CIRIA Guide 2  Design of Deep Beams in R.C
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Design of Deep Beams in R.C...
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CIRIA GUIDE 2
The design of deep beams in reinforced concrete aVE ARUP & PARTNERS
CIRIA PUBLICATION _ _ _ _ _ _ _ _ _ _ __
CIRIA
is the Construction Industry Research and Infonnation Association.
It is a nonprofitdistributing organisation carrying out research on behalf of its members. Membership includes all types of flnns and organisations which have an involvement with construction, including clients, designers, consultants, contractors and suppliers. The members collaborate in research aimed at improving the efficiency of design, construction and management, and the perfonnance and serviceability of building and civil engineering works. They initiate and take part in the research programme, and have preferential access to the results of research projects. CIRIA obtains cost effectiveness by contracting out the detailed and specialist aspects of research to the most suitable external bodies. The cost of research is met from member SUbscriptions and special contributions, supplemented in some cases by public funds. For further details and SUbscription rates apply to The Secretary, CIRIA, 6 Storey's Gate, Westminster, London SWIP 3AU. Telephone: 01222 8891.
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I SUMMARY
The design of deep beams in reinforced concrete Omstruction Industry Research and Information Association Guide 2, January 1977 The Guide provides simple rules for designing reinforced concrete beams of span/depth ratio below 2 for single spans or 2.5 for multiple spans. Supplementary rules deal with cases where load capacity of the beam may be affected by elastic instability, where loads are concentrated, or where holes in the beam may affect performance. The derivation of the rules is given. Appendices give typical stress distribution in deep beams, design examples, buckling strength, shear strength of toploaded deep beams, and an estimate of deflections in deep beams.
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SUMMARY
The design of deep beams in reinforced concrete Construction Industry Research and Information Association Guide 2, January 1977 The Guide provides simple rules for designing reinforced concrete beams of span/depth rati below 2 for single spans or 2.5 for multiple spans. Supplementary rules deal with cases whe load capacity of the beam may be affected by elastic instability, where loads are concentrat or where holes in the beam may affect performance. The derivation of the rules is given. Appendices give typical stress distribution in deep beams, design examples, buckling strengt shear strength of toploaded deep beams, and an estimate of deflections in deep beams.
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SUMMARY
SUMMARY
The design of deep beams in reinforced concrete Construction Industry Research and Information Association Guide 2. January 1977
The design of deep beams in reinforced concrete Construction Industry Research and Information Association Guide 2, January 1977
The Guide provides simple rules for designing reinforced concrete beams of span/depth ratio below 2 for single spans or 2.5 for multiple spans. Supplementary rules deal with cases where load capacity of the beam may be affected by elastic instability. where loads are concentrated, or where holes in the beam may affect performance. The derivation of the rules is given. Appendices give typical stress distribution in deep beams, design examples, buckling strength. shear strength of top loaded deep beams, and an estimate of deflections in deep beams.
The Guide provides simple rules for designing reinforced concrete beams of span/depth rati, below 2 for single spans or 2.5 for multiple spans. Supplementary rules deal with cases whe load capacity of the beam may be affected by elastic instability, where loads are concentrat or where holes in the beam may affect performance. The derivation of the rules is given. Appendices give typical stress distribution in deep beams, design examples, buckling strengt shear strength of toploaded deep beams, and an estimate of deflections in deep beams.
KEYWORDS (from Construction Industry Thesaurus)
KEYWORDS (from Construction Industry Thesaurus)
Architecture, Cracking; Cracks; Deep beams; Elastic analysis; Holes; Ughtweight aggregate concrete; Umit analysis; Loading properties; Reinforced concrete; Reinforcement; Shear strength; Steel; Structural engineering; Structural members; Structure; Testing
Architecture, Cracking; Cracks; Deep beams; Elastic analysis; Holes; Lightweight aggregate concrete; Limit analysis; Loading properties; Reinforced concrete; Reinforcement; Shear strength: Steel; Structural engineering; Structural members; Structure; Testing
READER INTEREST
READER INTEREST
Concrete designers
Concrete designers
KEYWORDS (from Construction Industry Thesaurus)
KEYWORDS (from Construction Industry Thesaurus)
Architecture, Cracking: Cracks; Deep beams; Elastic analysis; Holes; Ughtweight aggregate concrete; Limit analysis; Loading properties; Reinforced concrete; Reinforcement; Shear strength; Steel; Structural engineering; Structural members; Structure; Testing
Architecture, Cracking; Cracks; Deep beams; Elastic analysis; Holes; Lightweight aggregate concrete; Limit analysis; Loading properties; Reinforced concrete; Reinforcement; Shear strength; Steel; Structural engineering; Structural members; Structure; Testing
READER INTEREST
READER INTEREST
Concrete designers
Concrete designers
January 1977 (reprinted 1984)
Guide 2
The design of deep beams in reinforced concrete
Dve Arup & Partners
This document is intended to provide members with guidance on the design of deep beams in reinforced concrete. It is based largely on existing published information which has been assessed critically and presented in a form suitable for application in practice. Every effort has been made to ensure that the guidance given is based on the best available knowledge or experience at the time of presentation, but no responsibility of whatever kind for any injury, delay, loss or damage however caused reSUlting from the use of such guidance can be accepted by CI RIA or the contributors.
Price: £35 (£7 to CIRIA Members)
ISBN: 086017 081 0
ISSN: 03063267
CONSTRUCTION INDUSTRY RESEARCH AND INFORMATION ASSOCIATION 6 STOREY'S GATE, LONDON SW1 P 3AU
Telephone 01222 8891 Telex 24224, prefix 2063
This Guide has been prepared for CIRIA by:
Authors (Ove Arup & Partners) A. Stevens FICE FISructE (Team Leader) 1. Blanchard BSc(Eng) DIC MIStructE E. Booth MA MICE B. Corr PhD 1. Konopicky 1. Newey BSc MICE
Assessors R. E. Rowe MA SeD FICE FIStructE FIHE A. G. Senior MSc FICE FIStructE 1. A. Waller BSc ACGI DIC FICE FIStructE
Cement and Concrete Association Atkins Research and Development Oscar Faber & Partners
CIRIA acknowledge with thanks the assistance given by Dr F. K. Kong of Cambridge University to the authors at various stages of the work.
Contents
Summary
4
1.
5 5 5 5 5 5 7 8 8 8 9
Introduction 1.1 General 1.2 Design principles 1.3 Scope 1.4 Limitations 1.5 Multipurpose deep beams 1.6 Geometry and behaviour of deep beams 1. 7 Design ru les  general considerations 2. Simple rules for the analysis of deep beams 2.1 General 2.2 Geometry 2.3 Computation of forces 2.4 Strength: ultimate limit state 2.5 Holes through deep beams 2.6 Serviceability limit state 3. Supplementary rules for the analysis of deep beams 3.1 General 3.2 Geometry 3.3 Computation of forces 3.4 Strength: ultimate limit state 3.5 Holes through deep beams 3.6 Serviceability limit state 4. Behaviour of deep beams: an explanation of the rules 4.1 Elastic analysis 4.2 Elastic instability 4.3 Ultimate load behaviour 4.4 Flexural failure 4.5 Shear failure 4.6 Shear capacity: toploaded beams 4.7 Shear capacity: bottom and indirectly loaded and indirectly supported deep beams 4.8 Shear capacity: upper limits 4.9 Shear capacity: load combination 4.10 Bearing failure 4.11 Bursting tensions produced under concentrated loads 4.12 Holes 4.13 Deformation and deflection 4.14 Settlement of foundations 4.15 Reinforcement 5. References Appendices A. Typical stress distribution in deep beams B. Examples of deep beam design C. Buckling strength of deep beams D. Shear strength of reinforced, toploaded deep beams E. Estimate of deflections in deep beams List of illustrations List of tables Symbols, terms and units of measurement
10 13 19 23 25 25 25 26 26 34
37 37 37 39 39 39 40 41
44 44 45 45
46 47 48 48 48 48 51 71 105 115
117 123 129 129
3
Summary The Guide provides simple rules for designing reinforced concrete beams of span/depth ratio below 2 for single spans or 2.5 for multiple spans. Supplementary rules deal with cases where load capacity of the beam may be affected by elastic instability, where loads are concentrated, or where holes in the beam may affect performance. The derivation of the rules is given. Appendices give typical stress distribution in deep beams, design examples. buckling strength, shear strength of top loaded deep beams, and an estimate of deflections in deep beams.
4
1.
Introduction 1.1 GENERAL This Guide has been prepared by a team of designers and has been approved by a panel of assessors as an authoritative statement of the art and of good practice in designing reinforced concrete deep beams. It is based on an exhaustive study of published literature and of research reports on the subject and owes much to the work of Leonhardt and Kong l  B and the CEB FIP international recommendations. 9 The information is presented in a manner considered suitable for use by engineers who have a good general knowledge and understanding of reinforced concrete structures but little or no experience in the design or testing of deep beams. The Guide is intended for use as a design aid and is not therefore primarily concerned with minimum requirements which may be appropriate in regulatory documents. It should be used in conjunction with the British Standard Code of Practice for the structural use of concrete CPII 0 10.
1.2 DESIGN PRINCIPLES The methods used in the Guide conform with the limit state principles adopted in CP 110 10 and the CEBFIP recommendations 9 , and apply to deep beams made with normal or lightweight aggregates.
1.3 SCOPE The Guide provides simple rules for designing the simpler forms of reinforced concrete beams which are deep in relation to their span (Le. beams with a span/depth ratio of less than 2 for single span beams or less than 2.5 for multispan beams). For depths greater than the span, the beam is considered to behave as a beam having a depth equal to the span, with the remainder of the beam supported by it, and it is analysed and designed accordingly. Supplementary rules are provided for more complex cases in which load capacity may be affected by elastic instability or where loads are concentrated or where holes in the beam may Significantly affect performance. For some cases not covered by the rules, an indication is given on how to set about finding a solution to the problem. Design examples and information on stress patterns, buckling, shear strength and deformation are given in the appendices. In order to cover as broad a range of applications as possible, it has been necessary, in some cases where adequate data are not available, to rely on reasonable supposition and engineering judgement.
1.4 LIMITATIONS Problems which would require extensive analysis or further experimental evidence and also the design and analysis of nonplanar (curved, stepped or angled) beams are not considered. Because of the lack of adequate test data on the behaviour of prestressed deep beams, this form of construction has not been covered nor has the effect of lateral loading. In these more complex cases and where the beam serves more than one purpose, the designer must make his own assessment by suitable methods of extrapolation, superimposition, testing or analysis as he considers necessary.
1.5 MULTIPURPOSE DEEP BEAMS Deep beams may be required to perform additional functions. Typical examples are given in Figures 1 and 2. Figure 1 shows a case where the deep beam is also a retaining wall and, in addition, spreads column loads down to a continuous foundation.
5
Retaining wall under
Column load from building above
Deep beam
Basement slab and foundation
Figure 1
Typical mUltipurpose deep beam
Load bearing wall
Floor slab omitted to sOON load bearing wall
Figure 2
Deep beams as part of other structures 6
Figure 2 shows a case where deep beams form part of a load bearing wall. In such cases, the designer must judge for himself, the effect of interaction of the several functions. For this purpose he will need an understanding of the flow or pattern of forces within the beam. Characteristic patterns of principal stresses derived from elastic analysis of deep beams under a number of load conditions are given in Appendix A. From these typical stress plots, it is seen that, unlike shallow beams, the whole area of the member is within the zone of influence of the applied loads and support reactions. Consequently, deep beams must be designed to take account of both the vertical and horizontal stress components within the beam.
1.6 GEOMETRY AND BEHAVIOUR OF DEEP BEAMS The geometry of a deep beam is only rarely determined initially by strength and serviceability considerations related to its function as a deep beam. However, its geometry determines how the beam behaves and, therefore how it should be analysed and reinforced. In designing normal (shallow) reinforced concrete beams of the proportions more commonly used in construction, it is assumed that plane sections remain plane after loading. This assumption is not strictly true but the errors resulting from it do not become Significant until the depth of the beam becomes equal to, or more than, about half the span. The beam is then classed as a deep beam, and special methods of analysis and design have to be used. These methods take into account not only the overall applied moments and shears, but also the stress patterns and internal deformations within the beam. The general pattern of forces in a singlespan deep beam, after the concrete in tension has cracked, can be considered as analogous to a tied arch or truss as shown in Figure 3. However, for some dispositions of loads and supports, this analogy is not precise. The centre of the compression force in the arch rises from the support to a position which is a distance of approximately half the span above the tie. The tension force in the tie is roughly constant along its length because, although the bending moment reduces from the centre of the span to the support, so does the lever arm.
Arch zone
UnIformly distributed
bottom load
Figure 3
Tied arch analogy for single span deep beam action  uniformly distributed bottom load Although the structural system for a single span deep beam may be thought of as a simple tied arch, the load on a continuous beam is transferred to the supports by a combination of two distinct systems as shown in Figure 4. First, there is the tied arch system as described for single spans and, second, a suspension system centred over internal supports. These analogies are helpful in understanding the typical pattern of forces within the beam, but they do not lend themselves to the accurate prediction of the value and direction of principal stresses for different load and support conditions. A fuller explanation and discussion of the behaviour of deep beams is given in Section 4 for engineers who seek a better understanding of the subject.
7
Tension bands ~
1
Figure 4 Combined tied arch and
J
J
FM ...... """""""" """'"
suspension system analogy for mUltispan deep beams
Uniformly distributed
.
Tie
Tension" system. sharing load 0IIeI" an internal support
Compression system sinilar to that in the single span
bottom load
1.7 DESIGN RULESGENERAL CONSIDERATIONS The derivation of the design rules given in this Guide is discussed in Section 4. The rules are based on the observed behaviour and analysis of deep beams tested and reported in various research reports and publications, the most important of which are included in the references (Section 5). The rules are condensed as much as possible to Simplify their use. Information which does not. contribute directly to the process of checking the structural adequacy of a deep beam, or of selecting and arranging the reinforcement, has been omitted. It is recommended that the design sequence to be followed in applying the rules should
be: 1. establish significant dimensions 2. compute applied forces 3. check adequacy of concrete section 4. select reinforcement sizes and arrangement. Use of this simple sequence should minimise the design work, especially when it becomes necessary to alter the geometry of the beam during the course of the design process.
2.
Simple rules for the analysis of deep beams 2.1
2.1.1 Scope
GENERAL
The simple rules given in this section apply to the design of reinforced concrete deep beams over two or more supports, which satisfy the following conditions: 1. The beam is a flat plate. 2. Holes in the beam are not such as to interfere with the stress pattern Significantly (see Section 2.5).
8
3. There is no appreciable differential movement of supports under load (but see Appendix E). 4. There are no concentrated loads (see Terms, page 130, and Section 2.3.1). 5. There are no indirect loads or supports (see Terms, page 130). Conditions (1), (2) and (3) exclude structures to which the Guide does not apply. Conditions (4) and (5) exclude beams to which the simple rules are not intended to apply (i.e. they occasion the use of complex reinforcement). The simple rules are primarily intended for a uniformly loaded deep beam with insignificant holes. Where this is not the case, the above conditions should be checked, before design begins, and, if necessary, reference made to the provisions of the supplementary rules (Section 3).
2.1.2 Limit states
The design loads to be used with this Guide should be appropriate to the limit state under ~onsideration, in accordance with Clause 2.3 of CP 110. 10 Design recommendations are quoted at the ultimate limit state unless otherwise stated.
2.2 2.2.1 Effective span and height of a deep beam
GEOMETRY
1l1ese should be determined in accordance with Figure 5 as follows: Effective span (I)
=
10 + (the lesser of c 1/2 or 0.110) + (the lesser of C2 /2 or 0.110 )
Active height (h a ) = h when I = I when h
> >
h
1
Effective span I
Figure 5
Basic dimensions of deep beams
2.2.2 Minimum thickness of beam (see Figure 6)
For practical construction purposes, the thickness of the beam, b, should not be less than the sum of the six dimensions described below: 1. twice the minimum concrete cover to the outer layers of reinforcement.
2. a dimension allowing for four layers of web reinforcement, one vertical and one horizontal at each face of the beam, including bar deformations as appropriate.
9
3. space for the principal reinforcement bars if not accommodated in the web steel allowance. 4. space for the introduction and compaction of the concrete between the inner layers or reinforcement  say, a minimum of 80 mm. 5. space for the vertical reinforcement from the supports, if not accommodated within other allowances. 6. space for the Ubar anchorage at the supports, if this is required. The diameter of the Ubend depends on the force in the bars and the allowable bearing stress against the curve (see CP 110, Gause 3.11.6.8). As a first approximation the outside dimensions of such a U bar bend can be taken as 16 bar diameters. Thus the minimum thickness will only rarely be less than about 200 mm and may reasonably be assumed to be nearer 300 mm.
Allow for \lertical bars     , from supports. if necessary
Minimum CO\ler
Vertical web steel anchored
around principal bar reinforcement
, ~.: '~.Jr,.01

/
L.
Figure 6
Allow for diameter of bend in any necessary U bar anchorages
,,
Allow space to admit & compact concrete 80 mm min.
Lower bars of horizontal web reinforcement included with principal bars where necessary
l
Vertical section throug!J
Thickness of deep beams
2.2.3 Slenderness limits: elastic stability
bottool
of beam
The siinple rules assume no reduction of capacity due to the slenderness of the section or to lack of adequate restraint. This assumption is valid if every panel can be defined as a short braced wall in the terms of Gause 3.8.1.1 of CP 110. In the examination of this condition, the effective height of each panel is taken to be I. I x the shortest distance between cen tres of parallel lateral restraint (where there are effective lateral restraints at all four edges of the panel) or as 1.5 x the distance between the centres of the parallel lateral restraints (where one or two opposite edges of the panel are free). When both rotational and lateral movements are restrained, the effective height may be taken as the clear distance between restraints. Otherwise, refer to Section 3.2.2 and Appendix C.
2.3 COMPUTATION OF FORCES
2.3.1 Loading
Beams to which the simple rules apply will be subject to loads as follows: 1. Vertical forces applied above a line O.7Sha above the beam soffit are considered to be applied at the top of the beam and may be represented in their original form or by
IO
their static equivalent, which may be uniformly or variably distributed along the span or part of the span. Consideration should be given to the possibility of tension stresses being set up by the spread ofload (see Section 3.4.4). 2. Loads applied within the depth of the beam (i.e. below a line 0.75ha above the beam soffit) will be considered as hanging loads (e.g. beam selfweight, floor loads). 3. Loads applied to the lower edge of the beam will be considered as hanging loads. 4. There will be no indirect loads. 5. The intensity of any load will be less than 0.2fcu, and applied over a length which exceeds 0.21. More intense loads and those applied over shorter lengths are considered to be 'concentrated', in which case reference should be made to Section 3.
2.3.2 Forces: moments, shears
To estimate the bending moments, only the forces applied over the clear span, 10 , need be considered, assuming a uniform distribution of the reactions over the supports, as illustrated typically in Figure 7. The total force on the support must, of course, include the loads applied over the support width.
Static eq..ivaJent of loading applied to beam
Loading CHef actual support
if any. not accarted for in bending moments Moments based on span loading only
'+++""..'4I...L..." Reactions assumed uniformly
II II
M Effective
l
I I II II
support length
j; so;l;
l
E_~L
l'¥,'lAc=tualsupport~"'lengthjIIif' , Figure 7
M
distributed over effective support length
l_~L
'1
Support lengths and span moments The length of support at each end of the span may be taken as the actual length or 0.2/0 , whichever is less. Support moments in a continuous beam should be computed and distributed on the basis of effective spans, and on the assumption that beam stiffnesses in each span are proportional to bh!/l. The design envelope for shear forces should be that for simply supported or continuous spans, but beam shears at an end support should not be modified to account for support moments. The redistribution of moments is not permitted. Forces acting over a support apply an equivalent and additional shear force to the critical section of the beam (Le. at the support face). The additional force over an end support, Qe' is nw 2) (see Figure 8)
Cl (nWl 
and at Ii typical internal support, Qi> is C2
(n w 4

nw 3) (see Figure 9) 11
/  Load UDL.rlw,
Load UOL.rlw,
I
I
V I I
~ Addio;""" ""'" "" \hi, ""'"
!
E_ _ '
I
J1 .tLr +~~+¥
Figure 8
I
r   1
Additional shear at end support
I . End
suppor'~.
jEffectilie Sl4JPOI"t length e
1
Ie
jL ....
Load LOL. nW. ~ ,/
I
/
Load In.. nW3
  _ .   ...  
~
v/ I
I
/
Additional shear
on this
plane
I I
I I I
Figure 9
Additional shear at internal support
~
_ _ _ _ _ _..,_ _
~ SlQXlI1

~~
______ E_ffec__tive_span__!14
I } ~ Effective SlWQI1 length ~~~ on relevant span
C2
Thus the additional shear forces are: At the end support
Qe(h a  c.)/ha
At an intermediate support
Qj(h a 
2C2)/2ha
where ha is the beam height for the span adjacent to the support face under consideration.
12
Definition of concentrated loads Concentrated loads are defined as those applied over a length of beam less than 0.21 and those whose stress intensity is greater than 0.2fcu. The measures required to deal with concentrated loads are not included in the Simple Rules. Where loads are spread into the beam through a plate or slab thickness, the intensity of loading on the beam may be checked by reference to Figure 10. Ifloads are found to be concentrated as defmed above, reference should be made to the Supplementary Rules (Section 3). Concentrated load
Deep beam
Concentrated load Critical
Deep beam
Plate or floor slab permits
Figure 10
spread of
Spread of concentrated top and bottom loads
Critical
load Level of hanger anchorage
Top loading
Bottom loading
2.4 STRENGTH: ULTIMATE LIMIT STATE 2.4.1 Strength in
Limits
bending
If I/ha < 1.5, it is unnecessary to confirm the strength of the concrete in compression due to bending. The bearing stress, if critical, will govern. If l/ha > 1.5, the condition that the applied moment M < 0.12fcubh~, must be satisfied. Area of reinforcement
The area of reinforcement provided to resist positive and negative moment should satisfy the following condition: M
As>   0.87fyz
where M is the design moment at ultimate limit state and z is the lever arm at which the reinforcement acts. The use of compression reinforcement is not envisaged. The lever arm, z, may be assumed to be: for single spans, positive sagging moments: I/h < 2 z = 0.21 + O.4ha for multispan beams, mid span and support moments: l/h < 2.5
z = 0.21 + O.3ha Where adjacent span lengths differ, the lever arm over the support shall be related to the longer span, provided that the value for z does not exceed twice the shorter span.
13
Detailing: principal bending moment reinforcement Reinforcement: positive (sagging moment) Reinforcement is not to be curtailed in the span and may be distributed over a depth ofO.2h a • The bars must be anchored to develop 80% of the maximum ultimate force beyond the face of the support and 20% of the maximum ultimate force at or beyond a point 0.2/0 from the face of the support, or at or beyond the far face of the support, whichever is less (see Figure 11). The main reinforcement must be anchored so that the concrete within the area of support relied upon for bearing is adequately confined. Anchorage bond stresses and the effective dimensions of hooks, bends and Ubars should comply with Clauses 3.11.6.2 and 3.11.6.7 and 8, respectively, of CP 110. Bar spacing should comply with Section 2.6.2 of this Guide. It is not necessary to check local bond stresses. Reinforcement: negative (hogging reinforcement) About half of the negative reinforcement should extend over the full length of adjacent spans and may be counted as part of the minimum web reinforcement. The remaining bars may be terminated at a distance 0.4/, or an anchorage bond length, measured from the face of the support (see Figure 11).
j h, Pms _ _ 100As) _ bh,.
b XS h =
200Asv
b xSv
It is assumed that the web reinforcement represented by Pwh, Pwv is uniform over the height ha of the beam and over the critical lengths of the span.
TABLE 6
Main (sagging) steel shear stress parameter,
% main (sagging) steel (Pms)
Oear shear span/height
(x/h)
1.0 0.8 0.6 0.4 0.2 0
TABLE 7
0.2
0.4
0.6
0.8
1.0
0.20 0.24 0.29 0.34 0.38 0.39
0.39 0.48 0.57 0.67 0.75 0.78
0.59 0.71 0.86 1.01 1.13 1.17
0.78 0.95 1.15 1.34 1.50 1.56
0.98 1.19 1.43 1.68 1.88 1.95
Horizontal web steel shear parameter,
1.0 0.8 0.6 0.4 0.2 0
TABLE 8
0.30
0.35
0.4
0.6
0.8
1.0
0.12 0.15 0.18 0.21 0.23 0.24
0.15 0.18 0.22 0.25 0.28 0.29
0.17 0.21 0.25 0.29 0.33 0.34
0.20 0.24 0.29 0.34 0.37 0.39
0.29 0.36 0.43 0.50 0.56 0.59
0.30 0.48 0.57 0.67 0.75 0.78
0.40 0.60 0.72 0.84 0.94 0.98
vwv
(N/mm 2 )
% vertical web reinforcement (pwv)
Clear shear span/height
1.0 0.8 0.6 0.4 0.2 0
(N/mm 2 )
0.25
Vertical web steel shear parameter, (x/h)
vwh
% horizontal web reinforcement (Pwh)
Oear shear span/height
(x/h)
(N/mm 2 )
Vms
0.15
0.20
0.30
0.40
0.60
0.80
1.0
0.07 0.05 0.02 0.01 0.00 0.00
0.10 0.06 0.03 0.01 0.00 0.00
0.15 0.09 0.05 0.02 0.00 0.00
0.20 0.12 0.06 0.02 0.00 0.00
0.29 0.18 0.10 0.03 0.00 0.00
0.39 0.24 0.12 0.04
0.49 0.30 0.16 0.05
0.00
0.01
om
om
31
Combined ioading Under combined top, bottom and indirect loads or supports, the following conditions must be satisfied:
where
Vet is the shear capacity assuming top loads only Vcb
is the shear capacity assuming bottom or indirect loads or an indirect support only
Vat
is applied shear from top loads
Vab
is applied shear from bottom or indirect loads or with indirect supports
orV<
Vcb
Where these conditions cannot be satisfied, the reinforcement, geometry or loading of the beam should be revised accordingly.
3.4.3 Bearing capacity at supports and concentrated loads
The limiting bearing stress (set at O.4fcu in the simple rules) may be exceeded under the conditions described below, but in practice the capacity of the supporting columns may prove to be the critical factor. However, it may be possible to extend the column reinforcement, upwards into the deep beam, to improve load bearing capacity locally. At end supports, and at internal supports for continuous beams, the bearing stress limit may be increased to 0.6fcu and 0.8fcu, respectively, provided the stressed zone is adequately confined. In the plane of the beam, such confinement is afforded by the stress pattern itself, provided that at end supports the anchorage of the main bending reinforcement at the support is effective (e.g. properly proportioned Ubars). However, when bearing stresses exceed O.4fcu, transverse confining forces must be provided, normal to the plane of the beam, by reinforcement stressed at 0.87fy, located in transverse walls or slabs, or with some qualification, by anchored links or binding (see Figure 30). The transverse force, which can be taken as approximately equal to 1/6 of the total support reaction, may be distributed over a depth equal to the width of the beam, b, and must extend, at this intensity, over the area in which stress in excess of 0.4fcu exists.
.
~
II
,...
J
Binding extends CNer area where bearing stress of Oo4fcu is exceeded
.. ==j..
..
• I
.
~
Figure 30
J
Design force
b
Reinforcement over a highly stressed support
32
Vertical reactions Q
0/6
Under concentrated loads, the bearing stresses may rise to not more than D.8/cu , provided that adequate confining reinforcement is present. Where a concentrated load is closer to the end of a deep beam than three times the length of the load, the limiting stress should be O.6fcu. Restraint within the plane of the beam may not be afforded by the pattern of stress trajectories, and special reinforcement may be needed, as described in Section 3.4.4. Confining reinforcement normal to the plane of the beam may be designed as described for supports. It should be noted that the confinement provided, within and normal to the plane of a deep beam, by systems of links or by a mesh of welded bars, cannot be entirely effective across the full width of the beam. Moreover, the degree of compaction (and therefore the strength of the concrete within the binding) may be adversely affected by the congestion of reinforcement. It may, therefore, be found that the feasibility of satisfactorily anchoring, fixing and concreting round the reinforcement demanded by these provisions may limit the bearing capacity at supports and at concentrated loads.
3.4.4 Reinforcement for concentrated loads
Concentrated loads give rise to both compressive and tensile forces within the beam. Reinforcement for the compressive forces should be provided as described above for bearing stresses. Reinforcement for the tensile forces should be as described below (see Figure 31). When a concentrated load, Q, which has not been accounted for in the moment analYSiS, is applied over the support at the end of a deep beam, tensions develop near the top surface for which reinforcement must be provided. A limiting tension force of Q/4 may be assumed, when the length over which the load is applied is very short. The tension may reduce to about Q/6 where the applied load is spread over say O.2h a. Values may be interpolated. This force may be resisted by steel stressed at O.87fy arranged as indicated in Figure 31. Vertical stresses which may develop at about midspan should be controlled as suggested in Figure 31.
O·5l,
LoadJ"
f l.sI T~ ·I~
j Figure 31
Reinforcement for tensile stresses caused by concentrated top loads above an end support
Zone
Zone
T_ _ ... deIIekJp (about
. tiSi 1.___....,!..:::::::::::x::::1 ~~max~'::,~ iTUSt be controlled by compressiYe stresses from other loading
ical stress on horizontal section at midheiglt
ofbeam
..l
or ,eilfolcemeilt
I
Effectille span l,
..l
A concentrated load, Q, over an internal support (Figure 32) which will not have been accounted for in the moment analysis, will induce a bursting force which will vary from Q/4 to Q/6 for loaded lengths of O.OSha and O.Zh a, respectively. Values may be interpolated. Reinforcement to resist this force, stressed at O.87fy, should be located as shown on Figure 32.
33
Figure 32 Reinforcement for tensile stresses caused by concentrated top loads above an internal support A concentrated load applied in the span will produce similar tensions to those over a support when the ratio Ilh < 1.0. Reinforcement should be provided as shown on Figure 33. When Ilh > 1, the tension forces under concentrated loads in the span may be disregarded, provided that it can be shown that, by their location, they are neutralised by compressive bending stresses. Nominal (but not designed) web reinforcement may be assumed to contribute to the reinforcement required to satisfy these provisions. Bar spacing and minimum reinforcement areas should comply with the requirements of Section 2.6.2. O·3l,
L
_.l~3l,
    . . f  .,, '
Figure 33 Reinforcement for tensile stresses caused by concentrated top loads away from supports
I
I
l
I
l,
I
l
3.5 HOLES THROUGH DEEP BEAMS 3.5.1 Characteristic stress patterns and loading at holes
34
In the region round the hole the structural behaviour and conditions are assumed to be as described for the simple rules. Typical force band diagrams are given in Figures 34 to 37 for cases where there are uniformly distributed loads, indirect loads and indirect support reactions and where l/ha :!!: 1.
These diagrams may be used for the determination of principal stresses due to these loads, but for concentrated loads the stress plots given in Appendix A should be used. Principal stresses for the different load conditions may be combined to give envelope requirements for admissibility and reinforcement size and location.
Hanger tensions 
Centre of COl I!PI e55ion
Corrc:x assion
band approximately parabolic path
Indirect support web tensions Band widths for LO assessing. hole 6 admissibility    t 
Tension band Indirect support
Approximate direction of .....;.......""'1'
'+ Mail bending steel
stresses indicated thus
.
L
(}5h"a_ _,,..~
'I
Stress pattern in a single span beam with indirect supports and uniformly distributed bottom loads
Figure 34
UD Top load
Indirect support
L
1
t
I,
L
;
i
I
, L
Indirect support
Centre of COIl IPf8SSion Actual stress trajectories closer
Compression band
to this line
Band widths for assessilg hole admissibility Approxinate
cirection of pri1cipaI stress indicated
trus
•
Figure 35
•
l
Oo4h.
..l
Mail tension steel
Stress pattern in single span beam with indirect supports and uniformly distributed top loads 3S
t
r H f      .........""f Centre of compression
..~1_
Band widths for assessing hole admissibility
I I
 Indirect support web tension
+I I
Figure 36
s!
~i#;;f
10
6
Stress pattern in single span beam with indirect supports and top loads uniformly distributed over part of span
O·4h a
Indirect support ~ndirectlY applied load ~1
Main tension steel
Approximate directions of principal stress shown • • compression    .... tension
Compression band
Approximate direction of principal stress indicated thus .. ..
Band widths for assessing hole admissibility
.s! /;

0 l~ 0
 S.... .. fi I °08
Figure 63

• 009
0
~
0·19
I
~
~
"i
..
, CHl6
..
!O
M
I
I
~•
..,
1·~
fo~
,~
~
1·66
12:05L O'5L
1 I
Unit IoadfUnit width 0'09 0' O.
I
Figure 65
MUltispan
HIL
= 112
CIL
= 1120
Uniformly distributed top load (Stresses proportional to load)
~
0..
0·18
I
Figure 66
Multispan
HIL = 112
CIL = 1120
Uniformly distributed bottom load (Stresses proportional to load)
61
UnIt loeIJ
%
015
l' f_~~=O_29... ~ I
Figure 67
MUltispan HIL = 112 CIL = 1120 One top point load at midspan (Stresses proportional to loadlspan)
+
~ I
i
c)'SL
I
Unit loeIJ /2
Figure 68
MUltispan HIL = 112 CIL = 1120 Two top point loads at 114 span (Stresses proportional to unit loadl span)
62
~ 0028
T
(}oS
~
6 I
Figure 69 ~.O_'_S7~_~l ___~~
Multispan
HIL = 213
I
CIL
= 1120
Uniformly distributed top load (Stresses proportional to load)
1·00
O'SL
Figure 70
MUltispan HIL = 213 CIL = 1120 Uniformly distributed bottom load (Stresses proportional to load)
63
Unit Load
Figure 71
MUltispan HIL = 2/3 CIL = 1/20 One top point load at midspan (Stresses proportional to loadlspan)
Figure 72
Multispan HIL = 213 CIL = 1/20 Two top point loads at 114 span (Stresses proportional to unit loadl span) 64
I
I
()'09
N
t~~~~~~~~~~~~~~
Figure 73
MUltispan
HIL
= 1.0
CIL = 1120
Uniformly distributed top load (Stresses proportional to load)
J I
Figure 74
MUltispan
HIL = 1.0
CIL
= 1120
Uniformly distributed bottom load (Stresses proportional to load) 6S
HO
1008
1005
100
1ft
~.~~~~~~~~~~~~~
Figure 75
Multispan HIL = 1.0 CIL = 1120 One top point load at midspan (Stresses proportional to loadlspan)
!~~~.~~~~~~~~~ I
+
1 ~o
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109
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Steel c:ontirued ~ from column and tied 10
~
5000
Figure 93
I I ~
I
t
~
'iCaI sleel.
~~
I
l
;.
2000
Arrangement of reinforcement  Example 2 81
Material properties
2.1.2.
Reinforced concrete: dense aggregate,/cu = 30 N/mm 2 Steel reinforcement: hot rolled deformed bars,/y = 410 N/nun2 4700
300
500
3500
300
300
§
1 ~H~
Panel
A
I
~J~;~:A~ u'
1'
~
~
Indirect load
applied each side
1'
I~ 1200
Panels.
r Figure 94
I~
ncJirect~
®
6000
500
Proposed geometry Examp/e2
Floorsiab
250>
~B~
1050
u
1
@
Direct supports
1500
]81
7000
~ 1'
!
©
~
1000
Geometry (see Figure 94)
2.2
Effective span and active height of the beam
2.2.1
Panel A lAB = 6000 nun
= 500 nun = 1500mm
CA
cB
Effective support width is lesser of 0.2/0 (= 1200 mm) or c, so
lAB
= 6000
+ 250 + 600
= 6850 mm
Total height = 3000 + 3500 + 300 = 6800 mm Since h
< I,
ha = 6800 mm
Panel B
I Bc = 7000mm Cc = 1000 mm cB = 1500mm Effective support width is lesser of 0.210 (= 1400 mm) or c, so
I Bc = 7000 + 500 + 700 = 8200 mm
Total height = 3000 + 3500 + 300 = 6800 mm Since h
< I,
ha = 6800 mm
Minimum thickness of beam
2.2.2
Contribution to total thickness
82
Assume external conditions: 2 x 40 mm cover
80mm
2.2.2. (a)
Assume four layers of 16 nun deformed bars + 15% of diameter for defonnation = 4 x 16 X 1.1 5 = 73.6
7Smm
2.2.2. (b)
Space for principal bending steel allowed for in web steel total.
2.2.2. (c)
Space for compaction 80 mm minimum, but Vbar condition may govern.
2.2.2. (d)
Assume vertical steel from supports passes through Vbar and does not require any additional room
2.2.2. (e)
Space for Vbar say 10 diameters x 16 mm = 160 mm, assuming that the Vbar is not fully stressed
160mm
Total (minimum beam thickness)
315mm
2.2.2. (f)
Hence try an initial value of 350 mm
Slenderness limits
2.2.3
Check panels Bl , B2 , B3 Worst case is panel Bl Shortest distance between centres of parallel lateral restraint = 3000 mm The panel is restrained along four edges. The effective height Ie = 3000 x 1.1 = 3300 mm 3300 Slenderness ratio   = 9.43 350 Hence the panel may be considered as a short braced wall and the criteria for elastic stability are satisfied. Since B2 and B3 have a height of 2500 mm they can also be considered as short braced walls. Check panel A Slenderness ratio = 6400 X 1.1 = 20.1 350 Hence the panel must be checked under the Supplementary Rules. Before referring to the Supplementary rules the loading will be defined.
Computation of forces
2.3
The loading on panel A at level 10.50 m will be taken as due to two levels of brick wall and three concrete slabs and will be assumed to be 120 kN/m dead load and 60 kN/m live load. At level 10.50 m, the loading on the slab will be taken as heavy plant loading over a span of 10 m, giving a 110 kN/m dead load and 75 kN/m live load. The loading on panel B will be identical to that on panel A with an additional loading of 110 kN/m dead load and 75 kN/m live load at level 13.50 m, again due to heavy plant loading. There is an indirect load of 850 kN dead and 820 kN live load.
Loading (see Figure 95)
UDL live (kN/m)
dead (kN/m)
60."0* 75.0
120.0* 110.0 57.1
Panel A
1. Slab at level 16.80 m 2. Slab at level 10.50 m 3. Self weight Total Factored loads
(x 1.6)
135.0 216.0
287.1 (x 1.4) 401.9
83
12OkN/m dead load
+
6OkN/m live load
16·80
~============~~========~======~~ 14·00 ~~L~~L~~~~L~~U ~
Figure 95 tIOkN/m dead load
+ 75kN/m liYe load
57·1kN/m seH weight
+
10·50
~
Loading arrangement Example 2 Panel B
120.0· 110.0 110.0 57.1
60.0· 75.0 75.0
4. Slab at level 16.80 m 5. Slab at level 14.00 m 6. Slab at level 10.50 m 7. Self weight Total Factored loads Indirect load Factored load
(x 1.6)
210.0 336.0
397.1 (x 1.4) 555.9 850 kNt 1190 kN
820 kNt 1312 kN
Vertical forces applied above a level of 10.00 + 6.80 x 0.75 = IS.10 m will be considered as top loads
2.3.1. (a)
Top loads:
2.3.1. (a) UDL live (kN/m)
1. Panel A, Slab at level 16.80 m Factored loads
2. Panel B, Slab at level 16.80 m Factored loads
dead (kN/m)
60.0 (x 1.6) 96.0
120.0 (x 1.4) 168.0
60.0
120.0 (x 1.4) 168.0
(x 1.6) 96.0
3. Indirect load Length of indirect load acting above 0.7 Sha = 1700 mm Length of indirect load acting below O.7Sha = 1600 mm Hence 1700/3300 = O.SIS of the indirect load acts as top load.
Indirect load Factored load
422.4 kN 437.9 kN (x 1.6) 67S.8 kN (x 1.4) 613.0 kN 2.3.1. (b)
Loads applied within the beam: UDL live (kN/m)
dead (kN/m)
75.0
110.0 57.1
7S.0 (x 1.6) 120.0
167.1 (x 1.4) 234.0
Panel A
1. Slab at level 10.50 m 2. Self weight Total Factored loads ·Considered as top loads tPartial top load
84
Panel B
3. Slab at level 14.00 m 4. Slab at level 10.50 m
75.0 75.0
110.0 110.0 57.1
5. Self weight Total Factored loads
150.0
277.1
(x 1.6) 240.0
(x 1.4) 388.0
There are no loads applied to the lower edge of the beam. The loads from the lowest slabs have already been included under Section 2.3.1. (b).
2.3.1. (c)
There is an indirect load. This will be considered as a partial bottom load.
2.3.1. (d) and 3.4.2
Proportion of load below 0.75ha = 1600/3300 = 0.485 Hence additional bottom load: live Indirect load (bottom load) Factored loads
dead
397.6 (x 1.6) 636.2
412.1 kN (x 1.4) 577.0 kN
There are no concentrated loads Bearing stresses are checked later.
2.3.2. (e)
Force and moments on the beam
2.3.2.
Since the effective heights and thicknesses are identical, the flexural rigidities for each span are equal. The moments are: Maximum hogging moment at support B = 7615 kN m (Figure 96) Maximum sagging moment in span AB = 1827 kN m (Figure 97) Maximum sagging moment in span BC = 9182 kN m (Figure 98)
Figure 96
0'25
6·0
,jJ·6JH
(820 '1·6+850 x 1'4)kN 3·45 3'55 L
0·5
1 Loads for maximum hogging moment at support B
,il.,
AB
~~
~~
B
A
C 850kN
Figure 97
Loads for maximum sagging moment in span
(397'1 • 14 + 210 • 16) kN/m
(2871 '14 +135 '16)kN/m
0'25
6'0
,0·6,0'7
(287'1 • 1·4 +135 • 16) kN/m
L
.
3'45
.!~
C
B
A
0·25
6·0
0·5
397'1kN/m .!~
~~
Figure 98
3'55
,0·6 0'7
(820 x 1,6+850 x l4)kN I, 3·45 3'55
0·5
1 Loads for maximum sagging moment in span Be
287'1kN/m
(397'1 1·4 + 210 • 1'6)kN/m ~~
L~
B
A
L~
C
Defining shear forces Toploaded system, span AB For support A, assuming span AB to be simply supported (Figure 99) For support B, assuming span AB to be continuous at B (Figure 100)
85
60 16
Figure 99
120 14 kN/m
t
Top loaded system for top shear at support A
025 R.8325
60
025
600
R. 7515kN ,..(}fl ..07
(4377 14+4224 16)kN 345 355 L
05
1
Figure 100
60 x 16kN/m
60 x16kN/m
Top loaded system for top shear at support B, left hand side
t..
060
120 x14kN/m
~~
120 x14kN/m
I
Left hand shear force 11787kN
l.i~
.0
R.30592
R.  4052kN
Bottomloaded system. span AB For support A. assuming span AB to be simply supported (Figure 101) For support B, assuming span AB to be continuous at B (Figure 102)
Figure 101
75 '16 167 1  14 kN/m
r~0~25=____~6~00=______ o~l_!
Bottom loaded system for bottom shear at support A
R. 1OO76kN
R... 11163
Figure 102
60
025
..06 07
(3976 1 16 +4121 q4)kN 355 l, 345
05
1 Bottom loaded system for bottom shear at support B, left hand side
150 1 '6 kN/m
75"'6kN/m '671 1 ,4kN/m
771'14kN/m
I
Left hand shear force  16946kN I ~~
~~
f
R.=4997
R.=4294kN
1
Toploaded system, span BC For support B, assuming span BC to be continuous at B (Figure 103) For support C, assuming span BC to be simply supported (Figure 104)
025
Figure 103
60
60 16kN/m
Top loaded system for top shear at support B, right hand side
86
120 '14kN/m
t
..06,07
t
.i~
f
R.4052kN
R.30592
(4377  14 + 4224 161 kN 355 345
1
05
1" ,r
6O16kN/m 120  14kN/m
I Right hand shear force
188Q5kN
~~
(437·9 xl·4+ 422'4 xl'6)kN 3·55 3'45 L
0·70
(}50
..
Figure 104
(120 x 1·4 + 60 x1'6) kN/m ~~
Top loaded system for shear at support C
~~
1522'4kN
1614'6kN
c
B
Bottomloaded system, span Be For support B, assume support B to be continuous (Figure lOS) For support C, span AB is treated as simply supported (Figure 106)
Figure 105
Bottom loaded system for bottom shear at support B, right hand side
0·25
..0·6 0'7
6·0
(397'6 xl·6 +412·1 x14)kN L 3'55 3'45
.
75 xl'6kN/m
150 x 1·6kN/m
167'1 Q'4kN/m
nl x1·4 kNJm
~~
I l~ I Right hand shear force  3302·5kN
R.429·4kN
R.4997
(412·l x l·4+397·6 xl'6)kN I, 3·55 3·45
0'70
1~
0'5
f i I I
t
0·50
1
Figure 106
(150 x l'6+2nl x 14)kN/m
Bottom loaded system for shear at support C
l~
~~
2880· 3kN
2728·7kN
c
B
Summary of shear forces on beam Span AB VAat
VAab VBat VBab
833 kN 1116 kN 1177 kN 1693 kN
Span BC VBat VBab Veat Veab
= 1878 kN = 3301 kN = 1615 kN = 2880 kN
87
Buckling check Check on buckling capacity for Panel A in accordance with Appendix e, using the simplified single panel method, and compare with provisions of supplementary rules. Estimate maximum vertical stress from o. . th gh th b approx. reaction           (allowing a 45 dispersIOn rou e ottom support length at panel edge slab  see Figure 10).
Actual compressive stresses: support at end A; less than
support at end B; less than
618 x 3.5 (0.5 + 0.5) 618 x 3.6
= 2160 kN/m
= 2020 kN/m
(0.6 + 0.5) Estimate horizontal stresses from elastic distribution, Figure 83 horizontal at end B; max. at bottom =
30.4 x 7615 6.8 2
~
5000 kN/m
Similarly, at mid span, conservatively assuming span is simply supported (from Figure 53) max. at top = 0.25 x 618 ~ 160 kN/m Therefore equivalent panel, which includes these maximum stresses, is illustrated on Figure 107, where N v = 2160 kN/m; average horizontal stress Nh = 2600 kN
Therefore
Q
= 5000
1 = 0.92
2600
2160kN/m
Figure 107 5000kN/m
Equivalent panel maximum stresses
216OkN/m
From Figure 124 (page 110): for
,
N vcr : tP =
:. N'vcr 88
6.0
0.94,
6.4 1.75
X
Q
= 0
tI' = 0.480 EI' 6.0 2
,"2
for
6.4
N~cr: f/J

:.N'hcr
1.97 x
0.92
1T"'lEI'
0.540EI'
6.0 2 2160
Nv
R'2(R~)
4500
N~cr
0.480 El'
H'
Nh
2600
4815
Nhcr
0.540El'
EI'
R~(Rh) R~
= 1.07, Q
6.0
0.94
R~ (neglecting shear stress·)
From Figure 130 (page Ill):
R'I
modified critical stresses are Nvcr N hcr
= 0.515
Nv
Nh
N vcr
N hcr
check
MI M2
0.515 0.485
0.485 x 0.480 EI' = 0.233 EI' x 0.540 EI'
= 0.278 EI'
9300
(an arithmetrical check to test that, as intended, EI' horizontal and vertical critical loads are in the same ratio as applied loads)
:!2:  
6510mm and
Ie
=
.j1T2 El' /Nhcr
= 6000 mm
Crosssections are now analysed as slender columns in accordance with CP 110, using he and Ie as effective lengths and taking the applied loads as those of the equivalent panel. ." 6510 For vertlcal steel he/h =   = 18.6 350 for any vertical strip 1 m wide, N = 2160 kN
• f'or simpliCity the effect of shear stresses was neglected in the above. The following calculations are given to indicate how they would be allowed for. It should be noted that where the panel is the whole span of a beam the shear stress decreases steadily from one end to the other and changes sign near midspan. This is a very different distribution from the constant shear of Figure 125 (which is more appropriate where the panel is only part of the beam). Choice of the boundary shear stress T is therefore a matter of judgment but for Panel A it would be safe to take it as the average shear stress on the end vertical sections. Then
= 618 X 3.2/6 = 330 kN/m
T
4.71r2 From Figure 125, T~r =   El' 62
R~ =
330
256
1.29 EI'
EI'
As before
R;
4500
=EI'
:. from Figure 129
1.29 EI'
17.6 ~ = 0.98
(Note that the value of Ms is not used in this analysis. whereas it forms part of the design criteria of Reference 25 from which this figure is taken). :. N:;cr = 0.98 X 0.480 EI' = 0.470 EI' '" :. R2/R,
2160/2600 =   = 0.955 0.470
From Figure 1 30
= 0.49
0.540 M,
= 0.51
M2
X 0.470EI'
0.230 EI'
Nhcr = 0.51 X 0.540EI'
0.275 EI'
:. N vcr
0.49
so that the modification here due to shear was in fact negligible.
89
additional moment =
2160 x 0.35
x 18.62
X
(1  0.0035 x 18.6)
1750
= 140 kNm = 0.05 x 2160 x 0.35
initial moment
= 38 kN m
Assuming (conservatively) that modification factor K for additional moment is 1, total moment M t = 140 + 38 = 178 kN m 10 = 6.2 N/mm 2 ; ~ = 178 x 106 1000 x 350 bh 2 1000 x 3502
= 2160
N
bh
3
X
:. with d/h = 0.85, using Figure 65 of CP 110; Pt 2,
A~
1.45 N/mm 2
=0
6000 For horizontal steel Ie /h =   = 17.1 350 Consider lowest horizontal strip 1 m wide, 0.5 N = 5000  x 4800 = 4600 kN 6.0 4600 x 0.35
additional moment
x 17.12 x (1  0.0035 x 17.1)
1750 = 253 kNm = 0.05 x 4600 x 0.35 = 80 kN m
initial moment Try Asc
= 1.25%,
d/h
:.K =
= 0.9, then Nuz = 6042 kN, Nbal = 2425 kN 6042  4600
= 0.40
6042  2425 :. total moment, Mt = 80 + 0.4 x 253
181 kN m
3
Now N/bh
4600 x 10 = 13.1 N/mm2 1000 x 350 :. from Figure 64 ofCP 110: Pt 2,M/bh 2 = 1.7 N/mm2 1.7 x 1000 x 350 2 :. allowable Mu = .:: 210 kN m. This is satisfactory. 10 6
Consider horizontal strip between levels 1 m and 2 m above soffits. N = 5000 
~
x 4800 = 3800kN
6.0
additional moment = 253 initial moment
x 3800/4600
= 0.05 x
= 209 kN m
3800 x 0.35 = 67 kN m
Assuming Asc =0.55% and carrying out analysis as above gives K =0.49, and total moment M t = 67 + 0.49 x 209 = 170 kN m, and allowable Mu = 1.5 X 350 2 /1000 = 185 kN m. This is satisfactory . For comparison, using the supplementary rules of Section 3.2.2. Effective height = 1.1 x 6000 = 6600 mm Slenderness ratio = 6600/350 = 18.9 The maximum axial stress is horizontal. Therefore each horizontal strip 1 m wide would be designed to carry its maximum axial load together with an additional moment calculated from that axial load for a slenderness ratio of 18.9.
90
Each vertical strip 1 m wide would be designed to carry its maximum axial load, together with an additional moment equal to the greatest of the additional moments calculated for the horizontal strip. In this example, reinforcement will be provided as calculated above from Appendix C, for which, summarising: Horizontal reinforcement required for buckling is for lowest 1 m wide strip 1.25% 0.55%, elsewhere nominal next lowest 1 m wide strip Vertical reinforcement required for buckling, nominal. This reinforcement is to be provided throughout the length of the panel. It need not be provided as additional to any other reinforcement where that reinforcement is required to provide a horizontal tensile resistance.
Strength: ultimate limit state
3.4
Strength in bending
3.4.1
Design of main bending steel Maximum sagging moment in AB = 1827 kN m = 7615 kN m Maximum hogging moment at B = 9182 kN m Maximum sagging moment in BC For multispan beams; midspan and support moments
z = 0.21 + O.3ha ha = 6800 mm; lab = 6850 mm; Ibe = 8200 mm lever arm in span AB = 0.2 x 6850 + 0.3 x 6800 = 3410 mm As
1827
>
X
106
0.87 x 410 x 3410
= 1502mm1
1502
==
1141 mml/m
0.2 x 6.8
At the internal support B. Use I Be to calculate lever arm. lever arm z = 0.2 x 8200 + 0.3 x 6800 = 3680 mm
>
As
7615 x 106 0.87 x 410 x 3680
= 5801 mm 2
in upper band (Figure 12) 5801(8.2/6.8  1)/2 = 597 mm 2 over depth 0.2 x 6.8, i.e. 439 mm2/m: in lower band (5801  597)/(0.6 x 6.8) = 1275 mm2 /m At midspan BC
z = 0.2 x 8200 + 0.3 x 6800
>
As
9182 x 10 6 0.87 x 410 x 3680
= 3680 mm
= 6995 mm 2
==
6995 0.2 x 6.8
5143 mm 2 /m 3.4.2
Shear capacity of span AB
Shear capacity at support A Shear capacity at an indirect support V
For feu
30 N/mm V
< 0.75bhavu 2
,
from CP 110, Vu = 4.10 N/mm2
< 0.75 x
350 x 6800 x 4.10 x 103
= 7319 kN
Maximum shear force at indirect support A Vat
+
1949
Vab = 833
<
+ 1116 = 1949 kN (Figures 99, 101)
7319/3 = 2440 kN
91
Hence shear reinforcement in the form of an orthogonal mesh is required at the support. Vertical bars in shear mesh 1949 X 106 A = s 0.87 x 410 x 0.4 x 6800
2008 mm 2 /m
Minimum area of horizontal steel in shear mesh As =
0.8 x 1949 x 106 0.87 x 410 x 0.5 x 6800
1285 mm2/m
But the minimum reinforcement in tension zone is 0.8% 2.6.2 At support A, use 2800 mm 2/m vertically and horizontally, the minimum for tension zones. 2.4.2 3.4.2
Check on hanger steel Shear capacity of a bottom loaded span AB U.D. bottom load = 354 kN/m Hence minimum area of vertical steel As
>
354
X
103
0.87 x 410
= 992 mm 2 /m
Shear capacity at support B
The limiting bottomloaded shear capacity at support B
= 0.75
Veb
x 350 x 6800 x 4.10 x 10 3
= 7319 kN
Required vertical steel to support the bottom load on span AB = 992 mm 2 /m. A minimum of 2800 mm 2 /m must be provided for crack control. This will support a UDL of 2800 x 0.87 x 410 x 103 = 999 kN/m and give a shear capacity at support B of = 999 x 6 x (3
Veb
+ 0.25)/6.85 = 2843 kN
1693
=   = 0.596 <
Hence:
2843
1.0 2.4.2
Check toploaded shear capacity of span AB at support B. 6850 ... . xe = 1/4 =   = 1712 mm because load IS umformly dlstnbuted 4 % area ofSteel
=
1502 x 100 350 x 6800
= 0.063
Hence from Table 5, CP 110, ve = 0.35 N/mm2 Shear capacity Vet
= 350
X
6800
2
x 2 x 0.35 x 0.6 x 10
3
= 3968 kN
1713 This is less than the limiting shear capacity 350 x 6800 x 4.10 x 10 3 Shear at support B due to top load
=
1177 kN
<
3968 kN
Check combined loading at support B. (Vat/Vet + Vab/Veb ) 1177
1693
3968
2843
<
1
or V/Vcb
<
1
  +   = 0.89 < 1 Hence in span AB the shear at B is satisfactory under combined loading.
92
= 9758 kN
Shear capacity 0/ span BC
There are no indirect supports in span BC Distributed bottom load = 628 kN/m 628 x 103
Area of steel required
410 x 0.87
= 1760 mml/m
But a minimum of 2800 mm 2 /m must be provided Extra steel must be provided to support the concentrated hanging load of 1213 kN 1213 x 10 3 Area of steel =             = 1232 mml/m 0.87 x 410(0.30 + 0.30 x 8.20) supplied over a distance of (0.3 + 0.3 x 8.20)mm = 2760 mm
A total of 1760 + 1232 = 2992 mml/m and is to be supplied over a distance of 1230 mm each side of the indirect load. Bottomloaded shear capacity a/supports B and C
limiting V = 0.75 x 350 x 6800 x 4.10 x 10 3 (0.75bhavu) = 7318 kN VBab =
Hence
VBab
3301 kN
>
Veab
Veab =
> 7318/3
2880 kN
= 2440 kN
therefore diagonal steel must be provided. At support B VBab =
3301 kN
Splitting shear in vertical and diagonal reinforcement in proportion 40 : 60.
Vertical bars take a force of 1320 kN and diagonal bars a force of no less than 1981 kN. Establish the shear capacity of a web with reinforcement, 2800 mml/m, vertical and horizontal, in an orthogonal mesh. Capacity of vertical steel = 2800 x 0.87 x 410 x 0.30 x 8.20 x 10 3 = 2457 kN . fh' I I 2800 x 0.87 x 410 x 0.5 x 6.80 x 10Capaclty 0 onzonta stee = 0.8
3
= 4245 kN
so limiting shear capacity of orthogonal mesh is 2457 kN Area of diagonal steel Hence area of steel
0.8 x 1981 x 103 0.87 x 410
= 4443 mm
4443 x 10 3 2263 1963 mml/m (see Figure 108)
and total bottom load capacity is: (2457 + 1981) = 4438 kN VBcb =
4438 kN
> 3301 kN
At support C Veab
= 2880 kN
Vertical bars must take a force of 1152 kN Diagonal bars must take a force of 1728 kN Shear capacity of orthogonal mesh =2457 kN (as at support B). 93
Anchor over support
Inclined band of
Figure 108 Diagonal steel at
,l
2·720
1
supportS
3
Area of diagonal steel = 0.8 x 1728 x 10 0.87 x 410
= 3876 mm 2
This is spread over a distance of 2263 mm (Figure 108) Hence area ofsteel required = 1713 mm 2 /m
VCeb
= (2457
+ 1728)
= 4185 kN
Oleck on topload span BC Topload shear capacity at supports Band C
Vetlbh a = AI V x +
(J31 V rns
+
~2Vwh
+
~3Vwv)
For deformed bars, ~1 = ~2 = ~3 = 1.0
Vet = bha(A1 Vx + Pms
Vms
+
vwh
+ vwv )
x 6771 _ 0284 . 350 x 6800
= 100
The weighted xelha (from ~ Vrxr/~ v..) :C: 0.4 so, from Table 6, Vms = 0.47 N/mm 2 Assume 0.8% vertical steel over the length of beam and 0.8% horizontal steel over shear area with 0.3% nominal steel elsewhere. From Table 8, Vwv = 0.04 N/mm 2 • For horizontal web steel shear parameter take 0.3% over full beam. From Table 7, vwh = 0.25 N/mm 2 • From Table 4, Vx = 4.71 N/mm2 •
Vet = 350 x 6800
X
(0.44
X
4.71 + 0.47 + 0.25 + 0.04)
X
= 6741 kN
Check Kong limit
Vet
< 350
X
Hence Vet = 6741 kN
94
6800
X
1.3 x 0.44 x
V30 x
103
7456 kN
103
At support B VBat = 1878
< 6741 kN
and at support C Veat = 1615
<
6741 kN
Hence for the nominal reinforcement assumed span BC is satisfactory under top loading; Check combined loading at supports Band C
V;
Bat VBct
+
V;
Bab VBcb
<
1 or V
<
Vcb
Check combined loading condition at support B 1878
3301
6741
4438
1615
2880
6741
4185
  +   = 1.02
>
1
Satisfactory if nominal vertical steel is increased from 2800 to 2986 mml/m
<
1
Satisfactory
At support C
  +   = 0.93 Check on bearing stresses at support A
3.4.3
As support A is indirect there is no need to check the bearing stress Oteck on bearing stresses at support B
Take account of the spread of stress through the floor slab (see Figures 10, 109)
x 103 /(350 x 1700) = 4.83 N/mm 2 = 5180 x 103 /(350 x 1900) = 7.79N/mml 892 x 1.5 x 103
Bearing stress due to lefthand load Bearing stress due to righthand load
= 2870
= 1.53 N/mml
Stress due to load over support
350 x 2500
Total bearing stress in overlapping zone 14.15 N/mml This is greater than O.4leu, but the additional capacity can be provided by taking the column vertical steel up into the wall to carry about 1800 kN. This is less clumsy than providing reinforcement for lateral restraint.
<
)
""
""
Figure 109
Bearing lengths at support B
""
"'
'"1
!~
/
L
1200
~
1400
1500
/
/
/
f
/
j
t
Oteck on bearing stresses at support C
Total reaction on support C
1615 + 2880 = 4495 kN
Effective length of support including dispersion through floor slab
700 + 500 = 1200 mm
9S
4495
X
10 3
    = 10.7N/mm2
Hence bearing stress due to load
1200 x 350 Stress due to load over the support
892
x
1.0
x
103
=1500 x 350
Total stress = 12.4 N/mm 2 >O.4fcu, but the additional bearing strength can be provided by extending the vertical column steel up into the wall.
Check on admissibilities of hole A
Hole A may be considered as being in the top 0.1 of the beam. Width of horizontal tension force band over support = 0.6 x 6800 = 4080 mm Depth of hole = 500 mm; 0.2 x 4080 = 816 mm, so hole is admissible.
3.5.2
Maximum moment over support = 7615 kN m Hence force = 7615/3.4 = 2240 kN Hence intensity of stress over tension zone = 2240/4.08 = 549 kN/m The vertical loading will be taken as that on top of the beam, hence the loading on the equivalent hole becomes 264 kN/m (Figure 110)
2.5.2
264kN/m
55OkN/m
r
~
r
rTTTTT
Figure 110
500
j,
Stresses at hole A
In this position the principal stresses win be vertical and horizontal. For vertical stresses, consider a deep beam loaded as shown in Figure 111.
Max. bending moment: 66 x 0.3  264 x
0.25 2
2
= 11.6 kN m
Active height of beam above hole = 300 mm For single spans, positive sagging moments, Z
96
=0.2 x 600 + 0.4 x 300 =240 mm
2.4.1
264kN/m
Figure 111
lo·os 'l
Vertical/Dad at hole A
Q 0·05 l " s·
l
0·50
Hence area of steel required above hole 11.6 x 106 As>240 x 0.87 x 410 Active height of beam below hole = 600 mm. :. z = 0.2 x 600 + 0.4 x 600 = 360 mm
Hence area of bending steel required
As
>
11.6 x 106
= 90mm
0.87 x 410 x 360
2
The upper steel should be concentrated in an area 0.2 x 300 = 60 mm above the hole, so use two bars only. 2.4.2
Oteck top and bottom beams for shear Consider beams as toploaded beams Hence shear span = 1/4 = 150 mm %steelintopbeam= 135 x 100/(300 x 300)=0.15 % steel in bottom beam = 90 x 100/(600 x 300) = 0.05 Hence ve top = 0.35 N/mm 2 Ve
Table 5, CP 110
bottom = 0.35 N/mm 2 2 x 350
X
3002 x 0.35 x In x 10 3
Shear capacity of top beam
150 147 kN
Oteck bhavu = 350 x 300 x 4.1 0 x 10 3 = 431 kN So Vet' 147 kN > shear V, 66 kN. Satisfactory. Shear capacity of bottom beam
2 x 350
X
600 2 x 0.35 x 1.0 x 10 3 150
588 kN Oteck bhavu = 350 x 600 x 4.10
X
103
861 kN
Again, Vet' 588 kN > shear V, 66 kN. Hence the shear forces at supports At and Bt are satisfactory Oteck bearing capacity of supports At and Bt. Width of support = 0.2 x 500 = 100 mm
2.4.3
Bearing stress = 66 x 103 /(350 x 100) = 1.9 N/mm2 Hence the bearing stresses are satisfactory.
97
Figure 112 55OkN/m
Horizontal load at hole A
0'50
For horizontal stresses (Figures 110, 112)
2.4.1
Bending moment = 137.5 x 0.3  550 x
0.25
2
2 = 24.1 kNm
Area of steel required each side
A
s
>
24.1 x 10 6 0.87 X 410 X 0.6 x 600
= 187mm2
This is spread over a depth of 0.6ha = 360 mm (Tension stress field) Intensity of reinforcement = 187 x 1000/360 = 521 mm 2 /m 2.4.2
Check beams for shear 2 x 350
X
600 2
X
0.35 x 1.0 x 103
= 588 kN
150 350 x 600 x 4.10 = 861 kN 137.5
< 588 kN
Hence shear check is satisfactory Total direct 'tensile' force at supports = 137.5 kN
2.4.3
~ 500
500
~
Figure 113
Reinforcement around hole A
98
~
500
Hence area of steel required in bearing area = 137.5 over 0.2 x 600 = 120 mm at bottom of beam
x 103 /0.87 x 410 = 385 mm l
Hence intensity of steel = 3208 mm 2 /m The distribution of steel round the hole is subject to the requirement of 2800 mm2 1m in tension zones. The hole is already surrounded by Y16 at 125 EF =3216 mm2 /m. This reinforcement need not be augmented to satisfy the requirements round the hole (Figure 113).
Admissibility and stresses around hole B Stresses due to indirect load (Figure 114)
Width of tension band Width of hole in tension band
= 2300 mm
= 400 mm
< 2300
x 0.2 = 460 mm
Approximating from Figures 55 and 56, average compressive stress along band
2.0
x 2502
610 kN/m at about 10° to 12° to
8.2 the vertical. 0.1 x 2502 Average tensile stress across the band =     8.2
30 kN/m
Figure 114 Force bands at hole 8 indirect load
3·45m
Stresses due to bo ttom load (Figure 115)
The hole is subject under this system to a compressive stress, due to the arch strut, and tensions due to the vertical hanging reinforcement. Approximating from Figure 54 Average compressive stress across band
=628 x
1.25 = 785 kN at about 15° to the vertical
Average vertical tension stress = 628 x 0.2 = 126 kN at about 15° to the horizontal 99
Compressive force
Figure 115
Force bands at hole B bottom load
HangIng load 628 kN/m
264kN/m
Figure 116
Force bands at hole B top load
100
Stresses due to top load (Figure 116) The hole is subject under this system to compressive stresses due to the top load 'strut' and negligible minor stress. Vertical load across band Compressive component Load in strut at hole Band width at hole Average compressive stress
= 2 x 264 = 528 kN (a 2m wide strip) = 528/Cos 20° = 562 kN = 562 kN = 1100mm = 510 kN/m at about 20° to the vertical (approximately consistent with Figure 53)
Admissibility of hole B
3.5.
The reinforcement for hole B will be designed using idealised force diagrams to estimate the stresses on the hole. The loading can be broken down into three systems: 1. indirect load (Figure 114) 2. bottom load (Figure 115) 3. top load (Figure 116) Because an approximate method is being used to estimate the stresses, it is important to ensure that all estimates are conservative. From the force diagrams, we can summarise the forces on the holes as follows (Figure 117):
+30 kN/m Reference stress planes 610kN/m (a) Indirect load
+126 kN/m
785kN/m (b) Bottom load
Figure 117 510kN/m Loads around hole 8
(c) Top load 101
These stresses may be transformed into x,y coordinates either graphically or by using a formula. We must find the worst case for any possible combination of these loading conditions. The resolved forces become x
y
xy
(a)
11
 590
110
(b)
65
723
228
(c)
60
450
164
1900 kN/m
Figure 118
0·310
Loads on longest spanhole B
295kN
295kN
These stresses combine to give a compression principal stress at about 15° to the vertical of about  1900 kN/m and a minor tensile principal stress of about 150 kN/m. Consider a loading of 1900 kN/m on the longest possible clear span which is about 2SO(Cos 15° + Sin IS°) = 310 mm Bending moment
= 295
x 0.186  1900 x
0.31
2.4.1
2
8
= 32 kN m
Area of steel
0.87 x 410 x 0.6 x 372 = 402 mm 2
This steel is spread over a depth of O.2ha giving an intensity of
402 x 103
    = 5402 mm 2 /m
372
x 0.2
Check bearing stress 3
295 x 10 350 x 60
= 13.6
N/mm2
Bearing stress> O.4!cu, but, taking account of the restrained trimming steel to hole, is acceptable 2.4.3 Check beam for shear Consider compressive forces as top loads Take xe as 93 mm Shear capacity of toploaded beam
2 x 350
X
372
2
2.4.2
x 0.35 x 1.0
X 10 3
93 = 365 kN
Check 365
295
< 350 x 372 x < 365 kN
4.10
x 103
= 534 kN
Hence beam is satisfactory for shear. An orthogonal mesh of 5402 mm 2 /m is to be provided over the zone shown in Figure 119
2.5.4
102
~ N
250
250
~
250
~
Figure 119
Zone of reinforcement hole 8
This is equal to 1350 mm 2 over the area shown and is equivalent to 7Y16 within the zone (1410 mm2 ).
This is satisfied by the vertical steel which is spaced at less than 125 mm EF. However. the horizontal steel should be supplemented by one additional Y16 EF trimmer bar. Check on tension components Maximum tensile force ~ 150 kN/m Lever arm
=0.6ha = 0.6 x
Bending moment =
150
372 = 223 mm
x 32
= 2.53 kN m
1900
2.53 x 106
Area of steel =      
0.87 x 410 x 223
. sprea d over 06h' . an mtenslty . . 0 f 32 X 10 ThiSIS . a' gIvmg
3
0.6 x 372
= 142 mm 2/m
This steel is already supplied. Direct tensile force = 0.31 Area of steel required =
x 150/2 = 23.3 kN
x 103 0.87 x 410 23.3
= 65 mm 2
Area of steel provided within 0.2ha of face of hole
= 2Y16EF = 804 mm2 103
Area of bending steel required Area of direct tension Steel
= 402 mm l = 65 mm l
467 mml
Total
An area of 804 mm l has already been provided.
Notes on detailing
Vertical steel areas (see Figure 120) The beam must have a minimum of 2 800 mmllm vertical steel everywhere to deal with the tension stresses from the hanging loads. The vertical steel must be augmented to 2986 mmllm in the zone adjacent to support B in span BC. A minimum of 2938 mmllm must be provided at the indirect load. A mesh of Yl6 at 125 EF equivalent to 3208 mmllm has been used for vertical steel everywhere.

3876,","
4443mm' total for diagonal steel
.r
total for diagonal steel
,
I,
I I I
I I Vertical steel 2IilOOmm'(m nominal steel for hanging load
I
! i
\.
,J
I
I
I~V Hanging
Iindirect load
I
I I
I
!
/1
V
1 2938 mm'(m I
Figure 120
I
I
Vertical and diagonal steel areas  Example 2
875mm'(m nominal steel from CP110
f.f,
2800mm'/m support tension zone
I I
I
Horizontal steel areas Example 2 104
VertICal steel Increased to 2986mm'/m in zone adjacent to Sl4JPOI"t B
I
i 2800m 11 m tension zone due \0 ~ moment and SUI )rt tension zones
I ! I
'I
i
I
875mm'/m nominal steel from CP110
rI
r:;
2800mm2/m
I /Zone .~~support tension
I
1
·r·L~·2800mm'/m main bending steel
L,
I I I
Figure 121
~
I
I=~~
5143mm>(m main bending steel
IAppendix C 1
I
I
L
Horizontal steel areas (see Figure 121)
A minimum of 2800 mm 2 /m is required for 1. the tension zone due to the sagging moment in span AB 2. the tension zone due to the hogging moment over support B 3. the tension zones at supports A, Band C. A minimum of 4938 mm 2 is required in the
ten"IOlI
I.llne due to sagging lif.
Diagonal steel areas (see Figure 120)
Diagonal shear steel required is ) 963 mm2 /m at support B JJld 1712 mm 2 jill over approx. 2.6 m at support C. Reinforcement detailing
The complete reinforcement for Example 2 is shown on Figure 122.
Appendix C
Buckling strength of deep beams PROCEDURE IN PRINCIPLE The procedure described below provides a means by which the adequacy of a deep beam, against failure by buckling instability, may be checked, employing the standards for the stability of reinforced concrete columns given in Clause 3.5 of CP 110. The buckling stability of the web of a deep beam, considered as a reinforced concrete wall, differs from that of a normal wall, because the web is subject to biaxial in plane stresses which interact to reduce the reserve of buckling strength. It is assumed in this analysis that the distribuiion of stresses in the deep beam is known, either from Appendix A directly, or by interpolation, and by combination, or by specific elastic analysis. It is presumed that the elastic distribution of stress is sufficiently representative of ultimate design stresses for the purposes of a valid stability check. The definition of stress throughout this Appendix is force per unit width. The beam is assumed to be divided into panels by adequate restraints. If there are lateral restraints at the top and bottom edges only, then the whole beam forms one panel. Each panel is considered individually. The key parameter, sought by the procedure, is the effective 'column' height, Ie' as indicated in CP 110, Clause 3.5. The procedure determines safe values, of the effective height and the effective length, to be used in the manner described in CP 110, Clause 3.5 for calculating the additional moments associated with the vertical and horizontal stresses, respectively. The effective height and length are derived from theoretical elastic critical buckling stresses for rectangular plates, with inplane compressive or shear loads applied along the edges. Theory is available only for uniform or linearly varying edge loads. Where this is not the case it is necessary to use equivalent loads so distributed. These equivalent loads must be selected to give conservative values of effective lengths. The relative stiffnesses in the vertical and horizontal directions of the panel are also important in the selection of a safe effective height or length for a panel. The safe value for the effective height of a panel is influenced by both the horizontal and vertical stresses. The greater the horizontal compressive stress, the less stiff the horizontal restraint and, consequently, the greater the value which should be set on the effective height of the panel. However, the greater the vertical compressive stress, the less stiff the vertical restraint, and thus the relative stiffness of the horizontal restraint is increased, so reducing the value for the effective height. In other words, it is safe to overestimate the horizontal stresses but a lowerbound estimate of the vertical stresses should be used. Similar considerations apply when
105
the effective panel length is to be predicted. In this instance, for a safe estimate, the vertical forces must not be underestimated nor the horizontal forces overestimated. It is necessary to distinguish carefully between the equivalent loading used to predict effective length or height, and the axial loads used in the final check on column stability.
The resolution of these difficulties is found in two procedures; the 'simplified single panel' and the 'twopanel' methods. For the 'simplified single panel' method, the actual panel is replaced by a notional safe equivalent panel (see Figure 123). The equivalent panel comprises a rectangular plate, with its edges either simply supported or free, subjected to an equivalent load comprising linearly varying compressive stresses applied to the edges (equal and opposite load on opposite edges), and to a constant shear stress. The width,lp, and the height, hp, of the eqUivalent panel are taken as equal to the width and height of the actual panel at the points where the actual horizontal and vertical stresses, respectively, are at a maximum. The linear equivalent applied compressive stresses are chosen to produce compressive stresses within the panel that are at no point less than the actual stresses. The applied constant shear stress is taken as equal to the mean (with due account of sign) of the actual average vertical shear stresses applied at the ends of the panel. Where the actual stresses are tensile, they should be treated as if they were zero. When the effective column height and length have been derived by the means described below, cross sections of the reinforced web are analysed for the equivalent applied loads, and for additional moments derived from the axial stresses and from the effective height or length of the safe eqUivalent panel. The method is convenient, but may be too conservative where the actual stresses vary abruptly (e.g. when concentrated loads or reactions are applied). Point of max. horizontal stress
~ Maximum horizontal stress in panel
/
r
Safe equivalent
load
////
~1
bt'';~II~TOI
J~L~
Figure 123
Equivalent panel
~Ma"m"m ~ Safe
",rt'ca' " ....
'Safe equivalent' panel
~
,aneI,,",.!.....Lt
[f
equivalent load
In the 'two·panel' method, an eqUivalent panel is selected by the method described above for the 'simplified single panel', but the equivalent loads adopted to select the effective height and the effective length, respectively, differ in two panels. The effective height is calculated in the first panel on the basis of the eqUivalent horizontal load being an upperbound value, producing horizontal stresses everywhere equal to, or greater than, the actual stresses, and the equivalent vertical load being a lowerbound value. In the second panel the effective length is calculated on the basis of the equivalent horizontal stresses being lowerbound values and the equivalent vertical stresses upperbound values. The choice oflowerbound equivalent loading is frequently a matter of judgement. The natural definition of a lower· bound eqUivalent load would be 'the linearly distributed applied edge loading which would produce stresses within the panel at no point greater than the actual stresses', but in many panels this will be inappropriate. For example, for simply supported singlepanel beams, this definition would give a lowerbound value of zero for horizontal stresses (since the horizontal stresses all vanish at the ends) and hence an unrealistic value for effective length. Some recommendations for calculating lowerbound equivalent loads are given in the detailed description of the procedure.
106
u.
wI ~ N
~ 1;
j
u.
u.
II
It)
~
N
0
.c.
"l' N
=
co
I
I
V

"
1:
15Y16175 Ubars Length to suit

r
(M+3IY16125EF 3Y16175 Ubars
~tt
,
"\Q
~l
~
~
r r
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