ELEMENT DESIGN MITOITUS - AMK · RAKENNUSTEKNIIKKA Olli Ilveskoski 30.08.2006 rev2 10.01.2007 142 1. INTRODUCTION When designing a structure and its components, the designer must

RAKENNUSTEKNIIKKA Olli Ilveskoski 30.08.2006 rev2 10.01.2007

131

ELEMENT DESIGN

MITOITUS https://www.virtuaaliamk.fi/opintojaksot/030501/1132142124407/1133882367215/1136369180381/1136369611545.html.stx

Kuva: TRY ry:n / Risto Liljan Oppimisympäristön mitoitusperusteet

- Yleistä- Runkojärjestelmät- Palkit- Profiilipellit- Katto- ja seinäorret- Ohutuumapalkit- Ristikot- Pilarit- Jäykistys- Liitokset- Palosuojaus- Liittorakenteet- Tuotemallinnus ja suunnitelmat


132

Kuva: Määräykset ja ohjeetTRY ry:n / Risto Liljan Oppimisympäristö

Kuva: RajatilamitoitusTRY ry:n / Risto Liljan Oppimisympäristö


133

Kuva: KuormatTRY ry:n / Risto Liljan Oppimisympäristö



134


Kuva: MateriaaliominaisuudetTRY ry:n / Risto Liljan Oppimisympäristö


135

Kuva: PoikkileikkausluokatTRY ry:n / Risto Liljan Oppimisympäristö

Kuva: MitoitustaTRY ry:n / Risto Liljan Oppimisympäristö


136




137


Kuva: Palkin mitoitustaTRY ry:n / Risto Liljan Oppimisympäristö


138

Opiskelija perehtyy teräsrakenteiden mitoitukseen.ks ESDEP – oppimisympäristö: http://www.terasrakenneyhdistys.fisuomenkielinen versio

DESIGN OF ELEMENTS

Course Contents

ELEMENTS

Lecture 7.1 : Methods of Analysis of Steel Structures

Lecture 7.2 : Cross-Section Classification

Lecture 7.3 : Local Buckling

Lecture 7.4.1 : Tension Members I

Lecture 7.4.2 : Tension Members II

Lecture 7.5.1 : Columns I

Lecture 7.5.2 : Columns II

Lecture 7.6 : Built-up Columns

Lecture 7.7 : Buckling Lengths

Lecture 7.8.1 : Restrained Beams I

Lecture 7.8.2 : Restrained Beams II

Lecture 7.9.1 : Unrestrained Beams I

Lecture 7.9.2 : Unrestrained Beams II

Lecture 7.10.1 : Beam Columns I

Lecture 7.10.2 : Beam Columns II


139

Lecture 7.10.3 : Beam Columns III

Lecture 7.11 : Frames

Lecture 7.12 : Trusses and Lattice Girders


140

Workgroup Contents

Lecture 7.2 : Cross-Section Classification

Top

1. INTRODUCTION

2. REQUIREMENTS FOR CROSS-SECTION CLASSIFICATION

3. CRITERIA FOR CROSS-SECTION CLASSIFICATION

4. CONCLUDING SUMMARY

5. REFERENCES

6. ADDITIONAL READING


141

Previous | Next | Contents

ESDEP WG 7

ELEMENTS

Lecture 7.2: Cross-Section Classification

OBJECTIVE

To describe the classification of cross-sections and explain how this controls the application of the methods of analysis given in Eurocode 3 [1].

PREREQUISITES

Lecture 7.1: Methods of Analysis of Steel Structures

RELATED LECTURES

Lecture 7.3: Local Buckling

Lectures 7.5.1 & Lecture 7.5.2: Columns

Lectures 7.8: Restrained Beams

Lectures 7.9: Unrestrained Beams

Lectures 7.10: Beam Columns

Lecture 7.11: Frames

Lecture 14.10: Simple Braced Non-Sway Multi-Storey Buildings

RELATED WORKED EXAMPLES

Worked Example 7.1: Cross-Section Classification

SUMMARY

The analysis methods used are primarily dependent upon the geometry of the cross-section and especially on the width-thickness ratio of the elements which make it up.

The lecture describes how sections are classified as plastic, compact or semi-compact and gives the limiting proportions of the elements by which these classifications are made.


142

1. INTRODUCTION

When designing a structure and its components, the designer must decide on an appropriate structural model. The choice of model effects:

• the analysis of the structure, which is aimed at the determination of the stress resultants (internal forces and moments), and

• the calculation of the cross-section resistance.

Thus a model implies the use of a method of analysis combined with a method of cross-section resistance calculation.

There are several possible combinations of methods of analysis and methods of cross-section calculation, for the ultimate limit state, involving either an elastic or plastic design approach; the possible combinations are listed in Table 1.

Table 1 Ultimate Limit State Design - Definition of Design Models

Model Method of Global Analysis (Calculation of internal forces and moments)

Calculation of Member Cross-Section Resistance

I

II

III

IV

Plastic

Elastic

Elastic

Elastic

Plastic

Plastic

Elastic

Elastic Plate Buckling

Model I is related to plastic design of structures. Full plasticity may be developed within cross-sections, i.e. the stress distribution corresponds to a fully rectangular block, so that plastic hinges can form. These have suitable moment rotation characteristics giving sufficient rotation capacity for the formation of a plastic mechanism, as the result of moment redistribution in the structure.

For a structure composed of sections which can achieve their plastic resistance, but have not sufficient rotation capacity to allow for a plastic mechanism in the structure, the ultimate limit state must refer to the onset of the first plastic hinge. Thus, in Model II, the internal forces are determined using an elastic analysis and are compared to the plastic capacities of the corresponding cross-sections. For statically determinate systems, the onset of the first plastic hinge produces a plastic mechanism; both methods I and II should thus give the same result. For statically indeterminate structures, Model II, in contrast to Model I, does not allow moment redistribution.

When the cross-sections of a structure cannot achieve their plastic capacity, both analysis and verification of cross-sections must be conducted elastically. The ultimate


143

limit state, according to Model III, is achieved when yielding occurs at the most stressed fibre. Sometimes yielding in the extreme fibre cannot even be attained because of premature plate buckling of one component of the cross-section; in such cases, the above ultimate limit state should apply only to effective cross-sections (Model IV).

It is obviously not possible to have a model where a plastic method of analysis is combined with an elastic cross-section verification. Indeed, the moment redistribution which is required by the plastic analysis cannot take place without some cross-sections being fully yielded.

2. REQUIREMENTS FOR CROSS-SECTION CLASSIFICATION

In the previous section, the models are defined in terms of structural design criteria; these are actually governed by conditions related to stability problems. Plastic redistribution between cross-sections and/or within cross-sections can take place provided that no premature local buckling occurs, as this would cause a drop-off in load carrying capacity.

It must be guaranteed that no local instability can occur before either the elastic (Model III), or the plastic (Model II), bending resistance of the cross-section, or the formation of a complete plastic mechanism (Model I), is achieved.

Such a mechanism, as envisaged by Model I, can occur provided that the plastic hinge, once formed, has the rotational capacity required for the formation of a plastic mechanism.

To ensure sufficient rotation capacity, the extreme fibres must be able to sustain very large strains without any drop-off in resistance. In tension, the usual steel grades have sufficient ductility to allow for the desired amount of tensile strains; in addition, no drop-off is to be feared before the ultimate tensile strength is reached. With compressive stresses, however, it is not so much a question of material ductility, as of ability to sustain these stresses without instability occurring.

Table 2 gives a summary of the requirements for cross-sections in terms of behaviour, moment capacity and rotational capacity. As can be seen from this table, the limits are referred to cross-section classes, according to Eurocode 3 [1], each corresponding to a different performance requirement:

Class 1 Plastic cross-sections: those which can develop a plastic hinge with sufficient rotation capacity to allow redistribution of bending moments in the structure.

Class 2 Compact cross-sections: those which can develop the plastic moment resistance of the section but where local buckling prevents rotation at constant moment in the structure.

Class 3 Semi-compact cross-sections: those in which the stress in the extreme fibres should be limited to yield because local buckling would prevent development of the plastic moment resistance of the section.


144

Class 4 Slender cross-sections: those in which yield in the extreme fibres cannot be attained because of premature local buckling.

Table 2 Cross-section requirements and classification

M

M

M

Mpl

Mpl

Mpl

Mel

Mel

Localbuckling

Localbuckling

Localbuckling

φ

φ

φ

Model ofBehaviour

MomentResistance

Rotation Capacity Class

Plastic momenton full section

Plastic momenton full section

fy

fy

Elastic momenton full section

Elastic moment oneffective section

fy

fy

M

Mpl

1

1

1

1

Localbuckling

φφ

φ

φ

φ

φrot

M

M

M

M

φpl

φpl

φpl

φpl

φpl

Mpl

Mpl

Mpl

Mpl

Mal

Mpl

Sufficient

Limited

None

None

1

1

1

1

1

2

3

4

The moment resistances for the four classes defined above are:

for Classes 1 and 2: the plastic moment (Mpl = Wpl . fy)

for Class 3: the elastic moment (Mel = Wel . fy)

for Class 4: the local buckling moment (Mo < Mel).

The response of the different classes of cross-sections, when subject to bending, is usefully represented by dimensionless moment-rotation curves.

The four classes given above are recognised for beam sections in bending. For struts loaded in axial compression, Classes 1, 2 and 3 become one, and, in the absence of overall buckling are referred to as "compact"; in this case Class 4 is referred to as "slender".


145

3. CRITERIA FOR CROSS-SECTION CLASSIFICATION

The classification of a specific cross-section depends on the width-to-thickness ratio, b/t, of each of its compression elements. Compression elements include any component plate which is either totally or partially in compression, due to axial force and/or bending moment resulting from the load combination considered; the class to which a specified cross-section belongs, therefore, partly depends on the type of loading this section is experiencing.

a. Components of cross-section

A cross-section is composed of different plate elements, such as web and flanges; most of these elements, if in compression, can be separated into two categories:

• internal or stiffened elements: these elements are considered to be simply supported along two edges parallel to the direction of compressive stress.

• outstand or unstiffened elements; these elements are considered to be simply supported along one edge and free on the other edge parallel to the direction of compressive stress.

These cases correspond respectively to the webs of I-sections (or the webs and flanges of box sections) and to flange outstands (Figure 1).

b. Behaviour of plate elements in compression


146

For a plate element with an aspect ratio, α = a/b (length-to-width), greater than about 0,8, the elastic critical buckling stress (Euler buckling stress) is given by:

σcr = kσ (1)

where kσ is the plate buckling factor (see below),

υ Poisson's coefficient,

E Young's modulus.

The critical buckling stress is proportional to (t/b)2 and, therefore, is inversely proportional to (b/t)2. The plate slenderness, or width-to-thickness ratio (b/t), thus plays a similar role to the slenderness ratio (L/i) for column buckling.

In accordance with the definition of Class 3 sections, the proportions of the plate element, represented by the b/t ratio, must be such that σcr would exceed the material yield strength fy so that yielding occurs before the plate element buckles. The ideal elastic-plastic behaviour of a perfect plate element subject to uniform compression may be represented by a normalised load-slenderness diagram, where the normalised ultimate load:

= σu/fy

and the normalised plate slenderness:

p =

are plotted as ordinates and abscissae respectively (Figure 2).


147

For p < 1, = 1 which means that the plate element can develop its squash load

σu = fy. For p > 1, decreases as the plate slenderness increases, σu being equal to σcr. Substituting the Equation (1) value for σcr into the above and taking υ = 0,3 gives:

p = (2)

This expression is quite general as loading, boundary conditions and aspect ratio all influence the value of the buckling factor kσ .

The factor kσ is a dimensional elastic buckling coefficient, depending on edge support conditions, on type of stress and on the ratio of length to width (a/b), aspect ratio, of the plated element.

In general, the plated elements of a section have an aspect ratio much larger than unity and most of them are submitted to uniform compression. For such cases, Table 3 gives buckling factors for plated elements having various long edge conditions.

Table 3 Elastic buckling factor kσ

Support conditions at long edges

Clamped + clamped

Clamped + simply supported

Simply supported + simply supported

Clamped + free

Simply supported + free

Free + free

Buckling factor kσ

Various supportconditions

6,97

5,40

4,00

1,28

0,43

(b/a)2

a

b

a/b >> 1


148

When plated elements of sections are submitted to any kind of direct stress, other than uniform compression (e.g. webs of a girder in bending), the buckling factor kσ

has to be modified to take account of the stress gradient, given by the stress ratio, ψ.

Table 4 gives the buckling factors for different stress ratios ψ , for internal or outstand elements. In the latter case a distinction is made for elements with tip in compression or in tension.

Table 4 Buckling factors and stress distribution

I II III

σ1

σ1 1= maximum compressive stress and s is positive

σ1

σ2σ2σ2

σ1

ψ σ σ= /2 1 +1 1 > > 0ψ 0 0 > > -1ψ -1

σ2

σ1ψ =

Bucklingfactorkσ

Case Iinternalelement

Case IIoutstandelement

Case IIIoutstandelement

4,0

0,43

0,43

8,41,1 + ψ

0,578ψ + 0,34

0,57 - 0,21 +0,07

ψψ2

7,64

1,70

0,57

7,64 - 6,26 +10

ψψ2

1,7 - 5 +17,1

ψψ2

0,57 - 0,21 +0,07

ψψ2

23,9

23,8

0,85

c. Limit plate element slendernesses

The actual behaviour is somewhat different from the ideal elastic-plastic behaviour represented in Figure 2 because of:

i. initial geometrical and material imperfections,

ii. strain-hardening of the material,

iii. the postbuckling behaviour.

Initial imperfections result in premature plate buckling, which occurs for p < 1.

The corresponding limit plate slenderness p3, for Class 3 sections, may differ substantially from country to country because of statistical variations in imperfections and in material properties which are not sufficiently well known to be quantified accurately; a review of the main national codes shows that it varies from


149

0,5 to 0,9 approximately. Eurocode 3 [1,2] has adopted p3 = 0,74 as the limit plate

slenderness of Class 3 compression elements and p3= 0,9 for elements in bending where the yield strength may be reached in the extreme fibre of the cross-section.

For plate elements for which p < p3, no plate buckling can occur before the maximum compressive strength reaches the yield strength.

A Class 1 section must develop a resistance moment equal to the plastic capacity of the section and must maintain this resistance through relatively large inelastic deformations. In order to fulfil these conditions without buckling, the entire plate element must be yielded and the material must be strained in the strain-hardening region (see Table 2); this is only possible for elements with low reference

slendernesses ( p < p1), see Figure 2.

On the basis of certain theoretical approaches [3, 4, 5] values of p1 between 0,46 and 0,6, are proposed in various standards. The difference can be explained in the

choice of the amount of necessary rotation capacity. A value of p1= 0,6 corresponds to a limited rotation capacity which is estimated to be sufficient for usual plastic design (continuous beams, non-sway frames, etc.). In Eurocode 3 [1], the proposed value is:

p1 = 0,5

A Class 2 (or compact section) is one which can just reach its plastic moment resistance but has a rapid drop-off in resistances at that point (Table 2). The plate element is yielded and the material strained in the plastic range; it occurs for

elements with medium reference slendernesses p2 where:

p1 < p2 < p3

In Eurocode 3 [1], the proposed value is p1 = 0,6.

Using formula (2), and the appropriate values of p and kσ , the limiting b/t ratios can be calculated. Table 5 gives some limiting value of b/t for the elements of the cross-section of a rolled I-profile in compression or bending.


150

Table 5 Maximum slenderness ratios for the elements of a rolled section in compression or in bending

Class 3 cross-sectionElement Class 1

cross-section

Class 2

cross-section Formula kσ b*/t

or d/tw

Flange (1)

(b*/t)

9ε 10ε21ε

0,43 14ε

(1)

Web in compression d/tw

33ε 38ε21

1,0 42ε

Web in pure bending d/tw

72ε 83ε25,4ε

23,9 124ε

fy 235 275 355

ε = ε 1,0 0,92 0,81

(1) In practice, b, the half-width of the flange is considered instead of b*. For this reason, the values given in the "Essentials of Eurocode 3" is b = 15 ε > b*.

The most important limiting proportions of the elements of a cross-section, which enable the appropriate classifications to be made, are specified in Eurocode 3 [1]. Appendix 1 gives the limiting proportions for compression elements of Class 1 to 3.

The limiting values of the width-to-thickness ratio (b/t) of the plate elements of sections apply to members in steel of a specific yield strength. In order to cover all grades of steel, Eurocode 3 presents local buckling data non-dimensionally, in terms

of a reduction factor ε = , where 235 represents the yield stress of mild steel and fy that of the steel considered.


151

The various compression elements in a cross-section (such as a web or a flange) can, in general, be in different classes and a cross-section is normally classified by quoting the least favourable (highest) class of its compression elements.

It is important, particularly in plastic design, that the sections selected for various members should be, in all cases, appropriate for the assumed mode of behaviour.

When any of the compression elements of a cross-section fail to satisfy the limits given in Table 5 for Class 3, the section is classified as "slender" and local buckling shall be taken into account in the design. This may be done by means of the effective cross-section method which is discussed in detail in Lecture 7.3.


• The methods of analysis used are influenced by the geometry of the cross-sections and, more particularly, by the width to thickness ratios of the plate elements in compression.

• It must be guaranteed that no local instability can occur before a complete mechanism is achieved or before the plastic or elastic moment can be reached.

• Four cross-sectional classes are identified, each corresponding to a different performance requirement: plastic, compact, semi-compact and slender cross-sections.

• Limiting proportions for the elements of a cross-section, which enable the appropriate classifications to be made, are given in the lecture.

• When any of the compression elements of a cross-section fail to satisfy the limiting proportions for Class 3 (semi-compact), local buckling shall be taken into account in the design.


152

3. DESIGN OF MEMBERS3.1 Columns in Compression

As shown by the effective cross-sections 1 and 2, in Figure 4, the neutral axes of doubly-symmetrical cross-sections will not change with the formation of effective holes. Hence the compression load NSd that is central to the gross cross-sections will also be central to the effective cross-section.

The column buckling design check, therefore, is based on the non-dimensional

slenderness = , where Ncr is calculated on the basis of the gross cross-section and Npl is calculated using the effective cross-sectional area Aeff (Npl = Aeff . fy).

The design buckling resistance is given by:

NbRd = χ . Npl /γM

where χ is the reduction factor for the relevant buckling curve.

• For singly symmetrical cross-sections - type 3 in Figure 4 - or unsymmetrical cross-sections, the formation of effective holes may lead to a shift, eN, in the neutral axes position. The compression load, NSd, that is central to the gross cross-section will, therefore, be eccentric to the effective cross-section and hence will cause an additional bending moment M=NSd.eN. The member is now a beam-column and must be checked in accordance with Section 3.3.

3.2 Beams in Bending

Beams must be checked using the section modulus determined for the effective cross-sections, as given in Figure 5. In general the attainment of the yield strength at the compression face will limit the design bending resistance of effective cross-sections:

Mo,Rd = Weff . fy /γM1

For cross-sections similar to type 3 of Figure 5, Weff will be determined for the actual edge (e):

Weff =

and not the edge of the effective hole.


153

The lateral-torsional buckling check for beams is analogous to that for columns. The

non-dimensional slenderness LT = , is calculated with Mu=Weff.fy for the effective cross-section, and with Mcr calculated for the gross cross-sectional values. The design lateral-torsional buckling resistance is then given by:

Mb,Rd = χLT . Mu /γM1

where χLT is the reduction factor for the relevant buckling curve.

3.3 Beam-Columns

In the case of members that are subject to compression and monoaxial or biaxial bending (e.g. in the case of columns with monosymmetrical or unsymmetrical cross-sections) the design check is carried out using an interaction formula in which the checks for a centrally compressed column, for a beam with bending about the y-axis only, and for a beam with bending about the z-axis only, are combined.

If lateral-torsional bucking is prevented the interaction formula is as follows:

If lateral-torsional bucking can occur:

where eNy or eNz are the eccentricities due to the shift of the neutral axis for compression only. The resistances NbRd and Nbz.Rd are related to the case of central compression; Moy.Rd and Mby.Rd are related to bending about the y-axis only; and Moz.Rd is related to bending about the z-axis (weak axis) only.


• The design of members with Class 4 sections is carried out as for members with Class 3 sections (elastic analysis, elastic cross-sectional resistance limited by yielding in the extreme fibres) except that an effective cross-section (derived from gross sections with "effective holes", where buckles may occur) is used.

• The buckling checks for columns and the lateral-torsional buckling check for beams, requires the critical values, Ncr and Mcr, to be calculated using the gross cross-sectional data without considering "effective holes".


154

• In the case of columns with non-doubly symmetric cross-sections, the formation of "effective holes" may cause a shift in the neutral axis position resulting in eccentric compression and hence a beam-column problem.

• Beam-columns (compression and biaxial bending) are verified by using an interaction formula where the checks for the column in compression only, for the beam with bending about the y-axis only, and for the beam with bending about the z-axis only, are combined.

5. REFERENCES

[1] "Eurocode 3: "Design of steel structures" ENV 1993-1-1: Part 1.1, General rules and rules for buildings, CEN, 1992.

6. ADDITIONAL READING

1. "Eurocode 3: Part 1.3: "Cold formed thin gauge members and sheeting", CEN, (in preparation).

2. Eurocode 3: Background Document 5.5. (Justification of the design resistances, for buckling verifications)


155

3.

5.2 Basis of the ECCS Buckling Curves

From 1960 onwards, an international experimental programme was carried out by the ECCS to study the behaviour of standard columns [2]. More than 1000 buckling tests, on various types of members (I, H, T, U, circular and square hollow sections), with different values of slenderness (between 55 and 160) were studied. A probabilistic approach, using the experimental strength, associated with a theoretical analysis, showed that it was possible to draw some curves describing column strength as a function of the reference slenderness. The imperfections which have been taken into account are: a half sine-wave geometric imperfection of magnitude equal to 1/1000 of the length of the column; and the effect of residual stresses relative to each kind of cross-section.

The European buckling curves (a, b, c or d) are shown in Figure 14. These give the value for the reduction factor χ of the resistance of the column as a function of the reference slenderness for different kinds of cross-sections (referred to different values of the imperfection factor α).

The mathematical expression for χ is:

χ = 1/ {φ + [φ2 −€ 2]1/2} ≤ 1 (10)

where: φ = 0,5 [1 + α ( - 0,2) + 2] (11)

Table 1 gives values of the reduction factor χ as a function of the reference

slenderness .

The imperfection factor α depends on the shape of the column cross-section considered, the direction in which buckling can occur (y axis or z axis) and the fabrication process used on the compression member (hot-rolled, welded or cold-formed); values for α, which increase with the imperfections, are given in Table 2.


156

Curve a represents quasi perfect shapes: hot-rolled I-sections (h/b > 1,2) with thin flanges (tf ≤ 40mm) if buckling is perpendicular to the major axis; it also represents hot-rolled hollow sections.

Curve b represents shapes with medium imperfections: it defines the behaviour of most welded box-sections; of hot-rolled I-sections buckling about the minor axis; of welded I-sections with thin flanges (tf ≤ 40mm) and of the rolled I-sections with medium flanges (40 < tf ≤ 100mm) if buckling is about the major axis; it also concerns cold-formed hollow sections where the average strength of the member after forming is used.

Curve c represents shapes with a lot of imperfections: U, L, and T shaped sections are in this category as are thick welded box-sections; cold-formed hollow sections designed to the yield strength of the original sheet; hot-rolled H-sections (h/b ≤ 1,2 and tf ≤ 100mm) buckling about the minor axis; and some welded I-sections (tf ≤40mm buckling about the minor axis and tf > 40mm buckling about the major axis).

Curve d represents shapes with maximum imperfections: it is to be used for hot-rolled I-sections with very thick flanges (tf > 100mm) and thick welded I-sections (tf> 40mm), if buckling occurs in the minor axis.

Table 4 helps the selection of the appropriate buckling curve as a function of the type of cross-section, of its dimensional limits and of the axis about which buckling can occur. For cold-formed hollow sections, fyb is the tensile yield strength and fya is the average yield strength. If the cross-section in question is not one of those described, it must be classified analogously.

It is important to note that the buckling curves are established for a pin-ended, end loaded member; it is necessary carefully to evaluate the buckling lengths if the boundary conditions are different, see Lecture 7.7.

5.3 Equivalent Initial Bow Imperfection

To study a column using second order theory, it is necessary to choose geometrical imperfections (initial out-of-straightness and eccentricities of loading) and mechanical imperfections (residual stresses and variations of the yield stress). Eurocode 3 proposes values for a bow imperfection, eo, whose effect is equivalent to a combination of the two previous kinds of imperfections [1].

If the column is designed using elastic analysis, eo is as follows:

eo = α ( - 0,2) Wpl/A for plastic design of cross-sections

or,

eo = α ( - 0,2) Wel/A for elastic design of cross-sections (12)


157

If it is designed with an elastic-plastic analysis (elasto-plastic or elastic-perfectly plastic), the values of eo are functions of the buckling length L, and are given in Table 3.

3. DESIGN OF BEAMS FOR SIMPLE BENDING

For a doubly-symmetrical section or a singly-symmetrical section bent about the axis of symmetry, the basic theory of bending, assuming elastic behaviour, gives the distribution of bending stress shown in Figure 1.

Since the maximum stress σmax is given by:

σmax = (1)

where M is the moment at cross-section under consideration.

d is the overall depth of section.

I is the second moment of area about the neutral axis (line of zero strain).


158

it follows that limiting this to a fraction of the material yield stress gives a design condition of the form:

W ≥ M/fd (2)

where W =

fd is the limiting normal bending stress.

When M is taken as the moment produced by the working loads, this approach todesign is termed 'elastic' or 'permissible stress' design and is the method traditionally used in many existing codes of practice. In more modern Limit States codes fd is taken as the material strength fy possibly divided by a suitable material factor γM and M is taken as the moment due to the factored loads i.e. the working loads suitably increased so as to provide a margin of safety in the design. Equation (2), in this case, represents the condition of first yield. Values of W for the standard range of sections are available in tables of section properties.

Selection of a suitable beam therefore comes down to:

1. determination of the maximum moment in the beam,2. extraction of the appropriate value of fd from a suitable code,3. selection of a section with an adequate value of W subject to

considerations of minimum weight, depth of section, rationalisation of sizes throughout the structure, etc...

Clearly, sections for which the majority of the material is located as far away as possible from the neutral axis will tend to be the most efficient in elastic bending. Figure 2 gives some quantitative idea of this for some of the more common structural shapes. I-sections are most often chosen for beams because of their structural efficiency; being open sections they can also be connected to adjacent parts of the structure without undue difficulty. Figure 3 gives some typical examples of beam-to-column connections.


159


160

Utilisation of the plastic part of the stress-strain curve for steel enables moments in excess of those which just cause yield to be carried. At full plasticity the distribution of bending stress in a doubly-symmetrical section will be as illustrated in Figure 4, with half the section yielding in compression and half in tension.

The corresponding moment is termed the fully plastic moment Mpl. It may be calculated by taking moments of the stress diagram about the neutral axis to give:

Mpl = fy Wpl (3)

where fy is the material yield stress (assumed identical in tension and compression).

Wpl is the plastic section modulus.

Basing design on Equation (3) means that the full strength of the cross-section in bending is now being used, with the design condition being given by:

Wpl ≥ M/fyd (4)

where M is the moment at cross-section under consideration.

fyd is the design strength (material yield strength divided by a suitable material factor).

When M is due to the factored loads Equation (4) represents the design condition of ultimate bending strength used in Eurocode 3 for beams whose cross-sections meet at least the Class 2 limits [1]. It is usual in codes such as Eurocode 3 for the value of fyd to be taken as the material yield strength, reduced slightly so as to cover possible variations from the expected value.

For continuous (statically indeterminate) structures, attainment of Mpl at the point of maximum moment will not normally imply collapse. Providing nothing triggers unloading at this point, e.g. local buckling does not occur, then the local rotational stiffness will virtually disappear, i.e. the cross-section will behave as if it were a hinge, and the pattern of moments within the structure will alter from the original elastic distribution as successive plastic hinges form. This redistribution of moments


161

will enable the structure to withstand loads beyond that load which produces the first plastic hinge, until eventually collapse will occur when sufficient hinges have formed to convert the structure into a mechanism, as shown in Figure 5. Utilisation of this property of redistribution is termed "plastic design". It can only be used for continuous structures and then only when certain restrictions on cross-sectional geometry, member slenderness, etc.., are observed. The topic is covered in detail in Lecture 7.8.2 in the context of beams and in Lecture 7.11 in the context of frames.


162

4. DESIGN OF BEAMS FOR SHEAR

Although bending will govern the design of most steel beams, situations will arise, e.g. short beams carrying heavy concentrated loads, in which shear forces are sufficiently high for them to be the controlling factor.

Figure 6 illustrates the pattern of shear stress found in a rectangular section and in an I-section assuming elastic behaviour. In both cases shear stress varies parabolically with depth, with the maximum value occurring at the neutral axis. However, for the I-section, the difference between maximum and minimum values for the web, which carries virtually the whole of the vertical shear force, is sufficiently small for design to be simplified by working with the average shear stress, i.e. total shear force/web area. Since the shear yield stress of steel is approximately 1/√3 of its tensile yield stress, a suitable value for "permissible" shear stress when using elastic design is 1/√3 of the permissible tensile stress.

The ultimate shear resistance (based on plastic principles) used in Eurocode 3 [1] is fyd/√3; this is used in conjunction with a shear area Av, examples of which are:

rolled I-section, load parallel to web Av = A-2btf + (tw+2r)tf

plates and solid bars Av = A

circular hollow sections Av = 2A/π

In cases where high shear and high moment coexist, e.g. the internal support of a continuous beam, it may sometimes be necessary to allow for interaction effects. However, since the full shear capacity may be developed in the presence of quite large moments and vice versa (Eurocode 3 only requires reductions in moment resistance when the applied shear exceeds 50% of the ultimate shear resistance [1]) this will not often be required.


163

5. DEFLECTIONS

Excessive deflections, whilst not normally leading to a structural failure, may, nonetheless, impair the serviceability of the structure.

Deflection might, for example, cause:

• cracking of plaster ceilings.• misalignment of crane rails.• difficulty in opening large doors.

Since these affect the performance of the structure in its working conditions, it is usual to conduct this type of check at service load levels. Eurocode 3 suggests a limit of span/300 for the maximum deflection of beams supporting floors under service live load [1]. Such limits should, however, be regarded as advisory only, and smaller or larger values can be more appropriate in any given situation. Wherever possible, specialist advice should be sought; crane manufacturers, for example, should be able to give detailed guidelines regarding permissable deflection for gantry beams.

7.3 Sway Frames

Sway frames shall be analysed under those arrangements of the variable loads which are critical for failure in a sway mode. In addition, sway frames shall also be analysed for the non-sway mode.

The initial sway imperfections, and member imperfections where necessary, shall be included in the global analysis of all frames.

The allowance for imperfections in the analysis of sway frames is intended to cover effects such as lack of verticality, lack of straightness, residual stresses, etc. It is expressed in Eurocode 3 by means of a set of equivalent geometrical imperfections [1]. These imperfections are not actual construction tolerances but, because they are intended to represent the effect of a number of factors, are likely to be larger than such tolerances. The form specified in Eurocode 3 is:

• The effects of imperfections shall be allowed for in frame analysis by means of an equivalent geometric imperfection in the form of an initial sway imperfection φ determined from:

φ = kc ks φo

with φo = 1/200

kc = [0,5 + 1/nc]0,5 but kc ≤ 1,0

and ks = [0,2 + 1/ns]0,5 but ks ≤ 1,0

where nc is the number of columns per plane.


164

ns is the number of storeys.

• Columns which carry a vertical load NSd of less than 50% of the mean value of the vertical load per column in the plane considered, shall not be included in nc.

• Columns which do not extend through all the storeys included in ns shall not be included in nc. Those floor levels and roof levels which are not connected to all the columns included in nc shall not be included when determining ns.

• These initial sway imperfections apply in all horizontal directions, but need only be considered in one direction at a time.

• The possible torsional effects on the structure of anti-symmetric sways, on two opposite faces, shall also be considered.

• If more convenient, the initial sway imperfection may be replaced by a closed system of equivalent horizontal forces, see Figure 7.

• In beam-and-column building frames, these equivalent horizontal forces should be applied at each floor and roof level and should be proportionate to the vertical loads applied to the structure at that level, see Figure 8.

• The horizontal reactions at each support should be determined using the initial sway imperfection and not the equivalent horizontal forces. In the absence of actual horizontal loads, the net horizontal reaction is zero.


165

First-order or second-order analysis may be used. If the analysis is first-order, second-order effects may be allowed for in an appropriate way when designing the columns by using the results of a first-order analysis and either:

- using amplified sway moments, or

- using the sway-mode buckling lengths.

• When second-order elastic global analysis is used, in-plane buckling lengths for the non-sway mode may be used for member design.

• In the amplified sway moments method, the sway moments found by a first-order elastic analysis should be increased by multiplying them by the ratio:

where VSd is the design value of the total vertical load.

Vcr is its elastic critical value for failure in a sway mode.

• The amplified sway moments method should not be used when the elastic critical load ratio VSd/Vcr is more than 0,25.

• Sway moments are those associated with the horizontal translation of the top of a storey relative to the bottom of that storey. They arise from horizontal loading and may also arise from vertical loading if either the structure or the loading is asymmetrical.

• As an alternative to determining VSd/Vcr directly, the following approximation may be used in beam-and-column type frames:

where δ, h, H and V are as defined previously.

• When the amplified sway moments method is used, in-plane buckling lengths for the non-sway mode may be used for member design.

• When first-order elastic analysis with sway-mode in-plane buckling lengths is used for column design, the sway moments in the beams and the beam-to-column connections should be amplified by at least 1,2 unless a smaller value is shown by analysis to be adequate.

Rules for the application of plastic analysis procedures to sway frames are given in Clause 5.2.6.3 of Eurocode 3 [1].


166

Lecture 7.12: Trusses and Lattice Girders

OBJECTIVE/SCOPE

To introduce two-dimensional trusses: types, uses and principal design considerations.

PREREQUISITES

None.

RELATED LECTURES

None.

SUMMARY

This lecture presents the types and uses of trusses and lattice girders and indicates the members that are most often used in their construction. A discussion of overall truss design considers primary analysis, secondary stresses, rigorous elastic analysis, cross- braced trusses and truss deflections. The practical design of truss members is discussed.


167

1. INTRODUCTION

The truss or lattice girder is a triangulated framework of members where loads in the plane of the truss or girder are resisted by axial forces in the individual members. The terms are generally applied to the planar truss. A 'space frame' is formed when the members lie in three dimensions.

The main uses are:

• in buildings, to support roofs and floors, to span large distances and carry relatively light loads, see Figure 1.

• in road and rail bridges, for short and intermediate spans and in footbridges, as shown in Figure 2.

• as bracing in buildings and bridges, to provide stability where the bracing members form a truss with other structural members such as the columns in a building. Examples are shown in Figure 3.


168

The principle of a truss is simple. The structure is composed of top and bottom chords triangulated with diagonals in the webs so that each member carries purely axial load. Additional effects do exist but in a well designed truss these will be of a secondary nature.

A global moment on a truss is carried as compression and tension in the chords. A global shear is carried as tension or compression in the diagonal members. In the simplified case, where joints are considered as pinned, and the loads are applied at the panel points, the loading creates no bending moment, shear, or torsion in any


169

single member. Loads applied in such a way as to cause bending, shear, or torsion will usually result in inefficient use of material.

Trusses and lattice girders are classified in accordance with the overall form and internal member arrangement. Pitched trusses are used for roofs. Parallel chordlattice girders are used to support roofs and floors and for bridges, although in continuous bridges, additional depth is often required at the piers. In the past, proper names were given to the various types of trusses such as the Fink truss, Warren girder, etc.. The most commonly used truss is single span, simply supported and statically determinate with joints assumed to act as pins.

The Vierendeel girder should also be mentioned. It consists of rigid jointed rectangular panels as shown in Figure 1d. This truss is statically indeterminate and will not be further considered in this lecture, although it has a pleasing appearance and is often used in foot-bridges.

The saving over a plate girder is clear when the webs are considered. In a truss the webs are mainly fresh air - hence less weight and less wind force.

A truss can be assembled from small easily handled and transported pieces, and the site connections can all be bolted. Trusses can have a particular advantage for bridges in countries where access to the site is difficult or supply of skilled labour is limited.

2. TYPICAL MEMBERS

Truss, lattice girder and bracing members for buildings are selected from:

• open sections, primarily angles, channels, tees and joists.• compound sections, i.e. double angle and channels.• closed sections, in practice structural hollow sections.

For bridges, members are selected from:

• rolled sections.• compound sections.• built-up H, top hat and box sections.

Typical sections are shown in Figure 4.


170

The selection of members depends on the location, use, span, type of connection and the appearance required. Hollow sections are more expensive than open sections but are cheaper to maintain and have a better appearance. However, in exposed trusses corrosion can occur at the crevices which are formed at gusset positions. Angles are the sections traditionally used for small span truss construction.


171

3. LOADS ON TRUSSES AND LATTICE GIRDERS

The main types of loads on buildings are shown in Figure 5, namely:

1. Dead loads. These are caused by self-weight, sheeting, decking, floor or roof slabs, purlins, beams, insulation, ceilings, services and finishes. Dead loads for the construction to be used in any particular case must be


172

carefully estimated from material weights given in handbooks and manufacturers' literature.

2. Imposed loads. These are given in Eurocode 1 [1] for floors in various types of building and for roofs with or without access. The imposed load may cover the whole or part of the member and should be applied in such a way as to cause the most severe effect.

3. Wind loads. These are given in Eurocode 1 [1] and can be estimated from the location of the building, its dimensions and the sizes of openings on its faces. Wind generally causes uplift on roofs and this can lead to reversal of load in truss members in light construction. In multi-storey buildings, wind gives rise to horizontal loads that must be resisted by the bracing.

In special cases, trusses resist dynamic, seismic and wave loads. A careful watch should be kept for unusual loads applied during erection. Failures may occur at this stage when the final lateral support system is not fully installed.

For bridges, in addition to the dead loads and the vertical effects of live loads due to highway or railway loading, horizontal effects of live load have to be considered. These include braking and traction effects, centrifugal loads and accidental skidding loads. Temperature effects are significant in some bridges.

4. ANALYSIS OF TRUSSES4.1 General

Trusses may be single span, statically determinate or indeterminate, or may be continuous over two or more spans, as shown in Figure 6. Only single span, statically determinate, trusses are considered in this section.

A truss is usually statically determinate when:

m = 2j - 3,

where m is the number of members in a truss


173

j is the number of joints.

However compliance with this formula for the truss as a whole does not preclude the possibility of a local mechanism in part of the truss.

Manual methods of analysis for trusses, where the loads are applied at the nodes, are joint resolution, the method of sections and the force diagram. Joint resolution is the quickest method for analysing parallel chord lattice girders when all the forces are required. The method of sections is useful where the values of the forces in only a few critical members are required. The force diagram is the best general manual method. Computer programs are also available for truss analysis.

4.2 Secondary Stresses in Trusses

In many cases in the design of trusses and lattice girders, it is not necessary to consider secondary stresses. These stresses should, however, be calculated for heavy trusses used in industrial buildings and bridges.

Secondary stresses are caused by:

• Eccentricity at connections• Loads applied between the truss nodes• Moments resulting from rigid joints and truss deflection.

They are discussed in detail below:

1. Eccentricity at connections

Trusses should be detailed so that either the centroidal axes of the members or the bolt gauge lines meet at a point at the nodes. Otherwise, members and connections should be designed to resist the moments due to eccentricity. These moments should be divided between members meeting at joints in proportion to their rotational stiffnesses. Stresses due to small eccentricities are often neglected.

2. Loads applied between the truss nodes

Moments due to these loads must be calculated and the stresses arising combined with those due to primary axial loads; that is the members concerned must be designed as beam-columns. This situation often occurs in roof trusses where the loads are applied to the top chord through purlins which may not be located at the nodes, as shown in Figure 7. The manual method of calculation is first to analyse the truss for the loads applied at the nodes which gives the axial forces in the members. Then a separate analysis is made for bending in the top chord which is considered as a continuous beam. The ridge joint E is fixed because of symmetry, but the eaves joint A should be taken as pinned; otherwise, moment will be transferred into the bottom chord if the joint between the truss and column is assumed to be


174

pinned. The top chord is designed for axial load and bending. Computer analysis is mentioned below.

3. Moments resulting from rigid joints and truss deflection

Stresses resulting from secondary moments are important in trusses with short thick members. Approximate rules specify when such an analysis should be made. Secondary stresses will be insignificant if the slenderness of the chord members in the plane of the truss is greater than 50 and that of most of the web members is greater than 100. In trusses in buildings, the loads are predominantly static and it is not necessary to calculate these stresses. The maximum stresses from secondary moments occur at the ends of members and are not likely to cause collapse. However, where fatigue effects are significant, these secondary stresses have to be considered. The method of analysis for secondary moments is set out below.

4.3 Rigorous Elastic Analysis

Rigid jointed, redundant or continuous trusses or trusses with loads applied between the nodes can be analysed using a plane frame program based on the matrix stiffness method of frame analysis. The truss can also be modelled taking account of joint eccentricity. Member sizes must be determined in advance using a manual analysis. All information required for design is output including joint deflections.

It is important that a consistent approach is adopted for analysis and design. This means that if secondary moments are to be ignored then the primary axial forces to be used in design must be obtained from the simple analysis of the truss as a pin jointed frame. The axial forces obtained from a rigid frame computer analysis may be modified considerably by the joint moments.


175

5.4 Design Steps for Compression Members

To design a simple compression member it is first necessary to evaluate its two effective lengths, in relation to the two principal axes, bearing in mind the expected connections at its ends. Secondly, the required second moment of area to resist the Euler critical loads should be calculated to give an idea of the minimum cross-section necessary. The verification procedure should then proceed as follows:

• the geometric characteristics of the shape, and its yield strength give the reference slenderness (Equation (9)).

• χ is calculated, taking into account the forming process and the shape thickness, using one of the buckling curves and (Equations 10 and 11).

Buckling resistance of a compression member is then taken as:

Nb.Rd = χ A fy/γM1 (13)

the plastic resistance as:

Npl.Rd = A fy/γM0 (14)

and the local buckling resistance as:

No.Rd = Aeff fy/γM1 (15)

If these are higher than the design axial load, the column is acceptable; if not, another larger cross-section must be chosen and checked.

In addition, torsional or flexural-torsional buckling of the column must be prevented.


o Many different kinds of cross-sections are used as compression members; these include simple members, built-up, tapered and stepped columns.

o A stub column (with ≤ 0,2) can achieve the full plastic resistance of the cross-section and buckling does not need to be checked.

o If > 0,2, reduction of the load resistance must be considered because of buckling. Columns with medium slenderness fail by inelastic buckling and slender columns by elastic buckling.

o European buckling curves give the reduction factor for the relevant buckling mode depending on the shape of the cross-section, the forming-process, the reference slenderness and the axis about which buckling can occur. They take into account experimental and theoretical approaches and give reliable results.


176

LASKUESIMERKKEJÄ:Saarinen, Eero. Teräsrakenteiden suunnittelu, Eurocode. Sähköinen versio. Teräsrakenneyhdistys r.y. 2002. Julkaisematon

Saarinen, Eero. Teräsrakenteiden suunnittelu, Eurocode. Sähköinen versio. Teräsrakenneyhdistys r.y. 2002. Julkaisematon

Esimerkki 3.1Määritä kuvan 3.1 vedetyn sauvan kestävyys, kun teräs on S355 ja levyn paksuus t = 10 mm.On tutkittava leikkaukset I - I ja II - II:

AI = 10 mm ⋅ (120 - 18) mm = 1020 mm2

AII = 10 mm ⋅ (120 - 2 ⋅ 18 + 502/(4 ⋅ 60)) mm = 944 mm2

Tässä tapauksessa Anet = AII .γM0 = 1,1 (poikkileikkausluokka 1)γM2 = 1,25A = 10 mm ⋅ 120 mm = 1200 mm2

Npl.Rd = 1200 mm2 ⋅ 0,355 kN/mm2/1,1 = 387 kNNu.Rd = 0,9 ⋅ 944 mm2 ⋅ 0,510 kN/mm2/1,25 = 347 kN

⇒ Vetokestävyys Nt.Rd = 347 Kn


177

Kuva 3.8. Vapaasti tuettu vesikaton pääkannattaja

Esimerkki 3.2Mitoitettava vapaasti tuettu vesikaton pääkannattaja (kuva 3.8), jonka jänneväli L = 10 m. Kehäväli on 6,0 m ja orsiväli L1 = 2,0 m. Teräs S235J2G3.

.

Palkin kuormitus:kate + eriste 0,30 kN/m2

orret 0,10 “ pysyvä kuorma yhteensä 0,40 kN/m2

lumikuorma 1,80 kN/m2

Ominaislumikuorma orrelta palkille Fqk = 6,0 m ⋅ 2,0 m ⋅ 1,8 KN/m2 = 21,6 kN⇒ max Vqk = 2 ⋅ 21,6 kN = 43,2 kN

max Mqk = (43,2 kN ⋅ 5,0 m - 21,6 kN ⋅ (3,0 + 1,0)m) = 130 kNm

Arvioidaan palkin omaksipainoksi gk ≈ 0,7 kN/m

⇒ max Mk = 18

0 7 10 1302⋅ ⋅ +

, kNm = 139 kNm

Laskentakuormat: gd = 1,2 ⋅ 0,7 kN/m = 0,84 kN/mFd = 1,2 ⋅ 6,0 m ⋅ 2,0 m ⋅ 0,4 kN/m2 + 1,5 ⋅ 21,6 kN = 38,16

kN

max VSd = 12

⋅ 0,84 kN/m ⋅ 10 m + 2 ⋅ 38,16 kN = 80,5 kN


178

max MSd = (80,5 ⋅ 5 -12

⋅ 0,84 ⋅ 52 - 38,16 ⋅ 4) kNm = 239 kNm

S235 ⇒ fy = 235 N/mm2 (t < 16 mm)Poikkileikkausluokka 1 ⇒ γM0 = 1,1

⇒ Wpl ≥11 239 10

235

6, ⋅ ⋅mm3 = 1119 ⋅ 103 mm3

Taipuman rajatila on L/250 (vrt. taulukko 2.4), sillä kokonaistaipuman rajatila antaa pienemmän tuloksen jäyhyysmomentille. ⇒ k = 250

δf = 548

⇒ I ≥ 250548

130 102 1 10

100006

5⋅ ⋅⋅⋅

⋅,

mm4 = 16120 ⋅ 104 mm4 (kaava 3.24)

Vaaditun taivutusvastuksen ja jäyhyysmomentin perusteella valitaan IPE 400.

Poikkileikkausarvot: Iy = 23130 ⋅ 104 mm4

Wpl.y = 1310 ⋅ 103 mm3 (Wel = 1160 ⋅ 103 mm3)iz = 39,5 mmtw = 8,6 mmhs = (400 - 13,5) mm = 387 mmr = 21 mm (pyöristyssäde)gk = 0,663 kN/m < 0,7 kN/m (oletus)

Av = 8450mm2 - ( )2 180 13 5 8 6 2 21 13 5⋅ ⋅ + + ⋅ ⋅mm mm mm mm, , , = 4273 mm2

Leikkauskestävyys Vpl.Rd = 42730 235

11 32

2

mmkN mm

⋅⋅

, /,

= 527 kN > VSd (kaava 3.7)

λLT =

⋅ + ⋅

200039 5

1 1120

2000 39 5400 13 5

2

,

/ ,/ ,

= 47,3 (kaava 3.30)

λ1 = 93,9 ⇒ λ_

LT =47 393 9

,,

= 0,504 ( kaava 3.26 kun βw = 1)

⇒ χLT = 0,9229 (taulukosta 3.4 interpoloimalla)

⇒ Mb.Rd = 0 9229 1 1310 100 235

113 3

2

,, /

,⋅ ⋅ ⋅ ⋅mm

kN mm= 258 kNm > MSd

Yhteisvaikutusta ei tarvitse tarkistaa, koska normaalivoimaa ei esiinny ja taivutusmomentin sekä leikkausvoiman suurimmat arvot vaikuttavat eri poikkileikkauksissa. IPE 400 on siten sopiva profiili. Jos teräkseksi valitaan S355J2G3, voidaan profiiliksi valita IPE 360.


179

pgk2 0.70kNm

⋅:=Palkin op:n metrikuorma:

Fqk1 21.6kN=Fqk1 kk L1⋅ qk1⋅:=Orren pistekuorma lumesta: Fgk1 4.8kN=Fgk1 kk L1⋅ gk1⋅:=Orren pistekuorma rakenteista:

Voimasuureet:

KÄYTTÖRAJATILA:

kk 6 m⋅:=L1 2 m⋅:=L 14 m⋅:=

gk2 0.70kNm

⋅:=

qk1 1.8kN

m2⋅:=

gk1 0.40kN

m2⋅:=

Palkin kuormitus:kate + eriste 0,30 kN/m2orret 0,10 “ pysyvä kuorma yhteensä 0,40 kN/m2


palkin omapaino 0,70 kN/m2

Palkin geometria:jänneväli L = 10 morsiväli L1 = 2 mkehäväli kk = 6 m

Kuva: Vapaasti tuettu vesikaton pääkannattaja. kNm 1000N m⋅:=

kN 1000N:=

Esimerkki : Vapaasti tuettu vesikaton pääkannattajaMitoitettava vapaasti tuettu vesikaton pääkannattaja, jonka jänneväli L = 14 m. Kehäväli on 6,0 m ja orsiväli L1 = 2,0 m. Teräs S235J2G3


180

maxMsd 478.5kNm=

maxMsd 3 Fgd1 Fqd1+( )L2

⋅

Fgd1 Fqd1+( ) 5 m⋅ 3 m⋅+ 1 m⋅+( )⋅[ ]− pgd2L2

8⋅+

:=

maxVsd 120.36kN=maxVsd 3 Fgd1 Fqd1+( ) Lpgd2

2⋅+:=

Muuttuvien ja pysyvien kuormien aiheuttamat voimasuureet käyttötilassa:

pgd2 0.84kNm

=pgd2 γG pgk2⋅:=Palkin op:n metrikuorma:

Fqd1 32.4kN=Fqd1 γQ kk⋅ L1⋅ qk1⋅:=Orren pistekuorma lumesta: Fgd1 5.76kN=Fgd1 γG kk⋅ L1⋅ gk1⋅:=Orren pistekuorma rakenteista:

γQ 1.5:=γG 1.2:=

MURTORAJATILA

maxMk 333.95kNm=

maxMk 3 Fgk1 Fqk1+( )L2

⋅

Fgk1 Fqk1+( ) 5 m⋅ 3 m⋅+ 1 m⋅+( )⋅[ ]− pgk2L2

8⋅+

:=

maxVk 84.1kN=maxVk 3 Fgk1 Fqk1+( ) Lpgk2

2⋅+:=

Muuttuvien ja pysyvien kuormien aiheuttamat voimasuureet käyttötilassa:

maxMq1 259.2kNm=maxMq1 3 Fqk1⋅L2

⋅ Fqk1( ) 5 m⋅ 3 m⋅+ 1 m⋅+( )⋅−

:=

maxVqk1 64.8kN=maxVqk1 Fqk1 3⋅:=

Muuttuvan kuorman= lumikuorman aiheuttama voimasuureet käyttötilassa:


181

L1 2m=

b 210 mm⋅:=h 550 mm⋅:=

r 24 mm⋅:=tf 17.2 mm⋅:=tw 11.1 mm⋅:=iz 43.1 mm⋅:=

Wply 2787000mm3⋅:=

A 8450 mm2⋅:=

Vaaditun taivutusvastuksen ja jäyhyysmomentin perusteella valitaanIPE 550:

Poikkileikkausarvot:A = 8450 mm2Iy = 671200000 mm4Wpl.y= 1310 103 mm3 (Wel = 1160 103 mm3)iz = 43.1 mmtw = 11,1 mmtf = 17,2 mmhs = (400 - 13,5) mm = 387 mmr = 24 mm (pyöristyssäde)gk = 0,663 kN/m < 0,7 kN/m (oletus)h =550 mmb = 210 mm

Itarv 57977.431cm4=Itarv k δf⋅MkE

⋅ L⋅:=

Mk maxMk:=k 250:=δf 548

:=

Taipuman rajatila on L/250 (vrt. taulukko 2.4), sillä kokonaistaipuman rajatila antaa pienemmän tuloksen jäyhyysmomentille. k = 250

Taipuma

Wpltarv 2239.787cm3=WpltarvMsdfyd

:=Msd maxMsd:=PL1

Taivutus

E 2.1 105⋅N

mm2⋅:=

fyd 213.636N

mm2=fyd

fyγM

:=γM 1.1:=fy 235N

mm2⋅:=S235

LujuudetMITOITUS


182

Yhteisvaikutusta ei tarvitse tarkistaa, koska normaalivoimaa ei esiinny ja taivutusmomentin sekä leikkausvoiman suurimmat arvot vaikuttavat eri poikkileikkauksissa. IPE 550 on siten sopiva profiili.

>MSdMbRd 555.714kNm=MbRd χLT βw⋅ Wply⋅fy

γM⋅:=

χLT 0.933=χLT 1

φLT φLT2λLTmuunn2−+

:=

φLT 0.659=φLT 0.5 1 αLT λLTmuunn 0.2−( )⋅+ λLTmuunn2+ ⋅:=

αLT 0.21:=λLTmuunn 0.47=λLTmuunn( ) λLTλl

βw⋅:=

βw 1:=λl 93.913=λl πEfy

⋅:=

λLT 44.138=λLT

L1iz

C1( ) 1120

L1izhtf

2

⋅+⋅

:=

C1 1:=

> VSd

VplRd 276.599kN=VplRd Avfy

γM 3⋅

⋅:=

Av2 6349.2mm2=Av2 1.04 h⋅ tw⋅:=

Av 2242.52mm2=Av A 2 b⋅ tf⋅− tw 2 r⋅+( ) tf⋅+:=

Leikkaus


183

Esimerkki 3.3Kuvan 3.9 jatkuva palkki saa välipohjalta pysyvää kuormaa g1k = 15,0 kN/m ja oleskelukuormaa qk = 7,2 kN/m. Mitoita palkki teräksestä S355J2G3.

Saarinen, Eero. Teräsrakenteiden suunnittelu, Eurocode. Sähköinen versio. Teräsrakenneyhdistys r.y. 2002. Julkaisematon

Kuva 3.3. Esimerkin 3.3 jatkuva palkki.

Arvioidaan omaksipainoksi g0k ≈ 0,4 kN/m ⇒ gk = 15,4 kN/m

Laskentakuormat: gd = 1,2 ⋅ 15,4 kN/m = 18,5 kN/m qd = 1,5 ⋅ 7,2 kN/m = 10,8 kN/m

Pysyvä kuorma gd kuormittavaa koko palkkia. Liikkuvan kuorman qdvaarallisimmat kuorma-asennot eri voimasuureille on esitetty kuvassa 3.10 .

a) Lasketaan voimasuureet kimmoteorialla.Murtotilassa:max M0-1 = [0,080 ⋅ 18,5 + 0,101 ⋅ 10,8] ⋅ 6,02 kNm = 92,5 kNm(kuormitustap. 1)vast. M1 = - [0,100 ⋅ 18,5 + 0,050 ⋅ 10,8] ⋅ 6,02 kNm = - 86,0 kNmmin M1 = - [0,100 ⋅ 18,5 + 0,117 ⋅ 10,8] ⋅ 6,02 kNm = - 112 kNm(kuormitustap. 3)vast. M2 = - [0,100 ⋅ 18,5 + 0,033 ⋅ 10,8] ⋅ 6,02 kNm = - 79,4 kNmmax M1-2 = [0,025 ⋅ 18,5 + 0,075 ⋅ 10,8] ⋅ 6,02 kNm = 45,8 kNm(kuormitustap. 2)

max V0.1 = ( )12

18 5 10 8 6 086 06 0

73 6⋅ + ⋅ − =, , ,,,

,kN kN kN

min V1.0 = ( )− + ⋅ − = −12

18 5 10 8 6 01126 0

107, , ,,

kN kN kN

max V1.2 = ( )12

18 5 10 8 6 0112 79 4

6 093 3⋅ + ⋅ +

−=, , ,

,,

,kN kN


184

Käyttötilassa (kuormitustapaus 1):max M0-1 = [0,080 ⋅ 15,4 + 0,101 ⋅ 7,2] ⋅ 6,02 kNm = 70,5 kNmvast. M1 = - [0,100 ⋅ 15,4 + 0,050 ⋅ 7,2] ⋅ 6,02 kNm = - 68,4 kNm

δ f = ⋅ −⋅

=548

168 4

10 70 50 0941

,,

,

Poikkileikkauksen valinta:

Wpl ≥11 112 10

355

6, ⋅ ⋅mm3 = 347 ⋅ 103 mm3

I ≥ 250 0 094170 5 102 1 10

60006

5⋅ ⋅⋅⋅

⋅,,,

mm4 = 4739 ⋅ 104 mm4

Valitaan IPE 270 (painoltaan pienin valssattu I-profiili, joka täyttää vaatimukset).

Tarkistetaan leikkauskestävyys:Av = [4594 - 2 ⋅ 135 ⋅ 10,2 + (6,6 + 2 ⋅ 15) ⋅ 10,2] mm2 = 2213 mm2

Vpl.Rd = 2213 mm2 ⋅0 355

11 3

2, /,kN mm

⋅= 412 kN > VSd

Kiepahduskestävyyttä kentässä ei tartvitse tarkistaa, koska välipohjalaatta tukee ylälaipan sivusuunnassa, mutta välitukien kohdalla tarkistus on tarpeen. Kokeillaan aluksi onko kiepahduskestävyys riittävä, jos alalaippa on tukematon. Tällöin kuorman vaikuttaessa ylälaipan tasossa saadaan kiepahduskestävyys laskettua seuraavasti:

iz = 30,2 mm tf = 10,2 mm hs = (270 - 10,2) mm = 259,8 mm Wpl = 484 ⋅ 103 mm3

zg = 270

2mm

= 135 mm k = kw = 1,0 C1 ≈ 1,879 C2 = 0

λLT =

+

600030 2

1 879 1120

6000 30 2270 10 2

2

,

,/ ,/ ,

= 74,2

λ1 = π ⋅210000

355= 76,4 ⇒ λ

_ ,,LT = ⋅

74 276 4

1 = 0,971 ⇒ χLT = 0,685

⇒ Mb.Rd = 0,685⋅ 484000 mm3 ⋅0 355

11

2, /,

kN mm= 107 kNm

⇒ Tämän likimääräisen tarkastelun perusteella välituilla tarvittaisiin jäykisteet, mutta tarkemmin laskettaessa ne todennäköisesti olisivat tarpeettomia.


185

Yhdistetyt rasitukset välituella:

b) Voimasuureet plastisuusteorialla

Mekanismi murtotilassa on kuvan 3.11 mukainen. Työyhtälöksi saadaan

( ) ( ) ( )12

1 11

11

g q L ML L Ld d p+ = +

−+

−

ξ ξ ξ

josta ratkaisemalla seuraa täysplastiselle momentille Mp lauseke

Mp = ( ) ( )12

11

2g q Ld d+−

+ξ ξ

ξ(3.33)

Parametrin ξ arvo, jolla Mp saavuttaa suurimman arvonsa, voidaan ratkaista derivoimalla yhtälö 3.43 ja asettamalla derivaatta yhtäsuureksi kuin 0.

( )( )

∂

∂ξξ ξ

ξ

Mg q Lp

d d= +− −

+=

12

1 2

102

2

2

⇒ ξ = − ≈2 1 0 414,Vaatimukseksi taivutuskestävyydelle saadaan siten

Mc.Rd ≥ Mp = ( )12

18 5 10 8 6 02⋅ + ⋅, , , kNm ( )

⋅⋅ −0 414 1 0 4141 414

, ,,

= 90,5 kNm

Wp ≥90 5 10

355

6, ⋅mm3 = 255 ⋅ 103 mm3

⇒ IPE 220 riittäisi taivutuskestävyyden perusteella, mutta taipumarajatila on tässä tapauksessa määräävä. Sen vuoksi on valittava profiili IPE 270. Plastisuusteorian soveltamisesta ei siten tällä kertaa ollut saavutettavissa hyötyä. Palkin teräksen lujuusluokaksi riittäisi S235, mutta silloin alalaippa olisi jäykistettävä välitukien molemmin puolin.


186

Esimerkki 3.4Mitoitetaan kaksitukinen vapaasti tuettu suora kattopalkki, jonka jänneväli L = 25,0 m. Kehäväli on 8,0 m ja orsiväli L1 = 2,5 m. Teräs S355J2G3.

Katon kuormitus:kate + eriste 0,27 kN/m2

orret 0,12 “ pysyvä kuorma yhteensä 0,39 kN/m2


Arvioidaan omaksi painoksi 1,7 kN/m

Kuormat palkille:gk = 8,0 m ⋅ 0,39 kN/m2 + 1,7 kN/m = 4,82 kN/mqk = 8,0 m ⋅ 1,8 kN/m2 = 14,4 kN/m

Käyttötilan voimasuureet

max Vk = ( )12

14 4 4 82 25 0 240⋅ + ⋅ =, , / ,kN m m kN

max Mk = 18

19 22 25 0 15022 2⋅ ⋅ =, / ,kN m m kNm

Murtorajatilan voimasuureet

max VSd = ( )12

1 2 4 82 1 5 14 4 25 0⋅ ⋅ + ⋅ ⋅, , , , / ,kN m m = 342 kN

max MSd = ( )18

1 2 4 82 15 14 4 25 02 2⋅ ⋅ + ⋅ ⋅, , , , / ,kN m m = 2139 kNm

Poikkileikkauksen valinta

Koska rakennekorkeutta ei ole rajoitettu, valitaan optimikorkeuden perusteella (ks. Hitsatut profiilit käsikirja) mahdollisimman kevyt poikkileikkaus.

hopt = 7 011 2139 10

355

63,

,⋅

⋅ ⋅mm = 1315 mm

⇒ Valitaan palkki WI 1300-10-15x300Valitun palkin poikkileikkaus on esitetty kuvassa 3.19 ja sen poikkileikkaussuureilla on seuraavat arvot:

hf = hs = (1300 - 15) mm = 1285 mmhw = d = (1300 - 2 ⋅ 15) mm = 1270 mm (hoikkuutta määritettäessä ollaan

varmallapuolella kun hitsejä ei oteta huomioon)

A = (1270 ⋅ 10 + 2 ⋅ 15 ⋅ 300) mm2 = 21700 mm2

Iy = (10 ⋅ 12703/12 + 2 ⋅ 300 ⋅ 153/12 + 2 ⋅ 15 ⋅ 300 ⋅ 642,52)mm4 = 171954⋅104 m4


187

Kuva 3.19. Esimerkin 3.4 hitsatun palkin poikkileikkaus.Saarinen, Eero. Teräsrakenteiden suunnittelu, Eurocode. Sähköinen versio. Teräsrakenneyhdistys r.y. 2002. JulkaisematonTehollinen poikkileikkausKyseessä on poikkileikkausluokka 4, sillä uuma ei täytä poikkileikkausluokka 3:n ehtoa:

d/tw = 127 > 124 ε = 100

Pristetulle laipalle kσ = 4,0 ja ε = 0,81 (S355)

⇒ λ_ /

, ,p =

⋅ ⋅150 15

28 4 0 81 4= 0,217 < 0,673 ⇒ ρ = 1 ⇒ beff = bf = 300 mm

Uumalle ψ = -1 ⇒ kσ = 23,9 ⇒ λ_ /

, , ,p =

⋅ ⋅1270 10

28 4 0 81 23 9= 1,129 > 0,673

⇒( )

ρ =−1129 0 22

11292

, ,,

= 0,713 ⇒ deff = 0,713 ⋅ 635 mm = 453 mm

de1 = 0,4 ⋅ 453 mm = 181 mm ⇒ de2 = (453 - 181) mm = 272 mm⇒ tehoton alue a = (635 - 453)mm = 182 mm (ks. kuva 3.20)

Tehollinen pinta-ala: Aeff = 21700 mm2 - 10 mm ⋅ 182 mm = 19880 mm2

et = ( )119880

300 15 1292 5 181 10 1194 5 922 10 476 300 15 7 5⋅ ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅, , , mm = 624

mm⇒ ec = (1300 - 624)mm = 676 mmNeutraaliakseli laskee matkan e = (650 - 624)mm = 26 mmIeff =

( ) ( )2 3001512

1018112

1092212

300 15 676 7 5 181 10 676 105 53 3 3

2 2⋅ ⋅ + ⋅ + ⋅ + ⋅ ⋅ − + ⋅ ⋅ − +

, ,

+ ( ) ( ) ]922 10 624 476 300 15 624 7 52 2⋅ ⋅ − + ⋅ ⋅ − , mm4 = 517065 ⋅ 104 mm4

Weff = Wec = 517065 10

676

4⋅mm3 = 7649 ⋅ 103 mm3


188

Kuva 3.20. Tehollinen poikkileikkaus.

Tarkistetaan laipan taipumisesta johtuva lommahdus. Hoikkuussuhteen raja-arvo lasketaan kaavasta 3.37.

ct f

= = < ⋅ =15015

10 14 0 81 11 34, ,

⇒ puristettu laippa kuuluu poikkileikkausluokkaan 3 ⇒ k = 0,55dtw

≤ ⋅ ⋅⋅⋅

0 55210000

3551270 10300 15

, = 547

dtw

=127010

= 127 < 441

Taivutuskestävyys

Taivutuskestävyys Mc.Rd = 7649 100 355

113 3

2

⋅ ⋅mmkN mm, /

,= 2469 kNm > MSd

Kiepahduskestävyys

Iz = 2 1530012

12701012

3 3

⋅ ⋅ + ⋅

mm4 = 6761 ⋅ 104 mm4

Iw = 6761 ⋅ 104 mm4 ⋅1285

4

2 2mm= 27, 91 ⋅ 1012 mm6

epl =⋅ ⋅ + ⋅ ⋅15 300 642 5 635 10 317 5

10850, ,

mm = 452,3 mm

Wpl = 452,3 mm ⋅ 21700 mm2 = 9815 ⋅ 103 mm3

It = ( )13

2 15 300 10 12703 3⋅ ⋅ ⋅ + ⋅ mm4 = 10,98 ⋅ 105 mm4


189

Mcr = ⋅⋅ ⋅ ⋅ ⋅ ⋅

⋅+

⋅ ⋅ ⋅ ⋅⋅ ⋅ ⋅ ⋅

1 0

2 1 10 6761 102500

27 91 1067 61 10

2500 0 8 10 10 98 102 1 10 676 1 10

2 5 4

2

12

6

2 5 5

2 5 5,, ,

,, ,

, ,π

πNmm

Mcr = 934 ⋅ 104 kNm

λ πLT = ⋅ ⋅ ⋅⋅

⋅2 5

3

102 1 109815 10934 10

, = 1,476

λ π1

210000355

= ⋅ = 76,409

λ_ ,

,LT = ⋅1 476

76 40976499815

= 0,0171 < 0,2 tai λ_

LT =⋅ ⋅

⋅7649 10 355

934 10

3

10 = 0,0171 < 0,2

χLT = 1,0

LeikkauskestävyysLeikkauslommahduskestävyys on mitoittava. Sen määrittämiseksi käytetään yksinkertaista ylikriittiseen tilaan perustuvaa menetelmää.

S355 ⇒ ε = 0,81 ⇒ λ_ /

, , ,w =

⋅ ⋅1270 10

37 4 0 81 5 34= 1,814 (uuman jäykisteet vain tuilla)

⇒ τba = 0 9

1814355

3,

,⋅ N/mm2 = 101,7 N/mm2

⇒ Vba.Rd = 1270 mm ⋅ 10 mm ⋅0 1017

11

2, /,

kN mm= 1174 kN

VSd = 0,5 ⋅ 27,38 kN/m ⋅ 25,0m = 342 kN < Vba.Rd

Taipumarajatila

max δ = 548

1502 102 1 10 517065 10

25 106

5 2 4 42 6 2⋅

⋅⋅ ⋅ ⋅

⋅ ⋅Nmm

N mm mmmm

, /= 90,1 mm<

L200

=

125mm


190

Esimerkki 3.5

Molemmista päistään vapaasti tuettu kotelopoikkileikkauksen omaava välipohjapalkki (kuva 3.21 a) on mitoitettava hyötykuormalle qk = 2,0 kN/m2 , kun jänneväli L ≈ 5,75 m (pilariväli 6,0 m). Ontelolaattojen pituus palkin vasemmalla puolella 6,0 m ja oikealla puolella 8,4 m. Teräs S355. Välipohjan (ontelolaatat + pintabetoni) pysyvä kuorma gk = 5,0 kN/m2.

Valitaan pikavalintataulukon perusteella (ontelolaattojen paksuus 265 mm) WQ 265x5-15x240-12x450 (vrt. kuva 3.21 b). ⇒ g0k = 0,915 kN/m.

Kuva 3.21. Esimerkki 3.5:n palkki.

[Kuva 3.21 (alkuper. 3.37) korjattava alalaipan paksuus 15 → 12 kuvassa b)]

Poikkileikkaussuureet

Osa Ai (mm2) Ai ai (mm3) e2 - ai(mm)

Ai(e2 - ai)2

1 3600 970200 -150,6 81649 x 103

2 2650 382925 -25,6 1737 x 103

3 5400 32400 112,9 68831 x 103

Σ 11650 1385525 152217 x 103


191

e2 = 138552511650

mm = 118,9 mm ⇒ e1 = 158,1 mm

Iy = 152217 1024012

151012

26545012

123 3 3 3⋅ + ⋅ + ⋅ + ⋅

mm4 = 16786 ⋅ 104 mm4

Wc = 16786 10

158 1

4⋅,

mm3 = 1062 ⋅ 103 mm3

Wt = 16786 10

118 9

4⋅,

mm3 = 1412 ⋅ 103 mm3

Uuma:

( )ψ =−

− −= −

118 9 12158 1 15

0 747,

,,

⇒ kσ = 7,81 + 6,29 ⋅ 0,747 + 9,78 ⋅ 0,7472 = 17,97

⇒( )

λ_ /

, , ,p =

−⋅ ⋅

265 15 528 4 0 81 17 97

= 0,513⇒ ρ = 1

d/tw = 265 15

5−

= 50 < 72 ⋅ ε = 58,3 ⇒ uuma kuuluu poikkileikkausluokkaan

1

Puristettu laippa: b/tf = 240/15 = 16 < 33 ε = 26,7 ⇒ poikkileikkausluokka 1

Mitoitus taivutukselle

Kuorma palkille:

gk = 0,915 kN/m + ( )12

6 0 8 4⋅ +, , m ⋅ 5,0 kN/m2 = 36,9 kN/m

qk = ( )12

6 0 8 4⋅ +, , m ⋅ 2,0 kN/m2 = 14,4 kN/m

pk = 51,3 kN/m⇒ pd = (1,2 ⋅ 36,9 + 1,5 ⋅ 14,4) kN/m = 65,9 kN/m

max MSd = 18

65 9⋅ , kN/m ⋅ 5,752 m2 = 272 kNm

α d =

116502

240 15

2 5

− ⋅

⋅mm = 222,5 mm ⇒ (d - αd) = (265 - 222,5) mm = 42,5 mm

Wpl = ( )240 15 222 5152

222 5 2 5222 5

22 5

42 52

450 12 42 5 62

⋅ ⋅ −

+ ⋅ ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ +

, ,

, ,,

mm3


192

Wpl = 1292 ⋅ 103 mm3 ⇒ Mc.Rd = 1292 100 355

113 3

2

⋅ ⋅mmkN mm, /

,= 417 kNm >

MSd

Mitoitus leikkaukselled/tw = 50 < 69 ε = 55,9 ⇒ ei leikkauslommahdusvaaraaAv = 2 ⋅ 5 mm ⋅ (265 - 15) mm = 2500 mm2

⇒ Vpl.Rd = 2500 mm2 ⋅0 355

11 3

2, /,kN mm

⋅= 466 kN

max VSd = 12

65 9⋅ , kN/m ⋅ 5,75 m = 189 kN < Vpl.Rd

Mitoitus väännölle

Kuormitustapaus 1 (hyötykuorma oikealla):Ontelolaatan tukireaktion suurin mahdollinen etäisyys palkin painopisteestä (ks. kuva 3.21c) = 202,5 mm.

pvas = 12

6 0⋅ , m ⋅ 5,0 kN/m2 = 15,0 kN/m ⇒ pd.vas = 1,2 ⋅ 15,0 kN/m = 18,0 kN/m

pd.oik = 12

8 4⋅ , m ⋅ (1,2 ⋅ 5,0 + 1,5 ⋅ 2,0) kN/m2 = 37,8 kN/m

max TSd = ( )12

37 8 18 0⋅ −, , kN/m ⋅ 0,2025 m ⋅ 5,75 m = 11,5 kNm

A0 = 263,5 mm ⋅ 245 mm = 64558 mm2

VSd = ( )12

18 0 37 8 1 2 0 915⋅ + + ⋅, , , , kN/m ⋅ 5,75 m = 164 kN

τV

Nmmmm mm

=⋅

⋅ ⋅=

115 102 64558 5

17 86

2

,, N/mm2

( ) ( ) ( ) ( )βγ

τ

γV

Sd v

y M

v

y M

V A

f f= + =

⋅+

⋅

/

/ / / /

// ,

,/ ,3 3

164000 2500355 11 3

17 8355 11 30 0

= 0,448 < 0,5

⇒ leikkauskestävyys on riittävä, eikä taivutuskestävyyttä tarvitse pienentää.

Kuormitustapaus 2 (täysi kuorma):

pvas = 12

6 0⋅ , m ⋅ (1,2 ⋅ 5 + 1,5 ⋅ 2) kN/m2 = 27,0 kN/m

TSd = ( )12

37 8 27 0 0 2025 5 75 6 29⋅ − ⋅ ⋅ =, , / , , ,kN m m m kNm

τv = 6 29 10

2 64558 5

6, ⋅⋅ ⋅

N/mm2 = 9,74 N/mm2


193

⇒ ( ) ( )βV =⋅

+⋅

189000 2500355 11 3

9 74355 11 3

// ,

,/ ,

= 0,458 < 0,5

Taipumarajatila

max Mqk = 18

14 4 5 752⋅ ⋅, , kNm = 59,5 kNm

δq = 548

59 5 102 1 10 16786 10

5 75 106

5 42 6⋅

⋅⋅ ⋅ ⋅

⋅ ⋅,

,, mm = 5,81 mm < δadm =

5750350

mm = 16,4

mm

max Mk = 18

51 3 5 752⋅ ⋅, , kNm = 212 kNm

max δ = 548

212 102 1 10 16786 10

5 75 106

5 42 6⋅

⋅⋅ ⋅ ⋅

⋅ ⋅,

, mm = 20,7 mm < δadm = 5750

250mm

= 23,0

mm

WQ-palkin mitoittamiseksi on käytettävissä tietokoneohjelma, jonka avulla voidaan myös alustavasti tarkistaa ontelolaattojen leikkauskestävyys.


194

Esimerkki 3.6. Keskisesti kuormitettu pilari.Tarkista kuvan 3.40 mukaisen pilarin normaalivoimakestävyys, kun pilari pääsee nurjahtamaan poikkileikkauksen heikompaan suuntaan. Pysyvän kuorman osuus normaalivoimasta Ngk = 200 kN.

1. LähtöarvotTeräs S235 ⇒ fy = 235 N/mm2

Profiili HE200A ⇒ A = 5383 mm2

iz = 49,8 mmpoikkileikkausluokka 1, koskauuman d/tw = (190 - 2 ⋅ 10 - 2 ⋅ 18)/6,5 = 20,6 < 33ε = 33

laipan c/tf = 2002 10⋅

= 10 = 10ε

Nurjahduspituus Leff = γ ⋅ L = 0,8 ⋅ 4000 mm = 3200 mm

Koska profiilin h/b <1,2 ⇒ nurjahduskäyrä c⇒ α = 0,49

⇒ λπ

_

,,

=⋅

⋅⋅3200

49 81 0 235210000

= 0,6842

⇒ φ = 0,5 ⋅ [1 + 0,49 ⋅ (0,6842 - 0,2) + 0,68422] = 0,8527

⇒ χ =+ −

10 8527 0 8527 0 68422 2, , ,

= 0,7344

Nb.Rd = 0 7344 1 0 53800 23511

22, ,

,,

⋅ ⋅ ⋅mmkN

mm= 844 kN

NSd = (1,2 ⋅ 200 + 1,5 ⋅ 380) kN = 810 kN < Nb.Rd ⇒ profiili on sopiva.


195

Esimerkki 3.7. Keskisesti kuormitettu putkipilari.Määritettävä kuvan 3.41 putkipilarin nurjahduskuormat.

1. LähtöarvotTeräs S355J2H EN10219 ⇒ fy = 355 N/mm2

Profiili : suorakaideputki 300 x 200 x 6A = 5763 mm2

Iy = 7370 ⋅ 104 mm4 iy = 113,1 mmIz = 3962 ⋅ 104 mm4 iz = 82,9 mm

Toimivat leveydet

Lyhyempi sivu: b b t_

= − 3 = 200 mm - 3 ⋅ 6 mm = 182 mm

⇒ λ_ /

, ,p =

⋅ ⋅182 6

28 4 1 0 4= 0,534 < 0,673 ⇒ ρ = 1

⇒ lyhyt sivu toimii täysin tehollisena

Pidempi sivu: ( )b_

= − ⋅300 3 6 mm = 282 mm

⇒ λ_ /

,p =

⋅282 6

28 4 4= 0,8275 > 0,673 ⇒ ρ =

0 8275 0 220 82752

, ,,

−= 0,8872

⇒ beff = 0,8872 ⋅ 282 mm = 250mm ⇒ ∆b = (282 - 250) mm = 32 mmTehollinen poikkileikkaus (ks. kuva 3.42): Aeff = A - 2 ⋅ ∆b ⋅ t = (5763 - 2 ⋅ 32 ⋅ 6) mm2

Aeff = 5379 mm2


196

2. Nurjahduskuormat2.1 Nurjahdus y - y akselin suhteen (suuntaan z - z)nurjahduskäyrä b ⇒ α = 0,34Leff = 1,0 ⋅ L = 8000 mm

βA =53795763

= 0,9334 ⇒ λπ

_

,,

y =⋅

⋅⋅8000

11310 9334 355

210000= 0,8944

⇒ φy = 0,5 ⋅ [1 + 0,34 ⋅ (0,8944 - 0,2) + 0,89442] = 1,018

⇒ χ y =+ −

11 018 1 018 0 89442 2, , ,

= 0,6648

⇒ Nb.Rd = 0 6648 0 9334 57630 35511

22, ,

,,

⋅ ⋅ ⋅mmkN

mm= 1154 kN

2.2 Nurjahdus z - z akselin suhteen (suuntaan y - y)

λπ

_

,,

z =⋅

⋅⋅8000

82 90 9334 355

210000= 1,220

⇒ φz = 0,5 ⋅ [1 + 0,34 ⋅ (1,220 - 0,2) + 1,2202] = 1,418

⇒ χz =+ −

11 418 1 418 1 2202 2, , ,

= 0,4673

⇒ Nb.Rd = 0 4673 0 9334 57630 35511

22, ,

,,

⋅ ⋅ ⋅mmkN

mm= 811 kN

Pilarin kestävyys on siten = 811 kN.


197

Esimerkki 3.8.Mitoita kuvan 3.43 mukaisen kehärakenteen puristettu ja taivutettu pilari, kun pilarit on vaakaorsilla tuettu pienemmän jäykkyyden suunnassa.

Kuva 3.43. Esimerkki 3.8:n rakennemalli.

1. LähtöarvotTeräs S355 gk = 4,0 kN/m

q3k = 14,0 kN/m (lumikuorma)q1k = 2,6 kN/mq2k = 1,1 kN/mFwk = 3,0 kN

2. Voimasuureiden ominaisarvotPilarin arvioitu omapaino ≈ 7,0 kN (sisältää seinäorsilta tulevan kuorman)

⇒ Ngk = 7,0 kN + 10m ⋅ 4,0 kN/m = 47,0 kN Nqk = 10 m ⋅ 14,0 kN/m = 140 kN

Tuulikuorman aiheuttama 1.kertaluvun taivutusmomentti pisteessä A:

MAk = 18 21

2⋅ ⋅ + ⋅q HF

Hkh

Fh = ( )38 1 2⋅ + ⋅ +q q H Fk k wk

Fh = ( )38

2 4 1 7⋅ +, , kN/m ⋅ 7,0 m + 3,0 kN = 13,8 kN

⇒ MAk = 18

2 4 7 013 8

27 0 63 02 2⋅ ⋅ + ⋅ =, ,

,, ,kN m m kN m kNm

3. Kuormitusyhdistelmien aiheuttamat rasitukset

Koska pilari on tuettu heikommassa suunnassa seinäorsilla siten,että λ λ_ _

z y< ja

λ_

,LT < 0 4 , pilari nurjahtaa jäykempään suuntaan ei kiepahdusta tapahdu.

a) voimakas tuuli + samanaik. lumiNSd = 1,2 ⋅ 47,0 kN + 1,5 ⋅ 0,7 ⋅ 140 kN = 203 kN


198

kc = +0 51

18, = 0,745

k s = +0 211

, = 1,10⇒ φ = ⋅ ⋅ =0 745 1101

2001

244, ,

lisävaakavoima pilarien mahdollisesta vinoudesta = Nd/244

⇒ MAd = 1,5 ⋅ 63,0 kNm + 203

2447 0 100

kNm⋅ =, kNm

Valitaan alustavasti IPE300⇒ A = 5381 mm2

iy = 124,6 mmWpl.y = 628 ⋅ 103 mm3

Leff = 2,1 ⋅ 7,0 m = 14,7 m_

,,λ πy =

⋅⋅ =

14700124 6

355210000

1544

nurjahduskäyrä a (nurjahdus suuremman jäykkyyden suunnassa, kun h/b > 1,2)⇒ α = 0,21⇒ φ = 0,5 ⋅ [1 + 0,21 ⋅ (1,544 - 0,2) + 1,5442] = 1,833

⇒ χ y =+ −

11 833 1833 15442 2, , ,

= 0,3545

Tarkistetaan poikkileikkausluokka:

fN

mmd =355

11 2,= 322,7 N/mm2

Uuma: d = (300 - 2 ⋅ 7,1 - 2 ⋅ 15) mm = 248,6 mm

α = 12

203248 6 7 1 0 3227

+⋅ ⋅

, , ,

= 0,856

dtw

= = <⋅

⋅ −=

248 67 1

35 0456 0 81

13 0 856 136 5

,,

,,

,, ⇒ poikkileikkausluokka 2

laippa:ct f

=⋅

= < =150

2 10 77 01 10 8 1

,, ,ε ⇒ poikkileikkausluokka 1

MQ = 0,125 ⋅ 1,5 ⋅ 2,6⋅ 7,02 kNm = 23,9 kNm

∆M = 100 kNm +9

12815 2 6 72⋅ ⋅ ⋅, , kNm = 113,4 kNm

( )βMy = + ⋅ −1 823 9

113 41 3 1 8,

,,

, , = 1,69

Wel.y = 557 ⋅ 103 mm3

( )µy = ⋅ ⋅ − +−

1 544 2 1 69 4628 557

557, , = - 0,830


199

k y = +⋅

⋅ ⋅1

0 830 2030 3545 5380 0 355

,, ,

= 1,25

Yhteisvaikutusehto:203

0 3545 5380 0 32271 25 100

628 0 3227, ,,

,⋅ ⋅+

⋅⋅

= 0,947 < 1 (kaava 3.56)

b) runsas lumi + samanaik. tuuliNSd = 1,2 ⋅ 47,0 kN + 1,5 ⋅ 140 kN = 266 kN

MAd = 1,5 ⋅ 0,5 ⋅ 63,0 kNm + 266244

7 0 54 9kN m⋅ =, , kNm

2660 3545 5380 0 3227

1 25 54 9628 0 3227, ,

, ,,⋅ ⋅

+⋅

⋅= 0,771 < 1

IPE300 on täten painoltaan pienin I-profiili, joka soveltuu pilariksi.


200

Esimerkki 3.9. Teollisuushallin päädyn tuulipilarin mitoitus (ks. kuva 3.44).

Lähtöarvot:Koko kohdetta käsittelevästä rakenneselostuksesta saadaan:

teräs S235rakennemalli kuvassa 3.44bHEA-pilari on sivusuunnassa tukematon.Ngk = 8,0 kNNqk = 30,0 kNqw.ref = 0,331 kN/m2

Kuva 3.44. Esimerkin 3.9 tuulipilari.

Voimasuureet:wk = we + wi = 0,331 kN/m2 ⋅ 1,7 ⋅ (0,6 + 0,5) = 0,619 kN/m2

M1k = 18

5 0 0 619 6 02 2 2⋅ ⋅ ⋅, , / ,m kN m m = 13,9 kNm

Mfk = 9

1285 0 0 619 6 02⋅ ⋅ ⋅, , , kNm = 7,83 kNm

Mitoituksessa määräävä kuormitustapaus on omapaino + tuuli + samanaikainen lumi.⇒ NSd = (1,2 ⋅ 8,0 + 1,5 ⋅ 0,7 ⋅ 30,0) kN = 41,1 kN

M1d = 1,5 ⋅ 13,9 kNm = 20,9 kNmMfd = 1,5 ⋅ 7,83 kNm = 11,7 kNm

1. Alustava mitoitus

AN

N mm≥

4110040 2/

= 1028 mm2

Wel.y ≥⋅20 9 10

140

6

2

,/

NmmN mm

= 149 ⋅ 103 mm3

Kokeillaan profiilia HE 140A :A = 3140 mm2 Wel.y = 155 ⋅ 103 mm3

h = 133 mm Wpl.y = 173 ⋅ 103 mm3

b = 140 mm iy = 57,3 mm


201

tf = 8,5 mm iz = 35,2 mmtw = 5,5 mm r = 12 mm

2. PoikkileikkausluokkaUuma: d = (133 - 2 ⋅ 8,5 - 2 ⋅ 12) mm = 92,0 mm

α = +⋅ ⋅

12

4110092 5 5 235 2

Nmm mm N mm, /

= 0,846

dtw

= = <⋅

⋅ −=

92 05 5

16 7396 1

13 0 846 139 6

,,

,,

, ⇒ poikkileikkausluokka

1

Laippa:ct f

=⋅

= <140

2 8 58 24 10

,, ⇒ poikkileikkausluokka 1

3. HoikkuudetLaskettaessa hoikkuutta z-z akselin suhteen (heikompaan suuntaan)

käytetäännurjahduskäyrää c, sillä h/b < 1,2.⇒ α = 0,49Leff = 0,8 ⋅ 6000 mm = 4800 mm

⇒ λπ

_

,z =⋅

⋅⋅4800

35 21 235210000

= 1,452

⇒ ( )[ ]φz = ⋅ + ⋅ − +0 5 1 0 49 1 452 0 2 1 4522, , , , , = 1,861

⇒ χz =+ −

11861 1861 1 4522 2, , ,

= 0,331

Valitaan tekijäksi C1 = 1,2 (toinen pää kiinnitetty, toinen vapaasti tuettu)

⇒ λLT =

⋅ +

6000 35 2

1 2 1120

6000 35 2133 8 5

2

/ ,

,/ ,/ ,

= 59,1

λ π1

210000235

= ⋅ = 93,9

⇒ λ_ ,

,LT =59 193 9

= 0,629 ⇒ χLT = 0,8777

4. YhteisvaikutusehtoMQ = 20,9 kNm ∆M = (20,9 + 11,7) kNm = 32,6 kNm

( )βM LT. ,,,

, ,= + −1 820 932 6

1 3 1 8 = 1,48

⇒ µLT = ⋅ ⋅ −0 15 1 452 1 48 0 15, , , , = 0,172

⇒ k LT = −⋅

⋅ ⋅1

0 172 411000 331 3140 235

,,

= 0,971


202

f d =23511,

N/mm2 = 214 N/mm2

⇒41100

0 331 3140 2140 971 20 9 10

0 8777 173 10 214

6

3,, ,

,⋅ ⋅+

⋅ ⋅⋅ ⋅ ⋅

= 0,809 < 1

HE 140A on sopiva profiili, sillä jos valitaan HE 120A, saadaan seuraavat tulokset:

A = 2530 mm2 Wpl.y = 83,0 ⋅ 103 mm3 iz = 30,2 mm

⇒ λ_

z = 1,692 ⇒ φz = 2,298 ⇒ χz = 0,260

λLT = 55,4 ⇒ λ_

LT = 0,590 ⇒ χLT = 0,8934µLT = 0,226 ⇒ kLT = 0,940

⇒41100

0 260 2530 2140 940 20 9 10

0 8934 83 10 214

6

3,, ,

,⋅ ⋅+

⋅ ⋅⋅ ⋅ ⋅

= 1,53 > 1

Huom! Profiilin HE 120A taivutuskestävyys ei edes ole riittävä ( 83 ⋅ 0,214 kNm < 20,9 kNm ).


203

Esimerkki 4.1On laskettava kuvan 4.20 mukaisen liitoksen kestävyys. Teräs S355, ruuvit 8.8 M20, b = 160 mm, e1 = 50 mm, e2 = 40 mm, p1 = 70 mm, p2 = 80 mm, t1 = 25 mm ja t2 = 15 mm.

Valitaan Eurocode 3:n mukainen välys ⇒ d0 = 22 mmTeräs S355, t < 40 mm ⇒ fu = 510 N/mm2

8.8 ruuvit ⇒ fub = 800 N/mm2

M20 ⇒ A = 314 mm2 As = 245 mm2

Kuva 4.20. Esimerkin 4.1 liitos.

Ruuvien leikkauskestävyys:

FkN

mmmmv Rd. ,

,,

= ⋅ ⋅0 60 8001 25

24522 = 94,08 kN (kierteet leikkautumistasossa)

2-leikkeisiä ruuveja 6 kpl ⇒ ∑ Fv.Rd = 2 ⋅ 6 ⋅ 94,08 kN = 1129 kN

Reunapuristuskestävyys:e1 = 2,5 d ja p1 > 3,4 d = 68 ⇒ F kN mm mm mmb Rd. , , /= ⋅ ⋅ ⋅1 5 0 355 20 252 = 266,25 kNRuuveja 6 kpl ⇒ ∑ Fb.Rd = 6 ⋅ 266,25 kN = 1598 kN (keskimmäinen levy on mitoittava, sillä reunimmaisten levyjen paksuus on suurempi kuin puolet keskimmäisen paksuudesta, mutta rasittava voima vain puolet keskimmäisen arvosta).

Perusaineen kestävyys:( )Anet = ⋅ − ⋅ ⋅25 160 2 25 22 mm2 = 2900 mm2

N mm mmkN

mmpl Rd.

,,

= ⋅ ⋅25 1600 35511 2 = 1291 kN

N mmkN

mmu Rd. ,,,

= ⋅ ⋅0 9 29000 5101 25

22 = 1065 kN < Npl.Rd ⇒ liitos ei toimi sitkeästi.

Nettopoikkileikkauksen kestävyys on mitoittava.


204

Esimerkki 4.2Tarkistettava kuvan 4.21 palkkien välisen liitoksen kestävyys, kun Vd = 120 kN. Teräs S355, ruuvit 8.8 M16. e1 = 35 mm, e2 = 35 mm ja p1 = 40 mm sekä d0 = 18 mm.Koska leikkausvoiman Vd epäkeskisyys ruuviryhmän painopisteakselin suhteen on 45 mm, aiheutuu tästä momentti Md = 120 kN ⋅ 0,045 m = 5,4 kNm.IPE 300 ⇒ tw = 7,1 mm

Kuva 4.21. Esimerkin 4.2 palkkien välinen liitos.Ruuvin leikkauskestävyys:M16 ⇒ A = 201mm2, As = 0,78 ⋅ 201 mm2 = 157 mm2

F mmkN

mmv Rd. ,,,

= ⋅ ⋅0 6 1570 8001 25

22 = 60,3 kN

Reunapuristuskestävyys:F kN mm mm mmb Rd. , , / ,= ⋅ ⋅ ⋅1 0 0 355 16 7 12 = 40,3 kNPalamurtumiskestävyys:

Lv = 4 ⋅ 40 mm = 160 mm, L1 = a1 = 35 mm, L2 = ( )35 0 5 18510355

− ⋅, mm = 37,4

mm

( )L3 160 35 35 5 18510355

= + + − ⋅ mm = 201 mm ⇒ Lv.eff = 201 mm

Av.eff = 7,1 mm ⋅ 201 mm = 1427 mm2

⇒ VkNmm

mmeff Rd.

,,

=⋅

⋅0 355

11 314272

2 = 266 kN > Vd

Äärimmäisille ruuveille tuleva rasitus, kun sovelletaan kimmoteoriaa:( ) ( )Σ x y m mi i

2 2 2 2 2 22 0 04 0 08 0 016+ = ⋅ + =, , ,

FkNm m

mx =⋅5 4 0 08

0 016 2

, ,,

= 27,0 kN

FkN

y =120

5= 24,0 kN ⇒ Fv Sd. , ,= +27 0 24 02 2 kN = 36,1 kN < Fb.Rd = 40,3

kN⇒ Liitos kestää sille tulevat rasitukset.


205

Esimerkki 4.3Lasketaan esimerkin 4.1 liitos liukumisen kestävänä liitoksena käyttäen 4 kpl 10.9 ruuveja ja oletetaan, että pysyvän kuorman osuus vetovoimasta N on 40 %. Luokan C kosketuspinnat.

Standardin mukaiset reiät ⇒ ks = 1,0Luokan A kosketuspinnat ⇒ µ = 0,30

Liitos suunnitellaan luokkaan BF kN mm mmp Cd. , , /= ⋅ ⋅0 7 1 0 2452 2 = 171,5 kN

⇒ FkN

s Rd ser. . , ,,,

= ⋅ ⋅ ⋅1 0 2 0 31715

110= 93,5 kN

4 ruuvia ⇒ NSd.ser ≤ 4 ⋅ 93,5 kN = 374 kN (vetovoima käyttörajatilassa)⇒ murtorajatilassa NSd ≤ (1,2 ⋅ 0,4 + 1,5 ⋅ 0,6) ⋅ 374 kN = 516 kN

F mmkNmmv Rd. ,

,,

= ⋅ ⋅0 5 2451 0

1 252

2 = 98,0 kN

⇒ NSd ≤ 2 ⋅ 4 ⋅ 98,0 kN = 784 kN ⇒ käyttörajatila on mitoittava.

Jos kosketuspinnat käsitellään luokkaan A, muuttuu mitoitus seuraavasti:

Fs Rd ser. . ,,

,= ⋅ ⋅2 0 5

171511

kN = 156 kN

⇒ ( )N Sd ≤ ⋅ + ⋅ ⋅ ⋅1 2 0 4 15 0 6 4 156, , , , kN = 861 kN⇒ ruuvien leikkauskestävyys on nyt mitoittava.


206

Esimerkki 4.4

Laskettava kuvan 4.56 liitoksessa tarvittava pienahitsin a-mitta, kun teräs on S235, Lw = 80 mm, t = 15 mm ja FSd = 150 kN.

Sijoittamalla annetut arvot kaavaan 4.36 saadaan

akN

mm kN mm≥

⋅ ⋅ ⋅

⋅

150 0 8 1 25 280 0 360 2

, ,, /

= 7,4 mm ⇒ valitaan a = 8 mm


207

Esimerkki 4.5

Määritettävä kuvan 4.57 pienahitsin vaadittu a-mitta, kun teräs on S 275, t = 10 mm, Lw= 200 mm ja FSd = 300 kN.

Kaavasta 4.35 saadaan a ≥⋅ ⋅ ⋅

⋅300 0 85 1 25 3

200 0 430, ,

,mm = 6,4 mm ⇒ valitaan a = 7 mm


208

Esimerkki 4.6

Mitoitettava kuvan 4.60 mukaisen levykorvakkeen pienahitsit rakenneaineessa vaikuttavien jännitysten perusteella, kun teräs on S 355. Korvaketta kuormittavan voiman mitoitusarvo on FSd = 300 kN, ja voima vaikuttaa 45° kulmassa korvakkeen pituusakseliin nähden. Mitat on merkitty kuvaan.Voimasuureet ulokkeen tukipoikkileikkauksessa ovat:

NSd = 212,1 kNVSd = 212,1 kNMSd = 29,7 kNm.

Kimmoteorian mukaiset jännitykset pisteissä 1 ja 2 ovat seuraavat:

σ z

Nmm mm

Nmmmm mm1

6

2 2

21210020 200

6 29 7 1020 200

=⋅

+⋅ ⋅

⋅,

= 275,8 N/mm2

τyz.1 = 0

σ z

Nmm mm2

21210020 200

=⋅

= 53,0 N/mm2

τ yz

Nmm mm.2

32

21210020 200

= ⋅⋅

= 79,5 N/mm2

Vaadittu a-mitta pisteessä 1:

amm

≥⋅ ⋅

⋅⋅

0 9 1 25 202 510

2 275 82, ,, = 8,6 mm

Vastaavasti pisteessä 2 vaadittu a-mitta:

amm

≥⋅ ⋅

⋅⋅ + ⋅

0 9 1 25 202 510

2 53 0 3 79 52 2, ,, , = 3,5 mm

Valitaan edelläolevan perusteella a = 9 mm.


209

Esimerkki 4.7

Tutkitaan plastisuusteoriaa soveltamalla edellisen esimerkin korvakkeen hitsien kestävyys, jos a = 6 mm. Sovelletaan kuvan 4.61 rakennemallia.Hitsin leikkausjännitys leikkausvoimasta olettaen jännitys tasan jakautuneeksi koko hitsin korkeudelle:

τ|| = 212100

2 6 200N

mm mm⋅ ⋅= 88,4 N/mm2

Voidaan osoittaa (E.Niemi), että hitsin ns. vertailujännitys on laskentalujuuden (Eurocode

3: f u

Mwγ) suuruinen, jos hitsiin kohdistuva nimellinen poikittainen jännityskomponentti on

suuruudeltaan

σβ γ

τ

w

u

w Mwll

f

⊥ =

−

2

23

2(4.39)

Murorajatilassa hitsiin vaikuttava nimellinen poikittainen jännitys on tässä tapauksessa

σ w⊥ =⋅

− ⋅510

0 9 1 253 88 4

2

22

, ,,

N/mm2 = 301,7 N/mm2

Varataan hitsien keskiosa kantamaan normaalivoimaa. Tarvittavan alueen korkeus h on

hNa

Sd

w=

⊥2 σ=

2121002 6 301 7 2

Nmm N mm⋅ ⋅ , /

= 58,6 mm ≈ 59 mm

Liitoksen kestävyys taivutusmomentin suhteen lasketaan kaavasta( )

M aH h

mmmm

kN mmpl Rd w. , /= ⋅−

= ⋅ ⋅−

⋅⊥24

2 6200 59

40 3017

2 2 2 2 22σ =

= 33,1 kNm > MSd = 29,7 kNmLiitoksen kestävyys on siis riittävä ja plastisuusteoriaa soveltamalla riittää pienempi hitsin a-mitta kuin kimmoteorian mukaan.


210

Esimerkki 4.8Mitoitettava kuvan 4.62 pilarin kiinnityshitsit pohjalevyyn. Teräs on S 355, pilari HE 240A ja voimasuureiden mitoitusarvot ovat seuraavat:Kuormitustapaus 1: NSd = 55,0 kN

VSd = 30,0 kNMSd = 110 kNm

Kuormitustapaus 2: NSd = 245 kNVSd = 35,0 kNMSd = 160 kNm

Puristusjännitykset siirretään pohjalevyyn kosketuspaineen välityksellä. Hitsit mitoitetaan pilarissa vaikuttavien veto- ja leikkausjännitysten perusteella. Koska liitos kuuluu staattisesti määräämättömään rakenteeseen edellytetään liitokselta riittävää muodonmuutoskykyä. Tämän varmistamiseksi oletetaan vetojännitykseksi vähintään 80

% pilarin mitoituslujuudesta (σx ≥ 0,8 ⋅f y

Mγ 0).

HE 240A ⇒ Iy = 7763 ⋅ 104 mm4 A = 7680 mm2 hf /2 = 109 mm

Kuva 4.62. Pilarin kiinnitys pohjalevyyn.a) Pilarin laipan hitsitSuurin vetojännitys laipan keskellä esiintyy kuormitustapauksella 2 ja se on

σ x =⋅⋅

⋅ −

160 1077 63 10

1092450007680

6

6,N/mm2 = 193 N/mm2

Muodonmuutoskyvyn perusteella

σ xN

mm= ⋅0 8

35511 2,,

= 258 N/mm2

Kaavan 4.38 perusteella saadaan, kun τyz = 0, σz = 258 N/mm2 ja tf = 12 mm

a mmf ≥⋅ ⋅

⋅⋅

0 9 1 25 2582 510

12, ,

= 4,8 mm ⇒ valitaan af = 5 mm


211

b) Uuman hitsitLasketaan jännitykset pisteessä 2 (laipan ja uuman liittymiskohta):

σ x = ⋅ −

16077 63

1032450007680,

N/mm2 = 180 N/mm2

τ yxN

mm mm=

⋅35000

7 5 206,= 22,7 N/mm2

⇒ amm

w ≥⋅ ⋅

⋅⋅ + ⋅

0 9 1 25 7 52 510

2 180 3 22 72 2, , ,, = 2,1 mm

Muodonmuutoskyvyn perusteella saadaan taulukosta 4.6 aw = 0,403 ⋅ 7,5 mm = 3,0 mm⇒ valitaan uuman hitseiksi aw = 3 mm.


212

Esimerkki 4.9Tarkastellaan kuvassa 4.63 esitettyä IPE 220-profiilin nivelliitosta, joka on tarkoitettu standardiratkaisuksi. Mitoitetaan hitsit tasalujiksi ruuviliitoksen kanssa. Ruuviliitoksen kestävyys määräytyy tässä tapauksessa reunapuristuksen perusteella. Kahden M20 ruuvin yhteiseksi reunapuristuskestävyydeksi saadaan kaavasta 4.18 ∑Fb.Rd = 103 kN.

Tapa 1:Jaetaan liitos neljään osahitsiin kuvan 4.64 osoittamalla tavalla. Oletetaan jatkoslevyn kiertyvän murtorajatilassa pisteen P ympäri. Sijoitetaan osahitsien kestävyydet Fi niiden painopisteisiin kuvan 4.64 mukaisesti. Kestävyydet lasketaan yksinkertaisella menetelmällä kaavojen 4.29 ja 4.30 avulla. Liitoksen kestävyys tarkistetaan sitten sekä leikkausvoiman että taivutusmomentin suhteen.

Kuva 4.63. IPE-profiilin nivelliitos jatkoslevyineen.

Rasitusten mitoitusarvot : VSd = 103 kNMSd = 103 kN ⋅ 0,16 m = 16,5 kNm

Valitaan alustavasti a = 5 mm, jolloin hitsien kestävyyksiksi saadaan

f vw d. , ,=

⋅ ⋅360

0 8 1 25 3N/mm2 = 207,8 N/mm2

Fw.R1 = 0 2078 5 1002, /kN mm mm mm⋅ ⋅ = 103,9 kNFw.R2 = 0 2078 5 70, ⋅ ⋅ kN = 72,7 kN

Leikkauskestävyys, joka saadaan voimien Fw.R1 pystykomponenteista, on

Vpl Rd. ,= ⋅ ⋅25086

103 9 kN = 121 kN.

Kestävyys momentin suhteen onMpl.Rd = ( )2 103 9 86 72 7 35⋅ ⋅ + ⋅, , kNmm = 23,0 kNm.

Kestävyydet ovat riittäviä, joten 5 mm on riittävä hitsin a-mitaksi.


213

LASKUESIMERKKEJÄ B7 MUKAAN

Osiossa esitetään laskuesimerkkejä RakMK B7 mukaan. Tehtävänä on ratkaista tehtävät EC3 mukaan.


214

Wx 252000:=

VR 0.6 fd⋅ tw⋅ b⋅:= VR 2.378 105×= > V2d 5.375 104

×=

>MR fd Wx⋅:= MR 5.922 107×= M2d 5.375 107×=

Yhteisvaikutusyhtälö:

Md M2d:= Vd V2d:=

MdMR

0.63VdVR

⋅+

1.05= 1.38< OK

Käyttörajatilamitoitus:

Ix 27700000:= E 2.1 105⋅:=

ymaxgk qk+( ) L4

⋅187 E⋅ Ix⋅

:= ymax 6.32=L

40012.5= OK

ymax2 0.521 gk qk+( )⋅L4 10 2−⋅

E Ix⋅⋅:= ymax2 6.158=

SIVUTTAISTUETTU KAKSIAUKKOINEN PALKKIALUSTAVAN MITOITUKSEN MUKAAN / B7

Mitoita sivuttaistuettu kaksiaukkoinen valssattu teräspalkki alustavan mitoituksen kaavoilla, kun aukkomitta L= 5000 mm, kuormituksena ja tasainen kuorma qk=10 kN/m.

Alkuarvot: Teräs S235 IPE L 5000:= gk 1:= qk 10:=

fy 235:= γ 1.0:= fdfyγ

:=

Laskentakuormat:

gd 1.2 1⋅:= qd 1.6 qk⋅:=

M2d gd qd+( )L2

8⋅

:= M2d 5.375 107×=

V2d 0.625 gd qd+( )⋅ L⋅:= V2d 5.375 104×=

Kimmoteorian mukainen mitoitus poikkileikkausluokan 3 perusteella:

WtarpM2d

fd:= Wtarp 2.287 105

×= IPE 220 Wx=252 000 mm3

IPE 220- profiilin kimmoteorian perusteella määritetyt kestävyydet ovat:

fd 235= b 220 9.2−( ):= tw 8:= Aw b tw⋅:=


215

L 2.623 103×=L 2.7Efy

⋅ 1 0.5M2MR

⋅−

⋅ iy⋅:=M2 MR:=

E 2.1 105⋅:=MR 196.5106⋅:=

iy 65:=fy 235:=Teräs S235HE260

Alkuarvot:

Sivuttaistukien enimmäisetäisyys kimmoteorian mukaan:

iy 65:=fy 235:=Teräs S235HE260A

Alkuarvot:

Selvitä taivutetun HE260A- profiilin sivuttaistukien välinenenimmäisetäisyys, jotta kiepahduksella ei ole vaikutusta momentti-kestävyyteen.

SAUVAN SIVUTTAINEN TUKEMINEN / B7

KANTAVAN PROFIILIPELLIN ALUSTAVA VALINTATAULUKOIDEN AVULLA / B6

Valitse vesikaton kantava profiilipelti taulukoidenavulla seuraaviin rakenteisiin, lumikuorma qk=1.8 kN/m2.Ks. osoite www.rautaruukki.fi

a) kylmä autokatos, kantava profiilipelti yksiaukkoinen, aukonpituus 5m

b) lämpöisen teollisuushallin vesikatteen kantava, 2-aukkoinen profiilipelti, orsiväli 3000 mm, päällä lämpöeriste ja vedeneristys, paino 0.3 kN/m2.

c) lämpöisen teollisuushallin ulkoseinän verhouspelti, orret k/k 1200, tuulen paine 0.5 kN/m2.


216

KATTO- ja SEINÄORREN ALUSTAVA VALINTATAULUKOIDEN AVULLA B6

Valitse teollisuushallin vesikatto-orsi. Orsi on 2-aukkoinen,aukon pituus on 6m, k/k = 2400 mm.Ks osoite www.rautaruukki.fi , Teräsrakenne-oppimisympäristö

Valitse teollisuushallin seinäorsi. Orsi on 2-aukkoinen,aukon pituus on 6m, k/k = 1200 mm.Ks osoite www.rautaruukki.fi, Teräsrakenne-oppimisympäristö

HITSATTU OHUTUUMAPALKKITEOLLISUUSHALLIN PÄÄKANNATTAJANA ALUSTAVA TAULUKKOMITOITUS B7

Valitse teollisuushallin pääkannattajaksi hitsattu ohutuumapalkki.Hallin kehäväli on 6m ja jänneväli 20m, orsiväli 3m, pääkannattajan kattokaltevuus 1:16, teräs S355 ja tuentapa 1= leikkausjäykiste tuella. Kuormituksena on lumikuorma qk=1.8 kN/m2 ja omapaino gk= 0.5 kN/m2.

Valitaan TH-harjapalkki 1:16-600/20-6-12*270

Piirrä palkin valmistuspiirustus Auto-Cad:illä.


217

α 0.34:=

Muunnettu hoikkuus:

λkLcix π⋅

fyE

⋅:= λk 1.109=

Apusuure β:

β1 α λk 0.2−( )⋅+ λk

2+

2 λk2

⋅

:= β 1.032=

Reduktiokerroin:

χ β β2 1

λk2

−

−

:=χ 0.53=

Nrcx χ fd⋅ A⋅:= Nrcx 1.637 106×= Mrx fd Wx⋅:= Mrx 3.243 108×=

NR fd A⋅:= NR 3.088 106×= Nelx

NR

λk2

:= Nelx 2.512 106×=

C 1.0:=

NdNrcx

CMdMrx

⋅

1

1Nd Nrcx⋅

NR Nelx⋅−

⋅+ 0.766= Valittu poikkileikkaus OK.

KAKSOISSYMMETRISEN PURISTETUN JA TAIVUTETUNSAUVAN KESTÄVYYS B7Mastojäykistetyn teollisuushallin pääpilarin , L=6000mmm, laskentakuormat ovat Nd=200 kN ja Md=200 kNm. Valitse HE- profiili, kun teräs on S235.

Alkuarvot: Kokeillaan HE280B L 6 103⋅:=

Md 200 106⋅:= Nd 200 103⋅:= ix 121:= fy 235:=

A 131.4102⋅:= Wx 1380 103⋅:= fd 235:= E 2.1 105⋅:=

Alustava mitoitus:

Ndfd A⋅

Mdfd Wx⋅

+ 0.681= <1 jatketaan HE280B-profiililla

Tarkempi mitoitus yhteisvaikutusyhtälöllä:

fy 235=Nd 2 105×= Md 2 108×=

Lc 2.1 6000⋅:= ix 121=


218

λkLcix π⋅

fyE

⋅:= λk 0.518= α 0.21:=

Apusuure β:

β1 α λk 0.2−( )⋅+ λk

2+

2 λk2

⋅

:= β 2.489=

Reduktiokerroin:

χ β β2 1

λk2

−

−

:= χ 0.919=

Nrcx χ fd⋅ A⋅:= Nrcx 4.339 105×= Mrx fd Wx⋅:= Mrx 2.562 107

×=

NR fd A⋅:= NR 4.724 105×= NelxNR

λk2

:= Nelx 1.761 106×=

C 1.0:=

NdNrcx

CMdMrx

⋅

1

1Nd Nrcx⋅

NR Nelx⋅−

⋅+ 0.862= Valittu poikkileikkaus OK.

TUULIPILARIN MITOITUS B7Mastojäykistetyn teollisuushallin tuulipilarin, L=4000 mm, laskentakuormat ovat Nd=30 kN ja Md=20 kNm. Valitse IPE- profiili, kun teräs on S235.

Alkuarvot: Kokeillaan IPE160 L 4000:=

Md 20 106⋅:= Nd 30 103⋅:= ix 65.8:=

A 20.1 102⋅:= Wx 109 103⋅:= fd 235:=

Alustava mitoitus:

Ndfd A⋅

Mdfd Wx⋅

+ 0.844= <1 jatketaan IPE160-profiililla

Tarkempi mitoitus yhteisvaikutusyhtälöllä:

Nd 3 104×= Md 2 107×= fy 235:=

Lc 0.8 4000⋅:= ix 65.8= E 2.1 105⋅:=

Muunnettu hoikkuus:


219

PUTKIPALKKIRAKENTEIDEN KESTÄVYYS / B7

Selvitä seuraavien putkipalkkirakenteiden leikkaus-, taivutus- ja momenttikestävyydet vahvempaan suuntaan Putkipalkkikäsikirjan avulla:

a) 100x70x5

b) 200x100x5

c) pyöreä , halkaisija 101.6 x 5

PUTKIPALKKIRISTIKON ALUSTAVA SUUNNITTELU / B7

Suunnittele K- putkipalkkiristikko, mikä muodostuu yhteensä viidestä kolmiosta, joiden kateetit ovat 45 asteen kulmassa ja kolmion korkeus on 5m. Ristikon solmupisteitä rasittaa 5 kN laskentakuormat.

1. Selvitetään kuormat

2. Määritetään ristikon korkeus

3. Valitaan sauvat alustavasti laskemalla ristikon momentti olettamalla ristikko palkiksi.

Paarrevoimat saadaan yhtälöstä: No=Mmax/hUumasauvat valitaan leikkausvoiman maksimiarvon perusteella.Uumasauvan ja paarteen leveyden suhde on noin 0.7-0.8.

4. Lasketaan ristikon voimasuureet käsin tai es WINRAMI-ohjelmalla.

5. Lasketaan liitosten kestävyydet. Hitsit mitoitetaan putkipalkin vetokestävyyden perusteella.

6. Lasketaan taipuma ja verrataan sitä sallittuun arvoon.

7. Suunnitellaan ristikon poikittainen tukeminen ja orsien liitokset ristikkoon. Suunnitellaan ristikon asennusjatkokset ottaen myös kuljetus huomioon.


220

HALLIN RISTIKKOJÄYKISTYS B7

Vesikatto jäykistetään katon kehävälin jäykistysristikolla, jonka tukireaktiot hallin ulkoseinällä on 60 kN. Mitoita diagonaali vedolle ja puristukselle, kun seinän korkeus on sama kuin kehäväli =5m.

Tarkista myös kannakkeen kestävyys !!

<laipan hhvaad 144=hvaad 1.5d 3d+ 3d+ 1.5d+:=

d 16:=

Sijoitus yhteen riviin, laipan korkeus

3 pulttian 2.59=nQdFrv

:=Qd 200000:=

Pulttien lukumäärä:

Frv 7.721 104×=Frv frvd A⋅:=A π R2⋅:=R 8:=

frvd 384=frvd 0.6fyγm

⋅:=γm 1.0:=fy 0.8 800⋅:=

Leikkauskapasiteetti:

Valitaan M16 laatua 8.8

Pilarin HE280B ja palkin IPE160 välistä täysin nivelistä liitosta rasittaa leikkausvoima Qd=200kN. Suunnittele pilarin ja palkin välinen pulttiliitos.

LIITOKSET / B7


221

Md 8 107×=

FpMd250

:= Fp 3.2 105×=

Pultin vetokapasiteetti:

fy 640:= R 10:= Asp π R2⋅:=

frt 0.8 fy⋅:= Frt frt Asp⋅:= Frt 1.608 105×=

nFpFrt

:= n 1.989=

Pulttikehikkoon sijoitetaan 2+2 M20 pulttia

Pohjalevyn paksuus:fcd 11.7:= c 100:= md

12

fcd⋅ c2⋅:=

fd 235= t 6mdfd

⋅:= t 38.647=

Valitaan pohjalevyn paksuudeksi 50mm.

PILARIN LIITOS PERUSTUKSIIN B7

Mitoita alustavasti teräspilarin HE200B liitos aluslevyyn, aluslevyn paksuus ja pulttikehikon pultit. Pilarin Md=100 kNm, Nd=100 kN ja Qd=50 kN.

Pilarin liitos perustuksiin. Mitoitetaan hitsiliitos huomioimalla vain laippojen hitsauksen antama kapasiteetti. Hitsataan uuma samalla a-mitalla. Käytetään alustavaa ja pienahitsikaavaa.

Alkuarvot:

Md 80 106⋅:= l 2 200⋅:= fwd 190:=

Voimasuureet:

FdMd

200 30−( ):= Fd 4.706 105×=

Hitsin a-mitta

alustava aFd

l fwd⋅:= a 6.192=

pienahitsikaava:

fd 235:= β 0.7:= a 2 β⋅Fdl fd⋅

⋅:= a 4.956=

Pulttikehikon pultit, oletetaan pulttiryhmien voimanvarreksi z= 250 mm.

Pultin vetovoima:


222

TERÄKSEN PALOSUOJAUS

Suunnittele teräshallin pääpilarin HE280B palosuojaus Gyproc- levyjä käyttäen. Vaadittava palonkestoaika on 30 min ja kriittinen lämpötila on 450 C.

Profiilin F/V - suhde:

h 280:= b 280:= d 10.5:= t 18:=

A 2 h t⋅( )⋅ h 2 t⋅−( ) d⋅+:= A 1.264 104×=

2 h⋅ 4 b⋅+ 2 d⋅−( )A

0.131= F/V=131

Rakenteen kriittinen lämpötila Tkr=450 C on lämpötila, missä teräksenlämpötilan nousun aiheuttama lujuuden aleneminen on sitä luokkaa että rakenne ei enää pysty kantamaan edes palotilanteen kuormia, jotka ovat noin puolet normaalitilan kuormista.Kriittisen lämpötilan Tkr=450C ja F/V-suhteen avulla määrätään käyrästä tarvittava palosuojaus, kun vaadittava palonkestoaika on 30 min.


223

LIITTOLAATAN JA -PILARIN MITOITUS

18. Tehtävänä on omakotitalon kaksiaukkoisen välipohjarakenteen mitoitus liittorakenteisena Rautaruukin mitoitustaulukoita käyttäen. Laataston jännevälit ovat 6.0 m ja välipohjarakenteen leveys on 5m. Rakenteessa käytetään betonia K25-2 ja terästä A500 HW. Välipohjarakenteen kuormitus on OLI, palovaatimusta rakenteelle ei aseteta. Piirrä laatan raudoitus.

Valitaan raudoitus taulukosta: 6.1 T4, mikä tarkoittaa, että 6.1metrin jännevälillä laatan paksuuden ollessa 140mm ja liittolevyn paksuuden 0.7mm riittää, kun tuelle asennetaan raudoite T4, mikä vastaa T10k150- raudoitusta.

19. Koulun aulan pilari on nivelisesti tuettu molemmista päistään, pilarin kuormitus Nd=750 kN, pituus L=4m. Suunnittele pilari liittorakenteisena taulukoita käyttäen, kun palonkestovaatimus on 60 min.

Valitaan neliöpilari 250x250x12.5, teräs S355, betoni K40-2, teräsA500HW.Taulukon mukaan pilarin B+4T25 A60 normaalivoimakapasiteettion 4m nurjahduspituudella noin 800 kN. Pilari siis varustetaan palotilanteen varalta 4 kpl T25 pääteräksillä ja haoilla ja betonoidaan K40-2 betonilla.


224

0.3Efd

⋅ 8.968=

Puristettu laippa kuuluu poikkileikkausluokkaan 1

Uuma:b 360 2 12.7⋅− 2 18⋅−( ):= t 8:= λ

bt

:=

λ 37.325= < 2.4Efy

⋅ 71.744=

Taivutettu uuma kuuluu poikkileikkausluokkaan 1

Kimmoteorian mukainen poikkileikkauksen mitoitus:

Md 1.856 108×= Vd gd qd+( )L2

⋅

Pgd2

+:= Vd 8.08 104×=

IPE 360 Wx 904000:= b 360 2 12.7⋅−( ):= t 8= Aw b t⋅:=

σMdWx

:= σ 205.31= < fd 235=

τVdAw

:= τ 30.185= < fv 0.6 fd⋅:= fv 141=

SIVUTTAISTUETTU YKSIAUKKOINEN PALKKI B7

Mitoita sivuttaistuettu yksiaukkoinen valssattu teräspalkki, kun jänneväli L= 8000 mm, kuormituksena pistekuorma Pgk= 10 kN ja tasainen kuorma qk=10 kN/m.

Alkuarvot: Teräs S235 IPE L 8000:= gk 0.6:= qk 10:= Pgk 10000:=

fy 235:= γ 1.0:= fdfyγ

:=

Laskentakuormat:

gd 1.2 1⋅:= qd 1.6 qk⋅:= Pgd 1.2 20000⋅:=

Alustava arvio poikkileikkausluokan 3 perusteella:

Md gd qd+( )L2

8⋅

LPgd

4⋅+:= Md 1.856 108

×=

WtarpMdfd

:= Wtarp 7.898 105×= IPE 360 Wx=904 000 mm3

Poikkileikkausluokan määritys IPE 360- profiilille

Puristettu laippa:

b170 8− 2 18⋅−( )

2:= t 12.7:= λ

bt

:=

E 210000:= λ 4.961= <


225

fd 235= tw 8:= b 360 2 12.7⋅−( ):=

VR 0.6 fd⋅ tw⋅ b⋅:= VR 3.774 105×= > Vd 8.08 104

×=

MR fd Wx⋅:= MR 2.124 108×= > Md 1.856 108×=

Käyttörajatilamitoitus:

Ix 162700000:=

ymax5 gk qk+( )⋅ L4⋅

384 E⋅ Ix⋅

Pgk L3( )⋅

48E Ix⋅+:= ymax 19.668=

L400

20= OK

Uuman tarkistus lommahdukselle

Leikkausjännitykset:

b 360 2 12.7⋅− 2 18⋅−( ):= t 8:= a 8000:=

k 5.34 4.00ba

2⋅+:= k 5.346=

λp 1.05bt

⋅fyk E⋅

⋅:= λp 0.567= <0.9 OK

Taivutusjännitykset:

k 23.9:= λp 1.05bt

⋅fyk E⋅

⋅:= λp 0.268= <0.72 OK

Uuma toimii täysin tehollisena

IPE 360- profiilin kimmoteorian perusteella määritetyt kestävyydet ovat:

Documents

ELEMENT DESIGN MITOITUS - AMK · RAKENNUSTEKNIIKKA Olli Ilveskoski 30.08.2006 rev2 10.01.2007 142 1. INTRODUCTION When designing a structure and its components, the designer must