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28

CHAPTER 3

Analysis of thestructure at the

ultimate limit stateCHAPTER INTRODUCTION.............................................................................

A reinforced concrete structure is a combination of beams, columns, slabs and walls,

rigidly connected together to form a monolithic frame. Each individual member must

be capable of resisting the forces acting on it, so that the determination of these forces

is an essential part of the design process. The full analysis of a rigid concrete frame is

rarely simple; but simplified calculations of adequate precision can often be made if

the basic action of the structure is understood.

The analysis must begin with an evaluation of all the loads

carried by the structure, including its own weight. Many of

the loads are variable in magnitude and position, and allpossible critical arrangements of loads must be considered.

First the structure itself is rationalised into simplified forms

that represent the load-carrying action of the prototype. The

forces in each member can then be determined by one of the

following methods:

1. applying moment and shear coefficients

2. manual calculations

3. computer methods

Tabulated coefficients are suitable for use only with simple,

regular structures such as equal-span continuous beams

carrying uniform loads. Manual calculations are possible forthe vast majority of structures, but may be tedious for large or

complicated ones. The computer can be an invaluable help in

the analysis of even quite small frames, and for some

calculations it is almost indispensable. However, the amount

of output from a computer analysis is sometimes almost

overwhelming; and then the results are most readily inter-

preted when they are presented diagrammatically.

......................................

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Analysis of the structure 29

Since the design of a reinforced concrete member is generally based on the

ultimate limit state, the analysis is usually performed for loadings corresponding

to that state. Prestressed concrete members, however, are normally designed for

serviceability loadings, as discussed in chapter 11.

3.1 Actions

The actions (loads) on a structure are divided into two types: permanent actions, and

variable (or imposed) actions. Permanent actions are those which are normally constant

during the structures life. Variable actions, on the other hand, are transient and not

constant in magnitude, as for example those due to wind or to human occupants.

Recommendations for the loadings on structures are given in the European Standards,

some of which are EN 1991-1-1 General actions, EN 1991-1-3 Snow loads, EN 1991-1-4

Wind actions, EN 1991-1-7 Accidental actions from impact and explosions, and

EN 1991-2 Traffic loads on bridges.

A table of values for some useful permanent loads and variable loads is given in the

appendix.

3.1.1 Permanent actions

Permanent actions include the weight of the structure itself and all architectural

components such as exterior cladding, partitions and ceilings. Equipment and static

machinery, when permanent fixtures, are also often considered as part of the permanent

action. Once the sizes of all the structural members, and the details of the architectural

requirements and permanent fixtures have been established, the permanent actions canbe calculated quite accurately; but, first of all, preliminary design calculations are

generally required to estimate the probable sizes and self-weights of the structural

concrete elements.

For most reinforced concretes, a typical value for the self-weight is 25 kN per cubic

metre, but a higher density should be taken for heavily reinforced or dense concretes. In

the case of a building, the weights of any permanent partitions should be calculated

from the architects drawings. A minimum partition loading equivalent to 1.0 kN per

square metre is often specified as a variable action, but this is only adequate for

lightweight partitions.

Permanent actions are generally calculated on a slightly conservative basis, so that a

member will not need redesigning because of a small change in its dimensions. Over-

estimation, however, should be done with care, since the permanent action can oftenactually reduce some of the forces in parts of the structure as will be seen in the case of

the hogging moments in the continuous beam in figure 3.1.

3.1.2 Variable actions

These actions are more difficult to determine accurately. For many of them, it is only

possible to make conservative estimates based on standard codes of practice or past

experience. Examples of variable actions on buildings are: the weights of its occupants,

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30 Reinforced concrete design

furniture, or machinery; the pressures of wind, the weight of snow, and of retained earth

or water; and the forces caused by thermal expansion or shrinkage of the concrete.

A large building is unlikely to be carrying its full variable action simultaneously on

all its floors. For this reason EN 1991-1-1: 2002 (Actions on Structures) clause 6.2.2(2)

allows a reduction in the total variable floor actions when the columns, walls orfoundations are designed, for a building more than two storeys high. Similarly from the

same code, clause 6.3.1.2(10), the variable action may be reduced when designing a

beam span which supports a large floor area.

Although the wind load is a variable action, it is kept in a separate category when its

partial factors of safety are specified, and when the load combinations on the structure

are being considered.

3.2 Load combinations and patterns

3.2.1 Load combinations and patterns for the ultimate limit state

Various combinations of the characteristic values of permanent Gk, variable actionsQk,wind actionsWk, and their partial factors of safety must be considered for the loading of

the structure. The partial factors of safety specified in the code are discussed in

chapter 2, and for the ultimate limit state the following loading combinations from

tables 2.2, 2.4 and 2.5 are commonly used.

1. Permanent and variable actions

1:35Gk 1:5Qk

2. Permanent and wind actions

1:35Gk 1:5Wk

The variable load can usually cover all or any part of the structure and, therefore,

should be arranged to cause the most severe stresses. So, for a three-span continuous

beam, load combination 1 would have the loading arrangement shown in figure 3.1, in

order to cause the maximum sagging moment in the outer spans and the maximum

possible hogging moment in the centre span. A study of the deflected shape of the beam

would confirm this to be the case.

Load combination 2, permanent + wind load is used to check the stability of a

structure. A load combination of permanent + variable + wind load could have the

arrangements shown in figure 2.4 and described in section 2.4 of Chapter 2.

Figure 3.1

Three-span beam

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Figure 3.2 shows the patterns of vertical loading on a multi-span continuous beam to

cause (i) maximum design sagging moments in alternate spans and maximum possible

hogging moments in adjacent spans, (ii) maximum design hogging moments at

support A, and (iii) the design hogging moment at support A as specified by the EC2

code for simplicity. Thus there is a similar loading pattern for the design hogging

moment at each internal support of a continuous beam. It should be noted that the UK

National Annex permits a simpler alternative to load case (iii) where a single load

case may be considered of all spans loaded with the maximum loading of

(1:35Gk 1:50Qk).

3.3 Analysis of beams

To design a structure it is necessary to know the bending moments, torsional moments,

shearing forces and axial forces in each member. An elastic analysis is generally used to

determine the distribution of these forces within the structure; but because to some

extent reinforced concrete is a plastic material, a limited redistribution of the elastic

moments is sometimes allowed. A plastic yield-line theory may be used to calculate the

moments in concrete slabs. The properties of the materials, such as Youngs modulus,

which are used in the structural analysis should be those associated with their

characteristic strengths. The stiffnesses of the members can be calculated on the basis of

any one of the following:

1 the entire concrete cross-section (ignoring the reinforcement);

2. the concrete cross-section plus the transformed area of reinforcement based on themodular ratio;

3. the compression area only of the concrete cross-section, plus the transformed area ofreinforcement based on the modular ratio.

The concrete cross-section described in (1) is the simpler to calculate and would

normally be chosen.

Figure 3.2

Multi-span beam loading

patterns

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A structure should be analysed for each of the critical loading conditions which

produce the maximum stresses at any particular section. This procedure will be

illustrated in the examples for a continuous beam and a building frame. For these

structures it is conventional to draw the bending-moment diagram on the tension side of

the members.

Sign Conventions

1. For the moment-distribution analysis anti-clockwise support moments are positiveas, for example, in table 3.1 for the fixed end moments (FEM).

2. For subsequently calculating the moments along the span of a member, momentscausing sagging are positive, while moments causing hogging are negative, as

illustrated in figure 3.4.

3.3.1 Non-continuous beams

One-span, simply supported beams or slabs are statically determinate and the analysis

for bending moments and shearing forces is readily performed manually. For the

ultimate limit state we need only consider the maximum load of 1 :35Gk 1:5Qk onthe span.

EXAMPLE 3 .1

Analysis of a non-continuous beam

The one-span simply supported beam shown in figure 3.3a carries a distributed

permanent action including self-weight of 25 kN/m, a permanent concentrated action of

40 kN at mid-span, and a distributed variable action of 10 kN/m.

Figure 3.3 shows the values of ultimate load required in the calculations of the

shearing forces and bending moments.

Maximum shear force 54

2

195

2 124:5 kN

Maximum bending moment 54 4

4

195 4

8 151:5 k N m

The analysis is completed by drawing the shearing-force and bending-moment

diagrams which would later be used in the design and detailing of the shear and bending

reinforcement.

Figure 3.3

Analysis of one-span beam

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3.3.2 Continuous beams

The methods of analysis for continuous beams may also be applied to continuous slabs

which span in one direction. A continuous beam is considered to have nofixity with the

supports so that the beam is free to rotate. This assumption is not strictly true for beamsframing into columns and for that type of continuous beam it is more accurate to analyse

them as part of a frame, as described in section 3.4.

A continuous beam should be analysed for the loading arrangements which give the

maximum stresses at each section, as described in section 3.2.1 and illustrated in

figures 3.1 and 3.2. The analysis to calculate the bending moments can be carried out

manually by moment distribution or equivalent methods, but tabulated shear and

moment coefficients may be adequate for continuous beams having approximately equal

spans and uniformly distributed loads.

For a beam or slab set monolithically into its supports, the design moment at the

support can be taken as the moment at the face of the support.

Continuous beams the general case

Having determined the moments at the supports by, say, moment distribution, it is

necessary to calculate the moments in the spans and also the shear forces on the beam.

For a uniformly distributed load, the equations for the shears and the maximum span

moments can be derived from the following analysis.

Using the sign convention offigure 3.4 and taking moments about support B:

VAB

L wL2

2

MAB MBA 0

therefore

VABwL

2

MAB MBA

L 3:1

and

VBA wL VAB 3:2

Figure 3.4

Shears and moments in a

beam

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Maximum span moment Mmax occurs at zero shear, and distance to zero shear

a3V

AB

w 3:3

therefore

MmaxVAB

2

2wMAB 3:4

The points of contraflexure occur at M 0, that is

VABxwx2

2MAB 0

where x the distance from support A. Taking the roots of this equation gives

xVAB

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiVAB

2 2wMAB q

w

so that

a1VAB


2 2wMAB q

w 3:5

and

a2 L VAB


2 2wMAB q

w 3:6

A similar analysis can be applied to beams that do not support a uniformly distributed

load. In manual calculations it is usually not considered necessary to calculate the

distancesa1,a2and a3which locate the points of contraflexure and maximum moment

a sketch of the bending moment is often adequate but if a computer is performing thecalculations these distances may as well be determined also.

At the face of the support, width s

M0AB MAB VAB ws

4

s2

EXAMPLE 3 .2

Analysis of a continuous beam

The continuous beam shown in figure 3.5 has a constant cross-section and supports a

uniformly distributed permanent action including its self-weight ofGk 25 kN/m and avariable actionQk 10 kN/m.

The critical loading patterns for the ultimate limit state are shown infigure 3.5 where

the stars indicate the region of maximum moments, sagging or possible hogging.

Table 3.1 is the moment distribution carried out for thefirst loading arrangement: similar

calculations would be required for each of the remaining load cases. It should be

noted that the reduced stiffness of3

4

I

Lhas been used for the end spans.

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Figure 3.5

Continuous beam loading

patterns

Table 3.1 Moment distribution for the first loading case

A B C D

Stiffness (k)3

4:I

L

3

4:

1

6 0:125

I

L

1

4 0:25

3

4:I

L

0:125

Distr. factors 0:125

0:125 0:25

1=3

0:25

0:125 0:25

2=3 2=3 1=3

Load (kN) 292 135 292

F.E.M.0

292 6

8

135 4

12

292 6

8 0

0 219.4 45.0 45.0 219.4 0

Balance 58.1 116.3 116.3 58.1

Carry over 58.1 58.1

Balance 19.4 38.7 38.7 19.4

Carry over 19.4 19.4

Balance 6.5 12.9 12.9 6.5

Carry over

6.5

6.5Balance 2.2 4.3 4.3 2.2

Carry over 2.2 2.2

Balance 0.7 1.5 1.5 0.7

M(kN m) 0 132.5 132.5 132.5 132.5 0

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The shearing forces, the maximum span bending moments, and their positions along

the beam, can be calculated using the formulae previously derived. Thus for the first

loading arrangement and span AB, using the sign convention offigure 3.4:

Shear VAB

load

2

MAB MBA

L

292:5

2

132:5

6:0 124:2 kN

VBA load VAB

292:5 124:2 168:3 kN

Maximum moment, span AB VAB

2

2wMAB

where w 292:5=6:0 48:75 kN/m. Therefore:

Mmax 124:22

2 48:75 0 158:2kNm

Distance from A,a 3

VAB

w

124:2

48:75 2:55 m

The bending-moment diagrams for each of the loading arrangements are shown in

figure 3.6, and the corresponding shearing-force diagrams are shown infigure 3.7. The

individual bending-moment diagrams are combined infigure 3.8a to give the bending-

moment design envelope. Similarly, figure 3.8b is the shearing-force design envelope.

Such envelope diagrams are used in the detailed design of the beams, as described in

chapter 7.

In this example, simple supports with no fixity have been assumed for the end

supports at A and D. Even so, the sections at A and D should be designed for a hogging

moment due to a partialfixity equal to 25 per cent of the maximum moment in the span,

that is 158=4 39:5 kNm.

Figure 3.6

Bending-moment diagrams

(kN m)

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Continuous beams with approximately equal spans and uniform loading

The ultimate bending moments and shearing forces in continuous beams of three or

more approximately equal spans without cantilevers can be obtained using relevant

coefficients provided that the spans differ by no more than 15 per cent of the longest

span, that the loading is uniform, and that the characteristic variable action does not

exceed the characteristic permanent action. The values of these coefficients are shown

in diagrammatic form in figure 3.9 for beams (equivalent simplified values for slabs are

given in chapter 8).

Figure 3.7

Shearing-force diagrams (kN)

Figure 3.8

Bending-moment and

shearing-force envelopes

Figure 3.9

Bending-moment and

shearing-force coefficients

for beams

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The possibility of hogging moments in any of the spans should not be ignored, even if

it is not indicated by these coefficients. For example, a beam of three equal spans may

have a hogging moment in the centre span ifQk exceeds 0:45Gk.

3.4 Analysis of frames

In situ reinforced concrete structures behave as rigid frames, and should be analysed as

such. They can be analysed as a complete space frame or be divided into a series of

plane frames. Bridge deck-type structures can be analysed as an equivalent grillage,

whilst some form offinite-element analysis can be utilised in solving complicated shear

wall buildings. All these methods lend themselves to solution by computer, but many

frames can be simplified for a satisfactory solution by hand calculations.

The general procedure for a building is to analyse the slabs as continuous members

supported by the beams or structural walls. The slabs can be either one-way spanning or

two-way spanning. The columns and main beams are considered as a series of rigid

plane frames which can be divided into two types: (1) braced frames supporting verticalloads only, (2) frames supporting vertical and lateral loads.

Type one frames are in buildings where none of the lateral loads such as wind are

transmitted to the columns and beams but are resisted by much more stiffer elements

such as shear walls, lift shafts or stairwells. Type two frames are designed to resist the

lateral loads, which cause bending, shearing and axial loads in the beams and columns.

For both types of frames the axial forces in the columns can generally be calculated as if

the beams and slabs were simply supported.

3.4.1 Braced frames supporting vertical loads only

A building frame can be analysed as a complete frame, or it can be simplified into a

series of substitute frames for the vertical loading analysis. The frame shown infigure 3.10, for example, can be divided into any of the subframes shown infigure 3.11.

The substitute frame 1 in figure 3.11 consists of one complete floor beam with its

connecting columns (which are assumed rigidly fixed at their remote ends). An analysis

of this frame will give the bending moments and shearing forces in the beams and

columns for the floor level considered.

Substitute frame 2 is a single span combined with its connecting columns and two

adjacent spans, all fixed at their remote ends. This frame may be used to determine the

bending moments and shearing forces in the central beam. Provided that the central span

is greater than the two adjacent spans, the bending moments in the columns can also be

found with this frame.

Substitute frame 3 can be used tofind the moments in the columns only. It consists of

a single junction, with the remote ends of the members fi

xed. This type of subframewould be used when beams have been analysed as continuous over simple supports.

In frames 2 and 3, the assumption of fixed ends to the outer beams over-estimates

their stiffnesses. These values are, therefore, halved to allow for the flexibility resulting

from continuity.

The various critical loading patterns to produce maximum stresses have to be

considered. In general these loading patterns for the ultimate limit state are as shown in

figure 3.2, except when there is also a cantilever span which may have a beneficial

minimum loading condition (1:0Gk) see figure 7.21.

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When considering the critical loading arrangements for a column, it is sometimes

necessary to include the case of maximum moment and minimum possible axial load, in

order to investigate the possibility of tension failure caused by the bending.

EXAMPLE 3 .3

Analysis of a substitute frame

The substitute frame shown infigure 3.12 is part of the complete frame in figure 3.10.

The characteristic actions carried by the beams are permanent actions (including self-

weight)Gk 25 kN/m, and variable action,Qk 10 kN/m, uniformly distributed alongthe beam. The analysis of the subframe will be carried out by moment distribution: thus

the member stiffnesses and their relevant distribution factors arefirst required.

Figure 3.10

Building frame

Figure 3.11

Substitute frames

Figure 3.12

Substitute frame

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Stiffnesses,k

Beam

I0:3 0:63

12

5:4 103 m4

Spans AB and CD

kAB kCD 5:4 103

6:0 0:9 103

Span BC

kBC5:4 103

4:0 1:35 103

Columns

I0:3 0:353

12 1:07 103 m4

Upper

kU1:07 103

3:5 0:31 103

Lower

kL1:07 103

4:0 0:27 103

kU kL 0:31 0:27103 0:58 103

Distribution factors

Joints A and DPk 0:9 0:58 1:48

D.F.AB D.F.DC 0:9

1:48 0:61

D.F.cols0:58

1:48 0:39

Joints B and CPk 0:9 1:35 0:58 2:83

D.F.BA D.F.CD 0:9

2:83 0:32

D.F.BC D.F.CB1:35

2:83

0:48

D.F.cols0:58

2:83 0:20

The critical loading patterns for the ultimate limit state are identical to those for the

continuous beam in example 3.2, and they are illustrated in figure 3.5. The moment

distribution for the first loading arrangement is shown in table 3.2. In the table, the

distribution for each upper and lower column have been combined, since this simplifies

the layout for the calculations.

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Table

3.2

Momentdistributionforthefirstloadingcase

A

B

C

D

Co

ls.

(P

M)

AB

BA

Cols.

(PM)

BC

CB

Cols.

(PM)

CD

DC

Cols.

(PM)

D.F.s

0.

39

0.6

1

0.3

2

0.2

0

0.4

8

0.4

8

0.2

0

0.3

2

0.6

1

0.3

9

LoadkN

292

135

292

F.E.M.

146

146

45.0

45.0

146

146

Bal.

56

.9

89.1

32.3

20.2

48.5

48.5

20.2

32.3

89.1

56.9

C.O.

16.2

44.6

24.2

24.2

44.6

16.2

Bal.

6

.3

9.9

22.0

13.8

33.0

33.0

13.8

22.0

9.9

6.3

C.O.

11.0

5.0

16.5

16.5

5.0

11.0

Bal.

4

.3

6.7

6.9

4.3

10.3

10.3

4.3

6.9

6.7

4.3

C.O.

3.4

3.4

5.2

5.2

3.4

3.4

Bal.

1

.3

2.1

2.8

1.7

4.1

4.1

1.7

2.8

2.1

1.3

M

(kN

m)

68

.8

68.8

135.0

40.0

95.0

95.0

40.0

135.0

68.8

68.8

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The shearing forces and the maximum span moments can be calculated from the

formulae of section 3.3.2. For the first loading arrangement and span AB:

Shear VABload

2

MAB MBA

L

292:5

2

68:8 135:0

6:0 135 kN

VBA load VAB

292:5 135 157 kN

Maximum moment, span AB VAB

2

2wMAB

1352

2 48:75 68:8

118kNm

Distance from A, a3

VAB

w

135

48:75 2:8 m

Figure 3.13 shows the bending moments in the beams for each loading pattern;

figure 3.14 shows the shearing forces. These diagrams have been combined in

figure 3.15 to give design envelopes for bending moments and shearing forces.

A comparison of the design envelopes offigure 3.15 and figure 3.8 will emphasise the

advantages of considering the concrete beam as part of a frame, not as a continuous

beam as in example 3.2. Not only is the analysis of a subframe more precise, but many

moments and shears in the beam are smaller in magnitude.

The moment in each column is given by

McolX

Mcol kcol

Pkcol

Figure 3.13

Beam bending-moment

diagrams (kNm)

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Thus, for the first loading arrangement and takingP

Mcol table 3.2 gives

Column moment MAJ 68:8 0:31

0:58 37kNm

MAE 68:8 0:270:58

32kN m

MBK 40 0:31

0:58 21kN m

MBF 40 0:27

0:58 19kN m

This loading arrangement gives the maximum column moments, as plotted in

figure 3.16.

Figure 3.14

Beam shearing-force

diagrams (kN)

Figure 3.15

Bending-moment and

shearing-force envelopes

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EXAMPLE 3 .4

Analysis of a substitute frame for a column

The substitute frame for this example, shown in figure 3.17, is taken from the building

frame in figure 3.10. The loading to cause maximum column moments is shown in the

figure for Gk 25 kN/m andQk 10 kN/m.

The stiffnesses of these members are identical to those calculated in example 3.3,

except that for this type of frame the beam stiffnesses are halved. Thus

kAB1

2 0:9 103 0:45 103

kBC1

2 1:35 103 0:675 103

upper column kU 0:31 103

lower columnkL 0:27 103

Pk 0:45 0:675 0:31 0:27 103 1:705 103

fixed-end momentMBA 292:5 6

12 146kNm

fixed-end moment MBC 135 4

12 45kNm

Column moments are

upper column MU 146 45 0:31

1:705 18kNm

lower columnML 146 45 0:27

1:705 16kNm

The column moments are illustrated infigure 3.18. They should be compared with the

corresponding moments for the internal column infigure 3.16.

Figure 3.16

Column bending

moments (kN m)

Figure 3.17

Substitute frame

Figure 3.18

Column moments

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In examples 3.3 and 3.4 the second moment of area of the beam was calculated as

bh3=12 a rectangular section for simplicity, but where an in situ slab forms a flange tothe beam, the second moment of area may be calculated for the T-section or L-section.

3.4.2 Lateral loads on framesLateral loads on a structure may be caused by wind pressures, by retained earth or by

seismic forces. A horizontal force should also be applied at each level of a structure

resulting from a notional inclination of the vertical members representing imperfections.

The value of this depends on building height and number of columns (EC2 clause 5.2),

but will typically be less than 1% of the vertical load at that level for a braced structure.

This should be added to any wind loads at the ultimate limit state

An unbraced frame subjected to wind forces must be analysed for all the vertical

loading combinations described in section 3.2.1. The vertical-loading analysis can be

carried out by the methods described previously. The analysis for the lateral loads

should be kept separate. The forces may be calculated by an elastic computer analysis or

by a simplified approximate method. For preliminary design calculations, and also only

for medium-size regular structures, a simplified analysis may well be adequate.A suitable approximate analysis is the cantilever method. It assumes that:

1. points of contraflexure are located at the mid-points of all columns and beams; and

2. the direct axial loads in the columns are in proportion to their distances from thecentre of gravity of the frame. It is also usual to assume that all the columns in a

storey are of equal cross-sectional area.

It should be emphasised that these approximate methods may give quite inaccurate

results for irregular or high-rise structures. Application of this method is probably best

illustrated by an example, as follows.

EXAMPLE 3 .5

Simplified analysis for lateral loads cantilever method

Figure 3.19 shows a building frame subjected to a characteristic wind action of 3.0 kN

per metre height of the frame. This action is assumed to be transferred to the frame as a

concentrated load at each floor level as indicated in the figure.

By inspection, there is tension in the two columns to the left and compression in the

columns to the right; and by assumption 2 the axial forces in columns are proportional to

their distances from the centre line of the frame.

Figure 3.19

Frame with lateral load

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Thus

Axial force in exterior column :axial force in interior column 4:0P : 1:0P

The analysis of the frame continues by considering a section through the top-storey

columns: the removal of the frame below this section gives the remainder shown in

figure 3.20a. The forces in this subframe are calculated as follows.

(a) Axial forces in the columns

Taking moments about point s,P

Ms 0, therefore

5:25 1:75 P 6:0 P 10:0 4P 16:0 0

and therefore

P 0:135 kN

thus

N1 N4 4:0P 0:54kN

N2 N3 1:0P 0:135 kN

(b) Vertical shearing forcesFin the beams

For each part of the subframe, PF 0, thereforeF1 N1 0:54kN

F2 N1 N2 0:675 kN

(c) Horizontal shearing forcesHin the columns

Taking moments about the points of contraflexure of each beam,P

M 0, therefore

H1 1:75 N1 3:0 0

H1 0:93kN

Figure 3.20

Subframes at the roof

and 4th floor

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and

H1 H21:75 N1 8:0 N2 2:0 0

H2 1:70kN

The calculations of the equivalent forces for the fourth floor (figure 3.20b) follow asimilar procedure, as follows.

(d) Axial forces in the columns

For the frame above section tt0,P

Mt 0, therefore

5:253 1:75 10:5 1:75 P 6:0 P 10:0 4P 16:0 0

P 0:675 kN

therefore

N1 4:0P 2:70 kN

N2 1:0P 0:68kN

(e) Beam shears

F1 2:70 0:54

2:16kN

F2 2:70 0:68 0:54 0:135

2:705 kN

(f) Column shears

H1 1:75 0:93 1:75 2:70 0:543:0 0

H1 2:78kN

H2 12

10:5 5:25 2:78

5:1 kN

Values calculated for sections taken below the remainingfloors are

third floorN1 7:03kN N2 1:76kN

F1 4:33kN F2 5:41kN

H1 4:64kN H2 8:49kN

second floor N1 14:14kN N2 3:53kN

F1 7:11kN F2 8:88kN

H1 6:61kN H2 12:14kN

first floorN1 24:37kN N2 6:09kNF1 10:23kN F2 12:79kN

H1 8:74kN H2 16:01kN

The bending moments in the beams and columns at their connections can be

calculated from these results by the following formulae

beams MB F12

beam span

columns MC H12

storey height

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so that the roofs external connection

MB 0:54 1

2 60 1:6 k N m

MC 0:93 1

2 3:5 1:6 k N m

As a check at each joint,P

MBP

MC.

The bending moments due to characteristic wind loads in all the columns and beams

of this structure are shown infigure 3.21.

3.5 Shear wall structures resisting horizontal loads

A reinforced concrete structure with shear walls is shown in figure 3.22 . Shear walls are

very effective in resisting horizontal loads such as Fz in the figure which act in the

direction of the plane of the walls. As the walls are relatively thin they offer little

resistance to loads which are perpendicular to their plane.

The floor slabs which are supported by the walls also act as rigid diaphragms which

transfer and distribute the horizontal forces into the shear walls. The shear walls act as

vertical cantilevers transferring the horizontal loads to the structural foundations.

3.5.1 Symmetrical arrangement of walls

With a symmetrical arrangement of walls as shown in figure 3.23 the horizontal load is

distributed in proportion to the the relative stiffness ki of each wall. The relative

Figure 3.21

Moments (kN m) and

reactions (kN)

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stiffnesses are given by the second moment of area of each wall about its major axis

such that

ki h b3

where h is the thickness of the wall and b is the length of the wall.

The force P i distributed into each wall is then given by

Pi F kiP

k

EXAMPLE 3 .6

Symmetrical arrangement of shear walls

A structure with a symmetrical arrangement of shear walls is shown in figure 3.23.

Calculate the proportion of the 100 kN horizontal load carried by each of the walls.

Figure 3.22

Shear wall structure

Figure 3.23

Symmetrical arrangement

of shear walls

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Relative stiffnesses:

Walls A kA 0:3 203 2400

Walls B kB 0:2 83 346

Pk 22400 346 5492

Force in each wall:

PA kAP

k F

2400

5492 100 43:7 kN

PB kBP

k F

346

5492 100 6:3 kN

Check 243:7 6:3 100 kN F

3.5.2 Unsymmetrical arrangement of wallsWith an unsymmetrical arrangement of shear walls as shown in figure 3.24 there will

also be a torsional force on the structure about the centre of rotation in addition to the

direct forces caused by the translatory movement. The calculation procedure for this

case is:

1. Determine the location of the centre of rotation by taking moments of the wallstiffnesses kabout convenient axes. Such that

x

PkxxP

kxand y

PkyyP

ky

where kx and ky are the stiffnesses of the walls orientated in the xand y directions

respectively.

2. Calculate the torsional momentMt on the group of shear walls as

Mt F e

where e is the eccentricity of the horizontal force Fabout the centre of rotation.

3. Calculate the force Pi in each wall as the sum of the direct component Pd and thetorsional rotation componentP r

Pi Pd Pr

F kxP

kxMt

kiriPkiri2

where ri is the perpendicular distance between the axis of each wall and the centre

of rotation.

EXAMPLE 3 .7

Unsymmetrical layout of shear walls

Determine the distribution of the 100 kN horizontal force F into the shear walls A, B, C,

D and E as shown in figure 3.24. The relative stiffness of each shear wall is shown in the

figure in terms of multiples ofk.

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Centre of rotationPkx 20 5 5 30

Taking moments forkxabout YY at wall A

x

PkxxP

k

20 0 5 32 5 40

30

12:0 metresPky 6 4 10

Taking moments forky about XX at wall C

y

PkyyP

ky

6 0 4 16

10

6:4 metres

The torsional momentMt is

Mt F 20 x 100 20 12

800 kNm

The remainder of these calculations are conveniently set out in tabular form:

Wall k x ky r kr kr 2 Pd Pr Pi

A 20 0 12 240 2880 66.6 20.4 46.2

B 0 4 9.6 38.4 369 0 3.3 3.3

C 0 6 6.4 38.4 246 0 3.3 3.3

D 5 0 20 100 2000 16.7 8.5 25.2

E 5 0 28 140 3920 16.7 11.9 28.6P

30 10 9415 100 0 100

As an example for wall A:

PA Pt Pr F kAP

kMt

kArAPkiri

2

100 20

30 800

20 12

9415 66:6 20:4 46:2 kN

Figure 3.24

Unsymmetrical arrangement

of shear walls

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3.5.3 Shear walls with openings

Shear walls with openings can be idealised into equivalent plane frames as shown in

figure 3.25. In the plane frame the second moment of area Ic of the columns is

equivalent to that of the wall on either side of the openings. The second moment of area

Ib of the beams is equivalent to that part of the wall between the openings.

The lengths of beam that extend beyond the openings as shown shaded in figure 3.25

are given a very large stiffnesses so that their second moment of area would be say

100Ib.

The equivalent plane frame would be analysed by computer with a plane frame

program.

3.5.4 Shear walls combined with structural frames

For simplicity in the design of low or medium-height structures shear walls or a lift

shaft are usually considered to resist all of the horizontal load. With higher risestructures for reasons of stiffness and economy it often becomes necessary to include

the combined action of the shear walls and the structural frames in the design.

A method of analysing a structure with shear walls and structural frames as one

equivalent linked-plane frame is illustrated by the example in figure 3.26.

In the actual structure shown in plan there are four frames of type A and two frames

of type B which include shear walls. In the linked frame shown in elevation the four

type A frames are lumped together into one frame whose member stiffnesses are

multiplied by four. Similarly the two type B frames are lumped together into one frame

whose member stiffnesses are doubled. These two equivalent frames are then linked

together by beams pinned at each end.

The two shear walls are represented by one column having the sectional properties of

the sum of the two shear walls. For purposes of analysis this column is connected to therest of its frame by beams with a very high bending stiffness, say 1000 times that of the

other beams so as to represent the width and rigidity of the shear wall.

The link beams transfer the loads axially between the two types of frames A and B so

representing the rigid diaphragm action of the concrete floor slabs. These link beams,

pinned at their ends, would be given a cross-sectional area of say 1000 times that of the

other beams in the frame.

As all the beams in the structural frames are pressing against the rigid shear wall in

the computer model the effects of axial shortening in these beams will be exaggerated,

Figure 3.25

Shear wall with openings

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whereas this would normally be of a secondary magnitude. To overcome this the cross-

sectional areas of all the beams in the model may be increased say to 1000 m2 and this

will virtually remove the effects of axial shortening in the beams.

In the computer output the member forces for type A frames would need to be

divided by a factor of four and those for type B frames by a factor of two.

3.6 Redistribution of moments

Some method of elastic analysis is generally used to calculate the forces in a concrete

structure, despite the fact that the structure does not behave elastically near its ultimate

load. The assumption of elastic behaviour is reasonably true for low stress levels; but as

a section approaches its ultimate moment of resistance, plastic deformation will occur.

Figure 3.26

Idealised link frame for a

structure with shear walls and

structural frames

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This is recognised in EC2, by allowing redistribution of the elastic moments subject to

certain limitations.

Reinforced concrete behaves in a manner midway between that of steel and concrete.

The stressstrain curves for the two materials (figures 1.5 and 1.2) show the elastoplastic

behaviour of steel and the plastic behaviour of concrete. The latter will fail at a

relatively small compressive strain. The exact behaviour of a reinforced concrete

section depends on the relative quantities and the individual properties of the two

materials. However, such a section may be considered virtually elastic until the steel

yields; and then plastic until the concrete fails in compression. Thus the plastic

behaviour is limited by the concrete failure; or more specifically, the concrete failure

limits the rotation that may take place at a section in bending. A typical moment

curvature diagram for a reinforced concrete member is shown in figure 3.27

Thus, in an indeterminate structure, once a beam section develops its ultimate

moment of resistance,Mu, it then behaves as a plastic hinge resisting a constant moment

of that value. Further loading must be taken by other parts of the structure, with the

changes in moment elsewhere being just the same as if a real hinge existed. Provided

rotation of a hinge does not cause crushing of the concrete, further hinges will be

formed until a mechanism is produced. This requirement is considered in more detail in

chapter 4.

EXAMPLE 3 .8

Moment redistribution single span fixed-end beam

The beam shown in figure 3.28 is subjected to an increasing uniformly distributed load:

Elastic support momentwL2

12

Elastic span moment wL2

24

In the case where the ultimate bending strengths are equal at the span and at the

supports, and where adequate rotation is possible, then the additional loadwa, which themember can sustain by plastic behaviour, can be found.

At collapse

MuwL2

12

wL2

24 additional mid-span momentmB

where mB waL2=8 as for a simply supported beam with hinges at A and C.

Figure 3.27

Typical momentcurvature

diagram

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Thus wL2

12

wL2

24

waL2

8

Hence wa w

3

where w is the load to cause thefirst plastic hinge; thus the beam may carry a load of

1:33w with redistribution.From the design point of view, the elastic bending-moment diagram can be obtained

for the required ultimate loading in the ordinary way. Some of these moments may then

be reduced; but this will necessitate increasing others to maintain the static equilibrium

of the structure. Usually it is the maximum support moments which are reduced, so

economising in reinforcing steel and also reducing congestion at the columns. The

requirements for applying moment redistribution are:

1. Equilibrium between internal and external forces must be maintained, hence it is

necessary to recalculate the span bending moments and the shear forces for the loadcase involved.

2. The continuous beams or slabs are predominately subject to flexure.

3. The ratio of adjacent spans be in the range of 0.5 to 2.

4. The column design moments must not be reduced.

There are other restrictions on the amount of moment redistribution in order to ensure

ductility of the beams or slabs. This entails limitations on the grade of reinforcing steel

and of the areas of tensile reinforcement and hence the depth of the neutral axis as

described in Chapter Four Analysis of the Section.

EXAMPLE 3 .9

Moment redistribution

In example 3.3, figure 3.13, it is required to reduce the maximum support moment of

MBA 147 kN m as much as possible, but without increasing the span moment abovethe present maximum value of 118 kN m.

Figure 3.28

Moment redistribution,

one-span beam

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Figure 3.29a duplicates the original bending-moment diagram (part 3 offigure 3.13)

of example 3.3 while figure 3.29b shows the redistributed moments, with the span

moment set at 118kN m. The moment at support B can be calculated, using a

rearrangement of equations 3.4 and 3.1.

Thus

VAB

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiMmax MAB2w

pand

MBA VAB wL

2

L MAB

For span AB, w 48:75 kN m, therefore

VABffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

118 67 2 48:75 p

134 kN

MBA 134 48:75 6:0

2

6:0 67 140kNm

and

VBA 292:5 134

158:5 kN

Reduction inMBA 147 140

7 k N m

7 100

147 4:8 per cent

Figure 3.29

Moments and shears after

redistribution

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In order to ensure that the moments in the columns at joint B are not changed by the

redistribution, momentMBC must also be reduced by 7 kN m. Therefore

MBC 115 7 108 kNm hogging

For the revised moments in BC:

VBC108 80

4

195

2 105 kN

VCB 195 105 90kN

For span BC:

Mmax 1052

2 48:75 108 5 kNm sagging

Figure 3.29c shows the revised shearing-force diagram to accord with the

redistributed moments. This example illustrates how, with redistribution

1. the moments at a section of beam can be reduced without exceeding the maximum

design moments at other sections;2. the values of the column moments are not affected; and

3. the equilibrium between external loads and internal forces is maintained.

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