with equation (5-2), we find 0

A AV V= , 0BV VB= (5-3)

Obviously, the magnitudes of vertical reactions of the arch are the same as those of its corresponding simple beam.

Considering the entire arch as a free body and applying the horizontal force equilibrium condition ( ) to it, we obtain 0X =∑

A BH H H= =

That is, the horizontal reactions at supports A and B of the arch are equal in magnitude but opposite in direction, and we use alphabet H indicate the thrust. The direction of H shown in Fig.5.2 (a) is recognized to be positive.

In order to determine the thrust H, we employ the additional condition provided by hinge C, i.e., the bending moment at hinge C must be equal to zero ( 0CM = ), and if calculating CM from A to C we

obtain

[ ]1 1 1 0AV l Pd Hf− − =

in which, the first two terms (in square brackets) are the algebraic sum of the moments about hinge C of all the external loads and reactions acting on left side of the hinge C, which is equal to the magnitude of the bending moment on the corresponding section C of the corresponding simple beam. Indicating the bending moment by 0

CM , we write


01 1C A 1M V l Pd= −

Upon that, the moment about hinge C of the three hinged arch may be expressed as 0 0CM Hf− =

0CMH

f= (5-4)

It will be seen from this that under the action of vertical loads the thrust has no relationship with the form of the axial line of the arch; in fact, it is related to the locations of the three hinges of the arch; the magnitude of thrusts H is in direct proportion to 1/ f , i.e., the flatter the arch becomes the lager the thrust

will be. When the direction of a load is downward the thrust, imposed to the arch by the supports, is positive as shown in Fig.5.2 (a), and its direction is inward. In the limit case if the rise f vanishes, i.e., the thrusts H approach to infinite 0f →

value. As a matter of fact, if the three hinges lie on the same line the special system will be unstable and it can not keep in equilibrium.

For the three hinged arch with a tie shown in Fig.5.3 (a), there are only three reactions, which can be determined by the three equilibrium equations ( 0BM =∑ , 0AM =∑ and 0X =∑ ) of the entire arch.

The relationships ,0AH = 0A AV V= and 0

BV VB= remain usable.

The tensile force of the tie can be obtained by cutting the tie and the hinge, considering the left portion of the section C as a free body [Fig.5.3 (b)] and applying the additional condition that the algebraic sum of the moments about hinge C of all the external load, reaction and the tensile force acting on the side must be

1d1P

2P

A0AH = B

BVAV

C

1l 2l

(a) (b)

A

AV 1l

1d1PC

CV

AB

N

CH

Fig.5.3 A three hinged arch with a tie

5.3 Formula for Calculating Internal Forces of Three Hinged Arches under Action of Vertical Loads 105

equal to zero, i.e., . Thusly doing, we write 0CM =

[ ]1 1 1 0A ABV l Pd N f− − =

In which the first two terms (in square brackets) remain equal to the magnitude of the bending moment on the corresponding section C of the corresponding simple beam. Indicate the bending moment with symbol

0CM yet.

It can be seen from the analytical results that the magnitude of the tensile force of the tie of a three hinged arch with a tie is the same as that of the thrust H of a three hinged arch with no tie. The equation (5-4) is still usable for computing the tensile force of the tie. It can be rewritten as

0C

ABMN

f= (5-4＇)

5.3 Formula for Calculating Internal Forces of Three Hinged Arches under Action of Vertical Loads

A three hinged arch is composed of curved members. The cross-sectional dip ϕ , which is the angle

formed by the tangent to the centre line of the arch and the axis of abscissa, varies with the position of the cross section of the arch. When the centre line of an arch is given (i.e., the centre line function is known), the slope of a section of the arch may be determined by the first derivative, i.e., ta

( )y f x=n /dy dxϕ = ,

of the centre line function of the arch, sinϕ and cosϕ of the angle ϕ may be determined as well.

Internal force components over an arbitrary cross section of an arch may be the bending moment, shearing force (or simply shear) and axial force.

The sign convention of internal force components are regulated as following: a positive bending moment tends to cause tension in the inner fibers and compression in the outer fibers of the arch; a positive shear makes the section rotate clockwise and positive axial force makes the section in tensile state.

The method to calculate the internal forces of an arch is still the method of sections. The formulae for calculating the internal forces acting over an arbitrary cross section D [Fig.5.2 (a)] of an arch under the action of vertical loads will be derived now.

Pass an imaginary section perpendicular to the centre line of the arch at the point D where the internal forces DM , DQ and DN [see Fig.5.4 (a)] are desired and take the left portion of section D as a free

body. In order to facilitate the analysis the internal force components on section D are replaced by its horizontal and vertical force components DX and DY shown as in Fig.5.4 (b); the corresponding

bending moment and shear on the corresponding section of a corresponding simple beam, indicated by 0DM and 0

DQ [see Fig.5.4 (c)], are herein borrowed.

By applying the equilibrium conditions in the vertical and horizontal directions to the free bodies


shown in Fig.5.4 (b) and (c) respectively, we obtain

0Y =∑ ， 01D AY V P Q= − = D

H

0X =∑ ， DX =

The internal forces on section D will be expressed by the corresponding bending moment 0DM ，shear

0DQ of a corresponding simple beam and the thrust H of the arch in the following.

1. Formula for calculating bending moments Considering the free body shown in Fig.5.4 (a) or (b) and applying the equilibrium equation of

moment about the centroid of section D, we write

0DM =∑ ， [ ]1 1( )D A D D DM V x P x a Hy= − − −

Since 0A AV V= , the first two terms in square brackets on the right hand side of the equation are the

bending moment on the section D of the corresponding simple beam. The above equation can be thusly expressed as

0D D DM M Hy= −

Therefore, formula for calculating the bending moment of an arbitrary section will be 0M M H= − y (5-5)

In which 0M denotes the bending moment on the corresponding section in the corresponding simple

0AV 0

DQ

0DM

(c)

A1P

D

(d)H

Dϕsin DH ϕcos DH ϕ

(e)

Dϕ0 cosD DQ ϕ

0 sinD DQ ϕ

0DQ

Fig.5.4 Internal force calculation of a three hinged arch

AH

AV

(a)

1a 1P

D

DM

DQ

DNDϕ

Dy

Dx

0D DY Q=(b)

DX H=

Dϕ

AH

AV

1P

D M D


beam; y represents the distance between the centroid of the section and the line connecting the two hinges of the arch supports.

2. Formula for calculating shears and axial forces Obviously, the shear DQ and axial force DN on section D must be the algebraic sum of the

components in the directions perpendicular and parallel to the tangent to the centre line of the arch of forces H and 0

DQ acting on section D respectively. That is 0 cos sinD D DQ Q H Dϕ ϕ= −

0 sin cosD D DN Q H Dϕ ϕ= − −

Formulae for calculating the shears and axial forces of an arbitrary section thusly be 0 cos sinQ Q Hϕ ϕ= − (5-6)

0 sin cosN Q Hϕ ϕ= − − (5-7)

Where, ϕ denotes the cross-sectional dip, in a rectangular coordinate system if axis of abscissa is positive

to right and axis of ordinate is positive upward, 0dydx

> , 0ϕ > in the left half portion of an arch and

0dydx

< , 0ϕ < in the right half portion of an arch. denotes the shear on the corresponding section

in the corresponding simple beam.

0Q

It can be seen from above formulae that if the centre line of an arch is given the location and slope of a section (i.e., x, y andϕ ) of the arch may be determined; and then by employing the internal forces 0M

and on the corresponding section of a corresponding simple beam of the arch, the internal forces on

each section of the arch may be calculated directly by the above formulae (5-5) through (5-7).

0Q

3. Construct internal force diagrams Since an arch is composed of curved members the curves representing the internal force variation are

related with the ordinate y of the centre line and the cross-sectional dip ϕ of the arch. So the method of

constructing the internal force diagrams for straight members is not available for arches. When the internal force diagrams are required, sufficient number of cross sections must be considered. Based on the obtained numeric values of internal forces for these sections, ordinates may be laid off from the curved centre line of the arch or from the horizontal abscissa, and then the diagrams may be plotted.

Example 5-1

A three hinged arch and the loads acting on it is shown in Fig.5.5. The centre line of the arch follows a conic parabola given by the equation


2

4 ( )fy x ll

= − x

Determine the reactions and construct internal force diagrams.

Solution

1. Calculation of reactions Employing equations (5-3) and (5-4), we write

0 10 3 3 6 9 16kN( )12A AV V × + × ×

= = = ↑

0 3 6 3 10 9 12kN( )12B BV V × × + ×

= = = ↑

0 16 6 3 6 3 10.5kN4

CMHf

× − × ×= = =

2. Calculation of internal forces Once the reactions have been obtained the

internal forces on any section will be determined by using the formulae (5-5) through (5-7). In order to construct the internal force diagrams eight sections, located at the ends of segments (divided along the span by using identical horizontal projective length), are considered and the internal forces on these sections have to be calculated. As an example to explain the computing procedure we will evaluate the internal forces on the section D whose abscissa is 9m from the origin of the rectangular system shown in Fig.5.5.

(1) Geometric parameter of section D By the function of the axial line of the arch when , we determine 9mx =

2 2

4 4 4( ) 9(12 9) 3m12

fy x l xl

×= − = × − =

2 2

4 4 4tan ( 2 ) (12 2 9) 0.066712D

dy f l xdx l

ϕ ×= = − = − × = −

Therefore, we obtain sin 0.55Dϕ = − ， cos 0.832Dϕ =

(2) Internal forces of section D Based on formula (5-5), we write

0 12 3 10.5 3 4.5kN mD D DM M Hy= − = × − × = ⋅

We will employ formulae (5-6) and (5-7) to find the shear and axial force on section D. since there is a concentrated force at point D the corresponding shear 0

DQ in the corresponding simple beam will vary

abruptly from the left to the right of the section, and we must compute the shear value on the left and the

Fig.5.5 A three hinged arch of example 5-1

3kN/m

y

B

10kN

CD

x

A

6m 3m 3m12m


When implementing the concrete calculation tabulation (see table 5-1) may be utilized for clsimp

of a corre

internal force calculation of a three hinged arch

Geometric parameters of sections computation Shear computation Axial force computation

right sections of point D. We write

Table 5-1

Bending moment

SeN

octio

n .

x y tanϕ sinϕ cosϕ 0M Hy− M

0Q 0 cosQ ϕ sinH ϕ− Q 0 sinQ ϕ− cosH ϕ− N

0 0 1.333 0.800 0.600 0.00 0.00 0.00 16.00 9.58 -8.40 1.18 -12.8 -6.29 -19.090

1 1.5 1. 20. 75 1.000 0.707 0.707 63 18.38 2.25 11.50 8.13 -7.42 0.71 -8.13 -7.42 -15.55

2 3.0 3.0 0.667 0.555 0.832 34.50 31.50 3.00 7.00 5.82 -5.83 -0.01 -3.89 -8.74 -12.63

3 4.5 3.75 0.333 0.316 0.949 41.63 39.38 2.25 2.50 2.37 -3.32 -0.95 -0.79 -9.96 -10.75

4 6.0 4.0 0.000 0.000 1.000 42.00 42.00 0.00 -2.00 -2.0 0.00 -2.00 0.00 -10.50 -10.50

5 7.5 3.75 -0.333 -0.316 0.949 39.00 39.38 -0. -1.38 -2.00 90 3.32 1.42 -0.63 -9.96 -10.59

-2.00 -1.66 5.83 4.16 -1.11 -8.74 -9.85 6 9.0 3.0 -0.667 -0.555 0.832 36.00 31.50 4.50

-12. -4. -15.00 -9.98 5.83 16 -6.66 -8.74 40

7 10.5 1.75 -1.000 -0.707 0.707 18.00 18.38 -0.38 -12.00 -8.48 7.42 -1.06 -8.48 -7.42 -15.90

8 12.0 0 -1.333 -0.800 0.600 0.00 0.00 0.00 -12.00 -7.19 8.40 1.21 -9.60 -6.29 -15.89

0

0 0

cos sin ( 2) 0.832 10.5 ( 0.555) 4.16kN

sin cos ( 2) ( 0.555) 10.5 0.832 9.85kNDL DL D D

DL DL D D

Q Q H

N Q H

ϕ ϕ

ϕ ϕ

⎧ = − = − × − × − =⎪⎨

= − − = − − × − − × = −⎪⎩0 cos sin 12 0.832 10.5 0.555 4.16kNDR DR D DQ Q Hϕ ϕ= − × − × − −＝（－）（）＝

0 sin cos ( 12) ( 0.555) 10.5 0.832 15.4kNDR DR D DN Q Hϕ ϕ= − − = − − × − − × = −

arity and licity. By the numeric values listed in table 5-1 of the internal forces of the arch, bending moment M

and shear Q and axial force N diagrams are depicted in Fig.5.6 (a), (b) and (c) (which ordinates from the curve centre of the arch), or in Fig.5.6 (e), (f) and (g) (which ordinates from the horizontal abscissa).

In the interest of comparing an arch and its corresponding beam, the bending momentsponding simple beam (with the same span and carrying the same loads) is plotted in Fig.5.6 (d).

Comparing the bending moment diagram shown in Fig.5.6 (a) with that shown in Fig.5.6 (d), it will be seen that the bending moments in the arch (with the maximum bending moment of 4.5kN·m) are much smaller than those in a corresponding simple beam (with the maximum bending moment of 42.67kN·m). Obviously,


the reason that makes the bending moments in an arch decrease is the thrust of the arch. One of primary characteristics of arches is that the thrust will be developed at its supports under the action of vertical loads; consequently, the arch-typed structures are termed as compression structure.

An arch tends to flatten out under vertical loads and it is the horizontal thrust or the axial tension in the tie w

5.4 Stressing Performance of Three Hinged Arches

Foregoing discussion reveals that:

Fig.5.6 Internal force diagrams of example

15.40

9.8510.5910.5010.7512.6315.5519.09

ho works against the deviation between the supports. The primary internal force in an arch is axial compression, while the primary force in a beam is the bending moment.

15.90 15.89

A B

(g) diagram (kN)N

1.18

0.71 0.01− 0.95−2

A

C1.42 4.16

4.16− 1.06−1.21

(b) diagram (kN)QB

42.67

A B

C

0(d) diagram (kN m)M ⋅

41.6

3

34.5

42 39

3620.6

3

18

A B0

(e) diagram (kN m)M ⋅

2.25 3.0 2.25

0.38 0.380.38

4.53.02.25

C

1.5m

(a) diagram (kN m)M ⋅

1.5m1.5m1.5m1.5m1.5m1.5m1.5m

0.38

2.254.5

A B

A B0.01 0.95

4.16

1.06

(f) diagram (kN)Q1.18 0.71

2

1.42

4.16

1.21

C

10.50−

A B(c) diagram (kN)N

9.85−

10.75

19.09−

15.89−

15.55−

12.63−

− 15.40−10.59−

15.90−

5.5 Rational Axial Lines of Three Hinged Arches 111

1. Under the action of vertical loads, unlike a beam, which develops only vertical reactions at its supports, the arch gives rise to both horizontal thrust and vertical reactions at its supports.

2. It can be seen from the formula (5-5) that the bending moments in an arch are much smaller than those in a corresponding simple beam.

3. Under the action of vertical loads, unlike a beam, in which there exist no axial force, an arch yields axial compression, which is the dominated internal force in the arch.

4. Since the stress distribution on the sections of arches is more uniform than that of beams the capabilities of constructional materials can be effectively utilized in arch-typed structures. Moreover, because arches develop mainly compressive stresses when subjected to external loads the constructional materials which are stronger in compression but weaker in tension such as bricks, dimension stone and concrete, and the like, are suitable to build arch-typed structures.

However, any thing bearing two sides, when an inward thrust is provided by the supports for an arch which makes the bending moments relatively small in the arch, an outward thrust is as well imposed on the supports by the arch. Therefore, the three hinged arches with ties may be applied to the roof structures so as to reduce the action of the thrusts imposed to side walls or columns by the arches.

5.5 Rational Axial Lines of Three Hinged Arches

From the previous analysis, it is realized that although the dominant normal stresses acting on the sections of a three hinged arch are caused by axial compression, the magnitude of another kind of normal stresses induced by bending moments and distributed unevenly on the sections are considerable as well. In the interest of utilizing the capability of constructional materials sufficiently, the unevenly distributed normal stresses on the sections of an arch must be decreased as less as possible. If there exists only axial compression on the sections and the bending moments induced in an arch vanish, the normal stresses on the sections will be uniformly distributed and the dimension of the arch will be expected to have a minimum theoretical volume. Under the action of stationary loads the axial line of an arch that will induce the bending moments on the sections to vanish (i.e., the arch remains no bending-moment state) is referred to as rational or optimal axial or centre line of the arch.

5.5.1 Optimal centre line of three hinged arches under vertical loads

According to formula (5-5), it is recognized that under the action of vertical loads the bending moment on an arbitrary section of a three hinged arch ( )M x is equal to the bending moment of its corresponding

simple beam 0 ( )M x superposing the term ( )Hy x− , that is, 0( ) ( ) ( )M x M x Hy x= −

When the centre line of an arch becomes optimal the bending moments of sections of the arch are


equal to zero, that is,

，( ) 0M x ≡ 0 ( ) ( ) 0M x Hy x− ≡ Therefore, the function of a rational axial line ought to be

0 ( )( ) M xy xH

= (5-8)

In which 0 ( )M x is the expression of the bending moments of the corresponding simple beam (with the same span and carrying the same loads as those of the arch), and if depicting it as a figure the figure will be the bending moment diagram of the simple beam.

Obviously, under vertical loads the difference between the expression of the optimal centre of a three hinged arch and the expression of its corresponding simple beam is only a proportional constant H, that is, the ordinates of optimal centre line of the arch is directly proportional to those of the bending moment diagram of the simple beam.

With mastery of the concept of the optimal centre line of an arch, the rational form of an axial line of an arch will be selected when design the arch; the capability of its constructional materials will be thusly utilized sufficiently.

Example 5-2

In Fig.5.7 (a) a three hinged arch is under the action of uniformly distributed vertical load q, find the optimal centre line of the arch.

Solution

Locate the origin of the rectangular system at point A shown in Fig.5.7 (a), and the corresponding simple beam of the arch is shown in Fig.5.7 (b).

By formula (5-8), we write 0 ( )( ) M xy xH

=

The bending moment equation of the corresponding simple beam is

0 21( ) ( )2 2 2ql qxM x x qx l x= − = −

The thrust determined by equation (5-4) is 0 2

8CM qlH

f f= =

After some simple operation we obtain the function of the optimal centre line of the arch as follows


2 2ql l

( ) 42( ) ( )

qx l x fy x x l x−

8 f

= = − (5-9)

rabola. Accordingly, the arches with a

h is a curve of circular arc.

an in of the arc e body [Fig.5.8 (b)]. The angle between the two end sections is denoted by

It can be seen from equation (5-9) that under the action of uniformly distributed vertical load the function of the optimal centre line of a three hinged arch is a quadratic pa

par bola centre lines are commonly employed in the civil engineering structures.

q

C

5.5.2 Optimal centre lines of three hinged arches under uniformly distributed hydraulic pressure

The uniformly distributed hydraulic pressure is one sort of load which encounters frequently in water power engineering, pipework and the like. Following example will show the optimal centre line of a three hinged arch under uniformly distributed hydraulic pressure.

Example 5-3

A three hinged arch shown in Fig. 5.8 (a) is under the action of uniformly distributed hydraulic pressure q, prove that the optimal centre line of the arc

Solution

First, assume the arch is under zero bending moment state under the action of uniformly distributed hydraulic pressure.

(1) Take finitesimal element DE h as a fredϕ ; the curvature radi

respectus at point D and E are denoted by r and r dr+

ively; the length of the infinitesimal axial line will be ds rdϕ= .

B x

y

2ql

2ql

(b)

q

Fig.5.7 Arch and its corresponding beam of example 5-2

A

f

(a)2l l

x

l2


(2) Since the arch is under zero bending moment state there exists only axial force but no bending moment and shear on the sections D an . The moment equilib

ature of the arch, i.e., we write

，

d E rium equation may be set up by equating to zero the algebraic sum of moments about the centre of curv 0 0M =∑ ,

0 0M =∑ ( ) 0D EN r N r dr− + =

Neglecting the infinitesimal term , we obtain EN dr

D EN N=

This means that under the action of uniformly distributed hydraulic pressure if the arch is under zero bend

isector of the angl

ing moment state the axial force is a constant throughout the sections of the arch. It may be denoted by N.

(3) The force equation may be set up by equating to zero the algebraic sum of the projections of the forces in the direction of the b e dϕ . We write

sin sin 02 2D E

d dN N qrdϕ ϕ ϕ+ + =

Since N N= by2

dϕD E N= and dϕ is infinitesimal quantity we may replace sin

2dϕ

, and

then we obtain 0Nd qrdϕ ϕ+ =

N q= − r ， constantN=r

q= −

nt, the

curvature r of each section throughout the arch remains constant as well. Recalling from geometry, the

Because both the axial force N throughout the arch and the external load q remain consta

A B

Cq

DE

q

odϕ

O'

DNds

Dr

r+drE

d2ϕ

(a) (c)(b)

ND

d2ϕ

EN

NE

Fig.5.8 Arch and its infinitesimal free body of example 5-3


curve with a constant curvature is a circular arc. This means that the optimal centre line of a three hinged arch under the action of uniformly distributed hydraulic pressure is a circul

In conclusion, under the action of uniformly distributed hydraulic pressure the rational or optimal centre line of a three hinged arch is an arc of circle. This is the reason why the cross sections of structures unde

The earth pressure is one common load which encounters frequently in bridge engineering and tunneling. Following example will show the optimal centre line of a three hinged arch under earth pressure.

Example 5-4

A three hinged arch shown in Fig. 5.9 is under earth pressure. The upper surface of earth is a horizontal plane; the arch is thusly under the load of earth weight which may be expressed by

( ) cq x q y

ar arc.

r uniformly distributed hydraulic pressure, such as waterpipes, high-pressure tunnels and arch dams, are designed in annular shapes.

5.5.3 Optimal centre lines of three hinged arches under earth pressure

γ= +

Wher is the intensity of the distributed load at the crow e C,

e cq n hing γ is the weight of the earth per unit

volume. Find the optimal centre line of the arch under the earth pressure.

Solution

Since earth pressure ( )q x varies with respect to the ordinate y of the arch and y is an unknown

function now, we differentiate Eq. (5-8) with respect to x twice times and obtain the following differential equation

2 2 0

2 2dx H dx= (a)

Because the positive y axis is in downward direction, here Eq. (a) should be expres

1 ( )d y d M x

sed as 2 2 01 ( )d y d M x

= − (b)

ential relationship existing between the bending moment

2 2dx H dxBy using the differ 0 ( )M x and vertical load

e( )q x , w have 2 0

2

( ) (d M x q x) ( )cq ydx

γ= − = − + (c)

qc q(x)

x

A

y

C

B

Fig.5.9 A three hinged arch subjected to earth pressure


By a p sim le operation, Eq. (b) may be expressed as 2

2

1 ( )cd y q ydx H

γ= +

or 2

22 cq

dx H− =

1d y k y (d)

where 2k γH

=

It can be seen from Eq. (c) that both 0 ( )M x and ( )q x are related to the arch configuration. In this

situation, finding the solution of the governing differential equation (d), we obtain the optimal curve of the arch which is expressed by a hyperbolic function

1 2cosh( ) sinh( ) cqy C k x C k xγ

= + −

Where an are the undetermined constants of integration which may be determined by the

boun nditions of the a for

1C d 2Cdary co rch

0, 0,x dy dx= = hence 2 0C =

for 0, 0,x y= = thusly 1cqCγ

=

The function of the optimal centre line now becomes

(cosh 1)cqy kγ

= −x

In mathematics, the curve described by such equation is referred to as catenary line. In conclusion, under the action of earth pressure the rational or optimal centre line of a three hinged

arch is a catenary. This is the reason why the cross sections of structures under earth pressure such as highway and bridge tunnels are designed in catenoid shapes.

SUMMARY

The key points of the chapter are as followings: 1. Composing of curved mem and yielding a horiz

are the two pronounced characteristics of a three hinged arch, which are the difference between arches and beam o

bers ontal thrust under the action of vertical loads

s as well. The two characteristics make the analysis and stressing performance f three hinged arches quite different from those of other type of structures.

Summary 117

2. The formulae for calculating the reactions and internal forces of a three hinged arch under the action of vertical loads are summarized as followings:

0

0

0B B

V V

V V

⎫=

/

A A

CH M f

⎪= ⎬

⎪= ⎭

0M M Hy ⎫= −0

0

cos sinsin cos

Q Q HN Q H

ϕ ϕ

ϕ ϕ

⎪= − ⎬

⎪= − − ⎭

The above formulae also express the relationship between internal forces and reactions of three hinged arch

ing the internal force diagrams of curved members force diagrams are required, sufficient number of

g to the obtained numeric values of internal forces for these sections, ordinates may be lai tre line of the arch or from the horizontal abscissa,

th g moments greatly small (comparing with those of axial compression, the stress distribution on the

more uniform than that of beams. So the arch-typed structures may overpass longer span ger loads comparing with those of beams. Moreover, the capability of their constructional mate utili

For determined stationary loads, the centre line which makes the bending moments in an arch vanish is referred to as the optimal centre line of the arch. Under thload the function of the optimal centre line of a three hinged arch is a quadratic parabola; under the action

s an arc of circle; under the action of earth pressure the optimal centre line of a three hi

5-1 types are there of three hinged arches? 5-2 Are ties usually used to the three hinged arches applying in roof structures? Why? 5-3 What is the difference between the reactions of three hinged arches

es and those of their corresponding simple beams 3. It is the first time that the method of construct

is presented in the chapter. That is, when the internalcross sections must be considered; accordin

d off from the curved cenand then the diagrams may be plotted.

4. Since e horizontal thrust makes the bendinbeams) and the dominant internal force in an arch issections of arches is

and carry larrials can be effectively zed and the brittle materials such as bricks, dimension stone and concrete,

and the like, which are stronger in compression but weaker in tension, are suitable to build arch-typed structures.

5. e action of uniformly distributed vertical

of uniformly distributed hydraulic pressure the optimal centre line of a three hinged arch inged arch is a catenary.


How many

and those of beams under the


action of vertical loads? 5-4 How does the thrust H of an arch vary when

condition that the span and loads of the arch rem What is the difference between the internal forces of three hinged arches and those of beams under

the action of vertical loads? 5-6 How calculate

its rise f changes (increase or decrease) under the ain unchangeable?

5-5

, sinϕ ϕ and c osϕ of form5-7 Whether or not the formulae which are used to calculate the reactions and internal forces of three

inged arc es can be utilized to compute those of three hinged rigid frames? 5-8 ristics fo hing

5-10 What is the optimal centre line of a three hinged arch?

5-11 al centre lines of three hinged arches under the action of vertical loads? How determine the opaction of the loads shown in the figure [Fig.5.11 (a) and (b)]? Whether or not the rise of a three

ith a centre line of a quadratic parabola, of which the function is

ulae (5-6) and (5-7)?

h hWhat are the main stressing characte r three ed arches?

5-9 Why the capability of constructional materials constructed in arches can be utilized more efficiently than it can be done in beams?

P P P

What are the characteristics of the forms of the optimtimal centre line of a three hinged arch under the

hinged arch influences its optimal centre line?


5-1 The figure in problem 5-1 shows a three hinged arch w

2

4 ( ), 16m, 4fy x l x l fl

= − = = m

nd axial force N on sections D.

(1) Find reactions (2) Calculate the bending moment M, shear Q and axial force N on sections D and E.

5-2 The figure in problem 5-2 shows a three hinged arch with a centre line of a circle arc, Calculate the reactions and the bending moment M, shear Q a

2l

2l

3l

3l

3l

(a) (b)

igs. of problem f g 5-11F or reflectin


5-3 The figure in problem 5-3 shows a three hinged arch with a centre line of a quadratic parabola, of which the function is

2

4 ( )fy x ll

= − x

(1) Find reactions and internal forces in all two-force members. (2) Calculate the bending moment M, shear Q and axial force N on sections K.

5-4 For the three hinged arch shown in the figure of problem 5-1, (1) If change the height of the rise (set 8mf = ), how the reac(2) If change the height of the rise and the length of span (set

tions vary? 32ml = , 8mf = ) but the ratio /f l

remains unchangeable, how the reactions and bending moments vary?

D

5m

A B

C

4mf =D

E

4m

1kN/m

4m 4m 4m

10kN

y

x

5m 3m

20kN/m

5m

C

A B 030

Problem 5-1 Problem 5-1

C

K

BAED

10kN/m

3m1m

3

-3

m 3m 3m 3m 6m

Problem 5

CHAPTER 6 PLANE STATICALLY DETERMINATE TRUSSES AND

COMPOSITE STRUCTURES

The abstract of the chapter A truss is an assemblage of straight members connected at their ends by flexible connections

(hinged joints) to form a rigid configuration. The members are subjected only to axial forces under the action of joint loads. There are two sorts of methods used for analyzing plane trusses. They are the method of joints, considering joints as free bodies and applying two equilibrium equations at each joint, and the method of sections, taking a desired portion of the truss as a free body and applying three equilibrium equations to the portion. The analyzing sequence for internal forces should be the order which is the reverse of that of the geometric construction of the truss to avoid solving simultaneous equations. The composite structures are composed by flexural and two-force members, which should be distinguished clearly. Generally, the axial forces in two-force members ought to be analyzed first, then the bending moments and shears and axial forces in flexural members will be obtained favorably.

6.1 Characteristics and Classification of Trusses

Unlike beams and frames, of which the dominating internal forces are bending moments that make the stress distribution on the section are uneven [Fig.6.1 (a)] under the action of loads, a truss is an assemblage of straight members connected at their ends by frictionless hinges to form a stable system, under the action of joint loads whose members are subjected only to axial forces which develop uniformly distributed stress [Fig.6-1 (b)]on the sections and make the efficient utilization of capacity of the constructional materials of the truss. Because of their light weight and high strength, trusses are widely used, and their applications range from roofs of buildings (Fig.6.2) and supporting structures of bridges (Fig.6.3) to being support structures in space stations. Modern trusses are constructed by connecting members, which usually consist of structural steel or aluminum shapes or wooden struts, to gusset plates by bolted or welded connections (Fig.6.4 through Fig.6.5).

6.1.1 Assumptions for analysis of trusses and their internal force characteristics

For practical trusses, the connections between their members are almost never completely frictionless hinges. In fact, real trusses are constructed by connecting members to gusset plates by welded or bolted connections. Some members of the truss may even be continuous at the joints. The centroidal axes of

120



6-1 Analyze the geometric constructions and distinguish their types (simple, compound or complex) of the following trusses.

(b)

P

(c) P P

(a) Paaa

030

(d)P

(e)P 2P

(f)

P

problem 6-1 (contd)

150 Chapter 6 Plane Statically Determinate Trusses and Composite Structures

(a)

AB

C

D

EF

20kN

20kN

4 3=12m×

(b)B

P 4a

a3m

(g)

P

(h) P

problem 6-1

(c)

A

B

C

DE

F

2P 1P

4a2P 1P

a

F

CD

G H(d)

A B

E

4kN

4kN

3m 3m

4m4m

8kN

problem 6-3 (contd)


(e)

P P

6-2 Solve the following questions in the following problems: (1) point out zero-force members; (2) point out the characters of axial forces in the members (compression or tension) by considering the equilibrium of joints and depict out the clue of the transmission of forces. （a） The truss shown in the figure of problem 6-1 (c); （b） The truss shown in the figure of problem 6-1 (d); （c） The truss shown in the figure of problem 6-1 (e); （d） The truss shown in the figure of problem 6-1 (f).

6-3 Determine the member forces of the following trusses by using the method of joints. 6-4 Determine the axial forces in the indicated members of the following trusses by using the method of

sections. 6-5 Determine the axial forces in the indicated members of the following trusses by selecting either of

the two methods, which should be an optimal one for computation. 6-6 Determine the axial-force sign of the members and the maximum-force member of each of the

following two trusses. 6-7 Compute the axial forces in the two-force members and construct bending moment M, shear Q and

axial force N diagrams in the beam members of the composite structures shown in the following figures.

A B

C D

E

H

F

G

2m 4m 2m

4m4m

3m

B C D

F E

P

3m

(d)

A

4m

3m6m

problem 6-3


(a)10kN

10kN

10kN

4m

(b)

5kN10kN

10kN10kN

12 3

6 2=12m

1 23

3m 3m

3m

3m

×

BA

(d)P

P

1

2

34

3a

4a1

(c)

50kN 30kN

6 3=18m

2

3

4m

×

(e)

6 3=18m×P P P P

4m

1

2

3

problem 6-4 (contd)


1(b)

60kN

2

3

3m 3m 3m 3m

4m4m

(a)

20kN 20kN40kN

3m 6=18m×

2

13

4m

(f)

10kN 10kN 10kN 10kN 20kN1

23

6 4=24m

4m

×

problem 6-4

(c)

1kN

1kN

2kN 2kN1

2

3

44kN

3m3m2m2m

1.5m

1.5m

3m 3m

3m

(d)

4kN

1kN 2kN 2kN 2kN 1kN6 3=18m

4m

1

2

3

4

×

problem 6-5


(b)

d

AC D E F G

P P P P P

6 d

C D E

G5 d×

F H

d

P(a) f g bea c dA

B

×

problem 6-6

(a)1kN/m

3m 3m 3m 3m

3m

A B

C

D E

F G

(b)

A

C

DE

4 3m 12m

B

× =

20kN 30kN 40kN

3m4m

problem 6-7


members of a truss may not be perfectly straight lines, and the axes of members meeting at a joint may not all intersect at the common point. So it is almost impossible to make analysis for a truss according to its practical stressing state. In this circumstance, trusses have to be simplified or idealized, by throwing away some unimportant details and maintaining the very important features which is desired for analysis, as analyzing models prior to computation. Therefore, the following assumptions are made in order to simplify the analysis of trusses:

σ−

(b)

(a)

σ+

b

a

Fig.6.1 Simply supported beam and truss; (a) beam; (b) truss

13 600 600

120

6060

20 60 60(a)

(b)

150120 10×3 2450=7350×

90 8

90 8×

×

15000

110

2450

Unit: (mm)

Fig.6.2 A reinforced concrete truss and its computing model

(1) Truss members are connected together at their ends by frictionless hinges;


(2) The centroidal axes of the members are all straight lines, those meeting at a hinge all intersect at the centre of the hinge;

(3) External loads and reactions are all applied on the truss only at its joints

(b)

(a)

128m 64m

128m 64m

16m

16m

Fig. 6.3 The great bridge over Yangtse River in Wuhan (of China) and its computing model

Fig.6.4 A modern space truss structure

The reason for making these assumptions is to obtain an ideal truss, whose members are subjected only to axial forces. If all the members of a truss and the applied loads lie in a single plane, the truss is called a plane truss. The truss shown in Fig.6.6 (a) is a plane truss, Fig.6.6 (b) is its computing model which is idealized according to above assumptions. Each of the members is represented by its centroidal axis; the connections between members are all hinges; concentrated loads and reactions are all applied on the joints. Fig.6.6 (c) shows an arbitrary isolated member from the truss. Since member CD is connected at its ends by frictionless hinges [assumption (1)] with no loads applied between its ends

1 2, P P , A BY Y


[assumption (3)], the member would be subjected to only two forces at its ends. Since the member is in equilibrium, the two forces must be equal in magnitude but opposite in directions. The line of action of the two forces must coincide with the centroidal axis of the member CD. Thereby, the arbitrary member CD is subjected only to axial force. Because the members of a truss are subjected only to two forces they are named two-force members. As long as two-force members with straight axial lines are subjected only to axial forces, the internal forces in the members of a truss which satisfies above assumption are only axial forces.

Fig.6.5 Truss members made of aluminum shapes

The members of a truss are classified into two categories named as chords and web members

BA

C

D d

2P1P

h

AYBY

bottom chord

top chordvertical

diagonal

(a)

BD d

2P1P

h

AY BY

A

C

C

DN

N

(c)

Fig.6.6 A truss and its computing model

(b)


according to their arrangements and orientations [see Fig.6.6 (a)]. Chords indicate the members which are located along the top and bottom periphery of the truss, including top chords and bottom chords. Web members are arranged between top chords and bottom chords, including verticals and diagonals. The joint at which a web member is connected to chord is called a panel point, and the length d between two panel points on the same chord is called the panel length or simply the panel. The maximum distance between the top chords and the bottom chords is termed as the height of the truss, denoted by alphabet h.

6.1.2 Classification of trusses by their geometric constructions

A truss is an assemblage of straight members connected at their ends by frictionless hinges to form a stable system. The arrangement of the members of a statically determinate truss must obey the rules of geometric construction of a stable system with no redundant restraint. According to the geometric construction characteristics, trusses can be grouped into the following three classes.

1. Simple trusses Beginning with the foundation or a basic triangle which is composed by three members (bars)

connected by hinges pairwise at their ends, enlarge a simple truss by attaching two new members (or links) which do not lie on the same straight line and are pined together, or by adding a binary system, to the foundation or the basic triangle to form a new joint. The truss can be further enlarged by repeating the same procedure as many times as desired. Trusses constructed by this procedure are called simple trusses.

(a) (b)

12

3

4

5 6

7

8

9 10

11

12

CD

E F 1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

C

D

E

F

G

H

JA

I

B K

B

Fig.6.7 Simple trussesA

As shown in Fig.6.7 (a), beginning with the foundation, enlarge a simple truss by attaching two new members (1, 2) to the foundation to form joint A, further enlarge the new truss by attaching in turn two new members (3, 4), (5, 6), (7, 8) … (11, 12) to form joints B, C, D, E and F so as to construct a simple truss.


In Fig.6.7 (b), beginning with the basic triangle ABC, enlarge the triangle by attaching in turn two new members (1, 2), (3, 4), (5, 6), (7, 8) … (15, 16) to form joints D, E, F , …, J and K so as to construct a simple truss, which is internally stable with no redundant restraint. Finally, the truss is attached to the foundation by one hinged support and one roller support to form a simply supported truss.

2. Compound trusses Compound trusses are constructed by connecting two or more simple trusses to form a new truss

accoink BC, which does not

cros

3. Complex trusses classified as either simple or compound are referred to as complex trusses. An

exam

ate the tension as the positive axial force and compression as the negative. In

rding to the geometric construction rule 2 or rule 3 discussed in the Chapter 2. In Fig. 6.8, two simple trusses ABE and CDE are connected by hinge E and l

s hinge E, to form an internally stable truss with no redundant restraint, and then the truss is attached to the foundation by one hinged support and one roller support.

Trusses that cannot beple of complex truss is shown in Fig.6.9. Sign convention of axial forces and indication of internal

forces for inclined members It is customary to design

AB C

D

E

Fig.6.8 A compound truss Fig.6.9 A complex truss

Fig.6.10 Similar relation between the dimension and member force of an inclined member

(a)(b)

y

NB

Nl Yx

A α

N

Xα

xl

yl


calcu

he analysis of inclined members, the axial force N of an inclined member will be resol

lating process, it is recommended that unknown axial forces should be indicated at their locations in their positive directions. Based on this regulation, if consequent results of the unknowns are positive they are actually positive, this means the corresponding members are in tension; if some of the unknowns are negative they are actually negative, their direction indication are just in the opposite direction, it means the corresponding members are in compression. Therefore, you can avoid the ambiguity caused by directions of unknown axial forces.

In order to facilitate tved into its horizontal and vertical components, X and Y, in the calculation. The axial force N [see Fig.

6.10 (a)] of an inclined member AB and its horizontal and vertical force components X and Y form a triangle shown in Fig.6.10 (b), while the length l of the inclined member AB and its horizontal and vertical projections xl and yl forms another triangle [see Fig. 6.10 (a)]. Since the directions of axial forces in the members of truss c incide with their centroidal axes, the two triangles are similar to each other. There exists the following relation:

a o

x y

N X Yl l l= = (6-1)

Generally, xl and are given, so the length l will be determined. By using Eq. (6-1), once either of the f io r

6.2 The Method of Joints

In the method of joints, the axial forces in the members of a plane statically determinate truss are deter

ylorce project ns X o Y are determined another force projection and axial force N may be obtained

easily.

mined by considering the equilibrium of its joints. Since the entire truss is in equilibrium, each of its joints must also be in equilibrium. At each joint of the truss, the member forces and any applied loads and reactions form a coplanar concurrent force system, which must satisfy two equilibrium equations,

0X =∑ and 0Y =∑ , in order for the joint to be in equilibrium. These two equilibrium equations must d at eac f the truss. There are only two equations of equilibrium at a joint, so they cannot

be used to determine more than two unknown forces. In its practical application, the method of joints i

be satisfie h joint o

s suitable for analysis of simple trusses. Based on the geom

onstructing the truss, starting from stationary joint A and B, in turn

etric constructions of a simple truss, if we consider the joints successively in the order reverse through that it were set up, we could proceed from one joint to another, in such a way that there are never more than two unknown axial forces acting on the joint under consideration, to accomplish the calculation of all unknown axial forces of the simple truss.

As the truss shown in Fig.6.11, when cadd a binary system to fix a new joint in the order of C, D, E, F, G and H. when analyzing the truss,

6.2 The Method of Joints 127

the order of isolating joints should be the reverse of above constructing order. That is, we start with joint H at the right bottom end and proceed in succession to joints G, F and E, then to joints D and C. Apparently, if we calculate unknown axial forces in this sequence, there are never more than two unknown axial forces acting on the joint under consideration. For each joint, we can apply two equilibrium equations, and 0X =∑ 0Y =∑ to solve two unknown axial forces, so all of the unknown axial forces can be obtained successfully.

Example 6-1

Calculate the axial forces of every member of the truss shown in Fig.6.12 (a) by using the method of joints.

Solution

1. Calculation of reactions Considering the entire truss as a free body, we write

，

C

D

E

F

G

H

P P

1

23

4

5

67

8

9

10 11

12

Fig.6.11 A simple truss

A

B

0X =∑ 0AX =

， 0AM =∑ 12kN( )BY = ↑

, 0BM =∑ 12kN( )AY = ↑

2. Calculation of axial forces The truss shown in Fig.6.12 (a) is a simple truss. When constructing the truss, starting from a triangle

BEG, in turn add a binary system to fix a new joint in the order of D, F, C and A. when analyzing the truss, the order of isolating joints should be the reverse of the above constructing order. That is, we start with joint A at the left end and proceed in succession to joints C, F, D and E, then to joint G. Apparently, there are only two unknown axial forces acting on the each desired joint. For each joint, we can apply two equilibrium equations, 0X =∑ 0Y =∑ and to solve the two unknowns, so all of the unknown axial forces can be obtained successfully.

(1) joint A Taking joint A as a free body as shown in Fig.6.12 (c), depict reaction in its real magnitude and

direction, assume the unknown axial forces and being tensile forces, resolve the unknown axial force in the inclined member, whose length can be seen in Fig.6.12 (b), into components

AY

ACN AFN

ACN


ACX and along the horizontal and vertical directions, respectively. By writing the equation of vertical force eq m, we can immediately solve for

，

ACYuilibriu ACY

0Y =∑ 12 0ACY + =

By using proportional relation, Eq. (6-1), we find

12kNACY = −

12 2 16k1.5ACX = − × = − N

(a)

A B

C D E

FG

2m 1.5m 2m

1.5m (b)

A2

2 .5C D

F2 .1 2

1 .5

(c)

12kN

9kN 6kN 9kN

ACYACN (e)

16kN

3kN

FA

ACX

AFN

(d)

20kN12kN

16kNC

CFN

CDN

FDYFN D

FDX

FN G

12

16+

3

9kN 6kN 9kN

16− 16−

19+ 16+

3

12

4.24

16 16

6

D

33

3 3 4.24

Fig.6.12 The simple truss and its free bodies of example 6-1

9kN

(f) (g)

-20 -20-4.24 -4.24

unit: kN

1.5

1.5

1.5m


2.512 21.5ACN = − × = −

By writing the equation of horizontal force equilibrium

0X =∑ ,

0kN (compression)

, we obtain

0=+ AC

(tension)

.6.12 (e). In which given forces (the load and axial e real de and direction, unknown axial forces and

ibrium equations to the free body, we obtain

AF XN

16kNAF ACN X= − =

(2) joint C The free body diagram of joint C is shown in Fig

force of member AC) ar depicted in its magnitu

CFN are assumed to be tensile forces. Applying the equil

0X =∑ , CDN

CDN

16 0+ =

16kNCDN = − (compression)

0Y =∑ , 12 9 0CFN− − =

3kNCFN = (tension)

F nt F is shown in Fig.6.12 (d); the unknown axial force in inclined

mem b), is indicated by its horizontal and vertical components

(3) joint

FDNThe free body diagram of joiber FD, whose length can be seen in Fig.6.12 (

FDX and . By writing the on of solve for

FDY equati vertical force equilibrium, we can immediately

FDY

0Y =∑ ， Y 3 0FD + =

3kNFDY = −

By u

sing Eq. (6-1), we find 2.123 4.24kN1.5FDN = − × = − (compression)

1.53 3kN1.5FDX = − × = −

By applying the equation of horizontal force equilibrium, we obtain

X =∑ 16 0FD FGX N+ − =0 ,

19kNFGN = (tension)

etric istribal to

(4) Utilization of symmetry Since the truss and its loads are all symm al the d ution of the internal forces of the truss must

be symmetric o. So the axial forces of two members which lie on the symmetrical positions must be the


samen in ig.6.12 (f).

(5) Checking In order to check the correctness of above calculation, apply the equilibrium conditions of joint D,

which has not been utilized so far and whose free body diagram is shown in Fig.6.12 (g). Since the symmetry is considered the equilibrium condition

. Therefore, the axial forces of the other portion of the truss may be determined by symmetrical condition. The finally calculating results are show F

0X =∑ is automatically satisfied and the other equilibrium condition 0Y =∑ is needed to check as follows

, 0Y =∑ 6 3 3 0− + + =

The constructing sequence of the truss shown in Fig.6.12 (a) may be also considered as: starting from a triangle ACF, in turn add a binary system to fix a new joint in the sequence of D, G, E and B. So the sequence of isolating joints should be: starting with joint B at the right end and proceeding in succession to joints E, G, D and C, finally to joint F.

Special cases of the method of joints—Identification of zero-force members

Members with zero forces in them are named zero-force members. Because trusses are usually designed to support several different loading conditions, it is not uncommon to find members with zero forces in them when a truss is being analyzed for a particular loading condition. Zero-force members are also added to trusses to brace compression members against buckling and slender tension members against vibrating. The analysis of trusses can be expedited if we can identify the zero-force members by inspection instead of calculation. Three common types of member arrangements that result in zero-force members are the following:

1. If only two noncollinear members are connected to a joint that has no external loads or reactions applied to it [Fig.613 (a)], then the force in both mcentroidal line of member 1 as x axis of a rectangular coordinate system we can immediately obtain

(a) (b)(c)

embers are equal to zero. For instance, if taking the

A

1N 2N

1 2 0N N= =P

1N2N

2 10,N N P= = −

1N 2N

3N

1 2 3, 0N N N= =

Fig.6.13 Identification of zero-force members


0Y =∑ 2 0N = by applying the equilibrium equation an2. If two noncollinear members are connected to a joint [Fig.613 (b)] and the external loads of the

al line of one of the two members, then the axial force in the member whose cen al line does noequal to zero. For example, if we take the line of arecta te system we can immediately find

d vice versa.

joint are applied in the direction which coincides with the centroidtroid t coincide with the acting-lines of the loads is

ction of load P and axial force 1N as x axis of a

2 0N =ngular coordina by applyi e equilibrium equa

For inst nate system we can immediately find plying the e

ng thtion 0Y =∑ . 3. If three members, two of which are collinear[Fig.613 (c)], are connected to a joint that has no

external loads or reactions applied to it, then the force in the member that is not collinear is equal to zero. ance, if taking the connecting line of two collinear members as x axis of a rectangular coordi

by ap quilibrium equation 0Y =∑ . 3 0N =

Example 6-2

Identify the zero-force members of the truss shown in Fig.6.14.

1 2 3 4=

2. Identifying joints K, L, E, M and N, we find (arrangement 3)

5 6 7 8 9 0N N N N N= = = = =

and J (arrangement 3)

N N

P P

D E FG

C 1 3

Solution K L M N

1. Starting with joint C and G, we identify (arrangement 1)

0N N N N= = =

A B

2 45 6 812 13

H I J

7

910 11

14 15

Fig.6.14 The truss of example 6-2

3. Identifying joints H

010 11= =

4. Identifying joint I Since joint I lies on the symmetrical axis of the t

qual to each other, i.e.,

russ and members 12 and 13 are located at the symmetrical position, their axial forces 12 and 13 must be eagain since 0N = , by the equilibrium equation 0Y

N N 12 13N N= ; =∑ , we find 7 12 13 0N N+ = , i.e.,

This conflicts with the fact of , so two ai.e.,

n joints L and M, by considering the equilibrium of the two joints, we

12 13

= xial forces must be equal to zero, N N= − .

12 13N N the

12 13 0N N= = . 5. Focusing our attention oobtain

14 15 0N N= =


It can be seen finally that under the action of the two concentrated loads only the members depicted with zebralike lines are not equal to zero (see Fig.6.14).

6.3

e desiredrium of o ss. Each portion of the truss is treated as a

rigid body in equilibrium, under the act of any appliedthat

ugh more than three nonconcurrent and nonparallel members with unkn which there are only three m n imaginary section throu

The Method of Sections

The method of sections involves cutting the truss into two portions by passing an imaginary section through the members whose forces are desired. Th member forces are then determined by considering the equilib ne of the two portions of the tru

ion loads, reactions and the forces in the members have been cut by the section. The unknown member forces are determined by applying the three

equations of equilibrium to one of the two portions of the truss. There are only three equilibrium equations available, so they cannot be used to determine more than three unknown forces. Thus, in general, sections should be chosen that do not pass thro

own forces. The truss with parallel chords shown in Fig.6.15, in each panel of members (one top chord, one botto chord and one diagonal). If passing agh any one of the panels three members will be cut and the truss will be divided into two portions, the

three unknown forces can be determined by considering the equilibrium of any one of the two portions of the truss. For instance, if passing section Ⅰ-Ⅰ through panel BC, the unknown forces of members 1, 2 and 3 will be determined by applying equations of equilibrium 0BM =∑ , 0CM =∑ and 0Y =∑ to any one of the two portions of the truss.

0.5kN 0.5kN1kN1kN1kN

In the process of calculation, in order to avoid solving simultaneous equations it is preferable to select the types of equilibrium equations. The progress for applying the method of sections is now explained through the following example.

AB C D

E

a b c d e

1

2

3

4d

d

Ⅰ

Ⅰ

Fig.6.15 A truss with parallel chords

6.3 The Method of Sections 133

Ex

lying equ um equations to the entire truss, we find

ample 6-3

Calculate the axial forces of members 1 through 3 of the truss shown in Fig.6.16 (a) by using the method of sections.

Solution

1. Calculation of reactions App ilibri

0AM =∑ ， 1 (2 3 2 6 1 9) 1.5kN( )

18BY = × + × + × = ↑

0BM =∑ , 1 (1 9 2 12 2 15 1 18) )

18= × + × + × + × = ↑ 4.5kN(AY

Checking: 0Y =∑ , 1 2 2 1 4.5 1.5 0+ + + − − =

n in Fig.6.16 (b), a section m-m is passed through members 1, 2 and 3, cutting the truss into 2. Find axial forces of member 1 through 3 As show

21kN

kN 2kN 1kN

E G

D Hm

m I

1

3

2

(a)J

C

1N

A BF

6 3m=18m×

3m 1.

5m

(b)

(c)

G2N

3N

2X

3X 3Y

H

2Y J

B

I1kN

3.040.5

3G

I

G

H

2.53.91

truss of ex

F 3

1.5kN

Fig.6.16 The ample 6-2


two portions, ACGF and BJIH. For simplifying the calculation, used here

are assumed to be tensile and indicated by arrows pulling away from the corresponding joints on the diagram. The unknown member forces can be

m s of equilibrium to the free body. However, in order to avoid solving simultaneous equations we should select the types of equilibrium equations. It is helpful to select the inter

the right-hand portion, BJIH, will be to analyze the axial forces. The free-body diagram of the portion BJIH of the truss is shown in

Fig.6.16 (b). All three unknown forces, 1N , 2N and 3N

deter ined by applying the equation

secting point of two forces as the centre of moments to find the third unknown force, then to determine the other two unknown forces by applying equilibrium equations of force projections.

(1) Because 2N and 3N pass through point G their moments about point G are equal to zero. So we can use the equilibrium of the moments about G to calculate member force 1N as follows:

0GM =∑ ， 1 2.5 1 3 1.5 12 0N × + × − × =

1 6kNN = (tension)

(2) We calculate 2N by summing moments about point H, which is the point of intersection of the lines of action of 1N and 3N . In order to facilitate the calculation we resolve axial force 2N at point I into its horizontal and vertical components, 2X and 2Y , along the horizontal and vertical directions; the length of member 2 is shown in Fig.6.16 (c). Equilibrium of moments about H yields

0HM =∑ , 2 3 1.5 9 0X− × − × =

4.5kNX =2 −

By proportion relation, Eq. (6-1), we find

23.044.5 4.56kN

3N = − × = − (compression)

20.54.5 0.75kN3

Y = − × = −

(3) After the axial forces and have been determined, we may utilize the equilibrium equation of force projection in the horizo ction to fin . By summing the force components in horizontal direction of all forces imposed to the free body, we write

,

By proportion relation, Eq. (6-1), we obtain

1N 2Nntal dire d 3N

0X =∑ 2 1 3 0X N X− − − =

3 2 6 4.5 1.5kNX N X= − − = − + = −

33.911.5 1.96kN

3N = − × = − (compression)

6.3 The Method of Sections 135

32.51.5 1.25kN3

Y = − × = −

(4) Checking computations To check our computation, we apply the equilibrium equation 0Y =∑ which has not been utilized

so far to the free-body diagram. Thus

0Y =∑ 1.5 1 0.75 1.25 0− + − =, (satisfied)

Note that since the truss is a simple truss the axial forces of members 1, 2 and 3 can be determined by the method of joints. For doing so, however, the member forces connected to joints A, C, D and E have to be involved in the calculation before joints F and G are reached that can be analyzed for members 1, 2 and 3. Therefore, the method of sections enables us to determine forces in the specific members of trusses directly, without first calculating many unnecessary member forces, as may be required by the method of joints.

It is instructive to emphasize that if we consider the order of calculating axial forces in the members of a truss as the reverse order through that it were set up, we could accomplish the calculation of all unknown axial forces without encountering any trouble. Since compound and complex trusses are constructed by several simple trusses the method of sections is indispensable to analyzing the member forces of these types of trusses. For instance, the truss shown in Fig.6.17 (a) is a compound truss formed by connecting two simple trusses, AEFC and BGFD, by a common hinge F and a member CD. Thus, the order of calculating axial forces in the members of the truss ought to be: first calculating the member force of member CD; then the members FG and FI (or FE and FH) by using the method of sections [section aa or bb as shown in Fig.6.17 (a)]. Once the connecting member forces of CD, FG and FI (or FE and FH) are

known, we can focus our attention on the simple trusses and to find corresponding member forces by the method of joints. Apparently, following above mentioned analyzing sequence we will determine the unknown member forces of the truss successfully. The truss shown in Fig.6.17 (b) is a compound truss as well, after the connecting member forces of 1, 2 and 3 are determined by using the method of sections, the

(b)

1

2 3

(a)

BC D

E F Gb a

a b

IH

Fig.6.17 Compound trusses

A


whole member forces in the truss can be obtained without any difficulty. Generally, when sections are chosen that do not pass through more than three members with unknown

forces we may directly use the equations of equilibrium for a coplanar force system to determine the unknown forces. However, in some trusses, the arrangement of members may be such that by using sections that pass through more than three members with unknown forces, we can determine one or, at most,

two unknown forces. Such sections are employed in the analysis of only certain types of trusses. For instance, the truss shown in Fig.6.18 (a), the section m-m cuts five members, but four of which are concurrent to joint C. So by considering either portion of the truss (divided by section m-m) and summing the moments about C we can find the axial force in the member (numbered 1) not crossing the joint C. Another example is the truss shown in Fig.6.18 (b). The section m-m cuts four members, but three of which are parallel each other. So by considering either portion of the truss (divided by section m-m) and summing the force components in the direction perpendicular to the axial lines of the three parallel members we can obtain the axial force in the member (numbered 1) being not parallel to the other three members.

6.4 The Combination of the Method of Joints and the Method of Sections

Although the method of joints and the method of sections described in the preceding sections can be used individually for the analysis of trusses, the analysis of compound or some complex trusses can sometimes be expedited by using a combination of the two methods. For some types of compound or complex trusses, the sequential analysis of joints breaks down when a joint with more than two unknown forces be found. In such a case, the method of sections is then employed to calculate some of the member forces, thereby yielding a joint with two or fewer unknowns, from which the method of joints may be continued. This approach is illustrated by the following example.

Example 6-4

Find the axial forces of members 1 through 3 of the truss shown in Fig.6.19.

(a)

BC

m

m1 (b)

AA B

1m

m

Fig.6.18 Special trusses

6.4 The Combination of the Method of Joints and the Method of Sections 137

Solution

1. Finding reactions Considering the whole truss as a free body, we obtain

AX

Findi members 1, 2 and 3 cut into two portions. Taking the right

porti body diagram is omitted) and by summing the moments about point C of all f s applied to the portion, we find

12kN( ), 4kN( ), 0A BY Y= ↑ = ↑ =

2. ng the axial forces of (1) By using section m-m as shown in Fig.6.19 (a), the truss ison on the section m-m of the truss as a free body (the free-

orce

0CM =∑ 1 4 4 8 0N × − × = 1 8kNN =, ， (tension)

(2) By using section n-n as shown in Fig.6.19 (a), the truss is cut into two portions. Fothe analyzing, the axial force

r simplifying

2N of member 2 [whose length is shown in Fig.6.19 (c)] is resolved into a horizontal component 2X and a vertical componen . By taking the left portion on the section n-n of the truss as a free body [see Fig. (b)] and by summing the moments about point F of all forces imposed to the portion, we find

t 2Y

0FM =∑ 2 2(12 2) 4 4 2 1 2 8 2 0Y X− × − × − × − × − × = ，

By using the proportion relation of Eq. (6-1), we write

22 1.5

2YX = ×

Solving above two equations simultaneously, we obtain

2 6.4kNY =

22 2.5 8kN

2YN = × = (tension)

(3) With and now known, there are only two unknowns, and1N 2N 3N DHNody of jo

, at joint D. These forces can be determined by applying the two equations of equilibrium to the free b int D, as shown in Fig.6.19 (d). For simplifying the analyzing, the axial force of member 3 [whose length is shown in Fig.6.19 (c)] is resolved into its horizontal and vertical com nts,

3N pone 3X and . By summing the force

components in the vertical direction of all forces applied to joint D, we write

，

3Y

0Y =∑ 3 6.4 0Y + =

3 6.4kNY = −

3 5 72

N = −6.4 .16kN= − (compression)


It can be seen from the above example that when utilizing a combination of the method of joints and the method of sections an optimal method of the two methods may be selected according to the location of the member whose axial force is desired.

6.5 Form and Characteristics of Member Forces of Girder Trusses

used in pra

In order to make a clear underst about the argument tused girder trusses, which are parallel chord trusses, triunder the action of top-chord-joint loads (which are the sim

Since girder trusses are widely ctical engineering structures it is interest to discuss some of their form and member force characteristics. In fact, there is relation between the configuration and the member force characteristic of a truss, i.e., different configuration has different member force characteristic.

and he member force performance of three commonly angular trusses and parabolic trusses, are analyzed

plification of uniformly distributed loads). The bending moment 0M and shear 0Q of a corresponding simply supported beam (with the same span and subjected to the same loads) [see Fig.6.20 (b)] are utilized here to represent axial forces in a truss.

1. Parallel chord trusses [Fig.6.20 (a)] (1) Axial forces in chords When calculating the axial forces in chords of a parallel chord truss the method of sections may be

3

C

GF

nm

2kN

2kN

4kN

4kN

4kN(a)

4m 1.5

2.52

D

G(c)

BAED nH m1

2F

2m

5m

2m 2m 2m

3m52

D18m

(b)

F

A

2kN

2X

D

4kN

4kN

2

(d)

3Y3N 8kN

6.4kNN

2Y3X 4.8kN

DHN8kND

8kN

12kN

Fig.6.19 The truss and some its free body diagrams of example 6-4

6.5 Form and Characteristics of Member Forces of Girder Trusses 139

used. That is, passing a section through a panel cutting the truss into equilibrium equations to either of the two portions and by selecting the moment centres at the intersection poin mem

example, if we are interested in the axial force in bottom chord 1 as show

l force in top chord 2 as show

expressed as follows

two portions, by applying moment

ts of the chords and the webbers, the axial forces of chords

will be obtained successfully. For

n in Fig.6.20 (a) the joint f, the point of intersection of top chord 2 and web member Ef, should be chosen as the moment centre of the desired moment equation; if we desire to find the axia

n in Fig.6.20 (a) the joint E, the point of intersection of bottom chord 1 and web member Ef, should be chosen as the moment centre of the desired moment equation. The arms of the axial forces in chords are equal to the height h of the truss.

The formula to calculate the axial forces in chords of the truss may be

0' MN

h= ± (6-2)

in whic 0h, M is the bending moment in the corresponding simple beam on the section corresponding to the desired location of moment centre of the truss; h is the arm of the axial force; the positive sign is for the axial forces in the bottom chords under the action of tensile forces; the negative sign is for those in the top chords under the action of compressive forces. Since the height h of the parallel chord truss is a constant, the variation of the axial forces in chords is directly proportional to that of the bending moment diagram of the corresponding simple beam. i.e., the axial forces in the middle chords are larger than those in the end chords of the truss, or the axial forces in the chords would decrease from the middle to the ends of the truss.

(a)A

a

C D E F GB

6 d×

P P P P P/ 2P / 2Pc d e f g b2

1

h

/ 2P P P P P P / 2P

6 d×

AC D E F G

B(b)

3P 3P

(c)

(d)

0 diagramM

2.5Pd4.0Pd 4.5Pd

1.5P2.5P

Fig.6.20 A parallel chord truss and its corresponding simple beam

0 diagramQ0.5P

2.5P1.5P

0.5P


(2) Axial forces in web members When calculating the axial forces in the diagonals and the verticals of a parallel chord truss the method

of sections may be used as well. That is, passing a section through a panel cutting the truss into two portions, by applying vertical force projection equations of equilibrium to either of the two portions, the axial forces may be determined successfully. The formula to calculate the vertical components in the

truss into two portions), the sign for axial forces in the diagonals and the verticals must be accordingly changed as well. Apparently, the variation of the axial forces in web members is directly proportional to that of the shearing force diagram of the corresponding simple beam. i.e., the axial forces in the end-portion web mem

diagonals and the axial forces in the verticals of the truss may be written as follows

(6-3)

in which, is the shear in the corresponding simple beam on the section corresponding to the location of the desired panel of the truss; the positive sign is for the vertical components of axial forces in the diagonals under the action of tensile forces and the negative sign is for the axial forces in the verticals for the arrangement of the truss shown in Fig.6.20 (a). If exchanging the right-portion’s diagonal direction with

the left-portion’s diagonal direction as shown in Fig.6.20 (a) (the vertical line of symmetry of the truss partitions the

bers are middle-portion

xial

forces in chords of a triangular truss may be still expressed as follows

0Y Q= ±0Q

larger than those in theweb members of the truss, or the aforces in the web members would increase from the middle to the ends of the truss.

2. Triangular trusses [see Fig.6.21 (b)] (1) Axial forces in chords The formula to calculate the axial

(a)

12 1 1 1 1 1

12

2.5− 4.0− 4.5−

0 2.5 4.0

6 d×

3.0

−

3.54 2.5

− 2.21 1.5

− 0.71 1.0

− h

12

7.5 7.5 6.0 7.91−

6.32−

1 4.74−1

11

11

h

20 1.58− 5

+0.

1.8− 2.0+ (b)

6 d×

(c)

h

6 d×

12

112

1 1 11

4.5 4.5 4.50

0 00 0

5.15−

4.75− 4.53−

Fig.6.21 Three commonly used trusses

6.5 Form and Characteristics of Member Forces of Girder Trusses 141

0' MN

r= ±

in which, 0M is still the bending moment in the corresponding simple beam on the section corresponding to the de ation of the moment centre of the truss; r is the arm of the which is the distance between the desired chord and the centre of the moment, the variation of r is linear from ude of the arm r reduces more quickly than that of the b

sired loc axial force in the desired chord,

the middle to the ends of the truss; Since the magnitending moment 0M , the variation of the axial forces in chords increase from the middle to the ends

of the truss, or the axial forces in the middle-portion chords are smaller than those in the end-portion chords of the truss.

(2) Axial forces in web members c

verticals are tensile members. 3. Parabolic trusses

rabolic curve, the variations of the length r of the verticals and the coordinates of the bending mo

It is onvenient to determine the axial forces in the diagonals and the verticals of a triangular truss by means of method of sections, i.e., passing a section through a panel cutting the truss into two portions, by applying moment (about joint at support) equations of equilibrium to either of the two portions, the axial forces may be determined successfully. Apparently, the axial forces in the end-portion web members are smaller than those in the middle-portion web members of the truss, and the diagonals are compressional member and the

Since the panel points along the top chords locate on a pa0M ment of the corresponding simple beam obey the

he horizontal components of axial forces in top chords same parabolic function (the formula for calculating t

and 0MN = ± ) the axial forces in bottom chords are still

r, the horizontal component of axial forces in

the top chords and the axial forces in the bottom chords of each panel of the truss would be identical. Because the variations of the inclinations of the top chordsmay be considered as being nearly identical. All of the web members of the truss are the zero-force mem rea th ax

he en Fig.6.21 that are obtained in

the case By analyzing the distribution me argumen (1) Parallel chord trusses: The axial forces in the memb

The axial forces in the middle chords are larger than those in the end chords of the truss. From mechanical point of view, when constructing this type of truss the cross-secould be different or nonuniform, which would yield difficulty to connect the members together; if we

are small enough the axial forces of top chords

bers for the son that e magnitudes of the horizontal component of ial force in the top chord and the axial force in the bottom chord of each panel of the truss are equal to each other but in opposite sense (one is tension another is compression).

In the interest of comparing the member force characteristics of t above m ntioned commonly used three types of girder trusses, the distributions of their axial forces are shown i

h d= . , so ts are stated as follows: ers of a parallel chord truss are nonuniform.

ctional areas of the chords in each panel


adop

rmalizing web members and so forth, they are widely applied to engineering structures, especially to light-weighted structures for the reason that the same cross-sectional area of chord members will not cause too much waste of constructing materials. So the parallel chord trusses may be often appeared in long-span crane girders, supporting-structure systems and railroad bridge structures and the like.

(2) Triangular trusses: The axial forces in the members of a triangular truss are nonuniform as well. The axial forces in the end chords are larger than those in the middle chords of the truss, and the inclinations of end chords are too small to manufacture. However, since the inclined configuration satisfies the requirement of roof drainage triangular trusses are often applied to roof structures with a bit small span and large inclination, such as wooden roof structures, light-weighted-steel roof structures and composite (consisting of reinforced concrete inclined beams and steel tensile members) roof structures.

(3) Parabolic trusses: The axial forces in the members of a parabolic truss are by and large uniform. So their constructing materials would be utilized more economically. However, since the inclinations of top chord in each panel have different value the joint conformations of the truss are complicated so that it is quite difficult to manufacture this type of truss. In practical engineering structures, parabolic trusses may be often occurred in long-span (18m~30m) roof structures and bridge structures.

6.6 Composite Structures

6.6.1 Constitutes and types of composite structures

ve different stressing xial forces, thus called

two-

l t a

re named three-hinged rigid-frame-like structure, of which colu others are two-force members.

t the uniform cross-sectional area for each chord member, their constructing materials would be wasted. However, since the advantages in manufacture of parallel chord trusses, such as uniform joint structure, the identical length of chord members and no

A composite structure is composed by two types of members which haperformance. One type of members is the members which are only subjected to a

force members. The other type of members is the members whose dominating internal force is bending moment and additional internal forces are shearing forces and axial forces only in some cases, which are thusly named beam members or flexural members. Composite structures often come into your sight when you focus your attention on the roofs of buildings in industrial and civilian architectures, crane girders and supporting systems of bridges.

The computing model of a pentagonal below-supported roof structure is shown in Fig.6.22 (a), whose top chords are flexura members tha re made of reinforced concrete and bottom chords and web members are all two-force members which are made of steel shapes.

Fig.6.22 (b) shows a composite structumns are flexural members andFig.6.22 (c) shows an analyzing model of an arched bridge. It is a composite structure composed by

6.6 Composite Structures 143

stiffened beam and linked arch constructed by two-force members.

6.6.2 Analysis of composite structu

The most fundamental point for ancharacters of different members. It is essthree types of internal force componentswhen using the method of sections unknown internal forces acting on desired imaginary section should coincided with the stressin

(b) (c)

Stiffened beam

Fig.6.22 Three composite structures

(a)

res

alyzing a composite structure is to distinguish the different stressing ential that there exists only axial force in a two-force member, but (bending moment, shear and axial force) in a flexural member. So the the be

g characteristic of t m

ary rnal forces on the

section must be be ear and axial force in some cases; while a

mber is cut by the desired al

xial

he cut member. That is, if a beamember is cut by the desired imaginsection the unknown inte

nding moment, sh

two-force meimaginary section the unknown internforces on the section ought to be only aforce.

Similarly, when taking a joint as a free body it is essential to differentiate whether the joint is a perfect hinged joint, the intersection point of two-force members, or a nonperfect hinged joint, the intersection point of two-force members and flexural members. As shown in Fig.6.23 (a), joint C is a perfect hinged joint and joint D is a nonperfect hinged joint.

B(a) (b)B C

BN C

BCQD

BN D

A

(c) DN B

DM BDN C

DBQ

DDM A

DAQ

DN A

Fig.6.23 A composite structure and some of its free body diagrams


It should be noted that the conclusion about the method of joints discussed in the previous section of the c

load applied to joint B; whereas the member DC, which is connec

e

hapter is based on the fact that the joint is a perfect hinged joint. For a nonperfect hinged joint formed by flexural members and two-force members, the conclusion about the method of joints only being suitable to trusses must not be applied to it because there exist three internal force components in flexural members. As shown in Fig.6.23 (a), the member BDA is a flexural member, the member BD and BC, which are connected by joint B [Fig.6.23 (b)], are not zero-force members although there is no

ted to joint D [Fig.6.23 (c)], is not zero-force member as well since there xist axial forces DCN , DBN and DAN , shears DBQ and DAQ , the bending moments

DBM and DAM on the free-body diagram of joint D. It is worth of emphasizing again that the order to calculate internal forces in the members of a

composite structure should be the reverse order through that it was set up. Because the flexural members he internal forces in the members of a members, and then the internal forces in their diagrams may be constructed. The

cture shown in Fig.6.24.

have at least two internal force components the first step to find tcomposite structure is to determine the internal forces in two-force flexural members. When the desired internal forces are determined procedure is illustrated by the following example.

Example 6-5

Construct the internal force diagrams of the composite roof stru

Solution

1kN / m

(a) (b)

(c) (d)

A BC

D EGF

1kN/m

3m 3m 3m 3m

3m

Ⅰ

Ⅰ

AC

D

F

6kNDEN

CY

CX

A CF

6kN6kN6kN

6kN

CFN

CFQ

A

6kN6kN

AFQ

AFN6kN

Fig.6.24 The composite structure and some of its free-body diagrams of example 6-5

6.6 Composite Structures 145

(1) Find reactions By considering the equilibrium of the whole composite structure and utilizing the symmetry of the

e wstructure, w rite

0X =∑ ， 0AX =

1 11 12 6kN( )2 2A BY Y ql= = = × = ↑

(2) Calculate the axial forces in two-force members Geometric construction analysis: The structure may be considered as a system formed by two rigid

bodi ge, C, and one link, DE, which does not cross the hinge. So the struc

section [Fig.6.24 (b)], we obtain

es ADC and BCE by using one hinture is a stable system with no redundant restraint. Based on the geometric construction sequence of

the structure, its analyzing steps should be: first calculating the internal force in the link, member DE, and restraint forces of hinge C; then restraint forces at joints D, E, F and G; finally the internal forces in the beam member AFC or CGB. The approach is implemented as follows:

By using the method of sections (the imaginary section is I-I) and by applying moment equilibrium equation about hinge C to the left portion of the

0CM =∑ 6 6 1 3 6 3 0DEN× − × × − × = ，

6kNNDE = (tension)

By considering the equilibrium of joint D, we write

0X =∑ 6kNDAX = ，

6kNDAY =

6 2 8.48kNDAN = × = (tension)

0Y =∑ , 6kNDFN = − (compression)

(3) Internal forces of flexural members Take beam member AFC as a free body whose diagram is shown in Fig.6.24 (c) and select sections A,

F and C as the control sections. When focusing our attention on the equilibrium of joint A [Fig.6.24 (d)], we find

0X =∑ ， 6 0AFN + =

6kNN = −AF (compression) 0Y =∑ , 6 6 0AFQ + − =

0AFQ =

By considering the equilibrium of flexural member AFC [Fig.6.24 (c)], we write


0X =∑ ， 6 0CFN + =

6kNCFN = − (compression)

0Y =∑ , 1 6 6 6 6 0CFQ× + − − + =

0CFQ =

By using the method of sections and calculating from A to F, we find

1 3 1.5 4.5kN mFA

6 6 1 3 3kN mFAQ = − + − × = − ⋅

6kN mFAN = − ⋅ (compression)

Similarly, by using the method of sections and calculating from C to F, we find

4.5kN mFCM = − ⋅ (tension in upper fibers)

1 3 3kN mFCQ = × = ⋅

M = − × × = − ⋅ (tension in upper fibers)

s S AFC and CGB are subjected to uniformly distributed loads, when

constructing their internal force diagrams the superposition method may be used. The final bending nt in Fig.6.25 (a) through (c).

6kN mFCN = − ⋅ (compression)

Because the structure and its loads are all symmetrical the distribution of its internal forces must be symmetrical too. After the internal forces of the left portion AFC of the structure are determined, those in the right portion CGB of the structure should be determined according to the symmetry of the structure.

(4) Constructing bending moment M, shear Q and axial force N diagramince the beam members

mome M, shear Q and axial force N diagrams are shown

(a) (b)

6kN

AF C G

B AF C G B

3

3

3

3

diagram (kN)Q4.5kN m⋅4.5kN m⋅

Fig.6.25 The composite structure and its internal force diagrams of example 6-5

(c) diagram (kN)N

8.48kN 8.48

kN 6k

N− 6k

N−

6 6A F C G

B diagram (kN m) and diagram (kN)M N⋅

Summary 147

SUMMARY

The chapter was mainly discussed the analysis of plane statically determinate trusses. Its key points are as followings:

1. The member forces of a truss are only axial forces if the truss satisfies the assumptions made for a perfect truss. So readers should understand the conditions to simplify a practical structure to a perfect truss.

2. There exists relationship between the analyzing sequence and the order of the geometric construction for a truss. It is instructive to emphasize that if we consider the sequence of calculating axial forces in the members of a truss as the reverse sequence through that the truss were set up, we could accomplish the calculation of all unknown axial forces without encountering any trouble. The trusses may be classified into three kinds of trusses, simple trusses, compound trusses and complex trusses according to

eo aking, the method of joints is suitable to analyzing the member forces

tions usually proves to be more convenient when forces

he inactive members or zero-force members may be found ation, Eq. (6-1), between the member force components and

used. influence on the axial forces in a truss. By comparing the

sed girder trusses with their corresponding simple beams, the haracteristics of the trusses ought to be understood clearly.

ting of flexural and two-force members, which must be ferentiate whether the joint is a perfect or nonperfect hinged the members of a composite structure is: first the axial forces s in flexural members.

blems for Reflecting

for a perfect truss? Why there exist only axial forces in the

ation between axial force N and

their g metric constructions. Generally speof simple trusses while the method of sections will be adapted to compound and some complex

trusses. Since the method of joints and the method of sections are two fundamental methods to analyze trusses,

they must be soundly mastered. The method of joints is most efficient when forces in all or most of the members of a truss are desired. The method of secin only a few specific members of the truss are desired.

In the interest of simplifying the analysis, tbefore starting calculation, and proportional relprojections of member length should be neatly

3. The configurations will impose anstressing performance of three commonly udistribution of member forces and structural c

4. A composite structure is consisdistinguished accurately. It is essential to difjoint. The order to find the internal forces in in two-force members, then the internal force

Pro

6-1 What are the fundamental assumptionsmembers of a truss under the three assumptions?

6-2 How many types of geometric constructions of the trusses? How classify them? 6-3 Try to deduct equation (6-1) based on the trigonometric-function rel


its horizontal and vertical components X and Y in an inclined member. 6-4 How using the characteristics of geometric constructions of simple trusses to find their member

forces? 6-5 How to distinguish zero-force

members? 6-6 How to selection equilibrium

equations so as to avoid solving simultaneous equations when using the method of sections to find m er forces of a truss. embHow using the characteristics of geometric constructions of compound trusses to find their member forces?

6-8 Comparing the differences of member forces between the three

6-7

n in the problem figure. If the joint loads are

of the bottom chords, what are the variations of their member forces?

6-9 How to distinguish the two-force and non-two-force members in a composite structure? What are their stressing differences? Which attentions should be paid when analyzing them? Whether the members a and b in the truss shown in the problem figure are zero-force members?

6-10 What is the procedure to analyze a composite structure?

(a)

P P P P P /2P/2P

trusses show

removed to the corresponding position

AC D E F G B

a c d e f g b

h

6l d= ×

h

(b)

(c)

h

Problem for reflecting Fig.6-8

EC

a

b

F

P

B

D

A

Problem for reflecting Fig.6-9

CHAPTER 7 GENERAL REMARKS ON

STATICALLY DETERMINATE STRUCTURES

The abstract of the chapter The chapter is the summaries of the five preceding chapters treated of from the analysis of the

geometric constructions of structures to that of internal forces in statically determinate structures. Unlike the five preceding chapters, which are classified on the types of structures, the chapter, combining all of statically determinate structures together, will discuss the structural analysis in following three aspects, the methods for analyzing statically determinate structures, the general performance of statically determinate structures and the stressing characteristics of various types of statically determinate structures, so as to enhance and deepen the comprehension about statically determinate structures and their analytical methods.

7.1 Analytical Methods For Statically Determinate Structures

The analysis of statically determinate structures mainly treats of the determination of reactions and internal forces existing in statically determinate structures by means of equilibrium equations, and the constructions of the internal force diagrams. The basement of the analysis is the isolation of free-body diagrams, on which internal forces and reactions may be emerged and to which the equilibrium equations can be applied so as to determine the unknown forces.

Fig.7.1 (a) shows a statically determinate multispan beam, which has 9 unknown restraint forces [Fig.7.1 (b)], and three members to which 9 available equilibrium equations are able to apply. Obviously, the number of its restraint forces is just equal to that of available equilibrium equations. In theoretical mean, the unknown forces may be determined as a whole. However, in order to expedite the computation and avoid solving simultaneous equations, we should write the equilibrium equations so that each equation involves only one unknown. Based on this purpose, in the concrete analyzing process, the sequence of isolating the free bodies of the beam must be in the reverse order by which the beam was set up. Then apply equilibrium equations to each free-body diagram to determine the desired unknown restraint forces. In order to implement efficient analysis, some of the analytical key points are given as follows:

(1) Types of free bodies and unknown forces A free body isolated from a structure may be a joint, a member, a portion or the whole of the structure.

It has been observed from previous chapters that the method of joints used for analyzing trusses isolates

155

156 Chapter 7 General Remarks on Statically Determinate Structures

joints as free-body diagrams; in analyzing statically determinate multispan beams the members are isolated as free-body diagrams; in analyzing statically determinate rigid frames the members are often isolated as free bodies to determine the member-end moments and shears, and rigid joints are selected to calculate member-end axial forces; the method of sections used for analyzing trusses isolates portions of trusses as free bodies.

On a free-body diagram there exist given forces and unknown forces. The number of unknown forces is dominated by the kinds of restraints.

Fig.7.2 shows a simple example. Since the members AC and BC in Fig.7.2 (a) are two-force members subjected only to axial force, by isolating joint C as a free body and using the method of joints the axial forces and can be conveniently determined. 1N 2N

In Fig.7.2 (b), since member AC is a beam member and BC is a two-force member, by isolating member AC as a free body and considering its equilibrium, unknown forces AX , AY and may be obtained successfully. In Fig.7.2 (c), because members AC and BC are all beam members, by using the method similar to that used in three hinged arches, i.e., detaching the two members from their supports and considering the entire system as a free body and applying its equations of equilibrium and condition ( ), unknown forces

N

0CM = AX , AY and AX , may be found successfully. BY

P

BC D

E

(2) Number of equations There exists relation between the number of equations and geometric construction of a system. As

shown in Fig.7.2 (a), there are two degrees of freedom for joint C; on the other hand, two available equilibrium equations can be applied to the joint. In Fig.7.2 (b), member AC has three degrees of freedom, and three available equilibrium equations are able to apply to it. While in Fig.7.2 (c), the unstable system

A F

PC D

C

XX D

CY DY

AX

AY BYCX

CY

EY FY

DYA B E

DXC D

(a)

(b)

Fig.7.1 A multispan beam and some of its free-body diagrams

F

7.1 Analytical Methods for Statically Determinate Structures 157

ACB has four degrees of freedom, and the number of its available equilibrium equations ( ) is also four. Generally, for statically determinate structures, the number of degrees of freedom of a free body is just equal to that of its available equilibrium equations.

0, 0, 0, 0CX Y M M= = =∑ ∑ ∑ =

(b)

However, this argument is not always true for any free body. For instance, the free-body diagram of member CD shown in Fig.7.1 (b) indicates four unknown forces but only three available equilibrium equations can be applied to it. The three equilibrium equations may determine the unknown forces , CY

DY and the relation of , but not the magnitude of andCX X= D CX DX . Their magnitude will be determined after considering the equilibrium of member DF.

(3) Simplification of analysis and sequence of isolating free bodies When considering the equilibrium of a free body to determine its unknown forces several equilibrium

equations may be used. However, in order to expedite the computation and avoid solving simultaneous equations, we should write the equilibrium equations so that each equation involves only one unknown. For instance, when analyzing a truss by using the method of joints the commonly considered equilibrium equations are force projection equilibrium equations, but moment equilibrium equations may be also utilized to simplify the calculation; when analyzing a truss by using the method of sections the selection of

C

A B

P

(a) q

A B

C

q

A B

C

(c)

C P

2N1N

q

C

AXB

C

N

A

AY

A

BY

q

AY

AX X

Fig.7.2 Types of free-body diagrams(a) a joint as a free body; (b) a member as a free body; (c) a portion as a free body

B


the centre of moments selected for moment equilibrium equations and the axes of force projections used to sum force components in the directions of the axes impose very important effect on the simplification of the calculation.

The capability of identifying the stressing characters in the members of a structure may help us to simplify the calculation. As the structure shown in Fig.7.2 (a), if you can identify out that members AC and BC are all two-force members before implementing the calculation you may isolate joint C as free body and find the member forces by considering the equilibrium of joint C. If the two members are recognized as beam members the method similar to that used to analyze three hinged arches has to be utilized to find the internal forces in the members.

The utilization of symmetry of a structure may simplify the computation of reactions and internal forces. Here the symmetry of a structure means that the geometry of the structure (the member arrangement and the types of supports) is symmetrical with respect to a line, which is named the line of symmetry. It is a fact that if a symmetrical structure is under the action of symmetrical loads its reactions and internal forces are symmetrical too with respect to the line of symmetry. Thus, for a symmetrical structure only a half of its unknown forces are needed to determine. A symmetrical structure is shown in Fig.7.3 (a), since the

geometry of the structure and the applied loading are symmetrical about its vertical center line of the structure, its reactions will also be symmetrical with respect to the line, so only two reaction components at one support are needed to determine. Based on the symmetry, if taking member AC as a free-body diagram shown in Fig.7.3 (b), there is only horizontal restraint force but no vertical force at joint C (since point C lies on the line of symmetry and is a pair of antisymmetric forces).

CXCY

The most important way for simplifying the calculation of a statically determinate structure is the sequence of isolating the free bodies, which must be the reverse order by which the structure was built up.

For a multispan beam, the sequence follows thusly in this order: first the subsidiary portions, then the

(a) (b)

A B

C

H

V

H

V2l

2l

q

f

CX

A

C

V

H

q

Fig.7.3 Analysis of a symmetrical three hinged structure

7.1 Analytical Methods for Statically Determinate Structures 159

main portion.

For the analysis of a truss, generally, the zero-force members must identified by inspection before the calculation. On above argument pertinent to isolating free bodies, for a simple truss, the sequence of isolating the joints must be the reverse order by which the truss was constructed; for a compound truss, the axial force of the final two-force member which is used to connect simple trusses has to be determined first by using the method of sections, then other member forces of the truss may be obtained. It is observe that in order to select rational analytical sequence the geometric construction analysis has to be made. In the analysis of a structure, the essential problem is the isolation of free bodies, i.e., the problem how to disassemble a structure; whereas in the analysis of geometric construction of a structure, the main problem is the composition of the structure, i.e., the problem how to build a structure up. From the point of structural analytical view, there is relation between the disassembly and the build of a structure. If the sequence of disassembly of a structure is in the reverse order by which the structure was built up, i.e., the

BA C D

E

(b)

(a)F G

2P1P

Ⅱ ⅡⅠ

A B C D

E F G

2P1P

GEEX EX

EY EYG XX G

GY GY

AYDY

Fig.7.4 The analysis of a statically determinate multispan frame


sequence of isolating members as free bodies is in the reverse order in which the structure was composed by adding the members, the analysis of the structure will be simple and successful.

As the statically determinate rigid frame shown in Fig.7.4 (a), the sequence of constructing the frame is from the main portionⅠto the subsidiary portionⅡ, but the analytical sequence, i.e., the order of isolating free bodies must be the reverse order of the construction as shown in Fig.7.4 (b).

7.2 General Property of Statically Determinate Structures

The statically determinate structures and statically indeterminate structures are all stable systems, the difference between them are:

From the point of geometric-construction-analysis view, a statically determinate structure is a stable system with no redundant restraint, while a statically indeterminate structure is a stable system with at least more than one redundant restraint.

From the point of static-equilibrium view, the internal forces of a statically determinate structure can be determined uniquely by their equilibrium equations, whereas those of a statically indeterminate structure are not able to be determined by their equilibrium equations, the unique solution of the internal forces may be determined by considering both its some deformation conditions and equilibrium conditions.

One of essential characters of a statically determinate structure is that if the solution of its unknown internal forces of the structure is obtained only by the equilibrium equations the solution is a unique solution. Some of the undermentioned characters will be all deducted from the character.

(1) The effect on internal forces due to temperature changes, support settlements and manufacture errors

The temperature changes, support settlements and manufacture errors and the like will not induce internal forces in a statically determinate structure.

As the beam shown in Fig.7.5 (a), the settlement of support B may only make the beam rotate around hinged support A but no internal forces will be developed. The conclusion may be further explained by an imaginary experimentation: imagine removing the support B so as for the beam to be a mechanical system and let the beam rotate around support A an arbitrary angle until the support B reaches to a new position B′ , then attach the beam to the foundation with a roller support at the new position. Apparently, in the entire process, no internal force is yielded in the beam.

There exist construction errors in the three hinged arch shown in Fig.7.5 (b), but after attached the wrong members to the foundation, there is no internal force caused by the errors in the arch represented by the dashed lines.

In Fig.7.5 (c), the simple beam is assumed to undergo the action of a temperature change from the amount of at the top and the amount of ot C− ot C+ at the bottom of the beam. Since the beam can be

7.2 General Property of Statically Determinate Structures 161

bent freely there is no internal force induced by the temperature change in the beam as shown in dashed lines.

(2) Characters of local equilibrium of a statically determinate structure Under the action of a system of external loads, if a local portion of the statically determinate structure

can maintain the equilibrium with the loads by itself, the internal forces in the other portion of the structure will be equal to zero.

A statically determinate multispan beam is shown in Fig.7.6 (a). Since the beam AB is a stable portion it can maintain equilibrium by itself under the action of the loads imposed on it, so there is no internal force induced by the loads in subsidiary portion BC. As shown in Fig.7.6 (b), since the member AB of a statically determinate truss is subjected to an arbitrary balanced force system the internal forces in all other members except in member AB of the truss will be equal to zero.

In fact, since the above mentioned internal force states have satisfied all of the equilibrium conditions of each portion of the statically determinate structures, the solution of the unknown internal forces is its unique solution.

It should be pointed out that there is no special requirement about a local equilibrium system, which may be a stable or unstable system, but must keep in equilibrium under some kinds of loads. As the statically determinate truss shown in Fig.7.7 (a), which is subjected to a pair of compressive forces equal in

A B

B′

(a)

0t C+

A 0t C−

(c)

C

C′B

A B

(b)

Fig.7.5 Support settlement, manufacture error and temperature change

(a)

A

P

B C

P2

PP

B2(b)A

Fig.7.6 Character of local equilibrium of statically determinate structures


magnitude but opposite in direction at the ends of the bottom chords, it is able to remain in equilibrium only by its bottom chords (which consists of a local equilibrium system) as shown in Fig.7.7 (b), so the internal forces in other members must be equal to zero.

(3) Character of equivalence of loads on statically determinate structures When a statically equivalent transformation of loads acting on an internally stable portion of a

statically determinate structure takes place, the difference between the internal force solutions for the two loading cases exists only in the internally stable portion, and the internal forces in other portions of the structure for the two loading cases are keeping identical. Here, so called equivalent loads mean that the loads which distribute differently but have the identical resultant.

The concentrated load P as shown in Fig.7.8 (a) is statically equivalent to the two concentrated loads in magnitude of acting on the joints A and B. If we change the two acting types of loads, i.e., replace the load P acting at the middle point of member AB as shown in Fig.7.8 (a) with the two loads P acting at the ends of member AB as shown in Fig.7.8 (b), only the internal force in the member AB will be changed and the internal forces in the other members will remain identical.

/ 2P

/ 2

1 2P

The character of statically determinate structures under the action of equivalent loads may be further explained by the character of their local equilibrium. Assume that a certain stable portion of a statically determinate structure is separately subjected to two types of loads and , which are equivalent to each other; the internal forces corresponding the two types of loads are and respectively [see Fig.7.8 (a)

and (b), the stable portion is member AB]. By the principle of superposition, the internal force state in the

P

1S 2S

2P

B

A

P

(a)

1Load P 1Internal force S

2P

2P

AB

2Load P 2Internal force S

(b)

P

A(c) B2

P

Fig.7.8 Character of equivalent loads

P

(a) (b)

P P P

Fig.7.7 Bottom chords keep in equilibrium

7.2 General Property of Statically Determinate Structures 163

structure ought to be under the simultaneous action of and1S S− 2 1P 2P− . Since and1P 2P− compose a balanced force system which is imposing only on the stable portion, the internal forces (in magnitude of ) in all other members except in the stable portion (member AB) of the structure must be equal to zero, i.e., . Therefore, when the two types of loads and are imposed to the stable portion (member AB in Fig.7.8) of the structure separately, the corresponding internal forces and

in all other members except in the stable portion (member AB) of the structure must be identical.

1 2S S−1 2 0S S− = 1P 2P

1S2S

(4) Character of variable conformations of statically determinate structures The internal forces in the other portion of a statically determinate structure will remain identical when

a stable portion of the structure changes its conformation.

P

2P

P

A B(a) (b)

A B

The truss shown in Fig.7.9 (a) is assumed to change its top chord AB into a small truss as shown in Fig.7.9 (b). Apparently, the internal forces in all other members except in member AB of the truss must remain unchangeable. The argument may be further explained by portioning the truss into two portions, member AB and the others, which respectively keep in equilibrium under the action of their own loads and restraint forces and whose free-body diagrams are as shown in Fig.7.9 (c). Now assume changing the member AB into a small truss. If the internal forces in other members and the restraint forces between the

(c) (d)P

A B

ABN

A B

ABN ABNABN ABN

ABN ABN ABNA B A B2P

2P

2P

2P

2P

2P

2P

Fig.7.9 Character of variable conformations

P


small truss and the other portion of the truss remained unchangeable, the original equilibrium conditions existing in the other portion would be still unchangeable; while the small truss could unaffectedly keep in equilibrium under the action of original load and restraint forces. Therefore, this sort of internal force state existing in the members of the truss is the actual internal force state after the variation of conformation of the truss.

7.3 Stressing Characteristics of Various Types of Structures

From the preceding chapters, we have learned five types of structures. They are: beams, rigid frames, arches, trusses and composite structures. However, the types of structures may be classified in different ways. The commonly used two kinds of methods of classifying structures are adduced as follows.

First classifying method: The method classifies structures into two categories, structures with thrusts and structures with no thrust. Beams and girder trusses belong to the former, while three hinged arches and frames, arched trusses and some of composite structures belong to the latter.

Second classifying method: The method classifies the members of a structure into two-force members and flexural (or beam) members. The members of a truss are all two-force members; the members of a statically determinate multispan beam and the members of a statically determinate multispan rigid frame are beam members; for a composite structure, some of the members belong to two-force members and some of them are beam members.

In a two-force member, the axial force is its only internal force; there is no bending moment. Under no bending moment state, the distribution of normal stress on the cross-sectional area of a member is uniform, so the strength of the constructional materials of such a member is utilized in the most efficient manner.

A beam member is under the action of its bending moments. Recall from previous courses on statics and strength of materials that the bending (normal) stress varies linearly over the depth of a beam from the maximum compressive stress at the fiber farthest from the neutral axis on the concave side of the bent beam to the maximum tensile stress at the outermost fiber on the convex side. The material near neutral face of a beam member is thusly not utilized in efficient way. In order to utilize the material of a beam member most efficiently under the bending moments, the best way is to reduce the value of the bending moments or even eliminate the bending moments so as to make the member be under no bending moment state. Based on the argument, we will discuss the characteristics of some types of structures as follows.

(1) In a multispan beam or an overhanging beam, the value of maximum bending moment on the section of the middle span may be reduced by utilizing the negative bending moment on the sections of overhanging ends as shown in Fig.7.10.

(2) In a structure with thrusts, the value of maximum bending moment may be reduced by utilizing

7.3 Stressing Characteristics of Various Types of Structures 165

the actions of the thrusts. The conclusion may be further explained through the three hinged rigid frame shown in example 4-6, the three hinged arch shown in example 5-1 and the composite structure shown in example 6-5.

0 2 / 8CM ql=

BCA

2l

2l

(a) q

0.207l

BA

0 / 6CM 0 /6CM

0.207l

0 /6CM0.586l

(b)

0 /CH M f=

0 / 4CM 0 / 4CMA

C

B

(c)

f

2l

/ 6fA

C

B

0 / 6CM 0 / 6CM

0 / 6CM

0 /CH M f=

2l

(d)

f

A

C

B

/ 4l

0 / 24CM

0 / 24CM

1 5 /12f f=

2 7 /12f f=

0 / 24CM

/ 4l / 4l / 4l

(g)

f

0 /CN M h=

(f)

A B

C

h

(e)q

0 /CH M f=

f

A B

2l

2l

Fig.7.10 Comparison of stressing states of different structures(a) a simple beam; (b) an overhanging beam; (c) a three hinged triangular arch with tie;(d) a three hinged triangular arch with eccentric top chords; (e) a three hinged arch;(f) a parallel chord truss; (g) a composite structure with lower bracing members


(3) In a truss, the members of the truss may be subjected only to the axial forces with no bending moment by connecting the members with frictionless hinges and proper arrangement of them and by the load-transmission manner by joints. For three hinged arches, the arches with optimal centre lines are subjected only to axial forces with no bending moment too. From the point of mechanical view, the state of no bending moment is a rational stressing state for a structure. The above mentioned structures (trusses and three hinged arches with optimal centre lines) are all rational types of structures. In addition, in a composite structure, some of the members are stayed in the state of no bending moment as well.

In order to make a comprehensive comparison about the characteristics of various types of structures, the values of the dominated internal forces in some types of structures, with the same span and subjected to the same uniformly distributed loads, are shown in Fig.7.10.

The beam shown in Fig.7.10 (a) is a simple beam, of which the value of the bending moment on the section C, at the middle span, is

20

8CqlM = . If the section of the beam is rectangular (depth indicated by

h), the distribution of the bending (normal) stress over the sections of the beam looks like a triangularity. The magnitude of the resultant of the compressive stress and that of the tensile stress on the section C are all equal to

0

2 3CM

h.

The beam shown in Fig.7.10 (b) is an overhanging beam. In order to reduce the value of the positive maximum bending moment on the section at the middle of the span, the length of the overhanging is controlled in such a length that the value of the positive maximum bending moment and those of the negative on the sections at the two supports are equal to each other. Based on the condition, the length of

the overhanging is and the value of the maximum bending moment becomes 0.207l 016 CM .

The arch shown in Fig.7.10 (c) is a triangular three hinged arch with a tie, whose thrust is

01CH M

f= . Since the action of the thrust the maximum value to the bending moments in top chord

members is decreased to 014 CM . If the top chords become the chords as shown in Fig.7.10 (d), with an

eccentricity of 6fe = , the magnitude of the negative maximum bending moments, 01

6 CM , becomes just

equal to that of the positive maximum one in the top chord members.

The arch shown in Fig.7.10 (e) is a parabolic arch, of which the value of its thrust is still 0CM

f. Since

the centre line of the arch is an optimal one, it remains in the state of no bending moment. The truss shown in Fig.7.10 (f) is a girder truss, of which the values of the axial forces in the bottom


chords are 0CM

h. Since the joint loads are imposed to it the truss remains in the state of no bending

moment. The structure shown in Fig.7.10 (g) is a composite structure. In order to set the value of the negative

maximum bending moments and those of the positive in the top chords are equal to each other, the rises

must be: 15

12f f= and 2

712

f f= , and now the value of the maximum bending moment in the top

chords decreased to 0124 CM and the axial force of bottom chord becomes to

0CM

f.

It is observed from above comparison that for the structures with the same span and subjected to the same loads, the maximum value of the bending moments occurs in a simple beam; the bending moments in overhanging beams, statically determinate multispan beams, three hinged rigid frames and composite structures are quite small comparing with those in a simple beam; whereas trusses and the arches with optimal centre lines are in the state with no bending moment. In practical engineering, thereby, simple beams are used to short span structures; overhanging beams, statically determinate multispan beams, three hinged rigid frames and composite structures may be applied to the structures with quite long spans. When the span of a structure is very long trusses or arches with optimal centre lines may be selected. It should be thusly realized that when selecting a type of structure different types of structures posses different best-span ranges.

On the other hand, every type of structure has its own advantages and disadvantages. Although simple beams have foregoing mentioned disadvantages some of their advantages such as being constructed easily and facilitating application make the type of structure be widely used in practical engineering projects. While other types of structures have some advantages they have disadvantages too. For instance, a truss has too many members and its connections (joints) are difficult to manufacture; three hinged arches have a strict requirement on its foundation because of existence of its thrust or need a tie to undergo the thrust, and its curved shape increases the difficulty of construction as well.

It is observed from above statement that when selecting a type of structure one must, in all aspects not just their stressing states, analyze and compare the characteristics of a structure before make a decision.


7-1 Discuss the composition and stressing character of the truss shown in the problem figure and find the internal forces in members 1 and 2.

7-2 The problem figure shows two different trusses subjected to the same loads. What is the difference between their internal forces?


problem 7-1

E FD

A BCP P4P

1

2

8a

aa

(a) 112

1 1 1 1 12

problem 7-2

(b) 14

14

12

12

12

12

12

12

12

12

12

12

12

CHAPTER 8 INFLUENCE LINES

The abstract of the chapter In previous chapters, we discussed the analysis of statically determinate structures subjected to

stationary loadings. In this chapter, we study the analysis of statically determinate structures subjected to variable loads. The influence lines are the important tools and means used in the analysis of structures subjected to variable loads.

We begin this chapter by introducing the concept of influence lines. We next discuss the two kinds of methods, named static method and mechanismic (or virtual displacement) method, to construct influence lines. Finally, we consider the application of influence lines to determination of the maximum values of response functions for structures due to variable loads.

8.1 Concept of Influence Lines

In the previous chapters, we considered the analysis of structures subjected to loads whose positions were fixed on the structures. An example of such stationary loading is the dead load due to the weight of the structure itself and of other material and equipment permanently attached to the structure. However, structures generally are also subjected to loads (such as moving loads and movable loads) whose positions may vary on the structure. The hoist loads shown in Fig.8.1 (a) and (b), the carload shown in Fig.8.2 and pedestrian loads are the examples of moving loads; while the wind and snow loads imposed on buildings are the examples of movable loads. In this chapter, we study the analysis of statically determinate structures subjected to moving loads and movable loads, simply called variable loads.

Under the action of variable loads, the internal forces and reactions of a structure will vary with the variation of the loads. Consider, as an example, the bridge truss shown in Fig. 8.2. As the car moves across the bridge, the forces in the members of the truss will vary with the position x of the car. It should be realized that the forces in different truss members will become maximum at different positions of the car. For example, if the force in member AB reaches its maximum value when the car is at a certain position

1x x= , then the force in another member, for example, member FG, may reach maximum value when the car is at a different position 2x x= . The design of each member of the truss must be based on the maximum force that develops in that member as the car moves across the bridge. Therefore, the analysis of the truss would involve, for each member, determining the position of the car at which the force in the

169

170 Chapter 8 Influence Lines

member becomes maximum and then computing the value of the maximum member force.

From the foregoing discussion, we can see that the analysis of structures for variable loads consists of two steps: (1) determining the position(s) of the load(s) at which the response function of interest (e.g., a reaction, shear or bending moment on a section of a beam, or force in a truss member) becomes maximum and (2) computing the maximum value of the response function.

However, there are a variety of kinds of moving loads and a great deal of ways of distributions of movable loads. It is impossible to determine the each most unfavorable loading position of all variable loads which may impose on a structure. An

(a)(b)

Beam of hoist

Fig.8.1 Hoist loads(a) transverse section of the workshop; (b) longitudinal section of the workshop; (c) the computing model of the beam

Beam of hoist

P P

l

P PA B

x

G HTruss

Beam Deck

CD

E

Fig.8.2 Car loads

A

B

F

8.1 Concept of Influence Lines 171

efficient way to analyze structures under the action of variable loads is the utilization of influence lines. An influence line is a graph of a response function of a structure as a function of the position of a

downward unit load moving across the structure. When the influence line for a desired quantity (such as a reaction, a shear, an axial force and a bending

moment) is determined, the maximum value of the quantity due to the actual variable loads may be determined by the principle of superposition. In fact, an influence line for a desired quantity of a structure is the graphic expression of the quantity with respect to the position of a downward unit load moving across its possible range on the structure, which shows graphically how the movement of a unit load on the structure influences the magnitude of the quantity; whereas the actual magnitude of the desired quantity of the structure due to variable loads could be considered as that of the combination of a series of unit loads. The concept of influence lines will now be further explained by the following example.

Consider the simply supported beam shown in Fig. 8.3(a). The beam is subjected to a downward concentrated load of unit magnitude, which moves from the left end A of the beam to the right end B. The position of the unit load is defined by the coordinate x measured from the left end A of the beam, as shown in the figure. Suppose that we wish to draw the influence line for the vertical reaction at support B.

To develop the influence line for the vertical reaction BR of the beam, we determine the expression for BR in terms of the variable position of the unit load, x, by applying the equilibrium equation of the moment about point A, we obtain

0, 1 0A BM x R l= × − × =∑

(0 )BxR x ll

= ≤ ≤ (8-1)

From equation (8-1), it is obvious that BR is a linear function of x, with

at and at

0BR = 0= 1BR =xx l= .

Equation (8.1) represents the equation of the influence line for BR , which is constructed by plotting this equation with BR as ordinate against the position of the unit load, x, as abscissa, as shown in Fig.8.3 (b). Note that this influence line [Fig.8.3 (b)] shows graphically how the movement of a unit load across the length of the beam influences the magnitude of the reaction BR . As this influence line indicates, 0BR = when the unit load is located at the left support A of the beam (i.e., when 0x = ). As the unit load moves from A to B, the magnitude of BR increases linearly until it becomes 1.0 when the unit load reaches the

(b)

(a) A B

x

x

xy

l

1.0

1P =

RB

By R=

Fig.8.3 Determination of influence line for (a) a simple beam; (b) influence line for

BR

BR


right support B (i.e., when x l= ). It is important to realize that the ordinate of the influence line at any position x is equal to the magnitude of BR due to a unit load acting at the position x on the beam. For example, from the influence line for BR [Fig.8.3 (b)], we can determine that when a unit load is applied at a distance of from the end A of the beam, the magnitude of the reaction 0.25l BR will be 0.2 . Similarly, when the unit load is acting at

50.6x l= the magnitude of BR will be , and so on. 0.6

The foregoing discussion about the influence line for reaction BR has introduced the concept of an influence line. Generally, a graph indicating the variation of a response function such as one of reactions or internal forces of a structure, across which a downward unit load is moving, is termed as an influence line. For the graph, the position of the unit load, x, is abscissa, the value of the response function is ordinate. When constructing an influence line, the base line (as abscissa) represents the possible moving range of the unit concentrated load, and the positive value of the influence line is depicted on the upper side of the base line and vice versa. The dimension of the ordinate is equal to the dimension of the response function divided by the dimension of the unit load.

8.2 Influence Lines for Statically Determinate Single Span Beams by Static Method

There are two kinds of elementary methods, static method and mechanismic (or virtual displacement) method, to construct influence lines for the reactions or internal forces of a structure. The static method is illustrated by determining the influence lines for the reactions and internal forces of a single span beam as follows.

8.2.1 Influence lines for simple beams

(1) Influence lines for reactions The influence line for vertical reaction at right-hand support B of a simple beam is determined in the

preceding section [see Fig.8.3]. Now we determine and draw the influence line for vertical reaction at left-hand support A of the simple beam.

To develop the influence line for the vertical reaction AR of the beam, we determine the expression for AR in terms of the variable position of the unit load, x [see Fig.8.4 (a)], by applying the equilibrium equation of the moment about point B, we write

0BM =∑ ， 1 ( ) 0AR l l x× − × − =

Al xR

l−

= (0 )x l≤ ≤ (8-2)

From equation (8-2), it can be seen that AR is a linear function of x, with at 1AR = 0x = and at0AR = x l= . Actually, the linear function is just the influence line for the vertical reaction AR and

can be constructed by plotting this equation with AR as ordinate, being dimensionless, against the

8.2 Influence Lines for Statically Determinate Single Span Beams by Static Method 173

position of the unit load, x, as abscissa, as shown in Fig.8.4 (c).

x 1P =

C BA x

BR

x 1l

BA

x1−1l

bl

1

1

ARa b

l

(a)

(b)

(c)

(d) (e)

(f)

(g)

Fig.8.4 Influence lines for a simple beam(a) a simple beam; (b) influence line for ; (c) influence line for ; (d) and (e) free-body diagrams; (f) influence lin

B AR Re for ; (g) influence line for C CQ M

BA

A C

AR a

CM

CQ

CMB

CQ

C

bBR

A BC

al

abl

a

b

A BC


(2) Influence lines for shears The influence lines for shears can be developed by employing a procedure similar to that used for

constructing the influence lines for reactions. To develop the influence line for the shear at section C of the beam [Fig.8.4 (a)], we determine the expressions for . It can be seen from Fig.8.4 (a) that when the unit load is located to the left of point C—that is, on segment AC of the beam (

CQ0 x a≤ ≤ )—the shear at C can

be conveniently obtained by using the free body of the portion CB of the beam that is to the right of C [Fig.8.4 (e)]. Considering the downward external forces and reactions acting on the portion CB inducing positive shear on section C in accordance with the beam sign convention (Section 3.1), we write

C BQ = −R ( 0 x a≤ ≤ )

When the unit load is located to the right of point C—that is, on segment CB of the beam —it is simpler to determine by using the free body of the portion AC, which is to the left

of C [Fig.8.4 (d)]. Considering the upward external forces and reactions acting on the portion AC inducing positive shear on section C, we determine the shear as to be

1P =

(a x l≤ ≤ )

A

CQ

CQ R= ( a x l≤ ≤ )

Thus the equations of the influence line for can be written as CQ

(0 ) ( )

BC

A

R xQ

R a xal

− ≤ ≤⎧= ⎨ ≤ ≤⎩

(8-3)

Note that Eq. (8.3) expresses the influence line for in terms of the influence lines for the reactions

CQAR and BR . This equation indicates that the segment of the influence line for between

points A and C ( 0CQ

x a≤ ≤ ) can be obtained by multiplying the ordinates of the segment of the influence line for BR between A and C by -1. Also, according to this equation, the segment of the influence line for between points C and B ( a x ) is the same as the segment of the influence line for CQ l≤ ≤ AR between the same two points. The influence line for thus constructed from the influence lines forCQ AR and BR is shown in Fig.8.4 (f). It is usually more convenient to construct the influence lines for shears and bending moments (to be discussed subsequently) from the influence lines for reactions instead of from the equations expressing the shear or bending moment explicitly in terms of the position of the unit load, x. If desired, such equations for the influence line for in terms of x can be obtained by simply substituting Eqs. (8.1) and (8.2) into Eq. (8.3); that is,

CQ

(0 )

1 ( )

B

C

A

xR x alQxR a x ll

⎧− = − ≤ ≤⎪⎪= ⎨⎪ = − ≤⎪⎩

≤ (8-4)

The influence line for [Fig.8.4 (f)] shows that the shear at C is zero when the unit load is located CQ


at the left support A of the beam. As the unit load moves from A to C, the shear at C decreases linearly until it becomes when the unit load reaches just to the left of point C. As the unit load crosses point C, the shear at C increases abruptly to . It then decreases linearly as the unit load moves toward B until it becomes zero when the unit load reaches the right support B. The ordinate of is dimensionless.

/a l−1 /a l−

CQ(3) Influence lines for bending moments When the unit load 1P = is located to the left of point C [Fig.8.4 (a)], the expression for the

bending moment at C can be conveniently obtained by using the free body of the portion CB of the beam to the right of C. Considering the counterclockwise moments of the external forces and reactions acting on the portion CB yielding positive bending moment on section C in accordance with the beam sign convention (Section 3.1), we write

C BM R b= × (0 )x a≤ ≤

When the unit load 1P = is located to the right of point C, we use the free body of the portion AC to the left of C to determine CM . Considering the clockwise moments of the external forces and reactions acting on the portion AC yielding positive bending moment on section C, we obtain

C AM R= ×a ) (a x l≤ ≤

The equations of the influence line for CM can be thusly written as

( ) (0 )( ) ( )

B BC

A A

R b R l a x aM

R a R a a x l× = − ≤ ≤⎧

= ⎨ × = ≤⎩ ≤ (8-5)

It can be seen from Eq. (8.5) that the segment of the influence line for CM between points A and C ( 0 x a≤ ≤ ) can be obtained by multiplying the ordinates of the segment of the influence line for BR between A and C by ( ). Similarly, according to this equation the segment of the influence line for (l a− )

CM between points C and B ( a x ) can be obtained by multiplying the ordinates of the segment of the influence line for

l≤ ≤AR between C and B by a. The influence line for CM thus constructed from the

influence lines for AR and BR is shown in Fig.8.4 (g). The equations of this influence line in terms of the position of the unit load, x, can be obtained by substituting Eqs. (8.1) and (8.2) into Eq. (8.5); that is,

( ) (0 )

(1 ) ( )

B

C

A

xR b l a xlM

x

a

R a a a xl

⎧

l

× = − ≤ ≤⎪⎪= ⎨⎪ × = − ≤ ≤⎪⎩

(8-6)

The ordinate of CM is the dimension of length. Although the influence line for CM [Fig. 8.4(g)] resembles, in shape, the bending moment diagram

of the beam for a concentrated load applied at point C, the influence line for bending moment has an entirely different meaning than the bending moment diagram, and it is essential that we clearly understand


the difference between the two. A bending moment diagram denotes how the bending moment varies on all sections along the length of a beam for a loading condition whose position is fixed on the beam, whereas an influence line for bending moment denotes how the bending moment varies on one particular section as a unit load moves across the length of the beam. 1P =

It should be pointed out that the influence lines [see Fig. 8.4] for the reactions, shear, and bending moment of the simply supported beam consist of straight-line segments. We will show in the following sections that this argument is true for the influence lines for all response functions involving forces and moments (e.g., reactions, shears, bending moments, and forces in truss members) for all statically determinate structures.

8.2.2 Influence lines for overhanging beams

(1) Influence lines for reactions Now we determine and draw the influence lines for vertical reactions, AR and BR , at supports A and B

of the overhanging beam [Fig. 8.5(a)]. To develop the influence lines for the vertical reactions AR and BR of the beam, we determine the expressions for AR and BR in terms of the variable position of the unit load, x measured from the left support A and considered to be positive in the right direction, by applying the equilibrium equations of the moment about supports A and B, we write

0,B Al xM R

l−

= =∑ 1( l x l l2 )− ≤ ≤ + (8-7a)

0 ,A BxM Rl

= =∑ 1( l x l l2 )− ≤ ≤ + (8-7b)

The two equations of influence lines for vertical reactions, AR and BR , at the supports A and B of the overhanging beam is just the same as those for the reactions of a corresponding simple beam with the same span [Eqs.(8-1) and (8-2)], but the moving range of the unit load 1P = being extended. For the simple beam, the moving range of 1P = is 0 x l≤ ≤ , and for the overhanging beam the range becomes

. The influence lines for 1l x l l− ≤ ≤ + 2 AR and BR [shown in Fig.8.5 (b) and (c)] of the overhanging beam, which is obtained by plotting the equations, are the same as those of the simple beam in the segment AB; and in the overhanging portions, the influence lines may be just extended to the ends of the beam. It should be noted that when is located to the left of A the coordinate x becomes minus. 1P =

(2) Influence lines for shears and bending moments on segment AB The influence lines for the shear and bending moment on section C of segment AB of the overhanging

beam [Fig.8.5 (a)] is now determined as follows. First, we place the unit load at a variable position x to the left of point C—that is, on segment

AC of the beam—and determine the shear and bending moment at C by using the free body of the portion CB of the beam, which is to the right of C:

1P =


(0 )C B

C B

Q Rx a

M R b= − ⎫

≤ ≤⎬= ⋅ ⎭

Next, the unit load is located to the right of C—that is, on segment CB of the beam—and we use the free body of the portion AC, which is to the left of C, to determine the shear and bending moment at C:

1lll

+

2ll

2lll

+

1ll

bl

al

2lal

1lbl

2ll1l

1

1

1

1

(a)

(b)

(c)

(d)

A BC D EF

x

a b

A BCE

F

ABC E

F

A BCE

F

ABC

EF

a

a

(e)

1P =

d

1ll 2l

l

abl

Fig.8.5 Influence lines for an overhanging beam(a) an overhanging beam; (b) influence line for ; (c) influence line for ; (d) influence line for ; (e) influence line for

A

B C

C

RR QM


( )C A

C A

Q Ra x l

M R a= ⎫

≤ ≤⎬= ⋅ ⎭

It can be observed from above equations that the equations of the influence lines for and CQ CM between two supports of an overhanging beam are the same as those for the shear and bending moment of a corresponding simple beam with the same span; and if we extend the varying range of x to segment FA and BE, the above two expressions will remain unchangeable. Therefore, the influence lines for and CQ CM on the overhanging portions of an overhanging beam may be obtained by just extending the corresponding influence lines for and CQ CM of the corresponding simple beam to the range of the overhanging portions, as shown in Fig.8.5 (d) and (e).

(3) Influence lines for shears and bending moments on overhanging portions Now determine the influence lines for the shear and bending moment on an arbitrary section D, DQ

and DM , of the right overhanging portion of the overhanging beam [Fig.8.5 (a)]. First, we place the unit load at a variable position x to the left of point D—that is, on segment

FD of the beam—and determine the shear and bending moment at D by using the free body of the portion DE of the beam, which is to the right of D; since there is no load imposed on the portion we find:

1P =

00

D

D

QM

= ⎫⎬= ⎭

1 2( l x l l d)− ≤ ≤ + −

Where d is the distance between points D and E [see Fig.8.6 (a)]. Next, the unit load is located to the right of D—that is, on segment DE of the beam—and we use Eq.

(8-7) and the free body of the portion FD, which is to the left of D, to determine the shear and bending moment at D [see Fig.8.6(a)]:

2 2 2

1( ) ( ) [ ( )

D A B

D A B

Q R R]R l l d R l d x l l d

= + = ⎫⎬= − + − − − = − − + − ⎭

2 2l l d x l l+ − ≤ ≤ + ( ) M

The influence lines for DQ and DM of the overhanging beam are shown in Fig.8.6 (b) and (c). As all the ordinates of the influence lines on the beam except for the segment DE are equal to zero, this means that the shears and bending moments may be developed when the loads only applied on the segment DE of the beam.

Through constructing the influence lines for simple and overhanging beams we may summarize the following procedure for determining the influence lines for statically determinate structures by using the static method.

1. Select an origin from which the position of a moving downward concentrated unit load 1P = will be measured. It is usually convenient to assume that the unit load moves from the left end of the structure to the right end, with its position defined by a coordinate x measured from the left end of the


structure. 2. Determine the expressions for the reactions in terms of x by applying equations of equilibrium or

condition and construct the influence lines by plotting the expressions with the magnitudes of the reactions as ordinate against the position x of the unit load as abscissa.

3. It is generally convenient to construct the influence lines for shears and bending moments by using the influence lines for support reactions. Thus, before proceeding with the construction of an influence line for shear or bending moment at a section of the structure, make sure that the influence lines for all the reactions, on either the left or right side of the section under consideration, are available. Otherwise, draw the required influence lines for reactions by using the procedure described in the previous step. An influence line for the shear (or bending moment) on a section of the structure can be constructed as follows:

1P =

(1) Locate the unit load on the structure at a variable position x to the left of the section under consideration, and determine the expression for the shear (or bending moment). If the influence lines for all the reactions are known, it is usually convenient to use the portion of the structure to the right of the section for determining the expression for shear (or bending moment), which will contain terms involving only reactions. The shear (or bending moment) is recognized to be positive or negative in accordance with the beam sign convention established in Section 3.1.

1P =

(2) Next, place the unit load to the right of the section under consideration, and determine the expression for the shear (or bending moment). If the influence lines for all the reactions are known, it is usually convenient to use the portion of the structure to the left of the section for determining the desired expression, which will contain terms involving only reactions.

1P =

(3) If the expressions for the shear (or bending moment) contain terms involving only reactions, it is generally simpler to construct the influence line for shear (or bending moment) by combining the segments of the reaction influence lines in accordance with these expressions. Otherwise, substitute the expressions for the reactions into the expressions for the shear (or bending moment), and depict the resulting

F Dx

E

1

d

d

(a)

(b)

(c)

D E

D E

Fig.8.6 Other influence lines for an overhanging beam(a) an overhanging beam; (b) influence line for ; (c) influence line for

D

D

QM


expressions, which will now be in terms only of a variable position x, to obtain the influence line. (4) Repeat step (3) until all the desired influence lines for shears and bending moments have been

determined.

8.3 Influence Lines for Girders with Floor Systems

In the previous sections, we discussed the influence lines for beams that were subjected to a moving unit load applied directly to the beams. In most bridges and buildings, there are some structural members that are not subjected to live loads directly but to which the live loads are transmitted via floor framing systems. The framing system of a bridge is shown in Fig. 8.7(a). The loads caused by vehicle and the like are applied on beams called stringers, which are supported by floor beams, which, in turn, are supported by the girders. Thus, any live loads (e.g., the weight of the traffic), regardless of where they are located on the stringers and whether they are concentrated or distributed loads, are always transmitted to the girders as concentrated loads applied at the points where the girders support the floor beams.

1P =

To illustrate the procedure for constructing influence lines for shears and bending moments in the girders supporting bridge or building floor systems, consider the simply supported girder shown in Fig. 8.7 (a). As shown, a unit load moves from left to right on the stringers, which are assumed to be simply supported on the floor beams. The effect of the unit load is transmitted to the girder at points A through B, at which the girder supports the floor beams. The points A through B are commonly referred to as panel points, and the portions of the girder between the panel points (e.g., AC or CE) are called panels. Fig. 8.7(a) shows the stringers resting on top of the floor beams, which rest on top of the girder. Although such sketches are used herein to show the manner in which the load is transmitted from one structural member to the others, in actual floor systems, members are seldom supported on top of each other, as depicted in Fig. 8.7(a). Instead, the stringers and the floor beams are usually positioned so that their top edges are even with each other and are either lower than or at the same level as the top edges of the girders.

1P =

The characteristics and constructing method of influence lines for girders under panel-point loads are herein planning to be explained by the influence line for bending moment DM on section D of the girder shown in Fig.8.7 (a).

In order to determine the influence line for the bending moment DM on the section D, which is located in the panel CE, of the girder due to panel-loads, which means that a unit load will move from the left end A to the right end B on the stringers, we will consider three sorts of situation, the unit load 1P = is located to the left of point C (within stringer AC), to the right of point E (on stringers EF and FB) and within the stringer CE.

It should be noted that regardless of where the unit load is located, the influence lines for vertical reactions AR and BR at the supports A and B of the girder is identical as those for the reactions of a

8.3 Influence Lines for Girders with Floor Systems 181

corresponding simple beam with the same span. Rewrite them as follows:

1P =

2C

d xYd−

= E

x dYd−

=

34

d58

d

1516

d

58Cy d=

34Ey d=

2.54

1.54

14

12

2d

4l d=

(a)Floor beam Stringer

BA C D E F

2d Girder

x 1P =

(b)A

C E

C E

d

(c)

B

BA D

(d)BA C E

(e)

CE

BA

1

1 Fig.8.7 Influence lines under panel-loads(a) framing system of a bridge; (b) manner of load transmission; (c) influence line for under direct loading; (d) influence line for under p

D

D

MM anel-loading; (e) influence line for DQ


A

B

l xRl

xRl

− ⎫= ⎪⎪⎬⎪=⎪⎭

(0 )x l≤ ≤

Now let us consider the first sort of situation, that is, when a moving unit load is located to the left of point C (within stringer AC). In this situation, the equation of the influence line for bending moment DM on section D of the girder may be expressed as [reference to Eq. (8-6)]

3 3( ) ( ) (0 )2 2D Bd x dM R l l x d

l= − = − ≤ ≤

Next, consider the second sort of situation, that is, a moving unit load is located to the right of the point E (on stringer EF and FB). In the situation, the bending moment at D is given by

3 3(1 ) (2 )2 2D Ad x dM R

l= = − ≤ ≤d x l

Then, consider the third sort of situation, that is, a moving unit load is located within stringer CE, as shown in Fig. 8.7 (a). In this situation, the restraint forces and , which are exerted on the girder by the floor beam at C and E will be

CY EY

2C

d xYd−

=

Ex dY

d−

=

The force must be included in the expression for bending moment at D, so CY DM is given by 3 1 ( 2 )2 2 2 8D A Cd d dM R Y x d x= − = + ≤ ≤ d

Thus, the equations of the influence line for DM can be written as

5 3 ( ) (0 )2 2

3 1 ( 2 )2 2 2 8

3 3 (1 ) (2 )2 2

B

D A C

A

d x dR l xl

d d d

d

M R Y x d x

d x d

d

R d x ll

⎧ × = − ≤ ≤⎪⎪⎪= − = + ≤ ≤⎨⎪⎪ = − ≤⎪⎩

≤

(8-8)

The influence line, which is obtained by plotting the equation, is shown in Fig. 8.7(d). For comparison, Fig.8.7 (c) shows the influence line for the bending moment DM , which may be constructed by using the procedure discussed in subsection 8.2.1, of the girder to which a moving unit load is applied directly. It can be seen from Fig.8.7 (d) that the influence line for

1P =DM consists of three straight-line

8.3 Influence Lines for Girders with Floor Systems 183

segments; the two side straight-line segments, on segments AC and EB, are identical as those of the influence line for DM shown in Fig.8.7 (c). The only difference between the two figures [Fig.8.7 (c) and Fig.8.7 (d)] occurs within the panel CE containing the location of the response function DM under consideration; and the influence line for the response function within the panel is a straight line as well. Founded on the analysis, we summarize the procedure for constructing influence lines for girders due to panel-point loads as follows:

(1) Construct the influence line (plotted by dashed lines) for desired quantity of a girder to which a moving unit load is applied directly. 1P =

(2) Complete the influence line by connecting the all of previously computed ordinates of the panel points between two adjacent panel points by straight lines (plotted by solid lines) and by determining any remaining ordinates by using the geometry of the influence line.

By the procedure, the influence line for shear DO on section D of the girder shown in Fig.8.7 (a) is constructed as follows:

First, construct the influence line (plotted by dashed lines) for DO of the girder to which a moving unit load is applied directly, as shown in Fig.8.7 (e). 1P =

Then, complete the influence line DO by connecting the all of previously computed ordinates of the panel points between two adjacent panel points by straight lines (plotted by solid lines) as shown in Fig.8.7 (e).

In order to explain the characteristics of influence lines for shears of the girders due to panel-point loads. We write the equation of influence line for shear DO as follows (which may be obtained by using the similar procedure utilized in determining the equation of DM ):

- (0 )

31 ( 2 )4

1 (2 )

B

D A C

A

xR x dl

Q R Y x d xd

x

d

R d x ll

⎧ = − ≤ ≤⎪⎪⎪= − = − + ≤ ≤⎨⎪⎪ = − ≤⎪⎩

≤

(8-9)

It is observed that the expressions for shear DO do not depend on the exact location of a point within the panel; that is, these expressions remain the same for all points located within the panel CE. The expressions do not change because the loads are transmitted to the girder at the panel points only; therefore, the shear in any panel of the girder remains constant throughout the length of that panel. Thus for girders with floor systems, the influence lines for shears are usually constructed for panels rather than for specific points along the girders.


8.4 Influence Lines for Trusses by Using Static Method

The floor framing systems commonly used to transmit live loads to trusses are similar to those used for the girders discussed in the preceding section. Fig. 8.8 shows a typical floor system of a truss bridge. The deck of the bridge rests on stringers that are supported by floor beams, which, in turn, are connected at their ends to the joints on the bottom chords of the two longitudinal parallel chord trusses. Thus, any live loads (e.g., the weight of the traffic), regardless of where they are located on the deck and whether they are concentrated or distributed loads, are always transmitted to the trusses as concentrated loads applied at the joints. Live loads are transmitted to the roof trusses in a similar manner. As in the case of the girder floor systems [Fig.8.8 (c)], the stringers of the floor systems of the trusses are assumed to be simply supported at their ends on the adjacent floor beams [Fig.8.8 (b)]. Thus, the influence lines for trusses also contain straight-line segments between panel points.

To illustrate the construction of influence lines for trusses by using static method based on method of joints and method of sections, consider the parallel chord bridge truss shown in Fig. 8.8(a). A unit load

moves from left to right on the stringers [see Fig.8.8 (b)] of a floor system attached to the bottom chord AB of the truss. The effect of the unit load is transmitted to the truss at joints (or panel points) A through B, where the floor beams are connected to the truss. Suppose that we wish to draw the influence lines for the vertical reactions at supports A and B and for the axial forces in members de, DE, dE, cC and eE of the truss.

1P =

(1) Influence lines for reactions The equations of the influence lines for the vertical reactions, AR and BR , can be determined by applying the equilibrium equations:

A

B

l xRl

xRl

− ⎫= ⎪⎪⎬⎪=⎪⎭

(0 )x l≤ ≤

The influence lines are not plotted out. But note that these influence lines are identical to those for the reactions of a corresponding simple beam to which the unit load is applied directly.

(2) Influence line for axial force in top chord member de The expressions for the axial force can be determined by passing an imaginary sectionⅠ-Ⅰ

through the members de, dE, and DE, as shown in Fig. 8.8(a), and by placing the unit load to the left of joint E and by applying the moment equilibrium equation

deN

0EM =∑ for the right portion EB, we obtain the following expressions for . deN

0EM =∑ ， 3B deR d N h 0× + × =

8.4 Influence Lines for Trusses by Using Static Method 185

3 3= (0 3 )de Bd d xN R xh h l

= − − ≤ ≤ d

This indicates that the segment of the influence line for between A and E can be obtained by multiplying the corresponding the influence line for

deNBR of the segment by 3 /d h− . When the unit load is

located to the right of joint E, it is convenient to determine by using the free body of the left portion AD deN

3dh

B

B

(a)

3dh 3

2dh

43dh

12

13

23

16

56

B

h

(b)

(c) (d)

A

P=1D E F G

ⅠⅡ

Ⅱ Ⅰ

6l d=AR

BR

C

a c d f g be

A

C D E F GB

(e)A

DB

1

Truss Floor beams

Truss Stringer

EA B

(f)

1

DE

A B

1

(h)

(j)

C

D

CD(g) A B A

AB(i)

AE F

1 Fig. 8.8 Influence lines for truss(a) a parallel chord truss; (b) floor system of the truss bridge; (c) a substitute girder; (d) influence line for N ;(e) influence line for ;(f) influence line for ; (g) influence line for (through truss); (h) influence line for (deck truss); (i) influence line for (through truss); (j) influe

de

DE dE cC

cC eE

N Y NN N nce line for (deck truss)eEN


0EM =∑ ， 3 0A deR d N h× + × =

3 3= (1 ) (3 6 )de Ad d xN R d x lh h l

= − − − ≤ ≤ = d

This indicates that the segment of the influence line for between E and BdeN can be obtained by multiplying the corresponding influence line for AR of the ment by 3 /d h− . The influence line for

deN thusly constructed from the influence lines for A

seg R and BR is shown in .8 (d).

(3) Influence line for axial force in bottom cho emb DE Fig.8

rd m er y applying the moment equilibrium

equaBy considering the same sectionⅠ-Ⅰused for deN but btion 0dM =∑ , and by placing the unit load first to eft and then to the right of joint D, we obtain

the follow sions for the l

ing expres DEN :

0∑ dM = ， 4 0B DER d N h× − × =

4 4= (0 2 )DE Bd d xN R xh h l

= ≤ d≤

，0dM =∑ 2 0A DER d N h− × + × =

2 2= (1 ) (2 6 )DE Ad d xN R d x lh h l

= − ≤ ≤ d=

The influence line for DEN thusly obtained is shown in Fig.8.8 (e). e for

sidering again sectionⅠ-Ⅰand by applying the equi

(4) Influence lin force in diagonal member dE The expressions for dEN can be determined by con

librium equation Y 0=∑ to one of the two portions of the truss. When the unit load is located to the left of joint D, applica e equilibrium equation 0Ytion of th =∑ to the right portion EB of the truss yields

0Y =∑ ， (0 2 )hN Y x d= ≤ ≤ dE dE BdE

Rl

= −

When the load is located to the right of joint E, we write 1P =

0Y =∑ ， hN Y R= = (3 6 )dE dE AdE

d x l dl

≤ ≤ =

In which, and represent the length of the diagonal dE and the vertical component of. S d DdEl dEY dEN

respectively o the segments of the influence line for the vertical component of dEN between A an and between E and B thus constructed from the influence lines for BR and AR , res vely, are shown in Fig. 8.8(f). The ordinates at D and E are then connected by a straight e to plete the influence line for

dEY (reference to the procedure for constructing influence lines for girders due to panel-point loads in ion 8.3), as shown in the figure.

pecti lin com

sect

8.5 Virtual Displacement Method for Constructing Influence Lines for Statically Determinate Beams 187

vertical member cC By considering the sectionⅡ-Ⅱand by applying the equilibrium equation(5) Influence line for force in

0Y =∑ d considerin

to one of the two portions of the truss, and by placing the unit load first to the left of joint C an g the right portion DB of the truss, we obtain the following expressions for

，

cCN :

0Y =∑ (0 )cC BN R x d= ≤ ≤

When the load is located to the right of joint D, application of the equilibrium equation1P = 0Y =∑

to the left portion AC of the truss yields

， d0Y =∑ (2 6 )cC AN R d x l= − ≤ ≤ =

The segments of the influence line for between A and C and between D and B thusly constructed from the influence lines for

cCNBR and AR , respectively, are shown in Fig. 8.8(g). The ordinates at C and D

are then connected by a straight line to plete the influence line for as shown in the figure. (6) Influence line for force in vertical member eE The influence line for can be constructed by considering the equilibrium of joint e. By applying

the equilibrium equation to the free body of joint e, we determine that is zero when the load is located at joints A through B. The influence line for coincides with the base line as shown in Fig.8.8 (i).

If the load is placed to top panel points, the influence lines for chord and diagonal members will remain unchangeable but those for vertical members will have variation.

For example, when the load is located to the top panel points, the ordinates of the segment cb of the influence line for , obtained by using sectionⅡ-Ⅱand placing the unit load to the right of joint c and applying the eq brium equation

com cCN

eEN 0Y =∑ eEN

1P = eEN

1P =

1P =

cCNuili 0Y =∑ to the left portion, will become AR− and the

influence line is shown in Fig.8.8 (h). If the load 1P = ne fo

moves on the top panel points and when the unit load reaches to the joint e, the ordinate of influence li r is equal to -1, whereas the ordinates at a through d and b through f are zero. The influence line for tained by connecting these ordinates by straight lines, is shown in Fig. 8.8(j). As this influence li icates, the force in member will be nonzero only when the unit load is located in the panels de and ef of the truss.

It is observed from the above that there is difference between the influence lines for deck trusses, of which the moving unit load is applied on the top panel points, and those for through trusses, of which the moving unit load is applied on the bottom panel points.

8.5 Virtual Displacement Method for Constructing Influence Lines for Statically Determinate Beams

In the preceding section, we have discussed the method for constructing influence lines for statically

eEN

eEN , obne ind eEN


determinate structures. In fact, the construction of influence lines for the response functions involving forces and moments can be considerably expedited by applying virtual displacement method, a procedure developed by Heinrich Müller-Breslau in 1886. The procedure, which is commonly known as Müller-Breslau's principle, is based on the principle of virtual displacements for rigid bodies.

8.5.1 Principle of virtual displacements for rigid bodies

The principle of virtual displacements for rigid bodies (be further discussed in the next chapter) can be stated as follows:

If a rigid body is in equilibrium under a system of forces and if it is subjected to any infinitesimal virtual rigid-body displacement compatible with its restraints, the virtual work done by the external forces is zero.

The term virtual simply indicates that the infinitesimal displacement is not caused by the forces but by other causation. Or you may consider that the virtual displacement is an imaginary displacement, not a real one.

Consider the simple beam shown in Fig. 8.9(a). The free-body diagram of the beam is shown in Fig.8.9 (b), in which forc s 1P , e

2P , 1 R and 2R form a balanced force system. Now pose that the beam is given an arbitrary itesimal virtual rigid-body displaceme port settlement) from its initial equili m position [Fig.8.9 (b)] to another posit [Fig.8.9 (c)], as shown in Fig.

8.9(c). As the beam undergoes the virtual displacement from initial librium position to virtual displacement position, the forces acting on it perform work, which is cal irtual work. The total virtual work, , performed by the external forces acting on the beam can be expressed as the sum of the virtual work done by each force; that is,

(8-10)

To prove the validity of the principle of virtual displacement, consider the condition that in Fig.8.9, set and place at the middle span of the beam. The equation (8-10) will become:

, sup infin

nt (supbriuion equiled v

eW

1 1 2 2 1 1 2 2 0eW P P R c R c= Δ + Δ − − =

2 0P = , 2 0c = 1P

1 11 1 1 10,

2 2c PP R c R− = =

1Δ2Δ

(a)

(b)

(c)

1P 2P

1P 2P

1R2R1C

2C

Fig. 8.9 Principle of virtual work(a) a simple beam; (b) initial equilibrium position;(c) virtual displacement position


In th s quations of the simple beam. is ca e, equation (8-10) becomes one of equilibrium eFor general cases, the equation (8-10) may be expressed as:

0i i K KP R cΔ + =∑ ∑ (8-11)

in which, represents th s applied on the system, iP iΔe load represents tcorresponding to (when ion coin ith that of the produc

he virtual displacements its direct cides w t PiP iP i iΔ is positive and vice

versa), while Kc resents the virtual displacements that is consistent with the given restraint c rep ondition,

KR represents the restraint forces corr Kc (when its sense coincide with that of Kcesponding to the uct of prod K KR c is positive and vice versa) Fig.8.9 (b) and (c), the virtual restrain. In t displacements Kc

are support settlements, the correspondin t forces g restrain KR are support reactions.

8.5.2 Mechanismic (or virtual displacement) method

The construction of influence lines for the response functions in ving forces and momentsvol can be cons b tual displacement) method. The procedure of the meth ws:

r a force defled str g to

by giving the released structure a unit displacement (or rotation) at the location and in th ection

tions involving forces and mom bendin

flecti virtual

8.10(evio

idera ly expedited by applying mechanismic (or virod can be stated as folloThe influence line fo (or moment) response function is given by the ected shape of the

releas ucture obtained by removing the restraint correspondin the response function from the original structure and

e positive dir of the response function. This procedure is valid only for the influence lines for response funcents (e.g., reactions, shears, g moments, or forces in truss members), and it does not apply to

the influence lines for de ons. To prove the validity of the method of displacement, consider the simply supported beam

subjected to a moving unit load, as shown in Fig. a). The influence lines for the vertical reactions at supports A and B of this beam were developed in the pr us section by applying the equations of equilibrium (see Fig. 8.4). We now draw the influence lines for the same two response functions by using method of virtual displacement.

To construct the influence line for the vertical reaction AR , we remove the restraint corresponding to

AR by replacing the hinged support at A by a roller support, which can exert only a horizontal restraint to A, as shown in Fig. 8.10(b). Note that point A of the beam is now free to displace in the direction of AR . Although the restraint corresponding to AR has been rem he reaction oved, t AR still acts on the which remains in equilibrium in the ho al position (shown by the solid lines in the figure) under the

n

beam,rizont

actio of the unit load and the reactions AR and BR . Next, a virtual unit displacement 1δ = in the e direction of Apositiv R is given to poin of the released beam as shown by the dash n Fig.

8.10t A ed lines i

(b). Note that the pattern of virtual displacement applied is consistent with the support conditions of the


released beam; that is, points A and B cannot move in the horizontal and vertical directions, respectively. Also, since the original beam is statically determinate, removal of one restraint from it reduces it to a statically unstable beam. Thus, the released beam remains straight (i.e., it does not bend) during the virtual displacement. Since the beam is in equilibrium, according to the principle of virtual displacements for rigid

, t g through the virtual displacements must be

zero;

bodies he virtual work done by the real external forces actin

1P =x

A B

that is,

1 1 0AR y× − × =

from which

AR y= (8-12)

yδ 1=

BR

1P =

AR

1δ = y

1

A B

1P =x

1(e)

A B

A B

(a)

(b)

(c)

(d)

x

BA

Fig.8.10 Influence lines for a simple beam(a) a simple beam; (b) deflecte(c) influence line for ; (d) d

d shape diagram for influence line ; eflected shape diagram for influenc

A

A

RR e

line ; (e) influence line for B BR R


wn in Fig. 8.10(b). Equa any position x is equal to the magnitude of

where y represents the displacement of the point of application of the unit load, as shotion (8-12) indicates that the displacement y of the beam at

AR due to a unit load acting at the position x of the beam. Thusequal to the ordinate of the influence line for

, the displacement y at any position x is

AR at that position. Since the unit loa is moving from ti

ici e n Fi

d 1P =A to B, equa on (8-12) expresses the deflected shape diagram of the beam, it is the influence line diagram as well. For simpl ty, draw it in Fig.8.10 (c), which is the sam as that shown i g.8.4 (c) obtained by static method.

The influence line for the vertical reaction BR m y be determined in a similar manner, as sh n in Fig. 8.10(d) and (e).

Through above analysis, we summarize the procedure r determining the reaction

a ow

fo and internal force c mechanismic (or virtual displacement) method

as fo

se a corresponding force(s) to obtain the released structure.

lines for statically determinate structures consist only of strai

influen e lines for statically determinate structures by usingllows: 1. From the given structure, remove the restraint corresponding to the response function whose

influence line is desired, and impo2. Apply a unit infinitesimal displacement to the released structure at the location and in the positive

direction of the response function. Draw a deflected shape diagram of the released structure that is consistent with the support and continuity conditions of the released structure to obtain the general shape of the influence line. (Remember that the influence

ght-line segments.) 3. By using the geometric relationship in deflected shape diagram and the given value of 1δ = , we

can directly determine the numerical values of all the remaining ordinates of the influence line.

ow

4. The ordinates located above the base line are positive and those below are negative.

8.5.3 Influence lines for simple beams by using mechanismic method

The following example will further explain the construction of influence lines for the response functions involving shears and bending moments by using the mechanismic (or virtual displacement) method.

Example 8-1

Construct the influence lines for the bending moment and shear at section C of the simple beam sh n in Fig.8.11 (a).

Solution

(1) Influence line for bending moment CM To construct the influence line for the bending moment CM , we remove the restraint corresponding

to CM by inserting a hinge at C and impose a pair of couples CM at C, as shown in Fig. 8.11(b). The


bending moment is assumed to be positive in accordance with the beam sign convention. The portions AC and CB of the released beam are now free to rotate relative to each other. Next, a virtual unit rotation, 1θ = ,

1θ =

abl

1Δ =

bl

al

Cy

1θ 2θ

x

1Δ

2Δ

(a)

(b)

(c)

(d)

CA B

CA B

C B

1P =

a bl

y

C

1B

A

MCM

1

2C

1C

CQ

CQ

1

1

b

b

1

1

(e)

y

Fig.8.11 Influence lines for simple beam(a) a simple beam; (b) deflected shape diagram for in

r ; (d) deflected CM shape diagram for influence line for ; (e) influence line for C CQ Q

by mechanismic methodfluence line for ; (c) influence line CM

fo


is introduced at C [Fig. 8.11(b)] by rotating portion AC by 1θ counterclockwise and portion CB by

2θ clockwise, so that 1 2 1θ θ θ+ = = . Applying the principle of virtual displacements, we write

1 2 1 21 ( ) 1 (1) 1( ) 0e C C C CW M M y M y M yθ θ θ θ= × + × − × = + − × = − =

From which

CM y=

This indicates t CMhat the e of the beam [Fig. 8.11(b e influence line for deflected shap )] is th . The value h

he geometric relation of t e ordinate Cy can be directly determined by using the geometric relation in the deflected shape diagram. By applying t 1 2 1θ θ+ = , we obtain

Cabyl

=

The influence line diagram is shown in Fig. 8.11(c).

the restraint corre

(2) Influence line for shear CQ To construct the influence line for the shear CQ at section C of the beam, we removesponding to CQ by replacing the rigid connection at C by a two-vertical-roller connection, as shown

in Fig. 8.11(d). Note that sections C of the portions AC and CB of the released beam are now free to displace vertically relative to each other. Next, a virtual unit relative displacement, 1Δ = , in the positive direction of CQ is given at C of the released beam [Fig. 8.11(d)] by moving the end C of the portion AC downward by 1Δ and the , so that 1 2 1Δ + Δ = Δ =end C of the portion CB ard by upw 2Δ . The values of ition

1 and 2Δ depend on the requirement that the segments ACΔ 1 and BC2 in the displaced pos must be parallel to each other. Applying the principle of virtual displacements, we write

1 2 1e C CW Q Q y= ×Δ + ×Δ ± ×

1 2 ( ) 1CQ y= Δ + Δ ± ×

(1) 1( ) 0CQ y= ± =

From which

CQ y= ∓

The plus before y are, respectively, applied to portions AC or CB. The above equation indic

inates and can be directly determined by using the geometric relation g in the d ape diagram. Thusly, we obtain

sign minus orates that the deflected shape diagram of the beam [Fig. 8.11(d)] is the influence line for CQ . The

values of the ord 1Δ 2Δ existineflected sh

1Δ = 2, a bΔ =

inf

l lThe luence line diagram for CQ is shown in Fig.8.11 (e).


8.5.4 Influence lines for statically determinate multispan beams by using mechanismic method

It is very convenient to construct influence lines for statically determinate multispan beams by using mechanismic (or virtual displacement) method. The procedure of constructing the influence lines for the multispan beam is the same as that of constructing the influence lines for simple beams introduced in the preceding section. The only notation is the deflection shapes of the released structures from which the restraints corresponding to the desired response functions were removed. Generally, a statically determinate multispan beam is composed of a stable portion and subsidiary portion(s). It should be realized that which portion(s) would develop virtual displacement(s) and which one(s) would not yield virtual displacement(s) after the restraint(s) of the multispan beam was removed and a virtual displacement was given. If the influence line of a desired response function is for the force (reactions and internal forces) in the subsidiary portion(s) of a multispan beam a virtual displacement corresponding to the desired function will not make the stable portion of the beam develop displacement and the virtual displacement will occur only on the subsidiary portion(s). However, the virtual displacement corresponding to the force(s), for which the influence line(s) is desired, of the stable portion of a multispan beam will make the entire multispan beam develop virtual movement. The following example will explain the argument in detail.

Example 8-2

Construct the influence lines for the desired forces BM , FQ , BR , CQ and GM of the multispan beam shown in Fig.8.12 (a).

Solution

(1) Influence line for BM

BMTo determine the general shape of the influence line for , we remove the restraint corresponding to BM portion

by replacing the rigid joint at B by a hinged joint that allows relative rotation between the left of B and the right portion of B, and impose a pair of couple BM as shown in Fig.8.12 (b) to

obtain a released structure (a mechanism). Next, a unit relative virtual rotation in the positive sense of BM 1θ = is given to hinged joint B of

the released structure, and a deflected shape diagram of the beam is drawn as shown in Fig.8.12 (b). Note that the deflected shape is consistent with the support and continuity conditions of the released structure; the portion AB, which is a stable portion, remains its original position; also, points B and D are attached to the roller supports, therefore, they cannot displace in the vertical direction; thus, the portion BC rotates about B and portion CDE rotates about D as shown in the figure.

Then, by using the geometric relation in the deflected shape diagram of the released structure, the numerical values of the ordinate of the influence line can be determined. Since the virtual displacement is infinitesimal, is small enough to let the length of CC′1θ = be 1 2m× and the length of EE′ 1m .


Hence, the values of the ordinate of the influence line at C and E are 2m and 1m respectively. The influence line diagram is shown in Fig.8.12 (c).

(a)

3m

C D EB GF

3m 2m 2m 2m 2m

1P =

A

(2) Influence line for FQTo remove the restraint corresponding to the shear at section F, we cut the given multispan beam at F

to obtain the released structure shown in Fig.8.12 (d). Next, a infinitesimal virtual relative displacement, , in the positive direction of the shear is given to the released structure to obtain the deflected shape

diagram or the shape of the influence line for shown in the figure. Note that the deflected shape is 1Δ =

FQ

1θ =

(b)E′

C D

C

M B

AE

B′

(c)

2m

1mE′

C D

C

AEB

′(d)FQ

FQ

C DE

A12

1

BFE′

C′3m 3m 2m 2m 2m 2m

161

2 13

(e)C

D EA

B

F ′′E′

F ′C′

13

12

12

16

Fig.8.12 (contd.)(a) a statically determinate multispan beam; (b) deflected shape of ; (c) influence line for ; (d) deflected sh

B

B

MM ape of ; (e) influence line for F FQ Q

1Δ=


consistent with the support and continuity conditions of the released structure. Since point A is attached to the hinged support and points B and D are attached to the roller supports, the portions AF, FBC and CDE must rotate about A, B and D respectively. Also, the rigid portions AF and FBC must remain parallel to each other in the displaced configuration. Then, by using the geometric relation in the deflected shape diagram shown in Fig.8.12 (d), the numerical values of the influence-line ordinates adjacent to F are firstly determined, and then the ordinate values at C and E can be fined successively. The influence line diagram for is shown in Fig.8.12 (e). CQ

(3) Influence line for BR The roller support at B is removed from the given beam, and an infinitesimal virtual displacement,

in the positive direction of 1Δ = BR , is applied at B to obtain the shape of the influence line shown in Fig 8.12(f). Because of the hinged support at A and the roller support at D, the rigid portions ABC and CDE of the released beam must rotate about A and D respectively. By realizing that the influence-line ordinate at B is equal to 1, the ordinates at C and E are then determined from the geometric relation of the influence line. The influence line thusly obtained is shown in Fig. 8.12(f).

43

23

E′

C′′B

AB C D

E

1(f)

C DE

E′

C′

12

AB

(g)

C D

E

AB

E′

G′(h)

G

1m

1m

D′

1θ = 2m

1

Fig.8.12 Constructing influence lines for a statically determinate multispan beam (f) influence line for ; (g) influence line for ; (h) influence line for B C GR Q M

8.6 Application of Influence Lines 197

(4) Influence line for CQThe restraint at C is removed from the given multispan beam, and an infinitesimal virtual

displacement, in the positive direction of , is applied at C to obtain the shape of the influence line shown in Fig 8.12(g). Because of the portion ABC of the released structure forms a stable overhanging beam, it neither translates nor rotates as a rigid system; and the rigid portion CDE of the released beam can only rotate about D. By realizing that the influence-line ordinate at C is equal to 1, the ordinate at E is then determined from the geometric relation of the influence line. The influence line thusly obtained is shown in Fig. 8.12(g).

1Δ = CQ

(5) Influence line for GM An internal hinge is inserted in the given multispan beam at section G, and an infinitesimal relative

rotation 1θ = is applied at G to obtain the shape of the influence line shown in Fig. 8.12(h). It can be seen from the figure that the length of is 1 2DD′ m=2m× , the ordinates at G and E are then determined from the geometric relation of the influence line. The influence line is shown in Fig. 8.12(h).

8.6 Application of Influence Lines

In the preceding sections, we have learned how to construct influence lines for various response functions of structures. In this section, we consider the application of influence lines in determining the maximum values of the response functions at particular locations in the structures due to variable loads.

8.6.1 Response at a particular location due to moving concentrated loads

1. Response at a particular location due to a single moving concentrated load As discussed in the preceding sections, each ordinate of an influence line gives the value of the

response function due to a single concentrated load of unit magnitude placed on the structure at the location of that ordinate. Thus we can state the following:

(1) The value of a response function due to any single concentrated load can be obtained by multiplying the magnitude of the load by the ordinate of the response function influence line at the position of the load.

(2) To determine the maximum positive value of a response function due to a single moving concentrated load, the load must be placed at the location of the maximum positive ordinate of the response function influence line, while to determine the maximum negative value of the response function, the load must be placed at the location of the maximum negative ordinate of the influence line.

Consider, for example, an overhanging beam subjected to a moving concentrated load of magnitude P, as shown in Fig. 8.13(a). Suppose that we wish to determine the bending moment at B when the load P is located at a distance x from the left support A. The influence line for BM , given in Fig. 8.13(b), has an ordinate y at the position of the load P, indicating that a unit load 1P = placed at that position of P causes


a bending moment BM y= . By the principle of superposition, the load of magnitude P must cause a bending moment at B, whose value is P times as large as that caused by the load of unit magnitude. Thus, the bending moment at B due to the load P is BM Py= .

Next, suppose that our objective is to determine the maximum positive and the maximum negative bending moments at B due to the load P. From the influence line for BM [Fig. 8.13 (b)], we observe that the maximum positive and the maximum negative influence-line ordinates occur at points B and D, respectively. Therefore, to obtain the maximum positive bending moment at B, we place the load P at point B, as shown in Fig. 8.13 (c), and compute the magnitude of the maximum positive bending moment as

B

P

C D

BM Py= , where is the influence-line ordinate at B [Fig. 8.13(b)]. Similarly, to obtain the maximum negative bending moment at B, we place the load P at point D, as shown in Fig. 8.13 (d), and compute the magnitude of the maximum negative bending moment as

By

B DM Py= − .

BA

A B CD

Byy

(a)

(b)

(c)

(d)

x

C D

BA

CD

BA

Dyx

P

P

Fig.8.13 Response due to a single moving load(a) an overhanging beam; (b) influence line for ; (c) position of load for maximum positive ; (d) position of load for maxi

B

B

MM mum negative BM


2. Response at a particular location due to a series of moving concentrated loads As mentioned in Section 1.5, live loads due to vehicular traffic on highway and railway bridges are

represented by a series of moving concentrated loads with specified spacing between the loads. Influence lines provide a convenient means of analyzing structures subjected to such moving loads. In the following, we will discuss how the influence line for a response function can be used to determine the value of the response function for a given position of a series of concentrated loads.

Consider, for example, a simple beam shown in Fig. 8.14(a). Suppose that we wish to determine the shear at section C of the beam due to the three concentrated loads , and , as

shown in the figure. The influence line for the shear at C is also shown in Fig. 8.14(b). The influence line for , given in Fig. 8.14(b), has the ordinates , and at the positions of

the loads , and respectively, indicating

that a unit load placed at those positions of , and cause shears ,

1P 2P 3P

CQ

1y 2y 3y

1P 2P 3P1=P

1P 2P 3P ( ) 11CQ y=

( ) 22CQ = y and ( ) 3y3CQ = respectively.

Because the principle of superposition is valid, the load of magnitude , and must cause shears at C,

whose values are , and times as large as those caused by the load of unit magnitude respectively,

i.e., , and . Thus, the shear at C due to the three concentrated loads , and

are

1P 2P 3P

1P 2P 3P

1 1P y 2 2P y 3 3P y 1P 2P 3P

1 1 2 2 3 3CQ P y P y P y= + +

Generally, suppose that we wish to determine the value of a desired response function (reactions or internal forces), Z, at a particular location due to a given position of a series of concentrated loads ,

,…, . The ordinates of the influence line for Z at those positions of , ,…, are , ,…, respectively. Thus, the value of the response function Z is the algebraic sum of a series of product

, ,…, . That is,

1P2P nP 1P 2P nP 1y 2y ny

1 1P y 2 2P y n nP y

1 1 2 21

n

n n i ii

Z P y P y P y P y=

= + + + = ∑ (8-13)

3. Response due to a uniformly distributed live load at a particular location

bl

al

(a)

(b)

CA B

3P1P 2P

la b

1

1y

2y3y

1

Fig.8.14 Response due to a series of moving loads (a) a simple beam; (b) influence line for CQ


Influence lines can also be employed to determine the values of response functions of structures due to distributed loads. Consider, for example, a simple beam subjected to a uniformly distributed live load of intensity , as shown in Fig. 8.15(a). Suppose that we wish to determine the shear at C when the load is placed on the beam, from

qx d= to x e= , as shown in the figure. The influence line for is also

given in the figure. By treating the distributed load applied over a differential length of the beam as a concentrated load of magnitude , as shown in the figure, we can express the shear at C due to the load as

CQdx

dP qdx=dP

( ) ( )CdQ dP y qdx y= = (8-14)

Where y is the influence-line ordinate at x, where is the point of application of , as shown in the figure. To determine the total shear at C due to the distributed load from

dPx d= to x e= , we integrate equation

(8-14) between these limits to obtain

( )e e e

C d d dQ qdx y q ydx q da qA= ⋅ = = =∫ ∫ ∫ (8-15)

in which e

dA ydx= ∫ represents the area under

the segment of the influence line, which corresponds to the loaded portion of the beam. This area is shown as a shaded area on the influence-line for in Fig. 8.15(b). CQ

Generally, the equation for calculating the value of a desired response function (reactions or internal forces) Z due to a uniformly distributed load at a particular location should be

Z qA= (8-16)

Equation (8-16) indicates that under the action of a uniformly distributed load the value of the desired function Z at a particular location of a structure is equal to the load-intensity times the area A of the influence line for the response function corresponding to the loaded range of the

structure. Note that the sign of A must be taken into consideration when applying Eq. (8-16).

q

Example 8-3

For the simple beam shown in Fig.8.16 (a), determine the magnitude of shear at section C due to a uniformly distributed load with the intensity of applied on the whole span of the beam by

CQ10kN/m

bl

al

(a)

(b)

C D EA B

x

la b

dx

dq

e

dA

y A

Fig.8.15 Response due to a uniformly distributed load (a) a simple beam; (b) influence line for CQ


using the influence line for , which has constructed as shown in Fig.8.16 (b).

CQ

Solution

It is observed from Fig.8.16 (b) that the positive area 1A and negative area 2A of the influence line Q may be easily determined as

C

11 36 1.8m2 5

A = × × =

21 24 ( ) 0.8m2 5

A = × × − = −

By using equation (8-16), we obtain

1 2( ) 10 (1.8 0.8) 10kNCQ q A A= + = × − =

8.6.2 The most unfavorable arrangement of uniformly distributed moveable loads

The most unfavorable arrangement of a uniformly distributed moveable load, which may be distributed arbitrarily on a structure, means the arrangement of the uniformly distributed moveable load, by which the value of a desired response function of the structure will reach to the positive (or negative) maximum.

For an arbitrarily uniformly distributed moveable load, equation (8-16) also indicates that the desired response function of a structure will be maximum positive if the uniformly distributed load is placed over all those portions of the structure where the influence-line ordinates are positive and vice versa. From Fig.

35

25

(a)

(b)

CB

C

A

A B

10kN/m

4m 6m

Fig.8.16 (a) a simp

due to a uniformly distributed loadle beam; (b) influence line for CQ

C DBA

C DBA

C DA B

(a)

(b)

(c)

q q

q

max min

Fig.8.17 The most unfavorable arrangement of uniformly distributed moveable loads(a) influence line for ; (b) unfavorable position for ; (c) unfavorable position for Z Z Z


8.17(a), we can see that the ordinates of the influence line for Z are positive between the points A and B and between the points C and D, and negative between points B and C. Therefore, to obtain the maximum positive maxZ , we place the uniformly distributed load q over the portions AB and CD of the beam, as shown in Fig. 8.17(b). Similarly, to obtain the maximum negative minZ , we place the load over the portion BC of the beam, as shown in Fig. 8.17(c).

Example 8-4

Determining the maximum positive and negative shears on section C of the simple beam shown in Fig.8.16 (a) due to a uniformly distributed moveable load with the intensity of10 , which may be arranged arbitrarily over the span of the beam by using the influence line for , which has constructed as shown in Fig.8.16 (b)

kN/m

CQ

Solution

From Fig.8.16 (b), we observe that the ordinates of the influence line for are positive between the points C and B. Therefore, to obtain the maximum positive due to the uniformly distributed moveable load10 , we place the uniformly distributed load over the portions CB of the beam and compute the maximum positive value of due to this load by multiplying the load intensity by the area under the portion CB of the influence line. Thus

CQmaxCQ

kN/m

maxCQ

max 10 1.8 18kNcQ qA+= = × =

Similarly, since the ordinates of the influence line for are negative between the points A and C, the load must arranged over the portion AC to obtain the maximum negative shear at C. By multiplying the load intensity by the area under the portion AC of the influence line, we obtain

CQ

min 10 ( 0.8) 8kNcQ qA−= = × − = −

8.6.3 The most unfavorable arrangement of a series of moving concentrated loads

The most unfavorable arrangement of a series of moving concentrated loads herein means the arrangement of the loads, by which the value of the desired response function for a structure will reach to the maximum positive (or negative). Influence lines can also be used for determining the maximum values of response functions at particular locations of structures due to a series of moving concentrated loads.

Consider, for example, the beam shown in Fig. 8.18(a). Suppose that our objective is to determine the maximum positive shear at point B due to the series of four concentrated loads shown in the figure. The influence line for is shown in Fig. 8.18(b). Assuming that the load series moves from right to left on the beam, we can observe from these figures that as the series moves from the end C of the beam toward point B, the shear at B increases continuously as the ordinates of the influence line under the loads increase. The shear at B reaches a relative maximum when the first load of the series, the load 8kN, reaches just to

BQ


the right of B, where the maximum positive ordinate of the influence line is located. As the load 8kN crosses point B, the shear at B decreases abruptly by an amount equal to 8(0.6 0.4) 8kN− + = − . With the series of loads continuing to move toward the left, increases again and it reaches another relative maximum when the second load of the series, the load 10kN, reaches just to the right of B, and so on. Because becomes a relative maximum whenever one of the loads of the series reaches the maximum

positive influence-line ordinate, we can conclude that during the movement of the series of loads across the entire length of the beam, the (absolute) maximum shear at B occurs when one of the loads of the series is staying at the location of the maximum positive ordinate of the influence line for . Since it is not possible to identify by inspection which load placed at the maximum influence-line ordinate will cause the maximum positive , we use a trial-and- error procedure to determine the value of the maximum positive shear at B. As shown in Fig. 8.18(c), the series of loads is initially positioned on the beam, with its first load, the load 8kN, placed just to the right of B, where the maximum positive ordinate of the influence line is located. Noting that the slope of the influence-line segment for the portion BC is 1:10 [Fig. 8.18(b)], we compute the value for this loading position as

BQ

BQ

BQ

BQ

BQ

1m1.5m 1.5m

8kN 10kN 15kN5kN

B C

4m 6m

A

A B C

4m 6m

A B C

4m 6m

(a)

(b)

(c)

(d)

0.6

0.4B C10

10

1

1Influence line for A BQ

Loading position 1

8kN10kN15kN 5kN

1.5m 1m 1.5m

8kN 10kN 15kN5kN

Loading position 2

Fig.8.18 The most unfavorable arrangement of a series moving loads

1.5m 1m 1.5m


( )11 1 1 18 6 10 4.5 15 3.5 5 2 15.55kN

10 10 10 10BQ = × × + × × + × × + × × =

Next, the entire series of loads is moved to the left by 1.5m to place the second load of the series, the load 10kN, at the location of the maximum positive ordinate of the influence line, as shown in Fig. 8.18(d). The shear at B for this loading position is given by

( )2

1 1 1 18 2.5 10 6 15 5 5 3.5 13.25kN10 10 10 10BQ = − × × + × × + × × + × × =

By inspection, no other loading positions could cause larger positive values of than that obtained from the above two loading positions.

BQ

By comparing the values of determined from the above two loading positions, we conclude that the maximum positive shear at B occurs for the first loading position—that is, when the load 8kN is placed just to the right of B [Fig. 8.18(c)].

BQ

Maximum positive shear at B:

( )max15.55kNBQ =

From forgoing discussion, we may summarize that the determination of the most unfavorable arrangement of a series of moving concentrated loads or the determination of the most unfavorable position of a series of moving concentrated loads will follow the following procedure.

1. Determination of the critical loading positions on which the values of a desired response function, Z, of a structure will reach to extreme values (positive or negative).

2. By comparing the magnitudes of the extreme values determined for all the critical loading positions considered, obtain the maximum positive (or negative) value(s) of the response function so as to determine the most unfavorable position of a series of moving concentrated loads.

The characteristics and the principle of determination of the critical loading positions will be further explained by using one of polygonal influence lines as follows.

Consider, for example, a series of moving loads is shown in Fig.8.19 (a). The influence line for a desired response function Z of a structure is shown in Fig.8.19 (b), of which the shape is polygonal. Although the influence-line ordinates corresponding to the loads can be obtained by using the properties of the similar triangles formed by the influence line, it is usually convenient to evaluate such an ordinate by multiplying the slope of the segment of the influence line where the load is located by the distance of the load from the point at which the influence line changes its slope. The sign (plus or minus) of the ordinate is obtained by inspection. Therefore, the angles with respect to the horizontal coordinate axis x of each of the straight lines of the polygon [Fig.8.19 (b)] are represented by 1α , 2α and 3α ( 1α and 2α are positive, 3α is negative). The resultants of the loads applied over each segment corresponding to each straight line of the influence line are represented by 1R , 2R and 3R [Fig.8.19 (a)]. , and 1y 2y 3y


[Fig.8.19 (b)] are the ordinates of the influence line at positions corresponding to , and , while 1P 2P 3P1y , 2y and 3y [Fig.8.19 (b)] are the ordinates of the influence line at positions corresponding to

1R , 2R and 3R . For any arrangement of the loads, we may evaluate the magnitude of Z by adding the values of the

response function contributed by the individual loads together. Thus

1 1 2 2 6 6Z P y P y P y= + + +

Since the influence line for quantity Z is composed of individual straight lines, corresponding to each straight line the magnitude of Z may be expressed by the contribution of each resultant of the loads applied on that segment. For example, corresponding to firs segment, we find

1 1 2 2 1 1P y P y R y+ =

By the principle of superposition, we may express the magnitude of Z as the sum of all contributions of each resultant of the loads as follows

3

1 1 2 2 3 31

i ii

Z R y R y R y R y=

= + + =∑ (a)

It is evident that since the shape of the influence line for quantity Z is a polygon the response function Z must be a linear piecewise function with respect to x. With the knowledge of mathematics, we know that

(a)

1y

xΔ

(b)

1R2R 3R

1P 2P 3P 4P 5P 6P

y

x

1y2y

1 0α >

2 0>3 0α <

α

Fig.8.19 Determination of critical loading position for polygonal influence line(a) a given position of a series of moving concentrated loads; (b) influence line for Z

2y 3y

xxΔ Δ


if the quantity Z reaches to one of the extreme values at a loading position of the series, the increment of Z, ZΔ , must less and equal to zero with respect to the infinitesimal increment xΔ (right or left), around the

position of the loads [see Fig.8.20]. That is, the ZΔ must satisfy 0ZΔ ≤ (b)

from equation (a), we write

i iZ R yΔ = Δ∑ (c)

Suppose that all the concentrated loads are moved an infinitesimal distance xΔ (positive to right), the line of action of iR is moved the same distance xΔ as well. From the geometry, the increment of the ordinate of iy must be

taniy x iαΔ = Δ

Thus, the increment of Z is

taniZ x R iαΔ = Δ ∑

Therefore, when the quantity Z reaches to one of the positive extreme values the critical position of the loads must satisfy the following condition. That is,

tan 0i ix R αΔ ≤∑ (d)

The condition (d) has two situations:

when the loads move slightly to the right, 0, tan 0

when the loads move slightly to the left, 0, tan 0i i

i i

x R

x R

α

α

⎫Δ > ≤ ⎪⎬

Δ < ≥ ⎪⎭

∑∑

(8-17a)

Similarly, when the quantity Z reaches to one of the negative extreme values the critical position of the

0Zx

Δ>

Δ 0Zx

Δ<

Δ

0Zx

ΔZ Z =Δ

0Zx

Δ>

Δ( )Z f x=Z

O x x

Extreme Z Extreme Z

(a) (b)O

( )f x=

0Zx

Δ<

Δ

Fig.8.20 The character of critical position of loads (a) extreme point; (b) extreme domain


loads must satisfy the condition as follows:

when the loads move slightly to the right, 0, tan 0

when the loads move slightly to the left, 0, tan 0i i

i i

x R

x R

α

α

⎫Δ > ≥ ⎪⎬

Δ < ≤ ⎪⎭

∑∑

(8-17b)

Equation (8-17) indicates that when Z reaches an extreme value the sum of taniR iα∑ must change its sign with respect to xΔ , a given infinitesimal increment of the position of the loads. In other words, the response function Z reaches extreme values when a series of moving loads moves to their critical positions.

Now we will analyze when the sum of taniR iα∑ changes its sign. It is observed from forgoing sections that the shape of the influence line for a desired response function Z of a statically determinate structure is composed of straight lines. The slope, tan iα of any straight line, must be a constant. Only when the value of one of the resultant iR changes its magnitude that has the possibility to make the sum

taniR iα∑ change its sign. Next, in order to make the resultant iR of the loads change its magnitude, one of the series of moving loads must move through a position where the shape of the influence line for Z occurs a vertex. When the series of moving loads moves to the left, the load moves to the left of the position corresponding to the vertex as well, so that the resultant iR of the loads and the sum tani iR α∑ maybe change its magnitude and sign respectively; and vice versa. The load, which makes the sum

taniR iα∑ change its sign when it moves to the left or to the right of the position of a vertex of the influence line for Z, is termed as a critical load, indicated by ; while the loading position (i.e., putting

at the vertex) is called as critical position of the loads. crP

crPWhen the influence line for a desired response function, Z, is composed of two straight lines as shown

in Fig.8.21 (b), the critical loading position may be expressed more simply. The case in which the desired quantity Z reaches to its peak value is now discussed as follows.

By using equation (8-17a), now 1 21,2; , i α α α β= = = − , 1R and 2R represent the resultants of loads applied on the left portion and the right portion (segments a and segment b) respectively. In order to make the sum taniR iα∑ change its sign one of the series of moving loads, , must be placed at the position where the ordinate of the influence line reaches to its peak value. and represent the resultants of loads to the left of and to the right of respectively.

crPleftR rightR

crP crPFrom Eq. (8-17a), when the loads move to the right, we write

1 2tan tan 0R Rα β− ≤

tan ( ) tan 0left cr rightR P Rα β− + ≤

( )left cr rightcR P Ra b

0c− + ≤

thus, we obtain


left cr rightR P Ra b

+≤

When the loads move to the left, the equation (8-17a) will be written as

1 2tan tan 0R Rα β− ≥

( ) tan tanleft cr rightR P R 0α β+ − ≥

( )left cr rightc cR P Ra b

0+ − ≥

thus, we obtain

left cr rightR P Ra b+

≥

Therefore, for the influence line with a triangular shape, the necessary condition for the critical loading position of a series moving loads may be expressed as

left cr right

left cr right

R P Ra b

R P Ra b

+ ⎫≤ ⎪⎪

⎬+ ⎪≥ ⎪⎭

(8-18)

α β

leftR

crPnP

rightR

(a)

(b)

1P 2P

y

x

c

a b

Fig.8.21 Critical position of loads for influence line with a triangular shape (a) critical position of loads; (b) influence line with a triangular shape


Equation (8-18) indicates that the character of the critical loading position for the influence line with a triangular shape is that one of the series moving loads, , is located at the position where the ordinate of the influence line reaches to its peak value, and when is added to the left portion the load intensity of the left portion will be greater than that of the right portion and vice versa.

crPcrP

Example 8-5

For the crane beam and its loadings shown in Fig.8.22 (a), find the most unfavorable loading position of the loads and the maximum bending moment CM at section C. 1 2 3 4 82kNP P P P= = = =

Solution

(a)

(b)

(c)

CBA

A B

3.6m 8.4m

12m

3.5m 3.5m1.5m1P 2P 3P4P

0.07m

2.52m2.07m

1.02m

1P 2P3P 4P

C

1P 2P3P 4P

1 2 3 4 82kNP P P P= = = =

maxFig.8.22 Determination of due to the pressure of crane wheels (a) crane beam and its wheel pressures; (b) influence line for ; (c) the most unfavorable loading position

C

C

MM

(1) Construct the influence line for CM as shown in Fig.8.22 (b). (2) The possible critical load may be or . First, consider as a critical load and put

at the position where the ordinate of the influence line reaches to its peak value. From equation (8-18), crP 2P 3P 2P

2P


we write

The load move slightly to the left, 82 82 82 82 3.6 8.4+ +

>

The load move slightly to the right, 82 82 82 823.6 8.4

+ +<

It is observed that when is added to the left portion the load intensity of the left portion will be greater than that of the right portion and vice versa. So is one of critical loads.

2P2P

Next, consider as a critical load and put at the location where the ordinate of the influence line reaches to its peak value. Note that the load has moved off the span of the beam. By using equation (8-18), we obtain

3P 3P1P

The load move slightly to the left, 82 82 82 3.6 8.4+

>

The load move slightly to the right, 82 82 823.6 8.4

+>

Apparently, when is added to no matter what portion the load intensity do not satisfy equation (8-18). So is not a critical load.

3P3P

(3) From above analysis, when concentrated load is placed at point C, which is the most unfavorable loading position for

2PCM [Fig.8.22 (c)], the bending moment CM reaches to its maximum

value. By principle of superposition, the magnitude of CM is given by

( )max82 (0.07 2.52 2.07 1.02) 465.76kN mCM = × + + + = ⋅

Example 8-6

Determine the maximum axial force of member 1 of the girder truss shown in Fig.8.23 (a) due to the wheel loads of a group of trucks and find the most unfavorable loading position of the loads.

1N

Solution

(1) Construct influence line for as shown in Fig. 8.23(b). 1N(2) First, suppose the group of trucks move from right to left as shown in Fig.8.23 (a), now there are

only five concentrated loads applied on the truss. Consider the third load (numbered from the left), , as and put it at joint C. From equation (8-18), we write 120kNP = crP

The load move slightly to the left, 60 120 120 70 13018 27

+ + +>

The load move slightly to the right, 60 120 120 70 130 18 27+ + +

<

It is observed that when the third load is added to the right portion the load intensity of the right


portion will be greater than that of the left portion and vice versa. So the third load (numbered from the left) is one of critical loads and the axial force of member 1 corresponding to the arrangement of loads is given by:

( )1 max160 1.26 120 1.66 120 1.80 70 1.13 130 0.87 682.8kNN = × + × + × + × + × =

(3) Next, suppose the group of trucks move from left to right as shown in Fig.8.23 (c), now there are seven concentrated loads applied on the truss. Consider the fourth load [numbered from the right, see Fig.8.23 (c)], , as and put it at joint C. The equation (8-18) is written by 120kNP = crP

The load move slightly to the left, 130 70 120 120 60 130 7018 27

+ + + + +>

The load move slightly to the right, 130 70 120 120 60 130 70 18 27

+ + + + +>

It is evident that the arrangement of the loads is not a critical loading position. (4) By comparing the results of above calculation for the two loading positions, we conclude that

the magnitude of the maximum axial force that develops in member 1 is ( )1 max682.8kNN = tension.

C

(a)

(b)

60kN 120kN 70kN 130kN

4m 1.4m10m

4m

C

A B1

4 18md = 6 27md =

45m

1.261.66 1.8

1.130.87

A B

120kN

6

hm

=1

Fig.8.23 (contd.)(a) girder truss and trucks moving towards left; (b) influence line for ;N


(5) This maximum force occurs when the third load (numbered from the left) of the series (move from right to left) is placed at joint C of the truss, as shown in Fig. 8.23(b).

130kN 70kN

4m10m

1.4m

120kN120kN

4m

60kN 15m130kN 70kN

4m

(c)

(d)

C

SUMMARY

The influence lines and their applications for reactions and internal forces of plane statically determinate beams and trusses are discussed in this chapter. The key points are as followings:

1. It is essential that we clearly understand the difference between an internal force diagram and an influence line for an internal force at a given position. An internal force diagram denotes how the internal force varies on all sections along the length of a member for a loading condition whose position is fixed on the member, whereas an influence line for the internal force at the given position denotes how the internal force varies as a unit load moves across the length of the member 1P =

2. There are two kinds of methods for constructing influence lines, static (or equilibrium) method and mechanismic (or virtual displacement) method. If you use static method to construct influence lines for structures, it is generally convenient to employ the influence lines for their support reactions. Thus, before proceeding with the construction of an influence line for shear or bending moment at a point on the structure, make sure that the influence lines for all the reactions, on either the left or right side of the point

A B1

18m 27m

0.26 0.661.66

1.81.53

0.53 0.27 BC

6m

1

1

Fig.8.23 due to the wheel loads of moving trucks(c) girder truss and trucks moving towards right; (d) influence line for

N

A

N


under consideration, are available. Otherwise, draw the required influence lines for reactions by using equilibrium method.

3. Note that all influence lines for a statically determinate structure are composed of straight lines. 4. One of the characteristics of the influence lines for some structural members that are not

subjected to live loads directly but to which the live loads are transmitted via panel points is that the influence line between adjacent panel points is a straight line. By using this feature, the influence lines for statically determinate beams and trusses may be constructed conveniently.

5. The construction of influence lines for the response functions involving forces and moments can be considerably expedited by applying mechanismic (or virtual displacement) method. The key steps of the method is: (1) from the given structure remove the restraint and impose corresponding force, corresponding to the response function whose influence line is desired, to obtain the released structure; (2) apply a unit infinitesimal displacement to the released structure at the location and in the positive direction of the response function, and draw a deflected shape of the released structure to obtain the general shape of the influence line; (3) by using the geometric relations of the deflected shape diagram and the given unit infinitesimal displacement 1δ = , the numerical value of the influence-line ordinate cam be determined; (4) the ordinates above the base line are positive and those below are negative.

6. We discussed two aspects of the application of the influence lines. One is the determination of the magnitude of a desired response function by using its influence line. Another utilization of the influence line is the determination of the most unfavorable loading position of a series of moving loads, at which the desired response function will reach its maximum value.


8-1 What is the definition of an influence line? Why take a downward concentrated load of unit magnitude as the foundation of the construction of an influence line? What mechanical meanings the abscissa x and ordinate y of an influence line represent? What is the difference between an internal force diagram and an influence line?

8-2 If the beam shown in Fig.8.3 (a) is subjected to a concentrated load , whether or not be able to use the influence line for

PBR shown in Fig.8.3 (b) to determine the reactions due to the concentrated

load? 8-3 Try to compare the difference between two static methods (first draw free-body diagram and then

use equilibrium equation; directly write out the equations of the desired quantity by the method of sections) used to determine the influence line for a desired quantity Z?

8-4 When write the function of an influence line for the force of a beam why sometime the function of an influence line is defined in the entire span of a beam and why sometimes the function of an


influence line must be defined piecewise along the length of a beam? 8-5 Try to explain the meaning of the ordinate of the influence line at the point C for the bending

moment at section C of a simple beam and the difference between the influence line and the bending moment diagram of the beam due to the action of the unit load 1P = applied at the point C.

8-6 Try to explain the reason why the influence line for shear at section C is parallel between the left portion and the right one of section C, and why the influence line for shear at section C of a simple beam has an abrupt change and what is the meanings of the ordinates corresponding to the left and the right of the abrupt change.

8-7 Please simply state the procedure for constructing the influence line for a desired quantity Z due to the live loads transmitted via panel points.

8-8 What is the character of an influence line due to a unit load transmitted via panel points? 8-9 What is the character of constructing the influence line for trusses by using static method? 8-10 When constructing the influence lines for trusses why we have to make clear that the truss is a deck

truss or a through truss? Under what condition the influence lines for the two trusses are identical and under what condition the influence lines for the two trusses are different?

8-11 What is the procedure to construct influence lines by mechanismic method? 8-12 By using the character of the geometric construction of a statically determinate multispan beam,

explain why the shape of the influence line for a quantity in the subsidiary portion is coincided with the initial base line in the stable portion of the multispan beam when using the mechanismic method to construct the influence line?

8-13 How to determinate the most unfavorable loading position and a critical loading position? What is their difference?

8-14 Simply explain the meanings of sign change of the discriminant, taniR iα∑ , of a critical loading position ?


8-1 Construct the influence lines for reactions and internal forces on designated sections by using static (or equilibrium) method. (a) Construct the influence lines for reactions AR and AM on support A and internal forces CM

and on section C. CQ(b) Construct the influence lines for reactions AR at support A and at support B and internal

forces BR

CM and on section C andCQ AM , ( )A leftQ and ( )A right

Q on section A.

(c) Construct the influence lines for reactions AR at support A and at support B and internal BR



forces CM , and on section C. CQ CN(a) Construct the influence lines for reactions AxR and AyR at support A and internal forces CM ,

and on section C. CQ CN

x

8-2 Construct the influence lines for statically determinate beams due to unit load transmitted via panel points by using static method.

(a) Construct the influence lines for internal forces CM and on section C. CQ

(b) Construct the influence lines for reactions AR at support A and internal forces CM and

on section C, and on section D.

CQ

( )D leftQ ( )D right

Q

α

A

a bl

1P =

C

x 1P =

B

Problem 8-1 (a)

x 1P = B

C

a bl

Problem 8-1 (c)

D BA C

2m 3m 2m

Problem 8-1 (b)

B

C

1P =

αA

A x

a bl

Problem 8-1 (d)


8-3 onstruct the influence lines for axial forces of the designated members of the statically determinate

es for axial forces of the designated members 1, 2 and 3 of the deck

(b) nes for axial forces of the designated members 1, 2, 3 and 4 of the

Ctrusses by using static method. (a) Construct the influence lin

truss shown in the figure. Construct the influence lideck truss shown in the figure.

1P =

AD

C

B1

2

3

E4 2m×

2m

Problem 8-3 (a)

C D 1 2

34

E F

A B

1P =

2a 2a4a

a

Problem 8-3 (b)

AB

C DA

CB

D1P = 1P =

1m 2m 2m

E F G

3m 1m 1m 1m2m 2m 2m3m

Problem 8-2 (a) Problem 8-2 (b)

H

1m


(c) Construct the influence lines for axial forces of the designated members 1, 2 and 3 of the deck truss shown in the figure.

(d) Construct the influence lines for axial forces of the designated members 1, 2, 3 and 4 of the through truss shown in the figure.

C D1P =

1 E AA

8-4 Construct the influence lines for the designated quantities by using mechanismic (or virtual displacement) method. (a) The beam and its designated quantities have been pointed out in problem 8-1 (c).

(b) Construct the influence lines for internal forces EM on section E, ( )B leftQ and on

section B.

( )B rightQ

(c) Construct the influence lines for internal forces on section C, CY HM and HQ on section H

AE

1P =

BC

D

1m 1m 2m 2m 2m

Problem 8-4 (b)

AB C D E F

HG

1m 1m 1m 1m 1m2m 2m

Problem 8-4 (c)

1P =

32

B3 3m×

Problem 8-3 (c) Problem 8-3 (d)

2m1m

D E B

1P =

1 2 3

4 F C

3 4m 2 4m×

4m

×


and on section G. GQ(d) Construct the influence lines for internal forces AY on section A, on section B,BY EM ,

and ( ) on section E. ( )E leftQ E right

Q

C D E F1P =

BA

4m 1.5m 2.5m 1m 3m

Problem 8-4 (d)

8-5 Compute the magnitude of the designated quantities by using their corresponding influence lines. (a) Compute the magnitudes of shear ( )D left

Q on the left section of D and bending moment EM

on section E. (b) Compute the magnitudes of shear DQ on section D and bending moment EM on section E.

(c) Compute the magnitudes of shear and bending moment CQ CM on section C.

(d) Compute the magnitudes of reaction at support B, bending moment BY EM on section E and

shear on the left section of B. ( )B leftQ

20kN 20kN 5kN

A CB

3kN 2kN/m

8m2m2.5m

Problem 8-5 (c)

A

40kN 40kN20kN/m

E B CD

2m 2m 2m 2m 4m

Problem 8-5 (d)

AD E

BC

4m2m2m 2m

Problem 8-5 (a)

ABD E C

2kN/m

2m 2m 2m4m

Problem 8-5 (b)


8-6 Suppose the beams are subjected to arbitrarily uniformly distributed loads, determine the most unfavorable arrangement of the loads by using the influence line for designated quantity Z; and compute the maximum maxZ and the minimum minZ if the load intensity 20kN/mq = . (a) Compute bending moments ( , )maxCM ( )minCM , shears ( )maxCQ and ( )minCQ of the beam

shown in the figure of problem 8-1 (b). (b) Compute bending moments ( ) ,maxHM ( )minHM , shears ( )maxHQ and ( )minHQ of the beam

shown in the figure of problem 8-4 (c). 8-7 Determine the most unfavorable arrangement of the crane loads and compute the maximum and

minimum values of the designated quantities by using their corresponding influence lines. (a) Compute bending moment , shears( )maxCM ( )maxCQ and ( )minCQ of the beam shown in the

figure. (b) Compute reaction ( )maxBR of the beam shown in the figure, the loading is the same as shown in

problem 8-7(a). (c) Compute bending moment , shears( )maxCM ( )maxCQ and ( )minCQ of the beam shown in the

figure.

82kN 82kN 82kN 82kN3.5m 1.5m 3.5m

BB AAC

C

3m 6m

Problem 8-7 (a) Problem 8-7 (b)

9m 9m

1.15m

108.5kN 108.5kN 76kN 76kN4.4m 3.5m

BAC

5m 7m

Problem 8-7 (c)

CHAPTER 9

PRINCIPLE OF VIRTUAL WORK AND DISPLACEMENTS OF STRUCTURES

The abstract of the chapter The attention of the chapter is mainly focused on the calculation of the displacement of statically

determinate structures by using the principle of virtual work. In the analysis of statically determinate and indeterminate structures, the chapter plays a role of

the link between the preceding and the following of the text: it is not only the end of the analysis of statically determinate structures but also the beginning of the analysis of statically indeterminate structures.

First, we will discuss the principle of virtual work for rigid bodies and its two kinds of applications: determining the restraint forces and determining the displacement of structures.

Then, we mainly discuss the principle of virtual work for deformable bodies, the general equations for determining the displacement of structures, the calculation of the displacement of structures due to loadings, temperature changes and support settlements, and the method of graph multiplication and so on.

Finally, introduce the reciprocal laws for linearly-elastic deformable systems.

9.1 Introduction for Calculation of Structural Displacement

The calculation of the displacement of statically determinate structures is a very important portion of the content of Structural Mechanics, and it is also the foundation of the analysis for statically indeterminate structures.

The evaluations of the displacement for a structure have two sorts of purpose. The first purpose is for checking the stiffness requirement of the structure. In design stage of a structure, it is necessary that a structure has to satisfy not only the requirement of strength but the requirement of stiffness as well. That is, the deflection of the structure must not exceed the allowable value specified by design code (For example,

the allowable value of the beam of a building roof is equal to 1 1~200 400

of its span, while the allowable

value of a crane beam is equal to 1600

of its span). The second purpose is for the preparation of analysis

for statically indeterminate structures. Because in the analysis of statically indeterminate structures, we

220

9.1 Introduction for Calculation of Structural Displacement 221

have to consider not only static equilibrium conditions but the compatible conditions of the displacements of a structure as well.

There are three factors that cause a structure to yield displacements. They are: (1) loadings; (2) temperature changes and material expansion; (3) support settlements.

Under the action of above three factors, the members of a structure will be deformed, which is named the deformation of the structure. When a structure develops deformation, the translation of a point, the translation or rotation of a section of a member of the structure are referred to as the displacement corresponding to the point and the section of the structure, respectively.

There are two kinds of displacements for a structure. One is linear displacement or translation, which indicates the distance of movement of a point of the structure along a line. Another is angular displacement or rotation, which indicate the sectional rotation of a member of the structure.

As shown in Fig. 9.1(a), a simple beam is under the action of uniformly distributed loads; the deflection shape is shown in dashed line in the figure. The point C (at the middle of the span) in unloaded state moves to point in the deformed state, C′ CC ′ is the vertical displacement of C; and the rotation

Bθ at support B is the angular displacement of section B.

Fig. 9.1(b) shows a cantilever rigid-frame. Under the action of a concentrated load P at the cantilever end, the deformed shape of the rigid frame is shown in dashed line in the figure. As shown in the figure, the beam end, C, in unloaded position moves to C′ in deformed position and the displacement is the

translation of point C in the direction of CCCΔ

′ . The horizontal and vertical components of are

indicated by and , respectively, and the geometric relationship between them is shown in Fig.9.1

(b).

CΔ

CxΔ CyΔ

Cθ represents the rotation of section C and is named the angular displacement of the section.

CxΔ

Cθ

C yΔ

C xΔ

CΔ

BθC

C′

(a)(b)

A B

q

P

C′

CB

A

C

yΔ

Fig.9.1 Displacement of structures(a) displacement of a beam; (b) displacement of a rigid frame

222 Chapter 9 Principle of Virtual Work and Displacements of Structures

The chapter will pay attention to the calculation of the displacements of linearly deformable systems (or structures). The theoretical foundation of the calculation is the principle of virtual work. The method of the calculation is the method termed unit load (or unit force) method.

The so called linearly deformable systems implies that the displacements are proportional to the loads applying on them. That is, when the loads are removed the displacements are disappeared. The conditions which a linearly deformable system must be satisfied are:

(1) The materials of the system are linearly elastic, i.e., the relationship between stress and strain is linear.

(2) The deformation of the system is small enough to neglect the changes of originally acting effect of its loadings.

A linearly deformable system is also termed linearly elastic system. The principle of superposition can be applied to a linearly elastic system. Therefore, the calculation of displacements of a linearly deformable system can be implemented by using the principle of superposition.

9.2 Virtual Work and Principle of Virtual Work

9.2.1 Virtual work

As we have already known, the magnitude of the work done by a unchangeable force P is the product of the force P and the displacement of its point of application undergone in the direction of P. That is, Δ

W P= Δ (9-1)

Now, let us extend the definition. Let W represents virtual work, P indicates a generalized force, and Δ

identifies the generalized displacement corresponding to P. Here, the definition of work is still the product

of force P and its corresponding displacement Δ . The dimension or unit of work W is Newton multiplying

meter (i.e. N m⋅ ). (1) Generalized force and its corresponding displacement If P is a force, its corresponding displacementΔ will be a translation along the direction of P. as

shown in Fig.9.2 (a), a vertical force P is exerted at point C on the simple beam; when the force does work through the displacements shown in Fig.9.2 (c), its corresponding displacement is the vertical translation

at the same point C. ΔIf P is a moment, its corresponding displacementΔ will be an angular displacement about the same

direction of the moment P. As shown in Fig.9.2 (b), a moment M is exerted at right-hand end B of the simple beam; when the moment does work through the displacements shown in Fig.9.2 (c), its corresponding displacement is the rotation θ about the same direction of the moment M acting at the end section B of the beam.

9.2 Virtual Work and Principle of Virtual Work 223

Δ

(a) (b) (c)

C

P

C

P M

If a system of forces does work through their corresponding displacements, their work can be expressed in the form of equation (9-1) as well. The system of forces may be represented by symbol P and their corresponding displacements may be represented by symbol Δ . Here, P is termed a generalized force and is named a generalized displacement corresponding to P. For example, a pair of forces and shown in Fig.9.3 (a), equal in magnitude (

Δ 1P 2P1 2P P P= = ) but opposite in direction, can be recognized as a

generalized force P. Its corresponding displacement Δ ( 1 2Δ = Δ +Δ ) is the sum of the horizontal components of the displacements at point A and B in the sense of the connecting line of the two forces, that is, the relative horizontal displacement between point A and B of the rigid frame, as shown in Fig.9.3 (b). In this case, is just the work done by the two forces shown in Fig.9.3 (a). Again for example shown in Fig.9.3 (c), the two parallel force and are equal in magnitude ( ) but opposite in direction, can be considered as a generalized force P as well. Its corresponding displacement

1 2 1 1 2( )P P P PΔ = Δ +Δ = Δ + Δ2

1P 2P

1 2P P P= =Δ ( ) is the sum of the vertical components of the displacements

at points A and B in the sense of the two forces, that is, the relative vertical displacement between points A and B of the rigid frame, as shown in Fig.9.3 (d). In this situation,

1Δ = Δ +Δ2

21 2 1 1 2( )P P P PΔ = Δ +Δ = Δ + Δ is just the work done by the two forces shown in Fig.9.3 (c). Similarly, if a generalized force P is a couple of moments, then its corresponding displacement Δ will be the sum of the angular displacements about the same sense of the couple P. As shown in Fig.9.3 (e), a couple of moments 1M and 2M are equal in magnitude ( 1 2M M M= = ) but opposite in direction, acting at the left and right side of hinge C of the rigid frame, can be realized as a generalized force M. Its corresponding displacement θ ( 1 2θ θ θ= + ) is the sum of the slopes of the left and right sections adjacent to hinge C in the sense of the couple, the relative rotation between the left and right sections adjacent to hinge C, as shown in Fig.9.3 (f). In this condition, 1 2 1 1 2( )M M M M 2θ θ θ θ θ= + = + is just the work done by the couple shown in Fig.9.3 (e).

(2) Virtual work The term virtual is used here for the purpose to distinguish the traditional concept of real work and the

θ

A B A B A C

Fig.9.2 Forces and their corresponding displacements (a) a vertical force ; (b) a moment ; (c ) displacements corresponding to and

P MP M

B


new concept of virtual work. The so called virtual work simply means imaginary (not real) work, of which the force (or forces) has no relation with the displacement (or displacements) undergone by the force (or forces) during performing the work. That is, the displacement (or displacements) is not caused by the force (or forces) doing the work; or when performing the work through the displacement (or displacements), the force (or forces) is a constant value (or values). As shown in Figs.9.2 (a) and (b), the generalized force in the figure (a) does not depend on the generalized displacement in the figure (c).

1Δ2Δ

(a) (b)

(c) (d)

(e)

C

1θ 2θ(f)

C

1PA B

2PA B

1P 2P

A BA

B

2M1M

1Δ 2Δ

1 2 1

Fig.9.3 Generalized forces and displacements(a) two horizontal forces and ; (b) corresponding displacements of the two horizontal forces aP P P 2

1 2 1 2

1 2 1 2

nd ;(c) two vertical forces and ; (d) corresponding displacements of the two vertical forces and ;(e) a couple of moments and ; (f) corresponding rotations of the couple and

PP P P P

M M M M


9.2.2 Two kinds of applications of principle of virtual work for rigid body systems

If the strain of the materials composing a system does not take into consideration, only the movements of the members of the system are concerned during its moving process. The system is recognized as a rigid body system. Since the principle of virtual work for rigid body systems was introduced in the previous section 8.5.1, now two kinds of its applications will be discussed.

For work, there exist two states, one is force state and another is displacement state. Since the force state and the displacement state are independent to each other for the virtual work performed by the force state acting through the displacement state, an unknown force of a balance force system acting on a structure will be determined by imagining a likely displacement state for the structure; whereas an unknown displacement of a given displacement state of a structure will be obtained by imagining a likely balanced force state for the structure. Now separately discuss them as follows.

1. Determination of unknown restraint forces for statically determinate structures Fig.9.4 (a) shows an overhanging beam. The reaction X at roller support A is desired when the action

of a concentrated force P at the right end of the beam is considered. In order to make the beam develop a rigid body displacement, the restraint against the movement in

the direction of the unknown reaction X is removed and substituted by the reaction force X (now the force X has been an active force). The original system has thusly become an unstable system (or mechanism) with one degree of freedom. The rigid body ABC can rotate about the hinge B freely; make point A moves to 1A and point C to ; then a virtual (imaginary) displacement state has been obtained, as shown in Fig.9.4 (b).

1C

The force system shown in Fig.9.4 (a) is in equilibrium under the action of external force P and the reaction forces and , that is, the Fig.9.4 (a) shows a balanced-force-system state. , BX Y BX

Let the balanced-force-system state shown in Fig.9.4 (a) undergo the virtual displacement state shown in the Fig.9.4 (b) to perform virtual work, the expression for the virtual work can be written as

0X PP PΔ + Δ = (9-2)

In which, PΔ and are the virtual displacements along the directions of P and X, respectively. By the

geometric relation of the virtual displacement state, we obtain XΔ

X aϕΔ = , P bϕΔ = −

Since the direction of is the same as that of X, it has positive sign; the direction of XΔ PΔ is opposite to

that of P, it has negative sign. Then they relation will be

P

X

ba

Δ= −

Δ (a)

Substitute equation (a) into equation (9-2), we obtain


0X XbX Pa

Δ − Δ =

That is,

0bX Pa

− = (b)

So bX Pa

= (c)

It is observed from equation (a) that the ratio P

X

ΔΔ

does not vary with respect to the magnitude of

, then for the convenience of calculation, the virtual displacement in the direction of X may be

imagined as a unit value [shown as Fig.9.4 (c)]. The virtual displacement in the direction of P will thusly be XΔ

Xδ . Let 1Xδ = , by the geometric relation, we write

(a) B

PΔ

ϕ

ϕ(b)

1Xδ =

P

ba

δ = −(c)

AC

1

a

P

X B

A

XΔ

X BY b

C

1C

BA

BAC

Fig.9.4 Determination of support reaction by principle of virtual work (a) a balanced force system; (b) virtual displacement state; (c) simple indication of virtual displacement state


Pba

δ = −

Now the virtual-work equation and unknown reaction X become as

1 0PX P δ⋅ + ⋅ = , PbX P Pa

δ= − =

The eventually positive result of X tells us that the assumed direction of X (downward) is the same as it really is.

When the shear force X ( DX Q= ) at section D of the overhanging beam shown in Fig.9.5 (a) is desired to determine, remove the restraint against the movement in the direction of X, that is, replace the rigid connection at D by two infinitesimal-length links parallel to the beam axis, then replace the restraint for shear by a pair equal in magnitude but opposite in direction shear force DX Q= . Here, X is a generalized force, and the original beam has become a mechanism with one degree of freedom shown as Fig.9.5 (c). The rigid body DBC can rotate about the hinge B freely; make point D moves to and point C to ; since the two portions AD and DBC are connected by two infinitesimal-length links parallel to the beam axis, the portion

1D1C

2AD remains parallel to during any virtual displacements; the deflection of A is equal to zero because of the roller support at A. A virtual (imaginary) displacement state has been thusly obtained, as shown in Fig.9.5 (c). Let the balanced-force-system state shown in Fig.9.5 (b) undergo the virtual displacement state shown in Fig.9.5 (c) to perform virtual work, the equation of the virtual work can be expressed as

1D BC1

0X PX PΔ + Δ =

In which, is the virtual displacements along the directions of X, i.e. the relative deflection between

the immediate left and right sections of D. It is a generalized displacement. Let XΔ

1Xδ = , as shown in

Fig.9.5 (d), with the geometric relation of the virtual displacement state, we write

Pba

δ =

At now, the virtual-work equation will be

1 PX P 0δ⋅ + =

So

DbX Q Pa

= = −

The eventually negative result of DX Q= indicates that the assumed direction of X is opposite to its real

direction, i.e. the shear force DQ has negative sign.


The forgoing method used to determine reactions and internal forces of a structure is termed as unit displacement method.

Now we conclude the procedure for determination of a restraint force X (reaction or internal force) of a statically determinate structure by using unit displacement method as follows.

CD B

(1) Real system Remove the restraint against the movement in the direction of the desired force and substitute for it a restraint force X so as to make the structure become a mechanism with one degree of freedom, and the restraint force X becomes an active force. So the original force system of the structure and the force X compose a new balanced force system to make the mechanism remain in equilibrium.

(2) Virtual system Apply a unit infinitesimal displacement consistent with the restraint conditions to the mechanism along the action line in the positive direction of the restraint force X ( 1Xδ = ) so as to obtain a virtual displacement state, in which the virtual displacement corresponding to a force P is Pδ .

(3) Determine Pδ By using the geometric relation between Xδ and Pδ , the value of each virtual displacement corresponding to each force P will be determined.

(4) Determine Let the new balanced force system undergoes the virtual displacement state to X

PΔ

xΔ

1xδ =

P

ab

δ =

(a)

(b)

(c)

(d)

A

2a

2a b

P

P

CD BA

X

C

1DBA ϕ

ϕϕ

2D1C

A CD

0.5

0.5

B

Fig.9.5 Determination of internal force by principle of virtual work(a) an overhanging beam; (b) mechanism for determining ; (c) virtual displacement;(d) simple indication of virtual displ

Xacement state


perform virtual work, we obtain a virtual-work equation as follows.

1 PX Pδ 0⋅ + =∑

then

PX Pδ= −∑

The key steps in the procedure are removal of the restraint against the movement in the direction of a desired restraint force, the assumption of a unit virtual displacement along the action line in the positive direction of the desired restraint force, the construction of correct virtual displacement configuration and determination of Pδ by using the geometric relation.

Example 9-1

Determine the reaction at support C and the bending moment CR GM on section G for the statically determinate multispan beam shown in Fig.9.6 (a) by using the principle of virtual displacement.

Solution

(1) Determine the reaction CR① Real system Remove the roller support C and substitute for it a vertical reaction X ( ), a

mechanism as shown in Fig. 9.6 (b) is thusly obtained. CR X=

② Virtual system Apply a unit infinitesimal displacement consistent with the restraint conditions to the mechanism along the action line in the positive direction of X, i.e. 1Xδ = , so as to obtain a virtual

displacement state of the system as shown in Fig.9.6 (c). ③ Determine Pδ By the geometric relation of the virtual displacement state, we find

1 3Pδ = − , 2 1.5Pδ = (a)

The negative value of 1Pδ here means that the direction of 1Pδ is opposite to that of . 1P④ Determine Substitution of each corresponding virtual displacement will obtain the virtual

work equation as X

1 1 2 21 P PX P P 0δ δ⋅ + + = (b)

Substitute equation (a) into equation (b), we obtain 2 ( 3) 1.5 4.5X P P P= − − − × =

(2) Determine the bending moment GM on section G

① Real system Remove the restraint against the rotation in the direction of bending moment GM ,

i.e. replace the rigid-joint section G by a hinged joint, and substitute for it a couple of moments X equal in magnitude but opposite in direction. Then the couple of moments GX M= have been changed into a pair


of active moments from its original form of internal force, and a mechanism shown as Fig. 9.6 (d) is consequently obtained.

② Virtual system Apply a unit infinitesimal relative rotation consistent with the restraint conditions in the positive direction of the couple X to the mechanism, i.e., 1Xδ = , so as to obtain a

virtual displacement state of the system as shown in Fig.9.6 (e).

1Xδ = 1Pδ

2Pδ

Bδ

1Pδ

2Pδ1Xδ =

(a)

(b)

(c)

(d)

(e)

A B C

D

E

a

1 2P P=

F

G

2a a 2a 2a a

C

D

E

F

2P P=

1P 2PA B

X

A

C D

EF

B

A C

D

EF

1P 2PG B

X

BA C

D

EF

G

Fig.9.6 Determination of reactions and internal forces by principle of virtual work (a) a statically determinate multispan beam; (b) mechanism for determining ; (c) virtual displacement for determ

CRining ; (d) mechanism for determining ;

(e) virtual displacement for determining C G

G

R MM


③ Determine Pδ By the geometric relation of the virtual displacement state, we write

1 4P aδ = − , 2 2P aδ = (c)

④ Determine By Substituting each corresponding virtual displacement into the virtual work equation, we will consequently obtain

X

1 1 2 21 P PX P P 0δ δ⋅ + + = (d)

Substitute equation (c) into equation (d), we find 1 2 ( 4 ) 2 0X P a P a⋅ + × − + × = , 6X Pa=

2. Determination of displacements for statically determinate structures Fig. 9.7 (a) shows an overhanging beam, its roller support A has a vertical settlement in a distance of

. Now the vertical displacement on section C is desired to be determined. 1c ΔSince the displacement state shown in Fig. 9.7 (a) has been given, in order to use principle of virtual

work to determine , a balance force system has to be assumed. Then, a concentrated force P along the direction of the displacement is assumed to be imposed at point C so as to let it perform virtual work through the displacement . Consequently, the force P and its corresponding reactions ( ) compose a balanced force system as shown in Fig. 7(b), which is a virtual-force-system state.

ΔΔ

Δ /AR Pb= a

Let the balanced force system shown in Fig. 9.7(b) perform virtual work through the rigid-body displacement shown in Fig. 9.7(a), we find

1 0AP R cΔ+ = (a)

1 0bP P ca

Δ + = , 1b ca

Δ = − (b)

It can be observed from the equation (b) that Δ has no relation with the magnitude of the assumed force P. In the circumstances, in order to simplify calculation we may assume the force P to be a unit magnitude, i.e., 1P = as shown in Fig. 9.7(c). Then let the balanced force system formed by the unit force 1P = to perform virtual work through the rigid-body displacement shown in Fig. 9.7 (a), we immediately obtain

1 0b ca

Δ + =

1b ca

Δ = −

The minus sign of means that the direction of displacement Δ Δ should be in the opposite direction of the assumed force P, i.e., upward direction as shown in the dashed line in Fig. 9.7 (a).

We may find implication from above that if we apply principle of virtual work between assumed


force-system state [Fig.9.7 (b) or (c)] and a given displacement state [Fig.9.7 (a)] the real rigid-body displacement of a structure can be determined. The main point here is the imposition of a unit load 1P = , which should be imposed in the direction of the desired displacement. The desired displacement will be determined by establishing virtual-work equation between the force system formed by the unit load and the displacement state of the structure. Consequently, the method is also termed unit load method.

9.2.3 Principle of virtual work for deformable bodies

When a system undergoes a deforming process not only each members of the system develop rigid-body deformation, the materials making of the system yield deformation as well. The system belongs to the category of deformable systems. One of applications of principle of virtual work to deformable bodies is the principle of virtual forces and it can be stated as follows:

If a deformable structure is in equilibrium under the action of a virtual system of forces (and couples) and if it is subjected to any small real deformation consistent with the support and continuity conditions of the structure, then the virtual external work done by the virtual external forces (and couples) acting through the real external displacements (and rotations) is equal to the virtual internal work done by the virtual internal forces (and couples) acting through the real internal displacements (and rotations).

Δ

A

bR Pa

=

ba

(a)

(b)

(c)

1c

BA

C

a b

BA

C

BA

C

P

1

Fig.9.7 Determination of displacement of a statically determinate structure by principle of virtual work (a) a given displacement state; (b) a virtual force system; (c) a virtual unit force system


In this statement, the term virtual is associated with the forces to indicate that the force system is arbitrary and has no relation with the action causing the real deformation. The statement may be expressed mathematically as

eW Wi= (9-3)

In which, represents the virtual work done by external force system; represents the virtual work

done by internal force system. eW iW

It should be realized that the principle of virtual forces as described here is applicable regardless of the cause of real deformations; that is, deformations due to loads, temperature changes, or any other effect can be determined by the application of the principle. However, the deformations must be small enough so that the virtual forces remain unchangeable in magnitude and direction while performing the virtual work. Also, although the application of this principle in this text is limited to elastic structures, the principle is valid regardless of whether the structure is elastic or not.

Now we discuss how to obtain the expression of , virtual work done by internal forces. iWConsider a simple beam shown in Fig.9.8 (a). The simple beam is in equilibrium under the action of a

system of forces as shown. Fig.9.8 (b) shows a displacement and deformed state of the simple beam caused by other reason other than the system of the forces.

Take a differential segment of the beam as a discussing object as shown in Fig.9.8 (c). The internal forces acting on caused by the force system in Fig.9.8 (a) is shown in Fig.9.8 (c), and the relative deformation of due to the deformation of the state shown in Fig.9.8 (b) is shown in Fig.9.8 (d).

dsds

ds

The virtual work done by the internal forces shown in Fig.9.8 (c) acting through the deformation shown in Fig.9.8 (d) can be defined as

idW Nd Qd Mdλ η θ= + +

In which, and , N Q M are generalized forces; , d dλ η and dθ are generalized displacement corresponding to and , N Q M .

Consequently, the expression for the virtual work done by the internal forces of the simple beam will be expressed as

(B

i AW Nd Qd Md )λ η= + +∫ θ

)

For a framed structure, the expression will become

(iW Nd Qd Mdλ η= + + θ∑∫

Introducing the expression for external virtual work (see subsection 8.5.1), the Eq. (9-3) can be rewritten as

eW


( )i i K KP R c Nd Qd Mdλ ηΔ + = + + θ∑ ∑ ∑∫ (9-4)

Since

0, , d ds d ds d dsλ ε η γ θ κ= = = (9-5)

The Eq. (9-4) will become as

0(i i K K )P R c N Q M dsε γ κΔ + = + +∑ ∑ ∑∫ (9-6)

P(a)

(b)

(c)

(d)

A B

a ds

A B

N NQ Q M M

dθ

In which, and , N Q M are internal forces acting on the differential segment , i.e., axial force, shear force and bending moment;

ds, d dλ η and dθ are relative deformation corresponding to

and

, N QM , i.e., relative axial deformation, relative shearing deformation and relative rotation of the

dλ

dη

1Rk

=

Fig.9.8 Stress and deformation state of a simple beam(a) a given stress state; (b) a given displacement and deformed state; (c) internal forces acting on caused by the force system in figure (a)(d) relative deformation of due to the deformation of figure (b)

ds ds ds

dsds ds


differential segment ; ds 0, ε γ and are the relative axial, tangent and bending strains of the κ

9.3 General Equation and Unit Load Method for Computing Displacements 235

differential segment . The symbol means integral over the axial length of a member, while symbol

means summation over all the members.

ds ∫∑

Equation (9-6) is the mathematical expression of principle of virtual work, which is commonly termed virtual work equation for deformable bodies. There is a variety of application of principle of virtual work. However, the chapter will only discuss one of its applications to the calculation of displacements for various deformable structures.

9.3 General Equation and Unit Load Method for Computing Displacements

The unit load method for deformable bodies can be very easily extended by that for rigid bodies. Now assume a real deformed state of a structure has been given. Our purpose is to determine one of its

displacements of the structure by using Eq. (9-6). In order to let one equation to solve one unknown displacement, only one unit load

Δ1P = must be imposed on the position, where the displacement is

desired, in the direction corresponding to the desired displacement Δ

Δ so as to let the virtual work equation just involve in and no other unknown displacement. Then take the unit load Δ 1P = and reactions and internal forces caused by the unit load as a balanced force system, the real deformed state of the structure as a virtual displacement state; let the virtual balanced force system acting through the real displacement state perform virtual work. Consequently, the virtual work equation (9-6) will be established and the desired displacement will be determined by solving the equation. Δ

By substituting the unit load 1P = and its corresponding reactions KR and internal forces ( N , Q and M ) into Eq. (9-6), the general expression for computing a desired displacement Δ by using unit load method can be expressed as follow

01 ( ) K KN Q M ds R cε γ κ×Δ = + + −∑∫ ∑ (9-7)

It is worth of mentioning again that the displacement and the deformed state (Δ , Kc , dsε , 0dsγ , )

here are given by a real structure, while the reactions (

dsκ

KR ) and internal forces ( N , Q , M ) of the structure,

which compose a virtual balanced force system with 1P = , are caused by an assumed unit load 1P = . If the strains (ε , 0γ , ) and support settlements (κ Kc ) of each differential segments of a structure are

prescribed, the procedure to find a desired displacement Δ of the structure may be stated as follows: (1) Impose a unit load 1P = at the location and in the direction of the desired displacement . Δ(2) Determine the reactions ( KR ) and internal forces ( N ,Q , M ) of the structure caused by the

unit load 1P = according to the static equilibrium conditions. (3) Calculate the desired displacement Δ by using Eq. (9-7). In equation (9-7), the sign convention is: when the directions of internal forces ( N , Q , M ) and

reactions ( KR ) in the virtual force-system state are the same as those of the strains (ε , 0γ , ) and support κ


settlements ( Kc ) of the real structure, the products between force and deformation are positive (for instance, when M and make the fiber of the same side of a differential segment be in tension, the product of κMκ will be positive) and vice versa.

If a desired displacement determined by Eq. (9-7) has positive sign, it means that the direction of is the same as that of

ΔΔ 1P = and vice versa.

It should be realized that the equation (9-7) is a general formula to calculate a displacement Δ . It can be employed to determine a displacement of a structure regardless of the properties of its materials, the type of the deformation, the factors causing the deformation, and the type of the structure.

9.4 Calculation of Displacements Caused by Loads

9.4.1 Computing formula and procedure for determination of displacements

The subsection will discuss the calculation of displacements for statically determinate elastic structures caused by loads.

For elastic framed structures, the strains (ε , 0γ andκ ) in Eq. (9-7) can be determined by using Hook’s law learned in strength of materials. The axial strain ε due to axial force PN , average shearing strain 0γ due to shear force PQ and bending strain κ due to bending moment PM may thusly be expressed as follows

PNEA

ε = (9-8a)

0PQk

GAγ = (9-8b)

PMEI

κ = (9-8c)

In which, E and G are the elastic tension (or compression) and shearing modulus of the materials, respectively; A and I are the area and inertial moment of a cross section, respectively; EA, EG, and EI represent the axial, shearing and flexural rigidities of a cross section, respectively; k is a modified factor because of non uniformity of shearing stress over a cross section, it is associated with the shape of the cross section, and for rectangular shape, 1.2k = , circular shape 10 / 9k = , I shape (1/k A A= 1A represent the area of the web of the I shaped cross section).

Substituting Eq. (9-8) into Eq. (9-7), the general formula for calculating elastic displacements due to the action of loads will be written as

P PNN kQQ MMds ds dsEA GA EI

Δ = + +∑ ∑ ∑∫ ∫ ∫ P (9-9)

9.4 Calculation of Displacements Caused by Loads 237

It should be noted that in Eq. (9-9) there are two sorts of internal forces. N , Q and M are internal

forces caused by a assumed unit load 1P = , while PN , PQ and PM are the internal forces due to

loads acting on a structure. The sign convention about N and PN , Q and PQ , M and PM are

the same as those in the previous chapters, that is, N and PN are positive when they are tensile forces;

Q and PQ are positive when they make a differential segment rotate clockwise. The sign convention of

M and PM are regulated by their product, i.e., the product of M and PM are positive when both of

them make the fibers in the same side of a member be in tension, otherwise, it is negative. The following step-by-step procedure can be used to determine the displacements of a structure due to

action of external loads by unit load method. (1) Real system Write out the expressions expressing the internal forces PN , PQ and PM of

the structure due to its real external loads by static equilibrium equations, and then construct their diagrams. (2) Virtual system Remove all the given loads from the structure; then apply a generalized unit

load at the location and in the direction of the desired displacement Δ . Find out the expressions expressing the virtual internal forces N , Q and M of the structure due to the generalized unit load by static equilibrium equations, and then construct their diagrams.

(3) The desired displacement of the structure now can be determined by applying Eq. (9-9) by substituting the results from step (1) through (2) into Eq. (9-9).

Δ

9.4.2 Formulae for calculating displacements of various types of structures

The expression (9-9) is a general equation for computing elastic deflections of a statically determinate structure due to external loads. There are three terms on the right-hand side of the equation. The first term represents the effect of axial deformation; the second term indicates the effect of shearing deformation and the third term involves in the effect of bending deformation of the all members of the structure. Obviously, the ratio of the three aspects of the effects in the equation will vary with the characteristics of deformation and type of the structure. The different simplified equations of expression (9-9) will thusly be obtained by consideration, on which the main effect of deformation is taken and negligible effect of deformation is discarded, of characteristics of deformation of a structure.

(1) Beam and rigid frames For beams and rigid frames, their deformation is mainly caused by bending moments. The deformation

due to axial force and shear forces are small enough to be neglected (see examples 9-2 and 9-8). Therefore, the expression (9-9) can be simplified as

PMM dsEI

Δ =∑∫ (9-10)

(2) Trusses


For trusses, each of the members is subjected to only axial force. In addition this, the cross-sectional area, axial forces N and PN , and the elastic modulus remain unchangeable over the length of each of the members, generally. The expression (9-9) is thusly simplified as

PNN NNds lP

EA EΔ = =∑ ∑∫ A

(9-11)

(3) Composite structures For composite structures, beam-typed (or flexural) members are mainly subjected to bending moments,

and two-force members are mainly undergone axial forces. Therefore, the expression (9-9) may be modified as

P PMM Nds lNEI E

Δ = +∑ ∑∫ A (9-12)

The first summation is applied to flexural members and the second for two-force members. (4) Arches For a general arch with massive body, the effect of its curvature is commonly neglected, and the

deformation due to bending moment is just considered. In the circumstance, equation (9-10) may be used to calculate the deflection of this sort of arches. However, the deformations due to bending moments and axial forces have to be taken into consideration for arches with small rise and long span. Thusly, the expression will be simplified as

PMM NNds dsEI EA

Δ = +∑ ∑∫ ∫ P (9-13)

9.4.3 Examples for calculating deflections due to external loads

Example 9-2

Determine the deflection on the middle-span section C of the simple beam shown in Fig.9.9 (a), and compare the effect on the deflection due to the bending deformation with that due to shearing deformation. The shape of the cross section of the beam is rectangular and its area is .

Δ

b h×Solution

(1) Real system The internal force diagrams of the beam due to real external loads are shown in Fig. 9.9 (a), which are depicted out according to the following equations expressing (in terms of a position coordinate x) the variations of the internal forces.

2(2Pq )M lx x= −

( 2 )2PqQ l= − x


0PN =

(a) (b)

(2) Virtual system Remove all the given (real) loads from the beam; then apply a unit load 1P = at point C, where the deflection is desired, and in the direction of the desired deflection to form the virtual force system as shown in Fig. 9.9(b). The internal force diagrams of the beam due to the virtual load are also shown in Fig. 9.9 (b), whose equations are written, in terms of the same position coordinate x, out as follows

Δ

12

M x=

12

Q =

0N =

(3) Calculation of deflection Δ Substituting equations coming from steps (1) through (2) into every term of expression (9-9), we

obtain deformation caused by bending moments and shear forces, respectively, as follows

242

0

1( ) ( ) 52 22384

lP

M

qx lx xMM qlds dxEI EI EI

−Δ = = = ↓∫ ∫ ( )

2

8ql

12

C

1P =

C

qA BA B

14

2ql

2l

2l

x

diagramQ

diagramPM diagramM

diagramPQ

2l

2l

x2ql 1

2

2ql

2ql 1

2 12

Fig.9.9 Determination of deflection at middle point (a) a simple beam and its internal force diagrams;(b) virtual force system and its internal force diagrams

C


22

0

1( ) ( 2 )2 22 1.2 0.15 ( )

lP

Q

q l xQQ qlk ds dxGA GA GA

−Δ = = × = ↓∫ ∫

Note: for rectangular cross section, 1.2k = . The positive answers for MΔ and QΔ indicate that the directions of the two deflections coincide

with that of the unit load 1P = . Since the axial forces are equal to zero, there is no axial deformation. Consequently, the total

deflection is written as 4 25 0.15 ( )

384M Qql ql

EI GAΔ = Δ + Δ = + ↓

(4) Compare the effects on the deflection due to the bending deformation with that due to shearing deformation.

The ratio between bending and shearing deformation is 2

4 2

0.1511.52

5384

Q

M

qlEIGA

ql GAlEI

Δ= =

Δ

For a rectangular cross section, 2

12I hA= , assume Poisson ratio 1

3μ = , then 82(1 )

3EG

μ= + = .

Substituting them into above expression, we obtain 2

2.56Q

M

hl

Δ ⎛ ⎞= ⎜ ⎟Δ ⎝ ⎠

Obviously, when the ratio between depth and span of the beam ( hl

) is equal to 110

, the ratio between

shearing and bending deformation will be 2.56%Q

M

Δ=

Δ. This implies that the shearing deformation is less

than 3 percent of bending deformation. Accordingly, the shearing deformation can be neglected for a long-span beam and its displacements may be calculated by Eq. (9-10) straightly. However, when the ratio

between depth and span of a beam is more than 15

, the ratio between shearing and bending deformation

becomes to . Therefore, the shearing deformation cannot be neglected for a beam with quite high ratio between the depth and span when calculating the displacements of the beam.

10%

(a)

(b)

constantEA =

82.1 10 kPaE = × 212cmA =B

D

C

E

2m2m 2m2m

45kNP =45kNP =

1.5m

A

45 45

60


Example 9-3

Determine the vertical displacement of joint C located at the middle point of the bottom chords of the truss shown in Fig.9.10 (a).

Solution

(1) Real system The real system and its member axial forces PN due to real loads obtained by

using the method of joints are shown in Fig. 9.10 (b). (2) Virtual system The virtual system consists of a unit load 1P = applied in the vertical

direction at joint C, as shown in Fig. 9.10(c). The member axial forces N due to the virtual unit load are determined by applying the method of joints, as depicted in Fig.9.10 (c).

Table 9-1 calculation of CΔ

member N (kN)PN (cm)l 2(cm )A 2(kN/cm )E (cm)PNN lEA

AC 2/3 60 400 12 42.1 10× 0.063

AD -5/6 -75 250 12 42.1 10× 0.062

DE -4/3 -60 1 4002× 12 42.1 10× 0.063

DC 5/6 0 250 12 42.1 10× 0

=0.188 (cm)∑

(3) Vertical displacement at C ( ) CΔ

The formula to determine the displacement of a truss is Eq. (9-11). That is,

PNN lEA

Δ =∑

To facilitate the computation of the desired deflection, the real and virtual member forces are tabulated along with the length (l), the cross-sectional area (A) and the elastic modulus (E) of the members, as shown in Table 9-1. Note that because of the symmetry of geometry and loads acting on the truss, only half of the truss is computed. The final value of must be 2 times of the sum tabulated in the table. That is, CΔ

2 0.188 0.376cm ( )CΔ = × = ↓

Example 9-4

Determine the vertical displacement at point B for the curved beam shown in Fig. 9.11 (a), and compare the effects of displacement components due to shearing deformation and axial deformation. The centre line of the beam is a segment of arc; the shape of its cross section is rectangular; the central angle and radius of the arc is equal toα and R, respectively.


Solution

(1) Real system Determine internal forces due to the real loads. The equations expressing the variation of the internal forces of the real beam can be expressed in terms of a position coordinate x (located at a distance from right end B) as

212PM qx= −

sinPN qx θ= −

cosPQ qx θ=

(2) Virtual system The virtual system consists of a unit load 1P = applied in the vertical direction at end B, as shown in Fig. 9.11(b). The equations expressing the variation of the internal forces of the virtual beam can be expressed in terms of the same x position coordinate as that in step (1) as

M x= −

sinN θ= −

cosQ θ=

(3) Vertical displacement at B ΔIf neglecting the effects of the curvature of the beam, the equation determining the displacement Δ

will be

P PMM NN kQQds ds dsEI EA GA

Δ = + +∑ ∑ ∑∫ ∫ ∫ P

For comparison, the displacement components due to the bending moments, axial forces and shear forces represented separately by MΔ , NΔ and QΔ , are determined by following equations.

θdθ

α

x(a) (b)

B B

q

y

1P =

C x C

θA A

Fig.9.11 Determination of displacement of a curved beam(a) real system and its load; (b) virtual system and its unit load


3

2A A

PM B B

MM qds x dsEI EI

Δ = =∫ ∫

2sinA A

PN B B

NN qds x dsEA EA

θΔ = =∫ ∫

2cosA A

PQ B B

kQQ kqds x dsGA GA

θΔ = =∫ ∫

In order to identify the variables we select variable θ as unique variable. Then, sinx R θ= , (1 cos )y R θ= − , ds Rdθ=

Substituting them into above equations, we obtain 4

3

0sin

2MqR d

EIα

θ θΔ = ∫

23

0sinN

qR dEA

αθ θΔ = ∫

22

0cos sinQ

kqR dGA

αθ θ θΔ = ∫

Since

3 2 3

0 00

1 2 1sin (1 cos )sin cos cos cos cos3 3 3

d dα

α α 3θ θ θ θ θ θ θ α⎡ ⎤= − = − + = − +⎢ ⎥⎣ ⎦∫ ∫ α

2 3

00

1 1cos sin cos (1 cos )3 3

dα

α 3θ θ θ θ α⎛ ⎞= − = −⎜ ⎟⎝ ⎠∫

We write 4

32 1( cos cos2 3 3MqR

EI)α αΔ = − +

232 1( cos cos

3 3NqREA

)α αΔ = − +

23(1 cos )

3QkqR

GAαΔ = −

If , then 090α =4

3MqREI

Δ =


223NqREA

Δ =

2

3QkqR

GAΔ =

(4) Comparison of effects of displacement components due to bending, shearing and axial

9.5 Graph-Multiplication Method 245

deformations If we set , , 090α = / 1/1h R = 0 3/ 8 /E G = , and select a rectangular cross section

with (h indicates the depth of the cross section) and 2/ /I A h= 12 1.2k = , the ratios between the displacement components will be

2

2

2 1 16 6

N

M

I hR A R

Δ ⎛ ⎞= = =⎜ ⎟Δ ⎝ ⎠ 00

2

2

112 375

Q

M

kEI k E hR GA G R

Δ ⎛ ⎞= = =⎜ ⎟Δ ⎝ ⎠

The results of the ratios indicate that the displacement components caused by shearing and axial deformations may be negligible in some conditions.

9.5 Graph-Multiplication Method

Recall from previous section that when calculating the displacements of beams and frames, the following integral have to be carried out.

( ) ( )i kP M x M xMM ds dxEI EI

=∫ ∫

The expression involves with the integral of a product of two internal force functions. Directly performing the integral is often tedious and time consuming. However, for a straight member or a segment of a straight member with a constant flexural rigidity EI, the integral can be conveniently performed in terms of so called graph-multiplication method that can take advantage of the fact that the diagram of one of the two functions is composed by straight lines to facilitate the evaluation of the integral.

9.5.1 Calculating equations of graph-multiplication method

Consider two bending moment diagrams of a straight segment AB of a member, one of which (say iM ) is a straight-line diagram as shown in Fig.9.12. If the flexural rigidity EI is constant over the length of the segment, EI can be put outside the integral. Then we write

1B Bi ki kA A

M M dx M M dxEI EI

=∫ ∫ (a)

Since the diagram of iM is a straight line with a slope ofα , if orienting the intersecting point

between the line representing iM and the base line x as the origin, the ordinate of iM at an arbitrary x

position will be taniM x α= (b)


Substitute equation (b) into the integral of equation (a) and take the consideration of the fact that the slopeα is a constant, we write

tan tanB B B

i k k kA A AM M dx x M dx xM dxα α= =∫ ∫ ∫ (c)

M

The graphic meaning of the right-hand side of equation (c) is that kM dx is the differential area of

the diagram kM at an arbitrary position x, i.e., the dashed area shown in Fig.9.12; kxM dx is the

moment of the differential area with respect to coordinate axis y; the integral B

kA

xM dx∫ means the

summation of the moments of the all differential areas over the segment AB with respect to axis y. Recalling from the theorem of moment of an area, the integral equals to the moment of the area of the bending moment kM diagram (indicated by A) with respect to axis y. If let 0x represent the distance

between the centroid of the diagram kM and the axis y, we write

0

B

kAxM dx Ax=∫ (d)

Substituting equation (d) into equation (c) and considering the geometric relation of the bending moment diagram iM , we obtain

0tan ( )B

i kA 0M M dx Ax Ayα= =∫ (e)

Substituting equation (e) into equation (a), we find

01B i k

A

M M dx AyEI EI

=∫ (f)

×M k

Ak

α

dxA C

0

B

x

yA B x

y

0

0y iMMi

A

Fig.9.12 Graph-multiplication method


in which, represents the ordinate of 0y iM diagram at the location corresponding to the centroid of the

area A of kM diagram. Consequently, the expression for determining displacements in terms of

graph-multiplication method will be written as

01

P1MM ds Ay

EI EIΔ = =∑ ∑∫ (9-14)

2lhA =

(a)

Expression (9-14) is the equation of calculating displacements in terms of graph-multiplication method. It transforms the calculation of an integral with the product of two functions, one of which is a

23

A lh=

(c) (d)

13

A lh=

(e) (f)

ba

C×

( ) 3l a+ ( ) 3l b+

h

2l

h

2l

l

l

l

h3 8l5 8l

4l3 4l

l

h

4 5l 5l

h

triagle

(b)

2 3l

l

h

3l

triagle

2d-degree parabola

vertexvertex

vertex vertex

2d-degree parabola

2d-degree parabola

3d-degree parabola

2lhA

C×

C

=

×

23

A lh=

C×

C× C×

13

A lh=

Fig.9.13 Graphs and their areas and centroids


linear function, into the calculation of their graphic areas, locations of the centroids of the areas and their corresponding ordinates ( ). When applying the method, following points must be noted. 0y

(1) Implementing conditions The members must be straight; their flexural rigidity EI remain constant; one of bending moment function under the integral must be a linear function; the value of ordinate

must be measured on the bending moment diagram depicted by the linear bending moment function. 0y(2) Sign convention If the area A of a bending moment diagram (or a portion area of a bending

moment diagram) lies on the same side with its corresponding ordinate , the product of 0y 0Ay is positive and vice versa.

Fig.9.13 shows some of commonly used graphs and their area centroids to facilitate the calculation when applying graph-multiplication method. It should be noted that when employing equations to determine the areas of the parabolic graphs, the slopes at the vertexes of the graphs must be parallel to the base lines of the graphs, i.e. shear forces at the locations corresponding to the vertexes are equal to zero ( ), the graphs are called standard parabolas. 0Q =

9.5.2 Division of segments and superposition of graphs when using graph-multiplication method

When implementing graph-multiplication method, if one of the two graphs multiplied each other is a straight line graph and another has a curved line, the ordinate must be measured on the graph of straight line. If the two graphs multiplied each other are all straight lines, the ordinate may be

determined on any one, whose area is not concerned, of the two diagrams. If the following cases are encountered, the product of the two graphs must be carried out by dividing them into some simple portions,

0y0y

1A×2A× 3A×

1A×2A×

1I 2I

2y1y3y

2y1y

diagramkM

diagramkM

diagramkM diagramkM

Fig.9.14 Graph multiplication on segments of polygonal diagram

Fig.9.15 Graph multiplication on segments with varying rigidity


on each of which the graph-multiplication method can be applied, then superpose them together. (1) Division ① If one of the two diagrams is a polygon and another is a curved trapezoid, the calculation of the

integral i kM M dx∫ must be performed over several segments. For the diagram shown in Fig.9.14, the

calculation of the integral will be

1 1 2 2 3 3i kM M dx A y A y A y= + +∫

② If the flexural rigidity EI specified over the extent of the integral varies in different segments, the integral must be carried out over the segments divided by the variation of EI. For example, the diagram shown in Fig.9.15, the integral becomes

1 1 2 21 2

1 1i kM M dx A y A yEI EI EI

= +∫

(2) Superposition When the diagram of kM is quite difficult to determine its area an

divid diagrams and then use superposition method to perform the integral. s are

d centroid, the diagram may be ed into several simple① If the two diagram all trapezoid [Fig.9.16], the two trapezoids may be separated into two

triangles (or one rectangle and one triangle) to simplify the application of graph-multiplication method. This means

1 1 2 2i kM M dx A y A y= +∫ (a)

In which, ordinates and may be expressed as

D

1y 2y

1A×2A×

ba

2y1yc d

1A×

2A ×

a

c

b

d

2y

1y

BA

Cl diagramkM

diagramkM

diagramiM diagramiM

Fig.9.16 Graph multiplication on the same side

Fig.9.17 Graph multiplication on different side


1

2

2 13 31 23 3

y c d

y c d

⎫= + ⎪⎪⎬⎪= +⎪⎭

If the two diagrams are all straight lines and tshown in Fig. 9.17. The diagram of

he two diagrams have positive and negative portions as

kM may be treated as two triangles, one of which has positive area

(ADB) and another has negative area (ABC) as shown in Fig. 9.17. The equation (a) can still be used to calculate integral i kM M dx∫ , but ordinates 1y and 2y must be determined by following formulae.

12 13 3

y c d= − + (corresponding area 1 2laA = )

21 23 3

y c d= − (corresponding area 2 2lbA = )


② If the diagram of kM is the diagram of bending moment PM of a straight segment of a member

due to uniformly distributed loads and member end couples, recalling from the method of superposition used to construct bending moment diagrams, the PM diagram is composed by simple graphs. As shown

in Fig.9.18 (a), the bending moment diagram of PM over segment AB due to uniformly distributed load

is superposed by a graph with straight side (caused by member-end couples

q

AM and BM ) and a graph

with standard parabolic-curve side (caused by q). Therefore, the diagram may be decomposed into two simple graphs ( M ′ and 0M ) as shown in Fig.

9.18 (b) and (c), respectively. The graph of M ′ may be decomposed again into two triangles. Then, the graph-multiplication method will be easily used to the three simple graphs.

It should be noted that the superposition of bending moment diagram is actually the superposition of their ordinates. So at the same x position, the ordinates of 0M [Fig. 9.18 (a)] depicted on an inclined base line CD, and that of 0M [Fig.9.18 (c)] depicted on a horizontal base line, are identical. Consequently, the differential areas as shown in dashed lines in Fig. 9.18(a) and (c) are identical, and then the areas and their centroids are as well identical, respectively.

dxx

2

8ql

C

(c)

(a)

(b)

9.5.3 Examples for determining displacements by graph-multiplication method

Example 9-5

A B

D

M ′

AMdiagramPM

diagramM

0M

BM

′

A BBM

C

D

AM

1A

2A

0 diagramMA B0M

Fig.9.18 Decomposition, superposition and graph multiplication of diagram


Calculate the deflection at point C for the beam shown in Fig.9.19 (a) by graph-multiplication method.

Solution

(1) Real system The bending moment PM diagram due to real load is shown in Fig.9.19 (a).

(2) Virtual system The virtual system with a unit force 1P = applied at C and its moment diagram M are shown in Fig.9.19 (b).

(3) Deflection Δ at point C Due to the symmetry of the two diagrams, only half of them are computed. Since the left half of

PM diagram is a curved-line graph and M is a straight-line graph, the calculation of the area (A) and

ordinate ( ) used in graph-multiplication method must be performed in 0y PM and M diagrams separately

as 2 32

3 2 8 24l ql qlA = ⋅ ⋅ =

05 58 4 32

l ly = ⋅ =

Finally, 3 4

01 1 1 5 52 2

24 32 384Pql l qlMM dx Ay

EI EI EI EI( )Δ = = = ⋅ ⋅ ⋅ =∑∫ ↓

The final result is the same as that in example 9-2.


Example 9-6

Calculate the deflection at point B of the cantilever beam shown in Fig.9.20 (a) by graph-multiplication method.

Solution

(1) Real system The bending moment PM diagram due to real loads is shown in Fig.9.20 (b). (2) Virtual system The virtual system with a unit force 1P = applied at B and its moment

diagram M are shown in Fig.9.20 (c). (3) Deflection Δ at point B Since the slope at point B of PM diagram is not parallel to base line AB, the equation (e) used for

calculating area and centroid expressed in Fig.9.13 can not be used directly. The PM diagram is thusly discomposed into two simple graphs 1A and 2A , in which 2A is a standard parabola, as shown in Fig.9.20 (b). Their areas ( 1A and 2A ) and corresponding ordinates measured on M diagram are expressed as

2 31

1 12 2

A l ql ql= ⋅ ⋅ = ， 123

y l= ( and1y 1A lie on the same side)

58 4

l×

4l

(b)

2

8ql

×

(a)

C

constantEI

q

BAA

=2l

2l

C

1P=

BA

Fig.9.19 Determination of the deflection at the middle span of a simple beam(a) real loads and bending moment diagram; (b) virtual unit load and bending moment diagram

PMM


2 32

2 1 13 8 12

A l ql ql= ⋅ = ， 212

y l= ( and2y 2A lie on different sides)

Consequently, we obtain 3 3 4

1 1 2 21 1 1 2 7( ) ( )

2 3 12 2 24B Pql l ql l qlMM dx A y A y

EI EI EI EIΔ = = − = × − × = ↓∑∫ ( )

2ql

Example 9-7

Calculate the slope of section C for the beam shown in Fig.9.21 (a) by graph-multiplication method.

Solution

(1) Real system The bending moment PM diagram due to real loads is shown in Fig.9.21 (b). (2) Virtual system The virtual system with a unit couple (corresponding to the desired slope) 1M = applied at C and the bending moment (due to 1M = ) diagram M are shown in Fig.9.21 (c).

(3) Slope at section C ΔSince the M diagram is composed by two straight lines (AB and BC), the graph-multiplication

method has to be applied to the two segment separately. The PM diagram on segment AB decomposed into a triangle and a standard parabola as shown in Fig.9.21 (b). The areas and their corresponding ordinates are calculated as follows.

constantEI =

(a)

× ×

(b)

(c)

A B

1A

q

l

2ql2A

BA

l

A B1y

2y

1P=

Fig.9.20 Determination of the deflection at the end of a cantilever beam(a) real loads and bending moment diagram; (b) virtual unit load and bending moment diagram

PMM


11 4 2 42

A = × × = ， 123

y = ( and1y 1A lie on the same side)

22 4 6 163

A = × × = ， 212

y = ( and2y 2A lie on different sides)

31 1 2 12

A = × × = ， 3 1y = ( and3y 3A lie on the same side)

(a)

245kN mEI

B

Finally, we obtain

1 1 2 2 3 30

1 1 ( )

1 2 1 13 13 (4 16 1)3 2 3 3

0.096 (rad)

Ay A y A y A yEI EI

EI EI

Δ = = − +

= × − × + = − = −45×

= −

∑

( )

The minus means that the slope of section C is counterclockwise (opposite direction ofΔ 1M = ).

Example 9-8

×

×

×(b)

(c)

= ⋅

A C

2

l

1A

3kN/m

4m 1m

2kN

3AA CB

A2

1y2y

AB C

3

l

1M =y

Fig.9.21 Determination of the slope at the end of an overhanging beam (a) an overhanging beam and its loads; (b) bending moment diagram; (c) virtual bending moment diagram

PMM

１

６

５


Calculate the deflection at point D (the middle section of the cantilever) due to the bending moments for the cantilevered rigid frame shown in Fig.9.22 (a) by graph-multiplication method, and Compare the effect on the deflection between the deflection component due to the axial forces and that due to the bending moments. The cross-sectional area A and flexural rigidity EI are indicated in the figure as well.

Solution

(1) Real system The bending moment PM diagram due to the real load is shown in Fig.9.22 (b). (2) Virtual system The virtual system with a unit force (corresponding to the desired deflection)

applied at D and the bending moment (due to1P = 1P = ) diagram M are shown in Fig.9.22 (c). (3) Calculate deflection MΔ at section D due to bending moments The area of M diagram, which is divided into two portions indicated by and , respectively,

and the ordinates 1A 2A

1y and 2y , which are measured from PM diagram at the positions corresponding to

(a)

constant EIA b h== ×

(b)

real system and diagramPM

2l

(c)

C

DP

1

virtual system and diagramM

(d)

real system and diagramPN

(e)

virtual system and diagramN

A

B A1y

2y2A

2l

l

2l

A

B C

P

Pl

Pl

CD

1P=

2l

B

A

A

B C

P

P

A

B CD

1P=

1

Fig.9.22 Determination of the deflection at point of the rigid frame (a) a rigid frame and its load; (b) bending moment diagram; (c) virtual system and its bending moment d

P

DM

M iagram; (d) real axial force diagram; (e) virtual axial force diagramPN N

Pl2l

1P


the centroids of and , respectively, are calculated as follows. 1A 2A2

112 2 2 8

l l lA = ⋅ ⋅ = ， 156

y Pl= ( 1y and 1A lie on the same side)

2

2 2 2l lA l= ⋅ = ， 2y Pl= ( 2y and 2A lie on the same side)

Therefore, we find 2 2

30

1 1 1 5 29 ( )8 6 2 48M Pl lMM ds Ay Pl Pl Pl

EI EI EI EI⎡ ⎤

Δ = = = ⋅ + ⋅ = ↓⎢ ⎥⎣ ⎦

∑ ∑∫

(4) Compare the effect on the deflection between the deflection component due to the axial forces and that due to the bending moments

The axial force PN diagram due to real loads and the virtual axial force N (due to ) diagram are shown in Fig.9.22 (d) and (e), respectively.

1P =

The deflection component due to axial forces may be expressed as

P PN

NN NN Plds lEI EA EA

Δ = = =∑ ∑∫

The ratio between the deflection component due to axial forces and that due to the bending moments is thusly expressed as

3 2

4829 2948

N

M

plIEA

Pl AlEI

Δ= =

Δ (a)

Substituting the cross-sectional area A b h= × and inertial moment , we obtain 3 /12I bh=2

2

429

N

M

hl

Δ=

Δ (b)

It can be observed from equation (b) that if 110

hl= ,

2

2

1100

hl

= and 1725

N

M

Δ=

Δ. This implies that the

effect on the deflection due to axial deformation is much smaller than that due to bending deformation. So for a beam (or flexural) member, the effect to a displacement due to axial forces is generally neglected.

Example 9-9

Calculate the relative horizontal displacement between points C and D due to the hydrostatic pressures for the rigid frame shown in Fig.9.23 (a) by graph-multiplication method.

Solution


(1) Real system The bending moment PM diagram due to the real pressures is shown in Fig.9.23 (b).

(2) Virtual system The virtual system with a pair of unit forces opposite in direction (corresponding to the desired displacement) applied at C and D, and the bending moment diagram M (due to the two unit forces) are shown in Fig.9.23 (c).

(3) Calculate relative displacement between C and D ΔIt is noted that because the members AC and BD are subjected to the hydrostatic pressure distributed

triangularly along the axes of the two members, the functions of the PM diagram on the two members are standard cubical parabolas. The vertexes of the two parabolas lie on point C and D, respectively. The areas of PM diagram on the two members can be expressed as follows.

2 3

114 6 2

qa qaA a= × × =4

， 14 45 5

y a= × = a ( 1y and 1A lie on the same side)

The areas of PM diagram on the member AB can be written as 2 3

2 26 3

qa qaA a= × = ， 2y a= ( 2y and 2A lie on the same side)

23

32 223 2 3

qaA a q= × × = 3a ， y a= ( 3y and 3A lie on different sides)

Finally, we obtain

× ××

×

2

2qa

2

6qa

(b) (c)11

(a)

constantEI =

A B

C D

a1A

1y 2y2Aq

2a

a

A B

C D

3A

4A

A B

C D

a

a a3y 4y

2

6qa

Fig.9.23 Calculate relative displacement between and (a) a rigid frame and its loads; (b) bending moment diagram; (c) virtual system and its bending moment diagram

P

C DM

M


1 1 2 2 3 3

3 3 3

4

1

1 (2 )

1 4 2 224 5 3 3

1 4 ( )15

PMM dxEI

A y A y A yEI

qa a qa qaa aEI

qaEI

Δ =

= + −

⎡= × × + × − ×⎢ ⎥

⎣ ⎦

= − × →←

∑∫

⎤

The minus indicates that the relative horizontal displacement between points C and D is in the opposite direction of the pair of assumed unit forces, i.e. points C and D are closed to but not deviate from each other.

Δ

Example 9-10

Find the relative rotation between the left section and the right one of the hinge C of the three-hinged frame shown in Fig.9.24 (a) by graph-multiplication method.

Solution

(1) Real system The bending moment PM diagram due to the real load is shown in Fig.9.24 (b). (2) Virtual system The virtual system with a pair of unit couples opposite in direction

(corresponding to the desired relative rotation) applied at the left and the right of C, and the bending moment (due to the pair of unit couples) diagram M are shown in Fig.9.24 (c).

(3) Find relative rotation ΔBecause of the symmetry of the structure and the load, the bending moment diagram of the structure

2

8ql

××

4ql

2ql

(b)

1

(c)(a)

constantEI =

A B

CD

E

q

2l

2l

2l

A B

CD E2A

1A

A B

CD E

2

8ql

2

8ql 2

8ql

4ql

2ql

1 1 1 1

1

Fig.9.24 Find the relative rotation between the left and the right of (a) a three-hinged frame and its load; (b) bending moment diagram;(c) virtual system and its bending moment diagram

P

CM

M


must be symmetric. So the shear force at hinge C must be equal to zero, and then the bending moment diagrams on portion DC and CE are standard second-degree parabolas, whose vertexes at point C. Therefore, the areas of PM diagram can be expressed as follows (only half of them are needed).

2 31

1 1 13 2 8 48

lA ql ql= ⋅ ⋅ = ， 1 1y = ( 1y and 1A lie on different sides)

2 32

1 12 2 8 32l lA ql q= ⋅ ⋅ = l ， 2

23

y ( 2y and 2A lie on different sides) =

Consequently, we find

[ ]3 3

1 1 2 20

1 2 2 2148 32 3 12ql ql qlAy A y A y

3

EI EI EI⎡ ⎤

Δ = = − − = − × − × = −⎢ ⎥⎣ ⎦

∑ EI( )

The minus implies that the relative rotation between the left section and the right one of the hinge C of the three-hinged frame is in the opposite direction of the pair of assumed unit couples.

Δ

Example 9-11

Determine the vertical displacement at point C of the composite structure shown in Fig.9.25 (a). The two-force members BC and CD are I shaped steel of number 12 and the flexural member ADB is I shaped steel of number 18, their axial rigidity EA and flexural rigidity EI are shown in the figure as well.

C

(a)

23738 10 kNEA = ×23484kN mEI = ⋅

(b)23

43

53

−

(c)

y2m

4kN

1

1P =

A

C

D

6kN

4m

3m 3m

4kN

12kN m⋅

C

6kN

8kN

10kN

B BB

−

D

23

D

AA

6kN

Fig.9.25 Determine the vertical displacement at point of the composite structure(a) a composite structure and its load; (b) bending moment diagram and ;(c) virtual system and its bending mome

P P

CM N

nt diagram and M N

EA

EA

EI


Solution

When calculating the displacement, the axial deformation of two-force members and bending deformation of the flexural members are only taken into consideration.

(1) Real system The bending moment PM diagram and the axial force PN due to the real load is shown in Fig.9.25 (b).

(2) Virtual system The virtual system with a unit force (corresponding to the desired displacement) applied at point C, and the bending moment diagram M and the axial force N (due to the unit force) are shown in Fig.9.24 (c).

(3) Determine the displacement CΔ By equation (9-12) used for calculating the displacement for composite structures, the desired

displacement may be calculated as follows (the graph-multiplication method is only used to the flexural member).

0

2

3

1

1 1 2 1 4 5 48 128 250 ( 3 12) ( 2) 8 4 ( 10) ( ) 52 3 3 3 3484 3738 10 3

14.11 10 m( )

P P PMM NN NNdx l Ay lEI EA EI EA

EI EA−

Δ = + = +

+⎡ ⎤ ⎡ ⎤= × × × × + × × + − × − × +⎢ ⎥ ⎢ ⎥ × ×⎣ ⎦ ⎣ ⎦= × ↓

∑ ∑ ∑ ∑∫ ∫

9.6 Calculation of Displacements Caused by Temperature Changes

For a statically determinate structure, the change of its temperature will not cause internal forces, but the materials of the structure will have free expansion or contraction. The deformation and displacement of the structure will be thusly induced by the temperature change. The calculation of displacements of a structure caused by temperature changes is sometimes very meaningful for the design of the structure.

The structural displacements caused by their temperature changes are still determined by employing the expression of the unit load method as given by Eq. (9-7). When applying the equation to calculate the displacement induced by temperature changes, the main task will be the determination of the axial strainε , the shearing strain 0γ and the curvature κ elicited by the temperature changes.

The analysis of the deformation elicited by the temperature changes may be illustrated by reference to the beam subjected to a change in temperature as shown in Fig. 9.26 (a). Assume that the temperature rises are equal to and at the top and bottom side of the beam, respectively; and the rise of temperature varies linearly along the depth h of the cross section [Fig.9.26 (b)], i.e., after the thermal deformation, the cross sections of the beam still remain planar. Considering an element of length ds, the deformation of a cross section may be decomposed as an axial deforming component and a bending deforming component

01t C 0

2t C

du

dθ ; there is no shearing deformation. Then, the average rise of temperature at 0t

9.6 Calculation of Displacements Caused by Temperature Changes 261

the central line of the element is

0 11 (2

t t t= + 2 ) (when the cross section is symmetric to its central line, i.e., ) (9-15a) 1 2h h=

or (when the cross section is asymmetric to its central line, i.e., 1 2h h≠ )

2 1 1 2 2 10 1 1

t t t h t ht t hh h− +

= + = (9-15b)

Where h is the depth of the cross section; and are the distances between the central line and the top fiber and the bottom fiber of the cross section, respectively; the difference of the temperature rises between the top and bottom of the element is denoted by

1h 2h

tΔ

2 1t t tΔ = − (9-16)

If the coefficient of thermal expansion of the material of the beam is equal to α , the axial deforming component and the bending deforming component du dθ will be determined by

0du ds t dsε α= = ， 0tε α= (a)

2 1( )t td ds dh

sαθ κ −= = ， 2 1( )t t t

h hα ακ − Δ

= = (b)

Substituting equations (a) and (b) into expression (9-7) and setting 0 0, 0Kcγ = = , we will find the equation for calculating the displacements elicited by temperature changes as follows.

0tM ds N t ds

hα αΔ

Δ = +∫ ∫

For a framed structure, the equation will become

dsdθ

0t ds duα =

1t dsα

1t

2t

(b)(a)

ds 2t dsα

A B1t C+ °

2t C+ °

2h 1h

h

Fig.9.26 Deformation caused by temperature changes(a) a simple beam and its temperature changes; (b) deformation of an element ds


0tM ds N t ds

hα αΔ

Δ = +∑ ∑∫ ∫ (9-17a)

If and are constants along the members, the equation may be rewritten as 0t tΔ

0t Mds t Nds

hα αΔ

Δ = +∑ ∑∫ ∫ (9-17b)

In which, Mds∫ and Nds∫ represent the areas of the internal force diagrams M and N of each member, respectively; the summation symbol means to find the algebraic sum of all members.

The sign convention used for the virtual forces must be the same as that adopted for the real system; that is, if temperature increases was considered as positive in the real system, then the virtual tensile forces N must also be considered to be positive and vice versa; if the direction of the bending deformation elicited by the difference of the temperature changes was the same as that caused by the virtual bending moments M , the virtual bending moments M must be recognized to be positive and vice versa.

Example 9-12

Find the vertical displacement at point C for the rigid frame shown in Fig.9.27 (a) due to a rise in temperature by 15 degrees inside of the frame, 0 degree outside of the frame. The coefficient of thermal expansion 0.00001α = per degree; the each of the cross sections of the rigid frame is a rectangle with a depth of 40cm.

Solution

Real system The real system consists of the temperature changes given in the problem, as show

(1) n in Fig.9.27 (a).

(b) (c)(a)

4ma =

A

B

Ca

a0 C°

0 C° 15 C+ °

15 C+ °

A

B C

A

B C

1P = 1P =

4m

a= diagramM diagramN

Fig.9.27 displacement of rigid frame caused by temperature changes(a) a rigid frame and its temperature changes; (b) virtual system and its bending moment diagram;(c)

M virtual system and its axial force diagramN

a 1

1

9.7 Calculation of Displacements Caused by Support Movement 263

(2) Virtual system The virtual system consists of a unit force applied in the vertical direction at point C, and the bending moment diagram M and the axial force N due to the unit force are shown in Fig.9.27 (b) and (c), respectively.

(3) Calculation of the average rise and the difference of the temperature rise0t tΔ

01 20

0 15 7.52 2

t tt C+ += = =

(4) Determine the displacement

Substituting the above values into equation (9-17b), the desired displacement is obtained as follow

02 1 15 0 15t t t CΔ = − = − =

CΔ

0

15 1 ( ) 7.5 ( )23 7.5 ( 1)

Ct Mds t Nds

ha a a a a

haah

α α

α α

α

ΔΔ = +

= − × × + × + −

= − +

∑ ∑∫ ∫

Since the direction of the bending deformation elicited by tΔ was opposite to that caused by the virtual bending moments M , the first term must be negative; and because the temperature increases was considered as posi n the real system but the virtual forcestive i N is compression, the second term must be negative too. Substituting the numerical values into above expression, we obtain

The minus answer means that the vertical displacement elicited by the temperature change is actually upward, opposite to the direction of the unit force.

9.7 Calculation of Displacements Caused by Support Movement

Since a statically determinate structure is a stable system with no redundant restraint, only rigid body motion of the members would occur when the supports of the structure were subjected to movements. As the multispan beam shown in Fig.9.28 (a), when the support B has a settlement the member AC will rotate about hinge A; the member CD will rotate about support D; only rigid body motion occur to all the members as shown in dashed lines in the figure.

Consequently, the calculation of displacements due to support movements belongs to the displacement determination of rigid-body system. It may be calculated by principle of virtual work for rigid bodies. So when using the unit load method, derived from principle of virtual work, to determine the displacements, the calculating expression will be obtained from equation (9-7) as follow

0.93cm( )CΔ = − ↑

Bc


K KR cΔ = −∑ (9-18)

Where Kc represents the real support movement in the direction of K; KR indicates the reaction in direction of K due to unit load.

K KR c means the virtual work done by KR on movement Kc . When KR and Kc have the same direction their product is positive and vice versa.

In Fig.9.28 (a), if the deflection at hinge C is desired, a virtual system, which does not consider the support settlement, with a unit load applied at C in the direction corresponding to the deflection, must be assumed and the reactions corresponding the unit load have to be determined, as shown in Fig.9.28 (b). By equation (9-18), we find

5 5( ) (4 4B Bc cΔ = − − × = ↓ )

Here, since only support B has a movement and other supports have no movements, just the virtual reaction at support B performs virtual work and the other virtual reactions perform no work. In addition, because the reaction BR and settlement are opposite in direction, the virtual work or their product must have negative sign.

Consequently, the procedure to determine the displacements of a structure due to its support movements may be stated as:

(1) Assume a virtual system by applying a unit load at the location and in the direction of a desired displacement of the structure without considering its support movements.

(2) Find each reaction

Bc

KR corresponding to each support movement Kc due to the unit load. (3) Determine the desired displacement Δ by means of equation (9-18).

14

54 0

(a)

(b)

B C D

l

A

CB

4l 3

4l

C

D1P =

A B

Fig.9.28 displacement calculation caused by support movement(a) support has a settlement; (b) virtual system and its balanced force systemB

9.7 Calculation of Displacements Caused by Support Movement 265

Exa

po secf support A.

Solu

① The virtual system with a unit load applied at support 9.29 (b).

mple 9-13

Find the horizontal displacement at sup rt B and the slope of tion B for the rigid frame shown in Fig.9.29 (a) due to the settlement o

tion

(1) Find the horizontal displacement at support B B in horizontal direction is shown in Fig.

② Find each reaction KR corresponding to each support movement Kc due to the horizontal unit force [see Fig.9.29 (b)].

③ Determine the horizontal displacement Δ at support B. By equation (9-18), we find

( ) ( )R c aΔ = − = − × = − ←∑ K K l lHere, the direction of reaction

h ha

AR is the same as that of settlement Ac , so their product is positive. The minus Δ indicates that the horizontal displacement at support B is opposite to the direction of

the unit force, i.e., toward left. (2) Find the slope of section B ① The virtual system with a unit clockwise couple applied at section B is shown in Fig. 9.29 (c). ② Find each reaction KR corresponding to each support movement Kc due to the unit

clockwise couple [see Fig.9.29 (c)]. ③ Determine the slope Δ of section B. By equation (9-18), we write

a

(b) (c)(a) ΔB

hl

1l

M 1=

A

B

0

l

PB B1=

h

A1 A

hl

1l

d fryst

nce

Fig.9.29 Determination of displacement of rigi ame caused by support movement(a) a rigid frame and its support settlement; (b) virtual s em for the horizontal displacement at and i

Bts bala d force system; (c) virtual system for the slope of section and its balanced force systemB


1( ) K KR c aΔ = − = − ×∑ ( ) l

Here, the direction of reaction AR is the same as that of settlement , so their product is positive. The ne e s is ire

9.8

near elastic structures. They are: law of reciprocal work (E. Betti’s law in 1872), law of reciprocal displacements (James C. Maxwell’s law in 18

9.8.1 Law of reciprocal work

The law, initially developed by E. Betti in 1872, can be stated as follows: For a linearly elastic structure, the virtual work done by numberⅠsystem of forces and couples acting

through the deformation caused by numberⅡ system of forces and couples is equal to the virtual work of theⅡsystem acting through the deformation due to theⅠsystem.

To show the validity of this law, consider the beam shown in Fig.9.30. The beam is subjected to two different systems of forces, ⅠandⅡsystems, as shown in Figs.9.30 (a) and (b), respectively. In systemⅠ,let us assume that and

Acgative answer for Δ implies that th lope of section B opposite to the d ction of the unit

couple, i.e., counterclockwise.

Reciprocal Laws for Linear Elastic System

The section will introduce three reciprocal laws that play very important roles in the analysis of statically indeterminate li

64) and law of reciprocal reactions.

1,P 1,N 1Q 1M represent the system of forces and 1,Δ 1,ε 01γ and 1κ represent its displacements and strains. In systemⅡ,we use and 2 ,P 2 ,N 2Q 2M to denote the system of forces and 2 ,Δ 2 ,ε 02γ and to denote its displacements and strains. Now, let us assume that we subject the beam that has theⅠforces already acting on it [Fig.9.30 (a)] to the displacements caused by theⅡsystem of forces [Fig.9.30 (b)]. The virtual work equation can be written as

2κ

12 1 2 1 2 1 02 1 2

1 2 1 2 1 2 N N kQ Q M Mds ds dsEA GA EI

= + +∑ ∑ ∑∫ ∫ ∫Next, we assum that the beam with theⅡforces acting on it [Fig.9.30 (b)] is subjected to the

displacements cause

W P N ds Q ds M dsε γ κ= Δ = + +∑ ∑ ∑ ∑∫ ∫ ∫

e d by theⅠforces [Fig.9.30 (a)]. By equating the virtual external work to the virtual

internal work, we obtain

21W P2 1 2 1 2 01 2 1N ds Q ds M ds

2 1 2 1 2 1 N N kQ Q M Mds ds dsEA GA EI

ε γ κΔ = + +∑ ∑ ∑∫ ∫ ∫

Noti ate

=∑= + +∑ ∑ ∑∫ ∫ ∫

ng that the right-hand sides of above two equations are identical, we equ the left-hand sides to

9.8 Reciprocal Laws for Linear Elastic System 267

obtain

1 2 2 1P PΔ = Δ∑ ∑ (9-19a)

W12 21W= (9-19b)

9-19) represents the mathematical statement of reciprocal work (Betti's law).

9.8.2 Law of reciprocal displacements

n 1864, will be derived here as a special case of the more genera

For a liequal to th due to a un

TodisplacemFig.9.31. ThetheⅠandⅡsystems, consisting of the unit loads at points i and j, respectively, as shown

indic

Equation (

(a)

Δ 1c Δ 1d

Δ Δ2a2b

(b)

a bc d

1aP

a bc d

1bP

2cP 2dP

Fig.9.30 Law of reciprocal work (a) system ; (b) systems Ⅰ Ⅱ

The law, initially developed by James C. Maxwell il form of the law of reciprocal work. It can be stated as follows: nearly elastic structure, the displacement at a point i due to a unit load applied at a point j is

e displacement at j it load at i. prove the law of reciprocal

ents, consider the beam shown in beam is separately subjected to

(a) 1 1P =

in Figs.9.31 (a) and (b). As the figure ates, ijδ re resents the displacement

at i due to the unit load at j, p

whereas jiδ deno h e unit load Tunit fl

law of recip

tes t e displacement at j due to th at i. hese displacements caused by per

load are referred to as exibility coefficients. By employing the

rocal work, we obtain

21δ

(b) 2P

12δ

1=

ⅡFig.9.31 Law of reciprocal displacements (a) system ; (b) system Ⅰ


1 12 2 21P Pδ δ=

Since and , then the equation will become

1 21P = 1P =

12 21δ δ= ( 9-20)

ent of reciprocal-displacement law (Maxwell’s law). The reciprocal relationship remains valid between the rotations caused by two unit couples as well as ee ple and a unit force, respectively.

9.8.3

ture, the reaction corresponding to displacement due to a unit displacement is equal to the reaction corresponding to displacement due to a unit displ

This is the mathematical statem

betw n the deflection and the rotation caused by a unit cou

Law of reciprocal reactions

The law of reciprocal reactions is also a special case of the more general form of the law of reciprocal work. It can be stated as

2cFor a linear elastic struc1c = 1c1

acement 1c = . 2

To prove the law of reciprocal reactions, consider two different deformation systems of the multispan beam shown in Fig.9.32. In the first system [Fig.9.32 (a)], the reactions at supports 1 and 2 due to the support movement c1 1= at support 1 are denoted by and , respectively. In the second system [Fig.9.32 (b)], the reactions at supports 1 and 2 due to the sup ovement at support 2 are represented by and , respectively. Other reactions that do not perfosketched. By applying the law of reciprocal work, we write

11k 21kport m 2

rm virtual work are not 1c =

12k 22k

12 1 21 2k c k c=

11k 21k

(a) 1 1c

(b)

2

1

21

=

2 1c =

12k 22k

Fig.9.32 Law of reciprocal reactions (a) system ; (b) systemⅠ Ⅱ

Summary 269

Since and , then the equation will be written as 1 1c = 2 1c =

12 21k k=

This is the mathematical statement of reciprocal-reacotations as well as

between the force and the couple caused by a unit rotation and a unit deflection, respectively.

SUMMARY

The chapter has mainly discussed the determination of displacements for statically determinate structures by using the principle of virtual work. It is not only the requirement of the determination of displacements of structures, but the preparation of analysis of statically indeterminate structures. And it plays a role of the link between the preceding and the following of the text. Therefore, it has a very important significance for students to have a sound mastery for the chapter.

(1) The principle of virtual work is one of the fundamental principles in mechanics. It is the characteristics of virtual work and equation of virtual work that the forces and displacements through which the forces perform virtual work are independent each other. By means of the characteristics, the unknown

n ven structure under the action of a system of forces by assuming ua ure; in a similar way, the unknown displacements may be

deterssed in the chapter is actually

the ments of a structure is the method of unit load. The

expr quation (9-7), that is

(9-21)

tion law. The reciprocal relationship remains valid between the couples caused by two unit r

restrai t forces may be determined for a gia virt l displacement system of the struct

mined for a structure subjected to a given deformation by assuming a virtual force system for the structure. The method of determining the displacements of a structure discu

ethod of assuming a virtual force system. (2) The method for calculating the displacem

ession for calculating the displacements is e

0( ) K KQ M ds R cε γ κ+ + −∑

arate systems: ① a real system of loads (or other effects)

NΔ =∑∫The unit load method employs two sep

causing the deformation (strains 0, , ε γ κ and su t Kcpport movemen ) to be determined and ② a virtual point and in the direction of the desired displacement，

rnasystem consisting of a unit load applied at thewhose inte l forces and reactions are KR . , , N Q M and

(3) For a linear elastic structure under action of loads, the strain expression can be written as

PNEA

ε = , 0PQk

GAγ = , PM

EIκ =

to be used to determine the displacements of the structures The explicit expression for unit load methodunder action of loads is as follow:


P P PMMds ds+∑ NN kQQds

EIΔ = +∑ ∑∫ ∫ ∫

Here, there is two separate systems: ① a real sywhose internal forces are N Q M , and ② a virtual system consisting of a unit load applied at the

EA GAstem of loads causing the deformation to be determined,

, , P P P

point and in the direction of the desired displacement, whose internal forces are , , N Q M .

When applying equation (9-9), the explicit expressions to determine the displacements of trusses, s, and frabeam mes are as follows:

Beams and rigid frames PMM dsEI

Δ =∑∫

Trusses P PNN NNds lEA EA

Δ = =∑ ∑∫

A

rches P PMM NNds dsEI EA

Δ = +∑ ∑∫ ∫

Composite structures P PMM Nds lNEI E

Δ = +∑ ∑∫ A

(4) When calculating the displacements for a structure under action of loads, the graph-multiplication method is expedient to simplify the calculation of the integrals with the mastery of the skill of dividing segments and superposition.

(5) When calculating the displacements of a structure under action of temperature changes, the expressions for determining strains can be expressed as:

， t

hακ Δ

= 0tε α=

Substituting above two expressions into equation (9-7), the expression used to determine the displacements due to temperature changes can be obtained as equation (9-17).

(6) The calculation of displacements due to support movements belongs to the displacement determination of rigid-body system. It may be calculated by equation (9-18). The key point of the calculation is the determination of reactions KR .

(7) There are many sign conventions in the calculation of displacements of a structure. The main sign convention is ign of work. That is, the sig onventi of the product of a force and a displacement. When the force and the displacement have the same directions, their uct is positive and vice versa. In gra multip tion method, when the area A of a bending moment diagram and the ordinate

0

the s n c on prod

ph- licay of another bending moment diagram lie on the same side of a member, their product is positive and vice

versa.


(8) Reci ocal la another one of fundamental principlepr w is in mechanical analysis. The significance and the application of reciprocal displacements and recip

applied to either the analysis of statically


9-1 Why are the displacements of a structure desired? 9-2

splacements or corresponding

e force an placement used in the principle of virtual work? 9-7 tically

determinate structures? 9-8 For the beam shown in the figure of problems for reflecting 9-8 (a) and (b), how to assume virtual

rocal reactions in different deforming conditions must be understood. The reciprocal displacements may be determinate structures or that of statically indeterminate structures. But the reciprocal reactions is only used to the analysis of statically indeterminate structures.

What are the main reasons which caused the displacements of a structure? 9-3 How many categories of displacements are there for a structure? 9-4 What is the characteristic of virtual work? 9-5 How to understand that the force and displacement used in principle of virtual work are

corresponding to each other? Please depict out the corresponding diforces in the figure of the problem.

9-6 What are the requirements for th d disWhat are the two kinds of applications of principle of virtual work for rigid bodies to sta

system to determine the desired bending moment?

(a) (b)

Δ

(c)

θ

(d)

A

A

A

A

Problem for reflecting 9-5

272 Chapter 9 Principle of Virtual Work And Displacement of Structures

9-9 Please explain the common ground and distinction between the principle of virtual work for rigid bodies and that for deformable bodies.

9-10 Please explain how equation (9-6) of the chapter satisfies compatible conditions of the deformation. 9-11 Please write out the general expression for calculating a displacement of a structure by unit load

method. 9-12 Why are there general universality for the application of equation (9-7) of the chapter? 9-13 Please find the difference between the equation (9-7) and equation (9-9) of the chapter. 9-14 Please write out the explicit expressions to determine the displacements of trusses, beams, and

frames due to loads, and explain the meanings and sign convention of each term. 9-15 Please explain how to determine the real direction of a desired displacementΔ ? 9-16 What are the conditions of application of graph-multiplication method? Whether or not the method

can be employed when determining the displacements of beams with variable cross sections, or curved beams, or arches? Whether or not graph-multiplication method may be applied between the

shear forces PQ andQ or between the axial forces PN and N ?

9-17 How is the sign convention regulated for the equation (9-14) used for graph-multiplication method? 9-18 Whether or not the following calculations are correct? Why?

(1) 1 1 2 2i KM M dx A y A y= +∫ [see the figure (a)]

(2) 22 1(3 8P

lMM dx l ql= × ×∫ )4

[see the figure (b)]

(3) 21 3( )3 4PMM dx l ql l= × ×∫ [see the figure (c)]

(a) (b)A B

C

P

a ab b

l l

C

P

find A

BA

Mfind CM



(4) 1 1( )2 6P

lMM dx l Pl2

= × × ×∫ [see the figure (d)]

9-19 How to determine strains 0, , ε γ κ when calculating the displacements caused by temperature changes?

9-20 Under what condition the integral can be substituted by the calculation of the areas when applying the equation (9-17)? How is the sign convention of the terms in the equation (9-17) prescribed?

9-21 What are the significances of KR , Kc and symbol ∑ in the equation (9-18)? How to determine the sign of the virtual work done by the reactions?

9-22 Why always is there a minus sign in front of the symbol ∑ in the equation (9-18)? 9-23 Please use the following figure to explain the law of reciprocal displacements, and to depict out the

displacements 12δ and 21δ , and account for the dimension of 12δ and 21δ .

××

2

8ql

4l

2ql

×

2ql

34l

×

2l

12l

diagramM

diagramPM

diagramMdiagramM

diagramPM

diagramPM(a) (b)

(c) (d)A B

q P

l

1AdiagramiM

diagramkM

second-degree parabola

C

1y2y

2A

A B

2l

2l

PPl

AAAA BB3

4l

3l

1P =

A B

A C

1P =

Problems for reflecting 9-18

lB


(a) (b)

9-24 Whether or not the law of reciprocal reactions can be applied to statically determinate structures? What results will be obtained when use it?


9-1 Determine the designated restraint forces of the following multispan beams by means of the principle of virtual work.

9-2 Determine the deflections and slopes of section A for following cantilevers (neglect the effect of shear forces).

AB

l

constantEI =1 1P =

l

constantEI =A B2 1M =


(a)

(b)

find , , , C B C left C rightY M Q Q

find , , ,A D A CM M Q Y

AB

C

D E

F G H

2m 2m 1m 1m1.5m 1.5m 3m

20kN 10kN

2kN/m

ABD

C

1.5m 1.5m 3m

Problem 9-1


(a) (b)q P

9-3 Determine the deflections at the middle spans for the following simple beams shown in the figure (neglect the effect of shear forces) by integral method.

9-4 Determine the slope of section A for the following simple beam.

9-5 Determine the deflections at joint C for the following trusses (elastic modulus ; in the figure (b), the area of every member is indicated in the parentheses on the member, its dimension is ).

82.1 10 kPaE = ×

2cm9-6 Determine the horizontal displacement at joint C for the following trusses. The axial rigidity of every

member is the same, i.e., . constantEA =9-7 Determine the horizontal and vertical displacements AHΔ and AVΔ of the free end for the circular

cantilever beam shown in the figure. EI is constant.

9-8 Determine the horizontal displacement at point B BΔ for the curved beam. The centre line of the

A B

2l

EI

EI

2ll

A B

Problem 9-2

(a) (b)

A BEI

l

q

A BEI

2l

2l

P

Problem 9-3

M EI

l

BA

Problem 9-4


beam is a parabola 2

4 (f )y x l xl

= − . EI is constant. Both the axial and shear deformations may be

neglected. Assume that the beam is flat, we may consider . ds dx

(b)(a)

230cmA =

A B

C

10kN 10kN5m 5m3m 3m

4mA

C

5m 5m

Problem 9-5

(35)

(50)

(50)

(20)(20)

(35)

(20)

20kN

5m

B

(b)(a)

C DP

A B

a

a

aa

aa

a

BA

a

P

Problem 9-6

P

R

A

B

q

y

f

A B x

/2 ll/ 2



9-9 Solve the problems 9-2 to 9-4 by means of graph-multiplication method. 9-10 Find the deflection at point C and E. 9-11 Find the deflection and slope of section B for the cantilever beam.

9-12 Find the deflection at point D. EI is constant. 9-13 Find the deflection at point C for the beam shown in the figure. The loads 9kNP = , ;

and the moment of inertia of the cross section ; the elastic modulus .

15kN/mq =41660cmI =

82.1 10 kPaE = ×

9-14 Determine the vertical displacements at points A and D for the rigid frame shown in the figure. 9-15 Determine the horizontal displacement at point C for the rigid frame shown in the figure. 9-16 Determine the horizontal displacement at point E and the slope of section B for the three hinged rigid

frame shown in the figure. 9-17 Determine the relative horizontal and vertical displacements and the relative rotation between section

E and F , termed , vΔ hΔ and , respectively, for the pair of cantilevered rigid frames shown in the figure under the following conditions:

rΔ

(1) Neglect the effect of axial deformation. (2) Consider the effect of axial deformation.

2l

2l3

l 0.9m 1.5m 1.5m

qP P

A AB BD

CC


82.06 10 kPaE = ×4

1 6560cmI =

42 12430cmI =

AC E D

B

30kN 30kN

1I 2I3I

2m 2m 2m 2m


1EI 2EI

C B

P

l aA

a

l

−


9-18 Find the vertical displacement at point C due to the temperature changes for the three hinged rigid frame with rectangular cross sections shown in the figure. The coefficient of thermal expansion is 0.00001α = .

9-19 Find the horizontal displacement at point D due to the temperature changes for the rigid frame with I shaped steel sections shown in the figure. The coefficient of thermal expansion is 0.00001α = .

q

C D

1I 1I

2I =

constantEI = constantEI =

2l

2l

q

D C E

A B

a a

a

Problem 9-16

q

C

A

E F D

B

Problem 9-17

h

A D B2EI

Ca a

1.5a

Problem 9-14

q

A B

∞

Problem 9-15

a

43a

EI

constantEI =

depth of cross section 60cmh =depth of cross section 18cmh =

20 C+ °

20 C+ °20 C+ °

0 C°

0 C°0 C°

20 C+ °

20 C+ °

20 C+ °

10 C+ °

0 C°

0 C°A B

CD E

6m 6m

6m

Problem 9-18

BA

CD

4m

6m

Problem 9-19


9-20 Find the horizontal displacement for the rigid frame shown in the figure under the following conditions: (1) Support A move 1cm toward left, (2) support A has 1cm settlement, (3) support B has 1cm settlement.

9-21 Find the vertical and horizontal displacements and rotation at point K ( v ,u andθ ) for the rigid frame shown in the figure under the conditions: support A has horizontal, vertical and rotating displacements xΔ , yΔ and ϕΔ , separately.

9-22 Determine the vertical and horizontal displacements and the rotation at point D, termed , vΔ hΔ

and , separately, due to the movement of support A for the beam shown in the figure. rΔ

9-23 Determine the vertical and horizontal displacements at point C and the relative rotation of hinge C, termed , and vΔ hΔ rΔ , separately, due to the movement of support B for the three hinged rigid

frame shown in the figure.

K

ϕΔ

3aa

a

Problem 9-21

xΔ

0.01rad

2cm

AB C

D

4m 1m

1m

Problem 9-20

A

yΔA

AB C

D

3m3m 1m

1cmProblem 9-23

B

C

4m 4m 1cm

3m

A

CHAPTER 10

FORCE METHOD

The abstract of the chapter The chapter will discuss the analysis of statically indeterminate structures by using force

method (the method of consistent deformations). Its main attention will be focused on the fundamental concepts and principles of force method. The selection of basic unknowns, primary systems and development of compatibility equations of the method will be discussed in detail. As the application of force method, the analysis of statically indeterminate beams, rigid joint frames, bent frames, trusses and composite structures will be discussed, respectively; in addition, the analysis of internal forces of statically indeterminate structures due to temperature changes and support movements will also be evaluated, and the simplification of analysis for statically indeterminate symmetric structures will be discussed as well.

Finally, introduce the problems pertinent to the calculation of displacements and verification of analyzing results of statically indeterminate structures.

10.1 Statically Indeterminate Structures and Degrees of Indeterminacy

10.1.1 Statically indeterminate structures

In the previous chapters of this text, we have discussed the analysis of statically determinate structures. In this chapter, we begin focusing our attention on the analysis of statically indeterminate structures.

For a structure, if its support reactions and internal forces can be uniquely determined from the equations of equilibrium (including equations of condition, if any), the structure is referred to as a statically determinate structure. The simple beam shown in Fig.10.1 (a) is one of the examples of a statically determinate structure. However, if the support reactions and internal forces can not be completely and uniquely determined from the equations of equilibrium, the structure will be named a statically indeterminate structure. The two-span continuous beam shown in Fig.10.1 (b) is one of the examples of a statically indeterminate structure.

It can be seen, by analyzing the geometric constructions of the two structures, that the two beams are all stable. If the roller support B is removed from the simple beam shown in Fig.10.1 (a), the simple beam will become an unstable system. Note however, if the roller support C is removed from the two-span continuous beam shown in Fig.10.1 (b), the two-span beam will be converted into a simple beam, and it is

280

10.1 Statically Indeterminate Structures and Degrees of Indeterminacy 281

still stable. Therefore, from the viewpoint of geometric construction, the support C is excessive (or redundant) restraint. With extending the meaning of geometric construction of the two-span continuous beam, we will introduce the conclusion that a statically determinate structure is a stable system with no redundant restraint, while a statically indeterminate structure is also a stable system but with redundant restraint(s).

Generally speak, the essential characteristics of a statically indeterminate structure is that it has more support reactions and/or members than required for static stability, the equilibrium equations alone are not sufficient for determining the reactions and internal forces of such a structure.

10.1.2 Determination of indeterminacy

From static stability point of view, the degrees of indeterminacy indicate the number of redundants involved in a statically indeterminate structure. If after n restraints were removed from a given statically indeterminate structure the structure would become statically determinate, the given structure is nth degrees of indeterminacy. Thusly, we express

Degrees of indeterminacy = number of redundants = number of released restraints (which are required to convert a statically indeterminate structure into a determinate one) (a)

Since the given structure is stable, recalling from section 2.2.4 of chapter 2, the following expression will thusly be obtained.

Degrees of indeterminacy = -W (b) Where W is the degrees of freedom of a system.

From static analysis point of view, the degrees of indeterminacy are equal to the number of required supplementary equations which are needed to determine the reactions and internal forces of the structure. Consequently, we find

Degrees of indeterminacy = number of redundant unknown forces = number of unknown forces

– number of equilibrium equations (c) Figs.10.2 (a) through (d) are shown four kinds of statically indeterminate structures. Fig.10.3 is shown

(a) (b)

AX

AY

A B CA B

BY CY

AX

AYBY

Fig.10.1 Statically determinate and indeterminate structures(a) a statically determinate beam; (b) a statically indeterminate beam

282 Chapter 10 Force Method

the statically determinate structures which are converted by removing sufficient excessive restraints from the indeterminate ones. They degrees of indeterminacy are 2, 4, 6 and 3, respectively.

When using equation (a) to determine the degrees of indeterminacy, the main point is that the mastery of how to release a statically indeterminate structure into a determinate one. The common ways to release a statically indeterminate structure are as follows.

(1) Removing a roller support or cutting a link (or two-force member) is equivalent to release one restraint [Fig.10.3 (a) and (b)].

(2) Removing a hinged support or a single hinge is equivalent to release two restraints.

Fig.10.2 Original statically indeterminate structures(a) a beam; (b) a truss; (c) a rigid joint frame; (d) an arch

(a)

(c)

(d)

(b)

(a)

1X 2X

(b)

1X 2X 3X 4X

(c)1X

1X3X

2X

4X4X 6X

5X

(d)

1X

2X

3X

Fig.10.3 Corresponding statically determinate structures(a) a beam; (b) a truss; (c) a rigid joint frame; (d) an arch

10.2 Basic Concept of Force Method 283

(3) Removing a fixed support or cutting out a flexural member is equivalent to release three restraints [Fig.10.3 (c)].

(4) Changing a rigid joint of two member ends into a hinged joint is equivalent to release one restraint [Fig.10.3 (d)].

The following two points must be paid attention to: (1) Do not release a statically indeterminate structure into an unstable one. For instance, if replace

the left hinged support by a roller support for the three span beam shown in Fig.10.2 (a), the beam will become an unstable system.

(2) The all excessive restraints must be removed completely. As the structure shown in Fig.10.4 (a), if only one roller support is removed as shown in Fig.10.4 (b), the closed frame is still internally indeterminate and has three excessive restraints. Therefore, only when the closed frame is cut into the frame as shown in Fig.10.4 (c), the original structure will be converted into a determinate one. So the original structure has four redundant restraints.

10.2 Basic Concept of Force Method

The force method (or the method of consistent deformations) is an essential method used to analyze statically indeterminate structures. The method, which was introduced by James C. Maxwell in 1864, essentially involves removing enough restraints from the indeterminate structure to render it statically determinate so as to convert a statically indeterminate analyzing problem into a statically determinate one, about which the analytical methods have been familiar to readers since they have already been studied in forgoing chapters.

Since the independent variables or unknowns in the method of consistent deformations are the redundant forces (and/or moments), which must be determined before the other response characteristics (e.g., displacements and internal forces) can be evaluated, the method is termed force method.

(a) (b) (c)

Fig.10.4 Statically indeterminate structures with a closed frame(a) a statically external and internal indeterminate structure; (b) a statically internal indeterminate structure; (d) a statically determinate structure


10.2.1 Primary unknowns and systems of force method

When analyzing a statically indeterminate structure by means of force method, the enough restraints must be removed from the indeterminate structure to render it statically determinate. This determinate structure, which must be statically stable, is referred to as the primary structure. The excess restraints removed from the given indeterminate structure to convert it into the determinate primary structure are called redundant restraints, and the reactions or internal forces associated with these restraints are termed primary unknowns (variables) or redundants. The primary unknowns are then applied as unknown loads on the primary structure, and their values are determined by solving the compatibility equations based on the condition that the deformations of the primary structure due to the combined effect of the primary unknowns and the given external loadings must be the same as the deformations of the original indeterminate structure. When the primary structure is under combined action of primary unknown(s) and the given external loadings, it is referred to as a primary system.

To illustrate the concept of primary unknowns and primary systems of force method, consider the propped cantilever beam subjected to uniformly distributed load shown in Fig.10.5 (a). Since the beam is supported by four support reactions ( AX , AY , AM and ), the three equations of equilibrium ( , , and ) are not sufficient for determining all the reactions. Therefore, the beam is statically indeterminate. However, if removing roller support B and substituting for it a redundant unknown force , which is now actively applied on the cantilever as shown in Fig.10.5 (b), the propped cantilever will become a statically determinate cantilever that is subjected to original external load and the unknown force . Thus, if we can determine the unknown force , then the remaining three reactions can be obtained from the three equations of equilibrium. Here, the unknown force is named the primary unknown (or variable) and the statically determinate cantilever shown in Fig.10.5 (b) is termed primary system and the cantilever shown in Fig.10.5 (c) is called the primary structure of the propped cantilever shown in Fig.10.5 (a).

BY0X =∑ 0Y =∑ 0M =∑

1X

1X 1X1X

(a) (b) (c)

AB

1XAY

AM

BY

constantEI =

q q

BBX AA

Fig.10.5 Primary system and primary structure of force method(a) original structure; (b) primary system; (c) primary structure

A

10.2 Basic Concept of Force Method 285

10.2.2 Compatibility equations of force method

How to determine the unknown force shown in Fig.10.5? It is obvious that three equilibrium equations can not solve it. We have to find a new supplementary condition to determine it.

1X

To establish the new supplementary condition, we must consider the compatible condition that the vertical displacement at the free end B of the primary system [Fig.10.6 (a)] due to the combined effect of the unknown force 1X and the given uniformly distributed external load must be the same as the vertical displacement at the support B of the original indeterminate structure. That is, only if the vertical displacement at end B of the beam shown in Fig.10.6 (a) is equal to zero, the primary system will be the same as the original structure. The supplementary condition is thusly developed by using a compatibility equation based on the geometry of the deformation at support B of the beam. That is

1 0Δ = (a)

If the beam is linear elastic, by using principle of superposition, the total vertical displacement at end B of the primary beam due to the combined effect of q and can be conveniently expressed by superimposing (algebraically adding) the vertical displacements due to the external load q and the redundant acting individually on the beam; that is,

1Δ

1X

1X

1 11 1 0PΔ = Δ +Δ = (b)

In which, 1PΔ and represent, respectively, the vertical displacements at the end B of the primary beam due to the external load q and the redundant , each acting alone on the beam [see Figs.10.6 (b) and (c)]. Note that two subscripts are used to denote the vertical displacements

11Δ

1X1PΔ and of the

primary beam. The first subscript, 1, indicates the location of these deflections; the second subscript, P, is used to indicate that

11Δ

1PΔ is caused by the given external loading, whereas the second subscript, 1, of 11Δ implies that it is due to the redundant . Both of these deflections are considered to be positive if they occur in the direction of the redundant , which is assumed to be upward, as shown in Fig.10.6.

1X1X

For a linear elastic system, the displacement 11Δ may be expressed by means of principle of

= +

(a) (b) (c)

1XconstantEI =

q qB

AA BB AP

1X 1

Δ

11Δ

Fig.10.6 Linear superposition of primary system(a) primary system; (b) primary structure under action of external loads; (c) primary structure under action of unknown force

l


superposition. Since the redundant is unknown, it is convenient to determine by first evaluating the deflection at B due to a unit value of the redundant , as shown in Fig.10.7 (b), and then multiplying the deflection thus obtained by the unknown magnitude of the redundant. Thus

1X 11Δ

1X

11 11 1XδΔ = (c)

In which, 11δ denotes the vertical displacement at point B of the primary beam due to the unit value of the unknown force . It has units of displacement per unit force, is referred to as a flexibility coefficient. By substituting equation (c) into equation (b), we obtain the compatibility equation

1X

11 1 1 0PXδ + Δ = (10-1)

Eq. (10-1) is the basic equation in the force method which is referred to as the canonical equation for a structure indeterminate to the first degree.

The coefficient 11δ and the free term 1PΔ in equation (10-3) are actually the displacements of a statically determinate structure, so they can be calculated by unit load method. When 11δ and 1PΔ are determined the unknown can be solved by the equation. 1X

In order to calculate coefficient 11δ and the free term 1PΔ , first construct the bending moment diagrams of PM and M for the primary structure due to the external load and the unknown , each acting alone on the structure, respectively. Then by graph-multiplication method, we write

1X

31 1

111 2( )

2 3 3M M l l l ldx

EI EI EIδ ⋅

= = ⋅ =∫

2 41

11 1 3( )

3 2 4 8P

PM M ql l qldx l

EI EI EIΔ = = − ⋅ ⋅ × = −∫

Substituting 11δ and 1PΔ into equation (10-1), we find

(a) (b)

A

B

constantEI = l

1P

Δ

q

11

δ

BA constantEI =

l1 1X =

A B2

2ql

lB

A

1

1

Fig.10.7 Bending moment and diagrams (a) diagram; (b) diagram;

P

P

M M

M M

10.3 Canonical Equations of Force Method 287

3 4

1 03 8l qlXEI EI

− =

Finally, we obtain

138

X ql=

The positive answer indicates that the assumed direction of the unknown is the same as its real direction.

1X

Once the unknown has been computed, the reactions and all other internal forces of the beam can now be determined by applying the three equilibrium equations to the free body of the indeterminate beam. The computed results are shown in Fig.10.8.

1X

(c)(b)(a)

The bending moment M on an arbitrary cross section of the structure can also be determined by employing superposition relationships similar in form to the displacement superposition relationship expressed in Eq. (10-1). Thus, the bending moment M on an arbitrary cross section can be expressed as

1 1 PM M X M= + (10-2)

In which, 1M is the bending moment on an arbitrary cross section of the primary structure due to the unit load ; whereas 1 1X = PM is the bending moment on the arbitrary cross section of the primary structure due to original given external loads.

10.3 Canonical Equations of Force Method

As mentioned previously, the analysis of a statically indeterminate structure by means of force method is, in fact, the method by which the primary system is used as an analytical tool and the redundants of the structure are selected as the primary unknowns. The primary unknowns are then applied as unknown active forces on the primary structure, and their values are determined by solving the compatibility equations

AB

constantEI =

q

l 2l

2l

A B AB

2

8ql

2

8ql

2

8ql 5

8ql

58

ql38

ql

2

16ql 3

8ql

Fig.10.8 Reaction and internal force diagrams(a) reactions; (b) diagram; (c) diagram;M Q


based on the condition that the deformations of the primary structure due to the combined effect of the unknown forces and the given external loadings must be the same as the deformations of the original indeterminate structure. After the unknowns have been computed, the reactions and all other internal forces of the structure can be determined by applying the methods used to analyze statically determinate structures. Therefore, the key step of analyzing statically indeterminate structures is establishment of compatibility equation(s) or force-method equation(s) and determination of the unknowns existing in the equation(s).

10.3.1 Canonical equations for structures with two degrees of indeterminacy

Th force-method equations for a structure with second degrees of indetee rminacy can be easily deve

0⎬

loped by extending the method of consistent deformations developed in the preceding section for analyzing structures with a single degree of indeterminacy. Consider, for example, the rigid frame with two degrees of indeterminacy, as shown in Fig.10.9 (a). To analyze the frame, we must remove two excess restraints. Suppose that we select the support reactions at support B as redundants ( 1X and 2X ). The hinged support at B is then removed from the given indeterminate rigid frame to obtain th tatic determinate and stable primary rigid frame, as shown in Fig.10.9 (b). The two redundants are now treated as unknown active forces on the primary frame, and their magnitudes can be determined from the compatibility conditions that the deflections of the primary frame at the locations B of the redundants due to the combined effect of the known external load and the unknown redundants ( 1X and 2X ) must be equal to zero. This is because the deflections of the given indeterminate frame at the hinged support B are zero. Therefore, the compatibility conditions may be written as

1 0Δ = ⎫

e s ally

Δ2 = ⎭ (a)

In which, represents the displacement along the direction of (i.e., the vertical displacement at to1Δ 1X

point B) due the combined effect of the known external load and the unknown redundants, 1X and 2X ; whereas 2Δ represents the displacement along the direction of 2X (i.e., the horizontal displacement t point B) d to the combined effect of the known external load and th unknown redundants, 1X and 2X .

For a linear elastic system, in order to determine the displacements 1

aue e

Δ and 2Δ by un oad meit l thod, we s

PX Xδ δ ⎬Δ = + + Δ ⎭ (b)

Substituting Eq. (a) into Eq. (b), we obtain

ubject the primary frame separately to a unit value of each of the red ants and 2X [Figs.10.9 (c) and (d), respectively] and the external load [Fig.10.9 (e)]. The displacements and Δ in equation (a) will be thusly, by the principle of superposition, rewritten as follows:

1 11 1 12 2 1PX Xδ δΔ = + + Δ ⎫

und 1X1Δ 2

2 21 1 22 2 2


11 1 12 2 1

21 1 22 2 2

00

P

P

X XX X

δ δδ δ

+ + Δ = ⎫⎬+ + Δ = ⎭

(10-3)

(a)

This equation is the force-method equation of a structure with second degrees of indeterminacy. The meanings of the coefficients and free terms of equation (10-3) are explained as follows (note that the first subscript denotes the location of the displacement, and the second subscript indicates the location of the unit load causing the displacement):

11δ and 21δ represent, respectively, the displacements of the primary frame at the action points and in the directions of and due to unit value of the redundant alone, as shown in Fig.10.9 (c). 1X 2X 1X

12δ and 22δ identify, respectively, the displacements of the primary frame at the action points and in the directions of and due to unit value of the redundant alone, as shown in Fig.10.9 (d). 1X 2X 2X

1PΔ and 2PΔ denote, respectively, the displacements of the primary frame at the action points and in the directions of and due to the external loads alone, as shown in Fig.10.9 (e). 1X 2X

22δ12δ

(d) (e)

2 1X =

A B

C D1P

2P

1P

2P

(b)

C D1P

2P

XA B 2

X1

C D

21δ

(c)

1 1X =

C D C D

A B11δ

A A BB P

1Δ

2PΔ

1

Fig.10.9 A structure with two degrees of indeterminacy(a) original structure; (b) primary system; (c) primary structure under action of 1; (d) primary structure under actio

X =2n of 1; (e) primary structure under action of external loadingX =


Since the coefficient δ and the free term Δ existing in the force-method equations are the displacements of a primary structure that is statically determinate, they can be evaluated by using the methods discussed previously in this text, for example, by unit load method.

Once the redundants and have been determined by solving equation (10-3), the other response characteristics (reactions and internal forces) of the structure can be evaluated either by equilibrium equations or by method of superposition. For example, the bending moment on an arbitrary cross section of the rigid frame can be expressed as

1X 2X

1 1 2 2 PM M X M X M= + + (10-4)

Where PM denotes the bending moment on the arbitrary cross section of the primary rigid frame due to the known external loading alone; 1M and 2M represent, respectively, the bending moments on the arbitrary cross section of the primary rigid frame due to 1 1X = and 2 1X = alone.

Different primary structures and unknowns can be selected for the analysis of an indeterminate structure. The structure shown in Fig10.9 (a) can select the primary structures shown in Fig.10.10 (a) or (b) as its primary structure. Although the form of the canonical equation (10-3) is the same, the physical meanings of redundants ( and ) and compatibility equation (10-3) are different. As shown in Fig.10.10 (a), represents the moment provided by support A; while

1X 2X2X 2 0Δ = identifies that the slope

at support A of the original structure must be equal to zero. However, in Fig.10.10 (b), denotes the bending moment of the left and right sections adjacent to middle section E; whereas indicates that

the relative rotation of the left and right sections adjacent to middle point of the beam must be equal to zero. However, additional consideration has to be taken into that an instantaneously unstable system cannot be used as a primary structure. For example, the system shown in Fig.10.10 (c) cannot be employed as a primary structure because it is an instantaneously unstable system.

2X2 0Δ =

10.3.2 Canonical equations for structures with n degrees of indeterminacy

(b) (c)(a)

AB

C D

1X2X

A B

C D

1

X

X

2

C

D

1X

E

X 2

A B

Fig.10.10 Multiple choice of a primary structure


For a general case in which a structure is statically indeterminate to n degree, when employing force method to analyze the structure, the primary unknowns are n redundant forces, , ; the primary structure is a determinate structure that is obtained by removing n excess restraints from the original indeterminate structure to form a stable system; the compatibility equations are the consistent deformation conditions at the locations and in the directions corresponding to the n redundant forces of the original structure, that is, the displacements of the primary structure at the locations and in the directions of the n redundants due to the combined effect of the known external loading and the unknown redundants ( , ) must be identical to the actual displacements of the original indeterminate structure. For a linear elastic system, by principle of superposition, the n compatibility conditions may be expressed as

1X 2 , , nX X

1X 2 , , nX X

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

00

0

n n P

n n P

n n nn n nP

X X XX X X

X X X

δ δ δδ δ δ

δ δ δ

+ + + + Δ = ⎫⎪+ + + + Δ = ⎪⎬⎪⎪+ + + + Δ = ⎭

(10-5)

The equation (10-5) is the general form of force-method formulation for a structure with n degrees of indeterminacy, referred to as the canonical equation of force method.

In equation (10-5), the flexibility coefficient ijδ and free term iPΔ represent, respectively, the

displacements of a primary structure due to a unit load and the external loading. Note that the first subscript of ijδ or indicate the location and direction of the displacement, whereas the second subscript is

used to indicate that the displacement is due to the unit value of jth redundant or the external loading. That is,

iPΔ

ijδ represents the displacement of the primary structure at the action point and in the directions of

due to unit value of the redundant iX

jX alone; while iPΔ denote the displacement of the primary structure

at the action point and in the directions of due to the external loading alone. iXThe sign convention is that when the direction of ijδ or iPΔ is the same as that of its corresponding

redundant , iX ijδ or will have positive sign and vice versa. iPΔ

In equation (10-5), the flexibility coefficients located on the main diagonal such as 11 22, , nnδ δ δ

are termed main coefficients, which are always greater than zero. The other coefficients, (ij i j)δ ≠ , are

named secondary coefficients, which may be greater than, smaller than or equal to zero. Recall from the law of reciprocal displacements, the secondary coefficients ijδ and jiδ are identical.

That is,

ij jiδ δ=

By using the definition of matrix multiplication, Eq. (10-5) can be expressed in matrix form as


11 12 11 1

21 22 2 2 2

1 2

00

0

nP

n P

n nPn n nn

XX

X

δ δ δδ δ δ

δ δ δ

⎡ ⎤ Δ⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪Δ⎢ ⎥ + =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ Δ ⎩ ⎭⎩ ⎭ ⎩ ⎭⎣ ⎦

(10-6)

In which, the matrix composing of flexibility coefficients ijδ is called the flexibility matrix. Based on this

reason, the canonical equation (10-5) is also referred to as the flexibility equation, and force method as flexibility method.

After the flexibility coefficients and free terms have been determined, the redundants , can be evaluated by solving equation (10-5), then the internal forces and their diagrams of

the structure can be obtained either by equilibrium conditions or by principle of superposition. The equations used to determine, by principle of superposition, the internal forces can be expressed as

1X 2 , , nX X

1 1 2 2

1 1 2 2

1 1 2 2

n n P

n n P

n n P

M M X M X M X MQ Q X Q X Q X QN N X N X N X N

⎫= + + + +⎪

= + + + + ⎬⎪= + + + + ⎭

(10-7)

Where, iM , iQ and iN identify, respectively, the bending moment, shear and axial forces of the

primary structure due to alone; 1iX = PM , PQ and PN denote, respectively, the bending moment,

shear and axial forces of the primary structure due to only the external loading. An alternative way to construct the diagrams of shear force Q and axial force N of an indeterminate

structure when employing equation (10-7) is that first draw the bending moment diagram and then determine the shear and axial forces by using the equilibrium conditions of the free bodies of each of the members or joints of the structure.

10.4 Statically Indeterminate Beams, Rigid Frames and Bent Frames

10.4.1 Statically indeterminate beams and rigid frames

A common means used in analysis of statically indeterminate beams and rigid frames by force method is that the effect on displacements due to the axial and shearing deformations is neglected and only that due to the bending deformation is under consideration. In the circumstance, the flexibility coefficients and free terms existing in force-method equations can be expressed as

10.4 Statically Indeterminate Beams, Rigid Frames and Bent Frames 293

2i

ii

i jij

i PiP

M dxEIM M

dxEI

M M dxEI

δ

δ

⎫= ⎪

⎪⎪

= ⎬⎪⎪

Δ = ⎪⎭

∑∫

∑∫

∑∫

(10-8)

Example 10-1

Draw the bending moment M and shear force Q diagrams for the beam with two fixed supports and subjected to uniformly distributed load of intensity q as shown in Fig.10.11 (a) by force method.

Solution

(1) Select primary system This is an indeterminate beam with three degrees of indeterminacy. The restraints against rotation at

ends A and B and horizontal translation at end B of the fixed beam are then removed to obtain the simply supported primary beam shown in Fig.10.11 (b).

(2) Develop compatibility equations Noting that the slopes of the actual indeterminate beam at the fixed supports A and B and the

2

8ql

(a)

(d) (e)

(b) (c)

l

diagramPM

BCE

q

EI

q

A BD

(f)

1q 2XX

BAXA 3

l

l

2 1X2 diagramM

3 diagramM1 diagramM1 1X =

l

=AB AA B 3 1X =B

3 1N = −3 0M =

Fig.10.11 figures of example 10-1(a) a beam with two fixed supports; (b) primary system; (c) primary structure under action of externa

1 2

3

l loading;(d) primary structure under action of 1; (e) primary structure under action of 1;(f) primary structure under action of 1

X XX

= ==


horizontal translation at the support B are zero, we write the compatibility equations as

11 1 12 2 13 3 1 0PX X Xδ δ δ+ + + Δ = (a)

21 1 22 2 23 3 2 0PX X Xδ δ δ+ + + Δ = (b)

31 1 32 2 33 3 3 0PX X Xδ δ δ+ + + Δ = (c)

(3) Calculation flexibility coefficients and free terms Since the flexibility coefficients and free terms are all displacements of the primary beam (a simple

beam), when just bending deformation is considered, only the bending moment diagrams, PM due to the external loading, 1M due to , 1 1X = 2M due to 2 1X = and 3M due to of the primary beam are needed to construct, as shown in Fig.10.11 (d) through (f).

3 1X =

By graph-multiplication method, we write 2

111

1 1 2( 1)2 3

M ldx lEI EI EI

δ = = × × × =∫ 3

22

221 1 2( 1)

2 3 3M ldx lEI EI EI

δ = = × × × =∫

1 212 21

1 1 1( 1)2 3

M M ldx lEI EI EI

δ δ= = = − × × × = −∫ 6

When calculating 33δ , the effect on the displacement due to the axial deformation must be considered because of 3 0M = . Thus

2 23 3

33 0M N l ldx dxEI EA EA EA

δ = + = + =∫ ∫

1 313 31 0M M dx

EIδ δ= = =∫

2 323 32 0M M dx

EIδ δ= = =∫

321

11 2 1 1( )

3 8 2 24P

PM M qldx l ql

EI EI EIΔ = = × × × =∫

32

2 24P

PM M qldx

EI EIΔ = = −∫

3 33 0P P

PM M N Ndx dx

EI EAΔ = + =∫ ∫

Noting that the minus sign of 2PΔ is the reason that the diagrams of 2M and PM are located in


different sides of the beam. (4) Determine the unknown redundants By substituting the expressions for the flexibility coefficients and free terms into the compatibility

equations [equations (a) through (c)], we obtain 2

1 2

2

1 2

3

2 04

24

0

qlX X

qlX X

l XEA

⎧− + =⎪

⎪⎪− + − =⎨⎪⎪ =⎪⎩

0

Solving the equations, we find

21

112

X ql= − , 22

112

X ql= , 3 0X =

3 0X = implies that the horizontal reactions of an indeterminate beam with two fixed supports subjected to vertical loading must be zero. So to analyze this beam, we need to select only two of the remaining four reactions as the redundants. The compatibility conditions may become

11 1 12 2 1 0PX Xδ δ+ + Δ =

21 1 22 2 2 0PX Xδ δ+ + Δ =

(5) Construct internal force diagrams Once the redundants are determined the internal forces at the end sections of the beam can now be

determined in terms of superposition of the internal forces of the primary beam due to the external loading and each of the redundants. And then the internal force diagrams will be drawn.

① Bending moment diagram

By the equation of superposition for bending moments, we write

1 1 2 2 3 3 PM M X M X M X M= + + +

By taking the values of ABM and BAM as the ordinates and laying them off by connecting them in dashed lines on the tension side of the beam AB, and then superposing the bending moment diagram of a corresponding simple beam subjected to the uniformly distributed load, the final bending moment diagram can be obtained as shown in Fig.10.12 (a).

② Shear force diagram The shear forces at the ends of the beam can be fined by considering the equilibrium conditions of the

free body of the beam [Fig.10.12 (b)] because the end bending moments are now known, and shear force


diagram can be thusly depicted. Considering the equilibrium conditions of the free body of the beam, we write

0AM =∑ , 2 21 1 02 12 12 BAlql ql ql Q l× − + + × = ,

12BAQ = − ql

0BM =∑ , 12ABQ = ql

The shear force diagram is shown in Fig.10.12 (c).

When the internal force diagrams have been finished the elastic curve of the deformed beam may be sketched by referring the bending moment diagram as shown in dashed lines in Fig.10.11 (a). The zero points on the bending moment diagram are the inflection points of the elastic curve of the deformed beam. In the segment DE, because the lower fibers are in tension its elastic curve is concave upward; whereas in the segments AD and EB, since the upper fibers are in tension their elastic curves are concave downward.

Example 10-2

Draw the internal force diagrams for the indeterminate rigid frame shown in Fig.10.13 (a).

Solution

(1) Select primary system This is a single indeterminate rigid frame. The restraint against horizontal translation at hinged support

B is removed, which is done by replacing the hinged support by a roller support, to obtain the primary rigid frame shown in Fig.10.13 (b).

(2) Develop compatibility equation Noting that the horizontal translation at support B of the primary rigid frame due to the combined

effect of the known external loading and the unknown redundant must be zero, we write the compatibility equation as

1X

2

12ql 2

12ql

2

24ql

2

12ql 2

12ql

2ql

2ql

(a)

A B

ABQ

D

(c)

q

A BE

(b)

l

BA

BAQ

Fig.10.12 Bending moment and shear force diagrams(a) diagram; (b) free body diagram of beam ; (c) diagramM AB Q


11 1 1 0PXδ + Δ =

(3) Calculate flexibility coefficient and free term The bending moment diagrams PM due to the external loading and 1M due to of the

primary rigid frame are depicted out as shown in Figs.10.13 (c) and (d). 1 1X =

10kN mq =

C D

6C D

1

By using the graph-multiplication method, we find

11

1 1

1 2 1080( 6 45) 63

PP

M M dxEI EI EI

Δ = = − × × × = −∑∫

21

111 2 1

1 2 1 2 216 144(6 6 6) ( 6 6) ( 6)2 3

M dxEI EI EI EI EI

δ = = × × + × × × × = +∑∫2

XA B

(a)

(d)

(b)

(c)

1I

1 22I I=2I2I

6m

6m

q

BA

C D

1

q

C D

45

diagram(unit: kN m)

PM⋅

1 diagram

M

XBA BA

6 66

1

Fig.10.13 Figures of example 10-2(a) a redundant rigid frame; (b) primary system; (c) primary structure under action of external loading; (d) primary structure under action of 1X =


Since 1 22I I= , 11δ is simplified as

111

504EI

δ =

(4) Determine the unknown redundant Substituting the expressions for 11δ and 1PΔ into the compatibility equation, we determine the

redundant to be

11 1

504 1080 0XEI EI

− =

1 2.14kNX =

It is instructive to note from above calculations that although the flexibility coefficient and free term in compatibility equation are related to the absolute value of flexural rigidity EI of each of the members, the final value of the unknown is only related to the relative value of flexural rigidity EI of each of the members. Therefore, only the relative value of flexural rigidity EI of each of the members of a statically indeterminate structure is necessary when analyzing the structure.

(5) Construct internal force diagrams ① Bending moment diagram The equation of superposition for bending moments now is

1 1 PM M X M= +

By multiplying the values of 1M by 1 2.14kNX = and then superposing them with PM diagram, the final bending moment diagram can be obtained as shown in Fig.10.14 (a).

② Shear force diagram In a rigid frame, the shear forces at the ends of each of the members can be fined by considering the

equilibrium conditions of the free body of each of the members because the end bending moments of each of the members are now known, and shear force diagram can thusly be depicted.

Considering the free body of member CD as an example shown in Fig.10.14 (b), by the equilibrium conditions (when determining shear forces the axial forces do not need to take into account, so the axial forces do not draw out), we write

0DM =∑ , 6 10 6 3 12.86 12.86 0CDQ × − × × − + = , 30kNCDQ =

0CM =∑ , 30kNDCQ = −

Using the similar approach, we can find the shear forces of members AC and BD. When the shear forces of all the members have been determined the shear force diagram can be finished, as shown in Fig.10.14 (c).


③ Axial force diagram By using the known shear forces, the axial forces of each of the members can be calculated by the

equilibrium conditions of the free bodies of the associated rigid joints. After the axial forces are determined the axial force diagram of the rigid frame can be drawn out conveniently.

C D

(unit: kN m)⋅

Taking rigid joint C as an example, its free body diagram is shown in Fig.10.14 (d) (when determining axial forces the bending moments do not need to take into account, so the bending moments do not draw out). By the equilibrium conditions, we write

0X =∑ , 2.14 0CDN + = , 2.14kNCDN = −

(a)

(e)

A B

10kN m

(b)

(c)

A B

C D

12.86

12.86

12.86

12.86

45

32.14

2.14kN2.14kN

12.86 12.86

6m

CDQ DCQ

(d)

2.14kNC

30kN

CDN

NCA

2.14

30kN 30kN

30

2.14

30

2.142.14

(unit: kN)

CD

(unit: kN)

2.14 2.1430 30

BA

C D

30 30

Fig.10.14 Internal force diagrams of example 10-2(a) diagram; (b) free body diagram of member ; (c) diagram;(d) free body diagram of joint ; (e) diagram

M CD QC N


0Y =∑ , 30 0CAN + = , 30kNCAN = −

By isolating the free body diagrams of rigid joints in an appropriate order, all the axial forces will be determined by using two projection equations of equilibrium of each of the joints. The axial force diagram is shown in Fig.10.14 (e).

Noting that in shear and axial force diagrams signs must be identified. By referring to the bending moment diagram, the elastic curve of the deformed rigid frame may be

sketched as shown in dashed lines in Fig.10.13 (a). Since the member CD has two zero points on its bending moment diagram, there are two inflection points lying on the elastic curve of the deformed member. Because the bending moments of members AC and BD do not change their sign there is no inflection point on the elastic curves corresponding to the portions of the two members. The cross-sectional rotations of member ends meeting in the same joint (joints C and D) must be identical. In addition, because this rigid frame is a symmetric structure subjected to a symmetric loading its elastic curve must be symmetric with respect to the axis of symmetry of the frame as well.

Procedure for Analysis

Based on the foregoing discussion, we can develop the following step-by-step procedure for the analysis of indeterminate structures by using force method.

(1) Select primary system Select excess restraint force(s) as the primary unknown(s) (redundants). The choice of redundant(s) is

merely a matter of convenience, and any reaction or internal force(s) can be selected as the redundant(s), provided that the removal of the corresponding restraint(s) from the given indeterminate structure results in a primary structure that is statically determinate and stable. The sense(s) of the redundant(s) is not known and can be arbitrarily assumed. The actual sense(s) of the redundant(s) will be known after its magnitude(s) has been determined by solving the compatibility equation(s). A positive magnitude for the redundant(s) will imply that the sense initially assumed was correct, whereas a negative value of the magnitude will indicate that the actual sense is opposite to the one assumed initially.

(2) Develop compatibility equation(s) Write the compatibility equation(s) by setting the algebraic sum of the displacements of the primary

structure at the location(s) and in the direction(s) of the redundant(s) due to the external loading and the redundant(s) equal to the given displacement(s) of the restraint(s) of the actual indeterminate structure.

(3) Calculate flexibility coefficients and free terms a. First，draw internal force diagram(s) of the primary structure with only the external loading applied

to it. b. Next, draw, respectively, internal force diagrams of the primary structure with only the unit value of


each of the redundants applied to it alone. The unit force (or moment) must be applied in the positive direction of the redundant.

c. Finally, calculate the flexibility coefficients and free terms in the compatibility equation(s) by using graph-multiplication method

(4) Determine the redundant(s) Substitute the values of the flexibility coefficients and free terms computed in step 3 into the

compatibility equation(s), and solve for the unknown redundant(s). (5) Draw internal force diagram(s) First, draw the bending moment diagram of the structure by superposition of the bending moments of

the primary structure due to the external loading and due to the redundant(s). Then, draw shear and axial force diagrams by using appropriate equilibrium equations.

10.4.2 Bent frames

A bent frame is composed of its roof (or roof beam) and columns. Fig.10.15 (a) shows a sketch of a cut plane of the structure commonly used in one-storey precast workshops in industrial factories. The roof (or roof beam) is usually simplified as a link with infinite axial rigidity, which is hinged at the tops of the columns. The columns are usually with variable cross sections and fixed to their foundations. The analytical model of a bent frame is shown in Fig.10.15 (b). The degree(s) of indeterminacy of a bent frame is equal to its number of span(s). The primary system may be obtained by cutting the link(s) connecting the columns, and the redundant(s) is a pair (or pairs) of forces equal in magnitude but opposite in direction, which is actually the substitution of restraint(s) of the link(s), as shown in Fig.10.15 (c).

Since the axial rigidity of the link(s) is infinite, i.e., EA = ∞ , the effect to the displacements of axial deformation of the link(s) will be neglected when calculating the flexibility coefficients and free terms in

(b)(a)

C D

(c)

P C DP1X

C DP EA = ∞

BABAA B

Fig.10.15 Computing model and primary system of a bent frame(a) sketch of cross-sectional structure of a workshop; (b) computing model; (c) primary system


force-method equation(s). Thusly, the expressions for the flexibility coefficients and free terms are still expressed as form of equation (10-8).

Example 10-3

Analyze the two-span bent frame with different heights of columns as shown in Fig.10.16 (a) and draw its bending moment diagram due to the action of a horizontal braking force of a crane.

Solution

(1) Select primary system The bent frame is indeterminate to the second degree. The primary system is obtained by cutting the

two links and by substituting them by two pairs of corresponding redundants, as shown in Fig.10.16 (b). (2) Develop force-method equations (compatibility equations) Noting that the relative horizontal displacements at the two cuts of the primary system due to the

combined effects of the given external loading and the unknown redundants and must be zero, i.e., the axial displacements of the each of the two pairs of sections adjacent to the two cuts should remain consistent, so we write the force-method equations as

1X 2X

11 1 12 2 1 0PX Xδ δ+ + Δ =

21 1 22 2 2 0PX Xδ δ+ + Δ =

(3) Calculate flexibility coefficients and free terms Draw the bending moment diagrams, PM due to the external loading, 1M due to 1 1X = and

2M due to of the primary system as shown in Figs.10.16 (c) through (e). 2 1X =By employing the graph-multiplication method, we write

111 2

1 1 2 1 1 2 504( 6 6 6) ( 6 6 6)2 3 2 3EI EI EI

δ = × × × × + × × × × =2

221

2

2

2 1 2( 3 3 3)2 3

2 1 2 1 1 1 2 7 10 ( 10 3) 7 3 ( 10 3)2 3 3 2 3 3

2270 3

EI

EI

EI

δ = × × × × +

⎡ ⎤× × × × + × + × × × × + ×⎢ ⎥⎣ ⎦

=

12 212 2

1 1 2 1 1446 6 ( 10 4)2 3 3EI EI

δ δ ⎡ ⎤= = − × × × × + × = −⎢⎣ ⎦⎥

1 0PΔ =


13.492

1 diagram (m)M 2 diagram

(m)M

diagram (kN m)M

⋅ diagram (kN m)

PM⋅

(a)

(d)

(e)

A B C

D E

1I

20kNP =

92.540

2I

(b)

(c)

20.238

F G

EA = ∞

EA = ∞

3

(f)

1I 1I

2I2 16I I=

3m2m

6m 7m

2X

1X20kNP =

1 1X = 3

1010

20kNP =20

11.562 55.898

0.238

1

Fig.10.16 Figures of example 10-3(a) a bent frame with different heights of columns; (b) primary system; (c) primary structure under action of 1; (d) primary X = 2structure under action of 1;(e) primary structure under action of external loading; (f) bending moment diagram

X =

6 6

160

26.984

2 1X =


21

2

2

1 1 2 11 20 ( 3 2)2 3 3

1 1 2 1 1 1 2 ( 7 160 ( 10 3) 7 20 ( 10 3)2 3 3 2 3 3

14480 3

P EI

EI

EI

⎡ ⎤Δ = − × × × × + ×⎢ ⎥⎣ ⎦⎡ ⎤− × × × × + × + × × × × + ×⎢ ⎥⎣ ⎦

= −

(4) Determine the unknown redundants By substituting the expressions for the coefficients and free terms into the force-method equations, we

determine the redundants to be

1 2504 144 0X X− =

1 22270 14480144 0

3 3X X− + − =

1 1.927kNX = ， 2 6.746kNX =

(5) Construct bending moment diagram The superimposing equation for bending moments now is

1 1 2 2 PM M X M X M= + +

By multiplying the values of 1M by 1 1.927kNX = , 2M by and then superposing them with

1 6.746kNX =

PM diagram, the final bending moment diagram can be obtained as shown in Fig.10.16 (f).

10.5 Statically Indeterminate Trusses and Composite Structures

10.5.1 Statically indeterminate trusses

A truss is assemblage of straight members connected at their ends by flexible connections, which are modeled as frictionless hinges. The members of a truss are only subjected to axial forces when there is no load applied between the ends of every member. In the circumstance, the flexibility coefficients and free terms in force-method equations can be expressed as follows.

i iii

i jij

i PiP

N N lEA

N Nl

EAN N lEA

δ

δ

⎫= ⎪

⎪⎪

= ⎬⎪⎪

Δ = ⎪⎭

∑

∑

∑

(10-9)

10.5 Statically Indeterminate Trusses and Composite Structures 305

The superimposing equation of axial forces for each of the members will be

1 1 2 2 PN N X N X N= + + + (10-10)

Example 10-4

Determine the axial forces in the members of the truss shown in Fig.10.17 (a). The elastic moduli of the materials of each of the members are identical; the area of every member is tabulated in table 10-1.

Solution

(10)(9) (8)(7)(6)

(5)

(4)

(3)(2)(1)

(a)

(d)

3 3 9m× =

(b)

(c)

3m

30kNP

(1) Select primary system The truss is a single degree internally indeterminate truss. The primary system is obtained by cutting

the number 10 member and by substituting it by a pair of corresponding redundants, as shown in Fig.10.17 (b).

(2) Develop force-method equation (compatibility equation) By the consistence deformation condition of member 10, i.e., the relative displacement in the direction

of centroidal axis at the cut through the member must be equal to zero, we write the force-method equation as

11 1 1 0PXδ + Δ =

=

0

0 0

01 1X =

0.7−

0.7−11

0.7−

0.7−

Fig.10.17 Figures of example 10-4(a) an indeterminate truss; (b) primary system; (c) primary structure under action of external loading; (d) primary structure und 1er action of 1X =

10 20

20201030kN

(kN)PN

10

10−

014− 28−14− 30

X

P

1

1N


(3) Calculate flexibility coefficient and free term Draw the axial force diagram PN due to the external loading and 1N due to of the primary

system as shown in Figs.10.17 (c) to (d). The coefficient and free term can be evaluated by equations 1 1X =

21

11N lEA

δ ==∑

11

PP

N N lEA

Δ =∑

As shown in table 10-1

1189.5

Eδ =

11082

P EΔ = −

Table 10-1 calculation of 11δ , 1PΔ and axial forces N

member (cm)l 2 (cm )A

(kN)PN 1N

21N lA

1 PN N lA

1 1 (kPN N X N= +

(1) 300 15 10 0 0 0 10.0 (2) 300 20 20 -0.7 7.5 -210 11.5 (3) 300 15 20 0 0 0 -20.0 (4) 424 20 -14 0 0 0 -14.0 (5) 300 25 -10 -0.7 6 84 -18.5 (6) 424 20 -28 0 0 0 -28.0 (7) 300 15 10 -0.7 10 -140 1.5 (8) 300 15 30 -0.7 10 -420 21.5 (9) 424 15 -14 1 28 -396 -1.9 (10) 424 15 0 1 28 0 12.1

∑ 89.5 -1082

(4) Determine the unknown redundant By substituting the expressions for the coefficient and free term into the force-method equation, we

determine the redundant to be

11

11

1082 12.1kN89.5

P EXEδ

Δ= − = − =

(5) Calculate axial forces of each member The superimposing equation of axial forces for each member now is


1 1 PN N X N= +

The calculating results are tabulated in table 10-1.

10.5.2 Statically indeterminate composite structures

A statically indeterminate composite structure is a composition of flexural members and two-force members. In the structure, the two-force members are subjected to only axial forces, whereas the flexural members are undergone bending moments, shear forces and axial forces. However, the effects on displacements due to the shear and axial forces are, compared to that due to bending moments, negligible in the flexural members. Thusly, when evaluating the flexibility coefficients and free terms in force-method equation(s) of a composite structure, only axial forces in two-force members and bending moments in flexural members are taken into account. In the circumstance, the expressions calculating the coefficients and free terms can be written as

2 2i i

ii

i j i jij

i P i PiP

M Ndx lEI EAM M N N

dx lEI EA

M M N Ndx lEI EA

δ

δ

⎫= + ⎪

⎪⎪

= + ⎬⎪⎪

Δ = + ⎪⎭

∑ ∑∫

∑ ∑∫

∑ ∑∫

(10-11)

The superposing equations of bending moment for flexural members and axial force for two-force members will be written as

1 1 2 2

1 1 2 2

P

P

M M X M X MN N X N X N

⎫= + + + ⎪⎬

= + + + ⎪⎭ (10-12)

Example 10-5

Analyze the internally indeterminate trussed beam shown in Fig.10.18 (a). The inertial moment of the cross section of the beam is ; the cross sectional area of each of the two-force members is

; the elastic modulus of the materials of the structure is

4 41 10 mI −= ×3 21 10 mA −= × constantE = .

Solution

(1) Select primary system The trussed beam is a single degree internally indeterminate structure. Its primary system is obtained

by cutting the vertical two-force member and by substituting it by a pair of corresponding redundants, as shown in Fig.10.18 (b).

(2) Develop force-method equation (compatibility equation)


By the consistence deformation condition of the vertical two-force member, i.e., the relative axial displacement of the up and down sections adjacent to the cut must be equal to zero, we write the force-method equation as

11 1 1 0PXδ + Δ =

(3) Calculate flexibility coefficient and free term Draw the bending moment diagram and axial force, 1M and 1N due to , 1 1X = PM and PN

due to the external loading of the primary structure as shown in Figs.10.18 (c) to (d). The coefficient and free term can be evaluated as follows

(kN m), P PM N⋅

(kN m), (kN)M N⋅ when , (kN m), (kN)A M N→∞ ⋅

5 / 2− 5 / 2−

1 1(m), M N

(a)

(e) (f)

(b)

(c)

4m

4m 4m

A A2A

10kN/m

1.76m

I I 2m

10kN/m

1X

1 1N =

1 1X =2

10kN/m

0 0 080

0PN =(d)

9.8

15.4 11.350.2+50.2+

44.9N = −

44.9

−

1.5m 20 50N = −

55.9+ 55.9+ 50

−

1

Fig.10.18 figures of example 10-5(a) an indeterminate trussed beam; (b) primary system; (c) primary structure under action of 1;(d) primary structur

X =e under action of external loading; (e) bending moment diagram and axial force;

(f) bending moment diagram and axial force when A →∞


2 21 1

11 4

22

3

5 5 5

1 4 2 2(2 )1 10 2 3

1 1 2 5 2 ( ) 2 51 10 2 2

1 1 (1.067 10 0.122 10 ) (1.189 10 )

M N ldxEI EA E

E

E E

δ −

−

× ×= + = × ×

× ×⎡ ⎤×

+ × − ×⎢ ⎥× × ⎣ ⎦

= × + × = ×

∑ ∑∫2+

1 11

64

1 4 2 80 5 2 1 (2 ) 0 (5.333 10 )1 10 3 8

P PP

M M N Ndx lEI EA

E E−

Δ = +

× × ×= × × + =

× ×

∑ ∑∫×

(4) Determine the unknown redundant By substituting the expressions for the coefficient and free term into the force-method equation, we

determine the redundant to be 6

11 5

11

5.333 10 44.9kN1.189 10

PXδΔ ×

= − = − = −×

(5) Calculate internal forces of each member The superimposing equations of internal forces now should be

1 1 PM M X M= +

1 1 PN N X N= +

The bending moment diagram and axial forces of the composite structure are depicted out as shown in Fig.10.18 (e).

(6) Discussion It is noted from comparing the bending moment diagrams M [Fig.10.18 (e)] and PM [Fig.10.18 (d)],

which is actually the bending moment diagram of a simple beam, that because of the elastic restraint provided by two-force members the bending moment M in the beam of the composite structure has been reduced by a percentage of , from 80.75% 80kN m⋅ to 15.4kN m⋅ .

If we change the magnitude of the cross-sectional area A of the two-force members, the distribution of internal forces in the composite structure will be changed as well. When the area A is reduced, the positive bending moment of the beam will become greater, whereas its negative bending moment becomes smaller. When the area A vanish the bending moment diagram M is the same as PM . Contrariwise, when the area A is increased, the positive bending moment of the beam will become smaller, whereas its negative bending moment becomes greater. When the area A goes infinitely great the elastic restraint in the middle of the beam will become a rigid support and the bending moment diagram M of the beam will become the same as


that of a two-span continuous beam as shown in Fig.10.18 (f).

10.6 Analysis of Symmetric Structures

Many structures, because of aesthetic and/or functional requirements, are arranged in symmetric forms. If a symmetric structure is linearly elastic, the response (i.e., member forces and deformations) of the entire structure under symmetric loading can be obtained from the response of one of its portions separated by the axes of symmetry. Thus only a portion (usually half) of the symmetric structure needs to be analyzed. In the section, we discuss how to recognize structural symmetry and how to utilize it to simplify the computational effort required in the analysis of symmetric structures.

When employing force method to analyze a statically indeterminate structure, the higher the degrees of its indeterminacy, the more the required computational effort. The main calculation work is the solution of compatibility equations, i.e., the evaluation of flexibility coefficients and free terms in compatibility conditions and the solution of the simultaneous linear equations. In the circumstance, the key step to simplify the calculation work in force method is the simplification of force-method equations. In other words, if the approach making some or a part of the flexibility coefficients and free terms in canonical equations to be zero is possible, the solution of the equations will be simplified. As we know, since the coefficients located in the main diagonal in flexibility matrix [Eq. 10-6] are always positive and greater than zero, obviously the principle of simplification of force-method equations will come down to the way by which the secondary coefficients and free terms in flexibility matrix can be made to be zero as many as possible. In fact, there are a few of simplifying methods such as the utilization of symmetry of structures, the method of elastic centre and etc. The key points of the methods are the selections of reasonable primary systems and primary unknowns. The objective of the section is the discussion of the utilization of symmetry of structures.

In practical engineering, a lot of structures actually can be considered to be symmetric. The utilization of symmetry will help us to simplify the analysis of the structures.

10.6.1 Symmetric structures and symmetric loadings

(1) Symmetric structures A plane structure is considered to be symmetric with respect to an axis of symmetry in its plane if the

reflection (or mirror image) of the structure about the axis is identical in geometry (dimension and shape), supports, and material properties (axial rigidity EA, flexural rigidity EI and shear rigidity GA etc.) to the structure itself. Some examples of symmetric structures are shown in Fig. 10.19. For each structure, the vertical axis of symmetry is identified as y-y axis and horizontal axis of symmetry as x-x axis. Note that the reflection of each structure about its axis of symmetry is identical in geometry, supports and material properties to the structure itself.

10.6 Analysis of Symmetric Structures 311

(2) Symmetry of loadings Symmetric loadings: A loading is considered to be symmetric with respect to an axis in its plane if

the reflection (or mirror image) of the loading about the axis is identical to the loading itself. An example of symmetric loading is shown in Fig.10.20 (b).

Antisymmetric loadings: A loading is considered to be antisymmetric with respect to an axis in its plane if the negative of the reflection (or mirror image) of the loading about the axis is identical to the loading itself. An example of antisymmetric loading is shown in Fig.10.20 (c).

Decomposition of a general loading into symmetric and antisymmetric components: Any general loading [Fig.10.20 (a)] can be decomposed into symmetric [Fig.10.20 (b)] and antisymmetric [Fig.10.20 (c)] components with respect to an axis by implementing the following procedure:

axis of symmetry

axis of symmetry

axis of symmetry(a)

1EI

y

(b)

2l

2l

y

2a

2EI 2EI h

y

y

2EI 2EI1EI

2a

2b 2b

xx

1EI

Fig.10.19 Symmetric structure(a) a symmetric structure with one axis of symmetry;(b) a symmetric structure with two axes of symmetry

(a)

C

2PP

(b) (c)

a2P

2P

a a 2P

a a

C C

Fig.10.20 symmetric and antisymmetric loadings(a) any general loading; (b) symmetric loading; (c) antisymmetric loading


① Divide the magnitudes of the forces and/or moments of the given loading by 2 to obtain the half loading.

② Draw a reflection (or mirror image) of the half loading about the specified axis. ③ Determine the symmetric component of the given loading by adding the half loading to its

reflection. ④ Determine the antisymmetric component of the given loading by subtracting the symmetric

loading component from the given loading.

10.6.2 Analysis by using symmetric system

When analyzing a statically indeterminate symmetric structure, the utilization of a symmetric system can considerably expedite the analysis of the structure. Considering the portal rigid frame shown in Fig.20

(a), a symmetric system is obtained by cutting the frame through its axis of symmetry and by adding three pairs of redundants as substitution of the restraints at the cut, as shown in Fig.10.21 (a). Apparently, by the definitions of symmetric and antisymmetric loadings, both of the pair of bending moments and the pair of axial forces is symmetric loadings, whereas the pair of shear force is antisymmetric

1X2X 3X

loading.

1 diagramM

2 diagramM3 diagramM

(a)

(d)

P (b)

(c)

1X

2X2X

3X

1 1X =

2 1X =

3 1X =

1

2

Fig.10.21 Bending moment diagrams and deformed shapes of symmetric primary structure(a) primary system; (b) primary structure under action of 1; (c) primary structure under action of 1; (d) pr

XX

== 3imary structure under action of 1X =


By setting the algebraic sum of the relative rotation, relative horizontal displacement and vertical relat

00

P

P

X X XX X X

δ δ δδ δ δ

ive deflection at the location of the cut of the symmetric primary structure due to each of the redundants and the external loading equal to zero, we can write compatibility conditions as

11 1 12 2 13 3 1 0PX X Xδ δ δ+ + + Δ = ⎫⎪

21 1 22 2 23 3 2

31 1 32 2 33 3 3

+ + + Δ = ⎬⎪+ + + Δ = ⎭

(a)

Figs.10.21 (b) through (d) show, separately, the bending moment diagrams and deformed configurations due to 1 1X = , 2 1X = and 3 1X = . It can be observed that the bending moment diagrams of 1M , 2M ir c onding ing shapes are symmetric, while that of and the orresp deform 3M and its correspondi d ming shape are antisymmetric. Therefore, the secondary flexibility coefficie 13ng efor nts δ ,

31δ , 23δ and 32δ will be

1 313 31 0M M ds

EIδ δ= = =∑∫

2 323 32 0M M ds

EIδ δ= = =∑∫

Thusly, the force-method equations can be simplified as

P

P

X XX

δ δδ

= ⎫11 1 12 2 1PX Xδ δ+ + Δ

21 1 22 2 2

33 3 3

00

0

⎪+ + Δ = ⎬⎪+ Δ = ⎭

(b)

Obviously, the force-method equations are now decomposed into two sets. The first set only involves sym

eq n ( loa gs. cted to a

sym

metric unknowns 1X and 2X , whereas the second set only antisymmetric unknown 3X . The free terms in uatio b) can be also simplified according to the symmetry of dinSymmetric structures subjected to symmetric loadings: when a symmetric structure is subje

metric loading the bending moment diagram of the symmetric primary structure due to the external loading is symmetric as well. As shown in Fig.10.22 (a), the bending moment diagram of PM is symmetric, while the bending moment diagram of 3M is antisymmetric [Fig.10.21 (d)]. In the circumstance, the free term 3PΔ will be

33 0P

PM M ds

EIΔ = =∑∫

Recalling from equation (b), the antisymmetric unknown must be equal to zero, i.e., Only tri eeded t

3X 3 0X = . the symmetric unknowns 1X and 2X exerting on the symme c section [Fig.10.22 (b)] are n o

determine.


Symmetric structures subjected to antisymmetric loadings: For the case that a symmetric structure is subjected to an antisymmetric loading, the bending moment diagram of the symmetric primary structure due to the external loading is also antisymmetric. As shown in Fig.10.22 (c), the bending moment diagram of PM is antisymmetric, while the bending moment diagrams of 1M and 2M are symmetric [Fig.10.21 (b) and (c)]. Therefore, the free terms 1PΔ and 2PΔ will vanish

1M MΔ = =1 0P

P dsEI∑∫

22 0P

PM M ds

EIΔ = =∑∫

It is noted from the first two equations of expression (b) that the symmetric unknown and must be eq

s 1X 2Xual to zero, i.e., 1 0X = and 2 0X = . Only the antisymmetric unknown 3X exerting on e sy etric

section [Fig.10.22 (d)] ed to te. Generally, when a symmetric structure i

th mm is need evalua

s subjected to a loading that is symmetric with respect to the

diagramPM

diagramPM

(a)

(d)2P

(b)

(c)

1 2 0X X= =

2P

2P

A

2P

2P

1X

2X 2X

3 0X =

2P

2P

2P

3X

1 2

Fig.10.22 Symmetric primary structure under action of symmetric and antisymmetric loadings(a) symmetric primary structure under action of symmetric loading; (b) only symmetric unknowns and ; (c)

X X

3

symmetric primary structure under action of antisymmetric loading;(d) only antisymmetric unknown X


struc

metric with respect to the struc

t half a structure to analyze

o symmetric loadings

f a symmetric structure under action of a sym

to a loading that is symmetric with respect to the fram

ture's axis of symmetry, the response of the structure (i.e., reactions, internal forces and displacements) is also symmetric, with the point(s) of the structure at the axis of symmetry neither rotating (unless there is a hinge at such a point) nor deflecting perpendicular to the axis of symmetry [Fig.10.20 (b)]; thus, in the symmetric system, there is no antisymmetric unknown(s) on the symmetric axis and only symmetric unknowns are existed on the symmetric axis and they are needed to determine.

When a symmetric structure is subjected to a loading that is antisymture's axis of symmetry, the response of the structure (i.e., reactions, internal forces and displacements)

is also antisymmetric, with the point(s) of the structure at the axis of symmetry not deflecting in the direction of the axis of symmetry [Fig.10.20 (c)]; thusly, in this system, there is no symmetric unknown(s) on the symmetric axis and only antisymmetric unknown(s) is existed on the symmetric axis and it is needed to determine.

10.6.3 Selec

(1) Frames with odd-number spans ① Symmetric structures subjected tFor emphasis, we rewrite, in italic form, the performance o

metric loading. When a symmetric structure is subjected to a loading that is symmetric with respect to the structure's axis of symmetry, the response of the structure is also symmetric, with the points of the structure at the axis of symmetry neither rotating (unless there is a hinge at such a point) nor deflecting perpendicular to the axis of symmetry. Thus, to determine the response (i.e., member forces and deformations) of the entire structure, we need to analyze only half the structure, on either side of the axis of symmetry, with symmetric boundary conditions (i.e., slopes must be either symmetric or zero, and deflections perpendicular to the axis of symmetry must be zero) at the axis. The response of the remaining half of the structure can then be obtained by reflection.

Consider a symmetric portal rigid frame subjected e's axis of symmetry as an example, shown in Fig.10.23 (a). The deflected shape (elastic curve) of the

frame is also shown in the figure. It can be seen that, like the loading, the deflected shape is symmetric with respect to the axis of symmetry of the frame. Note that the slope and the horizontal deflection are zero at point C, where the axis of symmetry intersects the frame, whereas the vertical deflection at C is not zero. The response of the entire frame can be determined by analyzing only half the frame, on either side of the axis of symmetry. The left half of the frame cut by the axis of symmetry is shown in Fig.10.23 (b). Note that the symmetric boundary conditions are imposed on this substructure by supporting it at the end C by a double link support, which prevents the rotation and the horizontal deflection at the axis of symmetry but cannot prevent the vertical deflection along the axis. Once the response of the left half of the frame has been determined by analysis, the response of the right half can be obtained from that of the left half by


reflection (or mirror image). ② Symmetric structures subjected to antisymmetric loadings

For the same reason of emphasis, we rewrite, in italic form, the performance of a symmetric structure unde

ymmetric frame subjected to a loading that is antisymmetric with respect to the frame's axis of symm

r action of an antisymmetric loading. When a symmetric structure is subjected to a loading that is antisymmetric with respect to the structure's axis of symmetry, the response of the structure is also antisymmetric, with the points of the structure at the axis of symmetry not deflecting in the direction of the axis of symmetry. Thus to determine the response of the entire structure, we need to analyze only half the structure, on either side of the axis of symmetry, with antisymmetric boundary conditions (i.e., deflections in the direction of the axis of symmetry must be zero) at the axis. The response of the remaining half is given by the negative of the reflection (or mirror image) of the response of the half structure that is analyzed.

Take a setry as an example, shown in Fig.10.24 (a). It can be seen that, like the loading, the deflected shape of

(a)

A B

CD E

(b)q q

DC

A

C′

Fig.10.23 Odd-number-span rigid frame under action of symmetric loading(a) a portal rigid frame subjected to symmetric loading; (b) equivalence of half the frame;

(a)A B

CDE

P

(b)

C′

P D C

A

P

Fig.10.24 Odd-number-span rigid frame under action of antisymmetric loading(a) a portal rigid frame subjected to antisymmetric loading; (b) equivalence of half the frame;


the frame is antisymmetric with respect to the frame's axis of symmetry. Note that the vertical deflection is zero at point C, where the axis of symmetry intersects the frame, whereas the horizontal deflection and slope at C are not zero. The response of the entire frame can be determined by analyzing only half the frame, on either side of the axis of symmetry. The left half of the frame cut by the axis of symmetry is shown in Fig.10.24 (b). Note that the antisymmetric boundary conditions are imposed on this substructure by supporting it at end C by a roller support, which prevents the vertical deflection at the axis of symmetry but cannot prevent the horizontal deflection and rotation at C. Once the response of the left half of the frame has been determined by analysis, the response of the right half is given by the negative of the reflection of the response of the left half.

(2) Frames with even-number spans symmetric loadings

ected to symmetric loading, as shown in Fig.

supporting it at the end C by a fixed support. Once the response of the left half of the frame has been

ings to an antisymmetric loading as an

exam

① Symmetric structures subjected toConsider an even-number-span symmetric rigid frame subj10.25 (a). The left half of the frame with symmetric boundary conditions is shown in Fig. 10.25 (b). As

this figure indicates, if we neglect the axial deformation of column CD (in fact, the members located along the axis of symmetry will undergo only axial deformations without bending), all of the displacements at joint C have been restrained. Thus, the symmetric boundary conditions are imposed on this substructure by

determined by analysis, the response of the right half is obtained by reflection (or mirror image). However, it should be noted that there is an axial force in member CD (that is located along the axis of symmetry) and it can be determined by appropriate equilibrium equations.

② Symmetric Structures Subjected to Antisymmetric LoadTake a six-degree-indeterminate symmetric rigid frame subjectedple, as shown in Fig.10.26 (a). The structure contains a member along the axis of symmetry. If we

(a)

A B

C

D

E

(b)q

FE C

A

q

Fig.10.25 Even-number-span rigid frame under action of symmetric loading(a) a rigid frame subjected to symmetric loading; (b) equivalence of half the frame;


halve the properties of the member, cross-sectional inertia moment I and area A (if there are any concentrated loads and couples acting on the structure’s axis of symmetry, their magnitudes should be halved as well), as shown in Fig.10.26 (b), only a pair of shear forces equal in magnitude but opposite in direction should be imposed on the cut (at points 1C and 2C ). The half structure selected for analysis will be simplified as the structure shown in Fig.10.26 (c).Th ntisymmetric boundary conditions are thusly imposed on this substructure by supporting it at end 1C by a roller support, which prevents the vertical deflection at the axis of symmetry but cannot prevent th orizontal deflection and rotation at 1C .

Note that the roller support just affects the magnitude of axial force in member CD tha is al

e a

e ht ong the

axis

summ is of simplification for symmetric structures as follows: into two

sets;

of symmetry and cannot undergo any axial deformations and has no any axial forces. Therefore, the half structure selected for analysis may be further simplified as portal rigid frame shown in Fig.10.26 (d). Once the response of the left half of the frame has been determined by analysis, the response of the right half is given by the negative of the reflection (or mirror image) of the response of the left half. However, it should be mentioned that there exist bending moments and shear forces in member CD that is located along the axis of symmetry and the internal forces can be obtained by doubling the bending moments and shear force in member 1 1C D [Fig.10.26 (d)].

Now we will arize the analys(1) Select symmetric primary system, and then the primary unknowns will be decomposed

the first set only involves symmetric unknowns, whereas the second set only antisymmetric unknown(s). Then, the secondary flexibility coefficients relating to the above two sets are equal to zero, i.e.,

0ijδ = ; the subscript i denotes the location and direction of a symmetric unknown, while the subscript j

nts that of an antisymmetric unknown. Consequently, a set of multi-unknown linear equations (force-method equations) will become two sets of few-unknown linear equations. For different loading conditions, there exist following three cases:

① When loading is symmetric as w

represe

ell, only the symmetric unknowns are needed to determine (anti

ly the antisymmetric unknown(s) is needed to determine (sym

e loading may be decomposed into superposition of a symmetric load n

alf a structure to analyze. It should be mentioned again that the internal forces and displ

symmetric unknown(s) are equal to zero). ② When loading is antisymmetric, onmetric unknowns are equal to zero). ③ When loading is asymmetric, th

ing a d an antisymmetric loading. The response (reactions, internal forces and displacements) of the structure can be obtained by using principle of superposition. Or you may just employ symmetric primary system and symmetric primary unknowns, which will simplify the force-method equations, to analyze the structure directly.

(2) Select hacements in the member that is located along the axis of symmetry of a symmetric structure with odd


number of span(s) are different from those of a symmetric structure with even number of spans under action of symmetric or antisymmetric loadings. If the half a structure is selected to analyze, the loadings must be symmetric or antisymmetric. Any general loading or asymmetric loading can be decomposed into symmetric and antisymmetric components with respect to the structure’s axis of symmetry. Once the response of half the structure has been determined by analysis, the response of the entire structure can be given by using the principle of superposition.

(a)

Example 10-6

Analyze the portal rigid frame shown in Fig.10.27 (a) and draw the bending moment diagram for the frame and then discuss the relation of variation with the change of ratio k between the flexural rigidity of the beam and that of the columns.

Solution

(1) Analysis of symmetry

B

A

B

C

D

E

2D

(b)

(d)(c)

F

P PC′ E

A A

F

2l

2l

1C

1D

2C

P E

1D

2l

1C

2l

EP 1C

1DA

Fig.10.26 Even-number-span rigid frame under action of antisymmetric loading(a) a rigid frame subjected to antisymmetric loading; (b) halved portions of the frame; (c) equivalence of the left half of the frame; (d) final equivalence of the left half of the frame


The frame is symmetric with respect to the vertical axis passing through the middle of the beam and the loading is asymmetric, which can be decomposed into symmetric and antisymmetric components as shown in Figs.10.27 (b) and (c).

For the case shown in Fig.10.27 (b), if neglecting the axial deformation of the beam the two columns are inactive (zero force member), only the beam is subjected to an axial force . The stressing and deforming states not only satisfy the equilibrium conditions but also the compatibility conditions of the displacements, so the internal force state is the real stressing state. In the circumstance, the bending moment diagram of the entire frame is the same as that of the frame only subjected to the antisymmetric loading as shown in Fig.10.27 (c), which can be obtained by subsequent steps.

/ 2N P= −

(2) Primary system In this example, the symmetric primary system is planed to use. Under action of antisymmetric

loading, the symmetric primary system is shown in Fig.10.27 (d). Since bending moment and axial force on the cut (passing through the intersection point of the axis of symmetry and the beam) are symmetric unknowns, only antisymmetric shear force is imposed on the cut. 1X

(3) Force-method equation (compatibility condition) Noting that the relative vertical displacements at the two ends of the cut of the primary system due to

diagramPM

1 diagramM

(f)

(a)

(d) (e)

l

1EI

2Ph

P

(b) (c)

1EI

2EI

h2P

2P

1X

2P

2P

2P

2P

2l

2Ph

2l

2l

2l

Fig.10.27 Figures of example 10-6(a) a rigid frame subjected to asymmetric loading; (b) the rigid frame subjected to symmetric loading;(c) the rigid fr

1

ame subjected to antisymmetric loading; (d) symmetric primary system;(e) primary structure under action of external loading; (f) primary structure under action of 1X =

1 1X =

2P

2P


the combined effects of the given external loading and the unknown redundant must be zero, we write the force-method equation as

1X

11 1 1 0PXδ +Δ =

(4) Flexibility coefficient and free term Draw, respectively, the bending moment diagrams, PM due to the external loading and 1M due to

of the primary structure as shown in Figs.10.27 (e) and (f). By employing the graph-multiplication method, we write

1 1X =

2 3

111 2

1 12 ( ) ( )2 2 2 2 2 3 2 12l l l l l l h lh

EI EI EI EIδ

⎡ ⎤= × ⋅ × + × × × × = +⎢ ⎥

⎣ ⎦ 1 2

2

11 1

12( )2 2 2 4P

Ph l Ph lhEI EI

Δ = × × × =

(5) Solve the force-method equation By substituting the expressions for the coefficient and free term into the force-method equation, we

determine the redundant to be

11

11

66 1 2

P k PhXk lδ

Δ= − = −

+

In which,

2

1

I hkI l

=

(6) Construct bending moment diagram By using the superimposing equation for bending moments 1 1 PM M X M= + , we draw the bending

moment diagram as shown in Fig.10.28 (a). (7) Discussion ① When the flexural rigidity of the beam is much smaller than that of the columns, i.e.,

, their ratio k vanish, i.e., . It implies that the beam is so weak that it can not resist the rotation at the joints caused by the deflection of the columns. Consequently, the top ends of the columns behave like hinged ends. The inflection points, where the bending moment is equal to zero, are at the top of the columns. In the circumstances, the bending moment diagram is shown in Fig.10.28 (b).

2EI EI1 0k →

② When the flexural rigidity of the beam is much greater than that of the columns, i.e., , their ratio k become infinite great, i.e., . It means that the beam is strong enough to hold the rotation at the joints. The restraint conditions at the two ends of the column are identical, thus the inflection

2 1EI EIk →∞


point locates at the mid-height of the columns. In this situation, the bending moment diagram is shown in Fig.10.28 (d).

③ In general cases, as the bending moment diagram shown in Fig.10.28 (c) indicates, the inflection point will locate somewhere between the top end and the mid-height of the column. When , we can approximately realize that the inflection point is at the mid point of the column. At the inflection point, the bending moment is equal to zero and the shearing force is equal to .

3k =

/ 2P

Example 10-7

Analyze the statically indeterminate rigid frame shown in Fig.10.29 (a) and draw the bending moment diagram for the frame.

Solution

(1) Analysis of symmetry

66 1 4

k Phk

⋅+

(a)

(d)

P

(b)

(c)

6 26 1 4k Phk+

⋅+

2Ph

P

k vanish

P P

2019 4

Phi

1819 4

Ph⋅ 4

Ph

3k =

4Ph

becomes infinite greatk

Fig.10.28 Bending moment diagrams of example 10-6(a) bending moment diagram in a general case; (b) bending moment diagram in the case of 0;(c) bending moment diagr

k →am in the case of 3; (d) bending moment diagram in the case of k k= →∞


In this example, we are planed to show how selecting the half of a symmetric structure for analysis. As shown in Fig.10.29 (a), the frame is symmetric with respect to the vertical axis that coincides with the centroidal line of member BE, and its loading is also symmetric. The left half of the frame is selected for analysis, as shown in Fig.10.29 (b). Note that the symmetric boundary conditions are imposed on this substructure by supporting it at the end E by a fixed support for the reason that the axial deformation of member BE is small enough to be ignored.

(2) Primary system As shown in Fig.10.29 (b), the substructure is indeterminate to the second degree. The primary system

may be selected as the three hinged frame shown in Fig.10.29 (c). The primary unknowns, and , are end bending moments of member DE as shown in the figure.

1X 2X

(3) Force-method equations (compatibility conditions) Noting that the relative rotation at the two positions of the primary system due to the combined effects

of the given external loading and the unknown redundants and must be zero, the force-method equations will be written as

1X 2X

1 diagramM diagram (kN m)

PM⋅

diagram (kN m)

M⋅

2 diagramM

(a)

(d) (e) (f) (g)

(b) (c)

2kN

/m

2kN

/m

6m 6m

D E

I

2I

D E

2kN

/m

A B C

D E F

4mI I I

2I2I X X

1

2kN

/m 1 2

A A

2 1X =

1 1X =

D E D E

A A A

D E

2kN

/m

4 4 4

A B C

DE

F

2.72 2.72

1.281.28

2.56

2.562.562.56

2.56

1

Fig.10.29 Figures of example 10-7(a) original rigid frame; (b) equivalence of half the frame; (c) primary system; (d) primary structure under action of 1X = 2; (e) primary structure under action of 1;(f) primary structure under action of external loading; (g) bending moment diagram

X =

1

1


11 1 12 2 1 0PX Xδ δ+ +Δ =

21 1 22 2 2 0PX Xδ δ+ +Δ =

(4) Flexibility coefficients and free terms Draw, respectively, the bending moment diagrams, PM due to the external loading, 1M due to

and 1 1X = 2M due to of the primary structure as shown in Figs.10.29 (d) through (f). By using the graph-multiplication method, we write

2 1X =

11 12 227 1, ,

3 21

EI EIδ δ δ= = =

EI

1 216 , 0

3P PEIΔ = Δ =

(5) Solve the force-method equation By substituting the expressions for the coefficients and free terms into the force-method equations, we

determine the redundants to be

1 2.56kN mX = − ⋅

2 1.28kN mX = ⋅

(6) Construct bending moment diagram By using the superimposing equation for bending moments 1 1 2 2 PM M X M X M= + + , we draw

the bending moment diagram as shown in Fig.10.29 (g).

Example 10-8

Analyze the statically indeterminate rigid frame shown in Fig.10.30 (a) and draw the bending moment diagram for the frame.

Solution

(1) Analysis of symmetry The purpose of this example is further to show how selecting the half of a symmetric structure for

analysis. As shown in Fig.10.30 (a), the frame and its loading have two axes of symmetry, x-x and y-y. The one quarter of the frame is selected for analysis, as shown in Fig.10.30 (b). Note that the symmetric boundary conditions under the action of the symmetric loading are imposed on this substructure by supporting it at the ends B and D by a double-link support, respectively, for the reason that only the translations of points B and D along the axes of symmetry will occur.

(2) Primary system


As shown in Fig.10.30 (b), the substructure is indeterminate to the first degree. The primary system may be selected as the frame shown in Fig.10.30 (c). The primary unknown, , is end bending moment of member CD as shown in the figure.

1X

(3) Force-method equation (compatibility condition) Noting that the slope at support D of the primary system due to the combined effects of the given

external loading and the unknown redundant must be zero, the force-method equation is thusly written as 1X

11 1 1 0PXδ +Δ =

(4) Flexibility coefficient and free term Draw, respectively, the bending moment diagrams, 1M due to 1 1X = and PM due to the

external loading of the primary structure as shown in Figs.10.30 (d) and (e). By using the graph-multiplication method, we find

1 diagramM

21

1

1 2

1 2

8

PlEI

l lEI EI

⎛ ⎞+⎜ ⎟

⎝ ⎠

diagramPM

(a)

(d) (e)

(f)

(b) (c)

x

1EI2EI2EI

2EI

1EI 1EI PP xx

A

B

C D Ey

y

2

2l

1 2l 1 2l

2l

B

C D

2P

C

B2P

D

1X

1 1X =

1

11

1 2P 1

4Pl

y

y

A

B

C

D1 2l

Fig.10.30 Figures of example 10-8(a) a rigid frame with two axes of symmetry; (b) one quarter structure of the frame;(c) primary system; (d) primary structure unde 1r action of 1; (e) primary structure under action of external loading; (f) bending moment diagram of original frame

X =

x2

1 1 2

1 2

1 2

1 2

2

8

Pl Pl lEI EI

l lEI EI

+

⎛ ⎞+⎜ ⎟

⎝ ⎠


1 211

1 22 2l lEI EI

δ = +

21

1116P

PlEI

Δ =

(5) Solve the force-method equation (compatibility condition) By substituting the expressions for the coefficient and free term into the force-method equation, we

obtain 2

1

11

1 2

1 2

8( )

PlEIX l l

EI EI

= −+

(6) Construct bending moment diagram First, draw the bending moment diagram of one quarter of the frame, as shown in the top left corner in

Fig.10.30 (f) by using the superimposing equation for bending moments 1 1 PM M X M= + ; then, by employing the property of symmetry of the frame, the final bending moment diagram will be finished as shown in Fig.10.30 (f).

10.7 Statically Indeterminate Arches

Statically indeterminate arches are the types of structures which are widely adopted in engineering structures. Such as the Chinese Zhaozhou Bridge [Fig.10.31] that was made of squared stone and designed

by Chinese craftsman, Lichun, in AD 605, it is a classical example of utilization of arches in bridges. Nowadays, double arch bridges are often selected in bridge structures. Arches are also commonly used in roof structures of buildings. Fig.10.32 (a) shows a roof structure that is reinforced concrete arch with a steel tie. Its computing model is shown in Fig.10.32 (b). The lining of tunnels used underground engineering and water conservancy engineering as shown in Figs.10.32 (c) and (d) are sorts of arches as well.

In practical engineering, most of statically indeterminate arches are two-hinged arches or hingeless arches. Their analytical models are shown in Fig.10.33. Figs.10.33 (a) and (b) show, respectively, a two-hinged arch with no tie and a two-hinge arch with a tie; whereas Figs.10.33 (c) and (d) depict, respectively, a hingeless arch and a closed-loop arch that may be realized as a kind of a special hingeless

Fig.10.31 Zhaozhou Bridge

10.7 Statically Indeterminate Arches 327

arch.

(a)

arch ring

side wall

inverted arch

(d)

(b)

(c)

12φ

2L90 60 10× ×

15.6m

25cm

50cm

Fig.10.32 arch structures(a) an arched roof with a tie; (b) computing model of arches with ties;(c) arched roof of tunnels; (d) double shaped lining of a tunnelU

2.60

m

(a)

(d)

(b)

(c)

Fig.10.33 two-hinged arches and hingeless arches(a) a two-hinged arch; (b) a two-hinged arch with a tie;(c) a hingeless arch; (d) a ring shaped structure


10.7.1 Analysis of two-hinged arches

(1) Two-hinged arches ① Primary systems of two-hinged arches A two-hinged arch [Fig.10.34 (a)] is a single degree indeterminate structure, so if the horizontal

reaction at hinged support B is selected to be the redundant , the primary system will be obtained by replacing the hinged support of the given two-hinged arch by a roller support as shown in Fig.10.34 (b), which now is a curved beam.

1X

② Force-method equation (compatibility condition) Because the horizontal displacement at support B of the actual indeterminate arch is zero, the algebraic

sum of the displacements of the primary structure (a curved beam) at B due to the redundant and external loading must also be zero. Thus the compatibility equation can be written as

1X

11 1 1 0PXδ +Δ = (a)

③ Determination of flexibility coefficient and free term Since the simply supported arch is a curved member, so we cannot use graph-multiplication method

given previously to calculate the flexibility coefficient 11δ and free terms 1PΔ in the compatibility

ϕ

(a)

(d)

y

(b)

(c)

1XA A BB

ϕN

M

AB

ACy x

x0

1

0

1 1X =

C

1Q

0

1

Fig.10.34 a two-hinged arch(a) a two-hinged arch; (b) primary system; (c) primary structure under action of 1; (d) free body diagram of segment X AC=


equation. Therefore, we will use integrate method, discussed in section 9.4, for determining the coefficient and free term. Noting that the ratio between the depth (indicated by h) and the rise (represented by f) of an arch is generally less than 0.5, i.e., 1

2hf< , so the effect on the coefficient and free term due to axial

deformation is negligible (when 12

hf> , axial deformation must be considered). So in general, only

bending deformation is considered when calculating the free term 1PΔ , and bending deformation and

sometimes axial deformation are taken into account when determining 11δ . And the shear deformation is

ignored. Thus, we rewrite the following expressions: 2 2

1 111

11 0P

P

M Nds dsEI EA

M M dsEI

δ⎫

= + ⎪⎪⎬⎪Δ = = ⎪⎭

∫ ∫

∑∫ (b)

Assume that the origin of the rectangular coordinate system used here is oriented at point A, the abscissa and ordinate of an arbitrary cross section C of the arch are, respectively, denoted by x and y as shown in the figure. ϕ is the angle formed by the tangent to the centre line of the arch and the axis of abscissa, which is positive in the left portion and negative in the right portion of the arch. The bending moment M has positive sign when it makes the inner fibers of the arch in tension, and axial force N has positive sign when it is a tensile force.

The expressions for the bending moment 1M and axial force 1N due to [Fig.10.34 (c)] on an arbitrary cross section C [Fig.10.34 (d)] of the primary structure (a simply supported curved beam) can be written, by equilibrium conditions, as follows

1 1X =

1

1 cosM yN ϕ

⎫= − ⎪⎬

= − ⎪⎭ (c)

Under the action of vertical loading, the bending moment PM of the simply supported curved beam (the primary structure) is identical to that, denoted by 0M , of a corresponding simple beam covering the same span and carrying the same loading. That is,

0PM M= (d)

By substituting the expressions (c) and (d) into expression (b), we obtain 2 2

11

0

1

cos

P

y ds dsEI EA

M y dsEI

ϕδ⎫

= + ⎪⎪⎬⎪Δ = − ⎪⎭

∫ ∫

∫ (10-13)


④ Solve the force-method equation (compatibility condition) By substituting the expressions for the coefficient and free term into the force-method equation, the

horizontal thrust H or the redundant will be obtained. That is, 0

11 2 2

11 cosP

M y dsEIX H

y ds dsEI EA

ϕδΔ

= = − =+

∫

∫ ∫ (10-14)

⑤ Calculation of internal forces After the thrust H has been obtained the internal forces of a two-hinged arch can be determined by the

formulation used for determining the internal forces of a three hinged arch. Under the action of vertical loading, the equations for calculating internal forces are rewritten as follows

0

0

0

cos sinsin cos

M M HyQ Q HN Q H

ϕ ϕ

ϕ ϕ

⎫= −⎪

= − ⎬⎪= − − ⎭

(10-15)

If an internal force diagram is desired, it may be obtained by first dividing the curved axial centre line of the arch into several segments, then calculating the internal forces of each ends of the segments, finally taking the values of the internal forces as ordinates of the curved axial line and connecting them into a smooth curve.

⑥ Summary of analyzing formulation and the characteristics of internal forces of two-hinged arches a. From the viewpoint of force method, the analyzing procedure for a two-hinged arch is the same as

that for a two-hinged rigid frame except that the method used to calculate flexibility coefficient 11δ and free terms 1PΔ in the compatibility equation is different. In other words, for a two-hinged rigid frame, graph-multiplication method can be employed to determine 11δ and 1PΔ , while for a two-hinged arch, integrate method must be used to determine 1PΔ and 11δ ; whereas in computing 11δ the axial deformation has to be taken into account sometime.

b. From the viewpoint of characteristics of internal forces, the internal forces of a two-hinged arch are almost same as that of a three-hinged arch except that the horizontal thrust H is different. For a three hinged arch, the thrust H is determined by an equilibrium condition, whereas for a two hinged arch, the thrust H must be determined by a compatibility condition.

(2) Two-hinged arches with ties Two hinged arches with ties are often applied to roof structures of buildings as shown in Fig.10.35 (a).

The function of the tie, which must posses enough stiffness, of a two hinged arch is to resist the horizontal reaction induced by the arch so as to relieve the thrust acting on the wall or column(s) of a building, and to


reduce the bending moment in the arch. ① Primary systems of two-hinged arches As shown in Fig.10.35 (a), a two-hinged arch with a tie is a single degree indeterminate structure too,

so if the axial force of the tie is selected to be the redundant , the primary system will be obtained by cutting the tie of the given arch as shown in Fig.10.35 (b), which now is a curved beam.

1X

(a) (b)

1X

EI

1 1E A

EA

Fig.10.35 a two-hinged arch with a tie(a) a two-hinged arch with a tie; (b) primary system

② Force-method equation (compatibility condition) Since the horizontal relative displacement at the cut of the actual tie is zero, the algebraic sum of the

relative displacements at the left and right sections adjacent to the cut of the primary structure (a curved beam) due to the redundant and external loading must also be zero. Thus the compatibility equation can be written as

1X

11 1 1 0PXδ +Δ = (e)

③ Determination of flexibility coefficient and free term As shown in Fig.10.35 (b), the primary curved beam is a curved member, so we cannot use

graph-multiplication method to calculate the flexibility coefficient 11δ and free terms 1PΔ in the compatibility equation. Therefore, we will use integrate method for determining the coefficient and free term. Note that only bending deformation is considered when calculating the free term 1PΔ , bending deformation and sometimes axial deformation are taken into account when determining 11δ . Thus, we write the following expressions:

2 2 21 1 1

111 1

M N Nds ds dxEI EA E A

δ = + +∫ ∫ ∫ (f)

In which, the first two items are the integer with respect to the arch, while the last item is the integer with the tie. is the axial rigidity of the tie. 1 1E A

Since the axial force in the tie due to 1 1X = is equal to a unit magnitude, for a tie with a length of , the third integer of equation (f) will be equal to l


2 21

0 01 1 1 1 1 1

1l lN dx dx lE A E A E

= =∫ ∫ A (g)

Because expressions of the bending moment and axial force for a two hinged arch with a tie is identical to that for a two hinged arch with no tie, the expression for determining the coefficient 11δ can be written, by substituting equations (c) and (g) into equation (f), as

2 2

111 1

cosy lds dsEI EA E A

ϕδ = + +∫ ∫ (h)

As long as there is no axial force in the tie of the primary curved beam due to external loading, the expression for the free term 1PΔ can be written as

01

1P

PM M M yds ds

EI EIΔ = = −∫ ∫ (i)

④ Solve the force-method equation (compatibility condition) By substituting the expressions for the coefficient and free term into the force-method equation, the

redundant or the tensile force in the tie will be expressed as 0

11 2 2

11

1 1

cosP

M y dsEIX

y lds dsEI EA E

ϕδΔ

= − =+ +

∫

∫ ∫ A

(10-16)

⑤ Calculation of internal forces The expressions [Eq. (10-15)] used to determine the internal forces of a two hinged arch with a tie are

the same as that used to determine the internal forces of a two hinged arch with no tie if changing thrust H into the tensile force . That is 1X

01

01

01

cos sin

sin cos

M M X y

Q Q X

N Q X

ϕ ϕ

ϕ ϕ

⎫= −⎪

= − ⎬⎪= − − ⎭

(10-17)

It is observed, from comparing the expression used to determine the tensile force of the tie, [Eq.

(10-16)], with that used to determine the thrust H [Eq. (10-14)], that the item

1X

1 1

lE A

is added in the

denominator of Eq. (10-16). Eq. (10-16) will approach to Eq. (10-14) provided that the axial rigidity of the tie goes to infinitely great, i.e., . In the circumstance, the internal forces in the two kinds of

arches mentioned above are almost the same. In the case that the axial rigidity of the tie vanishes, i.e.,

1 1E A →∞


1 1 0E A → , the tensile force in the tie approaches to infinitely small, i.e. , . In this situation, the two

hinged arch is actually a simple curved beam.

1 0X →

Example 10-9

Determine the horizontal thrust of the two hinged arch due to the action of two concentrated loads as shown in Fig.10.36 (a), which possesses a constant cross section and a conic parabola centre line that may be expressed as

2

4 (f )y x l xl

= −

Solution

(1) Assumptions for simplification The example adopts two assumptions: ① Only bending deformation is taken into account whereas

the axial deformation is ignored. ② The curvature of the arch can be neglected for flat arch (in the case of 15

fl< ). That is, the differential length of a curved line ds is equal to that of a straight line dx i.e., .

The expressions for determining the coefficient and free term in the compatibility equation can be thusly written as follows

ds dx=

211 0

1 ly dx

EIδ = ∫

(a)

(d)

0 Fig.10.36 figures of example 10-9(a) a two-hinged arch; (b) diagram of a corresponding simple beam;(c) diagram of the two hinged arch; (d) diagram of

MM M a corresponding three-hinged arch

A BCD E

4l

Py

0.0034

(b)P

(c)

D C EA B

x

4l

4l

4l 4

P l4Pl

CD

0.042

0.0280.011 C

D E0.0156

0.06250.0156A B A B( )Pl× ( )Pl×


01 0

1 l

P M ydxEI

Δ = − ∫

(2) Primary system The primary system of the two-hinged arch is shown in Fig.10.34 (b), which actually is a curved

beam. (3) Calculation of flexibility coefficient and free term The bending moment expression, by employing the symmetric property of the arch, for the

corresponding curved beam can be expressed, separately in segment, as

Segment AD: 0 , 04lM Px x= < ≤

Segment DC: 0 , 4 4 2Pl l lM x= < ≤

After above preparation that is needed to determine 11δ and 1PΔ , we finally obtain 2 2

2 2 2 2 3 4 211 2 40 0 0

1 1 4 16[ ( )] ( 2 )15

l l l 8f f fy dx x l x dx l x lx x dxEI EI l EIl EI

δ = = − = − + =∫ ∫ ∫l

4 201 0 0 4

4 2

2 20 4

2

1 24

2 4 4 ( ) ( )4

19 128

l l l

P l

l l

l

PlM ydx Pxydx ydxEI EI

f f Plx l x Pxdx x l x dxEI l l

PflEI

⎡ ⎤Δ = − = − +⎢ ⎥⎣ ⎦⎡ ⎤= − − + −⎢ ⎥⎣ ⎦

= −

∫ ∫ ∫

∫ ∫

(4) Solve the force-method equation (compatibility condition) By substituting the expressions for the coefficient and free term into the force-method equation, we

obtain

1

11

0.278P PlHfδ

Δ= − =

(5) Calculate internal forces Taking the bending moment as an example, by superposing equation

0M M Hy= −

We draw the bending moment diagram as shown in Fig.10.36 (c). (6) Discussion

It can be observed from the computing result that the thrust ( 0.278Plf

) of a two hinged arch approaches


to that ( 0.25Plf

) of a corresponding three hinged arch, only 10% of relative difference, in the condition that

the axial deformation and curvature of a two hinged arch are ignored. Since the thrust of a two hinged arch is greater than that of a three hinged arch, the bending moment of the two hinged arch is smaller than that of the three hinged arch [Fig.10.36 (d)].

10.7.2 Symmetric hingeless arches

Symmetric hingeless arches are the structures which are commonly adopted in bridge structures. The computing model of this sort of structure is shown in Fig.10.37 (a), indeterminate to third degrees. If we select a symmetric primary system as shown in Fig.10.37 (b) the force-method equations may be expressed as

3X

11 1 12 2 13 3 1

21 1 22 2 23 3 2

31 1 32 2 33 3 3

000

P

P

P

X X XX X XX X X

δ δ δδ δ δδ δ δ

+ + + Δ = ⎫⎪+ + + Δ = ⎬⎪+ + + Δ = ⎭

(a)

The primary unknown is a pair of bending moments, a pair of axial forces and a pair of shear forces, so the following secondary coefficients will be zero in terms of the unknowns’ symmetric properties. That is,

1X 2X 3X

13 31

23 32

00

δ δδ δ

= = ⎫⎬= = ⎭

(b)

Substituting equation (b) into equation (a), we obtain the following two sets of equations

11 1 12 2 1

21 1 22 2 2

33 3 3

00

0

P

P

P

X XX X

X

δ δδ δ

δ

+ + Δ = ⎫⎪+ + Δ = ⎬⎪+ Δ = ⎭

(10-18)

(a) (b)

y

2X 2X1X 1X

3XD

Fig.10.37 symmetric hingeless arch (a) original structure; (b) primary system

x


In order to evaluate the coefficients in equation (10-18), the expressions of internal forces due to and must be first determined.

Assume that the origin of the rectangular coordinate system used here is oriented at the top of the arch, the abscissa and ordinate of an arbitrary cross section of the arch are, respectively, denoted by x and y as show

1 1X = , 2 1X = 3 1X =

n in Fig.10.37 (b). ϕ is the angle formed by the tangent to the centre line of the arch and the axis of , is pos the right portion and negaabscissa which in tive in the left portion of the arch. The sign

convitive

ention for internal forces is identical to that used previously. The expressions of internal forces due to

1 1X = , 2 1X = and 1X = are thusly expressed as follows [see Figs.10.38 (a) through (c)]. 3

31 2

1 2 3

1 2 3

10 cos sin0 sin cos

M xM M yN N N

Q QQϕ ϕ

ϕ ϕ

⎫⎫ ⎫ == =⎪⎪ ⎪

= =

When evaluate coefficients

− =⎬ ⎬ ⎬⎪ ⎪ ⎪= = =⎭ ⎭ ⎭

iiδ , ijδ and iPΔ , only bending deformation is taken into account

whereas the axial deformation is considered in th of e case 15

fl< (noting that f indicating the rise and l the

span of the arch). The expressions for determining the coefficients and free terms in the compatibility equa ll s tions can be thusly written as fo ow

21

11

1 212

1MEI EIM M dsδ

⎫

2 2 2 22 2

22

2 23

33

11

22

33

cos

P PP

P PP

P PP

ds ds

y dsEI EI

M N yds ds ds dsEI EA EI EAM xds dsEI EIM M Mds ds

EI EIM M M yds ds

EI EIM M M xds ds

EI EI

δ

ϕδ

δ

= = ⎪⎪⎪= = ⎪⎪⎪= + = + ⎪⎪⎪= = ⎬⎪⎪

Δ = = ⎪⎪⎪Δ = = ⎪⎪⎪Δ = = ⎪⎭

∫ ∫

∫

∫ ∫ ∫ ∫

∫ ∫

∫ ∫

∫ ∫

∫ ∫

(10-19)

∫


Example 10-10

Determine the horizontal thrust and bending moments on the sections of the top and skewback of the hingeless arc arch due to the action of uniformly distributed load as shown in Fig.10.39 (a).

Solution

etric primary system is selected as shown in Fig.10.39 (b), noting that the antisymmetric unkn

(1) Primary system The symmown (shear force) 3 0X = due to the symmetric loading. (2) Relation between the radius R and central angle 0ϕ and coordinates From the rectangular triangle [see Fig.10.39 (a)], we obtain O AE′

0ϕϕ

20kN/m

(a)

A B

C D(b)

E

8m8m

R0ϕ

4m

y

y

20kN/m

A B

R

xO′

2X2X1X 1X

O′

Fig.10.39 figures of example 10-10(a) original structure and loading; (b) primary system

constantEI =

x

1M

1N1Q

2M

2N2Q

(a)

1 1X =

(b) (c)

2 1X 3 1X ==

M 3

3N3Q

OOO

1 2 3

Fig.10.38 internal forces due to unit loa(a) under action of 1; (b) under action of 1; (c) underX X

ding action of 1X= = =


2 2( ) ( )2lR2

R f= + −

2 24l f+ 10m8

Rf

= =

02sin 0.8

lAEO A R

ϕ = = =′

， 0 0.6O E R fϕcosO A R′ −

= = = ′

Thusly

0 0.9273 radianϕ =

The relation between the radius R and angle ϕ and coordinates will be written as sinx R ϕ= ， y cosR R= ϕ−

Force-method equations Based on the primary system, we write the force-m

X X

(3)ethod equations as

011 1 12 2 1Pδ +δ +Δ

2 0PX X

=

21 1 22δ δ+ +Δ =2

coefficients (4) Calculation of flexibility 1 14 5

fl= > , only the bending deformation is concerned whDue to en calculating the flexibility

coefficients. So by equation (10-19), we write 02

11 1 002 2EI M ds Rd

ϕRϕ= = =∫ ∫ ϕ δ

[ ]

0

0 0

12 1 2 0

02

2 ( cos )

2 ( 2 2 20

0 0

cos ) 2 sin

2 ( sin )

EI M M ds yds R R Rd

R Rϕ

d R

R

ϕ

ϕ

δ ϕ ϕ

ϕ ϕ ϕ ϕ

ϕ ϕ

= −

= −

= = = −

= −

∫ ∫ ∫∫

0 0

0

2 2 2 3 222 2 0 0

3 20 0 0sin 2 )

40

2 2 ( cos ) 2 (1 2cos cos )

1 3 1 2 2sin ( sin 2 ) 2 ( 2sin2 4 2

I M ds y ds R R Rd R d

R R

ϕ

ϕ

ϕE δ ϕ ϕ ϕ ϕ ϕ

ϕϕ ϕ ϕ ϕ ϕ ϕ

= = = − = − +

⎡ ⎤= − + + = − +⎢ ⎥⎣ ⎦

∫ ∫ ∫ ∫

Substituting 0 0.9273 radianϕ = into above expressions, the coefficients can be evaluated as

11 1.855EI Rδ = ， 212 0.2546EI Rδ = ， 3

22 0.0619EI Rδ =

(5) Calculation of free terms


The bending moment expression of the primary structure due to the external loading can be expressed, by equilibrium equation, as

2 2 sin2 2Pq qM x R 2ϕ= − = −

quation (10-19), weBy using e find

0

03 1 sin 2

2 21 1 0

0

3 00

2 1 ( sin )2

2 41 ( sin 2

2 4

P Pq

)

EI M M ds R Rd

qR

ϕϕ

qRϕ

ϕ

ϕ ϕ

Δ = = × −

⎢ ⎥⎣ ⎦

= − −

∫ ∫ϕ ϕ⎡ ⎤= − −

0

0

0

2 22 2PEI M

0

4 2 2

0

4 3

0

4 300 0

2 ( cos )( sin )2

(sin sin cos )

1 1 sin 2 sin2 4 3

1 1 ( sin 2 sin )2 4 3

P PqM ds M yds R R R Rd

qR d

qR

qR

ϕ

ϕ

ϕ

ϕ ϕ ϕ

ϕ ϕ ϕ ϕ

ϕ ϕ ϕ

ϕ ϕ ϕ

= = − −

= − −

⎡ ⎤= − − −⎢ ⎥⎣ ⎦

= − − −

∫ ∫∫

Substituting

Δ = ∫

0 0.9273 radianϕ = into above expressions, the free terms can be written as 3

1 0.2237PEI qR 42 0.0530PΔ = − , EI qΔ = − R

(6) Solve the force-method equations By substituting the expressions for the coefficients and free terms into the force-method equations, we

write

1.855 0.2546 0.2237 0

0.

X RX qR ⎫+ − =21 2

2

⎪

1 22546 0.0619 0.0530 0X X qR⎬

+ − = ⎪⎭

Finally, we obtain

14.2kN mqR ⋅

) Calculation of bending moments on the sections of the top and skewback of the arch By equilibrium conditions, the bending moments on the sections of the t

will be evaluated as

21 0.0071X = =

2 0.827 165.4kNX qR= =

(7op and skewback of the arch


1 14.2kN mCM X= = ⋅

2 21 2 ( ) 14.2 165.4

2 2A Bq lM M X X f 20 164 ( ) 35.8kN m=

2 2= + − = +

If oth

rust of a three hinged arch with the same span and carrying the same loading, i.e.,

× − = ⋅

er internal forces are desired, they can be determined by equilibrium conditions as well. (8) Discussion Comparing the th

0 2M × 220 (16) 160kN8 8 4

C qlHf f

′ = = = =×

, with that of the hingeless arch, i.e., 1 165.4kNH X= = , it can be

observed that the two thrusts only have a relative difference of 3.4%, and they are very closed.

movemindeterminate structures and must b pport settlements, w

externally ions, can be alysis. And

the procedure for the analysis of structures subjected to support settlements and temperature changes is the same

l forces of the rigid frame shown in Fig.10.40 (a) due to a temperature change as

Solution

10.8 Internal Forces due to Temperature Changes and Support Settlements

Thus far, we have considered the analysis of structures with unyielding supports. However, support ents due to weak foundations and the like may induce significant stresses in externally

e considered in their designs. Unlike su hich affect only externally indeterminate structures, temperature changes may affect the stress conditions of and/or internally indeterminate structures. The force method, as developed in the preceding secteasily modified to include the effects of support settlements and temperature changes in the an

as used previously for the case of external loads. The only difference is that the primary structure is now subjected to the prescribed support settlements and temperature changes (instead of external loads) to evaluate its displacements at the locations of redundants due to these effects. The redundants are then determined by applying the usual compatibility conditions that the displacements of the primary structure at the locations of the redundants due to the combined effects of support settlements and temperature changes and the redundants must equal the known displacements at the corresponding locations on the actual indeterminate structure.

10.8.1 Temperature changes

As mentioned above, the procedure for the analysis of structures subjected to temperature changes is the same as used previously for the case of external loads. The procedure is illustrated by the following example.

Example 10-11

Determine the internaindicated in the figure.

10.8 Internal Forces due to Temperature Changes and Support Settlements 341

(1) Primary system The horizontal reaction at support B is selected to be the redundant. The primary rigid frame obtained

by replacing the hinged support by a roller support. Next, the primary rigid frame is subjected to the pres e changes and an unknown concentrated force so as to form a primary system as show

(2) Force-method equation (compatibility condition) The compatibility equation can be expressed as

cribed temperaturn in Fig.10.40 (b).

11 1 1 0tXδ +Δ = (a)

in which, 11δ denotes the horizontal displacement at support B due to a unit value of the redundant and denotes the horizontal displacement at support B of the primary rigid frame due to the temperature changes.

(3) Calculation of the flexibility coefficient and free term The approach for determining coefficient

1X

1tΔ

11δ rams

is the same as used previously for the case of external loads, so bending moment and axial force diag due to 1 1X = are drawn in Fig.10.40 (c) and (d). By graph-multiplication method, 11δ is determined to be

21

111 1 2 1 2 168( 6 6) 6 ( 8 6) 6

2 3 2 3M dsEI EI E

δ ⎡ ⎤= = × × × × + × × × × =⎢ ⎥⎣ ⎦∑∫

I

1

Fig.10.40 Figures of example 10-11(a) original structure and temperature changes; (b) primary system; (c) bending mement diagram due to 1; (d) axial force diagram due tX = 1o 1X =

050 C−

00 C050 C−

00 C2144000kN mEI = ⋅

0.00001α =

)(a

A

BC

8m

6m

050 C−

00 C

050 C− 00 C

(b)

A

BC 1X

1 diagram (unit: m)M

(c)6

6

A

B

C1 1X =

34

34

1 diagram(unit: kN)N

(d)

BC1 1X =1 1

A


1tΔ The approach for determining free term is different from that used previously for the case of external loads, but we can use Eq. (9-17b) in section 9.6 to determine it. Thusly, we rewrite

1 1 0tt

1M ds t N dsh

α αΔΔ = +∑ ∑∫ ∫ (b)

In which, represents the difference of the temperature changes between the inner and outer

fibers of

2 1t t tΔ = −

the members; 1 20 2

t tt += denotes the average changes of the temperature at the central line of

the members. They are evaluated to be 0 0 0

2 1 0 ( 50 ) 50t t t C C CΔ = − = − − =

0 0 00 1 2

1 1( ) (0 50 ) 252 2

t t t C C= + = − = − C

Finally, we find

1 1 0 1

50 1 1 3 ( 6 6 8 6) 25 (1 8 6)0.6 2 2 4

3187.5

tt M ds t N ds

hα α

α α

α

ΔΔ = +

= − × × + × × + ⋅ × × + ×

= −

∑ ∑∫ ∫

Note that the two terms in equation (b) represent virtual work done by the virtual force system, so the sign of each term will be determined by the sign of virtual work. Consequently, the first term in must be

negative because the virtual bending moment perform negative virtual work; whereas the virtual axial force performs positive work, so the second term is positive.

(4) Solve the force-method equation By substituting the expressions for the coefficient and free term into the force-method equation, we

find

1tΔ

1168 3187.5 0XEI

α− =

I1 18.97X E α= (c)

(5) Construction of internal force diagrams Since the primary structure is a statically determinate structure, the temperature changes will not

induce internal forces in the primary structure. All of the internal forces of the indeterminate rigid frame are thusly caused by its redundant. The bending moment and axial force equations can be written as

1 1

1 1

M M XN N X

⎫= ⎪⎬

= ⎪⎭ (d)


The i xial force diagrams are depicted as shown in Fig.10.41 (a) and (b).

thod equation(s) [see Eq. (a)] is symbolically the same as that formulated previously for the case of external loads, but the meaning of free term(s) is different, which now is the displacement(s), evaluated by Eq. (b), at the location(s) of redundant(s) due to the effect of temperature changes (instead of external loads).

(2) All of the internal forces of an indeterminate structure are caused by its redundant(s) [see Eq. (d)].

(3) It is observed from the computing result [see Eq. (c)] and internal force diagrams [see Fig.10.41] that the magnitude of the internal forces, caused by temperature changes, is not only in self-equilibrium but also in direct proportion to the absolute value of the flexural rigidity EI of the member. Therefore, selecting a large cross section for the member will not modify its stress condition caused by its temperature changes.

(4) It can be observed from Fig.10.41 (a) that the tensile stress due to temperature changes occurs on the side of the member where the temperature rise is lower. It is instructive to notice that the cracks may be occurred in the concrete member due to decrease of its temperature.

ements

p ate structures; they may, however, induce sign

i Fig he support B of the determinate beam undergoes a

small settlement , the beam moves as rigid body without bending—that is, they remain no stresses develop in the determinate beam. However, when the indeterminate beam of Fig.10.42 (b) is

bend ng moment and aThe characteristics of analyzing the internal forces of a statically indeterminate structure due to

temperature changes may be summarized, by considering the analytical procedure of above example, as follows

(1) The force-me

10.8.2 Support settl

Sup ort settlements do not cause any stresses in determinificant stresses in externally indeterminate structures, which should be taken into account when

designing externally indeterminate structures. Consider the determ nate and indeterminate beams shown in Fig.10.42. It can be seen from . 10.42 (a) that when t

BΔ straight. Thus,

diagram(unit: kN m)

M⋅

C BB

diagram(unit: kN)N

(a)A

(b)A

C

163.93

163.93

27.3227.32

20.49

20.49

Fig.10.41 internal force diagrams of example 10-11(a) bending moment diagram; (b) axial force diagram


subje ds,ike may induce

significant stresses in externally indeterminate structures and must be cSupport settlements, however, do not have any effect on the stress

ndeterminatcause such structures to displace and/or rotate as rigid bodies without changing their shapes. The force

the preceg procedure is illustrated by the following example.

Example 10-12

Determine the internal forces and draw their diagrams for the indeterminate beam shown in Fig.10.43

cted to a similar support settlement, it ben as shown in the figure; then, bending moments develop in the indeterminate beam. Therefore, support movements due to weak foundations and the l

onsidered in their designs. conditions of structures that are

internally i e but externally determinate. This lack of effect is due to the fact that the settlements

method, as developed in ding sections, can be easily modified to include the effect of support settlements in the analysis. The analyzin

A B(a) (b)A B

(a) due to a rotation at support A as shown in the figure. constantEI = .

Solution

1 diagramMθ

θ

diagramM diagramQ

(d) (e)

l

(a) (b) (c)

(f)

A BEI

EIA

B

A BEI

1X 1 1X =1

A

A B

lθ

B

3EIlθ

2l3EIθ

A B

θ

θ

θ

θ

1(a) rotation at support ; (b) first primary system; (c) primary beam under action of 1;(d) primary structure subjected

A X = to a support

(f) shear force diagram

Fig.10.43 figures of example 10-12

settlement; (e) bending moment diagram;

BΔ BΔ

1B 1B

Fig.10.42 effect due to support settlements(a) a determinate beam; (b) an indeterminate beam


(1) First primary system ① Primary system 1 The vertical reaction at support B is selected to be the redundant. The primary beam obtained by

removing the support forms a determinate beam. Next, the primary beam is subjected to the prescribed support rotation θ and an unknown concentrated force so as to form a primary system as shown in Fig.10.43 (b).

② Force-method equation (compatibility condition) By realizing that the deflection of the actual indeterminate beam at support B is equal to zero due to

the c inomb ing actions of the redundant and support settlement, we write the compatibility equation as

11 1 1 0cXδ +Δ = (a)

in which, 11δ denotes the vertical displacement at support B due to a unit value of the redundant 1X and

1cΔ denotes the vertical displacement at support B of the primary beam due to the support rotation θ . ③ Calculation of the flexibility coefficient and free term

aThe pproach for determining coefficient 11δ is the same as used previously for the case of external loads, so bending moment diagram due to 1 1X = is drawn in Fig.10.43 (c). By graph-multiplication method, 11δ is determined to be

2 31

111 1 2( )

2 3 3M ldx l l lEI EI EI

δ = = × × =∫

The approach for determining free term 1cΔ is different from that used previously for the case of external loads, but we can use Eq. (9-18) in section 9.7 to determine it. Thusly, we rewrite

1c RcΔ = −∑ (b)

Here, , R l c θ= = , thusly, equation (b) will be changed to be

1c lθΔ = −

The free term may be determined by the displacement shape formed by the given support rotation as shown in Fig.10. The direction of is opposite to , so it has a minus sign.

d equation

1cΔ

43 (d). 1cΔ 1X④ Solve the force-methoBy substituting the expressions for the coefficient and free term into the force-method equation, we

find

11 3 2

11

3/ 3

c l EIXl EI l

θ θδΔ

= − = = (c)

⑤ Construc of bending moment diagram Since the primary structure is a statically determinate structure, the support movement will not induce

tion


internal forces in the primary structure. All of the internal forces of the indeterminate structure are thusly caused by its redundant. The bending moment equation can be written as

1 1M M X= (d)

① Primary system 2 The moment at support A is selected to be the redundant. The primary beam obtained by changing the

fixed support of the indeterminate beam into a hinged support. Next, the primary beam is subjected to the prescribed support rotation

The bending moment diagram is constructed as shown in Fig.10.43 (e). ⑥ Shear force diagram With 1X determined, the shear force of the beam can be found by equilibrium conditions. The shear

force diagram is shown in Fig.10.43 (f). (2) Second primary system

θ and an unknown concentrated moment at support A so as to form a primary system as shown in Fig.10.44 (a).

② Force-method equation (compatibility condition) By realizing that the slope of the actual indeterminate beam at support A is equal to the rotation θ

due to the redundant, we write the force-method equation as

11 1Xδ θ= (e)

③ Calculation of the flexibility coefficient The approach for determining coefficient 11δ is the same as used above, so bending moment diagram

due to is drawn in Fig.10.44 (b). By graph-multiplication method, 1 1X = 11δ is calculated to be 21

111 1 2( 1)

2 3 3M ldx lEI EI EI

δ = = × × =∫

④ Solve the force-method equation By substituting the expression for the coefficient into the force-method equation, we obtain

13l XEI

θ= , 13EIX

lθ

= (f)

1 diagramM(a) A

X1 1X =1

AB B(b)θ

1

1

10-12nd primary system; (b) primary beam under action of 1X

Fig.10.44 other figures of example(a) seco =

10.9 Computation of Displacements of Statically Indeterminate Structures 347

⑤ Construction of internal force diagrams ystem 2 is different from primary system 1 their final bending moment and

shea

s external loads, as follows

s. (a) and (e)] may be different for different prim

(b) or by the geometric relation (if convenient) of the displacemewhich is actually a determinate structure.

The inforc

and (f)] and internal force diagrams [see Fig.1 e in

e absol

10.9 Computation of Displacements of Statically Indeterminat

of sated in terms of

principle of virtual work discussed in Chapter 9, can be appliedtically determ inate structures. Recalling for unit load

method, the equilibrium condition is the only condition thaso the unit force may be applied on any one of the possiblestruc n

ts c unit load method to an the analyzed indeterminate structure. The internal force com

ined by equilibrium equations alone. In fact, the basic consideration of force method used to analyze

selecting a proper statically determinate structure, named primary structure, obtained by removing enough redu hen determining internal forces of the original inde

Although the primary sr force diagrams of the indeterminate beam are identical as those shown in Fig.10.43 (e) and (f). The characteristics of analyzing the internal forces of a statically indeterminate structure due to

support settlement may be summarized, by comparing the analytical procedure due to support movements with that due to

(1) The form of force-method equation(s) [see Eqary system. That is, the right hand side of the equality may not be equal to zero. (2) There are two approaches to determine the free term in Eq. (a). It can be determined by Eq.

nt configuration of the primary structure,

(3) re are no external loads exerting on the determinate primary structure, so all of the ternal es are caused by the redundant(s) [see Eq. (d)]. (4) It is observed from computing result [see Eqs. (c) 0.43 (e) and (f)] that the magnitude of th ternal forces, caused by support settlements, is also in

self-equilibrium and in direct proportion to th ute value of the flexural rigidity EI of the member as well.

e Structures

The analytical elements of displacements tatically indeterminate structures are the same as that of statically determinate structures. In other words, the unit load method, which is formul

to the calculation of displacements not only for sta inate structures but also for statically indeterm

t has to be satisfied for the virtual force system, primary structures of a desired indeterminate

ture. Conseque tly, as soon as the internal forces of an indeterminate structure are obtained by the force method, any one of its displacemen an be conveniently determined by using the

y one of the possible primary structures of ponents due to the unit force, which are needed when applying unit load method, may be thusly

determa statically indeterminate structure is

ndant restraints from the given structure, and tterminate structure by the primary system gained by applying external loadings and the unknown


redundants to the primary structure. In addition the consideration, the consistence of deformation is also taken into account. That is, in order to make the deforming conditions of the primary system to meet the deforming conditions of the original structure

o aspects of consideration ensure system are identical to those of its

origi

redundants are active forces that are applied to the primary struct

Consider, for example, the beam shown inindeterminate beam shown in Fig.10.45 (a); while force the d However, the stressing and deforming conditions of th

i be .45 (a) but also the bending moment diagram of the determinate beam showdeflection at the middle span of the beam shown in Fig.10.45 (a) has the d shown inFig.10.45 (b).

statically indeterminate structure can transformed into the calculation of displacements for its statically determinate primary system due to the combining actio f external lo

, the conditions of consistence of deformation or compatibility conditions have been satisfied in force method. These twthat the g conditions of the determinate primary stressing and deformin

enal indet rminate one. However, the two states are represented in different forms. For the original indeterminate structure, the redundants are passive forces that are existed in corresponding restraints, but for t r m, thehe p imary determinate syste

ures.

Fig.10.45. Reaction BY is a passive force for the Y is an active force, now denoted by X , for B 1

eterminate primary system shown in Fig.10.45 (b).e two beams shown in the figures (a) and (b) are identical. As the bending moment diagram shown in

Fig.10.45 (c) s not only the bending moment diagram of the indeterminate am shown in Fig.10n in Fig.10.45 (b). Similarly, the

the same magnitude and direction of eflection at the middle span of the beamBy above discussions, we conclude that the calculation of the displacements for a

ns o adings and the unknown redundants. For

38B

qlY =

Δ

(a)

B

(b)q q

AC

BACΔ

diagramM diagramM

(d)

C

38B

qlY =

(c)

2l

2l

2l

1P =2

8ql

BB AA

Fig.10.45 displacement calculation of indeterminate structurem;(a) an indeterminate beam; (b) a primary syste

(c) bending moment diagram; (d) primary structure and its unit load


instance, if the deflection of the middle span of the beam shown in Fig.10.45 (a) is desired, the unit load may be applied to the middle point C of the determinate primary beam shown in Fig.10.45 (d), on which the bending moment diagram M due to 1P = is drawn. By graph-multiplication method, we obtain

1 1 2 1 1 1 2 1 32 2 2 ( )( ) ( )2 2 2 3 8 3 16 3 2 8 8 2

4 ( )

192

CMM dsEI

l l l lql ql qlEIql

EI

Δ = ∫

⎡ ⎤= × × × + × − × × × ×⎢ ⎥⎣ ⎦

= ↓

By unit load method, the general expression for determining any one of the displacements of a statically indeterminate structure can be expressed as

0MM NN QQ tds ds k ds M ds N t ds RcEI EA GA h

α αΔΔ = + + + + −∑ ∑ ∑ ∑ ∑ ∑∫ ∫ ∫ ∫ ∫ (10-20)

Equation (10-20) is symbolically the same as that used for determining any one of the displacements of a statically determinate structure. However, it should be noted that M, Q and N are herein the bending moment, shear force and axial force of a statically indeterminate structure due to external loadings; while M , Q , N and R are the bending moment, shear force, axial force and reaction of one of its statically determinate primary structures due to a unit load, named virtual system in Chapter 9.

The first three items are the displacements caused by external loads; the fourth and fifth items are the displacements induced by temperature changes; whereas the last or sixth item is the displacement resulted from the support settlements.

If only the external loads are applied, the expression for the calculation of the displacements of a statically indeterminate structure will be

MM NN Qds ds k dsQEI EA G

Δ = + +∑ ∑ ∑∫ ∫ ∫ A (10-21)

Example 10-13

Determine the deflection at the middle point C for the indeterminate beam shown in Fig.10.45 (a) by unit load method under the condition that the primary beam shown in Fig.10.45 (b) will not be used.

. constantEI =Solution

(1) Real system The real system is the given indeterminate beam shown in Fig.10.45 (a) and its bending moment


diagram is shown in Fig.10.46 (a) again. (2) Virtual system The virtual system selected here is a simple beam with a unit load applied at the point C, which is one

of the primary beams, obtained by replacing the fixed support A by a hinged support and then applying a unit load at the location and in the direction of the desired displacement, as shown in Fig.10.46 (b), on which the bending moment diagram M is also depicted.

218

ql

4l

diagramM(a) (b)

C1P =

C BABA

(3) Calculation of desired displacement CΔ If only bending deformation is concerned, the expression for determining the displacement can be

expressed as

CMM dsEI

Δ =∑∫

By graph-multiplication method, we obtain 4

2 21 1 1 2 1 5( ) 2 ( ) (2 4 16 3 2 8 8 4 192C

l l l ql ql qlEI EI

⎡ ⎤Δ = − × × + × × × × × = ↓⎢ ⎥⎣ ⎦)l

The positive answer for indicates that the direction of the deflection coincide with that of the unit load

CΔ1P = .

It is observed from the example that when determining the displacements of an indeterminate structure by unit load method the unit load can be applied to any one of the primary structures of the given indeterminate structure. Since a primary structure is a determinate structure the determination of the displacement will be simplified.

Example 10-14

Determine the horizontal displacement at the top point D of column BD for the indeterminate bent frame shown in Fig.10.47 (a) by unit load method.

2l

2l

Fig.10.46 figures of example 10-13(a) an indeterminate beam and its bending moment diagram;(b) virtual system and bending moment due to 1P =


Solution

(1) Real system The real system is the given indeterminate bent frame shown in Fig.10.47 (a) and its bending moment

diagram is also shown in the figure.

(2) Virtual system The virtual system is shown in Fig.10.47 (b), which is one of the primary structures, obtained by

cutting the two-force member CD and then applying a unit load at the location and in the direction of the desired displacement, as shown in Fig.10.47 (b), on which the bending moment diagram M is also depicted.

(3) Calculation of desired displacement DΔ If only bending deformation is considered, the displacement can be determined, by

graph-multiplication method, as

1 1( 3 17.7) 22

1 1 2 1 1 1 2 ( 6 17.7)( 3 9 ( 6 53.1) 3 9)6 2 3 3 2 3 353.1 1380.9 283.2 ( )

6

DMM dsEI EI

EI

EI EI EI

Δ = = × × × +

⎡ ⎤× × × + × + × × × + ×⎢ ⎥⎣ ⎦

= + = →

∑∫）（

The positive answer for DΔ indicates that the sense of the displacement is the same as that of the unit load 1P = .

By the way, if we apply the unit load 1P = at the point C, the same result will be obtained. It may be checked by readers.

diagram(unit: kN m)

M⋅

diagram(unit: m)

M

11.8 17.72.3

(b)(a) 1P =

20kN

A B

C D

86.9

6I 6I

I

BA

C D

6I 6I

I

3

9

I

53.1 6m

3m

2m

Fig.10.47 Figures of example 10-14(a) an indeterminate bent frame and its bending moment diagram;(b) virtual system and bending moment due to 1P =

I


10.10 Verification of Calculation of Statically Indeterminate Structures

In the process of analyzing a statically indeterminate structure, many steps and numeric operations must be carried out, so mistakes are prone to occur. Therefore, in order to ensure the rightness of the final analyzing results, it is instructive to do some checking work during the analyzing process.

The checking work involves the following three aspects. (1) Verification of analyzing process The analyzing process involves the following checking steps and each of the steps will not allow

making any mistakes. ① Whether or not the determination of the degrees of indeterminacy of an indeterminate structure

is correct? Whether or not the primary structure is a stable system? ② Whether or not the internal force diagrams or expressions due to the external loading and due to

a unit load are correct? ③ Whether or not the calculation of flexibility coefficients and free terms in compatibility

equations has no mistakes? Whether or not the areas of internal force diagrams, their corresponding ordinates at the position of their centroids, measured from another bending moment diagram with straight lines, and flexural rigidity EI are correctly determined?

④ Whether or not the solution of the force-method equations is correct? Whether or not the answers of redundants satisfy the force-method equations?

⑤ The final internal force diagrams must be checked by the equilibrium and deformation conditions of the structure.

(2) Verification of consistent deformations When analyzing a statically indeterminate structure, in addition that the equilibrium conditions must

be used, the conditions of consistent deformations or compatibility conditions must be considered. Especially in force method, the unknown redundants are determined by the compatibility equations, so the keynote checking work is the verification of deformation consistence of the structure.

The common way for checking the deformation consistence of a statically indeterminate structure can be stated as: ① select any one of the primary structures as a desired primary structure and any one of the unknowns as a desired redundant, say ; ② calculate the displacement at the location and in the direction of based on the final internal force diagrams or expressions determined by force method, named ; ③ check the displacement

iXiX

iΔ iΔ whether or not equal to the corresponding prescribed displacement of the original indeterminate structure. That is

iΔ = corresponding prescribed displacement (10-22) For more general case, i.e., the displacements are determined by Eq. (10-20), the equation (10-22) will be written as

10.10 Verification of Calculation of Statically Indeterminate Structures 353

0

corresponding prescribed displacement

MM NN QQ tds ds k ds M ds N t ds RcEI EA GA h

α αΔ+ + + + −

=∑ ∑ ∑ ∑ ∑ ∑∫ ∫ ∫ ∫ ∫ (10-23)

In which, M, Q and N are the bending moment, shear and axial force of the original indeterminate structure; while M , Q , N and R are the bending moment, shear force, axial force and reaction of the primary structure due to . 1iX =

If the corresponding prescribed displacement is equal to zero and only the external loads are applied on the indeterminate structure, the equation (10-23) will be

0MM NN QQds ds k dsEI EA GA

+ +∑ ∑ ∑∫ ∫ ∫ = (10-24)

If the indeterminate structure is a beam or a rigid frame with the external loadings only applied on it and merely the bending deformation is concerned, the Eq. (10-24) will further become

0MM dsEI

=∑∫ (10-25)

We can use Eq. (10-25) to check the rightness of the final bending moment diagram. For an indeterminate structure, if the value of one of its displacements is given, we may employ the bending moment diagram obtained by the force method to calculate the displacement again. The coincidence of the solution for the displacement with the given value may verify the correctness of the bending moment diagram.

C D

P

1111

C D

1 1X =

diagramM diagramM

(a) (b)

BAB

1

Fig.10.48 verification of displacement of a closed rigid frame (a) a rigid frame and its bending moment diagram; (b) virtual system and bending moment due to 1X =

1 1A

Consider the closed rigid frame shown in Fig.10.48 (a). By consistent deformation, we have already known that the relative rotation of any one of the cross sections is equal to zero. However, we can use the bending moment diagram obtained by force method as shown in Fig.10.48 (a) again to calculate the relative


rotation of the sections adjacent to the middle point of the beam CD. The virtual force system and its bending moment diagram due to are shown in Fig.10.48 (b). By Eq. (10-25), we write 1 1X =

0MM Mds dsEI EI

= =∑∫ ∫ (10-26)

The solution for the displacement shows that the final bending moment diagram [Fig. 10.48 (a)] obtained by force method can satisfy the known displacement condition.

It can be seen from the discussion that the sum of the area of MEI

diagram for members of a closed

frame that is only subjected to external loads should be equal to zero. (3) Verification of equilibrium conditions The final internal force diagrams obtained by force method must satisfy equilibrium conditions. In

14.4

diagram(unit: kN m)

M⋅

diagram(unit: kN)

Q diagram (unit: kN)N

(a)

(b) (c)

(d)(e)

(f)

4kN/m

14.4 14.4

14.4

3.6

7.2 7.2

18

3I =

1I = 1I =

6m

A B

C D

12

5.4 5.4

5.4 5.4

12

CD

A B A B

C D12 12

12 12

C

14.4kN m

5.4 5.4

4kN/m

⋅

14.4kN m⋅

12kN

12kN5.4kN

5.4kNC

12kN

5.4kN

7.2kN m⋅

12kN5.4kN

7.2kN m⋅A B

C D

4m

Fig.10.49 verification of internal force diagrams of a closed rigid frame(a) a rigid frame and its bending moment diagram; (b) shear force diagram; (c) axial force diagram; (d) balance of moment of joint ; (e) balance of forces of joint ; (f) balance of load and reactions of entire structure

C C

10.10 Verification of Calculation of Statically Indeterminate Structures 355

other words, any one of the free body diagrams such as the entire structure, a rigid joint, a member, or a portion of the structure must meet their corresponding equilibrium conditions

Example 10-15

Verify the correctness for the diagram of the bending moment M as shown in Fig.10.49.

Solution

The given indeterminate rigid frame shown in Fig.10.49 (a) is only subjected to external load, so its bending moment diagram must satisfy Eq. (10-26). That is,

0MM Mds dsEI EI

= =∑∫ ∫

However, 1 2 2 1 1( 14.4 6 6 18) ( 4 14.4 4 7.2) 86.4 52.8 33.6 03 3 1 2 2

M dsEI

= − × + × × + − × × + × × = − + = − ≠∫

It implies that the bending moment diagram do not meet the conditions of deformation of the structure though the free body diagrams shown in Figs.10.49 (d) through (f) satisfy their equilibrium equations, so the computing result of the example is wrong.

Example 10-16

Verify whether or not the bending moment diagram shown in Fig.10.50 (a) satisfying the conditions of consistent deformation.

Solution

388

Pa

1588

Pa

diagramM diagramM

( )a ( )b

B

A

1I

12I

C

a

P

1 1X

2a

=

2a

a

a

1

Fig.10.50 figures of example 10-16(a) a rigid frame and its bending moment diagram; (b) virtual system and bending moment due to 1X =


It is known by observation that the horizontal displacement at support A of the frame shown in Fig.10.50 (a) is zero. The given displacement condition can be used to check the correctness of the bending moment diagram, so a virtual force system with a unit load corresponding to the given displacement applied on one of the primary structures of the given indeterminate frame, and the bending moment diagram due to the unit load are all shown in Fig.10.50 (b). By Eq (10-25) and graph-multiplication method, we find

21 2 3( )2 3 88

1 1 3 2 1 15 1 ( ) ( ( )2 2 88 3 2 88 3 2 4 2

0

AHMM ads PaEI EI

Pa a Pa a Pa aa aEI

Δ = = × × +

⎡ ⎤× × + × × − × ×⎢ ⎥⎣ ⎦=

∑∫） a

It implies that the bending moment diagram meets the conditions of consistent deformation of the indeterminate frame. The equilibrium conditions may be checked by readers.

SUMMARY

(1) Principle of force method The force (flexibility) method or the method of consistent deformations is one of the fundamental

methods used to analyze statically indeterminate structures. In the chapter we have discussed the formulation of the force method. The method of consistent deformations involves removing enough restraints from an indeterminate structure to render it statically determinate. The determinate structure is called the primary structure, and the reactions or internal forces associated with the excess restraints removed from the indeterminate structure are termed primary unknowns or redundants. The primary unknowns are now treated as unknown loads together with the external loadings applied to the primary structure to form a primary system, and their magnitudes are determined by solving the compatibility equations based on the condition that the displacements of the primary structure at the locations (and in the directions) of the primary unknowns, due to the combined effect of the prescribed external loading and the unknown redundants, must be equal to the known displacements at the corresponding locations on the original indeterminate structure. Once the redundants have been determined, the analysis of the indeterminate structure will be transformed to the analysis of its primary determinate structure, and the other response characteristics of the indeterminate structure can be evaluated either through equilibrium conditions or by superposition of the responses of the primary structure due to the external loadings and due to each of the redundants.

Therefore, the key points of force method are that: ① determine primary unknowns; ② select primary structure; ③ develop force-method equations (compatibility equations).

Summary 357

(2) Determine primary unknowns and select primary structures The primary unknowns must be equal to the degrees of indeterminacy of the structure. Remove the

restraints corresponding to the redundants from the given indeterminate structure to obtain the primary (determinate) structure. It is possible to choose several primary structures for one indeterminate structure, but the best one will result in simplest analysis. Also, the redundants must be chosen so that the removal of the corresponding restraints from the given indeterminate structure results in a primary structure that is statically determinate and stable. Applying the unknowns and the external loadings (external loads, support settlements or temperature changes etc.) to the primary structure will form a primary system.

The senses of the redundants are not known and can be arbitrarily assumed. A positive answer for a redundant will imply that the sense initially assumed for the redundant was correct.

(3) Develop force-method (compatibility) equations Write a compatibility equation for the location of each redundant by setting the algebraic sum of the

displacements of the primary structure due to each of the redundants and the external loading equal to the known displacement at the corresponding location on the actual indeterminate structure. The total number of compatibility equations thus obtained must be equal to the number of redundants.

(4) Calculation of flexibility coefficients and free terms in force-method equations a. First, draw internal force diagrams, especially bending moment diagram, of the primary structure

with only the external loading applied to it when the graph-multiplication is desired, or develop the expressions of internal forces of the primary structure with only the external loading applied to it when the integral method is employed.

b. Next, for each redundant, internal force diagrams, especially bending moment diagram, of the primary structure with only the unit value of the redundant applied on it when the graph-multiplication is employed, or develop the expressions of internal forces of the primary structure with only the unit value of the redundant applied on it when the integral method is employed.

c. Finally, compute the flexibility coefficients and free terms involved in the compatibility equations by using the graph-multiplication method or by the integral method and by the application of Maxwell's law of reciprocal deflections. A displacement (or flexibility coefficient) at the location of a redundant is considered to be positive if it has the same sense as that assumed for the redundant.

(5) Calculation of internal forces and the construction of their diagrams Once the redundants have been determined, the internal forces of the indeterminate structure can be

evaluated either through equilibrium conditions or by superposition of the internal forces of the primary structure due to the external loadings and that due to each of the redundants.

For beams and rigid frames, the member end moment should be determined first, then draw the bending moment diagram by the method discussed in the text, and then draw shear and axial force


diagrams by any convenient method. (6) Utilization of symmetric property and simplification of symmetric structures When a symmetric structure is subjected to a loading that is symmetric with respect to the structure's

axis of symmetry, the response of the structure is also symmetric. Thus we can obtain the response of the entire structure by analyzing a half of the structure, on either side of the axis of symmetry, with symmetric boundary conditions; and by reflecting the computed response about the axis of symmetry.

When a symmetric structure is subjected to a loading that is antisymmetric with respect to the structure's axis of symmetry, the response of the structure is also antisymmetric. Thus, the response of the entire structure can be obtained by analyzing a half of the structure, on either side of the axis of symmetry, with antisymmetric boundary conditions; and by reflecting the negative of the computed response about the axis of symmetry.

The response of a symmetric structure due to a general asymmetrical loading can be obtained by determining the responses of the structure due to the symmetric and antisymmetric components of the asymmetrical loading, and by superimposing the two responses.

(7) The calculation of displacements of indeterminate structures and verification of consistent deformations

The displacements for statically indeterminate structures are the same as that for the corresponding statically determinate primary structures, so the unit load can be applied on any one of the possible primary structures of the indeterminate structure to form the virtual system.

For an indeterminate frame or beam, if the value of one of its displacements is given, we may employ the bending moment diagram obtained by the force method to check the consistent deformation conditions of the structure by calculating the displacement again. The coincidence of the solution for the displacement with the given value can verify the satisfaction of the consistent deformation conditions of the structure.


10-1 What is a statically indeterminate structure? What is the difference between a statically indeterminate structure and a statically determinate structure? What are the degrees of indeterminacy of an indeterminate structure? How to determine the degrees of indeterminacy of a structure? How many methods are there for removing excess restraints?

10-2 What is the principle of force method? What are the primary system and primary unknowns? Why do we have to determine the primary unknowns first when analyzing a statically indeterminate structure? What is the difference between a primary system and its original structure?

10-3 What is the physical meaning of a force-method equation? The figures (b) and (c) of the problem show two kinds of the primary structures of the single degree indeterminate rigid frame shown in


the figure (a), please lay off their corresponding primary systems and explain the physical meanings of two kinds of force-method equations corresponding to the two primary systems, and sketch the displacements of 11δ and 1PΔ on the two corresponding primary structures, respectively.

10-4 Please explain, by physical meanings, why must the main flexibility coefficients be garter than zero while the secondary coefficients may be greater, or less, or equal to zero?

10-5 Which deformation effects must be considered when calculating the coefficients and free terms in force-method equations derived from indeterminate beams, bent frames and rigid frames?

10-6 Why are the internal forces of members of a statically determinate structure not related to their flexural rigidity EI, but those of members of a statically indeterminate structure is related to their flexural rigidity EI? Why are the internal forcestructure due to external loads related to the relative value of their flexural rigidity EI and not related to the absolute value of EI? Which deformation effects must be considered when

s of the members of a statically indeterminate

10-7 calculating the coefficients and free

10-8 in the figure of the

terms in force-method equations derived from indeterminate trusses? Please lay off the bending moment diagrams of the structure shown problem under following two conditions: (1) the area of member CD vanishes, i.e., 0A→ ;

reflecting problem 10-8

A

B

C

D E

P

I I

l l

l

PBB

B

primary structure 1 primary structure 2

AA A(b)(a) (c)

Reflecting problem 10-3(a) an indeterminate rigid frame; (b) primary structure 1; (c) primary structure 2


(2) the area of member CD goes infinitely great, i.e., A→∞ . What is a symmetric structure? Whether or no ru10-9 t the ctures shown in the figure of the problem

10-10 Why are the antisymmetric unknown forces on the section lying on axis of symmetry of a

10-11 he structure is a symmetric and subjected to

10-12 stem to analyze for the structures that have two axes of

10-13

stare symmetric structures? Why?

(a) (b)


symmetric structure subjected to symmetric loadings equal to zero? Contrariwise, why are the symmetric unknown forces on the section lying on axis of symmetry of a symmetric structure subjected to antisymmetric loadings equal to zero? Why can only half a structure be analyzed when tsymmetric or antisymmetric loadings? How select the symmetric primary sysymmetry, as shown in Fig.10.30 (a) of example 10-8?

In example 10-8, if set 1 21 2

1 2l lw the bending moment diagram of the

llowing two conditions: (1)10-14 ma neglected when calculating the

10-15 t the optimal centre line of two hinged arch subjected to vertical

2 1, , EI EI

i i i k i= = = , ho

example will vary under fo 0k → ; (2) k →∞ ? Under what condition the shear and axial deformations y bedisplacements of an arch? Whether or not affirm tha

uniformly distributed loads is still a second-degree parabola 2

4 ( )fy x l xl

= − ? Whether or not the

stressing condition of a three hinged arch with a centre ined by function line determ

2

4 ( )fy x l x= − subjected to vertical distributed loads is the same as that of a two hinged arch with

ne and subjected the same loads? Please explain the function of the tie and the variatio

l

the same centre li10-16 n of the bending moment in a two hinged arch

when the axial rigidity of the tie varies?

A B C

D E F

EI

EI EI

2EI 2EI

l l

A B C

D E FEIEI

EI EEI I

l l

h h


10-17 is subjected to no external loads there is no internal forces in the structure true?

10-18 What are the difference between the force-method equations used to analyze a statically indeterminate structure subjected to external loads and those subjected to support movements and that subjected to temperature changes?

10-19 In which situation is only the relative value of flexural rigidity EI of members of an indeterminate structure required when analyzing the structure? In which situation is the absolute value of flexural rigidity EI of members of an indeterminate structure needed when analyzing the structure?

10-20 Consider the beams had support movements as shown in the figures (a) and (b). Whether or not the force-method equations are identical when selecting the figure (c) and figure (d) as a primary structure, respectively? What is the meaning of each force-method equation?

10-21 What is the difference between the determination of displacements for a statically determinate structure and that for a statically indeterminate structure?

10-22 Why can the unit load be applied on any one of the primary structures of a statically indeterminate structure when use unit load method to calculate the displacements of the structure?

10-23 Please explain the physical meanings of the condition of consistent deformations of Eq. (10-25),

i.e.,

When is the statement that if a structure

AA constantEI = BB

A BBA

1X

Δ

Δ

(d)

(a)

(c)

(b)

1X

Reflecting problem 10-20

0MM dsEI

=∑∫ .


10-1 Determine the degrees of indeterminacy for the statically indeterminate structures shown in the figure. And remove sufficient restraints to change each of the indeterminate structures into a stable and statically determinate one.


(e) (f)

Problem 10-1 (contd)

(h)(g)

(c) (d)

(a) (b)


10-2 Analyze the structures shown in the figure by force method and draw their bending moment and shear force diagrams.

10-3 Analyze the rigid frames shown in the figure by force method and draw their bending moment, shear and axial force diagrams.

10-4 Analyze the rigid frames shown in the figure by force method and draw their bending moment diagrams.

10-5 Analyze the bent frames shown in the figure by force method and draw their bending moment diagrams.

Problem 10-1

(i)

(j)

1

2

I kI=

in (d), How is the variation of the bending moment diagram when increase from small to largek

(b)(a)

(c) (d)

Problem 10-2

A B

C D

2l

2l

EI

PA B

EI

P

2l

2l

q

A B

1 2

10, 0.1I kI

k k=

= =① ②

2I

l l

A

B

C

E F

1I1I

2I

1I

2I

2l

2l

EA = ∞


Problem 10-3

constantEI =

(b)(a) 1kN/m 1kN/m

A B

C D

q

12m

1I 1I 6m

A

BC

l

l

2 14 I I=

(b)(a)

20kN/m EA = ∞

B

C D

A

12m

I I 6m

A B

C D

EA = ∞

I 4kN I

6I 6I 6m1m

2m

12m

Problem 10-5 (contd)

constantEI =

(a)

Problem 10-4

A B

CD E

60kN

3m 3m

3m3m

10kN/m

2.5m 5m 2.5m5m

(b)

A B

CD E

10m constantEI =


10-6 Analyze the trusses shown in the figure by force method. 10-7 The figure shows a composite crane beam. Please analyze the beam and calculate the tensile force

and draw the bending moment diagram for the beam.

1.5

7.5

1.5

7.5

Problem 10-5

(c)

A B C

D E

F G

EA = ∞

EA = ∞ 4kN/m 4.

5m 2.

1m

6.6m

2.4m

①

③

numbers enclosed within circles represent the relative values of EI

constantEA =

constantEA =(b)(a)

Problem 10-6

AB

C

D E

P P

2a 2a

a

P

B CP

A D

a

a

I20a 2 20φ 2 20φ

Problem 10-8Problem 10-7

21400kN mEI = ⋅

52.56 10 kNEA = ×

3.5m7.6kNP = 7.6kNP =

2m 2m 2m

A B

C D

E F

2m

A B C

P P4m 4m2m 2m

1.2m

20I a


10-8 The figure shows a two-span suspending crane beam. Please analyze the beam and calculate the tensile force and elongation of suspension rods by the consideration of the axial deformation, and draw the bending moment, shear force diagrams for the beam. Assume that , the area of the rods , the moment of inertia for the shaped steel is .

4.5kNP =23.14cmA = I20a 42370cmI =

10-9 Analyze the structures shown in the figure by using their property of symmetry and draw their bending moment diagrams. For problems (a) and (f), please draw shear and axial force diagrams as well.


constantEI =

(b)(a)

(c) (d)

(e) (f)

Problem 10-9

10kN/m

A B

C

D

E

P

q

4m 4m

3m3m BA

C

D E F

l l

l

A B

C

D

E F G H

I I I I h

qA

B

DI

2I 2I

l l

l

q

q

A B

l l

l

2I

2I

I I

I I

A B

D

E F6kN

9m

6m6m


10-10 Analyze the internally indeterminate structure shown in the figure and draw its bending moment diagram.

10-11 The members of the frame shown in the figure are of a rectangular cross section of . Please analyze the frame and draw its bending moment diagram by following two conditions: (1) neglect the axial deformation; (2) take the axial deformation into account. Finally, make a discussion by comparing the two solutions.

b h×

10-12 Please derive the expressions for calculating the internal force of the tie of the two hinged arch shown in the figure by neglecting the axial deformation and effect of curvature of the arch, i.e.,

. The centre line of the arch is a parabola of ds dx=2

4 ( )fy x ll

x= − .

10-13 Determine the internal forces of the arc arch shown in the figure. 10-14 Determine the internal forces at sections A and B of the circular tube shown in the figure.

P

constantEI =

Problem 10-10

l l

l

A B

C

Problem 10-11

l l lC

P

BA

q

A

constantEI =

Problem 10-13

C

P

O

R=

B

R R

f

Problem 10-12

constantEI =

l

y

fxBA

1 1E A


q

10-15 Draw the bending moment diagrams for the beams shown in the figure due to temperature changes. The coefficient of thermal expansion is α ; the depths of the beams are of h.

10-16 Draw bending moment and shear force diagrams of the beams due to support movements as shown in the figure.


2 1t t> 2 1t t>

Problem 10-15

(b)(a)1t+A B

l2t+

AB

l

1t+

2t+

αconstantEI = constantEI =


(b)(a)

(c) (d)

Problem 10-16

ABAB

AB

Δ

1B

AB

1B

1B

Δα

α

α

l

ll

l

B

AR

constantEI =

Problem 10-14q


10-17 Draw bending moment diagram of the frame due to support movements as shown in the figure, , , 30mma = 40mmb = 0.01 radianϕ = .

10-18 Calculate the horizontal displacement at point C of the frame shown in the figure of problem 10-3(a).

10-19 Calculate the horizontal displacement at point C of the bent frame shown in the figure of problem 10-5(a).

10-20 Calculate the rotation at end B of the beam shown in the figure of problem 10-15(a). 10-21 Please check the correctness of following bending moment diagrams by the designated conditions:

(1) the relative rotation of an arbitrary section of the beam AB shown in the figure of problem 10-2 (b) is zero; (2) the vertical displacement at support B of the frame shown in the figure of problem 10-3 (b) is zero.

C

ϕ213440kN mEI = ⋅

Problem 10-17

A B

D E

F

2m 2m a

2m2m

b

CHAPTER 11 DISPLACEMENT METHOD

In this chapter, we will study a classical formulation of the displacement (stiffness) method used for the analysis of statically indeterminate structures. We will first present the fundamental concepts and principles of displacement method, such as the selection of basic unknowns, primary structures and primary systems and development of fundamental (equilibrium) equations of displacement method. We then consider the application of the method to the analysis of continuous beams, rigid frames, bent frames and other indeterminate structures. Finally, we introduce the development of displacement-method equations by direct equilibrium method or slope-deflection method.

The abstract of the chapter

11.1 Fundamental Concepts and Analyzing Procedures of Displacement Method

The slope-deflection method for the analysis of indeterminate beams and frames was introduced by George A. Maney in 1915. The method takes into account only the bending deformations of structures. Nowadays, the method has been progressed a lot and can be applied to all of statically indeterminate structures; for generalization, the method is termed displacement method. Although displacement method is itself considered to be a useful tool for analyzing indeterminate beams and frames, an understanding of the fundamentals of this method provides a valuable introduction to the matrix displacement method, which forms the basis of most computer software currently used for structural analysis.

In Chapter 10, we considered the force (flexibility) method of analysis of statically indeterminate structures. Recall that in the force method, the unknown redundant forces are determined first by solving the structure’s compatibility equations of consistent deformations; then the other response characteristics of the structure are evaluated by equilibrium equations or superposition. Unlike the force method, in the displacement method the unknown displacements are determined first by solving the structure's equilibrium equations; then the other response characteristics are evaluated through member force-deformation relations.

The fundamental idea and analyzing procedures of displacement method may be presented by the following example, a rigid frame shown in Fig. 11.1 (a).

1. Determination of primary unknowns When the frame is subjected to the external loading, members AC and CB deform, as shown in the

figure using an exaggerated scale. If we ignore the axial deformations of the two members (i.e. elongation

370

11.1 Fundamental Concepts and Analyzing Procedures of Displacement Method 371

or contraction), only joint C has a rotation Cθ , the common rotation of member ends of CA and CB. If we take the rotation as a primary unknown, then develop a equation involving Cθ and external loads, and then manage to resolve the Cθ , the other response characteristics will be obtained through member force-deformation relations.

2. Selection of primary system Now let us discuss how to determine the rotation Cθ . The procedure is as follows. Artificially add a restraint at joint C against the rotation of Cθ as shown in Fig.11.1 (b) to form a

primary structure, i.e. lock the joint so as to make it have no rotation. Thusly, the frame composed of two members now becomes an assemblage of two individual single-span indeterminate beams. Next, let the assemblage of single-span indeterminate beams subject the given external load and the unknown displacement Cθ as shown in Fig.11.1 (c) to form a primary system. Note that the primary structure in

P

1

1

3EIl

1Cθ =

C B

A

CCk

Cθ

A

BC

2EI

1EI

1

2l

P

(a) (b) (c)

(e)

15

32Pl

CPM C

P13

16Pl

(d)A

Cθ

1

2l

2l0CM =

C B BCCθ

Cθ

A

2

2

4EIl

1Cθ =

B

2

2

2EIl

Fig.11.1 An indeterminate rigid frame analyzed by displacement method(a) original structure; (b) primary structure; (c) primary system; (d) primary structure under external load; (e) primary structure under 1Cθ =

372 Chapter 11Displacement Method

displacement method is an assemblage of various single-span indeterminate beams and the primary system is the assemblage of various single-span indeterminate beams subjected to given external loading and unknown displacement.

3. Development of displacement-method (equilibrium) equation To establish the displacement-method equation, let us consider the equilibrium condition of restraint

moments at the artificial restraint at joint C of the primary system due to the combined effect of the external loading and the unknown rotation Cθ . Since there is no external moment exerted at joint C of the original structure, the restraint moments yielded on the artificial restraint at joint C of the primary system due to the combined effects of the external loading and the unknown rotation Cθ must be also equal to zero [see Figs.11.1 (a) and (c)]. That is,

0CM = (a)

4. Determine primary unknown To determine the unknown Cθ by using the equilibrium condition as mentioned above, the total

restraint moment at the artificial restraint of the primary structure due to the combined effects of the external load and Cθ can be conveniently expressed by superimposing (algebraically adding) the restraint moments due to the external load and the rotation Cθ acting individually on the primary structure; that is,

C CP CCM M M= + (b)

In which CPM and CCM represent, respectively, the restraint moments on the artificial restraint of the primary structure due to the external load and the unknown rotation Cθ , each acting alone on the structure.

Since the unknown rotation Cθ is unknown, it is convenient to determine CCM by first evaluating the restraint moment on the artificial restraint due to a unit value of the unknown rotation Cθ , as shown in Fig.11.1 (e), and then multiplying the restraint moment thus obtained by the unknown magnitude of the rotation Cθ . Thus

CC CC CM k θ= (c)

In which denotes the restraint moment at the artificial restraint of the primary structure due to the unit value of the rota on C

CCkti θ . By substituting equations (b) and (c) into equation (a), we obtain the

equilibrium equation

0C CP CC CM M k θ= + =

The equation can be solved to express the unknown rotation Cθ in terms of the restraint moment CPM and the stiffness coefficient of the primary structure. That is, CCk

CPC

CC

Mk

θ = −

11.1 Fundamental Concepts and Analyzing Procedures of Displacement Method 373

Since the primary system is composed by two individual single-span beams, their end moments can be easily determined by force method previously studied in last chapter. Thusly, we find

and by force method, we write 1 1 2 23( / ) 4( / )CCk EI l EI= + l 13 /16CPM Pl= −

1

1 2

1 2

316

3 4

Pl

EI Ell l

θ =+

After the unknown rotation Cθ has been computed, all other response characteristics of the original structure can be also conveniently determined by employing principle of superposition. For instance, the bending moment diagram for the original structure can be obtained by superimposing the bending moment diagram of the primary structure due to the external loading and the bending moment diagram of the primary structure due to a unit value of the unknown rotation Cθ multiplied by the value of Cθ .

Based on the foregoing discussion, we can summarize the following step-by-step procedure for the analysis of a statically indeterminate structure by displacement method.

(1) Select primary unknowns. The primary unknowns (or variables) used in displacement method are joint displacements (joint rotations and joint translations) of the structure.

(2) Select primary system. Add artificial restraints against the primary unknown displacements to discretize the original structure as an assemblage of individual single-span indeterminate beams. Next, let the assemblage of individual single-span indeterminate beams subject their given external loads and primary unknown displacements, which must satisfy the compatibility conditions of consistent deformations, to form a primary system.

(3) Develop displacement-method (or equilibrium) equations. The displacement-method equations are the equilibrium equations of restraint forces associated with the primary unknowns. The equations can be developed by the consideration that the restraint forces induced at the artificial restraints of the primary system must be equal to the joint forces acting on the original structure at the locations and in the directions associated with the primary unknown displacements. The stiffness coefficients and free terms in the equations can be conveniently obtained by referring the tabulated results (see tables 11-1 and 11-2) of various single-span indeterminate beams calculated by force method learned in Chapter 10.

(4) Determine the primary unknowns. The primary unknown displacements can be determined by solving the displacement-method equations.

(5) Determine the response characteristics of original structure. Once the unknown displacements have been computed, the reactions and all other response characteristics of the original structure can be also conveniently determined by employing superposition relationships. For instance, the bending moment diagram for the original structure can be obtained by superimposing together each individual bending


moment diagram of the primary structure due to the external loading and due to a unit value of each unknown displacement multiplied by its value individually.

11.2 Member End Forces of Various Single-Span Indeterminate Prismatic Beams Due to Their End Displacements and External Loadings

It is known from last section that the primary structure used in displacement method is an assemblage of individual single-span indeterminate beams. So the relationships between the member end forces and end displacements and external loadings of various single-span indeterminate prismatic beams are the basis of displacement method and they must be discussed before the application of displacement method.

11.2.1 Member end forces of various single-span indeterminate prismatic beams due to their end displacements

To derive the relationship between the member end forces and displacements, let us focus our attention on an arbitrary prismatic member AB of an indeterminate structure shown in Fig.11.2. When the structure is subjected to external loads and support settlements, member AB deforms, as shown in the figure, and internal moments are induced at its ends. The free-body diagram and the elastic curve for member AB are shown using an exaggerated scale in the figure. As indicated in this figure, double-subscript notation is used for member end moments, with the first subscript identifying the member end at which the moment acts and the second subscript indicating the other end of the member. Thus, ABM denotes the moment at end A of member AB, whereas

BAM represents the moment at end B of member AB. Also, as shown in Fig.11.2, Aθ and Bθ denote, respectively, the rotations of ends A and B of the member with respect to the undeformed (horizontal) position of the member; Δ denotes the relative translation between the two ends of the member in the direction perpendicular to the undeformed axis of the member; and the angle ϕ denotes the rotation of the member's chord (i.e., the straight line connecting the deformed positions of the member ends) due to the relative translation Δ . Since the deformations are assumed to be small, the chord rotation can be expressed as

lϕ Δ=

The sign convention used in this chapter is as follows:

ϕAθ

Bθ

B′ABQ

A B

BAQ

BAMl

constantEI =

Δ

Fig.11.2 Member end forces and displacements

ABM

11.2 Member End Forces of Various Single-Span Indeterminate Prismatic Beams Due to Their End Displacements and External Loadings 375

Note that all the member end forces and displacements are shown in the positive sense in Fig. 11.2. As mentioned previously, the relationship between the member end forces and displacements is the

basement of displacement method. Here, we especially interest in the member end forces expressed by end displacements. The member end forces due to a unit value of a displacement are named stiffness coefficients. Since the stiffness coefficients are only related to the material’s property and the size and shape of the member, they are termed shape constants. The shape constants may be determined conveniently by force method. Now we will give them directly as follows:

The member end moments, end rotations, and chord rotation are positive when clockwise; the sign convention for shear forces is the same as that previously used in the text.

1. Member end forces (shape constants) of member AB due to a unit rotation of end A, i.e., 1Aθ = (1) When end B is fixed, i.e., 0Bθ = Δ =

Consider a single-span indeterminate beam AB with two fixed supports as shown in Fig.11.3 (a). When support A is subjected to a rotation Aθ provided that the other end displacements are prevented, its deformed shape is denoted in dashed lines using an exaggerated scale in the figure and its corresponding end forces determined by force method are given by

BAM

AB Ai θ

AθA B

ABM

BAQABQ

BAM ABconstantEI = constantEI = constantEI =

l l l

BA BA BA

(b)(a) (c)

A

B

Aθ

ABM

BAQABQ

ABM

BAQABQ

2 AB Ai

Aθ

θ

A Fig.11.3 Member end forces due to (a) a single-span indeterminate beam with two fixed supports;(b) a single-span indeterminate beam with one fixed support and one r

θ

oller support ;(c) a single-span indeterminate beam with one fixed support and one double-link support

AB Ai3 AB Ai θ4 AB Ai θ θ


42

6

AB AB A

BA AB A

ABAB BA A

M iM i

iQ Ql

θθ

θ

⎫⎪=⎪

= ⎬⎪⎪= = −⎭

(11-1)

In which, AB

EIil

= is named the linear stiffness of member AB. When 1Aθ = , the bending-moment shape

constants at ends A and B are, respectively, and , the shear-force shape constants at the two

ends are

4 ABi 2 ABi6 ABi

l− . They are actually the restraint forces at the member ends due to the unit rotation 1Aθ = ,

provided that the other end displacements are prevented. (2) When end B is hinged, i.e., 0BM = Consider a single-span indeterminate beam AB with fixed end A and hinged (or rolled) end B as shown

in Fig.11.3 (b). When support A is subjected to a rotation Aθ , its deformed shape is denoted in dashed lines using an exaggerated scale in the figure and its corresponding end forces determined by force method are given by

30

3

AB AB A

BA

ABAB BA A

M iM

iQ Ql

θ

θ

⎫⎪=⎪

= ⎬⎪⎪= = −⎭

(11-2)

It is observed that when 1Aθ = , the bending-moment shape constant at end A is , and the

shear-force shape constants at ends A and B are

3 ABi3 ABi

l− .

(3) When end B is restrained by a double-link support Consider a single-span indeterminate beam AB with fixed support A and double-link support B as

shown in Fig.11.3 (c). When support A is subjected to a rotation Aθ , its deformed shape is denoted in dashed lines using an exaggerated scale in the figure and its corresponding end forces determined by force method are given by

0

AB AB A

BA AB A

AB BA

M iM iQ Q

θθ

= ⎫⎪= − ⎬⎪= = ⎭

(11-3)

It is obvious that when 1Aθ = , the bending-moment shape constants at ends A and B are and ABi


ABi− , respectively; and the shear-force shape constants are equal to zero because the restraint property of the double-link support at end B.

2. Member end forces (shape constants) of member AB due to a unit relative displacement between ends A and B, i.e., 1Δ =

(1) When end B is fixed Consider a single-span indeterminate beam AB with two fixed supports as shown in Fig.11.4 (a).

When support A is fixed and support B is subjected to a translationΔ provided that the other end displacements are prevented, its deformed shape is indicated by dashed lines using an exaggerated scale in the figure and its corresponding end forces determined by force method are given by

2

6

12

ABAB BA

ABAB BA

iM Ml

iQ Ql

⎫= = − Δ⎪⎪⎬⎪= = Δ⎪⎭

(11-4)

Thusly, when , the bending-moment shape constants at ends A and B are 1Δ = 6 ABil

− , and the

shear-force shape constants are 2

12 ABil

.

(2) When end B is hinged Consider a single-span indeterminate beam AB with fixed support A and hinged (or rolled) support B

as shown in Fig.11.4 (b). When support A is fixed and support B is subjected to a translation , its Δ

MM AB BA

A

ABQ

BAQ

BB

A

3ilΔ

l l

diagramMdiagramM

constantEI = constantEI

6ilΔ

A BA B

=

(b)(a)

ABM

ABQ

BAQ

Δ

Δ

6ilΔ

Fig.11.4 Member end forces due to (a) a single-span indeterminate beam with two fixed supports;(b) a single-span indeterminate beam with one fixed support and one

Δ

roller support ;


deformed shape is denoted in dashed lines using an exaggerated scale in the figure and its corresponding end forces determined by force method are given by

2

3

03

ABAB

BA

ABAB BA

iMl

MiQ Ql

⎫= − Δ ⎪⎪

= ⎬⎪⎪= = Δ⎭

(11-5)

Table 11-1 shape constants of single-span indeterminate prismatic beams

Bending moments Shear forces Number Single-span indeterminate beam

ABM BAM ABQ BAQ

1

4i 2i 6il

− 6il

−1Aθ =

1Aθ =

l

EIA B

Two fixed end support

2

6il

−6il

− 2

12il

2

12il

3

3i 0 3il

− 3il

−

One end is fixed and one end is hinged

or rolled 4

3il

− 0 2

3il

2

3il

One end is fixed and one

end is supported by double links

5 i i− 0 0

1Aθ =

1Aθ =

l

BA

EI

EI

l

AB

1Δ=

1Aθ = EIA B

l1Aθ =

EI BA

l

=1Δ


3 ABil

−It is observed that when , the bending-moment shape constant at end A is 1Δ = , and the

shear-force shape constants at ends A and B are 2

3 ABil

.

To facilitate the application of displacement method, table 11-1 gives the shape constants (stiffness coefficients), denoted by ABM BAM ABQ BAQ, , and , for various single-span indeterminate prismatic members with different restraint conditions.

11.2.2 Member end forces of various single-span indeterminate prismatic beams due to typical loadings

As mentioned previously, the relationships between the member end forces and end displacements and external loadings of various single-span indeterminate prismatic beams are the basement of displacement method. Now we will discuss the member end forces of various single-span indeterminate prismatic beams due to typical loadings. For easy distinguish, the member end forces caused by external loadings are termed fixed-end forces. Since the fixed-end forces are related with the types of external loadings they are referred to as loading constants, which can be determined by force method discussed in last chapter.

To facilitate the application of displacement method, table 11-2 gives the loading constants, represented by F

ABM FBAM, , and , for various single-span indeterminate prismatic members with

different restraint conditions.

FABQ F

BAQ

Table 11-2 loading constants of single-span indeterminate prismatic beams

Bending moments Shear forces Number Single-span indeterminate beam F

BAM FBAQF

ABQFABM

1 8Pl

− 8Pl

+ 2P

+ 2P

−

2

2

2

Pabl

− 2

2

Pabl

+2

2

21Pb al l

⎛ ⎞+⎜ ⎟⎝ ⎠

2

2

21Pa bl l

⎛ ⎞− +⎜ ⎟⎝ ⎠

Two ends are

fixed

3 2

12ql

− 2

12ql

+ 2ql

+ 2ql

− A B

q

l

A B

P

a b

l

2l

2l

B

P

A


The continue of table 11-2


BAM FBAQF

ABQFABM

4 2

30ql

− 2

20ql

+ 320ql

+ 720ql

−

q

Two ends are

fixed 5

EI thαΔ EI t

hαΔ

− 0 0

6

316Pl

− 0 1116

P+ 5

16P

−

7

( )2 2

22Pb l b

l−

− 0 ( )2 2

3

32

Pb l bl−

+ ( )2

3

32

Pa l al−

−

8 2

8ql

− 0 58ql

+ 38ql

−

9 2

15ql

− 0 25ql

+ 10ql

−

One end is fixed and

another end is hinged

or rolled

10 27

120ql

− 0 940ql

+ 1140ql

−

P

la b

A B

B

2l

2l

P

A

1t

1 2t t tΔ = −

A B

l2t

B

l

A

B

q

A

l

A B

q

l

B

q

l

A




BAM FBAQF

ABQFABM

One end is fixed and

another end is hinged or

rolled

11

32

EI thαΔ 0 3

2EI t

hlαΔ

− 32

EI thlαΔ

−

12

2Pl

− 2Pl

− P+ leftB P+

0rightB

13 ( )22

Pa l al−

−2

2Pa

l− P+ 0

14

2

3ql

− 2

6ql

− ql+ 0

15 2

8ql

− 2

24ql

−2ql

+ 0

One end is fixed

and another end is

supported by double

links

16

2524ql

− 2

8ql

−2ql

+ 0

1 2t t tΔ = −

A B

l2t1t

B

P

Al

lA B

q

Al

B

q

AB

q

l

P

l

AB

a b



Bending moments Shear forces Number Single-span indeterminate beam

FBAM F

BAQ FABQ F

ABM

One end is fixed and another end is

supported by double

links

17

EI thαΔ EI t

hαΔ

− 0 0

B

11.3 Primary Unknowns and Primary Systems in Displacement Method

Recalling from section 11.1, when apply displacement method to analyze a statically indeterminate structure the primary unknowns and primary system must be first selected.

The primary unknowns are the variables by which the main displacements, used to determine member end forces, of a structure can be described completely. They are actually the joint translations and rotations of the structure. The primary system is obtained by first adding the artificial restraints against the primary unknowns to discretize a structure into an assemblage of single-span indeterminate members (or beams) so as to form a primary structure, and then subjecting the primary structure to the prescribed external loadings and the primary unknown displacements. In the following, the choices of primary unknowns and primary systems will be discussed in details.

11.3.1 Primary unknowns in displacement method

First, let us discuss how to select joint rotations as primary unknowns. Consider the continuous beam shown in Fig.11.5 (a). As shown in the figure, all joints (joints A, B, C and D) have no translation; the end A is fixed, so the rotation of end A is zero; the joints B and C are the rigid joint of two members, which will rotate when the beam is subjected to the external loading, thus we denote, respectively, the rotations as Bθ and Cθ ; the end D is hinged, so there is rotation, represented by Dθ , at the joint under the external loading. If we assume that the rotation at end B of member AB is BAθ , the rotation at end B of member BC is BCθ , the rotation at end C of member CD is CDθ , and the rotation at end D of member CD is DCθ , there are totally 5 rotations for the continuous beam. Note however, that if we consider the conditions of consistent deformations, only 3 of them are independent. That is,

1 2t t tΔ = −

A

l2t1t

11.3 Primary Unknowns and Primary Systems in Displacement Method 383

BA BC B

CB CD C

θ θ θθ θ θ

= = ⎫⎬= = ⎭

(a)

Therefore, if we consider the consistent deformation conditions and manage to determine the joint rotations Bθ and Cθ , the member end rotations BAθ , BCθ , CDθ and CBθ

ationscan be determined as well.

Also, the member end forces of the beam can be determined by the rel hip between end force and displacement. The joint rotations Bθ and Cθ , again denoted by unified symbols and , can be taken as the primary unknowns of the beam.

Since end D is hinged the member end moment is zero, i.e.,

1Δ 2Δ

0DCM = . The rotation Dθ is useless to determine member end forces and it may thusly not be employed to be a primary unknown.

It is observed from above discussion that there are two rigid joints for the structure and also there are two primary unknowns. It implies that the number of rotating primary unknowns of a structure is equal to the number of rigid joints of the structure.

Next, let us discuss how to select joint translations as primary unknowns. Consider the rigid frame shown in Fig.11.6 (a). As shown in the figure, there are two rigid joints (joints C and D) for the frame.

CDθ

CBθ

BAθ

BCθ

DCθ

B

Bθ Cθ

(a)

(b)

(c)

A

B

C D

P

P

A C DB

Fig.11.5 Joint rotations as primary unknowns(a) original structure; (b) primary system ; (c) primary structure


Reca ing from chapter 2 of the text, there are two degrees of freedom for a point, one horizontal displacement and one vertical displacem dinate system. There are thusly 4 translating displacements for the rigid frame. Based on the discussion, the bent frame shown in Fig.11.6 (b) has 6 tra of the ).

llent, in a plane rectangular coor

three joints (joints D, E and Fnslations

However, the computing task is directly proportional to the number of primary unknowns when use displacement method to analyze a statically indeterminate structure. In order to reduce analyzing working, the simplification of primary unknowns is necessary in some cases, especially for hand-oriented calculation. Thusly, the following assumption is presented:

(1) The axial deformation is ignored due to the axial forces of the members. (2) The member end rotation θ and the relative translation Δ between member ends are small

enough so that the chord rotation ϕ of the member satisfies the equality l

ϕ Δ= .

By the assumption (1), the straight length of a deformed member is equal to its original undeformed straight length of the member. According to assumption (2), the length of the chord of a deformed member is the same as that of the deformed member. By summarizing the above, we conclude that the distance between two joints of a deformed structure is the same as that of the two joints of the original undeform struc e. That is, the lengths of the members of a structure remain unch geable under external loadings.

Based of above assumption, let us discuss the number of ependent translations of a structure. Co er the rigid frame shown in Fig.6 (a). Since re is no vertical translation for the joints C and n according with the small deformation assumption, there are only horizontal translations for the joints C and D. However, because the length of member CD must remain unch

edtur an

indnsid the D i

angeable the joints C and D only have the same horizontal translation, herein denoted byΔ . Consequently, only one horizontal translation of the four translations of joints C and D is independent, which can be selected as one of the primary unknowns of the

C′ D′

A B

CD

P ΔΔ

CθCθ Dθ

Dθ

A B C

D E F

Δ Δ Δ

F ′E′D′P

(b)(a)

Fig.11.6 Translating primary unknowns in displacement method

11.3 Primary Unknowns and Primary Systems in Displacement Method 385

structure. Finally, only three joint displacements may be chosen as the primary unknowns. They are joint rotations at joints C and D ( Cθ and Dθ ) and the independent horiWith similar analysis, the bent frame shown in Fig.11.6 (b) has one primary un n, which is the independent translation of joints D, E and F, as shown in the figure.

Since we ignore the distance change between the two ends of a member in the an is of rigid frames, the number of independent joint translations of a rigid frame may be determined the method for analyzing geometric construction of a structure. In fact, the neglect of the distance ch a member will make the number of the independent joint translations of a structure unchangeable n atter whether the members are rigidly or pin connected at the joints. Therefore, if we change the rigi nts (include fixed supports) of a rigid frame into pin joints to form a pin-connected system, the num grees of freedom of the pin-connected system is the same as the independent translations of the origin id frame. In other words, the number of the restraints or links, needed to add to make the pin-connect stem to be a stable system with no redundant restraint, is equal to the number of independent translatio he original rigid frame.

sy n .7. Ill of the rigi de t

d

zontal translation Δ of joints C and D. know

alys by

ange of o m

d joiber of de

al riged syns of t

(b)(a)

ame b) c ond con Fig.11.7 Determination of translating primary unknowns(a) an original rigid fr ; ( orresp ing pin- nected system

Thusly, we may determine the independent joint translations of a rigid frame by analyzing the degrees of freedom of a corresponding pin-connected stem. Consider the rigid frame shown i Fig.11 n order to determine the independent translations of the structure, change a d joints (inclu hree fixed supports) into hinged joints to form a pin-connected system as shown in Fig.11.7 (b) by soli lines. The pin-connected system shown in Fig.11.7 (b) by solid lines is a mechanism, so we must add two restraints or links, shown in the figure by dashed lines, to make the pin-connected system be a stable system with no redundant restraint. The original rigid frame thusly has two independent translations.

In general, the primary unknowns in displacement method for a statically indeterminate rigid frame include joint rotations and independent translations of the frame. The number of joint rotations is equal to the number of rigid joints; the number of joint independent translations is the same as that of the degrees of


freedom of a corresponding pin-connected system obtained by changing the rigid joints of the original frame into hinged joints.

Noted from above discussion, that the conditions of consistent deformations (or compatibility conditions of displacements) of a structure have already been satisfied because the same rotations and the same translations of the member ends connected to a joint are guaranteed when determining the primary unknowns of the structure.

11.3.2 Primary systems in displacement method

As mentioned in section 11.1, the primary structure of a structure is an assemblage of individual single-span indeterminate beams, which is obtained by adding corresponding artificial restraints against each of the primary unknowns of the structure. The primary system is obtained by subjecting the assemblage of individual single-span indeterminate beams to their given external loads and primary unknown displacements, which must satisfy the compatibility conditions of consistent deformations.

For example, Fig.11.8 (a) shows a rigid frame with only one rigid joint D. Since there is no joint translation the structure only has one primary unknown, which is the joint rotation of joint D, Dθ . The primary structure can be obtained by adding artificially a restraint against the rotation Dθ as shown in Fig.11.8 (b). The primary system can be gained by subjecting the primary structure to the external loading and the primary unknown as shown in Fig.11.8 (c), which actually is the composition of individual single-span indeterminate beams under the combined actions of the external loading and the support movements.

Similarly, the primary system and primary structure of the structure shown in Fig.11.5 (a) are, respectively, the system shown in Fig.11.5 (b), which is the assemblage of three single-span indeterminate

A1P

1Δ

D

B

C1Δ

1( )DθΔ

2P 2P1P

A

B

CD

(a) (b)

D

1Δ

A C

B

(c) Fig.11.8 P mary system in displacement metho(a) original struct re; (b) primary structure; (c) primary ystem

ri du s

11.4 Displacement-Method Equations 387

beams under the combined actions of a concentrated force and two support rotations, and the structure shown in Fig.11.5 (c), which the ass blage of three single-span indeterminate beam

Again consider the rigid frame shown in Fig.11.9 (a). Since the frame has one rigid-joint rotation and one independent joint translation, the primary unknown for the frame

is em s.

is two, the rotation of joint C and the trans

It is observed from foregoing discussion that the primary system of a structure can be obtained by adding the artificial restraints against the primary unknowns on the original structure. If a primary unknown is a rigid-joint rotation, a restraint resisting the rotation, herein named rotative restraint, will be added; similarly, if a primary unknown is a joint translation, a restraint resisting the translation (or a link), termed translating restraint, will be affixed. Note that the two kinds of artificial restraints are independent each other, the rotative restraint only resists the joint rotation, whereas translating restraint just prevent the joint trans

lation of joints C and D. Therefore, we add a rotative restraint against the rotation of joint C and a translating restraint against the translation of joints C and D as shown in Fig.11.9 (c) to form the primary structure; and let the primary structure undergo the combined actions of the external loading and the primary unknown to form the primary system as shown in Fig.11.9 (b).

lation. The distinguish between the primary system and the original structure is that the artificial restraints against the primary unknowns have been affixed to the original structure so as to make the original structure to be an assemblage of single-span indeterminate beams. The following section will discuss how to develop the fundamental formulations in displacement method by using the primary system of a structure.

11.4 Displacement-Method Equations

How to let the response characteristics (reactions, internal forces and displacements) of the primary system of a structure due to the combined action of the external loadings and the primary unknowns be identical to those of the original structure? The transforming conditions are just the equilibrium conditions

B

C

1ΔP

D

A A B

C D

B

C

A

D

2Δ

P 1( )DθΔ

1( )DθΔ

2Δ2

(b)(a) (c)

Δ 2Δ

Fig.11.9 Primary system in displacement method(a) original structure; (b) primary system ; (c) primary structure


of restraint forces acting on the artificial restraints; they are actually the displacement-method equations. The equations can be developed by the consideration that the restraint forces induced at the artificial restr

in the following discussion.

tions

aints of the primary system must be equal to the joint forces acting on the original structure at the locations and in the directions associated with the primary unknown displacements. In order to make the expressions of the displacement-method equations have generality, all of the primary unknowns are denoted by unified symbol Δ

11.4.1 Development of displacement-method equa

To illustrate the development of displacement-method equations, consider the rigid frame shown in Fig.11.10 (a). Since the rigid frame has only one rigid joint and has no joint translation, the primary unknown of the frame must be the rotation of joint C, denoted by 1Δ . The primary system will be obtained by adding artificial restraint against the rotation to the original structure as shown in Fig.11.10 (b).

The displacement-method equation can be developed by the consideration that the restraint forces induced at the artificial restraint of the primary system must be equal to the joint forces acting on the original structure at the locations and in the directions associated with the primary unknown displacement. Since no joint moment is exerted at joint C, the restraint moment at the artificial restraint due to the combined effects of the external loading and the unknown rotation 1Δ must be also equal to zero. That is,

1 0F = (a)

the value of the restraint moment at the artificial restraint yielded by the external load is the same

Then the ment equilibrium condition at joint C of the primary system due to the combined effects of the external l and the unknown rotation

Apparently, the equation (a) is an equilibrium equation. It is can be also established by considering the reason that if

in magnitude but opposite in direction as the moment necessary to rotate the joint C a angle in the value of 1Δ , the restraint moment at the artificial restraint on the primary system must be equal to zero.

mooad 1Δ is the same as that of the original structure at joint C.

1F

A

BC

q

(b)A

BC

2I1Δ

1Δ

q

I

(a)

11F

A

BC

(d)A

BC

q1FP

(c)

= +

1Δ

1Δ1Δ

1Δ

Fig.11.10 Figures to show the development of displacement-method equation(a) original structure; (b) primary system ; (c) primary structure under external loading;(d) primary structure under support rotation


The restraint force 1F can be determined by principle of superposition. That is, the total restraint moment at the artificial restraint of the primary structure due to the combined actions of the external load and the joint rotation 1Δ can be conveniently expressed by superimposing (algebraically adding) the restraint moment due to the external load and the rotation 1Δ subjected individually to the primary structure; that is,

1 1 11PF F F= + (b)

In which 1PF and 11F represent, respectively, the restraint moments at the artificial restraint of the primary structure due to the external load and the unknown rotation 1Δ , each acting alone on the structure. Note th subscri used to denote the restraint moments, 1PFat two pts are and , of the primary structure. The first subscript 1 indicates the location of these restraint moments; the s d subscript P is used to indicate that

11Fecon

1PF is caused by the given external loading, whereas the second subscript 1 of implies that it is due to the unknown rotation . Both of these restraint moments are considered to be positive if they occur in the direction of the unk rotation

11F1Δ

nown 1Δ , which is assumed to be clockwise. Since the unknown rotation is unknown, it is convenient to determine by first evaluating the

restraint moment at the artificial restraint due to a unit value of the unknown ro tion , as shown in Fig.11.10 (d), and then multiplying the restraint moment thus obtained by the unknown m gnitude of the rotation . Thus

For a structure which has more than one primary unknown, the displacement-method equations may de tion. First, let us consider a structure with two primary

unknowns.

11F ta

1Δ

1Δa

1Δ

11 11 1 (c)

In which 11k denotes the restraint moment at the artificial restraint of the primary structure due to the unit value of the rotation 1Δ , which has the unit of moment per unit radian, is referred to as a stiffness coefficient. By substituting equations (b) and (c) into equation (a), we obtain the equilibrium equation

1 11 1 1 0PF k F= Δ + = (11-6)

It is observed that when use displacement method to analyze a statically indeterminate structure, one primary unknown matches one corresponding artificial restraint, and one artificial restraint will yield a equilibrium equation. Therefore, the total number of primary unknowns must be equal to the total number of equilibrium equations.

11.4.2 Canonical equations of displacement method

F k= Δ

be also veloped by the idea used in above subsec

1. Displacement method of two primary unknowns Now let us focus our attention on the rigid frame shown in Fig.11.11 (a) to present how


displacement-method equations for multiple primary unknowns are developed. (1) Determine primary unknowns The frame has one rigid-joint rotation (of joint C) and one independent joint translation (of joint D)

based on the assumption in subsection 11.3.1, so the frame has two primary unknowns, denoted by 1Δ and 2Δ as shown in Fig.11.11 (a).

(2) Develop primary structure and primary system Remove the external loading and add artificial restraints against each of the primary unknowns, thusly

one rotative restraint against joint rotation 1Δ is affixed at joint C and one translating restraint (or link) against joint translation is added at joint D, to form a primary structure, which is an assemblage of 3 single-span indeterminate beams, as shown in Fig.11.11 (b). The primary system can be obtained by subje

D

2Δ

cting the external loading and the primary unknown displacements, which now are the support movements of the single-span indeterminate beams, to the primary structure to form a primary system as shown in Fig.11.11 (c).

(3) evelop displacement-method equations The displacement-method equations are actually equilibrium conditions developed at the artificial

A B

CD

A B

CD

12F22F

2 1Δ =

A B

C D11F

21F

1 1Δ =

1Δ

B

C

A

D

2Δ

P

1F

2F

B

C

A

D

P

(a)

A B

CD

1PF2PF

P

(f)(d) (e)

(c)

1Δ

2Δ 2Δ

1Δ

1Δ

2 1Δ =

(b)

Δ 11 =

Fig.11.11 A rigid frame with two primary unknowns(a) original structure; (b) primary structure; (c) primary system ; (d) primary structure under external loa

1 2

d;(e) primary structure under =1; (f) primary structure under 1Δ Δ =


restraints. They can be established by usi the principle of superposition as follows: ngre

m w

d to it, as shown in Figs.11.11 (e) and (f). By the principle of superposition, the real restraint force at the artificial restraint against prima

① Determine the straint forces corresponding to the primary unknowns at each artificial restraint of the primary structure with only external loading applied on it.

② For each pri ary unknown, determine the restraint forces corresponding to all the primary unknowns at each artificial restraint of the primary structure ith only the unit value of the primary unknown subjecte

ry unknown iΔ due to the unit value of the primary

unknown jΔ , termed ij , must be

ij ij jF k

F

= Δ (a)

Where ijk represents the restraint force against primary unkno i

wn Δ due to the unit value of the

primary unknown jΔ . Note that two subscripts are used to denote the restraint forces ijF and ijk of the

primary structure. The first subscript indicates the location and direction of these restraint forces; the second subscript

i j is used to indicate the restraint rces caused by the unknown rotation jΔ . fo

③ Write a displacement- thod (equilibrium) equation for the location each artificial restraint by setting the algebraic sum of the restraint forces (or moments) of the primary structure due to the external loading and each of the primary unknowns equal to the known joint force (or moment) at the corresponding location the orig

me of

on inal indeterminate structure. The total number of equilibrium equations thus obtained must be equal to the number of the primary unkn

e

PF F F

owns. Since the example has two primary unknowns and no joint forc s exerted on it, there are two equilibrium equations which can be expressed, by principle of superposition, as

1 1 11 12 0PF F F F= + + = ⎫

2 2 21 22 0F ⎬= + + = ⎭(b)

In which 1F and 2F denote, respectively, the total restraint moment at artificial restraint 1 (rotative restraint at joint C) and restraint force at artificial restraint 2 (translating restraint at joint D). 1PF , 11F and

12F represent, respectively, the restraint moments at the artificial restraint 1 of the primary structure due to the external load, the unknown rotation 1Δ and unknown translation 2Δ , each actin the g alone onstructure. Similarly, 2PF , 21F and 22F denote, respec2 of the ary structure due to the external load, the unknown rot

tively, the s at the artificial restraint prim ation

restraint force

1Δ and unknown translation 2Δ , each acting alone on the structure.

By equation (a), we write equation (b) in detail as

11 1 12 2 1

21 1 22 2 2

00

P

P

k k Fk k FΔ + Δ + = ⎫

⎬Δ + Δ + = ⎭ (11-7)


The equation (11-7) is just the displacement-method equation for the indeterminate structure with two primary unknowns, which can be solved by the equation.

2. Displacement method of n primary unknowns For a structure with n primary unknowns, the canonical equations in displacement method may be

written as

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

00

0

n n P

n n P

n n nn n nP

k k k Fk k k F

k k k F

Δ + Δ + + Δ + = ⎫⎪Δ + Δ + + Δ + = ⎪⎬⎪⎪Δ + Δ + + Δ + = ⎭

(11-8)

In which, the coefficient , termed main stiffness coefficient, represents the restraint force (or moment) at

the location and in the direction of ith artificial restraint of the primary structure due to the unit value of primary unknown ; the coefficient , termed secondary stiffness coefficient, represents the restraint

force (or moment) at the location and in the direction of ith artificial restraint of the primary structure due to the unit value of primary unknown

iik

iΔ ijk

jΔ ; the free term iPF denotes the restraint force (or moment) at

the location and in the direction of ith artificial restraint of the primary structure due to the external loading. The meaning of each of the equations in expression (11-8) is that all the restraint forces exerted at the

associated artificial restraint of the primary structure due to the combined actions of the external loading and the primary unknown displacements remain in equilibrium. So the expression (11-8) is the equilibrium conditions of the primary system. The total number of equilibrium equations is equal to the number of the primary unknowns. Consequently, the n primary unknowns can be solved by the equations.

It should be explained that when develop the displacement-method equations all primary unknowns, are assumed to be positive. That is, positive joint rotation is clockwise; positive joint

translation makes the chord rotation to be clockwise too. The positive answer of a primary unknown implies that the direction of the primary unknown is coincident with the assumed one and vice versa.

1 2, , , nΔ Δ Δ

Obviously, the stiffness coefficient is the restraint force (or moment) at the location and in the

direction of ith artificial restraint of the primary structure due to the unit value of jth support movement associated with

ijk

jΔ of the primary structure. Recalling from previous discussion, it should be the shape

constant and can be obtained from table 11-1 listed in section 11.2. The free term iPF is the restraint force

(or moment) at the location and in the direction of ith artificial restraint of the primary structure due to the external loading, it should be loading constant defined in section 11.2 and may be determined by using table 11-2 listed in section 11.2.

Since is a main coefficient it is always great than zero; is a secondary coefficient, so it may iik ijk

11.5 Analysis of Statically Indeterminate Beams and Rigid Frames without Sidesway 393

be great than or less than or equal to zero. Recalling from the law of reciprocal reactions, we write

ij jik k= (11-9)

11.5 Analysis of Statically Indeterminate Beams and Rigid Frames without Sidesway

The following examples will illustrate the analyzing procedure for statically indeterminate beams and rigid frames without sidesway by using displacement method.

Example 11-1

Determine the internal forces for the continuous beam shown in Fig.11.12 (a) by displacement method. The flexural rigidity of the beam is constant, i.e., constantEI = .

Solution

(1) Primary unknowns Since the two-span continuous beam is connected by rigid joint B, the joint rotation will be chosen

as the primary unknown of the beam. 1Δ

(2) Primary structure and primary system Remove all external loads and add an artificial restraint against the primary unknown (the rotation of

joint B) at joint B to form a primary structure. The primary system will be obtained by subjecting the external loading and the joint rotation to the primary structure as shown in Fig.11.12 (b). 1Δ

(3) Develop displacement-method (or equilibrium) equation The equilibrium equation is based on the requirement that the restraint moment of the artificial

restraint induced by the combined actions of the external loading and the joint rotation must be equal to zero because there is no joint moment acting at the joint of the original structure. By the principle of superposition, we write

1Δ

11 1 1 0Pk FΔ + =

(4) Calculation of stiffness coefficient The coefficient actually is the restraint moment induced at the artificial restraint due to the unit

value of the joint rotation . Referring to the table 11-1, we find 11k

1Δ3BCM i= 4BAM i= 2ABM i=

The bending moment diagram of the primary beam due to 1 1Δ = is shown in Fig.11.12 (c). By the moment equilibrium condition due to as shown in Fig.11.12 (d), the restraint moment at the artificial restraint due to must satisfy the following equilibrium condition. That is,

1 1Δ = 11k1 1Δ =

0BM =∑ 11 4 3 7k i i i= + =


In which, 6EIi = , liner flexural rigidity of the beam.

(5) Calculation of free term The free term 1PF actually is the restraint moment induced at the artificial restraint due to the

external loading, termed fixed-end moment. Referring to the table 11-2, we find 2 22 6 6kN m

12 12F FAB BA

qlM M ×− = = = = ⋅

3 3 16 6 18kN m16 16

FBC

PlM × ×= − = − = − ⋅

The bending moment diagram of the primary beam due to the external loading is shown in Fig.11.12 (e). By the moment equilibrium condition due to the external load as shown in Fig.11.12 (f), the restraint moment, 1PF , at the artificial restraint due to the external loading must satisfy the following equilibrium condition. That is,

0BM =∑ 1 18 6 0PF + − = 1 12kN mPF = − ⋅

(6) Solve primary unknown

A B C2kN/m 16kN

3m6m 3m

AB C1 1Δ =

2i

4i

1 1Δ = A

6

B C

2kN/m

16kN6

AB

C2kN/m 16kN

1Δ

3i

11k

4i

(b)(a)

(c)

(d)

(e)

(f)18

1PF

6

1 diagramM diagramPM

1

11

Fig.11.12 Figures of example 11-1(a) original structure; (b) primary system; (c) bending moment diagram due to 1; (d) equilibrium condition of ;(e) bending k

Δ =

1

moment diagram due to external loading; (f) equilibrium condition of PF

18

3i15

(unit: kN m)⋅


Substituting the numeric values of stiffness coefficient and free term 11k 1PF , we solve the primary unknown to be 1Δ

11

11

12 11.7147

PFk i

Δ = − = =i

(7) Draw bending moment diagram By the superposing equation for bending moment, i.e., 1 1 PM M M= Δ + , the member end moments

are calculated to be

11.7142 2 6 2.57kF

AB ABM i M ii

⎛ ⎞= Δ + = − = − ⋅⎜ ⎟⎝ ⎠

N m

11.7144 4 6 12.86kN mF

BA BAM i M ii

⎛ ⎞= Δ + = + = ⋅⎜ ⎟⎝ ⎠

11.7143 3 18 12.86kN mF

BC BCM i M ii

⎛ ⎞= Δ + = − = − ⋅⎜ ⎟⎝ ⎠

Once the member end moments have been determined, the bending moment diagram can be drawn by superimposing method used previously in the text, as shown in Fig.11.13 (a). Mention again that the bending moment diagram must be drawn on the tensile side of the beam.

(8) Draw shear force diagram Once the member end moments are determined the member end shear forces can be determined by

consideration of the free body diagram of each member. By the free body diagrams shown in Fig.11.13 (b), we write

0BM =∑ 12.86 2 6 3 2.57 4.29kN

6ABQ − + × × += =

0AM =∑ 12.86 2 6 3 2.57 7.72kN

6BAQ − − × × += = −

0BM =∑ 16 3 12.86 5.86kN

6CBQ − × += = −

0CM =∑ 16 3 12.86 10.14kN

6BCQ × += =

After the member end shear forces have been determined, the shear force diagram can be drawn by the method used previously in the text, as shown in Fig.11.13 (c).

(9) Verification By considering the moment equilibrium condition of joint B, we write


12.86 12.86 0BM = + − =∑

Obviously, the moment equilibrium condition is indeed satisfied. By consideration of the equilibrium condition of vertical component projections of all the forces

acting on the entire beam, we write

0Y =∑ 4.29 17.86 5.86 2 6 16 0+ + − × − ≈

Apparently, the equilibrium condition is satisfied as well. The above verification implies that the calculation for the example is correct.

ending moment diagram for the rigid frame shown in Fig.11.14 (a) by displacement meth

on

ary unknowns of 5 members connected by rigid joints D and E as shown in the

figur

Example 11-2

Determine the bod.

Soluti

(1) PrimBecause the rigid frame consists e, the rotation of joint D, 1Δ , and the rotation of joint E, 2Δ , will be chosen as the primary

unknowns of the frame. (2) Primary structure and primary system

o cial restraints against the primary unknowns, the rotations of jo

Rem ve all external loads and add two artifiints D and E, at the two joints to form a primary structure. The primary system will be obtained by

subjecting the external loading and the joint rotations 1Δ and 2Δ to the primary structure as shown in Fig.11.14 (b).

17.57

AB

C

12.86

1.29

2.57 ( )9(24)

CA B

10.14 10.14

5.86 5.867.72

4.29

C

2kN/m 16kN(b)(a)

(c)

(unit: kN m)⋅

(unit: kN)

BCQ CBQ

BA

BAQABQ

Fig.11.13 Internal force diagrams of example 11-1(a) bending moment diagram; (b) free body diagrams ; (c) shear force diagram

2.5712.86


(3) Develop displacement-method (or equilibrium quationEach of th

) e s e equilibrium equations is based on the requirement that the restraint moments yielded at

each h xternal loading and the joint rotations of t e artificial restraints due to the combined effects of the e

1Δ and 2Δ must be equal to zero, because there is no joint moment acting at the joints for the original structure. By superposing the combined effects of the external loading and the joint rotations 1Δ and 2Δ ,

write t displacement-method equations to be

11 1 12 2 1

21 1

0Pk k Fk

we he

0Pk F22 2 2

Δ + Δ + =Δ + Δ + =

(4) Calculation of stiffness coefficients In order to calculate the stiffness coefficients, we first determine the linear flexural rigidity of each

mem ter, the internal forces of a statically indeterminate structure are o

ber of the frame. As discussed in last chapnly related to the relative flexural rigidity of each member of the structure. Thusly, we assume

04I 04IC

D

A B

E

20kN/m

06I

02I 03I

4m4m 6m

F

20kN/m

A B

C

1ΔD E

FΔ2

4m4m 6m

(b)

(a)

4m4m

Fig.11.14 Figures of example 11-2(a) original structure; (b) primary system


04 14CD EFEIi i= = = 06 1

6DEEIi = =

02 14 2DAEIi = = 03 3

4 4EI

EBi = =

Actually, the coefficient is the restraint moment induced at ith artificial restraint due to the unit

value of the joint rotation

ijk

jΔ . Referring to the table 11-1, we find the shape constants due to 1 1Δ = as

follows

3 3DC CDM i= = 14 4 2M i2DA DA= = × =

4 4DE DEM i= = 2 2ED DEM i= =

And the shape constants due to must be

2 1Δ =

A B

C

1 1Δ =

DE F

1 1Δ =4 DAi

3 DCi

4 DEi

2 DAi

11k

4 DAi

3 DCi 4 DEiD

21k

E

2 DEi

(b)(a) (c)

1 diagramM

1 1Δ =

(b)(a) (c)

A B

C

2 1Δ =

D E F2 1Δ =2 1Δ =

2 DEi4 EBi

3 EFi

2 EBi

22k

4 EBi

3 EFi4 DEi E12k

2 DEi

D

2 diagramM

1

1 11 21

Fig.11.15 diagram of example 11-2(a) diagram; (b) equilibrium condition of ; (c) equilibrium condition of

MM k k

2

1 12 22


Mk k

2 DEi

4 DEi

M


4 4ED EDM i= = 3 3EF EFM i= = 4 3EB EBM i= =

The bending moment diagram e primary s of th frame due to 1 1Δ = and due to are shown in Fig.11.15 (a) and Fig.11.16 (a), respectiv

2 1Δ =ely.

By the moment equilibrium conditions due to 1 1Δ = as show gs.11.15 (b) moments, 11k and 21k , at the artificial restraints 1 and 2 m y the fo equilibrium cond

n in Fi and (c), the restraint ust satisf llowing

itions. That is,

0DM =∑ 11 3 4 4 3 2 4 9CD DA DEk i i i= + + = + + =

0EM =∑ 21 2 2DEk i= =

By the mome ilibrium con due to 2 1Δ =nt equ ditions as shown in Figs.11.16 (moments, and , at the ar l restraints 1 cond

b) and (c), the restraint

12k 22k tificia and 2 must satisfy the following equilibrium itions. That is,

0DM =∑ 12 212 2DEk i k= = =

0EM =∑ 022 4 4 3 3 3 4 1DE EB EFk i i i= + + = + + =

(5) Calculation of free terms In fact, the free term iPF is the int moment induced at ith artificial restrai

loadi loading constant. Referring to the table 11-2, we find restra nt due to the external

ng, termed fixed-end moment or 2 220 6 60kN m

12 12F

DE EDqlM M ×

= − = − = − = − ⋅ F

221 1 20 4 40kN m

8 8FEFM ql × ×= − = − = − ⋅

40

E

2PF

60D

1PF

60

2F (b) (c)P

E

A B

CD

1F(a) P

606040

F

diagram (unit: kN m)

PM⋅

1 2


P

P P

MM F FP


The bending moment diagram of the primary frame due to the external loading is shown in Fig.11.17 (a). By the moment equilibrium conditions due to the external load as shown in Figs.11.17 (b) and (c), the restraint moments, 1PF and 2PF , at the artificial restraint 1 and 2 must satisfy the following equilibrium conditions. That is,

0DM =∑ 1 60 0PF + = 1 60kN mPF = − ⋅

0EM =∑ 2 40 60 0PF + − = 2 20kN mPF = ⋅

(6) Solve primary unknown Substituting the numeric values of stiffness coefficients and free terms into displacement-method

equations, we write

1 2

1 2

9 2 60 02 10 20Δ + Δ − =⎧

⎨ Δ + Δ + =⎩ 0

Thusly, the primary unknown and1Δ 2Δ are solved to be

1 7.442Δ = 2 3.488Δ = −

(7) Draw bending moment diagram By the superposing equation for bending moment, i.e., 1 1 2 2 PM M M M= Δ + Δ + , the member end

moments are calculated to be

112 2 7.442 7.44kN m2AD DAM i= Δ = × × = ⋅

114 4 7.442 14.88kN m2DA DAM i= Δ = × × = ⋅

232 2 ( 3.488) 5.23kN m4BE BEM i= Δ = × × − = − ⋅

234 4 ( 3.488) 10.46kN m4EB BEM i= Δ = × × − = − ⋅

13 3 1 7.442 22.33kN mDC DCM i= Δ = × × = ⋅

1 24 2 4 1 7.442 2 1 ( 3.488) 60 37.21kN mFDE DE DE DEM i i M= Δ + Δ + = × × + × × − − = ⋅

1 22 4 2 1 7.442 4 1 ( 3.488) 60 60.93kN mFED DE DE EDM i i M= Δ + Δ + = × × + × × − + = ⋅

23 4 1 ( 3.488) 40 50.46kN mFEF EF EFM i M= Δ + = × × − − = − ⋅

Once the member end moments have been determined, the bending moment diagram can be drawn by superimposition method used previously in the text, as shown in Fig.11.18.

11.6 Analysis of Statically Indeterminate Rigid and Bent Frames with Sidesway 401

11.6 Analysis of Statically Indeterminate Rigid and Bent Frames with Sidesway

It is realized from last section that for the structures whose joint rotations are chosen as the primary unknowns, the physical meanings of the corresponding displacement-method equations are the moment equilibrium equations of the joints. Similarly, for the structures whose joint translations are chosen as the primary unknowns, the physical meanings of the corresponding displacement-method equations are the force component projection equilibrium equations of the forces acting on the entire or a portion of the primary structure, which is isolated by the cut along the associated directions of the translations. Following examples will further illustrate the application of displacement method to the analysis of statically indeterminate rigid and bent frames with sidesway.

Example 11-3

Draw the internal force diagrams for the rigid frame shown in Fig.11.19 (a) by using displacement method.

Solution

(1) Primary unknowns By the similar analysis with that of the examples solved in last section, the rotation of joint C, denoted

by , and the translation of joint D, represented by 1Δ 2Δ , will be chosen as the primary unknowns of the frame.

(2) Primary structure and primary system Remove all external loads and add two artificial restraints against the primary unknowns (the rotations

of joint C and the translation of joint D) at joints C and D to form a primary structure. The primary system

diagram(unit: kN m)M

⋅A B

CD E

F( )40

40.93

60.93

22.3337.21

50.46

14.7710.4614.88

5.237.44

( )90

Fig.11.18 Bending moment diagram of example 11-2


will be obtained by subjecting the external loading and the joint displacements and 1Δ 2Δ to the primary structure as shown in Fig.11.19 (b).

(3) Develop displacement-method (or equilibrium) equations The first equilibrium equation is based on the requirement that the restraint moments yielded at the

artificial restraint added at joint C due to the combined effects of the external loading and the rotation of joint C and the translation of joint D, 1Δ and 2Δ , must be equal to zero, because there is no joint moment acting at the joint C for the original structure. The second equilibrium equation is based on the condition that the restraint forces yielded at the artificial restraint added at joint D due to the combined effects of the external loading and the rotation of joint C and the translation of joint D, 1Δ and , must be equal to zero, because there is no joint force acting at the joint D for the original structure. By superposing the combined effects of the external loading and the joint displacements

2Δ

1Δ and , we write the displacement-method equations to be

2Δ

11 1 12 2 1 0Pk k FΔ + Δ + =

21 1 22 2 2 0Pk k FΔ + Δ + =

(4) Calculation of stiffness coefficients In order to calculate the stiffness coefficients, we first calculate the linear flexural rigidity of each

member of the frame. Thusly, we assume

4AC BDEIi i= = = i

3 26CDEIi i= =

Referring to the table 11-1, we find the shape constants due to 1 1Δ = as follows

4 4CA CAM i= = i 2 2AC CAM i i= = 3 6CD CDM i i= =

And the shape constants due to are 2 1Δ =

B

C D

A6m

B

C

A

1Δ

D2Δ

3EI

EI EI

(b)(a)

4m

10kN/m



6 6 1.54

ACAC CA

CA

i iM Ml

= = − = − = − i

3 34

BDBD

BD

iM il

= − = −

The bending moment diagrams of the primary frame due to 1 1Δ = and due to are shown in Fig.11.20 (a) and Fig.11.21 (a), respectively.

2 1Δ =

By the equilibrium conditions of the free body diagrams shown in Figs.11.20 (b) through (e), the restraint moment and restraint force , at the artificial restraints 1 and 2 must satisfy the following equilibrium conditions. That is,

11k 21k

0CM =∑ 11 3 4 6 4 10CD CAk i i i i i= + = + = [see Fig.11.20 (b)]

0X =∑ 21CA DBQ Q k+ = [see Fig.11.20 (c)]

0AM =∑ 4CA AC CAQ M M 0× + + = [see Fig.11.20 (d)]

( ) (2 4 ) 1.54 4

AC CACA

M M i iQ i+ += − = − = −

0BM =∑ 4 0DB BDQ M× + = 0DBQ = [see Fig.11.20 (e)]

Thusly, we find

21 1.5k i= −

Note that the coefficient is obtained by considering the force component projection equilibrium equation of the forces acting on a portion of the primary structure, which is isolated by the cut through the top of the columns and along the associated direction of the translation of joint D, as shown in Fig.11.20 (c).

21k

By the equilibrium conditions of the free body diagrams shown in Figs.11.21 (b) through (e), the restraint moment and restraint force , at the artificial restraints 1 and 2 must satisfy the following equilibrium conditions. Thusly,

12k 22k

0CM =∑ 126 0

4ACik + = 12 211.5k i k= − = [see Fig.11.21 (b)]

0X =∑ 22CA DBQ Q k+ = [see Fig.11.21 (c)]

( ) ( 1.5 1.5 ) 34 4

AC CACA

M M i iQ i+ − −= − = − =

4 [see Fig.11.21 (d)]

( 0.75 ) 0.754 4BD

DBM iQ −

= − = − =4

i [see Fig.11.21 (e)]


Consequently, we obtain

223 0.754 4

i i+ = k 223.75

4k = i

(5) Calculation of free terms The fixed-end forces or loading constants can be determined by referring to the table 11-2. Thusly, we

write

A B

C

3 CDi

1 1Δ = D

1 1Δ =

4 CAi

2 CAi

11k

C

4 CAi

3 CDi

A

CCAM

CAQ

ACQ

ACM

DBQ

BBDQ

BDM

D

CAQ DBQC

D 21k(b)(a) (c)

(d) (e)

4m 4m

1 diagramM

1 11 21

Fig.11.20 and free body diagrams of example 11-3(a) diagram; (b) equilibrium condition of ; (c) equilibrium condition of ;(d) free body diagram of member ; (e) fre

M k kCA e body diagram of member DB

11k

12k

A

CCAM

CAQ

ACQ

ACM

BBDQ

DBQ

BDM

D

CD

CAQ DBQ22k

B

C

2 1Δ =

D

A 34BDi6

4ACi

64ACi

C 0

64ACi

12k

(d) (e)2 diagramM

(b)(a) (c)2 1Δ = 22k

4m 4m

2

2 12 22

Fig.11.21 and free body diagrams of example 11-3(a) diagram; (b) equilibrium condition of ; (c) equilibrium condition of ;(d) free body diagram of member ; (e) fr

MM k k

CA ee body diagram of member DB


221 1 10 4 20kN m

8 8FBDM ql × ×

= − = − = − ⋅

The bending moment diagram of the primary frame due to the external loading is shown in Fig.11.22 (a). By the equilibrium conditions of the free body diagrams shown in Figs.11.22 (b) through (e), the restraint moment 1PF and restraint force 2PF at the artificial restraints 1 and 2 must satisfy the following equilibrium conditions. That is,

0CM =∑ 1 0PF = [see Fig.11.22 (b)]

0X =∑ 2F FCA DB PQ Q F+ = [see Fig.11.22 (c)]

0FCAQ = [see Fig.11.22 (d)]

10 4 2 15kN4

FF BDDB

MQ + × ×= − = − [see Fig.11.22 (e)]

Finally, we find

2 15PF = −

Note that the restraint force 2PF is obtained by considering the force component projection equilibrium equation of the forces acting on a portion of the primary structure, which is isolated by the cut through the top of the columns and along the associated direction of the translation of joint D, as shown in Fig.11.22 (c).

(6) Solve primary unknown Substituting the numeric values of stiffness coefficients and free terms into displacement-method

1PF

C

0

0

B

C

A

D 2PF1PF

20A

C0

0

FCAQ

FACQ

FBDQ

C D

FCAQ F

DBQ

B

D

FBDQ

FDBQ

(d) (e) diagram (unit: kN m)

PM⋅

(a) (b) (c)

10kN/m

2F P

4m 4m

10kN/m

1 2

Fig.11.22 and free body diagrams of example 11-3(a) diagram; (b) equilibrium condition of ; (c) equilibrium condition of ;(d) free body diagram of member ; (e) f

P

P P

MM F F

CA ree body diagram of member DBP


equations, we write

1 2

1 2

10 1.5 03.751.5 15 0

4

i i

i i

Δ − Δ =⎧⎪⎨− Δ + Δ − =⎪⎩

Thusly, the primary unknown and1Δ 2Δ are solved to be

113.158i

Δ = 2121.05i

Δ =

The positive answer implies that the real directions of joint displacements and coincide with the assumed ones.

1Δ 2Δ

(7) Draw bending moment diagram By the superposing equation for bending moment, i.e., 1 1 2 2 PM M M M= Δ + Δ + , the member end

moments are calculated to be

1 26 42 2 3.158 21.05 25.26kN m

4 6AC

AC ACiM i= Δ − Δ = × − × = − ⋅

1 26 44 4 3.158 21.05 18.94kN m

4 6AC

CA ACiM i= Δ − Δ = × − × = − ⋅

13 6 3.158 18.95kN mCD CDM i= Δ = × = ⋅

23 3 21.05 20 35.79kN m

4 4FBD

BD BDiM M= − Δ + = − × − = − ⋅

Once the member end moments have been determined, the bending moment diagram can be drawn by superimposing method used previously in the text, as shown in Fig.11.23 (a).

(8) Draw shear and axial force diagrams

( )20

B

C

A

D

18.95

18.94

35.79

2.48

B

C

A

D

28.95

11.05 11.05

3.16 3.16

C

BA

D11.05

3.16

3.16 3.16

3.16

(b)(a) (c)

(unit: kN m) (unit: kN) (unit: kN)

Fig.11.23 Internal force diagrams of example 11-3 (a) diagram; (b) diagram; (c) diagramM Q N

⋅

11.05

25.26 11.05


After the member end moments are determined the member end shear forces can be obtained by consideration of equilibrium conditions of the free body diagram of each member; the member end axial forces can be obtained by consideration of equilibrium conditions of each joint. By doing thusly, we draw the shear and axial force diagrams as shown in Figs.11.23 (b) and (c).

Example 11-4

Draw the bending moment and shear force diagrams for the bent frame shown in Fig.11.24 (a) by using displacement method.

Solution

(1) Primary unknowns The bent frame consists of 3 columns hinged by two beams as shown in the figure. The translation will

occur at the top of the three columns under the action of the external load. However, if neglect the changes of member lengths the only independent joint displacement is the horizontal translation of joints D, E and F, denoted by . So the translation will be chosen as the primary unknown of the frame. 1Δ

(2) Primary structure and primary system Remove the external load and add one artificial restraint against the primary unknown at joint F to

form a primary structure. The primary system will be obtained by subjecting the external loading and the joint displacement to the primary structure as shown in Fig.11.24 (b). 1Δ

(3) Develop displacement-method (or equilibrium) equation The equilibrium equation is based on the condition that the restraint forces yielded at the artificial

restraint added at joint F due to the combined effects of the external loading and the translation must be equal to zero, because there is no joint force acting at the joint F for the original structure. By superposing the combined effects of the external loading and the joint displacement

1Δ

1Δ , we write the displacement-method equation to be

1Δ1Δ 1Δ

AB

C

D FEP

1I1h2I

3h2h

3I AB

C

D E FP

(b)(a)



11 1 1 0Pk FΔ + =

(4) Calculation of stiffness coefficient In order to calculate the stiffness coefficient, we assume the linear flexural rigidity of each member of

the frame to be

11

1AD

EIi ih

= = 22

2BE

EIi ih

= = 33

3CF

EIih

i= =

Referring to the table 11-1, we find the shape constants due to 1 1Δ = as follows

1

1

3AD

iMh

= − 2

2

3BE

iMh

= − 3

3

3CF

iMh

= −

The bending moment diagram of the primary frame due to 1 1Δ = is shown in Fig.11.25 (a). By the equilibrium condition of the free body diagram shown in Fig.11.25 (b), the restraint force

at the artificial restraint must satisfy the following equilibrium condition. That is, 11k

0X =∑ 11DA EB FCQ Q Q k+ + = [see Fig.11.25 (b)]

12

1 1

3ADDA

M iQh h

= − = [see Fig.11.25 (c)]

22

2 2

3BEEB

M iQh h

= − = [see Fig.11.25 (d)]

ADQADM

DAQ

B

BEM

E

BEQC

CFM

F

CFQ

D E F

DAQ EBQFCQ

AB

C

D E F 11k

3

3

3ih

1

1

3ih

2

2

3ih

1 diagramM

A

1 1Δ =1 1Δ = 1 1Δ =

(d) (e)

(a)

(c)

(b)

11k 1h

2h 3h

EBQ FCQD

1

1 11

Fig.11.25 and free body diagrams of example 11-4(a) diagram; (b) equilibrium condition of ; (c) free body diagram of member ;(d) free body diagram of member

MM k DA

E ; (e) free body diagram of member B FC


32

3 3

3CFFC

M iQh h

= − = [see Fig.11.25 (e)]

Thusly, we find to be 11k31 2

11 2 2 21 2 3

3( )ii ikh h h

= + +

(5) Calculation of free term Note that the concentrated force P cannot yield any internal forces in the columns of the primary

structure. So by the equilibrium condition of the free body diagram shown in Fig.11.26 (b), the restraint force 1PF at the artificial restraint must satisfy the following equilibrium condition. That is,

1F F FDA EB FCQ Q Q P F+ + − = P

Because of

0FDAQ = [see Fig.11.26 (c)]

0FEBQ = [see Fig.11.26 (d)]

0FFCQ = [see Fig.11.26 (e)]

Thusly, we find

1PF P= −

(6) Solve primary unknown Substituting the numeric values of stiffness coefficient and free term into the displacement-method

equation, we write

D

0

FDAQ

FADQ

B

E

0

FBEQ

FEBQ

C

F

0

FCFQ

FFCQ

D

E

FPFDAQ F

FCQFEBQ

AB

C

D

E

FP 1F P

(b)(a)1F P

A

(d) (e)(c)

1h 2h 3h

1

Fig.11.26 and free body diagrams of example 11-4(a) diagram; (b) equilibrium condition of ; (c) free body diagram of member ;(d) free body diagram of member

P

P P

MM F DA

E ; (e) free body diagram of member B FC


31 212 2 2

1 2 3

3( ) 0ii i Ph h h

+ + Δ − =

Thusly, the primary unknown is solved to be 1Δ

1 331 2

2 2 2 211 2 3

3( ) 3 m

m m

P Pii i i

h h h h=

Δ = =+ + ∑

(7) Draw bending moment diagram By the superposing equation for bending moment, i.e., 1 1 PM M M= Δ + , the member end moments

are calculated to be

1

1 11 3

12

1

3AD

m

m m

iPi hM

ihh=

= − Δ = −

∑

2

2 21 3

22

1

3BE

m

m m

iPi hM

ihh=

= − Δ = −

∑

3

3

3 31 3

21

3CF

m

m m

iPiM

ihh=

= − Δ = −

∑h

Once the member end moments have been determined, the bending moment diagram can be drawn by superimposing method as shown in Fig.11.27 (a).

(8) Draw shear force diagram After the member end moments are determined the member end shear forces can be obtained by

consideration of equilibrium conditions of the free body diagram of each member. Thusly, we calculate

12

13

21

3AD DA

m

m m

iPhQ Qih=

= = −

∑

122

3

21

3BE EB

m

m m

iPhQ Qih=

= = −

∑

123

3

21

3CF FC

m

m m

iPhQ Qih=

= = −

∑

The shape of the shear force diagram is shown in Fig.11.27 (b).

AB

C

D FEP

AB

C

D FE

(b)(a)

ADMM BE

M CF

Fig.11.27 Internal force diagrams of example 11-4 (a) diagram; (b) shape of diagram M Q


11.7 Analysis of Statically Indeterminate Symmetric Structures 411

11.7 Analysis of Statically Indeterminate Symmetric Structures

It is observed from last section that when employing displacement method to analyze a statically indeterminate structure, the higher the degrees of its indeterminacy, the more the required computational effort. The main calculation work is the solution of simultaneous linear equations, i.e., the evaluation of stiffness coefficients and free terms in equilibrium conditions and the solution of the simultaneous linear equations. However, many structures, because of aesthetic and/or functional requirements, are arranged in symmetric forms. If a symmetric structure is linearly elastic, the response (i.e., member forces and deformations) of the entire structure under any general loading can be obtained from the response of one of its portions separated by the axes of symmetry. Thus only a portion (usually half) of the symmetric structure needs to be analyzed. In the section, we discuss how to utilize symmetric property of a structure to simplify the computational effort required in the analysis of the structure.

As discussed in chapter 10, any one of asymmetric loadings may be decomposed into the superposition of one symmetric component and one antisymmetric component. When a symmetric structure is subjected to a symmetric loading the response characteristics of the symmetric primary structure due to the external loading is symmetric as well. Similarly, when a symmetric structure is subjected to an antisymmetric loading, the response characteristics of the symmetric primary structure due to the external loading are also antisymmetric. Thus, to determine the response (i.e., member forces and deformations) of the entire structure, we need to analyze only half the structure, on either side of the axis of symmetry, with symmetric or antisymmetric boundary conditions at the axis. The response of the remaining half of the structure can then be obtained by reflection. Following examples will illustrate the procedure of the analysis.

Example 11-5

Draw the bending moment diagram for the symmetric rigid frame shown in Fig.11.28 (a) by using displacement method.

Solution

(1) Primary unknowns Although the frame has 2 rigid-joint rotations and 1 joint translation, it is a symmetric structure under

symmetric loading. So only half of the structure shown in Fig.11.28 (b) may be selected to analyze. With the similar analysis discussed in last section, the rotation of joint C of the half structure, denoted by 1Δ , will be chosen as the primary unknown of the frame.

(2) Primary structure and primary system Remove the uniformly distributed external load and add one artificial restraint against the primary


unknown (the rotation of joint C) at joint C of the half structure to form a primary structure. The primary system will be obtained by subjecting the external loading and the joint displacement to the primary structure as shown in Fig.11.28 (c).

1Δ

(3) Develop displacement-method (or equilibrium) equation The displacement-method equation is based on the requirement that the restraint moment yielded at

the artificial restraint added at joint C due to the combined effect of the external loading and the rotation of joint C must be equal to zero, because there is no joint moment acting at the joint C for the original structure. Thusly, by superposing the combined effects of the external loading and the joint displacement

, we write the displacement-method equation to be 1Δ

11 1 1 0Pk FΔ + =

(4) Calculation of stiffness coefficient 11kBy referring to the results listed in table 11-1, the bending moment diagram of the primary frame due

to is shown in Fig.11.29 (a). Then, the stiffness coefficient can be determined by the moment equilibrium condition of the free body diagram shown in Fig.11.29 (b) to be

1 1Δ = 11k

113 44

3 4CE CAEI EIk i i= + = + = 2i

In which, assume EI i= . (5) Calculation of free term 1PF The fixed-end forces or loading constants can be determined by referring to the table 11-2. Thusly, we

A B

C D

6kN/m

6m

EI EI

3EI

6kN/m

1Δ

A

EC

6kN/m

3m

3EI

EI

A

EC

EI

3m

3EI

(b)(a) (c)

4m 4m 4m

Fig.11.28 Figures of example 11-5(a) original structure; (b) half of the original structure; (c) primary system


write 2

21 1 6 3 18kN m3 3

FCEM ql × ×= − = − = − ⋅

221 1 6 3 9kN m

6 3FECM ql × ×= − = − = − ⋅

The bending moment diagram of the primary frame due to the external loading is shown in Fig.11.29 (c). By the moment equilibrium condition of the free body diagram shown in Fig.11.29 (d), we find

1 18kN mPF = − ⋅


equation, we obtain

12 18 0iΔ − = 19i

Δ =

(7) Draw bending moment diagram By the superposing equation for bending

moment, i.e., 1 1 PM M M= Δ + , the bending moment diagram for the half of the structure can be obtained and another half of the bending moment diagram can be constructed by reflection, as shown in Fig.11.30.

C

4 CAi

CEi

A

EC11k

4 CAi

CEi

2 CAi

CEi1 1Δ =

1 1Δ =

A

EC

18

9

6kN/m

1PF

18

(d)(b) (c)(a)11k

1F P

C

1

1 11 1

Fig.11.29 and of half structure(a) diagram; (b) equilibrium condition for ; (c) diagram; (d) equilibrium condition for

P

P P

M Mk M F

(unit: kN m)⋅

M

( )27

18

99

9 9

4.5 4.5A B

C D

(unit: kN m)⋅

Fig.11.30 diagram of example 11-5M


Example 11-6

Draw the bending moment diagram for the symmetric rigid frame shown in Fig.11.31 (a) by displacement method.

Solution

(1) Primary unknowns Although the closed rigid frame has 4 rigid-joint rotations, it is a symmetric structure with two axes of

symmetry, indicated by x-x and y-y, under symmetric loading. Therefore, only one quarter of the structure shown in Fig.11.31 (b) may be selected to analyze. The rotation of joint A of the quarter of the structure, denoted by , will be chosen as the primary unknown of the closed frame. 1Δ

(2) Primary structure and primary system Remove the uniformly distributed external load and add one artificial restraint against the primary

unknown at joint A of the quarter structure to form a primary structure. The primary system will be obtained by subjecting the external loading and the joint displacement 1Δ to the primary structure as shown in Fig.11.31 (c).

(3) Develop displacement-method (or equilibrium) equation The displacement-method equation is based on the requirement that the restraint moment yielded at

the artificial restraint added at joint A due to the combined effects of the external loading and the rotation of joint A must be equal to zero, because there is no joint moment acting at joint A of the original structure. Thusly, by superposing the combined effects of the external loading and the joint displacement 1Δ , we

EI

EI

EI

EI

A

C

D

qy

x

qa a

B AD

C

q

a

D

C

q

a

A EIEI

EI EI

1Δ

y

(b)(a) (c)a

a a a

Fig.11.31 Figures of example 11-6(a) original structure; (b) one quarter of the structure;(c) primary system

x


write the displacement-method equations to be

11 1 1 0Pk FΔ + =

(4) Calculation of stiffness coefficient 11kBy referring to the computing results listed in table 11-1, the bending moment diagram of the primary

frame due to can be determined as shown in Fig.11.32 (a). Thus, the stiffness coefficient can be determined by the moment equilibrium condition of the free body diagram shown in Fig.11.32 (b) to be

11k1 1Δ =

11 2k i=

Where . /i EI a=

q

(5) Calculation of free term 1PF The fixed-end forces or loading constants can be determined by referring to the table 11-2. Thusly, we

write

213

FADM qa= − 21

6FDAM qa= −

The bending moment diagram of the primary structure due to the external loading is shown in Fig.11.32 (c). By the moment equilibrium condition of the free body diagram shown in Fig.11.32 (d), we find

21

13PF q= − a


equation, we obtain

A D

C

1 1Δ =

11k

i

iAA

D

C

2

3qa

(b)(a) (c) (d)

2

6qa

1PF2

3qaA

11k 1F P

1

1 11 1

Fig.11.32 and of one quarter structure(a) diagram; (b) equilibrium condition for ; (c) diagram; (d) equilibrium condition for

P

P P

M MM k M F

i ii

i


2

1 6qa

iΔ =2

1123

i qaΔ − = 0

(7) Draw bending moment diagram

1 1 PM M M= Δ +By the superposing equation for bending moment, i.e., , the bending moment diagram for the quarter of the structure can be obtained and other portion of the bending moment diagram can be constructed by reflections, as shown in Fig.11.33.

11.8 Development of Displacement-Method Equations by Direct Stiffness Method

In this section, we present an alternative approach to develop the displacement-method equations of a statically indeterminate structure by direct stiffness method. So called direct stiffness method means that when establish the displacement-method equations of a structure, the primary system used in previous section is not needed. The displacement-method equations may be directly developed by considering the equilibrium conditions of joints or a portion of the structure in terms of slope-deflection equations of associated members. The slope-deflection equations relate the moments at the ends of a member to the rotations and displacement of its ends and the external loading applied to the member.

11.8.1 Slope-deflection equations of prismatic members

In section 11.2, we have already discussed the shape constants and loading constants for various single prismatic members. That is, the member end forces of various single-span indeterminate prismatic beams separately due to their end displacements and external loadings. If the end forces of a member are caused by the combined effects of external loads and its end displacements, they may be determined by using principle of superposition, i.e., by superimposing corresponding loading and shape constants together, which have been tabulated in table11-1 and table 11-2 in section 11.2. The relationship between the end

2

6qa 2

6qa

2

3qa

2

3qa

2

6qa2

6qa

Fig.11.30 diagram of example 11-6M

11.8 Development of Displacement-Method Equations by Direct Stiffness Method 417

moments and the end displacements and the external loading applied on the member is named slope-deflection equation of the member, which is the basis of the section. Thusly doing, we can obtain the slope-deflection equations for three kinds of prismatic members as follows.

(1) Member with two ends fixed (or rigidly connected) Focus our attention on an arbitrary prismatic member AB with the two ends fixed as shown in

Fig.11.31. When the member is subjected to external load and support settlements it deforms, as shown in the figure, and internal moments are induced at its ends.

Note that all the moments and rotations are shown in the positive sense in the figure. The slope-deflection equations can be derived by relating the member end moments to the end

rotations and chord rotation by applying the principle of superposition to the computing results tabulated in table 11-1 and table 11-2. Thusly, we write

4 2 6

2 4 6

FAB AB A AB B AB AB

FBA AB A AB B AB BA

M i i i Ml

M i i i Ml

θ θ

θ θ

Δ ⎫= + − + ⎪⎪⎬Δ ⎪= + − +⎪⎭

(11-10)

(2) Member with one end fixed (or rigidly connected) and another end hinged or rolled Consider an arbitrary prismatic member AB with one end fixed (or rigidly connected) and another end

hinged or rolled as shown in Fig.11.32. When the member is subjected to external load and support settlements it deforms as shown in the figure, and internal moments are induced at its fixed end. Its slope-deflection equations can be derived by relating the member end moments to the end rotation Aθ and chord rotation by applying the principle of superposition to the computing results tabulated in table 11-1 and table 11-2. Therefore, we write

/ lΔ

ϕ

BθABQ

A B

BAQ

BAM

P

A

ABM M AB PA

θ

l

EI

A

θ EI

θ

Fig.11.31 A two fixed-end member

Δ Aθ Δ

ABQ

BAQl

Fig.11.32 A member with one end fixed and another end hinged


3 3

0

FAB AB A AB AB

BA

M i i Ml

M

θ Δ ⎫= − + ⎪⎬⎪= ⎭

(11-11)

(3) Member with one end fixed (or rigidly connected) and another end fixed by double links Imagine an arbitrary prismatic member AB with one end fixed (or rigidly connected) and another end

fixed by double links as shown in Fig.11.33. When the member is subjected to external load and support settlements it deforms as shown in the figure, and internal moments are induced at its ends. Its slope-deflection equations can be derived by relating the member end moments to the end rotation Aθ by applying the principle of superposition to the computing results tabulated in table 11-1 and table 11-2. Consequently, we write

FAB AB A AB

FBA AB A BA

M i M

M i M

θ

θ

⎫= + ⎪⎬

= − + ⎪⎭ (11-12)

11.8.2 Analysis of statically indeterminate structures by direct stiffness method

Actually, the displacement-method equations of a statically indeterminate structure are the equilibrium equations. A moment equilibrium equation corresponding to a joint rotation can be developed by the consideration that the algebraic sum of the end moments of the members connected to the joint must be equal to the joint moment acting on the original structure. A force equilibrium equation in the direction of the joint translation can be established by the consideration that the algebraic sum of the end force components of the members associated with the joint translation in the direction must be equal to the algebraic sum of the force components acting on the portion, associated with the joint translation, of the original structure in the same direction. When the primary system is used to establish the displacement-equilibrium equations, the stiffness coefficients and free terms in the equations can be

BAM

ABQ

ABM PEI

l

BAQ

ABAθ

Aθ

Fig.11.33 A member with one end fixed and another end fixed by two links

11.8 Development of Displacement-Method Equations by Direct Stiffness Method 419

conveniently obtained by the tabulated results of various single-span indeterminate beams, see tables 11-1 and 11-2.

Alternatively, the displacement-method equilibrium equations may be developed directly by using the slope-deflection equations of the members of the structure. The following examples will illustrate the employment of slope-deflection equations when analyzing the structure by displacement method or slope-deflection method.

Example 11-7

Draw the bending moment diagram for the rigid frame shown in Fig.11.34 (a) by direct stiffness method.

Solution

C

(1) Primary unknowns From Fig.34 (a), we can see the rigid-joint rotation is the rotation of joint C and independent joint

translation is the horizontal translation of joint C or D of the frame if the changes of the member lengths are neglected. So the rotation of joint C of the structure, denoted by 1Δ , and horizontal translation of joint C or D, denoted by , will be chosen as the primary unknowns of the frame. 2Δ

(2) Fixed-end moments By using the fixed-end moment expressions given in table 11-2, we evaluate the fixed-end moments

due to the external loading for each member: 2 210 4 20 (kN m)

8 8BD

F qlM ×= − = − = − ⋅ 0F F F

AC CA CDM M M= = =

Note that in accordance with the slope-deflection sign convention, the clockwise fixed-end moments are

B

C

A

D

6m

EI EI

3EI

CAM

D

A

C CAQ

ACQ

ACM

CAQ DBQ

BBDQ

DBQ

BDM

D

C M CD

CAM

(b)(a) (c)

(d) (e)

4m

10kN/m

10kN/m

Fig.11.34 Figures of example 11-7(a) original structure; (b) free body diagram of joint ; (c) free body diagram of the portion around beam ; (d) free body

CCD diagram of column ; (e) free body diagram of column AC BD


considered as positive, whereas the counterclockwise fixed-end moments are considered to be negative. (3) Chord rotations Since the rotation of joint C and the translation of joints C and D are selected to be the primary

unknowns, the chord rotation for each member to be:

220.25

4AC BDϕ ϕ Δ= = = Δ 0CDϕ =

(4) Slope-deflection equations To relate the member end moments to the primary unknowns, we write the slope-deflection equations

for the three members of the frame by applying Eqs. (11-10) and (11-11). Thus the slope-deflection equations for the members of the frame can be expressed as

2 21 1

2 21

1 1 1

2 2

34 6 42

32 6 22

3 3(2 ) 63 3 20

4

FCA CA CA CA

CA

FAC CA CA AC

CAF

CD CD CD

FBDBD BD

BD

iM i i M il

iM i i M il

M i M i ii iM Ml

θ

θ

Δ Δ ⎫= − + = Δ −

1

⎪⎪

Δ Δ ⎪= − + = Δ − ⎪

⎬⎪= Δ + = Δ = Δ ⎪⎪

= − Δ + = − Δ − ⎪⎭

(a)

4CA BDEIi i i= = =In which, we assume ,

3 26CDEIi i= = .

(5) Equilibrium equations The free-body diagram of joint C is shown in Fig.11.34 (b). Note that the member end moments,

which are assumed to be in a clockwise direction on the ends of the members, must be applied in the (opposite) counterclockwise direction on the free body of the joint, in accordance with Newton’s third law. By applying the moment equilibrium equation 0CM =∑ to the free body of joint C, we obtain the equilibrium equation

0CD CAM M+ = (b)

0X =∑To establish the second equilibrium equation, we apply the force equilibrium equation to the free body of the portion, associated with 2Δ , of the frame shown in Fig.11.34 (c), to obtain

0CA DBQ Q+ = (b)

In which, and CAQ DBQ represent the shears at the upper ends of columns AC and BD, respectively, as

shown in the Fig.11.34 (c). To express the column end shears in terms of column end moments, we draw the free-body diagrams of the two columns [Figs.11.34 (d) and (e)] and sum the moments about the bottom

Summary 421

of each column:

DB

0

10 Q2

AC CAA CA

AC

BDB B

BD

M MM Ql

MDM ql

l

+ ⎫= = − ⎪⎪⎬⎪= = − −⎪⎭

∑

∑ (c)

By substituting Eq. (c) into Eq. (b), we obtain

80 0AC CA BDM M M+ + + = (e)

By substituting the end moment expressions in Eq. (a) into Eq. (b) and Eq. (e), we write

1 2

1 2

310 02

6 3.75 60 0

i i

i i

⎫Δ − Δ = ⎪⎬⎪Δ − Δ + = ⎭

(f)

(6) Solution of primary unknowns Solving simultaneous Eq. (f) yields

113.16i

Δ = 2121.05i

Δ =

(7) Member end moments By substituting the expressions of and1Δ 2Δ into the slope-deflection equations expressed in Eq. (a),

we obtain

18.94kN mCAM = − ⋅ 25.26kN mACM = − ⋅

⋅

18.95kN mCDM = 35.79kN mBDM = − ⋅

(8) Internal force diagrams The bending moment and shear and axial force diagrams can now be constructed by using the methods

used previously in the text. These diagrams have already been shown in Fig.11.23. It is observed from above progress that the principle for developing the equilibrium equations by

slope-deflection equations is the same as that for establishing displacement-method equations by primary system. The only distinguish is the presenting form of the equations. In fact, the slope-deflection equations have already reflected the consideration of the combined effects of external loadings and primary unknown displacements, the same consideration as applied to the primary system.

SUMMARY

(1) The displacement method is another approach that can be used for analyzing indeterminate


structures (although it may be also used for analyzing determinate structures). The method is especially suitable for analyzing continuous beams and rigid frames with high degrees of indeterminacy. Also, an understanding of the fundamentals of this method provides a valuable introduction to the moment distribution method (including no-shear distribution method) and the matrix stiffness method, which forms the basis of most computer software currently used for structural analysis.

(2) The primary unknowns (or variables) used in displacement method are joint displacements of a structure, i.e., rigid joint rotations and independent joint translations of the structure. A clear understanding of the mechanical meanings of shape and loading constants of various single-span indeterminate beams helps readers to realize why rigid joint rotations and independent joint translations rather than any displacements (such as rotations of hinged joints) of a structure are selected to be the primary unknowns. Note that the sign convention for member end forces and displacements will also facilitate the analysis of a structure by displacement method.

(3) In displacement method, the displacement-method equations used to solve primary unknowns are the equilibrium equations. For a rigid-joint rotation, a corresponding moment equilibrium equation can be written; whereas a force equilibrium equation can be obtained for an independent joint translation. The number of equilibrium equations is just the same as that of primary unknowns.

(4) The development of displacement-method equations by employing primary system, which will relate the analyzing procedure of the displacement method to that of the force method discussed in last chapter, is helpful to further understand the two fundamental methods. The utilization of primary systems not only makes each item in displacement-method equations have a definite meaning but also echoes the matrix stiffness method, which is plan to discuss in Chapter 13.

(5) The analysis of symmetric structures by displacement method is the analysis of corresponding half structures. The keynote of the analysis is the choice of the half of the structures. The understanding of response characteristics of a symmetric structure under symmetric and antisymmetric loadings will help the choice of a half of the structure.

(6) The direct stiffness method is based on the slope-deflection equation:

2 (2 3 ) Fnf n f nf

EIM Ml

θ θ ϕ= + − + (11-13)

The equation relates the moments at the ends of a member to the rotations and displacements of its ends and the external loads applied to the member.

The procedure of the method essentially involves (1) identifying the unknown joint displacements (degrees of freedom for independent joint displacements) of the structure; (2) for each member, writing slope-deflection equations relating member end moments to the unknown joint displacements; (3) establishing the equations of equilibrium of the structure in terms of member end moments; (4)substituting


the slope-deflection equations into the equilibrium equations and solving the resulting system of equations to determine the unknown joint displacements; and (5) computing member end moments by substituting the values of joint displacements back into the slope-deflection equations. Once member end moments have been evaluated, member end shears and axial forces, and support reactions, can be determined through equilibrium considerations.


11-1 What is the sign convention about the internal forces and displacements of a member ends in displacement method?

11-2 What are the shape and loading constants for a single-span indeterminate beam? 11-3 Please derive equations (11-1) through (11-5) and keep the form of equation (11-1) in your mind. 11-4 Please calculate the loading constants of the single-span indeterminate beams numbered by 1, 3, 6,

8, 12 and 14 in table 11-2 and keep the constants in your mind.

ABM BAM11-5 What are the magnitudes of end moments and due to 1Aθ = for the beams shown in the figure of the problem?

ABM BAM 1Δ =11-6 What are the magnitudes of end moments and due to for the beams shown in the figure of the problem?

1Δ =

EI

l

MBA

ABEI

1Aθ =l

A

ABMABM

EI

1AθB

= l

1Δ =

ABMEI

l



(a) one end hinged and one end fixed (b) two ends hinged

(a) one end hinged and one end fixed (b) two ends hinged

AB

BAMA B

AB MM BA


11-7 How many kinds of primary unknowns are there for displacement method? 11-8 How to determine joint rotation? What is the basis of the determination of joint rotations? 11-9 How to determine the number of independent joint translations? What is the basic assumption for

determining the independent translations? Why the determination of the number of degrees of freedom of a hinged system can be applied to the determination of the number of the independent joint translations of a structure?

11-10 How to select primary structure and primary system when using displacement method to analyze a statically indeterminate structure? What is the difference between the consideration of selecting primary structure and primary system used in force method and that used in displacement method? As we have known in force method, there are several kinds of primary systems for a structure. Whether or not more than one primary system for a structure can be used in displacement method?

11-11 What is the individual function of the artificial rotative restraints and artificial translating restraints?

11-12 What are the meanings of , and iik iPFijk occurring in displacement-method equations?

11-13 Why can we say that the displacement-method equations are actually the equations for equilibrium?

11-14 Whether or not displacement method can be used to analyze a statically determinate structure? 11-15 Why can the relative stiffness values of the members of a structure determine the internal forces?

Whether or not the solutions of primary unknowns 1 1, , nΔ Δ Δ are their real values if the relative stiffness values of the members of a structure are used?

11-16 Why can the rotations of a hinged joint not be used as primary unknowns? 11-17 Whether or not the structure shown in the figure of

the problem can be analyzed as the approach used in example 11-4?

C D

1 1I h2 2

11-18 Why can only a half of the structure be selected to analyze for a symmetric structure under symmetric or antisymmetric loadings? Whether or not just a half of the structure can be selected to analyze for a symmetric structure under asymmetric loadings?

P I

A B

11-19 Whether or not the symmetric property of a structure can be used to simplify the analysis of the structure if the original structure instead of a half of the structure is selected as the primary structure? Why is the

h


11-20 direct stiffness method actually the same as the method in which the primary system is employed?


11-21 are the direct stiffness method with primary-system method. 11-22 Please compare the displacement method that uses primary system with force method.


11-1 Determine the number of primary unknowns used in displacement method for the statically indeterminate structures shown in the figures, and lay off the corresponding primary structures.

(b)

Please comp

E

F

A B C

D

G H

A

D

E

F

B C

(a)

0α ≠ 0α =(1) (2)

(d)J

G

I K

A B C D

HFE

αα

A

B

C

α

0α ≠ 0α =(1) (2)

(c)

problem 10-1 (contd)

B

E

A

C

D

F

EA

A B

CE F

D

EA = ∞EA ≠ ∞(1) (2)

(e) (f)


11-2 Lay off the corresponding primary system for each of the following structures shown in the figures and also construct the bending moment diagrams due to the unit value of each primary unknown and due to the external loadings.

D

A B C

E

F G

(h)

A C

F G

EEI = ∞

B

D

(i)

Problem 11-1

EA

A B

E

C D

(g)

EA =∞EA ≠ ∞(1) (2)

A

Bi DC

qql

i

i

AB

C

EI

2.5kN/m

DEI EI

10kNP =

(b)(a)

constantEI =

4m 4m

4m

2l l

2l



11-3 Draw the bending moment diagrams for the following continuous beams and rigid frames by displacement method. Note that constantEI = .

A BC

D

ql

F

E

q

2l

ql

ll

A

B

C D

E

2l =OM

4l =

1l =

1l =

ll

(c) (d)

Problem 11-2

constantEI = l 2l 2l

2l 2l

problem 11-3

10kN/m40kN

6m6m3m3m

A B C

(a)

constantEI =

DB CA D

10kN/m 10kN50kN

3m6m 2m3m

(b)

constantEI =

P

FE

DCB

ll2l

2l

A

(c)

l

constantEI

B C

=

A D

E

20m

2.4kN/m

15m

(d)

constantEI

15m

=


11-4 Lay off the corresponding primary system for each of the following two structures shown in the figures and also construct the bending moment diagrams due to the unit value of each primary unknown and due to the external loadings.

11-5 Draw bending moment diagrams for each of the following structures shown in the figures by displacement method.

11-7 or each of the following structures shown in the figures by using the properties of the structures and by displacement method

11-8 Draw bending moment diagrams for each of the following structures shown in the figures by using the properties of the structures and by displacement method.

11-9 Discuss how to use the property of the rigid frame to analyze it. 11-10 Analyze the structures shown in the figures of problem 11-3 (c) and (d) and in the figures of

problem 11-5 (c) and (d) by direct stiffness method.

11-6 Analyze the following bent frames shown in the figures by displacement method and draw their bending moment, shear and axial force diagrams. Draw bending moment diagrams f

problem 11-4

2l

2l

lA

B D

C

E F

q

ql

(a)

l l

constantEI =

l

A

B C

DE

q

(b)

l l

constantEI =


B CA

D

q

l

E F2I2I

I I I

l

A B C

FE

D50kN

6m6m

l

5m

7m

(a) (b)

constantEI =

A B C

H

D

6m6m

E F

G

EI EIbEI

2EI

= ∞

(d)

problem 11-5

A B

C

E

D8kN

6m

(c)

constantEI =

4m4m

4m4m

/m10

kN

bEI = ∞ bEI = ∞

2EI 2EI

problem 11-6

B

C

A

DEA→∞

12m

20

kN/m

q=

(a)

A B

C D

20kNP =

40kN/m

3i =4i =

6i =

4m

(b)

6m

2m2mEI EI


A B

CD

E

40kN/m

3I

3I

I

I I

I

F

6m

problem 11-8 (a)4m

4m

EI

A B

D50kN

10m

EI

EI

C

problem 11-8 (b)

12m

P

A B

E F

lD

2I

I I

I

II

I

l

C

problem 11-9

problem 11-7 (a)

B CA

D

l l

E F

l constantEI =constantEI =

q

A B

CD E F G

6m6m 6m

constantEI

6m

20kN/m

= 4m

problem 11-7 (b)

A

C

3m

40k /m

3m80kN/m

40kN/m 40kN/m

40kN/mN

80kN/m

B

DE

F

constantEI = 3m

problem 11-7 (c)

CHAPTER 12 METHOD OF SUCCESSIVE APPROXIMATIONS AND

INFLUENCE LINES FOR INDETERMINATE STRUCTURES

The abstract of the chapter In this chapter, we will study two methods of successive approximations, which are all based

on the principle of the general displacement (stiffness) method discussed in last chapter, called moment distribution method and no-shear distribution method, respectively.

As it will be seen in the later content of the chapter, that when analyzing a structure by the methods of moment distribution or no-shear distribution, a basic process of calculation, in which each joint of the structure is first locked and then unlocked, is repeated in succession. The process includes three steps: (1) evaluate the fixed-end moments at the ends of each member and restraint moments at each joint in terms of external loadings; (2) determine distribution moments by distribution factor; (3) calculate carryover moments by carryover factor.

The moment distribution method is the keynote of the chapter. In addition, shear distribution method, the construction of influence lines and envelop of internal forces for statically indeterminate continuous beams will be presented as well.

12.1 General Remarks on Methods of Successive Approximations

In this chapter we will learn another classical formulation of the displacement method, the moment distribution method. In fact, the moment distribution method is classified as a displacement method, and from a theoretical point of view, it is very similar to the displacement method.

Like the displacement method, the moment distribution method can be employed only for the analysis of continuous beams and rigid frames, taking into account their bending deformations only.

Recalling from previous two chapters, force and displacement methods, a sort of simultaneous equations associated with the equilibrium and compatibility conditions as well as member force and displacement relations must be developed and solved when analyze an indeterminate structure. However, in the moment distribution method the moment equilibrium equations of the joints are solved iteratively by successively considering the moment equilibrium at one joint at a time, while the remaining joints of the structure are assumed to be restrained against displacement. As it will be explained in detail later, that accuracy of the successive approximations may be carried out to any desired degree.

The moment distribution method, which was initially developed by Hardy Cross in 1924, was the most

431

12.4 Moment Distributions at Multi-Joints—Successive Approximations 447

Example 12-4

Determine the member-end moments by using the moment distribution method and draw the bending moment diagram for the continuous beam shown in Fig.12.8 (a). The relative flexural rigidity is shown in the figure too. Assume . 1EI =

Solution

It is usually convenient to carry out the moment distribution analysis in a tabular form, as shown in Fig.12.8 (b). Note that the table, which is sometimes known as a moment-distribution table, consists of six columns, one for each member end of the beam. All the computations for a particular member end moment are recorded in the column for that member end.

(1) Bending stiffnesses and distribution factors at each rotatable joint The first step in the analysis is to calculate the distribution factors at those joints of the structure that

are free to rotate. As discussed in Section 12.2 [Eq. (12-6)], the distribution factor for an end of a member is equal to the bending stiffness of the member divided by the sum of the bending stiffnesses of all the members connected to the joint. From Fig. 12.8(a), we can see that only joints B and C of the continuous beam are free to rotate. Their bending stiffnesses and distribution factors can be calculated as follows.

Bending stiffnesses and distribution factors at joint B:

14 4 0.666BA BAS i= = × = 7

1.54 46BC BCS i= = × =1

Thus,

0.667 0.40.667 1

0.4 0.6 11 0.6

0.667 1

BABA

Bjj

BjBC j

BCBj

j

SS

SS

μ

μμ

⎫= = = ⎪+ ⎪⎪ = + =⎬⎪= = +

+ ⎪⎪⎭

∑∑

∑

Bending stiffnesses and distribution factors at joint C:

1.54 46CB CBS i 1= = × =

23 36CD CDS i 1= = × =

Consequently,

1 0.51 1

0.5 0.5 11 0.5

1 1

CBCB

Cjj

CjCD j

CDCj

j

SS

SS

μ

μμ

⎫= = = ⎪+ ⎪⎪ = + =⎬⎪= = +

+ ⎪⎪⎭

∑∑

∑

The distribution factors are written above their corresponding ends within a square as shown in Fig.12.8 (b).

(2) Fixed end moments Next, by assuming that joints B and C are locked, we calculate the fixed end moments that develop at

the ends of each member. By using table 11-2, we obtain

1 1 80 6 60kN m8 8

FBCM Pl × ×= − = − = − ⋅

1 60kN m8

FCBM Pl= + = + ⋅

221 1 20 6 90kN m

8 8FCDM ql × ×

= − = − = − ⋅

The fixed end moments are recorded on the first line of the moment-distribution table as shown in Fig.12.8 (b).

(3) Unlocking joint B Since joints B and C are actually not clamped, we release them, one at a time. We can release either

joint B or joint C; let us begin at joint B because it has the maximum unbalanced moment. From Fig. 12.8 (b), we can see that there is a (counterclockwise) fixed-end moment at end B of member BC, whereas no moment exists at end B of member AB. As long as joint B is restrained against rotation by the clamp, the unbalanced moment is yielded at the clamp. However, when the imaginary clamp is removed to release the joint, the negative of the unbalanced moment, , will equivalently act at the joint, causing it to rotate in the clockwise direction until it is in equilibrium. The rotation of joint B causes the distributed moments to develop at ends B of members BC and AB, and which can be evaluated by multiplying the negative of the unbalanced moment(i.e., 60 ) by the

60kN m− ⋅

60kN m− ⋅60kN m⋅

BCDM BADMkN m⋅


distribution factors, respectively. Thus

0.6 60 36kN mBCDM = × = ⋅

0.4 60 24kN mBADM = × = ⋅

These distributed moments are recorded on line 2 of the moment-distribution table [Fig. 12.8 (b)], and they are underlined to indicate that joint B is now balanced. Note that the sum of the three moments above the line at joint B is equal to zero (i.e., ). 36 24 60 0+ − =

The distributed moment at end B of member BC induces a carryover moment at the farend C, which can be determined by multiplying the distributed moment by the carryover factor of the member. Since joint C remains clamped, the carryover factor of member BC is 1/2. Thus, the carryover moment at the end C of member BC is

1 36 18kN m2× = ⋅

Similarly, the carryover moment at the end A of member AB is computed as

1 24 12kN m2× = ⋅

These carryover moments are recorded on the same line of the moment distribution table as the distributed moments, with a horizontal arrow from each distributed moment to its carryover moment, as shown in Fig. 12.8 (b).

It can be seen from this table that joint B is now in equilibrium, because it is subjected to two equal, but opposite, moments. Joint C, however, is not in equilibrium, and it needs to be balanced. Before we release joint C, an imaginary clamp must be applied to joint B in its rotated position.

(4) Unlocking Joint C Joint C is now released. The unbalanced moment at this joint is obtained by summing all the moments

acting at the ends C of members BC and CD. From the moment-distribution table (lines 1 and 2), we can see that there is a fixed-end moment and a 60kN m+ ⋅ 18kN m+ ⋅ carryover moment at end C of member BC, whereas the end C of member CD is subjected to a 90kN m− ⋅ fixed-end moment. Thus the unbalanced moment at joint C is

60 18 90 12kN m+ − = − ⋅

When the imaginary clamp is removed to release the joint, the negative of the unbalanced moment, , will equivalently act at the joint, causing it to rotate in the clockwise direction until it is in

equilibrium. The rotation of joint C induces the distributed moments to develop at ends C of members BC 12kN m⋅

and CD, and , which can, as discussed previously, be evaluated by multiplying the negative of the unbalanced moment(i.e.,

CBDM CDDM12kN m⋅ ) by the distribution factors, respectively. Thus

0.5 12 6kN mCBDM = × = ⋅

0.5 12 6kN mCDDM = × = ⋅

These distributed moments are recorded on line 3 of the moment-distribution table, and a line is drawn beneath them to indicate that joint C is now balanced. One-half of the distributed moment are then carried

B

BA BC0.5 0.50.4 0.6

C

CDA D

CB

B

6m 6m

20kN/m

2EICA DEI

3m3m1.5EI

80kN

fixed end moments -60 +60 -90 0unlocking B 12 24 36 181 unlocking C

← → 3 6 6

unlocking B -0.6 -1.2 1.8 -0.9 2 unlocking C 0.23

⎧⎨ ←⎩

← − → 0.45 0.45

unlocking B -0.05 -0.09 0.14 -0.07 3 unlocking C 0.04 0.03

member end moments 1

⎧⎨ ←⎩

← − →⎧⎨⎩

1.35 22.71 -22.71 83.52 -83.52 0

distribution factors

B C

( )90( )120

11.35

22.71

66.89

83.52

48.24 53.08

(b)

(c)

(a)

2.304m

A D

(unit: kN m)⋅

Fig.12.8 Figure of example 12-4(a) original structure and its loads; (b) moment distribution and carryover; (c) bending moment diagram


over to the farend B of member CB, as indicated by a horizontal arrow on line 3 of the table. Joint C is then reclamped in its rotated position.

(5) Again unlocking joint B With joint C now balanced, we can see from the moment-distribution table (line 3) that, due to the

carryover effect, there is a unbalanced moment at joint B. Recall that the moments above the horizontal line at joint B were balanced previously. Thus we release joint B again and distribute the unbalanced moment to ends B of members BC and AB as

3kN m⋅

0.6 3 1.8kN mBCDM = − × = − ⋅

0.4 3 1.2kN mBADM = − × = − ⋅

These distributed moments are recorded on line 4 of the moment-distribution table, and one-half of these moments are carried over to the ends C and A of members BC and AB, respectively, as indicated on the table. Joint B is then reclamped.

(6) Again unlocking joint C The unbalanced moment at joint C (line 4 of the moment-distribution table) is balanced

in a similar manner. The distributed and the carryover moments thus computed are shown on line 5 of the table. Joint C is then reclamped.

0.9kN m− ⋅

(7) Thirdly unlocking joint B and C It can be seen from line 5 of the moment-distribution table that the unbalanced moment at joint B has

now been reduced to only . Another balancing of joint B produces an even smaller unbalanced moment of at joint C, as shown on line 6 of the moment-distribution table. Since the distributed moments induced by this unbalanced moment are negligibly small, we end the moment-distribution process. The final member end moments are obtained by algebraically summing the entries in each column of the moment-distribution table. The final moments thus obtained are recorded on line 8 of the table. Note that the final moments satisfy the equations of moment equilibrium at joints B and C.

0.23kN m⋅0.07kN m− ⋅

(8) Reactions, internal forces and their diagrams With the member end moments known, member end shears and support reactions can now be

determined by considering the equilibrium of the free bodies of the members and joints of the continuous beam, as discussed in previously. The bending moment diagram can then be constructed in the usual manner by using the beam sign convention as shown in Fig.12.8 (c).

Example 12-5

Determine the member-end moments by using the moment distribution method and draw the internal force diagrams and reactions for the rigid frame shown in Fig.12.9 (a). Assume 1EI = .

Solution

Application of moment distribution method for rigid frames without sidesway follows the same procedure as that given for continuous beams. However, unlike the continuous beams, more than two members may be connected to a joint of a rigid frame. In such cases, care must be taken to record the computations in such a manner that mistakes are avoided. Whereas some engineers like to record the moment-distribution computations directly on a sketch of the frame, others prefer to use a tabular format for such purposes. We will use the former for calculations, as illustrated in Fig.12.9 (b).

(1) Bending stiffnesses

2 16 3DCEIi = = 3DC DCS i 1= =

2 14 2DAEIi = = 4DA DAS i 2= =

3 16 2DEEIi = = 4DE DES i 2= =

3 16 2EDEIi = = 4ED EDS i 2= =

4 43 3EFEIi = = 3EF EFS i 4= =

2 14 2EBEIi = = 4EB EBS i 2= =

(2) Distribution factors At joint D:

1 0.21 2 2

DC DCDC

Dj DC DA DEj

S SS S S S

μ = = = =+ + + +∑

2 0.45

DADA

Djj

SS

μ = = =∑

2 0.45

DEDE

Djj

SS

μ = = =∑


1Djjμ =∑

10kN/m

D E

At joint E:

2 0.252 2 4

ED EDED

Ej ED EB EFj

S SS S S S

μ = = = =+ + + +∑

6m 3m6mBA

FC

4m

BA

CE

FD

DC0.2 0.4 0.5

DA DE ED EB EF0.25

2.25−

30.00−3.75− 7.50−

45.00 30.00+15.00−

4.50− 2.25−

7.50−

4.50−

0.11− 0.06−0.56+

2.25−BE3.75−

AD

2.31−

1.13+0.06−

0.02+ 0.03+

0.11+3.46−

0.28+ 0.56+0.11−

0.02+

0.28+0.06−

42.69+ 4.61− 38.08− 20.77+ 6.92− 13.84−

(b)

(a)

0.4 0.25

Fig.12.9 Figure of example 12-5(a) original structure and its loads; (b) processes of moment distribution and carryover

2 0.258

EBEB

Ejj

SS

μ = = =∑

4 0.58

EFEF

Ejj

SS

μ = = =∑

1Ejjμ =∑

( )45

(3) Fixed end moments

221 1 10 6 45kN m

8 8FDCM ql × ×

= = = ⋅

221 1 10 6 30kN m

12 12FDEM ql × ×

= − = − = − ⋅

( )45

BA

23.66

42.69

FE

CD

15.58

38.08

4.61

20.77

2.31

13.84

6.92

3.46 BA1.73

FEC

D

2.60

32.89

37.11

22.89

27.11

4.61 4.61

D E

BA

F

70.00

ED2.60

31.72

4.33

2.60 31.72

4.33

70.00C

BA

F

22.89

4.33

70

1.73

2.31

2.60

31.72

3.46

4.61

C

(b)

(c)

(a)

(d)

(unit: kN m)⋅ (unit: kN)

(unit: kN)unit: kN, kN m⎛ ⎞⎜ ⎟⋅⎝ ⎠

Fig.12.10 Internal force diagrams of example 12-5(a) bending moment diagram; (b) shear force diagram; (c) axial force diagram; (d) reaction diagram


221 1 10 6 30kN m

12 12FEDM ql × ×

= + = + = ⋅

(4) Moment distribution and carryover As long as joints D and E are actually free to rotate, we release them, one at a time. We can release

either joint D or joint E; let us begin at joint E because it has the maximum unbalanced moment so as to shorten the calculation process. Note that any unbalanced moment at joint D or E must be distributed to the ends D or E of the three members connected to it in accordance with their distribution factors. The processes of moment distribution and carryover are shown in Fig.12.9 (b)

(5) The final moments are recorded on the last line of the moment-distribution table and the moment diagram of the frame is depicted in Fig.12.10 (a) by superposition method as used in the previous chapters.

(6) Reactions, shear and axial forces and their diagrams With the member end moments known, member end shear and axial forces and support reactions can

now be determined by considering the equilibrium of the free bodies of the members and joints of the frame, as discussed in previously. The shear and axial force diagrams can then be constructed in the usual manner as shown in Fig.12.10 (b) and (c). Fig.12.10 (d) is also shown the reactions.

Example 12-6

Determine the member-end moments by using the moment distribution method and draw the bending moment diagram for the rigid frame shown in Fig.12.11 (a).

Solution

(1) Simplification of calculation Since the rigid frame shown in Fig. 12.11 (a) and the loading applied on it are all symmetric we can,

recalling from section 10.6.3, take one-half of the structure to analyze as shown in Fig.12.11 (b). In the figure, supports G and H are known as double link support that prevent the translation along the links and rotation but not translation perpendicular to the links.

(2) Bending stiffnesses and distribution factors Bending stiffnesses:

2 4

3 32

AGEI ES = =

I

43AC CAEIS S= =

2

3CHEIS =

43CEEIS =

Distribution factors at joint A:

0.5AGμ =

0.5ACμ =

Distribution factors at joint C:

0.2CHμ =

0.4CAμ =

0.4CEμ =

The distribution factors are recorded around their corresponding joint as shown in Fig.12.11 (b). (3) Fixed end moments The fixed end moment of member AB can be determined in terms of original or its half structure by

using table 11-2.

A0.5

0.5

7.50.75−

0.373.75

0.02

0.03−

2.36

7.11−0.19

0.04−

0.08−

1.50−

0.75−

7.11

15−7.5

0.38

0.02

1.50−0.08−

0.75−0.04−

1.58− 0.78−

0.79−

0.4

0.4

0.2C

E

G

H

(b)

A B

C

20kN/m

D

E F

3m

I2

I I

I

I I 3m3m

(a)

Fig.12.11 (contd)

A B

C D

E F

7.11

0.79 0.79

0.781.58

2.36

7.11

15.37

2.36

1.580.78

I I

(c)

unit: kN m⋅


(d)Joint G A C H E

Member GA AG AC CA CE CH HC EC

DF 0.5 0.5 0.4 0.4 0.2 FEM Dist

-7.5

-157.5

7.5

CO Dist

-7.5

3.75-1.5

-1.5

-0.75

CO Dist

0.37

-0.750.38

0.75

-0.75

CO Dist

-0.37

0.19-0.08

-0.08

-0.03

CO Dist

0.02-0.040.02

0.03

-0.04

M∑ -15.37 -7.11 7.11 2.36 -1.58 -0.78 0.78 -0.79

Fig.12.11 Figures of example 12-6(a) original structure and its loading; (b) one-half of the frame and process of the calculation;(c) bending moment diagram; (d) moment distribution table

In terms of original structure: 21 20 3 15kN m12

FABM = − × × = − ⋅

In terms of one-half of the structure: 21 20 1.5 15kN m3

FAGM = − × × = − ⋅

21 20 1.5 7.5kN m6

FGAM = − × × = − ⋅

(4) Unlocking joints in the sequence of A, C, A, C and A will carry out the analysis as shown in Fig.12.11 (b) or (d). Note that when columns for two ends of a member cannot be located adjacent to each other, then an overhead arrow connecting the columns for the member ends may serve as a reminder to carry over moments from one end of the member to the other. In Fig.12.11 (d), such an arrow is used between the columns for the end of member CE. This arrow indicates that a distributed moment at end C of member CE induces a carryover moment at the far end E. Also, the carryover factors for end G and H of members AG and CH are equal to -1.

(5) Bending moment diagram

By the method of superposition, the bending moment diagram will be drawn as shown in Fig.12.11 (c)

432 Chapter 12 Method of Successive approximations and Influence Lines for Indeterminate Structures

widely used method for analysis of structures from 1930, when it was first published, through the 1960s. Since the early 1970s, with the increasing availability of computers, the use of the moment distribution method has declined in favor of the computer-oriented matrix methods of structural analysis. Nonetheless, the moment distribution method is still preferred by many engineers for analyzing smaller structures, since it provides a better insight into the behavior of structures. Furthermore, this method may also be used for preliminary designs as well as for checking the results of computerized analyses.

12.2 Concepts and Terminology in Moment Distribution Method

12.2.1 Sign convention

We will establish the same sign convention as that adopted for the general displacement method and the slope deflection equations: clockwise moment that act on the member are considered positive, whereas counterclockwise moments are negative.

Since a clockwise moment at an end of a member must act in a counterclockwise direction on the adjacent joint, the foregoing sign convention implies that counterclockwise moments on joints are considered positive.

12.2.2 Distribution and carryover of joint moments

Consider the rigid frame with only one freely rotating joint A, on which a concentrated moment M is applied, as shown in Fig.12.1 (a). If we use displacement method to analyze the structure, primary unknown is the rotation of joint A, indicated by 1Δ ; and the primary equation will be written as

01111 =+Δ PFk (a)

The bending moment diagrams due to a unit rotation and due to the external load at joint A have been drawn in Fig.12.1 (b) and (c). From the figure (b), we can find that

43

AB AB

AC AC

AD AD

M iM iM i

⎫=⎪= ⎬⎪= ⎭

(b)

Consider the moment equilibrium condition of the free body of the restrained joint A, we write

ADACAB iiik ++= 3411 (c)

From the figure (c), by considering the moment equilibrium condition of the restrained joint, we can obtain

01 =+ MF P MF P −=1 (d)

12.2 Concepts and Terminology in Moment Distribution Method 433

Substituting Eqs. (c) and (d) into Eq. (a), solve out

ADACAB

P

iiiM

kF

++=−=Δ

3411

11 (e)

The substitution of in the superposition equation 1Δ 1 1 PM M M= Δ + will yield the bending moments of member ends as follows

1 1

1 1

1 1

444 3

334 3

4 3

ABAB AB AB

AB AC AD

ACAC AC AC

AB AC AD

ADAD AD AD

AB AC AD

iM M ii i i

i

M

M M ii i i

iM M i Mi i i

⎫= Δ = Δ =

M

⎪+ + ⎪⎪= Δ = Δ = ⎬+ + ⎪⎪= Δ = Δ = ⎪+ + ⎭

(f)

A

B

C D A

B

C D

11kM

A

B

C D

1 1Δ =

1PF M

ACi

ABi

ADi

(b)

(c)

(a)

1 1

Fig.12.1 Distribution of joint moment(a) a rigid frame subjected to a concentrated moment; (b) diagram due to 1; (c) diagram due external loadPM MΔ =


1 1

1 1

224 3

4 3

ABBA BA AB

AB AC AD

DADA DA DA

AB AC AD

iM M i Mi i i

iM M ii i i

⎫= Δ = Δ =

M

⎪+ + ⎪⎬− ⎪= Δ = − Δ =

+ + ⎪⎭

(g)

The final bending moment diagram can be obtained by multiplying the 1M diagram shown in Fig.12.1 (b) by . 1Δ

Forgoing discussion is the analyzing process for the rigid frame with a single free rotating joint, on which a concentrated moment is applied, by means of the primary system of displacement method. Now we like to provide the quantities appearing in the process with some new explanations and define some new terminology as follows.

(1) Bending stiffness ABSIf we write the first equation of Eq. (b) as

4AB AB ABM i S= =

ABS is defined as the bending stiffness of member AB at end A, which is the moment that must be applied at end A of the member to cause a unit rotation of the end. Actually, the bending stiffness reflects the resistance ability of the member end against its corresponding end rotation.

Recalling from the shape constants for various single-span indeterminate prismatic members with different restraint conditions in table 11-1 of Chapter 11, the magnitude of the bending stiffness of a member end is equal to the magnitude of the corresponding shape constant. From table 11-1, we obtain the bending stiffnesses of near end A of member AB for three kinds of prismatic members listed in numbers 1, 3 and 5 in the table, as shown in Fig.12.2.

Note that: ① the first subscript A of means that end A is the end at which a moment must be

applied to cause a unit rotation at the end; for simplicity, end A is named nearend and end B is named

farend, respectively. ② the magnitude of is related to both the linear bending stiffness

ABS

ABS EIiL

= ,

bending stiffness per length, or relative bending stiffness of the member and the support conditions of the farend. The different support conditions yield different . For example, the Fig.12.2 shows four kinds

of support conditions at the farend of member AB. Their corresponding bending stiffness at the nearend will be written as follows.

ABS

ABS

The farend is fixed, iSAB 4= (12-1)

The farend is hinged, iSAB 3= (12-2)

The farend is rollered, iSAB = (12-3)


The farend is free, 0=ABS (12-4)

(2) Distribution factor A jμ and rotational stiffness of a joint

Recalling from Eq. (f), when a moment M is applied to a rigid joint as shown in Fig.12.1, the connecting members will each supply a portion of the resisting moment proportional to its bending stiffness and necessary to satisfy moment equilibrium at the joint. The fraction of the total resisting moment supplied by the member is called the distribution factor. To determine what fraction of the applied moment M is resisted by each of the three members connected to the joint, we rewrite Eq. (f) again as

44 3

34 3

4 3

ABAB AB

AB AC AD

ACAC AC

AB AC AD

ADAD AD

AB AC AD

iM M Mi i i

iM M Mi i i

iM M Mi i i

μ

μ

μ

⎫= = ⎪+ + ⎪

⎪= = ⎬+ + ⎪⎪= = ⎪+ + ⎭

Generally,

A j A jM Mμ= (12-5)

In which,

4ABS EI l=B

A1

l

EI AABS EI l=

EI1

B

l

3AbS EI l=B

A1

EI A0AbS =

EI

1B

(b)

(c)(a)

(d)

Fig.12.2 Bending stiffness(a) far end is fixed; (b) far end is hinged; (c) far end is fixed by double links; (d) far end is free


A jA j

A jj

SS

μ =∑

(12-6)

A jμ is termed distribution factor at end A of member Aj. Obviously, the second subscript j can be B, C

and D. A jj

S∑ represents the sum of the bending stiffnesses of all the members connected to joint A.

The rotational stiffness of a joint is defined as the moment required causing a unit rotation of the joint. From Eq. (12-6), we can see that the rotational stiffness of a joint is equal to the sum of the bending stiffnesses of all the members rigidly connected to the joint.

Note that the distribution factors at the identical joint posses following relationship.

1AB AC ADAB AC AD

A A jj

S S SS

μ μ μ μ + += + + = =∑ ∑

(12-7)

(3) Carryover factor A jC

The carryover factor represents the ratio of the farend moment to the nearend moment when the nearend rotates.

Consider again the rigid frame shown in Fig.12.1. When the concentrated moment applies at joint A, the nearend of each connecting member supply a resisting moment, A jM , in the meanwhile, the farend of

each connecting member supply a resisting moment as well, whose magnitude is related to the support conditions of the farend. Thus, we can express carryover factor as

j AA j

A j

MC

M= (12-8)

From Eqs. (f) and (g), we can obtain

12

BAAB

AB

MCM

= = 0CAAC

AC

MCM

= = 1DAAD

AD

MCM

= = −

Therefore, we can summarize the magnitude of the carryover factor of a prismatic member for different farend support conditions as follows.

The farend is fixed, 21

=C (12-9)

The farend is hinged, 0=C (12-10)

The farend is fixed by double links, 1−=C (12-11)


From Eq. (12-8), we can express the farend moment by carryover factor as follows

j A A j A jM C M= (12-12)

Now reexplain the performance of the member-end moments for the rigid frame shown in Fig.12.1 as follows:

When a moment M applies at a joint, the moment will be distributed among the nearends of the various members connected to the joint and carryovered to the farends of the members. The distributed moment of each nearend is named nearend bending moment or distributed bending moment whose magnitude is equal to the multiplication of distribution factor and the moment M; whereas the carryover moment of each farend is known as farend bending moment or carryover bending moment equaling to the product of the carryover factor and corresponding nearend bending moment.

Forgoing discussion has shown how to use the concepts of moment distribution and carryover to analyze a structure subjected to concentrated joint moment. The following example will give a detail explanation.

Example 12-1

Determine the member-end moments for the rigid frame with a concentrated couple applied at joint A as shown in Fig.12.3 by using the moment distribution method.

100kN mM = ⋅

Solution

(1) Bending stiffnesses and distribution factors In terms of Eqs. (12-1) through (12-4), the bending stiffnesses of the members connected at joint A

will be evaluated as follows

AB

C

25 12.5 12.5

37.5

50

Di i

i

i

100kN mM = ⋅

E(b)(a)

Fig.12.3 Distribution and carryover of a moment(a ) a rigid frame subjected to a concentrated moment; (b) bending moment diagram

(unit: kN m)⋅


iiS ABAB 44 ==

iiS ACAC 33 ==

iiS ADAD ==

0=AES

By using Eq. (12-6), the distribution factors will be calculated to be

4 0.54 3

ABAB

A jj

S iS i i i

μ = = =+ +∑

3 0.3758

ACAC

A jj

S iS i

μ = = =∑

0.1258

ADAD

A jj

S iS i

μ = = =∑

0 08

AEAE

A jj

SS i

μ = =∑

=

(2) Member end moments By employing Eq. (12-5), the member end moments can be obtained to be

0.5 100 50kN mAB ABM Mμ= = × = ⋅

0.375 100 37.5kN mAC ACM Mμ= = × = ⋅

0.125 100 12.5kN mAD ADM Mμ= = × = ⋅

01000 =×== MM AEAE μ

In terms of Eqs. (12-9) through (12-11), the carryover moments of the members can be calculated as

1 50 25kN m2BA AB ABM C M= = × = ⋅

05.370 =×== ACACCA MCM

1 12.5 12.5kN mDA AD ADM C M= = − × = − ⋅

Under the action of a joint couple, the distribution and carryover moments of the members are their

12.3 Moment Distribution at A Single Joint—Moment-Distribution Process 439

final end moments. By the end moments, the bending moment diagram of the frame can be drawn as shown in Fig.13.3 (b).

12.3 Moment Distribution at A Single Joint—Moment-Distribution Process

Moment distribution is based on the principle of successively locking and unlocking the joints of a structure in order to allow the moments at the joints to be distributed and balanced. The best way to explain the concept of the moment distribution at a single joint may be by the following practical model.

Consider the continuous beam shown in Fig.12.4 (a), which is made of a thin steel strap. As shown in the figure, when the bobweight reaches to a magnitude of the beam will be deformed as shown in dashed line and the ends of the members will yield moments that we are desired to determine. The procedure to determine the member end moments due to a general load can be taken as following processes:

P

(1) Initially assume that joint B of the beam that is free to rotate is temporarily restrained (or locked) against rotation by a clamp applied to it so as to make the joint as a hypothetical fixed support ( 0Bθ = ). As we can see, the hypothetical clamp has made the original continuous beam becomes two independent

FABM F

BAM 0FBCM =

BM

P

A B C

BAM BCM

BθA B

P

CBθ

M B

BCM ′BAM ′ABM ′

A CB

BM FBA

0FBCM =

unlock joint

lock joint(b)

(c)

(a)

Fig.12.4 Concept of moment distribution method for a single joint (a) original structure; (b) locking joint ; (c) unlocking joint B B


single span indeterminate beams. Then let the external load apply to this hypothetical fixed structure. As shown in dashed lines of Fig.12.4 (b), only span AB is deformed and only the fixed end moments at the ends of member AB are yielded. Since member BC is subjected to no loading there is no deformation and no fixed end moment at the ends of the member. Obviously, the restraint moment at the clamp, indicated by

P

BM , is temporarily balanced by the fixed end moment at the end B of member AB, FBAM . The restraint

moment BM can be thusly determined by the moment equilibrium condition of joint B. We can see from Fig.12.4 (b) that since the fixed end moment at the end B of member BC is equal to

zero, i.e., , the restrain moment at the clamp can be determined to be 0FBCM =

F FB BA BC

FBAM M M M= + =

Apparently, the restraint moment at the clamp is equal to the algebraic sum of the fixed end moments of the member ends connected to the locked joint. The sign convention of the restraint moment is that the clockwise is positive.

(2) Since the joint B of the original continuous beam is free to rotate and there is no moment applied to it [Fig.12.4 (a)], i.e., , we must release joint B by removing the clamp, thereby allowing it to

rotate an original angle

0BM =∑Bθ under the action of the minus unbalanced moment, BM− , same in magnitude

but opposite in sense to the restraint moment. The rotation of joint B due to BM− will induce the

distribution moments at the ends of the members connected to the joint, indicated by BAM ′ and BCM ′ ,

respectively. The bending of member AB due to the distributed moment causes carryover moments to yield at the farend of the member, ABM ′ , as well.

(3) If we superimpose the cases shown in Fig.12.4 (a) and (b) together, the superimposing results for both the deformations and the member end moments are identical to the original case shown in Fig.12.4 (a). For instance, the moment at the end B of member AB will be

FBA BA BAM M M ′= +

Now we will briefly summarize the concepts and processes of moment distribution method due to the action of general loads as follows:

(1) Lock joint Lock the joint that connects the members of a continuous beam by a clamp to decompose the continuous beam into independent single span indeterminate beams. External loads are applied to these individual single span indeterminate beams, and fixed-end moments at the ends of the beams are computed, and unbalanced restraint moment is evaluated by taking the algebraic sum of the fixed-end moments at the ends connected to the locking joint.

(2) Unlock joint Release the locking by removing the clamp, allowing it to rotate under the action of the minus of the unbalanced moment, i.e., same in magnitude but opposite in sense to the restraint


moment. The rotation of the joint will induce the distribution moments at the nearends and carryover moments at the farends of the members connected to the joint.

(3) Superimpose results Superposing the results of above two steps will obtain the real member-end moments of the continuous beam.

Example 12-2

Determine the member-end moments for the two-span continuous beam shown in Fig.12.5 (a) by using the moment distribution method. The flexural rigidity is a constant, i.e., constantEI = .

Solution

(1) Lock joint B. A clamp is applied to joint B to restrain its rotation as shown in Fig.12.5 (b). The fixed-end moments, the clockwise is positive, due to the external loads can be evaluated by using table 11-2 as follows.

221 20 6 60kN m

12 12FABM ql ×= − = − = − ⋅

221 20 6 60kN m

12 12FBAM ql ×= + = + = + ⋅

3 3 32 6 36kN m16 16

FBCM Pl × ×

= − = − = − ⋅

The fixed end moments are written below their corresponding ends. The restrain moment at joint B,

BM , must be the algebraic sum of the fixed-end moments connected to the joint. Thusly, BM can be written as

60 36 24kN mBM = − = ⋅

(2) Unlock joint B. Release joint B by removing the clamp and apply a couple, equal in magnitude but opposite in sense, , at the joint as shown in Fig.12.5 (c). The distribution moments at the nearends and the carryover moments at the farends will be calculated as follows.

24kN mBM− = − ⋅

The linear stiffness (or relative stiffness) of the two members AB and BC can be defined as

EIil

=

The bending stiffnesses at end B for the two members will be written as

4BA ABS i 4i= =


3BC BCS i 3i= =

B C20kN/m

6m 3m3m

32kN

Then, the distribution factors will be

4 0.5717

ABAB

BA BC

S iS S i

μ = = =+

3 0.4297

BCBC

BA BC

S iS S i

μ = = =+

Checking: 1BA BCμ μ μ= + =∑The distribution factors written above their corresponding ends within a square are all right.

The distributed moments can be evaluated by multiplying the negative of the unbalanced joint moment

B

60 36

BM

EIA

24kN m⋅

60− 36−48+

B C

20kN/m 32kN

A

BCA

24kN m⋅0.571 0.429

13.70− 10.30−6.85−

(b)

(c)

(a)

Fig.12.5 Process of moment distribution for a single joint(a) original structure and its loads; (b) locking joint and evaluating fixed end moments; (c) unlocking joint and determine the distribute

BB

d and carryover moments

EI


by the corresponding distribution factors as follows. ' ( 24) 0.571 13.70kN mBAM = − × = − ⋅

' ( 24) 0.429 10.30kN mBAM = − × = − ⋅

The distributed moments are written below their corresponding ends underlined to indicate that the joint is release and balanced as shown in Fig.12.5 (c).

The carryover moment at a farend can be computed by multiplying the distributed moment by the member carryover factor (fixed farend, 1/2; hinged farend, 0). Thus,

' '1 1 ( 13.70) 6.85kN m2 2AB BAM M= = × − = − ⋅

' 0CBM =

The carryover direction of a carryover moment is indicated by an arrow as shown in Fig.12.5 (c). (3) Superpose member-end moments. Algebraically summing the member-end moments obtained in

steps (1) and (2) can yield final member-end moments.

BA BC0.571 0.429

BCA

distribution factors

In order to facilitate the calculation, the above first two steps can be combined together in terms of the form shown in Fig.12.6 (a). Note that the member end moment doubly underlined is its final result. A

66.8546.30

33.43 24.85

( )90( )48

BCA

fixed end moments -60 +60 -36 0distributed moments and -6.85 -13.70 -10.30 carryover moments

⎫⎪ ← →⎬⎪⎭

0

member end moments -66.85 46.30 -46.30 0

(b)

(a)

Fig.12.6 Calculating form for moment distribution method and diagram(a) calculating form for moment distribution method; (b) bending moment diagram

M

(unit: kN m)⋅


positive answer for an end moment indicates that its sense is clockwise, whereas a negative answer for an end moment implies a counterclockwise sense. These final moments must satisfy the equation of moment equilibrium at joint B. That is,

46.30 46.30 0Bjj

M = − =∑

The bending moment diagram of the beam is constructed in Fig.12.6 (b).

Example 12-3

Determine the member-end moments by using the moment distribution method and construct the bending moment diagram for the rigid frame shown in Fig.12.7 (a). The relative flexural rigidity is shown in the figure.

Solution

(1) Bending stiffnesses and distribution factors at joint A Bending stiffnesses:

3 3 2 6ABS i= = × =

4 4 1.5 6ACS i= = × =

4 4 2 8ADS i= = × =

6 6 8 20A jj

S = + + =∑

Distribution factors:

6 0.320

ABAB

A jj

SS

μ = = =∑

6 0.320

ACAC

A jj

SS

μ = = =∑

8 0.420

ADAD

A jj

SS

μ = = =∑

0.3 0.3 0.4 1A j AB AC ADjμ μ μ μ= + + = + + =∑


(2) Fixed end moments

221 15 4 30kN m

8 8FABM ql ×= = − = ⋅

2 2

2 2

50 3 2 24kN m5

FAD

PabMl

× ×= − = − = − ⋅

2 2

2 2

50 3 2 36kN m5

FDA

Pa bMl

× ×= = − = ⋅

(3) Distribution and carryover The processes for moment distribution and carryover are carried out in the form as shown in Fig.12.7

(b). Note that the columns for all member ends, which are connected to the same joint, are grouped together, so that any unbalanced moment at the joint can be conveniently distributed among the members connected

AB AC

AB D

C

AD0.30.3

3630 24−1.8− 1.8−

1.82.4− 1.2−

28.2 26.4−

0.9−

34.8

AB D

C3m

15kN/m50kN

2i = 2i =

1.5i =

4m 2m

4m

0.4

BD

30 60

1.8

26.4

29.4

28.2

15.9

0.9

34.8

(b)

(c)

(a)

(unit: kN m)⋅

Fig.12.7 Figure of example 12-3(a) original structure and its loads; (b) moment distribution and carryover; (c) bending moment diagram


to it. (4) The bending moment diagram The final member end moments are obtained by summing all the moments in each column of the

moment-distribution table shown in Fig.12.7 (b). Note that the final moments satisfy the equation of moment equilibrium at joint A. The bending moment diagram is shown in Fig.12.7 (c), which can be constructed by superposition method discussed previously.

12.4 Moment Distributions at Multi-Joints—Successive Approximations

The moment distribution method used to analyze the continuous beams with only one rotatable joint can also be employed to analyze both of continuous beams and rigid frames without sidesway which have more then one free rotatable joints by an iterative procedure, in which it is initially assumed that all the joints of the structure that are free to rotate are temporarily restrained against rotation by imaginary clamps applied to them. External loads are applied to this hypothetical fixed structure, and fixed-end moments at the ends of its members are computed. These fixed-end moments generally are not in equilibrium at those joints of the structure that are actually free to rotate. The conditions of equilibrium at such joints are then satisfied iteratively by releasing one joint at a time, with the remaining joints assumed to remain clamped. A joint at which the maximum moments are not in balance is selected. The joint is then released by removing the clamp, thereby allowing it to rotate under the negative of the unbalanced moment until the equilibrium state is reached. The rotation of the joint induces distributed moments at the ends of the members connected to it, whose values are determined by multiplying the negative of the unbalanced joint moment by the distribution factors for the member ends connected to the joint. The bending of these members due to the distributed moments causes carryover moments to develop at the farends of the members, which can easily be evaluated by using the member carryover factors. The joint, which is now in equilibrium, is reclamped in its rotated position. Next, another joint with the maximum unbalanced moment is selected and is released, balanced, and reclamped in the same manner. The procedure is repeated until the unbalanced moments at all the joints of the structure are negligibly small. The final member end moments are obtained by algebraically summing the fixed-end moment and all the distributed and carryover moments at each member end. This iterative process of determining member end moments by successively distributing the unbalanced moment at each joint is called the moment-distribution process.

With member end moments known, member end shears, member axial forces, and support reactions can be determined through equilibrium considerations, as discussed in the previous chapters.

The following examples will illustrate the application of moment distribution method to the analysis of the continuous beams and rigid frames without sidesway and with more then one free rotatable joints

458

Example 12-7

Analyze the symmetric rigid frame shown in Fig.12.12 (a) by using the moment distribution method and draw its bending moment diagram. . constantEI =

Solution

(1) Simplification of calculation Because the rigid frame shown in Fig. 12.12 (a) and the loading applied on it are both symmetric about the axes x and , we can, recalling from section 10.6.3, take a quarter of the structure to analyze as shown in Fig.12.12 (b). In the figure, supports E and G are termed double link supports that prevent the translation along the links and rotation but not translation perpendicular to the links.

y

(2) Bending stiffnesses and distribution factors

Bending stiffnesses (assume 61EI i= ):

23CG CGEIS i= = = i 3

2CE CEEIS i i= = =

2 0.42 3

CGCG

CG CE

S iS S i i

μ = =+ +

= 3 0.6

2 3CE

CECG CE

S iS S i i

μ = =+ +

=

(3) Fixed end moments By using table 11-2, we can find

221 24 3 72kN m

3 3FCG

qlM = − = − × × = − ⋅

3m

A B

C D

E Fx

G

3m

24kN/my

y

E

C G

3m

24kN/m

x

2m2m 2m

(b)(a)

Fig.12.12 (contd)


2

21 24 3 36kN m6 6

FGC

qlM = − = − × × = − ⋅

(4) Unlocking joint C once will carry out the process of the analysis as shown in Fig.12.12 (c). Note that the carryover factors for end E and G of members CE and CG are equal to -1.

(5) Bending moment diagram By using the symmetric property of the frame and the method of superposition, the bending moment

diagram of the frame will be constructed as shown in Fig.12.12 (d).

( )108

E

CG0.6 0.4

CE CG

36−72

Example 12-8

Analyze the continuous beam with a hangover as shown in Fig.12.13 (a) by using the moment distribution method and draw its bending moment diagram. Assume 1EI = .

Solution

(1) Bending stiffnesses and distribution factors At joint B:

14 44BC BCS i= = × =1

13 3 06BA BAS i= = × = .5

1 0.6671 0.5BCμ = =+

0.5 0.333

1 0.5BAμ = =+

−

43.2−

28.843.2

64.8

28.8−43.2−43.2

( )10843.2

43.2 43.243.2

43.243.243.2

43.2

64.8

(d)(c)

64.8− unit: kN m⋅

Fig.12.12 Figures of example 12-7(a) original structure and its loading; (b) a quarter of the frame and process of the calculation; (c) moment distribution process; (d) bending moment diagram

460

At joint C:

0CDS = 4CB CBS i=

0 0 0 4CD

CBiμ = =

+

4 14CB

CBi0CBiμ = =+

Note that since end D of member CD is free no moment is needed to rotate a unit value of joint C, i.e., . This implies that the moment at the joint is not needed to distribute at the end C of member CD

and t0CDS =

o carry over to end D.

60kN 10kN

BA C D

6m 2m2m 2m

B

AC

D0.6670.333

30

1 0

−

14.0−

10

14.0

20

14.020

43

( )60

BA

CD

→10− ←

3.33 6.67 →1.67− ←

0.56 1.11 →0.28− ←

0.09 0.19

(a)

(b)

(c)

20

(unit: kN m)⋅

Fig.12.4 Figures of example 12-8(a) original structure; (b) process of moment distribution and carryover; (c) bending moment diagram

→

20

1020−

3.33

0.560.56−

30

0.10

−

0

3.33− 0

0

0.10−0.05− ← 00.02 0.03

20−

12.5 No-Shear Moment Distribution Method 461

(2) Fixed end moments It can be seen that the fixed end moment at the end C of member CD can be directly obtained by the

method of cross section. It is

10 2 20kN mCD

FM = − × = − ⋅

Recalling from table 11-2, we find

60 4 30kN m8 8

FBC

PlM ×= − = − = − ⋅

60 4 30kN m8 8

FCB

PlM ×= = = ⋅

(3) Process of moment distribution and carryover The process of moment distribution and carryover is shown in Fig.12.13 (b). (4) Bending moment diagram The bending moment diagram is shown in Fig.12.13 (c). (5) Discussion As we can see from above process that the moments are distributed at two joints, could you distribute

them only at joint B?

12.5 No-Shear Moment Distribution Method

Thus far, we have applied moment distribution method to the analysis of structures in which the translations of the joints were zero. In this section, we apply the moment distribution method to analyze rigid frames whose joints may undergo both rotations and translations but whose columns’ shears can be statically determined by equilibrium considerations. Since there is no shear yielded in the process of moment distribution and carryover, the method is referred to as no-shear moment distribution method.

12.5.1 Applying condition of no-shear moment distribution method

As mentioned before, we will apply the moment distribution method to some rigid frames with sidesway. The applying condition is that the shear forces of all the members can be statically determined by equilibrium conditions except those without relative end translation.

Consider, for example, the rigid frame shown in Fig.12.14 (a), which is a commonly utilized primary system of a symmetric structure subjected to antisymmetric loads as discussed in Chapter 10. Both ends of each beam of the frame do not have any relative translation during the sidesway of the frame, i.e., there is no relative translation in the direction perpendicular to its axis, so the frame’s sidesway will not induce internal forces in the beams. These sorts of members are named members without relative end translation.

462 Chapter 12 Method of Successive Approximations and Influence Lines for Indeterminate Structures

Further more, the shear force of each column of the frame can be determined statically although the ends of the columns have sidesway. The shear force diagram of the columns is shown in Fig.12.14 (b). These kinds of members are known as members with statically determinate shears.

The columns AB and CD of the rigid frame shown in Fig.12.14 (c), however, are neither the members without relative end translation and nor members with statically determinate shears, so no-shear moment distribution method cannot be applied to the frame.

The detailed discussion about no-shear moment distribution will take place in the following subsections.

12.5.2 Fixed end moments for the members with statically determinate shears

Actually, the analyzing process of no-shear moment distribution method is the same as that of the moment distribution method discussed in the previous sections. The only difference between them is that the structures to be analyzed have different characteristics in internal forces and deformations.

For instance, when use no-shear moment distribution method to analyze the rigid frame shown in Fig.12.15 (a), the process includes following three steps: (1) Lock the rotatable displacement of joint A [Fig.12.15 (b)], which connects the two members, and determine the fixed end moments for each member. (2) Unlock the joint and evaluate the distributed and carryover moments of the member ends [Fig.12.15 (c)]. (3) Superpose the member end moments obtained from the first two steps to determine the final results of the member end moments.

Now let us concern the fixed end moments for the members of the frame shown in Fig.15 (a). Since the sidesway of the frame will not induce bending moments in beam AC, the fixed end moments of the member can be determined as before. This means that for the members without relative end displacement,

D

(a)(b) (c)

P

P

P

B

C

D ED

B

A

C

A

B

C

P

2P

3P

A

Fig.12.14 Rigid frame being able to use no-shear moment distribution method(a) half a portal frame; (b) shear diagram of columns; (c) rigid frame not being able to use no-shear moment distribution method


the fixed end moments can be evaluated as previous. However, the determination of the fixed end moments for the column will have some difference as we do previously. The difference is that when we lock the rotation of joint A, the joint still has horizontal translation. As long as the shear ABQ at the end A of member AB is determinate ( ), if we substitute the shear 0ABQ = ABQ for its corresponding restraint the column AB will be equivalent to the single span indeterminate member as shown in Fig.12.15 (d), whose fixed end moments can be conveniently determined by use the results in line 14 of table 11-2.

For the multistory rigid frame as shown in Fig.12.16 (a), which is composed of members without relative end displacement and members with statically determined shears, the method to evaluate the fixed end moments is the same as that mentioned above (used for one-story similar frame). If we substitute the shear of each top end of the two columns for their corresponding restraint the determination of the fixed end moments for the two columns will be equivalent to those of the two single span indeterminate members

A

B

CA

B

C A

B B

A0ABQ =

(a) (d)(b) (c)

Fig.12.15 Analytical process of no-shear moment distribution method (a) half a portal frame; (b) locking joint ; (c) unlocking joint ; (d) equivalent computing model of member

A AAB

(a) (c)

B

A

C

1P

2P

C

(b)

B

A1P

B1 2P P+

Fig.12.16 Fixed end moment determination for a two-story frame(a) locking joints and ; (b) equivalent computing model of member ;(d) equivalent computing model of member

A B ABBC


as shown in Fig.12.16 (b) and (c). It is observed from above that the steps to determine the fixed end moments of a member with

statically determinate shears can be concluded as: (1) determine the member end shear by a proper equilibrium condition; (2) evaluate fixed end moments by substituting the shear for its corresponding restraint of the member, which is a single span indeterminate member with one end fixed and another end doubly linked.

12.5.3 Distribution and carryover factors for the members with statically determinate shears

Next, let us focus our attention on the distributed and carryover moments for the members of the frame shown in Fig.12.15 (c) due to the unlocking of joint A. Unlocking joint A means releasing the joint by removing the clamp and applying a couple, equal in magnitude but opposite in sense, at the joint as shown in Fig.12.17 (a). Under the action of the couple, the internal force and displacement of column AB will be equivalent to a cantilever member shown in Fig.12.17 (b). It can be seen from the figure that there is no shear in the member, so bending moment is a constant. Thusly, this sort of member is named no-shear member or zero-shear member. Recalling from table 11-1 (in line 5), the distribution moments at the near ends and the carryover moments at the far ends will be calculated as follows.

AB ABM i Aθ= BA ABM M= −

Obviously, the bending stiffness of member AB is

AB AS i B= (12-13)

The carryover factor of the member is

(a) (b)

A

B

AθAθ

B

AABM

Aθ

Fig.12.17 Unlocking joint (a) performance of internal force and displacement of member ;(d) equivalent computing model of member

AAB

AB

0Q =0Q =


1ABC = − (12-14)

Similarly, the bending stiffnesses and carryover factors for the columns shown in Fig.12.16 can be determined as

BC BCS i= 1BCC = −

BA BAS i= 1BAC = −

It can be observed from forgoing discussion that the members with statically determinate shears are zero-shear members under the action of the joint moments. Their distributed and carryover moments are obtained under the condition of no-shear forces. The method is thusly referred to as no-shear moment distribution method.

Example 12-9

Determine the member-end moments by using no-shear moment distribution method and draw the bending moment diagram for the rigid frame shown in Fig.12.18 (a).

Solution

It can be seen from the figure that since the member BC is a member without relative end translation during the frame’s sidesway and the member AB is a member whose shear force can be determined statically, no-shear moment distribution method can be utilized to the frame. Recalling table 11-2, the fixed end moment will be determined as

6.61A

B C1.39

5.70

1.39

6.61−A

B C

1.281.39−

4 / 51/ 5BCBA

2.67

A

B C

1 4i =

2m

5kN

3.75− −

1.28−5.33−

1.395.14

2 3i =

2m

4m 1kN

/m

(a) (b) (c)

unit: kN m⋅

Fig.12.18 Figures of example 12-9(a) original structure and its loads; (b) computing process;(c) bending moment diagram


3 5 4 3.75kN m16

FBCM = − × × = − ⋅

221 1 4 2.67kN m

6 6FBAM ql ×= − = − = − ⋅

21 5.33kN m3

FABM ql= − = − ⋅

The bending stiffnesses and distribution factors of the members can be evaluated to be

4515

BC

BA

μ

μ

⎧ =⎪⎪⎨⎪ =⎪⎩

1

2

3 13

BC

BA

S iS i

= =⎧⎨ = =⎩

2

The carryover factor of member BA is equal to -1. The process of moment distribution and carryover is shown in Fig.12.18 (b). Fig.12.18 (c) shows the

bending moment diagram of the frame.

Example 12-10

Determine the member-end moments by using no-shear moment distribution method and draw the bending moment diagram for the rigid frame shown in Fig.12.19 (a).

Solution

The rigid frame shown in Fig.12.19 (a) is a symmetric structure subjected to an asymmetric loading. If we decompose the loading into a symmetric one and an antisymmetric one as shown in Fig.12.19 (b) and (c), the bending moments of the structure will be induced by the antisymmetric loading provided that we ignore the axial deformations of the two beams. If we use the symmetric property of the structure shown in Fig.12.19 (c), half its structure as shown in Fig.12.19 (d) can be analyzed, in which the flexural rigidity of the beams will be doubled as shown in the figure.

(1) Fixed end moments Since columns AB and BC are the members with statically determinate shears, their shear forces can

be determined to be

4kNABQ = 12.5kNBCQ =

From the equivalent computing models of the two columns shown in Fig.12.19 (e), their fixed end moments will be determined, by using table 11-2, to be


4 3.3 6.6kN m2

F FAB BAM M ×= = − = − ⋅

12.5 3.6 22.5kN m2

F FBC CBM M ×

= = − = − ⋅

(2) Distribution factors The bending stiffnesses of the members connecting to joint B can be written to be

3.5BA BAS i= =

5BC BCS i= =

3 3 54 162BE BES i= = × =

170.5B jj

S =∑

Thusly, the distribution factors at joint B can be obtained as

B

A

C

8.5kN

4kN 4kN

8.5kNB

A

C

17kN

8kN2.7i =

5 5

3.3m

3.6m

3.5

B

A

C

8.5kN

4kN

5

5.4

B

A

C

8.5kN

4kN 4kN

8.5kN 5.4

(a)

(d)

(b)

(c) (e)

4kN

B

A

3.5

B12.5kN

C

3.5

Fig.12.19 Figures of example 12-10(a) original structure and its loads; (b) symmetric structure subjected to symmetric loading;(c) symmetric structure subjected to antisymmetric loading; (d) half the structure; (e) equivalent computing models of columns


3.5 0.0206170.5BAμ = =

5 0.0293170.5BCμ = =

162 0.9501170.5BEμ = =

Similarly, we can determine the distribution factors at joint A, which is written in the square box shown in Fig.12.20 (a).

(3) Moment distribution and carryover The computing process is shown in Fig.12.20 (a), which adopted the order of B, A, B, A. Note that the

carryover factors of the columns are -1. The final bending moment diagram is drawn in Fig.12.20 (b).

21.64−

6.60−

A D

0.0211A

0.60−0.15 7.05

7.057.05−

0.0293

0.0206

C

6.60−0.600.15−

22.5−

0.00 0.01 0.14

27.650.85

22.5−0.85−0.01−23.36−

B

27.796.15−

E

23.36

B

C F

7.05

27.79

6.15

7.05

27.79

21.646.15

23.36

0.97

89 0.

9501

21.64

(a) (b)

unit: kN m⋅

Fig.12.20 Computing process and result of example 12-10 (a) computing process; (b) bending moment diagram

12.6 Shear Distribution Method 469

12.6 Shear Distribution Method

Up to now, we have discussed the application of moment distribution method to the analysis of structures without sidesway and no-shear moment distribution method to the analysis of some special structures with sidesway. In this section, we will discuss the shear distribution method and its application to the analysis of bent frames with sidesway.

12.6.1 Bent frames subjected to horizontal loads at their top hinged joints

Now let us reconsider the bent frame subjected to a concentrated horizontal force shown in Fig.12.21 (a), which was analyzed by displacement method in Chapter 11 (example 11-4). When one of the columns of the bent frame is subjected to a horizontal translation of a unit magnitude at its top, [Fig.12.21

(b)], the shear force yielded in the column will be

1Δ =

2

3iQh

= , which is actually the stiffness coefficient due

to the unit sidesway of the column. Since the each column has the same sidesway Δ , the shear forces of the columns induced by the sidesway will be written as

11 12

1

22 2

2

33 32

3

3

3

3

iQ dh

iQhiQ d

h

⎫

2d

= Δ = Δ ⎪⎪⎪= Δ = Δ⎬⎪⎪

= Δ = Δ ⎪⎭

(a)

2

3 jj

j

id

h=

( ) jj

j

EIi

l= jhIn which, , , represent the stiffness coefficient due to sidesway, relative flexural

rigidity and height of the jth column, respectively. Recalling from the horizontal force equilibrium condition of the free body of the top portion of the

bent frame [Fig.12.21 (c)], we write

1 2 3Q Q Q P+ + = (b)

Substituting Eq. (a) into Eq. (b), we find

3

1

jj

jj

dQ P

djμ

=

= =

∑ (c)

It can be observed from Eq. (c) that the shear forces of the columns are direct proportion to their stiffness


3

1

jj

jj

d

dμ

=

=∑

, which is now defined as shear distribution factor. coefficient due to sidesway,

Now we will explain the problem in the shear distribution point of view as follows. When the bent frame shown in Fig.12.22 (a) is subjected to a horizontal concentrated force, the force is distributed to the columns in the proportion of their shear distribution factor, i.e., the column’s shear is evaluated by multiplying the force by its shear distribution factor ( j jQ Pμ= ). Since the bending moments at the hinges

are equal to zero, the bending moments of the columns can be determined by their distributed shear forces.

P

1dΔ2d Δ 3d Δ

P

33

3

EIih

=

Δ 1Δ =P

2

3idh

=

11

1

EIih

= 22

2

EIih

= i h

(a) (b)

(c)

Fig.12.21 Analysis of a bent frame subjected to a horizontal load at its top(a) original structure and its loads; (b) stiffness coefficient due to sidesway;(c) free body diagram of the top portion of the bent frame

P

1 1Qh

P

3i1i

2i

1h2h

3h

1 1Q Pμ=

2 2Q h

3 3Q h

2 2Q Pμ= 3 3Q P=

(a)

(b)

μ

Fig.12.22 Shear distribution method of a bent frame(a) original structure and its loads; (b) shear distribution


Thusly, the method to analyze bent frames subjected to horizontal loads at their top is referred to as shear distribution method.

12.6.2 Shear distribution of rigid frames with beams of infinite flexural rigidity

Now let us discuss the application of shear distribution method to the rigid frame with a beam of infinite flexural rigidity subjected to a concentrated horizontal force as shown in Fig.12.23 (a).

Since the flexural rigidities of the beams are infinite, there is neither deflection nor rotation for the beams. Therefore, only the common horizontal translation of the top joints can be considered to be the unknown displacement of the frame. While the shear force induced in a column having no end rotation due

to a unit horizontal displacement as shown in Fig.12.23 (b) will be equal to 2

12iQh

= , the stiffness

coefficient due to the unit sidesway must be 2

12ih

, denoted as 2

12idh

= . As long as each column has the

same sidesway , their shears can be expressed as Δ

j jQ d= Δ (a)

In which, 2

12 jj

j

id

h= represents the stiffness coefficient of the column having no end rotation due to

sidesway. Recalling from the horizontal force equilibrium condition of the free body composed by the top

P

3i1i

2i

Δ 1Δ =bi = ∞ 0θ = P

P

1d Δ 2d Δ 3d Δ

2

12idh

=

h

i

(a) (b)

(c)

Fig.12.23 Analysis of a bent frame with beams of infinite flexural rigidities(a) a bent frame with beams of infinite flexural rigidities;(b) stiffness coefficient due to sidesway; (c) free body diagram of the top portion


portion of the rigid frame [Fig.12.23 (c)], we write

1 2 3Q Q Q P+ + = (b)

Substituting Eq. (a) into Eq. (b), we obtain

3

1

jj j

jj

dQ P

dμ

=

= =

∑ (c)

It can be observed from Eq. (c) that the shear forces of the columns of the rigid frame with beams of

infinite flexural rigidity are distributed by the distributed factor jj

j

dd

μ =∑

. In other words, when a rigid

frame with beams of infinite flexural rigidity as shown in Fig.12.24 (a) is subjected to a horizontal concentrated force at its top, the force is distributed to the columns by their shear distribution factor, i.e., the column’s shear is evaluated by multiplying the force by its shear distribution factor ( j jQ Pμ= ). Note

that the inflection point (the location of zero bending moment, referring example 10-6) is located at the middle of the column when subjecting the column having no end rotation to a sidesway . Therefore, the magnitude of the bending moments at the ends of the column can be determined by multiplying its

distributed shear by half the column’s height, i.e.,

Δ

( )2

jj j

hM Pμ= × . Once the bending moments of the

columns are known, the bending moments of the beams can be determined by equilibrium as well, and the bending moment diagram of the frame can be conveniently constructed as shown in Fig.12.24 (b).

12.6.3 Shear distribution of bent frames or rigid frames of infinite-rigidity beams subjected to no-joint loads

P

3i1i

2i

1h 2h3h

P

1Q2Q

3Q

1 1

2Q h

2 2

2Q h

3 3

2Q h

(a) (b)3 3

2Q h

2 2

2Q h

1 1

2Q h

bi = ∞

Fig.12.24 shear distribution of a rigid frame with beams of infinite flexural rigidities (a) original structure; (b) shear distribution and bending moment diagram


The forgoing discussed shear distribution method may also be applied to the analysis of the bent frames or rigid frames of infinite-rigidity beams subjected to no-joint loads. Consider, for example, the bent frame shown in Fig.12.25 (a), whose columns are subjected to no-joint loads. The following three steps will help us to apply shear distribution method to these kinds of structure.

(1) Lock the sidesway of the frame by adding an artificial corresponding restraint as shown in Fig.12.25 (b). Then, determine the shear forces of the ends (fixed end shears) of the columns subjected to external loads by using the results of table 11-2. By the fixed end shears, the restraint force corresponding to the artificial restraint, denoted by , will be determined. 1PF

(2) Unlock the sidesway by releasing the artificial restraint and apply the negative of the restraint force, on the frame at the location and in the direction corresponding to the artificial restraint as shown in Fig.12.25 (c). And then determine the shear forces distributed in the columns by shear distribution method.

1PF−

(3) Superposing the computing results obtained from steps (1) and (2) together will obtain the final shear forces of the columns of the frame.

Example 12-11

Determine the member-end moments by using shear distribution method and draw the bending moment diagram for the rigid frame shown in Fig.12.26 (a).

Solution

(1) Shear distribution factors of the columns ∑

=j

jj d

dμ

= +

P

Δ

=

1PFP

+q 1PF

q

(c)(a) (b)

Δ

1F P

Fig.12.25 Analysis due to no-joint loads(a) original structure; (b) locking sidesway; (c) unlocking sidesway and shear distribution

bi = ∞


1 31 0.25

1 2 1μ μ= = =

+ +

22 0.5

1 2 1μ = =

+ +

(2) Shears of the columns

1 3 0.25 10 2.5kNQ Q= = × =

2 0.5 10 5kNQ = × =

(3) The bending moments of the ends of the columns For the rigid frame of infinite-rigidity beams, the inflection points are located at the middles of the

columns, so the end moments of the columns will be determined to be

11 3 1

42.5 5kN2 2hM M Q= = − × = − × = −

22 2

45 102 2hM Q= − × = − × = − kN

10kNP=

10kN

5

By the equilibrium conditions of joint moments [Fig.12.26 (c) and (d)], the end moments of the beams can be evaluated to be

55

5

5kN m⋅ 510

10

1 2.5kNQ = 2 5kNQ = 3 2.5kNQ =

5

AABM

10

B

BCMBAM

BA C

2 2i =

bI =∞

1 1i = 4m3 1i =

FED

(d)(c)

(a) (b)

Fig.12.26 Figures of example 12-11(a) original structure and its loads; (b) shear distribution and bending moment diagram;(c) moment equilibrium condition of joint ; A (d) moment equilibrium condition of joint B


5kN mABM = ⋅

1 10 5kN m2BA BCM M= = × = ⋅

(4) Bending moment diagram Once the end moments of the members are known, the bending moment diagram will be constructed

as shown in Fig.12.26 (b).

Example 12-12

Analyze by using shear distribution method and draw the bending moment diagram for the rigid frame shown in Fig.12.27 (a).

Solution

the sidesway of the frame by adding an artificial corresponding restraint as shown in Fig.12

(1) Lock.27 (b). Then, the fixed end shears of the columns and the restraint force can be determined. By using

table 11-2, we write

5 4 10kN2 2

FAD

qLQ ×= − = − = −

0F F

BE CFQ Q= =

BA

6.66F

3.34

55kN m⋅

5

1.66

( )10

6.66kN m⋅

3.34

10kN

5

5

10

10

5

5

5kN m⋅

10

10

( )10

11.66

BA C

D F

Δ

2 2i =

bI = ∞

1 1i =

E

3 1i = 4m

5kN

/m

C

(d)

(a) (b)

(c)

D E

55 5

10kN

Fig.12.27 Figures of example 12-12(a) original structure and its loads; (b) primary system and fixed end moments and shears;(c) shear distribution and bending moment diagram; (d) bending moment diagram


0X =∑From the horizontal-force equilibrium condition of the beams, , the restraint force yielded at the

artificial restraint will be

1 10kNPF = −

(2) Unlock the si

desway by releasing the artificial restraint and apply the negative of the restraint force, on the frame at the location and in tshow

as sh

y employing the shear forces and the location of inflection points (at the midd

n orces of a statically graph of a response

func

es com

use the method of force method, discussed in Chap

1PF− he direction corresponding to the artificial restraint as n in Fig.12.27 (c). And then determine the shear forces distributed in the columns by shear distribution

method own in the figure. (3) Superposing the computing results obtained from steps (1) and (2) together will obtain the final

shear forces of the columns. Ble of the column), the bending moments of the members and the bending moment diagram of the

structure will be drawn as shown in Fig.12.27 (d).

12.7 Influence Lines for Forces of Statically Indeterminate Structures

I this section, we discuss the procedures for constructing influence lines for findeterminate structure. It may be recalled from Chapter 8 that an influence line is a

tion of a structure as a function of the position of a downward unit load moving across the structure. The basic procedure for constructing influence lines for forces of an indeterminate structure is the

same as that for determinate structures considered in Chapter 8. The procedure essentially involvputing the values of the response function of interest for various positions of a unit load on the

structure and plotting the response function values as ordinates against the position of the unit load as abscissa to obtain the influence line. Since the influence lines for forces and moments of determinate structures consist of straight-line segments, such influence lines were constructed in Chapter 8 by evaluating the ordinates for only a few positions of the unit load and by connecting them with straight lines. The influence lines for forces of an indeterminate structure, however, are generally curved lines. Thus the construction of influence lines for forces of an indeterminate structure requires computation of many more ordinates than necessary in the case of determinate ones.

Although any of the methods of analysis of indeterminate structures presented in the book can be used for computing the ordinates of influence lines, we will

ter 10, for such purposes. Once the influence lines for forces of an indeterminate structure have been constructed, they can be applied to determine the unfavorable loading position and absolute maximum response and the like of the structure.

12.7.1 Principle of constructing the influence lines for forces of statically indeterminate beams by static method

12.7 Influence Lines for Forces of Statically Indeterminate Structures 477

Consider the continuous beam shown in Fig.12.28 (a). The beam is subjected to a downward-moving rated load of unit xconcent magnitude, the position of which is defined by the coordinate measured from

the left end A of the beam, as shown in the figure. Suppose that we wish to draw the in e line for the vertical reaction

fluenc

1Z

is repla

at right-end support of the beam by using force method. Thusly, the support C is

removed and it ced by a vertical force

xB

A C

1Z . Now we will determine 1Z stem

1

by force method. Recalling from force method, the consideration of that the deflection of the primary sy [Fig.12.28 (c)] at C due to the combined effect of the external unit load and the unknown redundant Z must be equal to zero, will yield the compatibility condition as follows

11 1 1 0PZδ δ+ =

Consequently,

11

11

PZ δδ

= − (12-15)

Note that the primary structure is still an indeterminate beam [Fig.12.28 (b)], whose degrees of indeterminacy are equal to that the degrees of indeterminacy of the original structure subtract one (i.e.,

).

A

1n −

BA

C

xB

A C

1P =

1P =

1Z

BA

C1Pδ

BC

11δ

1 1Z =

x

(d)

(e)

(a)

(c)

(b)

x 1P =

1Pδ

Fig.12.28 Influence line for forces of an indeterminate structure(a) original indeterminate structure; (b) primary structure;(c) primary structure subjected to 1; (d) primary structure subjP = 1ected to 1Z =


In which the flexibility coefficient 1Pδ denotes the deflection of the primary beam at C due to the unit load at position 1P = x [Fig.12.28 (d)], whereas the flexibility coefficient 11δ denotes the deflection at C due to the unit value of the redundant 1 1Z = [Fig.12.28 (e)].

We can use Eq. (12-15) for constructing the influence line for 1Z by placing the unit load successively at a number of positions along the beam, evaluating 1Pδ for each position of the unit load, and plotting the values of the ratio 1 /P 11δ δ− . However, a more efficient procedure can be devised by applying law of reciprocal displacements (Section 9.8), according to which the deflection at C due to a unit load at 1P = x must be equal to the deflection at x due to a unit load of at C; that is, 1 1Z =

1 1P Pδ δ= . Thus equation (12-15) can be rewritten as

11

11

PZ δδ

= − (12-16)

In which 1Pδ denotes the deflection of the primary structure at position x due to the unit reaction at C [see Fig.12.28 (e)]. 1 1Z =

It can be observed that since the flexibility coefficient 11δ is a constant without respect to the unit load position x and the deflection 1Pδ varies with respect to x , 1Pδ is a function of x . Thusly, the expression for reaction 1Z in terms of the variable position x of the unit load can be expressed clearly as

1P =

11

11

( )( ) P xZ x δδ

= − (12-17)

Equation (12-17) implies that when the position x of the unit load varies, the shape of the function

1( )Z x is just the influence line for 1Z , whereas the variable shape of 1( )Z x is proportional to that of

1( )P xδ and the variable shape of 1( )P xδ is just about the deflection shape of the primary beam due to the unit reaction at C as shown in Fig.12.28 (e). 1 1Z =

Therefore, the deflection shape of the primary structure due to 1 1Z = can represent the shape of the influence line for 1Z . Eq. (12-17) also indicate that the influence line for 1Z can be obtained by multiplying the deflected shape of the primary structure due to the unit load by the scaling factor 1 1Z =

111/δ− . Note that the deflections 1( )P xδ and 11δ are considered to be positive when their senses coincide

with their corresponding unit loads. However, since the negative ratio 1 1/P 1δ δ represents the function of influence line for 1Z , the ordinate of the influence line for 1Z is positive in the upward direction (i.e., above abscissa x ).

Example 12-13


Construct the influence line for vertical reaction of the indeterminate beam shown in Fig.12.29 (a) by static method.

BY

Solution

(1) Selection of primary system The primary system will be obtained by substituting a vertical force at B, 1 BZ Y= , for its

corresponding support as shown in Fig.12.29 (b). (2) Bending moment diagrams The bending moment diagrams due to the unit value of the reaction and the unit load are

shown in Fig.12.29 (c) and (d), respectively. 1P =

(3) Flexibility coefficient 11δ and deflection 1Pδ By graph-multiplication method, the flexibility coefficient 11δ and deflection 1Pδ will be evaluated

to be

3

111 1 2

2 3 3ll l l

EI EIδ = × × × =

BA

1P =

l

x lα=

BA

1P =

1 BZ Y=

( )1 lα−

BA1 1Z =

l

1P =

40128

1

1

(d)

(e)

(a) (b)

(c)

x l=α

lα

x l=α

11128

81128

1

Fig.12.29 Figures of example 12-13(a) original indeterminate structure; (b) primary system; (c) diagram of primary structure; (d) diagram of primary structure; (e) influence lP

MM ine for BY


32

11 1 1 1 2 1 1(1 )

2 3 2 3 2Pll l l l l l

EI EI3

6δ α α α α α α α⎡ ⎤ ⎛= − × − × + × × = − −⎜⎢ ⎥

⎞⎟

⎣ ⎦ ⎝ ⎠

(4) Solution of 1Z Substituting the expressions for 11δ and 1Pδ into Eq. (12-17), the solution of 1Z will be obtained

as

2 3 211

11

( ) 3 1 1( ) (3 )2 2 2

P xZ x 3δ α α αδ

= − = − = −α (a)

(5) Construction of influence line for 1Z The ordinates of the influence line for 1Z can now be computed by applying Eq. (a) successively for

each position of the unit load . For example, when the unit load is located at , 1P = 0x = 14

x l= ,

12

x l= , 34

x l= and x l= (i.e., 1 1 30, , , , 14 2 4

α α α α α= = = = = ), the values of 1Z are given by

, 1 0Z = 111

128Z = , 1

40128

Z = , 181

128Z = and 1 1Z = , respectively. Connecting the points of the

influence line smoothly will obtain the influence line as shown in Fig.12.29 (e). It can be seen from the shape of the influence line that the main difference between the influence lines

for statically indeterminate and those for statically determinate is that the influence lines for forces of determinate structures consist of straight-line segments, while the influence lines for forces of indeterminate structures are generally curved lines.

12.7.2 Qualitative influence lines for continuous beams by principle of virtual displacement

In many practical design stages, such as when designing continuous beams or building beams subjected to uniformly distributed live loads, it is usually sufficient to draw only the qualitative influence lines to decide where to place the live loads to maximize the response functions of interest. As in the case of statically determinate structures (Section 8.5), mechanismic (or virtual displacement) method provides a convenient means of establishing qualitative influence lines for indeterminate structures.

Recall from Section 8.5.2 that virtual displacement method can be stated as follows: The influence line for a force (or moment) response function is given by the deflected shape of the

released structure obtained by removing the restraint corresponding to the response function from the original structure and by giving the released structure a unit displacement (or rotation) at the location and in the direction of the response function.

The procedure for constructing qualitative influence lines for indeterminate structures is the same as that for determinate structures discussed in Section 8.5. The procedure essentially involves: (1) removing


from the given structure the restraint corresponding to the response function of interest to obtain the released structure; (2) applying a infinitesimal displacement (or rotation) to the released structure at the location and in the positive direction of the response function; and (3) drawing a deflected shape of the released structure consistent with its support and continuity conditions. The influence lines for indeterminate structures are generally curved lines since the released structures are, generally, still an indeterminate structure.

Once a qualitative influence line for a structural response function has been constructed, it can be used to decide where to place the live loads to maximize the value of the response function. As discussed in Section 8.6, the value of a response function due to a uniformly distributed live load is maximum positive (or negative) when the load is placed over those portions of the structure where the ordinates of the response function influence line are positive (or negative). Because the influence-line ordinates tend to diminish rapidly with distance from the point of application of the response function, live loads placed more than three span lengths away from the location of the response function generally have a negligible effect on the value of the response function. With the live-load pattern known, an indeterminate analysis of the structure can be performed to determine the maximum value of the response function.

Example 12-14

Draw qualitative influence lines for the bending moments on support section C and section G, and the shear on section G of the five-span continuous beam shown in Fig.12.30 (a).

Solution

(1) Influence line for bending moment at support section C. To determine the qualitative influence line for the bending moment at C, we insert a hinge at C in the

actual beam and give the released beam an infinitesimal relative rotation in the positive direction of CM by rotating the portion to the left of C counterclockwise and the portion to the right of C clockwise, as shown in Fig.12.30 (b). The deflected shape of the released beam thus obtained represents the qualitative influence line for CM [Fig.12.30 (c)].

(2) Influence line for bending moment and shear at section G. The qualitative influence line for bending moment GM at section G is determined in a similar

manner and is shown in Fig.12.30 (d) and (e). To determine the qualitative influence line for shear at section G, we insert two links at G in the actual

beam and give the released beam an infinitesimal relative displacement in the positive direction of the shear by moving end G of the left portion of the beam downward and end G of the right portion upward, as

shown in Fig.12.30 (f). The influence line for is shown in Fig.12.30 (g). GQ

GQNote that to obtain a quantitative influence line for a force of an indeterminate structure, the restraint


corresponding to the force must be removed and a corresponding unit displacement must be applied to the released structure, and the deflection shape may be obtained by determining the deflection equations of the members of the structure. The further discussion about the quantitative influence lines can be fined in other special papers or books.

x 1P =

A B C D E FG

A B C

(a)

(b)D E FCθ

1Pδ

A B C D E F(c)

A B C D E FG

A B C D E F

G

A B C D E F

Gλ

A B C D E FG

(d)

(e)

(f)

(g)

Gθ

Fig.12.30 Influence line for continuous beam(a) original indeterminate structure; (b) deflection shape due to ;(c) influence line for ; (d) deflection shape due to ; (e)

C

C G

MM M influence line for ;

(f) deflection shape due to ; (g) influence line for G

G GQ Q

CM

MG

GQ

M

12.8 Most Unfavorable Arrangement of Live Loads and Envelops for Internal Forces of Continuous Beams 483

12.8 Most Unfavorable Arrangement of Live Loads and Envelops for Internal Forces of Continuous Beams

12.8.1 Most unfavorable arrangement of live loads on continuous beams

Continuous beams are commonly adopted structures in engineering projects. Generally, the loads applying on a continuous beam involve dead loads distributed in whole span and maintained in life, and live loads distributed arbitrarily and maintained from time to time. Thusly, the bending moment on some section of the beam due to the dead loads is unchangeable, while that due to the live loads is changeable. In order to guarantee sufficient strength of a continuous beam, the maximum bending moment of the beam due to both dead and live loads must be determined before design the beam. Actually, the determination of the maximum bending moment is the evaluation of the effect of the live loads, because once the maximum bending moment (positive or negative) due to the live loads is determined, the final maximum bending moment can be obtained by adding both the maximum bending moment due to live loads and that due to dead loads.

However, the determination of the maximum bending moment due to the live loads needs to know the unfavorable arrangement of the live loads on the continuous beam, whereas the estimation for the unfavorable arrangement of the live loads must resort to the help of influence line for the bending moment.

Consider, for example, the five-span continuous beam shown in Fig.12.31 (a). The qualitative influence lines for bending moments at support section B and section 2 are also shown in Fig.12.31 (b) and (e). It can be seen from the figure that to cause maximum positive bending moment at B, we place the live load over spans CD and EF, where the ordinates of the influence line for BM are positive [Fig.12.31 (c)]; contrarily, the placement of the live load over other spans will cause maximum negative bending moment at B, as shown in Fig.12.31 (d). Similarly, the arrangement of the live load to cause maximum positive and negative bending moments at 2 are shown in Fig.12.31 (f) and (g).

It can be observed from above analysis that (1) the live-load arrangement for maximum negative bending moment on the section near a support is that the load is placed over the adjacent and every second spans to the support; (2) the live-load arrangement for maximum positive bending moment on the section of a middle span is that the load must be applied on the span and every second spans to the section.

12.8.2 Envelops for internal forces

An envelop of the bending moments (or shears) of a continuous beam is a curved line whose ordinates are composed of the maximum bending moments (or shears) against their corresponding sections of the beam. The envelop of bending moment of the beam may be obtained by (1) evaluating the bending moments of the beam due to dead loads; (2) estimating, by considering the unfavorable live-load arrangements, the maximum (positive and negative) bending moments due to the live load; (3) superposing


the maximum (positive and negative) bending moments from the first two cases; (4) laying off the maximum bending moments over or below the axis of the beam; (5) connecting the maximum bending moments smoothly. Similarly, the envelop of shears can be also obtained by above five-step’s process. The following example will illustrate the construction of bending-moment envelop in detail.

A B C D E F

B C D E F2A

Minfluence line for B

live-load arrangement for positive maximum BM

A B C D E F

A B C D E F

live-load arrangement for negative maximum

(a)

(b)

(c)

(d)

2influence line for

M B

M

E FA B C D

2live-load arrangement for positive maximum M

2live-load arrangement for negative maximum M

A B C D E F

A B C D E F

(e)

(f)

(g)

Fig.12.31 Live-load arrangement for maximum bending moment of a continuous beam


The maximum (positive and negative) bending moment of each section might be obtained by method

Example 12-14

Draw the bending-moment envelop for the girder of the floor system shown in its cross section, Fig.12.32 (a).

Solution

(1) Analytical model of the girder The girder is the member of the floor system of an industrial workshop. Since the rigidity of the girder

is much stiffer than that of the columns and there is local deformations under ends of the girder, the bending moments of the girder will almost not transferred to the columns, and thusly the girder is considered to be supported by hinges and rollers. The analytical model of the girder will be simplified as shown in Fig.12.32 (b). In order to simplify analysis, the weight of the girder is considered as a portion of concentrated dead load which is measured by G as shown in Fig.12.33 (a).

(2) Bending moment due to dead loads The entire girder is finally simplified as a three-span continuous beam. The bending moment diagram

due to the dead loads can be obtained by using anyone of the method discussed in Chapters 11 through 12 for analyzing an indeterminate structure. Thusly obtained bending moment diagram of the beam is shown in Fig.12.33 (a).

(3) Bending moment due to the live load

b

girder

eamslab

column

3037

19 18730 750 2

A B

748 750 2

(a)

(b)

Fig.12.32 Figures of example 12-14(a) cross section of a floor system ; (b) computing model of the girder


of su

(4) Envelop of bending moment nts of the beam can be determined by superposing the bending

mom

perposition, that is: (1) draw the bending moment diagram due to the live loads arranged over each span of the beam individually, as shown in Fig.12.33 (b) through (d); (2) superpose the maximum (positive and negative) bending moments from the figures (b) to (d) for each section together.

The envelop of the bending momeent evaluated in cases (2) and (3). In other words, the ordinates of the envelop of the bending moments

can be obtained by adding the maximum (or minimum) bending moments due to the dead loads and those due to the live loads. For instance, the maximum and minimum bending moments at sections 1, 2 and B

AB C

D

G66kNG =

G G G G

42

33.333.3

131.6kN m

(a)

(b)

(c)

⋅

77.2121.2

3

1 5 6

AB

C D42

24.1

85.1

31

5 6

109.8kNP =P 145.8kN m⋅

177.3

24.5

12.1

225.9

36.5A B C D

42

165.0

31 5 6

PP109.5kN m⋅73.0

36.2

109.8kN

Fig.12.33 Determination of bending-moment envelop (contd)(a) bending moment diagram due to dead loads;(b) bending moment diagram due to live loads arranged over first span;(c) bending moment diagram due to live loads arranged over second span


M = − − − = −

Thusly obtained bending-moment envelop is shown in Fig.12.33 (e).

(d)

(e)

will be calculated as

1max 121.2 225.9 12.1 359.2kN mM = + + = ⋅

2max 77.2 177.3 24.1 278.6kN mM = + + = ⋅

1min 121.2 36.5 84.7kN mM = − = ⋅

2min 77.2 73.0 4.2kN mM = − = ⋅

Bmax 131.6 36.2 95.4kN mM = − + = − ⋅

B min 131.6 145.8 109.5 386.9kN m⋅

145.8kN m⋅

A

B C

D4

2

278.6

31 5 6

198.3

95.4

386.9kN m⋅

359.2

76.3

84.74.2

A B C D4

2

24.1

85.1

31

5 6

P 109.8kNP =

177.3

24.5

12.136.2

225.9

Fig.12.33 Determination of bending-moment envelop (d) bending moment diagram due to live loads arranged over third span;(e) bending-moment envelop


SUMMARY

In this chapter, we have studied two methods of successive approximations, which are two classical formulations of the general displacement (stiffness) method, called moment distribution method and no-shear distribution method. In addition, shear distribution method, the method of constructing influence lines for the forces of continuous beams, the arrangement to cause maximum bending moment and envelop of bending moments (or shears) of continuous beams are studied respectively.

1. The moment distribution method is suitable for the analysis of continuous beams and rigid frames without sidesway. The concept and procedure of the method are clear and simple as long as there is no need to develop the simultaneous equations of the unknown displacements, and the member end moments can be determined directly by simple algebraic operation. So the method is a widely used hand-oriented method. The analytical procedure of the method contains the following two steps:

(1) Lock the rigid joints. The step involves computing fixed-end moments due to the external loads by assuming that all the free joints of the structure are temporarily restrained against the rotations.

(2) Unlock the rigid joints. The step consists of balancing the moments at free joints by the moment-distribution process. In the moment-distribution process, at each free joint of the structure, the unbalanced moment is evaluated and distributed to the member ends connected to it. Carryover moments induced at the far ends of the members are then computed, and the process of balancing the joints and carrying over moments is repeated until the unbalanced moments are negligibly small. The final member end moments are obtained by algebraically summing the fixed-end moment and all the distributed and carryover moments at each member end.

2. No-shear distribution method is a special case of moment distribution method. Since there is no shear induced in the members whose shears are statically determinate and there is no

bending moment yielded in the members having no relative translation during the joint sidesway, the moment distribution can be applied to this kind of structure only consisting of members with statically determinate shears and those without relative translation due to the joint sidesway. Their distributed and carryover moments are obtained under the condition of no-shear forces.

3. Shear distribution method is suitable for the analysis of the structures without joint rotations, such as bent frames with hinged joints at their tops and rigid frames with beams of infinite rigidities. The horizontal forces will be distributed in the columns by their corresponding shear distribution factors.

4. There are two methods, static and mechanical (or virtual work) method, to construct influence lines for statically indeterminate structures. The formulation of static method is the principle of force method; the principle of mechanical method is actually the application of unit displacement method applied to determinate the deflection shapes of indeterminate structures.

The concept of envelop of bending moments (or shears) is very helpful in the design of bridges, beams


of building frames and the like.


12-1 What are the sign conventions for the fixed end moments and moments at the ends of the members of a structure?

12-2 What is the bending stiffness? How much are the bending stiffnesses for a member with both nearend and farend fixed and for a member with nearend fixed and farend hinged, respectively?

12-3 What is distribution factor? What is the difference between the distribution factors and bending stiffnesses? Why is the summation of the distribution factors at a joint equal to 1?

12-4 What is the carryover factor? How to determine a carryover factor? 12-5 How to determine the distributed and carry over moments due to a joint couple? 12-6 How many steps is the computing operation of the moment distribution method due to a general

loading? What is the physical meaning of each step? 12-7 What is a fixed end moment? How to determine the restraint moment at an artificial restraint? Why

must the restraint moment be added a minus sign when it is distributed? 12-8 How are the computing steps when analyze continuous beams and rigid frames without sidesway by

moment distribution method? How to determine the end moments of a member? 12-9 Why will the restraint moments at the artificial restraints vanish when implement moment

distribution method to the continuous beams and rigid frames? i.e., why is the analysis convergent? 12-10 Can the moment distribution method be used to analyze the internal forces due to the temperature

changes and support settlements? In what circumstance it can be used and in what cases it cannot be used?

12-11 What is no-shear distribution method? What are its application conditions? 12-12 Why can no-shear distribution method be only applied to single span symmetric rigid frames? 12-13 What problems will occur when no-shear distribution method implements to single span asymmetric

rigid frame? 12-14 What is shear distribution method? What are its application conditions? How to use shear

distribution method when the columns of a structure are subjected to no-joint loads? 12-15 What is the difference between the influence lines for statically indeterminate structures and those

for statically determinate structure? 12-16 Why must the primary structure be the beam from which a restraint corresponding to the force

whose influence line is desired is removed when determines the influence line for the force by static method?

12-17 Please explain the principle and summarize the steps of constructing the influence lines for


indeterminate structures by virtual displacement method.


12-1 Determine the bending moments at the ends of the members for the following continuous beams and rigid frames shown in the figure by moment distribution method.

12-2 Analyze the continuous beams shown in the figure by moment distribution method and draw their bending moment diagrams.

12-3 Analyze the rigid frame shown in the figure by moment distribution method and draw their bending moment diagram.

BEI

A CEI

M

6m6m

AB

C

EI

2EI2EI

60kN m⋅

4m 4m

3m

Dproblem 12-1 (a)

BA C

2m6m 2m

60kN m⋅ 40kN

BA C2m 4m

3EI =

20kN/mq =

1EI =

(a) (b)

=constantEI

problem 12-3

B

A

C3EI =

20kN/mq =

1EI =

6m

4m

problem 12-1 (b)

problem 12-2 (b)problem 12-2 (a)


12-4 Analyze the continuous beams shown in the figure by moment distribution method and draw their bending moment diagrams and determine their reactions.

12-5 Analyze the rigid frames shown in the figure by moment distribution method and draw their bending moment diagrams.

12-6 Analyze the rigid frames shown in the figure by moment distribution method and draw their bending moment, shear and axial force diagrams.

BA C

2m

20kN/m

2m2m6m6m

40kN 40kN

D6m 6m3m3m

3EI = EEI 41= I =

400kN 40kN/mBA C D

(a) (b)

=constantEI

(c)

BA C

2m

20kN/m

2m 1.2m

40kN 40kND

4m 4m4m

FE

problem 12-4

=constantEI


BA C D

FE

15kN/m80kN

6m3m 6m3m

BA C D

FE1m5m 5m

20kNP =10kN/mq =

4m6m

(a) (b)

=constantEI

4I 4I

2I 2I


(c)

AB C

D E F

80kN20kN/mG

6m 3m6m 3m

EI

2EI

EIEI

EI EI 6m2m

(d)

4m 4m

60kN

A

B C

D E

6m6m 20

kN/m

=constantEI

problem 12-5

4m 4m 4m 4m

10kN/mq =

A B C D E

4m(b)

constantEI =

A

B C

D

20kN/mq =

12m

6m

(a)

=constantEI

problem 12-6

EI

EI EI

EI

EI

EI

EI

EI60kN 60kN 60kN 60kN

20kN/m

4EI 4EI 4EI

2m 2m 2m 2m 2m 2m2m

4m4m

A B C D

I J K L

(d)

E F G H

2m

A B

1.2m 1.2m0.6m

12kN 12kN4kN4kN

1.5EI

EI EI

EI EI

0.5m 0.5m

4m 3m1kN/m

C D

E F

(c)


12-7 Analyze the rigid frames shown in the figure by moment distribution method and draw their bending moment diagrams.

12-8 Analyze the structures shown in the figure by shear distribution method and draw their bending moment diagrams. constantEI =

A

BC

D

20kN

12m6m

(a)

1EI = 1EI =

2EI =

problem 12-7

B

A

10kN/m

2 3i

C1 4i =

4m=

4m

(b)

D

A

B

C

E

F8m

0.9i =

3.8m

3.8m

0.9i =

0.395 0.395

0.395 0.395

(c)

19kN

38kN

problem 12-8

A B C DI

HF G

(a)

I I I

E

8m

EA = ∞ EA = ∞ EA = ∞

20kN

/m

A

C

12m(b)

6m

20kN

/m

bEI = ∞

I I

B

D


12-9 Draw the influence line for the bending moment AM of the beam shown in the figure and answer the question when the unit load reaches to where the magnitude of AM will get the maximum value.

12-10 Draw the influence lines for the reaction , bending moment BR DM and sear force DQ of the beam shown in the figure. Assume constantEI = .

A B

1P=

l

x x 1P=

A B CD

6m3m 3m


CHAPTER 13 MATRIX DISPLACEMENT METHOD

In this chapter, we begin with an introduction, discussing the process of preparing a computing model of the structure to be analyzed. We also define global and local coordinate systems and explain the concept of unknown displacements of a structure. Next we derive member force-displacement relations in local coordinates. We consider the transformation of member end forces and end displacements from local to global coordinates and vice versa and develop the member stiffness relations in global coordinates. We formulate the stiffness relations for the entire structure by combining the member stiffness relations and, finally, develop a step-by-step procedure for the analysis of continuous beams and frames by the matrix displacement (stiffness) method.

The abstract of the chapter

13.1 Introduction

In the previous of the text, we have focused our attention on the classical methods of structural analysis. Although a study of classical methods is essential for developing an understanding of structural behavior and the principles of structural analysis, the analysis of large structures by using these hand-oriented methods can be quite time consuming. With the availability of inexpensive, yet powerful, microcomputers, the analysis of structures in most design offices is routinely performed today on computers using software based on matrix methods of structural analysis.

The matrix structural analysis is the method which uses the principle of classical structural mechanics to formulate the analytical procedure of a structure by matrix algebra and compute the algebraic equations representing the structure’s response by a computer. In matrix analysis, all the calculations are carried out with matrix algebra. The use of matrix notations will uniformly express the formulations of structural analysis in a simple and elegant form.

Although both the force (flexibility) and the displacement (stiffness) methods can be expressed in matrix form, the displacement method is more systematic and can be more easily implemented on computers. Thus, most of the commercially available computer programs for structural analysis are based on the displacement method. In this chapter, we will consider only the matrix displacement (stiffness) method of structural analysis.

13.1.1 The fundamental principle of matrix displacement method

In fact, the fundamental principle of matrix displacement method is identical to that of the

495

496 Chapter 12 Matrix Displacement Method

displacement method discussed in chapter 11. Matrix methods do not involve any new fundamental principles, but the fundamental relationships of equilibrium, compatibility, and member force-displacement relations are now expressed in the form of matrix equations, so that the numerical computations can be efficiently performed on a computer. Like the displacement method, a structure is also considered to be an assemblage of straight members connected at their ends to joints in the matrix displacement method. A member is defined as a part of the structure for which the member force-displacement relations to be used in the analysis are valid in matrix algebraic form. By assembling the force-displacement relations of the members under the equilibrium and compatibility conditions, the force-displacement relation, expressed in matrix form, of the entire structure can be obtained.

Therefore, the analysis of matrix displacement method comprises of the same stages as those of displacement method presented in chapter 11. They are: (1) the discretization of a structure; (2) the establishment of force-displacement relations of members; (3) the development of force-displacement relation of the structure. All of these relations must be expressed in matrix algebraic form.

13.1.2 Computing model

(1) Discretization of a structure In matrix displacement method of analysis, a member is defined as a part of the structure for which the

member force-displacement relations to be used in the analysis are valid in matrix algebraic form. In other words, given the displacements of the ends of a member, one should be able to determine the forces and moments at its ends by using the force-displacement relations expressed in matrix algebra. A joint is defined as a structural part of infinitesimal size to which the member ends are connected. The members and joints of structures are also referred to as elements and nodes, respectively.

Before proceeding with the analysis, the structure must be discretized, or a computing model of the structure must be prepared. The model is represented by a line diagram of the structure, on which all the joints, the joints’ unknown displacements and members are identified by numbers. Consider, for example, the rigid frame shown in Fig. 13.1(a). The computing model of the frame is shown in Fig. 13.1 (b), in which the element numbers are enclosed within circles to distinguish them from the joint numbers. As shown in this figure, the frame is considered to be composed of four members and six joints for the purpose of analysis. Note that, since the member 2 and member 4 have, generally, different rotations at their joint, the joint are numbered by two different numbers.

(2) Global and local coordinate systems In the matrix displacement method, the overall geometry and behavior of the structure are described

with reference to a rectangular global (or structural) coordinate system. The global coordinate system used in this chapter is a right-handed x yz coordinate system, with the plane structure lying in the x y plane, as shown in Fig. 13.1(b).

13.1 Introduction 497

Because it is usually convenient to derive the basic force-displacement relations in terms of the forces and displacements in the directions along and perpendicular to members, a local (or member) coordinate system is defined for each member of the structure. The origin of the local x y z coordinate system for a member may be arbitrarily located at one of the ends of the member, with the x axis directed along the centroidal axis of the member. The positive direction of the y axis is chosen so that the coordinate system is right-handed, with the local z axis pointing in the positive direction of the global axis. In Fig. 13.1 (b), the positive direction of the

zx axis for each member is indicated by drawing an arrow along each

member on the line diagram of the structure. For example, this figure indicates that the origin of the local coordinate system for member 2 is located at its end connected to joint 2, with the 1x axis directed from joint 2 to joint 4. The joint to which the member end with the origin of the local coordinate system is connected is referred to as the beginning joint for the member, whereas the joint adjacent to the opposite end of the member is termed the end joint. For example, in Fig. 13.1 (b), member 1 begins at joint 1 and ends at joint 2, whereas member 2 begins at joint 2 and ends at joint 4, and so on. Once the local x axis is defined for a member, the corresponding y axis can be established by applying the right-hand rule. That is, for each member, if we curl the fingers of our right hand from the direction of the x axis toward the direction of the corresponding y axis, then our extended thumb points through the plane of the page, which is the positive direction of the global axis. z

(3) Unknown displacements or nodal displacement vector The unknown displacements of a structure are the independent joint displacements (translations and

rotations) that are necessary to specify the deformed shape of the structure when subjected to an arbitrary loading. Unlike in the case of the classical methods of analysis considered previously, it is usually not necessary to neglect member axial deformations when analyzing frames by the matrix displacement method. Consider again the plane frame of Fig. 13.1 (a). From Fig.13.1 (b), we can see that joint 1, which is attached to the fixed support, can neither translate nor rotate; therefore, it does not have any unknown

(a)

1 1 2 32 ( , , )Δ Δ Δ

3(b)

① ②

③ ④

x

y

4 5 64 ( , , )Δ Δ Δ

4 5 75 ( , , )Δ Δ Δ

6

Fig.13.1 Discretization of a structure(a) actual structure; (b) analytical model and unknown displacements


displacement. Since joint or node 2 of the frame is not attached to a support, three displacements—the translations and 1Δ 2Δ in the x and directions, respectively, and the rotation about the axis—are needed to completely specify its deformed position. Thus node 2 has three unknown displacements. Similarly, nodes 4 and 5, which are connected to a free hinged joint, have three unknown displacements each. However, since node 4 and node 5 have the same translations but different rotations, their translations are numbered in the same symbols,

y 3Δ z

4Δ and 5Δ . Finally, joint 3 and 6, which are attached to the fixed supports, can neither translate nor rotate; therefore, they do not have any displacement. Thus, the entire frame has a total of 7 unknown displacements. As shown in Fig. 13.1(b), the basic unknown displacements are defined relative to the global coordinate system, with joint translations considered as positive when in the positive directions of the x and axes and joint rotations considered as positive when clockwise. The unknown displacements of the frame can be collectively written in matrix form as

y

{ }1

2

7

Δ⎧ ⎫⎪ ⎪ΔΔ = ⎨ ⎬⎪ ⎪Δ⎩ ⎭

(13-1)

In which, { }Δ is termed the nodal displacement vector of the structure. As indicated in Fig.13.1 (b), the unknown displacements are numbered by starting at the lowest node

number and proceeding sequentially to the highest node number. In the case of more than one unknown displacements at a node, the translation in the x direction is numbered first, followed by the translation in the direction, and then the rotation. y

1 12 ( ) 2 33 ( , )Δ Δ Δ55( )

① ② ③

(a)

(b)2 44 ( , )

ΔΔ Δ

Fig.13.2 Discretization of a continuous beam(a) actual continuous beam;(b) analytical model and unknown displacements

In multispan or continuous beams subjected to lateral loads, the axial deformations of members are zero. Therefore, it is not necessary to consider the joint displacements in the direction of the beam's centroidal axis in the analysis. Thus a joint of a plane multispan or continuous beam can have up to two

13.1 Introduction 499

unknown displacements, namely, a translation perpendicular to the beam's centroidal axis and a rotation. For example, the two-span beam of Fig. 13.2(a) has 5 unknown displacements, the translation of node 3 and rotations of nodes 2，3 and 4, as shown in Fig. 12.2(b).

Since the joints of trusses are assumed to be frictionless hinges, they are not subjected to moments; therefore, their rotations are zero. Thus, when analyzing plane trusses, only 2 unknown displacements, namely, translations in the global x and directions need to be considered for each joint or node. For example, the truss of Fig. 13.3 (a) has 12 unknown displacements, as shown in Fig.13.3 (b).

y

(a) (b)

1 8①

②③ ④

⑤

⑥

⑦

⑧

⑨

⑩

⑫

⑬⑪

1 22 ( , )

(4) Sign convention of member end displacements and forces Fig.13.4 shows an arbitrary prismatic flexural member (or element) of a structure. When the

structure is subjected to external loads, the member deforms and internal forces are induced at its ends. As indicated in this figure, three displacements—translations in the

ⓔ

x and y directions and rotation about the z axis—are needed to completely specify the deformed position of each end of the member. Thus the member has a total of six end displacements or unknown displacements. As shown in the figure, the member end displacements are denoted by 1u , 1v , 1θ and 2u , 2v , 2θ [ Fig.13.4 (a)], and the

ⓔ 211u

1v

1θ

ⓔ 211X

1Y

1M

2u

2v

2θx

y

2X

2Y

2Mx

y

(a)

(b)

Fig.13.4 Member end displacements and forces(a) member end displacements; (b) member end forces

Δ Δ

3 43( , )Δ Δ

5 64( , )Δ Δ

7 85( , )Δ Δ

9 106( , )Δ Δ

11 127( , )Δ Δ

Fig.13.3 Discretization of a truss(a) actual truss; (b) analytical model and unknown displacements


corresponding member end forces are denoted by 1X , 1Y , 1M and 2X , 2Y , 2M . Note that these end displacements and forces are defined relative to the local coordinate system of the member, with translations and forces considered as positive when in the positive directions of the local x and y axes, and rotations and moments considered as positive when clockwise. As indicated in Fig.13.4, the member end displacements and forces are numbered by beginning at the member end 1, where the origin of the local coordinate system is located, with the translation and force in the x direction numbered first, followed by the translation and force in the y direction, and then the rotation and moment. The displacements and forces at the opposite end 2 of the member are then numbered in the same sequential order.

In matrix displacement method, the six member end displacements and six member end forces are arranged, in the order starting from the beginning end and ending in the ending end, in matrix form named member end displacement vector and member end force vector, respectively. They can be written as

{ }

11

1 2

31

2 4

2 5

2 6

uv

uv

θ

θ

⎧ ⎫Δ⎧ ⎫⎪ ⎪⎪ ⎪ Δ⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪ Δ⎪ ⎪⎪ ⎪ ⎪ ⎪Δ = =⎨ ⎬ ⎨ ⎬

⎪ ⎪ ⎪ ⎪Δ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪Δ⎪ ⎪ ⎪ ⎪

Δ⎩ ⎭ ⎪ ⎪⎩ ⎭

ⓔⓔ

ⓔ

{ }

11

21

31

2 4

2 5

2 6

FXFYFM

FX FY FM F

⎧ ⎫⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪ ⎪ ⎪= =⎨ ⎬ ⎨ ⎬

⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

ⓔⓔ

ⓔ (13-2)

The bars over the displacement and force symbols identify that the associated physical quantities are defined in local coordinates.

13.2 Member Stiffness Matrix in Local Coordinate System

In the matrix displacement method of analysis, the unknown displacements of a structure are determined by solving a system of simultaneous equations, which are expressed in the matrix form

{ } [ ]{ }P K= Δ (13-3)

In which denotes the unknown displacement vector, as discussed previously; represents the effects of external loads at the joints of the structure; and is named the structure stiffness matrix. As discussed in the latter sections, the stiffness matrix for the entire structure, { , is obtained by assembling the stiffness matrices for the individual members of the structure. The stiffness matrix for a member can be expressed as

{ }Δ { }P{ }K

}K

13.2 Member Stiffness Matrix in Local Coordinate System 501

{ } [ ] { }F k= Δⓔⓔ ⓔ

In which represents the stiffness matrix for a member in global coordinates. In local coordinates, the stiffness matrix for a member can be expressed as

[ ]k ⓔ

{ } { }F k⎡ ⎤= Δ⎣ ⎦ⓔⓔ ⓔ

(13-4)

The bars over symbols in Eq. (13-4) denote that the corresponding quantities are in local coordinates. The member stiffness matrix is used to express the forces at the ends of the member as functions of the

displacements of the member's ends. Note that the terms forces and displacements are used here in the general sense to include moments and rotations, respectively. In this section, we derive stiffness matrices for the members of plane rigid frames, continuous beams, and plane trusses in the local coordinate systems of the members.

13.2.1 Stiffness matrix for a general flexural member

To establish the stiffness relationships for a general flexural member of a structure, let us focus our attention on an arbitrary prismatic member with the length l, flexural rigidity EI and axial rigidity EA, respectively, as shown in Fig.13.5. Our objective here is to determine the relationships between the member end forces and end displacements. Such relationships can be conveniently established by subjecting the member, separately, to each of the six end displacements and by expressing the total member end forces as the algebraic sums of the end forces required to cause the individual end displacements. If we neglect the coupling influence between axial deformation and bending deformation, the stiffness relationships pertinent to axial deformation and those pertinent to bending deformation can be derived separately.

First, let us determine the relationships between the member end axial displacements and member end axial forces. From Fig.13.5, if neglecting the coupling influence between axial deformation and bending deformation and recalling from strength of materials, we can see that

ⓔ

, , ,E I A ly

x

M

1X2X

2Y

2M

1Y

1

1u2u

2θ

1θ

1 2

2ν

1ν

Fig.13.5 A general flexural prismatic member


1 1 2

2 1

( )

( )

EAX u ulEA

2X u ul

⎫= − ⎪⎪⎬⎪= − −⎪⎭

(13-5)

Note that the member end axial force 2X is obtained by applying the equation of equilibrium in the x direction.

Next, let us determine the relationships between the bending deformations and forces. In fact, the equations between member end displacements and member end shears and bending moments have already determined in chapter 11 (displacement method). By using the shape constants of single-span indeterminate prismatic members shown in table 11-1 and principle of superposition, we can conveniently obtain

1 1 1 2 23 2 3 2

1 1 1 2 22 2

2 1 1 23 2 3 2

2 1 1 2 22 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

EI EI EI EIYl l l lEI EI EI EIMl l l l

EI EI EI EIYl l l lEI EI EI EIMl l l l

ν θ ν θ

ν θ ν θ

2ν θ ν

ν θ ν θ

⎫= + − +

θ

⎪⎪⎪= + − + ⎪⎬⎪= − − + −⎪⎪⎪= + − +⎭

(13-6)

Note that the signs of bending moments 1M and 2M have been adopted the sign convention of this chapter, that is, they are positive when clockwise.

Combining equation (13-5) and (13-6), we can write them in matrix form as

1

1 3 2 3 2

1 2 2

2

3 2 3 22

2 22

0 0 0 0

12 6 12 60 0

6 4 6 20 0

0 0 0 0

12 6 12 60 0

6 2 6 40 0

EA EAX ul l

EI EI EI EIYl l l lEI EI EI EI

Ml l l l

EA EAX l l

EI EI EI EIY l l l l

EI EI EI EIM l l l l

⎡ ⎤⎧ ⎫ −⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪ −⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥−⎪ ⎪⎢ ⎥⎪ ⎪ =⎨ ⎬ ⎢ ⎥

⎪ ⎪ ⎢ ⎥−⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎢ ⎥

− − −⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎢ ⎥

−⎪ ⎪ ⎢ ⎥⎩ ⎭ ⎣ ⎦

ⓔ

1

1

1

2

2

2

v

u

v

θ

θ

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

ⓔ

(13-7)

The equation can be symbolically written as


{ } { }F k⎡ ⎤= Δ⎣ ⎦ⓔⓔ ⓔ

The above equation is the stiffness equations (or relationships) for a general flexural prismatic member. In which, [ ]k ⓔ is referred to as the element (or member) stiffness matrix, which is a square matrix. It is written again as follows.

6 6×

3 2 3 211 12 13 14 15 16

21 22 23 24 25 26

2 231 32 33 34 35 36

41 42 43 44 45 46

51 52 53 54 55 56

61 62 63 64 65 66

0 0 0 0

12 6 12 60 0

6 4 6 20 0

0 0

0

EA EAl l

EI EI EI EIk k k k k k l l l lk k k k k k EI EI EI EIk k k k k k l l l lk EAk k k k k k

lk k k k k kk k k k k k

−

−⎡ ⎤⎢ ⎥⎢ ⎥ −⎢ ⎥⎢ ⎥⎡ ⎤ = =⎣ ⎦ ⎢ ⎥ −⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

ⓔ

3 2 3 2

2 2

0 0

12 6 12 60

6 2 6 40 0

EAl

EI EI EI EIl l l lEI EI EI EIl l l l

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥

− − −⎢ ⎥⎢ ⎥⎢ ⎥

−⎢ ⎥⎣ ⎦

(13-8)

13.2.2 Property of element stiffness matrix

(1) Meanings of stiffness coefficients The element ijk in matrix [ ]k ⓔ represents the force at the location and in the direction of iF

required, along with other end forces, to cause a unit value of the displacement jΔ while all other end displacements are zero. These forces per unit displacement in matrix [ ]k ⓔ are referred to as stiffness coefficients. Note that a double-subscript notation is used for stiffness coefficients, with the first subscript identifying the force and the second subscript identifying the displacement. For instance, 63k represents the moment at the location and in the direction of 6F ( 2M ) required, along with other end forces, to cause a unit value of the displacement 3 1Δ = ( 1 1θ = ) while all other end displacements are zero.

The ith column in matrix [ ]k ⓔ denotes the end forces required to cause a unit value of the displacement iΔ while all other displacements are zero. For example, the second column of [ ]k ⓔ denotes the six end forces required to cause the displacement 2 1Δ = ( 1 1v = ).

(2) Symmetry of element stiffness matrix The element stiffness matrix [ ]k ⓔ is symmetric with respect to the main diagonal, that is

ij jik k i j= ≠ (13-9)

In fact, this property is the reflection of the law of reciprocal reactions (see subsection 9.8.3).


(3) Singularity of element stiffness matrix The element stiffness matrix [ ]k ⓔ is singular because the determinant of the matrix is equal to zero.

That is,

[ ] 0k =ⓔ (13-10)

Since the matrix [ ]k ⓔ is singular, there exists no reciprocal matrix of [ ]k ⓔ . In other words, by equation (13-4), the end forces of the member can be uniquely determined by the given end displacements [ ]Δ ⓔ . But under the action of a given end force [ ]F ⓔ , the end displacements [ ]Δ ⓔ cannot be uniquely determined.

13.2.3 Special elements

Equation (13-7) is the stiffness equation for a general flexural member of a structure, in which the six end displacements of the member are unknowns. However, in some cases, one or some of the end displacement of a member is designated or small enough to be neglected. For instance, the axial deformation effect may be neglected in the analysis of rigid frames and continuous beams. For this kind of special member, the stiffness relationships may be obtained by modifying the stiffness relations of a general element, i.e., equation (13-7).

(1) Stiffness equation for an element without considering axial deformation In the analysis of a rigid frame, the axial deformations of the beams and columns are often be

neglected because they are to small comparing with their bending deformations. So the unknown end displacements for one of the beams or columns remain only four, i.e., 1v , 1θ and 2v , 2θ ; the axial displacements vanish as shown in Fig.13.6 (a). That is,

1 2 0u u= = (a)

21

l

EIⓔ

ⓔ

21EI

lM

1

1Y

1M

10u =

1θ

10u =

1θ

2θ

2M

2 0u =

1 0v = 20v =

1ν

2ν 2

12 0u = M2

2Y

2θ ν

ν−

(a) (b)

Fig.13.6 Special elements(a) beam element with no axial deformation; (b) continuous beam element


Substituting equation (a) into equation (13-7), we automatically obtain

1 13 2 3 2

1 2 2

3 2 3 2 22

2 22

12 6 12 6

6 4 6 2

12 6 12 12

6 2 6 4

EI EI EI EIY vl l l lEI EI EI EIMl l l lEI EI EI EI

vY l l l lEI EI EI EI

M l l l l

θ

θ

⎡ ⎤⎧ ⎫ ⎧ ⎫−⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪−⎢ ⎥⎪ ⎪ ⎪ ⎪= ⎢ ⎥⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪− − −⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥−⎪ ⎪ ⎪ ⎪

⎩ ⎭⎩ ⎭ ⎢ ⎥⎣ ⎦

ⓔ ⓔ

(13-11)

Equation (13-11) is the stiffness equation of an element without considering axial deformation, whose stiffness matrix can be written as

3 2 3 2

2 2

3 2 3 2

2 2

12 6 12 6

6 4 6 2

12 6 12 12

6 2 6 4

EI EI EI EIl l l lEI EI EI EIl l l lk EI EI EI EIl l l lEI EI EI EIl l l l

⎡ ⎤−⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥

⎡ ⎤ = ⎢ ⎥⎣ ⎦⎢ ⎥− − −⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

ⓔ (13-12)

As a matter of fact, the special stiffness matrix can be obtained by deleting the 1st rows and columns and 4th rows and columns from the stiffness matrix (13-8), respectively.

(2) Stiffness equation for an element of continuous beams Since the axial deformations of the members of continuous beams subjected to lateral loads are

considered to be equal to zero and the beams are supported on the rigidity supports, we do not need to consider the end displacements in the axial and lateral directions of the members. Thusly, only two unknown end rotations 1θ and 2θ of a member as shown in Fig.13.6 (b) are needed to take into consideration; the other four end displacements can be recognized to be zero. That is,

1 1 2 2 0u v u v= = = = (b)

Substituting equation (b) into equation (13-7), we will obtain stiffness equation for an element of continuous beams as follows.


1 1

2 2

4 2

2 4

EI EIMl lEI EI

M l l

θ

θ

⎡ ⎤ ⎧ ⎫⎧ ⎫ ⎢ ⎥⎪ ⎪ ⎪ ⎪= ⎢ ⎥⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪

⎩ ⎭ ⎩ ⎭⎢ ⎥⎣ ⎦

ⓔⓔ

(13-13)

Here, the stiffness matrix for an element of continuous beams will be

4 2

2 4

EI EIl lkEI EIl l

⎡ ⎤⎢ ⎥

⎡ ⎤ = ⎢ ⎥⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦

ⓔ (13-14)

It can be seen from equation (13-14) that since we have introduced the support conditions in formulating the stiffness matrix of an element of continuous beams, the stiffness matrix is reversible. Therefore, the end forces [ ]F ⓔ of the member can be uniquely determined by the given end displacements [ ]Δ ⓔ ; on the contrary, the end displacements [ ]Δ ⓔ can be uniquely determined by the given end forces [ ]F ⓔ as well.

(3) Stiffness equation for an element of truss members A member of a truss is subjected to only axial forces, which can be determined from its end

displacements in the direction of the centroidal axis of the member. Thus only two unknown end displacements 1u and 2u of a member are needed to take into consideration; the other four end displacements can be recognized to be zero as shown in Fig.13.7. However, in order to facilitate the global

analysis, the stiffness equation for a truss member can be written as

21

1 1

1 1

22

22

0 0

0 0 0 0

0 0

0 0 0 0

EA EAX ul l

Y v

EA EA uXl l

vY

⎧ ⎫ ⎡ ⎤ ⎧ ⎫−⎪ ⎪ ⎢ ⎥ ⎪ ⎪⎪ ⎪ ⎢ ⎥ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢=⎨ ⎬ ⎨ ⎬⎢⎪ ⎪ ⎪ ⎪−⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎩ ⎭⎣ ⎦⎩ ⎭

ⓔ ⓔ

⎥⎥ (13-15)

l

EAⓔ

2Y1Y

1u

1X

2u2

X

1 0v = 20v =

Fig.13.7 Truss element

13.3 Member Stiffness Matrix in Global Coordinate System 507

Thus, the stiffness matrix for an element of a truss will be

0 0

0 0 0 0

0 0

0 0 0 0

EA EAl l

kEA EAl l

⎡ ⎤−⎢ ⎥⎢ ⎥⎢⎡ ⎤ =⎣ ⎦ ⎢−⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

ⓔ ⎥⎥ (13-16)

The forgoing stiffness matrix for truss members can be also obtained by deleting rows 3 and 6 and columns 3 and 6 and by letting the elements in rows 2 and 5 and columns 2 and 5 to be zero, from the general stiffness matrix (13-8), respectively.

13.3 Member Stiffness Matrix in Global Coordinate System

In last section, we formulate the stiffness matrix of an element in its local coordinate system for the purpose of simplicity. However, for a bit complex structure, the members of the structure are oriented in different directions. Thusly, it becomes necessary to transform the stiffness relations for each member from the member’s local coordinate system (denoted by x y ) to a common global coordinate system (denoted by x y ). The member stiffness relations in global coordinates thus obtained are then assembled to establish the stiffness relations for the entire structure.

In order to derive the stiffness matrix in global system of an element, we use the method of coordinate transformation. First, we discuss the transformation of member end forces from local to global coordinates or vice versa so as to obtain the member transformation matrix; then we implement the transformation of member stiffness matrix from local to global coordinate system, and vice versa.

[ ]k ⓔ

13.3.1 Element transformation matrix of coordinates

Consider an arbitrary general member as shown in Fig. l3.8 (a). The orientation of the member with respect to the global

ⓔ

x y coordinate system is defined by an angle α measured clockwise from the positive direction of the global x axis to the positive direction of the local x axis, as shown in the figure. The stiffness relations derived in the preceding section are valid only for member end forces [ ]F ⓔ and end displacements [ ]Δ ⓔ described with reference to the local x y coordinate system of the member, as shown in the figure.

Now, suppose that the member end forces and end displacements are specified relative to the global x y coordinate system [Fig.13.8 (b)] and we wish to determine the relations between the end forces in the two coordinate systems. As shown in Fig.13.8 (b), the member end forces in global coordinates are


denoted by 1X , , 1Y 1M and 2X , , 2Y 2M . These global member end forces are numbered by beginning at the member end 1, where the origin of the local coordinate system is located, with the force in the x direction numbered first, followed by the force in the y direction and then the moment. The forces at the opposite end 2 of the member are then numbered in the same sequential order.

By considering the projection relations of forces shown in Fig.13.8 (b) and (a), we will find the relation between the end forces in the global coordinate system and those in the local coordinate system as follows

1 1 1

1 1 1

1 1

2 2 2

2 2 2

2 2

cos sinsin cos

cos sinsin cos

X X YY X YM MX X YY X YM M

α α

α α

α α

α α

⎫= +⎪

= − + ⎪⎪= ⎪⎬

= + ⎪⎪= − + ⎪⎪= ⎭

(13-17)

Equation (13-17) can be written in matrix form as

1 1

1 1

1 1

22

22

22

cos sin 0 0 0 0sin cos 0 0 0 00 0 1 0 0 0

0 0 0 cos sin 00 0 0 sin cos 00 0 0 0 0 1

X XY YM M

XXYYMM

α αα α

α αα α

⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪⎢ ⎥−⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪= ⎢ ⎥⎨ ⎬ ⎨⎢ ⎥⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪−⎢ ⎥⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪⎣ ⎦ ⎩⎩ ⎭

ⓔ

⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭

ⓔ

(13-18)

ⓔ2

1α

α

O

y

xⓔ

2

1

1Y

αO

y

x

2Y

2X

2M

1M

1X

y x

1X

2X

2Y2

M1Y

1M

y

(a) (b)

α

x

Fig.13.8 Transformation relations of member end forces(a) end forces in local coordinate; (b) end forces in global coordinate


or symbolically as

{ } [ ]{ }F T F=ⓔ ⓔ

(13-19)

in which

[ ]

cos sin 0 0 0 0sin cos 0 0 0 00 0 1 0 0 0

0 0 0 cos sin 0 0 0 0 sin cos 0 0 0 0 0 0 1

T

α αα α

α αα α

⎡⎢−⎢⎢

= ⎢⎢

−

⎣

⎤⎥⎥⎥⎥⎥

⎢ ⎥⎢ ⎥⎢ ⎥⎦

(13-20)

is referred to as the transformation matrix of coordinates. Equation (13-19) is the transformation of end forces from global to local coordinate systems.

Like end forces, the member end displacements are vectors as well, which are defined in the same directions as the corresponding forces. Therefore, the transformation matrix developed for the case of end forces [Eq. (13.20)] can also be used to transform member end displacements from global to local coordinate systems:

[ ]T

{ } [ ]{ }TΔ = Δⓔ ⓔ

(13-21)

Next, we determine the transformation of member end forces and end displacements from local to global coordinates. From the projection relations of forces shown in Fig.13.8 (a) and (b), we will observe the relation between the end forces in the local coordinate system and those in the global coordinate system as follows

1

1

1

2

2

2

cos sin 0 0 0 0sin cos 0 0 0 0

0 0 1 0 0 0 0 0 0 0 0 0 0 0

XYMXYM

α αα α

−⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪ =⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

ⓔ1

1

1

2

2

2

cos sin 0sin cos 0

0 0 0 1

XYMXYM

α αα α

⎧ ⎫⎡ ⎤ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎨ ⎬−⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎣ ⎦ ⎩ ⎭

ⓔ

(13-22)

A comparison between equations (13-18) and (13-22) indicates that the transformation matrix in equation (13-22), which transforms the end forces from local to global coordinates, is the transpose of the


transformation matrix [ in equation (13-18), which transforms the end forces from global to local coordinates. Thus equation (13-22) can be written as

]T

{ } [ ] { }TF T F=ⓔⓔ

(13-23)

The matrix can also determine the transformation of member end displacements from local to global coordinates; that is,

[ ]TT

{ } [ ] { }TTΔ = Δⓔⓔ

(13-24)

Recall from matrix algebra that the matrix is of orthogonality. That is, [ ]T

[ ][ ] [ ] [ ] [ ]T TT T T T I= = (13-25)

In which, represents an identity (or unit) matrix with the same order as that of . [ ]I [ ]T13.3.2 Element stiffness matrix in global coordinates

In global coordinates, the relations between the member end forces and end displacements can be also expressed as

{ } [ ] { }F k= Δⓔⓔ ⓔ (13-26)

In which, is referred to as the member (or element) stiffness matrix in global coordinates. [ ]k ⓔ

By using the member stiffness relations in local coordinates and the transformation relations, we can now develop the stiffness relations for members in global coordinates.

The stiffness relation of an element in local coordinates is ⓔ

{ } { }F k⎡ ⎤= Δ⎣ ⎦ⓔⓔ ⓔ

(a)

Substituting equations (13-19) and (13-21) into equation (a), we will obtain

[ ]{ } [ ]{ }T F k T⎡ ⎤= Δ⎣ ⎦ⓔⓔ ⓔ

(b)

Premultiplying to the both sides of equation (b) and in the meanwhile introducing equation (13-25), we will write

[ ]TT

{ } [ ] [ ]{ }TF T k T⎡ ⎤= Δ⎣ ⎦ⓔⓔ ⓔ

(c)

A comparison between equations (c) and (13-26), we find


[ ] [ ] [ ]Tk T k⎡ ⎤= ⎣ ⎦ⓔⓔ T (13-27)

The forgoing equation is just the transformation relations of element stiffness matrices in the two coordinates. By the equation, if the stiffness matrix [ ]k ⓔ in local coordinates and the transformation matrix between the local and global coordinates of a member are determined, the stiffness matrix

in global coordinates will be conveniently obtained. [ ]T

[ ]k ⓔ

It can be seen from above that the element stiffness matrix referred to global coordinates not only has the same order as the stiffness matrix

[ ]k ⓔ

[ ]k ⓔ referred to local coordinates, but also has the similar properties with [ ]k ⓔ . The properties are:

(1) The element in the matrix [ ] is the force corresponding to coordinate due to a unit displacement of coordinate , which is also termed stiffness influence coefficient.

ijk k ⓔ ij

(2) [ ] is a symmetric matrix. k ⓔ

(3) The stiffness matrix [ ] for a general member is singular. k ⓔ

Example 13-1

Determine the stiffness matrix for the members of the rigid frame shown in Fig.13.9. Assume both the length and the size of the cross section for each member are the same. They are: length , axial rigidity and flexural rigidity

[ ]k ⓔ

5ml =71.5 10 kNEA = × 6 21.25 10 kN mEI = × ⋅ .

Solution

(1) Stiffness matrix in local coordinates [ ]k ⓔ The positive direction of the axis of abscissas x for each member is indicated by an arrow along the

axis. Since member 1 and member 2 have the same size, their stiffness matrices in local coordinates are same too, i.e., [ ] [ ]k k=① ② . By substituting the length 5ml = , axial rigidity and flexural rigidity into equation (13-8), we obtain

71.5 10 kNEA = ×61.25 10 kN mEI = × ⋅ 2

4

300 0 0 300 0 00 12 30 0 12 300 30 100 0 30 50

10300 0 0 300 0 00 12 30 0 12 300 30 50 0 30 100

k k

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−

⎡ ⎤ ⎡ ⎤= = ×⎢ ⎥⎣ ⎦ ⎣ ⎦ −⎢ ⎥⎢ ⎥− − −⎢ ⎥

−⎢ ⎥⎣ ⎦

① ②

(2) Stiffness matrix in global coordinates [ ] k ⓔ

Element : since ① 0α = , cos 1α = , sin 0α = . Therefore, [ ] [ ]T I= , i.e., no coordinate


transformation is needed. Thusly,

[ ]k k⎡ ⎤= ⎣ ⎦①①

Element : since , ② 090α = cos 0α = , sin 1α = . Thusly, the transformation matrix of element will be ②

[ ]

0 1 0 0 0 01 0 0 0 0 0

0 0 1 0 0 00 0 0 0 1 00 0 0 1 0 00 0 0 0 0 1

T

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥

= ⎢⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

⎥

By applying equation (13-27), we can determine the stiffness matrix in global coordinates for member as ②

[ ] [ ] [ ] 4

12 0 30 12 0 300 300 0 0 300 030 0 100 30 0 50

1012 0 30 12 0 300 300 0 0 300 030 0 50 30 0 100

Tk T k T

− − −⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−

⎡ ⎤= = ⎢ ⎥⎣ ⎦ −⎢ ⎥⎢ ⎥−⎢ ⎥−⎢ ⎥⎣ ⎦

②② ×

, , ,E I A l

, , ,E I A l

y

x①

②

5ml =

5m

l=

Fig.13.9 The figure of example 13-1

13.4 Global Stiffness Matrices of Continuous Beams 513

13.4 Global Stiffness Matrices of Continuous Beams

In previous sections, we have discussed the member analysis, involving with the establishment of the stiffness relationships and development of stiffness matrices for the members of a structure. Beginning from this section, we will discuss the global (or structure) analysis, comprising the establishment of the stiffness relationships and development of stiffness matrix for an entire structure. The section will discuss the development of global stiffness matrix for continuous beams.

The stiffness relations of a structure, which are used to express the joint forces as functions of the joint displacements of the structure by the structure (or global) stiffness matrix, can be established by displacement method. Generally, there are two kinds of methods to develop the stiffness relations of a structure; one is the classical displacement method discussed in chapter 11, another is the method so called element assembly method (or stiffness assembly method or direct stiffness method). One of the advantages of element assembly method is that it can be easily implemented on computers.

In order to make a comparison between the two methods, let us simply review the procedure of the development of stiffness relations of traditional displacement method through the following example.

For the continuous beam shown in Fig.13.10 (a), its primary system used in displacement method is shown in Fig.13.10 (b). As discussed in chapter 11, the primary unknowns of the beam are the joint rotations, , and , which consist of the joint displacement vector 1Δ 2Δ 3Δ { }Δ of the beam, that is,

(a)

(b)

②①

{ } [ ]1 2 3 TΔ = Δ Δ Δ

The joint forces, corresponding to the joint displacements 1Δ , 2Δ and 3Δ , are the restraint moments, , and , which compose of the joint force vector of the beam. It can be written as 1F 2F 3F { }F

22

2

EIil

=

2

11

1

EIil

=

Δ

① ②

1Δ

3Δ1F3F2F

Fig.13.10 Global analysis of a continuous beam(a) original structure; (b) primary system in displacement method


{ } [ ]1 2 3 TF F F F=

In traditional displacement method, the restraint moments , and can be obtained by superposing each of the restraint forces due to the joint displacements

1F 2F 3F1Δ , 2Δ and individually as

shown in Fig.13.11 (a) through (c). The results of superposition can be written as following matrix form. 3Δ

1 1 1

2 1 1 2 2

3 2 2

4 2 02 4 4 20 2 4

F i iF i i i iF i i

1

2

3

Δ⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢= +⎨ ⎬ ⎨ ⎬⎢⎪ ⎪ ⎪ ⎪⎢ ⎥

⎥ Δ⎥Δ⎩ ⎭ ⎣ ⎦ ⎩ ⎭

(13-28)

Or symbolically as

{ } [ ]{ }F K= Δ (13-29)

Equation (13-28) or equation (13-29) is referred functions) of the continuous beam, which represents tdispl

2

to as the structure (or global) stiffness relations (or he transformation relationship between the joint

acements and forces. The relationship is transformed by the matrix [ ]K , termed as structure (or global) stiffness matrix of the continuous beam. It can be expressed as

Δ

1 14iΔ 01 12iΔ

1Δ

3Δ

1 22i Δ ( )1 2 24 4i i+ Δ 2 22i Δ

0 2 32i Δ 2 34i Δ

(a)

(b)

(c)

① ②

①

②

① ②

1 2

3

Fig.13.11 Joint moments due to joint rotations(a) joint moments due to ; (b) joint moments due to ;(c) joint moments due to

Δ ΔΔ


[ ]1 1

1 1 2 2

4 2 02 4 4 2i i

K i i i i

2 20 2 4i i

⎡ ⎤⎢ ⎥= +⎢ ⎥⎢ ⎥⎣ ⎦

(13-30)

The following subsection will introduce element stiffness assembstiffness matrix of a structure.

13.4.

discussion that when determining the global stiffness equation (13-29) of a contribution to the joint force vector of each joint

displ

of the structure was considered indiv

e st

s for the joints of the struc

ly method so as to form the global

1 Concept of element assembly method

It can be seen from above structure by traditional displacement method, the { }F

acement of the structure was considered individually as shown in Fig.13.11; then the individual contributions were algebraically added so as to form the joint force vector.

If the global stiffness equation (13-29) of a structure is formed by using element assembly method, the contribution to the joint force vector { }F of each element (or member)

idually; then the individual contributions will be assembled in the sense of matrix algebra so as to form the joint force vector, which is th iffness relations (or equations) between joint displacements and forces of the structure. One of the characteristics of the method is that the global stiffness equations of a structure can be directly assembled by the global stiffness equations of its members.

In fact, once the member stiffness equations in global coordinates have been determined, the stiffness equations for the entire structure can be established by writing equilibrium equation

ture and by applying the compatibility conditions that the displacements of the member ends rigidly connected to joints must be the same as the corresponding joint displacements.

To illustrate this procedure, again consider the two-span continuous beam shown in Fig.13.12 (a), which indicates that the beam has three joint displacements 1Δ , 2Δ , 3Δ and three joint forces

here ar wo e f ure

On is

tions for elements and elem

(a)

1F , 2F ,

3F . The continuous beam is divided into two elements, element ① and element ② as shown in Fig.13.12 (b), whose numbers are enclosed within circles. T e t kinds of numbers in th ig :

e is global number such as joint numbers 1, 2 and 3 shown in Fig.13.12 (a); another local number which is enclosed within a bracket such as (1) and (2) shown in Fig.13.12 (b).

(1) Element stiffness equations For the elements shown in Fig.13.12 (b), the element stiffness equa ①

ent ② can be written as follows.Element① :

(1) (1)1 1

(2) (2)1 1

4 22 4

F i iF i i

Δ⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎢ ⎥ Δ⎣ ⎦⎩ ⎭ ⎩ ⎭

① ①


Element

②

(b)

Here, indicate the member end forces of element

② :

(1) (1)2 2

(2) (2)2 2

4 22 4

F i iF i i

Δ⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎢ ⎥ Δ⎣ ⎦⎩ ⎭ ⎩ ⎭

②

(1)Fⓘ , (2)Fⓘ ⓘ ; (1)Δⓘ , (2)Δⓘ represent the member end

displacements (or rotations) of element Note that for the elements of a continuous beam, their local and global coordinates are coincident, so

ment stiffn ates main unch

(2) Equilibrium equations By applying the equation of equilibrium,

ⓘ

the ele ess equations in their local and global coordin re angeable.

0M =∑ , to the free body of each of the joints as shown in Fig.13.12 (c), we obtain the equilibrium equations for the joints

F F

⎫1

2 (2) (1)F F F

=

3 (2)

F F⎪⎪= + ⎬⎪

= ⎪⎭

①(1)

②

(c)

If we express the equilibrium relation by the joint force vector of the structure, it will be the

2

① ②

Δ1Δ

3Δ1F 2F 3F21 3

1F

1

2F2

(a)

( )F①

1

(1)Δ①

(2)Δ① F①(2)

F①(1) F①

(2)

33F

( )F 2②

( )F 1②

( )F 2②( )Δ 2

②

( )Δ 1②

( )F②

1

(b)

(c)

Fig.13.12 Global analysis of a two-span continuous beam(a) original structure; (b) elements and ; (c) equilibrium conditions of joints① ②

① ②


superposition of the member end force vectors, which is extended to have the same elements as the joint force vector. That is,

0F FF F F

1 (1) ⎧ ⎫⎧ ⎫ ⎧ ⎫⎪

2 (2) (1)

3 (2)0F F

⎪⎪ ⎪ ⎪ ⎪= +⎨ ⎬ ⎨ ⎬ ⎨ ⎬

②①

(d) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭ ⎩ ⎭

Or symbolically as

{ } { } { }F F F= +① ② (13-31)

(3) Compatibility equations As mentioned previously, when developing the stiffness equations for the entire structure not only the

ns for the joints of the structure have to be considered but also the compatibility conditions that the displacements of the member ends rigidly connected to joints must be the same as the corresponding joint displacements must be taken into account. By coobserve lement is hinged to joint 1, the displacement of end (1) of this elem

equilibrium equatio

mparing Fig.13.12 (a) and (b), we can that since the left end of e

①

① ent, (1)Δ , must be the same as the displacement of joint 1, 1Δ . Similarly, the displacement of end (2)

of element ① is rigidly connected to joint 2, the displacement of end (2), (2)Δ① , must be the same as the

displacement of joint 2, 2Δ . Thus, the compatibility equations for element 1 are

1) 1Δ( = Δ① (2) 2Δ = Δ① (e)

In a s lar manner, the compatibility equations for element 2 are

(1) 2

imi

Δ = Δ② (2) 3Δ = Δ② (f)

By substituting the compatibility equations for element [Eq. (e)] into the element’s force-displacement relation [Eq. (a)] and by extending the element stiffness m ess the element force vector

①

force and displacement vectors and atrix to the same orders as those of the entire structure, we expr

① in terms of the joint displacement vector { }Δ as { }F

(1) 1 1 1

(2) 1 1 2

3

2 4 00 0 0 0

F i i4 2 0F i i Δ⎧ ⎫ ⎡ ⎤ ⎧ ⎫

⎪⎪ ⎪ ⎪⎢ ⎥= Δ⎨⎨ ⎬ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥ Δ⎩ ⎭ ⎣ ⎦ ⎩ ⎭

bolically as

①

Or sym


{ } [ ] { }F K= Δ①① (g)

In which,

i iK i i[ ]

1 1

1 1

4 2 02 4 00 0

⎡ ⎤⎢ ⎥= ⎢

0⎥

⎢ ⎥⎣ ⎦

①

element-contribution matrix, which may be recognized as the contribution to global stiffness matrix of the structure provided by element

Similarly, for element 2, substitution of Eq. (f) into Eq. (b) yields

2 2 3

0 0 0 00 4 2F i i

⎧ ⎫

is termed ① .

1Δ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪

(1) 2 2 2

(2) 0 2 4F i i

⎪⎢ ⎥= Δ⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥ Δ⎣ ⎦ ⎩ ⎭

②

Or sy

⎩ ⎭

mbolically as

{ } [ ] { }F K= Δ②② (h)

In which,

0 0 00 4 20 2

K i ii

[ ] 2 2

2

⎡ ⎤⎢ ⎥= ⎢

24i⎥

⎢ ⎥⎣ ⎦

②

element-contribution matrix, which may be considered as the contribution to global stiffness matrix of the structure provided by the element

(4) Global stiffness equation of a structure Finally, by substituting equations (g) and (h) into the

desired relationship (or equation) between the joint force vector and the joint displacement vector

is named ② .

equilibrium equation (13-31), we obtain the { }F

{ }Δ of the structure as

{ } [ ] [ ] { }(F K K )= + Δ

Or

① ②

{ } [ ]{ }F K= Δ (13-32)


Therefore, the global stiffness matrix of the structure will

(13-33)

It can be seen that the global stiffness matrix of a structure

The expanding form of equation (13-31) will be

1

2 2 3

4 2 0

0 2 4

F i i

i i

[ ]K be

[ ] [ ] [ ] [ ]K K K K= + =∑ ⓔ① ②

ⓔ

is equal to the sum of element-contribution matrices of all the elements of the structure.

Δ1 1 1

2 1 1 2 2 2

3

2 4 4 2F i i i iF

⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥= +

⎢ ⎥Δ⎨ ⎬ ⎨ ⎬⎢ ⎥

⎪ ⎪ ⎪ ⎪Δ⎩ ⎭

⎣ ⎦ ⎩ ⎭

mbly method, is identical to equation (13-28) that is the result derived by traditional displacement method.

It can be observed from forging discussion that the steps for determinaa structure by so called stiffness assembly method are two: (1) determine element-contribution matrix

by element stiffness matrix ; (2) superpose all of the element-contribution matrices to form glob

I e d element global stiffness equations. In the process, the element

glob ly arranging the elements at the new positions and n to form element-contribution matrix . The key point

tions of the elements in , i.e., where the elements in

lobanding b

numb ement other positions, lement-contribution matrix will be formed. n Fi

Note that the above equation, derived by stiffness asse

tion of global stiffness matrix of

[ ]K ⓔ [ ]k ⓔ

al stiffness matrix of the structure. We will further discuss the first step as follows.

13.4.2 Element orientation vector

n the preceding subsection, w etermined the element-contribution matrix by substituting the element compatibility equations into the

al stiffness matrix [ ]k ⓔ was extended by newcomplementing enough zero elements ifor developing [ ]K ⓔ is the new posi

[ ]k ⓔ [ ]K ⓔ

[ ]k ⓔ

[ ]k ⓔmust be placed in [ ]K ⓔ ? It can be seen from the discussion of last subsection that if we extend the matrix [ ]k ⓔ to the same

order as that of the g l stiffness matrix [ ]K of the structure, change the element local end displacement numbers into its correspo joint global displacement numbers compatibility equations, newly place the elements in the rows and columns corresponding to its joint global displacement

ers, add the zero el s at

y its

the e [ ]K ⓔ

To illustrate the technique, consider the two-span continuous beam shown i g.13.13. The corresponding relations between the end local displacement numbers and its joint global displacement numbers for each element are tabulated in table 13-1.

In order to facilitate the development of element-contribution matrix, we define the vector whose


components are composed by the joint global displacement numbers of a mem ermed orientation ber, tvector and indicated by { }λ ⓔ . Then, the relationship between end local displacement numbers and its joint global displacement numbers of the member will be determined by the vector. So the orientation vector is also known as number exchange vector. The orientation vectors of the continuous beam shown in Fig.1

Table 13-1 relationship between end local displacement numbers and its joint global displacement numbers

Element End local displacement

numbers Corresponding joint global

displacement numbers Orientation vectors

3.13 are shown in the fourth columns of table 13-1.

① (2)

① ②

( )

(1) 1 2

12

λ⎧ ⎫

= ⎨ ⎬⎩ ⎭

①{ }

② (1) (2)

2 3

23

λ⎧ ⎫

= ⎨ ⎬⎩ ⎭

②{ }

The pro r developing an contributio mariz(1) Set up square matrix has the same order a he numbers of the joint displ ments

of a structure and let the initial values of he elements in be zero; (2) exchange numbers between the end local displacements and joint global displaceme f an

element; that is, exchange the row numb (i) and column number (j) of the element in for

the n

cedure fo element- n matrix may be sum ed as follows: a [ ]K ⓔ that s t ace

all t [ ]K ⓔ tonts o

er ( )( )i jkⓔ [ ]k ⓔ

ew row number iλ and column number jλ to form a new element Ki jλ λ

in [ ]k ⓔ ;

ⓔ for each of the elements

(3) Add each of the elements i j

Kλ λ to the position where is in row iⓔ λ and column jλ in

[ ]K ⓔ .

( )2( )1① ②

1

(b)

(a)1 2 3

( )2

Fig.13.13 Relationships between local and global numbers(a) joint global displacement numbers; (b) end local displacement numbers


By the procedure the corresponding positi ns between the elements ⓔ and those in [ ]K ⓔ for ea

, o in the b m shown in Fig.13.13 will be as follows.

1 1

1 2 3(1) (2)

0 12 4 0 2

)0 0 0 3i iK

[ ]k

Element ① : orientation vector { } [1, 2]Tλ =①

[ ]1 1

1 1

1 1

4 2 4 2 (1)

2 4 (2

i ii i

k i i

⎡

[ ]⎤

⎡ ⎤ ⎢ ⎥⇒ == ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦

Element : orientation vector

① ①

② { } [2, 3]Tλ =②

[ ] [ ]2 22 2

2 22 2

1 2 3(1) (2)

0 0 1 4 2 (1)

0 4 2 2 2 4 (2)0 2 4 3

i ii ik Ki ii i

⎡ ⎤⎡ ⎤ ⎢ ⎥= ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦

② ②

13.4.3 Implementation of element assembly method on computers

In subsection 13.4.1, we used two steps to determine the global (or structure) stifThe two steps are: (1) substitute the element compatibility equations into the elemrelations, i.e., orient the elements’ positions of in the element-contribution matrix orientation vector

fness matrix [ ]K . ent global stiffness

[ ]k ⓔ [ ]K ⓔ by{ }λ ⓔ ; (2) substitute the resulting relationships derived from step (1) into the joint

to form . This process of deve

bers that trix

atrix oidi directly

by t

equilibrium equations, i.e., accumulate the elements in each [ ]K ⓔ [ ]Kloping three types of matrices can be quite tedious and time consuming for large structures. From the expansion of Eq. (13-31), we observe that the stiffness of a joint in a direction equals the

sum of the stiffnesses in that direction of all the mem meeting at the joint. This fact indicates the global stiffness ma ]K can be formulated directly by accumulating the elements of the member global stiffness matrices into their proper positions in the global m , thereby av ng the necessity of developing element-contribution matrices. In other words, we can develop the global matrix [K

[

]he elements in [ ]k ⓔ . The technique of directly forming a global stiffness matrix by assembling the

elements of the member global stiffness matrices is known as element assembly method. Put concretely, the process of directly forming a global stiffness matrix [ ]K by assembling the

elements in [ ]k ⓔ is the process that firstly calculate the numerical value of one of the elements in [ ]k ⓔ ; secondly determine the position of the element in [ ]K by orientation vector { }λ ⓔ ; thirdly accumulate the value to its newl ermined position in [ ]K ; the same three steps are repeated for next element until all of the elements in all [ ]k ⓔ is assembled. The process can be implemented on comp

y detuters by foll

stepsowing

.


(1) Develop an array representing [ ]K and let the initial value of its elements to be zero. (2) Determine the element’s positions from [ ] in [ ]K by { }k ① λ ① an umulate them to the

newly determined positions. By now [ ] [ ]K K= . (3) Determine the ent’s positions from [ ]k ② in [ ]K by { }

d acc

elem

①

λ ② and algebraically add them

to the elements located in the newly determined po [ ]K= +① ② . The same step is

then stif ss matrix K

mor be ness ded.

Determine e global stiffness matrix for the continuous beam shown in Fig.13.14.

(1) Numbers of global joint displacements From the computing model shown in Fig.13.14 (b), we observe that joints 1, 2 and 3 are free to rotate.

Thus the beam has 3 joint global displacements,

sitions. By now [ ] [ ]K K repeated for all of the next element fness matrices. Finally, we obtain the global stiffne

[ ] [ ]K= ∑ ⓔ

ⓔ

.

It should be mentioned again that when two or e mem r stiff coefficients are located in the same position in [ ]K , then the coefficients must be algebraically ad

Example 13-2

th [ ]K

Solution

21 031i 2i 3i

1Δ , 2Δ and 3Δ , which are unknown rotations of the three joints.

Note that at a fixed joint, the joint displacements are zero. So we let the value of a component of joint glob estraints of the joint be equal to zero. al displacements corresponding to one of the r

(2) Orientation vector { }λ ⓔ The vector’s components are composed by the int global dis acement numbers of the element. Then,

the three elem jo pl

ents’ orientation vectors can be written as follows.

21 031Δ 2Δ 3Δ 0 0Δ =

① ② ③

① ② ③

(a)

(b)

Fig.13.14 Figures of example 13-2(a) original structure ; (b) computing model


{ }1⎧ ⎫①

{ }2

2λ = ⎨ ⎬

⎩ ⎭ 3λ

⎧ ⎫= ⎨ ⎬⎩ ⎭

② { }

30

λ⎧ ⎫

= ⎨ ⎬⎩ ⎭

Stiffness matrix of element

4 2 (1)(2)

i i⎡ ⎤= ⎢ ⎥⎣ ⎦

①

According to

③

(3) Development of global stiffness matrix [ ]K ①

[ ] 2 4k i i1 1

1 1

(1) (2)

{ }λ ①

obal nu, local number (1) must be changed into global number 1; local number (2) must

be changed into gl mber 2. Then directly place the elements in to their corresponding positions in . By now, will have the form as

0 0 0 3

[ ]k ①

[ ]K [ ]K

[ ]1 1

1 1

4 2 0 1

2 4 0 2i i

K i i

1 2 3

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

Stiffness matrix of element

By

②

[ ] 2 2

2 2

(1) (2)4 2 (1)2 4 (2)i i

k i i⎡ ⎤

= ⎢ ⎥⎣ ⎦

②

{ }λ ② , al num

local number (1) must be changed into global number 2; local number (2) must be changed into glob ber 3. Then directly algebraically add the elements in to elements located in their newly determined corresponding positions in . Then, will

[ ]k ②

[ ]K [ ]K have the form as

[ ]1 1

1 1 2 2

4 2 0 1

2 4 4 2

2 2

1 2 3

2i0 2 4 3

i iK i i i i

i

⎡ ⎤⎢ ⎥= +⎢ ⎥⎢ ⎥⎣ ⎦

Stiffness matrix of element ③


[ ] 3 3

3 3

(1) (2)4 2 (1)2 4 (2)i i

k i i⎡ ⎤

= ⎢ ⎥⎣ ⎦

③

In accordance with { }λ ③ , local number (1) must be changed into global number 3; local number (2) must al n

obtai

4 2 0 1

2 4 4 2 20 2 4 4 3

i iK i i i i

i i i

⎡ ⎤⎢ ⎥= +⎢ ⎥⎢ ⎥+⎣ ⎦

The matrix is just the global stiffness matrix of the continuous beam.

per

global stiffness matrix ess coefficient, which identifies the

force

be changed into glob umber 0. Then directly algebraically add the elements in [ ]k to elements located in their newly determined corresponding positions in [ ]K . By now, [ ]K will be ned as

1 2 3

③

[ ]1 1

1 1 2 2

2 2 3

[ ]K

13.4.4 Pro ty of global stiffness matrix

(1) Meanings of stiffness coefficients inThe element ijk in global stiffness matrix [ ]K is termed stiffn

(per unit disp cement) at the location and he direction of iF required, along with other end

forces, to cause a unit value of the displacement

la in t

jΔ while all othe nd displacements are zero. The

element iik in the main diagonal of [ ]K is the s ness of joint i in i direction and its magnitude

equals th um of the stiffnesses in the di tion of all the members meeting at the joint. (2) Global stiffness matrix [ ]K is symmetric. The property may be proved by

react

r e

tiff

e s rec reciprocal law of

(3) If the support conditions of a structure are introduced when determine the global stiffness matrix

nonzero elements ed in a rather narrow band near the main diagonal.

ions presented in section 9.8.

1 1,n nF Δ,n nF Δ

[ ]K , the [ ]K will be reversible. (4) [ is a sparse and banded matrix because there are many zero elements in the matrix and the

in it are only cluster]K

21 32 2,F Δ 3 3,F Δ1 1,F Δ + +

n 1n +

ⓝ① ② ③

Fig.13.15 A continuous beam with n spans


The F 3.15 shows a continuous beam with n spans. If we use the approach discussed previously, the stiffness equation of the structure might be obtained as follows.

ig.1

re just distributed near the di l of It can be observed that there are many zero elements in the global stiffness matrix [ ]K , so [ ]K is

known as a sparse matrix. Since the nonzero elements in [ ]K a main agona[ ]K , [ ]K is also referred to as a banded matrix.

13.5 Global Stiffness Matrices of Rigid Frames 525

1 1 1 1

2 1 1 2 2 2

3 2 2 3 3

1 1

1 1

4 2 0 02 4( ) 2 00 2 4( ) 2 0

0 2 4( ) 2

0 2 4n n n n n n

n n n n

F i iF i i i iF i i i i

F iF i

− −

+ +

Δ⎧ ⎫ ⎡ ⎤ ⎧⎪ ⎪ ⎢ ⎥ ⎪+ Δ⎪ ⎪ ⎢ ⎥ ⎪⎪ ⎪ ⎢ ⎥ ⎪+ Δ⎪ ⎪ ⎢ ⎥ ⎪⎪ ⎪ ⎢ ⎥ ⎪⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨

⎢ ⎥⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪+ Δ⎢ ⎥⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪Δ⎩ ⎭ ⎣ ⎦ ⎩

⎪⎪⎪

⎪⎪⎪⎪⎪⎪⎭

3

i i ii

⎫⎪⎪⎪

⎬ (13-34)

13.5 Global Stiffness Matrices of Rigid Frames

In this section, we will discuss the development of global stiffness matrix for rigid frames. Comparing with the last section that discussed the establishment of global stiffness matrix for continuous beams, the fundamental principle and procedure to develop the global matrix are identical. However, the technique to develop the global stiffness matrix for a rigid frame is more complex than that for a continuous beam. The complexity occurs in following aspects:

(1) The joint displacement components for each joint of a rigid frame increase to three, the translations in x and y directions and the rotation of the joint, respectively.

(2) The members of a rigid frame are oriented in different directions, so it becomes necessary to transform the stiffness relations for each member from the member’s local coordinate system to a common global coordinate system.

(3) If hinged joints occur on a rigid frame, the relative rotations between the member ends meeting at the same hinged joint must be taken into consideration.

Unlike in the case of the classical methods of analysis considered previously, it is usually not necessary to neglect member axial deformations when analyzing rigid frames by the matrix displacement method. In fact, the case of neglecting member axial deformations will be discussed as a special example individually.

13.5.1 Element orientation vectors and assembly of element stiffness matrices

(1) Nodal global displacement numbers As we have known in last section, there is only one nodal displacement component of a node for a

continuous beam, the rotation of the node. However, there are three nodal displacement components of a node for a rigid frame, the translations in the x and y directions, respectively, and the rotation about the z axis.


Therefore, when numbering the displacements of the nodes of a rigid frame, all of the nodal displacement components must be uniformly numbered in accordance with the connecting manner of each node.

Consider the rigid frame shown in Fig.13.16 (a). We can see that node A has three displacement components—the translations and in the x and y directions, respectively, and the rotation Au Av Aθ about the z axis, so the nodal global displacement numbers must be [1 2 3]. Since joint B is attached to the fixed support, it can neither translate nor rotate; it has three known displacement components, the translations

, and the rotation 0Bu = 0Bv = 0Bθ = , so the nodal global displacement numbers must be [0 0 0]. Similarly, node C, which is connected to a hinged support, has two given displacement components, the translations , , and a unknown rotation 0Cu = 0Cv = Cθ , thus the nodal global displacement numbers must be [0 0 4].

Note that the unknown displacement components are numbered by starting at the lowest node number and proceeding sequentially to the highest node number. In the case of more than one unknown displacement components at a node, the translation in the x direction is numbered first, followed by the translation in the y direction, and then the rotation. The known nodal displacement components are numbered by zero.

Since the rigid frame has four unknown nodal displacement components, its nodal (or joint) displacement vector { }Δ will be written as follows.

{ } [ ] [ ]1 2 3 4T T

A A A Cu v θ θΔ = Δ Δ Δ Δ =

A

B

C

0

2

1

30

40

y

x

00

①

②

(a)

( )5( )6

( )2( )3

( )1 ( )4 x( )2( )3

( )1

( )6

(b)

①

②

( )4( )5

x

Fig.13.16 Global displacement number for planar rigid frame(a) nodal global displacement numbers; (b) element local end displacement numbers


Its corresponding nodal (or joint) force vector will be written as:

{ } [ ]1 2 3 4TF F F F F=

(2) Element orientation vectors The rigid frame shown in Fig.13.16 (a) has two elements, and . The arrows on the members

indic scissas ① ②

x ate the positive directions of the ab of their local coordinate systems. The six numbers within the brackets, (1), (2), (3), (4), (5) and (6) identify the element local end displacement components, the translation in the x direction numbered first, llowed by the translation in the fo y direction, and then the rotation [Fig.13.16 (b)].

As defined i subsection 13.4.2, the element orientation vector is composed by its nodal global displacement component

n numbers of the element.

ers for each element of the rigid frame.

and its nodal global displacement numbers

Element

Table 13-2 shows the corresponding relations between the end local displacement component numbers and its nodal global displacement component numb

Table 13-2 relationship between end local displacement numbers

Element② ①

local num erb global numbers Orientation vector b global numbers Orientation vectors→ local num ers→

(1) → 1 (2) → 2 (3) 3 →(4) 0 →(5) 0 →(6) 4 →

123004

λ

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

①{ }

(2) ⎪ ⎪⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

②{ }

(1) → 1 → 2

(3) 3 →(4) 0 →(5) 0 →(6) 0 →

1⎧ ⎫⎪ ⎪2

3000

λ

(3) Assembling process of element stiffness matrices The global stiffness matrix of element has been calculated in example 13-1. We write it again as

follo①

ws


[ ] 4

(1) (2) (3) (4) (5) (6)(1) 300 0 0 300 0 0(2) 0 12 30 0 12 30

(3) 0 30 100 0 30 5010

300 0 0 300 0 0(4)0 12 30 0 12 30(5)0 30 50 0 30 100(6)

k

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−

×= ⎢ ⎥−⎢ ⎥⎢ ⎥− − −⎢ ⎥

−⎢ ⎥⎣ ⎦

① (a)

Based on the orientation vector { }λ ① tabulated in table 13-2 and relationships between the end local displacement numbers and nodal global displacement numbers, place sequentially the element in row (i) and column (j) of to the position in row [ ]k ①

iλ and column jλ in matrix [ to form the first-stage result of global stiffness matrix of the frame as follow

]K

[ ] 4

(1) (2) (3) (6) 1 2 3 4

(1) 1 300 0 0 0

(2)2 0 12 30 3010

(3) 3 0 30 100 500 30 50 100(6) 4

K

⎡ ⎤⎢⎢ ⎥= ×⎢ ⎥⎢ ⎥⎣ ⎦

⎥

Note that the nodal global displacement component numbers corresponding to the element’s local numbers (4) and (5) are zero, so there are no positions in global stiffness matrix for the elements in row (4) and column (4), in row (5) and column (5) of .

[ ]K[ ]k ①

The global stiffness matrix of element has been also obtained in example 13-1. We write it again as follow

②

[ ] 4

(1) (2) (3) (4) (5) (6)(1) 12 0 30 12 0 30(2) 0 300 0 0 300 0

(3) 30 0 100 30 0 5010

12 0 30 12 0 30(4)0 300 0 0 300 0(5)30 0 50 30 0 100(6)

k

− − −⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−

×= ⎢ ⎥−⎢ ⎥⎢ ⎥−⎢ ⎥−⎢ ⎥⎣ ⎦

② (b)

Based on the orientation vector { }λ ② tabulated in table 13-2 and relationships between the end local displacement numbers and nodal global displacement numbers, place sequentially the element in row (i) and column (j) of [ ] to the position in row k ②

iλ and column jλ in matrix [ , which, at this time, has ]K


already added the stiffness coefficients of element , to form the final result of global stiffness matrix of the frame as follows

①

[ ]

(1) (2) (3) 1 2 3 4

(1) 1 300 (12) 0 (0) 0 ( 30) 0

(2) 2 0 (0) 12 (300) 30 (0) 30(3) 3 0 ( 30) 30 (0) 100 (100) 50

4 0 30 50 100

K

+ + + −⎡ ⎤⎢ + + +⎢=⎢ + − + +⎢⎣ ⎦

410⎥⎥×⎥⎥

Note that the nodal global displacement component numbers corresponding to the element’s local numbers (4), (5) and (6) are zero, so there are no positions in global stiffness matrix for the elements in row (4) and column (4), in row (5) and column (5), in row (6) and column (6) of .

[ ]K[ ]k ②

13.5.2 Treatment of hinged joints

In this subsection, we will discuss how to deal with the rigid frames with hinged joint. Consider the frame shown as in Fig.13.17, on which there is a hinged joint.

First, let us consider the joint global displacement numbers. As shown in the figure, joints B and D are attached to the fixed supports, i.e., the two joint displacements

are known to be equal to zero, so the three global nodal displacement numbers for the two joints are [0 0 0] each. Since node A is a rigid joint, it has three unknown displacement components. Thusly, its global nodal displacement numbers are [1 2 3]. Because joint C is a hinged joint, it has relative rotation between the two member ends. Therefore, it is necessary to consider the joint as two nodes, nodes and . The two nodes have the same translations but different rotation. In this circumstance, we use global nodal displacement numbers [4 5 6] to identify the displacement components of node , [4 5 7] of node

.

1C 2C

1C2C

Next, let us determine orientation vectors of the elements. From Fig.13.17, the orientation vectors of the elements , and , whose positive axial

directions of abscissas ① ② ③

x in the local coordinates are indicated by the arrows shown in the figure, can be written as follows

{ } [ ]1 2 3 4 5 6 Tλ =①

{ } [ ]1 2 3 0 0 0 Tλ =②

{ } [ ]4 5 7 0 0 0 Tλ =③


Finally, let us develop global stiffness matrix of the frame by use element assembly method in terms of the order proceeding from the element with the smallest number to the largest number.

Assume that the size and material of each member of the frame shown in Fig.13.17 are identical, which

Assembly of element ① :

are the same as those of the members of the frame in example 13-1.

iffness matrix has been determined as equation (a) in subsection 13.5.1. Base

The global element st [ ]k ① d on the orientation vector { }λ ① , pla e element in [ ]k ① to their proper positions in matrix [ ]K ,

which, at this time, has only zero nts, to form the first-sta sult of global stiffness matrix of the f e as follows

ce theleme ge re ram

Assembly of element

1

23

4

[ ]

(1) (2) (3) (4) (5) (6) 1 2 3 4 5 6 7

(1) 1 300 0 0 300 0 0 0(2) 2 0 12 30 0 12 30 0

(3) 3 0 30 100 0 30 50 0(4) 4 300 0 0 300 0 0 0(5) 5 0 12 30 0 12 30 0(6) 6 0 30 50 0 30 100 0

7 0 0 0 0 0 0 0

K

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−⎢ ⎥= − ×⎢ ⎥⎢ ⎥− − −⎢ ⎥

−⎢ ⎥⎢ ⎥⎣ ⎦

410

② :

A

B

2C1C

D

0

y

00

x

00

0

①

② ③

56

74

5

Fig.13.17 Global displacement number for the rigid frame with hinged joint

13.6 Equivalent Nodal Loads 531

The global element sti esffn s matrix has been obtained as equation (b) in subsection 13.5.1. By the orientation vector

[ ]k ② { }λ ② , accumulate the element in to their proper positions in above matrix

to form the second-stage result of global stiffness matrix of the frame as follow

Assembly of element The global element stiffness matrix is equal to . By the orientation vector

[ ]k ② [ ]K

[ ]

(1) (2) (3) 1 2 3 4 5 6 7

(1) 1 300 12 0 0 0 30 300 0 0 0(2) 2 0 0 12 300 30 0 0 12 30 0

(3) 3 0 30 30 0 100 100 0 30 50 04 300 0 0 300 0 0 05 0 12 30 0 12 30 06 0 30 50 0 37

K

+ + − −+ + + −− + + −

= −− − −

−

410

0 100 00 0 0 0 0 0 0

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥

×⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

③ : [ ]k ③ [ ]k ② { }λ ③ , add the

element of to their proper positions in matrix , which, by now, has already added the stiffness coefficients of two elements and , to form the final result of global stiffness matrix of the frame as follow

13.6 Equivalent Nodal Loads

In the previous two sections, we have discussed the global stiffness matrix and developed global stiffness equations of a structure in matrix form as follow

[ ]k ③ [ ]K① ②

[ ]

(1) (2) (3)1 2 3 4 5 6 7

1 312 0 30 300 0 0 02 0 312 30 0 12 30 0

3 30 30 200 0 30 50 0(1) 4 300 0 0 300 12 0 0 0 0 30(2)5 0 12 30 0 0 12 300 30 0 0

6(3) 7

K

− −−

− −= − + + −

− − + + − +

410

0 30 50 0 30 100 00 0 0 0 30 0 0 0 0 100

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥

×⎢ ⎥⎢ ⎥⎢ ⎥

−⎢ ⎥⎢ ⎥− + +⎣ ⎦

[ ]K


{ } [ ]{ }F K= Δ (13-35)

The global stiffness matrix , which represents the relationships between the global joint displacements and global joint forces (that is, the forces yielded on their corresponding artificial restraints), is yielded in terms of displacement-method’s primary system of the original structure. The relationships only reflect the rigidity properties but not involve the actual loads applying on the structure, so they are not the primary equations used to analyze the original structure by displacement method.

Recalling from chapter 11, when establishing the primary equations used to analyze a structure by displacement method, we have considered two states of the primary system:

① The primary structure was subjected only to external loads ( let the joint displacements

[ ]K{ }Δ { }F

{ } {0}Δ = ). In this situation, we used { }PF to denote the restraint forces yielded on the artificial restraints on the primary structure.

② The primary structure was subjected only to joint displacements { }Δ raints

( let the external loads equal to zero). In this case, the restraint forces yielded on the artificial rest on the primary structure were identified by

Superimposing above two cases, we obtained the primary equations as follows { } [ ]{ }F K= Δ .

{ } { } 0PF F+ =

or

[ ]{ } { } { }0PK FΔ + = (13-36)

(1) Concept of equivalent joint loads and equivalent nodal load vector The loads applying on a structure may be distributed loads (named non-joint loads afterwards), or joint

loads, or the combination of joint and non-joint loads. In order to facilitate the matrix analysis, we have to transform non-joint loads into equivalent joint loads, indicated by . The rule for transforming non-joint loads into equivalent joint loads is that the original non-joint lo their equivalent joint loads yield identical restraint forces on the artificial restraints at the joints of th ary system of the structure. That is, if the restraint forces on the primary system caused by the original are

{ }EPads and e prim

loads { }PF , the restraint forces on the primary system induced by their equivalent joint loads are { }EP { }PF as well. Therefore, we have following conclusion in function form as

{ } { }E PP F= − (13-37)

As a matter of fact, if there are joint loads applying on the original structure, representing them by { }JP , equation (13-36) must be written as


[ ]{ } { } { }P JK F PΔ + = (13-38)

Substituting equation (13-37) into equation (13-38), we obtain

[ ]{ } { } { } { } { } { }J P J EK P F P P PΔ = − = + =

Or simply as

[ ]{ } { }K PΔ = (13-39)

In which,

{ } { } { } { } { }J P JP P F P P= − = + E (13-40)

represents the effects of all external loads on the nodes of the structure. It is termed equivalent nodal load vector.

It is observed from equation (13-35) and equation (13-39) that if we change the nodal restraint force vect -35 into the equivalent nodal load vector , we will obtain the disp of a

uival t cas cal

o ends of an element, i.e. 6 restraints, the two ends of the element will be fixed. Under action of external loads, the two ends will yield 6 fixed-end forces. The matrix form of the six fixed-end forces,

{ }P

or { }F in Eq. (13 ) { }Placement-method’s primary equations structure. (2) Eq ent joint loads { }EP ⓔ of an elemenFirst, let us consider the e in lo coordinates. If we impose restraint to all likely moving directions

of the tw

{ }PF ⓔ , will be written as

{ } 1 1 1 2 2 2

T

P P P P P P PF X Y M X Y M⎡ ⎤= ⎣ ⎦ⓔ

(13-41)

Table 13-3 shows the fixed-end forces of seven typical loading forms for a prismatic member. If add minus signs in the fixed-end forces, we will obtain the element’s equivalent joint loads { }EP ⓔ in its lcoordinates as follows.

ocal

{ } { }E PP F= −ⓔ ⓔ

(13-42)

Next, let us consider the situation in global coordinates. By the transformation equations of coordinates, Eq. (13-22), we obtain

{ } [ ] { }TE EP T P=

ⓔⓔ (13-43)

joint loads of element in global coordinates. ⓔ The equation (13-43) is nothing else but the equivalent


{ }PF ⓔ Table 13-3 fixed-end forces du

pes o

e to non-joint loads in local coordinates

Num

ber Ty f load

End forces

Beginning end（1） Ending end（2）

1

P

P

P

X

Y

M

2 3

2 31

2a aqal l

⎛ ⎞

2 2

26 8 3

12qa a a

l l⎛ ⎞

0

− − +⎜ ⎟⎝ ⎠

− − +⎜ ⎟⎝ ⎠

3

1qa a⎛ ⎞2 2l l

3

4 312qa a

l l⎛ −⎜⎝ ⎠

0

− −⎝ ⎠

⎞⎟

⎜ ⎟

q

2

P

P

P

X

Y

M

2

2

21qb al l

⎛ ⎞

2

0

qabl

−

− +⎜ ⎟⎝ ⎠

2

2

21qa bl l

⎛ ⎞

2

2

0

qa bl

− +⎜ ⎟⎝ ⎠

3

P

P

P

X

Y

M

3

0

6

2

qab

−⎜ ⎟

3l

qb b⎛ ⎞l l⎝ ⎠

3

0

6

2

qab

l l

−

−⎜ ⎟⎝ ⎠

3l

qa a⎛ ⎞

4

P

P

P

X

Y

M

2 3

2 3

2 2

2

0

1.64

2 3 1.26

qa a al l

qa a al l

⎞− − +

32⎛⎜ ⎟⎝ ⎠⎛ ⎞

− − +⎜ ⎟⎝ ⎠

3

2

3

0

3 1.64

1 0.84

qa al l

qa al l

⎛ ⎞− −⎜ ⎟⎝ ⎠

⎛ ⎞−⎜ ⎟⎝ ⎠

5

P

P

P

X

Y

M

1 0.⎛ ⎞⎜ ⎟5 aqa

l− −

⎝ ⎠

0

0

2

0.5 qa−

0

0

l

xba

l

y

q

ba

l

x

y

q

xba

ly

ba

l

q

x

y

q

bax

l

y


6

P

P

P

X

Y

M

0

0

qbl

−

0

0

qa−

l

q

7

P

P

P

X

Y

M

2 3a b⎛ ⎞2

2

2

0

qal l lqab

l

+⎜ ⎟⎝ ⎠

−

2 3qa a b⎛ ⎞

2

2

0

l l lqa b

− +⎜ ⎟⎝ ⎠

2l−

their pr

(3) Equivalent joint loads { }P of a structure E

As mentioned in previous, the number of components of the equivalent joint load vector { }EP of a structure is the same as that of the joint displacement vector of the structure. The components of vector { }EP can be obtained by algebraically adding each of the elements in all { }P ⓔ to oper positions oriented by the

E

proper { }λ ⓔ { }P in E . { }P(4) Equivalent nodal load vector of a structure

In general sense, a structure might be subjected to both joint loads and non-joint loads. As known from previous discussion, we use vector { }JP represents the effect of joint loads and { }EP indicates the effect of non-joint loads. Therefore, the total effects of all external loads, known as equivalent nodal load vector

the nodes of the stru{ }P , on cture w expressed as follows. ill be

{ } { } { }J EP P P= +

3.18.

(1) Fixed-end force

(13-44)

Example 13-3

Determine the equivalent nodal load vector { }P for the rigid frame shown in Fig.1

Solution

{ }PF ⓔ in local coordinates Element ① : Referring to the row 1 in table 13-3 and substituting 4.8kN/mq = , 5ma l= = into

the equations tabulated in the row, we write the fixed-end forces of the element as follows

1

1

1

012kN

10kN mPM⎪ = − ⋅⎩

P

P

XY⎧ =⎪

= −⎨ 2 0PX

⎪2 12kN

kNPY

⎧ =

= −⎨

2 10 mPM⎪ = ⋅⎩

xba

ly

q

ba

l

x

y


Element in tabl② : Referring to the row 2 e 13-3 and substituting 8kNq = − , a b= = 2.5m into the equations tabulated in the row, we find that the fixed-end forces of the element will be

1

1

1

04kN5kN m

P

P

P

XYM

⎧ =⎪

=⎨⎪ = ⋅⎩

2 0P =

2

2

4kN5kN m

P

P

XYM

⎧

=⎨⎪

⎪

= − ⋅⎩

Consequently,

{ }

01210

012

10

PF

⎧ ⎫⎪ ⎪−⎪ ⎪⎪ ⎪−⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪−⎪ ⎪⎪ ⎪⎩ ⎭

① { }

045

045

PF

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪−⎩ ⎭

②

(2) Element’s equivalent joint loads in gl bal coordinates du to non-joint loa s Element As known from the figure, since

{ }EP ⓔ o e d① : 1 0α = ° , [ ] [ ]T I=① . Referring to equations (13-42)

and (13-43), we write the equivalent joint load in global system as follows s { }EP ①

y

x1Δ

4.8kN/m20kN

2Δ3Δ

4Δ

8kN

A

B

C

5m

2.5m

2.5m

①

②

Fig.13.18 Figure example 13-3 of


{ } [ ] { } [ ]{ } { }

01210

01210

TE P P PP T F I F F

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪= − = − = − = ⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪−⎩ ⎭

① ① ①①①

Element : It can be observed from the figure, that since② 2 90α = ° , by equations (13-42) and (13-43), we write the equivalent joint loads in global system as follows { }EP ②

0 1 0 0 0 0 0 41 0 0 0 0 0 4 0

1 00 0 0 0 0 1 5 5

−⎡ ⎤ ⎧ ⎫ ⎧ ⎫⎢ ⎥ ⎪

{ } [ ] { }0 0 1 0 0 0 5 50 0 0 0 1 0 0 40 0 0 0 4 0

TE PP T F

⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪= − = − =⎢ ⎥ ⎨ ⎬ ⎨ ⎬−⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪

−⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎣ ⎦

②②②

(3) Joint loads { }JP of the struct ureIt can be seen from the figure that only second component of the joint load vector { }JP is non-zero.

Thusly, the vector will be

0⎧

{ }2000

JP

⎫⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭

he s in the figure, the orientation vectors of the two elements will be written as

004

λ

(4) Equivalent nodal load vector { }P of t tructure As shown

1

{ }

23

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

① 000

λ{ }

12

⎧ ⎫⎪

3

⎪⎪ ⎪⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

②


As known from equation (13-44), the total effects of all external loads, called equivalent nodal load vector , on the nodes of the structure will be obtained by algebraically adding each of the elements in

to that in { }P

{ }EP { }JP . For the rigid frame, the equivalent joint load vector due to the non-joint loads will be obtained by accumulating the elements at the proper positions oriented by { }

{ }EPλ ① and { }λ ② in the

two and { together. Therefore, { will be written as { }EP ① }EP ② }EP

{ }

(1) 1 0 (1) 1 4 4(2) 2 12 (2) 2 0 12(3) 3 10 (3) 3 5 5(6) 4 10 4 0 10

EP

⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪= + =⎨ ⎬ ⎨ ⎬ ⎨−⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪− −⎩ ⎭ ⎩ ⎭ ⎩ ⎭

⎬

0

Finally, by equation (13-44) the equivalent nodal load vector of the structure will be obtained to be

{ }P

{ } { } { }

0 4 420 12 320 5 50 10 1

J EP P P

⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪= + = + =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪− −⎩ ⎭ ⎩ ⎭ ⎩ ⎭

13.7 Procedure for Analysis and Examples

Based on the discussion presented in the previous sections, we can develop the following step-by-step procedure for the analysis of structures by the matrix displacement method.

1. Prepare a computing model of a structure as follows: (1) Draw a line diagram of the structure, on which each node (or joint) and element (or member) must

be identified by a number. (2) Select a global x y coordinate system, with the x and y axes oriented in the horizontal (positive to

the right) and vertical (positive downward) directions, respectively. (3) For each element, establish a local x y coordinate system by selecting one of the joints at its ends

as the beginning joint and the other as the end joint. On the line diagram of the structure, for each element indicates the positive direction of the local x axis by drawing an arrow along the member pointing toward the end joint. For horizontal members, the coordinate transformations can be avoided by selecting the joint at the left end of the member as the beginning joint.

(4) Identify the unknown joint (or nodal) displacements, { }Δ , of the structure. The nodal displacements are specified on the structure's line diagram by drawing arrows at the nodes and are numbered

13.7 Procedure for Analysis and Examples 539

by starting at the lowest nodal number and proceeding sequentially to the highest nodal number. In the case of more than one nodal displacement at a node, the x translation is numbered first, followed by the y translation, and then the rotation. Recall that a joint of a plane frame can have up to three unknown displacements (two translations and a rotation); a joint of a continuous beam can have up to two unknown displacements (a translation perpendicular to the beam's centroidal axis and a rotation); and a joint of a plane truss can have up to two unknown displacements (two translations). Note that joint translations are considered as positive when in the positive directions of the x and y axes; joint rotations are considered as positive when clockwise.

2. Evaluate the structure stiffness matrix [ ]K(1) Compute the member stiffness matrix in local coordinates, [ ]k ⓔ . Expressions of [ ]k ⓔ for the

members of rigid frames, continuous beams and trusses are given in equations (13-8), (13-14) and (13-16), respectively.

(2) Compute the member's transformation matrix [ by using Eq. (13-20). For horizontal members with the local

]Tx axis positive to the right (i.e., in the same direction as the global x axis), the element

stiffness equations in the local and global coordinates are the same (i.e., [ ] [ ]k k=ⓔ ⓔ ); go to step (4). (3) Determine the element stiffness matrix in global coordinates, [ ] [ ] [ ] [ ]Tk T k T=ⓔ ⓔ (Eq. (13-27)).

The matrix must be symmetric. [ ]k ⓔ

(4) Identify the member's orientation vector { }λ ⓔ and store the pertinent elements of in their proper positions oriented by

[ ]k ⓔ

{ }λ ⓔ in the structure stiffness matrix . The complete structure stiffness matrix obtained by assembling the stiffness coefficients of all the elements of the structure must be symmetric.

[ ]K[ ]K

3. Form equivalent nodal load vector of the structure { }P(1) If the member is subjected to external loads, then evaluate its fixed-end force vector in local

coordinates, { }PF ⓔ , by using the expressions for fixed-end forces given in table 13-3. (2) Determine member’s equivalent joint loads { }EP ⓔ in local coordinates by using Eq. (13-42). (3) Identify member’s equivalent joint loads { in global coordinates by using Eq. (13-43). }EP ⓔ

(4) Calculate equivalent joint loads { of the structure by storing the pertinent components of in their proper positions oriented by

}EP{ }EP ⓔ { }λ ⓔ in { }. EP

(5) Find joint load vector { }JP of the structure whose component numbers are associated with those of the joint displacements { }Δ .

(6) By Eq. (13-44), add the corresponding components of { }JP and together to form equivalent nodal load vector of the structure.

{ }EP{ }P

4. Determine the unknown joint displacements { }Δ . Substitute { , and into the structure stiffness equations, (Eq. (13-39)), and solve the resulting system of simultaneous equations for the unknown joint displacements

}P [ ]K { }Δ{ } [ ]{ }P K= Δ

{ }Δ .


5. Compute member end displacements and end forces. For each member, do the following: (1) Obtain member end displacements in global coordinates, { }Δ ⓔ , from the joint displacements, by

using orientation vector { }λ ⓔ . (2) Determine member end displacements in local coordinates by using the relationship

{ } [ ]{ }TΔ = Δⓔ ⓔ (Eq. (13-21)). For horizontal members with the local x axis positive to the right, { } { }Δ = Δⓔ ⓔ .

(3) Compute member end forces in local coordinates by using the relationship

{ } [ ] { } { }PF k F= Δ +ⓔ ⓔ ⓔ ⓔ (13-45)

For trusses, { } {0}PF =ⓔ . 6. Determine support reactions by considering the equilibrium of the joints located at the supports of

the structure.

Example 13-4

Determine the internal forces for the continuous beam shown in Fig.13.19 (a) by using matrix displacement method. The flexural rigidity of each members is unchangeable, i.e., . constantEI =

Solution

own joint displacements ig.13.19 (b). From the model, we observe that joint 0 is fixed to the

supp

12kN/mq = 60kN100kN

8m4m 4m

210 3Δ

(1) UnknThe computing model is shown in Fort and joints 1, 2 and 3 are free to rotate. Thus the beam has 3 unknown joint displacements, 1Δ , 2Δ

and 3Δ , which are the unknown rotations of joints 1, 2 and 3, respectively. (2) Element stiffness matrix [ ]k ⓔ

ce the local Sin x axis is identical to global x axis, the element stiffness matrices in local and global coordinates are the same, i.e., [ ] ]k ⓔ [k . In addition, the flexural rigidity per unit length of each element, = ⓔ

Δ 21 3Δ

(a) (b)

① ② ③

4m 4m

x

y

Fig.13.19 Figure of example 13-4(a) original structure; (b) computing model


EIil

= , is unchangeable, by using Eq. (13-14) we write

[ ] [ ] [ ] 4 22 4i i

k k ki i

⎡ ⎤= = = ⎢ ⎥

⎣ ⎦

① ② ③

(3) Global stiffness matrix [ ] KFrom Fig.13.19 (b), we can see that the orientation vectors of the elements are as follows.

{ }01

λ⎧ ⎫

= ⎨ ⎬⎩ ⎭

① { }

12

λ⎧ ⎫

= ⎨ ⎬⎩ ⎭

② { }

23

λ⎧ ⎫

= ⎨ ⎬⎩ ⎭

③

Based on the orientation vectors, the pertinent elements of can be stored in their proper positions in the global stiffness matrix . The final will be

[ ]k ⓔ

[ ]K [ ]K

[ ]8 2 02 8 20 2 4

i iK i i

i ii

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

(4) Element’s equivalent joint loads { }EP ⓔ

Referring to table 13-3, the fixed-end forces { }PF ⓔ for each element will be obtained as follows.

{ } 1008100

8

P

ql

Fql

⎧ ⎫−⎪ ⎪ −⎧ ⎫⎪ ⎪= =⎨ ⎬ ⎨+⎩ ⎭⎪ ⎪+⎪ ⎪⎩ ⎭

①

⎬ { }

2

2

641264

12

P

ql

Fql

⎧ ⎫−⎪ ⎪ −⎧ ⎫⎪ ⎪= =⎨ ⎬ ⎨ ⎬+⎩ ⎭⎪ ⎪+⎪ ⎪⎩ ⎭

② { } 608

608

P

ql

Fql

⎧ ⎫−⎪ ⎪ −⎧ ⎫⎪ ⎪= =⎨ ⎬ ⎨+⎩ ⎭⎪ ⎪+⎪ ⎪⎩ ⎭

③

⎬

Since { } { }E EP P=ⓔ ⓔ , by using Eq. (13-37) we write element’s equivalent joint loads as

{ }100100EP+⎧ ⎫

= ⎨ ⎬−⎩ ⎭

① { }

6464EP+⎧ ⎫

= ⎨ ⎬−⎩ ⎭

② { }

6060EP+⎧ ⎫

= ⎨ ⎬−⎩ ⎭

③

(5) Equivalent nodal load vector { of the structure }PFrom the Fig.13.19, we can see that there is no joint load, i.e., { } {0}JP = . The equivalent joint loads of the structure can be obtained by storing the pertinent components of { } in their proper

positions oriented by { }{ }EP EP ⓔ

λ ⓔ in { . By Eq. (13-44), add the corresponding components of {}EP }JP and together to form equivalent nodal load vector of the structure. The final becomes as { }EP { }P { }P


{ }36460

P−⎧ ⎫⎪ ⎪= −⎨ ⎬⎪ ⎪−⎩ ⎭

(6) Solve equation { } to determine { }[ ]{ }P K= Δ Δ Substituting and [ into the equation, we write the equation in expanded form as { }P ]K

1

2

3

8 2 0 362 8 2 40 2 4 60

i ii i i

i i

Δ −⎡ ⎤ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥ Δ = −⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥ Δ −⎣ ⎦ ⎩ ⎭ ⎩ ⎭

By solving these equations simultaneously, we determine the joint displacements to be

1

2

3

5.84815.385

17.693i

Δ −⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪Δ =⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪Δ −⎩ ⎭ ⎩ ⎭

(7) Member end displacements { }Δ ⓔ and forces { } F ⓔ

The member end displacements in global coordinates, { }Δ ⓔ , can be obtained by simply placing the corresponding components, whose positions are oriented by { }λ ⓔ , of { }Δ to the proper positions in

. Then, the end displacement vectors are as follows. { }Δ ⓔ

{ }0 1

5.848 i⎧ ⎫

Δ = ⎨ ⎬−⎩ ⎭

① { }

5.848 15.385 i−⎧ ⎫

Δ = ⎨ ⎬⎩ ⎭

② { }

5.385 117.693 i

⎧ ⎫Δ = ⎨ ⎬−⎩ ⎭

③

Because of { } { }F F=ⓔ ⓔ , by applying equation (13-45), the member end forces for the elements of the continuous beam can be determined as follows:

{ }4 2 0 100 111.7112 4 5.848 100 76.62i i

Fi i i

− −⎡ ⎤ ⎧ ⎫ ⎧ ⎫ ⎧= + =

⎫⎨ ⎬ ⎨ ⎬ ⎨⎢ ⎥ − +⎣ ⎦ ⎩ ⎭ ⎩ ⎭ ⎩

①

⎬⎭

{ }4 2 5.848 64 76.6212 4 5.385 64 73.84i i

Fi i i

− − −⎡ ⎤ ⎧ ⎫ ⎧ ⎫ ⎧= + =

⎫⎨ ⎬ ⎨ ⎬ ⎨⎢ ⎥ +⎣ ⎦ ⎩ ⎭ ⎩ ⎭ ⎩

②

⎬⎭

{ }4 2 5.385 60 73.8412 4 17.693 60 0i i

Fi i i

−⎡ ⎤ ⎧ ⎫ ⎧ ⎫ ⎧= +

⎫=⎨ ⎬ ⎨ ⎬ ⎨⎢ ⎥ − +⎣ ⎦ ⎩ ⎭ ⎩ ⎭ ⎩

③

⎬⎭

(8) Internal force diagrams


Based on the member end forces, the bending moment and shear force diagrams are drawn as shown in Fig.13.20.

76.62

105.84

111.70

83.08

Example 13-5

Determine the internal forces for the rigid frame shown in Fig.13.21 (a) by using matrix displacement method. Assume the cross-section of each member is rectangular; the elastic modulus . Their areas and inertial moments are also shown in the figure.

1E =

Solution

(1) Unknown joint displacements From the computing model shown in Fig.13.21 (b), we observe that while joints C and D of the structure

can neither translate nor rotate, joints A and B are free to translate as well as rotate. Thus each of the two joints has three unknown displacements, translations in x and y directions, respectively, and rotation about z. The total unknown joint displacements of the rigid frame are six; the displacement numbers of joint A are indicated by [1 2 3], joint B by [4 5 6], and joints C and D by {0}, as shown in Fig.13.21 (b).

12m

6m

1k

N/m

q=

A B

C D

21 0.5mA =

41

1 m24

I =

21 0.5mA =

41

1 m24

I =

22 0.63mA =

42

1 m12

I =

(a)

A B

C D

{ }0{ }0y

x①

②

③

(b)

1Δ

2Δ

3Δ

4Δ

5Δ

6Δ

Fig.13.21 Figure of example 13-5(a) original structure; (b) computing model

20.77

73.84

47.65

54.38 39.23

45.62

48.35

20.77

diagram (kN m)M ⋅ diagram (kN)Q(a) (b)

Fig.13.20 Internal force diagrams of example 13-4


[ ]k ⓔ (2) Element stiffness matrix in local coordinates The pertinent quantities are calculated as follows:

31

1

2 13.9 10EIl

−= ×31

1

83.3 10EAl

−= × 3

1

1 1094.6 −×=l

EI Column:

31

1

4 27.8 10EIl

−= × 32

1

1 1094.66 −×=lEI 3

31

1 1031.212 −×=lEI

32

2

2 13.9 10EIl

−= ×32

2

52.5 10EAl

−= × 32

2

6.94 10EIl

−= × Beam:

3222

6 3.47 10EIl

−= × 3232

12 0.58 10EIl

−= ×3

2

2 108.274 −×=lEI

Elements and ③ : ①

3

83.3 0 0 83.3 0 00 2.31 6.94 0 2.31 6.940 6.94 27.8 0 6.94 13.9

1083.3 0 0 83.3 0 00 2.31 6.94 0 2.31 6.940 6.94 13.9 0 6.94 27.8

k k −

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−

⎡ ⎤ ⎡ ⎤= = ×⎢ ⎥⎣ ⎦ ⎣ ⎦ −⎢ ⎥⎢ ⎥− − −⎢ ⎥

−⎢ ⎥⎣ ⎦

① ③

Elements : ②

3

52.5 0 0 52.5 0 00 0.58 3.47 0 0.58 3.470 3.47 27.8 0 3.47 13.9

1052.5 0 0 52.5 0 00 0.58 3.47 0 0.58 3.470 3.47 13.9 0 3.47 27.8

k −

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−

⎡ ⎤ = ×⎢ ⎥⎣ ⎦ −⎢ ⎥⎢ ⎥− − −⎢ ⎥

−⎢ ⎥⎣ ⎦

②

(3) Element stiffness matrix in global coordinates [ ]k ⓔ

x 90α = ° axis of element (or ③ ) and x axis is ①Since the angle between , the direction cosines of the two elements are as follows

cos 0α = sin 1α =


Substitution of these values into equation (13-20) yields the transformation matrix [ for the element:

]T

[ ]

[ ]

[ ]

0 1 01 0 0 0

0 0 10 1 0

0 1 00 0 1

T

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥

= ⎢⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

0

⎥

To determine the two element stiffness matrices in global coordinates, , we substitute the matrix [ ]k ⓔ

[ ]k ⓔ and the into the relationship [ ]T [ ] [ ] [ ] [ ]Tk T k T=ⓔ ⓔ and carry out the necessary matrix multiplications to obtain

[ ] [ ] [ ] [ ] 3

2.31 0 6.94 2.31 0 6.940 83.3 0 0 83.3 0

6.94 0 27.8 6.94 0 13.9 102.31 0 6.94 2.31 0 6.940 83.3 0 0 83.3 0

6.94 0 13.9 6.94 0 27.8

Tk k T k T −

− − −⎡ ⎤⎢ ⎥−⎢ ⎥−⎡ ⎤ ⎢ ⎥= = = ×⎣ ⎦ −⎢ ⎥

−⎢ ⎥⎢ ⎥−⎣ ⎦

①① ③

x y x yBecause the local coordinate for element coincides with the global ② coordinate, no coordinate transformations are needed, or [ ] [ ]T I= ; i.e., the element stiffness relations in the local and global coordinates are the same. That is,

[ ]k k⎡ ⎤= ⎣ ⎦②②

(4) Assemble global stiffness matrix by element assembly method [ ]KFrom the computing model shown in Fig.13.21 (b), the orientation vector, reflecting the relations

between the local and global numbers of each element, can be written as follows:

{ } { }

123000

λ

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

①

123456

λ

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

② { }

456000

λ

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

③


By using the orientation vectors, { }λ ⓔ , the relevant elements of can be added into their positions in the global stiffness matrix . Thusly assembled will be as follows

[ ]k ⓔ

[ ]K [ ]K

[ ] 3

54.81 0 6.94 52.5 0 00 83.88 3.47 0 0.58 3.47

6.94 3.47 55.6 0 3.47 13.9 1052.5 0 0 54.81 0 6.940 0.58 3.47 0 83.88 3.470 3.47 13.9 6.94 3.47 55.6

K −

− −⎡ ⎤⎢ ⎥−⎢ ⎥− −⎢ ⎥= ×− −⎢ ⎥

− − −⎢ ⎥⎢ ⎥− −⎣ ⎦


{ }PF ①Referring to table 13-3, the fixed-end forces for element , only the element subjected to load, will be obtained as follows.

①

{ } { } [0 3 3 0 3 3 TE PP F= − = − − −① ① ]

x 90α = ° axis of element ① and x axis is Since the angle between , by using Eqs. (13-23) and (13-37) we write the element’s equivalent joint loads as { }EP ①

{ } [ ] { }[ ]

[ ]

0 1 0 0 31 0 0 0 3 0

0 0 1 3 30 1 0 0 3

0 1 0 0 3 00 0 1 3 3

TE EP T P

⎡ ⎤ ⎧ ⎫ ⎧ ⎫⎢ ⎥ ⎪ ⎪ ⎪ ⎪− −⎢ ⎥ ⎪ ⎪ ⎪ ⎪− −⎪ ⎪ ⎪ ⎪⎢ ⎥= = − =⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎪ ⎪ ⎪ ⎪

− −⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎣ ⎦

①①

(6) Equivalent nodal load vector { } of the structure P{ } {0}JP =From the Fig.13.21, we can see that there is no joint load, i.e., . The equivalent joint loads

of the structure can be obtained by adding the relevant elements of { into their proper positions determined by { }EP }EP ①

{ }λ ① in { . By Eq. (13-44), accumulate the corresponding components of }EP{ }JP and { together to form equivalent nodal load vector { of the structure. The final { } will be

}P P}EP

{ } [ ]3 0 3 0 0 0 TP = −



1

2

3 3

4

5

6

54.81 0 6.94 52.5 0 0 30 83.88 3.47 0 0.58 3.47 0

6.94 3.47 55.6 0 3.47 13.9 310 52.5 0 0 54.81 0 6.94 00 0.58 3.47 0 83.88 3.47 00 3.47 13.9 6.94 3.47 55.6 0

−

Δ− − ⎧ ⎫⎡ ⎤ ⎧ ⎫⎪ ⎪⎢ ⎥ ⎪ ⎪Δ−⎪ ⎪⎢ ⎥ ⎪ ⎪− − Δ −⎪ ⎪ ⎪ ⎪⎢ ⎥× =⎨ ⎬ ⎨ ⎬− − Δ⎢ ⎥ ⎪ ⎪ ⎪ ⎪

− − − Δ⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥− − ⎪ ⎪ ⎪Δ ⎩ ⎭⎣ ⎦ ⎩ ⎭ ⎪


1

2

3

4

5

6

8475.13

28.4

8245.1396.5

A

A

A

B

B

B

uv

uv

θ

θ

Δ⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪Δ −⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪Δ⎪ ⎪ ⎪ ⎪ ⎪ ⎪= =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪Δ⎪ ⎪ ⎪ ⎪ ⎪ ⎪Δ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪Δ ⎩ ⎭⎩ ⎭⎩ ⎭

{ }F ⓔ{ }Δ ⓔ(8) Member end displacements and forces in local coordinates { }Δ ⓔThe member end displacements in global coordinate, , can be obtained by simply placing the

corresponding components, whose positions are oriented by { } { }Δλ ⓔ , of to the proper positions in . Then, the end displacement vectors are as follows. { }Δ ⓔ

{ }

8475.13

28.4

000

⎧ ⎫⎪ ⎪−⎪ ⎪⎪ ⎪⎪ ⎪Δ = ⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

① { }

8475.13

28.4

8245.1396.5

⎧ ⎫⎪ ⎪−⎪ ⎪⎪ ⎪⎪ ⎪Δ = ⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

② { }

8245.1396.5

000

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪Δ = ⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

③

{ } [ ]{ }F T F=ⓔBecause of and { } [ ] { } { }PF k F= Δ +ⓔ ⓔ ⓔ ⓔⓔ , the member end forces for the elements of the portal frame can be determined as follows:

Element : ①

Member end force in global coordinate system:


{ } [ ] { } { }

3

2.31 0 6.94 2.31 0 6.94 847 30 83.3 0 0 83.3 0 5.13 0

6.94 0 27.8 6.94 0 13.9 28.4 3 10 2.31 0 6.94 2.31 0 6.94 0 30 83.3 0 0 83.3 0 0 0

6.94 0 13.9 6.94 0 27.8 0 3

PF k F

−

= Δ +− − − −⎡ ⎤ ⎧ ⎫ ⎧

⎢ ⎥ ⎪ ⎪ ⎪− −⎢ ⎥ ⎪ ⎪ ⎪− ⎪ ⎪ ⎪⎢ ⎥= × +⎨ ⎬ ⎨− −⎢ ⎥ ⎪ ⎪ ⎪

−⎢ ⎥ ⎪ ⎪⎢ ⎥− −⎪ ⎪⎩ ⎭ ⎩⎣ ⎦

①① ① ①

1.240.432.094.76

0.438.49

−⎫ ⎧ ⎫⎪ ⎪ ⎪−⎪ ⎪ ⎪−⎪ ⎪ ⎪=⎬ ⎨ ⎬−⎪ ⎪ ⎪

⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪⎭ ⎩ ⎭

Member end force in local coordinate system:

{ } [ ]{ } [ ]0.43 1.24 2.09 0.43 4.76 8.49 TF T F= = − − −① ①

Element : ②


{ } [ ] { } { }

3

52.5 0 0 52.5 0 0 8470 0.58 3.47 0 0.58 3.470 3.47 27.8 0 3.47 13.9 10 52.5 0 0 52.5 0 00 0.58 3.47 0 0.58 3.470 3.47 13.9 0 3.47 27.8

PF k F

−

= Δ +−⎡ ⎤

⎢ ⎥− −⎢ ⎥−⎢ ⎥= × −⎢ ⎥

− − −⎢ ⎥⎢ ⎥−⎣ ⎦

②② ② ②

0 1.245.13 0 0.43

28.4 0 2.09847 0 1.245.13 0 0.4396.5 0 3.04

⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪+ =⎨ ⎬ ⎨ ⎬ ⎨ ⎬−⎪ ⎪ ⎪ ⎪ ⎪ ⎪

−⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭ ⎩

⎭

{ } { }F F=②0α = ° ②, . The member end force in local coordinate system will be Since

{ } [ ]1.24 0.43 2.09 1.24 0.43 3.04 TF = − −②

Element : ③


{ } [ ] { } { }

3

2.31 0 6.94 2.31 0 6.94 8240 83.3 0 0 83.3 0 5.13

6.94 0 27.8 6.94 0 13.9 96.5 10 2.31 0 6.94 2.31 0 6.94 00 83.3 0 0 83.3 0 0

6.94 0 13.9 6.94 0 27.8 0

PF k F

−

= Δ +− − −⎡ ⎤ ⎧ ⎫

⎢ ⎥ ⎪ ⎪−⎢ ⎥ ⎪ ⎪− ⎪ ⎪⎢ ⎥= × +⎨ ⎬−⎢ ⎥ ⎪ ⎪

−⎢ ⎥ ⎪ ⎪⎢ ⎥− ⎪ ⎪⎩ ⎭⎣ ⎦

③③ ③ ③

0 1.240 0.430 3.00 1.20 0.40 4.3

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬−⎪ ⎪ ⎪ ⎪

−⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

4438


{ } [ ]{ } [ ]0.43 1.24 3.04 0.43 1.24 4.38 TF T F= = − − − −③ ③

13.8 Global Analysis of Rigid Frames without Considering Axial Deformation 549

(9) Internal force diagrams Based on the member end forces, the bending moment, and shear and axial force diagrams are drawn,

respectively, as shown in Fig.13.22.

13.8 Global Analysis of Rigid Frames without Considering Axial Deformation

In practical engineering, the rectangular shaped rigid frames composed of columns and beams have a variety of application. The axial deformations of the members of this sort of rigid frames are quite small in comparison with other deformations. In order to simplify the analysis, the member’s axial deformations of the rigid frames are often neglected. The section will discuss the matrix analysis for the plane rigid frames without considering the axial deformations of their members.

Consider the portal rigid frame shown in Fig.13.23. Following aspects must be concerned when analyze the frame by neglecting its member’s axial deformations.

(1) Nodal global displacement numbers From the frame shown in Fig.13.23, we can see that joints B and D are both attached to the fixed

supports, they can neither translate nor rotate; each of them has three known displacement components, the translations in x and y directions, and the rotation about z axis. They are all equal to zero, so the nodal

DC

A B

4.76

0.431.24

1.24B

CA

D4.38

2.09 3.04

8.49

(a) (b)

(c)

diagram (kN m)M

⋅

diagram (kN)

Q

diagram (kN)N

0.43

A B

C

0.43

1.24

D



global displacement numbers for each of the two joints must be [0 0 0]. Since the member’s axial deformations are neglected, the vertical displacements at nodes A , and must be equal to zero, then the corresponding global displacement numbers must be zero as well; in addition, the horizontal displacements at nodes A , and must be identical, thusly the corresponding global displacement numbers must be the same. Consequently, the global displacement numbers for node A are represented by [1 0 2]; the numbers for node by [1 0 3] and for by [1 0 4], as shown in the figure.

1C 2C

1C 2C

1C 2C(2) Element orientation vectors The rigid frame shown in Fig.13.23 has three elements, numbered by , and . The arrows

on the members indicate the positive directions of the abscissas ① ② ③

x of their local coordinate systems. For each element, six numbers are used to identify the element’s global end displacement component numbers, which consist of so called orientation vector. The orientation vectors for the three elements are thusly

written as follows.

{ } [ ]1 0 2 1 0 3 Tλ =①

{ } [ ]1 0 2 0 0 0 Tλ =②

{ } [ ]1 0 4 0 0 0 Tλ =③

(3) Assembly of element stiffness matrices and development of global stiffness matrix [ ]KAssume the member’s size and materials of the frame are the same as those of the frame in example

13-1. Thusly, the global stiffness matrix of element has been calculated in example 13-1. We write it again as follows

①

A

B D

2C1C

0

y

00

x

00

0

0 0

0

①

② ③

1Δ

3Δ2Δ

1Δ

1Δ

Δ4

Fig.13.23 The computing model of a rectangular shaped rigid frame


[ ]4

(1) (2) (3) (4) (5) (6)(1) 300 0 0 300 0 0(2) 0 12 30 0 12 30

10(3) 0 30 100 0 30 50(4) 300 0 0 300 0 0(5) 0 12 30 0 12 30(6) 0 30 50 0 30 100

k

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥×−

= ⎢ ⎥−⎢ ⎥⎢ ⎥− − −⎢ ⎥

−⎣ ⎦

①

Based on the orientation vector and relationships between the end local displacement numbers and nodal global displacement numbers, place sequentially the elements of

to the relevant positions in matrix to form the first-stage result of global stiffness matrix of the frame as follows

{ } [1 0 2 1 0 3]Tλ =①

[ ]K[ ]k ①

[ ] 4

(1)(4) (3) (6) 1 2 3 4

(1)(4)1 300 300 300 300 0 0 0 0 0

(3) 2 0 0 100 50 010

(6) 3 0 0 50 100 04 0 0 0 0

K

− − + + +⎡ ⎤⎢ ⎥+⎢ ⎥= ×⎢ ⎥+⎢ ⎥⎣ ⎦

Note that the nodal global displacement component numbers corresponding to the element’s local numbers (2) and (5) are zero, so there are no positions in global stiffness matrix for the elements in row (2) and column (2), in row (5) and column (5) of .

[ ]K[ ]k ①

The global stiffness matrix of element has been also obtained in example 13-1. We write it again as follows

②

[ ]4

(1) (2) (3) (4) (5) (6)(1) 12 0 30 12 0 30(2) 0 300 0 0 300 0

10(3) 30 0 100 30 0 50(4) 12 0 30 12 0 30(5) 0 300 0 0 300 0(6) 30 0 50 30 0 100

k

− − −⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥×−

= ⎢ ⎥−⎢ ⎥⎢ ⎥−⎢ ⎥−⎣ ⎦

②

Based on the orientation vector and relationships between the end { } [1 0 2 0 0 0]Tλ =②


local displacement numbers and nodal global displacement numbers, add sequentially the elements of to the pertinent positions in matrix [ , which, at this time, has already added the stiffness

coefficients of element , to form the second-stage result of global stiffness matrix of the frame as follows

]K[ ]k ②

①

[ ] 4

(1) (3) 1 2 3 4

(1) 1 0 12 0 30 0 0

(3) 2 0 30 100 100 50 010

3 0 50 100 04 0 0 0 0

K

+ −⎡ ⎤⎢ − +⎢ ⎥= ×⎢ ⎥⎢ ⎥⎣ ⎦

⎥

⎥

Note that the nodal global displacement component numbers corresponding to the element’s local numbers (2), (4), (5) and (6) are zero, so there are no positions in global stiffness matrix for the elements in row (2) and column (2), in row (4) and column (4), in row (5) and column (5), in row (6) and column (6) of .

[ ]K

[ ]k ②

Since the global stiffness matrix of element is equal to that of element , i.e., , by orientation vector { and relationships between the end local displacement numbers and nodal global displacement numbers, add sequentially the elements of to the proper positions in matrix , which, at this time, has already added the stiffness coefficients of elements and ② , to form the final result of global stiffness matrix of the frame as follows

[ ] [ ]k k=③ ②③ ②

} [1 0 4 0 0 0]Tλ =③

[ ]k ③

[ ]K①

[ ] 4

(1) (3) 1 2 3 4

(1) 1 12 12 30 0 0 30

2 30 200 50 010

3 0 50 100 0(3) 4 0 30 0 0 0 100

K

+ − −⎡ ⎤⎢ −⎢ ⎥= ×⎢ ⎥⎢ ⎥− +⎣ ⎦

Note that the nodal global displacement component numbers corresponding to the element’s local numbers (2), (4), (5) and (6) are zero, so there are no positions in global stiffness matrix for the elements in row (2) and column (2), in row (4) and column (4), in row (5) and column (5), in row (6) and column (6) of .

[ ]K

[ ]k ③

Example 13-6

Determine the internal forces for the rigid frame shown in Fig.13.21 by using matrix displacement


method. Neglect the member’s axial deformations and assume the member’s size and materials are the same as those used in example 13-15.

Solution

(1) Unknown joint displacements From the computing model shown in

Fig.13.24, if we neglect the member’s axial deformations the total unknown joint displacements of the rigid frame will become three, the translation of node A or B in x direction, and the rotations of nodes A and B about z axis, respectively; thus the displacement numbers of joint A are indicated by [1 0 2], joint B by [1 0 3], and joints C and D by {0}, as shown in Fig.13.24.

(2) Element stiffness matrix in local coordinates [ ]k ⓔ [ ]k ⓔ is the same as that of example 13-5. (3) Element stiffness matrix in global coordinates [ ]k ⓔ

[ ]k ⓔ is the same as that of example 13-5. (4) Assemble global stiffness matrix by element assembly method [ ]KFrom the computing model shown in Fig.13.24, the orientation vector of each element can be written

as follows:

{ } [ ]1 0 2 0 0 0 Tλ =①

{ } [ ]1 0 2 1 0 3 Tλ =②

{ } [ ]1 0 3 0 0 0 Tλ =③

By using the orientation vectors, { }λ ⓔ , the relevant elements of can be sequentially added into their proper positions in the global stiffness matrix . Thusly assembled will be as follows

[ ]k ⓔ

[ ]K [ ]K

[ ] 34.62 6.94 6.946.94 55.6 13.9 106.94 13.9 55.6

K −− −⎡ ⎤

⎢= − ×⎢ ⎥−⎣ ⎦

⎥


{ }EP ⓔ is the same as that of example 13-5.

A B

C D

{ }0{ }0

yx

Δ2

①

②

③

1Δ

3Δ

1Δ

Fig.13.24 The computing model of example 13-6


(6) Equivalent nodal load vector { of the structure }PThe equivalent joint loads of the structure can be obtained by adding the relevant elements of

(since only element ① subjected to load) into their proper positions determined by in { . Because of

{ }EP{ }EP ①

{ } [1 0 2 0 0 0]Tλ =① }EP { } {0}JP = , by Eq. (13-44), the final will be

{ }P

{ } [ ]3 3 0 TP = −


13

2

3

4.62 6.94 6.94 310 6.94 55.6 13.9 3

6.94 13.9 55.6 0

−Δ− − ⎧ ⎫⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥× − Δ = −⎨ ⎬ ⎨ ⎬

⎢ ⎥ ⎪ ⎪ ⎪ ⎪− Δ⎣ ⎦ ⎩ ⎭

⎩ ⎭


13

2

3

0.8380.0261 100.0979

A

A

B

uθθ

Δ⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪Δ = = ×⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪Δ ⎩ ⎭⎩ ⎭ ⎩ ⎭

{ }F ⓔ{ }Δ ⓔ and forces in local coordinates (8) Member end displacements { }Δ ⓔThe member end displacements in global coordinates, , can be obtained by simply placing the

corresponding components, whose positions are oriented by { } , of { }Δλ ⓔ to the proper positions in . Then, the end displacement vectors are as follows. { }Δ ⓔ

{ }

0.8380

0.0261000

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪

Δ = ⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

① { }

0.8380

0.02610.838

00.0979

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪

Δ = ⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

② { }

0.8380

0.0979000

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪

Δ = ⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

③

{ } [ ]{ }F T F=ⓔBecause of { } and [ ] { } { }PF k F= Δ +ⓔ ⓔ ⓔ ⓔⓔ , the member end forces for the elements of the rigid frame can be determined as follows:

Element : ①



{ } [ ] { } { }2.31 0 6.94 2.31 0 6.94 0.838 3

0 83.3 0 0 83.3 0 0 06.94 0 27.8 6.94 0 13.9 0.0261 3 2.31 0 6.94 2.31 0 6.94 00 83.3 0 0 83.3 0 0

6.94 0 13.9 6.94 0 27.8 0

PF k F= Δ +− − − −⎡ ⎤ ⎧ ⎫

⎢ ⎥ ⎪ ⎪−⎢ ⎥ ⎪ ⎪− ⎪ ⎪⎢ ⎥= +⎨ ⎬−⎢ ⎥ ⎪ ⎪

−⎢ ⎥ ⎪ ⎪⎢ ⎥− ⎪ ⎪⎩ ⎭⎣ ⎦

①① ① ①

1.250

2.093 4.7

0 03 8.4

−

5

1

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬− −⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪− −⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭


{ } [ ]{ } [ ]0 1.25 2.09 0 4.75 8.41 TF T F= = − −① ①

Element : ②


{ } [ ] { } { }52.5 0 0 52.5 0 0 0.8380 0.58 3.47 0 0.58 3.47 00 3.47 27.8 0 3.47 13.9 0.0261 52.5 0 0 52.5 0 0 0.8380 0.58 3.47 0 0.58 3.47 00 3.47 13.9 0 3.47 27.8

PF k F= Δ +−⎡ ⎤

⎢ ⎥−⎢ ⎥−⎢ ⎥= −⎢ ⎥

− − −⎢ ⎥⎢ ⎥−⎣ ⎦

②② ② ②

0 00 0.430 2.090 00 0.4

0.0979 0 3.093

⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪+ =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪

−⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭ ⎩

⎭

{ } { }F F=②0α = ° ②Since , . The member end force in local coordinate system will be

{ } [ ]0 0.43 2.09 0 0.43 3.09 TF = −②

Element ③ : Member end force in global coordinate system:

{ } [ ] { } { }2.31 0 6.94 2.31 0 6.94 0.838

0 83.3 0 0 83.3 0 06.94 0 27.8 6.94 0 13.9 0.0979 2.31 0 6.94 2.31 0 6.94 0 0 83.3 0 0 83.3 0 06.94 0 13.9 6.94 0 27.8 0

PF k F= Δ +− − −⎡ ⎤ ⎧ ⎫

⎢ ⎥ ⎪ ⎪−⎢ ⎥ ⎪ ⎪− ⎪ ⎪⎢ ⎥= ⎨ ⎬−⎢ ⎥ ⎪ ⎪

−⎢ ⎥ ⎪ ⎪⎢ ⎥− ⎪⎩ ⎭⎣ ⎦

③③ ③ ③

0 1.250 50 3.00 1.20 00 4.4

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪+ =⎨ ⎬ ⎨ ⎬−⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪

−

95

7⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭



{ } [ ]{ } [ ]0 1.25 3.09 0 1.25 4.47 TF T F= = − − −③ ③

(9) Internal force diagrams Based on the member end forces, the bending moment, shear and axial force diagrams are drawn,

respectively, as shown in Fig.13.25. in comparison with the diagrams shown in Fig.13.22, it is observed that the effect on the internal forces due to the member’s axial deformations are small enough to be neglected. In addition, since the member’s axial deformations are assumed to be zero, the axial force determined by the stiffness equations are zero as well. Then, the member’s axial forces must be identified by the equilibrium equations derived from shear force diagram. Consequently, the internal force diagrams for the rigid frame are shown in Fig.13.25.

13.9 Block Diagram and Computer Programs for Plane Framed Structures

Based on forgoing discussion about the basic principle and procedure of matrix displacement method, a block diagram summarizing the various steps involved in the method and the computer programs, edited in the Fortran 90 language, for the analysis of plane framed structures, such as rigid frames, continuous beams, trusses, bent frames and composite structures are presented as follows.

DC

A B

4.75

0.431.25

1.25B

CA

D4.47

2.09 3.09

8.41

(a) (b)

(c)

diagram (kN m)

M⋅

diagram (kN)

Q

diagram (kN)N

0.43

A B

C

0.43

1.25

D


13.9 Block Diagram and Computer Programs for Plane Framed Structures 557

13.9.1 Design of the block diagram

For each element: (1) evaluate direction cosines [Subroutine SCL]; (2) determine element’s orientation vector { )λ ⓔ [Subroutine

EJC]; (3) Set up element’s coordinate transformation matrix [Subroutine CTM]; (4) yield element stiffness matrix in local coordinates [ ]k ⓔ [Subroutine ESM]; (5) calculate element stiffness matrix in global

coordinates, [ ] [ ] [ ] [ ]Tk T k T=ⓔ ⓔ , and store [ ]k ⓔ into global stiffness matrix [K] by { )λ ⓔ

2

Form global nodal load vector { } { } { }J EP P P= + [Subroutine nlv]

3

Solve structural stiffness equation {P}=[K]{Δ} [Subroutine BGS] to determine {Δ}

4

For each member: (1) obtain { }Δ ⓔ from { }Δ ; (2) calculate { } [ ]{ }TΔ = Δⓔ ⓔ ; (3) evaluate { } [ ] { } { }PF k F= Δ +ⓔ ⓔ ⓔ ⓔ

[“Subroutine MQN” and “Subroutine EFX(w)”

5

Name of the program

Develop a module to define the types of variables to digitalize a structure [“module variable_type_module” and “Subroutine input_data”]

1

13.9.2 Computer program for plane framed structure

The program consists of two portions, the main program and sub program composed by modules. The main program shows how to inform computer to read and store data, and how to call relevant subroutine programs. The further explanation for each module of the sub program will be taken place in latter subsections.

13.9.2.1 Main program for plane framed frames

!======================================================================! PROGRAM plane_frame_analysis_programme use Frame_analysis_sub


implicit none character*60,dimension(11)::FORM print *,'input file name ' read *, input_file_name OPEN (5,file=input_file_name) print *,'output file name ' read *, output_file_name OPEN (6,FILE=output_file_name) CALL input_data CALL nlv CALL ssm CALL BGS CALL MQN CLOSE (5) CLOSE (6) STOP END PROGRAM plane_frame_analysis_programme !======================================================================!

13.9.2.2 Subroutine analytical programs for plane framed structures

!===================================================================! module variable_type_module implicit none integer(kind(1)),parameter::Ikind=kind(1),Rkind=kind(0.d0) integer(Ikind)::NE,NJ,N,NW,NPJ,NPF,J1,J2,I,J,I1,M,K,L,IND,JJ real(Rkind)::BL,SI,CO,S,C,G,G1,G2,G3,DX,DY,A,B,Q integer(Ikind),allocatable::JE(:,:),JN(:,:) integer(Ikind),dimension(:)::JC(6) real(Rkind),allocatable::EA(:),EI(:),X(:),Y(:),P(:)


real(Rkind),allocatable::PJ(:,:),PF(:,:),KB(:,:) real(Rkind),dimension(:)::KE(6,6),KD(6,6),T(6,6),F(6),F0(6),D(6) character*20::input_file_name,output_file_name end module variable_type_module !------------------------------------------------------------------------------------------------------------------! module Frame_analysis_sub use variable_type_module contains !-------------------------- input original data ---------------------------------------------------! Subroutine input_data read (5,*) NE,NJ,N,NW,NPJ,NPF allocate (JE(2,NE),JN(3,NJ),P(N),PF(4,NPF)) allocate (EA(NE),EI(NE),X(NJ),Y(NJ),PJ(2,NPJ),KB(N,NW)) read (5,*) (X(J),Y(J),(JN(I,J),I=1,3),J=1,NJ) read (5,*) ((JE(I,J),I=1,2),EA(J),EI(J),J=1,NE) IF (NPJ.NE.0) READ (5,*) ((PJ(I,J),I=1,2),J=1,NPJ) IF (NPF.NE.0) READ (5,*) ((PF(I,J),I=1,4),J=1,NPF) WRITE (6,'(/6X,"NE=",I5,2X,"NJ=",I5,2X,"N=",I5,2X,"NW=",I5,2X,& "NPJ=",I5,2X,"NPF=",I5)') NE,NJ,N,NW,NPJ,NPF WRITE (6,'(/7X,"NODE",7X,"X",11X,"Y",12X,"XX",8X,"YY",8X,"ZZ"/)') WRITE (6,'(1X,I10,2F12.4,3I10)') (J,X(J),Y(J),(JN(I,J),I=1,3),J=1,NJ) WRITE (6,'(/4X,"ELEMENT",4X,"NODE-I",4X,"NODE-J",11X,"EA",13X,"EI"/)') WRITE (6,'(1X,3I10,2D15.6)') (J,(JE(I,J),I=1,2),EA(J),EI(J),J=1,NE) IF (NPJ.NE.0) WRITE (6,'(/7X,"CODE",7X,"PX-PY-PM"/(1X,F10.0,F15.4))')& ((PJ(I,J),I=1,2),J=1,NPJ)


IF (NPF.NE.0) WRITE (6,'(/4X,"ELEMENT",7X,"IND",10X,"A",14X,"Q"/& (1X,2F10.0,2F15.4))') ((PF(I,J),I=1,4),J=1,NPF) end subroutine input_data !---------------------------------------------------------------------------------------------------! !------------- Calculation of direction cosines of elements -----------------------------! Subroutine SCL J1=JE(1,M) J2=JE(2,M) DX=X(J2)-X(J1) DY=Y(J2)-Y(J1) BL=SQRT(DX*DX+DY*DY) SI=DY/BL CO=DX/BL end Subroutine SCL !----------------------------------------------------------------------------------------------------! !------------- Set up element’s orientation vectors ----------------------------------------! Subroutine EJC J1=JE(1,M) J2=JE(2,M) DO I=1,3 JC(I)=JN(I,J1) JC(I+3)=JN(I,J2) ENDDO END Subroutine EJC !----------------------------------------------------------------------------------------------------! !------------- Set up element’s coordinate transformation matrix [T]-------------------! Subroutine CTM


T=0.D0 T(1,1)=CO T(1,2)=SI T(2,1)=-SI T(2,2)=CO T(3,3)=1.D0

DO I=1,3 DO J=1,3 T(I+3,J+3)=T(I,J) ENDDO ENDDO END Subroutine CTM

!----------------------------------------------------------------------------------------------------! !----------------------- Set up element stiffness matrix --------------------------------------!

Subroutine ESM

G=EA(M)/BL G1=2.D0*EI(M)/BL G2=3.D0*G1/BL G3=2.D0*G2/BL

KD=0.D0 KD(1,1)=G KD(1,4)=-G KD(4,4)=G KD(2,2)=G3 KD(2,3)=-G2 KD(5,5)=G3 KD(2,5)=-G3 KD(2,6)=-G2 KD(3,3)=2.D0*G1 KD(3,5)=G2


KD(3,6)=G1 KD(5,6)=G2 KD(6,6)=2.D0*G1 DO I=1,5 I1=I+1 DO J=I1,6 KD(J,I)=KD(I,J) END DO END DO END subroutine ESM !----------------------------------------------------------------------------------------------------! !---------------------- Assemble structural stiffness matrix ---------------------------------! Subroutine ssm KB=0.D0 DO M=1,NE

call SCL call CTM call ESM call EJC KE=matmul(transpose(T),matmul(KD,T)) DO L=1,6 I=JC(L) IF (I.NE.0) THEN DO K=1,6 J=JC(K) IF (J.NE.0.AND.J.GE.I) THEN JJ=J-I+1


KB(I,JJ)=KB(I,JJ)+KE(L,K) END IF END DO END IF END DO

END DO

END subroutine ssm

!----------------------------------------------------------------------------------------------------! !-------------------------- Set up global nodal load vector ------------------------------------!

Subroutine nlv P=0.D0

!---------------------calculate joint load vector IF (NPJ.NE.0) THEN DO I=1,NPJ L=INT(PJ(1,I)) P(L)=PJ(2,I) ENDDO ENDIF

!------------calculate non-joint node load vector

IF (NPF.NE.0) THEN DO JJ=1,NPF M=INT(PF(1,JJ)) CALL SCL CALL EFX(JJ) CALL CTM CALL EJC DO L=1,6 S=0.D0 DO K=1,6 S=S-T(K,L)*F0(K)


END DO F(L)=S END DO DO J=1,6 L=JC(J) IF (L.NE.0) P(L)=P(L)+F(J) END DO END DO END IF END Subroutine nlv !----------------------------------------------------------------------------------------------------! !-------------------------------Solve structural stiffness equation {P}=[K]{Δ}-----------! Subroutine BGS N1=N-1 !------------------------------------------------loop of elimination DO K=1,N1 IM=K+NW-1 IF(N.LT.IM) IM=N ! the maximum number of eliminating row, IM I1=K+1 !---------------------------------------- loop of eliminating row DO I=I1,IM L=I-K+1 C=KB(K,L)/KB(K,1) JM=NW-L+1 ! find the maximum column number of eliminating row JM !----------------------------- column’s loop of eliminating row DO J=1,JM JJ=J+I-K KB(I,J)=KB(I,J)-C*KB(K,JJ) END DO !----------------------------- column’s loop of eliminating row


P(I)=P(I)-C*P(K) END DO

!---------------------------------------- loop of eliminating row END DO

!------------------------------------------------loop of elimination !----------------------------------------------------------------------------------------------------!

P(N)=P(N)/KB(N,1) ! find nth displacement !------------------------------------------------back substitution !---------------------------------------------loop of back-substitution row

DO K=1,N1 I=N-K JM=K+1

IF (NW.LT.JM) JM=NW ! find the maximum column number of back-substitution row JM !----------------------------------------- column’s loop of back-substitution row

DO J=2,JM L=J+I-1 P(I)=P(I)-KB(I,J)*P(L) END DO

!----------------------------------------- column’s loop of back-substitution row P(I)=P(I)/KB(I,1) ! find ith displacement END DO

!---------------------------------------------loop of back-substitution row !------------------------------------------------back substitution

WRITE (6,'(/7X,"NODE",10X,"U",14X,"V",11X,"CETA")') DO I=1,NJ DO J=1,3 D(J)=0.D0 L=JN(J,I) IF (L.NE.0) THEN D(J)=P(L) END IF END DO WRITE (6,'(1X,I10,3D15.6)') I,D(1),D(2),D(3) ! output nodal displacements


END DO END subroutine BGS !----------------------------------------------------------------------------------------------------! !------------------ Calculate element’s end forces due to its displacements-------------! Subroutine MQN WRITE (6,'(/4X,"ELEMENT",13X,"N",17X,"Q",17X,"M")') DO M=1,NE call SCL call CTM call ESM call EJC DO I=1,6 L=JC(I) D(I)=0.D0 IF (L.NE.0) THEN D(I)=P(L) END IF END DO DO I=1,6 F(I)=0.D0 DO J=1,6 DO K=1,6 F(I)=F(I)+KD(I,J)*T(J,K)*D(K) END DO END DO END DO IF (NPF.NE.0) THEN DO I=1,NPF


L=PF(1,I) IF (M.EQ.L) THEN CALL EFX(I) DO J=1,6 F(J)=F(J)+F0(J) END DO END IF END DO

WRITE (6,'(/1X,I10,3X,"N1=",F12.4,3X,"Q1=",F12.4,3X,"M1=",F12.4,& /14X,"N2=",F12.4,3X,"Q2=",F12.4,3X,"M2=",F12.4)')& M,(F(I),I=1,6)

END IF END DO END subroutine MQN

!----------------------------------------------------------------------------------------------------! !-------------------------------Calculate element’s fixed-end forces-------------------------!

Subroutine EFX(w) implicit none integer(Ikind)::w

IND=INT(PF(2,w)) A=PF(3,w) Q=PF(4,w) C=A/BL G=C*C B=BL-A

F0=0.D0

!-----------------------------------due to transverse distributed loads select case(IND) case(1) S=Q*A*0.5D0


F0(2)=-S*(2.D0-2.D0*G+C*G) F0(5)=-S*G*(2.D0-C) S=S*A/6.D0 F0(3)=S*(6.D0-8.D0*C+3.D0*G) F0(6)=-S*C*(4.D0-3.D0*C) !--------------------------------------- - due to transverse concentrated force case(2) S=B/BL F0(2)=-Q*S*S*(1.D0+2.D0*C) F0(5)=-Q*G*(1.D0+2.D0*S) F0(3)=Q*S*S*A F0(6)=-Q*B*G !--------------------------------------- - due to transverse concentrated couple case(3) S=B/BL F0(2)=-6.D0*Q*C*S/BL F0(5)=-F0(2) F0(3)=Q*S*(2.D0-3.D0*S) F0(6)=Q*C*(2.D0-3.D0*C) !--------------------------------------- - due to transverse triangularly distributed loads case(4) S=Q*A*0.25D0 F0(2)=-S*(2.D0-3.D0*G+1.60*C*G) F0(5)=-S*G*(3.D0-1.6D0*C) S=S*A F0(3)=S*(2.D0-3.D0*C+1.2D0*G)/1.5D0 F0(6)=-S*C*(1.D0-0.8D0*C) !--------------------------------------- - due to axially distributed loads case(5) F0(1)=-Q*A*(1.D0-0.5D0*C) F0(4)=-0.5D0*Q*C*A !--------------------------------------- - due to axial concentrated force


case(6) F0(1)=-Q*B/BL F0(4)=-Q*C

!--------------------------------------- due to distributed couples case(7) S=B/BL F0(2)=-Q*G*(3.D0*S+C) F0(5)=-F0(2) S=S*B/BL F0(3)=-Q*S*A F0(6)=Q*G*B

!---------------------------------------- due to member end axial displacements case(8) L=INT(A) S=Q*EA(M)/BL F0(L)=S IF (L.EQ.1) F0(4)=-S IF (L.EQ.4) F0(1)=-S

!---------------------------------------- due to member end transverse displacements case(9) L=INT(A) F0(L)=12.D0*EI(M)*Q/(BL*BL*BL) IF (L.EQ.2) F0(5)=-F0(2) IF (L.EQ.5) F0(2)=-F0(5) F0(3)=0.5D0*BL*F0(5) F0(6)=F0(3)

!---------------------------------------- due to member end rotations case(10) L=INT(A) S=2.D0*EI(M)*Q/BL F0(L)=2.D0*S IF (L.EQ.3) F0(6)=S IF (L.EQ.6) F0(3)=S F0(5)=3.D0*S/BL


F0(2)=-F0(5) end select END subroutine EFX end module Frame_analysis_sub !--------------------------------------------------------------------------------------------------------- ----! !===================================================================!

13.9.2.3 Explanation of variable_type_module

In the module, some variable identifiers are presented, which identifies variables and arrays of double precision. They will be explained as follows:

(1) Integer variable identifiers NE—number of elements; NJ—number of joints; N—number of unknown nodal displacements NW—number of maximum band; NPJ—number of joint loads; NPF—number of non-joint loads IND—number of types of non-joint loads; M—serial number of elements J1, J2, I, J, I1, K, L, JJ are arbitrary integer variables. (2) Real variable identifiers BL—length of element; SI—direction sine of element; CO—direction cosine of element SI, S, C, G, G1, G2, G3, DX, DY, A, B, Q are arbitrary real variables. (3) Integer array identifiers JE(:,:)—nodal number of element ends (allocatable); JN(:,:)—nodal number of element ends

(allocatable) JC(6) —array used to store the components of element orientation vector. (4) real array identifiers EA(:)—axial rigidity of element; EI(:)—flexural rigidity of element X(:),Y(:)—coordinates of a joint (allocatable) P(:)—global nodal load vector (finally, it will be nodal displacement vector) (allocatable) PJ(:,:)—array used to store the loads applying at joints (allocatable) PF(:,:)—array used to store the loads applying on elements (allocatable) KB(:,:)—array used to store elements in global stiffness matrix (allocatable) KD(6,6) —array used to store elements in [ ]k ⓔ KE(6,6) —array used to store elements in [ ]k ⓔ

T(6,6) —array used to store elements in coordinate transformation matrix [ ]T

F0(6) —array used to store components of fixed-end forces in local coordinates F(6) —array used to store components of element equivalent joint loads in global coordinates (finally,


store the member end forces in local coordinates) D(6) —array used to store components of member end displacements in global coordinates

13.9.2.4 Explanation of module Frame_analysis_sub

The module consists of 10 subroutines. Each of them will be interpreted as follows. 1. Subroutine input_data The subroutine is used to let a computer to read the data of a desired structure. The following block

diagram will explain the design of the subroutine.

From the data file read following data: Control parameter: NE, NJ, N, NW, NPJ, NPF

Nodal coordinates: X(NJ),Y(NJ) Nodal number of element end: JE(2,NE)

Axial rigidity of element: EA(NE) Flexural rigidity of element: EI(NE)

loads applying at joints: J(2,NPJ) loads applying on elements : PF(2,NPF)

Let computer read the data file

Output the data for checking

Develop a data file

2. Subroutine SCL The subroutine is used to calculate the direction cosine of elements of a desired structure. The

following block diagram will give a further explanation of the subroutine.

2 2BL SQRT DX DY⇐ +（） l SI DY/BL⇐ sinα CO DX/BL⇐ conα

J1 JE 1 M⇐ （，） number of beginning node of element M J2 JE 2 M⇐ （，） number of ending node of element M

DX X J2 X J1⇐ −（）（） xΔ DY Y J2 Y J1⇐ −（）（） yΔ


3. Subroutine EJC The subroutine is employed to determine orientation vectors of the elements. The block diagram of

the subroutine will be drawn as follows.

J1 JE 1 MJ2 JE 2 M⇐ （，）

4. Subroutine CTM The subroutine is utilized to form the transformation matrixes of the elements. The design of the

subroutine is based on the following block diagram.

5. Subroutine ESM

⇐ （，）

JC I JN I J1JC I+3 JN I J2

⇐⇐

（）（，）

（）（，）

I=1 3，

T(I J) 0.0⇐，

T 1 1 C0 T 1 2 SIT 2 1 SI T 2 2 C0T 3 3 1.0

⇐ ⇐⇐ − ⇐⇐

（，），（，）

（，），（，）

（，）

T(I+3 J+3) T I J⇐，（，）

I=1, 3

J=1, 3


The subroutine is used to yield the element stiffness matrix in global coordinates. Following block diagram is the base of the design of the subroutine.

6. Subroutine ssm The subroutine is employed to form the global stiffness matrix. Its design is based on following block

diagram.

I1 I+1=

KD(J,I) KD(I,J)⇐

I=1 5，

J=I1 6，

KD(1 1 G KD 2 3 G2KD(1 4 G KD 2 6 G2KD(4 4 G KD 3 5 G2KD(2 2 G3 KD 3 6 G1KD(5 5 G3 KD 5 6 G2KD(2 5 G3 KD 3 3 2.0 G1KD(6 6 2.0 G1

⇐ ⇐⇐ − ⇐⇐ ⇐ −⇐ ⇐⇐ ⇐ −⇐ − ⇐ ×⇐ ×

，），（，）

，），（，）

，），（，）

，），（，）

，），（，）

，），（，）

，）

Form upper

triangular submatrix

Form below

triangular submatrix

G EA(M)/BLG1 2.0 EI(M)/BLG2 3.0 G1/BLG3 2.0 G2/BL

⇐⇐ ×⇐ ×⇐ ×

Basic Stiffness

Coefficient

[ ]0k⎡ ⎤ =⎣ ⎦ⓔ

KD I J 0.0⇐（，）


7. Subroutine nlv The subroutine is utilized to form the equivalent nodal load vector {P}. Its design is based on

L ，=1 6

TKE [T] [ ] [T]k⇐ ⓔ

K=1，6

I JC L⇐ （）

I=0？

J JC K⇐ （）

J=0 J Ior ？<

JJ J-I+1⇐

KB I JJ KB I JJ +KE L K⇐（，）（，）（，

yes (dischage zero elements)

Calculate [ ]k ⓔ

stor

e el

emen

ts o

f []

in [K

]kⓔ

yes

no

no

[ ] [ ]0K =KB(I J) 0.0⇐，

M=1，NE

CALL SCLCALL CTMCALL ESMCALL EJC


following block diagram.

P(I) 0.0⇐

NPJ=0？yes (no joint load)

no

F(L) S⇐

S 0.0⇐

S S T K L F0 K⇐ − ×（，）（）

M PF 1 I⇐ （，）

CALL SCLCALL EFXCALL CTMCALL EJC

L PJ 1 IP L PJ 2 I⇐

⇐（，）

（）（，）

NPF=0？

I=1 NPJ，

I=1 NPF，

L=1 6，

stor

e jo

int l

oads

in {

P}

yes (no element loads)

stor

e el

emen

t loa

ds in

to {

}Pⓔ

contd

no


L JC C⇐ （）

L=0？

P L P L +F L⇐（）（）（）

J=1 6，

asse

mbl

e {

} in

to {

}P

Pⓔ

yes

no

8. Subroutine BGS The subroutine is used to solve the structure stiffness equations { } , by Gauss

elimination method, so as to determine the unknown displacements [ ]{ }P K= Δ

{ }Δ . Its program design is referenced to following block diagram.

N1 N-1⇐

K=1 N

IM K+NW-1

⇐

N IM？﹤

IM N⇐

I1 K+1

，

elim

inat

ionyes

no

contd

⇐


JJ J+I-KKB I J KB I J C KB K JJ⇐

⇐ − ×（，）（，）（，）

P I P I C P K⇐ − ×（）（）（）

P N P N /KB N 1⇐（）（）（，）

L I K+1C KB(K L /KB K IJM NW-L+1

⇐ −⇐⇐

，）（，）

I N-KJM K+1⇐⇐

L J+I-1P I P I -KB I J P L⇐

⇐ ×（）（）（，）（）

P I P I /KB I 1⇐（）（）（，）

NW JM？﹤

IM N⇐

I=I1 IM，

K=1 N1，

J=1 JM，

elim

inat

ion

b

J=2 JM，

ack

subs

titut

ion

yes

no

contd


write title

I=1 NJ，

J=1 3，

D J 0.0L JN J I

⇐⇐（）

（，）

L=0？

9. Subroutine MQN The subroutine is employed to member end forces. Its program design is based on following block

diagram.

D J P L⇐（）（）

output nodal displacement

yes

no outp

ut n

odal

dis

plac

emen

ts {

}Δ

CALL SCLCALL ESMCALL CTMCALL EJC

write title

get p

rope

r dat

a of

ele

men

ts

contd

M=1 N，


F I 0.0⇐（）

F I F I KD I J T J K D(K⇐ + × ×( ) ( ) (， ) ( ， ) )

L JC ID I 0.0⇐

⇐（）

（）

D I P L⇐（）（）

M L≠ ？

NPF=0

L PF 1 I⇐ ( ， )

L=0？

I=1 6，

I=1 6，

I=1 NPF，

J=1 6，

K=1 6，

CALL EFX

yes (discarge zero component in orientation vector)

no

yes (no element load)

no

yes

no

contd

obta

in m

embe

r end

dis

plac

emen

ts de

term

ine

tota

l mem

ber e

nd fo

rces

find member end forces due to its displacements


10. Subroutine EFX(w) The subroutine is utilized to calculate member fixed-end forces. The following block diagram is its

design procedure.

F J F J +F0 J⇐( ) ( ) ( )

J=1 6，

Output member end

stop

IND PF 2 IA PF 3 I)Q PF 4 I)

⇐⇐⇐

（，）

（，

（，

F0(I) 0.0⇐

C A/BLG C CB BL-A

⇐⇐ ×⇐

SELECT CASE 1 2 3 4 5 6 7 IND（，，，，，，），

IND=1 2 3 4 5 6 7

I=1 6，

conc

entra

ted

mom

ent

tran

sver

se tr

iang

ular

ly

dist

ribut

ed lo

ad

axia

lly d

istri

bute

d lo

ad

axia

lly c

once

ntra

ted

load

dist

ribut

ed m

omen

t

tran

sver

se c

once

ntra

ted

load tr

ansv

erse

dis

tribu

ted

load


13.9.3 Example and its computing results

As previous discussion mentioned, when analyze a structure on computer, the structure must be digitalized. So a data file, commonly known as input data file, about the structure have to be make according to the relevant read statements before running the program. In addition, the computing results of the structure are also needed to output in a data file. Following example will illustrate the development of input and output data files.

Example 13-7

Determine the internal forces for the rigid frame shown in Fig.13.26 by using the computer program listed in the section. Assume the axial and flexural rigidities are and

, respectively.

64.0 10 kNEA = ×41.6 10 kN mEI = × ⋅ 2

Solution

(1) Input data file The form of the file is text on which all of the data are written in free format, and the name of the file

is defined by users. The data and their corresponding read statements of the file are written as follows.

3,5,8,7,1,2 read (5,*) NE,NJ,N,NW,NPJ,NPF

0.D0,0.D0,0,0,0 read (5,*) (X(J),Y(J),(JN(I,J),I=1,3),J=1,NJ) 0.D0,4.D0,1,2,3 0.D0,4.D0,1,2,4 4.D0,4.D0,5,6,7 4.D0,0.D0,0,0,8

y

x

15kN m⋅

( )2 1,2,3

( )1 0,0,0 ( )5 0,0,8

( )4 5,6,7( )3 1,2,4

25kN/m

①

②

③

2m2m

18kN

4m

Fig.13.26 Figure of example 13-7


1,2,4.0D+06,1.6D+04 read (5,*) ((JE(I,J),I=1,2),EA(J),EI(J),J=1,NE) 3,4, 4.0D+06,1.6D+04 5,4, 4.0D+06,1.6D+04

7.0,-15.0 READ (5,*) ((PJ(I,J),I=1,2),J=1,NPJ)

1.0,2.0,2.0, 18.0 READ (5,*) ((PF(I,J),I=1,4),J=1,NPF) 2.0,1.0,4.0, 25.0

(2) output data file

The form of the file is text on which all of the data are written in free format by computer, and the name of the file is defined by users. The computing results of the example are written as follows. NE= 3 NJ= 5 N= 8 NW= 7 NPJ= 1 NPF= 2 NODE X Y XX YY ZZ 1 0.0000 0.0000 0 0 0 2 0.0000 4.0000 1 2 3 3 0.0000 4.0000 1 2 4 4 4.0000 4.0000 5 6 7 5 4.0000 0.0000 0 0 8 ELEMENT NODE-I NODE-J EA EI 1 1 2 0.400000D+07 0.160000D+05 2 3 4 0.400000D+07 0.160000D+05 3 5 4 0.400000D+07 0.160000D+05 CODE PX PY PM 7. -15.0000 ELEMENT IND A Q 1. 2. 2.0000 -18.0000

Summary 583

2. 1. 4.0000 -25.0000 NODE U V CETA 1 0.000000D+00 0.000000D+00 0.000000D+00 2 -0.221743D-02 -0.464619D-04 -0.139404D-02 3 -0.221743D-02 -0.464619D-04 0.357876D-02 4 -0.222472D-02 -0.535381D-04 -0.298554D-02 5 0.000000D+00 0.000000D+00 0.658499D-03 ELEMENT N Q M 1 N1= 46.4619 Q1= 10.7119 M1= -6.8477 N2= -46.4619 Q2= 7.2881 M2= 0.0000 2 N1= 7.2881 Q1= 46.4619 M1= 0.0000 N2= -7.2881 Q2= 53.5381 M2= 14.1523 3 N1= 53.5381 Q1= 7.2881 M1= 0.0000 N2= -53.5381 Q2= -7.2881 M2= -29.1523

SUMMARY

In this chapter we have studied the basic concepts of the matrix displacement method for the analysis of plane framed structures. The matrix displacement method is another kind of structural analytical method by which a structure can be analyzed by a computer in terms of the principle of traditional displacement method and the knowledge of matrix algebra.

The matrix displacement analysis of a structure consists of three pivotal processes. They are: (1) Discretization of the structure The Discretization means a computing model of the structure must be prepared. The model is

represented by a line diagram of the structure, on which all the joints, the joints’ unknown displacements and members are identified by numbers. (2) Element analysis

Element analysis involves with the development of relations between the forces and displacements of the member's ends, known as element stiffness equations. The pivot of the equations is the element stiffness matrix. In this chapter, we mainly derived stiffness matrices for a general flexural member, while the stiffness matrices for other members such as the elements of continuous beams, plane trusses and those


without considering axial deformations can be recognized as special elements of the general flexural member. The advantage of this operation can facilitate the standardization and routinization of the analysis. Therefore, in the computer program given in the chapter, only general flexural members are programmed. However, the program can be used to analyze any plane framed structures.

It must be kept in mind that the comprehension of meanings of stiffness coefficients in stiffness matrix will expedite the element analysis.

(3) Global analysis The objective of global analysis of a structure is to express the joint forces as functions of the joint

displacements of the structure by the structure (or global) stiffness matrix. The relations can be expressed as following matrix form

[ ]{ } { }K Δ = P

In the process of establishing the stiffness relations of a structure, the pivotal work is the developments of stiffness matrix and equivalent nodal load vector . [ ]K { }P

When develop global stiffness matrix of the structure, the method so called element assembly method (or stiffness assembly method or direct stiffness method) is used. By the method, the elements in each are algebraically added to their proper positions, oriented by corresponding { }

[ ]K

[ ]k ⓔ λ ⓔ , in the global stiffness matrix . The process has two steps, orienting the element’s position and adding the value of the element to the value that has all ready stored in its proper position of , but the two steps are simultaneously implemented on computers.

[ ]K[ ]K

While determine the equivalent nodal load vector of the structure, the external loads applying on the members are equivalently transformed into so called equivalent joint loads { firstly; then algebraically add the components of each { by corresponding orientation vector { }

{ }P}EP ⓔ

λ ⓔ}EP ⓔ to the components of { }JP , the external loads applying on the joints of the structure.

In actually, there are two sorts of method to obtain the structure stiffness matrix for a structure. When the support conditions are considered before the determination of the global stiffness matrix, the method is termed preprocessing method, which has all ready presented in previous sections of the chapter; otherwise, the support conditions are concerned after the development of the global stiffness matrix, the method is known as postprocessing method, which can be referenced to in other pertinent books.


13-1 How to determine the positive sense of element’s local coordinates? What is the sign convention about the end forces and displacements of a member in matrix displacement method? What is the difference between the chapter’s sign convention and that of the traditional displacement method?

[ ] ? k ⓔijk13-2 What is the physical meanings of each coefficient in the element stiffness matrix


34k 42kPlease explain the meanings of the stiffness coefficients and in the element stiffness matrix [ ]k ⓔ (which is a stiffness matrix of a general flexural member in its local coordinate system) ?

13-3 Why the stiffness matrix of a general flexural member [see Eq. (13-8)] is singular? Whereas the stiffness matrix of the members of a continuous beam [see Eq. (13-14)] is nonsingular?

13-4 How to determine the angle α used in the transformation matrix [see Eq. (13-20)]? [ ]T13-5 How much property is there for an element stiffness matrix? 13-6 What is the orientation vector of an element? How determine the orientation vector? In the global

analysis, where is the orientation vector used? 13-7 What is the element assembly method? When assemble the global stiffness matrix [ , how are

the equilibrium and compatibility conditions of the structure satisfied? ]K

[ ? ]Kijk13-8 What is the physical meanings of each coefficient in the global stiffness matrix 13-9 Why do the elements located on the main diagonal of element and structure stiffness matrices

always possess positive signs? Whether or not the signs for the elements located on non-main-diagonal positions in element and structure stiffness matrices are always positive? Why?

13-10 How to deal with the hinged joints of a rigid frame when analyze the frame by matrix displacement method? Why are the rotations of the hinged joints of the frame taken as unknown displacements in matrix displacement method? Why do the rotations not used as unknown displacements in traditional displacement method?

13-11 Why must the external loads applying on the elements of a structure be equivalently transformed into joint loads in matrix displacement method?

13-12 How to evaluate element equivalent joint loads of a structure? Whether or not the equivalent joint loads are the same as the restraint forces of the primary system of the structure used in traditional displacement method?

13-13 Please try to derive the stiffness equations of a general flexural member without considering its axial deformation [see Eq. (13-12)].

13-14 Whether or not a statically determinate structure can be analyzed by the matrix displacement method?


13-1 Determine the joint rotations and member end moments of the continuous beams shown in the following figures by matrix displacement method.

13-2 Try to analyze the continuous beam shown in the figure by using matrix displacement method, and draw its bending moment diagram.


13-3 Determine the global stiffness matrix by element assembly method and write out the primary equations for the structure shown in the figure.

[ ]K

13-4 Try to find the global stiffness matrix by element assembly method for the continuous beam shown in the figure.

[ ]K

l l

2EI

EI

EI

2EI

2EI

EI

1 2 3

4 5 6

problem 13-3

l l

problem 13-4

ll l2EI EI

2 31 4

y

x

① ② ③

EI

1i1 50kN mM = ⋅

2 1i i=

6m

10kN/m

6m① ② ① ②

1i

2 1i i=

6m 6m

(b)(a)

problem 13-1

5kN/m5kN/m

problem 13-2

ll lA B C D

constantEI = 4ml =


13-5 Try to determine, neglecting the axial deformation, the global stiffness matrix [ by element

assembly method for the rigid frame shown in the figure. Assume that each member has the same size and elastic modulus. That is, the element’s length

]K

5ml = , cross sectional area ,

cross sectional inertial moment

20.5mA =41 m

24I = and elastic modulus . 43 10 MPaE = ×

13-6 In problem for solution 13-5, if assume that a downward uniformly distributed load is applied on element , please determine the internal forces and draw the internal force diagrams of the rigid frame.

4.8kN/mq =①

13-7 Try to write out the primary equations of the displacement method with considering the axial deformations of the members for the rigid frame shown in the figure. Assume that the elastic modulus and size of the members, i.e., E A I, and , are all constant.

13-8 Try to analyze the rigid frame shown in the figure by matrix displacement method without considering the axial deformations of the members.

problem 13-5

① ②

③

2 31

4 y

x

problem 13-8

2

3

65

EI EI

2EI

2EI

1.5EI 1.5EI

8m

10kN

20kN

4m 6m

1

4

problem 13-7

① ②

③

xy

2 31

4

2l

3P

l l

2P3P 2P

2l l l

2l

14 CHAPTER GENERAL REMARKS ON

STATICALLY INDETERMINATE STRUCTRES

The abstract of the chapter In this chapter, we will give a general discussion on statically indeterminate structures, starting

with the classification and comparison of the methods used to analyze the structures, then presenting a combined method by simultaneously applying moment distribution and displacement methods for the analysis of indeterminate structures with sidesway and the approximate analysis of statically indeterminate structures, also summing up the mechanical characteristics of the structures, finally ending with the further discussion about the computing models of the structures.

14.1 Classification and Comparison of Fundamental Method used to Analyze Statically Indeterminate Structures

In previous chapters, we have discussed the fundamental methods for analyzing statically indeterminate structures. In order to have an overall understanding about the methods, we now summarize them in the following table.

Table 14-1 Classification of methods for analyzing statically indeterminate structures

Type Generalized force method Generalized displacement method

Fundamental type Force method Displacement method

Matrix type (matrix force method) Matrix displacement method

(Successively approximate force method)

Moment distribution method, no-shear distribution method

Successively approximate type

Energy type (complementary energy method) (Potential energy method)

The contents in the table can be explained as follows: (1) The methods for analyzing statically indeterminate structures can be classified into two

categories in terms of the characteristics of the primary unknowns the methods adopted. They are: ① generalized force method, ② generalized displacement method.

588

14.1 Classification and Comparison of Fundamental Method used to Analyze Statically Indeterminate Structures 589

So called generalized force method means that if the method adopts force(s) as primary unknown(s), it will be termed force method or generalized force method, such as classical force method, complementary energy method and the like; whereas generalized displacement method indicates that the primary unknown(s) adopted by the method are displacement(s), such as classical displacement method, potential energy method, matrix displacement method, moment distribution method and no-shear distribution method presented in the book.

It should be pointed out that the methods enclosed in the brackets shown in table 14-1 were not discussed in the book; they are the subjects in some special books.

(2) The methods used to analyze statically indeterminate structures can be also categorized into two groups by means of the formulating way utilized by the methods. They are: ① equilibrium-geometry-physics method, ② energy method.

Recalling from the process formulating the classical force and displacement methods, three types of equations were employed, they are equilibrium equation(s), compatibility (or geometric) equation(s) and force-displacement (or physical) relation(s). So equilibrium-geometry-physics method means that the equilibrium equation(s), geometric condition(s) and physical properties of the material must be used to analyze an indeterminate structure. In static analysis, the method is called static method.

If we use energy method to formulating methods for analyzing statically indeterminate structures, the analytical method will be referred to as energy method.

Actually, in despite of equilibrium-geometry-physics method or energy method, their principles are identical. The only difference is their formulating way. In this context, force method is equivalent to complementary energy method; displacement method is equivalent to potential energy method. For the precise solution of a simple problem, it is hard to find the advantages of energy method over equilibrium-geometry-physics method. However, for the approximate solution of a complex problem, the energy method has remarkable advantages over equilibrium-geometry-physics method, because energy method formulate a complex problem as a mathematical problem for minimum value or extreme value, which is easy to solve in the viewpoint of mathematics. Energy method is widely applied in the dynamic and buckling analyses of structures.

(3) The methods for analyzing statically indeterminate structures can be as well grouped into two categories based on the calculating way the methods used. They are: ① hand-orientated methods, ② computer-oriented methods.

In hand-oriented methods, simplification of calculation is the objective of the methods. So much of the details about the members, connections and those have little effect on the desired characteristics of a structure must be discarded so as to simplify the calculations by hand. Thusly, much of the unknowns which have little effect on principal behavioral characteristics of the structure are neglected when use

590 Chapter 14 General Remarks on Statically Indeterminate Structures

hand-oriented method(s) to analyze the structure. For instance, when analyze a rigid frame the axial deformations of the members are ignored to reduce the number of unknowns so that the calculating work by hand is decreased. In that context, displacement method is advanced over force method; potential energy method is more powerful than complementary energy method, and vice versa.

In computer-oriented methods, routinization and automatization of the computing process for a structure are the objective of the methods. This is the reason why matrix algebra has been applied to structural analysis. Since the displacement method is more systematic and can be more easily implemented on computers, most of the commercially available computer programs for structural analysis are based on the method instead of force method.

(4) The methods for analyzing statically indeterminate structures can be also divided into two groups according to the way for solving the canonical equations (it is actually simultaneous linear equations) the methods adopted. They are: ① direct methods, ② successively approximate methods.

Classical force and displacement methods belong to direct methods; whereas moment distribution and no-shear distribution methods are successively approximate methods, which actually use the joint displacements as primary unknowns though the end moments are participated in the concrete process of operation. Since the convergent speed of the successively approximate methods based on force method are very slow, they are not applied to structural analysis.

(5) In the viewpoint of hand-oriented methods, there is a most favorable method for an indeterminate structure to be analyzed. Table 14-2 shows framed structures and their corresponding most favorable methods by which the structures can be conveniently analyzed.

Table 14-2 statically indeterminate structures and their corresponding most favorable methods

Types of structures Most favorable method

Indeterminate trusses and arches Force method

Continuous beams and rigid frames without sidesway Moment distribution method

Displacement method or no-shear distribution method or combined method

Rigid frames with sidesway

(6) In the viewpoint of computer-oriented methods, matrix displacement method is suitable for all kinds of structures. An all purpose computer program might be used to analyze many kinds of structures as the program shown in section 13.9.


=

14.2 Combined Method—Simultaneously Applying Moment Distribution and Displacement Methods to Frames with Sidesway

In practice, when solving a complex structural analytical problem, the complex problem might be subdivided into several simple sub-problems. For each of the sub-problems, a most favorable analytical method can be applied to solve it. Then, combining each of the solution will obtain the solution of the complex problem. By this combined way, a complex problem sometimes can be solved conveniently because the advantage of each analytical method is utilized properly.

There are many combined ways, such as the combination of force and displacement methods, the combination of moment distribution and displacement methods, and moment distribution and no-shear distribution methods. For various different problems, different combined methods can be applied. For an introduction, only combination of moment distribution and displacement methods are discussed in the following.

As we known in previous chapters, moment distribution method or no-shear distribution method alone cannot be applied to analyzing the rigid frames with sidesway. However, if we apply both moment distribution method and displacement method to the frames, they will be analyzed conveniently, since the effect of joint rotation(s) can be estimated by moment distribution method and the response of the sidesway (translation) can be valued by displacement method individually. Now we will illustrate the analytical procedure of combination of moment distribution and displacement methods by the consideration of the rigid frame with sidesway shown in Fig.14.1 (a).

(1) Evaluating the effect due to joint translation (sidesway) The response due to joint translation can be evaluated by displacement method, but the primary

structure used here is different from that used by traditional way. As shown in Fig.14.1 (b), on the primary structure only translation is restrained; in other words, only the horizontal translation is chosen as the primary unknown whereas the joint rotations do not chosen as the primary unknowns in this step.

+

A B C

q

A B C

q

1PF A B C

11 1k Δ

Fig.14.1 Sketch of combination of moment distribution and displacement methods (a) original structure; (b) primary structure under the action of external load; (c) primary structure under the action of sidesway


Consequently, by considering the restraint forces at the artificial restraint caused by the external load and the unknown translation, we can write displacement-method equation as follow

11 1 1 0Pk FΔ + = (14-1)

The expression for evaluating bending moments can be expressed as

1 1 PM M M= Δ + (14-2)

1PF PMIn Eqs. (14-1) and (14-2), and are the restraint force yielded at the artificial restraint and bending moment in the primary structure due to the external load; and 11k 1M are the restraint force yielded at the artificial restraint and bending moment in the primary due to the unit displacement 1 1Δ = .

(2) Estimating the effect due to joint rotations For the primary structure shown in Fig.14.1 (b) or (c), the moment distribution method can be

expediently applied to determine stiffness coefficient , restraint force 11k 1PF 1M, bending moments and PM . Once the quantities are obtained the translation (or sidesway) can be determined through Eq. (14-1) as to be

11

11

PFk

Δ = −

After is determined the bending moment in the frame will be evaluated by Eq. (14-2). 1ΔExample 14-1

Draw the bending moment diagram for the rigid frame shown in Fig.14.2 (a) by applying both displacement method and moment distribution method.

Solution

(1) Start with applying displacement method. Thusly, the roller support at D is replaced by a hinged support to prevent the frame’s sidesway as shown in Fig.14.2 (b). Since the primary structure shown in Fig.14.2 (b) has no joint translation, moment distribution method can be applied to it conveniently, and the final bending moment diagram obtained by the method is shown in Fig.14.3.

In order to find horizontal restraint force at D, the shears at the bottom sections of the two columns due to the external load can be determined as to be

3.45 1.7 1.29kN4EB

PQ += − = −

9.8 4.9 2.45kN6FC

PQ += =


0X =∑By considering the horizontal force equilibrium condition of the free body diagram for the entire frame, the horizontal restraint force at D can be determined to be

1 2.45 1.29 1.16kN( )PF = − = →

20kN/mq =

DA

B C05

(2) Draw the bending moment diagram 1M for the primary structure shown in Fig.14.2 (c) due to the unit horizontal displacement at D. Since now the support’s displacement is given, , we can employ moment distribution method to the structure to determine its bending moments. Thusly obtained bending moment diagram is shown in Fig.14.4.

1 1Δ =

I

F

E

03I 03I

04I 04I

5m4m 4m

6m

4m

20kN/mq =

DAB C05I

F

E

03I 03I

04I 04I

5m4m 4m

6m

4m

(a) original structure

Fig.14.2 Figures of example 14-1

(b) primary structure under action of external load


In order to find stiffness coefficient , the shears at the bottom sections of the two columns due to can be evaluated as

11k1 1Δ =

0.965 0.806 0.4434EBQ +

= =

0.467 0.436 0.1516FCQ +

= =

By considering the horizontal force equilibrium condition 0X =∑ of the free body diagram for the entire frame, the stiffness coefficient can be written to be

DAB C05I

F

E

03I 03I

04I 04I

5m4m 4m6m

4m

1Fig.14.2 (c) primary structure under action of 1=

1 1Δ =

0assume 1 EI =

Δ

Fig.14.3 diagramPM

1.29kN

4.9

1.7

AC D

1 1.16kNPF =

2.45kN

B

F

E

3.4518.2

43.446.9

24.414.6

9.8

(unit: kN m)⋅


→

11 0.443 0.151 0.594kN( )k = + =

(3) Determine the horizontal displacement 1Δ . By considering the displacement equation 11 1 1 0Pk FΔ + = , we can determine the horizontal

displacement to be

11

11

1.16 1.950.594

PFk

Δ = − = − = −

(4) Draw bending moment diagram. By using the superposition equation for bending moment, 1 1 PM M M= Δ + , we will draw the final

bending moment as shown in Fig.14.5. In the forgoing example, we illustrate the analytical procedure of combined method for the rigid frame

with only one joint translation. As to the rigid frame with more than one joint translation, similar procedure can be considered. For instance, the analytical procedure of the combined method for a rigid frame with two joint translations, and , can be written as follows: 1Δ 2Δ

(1) Attach two artificial restraints associated with 1Δ and 2Δ to the original structure to prevent the two translations so as to obtain the primary structure for displacement method.

(2) By using moment distribution method, determine the bending moment diagram PM , the restraint forces, 1PF and 2PF , yielded at the artificial restraints.

(3) Evaluate the bending moment 1M , stiffness coefficients and for the primary 11k 21k

A C DB

F

E0.443EBQ =0.965

0.096

0.318

0.806

0.340

0.436

0.467 0.154EBQ =

11 0.594K =

1Fig.14.4 diagramM

0.488 1 1Δ =


1 1Δ =structure due to the unit displacement . (4) Calculate the bending moment 2M , stiffness coefficients and for the primary

structure due to the unit displacement 12k 22k

2 1Δ = .

1Δ(5) Determine and by using displacement-method equation 2Δ

11 1 12 2 1

21 1 22 2 2

00

P

P

k k Fk k FΔ + Δ + = ⎫

⎬Δ + Δ + = ⎭

(6) Draw final bending moment diagram by employing the superposition equation

1 1 2 2 PM M M M= Δ + Δ + .

18.5B

A

F

14.3 Approximate Analysis of Statically Indeterminate Structures

The analysis of statically indeterminate structures using the force and displacement methods introduced in the preceding chapters can be considered as exact in the sense that the compatibility and equilibrium conditions of the structure are exactly satisfied in such an analysis. However, exact analysis of indeterminate structures involves computation of displacements and solution of simultaneous equations, so it can be quite time consuming without assistance of computers. If we discard some of the unimportant details of an indeterminate complex structure to simplify the analysis of the structure, approximate method(s) will be developed. In fact, approximate analysis proves to be quite convenient to use in the

E

C

42.8

47.8

47.8

5.0

3.6

4.0

26.7

14.843.7

D

8.9

(unit: kN m)⋅


14.3 Approximate Analysis of Statically Indeterminate Structures 597

planning phase of projects, when several alternative designs of the structure are usually evaluated for relative economy. The results of approximate analysis can also be used to estimate the sizes of various structural members needed to initiate the exact analysis. The preliminary designs of members are then revised iteratively, using the results of successive exact analyses, to arrive at their final designs. Furthermore, approximate analysis is sometimes used to roughly check the results of exact analysis, which due to its complexity can be prone to errors. Numerous methods have been developed for approximate analysis of indeterminate structures. In the section, two approximate methods, single-storey method and inflection-point method, will be introduced as follows.

14.3.1 Approximate Analysis of Multistory and/or Multibay Rigid Frames Subjected to Vertical Loads

Unlike the exact methods, which are general in the sense that they can be applied to various types of structures subjected to various loading conditions, a specific method is usually required for the approximate analysis of a particular type of structure for a particular loading. Now we will introduce so called single-storey method used to approximately analyze multistory and/or multibay rigid frames or rectangular building frames subjected to vertical loads. Based on the behavioral characteristics of the multistory and/or multibay rigid frame and the experience gained from the exact analyses of the types of the frames subjected to vertical loads only, two simplifying assumptions about the frames can be made for the method.

(1) Neglect the effect due to the sidesway of a frame so as to moment distribution method can be applied to the frame.

(2) Discard the effect of the loading acting on the girder(s) lying in the same storey on the other girder(s) lying in the different stories. The assumption can make the girders of a multistory frame to be analyzed storey by storey.

Consider, for example, the four-storey and three-bay rigid frame shown in Fig.14.6 (a). If we apply the two assumptions to the frame, its analytical model will become four single-storey rigid frames as shown in Fig.14.6 (b). Note that each of the columns except for the bottom ones belong to two adjacent stories, so the bending moments obtained from the two stories of the columns must be added together.

The rationality of the first assumption can be stood for by the small sidesway of a multistory and/or multibay rigid frame under vertical loading; whereas the rationality of the second assumption will be explained as that: ① Unbalance moments will be yielded at the joints of some storey under the action of its vertical loads. The unbalanced moments will be distributed to the nearends and carry over to the farends of the columns of the storey. ② The carryover moments will further influence the columns adjacent to the stories. However, the carryover moments have already gone a long way and become too small to affect their neighbor columns. So the carryover moments can be neglected. Obviously, this argument is the basis of the second assumption.


It must be pointed out that the farends of all single-storey frames except for bottom one are assumed fixedly (or rigidly) supported. In practice, the farends are not rigidly supported; they are elastically

supported. In order to reflect the characteristics of elastic supports, the relative rigidity, i.e. EIl

, will be

reduced by timing the relative rigidity by 0.9, and the carryover factor is changed from 13

12

to .

Ⅳ

Ⅲ

Ⅱ

Ⅰ

(a)

(b) Fig.14.6 Sketch of single-storey method(a) original structure; (b) single-storey frames

Ⅲ

Ⅳ

Ⅱ

Ⅰ

It is observed that the analyzed results of single-storey method will generally yield unbalanced moments at the rigid joints of the frame. However, the experiences of analyzers have proved that the unbalanced errors are generally very small. If necessary, the unbalanced moments can be distributed once again.

14.3.2 Inflection-Point Method for Analyzing Multistory and/or multibay Rigid Frames Subjected to Horizontal Loads

The behavior of multistory and/or multibay frames is different under lateral (horizontal) loads than under vertical loads, so different assumptions must be used in the approximate analysis for lateral loads than were used in the case of vertical loads considered previously. There are several methods used for approximate analysis of multistory and/or multibay frames subjected to lateral loads. This subsection only


presents inflection-point method which is a commonly used method for analysis of multistory and/or multibay frames with strong girders and weak columns subjected to lateral loads.

The fundamental assumption of inflection-point method is that the stiffnesses of the girders of the frames are infinitely great such that the shear distribution method can be applied to the frames.

In order to illustrate inflection-point method, consider the portal frame shown in Fig.14.7 (a) whose bending moment diagram has been drawn in Fig.14.7 (b). Because the ratio between the relative flexural

rigidity of the beam and that of columns is 3, i.e. 3b

c

ii= , the frame belongs to the frames with strong

beams and weak columns. So we can use inflection-point method to analyze the frame. Thusly, the flexural rigidity of the beam is assumed to be infinitely great, i.e. bi = ∞ . The frame will become into an ideal

frame as shown in Fig.14.8 (a), whose deflecting behavioral characteristic is that its rigid joints have sidesway but no rotation, and bending-moment characteristic is that the bending moments at the midpoints of the two columns are equal to zero as shown in Fig.14.8 (b), which has drawn by using shear distribution method. Based of the assumption and simplification, we can obtain that:

(1) The shear of each column is equal to 2P by consideration of the symmetric property of the

frame.

0θ =Δ

bi = ∞

ci

P

4Ph

4Ph

4Ph

P

h

2PQ =

2h 2h2

PQ =2PQ =

2PQ =

4Ph

b

c

ii= ∞

ci

Fig.14.8 A portal frame analyzed by inflection-point method (a) original structure; (b) diagramM

0.9474

Ph

0.9474

Ph

1.0534

Ph1.0534

Ph

P

/ 3b ci i =

Δ

biθ

hP

ci ci

Fig.14.7 A portal frame exactly analyzed (a) original structure; (b) diagramM


2 4h PhM Q= × =(2) The bending moments at the ends of the columns are equal to by consideration

of the inflection points lying at the midpoints of the columns. Thusly, the bending moment diagram can be drawn out as shown in Fig.14.8 (b).

(3) The bending moments of the beam can be determined by using the equilibrium conditions of the rigid joints, and then the bending moment diagram can be drawn as shown in Fig.14.8 (b).

It can be observed from comparison between Fig.14.7 (b) and Fig.14.8 (b) that only 5% error occurs when uses approximate method to analyze the frame. It implies that if the ratio between the relative rigidity

of beams and that of columns is great than 3, i.e. 3b

c

ii≥ , the inflection-point method will obtain a quite

exact analysis for the frames with strong beams and weak columns under lateral loads. Inflection-point method can be also applied to the frames with different columns. Consider, for

example, the frame shown in 14.9 (a), whose beam has infinitely great rigidity, the relative rigidity and height of left column are and , and those for right ones are and , respectively. Since the sidesways of the two columns are identical the shears of the two columns will be calculated to be [see Fig.14.9 (c)]

1i 1h 2i 2h

11 12

1

22 22

2

12

12

iQ dh

iQ dh

⎫= Δ = Δ ⎪⎪⎬⎪= Δ = Δ⎪⎭

(a)

2

12idh

=In which, is referred to as stiffness coefficient of the column due to its sidesway, i.e. the shear

induced by a unit sidesway at the top of the frame. By consideration of the equilibrium condition of horizontal forces of the frame, we write

1 2Q Q P+ = (b)

Substituting Eq. (a) into Eq. (b), the two shears will be determined to be

11 12

1

22 22

1

ii

ii

dQ Pd

dQ Pd

μ

μ

=

=

⎫= = P

P

⎪⎪⎪⎬⎪= =⎪⎪⎭

∑

∑

(c)


jj

i

dd

μ =∑

Where, , known as shear-distribution factor.

It is observed from above that the shear distributed to a column is directly proportional to its sidesway-stiffness coefficient . Thusly, the horizontal load is distributed to each column by means of its shear-distribution factor. Once the shears in columns are determined the bending moments will be determined by using the condition that the bending moments on the middle sections of the columns are equal to zero, and thus obtained bending moment is drawn in Fig.14.9 (b).

d P

Based on forgoing discussion, we can summarize some points used for approximate analysis of multistory and/or multibay frames by inflection-point method as follows:

(1) When the ratio between the relative flexural rigidity of beam(s) and that of column(s) is great

than 3, i.e. 3b

c

ii≥ , inflection-point method might be used to analyze multistory and/or multibay frames

subjected to lateral loads. (2) The fundamental assumption of inflection-point method is that the flexural rigidity of the beam(s)

of a frame is infinitely great so that the rigid joints of the frames only translate but not rotate. (3) When the columns in the same storey have the same sidesway, the shears in the columns induced

by the sidesway are directly proportional to their sidesway-stiffness coefficients. All the columns locating

Δ Δ

bi = ∞0θ =

1Q

1Q2Q

2Q2 2

2Q h1 1

2Q h

1 1

2Q h 2 2

2Q h

1h 2h

Δ

2

12iQ dh

= Δ = Δ

Q

h

1i

P P

2i

1 2h1 2h

2 2h

2 2h

(a)

(c)

(b)

Fig.14.9 A frame with different columns analyzed by inflection-point method (a) original structure; (b) bending moment diagram; (c) shear of a column


within some storey undertake all the lateral loads above the storey, which are then distributed to each column of the storey by means of its shear-distribution factor.

(4) The bending moments of the columns are induced by their sidesways, so the inflection points are located at the midpoints of the columns. Note that for multistory rigid frames, the inflection point of a column lying on bottom storey is assumed to be located at a height of 2/3 length of the column.

(5) The bending moments of the columns are determined by their shears and the locations of their inflection points. While the bending moments at the ends of the beams connecting to the side columns are determined by equilibrium of the corresponding rigid joints; the bending moments at the ends of beams connecting to the middle columns are the distribution of the unbalanced moments which are distributed by means of the bending stiffness of the beams connected to the joints.

It must be pointed out that when shear distribution method is applied to the frames with infinitely great beam(s), it is a sort of exactly analytical method; inflection-point method is a special kind of shear distribution method, and it is an approximately analytical method when it applied to the rigid frames with strong beams and weak columns.

Example 14-2

Analyze and draw the bending moment diagram for the rigid frame shown in Fig.14.10 (a) by inflection-point method. Note that the digits enclosed in the parentheses represent the corresponding relative flexural rigidities of the members.

Solution

Simplifying the frame In order to simplify the frame, we assume that all the inflection points are located at the midpoints of the columns. Note that although the inflection points of bottom columns should be located at the positions of 2/3 heights of the columns by the forgoing argument, the relative rigidities of the bottom-storey beams are more than three times of those of the bottom columns. Thusly, the assumption is rational for approximate analysis. If we pass imaginary sections through the inflection points the free body diagrams of the frame will be shown as in Fig.14.10 (b). Some calculations will be made as follows.

jj

ii

dd

μ =∑

(1) Determination of each shear distribution factor for the columns

The shear distribution factors for the columns on top storey:

2 0.2862 2 3GD IFμ μ= = =× +

3 0.4282 2 3HEμ = =× +

The shear distribution factors for the columns on bottom storey:


3 0.33 2 4AD CFμ μ= = =× +

4 0.43 2 4EBμ = =× +

(2) Evaluation of shears for the columns By using j jQ Pμ= , the shears can be evaluated as

0.286 8 2.29kNGD IFQ Q= = × =

0.428 8 3.42kNHEQ = × =

0.3 (8 17) 7.5kNAD CFQ Q= = × + =

0.4 (8 17) 10kNEBQ = × + =

(3) Evaluation of bending moments The following calculation will take joint E as an example. Bending moments at the ends of the columns

2 3.33.42 5.64kN m2 2EH HEhM Q= − × = − × = − ⋅

1 3.610 18kN m2 2EB EBhM Q= − × = − × = − ⋅

Bending moments at the ends of the beams The summation of the bending moments of the columns connected to joint E will be written as

I

D EF

BC

G H

A

E FD

AADQ

CCFQ

GDQ HEQ

BEQ

IFQ

B

G H

3.3m

3.6m

3.6m 4.5m

17kN

8kN

(12)

(3)

(2)

(3)

(3)

(2)

(4)

(12)

17kN

8kN

(15)

(15)

(a) (b)

Fig.14.10 Figures of example 14-2(a) original structure; (b) free body diagram

I


5.64 18 23.64kN mEH EBm M M= + = − − = − ⋅

The distributed bending moments at the ends of the beams connected to joint E will be evaluated as

12 23.64 10.51kN m27EDM = × = ⋅

15 23.64 13.13kN m27EFM = × = ⋅

Fig.14.11 shows the bending moment diagram of the frame. Note that the digits enclosed in the parentheses are the values of the bending moments evaluated by an exact method.

14.4 Properties of Statically Indeterminate Structures

In this section, we will focus our attention on the properties of statically indeterminate structures. During the discussion, some principle of force method and comparison between the indeterminate and their comparable determine will be employed.


I

D F

A B C

GH

E

3.78

3.78( )3.63

( )3.63

3.78

3.78

3.78

( )3.07( )( )6.61

( )13.63 ( )19.31 ( )13.90

( )12.11( )2.85 ( )18.41( )12.65

( )15.91( )11.52

( )13.42

( )14.95

( )6.14 ( )3.26

3.902.515.64

17.2810.51

13.5 18

13.13

13.5

13.5

13.518

5.64

17.28

8kN

17kN

( )3.533.13

(unit: kN m)⋅

14.4 Properties of Statically Indeterminate Structures 605

14.4.1 Redundant Restraints and Their Influence

As discussed previously, the support reactions and internal forces of statically determinate structures can be determined from the equations of equilibrium (including equations of condition, if any). However,

since the indeterminate structures have more support reactions and/or members than required for static stability, the equilibrium equations alone are not sufficient for determining the reactions and internal forces of such structures and must be supplemented by additional relationships based on the geometry of deformation of structures. These additional relationships, which are termed the compatibility conditions, ensure that the continuity of the displacements is maintained throughout the structure and that the structure's various parts fit together. The existence of redundant restraints, i.e., the excess support(s) and/or member(s), will change the stressing properties of a structure in the following aspects.

(1) Stronger defense abilities. Unlike a statically determinate structure of which if one of supports or members is collapsed the entire structure will be failed as well, a statically indeterminate structure, if properly designed, has the capacity for redistributing loads when certain structural portions become overstressed or collapse in cases of overloads due to earthquakes, tornadoes, impact (e.g., gas explosions or vehicle impacts), and other such events. So if a part (or member or support) of such a structure fails, the entire structure will not necessarily collapse, and the loads will be redistributed to the adjacent portions of the structure because indeterminate structures have more members and/or support reactions than required for static stability. (2) Uniformly internal force distributions and smaller stresses. The influence extent of local loading in a statically indeterminate structure is generally great than that in its comparable determinate structure, but the maximum stress in a statically indeterminate structure is generally lower than that in its comparable determinate structure. Consider, for example, the statically determinate and indeterminate beams shown in Figs.14.12 (a) and (b), respectively. The bending moment diagrams and their elastic deforming curves for the beams due to a concentrate load are also shown in the figure. It can be seen from the figures that the bending moment distribution extends in the whole indeterminate beam but in local portion of the determinate beam, and the maximum bending moment in the indeterminate beam is

P

Fig.14.12 Bending moment distributions and comparison of maximum bending moments (a) bending moment diagram of an indeterminate beam; (b) bending moment diagram of the comparable determinate beam

(a) P (b) P


significantly lower than in the determinate beam. This implies that the internal force distribution of an indeterminate structure is more uniform and the maximum bending stress in the structure is lower than those of its comparable determinate structure.

(3) Greater stiffness and stabilities. Statically indeterminate structures generally have higher stiffnesses than those of comparable determinate structures. From Fig.14.13, we observe that the maximum deflection of the indeterminate beam with one fixed support and one roller support is only 0.42 that of the simply supported determinate beam, and that the maximum deflection of the two-fixed-support indeterminate beam is only 0.2 that of the simply supported determinate beam.

Statically indeterminate structures generally have higher stabilities than those of comparable determinate structures. From Fig.14.14, we can see

that the critical force of the column with one fixed support and one axially glidingly fixed support is only 0.25 that of the simply supported column.

1P 2P

l

1P2P

Fig.14.14 Comparison of critical loads (a) a simply supported column; (b) an indeterminate column

40.0026 /f ql EI= 40.0054 /f ql EI= 40.013 /f ql EI=

EIl

q

EIl

q

EI

l

q

(a) (c)(b)

Fig.14.13 Comparison of maximum deflections(a) a two fixed-support beam; (b) one fixed-support and one roller-support beam; (c) simply supported beam


14.4.2 Influence due to Changes of Member Stiffness

Unlike a statically determinate structure whose stiffness changes of the members will not influence the internal force distribution of the structure, the internal force distribution of a statically indeterminate structure is strongly depended on its member’s relative stiffness. The more rigid is a member the more load will be distributed to it. This can be further explained by referencing to canonical equations of force method [see Eq. (10-5)]. Recalling from the equation that the flexibility coefficients and free terms vary with the member’s stiffnesses, if the member’s stiffnesses are changed the flexibility coefficients and free terms in the force-method equation are changed as well, subsequently the internal force distribution in the structure will be also changed. However, if only change the magnitudes of the member’s stiffnesses of an indeterminate structure but remain the proportions between the stiffnesses unchangeable, the proportions between the flexibility coefficients and free terms—subsequently the distribution of internal forces in the structure—will also unchangeable. Accordingly, the internal force distribution of an indeterminate structure due to external loads is only related to the proportions between the member’s stiffnesses but not the magnitudes of the stiffnesses. That is, the internal force distribution changes with the relative stiffnesses of the members but not the absolute stiffnesses of the members. This implies that when we analyze the internal forces of an indeterminate structure only the relative rigidities of the structure’s members are necessary.

Since the analysis of an indeterminate structure involves, in addition to the dimensions and arrangement of members of the structure, its cross-sectional and material properties (such as cross-sectional areas, moments of inertia, moduli of elasticity, etc.), which in turn, depend on the internal forces of the structure. The design of an indeterminate structure is, therefore, carried out in an iterative manner, whereby the (relative) sizes of the structural members are initially assumed and used to analyze the structure, and the internal forces thus obtained are used to revise the member sizes; if the revised member sizes are not close to those initially assumed, then the structure is reanalyzed using the latest member sizes. The iteration continues until the member sizes based on the results of an analysis are close to those assumed for that analysis. However, the design of a determinate structure will not need the iterative manner.

It is obviously that we can utilize the characteristic, i.e., the more rigid is a member the more load will be distributed to it, to adjust the relative stiffnesses of the members of an indeterminate structure so as to modulate the internal force distribution of the structure and to implement a rational structural design.

Consider, for example, the bending moment diagram of a portal frame shown in Fig.14.15 (a). If we increase the cross-sectional size of the girder and reduce the cross-sectional sizes of the columns—and subsequently the member’s relative stiffnesses will be changed—the bending moment diagram will approach the case shown in Fig.14.15 (b), where the bending moment diagram in the girder is close to that of a simply supported comparable girder, and the maximum bending moment occurred on the middle


section is too great to be properly adopted. Contrarily, if we decrease the cross-sectional size of the girder

and increase the cross-sectional sizes of the columns—and subsequently the member’s relative stiffnesses will be changed—the bending moment diagram will approach the case shown in Fig.14.15 (c), where the bending moment diagram in the girder is close to that of a comparable girder with two fixed supports, and the maximum bending moment occurred at the top ends of the columns is much greater than that on the girder’s middle-section. In practice, this kind of internal force distribution is as well not properly adopted. Obviously, it is wise to modulate the cross-sectional sizes of the columns and the girder to make the

1 2 1 2 1 2

Fig.14.15 Changes of bending moment diagrams with member's stiffnesses (a) 2 ; (b) ; (c) I I I I I I=

C D

6m

53.3

8m

20kN/mq =

160

106.7

53.3

BA

2

I1

I I 2

(unit: kN m)⋅

(a)

20kN/mq =

160

160≈

BA

0≈

2I 2I

1I

(b)

(unit: kN m)⋅

20kN/mq =

106.7≈

A B

53.3≈

2I 2I

1I

(c)

(unit: kN m)⋅

1600≈ 106.7≈


maximum bending moments occurred in both the girder and the columns are closed to each other.

14.4.3 Influence due to Temperature Changes and Support Settlements

The argument that if there is no load applying on a structure there will be no internal force yielded in the structure, is only true for a determinate structure but not for an indeterminate one. Actually, temperature changes and support settlements do not cause any stresses in determinate structures; they may, however, induce significant stresses in indeterminate structures, which should be taken into account when designing indeterminate structures. The reason is that the temperature changes and/or support settlements will cause deformations of members which are restrained by the redundant restraints of an indeterminate structure; consequently the restraint deformations will induce internal forces in corresponding members. The internal forces induced by the temperature changes and/or support settlements and/or the like other than external loads are referred to as self-induced internal forces in Structural Mechanics. Obviously, self-induced internal forces can be existed in an indeterminate structure but not in a determinate structure.

Furthermore, the self-induced internal forces are directly proportional to absolute values of the member’s stiffnesses of an indeterminate structure, i.e., the larger is the magnitude of a member’s rigidity the greater is the self-induced internal force in the member. The argument can be further explained by follows.

Consider the canonical equations of force method used for analysis of an indeterminate structure subjected to temperature changes and support settlements, which can be rewritten as

11 1 12 2 1 1 1

21 1 22 2 2 2 2

1 1 2 2

00

0

n n t c

n n t c

n n nn n nt nc

X X XX X X

X X X

δ δ δδ δ δ

δ δ δ

+ + + + Δ + Δ = ⎫⎪+ + + + Δ + Δ = ⎪⎬⎪⎪+ + + + Δ + Δ = ⎭

The corresponding superposition equations used for calculate internal forces will be written as

1 1 2 2

1 1 2 2

1 1 2 2

n n

n n

n n

M M X M X M XN N X N X N X

Q Q X Q X Q X

⎫= + + +⎪

= + + + ⎬⎪= + + + ⎭

Above equations are the general expressions used to express the self-induced forces in an nth indeterminate structure.

ijδIt can be seen in the canonical equations that the flexibility coefficients are inversely proportional

to the member’s rigidities ( , and ), and that the free terms, EI EA itΔ icΔGA and , have no relationship

with the rigidities because they are determined by the equations


0it i it M ds t N ds

hα αΔ

Δ = +∑ ∑∫ ∫

ic K KR cΔ =∑

In which, there is no quantity involved with the rigidities. It can be seen from above discussion that the self-induced internal forces (induced by temperature

changes and/or support settlements and/or etc.) are directly proportional to the absolute values of the member’s rigidities of an indeterminate structure, i.e., if increase the lateral dimensions of the members—actually increase the rigidities of the members—the self-induced internal force in the members will be increased as well. Obviously, it is not a wise measure to increase the lateral sizes of the members of an indeterminate structure in order to improve the ability to resist their temperature changes and/or support settlements and/or the like.

Based on above discussion, following two aspects must be paid attention to when designing an indeterminate structure.

(1) Avoid the occurrence of the self-induced internal forces so as to relieve or eliminate the disadvantageous influence caused by the self-induced internal forces.

Consider, for instance, the indeterminate rigid frame shown in Fig.14.16. If the entire structure settled a distance , i.e., the four columns A, B, C and D have an identical settlement as shown in Fig.14.16 (a), the settlement will not induce internal forces in the rigid frame. However, if the four columns have different settlements as shown in Fig.14.16 (b), the settlements will really yield internal forces in the frame.

In this context, uneven support settlements of indeterminate structures are the cause to induce internal forces in the structures, to which designer(s) must pay attention in practical design.

Δ Δ

BA C DC′ D′ Δ

BA C DC′A′ B′ D′ Δ

(a)(b)

Fig.14.16 Sketch of support settlements(a) even settlements; (b) uneven settlements

14.5 Complementary Discussion about Computing Models of Structures 611

Another consideration to avoid the occurrence of self-induced internal forces in a long building due to temperature changes is that temperature joint(s) is arranged in the building. In practice, the foundation of a building is buried under the ground; while its superstructure is exposed in the atmosphere. So the deformation of the foundation due to temperature changes is smaller than that of the superstructure, which will subject to larger temperature changes during its lifetime. Furthermore, the deformation—consequently temperature-induced stress—varies proportionally with the length of the building. Therefore, if a so-called temperature joint is arranged in the superstructure of a long building, which will separate the superstructure along its length into several portions, the temperature-induced stress will be greatly reduced in the building.

(2) Employ prestressed structures to modulate the internal forces of the indeterminate structures during their lifetime by means of the forces deliberately provided by the prestresses.

14.5 Complementary Discussion about Computing Models of Structures

In Chapter 1, we have already given some discussion about the development of the computing model of a structure. The section will present further complementary discussion about the establishment of computing models.

14.5.1 Simplification of Structural System

Although all practical structures are generally three-dimensional (or space) structures based on the viewpoint of member’s arrangement and their connections, many of them can be idealized as plane structures for simplifying the analysis by means of their constructions and stressing characteristics. The consideration made for simplifying analysis generally has following aspects.

(1) Take a plane element as the analytical model. For a long prismatic structure with constant cross

sections or a structure consisting of a series of identical plane structures, a plane element of them can be utilized as the analytical model.

Consider, for example, a tunnel whose cross sections remain constant along the direction of its length, are generally subjected to unchangeable loads along the axis of the tunnel. Thusly, an annular ring, obtained by two imaginary adjacent cuts with a unit distance and perpendicular to the axis of the tunnel, is developed as the computing model of the tunnel, as shown in Fig.14.17 for a circular tunnel.

Fig.14.17 Cross section of a circular tunnel

(2) Plane structures arranged in lateral and longitudinal directions are adopted as the computing models.


Now let us focus our attention on a multistory and multibay building frame shown in Fig.14.18. The frame is composed by the beams and the columns to form a space structure. However, the structural analysis of the three-dimensional frame adopts two kinds of simple plane rigid frames as its analytical models, as shown in Figs.14.18 (b) and (c). In practice, the loads applying on the frame are its weight (vertical load), likely wind load and seismic load (horizontal loads). Obviously, the lateral stiffness or ability of resisting sidesway due to horizontal loads is weaker than its longitudinal one. So the response of

the lateral plane rigid frame due to the external loads is more considerable than that of the longitudinal one. Thusly, in practice, the response of the lateral plane rigid frame shown in Fig.14.18 (c) due to horizontal and vertical loads is analyzed, and the response of the longitudinal plane rigid frame only due to the seismic loading is concerned.

(3) Take a presentative plane structure as a computing model of a structure. Some of the space structures are composed by several kinds of plane structures. Under some

conditions they can be simplified as a presentative plane structures to analyze. The establishment of a presentative plane structure needs following considerations.

① Summarize all kinds of substructures (or plane structures) composing a space structure as a presentative plane structure, and determine the loads distributed to the substructures.

② Analyze the substructures subjected to their distributed loads. Consider, for example, a tall building structure composed by two kinds of substructures—plane shear

4m4m 4m 4m 4m 4m 4m 4m 4m 4m 4m

3m6m

5m7m

7m5m

(a)

(c)(b)

Fig.14.18 Plane sections of a space rigid frame(a) arrangement of columns; (b) longitudinal rigid frame; (c) lateral rigid frame

late

ral d

irect

ion

longitudinal direction

column


wall and plane rigid frame—whose column-arrangement map is shown in Fig.14.19 (a). Now we intend to analyze the response of the building due to its horizontal loading (wind and/or seismic load). It can be seen from Fig.14.19 (a) that the horizontal loading is undertaken by 12 sets of plane substructures—two sets of plane shear wall structures and ten sets of plane rigid frame structures. For the purpose of simplifying analysis, two assumptions are made as follows. ⓐ The in-plane rigidity of a floor is infinitely great. The assumption ensures that the cross sections

of the shear walls and floor slabs lying on the same storey have an identical sidesway. ⓑ The plane shear walls and the plane rigid frames are arranged symmetrically. The assumption

assures that there is no torsion deformation happened in the structure under the action of symmetric horizontal loadings.

Based on above two assumptions, the analysis of the structure under the action of lateral loads may be simplified as the analysis of a presentative plane structure shown in Fig.14.19 (b), in which two sets of shear wall structures are composed as one set of the structure and ten sets of plane rigid frames are synthesized as one set of the structure as well; whereas the connections between the two presentative plane structures are rigid bars as shown in Fig.14.19 (b).

rigid frameshear wall

(c)

shear wall rigid bar rigid frame

(b)

Fig.14.19 Computing model for frame and shear-wall structure(a) arrangement map for columns and shear walls;(b) a presentative plane structure for frame and shear-wall structure

columnshear wall rigid frame

(a)


Furthermore, if we neglect the rigidity of connecting beams between the two shear walls shown in Fig.14.19 (b) they will be again synthesized as one shear wall as shown in Fig.14.19 (c).

When the computing model of the structure is idealized into the models shown in Fig.14.19 (b) or (c), the analysis of the structure will become the analysis for a plane structure, which will be conveniently analyzed by routine computer program for plane structures. Once the internal forces of the links between the shear wall and the rigid frame are determined, the internal forces in shear walls and in the members of the rigid frames can be obtained easily.

14.5.2 Simplification of Bars

In Chapter 1, some of the general rules are presented for the purpose of simplifying various bars and their connections. Followings are some complementary fundamentals about the simplification of general sense bars.

(a)

(b)

Fig.14.20 Computing model of a bent frame applied in industrial buildings(a) a bent frame applied in industrial buildings; (b) computing model of the bent frame

(1) Use straight bars to represent slightly curved or folded bars. Consider, for example, the bent frame applied to the structure of an industrial building shown in

Fig.14.20 (a). Since the shape of the roof will not effect the internal forces of the columns the roof can be idealized as a straight link, and although the central lines of the cross sections of the columns of the frame are not straight lines they can be idealized as the straight lines as shown in Fig.14.20 (b) for the purpose of simplifying the analysis, in which each column is composed by two different straight lines. The height of the column is the distance between the top of the foundation and bottom chord of the roof; the span of the bent frame is the distance between the central lines of the two bottom columns since the span will have no effect on the internal forces of the columns when the roof is hinged and the loads apply on the columns. If meeting the case that the roof is rigidly connected with the columns the distance between the two top columns will be fitted to be the span of the frame.


Another example is the portal frame shown in Fig.14.21 (a). We can see from the figure that the central lines of the cross sections of the two inclined beams are not straight lines and those of the two columns are not straight vertical lines. For simplifying the analysis, the straight line that cross the centroid of the cross section at the middle section of the beam and be parallel to the outer surface of the beam is adopted as the axial line of the beam; while the vertical line crossing the centroid of the bottom end of the column is used as the axial line of the column. Thusly obtained line diagram will form the computing model of the portal frame as shown in Fig.14.21 (b).

It must be noted that the internal forces obtained from the computing model are the internal forces on the cross sections lying on the axis of the computing model.

(a) (b)

Fig.14.21 Computing model of a portal frame(a) a portal frame; (b) computing model of the frame

(2) Substitute solid bar for latticed member. In the analysis for a complicated structure, a latticed member is often substituted by a solid bar. Again let us pay our attention on the bent frame shown in Fig.14.20 (a). As mentioned above that since

the shape of the roof has no effect on the internal forces in the columns due to the loads applying on the columns, when analyze the internal forces in the columns the latticed roof will be idealized as a solid bar (or link) as shown in Fig.14.20 (b). However, when the internal forces in the members of the latticed roof are interested, the computing model of the roof must be simplified as a simply supported truss subjected not only the roof joint loads but also the longitudinal forces acting along the bottom chords, which are obtained from the analysis of the bent frame shown in Fig.14.20 (b).

(3) Simplification of member’s stiffness The simplification of member’s stiffness has already utilized indirectly in the previous analysis. For

example, the axial deformations of the members of a rigid frame are neglected when evaluate the displacement(s) of the frame. Obviously, the simplification implies that the axial rigidities of the members of the rigid frame are assumed to be infinitely great. Another instance is that when we meet the case that the


girder’s rigidity is much greater than that of the columns of a rigid frame, the girder’s rigidity will be assumed to be infinitely great for the analysis due to horizontal loads. The utilization of the simplification can reference to subsection 14.3.2.

14.5.3 Simplification of Joints

Now let us have some complementary discussions about the simplification of the connections between the members of a structure.

(b)

Fig.14.22 Simplification of connections(a) analyze as a truss; (b) analyze as a rigid frame

(a)

The first consideration about the simplification of the connections is that except for the construction of the connections, the geometric construction of the structure must be taken into account. In practice, it is likely happened that in the viewpoint of geometric construction, the connections of the members of a structure are rigid, but the bending moments transferred through the connections are very small. In this context, the connections can be still idealized as hinged joints. Consider, for example, the structure shown in Fig.14.22 (a). Although the members are almost rigidly connected they can be simplified as hinged joints to form the structure as a simply supported truss because the bending moments yielded at the connections are negligible. However, the connections of the structure shown in Fig.14.22 (b) must be simplified as rigid joints to form the structure to be a rigid frame, insomuch as the bending moments are the principal internal force components on the cross sections of the ends of the members meeting at the joints.

As to the viewpoint of geometric construction of a structure, the difference between a truss and a rigid frame is that although the joints of the truss are hinged together the arrangement of the members ensure the truss still to be a stable structure; whereas the geometric stability of the rigid frame is guaranteed by the rigid joints, i.e., if the rigid joints are changed into hinged ones the system will be unstable. In practical engineering, although the constructions of the connections between steel trusses and reinforced concrete trusses are almost closed the internal forces of the trusses are quite different from those of rigid frames. Axial force is the dominated internal force component, whereas bending moment is only subsidiary internal force component in the members of the trusses. This is the reason why the connections of a truss are


generally idealized into hinged joints. Another consideration about the simplification of the connections is the effect on the member forces

caused by a connecting region through which the connection between members is implemented. In practice, every connection between members needs a connecting region. If the connecting region is small enough to have little effect on the forces in the members connected to it, the connection can be simplified as a joint, whose dimension is never concerned. However, if the connecting region is large enough (say 1/5 of a member’s length) to have considerable effect on the forces in the members connected to it, the effect must take into account. A roughly consideration about the effect of a connecting region is that the region is regarded as a rigid area and the portions of the members located within the rigid region are recognized as being rigid, i.e., they have infinitely great rigidities.

14.5.4 Simplification of Supports

In Chapter 1, we discussed the simplification about complete rigid supports. In the subsection, we focus out attention on the simplification of elastic supports, which not only produce reactions but also displacements. Generally, the ratios between the reactions and their corresponding displacements remain constant. The constant ratios are known as the stiffnesses of the elastic supports. In practice, when the rigidity of a structure approaches the rigidity of its supports the supports are generally fitted to simplify as elastic ones.

The concept of supports can be generalized. For instance, if the response of a portion of a structure is interested, the other portion(s) of the structure can be simplified (if conveniently) as the elastic support(s) of the desired portion, whose stiffness is related to the rigidity of the portion(s).

Consider, for example, the rigid frame shown in Fig.14.23 (a). Assume that only the internal forces in beam AB are desired. The beam AB can be idealized as the beam shown in Fig.14.23 (b), whose supports are elastic. The reaction moments AM BM and yielded at the two elastic supports A and B are proportional to the end rotations Aθ and Bθ , that is,

*A A AM S θ= *

B B BM S θ= *ASWhere and are the rotational stiffnesses of the two elastic supports. Actually, they are the

bending stiffnesses of the members meeting joints A and B except for member AB, i.e., the moments needed to let members AC and AD, BE and BF in Fig.14.23 (a) rotate a unit angle, respectively. They can be expressed as

*BS

*A AD ACS S S= +

*B BE BS S S= + F

Note that when the rigidity of the supports is greatly larger or greatly less than that of their supporting structure, the supports may be idealized as those discussed in Chapter 1.


SUMMARY

In this chapter, an all-around discussion has made about the fundamental analytical methods, the characteristics of statically indeterminate structures and the selection of computing models used for the analysis of the structures.

(1) The attention for the discussion about the fundamental analytical methods is mainly paid to the hand-oriented methods. The selection of the methods must be aimed at a concrete problem. This implies that the type of an analytical method is related to the type of loads applying on a structure as well as the type of the structures analyzed by the method in order to simplify the analysis. For the structures having too much unknowns, the matrix displacement method should be employed so as to utilize the assistance of computers.

(2) Because of the redundant restraint(s), a statically indeterminate structure has some special characteristics which its comparable statically determinate one does not have. The advantages of the special characteristics might be utilized and their disadvantages could be avoided if we have a complete cognition about them.

(3) The general rule for the establishment of computing models can be concluded into the two implementing principles: ① starting with practicality, ② distinguish the dominating and the subsidiary.

So-called ‘starting with practicality’ means that before the computing model of a structure is set up, the arrangement of the members, the geometric construction and the likely behavior of the structure must be soundly understood. While so-called ‘distinguish the dominating and the subsidiary’ implies that what are the main factors that strongly influence the results of the analysis, and what are the subsidiary factors which may have little effect on the results of the analysis, must be clearly known when establish the computing model.

F

ED

CBA

qA B

q

(a)

Fig.14.23 Simplification of elastic supports(a) an original structure; (b) computing model of beam AB

Aθ BθA A AM S θ∗= M SB B Bθ∗=

(b)



14-1 What is the difference between the primary system used in displacement method and that used in the combined method by moment distribution and displacement method?

14-2 What is the stiffness coefficient of sidesway? What is the physical meaning of the coefficient? 14-3 Please use the results given in figure (a) to derive the results shown in figure (b).

ΔΔ

hEI

Q

(a)

14-4 Why can a multistory and/or multibay rigid frame subjected to vertical loads be analyzed by single-storey method? Why?

14-5 In what condition(s) inflection-point method can be used? In which case(s) the results obtained by inflection-point method are exact solutions?

14-6 How many portions is a computing model composed? What are the principles to develop a computing model?

3

3Qh

EIΔ =

3

12Qh

EIΔ =

EI h

Q

(b)

reflecting problem 14-3(a) a cantilever column; (b) a column with bottom end fixed and top end fixed freely



14-1 Chose the proper method to analyze the structures shown in the figures and draw their bending moment diagrams as well.

14-2 Draw the bending moment and shear force diagrams and determine the reaction at support C for the structure shown in the figure.

14-3 Analyze the structure shown in the figure by using combined method of moment distribution and displacement method, and draw the bending moment diagram as well. constantEI =

14-4 Draw the bending moment diagram for the structure shown in the figure by inflection-point method.

problem 14-1

A B

D E

C

l

P

h h

2a

2a

P

D

H

B

A

F

EC

G

(a) (b)

constantEI =

constantEI =

hexagon

P

A

BC

D

E

30m

20m20m

4kN

8kN

constantEI =

10m

(c)

C D E

5m

F

15m

A

5m

15m

40m

20kN20kN

1kN/m

(d)

constantEI =

B


14-5 Analyze the rigid frame by shear distribution method; assume that the rigidity of the girder is infinitely great.

CBA

D E F

EI2

EI EI EI

2EI

6m

6m

6m

20kN/mq =

problem 14-2

problem 14-3

E

q

l lBA C D

2l

2l

G H

problem 14-4

E F

A B C

D

I

5m5m22 2

1 11

4i = 5

7100kN

50kN

4m 6m

A DIB

C

E

F

1=

11

12

2 2 2

22

l l

P

problem 14-5


14-6 If neglect the axial deformations due to the axial forces in the members of the structures shown in the figures, try to compare the internal forces between following three sorts of computing models.

P PP

(a) (c)(b)

problem 14-6

14-7 Discuss the stressing characteristics of member AB shown in the figures when the ratio of the

relative stiffnesses 2

1

0i

ki

= → 2

1

ik

i= →∞ or . If the computing model only for member AB is

desired, please draw it.

14-8 The middle 4l portion structure shown in the figure is solid and the other portions of the structure

are truss-liked form. Please establish a proper computing model and analyzing method for the structure.

A BD

q

1i

2i

h1i

B

l

A

qq h

1i

2i

1i

l

(a) (b)problem 14-7


8 a l× =

problem 14-8

14-9 Try to discuss the distribution of the joint load about the structures shown in the figure as to the questions as follows. Assume that the cross sections of the members are identical and square, whose area is .

P

h h×(a) How much load is respectively distributed to the cantilever beam AB and the simple beam CD? (b) How much load is respectively distributed to the beam AB and the column CD? (c) How much load is respectively distributed to the beam AB and the truss CDEF?

P

C

D

2l

2l

l

A B

(a)

A BC

D

P2l

2l

l

(b)

A B

C D

P

E

F

2l

2l

45°

(c)

problem 14-9


14-10 The figure shows a rectangular board subjected to the uniformly downward distributed vertical load, whose a pair of sides AB and CD are simply supported; the third side AD is fixed and the fourth side BC is free as shown in the figure. Please discuss the distributing way of the load and the computing model of the board according to the following two cases.

(a) when ; (b) when . 1l l2

2

1 2l l

14-11 Try to discuss the stressing characteristics and computing model of the rectangular cistern due to the pressure of the water for the following situations.

(a) shallow cistern ( ); (b) deep cistern ( ). / 1/h a < / 2h a >

A B

C

D

2l

1l

problem 14-10

ha

a

(a)

problem 14-11

(b)

h

a

a

Bibliography

1. 包世华主编，结构力学，上、下册，第二版，武汉理工大学出版社，2003 Bao Shihua (editor in chief), Structural Mechanics (volumes 1 and 2), Wuhan University of Technology Press, 2003

2. 包世华主编，结构力学，中央广播电视大学出版社，1996 Bao Shihua (editor in chief), Structural Mechanics, Central Broadcast Television University Press, 1996

3. 龙驭球、包世华主编，结构力学教程（Ⅰ、Ⅱ），高等教育出版社， 2000，2001 Long Yuqiu, Bao Shihua (editors in chief), Structural Mechanics (volumes 1 and 2), High Education Press, 2000, 2001

4. 龚耀清，胡必武，求超静定结构精确影响线的简捷方法，力学与实践 1990，12（1），26~30 Gong Yaoqing, Hu biwu, A simple and direct method to solve precise functions of influence lines for statically indeterminate structures, Mechanics and Practice, 1990, 12(1), 26~30

5. Aslam Kassimali, Structural Analysis, PWS Publishing Company, 1995 6. Bingchen Zhi, Structural Mechanics, Tsinghua University Press, 1996 7. Edwin C. Rossow, Analysis and Behavior of Structures, Prentice Hall Inc. 1998 8. Harry, H. West, Louis F. Geschwindner, Fundamentals of Structural Analysis, Second Edition, John

Wiley & Sons, Inc. 2002 9. James K. Nelson, Jack C. McComac, Structural Analysis Using Classical and Matrix Method, Third

Edition, John Wiley & Sons, Inc. 2003 10. Russell C. Hibbeler, Structural Analysis, Third Edition, Prentice Hall, Inc. 1997

书后语

本书是作者应武汉理工大学出版社之邀，为高等学校土木工程专业（即大‘土木’）编写的英文

结构力学教材。国家教育部要求高等学校在部分课程中施行‘双语’教学，清华大学土木工程系从上世纪 80 年

代中期开始就在结构力学课程中实行了‘双语’教学。当时是按以下两种方式进行的：一种是凡书

面作业（包括课堂板书及同学作业等）全用英文，课堂讲解以汉语为主；另一种则是连课堂讲授也

以英文为主，这样进行了很长时间，因教师的不同，分别采用不同的教学方式。近年则将修结构力

学的学生分为中文班、英文班，分开授课。中文班用中文教学，用自己编写的统一教材。英文班用

英文讲授，曾用过自编的英文讲义；现由任课教师选用相应的英文材料。此做法已正式订入教学计

划要求。本书第一作者自 80 年代中期开始，一直在清华大学土木系和水利系以上述第一种方式进行结构

力学‘双语’教学。本书第二作者近年在宁夏大学和河南理工大学也多次用这种方式讲授过结构力

学。本书的第一稿曾在河南理工大学经过了试用。本书是在积累了上述‘双语’教学经验的基础上

完成的。国内正缺合适的结构力学英文教材，所以我们乐意奉献此书给广大读者。我国土木工程专业有自己的教学计划，结构力学课程有自己的教学大纲和教学基本要求，多年

来也形成了自己的教学体系。这些与美、英等国家的差别是比较大的。原版英文教材，因无统一的

教学大纲与课程要求，各版本的体系与内容的差别极大。直接采用原版英文教材并不太适合我国高

校的教学状况。此外，无论是用中文教学，还是用英文教学，教学计划、教学大纲应当是一样的，

结构力学的体系与内容也应当是一样的。清华大学结构力学课程多年来‘双语’教学实践也是遵循

这一原则的。基于这样的认识，本书（英文版）在编写时是以我国自己的结构力学教学大纲和课程

基本要求为依据，用英文编写。也就是说，它与国内现通用的土木工程专业结构力学（中文版）在

体系与内容上是相同的。这样，使本书不仅完全适合于国内结构力学课程的基本要求，也方便国内

教学使用。另外，结构力学的内容较多，根据我国的教学大纲和教学基本要求，将内容分成为基本部分和

专题部分。基本部分为必修部分；专题部分为选修部分，且各校根据自己的情况常有特殊的安排。

据此，我们也将本书分册出版。基本部分（即本书）以 Structural Mechanics 为名，包含静定结构、

超静定结构和结构矩阵分析等内容。专题部分以 Advanced Topics of Structural Mechanics or Advanced Structural Mechanics 为名，包含能量原理、动力、稳定和极限荷载等内容，另册出版。各校根据自

己的安排，可以全用，也可以分别选用。顺便说一下，多数美英原版英文结构力学是不包含动力、

稳定和极限荷载等内容的，它们是放在后续的高等结构力学中的。在这一点上，我们的分册与它们

是一致的。本书是一本用英文编写的结构力学教材，内容包括了正文、思考题、习题及习题答案等，这些

均与国内中文教材是对应的。‘双语’教学的方式、方法是多种多样的，本书为‘双语’教学提供了

各种英文的书面材料。各校在结构力学‘双语’教学的实践中如何使用本书，可由各校教师根据自

己的情况决定。本书由包世华提出编写大纲和全部内容要求；龚耀清写出全书英文初稿，并设计了书中的图式

和版式，以及文字的录入工作；全书由包世华修改定稿。本书插图的绘制和公式的录入工作由龚云和何书峰等完成，特致谢意。由于英文不是作者的母语，难免有不恰当及错误之处，欢迎读者批评指正。清华大学河南理工大学包世华龚耀清

[email protected]

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