Reinforced concrete design to Eurocode 2 SIXT H ED ITI ON
•
BILL MOSLEY
FORMERLY NANYANG TECHNOLOGICAL UN IVERSITY, SINGAPORE AND DEPARTMENT OF CIVIL ENG INEER ING UN IVE RSITY OF LIVERPOOL
JOHN BUNGEY DEPARTMENT OF ENGIN EE RIN G UN IVERSITY OF LIV ERPOOL
RAY HULSE FORMER LY FACULTY OF ENG IN EERING AND COMPUTIN G COVENTRY UN IVERSITY
pal grave macmillan
!.. • W. H. Mosley and J. H. Bungey 1976, 1982, 1987, 1990 f ' W. H. Mosley, J. H. Bungey and R. Hulse 1999, 2007
All rights reserved. No reproduction, copy or transmission of th is publication may be made without written permission. No paragraph of this publication may be reproduced, copied or transmitted save with written permission or in accordance with the provisions of the Copyright, Designs and Patents Act 1988, or under the terms of any licence perm itting limited copying issued by the Copyright Licensing Agency, 90 Tottenham Court Road, London W1P 4LP. Any person who does any unauthorised act in relation to this publication may be liable to criminal prosecution and civil claims for damages. The authors have asserted their right to be identified as the authors of this work in accordance with the Copyright, Designs and Patents Act 1988. First published 2007 by PALGRAVE MACMILLAN Houndmills, Basingstoke, Hampshire RG21 6XS and 175 Fifth Avenue, New York, N.Y. 10010 Companies and representatives throughout the world PALGRAVE MACMillAN is the global academic Imprint of the Palgrave Macmillan division of St. Martin's Press, LLC and of Palgrave Macmillan ltd. Macmillan is a registered trademark in the United States, United Kingdom and other coun•rics. Palgrave is a registered trademark in the European Union and other countries. ISBN·13 978- 0- 230-50071- 6 ISBN-10 0- 230- 50071- 4 This book is printed on paper suitable for recycling and made from fully managed and sustained forest sources. Logging, pulping and manufacturing processes are expected to conform to the environmental regulations of the co~1ntry of origin. A catalogue record for this book is available from the British Library. A catalog record for this book Is available from the Library of Congress. Library Of Congress Catalogue Card Number - 2007023349 10 16
Properties of reinforced concrete 1.1 1.2 1.3 1.4 1.5 1.6
2
page viii
Composite action Stress-strain relations Shrinkage and thermal movement Creep Durability Specification of materials
Limit state design 2.1 2.2 2.3 2.4 2.5
Limit states Characteristic material strengths and characteristic loads Partial factors of safety Combination of actions Global factor of safety
3 Analysis of the structure at the ultimate limit state 3.1 3.2 3.3 3.4 3.5 3.6
Actions Load combinations and pattern s Analysis of beams Analysis of frames Shear wall structures resisting horizontal loads Redistribution of moments
4 Analysis of the section 4.1 4.2 4.3 4.4
Stress- strain relations Distribution of strains and stresses across a section in bending Bending and the equivalent rectangular stress block Singly reinforced rectangular section in bending at the ultimate limit stale 4.5 Rectangular section with compression reinforcement at the ultimate limit state 4.6 Flanged seclion in bending at the ultimate limit state 4.7 Moment redistribution and the design equations 4.8 Bending plus axial load at the ultimate limit state 4.9 Rectangular- parabolic stress block 4.10 Triangular stress block
Detailing requirements Span-effective depth ratios Calculation of deflection Flexural cracking Thermal and shrinkage cracking Other serviceability requirements Limitation of damage caused by accidental loads Design and detailing for seismic effects
7 Design of reinforced concrete beams 7.1 7.2
Preliminary analysis and member sizing Design for bending of a rectangular section with no moment redistribution 7.3 Design for bending of a rectangular section with moment redistribution 7.4 Flanged beams 7.5 One-span beams 7.6 Design for shear 7.7 Continuous beams 7.8 Cantilever beams and corbels 7.9 Curtailment and anchorage of reinforcing bars 7.10 Design for torsion 7.11 Servi ceability and durability requirements
Shear in slabs Span- effective depth ratios Reinforcement details Solid slabs spanning in one direction Solid slabs spanning in two directions Flal slab floors Ribbed and hollow block floors Stair slabs Yield line and strip methods
9 Column design 9.1 9.2 9.3 9..4 9.5 9.6
Loading and moments Column classification and failure modes Reinforcement details Short columns resisting moments and axial forces Non-rectangular sections Biaxial bending of short columns
Pad footings Combined footings Strap footings Strip footings Raft foundations Piled foundations Design of pile caps Retaining walls
Prestressed concrete 11.1 Principles of prestressing 11 .2 Methods of prestressing 11. 3 Analysis of concrete section under working loads 11 .4 Design for the serviceability limit state 11 .5 An alysis and design at the ultimate limit state
The design procedure Design of lhe steel beam for conditions during cons truction The composite section at the ultimate limit state Design of shear connectors Transverse reinforcement in the concrete flange Deflection checks at the serviceability limit state
The purpose of this book is to provide a straightforward introduction to the principles and methods of design for concrete structures. lL is directed primari ly at students and young engineers who require an understandi ng of the basic theory and a conci~e guide to design procedures. Although the detailed design methods arc generally according to European Standards (Curococles), much or the theory and practkc is of a l"undamental nature and should, therefore. be usel"ul Lo engineers in countries outside Europe. The search for harmonisation of Tech nical Standards across the E uropean Community (EC) hw; led LO the clevelopmeut or a seri es o r these SrrtrC/11./"(i/ Eurocodes which arc the technical documents intended for adoption throughout all the member states. The use of these common standards is intended to tower trncle barriers and enable companies to compete on a more equitable basis throughout the EC. Eurocode 2 (EC2) deals with the design of concrete structures, whit.:h has most recently been covered in the UK by British Standard BS811 0. B$8 11 0 is scheduled for withdrawal in 2008. Eurocode 2. which will consist of 4 parts. also adopts the limit state principles established in British Standards. This hook refers primarily to part I , dealing with general rules for buildings. curol"ode 2 must he used in conjunction with other European Standard:. including Eurocodc 0 (Basis of Oc!;ign) that deals with analysis and Eurocode I (Actions) that covers loadings on ~tructurcs. Other relevant Standards are Eurocode 7 (Geotechnical Design) and Eurocodc 8 (Seismic Design). .. Several UK bodies have also produced a range or supporting documents giving the requirements of the code. commentary and hackground explanation for some Further supporting documentation includes, for each separate country, the Nati01wl Annex which includes informat ion specinc to the incli vicluaJ member stotes and is l'upported in the UK by Lhe British Standards publicmion PD 66B7:2006 which provides huckground informnrion. Adcli tionn lly, the Briti sh Cement Association has produced "111<1 Concise Eurocode .fbr the DesigH of Concrere Buildinf:(S w hich cont:ains materi
or
viii
Preface and design is now based on concrete cylinder strength. wi th both of these changes incorporated in this edition. Changes in terminol ogy. arising partly from language differences. have resulted i n the introduction of a few terms that are unfamiliar to engineers who have worked with BS8110. The most obvious of these is the use of actions to describe the loading on strucwrcs and the use of the terms permanent and l'ariable actions to describe dead and imposed loads. Notwithstanding this, UK intluence in drafting the documcm has heen very strong and terminology is broadly the same as in existing British Standm·ds. Throughout this text. terminology has been kept generally in l ine with commonly accepted UK practice and hence. fo r example. loads and ac1ions are used interchangeably. Other ' new' terminology is identified at appropriate points in the text. The subject mlltter in thi s book has been arranged so that chapters I to 5 deal mostly with theory and analysis while the subsequent chapters cover the design and cletniling of various types of member ancl structure. Tn order to include topics that arc usually iJl an undergraduate course. there is a section on earth-retai ning stt·ucLurcs anti also chnpters on prestressed concrete nncl composite construction. A new section on seismic design has also been added. lmporlant equations thnt have been derived within the text are highlighted by an asterisk adjacent to t·he equation number and in the Appendix a summary of key equations is given. Where it has been necessary to indutlc material tl1a1 is not directl y provided by the Eurocodcs, this has been based on cun·cntly accepted UK good practice. Jn preparing this new edition Iwh ich replaces Reil!{orced Crmcrl'le Design to EC2 ( 1996) by the some authors]. the principal aim has been to retain the stru cture and features of the well-established book Reii!(Orced Concrcle De.1ign hy Mosley. Bungey and Hulse (Palgrave) which i~> based on British Standards. By comparing the books it is po~sible to see the essential difference~ between Eurocode 2 and existing British Standards and to contrast the different outcomes when stru ctures arc de~igned to either code. It should be emphasised that Codes of Prattice are always liable to be revised. and readers should cnst1rc that they arc usi ng the latest edition of any relevant standard. Finally, lhe authors would like to thank Mrs M ary Davison ror her hard work. patience and assistance with the prepnrntion or the manuscripl.
Acknowledgements Permission to reproduce EC2 Figures 5.2, 5.3, 6.7, 8.2. 8J, 8.7. 8.9. 9.4 nncl 9.9 and Tables A 1. 1 (EN 1990), 7 .4. 8.2 and 8.3 from BS EN 1992- 1- 1: 2004 is grnnted by l3S I, British Standards can he obtained from BS.I Customer Servi ces. 3~9 Chiswick lligh Road. London W4 4AL (tel. +44 (0)20 8996 900 1, email: cservices@ hsi-globnl.com). We would also like to acknowledge and thank ARUP for permission to reproduce the photographs shown in chapters 2 to 8, and 12. The photograph of The Tower. East Side Plaza, Portsmouth (cover unci chapter 1) is reproduced by courtesy of Stephenson RC Frame Contracror. Oakwood House. Gui ldford Road, Bucks Green, Horsham, West Sussex .
Dedicated to all ou r families for their encouragement and patience whilst writing this text
,
ix
...........................................
:.: :···
Notation
Notation is generally in accordance with EC2 and the principal symbols are listed below. Other symbols are defined in the text where necessary. The symbols c for strain and f for stress have been adopted throughout, with the general system of subscripts such that the first subscript refers to the material, c - concrete, s - steel. and the second subscript refers to the type of stress, c - compression, t - tension. t:: F
G I K
M N
Q T
v a b {/ d'
e II k II
1/ r s II
.r
Ac Ap As A~ A,.TCI)
moclu lus of elasticity loacl (action) permanent load second moment of area prestress loss factor moment or bending moment :txial load variable load torsional moment shear force deflection breadth or width effective depth of tension reinfo rcement depth to compression reinforcement ecct:ntricity overnll depth of section in plane of bending radius of gyration coeflicient length or span ultimate load per unit area curvature of a beam spacing of shear reinforcement or depth or ~ tress block th ickness punch ing shear perimeter neutral axis depth lever arm concrete cross-sectional area cross-sectional area of prestressing tendons cross-sectional area of tension reinforcement cross-sectjonal area of compression reinforcement cross-sectional area of tension reinforcement required al the ultimate limit state X
Notation As,p,.ov
A,w f ern E,
Gl lc
NhnJ ME.r Nlu Nb:d NEd
P0
Q11 Tec1 Vf',d
wk hw .f~k .f~,.,
j~'"'
·">k / yk
gl
k1 k2 / 11
/0
qk 0' Cl'e
't/J
'Yc 'Yr ')'n / 'Q
''
o
e 0'
(j;
cross-sectional area of tension reinforcement provided at the ultimate limit state cross-sectional area of shear reinforcemem in the form of links or bent-up bars secant modulus of elasticity of concrete modulus of elasticity of reinforcing or prestressing steel characteristic permanent load second moment of area of concrete moment on a column corresponding to the balanced condition design value of moment ultimate moment of resistance axial load on a colunm corresponding to the balanced condition design va lue of axial force initial prestress force characteristic variable load design value of torsional moment design value of ~hear force characteristic wind load minimum width of section characteristic cy linder strength of concrete mean cylindcr strength of concrete mean tensile strength of concrete characteristic yield strength of prestressing steel characteristic yield strength of reinforcement characterist ic permanent load per unit area average compressive stress in the concrete for a rectangular parabolic stress block a factor that relates the depth to the cenrroid of the rectangular parabolic stress block and the depth to the neutral axis lever-arm factor = z/ d effective height or column or wall chaructcristic variable load per unit area coel'licient or thermal expansion modular ratio action combinution facto r partinl sal'c ly factor fo r concrete strength parl'inl sn l'e l.y l'aclor for loads (actions), F parl'ial safety !'actor for permanent loads, G partia l safety l'ador for variable loads, Q partial safety !'actor for steel strength moment redistribution factor strain stress bar diameter
Notation for composite construction, Chapter 12 Au A"
h
Area of a structural steel section Shear area of a structural steel section Width of the steel flange
~
xi
xii
Notation bcrr d
Ea Ec.cff Ecrn
fctm
fY j~
II ha hr hp h,c
'"
l,,.:lll~r
L Me 11 Ill
PRd
R.:~
Rc~
R, R,l
'R" Rv
Rw Rwx tr fw Wpl,y X (~
t ')' I'Ed
'I
Effective width of the concrete fl ange Clear depth of steel web or diameter of the shank of the shear stud Modulus of elasticity of steel Effective modulus of elasticity of concrete Secant modulus of elasticity of concrete M ean value of the axial tensile strength of concrete Nominal value of the yield strength of the structural steel Specified ultimate tensile strength Overall depth: thickness Depth of structural steel secti on T hickness of the concrete flan ge Overal l depth of th e profiled steel sheetin g excluding embossments Overall nominal height of a shear stud connector Second moment of area ol' the structural steel section Second moment of area of the transfonnetl concrete area and the structural steel area Reduction factor for resistance of headed stud w ith profi led steel :;heeti ng parallel with the beam Reduction factor for resistance of headed stud w ith profi led steel sheeting trnnsverse to the beam
Length. span Moment of resistance of the composite secti on Modular ratio or number of shear connectors umber of shear connectors for full shear connection Design value of the shear resistance of a single connector Resistan ce of the concrete Aangc Resistance of the concrete above the neutral axis Rc.'i~tancc of the steel section Resistance of the steel flange Resistance of the steel flan ge above the neutral axis Resistance or the clear web depth Rcsistnnce or the overall web depth = Rs = 2R,r Resistnnce of the web above the neutral axis Thickness of the steel fl ange T hickness of the steel web Plastic secti on modulus of a steel stn.1ctural secti on Distance to the centroid of a section Lever arm Deflecti on at mid span 2 Constant equal to )235/.{y where.fy is in N/mm factor of safety Longitudinal shear stress in the concrete flange Degree of shear connection
1
CHAPTER .....................................
Properties of reinforced concrete CHAPTER INTRODUCTION
Reinforced concrete is a strong durable building material that can be formed into many varied shapes and sizes ranging from a simple rectangular column, to a slender curved dome or shell. Its utility and versatility are achieved by combining the best features of concrete and steel. Consider some of the widely differing properties of these two materials that are listed below. Concrete
Steel
strength in tension
poor
good
strength In compression
good
good, but slender bars will buckle
strength In shear
fair
good
durability
good
corrodes if unprotected
fire resistance
good
poor - suffers rapid loss of strength at high temperatures
It can be seen from this list that the materials are more or less complementary. Thus, when they are combined, the steel is able to provide the tensile strength and probably some of the shear strength while the concrete, strong in compression, protects the steel to give durability and fire resistance. This chapter can present only a brief introduction to the basic properties of concrete and its steel reinforcement. For a more comprehensive study, it is recommended that reference should be made to the specialised texts listed in Further Reading at the end of the book.
'1 2
Reinforced concrete design
1.1
Composite action
The tensile strength of concrete is only about 10 per cent of t11e compressive strength. Because of this, nearly all reinforced concrete structures are designed on the assumption that tile concrete does not resist any tensile forces. Reinforcement is designed to carry these tensile forces. which arc transferred by bond between the interface of the two materials. If this bond is not adequate. the reinforcing bars will just slip within the concrete and there will not be a composite action. Thus members should be detailed so that the concrete can be well compacted around the reinforcement during construction. In addition, bars are normally ri bbed so that there is an extra mechanical grip. 111 the analysis and design of the composite reinforced concrete section, it is assumed that there is a perfect bond, so that the strai n in the reinforcement is identical to the strain in the adjacent concrete. This ensures that there is whm is known as 'compatibility of strains' across the cross-section of the member. The coefficients of thermal expansion for steel and for concrete are of the order of I 0 X I 6 per "C and 7- 12 X 10- 6 per "C respecti vely. These values are sufficiently close that problems with bond seldom arise f'rom diff'ercntial expansion between the two materials over normal temperature ranges. Figure 1.1 illustrates the behaviour of' a simply supported beam subjected to bending and shows the position of steel reinl'orcement to resist the tensile force.~, while the compression forces in the top of the beam arc carried by the concrete.
o-
A
Fig ure 1. 1 Composite action
Load
Compression
Reinforcement
L~
JD Strain
Section A-A
Distri bution
A
Wherever tension occurs it is likely that cracking of the concrete wi ll l:ake place. This cracking, however, docs not detract from the safety of the structure provided there is good rei nforcement bonding to ensure that the cracks arc restrained from opening so that the embedded steel continues to be protected from corrosion. When the compressive or shearing forces exceed the strength of the concrete, then steel reinforcement must again be provided, but in thc.~e cases il is only required to supplement the load-carrying capacity of the concrete. For example. compression reinforcement is generally required in a column, where it takes the form of vertical bars spaced ncar the perimeter. To prevent rhese bars buckling, steel binders are used to assist the restraint provided by the surrounding concrete.
Properties of reinforced concrete
1.2
9
·3
Stress-strain relations
The loads on a structure cause distortion of its members with resulting stresses and strains in the concrete and the steel reinforcement. To carry out the analysis and design of a member it is necessary to have a knowledge of the relationship between these stresses and strains. Tlus knowledge is particularly impottant when dealing with reinforced concrete which is a composite material; for in this case the analysis of the su·esses on a cross-section of a member musr consider the equilibrium of the forces in the concrete and steel, and also the compatibility of the strains across the cross-section.
1.2.1 Concrete Concrete is a very variable material, having a wide range of strengths and stress- wain curves. A typical curve for concrete in compression is shown in figure 1.2. As the load is applied, the ratio between the stresses and strains is approximately l.inear Ht rirst and the concrete behaves almost as an elastic material with virtual ly full recovery of displacement if the loud is removed. Eventually, the curve is no longer linear und the concrete behaves more and more a~ a plastic material. II' the load were removed during the plasti.c rnnge the recovery would no longer be complete :mel n permanent deformation would remain. The ultimate strain for most structural concret·es tends to be a constant value of approximately 0.0035, although this is likely to rcdut:c for concretes with t:ubc strengths above nboul 60 N/mm 2 . BS EN I\>92 'Design of Concrete Structures' - commonly known as Eurocodc 2 (or EC2) recommends values for use in such cases. The precise shape of the stress-strain curve is very dependent on the length of time the load is applied. a factor which wi ll be further discussed in section 1.4 on creep. Figure 1.2 is typical for n short-term loading. Concrete generally increases its strength with age. This characteristic is illustrated by the graph in figure 1.3 which shows how the increase is rapid at first. becoming more gradual later. The precise relationship will depend upon the type of cement used. That shown is for the typical variation of nn adequately cured concrete made with commonly used class 42.5 Portland Cement. Some codes of practice allow the concrete strength
Strain
0.0035
Figure 1.2
Stress-strain CUIVe for concrete in compression
40 Figure 1.3
'E
E
£01 <:
Q)
t;
20
Q)
·~
!!! Q. E 0
u
/v V-
30 I--
z
10
v
Increase or concrete strength with age. Typical curve for a concrete made with a class 42.5 Portland cemenl wllh a 28 d
~
0 1
day
7 days
28 days
3
months
Age of concrete (log sc.a le)
year
5 years
f.l
~·
Reinforced concrete design used in design to be varied according to the age of the concrete when it supports the design load. European Codes, however, do not permit the use of strengths greater than the 28-day value in calculations, bm the modulus of elasticity may be modified to account for age as shown later.
ln the United Kingdom, compressive stress has traditionally been measured and expressed in terms of 150 mm cube crushing strength at an age of 28 days. Most other countries use 150 mm diameter cylinders which are 300 mm long. ror normal strength concretes. the cylinder strength is. on average. about 0.8 x the cube strength. All design calculations to EC2 are based on the characteristic cylinder strength K~ as defined in secti on 2.2.1 . Cube strengths may however be used for compliance purposes, with the ch..racteris tic strength identified as f~k. cube· Concretes wil l normal ly be specified in terms of these 28-day characterisllc strengths, !'or example strength class C35/45 concrete has a characteri sti c cylinder strength of 35 N/mm2 and a characteri stic cube stJength of 45 N/mm2 . It wi ll be noted tha t there is some ·round ing off' in these values. which are usual ly quoted i n multiples of 5 N /mm 2 for cube strength. Concretes made w ith l ightweight nggn;gatcs are identified by the prelix L C.
Modulus of elasticity of concrete lt is seen from the stress-strain curve for concrete that although elast'ic behaviour may be assumed for stresses below about one-third of the ultimate compressive strength, this relationship is not truly linear. Consequenlly it is ncces~ary to define precisely what value is to be taken as the modulus of elasticity. stress strain
=
A number of alternative definitions exist, lnu the most commonly adopted is E Ecm where Ecm is known as the secam or swrir modulus. This is measured for a particular concrete by means or a static test in which a cylinder is loaded to ju ~ t above one-third of tl1e correspontling mean control cube stress J~m. cube· or 0.4 mean cylinder strength. and then cycled back to zero stress. Thi!> removes the effec t or initial 'bedding-in' and minor stress redistributions in the concrete under load. The load is reapplied anclthc behaviour wi ll then be almost linear; the average slope of the line up to th e specilied stress is taken as the value for E<:m· The test is describecl in detnil in BS 188 1 and the resul t is generally known as the secan1 modulus '~/' elclSiicity. The dynamit modulus of elasticity, Ed. is sometimes ref erred to ~ i nce this is much easier to measure in the laboratory nnd t·here is u f<~ irl y wel l-cleftned relationship between /~em and E". The standard test is bnscd on determ in ing th e resonant frequency or u prism spct:imcn and is also described in BS l l:l8 l. It is also possible to obtain a good estimate of Ed !'rom ultrasonic measuring techniques, which may sometimes be used on site to assess the concrete in an actual structure. The stanclarcl test for fu is on an unstressed specimen. [[ can be seen from figure 1.4 that the value obtained represents the slope of the tangent at zero stress and Ed is therefore higher than Ecm· The relationship between the two moduli is often taken as Secant modulus Ecm
= (1.25£d
19) kN/mm 2
This equation is sufficiently accurate for normal design purposes. The actual value of E for a concrete depends on many ractor!> related to the mix, but a general relation ship is considered to exist between the modulu~ or da~ticity and the compressive strength.
Properties of reinforced concrete
~5
Figure 1.4 Moduli of elasticity of concrete
I I I .;
tangent or dynamic modulus
_ - - \ • secant or static modulus
I Strain
Typical vulues of Ecr11 for various concrete classes using gravel aggregates which are suitable for design arc ~hown in table 1.1. For limestone aggregates these vulues should be reduced by a !'actor of 0.9, or for basalt increased by a facLOr of 1.2. Thu magnitude of the modu lus of el asticity is required when investigating the de!lection and cracking or a structure. When considering short-tenn effects. member stjffness will be based on the static modulus Ecm defined above. If long-term effects are being considered, it can be shown that t·he effect of creep can be represented by modifying the value or Ec 111 to an effective value Ec,eff· and this is discussed in section 6.3.2. The elastic modulus at an age other than 28 days may be estimated from this table by using t·he anticipated strength value at that age. lf a typical value of Poisson's ratio is needed, this should be tuken as 0.2 for regions which arc not subject to tension cracking.
1.2.2
Steel
Figure 1.5 shows typical stress-strain curves for (a) hot rolled high yield steel. and (b) cold-worked high yield steel. Mi ld steel behaves as an clastic material , with the stmin proportionul 1o the stress up to the yield, ut whic:h point there is a sudden increase in strain with no change in stress. After the yield point, !his becomes a plastic material and the strain increases rapid ly up to the ultimate value. Iligh yield steel. which is most Table 1.1
Short-term modulus of elasticity of normal-weight gravel concrete
28 day characterlst:ic strength (N/mm 2) fck!fck. cube (cylinder/cube)
6 :·. Reinforced concrete design figure 1.5 Stress- strain curves for high yield reinforcing steel
0.2% proof stress
Strain
0.002
(a) Hot rolled steel
(b) Cold worked steel
commonly used for reinforcement. may behave in a similar m;mner or may, on the other hand, not have such a definite y ield point but may show a more gradual l:hange from elastil: to pla:-:tic behaviour and reduced ductility depending on the manufacturing prol:css. A ll mater.ials have a simi lar slope of the clastic region with elastic modul us Es 200 kN/mm 2 approximately. The speciricd strength used in design is based on either the y ield stress or a speci ried proof' stress. A 0.2 per cent proof stress is defined in 1'\gurc 1.5 by the broken line drawn parallel to the linear part of the stress-strain curve. Removal of the load within the plastic range would result in I he stress- strain diagram following a line approximately parallel to the loading pori ion - sec line BC in fi gure 1.6. The steel will be left wi th a permanent strain AC. which is known as ' t-.lip'. If the steel is again loaded. lhe su·ess-strai n diagram will follow the unlonding curve unti l it almost reaches the original stress at B and then it will curve i n the direction of the first loading. Thus, the proportional limit for the second loading is higher I han for the in itial loading. This action is rel'tm-ed tO as ·strain hardening' or 'work hardening'. The load deformntion of the steel is also dependent on the length of' time the load is applied. Under a constant stress the strains will gradually increase - this phenomenon is known as ·creep' or 'relaxation'. The mnount of creep that takes place over a period of ·time depends on the grade of steel and the magnitude of the stress. Creep of the steel is of little significance in normal reinforced concrete work, hut i1 is an important factor in prestres:-:ed concrete where the prestressing steel is very highl y stressed.
=
ll
A
C
Figure 1.6 Strain hardening
1.3
Shrinkage and thermal movement
As wncrete hardens there is a reduction in volu!lle. T his shrinlwge is liable to l:Huse cracking of the concrete, but it also has the benefic ial c!Tccl of strengthening the bond between the concrete and the steel reinforcement. Shrinkage begins to take place as soon as the concrete is mixed, and is cause(.] initially by the absorption of the water by lhe l:Oncrcte and the aggregate. Furrher shrinkage is caused by evaporation of the water which ri ses to the concrete smfnce. During the setting process the hy<.lrati on of the cement causes a great deal of heat ro be generated, and as the concrete cools, flllther shrinkage takes place as a result of thermal con traction. Even after the concrete has hardened, shrinkage continues as drying out persists over many months, and any subsequent wetting and drying can also cause swelling and shrinkage. Thermal shrinkage may be reduced by restricting the temperature rise during hydration, wh ich may be achieved by the following procedures:
1. Use a mix design with a low cement conterll or suitable cement replacement (e.g. Pulverised Fuel Ash or Ground Granulated Blast Furnace Slag).
Properties of reinforced concrete 2. Avoid rapid hardening and f inely ground cement if possible.
3. Keep aggregates and mixing water cool. 4. Use steel shuttering and cool with a water spray.
5. Strike the shuttering early to allow the heat of hydration to dissipate. A low water-cement ratio will help to reduce drying shrinkage by keeping to a minimum the volume of moisture that can be lost. rr the change in volume of the concrete is allowed to take place free ly and without restrnint, there will be no stress change within the concrete. Restraint of the shrinkage. on the other hand. wi ll cause tensile strains and stresses. The restraint may be caused extern ally by fixity with adjoining members or friction against an earth surface, and internnlly by the nction of the steel reinforcement. For a long wall or fl oor slab, the restraint from adjoining concrete may be reduced by constructing successive bays instead of alternate bays. T his allows the free end of every bay to contract before the next bay is cast. .Day-to-clay thermal expansion of the concrete can be greater than the movements caused by shrin kage. Thermal stresses and ~trai ns may be controlled by the correct positioni ng of movement or expansion j oints in a strut:turc. ror example, l'l1e joints should be placed at an abrupt change in cross-set:tion and they should, in generul , pass completely through the structure in one plane. When the tensi le stresses caused by shrinknge or thermal movement ext:cctl the !-.trength of' the concrete. cracking will occur. To control the crack widths. steel reinf'orcement must be provided close to the concrete surface: the codes of practice specif'y minimum quantities of reinforcement in a member for this purpose.
Calculation of stresses induced by shrinkage (a) Shrinkage restrained by the reinforcement The shrinkage stresses caused by rei nfort:ement in an otherw ise unrestrained member may he calculated quite simply. The member shown in figure 1.7 has u free shrinkage stmin of ec, if made of plain cont:rctc, but thi s overall movement is reduced by the int:lusion reinf'ort:ement, giving a compressive strnin e~c in the steel and t:ausing an ef'fectivc tensi le struin e~, the concrete.
or
Origina l member -
as cast
Plain concreteunresLrai ned
~ !"
~--~I EK _.)
l
Reinforced concrete unrestrained
'-Reinforced concrete fully retrained
Figure 1.7 Shrinkage strains
~
.::i
7
~fl Reinforced concrete design Thus ( 1.1 ) whcrc.f~ 1 is the tensile stress in com:retc area/\" ancl.f~c is the compressive !;tress in steel area !Is Equating forces in the concrete and steel for equilibrium gives
AJ~t
(1.2)
- A,f,c
therefore
A,
fct - - f -..: Ac
Substituting for j~ 1 in equation I. I e cs
I)
A~- + =.fsc. ( AcEcm Es
.. Es TllUS I 1 O:c = £em
I)
=f-c (AcE~ + E$ =h,c (aeA s+ I) cteA~
£c,
L,
1\c
Thercl'ore steel stress ( l.3)
(
EXAMPLE 1 . 1
Calculation of shrinkage stresses in concrete that is restrained by reinforcement only A member contains 1.0 per cent n:infun;cmcnt, and the Cree shrinkage strain C:c, of the concrete is 200 x 1()- 6. Por steel, Es = 200 kN/mm 2 and for concn:lc Ec111 = 15 kN/mm 2. Hence from equ:nion 1.3:
stress
,
.
111 relllforcement .f.~c
EcsC s
= I
11
T
et·eA-' c
200 X J0 6 X 200 X 1(}1 = ---..,2""0""'0;-----1,- lS X 0.01
= 35.3 N/mrn 2 compression 11,
stress in concrete ..fcr ::::A· ./~c c
= 0.01 X 35.3 = 0.35 N/nun2 tension
L~--------------------------------------~)
Properties of reinforced concrete
'
:::,:~
The stresses produced in members free from external resu·aint are generally small as in example 1.1, and can he ca~ily withstood both by the 'steel and tl1e concrete.
(b) Shrinkage fully restrained If the member is fully restrained. then the steel crumot be in compression since tl;A; = 0 and hence J,.c 0 (figure I. 7). In this case the tensile strain induced in the concrete tc1 must he equal to the free shrinkage strain £cs. and the corresponding stress will probably be high enough to cause cracking in immature concrete.
=
(EXAM PL E 1 .2
Calculation of fully restrained shrinkage stresses
rr
the member in example 1.1 were fu lly restrained. the stress in the concrete would be given by
where
ec1= ec,
= 200 X Io-
then
.f.:1 200
X
10
6X
15
X
10'
3.0N/mm 2
When cracking occurs. the uncracked lengths of concrete try to contract so thnt the embedded steel between cracks is in compression while the steel across the cracks is in tension. This feature is accompanied by localised bond breakdown, adjacent to each crnck. The equilibrium of the concrete and reinforcement is shown in figure 1.8 and calculations may be developed to relate crack widths and spacings to properties of the cross-section; this is examined in more detail in chapter 6, which deals with serviceabi lily requirements. Figure 1.8
Shrinkage forces adjacent toil crack
Thermal movement As the coefficients of thermal expansion of steel and concrete (ar. s and o:,., c) are similar. differential movement between the steel and concrete will only be very small and is unlikely to cause cracking. The dif ferential thermal strain due to a temperature change T may be calculated as
T(a-r.c- Ct,-.,) and should be added to the shrinkage strain Ecs if significant.
.~~_}~ Reinforced concrete design The overall thermal conu·action of concrete is. however, frequen tly effective in producing the first crack in a restrained member. since the required temperature changes could easily occur overnight in a newly cast member. even with good control of the heat generated during the hydration processes.
(
EXAMPLE 1.3 Thermal shrinkage
Find the faJJ in temperature required to cause cracking in a restrained member if ultimate tensile s trength of the concrete .f~t. cn - 2 N/mrn 2• 6'c111 = 16 kN/m m2 and 6 !'l:T, c = OT,, = !0 X 10- per °C. Ultimate tensile strain of concrete Cull
= I 6 X2 .103 =
125
X
10
6
Minimum lcmpcrature tlrop to cause cracking
It should be notctlthal full restraint, as assumed in this example, is unlikely to occur in practice: thus the temperature change required 10 cause cracking is increased. A maximum 'restraint factor' of 0.5 is nflen used. with lower va lue~ where external restrnint is likely to be smal l. The temperature drop required would then be given by the theoretical minimum divided by the ·reS!I'aint factor'. i.e. 12.5/0.5 = 25
1.4
Creep
Creep is the continuous clet'ormntion of a member under sustained loutl. Tt is " phenomenon associated with many materia ls, but it is particularly evident with concrete. The precise behaviour of a pnrticular concrete depends on t·he aggregates and the mix design as wel l as the ambient humidiry, member cross-section, nnd nge at (irst loading, hut the general pattern is illustrated by considering :1 member subjected to axial compression . For such a me1nber, a typical variation of deformation with rjme is shown by the curve in fi gure 1.9. The characteristics of creep are Short·IM>l t•anlc
1. The fi nal deformation of the member can he three to four times the shorHerm elastic deformation. 2. The deformation is roughly proportional to the intensity of loading and to the inverse of the concrete strength.
Figure 1.9 Typical increase of deformation with time for concrete
3. If the load is removed. only the instantaneous elastic deformation will recover- the plastic deformation will nor. 4. There is a redistribution or load between the concrete and any steel present.
Properties of reinforced concrete The redistribution of load is caused by the changes in compressive strains being transferred to the reinforcing steel. Thus th e compressive stresses i n the steel are increased so that the steel takes a larger propo1tion of the load. The effects of creep are particularly important in beams, where the increased deflections may cause tJ1e opening of cracks, damage to finishes. and the non-alignment of mechanical equipment. Redistribution of stress between concrete and steel occurs primarily in the uncracked compressive areas and has little effect on the tension reinforcement other than reducing shrinkage stresses in some instances. T he provision of reinfo rcement in the compressive zone of a nexural member. however. often helps to restrain the de flecti on~ due to creep.
1.5
Durability
Concreto Htructuros, properl y designed and constructed, 11re long lasting und should require little maintcnunt.;o. The durability of the concrete is inlluenced by
4. the cover to the rei nforcement; 5. the wid th of any cracks. Concrete can be exposed to a w ide range of conditions such as the soil. !.ea water. de-ici ng salts, stored chemicals or the atmosphere. T he severity or the exposure govern s the type of concrete mix required and the minimum cover to the reinforcing steel. Whatever the exposure, the concrete mix should be made from impervious and chemically inert a~grcgates. A dense, well-compacted concrete with a low watercement ratio is all important and for some soi l conditions it is advisable to usc a sulfateresiHti ng cement. A ir entrainment is usually specified where it is necessary to cater for repeated f'reezing and thawing. Ad equate cover is essential to prevent corrosive agents reachi ng the rei nforcement through cracks and pervious concrete. The thickness of cover required depends on the severity of' the exposure and the quality of the concrete (us shown in t.uhlc 6.2). T he cover is also necessary to protect the rein forcemenL against a rapid ri se in temperature and subsequent loss of strength during a fi re. Part 1.2 of EC2 provides guidance on t.his and other nspects of tire design. Durability requirements w ith rclu ted design culeulalions to check nnd control crack w idt hs and depths arc descr ibed in more dctuil in chapter 6.
1.6
Specification of materials
1.6.1
Concrete
T he selection of the type of concrete is frequently governed by the strength required, which in turn depends on tl 1e intensity of loading and the forn1 and size of the structural members. For example, in the lower columns of a multi-storey buildi ng a higherS!rength concrete may be chosen in preference to greatl y increasing the size of the column section with a resultant loss in clear floor space.
w.~
··<_:!2
12
~ Reinforced concrete design As indicated in section 1.2.1, the concrete strength is assessed by measuring the crushing strength of cubes or cylinders of concrete made from the mix. These are usually cured, and tested after 28 days according to standard procedures. Concrete of a given strength is identified by its 'class' - a Class 25/30 concrete has a characteristic cylinder crushing strength lfc~:) of 25 N/mm 2 and cube strength of 30 N/mm 2 . Table 1.2 shows a list of commonly used classes and also the lowest class normally appropriate for various types of construction. Exposure conditions and durability can also afTect the choice of the mix design and the class of concrete. A structure suhject to con·osive conditions in a chemical plant, for example, would require a denser and higher class of concrete than, say, the interior members of a school or office block. Although Class 42.5 Portland cement would be used in most structures, other cement types can also be used to advantage. Blast-furnace or sulfate-resisting cement may be used to resist chemical ullack, low-heat cements in massive sections to reduce the heat of hydration, or rapid-hardening cement when a high enrly strength is required. ln some circumstances it· mny be usefu l to replace some or the cement by materials such as Pulverised Fuel Ash or Grouud Granu lated Blast Furnace Sing which have slowly developing cernentitious propert'ies. These wi ll reduce the heat of hydration and may also lead to a smnlter pore structure and incn.:ased durability. Generally, naturul aggregates found locnlly <.~re preferred: however, manufactured lightweight material may be used when self-weight is importnnt. or " special dense aggregate when radiation sh ielding is required. The concrete mix may either be ci
Strength classes of concrete
Norma/lowest class for use os specified Plain concrete Reinforced concrete Prestressed concrele/Reinforced concrete subject to chlorides Reinforced concrete in foundations
~
Properti es of reinforced concrete
ch loride-induced con·osion). Detailed requirements for mix specification and compliance arc given by BS EN206 ·concrete - Performance. Production, Placing and Compliance Cliteria' and BS8500 'Concrete - Complementary British Srandard to BS E\'206'
1.6.2
Reinforcing steel
Table 1.3 lists the characteri stic design strengths of some of the more common types of 2 reinforcement currently used in the UK. Grade 500 (500N/mm characteri stic strength) has replaced Grade 250 and Grade 460 reinforci ng steel throughout Europe. The a bar is the diameter of an equivalent circular area. nominal si7.e Grade 250 bars arc hot-rolled m.ild-steeJ bars w hich usually have a srnooth surrace so tha t the bond w ith the concrete is by adhesion only. T his type or bar can be more readily bent, so they have in l he past been used where smal l radius bends arc necessary, such as l inks in narrow beams or colunUls, but plain bars arc not now recognised in the l~uropcan U nion and th ey nJe no longer avai lable for general use i n the UK. lligh-yicld bars are manufactured with a ribbed surfnce or in the form of n twisted square. Square twisted bnrs have inferior bond characteristics and have been used in the past, although they are now obsolete. Deformed bars have a mechanical bond w ith the conl:rctc, thus enlumcing ultimate bond ~ tresses ns described in section 5.2. T he bending or high-yield bars through a small radius is liable to cause tension cracking of the steel, and to avoid this t11e radius of the bend should not be less than two times the nominal bar size for small bars (~ 16 mm) or 3\12 times for larger bars (sec figure 5.11 ). The ductility of reinforcing steel is also classified for design purposes. Ribbed high yield bars may be classified as:
or
Class A - which is normally associated with small diameter ( < 12 mm) cold-worked bars used in mesh and fabric. This is the lowest ductil ity category and will include l imit~ on moment redistribution which can be applied (sec section 4.7) and higher quantities for fire resistance. Class B - which is most commonly used for reinforcing bars. Class C- high ducti lity which may be used in eart hquake design or simi lar situations. Floor slabs. walls, shells and roads may be reinforced with a welclecl fa bric of rcin forccmcnL, supplied in rolls and having a square or rectangular mesh. T his can give large economics in the detailing of the reinforcement and also in site labour costs of handling and fi xing. Prefabricated reinforcement bnr assemblies are also becoming increasingly popular for si mi lar reasons. Welded fabric mesh made of ribbed w ire greulcr than 6 mm dinmeter may be of any of the cluct.il ity cl
Table 1.3
Strength of reinforcement
Designation Hot-rolled high yield (854449) Cold-worked high yield (8$4449)
Normal sizes (mm)
Specified characteristic strength fyk (N/mm 1)
All sizes
500
Up to and including 12
500
•Note that BS4449 will be replaced by BS EN1 0080 in due course.
~;~~
q
~
Reinforced concrete design
The cross-sectional areas and perimeters of various sizes of bars. and the crosssectional area per unit width of slabs •are listed in the Appendix. Reinforcing bars in a member should either be straight or bent to standard shapes. These shapes must be fully dimensioned and listed in a schedule of the reinforcement which is used on site for the bending and fixing of the bars. Standard bar shapes and a method of scheduling are specified in BS8666. The bar types as previously described are commonly identified by the following codes: H for high yield steel, irres pective of ductility class or HA, HB, HC where a specific ducti lity class is required: this notation is generally used throug hout this book.
CHAPTER
2
······· ·· ·············•··············
Limit state design CHAPTER INTRODUCTION limit slate design of an engineerin g structure must ensure that (1) under the worst loadings the structure is safe, and (2) during normal working conditions the deformation of the members does not detract from the appearance, durability or performance of the structure. Despite the difficulty in assessing the precise loading and variations in the strength of the concrete and steel, these requirements have to be met. Three basic methods using factors of safety to achieve safe, workable structures have been developed over many years; they are 1. The permissible stress method in which ultimate strengths of the materials are divided by a factor of safety to provide design stresses which are usually within the elastic range. 2. The load factor m ethod in which the working loads are multiplied by a factor of sa fety. 3. The limit state method which multiplies the workin g loads by partial factors of safety and also divides the materials' l.lltimate strengths by further partial factors of safety. The permissible stress method has proved to be a simple and usefu l method but it does have som e serious inconsistencies and is generally no longer In use. Because it is based on an elastic stress d istribution, it is not really applicable to a semi-plastic materia l such as concrete, nor is it suitable when the deformations are not proportional to the load, as in slender columns. It has also been found to be unsafe w hen dealing with the stability of structures subject to overturning ~ forces (see example 2.2).
15
r~
16
Reinforced concrete design
~
In the load factor method the ultimate strength of the materials should be used in the calculations. As this method does not apply factors of safety to the material stresses, it cannot directly take account of the variability of the materials, and also it cannot be used to calculate the deflections or cracking at working loads. Again, this is a design method that has now been effectively superseded by modern limit state design methods. The limit state method of design, now widely adopted across Europe and many other parts of the world, overcom es many of the disadvantages of the previous two methods. It does so by applying parlial factors of safety, both to the loads and to the material strengths, and the magnitude of t he factors may be varied so t hat they may be used either with the plastic conditions in the ultimate state or with the more elastic stress range at working loads. This fl exibility is particularly important if full benefits are to be obtained from development of improved concrete and steel properties.
2.1
limit states
The purpose of design is to achieve acceptable probabilitie1> that a structure will not become unfit for its imendcd use- that is, that it will not reach a limit' state. Thus, ru1y way in wh ich a structure may cease to be fit for use will constitute a l imit stale and the design aim is to avoid any such condition being reached during the expected life of the structure. The two principal types of limit stale are the ultimate limit stntc and the serviceabi lity limit state.
(a) Ultimate limit state This requires that the structure must be able to withstnnd, with an adequnre factor of safety against coll apse. the loads for which it is dc:-;igncd to ensure the safety of the building occupants and/or the safety of' the structure itself. T he possibi li ty of buckling or overturning must also be taken imo uc<.:ount, as mu:-;t the possibility of' ucci.dental damage as c~ u sed , for example, by an intern al explosion.
(b) Serviceability limit states General ly the most important serviceabil ity limit st~tcs arc:
1. Deflection - the appearance or effi ciency of any part of the structure must not be adversely affected by detlection~ nor shou ld the com fort of the building users be adversely affected. 2. Cracking - local damage due to cracking and sp:~ ll ing must not affect the :~ppearancc. efficiency or durability of the struclUrc. 3. Durability- this must be considered in terms of the proposed life of the structure and its conditions of exposure. Other limit stares that may be reached include:
4. Excessive vibration - which may cau:-;e discomfort or alarm as well as damage. 5. Fatigue- must be considered if' cyclic loading is likely.
Limit state design
n
17
6. Fire resistance - this must be considered in terms of resistance to collapse. flame penetration and heal transfer. 7. Special ci rcumstances - any special requirements of the structure which are nol covered by any of the more common limit states, such as earthquake resistance, must be taken into account. The relative importance of each limit stale will vary according to the nature of the structure. The usual procedure is to decide which is the ciUcial limit state for a particular structure and base the design on this, although durability and fire resistance requirements may well innucnce initial member sizi ng and concrete class selection. Checks must also be made to ensure that all other relevant limit states arc satisfied by the results produced. Except in special cases, such as water-retaining structures, the ultimate limit staLe is generally critical for rei nfo rced concrete although subsequent serviceability checks may affect some of the details of the design. Prestressed concrete design, however, is generally based on serviceabi lity conditions with checks on the ult.imat:c limit stale. In assessing a particular limit state for a structure it is necessary to consider all the possible variable parameters such as· the loads, material strengths and all construct"ional tolerances.
2.2
Characteristic material strengths and characteristic loads
2.2. 1
Characteristic material strengt hs
The strengths of material:. upon which a design is based are, normally, those strengths below which results arc unlikely to fal l. These are called 'characteristic' strengths. ft i~-. assumed that for a given mmerial. the disuibution of sLrcngth will be approximately ·normal ', so that a frequency distribution curve of a large number of sample resu l t~> would be of the form shown in figure 2.1. The characteristic strength is taken as that value helow which it is unlikely that more than 5 per cent of lhe results wi ll fa ll. Th is is given by
Jk. =};,, -
1.64.1'
= characteristic strength,f;, = mean strength and s = standard deviation. The relnrionship between eharHcteristic and mean values aecounts fo r variations in results of test specimens and wi ll, therefore, reliect the method and cont.rol of rnonufncture, qual ity of const ituents, and nature of the materia l.
wherc.f~
Mean strength (f.n)
I
Figure 2.1 Normal frequency distribution
of strengths
Number of
test specimens
Strength
r Reinforced concrete design
]_!._
2.2.2
Characteristic actions
In Eurocode tenninology the set of applied forces (or loads) for which a structure is to be designed are called 'actions· although the terms ·actions· and 'loads' tend to be used interchangeably in some of the Eurocodes. 'Actions· can also have a wider meaning including the effect of imposed deformations caused by. for example. settlement of foundations. In this text we will standardise on the term 'actions· as much as possible. Ideally it should be possible to assess actions statistically in the same way that material characteristic strengths can be determined statistically, in wltich case characteristic action
= mean action ± 1.64 standard deviations
In most cases it is the maximum value or the actions on a structural member that is critical and the upper, positive value given by this expression is used: but the 10\vcr, minimum value may apply when considering stability or the behaviour or continuous members. These characteristic values represent the limits within which at least 90 per cenl of values will lie in practice. It is to be expected that not more than 5 per cent of cases will exceed the upper limit and not more than 5 per cent wi ll fa ll below the lower limit. They arc uesign values that take into account the aecmacy with which the structural loading can be predicted. Usually. however. there is insufficient stat istical data to al low actions to be treated in tl1is way. and in this case the standard loadings, such as those given in BS EN 199 1, Eurocode I - Actions on Strut.:tures, should be uscu as representing characteristic values.
2.3
Partial factors of safety
Other possible variations such as constructional tolerances are allowed for by partial factors of safely applied to the strength of the materials and to the actions. ll should theoretically be possible to derive values for these from a mathematical assessment of the probability of reaching each limit state. Lnck of adequate data. however, makes this unrealistic and, in practice, the values adopted arc based on experience anu simpl ified calculations. 2.3.1
Partial factors of safety for materials ('/'m)
. charact:cristic ~ trcn gt.h (f') Destan strenoth = . . . . ' "' "' parttal 'factor oJ· sal'ety b.,) The following factors are considered when selecting n suitable value for ~lm :
1. The strength of the material in an actual member. Th is strength wi ll differ from that measured in a carefu lly prcparcu test specimen and it is particularly true l'or concrete where placing, compaction anu curing arc so important to the strength. Steel, on the other hand, is a relatively consistent material requiring a small partial factor of safety. 2. The severity of the limit state being considered. Thus, higher values are taken for the ultimate limit state than for the serviceability limit state. Recommended vallLes for l m are given in table 2. 1 The values in the first two columns should be used when the structure is being designed for persistent design situations (anticipated normal usage) or transient design siruatiom (temporary
Limit state design
~~
Table 2. 1 Partial factors of safety applied to materials (/'m) to ed
Limit state
og
Persistent and transient Concrete
Reinforcing and Prestressing Steel
Concrete
Reinforcing and Prestressing Steel
1.50 1.50 1.50
1 .1 5 1 .15 1.15
1.20 1.20 1. 20
1 .00 1 .00 1.00
1.00
1.00
of
ie. iaJ
Accidental
Ultimate
Flexure Shear Bond
Serviceability IS
!f.
of ill
-;ituations such as may occur during construction) . The values in the last two columns ~ho uld be used when the structure is being designed for exceptional accidental design situations such as tl1e effects of tire or explosion.
e)
lg
m l.
IC
ttl
ld >i
'd
tl
II"
2.3.2 Partial factors of safety for actions (l't) Errors anti inaccuracies may be due to a number of causes:
1. design assumpt ions nnd inaccuracy of calculation: 2. possible unuf'ual increases in the magnirude of the actions; 3. unforeseen stress redistributions: 4. constructional inaccuracies. These cannot he ignore<.!, anti arc taken into account by applying a partial factor of ~afcty h f) on the ehaructcristic actions, so that design value of action = characteristic action x partial factor of safety (')r) The value or this factor ~ho u ld al ~o take into account the importuncc of the limit srate under considerat ion and reflects to some extent the accuntcy w ith which differclll rypes of actions cnn be predicted, and the probabili ty o r particular com bin ation.~ of actions occurrin g. It should be noted that design errors and constructional inaccuracies have similar effects and are thus sensibly grouped together. T hese factors will account udequutely for normal conditions alth ough gross errors in design or construction obviously cannot he catered for. Recommen<.lcd values or partial factors of safety are given in tables 2.2 and 2.3 accordin g to the di fferent categorisations of actions shown in the tnblcs. Actions arc cn tegori setl us either permanent (Od. such as the self-weight of the structure, or Pariab/e (Qk), such as the temporary imposed loading ari sing from the traffic of people, wind and snow lomli ng, anti the li ke. Va riable actions urc also cutegorised as leading (the predomi nant variable action on the structure such as an imposed crowd load - Q~ . 1) and accompanying (secondary vmiable action(s) such as the effect of wind loading. Qk, ;, where the subscript 'i' indicates the i'th action ). The rerms favourable and unfavourable refer to the effect or the action(s) on the design situation under consideration. For example, if a beam. continuous over several spans, is to be designed for the largest sagging bending moment it will have to sustain any action that has rhe effect of increasing the bending moment will be considered unfavourable wh ilst any action that reduces the bending moment will be considcrcu to be favourab le.
19
1·.
20
-
r
Reinforced concrete design
Table 2.2
Partial safety factors at the ultimate limit state
Persistent or transient design situation
Permanent actions
Leading variable action
(Gk)
(0...t)
Accompanying variable actions (Qk.i)
Unfavourable
Favourable
Unfavourable
Favourable
Unfavourable
Favourable
(a) For checking the static equilibrium of a building structure
1.10
0.90
1.50
0
1.50
0
(b) For the design of structural members (excluding geotechnical actions)
1.35
1.00
1.50
0
1.50
0
(c) As an alternative to (a) and (b) above to design for both situations with one set of calculalions
1.35
1.15
1.50
0
1.50
0
Table 2.3
Partial safety factors at t he serviceabili ty limit state
Design Situation
Permanent actions
Variable actions
All
1.0
1.0
Example 2. I shows how the panial safety factors at the ullirnate limit state from tables 2.1 and 2.2 are used to design the cross-sectional area of a steel cable supporting permanent and variable actions.
(
EXAMPLE 2.1
Simple design of a cable at the ultimate limit state Determine the cross-sectional area of steel rcquirecl for a cable which suppons a total charnctcti sti c permanent a<.:tion o f 3.0 kN and a characteristic variabl e action of 2.0 kN as shown in figure 2.2.
Figure 2.2 Cable design
Variable load - 2.0 kN (man ... eqtlipmenl)
Permanent load = 3.0 kN (platform + cable)
Limit state design
21
The characteristic yield stress of the steel is 500 N/mm 2 • Carry out the calculation using limit state design with the following factors of safety: IG = 1.35 for the pennanenl action. 1Q
= 1.5 for the variable action. and
1m
= 1.15 for the steel strength.
Design value = /G x pennanent action + 'YQ x variable action 1.35
For convenience, the p~utial factors of safety in the example arc the same as those recommended in EC2. Probably. in a practical design. higher factors of safety would he prefcrred for a single supporting cable, in view of the consequences of a failure.
___________________________________________)
~·
Example 2.2 shows the design of a foundation to resist uplift at the ultimate limit \late using the partial facto rs or safety from table 2.2. lt demonstrates the benelits or using the limit stare approach instead of the potential ly unsafe overal l factor of safety design used in part (b).
(E XAMPLE 2.2 Design of a foundation to resist uplift
Figure 2.3 shows a beam supported on foundations at A and B. The loads supported by the bC<1m arc its own uniformly distributed permanent weight of 20 kN/m and a 170 kN variable load concentrated at end C. Determine the weight of foundation required at A in order to resist uplift: (a)
hy applying a factor of safety of 2.0 to the reaction calculated for the working loads.
(b) by using an ultimate limit state approach with partial factors of safety of 'Yr. = 1.10 or 0.9 for the permanent action and I'Q = 1.5 for the variable action. Investigate the effect on these designs of a 7 per cent increase in the vatiablc action.
··~
22
Reinforced concrete design
Figure 2.3 Uplift calculation example beam
permanenlload 20 kN/m
. 1..
6m
(a)
2m
0.9 x permanent load
(b) Loading arrangement for uplifl at A at the ultimate limit state
(a) Factor of safety on uplift = 2.0 Taking moments about B Uplift
( 17()
R,, =
X
2
20 6.0
X
8 X 2) =:l :lJ kN ....
Weight of foundation required
= 3.33 x safety factor 3.33
X
=6.7 kN
2.0
With a 7 per cent incrc
·up1'lit,.
R _ ( l.07 lA -
X l 70 X 2- 20 X 8 X 2~
fi.O
= 7 .3 ''l\.1~"
1
T hus with a ~l i gh t increase in the vnri nble action there is u significant increase in the
..
upl ifl and the structure becomes unsafe.
(b) Limit state method - ultimate load pattern t\'1> this example includes a cantilever and also involves the requirement for static equi li brium a1 A, partia l factors of safety of 1.10 and 0.9 were chosen for the permanent actions as given in the lina row of values in table 2.2 The arrangement of tJ1e loads for the maximum uplift at A is shown in figure 2.3b. x 20 x 2
= I . I 0 x 20 x 2 = 44 kN
Design permanent action over AR - I<; x 20 x 6
= 0.9 x 20 x 6 = lORkN
Design permanent action over
BC
= IG = ~lo
Design v
x 170 = J .5 x 170
255 kN
Taking moments about B for the ultimate actions
.f R _ (255
UpI l t
A-
X2
+ 44 X 6.0
l - 108 X 3) _ }o kN -
0
T herefore weight of foundation required = 38 kN. A 7 per ccm increase in th e variable action will not endanger the structure, since the acl.u
Limit state design : 23 Parts (a) and (b) of example 2.2 illustrate how the l imit state method or design can ensure a safer result when the stability or strength of a structure is sensitive lo a smal l numerical difference between the effects of two opposing actions of a similar magnitude.
2.4
Combination of actions
Permanent and variable actions wi ll occur in different combina tions, all of which must be taken imo account in determi ning the most cri tical design situation for any structure. For example, the seU"-wcight of the structure may be considered in combination with the weight of furnishings and people, w ith or wi thout the effect of wind acting on the building (which may also act in more than one direction) In cal'es where actions are to be combined it is recommended that, in determining suituble design values, each characteri stic action is not only multiplied by the parti al factors of safety, as discussed above, but also by a further f act.or given th e symbol W. T his !"actor is generally taken as 1.0 other than wh ere described below:
(i)
Combination values of variable actions Where more than one variable action is to be considered (i.e a combination) then the variable actions should be multiplied by a value of IJ! (denoted as 'l'o) ns given in table 2.4. This ensures that the probability of a combination or a<.:tions being exceeded is approximately the same a<; that for a single netion. As can be seen in the table thi s is also dependent on the type of structure being designed. Combination values are used for designing for (i) the ultimate limit state and (i i) irreversible servicenbility limit states such as irreversible crucking due to temporary hut excessive overloading of the structure.
(ii)
Frequent values of variable actions Frequent combinations of actions arc used in the consideration of ( i) ultimate limit states involving accidental actions and (i i) reversible l imit states ~uch as the serviceabi lity limi t states or cracki ng and deflection where the m:tions causing these effects ure of a short transitory nature. Tn these cases the varinhle actions are multiplied by a vu lue of 1T1 (denoted as IJ!·1) as given in table 2.4. The values of 1T1 1 give an estim ati on of the proportion of the total var iable acti on that is li k<.:ly to be associrned w ith thi s parti cular combination of actions.
(i ii) Quasi-permanent values of variable action EC2 requires t.hat, in certain situations, th e c fTct:ts of ·quasi-permanent.' a<.:ti ons should be considered. Quasi-permanent (meaning 'almost.' pennnncnl) acti ons are those that may be su~ta i ned over a long period but arc not necessaril y as permanent as, say. the self-weight of the structure. A n example of such a loading wou ld be the effect of snow on the roofs of bui ldings at high altitudes w here the weight of the snow may have to be sustained over weeks or months.
)
Quasi-permanent combinations of actions are used in the consideration of (i) ultimate limil Slates involving accidemal actions and (ii) serviceability limit states attribmable to. for example, the long-term effects of creep and where the actions causing these effects. whilst variable, arc of a more long-tem1. sustained nature. In these cases the variable actions are multiplied by a value of IJ! (denoted as ili 2 ) as given in table 2.4. The values of 'T'2 give an estimation of the proportion of the total variable action that is l ikely to be associated with this particular combi nation of actions.
24
mReinforced concrete design
Table 2.4
Values of \II for different load combinations
Action
Combination
Frequent
Quasi-permanent
'1!,
IJiz
Imposed load in buildings, category (see EN 1991 -1-1) Category A: domestic, residential areas Category B: office areas Category C: congregation areas Category D: shopping areas Category E: storage areas Category F: traffic area, vehicle weigh t < 30kN Category G: traffic area, 30 kN < vehicle weight< 160 kN Category H: roofs
0.7 0.7 0.7 0.7 1.0 0.7 0. 7 0.7
0.5 0.5 0.7 0.7 0.9 0.7 0.5 0
0.3 0.3 0.6 0.6 0.8 0.6 0.3 0
Snow loads on buildings (see EN 1991 -1·3) For sites located at altitude /i > 1000 m above sea level For sites located at altitude H ~ 1000 m above sea level Wind loads on buildings (see EN 1991·1·4)
0. 7 0.5 0.5
0.5 0.2 0.2
0.2 0 0
0.7
Figure 2.4 Wind and imposed load acting on an office building - stability
X
1.5Q,
1.5Q,
HH
Ht
check
t
0.5
X
--
1.5W,
t
0.9G,
t
, 1.1 Gk
I /~
(a)
(b)
Figure 2.4 illustrales how the factors in table 2.2 and 2.4 can be applied when considering the stability of the office building shown for overturning abour point B. Figure 2.4(n) treats lhe wind load (Wk) as the lending variable action ancl lhe live load (Qk) on the roof a.o; the accompanyi ng variable action. Figure 2.4(b) considers the live load as the leading variable action and the wind as the accompanying variable action.
2.4.1
Design values of actions at the ultimate limit state
In general terms. for p('r.l'istent and transiem design situations the design value can be taken as: Design value (Ed) - (factored permanent actions) combined with (factored si ngle leading variable action) combined with (factored remaining accompanying variable actions) The 'factors' wi ll, in all cases, be the appropriate partial factor of safety hr) taken together with the appropriate value of iii as given in table 2.4.
Limit state design The design value can be expressed formalistically as: £d
\'ore that the ..1.. sign in this expression is not algebraic: it simply means ·combined with'. The L: symbol indicates the combined eiiect of all the similar action effects. e.g. 1Lj2:l/'G.jGk.j1 indicates the combined effects of all factored permanent actions, -.ummcd from the first to the 'j'th action. where there are a total of j permanent actions acting on the structure. Two other similar equations are given in EC2, the least tavourable of which can alternatively be used to give the design value. Jlowever. equation (2.1) will normally apply for most standard situations. For accide111al design silltations the design value of actions can be expressed in a -,imi Jar way with the permanent and variable actions being combined with the effect of the accidental design situation such as fire or impact. As previously indicated. such accidentnl design situations wi ll be based on the frequent or quasi-permcment values or actions with the load combinations calculated usiDg the appropriate 111 value(s) from table 2.4
2.4.2 Typical common design values of actions at the ultimate li mit state For the routine design or the members within reinforced concrete structures the sl'nndard design loading cases will oflcn consist of combinations of tJ1e permanent action with a -,ingle variable act ion and possibly with wind. If the single variable act ion is considered to he the leading variable action then wind loading will be the accompanying variable .lction. The reverse may, however. be true and both scenarios mu~t be considered. In ' uch cases the factors given in tnble 2.5 can be used to determine the design value of lhe actions. The value of 1.35 for unfavourable permanent actions is conservati ve, and used throughout this book for simplicity. Alternative equations indicated in 2.4.1 may, in some cases, give greater economy. Table 2.5
Combination of actions and load factors at the ultimate limit state
Permanent + Variable + Wind Either of these two cases may be critical - both should be considered
Variable action
1.50
Favourable
0 1.50 0
\1! 0 x 1.50 = 0.5(2) X 1 .50
1.35
1.00
IJ, 0 t3l x 1.50
0
-0.75 1.50
(1) For continuous beams with cantilevers, the partial safety factor for the favourable effect of the permanent action should be taken as 1.0 lor the span adjacent to the cantilever (see figure 7.21 ). (2) Based on the 'combination' figure in table 2.4 for wind (3) lflo to be selected from table 2.4 depending on category of building (most typical value = 0.7) (4) The partial safety factor for earth pressures may be taken as 1.30 when unfavourable and 0.0 when favourable
26
Reinforced concrete design
2.4.3
Design values of actions at the serviceability limit state
The design values of actions at the serviceability limit state can be expressed in a simi lar way to equation 2. L but taking account of the different combinations of actions to be used in the three different situations discussed above. ln the case of serviceability the partial factor of safety, ~1 r will be taken as equal ro 1.0 in all cases. (i)
Combination values of variable actions Ed =
I?:= Gk,jl +Qk. l +I?:= IJ!o.; Qk.il j>l
(2.2)
t> l
(ii) F r equent values of variable actions
Eo~ = IL Gk.jl +l
11
,> 1
t, l
Ql<.l +I?:= \]1 2. 1 Qk,il
(2.3)
t> l
(i ii) Quasi-permanent values of variable ac.tions
(2.4) Note that. as before. the ...1.. signs in these expressions arc not necessnrily algebraic: they simply mean 'combined with' . The terms in the expressions have the follow ing meanings: the combined effect of all the characteristic permanent actions where the subscript 'j' indicates that there could be between one and 'j' permanent actions on the structure the si ngle leading characteristic variable action mu ltiplied by the factor 1T1. where 111 takes the value of 1, \11 1 or 2 as appropriate from table 2.4. The subscript ·1· inclicmes that thi s relates to the single leading variable action 0 11 th e structure.
'r'
the combined cf'fect or nll the 'accompany ing' characteristi c variable actions each multipliecl by the J.a ' ct.or ~~ . where \]'1 takes the value of 111 0 or 1T1 2 as approprime l'rom table 2.4. The .subscript ' i' indicates that th ere could be up to ' i' variable actjons on t.he structure in ncldil'ion to the single leading varinble action
(
EXA MPLE 2.3
Combination of actions at the serviceability limit state A simply supported reinforced concrete beam forms part of a building within a shopping complex. It is to be designed for a characteristic pcrmanem action of 20 kN/m (its own self-weight and that of the supported structure) together with a characteristic, single leading variable action of 10 kN/m and an accompanying variable action of2 kN/m (both representing the imposed loading on the beam). Calculate each of the serviceability limit state design values as gi ven by equations (2.2) to (2.4).
Limit state design From table 2.4 the building is and 0.6.
w2 =
Combination value
Ed=
IL
GL,,, + Qu +
j~ l
IL Q~;) IL 'Vo.;
= 20 +
10 + (0.7 x 2)
= 0.7, 11! 1 = 0.7
= 31.4 kN/m
i> l
Frequent value
Ed =
classified as category D. Hence. Yio
I~ Gk.jl +\h I QL. I + J? l
qi2,i Qk,tl
= 20 + (0.7 X
10) + (0.6
X
2)
s> l
= 28.2 kN/m Quasi-permanent value
Ed =
~~ Gk,jl + ~~ W2,rQh,il = 20 + (0.6 x 10) + (0.6 x 2) = 27.2 kN/m
2.5 t Global factor of safety The use of partial faclors of safety on materials and aclions offers considerable flexibility. which may he used to allow for special conditions such as very high \tandards of construction and control or, ar tJ1e other extreme, where structural fai lure would be particularly disastrous. T he global factor of safety against a particular type of failure may be obtained by multiplying the appropriate partial factors of safety. For instance, a beam failure cau$Cd by yielding of tensile reinforcement would have a factor of ~(, 11
or
x "fr
1. 15 x 1.35 = L.55 for permanent loads only 1.15 x 1.5 = 1.72 for variable loads on ly
Thus practical cases will have a value between these two figures, depending on the 1.8 which was rclutivc load ing proportions, and this can be compared w ith the value the order of magnitude used by the load !'actor method prior to the introduction of l imit ~ late design.
or
Similarl y, failure by crushing of the concrete in the compression zone hns u f'w.:lor of
1.5 x 1.5 = 2.25 due t.o variable actions only, which reflects !he fact that such failure is generally without warning and may be very serious. Thus the basic values of partial factors chosen arc such !hal under normal circumstances the global factor of sn fety is similar to that used in earlier design methods.
27
CHAPTER
3
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 fo llowing meth ods: 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 inter~ preted 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. arc transient and not constant in magnitude, as for example those due to wind or to human occupants. Recommendations f'or the loadings on structures arc given in t·he European Swndarcls. some of which arc EN I 99 1-1- 1General actions, EN I 99 I -1-3 Snovv loads, EN 199 I - 1-4 Wind actions, EN I 99 I - I -7 Accidental actions from impact and explosions, and EN 199 I -2 Tral'lic 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 Stlllcture itself and all architectural components such as exterior cladding, partitions and ceilings. &Juipmcnt and static machinery, when permanent fixwres. arc also often considered as part of the permanent action. Once the sizes of all the structural members, and the details or the architectural requiremems and permanent lixtures have been established. the permanent actions cnn be calculated quite accurately; but, first or all , preliminary design calcul ations are generally rcquireu to estimate the probnble sizes and self-weights or the structural concrete clements. For most rcinf'orccd concretes, a typical value for the self-weight is 25 kN per cubic merre. bul a higher density should be taken for heavi ly reinforced or dense concretes. In the case of' a bu ilding, l.he weights of any permanent p
3.1.2
Variable actions
These actions arc more difficult to determine accurately. For many of 1hem. it is on ly possible 10 make conservative estimates based on standard codes of practice or past experience. Examples of variable actions on buildings arc: the weights of its occupants,
29
30
r
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 characte ri~l.ic va lues of permanent Gk, variable actions Qk, wind actions Wk, and thelr partial factors of safety must he considered for the loading of the structure. The partial factors of safety specified in the code arc discussed in chapter 2. and for the ultimate limit slate the followin g loading combinations from tables 2.2, 2.4 and 2.5 arc commonly used. 1. Permanent and variable nctions 1 .35G~
+ 1 .5Q~
2. Permanent and wind actions
The variable load can usually cover al l or any part or the structure and, therefore, should be ::IITanged to cause the most :;cvcrc stresses. So. for a three-span continuous beam, load combination I would huvc 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 stucly 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 stabi lity of a structure. A load combination of permanent + variable + wi nd load uould have the arrangements shown in figure 2.4 and described in section 2.4 of Chapter 2. Figure 3.1 Three-span beam
1.35G, + 1.50Q,
1.35Gk + 1.50Q,
t
A
l
1.3SG,
j
c
l
(a) Loading arrangement for maximum sagging moment at A and C
(b) Deflected shape
A nalysis of the st ruct ure
=i =i
1.3SC, + l.SOQ,
[
a
s
'·
1.35G, + 1.50Q,
1.35C, + 1.50Q,
F
1.35Ck + l.SOQ,
j"sc, j j'"c'l 1'%
(i) loading arrangements for maximum moments in the spans 1.3SC, + 1.500\
!'"~
1.35C,. + 1.500.
1.3SC, + l.SOQ,
I I l'"c·l
(ii) loading arrangements for maximum support moment at A 1.35G, + l .SOQ, 1.3SC,
~~
A
'
t
I
1.35Ck
t ~7
:.?~
t t t at the supports according to EC2
1 t (Iii) Load ing for design moments t
\
Note that w hen there Is a cantilever span the minim um load on the span adjacent to the can tilever should be 'I.OG, for loading pattern (i)
Figure 3.2 shows the patterns of vertical loading on a multi-span conl'inuous beam to cause (i) maximum design sngging moments in alternate spans and maximum possible hogging moments in adjacent spans. (ii) maximum clesign hogging moments at ~upport A, and (iii) the design hogging moment at support A as 11pccificd by the EC2 code for simplicity. Thus there is a similar loading pattern for the design hogging moment at each internal Mtpport of a continuous beam. It should be nmcd thai the UK :-.rational Annex permits a simpler alternative to load c.:ase (iii) where a <;ingle load ca~c may be considered of all spans loadecl with the maximum loading of
( 1.35(ik ·I 1.50Qd.
s
a
e
3.3
Analysis of beams
To design a Sltl teturc it is necessary ro know the bending moments. tor~ i onn l moments, 'hearing forces and axial forces i n each member. A n elastic analysis is generally used to determine the distribut ion of these forces within the strucmre; bu t hcc:.~ use - to some c,xtent - reinforced concrete is a plastic material, a l imited redistribution of the clasti c.: moments i ~ sometimes allowed. A plastic yield-line theory may be used tn calcu late the moments in concrete slabs. T he properti es of the materials, such as Young's modulus. which arc used i n the structural analysis should be those 11ssociatcd with thei r characteri stic: strengths. T he stiiTncsses of the members can be culcu latecl on the basis of any one the fo llowing:
or
the elllire 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 calculme and would normally be chosen.
Figure 3.2 Multi-span beam loading patterns
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-distrihution analysis anti-clockwise support moments arc positive as. for example, in table 3. 1 J'or the fixed end moments (FEM). 2. For subsequently calculating the moments along the span of a member, moments causing sagging nre 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 fo rces is readily performed manually. For the ultimate limit state we need only consider the maximum load of l.35Gk + 1.5Qk on the spun.
(
EXAMPLE 3. 1
Analysis of a non-continuous beam The one-span simply SLlpported hcam shown in rigure 3.3a carries a distributed permanent action including ~elf-weight of 25 kN/m, a permanent concentrated action of 40 kN at micl-spnn, and a distributed variable action of I 0 kN/m. Fig ure 3.3
Analysis of one-span beam (1.35 X 2S + 1.50 X 10)4 = 195 kN
,__
4.0 m
(a) Ultimate l0<1d
124.5 kN
27
27 (b) Shearing Force Dlngmm
'124.5 kN
(c) Bending Moment Diagram
Figure 3.3 shows the values ol' ultimate loacl required in the calculations of the sheru·ing forces and bending moments. Maximum shear force .
.
54
195
= 2 +T = 124.5 kN 54 x 4
Max1mum bend111g moment = -
195 x4
•
- -1 - - - = I:l l.5 kN m 4 8 The analysis is compleLccl by drawing the shearing-force and bcnding-momenL diagrams which would later be used in the design and detailing of the shear and bending reinforcement. )
l~----------------------------------------
Analysis of the structure icb
3.3.2 Continuous beams
be '5e
of
ive nts
as
·::.is the on
The methods of analysis for continuous beams may also be applied to cominuous 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 assumplion is not strictly t111e 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 cominuous beam should be anal ysed 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 curri ed out manually by moment distribution or equivalent methods. but tabulated shear and moment coefficients may be adequate fo r continuous beams having approximately equal spans anti uniformly distributed loads. For a beam or slab set monolithically into its supports, the design mornenl at th e support can be taken as the moment at the face of the support.
Continuous beams - the general case Hav ing tletermin cd the moments at the supports hy, 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 th e maximum 1-pan moments can be derived from the rollowing analysis.
Load = w/metre
Figure 3.4
Shears and moments in a beam
S.F
B.tvl
Using th e sign convemion of 11gure 3.4 and taking momems nhout suppon B:
therefore
{3.1 ) and
j
(3.2)
33
34
Reinforced concrete design
Maximum span moment
Mmax
occurs at zero shear, and distance to zero shear
V.,n
a3 = -
(3.3)
w
therefore VAB2
MmaJ<
= -2w- + MAB
(3.4)
TI1e points of conlraflexure occur at M
= 0. that is
wxl
VAB X -2+ MAo = 0
where x the distance from support A. Taking the roots of this equation gives .Y =
VAs± j(v~,~/ +2wMAo) ------'- -- -- - W
so that CIJ
=
\lAo - j(VAo 2 +2HMi\H)
(3.5)
and
(3.6) A similar analysis can be applied to beams that do not!>upport a uniformly distributed loud. In manual calculations it is usually not considered necessary to calculate the distances a 1• a2 and a3 which locate the points or eontrallexure and maximum moment a sketch of the bending moment is often adequate - hut if a computer is performing the culculations these distances muy us well be determined nlso. /\t the fnce of the support. width .v
M~u
(
MAn - ( VAo -
~~r) ~
EXA M PLE 3.2
Analysis of a continuous beam The continuous beam shown in fi gure 3.5 has fl constant cross-section and supports a uniformly distributed permanent action including its self-weight of Gk = 25 kN/m and a v;u·iable action Q~ - I0 k.N/m. The critical loading pallerns 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 noted that the reduced stiffness of ~Z has been used for the end spans.
F ...
Analysis of the structure : 35 G, = 25 kN/m Q, = 1 0 kN/m
6.0m
3.3) ( 1.35
X
6.0m
25 + 1.50 X 1 0)
1)
I
X
6 (1.35
= 292.5 kN
3.-l)
I
* ( 1.35
I
(2)
X
25
X
25
X
4,_)- -- - - - ,
I
292 .5 kN
*
I
=2o2 5 kN
(3)
X
135 kN
*
( 1.35 X 25 + 1.50 X 10)
6)
Figure 3.5 Continuous beam loading patterns
195 kN
*
*
292.5kN
195 kN
I
X4
202.5 kN
* 202.5 kN
* 202.5 kN
(4)
195 kN
I
292.5 kN
*
~,,:> ") _
Table 3. 1
Moment distribution for the first loading case A
3.6) uted the
3 I 4 'I 3 1 .,.. 4 ' 6 = 0.125
Stiffness (k)
Distr. factors
phe Load (kN)
0 Balance
1 4 = 0.25
2/ 3
- 292 X 6 8 - 219.4 + 58.1
da
Balance Carry over
+
ere
Balance Carry over
+
Balance
M (kNm)
'T
0
19.4 6.5 2.2 0.7
- 132.5
292
+135 ~ 12
+ 45.0 + 116.3 58. 1
+
be
1/ 3
2/ 3 135
Balance Carry over
Jar
= 0.125
0.25 0.'125 + 0.25
Carry over
~g.
3 I 4'I
I
292 0
D
L
0.125 0.125 + 0.25 - 1/ 3
~m
F.E.M.
c
B
+
-
38.7 19.4
X
45.0 116. 3
292 X 6 +--8 + 219.4 58. 1
19.4
38.7 1- 19.4
12.9 6.5
6.5
-
6.5
I
+
-
4.3 2.2
-
4.3 2.2
-
2.2
1
+
1.5
-
1.5
-
0.7
132.5
0
+ 58. 1
+ 12.9
+
0
- 132.5
+ 132.5
0
36
Reinforced concrete design
The sheruing forces. the maximum span bending moments. rutd their positions along the beam, can be calculated using the formulae previously delived. Thus for the first loading a1rangemem and span AB, using the sign convention of figure 3.4: Shear
VAR
load (MAR - MRA) = --..:..._----'--'2
vi3A
Maximum moment, span AB
'-
-292.5
132.5 -
= 292.5 -
124.2
?
- ---6.(}- L4. 2 = load - VAl/
2kN
= 168.3 kN
VAo2
= -2- + M,,R w
where w = 292.5/6.0 = 48.75 kN/m. Therefore: 2
Mm:1x
0 ? =?-X124.2 +0 = I5o._ kNm 48.7.5
['rom A, a,=VAI3 . D!Stance • 111 124.2
= 48.75 -
2.55 m
The bending-moment diagrams for each of the loading arrangements are shown in fi gure 3.6, and tl1e correspondi ng shearing-force diagrams arc shown in figu re 3.7. The individual bending-moment diagrams arc combined in figu re 3.8a to give the bendingmoment design envelope. Similarly. figure 3.8b is the l>hearing-forcc desig11 envelope. Such envelope diagrams arc used in the detailed de!>ign of the beams. as described in chapter 7. ln this example, simple supports with no fixity have been assumed for the end supports at A and 0. Even so. the sections at I\ and I) should be designed for a hogging momem clue to a partial fixity equal to 25 per ce11l of the maximum moment in the span, lhtll iN 158/4 Figure 3.6 Bending-moment dlagmms (kN 111)
= 39.5 kNm. 133
(1)
'133
Ld
~
"""-7
158
158
108 (2)
108
~
~ 103
~ 103
151
(3)
~
""=/ 151
1~
(4)
~
~
151
A
~
~ 151
Analysis of the structure 124 ( 1)
~
168 67.5
~
r::---... I
83
97.5
C>--.
~
c:::::::::::::J
121
<3> ~
~
119
1 1~ I
"'-J 171
85 (4) ['-:..,
119
"""""J 9 7.5
--=:::::::::::) 8 3
118~
I~
""""J 85
-==z::::J 8 s
171 ~
"""J
11 8
~
I
""'J
151 (a)
Shearing-force diagrams (kN)
.............. I '-..,J 124
~
SSt->,
11 0
""'J 121
151
Figure 3.8
~
Bending-moment and
.-::-----&'---====---->,~-----,., kN.m
\ ·-. ----/,i
11
~
" \: ;::.- - -- :
158
-:l
158
171
~
110
~
(b)
shearing-force envelopes
~
124
~
--- ~
kN
~~ 110 124 171
Continuous beams with approximately equal spans and uniform loading
or
The ultimate bending moments and shearing forces in continuous beams three or more approximately equal spans without canLilevers can be obtained using relevant coefficients provided that the spans c.liff'er 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 f'or beams (equi valent si mplified values for slabs are given in chapter R). End Span 0.11FL
_.-/1
(a) Bending Moments
~
Interior Spa11
0.10FL
./1
<:::::::::::7 0.07FL
0.55f
0.45 F
(b)
0.10FL
"'--
0.09H
Shearing Forces
r----__
r---._
-:::::::::]
---=::::::j
0.60F F = Total ultimate load on span = (1.35G 1 + 1.50~) kN L = Effective span
37
Figure 3.7
-=::::::J67 5 .
"'-J168
(2)
~
rnF.::l
0.55f
Figure 3.9 Bending-moment and shearing-force coefficients for beams
38
1
Reinforced concrete design The possibility of hogging moments in any of the spans should not be ignored, even if it is not i ndicated by these coeffici~nts. 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 equi valent gri llage. whi lst some form of finite-clement analysis can be utilised in solving complicated shear W
3.4.1
Braced frames supporting vertical loads only
A bui lding frame can be analysed as a complete frame, or il can be simplified into a series of substitute frames for the vertical loading analysis. The frame shown in figure 3. 10, for example. cw1 be divided i nto any of the sub frames shown in figure 3.11 . T he substi tu te frumc I in figu re 3. 11 consists of one complete lioor beam with its connectin g columns (which arc assumed ri gidly li xcd m th eir remote ends). A n analysis of thi s frame wi ll gi ve th e bending moments and shearin g forces in the beams and colu mns for Lhe lloor level consiclcrccl. Substitute frame 2 is a single span combined with its connectin g columns and two adjacent spans, all li xed at their remote ends. T his frame may be Ul'ed to determine the bending moments and shearing forces in the centrul beam. Provided that the central span is greater than the two adjacent spans, the bendi ng moments in the columns can also be found wi th this frame. Substitute f rame 3 can be used to fi nd the moments in the columns only . It consists of a si ngle j unction, with the remote ends of the members fixed. This type of subframe would be used when bean'ts have been analysed as continuous over simple supports. I n 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 maxjmum 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 wh ich may have a beneficial minimum loading condition (I.OGk) -sec figure 7.2 1.
Analysis of the structure 0
if
aay
Half stiffness
Half stiffness
l a<;
of (2)
ge. ear
~
my Ha lf sli[fncss
Half sliffn ess
lefS
r or gid
(3)
cal '/
.are
'lltS
the
1-1 1, 11 2=Sto rey Heig hts
1,7,?7;, 1,7,»; 1,7, »? '7i ~ Figure 3.10
Figure 3.11 Substitu te rram es
Building rrame
Ill!>. ~ if
., a
in I I. its ;oiS
lmd •\'0
When considering the critical loading arrangements for a column. it is sometimes necessary to include the case of maximum moment and minimum possible ax ial load, in order to investigate the possibility of tension fa ilure caused by the bending.
( EXA MPL E 3.3
Analysis of a substitute frame The substitute rrnme shown in figure 3.12 is part of the complete frame in fi gure 3.10. The characteristic actions carried by the beams are permanent actions (including selfweight) G~ = 25 kN/m, and variable action, Qk = 10 kN/m, uniformly distributed along the beam . The analysis of the sub frame will be carried out by moment distributiou: thus the member stiffnesses and their relevant distribution factors are fi rs! required.
' 0.58 ( 0 D. i'.cols = . = ).2 2 83 The critical loading pa!lcrns for the ultimate limit state are identical to those for lhe continuous beam in example 3.2, and they arc 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 ca.lculations.
D'
~·
ia
Table 3.2
~-
Moment distribution tor the first loading case A
D.F.s Load kN
Cols. (L;M)
AB
0.39
0.61
BA
Cols. {LM)
BC
0.32
0.20
0.48
292
Bal.
M (kN m)
0.32
-
44.6
24.2
X
DC
Cols. (l:M)
0.61
0.39
292
-
.,.
48.5
+
-
146 32.3
146
+
+
24.2
44.6
X
I
I
89.1
56.9
16.2
-
-
+
+
+
-
-
-
+
+
6.3
9.9
22.0
13.8
33.0
33.0
13.8
22.0
9.9
6.3
+
+
-
16.5
16.5
5.0 6.9
11 .0
+
-
11.0
5.0
-
-
+
+
-
-
-
4.3
6.7
6.9
4.3
10.3
10.3
4.3
+
3.4
+
+
5.2
5.2 4.1
-
3.4
3.4
c.o. Bal.
0.20
20.2
+ 20.2
c.o. Bal.
0.48
-
+ 32.3
X
CD
48.5
-
89.1
+
Cols. (LM)
45.0
-
16.2
CB
45.0
+ 146
56.9
c.o.
D
135
+ 146
F.E.M. Bal.
c
8
3.4
-
-
+
-
-
1.3
2.1
2.8
1.7
4.1
i"
.;-
-
135.0
40.0
95.0
95.0
-
+
68.8
68.8 ---
I
+
+
6.7
43
-
-
I
+
1.7
2.8
2.1
1.3
-
+
-
-
40.0
135.0
68.8
68.8 )> ::l til
'-< ;:;;· "'
0
:roT ,...,. "' 2 n
,...,.
c ..,
ro
~::;;;;;t
I~
lq
42 . Reinforced concrete design The shearing forces and the maximum span moments can be calculated from the f01mulae of section 3.3.2. For the' first loading arrangement and span AB: Shear VAs
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 wiU 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 arc smaller in magnitude. The moment in each column is given by ~
krol
Mcnl - LMcol X
"k
L- col
135
Figure 3.13 Beam bending-moment diagrams (kNm)
69
-v ~
( I)
J\{955
135
~~
99551
69
/1~1\
\7
A
118
118
147
(3)
79 114
(4)
~
147
102 80
6
~
A---;-&
67
~
\.1
79
"'---./
A
~ :;·:;
Analysis of the structure ·~ 43 Figure 3.14 Beam shearing-force diagrams (kN)
be (1)
135 97.5~
91 (2)
111
~
I"'-
[">. 133
(3) 92
n: lfl
(4)
133 147
147
Figure 3.15 Bending-moment and shearing-force envelopes
kNm
118
118
:2 Mcol table 3.2 gives
Thus. for the first loading arrangement and raking Column moment MAJ - 68.8 x
0.31 _ 0 58
= 37 kN m
0.27
= 32 kN
MAt:. = 68.8 X _ 0 58
0.31
m
MoK
= 40 x -0.58 = 21 kN
m
MoF
0.27 9 = 40 x - = I kN 0.58
m
This loading arrangement gives the maximum column moments, as plotted in figure 3.16.
44
~~
WReinforced concrete design
Figure 3.16 Column bending moments (kN m)
(
EXAMPLE 3.4
Analysis of a substitute frame for a column n1e substitute frame for this example, shown in fi gure 3.17, is taken from the building frame in figure 3. 10. The loading to cause maximum column moments is shown in the fi gure for Gk = 251cN/m and Qk = IOkN/rn. ~/,
Figure 3.17
___-..-
Substitute frame
E
~ l. lSG, <150Q. ~.35Gk~ ~
~A
=292.5 kN
f
~
=135 kN
B
~
~
C
~""
'"
E
<:!I
'"I
-
'
~ ,~ 6.0
Ill
.....
4.0 ll1
...
1
The stiffnesses of these members are identical to those calculated in example 3.3, except that for lhis type of l'rame the beam stillnesses arc halved. Thus kAu
= 2I X 0.9 X 10 J = 0.45 X )()'"3
kBc
= 'I
upper column ku
2
X
1.35
X ) ()- ) =
= 0.3 1 x 10
().675
X
J()- 3
3
= 0.27 x 10 3 I> = (0.45 -1- o.675 -1- 0.3 1 + 0.27) x w-3 = 1.705 x 6 fixed-end moment MsA = 292.5 x ? = .L46 kN m )_ lower column kL
fixed-end moment MHc 18 16 kN m
Figure 3.18 Column moments
10- 3
= 135 x 1~ = 45 kN m
Column moments are 45 ) x -0.31- = l 8. kN m 1.705 0.27 ( 146 - 45) x - - 16 kN m 1.75 0
upper column Mu
= ( 146. -
lower column 1\ll~_
=
=
The column moments are illustrated in figure 3.1 !\. They should be compm·ed with the corresponding moments for the internal column in figure 3.1 6. )
l ---------------------------------------~
Analysis of the structure ln examples 3.3 and 3.4 the second moment of area of the beam was calculated as
bfl 3/ 12 a rectangular section for simplicity, but where ah in situ slab forms a flange to the beam, the second moment of area may be calculated for the T-scction or L-section.
3.4.2 Lateral loads on frames
_)
mg the
Lateral loads on a structure may be caused by wind pressures, by retained catth or by ;;eismic forces. A horizontal force should also be applied at each level of a structure resulting from u notional incl ination oft11e vertical members representing imperfections. The value of this depends on building height and number of columns (EC2 clause 5.2). hut w ill 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 A n unbraced frame subjected to wind forcel:i must be analysed for all the vertical loading combinations described in section 3.2. 1. T he vertical-load ing analysis cun be carried out by the methods described previously . T he anal ysi s fo r the lateral loads ,hould be kept separate. The forces may be calculated by an elastic computer analysis or by a si mplified approx imate method. For preliminary design calculntions. and also only for med iu m-size regular structures, a simplified analysis may well be adequate. A suitable approximate anal ysis is the cantilever method. It assumes that:
1. points of con tra f lexure are located at· the mid-points of all colu mns 11ncl 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 al so usual to assume that ull the column~ in a storey arc of equal cross-sectional area.
It should he emphasised that these approximate methods may give quite inaccurate results for irregular or high-rise structures. Application of thi s method is probably best Illustrated by an example. as follows. 3.
( EXA MPLE 3.5
Simplified analysis for lateral loads - cantilever method Figure 3. 19 shows a building frame subjected to a charucteri stic wind ncti on of 3.0 kN per metre height of the f rame. This action is assumed to be tnmsfcrrecl to the frame as a concentrated loud at each floor level as indicated in the fi gure. By inspection, there is tension in the two columns w the left and compression in the columns to the ri ght; and by assumption 2 the axial forces jn columns arc proporti onal to their distances rrom the centre line of the frame.
--
5.25 kN
E .....
z
.><
0
.,.; II "0
"' !a ~ ;;;
.,
10 .5!!:!._
4th
10.5~
3rd
11 .2~
2nd
12 .OkN
1st
-"
he
_)
6.0 kN ~
~~
6.0
7l.r
"',.;
--I... j
...C!
--r ' C! ....
'?;~
4.0
I
~
--r '
'l/:(//
6.0
'
Figure 3.19 Frame with lateral load
f1 45
46
~
Reinforced concrete design
Figure 3.20 Subframes at the roof and 4th floor
I f,= 0.54
F1= O.HkN
F,= 0.675
t .------.:-----r---+.. J--,--.....,.... J t
5.25
-
tJ
J!
---, s' -
H,= 1.70
H;= 0.93
N,=4.0P = 0 .54
N 1= 1.0P
It;= 0 .93 N = l.OP 3
= 0.135
=0.135
(a) Roor
0.54
0.135 1.70
0.93
t
r:s
0.135
1.70
0.54
0.93
l'
5.1
2.70
5.1
0.68
0.68
7.78 2.70
(b) 4 th Floor
Thus Axial force in exterior column : axial force in interior column
= 4.0P : L OP
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 subl'rame arc calculated as follows.
(a) Axial forces in the columns
s, 2: M, - 0. therefore 6.0 - P X 10.0 - 4P X 16.0 ()
Taking moments about point
5.25
X
1.75
+p X
and therefore
P - 0. 135 kN thus
N1
= - N4 = 4.0P = 0.54 kN
N2 = - N3 =: I .OP =: 0. J 35 kN (b) Vertical shearing forces F in the beams For each part of the sub f rame,
L F = 0, therefore
F 1 = N 1 = 0.54kN
F2 = Nr
+ N2 =
0.675 kN
(c) Horizontal shearing forces H in the columns Taking moments about the poi nts of comraflcxurc of each heam,
Hr X 1.75 - Nr x 3.0 H 1 = 0.93kN
=0
L: M = 0, therefore
Analysis of the structure :md (HI
H1
+ H2) 1.75 -
N1 X 8.0- N2 X 2.0
=0
= 1.70kN
The calculations of the equivalent forces for the fourth floor (figure 3.20b) follow a '1milar procedure. as follows. d) Axial forces in the columns
I: M, = 0, therefore 5.25(3 X 1.75) + 10.5 X 1.75 + P X 6.0 - P X 10.0- 4P X P = 0.675 kN
For the frame above section tr',
16.0 = 0
therefore
= 4.0P = 2.70
kN N2 = l.OP = 0.68 kN N1
(e) Beam shears F,
= 2.70 - 0.54 = 2.16kN 2.70 I 0.68 - 0.54 - 0.135
F2
= 2.705 kN (f) Column shears H1 X
l-11
1.75 t 0.93 X 1.75
(2.70 - 0.54)3.0
=0
= 2.78 k.N
1 H2 = 2( 10.5 + 5.25)- 2.7!\
= 5.1kN Values calculated for sections taken below the remaini ng floors are third noor N1 = 7.03 kN F1 = 4.33 kN ll1 - 4.64kN second floor N1 = 14. 14 kN /··1 7. 11 kN
N2 = 1.76kN F2 = 5.41kN H2 = 8.49kN Nz = 3.53 kN
= 6.6 1kN
H2 = 12. 14kN
first floor N 1 = 24.37 kN F, = l0.23kN HI = 8.74kN
Fz = l2.79kN H2 = 16.01 kN
Ht
F2
= 8.88kN
N2 = 6.09k.N
The bending moments in the beams and columns m their connections can be calculated from these results by the following formulae beams
Mu - F x! beam span
columns Me
= H x i storey height
;:;* ~::~
P 47
48
Reinforced concrete design 1.6
Figure 3.21
1.4
1.6
1 .6
3.0
1.6
4.9
8.9
4.9
8.1
14.9
8.1
13.2
24.3
13.2
17.5
32.0
Moments (kN m) and reactions (kN)
17.5 24.4
6. 1
6.1
24.4
External Column
Internal
Beams
Column
so that the roofs external connection Mu
= 0.54 X 2I X 60 = 1.6kN m
Me
= 0.93 X ;;1 X 3.:1- = 1.6 kN 111
As a check at each joint. L: MR = 2:: Me. ThL: bL:nding moments due to characteristic wind londs in all the columns ancl beams of this structure are shown in fi gure 3.2 1.
3.5
Shear wall structures resisting horizontal loads
t\ reinforced concrete structure with shear walls is shown in fi gure 3.22. Shear walls are very effective in resisting horizontal loads such as P,. in the figu re which act in the direction of the plane of the walls. As the walls arc relatively thin they of!'er litl'le resistance to loads which arc perpendicular to their plane. The Aoor slabs which arc supported by the walls also act as rigid diaphragms which transfer and distribute the horizontal forces into the shear wa lls. The shear walls act us vertical cantilevers transferring the horizontal loads to the structural rounclations.
3.5.1
Symmetrical arrangement of walls
With a symmetrical an·angemcnt of walls as shown in figure 3.23 the horizontal load is distributed in proportion to the the relati ve stiffness k1 of each wall. The relative
Analysis of the structure
~
i~:~ 49
Figure 3.22 Shear wall structure
-tiffncsscs arc given by the second moment of area of each wall about its major axis -uch that
k, ~ h X b3 here h is the thickness of the wall and b is the length of the wall. The force P; distributed into each wal l is then given by P,
)
f x
i,kk·
( EXA MPLE 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 JOOkN horizontal load carried by each of the walls.
Figure 3.23
lre
Symmetrical arrangement of shear walls
:he
lle
ch
E
0 N
as
E
"'
1-
10m
_._
15m
'
50
~ Reinforced concrete design Relative stiffnesses: Walls A Walls B
kA
= 0.3 X 2()3 = 2400
kR = 0.2 X R3
= 346
I: k = 2(2400 + 346) = 5492 Force in each wall : 2400
kA
P.... = I;k
X
F = 5492
ks
P8 = - x F
l:k
X
100 = 43.7kN
346 =- x 100 = 6 .•3 kN 5492
Check 2(43.7, 6.3) = 100 kN = 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 wal l stiffnesses k about convenient axes. Such that
where k~ and k,. arc the stiiTncsses or the walls orientated in the x andy directions respectively. 2. Calculate rJ1e torsional moment M, on the group of shear walls as M,
=F x e
where e is the eccentricity of the horizontal force F ahout the centre of rotation. 3. Calcu late the force P, in each wall as the sum of the direct component Pc1 and tl1e torsional rotation component· Pr
P; = Pc~
+ Pr
kx k;r1 = F X I;k., ± M1 X I;(k;r; 2 )
where r; is the perpendicular distance between the axis of each wall and the centre of rotation.
(
EXAM PLE 3.7
Unsymmetrical layout of shear walls Determine the distribution of the I00 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 tenus of multiples of k.
Analysis of the structure
yt.
l1= 12.0m
1
Figure 3.24
20m
I
Centre of 1 rotation
A: ZOk
•' -
Unsymmetrical arrangement of shear walls
I
~ I'
'
E o\
'0
....
--- - ~-
E:5k
0: 5k
I
rf
C: 6k
J.'= 6.4m 32m
~Centre of
20m
20m
F = lOOkN
rotation
Taking moments
_
am I
= 20 + 5 + 5 = 30
L:;k,
L:;(kxx)
X =~=
for k.~
2()
X
about YY at wall A
0 +5 X 32 + 5 X 40 30
= 12.0 metres L_k,.= 6 H
10
Taking moments for k, about XX at wall C
l:(k,.y)
6 X 0 - 4 X 16
.)'---·-= l:k,. 10 = 6.4 metres The torsional moment /v/1 is !VI, = F X (20 - x)
100
X
{20 - 12)
= 800 kNm The remainder of these calculations are conveniently set out in wbular form: Wall
kx
A B
D E
20 0 0 5 5
6 0 0
E
30
10
c
ky
12 9.6 6.4 20 28
0 4
kr
J
pd
P,.
PI
240 38.4 38.4 100 140
2880 369 246 2000 3920
66.6 0 0 16.7 16.7
- 20.4 - 3.3 3.3 8.5 11.9
46.2 - 3.3 3.3 25.2 28.6
9415
100
As an example for wall A: PA
-
P,
+ Pr = F X
rkA k
M,
kArA
X
L (k;r; 2)
20 20 X 12 = 100 x 30 - soo x = 66.6 94 15
[f~
20.4
= 46.2 kN
0
100
!1';;
52
~~
Reinforced concrete design
Figure 3.25
Shear wall with openings
D
D D (a) Shear Wall
3.5.3
(b) Idealised Plane Figure
Shear walls with openings
Shear walls with openings can be idealised into equivalent plane frames as shown in figure 3.25. I n the plane frame the second moment or area lc of the columns is equivalent to that of the wall on either side of the Clpenings. T he second moment of area lb of the beams is equi valent to that part of the wall between the openings. The lengths of beam that extend beyond the openings as shown shaded in flgu re 3.25 are given a very large stiffnesses so that their second moment of area would be say
100/b. The equi valent plane frame would be analysed by computer with a plane f rame program.
3.5.4
Shear walls combined with structural frames
For simpl icity in the design or low or medium-height structures shear walls or a l ift shaft are usually considered to resist all of the horizontal load. With higher rise :-~ tructu res for reasons of stiffness and economy it often becomes necessaty to inc lude the combined action of' the shear walls and the structural frames in the design. A method of a n a l y~ing a structure with shear walls and structural frames as one equivalent linked-plane frame is illustrated by the example in figure 3.26. I n the actual structure shown in plan there arc fo ur frames of type A and two frames of type B which include shear walls. ln the linked frame shown in elevation the four type A frames arc lumped together into one frame whose member stiiTnesses arc multiplied by four. Similarly the two type B frames are lumped together into one frame whose member sti ffnesses arc doubled. T hese two equivalent frames arc then l inked together by beams pinned
1~ !:·~
Analysis of the struc ture .'.:: 53 A
B
A
B
A
A
Figure 3.26 Idealised link frame lor a structure with shear walls and structural frames
..
shear walls
t
l ateral Load
a) Plan of Structure p ins
s1 - I- s 2 - \..._b_-J
j.., ~-I-
s ~ ...1• sl ....1 very stiff beams
/
in is rca
..
shear walls
.25
say
me
lift
'?.~ 'J. ~
Ide
7?/?7-
4 N o, frames
::?,~::?,~
A~ 1
-
--
-
ll!e
leS
-
L ~nks
"?.~ 2 N o, frames B
7?/~
I
of large cross-sectional:rea pinned at their ends
(b) Elcva tfon or lin k-Frame Model
ur
!l'e
me .ed of
.he
whereas l:his would normally be of a secondary magni!llclc. To overcome th is the cross'>CCtional areas of nil the beams in the model may be increased say to 1000 m2 and this will virtually remove !he effects of axial shortenjng in the beams. In the computer output the member forces for type A frames would need to be divided by a factor or four and those for type 13 fra mes by a factor of two.
be :.0
3.6
Redistribution of moments
i!S.
be ID
~.
Some method of elastic analysis is generally used to calculate the forces in u concrete srmcturc, despite the fact that the strucmre docs not behave elastically near it!> ultimate load. The assumption of elastic behaviour is reasonably true for low stress level!>; but as a section approaches its ultimate moment of resistance, plastic deformation will occur.
54
Reinforced concrete design
Figure 3.27
Typical moment- curvature diagram
c
"' E ::E 0
Curvature
This is recognised in EC2, by allowing redistribution of the clastic moments subject to ccnain limitations. Reinforced concrete behaves in a manner midway between that or 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 exnct behaviour of a reinforced concrete section depends on the relative quantities nnd t·he iJ1diviclunl properties of the two materials. However, such a section mny be considered vi rtually elastic unti l the steel yields; and then plastic until the concrete fni ls in compres~ ion. Thus the plastic behaviour is limited by the concrete failure; or more specitically, the C()ncrcte fai lure limits the rotation that may tnke place at a section in bending. A typical momentcurvature diagram for a reinforced concrete member is shown in figure 3.27 Thus, in an indeterminate structure, once a beam section develops its ultimate momelll of resistance, M 0 , it then behaves as a plaMic hinge resisting a constant moment of that value. further loading must he taken by other parts of the structure. with the changes in moment elsewhere being just the same as if a real hinge eJtisred. Provided rotation of a hinge docs not cause crushing of the concrete. further hinges will be fonned until a mechanism is pro<.luced. This requirement is considered in more detai l in chapter 4.
(
EXAMPLE 3.8
Moment redistribution -single span fixed-end beam
The beam shown in figure 3.28 is subjected to an increasing uniformly distributed load: 2
Elaslic support moment Elastic span moment
wL =12
wL2
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 w9 , which the member can sustain by plastic behaviour, can be found. At collapse wL2 12
IWu= -
wL2 + additional mid-span moment IIIH 24 where mR = (waL2 )/ 8 as for a simply supported beam with hinges at A and C. = -
1 Analysis of the structure w/unit length load
Elastic BMO MA=Mc = Mu
Additional moments diagram (Hinges at A and C) Collapse mechan ism :: [0
Elastic BMD (Collapse loads} Final Collapse BMD
?le.
~,:,tic
1\1
=3
.-
'' the load to cause IJ1c first pl astic hinge; thu s the beam may cnrry a load of ''ith redistribution.
fn,m the design point of view. the elastic bending-moment diagmm can be obtained • r.:quircd ultimate loading in the ordinary way. Some of theNe moments may then ·e1ruccd; hut thi s will necessitate increasing others to maintain the static equilibrium 'tructurc. Usually it is the maximum support moments which arc reduced. so -w..ing in reinforcing steel and also reducing congestion at the columns. The emcnts for applying moment redistribution arc: l:qullihrium between internal and external forces must be mnintaincd, hence it is necessary to n.:calculate the span bending moments and the shear forces for the load ~••'e involved.
2. The cor11inuous beams or slabs are predom inately suhj t~c t to fkxure. 3. The rmio of adjacent sptms be in Lhe range of 0.5 to 2.
lid:
~. The column design moments must not be reduced.
T here arc other restriction s on the :.~mount of moment. redistribution in order to ensure uucti lity of the beams or slabs. T his entails limjtations on the grade of rei nforcing steel ..ncl or th e areas of tensile reinforcement and hence rhe depth or the n eutr:~l axis as described in Chapter Four -'Annlysis of the Section'.
ne he
( EX AMPLE 3.9 Moment redistribution In exnmplc 3.3. ligurc 3. 13, it is required to reduce the maximum upport moment of 147 kN m as much as possible. but without increasing Lhe span moment nbove the present max imum value of ll8 kN m. MRI\
=
Figure 3.28 Moment redistribution, one-span beam
55
56
Reinforced concrete design 147
Figure 3.29 Moments and shears after redistribution
114
(a) Original Moments (kN m) 140
WA1o8
67
~
80 102
4 LG 2 ~ / -;= ~
\
"'---../ 118
79
(b) Redistributed Moments (kN m) 1 34
92 (c) Shears (kN)
Figure 3.29a duplicates the origi nal bending-moment diagram (pan 3 of figure 3.13) of example 3.3 while figure 3.29b shows the redistributed moments, with the span moment set at l18kN m. The moment at support B can be calculated. using a rcan·angement of equations 3.4 and 3.1. Thus
V,,B
J[(A'ln~.x - MAu)2wj
and
Mp,,, ~ (v.. .o- ;L)L I MAo 1
= 48.75 kN m, therefore V,., 11 / [(J 18+67) X 2 X 48.75 j = l 34kN 48.75 X 6.()) 6.0 - 67 = 140 kN m MBA= ( 134 2
For span AB,
w
and lloA
= 292.5 -
134
= 158.5 kN Reduction in MoA = 147 - 140
= 7kNm 7 X 100
= - 147 -- = 4.8 per cent
~
Analysis of the structure In order to ensure that the moments in the columns at joint B arc not changed by the 'tribution, moment M 8 c must also be reduced by 7·kN m. Therefore
:'!C
'=~•
= 115 - 7 = 108 kN m hogging
Fnr the revised moments in BC:
l BC
= (l 08 -
4
1 co
F
80) + 195
=:
I05 kN
2
= 195 - 105 = 90 kN
r ~pan
BC: 105 2 x . - 108 2 48 75
.
= 5 kN m saggmg
Figure 3.29c shows the revised shearing-force diagram to accord with the moments. This example illustrates how, with redistribution
-~..!Nrihuted
the moments al a section of beam can be reduced without exceeding the maximum c.;,ign moments at other sections; " values of the column moments are not arrcclcd; and 3
3.13) pan ng a
the equilibrium between external loads and internal forces is maimaincd.
A satisfactory and econom1c design of a concrete structure rarely depends on a complex theoretical analysis It is achieved more by deciding on a practical overall layout of the structure, careful attention to detail and sound constructional practice. Nevertheless the total des1gn of a structure does depend on the analys1s and design of the individual member sections. Wherever possible the analysis should be kept simple, yet it should be based on the observed and tested behaviour of reinforced concrete members. The manipulation and juggling with equations should never be allowed to obscure the fundamental principles that unite the analysis. The three most important principles are
1. The stresses and strains are related by the material properties, Including the stressstrain curves for concrete and steel. 2. The distribution of strains must be compatible with the d1storted shape of the crosssection. 3. The resultant forces developed by the sect1on must balance the applied loads for static equilibrium. These principles are true Irrespective of how the stresses and stra1ns are distributed, or how the member is loaded, or whatever the shape of the cross-section This chapter describes and analyses the action of a member section under load. It derives the basic equations used in design and also those equations required for
--.
58
Analysis of the section
___,. the preparation of design charts. Emphasis has been placed mostly on the analysis a sociated with the ultimate limit state but the behaviour of the section withm the elastic range and the serviceability limit state has also been cons1dered Section 4.7 deals with the redistribution of the moments from an elastic analysis of the structure, and the effect it has on the equat1ons denved and the tiesign procedure. It should be noted that EC2 does not g1ve any explicit equations for the analysis or design of sections. The equations given in this chapter are developed from the principles of EC2 in a form comparable with the (Quations formerly given in BS 8110 .
.1
Stress-strain relations
n-term ~trcs~-strain curvc1-. for concrete and steel are presented in I:C2. These e~ arc in un idealised form wh1ch can be used in the anCIIysh. of memhcr :-.ecuons. .. 1. 1
Concrete
bl:ha"iour of '>tructural concrete (figure 4.1) i~ reprel>ented by a paruholic '>Ire~' m rclauon~hip. up to n strain -,·~ · from which point the 1-.tr:un increa-;e, '' hile the "' rcmam~ con'>tant. The ultimate design :.tress is gl\·cn hy 0.85f.l 1.5
- 0.567l~l rc the factor ol 0.1!5 allow~ for the dillcrence bet\\ecn the bcnd111g \trength and the 1 fer cru1-.h ing stn.:ngth of the concrete. and;'< = 1.5 ;, the w.uul partial \lllety factor 'le strength of concrete. The ultimate strain of feu~ = 0.0035 ~~typical for cla~'c' of CS0/60. Concrete cla:.sc:. < C50/60 will, un less otherwise stated. be
Figure 4.1 Parabolic-recta ngular stress -strain diagram for C011crete m compression
~
E
-
.z
Porabolic
0.851,.
e
,, 0.0020
0.0035 Strain
59
60
Reinforced concrete design
cono;idcred throughout this book a~ these are the classes mo~t commonly used in reinforced concrete construction. Also for concrete clas~cs higher than CS0/60 the delining propertie such a-, the ultimate \train . ,111 vary for each of the hjgher classes. Oel>ign equations for the higher cla\),CS of concrete can in general be obtained using similar procedures to those shO\\n in the tC\t with the relative properties and coefficients obtained from the Eurocodes.
4.1.2
Reinforcing steel
The representative !.hort-term design !>tre~s strain curve for reinforcement is given in tigure 4.2. The behaviour of the steel i~ identical in ten,ion and comprc!>sion, being linear in the clastic range up to the de1-ign yteld ~tress of f>kh, where jyk b the characteristic yield ~>tress nnd ), is the partial tnctQr of safely. Figure 4.2 Short-term design wess-stmin CUIVe for relnforcEC>ment
= 0.00217 It !.houlu be noted that EC:! permits lhc u~c of an ulternutive design strc!>s-strain curve to that shown in figure 4.2 with an inclined top branch and the maximum strain limited to a value which is dependent on the clth:-. of reinforcement. However the more commonly used curve shown in figure 4.2 \\ill be U!.Cd 111 th1s chapter and throughout the text.
4.2
Distribution of strains and stresses across a section in bending
The theol) of bending for reinforced concrete assumes that the concrete will crack in the regions of tcnllilc <;train~ and that. after cracking, all the tension is carried by the
Analysis of the section in the
-
as.~
if- P- JFs- 0.8x
E« ' - -
,_•~ ._,_;_d~
•
c
Figure 4.3 Section with strain diagram and stress blocks
n':utral axts
,.._
A,
• • mtoon
Strains
(a)
(b)
(c)
triangular
rectangular paraboloc
cquovalent rectangular
Stress blocks
ment. It i-; also ns~umed thar plane ~ections of n ~tructural member remuin atter ~trnining, so thnt ncros~ the section thl.!re must be a linear distribution of rc l3 .~ hows the cross-section of a member subjected w bending, and the t struin diagram. together with three different types ol stress distribution in the triangular ~tress tlistribution applies when the stre<.sc~> 01rc very nearly ' ntonalto the strain~. which generally (lt;curs ntthc loading h:\cl!<. cncour11crcd e \Hirl,ing condltiOn\ and is, therefore. used ut the scrvicenhJiitv limit ~tate. ed.tngular parabolic \trcs« hlocli. represents the dJstrihution at failure when the 1 essi\c 'trams are\\ ithin the plastic range. and it "associated '~ith the deS~gn I rt ultunatc limit state. Cimpht1ed altcrnall\e to the rectangulartx_,hc dl'trihuuon - I
:ion
mere ts cornpatihtht) of strain<. bet\\cen the reinforcement and the adjacent e. the ~tecl \truins ~ ·t 111 ten~ion und ...... tn compresston cun he determined from n dragram. The relntlonshtp). het ween the depth of neutrul u\1!-. ( ') and the m concrete \trurn (~
u'
(\ d') .\
61
(4.J)
1' the effective depth of the beam and d' is the depth olthe cumpre~:-.mn ment. n.., determined the \train,, we can evaluate the streS!>es in the reinforcement from ,_,train curve of figure 4.2, together wrth the equations de\ cloped 111
- d
12 '\I\ of u secuon with known steel strains, the depth of the neutral a \i\ can be r d b) rearranging equation 4.2 ac;
tl (4A )
62
Reinforced concrete design
At the ultimate limit !.tate the maximum comprel!sive strain in the concrete is taken as feu:!
= 0.0035 for concrete clas-.
~
C50/60
For higher classes of concrete reference should be made to EC2 Table 3.1 - Strength and deformation characteristics for concrete. For \tecl ''ilh.f;t = SOON/mm1 the y1eld strain from section 4.1.2 is f> = 0.002 17. Inserting these values for .:cu~ and :> mto equation 4.4:
\' =
d
0.00217 = 0 ·611d 1
+ 0.0035
Hence. to X~
en~ure
yielding of the tcn!.ion ..,tccl at the ultimate limit state:
0.6J7c/
At the ultimate limit state it JS important that member sections in Aexure should be ducti le and that failure should occur with the gradual yielding of the tension steel and not hy a sudden catastrophic compression failure of the concrete. Also, yielding of the reinforcement enables the formation of p l n~tic hinges \O that rt!distrihution of maximum moments can occur. resulting in a safer and more economical ~trut:ture. To en .. ure rotation of the plastic hinges with ~ufficient yielding of the ten!'ion ... tecl and also to allow for other factors such a~ the strain h:.mlcning of the steel. EC'2 limit~ the depth of neutral axis to 0.4Sd for concrdc da~., CS0/60. Thi' 1:. the hmiting ma\imum \aluc for 1 gl\cn b} I.C2 w1th no redistrihut1on applied to the momcntl> calculated b) an cla\liC anal):-1' ot the c;tructure. a.c; described in Chapter 3. When moment redi,tnbuuon '' apphcd these maxunum values of .t arc reduced a:. described m Section 4 7 The UK Annex 10 EC2 can gl\·e different limiting \alucs for 1. The EC2 value of \ OA5d is within the Aunex·~ rcqu1red limit' and 11 ensures thut a grudual tcn~ion fuilure of the steel occur:, at the ultimate limit \late. and not wdtlcn briulc fuilure of the concrete in compression.
4.3
Bending and the equivalent rectangular stress block
For most reinforced concrete ~tructure~ it is u~ua l to commence the design for the conditiom at the ult i mr~te limit state, fo llowed hy check~> to ensure that the structure i~ adequate for the serv1ceability limit state with(lut excessive dcncction or cracking of the concrete. For this reason the analy\i~ in thi~> chapter will lirst consider the ~implitied rectangular stress block wh1ch can he U\Cd for thc design at the ultimate limit state. The rectangular stress block us shm\n in ligurc 4.4 may be used in preference to the more rigorous rectangular-parabolic strc" block. This \imphfied stress distribution will fac1htate the analysi' and provide more manageable de.,ign equations. in particular \\hen dealing \\ith non-rectangular cro!>s-scctions or when undertaking hand calculation~.
It can be \ecn from figure 4.~ that the stress block doe" not extend to the neutral axis of the section but has a depth~ = 0.8.\. Th1s will result Ill the centroid of the stress block being sf2 OAOx from the top edge of the ~ection, which i~ very ncar!) the same location a' for the more precise rectangular- parabolic stre~., hhx:k. Abo the areas of the
'res of slre~s block arc approximately equal (see 11ection 4.9). Thu~ the momen1 of ,t.tnce of the section will be simi lar using calculations based on either of the two ,, hlods. -he tle11ign cquntions derived in sections 4.4 to 4.6 arc for tcro redistribULion of ncnt!>. When moment redistribution is uppl ied, reference ~hou l d be mudc to 1011 4. 7 which uc,cribe~ how to modify the design equmions.
uld be I and . .,f the mum eo~ure
'o to rpth of
4.4
Singly reinforced rectangular section in bending at the ultimate limit state
4 4.1 Design equations for bending
lfDhcd lkd in I arc
a.:
te of
Bendang of the \ection will mduce a resultant tcn,ilc force F, 1 m thc reinforcmg \lecl, 31ld a resultant compre"t''e force Ill the concrete /\, which act' thmugh the centrotd of the- effective arca nf concrete in compres~ion. a!> 1-hnwn in figure -1...1. f-ur eqtullbnum. the ultimate destgn moment. M. mu't he balanced by the moment of ,t,tance of the 'ection so thai
lerl'\1011
1-.<-:.
H
t the
F,,:
(-1.5)
hcrc ;: the luver arm bet ween the resultant forces F,( anti /· , 1 Mres~
f _.
r the ore is
1lthe "'tfieu .Je. 11he \\ill ... ular hand axic; lock -arne
, the
x areu of uction
o.567J~~ x JJ.1
)lock .md : '>O
1hat
d
s/2
(4.6)
sub~tiluting
in equation 4.5
M Jnd replacing
1
Jrom equation 4.6 gives
M - 1.13-l.fclh(d
;::);::
(4.7}
Rearranging und wbslttuting K = M fbd~f.l: (:/d)'- (:./d)
Soh ing
Lhi~
K 1.1 34
0
quadratic equation:
t~[o.s + j(o.2s- K/ 1.134))
63
(-1.8)*
64
K
Reinforced concrete design
M bd1 f..
0.05
0.06
0.07
0.08
0.09
0.10
0.11
0.12
0.13
0.14
0.15
0.16
0.167
0 954
0.945
0.934
0.924
0.913
0.902
0.891
0.880
0.868
0.856
0.843
0.830
0.820
1.00
Figure 4.5 Lever-~rm
maxomum value of z/d accordong to the Concose Code and previous UK practoce
curve
0.95
..,';:; " ~
Compressoon reonforcement required (al Mb•t)
0.90
i1
0.85
0.82 0
010
0.05 K
0 lS 0 167
1
M/bd /l,
The percentage values on the K axos mMk the limits for stngly reonforc.ed sections woth moment redtstrobutoon applted (see Sectoon 4.7 dnd Table 4.2)
tn equation 4.5
F,,
= V~ "' 1 \
"ith ~ ,
I 15
0.87J; ..A, lienee
(4.9) 4
Equations 4.8 <~nd 4.9 can he u~cd to de~ign rhe area of tension reinforcement in ~ singly reinforced concrete section to resi~t an ultimate moment. ;II. Equation 4.R fm the lever arm :can also he used 10 ~ct up a table and drnw o lever· arm curve as shown in ligure 4.5. This curve may h~.: U\CULO determine the lever :trm • • instead nf solving equation 4.8. The lower limi t of 'l. 0.82d in llgurc 4.5 oct:ur~ when the depth of the neutral ax11 equals 0.45d. This is the max imum value allowcc.J hy EC2 foro singly reinforced seclim wirh concrete class less than or equal to ('50/60 in order to provide a ductile section thu will have a gradual ten~ion type failure as already described in section 4.2.
=
4.4.2 The balanced section The concrete ~ection "irh the depth of neutral axi~ at the spectficd mted reach their ulumare -.rr;un~ at the same time. This occurs a the ma-
= 0.45d
Analysis of the section
0e depth of the stres:. block is
0.1 67 • 00820
OoR\h,J)
= 008 X 0.45d = 0036d
The force in the concrete stress block is
F ,,
= 00567/.1. x b~ -
00204/c;;bd
For cqUJhbnum Lhe force m the concrete Fccbal mu~t he balanced by the force F,rt>al in
So that
'reel.
Ool:l7.1)~A,".ll -
...,,
F"b.'' = Oo2
1'hc cfore \ o.tl
= 0.234f~k [)(/ /})~
that
23.4/:~ /vk
per cent
u:h is the steel percentage for a balanced section which ~hou l u not be excccucd for a ..:ulc ~ingly reinforced secuono fhuo.,, for example. with f~k 25 N/ mm2 anti f..~ 500 N/mm ~
=
25 J()(JA,"" 1 - '>J • .< bd - -· o"t 500
nt :
"3 04 x ~
~ 500
1.17 1)er cent
ullunnte moment of resil'-tancc of the balanced ~cc1 ion b M ...,1 1
ti
\( 2 - 0o8:!c/
h'tituting for \/1\;11
Fn~,, 1 :hJI where
r ..h..l
anti ::
1
Ool6 7J~~..bd
(4010)
nd 1/J ~hd
(4.9)*
0.167
K NI
.
ment in u
a lever.:r arm. ::
• rral axis eJ ...ection ~uon that
M,t
\\hen the de~1gn moment M11 1~ such thm 0 b , Kbo•l 00167 then the sc~.:ltOn cannot fcl.. (1• 'mgly reinforced and compression reinforcing steel i:-. requino:tf in the compn.:s,ion nl' of the ~ection. Thi~ is the limiting value of K Oo l 67 mar~ctl on the horiz.ontal a.' ' of the lever arm curve l'hown in figure 4.5 .
,. EXA MPLE 4.1
Design of a singly reinforced rectangular section
Jeplh of <;tate the ;ecur' at on with
':'"he ultimate dc~ign moment 10 be resi sred by the sec1ion in figure 4.6 i~ I R5 kN mo Determine the area of ten:>ion remforcement (t\,) required given the charactcnMic 11aterial ~uocngths are fu = 500 :-..tmm1 and f., = 25 N/mm'
"
M
A,
• •
hd'J~ ~
185 :!6()
10"
4401
X
25
00147 < 00 167
tncrefore comprc~~ion :.teel is not required.
Figure 4.6 Design example - singly reinforced section
6
66
Reinforced concrete design
Lever arm:
~ = d{ 0.5
T
/
(
0.25-
).~14)}
=-wo{o.5~ /(o.:?5 O.l·H)} 1.134 = 373mm cor allcmalively. the value ol
~
= f.,d
he obtained from the lever-arm diagram,
figure 4.5.)
M
A,== 0.87}~~ :: 185
X
106
= 0.87 X 500 X 373 I 140 mrn 2
Analysis equations for a singly reinforced section 'I he following equations may be used to calculate the moment of resistance of a given
:-.cction \\ ith a kno'' n area of \lee! retnforcement. For equilibrium of the compres-.tve force 111 the concrete and the tensile force in the \tcl'l tn figure 4.4: or
= 0.87J~vl Therefore depth or ~lfC).S block. ·~ 0.567J~~b
~
0.87/yk/\,
(4. 11 )
0. 567~~b
and x = s/0.80 Therefore the moment
or resi\lalll:e of the section
i~
P,1 X::
M
-
0.8~~vl,(d -
.
=- o.87}yiA
(
.1/ 2)
O.!l7{1 ~t\,) d - u3.iJ~~..b
(4.12)
The~e
equations assume the tension reinforcement ha'> >ielded. which will be the case if ca~c. the problem would require solving by trying -.uccessh·e value' of..\ unttl
.\ < 0.617d. If thi., i-. not the
with the Meel Mrains and hence '>tresses being determtncd from equations 4.2 and 4.1. to be u\cd in equation 4.12 instead of 0.87f~k·
Analysis of the section EXA MPLE 4.2
Analysis of a singly reinforced rectangular section in bending Determine the uJlimatc moment of resistance of the cm'>s-~ecuon ~ho\\n in fiuure 4.7 ~I\ en that the chamctcri!>tic strengths are i)l = 500 Nlmm1 for the rcinforcen;ent and :!5 N/mm2 for the concrete. 0.567f,
agram,
Figure 4.7
- ,0
- ---
:;:;
•••
_____)
Analysis example - singly reinforced sect1on neutral axi~
A, .. t470 mm1
For equilibrium of the compressive and
tcn~ite
forces on the
~cct ion
Frc - /·,1 t
a gi\'en
tree m the
0.567/..lb.l' O.X7/;kA 1 0.567 < :!5 3()() X I= 0.87
X
5(X)
X
1470
therefore 150mm .1110 .1
t4.ll)
.1/ 0.8 15CJ; O.X lHH mrn
fhi~ value of .1 is l c~s than the value of 0.6l7d derived froml>Cction 4.2. and therefore the steel has yielded and /,1 = O.X7/y~ a~ a~"umed. Moment ol' re1>istance of the section is
0.~7/yki\,(d
,1'/2)
O.H7 x 500 x 1470{520- 150/2) x lO
6
284 kN m
-l.J2)
case if trying
.ru -l.l, to
4.5
Rectangular section in bending with compression reinforcement at the ultimate limit state
(a) Derivation of basic equations It -.houhJ be noted that the equauons in thb !-.Cc.:tion ha\'e been deri\ed for the case ot zero moment rcdiMrihution. When this is not the ca"e, reference should he made to 'ection 4.7 '' hich deal' with the effect of moment rcdbtrihut1on
68
Reinforced concrete design
Figure 4.8 Section w1Lh compress1on remforcemenl
~
0.567(,,
0 0035
b
..... 1
.--- ---,_j_d
1 X~ 0 45d
• A,' •
neutral ax1s
d
.
lbll
A, .
Section
Strains
Stress Block
from the ~ection deal ing with the analysis uf u 1>ingly reinforced section nnd for concrete class not grcall:r than CS0/60 when M
> 0.167fckbd~
the design ultimate moment exceeds the moment of' re~il>turu.:c of' the concrete(Mbal) and therefore compression reinforcement i~ required. For 1his condition the depth of neutral axis, 1 .... 0.45c/. the maximum value allowed hy the code in order to en&ure a tension failure with a ductile ~cc1ion. Therdorc
::t-111 = d- !it-aJ/'2 = d O.Rlh:ll/2
= d - O.t{ = 0.8:!d
0.45cl,2
For equilibrium of the section in ligurc 4.K
~o
that Wtth the reinforcement at yteld O.R?f)~l\~
= 0.567j~~bs
O. K~f~~~~:
or with
0.8
S
X
0.!:\7}~1.A,
0.45d
0.36d
(4. 13)
0.204fdbd 1- 0.8~/ykA~
and taking moments about the centroid of the tension l.lecl.
M - F.c x ;:""' I F\<;(d - d') = 0.204/.kbd X 0.82c/ + 0.87/;kA:(d - d')
= 0.167fckhcP + 0.87}y~,A' (d- d')
(4.14)
rrom cqunuon 4.14 M- O.l67hbd1 A, = O.R7J;dd- d' ) I
(4. 15)*
Muluplying both ;,ide;, or equation 4.13 by : A
'
=
O.l67fdhd1
0.'07};1.
\\ ith ;:,.,
X
= 0.82d.
.:ba1
.... 1\
I
'
=0.!!2d and rearranging gi vel> (-U6)*
Analysis of the section .. areas of compression steel. A~. and tension l>tecl, A,, can he calculated from 5 and 4.16. 1 1g Arut 0.167 and K = M I bd7cl into these equations would convert
(4.17 )* (4.18)* a }sts it hru, been assumed that the compressiOn steel ha' yielded so that the 'J ... = 0.87}yl.· From the proportions of the strain distribution diagram: 0.0035
and tor
(4. 19)
r ) and neutr.d
a ren ...ion 0.0035
=
\\ith };1. 500 N/mm~. the steel strain l l the compresston o,teel () 002 17 <, 0.38
Therefore for
=0...15tl
() 171
-+.13)
= ~) = 0.00217.
(4.20)•
0.0035 ~
"'
(4.21 )
rnttu or d' / d for the yielding of other grades of steel can be determined by u. . ing tclu ~train in equmion 4.19. but for values of ]yl. les& thun 500N/mrn'. the <. tton or equation 4.2 1 will provide an adequate safe check. II I' cl .. 0. 17 1. then it is necessary to calculate the strain f ,~ from equation 4. 19 and determine J~~ l'rom /:', X • 'c
200000 "' -l.14)
- 15
\,tlue of stress for the compressive 'tccl mu~t then he u~cd in the denominator of ton 4.15 in place of 0.87/yL in order to calcu late the area A: of compression steel. Tbe area of tension steel is calculated from a modified equation 4.16 such that
~
A
- 16
= 0.167/..kbc/~ -A 0.87/yL;J,at '
1
/,..;
0.87Al
ne above equauon' apply for the calo.e \\here the concrete clu"' "' Ic-.-. than or equal ( 50/60. l·or concrete cla!>ses greater than CS0/60 ~imilar equation ... \\llh different 'tam~. can be derived based on the EC2 requtrement for the-.c classes. The constant11 r concretes up to clalo.s CS0/60 arc tabulated in table 4.1.
69
70
Reinforced concrete design Table 4.1
Limiting constant values Concrete class~ CS0/60
Limiting xt...1 d Maximum Zbaf Ktx~l - limiting K Limiting d' d
0.45 0.82d
0.167 0.171 23.41, 0 , fv<
Maximum percentage steel area 1 OOAoa1, bd
(b) Design charts The equations for the design charts arc obtained by taking moments about the neutral axis. Thus
M = 0.567fck0.8x(.x- O.Sx/2) +J~cA~(,I
d') + f,.A,(d- .1)
This equation and 4.13 may be written in the form
A,
.
A~
Y
f.,, bd - 0.45·+fd d t .f-.: hd \-~
M
'
bl l-
= 0.454/,~ -,, (I t-
d') d
0.40)
A,( (~)
f.,, bd I
For ~pccificd ratios of A: / lu/ . .1jd and d' / d. the two non-dimcn:-.ional cquations can be '>olvcd to give \aluc~ for A j btl and M ' /}{/' \o that a ~ct of dc.,1gn charts such as the one 'hm' n in figure 4.9 ma> be plotted. Before 1hc equation' can be ~olved. 1he steel stresse~ and./~ f11U'1,l be calculmed for each value of 1/ d. Thi~ i!> achieved by first determining the rclc\ ant strain:- from 1he strain dwgram (or by applyu1g cquauons 4.2 and 4.3) and 1hen by e\aluating the stresse:-, from the Mres., ·Slram curve of figure 4.2. Values of t:/d below 0.45 will apply when mome111s are rcdi~tnbllled. ltl>hould be noted that EC2 does not give design charts for bending. Hence although 1t 1~ po'~1ble to derive cham; a<. indicated. it may be Simpler to u~e the equations derived earlier 111 thi~ chapter or simple computer program~.
Design of a rectangular section with compression reinforcement 'lO moment redistribution) "i'le ~ec lion !thown in figure 4.10 is to resist an ultimate design momcm of 2g5kNm. The c racterislic material strengths are fyk = 500 N/mm 2 and };:~ = 25 N/mm 2. Determine e areas of reinforcement required. M bd~fck
285 ( l6 260 X 440~ X 25 - ·--
A,
• •
> 0.167
aeutral
~refore compre~sion
d'!r/=50/440
as
111 ~.:quation
Figure 4.10 Design example with compression reinforcemen l,
steel is required
0. 11 <... 0.17 1
no moment redistribution
4.21 and the compres:.ion
!.~eel
will have y1elded.
C mpre 'l>ion :.tecl: (K - K~>•l }j;,bd' L
0.87.1)dd d') (0.226 0.167)25 ) 260 X 4402 =.;.._~~:--0.87 X 500(440 5())
Analysis of a doubly reinforced rectang ular section Determine the ultimate momem ol resistance of the cross-section shown in figure 4.1I that the characteristiC \trengths are f,k = 500 N/mm 2 for the reinforcement and J, = 25 N/mm 2 for the concrete. For equilibrium of the tcn~ i le and compressive forces on the .o,ection: ,_!IV en
F,1 =
F cc
\~suming
+ F,_
initially that the steel
0.87/ykAs
= 0.567fcL/H
~tresses
O.X7})LA~
fs1 and fs,
arc the design yield values, then
72
Reinforced concrete design
Figure 4.11
b~
I•
Analysis example, doubly reinforced section
t
280
0.5671,,
r ..
.. I
•A."= 628•
'
F,.
d =50
s = 0.8~
£ .,..
'
..,
-
.4.~2410
• •
F.,
Section
Stress Block
therefore 0.87/vk(/\,- A~)
.~
0.567f~kb
=
().H7 x 500(2410 628} 0.567 X 25 X 2SO 195 mm
s/0.8 = 244 mm 1/d - 244/510 0.48 < 0.617 1 -
!>O
(~ee ~ection
4.::!)
the tension steel will haw }icldcd. Also d' j.1
= 50/ 225
0.22
<..
0.38 (~cc equation 4.20)
so the compres<;ion steel \\Ill abo have ) tcltled. as Taking moments about the tcn-.iun steel M
= F,,(cl -
~ 2)
r "'( d
d' )
0.567f,kb:.(tl - 1/ 2) ~ O. H1j>~A~ (d ·0.567 x 25 x 280 x 195(510
= 319 -
124
as~umcd.
d')
195/ 2)+0.!!7
>
'i00
:><.
620(510
50)
X
lO
11
443 !...N m
If the depth of neutral axis was such that the comprc~>sive or tensile yielded. it would have been nccc1.sary to try successive vnluc:, of .r unlil
~teel
had not
F" = Fe,· + F.., balances with the &teel 4.1. The btcel
~;tresses
~trains
and
).Lrcs~es
being c.:ulculaled from C4m11ions 4.2, 4J and
at halancc would !hen be used to calculate lhe moment of
resistance.
4.6
Flanged section in bending at the ultimate limit state
T-sections and L-~ections which have their flanges in comprc~~•on can botl1 be designed or analy:-ed in a similar manner. and the equations \\ hich arc derived can be applied to either type or cross-section. As the Range~ gene rail} pro\ ide a large compressiYe area. it j.., usually unnecessary to cons1der the ca~c where compressiOn steel i:-, required; if it .should be required. the design would be ba~cd on the principles derived in ..,ection 4.6.3
Analysis of the section
73
lbc '1ngly reinforced section it is necessary to consider two conditions: --e'~
block lies within the compression tlonge, and .:'' block extends beiO\\ the flange.
Flanged section - the depth of the stress block lies within the :ge s hr (figure 4.12) acpth of stress block. the heam can be considered as an equivalent rectangular 1 llreadth bt equal to the flange width. Thi~ is because the non-rectangular belm the neutral axb i' in tcns10n and b. therefore, con'>idered to be cracked :U\c. rhus K = M /btd~f..~ can be calculated and the lever arm determ1ncd from a m curve of figure 4.5 or equation 4.H. The relation between the lever arm. ;:. \.of the neutral axis is given by d
I
2
- d- ::) O..S671,,
"1 •
, - --n-eu-tr_a_la-xi-5-~ X
Figure 4. 12
1r-1 -~ ·8rx - "F.,
T-~cctlon,
-r· r --S
stress block with1n the rldnge, s h1
s/2
l
, F.,
10 Section
Stress Block
e" than the flange thtd..nes-; Chr). the we~s block ....._.._..___...... 1nd the area of retnforccment is given hy
doc~
lie wtthtn the nange as
M ll S~fvk:.
de,ign of aT-section beam is described further in section 7.2.3 with a worked e.
It
E.
MP LE 4.5
~lysis
of a flanged section
n nc the ultimate moment of resistance of the T-section ~hown in figure 4.l3. dl.tractcri<.tie material ~trength~ arc [ 11. - 5()() N/mm1 and /d -= :!5 N/mm1. '\.5 ume initially that the Mrc~~ block depth lies wi thin the Oang~: and the ~:e mcnl is strained to the yield. ~>O thal /.11 0.87 }yk·
74
Reinforced concrete design 0.567f,,
b, = 800
Figure 4.13 Analysis example or a
T-section, s < llf
~ '
---~~ral~i~- -
- • • •
h1 =150
r-t-=fF-
_j
A,= 1470 mm<
F,.
Section
s/ 2
.
Stress Block
For equilibrium of the \ecuon
Fe,
= F,,
therefore
and
~olving
.\ -
for the depth ol stress
h l o~.;k
0.87 X 5()(} X 1470 0.567 X 25 X 8()()
= 56mm < 111 .\ = 1/ 0.8
150mm
70mm Hence the stress blocl. does lie within the Aangc ami with thi' depth of neutral axi~ the t->lcel wil l have yielued a~ ns:-.umec.J. Lever arm: ~
s/ 2 - 420 - 56/ 2 '392 nun d
Ta~ing moment~
M
about the centmid of the reinforcement the moment of resiMance il>
F" x :.:
;::: 0.567}~kbrs:.: 0.567
X
25
X
8()()
X
56
X
392
X
10
1 '
249 kN m If in the analysio; it had been found that s > h1 • then the procedure would have been :.unilar to that in example 4.7.
4.6.2 Flanged section -the depth of the stress block extends below the flange, s > h, For the design of a nanged section, the procedure dcscribt.:d in section 4.6.1 wil l check if the depth of the stress block extends below the nangc. An alternative procedure is to calculate the moment of resi tance. M1, of the section with s = lt1, the depth of the
Analysis of the section
75
~e ~h
(see equation 4.22 of example 4.6 following). Hence if the de~ign moment. Md. is that
\/d > Mr '1
the <;tress block must extend
beiO\\o-
the flange. and
ltr
ca\c the design can be carried out b) either:
ht\
w.. ing an exact method to determine the depth of the neutral axis. as in example 4.6
a
or for the conservative condition of x = 0.45d. which ;, the maximum value of r for a ~o. ing l y reinforced section and concrete class ~ C':'i0/60.
tle~igning
r EX AMPLE 4 .6
Design of a flanged section with the depth of the stress block below the flange ~I-tt:
T-section heam )hown in figure 4.14 is required to rcsi~t an ultimate design of IXO kN m. The churacteri~tic material strengths arc .f>l :'iOO N/mm 2 and 2 25 N/mm • Calcu late the area of reinforcement required.
~1 ment
1
Figure 4.14 Design example of a T·sewon
0.5671,,
'the
j
I
h,. 100
'
X
axo}
A
• •
·r-
s I
l
· -~
'""
j
lo
ll
F,.
l~b,. 20'L.
Section
Stress Block
In figure 4. 14 Fe~
is the l'orcc developed in the flange
I c"' is the force developed in the area of web in comprcsl.ion
been
_)
below
. eck if -e
IS
to
f the
\lomcnt of
re~istance.
Mr. of the flange is
Ml
F.~ X ;:1
Jfr
0.567}ckbl ltr(d
ur
0.567
25
= 170 I..N m <
X
4()()
111 / 2) X
100(350- 100/ 2) X
(4.22)"' JO -h
180 k~ m. the design moment
Therefore, the :..tress blod. mu~t extend belo'' the flange . It is now nece~saf) to determine the depth. s,. . of the web in compres,ion. where
h.
76
Reinforced concrete design
For equilibrium: Applied moment IHO
/·~r x
= 170
..L
::1 -'- F," x
~!
0.567f.:~ b"s"
- 170-0.567
X
25
;:2
v
200l., (250- 1" / 2)
X
170 + 2835sw(250
~w /2)
X
I()
6
10- h
X
Thi 'i equation can be rearranged into ~ .. 2 - soo~.. t
1.os
< 10
1
=o
Solving this quadratic equauon
v.. - 15 mm ~o
that Lhc depth of neutral axis
(/tr + sw)/0.8
X
= I.Wmm
(100
j
15)/0.8
0.4ld
0.45d compn!~~•on remforcement For the equilibrium of the section
A\ \
i~
not required.
F,1 - F,1 I Few
or 0.87/>kA, - 0.567j;.b,h
+ 0.567 /.:~ b,.. s,.
0.87 < 5(X) X tl ,
X
0 567
25(400
X
100 -r 200
15) = 610
X
101
Then:fore 610
X
10 1
0.87
X
5()()
1402mm~
(
EXAMPLE 4.7
Analysis of a flanged section Detcnnmc the ultimate moment of re~istance of the T-bcam section 'ho\\ n in figure 4.15 g1vcn h~ - 500 N/mm 2 and fck = 25 N/mm 2. rhc compressive force in the nange is F.r
0.567 f.:kb, It,
o 567 x 2s x 450 Then F·,1
tcn~ile
ISO x
Io-3
957 kN
force in the remlorcmg <;teel, a-.suming it has yielded. i'
0.87/ykA, 0.87 x 500
X
2592 x 10- 3 = I 128 kN
Analysis of the section 0.567fc,
/). =450
Figure 4.15 Analysis example of a T-section, s > /If
•ht =150 c
Section
cfore 1-', 1
f "
> F"
Stress Block
so that s > lit and the force in the web is
0.567/dbw{.V - fir) 0.567
X
25
X
3QO(s - 150)
X
lQ
3
4.25(.1 - 150) equi librium
F.
F,1
4 ~5 ( 1
F"
150)..: I 128 - 957
... '-e
= 190 nun 1 10.R
231:1 mm
= 0.43d
\\nh thr\ depth of ncutml a\is the reinforcement has yielded, F, ..
_)
4.25(190
150)
:I\
assumeu. and
170kN
> F,,, the the strc~~> block wou ld not extend beyond the llange and th~: ~cction he analysed Ul. in cxumplc 4.2 for a rectanguiLlr ~>ection of dimension!~ b1 x d.) rng momcms about th~.: centroid of the reinforcement
It f
1
~ld
l
\I
F~ r (d
flt /2) I f· cw(d
[957(550
~/2
hr /2)
150/2) I 170(550- 190/2- 150/2)] x 1() - 1
5.19kN m
!-+.15 ( EXA MPLE 4.8 Design of a flanged section with depth of neutral axis x - 0 4Sd
,Jfc but con:-.crvmive desrgn tor a flanged section with f > Jr1 can be achieved hy ung the depth of neutral axis to .r = 0.45cl, the maximum depth allowed in the code. [l~, rgn equations can IX! derrved for this condition a:. follows. Depth of Mres~ blocl... s
0.8x = 0.8 x OA5d = 0.36d
77
78
Reinforced concrete design O.S67fc~
r-
Figure 4.16 Flanged section with depth of neutral axis x 0 45d
=
x =OASd s" 0.8x
'
_j_
d aXIS
A,
• •
f,,
Section
Stress Block
Divide the flanged ~ection within the depth uf the ~tn:s~ block into areas l nnd 2 ns shuwn in figure 4.16. so that Area I
= bw X s = 0.36b,,,d
Area 2
(br
hw)
ht
X
and the comprc:-.sion forces developed hy these area~ urc
Fd
0.567.f.:~
Fc2 -
0.567.1~~ ~~1 (b, - /J" )
x 0.36b,.d
0.'1f..,b,.d
Taking moment~ about Fr' at the centroid of the flange
M
F,,(d
h1 /2)
Fd (.\j2
h, /2)
= 0.87/ikA,(d- h / '1 )- 0.2/..lb,.d(O. \()(/ 1
11,)1 2
Therefore
!:' + 0.1 /.:lbv. d(0.36d cunr~~ (d -
ht)
(4.23)*
0.511, )
Thi~ equation should not
he u~ed when 11 1 Applying thi::. equation to example 4.6: 180
A,
X
I 06 -
+ 0.1
X
(J.ln
:!5 X
0.36d.
X 200 X 350(0.36 X 15() 500(350 - 100/2)
l~ l 4mm 1 (compared w ith 1407
111111
2
I00)
in cxamrlc 4.6)
Before using equation 4.23 for calculating A,, it i\ necessary to confirm thnt compre!>sion reinforcement i~ not required. Thi~ is achil!vcd by using equation 4.24 to chcc.:k that the moment of rcsi~tancc ot the concrete. M 11ul. i~ greater than the design moment, M. )
l---------------------------------------~ 4.6.3
Flanged section wit h compression reinforcement
\V1th x OA5d m figure 4.16 and taking moment\ about A_. the max1mum res1stance moment of the concrete is
Mba!
= Fc1 X ~I + Fe'! X :'2 = 0.167fc..b,.d 2 - 0.567fc~ (bc- bw )(cl -
hr/ 2)
(4.24)
(Note that the value of 0.167 was derived in equation 4.10 for the rectangular section.)
,e the area of comprc)o~ion steel can be calculated from M - Mt\Jt 0.87J;dd d') coo~ tdcring
(4.26)
the equilibrium of forces on the section
r the nrco of tens10n steel is
0.87 ;;.~ n. t1
.7
1 /.\
~
(4.27)
< 0.18, otherwi11e the de~ign compre~-o~ive steel ~tre~s i~ less than 0.87 ;~~, .
Moment redistribution and the design equations
pht\tic hcha,·iour of reinforced concrete at the ultimate limn state affects the huuon of moment' 111 a \tructurc. To allow for this. the moment!> denved from an 111••tnal)l>i' may be redi-.trihuted based on the a~sumplion that plastic hinge-. have ed at the section<, \\ tth the largest moment' (see secuon 3.6). 1 he fonnatton ol 1 ~ htnge' require\ relnth ely large rotations with )'ielding of the ten \ion •rcement To ensure large \tram-; in the tension Meel. the code of practice n.:\tncts epth ol the neutral axi~> according to the magnitude of the moment redi,tribution cd out. Tne equation~-. for thi~. gtven by EC2 for concrete ci
i'
2: "I I
k
,\'b,ll
2
d
(4.28) •here
)
moment at section after redistribution ---< 1. 0 moment m section hefore redistribution .nd ~ ~ ore con .. tant~ from the EC2 code and the UK Annex and ,.h•l is the max1mum ah I! of the depth of the neutral axi~ which will take the limiting value of the equality of equation (4.28) but should be less than OA5d for concrete cia<;" < CS0/60. Tht: depth of the 1>trcs' block is
= 0.8.\'t>.l an the le\cl arm ts
(4.29)
79
80
Reinforced concrete design The moment of resistance of the concrete in compression is
Mt>al - Fcc
X
Zt>>l
=- 0.567 fc~bSt->al
X ;:bat
and
Koot = Mbal/bd~fck = 0.567.1hat X :.~>at/d~ This equation for K""1 and the previous equations from 4.28 to 4.29 can be arranged to give
Kt>al - 0.454(/' - k!) j k2 - 0.182"(6- ~l)/k2j~
(4.30)
or alternatively
0.454c~;t') c:')
Kt>ul
From EC2 clause 5.5 the constnnts k, anu k2 arc given as: k 1 0.44 and k2 - 1.25, but from the UK A1u1ex to EC2 k 1 = 0.4 and k2 1.0. The relevant values of Xtll'" Zbal and Kbnl ror varying pcrccntagcs of momem redistribution and concrete c;luss < C50/60 are shown in table 4.2. When the ultimate design moment i~ ~uch that M
> KhaJbd 2f.:l
K > Kbal
or
then compre:.sion !)teel i!> required '>Uch that I
Kto.tift~hd 1 0.87/ydd- d')
(K
A - .;_,..,..--.;_..;;..;.._--.,...
•
(4.31 t•
and
(4.32) where
A
Mto.. t
(4.33)
bd 2./,l
These equations arc iucnlical in form to tho~c derived previou~ly for the design of a ~ection with compre~sion reinforcement untl no moment rcdi:-.tributton. If the value ot
d' /d for the section exceeds that :-.hown in table 4.2, the compression steel will not have yieluetl and the compressive stress will be l cs~ thun 0.87 /y~· In such cases. the compressive stress _he will be E~!~c where the strain 1, i:-. ohtuincd from the proportion) or the strain dingram. This vnlue of ./~c shou ld replace 0.87/Yk in equation 4.31, and equation 4.32 becomes
A, :;:
Kbat/~k bd ~ 0.87 .1) ~ <-t>al
A1
+ '
./~c
X ---
0.87 /vl
It should be noted that for a singly reinforced ~ection (K < Kbatl. the lever arm i~ calculated from equution 4.8. For a l.ection requiring compre~sion 'teel, the lever arm can be calculated lrom equation 4.29 or by U'>ing the equution
-: = d[o.s- )(0.25- Koot/1.13-t)] "hich il> similar to equation 4.8 but with Krut replacing K.
(4.3-t
Analysis of the section
81
Table 4.2 Moment redistribution design factors x001/d
{o
l=-ed to ~JO )
t5. but mem
Zt>al/d
Kbal
d'/d
0.821 0.853 0.869 0.885 0.900 0.917
0.167 0.142 0.129 0.116 0.101 0.087
0.171 0.140 0.125 0.109 0.094 0.079
According to EC2, UK Annex, k1 - 0.4 and k; - 1.0 0 1.0 0.45 0.82 10 0.9 0.45 0.82 0.82 15 0.85 0.45 0.40 0.84 20" 0.8 0.75 0.35 0.86 25 30b 0.70 0.30 0.88
0.167 0.167 0.167 0.152 0.137 0.120
0.171 0.171 0.171 0.152 0.133 0.114
According to EC2, k, - 0.44 and k2 = 1 25 0 1.0 0.448 10 0.368 0.9 15 0.85 0.328 20. 0.8 0.288 0.248 25 0.75 30b 0.70 0.206
• Maximum perm1tted redistribution for class A normal ductility ste~l b Max1murn perrnltte?d redistribution lor class Band C h1gher ductility ~teel, see section 1.6.2
/EX AMPLE 4 . 9
Design of a section with moment redistribution applied and to
112)*
I 3.3)*
"t n e of IJVC
~
the ons
and
08
The -;cellon 'hown 1n figure 4.17 i::. subject to an ultimate design moment of 230 k.. m niter a 20'/r reduction due to momem redistribution. The characten,tit matcnal Mrcngths Jre ~~~ 'iOO t\/mm 2 and /.;~ = 25 N/mm 2• Determmc the area' of reinforcement required Ul>ing the con,tants ~~ ond ~1 from (a) the EC2 and (b) the UK unncx to EC2.
b e 260
d'• l
so
• • A,'
II
(a) Using EC2
•
(1) From first principles (b - k,)d/k~
Limiting neutral axis depth,xhJI I rum EC2 clau!>e 5.5 therefore StI'C!.S blocI,. depth Lever arm
~1 .thul
~hal :h•l
0.44 and k~ - 1.25, (0.8 - 0.44)490/1.25 = 14 1mm = ox~hJI 11 3 mm = d - .lbal/2 490 l l3/2 = 434mm
'-'1omcnt of rc~btance of the concrete MhJI
F,.,
:t>al
from
0.567 f~~hl'bal
X Zbal
0.567
260
X
25
X
X
113
X
-l34
X
10
II
181kNm <. 230 kN m, the applied moment
therdorc
d'jxt,.al
0
..."'
compre~<..ion
'oteel is required.
= S0/ 141- 0.35 < 0.38
therefore the compression steel has yielded.
( ee equation 4.20 in :.ection 4.5)
A,
"'
·-
Figure 4.17 Dc~lgn example with moment redistnbution, ~ = 0.8
82
Reinforced concrete design Compression steel:
M - MOO!
I
A,
= 0.87[f.:{d -
d')
(230 -181 ) X l(f 50)
= 0.87 X 500(490 256mm~
-
Tension steel: M t>al
A, -
1
0.87 /,LZbJJ J8J
-:-A,
106
X
= 0.87 X 500 X 434 -256
=959 + 256
12 15 mm2
(ii) Alternative solution applying equations developed in Section 4.7 From equations 4.30 to 4J4: Khal
= 0.454(0- k1 )/k1
0.1821(~
- 0.454(0.X 0.44 )/ 1 25 0.131 -0.015 = 0 116
At J/k1':
O.llUI(O.H - 0.44 )/ 1.25/
which agree<. with the value given in tahlc 4.2. M
Bending plus axial load at the ultimate limit state
.1pplled axial force may be ten\tle or compressive. In the analy,i'i that follows. a nn:sstve for<.:e 1:-. considered. For a ten~ile load the ~arne ba~1c principles ot lthrium. compatibility or strain!., and stre),s- strai n rcltttionships would apply, hut it td he necessary to change the 'ign of the upplied load (N) when we consider the 1 hrium of force-. on the cro-.-. ·~ection. (The area of concrete m compression ha~ nut reduced to allow for the concrete di~placed hy the comprc:-.sion steel. This could he en tnto uccount by reducing the stress.f,c in the comrression steel by an :unount cquul 56~~~~.)
cure 4.18 represent-. the cross-~ection of a member with typical struin and strcs' butions for varying position& of the neutral axi~. The cross-~ccuon ~~ subject to a ..,ent M and an nxial compressive force N. and in the figure the direction of rhc -nent i~ such a., to cause compre%ion on the upper part of the section and tcn~ion on •wer part. For ca~e!> ,.,here there is tension in the section (figure 4.18a) the limumg • ete !;train is taken as 0.0035 - the value used in the de,ign and analysts of !>ccuons ~ nding. However for cases where rherc is no tension in the section (figure 4.1 Rh) I miting strain i~ taken as a value of 0.002 at the level of 317 of the depth of the
83
84
Reinforced concrete design
Figure 4.18 Bendmg plus axial load w1th varying pos1tion of the neutral axis
r •
A,'
•
d •
d
h
A.
• •
·if--
s/2 neutral aXIS
t,
(a) S= O.Bx< h
r
0.567fc~
.....
0.0035
b
F,
0.5671<,
b
•
H
1 '
A,'
· -~ -
Ltc
d
h
• • A,
Section (b) s
Stral11s
Stress Block
=h: 0.8x > h
Let I·".:~
he the comprcS\IVC force developed in the concrete and acting through the cemroid of the <.tre-;<. block F.... he Lhe compre~~ive force in the reinforcement area and acting through centroid r , be the tensile or compres~i\e force in the rcmforccment area A~ and act • through it!> centroid.
A:
(I) Basic equations and design charts The applied force (N) mu!it be balanced by the forces developed within the cro section, therefore
N
= Fe~ + F + F~ M:
In thi!t equation, /·'_ wi ll be ncgutivc whenever the pn~ition of the neutral axis i~ Stk that the reinforcement/\, i~ in tcn!tion, at-. in ogure 4.18a. Sub1-tituting into thi~ equau the term), for lhe stresses and areas
{4.35
A;
where /.... i~ the comprec;sivc ~tress m reinforcement and .!- is the tensile compressive stress in reinforcement A . The de~ign moment M must he halunccd hy the moment of resi~tance of the fore" developed within the cro<,s-,cction. Hence, t:1king moments about the mid-depth oft" ~>ection
Analysis of the section
85
(·U6)* he depth of neutral axis is such that 0.8x ~ It, as in part (b) of ligure 4.18, then lc concrete 'cctton 11> 'uhject ton uniform compressive stres<> of 0.567./~... fn thi<> ~ concrete pmvide<; no contribution to the moment or rcsi•aancc and the lirst the right ~ide of the equation 4.36 di:-.appenr1-.. ') mmerricnl ammgement of reinforcement (A: =A, A,.j'2 and d' - It - d). ' -U5 and 4.36 can be re\\ riuen m the folio\\ ing form
0.567.1 f..c A, /~ A, --1·--+· -
!\
It
-l
}~~ bh
1".:1
0.567.1 ( 0.5- ~) It 2/t
1 "Ji.
bit
!-. ~
C'
fckbh II
0.5)
f..
t\,
.fck bh
c0.5) It
(4J7) (4J8)
cquauon' the Meel \tram'>, ami hence the 'tre'se' f, and f,. \ ary "11h the depth m:utral ax1' (\). Thu~ N / bl!f.:• and M j bh 1f,~ can he calculated for 'pccified ratios bh and r/ h so that culumn de))ign charts for a ~ymmcllicul arrangement of nrc ·ment l>Uch a~; the one !>hown 111 hgure 4.19 can he plotted. cltrcct solutiOn of equauonl> -1.37 and 4.38 for the dcs112-n ol column re1nforceml.!nt bl.! very tedious and, therefore, a set of design charts lor the usual case of trical section~ i~ available in publicatiOn!> :.Lu.:h a~ The IJeli[.lnt•r.\ Guide (rd. 20). pie' 'hO\•ing the dc~ign of column 'teel arc gl\:en 111 chapter 9.
\fodes of failure ~·
ltlve magnnude of the moment tM) and the ax1al load CN) governs whether the will fail in tension or in compression. With Iorge effective eccentricity ./ N) a tcn~i lc failure i~ liJ...ely. but\\ ith n '>mall cccentncity a comprc~~ive railure likely. Thi.! magmtude of the ecccntncll) affcch the po~11ion ol the neutral axi:-. hence the stmin~ and !>lrcs,cs in the re1nforceml.!nt.
Figur!! 4.19
TypiCal column design chart
0
01
0.2
0.3
OA
86
Reinforced concrete design
Let
A:
be the compressive strain in reinforcement be the tensile or compressi\'e o;train in reinforcement As be Lhe tensile yield strain of steel as -;hown in the stres\-stmin curve
£..., £,
£l
figure 4.2. From the linear strain distribution of figure 4.18(a) = 0.0035
E:-.:
c- d') X
(4.39
and
X)
d e:, = 0.0035 ( -x-
For values of x greater than h, when the neutral axis extends below the section, as shown in figure 4.18b, the steel !.trains are given by the allernmive expressions: 7(x d')
:..c = 0.002 (?x _ j fl)
and 7(x - d) 0.002 {7x 3h)
£, -
The ~lee! ~tres~es and strain'i arc then related according to the stres!'.- Strnin curve 0 1 figure 4.2. Consider the foliO\\ ing mode.., of f:ulure of the \CCtton a~ shown on the interaction dtagram of figure 4.20. (a) Tension failure,
£s
> "'y
Tim l)pc of failure is a-.~ociated with large eccentrictlles ((')and 'imall depth~ of ncutr... ax1~ (.\). l·ailure begms with yteldmg of the teno;Jic retnforcement, followed by cru'ihin,.. of the concrctl! as the tensile stratns raptdly 111crea'ie.
"s - .:y, point b on figure 4.20 When fat lure occurs with yielding or the tcn-.ion !.led und cru'ihing of the concrete at the
(b) Balanced failure,
sume inslant it is described us a ·oalanccd· ft1ilurc. WiLh ~ , d .I = ~"hQI
Figure 4.20 Bending plus axial load chart with modes of failure
e of
b_.cf .. 0
;;;
c
~
::l
"'=
~ , .!:!
M
nt h on the intenmion diagram of figure 4.20. N Nhut. M M"·'' and -0 87 ~~~· When the design load N > Nhul the section wi ll fail in wmpression, there will he ao initial tensile fai lure, wit h yie lding uf
Otroression failure c."e
1
'''"'
and N > N~-o.1 • The t:hange in slope at point r in ligurc 4.20 occurs
rune of uon
fror'l equation 4.39
0.0035d' (0.0035
.:) )
2.63d' for grade 500 l'tccl n
r "ill ot:t:ur tn the tension failure Lone of the imcrm:tton dtagrJm if .11 1 d
< '"·'''
_ O.H7 }~ ... and tcn,ile n
eat the 4.39
d
1
()
'hen
1
f, •
O.R7.f~k
•
d
nnd t:mnpressive
.. n r become:-. very large and the ~ec tion approac hc~ a .pression
0.00217
y
Mate
of uniform axial
lor grade 500 steel
stage, both layer!> of <,tee) will have yielded :md there will he 7ero moment of t-tam:c with a \ymmetrical section, 1.0 that
1 thl\
4.40)
~
0.567}~lbh
87
0.87J;K(A:- A,)
c\t the stage where the neutral a>. is cmncidcs with the bottom of the sectton the :.train ~ram changes from th:ushO\\ n in figure 4.18a to the alternmh c 'itratn diagram -;hown hgure 4.18b. To calculate N and M at this stage, corrcspondtng to potnt s in ,.urc 4.20. equation!> 4.35 and 4.36 should be used, taking the neutral ax.t!'l depth equal the overall section depth. h.
88
Reinforced concrete design
Such M-N interaction diagram~ can be constructed for any o;hape of cross-section vv hich has an axis of symmetry h) applying the ba!>ic equilibrium and strain compatibility equations vv ith the :-.tn.:~!>-~train rclauon\, a\ demonstrated in the foUowmg examples. These diagram., can be very u~cful for design purposes.
(
'\
EXA MPLE 4 .1 0
M-N interactio n diagram fo r a no n-symme trical section
Construct the interaction diagram for the ~cction :.hown m figure 4.21 wirh 1:~ = 25 Nlmm 2 and J, . . = 500N/mm2 The bending causes maximum compression on the face udjm:cnt to the steel area tl~. For a <.,ymmetrical cro~s-scction, taking momenb about the centre-line of the concrete section will give M = 0 with N = N11 and both area~ of stt.:cl ut the yield mess. Thi-; ' nn longer true for unsymmetrical steel area., a~ /• " I F~ at yidd therefore, Lheoreticnll~ moments should be calculated about an axi~ rcfcrn.:d to as tht.: 'plastic centroid'. The ultimate axial lond No acting through the pla~>lic ce111rt1id cau~e<. a unifnrm ~train ucru~, the section with compres~ion yielding of all the reinf(lrccment. and thu\ there i~> zcr mwm:nt of resistance. With uniform :.train the neutral-axis depth. x. is at infinity. Figure 4.21 Non ~ymmetr1eal ~l'Ction M ·N lnteraclton examplt'
~
A
lu
I~
.,"' II
II
-'=
...
A
•A,': 1610• •d'= 60
~
0
"'•
b 3SO ---
neutral ax1s A, • 982
' • '--
•
Section
Stratn
Otagr~m
The locution of the plasttc centroid i~ determined by w~ing moment~ of all the stre rewltants about an arhitrary axil. :.uch "' AA in figure 4.21 so that 'LJFccfl /2 I F,cd' I F,d} f' F)
.\p =~( f. L. 'cc
+ ·~r + '
= 0.567J~~A •• X 450/2 + O.S7.{y~A: >< 60
O.X7 J)~A, X 390 0.567 /.:kAcr + 0.87/y~A: 1 0.87(~k/\, 0.567 X ::!5 X 350 X 45t)a /2 I ().87 X 500{ l6JQ X 60 + 982 X 390) 0.567 25 X 350 X 450 + ().1{7" 500( 161() l 982) - -
=::! l::!mm frumAA The fundamental equation!> for calcul..ning points on the interaction diagram varying depth~ of neutral a>.b nre: (I)
'"th
Compatibility of stram~ (used in table 4 3. columns 2 and 3): E"
£,
= 0 0035 (" ~·
d') 0.0035 (d ~ ')
(4.41
Analysis of the section ---ection .. strain • n the
or when the neutral axis depth extends below the bottom <>f the section _
., 7{.t
d')
.
,.
c,, - 0.00. (?.I·_ 311) ,tnd ~' t ii 1 Stres~ \train relation~
~ ~ "Y
= 0.00217
=
7(x
0.002 (7.l _
d) 311
(x > II):
)
for the ~tee! (table 4.3. columns 4 and 5):
J - 0.87/yk f = l:: xc
(4.42 )
(Iii) Equil1hrium (t:lble 4.3. column~ 6 and 7):
0.!<1
"
0.8.1
,
N = Fe, ~ r . . + /', N - 0.567 J.~h O.lh f..._ A: t J A N 0.567l~ bh -f.,.,\: ~ f,,\,
'I ul.mg moment~ .tbout the pla,tic centroid O.!ll
h
M
().1:!.\
,
M
= / "( ' p- (Uh/ 2)
t F.,.( lr - d') - 1 (c/ lr ) l._ (.ir- h/ 2) + F,,:(.rp tl') f',(tl .lp)
F, i., negative when{, i\ a tensile
~Ire:.:..
Thc~c equation~ have been applied 10 provide the vo l ue~> in Utblc ·U lor a range of key \ in tahlc 4.3 us a 1-erie~ of straight line!>. or courr.e. N and '"' could have been calculated for imcrmediulc val ue!> x to provide u more accurate curve.
or
N(kN)
Figure 4.22
(0, 336 1)
M-N interaction diagram for
3000
\\ ith
a non-symmelrical sec11on
2000 1000
0
-lAL )
M (kNm)
90
Reinforced concrete design
(
EXAMPLE 4 . 11
M-N interaction diagram for a non-re ctangular section Con~truct
the interaction diagram for the equilateral triangular column section in figure -l.23 ''ith fck = ~5 N/mm 2 and /)k 500 Nlmm 2. The bending is about an axis parallel to lhe side AA ami cau~e' max1mum compres!-tion on the corner adjacent 10 the tee I an:a
,- -.
Figure 4.23 Non-rect~ngular 1nter~ction
i\:.
.•
GO
,..,II
M N
exam pip
"'.......
0
'
.,II
,..
2~3s
3
II
""
31125 bars A
~
•
A,
--
400
•
A
I or thi~ triangular 'ection. the plastic centroid i~ at the ~ame location a~ the geometric centrOid, 'ince the momcnt off"' equals the moment of r, ubout thi~ axi1. when aU the bur<. have yielded in compre.,~ion. The fundamental equation' lor ~tmin compatihility and the steel's 1.trcss stram relauon~ are as presented 111 e\amplc 4.10 and arc u~cd aga111 m th.b example. The equilihrium equauon!'t for the tnangular secuon hecomc ,\'
I"
+ F ,.
r,
or
0.8.1 <.. " 0.81 > h 0.!\.l
<,
0.8.1;::"
= 0.567}~k-\l /2 -l j-.._A:
I /.A 0.567./~kll > 400/2 -+ j',.A: f f,A, M 1··~, 2 ( 11 - O.XI')/1 f /- ,,(211/1 t!')- F,(cl M = F-. (211 /3- tl' ) /·',(c/ 211/])
N N
211/3)
F, is negative when f, b u tensile stress. and l'rom the geometry of figure 4.23 the
width or the section at depth ~
?
0.8..1 i~ I - ~ .vJ3.
Table 4.4 M-N Interaction values for example 4. 1 1 f!( (N/mm 1)
luhlc 4.4 has been calculated using the fundamental equation~ with the va lue~ of \ 'hown. The intcntction diagram b ~hown constructed 111 figure 4.:!4. With a non-rectangular section. it could he advisable to con~truct a more accurate ntcractinn dwgram U\ing other mtcrmcdJate value~ of r. Th1' would ceJtainly he the ~a'e "ith. say, a flanged \Cction \vherc there Js ~udden change in breadth.
c re111forccd concrete section. The stress block i\ suniln.r in ~hape to the ~trc!'.s ·!.train • trve for concrete in figure 4.1, having a maximum Mres' of 0.567 }~kat the ultJJnate r.tin of 0.0035. In ligure 4.25 I) 11•
the concrete wain at the end of the parabolic section the di~tancc !'rom th e neutral nxis to Mrain ec2 depth of the neutral axis
k1 J.1.r
the mean concrete stres~ depth to the centroid ol the ~tress hlock.
a) To determine the mean concrete stress, k1 om the strain diagram II - ---:
0.0035 acfore
0
92
Reinforced concrete design
Figure 4.25 Section in bending With a rectangular parabolic stress block
d
0 .567(,_
0.0035
b
neutral
-
Centroid of stress block
A-
• • Section
Sub~tituling II' =
for 0.571.\
For the
Strains
-"~
Stress Block
- 0.002
~ tre&~
hlot·k area or 'tress hlock r
aren pqrs - area rst X
Tim~.
using the area
,
0.567j~~x -
"' =
propcrlic~
of a parahol.1 "' \hown in figure 4.26. we have
0.567f..,.11·/3 t
Sub,tituting for '' from equation 4.43 !.!I"C'
J.. ,
= 0459},
(4.44)*
(b) To determine the depth of the centroid k2 x /.. ~ ~~ tktcnnincd lor a rectangular 'cellon by tal-..111g area moment!> of the stress blou about th~: m:utral axis see figun.:' 4.25 nnd 4.::!6. Thu-;
(.1 - /.. 1x)
nren pqrs x r/2 area r:-.t x 11-j ~ area of strl.!s~ block (0.567.f". ..1).1/2 - (0.567./~ ... 11'/l)ll-/4
=----
-
Figure 4.26 ProperliN of a parabola Area~:
w
'-
Pos1Uon of centroids: a1
s..
Analysis of the section Suh,liluting for (
t.
·)
\ - "2·'
u
from equation 4.-+3
rQ.)_- --.,0.571 ~] = 0.567k 1fcL·r x l L
I _ 0.168[., kt
= I _ 0.268/cL = 0.4! 6 0.459/..L
(4.45)"'
Once we 1-nO\\ the properties of the stress block. the magmtude and position of the e'ultnlll compres-.h e force in the concrete can he determtned. and hence the moment of e-.istance of the -.ection calculated u.~ing procedures stmtlar to tho:-l' for the rcctangulnr ,tress block. Comparison oJ' the rectangular-puraholic and the rectangulllr WC\1> hloc~s provrde~ (4.43)
(tl
Stress re!\u ltant. Fer rcctangular-paraholic: k1bx::::::: 0.459/;.kh.l rectangular: 0.567 f.·k x 0.8bx::::: 0.454/~kbx
(til
Lever arm.:. recwngu lar parabolic: d
k~.\
::::: tf - 0.416.1
rectangular: d - ~ x 0.8.1 = d- 0.4lh \o hoth \lrc'' hlocks ha\e almost the same moment Cll rc\i\tancc. I,., ' •. \howrng Itt\
lil'quate to u'e the ''mpler rectangular ... rres~ bloc!- for de\lgn culculatintl\.
4.44)*
4.10
' 'bloc!-
Triangular stress block
The triangular !'ltre'1> hlocl- applies to ela,tic condition' during the ),ervtceahiiH) limit ,t,Hc. In practice it i' nllt generally used in design calcultHion), cxcept for lrqllldetaming :-.tructure1>. or ror the calculations of cruel- width' :tnd dcflectiorh a' dc1-.crihed n chupter 6. With the triangular strc1-.s h loc~. the cro~s·M!Ct ton can he con,idcrcd as (t ) cradcd in the tcn,ion 1.onc, or (i t)
uncrackcd with the concrete re~isting a .~mn ll tlll\Ount or tension.
4.10.1
Cracked section
\ crad.cu \Cction i~ \hown in figure 4.27 with a stre)>s resultant /·,1 at'ting through the of the )>\eel and /·_. acting through the centrotd of the triangular \II\!\~ block. r:or cquilihrium of the ),ection
(I) Analysis of a specified section The t.lcpth of the ncutrnl axis, x. can he determ ined by converJ ing the section into nn ·equivalcnl' area concrclc as shown in fi gure ~.2X. where O'c £,/ Ec. the modulur
or
rnlio. b
Figure 4.28 f.qUJv~len t trc~mlorml'd
sect1on with the concr~tl' cr.1cked
d
h
Transformed • C,A, • rr..A steel arta f.
(,tl..ing the area momcnh about the upper edge: ,I
}JAI )
2.:::\
Then.: fore
or
Sofvtng thi~ quaurallc equation gives
(4A8)* f-.quation 4.48 may be \ohed U\tng a chart 'uch n~ the one ~hown in figure ~.29. Equattons ~.46 to ~.48 can be u~cd to analy).c a ~pectficd reinforced concrete section.
Analysis of the section a.A.'Ibd
Figure 4.29 Neutral-axis depths tor cracked rectangular sections elastic behaviour
0 30
-+
~
~
0 20
0.10
mto an mndular 0
0.2
0.4
0.6
x/d
(ii) Design of steel area, As, with stresses fn and fcc specified 1 e depth of the neutral axi~ can abo be expres~cu in tcnm. ollh~.: ~tr~uns and sucsses ol tl ~ concrete anu \!eel. I rom the ltnear Mrain distribution of figure 4.27: ~--·
\
/.;./£,
I = ~ - .t;.f '~" -1 f," £, J1,~. refore
1 ; -J.., --
"
oJ,.,.
Ec uations 4.47 and 4.49 may be used to design the area of tcn:-.ton \ tccl required, at a cified ~'>Ires::.. in order to resist a given moment.
, EX AMPLE 4. 12
Analysis of a cracked section using a triangular stress block
or the 1\ection ~how n in figure 4J O, delenninc the concrete and srcel slrc\scs caused hy
4.48)*
momcnl of 120 kN m, asl\uming a cracked section. TaJ.. c J:: .j t., 15 X 147() 0.16 'OO > 460 l '111g Ihe chart of figure 4.29 or equation .! 4X gi\'C\ ' - 197 mm. hom equation 4.47
.! 29.
e ...ection.
\(
~b,\ fc, ( d
9
nt
I..
15.
-
b- 300
Figure 4.30 Analysis example with triangular stress block 0
N .,..
•
j)
.c:
3H2S
'
1470 mmz
•••
L----l -
'
96
Reinforced concrete design
therefore 1:!0
10"
X
= 21 X 3000 X
197
X
fcc ( 46() 197) 3
therefore 10..1 N/mm 2
fcc
From equation 4.46
J~1A, = ~II.\ .t~~ therefore 300
10.3
I
197 '( -:;- x 1470
><
= 207 N/mm·,
4.10.2 Triangular stress block - uncracked section The concrete may be con~idered to resi~t a :-.mall amount of tension. In thts ca'>c a ten'>Jic stre% resultant Fc1 acts through the centroid of the triangular stress hloek 111 the tension Lone as ~hO\\ n in figure 4.31 Fnr equil ihrium of the section
F"
(4.50)
F,1 -+ F,t
= 0.5bx J,~ F" = 0.5b(ll r =A, f.,
I ,,
''here and
rl.(t
I
Taking moment!. about F._, the moment of resistance nf the section is given by
,
~ (It
M
(4.51 }
The depth ol the neutral axi~. '· can he determined by taking area moments about the upper edge /\A of the equivalent concrete section ~h(m n tn figure 4.32, :-uch that L(A1)
Figure 4.32 Equivalent transformed section witil the concrete uncracked
d -
Transformed= fA steel area E.
=M
Therefore v · -
u
X
11/ 2 + OcA,
X
d
bll ~ o. A, h +2ncn/
(4.52) "'
2 I 2ncr A,/bit
r n
bh
!-rom the linear
proportion~
of the strain diagram in figure 4J I:
.\'
fl,
= 1-1 -
r
d
\'
"
C',,
£,.
j,,
J~
(4.53 )•
·XC',,
\
a~
Thtrcfore
X C'ct
'tres<,
£
strain:
Cl
.\
=, - - x j"
(4 54)
I - ,\
d- \
l •
),1
"-
= - - x nJ.1 ,\
1-: <:nee tf the maximum lcn~ilc :-.tra•n or stre~s
1s spccihed. it is po:-.sihle to calculate the c •rre:-.ponthng concrete comprcs~ivc und steel tensile srre~l>CS from cquntinns U4. !'he equations denved can be u:.ed to analyse ~~ g1ven CI'Oi>s-scction in order to .:tt•rminc the moment ol rc~i~t:HlCC or the uncrad.cd 1-ection.
( EX AMPLE 4 .13
Analysis of an uncracked section lw the section shown in figure 4.30, calculate Lhc serviceability momcm of rc~istancc \\ ith nn cracking or the concrete, given ,k, - 3 Nlmm ~ . He - 30 I.Nimm' and l:.'s 200 I.Nimm ~ .
,.
A, bh 1470
300
"·
520
£, !:.",;
200
- 30
6.67
0 .()()94
97
98
Reinforced concrete design
,l
11 + '2o..,rd 2 + 2n.,r = 520 + 1 X 6.67 =-:-~-
2+2
f, = G~
X
X 0.009-f X 460 6.67 < 0.0094
= '2?2 mm
=:) uJ..,
= (460- 272)6.67 x 3 (420- 272 )
. N/mm1 15 2
M = il~!-, (d-~)-~h(h-Jlfc,c .l ~(II 1470 X 15.2 (460X
G X
272
i
272 3
~ (520
= 8.3 + 38.7 = 47 kN m
) J(}
t))
·I>+ ~~ X 30()(520 -
272)) 10 ,,
272)
X
3
CHAPTER
5
Shear, bond and torsion CHAPTER INTRODUCTION This chapter deals with the theory and derivation of the design equations for 5hear, bond and torsion. Some of the more practic
99
100
Reinforced concrete design
5.1
Shear
Figure 5.1 represents the distributton of principal -.trc\~es ac.:ro~s the o;pan of a homogcneou~ concrete beam. The din.!ction of the princtpal comprcssi'c <,tresses takes the fom1 of an arch. '' hile the tensile \tre~ ...es have the cun e of a catenary or sm.pemled chain. Toward~ mid-span. \\here the \hear i'> lm\ andthc bending Mrc~~c., arc dominant. the direction of the !>tresses rends to be parallel 10 the beam axi'>. Ncar the support.'>. where the shearing forces are greater. the principal :.tres!>es become inclined and the greater the -;hear force the greater the angle ot mcilnation. The tensile stresses due to l)henr are liable to cau-;e diagonal cracktng of the concrete near to the support so that 1>hear reinforcement must be provided. this reinforcement i~ either in the form of (I) 'ltirrups. or (2) inclined har-. (used in conjunction with Min·ups) as shown in ligure~ 5.4 ant.l 5.5. The :.tcel stirrup~ urc al~o often referred to as link~. Figure 5.1 Pnncipal \ltl'I\C'I in d lJ(ldm
Load
Comprtmlon
D1agonal tension cracks
The concrete uc;elf can rest\1 'hear hy a combinatton of the un-cracked concrete tnthe wmpresston zone. the dowelltng action of the bendtng remforcement and aggregate mterlocJ... across ten,ion cracks but. becau-;e concrete ts weal tn ten'>ton. the shenr remforcement is destgned to restM all the tetl'ille stn~"'e' cau<;ed by the ~hear force'> E\cn \\here the l>hear force!) are smull near the centre ol ~opan of a beam a minimum amount of shear retnforcement tn the form of ltnk'> mu..,t be provtded 111 order to form a cage supporting the longttUlltnal reinforcement and to resiM any ten~ilc ~>tresses due to factors ~uch hear tlc~> ign i~ prese111cd which will be unfamiliar to t ho~e designer:-. who have been usccl to design methods lxl'>cd on previous British Standard design codes. Thi!> method b known a\ The Variable S1r111 lncfinlllion Mellwd. The use of thi~> method allows the designer to seck out economic!. in the amount of shear rcinlorcement pro\ tded. but rccogni~tng thnt any economy achieved may be at the expense of having to prO\'tde :tdditional curtailment and anchorage lengths to the lcnMon steel O\'er and above that normally requtred for re,i-.tanc:e to bending a-. dcscnbed 10 'ecuon 7.9.
5.1.1 Concrete sections that do not require design shear reinforcement The concrete o;ectiom. that do not require -.hear rcmlorccmcnt are matnl) lightly loaded floor ~lah~ and pad foundation~. Beam!'. arc genera II) more hea\ il> loaded and ha\'e a smaller cross-.,cction \o that they ncar!) ah\3)''> require ~hear remforcement. Even
Shear, bond and torsion
p:m of a rakes lbpended "lmant. upport!-.. .md the ' due to 'o that torm of O\\ n in
iSe'
1ghtly loaded beams are required to have a minimum amount of shear link!.. The only e\ceptions to this are very minor beams ~uch a!. short span, lightly loaded linteb over "im.low\ and doors. Where shear forces are small the concrete section on it'> own rna} have l>Ufficient ,hear capacity hear remforcement will usually be pro\ ided. In those l>ectiono; where Vtd ::::; VRd, then no calculated .,hear reinforcement is requ1red. The shear capacity of the concrete. VRtt c. in such situations i:. given by an empirical expres~ion :
(5.1) with a m1nlrnum value of: VRd c =
[0.035k 3 1~f~~ 1/~] hwd
(5.2)
where:
I'Rd,
the dc.,ign shear rc!tistancc of the section without
(I +\ ~10)
/..
/-;;-
1
A ~,
form a 11! due w ICU it tO
IC\tigndopCU. :> tho:-c tndard The use • shear e at the to the :nding a<:
= the area of ten.,ilc con~•dercd
Some typ1cal values ot the corresponding ~he
Even
(I'Rd,
VRo1
,fb.,d)
5.1.2 The variable strut inclination method for sections that do require shear reinforcement In order to derive the design cquutionl- the action of a reinforced concrete betlln in ~hear is reprc:-.cmcd by an analogous truss a~> shown in figure 5.2. The concrete acts u~ the top
1.,.
d
loaded have a
::::: 2.0 with d cxpre!tscd in mm
reinforcement that extend'> beyond the \CCtion he1ng by at Jca.,l a full anchorage length plus one effective depth (d) b., - the \mallc~t \\idth of the l>ection m the tensile area (mm)
.:.:regate e 'hear orces.
ll1lliiUUIU
reinforcement
tl,, <. 0.02 b..,d -
{II
!ek mthe
~hear
'
b
---1
0 Section
Figure 5.2 Assumed truss model for t variable strut Inclination method
102
Reinforced concrete design
compression member and as the diagonal comprcs~10n member!. inclined at an angle B to the horizontal. The bottom chord is the horilCmtal tension steel anu the vertical links are the transverse tension members. 11 should be noted that in this model of !lhear bcha,iour all -;hear will be resisted by the prO\ i~ion of links 1rit/tno tlirect colllribution from tlte ~hear capacity of the concrete itself The angle (} increases with the magmtudc of the ma,amum shear force on the beam and hence the compressive forces m the diagonal concrete members. It is set by EC2 to h:t\C a value between 22 and 45 degrees. For mo\t cases of predominately unifonnl) distributed loading the angle 0 will be 22 degree~ but for heavy and concentrated Joads it can be higher in order to resist crul-lhing of the concrete diagonal mcmberl-.. The analysis of the truss to derive the de~ign equntions \.\ill be carried om in the fo llowing order: 1. Con),iderution of the compressive strength of the diagonal concrete strut and it1unglc 0;
2 . Calculation of the required shear reinfon:cmcnt A,w/~ for the vertical tics: 3. Calculm ion of the nuditionaltcn~ion stccl/1,1required in the bottom chord member. The following notation is used 111 the equations for the
~hear
de1-ign
= the cross-~ectional area of the two lege, of the lin~ 1 = the ~pacing of the linb
A\w
the lever arm between the upper anu lower chord members of the analogou:- tru'~ the dcl.ign yield strength of the link reinforcement
J;\\oJ 1~~
VEd V~: 1
v..
,j
VKu , VKd
rnu~
= =
the the the the
charactensttc strength of the link rctnforcement shear force due to the actionl-1 m the ultimate limit :-.tate ultimate l>hcar force at the face ol the suppon \hear force in the link
the shear rcsiMance of the ltnk\ the maximum design value com.:rcte strut
or the
shear which can be resisted by the
( 1) The diagonal compressive strut and the angle 0
The shear force applied to the sectton must be limite<.! so that excessive compressive ~trcssc:-. do not occur in the diagonal compressive struts, lea<.ling to compressive failure of the concrete. Thus the maximum design shear force VK.J. 111"' is limitcu hy the ultimate cm~hing !>trength of the diagonal concrete member tn U1e analogou~ trus), and its vertical component. With reference to figure 5.2, the effective cross sectional area of concrete acting a~ the diagonal ~trut il> taken a' b.., x ::co~o and the de!.ign concrete \trc~s.f.:.J f,k/ 1.5. fhe ultimate '>trength of the srnu
= ultimate dc~1gn \tres~ x cros\-sectional area = lf.l/1.5) x (b" x ::cos B)
and its venical component
- lf.l/1.5)" (b.. x :-cos B)) x sinO
!>o
that
v,td mu•
i.kb .... : coso sine1t.5
Shear, bond and torsion t.
angle 0
which by conversion of the trigometrical functions can also b~.: expressed as
~~·links
}~kb.. :
• -.hear
1.5(COLO-'- tan 0)
rbution
e beam EC2 to lformly I oads it
In EC2 this equation ts modified by the indus ton of a Hrengllt redu('tivn factor (r 1) for concrete cracked m shear. Thus VRu
• m
the
Jnd its
m.t~
J.~b.. ;:l't
= 1.5(cot0+tan8
)
where the strength reduction factor takes the value of putting ;: - 0.9d. equation 5.3 becomes
0.9t!
VRd. mn~ = -
X
(S. 3) 1• 1
0.6( I - j~~ /250) and.
b.. X 0.6( I - /.;~/250K~ l .5(cot8 + tan B)
0.:16bwrl( I - fck/250)fc~
(cotB+tanB)
111emher.
(S.4)*
and to c n ~ur~.: that there is no crushi11g of the diagona l compres~ivc 1-trut: (5.5)
~
of the
Thi:-. must he checked for the maximum value of ~hear on the beam, whid1 i~ u~ually taken !I)> the 'hear force. \'tt· at the face of the beam·~ 'upport~ \O that A~
pre\iou,ly noted EC2 hmitr. B to a 'aluc het\\Cen 22 and 45
(i) With 0
degree~.
22 degrees (this is the usual case for uniformly distributed loads)
1-rom equation 5.4: VRd , ma\(21J -
0. l24b~~od( l - fck/'250)fck
(5.6) *
It v~" mu\r!l , < v, r then a larger value of the angle() mu~t be u~ed ~o thm the diagonal concrete Mrut ha5 a larger vertical component to balance Vcd·
(ii) With
e
45 degrees (the maximum value of 0 as allowed by EC2)
From equation 5.4: ressive
• latlure •a mate enicnl ct ng as
1.5.
area
VRJ
mnx !4.~1 - 0. IBb,.d( I - ./~k/250lt~~
{5.7)*
which is the upper limit on the compres~ive strength of the concrete diagonal member in the analogous tnt).~. When \1~;1 > VRd. 013 ~t~ 5 ,, from equation 5.7 the diagonal ~trut wi ll be over ~tre~scd and the beam's dimensions must be mcrensed or a higher clas~ of concrete be u~ed. (iii) With 9 between 22 degrees and 45 degrees
The requtred value for 0 can be obtained by equating Vc.t to VMcl•• ,, and solvang for(} in equation 5.4 a~ follows: \' "' -
= sine X cos e - 0.5 sin 20 (sec proof in the Appendix)
therefore b} substitution
B=O.Ssin-'{
V1-.J
o.l8b"d( 1 -
fc~o./
..,
_so}f.;l
} S 4S
(5.8a)*
which alternatively can be expressed as: 0 = 0.5 sin-' { V VEr
}
< 45'
(5.8b)
Rd mu,l-l'il
where Vr:1 is the shear force at the face of the ~upport und the culculutcd vulue or the angle(} can then he used to uctcrmim: col 0 and cak:ulate the !<>hear reinforcement A~w/s from equation 5.9 hclow (\\hen 12 < 0 <- 45 ). (2) The vertical shear reinforcement A~ previously noted, all ~hear will be resisted hy tiK· provision of linl--1. with 110 direCt co11tri/.Jurion .fmm rlze 5I! ear cupacity c~( the COIIcrl'ft' it.1e(( U'ing the method of !>ections 11 can he seen that. at \ection X-X in figure 5.2. the force in the vertical link member
n·,",) mu~t equal the shear force (\'Ed). that i' \ '"d = Vt:.J - /w.
=115 = 0.87f1 ~A"'
If the hnks are spaced at a distance proporllonatel) and is given hy v,..~
1
apan. then the force
111
each link is reduced
-s- = o.87/ykA,. .
:;cot 8
or Vwu
= VE,t
= O.R7 ! l.1'sw ~fvl cot 0 J
0.87 A,w 0.9t/}~k COl 8
s
thus rearranging A,"'
I'EJ 0.78d/yk cot8
(5.9)"'
EC2 spec1fies a minimum value for A\v./:. such that
A'"
rnrn
0. 08fckO.S b,. f)~
(5. 10)*
Equation 5.9 can be used to determine the amount and '>pacing of the shear links and Will depend on the value Of (} USed Ill the de!o.ig_n. f-Or mo!-.t Ca\CS Of beams With
Shear, bond and torsion predominately unifonnJy distributed loads the angle B will be 22 degrees with cot 0 2.5. Otherwise the value for 0 can be calculated from equation 5.8. EC2 al~o ~pecifies that. for beams with predominately uniformly distributed loads. the design shear force VEd need not be checked at a distance less than d from the face of the ~upport hut the shear reinforcement calculated must be continued w the support. Equation 5.9 can be rearranged to give the shear restMancc \' Rd , of a gi\en arrangement of hnks Aw. / :.. Thus:
{5.11 )*
(3) Additional longitudinal force When using this rneU10d of shear design it i~ necessary to allow for the additional longitudinal force in the tcn::;ion steel cuused by the shear VEd· This longitudinnl tensile force t::.F1d i:- caused by the horizontal component requ ired to bulance the comprer-sive force in the inclined concrete strut. Resolving forces hol"iz.ontnlly in the section YY shown in figure 5.2. the longitudinal component of the force in the compressive ~lrul is given hy
fCI
Longitudinal force
= (Vr:.d / ~in 0) x cos() -
' '•J
cot 0
It •~ a'~umcd that half of tht~ force is carried by the reinforcement 111 the ten~ion wne of the beam then the additional tcll'.tle force to be pro\ided in Lhc tcn~llc 1one ts gi\en by
J./·111
• th
0.5\'htCOtO
(512 )
ro pnmdc for tlu~ longitudinal forl:c, at any crO\S-~>ecuon ll i~ necc,,al) to pro\ tde longlludtnnl reinforcement additional to thut required at that ~ecuon to re,i'it bending. In practice. increasmg the cunailmcnt length-. of the bottom-face tcn~ion reinforcement cun U\Uttlly prO\ tde the required force. Thi~> reinforcement provide~ hen<.ling re~iMancc in \Cction~ of lugh &ngging bending moment and then. "hen no longer reqUtrcd 10 rest<.! bending. can provtde the add itional tensile force to re1-1st shear in those sections away from mid ~ran nnd townrds the supports where ~agging bending moment~ reduce hut shcur forces increase. Tim, is discussed further and illuqruted in section 7.9 (Anchorage and curtailment of reinforcement) The lolul force given hy Mt•d/:. -1- !:::.F1d shou lclnm be taken us greater thnn MHcl, 11111 ,/<., where Mr,l,""'' is the ma>.imum hogging or mnximum sagging moment along the beam. Equntions 5.3 to 5. 12 can he Ul'ed together to design a i>ection for ~hear with n value of (I chosen by the de~igner within limits of 22 anu -+5 degreef. as sped lied in EC2. I· rom these equn ti on~ it i~ obvious that the steel ratio (A," /~) IS a function nf the inver~e or cot (J (equation 5.9), the ma>.imum ~hear force governed by diagonal compression failure is a function ot the inver~e of (cot 0 +tan 0) Cequmion 5.4) and the additional longitudinal tensile force. j.f 1cJ varies with cot 0 (equation 5.1 J). Figure 5.3 ~how~ the variation of the~c funl:tion~ from which it can be ~ecn that a~(}~~ reduced le'' ~henr reinforcement IS required, but thi.., is compen<;nted for by an increa'e in the ncce..,<;ary longitudtnal re111forcemcnl. At 'alue<., of 0 greater or les~ than 45 the o,hcar capacny of the secuon. based on compressive failure in the d1agonal ~trut1-. i.., alo,o reduced. 1 he de~1gner should calculate the -.alue of 0 but. a-. previously ~tated. for practical reasons I-.C2 places a lower and upper limit of 1.0 and 2.5 re~pectively on the \alue of em 0. This corresponds to limiting B ro 45 and 22 respectively.
105
106
Reinforced concrete design 2.0
Figure 5.3 Variation of VRd , •., ~;,d and A"" fS
1.5 c
g
...c
1.0
~
0.5
10
20
30
40
so
60
70
80
90
Angle (0)
Summary of the design procedure with vertical/inks 1. Calculate the ultimate design shear forces \lEd nlnng the beam's span. 2. Check the crushing strcnglh VRd.mo\ of the concrete diugonal strut at the l;CCtion of maximum shear, usually \11-1 at the face of the beam's support. For most cases the angle of inclination of the ~lrut i~ 0 2:! . with cot()= 2.5 and tan f) = 0.4 so that from cqumion 5.4
0.36b.,.d( I VRII ma\
=
/..k/250)J;k
(cot()+ tan())
and if \ 'Rd .max ?: Vn with 8 22 and cotO 2.5 then go directly to ~tep (3). However. if VRd m.L\ < V1 r then 0 22 and therefore B mu)>t he calculated from equauon 5.8 as:
0 5 -;in 1{ \ 'EI } .. 0. 18b.,.d(l ./..d250)/~~
()
<
45
.
Tf thi~ calculation give~ a value of B greater than 45 then the beam should be rc-"ized or a higher cla"1-. of concrete cou ld he u'cd.
3. The 1.hcar linb required can be calculated from equation 5.9 A"'
\lr:d
s
0. 7Rdf)'k cot (I
where Asw is rhe cross-sectional nrca of the legs of the links (2 x rrd>2/4 for single stirrups). For a predominately uniformly dislributcd load the \hear VE.t.J should be enlculnted at a di~tance d from the face of the suppoll ami the ~hear reinforcement should continue to the face of the support. The shear re~i!>tance for the links actually specified is
A"'
Vmm-- X
s
0.78df.,k cot 0
.
and this value \\ill be used together wnh the ~he:1r force envelope to detenninc the cunailment position of each "et of dc~igned hnk\ 4. Calculate the mmimum linh reqUired by EC2 from
A"' ·'"'"
0.08{.,~ h..
05
s
i)k
Shear, bond and torsion 5. Calculate the additional longitudinal tensile force caused by the shear
= 0.5 \l~..d cot 0
J.f,d
Thi~ addillonal tensile force can usually be allowed for by increa~ing the curtailment length of the ten!>ion bars as described in section 7.9. Example., illustrating the design of shear reinforcement for a beam arc given in Chapter 7.
( EXAMPLE 5. 1
Shear resistance of a beam
:tJon of
I
The beam in figure 5.4 spans 8.0mcLres on 300 mm wide :-.uppons. It is requin:d to ~uppon a uniformly distributed ultimate load, ll'u of 200 kN/m. The c hara~.:tcrist i c material strengths arc hk 30 N/mm 2 for the eontrcte and ./y~ 500 N/mm • for the steel. Check if the shear reinf<>rcement in the fonn of the vert ical links shown can supron, in shear, the given ultimate load.
= 2.5 •
H~ups
b: 350
at 17S spcg ...
[dII [1 11111 1 fO ~: 1
p (3).
' lrom
Hl2
Sectron 2H2S. A,
982mmz
MJid be Total ultimate load on beam Support renetion Sheur. V1 1 ut face of ~upport
= 200 "
8.0 1600 kN = 1600/ 2 HOO kN
- ROO - 200 X 0.3/ 2 770 kl"\ Shear. VC
1. Check the cru~hing strength VRd. mnxOf tlw concrete diagonal :-.trul at the face of the hcam!) ~uppmt. ulaled ,hould
From equation 5.6 with B= 22 '' vl{d
!n:t\ f !~J- 0.124h,.c/(1
0.124
X
.fc~ /250lfc~
350
X
- 7-l5 kN ( < From equation 5.7 with 0 IIRdmJ~ -1~1
650( J - 30/ 250)30
liEf
= 770 I.N)
45
0.18b"d( l-/d../ 250}kk = O.JK ) 350
X
650( 1
= 108 1 kN ( > VH Therefore: 22 < fJ < 45 .
30/ 250)30
770J..N)
Figure 5.4 Beam wrth stirrups
1 0/
108
Reinforced concrete design
2. Determme angle B From equation 5.8(a) B=0.5sin-'{
\IEf
0.18b_.d( I - f.:k 250}f.k
} <45
-
or aJternatively from equation 5.8(b) IJ = 0.5sin
'{v
VEt
Rtl
From which wt 0 3. Determine
~hear
}
=0.5stn '{
OM\{4~ '
2.39 and tan 0
s
} =:!2.7 108 1
0.42.
resistance of the linl-..s
The cross-1.cctional arca ""'' of a 12mm httr the link and a :.pacing of 175 mm ;\~II' _
°
77
11 3 mm 2• Thus for the two legs of
2 X 11 3 _ 1. 29 175
(or alternatively the value could have been obtained from Lahlc A4 in the Appendix) From equation 5.1 I the shear resi~tancc , VRJ , of the link~ il> given by \ 1Rd,,
A,w = - x 0.78l({)k cot() .\
= 1.29 X 0.78 > 650
500
><
:!.39
10
1
781 1-..N
Therefore shear rc-.istam:c of lin!-.-. 781 kN. Oe,ign shear. VEt! diMancc d from the face of the -;upport - 640k\ (< 78 1 k.N). Therefore. the beam can 'upport. in 11hcar. the ultimate load of 200 k'\/m. 4. Additional longitudinal ten:.ile force in the ten,ion '>tccl
It is nece:.sary to checlo.. that the bottom tension steel hu-, a suffictcnt length or curtailment and anchornge to resist the additional hori1.ontal tcn,ion oF1" cuused by the design shear. These addi tional tension forces arc calcu lutcd from equation 5. I2. Therefore D.Fcd
0.5Vtzd cot() X 640 X 2J9
0.5
765 kN
This force is added to the MEd/: dingrom. us dc!,crihed in section 7.9, to ensure there is sufficient curtailment of the ten<.ion reinf'mcemcnt and it~ anchorngc bono length at the suppons, as described 111 section 5.2.
l~--------------------------------------~) 5.1.3
Bent-up bars
To restst sheanng forces. longintdmal tension bars rna) be bent up near to the support• a~ shown in figure 5.5. The bent-up bar~ and the concrete 111 compress1on are considereo to act as an analogous lantce girder and the \hear resistance of the bars is determined b. taking a section X- X through the girder.
Shear, bond and torsion Anchorage length
X
Figure 5.5 Bent up bars
~I
-' s =0.9cl(cot u
~
+ cot 0)
(a) Single System
F' of
l/ZZZ?J l
the
(b) Multiple System
hom the geometry of part (a) of figure 5.5. the ~pacing of the bent-up bars is:
O.lJd(COl n t cot 0)
.1
and at the '>Cctaon X X the <.hear rel>i'itancc of a single bent-up bar (\l"d ) mw.t equal the '>hear force (I 1.a).
where A,.. i' the cross-.,ectional area of the bent-up bar. ror a multaplc 'Y'tem of bent-up bar~. a' in part (bJ of ligurc '\.5. the shear rc~i,tum:c " ancrca,cd proporttonately to the ~pacing, ~. Hence: 1
of
Va d
·~d
rom
. . ().!l7fvkA,w Sill 0
.
0. 9d(cmn
X --
cot 0)
s
or
-A,w = 0. 7Rrl/yk (col Vt-.a --:-:--:-·'' n + col 0}sin n urc nd
_)
(5.13)
Thi~ equation ic; ana logou~ to equation (5.9) for the shear rc!.istanc:e of shcua· lin i-s. In a !.imi lur way it can be ~hown that, bused on cru~hing of the concrete in the compressive ~truls, the analogous equation to (5.4) is given by:
(cot 0 + cotn ) VRdma~ < 0. 36h,.d ( l - fcl./250l.fck ;< ( I ' (J) ·
I cot·
-
(5.14)
und the additional tensile force to be provided by the provi!.ion of additional tension \lee I j.., gaven by a modtficd 'crsion of equation 5.12:
•11., ~ered
""by
~Frd
= 0.5va.J(cotfJ
cotn )
(5.15 )
EC2 abo require., that the max1mum longitudinal spacmg of bent-up ban, as hmJted to 0.6d( I + cot o ) and '>pecifies that at least 50 per cent or the rcquared shear reinforcement should be in the form of shear link'>.
109
11 0
Reinforced concrete design
5.1.4
Shear between the web and flange of a flanged section
The provision of shear links to resist vertical ~hear in a nangcd beam i., identical to that previously de cribed for a rectangular section. on the as!>umption that the web carries aJJ of the vertical shear and that the web Width. b..,. is used as the minimum width of the \Cction in the relevant calculations. Longitudinal complemental) shear stresses also occur in a nangcd !>ection along the interface between the weh and flange as shown in tigurc 5.6. Thi~ i~ allowed for b) providing transverse reinforcement over the width of 1he flange on the a:.sumption that thi~ reinforcement actl> a!> tic!> combined \\ ith compre.,s1ve stnlls in the concrete. It i., necesl>ary to check the posJ.ibility or fai lure by cxce. .sive compressive stresses in the ~truts and to provide sufficient steel area to prevent tensile fai lure in the ties. The variable strut inclination method is used in a similar munncr to that for the design to resist vertical shear in a beam described in 5CClion 5. 1.2. The design is divided into the following ~tngcs: 1. Caleul::~te lhe longitudintll design shear stres1.cs, ,.,~ ut the web-flange interface.
The longitudinal she:lf stresses arc ut a maximum in the region~ of the maximum changes in bending stresses that, in turn. occur ut the stccrest part<; of the bending moment diagram. These occur al the length~ up to the maximum hogging moment nver the supports and at the length~ awuy from the zero ~agging moments in the ~pan of the beam. rhe change in the longitudinal force .J.Fd in the llangc OUtl.tand at a section i' obtained from .J.Fd
= (d
.J.M X bto h, / 2) b,
b1 = the effective breadth of the flange
where
bro = the breadth of the OLII'>tand of the flange b.., the breadth of the weh h1
and
6M
the
thicknc),~
b.,. )/2
(b1
of the flange
the change in moment over 1he distance
~~
Therefore 6M
---...,.X (d hr/2)
(br - b-.)/2 hr
Figure 5.6 Shear between flange and web
•h
Shear, bond and torsion
that all the the r by
that i~
I
the The l.to
The longitudinal <;hear stress. I'Ed· at the vertical section between the outstand of the Hange and the web is caused by the change in the longitudinal force, ..:':J.Fd. which occur., over the di.,tance ~'· ~o that (5.16)
The maximum value allowed for ...lx is half the dtstance between the \ection with tcro moment and that where maximum moment occurs. Where point loads occur ..:':!..' should not exceed the distance between the loads. If ~'t::J is less than or equal to 40 per cent of the design tensile crad.ing \trength of the concrete. j~ 1J, t.e. I'F~ ~ OAfcw O..+Jc,,j 1.5 = 0.2?fc,~. then no ),hear reinforcement is required and proceed directly to step 4. 2. Ched.. the shear strel>ses in the inclined stmt
As before, the angle 0 for the inclination of the concrete strut is restricted to :J lower and upper value and EC2 recommends that. in this case:
:e.
num
2o.s · , e, _ 45 ·
bg
3R.6 ::;: 01
nent
the
ID
::;
i.e 2.0 > cot 0, ~ I .0 for fl ange~ in compre~~iu n i.e 1.25 > cot 01 > 1.0 for flanges in tcn~>ion.
45 '
To prevent crushing of the concrete in the compressive struts the longitudinal shear \Ires~ i~ limited to:
is l't:d
I ' Lf~L
<
- 1.5(cot o,
-
(S 17 )
tan e,)
where the ),treng.th reduction factor,., =0.6( l - }d/250). The lower value of the angle 0 i~ fiN tried and if the :\hear -;tre~se~ are too high the angle J\ calculated from the following equation:
e
0.5 \in
t{
0.1( I
3. Calculate the trunsversc shear reinforcement required The required transver~e reinforcement per unit length, A,1/.~ 1 • may he calculated lrom the equtltion:
A,, >
V[d h t
1r -
O.R?f~k cot Or
which is derived by considering the tensile force 4. The reqUirements of tran~vcr~c ~teel.
(5. 18) 111
each tic.
EC2 require' that the area of transverse steel should he the greater of (a) that given by equation 5. 18 or (b) half that given by equation 5.1 R plw. the area of steel required by transverlle bending of the flange. The minimum amount of tranwerse steel required in the flange is 2 1\, nun 0.26bdt.ifc~m/J).-.. (> 0.0013bdr) mm /m. where h - 1000 mm (),Ce table 6.8). Example 7.5 (p. 184) illustrates the approach to calculating reinforcement in flanged beam~.
tran~ver~e
shear
111
11 2
Reinforced concrete design
5.2
Anchorage bond
The reinforcing bar subject to direct ten'>ion 'ihown in figure 5.7 mu'>t be firmly ancll· • if it is not to be pulled out of the concrete. Bar.. suhject to forces induced by flexure rr. be ~imilarly anchored to develop their de),ign \tresses. The anchorage depend<; on bond between the bar and rhe concrete, the area of contact and \\ hethcr or not the b;.; located in a region where good bond condition'> can be expected. Let:
=
/b r
basic required anchorage length to prevent pull out bar size or nominal diUmeter
.ft>J = ultimate anchorage bond stress
.r:.
= the direct tenllilc or compres~ivc strC!.ll in the bar.
Figure 5.7 Anchorage bond
.I
Considering the forces on the bar: Tensile pull-out force - cross-sectional area of bar
direct '>treso.,
= m:rf'
-4 '
anchorage force
= comact area
.tnchorage hond '>trcll.,
- (lt> ryd7r
therefore
hence
k.P
= 'ifbtJ when h =/yd•
/h,rqd
and the de~ign yteld strength of the rcinrorccmcnt (=/yk/ 1.15) tl-nnchoruge length is given by /h.rtjd
= (¢/ 4) ([l;k/ 1.15]/!hd)
/h .rqd
= (/yl/-+.6/t>tJ)¢
(5.19)•
Basic anchorage length Equation 5.19 may be used to determine the /)(me anchorage lmgrlt of harl. which arc either 111 tension or compre:.sion. For the calculation of anchorage length~. design ,·alucs of ultimate anchorage bond stresses are specified according to whether the bond conditions are good or otherwise.
Shear, bond and torsion Figure 5.8 Definition of good and poor bond conditions
direction of concreting
t "chored -·e muc;t 'on the l~e har is
~
45°
h> 2SOmm
It< 250 mm
h> 600 mm
I
Good bond condition) for all bars
11
Good bond conditions in unhatchPd zone Poor bond conditions 1n hatched zone
Good hontl cond iti on~ are t:onsidered to he when (a) bors are inclined at an angle of hctwccn 45 and 90 to the hori.amlal or (h) zero to 45 provided thnt in thi:-. second case additional requirements arc mel. These additional condition~ nre that bar~ Ute
1. either placed in members whofte depth in the <.lircction or wncrcttng <.Inc~ not exceed 250mm or
2. embe<.lded in members With a depth greater than 250 mm an<.! arc ctther in the lower 250 mm of the member or at least 300 mm from the top ,urfm:c when the depth exceed~ 600 mm. Thc'c condlti<>n' ure Jllu,trated in figure 5.R. When hond condition' arc poor then the be reduced by a factor of 0.7. The dc,tgn \alue nf the ultimate hood ~tres<; 1s also dependent on the bar ~itc. For all har site' (o) greuter than :12 mm the hon<.l Mre:-.s 'hould addittonally he mulliplied by n fm:tor ( IJ2 ¢)/ I 00. Tahle 5. 1 give., the <.Jc),ign value~ of ultimate hond ),trc,~c., for 'good' conchtions. Thc.~c depend on the cln..,:. of concrete and arc obtained from the equation ./t-.1 1 . 50}~ 1 ~ where .f~r~ if> the characteristic tent.ile strength or the concrete. -.~ctfied ultimate bond ~trc'i\CS 'ihould
Design anchorage length
~)
the
'I he baviC' unchorage length discussed above lll\ISt be further JllOUiticd to give the minimum design anchorage lenglh taking into account factor~ not directly covered by tahlc 5. 1. Table 5. 1
Design values of bond stresses fbd (N/mm 7)
5.19)* f,k N/mm1
Kh are \alues e bond
Bars 32 mm diameter and good bond conditions Bars 32 mm diameter and poor bond conditions
12
16
20
25
30
35
40
45
50
ss
60
1.6
2.0
2.3
2.7
3.0
3.4
3.7
4.0
4.3
4.5
4.7
1.1
1.4
1.6
1.9
2.1
2.4
2.6
2.8
3.0
3.1
3.3
114
Reinforced concrete design
Table 5.2 Value of (I o1
n2
Coefficients a
(1
allows for the effect of:
Type of anchorage
The shape of the bars
Concrete cover to the reinforcement
Reinforcement in Tension
Compression
Straight
1.0
1.0
Other than straight
0.7 if
1.0
Straight
- 0. 15 ( c~
-Ol/o 0 7 and <:. 1.0
1.0
1 - 0.15(cu - 3o)to < 1.0
1.0
1
but Other than straight
but > 0.7 and OJ
Confinement of transverse reinforcement not welded to Lhe main reinforcement
All types of reinforcement
Confinement of transverse reinforcement welded to the main reinforcement
All types, position and sizes of reinforcement
Confinement by transverse pressure
All types of reinforcement
NOll': tht' product
"J >< "J •
O)
K>. but
>
0.7 and
1.0 ~
1 .0
0.7
but
0.7
0.04p 0.7 and • 1.0
should be greater than or equal to 0.7
The required minimum anchorage length (/~><~) " given hy
ft..t
r1t. n~. o,. o~ . o~/h ""11\, ,,.4/A, rrn'
where A,
(5.20
= area of reinforcemenl required nod 1"0\ idcd at that section o ( 1 to 5) = set of coeflicients a~ given in Table 5.2 n:,1•
A,. 1110,
In Table 5.2:
c0 K
-
concrete cover coefficient us ~hown in figure 5.9 values as shown in figure 5. 10
)., = (L:A,,- l:A,I,min)/A, LA" = the cross-sectional nrcn of the trnn!>vcrsc rcinfon:cment along the design anchorage length l:A,~.mon A~
= the
cross-sectional nrea of the minimum tnmsvcr~c rcinforccmcm ( = 0.25A, for beams nnd Lero for ~labs)
= the area of a single anchored bar with maximum har diameter
Thi!-. minimum design length must not be for tension bars:
le!>~
than:
0.3/b.rqd
for compression bars:
0.6/tuq.J
In both ca~es the mmimum value mu~t also exceed both lO bnr diameters and 100 mm. Anchorage~ may also be prm i<.lc<.l by hook' or bends in the remforcemem. Hooks and bends are con-;idere<.l adequate form' of anchorage to the main reinforcement if they
Shear, bond and torsion
115
Figure 5.9 Values of Cd for beams and slabs (see table 5.2)
Bent or hooked bars
Straight bars '• .. min (o/2, c,, c)
A,
looped bars Ceo= c
'• =min (o/2. c,)
o,A.,
A,
~A,.
A,
o,A,,
Figure 5.10 Values of K lor beams and slabs (see table 5.2)
o' K
0.1
Ka 0.05
Figure 5.11
-
:
Straight bar
•
Equivalent anchorage length~ for bends and hooks
90" < u < 1so• + r I
ft
'
Bend
loop Minimum Intern~ I radius of a hook, bend or loop = 2o or 3.5\'l for Q >16 mm
;n
~m is l'y the ITI1n1111UI11 uime n ~ i on s ~how n in figure 5. 1J. Bends and h ook~ ure IIOl recommended l'or u~e a!> compression anchorages. In the ense the hook~> and bend&
or
shown in ligure 5. II the anchorage length (shown m. h,.~q) which is l'quiva lcnl to that required by the straight bur can be simply calcula1ed from the ex pression: /b eq n 1/b,rqd where n 1 is Iuken us 0.7 or I .0 depending on I he cover co nd iti o n ~ (">ee table 5.2). The internal diameter of any bent bar (rcfemd to as the mandrel size) is limited to avoid damage to the bar when bending. For bar~ less than or equal to 16mm diameter the mternal diameter of any bend should be a minimum of 4 time~ the har dtametcr. For larger bar '>t7es the limJt as 7 umes Lhe bar diameter. To give a general idea of the full anchorage lengths required forf.:k 30 N/nun1 and 2 f..l 500 N/mm • with bar diameters, ¢ < 32 mm. /b ~ can vary between 25 bar diameter' (25¢) and 52 bar dJameten, (52o). depending on good and poor bond condi tion~>. and the value of the coefficiems Ct from rable 5.2.
=
116
Reinforced concrete design
(
EXAMPLE 5. 2 Calculations of anchorage length
Determine the anchorage length required for the top reinforcement of 25mm bar~> in the beam at its jum:lion with the external column as shown in figure 5. 12. The reinforcing bars are in ten~ioo resisting u hogging moment. The characteristic material ~trcngths are .fcL = 30 N/mm 2 and / ) k 500 N/mm~. Figure 5.12
Anchorage for a beam framing into an end column
H25 bars
100 = 4Q
~vespan
Assuming there is a con~Lruc t ion joint in the column ju&l above the beam :~nd, as the bars arc in the top of the hcum. from ligurc 5.8 the bond conditions arc poor nnd from table 5.1 the ultimate anchorage hond -.tress i' 2.1 N/mm ' . A~ the bar~ :~re hem mto the column and the concrete co,er coefficient. <'d (figure 5.9) is equivalent to 4o. which i' greater than 36, from table 5.2 cocfticiento 1 i-, 0.7. Abo from tahlc 5.2, codticicnl n' - I 0. 15ktt Jr11)/
/. l ) 0
- ( 4.6/bd
Cl jll'
= 0.7 x () 85 ( = 31
25 =
500
t:.
-
4.\) X 1_, I
77~
) ri1 = 31 ~'>
mm.
Sec ahn tahlc A.6 m the \ppcndix lor tahulated \aluc' of anchorage length'>.
l~------------------------------------~) 5.3
laps in reinforcement
Lapping of remforccmcnt is often necessary to transfer Lhe fo rce:. from one bar to another. Laps between bars should be ~taggcred and should not occur in regions of high stress. The length of the lap should be ba~ed on the minimum anchorage length moditied to take into account factors \Uch as cover. etc. The lap length /0 required 1~ g1ven hy
fu
ftvqJ X 0:1 X 01 X Cl~ X (l~ X f\(1
(5.21 )*
where e~ 1 • n:!, o,, and ns, arc obtained from table 5.2. n 6 (!1 1/25) .5 (with an upper and lower limit of 1.5 and 1.0 respectively) and p 1 ts the percentage of reinforcement lapped \\ithin 0.65/0 from the centre of the lap length hcing considered. Values of 0:0 can be comcniemly taken from table 5.3. 0
Shear, bond and torsion Ta ble 5.3
Values of the coefficient a 6
Percentage of lapped bars relative to the tolal cross-sectional area of bars at the section being considered
<25%
33%
50%
>50%
1.15
1.4
1.5
lntermedoate values mc~y be onterpolated from the table
Figure 5.13
·Jl11
Transverse reinforcement for lapped bars
m:i:,~~~ ~~J
14
(a) tension lap -~~ .
1, .1s the
1,/3
~from ~e
r
-
5.9)
- Al\0
=!.85.
(b) compresston lap
Notwithl-llandtng the abmc rcquirements. the absolute lllllltnlum lap length~~ gi\cn us
_
__,)
ne bar to .. of high modified _··en by 5.21 )"'
.m upper -cement ~ .. of n6
In mlu
0.3nr,ll, r 1ol > 15 diameter~ , 200 mm 1:.
(5.22)
'l runsvcrse rcinl'orccmcnt mu~t be providcd around lap), un lcs~ the hipped bars nrc less thon 20 mm tl iumctcr or there is lc~.), than 25 per cent lappcd bar)\. In thcsc case~ minimum trun~ver~e reinforcement provided J'or other purposes )>Uch a~ shcar links will he atlcyuate. Otherwi!>c transvcr~e reinforccmcm mu~t be prm 1ded. a:-, 1\hown in figure 5.13, having a total area of not less than the urea of one -.pliccc.l har. The arrangement of lapped bar~ must al~o confom1 to figure 5. 14. 'I he clear ~pace bet11 ecn lapped ha~ 'hould not he greater than ole;) or 50mm other11 i!>c an additional lap length e4ual to the clear <.pace must be provided. In the ca~e ol atljm.:ent lap~ the dear dtstancc hctween adJacent ha~ ~hould not be greater than 2o or 20 mm. The longirutlinal dl\tance between Lwo adjacent lap~ should he greater than 0. ""' If allthe!-.c condttion' arc compiled '' Hh then I00% of all tension bars 111 one layer at any o;ecuon may be lapped. mhemi'>e. where bar' arc in \everallayers. th1~ llgure \lmuld he reduced to sorr.. In the CU\C of compress tOn !.!eel. up to I0011- of the reinforcement at a ~ectton may be lapped.
11
118
Reinforced concrete design ~0.31.
Figure 5.14 Lapp1ng of adjilcent bars
,~
r--
'·
__..
<40or 50 mm
5.4
Analysis of section subject to torsional moments
5.4.1
Development of torsional equations
Torsional moments produce shear stresse~ thion reinforcement in the form of closed lin~ mU'.t hi! pro.. ided to rest\t the lullwn.ional moment. The cquauon!- for tor.;iorlJI destgn arc developed from a !>tructural model where it b :Mumed that the concrete beam m tor~ron beha1c-. in a <;rmilar fashiOn to a thin \\:tiled ht)\ .,cction. The box'' rdnl'orced wrth longitudmal bar:- 111 each corner v.itJl clo~edloop ... urrup' a-. tran-.1er~e tcn,ron uc-. and the concrde prm iding dmgonal cornprCl.loiOn .,trut,. It i., a..wmed that the concrete cannot provrde any tcn,tle resistance. EC2 gi\'C!> the princtple' .llld .,orne lunucd design equations for a gcnerali~ed .,hape of •1 hollow bo\ .. cction. In thi' se1.tron ot the text we\\ til develop the cquauons that can be used lor the dcsrgn and analy~>i' or a typic<~! solid nr hollow rcctungle hox section. Consider ligure 5.16a. The applied tmquc (Tt.r) at the far end or the section produce~ a .1hearjlnll' (q) around the perimeter of the box ~cction m the ncar end of the diagram. rhe sheur now is the product ol' the <;hc
T === 2A~q wher·c Ak ts the area enclosed within the centre \me of the hollow box section, hence (5.23) Figure 5.15 Tor.1onal reinforcement
l
Shear, bond and torsion
119
Figure 5.16 Structural model for torsion
(a) Compression slruts I
Torsional shear Ioree e'~es
·~ng
Tension In longitudinal steel
r.1cks Triangle of
·om
!.'JOg ~
•\ ill •mg
car
h
•
the
[
u
.
~
:[~, ,<'~. , qh
qh
L t
hcotU
is
o~.lled
for~Ps
(b) Forces actfng on whole body (one face shown representative of all four faces)
h
•
...
(c) Forces actang on one lace of the sectaon
l 10p
""'ton ~of
.
~be "\.
uccs
A~ q i'> the ~hrur force per unit length of the circumference of rhe bo\ :.l'ctton. the forr£' produced hy the !>hear llow is the product of q unci the circumference {u~) of the area A~. lienee. if it u<;:.umed thai this force i'> resi .. tcd hy the trus~ action of the ~:oncrerc compressive ~ trul\ acting m an angle. 0. together with tension in the longitudinal .-.tccl, from figure 5. 16b the force {F,) in the longitudinal tension Meclts given by
mm.
cos() l> in ()
(/Ilk
IIow ,hear
(jill.
tan()
Tu~
2/\k ton
B
(5.24)
The required urea of longi tudinul tcnl'ion steel lo rcsi .. t torsion (/\, 1 ), acting at its dc:.ign slrength ({y~/ 1.15), IS 1herel'orc given by ·nee
A,Jyt~
lit~
1.15
2A1. tan()
Tu1. cot 0 2Ak
(5.25)
In the ::tbO\C equation tJ1c torque. T, is the max11nurn that can he re~i,ted by the longirudinal rcinfnrcernent and i~ therefore equivalent to the de-;ign ultunare tor'>tonal moment. TCd· lienee A,J~ tl
1.15
(5.26)
The required cross-secrional area of torsional link, can be determ1ned b) considering one tace of the hox section m• .-.hown in figure 5.16c. If ir is assumed that the urea of one
120
Reinforced concrete design leg of a link (Asw) is acting at it)> design yield strength([.,~ / 1.15) the force in one link is given by
AsJ;i../ 1.15 = q
X
h
H1mc'"er if the linh arc spaced at a distance proportionately and i~ given by
A,.J,l I s · =qxtx--
h cot 0
1. 15
~
apart the force
10
each link b reduced
ql cOL ()
---
T&t-"
(5.27)*
2A• cot B
Equations 5.26 and 5.27 can be used w de~ign a section to res1st torsion and an example of their u~e i~ given in chapter 7. The calculated amount oJ reinforcement must he provided in addition to the full bending and ),hear reinforcement requirements for the ultimate load comhinalions corresponding to the tor:.ionul lofld cnse considered. Where longitudinal bending reinforcement is required the ndditional torsional steel nrea ma) either be provitll:d by increasing the size of the bars. or by additionul bnrs. Torsional l ink~ must consist of tully anchored clo~>cd links spaced longitudinally no more than 11~ /H apart. The longitudinal steel must con~i'>t of at lea'>! nne har in each corner of the :-.cction with other bar5 di\trihutcd around the 111ncr periphery of the links nt no mor~ than 350 mm centres. Where the reinforcement I!) known equation~ 5.26 and 5.27 can lx rearranged for :malysis purpo).es to gi\c TEd and 0 a-; follows:
rh,
A.,..
2A~o. ( --0.87/,~. \
.
/\,1 IlL
.
o.87J,,~.
5
[,~.
11
(5.28
•
• = (" '" )I (As-;;;:f)'"
tan· (}
) ~
I
(5.29
)
The use of ullthe above equations a~:-.t1mcs that the seclion i!> replaced by an equivalcn hoiiO\\ bo\ section. To determine the thickness of the !'>Cction an equivalent thickne~ Urtl 1-; used. detinet! a~ equul to the towl area of the cro!\)o-l'>CCtion dtvided hy the oute circumfen:ncc of the ~ection. In the case of un aduul hollow section the cro~:.-~cctio aren woulcl include uny inner hollow a rea~ and the culcu latcd thicknel.s should not be taken u~ grcnter than the actual wollthickncs!'l. In no ca~c :-.hould the thickness be take'" as lc:-.s than twice the cover 10 the longitudinal hm~. When analysing or dc)oigning a ~ectinn it is abo necess~1ry to check that excess"" cnmpresl>ive stresses do nol occur in the diagonnl compre~sivc strut!>, leading possibly t compre~sJve failure of the concrete. With reference to figun: 5. 16c and taking th.. lim1ting torsional moment for strut comprc~sive l"adure as '/~J 11111, : rOI'CI!
in ~trut
= (q X h )/~i n(;J
An•a of strut = r~ 1 x (II co~ 9)
Stri!H in
~trut =
rorcc; Area
= --. CB/ J,~.; l5 let Sill C0\11 II -
"here f.:~o. IS the characteristic comprc-.si\ e stre~ in the concrete. As q then the above equauon can be e\pre~scd a-.
~ma.•/(2AI.)
tcr sin0cos8 -
,1.
/ 1.5
= TRd ma,I(:!J\
Shear, bond and torsion K'
Link is
or TRd "''"
$ 1.33}ckldAksin8cos8)
which can ai\O be expressed as reduced
/Rdm~' <. l.33f,~tcr\~ /( cot0-tan8 )
(5.30)
In EC2 th1s equation j, modified b} the inclusion of a rtrength reduction facwr (1· 1) to gi\C
5 27)* and an
e,t mu't , for the Where ca may r~ional
e thnn tr of the more - l311 be
TMtl
m." < 1.331'lfcltcrAl/(cot8 +tan 0)
(5.31 )*
where the strength reduction factor takes the value of 0.6( I - f~l /250 ). In using the above C(JU on the value~ of(} that can he u~ed and EC2 recommends that 1.0 < cOLO ~ 2.5 rcpre~enting limiting values of 0 of -15 and 22 respectively. The arproach to design for torsion is therefore: (a)
Ba~ed on the calculated ultimate tor~ional moment (7'1c~). t:heck the maximum tor"onalmomentthat can be carried hy the ~ection (TRJ 111,,.) whit:h ~~ govemed hy compre~\ion in the concrete struts. w, given hy equation 5. 31:
7j ... <. /'~,l. ona' - 1.331·1f~ ~~~rA~ /(cot 0 t tan H)
: .28)* (h) Calculate the tor-.ional rcmforcemem requ1red from equation 5.17·
A,. , 1
~~here A " I'> the area
5.29)* (l:)
Jlent
~ection
not be
of one lt•g of a hnk.
Calculme the additmnal longitudinal reinforcemcnt (t\, 1) from equation 5.26: 1\, 1
\.'l.nes~
c outer
1,,1'(2. \~0 87}>k cot B)
('f',. 1 u~ / 2Ad cot 0/( 0.87{> 1d
Further infonnation on the pmctieal details of design fo1 tor~ion and a de11ign example arc given in charter 7.
>e taken ce~sivc
'1hly to • ng the
5.4.2
Torsion in complex shapes
A ~et:tion consi~ting of a T. Lor I :,hapc should be divided into component rectangles and each component h. then de~igncd separately to C
T,
lj Jk, (limon 31lma,),
(5.32 )•
L (Ahmon'lima\}
''here limon and limo\ arc the minimum and maximum dimcn'>lon of each ~ection. K i-. the 5t Venant'1 torlimwl cml\tallt that varies according to the ratio hm."/ llm111 : typical values of '' hich arc 'hOI\ n 111 table SA. The subdh ision of a ~hupc into Its component rectangles should be done in order to maximise the stiiTne~s expres~ion
L (Kh
111 ,,
~"""") .
121
122
Reinforced concrete design Table 5.4 St Venant's torsional constant K hmax hmon
K
hm••'hmon
K
1.0 1.2 1.5 2.0 2.5
0.14 0.17 0.20 0.23 0.25
3.0 4.0 5.0 10.0 >10
0.26 0.28 0.29 0.31 0.33
5.4.3 Torsion combined with bending and shear stresses Torsion is ~>eldom present alone, ttnd in most practh;a l cases will be combined ' shear and henuing ~lre~ses.
(a) Shear stresses Diagonal cracking will start on th.: ~iuc face ol a member where torsion and ' arc addi11ve. Pigurc 5.17 .,how' a typical ultimate torsion and ultimate ' intcrat.:tion diagram from wh1ch 11 can he ~ecn that the hcmn'., resistance to comh shear and torsion i' lc~s than that -when \UbJCCI to either effect alone. The effe~t comhined hear stres~ and tor~10nal ~trc ...s may therefore need to be considered a th1' ca~e both shear and tor.,ion are calculated on the ha.,i:-. of the <>arne equivalent "ailed <,ecuon pre\ iou:-.1} de~cnbed for tor\! onat de\lgn. The recommended ...rmpllfied approach to de,1gn i' to cnl>urc that the ultimate 'ht force (VEd l and the ultimate torl>IOnal moment (/ 1,1) ~o.ati•.f) the interaction formula ~tres.,cs
FigureS 17
Combined shear and tonton
(5. ~' ''here TRd ,..,
the dc\ign tor-.JOnttl
I'Rd ma~
the design shear
resi~tancc
resi~tancc
(equation 5.31)
(equation 5.8)
I r thi!o. imcraction equation i~ Mlti ~lied. the de~ign of the ~hear andLOr!.ionnllinks can b.. carried ()Lit separately providing th:.H the assumed angle of the compressive su·ll!s (0) ' the same for both tor:-.ionul and lihe;lr design. However for a solid rectangular sectiou, ~>uhj cctlo rcluti vcly small torsional and shea s trcl>se~. neither shear nor torsional reinlon:cmcnl i.., necessary if (5.3~
where VRd ~ is the ~hear capacity nf the concrete a:. given by equation 5. 1. TR.J.c i~ th~ torsional crad.ing momem \\hich can he calculated from equation 5.23 for a shear me, equal to the dc!>ign tens1le strc..,..,, /
T
q=21\l
where q T
shear force per u1111 length
= shear stre~' x tc~ x 2A~
:.hear ..,tress x
(It~
x I) or
Shear, bond and torsion '0
\\hen the concrete reaches its design tensile cracking strength. .f~1d
'lkd,c
=}~uJ X let -
26 28
.l<~k
- l.Sfet
2A
X
2A~
k
1.33J~~~~e~Ak
Z9 31
B
ned \>.ith
1.
(5.35)
It should abo be noted that the calculation for A"' for shear !equation 5.3) gives the required cross-sectional area or both leg~ of a link whereas equation 5.27 for tor~ion gives the required cross-sectional area of a single leg of a link. Thi~ needs to be taken mto con-;ideration when dctermming the total link requirement as 1n c'heur design (equation 5. 12) must be provided in the tension zone of the heam. whereas the udditional longitudinal reinforcement for torsion (equation 5.26) must be distributed Jround the inner periphery of the links.
(b) Bending stresses d \ hear
e 'hear mhincd ..!feel of c:J ,tnd in crll thin
~
te -,hcnr
nula -:i .J:I}*
' can be h (8) is
(5.34)
is the .tr stress
When a bending moment b pre~ent. dtagonal cracl,s will u~ually develop from the top of Aexurnl cracks. The nexural crac.:k~ them:-elve~ on ly slightly reduce the tor~ional 1-ttillness provided that the diagonal crack\ clu not dew lop. The ltnalmodc of failure "ill depend on lhc di-,tribution and quantity nt reinforcement present. Figure 5.18 shows a typical ultimate moment and ultimate torsion iutcrur.:tion curve for a ~>c~tion. As cun be seen. tor moments up to approximately RO per cent of the ultunate moment the section cun resbt the lull ultimate ltonal moment Hence no calculations tor tor~iOll OfC generally llCC.:CS~ttry for the tlltitnti!C limit Stale Of bending of reinl'orced concret~o: unless tor~ion has been iucludccl in lht.: original analysis or is required for equi lihriurn. When combined n~.:xure nnd lor~ion il> COll\tdered the longttuclinal ~lecl tor both C:l'\C:. can be determined :-.cparately. In the flexura l tension tone the longitudinal .~tet: l required for both cases can be added. llowever in the flexural compressive ,.:one no additional tor'lonul longitudinal 'lee! i'> ncces<;ary 1f the longitudinal force due to tor~ion i' le% than the concrete cnmpresstve force due 10 Ocxurc.
\
\ I 0.8M,. M, Figure 5. 18 Combined bend1ng and tor~1on
6
CHAPTER ......................................
Serviceability, durability and stability requirements CHAPTER INTRODUCTION
SeNiceability, durability and stability requirements
determination of cover lo reinfordng bars, as well as selection of suitable materials for the exposure conditions which are expected. Good construction procedures including adequate curing are also essential if reinforced concrete is to be durable. Simplified rules governing selection of cover, member dimensions and remforcement detailing are given in sections 6.1 and 6.2, while more rigorous procedures for calculation of actual deflection and crack widths are described 1n sections 6.3 to 6.5. Durability and fire resistance are discussed 10 $eCtion 6.6. The stability of a structure under accidental loading, although an ultimate limtt state analysts, will usually take the form of a check to ensure that empirical rules, designed to give a minimum reasonable resistance agatnst misuse or accident are satisfied. Like serviceability checks, this will often mvolve detailing of reinforcement and not affect the total quantity provided. Slclbilily requirements are discussed in section 6.7 and considered more fu lly for seismic effects In section 6.8.
6.1
,.
...... d for
11
"cal be and The :on of
Detailing requirements
Thc\C reqturemcntl> cn\urc that a ~rructurc hal> 'att~factory uwahiltty and 'cnu.:cuhility performan ce under normal circumstances. EC2 recommend~ \tmplc rule~ conccrmng the concrete llll\ and cm.cr to reinforcement, minimum memhc1 dm1cn'IOil'-. and hmlt' to n:infon.:emem quantilic!-., \pacmgs and bar diameter!> \\ hil:h mu\t be tuJ..en into account at the member \t.!ing and remforcement dctathng \tagc. In \OillC ca~e' tabulated value' arc pronded for I) p1cal common ca~c)>. "h1ch are ba,cd on more complc\ formulae gi\en 111 the code of practtce. Reinforcement detathng may al~o he aflcch.:d hy \tahi lity ctm~tderattOnll as tlc!>crihed in sectton 6.7. a~ well a~ rull!~ concerning anchorage and lapptng of bars which have been dbcu,~>cd in section' 5 2 and 5.3.
6. 1.1
Minimum concrete mix and cover (exposure conditions)
rhese requirement~ arc interrelated and. although not rully dcwiled in
EC'2, EN 206 Pt't.fnmtance, Prod11clirm. Plal'ing and Complirmce Criteria and the comrlememory Briti~>h SIUndard BS 8500 give more detailed gu idance on minimum Concre/1!
e in ~oped
s 1kely
e sure iO..Il its ~u of
and
........
comhinattons of thickness nf covel' anu mix chan.tcteris tics for variou~ clas~>cs of cxpthurc. It ~hould be noted that the UK national Annex to F.C2 (and B~ 8500) induuc 'igni licant modification~ to EC2 itself. The mixes arc CXJWC~'eu in tcnm of minimum cemen t content. maximum free water/cement ratio and corre~ponding lnwe~t concrete ~ln:ngth cl:l\S. bxpo~urc clal>sification~ are given intahlc 6.1 whtch then define~ the mix und cmcr reqUirements and so on "hich must be complied \\ith. Con~r to re111forcement 1s specified and ~hown on drawing~ a~ a nominal \:tlue. This ~~ obtaincu from i'""m -
~cdc,
"here ~cd.- b an allowance for construction dcnation<, and b normally taken a~ I 0 mm except where an appro\'ed qualtry control \)'Mcm on cover (e.g. 1111>1tu meawrcmenL~) i' specified in which ca.\C it can be reduced lo 5 mm.
=
126
Reinforced concrete design
Table 6.1
Exposure class designation
Oass designatton XO
XC -1
-2 -3 -4 XD -1
Description
Examples of environmental conditions
No risk of corrosion - Very dry
Unreinforced concrete (no freeze/thaw, abrasion or chemical attack) Reinforced concrete buildings with very low hum1d1ty
CarbonatiOn-induced corrosion risk Dry or permanently wet
Reinforced and prestressed concrete: - inside structures (except high hum1dity) or permanently submerged (non-aggressive water) completely buried in non-aggressive soil external surfaces (mcluding exposed to rain) exposed to alternate wetting and drying
Wet - rarely dry Moderate humidity Cyclic wet and dry Chloride-induced corrosion risk (not due Lo seawater) Moderate humidity
-2
Wet, rarely dry
-3
Cyclic wet and dry
xs -1
·2 -3 XF
·1
2 -3 -4
XA -1
-2 -3
.
Chlondc-induced corrosion risk (sea water) Exposed to airborne salt but not 1n direct water contact Perm.mently submerged Tidal, splt~sh and spray zones Freeze/thaw attack whilst wet Moderate water saturation - without dt>-icing agent Moderate water saturcll!On with de-Icing agent lligh water saturation -without de-icing agent High water saturation - with de-icing agcnL or sea water
Chemical auack Slightly aggressive Moderately aggressive Highly aggressive
Reinforced and prestressed concrete: exposed to airborne chlorides, bridge parts away from direct spray containing de-icing agents, occasional/slight chloride exposure - totally immersed in water containing chlorides (swimming pools, industrial waters) exposed to de-•cing salts and spray (bridges and adjacent structures, pavements, car parks) Reinforced and prewessed concrete: external 1n coastal areas - remaming saturated (e.g. below mid-tide level) - in upper tidal, splash and spray zones Concrete surfaces exposed to freezing: vertical exposed to rau1 vertical (road structures) exposed to de-Icing agents as spray or run -off horitontal exposed to ra in or water accumulation - horizontal exposed to de-icing agents directly or as spray or run -off. Others subject Lo frequent splashing
- Defined in specialist literature
Serviceability, durability and stability requirements
or
ter)
a 1ay
12
The cover is necesr.ary to provide 1. c;afe tran!>fcr of bond forces: 2. adequate durabilit)< 3. lire reSIStance. The value of r mon \hould not be le~s than the bar diameter (or cqui\ alent diameter of bundled bar:.) to ensure satisfactory bond performance. Tile value of <"nun lO cnwre adequate durability i'> inOuenced by the exposure das<;ificauon, mtx charactcri:.lb and intended design life of the structure. Thi~ i1> a potentially complex proces!> since there may commonly be a combination of expo·wre cla;;scs rduting to nuack ot the reinforcement. whil5t freeze/thaw and chemical auad, apply to the concrete rather tiMn the steel. The range of relevanr mix parameters include maximum water/cement ratio. minimum cement content, cement type. aggregate size and air-entrainment. Tuhle 6.2 show!' typical combinations of cover and mix dctui ls for commonly occurring situations. If any of the parameters of cement type. uggn:gatc ,;,c or de1.>ign life change, then adjuMmcnts will be nece.~sary. Design ~ hou l d he hu:-.cd on the most severe exposure clu~sifkation if more than one nre combined. Minimum com:rcle mix requirement:. f'or ca~>es where freeze/thaw apply are summnriseu in luble ().3, wh ilst exposure to chemical attack (class XA exposure) may place fun her limit), nn mix dctalls nnd mny nl))o require additional protective mea:-.urc),. Reference should he mode to the appropriate documentation in !>uch cases (e.g. BS R.'i00).
and
Jtlon
I or ~t
Table 6.2 Cover to reinforcement ~50-year des1gn life, Portland cement concrete with 20mm maximur aggregate size) [Based on UK Nationa Annex]
Exposure class
Nomma/ Cover (mm)
xo XCl XC2 XC3/4
Not recommended for reinforced concrete
25 35 45
35 40
---~
1
XDI XD2 XD3
45 50 2
35
35
40 1
40 45 1
45'
XS2 XS3 Maximum free water/cement ratio Minimum cement (kg/m 1) Lowest concrete
50
0.70
0.65
0.60
45
1
0.55
35 1
35 40 50 1
35 40 50 40 40
60 1
401 551
so 1
45 1
1
40 2 60 2
45 40 55 1
0.45
0.35
0.35
XSl 2
30
45
0.50
55
240
260
280
300
320
340
360
380
C20/25
C25/30
C28/35
C32/40
C35/45
C40/50
C45/55
CS0/60
Notes. l. C~mt>nt contt>nt should be mneased by 20 kgtm 1 abolll' thE' values shown in the tdble. 2 Cement contem should be Increased by 40 kg/m 1 AND wat4'f 'E'ment rauo reduced by 0.05 compdred woth the value$ shown in I he table.
Ctnerol Notes These v,llues may be reduced by S mm 1f an approved quality control sy~tcm os spcclfoed . Cover ~hould not be less than the bar diameter + 10 mm to ensure adequate bond perform<~ nee.
128
Reinforced concrete design
Table 6.3 Minimum concrete mix requirements for concrete exposed to freeze/thaw (Exposure Class XF) - 20mm aggregates Closs
Minimum member dimensions and cover (fire resistance)
In order that u reinforced concrete member is capable of withstanding tire for :1 specified pcriou or time. it is necessary to cn~ure the provbion of minimum dimensions and cover (here delincu as nominal minimum concrete !illrfncc to muin har axis dimension as illustrated m ligure 6.1) for various types of concrete mcmher. In many practical sli\Hltiono;, with modest fire resistance pcrioc.l\. the cover provided for durability wtll govern the design. Structural fire design in coni-.idcrcd h> part 1-2 of l:.C2. wh1ch gi-.cs several po~sible methods rang1ng from detailed calculations to 'impltried tabk:-. as presented here. Hre effect\ are con,idered further m ~ection 6.6.2 and dc).1gn 1~ based on ~ati1>fying load heanng (R). 1ntegrit} (E). and msulatmg (II performance as appropnate. !'he approach offers the destgner permts~lhlc comhtnallon~ ol member dimcn~iun and a-.:1' dt,tance a" indicated 111 Tables 6.-t to 6.C1. for a range of \tandard lire resi~Lunce period' (minutes). These '' 111 generally appl) ''hen nonnul deta11tng ntleo;, have been follm' cd und ''hen moment redt,tribuuon doc:-. not C\cecd 15q f-or benms nnd !.labs.
Table 6.4 Minimum dimensions and ax1s distance for RC beams for fire resistance Minimum dimensions (mm)
Standard fire n•slstance
Possible combinations of a and bm1n where a is the overage axis distance and brn 111 is the width of the beam Cotllinuou~
Simply supported
A
R60 R90 R120 R240
bll111l
a bmon a b,n•na bm~n-
120 40 150 55
200 65 280 90
B 160 35 200 45 240 60 350 80
c 200 30 300 40 300
D 300 25 400 35 500
55
50
500
700 70
75
c
r
120 25 150 35 200 45 280
200 12 250
75
G
H
450 35 650 60
500 30 700 50
25
300 35 500 60
Note: The oiXt) do~tdnce o,1 from the sidt' of a beam to the comer bar should be o+ IOmm except whtrl.' bn,., tsgreater lhdn the values in
cofumm C and f
Serviceability, durability and stability requirements
1~
further detailing requiremenb may appl} for higher fire periods. \\htl<;t effecti\e length and nxial load (relative to design capacity) may need 10 be specili<.:ally considered for columns. Table 6.5
Minimum dimensions and axis distance for RC slabs for fire resistance
Standard fire resistance
Minimum dimensions (mm) One-way sponmng
/1, -
REI 60
80 20 100 30 120 40 175 65
a-
11,
REI 90
a-
REI 120
hi
REI 240
oh, = a-
vet
Two-way spanning ly I.
<
--1.5 < ly, l. 80 15 100 20 120 25 175 50
7.5
80 10 100 15 120 20 175 40
Ribs in two-way spanning ribbed slob
2.0
bmln-
obman -
a=
b,,,ll ... abmn a-
100 25 120 35 160 45 450 70
120 15 160 25 190 40 700 60
20C 10 250 15 300 30
Notrs:
hie I 1re
1ad o~nd
I. The! sl"b thlckne~~ 11. 1S the 111111 ol the slab th1cknes~ ond tht' thlckncs~ ol ''"Y non·combultlbh• lloo11ng 2. In two-way siJbs the axis ref€'rs to the lower I,Jyfr of re1nlorcrmfnt 3. Th€' term 'two-way Sldbs' rcl<~l~s to slabs supported .11 cJIIfour edgPs, if this •s not the Cd5e they shmold be lteo~ttd .u one-wJy sp.mmng 4 For two-wc~y r bb od llabs. (a) Th~ axis d11l.10CP measuretlto the iatpr,ll surfctce oftht' rib should be •ltleu~t to 10J. (b) The values ~pply whPre thtorc h predomli1cH1tly unitormty clhtrllluted loatllllq (c) There \hO\Jid llc at le~\t olll' rrstrained t> (d) The top remfor<.cment should be pi.Krd in lht' upper h'll of the ll.snge
Table 6.6
Min1mum dimensions and axis d1stance for RC columns and walls ror fire res1stance
Standard fire rc1lstance
Minimum dimem1ons (mm)
--
thicknes~/oxis
Column width bmnlaxis distance, a, of the main bars
Wol/
Columns exposed on more than one side
Wall exposed on one side
Columns exposrd on one side
distance, a, of the main bars Wall exposed sldel
011
R60
250/46 350/40
155/25
130/10
140/10
R90
350/53 450/40 2
155/25
140/25
170/25
R120
350/571 450/51 2
175/35
160/35
220/35
R240
600/70
295170
270/60
350/60
Nore1. 1 Based
on the
ro~ho
conservat&v~>ly t~ken
two
of the deSJgn cJxtcJI lo.1d under lin: cond1Uom to thl! dPs1gn reSJStanct> at normal tempt>rature cond•llons as 0 7
2. Mmlmum of 8 bars reqwed
130
Reinforced concrete design
6.1.3
Maximum spacing of reinforcement
Crad..ing of a concrete member can result from the effect of loading or can an~ hccau~e of re~tmint to \hrinkage or thermal mo\l.:ment. In addition to providing minimum area of bonded reinforcement (see sectton 6.1.5). cracking due to loading minimised b} ensuring that the maxtmum clear 'pacing-. hetween longirudm~ reinforcing hars in beams i<. limited to thm gi\Cll in table 6.7. Thi., \\ill en~ure th..t the ma\imum crack \\ idths m the concrete do not exceed 0.3 mm. It can he seen that li c :-pacing depend~ on the •arcss in the reinforcement which should be taken as the !.trc,, under the aclion of the qua.11-pemwnem loadings. The qumi·perma11ellf loading i~ tal-.el' a' the permanent load. Gk. plu~ a propmtion of the variahle load. Qk, depending on the t) pe of su-ucture. The calculation or the f>tn.:.,!-> level (j~) can be complicated and an acceptable approximalion is to take l as
.f..
(6.1
I'm orlicc and domestic ~ituations (sec fable 2.4 for other circum&tances), where .r;k j, the characteristic )>trcngth of' the reinl'orccment. h will hove a value of 1.0 unle' mnml.!nt reubtribution hm. been earned out. in whil.'h ca~e t i~ the ratio of the distribute.. moment to the undi~tributcd moment at th~: ~ection at the ultimate Iunit. Table 6.7 Maximum clear bar spacings (mm) for high bond bars 1n tension caused predom~nantly by loading
Steel stress (N/mm 1)
160 200 240 280 320
360
Maximum bar spacing (mm)
-----
300 250 200 150 100 50
Thcl'te ~racing rule~> do not apply to ~ l 11hs with u11 overall thickness of200 mm or lc"' In this cu!.c the srmcing of longitudinal rcinforcemem '>hould be no greater than thrt:o; times the ovcrull ~luh dcpth or . wo mm, whichever is the le!>ser, and secondar: reinforcement three-and-a-half time' the ucpth m 450 mm gt:nerally. In arens o concentrated loadi't or maximum mo111cnh these shou ld he rl!duccd to 2h < 250111111 al1' 311 400 mm respectively.
6.1.4
Minimum spacing of reinforcement
To permit concrete llO\\. around reinforcement during con . . truction. the clear di~tan~.-~ between bar~ should not be less than (t) the maximum bar siLc. (ii) 20mm. or (iii) the maximum aggregate size plu~ 5 mm. \\h1chever ts the greater figure.
Serviceability, durability and stability requirements
6. 1.5
~on
the and an
6. 1
Minimum areas of reinforcement
For most purpo~es. thennal and shrinkage cracking can be controlled \~ithin acceptable mit<; by the use of minimum reinforcement quantities. The princtpal requirements, to "e checked at the detailing stage. are as specified in table 6.8. Requirement-; for \hear nks are gtven in sectiOns 7.3 and 9.3. In addttton to the requuements of table 6.8 a tmntmum \tecl area, \, mon· mu!'tt he " ovided tn all ca es to control cracking. The proviston of the minimum \!eel area .. n,ures that the remforcement does nor yield when the concrete in the ten~ion Llme .. mcks with a 1oudden transfer of stress LO the reinforcement Tim, could caUM! the tncontrolled development of a few wide cracks. Whenever thi'l minimum area is provided, then yield should not occur and crad.ing will then he diMributed throughout the !.Cction with a greater number of cracks but of lc%er width. A, min i~ given by the c\rn.:ssion J..~J..fct.el l Act/.fyk
A,,""" \\ here A,.m 111
(6.2)
minimum area of reinforcement that must be provided within the tensile zone Ac~ - area of concrete within tensile 70ne det1ned a., that area which is in ten ion JUSl before the initiation of the fir..,t crack .f,1 ctt ten<.tle strength of concrete at ttme of cracking \\'llh a 'ugg~"tcd minimum of 3 /mm 2 otherwi!.e obtained from tahle 6.11 u~;tng a concrete ~trcngth cla\s appropriate to the anticipated time of cracking J.., wess dt~tributton coeffictenl ( 1.0 for pure tenston. 0.4 for llc\ttre) non-linear ~tress dt!>tnbutton coeffictent leading to a rcduclton 111 restrattll force 1.0 for web less than 300 nun deep or 11ongcl> lcs~ than 100 mm \\ide 0.65 for webs greater than ROO nun deep or flange' grcatcr than ROO mm '' ide (intcrpohttc for imermeduue values) -
Table 6.8 Minimum areas of reinforcement Tension reinforcem ent In beams and slabs 0.26 fctm
(
0.00131
fyk
Secondary reinforcement
Concrete claS$ (fyh
-----------------C25/30 (30!35 0.0013
0.00 15
> 20% main reinforcement
Longitudinal reinforcement in columns A '"'" 0 1ON..., 0 87 f>~ 0 002Ac where N!d is the axial compression force Vertical reinforcement in walls A, m11• 0 002-A. Note· b, es the ml'dn wodth of the tenseon zone.
500N/mm 2 ) C40/50
CS0/60
0.0018
0.0021
13
132
Reinforced concrete design
6.1 .6
Maximum areas of reinforcement
The~e
are determined largely from th~ pracllcal need to achic\e adequate compaction of the concrete around the reinforcement. The limns specified arc a.-. follows
(a) For a
~lab
or beam, ten'>ion or compressiOn reinforcement
IOOA, A, < -l per cent other than at lap.., (b) For a column IOOA,/ Ac ::; 4 per cent other than at laps and 8 per cent at laps (c)
For a wall. verticnl reinforcement lOOA, j A, ::; 4 per cent
6.1.7
Maximum bar size
Sl·ction 6.1.3 dc\crihed the.: limitations on har \ ptlcing to cn:-.un.: that crack widths due to loading nrc h.epl within acceptable limi t\, When cnnsidcring lmu..l-induccd cracking bar diameters may be rc~ t rit'tcd a' indicated 111 tahlc 6.9 which is ha~ed on ('30/.n concrete nnd 25mm cover as an altcmativc to limiting -;pac1ng. Jn calculatmg the ~!ee l stre~s. the approximation given in equation 6. 1 may he U\ed. Table 6.9
Maximum bar diameters (0.3 mm crack width)
Steel stress (Nimm')
160 200 240 280 320 360 400 450
Mox1mum bar lize (mm)
32
----
15
16 12 10
8 6 5
When cracldng occurs as u re~ult of re:-.truint to ~ hrinh.nge or thcrmul effects then the hur si;ws nutst be limited us indicated in table 6.9. but the maximum ~pucing lim it~ of tnbk 6.7 do not need to be applied. l'he \Ice! ~tre!>s to he used in table 6.9 can be cnlculated from equation 6.3 where A, P''" is the steel area pmvidcd at the ~cc ti on under con~iderntion and A, nun i5. given in equation 6.2. (6.3 )
6.1.8
Side face and surface reinforcement in beams
In beams over I m deep addiuonal remforcemcnt mu ... t he prO\ 1ded in the 'Ide faces to control crad.ing as indicated 111 figure 6.21al Thts reinforcement ~hould be distnbuted evenly between the main tension steel and the neutral a\.t\ and within the \tirrups. The
Serviceability, durability and stability requirements I
Figure 6.2 Side-lace and surface re1nlorcement
I.
N/A
t(d - x)
< 600mm
Side-face reinforceml'nt (a)
rcmforcemcnt (b)
m111imum area of this reinforcement can be cakuluted from equation ()_2 with k taken as 0.5. In assessing the maximum spacing nnd ~itc of lhi:. reinforcement from tables 6.7 and 6.9 a -.tress -.alue equal to one half of that calculated for the main tenc;tle reinforcement may be used and it mny he a11sumcd that the side face reinforcement is in pure tension. In addauon to the ahove requirement, EC:? requires that surface reinforcement ~~ provided where it is necessary LO control spalling of the concrete due to tire (axis distance> 70 mm) or where hundled hun, or har~ greuter than -10 nun diameter are u~cd U\ main rcanforccment. In the Ul\, however. thi'> " not adopted due to practical dirtkullie~ in providing such relJ1fnrccmenl. !·or high covers il i~ n:conunencled that udditional fire protection i' provided and crack wiuth calculations are recommended with largl! daametcr har~. The sua·face reinforcement, if provided, :.hou ld consist of' welded mesh or small diumeter high bond bars located 0111\iclt• the lin"-' as indicated in lagurc 6.2(b). Cover to thi' reinforcement mu't comply\\ llh the requirements of )o.ectton 6. 1.1 and the mimmum area of longitudiawl ~urfacc reinforcement should be I per ecnt the area ol the cnncrete ouhide the link.!> and in the tension zone below the neutral axb: shO\\ n as the ~l1.1ded nre.t in figure 6.2(b). The ..urface reanfon.:cment bar~ 'hould he ~puccd no further than 150 mm ~tpnrt und if properly nm:hored can he taken into account as longitudinal bending and shear reinforcement.
or
6.2
Span-effective depth ratios
The appearance and function of a reinforced concrete heum or skah may he impaared if the deflectaon under 'e" iccahiltty Joadang ts c.\ce~stve. Deflections can he calculated il~ indicated in ~ection 6.3 but it is more usual to control dellcctions hy placing a limit on the rutio of the spun to the effective depth of the hcam or ,Jah. EC2 specilie' equations to calculnte ha,ic span-effecta\e depth ratio~. h> comrol deflection' to n maxamum of span/250. Some typical values are given in table 6.10 for rcct::tngular sections of cluss CJ0/35 concrete and for grade 500 steel. The ratios can abo he U~>ed for Hanged sections except where the rtltl() of the width of flange to the \\idth of web exceeds 3 when the h(l~ie values shou ld be multiplied by 0.8. For two-way ~>panni ng slabs. the check for the hn~;ic -;pau effective depth ratio ~hould be based on the \horter span whereas for nat 'lab~ calculations \hould be ba~cd on the longer span. The two columns given in table 6. 10 correspond to levels concrete !'>trcs!. under "erviceability conditaons: highly Slrc-,sed when the steel ratio p exceeds 1.5 per cent and
Basic span-effective depth ratio Factor for structural system K
Concrete highly stressed (p- 1 5%)
Concrete lightly stressed (p- 0.5%)
1.0
14
20
2. End span of continuous beam or one-way continuous slab or two-way slab continuous over one long side
1.3
18
26
3. In tenor span of continuous beam or one-way or two-way spanning slab
1.5
20
30
4. Slab on columns without beams
1.2
17
24
0.4
6
8
Structural system
1. Simply supported beam or one/two-way spanning simply supported slab
(!lilt slab) based on longer span
5. Cantilever
lightly strc:.\ed when p equal:. 0.5 per cent. pi-. gtven hy IOOA, rr.: ho reinforcement rntio and concrete &trcngth it may be more convenient to use the char in rigurc 6.3 \Vhich is for a simply supported '>pan with no compression steel togcthc with a modification factor K (as -;hO\\ n 111 table 6.1 0) accord111g to member type. T1 upprouch is basl.!d on the same ha~ic equations and offer!> greatl.!r flexibility than rchnn~ • placet! on tabulated values.
Figure 6.3
Graph ot basic ~pdn effective depth ratios for different classes or concrete
N 36
v"'
.,
~
b c 0
....vc
32
.
28
~
24
~
:t: -Q 0 Zl
~
-5
20
"' "'2:
16
CL
"Q
~ 'I'
::::
c
"'
Q.
"'
100A, .,1 12 0.40%
0.80%
1.2%
1.6%
2.0%
bd
Serviceability, durability and stability requirements
:-te)
The basic rutios are modillcd in particular cases a:- follows: (a)
For spans longer than 7 m (except tlat slabs) and where it is necc1-sary to limit dellcr.;tions to ensure that finishei>, such as partitions. are not damaged. the basic values -,hould be multiplied by 7/ span.
tb) For llat ,Jabs with span-, in e'
= 3 10N/mm~
where
[yk
( EXAM PL E 6 . 1
Span-effective depth ratio A rcctungul:lr
continuou~
hL·am
or cla:,s C25/30 concrete spans 10 m. If the breadth is
300 mm, check the acr.;eptahi lity of an eJTer.:ti vc <.h:pth of 600 mm when high yield reinfon.:ement, ./y~ = 500 N/mm'. is used. At the ultimate limit :.tate it is determined that 1250 mm1 nt tension Mecl ill needed and 3 No. 25 mm diameter reinforcing b
IOOA, "'
X
600)
0.7 per cent. From table 6.1 0, tor nn intcnm 11pan K = 1.5 Ba~tc
-.pan cffecuvc depth rn11o (figure 6.3)
Thercfure for an anterior ~pan.
ba~ir.;
= 16
srxan-etTcctivc depth ratio= L.5 x 16
:!~
T<> avoid damage to finishCll fllr span greater than 7 rn: Modified ratao = 24 x
1~
16.X
Ylodthcation for :.tccl area provaded: 1470 - 19.8 1250 . depth ratao . provJ'dcd = I 0 X I () l Span c1..lecuvc 600 Modified rmio
16.8 x
= 1,1. 7 t.
which tl> lcl>S than the allowable upper limiL thus deflection requirements nrc likely to be sati~tied.
135
136
Reinforced concrete design
6.3
Calculation of deflection
The general requirement i<. that neither the efficiency nor the appearance of a structure harmed hy the deflections that will occur during its life. Deflections must thu., considered at ntriou~ stages. The limitation~ neces ·ar) to satisfy the requirement' vary con.,iderably according to the nature of the structure and its loadiJ1gs, but reinforced concrete the folio"' ing ma) be considered as rea~onablc guides:
1. the final deflection of a beam. slab or cantilever should not exceed 'pan/ ::!50 2. that part of the dellection which takes place after the application of fini!>he~ or tl of partition., should not exceed span, soo to a-.osd damage to fixture~ and tutm the creep cflect~. Lateral delkction mu't not be ignored. especially on tull ~ lender structures. and limitation:-. in these case~ mu~t be Judged b) the engineer. It i~ smpot1ant to realise th. l there arc many factors which may have \ignificant effects on dcOcclion~. anti me difficult to allow for. Thus any calculated values mu-.t be regarded a' an cMtmatc onh The mo.,t unportunt of these factors arc: 1. suppon re-,traint mu\t he c:-.timatcd on the ham ol si mpllticd as~ umpti ons. whll \\ill have varying degree~> of aceur:tl.); 2. the prl'ci'e loadmg cannot he predicted and crrnrs in permanent loading may have a signsflcalll l'lfect, 3. a cr;K~cd rneml1t:1 will heha-.c differently from one that is uncracked - thi<. may be problem in lsghtly resnlnrccd mcmher~ wher1.· the v.orksng load may be ncar to th~ crac~sng 1tnu1. 4. the effects ul floor 'creed~. finishes und partition:-. urc very tlsfficult to as~>eS<> frequemly these are neglected dcl.pllc thesr ·.,uffemng' effect. It may he po~-.iblc to allow fur the'c factors hy averaging maximum and minimut estsmalcd effects and, provided that this is done. there :m.: a number of culculattor methods available \\ htch \\ill give reasonable rc.,ults. The method adopted by LC2 1 ha<.ed on the calculation of curvature oJ sections ~ ubjectcd to the nppropriatc moment' wtth allowance for creep and 'hrinl...age effect~ where necessary. Deflections arc the' calculated !rom Lhe~e curvature:.. A rigorou~ approach to dellcction is to calculate th( curvature at intervab along the ~pun .md then use numerical Integration technique' t\ estimate the critical dellections. ta~ing into account the fact that ~ome sections along tht span will be crac~cd under load and other:.. in rcgson' of lesser moment, \HII ill; uncracked. Such an upproach is rarely justified and the approach adopLed below. based on EC2, assumes that it i~ acceptahly accurate to calcu late the curvaltlre of the beam o slab ba,cd on both the cracked and uncracked ~eclion' and then l o U'\C an ·average· value in estimating the final deOection using standard deflection formulae or simpl" numerical mtegration based on clasric theor). The procedure for estimating deflcctionl> involves the following stages which :lrl slfustrated m example 6.2.
Serviceability, durability and stability requirements
137
6.3.1 Calculation of curvature Curvature under the action of the quasi-permanent load should be calcul::ued hased on both the cracked and 11/ICracked ~cction11. An estimate of an ·average' value of curvature can then lh: olnained usmg the formula:
1/r
~(1/r)~., +( I
(){ l fr)u,
(6.4)
where
lfr
average cunature
( 1/r)u, } _ (1/r)., -
value~ of curvature calculated for the uncracked Calle and cracked case respectively
coefficient g1ven by I - {1((1s,fa~l allowing for tcn1.ion stiffening loud duration factor (I for a single short-term load; 0.5 for load~ or cyclic loading) strc~1>
~tustained
in the tension ~tecl for the cracked concrete ~ection
stresf> in the tension steel calculated on the hasis of a cracked section 1111der the loadinp, thar ll'ill just cause rrar-king at the secllml heing c·nll\idered. Appropriate value\ of concrde tensile strength to he u-,cd in the cnlculation of 11,1 cnn he obtained from table 6.11. In calculaung ~. the rmio (a,,/11,) can more convcmcntly be replaced hy (Mc 1 /M) \\hen: M,., 1s the moment that \\ill ju't cau<,e cracl-ing of the '>ection and M i' the del>Jgn moment for the calculmion of curvature und deflection. In order to calculate the 'average' curvature. separate calculation' have to be earned out lor both the cracked and uncracl-cd ca'>e\.
Uncracked section fhe :h1-Um.:d clu-.tic stram and ~.trC\1-> di.,trihution for an uncraci-cd ~ec11on ts shown in tigure 6.4. For n given moment. M. and Irom elastic bending thenry. the curvature of the ~t:ctiou. ( I/ r).,,. is g1ven by M
(6.5)
where H~.cf1 i1. the ciTective clastic modulu~ of the concrete allowing for crccr effects anti ''" is the second moment of arcH nf the uncracl-ed concrete section.
Table 6. 11
Mean tensile strengths of concrete and secant modulus of elasticity
Strength class
fcrm Ecm
(N/mm 2) (kN/mm 2)
20/25
(25/30
00/37
05/45
(40/50
(45/55
CS0/60
2.2
2.6 31
2.9
3.2 34
3.5 35
3.8 36
4.1
30
33
37
138
Reinforced concrete design b
Figure 6.4 Uncracked section strain and stress distnbutiOn d
-
--I
-. . -f-
A.
Section
Strains
Stress
b
Figure 6.5 Cracked section - strain and stress distribution
x/3
neutral nxls
(d- x/3)
- -
F,,
A,
• • Equ1vaient transformed sect1on
Section
Strain
Stress
Cracked section The a'~umed ela~ttc strain and 'iti'C':- di•MihUiion tor a cracked section is ~ho\vn in figure 6.5. Thi-; IS identical to that shown in ligures -1.27 and -1.21:!. and equation 4.48 or figure -1.29 can be u:-ed to determine the neutral·a>.i' depth. Alternativcl). momenb ol area can be taken to c'tabli~h the neutral-a>;~). depth d1rectly. 'I he l>Ccond moment ol area of thc cracked section can then be determined hy laking <>econd moments of area about the neutral axi'
(6.6 where a~ is the modultlr nttio equal to the ratio of the cla!>tic modulus of the reinforcement to that of the concrete. For a given moment, M , and from clastic bending theory, the curv:uurc of the cracked section. { I / r).:,·• is therefore given by
M ( 1/rlcr=-E I
(6.7)
~. elf ~r
6.3.2
Creep and shrinkage effects
Creep The effect of creep will be to increase ddlccuons v.ith t1me and thu' 'hould be allowed for in the calculations by using an cffccthe modulus. /.:., ell· u-.ing the equatton t.~ cfl
where tt>
= Ecm/ ( I IS
- 6( ~. ro))
{6.8
a creep coefficient equal to the ratio of creep strain to initial elastic strum
Serviceability, durability and stability requirements Table 6.12
Final creep coefficient of normal weight concrete (Class C25/30)
Age at
Notional size (2fl.cfu) mm
loading (days)
100
200
300
500
Dry atmosphere (inside: 50% RH) 1
5.5
3 7
4.6 3.8 3.0 2.7
28 100+ Note:
139
A-
5.0 4.0 3.5 2.8 2.5
cross-~eclional
area ot concrete,
4.7 3.8 3.2 2.6 2.3 u
100
200
4.3 3.6 2.9 2.3 2.1
3.8 3.1 2.6 2.1 1.9
3.5 2.9 2.4 2.0 1.8
perimeter of that area exposed to drymg.
The value of rJ>. \\ hile oeing affected oy aggregate properties, mix design and curing conditions. is also governed by uge at lir~t loatling. the durm.ion of loatl and the section dunenl>ions. Table 6.12 gives some rypit:al long-term value~ of q'( XJ, to) a)< ~>ugge:-ted by EC2 for a class ('25/30 concrete made with a type N cement. The~e arc valid if the concrete i~ not subjected to a compressive ~Ires~ greater than 0.4~f...~ 1 1 ,, , at age fn (age at time of loading) and wi ll reduce as the concrd<.: strength incrc.:a~es . Equations and charts arc ~ tveo in EC2 for a range of cement type,. concrete cla,se~ . loadtng age' and nottonal member !>i7es. These equation~ includ~: the tlevelopmenL of creep with time anti .tdjustmcnts if the :.lre~s at loading exceeds thut indicated above. The ·notional ~ite' of the secttcm is taken U'\ I\\. tee the cm~s-!>ectwnal area divtded by the penmcter of the area e xpo~ed w drytng. An estimate the ela&tic moclttlus of concrete, l£,rn. can be obtained !rom taole 6.11 or from the cxpresl>ion:
or
22 [
f.:.'.:m
(!:\ ~ R)
r l
kN/mm 2
Shrinkage l'he ciTect or !>hrinkag~ nf the concrete will be to increusc the curvanm: and hence the deflection of the beam or 'ilab. The curvature due to shnnkage can be calculated using the equation
= E,,a,S/ 1
1/ t;;
(6.9)
where
L/ r.:, = rhc \hrinkage curvature
'".:, = free shrinkage strain
= fifl>t moment of area of reinforcement about the centrotd of the section I = second moment of area of section (cracked or uncracked a" appropriate)
S
lie -
300
500
Humid atmosphere (outside: 80% RH)
effective modular raho (l:. ,f Ec. cff )
Shrinl..age i~ influenced by many features of the mi\ and con~truction procedures hut for most normal weight concrete:. values of .,, may be obtatned from table 6.13. Shrinkage strains are affected by the ambiem humidity and t:lement dimensions. The total shrinkage strain can be considered a'i two component).: the drying shrinkage
3.4 2.8 2.3 1.9 1.7
3.3 2.8 2.2 1.9 1.6
140
Reinforced concrete design Table 6.13
Final shrinkage strains of normal weight concrete (10
Inside Out.side Note· A ..,
(Class C25/30
Notional size (2Ac tu) mm
Relative humidity
Location of member
6)
(%)
100
200
300
?::. 500
50 80
550 330
470 280
410 250
390 230
cross-sectional area of concrete, u
penmeter of that area expo!oed to drying
strain E:nt which develops slowly as water migrate~ through the hardened concrete and the autogenous shrinkage ~train. = cu which develop1. during hardening at early ages. Thus:
or the concrete from which the typica l long-term values in table o. l j have hecn d,:velopccl for a ciOS'I ('25/30 concrete. The total slui nknge wi ll tend to he fut lc:.s for higher strengrhs c'pccinlly a1 lower relative humidities. The ·average' shrinlage curvature can he calculmcd from equ:uion 6A having calculated the curvature based on both rhc 'cral.'kcd' and 'uncrad.cd' seclton ~. EC2 provide~> formu lae to evaluate lhese components al various ages
6.3.3 Calculation of deflection from curvature The total curvature can he determined b) adding the c;hnnkap.e curvature to the calculated cunmure due to the qua~i -permancnt load~. havmp. made aliO\Hmcc lor creep effect,. The dcflecuon of the beam or -.lab can be calculated from the IOta! cur\tic hcnthng theory which for smnll deflccuon~ 1~ b:1~ed on the cxprC'>'>ion d!,.
HI ~ tl\·-
M,
(6.10
whcre M~ ts the bending moment nt a section distun~.:e 1 rmm the origin 3); shown in ligure 6.6. l·or small deflections the term d'y/tl.t~ upproximalely equals the cu rvuturc which is the reciprocal of the radiu~ of curvalurc. Douhlc integnuion of equation 6.10 will yield un expression for deflection. This may be illu ~ t ratcd by considering the case of a pin-ended beam subjected to con.,lunl M throughout its length. ~o thttt M, - M. d2 1• El ., ·, u\-
(6.11 )
M
therefore
Figure 6.6
y
:::;
Pin-ended beam subji'Cl to a constdnl moment M
i. X
-
/
L
M
-, 8
"
Serviceability, durability and stability requirements ::25 30)
but if the ~lope ~ ~ zero at mid-span where x
L/2, then
C=- ML
2 am.l
d\· ML f1....:....-MI--
dl
2
.
Integrating agarn grves
M.1.:.
Ely
2
ML\
- -2+
D
0. y = 0.
•lit at support A whcn .t
D ncrete
Hence
()
ll\
1r a cia' ngth
M f,J
('"I2 - J..'·) 2
(6. 12)
ut any !>ection
The maxtmum dellcction in tht ~ cu~e will occur
m mid-'>pan, where
t
L/2.
in which
cac ,.ntJ\
but
=
M J2 £1 R
(6.1.3)
'"1c.:c at any uncracked section
H II
r
tne ma\imum dcllcction may be expressed a\ 6.10
Yrn l\
I ,I p, I. r
In general, the bending-moment di~tribution along a member wil l not be con~tant, hut Ill be a function or.\. The ha1.ic form of the result wi ll however he the l\ame. antlthc c:tlcction may be expres~cd as
the ca.~ I = \f
1
. mux .rmum del1cellon
a
= kL' · -I
rh
(o. I·W
6.11
k
n con~t unt , the value of which depends on the di~trihution nl bendrng moment~ in the member
L
the effective span
..!_ ,,, =-
the mid-span curvature for beams. or the support curvature for cantilevers
T)plcal valucll of k arc given in table 6.14 for varioull common shape!> of bendingment diagramll. If the loading is complex. then a value of k muM be estimated for the c :nplete load smce summing deflection" of simpler components will y1eld incorrect 1. ults.
141
142
Reinforced concrete design Table 6. 14
Typical deflection coefficients
Loading
B.M. diagram
(.M M ;;
!~d
w ....'aq-
~(1 -o)
t
~ w
~ wLl/8
Fi
-WaL
~~
~
k 0.125 4a 2 - Ba 1 1 ~a (if a- 0.5 then k 0.104
End deflection (if a - 1 then k
1--oL...
l
jltlw
wo1 L'/2
0.83)
End deflection
(if a
1 then k
at3 - a) 6
0 33)
-a(412 a) 0.25)
Although the derivation hn'> been on the bu-,h of an uncracked -;cction, the final cxprcl>sion is tn a form that will deal with u crad.cd ~ection ~imply by the :-.uh~titution of the appropriate curvature. Since the exrrel>~ ion involve11 the square of the lipan, it is important thut the true el'fccti vc span as dclincd in chapter 7 i~ used. particularly in the cu~c of cnmilevers. Dcllections of canti levers may al~o he im:rcascd by rotation of the ).Upporting member. and tim mu.,t he taken into :u.:count when the ~upporting structure 1s fairly 11cxible.
(
EXAMPLE 6.2 Calculation of deflection ~ - 30~
J ~
8,.... II
"'
II 't>
.• ....• Ll A,
5 No 25mm bars Figure 6.7
Deflection caltulalion example
E~llmatc the lung-term deflection of the he:.un ~h ow n in figun.: 6.7. It spans 9.5 metre~> nnd '·' designed to carry a unirormly di.~trihuted load giving rise to :1 qun~>i - pcrmuncm moment of 200 kNm. It i~ l:Onstructcd with class C25/30 concrete, is made of normal aggregate\ and the construction props are removed at 28 duys.
(a) Calculate curvature due to uncracked section I rom equation 6.5: ( I /r)""
M
where from table 6. Ll. Ecm - 31 1-.N/mm·. From table 6.12, a\suming loadtng at 28 day!. with indoor exposure, the creep cocflkienl ri> ~ 2.R hecau~c 2A~fu
(2 X [700
X
300])/2000
= 210
and hence from equation 6.X the effective modulus Ecctf
31/(l-2.8)
8.15k.N/mm~
i~
given by
Serviceability, durability and stability requirements lienee
(1/r)
2()()
-
uc -
8.15
X 101 X
2.86
X
X
106
(400
X
7003I 12)
10- 6 1 mm
• ote that in the above calculation lu..: has been calculated on the ba\i~ of the gross concrl.!tc \CCtional area ignoring lhe contribution of lhe remforcemcnt. A more accurate calculation could have been performed. as in example 4. 13 in chapter 4, but such accuracy i~> not JUstified and the 111mpier approach imlicatct.l will be ~ufficient l y accurate.
(b) Calculate curvature due to cracked section To calculate the curvmure of the crackct.l section the I value or the tra n~formcd concrete section must he calculated. With reference lO fi gure 6.5 the calcul:llionl> can be ~et out as
below. (i) Calculate the neutral axis position Taking arcu moments about the neutral axis:
b
X .1" X
300 nat
of
\~hich
.r/2
<.\' /
llcA,(d - t)
:wo .,
.
8.15 ( .450(600- .I J
2
has the \Oiuuon
x- 329mm (ii) Calculate the second moment of area of the cracked section 1 /" b.1 f 3 t a~A,(tl 1), \{)()X
3
329'
=7976 X
200 ()(60() 245 8 .15 1 I 0 ' mm 4
_
329
)2
(iii) Calculate the curvature of the cracked section From equation 6.7 ( 1 /r)~,
200 x J06 8. 15 x 101 x-7-9-76_ x_l_Q6 3.08 x 10
6
f mm
(c) Calculate the 'average' of the cracked and uncracked curvoture From equation 6.4:
1/ r
W/r)<, +(I
'"here
{= I -
i(a"fa,)
2
= I - 3(Mcr/M)1
-{)( 1/ r )oc
143
144
Reinforced concrete design (i) Calculate Mer
From lltble 6.11 the cracking ~trengr:h of the concretc.fctm· is given a~ 2.6 Nlmm2 • Hence from el~tic bending theory and considering the uncracked concrete section, the moment that will just cause cracking of the section. Mc1· , is given by M,, :: .fc~rn X (bwhzl6) = 2.6
and I rom table 6.13. · ,., ::::: 470 x 10 Therefore
470 X 10- 1'(200/ 8.15 )664
llr,., :: - - -w,6xlQO-
= 0.96 x 10
11
(becau:-e 211, lu
11
X
103
--
/ mm
(ii) For the uncracked section
1I rc,
= c, a~S I fuc
where S
A,(tf
x)
= 2450(600- 70012) =- 61 2.5 x 10~ mm 1 Therefore l I ,.,,
::::
470
X
10 11 (20018.15)612.5 300 X 7003 I 12
=0.82 x 10-f> j mm
X
1()3
210.
a.~
in pan (a)).
Serviceability, durability and stability requirements
145
(iii) Calculate the 'average' shrinkage curvature
~( l f r ),r- (1 - {)( 1/ r)IK
1/ r.,
= 0 95
X
= 0.95 x
0.96 10
11
X
10
6
( I - 0.95)
X
0.82
X
10
6
j mm
(e) Calculate deflection
Curvature due to loading = 3.07 x lO
6
j mm
Curvature due to !.hrinkage = 0.95 x lO~ / mm Therefore tottll curvature = 4.02 x 10
6
/ mm
l·or a 11imply supported ~.>pan subjected to a uniformly distributed load. the maximum mid-!>pan deflection is given by Deflection . 0. 104L2 (1 / r) 0.1()4 X 95()02 X 4.02 X l Q (>
37.X mm Thi~ value almost exactly matt.:hcs the allowable value of \pan /250 (9500/ 250 = 3X mm) and would be con~ic.Jercd acceptable noting the tnhercnt um:crtainty ol ~ome of
l
___________________________________________)
thc parameter'
~.
u~ed 111
the
calt.:ulation~.
6 .3.4 Basis of span-effective depth ratios The cnlculatmn ol tkOcction ha' been shO\\ n to be a tediou<, operation. llowevcr. for general u~c. ntb ba,cd on limiting the span-effecuve depth ruuo of n memhcr arc adequate to cn1>urc that the deflections are not excessive. The application of th11. method was described 10 \ection 6.2. 1 he rclauonsh•p between the deflection and the span effective depth ratio ol a memher can be derived from equation 6.14: thus dcflccliou r1
1-
1 l'h
L~
rb=-= I'll
"'em
+ " till d
where cc,rn." "'rm
1-
maximum compressive !>train in the concrete ten),ile \train in the reinforcement a factor whtch depends on lhe pattern of loading.
therefore 1>pan effective depth
-
tl
= al - - - - Lk (~c. rna\ - !'rn~)
146
Reinforced concrete design
Figure 6.8 Curvature and strain distribution
,
The strains in the concrete and temile reinforcement depend on the areas o• reinforcement provided and their ~tresses. Thus l'or a particular member section and ., pattern of loading. it i~ possible to determine a span-ciTective depth ratio to satisf) .. particular a/Lor deAecrion/span li mitation. The span- effective depth ratios obtained in section 6.2 are bnsecl on limiting Lhe tot •• dellection to !.pan/250 for a uniformly di1-trihuted loading and are presented fc· dirft:rent stress level~ depending on whether the concrete i~ highly or lightly stressed Thi~ in turn depends on the percentage or ten~ion reinforcement in the section. For span of le~!> than 7m this shoulcl abo ensure that the limit~ of ~pun /500 after application o· linishes are met but. for span~ over 7 m when: avoidance of damage to finishes may be important, the basic ratios of tnble 6.10 should he factored by 7 ; ...p
= baste rauo )
250
1
Tn cal>cs where the bas1c rntio ha!'l been modified for 1-ptm~ greater than 7 m, maximu deflections arc unlikely to exceed span/ 500 after con~truction of partitionl- and tini,h!: When another deflection limit ill required, the ratio' given ~hould he multiplied t'l. 500/ a where a is the propo~ed ma'\imum dellection.
(
EXAMPLE 6.3
Adjustme nt of basic span to effective depth ratio 20kN 10kN UDL
~
~ I I I I I II !
Determine the appropriate basic ratio for a cantilever beam supporting a uniform load and a concentrated point load at its tip a~. !.hown in figure 6.9. A:-.sume that Lhe concrete is C30/35 and is highly stressed. Basic ratio from table 6.10
effective span L Figure 6.9 Point load on a cantilever example
= 6.0 for u.d.l.
From table 6.14: /.. for cantilever with u.d.l. over full length /.. for cantilever with po1nt load at tip
0.25
= 0.33
Thu!>, for Lhe point load only. adjul>ted ba~ic ratio equal!> 0.25 - 4-
60 •
X
0.J3- ..:>
SeNiceability, durability and stability requirements An adjusted ba!>iC ratio to account for both loado; can be obtained hy factoring lhe moment due to the poim load by lhc ratiO of the J.. value:. as follow!> Mudl
= 10 X L/ 2 =
MP<,tnt
5/,
= 20L
. d balae . rauo . AdJUSte
. rauo . = Bastc
_ 6.()(5
(Mu
X
ku~t /kfl"'"')
Mud!+ M~mnt
20
X
0.25/0.33)
s 1 ::m
= 4.8 em. of
:rued for ..esscd.
f
r'pan~
~:o~Ho n of
11ay be \Cn by 1wn in
lThu~
6.4
it can be !teen that the effed of the point load dominates.
Flexural cracking
\1embcr~ ),UhJeCt to bending generally cxh1bit a ~cries of di::.trihuted fle.xural cracks. even at wQrking l ond~. The:-.c cracks ore unobtrusive und harmless unle!>s the width hecomcl> excessive. in which ca~e appearance and durability suffer a' the reinfon:cment " expo~ed to corrO!>JOn. I he actual widths of cracks in a reinforced concrete 'truclllre w11l vary between wide limits and cannot he precisely e!-.timated. thu:- the limitutg rcqui remclll to be sflt i~>licd is that the probability of the maximum width exceeding a !>atisfactory value i' <>mall. The ma\lmum acceptable "alue 'ugge,ted b) EC2 1\ 0.3 mm for all expo~ure classe'> under thi.! action ol the qua~i-permcmmr comtuna11on of load~. Other codes of pract1cc may rccommend lower values ol crnck wi<.lths for important member:-. and req u ireme n t~ for ~pedal cases. such a:-. water-retaining ~tructurell. may he even more stringent. Flexural cracking i-. generall)' controlled by providing a minimum area of ten.,ion rcinlorccment hection 6.1.5) and limiting bar ~pacing., (\ection 6.1.3) or limiting bar s1zes (section 6. 1.7). If culculmion::. lo estimate maximum crnck widths arc performed, they are based on the qua.1i f!t'mlwzelu combination of l oad~ and an effective modu lus of elasticity of the concrete \hould be u!>cd to aiiO\\ for creep effect,,
6.4. 1
M echanism of flexural cracking
Thi-; can be illu\trated by COihidcring the hcha' 1our of a mi.!mher subject to a uniform moment. A length or beam a~ shown in ligure 6.1 u Wi ll initia lly behave elastically throughout, as the urplied moment M i~> increased. When the limi ting tensile stra1n t'or the concrete i~ reached. a crnck \\ill form and the adjacem tensile ;one will no Iunger be acted on by direct tens1on force!'.. The curvature of the beam. hO\\.C\'Cr, cnusc~ further direct tcn~ion stres!-.cs to develop at some diswncc from the original eruck to maintain equi librium. This in turn cau~es further cracks to form, and the process continue1> until the distance between cracks does not permit sufficient tcn~ile strl.!sscs to develop and cauf.e further cracking. The-.e tnt11al crach are called 'pnmary crack•;. and lhe average spacing in a region of constant moment is largely independent of reinforcement detai ling. A"~> the applied moment is increased beyond this point. Lhc development of cracks is governed to a large extenL hy the reinforcement. Ten-.ilc stresses in the concrete
1·
148
Reinforced concrete design
Figure 6.10 Bending of a length of beam
Strain
surrounding reinforcing bar::. are caused hy homl a~ the strain rn the rcinforceme • increa:.es. These !:>tresscl. increase with distam:c from the primary cracks and m... eventually cause further cracks to form approximately mid-way between the primar cracks. This action may continue with increal.ing moment unti l the bond betwee concrete and steel i~ incapable of developing 1.uflicierH tension in the concrete to cau"t; further cracking in the length between existing crnc~s. Since the development of th~ tensile stresses is caused directl y by the prc),ence of reinforcing bars. the spacing o cracks wi ll he innuenced by the spacing of the rei nforcement. If bar<; are sufficienlly close for their 'tones of inAuem:c' to overlap then ~ecnnda~ eracb wil l join up across the member, while otherwif..e they will form only adjacent! individual bar~. . According to EC2 (sec 'ection 6.4.2) the average crack spacing 111 Hexural member depends in part on the effictency of hond, the diameter of reinforcin: bar used and the quantity and location of the reinforcement in relation to the tensto face of the <,cction.
6.4.2
Estimation of crack widths
If the behaviour of the member m figure o.ll i-. exammed. 11 C
where e., is the avcmgc Mrnin in the main reinforcement over the length considered, a may be as~umcd to be equal to E, where is the ~~tee I ).\res' ttl the cracked ~cctior 1 1 is the strain at level y which hy definition is the extension over the unit length ott. member. Hence, assuming any tensi le ~trnin of concrete hctween cracks is ~mal l. ~in~ !'ull bond is never devek1ped. the totnl width of al l craeb over this unit lenglh \\ equate tn the extension per unit length, that is
rrJ
y 0', (d - x) E,
Et = -
rr,
~
= L._. W
L
where 11' the sum of all cruc~ widths at level ."· The actual width of indJVrdual crack!> Will depend on the number of cracks in this u lcnglh, the average bemg given by unit length/a">erage 'pacrng (snn)· Thus average crack width
w., = - -L:\1' --av. number of crac~-. ~.
= ( If.\nn)
Serviceability, durability and stability requirements Figure 6.11 Bending strains
1
IX
d
The designer is concerned however with the maximum crnt.:k width which ha1-. an acceptably IO\\ probabilil) o1 betng exceeded. l·or dc!>1gn purpo~c1-. the design maumum crack width, ll'k, c~1n be hu~o.ed on the max1mum ~pacing. sr.rn•• ~· lienee the design crm:k \\ idth at an) level defined hy ,. in a member will thu'> t>e given b) "~cause
of the ·mg of
.... mdury ·em ttl
The exprc,sJOn for the dc~ign crack \\idth g1vcn in EC2 given a<;
tS
(<1.15)*
'"here H'k •
1rcing en~ ion
the dc.,.igu crack width
c\ 111
the maximum crack 'pal'ing the mean .llrain in thl! reinfiwn•nu'lll allowing for the effects of ten-.ion suiTcning of the concrete. ~hnnkagc etc.
~-."'
thc mean -.train in the concrete between c.:rads
.lr ma\
The mean ~tram. ,111 , will he less than the appun:nt value t 1 and (e\m the expression .fcL d l
/...,- (I ~ell
£,
.. uons. or the ,ince h will
of tlu: above form ami
em)
• 1g in a
l\'er!lll
J'
E'.:m) i~ given by
+ 1\dlp 0.6"" H,
(6.16)*
i-. the stres' in the tcn~mn steel calculated u~111g the cracked concrete sectl()n . factor that accounts for the dura!IOll of loading (0.6 for ~hort-te1m loud. 0.4 for long-term load). The maximum crack spacing•.11, 11,.,. ~~given hy the empirical formula where
/..: 1
rr,
i~ u
(6.17)*
,., unit
where is the bar 11i1e in mm ur an average bnr '11e \\here a mixture of different \i7es have been u~cd and c is the CO\Cr to thL' longllutlinal re111forccmen1. /q ~~ a coeffic1elll accounting fm thc honcl pmpettics of the reinforcemelll (0.8 for high hontl. 1.6 for plain bnr... ) und f..:, i' a coefficient accouming for the nature ol the \tram dJ~trihut1on "'htch for cracking due to llcxure cnn he taken as 0.5. l'p.cfl is the cffet.:tivc reinforcement ratio. -\,f A, efl· \\here A, i'i the area of reinforc.:ement within an effccllw ten\ion area of concrete Ac.en. as 11hown in figure 6.12. The effective ten~ion area io; that area of the concrete cros\-~Cction which \\ill crack due to the ten)>ion developed in bending. Thi~ the crncking which will he controlled hy the presence ol tm appropriate type. amount and distribution of reinforcement. Generally the dfecthc tensiOn area should be taken a-, having a depth equal to 2.5 tunes the distance from the tcn~>ion face of the concrete to the centroid of the reinforcement.
1'
14
150
Reinforced concrete design
Figure 6.12 Typical examples of effective concrete tension area Effecliv4!
h
Effective
tension area for thas lace
EffectiVe
d
tension area
-- -,
!1'".
. . .• • : q.:
ht,«ff
Seam Effecllv4!
tension area
{. • • • , • .~ . i'!
Member in tensaon
he, ttl
Slab
h, .~ ~
lesser of 2 5(h d), (h - x)/3 or h/2
although for slab~ the depth of this effective art:a should be limned to (h - \ )/3. A 1 ovcrnll upper dcplh Jimil of h/'2 al~o applies. Although not directly incorporared into the above formulae. it should be noted th.u crack wiuth11 may vary acros~ rhc width of the ~ol'lit of a beam and urc generally lil-.cl. to he greater at po,iliom, mid-way hclwccn longitudinal rcinforciJJg bar~ and at the corner-. nl the beam. Where the ma\lmum crack l>pacmg cxcccdf. 5(,· +- (;~/2) then o~. up)'ler hound to cracl-. "1dth can he estimated b) usmg ~r ma' 1.3(11 t).
6.4.3
Analysis of section to determin e crack widths
To uM: till' formula of EC2 i t i~ m.:ccs~ury to carry out an clastic analysis of the cracl-.eo conm.:tc ~ectinn u.,ing nn effectiw mmlulu~ £, ell· a~ given in equation (l.R to allow for t-rccp ette~·t~ The method.., d1scu~c;ed 10 \Cct1on 4.10.1 :-hould he used to find the neutral axi po),iuon. \, ant! hence the ~trc~~es. u, and rT,,. in rhe ten ile reinforcement from which "" (equation 6.16) can he obtained.
(
EXAM PLE 6.4 Calculation of flexural crack widths Calculntc rhe design flexural crack width.., for the beam 'hown in figure 6.13 when ~ubject to u qua~i-permancnt moment of 650 kN m. The concrete is da\s C25/30 and the reinforcement i s high bond with a total cross-secrional aren of 3770 mm2• (a) Calculate the mean strain f sm Fmm t.1ble 6.11. Ecm 31 kN/mm1. hom table 6.12. a~suming loading at28 days \\ith indoor exposure. the creep coefticicm o ~ 2.63 (hecau\e '21\J u = 2 x II000 x 4001/2800 = 285) and hence the effccri vc modulus is gi ven hy equation 6.8 a~
Ec,ttl - 31/( I + 2.63)
8.54 J...N/mm2
Serviceability, durability and stability requirements
-
b= 400
...
Figure 6.13 Crack width calculation example
X
_ y neutral axis
•J •
'-
3 No 40mm bars Section
Stress
(i) Calculate the neutral axis depth of the cracked section Taking moments about the neutral axis:
h x ,, x x/ 2 - acA,(d x) 1 00 400 X .r2 /2 -- - X 3770(93() 8.54 ~.
(iv) Calculate the maximum crack spacing (s, max) ~,rna\= 3.4c - 0.425klk~O/Pptff
where: c = cover = 1000 - 930 1.: 1 0.8 for ribbed bar;
401 2 = 50 mm to main bar-.
0.5 for flexure o - bar diameter = 40 mm
k1
hence
.I,
- 34 ln.l.l< -
X
•
50
+
0.425
X
296mm (which is
O.R X 0.5 0.0539
le~s
X
40
than S(r I r&/2)
350mm)
(v) Calculate crack width 1\'k
().()Ql
X
296
=0.30 mm which ju..,t satisfies the recommended limit. l_______________________________________ ~J
6.4.4 Control of crack widths II I'- ~1pparent from the expre:-.,ion'> derived ahove that there arc four fundamental \\'8) 111 \\ h1ch .,urfacc crack '' idth' may he reduced: (I) reduce the .. tre~' in the retnlorcement (a ) "h1ch w1ll hence reduce "": tiil rc.!duce the har dtamctcr' (ol \\h1eh \\Ill reduce b:u -;pacmg and ha\c the effect d reducing the crud. spacing (1, 111a,): (iii l tncrease the effective rcint'orccment rnrio (Pp.rtt ); (iv) U\C high bond rather thtm plain bar~. The u:-,c of steel at reduced 1>tn:s~c~ i:. generally uneconomicul and, ulthough ll approach is used in the design of watcr-rctllining ~trucwre~ where crading mu\t ofl~ he avoided altogether, it is generally easier to limit the bar diumcter11, incrcal!c f!p,dl u~c high bond bars in prclcrcncc to plnin bnn•. To increase Pr .r1 the effective ten~ ton area A ~.e l'f' should he made a~ small .:. possible. Thil' i~ hest achieved by placing rhc reinforcement close to the tension fal~ &uch that the depth of tension area {2 ..5(11 d)} i~ mudc as ~;mall as possibrecogni . . ing, nevertheless, that duruhility requirements limit the minimum value c. cover. The calculation of the design crack width~ indicated above only applie~ to region• '' ithin the effective tension zone. Since cracking can also occur in the side face of beam it j., also good practice to con!>idcr the provbion ot longitudinal !>tee! in the siti.. faces of hcam\. The critical po<.ition for the \ddth of ~uch crack\ is likely to tx approximately mid-way bet\\ecn the main ten!\ion steel and neutral axk Recommend .. tions regarding rhis and requirements for the matn remtorcement arc discus'>ed t• 'ection 6.1. u· these recommendation" are foliO\\ ed. 11 ts not nece'>'>ary to calculate cract width" except ll1 unu,ual circunmances. RelllfOrc.!ement detailing. however. has beer \hO\\ n to have a large effect on flexural crack1ng. and mu-.t 1n practice be a compromt'e hct\\ccn the requiremenL\ of crack.ing, durab1ltty and con.,tructional case and costs.
Serviceability, durability and stability requirements
6.5
153
Thermal and shrinkage cracking
Thermal and !>hnnkage effects. and the :.lrc~ses developed pnor to cracJ...ing of the concrete. were dtscuc;sed m chapter I. The rules for providing minimum areas of remforccment and ltmiung bar l.i.tes to conLrol thermal and c;hrinJ...age cracJ...ing were Ji~cu~scd in sections 6.1.5. 6.1.7 and 6.1.8. In this section. further con-.idcration "ill be given to the control of such crncJ...ing and the calculations that can be performed. tl ncccs~ary. to cakulate design cracJ... \vidth!.. Consider the com:rctc section of figure 6.14 which i1> in a state ot stress owing w thermal contraction and concrete shrinkage and the effects of external restrnmt. After cracking, the equilibrium of concrete adjacent to a cracJ... is lll> illustrateJ. Figure 6.14
ForcE's
l::.quaung tcthion and comprc-.sion forces. t\ ,/:t
A.J~L
- A,J-..,
If the condllton J), con-.idcrcd when steel and concrete \tmultancou,Jy reach the1r limiung values in tension. that "' /.. 1 /}~ and j~, =f(t.etr = the ten\llc strength ol concrete at the time \\hen cmcJ...ing i' expected to develop (usuall) taken a-. thn.!e days). then AJ)~
ll,tf.. ~ ..t(
-
A..J-..:
The value of 1~. can be calculated hut is generally very lomall and may he taJ...en a-. 7ero without llllroducing undue inaccuracy. Hence the criucal value of steel area i'
If the ~teel ureu isles!) than this amount then the steel will yield intension. resulting in a few wide cracks: however, if' it is greater, then more crach will be lormcd hut of narrower width. In EC2 this l'ormulu is modi lieu hy the inclusion ol a f..lrcss tlislrihution coefficient (kc) wJ..:en a~ 1.0 !'or pure tcn!'ion and a furt.hcr coefficient (/..) thnl accou nts for non-linear stress distribution wi thin the section. For thermal and shrinkage cl'fcc l ~. k can range from 1.0 for web~ where II 300 mm or nangc~ with width 300 mm tc> 0.65 for webs with II > ROO mm or flanges XOO mm interpolating accordingly. lienee the recommended minimum steel area required to control thermal and shrinJ..:agc cracking i\ given by A, mon
= /..).A,/.;L ell//~~ - kAJ,,
6.5.1
c~r/hl
(6.18)"'
Crack width calculation
The expression for the design cruel.. wtdth gtven m EC2. and di1>cusscd in 'ection 6.4.2 for the Calle of ncxural crackmg, can be used for the calculation of thermal and
t~djacent
to a crack
154
Reinforced concrete design Table 6 .15
Restraint factor values
R
Pour configuration
Thin wall cast onto massive concrete base Massive pour cast onto blinding Massive pour cast onto existing concrete Suspended slabs lnfill bays, i.e rigid restraint
0.6 0.1 0.1 0.3 0.1 0.2 0.8
to to to to to to to
0.8 at base 0.2 at top 0.2 0.4 at base 0.2 at top 0.4 1.0
;;hrin!..age cracking with some minor modilicationl-. The cruck widlh equation 6.1 S by ll'k
~r. max(E",m
i~
given in
f rm)
where wk is th~: design crack width..1,.""" is the maximum crack spacing and c~m is the mean 1-lrain in the ~eclion. For steel areal. greater than the minimum requir~:d value as given by equation 6.18. and when the totnl contraction exceed~ the ultimate tcnsi l~: ~train for the concrete. the shrint..age and thermal movement will be accommodated by cuntrollcd cracking uf the concrete. Any tensile strain in the concrete between crack~. em· as ~mall and Lhe effect may be approximated for buildang \tructure.., by u~ing the e>..pression (- .n1 - - cm l = 0.8R£" 1mp• \\here 1mp " the -.um ot the free shrinkage and thermal -.tralll~. That is {6.19)•
where " i' the "hrinkagc ~train. I i-. the rull in ll:mpcruturc rrom the hydration peak and oa • il. the coefficient of thermal expan~mn of concrete often taken as half the value for mature concrete to allO\\ for creep ctlcct~. The rcl.traint factor. R. i~ w altO\\ for differences in restraint according to pour conliguration. and typical value' are given in table 6.1 5. In prnctice. vnrintion~ in re~traim conditions cau~e large variations within members. and hetween otherwi~e :-.imilnr mcmhcr1., with ·full' re~ t rn 1 nt seldom occurring as indicmed in wble 6. .I 5. Cracking behaviour thus depend~ considerably on the degree and noture of the restraint and temperatures at the time of casting. CIRI A Guide C660 (ref. 25) offers further guidance on early-age crack control. The maximum cruck ~pacing. 11 mu' is given hy equation 6. 17 with fuctor k2 taken a:-. 1.0. Hence !'or rihheJ han. : (6.20) C'alculntion~ of crack width:!> ~hould there tore be considered as realistic 'estimates' only and engineering judgement may need to be applied in interpreting such results.
(
EXAMP LE 6 . 5
Calculation of shrinkage and thermal crack widths
A :-.cction of reinforced concrete wall i'> ISO mm Lhick and i:. cast onto a massive concrete base. A drying :.hrinkage Mrain of 50 mu:ro~tnun ( c.. l j<, anticipated together with a temperature drop (T) of 20 C after setting. Determine Lhe minimum
Serviceability, durability and stability requirements emforccmcnt ro control cracking in the lower prut of the wall and calculate the design rae~ width and maximum spacing for a suitable reinforcement arrangement. The ollowing dc~>ign parameters shou ld be used: Three-day tensile strength of concn.!te (/.: ctt )
= J.5 N/mm2
Effective modulus of elaf.ticity of concrete (Ecerd
10 k /mm~
Coefficient of thennal expansion for mature concrete (liT,)
= 12 microstram/ C
Cbaractcrbtic yield strength of reinforcement (j}l) - 500 N/mm 2 Modulus of ela~o.ticity of reinforcement
= 200 1-N/mm~
Mini mum steel area to be provided. from equation 6.18:
\,.mm- I.OA,Jc, tta/J;~ .N,uming n value of 1.0 for factor k in equation 6.18 If hori7ontal steel is to he placed in two layers the area of concrete within the tensile lOne, A.:t· can be taken as the full wall thickness multiplied by a one metre height. Hence 1.0(150
A s,nun
X
1000)
X
1.5/ 500
450mm'/m Thi1> could be convenient I> pro' ided a., I0 mm bar~ at 300 mm centres m each face of the membet (524 mm 2/m). For thb reinforcement and :1ssuming 35 mm cover, the crack spacing is given by equation 6.20 :ll> JA<·
Sr. mu'
+0.425 X 0.8 X
l.O
oil
"'here
JOmm
¢ fir ~rr
= t\, fA~.•rh = 524/( 150
x
HXlO} = 0.0035
therefore Sr. 1111,,
3.4 x 35 -1 0.425 x O.H x 1.0 x I0/0.0035
The imposed Mra1n in the section f 1mp=(.
=(50
i~
I090 mm
gtven by equuuon 6.19:
Tnr. )
20{12/2}}
X ]()
h
170 micro!>! rain
The ultimate tensile
Mrr~in
J'or the com:rclt'
= 1.5! ( J(X)()()) 150 rmcro~trai n Therefore the section can be considered as cracked. The design crack width is given W~ =
Sr nr.1x X
O.XRc,mp
Thus tiling U II'~
CUI (table 6.15)
= 1090 X 0.8 X 0.8 X = 0.12 mm
170 ( J0
h
n~
155
156
Reinforced concrete design
6.6
Other serviceability requirements
The 1wo principal other seniceability considerations are those of durability and re~i,tancc to fire. although occasionally a situauon an es ut \\hich some other factor may he of importance to ensure the proper performance of a stnlctural member in 'en icc. This ma) include fatigue due to moHng loads or machinery. or specific thermal ru1d sound in~ulation properues. The method\ of dealing with c.uch requirements rna} range from the use of reduced working \tresses in the material.... to the u\e of special concretes. for example light\\eJght aggregates for good thermal re~btilllce.
6.6.1
Durability
Deterioration will generally he associated with water permeating the concrete, and the opportunttie!> for this to occur shoult.l he minimiset.l us far a:- possihle hy providing gotld nrchitectural dewils with adequate drainage and prmcction to the com:rctc surface. Permeahility is the principal churactcriqi<.: or the com:n:te whi<.:h affects t.lurahility. although in some situation~ it is necc~sary to <.:Oilstdl'l al~o phy~icul und chemical effects which may cause the concrete lo dccuy. rOI rdnforcet.l <.:Oncretc. U further imp()rttllll uspe<.:t of durability i~ the degree of pmtcction which is given to the reinforcement. Cnrhonation by the rnmosphere will. in tim~.:, de,troy the alkalinity of the ~urfnce zone concrete, antl il thh reaches the Je,·el of the reinforcement will render the '-tee! vulnerahle to co1n1~10n 111 the pre~;ence of mm~ture and oxygen. II a concrete 15 made w11h a . .ound mert aggregate. dctennratmn \\' Ill not occur 10 the absence of an external mfluen<.:e. Smce concrete '' a h1ghly all-.alinc matenal. it' rc,Jstance to other alkah.., 1s good. hut Ill' hm\e\er \Cr) ,u..,ceptihlc to auad hy acid' or \Uhstnnces \\h1ch en,ily decompose w produce uc1d .... Concrete made \\ith Portland cemclll I'> thus not <.uttahle for U\C in ~itua11ons \\here 11 ~:nme' into contuct with ~uch material:-. \\ htch include hcer. mill. tmc.J futs . Some neutral \a Its may also attack ~:nncrete. the t\\o rno't notahk hc.:ing calciUm chlondc.: unt.l !>oluhlc \Uifulc.:!>. These react \\ l£h a mtnor wn,tituent of the hydration prmlucts in different \\aY'· The chloride mu~t he Ill t'lliH:Cntratcd \Oillllllll. \\ hl.!n 11 hll\ in 'ea water. The mallcr of exposure da:o.\itication' related to erwironmental condillon!. is dealt \\ ith Ill detail in E:-\ 206 and BS 8500 together \\'llh the provi~1on of appropnate concrete matenuk BS 8500 includes the use ol n \)'stem of clas~iticauon of a wide range of chemtcully aggrcr.si\e em·1ronments based on recommendauons of the UK Building Research E:-.tabli!>hment (BRE Spec1al Digest I). In some ca~es linble Ill aggressive
Serviceability, durability and stability requirements chemical attack Adtlitional Protective Measures (APMs) such a~ controlled penneahility formwork. smface protection. sacnficial layers or site drainage may he recommended. Phy~ical ~mack of the concrete must also be coni>itlered. This may come from nbra,ion or anriuon as mny be caused h> sand or shingle. anti by alremate \\cUing and df}ing. The Iauer effect I'> panJcularl) important in the ca.-.c of marine structure). nenr the \\ater surface, and cause strco,~e' to develop if the movements producetl arc re),traincd. It ~~ also JX>ssihle for crystal gro\\ lh to occur from drying out of sea water in cracks and pores. and this may cause further mtcmal stresses. leading 10 cracking. Alternate free7mg anti thawing is another maJOr cau~e of physical damage. particularly 111 road anti runwa) slabs and other ~>ituations \\here water in pores and cracks can frccLc und expand, thus lead111g to spalling. It hu~ been found that the cntrainmem of n .small percentage of air 111 the concrete 111 the form of small dllicrete bubble'> offer::. the mo<.t effective protection againl>t thi~ form or attack. Although tlus reduces the strength or the com:rcte. it is recommended that between 4.0 and 6.0 per cent by volume of entrained air llhould be included in concrete subjected to rcgulur welting and drying combined with ~cvcre lrost. All the:>e form~> o f attack may be minimised hy the production of a dense, wellcompacled and well cured concrete with low pcnncahility. thus rc1-.tricting damage to the surfucc 1011e of the rncmhcr. Aggregates which are likely to react with the a l~ ali matrix ~hou ld be avoidetl (or the alkah levels of the cement carefully limited). a:- must tho~e which exhibit unu:-.ually high -.Jmnknge charactemtic~. If thi\ i:-. done, then pcrmcahili ty, and hence durability. i~ affected by
1. aggregate type and dcn,lly; 2. water cement ratto: 3. degree of hytlr Coupled with this I' the need fnr n nn-porou~ aggregates \\ hich arc hard enough to re,ist any allntion, and lor thorough compuction. It is e~~entiul that the mix i' designed to have adequate workability J'or the situnlion in which it is to be used. thUll the cement content of the mix must be reasonably high. EN 206 1>pecifics minimum cement content:- for vnriou:- expo~ure contlitions according to cement ty pe~. ns well as minimum strength and maximum wmer t·em~·111 ratio which can ulso be related to minimum cover requirement~ a~ dc~crihcd in ~ce ll on 6. 1.1. The con~e4ucnccs of thermal effects on duruhility must not be ove1lookcd, and very high cement content~ ~hould only be used in conjunction with u detai led cmcking 1 u~~es~rnent. A cement content of 550kg/m i!. oJ'tcn regarded a~ an upper ]unit fnr general u~e. Provided thnt ~uch mcallure~ ure taken. and that adequutc cover ol \Ound concrete ~~ gtven to the remforcement, deterioration of reinforced concrete i~ unlikel> Thus although the ~urfnce concrete mu} be affected. the rcinfor'-111!! \ICC] \\Ill remain protected by nn alkaline concrete matrix which hn~ not hcen cnrbonated h) the atmosphere. Once tht!> CO\ er hrcaks dlm n and '' ater and po!>sihl) chem1cab can reach the Mecl, ru-;ting and con\cquent cxpan,ion lead rapid!) to cracking nnd ~pallmg of the cover concrete and /.evert: dnmage - \ i~uully and sometimes :-.tructurally.
157
158
Reinforced concrete design
6.6.2
Fire resistance
Depending on the type of structure under cono,idcration. it may be neces~ary to constde"" the lire reststance of the individual concrete members. Three conditions must examined: 1. effects on structural strength 2. name penetration re:>Jstance
3. hc
in the ca~c of dindmg memhcrll \UCh as \\'all } am! slabs
Concrete and steel in the form of reinforcement or prestressing tendon~ exhib reduced wcngth after being subjected to high tcmpcruturc.... Although concrete has k thermal conductivity. and thu1. good resistance to temperature rise. the strength begin~ • drop significamly at temperatures above 300 C and it haf> n tendency to spall at hi: temperatures. The extent of this spalling is governed hy the type of aggregate. \\ 1 si liceous materials being particularly susceptible while cn lcurcous and lightwetg aggregate concrete& suffer very little. Reinforcement will retain about 50 per cent of normal strength after reaching uhout 550 C. while for prestressing tendon~ tl-! corresponding temperature is only 400 C. I hu:- as the tcmpcralllre rise' the heat i\ tranF-ferred to the interior of a concrt : member. w11h a thermal gradient cstahlished in the concrete. Thi~ gradient will affected by the area and ma~~ of the member 111 addition to the thermal properties of the concrete. and may lend to e\pan,ion and los' of strength. Dependent on the thid.n and nature of cmer. the ~tccl ''ill ri~e 111 temperature and lo~c <;trength, thus leadin., dellecuom. and eventual \tructllral failure of the mcmher II the \tccl temper: • ~ becomes e\cc,.,i,·e. De~rgn mu\t therefore he aimed at prm tdtng and maintaining~.., cover of concrete as a protcction. thu., delayrng the tempcratllre nse m the o.;teel. 'The presence of plaster. ~creeds and other norH.:omhll'.tihlc lint~he~ a\,i<.ts the cover protecting the retnforcement and may thtl' he allowed for 111 the design. £~C2 givc~o tabulated values ot mi111mum dimen.,ion\ and cover' for variouf> types concrete member which arc neces . .ary to permit the member LO withstand fire for <;pecified period of time. These values. which have been ..,urnmari~ed in tables 6..+. 65 and 6.6 for siliceous aggregates may be considered ndequute for mo~t normal purpo'c' More detailed information concerning design for fire rcsi,tum:e i:o. given tn Part 1.2 l Eurococle 2 including concrete type, member type nnd detai l:. ol finishes. The periw that n member i)o required to ~urvive. both in respect of 1.1n.:ngth in relnt ion to worldn., londs and the containment of lire, will depend upon the Lypc and usage of the strucwre and minimum requirements arc generally ),pccilied by building regu lation~. Prcstrcs~ca concrete beams must be eon~idered ~epar:llc ly in view of the incrcu,cd vulnerability or the prestrcs~ing steel.
6.7
Limitation of damage caused
by accidental loads
While n would he unreasonable to expect u '>tntclllre to '' nh,tand cxtrcmcl. ot accidental loading as may be caused by colli'>ion, c\plo.,ion or stmtlar, n i' important that resulting dumagc 'hould not be di~proportionate to the cau-.e. It follm'' therefore that a major <;truclllral collapse mu~;t not be allowed to be cau ..cd h> a rclathely minor mi,hap which ma) ha\'e a rea~onahl) high probabihry of huppcnmg in the anticipated lifetime of the ~tructure.
Serviceability, durability and stability requirements The po~sihilities of a structure buckling or overturning under the 'design· loads will have been con~idered as part of the ultimate limit ~tate analysis. However. in ~orne instances a ~tructure will not have an adequate lateral -;trength e\en though 11 hal> hcen designed to rc~ist the speci fied combinations of wind load and vertical load. Thi' could be t11e case if there IS an explo~ion or a slight earth tremor. since then the lateral load\ are proportional to the ma~s of the '\tructure. Therefore it is recommended that at any floor level. a ~tructure should alwayl> he capable of rcsio;ting a minimum lateral force al> detailed 111 section :\.4.2. Damage and pOl>sihle in,tability should also he guarded again-.L \\hercvcr possible. for example vulnerable load-bearing member~ '\hould be prmected from collision by protective features iluch as hanJ..s or barrier),.
6.7.1 Ties l.n addition to the&c precautions, the general stabi lity and robustness or a bu ilding structure can be increased by providing reinforcement acti ng a~ tics. These ttes ~ohould act both vertically hetween roor and foundations. ami horizontally around anti m:rw.s each noor (figure 6.15). and all external vertical load-bearing mcmhl!rs should he anchored to the noor~ nnd beam~. If a building i~ divided hy expansion joint~ into ~lructurn ll y independent sections. then each section should have an independent tying ~y~tem.
Internal ties
Column ties
Penpheral lie
Vertical ties
Vertical ties Vertical tics are not generally neces~ary in ~tructurc~ of lc),s than f1vc ~torcy~ hut in higher hui l d 1 ng~ should be provided by reinforcement, effect1vely continuou~ from roof 10 foundation by means of proper laps. running through all vcnical load-hcruing member~. 'I hi~ \ tee! ~hould he capahle of resisting a tcnsih: force equal to the maximum tlc~ign ultimate load carried hy the column or \\all from any one ~torcy or the roof. Although the accidental load ca.\c io, an ultimate limit ~talc, the ultimate load u~ed hould reflect the loads hkely to he acting at the umc and the qut1.1i-pemwnen1 value would normally he taken. The mm i~ to contribute lO n bridging sy:-.tem in the event of loss of the member at a lower Jc,cl. In in silu concrete. thi~o. requtrement i~ almost invarinbly satisfied hy a normal design, but joint detailing may he affected in preca<;t work.
Figure 6.15 T1e forces
159
160
Reinforced concrete design
Horizontal ties HoriLontal ties '>hould be provided for all bulldmgs, irrespective of heighL in three \\'3}11:
1. peripheral Lies: 2. internal ties: 3. column and wall Lies. The re~istance of thc:,e ties when strcs!>ed to their charm:tcril>LiC strength is given m terms of a force F,. where F1 60 I. X or (:!0 + 4 x numhcr of storey~> in structure) k.\1. \\ hichever is less. Thil> expression tal.cs into account the increased risk of an accident in a large building and the seriousncs:-. of the collarsc of u tall structure.
(a) Peripheral ties The rcripheral rie must be provided. hy reinforcement which is effectively continuous. around Lhe perimeter of the building at each lloor and roof level. This reinforcement mu~l lie within 1.2 m of the outer edge and at it~ characterist ic strc!>s be capahle of resisting a force of at least /• 1•
(b) Internal ties lntcmal tie ... should abo h~ provided at each noor in two perpendicular directions anu he anchored at each end. c1thcr to the peripheral tic or to the continuou~ column or \\all tiCS The~e ues mu~t be effect!\ cl} contmuous and they may either be ')pread even I) acros'> a floor. or grouped at beam~ or wall~ a-. cun,en1ent. Where walls are used. the tie reinforcement must be withm 0.5 m of the lop or hottom of the floor -,lab. The resi•ilance required h. related to the :-.pan and loading. lntemal uc:-. mu~t be capable of rc~i-,ting a force ol F1 kN per metre w1dth or lltCIIl + lJk )/7.5 1r/5l.N per metre width. if this 1' greater. Ln th1s expression. lr ~~the greatest homontal di~tance 111 the d1rcction ot the tic between the centre!'> of vertkal load-bearing m~.:mbcr . Tht.. load1ng (Rl - q!..) I.N/m! is the avcrag~.: characteristic load on unit ~rca of the floor con~'>ldered. If the tie:-. are grouped their maximum spacing should he limited to 1.5/r.
(c) Column and wall ties Column and wnll tics must be ahle ll) resh.t n Coree of ut lenst ~ per ~.:cnt of the total verlil:al ultimate load at that level for which the member has been designeu Additionally, the resistance provided must not he li.:sl. than the smaller of 2F1 or F1IJ2 .5 kN where/, is the floor to ceiling he1ght in metre:.. Wullti~.:s ure nssesscd on tht basi' ()['the ahovc forces acting per metre length of the wall. while column tics are ~.:onccntrated within lm of ci th~.:r ~1de of the column cemrc lim.:. Part1cular care ~houla be taken with comer column' to cnr,ure they urc tied 111 two perpendicular directions In eon~idering the structure subjected w accidental loading it i., assumed that no othe force:-. are acting. Thu reinforcement pro\'ided lor an} other purpmes may also act a tics. Indeed, peripheral and intcmalt1es mu} abo be con~1dcred to be acting a:-. columr or ""all tie'>. A<, with Ycrtical tiC'>, the provision of horiLontal tiC~ for ;, \itu construction wi I -,cldom affect the amount of reinforcement promled. Detailing of the reinforcemcn may hm,cvcr be affected. and particular auenuon mu~t be paid to the manner in whic.. internal tics are anchored to penphcral tics. Typical detail"> for the anchorage of intern...
Serviceability, durability and stability requirements
,
Full anchorage length '
three
• •
~;-
-
en
•
.....-
-:1
111
Figure
16
6.16
Typical anchorage details fo1
internal tie~
•
•••
•••
•••
(a)
(b)
(c)
~.kN. l'lt in
1US,
tie~ are strel\se1.
illu~truted
in figure 6.16. If tull anchorage i-. not posc;iblc then the a~sumcd the ties muM he reduce<.! appropriately. Prcc~tst concrete construction. however. present~ " more l.eriou~ problem ~ince the requirements of tie forces anti ~imple easil> com.tructed joum are not always compatible. Unle!.!. the rctluir~:d ue force~ can be provided WJLh the bar!> anchored by hook!\ and bend~ in the case of column and wall tie~>. an analysis of the structure must be performed to as-.e~s the remaining stabilit) after a 'f>'!Ctfied degree of structural damage. tn
'
( EXAMPLE 6 .6 Stability ties
Calculate the \lability tiel> required in an eight-storey buildtng of figure 6.17: Clear storey height under lk:ams
= :! 9 m
rloor to ceiling height ({,)
3.4 m
Characteristic permanent load C.~d
=6 kX/m'
Characteristic variable lo
\ I..N/m ~
Chanl<.:tcn-;tie steel strength (/~ d
500 N/mrn'
f1
(20 + 4
= 20 I 4
plt~n
area 'ho\\11 in
X number Of sLOrcy~ )
8 - 52kN
60k'i
(a) Peripheral ties
Force to he r..:,i-.tcd - F, 52 kN 52 X J() 1 lhr area required I04 mm' 500 Thb. could be provided by one H12 bar. Precast floor slab
long1tud~nal
\
t
bedm
'""'~"' i•-----1----------11---.. . b,.m,
14bays® 6 Sm -
~ -- --
26m
----r
Figure 617 Structure ldyout
162
Reinforced concrete design (b) Internal ties
.
.
l·orce to be rcl>J~ted
FrCI?l -1 qk) 7.5
=
I,
" -::-) k p.:r metre
(1) Transverse direction
Force
5"'(6 = ---7.5
..L
3)
7 = 87.4 J...N/m > F1 5
x-
Force per bay= 87.4 x 6.5
= 568.1 kN Therefore. bnr area required in each mmwcrse interior beam i..,
56H.~= 500
113"
. \)mm
2
This could be provided hy 4 ll20 burs.
(2) Longitudinal direclion . f·orce
6.5 = -5::!(6_ ~~- --:-X1.1 J...N/m 7.::. )
1'1
rhe•o:forl' force alnng kngth ol huikhng required in each lnnglluc.hnal heam ., 567 7 2
IO'
_
XI. I x 7
567 7kN. hence bar area
,
--=::.67mm·
500
rtm could he prm 1dcd h) 2 H20 har,,
(3) Column ties
l·mcc
10
he designed for i!'.
(L1 )F~ G:~)52 = 1o.n~ < 2F, or 3 per cent of ultimate lloor loud on a column 8 [~ ( 1.35 x 6 + 1.5 x 3) x 6.5 x 10()
~]
2
i~
- 69 kN al ground lcvd
To allow for 3 per cent of' column self-weight. taJ...c design forc.:c tt> be 72 kN. say, at ground level. Area of ties. rcquire·u1
= 72 50010' I(
144mm~
Thi' would be provided by I IT20 bar and incorporall:d with the internal tics. At higher floor lcwb a design force ol 70.7 kN \\oultl hi! u~cd giving a similar practical remforccmcnt requirement. (c) Vertical ties As~ume quasi-pemumenr
=
loading \\ith 111 ,
1.0 > 6 1 0.6x3=7.8k~/m~.
0.6. Thull the ultimate dcllign load
Serviceability, durability and stability requirements
...,unum column load from one storey 7.S
X
3.5
X
6.5
i~
approximately equal to
177.5 kN
nerefore bar arc
500
,
=.355 mnr
' would be provided by 4 Ill 2 bar~.
6.7.2
Analysis of 'damaged' structure
I h1~ must be undertaken when a -.tructure ha\ five or more storeys and doe!> not comply nh the vertical-tie requirement~. or when every precast lioor or roof unit docs not have ,ufficient anchorage to resi~t a force equal to F1 kN per metre width acting in the direction of the 1-pan. The an:1ly:-.i1> mu~t ~>how that each key load-bearing m~.:mhcr. it~ ~:onnections. and the horizontal mcmher:-. '' hich provide lateral support. arc able to 1\ ithstand a sped lied loading from any direction. If this cannot he satt:,.hed. then the maly:,.is must demonstrate that the rcmm al of an} single \ en1cal load-hearing element. Hhcr than key membcrs, at each store) 111 turn \dll not re~uh 111 collapse of a !.igniticant part of the 1>1ructun.:. Thc minimum loading that m;,y m·t from any direction tlll a key member is recommended us 34 J..N/m1. The decision as to what load:. should he considered acttng is left to the engineer, but will generally he on the basis of permanent and reali~tie variable OOmewhat arbnral) value. The 'pressure' method wi ll generally be suitable for npplicatinn to column:- in preea~t tmmed structure~; however, where precast load-bearing panel con~truction is being used an approach incorporating the removnl of individual element.-. muy be more uppropriatc. ln this case. \ c111Cal loading/> -;hould be U'>:;e-..,cd a' described. and the ~tructure mve:-.tigated to dcterr111ne '' hcther it i-. able to remain ~tand1ng b) a d1fferent ,true tum I Jet ion. Tills actton may 1nclutle parts of the damaged :.tructurc bchtl\ 111g a. . a canttlcver or a catenary. and it ma> also be necc~sary to con~idcr the strength of non-load heanng pur1itions or cluoding. Whichever upprouch is adopted, such analyses arc tedious. and the provi~ion of ~o:lfeetive tie forces within the structure ~hould be regurded as the preferred solution hmh from the poun of view of design and performance. Continuity reinforcement and good detailing '' 111 greatly enhance the overall ltre resistance of a \tructure with respect to collapse. A tire-damaged ~tructurc with reduced member ~trength may even be ltkened to u ~tructure subjected to accidental overloud, <111d analysed nccordingly. 1\
6.8
Design and detailing for seismic forces
Earthquakes are cau:-.ed hy movement of the earth's cru~t along faults or slip planes. TI1ese movements re!>ult in horizontal und vertical force~ with vibrations of varying lrequcncy, amplitude nnd duration to act on ~tructures within the earthquake zone.
163
164
Reinforced concrete design
The earth· cmst i~ not one contmuous outer layer but cons1sts of seven major p number of minor tectonic plates as shown in figure 6. I8. These plate~ bear ag~:ale . lnten~itie:- up to ~ in magrutudt: are generally con!>idcred to motfl'rate but higher intensitie~ oJ' 0\·er 0 are \eVCI'e. i nlCJl',iliC~ as high as 9.5 have lk~ mea~ured.
In many parts or the world, such a.~ Turl-..cy, Japan and California. where earthqual-..t. can be se\erc. rc~•~tance 10 ~eism1c force' form' a cnllcal part of the structuml dest~ In other area:-. of the world. such a~ the British Isles. earthquakes arc less common an~.. not nearl) so -,e\ere so that the de~ign for \\ind loadtng. or the requirements for thl!' 'truclltrc to be able to resist a minimum hnri.wnwl force, plu' the prm ision of continuity \tcclthroughout the structure according to the requirements in section 6.7 are generally adequate. (l;evcnhele''· \\ tth Important structure~, -,uch a~ rmtJOr darm Jt nuclear pO\\Cr stations where l':ulure or damage can hove cnta!-.lrnphic cl'fects. the re~•~tunce to set'illllC diMurhance~ nm\1 .tbo he consrdcred. C\Cil in the Bntbh J,b The nmurc of the vibrations and the lorces induced by an cnrthtiLtnke arc complt. phenomena, as ,., the dyn.tmlc rcspon'e of a highl) 111detcm1inate concrete ~tructure This has ltd to the development of computer programs to carry out the analysi,, \Otllctimes rdem.:d to a~ a multi-modal rc,ponse 'pectrum analy\i~. A .,jrnplcr approacl'l i~ the equrvalcnt ~tallc analysis tn which 1he ba~e ),henr at the fool of the 'tru~;ture
Figure 6.18 TectoniC plaitS
Pacific Plat!'
Serviceability, durabilily and stability requirements
a
calculated and distributed as hori1ontal forces at each floor level according to certain defined criteri:t. Thil> approach is allowed in many national codes of practtce for the design of approximately regular and symmetrical structures. 1-:.urocode 8 provides guidance relev:tm to countries within the European Cnion. The full numerical dc'>ign requirements of the coder. of practtce are beyond the <;cope of tht<; book but it is hoped that highlighting some of the import:tnt principles and requirement<; tor the overall design and detailing may be of ~orne help in the design of safer structure~.
6.8.1
!!'ell
01
are
or the
165
Construction and general layout
Jt is particularly important that good quality material<;, including high ductility reinforcing )tccl, arc used together with rigorous tc~ting and control procedures. Design shou ld ensure thut sudden bnrtle shear or comprcs1;ivc fai lure b nvoit.lcu with emphasis on energy dissipmion. Good construction practices, inc.:luding steel lixing. c.:ompaction, curing and inspection arc also essential if n Stntcturc is to perform satl~fat.:tori l y under seismic loading. Foundntions should be designed to provide a regular lavout in plan and elevmion und to he aprroximntely symmetrical about both orthogonal axes in plan with no ~uddcn and major change in layout or con:.trul:tion. It is important that there i' adequate bending and tor~ional rcsi,t:tncc about both axes of the ~tructure . Some illu,trattve examples or good and poor prnc!lce are shown in figure~ 6.19 and 6.20. Sway effects under horizontal motions should be minimi'>cd hy ensuring approximately equal loading at each floor le\el with no hea\y lond' in the higher \tore}~. Effort' 'hould he made to pro' ide a highly indetermmate !.tructure that '' well
Figure 6.19 Examples of good and poor plan lc~youts
Good plan layout
D===O Poor piJn layout
Figure 6.20 Examples of good and poor elevations
Good elevat1on des1gn
( Poor elevat1on des1gn
166
Reinforced concrete design
Figure 6 .21 One-storey building
I
I
tied together with continully reinforcement so that the loading can be rcdi~tributed ano alternative :.tructurul actions may develop if ncce~sary. The principle~ d1~cussed m lleCLion 6.7 are relevant to this. Slah!. can provide rigid diaphragm~ to transfer loads at the roof" and cm:h floor Figure 6.21 :;hows hO\\. in a one-storey building, a rigid hori7ontal slab or hrucing roof le' el enables the -.tructurc to act us a closed box gi' ing more rigidit) and streogt to resist cracklllg.
6.8.2
Foundations
In addition to a regular and ~ymmetncal layout in plan a~ discuiosed above. it preferable Lhtll on ly one type or l'ound:llion is u!>ed throughom n :.tructure and thutn con~tructetl on a level ground base and tied together with strong ground beam11 to lin relati\e mu\'ement. Tim. i~ illw,trated in ligure 6.22. Land~lidc~ arc a common feature of earthquakes and they cause much 11tructu damage and loss ol lile. f'ht.:rcfore ~tructurcs ~houlu nm be built on steep ~lopes. 111 ncar gullcy~ or nct~r cl ifl\ II mu~t al:-.o be recognised thm vibrations during ca11hquake can cau).e liqucf.tction of ~m11c sot!~. ~uch as sandy or silty ~oils, cau-. lo~s of bearing !>trcngth. excessive seulcment and failu re. Figure 6.22 Examples of good and foundation design
poor
Good deslg~
firm
Serviceability, durability and stability requirements
6.8.3
167
Shear walls
Shear wal11> provtde a strong reSiStance to the lateral force~ from an earthquake and they should continue down to. and be anchored into. foundations. They should never be ~upported on beams. 'lah" or columns If coupling beam-, are required these should be reinforced '"ith diagonal cage~ of Mccl bar& and diagonal reinforcing bars should also be provided to resist hori7ontal sliding at constntction joint1> in the wall. These bars should ha\·e at lca't a tension anchorage on either ~ide of the con,truction JOint. Some typical reinforcing !.lccl detail" are given in I:C!! and a typtcal detail for a coupling beam is shown in figure 6.23. X
coupling beam
---
I
~
Figure 6.23
- ·v
TypiCal reinforcement detail for a coupling beam
~hear wall
diagonal cage of rCinforcement
Section X -X
6.8.4
\
opening X _.
v
Columns
Column' and their connection' to beaml> are critical parts of a 'tructurc. ratlurc of a column in a huilding can be cata~trophic leading to a progrcs-.ivc collapse. and the formation of plastic hinges 111 columns above the ba~c of u building 'hould be avoided. llorimntul hoops of helical reinforcing bars have been found to give a •Monger containment to the longitudinal vertical bars than thot provided by rectangu lar links and at a beom-to-cotumn joint horizontal steel reinforcement hoops not les.~ than 6mm diameter are udvisahlc with in the depth of the beam. At extetnal columns the longitudinal reinforcement of beam:-. should he well anchored within the colu mn. Thi!-. may rcquirc special mea~ures ~u ch as the provision of henm hounches or anchorage plate~ and some typical examples of details ure given in hC8.
6.8.5
Beams and slabs
Beam~ ~hould he ducule so that plastic hinges can form. and thc~e should be dtstributed throughout a \tructure. uvotding 'soft' sroreys. This will provide a gradual type of failure und not a 'udden catastrophic frulure such a:; that U!)suciatcd with shear or brittle compre~stve failure. The formotion of plastic hinges also allows the ma\imum moments to he rcdi,trihutcd to other pans of the statically indelenninatc \lructurc. thu)) provtding more overall Mtfcty. The fir:-t pluMic hinges arc likely to form in the sections of the beam ncar the column \\here the maximum moments are hogging, causing compression on the lo""cr fibres i.O that the section acts cffecuvely as u rectangular section. Plastic hinges which form later
168
Reinforced concrete design
at mid-span will have compression on the upper fibre!. -,o that the '\ection is effective!} .. T-section \\ith the slab acting a!-. the flange and there i' a large area to resist the compress10n. Further discu~sion of the de,ign of ducule \ectJOO'> i~ given in Section~ 4.2. 4.4 and 4.7. The beam section<; near the .,upport \hould he reinforced by dm.ed :-.tecl link!>. close!) 'paced to resist the c;hear and to prmide greater compre...sive resi,tancc to the cnclo:.t'd concrete The provision of comprc\sivc steel reinforcement abo cn\ures a more ductile ~ection.
The \lab' in a building act :1!> rigid horitontal diaphragms to stiffen the ~tructure against tor11ion during ~ei:.mic di~turbances and nlso tran~fer the hori7ontal forces into the columnl> and shear walls. The l>labs ~hould be well ned into the columns. the ::.hear walls and the perimeter beams wtth contimuty reinforcement a~ indicated previously. When precast concrete slabs nre used they ~hould have good length' of hearings onto the :-ttpporting beams and sheur walls ~hou l d also he provided with continuity 'ltcel over their supports so that they can net :\)., cont inuous indeterminate members. In this way they can also develop their full ultimate reserve of 'llrcngth hy enabling a tensi le (:!liCJHiry action.
CHAPTER
7
Design of reinforced concrete beams CHAPTER INTRODUCTION Reinforced concrete beam design consists primarily of producing member details which will adequately resist the ultimate bending moments, shear forces and torsional moments. At the same time serviceclbility requirements must be considered to en~ure that the member will behave satisfactorily under working loads. It is difficult lo separate these two critf'ria, hence the design procedure consists or a series of interrelated steps and checks. These steps are shown 1n detail in the flow chart in figure 7.1, but may be condensed into three basic design stages:
2. 3.
preliminary analysis and member sizmg; detailed analysis and design of reinforcement; serviceability calculations.
Much of the material in this chapter depends on the theory and design specification from the previous chapters. The loadIng and calculation of moments and shear forces should be carrit.>d out using the methods descnbed in chapter 3. The equations used for calculating the areas of reinforcement were derived in chapters 4 and 5. Full details of serviCeability requirements and calculations are given in chapter 6, but it IS normal practice to make use of simple rules which are specified
__..
169
-1 70
Reinforced concrete design
EC2 Section
Y4ttable actiOns
2.3.1
Concrete class
D
D
Estimated self-weight
Concrett' cover
D
Mtnimum sectton
D 5
4.4.1
D
Permanent actions
2.3.1
3.1
D
PRELIMINARY ANALYSIS
Durab1hty and fire re~istance
4.3 and 4.4
D Trial b
c
·;cOl 0
Esumate d from
~
c .E"'
~
...
6.1 singly reinforced doubly reinforced
r--
.[~ 6.2 7.4.2
m~ximum allow~ble?
Vw
ChPc~
basic 1pan-crfrcttVe depth ratoos _ __
.;L
Select II
.,
-<..>DETAILED ANALYSIS & DESIGN 8endong moment and shear force envPiclpcs
,J. 6.1
Bend1ng re1nforcement design
84
Anchor~ge
,l.. ..:.'"J..
(
8 and 9.2
~
Bending relnforcPmcnl details r-1
../....?
7 4.2
Check
span-cff~tovc
depth rauo
D 6.2
She~r
rconforcement deso9n
D 7.3
C~lculate
crack wodths (of rtQuored)
D 7.4
Calculate denectlom (if required)
D FINISH
Figure 7.1 Beam design flowchart
c
0'
·~
0
;;
c u:
"0
...c
~ ·~
c; 0
Design of reinforced concrete beams
in the Code of Practice and are quite adequate tor most situations. Typical of these are the span-effect1ve depth ratios to ensure acceptable deflections, and the rules for maximum bar spacings, maximum bar sizes and minimum quantities of reinforcement, which are to limit cracking, as described in chapter 6. Design and detailing of the bending reinforcement must allow for factors such as anchorage bond between the steel and concrete. The area of the tensile bending reinforcement also affects the subsequent design of the shear and torsion reinforcement. Arrangement of reinforcement is constrained both by the requirements of the codes of practice for concrete structures and by practical considerations such as construction tolerances, clearance between bars and available bar sizes and lengths. Many of the requirements for correct detailing are illustrated in Lhe examples which deal with the design of typical beams. All calculations should be based on the effective span of a beam which is given by
lrtt
In · a, ~ a2
where
111 is the clear distance between the faces of the supports; for a cantilever In is its length to the face of the support o1, o2 are the lesser of half the width, 1, of the support, or half the overall depth, h, of the beam, at the respective ends of the span
7.1
Preliminary analysis and member sizing
The luynut anti -.i;c of member\ nrc very often controlled hy archttectural tletail1-. and clearances for machtnery nnd equipment. The engineer mu<>t either chcc'- that the hcam ~i.Ge!> arc adequate to carry the londing. or altcmutively. decide on !lite:-. that arc adequate. The preliminary analy..,i~ need only provide the maximum moment~ and ~heors in ordet to asccrtuin reasonahle t.limensions. Beam dimension~ required arc
1. cover to the n.:inlorccmcnt 2. breadth (b) 3. effective depth (c/) 4. overnll depth (h)/ Adequate concrete cover i-. required to ensure adequate bond and to protect the reinforcement from corrosion and t.lamagc. The necessary CO\'er depend~ on the dass of concrete. the exposure of the beam. and the required fire re!.i~tnnce. Table 6.2 give~ the nominal cover that 'hould be provided to all reinforcement. including lin~s. Thi' cover may need to he increased to meet the fire resistance requtrements of the (ode of Practice. The -.trength of a benm is affected considerably more hy its depth th:ln th breadth The spun-depth rauos usually \'ary between say 14 and 30 hut for large spans the ratio\ can be greater. A \ttitahle breadth may be one-third to one-half of the depth: but it mu) be much les:-. for a deep heam. At other times wide shnllow hcam'i are u~ed to conserve
171
1 72
Reinforced concrete design
headroom. The beam should not be too narrow: if it is much less than 200 mm wide there may be difficult)' in providmg adequate side cover and c;pace for the reinforcing bare;. Suitable dimensions forb and d can be decided by a few trial calculations as follows:
1. Por no compression reinforcement K = M/bd~fd ~
Khut
where
0.167 for fc~:
Kt-..t
< C50
With compression reinforcement 11 can he shm' n that M / bd1fck
< 8/.f<~
approximmely. if the urea of hcnding reinforcement is not to be excessive. 2. The maximum design sheur force Vblmux !>hould not be grculcr than VRu m.u = 0.1 ~b"d( I - .f,~/2501fd· To avoid congested shear reinforcement. Vl'u 11 ~;1x should preferably be somewhat clo~cr to half (or less) of the maximum allowed. 3. The span -effective depth ratiO for 'pun' not exceeding 7 m should he within the hNc \'aluc' given in table 6.10 or figure 6.3. For ~pans greater than 7 m the basK ratio' arc multiphcd by 7/..,pan.
b
hold
4. The overnll depth of the beam 1s g1ven II
I , covN I
Figure 7.2 Bcdm dimensions
(
i
h)
d +cover + t
where 1 cslimnled distance from the out~> ide of the link to the centre of the tension bnr~ (:-.ec figure 7.2). For exumple. with nominal sited 12 111111 link!- unJ one layer ot 32 mm tension barli. 1 2~. mm approximately. It will. in fact, be slightly larger than this with deformed har~ a~ they have a larger overall dimension than the nominal bur :-.ite.
EXAMPLE 7.1 Beam sizing
A concrete lintel with an ciTcctive ~pa n of 4.0 rn supporLs 3 230 mm hrick wall as shown in ligurc 7.3. The loads on the lintel arc G~ lOOkN nnd Q~ 40kN. Determine ~uitable dimens1ons for the lintel If class C25/30 concre1e i~ u~cd.
I Figure 7.3 Lintel beam
Assumed load
Design of reinforced concrete beams
m wide rcing llows:
The beam breadth b will match the wall thieknl!ss so that
b;;;;: 230mm Allo" ing. say. 14 I..N for the \\eight of the beam. gives the ultimate loac.l /·
1.35
114
X
1.5
X
40
;;;;: 214k.\
Therefore maximum design shear force VFd
= 107 J..l\
Asc;uming a triangular load dbtribution for the preliminary nnalybis. we have
M _
r x span= ::!14 x 4.0 6
6
143 "'-N m For 1>uch a rclati vely minor beam the case with no compre~sion steel ~ h ould be con:-idered K
t n the le basic
M ,}. I )( / ' Ll
< Khu1
therefore 141 ( 101' 230 d' )< 25
.:11\tOn
er of arger ' n the
0.167
0.167
Rcammging, d .;> 386 mm. A\\umc a com: rete cover of 25 mm to the reinforcing steel. So for I0 mm link~ and . s.t}.
n mm han.
O"crall beam c.lcpth h
d + 25 + 10 =d+ 51
32/2
'll1crelorc make II - 5::!5 mm as an integer numher of bril:J.. courses. So that d
~7~0 = 8.35 < ::::: 20 (for a lightly ~trc!.scd heam in C25
concrete tabll! 6. 10) A heam siLc of 230 mm by 525 mm deep would he 'uitablc. Weight of beam
0.23 x 0.525 x 4.0 x 25 - 12.1 kN
l" hieh i!> 'uffic•entl} dose to the assumed value.
1 74
Reinforced concrete design
7.2
Design for bending of a rectangular section with no moment redistribution
The calculation of main bending reinforcement is performed using the equations and charts derived in chapter 4. In the ca~e of rectangular sections which require onl~ tension steel. the lever-arm curve method i~ probably the simplest approach. Whert compression :,teel is required. either design charts or a manual approach with the simplified design formulae may be u'cd. When design charts arc not applicable or nll available. as in the case of non-rectangular sections. the formulae balled on the equivalent rectangular stress block will !>implify calculations considentbly. The grade and ductility class of reinforcing steel to be u~ed must he decided initial -;ince this. in conjunction wtth the chosen concrete class, will affect the areas requirt.-. and also influence such factor' a:, bond calculation\. In moM ctrcumstance~ one of th.. available types of high-yield hars will be used. Areas of reinforcement are calculated the critical M.:ctions with maximum moment~ . antl1-uitable bar si1.c:-. ~-oc lectecl. (Table~ c bar areas arc given in the appendix.) 'I hi ~> permit~ anchorage calculations to bt performed und dewil\ of har arrangement to be produced, tak.ing into account t ~ gutdance gl\cn b) the C'otb of Practice. An exl:e.,~ tv e amount ot n:111forcemcnt u'ually indicate'> thUl a member is undcr!>iZt and it may also cause tid llculty in lixing the bars and pouring the concrete. Therefor~: the code ... tipulates that lOOt\ 1 \, _ 4Cf· except at
lap~.
On the other hand too ltttlc reinforcement i' abo
I00/\, I b1rI where: A,
i~
.>
unde~1rable
therefore
;. .f~trll 2of. - per cent and nol less than 0.I.,.., per cent y~
the area of concrete
= b x II for a n:ctangular ~cctmn
lJ 1 b the mean width of the beam's tension .wnc ~·, 1rn .ts ,'I1e concrete ' s mean ax .to I tenst'Ie MrengtI1 0.•"1 x.f'-d. :. 13 10r. r j'
C50
Value'> for tltfferent concrete .,trengths are given in table 6.8 To avotd exces,he detleetion' it i\ abo nece,~ary to check the span to cllecuve dcp ratio tt!-. outlined in chapter 6. It :-.hould be nmed thm t~e equations derived in th i:-. chapter are ror concrete class~ less than or equal w C5016d·. l'he equations for higher cla\!.es of concrete can be deri\( ut.ing '>imilar procedure., but using the ultimate concrete '>tratn' and constant~ for c•• ~ clas~ of concrete from EC2 and it~ attonul Annex.
7.2.1 Singly reinforced rectangular sections, no moment redistribution A beam section need<; reinforcement onl> in t11e tenstle tone \\hen K
M u < Kh.•t ld ) ./Ck
= 0.167
The singly reinforced section considered ;., \hown in figure 7.4 and it is subjected t< sagging de\tgn moment M at the ultimate limit state. The design calculntion'> for l longimdtnal \teet can be summarised a-. follows:
2. Determine the levcr-am1, <., from the curve of figure 7.5 or from the equation ., e' of to be 1t the
k
'ILCd
'le1"Cimc
" r~[o.s + ,j(0.25 -
K/1.134)]
(7. I)*
3. Calculate the area of tension steel required from M O.R7fvk:
(7.1 }*
4. Select ~Uitablc bar 'i7e~.
5. Chcd that the urea
or -.tccl actually provtded i1. within the limit~ rcqutred hy the
code. that ss 100 A, m.o, < 4.W'f bh
•
und '> whcrej~ 1111
depth
~\
f.:om
26 -% and not less than 0.13~ J~~
3
0.3 X /~~ for fc~
< C50
1.00 Figure 7.5 Lever-arm curve
maximum value of z/d according to the Concise Code Dnd previou s UK practice
du:-.~e1.
demcd
-I
0.95
~each
~
I
Compression reinforcement requ1red (at M0 .,)
II
~
0.90
0.85 0.82
L.._ _ __
0
ed to a
f1 •r the
L __
oo.s
_
17
_ _..L.._.::.:_
0.10 K = M/bd1fa
_
_
_
0 .15 0.167
The percentage values on the K axis mark the limits for singly reinforced section with moment redistribution applied {see section 4.7 and table 4.2)
176
Reinforced concrete design
(
EX AMPLE 7. 2
Design of tension reinforcement for a rectangular section, no moment redist ribution
The beam section shown in rtgure 7.6 h a~ charaetcristic material strengths of .fck = 25 Nlmm 2 for Lhe concrete and[yk = 500 N/mm 2 for the steel. The de~ign moment at Lhe ultimate limit state i'> 165 kN m which cause\ sagging ol the heam.
b .. 230
0
""
0 .,., .,.,
"' II "b
II .<:;
••• A, 3·H20
Figure 7.6 Singly reinforced beam example
1. K
_!!_ __165 X 106
p
-Q
bcf2/ck - 230 X 4902 X 25 -
.
~
Thh. is less than Kb•l -= 0. 167 therefore compression steel 2. From the lc\er-arm cur\C of figure 7.5 1.
0.88 ( 490
j.,
not required.
0.88. therefore lever ann := lad =
= 431 mm und
M 165 X 1011 1 3 ' A~ - 0.87/.l;: = 0.87 x 500 x 43 1 = RHO mm·
4. Prm ide three H20 bar... area
5. l-or thl!
~teel
.
= 943 mm' .
( ()()/\ ,
( (){) X
prov1ded
-c;;-- 230
943 )( 550
0.75 (
J<
94J 490
= 0.8.:1
( > 0.11%)
and IOOtl ,
100 ~~()
therefore the
~tccl
4.011)
perccntagl! is wuhin the
l imit~
l>pecilicd h) the code.
l~----------------------------------------~) 7.2. 2 Rectangular sections with tension and compression re info rcement, no moment redistribution Comprc!..,ion steel i ~ required whenever the concrete in cn mprc~sion , by i t~c l f. i!. lln::tbk to develop the ncce~~ary moment of resistance. Des1gn charts <;uch Ui> the one 1 figure 4.9 may he used to determine the \tee\ area~; hut the "mphtied equation-. based < the equivalent rectangular Mrcs' block are quick to appl} . The arrangement of the reinforcement to rc!>ist a sagging moment h shown in ligure 7.7 In order to have a ductile section so avoiding u sudden compressive i'ailure or the concrete it i'> generally required that the maximum depth of the neutrnl axis I' Figure 7.7 Beam doubly reinforced to resist a sagging moment
Sect1on
Strams
Equivalent rectangular stress block
Design of reinforced concrete beams Xt>al
177
= 0.45d and this is the value used in the design of a section with compression steel.
The de~ign method and equatiom are those derired in Chapter .J for fertions lllbject to bending. The design ot ~m
step~
arc:
1. Calculate K = f. M , ff K > Kt>al
=
. dbd· 0.167 compressiOn reinforcement is required and .I
=
.l~>;aJ
=0.45d.
2. Calcu late the area of compression steel from I
A, :..
(M - Kbalf~.bd~) .f-.(d dl)
(7.3)*
where .f~c i ~> the compressive stress in the steel.
If t/1 /x < 0.3S the compression steel hal> yielded and j~. O.S7jy~ If d1/r > 0.3S then rhc strain e,c in the CC>mpressi ve steel mu!>t be calculated fro m the proponion1) of the ~trai n diagrum and .f..L £~'"' '• 200 x 10 1e-". 3. Calculate the area of tension ~tee ! required from Ktmtf.:l. hd2
A,
Cl.S7f)•'
f~r lhe areas of \tee! required and the areas provided thai
(.\: """ A: rc") ~ (A, pro' - A, rtq}
_)
(7.4)*
= 0.82d.
with le'ver arm ::
4. Check
I f.,< 'O.S7],1.
11\ - -
(7 .5)
Thi' i'> w en~ure rhar the depth of rhe neutral axi~ has nnl exceeded the maximum value of 0.45d by providmg an over-excess of tensile rcinforcemem.
5. Check thm the area of 1>lecl actual!) provided i' within the ltmil'> required by rhe Code of Practice. ble Ill
,edon
the
(EXA MPLE 7.3 Design of tension and compression reinforcement, no moment redistribution
f the
,,
The beam section shnwn 1n figure 7.R has chnracteriM ic matcril1l :-.lrenp,lhs or f.k 25 N/mm2 and .f)k 500 N/mm1. The uliirnare design moment is 105 kN m, caul>ing hogging uf the heam:
1.
\O
2.
165
bd~J:.~
230 " 3302 0.26 ,> K~al
X
• b:: 230, 1
1011
M
X
15 0.167
that comprc,.,ion 'tee! is required. x 0.45d- 0.45 330 = 148 mm
tl' I \ 50/ 14X 0.14 < ().3R therefore the compre~~ion \tccl ha~ yielded and
~I
' Figure 7.8 Seam doubly reinforced to resist a hogging moment
1 78
Reinforced concrete design From equation 7.3
. Compress1on steel
1\
1
=
(M - 0 167f.:•bd2} ) 0.87J;dd d'
(165 X 10~ 0.167 X 25 X 230 = - -0.87 X 500(330 - 50)
X
3302)
= 496mm 2 Pro\ ide two H20 bar'> for A~. area 3. From equation 7.4 Tem.ion steel A,
Q.J61f~kbd2
628 mm 2 • bottom 'iteel.
---::-~- ~ A
0.87fyl;. 0.167 X 25
().1{7 X 5()() X8H + 496
X X
1
' 2J0 X 33Q2 - - - + 496 0.82 X 330 1384 mm 2
Prmide three H25 har~ for A" area
I
I
1470mm". top ~teel.
4. Chccl. equatton 7.5 for the area'> ot '>tccl required and prO\ 1ded for the compre-;sll)n and tcn~ion remtorccment to ensure dw.:tility of the section {/\ ~ pnw- A ~.fCIJ) ~ (1\l.prnv- A,, ,cq )
That i.,
628
496 (= 132)
1470- 1384 (
86)mm1
5. The har areas provided arc within the upper and lower limit~> specified by the codl To rc!.train the comJWO!.sion l.>teel, at least 8 rnm links nt 300 mm centres should ~ provided.
7.3
Design for bending of a rectangular section with moment redi stribution
The redl\trihution of the moments obtained from the ela'>tiC analysis ol ;1 concrt strucrurc Hil-es account of the plastic1ty of the reinforced concrete as it upproachc~ uhimatc limit srate. In ul is set according to the amount of redistrihution 6. Por the EC2 code it ,.bal ::::
0.8(b- 0.44)d for hL < CSO
(7.L
where
h = moment at the section after reclistrihulion moment at the ~ecuon before
rcdi&~ribution
However the UK Annex to the EC2 modifies the limit to xro~ as Xbal
< (h 0.-l}d
(7.6b •
Design of reinforced concrete beams Table 7.1
Moment redistribution factors for concrete classes
• M.vumum permitted rediWibutton for
cld$S
A
norm~ I
~
CS0/60
ductility steel
" Maximum permitted redistribution for cl.m 8 and C hogher ductihty ~teel, ~ee sectoon I 6.2
In this chapter the examples will he ha~ed on the L K Annex·~ equation 7.6h. hut. because many of the de~igns in the UK arc for project~ uver~ea~ which may require the u'c or the 1!('2 1-opcci lic ati on~. example 4.9 part (a) was hascr.l on the usc of the EC2 ~4ll at for both the EC2 and the UK Annex equationl. so that e examples on mo111cnt redi!>tribution in thi:-. chapter can be readily amended for usc in .:rm~ of the EC2 equmton. The ratio d' fd in table 7. 1 o;ct ~ the limiting upper value for l e yield of the comprCS\1011 Meel. The moment redtMribuuon i'> generally carried out on the maximum momems along a m and the'c arc generally the hogging moment~ ttl the beam~ ~>upports. Example J.9 oment rediwihu tion 'hows how the hogging moment may be n:duccd without l!a,mg the maximum sagging moment in the bending moment envelope. Thus there n economy un the amount of steel rcinfon.:ement requ ired and <1 reduction of the ~e,ti on of' ~tccl hur~ at the beam-column connection. 'he equa tion~ usl!U in the design procctlurc1-. that lollow arc ha~ed on the equation ~ ed tn section 4.7.
- 3 1 Singly reinforced rectangular sections with moment cG·stribution L'tgn procedure
u ~ul:ne K ~e
u~mg
the equations based on the UK Annex to EC2 "
M / fu/ 1/..1.
Kbnt from table 7.I, or alternatively calculate
i..~al
0.454(b - 0.4) - O. IR2{b- 0.4)~ l'or.fc~ 5. C50
here b = moment after redistribution/moment hcfore redislrihution
a
check. that K
< Kb.oi· Therefore compres~ton
steel b not required.
1 7S
180
Reinforced concrete design
3. Calculate ~=d[0.5- j(0.25 4. Calculate A 5
K1 1.134)]
M
= - -.-
o.87A,:
5. Check that the area or steel provided is within the maximum and minimum lirnll rcquiJed.
7.3.2 Rectangular sections with tension and compression reinforcement with moment redistribution applied (based on the UK Annex to EC2) TI1e \teps in the design are :
1. Calculate xb.u < (b- OA)d 2. Calculnte K 3. Take
Khnl
Kh.il If A.
= M /bd2.fck
from table 7. 1 or alternatively culcLLiatc
0.454( f. - 0.4) - 0. 182(b - 0.4 )2 for [cl ~ C50 K~., 1 •
compres<>ion 'tccl j.., required.
4. Calculate the area of comprc"ion 'tccl from
(K - Kt..,,JJ~l hd J~.Jd - d') where .f,,. i.\ the
~tre!>s
(7.7)
in the cmnpres,ion steel
If d'/ r <" OJ8 the compre)>sion /ilCel has yielded and/,.~
O.R7(y~
I r d' / r > 0.38 then the strain c'" in the compressive ~>I eel must he calculated from th(! proportions of the )>train diagram nnd .h, £,_.,._- 200 x 10 1 .._..
=
5. Calculate the area of tension \Lccl from KNtfclbd: -I-A'~ O.R7ho.:. '0.87/yk \\here ;: - d
O.RxhJI/2.
6. Ch(!c(.. cquauon 7.5 fnr the
area~
(1\~-ll'"'- !\~.rcc 1 ) > (A,,pto•
ol
~tecl
required and the an.:u ... provided that
A,,rcq)
This i~ to ensure that the depth of the neutral axis hns not exceeded th(! maximum va[U(! of' Xb3 1 by providing Ull tlVCf-eXCC~S Of tensile reinforcement. 7. ChccJ., that the area of steel pr within the maximum and minimum limit' required.
(
EXAMPLE 7.4
Design of tension and compression reinforcement, with 20 per cent moment redistribution, 8 = 0.8 (based on the UK Annex to EC2) The beam section sho\\n in figure 7.9 has characreri~tic mutcrial ~trengths of r.~ 25 Nlmm 2 and f.,k 500N/mm 2 . The ultimate momt.:nl j., 370 "-N m, cau<>ing hogging of the beam.
Design of reinforced concrete beams
181
Figure 7.9 Beam doubly reinforced to res1st a hogging moment
Section
Strain s
1. A~ the moment reduction factor b = 0.80, the limiting depth (6 - 0.4)d
= 1683-'- 19-t- 1877mm1 Provide four H:?S bars for ;\. area= l960mm~. top ~tee!.
6 . Check equation 7.5 for the areas of !>teet required and provided for the compress10r nnd ren:-.ion reinforcement w l.m!.ure ductili ty of the section
(!\~ I""'
A:. ceq) ;::: (/\,,prov - A,,rc4 )
-
That is
628 - :?24 ( 40-t)
1960
1877 (- 83 ) mm 2
7. Thc:.e area!-. he within the rnax1mum and minimum limits specified by the code. T , re)>train the compre~~ion 1>teel, at teaM 8 mrn links m 300 mm centre!> ~hould hi: provided.
7.4
Flanged beams
sections through a T-benm and an L-hcam \\hich form part ol a and slab tloor with the slab spanning between the beam~ and the area~ the :-.lnb n~ting as the llnngcs of the beams ns f,hown in figure 7.1 1. When the heams .., rc~i!.llng ),agging moments. the slab acts a~ a compn.:ssion llonge and the members rn he de~igned as T- or L heams. With hogging moment' the slab will he in tensiOn a a~o,umed to he cracked. therefore the beam must then be des1gncd a!-. J rectangu \CCII on of \\ itlth b,. and overall depth h. At in tel mediate supports of conunuous hcums where hogging moments occur t 1owl area ten~Jon reinforcement ~hould be spread over the effective width of tr flnngc as shown in ligurc 7.10. Part of the reinforcement mny be conccntrnted over the weh \\idth. I he eltccttve flange '' idth bcu i-. specified by the following equation: Figure_ 7.10
~how~
concrete~enm
or
b.,. -
b d1
L b~n
1
\\here 1><'1 1
1
0.2b, I 0.1/o < 0.2/o tllld also b~rr 1 b1 2h1 is the clear distanec between the webs of adjacent beams /0
i., the diswncc hetween the tigure 7.11.
point~
of contranexure along the beam as sho\\
111
Figure 7.10
T·beam and L br<~m
d
..!!!!:.• _
bv.+b"'2 Section
1- b.. _ Section
I,
Design of reinforced concrete beams lz _
_ _ _ _ _l_, :_
Figure 7.11 Dimensions to be used in the calculation or effective nange widths
zs: lo= 0.851,
0.1 5(1, ... 12)
10 = 0.70/z
lo = 0.1511 +I,
Note: (i) the length ol the cantilever should be less than hall the length or the adJacent span (ii) tht rat•o of ad1acent span lengths should be between 0.67 and 1 .50
b., b•• ,
b_l __ _
So that for the intenor ~pan of a ~ymmetncal T-heam "-ith h 1 brrr
h.,.,
+2 0.2b' -1
0.071J
b"
lh
1/ and /0
=0.7/
+ 1 O.l -'1
For ~ugging moments the flange), uct a~ a large compre~'ive areu rherefore the \Ires~ bh:k for the Hanged hearn \Cction U\ually fall!> within the tlange tlucknc~'· l·or tht!> po~ition of the \trc'' hlod... the \Cction may be designed as an cqutvulent rectangular \Cclion of breadth h1 • Tran~ver11e retnfon:cmcnt .'lhould he placed aero~' the full width of the flange to re.,t\1 the 11hear developed hetween the web and the flange. a., dcscnbed 111 l>Cctton 5.1.4. Qutte often thi\ reinforcement is adequately provided for hy the top !lteel of the hcnding reinforcemen t tn the \lah supported by the beam.
Design procedure for a flanged beam subject to a sagging moment
1. Cu leulatc
~
hrd :fck equm ion 7.I lever arm ~
183
and determine lu !'rom the lever-ann cttrvc of figure 7.5 or from
/Jd and the depth of the stres~ block ~
= 2(d
- ~)
rr ~ <' h the strc~s block falls within the flange depth. and the design mny proceed a~ for a rectongu lar 1.ection. breadth h1• On the very few occasion~ that the ncutrnl axi~. doc:-. fall bclov. the flange, reference should be made to the methOlh dcscnbed in \Cction 4.6.2 for u full analysi),.
2. De!.t£!n tranwcro;e \teel in the top of the flange to rcsi~t the longitudmal \hear Mre,se~ at the flange- \\Cb interface (see section 5.1.-J). These longitudinal shear Mre,,es are a maximum where the 'lopes dM / d\ ol the hcnding moment envelope arc the greatest. That i~ (a) in the regton of tcro moment for the span .'lagging moments, and (b) the region of the maximum momenh for the hogging moment~ at the :,uppon....
184
Reinforced concrete design
(
EXAMPLE 7.5
Design of bending and transverse reinforcement for aT-section A simply supported beam has
T-beam
T
,. .. d'mg moment at mlu-,pan ..I v•a\lmum ....a.en
•
t!t
.w X-6~ M =8
I90o kNm
(1) Longttudinal reinforcement M
/1r rf2./~k
19X X 106 600 x 5301 x 25
0.047
From the lever-arm curve. ligurc 7.5. Ia = 0.95, therefore h:vcr arm ;: = l,,d 0.95 x 530 depth of strcsl. block v '2(d
503 mm ~)
2(530 - 503 )- 54 mm ( < lrr)
Thu' the qres-; block hcs "11lun the Oangc 1\ ,
M
0.87/yl;:
19R )( 101' 0.87 X 5()() X 5()3
905 mm ~ Provide 1wo H25 bar~. area IOOA, b.,d
100 x 982
= 250 X 53()
982 mm 2• For these bars
0.74 per cent > 0. 13
Thus the ~ted percentage " grcutcr than the minimum specified b) the Code of Pmctice. (2) Transverse steel in the flange
The design foliOWl> the
procedure~
anti equations set out in
(i) Calculate the design longitudinal shear
vEd
~>CCtion
5. 1.4
at the web-flange interface
For a 'lagging moment the longitudinal ~)hear stre.~ses arc the greatest over a distance of ~x mcu.~urcd from the point of :t.ero moment and .6..\ i:. tal...cn tl'> half the diswnce to the maximum moment at mid-:,pan. or~'= 0.5 x L/ '2. = L/ 4 = 1500mm.
Design of reinforced concrete beams TI1en.:forl.! the change in moment :::C.M over distance 6x =L/ 4 from the tero moment i~ ll 'u
XL 2
L
-->(-
4
ll'u X I.
L
3wuL2
4
8
32
3
- - X -= - -
X
44
X 6~
32
= 1491-.:--!m
The change in longttudtnal force .J.F at the Hange- web interface ''
~Fd
=
.J.Af
(d - hr/2)
>< bto
b,
where b1., is the breadth of Hange outstanding from the web. Thus
(ii) Check the strength of the concrete strut From equation 5.17. to prevent cru~hing of the compres~ive ~trut in the !lange 0 6( I j~lf''!.50 )J.~ - I 5(cot 01 tan fl1)
l' t d <
fhc moment... arc 'aggmg 'o the nange
26.5 with Or
o,
i,}
in compre,~ton and the lun11' tor 81 an:
45
th~.: mimmum value
0.6( I
.'. l'ftll mto\1 -
of 26.5
25/ 250) x 25 (2.0 0.5)
= 5.4
( > O.•H N/mm1)
and the concrete ~ trut has ~ufh cient ' trength with () 26.5 (lor a nungc in ten&it1n the limit~ on{) arc 45' > 0 > 38.6 · or 1.0 cot (J :S 1.25).
(Iii) Design transverse steel reinforcement Transver~e shcllr reinforcement is rcquirl.!d if ,., d ;> 0.27fc 1 ~ where ./~ 11, i!. the charac1eriM ic axial tensile Mrength of concrete.! (= l.XN/mm1 for clu~>s 25 concrete). The maximum allowable va lue of ''Ed - 0.271_.~ = 0.27 o< I.H 0.49 N/mm2 ( • 0.43) and transvur!.c ~hear reinforcement i~ therefore not reqlllrctl. A minimum area of 0.13t:f, of tranwerse steel should he provided :h given in table 6.8 or in tahle A3 1n the Appendix. lienee
A,1
= 0 13bh 1 100 1
0. 13
1000 x 150j 100
=
195 mm1/m
Pro' ide 1110 har' at 300 mm centre' 262 mm 2/m (see table A.3 111 the Appendix ). Longitudinal remforcemem :.hould also be provided in the flange a\ !>ho'' n in figure 7.12.
185
186
Reinforced concrete design
7.5
One-span beams
The following example <.bcribe~ the calculations for designing the bending reinforcement for a simply supported beam. II b1ings together many of the items from the prcvioul> sections. The :,hear rcinfon.:cmcnt for this beam is de:;igned later in
example 7. 7.
(
EXAMPLE 7.6 Design of a beam - bending reinforcement
The beam shO\\n in tigure 7. 13 :-.upports the foliO\\ing uniform!) pennllncnt ;oad Kl ()0 1-N/m. including self-weight v:1r1ahl ~ load
The /~1
di~tributed
= I RkN/m
lfk
eharact~-istic
\lrengrhs of the concrete and :.tcel arc ./:k 30 N/mrn2 ami 500 N/mm". Effective tkpth. d 540 rnm anti breadth, /1 ~()0 mm.
Figure 7.14 Simplified rules for curtailment of bars in beams
50%
0.08l_
Simply supported
where. from the lever arm curve of figure 7.5 /3
-
0.82. Thus
0.167 X '\() X 30() ~ 5~0~ + 2')'l __ 0.87 X 500 X (0.82 ~ 54())
A~ =--
2275
+ 222
2497 mm 2
Provide two 1132 har~ and two H25 bar:.. area
2592 mm 2 , IOOA,/bd
1.6 > 0. 15.
(c) Curtailment at support
The rension reinf'on:cment shnuh.l he anchored over the supportll with a hend as shown in figure 7. 14 which i1. ba~cd on past UK practice. Two b:m, may he curtai led ncnr to the su pport~>.
(d) Span- effective depth ra tio
100. 1, ......41 /bd
f1
( 100 x 2~97)/(300
From tuhle 6 10 or figure 6.3
ba~ic
540)
U4 per cent
span-effective depth ratio
l..f
Modification fur !'ltccl area prov1ded: Modified ratio
1-Ul x
2592 2497
= 14 5
Span cffcctin: depth rauo provided
= 6000 = 11 .1 540
which Ill lcl-\ than the allowable upper limit. thu). deflection rcqturement-, arc likely to he sati.,licd.
7.6
Design for shear
Th~; theory and design requirements for :.hc(ll' were covered in chapter 5 and the relevam design equation!> were derived bused on the requirements of EC2 u~ ing the Variable Strut Inclination MetJ10u. The ~ heur reinforcement wi ll usually take the form of vertical lin~~ nr a combination of l i n~ s and bent-up bars. Shear reinforcement may not he required in very min uno light loads. The following notation ~~ U!>ed in the equation~ for the \hear de,ign
A\,.
,\ :: f~.., 0
the cro,:.-~ectionul area of the two legs of the slirrup the -,pacing of the \tirrups the lever arm between the upper and lower chord analogou-; tn1ss the design yteld
~trength
187
of the stirrup reinforcement
member~
or
the
188
Reinforced concrete design
f.11
the cbaractcri),tic strength of the stirrup reinforcement
VEd - the !>hear force due to the action!> at the ult1mate limit state
= the shear force in the stirrup = the shear resistance of the stirrups
v\\J VRtl ,
the maximum design value of' the shear which can be resisted by concrete strut
VRd.ma\
th~
\I 7.6.1
Vertical stirrups or links
The procedure for
dc~igning
the shear links
i~
as follows
1. Calculate the ultimate de~ign shear force' Vw along the beam ·s span. 2. Check the crushing .,trength VR11. m:u of the concrete diagonal 'itntt at the ~ection l maximum shear, usually at the face of the heum support. For most c;.u;cs the angle of inclination of the strut is () 22°, with col() = 2.3
and tan 0 ~I!J
0.4 so that from equation 5.6:
= 0.124/>.. d( l - / ck/ 250)/ck
111 ,
(7.9 •
and if VRJ , m11 , 2: Vr'd then go to Mcp (3) with fJ 22 nnd cot 0 - 2.5 but if VKJ mnx < V1" then () > 22 and 8 mu~t be calculated from equation 7.10 .. fJ
V&~ } < 45 O. !Mb.. tlj,d l - .f, k/ '250) -
() 5 ,in 1 { -
(7.10 •
3. The shear links required can be C
Vtu 0.7Hdf~k cot 0
(7. 11 •
\\here A"' 1' the cro~l>-ilecuonul area of the leg!. of the surrup' (2 x 7HJ 2/4 for ~ing stirrups) ror a predominately uniformly distributed loatlthc shear vl(d should be tuken at " distanced from the face of the support and the shear remforccment ~hould continu~ w the lace of the supp<.>rt. 4. Culculate the minimum ltnJ.. ... requtred by EC2 from A,"
O.O~fc~· 5 b.,..
mlu
- .\'
nnd the \1111111
;--}~k ~hear
resistance for the !Jnks actually specifred
A," o · n - X 0 . 7od/v~ COtv .$
(7. 12 •
•'
(7. 13 .
Thi~
value 'hould he marked on the shear force envelope to show the extent of the~ link!:. a' 'hov.n in figure 7.16 of example 7.7. 5. Calculate the additional longitudtnul ten~ile force cuusetl by the shear force 6F1<1
= 0.5\IE.JCOtB
(7.14)•
This additional tensile force increases the curtailment length of the tension bars a' shown in section 7.9.
Design of reinforced concrete beams Figure 7.15 Types of shear link
~by the
Open hnk
Closed link
Multiple lonk
/
The minimum spacing of the links is governed by the requirements of placing and compacting 1he concrete and should not normally be less than about 80 mm. EC2 gives the following guidance on the maximum link spacmg: (a) Maximum longitudinal ~pacing berween shear lin~s in a serie' of link~ S1 mox ~ction
of
ote = 2.5
(7.10)*
(7. 11 )*
I cot a)
where (t is the inclination of the shear reinforcement to the longitudinal axis of the beam. Mnximum transverse spacing between legs in a serie1. of shenr links
(h)
.111 mu~
(7 .9)* m 7.10 a!>
= 0.75d(l
0.75d ( ::; 600mm)
Types ol links or &tirrups arc shown in figure 7. 15. t he open link" an.: u"ually u~ed in the span of the beam with longitudinal '\teet wn~i~ung of top hanger hars and hotlom tensile remforcemcnt. The do~ed links arc u'ed to enclo'e top and bottom reinforcement such as lhal ncar to the -;upports. Multiple linh nrc u!-.ed when there arc high shear forces to be re~isted.
( EXAMP LE 7.7
· 1r 'iingle
Design of shear reinforcement for a beam
ai.;en at n
Shear reinforcement b to be designed for the one-span hearn of example 7.6 as 'hown in figures 7.11 nnd 7.16. The total ultimate load i1> 10~ kN/metrc und the characteristic ~ Lrcngths of the concrete and Mcel arc ./~l 30 ~lmm~ und ./y< 500 N/mm'.
I ~ont inue
=
1.45m ~ SR nofllln~t lmks
151kN I
(7. 13)*
308 kN [
-
~
..,..- ~
1 ~51kN
.-
~
I
SF d1agram
1 of these
7. 14)* m bars a:.
Figure 7.16
~I
p. l2)*
9
H8 It 200 HS links@ 350
9- H8 Cill 200
308kN
Non·contlnuous beam-she
relnforcement
190
Reinforced concrete design (a) Check maximum shear at face of support
Maximum design l>hear = '''u x effective span/ 2 - 108 x 6.0/ 2 = 324 k.N Design shear at face of ~upport \'Ed= 324 - 108 x 0.15 = 3081-.N Crushing ~trergth
VRd rna'
of diagonal strut, assuming angle
(J
= 22 . COl B= 2.5 is
OJ?4b"'d{1 - f..l /250}(r~
VRd. max
= 0 124 x 300 ) 5-tO(I - 30/ 250)
530 kl'\ ( > \'F.t
X
30 x 10--'
308 kN)
Therefore angle B = 22' and cot B = 2.5 as n'sumcd. (b) Shear links At distance d from face of support the de~ign ~hear i~ \'F..s 308 - 108 x 0.54 =- 250 kN
308 - '''ud =
ll~,;o
A,w
0.7Rd(,~
.1
cot() 250 10 1 540 x 500
0.7!\
0.475
2.5
Using table !\.4 tn thl! Appendix Provide 8 mm links at 200 mm l:Cntres. A "/ 1 = 0.503. (c) Minimum links
0.0~(1t 5 hw
- ,,k0 ()!{
X
3()0 <
X
100
500
0.26
Provide 8mm links at350mm centres. A,w/ 1 0.287. The shear rc~istance of the links actually specified i~ \fm•n -
' '" ~
v
0.78t/J;. k cot lJ
= 0.287 x 0.78 x 540
500 x 2.5
Ill' = 1511-.N
(d) Extent of shear links
Shear link\ are required at each end of the hcam from the face of the support to the point 11here the de~tgn ~hear force '" 1'm m = 1511..'1 as shown on the \hear force diagram of figure 7.16. From the face of the support dt~tance 1 ll'u
308
LSI 108
1.45 metre~
Therefore the number of H8 links at 200 mm centres required at each end of the beam is
1 -t( \ 1)
1+ ( 1450/ 200) =9
!>paced over a dbtance of (9 - 1)200 = 1600mm.
Design of reinforced concrete beams (e) Additional longitudinal tensile force
.J.F,d
= 0.5\lld COl 0 0.54'308
J(
2.5
=385kN This additional longitudinal ten~ile force is provided for by extending the curtailment l pomt of the mid-1.pan longitudinal reinforcement a~ di,cu!.sed in ).ection 7. 9.
7.6.2 =
Bent-up bars to resist shear
In regions of high ~>hear forces it may be found that the usc of links to carry the full force wil l eist bending forces but can be so used to re!.ist part of the ~>hear. The eq uation~ for dc11igning this type of shear reinforccml.!nt and thi.! additional longitudinol tension fore!.! were derived in chapter 5 and arc given below A,,.
~~~~~
0.78tif..,dcot o + cot 0) sin n ~Fu1
0 5 V&~ ( COl 0 - cot ll )
"hl.!rc o 1s the angle ol tnclinallon "ith the horitontal olthe belli up har. Bent up har-; muM be fully anchored paM the point at'' hich they arc acttng a' ten-.ion mcmhcrs. a'> W
7.7
1"101111 _ ~ of
-mts
Continuous beams
Beams. ~ labs and columns of :1 ca~t in .1i1U structure nil m:t together to fnrm a Cl>ntinuou~ lond·bcaring strtH.:turc. The rein forccment in a conti 11t1ous beam mu~t be de:-.igned and derailed to maintain lhi~ continuity by connecting adjacent sp11ns and tying together the benm nnd its supporting column!>. There must al:-.o be trnnsvcr'c rcinf'orcement to unite the ~ lnb nnd the benm. The bending-moment envelope is generally a 'cries of sagging moment). in the spans and hogging moments at the supports as in figure 7.17, hut occasionally the hogging moments may extend completely 0\·Cr the ).pan. Where the snggmg moments occur the hcam and -;lab act together. and the beam can he des1gned a~ a T-scdion. At the supports. the heam must be designed a.s a rectangular ~ec11on because the hogging moment~ cuu-;e ten~ton 111 the slab. The moment of reo;1stance of the concrete T-bcam 'ection 1s l>Ome\\hat greater than that of the rectangular concrete section at the support\. Hence it I') often advantageous to rcdbtrihu tl.! the support moments as described in chapter 3. B) this means the de~ign !>upport moment). can be reduced and the de-.ign -.pan moments possibly increased.
19
192
Reinforced concrete design
,---
Design of the beam follows the procedures and rule~ set out in the previous section, factors v. hich have to be con),idered in the detailed design are as follow\:
-o~er
1. At an exterior column the beam reinforcing bars which resist the design moment' must ha\e an anchorage bond length within the column. 2. In monolithic construction where a l-imple ~upport has been lll.!>umed in the structural analysis. partial fix.iry of atlcast 25 per cent of the ~pan moment should be allowed for in the design. 3. Reinforcement in the top of the \lao must pa<;\ over the beam steel and still have the necessary coYer. Th1' must be con\idered \\hen detailing the beam reinforcement and when deciding the effective depth of the heum at the :-.upport sections. 4. rhe column and beam reinforcement must be carefully detnilcd so that the bars can pa.\s through the junction' without interference. Figure 7.17 illustrate!:> n typical armngement of the bending rcmforcement torn twocontinuous beam. The rcinl'orcemcm hal> been arranged with reference to the bending-moment envelope and in accordance with the rules for anchorage ru1d
~>pnn
Figure 7.17 Continuous beam arrangement or bending remforcement
B.M Envelop!!
Design of reinforced concrete beams
the uld be
10
cment
curtailment dc....cnbed in section 7.9. The application of these rule!'. establishes the cutoff point~ beyond wblch the bars must extent! at leal>t a curtailment anchorage length. It should be noted that at the external columns the reinforcement ha-. been bent to gi\'e a ful l anchorage hond length. The shear-force envelope and the arrangement of the !.hear reinforcement for the same continuou~ beam are shown is figun.: 7.18. On the shear-force envelope the resistance of the minimum stirrups has heen marked and this show~ the length~ of the beam which need shear reinforcement. When de~igning the shear reinforcement. reference should be made l
·..,can ( EXAMPLE 7. 8
and
Design of a continuous beam
=
The be41m hal> a Width, bw 300 mrn and an overall depth, /1 660 mm with three equnl . . pans. L 5.0. In the tranwcr'c direction the beam'i ~pacing'> arc 8 4.0 m centre' \\Jlh a slah th1ckness. lrr = 180 mm. a!> :.hown in ligures 7.19 and 7.20. The ~uppm1~> have a width of 300 mm.
=
The unifonnl} distributed ulttmutc dl.!~ig.n load. 11" = 190 kN/m. The ultirnatl.! design and ),hcurs near mid-~ pan und the support~ nrc shown in 1igurc 7.19. The characterbtic 'trcngth~ of the concrete and ~ted arc ./,k = 'O '-'/mm' and
mome nt~>
fy~
500 N/mm,.
I
Shear V(kN)
428 427
428
333
570 522
8
At
0
523
523
0
Moment M (kNm)
1. s.om
t
522
S.Om
F• 1.3SG,
~
570
C l
427
1
ol 5.0m
L
I
1 SOQ,
Total ultimate loud em each ),pnn b
r
950kN
190 x5.0
Design for bendmg
(a) Mid-span of 1sl and 3rd end spans - design as a T-seclion Moment .[!8 k.N m sngging Lffectivc width of Hange ben
hw + 2t0.2b' + 0. 1 x 0.851.1
= 300 bw
( :5 ""' + 2[0.2 x O.X5Lj) (see fi gure 7. 11 )
:! 1(0.2 > (2000- 300, 2)) t (0.085 x 5000)J - 1890 mm
so that the ~tress block mu~t lie within the 180 mm thick flange and the eel ion i designed as a rectangular section with b = b1• M
A,
= 0.87}~k;: 6
X J0 = 0.!!7428 X 50() X 570
,
l?Jomm·
Provide lhn:c TI25 bars and two Hl 6 hnr~. area = IH72mm2 (hotlom ~teel). (b) Interior supports - design as a rectangular section M
523 ~ m hogging A/ btf]}, k
523 X (()(I 3()() !\ 5802 X 30
0.173 > 0.167
Therefore. compres<;ton steel i' required.
(K K'lf-~btl' 0.87j;dcl - d') (0.173 - 0.167) '< 10 100 x =- -O.H7 X 500(580 50)
sxo'
Thb :.mall area of reinforcement can be prt)\ tdcd h) e\tcnding the bottom :.pan bcyoml the i merna I supports. From the le\er arm curve (Jf figure 7.5 1., 0.82. therefore: 0.167fckbtf1
--::-~:--0.87/y~~
bar~
1
+A
• 0. 167 X 3() X ]()() X 580~ O.X7 X 500 x (0.82 x 580) 24-l4 I 79 - 2523 mmz
+
79
Provide four H25 bars plu~ two IT20 bar~. :1rea 2588 mm 2 (top !>ICe(). The arrangement or the reinforcement is shown in figure 7.20. At end support A two 1125 har~ have been provided a' top continuity ~ted to meet the requirement ol item (2) in section 7.7. (c) Mid-span of interior 2nd span BC - design as aT-section
M
= 333 kN m. sagging From figure 7 . II. cffccti\'c flange \\ idlh bttr
=- b.., +2[0.:!b' + 0.1 x 0.70L]( < bv. + 2[0 2 v 0.70/, ) =- 300-:! (0.2 x (2000 300/ 2)) (0.07 x 5000)1 1740 mm h.., I :![0.2 x 0.7L] = 300 + 2[0.:! > 0.7 ) 5000] = 1700mm
Design of reinforced concrete beams !.--
H8 @ 200
H8 @ 300
H10@ 200 Figure 7.20
2-H25
4-H25 >
II
2· H20
End-span remforcement details
II
...
3-H25 2-H16
ls5.0m_ __
2525 25
..
252525
1~ 300 Sections
rn1dspan
near the interior support
Calculating M /( brcPf
333 X 106 0.87 X 500(0.95 >. 600)
M
A,=--.0 87})~ ~
0.95
13-B mm'
Provtde three 112'5 har.... area = 1470 mm2 (hottom ~tccl ).
Design for shear (a) Check for crushmg of the concrete strut at the maximum shear force N'laxtmum !>hear i\ 111 spans AB and CD at supports D and (', At the face ot the support~ V~.J
570 570
ll'u X
support Width/ 2
190 x 0.15
= 542 kN
Cru ~hi 11g ~trenglh of diagonal ~t1111 i~
IlK~. "~"~ - 0. 124bwd( I
.kk/ 250Kk assuming angle {) 22' , cot() = 0.124 x 300 x 600( I - 30/ 250) x 30 = 589 ~N ( > Vm
nt
been
Therefore angle ()
22 and cot fJ
2.5
542 kN)
::.?..5 for all the ~hear calculation ~.
(b) Design of shear links
(i) Shear li11k.l in end rpan.s at
~ upporrs
A and D
Shear distanced from face of suppon b Vl:d A,,. ~
= 427 -
190 x (0.1 5 + 0.6)
VF..s ...,.:.:..:~~ 0.78df,l cotO > 10' = 0.78 X 2R5 - -,-=0.49 600 X 5()() X -·5
Provide 118 linb at 200 mm centres. Aw. j s
0.50 (Table A4 in Appendix)
285 kN
19~
196
Reinforced concrete design
Additional longitudinaJ tensile force is ~/·,d
=0.5VEdcot B - 0.5 X 285 X 2.5
=3561-.N This additional longttudinal tcn~ile force is prO\ ided for by extending the curtailment point of the longitudinal reinforcement, as discus~ed in !.CCtion 7.9 (ii) Shear lin/.. f in end 1pans at 1'11pports 8 and C Shear Vhu distance d fmm face of support is
570
VEd
190(0.15 - 0.58)
= 431 kN Therefore: I\"'
Vtu
0. 78t(fv~ col 0 43 1 X [() -' 0.78 < 580 ')()() 0.762
-- - - - -2.5 I(
Provide 1110 link~ a1 200 mm ccmrc~. A .,. /s =0 762 (Table A4 in Appendix) Additional longitudinal ten~1lc force 1s ~~ ~~~
0.5 \'1.1 cot U 0.5 ~ 431 "2.5
539 ~N I hb additional longiludinal tcn~de force is provided for by extending the curnulment point of the longitudinal reinforcement. a~ di~cus\ed in section 7.9.
(iii) Shear fink.1 in middle span IJC til mpport.\ B and C
Shear d"tance d from the face of ~uppurt = 522- 190(0. 15 I 0.6) 380kN. The calculatiOnS for the shear links \o\OUid be ~imilar 10 thol>c for the other supports in sections (i) and (ii) giving IOmm link'> at 225mm centres. fhe :~ddition:~l longitudinal ten,de force. F1.t 0.5 3RO 2 5 475 kK (i1•) Minimum 1/tear lin/.. v
O.O~f:~ ~ h"" --=-y;:-
A,.,., mm
o.ox 30°5 ---
300
500 = 0.:!63
Provitle HR Shear
Vmon
link~
m 300 mm spadng. A,w Is of links prm ided
= 0.335 (Table A4 in Appendix).
resi~tancc
Asw
=-X
s
0.78d/~1 . ' cotB
=0.335 X 0.78 X 600 X 500 X 2.5 X l96k~
)()
"\\ I
Design of reinforced concrete beams
197
( 1') £trem of vhear links
Links to re:.i't ~hear are required over a dististed by Vn1m 196 k~. a.., provided by the mimmum linl.s. For the face of the end supports A and D the distance r 1 i!>
=
VF.1-V
~ ~-0.15 ll'u
=
427
196
190
For the interior 570 .r2
-0.15=l.07m
~upports B
196
190
and C of the I st and 3rd spans
0. 15 = I.S2m
For rhe links nL ~upporti> 8 and C in the middle spun .\~
522
196 190
- 0.15- 1.57m
Based on these dimcm.ions the link!. are arranged
as shown in figure 7.20.
l ------------------------------------------~) --
7.8
Cantilever beams and corbels
The effective .,pan of a cantilever JS e1ther Ca) the length to the face of the -;upport plm, half the O..:am's overall depth, h or (b) the distance to the centre of the ~uppon 1f the beam " contlnuou.... The moment-., 'hear:. and deflections for a cantilever beam arc l>Uh'ltanually grcatc1 than tho~c for a beam that 1s l>Upponed at both ends \\.ith an equivalent load. Aho the moment:-. in a canlllcvcr can ne\'er be red1'ilributed to other part\ of the 'ltructurc the beam mullt alway!. he capable of resisting the full static moment. Occausc of these factor~. and the pmhlcms thut often occur with increased detlections due to creep, the design and detailing of a cuntilcvcr Ileum !.hould be done with cnre. Particular aucntion ~hnu ld he paid to the anchorage into the support of the top tcn~ion reinforcement. The steel should he anchored ut rhe support by. nt the very leust, a full maximum anchorage length hcyond the end of its effective spnn. Some de~ooign orli<.:cs 11pecify nn anchorage length equal LO the length of the <.:anlilcver, mostly to avoid !>lecl lixing error) on site. Load~ on n cantilever can cause the adj
1 35C + 1 SQ,
~endtng
Figure 7.21 Cantilever loadtng pattern
moments
198
Reinforced concrete design
7.8.1
Design of corbels
A corbel. as shown in figure 7.22 is cons1dered to be a ~hort cantilever when OAI!, :::; a, :::; he where h, is the depth of the corbel at it\ junction with the column and a" b tile distance from the face of the column to the hearing of the vertical force. F bJ· When the \·ertical load has a stijj' bearing a, may he measured to the edge of the bearing but where a flexible beanng 1<; u~ed a, i' mea,urcd to the vertical force. Corbels can be de~igned as a stnll-and-ue 'Y'tem illu!ttrated tn figure 7.'22. In the figure the \'erticalload FEd at point B 1s resi!.ted by the force Fed in the inclined concrete stnlt CB and the force F1d Ill the horilllntal steel tie AB. The design and de1ailing of a cnrhel has the following requirements:
1. The bearing stress of the load on the corhel directly under U1e lond should nut exceed 0.*8( I - j~~ /250)/~~. 2. A horizontal force H~:.d 0.2Fr J must nlso be re:-isted. Thii. force acts at rhe level ol the Wp of the bearing, a diMance ar r nbove the horitonlul tic. 3. The main tension steel. A, "'''1" must be fully :111chored into the column and the other end of these bars must he welded to an anchorage device or loopfl of reinforcing hars. 4. The ungle of inclination, (J of the compression \trut must he within the limn, 22 ~ 0 :::; 45°. or 2.5 ~ cotiJ ~ 1.0. 5. The dc-,ign ~tress.f~d of the c.:oncrctc strut mu~t not exceed (nu.kkhc)l•r where: ''r - 0.6( I - /..1../'250) o,..
= 0.85
-., = 1.5. the part1al factor ol ~al ety
for concrete in compression.
Therefore f~<~ mu~t not exceed 0 1-lf,d I ./,L/'150) 6. llorizontal linf.s of total area A hnk ~hould he prm idcd to confine the concrete in the compres'>ion strut and I: A, ltn~ 0.5/\, mJ"' . The strut and tie system of design
Tht.: forces on a corhcl produce n complex comhinutinn or stre~ses due to bearing. shcnr. direct compression, direct tension nnd hending concentrated into a small urea. The strut and tie system combined with good detailing i:-. able to simplify the design 10 produce a workable and safe design. Figure 7.22 shows the corbel with the inclined strut RC at un angle fJ to the horizontal tie AB. The force in the strut is Feu and F,d in the horizontal tic respectively. Point B i' distance a' (ac -r 0.2aH ) from the face of the column because of the effect of 1hc honzontal force. HEd ( = 0.2,..,~). From the geometry of the triangle ABC. the lever ann depth is given b) ;; = (a, -t 0.2an) tan 0. (a) Force in the concrete strut, F,d
The dc~1gn stress for the concrete strut IS/.:.r 0 . 34/..~ ( I f,~ /250 ) . From the gcometl) of figure 7.22 the width of the concrete strut mca.,urcd vertically is 2(d ;;). Hence. the width uf the strut measured at right angles to ih axis b given by '''wu1 2(d- ;;) cos (J
Design of reinforced concrete beams a'
-v··
Figure 7.22 Strut and tie system in a corbel
l 2(d
'
2(d- L)cosO '
,,
v Thus the force Fn1 ill the concrete strut is /·~~~
= }~d
X ~1 \uul X
= }~d
x 2{d
bw ;:) x bwcosO
(7. 15)
where b" is the width of the corbel.
(b) Angle of inclination, £1 of the concrete strut Rc~oh ing
vertically at point B:
/~oJ \in 0 - /cd )(
2(d
z) x b.. > cosO x -.inO
.: kiJ X (t/ a' tan O)b,. Stn 20
Re:urangtng I· F..t
(7. 16)
f~udbv.
Thi!- equation cannot be solved directly for f) but table 7.2 (overleaf), which has been developed directly !'rom equation 7.16, can be used.
(c) Main tension steel, As,maln Re~o lving
F,d
horiwntally at B. the force F,J in the steel tie is given hy
I· c1t cos 0
Fc.J ~;os ()/ ~in 0
FFJ cot B
The towf force F~ in the steel tie. including the effect of the hori1ontal force of0.2Ft.d• io; given by
F;d
FEdcot8 t 0.2fht = FEd(COl 0 t 0.2)
(7. 17)
The area of main tension steel, A, m.un• i~ given by
A,. maon
199
= F;d /O.R~f)l
(7.18)
200
Reinforced concrete design Table 7.2 Values of 0 to satisfy equation 7.16 H
EXAMPLE 7.9 Design of a corbel l)esign the reinforcement for the corbel ~hown in figure 7.23. The cor·hcl ha~ u breadth h 350 mm and supports an ultimate load of Vhl 400 1--N at a tli!.lancc ac - 200 mm from the face of the column. The bearing is ncxihlc and at u di~>I
Check the bearing stress Safe bearing Mrcss
0.48( I - !d/250)/.;k
Actual bearing wess = 400 x I0'/(350
0.48( I
120)
Concrete strut The effecti \'e depth of the corbel b d Di:.tance. u'
= (200 + 0.2 x 75 ) 215/ 550 = 0.40
Therefore a'/d
550 111111
215mm
30/ 250)
9.6 ~/mm 2
30 = 12.7 N/mm 2 12.7 N/J11Jn 2
Design of reinforced concrete beams Figure 7.23
I .fed
Corbel example
OJ4,t;.~ (I .f,;k/250)
Fr.d ./~ddllw
0.34 x 30 x ( I 400 x 1000 = O132
8.98 X 55()
lienee, from table 7.2. 0
X
35()
30/250)
8.98 N/mm 2
·-·
35.5 .
Main tension steel The force 1n the ma111 tcnc,10n steeiJS 400(cot 35.5
MO >< to' 0.87 500 Pro\ltlc two II 12 hal".. area
+ 0.2) = 400( 1.40 + 0.2) = 6-tO kN
1471 mm 2
1610 mm 2•
Horizontal/inks A,
hnb
•
0 5A, """"
735 mm,
Pro\itlc four 1116 linb. A, 11 11 ~, 804mm~. Figure 7.2-1 :-,how' the detailing of the reinforcement in the corbel. Figure 7.24
Reinforcement in corbel
2H32
201
202
Reinforced concrete design
7.9
Curtailment and anchorage of reinforcing bars
As the magnitude of the bendtng moment on a ~am decreases along its length so rna~ the area of bending reinforcement be reduced by curtailing the bars since they are no longer required. as shown in figure 7.25. It should he recognised though that because oi the approximations and assumption<, made for the loading. the qrucrural analyst:. and the behaviour of the reinforced concrete, the curtailment cannOt be a particularly precise procedure. In addition the curtailment length\ are in man} case11 supe!'lleded by the requirements for ser\'iccability, durability and detailing. such n~ maximum bar l>pacing minimum bar numbers and curtailment beyond the critical sections for shear. Each curtailed bar should extend a full anchorage length beyond the point at which 1 is no longer needed. The equations for an anchorage length were derived in !>ection 5.2 The equation for the design anchorage length, /h
fbd = - - ¢ X 4Ni,d
rtn
where is a series of coefficients depending on the unchoruge conditions r/1 is the bar diameter .fbd is the design bond strength which, for a beam. depends on the concrete >trer: and the bar size and whether the bar is in the top or hot tom of the beam. The honds better wnh the compacted concrete in the hottom of the beam.
For a straight bar with o _ 32 mm. the order of anchorage lengths are /bd = 52c 1 r top bar and /bd = 36¢ for a bonom bar \'wllh cla'~ C30 concrete. The curtailment of the tens1on reinforcement i' ha,ed upon the envelope of ten force~. F\. derived from the bending moment envelope a~ 'hown 1n figure 7.25 such tmr at any location along the -;pan F,- Mf.1/:. + 6.F1o
where MEd
il> the design bending moment from the moment envelope
z i~ the lever arm !:J.F1d i~ the additional tensile force obtuined from the design for ~hear 1ot:
~F1d is a maximum where the shear force is a maximum at sections of zero moment, and ~Ftd b zero at the maximum moment near to mid-span and the interior ~upp011. For members where shear reinforcement is not required the tenc;ile force envelope may be estimated by snnply 'shifting· the bending moment envelope diagram honzontally by a dt!>tance a1 (=d) as shown in figure 7.15. To determme the cunailmenr positions of e~ch reinforcing bar the ten\ile force em·elope i" di-.ided into secuons as shown. in proponion to the area or each har. In figure 7.25 the three haro; provided for the sagging envelope and the four for the hogging cm·clopc arc <.:onsidcrcd to be of equal area so the em elope i" div1ded 11110 three equal ~CCtions for the sagging part of the Cll\•clope and four for the hogging pan. When considering the curtailment the following rules must abo he applied:
1. At lea~ t one-quarter of the bottom reinforcement should extend to the \Upports 2. The hollom reinforcement at an end support should he anchored into the ~up pon n~ :-.hnwn in ligun; 7.26. 3. At un end ~upport where there i:, little or no fi xity the holtom steel should be designed to rc1.ist a lcn~i l e l'o rce of 0.5V~:c~ to allow for the tension induced by the l>hcur with a minimum requirement of 25% or the reinfon:emcnl rrovided in the ~pa n.
4. /\tan end :-.upporl when: there is f'ixity but it has been analysed a\ a ~imp l e ~upport. top steel 1>hould he tle~igncd and anchored lo resist at leu't 25 per cent of the maximum :-.pnn moment. 5. At internal ~upporh the houom ~tcel !.hould extend at lca~t 10 har diumcter:-. ~~ beyond the face of the \upport. To achieve continuity and re~i,tuncc to '>uch factor\ a:-. acctdcntal uamage or '>ei-.mic force!>. plice bar~ 1>hould be pm\tdcd acrm., the '>Uppon with a lull anchorage tap on each stde as shO\\Iltn ligure 7.27. 6. Where the load~ on a heam are $Ubstamially uniformly ut'-.tributed, 'implified rules for curtailment rna) he used. These rules only apply to contmuou:-. hcam" if the chnrncteri~tic variable load does not exceed the charncteristic permanent load and the l>ran'> arc approxi'lnntely equal. Figure 7.28 show~ the rule'> in diagrammatil.: form. llowcvcr 11 -.houlcl be noted that these rule:) do not appear in EC2 and arc based on previou& eMablished UK practice. Figure 7.26 Anchorage of bottom remforcement at end supports
( 1) Beam supported
on wall or column
(2) Beam mtcrsecting
another supporting beam
Figure 7.27 Anchorage ill intermediate supports
204
Reinforced concrete design
!'~ ;I
Figure 7.28 Simplified rules for curtailment of bars in beams
',
0.08L
100%
I
~008L
'
Simply supported
c .. 0.25~ c = 0 1St
60, ;;
20% 100%
30%
Continuous beam
7.10
Design for torsion
The theory and design requirement~ for torsion were covered in section 5.4. The design procedure consiw. of calculations to determme addit1onal arcus of lin~s and longiiUdinal reinforcement to re&ist the torsional moment. u~ing an equivalent hollow box section U),uully 1t is not necessary to design for tor!-.ion m \latically indetcnninate sLructure' \\here the tor<.ional forces are often only a 'econdary effect and the mucture can be m cquilihrium even if the tor,ion i' neglected. When the equilibrium depend~ on thl toNtmal re<,t~tance the effect\ ol tor... ,on nlll\t he cnn~tdered. 7.2.1
Design procedure for torsion combined with shear
Notation 71,J Design torsion moment I'Ku nw• Maximum torsional moment of re'iistance v,,,, Design shear force VRd ,m.. ~
Maximum shear resistance based on crushing of the concrete
The following section outline~ the procedure for de~tgning for tor~ion and explains ho\\ torsional de~ign must he considered together wtth the de~ign for shear. (1) Design for shear using the Variable Strut Inclination Method The procedure for thts is descrihed in 'cctiom. 5.1.2 and 7.6 and illu-.tratcd with examples 7.7 and 7.8. Usc the procedures previou~ly dc~cnbcd to determme the angle of inclination 0 of t11e concrete compressive strut and the stirrup remforcement to resi).t the shear forces. Also required is Lhe additional horiLOntal tensile force .:lf1.1. The angle 8 should range between 22 and 45 so that cot 0 i'> between 2.5 and 1.0. The \alue determined for 8 should be used throughout the \uhsequent sections of the design.
Design of reinforced concrete beams
(2) Convert the section into an equivalent hollow box section of thickness t !See figure 7.29b.) Area of the section Perimeter of the section ~o
A 11
that for a rectangular section b x lz
bll
2(/h- /r) Calculnte the nrea AL within the centreline of the equivalent hollow box section. For u rectangular section A~
= (b- t)(h
t)
and the peri meter of the centreline is IlL
2(b
~
lz - 2t)
(3) Check that the concrete section is adequate to resist the combined shear and torsion using the Interaction condition , , ...d
-+~ < 1.0
7Rd . mu
VRd
ma~ -
where
Ill:
and l't
0.6( I
}~L/250)
(4) Calculate the additional stirrup reinforcement required to resist torsion
A,w ~
Tt:.u 21\kO.R?{y~
col()
The ).pacing s of the lltin·ups should not exceed the bser of (a) llk/8, (b) 0.75d or (c) the leust dimension of the beam's cross-section. The stirrups should be of the closed type fully anchored by means of laps.
(5) Calculate the total amount of stirrup rein(orcement Asw/s This is the sum of the &I irrup reinforcement for shear and tor!.ion from ~teps ( I ) and (4).
(6) Calculate the area As1 of the additional longitudinal reinforcement Trc<~ llk cot 0
2Ak0.87/ytL Thts reinforcement should be arranged so that there is ar least one bar at each corner with the other bar~ dic;tributed equally around the inner penphery of the link'> ,.,paced at not more than 350 mm centres.
205
206
Reinforced concrete design
(
'
EX AMPLE 7.9 Desig n of torsional reinforcement Tor~ional
reinforcement i~ to be des1gned for the beam of examples 7.6 and 7.7 which 1 also ::.ubjecr to an ultimate torsional moment of T~e..t 14.0 kNm in addition to th. uniformly distributed loading of 108 kN/mctrc already considered in the previou example'>. The beam ~.:ros~-::.ection i.-. ~hown in figure 7.29a. The step<, in the ca lculation~ an numbered a~ outlined in the pre\ iou::. descriptioo of the dc.,ign procedure. Th. . charactcrh.tic strength-. of the concrete and steel arc .f..~ 30 '/mm~ and }}~ 500 N/mm~.
1. l)e.,ign for 'hear u~ing the Variable Stntt Inclination Method (~ee the de'i1g· calculations of example 7 7> From example 7.7: \'hi = 308 kN and Vku. 111 a~ 530 kN The angle of inclinntion of the concrclc strut i.-. 0 = 2:! with cot 0 1.5 a!ld tanH 0 4 For the 'hear
link~
A"
0.475 (required)
I
The additional longitudinal ten.-.de force
~F1u
3!-!5 J..N
2. Con\crt the rectangular M!Ction to an equivalent hollow hox !.ection figure 7.29h) ·n,i cknes~
A
of box section
/(
600
X
30()
= 2(600 + 300)
lOOmm
Area w1th111 centreline
Perimeter of centreline
A~
u~
(I>
t)(h
I)
200
500
100
10' mm~
2(h I h -1t ) 1400 mm
Figure 7.29 Torsion example
1- . - - - - - - ,
H16
H16 H8 lit 12S
••••
H16
H16 TI
Hl6
Hl6
... b = 300 ...J (a) Cross·section
(b) Equivalent box section
(c) Reinforcement details
1~--e
Design of reinforced concrete beams 3. Check if concrete section
i~
adequate
Tt:.u vlid - - + - - < 1.0 \'Rd. m;a~
TRd."'"'
-
where 1.331'dc~lcrAl
. [ Rd.m•~
= (cot 0 ~ tan B)
(~ee
equation 5.31)
tor~ion .
(Note that A__.
with
0.6( 1 - f<~ /'250)
r1
= 0.6( I -
30 t250) = 0.528
Therefore TRd
rna'-
1.33
0.5'28
X
X
30
X
100
X
100
X 1()- l
(2.5 + 4.0)
72.6 kN ( > Tt:d
24.0 kN)
and
24.~
308
72.6 I 530
= 0.•.3~ + ()·58 = 0.91
Therefore tlu: concrete
1>~c.:tion
10
is adequate.
4. Calculate the additional link r·einforc.:emcnt required to resist is for one leg onl})
r,,d
A~,.
2/\l O.!l7}yk cot 0 24.o x to<> 2 100 X 10' X O.!l7 X 500 X 2.5
.1
5. Therefore for shear plus
tor~ion
and
ba~cd
= o. I(()
on the area ot two legs
0.475 I 2 X(), ll O 0.695 !·or 8 rnm '>tirrup., at I '25 mm cemrcl> A... Spacing v
125mm ( .. u~ /8
.1
=0.805 (!>ee Appendix table A4)
175mm)
6. ('olculate the area A, 1 of' the udditional longitudinal reinforcement required for torsion
JJ;
=
11
(24 x 10 2 X 100
Thi~
X
10'
1400) " 2.5 = mm 2 966 X 0.87 x 500
additional longitudinal Meet can be pro\ ided for by ~ix H 16 bars. one in each corner and one in each of the side faces as 'hown in tigure 7.29(c). llte addttionallongrrudinal tensile force of 385 k.N rc:-.ulting from the design for shear will he catered for by appropriate curtailment Of the main tensile rcinforccmcnr as previously disCUS),Cd in section 7.9.
207
( 208
Reinforced concrete design
7.11
Serviceability and durability requirements
The requirements for the ser\'iceability and durability limit \Late~ have been co, em:_ cxtcru.ively in Chapter 6 -.o thi' section i' a 'hort review of the factors lhat appl} t de\ign and detailing of beam'>. Ahhm1!!h thh \CCllon " a :-.hort rc\ icw at the end <:.. chapter it sbould be empha\ISed that the de~ign for the :-.cniceabilit) and durabilit) I ~tates if> JUSt as important a-. the design for the ultimate limit '>Late. Failures of stru~ ... at the ULS are fortunately quite rare hut can get a lot of publicity. \\hereas failure-. to durabiliry and 'crviccability an; much more common during the life of a structure they can quite ea-,ily lead eventually to a :-.tructural failure or be one of the pn ~ causes of a fa.ilurc. Also poor detnal1ng and con\tructaon can be the cause of ' prob l em~ as lcaldng roofs nnd basements and dasfigurement of the structure 1\ const:qucnt high maintenance col!ts and reduced working life. Adequate concrete cover to nil the reinforcement bar.\ is all-important to pre ingress of moisture and corrosion of the steel hnr:- with resu ltant staining and spallin: the concrete. Cover of the concrete is u l ~o required for fire rcsi~>tigned concrete mixes. ~ccurc fixing the formwork and reinforcing stc~.: l with adi.!4U11tc placement, compaction and curing the concrete.
CHAPTER
8
Design of reinforced concrete slabs CHAPTER INTRODUCTION Remforced concrete slabs are used in floors, roofs and walls of buildings and as the deck of bridges. The floor system of a structure can take many forms such as m situ solid slabs, ribbed slabs or precast units. Slabs may span in one direction or in two directions and they may be supported on monolithic concrete beams, steel beams, walls or d1rectly by the structure's columns. Continuous slabs should in principle be designed to Withstand the most unfavoLJrable arrangements of loads, in the same manner as beams. As for beams, bcnd1ng moment coefficients, as given in table 8.1, may be used for one-way spanning slabs. These coefficients are comparable to those 111 ngure 3.9 for con tinuous beams and are based on UK experience. If these coefficients are used Lhe reinforcement must be of ductility class B or C and the neutral-axis depth, x, should be no greater than 0.25 of the effective depth such that the lever arm, z ( d O.Sx/2), is not less than 0.9d to allow for moment redistribution Incorporated in the values given (which may be up to 20 per cent). In addition, as for beams, table 8.1 should only be used when there are at least three spans that do not differ in length by more than 15 per cen t , and Q. should be less than or equal to 1.25Gk and also less than 5 kN/m 2 .
--+209
21 0
Reinforced concrete design
The moments in slabs spanning in two directions can also be determined using tabulated coefficients. Slabs which are not rectangular in plan or which support an irregular loading arrangement may be analysed by techniques such as the yield line method or the Hilleborg strip method, as described rn section 8. 9. Concrete slabs are defined as members where the breadth is not less than 5 times the overall depth and behave primarily as flexural members with the design similar to that for beams, although in general tl is somewhat simpler because:
1. the breadth of the slab is already fixed and a unit breadth of 1 m is used tn the calculations;
2.
3.
the shear stresses are usually low in a slab except when there are heavy concentrated loads; and compression reinforcement is seldom required.
Minimum thicknesses and axis d istances for fire resistance are given in table 6.5 but derleclion requirements will usually dominate.
Table 8.1 Ultimate bending moment and shear force coefficients in one-way spanning slabs
End support conditiOn Pmned
Moment Shear
Outer support
Near middle of end span
0
0.086FI
Continuous End St.Jpport
End span
At first interior st.Jpport
0.075FI
At mrddle At mterior of interior Sl.lpports spans 0.063F/
0 04FI
0.086F/
0.063Ff
0.46F
0 6F
O.SF
0 4F
I crrcct•ve spdn. Area or each b<~y ,:: 30 m1 • (A bay I$ a strip or sl,lb ocross thl! 1tructurl' bl'lween adjacent rows of column$.) F total ultimate load 1.35Gf + l .SOQ, ,
8.1
Shear in slabs
The ~hear rc~iMance of a bOiid :.lab may he calculated hy the procedure~ given ir> chapter 5. Experimental work ha~ indicated that. compared wtlh beam\, shallow slat.' fall at slightly higher shear stresses and thb 1~ incorporated into the values of th~ ultimate concrete shear resistance. VRd , • as given by equation~ 5.1 and 5.2. Calculation' arc usually based on a strip of slab I m wtde. Stnce shear stresses in slabs subJeCt to umtormly dt~tributed loads are general!) -.mall. shear reinforcement will seldom be requtred and 11 would be usual to design tbe slab such that the design ultimate shear force. \'Ed· ic; leo;s than the shear strength of tbe unreinforced ection. \'Rd..:· In thic; ca'e it " not neces<;ary to provide any shear reanforcement. This can conveniently he checked us1ng Table !!.2 which has been
Design of re inforced concrete slabs Table 8.2
I using ppo:t iiS~ Q
L~n
5 rdeSI
avy
fll
Shear resistance of slabs without shear reinfo rcement
A, j bd
N/mm 2 (Class C30/35 concrete)
Effective depth, d (mm)
s 200
225
250
300
350
400
500
600
750
0.25% 0.50% 0.75% 1.00% 1.25% 1.50% 2.00%
0.54 0.59 0.68 0.75 0.80 0.85 0.94
0.52 0.57 0.66 0.72 0.78 0.83 0.91
0.50 0.56 0.64 0.71 0.76 0.81 0.89
0.47 0.54 0.62 0.68 0.73 0.78 0.85
0.45 0.52 0.59 0.65 0.71 0.75 0 .82
0.43 0.51 0.58 0.64 0.69 0.73 0.80
0.40 0.48 0.55 0.61 0.66 0.70 0.77
0.38 0.47 0.53 0.59 0.63 0.67 0.74
0.36 0.45 0.51 0.57 0.61 0.65 0.71
k
2.000
1.943
1.894
1.816
1.756
1.707
1.632
1.577
1.516
Table 8.3
Concrete strength modiiicalion !actor
f,~ (N/mm1 )
one-way
vRd,c
Modification factor
25
30
35
40
45
50
0.94
1.00
1.05
1.10
1.14
1.19
derived from Equation!-. 5.1 and 5.2 for cia~~ C30 concrete on the basi~ that the allowable ~hear ~tress in the unreinforccd 1-lah i-; given hy ~
nterior supports
l'fld.<
\'RtJ < hd
In this cr"e, the applied ulttmntc ~hear Mres~ 0 063FI
vhll
I'J·.d
bd< -
~·R·'
uU
0 .5F rows
of
..!1\Cn Ill w 'lahs ., of the .ulation~
generally k:.tgn the ='h or the } shear ..as been
21
Tuhlc 8.2 also clearly illu!-.trute~ rhe cf'f'cct of increasing :-.lah ttllcknc!'l~ on the depth related factor k. as noted abo\oc. Where different concrete strength' are U\Cd. the values in table 8.2 may he modtfied hy the factors 111 tahlc lU provided fit 0.4~. As for beam~. the secuon should also he checked tn ensure that Vt'.d docs not exceed lhe maximum permissible shear force. Vlht, 111,,\· If 1ohcar reinfor~cmenl is required then the methods given in char>ter 5 can he u~cd , nlthough practical dif'lit:ulties conccmed with hending and fixing shear reinforcement make it unlikely thm ~hear reinforcement could he provtdcd in 'lah' le'\:. than 200 mm tluck. Localised 'punching' actions due to hca\ y concentrated loads ma). hO\\ever, cau~c more critical condition!\ as -,hown tn the following <;ections .
8.1.1 Punching shear - analysis A concentrated load on a slah causes shearing \tres~es on a section around the load: this effect ill n:ferred to as punching :-.hear. The critical \urface for chccktng punching '>hear ts shown ao; the perimeter in ligure 8.1 which i' located at 'l.Od from the loaded area. The maximum force that can be carried by the slah wtthout shear reinforcement (VK11,") can be obtained using the values of ''Rd , c given in table 8.2 based on equations 5.1 and 5.2 for normal shear in beams and &labs. where Pt J{pyfJ1 ) where /Jy and p, arc
212
Reinforced concrete design
Figure 8.1
Punching shear Critical Section -
Loaded area
I I
/ Plan
JJ, JJ Load
~1~r~===·===~·===~·~::J,·t ~!-~t Elevation
the reinforcement ratio~. A,/bd in the two mutually perpendicular directions (see table A.3 in the Appendtx for A,) then \/Rd c
= I'RJ.~ dll
(R.I )
where d = ciTecuve depth of sectton !average directions
(d)~-, d,)l
or the
two 11teel layer1- in perpcndicu..;
u = length of the punching shear perimeter. If there are axial loads in the plane ot the slah an addiuonalterm is added to VRd c to allow for the effect of these axial loads. This term is ..,.0.111er where 11cp is the nverage of the normal compre)\~ive :-.tre~~.es acting in the y and :- direction!> (from prestre!l~> or external forces). Such compre"ive \tresse:. thus increa~e the punching shear resistance whilst conversely tcn!>ilc stresses reduce the ~hear capacity. Checl>.~ must alsu be undertuken to ensure that the maximum permissible shear force ( VR,t max) i~ nnt exceeded at the race of the loaded nrca. The mnximum permissible shear for~.:c is given by VRu, mu• = 0.51'tf.:t~ud 0.51'tlf.~ / 1.5 )ud where 11 ill the pertmeter of the loaded area and v1 , the strength reduction factor = 0.6( I )~1../250).
(
EXAMPLE 8 .1
Punching shear A ~lab. l75 mm lhick. uverage effcctivc depth 145 mm is constructed with C25/30 concrete
(2a + 2/J -'- 2;r X 2d) = 2(a -1- b) -1 4r.d = 2(300 ~00 ) + 47i X 145 =- 3222mm
It 1 ::
Design of reinforced concrete slabs
21
hence from equation 8. J \!Kd c
= ~"Rd .c X 3222 X J45 = 467 190vRd. c
Average steel ratio fit
VPy
X
p,
where 754/( 1000 X 145) = 0.0052 393/( 1000 X 145) = 0.0027
P> {11
hence "n~
(see
(8. J)
)(0.0052
/It -
X
0.0027)
0.0038
= OJRl;f,
Thus from Iable 8.2. for a 175 mm slab, ~'Rd, c = 0.56 N/mm" for a cl:m C30 concrete and from table 8.3 for cia!>~ C25 concrete, a~ used ht:rc, modification factor O.
r~ndicu lar
itO\ Rd
\!Rd ,r
t
tO
erage of re'tress or -e,tstance
X
467
Jl)()
> 10 J
The maximum permi~~iblc 11hcar force ba~ed on the lace of the lnadcd arl'a i~ gf\ en b) the maximum !>hear re' t ~tancc \ ' Rd ""''
0.5ud[o.6(1 0.5
car force 'ble shear "!leter of
C25/30 m) and mtne the
0.94 >< 0.56 246lN
{;~,))f~
2(_1()() ..,.. 400)
X
145
X
[0.6 (I
25 )] 25 250 T5
) 10
1
914kN
v.hkh clearly exceeds the value VRd r based on the first critical perimeter. I fence the lmaximum load thar the ~l ub can carry is 246 kN. ;
8.1.2
Punching shear - reinforcement design
If reinforcement i~ required to resist shear around the contrnl perimeter indicatcd in Figure 8.1, il ~hould he placed between not more thnn 0.5d from the loaded area and a distance 1.5d inside the outer control perimeter m \Vhich shear reinforcement i~ no longer required. The length of this is given hy u0111 tr = II,.J/( ' 'Rd. eel) from wh1ch the nece~sary dio;tance from the loaded area can he calculated. If thi~ i' Je~, than .1d from the face of the loaded area. then reinforcement should be placed in the zone hetween 0.3d and l .5d from this face. Vertical link\ wtll nonnally be used and provtdcd around m lcao;t two perimeter<; not more than 075d apart. Link spacing around a perimeter wuhin 2d of the face of rhe loaded area ~hould not be greater than l.5d. incrcuo;ing to a limit of 1.0d at greater
214
Reinforced concrete design
perimeter-.. Provided that the slab is greater than 200 mm thick overaiJ then the amou of reinforcement reqlllred is given by: •
A ~w SJI1 0' ~
c' - 0. 751'Rd.l" d X r 1.5 J)'l'd.cf
IRd
.lr X 111d
where A," 1·,
1s
the total area of shear reinJorccment in one perimeter (mm 2)
is the radial spacing of perimeters of shear reinforeemenl
.f)wd,cr is the effective de~ign ~trenglh of the :.hear reinforcement and i:. gl\·en f,"cJ lab. and for vertical links can be e\prcssed a\: l'l(cJ _.
0.751'Rd.c (.--)
/\.-, ~-l.5 ./y"u d .I rill
where the requircu
l'lhl,c'
would be given hy
v~..,.
llj (
A ched, mu"t al:.o be made requirement that:
th<~l
the calculated reinforcement sausfie-. the minimu
0.053.j1,7(sr. v,) i>~
where ·'l I)> the spncing of link'i around the perimeter and A,11 mrn i~> the area of m indil'idt/(/1 fin/... h•g. Similar procedure:- must be applied to the regums of Om 'ilab' which arc close t<1 !.upponing. columns, hut allowances must he macte for reduced criw.:al perimeter~ near sinh edge' and the effect of moment tran~fer from the column11 a~ described 1n -.cction 8.6.
(
EXAMPLE 8 .2 Design of punching shear reinforcement
A 260 mm thid. slah ol class C25/30 concrete i~ reinforced by 12 mm high yield har~ m 125 mm centre!> in each direction. n1e 'lab is "UbJCCL to a dry environment and must be able to corry a localised concentrated ultimate load of 650 J..N over a square area of 100 mm :,Ide. Detcm1ine the -.hear reinforcement required for f>l = 500 N/mm 2• For exposure class XC-I, cover required for a C25/30 concrete is 25 mm. thu:. average cffecti\e depth for the two layer' of steel and allowing for 8 mm links i!. equal to 260- (25 +X+ 12 ) = 215mm
Design of reinforced concrete slabs (i) Check maximum permissible force at face of loaded area \11aximum shear resistance: \'Rd.
~t =- 0.5ud [0.6 (I - {;~)] ~~ - 0.5(4" 300)
= 1161 kJ\
X
215
X
25 )] 1. 25 [ 0.6 ( 1 - ::?SQ 5
.
X 1()_,
( > Vut =- 650 kN>
(ii) Check control perimeter 2d from loaded face Perimeter
11 1 -
2(a f b) t 4rrd
2(300 -1- 300) + 4r. x '215 an.:: I for
hence for concrete without ~hc~u· reinforcement the ~hear capacity is given hy:
Vnct,c
orced
= 3902 mm
l'fttt ,r X
3902
X
'215
= 838 930vRd .c
bending stc~·l ratio
A, -btl = I 000905:.. .,~1.5 = 0.0042
( ( > 1.40 per cent}
hence from tahlc 8.2. I'Rd , 0.5n fnr eta~~ C30 concrete and. Irom tahlc !U, modification factor for da-.s C25 concrete O.lJ-1 then \ 1Rd .,·
!OX 930
= 442 kN 1murn
0.56
X
<< I'&~ -
0.94 x 10-' 650 k:-.l)
and punching :.hear rc111forcemcnt 1:. required. (iii) Check outer perimeter at which reinforcement is not required
Vr:d ,d
llt~ut,tl
650 " 10' 0.56 > 0.94 X 215
I' Rtl
01 (/fl
"t
This wtll occur at n diMance .rd from the face of the lmtdcd area. such thlll
2( \()() I 300) I 2;; X 215
5743
fo,e w terS ncar 1bed in
= 57 ,3 mm
and
.1.36 (
1
X .I
3.0)
(iv) Provision or reinforcement Shear n:in l'orccment should thus he provided within thi.! tnnc extending from a di~t:111cc not gretttcr thnn 0.'id and less than (3.36 - I .5 )d 1.86d from the loaded l'ncc l·ot peritm:ters < 0.75d apar1. 3 perimerers of steel will thus h.: adequate located at 0.4d, 1.15d and 1.9d. i.e. l\5. 245 and ~00 111111 from the fucc of the loaded area (i.e ..11 ~0.75d loOmm apurt). Since all perimeter' lie within 2d (=- 430 mm) of the lnad and maximum lin I. spacing, (~ 1 ). is limited to 1.5d ( 323 mm). The minimum link leg area i), therefore given hy:
=
A,w,rnm = \erage
0.05 3..j]d ( ~,. \1 ) r J)l
0.053 /3( 16() X 323) 500
- 27.3 mm= \lhich tl> l>atil>ficd by a 6 mm diameter har (28.3mm'> Hence the as~umed Hmm lin"-' ''ill be adequate.
215
216
Reinforced concrete design
The area oJ steel required/perimeter is thus given by: A ~~~
> I'Rd ,, -
0.751'Rd.•
-
1.5
(f
/y..,d.tl
)
Sr X IIJ
where. for the outer perimeter 650
VF,t
\'Rd"
= -lltl1 - 3,.,)_ •v.,
~
= 0.94 x 0.56
l'l{d
!.v.d d = 250 1- 0.25 UIIU
Vr -
X
103 ~ _l)
,
= 0.77511.1/mm-
0.526 N/mm2 (u~> above) x 215
= 103 N/mm'
( ~ 500)
160111111
thu~
(0.775 - 0.75
X
0.526)
X
160
X
J9()2
A" ' > - -- -1.5 X303- - -
= 523mm2 (v) Number of links The nre:1 of one leg of nn Xmm link i~ 50.3 mm 2 . lienee the number of II nk-lcgs required 523/50.3 11 on the outer pcrimeler. Thl: same number of link::. cnn conveniently be pro\ tded around each of the 3 proposed perimeter' as 'ummarised in the table hclo\\ The table tndicate' the numhct of ~tng l c-leg g mm diameter link~'. (urea = 50.3 mm~ J proposed for each or lhc three n:inforcemcnt perimeter' taking account of the maximum requm:d spacing and practical h\ing cnn,idcrauon,. Bendtng retnforccment i' 'paced at I25 mm centres in hoth directions: hence link spacing is ~ct at multiple!> or thb value.
--Dl5tance from load face (mm)
Length of perimeter (mm)
85
1734 2739 3713
245 400
Reqtlired ltnk spacing (mm)
Proposed link spacing (mm)
Propo5ed number of finks
158
125 250 250
11
249 323•
14 15
• muxir'num allowed
8.2
Span-effective depth ratios
Excc~sivc <.lellection' of slabs will cause damage to lhe ceiling, 11om finishes or other archttccrural hni-.hc~. To a\Oid thi!>. hmil'> arc set on the -.pan-depth ratio. These limit~ arc exactly the ~arne ns those for hcarns a~ described in section 6.2. As n slab i'> usually a !.lender member. the restrictiom; on the ~pan-deplh ratio hecomc more important and this can often comrol the depth of slah required. ln tenm of the 'pan- effective depth nlliO. the depth Of slah IS gtVCI1 by . span mtnimum effective depth = -.--. -.--.- baste rauo x correcuon Iactor:-
Design of reinforced concrete slabs
The correction factors account for slab type and ~upport conditions a~ well as cases of span!> greater than 7 metre~ and for flat slabl> greater than R.5 metres. The ba:-.ic ratio may also be corrected to account for grades of sreel other than grade 500 and for" hen more reinforcement 1\ provided than that requlfed for de~ign at the ult1mate limit -.tate. Initial values of ba~ic ratio may be obtained from tables (e.g. table 6. 10) but thc-.e arc concrete strength dcj)l.!ndcnt. It may normally be a:.~umed that in usUlg such table-., ,Jab., are hghtl) Mres\ed although a more exact determination can he made from ligurc 6.3 when the percentage of tcnllion reinforcement is known. It can he seen 1hat the ba,ic ratio can he mcrcased hy reducing the stress condition in the concrete. The concrete \trel>' may be reduced by providing an area of tension reinforcement grcall.:r than that required to rc1>i1-.t the design moment up (() a maximum of 1.5 x lhm required. In the case of two-way spanning slabs, the check on the span effective depth ratio r.hould be hased on the sltorler span length. This doc!' not apply to llat slabs where the longer span should be checked.
8.3
Reinforcement details
To resist cracking of the concrete. codes of practice ~pccify dct:ul~ ~uch as the minimum area oJ reinforcement required in a ~ection and limit~ to the maxnnum anu min1mum spacmg oJ bar\. Some of thc\e rules are a~ follow~.
re'luircd ~tly be : bcJO\\. .3 mm1 ) imum I!P3CCd at \Ulue.
(a)
M111imum area\ of rdnforcemenr minimum area
0.26f.:1mb1d/f)1.
">
0 0013b1d
in hmh direcuons. where b is the mean '' idth of the terNie tone or scct1on. The m1nimum remforcement provision for crack control. a' spec1hed 111 'cction 6.1.5 may also huve to t>e cons1dered ''here the ,Jut> depth exceeds 200 mm. Secondary tran!>vcrsc re1nforcement should not be less than 20 per cent olthe minimum mam rcinfon.:cment requirement in one way <.lahll. (b) Maximum areas of longtnldinal and trun!>verse reinforcement muximum urea - 0.04A,
bsed
' of links
__
)
. or other :e'e limittu,ually :.1 1ant and 1 e depth
where A, il' the gross cross-sectional area. This limit applic1-. to sections away from arcus of bnr lapping. (c) Maximum spacing of bar!i For slabs not exct:eding 200 mm thicknc~s. hur spacing shou ld not exceed three timet- the overall depth of slab or -100 mm whichever i:-. the l es~er for main reinforcement, and 3.511 or 450 mm for secondary rellllmceml'nt. In areas of concentrated load or maximum moment, thc1-c value~ arc reduced to 211 < 250 mm and 111 <.. 400mm respectively. (d) Reinforcement in the flange of aT- or L-beam 'I his i' described in deuul in section 5.1.4. (c) Curtailment and anchorage of reinforcement The general rules for curtailment of ban. in nexural members \\ere discus'icd in section 7.9 Simphtied rules for curtailment in different I} pes of c;lab arc IIJu,trated in the sub,equent o;ecuons of this chapter. At a \imply \UPI10rted end. at lea!-.t half thl' span reinforcement should be anchored a~ ~pcc1ficd in figure 7.26 and at an unsupported edge U bar\ with leg length at leaM 2h llhould t>e provided. anchored h) top and bottom transverse bars.
2
218
Reinforced concrete design
8.4
Solid slabs spanning in one direction
The -.labs are de-.igned as if they consist of a serie-. of beams of I m breadth. The ma1 ~teel is in the direction of the span and secondary or distribution steel i~ required in tht transverse direction. The main steel should fonn the oUier layer of reinforcement to giH it the maximum lever ann. The cakulations for bending reinforcement foliO\\ a similar procedure to that used ir hcum design. The lever arm curve ol fi gure 4.5 is used to determine the lever arm (· and the area of tension reinforcement i~ then gi\cn by A, -
M -0 87f.,L:
~·or solid slabs spanning one-way the simplified rules for cwtuiling hars as shown 1 hgure 8.2 may be used pro\ 1ded that the load' are uniformly distributed. With continuous sluh it i-. also necessary that tht: spans arc approximately equal. The,, simplihed rule\ arc not gi,en in EC2 but arc recommended on the basis of prO\t satisfactory performance established in previous codes of practice.
8.4.1 Single-span solid slabs The ha-;1c span effective depth ratio lor thl\ type of slah i' 20: I on the bal>l'i that it ' lightly stressed' and that grade 500 steel is u~ctl in the design. For a start-point de~ign a \alue abo\'e thi~ can U\uall} he c),timatcd (uniCl>S the slah ic, known w bhcavily loaded) and subsequently checked once the main tcn~ion n.:mforccmcnt h been designed. fhe effective 1>pan ol the ~lah may he taf..en ~~s the clear d1stance hctween the face l lhc supports plus a distance at both ends taf..en a), the le~scr of (a) the di ~t:Jncc from tl, face of the support 10 11\ centreline and (b) nne-half of the menlll depth of the l!lab.
rlgure 8.2 Slmpliht-d rules for curtailment or ba~ ir1 slab }panning In one direCtion
100% < SO%
~
~
O.JJ. Stmpty Supported 25% or
' • 0.3L
m1d-span SLc!!l
c- 0.20L
'" 0.1SL ~
'-
100%
50% <
SO%
0.21
_L Continu ous Slab
100%
Design of reinforced concrete slabs
( EX AMPLE 8 . 3
Design of a simply supported slab The :-lab \ hown in figure 8.3 ts to be designed lO carry a variable load of 3.0 kN/m' plus floor fini~hc11 and cetling loads of 1.0 kN/m!. The charactcrhtic matcnnl ~trength<; are fc1.. = 25 Nlmm 2 nnd J,~ - 500N/mm2 • Baste span effecti,·c depth ratto = 19 for a lightly strc~~cd i-lab from rigure 6.3 for class C25/.10 concrete and p O.Y'f. For \implicity. take the effective span to be 4.5 m hetween centreline' of supports. HlO
10\\ n
in lth a These p m en
• fa~e of m the r ,Jab.
.
I~
300
,
:.
Figure 8.3 Simply supported slab example
HlO - 150
~0~ ,...
4 .5 m
(7) First design solution
Eslimate of slab depth
Try a baste span-depth ratio of 27 (approx. · ·
d
f
~OIX
ahovc value from ligurc 6.1)·
~pan
h
mnumum c tccuvc ept =- 27 x correction factor~ (C.f.l 4500
167 c.f. A,.,. hi~h ) icld '>tee I t'> lx:ing used and the span •~ le'' than 7 m the correction tactor:. can be taken a' unny. Try an effective depth of 170 nun. For a cia'' XC I cxpm.ure the co.,cr 25 mm. Allowing. ~>ay. 5 mm tl" half the har diameter of the rcinlorctng har: O\erall uepth of \lab 170 f 25 f 5 200 mm ').7 X c.f.
Slab loading ~tel f weight ol slah
200 x 25
x 10- 1 = 5.0 kN/m 2
1.0 1 5.0 - 6.0 kN/m2
tolul pcrmancttl load Fm a I tn width of slab:
ultimate load
( 1.35Rk I 1.5qk)4.5
= (J.35 X 6.0 + 1.5 X 3.0)4.5 M
56.7 x 4.5/R
56.H N
31.9 kNm
Bending reinforcement ,\.f
bd2}~L
31.9 X ~= 4 I()(X) X 170 2 x 25 0 .()4
From the lever-arm curve of ligure -t5. I~ = 0.96. Therefore adopt upper limtt of 0.95 and lever-arm ;: -lad- 0.95 x 170 = 161 mm: 6
A, =- -M -.- = 31.9 X 10 O.l:!7})kZ 0.87 X 500 X 161
= ..t55 mm '!m
Provide HI 0 bar.. at 150 rnm centres. As = 523 mm 2/m (al> shown in tahle A.3 in the Appendtx}.
21
220
Reinforced concrete design Check span-effective depth ratio I
I
= IOOA, n:q = I()() X 455 bd 1000 X 170
0.268''~ (>0. 13'* minimum requirement)
!-rom figure 6.3. thi~ corre~ponds to a baste span-eftectt\'e depth mtio of 32. The actual ratio= 45001 170 = 26.5: hence the chosen effecuve depth ts acceptable. Shear
At the face of the !>uppon Shear V~::d
55.5 (2.25 - 0.5 X O.J) . =T 2 25 I(){) X
..,
k
=- 59 . N
523
PI = IOOO x t70=0.J I VR~ . . - t'Rd,,bd where vR11., from table 8.2 adjustment since fiJ < 0.4%). Thus:
VKu , - 0.55
X
I 000
X
170
=0.55
(note: no concrete strength
93.5 kN
as VL:d b less than VRu, then no
~hear
reinforcement
i~
required.
End anchorage (figure 7.26)
From the tahle of anchorage lengths in the Appendix the tenswn anchorage length
= 40c•,
40 x 10 = 400 mm.
Distribution steel
Pm\ldc minimum
0.00 JJbd - O.O
221 nun 2/m.
Prm ide 1110 at 300 mm centres (262 mm 2/m) \\htch \llll'lhe' ma>:imum bar spacing It mit,. (2) Alternative design solution
The 1-.ccond part ot thl' exumple lllu<;tralel> how a ~muller depth of lab is adequate provided it b reinforced with Mecl in exec:-\ or !hat required for bending thus working at a lower ~tre5s in service. Try a thid.nc~s of sloh, II 170 mm and d = 140 mm:
or !.lab = 0.17 25 Iota! permanent load = I .0 + 4.25 ~elf-weight
X
ultimate load
4.25 l-.N/m~ 5.25 kN/ml
( l.35gk + 1.5q~ ) =- ( 1.35 X 5.25 1- 1.5
X
3.0)4.5
52. I kN
Bending reinforcement
4.5
= 52.1 X 8 !:!..__ = 29.3 )( M
bd2J'ck
1()0() X
29.3 kN Ill
106 140! X 25
= 0.060
From the le,er-arm curve of figure 4 5. 1.
=0.945. Therefore,
0945 x 140 = 132mm:
M 29.3 10~> A,= = . Q.87f}k: 0.87 ) 5()0 X J32
510mm2/m
Provide H 10 han. at 150 mm centre!>. A,
523 mm~/m.
lever-arm ;:
l.,d
=
Design of reinforced concrete slabs Check span-effective depth ratio IOOA, l00 x 510 p bd 1000 )( 140 = 0 '364%
=
f-rom figure 6.3 this correspond~ to a ba~ic span-effective depth rat1o of 24.0. Acrual
Span
Eff. depth
45 000 = 32.1 140
Thi!> is inadequate but can he overcome by increasing the Mccl area. Span . . A,. p01 , Ef ' d th = b
Limiting
Try IOmm bar~> at I 00 mm centres, A~ prov £Ience:
;th
A,.rrnv 1\,, ,'C~
Upper limit
785 5 10 = 1.54
w torrection
lienee al lowuhle Therefore d
n..,
= 785 mm 2/m.
factor (UK National Annex)
span . . efleetiVC depth
= 24 x 1.5 -
36 which
I 5. i~> greater than that provided.
140 mm is adequate.
8.4.2 Continuous solid slab spanning in one direction f-or a conllnuous slab, bottom reinforcement j., required \\lthlll the 'pan and top reinforcement over the support~. The effective span i-. the diStance hct\\ecn the centreline of the ..upport<> and the ba.<.ic l>pan-cffettivc depth ratio of an 1ntcnnr 1-pun i' 10.0 for 'lightly ~tressed' where grade 500 steel and clai-~ C'30/35 concrete arc U\Cd. The cnrre~>ptmding limit for an end span is 26.0. If the conditions given on page 209 are met. the bending moment and ~heur Ioree coefficient~ given in table R.l mt1y he used.
(EXAMPLE 8. 4
Design of a continuo us solid slab The four-spun slab shown in figure H.4 ~urports a variable load of 3.0 ~N/m 2 plus lloor fini\hes and a ceiling load of 1.0 kN/m 1. TI1c characteristic material ~trcngths arc .f.l 25 N/mm2 and f>k 500 N/mm2.
Estimate of slab depth
=
A~ the end 'pan i-. more critical th
221
L
222
Reinforced concrete design
figure 8.4 Continuou~
slab E'
:I
co
E
~,
-
I I
I I
'E'
'E'
1:1
Span
'""'
Span
I I , E'
1~1
1~1
Span
' ""'
'E
Span
IQ': . co
I 1---- 1 ---I I I I H H
I I I I H
Pldrl
LJ
u 4Sm
u
J
u
4.5m
J_
4.Sm
4.5m
Elevation
A"' high yield \tecl is being u ~cd and th\: span is lcs1. Lhnn 7 m the correction factor can be ta~en
as unity. Try rtn effective depth of 140mm. For a class XC-I exposure U1e 25 111m. AIIO\\ mg. 'ay. 5 mm as half the bar diameter of the reinforcing bar:
cover
overall depth of slab
140 + 25 + 5 170mm
Slab loading ~elf-weight of slab totul permanent load
170 x 25 I .0
4.25
I0 '
4 25 kN/m'
= 5.25 ~N/m ~
For a I m w1dth of slab
= (1 .35,11~ - 1.5q, )4.5 = ( I J5 x 5.25 I 1.5 x 3.0)4.5
ulumme loud. F
52.14 ~N
lhmg the codficiems of lrthle R.I. as~uming the end support i' pinned, the moment at the m1ddlc of the end ~pan j.., gi\cn by
M
0.086F/
0.086 x 52.14 x 4.5
= 20. 1XkN m
Bending reinforcement 20. 1X X 1011
M bd~f~·
JO()() X f.tQ1
25
0 04 12
From the lcvcr-rtrrn curve of figure 4.5, 0.95 x 140 133mm:
M 0.87!~.;:
/~
0.96. Therefore. lever-arm •.
20.18 X 101'1 -0.!!7 X 500 X 133
= 349 111111 2/m Provide
nI 0 har)o at 200 mm cemrc:.. A, = 3'>3 mm2/m.
Check span-effective depth ratio
IOOrl,,,.,~ IJd
= 100
X 349 = 0.2-19 I 000 x I 40
l.,d
Design of reinforced concrete slabs Hl0-400 Hl0-250
. s · ; , lfr.· HlO- 400
HlO
•
jk
200
• >
•
Figure 8.5 Reinlorcement in a continuous
slab
9
I
Hl0-200
Hl0 - 400
Hl0 - 250
Hl0-250
f-rom figure 6.3 thb corre1>pond~ to a basic span-effective depth ratio m excess of 12 x 1.3 (for an end ~pan) 41. The acrual ratio 4500/ 140 = 32.1; hence the chosen effective depth is acceptable. Similar calculations for the supports and the intelior ~pan give the steel area~ l.hown in figure 8.5. At the end ~upports there ill a monolithic connection between rhe slnb nnd the beam, therefore rop steel should be provided to resist t~ny negative moment. The momcnl w he de~igned fori~ rt minimum of 25 per cent of rhe span moment. I haL is 5.1 kN m. In fact, to rrovide a m1nimum of 0. 13 per cent of steel, I-l l0 ban m 400 mm centres have b~.:cn ~pecitied. The layout or the reinfmcemen\ in 11gure 8.5 i.., accmding to the ..,implit1ed rub for curtailment of bar~ in slabs as illuwared in llgure 8.2. fran~versc
remforcemcm = 0.00!3bd 0.0013 x 1000 " I~0 182mm ~/m
Provide Ill 0 m 400 mm centres top and bollom. where,·er there
t\
matn rcinforcem~.:nt
( 190 mm'lm).
lit at
=
8.5
223
Solid slabs spanning in two directions
When a .~olah is ~upportcd on t~ll four of it~ :.ides it effectively spun..; in both dir~.:c:tion~o. and it is sometimes more cconomic~1l to design the slab on this ha~is. Th~.: amount of bending in cuch direction will depend on the nttin of the two 11pan:-. and the conditions of remain! at each support. If the slab i~ square and the rcr.traints are similar along ti\C l'our sidus then the load will s.pon c4ually in both directions. If the slah i~ rccrungu lar I hen more than t)ne-half of the load will be c:urried in the stiffer, shorter direction and less in I hi! longer direction. If one span i~o much longer than the other. a large proportion nf the load will he carried in the ~hort direction and the slab may as well be designed as spanning in only one tlircction. Moment!> in eac:h direction of span arc generally calculated using tabulated coefficient<;. Areas of rcinforcemem ro resist the moments arc dctcrmmed mdi!pcndently for each direc11on of lipan. The slab is reinforced with bars in hoth d1rections parallel to the span" \\ith the steel for the shorter span placed funhest from the neutral a xi' to give it the grcatcr effective tlcpth. Thi! -.pan-etlecuve depth ratios are based on the -.horter :.pan and the percentage of reinlorccmcnt in that direction. With a uniformly chstributetl load lhe loads on the 'iUpportlllg beam' may gcn~.:rall) be apportioned a.' ~hown in figure 8.6.
224
Reinforced concrete design Beam A
Figure 8.6 Loads carried by supportmg beams
0
u E
E ..,
... ""
~
"" Beam B
8.5. 1
Simply supported slab spanning in two directions
A slab simply -.upponed on its four ~-.ides will c.lenccl uhmn both axe~ under load and th~ ~:omen; will tend to lift and c.:urlup l'rom the supports, c.:ausing torsional momems. Wher> no provision has been made w prevent th1s lifting or to resist the torsion then the moment cocffidents of tahle 8.4 may be U1'>Cd and the maximum moments are given o~
M,x and
= a,,nl;'
in direction of span f,
,
M,y = a,ynl~ in direction of
~pan
/).
where
M" and M,) are the moments at mid \pan on \trips of unit width with c;pans /, and respectively 11 - ( 1.35g~ + l.Sql.). that b the total ultimate lond per unit urea
I>
= the length of the longer side
I, - the length of the shorter side
a,, and a,> arc the moment coefficients from mhlc 8.4. The area of reinforcement in directions/, and /\ respectively arc 1\,~
M,x . = - 'f, . per metre w1dth 0.87 y~:
nnd
ll,y
M,y =-0.87/ykZ
per metre width
111c s!ah should be reinforced uniformly m:ro:-.s 1he full width, in each direction. The effective depth d used in calculating/\,~ should he le'' than that for A,, because Of the different depth~ of the tWO layer~ Of reinforcement. Table 8.4 Bending-moment coefficients for slabs spanning in two directions at right angles, simply supported on four sides 1.0
0.062 0.062
17
0.074 0.061
12 0.084 0.059
1.3
I 4
1.5
1.15
2.0
0.093 0.055
0.099 0.051
0.104 0.046
0.113 0.037
0.118 0.029
Design of reinforced concrete slabs F.stabli hed practice '>uggcsts that at least 40 per cent of the mid-span reinforcement <,hould extend to the supporL<; and the remaining 60 per cent should extend to within 0. 11, or 0.1/y of the appropriate support. It should be noted that the above method is not specially mentioned 111 EC2: however. as the method was deemed acceptable in B$8110. Its continued use should be an acceptable method of analysing this type of slab.
( EX AMPLE 8 .5 Design the reinforce ment fo r a simply supported slab The :,lab i~ 220 rnm thick and spans in two direction:.. ThL: effective span in each direction is 4.5 m and 6.3 m and the :-.lah supports a variable load of 10 kN/m 2• The churucteriM ic material su·engtb:, arc fck 25 N/mm2 and /yk 500 N/mm1 .
= J .4 From lahle ~.4. a~~ = 0.099 and a,) fy/1,
6.3/4.5
0.051.
Self-weight of !-lab = 220 x 25 x 10 ' ultimate toad
3IId 1
Bending
= 5.5 t..N/m2
= 1.35gk + 1.5qk = 1.35 X 5.5 t 1.5 X
10.0 - 22..431-N/m'
short span
With clas'> XC- I cxpo:-.urc conditions take d - 1R5 mm M"
0.099 x 22.43 x 4.52
a,,nt;
45 OkNm M,~
45.0 X 106
185~
I 000
bcl'.f..k
X
-
15
= 0.0::>3
From the lever-ann curve. figurc 4.5. lever-arm
~
I~
-= 0.95. TI1crefore
- 0.95 x 185 - 176 mm
and O.R7
X
500
176
X
2
588 rnm /m
n. ·au~e
A,= 646rnm2/rn.
Provide 11 12 at 175 mm centres,
Span-effective depth ratio ~.right
1
= IOOAueq
It
bd
=
100 X 588 1000 > 185
= 0.318
From figure 6.3. tht'> correl>pondl> to a ba'iic 'pan-effective depth ratio of 28.0:
20 118 ~
actual
029 Thu'i d
-;pan cffecti vc depth
4500 185
1R5 mm b adequate.
?4 3 - ..
225
226
Reinforced concrete design Hl0-200
Figure 8.7 Simply supported slab
I
spanning in two directions
'
H12-17S
4.Sm
Bending - long span
,
a,~nl;
M,) -
= 0.051 X 22.43 X 4.5! = 23.16kN m
Since the rei nforcement for this span will have a reduced effective depth. take
;: = 176 A -
' -
12 = 164mrn. Therefore M,y
0.87.J;~::-
:n. l6 x
11Y·
0.87 )( 500
164
325 mm~lm
Prm ide H10 at 200 mm centres. A, = 393 mm '/m lOOA,
100
.193
-;;;/ =- (()()() v
164
= 0.24 "h1ch i' greater than 0.13, the mm1mum for tranwer'e 'teel, wuh cia'" C25/30 concrete. rhc arrangement of the reinforcement ''shown 10 hgurc 8.7.
l~--------------------------------------~) 8.5.2
Restrained slab spanning in two directions
When the slabs have fixity at the supports and reinl'orccment is added to resist torsion and to prevent the corm:rs of the l>lab from lifting then the maximum moments per unit width are given hy
,
d,,111~
in direc1inn of span I,
= 3,>111~
in direction ol span ly
M,, =
and M,)
where J~\ and d,y are the moment coefficients given in table 8.5. hased on previou' experience. for the specified end condition~. and n - ( 1.35gl l.Sqk ), the total ultimate load per unit area. The slab is di\ided into middk and edge !.trips as \.hown in figure 8.8 and reinforcement is required in the mtddle smp-. lO re'>ist ,H,x and M,y. The arrangement this remforcement should tal..c i-. illustrated in figure 8.2. In the edge strip~ only nominal reinforcement i:-. ncccssaJ). such that A,/ bd = 0.26f.:1m/f}l > 0.0013 for high yield steel.
Design of reinforced concrete slabs Table 8.5
227
Bending moment coefficients for two-way spanning rectangular slabs supported by beams
Short span coefficients for values of ~ 1 1~ Type of panel and moments considered
1.0
7.25
1.5
1.75
2.0
Long-spon coeffidents for all values of ly , lx
0.053 0.040
0.059 0.044
0.063 0.048
0.032 0.024
0 063 0.047
0.067 0.050
0.037 0.028
0.083 0.062
0.089 0.067
0.037 0.028
0.093 0.070
0.045 0.034
Interior ponels Negative moment at continuous edge Positive moment at midspan
0.031 0.024
0.044 0.034
One short edge discontinuous Negative moment at continuous edge Positive moment at midspan
0.039 0.029
0.050 O.Q38
0.058 0.043
One long edge discontinuous Negative moment at continuous edge Positive moment at midspan
0.039 0.030
0.059 0.045
0.073 0.055
Two adjacent edges discontinuous Negative moment at continuous edge Positive moment at midspan
0.047 0.036
0.066 0.049
0.078 0.059
0.087 0.065
- - ---
..
... I Q.
51 .,... .
/,
"
;;
:;; "'
.g'l ....
tte
_)
'2
:I
I i
31.. 4
1..
(cl)
ion unit
1- _ _
I
...i ~
ror span I,
I,
-'
Edge slnp
Mtddle strip
8
3/,
T
I;§'
- · - Edge Slrlp · -
l
I,
...
~
•
l I, 8
8 (b) For span ~
In audition, torsion reinforcement is provided ut c.lbwnrinuous corner~ nntl it ~hould:
1. consist of top and bottom mats, each hoving bar~ in both directions uf ~-opan; 2. extend from the edges n minimum diE.tunce 1, / 5: 3. ut a corner where the slab is discontinuou!l in both directions hnve an area of steel in each of the four layers equal to tJ1ree-quartcr~ of the urea required I'm the muximum mid-span momem; OU\ <"~tal
and
e.,cnt minaI ~
eld
4. at a corner where the slnb is discontinuous in one direcrton only, have an area of tor~ion reinlorcement only half of that !.pecified in rule 3.
Torsion reinforcement il> not. however. nccc~;~ary at an} corner \\here the 1>lah ;.., continuous in both directions. Where /) fl, > 2. the slabs should be de~igned as ~panning in one dtrecllon only. lr ~>hould be noted thatrhe coefficients for both shear and momen111 can only be U\Cd if d w., B or C ductility reinforcement i~o .\pccificd and the mrio .1j d is limited 10 0.25.
figure 8.8 D1v1sion of slab into m1ddle and edge strips
228
Reinforced concrete design 1,= S.Om
Figure 8.9 Continuous panel spanning in
J
two direction~ support
--. .,E
a
b
Disc;ontinuous supported edge
'C
N
..;
"
&. a.
~
a
d
..L
support
(
\
EXAMPLE 8.6 Moments In a continuous two-way slab The panel con~tdered i~ an edge panel, U\ ~hO\~n in figure 8.9 and the uniform!~ U.i~>trihutctl lontl, n ( 1.35gk + l.Sqk) I 0 kN/m 2• The moment coefficient"\ nre taken from table 8.5. ~
I,
6.2) 5.0
I ::!5
Positive momenls at mid-span M,,
,J""'~
:;: ; ().().l5 X
5~
10
II :!5 kl\ m tn dtrc~.:uon I,
M,> - d,)111:
0.028
10
"s'
7.0kN min direction ly
Negative moments Support ad. M, Support<, nb (llld de, M~
0.059 x 10 )( 5'
14.75 kl\ m
o.o:n X
9.25 kN m
10
X
s'
The moment~ calculated arc for a metre width of slab. The de~tgn of reinforcement to resist these momenb would follow the usual procedure. Tor,ion reinforcement, accordtng to rule 4 1~ requtrcd at corners h and c. A check would aJ,o be required on the ~pan effective depth ratio of the slab.
l~--------------------------------------~) 8.6
Flat slab floors
A llat slab noor is a reinforced concrete sltth supported directly hy concrete column' without the usc of Lntermctliary be:um. T he ~l ab moy be of constant lh ickne~-. throughout or in the area of the column it may be thickened as a drop panel. The column may aJ<.o be of constant '>Cction or it may be llared to form a column hend or capital. These vanoul> forms of con~truction arc illu11trated in figure R.l 0.
Design of reinforced concrete slabs
rr Floor without drop panel or column head
nnl~
Figure 8.10 Drop panels and column
heads
v
Floor with column head but no drop panel
Floor wllh drop panel and column head
The drop panels ore effective in reducing the shearing stre:-.sc:-. where the column i11 liable to punch through the slab. and they also provide an increase<.! moment or re!.istance where the negative moments arc greatest. The llat slnb floor has many a<.lvantages over the beam and ~lob lloor. The simpl ified rormwork ar1d the reduced ~Lorey heights make it more economical. Win<.lows can extend up to the undcr~idc of the slab, and there are no beams to ob~ truct the light and the circulation of air. The absence of sharp corners gtves greater fire resistance as there rs less danger of the concrete spalliog and expo~ing. the reinforcement. Deflection requirements will generally govern slab thicknesl> which should not normally he le:,s than 180 mm fur lire rcl>istance as indicated tn table 8.6. l'hc analy\is of a nut 5lab strucrure may be carried out hy divtding the ~tntctun.: into a \eric' of equivalent frames. The moments in these frame~ may he determine<.! hy: (a)
a method of frame analy:,is '>uch ru. momelll dtsmbution. or the "iffne!-ts method on a computer:
(b) a stmpltlted method using the moment and \hear cocflictent~ ol table K.l subjeCt to the followtng requirements:
1 ,ual
i~
(i}
the lateral stabrltty
(ii)
the condition~ for using table tU <.lcscrihed on page 209 arc sati:,ricd:
(iii)
there nrc at lenst rhree rows of panel:, of tlpproximately equal span in the direction being considered;
(iv)
the hay si~c exceeds 30m 2
Table 8.6
not dependent on the slab-column connections:
Minimum dimensions and axis distance for Flat slabs for fire resistance
.. A
_)
Standard fire res1stance
Minimum diml'nslom (mm)
-------
Slab thickness, h1
REI60 REI 90 REI 120 REI 240
180 200
200 200
Axis
di~tance,
a
15
25 35 50
umn<.
·ness • umn ~ pttaJ.
Nott: 1. Redi~tnbuuon of moments not to exceed 15%. 2. for hre resi~tanct> R90 and above, 20% of the total top reinforcement in each dirtction over mtermedidte supports should be continuoU5 over the whole spJn cmd pl.lced '" the column strip.
22'
230
Reinforced concrete design
Figure 8.11 Flat slab divided into strips
Position of maximum - -- -- - - --:.._-_---~ negaL1ve moment
,
, Pos•tion of maximum - - 1 - - - pos1L1ve momcnl
W1dthof half column strip= //4 with no drops or = half drop width when drops arc used
Interior panels of the Aat sluh ~hoult.l be t.livit.led as shown in ligure g.l l into colur and middle strips. Drop panel~ should be ignored if I heir smaller dimension is less th~ one-third of the smaller panel dimension JK . If a panel is not square, strip widths in hv directions are ba.'>ed on l,. Moment~ dctennined from a structural anuly~i~ or the coefficient1. of table 8.1 ...or distributed between the strip~ as shown 111 table !.0 such that the negative and poslll ~ moments resi~tcd by the column and middle ~trips totul I 00 per cent in each ca~e. Reinforcemenl designed to re i't these \lab moments may be detailed according t.o the simplified rules for slabs, and c;atisfying nom1al spucing limtl!>. This should be spread aero~!. the respective strip hut. in <,olid 'lab' \\ithout drop,, top steel to re'l negati\e moments in column strip-. 'hould ha' e one half of the area located in 1 centml 4uartcr-strip width. If the column Mrip ts narrower because of drop,, moments rcsbted b} rhe column and middle \trip' ~hould be adJU~ted proponional _ illu,truted in example 8.7. Column moments can be calculated from the unaly~is of the equivalent fr.. Particular care is needed over the tran~;fer of moments to edge column<;. This b to en~ure that there is adequate moment capacity wtthin the slah udjucem to the column since moments will only be able to be transferred to the edge column by a strip ot sinh con~iderably narrower than the normtll internal panel column sttip width. As seen in table ~.7. a limit i~ placed on the negative moment transfcn·ed to an edge column, and ~ lab reinforcement 'ihou ld bl.! concentrated within width be a'i defined in figure R.l2. If exceeded the moment shoult.l be limited to this value and the positive moment increased to maintain equilibrium.
Table 8.7
Division of moments between strips Column stnp
Negative moment at edge column Negative moment at internal column Positive moment in span
b.• width ol edge strip.
100% but not more than 0 17bpd2 f,k 60-80% 50-70%
Middle strip 0 40-20% 50-30%
Design of reinforced concrete slabs c, Figure 8.12 Definition of be slab edge
C,
c,
I
~-----
* 4~ !.
y
j·
inner face of column
b. =l
'
•
y/2
..j-
~~
Note: All slab reinforcement perpendicular to a free edge transferring moment to the column should be concentrated within the width be
(a) Edge column
{b) Corner column
The reinforcement for :1 Rat slab should generally be arranged according to the rule!. lllui>tratec.J in Jigurc 8.2, but at least 2 bottom bars in each orthogonal c.Jirection shou lc.J pas~ through internal columns 10 enhance robustness. Important feature~ in the design of the slabs are the calculations for punching shenr nt the hend the column~ and at the change in depth of the slnb, if drop puncl~ are used. The design !'or ~hear .\hould follow the procedure described in the previous sect ton Oil punching ~henr except thm EC2 requires that the design shear force be incrca,cc.J ahnvc the calculated value by 15 per cent for internal columns. up to 40 per cent lor edge columns and 50 per cent for corner co lumn~. to allow for the effect~ of moment transfer. The~e 'iimpltfied rules only apply to braced :-.tructurc~ where adjacent 'pans do not differ by more than 25~. In constdenng punchtng shear. EC2 plaeei> additional requirements on the amount and di-;tnbuuon ol re111forcement around column head\ to en,urc that full punching 'hear capacuy io; de\'eloped. The U'iUlll basic \pan effective depth ratios may be u...cd hut where the greater '>PliO exceed' R.5 m the basic rntio should be multiplied by 8.5/span. For flm ~labs the ~pun effective depth calculation should be based on the longer span. Reference ~hould he made to codes of practice for further detatled informntion describing the requirements for the analysis and design of Oat ~labs. including the u~>c ol bent-up har~ to provide punching shear reststance.
or
(EXA MPLE 8. 7
Design of a flat slab The co l umn~ arc at 6.5 m centres in each direcUon and the slab support~ a variable lond 2 of 5 kN/m • I he characteristic material strenglhs are ~k 25 N/mm2 for the concrete, and /y~ 500 N/mm 2 for the reinforcement. 1t i~ decided to use a floor ...tah us ~hown in figure 8.13 wi1h 250 mm overall depth of slab. and drop panels 2.5 m square hy I 00 mm deep. The column heads arc to he made 1.2 m dwmetcr.
=
Permanent load Weight of slab = 0.25 x 25 x 6.51 = 264.1 kl\ Weight of drop - O.l x 25 \( 2.51
=
15.6 kl\
Total= 279.71<1'\
231
232
Reinforced concrete design
Figure 8.13
Flat slab example
6.5 m column centres each way
~
Variable load
Total= 5 x 6.5 2
= 211.3kN
Therefore ultimate load on the Ooor. F
= 1.35 x 279.7
1 1.5 x 211 .3
- 695 kN per panel
and equivalent distributed load. n
16.4 l..~fml
695 6.5 2
ll1c effective span.
L - clear
~pan
(6.5
between column head') +
1.2)
350 "'
., X-
10
~lab thicl..nc~~
2
at either end
l
5.65m A concrete cover of 25 mrn has been allowed, and where there urc two equal layer~ of rctnforcement the effect1ve depth has heen taken as the mean depth of the two l ayer~ in calculating the reinforcement area~. (d - 205 mm in ~>pan and 305 mrn ut suppOI'l!>.) The drop d1 men~ion is greater than one-th 1rd of the panel dimension. therefore the column strip is taken as the width of the dmp panel (2.5 m). Bending reinforcement
Since the variable load is less than the pl.!rmancnl load anti hay size
c
6.5 x 6.5
42.25 m 2 ( 30 m2 ), from table 8. I: 1. Centre of interior span Po~itivc
moment = 0.063/-'/
= 0.063 X 695 X 5.65
247 kN m
~trip i!'. (6.5- 2.5) = 4m which is greater than half the panel dimension, therefore the proponion ol thi ~ moment taken by the middle strip can be taken as 0.45 from table 8.6 adju<;ted as sho" n.
'I he width of the middle
0.45
X
6
-t
I" = 0.55
.5 -
Thus middle strip positi' e moment = 0.55 '>' 247 = 136 I..N m. The column strip poi>itivc moment
=( I -
0.55) x 247
llllli m.
Design of reinforced concrete slabs
(a) For the middle strip
_!!_ = bd2fck
I 36 X 106 4000 X 205~ X 25
= 0.032
From the lever-arm curve. figure 4.5. Ia = 0.97. thcrcfon: M 136 X let A - - - - ~=---=-=-=---::-:::-:---::-::-= ' - O.K7i)l~ - 0.87 X 500 X 0.95 X 205
1605 mrn 2 bottom steel Thu\ provide ,\ixteen HI 2 bar~> (A, 1809 mm1 ) each way in the ~pan. di~tributed evenly acros~ the 4 m width of the middle ~trip (spacing 250 mrn maximum allowuble for a slab in an area of maximum moment). (h) The column ~trip morm:nts will require 1310mm2 houom ~tcel which can be provided a~-. twelve H 12 bar~ (/\, 1356 mm 2) in the :.pan diMrihuted evenly across the 2.5 m width of the column strip (spacing approx 2 10 mm).
2. Interior &upport Ncg:uivc moment
= - 0.063FI 0.063
and
thi~
X
695
X
5.65
= 247 kN m
can also be divided into ()....') 5 X - 4 6.512
mrddle <;trip
X
.,_4 7
= 77k:"Jrn and column lltrip =( I - 0.31 ) x 247
0.31
X
247
= 0.69 x 247 = 170 kl\ m
tal For the middle wip M
btf'}~l
77 X lOt'> - OJ 4000 > :2()5Z X 25 - O. S
f-rom the lever-arm curve. rigure 4.5. /., M
A,
909mm
=
77 x
0.!\((vl::
0.87
X
500
X
= 0.98 (
0.95 ). therefore
106
0.95
X
205
2
Provide ~.: I even evenly ~paced Ill2 burs 400 mm maximum spacing limit.
::1:.
top steel (A 1
1243 mm 2 ) lo !-.ati:.fy
(b) For the column strip M
btf""iJ..l
6
170 X 10 = 0.0 29 2500 X 3052 X 25
From the lever-arm curve. figure 4.5. 1.
= 0.97 ( > 0 95). therefore
6
M
A = --= ' 0 X7f,l:_ 0.87
X
170 X 10 500 X 0.95
X
305
1
- 1349mm
Provrde H 12 bars a\ top steel at 200 centres. This ill cqui-.alcnt to fourteen bar~ (A, 1582 mm1 ) over the full 2.5 m width of the column Mrip. The bending reinforcement requirements arc summarhed in figure 8.14.
233
234
Reinforced concrete design t Column
figure 8.14
{Column
I
Detail$ of bending reinforcement
11H12- 400
11 H12- 400
, >
16H12-250each way
(a) Middle stnp4.0m wide 14H12-200e.w
14'--<-'-->
14H12- 200 e.w
• 12HI2 -210
(b) Column strip 2.5 m wide
Punching shear
1. At the column hC
= 1r x diameter of column head - ;r
x 1200 = J770mm
'hear force Vw - F
7r
4
,
J.2·n
695
676.4kN incrca~cd hy
To aiiO\\ for the effects of moment tran,fcr. \ i' intcmal column. thu' \'fd rtt
= l 15
676.4
15 per cent for an
778 J..N
Ma\imum permissible \hear furce.
vRd
!nO\
d[o6(1- 25(} 1-~)].1~~ 1.5
()5 • Ll
,
= 0.5 X J770 X 305 X
25 )] 25 0 , 250 1.5 x I
[().() (I
thus Vnd,rtf is signifieun()y lc:.s than
5174kN
VR11, ,,," .
2. The nr~t critical section for ~henr i~ 2.0 X effective depth from the face of the column head. that is. a section of diameter 1.2 t 2 x 2.0 x 0.305 2.42 m. (i.e. within the drop panel). Thus the length of the perimeter
11 1
rr
><
2420
Uhimate shear force, \!Ed= 695- ~ x 2.42 2 VU!cft
= 1.15 X 620
l·or the unrcinforced section 7602 X 305 !()() X 1582 ? With Pt = Py = Pz = x = 0.-1~ 2500 305 VRd.r = I'Rd cll td
=
I'Rd c X
1<
7602 mm
16.4
713kN
620k'l
Design of reinforced concrete slabs thus from table 8.2, VRd c
vRd.c
= 0.47 Nlmm~; therefore
=0.47 X 7602 X 305 X
10-~ = 1090k..~
A~ VFA er1 is less than I'Rd.c the section 1s adequate. and shear reinforcement i!. not
needed. 3. At the dropped panel the critical section i~ 2.0 x 205 a perimeter given by II
=(2a + 2b 2;r X 2d) = (4 x 2500+27:" x 410) =
= 410mm from the panel with
12576mm
The area within the perimeter is given by (2.5 I Jd}
2
(4
- r.}(2.0
X
0.205) 2
(2.5 -J 3 X 0.205) 2 - (4- r.)(0.41Q} 2
9.559 m2 Ultimat·e
~henr
force.
695 - 9.559
Vrtl
X
16.4 = 538 J..N
Ml
Vw<11 = 1.15 VRd,
PI
538=6191-.N
= I'Rd.c lld JOQ
where u 12576mm and tl 205mm J582 , 1-0.5 = 031%. thus from table 8.2 ''Rd ,c ?' 0.55 N/mm·
X
'l
_suo
hence \'Kd ~
0.55
12, 576 > 205
14181-.~
A~
I Ld c11 is lcs' than VR,, , the section b adequate.
INote in the above calculation
p 1 has been ba~ed on column \trip reinforcement at the support. Since the critical tone will lie partially in the middle strip. thi~ value will be a minor over-estirnme buL is not significam in thi~ ca~e.l
Span-effective depth ratios
At the centre
e
01
-1 m.
I OOA ~. •eq /it/
or the ~pan JO() x 1605 4(){)0 X 205
= 0. 20
Prom ligure 6.3 the limiting basic ~pan-effective depth ratio i., 32 for clas1- C25 concrete and thi~ i~ multiplictl by a K factor of 1.2 for n flat slab (!.ce table 6.1 0) giving J2 X 1.2 384. actual
~pan
effective depth ratio
6500/ 205
= 31 7
Hence the slab effetl!\C depth i~ acceptahle. To take care of '>tabilit} requ1remcnt~. cxLra reinforcement may be ncccs'ary in the column stnps to act as a tic hetween each pair of columns ~ee section 6.7, and the requirement for at lca~t two bottom bar.. to pa~s through each column will be satisfied by the spacings calculated above and sho" n in figure 8.14.
23!
236
Reinforced concrete design
8.7
Ribbed and hollow block floors
Cross-sections through a ribbed and hollow block floor slab are shown in figure 8.15. The ribbed floor is formed u~ing temporary or permanent shuttering while the hollow block floor IS generally com.tructed with blocks made of cia> tile or '' ith concrete containing a lightweight aggregate. If the blocks arc suitably manufactured and have adequate strength they can be considered to contribute to the strength of the slab in the design calculations. but in many design~ no such allowance is made. The principal advantage of the~c floors is the reduction in weight achieved by removing part or the concrete below the neutral ax i), and, in the ca'e of the hollow block lloor. replacing il with a lighter form of construction. Ribbed and hollow block lloors are economical for buildings where there are long spans. over about 5 m, and light or moderate live load,, such a<; in hoi>pital wards or apartment buildings. They would not he suitable for Slrut:turcs having a heavy loading, such as warehouses and garages.
Figure 8. 15 Sections through ribbed and hollow block floor~. and waffle slab (a) Sectton through a rtbbed lloor
II I
=s==-1! I
Ill E 21 I ·f il l ~IiI "' II
I· I
t=J;
Supporting beam
Solid end se
r - 1r- 1 r- 1 r 1 I
I
--d --, 1
-,
-l
r -
I I
I I
I I
II
I I
I I
-- -- r--- 1 r----- 1r
r - 1r - ' r I
I I
r -, r - 1 r - 1 r - 1 r - 1 r 1 ~- ~ -- ~ ~- ~ ~-~ '-S I I
I I
I I
I I
I I
~-'-'- 11 -'\),..--1-1~ I I light mesh
o ..M. o ..o ..o ..n
(b) Parttal plan of Clnd section through a waffle slab
(c) Section through a hollow b lock floor
Design of reinforced concrete slabs :\car to the supports the hollO\\ blocks are Mopped off and the slab i~ made solid. This .. 1ne to achieve a greater shear ~trength. and if the slab i~ supported by a monolithic ,crete beam the solid section acts as the tlange of a T-section. The ribs should he .ecked for shear at their junction with the solid ~; lab. lt is good practice to stagger the 'Ih of the hoUow blocks in adjacent rows so that. as they arc stopped off. there is no ::11pt change in cross-section extending acros~ the slab. The slabs arc usually made hJ under partitionl> and concentrated loads. During construction the hollow tiles should be well soaked in water prior to placing .: concrete. otherwise shrinkage cracking of the top concrete flange 1~ liable to occur. The thickness of tJ1e concrete flange should not be less than:
1. 40 mm or one-tenth of the clear distance between ribs. whichever is the greater. fnr slabs with permanent blocks: 2. 50 mm or one-tenth of the clear distance between nbs. whichever ll> the greater. for slabs without pcnnanent blocks. these requirements are not met. than n check of longitudinal shear between web und i ...ngc should be mude to see if additional tran~verl'e steel il> needed The rib width will be governed hy cover. bar-spacing ami lire resi~tunce (!.ection 6.1 ). The rihs !)hould be ~raced no further apart than 1.5 m and their depth below the flange 'hould not be greater than four times their width. Transver~c rihs should he provided at 'lacings no greater than ten tune~ the overall slab depth Provided that the above dimen"onal rcqu1remt!nt~ arc met. nbbcd ~lab~ can he treated or annly!>it-. as solid ~lah1-. and the clcl>ign rcquiremt!nts can be based on tho~e of a solid ,Jab. Calculaticms of reinforcement will require evaJualion of effective llangc breadths ;.Nng the procedurel> de1-crihcd for T-bcam-. in Chapter 7. Ribbed :-.labs will he del>igned for ~hear u-;ing the approach described previously With ?, taken us the breauth of the rib. Although no &pccific guiuance ~~ given in EC2, previou!> practice suggests that. where hollow block!-i urc used, the rib width may be mcrea'\ed by the wall thickncs~-o of the block on one sitlc of the rib. Span effective depth ratioo; \\Ill be based on the 'horter span \\ ith the ba:-.ic values 21ven 111 ligurc 6.3 multiplied by 0.8 where the ratio of the Oange width to the rih width exceeds 3. Again, no specifi~.: guidance i~> given in the Code but previous practice 'uggcsts that the thidness of the rih width may include the thickness of the two adjacent bloc!. walls. At least 50 per cent of the tensile reinforecmcm tn the ~pun ~hould continue to the supports and be anchored. In some instance~ the slabs are i>llpported by steel henrm and an! de:-.igncd as s1mply supported even though the topping i\ cominuou.:;. Reinforcement -.hould he provtded over the suppOrt!> to prevent cracl.1ng 111 these ca~e~. This top !>tee! <;hould be determtned on the basts or 25 per cent or the lllld-span moment and ~hould extend at lca~t 0.15 of the cleur span into the adjoining span. A light reinforcing mesh in the topping nange can give added 11trcngth and durubility 10 the sluh, particular!> if there are concentrated or moving loads. or 1f cracking due to ~hrinkage or thermal movement t<; likely. The m1111mum area of reinforcement reqUired to control shrinkage und thermal cracking can be calculated, as given in chapter 6, but established practice suggests that an area of rnesh equivalent to 0. 13 per cent of the topping ftangc will be adequate. Waffle slab!. arc designed as ribbed slab~ and thetr de~ign moment~ each way arc obtained from the moment coefficients tabulated in table 8.5 for two-way spanning slahs.
23i
238
Reinforced concrete design
(
EXAMPLE 8 .8 Design of a ribbed floor
The ribbed floor is coru.tructcd with permanent fibrcglasJo. moulds: it b continuous over several spans of 5.0 m. The charJcteristic material Mrcngths are fcL = 25 Nlmm2 and f>l = 400 N/mm 2 . An effective section. as shO\\n in figure lU6. which satisfie, requirements for .. 60 minute fire resistance {see table 6.5) is to he tried. The characteri.,tic permanent load including self-weight and finishe~ 111 4.5 I.. N/m~ and the charactcriMic variable load i~ 2.5 1..N/m 2• The cuJculaLions are for an end ~pan {which wi ll be most critical) for which the moments and shears can be determined from the ~.:oefficient::. in tnble 8.1. Considering a 0.4 m width of floor ns supportctl hy each rib: Ultimate loud = 0.4( 1.35g~ + l.5qd
= 0.4( 1.35 X 4.5
I 1.5
X
2.5)
3.93 kN/m Ultimate load on the span. F = 3.93
><
5.0
19.65 kN
Bending (I)
At
mid-~pan
de:.ign
a~
a T-sl.!ction:
M ::: 0.086F/ = 0.086 x 19.65 x 5.0 The dfl.'Cti H~ brcatlth uf nange
which both cxccetl the rih spacing of 400mm, whidl 6
\CCII On
Design of reinforced concrete slabs
239
hom the lever-ann curve. figure 4.5. la = 0.98 ( > 0.95). Thus the neutral axis depth lies within the flange and 6
A,
M X 10 = 0.87i)kfad - 0.87 X 8.45 5()() X 0.95 X J6Q
12R mm~
Pro,ide t\\0 HIO ban. in the ribs. A,= 157mm~.
2. At the end interior o;upport de~ign as a rectangular M
~ection
for the 'olid
~lah:
0.086 x 19.65 x 5.0- 8.45 k:-.1 m as in ( I )
0.086FI
128 mm 2 as at mid-span
and /\,
Provide two H 10 bars in each 0.4 m width of slab. A1
= 157 mm!.
3. At the section where the ribs terminate: this occur~ 0.6 m from the centreline of the .~upport and the moment may be hogging so that 125 mm rib:. mu~t provid<.: the concrete area in compression to resist the design moment. The maximum moment or rei.iswnce of the concrete is (), 167/..~/u/2
M
0.167
X
25
X
125
X
1 60~ X 10 -~
= 13.36 kNm which mu"t be greater than the moment at thi~ section. therefore compre.,~ion not rcquirctl.
~teeI
is
Span-effective depth ratio
At the centn: of the I
= I OOA
I
lt.9_
~pan
= I()()
X
400
hd
128 = 0 10% 160 ·-
f-rom figure 6.3 and table 6.10 the limiting basic "pan-effective depth ratio (p = 0.3%) for an end 'pan 1s 32 I J 41.6. For H T-section with ll nangc \\ idth gre
~<.pan
Thus d
effective depth ratio = 5000/ 160 = 31.3
160 mm is adequate.
Shear
Maximum V1 11
~h<.:11r
0.6F
A,
PI=-
bd
in the rih O.
0.6 X 3.93 0.6 X 19.65 157 0.0079 125 X J60
0.6
X
3.93
9.43kN
hom tublc 1!.2. the ~hear resistance without reinforcement VRd ..: = "Rd.,bd where 0.6R N/mm 2 and, from table 1!.3. the Mrength modification factor= 0.94. llcncc:
I'Rd.
Vkd c A~
0.94 < 0.68
X
1:!5
X
160 = 12.78k
VRo, i' grcatcr than \'1:<.1 then no l>hcar n:inforccmcnt in the rib~ arc ~ccurcly located during conl>truction.
Design of a waffle slab Del>ign a waffle
ultimate load
= (l35gk + I .5ql)
= (1.35 x 6.0)
f (1.5 x 2.5)
= I t.RS kN/m~
A:, the slab ha~ the same span in each uircction the moment coefficients. .J", .d~y arc taken from table 8.5 with fy/1~ = l.O. Calculations nrc given for n single 0.4 m wide beam section and in both clirct.:tll.ln~ of !>pan. Bending
1. At mid-span: design a:. a T -sect ion. Positive moment at mid-~pan
, 'i,~nf~
111"
0.024 Moment carried by each rih
v
I I .85 < 62
I0.24 k:'-lm/m
11.-l x 10.24 = 4. 10kNm
4.10 X 10~ bt/'fd = 4(XJ X 160~ X 25 M
0.016
where the effecth e breadth 1s 400 mm as 111 the pn!\JOU' c\nmple. Prom the lever-arm cur\C. f1gurc t5./ flange and O.S7f~k;:
to~>
-tto
M
A, =----=-
0.87
X
500
0 95. Thu~ the neutral
0.95
X
160
U\Jl>
lie:. \\ith.in the
62 rnrn'
Pro ville two HI 0 bars 1n cat.:h r1h at the bottom nt the beam, A, 157 mmZ to satisty minlfllum requirement of0.13bd% 0.0013 x 400 x 160 H3mm 2/rib. Note that ~incc the service ~>lre~>s in the ~tee l wi lt be reduced, thil\ wi ll kutl to a higher spuneffective ch.:pth ratio thus ensuring that the spon- effet.:tive depth ratio of the slab is kept wi thin acccptubll: limit~. 2. At the support: design as a rectangular ~>Cction for the \Oiid slab. Negative moment at support m,,
0.01 I x I I.R5 x b~
!1"111;
= 13.22 kN m/m Moment carried by each 0.4 m width
5.29
M
btJ2fcl.
4()()
X
X
101>
160~
X
25
0.87.1;~,::
5.29 0.87
y
500
5.29 kN m
2 O.Q l
hom the lever-arm curve, figure 4.5 1. M A, = - - - -
0.4 < I 3.22
X
= 0.95. Thu.,
1011 0.95 ) 160
80mm 2
Provtde two HI 0 ha~ in each OA m v. idth of 'lab . ./\,
157 mm 2 .
Design of reinforced concrete slabs
1
241
3. At the l>cttion where the ribs terminate: the maximum hogging moment of rc'istance of the concrete ribs is 13.36 kJ.'l m. as in the prcviou!> example. Thi~ b greater than the moment at this section. therefore compression :.tecl is not required.
Span-effective depth ratio
IOOA ~ req be/
100 x 62 = 0.096~
_
4()()
160
X
hence from tigure 6.3, limiting basic span depth ratio J2 )( 1.5 (for interior 1-~pan) X 0.8 (for nange > .1 X web thickness) when p < O.JCJr.
Thw. allowable ratio
= 32 x 1.5 x 0.8 = 38.4 = 6000 = 37.5
actunl span effective depth
160
=
Thus d 160 mm is just adequate. It ha:-. not been necesl-tilry here to allow for the increased ~pun/effective tlcrth resulting from providing an incn.:ascd steel aren. t h u ~ con~iclenu ion could be given to reducing the rib reinforcement to two II X bars ( I 0 I mm 2) which :-till sat i!>lie!. nominal rcquircml.!nts.
Shear
From Tahle 3.6 in the Desifollll!rs Guide (ref. 20) the shear force cocflicicnt for a wntinuous edge suppol1 is 0.33. I lence, for one rib, the shear at the 'upport 1',,
~"11/,
Ma~1mum \;Ld
v /)-
11 .85
)I
6
X
X
0.4- 9.38k:--J
\hear 111 the rib 0.6m from the centre-line''
9.18
0.6
Atthl\ po~IIIOil.
><
l'l!oJ.,
0.4
= 6.5-tk:-1
125
X.
160 and
100 )( 157 125 X 16()
btl
hence from table 8.2, .'. VRrlr
1185
)I
vi(oJ'
100/\ PI
0 33
I'Rd,
= 0.68 N/mm1 and from table 8.3 K
0.68> 0.94x 125x 160 x 10
1
0.94
12.8 J..N
Therefore the unrcinl orccd ~>Cellon i~ adcqume 111 shear, and no linJ..~ arc rctJuircd provided thut the bars in the ribs arc 1->Ccurcly located during const ruction.
The U\ual form of stam. can be classified into two type:-.: ( I ) tho'c :-.pannmg horiLontall) in the tran~ver<;e direction, and (2) those spanning longitudinall).
242
Reinforced concrete design
8.8.1
Stairs spanning horizontally
Stairs of this type may be supported on both sides or they may be cantilevered from a 'iupporting wall. Figure 8.17 shows a starr supported on one -;ide by a wall and on lhc olher b) a 'tringer beam. Each step is usually designed a<; ha\'10£ n breadlh b and an effective depth of d = D/2 as '>hown 10 the figure: a more ngorous annly&i& of lhc section i' rarely justified. Distribution steel in the longitudinal dtrection is placed above the main reinforcement. Detaill> of a cantilever stair are shown in figure R.l8. The effective depth of the member i~ taken as the me
.... o
Distribution steel
Stringer be~m
'
Waost
Figure 8.18 Cantilevered
Section A A
....
s-1
SltllrS
:= : = : = : = : = : ~
·- -- --·- ·- -- ,
r
•
I
I
I
•
II
Mdin
rconforcement
I
Ligh t mesh
8
~ ~
8.8.2
# •
Light mesh
td
Section B-B
_SP-~n
Stair slab spanning longitudinally
The l>tair -;lah may span into landings which <.pan at right angles to the stairs as in 11gure R.l9 or it may span between supporting beam\ as in ligure R.20 of example 8.1 0. The permanent load is calculated along the slope length of the stain. hut the variable load il> b~bcd on the plan area. Loads common on two spans which intersect at right angles and surround an open \\CII may he assumed to be dtvided equally between lhe !.pan&. The effective span (/) i'> mea~urcd hori7ontally between the centres of Lhe :.upports and lhe thickness of lhc wai'>t (h) i~ taken a-; the slnb thickness.
Design of reinforced concrete slabs
243
Figure 8.19 Stairs spanning into landings
Stair ~lab!. which arc continuous and constmcted monolithit:ally \\'ith their supporting <.bign~.:d for a bending moment of say Fl I 10. where F ~~the total Ill mate load. llowcvcr. in many in~Lances tht: ~lain.. are preca~t or con~tructed after the ma111 ~tructure. pod.et!. with dowel!> being left in the )Upporting beam~ to retch c the tatr.... and with no apprectablc end re~traint the dc~ign moment l>hould he 1'1/R. ah~ or beam~ cun
oc
( EXAMPLE 8 . 10
Design of a stair slab The 'tair' arc of the type ~ho~n in figure 8.20 11panning longitudinally and :.et into pudet' in the two ~upporting beam~. The effective lipan i., 3m and the ri~e of the Main.. '' l .S m with 260 mm treads and I 50 mm ri,ers. The variable load is lO k N/m ' and the chnrncteri&lic material ~trcnglhs arc .fc~ 30 N/mm~ and .h·~ 500 N/mml. Try a 140 mm thick waist, cffc<:tivc depth. d - 11 5 mm. Thb would give an initinl e~ timnte of the &pnn-cffcctivc depth ratio of 26. 1 (3000/ II 5) whi<:h. from table 6. 10, lies n lillie above the basic value for a 'lightly stressed' simply :-.uppmlctl slah. H12 - 400
Figure 8.20 Stairs supported by beams
l 'o
.,E
EHective depth, d = 1 I 5 H12-400
Span 3.0m _
244
Reinforced concrete design
Slope length of stairs= ~- 5}) = 3.35 m Consider a I m width of stairs: Weight of waist plus step1oo = (0.14 x 3.35
+ 0.26 x 1.5/ 2)25
= 16.60kl'\
3.0
Variable load Ultimate load, F
3
>.
= 1.35 x
9.0 k, 16.60
1.5 x 9.0 - 35.9li..N
With no effective end restraint: 35.9 1 X 3.0 _
Fl
M
if
8
146
k.N
m
' -·
Bending reinforcement M
bdlfck
13.46 X 106 _ 34 1000 X 1152 X 30 - O.O.
From the lever-arm curve, figure 4.5, /J = 0.95 (the maximum nonnnlly adopted in practice). therefore M 13.46 x 10" A - ---• - O.X7/yk<: - O.R7 X 500 X 0.95
Maximum allowable spacing is 311
X
283 mm 1/m
115
420 mm with an uppl!r limit of 400 rum.
3 < 140
Provide Hl2 bars at 300 mm cen tre~. area
377 mm1/m.
Span-effective depth ratio
At the centre of the 'pan
1001\, '"'" bel
= 100 " 377
- () l'\
I 000 x I 15
which i:.. greater than the minimum reqUirement of 0.15 for clas~ CJO concrete (sec Table 6.8). f'rom table 6.10 the basic span-effective depth ratio ror u simply supported span with f1rn1 0.5% is 20. Allowing for the actual steel area provided:
20 x A,
limiting span-effective depth ratio
pnlV
I A,. rc4
= 2() X 377/283 actual spun-effective depth ru1io
3000/ 11 5
26.6
26.09
lienee the slab effective depth is acceptuhle. (Note that the allowable nuio will actually be greater than estimated above since the required ~led ratio is lcs., than the 0.5% used with l!lblc 6.10.) Secondary reinforcement
Tranwerse diStribution ~tecl 2: O.::!A, nun
= 0.2
'X
377
75 .-l mm'tm
l11is i~ very small. and adequately co,crcd by II I0 bar~ at the maxtmum allowable ),pacing of 400 mm centres. area 174 mm 'lm. Continuity bars at the top and bottom of the l>pan :.hould be provided and. wherea<. about SO per cent of the main Mccl would be reasonable, the max1mum <,pacing i!> limited to 400mm. Hence provide, ~ay. II I::! bars at 400 mm cemres a' continuity Meel. )
l~--------------------------------------------
Design of reinforced concrete slabs
8.9
245
Yield line and strip methods
•r cac;cs \\hich are more complex a<, a result of sbapc. suppon condition!-.. the pre~ence · ilpening<,, or loading conditionl> it rna} be worth\\ hilc adopung an ullunate analy~i' .ethod. The two pnncipal approaches are the yield line method. \\hich i~ pantcularly table for -.Jab\ wuh a complex shape or concentrmcd loading. and the ),lnp method 11ch ts valuable \\ohere the slab contains opening<;. l'hese method\ have heen the subject of research. and arc \\ell documented although C) are ol a relati\cly specwltsed nature. A tmef tntroduction is mcludl'tl here to u'tratc the gem:ral principle~ and features of the methods. \\hkh nrc rarticularl) uahlc in ass1Ming an understanding or failure mechanism\. In practical tlc,lgn uatinns care must he taken to allov..· for the effect!-. of tie-down fnrcl..'s at cmnl..'t s and '1on ut free edges of slabs.
8.9. 1
Yield line method
T' e capacity of reinforced concrete to sustain pla;tic tlefomlation hn~ bcl.!n clc~crihed in lion 3.6. l·or an under-reinforced section the capacity to develop curvature'> between hr<.t yield of reinforcement and failure due to cntshing of concrete i' mnsiderable. Fe .t '>lab which is suhJected to tncreasing load, c.:mcktng and reinforcement yic.:ld \\' Ill t occur Ill the mo't highly ~tres~ed Lone. Th1' wtll then aLt a' u pla,ttc hinge "' ''equcnt load' arc distrihutcd to other region\ oJ' the lab~ Je,tgncd h) the yield line methou nut't h.: rc nforccu \\ llh Cia" B or(' (medium or IHghl ducltllly Meel nnd thc ratio 1/ cl 'hould •I exceed 0.25 lor concrete up to Clas~ C50/60. hn continuou!> ~>lab:-. the mtcn11cdiate suppon moment 'hould al'o lie between hall' 1J t\.\ icc the mugllttudc of the ~pun moments. It 1:-. a'Mnnl.!d that a pallern of yield line~ can be ~uperimpo,cd on the !-.lah. which w1ll IU'c a collupsc mcchani~m. and that the.: regions between yield line' remain rigid anu 'lCrtld.ed. Pigurc lU I ~>how:-. the yield line mechnnii>m which wi ll occur lor the ~implc 1'e of n lixeu endl.!u ,,lah f>panning in one direction with n uniform load. Rotation aloug •e yield lines wi II occur at a constnnt moment equal t.o the ull imarc.: moment 111 Yield line) Figure 8.21
Development of yield lmes FIXI'd
supports
Plastic hinges
246
Reinforced concrete design
resistance of the section. and will absorb energy. This can be equated to the energy expended by the applied load undergoing a compatible displacement and is known as the virtual work method. Considerable care must be taken over the selection of likely yield line patterns. since the method wi ll give an ·upper hound· 1.olmion. that is. either a correct or unsafe solution. Yield lines will form at right
1. Yield lines are usually \traight and end at a slab boundary. 2. Yield lines \\ill lie along axes of rotation. or pa'" through thelJ' points of mter!-cction. 3. Axes t)f rotation lie along 'upportcd edges. pass over column-. or cut unsupPQrtcd edge\. In !-implc cases the alternative paltcms to be considered will be readily determined on the ba1-.i~ of common ~en'>c, while for more complex cases difTerenlial calcu l u~ may be used. The danger of mis~ing the critical layout of yield lines, and thus obtaining an incorrect 'olution. mean" that the method can only be used wi1h confidence b) c\petienced dc.,igners. A number of typical panern~ urc 'hown in figure 8.22. Simple supports -
Figure 8.22
Exdmples of yield line patterns
~
!
Fixed support Positive yield line
- -_ ~.
......
....
Axes or rotation
Negattve yteld line Axe) ol rotation
'
Column rr~e
edge
A yield line caw;ed hy a sagging moment is generally referred to a-. a ·positive' yield line and i~ represented hy n full line, while a hogging momcm causing cracking on lhc top surface of the ~lab cuuses a 'negative' yield line shown by :.1 broken line. The basic approach of the method i:; illustrated for the ~ imple case of a one-way spanning slab in example B. II
The ~lab ~>h0\\11 in figure 8.23 is ~ubjected to a uniformJy distributed load'' per unit area. Longitudinal reinforcement is provided as indicated giving a uniform ultimate moment of resi~tance m per unit \\idth. Lonttudinal retn orc;ement
~
'
I
I~
m --
line
~
Fig ure 8.23 One-way spanntng slab
~
'
Yield
el ~
~
Pldn Hinge
Collapse mechanism
The ma"pan and a poltlll\c } ield line can thus be 'upenmpo:.ed a~ ~>h
area x load x average dbtanee moved for each rigid half nf the slah
=;(nLx~) xu· x~ therefore total
I
2
,
n£,· ~t·~
Jnternul energy absorbed by rotation along thl; yield line 1s moment x rotation x length - mdJoL where
G>~2(~) 0.5£,
4~
L
hence imcmal energy
247
4mo ~
thu~ equating mtemal energy abi>orbed \\ilh external work done
I ' 2 n /' "1\'..l or m
a:. anticipated
248
Reinforced concrete design
Since the displacement ~ is eliminated. this will generally be set to unity in calculations of this type. In the simple case of example 8.11. the yield line cro<.sed the reintorcement at right angle" and transverse steel wa~ not imoh ed in bending calculations. Generally. a yield line willl1e at an angle 0 to the orthogonal to the main reinforcement and will thus also c:ro~s tranwer~e <>teel. The ultimate moment of rc i~tance developed is not easy to define. but Johan~en·~ stepped yield criteria i~ the mo~t popular approach. Th1s a~'iumes that an inclined y1eld line consis~ of a number of Meps. each orthogonal to a reinforcing bar :b shown in figure 8.~4. Yield line
Figure 8.24
... ;
ray
Stepped yield line
~
I
m,
- I / - /
-
-/
,
14]: tJ
..."C3
€
m1 bm0
l11nH
"""
Remforcement
Stepped Yield lin<'
Movem<'nt Vectors
II the ultimate moments of re~IStance pro\ 1ded hy ma111 and tranwcrsc steel arc m1 and 111~ ~r unit width. it foliO\\' that for cquihhnum of the vector. :-.h0\\0. the ultimate momcnr of re~i~tance normal to the } 1eld Ime llln p.:r unit length is given by
hence
In the extreme ca~e of n 0. this reduce.-. to 1110 111 1, and ''hen m1 - m2 - m, then 111 for any value of 0. Thh. Iauer ca:-.c oJ an orthotropically reinforced sluh (rcin!'orcement mutually perpendicular) with equal moment' of re~>iswnce i,-. said to be isotrop1cally reinforced. When applying thi~ approach to complex situations it i~ often difficult to calculate the lengths and rotations of the yield lines. and a !.imple vector notation can be used. The total moment component mn can be resolved vectorial ly in the .\ and v direction~ :mel since internal cncrg)' dissipation along a y1elcl line is given by moment x rotation x length it follow~ that the energy di,s1pated hy rotation of yield line~ bounding any rigid area is given by
111 11
\\here m, and m, are }ield momenh 111 dm:ction~ x Jnd 1. I, and /> art: projections of ) 1eld line!'> along each axi'>. and ~~, and ¢) arc rotation~ about the axe~. Th1~ 1~ illustrated in example 8.12.
Design of reinforced concrete slabs
249
J
( EXAMPLE 8. 12
Slab simply supported on three sides The ~lab !>hown in figure 8.15 supports a uniformly distrihuted load (u.d.J.) of,,. pa umt
area. IlL
Figure 8.15 Slab supported on three sides
y
X
m1===:: .....::
mJIII
8
Internal energy absorbed (E) for unit displacement at points X and Y Area A
EA \\here
m,l,t'J,
+ m) l yr:>y
c:1, - 0: hence I
m 1nL x JL
L,
n
m1 J
Area B m,lyO~
m,l,¢,
L'll
\\here m} - 0: hence 21111
·
'JL X
I ctL
hence total for all rigid an.:a)> i~
External work done
This can also be calculated for each region separately I
Wa ' of :rated
_)
.
I
I
3
6
-(nLxdL)u•x -
W"
2
,
wodL·
= [~ 11'<1 3/} +of.(~- Jf.) I ' X~]
there lore total
21\'A
+ IVo
I ' =6o(323)hr
X1
250
Reinforced concrete design
Hem:c equaling internal and external work, the max1mum u.d.l. that the slab can !>Ustain is given by ..,- ( 6 1 J1) x _ J?( - llltO ~ -1112 _,..J1) -mta -nh ,- , , ai o(3 - 2 3)1.o·L-(311- 2 1·) ltts clear that the rc!>ull will var) according to the \alue of J. The maximum value of may be obtained by trial and error using se,·eral values of 3. or alternatively. b) di ftcrentiation. let m2 = Jllllt. then 11·
II'
12m d n 2
11,12 ) n~L2(3;J- 2J2)
and tl{ml /w) d,8
=0
will give the aitical value of I'J
hence JrtrJ2 + 4o 2 ;~- 3n2 =o
and
~] ,\ m:gah\e value
1~
1mpossihlc. hence the critical \aluc of ,j for usc in the analyc;il> ''
gi' en b} the posll1ve mot.
8.9.2
Hilleborg strip method
Thi!-t is based on the 'lower hound' concept of plu-,tic theory which suggests that it J ~trcil~ c.Ji~tribuuon throughout a Mructurc cnn be found "'hich sati'>fies all eqUJiibnum t'onditions without violating yield critcriu, then the structure is safe for the corrc~pond1ng system of external loath. Although safe. the '>Lructure will not necessaril) be ~erv1ceahle or economic. hence considerable sktll is required on the part of the engineer in selecting a suitable distribution of bending moment~ on which the ucsign cnn be ba~ed. Detailed analysi~ or a slab designed on thi~ basi~ is not necessary. but the cle~igner\ structural sense and 'feel' for tile way loads are transmitted to the support~ are of prime imporwncc. Although this method for design of slabs wa~ proposed hy ll il lcborg in the 1950s. developments by Wood and Armer in the 1960s havc pmduced it~ currently used form. The method can be applied to slab~ of any shape, and a~sume:-. that at fnilure the load will he carried by bending in either the x or y direction separately with no twisting action. llcncc the title of ·strip method'. Considering a rectangular ~lah ~1mply ~upported on four ~ides and carrying a uniformly di!.tributed load. the load may be expected to he di'itributed to the supports in the manner shown in figure R.26. Judgement will be required to detcnnine the nngle c\. but 11 can be !'teen that if n 90 the slab w11l be a'>sumed to be one-way spanning and. although safe. is unlikely to be servtceable because of cracking near the upports along the r axis. Hillcborg '>Uggcsts that for such a !>lab. n should be 45 . The load dtagrams causing bending moments along typical <,lrip-. spanning each direction are also shown. It will be seen that
It!!
'lie
Design of reinforced concrete slabs -o c "O"" -o 'Q.
.. -"".,.. 0
~·::: ·~
251
Figure 8.26 Assumed load distributions
2-
(8 Load Load and B.M.O Strip A
he ultcrnalive pallern. :-.uggcsled by Wood and Armer, in ligur~; R.27 will .~imp l ify the design, and in this case live :-.trips in each direction may be conveniently used as ~hown. Each of these will be designed in hending for ils particular lonuing, as if it were one-way 'panning using rhe method~ of ~ection BA. Reinforcement will he arrunged uniformly across each strip. to produce an overall pattern of reinforcement bands in two dirct:tion-.. Suppon reactions can also be obtained very simply from each strip. The approach 1s particularly suitable for ltlabs \\ith openings. in \\hich ca~c 'trengthcncd hamh can be provided round the opening-. with the remainder olthc \lah divided 1nto ~trips ttlo. appropriate. A typical pattern of thi-. type I'> -;ho\\ n in hgure H.2R. Suggl!)ted Strips Figure 8.27 lC
$'.:.
::. I ~ "'~ I I
!
E
VI
01 01 :I
load distribution according to Wood and Armer
j
Q.
V>
~.
N
Strips x, andx, y andy, etc are ldentlcal
t
"'
~ !-
Note
/
45" .
Stiffened bands Figure 8.28 Strong bands around openings
The columns in a structure carry the loads from the beam~ and slabs down to the foundations, and therefore they are primarily compresston members, although they may also have to resist bending forces due to the continUity of the structure. The analysis of a section subjected to an axial load plus bend1ng 1s dealt with in chapter 4, where it is noted that a direct solution of the equat1ons that determine the areas of reinforcement can be very laborious and impractical. Therefore, des1gn charts or computers are often employed to facilitate the routine des1gn of column sections. Destgn of columns is governed by the ultimate limit state; deflections and cracking during service condit1ons are not usuc~lly a problem, but neverth eless correct detailing of the reinforcement and adequate cover are important. Many of the principles used in this chapter fo r the design of a column can also be applied in a similar manner Lo oth er types of members that also resist an o;~xial load plus a bending mom ent.
252
Column design
9.1
25 3
Loading and moments
The loading arrangcmcnL'> and the analy!>is of a 'tructuml frame have been described with example~ in chapter 3. Tn the anal}sis it was necessal) to clas~ify the ~tructure mto one of the following type~:
1. hraced - \\here the lateral loads are resio;ted b) shear wall;., or other form~ of bractng capahle of transmitting all horizontal loading to the foundations. nnd
2. unbraced - where horizontal loads arc resisted h) the frame nction of rigid!) connected column~>. beam1. and slabs.
I!
5
If
With a braced structure the axial forces and moments in the column-; are cau~ed by the vcnicul permanent and variable actions only. whereas with an unhraced ~>tructurc the loading arrangements which include the effects of the lateral load~ must also be considered. Both braced and unhraccd strw.:turcs can be further classilied th swuy or non-swny. ln a sway structure sideswoy is likely to significantly increase the magnitude of the bending moments in the columns whereas in a non-sway ~>tructun: thil'l effect i~> let.s significant. This increase of moments due ro sway. known ns n \econd order' effect. is not considered to be significant if there is less than n 10 per cent im:rcasl! in the normal ('fir~t order') design moments as n result of the sidesway dbplacement ~ ol' the ~tructurc Suh,tunt1ally hraced ~tructures can normally be considered to be non-:-.wny. EC'2 gives further guidance concerning the classification of unhraeed structure'>. In thi' chapte1 only the de.,1gn of hraccd non-~wa} 'lructures will he considered. For a braced \trueturc the critical arrangement of the uhnnatc loud 1s u\uall> that which cau ... e~ the largc... t moment in the column. together" 1th a large axial load. \<;an example. figure 9.1 'ohO\\' a building frame "ith the critical loading an·angement for the design of it~ centre column lit the fir~t-floor k\el and al~o the left hund column at all floor levels. When the moment!-. in column~ are !urge und particularly "llh unbraced columm. 11 may nbo be ncces\at) to check the ca~e of ma:-.1mum moment combmed w11h the mmimurn axial lond. ln the case of braced lrames. the axial column force!- due to the vertical louding may he calculuted as though the beam11 and slabs are simply !'>upported. provitll!d that the span!. on either s1de of the column differ by no mnre than 30 per cent and therc i:-. not~~ cantilever spun. In M>me ~otructures it is unlikely that all the noon, ol u huilding will carry the full imposed load ut the same instant, therefore a reduction i!-. allowed in the tolll l impose(.) load when design ing columns nr foundotion~ iu buildings which urc greater than two storeys in height. Furthl!r guidun~o:c on thi), can ht! found in OS EN 1991 1-l (Act ion), on structurl!s) Figure 9.1 A crilicatloading arrangement
1.3SG, + l .SQ,
1.3SG,
+
1,5Q,
1st t _3SG, .. 1.50\ Floor 1 - - - -
1.35G,
254
Reinforced concrete design
9.2
Column classification and failure modes
(7 ) Slenderness ratio of a column The
~tenderness
A=~=
ratio A of a column bent about an
axi~
i-. given by
lo j(I / A)
(9.1
where / 0 r.~
the effective height of the column
i is the radiu5 of f:yrarion about the axis considered l is the second moment of urea of the section about the A is the cross-sectional area of the t:olumn
axi~
(2) Effective height 10 of a column
The effective height of a column, 111 • i~ the height of u thcoretical column of equivalent section but pinned at both ends. Thi~> depends on the degree ol' fixity at each end of the column. which itself depends on the rdmivc ~tiffnesse~ of the columns and beam-. connected to either end of the column under con:.idcration. EC2 give' two formulae for calculating the effective height: For braced member-;:
(9.2) For unbraced members the larger of:
lu
I
/-(1+ lOk' A2)
(9.3.a)
X
\
k,
+ k:
and lu
I( I + I
~ kJ (1+ I ~\J
(9.3.b)
In the above formulae, k1 and k1 arc the relative llexihilitics of the rotational restraints a1 ends ·1· and '2' of the column rcl.pectively. At each end /q and k2 can be taken as: column stillness _ (£/ / l),..rumn
(/ / /),ulurnn
L beam stiffness - L 2(£1 /1)~,: L 2(/ //)!> in the beam. Hence. for a typical column in a symmetrical frame with -;pans of approximate!} equal length, as shown in figure 9.2. k1 and J..~ can be calculated a<.,: kl = k~ =A
= COlUmn Sti.ffnel>!l
y/ /)cCllumn -
2:: beam str ffncss "£ 2(/ / I) t.:.m
(/ / llcolumn 2 x 2(/ / /) btam
I (/ / l)cnlumn 4 (//I) t-eam
Column
non-falling column
-
I End 1
End 2
9.1
non-fatling column
Note: the effective contnbution of the non-failing column to the JOint sttftness may be •gnored
Table 9.1
Column effective lengths
1 (f / l(olvrnn) 4 (Tj/btmnl
0 0.0625 0.125 (fixed end)
k
lo braced (equation 9.2) {x/} lo unbraced (equation 9.3(a) and 9.3(b)). Use greater value {x/}
0.25
0.50
1.0
/.5
2.0
0.5
0.56
0.61
0.68
0.76
0.84
0.88
0.91
1.0
1.14
1.27
1.50
1.87
2.45
2.92
3.32
1.0
1.12
1.13
1.44
1.78
2.25
2.56
2.78
Thus. for thic, '>ituat1on typical values of column efleeti\C kngth' c~tn he tahulated using equation~ 9.2 and 9.3 a\ ~hown in table 9.1.
(3) Limiting slenderness ratio - short or slender columns
113.a
EC2 place<; an upper limit on the ~lenderness rauo of a single memhcr below which second order effects may bc ignored. This limit i~ given by: A11 111
20
X
A
X
8
X
C/ VII
(9.4)
where:
13 b at
II
I /( I I· 0.2qlel )
lJ
v'Tf'2w 1.7 rm
C '/let w
255
Figure 9.2 Effective length calculation for a column in a symmetrical frame
beam failing column
design
effective creep ratio (if not known A can be taken a~ 0.7)
ll,f.,.J/(IIJ"") (if not known B can be taken
a~ 1.1)
.~d
the design yield strength of the reinforcement
f..t~
the de~ign compressiYe strength of lhc concrete
= the total area of longitudinal reinforcement II = Nr..J/(AJ'cd )
A,
N&~
- the dc~ign ultimate axial load in the column r 111 - M01 / Mo• (if rm not known then C can be taken"" 0.7) Mnt- Mm are the first order moments at the end of the column with IMu1
~
IMm
256
Reinforced concrete design The following condition~ apply to the value of C:
If the end momems. Mot and Mu~. give ri'\c to ten~ion on the swnc side of the column r01 should be taken as po~itive from which it follow" that C :5 1.7.
(a) (b)
If the con,erse to (a) is tn1e. i.e the column i' 111 a state of double curvature. then ~hould be taken as negative from which it folio'"' that C > 1.7.
rm (c)
For braced members in which the first order moment<. ame only from tmnsver~e loads or imperfections: C can be taken a<; 0.7.
(d)
For unbraced members: C can he taken a' 0.7.
For an embraced column an approximation to the limiting value of A wJII be given by:
At11n = 20
X
A
X
8
X
Cj [n
20
X
0.7 x I. I
X
0.7/ ../Nr:
- 10.8/ /NEc~/(A.:./~d) The limiti ng value of A Cor a lmtted column will depend on the relati ve value of the column 's end moments that will normally act in the ~-tame clockwi'\c or anti-clockwi se direction as in case (b) above. If th ese moments :trc of arproximatcly equtll value then r111 I , C 1.7 ~ J 2.7 and a typiml, approximate limit on A wi ll he given by:
=
Ahm
= 20 X A X 8 X Cj[ii = 20
0.7
X
I. I
x
2.7/v'Nt-tt/(AJcd)
41.6/ .fNt.J/{A.J~d) Alternatively for a bracl'd column the minirm1111 limitmg value of A will be given b) tal-.ing C 1.7. Hence:
=
\hnl
20
A X B '( c;../ii
20
X
0.7
>.
1.1
X
I 7, VNtd (AJc.J)
'26.2/ \ ' NEcJ (Ac.f~,t ) It the actual <.knderness milo is le'" than lh~ calculmed Htluc of \ 1101 1hcn 1he column can be trentcd a.' l>hon. Other'" 1~e 1he column mu'it be treated a-. slender and second order effec:l~-o mu-;t be accoumed for 111 the de,ign nf the column.
(
EXAMPLE 9. 1 Short or slender column Determine il' the column in the lm1ccd frame shown in figure 9.3 ii> short or slender. The concrete ~trcngth .1~·• 25 N/mm 2, and the ultimate axial load 1280 kN . It can be ~een thm the column will have the hi ghe~t s lendcrn c~!' ratio for bending about axes YY where II ~00 mm and abo the end restraint~ ~1re the les~ stiff 300 x 500 heams.
=
Effective column height /0 f u>t
= 4()() X 300'/J:!
90()" 10~ Innl~ 1
4
'"""'" - 300 ,.. 500' 1 12 - 3125 x I0 ' mm fcotfl~ol 900 >< 1011 / 3.0 101 k, =h ==-':..::..:._....:.;.;-~ - "iJ11tx-.llll f ltx:...,) 2(2 3125) HY'/4.0 X 101 ) 0.096 From table 9.1 and b) interpolation; effective column height/o
0.59 >. 3.0
= l.77 m.
Column design
Figure 9.3 Column end support detail
_ 3oo _
length= 4.0m each sid~ \
2
'
Beam
~ 400 -
z
Beam Note: the beams are continuous In both directions
I /·. I \(
Slenderness ratio ,\
"
Rudius of gymtion. i Slendernc's ratio \
3.46 111 /i
= 1.77 x l0'/86.6
86.6 mm
20.4
ror a bmced column the mmimum limiting value of \ will be given hy \h.,
26.2t v Nt;J /(AJ..,J)
''here: N~:t~ /(AJ~'Cl)- 1280 ( 10 /(400 1
X
300
25 / 1.5)
0.6-t
thu\ \1uu -
26.2/ V0.64
32.7
(> 20.4)
llcncc, compurl!d with the mwinwm limiting value of,\ the column
l order moment effect!> would not have to be wf...cn into uccount.
i~
shon and
~ccond
Load N
:r The mg
500
Moment M
(4) Failure modes
Short eolumns usually fail by crushing but a slender column b liable to foil by buckling. The end moments on a slender column cause it to deflect ~ideways and thuN bring into play an additional moment Ne114d as illustrnted in figure 9.4. TI1e moment Nendd cou~e:-. u further lmeral deflection and if the axial load (N) exceed~ a critical value thi~ deli eel ton, and the additional moment become self-propagating until the column hud.les. Euler denved the critical load for a pin-ended strut a~ M
The CfU\hing load Nu.t or a trul) axially loaded column may he taken a-.
Nu.J
= 0 567j~k Ac + 0.87AJ;..
where Ac ill the area of the concrete and A, is the area or the longitudinal
~~teet.
Figure 9.4 Slender column with lateral defle
258
Reinforced concrete design 10
Figure 9.5 \ Column fatlure m~
5 Buckling
35
70
175
105
210
I
/i
Value~ of N~n 1 /NuJ anu I have been calculated and plotted in figure 9.5 for a typical column crms-\cction. The rutin of Ncn 1/ Nud in ligurc 9.5 determine~ the type of failure of the column. With I/ i le~~ than, '>ay. 50 the load wtll probuhly cause cntshing, NuS than Nc111 , the load thm call'•C\ huckling and therefore a budding failure will not occur. This i~ not true with highl·r value of 1/ i nnd '>O a buckling failure is pos~ible. depending on such factor~ "' th~.: initit~l curvature of the column and the actual eccentricity of the load. When ! 1 1 is greater than 110 then N.,11 i~ les\ than Nud and in thi~ ca~c a huckling failure \\Ill oc<.·ur for the column considered. 'I he mode of l':ulurc of a column can be one ot the foliO\\ ing: ft~ilurc with negligible lateral dcllection. wh1ch usually occurs with ~hort column:; hut can ui'>O occur ''hen there arc large end momentf> on a column with an intermediate \lendcrnc's ratio.
1. Material
2. Material failun: inwnsificd hy the lmcral deflection anu the additional moment. Thts type of failure is typtcal of intermediate column\,
3.
ln~tahility l:ulure \\hich occur \\ ith l>lender column~ and is liable to he preceded by cxce,~ive deflections.
9.3
Reinforcement details
The rule., govern ing the mintmum and maximum amount~ of reinforcement in a load hcarmg column ure as follOW\.
Longitudinal steel 1. A mini mum of four bars i'i required in a rectangular column (one bar in each corner) and six bar., in a circul11r column. Bar diameter should not be less than 12 mm. 2.
fhe minimum area of steel is given by
A,
= O.IONEd > OOO"k 0.87/l~ - . - '
Column design
25
3. The rnaJtimum area of steel, at laps is given by 1\,
ma~ < 0.08
A,
\\.here/\, is the total area of longitudinal steel and Ac is the cro ..s-sectional area of the column. . .m regtons . As.nu\ Oh t erwtse. away f rom Iaps: -
A,
< 0"' . u-t .
Links 1. Minimum si7e = ~ x size of the compre11~ion bar hut not less than 6 mm. 2. Maximum spacing should not exceed the lesser of 20 x size of the smallest compression bar or the least lateral dimension of the column or 400 mm. Thi& ~racing shou l(l be reduced by a factor of 0.60. (a)
for a distance equal to the larger lateral dimension or the column above and below n beam or slab, and
(b)
at lapped joint!. of longiwdinal burs> 14 mm diameter.
~p1C3.1
Wtlh
o \m .
Ilu ''
uch
~ load
at lure
3. Where the direction of the longitudinal reinforcement changes, the ~pncing of' the links should be calculated, while taking account of the lateral force~ Involved. If' the change 111 direction b lesl. than or equal to I in 12 no calculation is nece~sary . 4. Every longitudinal bar placed in a corner ~h oulc.J he held by trans\ersc remforcemcnt.
5. No compression bar should be further than 150 mm from a rcstra111ed bar.
1
hon
I
lh 30
Although links are popular in lhe United Kingdom, helical remtorcemcnt '' popular Mlmc part<; of the world and provides added '>trcngth 1n nddtllon to added prutecuon agmn~>t sCI~>mtc load1ng. Si1ing and spacing of helical reinforcement ~hould he similar to linb. f-igure 9.6 shows po'isible nrrangements of reinforcing. har~ nt the junction of two column~ and u floor. In figure 9.6a the reinforcement in the lower column il> cranked Ml that it will lit within the :-.muller column above. The cr~,.,~ in the rciuforcemcnt11hould, if po:-siblc, commence uhove the soffit of a beam so that the moment of re~i~tance of the column i~ not reduced. For the same reason, the bnrs in the upper column ~ohou l d be the in
Figure 9.6 Detdils of splices in column reinforcement
load
omen ::m. Beam Soffot
(a)
(b)
(c)
260
Reinforced concrete design one., cranked \\hen both columns are of the same si;e., a:; in figure 9.6b. Links should be provided at the point<; ''here the bars are cranked in order to re~i:.t buckling due to horizontal components of force in the inclined lengths of bar. Separate dowel bars as in figure 9.6c may al~o be useu to provide continuity between the two lengths of column The column-beam junction <;bould be detailed !.ll that there is adequate space for both the column \ted and the beam 'tee I. Care lui attention to detail on thb point \\ill great!) as~ist the fixing of the :.teel during construction.
9.4
Short columns resisting moments and axial forces
The area of longitudinal steel for these columns is determined by: 1. u'ing design chan-. or con tructing M- N interaction diagram)> 2. a ~olution or the ba!'ic dc .. ign equations. or 3. an approximate method
a~
in chapter 4.
De'>ign cham are u~ually used for columns ha\ ing a rectangular or circular cross1>cction nnd a :.ymmetrical arrangement of reinforcement but internctton d i agrom~ can be constructed for any arrangement of cross-section us illustrated in examples 4. 10 and 4.11. The bu,ic equations or the approximate method can be u~ed when an unsymmetrical arrangement of rcinforccmenl i<. required. or ,.,hen the cro:.s-~ection i'> 1101\·rectangular as ucscnbetl in 5CCllOil 9..'i. Whichever design method is used, a column shou ld not be designed for a moment lcs~ than N1..t 1'111111 • where 1'111111 ha~ the lc\~er value of It / 30 or :!0 mm. Thh is to allow fm tulerancc' tn con,tructton The dtmcn,ton h i' the o' erall '"c or the column crosssection tn the pl
9.4.1
Design charts and interaction diagrams
The design or a section !>Uhjl.!ctcd to bending plw. axinl lo
(9.5)
M~ct - Fn
NfA
,
(
(9.6)
2
dc'>tgn ultimate axial load
Mr.d = destgn ultimate moment v = the depth of the stress hlod
0.8.\
A: = the area of longitudinal reinforcement in the more highly compre:..scd race 1\,
f,c
J,.
= the area of reinforcement in the other face the stress in reinforcement A~ the ~tre..,s in reinforcement A,. negative when tensile.
Column design 0.0035
0.567fk
r1
b yd'
• • • A,
d A,
• ••
f· -
Column section
aXIS
r,
Sec lion
Figure 9.7
- -j
F,
Stress
Stra1n
1.4
Figure 9.8
13
Rectangular column
e
1.2
(d'flt
A, • 2
1I 1.0
"'
0.9
0.6 N
bhf, , 0.7
A, 2
• • b_
'
d'
T
06
o.s 0.4 03 02 0 I
.... 0
0 OS
010
0.15
0.20
0 25
0 30
0 35
0 40
045
0.50
M
bh2 fc;
These equation~ urc not suitable for direct solution und 1he de!tign of n column with symmetrical n:infon.:ement in each ftH.:e is hesl curried OLll u~ing dc~>ign chart~ a:; illu:-.tnlled in figure 9.8. Set~ of these chart!. can he found in the Concise Eurocode (ref. 21), the Manual for the Design of' Concrete Structures (ref. 2~) and the web~> it e www .eumcode2.in ro.
6
(EXAMPLE 9.2
Column design using design charts Figure 9.9 show~ a frame of a heavily loaded industrial structure for whid1 the centre along lme PQ are to be designed in t11is example. TI1e frames at 4 m centres, arc braced again~t lateral forces. and <;upport the following floor load~: column~
permanent actum ..:~
I 0 kl\/m1
vanahle action q~ - 15 f...l\/m2 CharacterilillC material ~trengths are N/mm2 for the steel.
fck
= :!.5 'J/mm1
for the concrete and /y~
= 500
0.20)
261
262
Reinforced concrete design _-_-_-_=:._-.._=:_-._-
Figure 9.9 Columns in an industrial
_-.._*-._=:_-._-_---: I I
structure
1 I
Plan
I I I I I I I I
1 I
_-._=:_-._-----==----_-_-~-- _=-_- _3-:-:_1 1
I
p
beams 300x 700dp
1st floor
300 x 400 columns ground floor
6.0m Sl'cllon through the frame
·t.O( 1 .3511~ i I 5qd per metre length of beam 4( U5 X 10 I 1.5 15)
Ma\imum ultimme load at each noor
- l441..N/rn Mintmum ultimate load at each tloor 4.0 x 1.35gk - 4.()
X
1.35
X
I()
- 54 I..N per metre length of beam Consider fir~t the dc!tign or the CeJI!re column ut the llllUCr!o.idc (u.s.) of the first tloor. The critical arrangement of load lhat wil l cause the maximum moment in the column i~ shown in figure 9. 10a.
Column loads Second and third floors = 2 x 144 x I 0/ 2
first floor = 144 x 6/ 2 Column
~elf-weight,
54 x 4/ 2 say :! x 14
- 1440kN 540 :!8
20081..N Similar arrangcmenLS of load will give the axial load in the column at the under 1de (u.s.) and top side (t.s.) of each floor level and these value~ of Nw arc sho\',.n in table 9.2.
Column design 1.35G, + 1.5Q,
Figure 9.10 Substitute frame for column design example
1.35Gk + 1.50..
1 35G• + 1 .50..
1st Floor
Jc..oum X
1.35Gk
c
B
~''..,~ Cntlcalload1ng arrangement for centre columns at 1st floor
(~)
144
A
:.0.:..
6 864kN
A ~
'-
B
k.a
P"
54 x 4 =216kN
,
'
k,o ull•"'
2
c
A
I+432
-432 +72 -72 ' B
kNm
c
kac
2
(b) Subst1tute frame
(c) Fixed end moments
Table 9.2 Floor
3rd u.s.
beam
2nd l.S. 2nd u.s.
1st t.s. 1st u.s.
oor. nnb
A,
Nfd
Mld
Nrd
Mfd
(kN)
(kNm)
bhfck
bh 1 fck
A/yk bhfck
540 734 + 540 1274 1468 + 540
82.6
0.18 0.24
0.07 0.06
0 0
240
68.4 68.4 68.4
0.42 0.49
0.06 0.06
0 0.10
240
600
2008
68.4
0.67
0.06
0.30
1800
(mm 1)
240
Column moments
The loading arrangement and the substitute frame for determining lhc t;Oiumll moments at the lir~>l and second floors arc shown in figure 9.10(c). Membc1 stil'l'nc~scs arc I 2
kAB
- X
2
the
.. \\0
0.3 X I 0. 73 12 X 6
0. 71
X
I0 3
:1
2 '"1
12/.Afl
I 2
-- X
I 0.3 X 0.7' O O .,x., - = 1.7xl · 4
AR(
k
blr'
-
!_X
- 0 ·3 x OA' 12
X
3.0
= o53 .•
X
I
263
o-3
therefore
2) = (0.7 1 + 1.07 + 2 X 0.53) 10- 3 = 2.8-l X 10
l
264
Reinforced concrete design and
. "buuon . f acLOr 10r r the co Iumn d1qn
kc<'l = 0.53 = 0.19 = L-k '\' 2.84
Fixed end moments at B arc F.E.M ·BA
• :
=
144 X 62 -, I_
_ 54
F.f. .M ·UC - -
X
2
4
-
_
432 kN m ? .
- - - 7... k~ Ill 12
Thus column moment Mf.d
= 0.19(432- 72) = 68.4 k~ m
At the Jrd tloor
+ 1.07+0.53) 1o- '
'[)- (0.71
= 2.31
X
10-J
and column moment MEd
= 0.53 (432 2.31
72)
= 82.6 kN m
The area' of remforcemcnt in table 9.2 arc determined by u...ing the de~ign chart of figure 9.8. Section' through the column ore shO\\ll in figure 9.11.
Figure 9.11 Column sections In example
.-
design
8..,.
1-
300
HS at 300
[OJ [OJ 4H2S
(~)
HhL 300
I
Ground to
1st Floor
.H16
(b) 1st to 3rd Floor
NotP: thl' link spacing Is reduced to 0.60 x these values for 400mm above dnd below each floor level and Dt laps below 1sl floor level
('over for the reinforcement i~ token as 50 mrn and d' /It f!0/400 m1n1murn area of reinforcement allowed 111 the ~ection is given hy:
,\,
0.002blt
= 0.002 x 300 x .WO •
0.2. The
240 mm1
and the maximum area is t\, - 0.08 x 300 x 400
9600 mm'
and the reinforcement provided il> within these lirniU.. A lthough EC2 permits the usc of 12 mm main Meel, L6 mm bars have been u~ed to ensure adequate rigidiry of the rdnforcing cuge. A smaller column section could have been used nbove the first floor but thb would have involved change:- in formwork and possibly also increa,cd areas of reinforcement.
l_________________________________________)
Column design
9.4.2
2t
Design equations for a non-symmetrical section
The symmetrical amlJlgernent of the reinforcement with A~ = A, i~ ju\tifiablc for the columns of a butldtng where the axial loads are the domtnant force~ and where any moment\ due to the wind can be acting in either direction. But some member\ arc required to resist axial forces combined with large bendtng moment!-. !-.O that it is not economical to ha\'e equal areas of steel in both faces. and 1n these ca~cs the u~ual de~ign chart!. cannot be applied. A rigoro~ design for a rectangular section a\ shown in figure 9.12 invo l ve~ the following iterative procedure: 1. Select a depth of neutral axis, .\ (for thi1-. dc1>ign method where the moments are relatively large . .\ would generally be lesll than h). 2. Determine the ~tccl "trains f-;.; and ~, from the ~ train distribution. 3. Determine the steel stresses .he andj~ from the equations relating to the stress-strain curve for the reinforcing bars (see section 4. 1.2). 4. Taking moments about the centroid of As
Nl'cJ(e ..- ~
d2)
-: 0.56~/~~bs(d - .v/2) +.t~~A~(d
(9.7}
d' )
where s = 0 K1. This equation can he solved to give a value for A ~
of
5. A, is then determined from the equilibrium of the axial forces , that
N1 <1
i~
0.567/,lb.l / -.:A: l f. A,
(Y.S)
6. l·urther values of.\ may he selected and \tCpll {I ) to (5) repeated until a min1mum \'alue for A' A, 1' obtained. The term};. tn the equauons may be modified to (J"' 0 567/,-..) to allo\\ for the an~a ol concrete di'iplaced by the remforcement Stre"" f, ha'i a negative stgn whenever 11 b ten~ile .
A:.
N14 : Normal to the ~ect•on
('
h/2
The
/112
'
''
• A'•, • d'' A,
• • 1..
b Section
t dz
i
d
neutral axis
~~· -
0.567f,.bs
-f,A,
Stress Block
(EXAMPLE 9 . 3 d to
\e
_)
Column section with an unsymmetrical arrangement of reinforcement The column :-.cction shm' n in figur~ 9.13 res.i:.~ an axial loud or 1100 ~and a moment of 230 I.Nm at the ultimate limit state. Determine the areal> of reinforcement required if the chttracteri~tic material strengths are f.,l = 500 N/mrn~ and }~l = 25 N/mm1.
Figure 9.12 Column with a non-symmetrical arrangeme of reinforcement
266
Reinforced concrete design
0
8...
-
0.0035
.I
300
Figure 9.13 Unsymmetrical column design example
neutral axis
~ -- r----
As
• • • '
I
d,=60
'
Section
Strains
axi~ .
1. Select n depth of neutral 2. From the strain diagram . steel stnun
r = I90 mm.
0.0035 =- (x- d') X
E:,c
0 0035 =- '190 ( 190 - 80)
0 .00"-l)3.
and .
steel stram !,
0.0035 =- (d X
r)
0.0035 (340 190 '
190)
= 0.00276
3. hom the M rc,~-Mrain curve and the relevant equations of section 4. 1.2 yield strain. 0.00217 for grade 500 steel t,
> 0.002 17: therefore /....
435 ~/mm 2
500/1.15
and E..:.< 0.00217;
therefore
!...:
£,.,,
200 x 10 1 x 0.00203
4061'!/mm2• compres~>ion. 4. In equation 9.7
Nt::d
(e + ~- d2) = 0.56~(~~/n(d
e
MEd Nt;d
~
230 X 106
11 00 x 103
= 0.8.r = 0.8 x 190
= 209
r f...:A~(d d' )
.1'/2)
mm
= 152 mm
To allow for the area of concrete displaced
I-.e become!> 406 0.567fck = 406 - 0 567 x 25
= 392~/mm
2
and from equarion 9.7 1100 X 103 (209 J40) 0.567 X 25 > 300 X 152(14()- 152/2) 1 A,=------~~----~~~~~~~----~------~~ 392(340
= 2093mm~
RO)
Column design 5. !·rom equati on 9.8
0.567/c.. bJ - /scA: ...L /.,As
Nr..d
(0.567
25
X
X
300
X
152 )- (392
A,
X
2093)- ( JJOO
X
101 )
435
= 843mm2 Thu'
A: +A,
2936 mm 2 for
x = 190 mrn
6. Values of II~ f-A, calculated for other depths of neutral axis.
A. arc ploncd in figure 9. 14. h om this tigure the minimum area of reinforcement required occurs with x ~ 2 10 mm. Using thi s depth of neutral axis. step~> 2 to 5 arc repeated giving
C:sc
fr~hm
Ji., so
0.00217. e,
= 0.00217
435 N/rnm2 and f., = 435 N/nun 2 tcn~ion
that
A:
= 1837 mm~ and /\,
R9 1 mm 2
(Alternati vely separate values of A~ and A, as calcu lated for each value of x could have al~o have been plotted against x and their values remJ from the graph at r 2 10mrn.) This area would be provided \\ith
A:
three
H25 plus two lJ20 ban.
2098mm' and
A,
one 1125 plu' l\\ o 1120 bars - 1119mm 2
With n symmetrical arrangement of reinforcement the area from the design chart of ligure 9.X would he A~ A, ~ 3 120 mm 2 or 14 per cent grl!ater than the area wi th an unsymmetrical arrangement, and including no allowance for the area of concrete displaced by the ~ooteel.
r
Figure 9.14 Design chart tor unsymmetrical column example
2900 2800 2700 180
190
200
2t0
220
230
Depth of neutral axis, x
These types of iterative ealculalionJ> arc readily programmed for solution by computer or using spread~heets that could find the oplimum l>tecl areas without the necel>~ity of ploning a graph.
267
268
Reinforced concrete design
9.4.3 Simplified design method As an alternatiYe ro rhe previous rigoroul> method of design an approximate method ma:r be used when the eccenuiciry of loading. e ;., not lese; than {h/ 2- th). Figure 9.1S Simplified design method
M. = M + N(h/2 - d1 )
M
? A', A,
A', A,
\:-
\/
The moment MEd and the axial force when.:
NEd
arc replaced hy an Increased moment Ma
(9.9 l plus a compressive force NFA acting through the tcn~ilc 'teel A a~ shown tn figure 9.15. Hence the destgn of the reinforcement t'> carried out 111 two pal1!-..
1. The member is designed as a douhly rcmforced :-.ectton to rc,i:-.t Mu acting hy
it~elf.
The equation., for calculating the area' of rctnforcement to re<,i!,t M~ for grade:- C50 concrete (or below) arc given in 1>Cctinn 4.5 a~: /VI~
O.l67fc~bd~ I 0.87J;v'\:(d
d' )
O.R~/~~11, - 0.2CJ4(dbd I 0.8~()~/\:
2. The area of A, calculated in the
11r~t pun is reduced by the amount
(9.10 (9.11
Nf!.d /O.'d7f)·k·
This preliminary design method i~> prohahly most usc l\ 11 for non-rccttlllgulur column sections as shown in example 9.5, hut the procedure i& hrc;t clemonstn.1ted with a rectangu lar cross-section in the following example.
(
EXAMPLE 9 .4
Column design by the simplified method Calculate the area of steel requtrcd in 1hc 300 x .tOO column of figure 9.1.l = II 00 k . MF.d = 230 k m. /.k 251\/mm, and />l 500 Nlmm 2•
=
NF~
. .
23()
X )()6
I:.cccnlncll) e = ---...,.1
1100 X 10
= 209 mm >
G-(/2)
Column design 1. lncrea ed moment
=MrA+NwG-d1)
Ma
= 230 + 1100(200 - 60) 10-3 = 384k.J"l m The area of <;teel to resist this moment can he calculated u~ing formulae 9.10 and 9.11 for the de:;tgn of a beam with comprel>sive reinforcement. that •., 0.167fckblP
MJ
+ 0.87J;.k11~(d
d' )
and 0.8~(,lA,
= 0.204/-~bd + 0.87/y~A:
therefore 38-+ x 1o6
0.167 x 25 x 3oo x 3402 ~ o.s7 x 500A~ ( 340 - 80}
-
so lhat A~ = 2115mm~
and ().!{7 \,
99
J(
500
331 I
J<.
1\, = 0.204
X
25
X
3(){)
X
340 + Q,g7
X
5()() ' 21 15
mm 2
2. Rcductng thi!> an:a by Nw /O.R7j~ ,
A,
9 1"
I 100 " 10' 311 I - ..,.-...,_--,-0.87 x 5(Xl
- 782 mm 2
self C50
910
rhis compare' wtth A~ = 1837 mm~ and A, 891 mm~ with the de,ign method of example 9.1. (To give a truer comparison the WC)o\ in the comprc,~ive reinforcement 11hould have hcen modtfied to allO\~ for the an:a of concrete di,pluccd, as was done in exurnplc 9. 3. l
911
9.5 a
Dc~ign chart\ arc not usually uvuilttble for columns of other than a rectnngular or a circulur cross-section. Therefore the dc~ign of a non-rectangular ~ection entails either ( I) un iterative solution of design equations, (2) a ~imp lified form of design, or (3) con~truction of M N interaction diagrams.
9.5.1
3
Non-rectangular sections
Design equations
For a non-rectangular ~ecuon tt IS much simpler to consider the equivalent rectangular ~tre~~ block. Determination of the reinforcement areal. follow-. the same procedure a\ de.,cribcd for a rectangular column 111 section 9.4.2, namely
1. Select a depth ol nemral axis. 2. Determine the corresponding Mccl !>trains. 3. Determtne the steel stressc!>.
26~
270
Reinforced concrete design
.
Nld
Figure 9.16 Non-rectangular column section
normal to the section
!- · ~r ,
.!1 2
d
h 2
_
__
_
(K
-
~
p 0.567f,,
0.0035
neutral •
axts
•_
' Section
Stratns
Stress Block
4 . Tnl\t: moment!> about A, so that with reference to figure 9. 16:
NEd ( e-'-
~- d2) 0.567}~kAl., (d
x) I
f~cA: (d -
d')
Solve th1 ~ equation to give A~
5. For no resultant force on the sec11on Nw - 0.567..f..kAcc
+ f<.eA~ +./~A ,
Solve thi\ equation to give A,. 6. Repeal the previous step~ for different values of In
~tepl>
1 to
lind a minimum (A:+ A, ).
(..J ) and (5)
A., b the mea of concrete 111 comprcssmn shown \haded
.\ i:-. the di\tance from the centroid of '1,< to the extreme lihrc m compression j~ l'i the Mre~c; in reinforcement
A" negative 11' ten~ile.
The calculation for a particular cro~s-~cction would be very similur to that described 111 example 9.3 except when using the design equat1ons it would he necessary to determine A,c and x for cuch position of a neutral ax1~.
9.5.2
Simplified preliminary design method
The procedure is similar LO that de~cribed for a column with n rectangular section as described in section 9.4.3 and figure 9. 15. The column is designed to resist a moment M0 only. where
M,
MI\U I
NF.d
G-
(9. 12)
d2)
The slecl area required to resist this moment ~:w1 he calculalcd from
M,.
0.567/.kAcc(d- i } ~ 0.87/.;~A~ (d
d')
(9.13)
and
(9.14) where 11,. i'> the area of concrete in compre~~ion \\ ith 1 0.4Sd for concrete grades CSO and below and x ill the distance from the cenLroid of A•• to the extreme fibre in compression.
Column design
The area of tension reinforcement. A,, as given by equation 9.14 is then reduced by an amount equal to N~-.d/0.87/>1.. Thi~ method should not be used if the eccentricity, e, is less than (h/ 2 - d2) .
9.5.3 M- N interaction diagram These diagram~ can be con!>tructcd using lhc method described in sec11on 4.8 \\ ith example5 4.10 and 4.11. They are particularly u~eful for a column in a mult1-storey building where the moments and associated axial forces change at each ~torey. The diagrams can be constructed after carrying out the approx1mate design procedure in section 9.5.2 to obtain suitable arrangements of reinforcing bars.
(EXAMPLE 9.5
Design of a non-rectangular column section De~ign
the reinforcemeut for the non-rectangular section shown in figure 9. 17 given NEI.J = 1200 kN at the ultimate limit state and the characteristic material ~trenglhs are .fck = 25 N/mm 2 and ./)·• = 500 N/mm 2 •
= 320 ~N m.
MFd
11
t'
Ml.:d
Nl:.d
-
320 X 10 1200 X lQl
'?67
-
- -
lm:rca~cd moment M.
mm >
Mw I Nw
(II2 -d)-'
G
320 t 1200{200
d2)
80) I0
1
=464k..\lm With , 0.45d 144 mm. limit of the Mrcss hlocl.
b,
300
200{400
~
0.8.\
115 mm and the width (11 1) of the 'ection at the
115)
400
443mm A.,~
x(l> -1- "') 2 11 5{500 1443} 2 54 223 rnm 2
L l d'
.3
80
'
0
N
'
M
"
Figure 9.17 Non-rectangular section example
271
272
Reinforced concrele design
The depth of the centroid of the trapezium is given by _
s(b ...L 2bJ)
X=-'---,.;3(b T bJ)
= 115
(500 ..L 2 X 443) _ ) =56.3mm ( 3 500 443
Therefore ubstituting in equation 9.13 4(i.l x 106
0.567
25
X
X
54223(320
56.3) f 0.H7 x 500A:(320- 80)
hence
A: = 2503 mrn~ 2XI2 rnm 2•
Prnviue three H32 plus two H 16 bars, area From equation 9.14 lUl~/ykA,
0.567
25
X
X
54 223 I 0.87
X
500
X
2503
thcr~rorc
= 4269 mm2
A,
R~ducing
i\
'
A, by Ncd/0.87f,k gives
4'~-69
120()
0.87
X X
103 500
1510mm~
l
~.
Provtde one HI 6 plus two H32 hat·~. area IHII mm 2 • The total area of n:inforccmcnt prm ided 4623 mm 2 "hich i~ le~s than the 8 per cent allowed. An M-N mtcraction dtagram could nm' he con-;tructcd for tht'> ~teel arrangement. a~ 1n \CCtton 4.8. to provide a more rigurou~ dc~ign
___________________________________________) 9.6
Biaxial bending of short columns
For mol>t columns, biaxial bending wi ll not govern the design. The loadi ng pattern~ necessary to cause biaxial bending in a building's inLernal and edge columns will not usually cause large moments in borh directions. Corner eolumnf. muy have LO resist signilicant bending about bmh axes, hut the axial loads are u<,ually small and a design similar to the adjacent edge columns i~ generally adequate. A design for biaxial bending based on a rigorous analysts of the crol!.s-&ection and the ~train and ~tres., distribution~ would he done a~:cording ro the fundamental principles of chapter 4. For memhcrl> with a rectangular cros'>-,ecrion. \Cparate check<. in the two principal plane<, are perrnio,~ible if the ratio of the corre,ponding eccentricities satisfie~ one of the following conditions: etther
~; / ~ < 0.2
or
e>j~ < 0.2 b
b-
Column design
273
y
Figure 9.18 Section with biaxial bending
z
where ey ond e, arc the lirst-order eccentricitie:-. in the direction or the section dimensions b nnd It respectively. Where Lhc:-.c conditions are not fuiJillcd biaxial bending mu~t be accounted for and EC2 presents an interaction equation, relating the moment~ ubmu the two axes to tl1e moment or resistllncc about the two axes, whlch must he :-.ati~Hed. However, the given formula cannot be u~cd directly to design a column subject to biaxinl bending hut rather to check it once desig.m:tl. In the absence of ~pccilic dcstgn gutdance it would be acceptable in tl1e UK !hal the wlurnn he de~igned using the method previously presented in BS 8110. This approximate method specifies that a column subjected to an ultimate load N1, 1 and moment!> M, and M) in the direction of the ZL and YY axe!> rc~pccti\cly (sec figure 9.18) may be dc:-.igncd for a -;inglc axis bending but with an 111crca,ed moment and ~ubjcct to the follm' ing condition~:
. M, tl ,,
(3)
M, h'
then the tncreased smgle axi<> desrgn moment rs
h'
M, ~
j b' X M)
if M, ..._ AI/~ ,, ll
(h)
then the increased single oxis design moment is M;
My +
I/
lj;,, X M,
The dimension.., II' anull are defined in figure 9. I8 and the coefficient ;:J is ~pecihed in table 9.3. The coefficient~ in table 9.3 ore obtained from the equation d=l
Table 9.3
Ned bhfc• j
Nht
bhf,;k Values of coefficient 1 for biaxial bending
0
0.1
0.2
0.3
0.4
0.5
0.6
1.0
0.9
0.8
0.7
0.6
0.5
0.4
07
0.3
274
Reinforced concrete design
(
EXAMPLE 9 . 6 Design of a column for biaxial bending
The column section shown in figure 9.19 is to be dc~igncd to res1st an ultimate axial load of 1200 k · plus moments of M, = 75 k'J m and M> - 80 kN m. The chamcteristic material strengths are fck = 25 , /mm~ and f.,k = 500 l\/mm2. Figure 9.19 Biaxial bending example
t 0 .....
t
M,
e
I
z
~
Z -·- ·-·-·-
75kNm
l
""
60 I
y
106
M2
75
My
80 x I if
X
c•,
= -NF.d = 1200 X 1()3
C'y
= NFd = 1200 X lO'
62.5 mm 66.7 mm
lhU\
~/('> "
"
= 62.5/66.7 350
0 o.x > '2
300
and l'y/f!( - 66.7/62.5 . - -- - 1.24 /1 It 300 350
> 0.2
lienee the column must he designed for binxiul bending. Mz _ 75 _ O 168 '" (350 - 70) - ·Mv 80 0 333 b' (300 - 60) • M, My II' < b'
therefore the increased M;
M)
Nwfblifcl
~ing lc
axis design moment b
b'
.,..JWx M7 1200 x IO't(300
350 x 25)
h om tahle 9.3, 3 = 0.54 M'
~
X()..,.. 0.54
240
X -
X
280 .
75
= 114.7 kt\ m
0.46
Column design thus
Mtd bh2_{ck
-- -
I 14.7 X 106 2 350 X (300) X 25
=
0 .15
From the de!>ign chart of figure 9.8 AJ)k
blifck
= 0.47
Therefore required A,
2467 mm 2•
Su provide four 1-1 32 bars.
9.7
Design of slender columns
As specified in section 9.2. a column is da:-.silicd as slender if the slcndemess rutio uhout either ax is exceeds the value of AJim· lf A::; AJim then the column may be cla~l>ilicd a~ shmt and the slenderness effect may be neglected. A slender column with A > Alim must be designed for an additional moment cau~ed hy its curvature at ultimate conditions. EC2 identi!ie~ four different apprnachc~ to dc!tigning slender column~: 1. A general method based on a non-linear analys1~ of the 't1111:ture and allm~ing for ~econd-order effect~ that necc~~itates the u'c of computer analy"'·
2. A second-order analy1>i~ based on nominal st1ffness values of the beam' and column' that. agam, require~ computer analysi" using a proces~ of Iterative analy!.i!-1.
3. The ' moment magnification· method \\>here the design momenh arc obtained by factonng the fiN-order moments.
4. The 'nominal curvature' method where second-order momentl- arc dctermmed from an estimation of the column curvature. These second-order momentl- are added to the first-order moment:- to give the tmal column des1gn moment. Only the fourth method. as given above. will be detailed here ll). this method is not greatly dissimilur to the approach in the previous Briti~h Standard for concrete design, BS RI I 0. Fur1hcr information on the other methods can be found in specialist literature. T he expressions given in EC2 for lhc additional moments were derived hy ~ t udying the moment/curvuture behaviour for a member subject to bending plus uxiul loud. The equa t ion~ for calr..:ulating the design moments arc only applicable w colu mn~. of a rectangular or circular section with symmetricul reinfnrcemcnt. A slcnuer column ~hou ld be designed for an ultimate axial load (NCd) plu1. on im:reascd moment given by Mr
Nhll'rur
where
=
l'tor eo + eJ + e~ eo IS an equivalent fif).t-ordcr cccenlrieity
e0
•~ an accidental eccentricity which accounts for geometric Jmperfecuonl- in the column
e: is the
~econd-order
eccentricity.
275
276
Reinforced concrete design The equtvalent eccentricity eo is given by the greater of 0.6eo2 - 0.4eoJ or
0.4eo~
where e01 and eo:>. are the first-order eccentricities at the two ends of the column a.' described above, and le02l is greater than Jl'ot . The accidental eccentriciry is given by the equation In
ea-- ~· 2
where /0 is the effective column height about the axis considered and I
l
100/i
200
1'=-->where 1 is the height of the column in metres. A conservative estimate of e0 can be given hy: /o 1 /o lo 11-=-X-=-
2
200
2
400
The !.econd-order eccentricity e2 is an estimate ol the deflection of the column at failure and i::. given by the equation c~
15 ( -C'v
where
K1 = 1 -L. ( 0.35 + ,.\
{~ 1 ~0};;<,
;?::
I
slendeme:-s ratio
oc~
-effective creep ratto- c1( • to) Mor:,1p/ MoEAJ ¢( , to)= tinal creep coefficient M01 , 11, = the bending moment in the quu'i permanent load combtnation at the Sl Mm 11 the bending moment tn the design loud combination at the ULS h In most practical
ez
ca~e!.
the above equation may be
~>imp lili ed
to
K1K2lfi(yk rr2 x I 03 SOOd
r
where is ~omctimc!. approximnted to a value or 10. The coefficient K1 is a reduction factor to allow for the fact that the dellection mu ~t be less when there i!i a large propor1ion of the column ~cction in compre~sion. The value for K1 i~ given by the equation Nud- NFJ.l < I 0 NuoJ - Nbal .
owhere Nu.t
Nud i~
(9.15 1
the ultimate axial load such that
= 0.567fckA..: -r O.R7h~/\-..:
and Nrn.1 i:-. the axial load at balanced failure defined in section 4.8 and may he taken approximately Nbat = 0.29/cvl, for 1->ymmetricul reinforcement.
01>
Column design
In order to calculate K~. the area A, of the column reinforcement must he known and hence a trial-and-error approach is necessary. taking an initial conservative value of K~ 1.0. Values of K2 arc also marked on the column design chart~ a~ ~hown in figure 9.8. ( EXAMPLE 9.7 Design of a slender column
A non -~W:l) column of 300 x 450 cross-~ection rc~ists. m the ulurn:uc limit ~tnte. an axial load of 1700 k r and end moments of 70 I..N m and I 0 I..N m eau~ing double curvmure tlbour rhe minor axis YY as ~bown in figure 9.20. The cnlumn \ effective height!> arc l~) 6. 75 m and leL - 8.0 m and the charactcri~-.r ic material Mrengths J,k = 25N/mm2 and/yk = 500N/mm 2. The effective creep ratio 9'c~ 0.87. Eccentricities are t'u 1
M1 Nl!.J
M~ ('()2
Nw
10
X I()~
= 5.9nun 1700 70 101 I< = - 41 ::! mm 1700
where t'tP is negative since the column is bent in douhle curvature. I he limiting :-lcndcrne~s ratio can be calculated from equauon 9.4 where:
+ (0.2 x 0.87))
A
I /( I l ().2~)c
8
the default 'alue of I. I
C
1.7
) -
M111 / M11:
II( I
1.7- (- 10/70)
::W A" lJ x C//ii
)c
= 1.84 0 85
1700 X 10 3
II
.'. Alorn
20
(3()() 14.41
X
450)
X
0.567
O.R5
X
25
X
1.1
X
r.: 1.84I yl/
=J4.41 --r:'"" \Ill
0.89
.36.47
v'o !!9 l
N t d ,.
1700kN
Figure 9.20
Slender column example
z (a) Section
(b) Ax1alload and initial moments
277
278
Reinforced concrete design
Actual slenderness rat1os are
,__
\ = 7I~
-
/\ \ 1 -
6.75
~
= -() ~ X .>.46 = ..>
'",,
77.85 > 36.47
S.O
0.4 X 3A6 = 61.55 > 36.47 5
,
Therefore the column b ~lender. and ,.\ i<> critical. Equivalent eccentricity = 0.6eo2 + O.·kot > 0At·0~
0.6eo2
OAeo1 = 0.6 x 41.2
0.4eu:
+ 0.4 x ( 5.9)
22.15 mm
= 0.4 x 41.2 = 16.47 mm
Thcrl!fore the equivalent eccentricity ec 22.35 mm. Taking ,. a~ l / 200 the accidental eccentricity i!. ea
fry
=
I'
=
2
I
200
6750 x T - l6.88mm
The second-order eccentricity is
Kt K!lij{1 ~
('"t=
11. 1
-·-
7!"1 X
-
I03 500tl
= 1-
(oJs 1 200~~
I~O)Oet = I + ((U5
1
096 ( X 6750' ;;· " 103 500
I >- I ••• t'~
:r~
I 03 :'iOOd
>(
25
:wo
77 !15) 150
0.87
I)
5(Xl 240
92.92 111111 \\ tth K' 1.0 for the initial value. l-or the lir:.t iteration the towl eccentricity
+ 16.88 + 92.92
22.35
i~
132. 15 mm
and the total moment is M,
NL!det(lt
Nt:u 1111/..k
1700
X
1().1
= 45() X :100 X 25
M1 =
1Jh1fck
= 1700 x 11
225 x ~() .f5() X 300· X
'"/~ _;,
132. 15 x 10 ' - 225 kN rn
0.504
= 0. 222
From the dc~ign chart of figure 9.&
Ah = 0.80
blrJ..~
and K' -
= 0.78
This new value of K1 i<; used to calculate t'2 and hence M1 for the 'econd iteration. The de,ign chart b again u<>ed to determine AJ.\k/ bhf•• and a new value of K2 as shown in tahle 9.4. The iterations arc continued until the value of K2 in column~ ( l ) and (5) of the
Column design
279
Table 9.4 (7) K1
(2)
(3)
M
Mt bhZfck
(4) A,fyk bhf-;;
(5) K1
1.0 0.78
225 190
0.222 0.187
0.80 0.6
0.78 0.73
table arc in reasonable agreement. \\hich in th1s de!>ign o<.:cur,\ alter two itcrutinn~. So lhat the :-~eel area required is A,
0.6bflj~~ _
---
0.6
X
450
X ~()() X
soo
r}~
25 _
- - - - 4050 mm
~
and K~ 0.7.+. A\ a check on the final value of K2 interpolated from the design chart: Nj,JI
\\hkh agrees \\ith the linal value in column 5 of table 9.-1.
~·
9.8
Walls
Wall11 may tul.c the form of non-~tructural dividing element:-. in which ca~L· their thic l.n es~> will often rellcct sound insulation and fire resi~lllm:c requirement)>. Nominal
The :nm
lhc
reinfon.:ement will he used to control cracl.ing in such t:a,\e).. More commonly, reinforced cont:rctc wall~ will form part of a structural frame and wil l he designed rnr vertical and horitontal forces and moments obwincd hy normal unulysis methods. In thi ~ situ:ltion a wall is dl!fi ned as being a vertical l oad- hc~1ring member whose length is not less than four timc11 its lhickncss. Where several walls arc connected monolitlucally so that they behave a~ a llllll, they urc de~cnhed as a wall system. Sometimes horilontal f'orl:cs on a strucwrc are rest~ted hy more than one wall or \)Stem of wall~. in \vhich case the dt ... tribution of forces hctwcen the walls or sy<.,tcms will he assumed to be in propon1nn to thctr st iffncs-;e).. It j, normal practice to consider a wall as a 5eries of \Crtical stnps \vhen designing vcn1cal re111forcement. Eat:h strip i-; then designed as a wlumn ~ubject to the appropriate venical load and Lranwerse momentl) at its top and honom. Slcmlcrncs<; effects must be constdered \\here ncccl.'\ary. as for columns. If a wall 11. ).Uhjc<.:t predominant!) to lateral bending. the design and detailing \\ill be undertaken as if it were a 'lab, hut the wall thickness \\ill usually be governed by slenderness hmitauons. tire resiMancc requirements and construction practicalities.
For a wall designed either as a senes of column' or a' a ~lab. the area of \'enJca:. reinforcement should lie between 0.002A, and O.o.t, \, and thi~ "ill normally be equ.J} di\ ided het\\een each face. Bar -;pacing along the length of the wall should not e>;c~ the lesser of 400 mm or three time'> the \~all thickncs,, Homontal bar~ !.hould haYe a d1ametcr of not bs than one-quarter of the \'ert ,.. bar , and w1th a total area of not les' than 25'1 of the vertical ba~ or 0.00 lAc whiche a " greater. The horizomal har<; should lie between the Yerllcal bars and the con~.n:te surface. \\ith a spacing which i.., not greater than 400 rnm. If the arc
10
A building 1s generally composed of a superstructure c1bove the ground c1nd a substructure which forms the foundations below ground The foundations transfer and spread the loads from a structure's columns and walls into the ground. The safe bearing capac1ty of the soil must not be exceeded otherwise excessive settlement may occur, re~ulting in damc1ge to the buildmg and its service facilities, such as the Welter or gas mams Foundation failure can also affect the overall stabihty of a structure so that 1t is liable to slide, to lift vertically or even overturn. The earth under the foundations is the most variable of all the materials thdt are considered in the design and construction of an engineering structure. Under one small building the soil may vary from a sort clay to a dense rock. Also the nature and properties of the soil will change with the seasons and the weather. For example Keuper Marl, a relatively common soil, is hard like rock when dry but when wet it can change into an almost liquid stole. It is Important to have an engineering survey made of the soil under a proposed structure so that variations In the strata and the soil propertie~ can be determined. Drill holes or trial pits should be sunk, In situ tests such as the penetration Lest performed and samples of the soil taken to be tested in the laboratory. From the information gained it is possible to recommend safe bearing pressures and, if necessary, calculate possible settlements of the structure The structural design of any foundation or retaining wall will be based on the general principles outlined in previous chapters of this book. However where the foundation interacts w1th the ground the geotechnical
--+-
281
282
Reinforced concrete design
design of the foundation must be considered i.e. the ability of the ground to resist the loading transferred by the structure. Geotechnical design is in accordance with BS EN 1997: Eurocode 7. This code classifies design situations into three types: (i) category 1 - small and simple structures (ii) category 2- conventional with no difficult ground or complicated loading conditions and (iii) category 3 all other types of structures where there may be a high risk of geotechnical failure. The expectation is that structural engineers will be responsible for the design of category 1 structures, geotechnical engineers for category 3 and e1ther type of engineer could be responsible for category 2. This chapter will only consider foundation types that are likely to fal l within the first two categories.
General design approach Although EC7 presents three ahernativc t.bign approaches the UK National Annex a111m~ for only the hN ol thC\C. In this UC'-1gn approach, two \CI<. of load combination' (referred to as comhinat1on' I and 2 111 table 10.1) nuN be con~1dered at the ultimate linut -.t.lte. Thc'c two comhinallon~ w11l he u'ed for con~1derat1on of hoth structural fa1lure. SI"R (e\cc"ivc deformation. crackmg or fmlure of the 'tructun.:). and geotechnical fa1lurc. GEO (exces'>IVe deformation or complete failure ot the i>Upponing ma'' of earth). A th1rd combination must be taken when considenng po\\lhlc lo-,, of equilibrium (f'QL J of the l-tructure !\UCh as overturning. I he partial safct) factori> to be used for these three combinations are g1ven 1n mole I 0.1. Table 10.1
Partial safety factors at the ultimate limit state
Per\istent or transient design situation
Permanent actions
Leading variable action
(Gk)
(Qk ,)
Unfavourable
Favourable Unfavourable
Accomponymg variable action (Qk I)
Favourable
Unfavourable
Favourable
(a) lor consideration of structural or geotechnical failure: combination 1 (STR) & (GEO)
1.35
1 .00*
1.50
0
1.50
0
(b) for consideration of structural or geotechnK,ll failure: combination 2 (STR) & (GEO)
1.00
1.oo·
1.30
0
1 30
0
(c) for checking ~tatic equilibrium (EQU)
1.1
0.9
1.50
0
1.50
0
• To ~ ~pplted to be.mng, sliding dnd earth resistance fort6
Foundations and retaining walls
ode ed
ln determining lhe design values of actions to be used at the ultimate limit state the characteristic loads should he multiplied by a partial safety factor. Appropriate values of p such as the type shown in figure l0. l. where it ma) he necessary to check the possibility of uplift to the foundations and the stability of the structure when it is suhjcctcd to lateral loads. The critical loading lUTangement is usually the cmnbinallon of maximum lateral load with minimum permanent load and no variable load, that i~ l .5Wk I 0.9Gk. Minimum permanent load can sometimes occur during crcclion when many of the imerior finishes and fixtures may not have heen installed. At the same time us the dc:-ign values of actions urc determined. u:. above. the soil parameters u~ed in the geotechnical a!>pccts of the design are rnuliiplied by the parual factors of -;afety, appropn:ue to the load comh1nat1on under considcratton. e factors will not be developed further 111 lim text hut are g1vcn for completenes~. For ~irnplc spread foundation<; such U\ ~trip and rnu footings i:.C7 gives three altcmativc methods ol design: 1. The 'Direct Method' where calculation'> arc required for each It mit state u-;ing the
partial factor:. of ~afety a~ appropriate from tahlc~ 10. I and 10.2 2. The 'Indirect Method' which allows for a simultaneou.; hlending of ullimatc limit slate and <;erviceability limit state procedures 3. The 'Prescriptive Method' where an as~ u111ed ~arc bearing prc:.sure is used to si7.c the foundations based on the serviceability limit ~talc followed by dctuiled structural de~ign based on the ultimate limit Mate In the Prt•lcriptit·e Mt>rlwtlthe traditional UK approach to the sinng ol foundation~ t!\ effectively retained ~uch thnt a suitable base ~tLC may he determined based on the :.crviceability limit slate values for action~ nnd an assumed ullowuble ~ufc bearing pressure (see tnble IOJ). In !his way settlement will he controlled. with the exception that for foundations on soft clay full sculemcnt calculutions mw.t he t:arried out.
Table 10.2
Partial safely factors applied to geotechnical material properties Angle of shearing resistance '\
Combination 1 Combination 2
1.0 1.25
Effective cohesion 1<
1.0
1.25
Undrained shear strength
Unconfined strength
,\U
.'IJU
1.0 1.4
1.0 1.4
Bulk
density ..,._
1.0 1.0
Figure 10.1
Uplift on footing
28;
284
Reinforced concrete design Table 10.3 Typical allowable bearing values Rock or soil
Typical bearing value (kN/mz)
Massive igneous bedrock Sandstone Shales and mudstone Gravel, sand and gravel, compact Medium dense sand Loose fine sand Very st1ff clay Stiff clay
10000 2000 to 4000 600 to 2000 600 100 to 300 less than 100 300 to 600 150 to 300 75 to 150 Less than 75
Firm clay
Soft clay
When.: the foundation~ arc ~ttbJeCt to hoth vertical and hori7.ontal load~ the followine rule can be applied:
v P, when: I
the vertical load
H
the horizontal load
/'_
the allowable vertical loud
/'11
the allowable hori;ontal loud.
The ullownblc hnri7ontul lo:1d wou ld take account of the pa~siw resistance of the ground in contact with the vertical fac~: or the foundation plu' the friction and cohesion along th~: bal'>e. The cakulation~ to detcmune the 'tructural ~trength of the foundation~. that i~ the thu.:knc~' of the ba,cs and the :lrl!a~ of reinforcement, should he ha~ed on the loading~ ;md the rc,ultant ground prcs'iurc' corre~ponding to the ulumate ltmit \late and con,idenng the \\ON of the cnmhinution' I and 2 for the action' p1cviou\ly noted. combina11on I \\til usually govern the 'itructural tbign. For most designs a linear distribution or ~od prC!>SUre aero~~ the ba~o.e of the rooting 1\ :t'iMtllled :11, shown in figure 10.2(a). This assumption must be ba~ed on the soil acting as an elt1~tlc material and the foming having infinite rigidity. In fact. not on ly do most soi ls exhibit some pln~tic bchmiour and all footings have a finite stiffne~:-. but also the t.hstributton of 'ioil pre!>surc varies with time. 'I he aclltal diwibution of hearing pres~ure many moment may take the form shown in ligure 10.2(b) or (c), depending on the type of ~od and the sriffne'~ of the ha'ic and the structure. But as the behaviour ol foundation\ mvolves many uncenamue~ regarding the actton ol the ground and the loadmg. 11 " u~ually unrealbtic to con\ldcr an analys1<; that ts too soplusticated. Figure 10.2 Pressure d1~tribullons under foolings
6
ttttttt (a) Un1form distribution
(b) Cohesive soil
(c) Sandy soil
Foundations and retaining walls roundation~ should be constructed so that the undersides of the bases are below fro~t level. A'l. the concrete is subjected to more severe exposure conditions a larger nominal cover to the reinforcement i~ required. Despite the values -.uggested in tables 6.1 and 6.2 Cl\tabli.,hed practice in the UK would be to recommend that the mintmum cover should be not les~ than 75 mm \\·hen the concrete is cast against the ground. or less than 50 mm when the concrete is ca.~t against a layer of blinding concrete. A concrete class of at least C30/37 is required to meet durability requirement<;.
10.1
Pad footings
The tooting for a single column may be made square in plan. but \\hen.· there is a large moment acting about one axis it may be more economical to hnvc a n:~.:wngular base. Assuming there is a linear distribution the bearing pressure\ across the base will take one of lhe three forms shown in figure 10.3, according to the relative magnitudes of lhc axia l load N and the moment M acting on the base. 1. In figure 10.3(u) there is no moment ~tntl the pressure is uniform
N
p
( 10. 1)*
IJD
2. With a moment M acting as ~hown. the pressure~ are given by the equation for axial load plus bending. Thi~ ~~ provided there is positive contact hetwccn the ba~c and the gmund along the complete length D of the footing, a' 111 figure 10.3(b) ~o that N M\' IJIJ i I
p
where Its the second moment area of the ba~e about the nxt), of bendtng and .1 i' the dtstance from the ax1 to \\here the pres ure il) being calcu lated.
Breadth of footing
r
ting j, ng a~
'o the
r l"'~ure
~-
8
6
' \Oib
c type 1our of and the
~
r
8
'
r
8 i
D
l t ttf t t t 14
p
Eccentricity (e) ,. MIN
~c O
e <0,6
t >D 6
N P • BD
N 6M P- BD± B[)l
p ., 2N
BY where: Y=
(a)
(b)
3(1 e) (c)
Figure 10.3 Pnd-foot1ng pressure distributions
2f
286
Reinforced concrete design
Substituting for I
13D3/ 12 and y
D/2, the maximum pressure is
N 6M IW- BD'l
PI
(10.2)*
and the minimum pressure is p~
N
= BD
6M BD1
( 10.3)*
There tS positi\c contact along the base if p 2 from equation 10.3 is positive. When pressure p~ just equals zero
N
6M
/JD - BD1 - 0 or
M D N =(;
So that for Pl alway!. to be positive. M/N or the effective eccentricity, I ' - must never he greater than D/6. In thc\e cases the eccentrtCilY of loading is ~a id to lie within the 'middle third· of the hasc. 3. When the eccentricity, e is greater than D/6 there is no longer a po!>iuve pres~urc along the length D and the pres~urc diagram is triungulur ns shown in ligure IOJ(c). Bulancing the downward load and the upward prcs!-.urcs I .,pLJ}
=N
therefore maximum
pres~urc I'
2N
BY
''here Y "the length of positive conwct. The cemroid of the pres~ure diagram must coinc1de with the eccentricity of loading in order for the load and reaction to he equul tmd opposi re. J'hus
y
f)
3
or
therefore 1n the case of c• > D/ 6 2N B(D/ 2 _ c7} maximum pressure I' 3
( 10.4)*
A typ1cal arrangement ol the reinforcement in a pad footing i!-1 shown in figure 10.4. With a 'quare base the n!tnlorcemcnt to resist hcnding ~hould he distnhuted uniformly across the full wtdth of the footing. For n rectangular base the reinforcement in the shon direction ~hnuld he di~Lnbuted with a closer ~pacing in the region under t~nd near the t:nlumn. to allow for the fact that the transvcr!-.C momems must be greater nearer the column. It 1~ recommended that at least two-thirds of the reinforcement in the short dtrection !-.hould be concentrated in a band \\idlh of (r 1 3d) \\here c is the column dimension Ill the long d1rection and d i'i the effecuve depth. Lf the footing ~;hould be
Foundations and retaining walls
'
lap length
t
A,
8
J_
'-
L - - - - - - - ---'
bubjectc<.lto u large overturning moment so that there is onl y par1ial hearing, or i f there i~ a re:.ultant uplift for(;c. then reinforcement may :l111o he require<.! in the top face. Do\\eb or !'.tarter bar~ ~hould exten<.l from the looting into the column in order to provtde continutty to the reinforcement. The'e dowel\ should be cmheddec.J into the footmg and extend tnto the column~ n fulllCctions through the ha.,e for chcckmg '>henr. punchmg )~hear and bending arc shown in figure 10.5. The shcan ng force and hcndmg moment~ an.: cmt~ed hy the ultimute load!'~ from the column 1111d the wetght of 1he base ~hould not he included in thc:-c calculations. The thtcknes' of the base i~ oflcn governed h) the requirement' tor 'hear re~•stancc. Following the Prescriptiw! Ml!tlwd the princqxll 'tep~ in the dc:.tgn calculation' arc a!> follows:
1. Calculate 1he plan size of the footing using the permi&sible hcming pressure amlthe critical loading arrangement for the serviceability limit state.
2.
Calculate the heanng prcs'>urc., a<;sociatcd \\ith the critical loadtng arrangement at the ultimate l im1t state.
3.
Assume u suitable value for rhe thickness (II) and efi'cctive depth (d). Check thnt the shear force at the column fuce is less thun 0.51·Lf~JIId 0.5,• 1ifcl/ 1.5)ud where 11 is the pcnrneter of the column and ,., is the ~>trcngth reduction factor
0.6( I - .f,l( l50). 4. Carry out u preliminary check for pum:hing shear w cn~urc that the footrng thickness gives a punching shear strC's~ whil.:h is within the likely range of acceptable pctfonnance. 5. Detenninc the reinforcement reqUired to re!-.1).1 bending. 6 . Make a final check for the punching \hear. 7. Check the ~hear force at the critical ~rction~>. 8. Where upplicablc, both foundations and the <;tructure should be checked for overall stability at the ultimate li mit state. 9. Reinforcement to rc~ist bendtng in the bottom of the ha~e should extend at lea:.t a full tension anchorage length beyond the crttJcal !\CCtion of bending.
287
Figure 10.4 Pad footing rerntorcemenl details
288
Reinforced concrete design
Figure 10.5 Critical sectiom for design
Maximum shear
_
Punching shear penmeter column perimeter + 4n:d
2.0d _
I' --~~
I
\
i"" -
I
......
;
Shear
Bendmg
l.Od
(
EXAMPLE 10. 1
Design of a pad footing
I he fooun g t figure IO.(l) i~ required to resist characterii.ttc ax tal loud~ of 1()()() 1-N .u1d 350 1-.N 'ariahk from a 400 mm ~quare column. The sate bearing prcs\urc un the ~oil 1' 200 kN/m: and the charactemt ie material wcngths are .f..~ JO N/mm' and }, 1 SIX I :"J/mm~. As~umc a l'ooti11g weight or 150 1-.N so that the total permanent load i:. 11 50 kN and ha-.c the dcstgn on the Pn·1cnpt11'l' Method. ~nnancnt
1. l-or the
~erviceahility
limit 'tate
TotuJ dc..,ign axinl load
l .OG~
. 1500 Rcqwcd ba~c nrea - :wo
+ I.OQk
11 50 ..1... 350 = 1500 kN
' 7.5 m
Provide u ba-.e 2.8 m square = 7.8 m2• 2. l·or the ultimate limtt l>tatl: From table 10.1 it j.., apparent load combination I will give the largest set of al:lions for thi~ simple wucturc. Hence, u... ing the partial c;afety factor:. for load combination I: Column design axial lotld. Nht
U5G~
+ 1.5Qk
= 1.15 x 1000 Eanh pre,sure 400 sq
Figure 10.6 Pad foot1ng example
1---
•I 8 .-
,.-----......l!dl-----,
'4)
!J 12H16@ 225 e.w. 1~ -
2.8m.sq
--J
1.5 ll 350- 1875 kN 1875 - 119 kN/m 2 2.1F -·
Foundations and retaining walls 3. Assume a 600 mm thick footing and with the footing con~tructcd on a hl1nding layer of concrete the minimum cover is taken as 50 mm. Therefore take mean effecuve depth = d = 520 mm. At the column face Ma\imum
!>~hear res1~1ance. I'Rd
m;u
0.5ud [o.6(J - f, L ) ] h l 250
1.5
30 ) ] = 0.5(4 X 400) X 520 X [0.6 ( l - ?-50 = 43931-..N ( > N~-cJ IX75 kN)
l'w = -::---:-. __;;;;;._~ Pcnmcter d 626 X 10' N ' 8134 x 520 - 0' 15 • /mm
Tim, ultimate !.hear -.trc11s i' not excc~'ivc, (sec table 8.2) therefore It he u suiwhle e-.timate. 5. Bending rclnforccmenr see figure 10.7(a)
6(XJ mm will
At the columu face which i' the critical section
Mr:.
(2J9
><
2.8
X
1.2)
I ~
X
~-
482 kN m
0.68m 1..
...
l.Od= 0.52m
D (a) Bend1ng
(b) Shear
Figure 10.7 Cntical sections
289
290
Reinforced concrete design
For lhe concrete 0.167.f.:lbd1
Mhal
= 0.167 >< 30 X 2800 X 510~ X
3793 kN m ( > 4~2)
J0-1>
M&l A-= - -
,
0.87/yk.:
From lhe lever-arm curve. figure 4.5. /., -
482 )( 10''
A, - 0.87
= 0.95. Therefore:
- .,.,
~ 500 x (0.95 x 520) - --43 mm
~
2412mm 2 • Therefore
Pro\idl! tw·ehe lll6 bar:. at 225mm centres. A,
IOOA,
100
bd
28()()
2412
X
X
520
0.165 ( > 0.15
sec table 6.X)
that is. the minimum steel area requi rement is satil'fied. Maximum bar size The ~lecl stress should be calculated unde1 the action of the quasi-permanent loading which can he e~timated from equation 6.1 a~ follow~:
t;kl Gk + IUQd 1.15( 1.35Gl I 1.5Qd 5001 )()()() i () 3 ' ·'50) .< 1000- 1.5 350)
t !51 1.35
256 '1/mm '
Theretore lrom table 6.1) the ll1<1\lmum .llltmablc bar ~1te j-, 16mm. lienee. mtntmum area ami bar 'llC rcqutrcmcnt' "' :.pccthcd h} the code for the purpo!>e:> of crack control arc met. 6. Fuwl check of punching :.hear The 'hear re-,i~tance of the concrete without 'hear reinforcement can he obtained fwm table 8.2 where /'1 cnn he taken as the average of the steel rauo~ 111 both direction-, 2412
A,
= bd = 2800 X 520
0.0017 (
(), t 7~',i. < 2~)
hcm:c from table 8.2 I'Rd ,, = 0.4 N/mm 2 . Therefore the shear resistance of the concrete, lftlll,~
I'Rd
clld
0.40 x 8134 x 520 x I 0
1
VRd.•
il'> given by:
169 I kN ( .-. Vud
= 626 kN)
7. Maximum Shear Force - sec figure 10.7!b)
AI the critical section for shear. I.Od from the column face: De~ign
shear """
A' before. .', \'KJ
r
I'Kd.,
=
239 x 2.8 x 0.68 = 455kN
= OAON/mm~
I'Rd. cbd
= 0.40 X 2800 "
520
X
10- 3
-
582 kN ( ~
Therefore no shear reinforcement b required.
\'hi
455 kf':)
Foundations and retaining walls
291
Instead of as~uming a footing weight of 150kN at the stan of this example it i~ possible to allow for the weight of the footing by using a net safe bearing prc:-.surc flnct where P.wt
= 200 = 200
II x unit \\eight of concrete 0.6 >< 25
= I85.0kN/m2
Therefore Required base area
l~
1.0 x column load
1000 + 350
r>net
IR5.0
7 . 30m~
It should be noted that the self-weight of the footing or ih effect nw~t be included in the calculations at serviceability for determining the area of the base hut at the ultimate limit stme the self-weight should not he included.
_______________________________________)
Example 10.1 shows how w design a pad footing with a centrally located set of actions. If the action~ arc eccentric.: to the c.:entroidal axis or the base then in the chccking of punching shcor the maximum shear stress. ' '&J. i!- multirlied by an enhancement factor ,1i ( l ). This factor nccount~ for the non-linear tli:-.lrihution of strc'" urnund the critical perimeter due to the eccentricity of loading. Refcrcncc should he mudc to EC! Clause 6.4.3 for the details of this dcs1gn approach.
10.2
Combined footings
Where two column-; are clo'e together it ·~ sometime!> necessnl) or convement to combine thetr footing1> to form a continuous base. The uimen,ions of the l'oottng ~hould he chosen so that the resultant loud pa,~cs through the centroid of the ha'e area. This may be a~!.lllned to gne a uniform bearing pres:o.urc under the foming and help to prevent differential settlement. For most ~tructures the ratios of permanent wtd variable loads carried hy each column arc 11imilar so that if the t·c~uhnnt pa~~es through the centroid lor the serviceability limit ~Late then this will also he true or very neurly at the ultimate limit ~tate. and hence in these cu:-.es u uniform prc:-.sure dt.,tribution may be considered for both limit ~tates. The ~hape of the footing may be rectangular or trapezoidal u~ ~hown in figure I O.R. The trape:wi(lol base has the tlisadvant:~ge of tlt:ttrengthen the hase and economi~c on concrete a beam is incorporated bctweeu the two columns :-;o that the bose i~ designed as an inverted T-scction. Centroid of base and resultant load coincide
Figure 10.8 Combined bases
-"
--m
+'
o-.-
Rectangular
1--
Trapezoidal
292
Reinforced concrete design
The proportions of lhe footing depend on many factors. u· it i!> too long. there will be large longitudinal moments on the length~ projecting beyond the column!>, 'Whereas a ~hort base ''ill ha-.e a larger span moment between the column<, and the greater width ''ill cau-;e large transverse moments. The thickness of the footing mu~l be such that the c.,hear stres-;es arc not excessive.
(
EXAMPLE 10. 2
Design of a combined footing
The footing supports two columns 300 mm squurc ami 400 mm square with ch:.ml(.:tcrislic permanent and variable loads a~ shown in figure 10.9. The safe bearing pressun: is 300 kN/m~ and the churacteri stic material ~tn:ngth~ are}~~ = 30 N/mm1 and ./y~ - 500 N/mm 2• Asstune a ba~c thickne., s II X50 mm.
1. Base area (calcu lated at servicenhility limit
~tate.
basi ng the design on the
/Jrt'scriptil'l' Method1
Net safe bearing
prc'~sure
Pne1 = 300
'2511
300
25 x O.H5
= 27H.!l kN/m 2 Total load
= 1000 + 200
1400 + 300
- 2900kN 2900 278.8
Area of ha'c reqtmed
Pro' ide a re~tangular ba,e. 4.() m x 2J 111. area 1.24m
2.3m
Figure 10.9
G,
G, • lOOOkN Q, • 200kN
Combined footing cx.1mple
Q,
1400kN 300kN
l 0
"' 00
'
H16 ®150
(transverse)
E ,....
"'
300sq
400sq
column
column
E@-
-0
3.0m
0.54m
4.6m
1.06m
10 5R m'
Foundations and retaining walls be
2. Resultant of column loads and centroid of base: taking moment~ about the centre line of the 400 mm ~quare column
x=
1200 ). 3 - 1 ., f
1200-1700- · -~m
The base is centred on thrs position of the resultant of the column load a~ sho'' n in figure 10.9. 3. Bearing prc~surc at the ultimate limit qare (Load combination I):
Column load:.
= 1.35 x
1000- 1.5 x 200
1650 I 2340
1.35
1400 + 1.5 x 300
= 3990 k;\1
therefore earth
pres~ure
= 4.63990 = 377 k.N/nr X 2.3 1
4. A1>suming d 790 mm fur Lh~.: longitudinal harl' und with a mean d punching ~ hear calculations: J\t the column fuw Maximum shear resi~tance, I'Ru ma'
0.5wl
-
7RO mm for
fo.o (I
l-or 300 mm square column
o)ud (o.6 (I ;;~) ~ ~~~
VRu "'"'
os
1so [o.6 ( 1 - _so ;o ) l 1.•5 30
1200
4942lN (Nf.AJ I or 400 mm
square column
\IRd """
0.5ttt1 .6 ( I -
= 1650 l:
[o
0.5
>.
160()
X
)
1.~ ) ] T:5 .r..~
250
780 [0.6 ( 1 -
;~~)] [~.~~ X
[()
1
- 6589 kN (Nr,t - 2140 k~)
5. Longitudinal moments and 1\heilr forces: the shcnr-force and bending-moment diagrnms at the ultimate limit state and for a net upward pre~surc of 377 kN/m~ are shown in fi gure 10. 10 overleal. 6. Longiluuinal bending Maximum moment is at mid-span between the columns A, =
6
679 x 10 = -0.87 x 500 ----0.!!7/~l.: X 0.95 X 790 Mw
From tahle 6.8 - O. l 5bad -- 0.00 I)t\, nun -10()'"
.,1()()
x -·
><
2080 mm 2
790 - "~7"~6 ~ - mm ~
Pmvidc nine 1120 at 270 rum centres. area minimum area requirements.
= 2830 mm2• top and bouorn to meet the
293
294
Reinforced concrete design
Figure 10.10 Shear-force and bendingmoment diagrams
0.54m
3.0m
1.06m
23<0kN~
• ft
t t w: 377 x 2.3 = 867kN/ m
1180~
920~
I~
""'J 470
~
~S.FkN
I 1
~ 1420
679
~\ 7
B.MkNm
<:::J126
4s7V 7. Transverse bencling
Mnd = 377
1.152 Y
2
M&l A, - 0.87/.k::
= 2491-Nm/m
249
><
= 0.87 X 50() X
106 0 95
X
770
783mm' /m
But IVtmtmum A,=
0 15bd 0.15 " --wo= 100
I()()() X
Provide 1116 bar:. at 150 mm centre~. areu
770
= 1155 mm·, /m
1140 mm2 per metre.
The tran!>ver::.c reinforcement should be Jllm:ed a1 clo~er centre~ under the columns to allow for greater moment~ in those regions. For the purro~e~ of crack control. the ma"imum bar site or maximum bar 11pacing ~hould al~o be checked as in example I 0. I.
8. Shear Punching shear cannot be checked. since the critical peri meier 2.0d from the column lace lies outside the base area. The critical sec1ion for shear is tn~c n I.Od from the column face. Therefore with d 780 mm. Design ~hear Vr:d
= 1420 -
377 x 2.3(0.7K 1 0.2)
570 kN
The ~hear resistance of the concrete without shear reinforcement can be obtained from tabll: 8.2 where p 1 can be talcn as the average of thl: steel ratio' in hmh direction~
"""" As [ 2830 I340 ] =0.5 Lbd= 0·5 2300 x 790 + 1000 < 770
°·0016 (
0 · 16t;f< 2%)
=
hence from table 8.2 ' 'Rd , 0.36 N/mm2• Therefore 1he shear resistance ot the concrete. VRd 1s g1\Cn by: VR
I'Rd.o;bd
= 0.36 X 2300 X 780 'I<
10 ~
Therefore shear reinforcement is not required.
645 kN ( > \1~~~
570 kN)
l~------------------------------------~)
Foundations and retaining walls
295
Strap footings
10.3
Strap footings, as shown in figure I0.11. are used where the base for an exterior column must nor project beyond the property line. A strap beam is constructed between the exterior footing ami the adjacent interior footing - the purpose of the strap is to restrain the overturning force due to the eccentric load on the exterior footing. The base areas of the footings are proportioned ~o thai the bearing pressures are uniform and equal under both ba1.cs. Thus it is necessury that the resultant of the loads -::>n the two footing!'! should pass through the cemrotd of the areas of the two bases. The 'trap beam between the footings should not bear agatnst the soil, hence the ground directly under the beam \hould be IOO!.cncd and lefl uncompacted. As well as the loadings indicated in figure 10.11 EC2 recommend& that. where the action of compaction machinery could affect the lie beam. the beam should he designed for a minimum downward load <.lf 10 kN/m.
I-
r
0
1
Centroid of bases lo coincide with resultant of N, and N2
8
Figure 10. 11 Strap fooling with sheanng force and bending moments for the strap beam
5
s
+
'- L.,L---...J _ _ _ _...,;;._ Loads at the ult. limit state
N,
1.3S W,
-
0/2 R, ~
'
' A~
v
v
Shear Forces
I,
N,(r- f) - p.Brl
2
S (NI+l.35W1-RJ)-
c~
Pu =n:l upward pre?ssura at the ultimate ""(?' limit state kndmg Moments
2
296
Reinforced concrete design
To achieve suitable sizes for the footings sc\craltrial designs may be necessary. With reference to figure 10.11 the principal steps in the de$ign are as follows.
1. Chsc a trial width D for the rectangular outer footing and a.-,sume weigh~ W1 and W~ for the footings and Ws for the <;trap beam. 2. Take moments about the centre hne of the inner column in order to determine the reaction R 1 under the outer footing. Tile loadings -.hould be thmc required for the ~crviceahilit) limit state. Thus
(RI -
W Jl(L--•-f- ~) -NIL- w, ~
0
(10.5}
and solve for R 1• The width 8 of the outer footing is then given by
8 =!!.:.._ pD
where Jl i~ the safe heari ng pressure. 3. Equate the vertical loads and reaction!. to determine; the re
+ W2 + W,)
0
(10.6)
und solve for R:_. The \ite S of lhe square inner fooung is then given by
s
f
?2
\ p
4. Check that the rcl'lultant of all the loads on the flx>tmg!-. pas-;es through the cemro1d of the area., of the two ba~es. If the rc!-.ultant i~ too far away from the centrotd then \LCps (I) to (4) mu<;t be repeated until there " adequate agreement. 5. Appl) the load111g as.,ocmtcd with the ultunatc limn \tate. Accordingly. reqse equation\ 10.5 and 10.6 to determine the new value., for R1 and R·. Hence calculate the beanng pre!)sure Pu for thts lun1t l!tate. It lllll) he a!-.sumed that the bearing pressures for thil> Cal>e are abo equal and uniform, prov1ded the ratios of dead load to 1mpo!led load are similar for both columns. 6. Design the inner footing ns a :-quare base with bending in both directions. 7. Design the outer footing ns a base with bending in one direction und .. upported b) the strap beam. 8. Design the strap beam. The maximum hcnding moment on the hcam occurs nt the point of t.ero shear as shown in figure 10. 11. The !o.hcar on the beam is virtuall) constnnt, the slight decrea!>e being cau!lcd by the beam's self-weight. The stirrup' should be placed at a con&tant ~pacing but they should extend into the footings over the supports so as to give a monolithic foundation. The main tension steel i., required at the top of the beam but reinforcement \hould al1.o be provided in the bottom of the beam so as to cater for any differential settlement or downward load-; on the beam.
10.4
Strip footings
Strip footings arc u~cd under \\all, or under a line of clo,cly spaced columns. l:.ven \\here it i' possible to have indi .. idual bases, it is often 'iimpler and more economic to excavate and construct the formwork for a continuou' ha\c.
Foundations and retaining walls
297
lap< 2h Rgure10.12 Stepped footing on a sloping site
rr fTt ;ttfftt Uniform pressure
ttlfJ_ ~_LIJ_ t Non-unllorm pressure
On a sloping site the foundations ~hould be c.;onstructed on n horitontal hearing and where neces~nry. At the ~teps the footings 'hould he lapped as shm\n 111 figure I 0 12. The footings arc analysed and destgned 3!. an im crted continuous beam '>uhjcctcd to the ground hcanng pre~sure~. With a thick rigid footing and 3 hrm -.oil. J linear dtstributmn of bearing pre~o.sure is considered. If the column' are equally '>paced and equally loaded the prcso;ure IS untforrn.ly distributed but if the loading t)> not ')mmctrical then the base i' !'.Uhjectcd to Wl eccentric loau and the bearing pres.,ure \'ancs a' shown 10 figure IO.ll The bearing pre"urcs will not be linear when the footing i' not very ngid and the soil " soft and compressible. In these cases the bending-moment diagram would be quite unlike that tor a continuous beam with lirmly held support' and the moment~ cuuld be qutte large. particularly if the loading is unl.ymmetrical. For a large foundation it may be necessary to have 11 more dctai lctl inveqigution of the soil pressures under the base in order to determine the bending moments and ~hearing forces. Reinforcement is required in the b01tom of the base to resi~t the transverse hencling moments in addition to the reinforcement required for the longitudinal bending. Footings which support heavily loaded columns often require !>lirrups and bent-up bars to resist the shearing forcet-. ~tcppcd
(EXAMPLE 10.3 Design of a strip footing
Design a 'tnp fooung to carry 400 mm square column-. equally ~raced at '3.5 m centre.... On each column the characteristic load~ are I000 k:-.1 permanent and J50 kN \Uriable. The safe beanng preso;ure is 200 k;\/m2 and the characteristic material \trength~ aref.L JON/mm~ and};,= 500 •tmm2. Bal>c the dcl-tgn on the Prt•lcriptil'l' Method.
Figure 10.13 linear pressure distribution under a rigid strip footing
298
Reinforced concrete design
1. Try a thickness of footing - 800 with d reinforcement. Net bearing pressure. p,.,1 = 200 - 2511
200
740mm for the longitudinal 25 x 0.8
1
- 180.0 k '/m . rcqui!C . d= \v1'dlh of ..tooung
Provide a
~trip
1000 -l 350 X 3.4
180.0 footing 2.2 m wide.
2.14m
At the ultimate limit state column load, Nb.J 1.35 x 1000 . 1875 bcanng pressure= . x . 22 35 = 244k.N/m 2
1.5 x 350
1875 kN
2. Punching .\'hear at the C()lumn face Maximum shear resistance.
VKd,m.l\
= o.swt [o.6 ( 1- ;;~>) J {::~ 0.5(4
X
400) "740
X
30 )1301 '\ ( 10 \ [o.6( I 250
6251 k:'-l 8} in-.pection, the normal 'hear on a ~ection at the column face will be ~ignificantly le~~ \C\'crc than thi' value.
3. umxiTtulmal r~infurceme/11 L -;ing the moment and shear coeflicients for an equal-1.pan continuous beam (figure 3.9). for an interior pan
moment at the column1-. M~o.c~
244 x 2.2 x 1.5] x 0. 10
= 605 k
Ill
therefore
A1
665
X
106
= 0.87 X 5()() X 0.95 X 740
2175mm 2
From table 6.8
O. ;~~d = 0.0015
A, m'"
- 2442mm
X
2200
X
740
1
Provide eight H20 bar~ at 300 mm centre~. area
2510 mm,. boltom steel.
In the <,pan MFd
244
X
2.2
X
3.52
X
0.07
=460kNm Therefore. ~ in the bottom face, provide eight H20 bars at 100 mm centre!.. area= 2510mm2 • top steel (figure 10.14).
Foundations and retaining walls
.
3.Sm centres
ill
·~
.:
B
8H20
,..
·~
0 0
..Loa
6
Figure 10.14 Strip fooling with bending reinforcement
H20 @ 2SOctrs 2.2m _
8H20
299
I
4. Tran\Tene reinforcemenl In the transvcr~c direction the max.imum moment can he calculmcd on the assumption that the 2.2 m wide rooting is acting a~ a I.! m long cantilever for the purposes or calcutaLing lhe design moment:
1.] 2
244 x T
Mt!d
= 148 kNm/m
148 X L06 2 A1 = 0.87 X 500 X 0.95 X 720 4l)7 mm j m . . 0. 15hlt/ 720 ' 0.15 x iOOO x 1080mm / m MtmmumA$= 1oo 100 Prov tde 1120 bar~ at 250mm centre~. area 1260mm!/m. bottom 'lecl.
5. Normal1·11ear will govern a~ the punching perimeter i' ouhttlc the looting. The crttical '>CCtton for ~hear tS taken I J)d from the column face. Therefore wrth d 740mm Oe~tgn ~hear
244 > 2.2(3.5 x 0.55
Vrtt
0.74 - 0.2}
= 529kN (The cnefftcient of 0.55 is from figure 3.9.) The shear resistance of the concrete without shear reinforcement can be obtained from tahle 8.2 where p 1 can he taken as the average ot the ~teet ratio., in hmh directions
~A, = 0'5 ~bel
s[
25 10 _ ~] _ O.. 2200 X 740 I 1000 X 720 - 00 ' 0165 (
() 165 ~ <... 2%)
=
hcnt:e from table 8.2 "Rd, c 0.36 N/mmz. Therefore the shear resistance of the concrete. VRd. c is gi ven hy: VKd .c
VRd
cbd
0.36
X
2200
X
740
X
10 ~
586 kN ( '>
VE.t
52() kN)
Therefore shear reinforcement is not requ ired. l_________________________________________ )
10.5
Raft foundations
A raft foundmion tran~mits the loads to the ground hy means of a reinforced concrete -.Jab that tSconttnuous O\'Cr the base of the stmcturc. The raft is able to span any area\ of weaker sotl and it spread~ the loado; O\'er a wide area. Heavily loaded structures are often pro\ idcd with one continuou~ base in preference to many closely-:.paccd. separate fomings. Also where settlement is a problem. bccau~c of mining sub\itlence. It is
300
Reinforced concrete design
Figure 10.15 Raft foundations
Il~
. l .....
Figure 10.16 Raft foundation subject to uplift
11]
1 ~
-;• t;r,,p'
(a)
Pedertal
~~ ~11 ·
flat slab
(b) Downstand beam
~I
t ! +++ t ! ++ + !
l
~ .,""~..::,~
~
(c) Upstand beam
llf: !
table
Upward pressure
common practice to use a raft foundation in conjunction with a more flexible Stlpcr:-.lructure. The simplest type of raft is a nat slah of uniform thickncs' supporting the colunm~. Where punching :-.hear~ arc large the column~ may be provided with a pedestal at the base ns shown in figure I0.15. The pedestal ),Crves a similar function to the drop panel in :1 flat slab Aoor. Other, more heavily loaded raft~ require the l'oundmion to be ~trengthcncd by hcrun:- to form a ribbed consu-uction. The beams may be dowostnnding. projecting hclow the slab or they may be up~tnnding n~ shown in figure 10.15 Down standing bcaml> have the dJ&advantage of tilsturbu1g the ground below the slab and the excavated trenches are often a musance dunng conc;tructlon. while upManding hcams interrupt the clear floor area ahove the ~lah. To overcome th1'\. a o;econd <;lah 1' l>omctimes cast on top of the beam~. so formmg a cellular raft. Rafts ha' ing a unifonn slab. and without strengthen mg. bemm. arc generally anal) sed and de~igned as an mverted flat slah floor l.uhjected to earth hearing prc:.sure\. With regular column spacmg and equal column loading, the coefficients tahulatcd in section 8.6 for ftat slab floors are used to calculate the hending moment~ in lhc raft. The '>lab muM be checked for punchmg shear around the column' and urountl pedestals. rf they are used. A raft with strengthening beams is dc~>igned a~ un inverted beam and slab lloor. The slab ~~designed w span in t\\O directions where there arc supporting beams on all four side!). The hcam~ arc often subjected to high shearing forces which need to be resisted hy u combination or stirrups and bent-up bars. Raft foundations which arc below the level of the water table. as in Rgure I0.16. shou ld he checked to ensure that they are able to res1st the uplift forces clue to the hydrostatic pressure. This mny be criticn l during con~truction before the weight of the superstructure is in place, and it may be necessary to provide exltlt weight to the raft and lower the water table by pumping. An ullernutivc method is to anchor the slab dov. n with ~hon tension piles.
10.6
Piled foundations
Pilcl> arc used where the soil conditions arc poor and it is uneconomical. or not posc;rble. to provide adequate spread foundation\. The piles mul.t extend do\\ n to firm sotl so lh.. the load is carried by either (I) end bearing, (2) friction. or (3) a combination of bott end bearing and friction. Concrete piles rna) be precast and driven rnto the ground. ur they may be the cast-in-situ type which are bored or excavated.
Foundations and retaining walls I~
"el'FTEI
Figure 10.17 Bulbs of pressure
I
I
I
Bulb of
\
I I
I""
I
I
pressure Gravel
, •
.....,
_,~-
_
T!!::;w.
Soft clay
' .... Pile group
Single ptle
A soils survey of a proposed site should be carried out to determine the depth to lim1 and the properties of the 1-.oil. This information will provide a guide to the lengths of pile required and the probable ~arc load capacity of the pile!.. On a large contract the ~afc loads are often determined from full-~cale load tests on typical piles or groups of piles. With driven piles the safe load cun he calculated l'rom equation~ which relate the resi~tancc of the pile to the measured :-et per biO\\ and the dri\ ing force. The loud-carr}tng capacity of a group of piles 1<. not nccessanl} a multiple of that for a single ptle it is often con~iderably le!-.s. For a I:Jrgt: group of clmely spaced fricttoJJ piles the rcc.luerion can be of the order of one-third. In contrast. the load capacity of a group of cnd hearing pilcs on a thicl- stratum of rod. or compact sand gravel is suh)>tantially the :-urn total of the resi~tant:c of each 1ndl\ idual ptle. Figure I 0. I 7 shows the bulbs or pressure under piles and IllUstrates why the settlement of a group ol piles is th.:pcndent on the soil propl:T'lie:- at a greater depth. The minimum :-pacing pilel>. centre to l:Cilli'C, ~hould not be les~ lhan (I) the pile perimeter for friction pile~. 01 (2) tw1ce the lca!>t \\idth of the pilc for end hearing pilel>. Bored piles 3rc ~mnetime~ enlarged at the btl\e ~o that they have a larger bearing area or u greater resistance to uplift. A pile is designed as a ~ho1t column unle% it i~ ~>lender and the :-urrounding :-oil is too weak to provide rcstra1nt. Prcca\t pile<, muM also he del>igncd to resi~t the bending moments cau<>ed hy lifting and ~tacking, nnd the hcud of the pile must be reinforced to withstand the impact of the driving hammer. Jt i~ very difficult, if not impOSSible. to determine the true distribution of lond of ll pile group. Therefore. in general. it is more reali'>IIC to u-.c method-. that are Mmplc but log1cal. A \erucal load on a group of vcnical p1b wnh an ax1' of '>ymmetry is considered to he distributed according to the following equation, which is similar in form to that for an eccentric load on a pad foundation: ~oil
or
Pn
= -N ± Ne" Yn ± Ne» 11
'"
\n
In
where
Pn is the axial load on an individual p1le N IS the vertical load on the pile group n i~> the number of piles e.. and e,y arc the eccentricities of the load N about the centroitlul axes XX and YY of the pile group fu and In arc the second moment\ of area of the pile group about axes XX anti YY r 11 and .l'n arc the distances of the inclividu:.~l pile [rom axes YY und XX, re~>pectively.
301
302
Reinforced concrete design
(
EXAMPLE 10.4
Loads in a pile group Determine the distribution between the individual pile~ of a 1000 kN vertical load acting at the position ~hown of the group of vertical piles shown in figure 10.18. To determine the centroid of the pile group take moments about line T- T.
- LY
' ' =f/- =
.
2.0 - 2.0 .L 3.0 + 3.0 l 67 = . m 6
where 11 IS the number of piles. Therefore the eccentricitic~ of the load about the XX and YY centro1dal axis are e._ - 2.0 - 1.67
= 0.33 m
nnd ~'Y>'
-
1,~
= LY~
0.2m
with respect to the centroida l axis XX
= 2 X 1.672 + 2 X 0.332 + 2 X 1.33 ~ 9.33 Similarly 1,\
= I>~ = 3 X
1.0! + 3
X
1.0'
6.0
therefore N -
Nl'n I Yn
II
"
1000
= -6- J:
=Ne,, I .
tn
t)
1000 X 0.33
JOQO
9.33 _In± 166.7 ± 35.4yn ± 33.3.\n
0.2
6.0
111
y
Figure 10.18 Pile loading example
... l .Om +
l.Om _
I T &
@1
2®
I I
E
T ~
.....
10 II
X- · -
:t
E 0 ...;
1000kN
1>-
-@3 \ -·-
-
---
4~ ,
E
..... ..... 0
..= II
E
e-s -
O.Sm
·- X
60 -e,., ...
y
0.2m
~
Foundations and retaining walls Therefore.
166.7 - 35.4
P1
P2
X
for Yn and
Xn
1.67 t- 33.3
X
= 166.7-35.4 X 1.67-33.3 X
1.0 = 140.9!w'l 1.0 = 74Jk'
X
1.0 = 211.7 k.'l
= 166.7-35.4 X 0.33-33.3 X
1.0 = 145.1 k:\l
166.7
P1
p~
~ub~tituting
35.4
P5 - 166.7- 35.4
Pt.
166.7
X
X
0.33 r33.3 1.33- 33.3
35.4 x 1.33
X
1.0 = 247.1 k.\f
33.3 x 1.0 = 180.5 kN Total = 999.6::::: IOOOkN
When a pile group is unsymmetrical about both co-ordinate axes it is nece&sary to consider the theory of bending about the principal axes which is dealt with in mo11t tcxtboo"-~ on strength of materials. Ln thjs case the formulae for the pile l oad~ are f>u
N II
.L 1\•\'n .L IJ ru
where
N (e\~ 2::.>~- <'))
L:>~ L.\ 1~
L.An.\'n)
(l: \'n,l'n) 2
and
8
kn L \'~ I'" L .\n."n) L I~ L \'~ (L 1n,ln}2
N
l\ote that t'n 1s the eccentricity about the XX a\IS, while ~'n is the eccentricity about the YY a\is. a\ 111 figure I 0.18. Piled foundnllons are somet1mes required to res1st homontal force!. 10 addition to the vertical loads. II the horizontal forces are small they can often be resi'ited by the pass1ve pressure of the soil against verucal piles. otherwise if the forces are not ~mall then ra"-ing piles mu:>l be provided as shown in figure IO.l9(a). To determine the load in each pi le either a \latic method or an cla~tic method i!, avai lable. The static method b ~imply a graphical analysis U\ing Bow·~ notation as . illu:-.trated in Hgure IO.l9(h). This method assumes that the piles arc pinned at their ends so that the induced load~ lire axial. The dustic method takes into account the displtu.:cments und rotations of the piles which may be considered pinned or fixed ut their end~. The pile foundation is analysed in a similar manner to a plane frame or ~pacc fntmc and available cumputcr p)'()grams arc commonly used. Figure 10.19
Forces in raking piles
b
(a)
(b)
303
304
Reinforced concrete design
10.7
Design of pile caps
The pile cap must he rigid and capable of transferring the column loads to the piles. It l>hould ha"c sufficient thicknes<; for anchorage of the column dowels and rhe pile reinforcement. and it must be checked for punching shear. diagonal shear. bending and bond. Pile!, are rarely positioned at the exact location~ shov.n on the drawings. therefore thi'i mu:-.t be allowed for when designtng and tlctailing the p1le cap. Two method.<. of design are common: des1gn using beam theory or design using J truss analog) approach. In the former case the pile cap i:-. treated a!> an invcned beam and is designed for the u~ual conditions of bending and shear. The tru:-.s analogy method is used w determine the reinforcement requirements where the span-to-tlepth ratio I ' less than 2 such that beam Lheory is nul approprimc.
10.7.1
The truss analogy method
In the truss analogy the force from the suproncu column is asllumed to be transmitted hy a triangular trus), action with concrete proviuing the compres~>ive members of the I russ
and steel reinforcement providing the tensile tie force as 'hown in the two-pile cap in figure I0.20(al. The upper node of the ln1ss is locatctl ut the centre of the loaded are.. and the lower nodes at the inter~ection of the ten:.ile reinforcement with the centreline' of the pile~. Where the piles arc .,paced at a distance greater than three times the p1le diameter only the reinforcement within u d~'>t
Truss model lor a two-p1le cap d
(a)
(b)
Prom the geometry of the force diagram in rigurc I0.20h: T
N/2 21
d
therefore I
Nl 2c/
Hence required area of reinforcement Figure 10.21
Four-pile cap
= 0 87T{,~
N '!.d
X
I
0.87/,~
( 1 0. ~
Where the pile cap is supported on a four-pile group. a.-. ~hmtrted equall> by pnrallel pairs of trus:.es. such ..,
Foundations and retaining walls Table 10.4 Number of piles
2
Group arrangement
•
A {.
Tensile farce
~8
T
21
3
I
A8
A
21
•
'c 0 ~
4
2/
....
r ;
TAB- Tsc- TAc;
2NI 9d
TAs - Tac - Teo
Nl TAo - -
8
--1
A
'
_ Nl 2d
B
0
®
2/
' D®
4d
• •.._:~ c
21
AB and C'D. and equation 10.7 can be modified to gi\c . d f . r . h requ1re area o reJn.orccment m eac truss
T/ 2 = O.SJ./)k
( 10.8 )
and thi., rcinforcemem should be provided in both directions 111 the bottom face of the pile-cap The tru'' theory may be extended to give the ten~ile force 111 pile caps \\ ith other configuration~ of pile group~. Table 10.4 gwe~ the force for ~ome common ca~e~ .
10.7.2
Design for shear
The !.hear capac.:ity of a pile cap should be checked m the crilicul ~-.cction tnken w be 20 per cent or the pile ditlmetcr inside the face of the pile, as ~hown in ligure 10.22. In determining the shear resi:-.luncc, shear enhnncemenl may he considered such that the shear force, VEJ, mny be ckcreascd hy a,.j2d where a, i~> the distance from the face of the column to the critical scc.:lion. Where the ~pacing of the pile' is bs than or equal to ~/5 .,.II_
Figure 10.22 Critical sections for shear checks
Punching ~hear perimeter
305
306
Reinforced concrete design three times the pile diameter, thi~ enhancement may be applied across the whole of the critical !.ection; otherwise it may only be applied to strips of width three time!. the pile diameter located central to each pi Ie.
10.7.3
Design for punching shear
Where the spacing of the pilel> cxcechown in figure I0.22. The shear force at the column face should be checked LO ensure that it is less than 0.51'l/cdlui = 0.5,· 1(Jc~/ l.5}ud where u is the perimeter of the column and the strength reduction factor. 1· 1 = 0.6( I - /c~/250).
10.7.4 Reinforcement detailing As for al l member~. normal dct:Jiling requirements must be checked. These include maximum and minimum steel areas. har spucings, cover to reinforcement and anchorage lengths of the tension steel. The main tension reinforcement should eontimre past each pi le and ~>hould be bent up vertically to provide u full anchorage length beyond the centreline of each pi le. In orthogonal directions in the top and bottom faces of the pile cap a minimum steel area of 0.26(f..11n/f)dhd ( > 0.0013/}(/) ~hould be provided. It I' normal to provide fully lapped horiL.ontal links of si1.c not le'>~ than 12 mm and a ~pacings of no greater than 250 mm. a~ sho" n in figure I0.2:\(b). The piles ),hould be cu otl ~o that they do not extend Ulto the pile cap beyond th~.: lower mat of reinforcing bar. otherwise the punching shear strength may be reduced.
10.7.5 Sizing of the pile cap In determrntng a suitable depth of pile cap table I0.5 may be used as a guide ''hen thert are up to six piles in the pile group. Table 10.5
Depth of pile cap
Pile size (mm) Cap depth (mm)
(
300 700
350 800
400 900
450 1000
500 1100
550 1200
600 1400
750 1800
EXAMPLE 10. 5
Design of a pile cap A group of four piles supports a 500 mm ' an ultimate axial load of 5000 kN. The pile~ are 450 mm diameter and arc 'paced at 1350 mm centre~ a~ ~ho" n. Design the pile cap for }~L = 30 l'\/mm 2 and/>~ 500 N/mm~. (a) Dimensions of pile cap
Try an O\erall depth of HXX> mm and an a\·erugc etfecu,·e depth of 875 mm. Allow t11e pile cap to c.xtend 375 mm either ~ide to give a 2100 mm .,quare cap.
Foundations and retaining walls 2100 Figure 10.23 Pile-cap design example
9H20e.w.
column starter bars /'
_soo-l ~
T"T r;;=;::::;:=lF;:~-;;:::~
To 0
./'
~I
g 0
,.,.,
01)
~~==t:=F~
\
\.. ) 15H20e.w. 290 (a) Plan
(b) Relnrorcement details
(b) Design of main tension reinforcement 1-rom equation 10.8, the required area of reinforcement in each truss i ~
T/2
NxI =-----
0.87/yk 4tf X 0.87.f;•k 5000 X I01 X ( 1350/2) 4 I( 875 X O.ll7 X 500 2216mm' The total area of reinforcement required in each direction 2 x A, 2 x 22 16 4432 mm1. As the pllec; are l>paced at three time~ the pile diameter thi' retnforcement rna) he dt<.tributed uniformly acro~s the secuon. Hence prm ide fifteen 1120 bars. area
4710mm'. at 140mm centre!> in bnth directions:
IOOA,
100 x4710
lui
2100xR75
( > 0.26J~Im J;~
0.26
=0. 15)
(c) Check for shear
5000/2
Shear force, Vt::LI· along critical section enhancement thi ~ may be reduced 10:
2500 VRd
L
llv
X
tf 2
0.121.( IOOf!fck)'/.1
290 2 875
X
X
l'tld
and 10 allow for shear
= 414kN
(~ 0.035kl.5./~(t 5 )
I -t jWojd = I -t / : = 1.49
where: k
(< 2} and fl = 0.0026
~ = 0.12k(100p.hd 1 ~
=0.12 and
2500
= 2500 ~N
I Rchfmin\
X
1.49
X (
100
X
= 0.0351. 1 %'~ = 0.035 X 5
= 0.35 N/mm1.49 15 X 30° s = 0.35 N/mm'
0.0026
X
30) I '1.
'
therefore the 'hear resistance of the concrete. VRd c il> given by: \ ' lld
c
l'Rcl
.hd
- 0.35 "2 100 x 875 X 10)- 643kN (> V~.cd
414k:-.i)
307
308
Reinforced concrete design (d) Check for punching shear A~ the pile spacing is at three times the pile diameter no punching shear check necessary. The shear at the column face should he checked:
Su~.:h wal l ~ ore usually required to re~ist u combination of earth and hydrostatic loadings. The fundamental requirement is that the wall is capable of holding the retained material in plnce without undue movement arising from deflection, overturning or ).liding.
10.8.1 Types of retaining wall Conc.;rctc retaining walb ma) be constdered in terms of three ba~tc categories: (I) gruvity. (2) counterfort, and (3) cantilever. Within these group~ many common variations extsl. for C\ample cantilever wall~o may have additional surporting tiel> into the rctnined material. The su·uctural action or each type 11-. fundamentally different, but the techniques used in analysis, design and detailing nrc those normally u~ed for concrew structures.
(i) Gravity walls The~;e
arc uwall) con,tructcd of nw. ~ concrete. with reinforcement mcluded tn the face~ to rc-.trict thermal and ..,hrink<~gc cracktng. A\ tllu•arated in figure 10.24. reliance b placed on 'cll-\\etght to :-.au'f) l>tUblhty requirements. both in respect of overturning and \lldmg. It 111 generally taken a~ ultnnt of the ~ell'-weight and overturning forces mu~t lie within the middle third at the interface of the busc and soil. This en,ures lhnt uplift is avoided Ht this inte1fac:c, us described in !>Cction I0.1. Friction effects which resi~t 'liding are thus maintained across the entire base.
Figure 10.24 Gravity wall
Ioree
Foundations and retaining walls
Bending, ~hear. and deflection~ of such walls are u~ually insignificant in view of the large effective depth of the section. Distribution steel to conrrol them1al cracking i:. necessary. however, and great care mullt be taken to reduce hydration temperatures by mix destgn. construction procedures and curing technique~. (ii) Counterlort walls
This type of cot1\lructinn will probably be u:.ed where the overall height of the wall is too large to be con~lructed economically either in mass concrete or as a cantilevct. The basis of design of counterfort walls is that the earth pref.~urcs act on a thin wall which :-.pan~> hmizontally between the massive counterforts (tlgure 10.25). These mu~>t be ~ u fficient l y large to provide the necessary permanent load for !>Lability requirements. pussibly with the aid of the weight of backfill on an enlarged hase. The countcrfort~> muM he designed with rcinforl'cment to act as cantilevers to resist the considerable bending moments tiH\1 arc concentrated at these points. Countcrfort Figure I 0.25 Countcrfort wall
JllURl l
Cro~~-sewon
Span Plan
The i>paclllg of countcrfort'i \\til be go\cmcd by the abO\c factor:-.. coupled with the need to rnatntatn a l>atisfactory span-depth ratio on the wull "lab. \\ hich mu'it he del>tgned for bending us a continuou~ slab. The athalllage ol Lhtll form of con~truction i'i thut the volume of concrete involved is con.,idcrahly reduced. thereby removmg many of the problems of large pours. and reducing the quuntJfles of cxcavatton. Bulanc.:d aguin'it thi.s mu)t be con11idcrcd the generally increased -;htHtering complication anti the probable need for increased reinforcement.
(iii) Cantilever walls These arc tlc.signetl as vertical cantilevers ~panning !'rom a large ngicl ha~o.c which often t'Ci les on the weight of backfill on the hato.c to provide ~labi l ity . Two fo ll11l> of thito. rlgure 10.26
Cant1lever walls
~ -- H,
'
-- Heel beam
c. (a)
(b)
309
31 0
Reinforced concrete design
con.,Lruction arc illustrated tn figure 10.26. In both ca.c,es. stability calculations foliO\\ similar procedures to those for gravity walls to ensure that the re1-oultant force lie within the middle third of the base and that overturning and sliding requirements are met.
10.8.2 Ana lysis and design The design of retaining walls may be split into three fundamental !>tttges: (I) Stability analysis- ult1mtlle limit \tate (EQU and GEO), (2) Bearing pressure analysis - ultimate lunll '>tate (GEO). and (3) Member design and detmling - ultimate limit state (STR) and c,erviceability limit states.
(i) Stability analysis l~ nder the action of the loads corresponding to the ultimate limit state (EQU). a retaining wr~ ll must be stable in term!-- of resistance to Ol'errurning. This is clemon~lrated hy the simple case of a gravity wall as shown in figure 10.27. TI1e critical condition' for overtummg are when a maximum horilOntal force t~ch with a minimum verucal load. To guard again\! fmlure by overturning, it ic, u~ual to upply conservative fr~ctorl-1 of safety to the forces and loads. Tahle IO.I(c) gives the factors that arc relevant to these calculation1.. A panjal factor of safety of ~·c ; = 0 9 i' applied to the permanent load Gl if 11<; effect " ·favourahle'. and the ·unfavourable' effect- of the permanent earth prel>Sure lmtding at the n:ar laL'c or the wall arc multiplied hy a partial !'actor or safety of 'Yt 1.1. The ' unfti\'Ottrahk' et'fects of the variable ~urchargc loading, if any, arc multiplied by a partial facto1 of ,afety of ~ 1 1.5. h1r resi.,tance to overturning. moments would normally he taken ahout the toe of the ha~e. point A on figure 10.27. Thus the requirement is that 0.9G~x ,;;> 1 1 //~v
Resistance to .1/irlinf: i' provided by friction between the unders1de of the ba-;c and the ground. ami thus is also related to total self-weight Gk. Resiswncc providi.!d by the passive earth pressure on the.: front face of lhc base may make some contrihution. but \lllCe thi-. material is often backfilled against the face, th1s re)>istanee cannot hi! guaranteed and i-. usually ignored. Failure by \licl ing is considered under the acuon of the load), corresponding to the ultimate limit swte of GEO. Tahlc lO.I gives the factors that are relevant to these calculation.-,. Figure 10.27 Forces and pressure~ on a
grav1ty wall
Fnction force ~G,
~urcharge pressure
Foundations and retaining walls
A partial factor of safety of ~1G = 1.0 is applied to the permanent load Gk 1f its effect is ·favourable' (i.e. contrihutes to the sliding resistance) and the 'unfavourable' effects of the permanent earth pressure loading at the rear face of the wall arc multiplied by a partial factor of safety of I t = 1.35. The ·unfavourable' effecu, of the variable surcharge loading are multiplied by a partial factor of safety of I t I .5. Thus, if the coefficient of friction between base and :-oil IS JL, the total friction force will be given by JLGk for the length of the wall of weight GL: and the requirement is that
l.o,,ck ?: 1r Hk
2
·ed
where H1. is the hori7onwl force on this length of wall. If this criterion i~ not met. a heel bc is laken into account in the stahility unaly~is. The force~ a(;ting in this (;a~e are shown in figure 10.28. In addition to Gk and Hk there is an additional vertical load Vk due to the material above the base actmg a distance q from the toe. The worst condition for stability will be when thi<> is at a minimum; therefore u partial loa<.l fnctor ")r 0.9 b used for consideration of O\Crturnmg and 1.0 for con~i<.lcration of ~liding. The '>t:thility requirement~ then become 0.9GL.\ t 0.9Vllf ~ ")rlh\'
11( t.OGL t 1.0\'L) ?: / r Hl
for overturning for ~liding
( 10.9) ( 10.10)
When a heel hcam is prov1<.lcd the additional pa,s1ve rcsisloil mechanics. or to usc increa.~ed factorl> of sufety. L~r.:.:m.
I I - ~
D/2
Beanng pressures
Re;ultant force Hk
D/ 2
Figure 10.28 Forces on a cantilever wall
311
312
Reinforced concrete design
(ii) Bearing pressure analysis The bearing pressures underneath ret:Hntng walls arc a:-.sessed on the basis of the ultimate limit state (GEO) when determining the size of base that is required. The anaJy,is will be 'imilar tc> thai dJscussed in ~ection 10.1 with the foundation being subject to the combined effects of an eccentric verucal load, coupled with an overturning moment. Considering a unit length of the cantilever wall (figure 10.28) the resultant moment about the centroidal axis of the base i'> ( 10.11 )
and the vertical load is ( 10.12)
where in lhi~ ca:-.c ol the STRand GEO ultimute limit state~ the part ial factors of safely are gi\'cn in Tahlc 10.1: l-or load comhinmion I· " 1
-
1.35 and 'irz
For load comhinm ion 2: 1r1
= l 'r:!
")'f'
'in
= 1.0
= l.O
as,uming that. for load comh1nat10n I. the effect of the moment due to the hori7ontal load on the maxtmum hcanng pressure ut the toe ot the wall at A 1s ·unfavourahlc' whi lst the moments of the ~te l f-wcight of the wall and the earth acting on the heel or the wall act in the opposite sense and are thus ·ravourahle'. This assumption may need chcck1ng in mdi\1dual ca~es and the appropriate partwl ft~ctor:-. applied depending on whether the cllccl of the load can be con:,1dcrcd to be f~1vourable or unfavourahlc. The distribution of bearing prcssure11 wi II be os ~hown in tigurc 10.28. provided the cl'fcctive eccentricity lies within the 'middle third' of 1hc base. that i~
M
D
N
6
The ma'\imum hcanng
prc~),ure
1s then £1\en by
N M D
(11
D+ I
"2
Therefore N
D-t
PI
6M 1)2
( 10.13)
and N
"~
=
o
6M D~
( 10.14)
Foundations and retaining walls
(iii) Member design and detailing A<> with foundations, the design of bending and ~hear remforccment " ba\ed on an analyst<; of the load~ for the ultimate limit ">Late (STR) . wtth the corrc\pondmg beanng pres~ures. Gra' ity ''all~ will o;eldom require bending or shear :.tccl, \\hile the ''alb m counterfort and cantile,er constmction "ill he designed all lllabs. 1l1e de\ign or counterfons "ill generally be similar to that of a cantiJe,er beam unles:. they arc mas<;tve. With a cantile\cr-typc retaining wall the stem ill designed to re1:1iM the moment caused by the force It H1• with lr value!> taken for load combination I tf this load combinatton is deemed to be critical. For preliminary si1ing, the thickne~~ of the wall may be taken as 80 mm per metre depth of bacl..lill. The thidnc~~-. of the base b u~ually of the same order a~ that of the ~tern. The heel and toe mullt be designed to resist the moments due to the upward earth hearing presMtre~ and the downwm·d weight or soil and base. The soil bearing prc~sures ure CHJcuJalctl from equation~ lO.lJ 10 l0.14, provided the resultant of the horitontaJ and vertical force~ lies within the 'middle third'. Should the resultant lie outside the 'midd le third', then the bearing pressures should be calculated using equation I 0.4. The partial factors or ~afcty "'til· it:! and In sbould be taken to provide a combtnation which give~> the critical dc~ign condition (the worst of combinations I and ::!). Reinforcement detailing mu~t follow the gL•nera l ntle' for ~lah' und beam~ a~ uppropriate. Particular care must be given to the <.letading of relllforcement to hmit ~hnnf..uge and thermal cracf..ing. Gravity wall<. are particular!) vulncr.thle bccau'c of the large concrete pour~ thOil: thu~ a llliding layer 1' not po"ihlc. Re111lorccment tn the ba'e~ mu't therefore be adequate to control the cracf..ing cau,ed hy a !ugh degree of re.,lraint. Long wall~ re.,traincd by rigid ha'e" are parucularly ~>uscepllblc to crading during thermul mo,ement due to lo.,ll of hydranon heat, und detailing mu\t attempt to <.lbtnbutc these crncf..s to en . . ure ueceptable wi<.lth~. Complete vct1tcal movement jointr-. must be provided. '11tcse jomts will often incorporate a ~hear f..ey to prevem differential movcmem of udjaccnt r-.ections of wall. and watcrhar~ and scaler' should he used. The bucf.. races of retaining walls will usually be ~ubject to hydrostutic forces l'mm groundwater. These may be reduced by the provision of n drninage path at the J'ncc of the wall. lt is usual practice to provide such a dntin hy a layer of rubble or porous blocks as shown ln ligurc l0.29. with pipes to remove the wuter. often through the fron t of the wall. In addition to n:ducing the hydrostatic pres~urc on the wnll, the likelihood of leakage through the wall is reduced, and warer i~ abo lc'~ likely tn reach and damage the ~oil henemh the foundation~ of the wall. Figure 10.29
P1p~ ca~t
Dra1nage layer
mto wall
Dramdge
Porous pipe la1d to fall
cha~nnel
mm
'
"
31
314
Reinforced concrete design
(
EXAMPLE 10.6 Design of a retaining wall The cantilever retaining wall shown in figure 10.30 supporL~ a granular material of ~aiUrated density L700kg/m2• It is required to:
2. determine the bcanng pressure:. at the ultimate limit
3.
= 500 kN/mm2
and
(1) Stability Horizontal force
=
It b a:>sumed that the coefficient of active pressure K4 0.33. \\hich i<; a typical value for a gr:Jnular material. So the earth pressure is gi vcn by
Pu
Kupgh
where p j, the density of the backfill and h i~ the depth con ... idered. Thu\. at the base 1
[1u = 0.33x 1700x 10
x9.8 1 x 4.9
= "7.0 kNim' Allowing for the minimum required surcharge of 10 ~/m 2 an additional hori7.
nf
= Aa
p
'.J ~ \J/m'
I0
liCll' uni lmmly over the wholl: depth
h.
300
Flgure 10.30
... I·
Retain1ng wdll design example
165.1 .. 22.0
~.... I
H10·200ew
I Ground level I I
-
16 2kN
active earth
I p. ~ 27.0kN/m' A
'
H12-200 t
39.4kN
_.,.1 IJ•
passive
34.0kN
~800...j.!OT~200
P•
U ~
_
beanng p1 pressures
l
3.4m
_
I 1
3.3kN/ml
Foundations and retaining walls Therefore the horit.ontal force on I m length of wall i~ g1ven by: Hk lcanh
= 0.5pah = 0.5 x 27.0 x 4.9 =
66.1 kN from the active earth
pre~)ourc
and //l 1,ur)
3.3 x 4. 9 = 16.2 k.:-1 from the surcharge
p)r
pres~urc
Vertical loads (a) permanent load:.
..~ {0.4 + 0.3) x 4.5 x 25
wall
= 39.4
0.4 x 3.4 x 25
base earth
= 2.2 X 4.5 X
1700
X
10
= 34.0
3 X
9.81
165.1
Total = 238.5 kN (h) vnriahle loads
Nurchargc
2.2 x 10
= 22.0J...N
The paninl foetor~ of safety a1> given in table 10.1 will be usctl. (i)
01•ertrtminR: taking moments about point A at the edge of the toe.:. at th1.: ultimate limit state (l!QU). l·or the overturning (unfavourable) moment a factor of 1.1 is applied to the earth pn.:\surc.: and a factor of 1.5 to the ~urcharge pressure overturning moment
A
11
Hk lc~· h/3
+
A
11 H~
,ur/r/2
={ l.l ><66.1 x 4.9t3)+{1.5x 16.2 x 4.9/2)
178 J...l\ m ror the rc\tratning (favourable) moment a factor of 0.9 is applied to the pcnnancnt loads and () to the vnriahle ~urcharge load rcsmun1ng moment - -; 1(.39.4 x 1.0 + 34.0 x 1.7 + 165.1 x 2.3)
= 0.9 X 476.9 429kN m Thus the criterion for overtuming is sntisfied. (ii) Slidi11~: from equat ion 10.10 it i ~ necessary that
I'( I .OG~ -t I.OVk) >'''II Hk
for no heel heam
For the sliding (unfavourable) effect a factor of 1.35 i1> upplicd Lo the cunh pressure and a factor of 1.5 to the ~urcharge pressure sliding force
1.35 x 66.1
+ 1.50 x 16.2
113.5 kN For the re.,traming (favourable) effect a factor of 1.0 i~ applied to the permanent loads and 0 to the variable surcharge load. As~umu1g a value of coefficient of friction Jt 0.45 frictional rcl>istmg force
0.45 x 1.0 x 238.5 = 107.3kN
315
316
Reinforced concrete design Since the sliding forte exceed~ the frictional force, resistam:c mu\t aho be provided by the passive earth pre~sure acting against the heel beam antltht!> Ioree given by lip -
/1 X
0.5Kppgt?
when! Kp 11- the coefficient of pa~sive pre~:.urc. a<;!'tumetlto be 1.5 for rhi~ granular material and a b the depth of the heel below the 0.5 m 'trench' allowance tn fmnt of lbe ba..,e. Therefore 1.0
/lp
X
0.5 >< 3 5
X
1700
X
1
10
X
9.~1
X
()j~
= 7.3 kN
Then.:fore total rcsiqiog forte i' 107.3.,. 7.3 - 114.6kN '~hith
marginally cxteeds the
~liding
force.
(2) Bearing pressures al ultimate limit state (STR & GEO) Consider load combination l a" the critical combination that will gtvc the max1murn bearing prc.:s,ure at thc.: toe of the wall bee whle 10.1 ), although in practice lo..u.i combtnallon 2 ma) haH~ to he chcdcd to determine if 11 gives a \\Orse effect. Note that thL ''c.:ight ot the earth and the 'un:hargt· loa<.ling exerts a moment about Lh~.: ba'~: ccntrd111e that" ill r<'diiC<' the maximum p1cs<.urc ut the toe of the wall. lienee the cfteL ot the weight ol the ca1th i~ talo..en us a fttt'Ottrnbh• effect (/1 = I) and the weight of the 'un:hargc loud ~~ al,o tab.en as a jal·ourablt• ellect ("11 0) within the calcu lation' hl'lm' The tmjm·ourablc effect!. of the luteral earth prc.:s~urc and the lateral ~urcharg.e prc.,~urc an: mullipl1cd hy factors of r 1 1.35 and 1 1 = 1.50. rcspccth el) f·rom equall(li1S 10. 13 and 10.14 the bearing. prc~sures are given by
N J 61\1 0 D' ''here H i~ the moment about the base centreline. Thc.:refore 1'
M
~I (66 1 X 4 9/3)
+ . . , ( 16.2 X 4.9/2)
+")I X
19.4(1 .7
1.0)
-')r X
1.35
107 9 I 1.5 x 39.7
X
= 145.7 -f
59.6 + 37.3 - '.19.1
Thcrc!nrc. hearing pres,ure ( 1JS
X
+ 1.35 X
tiL
1.0
--
(a\
165.1)
6x l43.5
--±--3.·-1'
'hO\\ n in figure I0.1 1)
__.11. ~
Figure 10.31
m? ~: : : : : :
''ffi""'""'" ... '"'
_...
Ul
_2200
bearing
--
p,
pressure~
6 P1 = 3.2 __.; 1
152.2
1
~
34m
99.1
tm: and heel of wall
77.7- 74.5
- I52 2, 3.2 kN/m'
X
143.5 b.N m
{39.4 -f 34.()) I 1.0
- - - - -3.4
27.6
- - - - - -"'1
165.1 X (2.3 - 1.7)
Foundations and retaining walls (3) Bending reinforcement (i)
Wall Hon zontal force
=110.5Kopgl,Z + -.,p ,h = 1.35 X 0.5 X 0.33 X 1700 X 75.2 I 22.3 con~idenng
10 -~
X
4.51
+ 1.50 X
the effective 1>pan. the maximum moment X 106
l !l2.5 33()2
Mtu bd~f~k
X
3.3
X
4.5
97.5 kN
75.2 X (0.2 + 4.5/ 3) I 22.3
Mb.l
9.81
JO()() X
X
_
30
X
(0.2
i~
4.5/ 2)
182.5 kN m
O.056
for which 1., - 0.95 (figure 4.5). Therefore
~~~
=
6
182.5 X 10 0.95 X 330 X 0.87
X
500
= 1338 mm 2/m Provide 1120 bars at 200 mm centres (A, - 1570 mm ' ). (ii)
Base The hearing prc~~urc~ at the ultimate limit \late arc obtai ned from part (2) of the ~e calcu lation\ . Using the figurclt from part ( 2): prc~c;urc Pt - 152.2 kN/m 2 fh
and
111
= 3.2 J..N/m1
figure 1Cl.31 :
3.2 - ( 152.2 - 3.2)2.2/ 3.4
p1
= 99.6 J..N/m'
/feel: tal-ing moments about the qcm centreline for the vertical hearing pres, ure~ l't X
34.0
)I
3.4 (T
1.0 ) + "Yt
-
X
165 .1 X IJ
(99.6 1.35
X
23.8 + 1.0
X
2 14.6
9.2
l oad~
J.2 X 2.2 X l.J
3.2)
X -~"' X ') ')
9
139 kN m therefore Mw
btf2j;k
6
=
139 x 10 1()()(.)
for which Ia
X
3302
X
30
= 0.043
0.95 (figure 4.5). Therefore
139 X IOh () 95 X 33() X 0.87 1019mm1/m
X
5()()
Pro\'ide H20 bars at 250mm centres (A,
and the
1260mm 2/m). top steel.
("' 0.2) ') ~~-
317
318
Reinforced concrete design Toe: taking moments about the stem centreline
MFA
~ ~~~ X
34.0
~
1.35
X
~
-67k. m
X
0.6
0.8 - 152.2 3.4
X -
X
0.8
X
0.6
4.8-73.1
therefore 67
MFd
X
106
bcf-J,~ - 1000 x 330~ for which Ia
.,
X
30 - O.(L I
= 0.95 (figure 4.5). Therefore: 67
As= 0.95 x
:no
106 , x 0.87 x 500 = 491 mm·fm
X
The minimum area for this, and for longitudinal dblribution steel which is also required in the wall and the base. is given from wblc 6.8: A,,min
=
0. 15/Jtd
loO =
0.0015 x 1000
X
330
495 mm
2
Thus. provide H 12 bars at 200 rnm centre~ (A~ )()6 mm 2/m) • bottom and dbtribution steel. Also ~teel should be provided in the compres,ion face of the wall in order to prevent cracking - ~ay. IUO bars at 200 mm centres each way. Bendtng reinforcement i& requtred 111 the heel heum to n:~bt the moment due to the pa~'i'c earth pressure. Tht<. reinforcement would probabl) be in the form of dosed link~.
The analysis and design of prestressed concrete is a specialised field wh1ch cannot possibly be covered comprehensively in one chapter This chapter concentrates therefore on the basic principles of prestressmg, and the analy~is and des1gn of statically determinate members in bending for the serviceability and ultimate limit states. A fundamental aim of prestressed concrete is to limit tensile stres,es, .:md hence flexural cracking, in the concrete under working conditions. Design is therefore based initially on the requirements of the serviceability limit state. Subsequently considered are ult1mate limit state criteria for bending and shear. In add1lion to the concrete stresses under working loads, deflections must be checked, and attention must also be paid to the construction stage when th e prestress force is first applied to th e immature concrete. This stage is known as the transfer condition. The stages fn the design of prestressed concrete may therefore be summarised as:
1. design for serviceability- cracking 2. check stresses at transfer 3. check deflections 4. check ultimate limit state - bending 5. des1gn shear reinforcement for ultimate l1 mit state. They are illustrated by the flow chart in figure 11.1.
319
320
Reinforced concrete design
___. When considering the basic design of a concrete section subject to prestress. the stress distribution due to lhe prestress must be combined with the stresses from the loading conditions to ensure lhat permissible stress limits are satisfted Many analytical approaches have been developed to deaf with this problem however, it is considered that the method presented offers many advantages of simplicity and ease of mampulat1on in design.
EC2 Section 2.3 1
5. 10.2
Cdlculate moment Vdltdtion (non-permilnenl actions & finishes) Mv
D
Structur~
{ ConcrelP u)age
Slr(.')S limits
cl~ss
D Min, section moduli D cover, ,.._ { Shap!'. Trial !<'Ction IO)) allowance D Self-weight t d~pth,
~tc
2.3 1
~rmanent
actton momPnt
Ql
~
D D Draw McJgnel dtagram lor cnucal section D Select prestress force Jnd ecccrllrtetly D
-'
~
Total momt>nt
~
:.0 II)
Ql
v
·~
~
DPtPrmlnl' tendon proftle
5. 10.4-5.1 0.9
D D Check nnal stresses and wesst>s Ccllculate lOlSe~
under quasl·permanent lo.td~
D 7.'1
8.10.3 6.1. 5.10.8
6.2
8.10.3 Figure 11 .1
Prestressed concrete destgn flow chart
Check delle( lion~
D D Ultimate moment of reslsta11ce D Untenstoned reinforcement Design end block
D Shear remforcPmPnl des.gn D Check end-block (unbonded) D FINISH
Pr~strtss
lystem
, . . _ Ultimate moment
~
~ -'
, . . _ Ultimate shear lore~
~
~
§"' 5
Prestressed concrete
Principles of prestressing
11.1
In the dei>ign of a retnforced concrete beam subjected to bending it i1- accepted that the concrete Ill the ten~ile .tone !!> cracked, and that all the ten!>ilc rc-.i'>tancc " prov1ded hy the reinforcement. The '>trel>i> that rna) be pcrmiued in the reinforcement i-. limited by the need to keep the cracls in the concrete to acceptable '' idth~ under working condillons. thus there is no advamage to be gained from the u~c of the very high Mrength steels wh1ch are available. The design is therefore uneconomic in two rc)!pccts: (I) dead \\eight includes 'uselesl> · concrete in the tensile 7one. and (2) economic uM: of Mccl resource~ is nm po~'ihle. ·Prestressing' mean:- the artificial creation of Mres.,es in a structure before loading, l>O that the stre~ses which then exist under load arc more favour which arc anchored against the concrete nnd, since the stress in thi<. Meet is not an imrortant factor in the behaviour of the beam but merely a means of apply1ng the appropriate fwce. full advantage may be taken of very high strength steel~. The way in wh1ch the c;tresc;es due 10 bending and an applied compre"IH' force may be comb1ned I\ demonstrated 111 figure 11 ..2 for the ca~e ot an a\Wll> applied force acting over the length of a beam. The stres!-. & .. tributiun at any ~;ecuon \\Ill equal the 'um of the compression and bcndmg ..,trcsse\ if it i., avmmcd that the concrete behave-. elastically. 'I hu., 11 is poss1hle to determine the applied force 'o that the combmed stresses arc alway!> comprel>SI\·e. By appl) ing the compre11-.ive force eccentrically on the concrete cross-section. a funhe1 Mrc"s distribution. due to the bending effects of the couple thu' created, 1s added lC) lho\e 'hown in figure 11.2. rhi'l effect i:-, illu<,trated in figure 11.1 and otters further udvuntages when aucmpting to produce W~c.., \\ ith1n required limm.
2-+ .~-~!.~~~>I·~·1
p
c
c
c
c
! 0![7 c
c
T Bending strain dlstnbution
Prestress
Seclion 8-B
Stress distribution - Section 8-8
T
Bending
Total
Figure 11.2 Elfecls of axial preslre~s
321
322
Reinforced concrete design
Figure 11.3 Effects of eccentric prestress
t t
+
t '
' e 1
t
•
l_s C
C
T
C
Of\· [7 C
Axial
C
T
Bending
Eccentricity
C
or
Total
prestress prestress Stress distribution- Section B B
Early
attempt~ to
achieve this effect were hampcn:d both by the limited steel availahk and by shrJllJ..agc and acep ol the concrete under sustained romprcs•.1on. cnuplco with rela\atmn of the '>tccl. This meant that the steel lo~t a large part nl Its innial prcten~ion ano as a n:~uh re~idual stre~~e~ were so small a~; to be t"cle". It is no\\ posSible. howc\er. to produce stronger conere::tes which have good cn:ep propcrttcs, and \er} htgh stn.:ngth '>teclc. whtch can he stressed up to a lugh pcn.:cntage of their 0.2 per cent prool -.tres'> arc al-;o avatlable. l·or example. hard-drawn ''ires m.l) c:arr) stre:-.seo.; up to ahout three tunes tho-.e fXl''-ihle in grade 500 remforcmg \tccl. This nut onl} result\ in saving-. Oll>tcel quantity. hut al'o the cfli!ct!) ot shnnkage and creep become relamcly ~mailer and rna) t}ptcally amount til the loss of only about 25 per cent of the tmtial applied force. Thu,, modern material-, mean that the pn:strcssing of concrete i' a pracllcal propo~Jtton. with the force' being provided b} 'teel p
11.2
Methods of prestressing
l'wo hn,ic techniques arc common ly employed in the con~>trtl Ction of prcstrusscd concn:te. their chief difference being whether the !-tleel tCIISlOtllng proces~ is performed before or niter the hardening of the concrete. The choice of method will be governed largely by the type ano size of member coupled w1th the need for prcc:ust or in situ con~tructJon.
11 .2.1
Pretensioning
In tlw• method the steel wire~ or ~trand~ are '>!retched tn the required ten'>ion and anchored 10 the ends of the moulds for the concrete. l'hc concrete is cast around the Lc!n,toned \lee!. and \\hen it ha' reached o;;uffictent 'trength. the anchors are released and the force in the c;teel ts tmn,ferred 10 the concrete hy hond. In addition to long-term lo,ses due to creep. -,hrinl-uge and relaxation. an 1mmediate drop 111 pre,tress force occurs due to elastic shortening of the concrete. TheM! features are illustrated in figure II A.
Prestressed concrete Beam with pretensioned tendons
Figure 11.4 Tendon stres~es pretensionmg
, Before transfer After transfer and losses
-I
bond length
length
Becau))e or the dependence on bond. Lhe tendon~ for this form of c:on.~truction generally conflisl of small diameter wires or small strands which have good bond characteristics. Anchorage near the ends of these wires is often enhanced by the provision of small indentations in the surface of the wire. The melhml is ideally suited for factory production where large number\ or identical unit~ can he economkal ly made under comrolled conditJon~. a development of thi'> being the 'long line' :-y~tem where several unit\ can be cast at once end to end - and the tendon~ merely cut between each unit after rclea\e of the anchorage~. An adv:mtage of factory production of prestrc~scd units i~ that ~pcciali)..ed curing technique~ ~uch a~ stenm curing cnn he employed to increase the rate of hardening of the concrete nnd to enable earlier 'transfer' or the stress to the concrete. Thi~ is particularly important where re-use of mould!. t\ reqtured. but 11 t~ essential that under no circum:,tancc., rnu.,l cnlcium chloride be U'>ed ac; an ncceleraLOr becnuse of its severe corrosi' c acuon on small diameter ~teel wtres. One maJor hmlfallon of this approach is that tendons mu~t be straight. whtch may cau~e difficulue~ when attempting to produce acceptable final !.lre:,:, level\ throughout the length of a member. It may therefore be necessary 10 reduce either the pre~tress or eccentnctty of force nenr the ends of a member. in which cu~c tcntlon-; muM either be 'dcbondcd' or 'denccted'. 1. Deboncling consist~> of applying a wrapping or coating to the Mcel to prevent bond developing with the surrounding concrete. Treating some of the wires in thi~> way over pun of their length allows the magnitude of effective prc:-tn:s~ force to be varied along the length of a member. 2. Dellecting tendons is a more complex operation and i~ U)..Ulllly restricted to lurge member!'., such as bridge benms. where the individual member:- may he required to form purl of a continuous structure in conjunction with in .1it11 concrete ~labs and ~ill beam!.. A typicnl arrangement for det1ecting tendons is shown in figure I 1.5, but it must be appreciated that substantial ancillary equipment is rc4uircd w provide the necessary reactton~. Deflection supports (cut off after tr.1nsfer)
Concrete
l
~
Tojacks~ ~
l
Prestressed tendons
~
- ~,
Y"" 7 - \;..,
I
.
22 , 7 2 2 ,~Tofacks
Figure 11.5 Tendon deflection
323
324
Reinforced concrete design
11.2.2
Post-tensioning
This method. wtuch is the most suitable for in .1i111 con~truction. involves the stressing again~! the hardened concrete of tendon~ or steel bar\ which are not bonded 10 the concrete. The tendons are passed through a flexible ~heathing. wh1ch is cast into the concrete in the correct position. They arc tcm.ioncd by Jacking against the concrete. and anchored mecharucally by mean\ of \tccl thrust plate:. or anchorage blocks at each end of the member. Alternatively. steel bar!> threaded at their ends may be tensioned against hearing plates by means of tightening nms. It i~ of cour e usually necessary to wait a con~iderable time between casting and stre~sing to permit the concrete to gain sufficient strength under in siw conditions. The use of tendons cons1sting of u number of ~trunds pas:-.ing through tlexible sheathing offer~ considerable advantage~ in that curvet.! tcnt.lon protiles may be obtarned. A post-tensioned strucruml member may be constructed from an n~semb l y of sep:~rme pre-cast units which [Ire constrained to act LOgcther by means of tensioned cnb l e~ which ~u·e often curved as illu~tntted in figure 11 .6. Alternntively, the member may he cast a:-. one uni1 in rJ1e normal way but a lighl cage of untensioned reinforcing steel i~ nl:ccssary to hold the ducts in their correcl position during concreting. Af1er strcs~ing. the remaining ~puce in the ducts may be left empty ( 'unbondcd· con~>truction). or more usually will be filled with grout under high prcs~urc ("bonded" con~truction). Although this grout a~sistc; 111 tmnsm1l1111g forces between the steel and eom:rcte under live loads, and improves the ultrmate ~trength of the member. the principal use rs to protect the highly stres-;ed ~trands from corro~ton. The quality of ~orl..manship of grouting i!> thu!:. criucal to avoid air pocl..cts which may permit corro~ion. The honding of the highly stressed o;tecl \\llh the 'urrount.ling concrete beam al'o greatly a''ist-. demolition. since the hcam may th~.:n \afcly be 'chopped-up· into 'mall length' wrthout rclea!)rng the energy ~torcd 111 th~.: -.tccl. Par;~bohc
Figure 11.6 Po~t·tensloned
tendons
s!'gmental
construction
Precast segments
11.3 Sine~:
Analysis of concrete section under working loads
the object of prestressing is to marntain favourable ~tress conditions 111 a concrete member under load. the ·worJ,.ing load' for the mcmhl:r must be considered 111 terms of hmh maximum and minimum values. rhu' at any . .ection. the stre!>se~ produced by the prcl>tress force mu:-.t he considered in conjunction with the \tres~es cau~ed by maximum and rmnimum values of applied moment. Unlike rernforccd concrete, the primary analysi\ of pre'>tressed concrete is b:t\ed on serv1ce condition'>. and on the U\)..Umption thai stres:-.es in the concrete are limited to values which will correspond to elu~tic behaviour. ln this l>Cction. the following assumption-. arc made in analy~'"·
Prestressed concrete lo= -
I
rr___
y,
b---, -
e-ve
h
- - - -T - f-
t a
f'!C:d tiber
JOg
1. Plane
and
the ) of
mm
section~ remain plane.
2. Stress strain rclation1>hip1. arc linear.
3. Bending occurs ahout a principal axb. 4. The prc1-trcssing force is the value remaining al'tcr all
5. Changes in tendon stress due to applied
lo~l.es have occurred .
load~ on the member have negligible effect
on the behaviour of the member.
l~tf
ded'
t+ve Bottom fibre Prestressing tendon
_,__ L - -- - --' _ _
Compressive stresses +Ve Tensile stresses - ve
nent t1ble be of
_.___
- - axis
Q
I =-;;;: ,.
' Centroidal --
e
)'b
lb
Figure 11.7 Sign convention and notation
Top fibre
y,
6. Section properlies are generally based on the
gro~s wncrete eros~ ~ection.
The ~tre~~ in the ~tecl is unimportant in the analy'>i'> of the concrete ~ection under wor~ing condition~. it being the force provided by the 'tccl that i., con.,tdered 111 the analy'>i~.
The sign convention!> and nmalion~ used for the analy'•' are indicated in ligurc 11.7.
mto
11 .3.1
Member subjected to axial prestress force
If secuon BB ol the member shown in figure II.X i\ subjected to moment' ranging between Mm 3 , and M111111 • the net \lrc~sc:. at the outer fibres of the beam urc given hy
{!
, ---+--P M~, A ;:,
fo
p
Mrnu'
A
~b
p
under M,,;n
{fi ./i.
:Is of
-- I A
Mmiu
at the top
(II. I)
ut the bottom
( 11.2)
at the top
(11.3)
at the bottom
( 11 .4)
:::c
p
Mmln
;\
Zh
where Zh nnd ::1 are the ela&tic section moduli and P b the final pre~tre!>s force. The critical condition for tension in the beam is g1ven by equation J1.2 which for no tens ton, that is fh = 0, becomes
or
p
. . r . d -Mm;,,i\ -= mmtmum prestress •orce requtre <:t>
325
326
Reinforced concrete design
Figure 11.8 Stresses in member with axial prestress force
PIA
Mlz,
f,
0![7
PIA Prestress
Mlzb
f.,
Bending
Total
Stress distribution- Section B-B
For thi~ value of pre!>trcs!> force. subl>titution in the other equation~ wi ll yield the ~trcs~e!> in the beam undt:r maximum load and also under minimum load. Similarly the stressc~ immediately after prestressing, before losses have occurred, may be calculated if the value or los~es i!. known. For example. the maximum stress in the top of the member is given by equation I l.J
\\here Mma\ A P--~~
therefore
,t;
p
p ~h
A
;\ <:t
p
;\
cb ~z') q
lt cun be seen from the stress distribution~> in figure 11 .8 that the top tlbre is generalI) n
consiucrnble compression. while the bottom f1hrc is generally at lower strcssc~. Muc better u~e of the concrete could be made if the ~tresses at both top and bottom can ~ caused to vary over the ful l range of permissible ~tresses for the two extreme loadm; conditions. This may be achieved by providing the force at an eccentricity e from 1 te centroid.
11.3.2
Member subjecte~ to eccentric prestress force
The ~tres!> distributions will be similar to those in section 1I .3.1 but with the add1tion the term ±Pe/<. due to the eccentricity e of the prestressmg force. For the (Xl'-lt. , shO\\ n in figure 11.9. e will have a positive value. So that
Prestressed concrete at the top
( 11 .5)
under Mm..., a1
the bottom
( 11.6)
at the lop
( 11.7)
at the bottom
( 11.8)
under Mm10
'ote that. as the prestressing force lies bel(lW the neutral axil-., it hu~ the effect of caul>ing hogging moments in the section. The critical condition for no tension in the hottom of the heam il. again given by equation 11 .6, which becomes
P A
Mm.,,
Pe
::b
Zb
0
or p
minimum prestrc~o.s force required for no tension in bottom ntm:
'-b +<' ) (1\
Thus for a given value of prestrc:-.~ force P. the beam may can·y a maximum moment of Mll\ii\ -
::h + l' ) p (A
=
When compared with Mm.u P::b/A for an axwl prc~tre~" force it md1catel> an increase in moment carrymg capacity of Pe. The maxunum stress 1n the Lop of the beam b gi\'en hy equat1on 11.5 a\
!.
J> t A
MIIIJ\
::,
Pt• :;,
where
n D! \ .[} rl~:-
11
l .... B
PIA
PIA
Mlz1
Ml zb
Pelz,
Ptl lo
Axial Bending Eccentricity of prestress prestress Stress diStribution - Section 8-B
11
'• Total
Figure 11.9 Stresses in member with eccentric prestress force
327
328
Reinforced concrete design
which is the same a~ lhat obtained in l.ection l I .3. I for an axially prestressed member. Thus lhe advantages of an eccentric prestre s force with rcl-pect to the maximum moment.carrying capacity of a beam are apparent. If the stress distributions of figure 11.9 are further examined. it can be ~ecn that lhe differences in the net stress diagrams for the extreme loading cases are solely due to the differences between lhe applied moment terms Mm1u and Mm.n· It follow!; that by increasing the range of the strc~se by the usc of an eccentric prestres' force the range of applied moments that the hcam can carry is also increased. The mtnimum moment Mmm that can be resisted is generally governed by the need to avoid tension in the top ot the beam, as indicated in equation 11.7. In lhc design of prestressed beams it i' important that the minimum moment condition •~ not merlooked. c'~c1ally \\hen \tr:ught tendon' arc employed. a' streJ..!>C!> ncar the ends of beams \\here momenb are ~mall may often exceed lho'>e at section~ nearer mid-span. Thi~ fcmure i11 illustrated 11y the result~> obtained in example II. I.
(
EXAMPLE 11 . 1
Calcu lation of prestress force and stresses A rectangu lar heam 300 x 150 mm is simply ~ upporled ovcr :1 4 rn ~pan. und supports a live load of 10 t..N/m. I f a straight tendon is provided at un eccentricity of 65 mrn below the centroid uf the section. find the minimum prc,tre...:. force nece~sary for no tension under li\'e load at mtd-~pan. Calculate the corrc~ ponding ~trcs\e~ under self-weight only at mid-span and at the endl> of the member. (a) Beam properties
Self-weight = 0.15 > 0.3 < 25 = 1.12t..N/m Area = 45 x 10' mm 2 Section moduli Z1 =Zh = :. (b) Loadings (mid-span)
1.12) X 4 2 22.2L m 8 1.12 X 42 k.N
( 1()
Mm.., = -
Mmm
8
= 22. ·
m
(c) Calculate minimum prestress force
For no tcnston
ut
~ _ Mmox -1 Pt! A .: =: where
e=65mm
the bonom under Mnn,
=O
Prestressed concrete hence Mm3 , (5;_
p_
\A
_
+e) -
22.2 X 106 X 10 3 :1.25 106 45 X 101 + 65
x
= 1931<1\ (d) Calculate stresses at mid-span under Mlliill Stresc; at top};
= -P + -Mmon _- - Pe ~ A
,
'·
where p
-
193
x 101
4.3 N/mm2 45 X 10 1 2.2 x 1011 Mmlu 1.0 N/mm2 2.25 X 101> '· Jle l 93 X I ()~ X 65 _ N/ = '>-·25 X 1Q6 5.6 mm2 (. A
~
lienee Stres1. nt top};
= 4.3 + 1.0 -
5.6 = - 0.3 l\lmm 2 (tension)
and
stri!Sl> at hottom ft> .
M nun = -AP - -+ -Pe :: ::
::: 4 10 - I 0 .._ 5.6 = - 8.9 N/mmz The calcula11on ~howo, that '"ith minimum load it is pos~ihle for the benm to hog with ten-.1le 1>tn!~ses Ill the top libre~. TI1is is paniculatl) likely at the iniunl transfer of the prestress force to the unloaded beam.
(e) Calculate stresses at ends In this situation M 0. Hence 4.3
A
5.6 - - 1.3 N/mm 2
and I' 1'1' jj, - A+ 7
'·
11.4
= 4.3 + 5.6 = 9.9N/mm2
Design for the serviceability limit state
The destgn of a pre:.tressed concrete member is based on maintaining the concrete '> tresses within specified limits at all stages of the life of the member. Hence the primary design ,., bac;ed on the serviceabilicy limit state. with the concrete ~tress limits based on the acceptnble degree of flexural cracking. the necessity to pre\ ent e\cc~\ive creep and the need to eno,ure that excessive compressiOn does not resuh in longitudinal and micro cracking.
329
330
Reinforced concrete design
Guidance regarding rhe allowable concrete compressive stress in bending is given in EC2 as limited to: (i) 0.6/c~ under the action of characteristic loads
and (ii) 0.45/ck under the action of the
qua.~i-pen11anem load~ .
The qzwJi-permwzemloads are the permanent and preme~~ing load. Gl + Pm.t· plus a proportion of the characteristic variable impo ed load. This proportion is taken as 0.3 for dwellings. offices and stores. 0.6 for parking areas and 0.0 for snow and wind loading. Jf the tensile stress in the concrete is limited to the values offc1m given in table 6.11 then all stresses can be calculated on the a~sumption that the ~ection is uncracked nnd the gross concrete section is re~isting bending. If th11> is not the case then ca l cu l ation~ may have to be based on a cracked section. Limited cracking is permissible depending on whether the beam is pre- or pol>t· tensioned and the appmpriate exposure class. Generally for prestressed members with bonded tendons crack witlths shoultl be limited to 0.2 mm under the action of the frequent loading combination taken as the permanent characteristic and prestre~sing load. Gk Pm I· plus a rroportion of the characteristic variable imrosed load as given by equation 2.3 and table 2.4. ln some, more aggre!.sive exposure contlitions, the pos~>ibility of decompre~sion under the quasi pemument load conditions may need to be consitlered. At initial tran1.fer of prestrest. to the concrete. the pre,tress force will be considembly higher than the 'long-term· value a' a result of sub~equent losses which are due to a number of causes including elastic shortening, creep and shrinkage of the concrete member. Esumation of losses is dc).cribed in section II .4. 7. Since these lossc' commence unmedtately. the condition at tran.,fcr represent\ a trunsitory stage in the life of a member and further con<;ideration 'hould be given to limiting ooth compre,sive and tensile stresses at this stage. In adtlition. the concrete, at thll- \tage. i' usually rclativcl) immature and not at fulll.trength and hence tran1>fer i). u critical stage which 'hould be considered carefully. The comprcs\ive stress at transfer \hould be limited to 0.6/,~ wherefc• is based on the Mrcngth on the concrete at transfer. The tcn1.ile stress should be limited to I N/mm 2 for sections designed not to be in tension in service. Where limited Aexural 'tress under service loads h. permillcd, some limited tensile stress is permitted at transfer. The choice of whether to permit cracking 1.0 take place or not wil l depend on a number of factors which include condllions of exposure nnd the nature of loading. Ira member consists of precast segments with monnr joint~. or if it is essential that cracking ~>hould not occur. then it will be designed to he in comprcs~ion under ulllond condition~. However a more efficient use of materials can be made if the tensile strength of the concrete. f~tm· given in table 6.11 is utilised. Provided thet--c ~trcs~c~ arc not exceeded then the section can be designed, ba!-.cd on the gm1>s uncrad.ctl section. Unless the section is designed to be fully in comprcs~ion untlcr the characteristic loads. a minimum amount of bontlcd reinforcement ~hould be provided to conrrol cracking. Thio; j., calculated in an identical manner to the minimum reqULrement for reinforced concrete (sec ~ection 6.1.5) with the allowance that a percentage of the prestressing tendons can be counted tO\\ ards this m111imum area. The de\ign of prestressing requirements ts based on the manipulation of the four basic exprc!.1>ions given in section 11.3.2 descnbing the stress distribution acros~ the
Prestressed concrete m
3 31
concrete section. These are used in conjunction with permissible ~tresses appropriate to the type of memher and covering the following conditions:
1. Initial tran~fer of prestress force with the associated loading (often JUSt the beam's sel f-weight); 2. At ~ervicc. after prestress losses. with minimum and maximum characteri~tic loading; 3. At !>ervice with the quasi-permanent loading.
The loadings mu~t encompass the full range that the member will encounter dunng its life, and the minimum values will thus be governed by the construction techniques u~ed. The partial factors of safety applied to these loadl> will he those for serviceability limit ~o.tat e, that is 1.0 for both permanent and variable load!-.. The qunsi-permancnt loading situation is considerc<.l with only a proportion of the characteristic variable load acting. For a beam with a cantilever span or a continuous beam it is necessnry to con~ider the loading patterns of the li ve loads at service in order to determine the minimum and maximum moments. For a singlc-:-.pan, simply ~upported beam it is u~ua ll y the minimum moment at tran~fer and the maximum moment at f>Cf\ 1cc that wil l govern. as shown in figure I 1.1 0. From tigure 11.1 0 the govcmi ng equauon~ for a ~i nglc-span beam arc:
At
tran~fer
Po
Pue
A
:.
Po
Po<'
M mon
A
;:h
;:h
At
= f'1 > J.'mm
M mon
.......
( 11.9)•
.f~ ~f/na~
( II 10)
~ervice
KPn
KPoe
A
~(
KPo
KPoe
A t
Zh
Mmnx -;·
onux
( I 1.1 I)*
M,n~• - jih >J,. - mon
(11 .12)*
-r - - ~l
I
Zh
where }~ 11 ~, ;;~in· j;11 u~ nnd };nln arc the appropriate permissible ~ trcssc~ ut tram, fer anti conditions. Po is the prc:-.tressing force at transfer and K is a loss factor that account~ for the prestress losses - for example, K = 0.8 for 20 per cent loss. ~erviccnbi lity
w ,.,·ln
f f f +
'
+ +
Transfer w"'...
f f f f f f
Figure 11 .10
r•, > r;...,.
I'• -~ I'm., f,c;f.,.,
[7 Service
Prestressed beam al lransfer and servoce
332
Reinforced concrete design
11.4.1
Determination of minimum section properties
The two pairs of expres~ions can be combined as follows: 11.9 and 11. I I ( 11.13)
II.!Oand 11.12 (Mmu~ - KMnun ) :::; (Kfm:tx -
Hence. tf (Mmal
.fmm):Oh
(11.1 4)
KMmm ) is written as M,. the moment variation
M,
•,>---.. - ifrru, KJ~,n)
( 11.1 5)
and
M,
( ll. J6)
Zb> - - - . - ( KJ~., - };mn)
In equations 11.15 and 11.16, for~~ and ::11 1t can be assumed with ~uftic.:ient accuracy, for preliminary siting that M10. , will depenu on both the imposed and dead (self-weight) load ami Mnun will depend on the dead (self-weight) load only. ~o that in effect the c.:n lculations for Mv bec.:ome independent of the self-weight of the beam. These minimum \ulues of section moduli muM be ~atistied by the cho,cn section in order that a prestress force and eccentricity exist which will permit the stress limits to be met; but to ensure that practical con~iderntion~ arc met the chosen section must have a margin above the minimum values calculated above. The equations for minimum moduli depend on the difference between maximum and minimum values of moment. The ma\imum moment on the section has not directly been included in these figure ... thu~> it i!. possible that the rc:-.ulttng prestresl> force may not he economic or practicable. However. it is found in the majority of cases t·hat if a section is cho,en which satislles thel>c minimum requirements. coupled wtth any other speci fied requirement!. regarding rhc shape or the section, then a snw;fuctory dc~>ign is usually possible. The ratio of acceptable '>pan to depth for a prcstres~cd beam cannot he categorised on the haiotis of dcnccttons as ea~tly as for reinforced concrete. In the ab~ence of any other criteria, the following formulae may be used as a guide und will generally produce reusonably consen ative de,ignc; for post-tensioned memhcr~. span< 16m
I1=
span > 36m
II
spnn
.,-+ _s 0.lrn
span m 20
:::.--
In the case of :-.hort-span members it may be possible to use very much greater spandepth ratios quite satisfactorily. although the resulting prestress forces may become ver) high. OU1er factors which must he consic.lcred at this stage include the slencleme~>s ratio or beams. where the same criteria apply as for reinforced concrete. and the possibility of web and flange splitting 111 tlanged members.
Prestressed concrete
(EXAMPLE 11 . 2
Selection of cross-section Select a rectangular <,ection for a posHensioncd beam to carry. in addition to 1ts 0\\11 a uniformly di:.lributed load of 3 k:-.f/m O\'er a simply l.upponed span of I0 m. The memher is to be designed with a concrete strength cla~s C40/50 and i:. restrained aga111~t torsion at the ends and at mid-span. Asc;ume 20 per cent lo~~ of
:n.5 x 106 ,, , :l ~ (J, = ("> 4 _ 0 S{-l}) =- 1.50 > 10 mm· ma~ mtn · M, 37.5 X 10" ., ft \ _ O) =- .93 X 10 111111' ;:~ > (O (KJ;:,.,-.f.n.8 16 0. 111 ) M, _ Kj'' )
21X) mm. I Jcnce
Take b ~
2()()11~ / 6
2 93 )' 10(>
Therefore h ' /(2.93 x 106 x 6/200) - 297 mm
The minimum depth ol' beam b therefore 2'->7 mm and 10 allow a mnrgw 111 subsequent det<\i lcd design u depth of 350 mm would be appropriate 11s ll first attempt. To pn:vl.!lll lateml buckling 8C2 ~pccili cs a maximum 'pan/hrcadth rat111 rcquin.:mcnt: lm /)
50 (11/1>) 1/3
where /.,1 Actual
the diswncc between torsional
I
~
of
5000 -200
!,., b
rna\imum 0
with h/ b ~ 2.5
1,~
b
re~traint'
= 5.0
Ill
in this
cx~1mplc.
= '?5 -
50 - - =41.5 (350/ 200) I J
--
hence the cho\cn dimension), arc satisfactory a~ an initial c~timatc of the required beam SI7C
3:
334
Reinforced concrete design
11.4.2
Design of prestress force
The inequalities of equation\ 11.9 to 11.12 may be rearranged to give expression)> for the minimum required prestrc~s force for a given eccentricity: p
0
p
0
p
< (-.Jma' - .Mill;J, ) -
( 11.17)
e)
K(<.t/ A
> {~J~lin- Mmao) -
( 11.18)
(~/A - e)
> {;:.hfrniu + Mrnax) o_ K(<.t./ A + e)
P0
<
( ;:tf~.,
-1
Mmin )
-- -
(~h/A
( 11.1 9)
+ e)
( 11.20)
-
:'\ote that in cquatton~ 11.17 and 11.18 it is pos),ible that the denominator term, (-. 1/A e). might be negative if e > ::1/A. Tn this Cideratton. Such limitil ~ill include consideration of lhc required minimum cover to the prcsLressing tendons which will depend on the e\po~urc and ~tructural cia~' assumed for the design. The ciTect of this limitation will he moM severe when considcnng the maximum moment' acti11g on the :-.cction. thnt is, the inequalities of equations II . II and 11.1 2. If the !uniting value for maximum eccentricity e""''' depend~ on cover requirements. equation I 1.11 becomes
Mma' $ };nax:.l
KPoC · ema~)
( 11.21 )
nnd equation 11.12 becomes
Mm." $ KPo e~ t
<'m•') fmm'-h
( 11.22)
These rept·esent linear relutionshipl- between MrTWJo. ond P0 . Por the case of n beam suhjcct to sagging moments enux will generally be positive in value, thu\ equation 11 .22 is of positive slope and represents a lower limit tO P0 . It can al..,o be shown that for moM practical cases !(;:1/ A) - ern•~ I < 0, thus equation 11 .21 is similarly a lower limit of positive. though 'imaller c,lopc. Figure 11.11 represents the general form of these cllpresston:-., and it can be seen clearly that providing a pre~·.tress force in excess of Y' produces only ~mu ll benelits of additional moment capacity. The value of )' ' is given by the intersection of the'>C two expressions, when
KPo (~ + ema,)
fmm:.t>
=fma,:.l KPoC elll3,)
Prestressed concrete 22
21
Figure 11 .1 1 Maximum moment and prestress force relationship
h
Max. moment lnequahttes satisfied In thts zone
'--'+---___;___;_ _ _ __ Po Y'
thus
p _};nax";t +};n.n;.h
()- K (- A<-~ +-)
( 11.23)
•·b
Thus the value of prestress force Po Y' may he conveniently considered a~ a maximum economic value beyond whi~.;h any incrca:--c in prestress force would be matched hy a diminishing rule of increase in moment-carrying capacity. If a force larger than thi~o. limit i ~ required for a given section il may be more economic to increase the siLc of this section.
(EXAMPLE 11.3
Calculation of prestress force The 10 metre ~pan heam in example 11.2 was determtned to have a breadth of :!IXlmm and a depth ot 150 mm (~ ~• 4.0!! x 106 mm3 ). Determine the min1mum 1nit1al prestress force requtred for an assumed maximum ccccntnclt) of 75 mm From example 11.2:
0.2 x 0.35 x 25 - 1.75 kN/m
Self-weight of heam 2
1.75 X 10 / H
M mm
Mm,1, "" 3.0
X
102/8
21.9kNm
+ 2 1.9- 59.4 kN m
(a) l·rom equation 11 .17: p
11
< (~tfma\ Mmux) -
K(:.,/A
-
l.'max )
106 X 24 59.4 X 10'') 0.8(4.08 X I/ 70000- 75)
< (4.08
X
X
lO
3
and allowing for the division b) the negative denominator Po~
- 2881 kN
335
336
Reinforced concrete design
Similarly from equations 11.18 to 11.20: Pn
s;
+1555kN
Po;::: +557kN Po s; -r65~k ·
( h)
The minimum 'alue of prc~trc~s force i!> therefore 557 kl'\ '' ith an upper limit of 654 kN. Check the upper economic limit to prc.:~trc.::-.' force From equation 11.~3: p
0
< fma,'l.t -1 /min:.h = -
Kc:b; : ,)
24z;\
s; 12A/ K s; 12 X (350 X 200) <;,.
2K~ X
10 ~ /0.8
1050kN
Sinc.:e thi~> b greater than the urpcr limit already eslabli~hcd from c.:quatiun 1 I .20 a design with an initial prc~tressing force between 557 lN and 654 kN will be acceptable.
11.4.3 Stresses under the quasi-pe rmanent loading The calculauon in example 11.3 " balled on the characteristiC load\. Once a value of pre~..trC'>S force lying between the minimum and upper limit value '' cho en. the cornpre.,.,ive ~tress at the top of the '>ection under the quu,i-pcnn~Ulent load' should also be calculnted and compared \\ith the les'ier allowahle value of 0.45!.:1 • If thi' proves to be critical then the section may have to be redesigned w~ing the quasi-permanent load conditiOn a~ more critical than the charactcmtic load condition.
(
EXAMPLE 11.4
Stress under quasi-permanent loads ror the previous example. using minimum preMress force or 557 kN. check the su·e~s condition under the quasi-permanent loading c.:ondition. Assume thtll the 3 kN/m imposed lo:~d consist~ of a permanent load ol'2kN/m as Jinishe~ and I.OkN!tn variable load. Tale 30 per cent of the variable load comributing to the quasi-pcnnancnt load. From the previou~ example: Moment due to ~elf-weight Moment due to finishes
-= :! 1.9 kN m
= 2 >.. 10'/8 = 25.0\.:N m
Moment due to variublc load
I < I0' 1 8
= 1~ .5k~m Quas1-pennanent moment
21.9 I 25.0 I (0.3 x 12.5)
= 5065kNm
Prestressed concrete
337
Strcs'> at the top of ~cclion is given by:
j;
= KP0 _
KPoe + M ~~
A
=
0.8
J(
6.37
557
~, J(
103
70000 8.19
0.8
J(
557
X
10'
X
75
4.08 X 1()6
50.65
+ 4.08
X
106 1
12.41
I 0.59 ~/mm~ Allowahlc compres,ive ~tress 0.45/cl. - 0.45 x 40 - I H "'/mrn 2 lienee the maxrmum compre,sive stress is Jess than the allowahle figure.
Equmlor1s l J . 17 to 11 .20 can be used ro determine u mnge possihlc vulucs of prestress force for n given or nssumed eccentricity. For different ussumcu value'> of eccentricity further limit& on the prestress force cnn be determined in an iucntknl manner although the ca l culation~ would he tedimt' and repetitive. In uddition. it is po~~ihle to U\sumc values of eccentricity for which there is no '\olution for the prestreso; force a~ the upper und lower limit~ wuld overlap. A much more u,cful approach to design can be developed if the equations arc treated graphicully a~ follow~. Equal ron-. 11.9 to II. I 2 can be rearranged into the following
fonn: K( I 1 A -
t•/':.1)
{ equauon 11.1 I }
( 11.24)
{ equauon I 1.9}
( 11.25)
K ( 1/ A e/':..h) (/;,"' I Mmu\/":.t>)
{equation 11.12}
( 11.26)
I ( l /A +e/:.h) ">-Pn • ({~,.., I M1111n/ /.h)
{equntion 11.10}
( I I .27)
(fmJ\ - Mm." /~.)
Pn Pu
.
t•f:.,)
({~1ln- M111u1 /~l)
<
Po
( 1/A-
These equations now express li near relationships hctwcen 1/Pu and''· Note that in equation 11 .25 the ~l.!nse or the inequality has been reversed to account for the fact that the dcnnminutor is negative (/~1 ; 11 i~ negative according 10 the chosen !.ign convention). The relationships can he ploued as shown in figure I 1.12(a) and (b) and the area of the graph to one ~ide or each line. as deli ned by the inequality. can be climtnated, resulting in an area of graph withtn which any combination of force and eccentricity will '>imultaneously ~atisly all four inequalities and hence will provide a satisfnctory destgn. The ltncs marked I to 4 correspond to equations 11.2~ to 11.27 re~pectively. This form of con~truction is known as a Magnet Diagram. The additional line (5) ~hown on the diagram correspond-; to :.1 po.,~ihle physical limitauon of the max1mum eccenmcicy allowing for the over.tll depth of secuon. cover 10 the preMres~ing tendons. provision of shear link:- and so on. Tv.o 11cparutc ligures are shown :t\ tt "po'>!.tble for line I. derived from equation 11.2~. to have either a po~itive or a negative '>lope depending on whether fm.t, is greater or less than Mm., /:.1 •
338
Reinforced concrete design
Figure 11.12 Magnel diagram construction
Q)
Po
--~~-L----LL--~• e
r
z.IA
zJA
· I· · I (b)
The Magnel diagram is a powerful design tool as it cover<; all possible solutions of the inequality equ:11ions and enables a range of prestress force and eccentricity values to be investigated. Values of minimum und maximum prestress force can be readi ly reutl from the diagram as can intermediate values where the range of possible eccentricities for a chosen force can he easily determined. The dwgram also 'how~ that the minimum prcstrcl>s force (largest value of 1/ Po) corresponds to the maximum eccentricit). and as the eccentricity is reduced the prestress force mu11t be increased to compensate.
(
EXAMPLE 11 .5 Construct ion of M agnel diagram
Construct the Magnel diagram for the beam given in example 11.2 and determine the m1nunum and maximum po~'ihlc valuct:. of pre~tre's force. Assume a maximum po!>sible eccentricity of 125 mm allo'' mg for co' er etc. to the tendon,. From the previou' examples:
-"~"'
= 16 N/mm2 };n,~ /~'" = - I .0 N/mm 2 /m n 1
Mm1n - 21.9 k\1 m K !\
0.8
Mm." Zb
24 N/mm
2
0 0 N/mm
2
59.4 kN m
= ~~
4.08 x 1011 mm 1
= 70 000 mm1
From equation 11 .24:
11
59.4 X 10 4.08
which can be re-arranged to give: 106 > 1210 Po -
20.77e
X
1()6
)
Prestres5ed concrete and similarly from the other three inequalities. equation~ 11.25 ro 11.27:
lOa > 2243 + 38.50e Pu t011 Po
-
tot.
-
Pn
< 785 -r l3.5e > 669 +- 1l.5e
These inequalities arc plotted on the Magnet diagram in figure 11.13 and the zone bounded by the four lincll defines an area in which all possible de!\ign solutions lie. The line of maKimum possible cccemricity is also plotted but, as it lies outside the .wne hounded by the four inequalities, does not place any restriction on the possible ~olulion~. From figure l 1.13 it can he ~ecn that the maximum and minimum values of prestn.:ss force tire given by:
TI1e intersection of the two lines at position A on the diagram cone~pond~ to a value of Po I 050 kN. established in example II '\ a' the max1mum econnm1cal value of prcstre~), force for this section (sec equation 11 23). Hence the inter,ection of the'e tWtl line~ 'ihould he taken as the maximum pre,tres' force and. force for the g1ven eccentricity.
10• Po
Minimum --~-
Figure 11 .13 Magnet diagram for example 11.5
Po " 414kN
-
Permissible
zone
Maximum economtc prestress Ioree
.,.. ~ ·
~I II'
"I - 60
--40
-20
20
40
-r 60
80
N ~
II
j I!
100
120
33~
340
Reinforced concrete design
Design of tendon profiles
11 .4.5
Having obtained a value of prcstres)> force which will permit all stress conditions to be :-.ati~fied at the critical section, it is necessary to determine the eccentricity at which this force must be provided. not only at the critical section but also throughout the length of the member. At any section along the member. e is the only unknown term 111 the four equations 11.9 to 11.12 and these will yield rwo upper and two 10\\er limits '' hich mu~t all be simultaneously satisfied. This reqturemenr must be met at all sections throughout the member and \\ill reflect both \'ariations of moment. prc:.trcs~ force and section properties along the member. The design expressions can be rewritten as: At transfer ( 11.28) (11.29) At service
fmJ,:t] -r- NJ_ma\
e>
[x
l' ~
[-~
KPu
( 11.30)
!\Po
fmm::t>] + Mm3.\ KPo
(11.3 1)
KPn
Equauons 11.28-11.31 can be C\'aluated at any 'cct1on to dctcrmmc the range of ccccntncltles "11h111 which the resultant force Po mu't lie. The moment-. Mn 1a~ and Mmm arc those relating to the secuon be111g con\ldcrcd. For a memher of constant croso,-~ection, if minor change., in prc~tress force along the length are neglected. the terms in hrackct~ in the uhovc expre:.s10ns are constant Therefore the LOne within which the centroid mu:-.t lie is governed by the :.hape of the bending moment envelopes. a~> ~ohown in ligure 11.14. In the case of uniform loading the bending moment envelopes arc p
Figure 11 .14 Cable zone limili
,_
t ·-
--- .- ----
Equation 31 Centroidal axis Equat1on 28
z, [+A
f._z,
- -;;;;-
1L---- - - ---....----__; M,... Po
Prestressed concrete
(EXAMPLE 11 .6
115
o be
llCh thi~
lngth of [UJIIOn'
~ all be lou• the secuon
Calculation of cable zone
Determine the cable zone limits at mid-span and ends of the member designed in examples 11.2 to ll.5 for a constant initial prestrcs~ force of 700 kN. Data for this question arc given in the previous example~. (a) Ends of beam
Limits to cable eccentricity are given by equation 11 .29. which at 1he end sec1ion can be readily shown, for this example. to be more cri1ical than equation 11 .28:
(' < [ ~~ +f~11 xZh] -
A
-1 Mmin
Pu
Po
and equation 11.31: II 2S)
, > [- ~ +/,ninZh] I
1
A
-
II 29)
MnM'
!\Po
KPn
As there are no moment\ due to c.xtcrnalloadmg at the end of a 'unpl} 'upported hcam equation 11 .29 bccnme~
< [- 4.0H X 10~ -~)( 4.08 (350
(!.
200)
X
-58.2!i
7()()
X
10"] t () 10 1
93.25
c. 35 111111
Similarly equation 11.31 become:.
4.08
I'
> [ • (350
106
v
]
200) I O I· ()
>-58 29mm At the ends of the beam whcre the moments are lero. and for : 1 ;:h. the im:quality cxpre~sions can apply with the tendon eccentricities above or below the neutrul axis (e po,itive or negative). So that e must lie within the range ±35 mm. (b) Mid-span Rq ur~lion
(!
11 .2H hecomcs:
4.08 X 1011 (- I )4.0X )' 10~] [ ~ (350 200) -]()() I~
21.lJ
y
+ 700 X
1 0~
JO'
< 64. 1 i 31
< 95. 1mm Equation 11.29 mighl be more critical than cquauon ll.2X checked. From equation 11 .29: 4.0!-l x 106
C'
16 x 4.08 x 106 ] 2J.9 ~ [ - (35() X 2()()) +-700 )I 103 f 700 ~ ~
X
r~nd
-.hould be ubo
1()6
10 1
58.3 + 93.3 + 31
66 mm
Hence equation 11 .29 is critical and the eccentricity mu-;t be less than 66 mm.
J
342
Reinforced concrete design
Equation 11.31 gives e
> - 4.0H X JW -0] - [ (350 X 200)
0.8
X
700
X
1() 1
-58.3 + 106.1 ~ 47.8mm ~
Hence at mid--.pan Lbe resultant of Lbe tendon force mu!>t lie at an eccentricity in the range of 47.H to 66 rnm. Provided that the tendons can be arranged Ml that their resultant force lies within the calculated l imit~ then the dc11ign will be acceptable. If a Magnet diagram for the ~tress conuition at mid-span had been drawn. a:. in example 11.5. then the eccentncity range could have t>een determined directly from the dtagram without further calculation. For tendon<, ,.. ith a combined prestrcs\ force at transfer of Po= 700kN (10t./ Pn = 1428). plotting this value on the diagram of figure 11.13 will give the range ol possible eccentricity between 48 mm and 66mm. )
rrom the Magncl diagram of ligure 11 . 13 it Cllll be seen that for any chosen value of prcstre~s fon.:c there i~ an ccccntncity range within which the resultant tendon force must lie. 1\' the Ioree approache' a value corresponding to the top and bottom limits of the diagram the width of the available cahlc ;.one tlimim11he~ until at the very extremitic11 the upper and lower limits of eccentricity coincide. giving :!ero width of cable zone. Practically. therefore. a prestre's force will he chosen which ha' a value in between the upper and lower limits of permissible pre-.trcs~ force v.hil\t. at the same time. ensuring that. for the cho,en force, a rea•.onahlc width of cuhlc lOne ext~!\. The prcstrcs~ing cahlc~ must also sati,fy requirements of cover. minimum spacing hctwccn tcn<.lon:.. avnilahle size of tendons and so 0 11. A number of alternative l.cn<.lon combination' and configurations are li~ely to be tried so that all requirement~ are 'imultaneou,ly met. Tile athantage of the Magnet dtagram i-. that a range of alternatives can be quid I) con . . idercd wtthoutthe nece:.~IIY tor any further calculation. as tllustrated at the end ol example I 1.6. 11.4.7
Prestress losses
From the tune that the prestre-..,ing force ts fir.,t applied to the concrete member. los<,cs olthi' force will take place bct:aU'.c of the following cause':
1. Elastic shortening of the concrete. 2. Creep of the concrete under su~tained compression. 3. Rclaxauon of the
pre~.trc"ing
steel under !.U\truned ten,ion
4. Shrinkage of the concrete. These los!.es will occur whichever fonn or construction is used, although the effccLs of elastic 'hortening will generally he much reduced when post-tensioning i!; u~cd. Thi~ is becaU'. e stresc;mg i<> a <;equential procedure. and not tn<;tantancous as with pretensioning. Creep and shrinkage losses depend to a large extent on the propertie~ of the
Prestressed concrete concrete wirb particular reference to the maturity at the time of stressing. ln pretensioning, where the concrete is usually relatively immature at transfer. these los<;es may therefore be expected to be higher than in post-tensioning. [n addition to lo&ses from these causes. which will generally total between 20 and 30 per cent of the initial prestress force at transfer. further losses occur in post-tensioned concrete during the ~trcssing procedure. These are due to friction between the :.trands and the duct. especially where curved profiles arc used. and to mechanical anchorage slip during the stressing operation. Both these factors depend on the actual ystem of ducts. anchorages and stressing equipment that are used. Thus although the basic losses are generally highest in pre-tensioned members. 10 ~>ome t nstancc~> ovcrull losses in post-tensioned members may he of ~imilar magnitude.
Elastic shortening The concrete will immediately shorten elastically when subjected to comprc~s i on , and the steel wi ll generally shorten by a similar amount (as in pre-tensioning) with u corrcsp1mding Joss of prestress force. To calcu late this it is neces~ury to obtain the compre!>sive ~
P' and the
Po - loss in fore~.: corrc~ponding
P' A
+
stress in the concrete at the level ol the tendon
(P'r) x (' I + ITcg
where <1,, ts the !.tress due to self-we1ght which \\ill he relatively small \\hen averaged over the length of the mcmher and may thus be neglected. lienee l1cp
P' 1\
(I t £'2'\) I
and concrete <.train
ITer / t;cm· thu~ reduction in steel strain
reduction in steel stress =
(~·P) £, fcm
rhus with A 11
area of tendons
loss in prestress force
= C~eact>Atl Ap '( I = CXeAP
hence
P' - Po so that
A,
rlcA
( +-' )
p' I
e-A 1
(l-c
a c11 / l:.~ rn unci
343
344
Reinforced concrete design
detailed calculation could be undertaken it is normally adequate to of the abo,·e losses. In thi~ case the remaimng presrres<; force is
P' =
Pu
a~sume
50 per cent
,
·1-0.5c.l/: (1 + e~A)
and it is this \'alue which applies to sub~cquent loss calculation-;. In calculating Oe, "-~m may be taken from table 6.ll where /.• ~hould be taken a.-. the tran~fer strength of the concrete.
Creep of concrete The ~ustained compressive stress on the com;n:te will abo cnu~e a long-term shortening due to creep. which will similarly reduce the prc~tress force. As above, it is the stress in the concrete at the level of the steel which il> important. that i'i
and lo~~
of steel stress
E,rrq,
specdic nccp .,tr:un
then L', Ap A
p'(l
/'~") 1
x ... pcc.:if1c
creep '>tratn
1l1c \aluc of ~pecific creep used m thi~ calculation \\'Ill be mflucnced by the facto~ di\CU'>~cd m <~ection 6.3.2. and ma) he ohtaincd from the \uluc'> of the final creep coefficient o( :x.. tn) given in table 6.12 in chapter 6 ll'-lng the rclation,hip •
•
Spec1ftc creep Mram
,In)/ , = -1."1(1. 05 c.:m r: N/mm ·
Tobie 6.12 may be used where the concrete MI'C'>'> doc~ not exceed 0.45/.:• m transfer. where f<~ relate~ ro the concrete strength at tran.,rer.
Relaxation of steel Despite developments in prestrcs~ing steel manufacture. rclaxutlon of the wire or strand under su~tai ncd tension may Mill be expected to he a signilicant factor. The precise value w11l dcpe11d upon whether pre-ten~ioning or po~t-tcn:.iolllng i!-. used and the charncteri~tics of the steel type. Equntions allowing for method of construction are given in EC:! section 3.3.2(7) which should he applied to 1000-hour relaxation values provided hy the manufacturer. The amount of rclaxmion will also depend upon the initial tendon load relative to ib breaking load. In most practical situations the transfer -.tee! stre\'> i~ about 70 per cent of the characteristic -.trength und relaxation los\cs are hkely to be approximately 4-10 per cent of the tendon load remainmg after ti".Ul'ifer.
Shrinkage of concrete Thi'> I'> based on cmptrical figures tor hnnkugc/unit length of concrete (£C>) for particular curing condjuono; and transfer malllrity a... di~CU\ ed 1n chapter 6. Typical values range from 230 x 10 6 for UK outdoor expo~ure (SOtq relative humidil}) to
346
Reinforced concrete design
(
EXAMPLE 11 . 7 Estimation of prestress losses at mid-span
A po~t-tensioned beam shown in figure 11 .15 IS stressed by two tendons\\ ith a parabolic proille and having a total cross-<;ectional area Ar = 7500 mm2. The total initial prestress force IS Po= 10500kN and the tom! charactensllc strength l'i Pp.._ = 14000k~. Figu re 11.15 Post-tensioned bec1m
The tendon supplier specifics clas.., 2 strands \\ ith a I 000 hour relaxation cent at 70 per cent of the charactcri~tic l>Lrength.
lo\~ of 2.5
per
(1) Friction
The equation of the pnrabola is I'= c~ and with the ongin at mid-span when 15000. y = 640. so that C = 640/15(}()(!1 2.844 x 10 11 The gradient (J at the ends i~ given hy
0 dy/ dt
2Cr
2
X
2.84-t
X
10
6 X
15000
= 0.0853 radians At mid-span
los!> ~P(x)
= Po ( 1 - e_,,(OH•)) = Po( 1 _ e - o111 (UI~Sl+U.tll x 15)) 0.0-MPo
= 460k..~ = 4.-t per Cl!nt
(2) Elastic shortening for post-tensioned construction
P'
ra~c
und
p
= ----.,.----- -
l - 0.5nc ~ (I e~ y)
lht: 3\erage eccentriClly for the parabolic tendon a:o. 5/8e, = E-/E,m = 205/32 = 6.41.
Oe
5/8 x 640
= 400 mm
I~
Prestressed concrete
Po
P'
-
1 + 0.5 0.968P0
3 X 10 ( X 1.05 X 1Q6
7.5
X
6.41
,
I
1.05
X
+ ..f()()· 0.36 X
IOIJ )JO I2
= 10 160kN
Los ~J = 10500 lO 160 340kN = 3.2 per cent Total l)hort-term los~es = 460 -'- 340 = 800 k.~ P0
p'
-
short-term losses
= 10500 (3) Creep
800
9700kN
m F.,Ap (I+e2 ~)I P' ' ( 1.05£.m)A
Los!. t:.P
6
205
1
10 7.5 X 10] ( ' 1.05 X IO'') - I , l ( 1.05 x 32) x I 03 x 1.05 X 1()1> I + ..JOO OJ6 Xi(iTi 9700 X
992 kN ( = 9.-1 per cent of Pn) (4) Shrinkage Los~
.:::,p = •c,c',Ar - JJ()
J0
1 ' X
2()5
X
7.5 X 10'
507 kN ( = ..J .8 per cent of Pu) (5) Relaxation
Long-tc.:nn rda\atJOn IO\S !actor = 2.5 for cia<;, 2 \trand estimated rrom equation 3.29 of EC2 IO~\ ~,,
(2.5
X
2.5/ IOO)P'
= 0.0625 X 9700 606 k.N ( = 5.8 per cent of flo)
Total c~timatc<.l losse!.
800 + 992 ,.. 507 l 606
= 2905 !..N
= 28 pt.:r cent of Po
11.4.8
Calculation of deflections
The anticipated <.lcncction of a prestressed member mu<;t alway'> be checked since <;pandfeetive depth rutios arc not specified in the code for prestrc.:ssed concrete member~. The deflection due to the eccentnc prestress force must be evaluated and aducd to that from the normal pennunent and variable load on the member. In the maJonty of ca~cs. particularly where the member is designed to hi! uncracked under full load. a simple linear elastic analy~i~ ha,ed on the gross concrete \Cction ''ill be ~ufficient to gi'e a reasonable and realilltJC e~timatc of deflections. Where the member ic; de<;igncd ~uch that, under the characten ... uc loads. the tensile 'ltrength c\ceeds the cracking \trength of the concrete, /.1m. it may be necessary to base the calculation of detlecLion on the cracked concrete section and reference o;hould be made to the Code tor the method of dealing with th1s ~ituation.
34;
348
Reinforced concrete design
The basic requirements which should generally be sausfied in respect of deflections arc ~imilar to tho~e of a reinforced concrete beam (~ection 6.3) which are: 1. Deflection under the action of the qumi-permanent load ~ span/250 measured
bekl\\ the level of the ),Upports: 2. Span/500 maximum movement after other element'>, \vhich are susceptible to damage by movement, are applied. The evaluation of dellections due to integration of the expresston
prc~trcss
loadtng can be obtained by double
d2v
M,=Pe,=Eid·~
t·
over the length of lhe member, although this calculation can prove tedious for complex tendon profiles. The simple case of straight tendons in a uniform member however, yields M = - Pe a wnstant. which is the situation evaluated in 5CCtion 6.3J to yield a rna xi mum mid-~ pan dellection of ML1fHHI Pe/.2 /81:.'1. H the cables lie below the centroidal axi'>. I! is po~itive. and the dellection due to pn.:we!>s is then negative. that is upwards. Another common ca~c of a ~ymmctrical pambolic tendon pronlc in a beam of coll\tant ~ection can also be evaluated quite simply by considering the bending-moment di'>lrihution in terms of an equivalent uniformly distributed load. Fm the beam in figure 11.16 the moment due to prestre-., loading 31 any section is A1, -Pe, but 'Iince e~ t puroholic, the prestre'l-. loading may be IJkened to a umfonnly dt<;trihutcd load lie on a 'tmpl} -.upponed beam; then mtd-span moment M
lleL~
- - = -Pe, 8 XPe,
~''c
= --, 1.·
Bul since the mid-span deflection Juc given by y
5 wL4 384 !:..'/
the deflection due w H'e i::. 2
5 (Pec) L ----+8 £1 Figure 11.16 Parabolic tendon profile
L/2
10
a uniformly dimihutcd load
w
over a span L i&
Prestressed concrete
34~
Figure 11.17 Parabolic tendon profile eccentric at ends of beam
[0
le
Jf the prel>tress force does not lie at the centroid of the section at the end' of the beam. but at an eccentricity eo as shown in figure 11.17, the expression for deflection mu:.t be modified. ll can be shown that the deflection is the same as thm cau,ed by a force P acting at a constant eccentricity eo throughout the lengt.h of the member, plu' n force P following n pnrabolic profile with mld-span eccentricity t< as shown in figure 11.17. The mid-::.p
2
(Peo)L 5 (P<)L ,.. -- ----. Rt:t 4R £1
11 '
f
IS
a
DcHcctions due to more complex tendon profile~ are most conveniently estimatcu on the basis of coeftic1ents wh1ch can be evaluated for commonly occurnng arrangements. These are on the ba i' y = ( Kt})jEl where K incnrporate~ the varwtion'> of curvature due to prestre~s along the memher length. There arc three pnncipal 'otages in the life of a prestrcs\Cd member m whtch deflections may he critical and may need to be as\e:o..,ed.
1. At tranllfcr - a chcd. of actual dcnection a1 tran~fcr for compari,on "ith estimated values i-. a useful guide that a pre. tres. ed beam hal> been correctly wnstructcd. 2. Under dead load, before application of lini!>he!-. - dcOecuons muo;t he evaluated to permit :-.ubscqucnt movement and possible damage to he csumated. 3. Long-term under full quasi-permanent action:-. deOect1ons are reqlllrcd. both to determine the sub~equent movement and al!lo to a~sess the arpearuncc of the final structure. Short-term dellections will be based on materials rroperrie!> u!.socimcd with characteristic strengths hm = I ) and wi th actual loading ("}1 I ). Long-term ns~essment however must not only take intn account loss in prestrcs~ force, but al!.o the effects of creep both on the applied louding and the prestress loading component~ of the del~ection. Creep is allowed for hy using an cffectivl! mmJulus of elusticity for the concrete, us discussed in section 6.3.2. Thu~ if f:..'.:r ~ 1 is the in~tantancou~ value. the effective vnlue after creep is given by
where the value of ¢{-x:.r0 ) , the creep coefficient can be obtained from table 6. 12 It can be ~hown in -.omc instances that when net upward deflections occur, the~e often increa.,c bccau~c of creep. thus the most criucal downward deftectton mny well be before creep IO!>~es occur. while the most critical upward dcftecuon may he long-term. This further complicates a procedure which alread) ha:-. many uncertainties as di)o.Cll~\ed in chapter 6: thus deflection!. must always he regarded as e~;ltmute-. only.
350
(
Reinforced concrete design
EXAMPLE 11 . 8
Calculation of deflection
Estimate transfer and long-term deflections for a 200 x 350 mm beam of I0 m span. The pre,tressing tendon has a parabolic profile with mid--.pan eccentricity 75 mm and the end eccentricity = 0 at both ends. The initial prestre:,-. force at traru.fer. Po. i~ 560 kl and there arc 20 per cent losses. The imposed load consbts of 2.0 1-N/m finishes and L.O kN/m 'ariable load. Ecm = 35 kl\/mm~ and the creep factor ~( oo.t0 ) = 2.0. Self-\\ eight= 0.2 x 0.35 x 25 = 1.75 kN/m bh3
1 =-= 12
(a)
2()() X 35()3 12
715x 106 mm4
At transfer 5 ~~'minL4 5 (Poer)L2 DeflccLtOn Yn = 384 Ecml -48 E,.ml .
5 1.75 X JO~ X 10 12 - 384 35 X l 03 X 715 X I 06 9.1 17.5 = -8 mm
5 560 48
X
35
103
X
75 X 102 X 106 7 15 X l 06
X I 01 X
(upward~)
(h) At application of ftni~hes A~sume
that only a small proportion of pn:strcss lo..sc-. have occurred:
Weight of
fini~hes
- 2 0 kN/m
therefore 5 ~ 2.0 : 10~ 10 1 ~ · -.a 384 x 35x 101 x 715 x I(J<' - R 1 IOmm 2mm (downwards) \'~ - ,.-
(c)
pre~>~ res~
force after losses
Assuming 30 per cent of the variable load contribute), to the acrion:
qua~i-pcrmancnl
In the long term due to the quasi-permanent action plus
Quasi-permanent action
self-weight I finishes I OJ x vuriablc load
= 1.75 + 2.0 + 0.3 x 1.0 Prestres~
forces after losses
4.05 kN/m
= 0.8Po
0.8 x 560 44l'l kN 35 , Ecetf=( ( )) =( 1- ., ) = 11.7kN/mm· l + cJ> oo. ru i -· 0 5 4.05 X 104 X 10 11 5 448 X 10 1 Y 75 X 102 X (()6 ."c = 38411.7 X 103 X 715 X !Qfl 48 11.3 X t01 X 715 ~ ,
Ecm
63.0- 43.3 = 20mm (downwards) < span/ 250 (d)
40 mm
1l1eretore sat.J\factory. \t1ovement after application of finishes )'d
y., -
)'b
= 20- 2 = 18 mm $ spnn/ 500 = 20 mm ('>allsfactory).
Prestressed concrete 11.4.9
End blocks
In pre-tensioned members. the prestress force is transferred to the concrete by bond over a definite length at each end of the member. The transfer of !>tre~s to rhe concrete is thus gradual. In post-tensioned members however, the force is concentrated over a small area at the end faces of the member. 11nd this leads to high-tensile forces at light angles to the direction of the compression force. This effect wi ll extend some distance from the end of the member until the compression has distributed itself aero:.' the full concrete cross~ection. Thi~ region is known as the 'end block' and muo;t be heavily reinforced by steel to resist the bursting tension forces. End block reinforcement will generally wnsist oJ closed links which surround the anchorages. and the quantities provided arc u~uaily obtained from empirical methods. Typical 'How line~· of compressive stre~~ arc sho\\n in figure 11.18. from \\hich it can be seen that ~hmever type of anchorage il> u~ed. the required di:.tribuuon can be expected to have been attained at a distnnce from the loaded face equal to the lateral dimension of the member. Thil> is relatively independent of the anchorage type. In dc:.igning the end block it i~> ncccs~ary to check that the bearing strcsl-. behind the anchorage plate due to the prestres:.ing force doc~ not e'\cecd the limiting ,\Ires\. /ll.du· given by /Rdu
= 0.67f._k(Ad/Aco) 0'5 "' 2.Qf:k
where b the loaded urea of the anchorage plate A.t is the maximum aren. having the ~ame ~hape a<. Ad1 wh1ch can be in the total area A., as ~hown in figure 11.19(a)
A<"
in~cribed
The lateral tcn~ile bursting fmces can be established hy the usc of a ~ tatil.:all y dcterminnte strut and tie model where it is a~sumed that the load is carried by a truss consisting of concrete !>truts and linJ..s of reinforcement acting a' steel tics. In carrying out these calculations a part1al factor of safet> of ') p = 1.2 i:-. applied to the prestre~sing Figure 11.18 Stress dtstributton In end blcxks
Flat plate anchorage
Contcal anchordge O.SP
_l _ O.SP
LJ
(a) Anchorage LOne (end vtew)
(b) Strut and tie model of load dtspersion
Figure 11.19 Sursttng tensile force In end blocks
35
352
Reinforced concrete design
force. EC2 suggests that in determining the geometry of thi!> truss the prestressing force can be a!>sumed to disperse at an angle of 33.7 to the longitudinal axis of the beam as shown in figure 11.19(b). The comprelosivc sLrc:.sc~ in the assumed struts should not exceed 0.4 ( 1 -
{;~))i~k and the reinforcement is designed to act at a design strength of
OJn(yk· ltowever if the ~tress in the reinforcement is limited to 300 N/nm1 2 then no checks on crack widths are necessary. This reinforcement, in the form of closed links, is then distributed over a length of the end-hlock equal to the greater lateral dimension of the block. this length being the length over which it is assumed that the lateral tensile strel.sel. are actmg.
(
EXA MPLE 11 .9
Design of end block reinforcement
The hcum in ligure 11 .20 is stres~cd by four identical 100 mm diameter conical anch(lrage<, located :lS shown. with a jacking force of 250 kN applied to each. The area may he ),Ubdivided into four equal end tones of 200 x 150 mm each. Determine the reinforcement required around the anchorages:!.:~ = 40 N/mm 2• J,. • 500 N/mm 2 . Consider one anchor. (a)
Ched: hearing Lress under the anchor Actual bearing strl!ss
~rcs tre~s10g
force
l.oadcd urea l.2 X 25() X 1() 1 - - - ; . >( 1002/ 4= 38.2 N/mm
\llowable bean ng 'lrc" .f~tdu
1
0.67.f..k (A.t / A<"~·d' 5
o.67 <.~o(""
150~/4 )
7TX !()()1j4
-10.2 N/mm2 ( > 38.2) Figure 11 .20 End block reinforcement example
8....
• ~l lOOOkN
~,
400 (a) End section. four anchorages
(b) Area for combined anchorage
0
~
Prestressed concrete
353
(b) Reinforcement Front figure ll.l9b. the tensile force in the tie of the equivalent trUl>S is given by
T = 0.33
X
1.2
X
250
lOOk'
Area of tensile steel required (assuming strc~s in the \tee I is limited to 300 !'\/nun') 10
A, -
I()() X 103 300 330 mrn 2
Thi), can he provided by Lhrcc 10 mm closed links (471 mm 2 ) at, say. 50. 125 ami 200 mm from the end race: that i~. djstrihuted over a length equal to the largest uimcnsion of the anchorage bloc~ (200 mrn). Note that in each direction there arc two legs of each link acting to rcsi~t the tensile force. (c) Check cornpres!>ivc stress in the stmts Allowable compressive stress
0.4( 1 - ./~~ /250)j~k 0.4( I - 40/250}40
Actual
.
~tres~ 1n
13.44 N/mm 2
Force in strut . 1 0.60 X 1.2 X 25() X JO~ =.,.,...,.-:----:-(200 X 150 X CO\ 33.7 )
strut = -C
ross-~ecl!ona an::~
= 1.21
m11D
2
'Jhe effect ol the combined anchorage can he con'>Jdered by con\idcnng the total
pre'> tress Ioree of I000 k • acting on an cffecuve end block of 400 The ten,Jie force in the tic of the equivalent tru'\s i\ given by O.B x 1.2
I()(X)
400 mm.
= 4001..N
Area of tensile \tccl rc4uircd 4()0 X A.\
)() I
300 1333 mm 2
This can be provided by six 12 mm closed links (1358 mm 2 J distributed over cqual to the largest dimension of the anchorage block, thm k 400 mm.
ll
length
___________________________________________)
l-·
11 .5
Analysi s and design at the ultimat e limit state
After a pn:~trc~'ed member has been designed to sali~fy \crviceability requirement~. a check must be carried out to ensure that the ultimate moment of resistance and shear resi~tancc arc adequate to satisfy the requirement~ of the ultimate Iunit '> tate. The partial factors of l>afct) on loads and materials for this analysis arc the normal values for the ulumme limit state which are given in chapter 2. However. in con:-.iderauon of the effect of the prestress force this force should be multtplled by a partial factor of safety. ir· of 0.9 (UK l\ational Annex) when the prestress force is considered to be, as '' usual. a ·favourable effect·.
354
Reinforced concrete design
11.5.1 Analysis of the section A~ the load~ on a prestressed member increase above the working values. cracking occurs and the pre.,trcssing steel begin~ to behave as conventional reinforcement. The behaviour of the member at the ultimate limit state is exactly as that of an or<.linary reinforced concrete member except that the initial strain in the steel must be taken into w::cuunt in the cak:ulations. The 'ection may easily be analy:.cd by the u~c of the equi\'alent rectangular stress hlock <.lescribed m chapter 4. Allhough illu~Lrated by a simple example thi., method rna) be applied to a crosssection of any ~hape which may have an arrangement of prestressing wires or tendons. Use is ma<.lc of the l.tres~-sLruin curve for the pn:stressing steel shown in figure l 1.21 to calcu late tension forces in each layer of steel. The total steel <;train is that due to bending added to the mitial ~train in the steel re~ulting from prestress. For a series ol al.sumed neutral axi~ posn1on". the total tension capac1ty 1s compared with the compressive force developed by a unifonn ~trel>~ of 0.567f
Figure 11.21 Stress-strain curve for prestressing ~teet
Ym
lz 205kN/mm'
Str.lln
(
EXAMPLE 11.10
Calculation of ultimate moment of resistance The section of n pn.:tcnsioned beam shown in ligure I 1.22 is :-trcs~-.cd by ten 5 mm wires of 0.1% proof ~lri!~S fPJ 1, 1600l'\/mm2• If these wires are initially ~trc~~>ed to I 120 .:--l/mm~ and 30 per cent ((h~C~ are anticipated. estimate the ultimate moment of n.:l>istance of thl! ~cction it clas~ C35/45 concrete ~~ used. The ~tre~s-!>Lrain curve lor prcl>Lre~sing wirc i~ ~hown in tlgure 11.23. Area of 5mm wire
rr x 52 /4
Stress in 'tccl after los.,cs
)p :>-.
19.6 mm 2 1120 x 0.7
=0.9 x
thcrdore . .111 'tcel alter . S • rra111
lossc~
= ·£(,., -
., -705
- 05 X I 01
wh1ch is les' than _ , the yield l>train.
= ()·{)<)34
11 20 x 0.7
= 705 N/mm2
Prestressed concrete b= 120
lee=
Figure 11 .22
-1
1
neu.t~a!___~
_j 8.....
0.0035
_ __
ax~
• ••••
(~O,Ih Ym
:
l
'
F,
Bending Strains
Ultimate moment of resistance example
1
~
•••••
Section
Stress Block
'1600 :: 1390
Figure 11.23 Stress-stra!11 curve For
1.15
prestressing wire
205kN/mm 7
Stram
0.00678
A depth 1 ot neutral axis must be found for which the compre-;,tve force /~ in the conm:te ~~ halanced b) the tensile force in lhe steel. Then the ultimate moment rcsi~~tuncc i' given by
r:
or
( 11.33) where :: is the lever ann between Fe and F,. As a first attempt try x 130 mm, approximmely cqmtl to 0.5tl.
(In cnlculating < the initial concn.:te strain due w prc!.trcs~ cnn be ignored without undut: error). Top layer
'"~a - 0.0034 I E~a therefore t~
r) = 0.0034 + .._ ( 175ec-c .I
= 0.003., 1
0.0046
(
175
( I 1.34)
130)0 003~
130
.
.)
355
356
Reinforced concrete design Bottom layer
= 0.0034,.. e:~b
!:,IJ
= 0.0034 -r
(275 -x)
e,.~
( 11.35)
X
-- 0.0034 I (
275
130 ) (. ) 0035 130 •.
- 0.()()73 (b) Steel stresses hom the :mess-~train curve the corresponding \tccl ~trc1.ses are: Top layer
J..3 =. 3 )( £,
=0.0046
)I
( 11.36) 205
- 943N/mm
)I
3
10
2
and 1390 ;-.Jfmm 2
f..b
as the str;Hn in th~.: bollom \tee I cx~.:ced~ the yield strum (c ~
0.00678).
(c) Forces in steel and concrete
Stcd tensile force /·,=I>~;\, (94~
229
(f,a
+ I WO} 1<
-l X
( 11.37 )
f,h)5 x 19.6 9X
10'
W1th a rectangular stre's block Concrete compressive force "'~
=0.567f.•b x 0.8x
(11.38)
0.567 > 35 ')( 120
X
0.8
X
130
= 248 < 10' N The force f· c in the concr~w i~o. lurge1 than the lnrt:e F, in the steel. therefore a :.mallcr depth of neutral axis mu~t hi! tried. Tahlc ll.l shows the re),ults of t:lllculation11 for further trial depth~ of neutral ax•'· For .\ II 0. F, became 'mallcr than F , therefore \ 120 and 123 were tried and it wa<, then found that /•~ = I• c. Table 11 .1 X
(mm)
130 110 120 123
,. 4.6 5.5 5.0 4.9
( >< I 03)
Forces
Stresses
Strains >b
7.3 86 79 7.7
f... (N/mm1 )
943 1121 1026 1000
'~b
1390 1390 1390 1390
F~
Fe (kN)
229 246 237 234
248 210 229 234
Prestressed concrete
357
In terms of the tensile force in the steel. the ultimate moment of resi tance of the ::.cction is given by
Mu
= F,:. = L lf-A,(d- 0.4x).
( I 1.39)
5 Y 19.6[1000( 175 - 0A x 123)+ 1390(275-0A x 123 ) = 43.1 x IOfl 'mm
H x had been incorrectly chosen as 130mm then u&ing equation I 1.39 M. would equal 42.0 kN m, or in terms of the concrete 0.56~/~\b X 0.8x~
Mu
~ 0.567 ~
X
35
X
120
0.8
X
X
130(225 -0.4
X
130)
X
10
I>
43kNm
Comparing the average of these two val ue~> of Mu ( 42.5 kN m) with the correct it can be seen !hat a 1>light error in the position of the neutrul axis does not have any significant effect on the calculated moment of rc~>i ~>tance. an~o.wcr,
_____________________________________________)
l
11.5.2
Design of additional reinforcement
If it i~ found that the ultimate limit -.tate requirements arc nm met, additional untcnsioncd or partially tensioned Mccl may be added to increase the ultimate moment nf rc~>i,tancc.
(EXA MPL E 11 .11
Design of untensloned reinforcement
=
o c.. ign UntCll\ionc<.l high yield reinforcement rjyl 500 N/mm 7 ) for the rectangular benm::.cction ::.hown in figure 11.24 which is stres~u by live 5 mm wire,, if the ultimute moment of resistance is to exceed 40 kN m for cl a~s 40/50 concrete. The characteri ~tit strength of te n~ toned steel.jj,o t\ 1600N/mm 2•
=
(a) Check ultimate moment of resistance Maximllm tensile force if
=0.9 X [5
X
19.6 X
prestre~sing
1600
1.15
]
X
10
3
steel yielded
= 123kN
Concrete compres:-.ive area to balance =
123 x 101 )( - 0.8 . 0 567 40
1:!01
thus ncutral-axi!. tleplh x = 56 mm. 1\~~uming prcstrain as caJculated in example 11.10 tollll
~tccl
\train
prestrain +bending strain
0.0034 ~ - 0.0034 + Lever arm
275
~X
0.0035
X
219
56 x 0.0035 =
0..+0 x 56
253 mm
0.0171 ( > yield)
358
Reinforced concrete design
-
Figure 11.24 Ultimate moment of resistance example
Lee=
0.0035
0.5671"
120
I
0.8x~ -- i.
X
~
neutral axis
"' ~ - ----
N
•
•
-
2Hl0
'
F,
•••••
Sect1on
Bend1ng Strains
Stress Block
hcm;c ultimate moment of resistance
=
253 x 123 x 10 1
31.1 kN 111
Unten~ioned steel is therefore required 1t1 permit the beam to support an ultimate moment of 40 kNm.
Additional moment capacity to be provided
Effective depth of additional steel
40
3 1.1
8.9 kN m
245 m111
then
IC\'Cr arm to additional steel :::: 220 m111 and . I add .lll()na
. r tcn~ton
.orce required - 8900 220
c 'N ..'(l _,"
lhll' CMtmatcd arcu of untensioned
&~eel
required at its yteld stress
40500 ' 0.87 x 500 = 93 mm· 'lry two lOmm diameter bars ( 157mm\ (b) Check steel strain
{(' additional steel has yielded, force in lwo 1110 burs 68.3 kN. therefore total
ten~ilc
force if all the steel has yielded
= 123 -+ 68.3 191.3 kN
thus
, , , depth ot neutral axts at ulnmate
191.3 X 103 x O.S 120 88mm
= 0. 567 x 40 x
TI1ereforc
. !>teeI SlraJO . esb pn!\ti'Cl>l>lllg
=
275 - 88 gg
X
00035 •
= 0.0108 (>yield)
0003 · 4
157
X
500
X
10 1/1.15
Prestressed concrete
and . d !Ilee! stram . :..., = 245 - 88 x 0.I)()35 untenswne 88 0.0062 Thi~
value b greater than the yield strain of 0.00217 from
= 43.!lkNm If it had been round in (b) thtll either the prestressing l>tecl or untcnsioned i>li.!CI had not yielded. then a trial and error approach !>i milar to example ll.l 0 would have been nc;cessary.
11 .5.3 Shear Shc;ar in pre~lre,l.cd concrete is considered at the ultimate limit ~late . Design for shear therefore ill\ olve., the rno~t 5everc loading condilion1-.. with the u~ual panial factors of safet)' being applied to the actions for the ultimate limit state being con.,idcrcd. The respon.,e of a member 111 remting shear is ~imilar to that for reinforced concrete. hut with the additional effects of the compression due to the prer.~res!lmg force. TI1is will incrca:.e the 1-.hcar resistance con.,iderahly and thi-. is taken into account in EC2 by enhancing the equnuon for the shear capacity ( VRLI J of the section without shear reinforcement. With a few slight modifications. the Code gives an almost 1dentical approach. ba'icd on the Variable Strut Inclination Method of shear dc.~ign. in prestressed l.ections as i'l U'led 111 reinforced wncrclc sections as outltned in Chapter 5. In calculating the design shear force, V!Zd. it is permissible to take into account the verucal component of rorcc in any inchned tendons which will tend tO uct in a direction thm resists ~heur. thu~ enhancing the shear capacity of the section. In ~o.uch a case the prestressing force .~hou l d be mu ltiplied by the partiul factor of safety, / p 0.9. Sections that do not require designed shear reinforcement
In regions of prcstrcs!>cd beams where shear forces arc small and. taking into account any beneficial effect of forces attributable to inclined prestrc~~ing tendon~. the concrete ~cction on it~ own may have sufficient shear capacity {VRd c) to resist the ultimate ~hear force ( VhJ). Notwithc;tanding this it is usual to provide a minimum amount of shear links unless the beam IS a minor memher \Uch as a short-span, lightly loaded lintel. The concrete shear strength (\' Kd.c ) is given by the empirical expressiOn: ( 11.40)
\\ith a minimum value of: VRu..:
rl0.035k3' •t,k ? 11 +0.15u, p] b... d
( llA l)
359
360
Reinforced concrete design
where: VRd.c
=the design shear resistance of the secuon without shear reinforcement
k= (1
_~_ \~) < 2.0 with d expressed tn mm
A, t Pt - b.,.d ~ 0.02 A~ 1
= the area of tensile reinforcement that extend~ beyond the section being considered by at lea~t a full anchorage length plus one effective deprh (d) b,. = the smallest width of the section in the tensile area (mm)
a,r = axial stress in section due to prestress (')1,KI'o/ M ( <
0.13~/~~)
It can be seen rhat equations I I .40 and 11 .4 1 are pruc:ticully identical to equations 5. 1 and 5.2 for 1-hear in reinforced conc:rctc sections. The additional term of 0.15crcp indicates rhalthc effect or the prestress i~ to enhance the ~->hear capacity of the section hy 15% of the longitudinal ~tress due to prc)otre~~>ing.
Shear strength without shear reinforcement - regions uncracked in bending (special case) For the spedal ca.-;e of a .1i11Kie Sf/all beam. in regiom. which arc uncracketl in bending (t.e where sagging moments arc relati\ely l>lllallnear to the 'upports). the 11hear strength of the concrete :.cction could be governed by the development of excel>\hc tensile 'ilrc~,c~ in the concrete. These regions are defined as "here the llcxural ren..,ile stress in the uncrad.ed <,ection does not exceed f ctk he· where }~ 1 ~ i:. the charactemtic axial tcn'iilc ~ l rength of the concrete. Tile applicable equariow; tn EC2 can be developed as follow~.
At an uncracked section. a Mohr'~ circle analyM' of a beam element shown in figure 11.25 \\>hich b subject to a longitudwal compressive stress.}~ and a shear stress ''cu give~ the principal tensile stress as:
Figure 11 .25
Stress ln uncracked section
This can be re-nrranged to give the shear stress
"'"- Jw -JJt) The actual shear stress at any level of a beam subjecr
10
a shear force. \1, can be
shO\\ n to be:
where Ay tS the first moment of area of the part of Lhc :-ection ahove the level considered about the centroidal axis of the beam, 3!> :.hown in figure 11.26, b i~ the breadth of the section at the level considered and I h. the second moment of area of the whole section about its centroidal axis.
Prestressed concrete Figure 11.26 Shear stress distribution
Shear stress Yeo
Cross-section
Hence if }~cJ IS the limiting value of principal tensile stress. the ultimate shear resistance VRd,c of the uncracked sectjon becomes:
bJ Ay Th i~
equation as:
rorm~
the
bu~i~
or the design equation given in 13('2 which is
cxprc~>:-.cd
( 11.42) where: ~tres~
in <.,ection due to prestress C!pKPot A)
11cp
axial
/.:cd
the de,ign tcnl>ilc ~trcngth of the concrete (- l~c~ h.>
n1
I for poo.,t -tcn~ioned
<
tendon~;
I for pretens1oned tendon!-. and in thi:-. ca!'te the value of n 1 is given 10 t.C2 according to the di,tancc of the !lection being considered Ill relation to the transnms1on length of the tendon.
~>tate~> that. for lhc l!pccial case of n simply supported beam. equation II .42 be used in thO~>C regionl> where the flexuralten~ile Mres~ in the uncrackccl section docs not exceed}~ckhc and where the beam i~ cr:1cked in bending equation 11.40 1>hould be used. Determming where the beam is uncracked al the ultimate limir ~;tate is not straight-l'orwarcl and. in practice, both these equations ~hou ld he upplied at each section considered and the lowest of the two va lues calcu lated then taken as the t-hcar caracity of the section.
EC'2
~hould
The variable strut Inclination method for sections that do require shear reinforcement A~ prcv10u~ly noted the design for shear and the prov1ston of shear reinforcement in prel>tre,!>cd concrete is practically idenucal to that for reinforced concrete and is ~ummarbcd below.
(7) The diagonal compressive strut and the angle 0 The maximum design shear force that a section can carry
361
362
Reinforced concrete design
compreo;sive struts of lhe assumed truss. leading possibly to compressive failure of the concrete. The maximum shear force is given by: Ctcwbw'Z.V!}~~
V Rd. max -
1.5 (cotB+tanB )~
where ;: = 0. 9d and VRd max
1• 1
= 0.6( I -
.fc~ /250).
Hence:
I - }~k/250).f~k < -ll'cwbw0.9d0.6( - ,..---,:-:-- ~--..::....:.::.:,..,..,--
-
II .S (c01 0
< C\cw0.36b.,..d( J
I tan B)j
( 11.43)
.fck /250lfc~
[col() + tan OJ
-
This equation i~ practically identical to equation 5.4 in Chapter 5 except that it includes a coefficient ac.,.. given by:
= I 1- l .5rJcp//c~ l~cw = 1.25 o •.,. = 2.5( I - 1.5rJcp/.fcd (~cw
for 0 < CTcp < O.l67fck for 0.16 7./:~ < 17'cp ~ 0.333/cL for 0.333,/;;~ < O'cp <... 0.667fcL
where
VRd ma~(22J
2.5:
1.0:
and with col I)
= ncw0.1241>\.,t!( I - .fck/250lJ~L
VRd. mux '4~1 - Ocw0.18h.,.d( I - }~k/250)f..~
( t 1.44) ( 11.45)
and ror values of Bthat lie between these two limiting values the required value of 0 can he obtained hy equating V~:d to VR
· 0.5 SIO
I { -
--
l'~.t- . -
,
Hcw0.18h.,..t/{I /ck/ 1_50lJ~k
} < 4.5
-
( 11.46a)*
which alternatively can be expressed as:
0
0.5 Sill
I
El } ~ 45 { VRd.vma~(4~ )
( 11.46b)
where Vnr is the shear force at the section being con:.idercd and the calculated value of the angle Bcan then be u\ed to determine cot Band to calculate the ~hear reinforcement A,.,. j s at thatl!ecuon fmm equation 11.47 below ('"hen 22 < 8 < 45 ). If the web of the section contains grouted ducts with diameter greater than one-eighth of the web thickne%, in the calculation of VRd ma•· the web thickness should be reduced by one-half of the sum of the duct diameterc; measured at the most unfa\(>urable o;ection of the web. For non-grouted duct~. grouted plastic duct~ and unbonded tendons the web thickness shou ld be reduced hy 1.2 times the !-Um of the duel diameters. u· the de:.ign ~hear force exceeds VRd m3~ then it will be necessary to increase the !tiLe of the ~cct ion. (2) The vertical shear reinforcement
As in reinforced concrete. ~hear reinforcement musl he provided to resist the shear force if it can not be sustained by the concrete '>ection including tbe enhanced shear resistance
364
Reinforced concrete design 4. The shear links required can be calculated from equation 11.47
A"'
VEd
s
0.78df~L cotB
where A~" is the cross-sectional area of the legs of the links (2 x 1r9~ / 4 for 1>ingle stirrups). For a predominate!) unifonnly distributed load the maximum shear Vw can be calculated at a distance d from the face of the ~upport and the shear reinforcement should continue to the face of the support. The shear resistance for the links actually specified is Vmin
Asw
-
s
x 0.78d/yL cot f)
and this value will be used together with the shear force envelope to determine the cunailment position of each set of designed links. 5. Calculate lhe minimum links required by EC2 Ji·om A,w
ndn _
0.08fc~ ~bw
.r-~ 6. Calculate the additional longitudinal ten<>ile force
s
ct~uscd
hy the shear
:::.f td = 0 5 \ '~"A~ cot 0 The al.1o'e procedure ~hould be repeated at different allustrated in the following example
(
~cctions
along the beam. as
EXAMPLE 11 . 12 Design of shear reinforcement
l hc beam cross-section l>hO\\-n in figure 11 .27 1s constant over a 10m :-.imply supported l)pan with a parabolic tendon prolile and an eccentricity varying between 300 mm at the ends and 750 mm at mid-span, measured hclt>w the ncutrulllxis in both cases. The beam 1.upports an ultimate uniformly distributed loud of 40 kN/rn and j~~ 35 N/mm 2•
Figure 11.27 Shear reinforcement example
1000
0
0\ "'
II 0
"'""+ 0
~ I
8
___._,"'" location of tendons at the supports
Prestressed concrete Given data: Prc...,tres~
force after losses
2590 kN
= 145 106 x A = 500 X I03 mm 2 Ar = 3450mm2 /yL = 500 1':/mm~ for the shear links fctk = 2.2 N/mm~ 106 mm4
I
The calculation~ will he presented for a section at the ~upport nnd then repeated and tnbulntccl at 3m intervals along the span. (1) Calculate shear force at the section
Although th e m:~ximum shear force can be taken at the fac;c or the ~upport , in thi~ example we wi ll, for illustrative purpo~e:., luke the 'ection nt the middle of the support itself. llcncc: 40 x 30/2
Vrd
600 kN
(2) Check if shear reinforcement is reqwred
rrom equation I 1.40 the concrete ~hear \trength i., gi"cn hy: I'Rd, = [o. I2k( lOOpJ.~)
1
'+ O.ISa,"• b,.d
\\here:
cl :: 1.5
(1 -1
1.
e
0.85
~2~))
= "'fq,KPn/11
= 0 95 m at the ~uppurt
~ f:o) = 1.46
(1
3450 150 X 950
rr,p
0.85 + 0.3
1.5
(::; 2.0)
( > 0.02) ... PI - O.D2
0.0242
- 0.9 x 2590 x 10 1/(500 x 101 )
4.66 N/mm 2
{5
0.133.f.·L = 0.133
X
35
4.66 OK}
lienee:
VRLI c
[o. I2k( IOOpt/~k) l/l + 0. I5crrp] b,..d [0.12
X
1.46(100
X
0.02
X
35)
111
-j
0.15
X
4.66] 150
X
950
X
JO-J
202kN Noll!: a check on equation 11.41 will show that the minimum value of VKd, as given by
equation I 1.41 1., not cruical in this case. As thb is u \imply supponed beam equation 11.42 should also he used to check the l>hear capac it) of the concrete ~ection. From equation 11.42: b~J
~
VMt~c= AJ vU~id+nlacJc•d
)
36:
366
Reinforced concrete design
where:
= axial stress in section due to prc:-.tress = 4.66 Nlmm2• as before
rr,p
= the design
tensile strength of the concrete = 2.2/ 1.5 o 1 = I for post-tensioned tendonl>.
.f.ld
= 1.47 N/mm2
Hence by reference to the dimensions shown in figure 11.25: bwf . j ,
VRd .c
= 1\y V (fc~d +
(I
I rTq/tld)
150 X 145 106 X 106 f. , = ~;{-IO_OO_x_l-75-x-56-2-.5-)_t_( 150 x 475 x 237.5 )) V ( l.47- + I x 4·66 x 1.4?
-3 X
lO .
S66kN This is considerably greater than the figure of 202 k~ calculated from equation II .40. We will take the lower value of202 kN as representing the shear capacity of the concrete section The effective resistance of the section is the sum of the shear resi~tnncc of the concrete. VRd , . plus that of the ,·crlical 'hear rclobtance of the inclined tendons. Shear strength Including the shear resistance of the inclined tendons
The vertical component of the prestress force is P sin 11 where r3 tendon slope. The tendon profile" y - c,.:! with the origin of the cable prohlc taken at mid-lipan; hence at .1 15000. ~ 750 - 3CXl 450 and
c x 15 ooo2
,150
c
2.0 . . . 10 (, 2.0 x 10· 11 rl and tendon :.lope
Therefore the tendon profile '' .'
dy/dt
2C\ .
At end
dyj dx
= 2 X 2,() X
10· <>
X
15 000
= ().()60 -
tan i
lienee, ,J
3.43
und 'in ;3::::: tan ii
= 0.06
Therefore vertind component, V1• of prc~trcs~ force 111 259!1:-in i = 2590 x 0.06 1551..1'\
at
the suppo11s
i~:
and the total shear capacit) i-,: VRd. c + V1
202 I / p X 155 202 + 0.9 142 kN at the support!-..
X
155
At the end ot the beam the design ~hear rorce is (40 >< 10/ 2) = 6001.:-1 and hence the shear capacity uf the concrete &ection i~ inudequatc and ~hear reinforcement mu~t be provided.
(3) Check the crushing strength
VRd max
of the concrete diagonal strut
A check must be made to cnwrc that the !-hear force doc:-. not cause excessive cnmpression to develop in the diagonal strut:. of the assumed Lrus~. From equation 11 .4-l (COle - 2.5):
VRu mn.t1 2~
= U<',O.I24b,.t/{l
/.k / 250lf~~
where the value of n,w depends on the magnitude of I'Tc~ given by: rTql
As the ~hear force at the end of the beam is 600 kN then the upper li mir to the shear force i~ not exceeded. (4) Calculate the area and spacing of links
Where the shear force exceeds the capacity of the concrete section, allowing for the enhancement from the inclined tendon force. shear reinforcement must he provide to resist the net shear force taking into account the beneficial effect of the inclined tendons. From equation I I A~ thi~> is given by:
A,w S
~ = (600 - 0.9 X 155) X 103 1.95dfyk 1.95 X 950 X 500
= 0.497
(5) Calculate the minimum link requirement
2
0.08fc~ b" = 0.08 X 35 1 ~ J..... 500
A,"•'"'" -
}
150 = O.l 4
X
Therefore provide I 0 mm links at 300 mm centres rcsi,tance of the linb actually ~pecified is: A,. Vnun -
I
(I\ ,..
h
0.523) \Uch that the 'hear
X 0.18dfy~ COL 0
0.523
X
0.7!!
X
950
X
500
X
2.5
X
10
J
4X4 kN
(6) Calculate the additional longitudinal force
The additional longitudinal tensile force is:
AF1d
0.5Vt:ucot()
0.5 x (600
0.9 x 155) x 2.5
575 kN
Hence:
575 X 101 0.87 X 500
- - - - = 1322 mm
1
Thi!. additional longitudinal steel can be provided for by four untensioned II25 burs (1960 mm 2) located at the bottom of the beam's cross-seclion and fully anchored pa~t the point required using hooks and bends as nece~sary. Umen ioned longitudinal reinforcement mu~t be provided at every cross-~ection to resist the longitudinal ten~ile force due to ~hear and the above calculation must be repeated at each :.ection to determine the longitudinal ~teel requirement. All of the above calculations can be repeated ar other cross-section)> and are tabulated in table 11 .2 from wh1ch it can be seen that, from mid-span to n liection approximately 9 111 from mid-1.pan, nominal shear reinforcement is rcquin.:d and in the outer 6 m of the !-.pan fully designed shear reinforcement is required. This can he provided a!.
367
368
Reinforced concrete design
Table 11 .2
Shear calculations at 3m intervals
Prestress
Mid-span
0 3 6
9
12 End-span 15
v,
(1)
(3)
(2)
(4) A,wfs
(kN) 1
(kN)
(kN)
(kN)
""'~ (kNi
1400 1382 1328
281 278 270
0 28 56
0 120 240
281 306 325
941 928 892
Minimum reinforcement only
1238 1112 950
255 234 202
84 112 140
360 480 600
339 346 342
832 747 638
Reinforcement 0.229 carries all the 0.339 shear force 0.497
D (m) (mm) X
'Yp
VRd.c
VEd
VRd c
~Pvt
VRd
(S) Asw/S m1111
(6) ~Ftd
(kN)
- -0 0.14 0.14 0.14
115 230 345 460 575
1 Equation 11 .40. 2 Equation 11.44.
Figure 11.28 Shear resistance diagram
z
.. 2 .."'
Ultimate sheM force
600
~
...c
~ ~
v,,.
Concrete plus tendon ~hear resi5Lance
t:
~
400
..c
:::?
V'l
200
0
Q.
:2
Q.
::>
V'l
' 6
9
12
15
Distance alonq IPMI (m)
(figure 11.28) 10 mm link::. at 300 mm centres in the outer 3 metres (A,..., / .1 0.523) changing to I 0 mm link' Jt 450 mm centres (A "/1 - 0 ~49) between 3 and 6 m from the end of the heam and then 8 mm at -150 centres (A ,,J 1 0 223) throughout the rc~t of the ::.pan.
CHAPTER
12
Composite construction CHAPTER INTRODUCTION
)
Many buildings are constructed with a steel framework composed of steel beams and steel columns but mostly with a concrete floor slab. A much sliHcr and stronger structure can be ach1eved by ensuring that the steel beams and concrete slabs act together as composite and so, effectively, monolithic units. This composite behaviour is obtained by providing shear connections at the interface between the steel beilm and the concrete slab as shown in figure 12.1. These shear connect1ons reSISt the horizontal shear at the interface and prevent slippage between the beam and the slab. The shear connectors are usually in the form of steel studs welded to the top flange of the bec~m and embedded in the concrete slab. The steel beam will usually be a universal 1-be.:~m. Other .:~ ltern
369
370
Reinforced concrete design
Figure 12.1 Composite beam sections
Shear stud connectors
Ribs parallel to beam Ribs perpendicular to beam Composite beam with ribbed slab
Figure 12.2 Composite floor beams
t~www~.·p ']~ ~ ~~ k ~ '
(a) Composite la ttice gird er
~~J (b) Compos1te castellated beam
Secondary \teet be.1m
figure 12.3 Compos1te floor systems
Steel beam (a) Typical sub-girder system
Steel beom Concrete slab Profiled sheeting
(b) Compos1te profiled slim deck system
Composite construction
The Sllmdec system shown in Figure 12.3 is manufactured by Corus. The special steel beams have a patterned tread on the top flange that provides an enhanced bond with the concrete slab so that a composite act1on can be developed without the use of shear studs. Deep ribbed profiled sheeting 1s used to support the slab with the deep ribs resting on the bottom flange of the beam. With this arrangement the steel beam is partially encased by the concrete which provides it with better fire resistance. Openings for services can be cut in the web of the beam between the concrete ribs. The concrete slab itself can also be constructed as a composite member using Lhe profiled steel decking on the soffit of the slab as shown in figure 12.4. The steel decking acts as the tension reinforcement for the slab and also as permanent shuttering capable of supporting the weight of the wet concrete. It is fabricated with ribs and slots to form a key and bond with the concrete. Properties of the steel decking and safe load tables for the decking and the compos1te floors are obtainable from the manufacturing companies. Many composite beams are designed as simply supported non-continuous beams. Beams that are continuous require moment resisting connections at the columns and additional reinforcing bars in the slab over the support. The method of construction may be either: • Propped • Unpropped With propped construction temporary props are placed under the steel beam dunng construction of the floor and the props carry all the construction loads. After the concrete has hardened the props are removed and then the loads are supported by the composite beam. The use of temporary props has the disadvantage of the lack of clear space under the floor during construction and the extra cost of longer construction t1mes. Unpropped construction requires that the steel beam itself must support the construction loads and the steel beam has to be designed for this condition, which may govern the size of beam required. The beam can only act as a composite section when the concrete in the slab has hardened. This also means that the deflection at setvice is greater than that of a propped beam as the final deflection is the sum of the deflection of the steel beam during construction plus the deflection of the composite section due to the additional loading that Lakes place after construction. The calculalions for this are shown In example 12.4 which ~ets out lhe serviceability checks for an unpropped beam. As there are differences in the design procedures for Lhese two types of construction it is important that the construction method should be established at the outset.
Shear studs -
___
-_---------
Profile steel decking
figure 12.4 Compos1te slab with steel deck1ng
371
372
Reinforced concrete design
12.1
The design procedure
The design procedure for composite beams follows the requirements of: (a) EC2. (E:\ 1992-1-1) for lhe design of concrete structures. (o)
EC3 (E:\ 1993-1-1) for the design of Mecl!ltructures, and
(c) EC4 (E:-.=1994-1-1) for the dc!.ign of compo:.itc Mecl and concrete structures. At the tune of writing this chapter the CK National Annex for EC3 and EC4, and the Concise Eurocodes are not available. Parh of these code~ are quite complex: for example the li~>l of symbols for the three codes extends to 21 pages. Tt is intended in this chapter to try and simplify many of the complications and enable the reader to gain a grasp of the basic principles of the design or composite heams.
12.1.1
Effective width of the concrete flang e (EC4, cl 5.4.1.2)
An early step in the design of the composite benm section is to determine the effective breatlth bcrr or the concrete flange. For builtling structures at mid-span or an internal support
bru
= L:b.,
where be~ i!> the effective width of the concrete flange on each side of the steel web and i'> taken as 4 /8. but not greater than half the dist:lnce to the centre of the adjacent beam. The length 1-t is the approximate di!-.tance ~tween points of cero bending moment "hich can be taken as L/ 2 for the mid-.,pan of a continuou:. hcam, or L for a one-span simply supported beam. The length L is the \pun of the hcam being conl>idered. For example. for a continuous hcam with a span oft 16m and the adjacent beams being at 5 m centre to centre the effective hrcatlth. hd1 , of the concrete flange is
bell
2 >. Lc/8 = 2 X 0.5 x 16/ 8
If the beam be 4.0 m.
wa.~
a nne-span
~imply
2.0 m supported beam the effective breadth. bttt. would
12.1.2 The principal stages in the design These
stage~
arc listed with brief description~
u~ fo ll ow~>:
(1) Preliminary sizing
The depth of a universal steel beam mny be taken as approximately the ~>pan/20 for a simply l'lupported span and the span/24 for a continuous heum. The yield strength. j~. and the section classification of the steel heam ~houltl he determined. (2) During construction (for unpropped construction only)
The loading i:-. taken ac; the self-we•ght of the steel beam with any 'buttering or ~lecl deckmg. the \\eight of the wet concrete and an unposed con~truction load of at lea1.t 0.75 kN/m~. The following design checkc; are reqUired: (a) At the ultimate limit state
Check the strength of the steel !>ection in bending untl 'hear.
Composite construction
(b) At the serviceability limit state Check the deflection of the ~tccl beam.
(3) Bending and shear of the composite section at the ultimate limit state Check the ultimate moment of rcsio;tanee of the composite section and compare 1t v. ith the ultimate design moment. Check the shear strength of the ~tccl beam.
(4) Design of the shear connectors and the transverse steel at the ultimate limit state The shear connecters are required to re~>iM the horizontal shear at the interface of the ~>teel and the concrete so !hut the steel beam and the concrete llange act as u cnmposih.: unit. TI1e shear connectors can be either a full shear connection or a partial f>hcur connection depending on the design and dewiling requirements. Tmnwerse reinforcement is required to re~>i:-.t the longitudinal 'hear in the concrete flange and to prevent cracking of the concrete in the reg1on of the ~hear connectors.
(5) Bending and deflection at the serviceability limit state for the composite beam The deflection of the heam i~> chcc"-ed to ensure it is not exces!.IVC and so cau:-.ing crac"-ing of the architectural hni\he~.
12.2
Design of the steel beam for conditions during construction (for unpropped beams only)
The steel beam muM be des1gned to support a dead load of its e~timated self-\\ eight. the weight of wet concrete ond the weight of the proflled ~teel dec"-ing or the formwork. plus a construction li ve loud or til leu:-.t 0.75 k.N/m 2 covering the lloor urt:a. A preliminary depth for the siling of the Meel beam can be raken m. the '-pan/ 20 for a one-::.pun simply ~.upported heam.
(a) At the ultimate limit state (i) Bending The plastic section modulu~ Wpt y· for the steel beam may he calculated from
IVpt.y
= MrcJ r Jy
( 12 I)
where MP.t i~ the ultimate de~rgn moment
J;
i~ the dc,ign strength of the steel as obtained from EC3. tahlc
3.1
1l1is as.,ume!. that the compre1.sion flange of the steel beam is adequatel> n!strallled against bud.. ling by the steel decktng for the ~lab and the !.tccl sect1on used can be classified a~ a plastic or compact section a~ defined in EC3. sections 5.5 and 5.6.
· 373
374
Reinforced concrete design
(ii) Shear The <>hear iJ. considered to he carried by the steel beam alone at the con~Lruction stage and also for the final composite beam. The ultimate shear strength of a rolled 1-beam is based on the following shear area. A,. of the section (122)
where Aa is the cros!.-scctional area of the ~tecl hcam and h.,. is the overall depth of the web. IJ can he taken as 1.0. The other dimensions of the cross-section arc detined 1n figure 12.5. Figure 12.5 Dimensions for
1-sectlon beam
1-I ~
an
b"----'
-
h.,
fw
d
'
-
/'
'
r''
r " radius ol roof f1llet
For class I and cla-.r. 2 1-beams with a predominately uniformly di~trihutcd load the design shear ~tre~se:-. arc seldom ex~o:c),~ivc and the shear area. A, may be con~c rvatively tal.cn a<; the \\Ch area so that ( 12.3)
where d i:-. the depth of the \tntight port10n of the \\Ch. The dc,ign plastic shear re~btance \ 'r1 Rd of the ~cction is given by: A,f> VpiRd
\\hCrC
') \10
-
( 12.4)
-
'i\loJ3
1.0 is the material parti
1->lCCI.
(b) At the serviceability limit state The deflection f, at m1d-span for a uniformly by:
di~trihutcd
loud on n steel beam is given
( 12.5) where
w is the ~crviceability load per metre at conMruetion LIS the beam's span Ea is the cla.... tic moduluo; of the <.tee] = 210 k\l/mm 1 1. is the second moment of area of the steel ~ecuon fhc deflections at the consLruction -.tagc due to the permanent loads arc locked into the beam a'> the concrete hardens.
Composite construction
375
I
(EXAMPLE 12. 1
Design of steel beam for construction loads Figure 12.6 show!> the ~ection of an unpropped composite beam. Check the ~trcngth of the universal 457 191 x 74 kg/m steel beam for the loading at con~truction. The Mccl •s grade S355 with ./~ = 355 N/mm 2 and the plastic modulus for the ~tee! section i::. Wr1 ~ 1653 em 1• Th~ one-span simply supponed beam spans 9.0 metre and the width of loading on the concrete flange i~ 3.0 metres. Profiled steel decking
Figure 12.6 Comlruclion design example
J
457
X
191
X
74
Universdl Be~m
' I
Section
Span L
9 Om
....
Elevation
(a) Steel strength and classrfication of the steel beam (see EC3, tables 3, 7 and 5.2)
The web thid.nel>~. t.., = 9 0 mm and tJte flange tluckne~~. t1 I t4 mm. and both arc h!\l> than 40 mm. Therefore from EC3. l>CCtion 3.2. table 3.1 the yield -.trength. f> 355 N/mm 2 . From EC3, sccuon 5.6, table 5.2 E
(235
O.R I
Vh t!
407.6
lv.
l) .0
45.3
< 72
X f -
58.3
therefore the steel ~cction i:. class !. (b) Loading at construction
and rib!>= 90 + 50/ 2 = 115 mm 0.115 x 25 x 3 8.62 J..N/m
Average depth of concrete Weight of concrete
~l ab
Steel deck
= 0.15 x 3
Stee I beam
= 74 X 9.8 1 X
Tmal dead load
= 0.-15 ( ()-)
= 0.7] = 9.8kN/m
Jmpo~ed con~truction
Ultimate load
load = 0.75 x 3 = 2.25 kN/m U5Gk I 1.5QL = (1.35 x 9.8 + 1.5 x 2.25) =
16.6 k '/merre
376
Reinforced concrete design (c) Bending
= (16.6 x 9~) /8 = 168 kN m Moment of re~istance of Meel section = Wpl J, 1653 x 355 x 10 = 587kNm > 168kNm OK Maxunum bending moment = wL~ /8
3
(d) 5heor
Maximum
~>hear
force V = lrL/2- 16.6 x 9/1
Shear Cl!sbtancc of M:clion -
vpl Rd -
74.7 kl'\
AJ)!'> l\10 v 3
For the steel ~>eclion tile web depth. d = 407.6 mm and the weh thic\..ness U),ing the conservative value of dt.,.,
1\,
Shenr
407.6 x 9.0
rc&i~tnnce
of section
1
= 9 mm.
3.67 x 101 mm 1
= Vpl. Rd
1\ v.f~
"/Mo/3
3.67
X
101
X
355
_,
- - --;:::-- " I 0 ·'
1.0
7'i2 I..N
X
/3
74.7 kl\1
hom the calculation!- for bending and shear il can he seen lhe loading on the beam during l'OINrul·tion is rdati\'cly IO\\ wmpan:d to the '-lrength of the beam. Abo, U1c 'reel th:ckmg "uh the COJTuga!lon~ Jl right angk' to I he 'pan g1vcs lateral and torsional re,lr:Jilll to the 'teeI heam. For the~e n!n,on' it i' con,idcn:d unnecc,sary to carry out tile Jmnht.>d calculatinn' for lateral and tor,ional stabllit} which arl! descnhcd in EC3. De,1gn ol ()reel Structure' I or the calculation of the deftccuon of the ~reel be;1m during construcllon at the M!I'\ICCahllll) hmll state see Example 12.4.
At the ultimate limit state it h. necc11sllry lll check the compo~ite section tor its moment cupacily and its shear strength, and compurc them aguin~t the maximum design ullimate moment and shear.
12.3.1 Th~:
Moment capacity with full shear connection
moment capacity M, of the composite sed ion i ~ derived in terms of the tensile or of the variou~ clement~ of the section as follows:
the concrete flange the steel sccti(m the steel flange overall web depth clear weh depth the concrete above the neutral :ms the ~teeI flange above the neutral axts the \\eh O\'er dl'>tance X•
R" R, R,1 R"
0.56~f,lhcrr(hf~A·
J; btr R\
2R,r
R, J.,dt., R., 0.567/.kbcnx R x -/>b.\ 1 R.,.., =f-,tv."<1
The dimensions used tn these expressiOn!!> are defined in ligures 12.7 and 12.8.
h11 )
n Composite construction b.t~
,-h
I
j_ h
'hp
37i
Figure 12.7 Composite section dimension
J -
-, c,
Note: dis the distance between the fillets of the steel section
ft is important to note in the figures Lhm the SlrC!-!> hlock for the concrete extends to the depth of the neutrnJ nxis as specified in EC4 for compo~>ite design. There ttre three pos~ible locations or the neutral axis a~ shown in figure 12.8. These are:
(a) The neutrul axis in the concrete flange: (b) The neutral axis in the steel t1ange: (c)
The nculrttl uxis in the steel web.
Figure 12.8 Stress blocks at the ultimate limtt state
A
(a) Neutral axis In the concrete flange X< h: Rc~> R,
b,,
c, '
,, •
!!I
.... b ...1
(b) Neutral axis In the steel flange II < x < h
1
t, and R, > Rtt > Rw
(,
d
It
(c) Neutral axis In the steel web x > h + t,: R.:t < R.,.
378
Reinforced concrete design
The location of the neutral axis is determined from the equilibrium equation of the forces R at the section.
rel>i~tance
'E,R = 0
i.e.
The moment of resistance at the section is then obtained by taking moments about a convenient axis such as the centreline of the ~teel section. 1>0 that where : is the lever arm about a chosen a xi'\ for the resistance R. For Calle~ (bland (c) the analysis i' facilitated by considering an equivalent system of the re~istaoce forces a shown in the relevant diagram~. (a) Neutral axis in the concrete flange, x
< h Figure
12.8(a)
This condition occur" when Rcr > R,. Then the depth x of the neutral axi1. is given by Rex = 0. 567]~kVettX- Rs . R, R$(11 - hp) Therefore x ::.. ," b 1?d 0 ·567J~~ eff The moment of resistance is
= R,:.
M,
where the lever arm (11.12
~
i\
+"- ,,2)
Therefore ( 12.6) (b) Neutral axis in the steel flange, h < x
This condition occurs when Rs > Rd
<
h + t1 (Figure 12.8(b))
> R..,.
l·or the equilibrium of the rcsiMancc forces Ret I 2/?,,
2R,x
i.e. and
Xt
where b i!. the breadth and
ft
=
1?,
= 2Jyi)),1
1?,
(R,- R, t )
I? cf
(!?,
~{yb
l?d)lt
21?,,
is the thickness of the
~tee t
nange.
The moment of resistance is given by M,
R.r:.t
+ 2 Rs~Z~
l?d::.t
+ (R\- R,r):.1
,.. here :. 1 and :.2 are the tc.. cr arms as shown :1 ;:z
(h4-" + hp)/2 :rt)/ 2
= (ha
tn
figure 12.R(b) and
Composite construction ;he
Therefore substituting for z1• z~ and
x1 and
37
rearranging
( 12.7)
ow a (c) Neutral axis in the steel web, x
of
> h + t1 ( Figure 72.8(c))
Thi\ condition occurs when Rcr < R-.. and is mostly associated \\tth hutlt-up beam !>Cctions wtth small top flanges and larger bottom flanges and also with ~tiffcncd weh~ to avoid wch buckling. For equilibrium of the equivalent arrangement of the rcsi~tance force~
2R-..x
= 2/y lwX2 = Rcr
Therefore
where x2 is the distance between the neutral axil> and the centreline of the steel section . The moment of re~ismnce of the composite ~ection is the moment of the two couples produc.:cd hy R, and Rc~ with 2Rwx so that
Me
\4 I
'
-f
"
R.r(h1
hp
X2)
2
or
A{
I
A' -~o
Rc (h.
+ h + 1111 ) R~rd 2
( 12.8)
4R,
':!6
(EXAMPLE 12.2 Moment of resistance of a composite section Determine the moment or resistance or the tompo).itc scctinn shown in Jigure 12.9. The universal 457 x 19 1 x 74kg/m steel beam has a cross-sectional area ()fAn 94.6em 2 and is grade S355 t'. tcel with .fy 355 N/mm2• Usc concrete clnss C25/30 with characteristic cylinder strength J:k 25 Nlmm 2.
=
btl1 :
0.5671,,
3000
Figure 12.9
Moment of resistance examp
h,- 50
h.= 457
py
457 x 191 x 74 UB
Section
Stress Blocks
380
Reinforced concrete design (a) From first principles Resi~tancc
0.567/..~berr(h
of concrete flange Ret -
hp )
= 0.567 X 25 X 3000 X ( 140- 50) X 10 J = 3827 kN Resistance of Meel beam
R~
/yAu
355
X
9460 x 10- 3 = 3358 kN
As R, < R., the neutral axis is within the concrete flange. Determine the depth of neutral axis: 0.567AkbeffX = R, Therefore
3358 1...1\
3358 X 101 x= 0.567 X 25 X 3()()()
79.0 mm
Moment of rcsi~Lance: Lever mm z to the centre of the steel section is (/lu /2 I h ~/2) = 457/ 2 -f 140 - 79.0/2 M,
R,z
1
3358 x 329 x 10
-
329 mm
1105 kN m
(b) Alternatively using the design equat1ons derived
From pari (a) the neutral axi'> i~ \\ Hhlll the concrete flange therefore. from equation the moment ot resi~tance of the o;ect1on 11- g1vcn by M,
1~ .6.
1?,{';" .. h - :c't (h ~ hr) } BSR{457 140 _ 3358 ( 140 50)} IO 3 = ( 5 k 3827 2 I l J Nm ... 2 t
12.3.2 The shear strength
VRd
of the composite section
f·or the compoc;ite section. as for the con~tmction ~tage. secunn 12.2a(u). the shear is rclliMcd by the shear area A, of the ~teel beam and the shear resistuncc VR.J ic; given by A..(y
VRll
Vpi , Rd
= ----r,; '}MO V 3
(12.9)
and the !\hear (!rca A,. is given hy ( 12.10)
where A1 ll. the cross-sectional area of the steel beam and the othe1 dimension:\ of the cro!ls secuon are defined in figure 12.5. For class I and cia.,~ 2 I-beams with a predominately uniformly dt!>tributed load the des1gn shear stress is seldom exces.,ive and the !.hear area, A,. may be safely taken conservattvely as the area of the web. 1\. - d
X fw
For heams where high shear force:- and moments occur at the same ~eel ion such Uu11 > 0.5VRd it is necessary LO usc a reduced moment capacity for the composite section by reducing the bending stress in the steel weh a~ de~cribed io EC4, c;ection 6.2.2.4. VFAI
Composite construction
12.4
381
Design of shear connectors
The shear connectors are required to prevent slippage between the concrete nange and the Mcel beam thus enabling the concrete and steel to act a' a composne unil. Stud 'hear connector~; welded to the steel nange are the most corrunon t} pc u'ed. The head on the <;tud acts to prevent the vertical lifting or pri'>ing of the concrete a\\:.t) lrom the \teet beam. Figure 12.1 O(a) shows the slippage that occur'> '' tthout 'hear connector.... The "li ppage is a maximum at the ~up ported end or the beam ''here the shear I' and the rate of change or moment dM / d1 are a maximum. The 11lippage reduce:. to tcro at mtd-span where the moment is a maximum and shear V = 0 for a uniformly tlistrihutcd load The connectors restrain the slippage by resisting the horizontal shear at the interface of the concrete nnd the ~tccl. The design is carried out for the conditions at the ultimate limit state. The design shear resistance, PHu· of a hended stud automatically welded i~ given hy I ~C4 as the lesser value of the following two equation;,: plld :
0.&/~ 1Td~ /4
(12.11 a)
or:
0.29nd~ ../J..J.Ecm
n-1
)
for ll..,j d
( 12. 11h)
>~
where -,, d
j;,
t~o.
the ptU1tnl safety factor
1.25
the diameter of the shun~ of the stud, between 16 mm and 25 mm 1s the ullimutc tensile ~o.trcngth of the stud 500 t'oilmm 2 or 450 N/mm, when equation 12.12b npplie~ 11o.
fck is the cylinder c:haraclcrislic compressive strength of the concrete £,111 is the
~cctmt
modu lus of eluMiciry of the concrete. sec table 1.1
A further reduction factor. k1 or k, is w he applied to PKtl ~~~ ~rcctticd in EC'4, section 6.6.4 and its value depend), on whether the ribs of the profiled ~heetulg arc parallel or tranwer~e to the supporting beam.
Horizontal shear
-.
/.
(a) Slippage
(b) Horizontal shear at beam-slab interface
figure 12.10 Slippage <1nd horizontal shear
382
Reinforced concrete design
(i) For ribs parallel to the supporting beam l..t
= 0.6 bo ( 11'"' hp
lip
~
1)
( 12.l2a)
1.0
(ii) For ribs transverse to the supporting beam
/.. _ !:.:!.._ bo I -
yll,hp
(h"'/tp _1)
( 12.l2b)
n, is rhe number of stud connectors in one nh at a beam connection, < 2 /1 5, is the overall nominal height of the srud hp is the overnll depth of the pronicd steel ~heeting bo is the mean width of the concrete rib
Al~o in this ~econd case the ulumare tensile -.tcel stress of the studs};,
< 450 N/mm2 •
There is an upper limit k1,,11111 for the retluction factor k, which is given in EC4, sc<.:tion 6.6.4, table 6.2.
12.4.1
Full shear connection
The change in hori7ontal shear between zero and maximum moment is the Jesser of the rcsi~ranct: R, of the ~ted ~ection und R. the resistance of the con~.:n:tc 11ange. Thus. to ue"elop tht: full hcnuing \trength of the composite section the number of shear connector.. 111 required O\'er half the ~pan r~ the lc~ser of
R, II( = -
k,Prhl
R \ll
llr -
k,PKd
( 12.13)
!'or a full ~hear connection. where P1<.t rs the effectrve '>tren!•th of a ...hear ... tutl k1 is the reduction tactor applied ro the characteristic Mrcngth PRu
Rc
0.567./',;kb,.lr(h
R,
.fA1
lip)
hrll scale tcsh \\ 1th unilom1ly distributed loading have shown thm "llh plu~tic ~.:ondition~ during the ultimate limit ~tate the ~>hear stutl-. can develop their full strength
when spaced uniformly ulong the ~ran of the hetun.
12.4.2 Partial shear connection In ~ome callc~ it i~ not necessary to have a full -.hear connection in order to resist an ultimate dc~ign momenr that is somewhat less than the ful l moment capacity of the composite r.ect ton. Also. u~ing fewer shear stutb can often provide a simpler detai l for the layout of the '>tud connector!>. For panial !.hear connection the degree of '>hear connection tJIS defined ac;
II
II
( 12.14)
llj
where 111 is the numher of shear connectors for full shear connection over a length of a hcam and 11 b the number ot shear connector~ prO\•ided 10 that length.
Composite construction
2
l~a)
EC4 provides limits to lhe degree of shear connection TJ hy two alternative equations accor<.ling to the distance L, for steel sections with equal flange .
1. 1l1e nominal diameter d of lhe shank of the headed stu<.l i~ within the runge: I6 mm $ d < 25 rnm and lhe overall length of the !>tltd after welding i'> ;::: 4d: ~
1
1.~ < 25
2b)
TJ ;: : I-(~ 5 ) (0.75-0.034)
11
2. The nominal diameter d of the shank or the headed ~tu<.l ic; d overall length of the stud after welding ;::: 76 mm:
< 25
1-'t
T
17 ;::: I - ( 355) (1.00 - 0.044)
1)
( 12.15a)
0.4
> 0.4
= 19 mm
and the
( 12. 15h)
where 4 i ~> the distance in sagging between the pomt:, or t.cro moment in mctre~>.ln both cases, where /,~ is greater than 25 m, the factor ~houlcl be greater than I . There urc also a number of other conditiont-. us listed in EC4 section 6.6. 1.2. The ultimate moment resistance of lhe composite section with rm1ial shear connection is delive<.l from lhe nnaly~is of the stress bloc!-. system)) shown in figure 12. 11. In the annlysi~ the depth of the concrete stre~~ blod. 1· i\ 4 fthe
_
,..to
0.56~f.:kb,.u
shi:ar
RQ
where Rq i!. the ~hear resistance of the '>hear <;tud~ prm ided. A'> pre\'iOU\Iy o,hown in ection 12.3 lhc depth of the 'ecuon·, neutral aX I\ I\ obtamed hy con'>idering the cquihhnum of the material remtanceo, R. The moment of re~t\tance M, i-; obtained by laking moment!'. about a con,enient aw. 'uch a' the centreline olthc 0.5671,,
IJ h,
j
I
d
't
I,
~ b -1
p.,
(a) Neutral axis in the sLeel flange h < x < h + 11: R, > Rq > R,
b,,,
(b) Neutral axis in the steel web x > h + t, : Rc < R.,
Figure 12.11 Stress blocks for part1al sh connect1on
384
Reinforced concrete design steel section. followed by some rearrangement of the equations. The diagrams for this analyo;is are shown in figure 12.11 for the two pos~ih1e ca~cs of: the neutral axis in the steel flange R4
(a)
RJ1a Me = - 2
-llr)] + Rq lrlI - Rq (ll2Rct -
> R... (R, - Rq )~tc -4R,I
--
(12.16a)
the neutral axis in the steel \\eb Rq < R,.
(b)
Me = M, -1
R q
["a2 -
I1
Rq (ll- lip)] -
2/?d
dR~
- 4R,
(12.16b)
Figure 12.12 ~haws the interaction diagram for the moment of resistance of the composite ~ection against the degree of shear connection 11 where II
'I =-
{equation 12.14)
llj
The curved interaction line (a) is based on the Mress blod. equations of 12.16a and 12.16b whil:h give the more precise resu lt'>. The ~traight interaction line (b) represen ts a lincnr relation between the moment capacity anti 11 which provide:- n simpler and safer hut Jess economic ~o lutinn.
12.4.3
Shear connection for concentrated loads
When the beam !-.Lipp011~ concentrated load~ the \lope dM jdx of the bending moment i!> greater and the "hear is more inten<.e. Thi" mean'> that the shear eonnecto~ have to be 'paced clo,er together between the concentrated load and the adjacent support. The d1:-.tnbuuon of the shear connectors l'i then ~pcc1 hed hy the equation.
N,
N,(M, M,) (Ale- M, )
( 12.17)
where
N, is the number of <;hear connector'> between the conccmrutcd load and the adjacent support N, "the total number of '>hear connector~ required between the support and the poinr of maximum moment (M 01,") M, IS the bending moment at the eonccnlrmed load M, is the moment capacity of the steel member M, is the moment capacity of the composi te section Figure 12.13 shows n beam !.upporting conccntraleu loncls and the distribution of the ~hear connector'>. Fully composite sectoon M,
Figure 12. 12
(a) Stress Block Method
Interaction diagram for partial shear connection
(b) Lonear Interaction Method
Steelloection M,
0
1.0 0.4 Degree or shear connection 11
Composite construction N I -1 NzLI -L - 1
---
, ~- ' ]
'
w.
N, (L
--- -L 1 W1
2-
L) 1
t
Figure 12.13 Distribution of shear connectors with concentr; loads
6a
~ Mz
B.M Oiagrdm (all loads including thE' pomt load~ shown)
M.,,.,
6h
lhe 1:!)
:md
t, a
12.5
Transverse reinforcement in the concrete flange
Transverse reinforcement is required to re1.iM the longitudinal sheur in the concrete Range. Thb shear ucts on vertical planes either side of the shear connectors as shown in figure 12.14.
er
fransversc reinforcement
b
Potential failure planes
be The
.nt
Figure 12.14 Transverse reinforcement l the concrete flanges
b
Ribs parallel to beam
Ribs perpendicular to beam
Tile analy'>i\ uno de!>ign for the trallS\'Cr!>C rcinron:cment to rt.!\i\1 the longiludma) shear in a flanged hcum follo\\s the variable strut inclination method all required 111 PC:! and dellcrihcd in thi' hooJ.. in ~ections 5.1.4 anti 7.4. in COilJI.IIlt:lion with C\Hmple 7.5
part (2). nt
F...C2 ~peci lie' a mini mum of tranwerse ~teet area equal to (0.1311 1 > l 000/ I 00) mm 1 per metre Width. The method of de~igning the tranwer~e steel for u composite beam is ~>hown in example 12.3 part (b) Tra11sverse reinforcellll!l/1.
(EXAMPL E 1 2.3 Shear coonectors and transverse reinforcement The compo~>itc beam of example 12.2 and ligurc 12.9 srans 9.0 m..:1rcs ant! ill provided with 80 ~hear stutl connectors in pair~ at 225 mm ccntn:s. The stud' arc 19 mm tliumcter and of IOOmm hc1ght. The plal>lic -.cction of modulus of the steel ~ection is IV111 ) 1653 N/mm2 ant! the 1 tlestgn 'trc\~ of the \teet. J., - 355 "lmm • The characteristic material strengths are /..-L 251'\/mm, for the con~rctc and frk 500 Nlmm 1 for the rem forcing bal'\.
=
=
(a)
Calculate the degree of shear rcsiswncc and the mom..:nt of rcsi,tance ot the compo,ite heam based upon the l>hear conncctorll proVIded
(b)
Dcllign the transverse reinforcement required to rcsi~t the tranwcrsc llhear in the concrete tlange.
386
Reinforced concrete design (a) Degree of shear connection and moment of resistance
The design :.hear resio;tance. PRd· of each shear 'tud i~ the le~ser value obtained from equations 12. ll a and 12.llb wilh/0 • the ulttmate ten~lle \trcngth of the steel. equal to 450N/mm~. Us~ng
these equations it is found from equauon 12.1I a th
P Rd =-~--=
0.8
X
45()
,,
X ii X
.., 1.- 5
192 / 4
x
w- 1
Sl.7k
A reduction tactor. k1• is calculated from equation 1:!. 12b with an·uppcr limit taken from table 6.2 of EC4. from equatton 12.12b. with reference to EC4. figure 6.13 and taking dimension h11 80 mm for the profiled ~teet sheeung
k,
= ~ ho (11,.,.. _ Jiir111,
flp
The upper limit of k,
t) = 0.7J2
x HO ( 100
50
50
1)
0 _79
= 0.8 !"rom EC4 tuhle 6.2.
lienee the design !.hear resistance,
P~ct. of
a stud
0.79 x 8 1.7
64.5 kN.
For full shear w nnection the number of <;t ud~ required over half the spnn 111
i~
R, = 3358 = 52
64.5
pll.d
CR, = i )-A. 3358k.i'\, i~ the resi~tance of the stee l beam as obta1ned from example 12.2.> Hence tor full ~hear connection the total numhcr ol .,tud . . rl.!qUtred over the whole 'pan I o.t The degree of shear connection. 'I· i<. 80
0.77
IJ- J{)4
I he lower lim1t lor q i~ calculated from equation 12.15a IJ ?:;.
I-
355) . (0.75 - 0.031...: ) ( ./)
355 ( 355 ) (0.75
:.1\
0.03
X
9)
0.52 < 0.77 OK
Th~
moment. of the rc!>iSLance A·1r of the compo~i t e heum hascd on the panial ~hear n:sistancc can he obtained using the linear interaction method of ligurc 12. 12. From the proportion!' of the straight line rclation:-.hip M11
tJ(Mc
M,)
+ M,
where Me is the moment capacity of the compo~ lle M!ction with full shear connection frum example I 2.2. M, i~ the moment capaci ty of the steel beam where M,
l¥p1
/ y = 1653 x 355
w-·' - 587 ~N m
Therefore Mr
IJ(M( - M,) - M,
= 0.77 x (1105- 587 )- 587 -
986kl'\ m
(b) Transverse reinforcement in lhe concrete flange
The de~ign follow:. Lhe example 7.5(2).
procedure~
and equauons \Ct out in
~ection 5. I .4
and
Composite construction (i) Calculate the design longitudinal shear vg.1 at the web-flange interface
irom hl to
For a 1>agging moment the longitudinal ~hear stresses arc th~.: greatest O\er a di~tance of .l.1 measured from the point of zero moment and u.x is talo.en all half the di~tancc to the maximum moment at mid-~pan. thus ~X =
0.5
L/ 2
X
= 0.5 x 9 x 103/ 2 m
2250mm
For a one '>pan ~imply ~upported beam wtth a uniformly di!.trihuted load the change in moment. uM over di~tance .6.x = L/ 4 from the 7ero moment at the ~upport is
ling I:::..M
=
ll'u X
f.
f. _
X
2
ll'u
4
X L X ~ = 3~1· 11 [} _3 (11\,L~) --X -8 32 4 8
4
Therefore 0.75 x 9S6 = 740k.N m
t::..M
The change in longitudinal force t::..Fd in th\: <:on~:re tc llangc fi gure 12. 14 is A
O.M
,
;:
u/'d -
om le
X
lta / 1
section h b in
0.5(belC - b) h.~,
where the lever arm ~ is taken as thc di,tance lrom the centre of the concrete flange. so that :
nl
cclllr~:
ot the 'tecl bc
10
the
h, /~
II
457/ 2 + 140 - 90/2
= 324 mm
Therefore 74Cb 10 1 324
I hc longttudinal ' hear
OK
0.5(3000 - 191 )
X --'---:-::-::-::--
3000
~trcl!~
indu<:ed.
..;..
10691-..
l'f.d·
i!>
.6.Fti I'I;J
(/11 X ~1 ) 1
bear
X 10 = 1069 ()0 X 2250
uhe
5.3 N/mn1 2
(II) Check the strength of the concrete strut
Prom equntion 5.17. to prevent crushing ol' the concn.·tc in the eotnprc:-.sivc strut in the
llangc I'UJ
Thc
0.6( l - ftd250)fck < - 1.5(cot Or tan Ot)
moment~
26.5 < o,
are sagging so the Oa11gc is in compression and the limit' tor 81 arc 45
With 81 = the minimum value of 26.5 0.6( I - 25; 250) x 25 _ , N/ , ., • , _..:........,-::-:::--'---=--::-:-- = ::> .... 1 mm- ( > 5··' /mm-) (2.0 t 0.5) and the concrete strut has sufficiem strength with Or = 26.5 .
388
Reinforced concrete design (iii) Design transverse steel reinforcement Tran~vcrse ~hear reinforcement is required if \'w > 0.27fc1~:. where f~ck is the characteristic axial tensile strength of concrete = I 8 l'\/mm1 for class 25 concrete. Therefore
''t-..J "''"
= 0.27f<~
0.17 x 1.8 = OA9 N/mm1
(
< 5.3 N/mm1 )
and tn.msverse shear reinforcement i\ required. The area required is given by: A,, sr
''~o<~
A
0.87/vL
A
hr cotO,
5.3 x 90 0.87 x 500 x 2.0
= 0.55
If bar:. are provided at (say) 175 mm cemres then A,1 = 0.55 x 175 % mm 2. lienee provide 12 mm bar~ (A, 113 mm 2 for one bar). This ~t eel area ~ati sfies the minimum requirement of 0.13% 11 7 mm 2/m. Longitudinal reinforeement shoultl also be provided in the Range.
l~--------------------------------------~) Deflection checks at the serviceability limit state
12.6
At the ~erviceabiluy limit \late it i~ hcam for the followmg condit1on~: (a)
nece~~ary
to check the maxunum deflections of the
Dunng con~truction \\hen the concrete nangc ha\ not hardened and the 'tee I beam alone has to carry all the load' due to the permanent and vanable actions at that time. ~ection
{b) At 'er\1ce when the concrete ha' hardened and the comrx,.,ite steel and concrete section carrie' the additional pem1ancnt and \'ariahlc load-..
12.6.1
Deflections during construction
The dencction h at mid-span for a uniformly
di~trihuted
loau if>
511•/} 384£010
( L2. L8)
where is the !.ervicenbiliry load per metre at construction Lis the beam's span E~ i\ the ela,tic modulus of the steel 210 kN/mm, 1. i~ the second moment of area of the steel section 1v
The deflection due to the permanent ac11on or dead load concrete harden~.
12.6.2
1s
locked into the beam as the
Deflections at service during the working life of the structure
At this stage the concrete has hardened and form' a cnmpo-.ite section together .,.. ith the -.tccl beam and the shear connectors.
Composite construction
The composite ~ection is convened into a transformed ~ection ~o lhat the area of concrete in compression is transformed into an equivalent ~tccl area wnh a flange width U!> ~hown in figure 12.15. where II
E,tcd . h _.~ I . 11 = -- ts t e mvuu ar ratiO Ec dl
and
For building~. EC4 states that Ec.eft may be taken as E,m/ 2 \\here F.,m il> the llecant modulus of elasticit) for concrete (see table 1.1 ). It i~> al~o staled in EC4 lhat for calculating dencctions at llerviec the effect~ of partial :-.hear Ctlllnectton can be ignored provided that the degree of shear connection. 'I ~ 0.5 and other practicnl requirements are satisfied.
The transformed composite section For two an.:a~ A, nnd ,h the position of their neutral nxi~ may he found by taking aren moments about the centroidol ax is of A 1 su.:h that A2.1
te
l ---
+ A~
A,
where ' is the di~tance to the neutral axis from the ccnrro1d o1 A 1 1 I!. the di<.tance between the centroid~ of A 1 and \ 1 . The '>ccond moment of area of the total sect ton about the neutral a·w, of A 1 und A2 combllled can he calculated from lr
-
)'
I'
/1
... A, ~ 2r
__.:.__;;__
At+ 1\2
where 1, nnd '=ure the 11econd moments of areas of A 1 and A~ rc'>pecuvely nboutthe1r cemroidnl axes. So with reference w figure I2.7 and taking A1 u~ the steel nreu A,, for tile u·an1.formcd composite scclion .1 (II,, f h I /tp)/2 nnd the equations for rand I become A,,n(h, I h I hp) +- brrr(h 111,)}
( 12. 19)
2~Anll fu,on~r
/J I
berr(h
1211
h1.)"
+
h el l (h
- hr )(h. - h t hp ).r
211
---
( 12.20)
~here
A. IS the area of the \!eel ~ection and Ia b the c;econd moment of area of the steel 'icction.
TI1e depth of the neutral ax1s. x1 from the top of the concrete nange ~~
ture u'le
.\,
,\ -I-
1f.\
< (/z
~emee
lz
lzr
-2-
lzp) then the neutral axi~ is \\ ithin the nange and the concrete is cracked at and these cquntion.;, cannot apply.
Figure 12.15 The transformed sectton
service
390
Reinforced concrete design
Deflection at service due to the permanent and variable loads The deflections are calculated for the unfactorcd acllon\. The second moment of area of the composite section is used in the calculation<,. For lhe unpropped case the total dcncct1on i':
\\here
(
t,on~rr
is the deflection of the Mecl beam due to the permanent load at construction
bl"'"l''"~'t
i-; the deflection of the compo.,itc beam due to the quasi-permanent load which i~ the additional permanent load plu-. a proportton or the variable load depending on the type of ~tructure. (See section 2.4 and table 2.4 in chapter 1.)
EXAMPLE 12. 4
Serviceability checks for deflection
For the compo~itt> hcam of the prcviou~ cxamph:~ dctcnninc the tlellection~. at ~ervice. The relevant \cctional properties for the -l57 x 191 x 74 J...g/m Untver~al Beam arc: Cr(l,,.,cctinn:'ll area A,,
94 (H:m'
'\ccond moment of area 1., - 33 300cm4 A''umc the beam i., pan of a huildtng floor ')'tcm and " unproppcd during Cllll\lruCI!Oil.
The uniform!) di-.trihuted charactcm.w.: uc11on' arc During com.truction - permanent load 9.R 1.. "1/m.
not hardened so that the :-tcel beam supporh the load
or the
wet
COilt'l'clC.
The tknection at mid-span of u beam with u unifmmly di:-tributc.:d loud i!-. given by 5wTJ
Figure 12.16 Transformed section exdmpl('
1..
,
..
b••• = 3000 - 222 n 13.5 l
h,
457
'---
457 X 191 X 74 UB
Se<:tion
x, .. 149
Composite construction Due to the pcnnanenr load. wo and the variable load, steel heam i~:
= (9.8
2.25)
12 -3
384
= 15 mm -
21()
X
X
li'Q
the mid-span deflection for the
JJ )()()
~p:tn/600
Thl.! 12 mm deflection due to the permanent load at thh :-.tage i~ locked into the beam a~ the concrete hardens after constntclion. (b) At
service
The composite 1.ection is transformed into an equivalent :-.lcel section ~~s shown in figure 12. 16. For u class C25/30 concrete the scctLnt modulus uf elast icity of the concrete. Etm 3 1kN/mm ~ (&cc tahlc 1.1). Take the modular ratio II
Eu 210 0.5£,m = 0.5 ,. . 31
13.5
The po~ition of the centr
A.n(ha + h l hr) 2{A.n + /1,11 (II hrl}
therefore the concrete is not cracked. lhc second moment ot area of the compo:-.tte ligun!
j,
111
90 mm
given h) equation 12.20 a\
i
'••J"''
I.
+
333
bttdh fir ) h.u (II - hp)(lt. I II +-lip) 1 I,~n I ,.,-11
X
1000 >. ( 140 - 50)
1011 1
1
12 X 13.5
+
.3000 x ( 140 - (333 1 14
= 1019 x
50) X (457 I 140 I 50) x IO·I 2 X 13.5
6n) " 10(\ mm 4
1011 mm 1
Defleclions at service
= 11.0 -9.8 18 0 kN/m. thus deflection
At 'ervice lh~: addittonal permanent load
pcnnanent variable load {I= (II'G
= (1.
2
+11'()) +
= I 17
18
5L~ I(
184£ala 'i
)
X
38~ X
9~
103 210 " 1019
Xmm=~pan/ 11 25
1.2 kN/m and the qua'i
391
392
Reinforced concrete design
Therefore the total final deflection including that at the construction stage is t
12 + 8 = 20rnm = span/ 450
At all !>tagcl> the deflection i~ well within the normally acceptable limits of span/ 250. Deflection!> are seldom a problem for eompostte heams in buildings. but for long spans the beaml> could be pre-cambered for a proportton of the permanent load to avoid vi~ible c;tgn\ of the beam ·~agging·.
l~------------------------------------~)
...............................................
Appendix
;5D.
_) Typical weights and live loads l kg
;;;; 9.8 1 N force
1 lb;;;; 0.454 kg ;;;; 4.448 N force I lb/ft 2 ;;;; 4.88 kg/m~ 47.9 N/m2
I lb/ft 1
=
=16.02 kg/m3 = 157 N/m
Weights Aluminium, cost Asphalt paving Brick~.
common
Brick!., prcs~cd Clay, dry Cloy. wet Concrete. re111forccd Gin.,.,. plate Lead OaJ... Pme, wlute Sand, dry Sand. wet Steel Water
Shear reinforcement Table A.4 Asw l s ror varying stirrup diameter and spacing Stirrup diameter (mm)
Stirrup spocmg (mm)
8 10 12 16
85
90
100
125
150
175
200
225
250
275
1.183 1.847 2.659 4.729
1.118 1.744 2.511 4.467
1.006 1.57 2.26 4.02
0.805 1.256 1.808 3.216
0.671 1 047 1.507 2.68
0.575
0.503 0.785 1.13 2.01
0447 0.698 1.004 1.787
0.402 0.628 0.904 1.608
0.366 0.571 0.822 1.462
0.897 1.29 1 2.297
30t
0.3?
0.52 0.75
1.34
Not£'. A,., I~ based on Lhc cross-sectional <~rca ol two leg~ of Lhe stirr
Wire fabric Table A.S
Secttonal areas for different rabric types
Fabflc reference
i!
Longitudinal wtres Wire size (rnm)
Cross wires
Pitch (mm)
Area (mm 1/m)
200 200 200 200 200
393 252 193 142 98
10 8 7 6
1131 785 503 385 283 196
8 8 8
5
100 100 100 100 100 100
10 9 8 7 6
100 100 100 100 100
785 636 503 385 283
6 6 5 5 5
5
200 100
98 49
2.5
Wife size (mm)
Pitch (mm)
Area (mm 1/m)
Square mesh
A393 A252 A193 A142 A98
10 8 7 6 5
5
200 200 200 200 200
393 252 193 142 98
200 200 200 200 200 200
252 252 252 193 193 193
Strucwral mesh
811 31 8785 8503 8385 8283 8 196
12 10 8 7
6
7
7 7
Long mesh
C785 C636 C503 C385 C283
400 400 400 400 400
70.8 70.8 49 49 49
200 100
98 49
Wrapping mesh
098 049
2.5
5
Appendix
396
Anchorage and lap requirements Table A.6
1 Anchorage and lap length coefficients (lenglh L = KA x bar size) for good bond conditions
KA for concrete strength, fck (N/mm1)
20
25
30
35
40
45
50
Straight bars Anchorage in tension and compression
47
40
36
32
29
27
25
Curved bars Anchorage in tension 3 Anchorage in compression
33 47
28 40
25 36
22 32
20 29
19 27
18 25
47 54 66 71
40 46 56 60
36 42 51
32 37
54
48
29 33 41 44
27 31 38 41
25 29 35 38
% of bars lopped at section Compression and tension laps~
<25% 33% 50% >50%
45
NOI£'5:
1. For poor bond cond•t•ons (~~ llgurc 5.8) dev1de the coefficient~ by 0.7. 2. lor bars greJter than 32mm d1vide the coeffiCient\ by 1!132 •1>)100 where •I• Is the banile. 3. For.., curved bM in tension the anchorage length 11 qenerally thJt or a straight bo~r II'C
0 7 but also depends on the cover conditions-
ltlble 52
4. ThP.sP ligures dpply for"' " I .. "' n 1 dch11ling \pPclllcd 1n section> S 2 .md 5.3
'I (src Ldble 52). Abo 1ee the nddltlono~l requirement' for minln,um lap lengths ~f'd
Maximum and minimum areas of reinforcement Table A.7
Maximum areas of reinforcement
(a) For a slab or beam, tem1on or compression reinforcement
·1OOA, I Ac
4 per cent other than at laps
(b) ror a column 1OOA,. A.:
_4
per cent other than at laps and 8 per cent at laps
(c) For a wall, vertical reinforcement
1OOA,/A1
c.
4 per cent
Appendix Table A.8
Minimum areas of reinforcement Concrete closs (fy. ~ 500N/ mm 1 )
Tension reinforcement in beams and slabs
\~"'
>
397
0.26
(
'~=
0 001 3)
C25/ 30
C30/ 35
C40/ 50
CS0/60
0.0013
0.0015
0.0018
0.0021
Secondary reinforcement > 20% main reinforcement Longitudinal reinforcement in columns A, min .;. 0 1ONKJ / 0.87 fyk ., 0.002Ac where NKl is the axial compression force Vertical reinforcement In walls A,,rnln "> 0.002At Note: t>. os the ml'an width
or the tension zone.
300
32
E' E'
250
g'
200
2
150
.§,
25
.§,
~
E
?;
bar spacmg
20
ti
sE E
16
E i(
Maximum bar size and spacing for crack control
'2
::>
"' E
£
Figure A.1
100
12
so
10
M
0
-~
2 E ::> E
sE
6
100
150
200
250
300
350
400
Stress In reinforcement under quasi-permanent load (see secuons 6.1.3 and 6.1.7)
(N/mm~)
Span-effective depth ratios 36
32
L
~ - ---+---1---
1I
- i- - t - - t - -
Figure A.2
---i--11--t--
Graph of basic span-effective depth ratios for different classes of concrete
---;---+-+
4
K • 1 0 for a som ply supported span
100A,,... 12 ~------~----~--~--~~~-----
0.40%
080%
1.2%
1 6%
2.0%
bd
398
Appendix
Summary of basic design equations for the design of reinforced concrete (a) Design for bending (see chapters 4 and 7) For a swgl) reinforced section: A,=
When moment rcdi),tribution modified - sec 1ablc -U. Table A.9
ha~
been applted lhl'n
th~.: above equation~
Limiting constant values
Concrete class
--------------------------------xn.)ljd Limit1ng Maximum z1111, Khlol limiting K Um1tmg d'td Maximum percentage steel area 100Ab,,1/bd
0.95
Compression rcmtorccment requ1red (al M1.,,)
II
~
0.90
0.85 082 0
0.05
CS0/60
0 45 0.167 0.171 23 Afckl fyk
m3x1mum valuf.' of 1/d according lo Lh~ Concise Codf.' and previous UK prt~clicc
~
<
0.82d
1.00
Figure A.3 Lever-arm curvf.'
must be
0 .10 K= M!bd1f(,.
The percentage values on the K axis mark the limits tor singly re~nforced secttons w1th moment red1slnbulion apphed (see section 4.7 and !able 4.2)
Appendix
(b) Design for shear (see chapters 5 and 7) VRd ma\ (21
= O.l24b"d( I
I Rd m3, f-15
-
/ck/ 250}f..:l
0.18h" d ( 1 - f~t , 250)/ck
0 = 0.5~>in- '{
l'f:.d
0.1 8b" elf,~ (1 -
.,-
fekf _)Q)
}
~ 45
I'Ed
A," -
0.78c/f}l COl() O.OSJ;.'~ 5b"
!1,,.,,111n
!>k A,w
.
0.7'6c/}yk COl
\ln1in - - X .I
~Ftd
()
= 0.511~-;.,cotO
(c) Design for torsion (see chapters 5 and 7) t=
for
rt
Area ol the section Perimeter of the section
II u
rectangular section h x h bll Ilk -
2(hhJ
2(b
"
:!t)
' llr..s_ < 1.0
~1M m.l\ -
1 . 33lj~kA~t
cot 0 I tan (}
.,,..,
2Al0.8~/y1.. COl(}
Tt:..tllk cot
n
211 ~.,0.8~/ylk
(d) Design for punching shear in slabs (see chapter 8) VRtl , mJx 11 1
2(a
.r..')].fck1.5 0.6 (I 250
0.5ud [ /J)
~
47Td
0.053 J.Tc.(.l, ..l·, )
r.,.. . I'Rd c'
A," > -
-
0.751'Rt.f c
15/;"t.l d lr
llf
-
3
400
Appendix
Proof of equation conversions used in Chapter 5 section 5.1.2 To prove that !>in 0 cos() =
1 : () tan -cot 8
Con1>ider a right-angled triangle with sides length a, band ft. where h b the length of the hypotenuse and () is the angle between :.ides of length a and b, and use the theorem of Pythagoras where ~~~ = a 2 + b~. l>in ()COS()
-
abab h X II = h1
ab
=a1 +
I b2 = a2
!J2
tan 8 +COl (J
tlb -r- ab
A reason for using this type of conver~ion in the equations for the analysis for shear is that it facilitates the selling up of quadratic equations which can he more readily solved.
Sted for tilt reinforn'IIWIII oj nmcn•tc f'n•.\trt'.HIIIJ.I 11ed1
STeel ./i1r Tltt• remfon·l'llll'llf of cmwn•t1. ll'l'ldablt reinforcmg .1/et'l. hw. n11/ and decoilecl pmduo' Cold reducetl 11m· far the l't'il!/on·emc•w of concrete
13 13S 'i057
ClliiCI'CII!
14
UK Nmional
'lA to F.N 19Y2
15 13S PD 66H7:2006
atfmi\1111'1'1 1111111' 1 to
l.umrodc• 2
lJadlvound flliJJN to tht• UA Narimwl Awtl' lt'.l to US I:.N I tJIJ2 -I
(b) Textbooks and other publications 16 J. 13ungcy, S. Mi llard. M. Gran1ham. 'l'hl' 1'1•s1in~ of Collc'/'1'11' l'uylo1 & Franci~. London. 200o
111
I7 11. Gu lvanc\Sian. J.A Cal gam. M. Holic.:~y. /k1i~nf!l'.l' Guit/1' 1\:lford. London 2005
Stl'll<'llll'fl'. 4th cdn.
to f:'N /WI!. Thomas
18 M.K. llurst. Prt•,trt'l'.led rrml'reu· D!!'it:n. ~nd cdn. Chapman & Hall. London, I
19 W. II. Mo'h:). J.H Bungey, R Hulse, Rem{or('('c/ (oncrt'le Londun. llJlJlJ
Dc•.1i~n.
'ith cdn.
Palgran~.
20 R.S J\ar,tyanan. A Beehy. De.11gnen Guulr 10 /:..;\ II.JCJ2-I · I and F V/99::!-1-::!. Thoma\ Telford. London, 2!Xl5 21
R.S Narayanan. C.H Goodchild. 77/(• Cmu·llt' f.umcodt ::!. The Concre1e Centre. Surrey. 2006
403
404
Further reading 22 A.M. Neville. Properties of Concrete, 4th Edn. Pearson Education Ltd. E~scx. 2000 23 Manual for the Design of Concrete Building Stmctttrer to Eurocode 2, The Institution of Structural Engineers, London. 2006 24 Swndard Method of Detailing Structural Concrete. 3rd cdn, The lmtitution of Structural Engineers. London, 2006 25 Early-age 111ermal Crack Comrol in Concrete, Guide C660. ClRlA. London. 2007
(c) Websites 26
Eurocode~
Expert- http://www.curocodcs.eo.uk/
27 The Concrete Centre website - htlp://www.concrctcccntrc.com
............. .. .. .. .................
Index
Action~
characteristic I!l combination 2:1. 30, 2!l2. :II 0 I I. 313 design values 24-7 frequent 2:1, permanent 19. 21) qua::.i-pcnnam:nt 23 typical value' 393 variable 2:1, 29 Age factors S Analy!.is of Mructurcs beams 31 7 column moment .f4, 4tl damaged 'tructurc ISS. 163 frame~
38-48
lateral load' 4.5 reta1n111g wall\ JOR 12 Analy .Vl'l' Foul i ng~ Beam' analysi~ of moments and <,he:m 31 anulysi~ or sections 58- 9!1 cantilever 197 continuous 3.1- 7. 191- 7 deflection' 136-45 tk\ign 63. 16l) 20!! dc::.ign cha1h M. 70 doubl} reinfon:ed 67. 176-8, 180--2 effect!\ e 'pan' 171 one 1>pan 12. 176-!17 prestressed 119- 6!1 reinforcement dctai Is 192-7. 202-4
Index Combined footings 291-4 Composite construction dc~ign
36Y-92
serviceability limit state 373. 374, 388-92 ~hear connector~
381-5
trans,crsc reinforcement 385 types 369-7 J ultimate limit state 372-4, 376-80 Compression reinforcement 67-72. 176-7 Concrete age factor 5 characteristic ~trength 12 clu~s 12, 126, 285 cover 127-9. 285 cracking 7- 10, 147-53 creep I0. 138-9. 344 durabili ty I I, 208 dnstic modu lu ~ 4. shrinh.agc 6. 153. 344 stn.:ngth dN 12, 126, 285 stress-strain curve 3. 59. thermal expan~ion 9. 153-4 Continuous beam; analy,is 33-7 cunmlmcnt of bar; 202-4 tbign 191-7 eO\ elope<, 37-43 loading arrangement' 30 moment and ~hear coefficient:. 37 Corbel~ 198-20 I Counterfon retaimng wall~ 309 Cover to rctnforccmcnt I27- 9. 285 Cracking control 152 nc.xurol 147- 52 thermal and shrinkage 153 Creep 10. 138. 344 Creep cocfricients 139 Critical section I06. 212, 28~. 305 Curtailment of bar' 202-4 Curvtllun:s 137, 140
Earth-bearing pressures 284-6, 312 Effective depth 61. 171 Effective Oange width 182-3 Effective height of a column 25+-5 Effective span 171 Elasuc analy~i~ of a ~ection 93. 324-8 Elastic modulu<. concrete 4 ~tccl 5 End block' 351 ·3 Envelope~. bending moment uno ~hear force 37. 43 Equivalent rectangular wes~ block 61 Factor~
of safety glohnl 27 partial 18. 26, 282, 283 Fire re~lstance 128, 158, 229 Flungccl~>cc t ion .vee T-he:nm Flat ~lab 228-35 Floms l'ee S l ab~ Footing~
