structural analysis

3

CHAPTER ......................................

Analysis of the structure at the ultimate limit state CHAPTER 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 all possible 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 for the 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 interpreted when they are presented diagrammatically.

28

Analysis of the structure

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 structure’s 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 can be 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. Overestimation, however, should be done with care, since the permanent action can often actually 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,

29

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 or foundations 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 actions Qk , wind actions Wk , 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 þ 1:5Qk 2. Permanent and wind actions 1:35Gk þ 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

Analysis of the structure Figure 3.2 Multi-span beam loading patterns

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 þ 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 Young’s 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 the modular ratio; 3. the compression area only of the concrete cross-section, plus the transformed area of reinforcement based on the modular ratio. The concrete cross-section described in (1) is the simpler to calculate and would normally be chosen.

31

32

Reinforced concrete design 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 positive as, for example, in table 3.1 for the fixed end moments (FEM). 2. For subsequently calculating the moments along the span of a member, moments causing 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 þ 1:5Qk on the span.

E X AM P L E 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 Analysis of one-span beam

Figure 3.3 shows the values of ultimate load required in the calculations of the shearing forces and bending moments. 54 195 ¼ 124:5 kN Maximum shear force ¼ þ 2 2 54  4 195  4 þ ¼ 151:5 kN m Maximum bending moment ¼ 4 8 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.

Analysis of the structure

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 no fixity with the supports so that the beam is free to rotate. This assumption is not strictly true for beams framing 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. Figure 3.4 Shears and moments in a beam

Using the sign convention of figure 3.4 and taking moments about support B: VAB L 

wL2 þ MAB  MBA ¼ 0 2

therefore VAB ¼

wL ðMAB  MBA Þ  2 L

ð3:1Þ

and VBA ¼ wL  VAB

ð3:2Þ

33

34

Reinforced concrete design Maximum span moment Mmax occurs at zero shear, and distance to zero shear a3 ¼

VAB w

ð3:3Þ

therefore Mmax ¼

VAB 2 þ MAB 2w

ð3:4Þ

The points of contraflexure occur at M ¼ 0, that is VAB x 

wx2 þ MAB ¼ 0 2

where x the distance from support A. Taking the roots of this equation gives qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ffi VAB  VAB 2 þ 2wMAB x¼ w so that a1 ¼

VAB 

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ffi VAB 2 þ 2wMAB w

ð3:5Þ

and a2 ¼ L 

VAB þ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ffi VAB 2 þ 2wMAB w

ð3:6Þ

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 distances a1 , a2 and a3 which locate the points of contraflexure and maximum moment – a sketch of the bending moment is often adequate – but if a computer is performing the calculations these distances may as well be determined also. At the face of the support, width s  ws s 0 MAB ¼ MAB  VAB  4 2

E X AM P L E 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 of Gk ¼ 25 kN/m and a variable action Qk ¼ 10 kN/m. The critical loading patterns for the ultimate limit state are shown in figure 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 the first loading arrangement: similar calculations would be required for each of the remaining load cases. It should be 3I noted that the reduced stiffness of has been used for the end spans. 4L

Analysis of the structure

35

Figure 3.5 Continuous beam loading patterns

Table 3.1 Moment distribution for the first loading case A 3 I : 4 L 3 1 ¼ : ¼ 0:125 4 6 0:125 0:125 þ 0:25 ¼ 1=3

Stiffness (k)

Distr. factors

Load (kN) F.E.M.

B

C I L

0 Balance

3 I : 4 L

1 ¼ 0:25 4 0:25 0:125 þ 0:25 ¼

¼ 0:125

2=3

2=3

292 0

D

1=3

135 292  6 8  219.4 þ 58.1

Carry over



292

135  4 12

292  6 8 þ 219.4  58.1

þ

þ 45.0 þ 116.3

 45.0  116.3

 58.1

þ 58.1

Balance Carry over

þ 19.4

þ 38.7  19.4

 38.7 þ 19.4

 19.4

Balance Carry over

þ

6.5

þ 12.9  6.5

 12.9 þ 6.5



6.5

Balance Carry over

þ

2.2

þ 

4.3 2.2

 þ

4.3 2.2



2.2

Balance

þ

0.7

þ

1.5



1.5



0.7

M (kN m)

0

 132.5

þ 132.5

 132.5

þ 132.5

0 0

0

36

Reinforced concrete design 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 of figure 3.4: load ðMAB  MBA Þ  2 L 292:5 132:5  ¼ 124:2 kN ¼ 2 6:0 ¼ load  VAB ¼ 292:5  124:2 ¼ 168:3 kN

Shear VAB ¼

VBA

Maximum moment, span AB ¼

VAB 2 þ MAB 2w

where w ¼ 292:5=6:0 ¼ 48:75 kN/m. Therefore: 124:22 þ 0 ¼ 158:2 kNm 2  48:75 VAB Distance from A, a3 ¼ w 124:2 ¼ 2:55 m ¼ 48:75

Mmax ¼

The bending-moment diagrams for each of the loading arrangements are shown in figure 3.6, and the corresponding shearing-force diagrams are shown in figure 3.7. The individual bending-moment diagrams are combined in figure 3.8a to give the bendingmoment 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 partial fixity equal to 25 per cent of the maximum moment in the span, that is 158=4 ¼ 39:5 kNm. Figure 3.6 Bending-moment diagrams (kN m)

Analysis of the structure

37

Figure 3.7 Shearing-force diagrams (kN)

Figure 3.8 Bending-moment and shearing-force envelopes

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 3.9 Bending-moment and shearing-force coefficients for beams

38

Reinforced concrete design 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 if Qk 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 of finite-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 vertical loads 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 in figure 3.10, for example, can be divided into any of the subframes shown in figure 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 to find the moments in the columns only. It consists of a single junction, with the remote ends of the members fixed. This type of subframe would 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.

Analysis of the structure

Figure 3.10 Building frame

Figure 3.11 Substitute frames

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.

EX AM PLE 3.3 Analysis of a substitute frame

The substitute frame shown in figure 3.12 is part of the complete frame in figure 3.10. The characteristic actions carried by the beams are permanent actions (including selfweight) Gk ¼ 25 kN/m, and variable action, Qk ¼ 10 kN/m, uniformly distributed along the beam. The analysis of the subframe will be carried out by moment distribution: thus the member stiffnesses and their relevant distribution factors are first required. Figure 3.12 Substitute frame

39

40

Reinforced concrete design Stiffnesses, k

Beam I¼

0:3  0:63 ¼ 5:4  103 m4 12

Spans AB and CD kAB ¼ kCD ¼

5:4  103 ¼ 0:9  103 6:0

Span BC kBC ¼

5:4  103 ¼ 1:35  103 4:0

Columns I¼

0:3  0:353 ¼ 1:07  103 m4 12

Upper kU ¼

1:07  103 ¼ 0:31  103 3:5

Lower kL ¼

1:07  103 ¼ 0:27  103 4:0

kU þ kL ¼ ð0:31 þ 0:27Þ103 ¼ 0:58  103 Distribution factors

Joints A and D P k ¼ 0:9 þ 0:58 ¼ 1:48 0:9 ¼ 0:61 D.F.AB ¼ D.F.DC ¼ 1:48 0:58 D.F.cols ¼ ¼ 0:39 1:48 Joints B and C P k ¼ 0:9 þ 1:35 þ 0:58 ¼ 2:83 0:9 ¼ 0:32 D.F.BA ¼ D.F.CD ¼ 2:83 1:35 D.F.BC ¼ D.F.CB ¼ ¼ 0:48 2:83 0:58 D.F.cols ¼ ¼ 0:20 2:83 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.

 68.8

M (kN m)

 135.0

þ 68.8

 1.3

Bal.

 4.3

 6.3

 56.9

 3.4 þ 2.8

C.O.

Bal.

C.O.

Bal.

C.O.

Bal.

F.E.M.

þ 3.4  2.1

0.32

 5.0 þ 6.9

292

BA

þ 11.0  6.7

0.61

AB

 146 þ 32.3  44.6 þ 22.0

Cols. P M) 0.39 þ 146  89.1 þ 16.2  9.9

D.F.s Load kN

(

A

Table 3.2 Moment distribution for the first loading case

(

þ 40.0

þ 1.7

þ 4.3

þ 13.8

þ 20.2

Cols. P M) 0.20

B

þ 95.0

 5.2 þ 4.1

 16.5 þ 10.3

þ 45.0 þ 48.5  24.2 þ 33.0

0.48

BC

135

 95.0

þ 5.2  4.1

þ 16.5  10.3

 45.0  48.5 þ 24.2  33.0

0.48

CB (

 40.0

 1.7

 4.3

 13.8

 20.2

Cols. P M) 0.20

C

þ 135.0

þ 3.4  2.8

þ 5.0  6.9

þ 146  32.3 þ 44.6  22.0

0.32

CD

292

 68.8

 3.4 þ 2.1

 11.0 þ 6.7

 146 þ 89.1  16.2 þ 9.9

0.61

DC

D

þ 68.8

þ 1.3

þ 4.3

þ 6.3

þ 56.9

Cols. P ( M) 0.39

Analysis of the structure 41

42

Reinforced concrete design 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: load ðMAB  MBA Þ  2 L 292:5 ð68:8 þ 135:0Þ  ¼ 135 kN ¼ 2 6:0

Shear VAB ¼

VBA ¼ load  VAB ¼ 292:5  135 ¼ 157 kN Maximum moment, span AB ¼ ¼

VAB 2 þ MAB 2w 1352  68:8 2  48:75

¼ 118 kN m VAB 135 ¼ 2:8 m Distance from A, a3 ¼ ¼ 48:75 w 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 of figure 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 X kcol Mcol  P Mcol ¼ kcol Figure 3.13 Beam bending-moment diagrams (kNm)

Analysis of the structure Figure 3.14 Beam shearing-force diagrams (kN)

Figure 3.15 Bending-moment and shearing-force envelopes

Thus, for the first loading arrangement and taking

P

Mcol table 3.2 gives

0:31 ¼ 37 kN m 0:58 0:27 ¼ 68:8  ¼ 32 kN m 0:58 0:31 ¼ 21 kN m ¼ 40  0:58 0:27 ¼ 19 kN m ¼ 40  0:58

Column moment MAJ ¼ 68:8  MAE MBK MBF

This loading arrangement gives the maximum column moments, as plotted in figure 3.16.

43

44

Reinforced concrete design

Figure 3.16 Column bending moments (kN m)

E X AM P L E 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 ¼ 25 kN/m and Qk ¼ 10 kN/m. Figure 3.17 Substitute frame

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 1 kAB ¼  0:9  103 ¼ 0:45  103 2 1 kBC ¼  1:35  103 ¼ 0:675  103 2 upper column kU ¼ 0:31  103 lower column kL ¼ 0:27  103 P k ¼ ð0:45 þ 0:675 þ 0:31 þ 0:27Þ  103 ¼ 1:705  103 6 fixed-end moment MBA ¼ 292:5  ¼ 146 kN m 12 4 fixed-end moment MBC ¼ 135  ¼ 45 kN m 12 Column moments are 0:31 ¼ 18 kN m 1:705 0:27 ¼ 16 kN m lower column ML ¼ ð146  45Þ  1:705

upper column MU ¼ ð146  45Þ  Figure 3.18 Column moments

The column moments are illustrated in figure 3.18. They should be compared with the corresponding moments for the internal column in figure 3.16.

Analysis of the structure 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 to the beam, the second moment of area may be calculated for the T-section or L-section.

3.4.2 Lateral loads on frames Lateral 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 the centre 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. EX AM PLE 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

45

46

Reinforced concrete design

Figure 3.20 Subframes at the roof and 4th floor

Thus Axial force in exterior column : axial force in interior column ¼ 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

P Taking moments about point s, Ms ¼ 0, therefore 5:25  1:75 þ P  6:0  P  10:0  4P  16:0 ¼ 0 and therefore P ¼ 0:135 kN thus N1 ¼ N4 ¼ 4:0P ¼ 0:54 kN N2 ¼ N3 ¼ 1:0P ¼ 0:135 kN (b) Vertical shearing forces F in the beams

For each part of the subframe,

P

F ¼ 0, therefore

F1 ¼ N1 ¼ 0:54 kN F2 ¼ N1 þ N2 ¼ 0:675 kN (c) Horizontal shearing forces H in the columns

Taking moments about the points of contraflexure of each beam, H1  1:75  N1  3:0 ¼ 0 H1 ¼ 0:93 kN

P

M ¼ 0, therefore

Analysis of the structure and ðH1 þ H2 Þ1:75  N1  8:0  N2  2:0 ¼ 0 H2 ¼ 1:70 kN The calculations of the equivalent forces for the fourth floor (figure 3.20b) follow a similar procedure, as follows. (d) Axial forces in the columns

For the frame above section tt0 ,

P

Mt ¼ 0, therefore

5:25ð3  1:75Þ þ 10:5  1:75 þ P  6:0  P  10:0  4P  16:0 ¼ 0 P ¼ 0:675 kN therefore N1 ¼ 4:0P ¼ 2:70 kN N2 ¼ 1:0P ¼ 0:68 kN (e) Beam shears

F1 ¼ 2:70  0:54 ¼ 2:16 kN F2 ¼ 2:70 þ 0:68  0:54  0:135 ¼ 2:705 kN (f) Column shears

H1  1:75 þ 0:93  1:75  ð2:70  0:54Þ3:0 ¼ 0 H1 ¼ 2:78 kN 1 H2 ¼ ð10:5 þ 5:25Þ  2:78 2 ¼ 5:1 kN Values calculated for sections taken below the remaining floors are third floor N1 ¼ 7:03 kN

N2 ¼ 1:76 kN

F1 ¼ 4:33 kN

F2 ¼ 5:41 kN

H1 ¼ 4:64 kN

H2 ¼ 8:49 kN

second floor N1 ¼ 14:14 kN

N2 ¼ 3:53 kN

F1 ¼ 7:11 kN

F2 ¼ 8:88 kN

H1 ¼ 6:61 kN

H2 ¼ 12:14 kN

first floor N1 ¼ 24:37 kN

N2 ¼ 6:09 kN

F1 ¼ 10:23 kN

F2 ¼ 12:79 kN

H1 ¼ 8:74 kN

H2 ¼ 16:01 kN

The bending moments in the beams and columns at their connections can be calculated from these results by the following formulae beams

MB ¼ F  12 beam span

columns MC ¼ H  12 storey height

47

48

Reinforced concrete design

Figure 3.21 Moments (kN m) and reactions (kN)

so that the roof’s external connection 1 MB ¼ 0:54   60 ¼ 1:6 kN m 2 1 MC ¼ 0:93   3:5 ¼ 1:6 kN m 2 P P As a check at each joint, MB ¼ MC . The bending moments due to characteristic wind loads in all the columns and beams of this structure are shown in figure 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

Analysis of the structure Figure 3.22 Shear wall structure

stiffnesses are given by the second moment of area of each wall about its major axis such that k i  h  b3 where h is the thickness of the wall and b is the length of the wall. The force Pi distributed into each wall is then given by ki Pi ¼ F  P

k

EX AM PLE 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 3.23 Symmetrical arrangement of shear walls

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50

Reinforced concrete design Relative stiffnesses: Walls A

kA ¼ 0:3  203 ¼ 2400

Walls B

kB ¼ 0:2  83 ¼ 346 P k ¼ 2ð2400 þ 346Þ ¼ 5492

Force in each wall: kA 2400  100 ¼ 43:7 kN PA ¼ P  F ¼ 5492 k kB 346 PB ¼ P  F ¼  100 ¼ 6:3 kN 5492 k Check 2ð43:7 þ 6:3Þ ¼ 100 kN ¼ F

3.5.2 Unsymmetrical arrangement of walls With 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 wall stiffnesses k about convenient axes. Such that P P ðky yÞ ðkx xÞ x ¼ P and y ¼ P kx ky where kx and ky are the stiffnesses of the walls orientated in the x and y directions respectively. 2. Calculate the torsional moment Mt on the group of shear walls as Mt ¼ F  e where e is the eccentricity of the horizontal force F about the centre of rotation. 3. Calculate the force Pi in each wall as the sum of the direct component Pd and the torsional rotation component Pr Pi ¼ Pd þ Pr kx ki ri ¼ F  P  Mt  P kx ðki ri 2 Þ where ri is the perpendicular distance between the axis of each wall and the centre of rotation.

E X AM P L E 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 of k.

Analysis of the structure

51

Figure 3.24 Unsymmetrical arrangement of shear walls

Centre of rotation P kx ¼ 20 þ 5 þ 5 ¼ 30 Taking moments for kx about YY at wall A P ðkx xÞ 20  0 þ 5  32 þ 5  40 x ¼ P ¼ 30 k ¼ 12:0 metres P ky ¼ 6 þ 4 ¼ 10 Taking moments for ky about XX at wall C P ðky yÞ 6  0 þ 4  16 y ¼ P ¼ 10 ky ¼ 6:4 metres The torsional moment Mt is Mt ¼ F  ð20  xÞ ¼ 100  ð20  12Þ ¼ 800 kN m The remainder of these calculations are conveniently set out in tabular form: Wall

kx

ky

r

kr

kr 2

A B C D E P

20 0 0 5 5

0 4 6 0 0

12 9.6 6.4 20 28

240 38.4 38.4 100 140

2880 369 246 2000 3920

30

10

9415

Pd 66.6 0 0 16.7 16.7 100

As an example for wall A: kA kA rA P A ¼ P t þ P r ¼ F  P  Mt  P k ðki ri 2 Þ ¼ 100 

20 20  12  800  ¼ 66:6  20:4 ¼ 46:2 kN 30 9415

Pr

Pi

 20.4  3.3 3.3 8.5 11.9

46.2  3.3 3.3 25.2 28.6

0

100

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Reinforced concrete design

Figure 3.25 Shear wall with openings

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 rise structures 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 the rest 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,

Analysis of the structure

53

Figure 3.26 Idealised link frame for a structure with shear walls and structural frames

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

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Reinforced concrete design

Figure 3.27 Typical moment–curvature diagram

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 stress–strain 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.

E X AM P L E 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 moment ¼

wL2 12

Elastic span moment

wL2 24

¼

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 load wa , which the member can sustain by plastic behaviour, can be found. At collapse wL2 12 wL2 ¼ þ additional mid-span moment mB 24

Mu ¼

where mB ¼ ðwa L2 Þ=8 as for a simply supported beam with hinges at A and C.

Analysis of the structure Figure 3.28 Moment redistribution, one-span beam

wL2 wL2 wa L2 ¼ þ 12 24 8 w Hence wa ¼ 3

Thus

where w is the load to cause the first 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 load case 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’.

EX AM PLE 3.9 Moment redistribution

In example 3.3, figure 3.13, it is required to reduce the maximum support moment of MBA ¼ 147 kN m as much as possible, but without increasing the span moment above the present maximum value of 118 kN m.

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Reinforced concrete design

Figure 3.29 Moments and shears after redistribution

Figure 3.29a duplicates the original bending-moment diagram (part 3 of figure 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 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi VAB ¼ ½ðMmax  MAB Þ2w and

  wL L þ MAB MBA ¼ VAB  2

For span AB, w ¼ 48:75 kN m, therefore pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi VAB ¼ ½ð118 þ 67Þ  2  48:75 ¼ 134 kN   48:75  6:0 6:0  67 ¼ 140 kN m MBA ¼ 134  2 and VBA ¼ 292:5  134 ¼ 158:5 kN Reduction in MBA ¼ 147  140 ¼ 7 kN m 7  100 ¼ 4:8 per cent ¼ 147

Analysis of the structure In order to ensure that the moments in the columns at joint B are not changed by the redistribution, moment MBC must also be reduced by 7 kN m. Therefore MBC ¼ 115  7 ¼ 108 kN m hogging For the revised moments in BC: ð108  80Þ 195 þ ¼ 105 kN 4 2 ¼ 195  105 ¼ 90 kN

VBC ¼ VCB

For span BC: Mmax ¼

1052  108 ¼ 5 kN m sagging 2  48:75

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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