Administrator

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ReynoldsrsquosReinforcedConcreteDesignerrsquosHandbook

Reynoldsrsquos Reinforced Concrete Designerrsquos Handbook has beencompletely rewritten and updated for this new edition to takeaccount of the numerous developments in design and practiceover the last 20 years These include significant revisions toBritish Standards and Codes of Practice and the introduction ofthe new Eurocodes The principal feature of the Handbook is thecollection of over 200 full-page tables and charts covering allaspects of structural analysis and reinforced concrete designThese together with extensive numerical examples will enableengineers to produce rapid and efficient designs for a large rangeof concrete structures conforming to the requirements of BS 5400BS 8007 BS 8110 and Eurocode 2

Design criteria safety factors loads and material propertiesare explained in the first part of the book Details are then givenof the analysis of structures ranging from single-span beamsand cantilevers to complex multi-bay frames shear walls

arches and containment structures Miscellaneous structuressuch as helical stairs shell roofs and bow girders are alsocovered

A large section of the Handbook presents detailed informationconcerning the design of various types of reinforced concreteelements according to current design methods and their use insuch structures as buildings bridges cylindrical and rectangulartanks silos foundations retaining walls culverts and subwaysAll of the design tables and charts in this section of the Handbookare completely new

This highly regarded work provides in one publication awealth of information presented in a practical and user-friendlyform It is a unique reference source for structural engineersspecialising in reinforced concrete design and will also be ofconsiderable interest to lecturers and students of structuralengineering

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Also available from Taylor amp Francis

Concrete Pavement Design GuidanceG Griffiths et al Hb ISBN 0ndash415ndash25451ndash5

Reinforced Concrete 3rd edP Bhatt et al Hb ISBN 0ndash415ndash30795ndash3

Pb ISBN 0ndash415ndash30796ndash1

Concrete BridgesP Mondorf Hb ISBN 0ndash415ndash39362ndash0

Reinforced amp Prestressed Concrete 4th edS Teng et al Hb ISBN 0ndash415ndash31627ndash8

Pb ISBN 0ndash415ndash31626ndashX

Concrete Mix Design Quality Control and Specification 3rd edK Day Hb ISBN 0ndash415ndash39313ndash2

Examples in Structural AnalysisW McKenzie Hb ISBN 0ndash415ndash37053ndash1

Pb ISBN 0ndash415ndash37054ndashX

Wind Loading of Structures 2nd edJ Holmes Hb ISBN 0ndash415ndash40946ndash2

Information and ordering details

For price availability and ordering visit our website wwwtandfcoukbuiltenvironment

Alternatively our books are available from all good bookshops

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ReynoldsrsquosReinforcedConcreteDesignerrsquosHandbookELEVENTH EDITION

Charles E ReynoldsBSc (Eng) CEng FICE

James C SteedmanBA CEng MICE MIStructE

and

Anthony J ThrelfallBEng DIC

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First edition 1932 second edition 1939 third edition 1946 fourth edition 1948revised 1951 further revision 1954 fifth edition 1957 sixth edition 1961revised 1964 seventh edition 1971 revised 1972 eighth edition 1974 reprinted1976 ninth edition 1981 tenth edition 1988reprinted 1991 1994 (twice) 1995 1996 1997 1999 2002 2003

Eleventh edition published 2008by Taylor amp Francis2 Park Square Milton Park Abingdon Oxon OX14 4RN

Simultaneously published in the USA and Canadaby Taylor amp Francis270 Madison Ave New York NY 10016 USA

Taylor amp Francis is an imprint of the Taylor amp Francis Groupan informa business

copy 2008 Taylor and Francis

All rights reserved No part of this book may be reprinted or reproduced or utilised in any form or by any electronic mechanical or other meansnow known or hereafter invented including photocopying and recordingor in any information storage or retrieval system without permission in writing from the publishers

The publisher makes no representation express or implied with regard to the accuracy of the information contained in this book and cannot accept any legal responsibility or liability for any efforts or omissions that may be made

British Library Cataloguing in Publication DataA catalogue record for this book is available from the British Library

Library of Congress Cataloging-in-Publication DataReynolds Charles E (Charles Edward)

Reynoldsrsquos reinforced concrete designers handbook Charles E ReynoldsJames C Steedman and Anthony J Threlfall ndash 11th ed

p cmRev ed of Reinforced concrete designerrsquos handbook Charles E Reynolds

and James C Steedman 1988Includes bibliographical references and index1 Reinforced concrete construction ndash Handbooks manuals etc

I Steedman James C (James Cyril) II Threlfall A J III ReynoldsCharles E (Charles Edward) Reinforced concrete designerrsquos handbookIV Title

TA6832R48 200762418341ndashdc22 2006022625

ISBN10 0ndash419ndash25820ndash5 (hbk)ISBN10 0ndash419ndash25830ndash2 (pbk)ISBN10 0ndash203ndash08775ndash5 (ebk)

ISBN13 978ndash0ndash419ndash25820ndash9 (hbk)ISBN13 978ndash0ndash419ndash25830ndash8 (pbk)ISBN13 978ndash0ndash203ndash08775ndash6 (ebk)

This edition published in the Taylor amp Francis e-Library 2007

ldquoTo purchase your own copy of this or any of Taylor amp Francis or Routledgersquoscollection of thousands of eBooks please go to wwweBookstoretandfcoukrdquo

ISBN 0-203-08775-5 Master e-book ISBN

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List of tables viPreface to the eleventh edition ixThe authors xAcknowledgements xiSymbols and abbreviations xii

Part 1 ndash General information 11 Introduction 32 Design criteria safety factors and loads 53 Material properties 144 Structural analysis 285 Design of structural members 446 Buildings bridges and containment structures 547 Foundations ground slabs retaining walls

culverts and subways 63

Part 2 ndash Loads materials and structures 738 Loads 759 Pressures due to retained materials 86

10 Concrete and reinforcement 9511 Cantilevers and single-span beams 10512 Continuous beams 11113 Slabs 12814 Framed structures 15415 Shear wall structures 16916 Arches 17517 Containment structures 18318 Foundations and retaining walls 195

19 Miscellaneous structures and details 20620 Elastic analysis of concrete sections 226

Part 3 ndash Design to British Codes 23721 Design requirements and safety factors 23922 Properties of materials 24523 Durability and fire-resistance 24924 Bending and axial force 25625 Shear and torsion 28326 Deflection and cracking 29527 Considerations affecting design details 31228 Miscellaneous members and details 322

Part 4 ndash Design to European Codes 33329 Design requirements and safety factors 33530 Properties of materials 33831 Durability and fire-resistance 34232 Bending and axial force 34533 Shear and torsion 36234 Deflection and cracking 37135 Considerations affecting design details 38136 Foundations and earth-retaining walls 390

Appendix Mathematical formulae and data 395

References and further reading 397

Index 399

Contents

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21 Weights of construction materials and concrete

floor slabs22 Weights of roofs and walls23 Imposed loads on floors of buildings24 Imposed loads on roofs of buildings25 Imposed loads on bridges ndash 126 Imposed loads on bridges ndash 227 Wind speeds (standard method of design)28 Wind pressures and forces (standard method

of design)29 Pressure coefficients and size effect factors

for rectangular buildings210 Properties of soils211 Earth pressure distributions on rigid walls212 Active earth pressure coefficients213 Passive earth pressure coefficients ndash 1214 Passive earth pressure coefficients ndash 2215 Silos ndash 1216 Silos ndash 2217 Concrete cements and aggregate grading218 Concrete early-age temperatures219 Reinforcement general properties220 Reinforcement cross-sectional areas of bars

and fabric221 Reinforcement standard bar shapes and method of

measurement ndash 1222 Reinforcement standard bar shapes and method of

measurement ndash 2223 Reinforcement typical bar schedule224 Moments shears deflections general case for beams225 Moments shears deflections special cases for beams226 Moments shears deflections general cases for

cantilevers227 Moments shears deflections special cases for

cantilevers228 Fixed-end moment coefficients general data229 Continuous beams general data230 Continuous beams moments from equal loads on

equal spans ndash 1231 Continuous beams moments from equal loads on

equal spans ndash 2232 Continuous beams shears from equal loads on

equal spans233 Continuous beams moment redistribution234 Continuous beams bending moment diagrams ndash 1

235 Continuous beams bending moment diagrams ndash 2236 Continuous beams moment distribution methods237 Continuous beams unequal prismatic spans and loads238 Continuous beams influence lines for two spans239 Continuous beams influence lines for three spans240 Continuous beams influence lines for four spans241 Continuous beams influence lines for five or

more spans242 Slabs general data243 Two-way slabs uniformly loaded rectangular panels

(BS 8110 method)244 Two-way slabs uniformly loaded rectangular panels

(elastic analysis)245 One-way slabs concentrated loads246 Two-way slabs rectangular panel with concentric

concentrated load ndash 1247 Two-way slabs rectangular panel with concentric

concentrated load ndash 2248 Two-way slabs non-rectangular panels (elastic

analysis)249 Two-way slabs yield-line theory general information250 Two-way slabs yield-line theory corner levers251 Two-way slabs Hillerborgrsquos simple strip theory252 Two-way slabs rectangular panels loads on beams

(common values)253 Two-way slabs triangularly distributed load (elastic

analysis)254 Two-way slabs triangularly distributed load (collapse

method)255 Flat slabs BS 8110 simplified method ndash 1256 Flat slabs BS 8110 simplified method ndash 2257 Frame analysis general data258 Frame analysis moment-distribution method

no sway259 Frame analysis moment-distribution method

with sway260 Frame analysis slope-deflection data261 Frame analysis simplified sub-frames262 Frame analysis effects of lateral loads263 Rectangular frames general cases264 Gable frames general cases265 Rectangular frames special cases266 Gable frames special cases267 Three-hinged portal frames268 Structural forms for multi-storey buildings

List of tables

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List of tables vii

269 Shear wall layout and lateral load allocation270 Analysis of pierced shear walls271 Arches three-hinged and two-hinged arches272 Arches fixed-ended arches273 Arches computation chart for symmetrical

fixed-ended arch274 Arches fixed-ended parabolic arches275 Cylindrical tanks elastic analysis ndash 1276 Cylindrical tanks elastic analysis ndash 2277 Cylindrical tanks elastic analysis ndash 3278 Rectangular tanks triangularly distributed load

(elastic analysis) ndash 1279 Rectangular tanks triangularly distributed load

(elastic analysis) ndash 2280 Rectangular containers spanning horizontally

moments in walls281 Bottoms of elevated tanks and silos282 Foundations presumed allowable bearing values

and separate bases283 Foundations other bases and footings284 Foundations inter-connected bases and rafts285 Foundations loads on open-piled structures286 Retaining walls287 Rectangular culverts288 Stairs general information289 Stairs sawtooth and helical stairs290 Design coefficients for helical stairs ndash 1291 Design coefficients for helical stairs ndash 2292 Non-planar roofs general data293 Shell roofs empirical design method ndash 1294 Shell roofs empirical design method ndash 2295 Bow girders concentrated loads296 Bow girders uniform loads ndash 1297 Bow girders uniform loads ndash 2298 Bridges299 Hinges and bearings2100 Movement joints2101 Geometric properties of uniform sections2102 Properties of reinforced concrete sections ndash 12103 Properties of reinforced concrete sections ndash 22104 Uniaxial bending and compression (modular ratio)2105 Symmetrically reinforced rectangular columns

(modular ratio) ndash 12106 Symmetrically reinforced rectangular columns

(modular ratio) ndash 22107 Uniformly reinforced cylindrical columns

(modular ratio)2108 Uniaxial bending and tension (modular ratio)2109 Biaxial bending and compression (modular ratio)31 Design requirements and partial safety factors

(BS 8110)32 Design requirements and partial safety factors

(BS 5400) ndash 133 Design requirements and partial safety factors

(BS 5400) ndash 234 Design requirements and partial safety factors

(BS 8007)35 Concrete (BS 8110) strength and deformation

characteristics36 Stress-strain curves (BS 8110 and BS 5400) concrete

and reinforcement

37 Exposure classification (BS 8500)38 Concrete quality and cover requirements for durability

(BS 8500)39 Exposure conditions concrete and cover requirements

(prior to BS 8500)310 Fire resistance requirements (BS 8110) ndash 1311 Fire resistance requirements (BS 8110) ndash 2312 Building regulations minimum fire periods313 BS 8110 Design chart for singly reinforced

rectangular beams314 BS 8110 Design table for singly reinforced

rectangular beams315 BS 8110 Design chart for doubly reinforced

rectangular beams ndash 1316 BS 8110 Design chart for doubly reinforced

rectangular beams ndash 2317 BS 8110 Design chart for rectangular columns ndash 1318 BS 8110 Design chart for rectangular columns ndash 2319 BS 8110 Design chart for circular columns ndash 1320 BS 8110 Design chart for circular columns ndash 2321 BS 8110 Design procedure for columns ndash 1322 BS 8110 Design procedure for columns ndash 2323 BS 5400 Design chart for singly reinforced

rectangular beams324 BS 5400 Design table for singly reinforced

rectangular beams325 BS 5400 Design chart for doubly reinforced

rectangular beams ndash 1326 BS 5400 Design chart for doubly reinforced

rectangular beams ndash 2327 BS 5400 Design chart for rectangular columns ndash 1328 BS 5400 Design chart for rectangular columns ndash 2329 BS 5400 Design chart for circular columns ndash 1330 BS 5400 Design chart for circular columns ndash 2331 BS 5400 Design procedure for columns ndash 1332 BS 5400 Design procedure for columns ndash 2333 BS 8110 Shear resistance334 BS 8110 Shear under concentrated loads335 BS 8110 Design for torsion336 BS 5400 Shear resistance337 BS 5400 Shear under concentrated loads ndash 1338 BS 5400 Shear under concentrated loads ndash 2339 BS 5400 Design for torsion340 BS 8110 Deflection ndash 1341 BS 8110 Deflection ndash 2342 BS 8110 Deflection ndash 3343 BS 8110 (and BS 5400) Cracking344 BS 8007 Cracking345 BS 8007 Design options and restraint factors346 BS 8007 Design table for cracking due to temperature

effects347 BS 8007 Elastic properties of cracked rectangular

sections in flexure348 BS 8007 Design table for cracking due to flexure

in slabs ndash 1349 BS 8007 Design table for cracking due to flexure

in slabs ndash 2350 BS 8007 Design table for cracking due to flexure

in slabs ndash 3351 BS 8007 Design table for cracking due to direct

tension in walls ndash 1

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List of tablesviii

49 EC 2 Design chart for doubly reinforcedrectangular beams ndash 1

410 EC 2 Design chart for doubly reinforcedrectangular beams ndash 2

411 EC 2 Design chart for rectangular columns ndash 1412 EC 2 Design chart for rectangular columns ndash 2413 EC 2 Design chart for circular columns ndash 1414 EC 2 Design chart for circular columns ndash 2415 EC 2 Design procedure for columns ndash 1416 EC 2 Design procedure for columns ndash 2417 EC 2 Shear resistance ndash 1418 EC 2 Shear resistance ndash 2419 EC 2 Shear under concentrated loads420 EC 2 Design for torsion421 EC 2 Deflection ndash 1422 EC 2 Deflection ndash 2423 EC 2 Cracking ndash 1424 EC 2 Cracking ndash 2425 EC 2 Cracking ndash 3426 EC 2 Early thermal cracking in end restrained panels427 EC 2 Early thermal cracking in edge

restrained panels428 EC 2 Reinforcement limits429 EC 2 Provision of ties430 EC 2 Anchorage requirements431 EC 2 Laps and bends in bars432 EC 2 Rules for curtailment large diameter bars

and bundles

352 BS 8007 Design table for cracking due to directtension in walls ndash 2

353 BS 8110 Reinforcement limits354 BS 8110 Provision of ties355 BS 8110 Anchorage requirements356 BS 8110 Curtailment requirements357 BS 8110 Simplified curtailment rules for beams358 BS 8110 Simplified curtailment rules for slabs359 BS 5400 Considerations affecting design details360 BS 8110 Load-bearing walls361 BS 8110 Pile-caps362 Recommended details nibs corbels and halving joints363 Recommended details intersections of members41 Design requirements and partial safety factors

(EC 2 Part 1)42 Concrete (EC 2) strength and deformation

characteristics ndash 143 Concrete (EC 2) strength and deformation

characteristics ndash 244 Stressndashstrain curves (EC 2) concrete and

reinforcement45 Exposure classification (BS 8500)46 Concrete quality and cover requirements for durability

(BS 8500)47 EC 2 Design chart for singly reinforced

rectangular beams48 EC 2 Design table for singly reinforced

rectangular beams

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Since the last edition of Reynoldsrsquos Handbook considerabledevelopments in design and practice have occurred These includesignificant revisions to British standard specifications and codesof practice and the introduction of the Eurocodes Although cur-rent British codes are due to be withdrawn from 2008 onwardstheir use is likely to continue beyond that date at least in someEnglish-speaking countries outside the United Kingdom

One of the most significant changes has been in the systemfor classifying exposure conditions and selecting concretestrength and cover requirements for durability This is now dealtwith exclusively in BS 8500 which takes into account theparticular cementcombination type The notation used todefine concrete strength gives the cylinder strength as well asthe cube strength For structural design cube strength is usedin the British codes and cylinder strength in the Eurocodes

The characteristic yield strength of reinforcement has beenincreased to 500 Nmm2 (MPa) As a result new design aidshave become necessary and the Handbook includes tables andcharts for beams and columns (rectangular and circular)designed to both British and European codes Throughout theHandbook stress units are given as Nmm2 for British codesand MPa for European codes The decimal point is shown by afull stop (rather than a comma) in both cases

The basic layout of the Handbook is similar to the previousedition but the contents have been arranged in four separateparts for the convenience of the reader Also the opportunityhas been taken to omit a large amount of material that was nolonger relevant and to revise the entire text to reflect moderndesign and construction practice Part 1 is descriptive in formand covers design requirements loads materials structuralanalysis member design and forms of construction Frequentreference is made in Part 1 to the tables that are found in therest of the Handbook Although specific notes are attached tothese tables in Parts 2 3 and 4 much of the relevant text isembodied in Part 1 and the first part of the Handbook shouldalways be consulted

Part 2 has more detailed information on loads materialproperties and analysis in the form of tabulated data and chartsfor a large range of structural forms This material is largelyindependent of any specific code of practice Parts 3 and 4 cover

the design of members according to the requirements ofthe British and European codes respectively For each code thesame topics are covered in the same sequence so that the readercan move easily from one code to the other Each topic isillustrated by extensive numerical examples

In the Eurocodes some parameters are given recommendedvalues with the option of a national choice Choices also existwith regard to certain classes methods and procedures Thedecisions made by each country are given in a national annexPart 4 of the Handbook already incorporates the values given inthe UK national annex Further information concerning the useof Eurocode 2 is given in PD 6687 Background paper to theUK National Annex to BS EN 1992ndash1ndash1

The Handbook has been an invaluable source of reference forreinforced concrete engineers for over 70 years I madeextensive use of the sixth edition at the start of my professionalcareer 50 years ago This edition contains old and new infor-mation derived by many people and obtained from manysources past and present Although the selection inevitablyreflects the personal experience of the authors the informationhas been well tried and tested I owe a considerable debt ofgratitude to colleagues and mentors from whom I have learntmuch over the years and to the following organisations forpermission to include data for which they hold the copyright

British Cement AssociationBritish Standards InstitutionCabinet Office of Public Sector InformationConstruction Industry Research and Information AssociationPortland Cement AssociationThe Concrete Bridge Development GroupThe Concrete Society

Finally my sincere thanks go to Katy Low and all the staff atTaylor amp Francis Group and especially to my dear wife Joanwithout whose unstinting support this edition would never havebeen completed

Tony ThrelfallMarlow October 2006

Preface to theeleventh edition

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Charles Edward Reynolds was born in 1900 and received hiseducation at Tiffin Boys School Kingston-on-Thames andBattersea Polytechnic After some years with Sir WilliamArroll BRC and Simon Carves he joined Leslie Turner andPartners and later C W Glover and Partners He was for someyears Technical Editor of Concrete Publications Ltd and thenbecame its Managing Editor combining this post with privatepractice In addition to the Reinforced Concrete DesignerrsquosHandbook of which almost 200000 copies have been soldsince it first appeared in 1932 Charles Reynolds was the authorof numerous other books papers and articles concerningconcrete and allied subjects Among his various professionalappointments he served on the council of the Junior Institutionof Engineers and was the Honorary Editor of its journal at hisdeath on Christmas Day 1971

James Cyril Steedman was educated at Varndean GrammarSchool and first was employed by British Rail whom he joinedin 1950 at the age of 16 In 1956 he began working for GKNReinforcements Ltd and later moved to Malcolm Glover andPartners His association with Charles Reynolds began whenafter the publication of numerous articles in the magazine

Concrete and Constructional Engineering he accepted anappointment as Technical Editor of Concrete Publications apost he held for seven years He then continued in privatepractice combining work for the Publications Division of theCement and Concrete Association with his own writing andother activities In 1981 he set up Jacys Computing Servicessubsequently devoting much of his time to the development ofmicro-computer software for reinforced concrete design He isthe joint author with Charles Reynolds of Examples of theDesign of Reinforced Concrete Buildings to BS 8110

Anthony John Threlfall was educated at Liverpool Institute forBoys after which he studied civil engineering at LiverpoolUniversity After eight years working for BRC Pierhead Ltdand IDC Ltd he took a diploma course in concrete structuresand technology at Imperial College For the next four years heworked for CEGB and Camus Ltd and then joined the Cementand Concrete Association in 1970 where he was engagedprimarily in education and training activities until 1993 Afterleaving the CampCA he has continued in private practice toprovide training in reinforced and prestressed concrete designand detailing

The authors

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The publishers would like to thank the following organisationsfor their kind permission to reproduce the following material

Permission to reproduce extracts from British Standards isgranted by BSI This applies to information in Tables 21 2324 27ndash210 215 216 219ndash223 242 243 245 255256 2100 31ndash311 321 322 331ndash345 353ndash361 41ndash46415ndash425 and 428ndash432 British Standards can be obtainedfrom BSI Customer Services 389 Chiswick High StreetLondon W4 4AL Tel 44 (0)20 8996 9001 emailcservicesbsi-globalcom

Information in section 31 and Tables 217ndash218 is reproducedwith permission from the British Cement Association andtaken from the publication Concrete Practice (ref 10)

Information in section 62 is reproduced with permissionfrom the Concrete Bridge Development Group and taken

from the publication An introduction to concrete bridges(ref 52)

Information in section 72 is reproduced with permissionfrom The Concrete Society and taken from TechnicalReport 34 Concrete industrial ground floors ndash A guide todesign and construction (ref 61) Technical Report 34 isavailable to purchase from The Concrete Bookshop wwwconcretebookshopcom Tel 0700 460 7777

Information in Chapter 15 and Table 270 is reproduced withpermission from CIRIA and taken from CIRIA Report 102Design of shear wall buildings London 1984 (ref 38)

Information in Tables 253 and 275ndash279 is reproduced withpermission from the Portland Cement Association (refs 32and 55)

Information in Tables 25 26 and 312 is reproduced withpermission from HMSO

Acknowledgements

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The symbols adopted in this book comply where appropriatewith those in the relevant codes of practice Although these arebased on an internationally agreed system for preparing nota-tions there are numerous differences between the British andthe European codes especially in the use of subscripts Whereadditional symbols are needed to represent properties not usedin the codes these have been selected in accordance with thebasic principles wherever possible

The amount and range of material contained in this bookmake it inevitable that the same symbols have to be used for

different purposes However care has been taken to ensure thatcode symbols are not duplicated except where this has beenfound unavoidable The notational principles adopted for con-crete design purposes are not necessarily best suited to otherbranches of engineering Consequently in those tables relatingto general structural analysis the notation employed in previ-ous editions of this book has generally been retained

Only the principal symbols that are common to all codes arelisted here all other symbols and abbreviations are defined inthe text and tables concerned

Symbols andabbreviations

Ac Area of concrete sectionAs Area of tension reinforcementAs Area of compression reinforcementAsc Area of longitudinal reinforcement in a columnC Torsional constantEc Static modulus of elasticity of concreteEs Modulus of elasticity of reinforcing steelF Action force or load (with appropriate

subscripts)G Shear modulus of concreteGk Characteristic permanent action or dead loadI Second moment of area of cross sectionK A constant (with appropriate subscripts)L Length spanM Bending momentN Axial forceQk Characteristic variable action or imposed loadR Reaction at supportS First moment of area of cross sectionT Torsional moment temperatureV Shear forceWk Characteristic wind load

a Dimension deflectionb Overall width of cross section or width of flanged Effective depth to tension reinforcementd Depth to compression reinforcementf Stress (with appropriate subscripts)fck Characteristic (cylinder) strength of concretefcu Characteristic (cube) strength of concretefyk Characteristic yield strength of reinforcementgk Characteristic dead load per unit areah Overall depth of cross section

i Radius of gyration of concrete sectionk A coefficient (with appropriate subscripts)l Length span (with appropriate subscripts)m Massqk Characteristic imposed load per unit arear Radius1r Curvaturet Thickness timeu Perimeter (with appropriate subscripts)v Shear stress (with appropriate subscripts)x Neutral axis depthz Lever arm of internal forces

Angle ratioe Modular ratio EsEc

Partial safety factor (with appropriate subscripts)c Compressive strain in concretes Strain in tension reinforcements Strain in compression reinforcement Diameter of reinforcing bar Creep coefficient (with appropriate subscripts) Slenderness ratio Poissonrsquos ratio Proportion of tension reinforcement Asbd Proportion of compression reinforcement As bd Stress (with appropriate subscripts) Factor defining representative value of action

BS British StandardEC EurocodeSLS Serviceability limit stateUDL Uniformly distributed loadULS Ultimate limit state

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Part 1

General information

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A structure is an assembly of members each of which under theaction of imposed loads and deformations is subjected tobending or direct force (either tensile or compressive) or to acombination of bending and direct force These effects may beaccompanied by shearing forces and sometimes by torsionImposed deformations occur as a result of concrete shrinkageand creep changes in temperature and differential settlementBehaviour of the structure in the event of fire or accidentaldamage resulting from impact or explosion may need to beexamined The conditions of exposure to environmental andchemical attack also need to be considered

Design includes selecting a suitable form of constructiondetermining the effects of imposed loads and deformationsand providing members of adequate stiffness and resistanceThe members should be arranged so as to combine efficientload transmission with ease of construction consistent withthe intended use of the structure and the nature of the siteExperience and sound judgement are often more important thanprecise calculations in achieving safe and economical structuresComplex mathematics should not be allowed to confuse a senseof good engineering The level of accuracy employed in thecalculations should be consistent throughout the designprocess wherever possible

Structural design is largely controlled by regulations or codesbut even within such bounds the designer needs to exercisejudgement in interpreting the requirements rather than designingto the minimum allowed by the letter of a clause In the UnitedKingdom for many years the design of reinforced concretestructures has been based on the recommendations of BritishStandards For buildings these include lsquoStructural use ofconcretersquo (BS 8110 Parts 1 2 and 3) and lsquoLoading on build-ingsrsquo (BS 6399 Parts 1 2 and 3) For other types of structureslsquoDesign of concrete bridgesrsquo (BS 5400 Part 4) and lsquoDesign ofconcrete structures for retaining aqueous liquidsrsquo (BS 8007)have been used Compliance with the particular requirements ofthe Building Regulations and the Highways Agency Standardsis also necessary in many cases

Since the last edition of this Handbook a comprehensiveset of harmonised Eurocodes (ECs) for the structural andgeotechnical design of buildings and civil engineering workshas been developed The Eurocodes were first introduced asEuronorme Voluntaire (ENV) standards intended for use inconjunction with a national application document (NAD) asan alternative to national codes for a limited number of years

These have now been largely replaced by Euronorme (EN)versions with each member state adding a National Annex(NA) containing nationally determined parameters in order toimplement the Eurocode as a national standard The relevantdocuments for concrete structures are EC 0 Basis of structuraldesign EC 1 Actions on structures and EC 2 Design of con-crete structures The last document is in four parts namely ndashPart 11 General rules and rules for buildings Part 12Structural fire design Part 2 Reinforced and prestressed con-crete bridges and Part 3 Liquid-retaining and containingstructures

The tables to be found in Parts 2 3 and 4 of this Handbookenable the designer to reduce the amount of arithmetical workinvolved in the analysis and design of members to the relevantstandards The use of such tables not only increases speed butalso eliminates inaccuracies provided the tables are thoroughlyunderstood and their applications and limitations are realisedIn the appropriate chapters of Part 1 and in the supplementaryinformation given on the pages preceding the tables the basisof the tabulated material is described Some general informa-tion is also provided The Appendix contains trigonometricaland other mathematical formulae and data

11 ECONOMICAL STRUCTURES

The cost of construction of a reinforced concrete structure isobviously affected by the prices of concrete reinforcementformwork and labour The most economical proportions ofmaterials and labour will depend on the current relationshipbetween the unit prices Economy in the use of formwork isgenerally achieved by uniformity of member size and the avoid-ance of complex shapes and intersections In particular casesthe use of available formwork of standard sizes may determinethe structural arrangement In the United Kingdom speed ofconstruction generally has a major impact on the overall costFast-track construction requires the repetitive use of a rapidformwork system and careful attention to both reinforcementdetails and concreting methods

There are also wider aspects of economy such as whetherthe anticipated life and use of a proposed structure warrant theuse of higher or lower factors of safety than usual or whetherthe use of a more expensive form of construction is warrantedby improvements in the integrity and appearance of the structureThe application of whole-life costing focuses attention on

Chapter 1

Introduction

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Introduction4

whether the initial cost of a construction of high qualitywith little or no subsequent maintenance is likely to be moreeconomical than a cheaper construction combined with theexpense of maintenance

The experience and method of working of the contractor theposition of the site and the nature of the available materials andeven the method of measuring the quantities together withnumerous other points all have their effect consciously or noton the designerrsquos attitude towards a contract So many andvaried are the factors involved that only experience and acontinuing study of design trends can give reliable guidanceAttempts to determine the most economical proportions for aparticular member based only on inclusive prices of concretereinforcement and formwork are likely to be misleading It isnevertheless possible to lay down certain principles

In broad terms the price of concrete increases with thecement content as does the durability and strength Concretegrades are often determined by durability requirements withdifferent grades used for foundations and superstructuresStrength is an important factor in the design of columns andbeams but rarely so in the case of slabs Nevertheless the samegrade is generally used for all parts of a superstructure exceptthat higher strength concrete may sometimes be used to reducethe size of heavily loaded columns

In the United Kingdom mild steel and high yield reinforce-ments have been used over the years but grade 500 is nowproduced as standard available in three ductility classes A B andC It is always uneconomical in material terms to use compressionreinforcement in beams and columns but the advantages gainedby being able to reduce member sizes and maintain the samecolumn size over several storeys generally offset the additionalmaterial costs For equal weights of reinforcement the combinedmaterial and fixing costs of small diameter bars are greater thanthose of large diameter bars It is generally sensible to use thelargest diameter bars consistent with the requirements for crackcontrol Fabric (welded mesh) is more expensive than barreinforcement in material terms but the saving in fixing time willoften result in an overall economy particularly in slabs and walls

Formwork is obviously cheaper if surfaces are plane and atright angles to each other and if there is repetition of use Thesimplest form of floor construction is a solid slab of constantthickness Beam and slab construction is more efficient struc-turally but less economical in formwork costs Two-way beamsystems complicate both formwork and reinforcement detailswith consequent delay in the construction programmeIncreased slab efficiency and economy over longer spans maybe obtained by using a ribbed form of construction Standardtypes of trough and waffle moulds are available in a range ofdepths Precasting usually reduces considerably the amountof formwork labour and erection time Individual mouldsare more expensive but can be used many more timesthan site formwork Structural connections are normally moreexpensive than with monolithic construction The economicaladvantage of precasting and the structural advantage of in situcasting may be combined in composite forms of construction

In many cases the most economical solution can only bedetermined by comparing the approximate costs of differentdesigns This may be necessary to decide say when a simplecantilever retaining wall ceases to be more economical thanone with counterforts or when a beam and slab bridge is moreeconomical than a voided slab The handbook Economic

Concrete Frame Elements published by the British CementAssociation on behalf of the Reinforced Concrete Councilenables designers to rapidly identify least-cost options for thesuperstructure of multi-storey buildings

12 DRAWINGS

In most drawing offices a practice has been developed to suitthe particular type of work done Computer aided drafting andreinforcement detailing is widely used The following observa-tions should be taken as general principles that accord with therecommendations in the manual Standard method of detailingstructural concrete published by the Institution of StructuralEngineers (ref 1)

It is important to ensure that on all drawings for a particularcontract the same conventions are adopted and uniformity ofsize and appearance are achieved In the preliminary stagesgeneral arrangement drawings of the whole structure are usuallyprepared to show the layout and sizes of beams columns slabswalls foundations and other members A scale of 1100 isrecommended although a larger scale may be necessary forcomplex structures Later these or similar drawings are devel-oped into working drawings and should show precisely suchparticulars as the setting-out of the structure in relation to anyadjacent buildings or other permanent works and the level ofsay the ground floor in relation to a fixed datum All principaldimensions such as distances between columns and walls andthe overall and intermediate heights should be shown Plansshould generally incorporate a gridline system with columnspositioned at the intersections Gridlines should be numbered 12 3 and so on in one direction and lettered A B C and soon in the other direction with the sequences starting at thelower left corner of the grid system The references canbe used to identify individual beams columns and othermembers on the reinforcement drawings

Outline drawings of the members are prepared to suitablescales such as 120 for beams and columns and 150 for slabsand walls with larger scales being used for cross sectionsReinforcement is shown and described in a standard way Theonly dimensions normally shown are those needed to positionthe bars It is generally preferable for the outline of the concreteto be indicated by a thin line and to show the reinforcement bybold lines The lines representing the bars should be shown inthe correct positions with due allowance for covers and thearrangement at intersections and laps so that the details onthe drawing represent as nearly as possible the appearanceof the reinforcement as fixed on site It is important to ensurethat the reinforcement does not interfere with the formation ofany holes or embedment of any other items in the concrete

A set of identical bars in a slab shown on plan might bedescribed as 20H16-03-150B1 This represents 20 numbergrade 500 bars of 16 mm nominal size bar mark 03 spaced at150 mm centres in the bottom outer layer The bar mark is anumber that uniquely identifies the bar on the drawing and thebar bending schedule Each different bar on a drawing is givena different bar mark Each set of bars is described only once onthe drawing The same bars on a cross section would be denotedsimply by the bar mark Bar bending schedules are prepared foreach drawing on separate forms according to recommendationsin BS 8666 Specification for scheduling dimensioning bendingand cutting of steel reinforcement for concrete

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There are two principal stages in the calculations requiredto design a reinforced concrete structure In the first stagecalculations are made to determine the effect on the structureof loads and imposed deformations in terms of appliedmoments and forces In the second stage calculations are madeto determine the capacity of the structure to withstand sucheffects in terms of resistance moments and forces

Factors of safety are introduced in order to allow for theuncertainties associated with the assumptions made and thevalues used at each stage For many years unfactored loadswere used in the first stage and total factors of safety wereincorporated in the material stresses used in the second stageThe stresses were intended to ensure both adequate safety andsatisfactory performance in service This simple approach waseventually replaced by a more refined method in which specificdesign criteria are set and partial factors of safety are incorpo-rated at each stage of the design process

21 DESIGN CRITERIA AND SAFETY FACTORS

A limit-state design concept is used in British and EuropeanCodes of Practice Ultimate (ULS) and serviceability (SLS)limit states need to be considered as well as durability and inthe case of buildings fire-resistance Partial safety factors areincorporated into loads (including imposed deformations) andmaterial strengths to ensure that the probability of failure (notsatisfying a design requirement) is acceptably low

In BS 8110 at the ULS a structure should be stable under allcombinations of dead imposed and wind load It should also berobust enough to withstand the effects of accidental loads dueto an unforeseen event such as a collision or explosion withoutdisproportionate collapse At the SLS the effects in normal useof deflection cracking and vibration should not cause thestructure to deteriorate or become unserviceable A deflectionlimit of span250 applies for the total sag of a beam or slabrelative to the level of the supports A further limit the lesser ofspan500 or 20 mm applies for the deflection that occurs afterthe application of finishes cladding and partitions so as to avoiddamage to these elements A limit of 03 mm generally appliesfor the width of a crack at any point on the concrete surface

In BS 5400 an additional partial safety factor is introducedThis is applied to the load effects and takes account of themethod of structural analysis that is used Also there are moreload types and combinations to be considered At the SLSthere are no specified deflection limits but the cracking limits

are more critical Crack width limits of 025 015 or 01 mmapply according to surface exposure conditions Compressivestress limits are also included but in many cases these do notneed to be checked Fatigue considerations require limitationson the reinforcement stress range for unwelded bars and morefundamental analysis if welding is involved Footbridges areto be analysed to ensure that either the fundamental naturalfrequency of vibration or the maximum vertical accelerationmeets specified requirements

In BS 8007 water-resistance is a primary design concernAny cracks that pass through the full thickness of a section arelikely to allow some seepage initially resulting in surfacestaining and damp patches Satisfactory performance dependsupon autogenous healing of such cracks taking place within afew weeks of first filling in the case of a containment vesselA crack width limit of 02 mm normally applies to all cracksirrespective of whether or not they pass completely through thesection Where the appearance of a structure is considered to beaesthetically critical a limit of 01 mm is recommended

There are significant differences between the structural andgeotechnical codes in British practice The approach to thedesign of foundations in BS 8004 is to use unfactored loadsand total factors of safety For the design of earth-retainingstructures CP2 (ref 2) used the same approach In 1994 CP2was replaced by BS 8002 in which mobilisation factors areintroduced into the calculation of soil strengths The resultingvalues are then used in BS 8002 for both serviceability andultimate requirements In BS 8110 the loads obtained fromBS 8002 are multiplied by a partial safety factor at the ULS

Although the design requirements are essentially the samein the British and European codes there are differences ofterminology and in the values of partial safety factors In theEurocodes loads are replaced by actions with dead loads as per-manent actions and all live loads as variable actions Each vari-able action is given several representative values to be used forparticular purposes The Eurocodes provide a more unifiedapproach to both structural and geotechnical design

Details of design requirements and partial safety factors to beapplied to loads and material strengths are given in Chapter 21for British Codes and Chapter 29 for Eurocodes

22 LOADS (ACTIONS)

The loads (actions) acting on a structure generally consist ofa combination of dead (permanent) and live (variable) loads

Chapter 2

Design criteria safetyfactors and loads

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Design criteria safety factors and loads6

In limit-state design a design load (action) is calculated bymultiplying the characteristic (or representative) value by anappropriate partial factor of safety The characteristic value isgenerally a value specified in a relevant standard or code Inparticular circumstances it may be a value given by a client ordetermined by a designer in consultation with the client

In BS 8110 characteristic dead imposed and wind loadsare taken as those defined in and calculated in accordancewith BS 6399 Parts 1 2 and 3 In BS 5400 characteristicdead and live loads are given in Part 2 but these have beensuperseded in practice by the loads in the appropriateHighways Agency standards These include BD 3701 andBD 6094 and for the assessment of existing bridgesBD 2101 (refs 3ndash5)

When EC 2 Part 11 was first introduced as an ENVdocument characteristic loads were taken as the values given inBS 6399 but with the specified wind load reduced by 10 Thiswas intended to compensate for the partial safety factor appliedto wind at the ULS being higher in the Eurocodes than in BS8110 Representative values were then obtained by multiplyingthe characteristic values by factors given in the NAD In theEN documents the characteristic values of all actions are givenin EC 1 and the factors to be used to determine representativevalues are given in EC 0

23 DEAD LOADS (PERMANENT ACTIONS)

Dead loads include the weights of the structure itself andall permanent fixtures finishes surfacing and so on Whenpermanent partitions are indicated they should be included asdead loads acting at the appropriate locations Where any doubtexists as to the permanency of the loads they should be treatedas imposed loads Dead loads can be calculated from the unitweights given in EC 1 Part 11 or from actual known weightsof the materials used Data for calculating dead loads are givenin Tables 21 and 22

24 LIVE LOADS (VARIABLE ACTIONS)

Live loads comprise any transient external loads imposed on thestructure in normal use due to gravitational dynamic andenvironmental effects They include loads due to occupancy(people furniture moveable equipment) traffic (road railpedestrian) retained material (earth liquids granular) snowwind temperature ground and water movement wave actionand so on Careful assessment of actual and probable loads is avery important factor in producing economical and efficientstructures Some imposed loads like those due to containedliquids can be determined precisely Other loads such as thoseon floors and bridges are very variable Snow and wind loadsare highly dependent on location Data for calculating loadsfrom stored materials are given in EC 1 Part 11

241 Floors

For most buildings the loads imposed on floors are specified inloading standards In BS 6399 Part 1 loads are specifiedaccording to the type of activity or occupancy involved Datafor residential buildings and for offices and particular workareas is given in Table 23 Imposed loads are given both as

a uniformly distributed load in kNm2 and a concentrated loadin kN The floor should be designed for the worst effects ofeither load The concentrated load needs to be considered forisolated short span members and for local effects such aspunching in a thin flange For this purpose a square contactarea with a 50 mm side may be assumed in the absence of anymore specific information Generally the concentrated loaddoes not need to be considered in slabs that are either solid orotherwise capable of effective lateral distribution Wherean allowance has to be made for non-permanent partitions auniformly distributed load equal to one-third of the load permetre run of the finished partitions may be used For offices theload used should not be less than 10 kNm2

The floors of garages are considered in two categoriesnamely those for cars and light vans and those for heaviervehicles In the lighter category the floor may be designedfor loads specified in the form described earlier In the heaviercategory the most adverse disposition of loads determinedfor the specific types of vehicle should be considered

The total imposed loads to be used for the design of beams maybe reduced by a percentage that increases with the area of floorsupported as given in Table 23 This does not apply to loads dueto storage vehicles plant or machinery For buildings designed tothe Eurocodes imposed loads are given in EC 1 Part 11

In all buildings it is advisable to affix a notice indicatingthe imposed load for which the floor is designed Floors ofindustrial buildings where plant and machinery are installedneed to be designed not only for the load when the plant is inrunning order but also for the probable load during erectionand testing which in some cases may be more severe Datafor loads imposed on the floors of agricultural buildings bylivestock and farm vehicles is given in BS 5502 Part 22

242 Structures subject to dynamic loads

The loads specified in BS 6399 Part 1 include allowancesfor small dynamic effects that should be sufficient for mostbuildings However the loading does not necessarily coverconditions resulting from rhythmical and synchronised crowdmovements or the operation of some types of machinery

Dynamic loads become significant when crowd movements(eg dancing jumping rhythmic stamping) are synchronisedIn practice this is usually associated with lively pop concertsor aerobics events where there is a strong musical beat Suchactivities can generate both horizontal and vertical loads Ifthe movement excites a natural frequency of the affected partof the structure resonance occurs which can greatly amplify theresponse Where such activities are likely to occur the structureshould be designed to either avoid any significant resonanceeffects or withstand the anticipated dynamic loads Somelimited guidance on dynamic loads caused by activities such asjumping and dancing is provided in BS 6399 Part 1 Annexe ATo avoid resonance effects the natural frequency of vibrationof the unloaded structure should be greater than 84 Hz for thevertical mode and greater than 40 Hz for the horizontal modes

Different types of machinery can give rise to a wide range ofdynamic loads and the potential resonant excitation of thesupporting structure should be considered Where necessaryspecialist advice should be sought

Footbridges are subject to particular requirements that willbe examined separately in the general context of bridges

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243 Parapets barriers and balustrades

Parapets barriers balustrades and other elements intended toretain stop or guide people should be designed for horizontalloads Values are given in BS 6399 Part 1 for a uniformlydistributed line load and for both uniformly distributed andconcentrated loads applied to the infill These are not takentogether but are applied as three separate load cases The lineload should be considered to act at a height of 11 m above adatum level taken as the finished level of the access platformor the pitch line drawn through the nosing of the stair treads

Vehicle barriers for car parking areas are also includedin BS 6399 Part 1 The horizontal force F as given in thefollowing equation is considered to act at bumper heightnormal to and uniformly distributed over any length of 15 m ofthe barrier By the fundamental laws of dynamics

F 05mv2(b c) (in kN)

m gross mass of vehicle (in kg)v speed of vehicle normal to barrier taken as 45 msecb deflection of barrier (in mm)c deformation of vehicle taken as 100 mm unless better

evidence is available

For car parks designed on the basis that the gross mass of thevehicles using it will not exceed 2500 kg (but taking as arepresentative value of the vehicle population m 1500 kg)and provided with rigid barriers (b 0) F is taken as 150 kNacting at a height of 375 mm above floor level It should benoted that bumper heights have been standardised at 445 mm

244 Roofs

The imposed loads given in Table 24 are additional to allsurfacing materials and include for snow and other incidentalloads but exclude wind pressure The snow load on the roofis determined by multiplying the estimated snow load on theground at the site location and altitude (the site snow load) byan appropriate snow load shape coefficient The main loadingconditions to be considered are

(a) a uniformly distributed snow load over the entire rooflikely to occur when snow falls with little or no wind

(b) a redistributed (or unevenly deposited) snow load likely tooccur in windy conditions

For flat or mono-pitch roofs it is sufficient to consider thesingle load case resulting from a uniform layer of snow asgiven in Table 24 For other roof shapes and for the effects oflocal drifting of snow behind parapets reference should bemade to BS 6399 Part 3 for further information

Minimum loads are given for roofs with no access (other thanthat necessary for cleaning and maintenance) and for roofswhere access is provided Roofs like floors should be designedfor the worst effects of either the distributed load or theconcentrated load For roofs with access the minimum loadwill exceed the snow load in most cases

If a flat roof is used for purposes such as a cafeacute playgroundor roof garden the appropriate imposed load for such a floorshould be allowed For buildings designed to the Eurocodessnow loads are given in EC 1 Part 13

245 Columns walls and foundations

Columns walls and foundations of buildings are designed forthe same loads as the slabs or beams that they support If theimposed loads on the beams are reduced according to the areaof floor supported the supporting members may be designedfor the same reduced loads Alternatively where two or morefloors are involved and the loads are not due to storage theimposed loads on columns or other supporting members maybe reduced by a percentage that increases with the number offloors supported as given in Table 23

246 Structures supporting cranes

Cranes and other hoisting equipment are often supported oncolumns in factories or similar buildings It is important that adimensioned diagram of the actual crane to be installed isobtained from the makers to ensure that the right clearances areprovided and the actual loads are taken into account For loadsdue to cranes reference should be made to BS 2573

For jib cranes running on rails on supporting gantries theload to which the structure is subjected depends on the actualdisposition of the weights of the crane The wheel loads aregenerally specified by the crane maker and should allow forthe static and dynamic effects of lifting discharging slewingtravelling and braking The maximum wheel load underpractical conditions may occur when the crane is stationary andhoisting the load at the maximum radius with the line of the jibdiagonally over one wheel

247 Structures supporting lifts

The effect of acceleration must be considered in addition to thestatic loads when calculating loads due to lifts and similarmachinery If a net static load F is subject to an accelerationa (ms2) the resulting load on the supporting structure isapproximately F (1 0098a) The average acceleration ofa passenger lift may be about 06 ms2 but the maximumacceleration will be considerably greater BS 2655 requiresthe supporting structure to be designed for twice the loadsuspended from the beams when the lift is at rest with anoverall factor of safety of 7 The deflection under the designload should not exceed span1500

248 Bridges

The analysis and design of bridges is now so complex thatit cannot be adequately treated in a book of this nature andreference should be made to specialist publications Howeverfor the guidance of designers the following notes regardingbridge loading are provided since they may also be applicableto ancillary construction and to structures having features incommon with bridges

Road bridges The loads to be considered in the design ofpublic road bridges in the United Kingdom are specified in theHighways Agency Standard BD 3701 Loads for HighwayBridges This is a revised version of BS 5400 Part 2 issuedby the Department of Transport rather than by BSI TheStandard includes a series of major amendments as agreed bythe BSI Technical Committee BD 3701 deals with both perma-nent loads (dead superimposed dead differential settlementearth pressure) and transient loads due to traffic use (vehicular

Live loads (variable actions) 7

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pedestrian) and environmental effects (wind temperature)The collision loads in BD 3701 may be applicable in certaincircumstances where agreed with the appropriate authority butin most cases the requirements of BD 6094 The design ofhighway bridges for vehicle collision loads will apply

Details of live loads due to traffic to be considered in thedesign of highway bridges are given in Table 25 Two typesof standard live loading are given in BD 3701 to representnormal traffic and abnormal vehicles respectively Loads areapplied to notional lanes of equal width The number ofnotional lanes is determined by the width of the carriagewaywhich includes traffic lanes hard shoulders and hard stripsand several typical examples are shown diagrammatically inBD 3701 Notional lanes are used rather than marked lanesin order to allow for changes of use and the introduction oftemporary contra-flow schemes

Type HA loading covers all the vehicles allowable under theRoad Vehicles (Construction and Use) and Road Vehicles(Authorised Weight) Regulations Values are given in terms ofa uniformly distributed load (UDL) and a single knife-edgeload (KEL) to be applied in combination to each notional laneThe specified intensity of the UDL (kNm) reduces as the loadedlength increases which allows for two effects At the shorterend it allows for loading in the vicinity of axles or bogies beinggreater than the average loading for the whole vehicle At thelonger end it takes account of the reducing percentage of heavygoods vehicles contained in the total vehicle population TheKEL of 120 kN is to be applied at any position within the UDLloaded length and spread over a length equal to the notional lanewidth In determining the loads consideration has been given tothe effects of impact vehicle overloading and unforeseen changesin traffic patterns The loading derived after application ofseparate factors for each of these effects was considered torepresent an ultimate load which was then divided by 15to obtain the specified nominal loads

The loads are multiplied by lane factors whose valuesdepend on the particular lane and the loaded length This isdefined as the length of the adverse area of the influence linethat is the length over which the load application increases themagnitude of the effect to be determined The lane factors takeaccount of the low probability of all lanes being fully loaded atthe same time They also for the shorter loaded lengths allowfor the effect of lateral bunching of vehicles As an alternativeto the combined loads a single wheel load of 100 kN appliedat any position is also to be considered

Type HB loading derives from the nature of exceptionalindustrial loads such as electrical transformers generatorspressure vessels and machine presses likely to use the roads inthe neighbouring area It is represented by a sixteen-wheelvehicle consisting of two bogies each one having two axleswith four wheels per axle Each axle represents one unit ofloading (equivalent to 10 kN) Bridges on public highwaysare designed for a specific number of units of HB loadingaccording to traffic use typically 45 units for trunk roads andmotorways 375 units for principal roads and 30 units for allother public roads Thus the maximum number of 45 unitscorresponds to a total vehicle load of 1800 kN with 450 kNper axle and 1125 kN per wheel The length of the vehicle isvariable according to the spacing of the bogies for which fivedifferent values are specified The HB vehicle can occupy anytransverse position on the carriageway and is considered to

displace HA loading over a specified area surrounding thevehicle Outside this area HA loading is applied as specifiedand shown by diagrams in BD 3701 The combined loadarrangement is normally critical for all but very long bridges

Road bridges may be subjected to forces other than those dueto dead load and traffic load These include forces due to windtemperature differential settlement and earth pressure Theeffects of centrifugal action and longitudinal actions due totraction braking and skidding must also be considered as wellas vehicle collision loads on supports and superstructure Fordetails of the loads to be considered on highway bridge parapetsreference should be made to BD 5293 (ref 6)

In the assessment of existing highway bridges traffic loadsare specified in the Highways Agency document BD 2101 TheAssessment of Highway Bridges and Structures In this casethe type HA loading is multiplied by a reduction factor thatvaries according to the road surface characteristics traffic flowconditions and vehicle weight restrictions Some of the contin-gency allowances incorporated in the design loading havealso been relaxed Vehicle weight categories of 40 38 25 1775 and 3 tonnes are considered as well as two groups offire engines For further information on reduction factors andspecific details of the axle weight and spacing values in eachcategory reference should be made BD 2101

Footbridges Details of live loads due to pedestrians to beconsidered in the design of footcycle track bridges are given inTable 26 A uniformly distributed load of 5 kNm2 is specified forloaded lengths up to 36 m Reduced loads may be used for bridgeswhere the loaded length exceeds 36 m except that specialconsideration is required in cases where exceptional crowds couldoccur For elements of highway bridges supporting footwayscycle tracks further reductions may be made in the pedestrian liveload where the width is greater than 2 m or the element alsosupports a carriageway When the footwaycycle track is notprotected from vehicular traffic by an effective barrier there is aseparate requirement to consider an accidental wheel loading

It is very important that consideration is given to vibrationthat could be induced in footcycle track bridges by resonancewith the movement of users or by deliberate excitation InBD 3701 the vibration requirement is deemed to be satisfiedin cases where the fundamental natural frequency of vibrationexceeds 5 Hz for the unloaded bridge in the vertical direction and15 Hz for the loaded bridge in the horizontal direction Whenthe fundamental natural frequency of vertical vibration fo doesnot exceed 5 Hz the maximum vertical acceleration shouldbe limited to 05radicfo ms2 Methods for determining the naturalfrequency of vibration and the maximum vertical accelerationare given in Appendix B of BD 3701 Where the fundamentalnatural frequency of horizontal vibration does not exceed15 Hz special consideration should be given to the possibilityof pedestrian excitation of lateral movements of unacceptablemagnitude Bridges possessing low mass and damping andexpected to be used by crowds of people are particularlysusceptible to such vibrations

Railway bridges Details of live loads to be considered in thedesign of railway bridges are given in Table 26 Two types ofstandard loading are given in BD 3701 type RU for mainline railways and type RL for passenger rapid transit systemsA further type SW0 is also included for main line railways

Design criteria safety factors and loads8

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The type RU loading was derived by a Committee of theInternational Union of Railways (UIC) to cover present andanticipated future loading on railways in Great Britain and on theContinent of Europe Nowadays motive power tends to be dieseland electric rather than steam and this produces axle loads andarrangements for locomotives that are similar to those for bogiefreight vehicles (these often being heavier than the locomotivesthat draw them) In addition to normal train loading whichcan be represented quite well by a uniformly distributed loadof 80 kNm railway bridges are occasionally subjected toexceptionally heavy abnormal loads For short loaded lengths itis necessary to introduce heavier concentrated loads to simulateindividual axles and to produce high shears at the ends Type RUloading consists of four concentrated loads of 250 kN precededand followed by a uniformly distributed load of 80 kNm For acontinuous bridge type SW0 loading is also to be considered asan additional and separate load case This loading consists oftwo uniformly distributed loads of 133 kNm each 15 m longseparated by a distance of 53 m Both types of loading whichare applied to each track or as specified by the relevant authoritywith half the track load acting on each rail are to be multipliedby appropriate dynamic factors to allow for impact lurchingoscillation and other dynamic effects The factors have beencalculated so that in combination with the specified loading theycover the effects of slow moving heavy and fast moving lightvehicles Exceptional vehicles are assumed to move at speeds notexceeding 80 kmh heavy wagons at speeds up to 120 kmh andpassenger trains at speeds up to 200 kmh

The type RL loading was derived by the London TransportExecutive to cover present and anticipated future loading onlines that carry only rapid transit passenger trains and lightengineersrsquo works trains Passenger trains include a variety ofstock of different ages loadings and gauges used on surfaceand tube lines Works trains include locomotives cranesand wagons used for maintenance purposes Locomotives areusually of the battery car type but diesel shunt varieties aresometimes used The rolling stock could include a 30t steamcrane 6t diesel cranes 20t hopper cranes and bolster wagonsThe heaviest train would comprise loaded hopper wagonshauled by battery cars Type RL loading consists of a singleconcentrated load of 200 kN coupled with a uniformly distrib-uted load of 50 kNm for loaded lengths up to 100 m Forloaded lengths in excess of 100 m the previous loading ispreceded and followed by a distributed load of 25 kNm Theloads are to be multiplied by appropriate dynamic factors Analternative bogie loading comprising two concentrated loadsone of 300 kN and the other of 150 kN spaced 24 m apart isalso to be considered on deck structures to check the ability ofthe deck to distribute the loads adequately

For full details of the locomotives and rolling stock coveredby each loading type and information on other loads to be con-sidered in the design of railway bridges due to the effects ofnosing centrifugal action traction and braking and in the eventof derailment reference should be made to BD 3701

249 Dispersal of wheel loads

A load from a wheel or similar concentrated load bearing on asmall but definite area of the supporting surface (called thecontact area) may be assumed to be further dispersed over anarea that depends on the combined thickness of any surfacing

material filling and underlying constructional material Thewidth of the contact area of a wheel on a slab is equal tothe width of the tyre The length of the contact area depends onthe type of tyre and the nature of the slab surface It is nearlyzero for steel tyres on steel plate or concrete The maximumcontact length is probably obtained with an iron wheel on loosemetalling or a pneumatic tyre on an asphalt surface

The wheel loads given in BD 3701 as part of the standardhighway loading are to be taken as uniformly distributed overa circular or square contact area assuming an effective pressureof 11 Nmm2 Thus for the HA single wheel load of 100 kNthe contact area becomes a 340 mm diameter circle or a squareof 300 mm side For the HB vehicle where 1 unit of loadingcorresponds to 25 kN per wheel the side of the square contactarea becomes approximately 260 mm for 30 units 290 mm for375 units and 320 mm for 45 units

Dispersal of the load beyond the contact area may be takenat a spread-to-depth ratio of 1 horizontally to 2 vertically forasphalt and similar surfacing so that the dimensions of thecontact area are increased by the thickness of the surfacingThe resulting boundary defines the loaded area to be used whenchecking for example the effects of punching shear on theunderlying structure

For a structural concrete slab 45o spread down to the levelof the neutral axis may be taken Since for the purpose ofstructural analysis the position of the neutral axis is usuallytaken at the mid-depth of the section the dimensions of thecontact area are further increased by the total thickness of theslab The resulting boundary defines the area of the patch loadto be used in the analysis

The concentrated loads specified in BD 3701 as part of therailway loading will be distributed both longitudinally bythe continuous rails to more than one sleeper and transverselyover a certain area of deck by the sleeper and ballast It maybe assumed that two-thirds of a concentrated load applied toone sleeper will be transmitted to the deck by that sleeper andthe remainder will be transmitted equally to the adjacent sleeperon either side Where the depth of ballast is at least 200 mmthe distribution may be assumed to be half to the sleeper lyingunder the load and half equally to the adjacent sleeper oneither side The load acting on the sleeper from each rail maybe distributed uniformly over the ballast at the level of theunderside of the sleeper for a distance taken symmetricallyabout the centreline of the rail of 800 mm or twice the distancefrom the centreline of the rail to the nearer end of the sleeperwhichever is the lesser Dispersal of the loads applied to theballast may be taken at an angle of 5o to the vertical down tothe supporting structure The distribution of concentrated loadsapplied to a track without ballast will depend on the relativestiffness of the rail the rail support and the bridge deck itself

25 WIND LOADS

All structures built above ground level are affected by thewind to a greater or lesser extent Wind comprises a randomfluctuating velocity component (turbulence or lsquogustinessrsquo)superimposed on a steady mean component The turbulenceincreases with the roughness of the terrain due to frictionaleffects between the wind and features on the ground such asbuildings and vegetation On the other hand the frictionaleffects also reduce the mean wind velocity

Wind loads 9

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Wind loads are dynamic and fluctuate continuously in bothmagnitude and position Some relatively flexible structuressuch as tall slender masts towers and chimneys suspensionbridges and other cable-stayed structures may be susceptible todynamic excitation in which case lateral deflections will be animportant consideration However the vast majority of build-ings are sufficiently stiff for the deflections to be small inwhich case the structure may be designed as if it was static

251 Wind speed and pressure

The local wind climate at any site in the United Kingdom can bepredicted reliably using statistical methods in conjunction withboundary-layer wind flow models However the complexity offlow around structures is not sufficiently well understood toallow wind pressures or distributions to be determined directlyFor this reason the procedure used in most modern wind codesis to treat the calculation of wind speed in a fully probabilisticmanner whilst continuing to use deterministic values of pressurecoefficients This is the approach adopted in BS 6399 Part 2which offers a choice of two methods for calculating wind loadsas follows

standard method uses a simplified procedure to obtain aneffective wind speed which is used with standard pressurecoefficients for orthogonal load cases

directional method provides a more precise assessment ofeffective wind speeds for particular wind directions which isused with directional pressure coefficients for load cases ofany orientation

The starting point for both methods is the basic hourly-meanwind speed at a height of 10 m in standard lsquocountryrsquo terrainhaving an annual risk (probability) of being exceeded of 002(ie a mean recurrence interval of 50 years) A map of basicwind speeds covering Great Britain and Ireland is provided

The basic hourly-mean wind speed is corrected according tothe site altitude and if required the wind direction seasonand probability to obtain an effective site wind speed This isfurther modified by a site terrain and building height factorto obtain an effective gust wind speed Ve ms which is used tocalculate an appropriate dynamic pressure q 0613Ve

2 Nm2Topographic effects are incorporated in the altitude factor for

the standard method and in the terrain and building factor for thedirectional method The standard method can be used in hand-based calculations and gives a generally conservative resultwithin its range of applicability The directional method is lessconservative and is not limited to orthogonal design cases Theloading is assessed in more detail but with the penalty ofincreased complexity and computational effort For furtherdetails of the directional method reference should be made toBS 6399 Part 2

252 Buildings

The standard method of BS 6399 Part 2 is the source of theinformation in Tables 27ndash29 The basic wind speed andthe correction factors are given in Table 27 The altitudefactor depends on the location of the structure in relation tothe local topography In terrain with upwind slopes exceeding005 the effects of topography are taken to be significant for

certain designated zones of the upwind and downwind slopesIn this case further reference should be made to BS 6399Part 2 When the orientation of the building is known the windspeed may be adjusted according to the direction under consid-eration Where the building height is greater than the crosswindbreadth for the direction being considered a reduction in thelateral load may be obtained by dividing the building into anumber of parts For buildings in town terrain the effectiveheight may be reduced as a result of the shelter afforded bystructures upwind of the site For details of the adjustmentsbased on wind direction division of buildings into parts and theinfluence of shelter on effective height reference should bemade to BS 6399 Part 2

When the wind acts on a building the windward faces aresubjected to direct positive pressure the magnitude of whichcannot exceed the available kinetic energy of the wind Asthe wind is deflected around the sides and over the roof of thebuilding it is accelerated lowering the pressure locally onthe building surface especially just downwind of the eavesridge and corners These local areas where the acceleration ofthe flow is greatest can experience very large wind suctionsThe surfaces of enclosed buildings are also subjected to internalpressures Values for both external and internal pressures areobtained by multiplying the dynamic pressure by appropriatepressure coefficients and size effect factors The overall force ona rectangular building is determined from the normal forces onthe windward-facing and leeward-facing surfaces the frictionaldrag forces on surfaces parallel to the direction of the wind anda dynamic augmentation factor that depends on the buildingheight and type

Details of the dimensions used to define surface pressuresand forces and values for dynamic augmentation factors andfrictional drag coefficients are given in Table 28 Size effectfactors and external and internal pressure coefficients for thewalls of rectangular buildings are given in Table 29 Furtherinformation including pressure coefficients for various roofforms free-standing walls and cylindrical structures such assilos tanks and chimneys and procedures for more-complexbuilding shapes are given in BS 6399 Part 2 For buildingsdesigned to the Eurocodes data for wind loading is given inEC 1 Part 12

253 Bridges

The approach used for calculating wind loads in BD 3701 is ahybrid mix of the methods given in BS 6399 Part 2 The direc-tional method is used to calculate the effective wind speed asthis gives a better estimate of wind speeds in towns and for sitesaffected by topography In determining the wind speed theprobability factor is taken as 105 appropriate to a return periodof 120 years Directional effective wind speeds are derived fororthogonal load cases and used with standard drag coefficientsto obtain wind loads on different elements of the structure suchas decks parapets and piers For details of the proceduresreference must be made to BD 3701

26 MARITIME STRUCTURES

The forces acting upon sea walls dolphins wharves jettiespiers docks and similar maritime structures include those dueto winds and waves blows and pulls from vessels the loads

Design criteria safety factors and loads10

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from cranes roads railways and stored goods imposed on thedeck and the pressures of earth retained behind the structure

For wharves or jetties of solid construction the energy ofimpact due to blows from vessels berthing is absorbed by themass of the structure usually without damage to the structureor vessel if fendering is provided With open constructionconsisting of braced piles or piers supporting the deck in whichthe mass of the structure is comparatively small the forcesresulting from impact must be considered The forces dependon the weight and speed of approach of the vessel on theamount of fendering and on the flexibility of the structureIn general a large vessel will approach at a low speed and asmall vessel at a higher speed Some typical examples are a1000 tonne vessel at 03 ms a 10 000 tonne vessel at 02 msand a 100 000 tonne vessel at 015 ms The kinetic energy of avessel displacing F tonnes approaching at a speed V ms isequal to 0514FV 2 kNm Hence the kinetic energy of a2000 tonne vessel at 03 ms and a 5000 tonne vessel at 02 msis about 100 kNm in each case If the direction of approachof a vessel is normal to the face of a jetty the whole of thisenergy must be absorbed on impact More commonly a vesselapproaches at an angle with the face of the jetty and touchesfirst at one point about which the vessel swings The energythen to be absorbed is 0514F[(Vsin13)2 ()2] with 13 theangle of approach of the vessel with the face of the jetty theradius of gyration (m) of the vessel about the point of impactand the angular velocity (radianss) of the vessel about the pointof impact The numerical values of the terms in the expressionare difficult to assess accurately and can vary considerablyunder different conditions of tide and wind and with differentvessels and methods of berthing

The kinetic energy of approach is absorbed partly by theresistance of the water but mainly by the fendering elasticdeformation of the structure and the vessel movement of theground and also by energy lsquolostrsquo upon impact The relativecontributions are difficult to assess but only about half ofthe total kinetic energy of the vessel may be imparted to thestructure and the fendering The force to which the structure issubjected is calculated by equating the product of the force andhalf the elastic horizontal displacement of the structure to thekinetic energy imparted Ordinary timber fenders appliedto reinforced concrete jetties cushion the blow but may notsubstantially reduce the force on the structure Spring fendersor suspended fenders can however absorb a large proportion ofthe kinetic energy Timber fenders independent of the jetty aresometimes provided to protect the structure from impact

The combined action of wind waves currents and tides on avessel moored to a jetty is usually transmitted by the vesselpressing directly against the side of the structure or by pullson mooring ropes secured to bollards The pulls on bollardsdue to the foregoing causes or during berthing vary with thesize of the vessel For vessels of up to 20 000 tonnes loadeddisplacement bollards are required at intervals of 15ndash30 m withload capacities according to the vessel displacement of 100 kNup to 2000 tonnes 300 kN up to 10 000 tonnes and 600 kN upto 20 000 tonnes

The effects of wind and waves acting on a marine structureare much reduced if an open construction is adopted and ifprovision is made for the relief of pressures due to water and airtrapped below the deck The force is not however relateddirectly to the proportion of solid vertical face presented to

the action of the wind and waves The pressures imposed areimpossible to assess with accuracy except for sea walls andsimilar structures where the depth of water at the face of the wallis such that breaking waves do not occur In this case the forceis due to simple hydrostatic pressure and can be evaluated forthe highest anticipated wave level with appropriate allowancefor wind surge In the Thames estuary for example the lattercan raise the high-tide level to 15 m above normal

A wave breaking against a sea wall causes a shock pressureadditional to the hydrostatic pressure which reaches its peakvalue at about mean water level and diminishes rapidly belowthis level and more slowly above it The shock pressure can beas much as 10 times the hydrostatic value and pressures up to650 kNm2 are possible with waves 45ndash6 m high The shape ofthe face of the wall the slope of the foreshore and the depthof water at the wall affect the maximum pressure and thedistribution of the pressure For information on the loads to beconsidered in the design of all types of maritime structuresreference should be made to BS 6349 Parts 1 to 7

27 RETAINED AND CONTAINED MATERIALS

The pressures imposed by materials on retaining structures orcontainment vessels are uncertain except when the retainedor contained material is a liquid In this case at any depth zbelow the free surface of the liquid the intensity of pressurenormal to the contact surface is equal to the vertical pressuregiven by the simple hydrostatic expression z wz where w

is unit weight of liquid (eg 981 kNm3 for water) For soilsand stored granular materials the pressures are considerablyinfluenced by the effective shear strength of the material

271 Properties of soils

For simplicity of analysis it is conventional to express the shearstrength of a soil by the equation

c n tan

where cis effective cohesion of soil is effective angle ofshearing resistance of soil n is effective normal pressure

Values of cand are not intrinsic soil properties and canonly be assumed constant within the stress range for which theyhave been evaluated For recommended fill materials it isgenerally sufficient to adopt a soil model with c 0 Such amodel gives a conservative estimate of the shear strength of thesoil and is analytically simple to apply in design Data takenfrom BS 8002 is given in Table 210 for unit weights of soilsand effective angles of shearing resistance

272 Lateral soil pressures

The lateral pressure exerted by a soil on a retaining structuredepends on the initial state of stress and the subsequent strainwithin the soil Where there has been no lateral strain eitherbecause the soil has not been disturbed during construction or thesoil has been prevented from lateral movement during placementan at-rest state of equilibrium exists Additional lateral strain isneeded to change the initial stress conditions Depending on themagnitude of the strain involved the final state of stress in the soilmass can be anywhere between the two failure conditions knownas the active and passive states of plastic equilibrium

Retained and contained materials 11

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The problem of determining lateral pressures at the limitingequilibrium conditions has been approached in different waysby different investigators In Coulomb theory the force actingon a retaining wall is determined by considering the limitingequilibrium of a soil wedge bounded by the rear face of thewall the ground surface and a planar failure surface Shearingresistance is assumed to have been mobilised both on the backof the wall and on the failure surface Rankine theory givesthe complete state of stress in a cohesionless soil mass whichis assumed to have expanded or compressed to a state of plasticequilibrium The stress conditions require that the earthpressure on a vertical plane should act in a direction parallel tothe ground surface Caquot and Kerisel produced tables ofearth pressure coefficients derived by a method that directlyintegrates the equilibrium equations along combined planar andlogarithmic spiral failure surfaces

273 Fill materials

A wide range of fill materials may be used behind retainingwalls All materials should be properly investigated and classi-fied Industrial chemical and domestic waste shale mudstoneand steel slag peaty or highly organic soil should not be usedas fill Selected cohesionless granular materials placed in acontrolled manner such as well-graded small rock-fills gravelsand sands are particularly suitable The use of cohesive soilscan result in significant economies by avoiding the need toimport granular materials but may also involve additionalproblems during design and construction The cohesive soilshould be within a range suitable for adequate compactionThe placement moisture content should be close to the finalequilibrium value to avoid either the swelling of clays placedtoo dry or the consolidation of clays placed too wet Suchproblems will be minimised if the fill is limited to clays with aliquid limit not exceeding 45 and a plasticity index notexceeding 25 Chalk with a saturation moisture content notexceeding 20 is acceptable as fill and may be compacted asfor a well-graded granular material Conditioned pulverizedfuel ash (PFA) from a single source may also be used it shouldbe supplied at a moisture-content of 80ndash100 of the optimumvalue For further guidance on the suitability of fill materialsreference should be made to relevant Transport ResearchLaboratory publications DoT Standard BD 3087 (ref 7)and BS 8002

274 Pressures imposed by cohesionless soils

Earth pressure distributions on unyielding walls and on rigidwalls free to translate or rotate about the base are shown inTable 211 For a normally consolidated soil the pressure on thewall increases linearly with depth Compaction results in higherearth pressures in the upper layers of the soil mass

Expressions for the pressures imposed in the at-rest activeand passive states including the effects of uniform surchargeand static ground water are given in sections 911ndash914Charts of earth pressure coefficients based on the work ofCaquot and Kerisel (ref 8) are given in Tables 212ndash214These may be used generally for vertical walls with slopingground or inclined walls with level ground

275 Cohesive soils

Clays in the long term behave as granular soils exhibitingfriction and dilation If a secant value (c 0) is used theprocedures for cohesionless soils apply If tangent parameters(c ) are used the RankinendashBell equations apply as given insection 915 In the short term if a clay soil is subjected torapid shearing a total stress analysis should be undertakenusing the undrained shear strength (see BS 8002)

276 Further considerations

For considerations such as earth pressures on embedded walls(with or without props) the effects of vertical concentrated loadsand line loads and the effects of groundwater seepage referenceshould be made to specialist books and BS 8002 For the pres-sures to be considered in the design of integral bridge abutmentsas a result of thermal movements of the deck reference shouldbe made to the Highways Agency document BA 4296 (ref 9)

277 Silos

Silos which may also be referred to as bunkers or bins aredeep containers used to store particulate materials In a deepcontainer the linear increase of pressure with depth found inshallow containers is modified When a deep container is filleda slight settlement of the fill activates the frictional resistancebetween the stored material and the wall This induces verticalload in the silo wall but reduces the vertical pressure in thematerial and the lateral pressures on the wall Janssen devel-oped a theory by which expressions have been derived for thepressures on the walls of a silo containing a granular materialhaving uniform properties The ratio of horizontal to verticalpressure in the fill is assumed constant and a Rankine coeffi-cient is generally used Eccentric filling (or discharge) tends toproduce variations in lateral pressure round the silo wall Anallowance is made by considering additional patch loads takento act on any part of the wall

Unloading a silo disturbs the equilibrium of the containedmass If the silo is unloaded from the top the frictional load onthe wall may be reversed as the mass re-expands but the lateralpressures remain similar to those during filling With a free-flowing material unloading at the bottom of the silo from thecentre of a hopper two different flow patterns are possibledepending on the characteristics of the hopper and the materialThese patterns are termed funnel flow (or core flow) and massflow respectively In the former a channel of flowing materialdevelops within a confined zone above the outlet the materialadjacent to the wall near the outlet remaining stationary Theflow channel can intersect the vertical walled section of the siloor extend to the surface of the stored material In mass flowwhich occurs particularly in steep-sided hoppers all the storedmaterial is mobilised during discharge Such flow can developat varying levels within the mass of material contained in anytall silo owing to the formation of a lsquoself-hopperrsquo with highlocal pressures arising where parallel flow starts to diverge fromthe walls Both flow patterns give rise to increases in lateralpressure from the stable filled condition Mass flow results alsoin a substantial local kick load at the intersection of the hopperand the vertical walled section

Design criteria safety factors and loads12

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When calculating pressures care should be taken to allow forthe inherent variability of the material properties In generalconcrete silo design is not sensitive to vertical wall load sovalues of maximum unit weight in conjunction with maximumor minimum consistent coefficients of friction should be usedData taken from EC 1 Part 4 for the properties of storedmaterials and the pressures on the walls and bottoms of silosare given in Tables 215 and 216

Fine powders like cement and flour can become fluidisedin silos either owing to rapid filling or through aeration tofacilitate discharge In such cases the design should allow forboth non-fluidised and fluidised conditions

28 EUROCODE LOADING STANDARDS

Eurocode 1 Actions on Structures is one of nine internationalunified codes of practice that have been published by the

European Committee for Standardization (CEN) The codewhich contains comprehensive information on all the actions(loads) normally necessary for consideration in the design ofbuilding and civil engineering structures consists of ten partsas follows

1991-1-1 Densities self-weight and imposed loads1991-1-2 Actions on structures exposed to fire1991-1-3 Snow loads1991-1-4 Wind loads1991-1-5 Thermal actions1991-1-6 Actions during execution1991-1-7 Accidental actions due to impact and explosions1991-2 Traffic loads on bridges1991-3 Actions induced by cranes and machinery1991-4 Actions on silos and tanks

Eurocode loading standards 13

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The requirements of concrete and its constituent materialsand of reinforcement are specified in Regulations Standardsand Codes of Practice Only those properties that concern thedesigner directly because they influence the behaviour anddurability of the structure are dealt with in this chapter

31 CONCRETE

Concrete is a structural material composed of crushed rockor gravel and sand bound together with a hardened paste ofcement and water A large range of cements and aggregateschemical admixtures and additions can be used to producea range of concretes having the required properties in boththe fresh and hardened states for many different structuralapplications The following information is taken mainly fromref 10 where a fuller treatment of the subject will be found

311 Cements and combinations

Portland cements are made from limestone and clay or otherchemically similar suitable raw materials which are burnedtogether in a rotary kiln to form a clinker rich in calciumsilicates This clinker is ground to a fine powder with a smallproportion of gypsum (calcium sulphate) which regulates therate of setting when the cement is mixed with water Overthe years several types of Portland cement have been developed

As well as cement for general use (which used to be knownas ordinary Portland cement) cements for rapid hardeningfor protection against attack by freezing and thawing or bychemicals and white cement for architectural finishes are alsomade The cements contain the same active compounds but indifferent proportions By incorporating other materials duringmanufacture an even wider range of cements is made includingair-entraining cement and combinations of Portland cementwith mineral additions Materials other than those in Portlandcements are used in cements for special purposes for examplecalcium aluminate cement is used for refractory concrete

The setting and hardening process that occurs when cementis mixed with water results from a chemical reaction known ashydration The process produces heat and is irreversible Settingis the gradual stiffening whereby the cement paste changesfrom a workable to a hardened state Subsequently the strengthof the hardened mass increases rapidly at first but slowinggradually This gain of strength continues as long as moisture ispresent to maintain the chemical reaction

Portland cements can be either inter-ground or blendedwith mineral materials at the cement factory or combined withadditions in the concrete mixer The most frequently used ofthese additional materials in the United Kingdom and therelevant British Standards are pulverized-fuel ash (pfa) toBS 3892 fly ash to BS EN 450 ground granulated blastfurnaceslag (ggbs) to BS 6699 and limestone fines to BS 7979 Otheradditions include condensed silica fume and metakaolin Theseare intended for specialised uses of concrete beyond the scopeof this book

The inclusion of pfa fly ash and ggbs has been particularlyuseful in massive concrete sections where they have been usedprimarily to reduce the temperature rise of the concrete withcorresponding reductions in temperature differentials and peaktemperatures The risk of early thermal contraction cracking isthereby also reduced The use of these additional materialsis also one of the options available for minimising the risk ofdamage due to alkalindashsilica reaction which can occur withsome aggregates and for increasing the resistance of concreteto sulfate attack Most additions react slowly at early stagesunder normal temperatures and at low temperature the reac-tion particularly in the case of ggbs can become considerablyretarded and make little contribution to the early strength ofconcrete However provided the concrete is not allowed to dryout the use of such additions can increase the long-termstrength and impermeability of the concrete

When the terms lsquowater-cement ratiorsquo and lsquocement contentrsquoare used in British Standards these are understood to includecombinations The word lsquobinderrsquo which is sometimes used isinterchangeable with the word lsquocementrsquo or lsquocombinationrsquo

The two methods of incorporating mineral additions makelittle or no difference to the properties of the concrete but therecently introduced notation system includes a unique code thatidentifies both composition and production method The typesof cement and combinations in most common usage are shownwith their notation in Table 217

Portland cement The most commonly used cement wasknown formerly as OPC in British Standards By grinding thecement clinker more finely cement with a more rapid earlystrength development is produced known formerly as RHPCBoth types are now designated as

Portland cement CEM I conforming to BS EN 197-1

Chapter 3

Material properties

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Cements are now classified in terms of both their standardstrength derived from their performance at 28 days and at anearly age normally two days using a specific laboratory testbased on a standard mortar prism This is termed the strengthclass for example CEM I 425N where 425 (Nmm2) is thestandard strength and N indicates a normal early strength

The most common standard strength classes for cements are425 and 525 These can take either N (normal) or R (rapid)identifiers depending on the early strength characteristics ofthe product CEM I in bags is generally a 425N cementwhereas CEM I for bulk supply tends to be 425R or 525 NCement corresponding to the former RHPC is now producedin the United Kingdom within the 525 strength class Thesecements are often used to advantage by precast concretemanufacturers to achieve a more rapid turn round of mouldsand on site when it is required to reduce the time for which theformwork must remain in position The cements which gener-ates more early heat than CEM I 425N can also be useful incold weather conditions

It is worth noting that the specified setting times of cementpastes relate to the performance of a cement paste of standardconsistence in a particular test made under closely controlledconditions of temperature and humidity the stiffening andsetting of concrete on site are not directly related to thesestandard setting regimes and are more dependent on factorssuch as workability cement content use of admixtures thetemperature of the concrete and the ambient conditions

Sulfate-resisting Portland cement SRPC This is a Portlandcement with a low tricalcium aluminate (C3A) content forwhich the British Standard is BS 4027 When concrete madewith CEM I cement is exposed to the sulfate solutions that arefound in some soils and groundwaters a reaction can occurbetween the sulfate and the hydrates from the C3A in thecement causing deterioration of the concrete By limitingthe C3A content in SRPC cement with a superior resistanceto sulfate attack is obtained SRPC normally has a low-alkalicontent but otherwise it is similar to other Portland cements inbeing non-resistant to strong acids The strength properties ofSRPC are similar to those of CEM I 425N but slightly lessearly heat is normally produced This can be an advantage inmassive concrete and in thick sections SRPC is not normallyused in combination with pfa or ggbs

Blastfurnace slag cements These are cements incorporatingggbs which is a by-product of iron smelting obtained byquenching selected molten slag to form granules The slag canbe inter-ground or blended with Portland cement clinker atcertain cement works to produce

Portland-slag cement CEM IIA-S with a slag content of6ndash35 conformimg to BS EN 197-1 or more commonly

Blastfurnace cement CEM IIIA with a slag content of36ndash65 conforming to BS EN 197-1

Alternatively the granules may be ground down separately to awhite powder with a fineness similar to that of cement and thencombined in the concrete mixer with CEM I cement to producea blastfurnace cement Typical mixer combinations of 40ndash50ggbs with CEM I cement have a notation CIIIA and at this levelof addition 28-day strengths are similar to those obtained withCEM I 425N

As ggbs has little hydraulic activity of its own it is referredto as lsquoa latent hydraulic binderrsquo Cements incorporating ggbsgenerate less heat and gain strength more slowly with lowerearly age strengths than those obtained with CEM I cementThe aforementioned blastfurnace cements can be used insteadof CEM I cement but because the early strength developmentis slower particularly in cold weather it may not be suitablewhere early removal of formwork is required They are amoderately low-heat cement and can therefore be used toadvantage to reduce early heat of hydration in thick sectionsWhen the proportion of ggbs is 66ndash80 CEM IIIA and CIIIAbecome CEM IIIB and CIIIB respectively These were knownformerly as high-slag blastfurnace cements and are specifiedbecause of their lower heat characteristics or to impart resis-tance to sulfate attack

Because the reaction between ggbs and lime released by thePortland cement is dependent on the availability of moistureextra care has to be taken in curing concrete containing thesecements or combinations to prevent premature drying out andto permit the development of strength

Pulverized-fuel ash and fly ash cements The ash resultingfrom the burning of pulverized coal in power station furnaces isknown in the concrete sector as pfa or fly ash The ash whichis fine enough to be carried away in the flue gases is removedfrom the gases by electrostatic precipitators to prevent atmos-pheric pollution The resulting material is a fine powder ofglassy spheres that can have pozzolanic properties that iswhen mixed into concrete it can react chemically with thelime that is released during the hydration of Portland cementThe products of this reaction are cementitious and in certaincircumstances pfa or fly ash can be used as a replacement forpart of the Portland cement provided in the concrete

The required properties of ash to be used as a cementitiouscomponent in concrete are specified in BS EN 450 withadditional UK provisions for pfa made in BS 3892 Part 1 Flyash in the context of BS EN 450 means lsquocoal fly ashrsquo ratherthan ash produced from other combustible materials and fly ashconforming to BS EN 450 can be coarser than that conformingto BS 3892 Part 1

Substitution of these types of cement for Portland cement isnot a straightforward replacement of like for like and thefollowing points have to be borne in mind when consideringthe use of pfa concrete

Pfa reacts more slowly than Portland cement At early ageand particularly at low temperatures pfa contributes lessstrength in order to achieve the same 28-day compressivestrength the amount of cementitious material may need to beincreased typically by about 10 The potential strengthafter three months is likely to be greater than CEM I providedthe concrete is kept in a moist environment for example inunderwater structures or concrete in the ground

The water demand of pfa for equal consistence may beless than that of Portland cement

The density of pfa is about three-quarters that of Portlandcement

The reactivity of pfa and its effect on water demand and hencestrength depend on the particular pfa and Portland cementwith which it is used A change in the source of either materialmay result in a change in the replacement level required

Concrete 15

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When pfa is to be air-entrained the admixture dosage ratemay have to be increased or a different formulation thatproduces a more stable air bubble structure used

Portland-fly ash cement comprises in effect a mixture ofCEM I and pfa When the ash is inter-ground or blended withPortland cement clinker at an addition rate of 20ndash35 themanufactured cement is known as Portland-fly ash cementCEM IIB-V conforming to BS EN 197-1 When this combina-tion is produced in a concrete mixer it has the notation CIIB-Vconforming to BS 8500 Part 2

Typical ash proportions are 25ndash30 and these cements canbe used in concrete for most purposes They are likely to havea slower rate of strength development compared with CEM IWhen the cement contains 25ndash40 ash it may be used toimpart resistance to sulfate attack and can also be beneficial inreducing the harmful effects of alkalindashsilica reaction Wherehigher replacement levels of ash are used for improved low-heatcharacteristics the resulting product is pozzolanic (pfa) cementwith the notation if manufactured CEM IVB-V conforming toBS EN 197-1 or if combined in the concrete mixer CIVB-Vconforming to BS 8500 Part 2

Because the pozzolanic reaction between pfa or fly ash andfree lime is dependent on the availability of moisture extra carehas to be taken in curing concrete containing mineral additionsto prevent premature drying out and to permit the developmentof strength

Portland-limestone cement Portland cement incorporating6ndash35 of carefully selected fine limestone powder is knownas Portland-limestone cement conforming to BS EN 197-1When a 425N product is manufactured the typical limestoneproportion is 10ndash20 and the notation is CEM IIA-L or CEMIIA-LL It is most popular in continental Europe but its usageis growing in the United Kingdom Decorative precast andreconstituted stone concretes benefit from its lighter colouringand it is also used for general-purpose concrete in non-aggressiveand moderately aggressive environments

312 Aggregates

The term lsquoaggregatersquo is used to describe the gravels crushedrocks and sands that are mixed with cement and water to pro-duce concrete As aggregates form the bulk of the volume ofconcrete and can significantly affect its performance the selec-tion of suitable material is extremely important Fine aggregatesinclude natural sand crushed rock or crushed gravel that is fineenough to pass through a sieve with 4 mm apertures (formerly5 mm as specified in BS 882) Coarse aggregates compriselarger particles of gravel crushed gravel or crushed rock Mostconcrete is produced from natural aggregates that are specifiedto conform to the requirements of BS EN 12620 together withthe UK Guidance Document PD 6682-1 Manufactured light-weight aggregates are also sometimes used

Aggregates should be hard and should not contain materialsthat are likely to decompose or undergo volumetric changeswhen exposed to the weather Some examples of undesirablematerials are lignite coal pyrite and lumps of clay Coal andlignite may swell and decompose leaving small holes on thesurface of the concrete lumps of clay may soften and formweak pockets and pyrite may decompose causing iron oxide

stains to appear on the concrete surface When exposed tooxygen pyrite has been known to contribute to sulfate attackHigh-strength concretes may call for special propertiesThe mechanical properties of aggregates for heavy-dutyconcrete floors and for pavement wearing surfaces may have tobe specially selected Most producers of aggregate are ableto provide information about these properties and referencewhen necessary should be made to BS EN 12620

There are no simple tests for aggregate durability or theirresistance to freezethaw exposure conditions and assessmentof particular aggregates is best based on experience of theproperties of concrete made with the type of aggregate andknowledge of its source Some flint gravels with a white porouscortex may be frost-susceptible because of the high waterabsorption of the cortex resulting in pop-outs on the surface ofthe concrete when subjected to freezethaw cycles

Aggregates must be clean and free from organic impuritiesThe particles should be free from coatings of dust or clay asthese prevent proper bonding of the material An excessiveamount of fine dust or stone lsquoflourrsquo can prevent the particles ofstone from being properly coated with cement and lower thestrength of the concrete Gravels and sands are usually washedby the suppliers to remove excess fines (eg clay and silt) andother impurities which otherwise could result in a poor-qualityconcrete However too much washing can also remove allfine material passing the 025 mm sieve This may result in aconcrete mix lacking in cohesion and in particular one that isunsuitable for placing by pump Sands deficient in fines alsotend to increase the bleeding characteristics of the concreteleading to poor vertical finishes due to water scour

Where the colour of a concrete surface finish is importantsupplies of aggregate should be obtained from the one sourcethroughout the job whenever practicable This is particularlyimportant for the sand ndash and for the coarse aggregate when anexposed-aggregate finish is required

Size and grading The maximum size of coarse aggregateto be used is dependent on the type of work to be done Forreinforced concrete it should be such that the concrete can beplaced without difficulty surrounding all the reinforcementthoroughly and filling the corners of the formwork In theUnited Kingdom it is usual for the coarse aggregate to havea maximum size of 20 mm Smaller aggregate usually with amaximum size of 10 mm may be needed for concrete that is tobe placed through congested reinforcement and in thin sectionswith small covers In this case the cement content may haveto be increased by 10ndash20 to achieve the same strength andworkability as that obtained with a 20 mm maximum-sizedaggregate because both sand and water contents usually haveto be increased to produce a cohesive mix Larger aggregatewith a maximum size of 40 mm can be used for foundationsand mass concrete where there are no restrictions to the flowof the concrete It should be noted however that this sort ofconcrete is not always available from ready-mixed concreteproducers The use of a larger aggregate results in a slightlyreduced water demand and hence a slightly reduced cementcontent for a given strength and workability

The proportions of the different sizes of particles makingup the aggregate which are found by sieving are known asthe aggregate lsquogradingrsquo The grading is given in terms of thepercentage by mass passing the various sieves Continuously

Material properties16

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graded aggregates for concrete contain particles ranging in sizefrom the largest to the smallest in gap-graded aggregatessome of the intermediate sizes are absent Gap grading may benecessary to achieve certain surface finishes Sieves used formaking a sieve analysis should conform to BS EN 933-2Recommended sieve sizes typically range from 80 to 2 mm forcoarse aggregates and from 8 to 025 mm for fine aggregatesTests should be carried out in accordance with the proceduregiven in BS EN 933-1

An aggregate containing a high proportion of large particlesis referred to as being lsquocoarselyrsquo graded and one containing ahigh proportion of small particles as lsquofinelyrsquo graded Overallgrading limits for coarse fine and lsquoall-inrsquo aggregates arecontained in BS EN 12620 and PD 6682-1 All-in aggregatescomprising both coarse and fine materials should not be usedfor structural reinforced concrete work because the gradingwill vary considerably from time to time and hence from batchto batch thus resulting in excessive variation in the consistenceand the strength of the concrete To ensure that the properamount of sand is present the separate delivery storage andbatching of coarse and fine materials is essential Graded coarseaggregates that have been produced by layer loading (ie fillinga lorry with say two grabs of material size 10ndash20 mm andone grab of material size 4ndash10 mm) are seldom satisfactorybecause the unmixed materials will not be uniformly gradedThe producer should ensure that such aggregates are effectivelymixed before loading into lorries

For a high degree of control over concrete production andparticularly if high-quality surface finishes are required it isnecessary for the coarse aggregate to be delivered stored andbatched using separate single sizes

The overall grading limits for coarse and fine aggregates asrecommended in BS EN 12620 are given in Table 217 Thelimits vary according to the aggregate size indicated as dD inmillimetres where d is the lower limiting sieve size and D isthe upper limiting sieve size for example 420 Additionally thecoarsenessfineness of the fine aggregate is assessed againstthe percentage passing the 05 mm sieve to give a CP MPFP grading This compares with the C (coarse) M (medium)F (fine) grading used formerly in BS 882 Good concrete canbe made using sand within the overall limits but there may beoccasions such as where a high degree of control is requiredor a high-quality surface finish is to be achieved when it isnecessary to specify the grading to even closer limits On theother hand sand whose grading falls outside the overall limitsmay still produce perfectly satisfactory concrete Maintaining areasonably uniform grading is generally more important thanthe grading limits themselves

Marine-dredged aggregates Large quantities of aggregatesobtained by dredging marine deposits have been widely andsatisfactorily used for making concrete for many years Ifpresent in sufficient quantities hollow andor flat shells canaffect the properties of both fresh and hardened concrete andtwo categories for shell content are given in BS EN 12620 Inorder to reduce the corrosion risk of embedded metal limitsfor the chloride content of concrete are given in BS EN 206-1and BS 8500 To conform to these limits it is necessary formarine-dredged aggregates to be carefully and efficientlywashed in fresh water that is frequently changed in order toreduce the salt content Chloride contents should be checked

frequently throughout aggregate production in accordance withthe method given in BS EN 1744-1

Some sea-dredged sands tend to have a preponderance ofone size of particle and a deficiency in the amount passing the025 mm sieve This can lead to mixes prone to bleeding unlessmix proportions are adjusted to overcome the problemIncreasing the cement content by 5ndash10 can often offset thelack of fine particles in the sand Beach sands are generallyunsuitable for good-quality concrete since they are likely tohave high concentrations of chloride due to the accumulation ofsalt crystals above the high-tide mark They are also oftensingle-sized which can make the mix design difficult

Lightweight aggregates In addition to natural gravels andcrushed rocks a number of manufactured aggregates are alsoavailable for use in concrete Aggregates such as sintered pfaare required to conform to BS EN 13055-1 and PD 6682-4

Lightweight aggregate has been used in concrete for manyyears ndash the Romans used pumice in some of their constructionwork Small quantities of pumice are imported and still used inthe United Kingdom mainly in lightweight concrete blocksbut most lightweight aggregate concrete uses manufacturedaggregate

All lightweight materials are relatively weak because of theirhigher porosity which gives them reduced weight The resultinglimitation on aggregate strength is not normally a problemsince the concrete strength that can be obtained still exceedsmost structural requirements Lightweight aggregates are usedto reduce the weight of structural elements and to giveimproved thermal insulation and fire resistance

313 Water

The water used for mixing concrete should be free fromimpurities that could adversely affect the process of hydrationand consequently the properties of concrete For examplesome organic matter can cause retardation whilst chloridesmay not only accelerate the stiffening process but also causeembedded steel such as reinforcement to corrode Otherchemicals like sulfate solutions and acids can have harmfullong-term effects by dissolving the cement paste in concreteIt is important therefore to be sure of the quality of water If itcomes from an unknown source such as a pond or boreholeit needs to be tested BS EN 1008 specifies requirements forthe quality of the water and gives procedures for checking itssuitability for use in concrete

Drinking water is suitable of course and it is usual simplyto obtain a supply from the local water utility Some recycledwater is being increasingly used in the interests of reducing theenvironmental impact of concrete production Seawater hasalso been used successfully in mass concrete with no embeddedsteel Recycled water systems are usually found at large-scalepermanent mixing plants such as precast concrete factories andready-mixed concrete depots where water that has been usedfor cleaning the plant and washing out mixers can be collectedfiltered and stored for re-use Some systems are able to reclaimup to a half of the mixing water in this way Large volumesettlement tanks are normally required The tanks do not needto be particularly deep but should have a large surface area andideally the water should be made to pass through a series ofsuch tanks becoming progressively cleaner at each stage

Concrete 17

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314 Admixtures

An admixture is a material usually a liquid which is added toa batch of concrete during mixing to modify the properties ofthe fresh or the hardened concrete in some way Most admix-tures benefit concrete by reducing the amount of free waterneeded for a given level of consistence often in addition tosome other specific improvement Permeability is therebyreduced and durability increased There are occasions when theuse of an admixture is not only desirable but also essentialBecause admixtures are added to concrete mixes in smallquantities they should be used only when a high degree ofcontrol can be exercised Incorrect dosage of an admixture canadversely affect strength and other properties of the concreteRequirements for the following main types of admixture arespecified in BS EN 934-2

Normal water-reducing admixtures Commonly knownas plasticisers or workability aids these act by reducing theinter-particle attraction within the cement to produce a moreuniform dispersion of the cement grains The cement paste isbetter lsquolubricatedrsquo and hence the amount of water needed toobtain a given consistency can be reduced The use of theseadmixtures can be beneficial in one of three ways

When added to a normal concrete at normal dosage theyproduce an increase in slump of about 50 mm This can beuseful in high-strength concrete rich in cement which wouldotherwise be too stiff to place

The water content can be reduced while maintaining the samecement content and consistence class the reduction in watercement ratio (about 10) results in increased strength andimproved durability This can also be useful for reducingbleeding in concrete prone to this problem and for increasingthe cohesion and thereby reducing segregation in concrete ofhigh consistence or in harsh mixes that sometimes arise withangular aggregates or low sand contents or when the sand isdeficient in fines

The cement content can be reduced while maintaining thesame strength and consistence class The watercement ratiois kept constant and the water and cement contents arereduced accordingly This approach should never be used ifthereby the cement content would be reduced below theminimum specified amount

Too big a dosage may result in retardation andor a degree ofair-entrainment without necessarily increasing workabilityand therefore may be of no benefit in the fresh concrete

Accelerating water-reducing admixtures Acceleratorsact by increasing the initial rate of chemical reaction betweenthe cement and the water so that the concrete stiffens hardensand develops strength more quickly They have a negligibleeffect on consistence and the 28-day strengths are seldomaffected Accelerating admixtures have been used mainlyduring cold weather when the slowing down of the chemicalreaction between cement and water at low temperature couldbe offset by the increased speed of reaction resulting fromthe accelerator The most widely used accelerator used to becalcium chloride but because the presence of chlorides even insmall amounts increases the risk of corrosion modern standardsprohibit the use of admixtures containing chlorides in all concrete

containing embedded metal Accelerators are sometimesmarketed under other names such as hardeners or anti-freezersbut no accelerator is a true anti-freeze and the use of anaccelerator does not avoid the need to protect the concrete incold weather by keeping it warm (with insulation) after ithas been placed

Retarding water-reducing admixtures These slow downthe initial reaction between cement and water by reducingthe rate of water penetration to the cement By slowing down thegrowth of the hydration products the concrete stays workablelonger than it otherwise would The length of time during whichconcrete remains workable depends on its temperature consis-tence class and watercement ratio and on the amount of retarderused Although the occasions justifying the use of retarders in theUnited Kingdom are limited these admixtures can be helpfulwhen one or more of the following conditions apply

In warm weather when the ambient temperature is higherthan about 20oC to prevent early stiffening (lsquogoing-offrsquo) andloss of workability which would otherwise make the placingand finishing of the concrete difficult

When a large concrete pour which will take several hours tocomplete must be constructed so that concrete already placeddoes not harden before the subsequent concrete can bemerged with it (ie without a cold joint)

When the complexity of a slip-forming operation requires aslow rate of rise

When there is a delay of more than 30 minutes betweenmixing and placing ndash for example when ready-mixed concreteis being used over long-haul distances or there are risks oftraffic delays This can be seriously aggravated during hotweather especially if the cement content is high

The retardation can be varied by altering the dosage a delayof 4ndash6 hours is usual but longer delays can be obtained forspecial purposes While the reduction in early strength ofconcrete may affect formwork-striking times the 7-day and28-day strengths are not likely to be significantly affectedRetarded concrete needs careful proportioning to minimisebleeding due to the longer period during which the concreteremains fresh

Air-entraining admixtures These may be organic resinsor synthetic surfactants that entrain a controlled amount of airin concrete in the form of small air bubbles The bubbles needto be about 50 microns in diameter and well dispersed Themain reason for using an air-entraining admixture is that thepresence of tiny bubbles in the hardened concrete increases itsresistance to the action of freezing and thawing especiallywhen this is aggravated by the application of de-icing saltsand fluids Saturated concrete ndash as most external paving willbe ndash can be seriously affected by the freezing of water inthe capillary voids which will expand and try to burst it If theconcrete is air-entrained the air bubbles which intersect thecapillaries stay unfilled with water even when the concrete issaturated Thus the bubbles act as pressure relief valves andcushion the expansive effect by providing voids into which thewater can expand as it freezes without disrupting the concreteWhen the ice melts surface tension effects draw the water backout of the bubbles

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Air-entrained concrete should be specified and used for allforms of external paving from major roads and airfieldrunways down to garage drives and footpaths which are likelyto be subjected to severe freezing and to de-icing salts The saltsmay be applied directly or come from the spray of passingtraffic or by dripping from the underside of vehicles

Air-entrainment also affects the properties of the freshconcrete The minute air bubbles act like ball bearings and havea plasticising effect resulting in a higher consistence Concretethat is lacking in cohesion or harsh or which tends to bleedexcessively is greatly improved by air-entrainment The riskof plastic settlement and plastic-shrinkage cracking is alsoreduced There is also evidence that colour uniformity isimproved and surface blemishes reduced One factor that has tobe taken into account when using air-entrainment is that thestrength of the concrete is reduced by about 5 for every 1 ofair entrained However the plasticising effect of the admixturemeans that the water content of the concrete can be reducedwhich will offset most of the strength loss that would otherwiseoccur but even so some increase in the cement content is likelyto be required

High-range water-reducing admixtures Commonlyknown as superplasticizers these have a considerable plasticizingeffect on concrete They are used for one of two reasons

To greatly increase the consistence of a concrete mix so thata lsquoflowingrsquo concrete is produced that is easy both to placeand to compact some such concretes are completely self-compacting and free from segregation

To produce high-strength concrete by reducing the watercontent to a much greater extent than can be achieved byusing a normal plasticizer (water-reducing admixture)

A flowing concrete is usually obtained by first producing aconcrete whose slump is in the range 50ndash90 mm and thenadding the superplasticizer which increases the slump to over200 mm This high consistence lasts for only a limited periodof time stiffening and hardening of the concrete then proceednormally Because of this time limitation when ready-mixedconcrete is being used it is usual for the superplasticizer to beadded to the concrete on site rather than at the batching ormixing plant Flowing concrete can be more susceptible tosegregation and bleeding so it is essential for the mix designand proportions to allow for the use of a superplasticizer As ageneral guide a conventionally designed mix needs to bemodified by increasing the sand content by about 5 A highdegree of control over the batching of all the constituents isessential especially the water because if the consistence of theconcrete is not correct at the time of adding the superplasticizerexcessive flow and segregation will occur

The use of flowing concrete is likely to be limited to workwhere the advantages in ease and speed of placing offset theincreased cost of the concrete ndash considerably more than withother admixtures Typical examples are where reinforcement isparticularly congested making both placing and vibrationdifficult and where large areas such as slabs would benefitfrom a flowing easily placed concrete The fluidity of flowingconcrete increases the pressures on formwork which should bedesigned to resist full hydrostatic pressure

When used to produce high-strength concrete reductions inwater content of as much as 30 can be obtained by using

superplasticizers compared to 10 with normal plasticizers asa result 1-day and 28-day strengths can be increased by as muchas 50 Such high-strength water-reduced concrete is used bothfor high-performance in situ concrete construction and for themanufacture of precast units where the increased early strengthallows earlier demoulding

315 Properties of fresh and hardening concrete

Workability It is vital that the workability of concrete ismatched to the requirements of the construction process Theease or difficulty of placing concrete in sections of varioussizes and shapes the type of compaction equipment neededthe complexity of the reinforcement the size and skills of theworkforce are amongst the items to be considered In generalthe more difficult it is to work the concrete the higher shouldbe the level of workability But the concrete must also havesufficient cohesiveness in order to resist segregation andbleeding Concrete needs to be particularly cohesive if it is tobe pumped or allowed to fall from a considerable height

The workability of fresh concrete is increasingly referred toin British and European standards as consistence The slumptest is the best-known method for testing consistence and theslump classes given in BS EN 206-1 are S1 (10ndash40 mm)S2 (50ndash90 mm) S3 (100ndash150 mm) S4 (160ndash210 mm) Threeother test methods recognised in BS EN 206-1 all with theirown unique consistency classes are namely Vebe timedegree of compactability and flow diameter

Plastic cracking There are two basic types of plastic cracksplastic settlement cracks which can develop in deep sectionsand often follow the pattern of the reinforcement and plasticshrinkage cracks which are most likely to develop in slabsBoth types form while the concrete is still in its plastic statebefore it has set or hardened and depending on the weatherconditions within about one to six hours after the concrete hasbeen placed and compacted They are often not noticed until thefollowing day Both types of crack are related to the extent towhich the fresh concrete bleeds

Fresh concrete is a suspension of solids in water and after ithas been compacted there is a tendency for the solids (bothaggregates and cement) to settle The sedimentation processdisplaces water which is pushed upwards and if excessiveappears as a layer on the surface This bleed water may notalways be seen since it can evaporate on hot or windy daysfaster than it rises to the surface Bleeding can generally bereduced by increasing the cohesiveness of the concrete This isusually achieved by one or more of the following meansincreasing the cement content increasing the sand contentusing a finer sand using less water air-entrainment using arounded natural sand rather than an angular crushed one Therate of bleeding will be influenced by the drying conditionsespecially wind and bleeding will take place for longer on colddays Similarly concrete containing a retarder tends to bleed fora longer period of time due to the slower stiffening rate ofthe concrete and the use of retarders will in general increasethe risk of plastic cracking

Plastic settlement cracks caused by differential settlementare directly related to the amount of bleeding They tend tooccur in deep sections particularly deep beams but they may

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also develop in columns and walls This is because the deeperthe section the greater the sedimentation or settlement thatcan occur However cracks will form only where somethingprevents the concrete lsquosolidsrsquo from settling freely The mostcommon cause of this is the reinforcement fixed at the top ofdeep sections the concrete will be seen to lsquohang-uprsquo over thebars and the pattern of cracks will directly reflect the layout ofthe reinforcement below Plastic settlement cracks can alsooccur in trough and waffle slabs or at any section where thereis a significant change in the depth of concrete If alterationsto the concrete for example the use of an air-entraining orwater-reducing admixture cannot be made due to contractualor economic reasons the most effective way of eliminatingplastic settlement cracking is to re-vibrate the concrete afterthe cracks have formed Such re-vibration is acceptable whenthe concrete is still plastic enough to be capable of beinglsquofluidizedrsquo by a poker but not so stiff that a hole is left when thepoker is withdrawn The prevailing weather conditions willdetermine the timing of the operation

Plastic shrinkage cracks occur in horizontal slabs such asfloors and pavements They usually take the form of one ormore diagonal cracks at 05ndash2 m centres that do not extendto the slab edges or they form a very large pattern of mapcracking Such cracks are most common in concrete placed onhot or windy days because they are caused by the rate ofevaporation of moisture from the surface exceeding the rateof bleeding Clearly plastic shrinkage cracks can be reducedby preventing the loss of moisture from the concrete surface inthe critical first few hours While sprayed-on resin-based curingcompounds are very efficient at curing concrete that has alreadyhardened they cannot be used on fresh concrete until the freebleed water has evaporated This is too late to prevent plasticshrinkage cracking and so the only alternative is to protect theconcrete for the first few hours with polythene sheeting Thisneeds to be supported clear of the concrete by means of blocksor timber but with all the edges held down to prevent a wind-tunnel effect It has been found that plastic shrinkage crackingis virtually non-existent when air-entrainment is used

The main danger from plastic cracking is the possibility ofmoisture ingress leading to corrosion of any reinforcement Ifthe affected surface is to be covered subsequently by eithermore concrete or a screed no treatment is usually necessaryIn other cases often the best repair is to brush dry cement(dampened down later) or wet grout into the cracks the day afterthey form and while they are still clean this encourages naturalor autogenous healing

Early thermal cracking The reaction of cement with wateror hydration is a chemical reaction that produces heat If thisheat development exceeds the rate of heat loss the concretetemperature will rise Subsequently the concrete will cool andcontract Typical temperature histories of different concretesections are shown in the figure on Table 218

If the contraction of the concrete were unrestrained therewould be no cracking at this stage However in practice thereis nearly always some form of restraint inducing tension andhence a risk of cracks forming The restraint can occur due toboth external and internal influences Concrete is externallyrestrained when for example it is cast onto a previously castbase such as a wall kicker or between two already hardenedsections such as in infill bay in a wall or slab without the

provision of a contraction joint Internal restraint occurs forexample because the surfaces of an element will cool fasterthan the core producing a temperature differential When thisdifferential is large such as in thick sections surface cracksmay form at an early stage Subsequently as the core of thesection cools these surface cracks will tend to close in theabsence of any external restraints Otherwise the cracks willpenetrate into the core and link up to form continuous cracksthrough the whole section

The main factors affecting the temperature rise in concreteare the dimensions of the section the cement content andtype the initial temperature of the concrete and the ambienttemperature the type of formwork and the use of admixturesThicker sections retain more heat giving rise to higher peaktemperatures and cool down more slowly Within the coreof very thick sections adiabatic conditions obtain and abovea thickness of about 15 m there is little further increasein the temperature of the concrete The heat generated isdirectly related to the cement content For Portland cementconcretes in sections of thickness 1 m and more the temper-ature rise in the core is likely to be about 14oC for every100 kgm3 of cement Thinner sections will exhibit lowertemperature rises

Different cement types generate heat at different rates Thepeak temperature and the total amount of heat produced byhydration depend upon both the fineness and the chemistry ofthe cement As a guide the cements whose strength developsmost rapidly tend to produce the most heat Sulfate-resistingcement generally gives off less heat than CEM I and cementsthat are inter-ground or combined with mineral additions suchas pfa or ggbs are often chosen for massive constructionbecause of their low heat of hydration

A higher initial temperature results in a greater temperaturerise for example concrete in a 500 mm thick section placedat 10oC could have a temperature rise of 30oC but the sameconcrete placed at 20oC may have a temperature rise of 40oCSteel and GRP formwork will allow the heat generated to bedissipated more quickly than will timber formwork resultingin lower temperature rises especially in thinner sectionsTimber formwork andor additional insulation will reduce thetemperature differential between the core and the surface ofthe section but this differential could increase significantlywhen the formwork is struck Retarding water-reducers willdelay the onset of hydration but do not reduce the total heatgenerated Accelerating water-reducers will increase the rate ofheat evolution and the temperature rise

The problem of early thermal cracking is usually confined toslabs and walls Walls are particularly susceptible becausethey are often lightly reinforced in the horizontal directionand the timber formwork tends to act as a thermal insulatorencouraging a larger temperature rise The problem could bereduced by lowering the cement content and using cementwith a lower heat of hydration or one containing ggbs or pfaHowever there are practical and economic limits to thesemeasures often dictated by the specification requirements forthe strength and durability of the concrete itself In practicecracking due to external restraint is generally dealt with byproviding crack control reinforcement and contraction jointsWith very thick sections and very little external restraint ifthe temperature differential can be controlled by insulating theconcrete surfaces for a few days cracking can be avoided

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Typical values of the temperature rise in walls and slabs forPortland cement concretes as well as comparative values forconcrete using other cements are given in Table 218 Furtherdata on predicted temperature rises is given in ref 11

316 Properties of hardened concrete

Compressive strength The strength of concrete is specifiedas a strength class or grade namely the 28-day characteristiccompressive strength of specimens made from fresh concreteunder standardised conditions The results of strength tests areused routinely for control of production and contractual confor-mity purposes The characteristic strength is defined as that levelof strength below which 5 of all valid test results is expectedto fall Test cubes either 100 mm or 150 mm are the specimensnormally used in the United Kingdom and most other Europeancountries but cylinders are used elsewhere Because theirbasic shapes (ratio of height to cross-sectional dimension) aredifferent the strength test results are also different cylindersbeing weaker than cubes For normal-weight aggregates theconcrete cylinder strength is about 80 of the correspondingcube strength For lightweight aggregates cylinder strengths areabout 90 of the corresponding cube strengths

In British Codes of Practice like BS 8110 strength gradesused to be specified in terms of cube strength (eg C30) asshown in Table 39 Nowadays strength classes are specified interms of both cylinder strength and equivalent cube strength(eg C2530) as shown in Tables 35 and 42

In principle compressive strengths can be determined fromcores cut from the hardened concrete Core tests are normallymade only when there is some doubt about the quality ofconcrete placed (eg if the cube results are unsatisfactory) orto assist in determining the strength and quality of an existingstructure for which records are not available Great care isnecessary in the interpretation of the results of core tests andsamples drilled from in situ concrete are expected to be lowerin strength than cubes made cured and tested under standardlaboratory conditions The standard reference for core testingis BS EN 12504-1

Tensile strength The direct tensile strength of concrete asa proportion of the cube strength varies from about one-tenthfor low-strength concretes to one-twentieth for high-strengthconcretes The proportion is affected by the aggregate used andthe compressive strength is therefore only a very general guideto the tensile strength For specific design purposes in regard tocracking and shear strength analytical relationships betweenthe tensile strength and the specified cylindercube strength areprovided in codes of practice

The indirect tensile strength (or cylinder splitting strength) isseldom specified nowadays Flexural testing of specimens maybe used on some airfield runway contracts where the methodof design is based on the modulus of rupture and for someprecast concrete products such as flags and kerbs

Elastic properties The initial behaviour of concrete underservice load is almost elastic but under sustained loading thestrain increases with time Stressndashstrain tests cannot be carriedout instantaneously and there is always a degree of non-linearityand a residual strain upon unloading For practical purpose theinitial deformation is considered to be elastic (recoverable

upon unloading) and the subsequent increase in strain undersustained stress is defined as creep The elastic modulus onloading defined in this way is a secant modulus related to aspecific stress level The value of the modulus of elasticity ofconcrete is influenced mainly by the aggregate used With aparticular aggregate the value increases with the strength of theconcrete and the age at loading In special circumstances Forexample where deflection calculations are of great importanceload tests should be carried out on concrete made with theaggregate to be used in the actual structure For most designpurposes specific values of the mean elastic modulus at28 days and of Poissonrsquos ratio are given in Table 35 forBS 8110 and Table 42 for EC 2

Creep The increase in strain beyond the initial elastic valuethat occurs in concrete under a sustained constant stress aftertaking into account other time-dependent deformations notassociated with stress is defined as creep If the stress isremoved after some time the strain decreases immediately byan amount that is less than the original elastic value becauseof the increase in the modulus of elasticity with age This isfollowed by a further gradual decrease in strain The creeprecovery is always less than the preceding creep so that thereis always a residual deformation

The creep source in normal-weight concrete is the hardenedcement paste The aggregate restrains the creep in the paste sothat the stiffer the aggregate and the higher its volumetricproportion the lower is the creep of the concrete Creep isalso affected by the watercement ratio as is the porosity andstrength of the concrete For constant cement paste contentcreep is reduced by a decrease in the watercement ratio

The most important external factor influencing creep is therelative humidity of the air surrounding the concrete For aspecimen that is cured at a relative humidity of 100 thenloaded and exposed to different environments the lower therelative humidity the higher is the creep The values are muchreduced in the case of specimens that have been allowed todry prior to the application of load The influence of relativehumidity on creep is dependent on the size of the memberWhen drying occurs at constant relative humidity the largerthe specimen the smaller is the creep This size effect isexpressed in terms of the volumesurface area ratio of themember If no drying occurs as in mass concrete the creep isindependent of size

Creep is inversely proportional to concrete strength at the ageof loading over a wide range of concrete mixes Thus for agiven type of cement the creep decreases as the age andconsequently the strength of the concrete at application of theload increases The type of cement temperature and curingconditions all influence the development of strength with age

The influence of temperature on creep is important in the useof concrete for nuclear pressure vessels and containers forstoring liquefied gases The time at which the temperature ofconcrete rises relative to the time at which load is appliedaffects the creepndashtemperature relation If saturated concrete isheated and loaded at the same time the creep is greater thanwhen the concrete is heated during the curing period prior to theapplication of load At low temperatures creep behaviour isaffected by the formation of ice As the temperature falls creepdecreases until the formation of ice causes an increase in creepbut below the ice point creep again decreases

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Creep is normally assumed to be directly proportional toapplied stress within the service range and the term specificcreep is used for creep per unit of stress At stresses aboveabout one-third of the cube strength (45 cylinder strength)the formation of micro-cracks causes the creepndashstress relationto become non-linear creep increasing at an increasing rate

The effect of creep is unfavourable in some circumstances(eg increased deflection) and favourable in others (eg reliefof stress due to restraint of imposed deformations such asdifferential settlement seasonal temperature change)

For normal exposure conditions (inside and outside) creepcoefficients according to ambient relative humidity effectivesection thickness (notional size) and age of loading are givenin Table 35 for BS 8110 and Table 43 for EC 2

Shrinkage Withdrawal of water from hardened concretekept in unsaturated air causes drying shrinkage If concretethat has been left to dry in air of a given relative humidity issubsequently placed in water (or a higher relative humidity)it will swell due to absorption of water by the cement pasteHowever not all of the initial drying shrinkage is recoveredeven after prolonged storage in water For the usual rangeof concretes the reversible moisture movement representsabout 40ndash70 of the drying shrinkage A pattern of alternatewetting and drying will occur in normal outdoor conditionsThe magnitude of the cyclic movement clearly depends uponthe duration of the wetting and drying periods but drying ismuch slower than wetting The consequence of prolonged dryweather can be reversed by a short period of rain More stableconditions exist indoors (dry) and in the ground or in contactwith water (eg reservoirs and tanks)

Shrinkage of hardened concrete under drying conditions isinfluenced by several factors in a similar manner to creep Theintrinsic shrinkage of the cement paste increases with thewatercement ratio so that for a given aggregate proportionconcrete shrinkage is also a function of watercement ratio

The relative humidity of the air surrounding the membergreatly affects the magnitude of concrete shrinkage accordingto the volumesurface area ratio of the member The lowershrinkage value of large members is due to the fact that dryingis restricted to the outer parts of the concrete the shrinkage ofwhich is restrained by the non-shrinking core Clearly shrink-able aggregates present special problems and can greatlyincrease concrete shrinkage (ref 12)

For normal exposure conditions (inside and outside) valuesof drying shrinkage according to ambient relative humidity andeffective section thickness (notional size) are given in Table 35for BS 8110 and Table 42 for EC 2

Thermal properties The coefficient of thermal expansionof concrete depends on both the composition of the concreteand its moisture condition at the time of the temperaturechange The thermal coefficient of the cement paste is higherthan that of the aggregate which exerts a restraining influenceon the movement of the cement paste The coefficient of thermalexpansion of a normally cured paste varies from the lowestvalues when the paste is either totally dry or saturated to amaximum at a relative humidity of about 70 Values for theaggregate are related to their mineralogical composition

A value for the coefficient of thermal expansion of concreteis needed in the design of structures such as chimneys tanks

containing hot liquids bridges and other elevated structuresexposed to significant solar effects and for large expanses ofconcrete where provision must be made to accommodate theeffects of temperature change in controlled cracking or byproviding movement joints For normal design purposes valuesof the coefficient of thermal expansion of concrete accordingto the type of aggregate are given in Table 35 for BS 8110 andTable 42 for EC 2

Short-term stressndashstrain curves For normal low to mediumstrength unconfined concrete the stressndashstrain relationship incompression is approximately linear up to about one-third ofthe cube strength (40 of cylinder strength) With increasingstress the strain increases at an increasing rate and a peakstress (cylinder strength) is reached at a strain of about 0002With increasing strain the stress reduces until failure occurs ata strain of about 00035 For higher strength concretes the peakstress occurs at strains 0002 and the failure occurs atstrains 00035 the failure being progressively more brittle asthe concrete strength increases

For design purposes the short-term stressndashstrain curve isgenerally idealised to a form in which the initial portion isparabolic or linear and the remainder is at a uniform stress Afurther simplification in the form of an equivalent rectangularstress block may be made subsequently Typical stressndashstraincurves and those recommended for design purposes are givenin Table 36 for BS 8110 and Table 44 for EC 2

317 Durability of concrete

Concrete has to be durable in natural environments rangingfrom mild to extremely aggressive and resistant to factors suchas weathering freezethaw attack chemical attack and abrasionIn addition for concrete containing reinforcement the surfaceconcrete must provide adequate protection against the ingressof moisture and air which would eventually cause corrosion ofthe embedded steel

Strength alone is not necessarily a reliable guide to concretedurability many other factors have to be taken into accountthe most important being the degree of impermeability This isdependent mainly on the constituents of the concrete in partic-ular the free watercement ratio and in the provision of fullcompaction to eliminate air voids and effective curing toensure continuing hydration

Concrete has a tendency to be permeable as a result ofthe capillary voids in the cement paste matrix In order for theconcrete to be sufficiently workable it is common to use farmore water than is actually necessary for the hydration of thecement When the concrete dries out the space previouslyoccupied by the excess water forms capillary voids Providedthe concrete has been fully compacted and properly cured thevoids are extremely small the number and the size of the voidsdecreasing as the free watercement ratio is reduced The moreopen the structure of the cement paste the easier it is for airmoisture and harmful chemicals to penetrate

Carbonation Steel reinforcement that is embedded in goodconcrete with an adequate depth of cover is protectedagainst corrosion by the highly alkaline pore water in thehardened cement paste Loss of alkalinity of the concrete canbe caused by the carbon dioxide in the air reacting with and

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neutralising the free lime If this reaction which is calledcarbonation reaches the reinforcement then corrosion willoccur in moist environments Carbonation is a slow processthat progresses from the surface and is dependent on thepermeability of the concrete and the humidity of the environ-ment Provided the depth of cover and quality of concreterecommended for the anticipated exposure conditions areachieved corrosion due to carbonation should not occur duringthe intended lifetime of the structure

Freezethaw attack The resistance of concrete to freezingand thawing depends on its impermeability and the degreeof saturation on being exposed to frost the higher the degree ofsaturation the more liable the concrete is to damage The useof salt for de-icing roads and pavements greatly increases therisk of freezethaw damage

The benefits of air-entrained concrete have been referred toin section 314 where it was recommended that all exposedhorizontal paved areas from roads and runways to footpathsand garage drives and marine structures should be made ofair-entrained concrete Similarly parts of structures adjacent tohighways and in car parks which could be splashed or comeinto contact with salt solutions used for de-icing should alsouse air-entrained concrete Alternatively the cube strength ofthe concrete should be 50 Nmm2 or more Whilst C4050concrete is suitable for many situations it does not have thesame freezethaw resistance as air-entrained concrete

Chemical attack Portland cement concrete is liable toattack by acids and acid fumes including the organic acidsoften produced when foodstuffs are being processed Vinegarfruit juices silage effluent sour milk and sugar solutions can allattack concrete Concrete made with Portland cement is notrecommended for use in acidic conditions where the pH valueis 55 or less without careful consideration of the exposurecondition and the intended construction Alkalis have little effecton concrete

For concrete that is exposed to made-up ground includingcontaminated and industrial material specialist advice shouldbe sought in determining the design chemical class so that asuitable concrete can be specified The most common formof chemical attack that concretes have to resist is the effect ofsolutions of sulfates present in some soils and groundwaters

In all cases of chemical attack concrete resistance is related tofree watercement ratio cement content type of cement and thedegree of compaction Well-compacted concrete will always bemore resistant to sulfate attack than one less well compactedregardless of cement type Recommendations for concreteexposed to sulfate-containing groundwater and for chemicallycontaminated brownfield sites are incorporated in BS 8500-1

Alkalindashsilica reaction ASR is a reaction that can occur inconcrete between certain siliceous constituents present in theaggregate and the alkalis ndash sodium and potassium hydroxide ndashthat are released during cement hydration A gelatinous productis formed which imbibes pore fluid and in so doing expandsinducing an internal stress within the concrete The reactionwill cause damage to the concrete only when the followingthree conditions occur simultaneously

A reactive form of silica is present in the aggregate in criticalquantities

The pore solution contains ions of sodium potassium andhydroxyl and is of a sufficiently high alkalinity

A continuing supply of water is available

If any one of these factors is absent then damage from ASRwill not occur and no precautions are necessary It is possiblefor the reaction to take place in the concrete without inducingexpansion Damage may not occur even when the reactionproduct is present throughout the concrete as the gel may fillcracks induced by some other mechanism Recommendationsare available for minimising the risk of damage from ASR innew concrete construction based on ensuring that at least oneof the three aforementioned conditions is absent

Exposure classes For design and specification purposes theenvironment to which concrete will be exposed during itsintended life is classified into various levels of severity Foreach category minimum requirements regarding the qualityof the concrete and the cover to the reinforcement are givenin Codes of Practice In British Codes for many years theexposure conditions were mild moderate severe very severeand most severe (or in BS 5400 extreme) with abrasive as afurther category Details of the classification system that wasused in BS 8110 and BS 5400 are given in Table 39

In BS EN 206-1 BS 8500-1 and EC 2 the conditionsare classified in terms of exposure to particular actions withvarious levels of severity in each category The followingcategories are considered

1 No risk of corrosion or attack

2 Corrosion induced by carbonation

3 Corrosion induced by chlorides other than from seawater

4 Corrosion induced by chlorides from seawater

5 Freezethaw attack

6 Chemical attack

If the concrete is exposed to more than one of these actions theenvironmental conditions are expressed as a combination ofexposure classes Details of each class in categories 1ndash5 withdescriptions and informative examples applicable in the UnitedKingdom are given in Tables 37 and 45 For concrete exposedto chemical attack the exposure classes given in BS EN 206-1cover only natural ground with static water which represents alimited proportion of the aggressive ground conditions found inthe United Kingdom In the complementary British StandardBS 8500-1 more comprehensive recommendations are providedbased on the approach used in ref 13

On this basis an ACEC (aggressive chemical environmentfor concrete) class is determined according to the chemicals inthe ground the type of soil and the mobility and acidity of thegroundwater The chemicals in the ground are expressed as adesign sulfate class (DS) in which the measured sulfate contentis increased to take account of materials that may oxidise intosulfate for example pyrite and other aggressive species suchas hydrochloric or nitric acid Magnesium ion content is alsoincluded in this classification Soil is classified as natural orfor sites that may contain chemical residues from previousindustrial use or imported wastes as brownfield Water in theground is classified as either static or mobile and according toits pH value

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Based on the ACEC classification and according to the sizeof the section and the selected structural performance level therequired concrete quality expressed as a design chemicalclass (DC) and any necessary additional protective measures(APMs) can be determined The structural performance level isclassified as low normal or high in relation to the intendedservice life the vulnerability of the structural details and thesecurity of structures retaining hazardous materials

Concrete quality and cover to reinforcement Concretedurability is dependent mainly on its constituents particularlythe free watercement ratio The ratio can be reduced and thedurability of the concrete enhanced by increasing the cementcontent andor using admixtures to reduce the amount of freewater needed for a particular level of consistence subject tospecified minimum requirements being met for the cementcontent By limiting the maximum free watercement ratio andthe minimum cement content a minimum strength class can beobtained for particular cements and combinations

Where concrete containing reinforcement is exposed to airand moisture or is subject to contact with chlorides from anysource the protection of the steel against corrosion depends onthe concrete cover The required thickness is related to theexposure class the concrete quality and the intended workinglife of the structure Recommended values for an intendedworking life of at least 50 years are given in Tables 38 and 46(BS 8500) and 39 (prior to BS 8500)

Codes of Practice also specify values for the covers neededto ensure the safe transmission of bond forces and provide anadequate fire-resistance for the reinforced concrete member Inaddition allowance may need to be made for abrasion or forsurface treatments such as bush hammering In BS 8110 valuesused to be given for a nominal cover to be provided to all rein-forcement including links on the basis that the actual covershould not be less than the nominal cover minus 5 mm In BS8500 values are given for a minimum cover to which anallowance for tolerance (normally 10 mm) is then added

Concrete specification Details of how to specify con-crete and what to specify are given in BS 8500-1 Threetypes ndash designed prescribed and standardised prescribedconcretes ndash are recognised by BS EN 206-1 but BS 8500adds two more ndash designated and proprietary concretes

Designed concretes are ones where the concrete producer isresponsible for selecting the mix proportions to provide theperformance defined by the specifier Conformity of designedconcretes is usually judged by strength testing of 100 mm or150 mm cubes which in BS 8500 is the responsibility ofthe concrete producer Prescribed concretes are ones where thespecification states the mix proportions in order to satisfyparticular performance requirements in terms of the mass ofeach constituent Such concretes are seldom necessary butmight be used where particular properties or special surfacefinishes are required Standardised prescribed concretes that areintended for site production using basic equipment and controlare given in BS 8500-2 Whilst conformity does not depend onstrength testing assumed characteristic strengths are given forthe purposes of design Designated concretes are a wide-ranginggroup of concretes that provide for most types of concreteconstruction The producer must operate a recognized accreditedthird party certification system and is responsible for ensuring

that the concrete conforms to the specification given inBS 8500-2 Proprietary concretes are intended to provide forinstances when a concrete producer would give assurance of theperformance of concrete without being required to declare itscomposition

For conditions where corrosion induced by chlorides doesnot apply structural concretes should generally be specifiedas either designated concretes or designed concretes Whereexposure to corrosion due to chlorides is applicable only thedesigned concrete method of specifying is appropriate Anexception to this situation is where an exposed aggregate ortooled finish that removes the concrete surface is required Inthese cases in order to get an acceptable finish a special mixdesign is needed Initial testing including trial panels shouldbe undertaken and from the results of these tests a prescribedconcrete can be specified For housing applications both adesignated concrete and a standardised prescribed concrete canbe specified as acceptable alternatives This would allow aconcrete producer with accredited certification to quote forsupplying a designated concrete and the site contractor or aconcrete producer without accredited certification to quote forsupplying a standardised prescribed concrete

32 REINFORCEMENT

Reinforcement for concrete generally consists of deformedsteel bars or welded steel mesh fabric Normal reinforcementrelies entirely upon the alkaline environment provided by adurable concrete cover for its protection against corrosion Inspecial circumstances galvanised epoxy-coated or stainlesssteel can be used Fibre-reinforced polymer materials havealso been developed So far in the United Kingdom thesematerials have been used mainly for external strengthening anddamage repair applications

321 Bar reinforcement

In the United Kingdom reinforcing bars are generally specifiedordered and delivered to the requirements of BS 4449 Thiscaters for steel bars with a yield strength of 500 MPa in threeductility classes grades B500A B500B and B500C Bars areround in cross section having two or more rows of uniformlyspaced transverse ribs with or without longitudinal ribs Thepattern of transverse ribs varies with the grade and can beused as a means of identification Information with regardto the basic properties of reinforcing bars to BS 4449 whichis in general conformity with BS EN 10080 is given inTable 219

All reinforcing bars are produced by a hot-rolling process inwhich a cast steel billet is reheated to 1100ndash1200oC andthen rolled in a mill to reduce its cross section and impart therib pattern There are two common methods for achievingthe required mechanical properties in hot-rolled bars in-lineheat treatment and micro-alloying In the former method whichis sometimes referred to as the quench-and-self-temper (QST)process high-pressure water sprays quench the bar surface as itexits the rolling mill producing a bar with a hard temperedouter layer and a softer more ductile core Most reinforcingbars in the United Kingdom are of this type and achieve class Bor class C ductility In the micro-alloying method strengthis achieved by adding small amounts of alloying elements

Material properties24

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during the steel-making process Micro-alloy steels normallyachieve class C ductility Another method that can be usedto produce high-yield bars involves a cold-twisting process toform bars that are identified by spiralling longitudinal ribs Thisprocess has been obsolete in the United Kingdom for sometime but round ribbed twisted bars can be found in someexisting structures

In addition to bars being produced in cut straight lengthsbillets are also rolled into coil for diameters up to 16 mm Inthis form the product is ideal for automated processes suchas link bending QST micro-alloying and cold deformationprocesses are all used for high-yield coil Cold deformation isapplied by continuous stretching which is less detrimental toductility than the cold-twisting process mentioned previouslyCoil products have to be de-coiled before use and automaticlink bending machines incorporate straightening rolls Largerde-coiling machines are also used to produce straight lengths

322 Fabric reinforcement

Steel fabric reinforcement is an arrangement of longitudinal barsand cross bars welded together at their intersections in a shearresistant manner In the United Kingdom fabric is producedunder a closely controlled factory-based manufacturing processto the requirements of BS 4483 In fabric for structural purposesribbed bars complying with BS 4449 are used For wrappingfabric as described later wire complying with BS 4482 may beused Wire can be produced from hot-rolled rod by eitherdrawing the rod through a die to produce plain wire or coldrolling the rod to form indented or ribbed wires In BS 4482provision is made for plain round wire with a yield strength of250 MPa and plain indented or ribbed wires with a yieldstrength of 500 MPa

In BS 4483 provision is made for fabric reinforcement tobe either of a standard type or purpose made to the clientrsquosrequirements The standard fabric types have regular mesharrangements and bar sizes and are defined by identifiablereference numbers Type A is a square mesh with identical longbars and cross bars commonly used in ground slabs Type B isa rectangular (structural) mesh that is particularly suitable foruse in thin one-way spanning slabs Type C is a rectangular(long) mesh that can be used in pavements and in two-wayspanning slabs by providing separate sheets in each directionType D is a rectangular (wrapping) mesh that is used in theconcrete encasement of structural steel sections The stock sizeof standard fabric sheets is 48 m 24 m and merchantsize sheets are also available in a 36 m 20 m size Fulldetails of the preferred range of standard fabric types are givenin Table 220

Purpose-made fabrics specified by the customer can haveany combination of wire size and spacing in either direction Inpractice manufacturers may sub-divide purpose-made fabricsinto two categories special (also called scheduled) and bespoke(also called detailed) Special fabrics consist of the standardwire size combinations but with non-standard overhangs andsheet dimensions up to 12 m 33 m Sheets with so-calledflying ends are used to facilitate the lapping of adjacent sheets

Bespoke fabrics involve a more complex arrangement inwhich the wire size spacing and length can be varied within thesheet These products are made to order for each contract as areplacement for conventional loose bar assemblies The use of

bespoke fabrics is appropriate on contracts with a large amountof repeatability and generally manufacturers would require aminimum tonnage order for commercial viability

323 Stressndashstrain curves

For hot-rolled reinforcement the stressndashstrain relationship intension is linear up to yield when there is a pronounced increaseof strain at constant stress (yield strength) Further smallincreases of stress resulting in work hardening are accompaniedby considerable elongation A maximum stress (tensile strength)is reached beyond which further elongation is accompanied bya stress reduction to failure Micro-alloy bars are characterisedby high ductility (high level of uniform elongation and high ratioof tensile strengthyield strength) For QST bars the stressndashstraincurve is of similar shape but with slightly less ductility

Cold-processed reinforcing steels show continuous yieldingbehaviour with no defined yield point The work-hardeningcapacity is lower than for the hot-rolled reinforcement withthe uniform elongation level being particularly reduced Thecharacteristic strength is defined as the 02 proof stress(ie a stress which on unloading would result in a residualstrain of 02) and the initial part of the stressndashstrain curve islinear to beyond 80 of this value

For design purposes the yield or 02 proof condition isnormally critical and the stressndashstrain curves are idealised to abi-linear or sometimes tri-linear form Typical stressndashstraincurves and those recommended for design purposes are givenin Table 36 for BS 8110 and Table 44 for EC 2

324 Bar sizes and bends

The nominal size of a bar is the diameter of a circle with an areaequal to the effective cross-sectional area of the bar The rangeof nominal sizes (millimetres) is from 6 to 50 with preferredsizes of 8 10 12 16 20 25 32 and 40 Values of the totalcross-sectional area provided in a concrete section according tothe number or spacing of the bars for different bar sizes aregiven in Table 220

Bends in bars should be formed around standard mandrels onbar-bending machines In BS 8666 the minimum radius ofbend r is standardised as 2d for d 16 and 35d for d 20where d is the bar size Values of r for each different bar sizeand values of the minimum end projection P needed to formthe bend are given in Table 219 In some cases (eg wherebars are highly stressed) the bars need to be bent to a radiuslarger than the minimum value in order to satisfy the designrequirements and the required radius R is then specified on thebar-bending schedule

Reinforcement should not be bent or straightened on site ina way that could damage or fracture the bars All bars shouldpreferably be bent at ambient temperature but when the steeltemperature is below 5oC special precautions may be neededsuch as reducing the speed of bending or with the engineerrsquosapproval increasing the radius of bending Alternatively thebars may be warmed to a temperature not exceeding 100oC

325 Bar shapes and bending dimensions

Bars are produced in stock lengths of 12 m and lengths upto 18 m can be supplied to special order In most structures

Reinforcement 25

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bars are required in shorter lengths and often need to be bentThe cutting and bending of reinforcement is generally specifiedto the requirements of BS 8666 This contains recommendedbar shapes designated by shape code numbers which areshown in Tables 221 and 222 The information needed to cutand bend the bars to the required dimensions is entered into abar schedule an example of which is shown in Table 223 Eachschedule is related to a member on a particular drawing bymeans of the bar schedule reference number

In cases where a bar is detailed to fit between two concretefaces with no more than the nominal cover on each face (eg linksin beams) an allowance for deviations is required This is tocater for variation due to the effect of inevitable errors in thedimensions of the formwork and the cutting bending andfixing of the bars Details of the deductions to be made to allowfor these deviations and calculations to determine the bendingdimensions in a typical example are given in section 1035with the completed bar schedule in Table 223

326 Stainless steel reinforcement

The type of reinforcement to be used in a structure is usuallyselected on the basis of initial costs This normally results inthe use of carbon steel reinforcement which is around 15of the cost of stainless steel For some structures however theselective use of stainless steel reinforcement ndash on exposedsurfaces for example ndash can be justified In Highways Agencydocument BA 8402 it is recommended that stainless steelreinforcement should be used in splash zones abutmentsparapet edges and soffits and where the chances of chlorideattack are greatest It is generally considered that where theconcrete is saturated and oxygen movement limited stainlesssteel is not required Adherence to these guidelines can meanthat the use of stainless steel reinforcement only marginallyincreases construction costs while significantly reducing thewhole-life costs of the structure and increasing its usable life

Stainless steels are produced by adding elements to iron toachieve the required compositional balance The additionalelements besides chromium can include nickel manganesemolybdenum and titanium with the level of carbon beingcontrolled during processing These alloying elements affectthe steelrsquos microstructure as well as its mechanical propertiesand corrosion resistance Four ranges of stainless steel areproduced two of which are recommended for reinforcement toconcrete because of their high resistance to corrosionAustenitic stainless steels for which chromium and nickel arethe main alloying elements have good general propertiesincluding corrosion resistance and are normally suitable formost applications Duplex stainless steels which have highchromium and low nickel contents provide greater corrosionresistance for the most demanding environments

In the United Kingdom austenitic stainless steel reinforcementhas been produced to the requirements of BS 6744 which isbroadly aligned to conventional reinforcement practice Thusplain and ribbed bars are available in the same characteristicstrengths and range of preferred sizes as normal carbon steelreinforcement Traditionally stainless steel reinforcement hasonly been stocked in maximum lengths of 6 m for all sizesBars are currently available in lengths up to 12 m for sizes upto 16 mm For larger sizes bars can be supplied to order in

lengths up to 8 m Comprehensive data and recommendationson the use of stainless steel reinforcement are given in ref 14

327 Prefabricated reinforcement systems

In order to speed construction by reducing the time needed tofix reinforcement it is important to be able to pre-assemblemuch of the reinforcement This can be achieved on site givenadequate space and a ready supply of skilled personnel Inmany cases with careful planning and collaboration at an earlystage the use of reinforcement assemblies prefabricated by thesupplier can provide considerable benefits

A common application is the use of fabric reinforcement asdescribed in sections 323 and 1032 The preferred range ofdesignated fabrics can be routinely used in slabs and walls Incases involving large areas with long spans and considerablerepetition made-to-order fabrics can be specially designed tosuit specific projects Provision for small holes and openingscan be made by cutting the fabric on site after placing thesheets and adding loose trimming bars as necessary Whilesheets of fabric can be readily handled normally they areawkward to lift over column starter bars In such cases it isgenerally advisable to provide the reinforcement local to thecolumn as loose bars fixed in the conventional manner

A more recent development is the use of slab reinforcementrolls that can be unrolled directly into place on site Each made-to-order roll consists of reinforcement of the required size andspacing in one direction welded to thin metal bands and rolledaround hoops that are later discarded Rolls can be produced upto a maximum bar length of 15 m and a weight of 5 tonnes Thewidth of the sheet when fully rolled out could be more than50 m depending upon the bar size and spacing The full rangeof preferred bar sizes can be used and the bar spacing andlength can be varied within the same roll For each area of slaband for each surface to be reinforced two rolls are requiredThese are delivered to site craned into position and unrolled oncontinuous bar supports Each roll provides the bars in onedirection with those in the lower layer resting on conventionalspacers or chairs

The need to provide punching shear reinforcement in solidflat slabs in the vicinity of the columns has resulted in severalproprietary reinforcement systems Vertical reinforcement isrequired in potential shear failure zones around the columnsuntil a position is reached at which the slab can withstand theshear stresses without reinforcement Conventional links aredifficult and time-consuming to set out and fix Single-leggedlinks are provided with a hook at the top and a 90o bend at thebottom Each link has to be hooked over a top bar in the slaband the 90o bend pushed under a bottom bar and tied in place

Shear ladders can be used in which a row of single-leggedlinks are connected by three straight anchor bars welded toform a robust single unit The ladders provide the required shearreinforcement and act as chairs to support the top bars The sizespacing and height of the links can be varied to suit the designrequirements Shear hoops consist of U-shaped links welded toupper and lower hoops to form a three-dimensional unit Byusing hoops of increasing size shear reinforcement can beprovided on successive perimeters

Shear band strips with a castellated profile are made from25 mm wide high-tensile steel strip in a variety of gauges to

Material properties26

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cater for different shear capacities The strip has perforatedholes along the length to help with anchorage and fixing Thepeaks and troughs of the profile are spaced to coincide withthe spacing of the main reinforcement Stud rails consist of arow of steel studs welded to a flat steel strip or a pair of rodsThe studs are fabricated from plain or deformed reinforcingbars with an enlarged head welded to one or both ends Thesize spacing and height of the studs can be varied to suit theshear requirements and the slab depth

The use of reinforcement continuity strips is a simple andeffective means of providing reinforcement continuity acrossconstruction joints A typical application occurs at a junctionbetween a wall and a slab that is to be cast at a later stageThe strips comprise a set of special pre-bent bars housed in agalvanised indented steel casing that is fabricated off-site ina factory-controlled environment On site the entire unit is castinto the front face of the wall After the formwork is struck thelid of the casing is removed to reveal the legs of the barscontained within the casing The legs are then straightenedoutwards by the contractor ready for lapping with the mainreinforcement in the slab The casing remains embedded in thewall creating a rebate into which the slab concrete flows andeliminating the need for traditional joint preparation

328 Fixing of reinforcement

Reinforcing bars need to be tied together to prevent their beingdisplaced and provide a rigid system Bar assemblies and fabricreinforcement need to be supported by spacers and chairs toensure that the required cover is achieved and kept duringthe subsequent placing and compaction of concrete Spacersshould be fixed to the links bars or fabric wires that are nearestto the concrete surface to which the cover is specifiedRecommendations for the specification and use of spacers andchairs and the tying of reinforcement are given in BS 7973Parts 1 and 2 These include details of the number and positionof spacers and the frequency of tying

33 FIRE-RESISTANCE

Building structures need to conform in the event of fire toperformance requirements stated in the Building RegulationsFor stability the elements of the structure need to providea specified minimum period of fire-resistance in relation to astandard test The required fire period depends on the purposegroup of the building and the height or for basements depth ofthe building relative to the ground as given in Table 312Building insurers may require longer fire periods for storagefacilities where the value of the contents and the costs ofreinstatement of the structure are particularly important

In BS 8110 design for fire-resistance is considered at twolevels Part 1 contains simple recommendations suitable formost purposes Part 2 contains a more detailed treatment with

a choice of three methods involving tabulated data furnacetests or fire engineering calculations The tabulated data is inthe form of minimum specified values of member size andconcrete cover The cover is given to the main reinforcementand in the case of beams and ribs can vary in relation to theactual width of the section The recommendations in Part 1 arebased on the same data but the presentation is different intwo respects values are given for the nominal cover to all rein-forcement (this includes an allowance for links in the case ofbeams and columns) and the values do not vary in relation tothe width of the section The required nominal covers to allreinforcement and minimum dimensions for various membersare given in Tables 310 and 311 respectively

In the event of a fire in a building the vulnerable elementsare the floor construction above the fire and any supportingcolumns or walls The fire-resistance of the floor members(beams ribs and slabs) depends upon the protection providedto the bottom reinforcement The steel begins to lose strengthat a temperature of 300oC losses of 50 and 75 occurringat temperatures of about 560oC and 700oC respectively Theconcrete cover needs to be sufficient to delay the time takento reach a temperature likely to result in structural failure Adistinction is made between simply supported spans wherea 50 loss of strength in the bottom reinforcement could becritical and continuous spans where a greater loss is allowedbecause the top reinforcement will retain its full capacity

If the cover becomes excessive there is a risk of prematurespalling of the concrete in the event of fire Concretes madewith aggregates containing a high proportion of silica are the mostsusceptible In cases where the nominal cover needs to exceed40 mm additional measures should be considered and severalpossible courses of action are described in Part 2 of BS 8110The preferred approach is to reduce the cover by providingadditional protection in the form of an applied finish or a falseceiling or by using lightweight aggregates or sacrificial steelThe last measure refers to the provision of more steel than isnecessary for normal purposes so that a greater loss of strengthcan be allowed in the event of fire If the nominal cover doesexceed 40 mm then supplementary reinforcement in the formof welded steel fabric should be placed within the thicknessof the cover at 20 mm from the concrete surface There areconsiderable practical difficulties with this approach and it mayconflict with the requirements for durability in some cases

For concrete made with lightweight aggregate the nominalcover requirements are all reduced and the risk of prematurespalling only needs to be considered when the cover exceeds50 mm The detailed requirements for lightweight aggregateconcrete and guidance on the additional protection provided byselected applied finishes are given in Table 310

EC 2 contains a more flexible approach to fire safety designbased on the concept of lsquoload ratiorsquo which is the ratio of theload applied at the fire limit-state to the capacity of the elementat ambient temperature

Fire-resistance 27

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Torsion-less beams are designed as linear elements subjectedto bending moments and shear forces The values for freelysupported beams and cantilevers are readily determinedby the simple rules of static equilibrium but the analysis ofcontinuous beams and statically indeterminate frames is morecomplex Historically various analytical techniques have beendeveloped and used as self-contained methods to solve partic-ular problems In time it was realised that the methodscould be divided into two basic categories flexibility methods(otherwise known as action methods compatibility methods orforce methods) and displacement methods (otherwise known asstiffness methods or equilibrium methods) The behaviour ofthe structure is considered in terms of unknown forces in thefirst category and unknown displacements in the secondcategory For each method a particular solution obtained bymodifying the structure to make it statically determinate iscombined with a complementary solution in which the effectof each modification is determined Consider the case of acontinuous beam For the flexibility methods the particularsolution involves removing redundant actions (ie the continuitybetween the individual members) to leave a series of discon-nected spans For the displacement methods the particularsolution involves restricting the rotations andor displacementsthat would otherwise occur at the joints

To clarify further the main differences between the methodsin the two categories consider a propped cantilever With theflexibility approach the first step is to remove the prop andcalculate the deflection at the position of the prop due to theaction of the applied loads this gives the particular solutionThe next step is to calculate the concentrated load needed at theposition of the prop to restore the deflection to zero this givesthe complementary solution The calculated load is the reactionin the prop knowing this enables the moments and forces in thepropped cantilever to be simply determined If the displacementapproach is used the first step is to consider the span as fullyfixed at both ends and calculate the moment at the propped enddue to the applied loads this gives the particular solution Thenext step is to release the restraint at the propped end and applyan equal and opposite moment to restore the rotation to zero thisgives the complementary solution By combining the momentdiagrams the resulting moments and forces can be determined

In general there are several unknowns and irrespectiveof the method of analysis used the preparation and solution ofa set of simultaneous equations is required The resulting

relationship between forces and displacements embodies a seriesof coefficients that can be set out concisely in matrix formIf flexibility methods are used the resulting matrix is built up offlexibility coefficients each of which represents a displacementproduced by a unit action Similarly if stiffness methods areused the resulting matrix is formed of stiffness coefficients eachof which represents an action produced by a unit displacementThe solution of matrix equations either by matrix inversionor by a systematic elimination process is ideally suited tocomputer technology To this end methods have been devised(the so-called matrix stiffness and matrix flexibility methods)for which the computer both sets up and solves the simultaneousequations (ref 15)

Here it is worthwhile to summarise the basic purpose ofthe analysis Calculating the bending moments on individualfreely supported spans ensures that equilibrium is maintainedThe analytical procedure that is undertaken involves linearlytransforming these free-moment diagrams in a manner that iscompatible with the allowable deformations of the structureUnder ultimate load conditions deformations at the criticalsections must remain within the limits that the sections canwithstand and under service load conditions deformationsmust not result in excessive deflection or cracking or both Ifthe analysis is able to ensure that these requirements are met itwill be entirely satisfactory for its purpose endeavouring toobtain painstakingly precise results by over-complex methodsis unjustified in view of the many uncertainties involved

To determine at any section the effects of the applied loadsand support reactions the basic relationships are as follows

Shear force (forces on one side of section) rate of change of bending moment

Bending moment (moments of forces on one side of section) (shear force) area of shear force diagram

Slope (curvature) area of curvature diagram

Deflection (slope) area of slope diagram

For elastic behaviour curvature MEI where M is bendingmoment E is modulus of elasticity of concrete I is secondmoment of area of section For the purposes of structural

Chapter 4

Structural analysis

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analysis to determine bending moments due to applied loadsI values may normally be based on the gross concrete sectionIn determining deflections however due allowance needs to bemade for the effects of cracking and in the long term for theeffects of concrete creep and shrinkage

41 SINGLE-SPAN BEAMS AND CANTILEVERS

Formulae to determine the shearing forces bending momentsand deflections produced by various general loads on beamsfreely supported at the ends are given in Table 224 Similarexpressions for some particular load arrangements commonlyencountered on beams either freely supported or fully fixedat both ends with details of the maximum values are given inTable 225 The same information but relating to simple andpropped cantilevers is given in Tables 226 and 227 respectivelyCombinations of loads can be considered by summing theresults obtained for each individual load

In Tables 224ndash227 expressions are also given for the slopesat the beam supports and the free (or propped) end of a cantileverInformation regarding the slope at other points is seldomrequired If needed it is usually a simple matter to obtain theslope by differentiating the deflection formula with respect to xIf the resulting expression is equated to zero and solved toobtain x the point of maximum deflection will have been foundThis value of x can then be substituted into the original formulato obtain the maximum deflection

Coefficients to determine the fixed-end moments producedby various symmetrical and unsymmetrical loads on beamsfully fixed at both ends are given in Table 228 Loadings notshown can usually be considered by using the tabulated casesin combination For the general case of a partial uniform ortriangular distribution of load placed anywhere on a membera full range of charts is contained in Examples of the Design ofReinforced Concrete Buildings The charts give deflection andmoment coefficients for beams (freely supported or fully fixedat both ends) and cantilevers (simple or propped)

42 CONTINUOUS BEAMS

Historically various methods of structural analysis have beendeveloped for determining the bending moments and shearingforces on beams continuous over two or more spans Most ofthese have been stiffness methods which are generally bettersuited than flexibility methods to hand computation Some ofthese approaches such as the theorem of three-moments and themethods of fixed points and characteristic points were includedin the previous edition of this Handbook If beams having twothree or four spans are of uniform cross section and supportloads that are symmetrical on each individual span formulaeand coefficients can be derived that enable the support momentsto be determined by direct calculation Such a method is givenin Table 237 More generally in order to avoid the need to solvelarge sets of simultaneous equations methods involving succes-sive approximations have been devised Despite the general useof computers hand methods can still be very useful in dealingwith routine problems The ability to use hand methods alsodevelops in the engineer an appreciation of analysis that isinvaluable in applying output from the computer

When bending moments are calculated with the spans takenas the distances between the centres of supports the critical

negative moment in monolithic forms of construction can beconsidered as that occurring at the edge of the support Whenthe supports are of considerable width the span can be taken asthe clear distance between the supports plus the effective depthof the beam or an additional span can be introduced thatis equal to the width of the support minus the effective depth ofthe beam The load on this additional span should be takenas the support reaction spread uniformly over the width of thesupport If a beam is constructed monolithically with a verywide and massive support the effect of continuity with the spanor spans beyond the support may be negligible in which casethe beam should be treated as fixed at the support

The second moment of area of a reinforced concrete beamof uniform depth may still vary throughout its length due tovariations in the amount of reinforcement and also becausewhen acting with an adjoining slab a down-stand beam maybe considered as a flanged section at mid-span but a simplerectangular section at the supports It is common practicehowever to neglect these variations for beams of uniformdepth and use the value of I for the plain rectangular section Itis often assumed that a continuous beam is freely supportedat the ends even when beam and support are constructedmonolithically Some provision should still be made for theeffects of end restraint

421 Analysis by moment distribution

Probably the best-known and simplest system for analysingcontinuous beams by hand is that of moment distributionas devised by Hardy Cross in 1929 The method whichderives from slope-deflection principles is described briefly inTable 236 It employs a system of successive approximationsthat may be terminated as soon as the required degree ofaccuracy has been reached A particular advantage of this andsimilar methods is that even after only one distribution cycleit is often clear whether or not the final values will be acceptableIf not the analysis can be discontinued and unnecessary workavoided The method is simple to remember and apply andthe step-by-step procedure gives the engineer a lsquofeelrsquo for thebehaviour of the system It can be applied albeit less easily tothe analysis of systems containing non-prismatic members andto frames Hardy Cross moment distribution is described inmany textbooks dealing with structural analysis

Over the years the Hardy Cross method of analysis begotvarious offspring One of these is known as precise momentdistribution (also called the coefficient of restraint method ordirect moment distribution) The procedure is very similar tonormal moment distribution but the distribution and carryoverfactors are so adjusted that an exact solution is obtainedafter one distribution in each direction The method thus hasthe advantage of removing the necessity to decide when toterminate the analysis Brief details are given in Table 236 andthe method is described in more detail in Examples of theDesign of Reinforced Concrete Buildings (see also ref 16)

It should be noted that the load arrangements that producethe greatest negative bending moments at the supports are notnecessarily those that produce the greatest positive bendingmoments in the spans The design loads to be considered inBS 8110 and EC 2 and the arrangements of live load that givethe greatest theoretical bending moments as well as the lessonerous code requirements are given in Table 229 Some live

Continuous beams 29

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load arrangements can result in negative bending momentsthroughout adjacent unloaded spans

422 Redistribution of bending moments

For the ULS the bending moments obtained by linear elasticanalysis may be adjusted on the basis that some redistributionof moments can occur prior to collapse This enables the effectsof both service and ultimate loadings to be assessed without theneed to undertake a separate analysis using plastic-hinge tech-niques for the ultimate condition The theoretical justificationfor moment redistribution is clearly explained in the Handbookto BS 8110 Since the reduction of moment at a section assumesthe formation of a plastic hinge at that position prior to theultimate condition being reached it is necessary to limit thereduction in order to restrict the amount of plastic-hinge rotationand control the cracking that occurs under serviceabilityconditions For these reasons the maximum ratio of neutralaxis depth to effective depth and the maximum distancebetween tension bars are each limited according to the requiredamount of redistribution

Such adjustments are useful in reducing the inequalitiesbetween negative and positive moments and minimising theamount of reinforcement that must be provided at a particularsection such as the intersection between beam and columnwhere concreting may otherwise be more difficult due to thecongestion of reinforcement Both BS 8110 and EC 2 allowthe use of moment redistribution the procedure which may beapplied to any system that has been analysed by the so-calledexact methods is described in section 123 with an illustratedexample provided in Table 233

423 Coefficients for equal loads onequal spans

For beams that are continuous over a number of equal spanswith equal loads on each loaded span the maximum bendingmoments and shearing forces can be tabulated In Tables 230and 231 maximum bending moment coefficients are given foreach span and at each support for two three four and five equalspans with identical loads on each span which is the usualdisposition of the dead load on a beam Coefficients are alsogiven for the most adverse incidence of live loads and in thecase of the support moments for the arrangements of live loadrequired by BS 8110 (values in square brackets) and by EC 2(values in curved brackets) It should be noted that the maximumbending moments due to live load do not occur at all thesections simultaneously The types of load considered are auniformly distributed load a central point load two equal loadsapplied at the third-points of the span and trapezoidal loads ofvarious proportions In Table 232 coefficients are given for themaximum shearing forces for each type of load with identicalloads on each span and due to the most adverse incidence oflive loads

424 Bending moment diagrams for equal spans

In Tables 234 and 235 bending moment coefficients forvarious arrangements of dead and live loads with sketches

of the resulting moment envelopes are given for beams oftwo and three spans and for a theoretically infinite systemThis information enables appropriate bending moment diagramsto be plotted quickly and accurately The load types consideredare a uniformly distributed load a central point load and twoequal loads at the third points of the span Values are givenfor identical loads on each span (for example dead load) andfor the arrangements of live load required by BS 8110 andEC 2 As the coefficients have been calculated by exactmethods moment redistribution is allowed at the ultimate statein accordance with the requirements of BS 8110 and EC 2 Inaddition to the coefficients obtained by linear elastic analysisvalues are given for conditions in which the maximum supportmoments are reduced by either 10 or 30 as described insection 1233 Coefficients are also given for the positivesupport moments and negative span moments that occur undersome arrangements of live load

425 Solutions for routine design

A precise determination of theoretical bending moments andshearing forces on continuous beams is not always necessary Itshould also be appreciated that the general assumptions ofunyielding knife-edge supports uniform sectional propertiesand uniform distributions of live load are hardly realistic Theindeterminate nature of these factors often leads in practice tothe adoption of values based on approximate coefficients InTable 229 values in accordance with the recommendationsof BS 8110 and EC 2 are given for bending moments andshearing forces on uniformly loaded beams of three or morespans The values are applicable when the characteristicimposed load is not greater than the characteristic dead loadand the variations in span do not exceed 15 of the longestspan The same coefficients may be used with service loads orultimate loads and the resulting bending moments may beconsidered to be without redistribution

43 MOVING LOADS ON CONTINUOUS BEAMS

Bending moments caused by moving loads such as those due tovehicles traversing a series of continuous spans are most easilycalculated with the aid of influence lines An influence line is acurve with the span of the beam taken as the base the ordinateof the curve at any point being the value of the bending momentproduced at a particular section when a unit load acts at thepoint The data given in Tables 238ndash241 enable the influencelines for the critical sections of beams continuous over twothree four and five or more spans to be drawn By plotting theposition of the load on the beam (to scale) the bending momentsat the section being considered can be derived as explained inthe example given in Chapter 12 The curves given for equalspans can be used directly but the corresponding curves forunequal spans need to be plotted from the data tabulated

The bending moment due to a load at any point is equal tothe ordinate of the influence line at the point multiplied by theproduct of the load and the span the length of the shortest spanbeing used when the spans are unequal The influence lines inthe tables are drawn for a symmetrical inequality of spans Thesymbols on each curve indicate the section of the beam andthe ratio of span lengths to which the curve applies

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44 ONE-WAY SLABS

In monolithic building construction the column layout oftenforms a rectangular grid Continuous beams may be provided inone direction or two orthogonal directions to support slabs thatmay be solid or ribbed in cross section Alternatively the slabsmay be supported directly on the columns as a flat slab Severaldifferent forms of slab construction are shown in Table 242These are considered in more detail in the general context ofbuilding structures in Chapter 6

Where beams are provided in one direction only the slab isa one-way slab Where beams are provided in two orthogonaldirections the slab is a two-way slab However if the longerside of a slab panel exceeds twice the shorter side the slab isgenerally designed as a one-way slab A flat slab is designedas a one-way slab in each direction Bending moments andshearing forces are usually determined on strips of unit widthfor solid slabs and strips of width equal to the spacing of theribs for ribbed slabs

The comments in section 425 and the coefficients for theroutine design of beams given in Table 229 apply equally toone-way spanning slabs This is particularly true when elasticmoments due to service loads are required However lightlyreinforced slabs are highly ductile members and allowanceis generally made for redistribution of elastic moments atthe ULS

441 Uniformly distributed load

For slabs carrying uniformly distributed loads and continuousover three or more nearly equal spans approximate solutionsfor ultimate bending moments and shearing forces accordingto BS 8110 and EC 2 are given in Table 242 In both cases thesupport moments include an allowance for 20 redistributionbut the situation regarding the span moments is somewhatdifferent in the two codes

In BS 8110 a simplified arrangement of the design loadsis permitted where the characteristic imposed load doesnot exceed 125 the characteristic dead load or 5 kNm2excluding partitions and the area of each bay exceeds 30 m2Design for a single load case of maximum design load on allspans is considered sufficient providing the support momentsare reduced by 20 and the span moments are increasedto maintain equilibrium Although the resulting moments arecompatible with yield-line theory the span moments are lessthan those that would occur in the case of alternate spans beingloaded with maximum load and minimum load The implicitredistribution of the span moments the effect of which on thereinforcement stress under service loads would be detrimentalto the deflection of the beam is ignored in the subsequentdesign In EC 2 this simplification is not included and thevalues given for the span moments are the same as those forbeams in Table 229

Provision is made in Table 242 for conditions where aslab is continuous with the end support The restrainingelement may vary from a substantial wall to a small edgebeam and allowance has been made for both eventualitiesThe support moment is given as 004Fl but the reducedspan moment is based on the support moment being no morethan 002Fl

442 Concentrated loads

When a slab supported on two opposite sides carries a loadconcentrated on a limited area of the slab such as a wheelload on the deck of a bridge conventional elastic methods ofanalysis based on isotropic plate theory are often used Thesemay be in the form of equations as derived by Westergaard(ref 17) or influence surfaces as derived by Pucher (ref 18)Another approach is to extend to one-way spanning slabs thetheory applied to slabs spanning in two directions For examplethe curves given in Table 247 for a slab infinitely long in thedirection ly can be used to evaluate directly the bendingmoments in the direction of and at right angles to the spanof a one-way slab carrying a concentrated load this methodhas been used to produce the data for elastic analysis givenin Table 245

For designs in which the ULS requirement is the maincriterion a much simpler approach is to assume that a certainwidth of slab carries the entire load In BS 8110 for examplethe effective width for solid slabs is taken as the load widthplus 24x(1 xl) x being the distance from the nearer supportto the section under consideration and l the span Thus themaximum width at mid-span is equal to the load width plus06l Where the concentrated load is near an unsupported edgeof a slab the effective width should not exceed 12x(1 xl)plus the distance of the slab edge from the further edge of theload Expressions for the resulting bending moments are givenin Table 245 For ribbed slabs the effective width will dependon the ratio of the transverse and longitudinal flexural rigiditiesof the slab but need not be taken less than the load width plus4xl(1 x l) metres

The solutions referred to so far are for single-span slabs thatare simply supported at each end The effects of end-fixity orcontinuity may be allowed for approximately by multiplyingthe moment for the simply supported case by an appropriatefactor The factors given in Table 245 are derived by elasticbeam analysis

45 TWO-WAY SLABS

When a slab is supported other than on two opposite sides onlythe precise amount and distribution of the load taken by eachsupport and consequently the magnitude of the bendingmoments on the slab are not easily calculated if assumptionsresembling real conditions are made Therefore approximateanalyses are generally used The method applicable in anyparticular case depends on the shape of the slab panel theconditions of restraint at the supports and the type of load

Two basic methods are commonly used to analyse slabsthat span in two directions The theory of plates which isbased on elastic analysis is particularly appropriate to thebehaviour under service loads Yield-line theory considersthe behaviour of the slab as a collapse condition approachesHillerborgrsquos strip method is a less well-known alternative tothe use of yield-line in this case In some circumstances itis convenient to use coefficients derived by an elastic analysiswith loads that are factored to represent ULS conditions Thisapproach is used in BS 8110 for the case of a simply supportedslab with corners that are not held down or reinforced fortorsion It is also normal practice to use elastic analysis for

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both service and ULS conditions in the design of bridge decksand liquid-retaining structures For elastic analyses a Poissonrsquosratio of 02 is recommended in BS 8110 and BS 5400 Part 4In EC 2 the values given are 02 for uncracked concrete and 0for cracked concrete

The analysis must take account of the support conditionswhich are often idealised as being free or hinged or fixed andwhether or not the corners of the panels are held down A freecondition refers to an unsupported edge as for example the topof a wall of an uncovered rectangular tank The condition ofbeing freely or simply supported with the corners not helddown may occur when a slab is not continuous and the edgesbear directly on masonry walls or structural steelwork If theedge of the slab is built into a substantial masonry wall or isconstructed monolithically with a reinforced concrete beam orwall a condition of partial restraint exists Such restraint maybe allowed for when computing the bending moments on theslab but the support must be able to resist the torsion andorbending effects and the slab must be reinforced to resist thenegative bending moment A slab can be considered as fixedalong an edge if there is no change in the slope of the slab atthe support irrespective of the incidence of the load A fixedcondition could be assumed if the polar second moment of areaof the beam or other support is very large Continuity over asupport generally implies a condition of restraint less rigid thanfixity that is the slope of the slab at the support depends uponthe incidence of load not only on the panel under considerationbut also on adjacent panels

451 Elastic methods

The so-called exact theory of the elastic bending of platesspanning in two directions derives from work by Lagrangewho produced the governing differential equation for platebending in 1811 and Navier who in 1820 described the useof a double trigonometric series to analyse freely supportedrectangular plates Pigeaud and others later developed theanalysis of panels freely supported along all four edges

Many standard elastic solutions have been produced butalmost all of these are restricted to square rectangular andcircular slabs (see for example refs 19 20 and 21) Exactanalysis of a slab having an arbitrary shape and supportconditions with a general arrangement of loading would beextremely complex To deal with such problems numericaltechniques such as finite differences and finite elementshave been devised Some notes on finite elements are givenin section 497 Finite-difference methods are considered inref 15 (useful introduction) and ref 22 (detailed treatment)The methods are suited particularly to computer-based analysisand continuing software developments have led to the techniquesbeing readily available for routine office use

452 Collapse methods

Unlike in frame design where the converse is generally trueit is normally easier to analyse slabs by collapse methods thanby elastic methods The most-widely known methods ofplastic analysis of slabs are the yield-line method developedby K W Johansen and the so-called strip method devised byArne Hillerborg

It is generally impossible to calculate the precise ultimateresistance of a slab by collapse theory since such elements are

highly indeterminate Instead two separate solutions can befound ndash one being upper bound and the other lower boundWith solutions of the first type a collapse mechanism is firstpostulated Then if the slab is deformed the energy absorbedin inducing ultimate moments along the yield lines is equal tothe work done on the slab by the applied load in producing thisdeformation Thus the load determined is the maximum thatthe slab will support before failure occurs However since suchmethods do not investigate conditions between the postulatedyield lines to ensure that the moments in these areas do notexceed the ultimate resistance of the slab there is no guaranteethat the minimum possible collapse load has been found Thisis an inevitable shortcoming of upper-bound solutions such asthose given by Johansenrsquos theory

Conversely lower-bound solutions will generally result in thedetermination of collapse loads that are less than the maximumthat the slab can actually carry The procedure here is to choosea distribution of ultimate moments that ensures that equilibriumis satisfied throughout and that nowhere is the resistance of theslab exceeded

Most of the literature dealing with the methods of Johansenand Hillerborg assumes that any continuous supports at the slabedges are rigid and unyielding This assumption is also madethroughout the material given in Part 2 of this book Howeverif the slab is supported on beams of finite strength it is possiblefor collapse mechanisms to form in which the yield lines passthrough the supporting beams These beams would then becomepart of the mechanism considered and such a possibility shouldbe taken into account when using collapse methods to analysebeam-and-slab construction

Yield-line analysis Johansenrsquos method requires the designerto first postulate an appropriate collapse mechanism for the slabbeing considered according to the rules given in section 1342Variable dimensions (such as ly on diagram (iv)(a) in Table 249)may then be adjusted to obtain the maximum ultimate resistancefor a given load (ie the maximum ratio of MF) This maximumvalue can be found in various ways for example by tabulatingthe work equation as shown in section 1348 using actualnumerical values and employing a trial-and-adjustment processAlternatively the work equation may be expressed algebraicallyand by substituting various values for the maximum ratio ofMF may be read from a graph relating to MF Anothermethod is to use calculus to differentiate the equation and thenby setting this equal to zero determine the critical value of This method cannot always be used however (see ref 23)

As already explained although such processes enable themaximum resistance for a given mode of failure to be foundthey do not indicate whether the yield-line pattern considered isthe critical one A further disadvantage of such a method is thatunlike Hillerborgrsquos method it gives no direct indication of theresulting distribution of load on the supports Although it seemspossible to use the yield-line pattern as a basis for apportioningthe loaded areas of slab to particular supports there is no realjustification for this assumption (see ref 23) In spite of theseshortcomings yield-line theory is extremely useful A consid-erable advantage is that it can be applied relatively easily tosolve problems that are almost intractable by other means

Yield-line theory is too complex to deal with adequately in thisHandbook indeed several textbooks are completely or almostcompletely devoted to the subject (refs 23ndash28) In section 134

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and Tables 249 and 250 notes and examples are given on therules for choosing yield-line patterns for analysis on theoreticaland empirical methods of analysis on simplifications that canbe made by using so-called affinity theorems and on the effectsof corner levers

Strip method Hillerborg devised his strip method in orderto obtain a lower-bound solution for the collapse load whileachieving a good economical arrangement of reinforcement Aslong as the reinforcement provided is sufficient to cater for thecalculated moments the strip method enables such a lower-boundsolution to be obtained (Hillerborg and others sometimes referto the strip method as the equilibrium theory this should nothowever be confused with the equilibrium method of yield-lineanalysis) In Hillerborgrsquos original theory (now known as thesimple strip method) it is assumed that at failure no load isresisted by torsion and thus all load is carried by flexure ineither of two principal directions The theory results in simplesolutions giving full information regarding the moments overthe whole slab to resist a unique collapse load the reinforcementbeing placed economically in bands Brief notes on the use ofsimple strip theory to design rectangular slabs supportinguniform loads are given in section 135 and Table 251

However the simple strip theory is unable to deal withconcentrated loads andor supports and leads to difficultieswith free edges To overcome such problems Hillerborg laterdeveloped his advanced strip method which involves the use ofcomplex moment fields Although this development extendsthe scope of the simple strip method it somewhat spoils thesimplicity and directness of the original concept A full treat-ment of both the simple and advanced strip theories is givenin ref 29

A further disadvantage of both Hillerborgrsquos and Johansenrsquosmethods is that being based on conditions at failure onlythey permit unwary designers to adopt load distributions thatmay differ widely from those that would occur under serviceloads with the risk of unforeseen cracking A development thateliminates this problem as well as overcoming the limitationsarising from simple strip theory is the so-called strip-deflectionmethod due to Fernando and Kemp (ref 30) With this methodthe distribution of load in either principal direction is notselected arbitrarily by the designer (as in the Hillerborg methodor by choosing the ratio of reinforcement provided in eachdirection as in the yield-line method) but is calculated so as toensure compatibility of deflection in mutually orthogonal stripsThe method results in sets of simultaneous equations (usuallyeight) the solution of which requires computer assistance

453 Rectangular panel with uniformlydistributed load

The bending moments in rectangular panels depend on thesupport conditions and the ratio of the lengths of the sides ofthe panel The ultimate bending moment coefficients given inBS 8110 are derived from a yield-line analysis in which thevalues of the coefficients have been adjusted to suit the divisionof the panel into middle and edge strips as shown in Table 242Reinforcement to resist the bending moments calculated fromthe data given in Table 243 is required only within the middlestrips which are of width equal to three-quarters of the panelwidth in each direction The ratio of the negative moment at

a continuous edge to the positive moment at mid-span has beenchosen as 43 to conform approximately to the serviceabilityrequirements For further details on the derivation of the coef-ficients see ref 31 Nine types of panel are considered inorder to cater for all possible combinations of edge conditionsWhere two different values are obtained for the negativemoment at a continuous edge because of differences betweenthe contiguous panels the values may be treated as fixed-endmoments and distributed elastically in the direction of spanThe procedure is illustrated by means of a worked example insection 1321 Minimum reinforcement as given in BS 8110is to be provided in the edge strips Torsion reinforcement isrequired at corners where either one or both edges of the panelare discontinuous Values for the shearing forces at the ends ofthe middle strips are also given in Table 243

Elastic bending moment coefficients for the same types ofpanel (except that the edge conditions are now defined as fixedor hinged rather than continuous or discontinuous) are givenin Table 244 The information has been prepared from datagiven in ref 21 which was derived by finite element analysisand includes for a Poissonrsquos ratio of 02 For ratios less than 02the positive moments at mid-span are reduced slightly and thetorsion moments at the corners are increased The coefficientsmay be adjusted to suit a Poissonrsquos ratio of zero as explainedin section 1322

The simplified analysis due to Grashof and Rankine can beused for a rectangular panel simply supported on four sideswhen no provision is made to resist torsion at the corners orto prevent the corners from lifting A solution is obtained byconsidering uniform distributions of load along orthogonalstrips in each direction and equating the elastic deflections atthe middle of the strips The proportions of load carried by eachstrip are then obtained as a function of the ratio of the spansand the resulting mid-span moments are calculated Bendingmoment coefficients for this case are also provided in Table 244and basic formulae are given in section 1322

454 Rectangular panel with triangularlydistributed load

In the design of rectangular tanks storage bunkers and someretaining structures cases occur of wall panels spanning in twodirections and subjected to triangular distributions of pressureThe intensity of pressure is uniform at any level but verticallythe pressure increases linearly from zero at the top to a maxi-mum at the bottom Elastic bending moment and shear forcecoefficients are given for four different types of panel to caterfor the most common combinations of edge conditions inTable 253 The information has been prepared from data givenin ref 32 which was derived by finite element analysis andincludes for a Poissonrsquos ratio of 02 For ratios less than 02 thebending moments would be affected in the manner discussed insection 453

The bending moments given for individual panels fixed atthe sides may be applied without modification to continuouswalls provided there is no rotation about the vertical edges Ina square tank therefore moment coefficients can be takendirectly from Table 253 For a rectangular tank distribution ofthe unequal negative moments at the corners is needed

An alternative method of designing the panels would be touse yield-line theory If the resulting structure is to be used

Two-way slabs 33

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to store liquids however extreme care must be taken to ensurethat the adopted proportions of span to support moment andvertical to horizontal moment conform closely to those givenby elastic analyses Otherwise the predicted service momentsand calculated crack widths will be invalid and the structuremay be unsuitable for its intended purpose In the case of struc-tures with non-fluid contents such considerations may be lessimportant This matter is discussed in section 1362

Johansen has shown (ref 24) for a panel fixed or freelysupported along the top edge that the total ultimate momentacting on the panel is identical to that on a similar panel withthe same total load uniformly distributed Furthermore as in thecase of the uniformly loaded slab considered in section 1346a restrained slab may be analysed as if it were freely supportedby employing so-called reduced side lengths to represent theeffects of continuity or fixity Of course unlike the uniformlyloaded slab along the bottom edge of the panel where the load-ing is greatest a higher ratio of support to span moment shouldbe adopted than at the top edge of the panel If the panel isunsupported along the top edge its behaviour is controlledby different collapse mechanisms The relevant expressionsdeveloped by Johansen (ref 24) are represented graphically inTable 254 Triangularly loaded panels can also be designed bymeans of Hillerborgrsquos strip method (ref 29) shown also inTable 254

455 Rectangular panels with concentratedloads

Elastic methods can be used to analyse rectangular panelscarrying concentrated loads The curves in Tables 246 and 247based on Pigeaudrsquos theory give bending moments on a panelfreely supported along all four edges with restrained corners andcarrying a load uniformly distributed over a defined area sym-metrically disposed upon the panel Wheel loads and similarlyhighly concentrated loads are considered to be dispersedthrough the thickness of any surfacing down to the top of theslab or farther down to the mid-depth of the slab as describedin section 249 The dimensions ax and ay of the resultingboundary are used to determine axlx and ayly for which thebending moment factors x4 and y4 are read off the curvesaccording to the ratio of spans k lylx

For a total load F acting on the area ax by ay the positivebending moments per unit width of slab are given by theexpressions in Tables 246 and 247 in which the value ofPoissonrsquos ratio is normally taken as 02 The curves are drawnfor k values of 10 125 radic2 ( 141 approx) 167 20 25 andinfinity For intermediate values of k the values of x4 and y4

can be interpolated from the values above and below the givenvalue of k The use of the curves for k 10 which apply to asquare panel is explained in section 1332

The curves for k infin apply to panels where ly is very muchgreater than lx and can be used to determine the transverse andlongitudinal bending moments for a long narrow panel sup-ported on the two long edges only This chart has been used toproduce the elastic data for one-way slabs given in Table 245as mentioned in section 442

For panels that are restrained along all four edges Pigeaudrecommends that the mid-span moments be reduced by 20Alternatively the multipliers given for one-way slabs could beused if the inter-dependence of the bending moments in the

two directions is ignored Pigeaudrsquos recommendations for themaximum shearing forces are given in section 1332

To determine the load on the supporting beams the rulesin section 46 for a load distributed over the entire panel aresufficiently accurate for a load concentrated at the centre ofthe panel This is not always the critical case for live loads suchas a load imposed by a wheel on a bridge deck since themaximum load on the beam occurs when the wheel is passingover the beam in which case the beam carries the whole load

Johansenrsquos yield-line theory and Hillerborgrsquos strip methodcan also be used to analyse slabs carrying concentrated loadsAppropriate yield-line formulae are given in ref 24 or themethod described in section 1348 may be used For detailsof the analysis involved if the advanced strip method is usedsee ref 29

46 BEAMS SUPPORTING RECTANGULAR PANELS

When designing beams supporting a uniformly loaded panelthat is freely supported along all four edges or with the samedegree of fixity along all four edges it is generally accepted thateach of the beams along the shorter edges of the panel carriesload on an area in the shape of a 45o isosceles triangle whosebase is equal to the length of the shorter side for example eachbeam carries a triangularly distributed load Each beam alongthe longer edges of the panel carries the load on a trapezoidalarea The amount of load carried by each beam is given bythe diagram and expressions in the top left-hand corner ofTable 252 In the case of a square panel each beam carries atriangularly distributed load equal to one-quarter of the totalload on the panel For beams with triangular and trapezoidaldistributions of loading fixed-end moments and moments forcontinuous beams are given in Tables 228 230 and 231

When a panel is fixed or continuous along one two orthree supports and freely supported on the remaining edges thesub-division of the total load to the various supporting beamscan be determined from the diagrams and expressions on theleft-hand side of Table 252 If the panel is unsupported alongone edge or two adjacent edges the loads on the supportingbeams at the remaining edges are as given on the right-handside of Table 252 The expressions which are given in terms ofa service load w may be applied also to an ultimate load n

For slabs designed in accordance with the BS 8110 methodthe loads on the supporting beams may be determined from theshear forces given in Table 243 The relevant loads are takenas uniformly distributed along the middle three-quarters of thebeam length and the resulting fixed-end moments can bedetermined from Table 228

47 NON-RECTANGULAR PANELS

When a panel that is not rectangular is supported along all itsedges and is of such proportions that main reinforcement intwo directions seems desirable the bending moments can bedetermined approximately from the data given in Table 248The information derived from elastic analyses is applicable toa trapezoidal panel approximately symmetrical about one axisto a panel that in plan is an isosceles triangle (or nearly so) andto panels that are regular polygons or circular The case of atriangular panel continuous or partially restrained along threeedges occurs in pyramidal hopper bottoms For this case

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reinforcement determined for the positive moments shouldextend over the entire area of the panel and provision must bemade for the negative moments and for the direct tensions thatact simultaneously with the bending moments

If the shape of a panel is approximately square the bendingmoments for a square slab of the same area should be usedA slab having the shape of a regular polygon with five or moresides can be treated as a circular slab with the diameter takenas the mean of the diameters obtained for the inscribed andcircumscribed circles for regular hexagons and octagons themean diameters are given in Table 248

For a panel circular in plan that is freely supported or fullyfixed along the circumference and carries a load concentratedsymmetrically about the centre on a circular area the totalbending moment to be considered acting across each of twomutually perpendicular diameters is given by the appropriateexpressions in Table 248 These are based on the expressionsderived by Timoshenko and Woinowski-Krieger (ref 20) Ingeneral the radial and tangential moments vary according to theposition being considered A circular panel can therefore bedesigned by one of the following elastic methods

1 Design for the maximum positive bending moment at thecentre of the panel and reduce the amount of reinforcementor the thickness of the slab towards the circumference If thepanel is not truly freely supported at the edge provide forthe appropriate negative bending moment

2 Design for the average positive bending moment across adiameter and retain the same thickness of slab and amountof reinforcement throughout the entire area of the panel Ifthe panel is not truly freely supported at the edge providefor the appropriate negative bending moment

The reinforcement required for the positive bending momentsin each of the preceding methods must be provided in twodirections mutually at right angles the reinforcement for thenegative bending moment should be provided by radial barsnormal to and equally spaced around the circumference or bysome equivalent arrangement

Both circular and other non-rectangular shapes of slab mayconveniently be designed for ULS conditions by using yield-line theory the method of obtaining solutions for slabs ofvarious shapes is described in detail in ref 24

48 FLAT SLABS

The design of flat slabs that is beamless slabs supporteddirectly on columns has often been based on empirical rulesModern codes place much greater emphasis on the analysis ofsuch structures as a series of continuous frames Other methodssuch as grillage finite element and yield-line analysis may beemployed The principles described hereafter and summarisedin section 138 and Table 255 are in accordance with thesimplified method given in BS 8110 This type of slab can beof uniform thickness throughout or can incorporate thickeneddrop panels at the column positions The columns may be ofuniform cross section throughout or may be provided with anenlarged head as indicated in Table 255

The simplified method may be used for slabs consisting ofrectangular panels with at least three spans of approximatelyequal length in each direction where the ratio of the longer tothe shorter side of each panel does not exceed 2 Each panel is

divided into column and middle strips where the width of acolumn strip is taken as one-half of the shorter dimension of thepanel and bending moments determined for a full panel widthare then distributed between column and middle strips as shownin Table 255 If drops of dimensions not less than one-third ofthe shorter dimension of the panel are provided the width of thecolumn strip can be taken as the width of the drop In this casethe apportionment of the bending moments between columnand middle strips is modified accordingly

The slab thickness must be sufficient to satisfy appropriatedeflection criteria with a minimum thickness of 125 mm andprovide resistance to shearing forces and bending momentsPunching shear around the columns is a critical considerationfor which shear reinforcement can be provided in slabs not lessthan 200 mm thick The need for shear reinforcement can beavoided if drop panels or column heads of sufficient size areprovided Holes of limited dimensions may be formed in certainareas of the slab according to recommendations given in BS8110 Larger openings should be appropriately framed withbeams designed to carry the slab loads to the columns

481 Bending moments

The total bending moments for a full panel width at principalsections in each direction of span are given in Table 255 Panelwidths are taken between the centrelines of adjacent bays andpanel lengths between the centrelines of columns Momentscalculated at the centrelines of the supports may be reduced asexplained in section 1383 The slab is effectively designedas one-way spanning in each direction and the commentscontained in section 441 also apply here

At the edges of a flat slab the transfer of moments betweenthe slab and an edge or corner column may be limited by theeffective breadth of the moment transfer strip as shown inTable 256 The structural arrangement should be chosen toensure that the moment capacity of the transfer strip is at least50 of the outer support moment given in Table 255

482 Shearing forces

For punching shear calculations the design force obtained bysumming the shear forces on two opposite sides of a column ismultiplied by a shear enhancement factor to allow for theeffects of moment transfer as shown in Table 256 Criticalperimeters for punching shear occur at distances of 15d fromthe faces of columns column heads and drops where d is theeffective depth of the slab or drop as shown in Table 255

483 Reinforcement

At internal columns two-thirds of the reinforcement neededto resist the negative moments in the column strips should beplaced in a width equal to half that of the column strip andcentral with the column Otherwise the reinforcement neededto resist the moment apportioned to a particular strip should bedistributed uniformly across the full width of the strip

484 Alternative analysis

A more general equivalent frame method for the analysis offlat slabs is described in BS 8110 The bending moments and

Flat slabs 35

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shearing forces are calculated by considering the structure asa series of continuous frames transversely and longitudinallyThe method is described in detail in Examples of the design ofreinforced concrete buildings For further information on bothequivalent frame and grillage methods of analysis of flat slabstructures see ref 33

49 FRAMED STRUCTURES

A structure is statically determinate if the forces and bendingmoments can be determined by the direct application of theprinciples of equilibrium Some examples include cantilevers(whether a simple bracket or a roof of a grandstand) a freelysupported beam a truss with pin-joints and a three-hinged archor frame A statically indeterminate structure is one in whichthere is a redundancy of members or supports or both andwhich can be analysed only by considering the elastic defor-mations under load Typical examples of such structures includerestrained beams continuous beams portal frames and othernon-triangulated structures with rigid joints and two-hinged andfixed-end arches The general notes relating to the analysis ofstatically determinate and indeterminate beam systems given insections 41 and 42 are equally valid when analysing framesProviding a frame can be represented sufficiently accurately byan idealised two-dimensional line structure it can be analysedby any of the methods mentioned earlier (and various othersof course)

The analysis of a two-dimensional frame is somewhat morecomplex than that of a beam system If the configuration ofthe frame or the applied loading (or both) is unsymmetricalside-sway will almost invariably occur making the requiredanalysis considerably longer Many more combinations of load(vertical and horizontal) may need to be considered to obtainthe critical moments Different partial safety factors may applyto different load combinations The critical design conditionsfor some columns may not necessarily be those correspondingto the maximum moment loading producing a reduced momenttogether with an increased axial thrust may be more criticalHowever to combat such complexities it is often possible tosimplify the calculations by introducing a degree of approxi-mation For instance when considering wind loads acting onregular multi-bay frames points of contra-flexure may beassumed to occur at the centres of all the beams and columns(see Table 262) thus rendering the frame statically determinateIn the case of frames that are not required to provide lateralstability the beams at each level acting with the columns aboveand below that level may be considered to form a separatesub-frame for analysis

Beeby (ref 34) has shown that if the many uncertaintiesinvolved in frame analysis are considered there is little tochoose as far as accuracy is concerned between analysing aframe as a single complete structure as a set of sub-frames oras a series of continuous beams with attached columns Ifthe effect of the columns is not included in the analysis of thebeams some of the calculated moments in the beams will begreater than those actually likely to occur

It may not always be possible to represent the true frame asan idealised two-dimensional line structure and analysis as afully three-dimensional space frame may be necessary If thestructure consists of large solid areas such as walls it may notbe possible to represent it adequately by a skeletal frame

The finite-element method of analysis is particularly suited tosolve such problems and is summarised briefly later

In the following pages the analysis of primary frames by themethods of slope deflection and various forms of momentdistribution is described Rigorous analysis of complex rigidframes generally requires an amount of calculation out ofall proportion to the real accuracy of the results and someapproximate solutions are therefore given for common casesof building frames and similar structures When a suitablepreliminary design has been justified by using approximatemethods an exhaustive exact analysis may be undertaken byemploying an established computer program

491 Building code requirements

For most framed structures it is not necessary to carry out afull structural analysis of the complete frame as a single unitand various simplifications are shown in Table 257 BS 8110distinguishes between frames subjected to vertical loads onlybecause overall lateral stability to the structure is provided byother means such as shear walls and frames that are requiredto support both vertical and lateral loads Load combinationsconsisting of (1) dead and imposed (2) dead and wind and(3) dead imposed and wind are also given in Table 257

For frames that are not required to provide lateral stabilitythe construction at each floor may be considered as a separatesub-frame formed from the beams at that level together withthe columns above and below The columns should be taken asfixed in position and direction at their remote ends unless theassumption of a pinned end would be more reasonable (eg ifa foundation detail is considered unable to develop momentrestraint) The sub-frame should then be analysed for therequired arrangements of dead and live loads

As a further simplification each individual beam span maybe considered separately by analysing a sub-frame consisting ofthe span in question together with at each end the upper andlower columns and the adjacent span These members areregarded as fixed at their remote ends with the stiffness of theouter spans taken as only one-half of their true value This sim-plified sub-frame should then be analysed for the loadingrequirements previously mentioned Formulae giving bendingmoments due to various loading arrangements acting on thesimplified sub-frame obtained by slope-deflection methods asdescribed in section 1421 are given in Table 261 Since themethod is lsquoexactrsquo the calculated bending moments may beredistributed within the limits permitted by the Codes Themethod is dealt with in more detail in Examples of the designof reinforced concrete buildings

BS 8110 also allows analysis of the beams at each floor as acontinuous system neglecting the restraint provided by thecolumns entirely so that the continuous beam is assumed to beresting on knife-edge supports Column moments are thenobtained by considering at each joint a sub-frame consistingof the upper and lower columns together with the adjacentbeams regarded as fixed at their remote ends and with theirstiffness taken as one-half of the true value

For frames that are required to provide lateral stability to thestructure as a whole load combinations 1 and 3 both need to beconsidered For combination 3 the following two-stage methodof analysis is allowed for frames of three or more approxi-mately equal bays First each floor is considered as a separate

Structural analysis36

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sub-frame for the effect of vertical loading as describedpreviously Next the complete structural frame is consideredfor the effect of lateral loading assuming that a positionof contra-flexure (ie zero bending moment) occurs at themid-point of each member This analysis corresponds to thatdescribed for building frames in section 4113 and the methodset out in diagram (c) of Table 262 may thus be used Themoments obtained from each of these analyses should thenbe summed and compared with those resulting from loadcombination 1 For tall narrow buildings and other cantileverstructures such as masts pylons and towers load combination2 should also be considered

492 Moment-distribution method no sway

In some circumstances a framed structure may not be subjectto side-sway for example if the frame is braced by other stiffelements within the structure or if both the configuration andthe loading are symmetrical Similarly if a vertically loadedframe is being analysed as a set of sub-frames as permitted inBS 8110 the effects of any side-sway may be ignored In suchcases Hardy Cross moment distribution may be used to evaluatethe moments in the beam and column system The procedurewhich is outlined in Table 258 is similar to the one used toanalyse systems of continuous beams

Precise moment distribution may also be used to solvesuch systems Here the method which is also summarised inTable 258 is slightly more complex to apply than in theequivalent continuous beam case Each time a moment iscarried over the unbalanced moment in the member must bedistributed between the remaining members meeting at the jointin proportion to the relative restraint that each provides Alsothe expression for the continuity factors is more difficultto evaluate Nevertheless the method is a valid alternative tothe conventional moment-distribution method It is describedin more detail in Examples of the design of reinforcedconcrete buildings

493 Moment-distribution method with sway

If sway occurs analysis by moment distribution increases incomplexity since in addition to the influence of the originalloading with no sway it is necessary to consider the effect ofeach degree of sway freedom separately in terms of unknownsway forces The separate results are then combined to obtainthe unknown sway values and hence the final moments Theprocedure is outlined in Table 259

The advantages of precise moment distribution are largelynullified if sway occurs but details of the procedure in suchcases are given in ref 35

To determine the moments in single-bay frames subjected toside sway Naylor (ref 36) devised an ingenious variant ofmoment distribution details of which are given in Table 259The method can also be used to analyse Vierendeel girders

494 Slope-deflection method

The principles of the slope-deflection method of analysing arestrained member are given in Table 260 and section 141together with basic formulae and formulae for the bending

moments in special cases When there is no deflection of oneend of the member relative to the other (eg when the supportsare not elastic as assumed) when the ends of the memberare either hinged or fixed and when the load on the member issymmetrically disposed the general expressions are simplifiedand the resulting formulae for some common cases of restrainedmembers are also given in Table 260

The bending moments on a framed structure are determinedby applying the formulae to each member successively Thealgebraic sum of the bending moments at any joint must equalzero When it is assumed that there is no deflection (or settle-ment) a of one support relative to the other there are as manyformulae for the end moments as there are unknowns andtherefore the restraint moments and the slopes at the endsof the members can be evaluated For symmetrical frameson unyielding foundations and carrying symmetrical verticalloads it is common to neglect the change in the position of thejoints due to the small elastic contractions of the members andthe assumption of a 0 is reasonably correct If the founda-tions or other supports settle unequally under the load thisassumption is not justified and the term a must be assigned avalue for the members affected

If a symmetrical or unsymmetrical frame is subjected to ahorizontal force the resulting sway causes lateral movementof the joints It is common in this case to assume that there isno elastic shortening of the members Sufficient formulae toenable the additional unknowns to be evaluated are obtainedby equating the reaction normal to the member that is theshear force on the member to the rate of change of bendingmoment Sway occurs also in unsymmetrical frames subjectto vertical loads and in any frame on which the load is notsymmetrically disposed

Slope-deflection methods have been used to derive bendingmoment formulae for the simplified sub-frames illustratedon Table 260 These simplified sub-frames correspond tothose referred to in BS 8110 as a basis for determiningthe bending moments in the individual members of a framesubjected to vertical loads only The method is describedin section 142

An example of applying the slope-deflection formulae to asimple problem of a beam hinged at one end and framed intoa column at the other end is given in section 141

495 Shearing forces on members of a frame

The shearing forces on any member forming part of a frame canbe simply determined once the bending moments have beenfound by considering the rate of change of the bendingmoment The uniform shearing force on a member AB due toend restraint only is (MAB MBA)lAB account being taken ofthe signs of the bending moment Thus if both of the restraintmoments are clockwise the shearing force is the numerical sumof the moments divided by the length of the member If onerestraint moment acts in a direction contrary to the other theshearing force is the numerical difference in the momentsdivided by the length of the member For a member with end Bhinged the shearing force due to the restraint moment at A isMABlAB The variable shearing forces caused by the loadson the member should be algebraically added to the uniformshearing force due to the restraint moments as indicated fora continuous beam in section 1112

Framed structures 37

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496 Portal frames

A common type of frame used in single-storey buildings is theportal frame with either a horizontal top member or twoinclined top members meeting at the ridge In Tables 263 and264 general formulae for the moments at both ends of thecolumns and at the ridge where appropriate are given togetherwith expressions for the forces at the bases of the columnsThe formulae relate to any vertical or horizontal load and toframes fixed or hinged at the bases In Tables 265 and 266corresponding formulae for special conditions of loading onframes of one bay are given

Frames of the foregoing types are statically indeterminatebut frames with a hinge at the base of each column and one atthe ridge that is a three-hinged frame can be readily analysedFormulae for the forces and bending moments are given inTable 267 for three-hinged frames Approximate expressionsare also given for certain modified forms of these frames such aswhen the ends of the columns are embedded in the foundationsand when a tie-rod is provided at eaves level

497 Finite elements

In conventional structural analysis numerous approximationsare introduced and the engineer is normally content to acceptthe resulting simplification Actual elements are considered asidealised one-dimensional linear members deformations due toaxial force and shear are assumed to be sufficiently small to beneglected and so on

In general such assumptions are valid and the results of theanalysis are sufficiently close to the values that would occurin the actual structure to be acceptable However when themember sizes become large in relation to the structure theyform the system of skeletal simplification breaks down Thisoccurs for example with the design of such elements as deepbeams shear walls and slabs of various types

One of the methods developed to deal with such so-calledcontinuum structures is that known as finite elements Thestructure is subdivided arbitrarily into a set of individualelements (usually triangular or rectangular in shape) which arethen considered to be inter-connected only at their corners(nodes) Although the resulting reduction in continuity mightseem to indicate that the substitute system would be muchmore flexible than the original structure this is not the case ifthe substitution is undertaken carefully since the adjoiningedges of the elements tend not to separate and thus simulatecontinuity A stiffness matrix for the substitute structure cannow be prepared and analysed using a computer in a similarway to that already described

Theoretically the pattern of elements chosen might bethought to have a marked effect on the validity of the resultsHowever although the use of a smaller mesh consisting ofa larger number of elements can often increase the accuracyof the analysis it is normal for surprisingly good results to beobtained by experienced analysts when using a rather coarsegrid consisting of only a few large elements

410 COLUMNS IN NON-SWAY FRAMES

In monolithic beam-and-column construction subjected tovertical loads only provision is still needed for the bending

moments produced on the columns due to the rigidity of thejoints The external columns of a building are subjected togreater moments than the internal columns (other conditionsbeing equal) The magnitude of the moment depends on therelative stiffness and the end conditions of the members

The two principal cases for beamndashcolumn connections areat intermediate points on the column (eg floor beams) and atthe top of the column (eg roof beam) Since each member canbe hinged fully fixed or partially restrained at its remote endthere are many possible combinations

In the first case the maximum restraint moment at the jointbetween a beam and an external column occurs when theremote end of the beam is hinged and the remote ends of thecolumn are fixed as indicated in Table 260 The minimumrestraint moment at the joint occurs when the remote end ofthe beam is fixed and the remote ends of the column are bothhinged as also indicated in Table 260 Real conditions inpractice generally lie between these extremes and with anycondition of fixity of the remote ends of the column themoment at the joint decreases as the degree of fixity at theremote end of the beam increases With any degree of fixity atthe remote end of the beam the moment at the joint increasesvery slightly as the degree of fixity at the remote ends of thecolumn increase

Formulae for maximum and minimum bending moments aregiven in Table 260 for a number of single-bay frames Themoment on the beam at the joint is divided between the upperand lower columns in the ratio of their stiffness factors K whenthe conditions at the ends of the two columns are identicalWhen one column is hinged at the end and the other is fixedthe solution given for two columns with fixed ends can still beused by taking the effective stiffness factor of the column withthe hinged end as 075K

For cases where the beamndashcolumn connection is at the top ofthe column the formulae given in Table 260 may be used bytaking the stiffness factors for the upper columns as zero

4101 Internal columns

For the frames of ordinary buildings the bending moments onthe upper and lower internal columns can be computed from theexpressions given at the bottom of Table 260 these formulaeconform to the method to be used when the beams are analysedas a continuous system on knife-edge supports as describedin clause 32125 of BS 8110 When the spans are unequal thegreatest bending moments on the column are when the value ofMes (see Table 260) is greatest which is generally when thelonger beam is loaded with (dead live) load while the shorterbeam carries dead load only

Another method of determining moments in the columnsaccording to the Code requirements is to use the simplifiedsub-frame formulae given on Table 261 Then consideringcolumn SO for example the column moment is given by

where DSO DST and DTS are distribution factors FS and FT arefixed-end moments at S and T respectively (see Table 261)This moment is additional to any initial fixed-end momentacting on SO

DSO2DTSFT 4FS

4DSTDTS

Structural analysis38

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To determine the maximum moment in the column it may benecessary to examine two separate simplified sub-frames inwhich each column is embodied at each floor level (ie thecolumn at joint S say is part of two sub-frames comprisingbeams QR to ST and RS to TU respectively) However themaximum moments usually occur when the central beam ofthe sub-frame is the longer of the two beams adjoining thecolumn being investigated as specified in the Code

4102 End columns

The bending moments due to continuity between the beams andthe columns vary more for end columns than for internalcolumns The lack of uniformity in the end conditions affectsthe moments determined by the simplified method describedearlier more significantly than for internal columns Howevereven though the values obtained by the simplified methodsare more approximate than for internal columns they are stillsufficiently accurate for ordinary buildings The simplifiedformulae given on Table 260 conform to clause 32125 ofBS 8110 while the alternative simplified sub-frame methoddescribed for internal columns may also be used

4103 Corner columns

Corner columns are generally subjected to bending momentsfrom beams in two directions at right angles These momentscan be independently calculated by considering two frames(also at right angles) but practical methods of column designdepend on both the relative magnitudes of the moments andthe direct load and the relevant limit-state condition Thesemethods are described in later sections of the Handbook

4104 Use of approximate methods

The methods hitherto described for evaluating the columnmoments in beam-and-column construction with rigid jointsinvolve significant calculation including the second momentof area of the members Often in practice and especially inthe preparation of preliminary schemes approximate methodsare very useful The final design should be checked by moreaccurate methods

The column can be designed provisionally for a direct loadincreased to allow for the effects of bending In determiningthe total column load at any particular level the load from thefloor immediately above that level should be multiplied by thefollowing factors internal columns 125 end columns 15 andcorner column 20

411 COLUMNS IN SWAY FRAMES

In exposed structures such as water towers bunkers and silosand in frames that are required to provide lateral stability to abuilding the columns must be designed to resist the effects ofwind When conditions do not warrant a close analysis of thebending moments to which a frame is subjected due to wind orother lateral forces the methods described in the following andshown in Table 262 are sufficiently accurate

4111 Open braced towers

For columns (of identical cross section) with braced cornersforming an open tower such as that supporting an elevated

water tank the expressions at (a) in Table 262 give bendingmoments and shearing forces on the columns and braces dueto the effect of a horizontal force at the head of the columns

In general the bending moment on the column is the shearforce on the column multiplied by half the distance between thebraces If a column is not continuous or is insufficiently bracedat one end as at an isolated foundation the bending moment atthe other end is twice this value

The bending moment on the brace at an external column isthe sum of the bending moments on the column at the points ofintersection with the brace The shearing force on the brace isequal to the change of bending moment from one end of thebrace to the other end divided by the length of the braceThese shearing forces and bending moments are additional tothose caused by the dead weight of the brace and any externalloads to which it may be subjected

The overturning moment on the frame causes an additionaldirect load on the leeward column and a corresponding relief ofload on the windward column The maximum value of thisdirect load is equal to the overturning moment at the footof the columns divided by the distance between the centres ofthe columns

The expressions in Table 262 for the bending moments andforces on the columns and braces apply for columns that arevertical or near vertical If the columns are inclined then theshearing force on a brace is 2Mb divided by the length ofthe brace being considered

4112 Columns supporting massivesuperstructures

The case illustrated at (b) in Table 262 is common in silos andbunkers where a superstructure of considerable rigidity iscarried on comparatively short columns If the columns arefixed at the base the bending moment on a single column isFh2J where J is the number of columns if they are all of thesame size the significance of the other symbols is indicated inTable 262

If the columns are of different sizes the total shearing forceon any one line of columns should be divided between them inproportion to the second moment of area of each column sincethey are all deflected by the same amount If J1 is the numberof columns with second moment of area I1 J2 is the number ofcolumns with second moment of area I2 and so on the totalsecond moment of area I J1I1 J2I2 and so on Then onany column having a second moment of area Ij the bendingmoment is FhIj2I as given in diagram (b) in Table 262Alternatively the total horizontal force can be divided amongthe columns in proportion to their cross-sectional areas (thusgiving uniform shear stress) in which case the formula for thebending moment on any column with cross-sectional area Aj isFhAj2A where A is the sum of the cross-sectional areas ofall the columns resisting the total shearing force F

4113 Building frames

In the frame of a multi-storey multi-bay building the effect ofthe wind may be small compared to that of other loads andin this case it is sufficiently accurate to divide the horizontalshearing force between the columns on the basis that an endcolumn resists half the amount on an internal column If in the

Columns in sway frames 39

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plane of the lateral force F Jt is the total number of columns inone frame the effective number of columns for the purpose ofcalculating the bending moment on an internal column is Jt 1the two end columns being equivalent to one internal columnsee diagram (c) in Table 262 In a building frame subjectedto wind pressure the forces on each panel (or storey height)F1 F2 F3 and so on are generally divided into equal shearingforces at the head and base of each storey height of columnsThe shearing force at the bottom of any internal column istoreys from the top is (F Fi2)( Jt 1) where F F1 F2 F3 Fi 1 The bending moment is then the shearingforce multiplied by half the storey height

A bending moment and a corresponding shearing force arecaused on the floor beams in the same way as on the braces ofan open braced tower At an internal column the sum of thebending moments on the two adjacent beams is equal to the sumof the moments at the base of the upper column and the head ofthe lower column

The above method of analysis for determining the effects oflateral loading corresponds to that described in section 491and recommended in BS 8110 for a frame of three or moreapproximately equal bays

412 WALL AND FRAME SYSTEMS

In all forms of construction the effects of wind force increasein significance as the height of the structure increases Oneway of reducing lateral sway and improving stability is byincreasing the sectional size of the component members ofsway frames However this will have a direct consequenceof increasing storey height and building cost

Often a better way is to provide a suitable arrangement ofwalls linked to flexible frames The walls can be external orinternal be placed around lift shafts and stairwells to form corestructures or be a combination of types Sometimes core wallsare constructed in advance of the rest of the structure to avoidsubsequent delays The lateral stiffness of systems with acentral core can be increased by providing deep cantilevermembers at the top of the core structure to which the exteriorcolumns are connected Another approach is to increase theload on the central core by replacing the exterior columns byhangers suspended from the cantilever members at the top ofthe building This also avoids the need for exterior columns atground level and their attendant foundations As buildings gettaller the lateral stability requirements are of paramount impor-tance The structural efficiency can be increased by replacingthe building facade by a rigidly jointed framework so that theouter shell acts effectively as a closed-box

Some different structural forms consisting of assemblies ofmulti-storey frames shear walls and cores with an indicationof typical heights and proportions taken from ref 37 areshown in Table 268

4121 Shear wall structures

The lateral stability of low- to medium-rise buildings is oftenobtained by providing a suitable system of stiff shear walls Thearrangement of the walls should be such that the building is stiffin both flexure and torsion In rectangular buildings externalshear walls in the short direction can be used to resist lateralloads acting on the wide faces with rigid frames or infill panels

in the long direction In buildings of square plan form a strongcentral service core surrounded by flexible external framescan be used If strong points are placed at both ends of a longbuilding the restraint provided to the subsequent shrinkageand thermal movements of floors and roof should be carefullyconsidered

In all cases the floors and roof are considered to act as stiffplates so that at each level the horizontal displacements of allwalls and columns are taken to be the same provided the totallateral load acts through the shear centre of the system If thetotal lateral load acts eccentrically then the additional effectof the resulting torsion moment needs to be considered Theanalysis and design of shear wall buildings is covered in ref 38from which much of the following treatment is based Severaldifferent plan configurations of shear walls and core units withnotes on their suitability are shown in Table 269

4122 Walls without openings

The lateral load transmitted to an individual wall is a functionof its position and its relative stiffness The total deflection of acantilever wall under lateral load is a combination of bendingand shear deformations However for a uniformly distributedload the shear deformation is less than 10 of the total forHD 3 in the case of plane walls and HD 5 in the case offlanged walls with BD 05 (where B is width of flange D isdepth of web and H is height of wall) Thus for most shearwalls without openings the dominant mode of deformation isbending and the stiffness of the wall can be related directly to thesecond moment of area of the cross section I Then for a totallateral load F applied at the shear centre of a system of parallelwalls the shearing force on an individual wall j is FIjIj

The position of the shear centre along a given axis y can bereadily determined by calculating the moment of stiffness ofeach wall about an arbitrary reference point on the axis Thedistance from the point to the shear centre yc Ijyj Ij

If the total lateral load acts at distance yo along the axis theresulting horizontal moment is F(yondashyc) Then if the torsionstiffness of individual walls is neglected the total shearingforce on wall j is

Fj FIj Ij F(yo yc)IjyjIj (yj yc)2

More generalised formulae in which a wall system is related totwo perpendicular axes are given in Table 269 The aboveanalysis takes no account of rotation at the base of the walls

4123 Walls containing openings

In the case of walls pierced by openings the behaviour ofthe individual wall sections is coupled to a variable degree Theconnections between the individual sections are provided eitherby beams that form part of the wall or by floor slabs or by acombination of both The pierced wall may be analysed byelastic methods in which the flexibility of the coupling elementsis represented as a continuous flexible medium Alternativelythe pierced wall may be idealised as an equivalent plane frameusing a lsquowide columnrsquo analogy

The basis of the continuous connection model is described insection 152 and analytical solutions for a wall containing asingle line of openings are given in Table 270

Structural analysis40

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4124 Interaction of shear walls and frames

The interaction forces between solid walls pierced walls andframes can vary significantly up the height of a building asa result of the differences in the free deflected shapes ofeach structural form The deformation of solid walls is mainlyflexural whereas pierced walls deform in a shearndashflexure modeand frames deform in an almost pure shear manner As a resulttowards the bottom of a building solid walls attract load whilstframes and to a lesser extent pierced walls shed load Thebehaviour is reversed towards the top of a building Thusalthough the distribution of load intensity between the differentelements is far from uniform up the building the total lateralforce resisted by each varies by a smaller amount

As a first approximation the shearing force at the bottom ofeach load-resisting element can be determined by considering asingle interaction force at the top of the building Formulae bywhich the effective stiffness of pierced walls and frames can bedetermined are given in section 153

413 ARCHES

Arch construction in reinforced concrete occurs sometimes inroofs but mainly in bridges An arch may be three-hingedtwo-hinged or fixed-ended (see diagrams in Table 271) andmay be symmetrical or unsymmetrical right or skew singleor one of a series of arches mutually dependent upon eachother The following consideration is limited to symmetrical andunsymmetrical three-hinged arches and to symmetrical two-hinged and fixed-end arches reference should be made to otherpublications for information on more complex types

Arch construction may comprise an arch slab (or vault) or aseries of parallel arch ribs The deck of an arch bridge may besupported by columns or transverse walls carried on an archslab or ribs when the structure may have open spandrels or thedeck may be below the crown of the arch either at the level ofthe springing (as in a bowstring girder) or at some intermediatelevel A bowstring girder is generally regarded as a two-hingedarch with the horizontal component of thrust resisted by a tiewhich normally forms part of the deck If earth or other filling isprovided to support the deck an arch slab and spandrel walls arerequired and the bridge is a closed or solid-spandrel structure

4131 Three-hinged arch

An arch with a hinge at each springing and at the crown isstatically determinate The thrusts on the abutments and thebending moments and shearing forces on the arch itself arenot affected by a small movement of one abutment relative tothe other This type of arch is therefore used when there is apossibility of unequal settlement of the abutments

For any load in any position the thrust on the abutmentscan be determined by the equations of static equilibrium Forthe general case of an unsymmetrical arch with a load actingvertically horizontally or at an angle the expressions for thehorizontal and vertical components of the thrusts are givenin the lower part of Table 271 For symmetrical arches the for-mulae given in Table 267 for the thrusts on three-hinged framesapply or similar formulae can be obtained from the generalexpressions in Table 271 The vertical component is the same asthe vertical reaction for a freely supported beam The bending

moment at any cross section of the arch is the algebraic sum ofthe moments of the loads and reactions on one side of thesection There is no bending moment at a hinge The shearingforce is likewise the algebraic sum of the loads and reactionsresolved at right angles to the arch axis at the section and actingon one side of the section The thrust at any section is the sumof the loads and reactions resolved parallel to the axis of thearch at the section and acting on one side of the section

The extent of the arch that should be loaded with imposedload to give the maximum bending moment or shearing forceor thrust at a particular cross section can be determined byconstructing a series of influence lines A typical influence linefor a three-hinged arch and the formulae necessary to constructan influence line for unit load in any position are given in theupper part of Table 271

4132 Two-hinged arch

The hinges of a two-hinged arch are placed at the abutmentsso that as in a three-hinged arch only thrusts are transmitted tothe abutments and there is no bending moment on the archat the springing The vertical component of the thrust from asymmetrical two-hinged arch is the same as the reaction fora freely supported beam Formulae for the thrusts and bendingmoments are given in Table 271 and notes in section 162

4133 Fixed arch

An arch with fixed ends exerts in addition to the vertical andhorizontal thrusts a bending moment on the abutments Like atwo-hinged arch and unlike a three-hinged arch a fixed-endarch is statically indeterminate and the stresses are affected bychanges of temperature and shrinkage of the concrete As it isassumed in the general theory that the abutments cannot moveor rotate the arch can only be used in such conditions

A cross section of a fixed-arch rib or slab is subjected to abending moment and a thrust the magnitudes of which have tobe determined The design of a fixed arch is a matter of trial andadjustment since both the dimensions and the shape of the archaffect the calculations but it is possible to select preliminarysizes that reduce the repetition of arithmetic work to a minimumA suggested method of determining possible sections at thecrown and springing as given in Table 272 and explained insection 1631 is based on first treating the fixed arch as ahinged arch and then estimating the size of the cross sectionsby greatly reducing the maximum stresses

The general formulae for thrusts and bending moments on asymmetrical fixed arch of any profile are given in Table 272and notes on the application and modification of the formulaeare given in section 163 The calculations necessary to solvethe general and modified formulae are tedious but are easedsomewhat by preparing them in tabular form The form givenin Table 272 is particularly suitable for open-spandrel archbridges because the appropriate formulae do not assume a con-stant value of aI the ratio of the length of a segment of the archto the mean second moment of area of the segment

For large span arches calculations are made much easier andmore accurate by preparing and using influence lines for thebending moment and thrust at the crown the springing and thequarter points of the arch Typical influence lines are given inTable 272 and such diagrams can be constructed by considering

Arches 41

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the passage over the arch of a single concentrated unit load andapplying the formulae for this condition The effect of the deadload and of the most adverse disposition of imposed load canbe readily calculated from these diagrams If the specifiedimposed load includes a moving concentrated load such as aKEL the influence lines are almost essential for determiningthe most adverse position The case of the positive bendingmoment at the crown is an exception when the most adverseposition of the load is at the crown A method of determiningthe data to establish the ordinates of the influence lines is givenin Table 273

4134 Fixed parabolic arches

In Table 274 and in section 164 consideration is given tosymmetrical fixed arches that can have either open or solidspandrels and be either arch ribs or arch slabs The method isbased on that of Strassner as developed by H Carpenter andthe principal assumption is that the axis of the arch is made tocoincide with the line of thrust due to the dead load This resultsin an economical structure and a simple calculation methodThe shape of the axis of the arch is approximately that of aparabola and this method can therefore be used only when thedesigner is free to select the profile of the arch The parabolicform may not be the most economic for large spans althoughit is almost so and a profile that produces an arch axis coinci-dent with the line of thrust for the dead load plus one-half of theimposed load may be more satisfactory If the increase in thethickness of the arch from crown to springing is of a parabolicform only the bending moments and thrusts at the crownand the springing need to be investigated The necessaryformulae are given in section 164 where these include a seriesof coefficients values of which are given in Table 274 Theapplication of the method is also illustrated by an examplegiven in section 164 The component forces and momentsare considered in the following treatment

The thrusts due to the dead load are relieved somewhat by theeffect of the compression causing elastic shortening of the archFor arches with small ratios of rise to span and arches that arethick in comparison with the span the stresses due to archshortening may be excessive This can be overcome by intro-ducing temporary hinges at the crown and the springing whicheliminate all bending stresses due to dead load The hinges arefilled with concrete after arch shortening and much of theshrinkage of the concrete have taken place

An additional horizontal thrust due to a temperature rise ora corresponding counter-thrust due to a temperature fall willaffect the stresses in the arch and careful consideration mustbe given to the likely temperature range The shrinkage of theconcrete that occurs after completion of the arch produces acounter-thrust the magnitude of which is modified by creep

The extent of the imposed load on an arch necessary toproduce the maximum stresses in the critical sections can bedetermined from influence lines and the following values areapproximately correct for parabolic arches The maximumpositive moment at the crown occurs when the middle third of thearch is loaded the maximum negative moment at a springingoccurs when four-tenths of the span adjacent to the springing isloaded the maximum positive moment at the springing occurswhen six-tenths of the span furthest away from the springing

is loaded In the expressions given in section 1644 the imposedload is expressed in terms of an equivalent UDL

When the normal thrusts and bending moments on the mainsections have been determined the areas of reinforcement andstresses at the crown and springing can be calculated Allthat now remains is to consider the intermediate sections anddetermine the profile of the axis of the arch If the dead loadis uniform throughout (or practically so) the axis will be aparabola but if the dead load is not uniform the axis must beshaped to coincide with the resulting line of thrust This canbe obtained graphically by plotting force-and-link polygonsthe necessary data being the magnitudes of the dead load thehorizontal thrust due to dead load and the vertical reaction(equal to the dead load on half the span) of the springing Theline of thrust and therefore the axis of the arch having beenestablished and the thickness of the arch at the crown and thespringing having been determined the lines of the extradosand the intrados can be plotted to give a parabolic variation ofthickness between the two extremes

414 PROPERTIES OF MEMBERS

4141 End conditions

Since the results given by the more precise methods of elasticanalysis vary considerably with the conditions of restraint atthe ends of the members it is important that the assumedconditions are reasonably obtained in the actual constructionAbsolute fixity is difficult to attain unless a beam or column isembedded monolithically in a comparatively large mass ofconcrete Embedment of a beam in a masonry wall representsmore nearly the condition of a hinge and should normally beconsidered as such A continuous beam supported internallyon a beam or column is only partly restrained and where thesupport at the outer end of an end span is a beam a hinge shouldbe assumed With the ordinary type of pad foundation designedsimply for a uniform ground bearing pressure under the directload on a column the condition at the foot of the column shouldalso be considered as a hinge A column built on a pile-capsupported by two three or four piles is not absolutely fixed buta bending moment can be developed if the resulting verticalreaction (upwards and downwards) and the horizontal thrust canbe resisted by the piles The foot of a column can be consideredas fixed if it is monolithic with a substantial raft foundation

In two-hinged and three-hinged arches hinged frames andsome bridge types where the assumption of a hinged joint mustbe fully realised it is necessary to form a definite hinge in theconstruction This can be done by inserting a steel hinge (orsimilar) or by forming a hinge within the frame

4142 Section properties

For the elastic analysis of continuous structures the sectionproperties need to be known Three bases for calculating thesecond moment of area of a reinforced concrete section are gen-erally recognised in codes of practice as follows

1 The concrete section the entire concrete area but ignoringthe reinforcement

Structural analysis42

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2 The gross section the entire concrete area together with thereinforcement on the basis of a modular ratio (ie ratio ofmodulus of elasticity values of steel and concrete)

3 The transformed section the concrete area in compressiontogether with the reinforcement on the basis of modular ratio

For methods 2 and 3 the modular ratio should be based on aneffective modulus of elasticity of concrete taking account ofthe creep effects of long-term loading In BS 8110 a modularratio of 15 is recommended unless a more accurate figure can bedetermined However until the reinforcement has been deter-mined or assumed calculation of the section properties in thisway cannot be made with any precision Moreover the sectionproperties vary considerably along the length of the member asthe distribution of reinforcement and for method 3 the depthof concrete in compression change The extent and effect ofcracking on the section properties is particularly difficult toassess for a continuous beam in beam-and-slab construction inwhich the beam behaves as a flanged section in the spans wherethe bending moments are positive but is designed as a rectan-gular section towards the supports where the bending momentsare negative

Method 1 is the simplest one to apply and the only practicalapproach when beginning a new design but one of the othermethods could be used when checking the ability of existingstructures to carry revised loadings and for new structureswhen a separate analysis for the SLSs is required In all cases itis important that the method used to assess the section propertiesis the same for all the members involved in the calculationWhere a single stiffness value is to be used to characterise amember method 1 (or 2) is likely to provide the most accurateoverall solution Method 3 will only be appropriate where thevariations in section properties over the length of membersare properly taken into account

415 EARTHQUAKE-RESISTANT STRUCTURES

Earthquakes are ground vibrations that are caused mainly byfracture of the earthrsquos crust or by sudden movement along analready existing fault During a seismic excitation structuresare caused to oscillate in response to the forced motion of thefoundations The affected structure needs to be able to resistthe resulting horizontal load and also dissipate the impartedkinetic energy over successive deformation cycles It would beuneconomical to design the structure to withstand a majorearthquake elastically and the normal approach is to provide itwith sufficient strength and ductility to withstand such an eventby responding inelastically provided that the critical regions

and the connections between members are designed speciallyto ensure adequate ductility

Significant advances have been made in the seismic designof structures in recent years and very sophisticated codes ofpractice have been introduced (ref 39) A design horizontalseismic load is recommended that depends on the importanceof the structure the seismic zone the ground conditionsthe natural period of vibration of the structure and the availableductility of the structure Design load effects in the structureare determined either by linear-elastic structural analysis forthe equivalent static loading or by dynamic analysis When alinear-elastic method is used the design and detailing of themembers needs to ensure that in the event of a more severeearthquake the post-elastic deformation of the structure willbe adequately ductile For example in a multi-storey framesufficient flexural and shear strength should be provided in thecolumns to ensure that plastic hinges form in the beams inorder to avoid a column side-sway mechanism The properdetailing of the reinforcement is also a very important aspectin ensuring ductile behaviour At the plastic hinge regions ofmoment resisting frames in addition to longitudinal tensionreinforcement it is essential to provide adequate compressionreinforcement Transverse reinforcement is also necessary toact as shear reinforcement to prevent premature buckling ofthe longitudinal compression reinforcement and to confine thecompressed concrete

Buildings should be regular in plan and elevation withoutre-entrant angles and discontinuities in transferring verticalloads to the ground Unsymmetrical layouts resulting in largetorsion effects flat slab floor systems without any beams andlarge discontinuities in infill systems (such as open groundstoreys) should be avoided Footings should be founded at thesame level and should be interconnected by a mat foundationor by a grid of foundation beams Only one foundation typeshould in general be used for the same structure unless thestructure is formed of dynamically independent units

An alternative to the conventional ductile design approach isto use a seismic isolation scheme In this case the structure issupported on flexible bearings so that the period of vibration ofthe combined structure and supporting system is long enoughfor the structure to be isolated from the predominant earthquakeground motion frequencies In addition extra damping isintroduced into the system by mechanical energy dissipatingdevices in order to reduce the response of the structure to theearthquake and keep the deflections of the flexible systemwithin acceptable limits

A detailed treatment of the design of earthquake-resistingconcrete structures is contained in ref 40

Earthquake-resistant structures 43

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51 PRINCIPLES AND REQUIREMENTS

In modern Codes of Practice a limit-state design concept isused Ultimate (ULS) and serviceability (SLS) limit-states areconsidered as well as durability and in the case of buildingsfire-resistance Partial safety factors are incorporated in bothloads and material strengths to ensure that the probability offailure (ie not satisfying a design requirement) is acceptablylow For British Codes (BS 8110 BS 5400 BS 8007) detailsare given of design requirements and partial safety factors inChapter 21 material properties in Chapter 22 durability andfire-resistance in Chapter 23 For EC 2 corresponding data aregiven in Chapters 29 30 and 31 respectively

Members are first designed to satisfy the most critical limit-state and then checked to ensure that the other limit-statesare not reached For most members the critical condition to beconsidered is the ULS on which the required resistances of themember in bending shear and torsion are based The require-ments of the various SLSs such as deflection and crackingare considered later However since the selection of an adequatespan to effective depth ratio to prevent excessive deflection andthe choice of a suitable bar spacing to avoid excessive crackingcan also be affected by the reinforcement stress the designprocess is generally interactive Nevertheless it is normal tostart with the requirements of the ULS

52 RESISTANCE TO BENDING AND AXIAL FORCE

Typically beams and slabs are members subjected to bendingwhile columns are subjected to a combination of bending andaxial force In this context a beam is defined as a member inBS 8110 with a clear span not less than twice the effectivedepth and in EC 2 as a member with a span not less than threetimes the overall depth Otherwise the member is treated as adeep beam for which different design methods are applicableA column is defined as a member in which the greater overallcross-sectional dimension does not exceed four times thesmaller dimension Otherwise the member is considered as awall for which a different design approach is adopted Somebeams for example in portal frames and slabs for example inretaining walls are subjected to bending and axial force Insuch cases small axial forces that are beneficial in providingresistance to bending are generally ignored in design

521 Basic assumptions

For the analysis of sections in bending or combined bendingand axial force at the ULS the following basic assumptionsare made

The resistance of the concrete in tension is ignored

The distribution of strain across the section is linear that issections that are plane before bending remain plane afterbending the strain at a point being proportional to its distancefrom the axis of zero strain (neutral axis) In columns ifthe axial force is dominant the neutral axis can lie outsidethe section

Stressndashstrain relationships for concrete in compression andfor reinforcement in tension and compression are thoseshown in the diagrams on Table 36 for BS 8110 andBS 5400 and Table 44 for EC 2

The maximum strain in the concrete in compression is 00035except for EC 2 where the strains shown in the followingdiagram and described in the following paragraph apply

h

0

0

c2

c2

cu

(37)h

For sections subjected to pure axial compression the strain islimited to c2 For sections partly in tension the compressivestrain is limited to cu For intermediate conditions the straindiagram is obtained by taking the compressive strain as c2 at alevel equal to 37 of the section depth from the more highlycompressed face For concrete strength classes C5060 thelimiting strains are c2 0002 and cu 00035 For higherstrength concretes other values are given in Table 44

Strain distribution at ULS in EC 2

Chapter 5

Design of structuralmembers

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In all codes for sections partly in tension the shape of thebasic concrete stress-block is a combination of a parabolaand a rectangle In EC 2 a form consisting of a triangle anda rectangle is also given In all codes a simplified rectangularstress distribution may also be used If the compression zoneis rectangular the compressive force and the distance of theforce from the compression face can be readily determined foreach stress-block and the resulting properties are given insection 241 for BS 8110 and section 321 for EC 2

The stresses in the reinforcement depend on the strains in theadjacent concrete which depend in turn on the depth of theneutral axis and the position of the reinforcement in relationto the concrete surfaces The effect of these factors will beexamined separately for beams and columns

522 Beams

Depth of neutral axis This is significant because the valueof xd where x is the neutral axis depth and d is the effectivedepth of the tension reinforcement not only affects the stress inthe reinforcement but also limits the amount of moment redis-tribution allowed at a given section In BS 8110 where becauseof moment redistribution allowed in the analysis of a memberthe design moment is less than the maximum elastic momentthe requirement xd (b 04) should be satisfied whereb is the ratio of design moment to maximum elastic momentThus for reductions in moment of 10 20 and 30 xdmust not exceed 05 04 and 03 respectively In EC 2 asmodified by the UK National Annex similar restrictions applyfor concrete strength classes C5060

and xd 0456 and dx 043 for BS 5400 For design toEC 2 considerations similar to those in BS 8110 apply

Effect of axial force The following figure shows a sectionthat is subjected to a bending moment M and an axial force Nin which a simplified rectangular stress distribution has beenassumed for the compression in the concrete The stress blockis shown divided into two parts of depths dc and (h 2dc)providing resistance to the bending moment M and the axialforce N respectively where 0 dc 05h

Resistance to bending and axial force 45

Section Strain diagram

Section Forces

The figure here shows a typical strain diagram for a sectioncontaining both tension and compression reinforcement Forthe bi-linear stressndashstrain curve in BS 8110 the maximumdesign stresses in the reinforcement are fy115 for values of s

and s fy115Es From the strain diagram this gives

and

In BS 5400 the reinforcement stressndashstrain curve is tri-linear withmaximum design stresses of fy115 in tension and 2000fy (2300 fy) in compression These stresses apply for values ofs 0002 fy115Es and s 0002 giving

With cu 00035 fy 500 Nmm2 and Es 200 kNmm2 thecritical values are xd 0617 and for BS 8110dx 038

dx (cu0002) cu

x d cu (cu 0002 fy 115Es) and

dx (cu fy 115Es) cux d cu (cu fy 115Es)

The depth dc (and the force in the tension reinforcement) aredetermined by the bending moment given by

M bdc(d 05dc) fcd

Thus for analysis of the section axial forces may be ignoredfor values satisfying the condition

Nb(h 2dc) fcd

Combining the two requirements gives

Nbhfcd 2M(d05dc)

In the limit when dc 05h this gives

Nbhfcd 2M(d025h)congbhfcd 3Mh

For BS 8110 the condition becomes N 045bhfcu 3Mhwhich being simplified to N 01bhfcu is reasonably valid forMbh2fcu 012 For EC 2 the same condition becomesN 0567bhfck 3Mh which may be reasonably simplified toN 012bhfck for Mbh2fck 015

Analysis of section Any given section can be analysed by atrial-and-error process An initial value is assumed for theneutral axis depth from which the concrete strains at the rein-forcement positions can be calculated The correspondingstresses in the reinforcement are determined and the resultingforces in the reinforcement and the concrete are obtained If theforces are out of balance the value of the neutral axis depth ischanged and the process is repeated until equilibrium isachieved Once the balanced condition has been found theresultant moment of all the forces about the neutral axis or anyother suitable point is calculated

Singly reinforced rectangular sections For a section thatis reinforced in tension only and subjected to a moment M aquadratic equation in x can be obtained by taking moments forthe compressive force in the concrete about the line of actionof the tension reinforcement The resulting value of x can beused to determine the strain diagram from which the strain in

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the reinforcement and hence the stress can be calculated Therequired area of reinforcement can then be determined fromthe tensile force whose magnitude is equal to the compressiveforce in the concrete If the calculated value of x exceeds thelimit required for any redistribution of moment then a doublyreinforced section will be necessary

In designs to BS 8110 and BS 5400 the lever arm betweenthe tensile and compressive forces is to be taken not greater than095d Furthermore it is a requirement in BS 5400 that if xexceeds the limiting value for using the maximum designstress then the resistance moment should be at least 115MAnalyses are included in section 2421 for both BS 8110 andBS 5400 and in section 3221 for EC 2 Design charts basedon the parabolic-rectangular stress-block for concrete withfy 500 Nmm2 are given in Tables 313 323 and 47 forBS 8110 BS 5400 and EC 2 respectively Design tables basedon the rectangular stress-blocks for concrete are given inTables 314 324 and 48 for BS 8110 BS 5400 and EC 2respectively These tables use non-dimensional parameters andare applicable for values of fy 500 Nmm2

Doubly reinforced rectangular sections A sectionneeding both tension and compression reinforcement andsubjected to a moment M can be designed by first selecting asuitable value for x such as the limiting value for using themaximum design stress in the tension reinforcement or satisfy-ing the condition necessary for moment redistribution Therequired force to be provided by the compression reinforcementcan be derived by taking moments for the compressive forcesin the concrete and the reinforcement about the line of actionof the tensile reinforcement The force to be provided by thetension reinforcement is equal to the sum of the compressiveforces The reinforcement areas can now be determined takingdue account of the strains appropriate to the value of x selected

Analyses are included in section 2422 for both BS 8110and BS 5400 and in section 3222 for EC 2 Design chartsbased on the rectangular stress-blocks for concrete are given inTables 315 and 316 for BS 8110 Tables 325 and 326 forBS 5400 and Tables 49 and 410 for EC 2

Design formulae for rectangular sections Designformulae based on the rectangular stress-blocks for concreteare given in BS 8110 and BS 5400 In both codes x is limitedto 05d so that the formulae are automatically valid for redistri-bution of moment not greater than 10 The design stress intension reinforcement is taken 087fy although this is onlystrictly valid for xd 0456 in BS 5400 The design stresses inany compression reinforcement are taken as 087fy in BS 8110and 072fy in BS 5400 Design formulae are given in section2423 for BS 8110 and BS 5400 Although not included inEC 2 appropriate formulae are given in section 3223

Flanged sections In monolithic beam and slab constructionwhere the web of the beam projects below the slab the beam isconsidered as a flanged section for sagging moments Theeffective width of the flange over which uniform conditionsof stress can be assumed is limited to values stipulated in thecodes In most sections where the flange is in compressionthe depth of the neutral axis will be no greater than the flangethickness In such cases the section can be considered to berectangular with b taken as the flange width If the depth of

the neutral axis does exceed the thickness of the flange thesection can be designed by dividing the compression zoneinto portions comprising the web and the outlying flangesDetails of the flange widths and design procedures are given insections 2424 for BS 8110 and 3224 for EC 2

Beam sizes The dimensions of beams are mainly determinedby the need to provide resistance to moment and shear In thecase of beams supporting items such as cladding partitions orsensitive equipment service deflections can also be criticalOther factors such as clearances below beams dimensions ofbrick and block courses widths of supporting members andsuitable sizes of formwork also need to be taken into accountFor initial design purposes typical spaneffective depth ratiosfor beams in buildings are given in the following table

Design of structural members46

Spaneffective depth ratios for initial design of beams

Span conditionsUltimate design load

25 kNm 50 kNm 100 kNm

Cantilever 9 7 5Simply supported 18 14 10Continuous 22 17 12

The effective span of a continuous beam is generally taken asthe distance between centres of supports At a simple supportor at an encastre end the centre of action may be taken at adistance not greater than half of the effective depth fromthe face of the support Beam widths are often taken as half theoverall depth of the beam with a minimum of 300 mm If amuch wider band beam is used the spaneffective depth ratiocan be increased significantly to the limit necessitated bydeflection considerations

In BS 8110 and BS 5400 to ensure lateral stability simplysupported and continuous beams should be so proportioned thatthe clear distance between lateral restraints is not greater than60bc or 250bc

2d whichever is the lesser For cantilevers inwhich lateral restraint is provided only at the support the cleardistance from the end of the cantilever to the face of the sup-port should not exceed 25bc or 100bc

2d whichever is the lesserone In the foregoing bc is the breadth of the compression faceof the beam (measured midway between restraints) orcantilever In EC 2 second order effects in relation to lateralstability may be ignored if the distance between lateralrestraints is not greater than 50bc(hbc)13 and h 25bc

523 Slabs

Solid slabs are generally designed as rectangular strips of unitwidth and singly reinforced sections are normally sufficientRibbed slabs are designed as flanged sections of width equalto the rib spacing for sagging moments Continuous ribbedslabs are often made solid in support regions so as to developsufficient resistance to hogging moments and shear forcesAlternatively in BS 8110 ribbed slabs may be designed as aseries of simply supported spans with a minimum amountof reinforcement provided in the hogging regions to controlthe cracking The amount of reinforcement recommended is

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25 of that in the middle of the adjoining spans extending intothe spans for at least 15 of the span length

The thickness of slabs is normally determined by deflectionconsiderations which sometimes result in the use of reducedreinforcement stresses to meet code requirements Typicalspaneffective depth ratios for slabs designed to BS 8110 aregiven in the following table

contain a modification factor the use of which necessitates aniteration process with the factor taken as 10 initially Details ofthe design procedures are given in Tables 321 and 322 forBS 8110 Tables 331 and 332 for BS 5400 and Tables 415 and416 for EC 2

Analysis of section Any given section can be analysed by atrial-and-error process For a section bent about one axis aninitial value is assumed for the neutral axis depth from whichthe concrete strains at the positions of the reinforcement can becalculated The resulting stresses in the reinforcement aredetermined and the forces in the reinforcement and concreteare evaluated If the resultant force is not equal to the designaxial force N the value of the neutral axis depth is changed andthe process repeated until equality is achieved The sum of themoments of all the forces about the mid-depth of the section isthen the moment of resistance appropriate to N For a section inbiaxial bending initial values have to be assumed for the depthand the inclination of the neutral axis and the design processwould be extremely tedious without the aid of an interactivecomputer program

For design purposes charts for symmetrically reinforcedrectangular and circular sections bent about one axis can bereadily derived For biaxial bending conditions approximatedesign methods have been developed that utilise the solutionsobtained for uniaxial bending

Rectangular sections The figure here shows a rectangularsection with reinforcement in the faces parallel to the axisof bending

Resistance to bending and axial force 47

Spaneffective depth ratios for initial design of solid slabs

Span conditionsCharacteristic imposed load

5 kNm2 10 kNm2

Cantilever 11 10Simply supported

One-way span 27 24Two-way span 30 27

ContinuousOne-way span 34 30Two-way span 44 40

Flat slab (no drops) 30 27

In the table here the characteristic imposed load should includefor all finishes partitions and services For two-way spans theratios given apply to square panels For rectangular panelswhere the length is twice the breadth the ratios given for one-wayspans should be used For other cases ratios may be obtainedby interpolation The ratios apply to the shorter span for two-wayslabs and the longer span for flat slabs For ribbed slabs exceptfor cantilevers the ratios given in the table should be reducedby 20

524 Columns

The second order effects associated with lateral stability are animportant consideration in column design An effective height(or length in EC 2) and a slenderness ratio are determined inrelation to major and minor axes of bending An effective heightor length is a function of the clear height and depends upon theconditions of restraint at the ends of the column A clear distinc-tion exists between a braced column with effective height clear height and an unbraced column with effective height clearheight A braced column is one that is fully retrained in positionat the ends as in a structure where resistance to all the lateralforces in a particular plane is provided by stiff walls or bracingAn unbraced column is one that is considered to contribute tothe lateral stability of the structure as in a sway frame

In BS 8110 and BS 5400 a slenderness ratio is defined asthe effective height divided by the depth of the cross section inthe plane of bending A column is then considered as eithershort or slender according to the slenderness ratios Bracedcolumns are often short in which case second order effects maybe ignored In EC 2 the slenderness ratio is defined as theeffective length divided by the radius of gyration of the crosssection

Columns are subjected to combinations of bending momentand axial force and the cross section may need to be checked formore than one combination of values In slender columns theinitial moments obtained from an elastic analysis of the structureare increased by additional moments induced by the deflectionof the column In BS 8110 and EC 2 these additional moments

Resolving forces and taking moments about the mid-depth ofthe section gives the following equations for 0 x h

N k1bxfc As1 fs1 As2 fs2

M k1bxfc (05h k2x) As1 fs1 (05h d) As2 fs2 (d 05h)

where fs1 and fs2 are determined by the stressndashstrain curves for thereinforcement and depend on the value of x Values of k1 and k2

are determined by the concrete stress-block and fc is equal to fcu

in BS 8110 and BS 5400 and fck in EC 2For symmetrically reinforced sections As1 As2 Asc2

and d h d Design charts based on a rectangular stress-blockfor the concrete with values of fy 500 Nmm2 and dh 08and 085 respectively are given in Tables 317 and 318 forBS 8110 Tables 327 and 328 for BS 5400 and Tables 411 and412 for EC 2 Approximate design methods for biaxial bendingare given in Tables 321 331 and 416 for design to BS 8110BS 5400 and EC 2 respectively

Section Forces

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Circular sections The figure here shows a circular sectionwith six bars spaced equally around the circumference Six is theminimum number of bars recommended in the codes and solu-tions based on six bars will be slightly conservative if more barsare used The arrangement of bars relative to the axis of bendingaffects the resistance of the section and it can be shown that thearrangement in the figure is not the most critical in every casebut the variations are small and may be reasonably ignored

braced structures are typically square in cross section withsizes being determined mainly by the magnitude of the axialloads In multi-storey buildings column sizes are often keptconstant over several storeys with the reinforcement changingin relation to the axial load For initial design purposes typicalload capacities for short braced square columns in buildings aregiven in the following table

Design of structural members48

The following analysis is based on a uniform stress-block forthe concrete of depth x and width hsin at the base (as shownin the figure) Resolving forces and taking moments about themid-depth of the section where hs is the diameter of a circlethrough the centres of the bars gives the following equationsfor 0 x h

N [(2 sin2)8]h2fcd (Asc3)( fs1 fs2 fs3)M [(3sin sin3)72]h3fcd 0433(Asc3)( fs1 fs3)hs

where fs1 fs2 and fs3 are determined by the stressndashstrain curvesfor the reinforcement and depend on the value of x Values of fcd

and respectively are taken as 045fcu and 09 in BS 811004fcu and 10 in BS 5400 and 051fck and 08 in EC 2

Design charts derived for values of fy 500 Nmm2 andhsh 06 and 07 respectively are given in Tables 319and 320 for BS 8110 Tables 329 and 330 for BS 5400 andTables 413 and 414 for EC 2 Sections subjected to biaxialmoments Mx and My can be designed for the resultant moment

Design formulae In BS 8110 two approximate formulae aregiven for the design of short braced columns under specificconditions Columns which due to the nature of the structurecannot be subjected to significant moments for example columnsthat provide support to very stiff beams or beams on bearingsmay be considered adequate if N 040fcuAc 067Asc fy

Columns supporting symmetrical arrangements of beamsthat are designed for uniformly distributed imposed load andhave spans that do not differ by more than 15 of the longermay be considered adequate if N 035fcuAc 060Asc fy

BS 5400 contains general formulae for rectangular sectionsin the form of a trial-and-error procedure and two simplifiedformulae for specific applications details of which are given inTable 332

Column sizes Columns in unbraced structures are likely tobe rectangular in cross section due to the dominant effect ofbending moments in the plane of the structure Columns in

M (M2x M2

y)

Concrete Column Reinforcement percentageclass size

1 2 3 4

C2530 300 300 1370 1660 1950 2240350 350 1860 2260 2650 3050400 400 2430 2950 3470 3980450 450 3080 3730 4390 5040500 500 3800 4610 5420 6230

C3240 300 300 1720 2010 2300 2580350 350 2350 2740 3130 3520400 400 3070 3580 4090 4600450 450 3880 4530 5170 5820500 500 4790 5590 6390 7190

C4050 300 300 2080 2360 2650 2930350 350 2830 3220 3600 3990400 400 3700 4200 4710 5210450 450 4680 5320 5960 6600500 500 5780 6570 7360 8150

Ultimate design loads (kN) for short braced columns

In the foregoing table the loads were derived from the BS 8110equation for columns that are not subjected to significantmoments with fy 500 Nmm2 In determining the columnloads the ultimate load from the floor directly above the levelbeing considered should be multiplied by the following factorsto compensate for the effects of bending internal column 125edge column 15 corner column 20 The total imposed loadsmay be reduced according to the number of floors supportedThe reductions for 2 3 4 5ndash10 and over 10 floors are 1020 30 40 and 50 respectively

53 RESISTANCE TO SHEAR

Much research by many investigators has been undertaken in aneffort to develop a better understanding of the behaviour ofreinforced concrete subjected to shear As a result of thisresearch various theories have been proposed to explain themechanism of shear transfer in cracked sections and provide asatisfactory basis for designing shear reinforcement In theevent of overloading sudden failure can occur at the onset ofshear cracking in members without shear reinforcement As aconsequence a minimum amount of shear reinforcement in theform of links is required in nearly all beams Resistance to shearcan be increased by adding more shear reinforcement but even-tually the resistance is limited by the capacity of the inclinedstruts that form within the web of the section

531 Members without shear reinforcement

In an uncracked section shear results in a system of mutuallyorthogonal diagonal tension and compression stresses Whenthe diagonal tension stress reaches the tensile strength of the

Section Forces

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concrete a diagonal crack occurs This simple concept rarelyapplies to reinforced concrete since members such as beamsand slabs are generally cracked in flexure In current practiceit is more useful to refer to the nominal shear stress v Vbdwhere b is the breadth of the section in the tension zone Thisstress can then be related to empirical limiting values derivedfrom test data The limiting value vc depends on the concretestrength the effective depth and the reinforcement percentageat the section considered To be effective this reinforcementshould continue beyond the section for a specified minimumdistance as given in Codes of Practice For values of v vc noshear reinforcement is required in slabs but for most beams aspecified minimum amount in the form of links is required

At sections close to supports the shear strength is enhancedand for members carrying generally uniform load the criticalsection may be taken at d from the face of the support Whereconcentrated loads are applied close to supports in memberssuch as corbels and pile-caps some of the load is transmittedby direct strut action This effect is taken into account in theCodes of Practice by either enhancing the shear strength of thesection or reducing the design load In members subjected tobending and axial load the shear strength is increased due tocompression and reduced due to tension

Details of design procedures in Codes of Practice are givenin Table 333 for BS 8110 Table 336 for BS 5400 andTable 417 for EC 2

532 Members with shear reinforcement

The design of members with shear reinforcement is based on atruss model in which the tension and compression chords arespaced apart by a system of inclined concrete struts and uprightor inclined shear reinforcement Most reinforcement is in theform of upright links but bent-up bars may be used for upto 50 of the total shear reinforcement in beams The trussmodel results in a force in the tension chord additional to thatdue to bending This can be taken into account directly in thedesign of the tension reinforcement or indirectly by shiftingthe bending moment curve each side of any point of maximumbending moment

In BS 8110 shear reinforcement is required to cater for thedifference between the shear force and the shear resistance ofthe section without shear reinforcement Equations are givenfor upright links based on concrete struts inclined at about 45oand for bent-up bars where the inclination of the concrete strutsmay be varied between specified limits In BS 5400 a specifiedminimum amount of link reinforcement is required in additionto that needed to cater for the difference between the shear forceand the shear resistance of the section without shear reinforce-ment The forces in the inclined concrete struts are restrictedindirectly by limiting the maximum value of the nominal shearstress to specified values

In EC 2 shear reinforcement is required to cater for the entireshear force and the strength of the inclined concrete struts ischecked explicitly The inclination of the struts may be variedbetween specified limits for links as well as bent-up bars Incases where upright links are combined with bent-up bars thestrut inclination needs to be the same for both

Details of design procedures in Codes of Practice are givenin Table 333 for BS 8110 Table 336 for BS 5400 andTable 418 for EC 2

533 Shear under concentrated loads

Suspended slabs and foundations are often subjected to largeloads or reactions acting on small areas Shear in solid slabsunder concentrated loads can result in punching failures on theinclined faces of truncated cones or pyramids For designpurposes shear stresses are checked on given perimeters atspecified distances from the edges of the loaded area Where aload or reaction is eccentric with regard to a shear perimeter(eg at the edges of a slab and in cases of moment transferbetween a slab and a column) an allowance is made for theeffect of the eccentricity In cases where v exceeds vc linksbent-up bars or other proprietary products may be provided inslabs not less than 200 mm deep

Details of design procedures in Codes of Practice are givenin Table 334 for BS 8110 Tables 337 and 338 for BS 5400and Table 419 for EC 2

54 RESISTANCE TO TORSION

In normal beam-and-slab or framed construction calculationsfor torsion are not usually necessary adequate control of anytorsional cracking in beams being provided by the requiredminimum shear reinforcement When it is judged necessary toinclude torsional stiffness in the analysis of a structure ortorsional resistance is vital for static equilibrium membersshould be designed for the resulting torsional moment Thetorsional resistance of a section may be calculated on the basisof a thin-walled closed section in which equilibrium is satisfiedby a closed plastic shear flow Solid sections may be modelled asequivalent thin-walled sections Complex shapes may be dividedinto a series of sub-sections each of which is modelled as anequivalent thin-walled section and the total torsional resistancetaken as the sum of the resistances of the individual elementsWhen torsion reinforcement is required this should consist ofrectangular closed links together with longitudinal reinforce-ment Such reinforcement is additional to any requirements forshear and bending

Details of design procedures in Codes of Practice are givenin Table 335 for BS 8110 Table 339 for BS 5400 andTable 420 for EC 2

55 DEFLECTION

The deflections of members under service loading should notimpair the appearance or function of a structure An accurateprediction of deflections at different stages of construction mayalso be necessary in bridges for example For buildings thefinal deflection of members below the support level afterallowance for any pre-camber is limited to span250 In orderto minimise any damage to non-structural elements such asfinishes cladding or partitions that part of the deflection thatoccurs after the construction stage is also limited to span500In BS 8110 this limit is taken as 20 mm for spans 10 m

The behaviour of a reinforced concrete beam under serviceloading can be divided into two basic phases before and aftercracking During the uncracked phase the member behaveselastically as a homogeneous material This phase is ended bythe load at which the first flexural crack forms The cracks resultin a gradual reduction in stiffness with increasing load duringthe cracked phase The concrete between the cracks continues

Deflection 49

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to provide some tensile resistance though less on average thanthe tensile strength of the concrete Thus the member is stifferthan the value calculated on the assumption that the concretecarries no tension This additional stiffness known as lsquotensionstiffeningrsquo is highly significant in lightly reinforced memberssuch as slabs but has only a relatively minor effect on thedeflection of heavily reinforced members These concepts areillustrated in the following figure

assumptions made in their derivation provide a useful basisfor estimating long-term deflections of members in buildingsas follows

Details of spaneffective depth ratios and explicit calculationprocedures are given in Tables 340 to 342 for BS 8110 andTables 421 and 422 for EC 2

56 CRACKING

Cracks in members under service loading should not impairthe appearance durability or water-tightness of a structure InBS 8110 for buildings the design crack width is generallylimited to 03 mm In BS 5400 for bridges the limit variesbetween 025 mm and 010 mm depending on the exposureconditions In BS 8007 for structures to retain liquids a limitof 02 mm usually applies Under liquid pressure continuouscracks that extend through the full thickness of a slab or wallare likely to result in some initial seepage but such cracks areexpected to self-heal within a few weeks If the appearance ofa liquid-retaining structure is considered aesthetically critical acrack width limit of 01 mm applies

In EC 2 for most buildings the design crack width is generallylimited to 03 mm but for internal dry surfaces a limitof 04 mm is considered sufficient For liquid-retainingstructures a classification system according to the degree ofprotection required against leakage is introduced Where asmall amount of leakage is acceptable for cracks that passthrough the full thickness of the section the crack width limitvaries according to the hydraulic gradient (ie head of liquiddivided by thickness of section) The limits are 02 mm forhydraulic gradients 5 reducing uniformly to 005 mmfor hydraulic gradients 35

In order to control cracking in the regions where tension isexpected it is necessary to ensure that the tensile capacity ofthe reinforcement at yielding is not less than the tensile force inthe concrete just before cracking Thus a minimum amount ofreinforcement is required according to the strength of thereinforcing steel and the tensile strength of the concrete atthe time when cracks may first be expected to occur Cracks dueto restrained early thermal effects in continuous walls and someslabs may occur within a few days of the concrete being placedIn other members it may be several weeks before the appliedload reaches a level at which cracking occurs

Crack widths are influenced by several factors including thecover bar size bar spacing and stress in the reinforcement Thestress may need to be reduced in order to meet the crack widthlimit Design formulae are given in Codes of Practice in whichstrain calculated on the basis of no tension in the concrete isreduced by a value that decreases with increasing amounts oftension reinforcement For cracks that are caused by appliedloading the same formulae are used in BS 8110 BS 5400 andBS 8007 For cracks that are caused by restraint to temperatureeffects and shrinkage fundamentally different formulae areincluded in BS 8007 Here it is assumed that bond slip occursat each crack and the crack width increases in direct proportionto the contraction of the concrete

Deflection actual spaneffective depth ratio

limiting spaneffective depth ratiospan250

Design of structural members50

Schematic load-deflection response

In BS 8110 for the purpose of analysis lsquotension stiffeningrsquo isrepresented by a triangular stress distribution in the concreteincreasing from zero at the neutral axis to a maximum value atthe tension face At the level of the tension reinforcement theconcrete stress is taken as 1 Nmm2 for short-term loads and055 Nmm2 for long-term loads irrespective of the strain in thetension reinforcement In EC 2 a more general approach isadopted in which the deformation of a section which could bea curvature or in the case of pure tension an extension or acombination of these is calculated first for a homogeneousuncracked section 1 and second for a cracked section ignor-ing tension in the concrete 2 The deformation of the sectionunder the design loading is then obtained as

2 (1 )1

where is a distribution coefficient that takes account of thedegree of cracking according to the nature and duration ofthe loading and the stress in the tension reinforcement underthe load causing first cracking in relation to the stress under thedesign service load

When assessing long-term deflections allowances need to bemade for the effect of concrete creep and shrinkage Creep canbe taken into account by using an effective modulus of elasticityEceff Ec(1 ) where Ec is the short-term value and is acreep coefficient Shrinkage deformations can be calculatedseparately and added to those due to loading

Generally explicit calculation of deflections is unnecessaryto satisfy code requirements and simple rules in the form oflimiting spaneffective depth ratios are provided in BS 8110 andEC 2 These are considered adequate for avoiding deflectionproblems in normal circumstances and subject to the particular

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Generally for design to BS 8110 and EC 2 there is no needto calculate crack widths explicitly and simple rules that limiteither bar size or bar spacing according to the stress in thereinforcement are provided Details of both rules and crackwidth formulae are given in Table 343 for BS 8110 and BS 5400Tables 344 and 345 for BS 8007 and Tables 423ndash425 forEC 2 Additional design aids derived from the crack widthformulae are provided in Tables 346ndash352 for BS 8007 andTables 426 and 427 for EC 2

57 REINFORCEMENT CONSIDERATIONS

Codes of Practice contain many requirements affecting thereinforcement details such as minimum and maximum areasanchorage and lap lengths bends in bars and curtailment Thereinforcement may be curtailed in relation to the bendingmoment diagram provided there is always enough anchorageto develop the necessary design force in each bar at every crosssection Particular requirements apply at the positions wherebars are curtailed and at simple supports

Bars may be set out individually in pairs or in bundles ofthree or four in contact For the safe transmission of bondforces the cover provided to the bars should be not less thanthe bar size or for a group of bars in contact the equivalentdiameter of a notional bar with the same cross-sectional area asthe group Gaps between bars (or groups of bars) should benot less than the greater of (aggregate size plus 5 mm) or thebar size (or equivalent bar diameter for a group) Details ofreinforcement limits and requirements for containing bars incompression are given in Table 353 for BS 8110 Table 359for BS 5400 and Table 428 for EC 2

571 Anchorage lengths

At both sides of any cross section the reinforcement should beprovided with an appropriate embedment length or other formof end anchorage In earlier codes it was also necessary to con-sider lsquolocal bondrsquo at sections where large changes of tensileforce occur over short lengths of reinforcement and thisrequirement remains in BS 5400

Assuming a uniform bond stress between concrete and thesurface of a bar the required anchorage length is given by

lbreq (design force in bar)(bond stress perimeter of bar) fsd (24)fbd () ( fsdfbd)(4)

where fsd is the design stress in the bar at the position fromwhich the anchorage is measured The design bond stress fbd

depends on the strength of the concrete the type of bar and inEC 2 the location of the bar within the concrete section duringconcreting For example the bond condition is classified aslsquogoodrsquo in the bottom 250 mm of any section and in the top300 mm of a section 600 mm deep In other locations thebond condition is classified as lsquopoorrsquo Also in EC 2 the basicanchorage length in tension can be multiplied by severalcoefficients that take account of factors such as the bar shapethe cover and the effect of transverse reinforcement or pressureFor bars of diameter 40 mm and bars grouped in pairs orbundles additional considerations apply Details of design

anchorage lengths in tension and compression are given inTable 355 for BS 8110 Table 359 for BS 5400 and Tables 430and 432 for EC 2

572 Lap lengths

Forces can be transferred between reinforcement by lappingwelding or joining bars with mechanical devices (couplers)Connections should be placed whenever possible away frompositions of high stress and should preferably be staggeredIn Codes of Practice the necessary lap length is obtained bymultiplying the required anchorage length by a coefficient

In BS 8110 for bars in compression the coefficient is 125For bars in tension the coefficient is 10 14 or 20 accordingto the cover the gap between adjacent laps in the same layerand the location of the bar in the section In slabs where thecover is not less than twice the bar size and the gap betweenadjacent laps is not less than six times the bar size or 75 mm afactor of 10 applies Larger factors are frequently necessary incolumns typically 14 and beams typically 14 for bottom barsand 20 for top bars The sum of all the reinforcement sizes ina particular layer should not exceed 40 of the width of thesection at that level When the size of both bars at a lap exceeds20 mm and the cover is less than 15 times the size of thesmaller bar links at a maximum spacing of 200 mm arerequired throughout the lap length

In EC 2 for bars in tension or compression the lap coefficientvaries from 10 to 15 according to the percentage of lapped barsrelative to the total area of bars at the section considered andtransverse reinforcement is required at each end of the lap zoneDetails of lap lengths are given in Table 355 for BS 8110Table 359 for BS 5400 and Tables 431 and 432 for EC 2

573 Bends in bars

The radius of any bend in a reinforcing bar should conform tothe minimum requirements of BS 8666 and should ensure thatfailure of the concrete inside the bend is prevented For barsbent to the minimum radius according to BS 8666 it is notnecessary to check for concrete failure if the anchorage of thebar does not require a length more than 5 beyond the end ofthe bend (see Table 227) It is also not necessary to check forconcrete failure where the plane of the bend is not close to aconcrete face and there is a transverse bar not less than its ownsize inside the bend This applies in particular to a link whichmay be considered fully anchored if it passes round anotherbar not less than its own size through an angle of 90o andcontinues beyond the end of the bend for a length not less than8 in BS 8110 and 10 in EC 2

In cases when a bend occurs at a position where the bar ishighly stressed the bearing stress inside the bend needs to bechecked and the radius of bend will need to be more than theminimum given in BS 8666 This situation occurs typically atmonolithic connections between members for example junc-tion of beam and end column and in short members such ascorbels and pile caps The design bearing stress is limitedaccording to the concrete strength and the confinementperpendicular to the plane of the bend Details of bends in barsare given in Table 355 for BS 8110 Table 359 for BS 5400and Table 431 for EC 2

Reinforcement considerations 51

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574 Curtailment of reinforcement

In flexural members it is generally advisable to stagger thecurtailment points of the tension reinforcement as allowed bythe bending moment envelope Bars to be curtailed need toextend beyond the points where in theory they are no longerneeded for flexural resistance for a number of reasons butmainly to ensure that the shear resistance of the section is notreduced locally Clearly of course no reinforcement shouldbe curtailed at a point less than a full anchorage length from asection where it is required to be fully stressed

In BS 8110 and BS 5400 except at end supports every barshould extend beyond the point at which in theory it is no longerrequired for a distance not less than the greater of the effectivedepth of the member or 12 times the bar size In addition barscurtailed in a tension zone should satisfy at least one of threealternative conditions one requires a full anchorage length onerequires the designer to determine the position where the shearresistance is twice the shear force and the other requires thedesigner to determine the position where the bending resistanceis twice the bending moment The simplest approach is to complywith the first option by providing a full anchorage lengthbeyond the point where in theory the bar is no longer requiredeven if this requires a longer extension than is absolutelynecessary in some cases Details of the requirements are givenin Table 356

In BS 8110 simplified rules are also given for beams andslabs where the loads are mainly uniformly distributed and inthe case of continuous members the spans are approximatelyequal Details of the rules are given in Tables 357 and 358

At simple end supports the tension bars should extend foran effective anchorage length of 12 times the bar size beyondthe centre of the support but no bend should begin before thecentre of the support In cases where the width of the supportexceeds the effective depth of the member the centre ofthe support may be assumed at half the effective depth from theface of the support In BS 8110 for slabs in cases where thedesign shear force is less than half the shear resistance anchor-age can be obtained by extending the bars beyond the centre ofthe support for a distance equal to one third of the supportwidth 30 mm

In EC 2 the extension al of a tension bar beyond the pointwhere in theory it is no longer required for flexural resistance isdirectly related to the shear force at the section For memberswith upright shear links al 05zcot13 where z is the lever armand 13 is the inclination of the concrete struts (see section 3512)Taking z 09d al 045dcot13 where cot13 is selected by thedesigner in the range 10 cot13 25 If the value of cot 13 usedin the shear design calculations is unknown al 1125d can beassumed For members with no shear reinforcement al d isused At simple end supports bottom bars should extend for ananchorage length beyond the face of the support The tensileforce to be anchored is given by F 05Vcot 13 and F 125Vcan be conservatively taken in all cases Details of the curtailmentrequirements are given in Table 432

58 DEEP BEAMS

In designing normal (shallow) beams of the proportions morecommonly used in construction plane sections are assumed toremain plane after loading This assumption is not strictly true

but the errors resulting from it only become significant whenthe depth of the beam becomes equal to or more than abouthalf the span The beam is then classed as a deep beam anddifferent methods of analysis and design need to be usedThese methods take into account not only the overall appliedmoments and shears but also the stress patterns and internaldeformations within the beam

For a single-span deep beam after the concrete in tensionhas cracked the structural behaviour is similar to a tied archThe centre of the compression force in the arch rises from thesupport to a height at the crown equal to about half the span ofthe beam The tension force in the tie is roughly constant alongits length since the bending moment and the lever arm undergosimilar variations along the length of the beam For a continuousdeep beam the structural behaviour is analogous to a separatetied arch system for each span combined with a suspensionsystem centred over each internal support

In BS 8110 for the design of beams of clear span less thantwice the effective depth the designer is referred to specialistliterature In EC 2 a deep beam is classified as a beam whoseeffective span is less than three times its overall depth Briefdetails of suitable methods of design based on the result ofextensive experimental work by various investigators are givenin ref 42 and a comprehensive well-produced design guide iscontained in ref 43

59 WALLS

Information concerning the design of load-bearing walls inaccordance with BS 8110 is given in section 618 Retainingwalls and other similar elements that are subjected mainly totransverse bending where the design vertical load is less than01fcu times the area of the cross section are treated as slabs

510 DETAILS

It has long been realised that the calculated strength of areinforced concrete member cannot be attained if the details ofthe required reinforcement are unsatisfactory Research by theformer Cement and Concrete Association and others has shownthat this applies particularly at joints and intersections Thedetails commonly used in wall-to-base and wall-to-walljunctions in retaining structures and containment vessels wherethe action of the applied load is to lsquoopenrsquo the corner are notalways effective

On Tables 362 and 363 are shown recommended details thathave emerged from the results of reported research The designinformation given in BS 8110 and BS 5400 for nibs corbels andhalving joints is included and supplemented by informationgiven elsewhere In general however details that are primarilyintended for precast concrete construction have not beenincluded as they fall outside the scope of this book

511 ELASTIC ANALYSIS OF CONCRETE SECTIONS

The geometrical properties of various figures the shapes ofwhich conform to the cross sections of common reinforcedconcrete members are given in Table 2101 The data includeexpressions for the area section modulus second moment ofarea and radius of gyration The values that are derived fromthese expressions are applicable in cases when the amount of

Design of structural members52

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reinforcement provided need not be taken into account in theanalysis of the structure (see section 141)

The data given in Tables 2102 and 2103 are applicable toreinforced concrete members with rectilinear and polygonalcross sections when the reinforcement provided is taken intoaccount on the basis of the modular ratio Two conditions areconsidered (1) when the entire section is subjected to stressand (2) when for members subjected to bending the concretein tension is not taken into account The data given for the

former condition are the effective area the centre and secondmoment of area the modulus and radius of gyration For thecondition when a member is subjected to bending and theconcrete in tension is assumed to be ineffective data giveninclude the position of the neutral axis the lever-arm and theresistance moment

Design procedures for sections subjected to bending andaxial force with design charts for rectangular and cylindricalcolumns are given in Tables 2104ndash2109

Elastic analysis of concrete sections 53

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The loads and consequent bending moments and forces onthe principal types of structural components and the designresistances of such components have been dealt with in thepreceding chapters In this chapter some complete structurescomprising assemblies or special cases of such componentsand their foundations are considered

61 BUILDINGS

Buildings may be constructed entirely of reinforced concreteor one or more elements of the roof floors walls stairs andfoundations may be of reinforced concrete in conjunction witha steel frame Alternatively the building may consist of interiorand exterior walls of cast in situ reinforced concrete supportingthe floors and roof with the columns and beams being formedin the thickness of the walls Again the entire structure or partsthereof may be built of precast concrete elements connectedtogether during construction

The design of the various parts of a building is the subjectof Examples of the Design of Buildings That book includesillustrative calculations and drawings for a typical six-storeymultipurpose building This section provides a brief guide tocomponent design

611 Robustness and provision of ties

The progressive collapse of one corner of a London tower blockin 1968 as a result of an explosion caused by a gas leak in adomestic appliance on the eighteenth floor led to recommen-dations to consider such accidental actions in the design of allbuildings Regulations require a building to be designed andconstructed so that in the event of an accident the buildingwill not collapse to an extent disproportionate to the causeBuildings are divided into classes depending on the type andoccupancy including the likelihood of accidents and thenumber of occupants that may be affected with a statementof the design measures to be taken in each of the classes TheBS 8110 normal requirements for lsquorobustnessrsquo automaticallysatisfy the regulations for all buildings except those wherespecific account is to be taken of likely hazards

The layout and form of the structure should be checked toensure that it is inherently stable and robust In some cases itmay be necessary to protect certain elements from vehicularimpact by providing bollards or earth banks All structures

should be able to resist a notional ultimate horizontal forceequal to 15 of the characteristic dead load of the structureThis force effectively replaces the design wind load in caseswhere the exposed surface area of the building is small

Wherever possible continuous horizontal and vertical tiesshould be provided throughout the building to resist specifiedforces The magnitude of the force increases with the numberof storeys for buildings of less than 10 storeys but remainsconstant thereafter The requirements may be met by usingreinforcement that is necessary for normal design purposes inbeams slabs columns and walls Only the tying forces needto be considered and the full characteristic strength of thereinforcement may be taken into account Horizontal ties arerequired in floors and roofs at the periphery and internally intwo perpendicular directions The internal ties which may bespread uniformly over the entire building or concentrated atbeam and column positions are to be properly anchored atthe peripheral tie Vertical ties are required in all columns andload-bearing walls from top to bottom and all external columnsand walls are to be tied into each floor and roof For regulatorypurposes some buildings are exempt from the vertical tyingrequirement Details of the tying requirements are given inTable 354

For in situ construction proper attention to reinforcementdetailing is all that is normally necessary to meet the tyingrequirements Precast forms of construction generally requiremore care and recommended details to obtain continuity ofhorizontal ties are given in the code of practice If ties cannotbe provided other strategies should be adopted as described inPart 2 of the code These strategies are presented in the contextof residential buildings of five or more storeys where eachelement that cannot be tied is to be considered as notionallyremoved one at a time in each storey in turn The requirementis that any resulting collapse should be limited in extent withthe remaining structure being able to bridge the gap causedby the removal of the element If this requirement cannot besatisfied then the element in question is considered as a keyelement In this case the element and its connections need tobe able to resist a design ultimate load of 34 kNm2 consideredto act from any direction BS 8110 is vague with regard to theextent of collapse associated with this approach but a moreclearly defined statement is given in the building regulationsHere a key element is any untied member whose removalwould put at risk of collapse within the storey in question

Chapter 6

Buildings bridges andcontainmentstructures

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and the immediately adjacent storeys more than 15 of thearea of the storey (or 70 m2 if less)

In EC 2 similar principles apply in that structures notspecifically designed to withstand accidental actions should beprovided with a suitable tying system to prevent progressivecollapse by providing alternative load paths after local damageThe UK National Annex specifies compliance with the BS 8110requirements as given in Table 429

612 Floors

Suspended concrete floors can be of monolithic constructionin the form of beam-and-slab (solid or ribbed) or flat slab(solid or waffle) or can consist of precast concrete slab unitssupported on concrete or steel beams or comprise one ofseveral other hybrid forms Examples of monolithic forms ofconstruction are shown in the figure on Table 242

Two-way beam and solid slab systems can involve a layoutof long span secondary beams supported by usually shorterspan main beams The resulting slab panels may be designed astwo-way spanning if the longer side is less than twice theshorter side However such two-way beam systems tend tocomplicate both formwork and reinforcement details with aconsequent delay in the construction programme A one-waybeam and solid slab system is best suited to a rectangular gridof columns with long span beams and shorter span slabs If aribbed slab is used a system of long span slabs supported byshorter span beams is preferable If wide beams are used thebeam can be incorporated within the depth of the ribbed slab

In BS 8110 ribbed slabs include construction in which ribsare cast in situ between rows of blocks that remain part of thecompleted floor This type of construction is no longer used inthe United Kingdom although blocks are incorporated in someprecast and composite construction The formers for ribbedslabs can be of steel glassfibre or polypropylene Standardmoulds are available that provide tapered ribs with a minimumwidth of 125 mm spaced at 600 mm (troughs) and 900 mm(waffles) The ribs are connected by a structural concrete toppingwith a minimum thickness of 50 mm for trough moulds and75 mm for waffle moulds In most structures to obtain thenecessary fire-resistance either the thickness of topping has toexceed these minimum values or a non-structural screed addedat a later stage of construction The spacing of the ribs may beincreased to a maximum of 1500 mm by using purpose-madeformers Comprehensive details of trough and waffle floorsare contained in ref 44

BS 8110 and EC 2 contain recommendations for both solidand ribbed slabs spanning between beams or supported directlyby columns (flat slabs) Ribs in waffle slabs and ribs reinforcedwith a single bar in trough slabs do not require links unlessneeded for shear or fire-resistance Ribs in trough slabs whichare reinforced with more than one bar should be provided withsome links to help maintain the correct cover The spacing ofthese links may be in the range 10ndash15 m according to the sizeof the main bars Structural toppings are normally reinforcedwith a welded steel fabric

Information on the weight of concrete floor slabs is given inTable 21 and details of imposed loads on floors are givenin Table 23 Detailed guidance on the analysis of slabs isgiven in Chapters 4 and 13 More general guidance includinginitial sizing suggestions is given in section 523

613 Openings in floors

Large openings (eg stairwells) should generally be providedwith beams around the opening Holes for pipes ducts andother services should generally be formed when the slabis constructed and the cutting of such holes should not bepermitted afterwards unless done under the supervision of acompetent engineer Small isolated holes may generally beignored structurally with the reinforcement needed for a slabwithout holes simply displaced locally to avoid the hole

In other cases the area of slab around an opening or groupof closely spaced holes needs to be strengthened with extrareinforcement The cross-sectional area of additional bars to beplaced parallel to the principal reinforcement should be at leastequal to the area of principal reinforcement interrupted by theopening Also for openings of dimensions exceeding 500 mmadditional bars should be placed diagonally across the cornersof the opening Openings with dimensions greater than 1000 mmshould be regarded as structurally significant and the area ofslab around the opening designed accordingly

The effect of an opening in the proximity of a concentratedload or supporting column on the shearing resistance of theslab is shown in Table 337

614 Stairs

Structural stairs may be tucked away out of sight within a fireenclosure or they may form a principal architectural feature Inthe former case the stairs can be designed and constructed assimply and cheaply as possible but in the latter case much moretime and trouble is likely to be expended on the design

Several stair types are illustrated on Table 288 Variousprocedures for analysing the more common types of stairhave been developed and some of these are described onTables 288ndash291 These theoretical procedures are based onthe concept of an idealised line structure and when detailingthe reinforcement for the resulting stairs additional bars shouldbe included to limit the formation of cracks at the points ofhigh stress concentration that inevitably occur The lsquothree-dimensionalrsquo nature of the actual structure and the stiffeningeffect of the triangular tread areas both of which are usuallyignored when analysing the structure will result in actual stressdistributions that differ from those calculated and this mustbe remembered when detailing The stair types illustrated onTable 288 and others can also be investigated by finite-elementmethods and similar procedures suitable for computer analysisWith such methods it is often possible to take account of thethree-dimensional nature of the stair

Simple straight flights of stairs can span either transversely(ie across the flight) or longitudinally (ie along the flight)When spanning transversely supports must be provided on bothsides of the flight by either walls or stringer beams In this casethe waist or thinnest part of the stair construction need be nomore than 60 mm thick say the effective lever arm for resistingthe bending moment being about half of the maximumthickness from the nose to the soffit measured at right anglesto the soffit When the stair spans longitudinally deflectionconsiderations can determine the waist thickness

In principle the design requirements for beams and slabsapply also to staircases but designers cannot be expected todetermine the deflections likely to occur in the more complex

Buildings 55

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stair types BS 8110 deals only with simple types and allows amodified spaneffective depth ratio to be used The bendingmoments should be calculated from the ultimate load due to thetotal weight of the stairs and imposed load measured on plancombined with the horizontal span Stresses produced bythe longitudinal thrust are small and generally neglected in thedesign of simple systems Unless circumstances otherwisedictate suitable step dimensions for a semi-public stairs are 165mm rise and 275 mm going which with a 25 mm nosing orundercut gives a tread of 300 mm Private stairs may be steeperand those in public buildings should be less steep In eachcase optimum proportions are given by the relationship(2 rise going) 600 mm Different forms of constructionand further details on stair dimensions are given in BS 5395

Finally it should be remembered that the prime purpose of astair is to provide safe pedestrian access between the floors itconnects As such it is of vital importance in the event of a fireand a principal design consideration must be to provide adequatefire-resistance

615 Planar roofs

The design and construction of a flat reinforced concrete roofare essentially the same as for a floor A water-tight coveringsuch as asphalt or bituminous felt is generally necessary andwith a solid slab some form of thermal insulation is normallyrequired For ordinary buildings the slab is generally built leveland a drainage slope of the order of 1 in 120 is formed byadding a mortar topping The topping is laid directly onto theconcrete and below the water-tight covering and can formthe thermal insulation if it is made of a sufficient thickness oflightweight concrete or other material having low thermalconductivity

Planar slabs with a continuous steep slope are not commonin reinforced concrete except for mansard roofs The roofcovering is generally of metal or asbestos-cement sheeting orsome lightweight material Such coverings and roof glazingrequire purlins for their support and although these are often ofsteel precast concrete purlins are also used especially if theroof structure is of reinforced concrete

616 Non-planar roofs

Roofs that are not planar other than the simple pitched roofsconsidered in the foregoing can be constructed as a series ofplanar slabs (prismatic or hipped-plate construction) or assingle- or double-curved shells Single-curved roofs such assegmental or cylindrical shells are classified as developablesurfaces Such surfaces are not as stiff as double-curved roofsor their prismatic counterparts which cannot be lsquoopened uprsquointo plates without some shrinking or stretching taking place

If the curvature of a double-curved shell is similar in alldirections the surface is known as synclastic A typical case isa dome where the curvature is identical in all directions Ifthe shell curves in opposite directions over certain areas thesurface is termed anticlastic (saddle shaped) The hyperbolic-paraboloidal shell is a well-known example and is the specialcase where such a double-curved surface is generated by twosets of straight lines An elementary analysis of some of thesestructural forms is dealt with in section 192 and Table 292but reference should be made to specialist publications for

more comprehensive analyses and more complex structuresSolutions for many particular shell types have been producedand in addition general methods have been developed foranalysing shell forms of any shape by means of a computerShells like all statically indeterminate structures are affectedby such secondary effects as shrinkage temperature change andsettlement and a designer must always bear in mind the factthat the stresses arising from these effects can modify quiteconsiderably those due to normal dead and imposed load InTable 281 simple expressions are given for the forces indomed slabs such as are used for the bottoms and roofs of somecylindrical tanks In a building a domed roof generally hasa much larger rise to span ratio and where the dome is partof a spherical surface and has an approximately uniform thick-ness overall the analysis given in Table 292 applies Shallowsegmental domes and truncated cones are also dealt with inTable 292

Cylindrical shells Segmental or cylindrical roofs are usuallydesigned as shell structures Thin curved slabs that behave asshells are assumed to offer no resistance to bending nor todeform under applied distributed loads Except near edge andend stiffeners the shell is subjected only to membrane forcesnamely a direct force acting longitudinally in the plane of theslab a direct force acting tangentially to the curve of the slaband a shearing force Formulae for these membrane forces aregiven in section 1923 In practice the boundary conditionsdue to either the presence or absence of edge or valley beamsend diaphragms continuity and so on affect the displacementsand forces that would otherwise occur as a result of membraneaction Thus as when analysing any indeterminate structure(such as a continuous beam system) the effects due to theseboundary restraints need to be combined with the staticallydeterminate stresses arising from the membrane action

Shell roofs can be arbitrarily subdivided into lsquoshortrsquo (wherethe ratio of length l to radius r is less than about 05) lsquolongrsquo(where l r exceeds 25) and lsquointermediatersquo For short shellsthe influence of the edge forces is slight in comparison withmembrane action and the stresses can be reasonably taken asthose due to the latter only If the shell is long the membraneaction is relatively insignificant and an approximate solutioncan be obtained by considering the shell to act as a beam withcurved flanges as described in section 1923

For the initial analysis of intermediate shells no equivalentshort-cut method has yet been devised The standard method ofsolution is described in various textbooks (eg refs 45 and 46)Such methods involve the solution of eight simultaneousequations if the shell or the loading is unsymmetrical or four ifsymmetry is present by matrix inversion or other means Bymaking certain simplifying assumptions and providing tables ofcoefficients Tottenham (ref 47) developed a popular simplifieddesign method which is rapid and requires the solution of threesimultaneous equations only J D Bennett also developed amethod of designing long and intermediate shells based on ananalysis of actual designs of more than 250 roofs The methodwhich involves the use of simple formulae incorporatingempirical coefficients is summarised on Tables 293 and 294For further details see ref 48

Buckling of shells A major concern in the design of anyshell is the possibility of buckling since the loads at which

Buildings bridges and containment structures56

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buckling occurs as established by tests often differ from thevalues predicted by theory Ref 49 indicates that for domessubtending angles of about 90o the critical external pressure atwhich buckling occurs according to both theory and tests isgiven by p 03E(hr)2 where E is the elastic modulus ofconcrete and h is the thickness and r the radius of the domeFor a shallow dome with spanrise cong 10 p 015E(hr)2 Afactor of safety against buckling of 2 to 3 should be adoptedSynclastic shells having a radius ranging from r1 to r2 may beconsidered as an equivalent dome with a radius of r radic(r1 r2)

For a cylindrical shell buckling is unlikely if the shell isshort In the case of long shells p 06E(hr)2

Anticlastic surfaces are more rigid than single-curved shellsand the buckling pressure for a saddle-shaped shell supportedon edge stiffeners safely exceeds that of a cylinder having acurvature equal to that of the anticlastic shell at the stiffenerFor a hyperbolic-paraboloidal shell with straight boundariesthe buckling load obtained from tests is slightly more than thevalue given by n E(ch)22ab where a and b are the lengthsof the sides of the shell c is the rise and h the thickness this isonly half of the value predicted theoretically

617 Curved beams

When bow girders and beams that are not rectilinear in planare subjected to vertical loading torsional moments occur inaddition to the normal bending moments and shearing forcesBeams forming a circular arc in plan may comprise part of acomplete circular system with equally spaced supports andequal loads on each span such systems occur in silos towersand similar cylindrical structures Equivalent conditions canalso occur in beams where the circle is incomplete provided theappropriate negative bending and torsional moments can bedeveloped at the end supports This type of circular beam canoccur in structures such as balconies

On Tables 295ndash297 charts are given that enable a rapidevaluation of the bending moments torsional moments andshear forces occurring in curved beams due to uniform andconcentrated loads The formulae on which the charts arebased are given in section 193 and on the tables concernedThe expressions have been developed from those in ref 50 foruniform loads and ref 51 for concentrated loads In both casesthe results have been recalculated to take into account values ofG 04E and C J2

618 Load-bearing walls

In building codes for design purposes a wall is defined as avertical load-bearing member whose length on plan exceedsfour times its thickness Otherwise the member is treated as acolumn in which case the effects of slenderness in relation toboth major and minor axes of bending need to be considered(section 524) A reinforced wall is one in which not lessthan the recommended minimum amount of reinforcement isprovided and taken into account in the design Otherwise themember is treated as a plain concrete wall in which case thereinforcement is ignored for design purposes

A single planar wall in general can be subjected to verticaland horizontal in-plane forces acting together with in-planeand transverse moments The in-plane forces and moment canbe combined to obtain at any particular level a longitudinal

shear force and a linear distribution of vertical force If thein-plane eccentricity of the vertical force exceeds one-sixth ofthe length of the wall reinforcement can be provided to resistthe tension that develops at one end of the wall In a plain wallsince the tensile strength of the concrete is ignored the distrib-ution of vertical load is similar to that for the bearing pressuredue to an eccentric load on a footing Flanged walls and coreshapes can be treated in a similar way to obtain the resultingdistribution of vertical force Any unit length of the wall cannow be designed as a column subjected to vertical loadcombined with bending about the minor axis due to anytransverse moment

In BS 8110 the effective height of a wall in relation to itsthickness depends upon the effect of any lateral supports andwhether the wall is braced or unbraced A braced wall is onethat is supported laterally by floors andor other walls able totransmit lateral forces from the wall to the principal structuralbracing or to the foundations The principal structural bracingcomprise strong points shear walls or other suitable elementsgiving lateral stability to a structure as a whole An unbracedwall provides its own lateral stability and the overall stabilityof multi-storey buildings should not in any direction dependon such walls alone The slenderness ratio of a wall is definedas the effective height divided by the thickness and the wall isconsidered lsquostockyrsquo if the slenderness ratio does not exceed15 for a braced wall or 10 for an unbraced wall Otherwise awall is considered slender in which case it must be designed foran additional transverse moment

The design of plain concrete walls in BS 8110 is similar tothat of unreinforced masonry walls in BS 5328 Equations aregiven for the maximum design ultimate axial load taking intoaccount the transverse eccentricity of the load including anadditional eccentricity in the case of slender walls The basicrequirements for the design of reinforced and plain concretewalls are summarised in Table 360

62 BRIDGES

As stated in section 248 the analysis and design of bridges isnow so complex that it cannot be adequately covered in a bookof this type and reference should be made to specialist publi-cations However for the guidance of designers who may haveto deal with structures having features in common with bridgesbrief notes on some aspects of their design and constructionare provided Most of the following information is taken fromref 52 which also contains other references for further reading

621 Types of bridges

For short spans the simplest and most cost-effective form ofdeck construction is a cast in situ reinforced concrete solid slabSingle span slabs are often connected monolithically to theabutments to form a portal frame A precast box-shaped rein-forced concrete culvert can be used as a simple form of framedbridge and is particularly economical for short span (up toabout 6 m) bridges that have to be built on relatively poorground obviating the need for piled foundations

As the span increases the high self-weight of a solid slabbecomes a major disadvantage The weight can be reduced byproviding voids within the slab using polystyrene formersThese are usually of circular section enabling the concrete to

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flow freely under them to the deck soffit Reinforced concretevoided slabs are economical for spans up to about 25 m Theintroduction of prestressing enables such construction to beeconomical over longer spans and prestressed voided slabswith internal bonded tendons can be used for spans up toabout 50 m If a bridge location does not suit cast in situ slabconstruction precast concrete beams can be used Severaldifferent types of high quality factory-made components thatcan be rapidly erected on site are manufactured Precast beamconstruction is particularly useful for bridging over live roadsrailways and waterways where any interruptions to trafficmust be minimised Pre-tensioned inverted T-beams placedside-by-side and then infilled with concrete provide a viablealternative to a reinforced concrete solid slab for spans upto about 18 m Composite forms of construction consistingtypically of a 200 mm thick cast in situ slab supported onpre-tensioned beams spaced at about 15 m centres can be usedfor spans in the range 12ndash40 m

For very long spans prestressed concrete box girders are theusual form for bridge decks ndash the details of the design beingdictated by the method of construction The span-by-spanmethod is used in multi-span viaducts with individual spans ofup to 60 m A span plus a cantilever of about one quarter thenext span is first constructed This is then prestressed and thefalsework moved forward after which a full span length isformed and stressed back to the previous cantilever In situ con-struction is used for smaller spans but as spans increase so alsodoes the cost of the falsework To minimise the cost the weightof the concrete to be supported at any one time is reduced bydividing each span into a series of transverse segments Thesesegments which can be cast in situ or precast are normallyerected on either side of each pier to form balanced cantileversand then stressed together Further segments are then addedextending the cantilevers to mid-span where an in situ concreteclosure is formed to make the spans continuous During erectionthe leading segments are supported from gantries erected on thepiers or completed parts of the deck and work can advancesimultaneously on several fronts When the segments are precasteach unit is match-cast against the previous one and thenseparated for transportation and erection Finally an epoxyresin is applied to the matching faces before the units arestressed together

Straight or curved bridges of single radius and of constantcross section can also be built in short lengths from one orboth ends The bridge is then pushed out in stages from theabutments a system known as incremental launching Archbridges in spans up to 250 m and beyond can be constructedeither in situ or using precast segments which are prestressedtogether and held on stays until the whole arch is complete

For spans in excess of 250 m the decks of suspensionand cable-stayed bridges can be of in situ concrete ndash constructedusing travelling formwork ndash or of precast segments stressedtogether For a comprehensive treatment of the aestheticsand design of bridges by one of the worldrsquos most eminentbridge engineers see ref 53 Brief information on typicalstructural forms and span ranges is given in Table 298

622 Substructures

A bridge is supported at the ends on abutments and may haveintermediate piers where the positions of the supports and the

lengths of the spans are determined by the topography of theground and the need to ensure unimpeded traffic under thebridge The overall appearance of the bridge structure is verydependent on the relative proportions of the deck and itssupports The abutments are usually constructed of reinforcedconcrete but in some circumstances mass concrete withoutreinforcement can provide a simple and durable solution

Contiguous bored piles or diaphragm walling can be used toform an abutment wall in cases where the wall is to be formedbefore the main excavation is carried out Although the cost ofthis type of construction is high it can be offset against savingsin the amount of land required the cost of temporary works andconstruction time A facing of in situ or precast concrete orblockwork will normally be required after excavation Reinforcedearth construction can be used where there is an embankmentbehind the abutment in which case a precast facing is oftenapplied The selection of appropriate ties and fittings is partic-ularly important since replacement of the ties during the life ofthe structure is very difficult

Where a bridge is constructed over a cutting it is usuallypossible to form a bank-seat abutment on firm undisturbedground Alternatively bank seats can be constructed on piledfoundations However where bridges over motorways aredesigned to allow for future widening of the carriageway theabutment is likely to be taken down to full depth so that it canbe exposed at a later date when the widening is carried out

The design of wing walls is determined by the topography ofthe site and can have a major effect on the appearance of thebridge Wing walls are often taken back at an angle from theface of the abutment for both economy and appearance Castin situ concrete is normally used but precast concrete retainingwall units are also available from manufacturers Concrete cribwalling is also used and its appearance makes it particularlysuitable for rural situations Filling material must be carefullyselected to ensure that it does not flow out and the fill mustbe properly drained It is important to limit the differentialsettlement that could occur between an abutment and its wingwalls The problem can be avoided if the wing walls cantileverfrom the abutment and the whole structure is supported onone foundation

The simplest and most economic form of pier is a verticalmember or group of members of uniform cross section Thismight be square rectangular circular or elliptical Shaping ofpiers can be aesthetically beneficial but complex shapes willsignificantly increase the cost unless considerable reuse of theforms is possible Raking piers and abutments can help toreduce spans for high bridges but they also require expensivepropping and support structures This in turn complicates theconstruction process and considerably increases costs

The choice of foundation to abutments and piers is usuallybetween spread footings and piling Where ground conditionspermit a spread footing will provide a simple and economicsolution Piling will be needed where the ground conditionsare poor and cannot be improved the bridge is over a river orestuary the water table is high or site restrictions prevent theconstruction of a spread footing It is sometimes possible toimprove the ground by consolidating grouting or applying asurcharge by constructing the embankments well in advance ofthe bridge structure Differential settlement of foundations willbe affected by the construction sequence and needs to becontrolled In the early stages of construction the abutments

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are likely to settle more than the piers but the piers will settlelater when the deck is constructed

623 Integral bridges

For road bridges in the United Kingdom experience has shownthat with all forms of construction continuous structures aregenerally more durable than structures with discontinuous spansThis is mainly because joints between spans have often allowedsalty water to leak through to piers and abutments HighwaysAgency standard BD 5701 says that in principle all bridgesshould be designed as continuous over intermediate supportsunless special circumstances exist The connections betweenspans may be made to provide full structural continuity or inbeam and slab construction continuity of the deck slab only

Bridges with lengths up to 60 m and skews up to 30o shouldalso be designed as integral bridges in which the abutmentsare connected directly to the deck and no movement joints areprovided to allow for expansion or contraction When the designerconsiders that an integral bridge is inappropriate the agreementof the overseeing organisation must be obtained HighwaysAgency document BA 5701 has figures indicating a variety ofcontinuity and abutment details

624 Design considerations

Whether the bridge is carrying a road railway waterway or justpedestrians it will be subject to various types of load

Self-weight and loads from surfacing parapets and so on

Environmental (eg wind snow temperature effects)

Traffic

Accidental loads (eg impact)

Temporary loads (during construction and maintenance)

Bridges in the United Kingdom are generally designed to therequirements of BS 5400 and several related Highways Agencystandards Details of the traffic loads to be considered forroad railway and footbridges are given in section 248 andTables 25 and 26 Details of structural design requirementsincluding the load combinations to be considered are given insection 212 and Tables 32 and 33

The application of traffic load to any one area of a bridgedeck causes the deck to bend transversely and twist therebyspreading load to either side The assessment of how much ofthe load is shared in this way and the extent to which it isspread across the deck depends on the bending torsion andshear stiffness of the deck in the longitudinal and transversedirections Computer methods are generally used to analysethe structure for load effects the most versatile method beinggrillage analysis which treats the deck as a two-dimensionalseries of beam elements in both directions This method canbe used for solid slab beam and slab and voided slabs wherethe cross-sectional area of the voids does not exceed 60 ofthe area of the deck Box girders are now generally formed asone or two cells without any transverse diaphragms These areusually quite stiff in torsion but can distort under load givingrise to warping stresses in the walls and slabs of the box It isthen necessary to use three-dimensional analytical methodssuch as 3D space frame folded plate (for decks of uniformcross section) or the generalised 3D finite element method

An excellent treatment of the behaviour and analysis of bridgedecks is provided in ref 54

It is usual to assume that movement of abutments and wingwalls will occur and to take these into account in the designof the deck and the substructure Normally the backfill used isa free-draining material and satisfactory drainage facilities areprovided If these conditions do not apply then higher designpressures must be considered Due allowance must be madealso for the compaction of the fill during construction and thesubsequent effects of traffic loading The Highways Agencydocument BA 4296 shows several forms of integral abutmentwith guidance on their behaviour Abutments to frame bridges areconsidered to rock bodily under the effect of deck movementsEmbedded abutments such as piled and diaphragm wallsare considered to flex and pad foundations to bank seats areconsidered to slide Notional earth pressure distributionsresulting from deck expansion are also given for frame andembedded abutments

Creep shrinkage and temperature movements in bridgedecks can all affect the forces applied to the abutments Piersand to a lesser extent abutments are vulnerable to impact loadsfrom vehicles or shipping and must be designed to resistimpact or be protected from it Substructures of bridges overrivers and estuaries are also subjected to scouring and lateralforces due to water flow unless properly protected

625 Waterproofing of bridge decks

Over the years mastic asphalt has been extensively used forwaterproofing bridge decks but good weather conditions arerequired if it is to be laid satisfactorily Preformed bituminoussheeting is less sensitive to laying conditions but moisturetrapped below the sheeting can cause subsequent lifting Theuse of hot-bonded heavy-duty reinforced sheet membranes ifproperly laid can provide a completely water-tight layer Thesheets which are 3ndash4 mm in thickness have good punctureresistance and it is not necessary to protect the membrane fromasphalt laid on top Sprayed acrylic and polyurethane water-proofing membranes are also used These bond well to theconcrete deck surface with little or no risk of blowing or liftingA tack coat must be applied over the membrane and a protec-tive asphalt layer is placed before the final surfacing is carriedout Some bridges have depended upon the use of a dense highquality concrete to resist the penetration of water without anapplied waterproofing layer In such cases it can be advanta-geous to include silica fume or some similar very fine powderedaddition in the concrete

63 CONTAINMENT STRUCTURES

Weights of stored materials are given in EC 1 Part 11 and thecalculation of horizontal pressures due to liquids and granularmaterials contained in tanks reservoirs bunkers and silosis explained in sections 92 and 93 in conjunction withTables 215 and 216 This section deals with the design ofcontainment structures and the calculation of the forces andbending moments produced by the pressure of the containedmaterials Where containers are required to be watertight thestructural design should follow the recommendations givenin either BS 8007 or EC 2 Part 3 as indicated in sections 213and 294 respectively In the following notes containers are

Containment structures 59

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conveniently classified as either tanks containing liquids orbunkers and silos containing dry materials

631 Underground tanks

Underground storage tanks are subjected to external pressuresdue to the surrounding earth in addition to internal waterpressure The empty structure should also be investigated forpossible flotation if the earth can become waterlogged Earthpressure at-rest conditions should generally be assumed fordesign purposes but for reservoirs where the earth is banked upagainst the walls it would be more reasonable to assume activeconditions Storage tanks are normally filled to check for water-tightness before any backfill material is placed and there isalways a risk that such material could be excavated in the futureTherefore no reduction to the internal hydrostatic pressure byreason of the external earth pressure should be made when atank is full

The earth covering on the roof of a reservoir in its final stateacts uniformly over the entire area but it is usually sensible totreat it as an imposed load This is to cater for non-uniformconditions that can occur when the earth is being placed inposition and if it becomes necessary to remove the earth formaintenance purposes Problems can arise in partially buriedreservoirs due to solar radiation causing thermal expansion ofthe roof The effect of such movement on a perimeter wall willbe minimised if no connection is made between the roof and thewall until reflective gravel or some other protective materialhas been placed on the roof Alternatively restraint to thedeflection of the wall can be minimised by providing a durablecompressible material between the wall and the soil Thisprevents the build-up of large passive earth pressures in theupper portion of the soil and allows the wall to deflect as a longflexible cantilever

632 Cylindrical tanks

The wall of a cylindrical tank is primarily designed to resist ringtensions due to the horizontal pressures of the contained liquidIf the wall is free at the top and free-to-slide at the bottom thenwhen the tank is full the ring tension at depth z is given bynzr where is the unit weight of liquid and r is the internalradius of the tank In this condition when the tank is full novertical bending or radial shear exists

If the wall is connected to the floor in such a way that noradial movement occurs at the base the ring tension will be zeroat the bottom of the wall The ring tensions are affectedthroughout the lower part of the wall and significant verticalbending and radial shear occurs Elastic analysis can be usedto derive equations involving trigonometric and hyperbolicfunctions and solutions expressed in the form of tables areincluded in publications (eg refs 55 and 56) Coefficients todetermine values of circumferential tensions vertical bendingmoments and radial shears for particular values of the termheight2(2 mean radius thickness) are given in Tables 275and 276

The tables apply to idealised boundary conditions in whichthe bottom of the wall is either hinged or fixed It is possible todevelop these conditions if an annular footing is provided at thebottom of the wall The footing should be tied into the floor ofthe tank to prevent radial movement If the footing is narrow

there will be little resistance to rotation and a hinged conditioncould be reasonably assumed It is also possible to form a hingeby providing horizontal grooves at each side of the wall so thatthe contact between the wall and the footing is reduced to anarrow throat The vertical bars are then bent to cross over atthe centre of the wall but this detail is rarely used At the otherextreme if the wall footing is made wide enough it is possibleto get a uniform distribution of bearing pressure In this casethere will be no rotation and a fixed condition can be assumedIn many cases the wall and the floor slab are made continuousand it is necessary to consider the interaction between the twoelements Appropriate values for the stiffness of the memberand the effect of edge loading can be obtained from Tables 276and 277

For slabs on an elastic foundation the values depend on theratio rrk where rk is the radius of relative stiffness defined insection 725 The value of rk is dependent on the modulus ofsubgrade reaction for which data is given in section 724Taking rrk 0 which corresponds to a lsquoplasticrsquo soil state isappropriate for an empty tank liable to flotation

633 Octagonal tanks

If the wall of a tank forms in plan a series of straight sidesinstead of being circular the formwork may be less costly butextra reinforcement and possibly an increased thickness ofconcrete is needed to resist the horizontal bending momentsthat are produced in addition to the ring tension If the tankforms a regular octagon the bending moments in each side areq l212 at the corners and q l 224 at the centre where l is thelength of the side and q is the lsquoeffectiversquo lateral pressure at depthz If the wall is free at the top and free-to-slide at the bottomq z In other cases q nr where n is the ring tension atdepth z and r is the lsquoeffectiversquo radius (ie half the distancebetween opposite sides) If the tank does not form a regularoctagon but the length and thickness of the sides are alternatelyl1 h1 and l2 h2 the horizontal bending moment at the junctionof any two sides is

634 Rectangular tanks

The walls of large rectangular reservoirs are sometimes built indiscontinuous lengths in order to minimise restraints to theeffects of early thermal contraction and shrinkage If the wallbase is discontinuous with the main floor slab each wall unit isdesigned to be independently stable and no slip membrane isprovided between the wall base and the blinding concreteAlternatively the base to each wall unit can be tied into theadjacent panel of floor slab Roof slabs can be connected to theperimeter walls or simply supported with a sliding jointbetween the top of the wall and the underside of the slab Insuch forms of construction except for the effect of any cornerjunctions the walls span vertically either as a cantilever orwith ends that are simply supported or restrained depending onthe particular details

A cantilever wall is statically determinate and if supportinga roof is also isolated from the effect of roof movement Thedeflection at the top of the wall is an important consideration

q12l1

3 l23h1

h23 l1 l2h1

h23

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and the base needs to be carefully proportioned in order tominimise the effect of base tilting The problem of excessivedeflection can be overcome and the wall thickness reduced ifthe wall is tied into the roof If the wall is also provided with anarrow footing tied into the floor it can be designed as simplysupported although considerable reliance is being put in theability of the joint to accept continual rotation If the wallfooting is made wide enough it is possible to obtain a uniformdistribution of bearing pressure in which case there will be norotation and a fixed condition can be assumed In cases wherethe wall and floor slab are made continuous the interactionbetween the two elements should be considered

Smaller rectangular tanks are generally constructed withoutmovement joints so that structural continuity is obtained inboth horizontal and vertical planes Bending moments andshear forces in individual rectangular panels with idealisededge conditions when subjected to hydrostatic loading aregiven in Table 253 For a rectangular tank distribution of theunequal fixity moments obtained at the wall junctions isneeded and moment coefficients for tanks of different spanratios are given in Tables 278 and 279 The shearing forcesgiven in Table 253 for individual panels may still be used

The tables give values for tanks where the top of the wall iseither hinged or free and the bottom is either hinged or fixedThe edge conditions are generally uncertain and tend to varywith the loading conditions as discussed in section 172 Forthe horizontal spans the shear forces at the vertical edges ofone wall result in axial forces in the adjacent walls Thus forinternal loading the shear force at the end of a long wall isequal to the tensile force in the short wall and vice versa Indesigning sections the combined effects of bending momentaxial force and shear force need to be considered

635 Elevated tanks

The type of bottom provided to an elevated cylindrical tankdepends on the diameter of the tank and the depth of water Forsmall tanks a flat beamless slab is satisfactory but beams arenecessary for tanks exceeding about 3 m diameter Someappropriate examples which include bottoms with beams anddomed bottoms are included in section 174 and Table 281

It is important that there should be no unequal settlement ofthe foundations of columns supporting an elevated tank and araft should be provided in cases where such problems couldoccur In addition to the bending moments and shear forces dueto the wind pressure on the tank as described in sections 25and 83 the wind force causes a thrust on the columns on theleeward side and tension in the columns on the windward sideThe values of the thrusts and tensions can be calculated fromthe expressions given for columns supporting elevated tanks insection 1742

636 Effects of temperature

If the walls of a tank are subjected to significant temperatureeffects due to solar radiation or the storage of warm liquids theresulting moments and forces need to be determined by anappropriate analysis The structure can usually be analysedseparately for temperature change (expansion or contraction)and temperature differential (gradient through section) For awall with all of the edges notionally clamped the temperature

differential results in bending moments causing compressionon the warm face and tension on the cold face given by

where E is the modulus of elasticity of concrete I is secondmoment of area of the section h is thickness of wall is thecoefficient of thermal expansion of concrete 13 is temperaturedifference between the two surfaces is Poissonrsquos ratio Forcracked sections may be taken as zero but the value of I shouldallow for the tension stiffening effect of the concrete The effectof releasing the notional restraints at edges that are free orhinged modifies the moment field and in cylindrical tankscauses additional ring tensions For further information onthermal effects in cylindrical tanks reference can be made toeither the Australian or the New Zealand standard Code ofPractice for liquid-retaining concrete structures

64 SILOS

Silos which may also be referred to as bunkers or bins aredeep containers used to store particulate materials In a deepcontainer the linear increase of pressure with depth found inshallow containers is modified Allowances are made for theeffects of filling and unloading as described in section 277The properties of materials commonly stored in silos andexpressions for the pressures set up in silos of different formsand proportions are given in Tables 215 and 216

641 Walls

Silo walls are designed to resist the bending moments andtensions caused by the pressure of the contained material If thewall spans horizontally it is designed for the combined effectsIf the wall spans vertically horizontal reinforcement is neededto resist the axial tension and vertical reinforcement to resist thebending In this case the effect of the horizontal bendingmoments due to continuity at the corners should also beconsidered For walls spanning horizontally the bendingmoments and forces depend on the number and arrangement ofthe compartments Where there are several compartmentsthe intermediate walls act as ties between the outer walls Forvarious arrangements of intermediate walls expressions forthe negative bending moments on the outer walls of the silosare given in Table 280 Corresponding expressions for thereactions which are a measure of the axial tensions in thewalls are also given The positive bending moments can bereadily calculated when the negative bending moments atthe wall corners are known An external wall is subjected tothe maximum combined effects when the adjacent compartmentis full An internal cross-wall is subjected to the maximumbending moments when the compartment on one side of thewall is full and to maximum axial tension (but zero bending)when the compartments on both sides are full In small silosthe proportions of the wall panels may be such that they spanboth horizontally and vertically in which case Table 253 canbe used to calculate the bending moments

In the case of an elevated silo the whole load is generallytransferred to the columns by the walls and when the clear spanis greater than twice the depth the wall can be designed as ashallow beam Otherwise the recommendations for deep beamsshould be followed (see section 58 and ref 43) The effect of

M EI13(1v)h

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wind loads on large structures should be calculated The effectof both the tensile force in the windward walls of the empty siloand the compressive force in the leeward walls of the full siloare important In the latter condition the effect of the eccentricforce on the inside face of the wall due to the proportion of theweight of the contents supported by friction must be combinedwith the force due to the wind At the base and the top ofthe wall there are additional bending effects due to continuityof the wall with the bottom and the covers or roof over thecompartments

642 Hopper bottoms

The design of sloping hopper bottoms in the form of invertedtruncated pyramids consists of finding for each sloping sidethe centre of pressure the intensity of pressure normal to theslope at this point and the mean span The bending momentsat the centre and edge of each sloping side are calculated Thehorizontal tensile force is computed and combined with thebending moment to determine the horizontal reinforcementrequired The tensile force acting along the slope at the centreof pressure is combined with the bending moment at this pointto find the inclined reinforcement needed in the bottom of theslab At the top of the slope the bending moment and theinclined component of the hanging-up force are combined todetermine the reinforcement needed in the top of the slab

For each sloping side the centre of pressure and the mean spancan be obtained by inscribing on a normal plan a circle thattouches three of the sides The diameter of this circle is the meanspan and its centre is the centre of pressure The total intensityof load normal to the slope at this point is the sum of the normalcomponents of the vertical and horizontal pressures and the deadweight of the slab Expressions for determining the pressures onthe slab are given in Table 216 Expressions for determiningthe bending moments and tensile forces acting along the slopeand horizontally are given in Table 281 When using thismethod it should be noted that although the horizontal span ofthe slab reduces considerably towards the outlet the amount

of reinforcement should not be reduced below that calculatedfor the centre of pressure This is because in determining thebending moment based on the mean span adequate transversesupport from reinforcement towards the base is assumed

The hanging-up force along the slope has both vertical andhorizontal components the former being resisted by the wallsacting as beams The horizontal component acting inwardstends to produce horizontal bending moments on the beam atthe top of the slope but this is opposed by a correspondingoutward force due to the pressure of the contained material Thelsquohip-beamrsquo at the top of the slope needs to be designed both toresist the inward pull from the hopper bottom when the hopperis full and the silo above is only partly filled and also for thecase when the arching of the fill concentrates the outward forcesdue to the peak lateral pressure on the beam during unloadingThis is especially important in the case of mass-flow silos(see section 277)

65 BEARINGS HINGES AND JOINTS

In the construction of frames and arches hinges are needed atpoints where it is assumed that there is no bending moment Inbridges bearings are often required at abutments and piers totransfer loads from the deck to the supports Various types ofbearings and hinges for different purposes are illustrated inTable 299 with associated notes in section 1941

Movement joints are often required in concrete structures toallow free expansion and contraction Fluctuating movementsoccur due to diurnal solar effects and seasonal changes ofhumidity and temperature Progressive movements occur due toconcrete creep drying shrinkage and ground settlementMovement joints may also be provided in structures wherebecause of abrupt changes of loading or ground conditionspronounced changes occur in the size or type of foundationVarious types of joints for different purposes are illustrated inTable 2100 with associated notes on their construction andapplication in section 1942

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71 FOUNDATIONS

The design of the foundations for a structure comprises threestages The first is to determine from an inspection of the sitetogether with field data on soil profiles and laboratory testing ofsoil samples the nature of the ground The second stage is toselect the stratum on which to impose the load the bearingcapacity and the type of foundation These decisions dependnot only on the nature of the ground but also on the type ofstructure and different solutions may need to be consideredReference should be made to BS 8004 Code of Practice forfoundations The third stage is to design the foundation totransfer and distribute load from the structure to the ground

711 Site inspection

The objective of a site inspection is to determine the nature ofthe top stratum and the underlying strata in order to detect anyweak strata that may impair the bearing capacity of the stratumselected for the foundation Generally the depth to whichknowledge of the strata is obtained should be not less than oneand a half times the width of an isolated foundation or thewidth of a structure with closely spaced footings

The nature of the ground can be determined by digging trialholes by sinking bores or by driving piles A trial hole can betaken down to only moderate depths but the undisturbed soilcan be examined and the difficulties of excavation with theneed or otherwise of timbering and groundwater pumping canbe determined Bores can be taken very much deeper than trialholes and stratum samples at different depths obtained forlaboratory testing A test pile does not indicate the type ofsoil it has been driven through but it is useful in showing thethickness of the top crust and the depth below poorer soil atwhich a firm stratum is found A sufficient number of any ofthese tests should be taken to enable the engineer to ascertainthe nature of the ground under all parts of the foundationsReference should be made to BS 5930 Code of practice for siteinvestigations and BS 1377 Methods of test for soils for civilengineering purposes

712 Bearing pressures

The pressure that can be safely imposed on a thick stratum ofsoil commonly encountered is in some districts stipulated in

local by-laws The pressures recommended for preliminarydesign purposes in BS 8004 are given in Table 282 but thesevalues should be used with caution since several factors cannecessitate the use of lower values Allowable pressures maygenerally be exceeded by the weight of soil excavated down tothe foundation level but if this increase is allowed any fillmaterial applied on top of the foundation must be included inthe total load If the resistance of the soil is uncertain a studyof local records for existing buildings on the same soil can beuseful as may the results of a ground-bearing test

Failure of a foundation can occur due to consolidation of theground causing settlement or rupture of the ground due toshearing The shape of the surface along which shear failureoccurs under a strip footing is an almost circular arc startingfrom one edge of the footing passing under the footing andthen continuing as a tangent to the arc to intersect the groundsurface at an angle depending on the angle of internal frictionof the soil Thus the average shear resistance depends onthe angle of internal resistance of the soil and on the depthof the footing below the ground surface In a cohesionless soilthe bearing resistance not only increases as the depth increasesbut is proportional to the width of the footing In a cohesive soilthe bearing resistance also increases with the width of footingbut the increase is less than for a non-cohesive soil

Except when bearing directly on rock foundations for all butsingle-storey buildings or other light structures should betaken down at least 1 m below the ground surface in order toobtain undisturbed soil that is sufficiently consolidated In claysoils a depth of at least 15 m is needed in the United Kingdomto ensure protection of the bearing stratum from weathering

713 Eccentric loads

When a rigid foundation is subjected to concentric loadingthat is when the centre of gravity of the loads coincides withthe centre of area of the foundation the bearing pressure on theground is uniform and equal to the total applied load divided bythe total area When a load is eccentrically placed on a base ora concentric load and a bending moment are applied to a basethe bearing pressure is not uniform For a load that is eccentricabout one axis of a rectangular base the bearing pressure variesfrom a maximum at the side nearer the centre of gravity of theload to a minimum at the opposite side or to zero at some inter-mediate position The pressure variation is usually assumed to

Chapter 7

Foundations groundslabs retaining wallsculverts and subways

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Foundations ground slabs retaining walls culverts and subways64

be linear in which case the maximum and minimum pressuresare given by the formulae in Table 282 For large eccentricitiesthere may be a part of the foundation where there is no bearingpressure Although this state may be satisfactory for transientconditions (such as those due to wind) it is preferable for thefoundation to be designed so that contact with the ground existsover the whole area under normal service conditions

714 Blinding layer

For reinforced concrete footings or other construction wherethere is no underlying mass concrete forming an integral part ofthe foundation the bottom of the excavation should be coveredwith a layer of lean concrete to protect the soil and provide aclean surface on which to place the reinforcement The thicknessof this blinding layer is typically 50ndash75 mm depending on thesurface condition of the excavation

715 Foundation types

The most suitable type of foundation depends primarily on thedepth at which the bearing stratum lies and the allowable bearingpressure which determines the foundation area Data relatingto some common types of separate and combined pad founda-tions suitable for sites where the bearing stratum is found closeto the surface are given in Tables 282 and 283 Several typesof inter-connected bases and rafts are given in Table 284 Inchoosing a foundation suitable for a particular purpose thenature of the structure should also be considered Sometimes itmay be decided to accept the risk of settlement in preference toproviding a more expensive foundation For silos and fixed-endarches the risk of unequal settlement of the foundations mustbe avoided at all costs but for gantries and the bases of largesteel tanks a simple foundation can be provided and probablesettlement allowed for in the design of the superstructure Inmining districts where it is reasonable to expect some subsidencea rigid raft foundation should be provided for small structuresto allow the structure to move as a whole For large structuresa raft may not be economical and the structure should bedesigned either to be flexible or as several separate elementson independent raft foundations

716 Separate bases

The simplest form of foundation for an individual column orstanchion is a reinforced concrete pad Such bases are widelyused on ground that is strong and on weaker grounds wherethe structure and the cladding are light and flexible For basesthat are small in area or founded on rock a block of plain ornominally reinforced concrete can be used The thickness ofthe block is made sufficient for the load to be transferred to theground under the base at an angle of dispersion through theblock of not less than 45o to the horizontal

To reduce the risk of unequal settlement the column basesizes for a building founded on a compressible soil should be inproportion to the dead load carried by each column Bases for thecolumns of a storage structure should be in proportion to the totalload excluding the effects of wind In all cases the pressure onthe ground under any base due to combined dead and imposedload including wind load and any bending at the base of thecolumn should not exceed the allowable bearing value

In the design of a separate base the area of a concentricallyloaded base is determined by dividing the maximum serviceload by the allowable bearing pressure The subsequentstructural design is then governed by the requirements of theultimate limit state The base thickness is usually determined byshear considerations governed by the more severe of two con-ditions ndash either shear along a vertical section extending acrossthe full width of the base or punching shear around the loadedarea ndash where the second condition is normally critical Thecritical section for the bending moment at a vertical sectionextending across the full width of the base is taken at the faceof the column for a reinforced concrete column and at the cen-tre of the base for a steel stanchion The tension reinforcementis usually spread uniformly over the full width of the base butin some cases it may need to be arranged so that there is aconcentration of reinforcement beneath the column Outsidethis central zone the remaining reinforcement must still con-form to minimum requirements It is also necessary for tensionreinforcement to comply with the bar spacing limitations forcrack control

If the base cannot be placed centrally under the column thebearing pressure varies linearly The base is then preferablyrectangular and modified formulae for bearing pressures andbending moments are given in Table 282 A base supportingfor example a column of a portal frame may be subjected to anapplied moment and horizontal shear force in addition to avertical load Such a base can be made equivalent to a base witha concentric load by placing the base under the column with aneccentricity that offsets the effect of the moment and horizontalforce This procedure is impractical if the direction of theapplied moment and horizontal force is reversible for exampledue to wind In this case the base should be placed centrallyunder the column and designed as eccentrically loaded for thetwo different conditions

717 Combined bases

If the size of the bases required for adjacent columns is suchthat independent bases would overlap two or more columnscan be provided with a common foundation Suitable typesfor two columns are shown in Table 283 for concentrically andeccentrically loaded cases Reinforcement is required top andbottom and the critical condition for shear is along a verticalsection extending across the full width of the base For someconditions of loading on the columns the total load on the basemay be concentric while for other conditions the total load iseccentric and both cases have to be considered Some notes oncombined bases are given in section 1812

718 Balanced and coupled bases

When it is not possible to place an adequate base centrallyunder a column owing to restrictions of the site and when forsuch conditions the eccentricity would result in inadmissibleground pressures a balanced foundation as shown in Tables 283and 284 and described in section 1813 is provided A beamis introduced and the effect of the cantilever moment causedby the offset column load is counterbalanced by load from anadjacent column This situation occurs frequently for externalcolumns of buildings on sites in built-up areas

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Sometimes as in the case of bases under the towers of atrestle or gantry pairs of bases are subjected to moments andhorizontal forces acting in the same direction on each base Insuch conditions the bases can be connected by a stiff beam thatconverts the effects of the moments and horizontal forces intoequal and opposite vertical reactions then each base can bedesigned as concentrically loaded Such a pair of coupled basesis shown in Table 283 which also gives formulae for thereactions and the bending moments on the beam

719 Strip bases and rafts

When the columns or other supports of a structure are closelyspaced in one direction it is common to provide a continuousbase similar to a footing for a wall Particulars of the designof strip bases are given in Table 283 Some notes on thesebases in relation to the diagrams in Table 284 together with anexample are given in section 1812

When the columns or other supports are closely spaced intwo directions or when the column loads are so high and theallowable bearing pressure is so low that a group of separatebases would totally cover the space between the columns asingle raft foundation of one of the types shown at (a)ndash(d) inTable 284 should be provided Notes on these designs are givenin section 1814

The analysis of a raft foundation supporting a set of equalloads that are symmetrically arranged is usually based on theassumption of uniformly distributed pressure on the groundThe design is similar to that for an inverted floor upon whichthe load is that portion of the ground pressure that is due to theconcentrated loads only Notes on the design of a raft for whichthe columns are not symmetrically disposed are also includedin section 1814 An example of the design of a raft foundationis given in Examples of the Design of Buildings

7110 Basements

The floor of a basement for which a typical cross section isshown at (e) in the lower part of Table 284 is typically a raftsince the weights of the ground floor over the basementthe walls and other structure above the ground floor and thebasement itself are carried on the ground under the floor ofthe basement For water-tightness it is common to construct thewall and the floor of the basement monolithically In mostcases although the average ground pressure is low the spansare large resulting in high bending moments and a thick floorif the total load is taken as uniform over the whole area Sincethe greater part of the load is transmitted through the wallsand any internal columns it is more rational and economicalto transfer the load on strips and pads placed immediatelyunder the walls and columns The resulting cantilever actiondetermines the required thickness of these portions and theremainder of the floor can generally be made thinner

Where basements are in water-bearing soils the effect ofhydrostatic pressure must be taken into account The upwardwater pressure is uniform below the whole area of the floorwhich must be capable of resisting the total pressure lessthe weight of the floor The walls must be designed to resist thehorizontal pressures due to the waterlogged ground and thebasement must be prevented from floating Two conditions needto be considered Upon completion the total weight of the

basement and superimposed dead load must exceed the worstcredible upward force due to the water by a substantial marginDuring construction there must always be an excess ofdownward load If these conditions cannot be satisfied oneof the following steps should be taken

1 The level of the groundwater near the basement should becontrolled by pumping or other means

2 Temporary vents should be formed in the basement floor orat the base of the walls to enable water to freely enter thebasement thereby equalising the external and internalpressures The vents should be sealed when sufficient deadload from the superstructure has been obtained

3 The basement should be temporarily flooded to a depth suchthat the weight of water in the basement together with thedead load exceeds the total upward force on the structure

While the basement is under construction method 1 normallyhas to be used but once the basement is complete method 3 hasthe merit of simplicity Basements are generally designedand constructed in accordance with the recommendations ofBS 8102 supplemented by the guidance provided in reportsproduced by CIRIA (ref 57) BS 8102 defines four gradesof internal environment each grade requiring a different levelof protection against water and moisture ingress Three types ofconstruction are described to provide either A tanked or Bintegral or C drained protection

Type A refers to concrete or masonry construction whereadded protection is provided by a continuous barrier system Anexternal tanking is generally preferred so that any externalwater pressure will force the membrane against the structureThis is normally only practicable where the construction is byconventional methods in excavation that is open or supportedby temporary sheet piling The structure should be monolithicthroughout and special care should be taken when a structureis supported on piles to avoid rupture of the membrane due tosettlement of the fill supported by the basement wall

Type B refers to concrete construction where the structureitself is expected to be sufficient without added protection Astructure designed to the requirements of BS 8007 is expectedto inhibit the ingress of water to the level required for a utilitygrade basement It is considered that this standard can also beachieved in basements constructed by using diaphragm wallssecant pile walls and permanent sheet piling If necessary theperformance can be improved by internal ventilation and theaddition of a vapour-proof barrier

Type C refers to concrete or masonry construction whereadded protection is provided by an internal ventilated drainedcavity This method is applicable to all types of constructionand can provide a high level of protection It is particularlyuseful for deep basements using diaphragm walls secant pilewalls contiguous piles or steel sheet piling

7111 Foundation piers

When a satisfactory bearing stratum is found at a depth of15ndash5 m below the natural ground level piers can be formedfrom the bearing stratum up to ground level The constructionof columns or other supporting members can then begin on thetop of the piers at ground level Such piers are generally squarein cross section and most economically constructed in plain

Foundations 65

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concrete When piers are impractical by reason of the depth atwhich a firm stratum occurs or due to the nature of the groundshort bored piles can be used

7112 Wall footings

When the load on a strip footing is distributed uniformly overthe whole length as in the general case of a wall footing theprincipal effects are due to the transverse cantilever action ofthe projecting portion of the footing If the wall is of concreteand built monolithically with the footing the critical bendingmoment is at the face of the wall If the wall is of masonry themaximum bending moment is at the centre of the footingExpressions for these moments are given in Table 283 If theprojection is less than the thickness of the base the transversebending moment may be ignored but the thickness should besuch that the shear strength is not exceeded Whether or not awall footing is designed for transverse bending longitudinalreinforcement is generally included to give some resistance tomoments due to unequal settlement and non-uniformity ofbearing In cases where a deep narrow trench is excavated downto a firm stratum plain concrete fill is normally used

7113 Foundations for machines

The area of a concrete base supporting a machine or enginemust be sufficient to spread the load onto the ground withoutexceeding the allowable bearing value It is advantageous if thecentre of area of the base coincides with the centre of gravityof the loads when the machine is working as this reducesthe risk of unequal settlement If vibration from the machine istransmitted to the ground the bearing pressure should beconsiderably lower than normally taken especially if the groundis clay or contains a large proportion of clay It is often importantthat the vibration of a machine should not be transmitted toadjacent structures either directly or via the ground In suchcases a layer of insulating material should be placed betweenthe concrete base carrying the machine and the groundSometimes the base is enclosed in a pit lined with insulatingmaterial In exceptional cases a machine base may stand onsprings or more elaborate damping devices may be installed Inall cases the base should be separated from any surroundingarea of concrete ground floor

With light machines the ground bearing pressure may not bethe factor that determines the size of the concrete base as thearea occupied by the machine and its frame may require a baseof larger area The position of the holding-down bolts generallydetermines the length and width of the base which shouldextend 150 mm or more beyond the outer edges of the holes leftfor the bolts The depth of the base must be such that the bottomis on a satisfactory bearing stratum and there is enough thick-ness to accommodate the holding-down bolts If the machineexerts an uplift force on any part of the base the dimensions ofthe base must be such that the part that is subjected to uplift hasenough weight to resist the uplift force with a suitable marginof safety All the supports of any one machine should be carriedon a single base and any sudden changes in the depth and widthof the base should be avoided This reduces the risk of fracturesthat might result in unequal settlements which could throw themachine out of alignment Reinforcement should be providedto resist all tensile forces

Advice on the design of reinforced concrete foundations tosupport vibrating machinery is given in ref 58 which givespractical solutions for the design of raft piled and massivefoundations Comprehensive information on the dynamics ofmachine foundations is included in ref 59

7114 Piled foundations

Where the upper soil strata is compressible or too weak tosupport the loads transmitted by a structure piles can be usedto transmit the load to underlying bedrock or a stronger soillayer using end-bearing piles Where bedrock is not located ata reasonable depth piles can be used to gradually transmit thestructural loads to the soil using friction piles

Horizontal forces due to wind loading on tall structures orearth pressure on retaining structures can be resisted by pilesacting in bending or by using raking piles Foundations forsome structures such as transmission towers and the roofs tosports stadiums are subjected to upward forces that can beresisted by tension piles Bridge abutments and piers adjacentto water can be constructed with piled foundations to counterthe possible detrimental effects of erosion

There are two basic categories of piles Displacement pilesare driven into the ground in the form of either a preformedsolid concrete pile or a hollow tube Alternatively a void canbe formed in the ground by driving a closed-ended tube thebottom of which is plugged with concrete or aggregate Thisallows the tube to be withdrawn and the void to be filled withconcrete It also allows the base of the pile to be enlarged inorder to increase the bearing capacity Non-displacement orlsquocast-in-placersquo piles are formed by boring or excavating theground to create a void into which steel reinforcement andconcrete can be placed In some soils the excavation needs tobe supported to stop the sides from falling in this is achievedeither with casings or by the use of drilling mud (bentonite)For further information on piles including aspects such aspile driving load testing and assessment of bearing capacityreference should be made to specialist textbooks (ref 60)

7115 Pile-caps

Rarely does a foundation element consist of a single pile Inmost cases piles are arranged in groups or rows with the topsof the piles connected by caps or beams Generally concrete ispoured directly onto the ground and encases the tops of the pilesto a depth of about 75 mm The thickness of the cap must besufficient to ensure that the imposed load is spread equallybetween the piles For typical arrangements of two to five pilesforming a compact group load can be transmitted by dispersionthrough the cap Inclined struts extending from the load to thetop of each pile are held together by tension reinforcement inthe bottom of the cap to form a space frame The struts areusually taken to intersect at the top of the cap at the centreof the loaded area but expressions have also been developedthat take into account the dimensions of the loaded areaInformation regarding the design of such pile-caps andstandardised arrangements and dimensions for groups of twoto five piles are given in Table 361

The thickness of a pile-cap designed by dispersion theory isusually determined by shear considerations along a verticalsection extending across the full width of the cap If the pile

Foundations ground slabs retaining walls culverts and subways66

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spacing exceeds three pile diameters it is also necessary todesign for punching shear In all cases the shear stress at theperimeter of the loaded area should not exceed the maximumdesign value related to the compressive strength of the strutsThe reinforcement in the bottom of the pile-cap should beprovided at each end with a full tension anchorage measuredfrom the centre of the pile Pile-caps can also be designed bybending theory but this is generally more appropriate where alarge number of piles are involved In such cases punchingshear is likely to be a critical consideration

7116 Loads on piles in a group

If a group of n piles is connected by a rigid pile-cap and thecentres of gravity of the load Fv and the piles are coincident eachpile will be equally loaded and will be subjected to a load FvnIf the centre of gravity of the load is displaced a distance e fromthe centre of gravity of the piles the load on any one pile is

where is the sum of the squares of the distance of each pilemeasured from an axis that passes through the centre of gravityof the group of piles and is at right angles to the line joining thiscentre of gravity and the centre of gravity of the load and a1 isthe distance of the pile considered from this axis (positive if onthe same side of the axis as the centre of gravity of the load andnegative if on the opposite side) If the structure supported onthe group of piles is subjected to a bending moment M whichis transmitted to the foundations the expression given for theload on any pile can be used by substituting e MFv

The total load that can be carried on a group of piles is notnecessarily the safe load calculated for one pile multipliedby the number of piles Some allowance has to be made for theoverlapping of the zones of stress in the soil supporting thepiles The reduction due to this effect is greatest for piles thatare supported mainly by friction For piles supported entirely oralmost entirely by end bearing the maximum safe load on agroup cannot greatly exceed the safe bearing load on the areaof bearing stratum covered by the group

7117 Loads on open-piled structures

The loads and forces to which wharves jetties and similarmaritime structures are subjected are dealt with in section 26Such structures can be solid walls made of plain or reinforcedconcrete as are most dock walls A quay or similar watersidewall is more often a sheet pile-wall as described in section 733or it can be an open-piled structure similar to a jetty The loadson groups of inclined and vertical piles for such structures areconsidered in Table 285

For each probable condition of load the external forces areresolved into horizontal and vertical components Fh and Fvthe points of application of which are also determined If thedirection of action and position are opposite to those shown inthe diagrams the signs in the formulae must be changed It isassumed that the piles are surmounted by a rigid pile-cap orsuperstructure The effects on each pile when all the piles arevertical are based on a simple but approximate method ofanalysis Since a pile offers very little resistance to bending

a2

Fv1n

ea1

a2

structures with vertical piles only are not suitable when Fh isdominant In a group containing inclined piles Fh can beresisted by a system of axial forces and the bending momentsand shear forces in the piles are negligible The analysis used inTable 285 is based on the assumption that each pile is hingedat the head and toe Although this assumption is not accuratethe analysis predicts the behaviour reasonably well Three designsof the same typical jetty using different pile arrangements aregiven in section 182

72 INDUSTRIAL GROUND FLOORS

Most forms of activity in buildings ndash from manufacturingstorage and distribution to retail and recreation ndash need a firmplatform on which to operate Concrete ground floors arealmost invariably used for such purposes Although in manyparts of the world conventional manufacturing activity hasdeclined in recent years there has been a steady growth indistribution warehousing and retail operations to serve theneeds of industry and society The scale of such facilities andthe speed with which they are constructed has also increasedwith higher and heavier racking and storage equipment beingused These all make greater demands on concrete floors Thefollowing information is taken mainly from ref 61 where acomprehensive treatment of the subject will be found

721 Floor uses

In warehouses materials handling equipment is used in twodistinct areas according to whether the movement of traffic isfree or defined In free-movement areas vehicles can travelrandomly in any direction This typically occurs in factoriesretail outlets low-level storage and food distribution centresIn defined-movement areas vehicles use fixed paths in verynarrow aisles This usually occurs where high-level storageracking is being employed and distribution and warehousefacilities often combine areas of free movement for low-levelactivities such as unloading and packing alongside areas ofdefined movement for high-level storage The two floor usesrequire different tolerances on surface regularity

722 Construction methods

A ground-supported industrial floor slab is made up of layersof materials comprising a sub-base a slip membranemethanebarrier and a concrete slab of appropriate thickness providinga suitable wearing surface Various construction methods can beused to form the concrete slab

Large areas of floor up to several thousand square metres inextent can be laid in a continuous operation Fixed forms areused up to 50 m apart at the edges of the area only Concrete isdischarged into the area and spread either manually or bymachine Surface levels are controlled either manually using atarget staff in conjunction with a laser level transmitter or bydirect control of a laser-guided spreading machine After thefloor has been laid and finished the area is sub-divided intopanels typically on a 6 m grid in both directions This is achievedby making saw cuts in the top surface for a depth of at leastone-quarter of the depth of the slab creating a line of weaknessin the slab that induces a crack below the saw cut As a result ofconcrete shrinkage each sawn joint will open by a small amount

Industrial ground floors 67

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With such large-area construction there are limitations on theaccuracy of level and surface regularity that can be achievedand the construction is most commonly used for free-movementfloor areas

The large-area construction method can also be employedwithout sub-dividing the area into small panels In this case nosawn joints are made but steel fibres are incorporated in theconcrete mix to control the distribution and width of the cracksthat occur as a result of shrinkage The formed joints at theedges of the area will typically open by about 20 mm

Floors can also be formed as a series of long strips typically4ndash6 m wide with forms along each side Strips can be laidalternately with infill strips laid later or consecutively orbetween lsquoleave-in-placersquo screed rails Concrete is poured in acontinuous operation in each strip after which transverse sawcuts are made about 6 m apart to accommodate longitudinalshrinkage As formwork can be set to tight tolerances and thedistance between the forms is relatively small the long-stripmethod lends itself to the construction of very flat floors and isparticularly suitable for defined-movement floor areas

723 Reinforcement

Steel fibres usually manufactured from cold-drawn wire arecommonly used in ground-supported slabs The fibres vary inlength up to about 60 mm with aspect ratios (lengthnominaldiameter) from 20 to 100 and a variety of cross sections Inorder to increase pull-out resistance the fibres have enlargedflattened or hooked ends roughened surface textures or wavyprofiles The composite concrete slab can have considerableductility dependent on fibre type dosage tensile strength andanchorage mechanism The ductility is commonly measuredusing the Japanese Standard test method which uses beams ina third-point loading arrangement Load-deflection curves areplotted as the load increases until the first crack and thendecreases with increasing deflection The ductility value isexpressed as the average load to a deflection of 3 mm dividedby the load to first crack This measure is commonly known asthe equivalent flexural strength ratio In large-area floors withshrinkage joints at the edges only fibre dosages in the order of35ndash45 kgm3 are used to control the distribution and width ofcracks In floors with additional sawn joints fibre dosages in therange 20ndash30 kgm3 are typically used

In large area floors with additional sawn joints steel fabricreinforcement (type A) can be placed in the bottom of the slabwith typically 50 mm of cover The proportion of reinforcementused is typically 01ndash0125 of the effective cross section bdwhich is small enough to ensure that the reinforcement willyield at the sawn joints as the concrete shrinks and alsosufficient to provide the slab with adequate rotational capacityafter cracking

724 Modulus of subgrade reaction

For design purposes the subgrade is assumed to be an elasticmedium characterised by a modulus of subgrade reaction ksdefined as the load per unit area causing unit deflection It canbe shown that errors of up to 50 in the value of ks have onlya small effect on the slab thickness required for flexural designHowever deflections are more sensitive to ks values and long-term settlement due to soil consolidation under load can be

Foundations ground slabs retaining walls culverts and subways68

Soil typeValues of ks (MNm3)

Lower Upper

Fine or slightly compacted sand 15 30Well compacted sand 50 100Very well compacted sand 100 150Loam or clay (moist) 30 60Loam or clay (dry) 80 100Clay with sand 80 100Crushed stone with sand 100 150Coarse crushed stone 200 250Well-compacted crushed stone 200 300

much larger than the elastic deflections calculated as part of theslab design

In principle the value of ks used in design should be relatedto the range of influence of the load but it is normal practice tobase ks on a loaded area of diameter 750 mm To this endit is strongly recommended that the value of ks is determinedfrom a BS plate-loading test using a 750 mm diameter plateand a fixed settlement of 125 mm If a smaller plate is used ora value of ks appropriate to a particular area is required thefollowing approximate relationship may be assumed

ks 05(103D)2k075

where D is the diameter of the loaded area and k075 is a valuefor D 075 m This gives values of ksk075 as follows

D (m) 03 045 075 12 infin

ksk075 20 14 10 08 05

In the absence of more accurate information typical values ofks according to the soil type are given in the following table

725 Methods of analysis

Traditionally ground-supported slabs have been designed byelastic methods using equations developed by Westergaard inthe 1920s Such slabs are relatively thick and an assessment ofdeflections and other in-service requirements has generallybeen unnecessary Using plastic methods of analysis thinnerslabs can be designed and the need to investigate in-servicerequirements and load-transfer across joints has become moreimportant The use of plastic analysis assumes that the slab hasadequate ductility after cracking that is it contains sufficientfibres or reinforcement as described in section 723 to give anequivalent flexural strength ratio in the range 03ndash05 Plainconcrete slabs and slabs with less than the minimum recom-mended amounts of fibres or reinforcement should still bedesigned by elastic methods

Westergaard assumed that a ground-supported concrete slabis a homogeneous isotropic elastic solid in equilibrium withthe subgrade reactions being vertical only and proportional tothe deflections of the slab He also introduced the concept of theradius of relative stiffness rk given by the relationship

where Ec is the short-term modulus of elasticity of concreteh is the slab thickness ks is the modulus of subgrade reaction

rk [Ech312(1v2)ks]

025

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and is Poissonrsquos ratio The physical significance of rk isillustrated in the following figure showing the approximatedistribution of elastic bending moments for a single internalconcentrated load The bending moment is positive (tensionat the bottom of the slab) with a maximum value at the loadposition Along radial lines it remains positive reducing to zero atrk from the load It then becomes negative reaching a maximumat 2rk from the load with the maximum negative moment (tensionat the top of the slab) significantly less than the maximumpositive moment The moment approaches zero at 3rk fromthe load

or internally as shown in Table 286 An externally stabilisedsystem uses an external structural wall to mobilise stabilisingforces An internally stabilised system utilises reinforcementsinstalled within the soil and extending beyond the potentialfailure zone

Traditional retaining walls can be considered as externallystabilised systems one of the most common forms being thereinforced concrete cantilever wall Retaining walls on spreadfoundations together with gravity structures support the soilby weight and stiffness to resist forward sliding overturningand excessive soil movements The equilibrium of cantileverwalls can also be obtained by embedment of the lower part ofthe wall Anchored or propped walls obtain their equilibriumpartly by embedment of the lower part of the wall and partlyfrom an anchorage or prop system that provides support to theupper part of the wall

Internally stabilised walls built above ground are known asreinforced soil structures By placing reinforcement within thesoil a composite material can be produced that is strong intension as well as compression A key aspect of reinforced soilwalls is its incremental form of construction being built up alayer at a time starting from a small plain concrete strip footingIn this way construction is always at ground level the structureis always stable and progress can be very rapid The result ofthe incremental construction is that the soil is partitioned witheach layer receiving support from a locally inserted reinforcingelement The process is the opposite of what occurs in aconventional wall where pressures exerted by the backfill areintegrated to produce an overall force to be resisted by the struc-ture The materials used in a reinforced soil structure comprisea facing (usually reinforced concrete) soil reinforcement (inthe form of flat strips anchors or grids made from eithergalvanised steel or synthetic material) and soil (usually a well-graded cohesionless material) Reinforced soil structuresare more economic than equivalent structures using externallystabilised methods

Internal soil stabilisation used in the formation of cuttingsor excavations is known as soil nailing The process is againincremental with each stage of excavation limited in depth sothat the soil is able to support itself The exposed soil face isprotected usually by a covering of light mesh reinforcementand spray applied concrete Holes are drilled into the soil andreinforcement in the form of steel bars installed and groutedWith both reinforced soil and soil nailing great care is takento make sure that the reinforcing members do not corrode ordeteriorate Hybrid systems combining elements of internallyand externally stabilised soils are also used

732 Walls on spread bases

Various walls on spread bases are shown in Table 286 Acantilever wall is suitable for walls of moderate height If thesoil to be retained can be excavated during construction ofthe wall or the wall is required to retain an embankment thebase can project backwards This is always advantageous asthe earth supported on the base assists in counterbalancing theoverturning effect due to the horizontal pressures exerted bythe soil However a base that projects mainly backwards butpartly forwards is usually necessary in order to limit the bearingpressure at the toe to an allowable value Sometimes due tothe proximity of adjacent property it may be impossible to

Retaining walls 69

Approximate distribution of elastic bending moments for an internal concentrated load on a ground-supported slab

As the load is increased the tensile stresses at the bottom ofthe slab under the load will reach the flexural strength of theconcrete Radial tension cracks will form at the bottom ofthe slab and provided there is sufficient ductility the slab willyield Redistribution of moments will occur with a reduction inthe positive moment at the load position and a substantialincrease in the negative moments some distance away Withfurther increases in load the positive moment at the loadposition will remain constant and the negative moments willincrease until the tensile stresses at the top of the slab reach theflexural strength of the concrete at which stage failure isassumed For further information on the analysis and designmethod with fully worked examples see ref 61

73 RETAINING WALLS

Information on soil properties and the pressures exerted bysoils on retaining structures is given in section 91 andTables 210ndash214 This section deals with the design of wallsto retain soils and materials with similar engineering propertiesIn designing to British Codes of Practice the geotechnicalaspects of the design which govern the size and proportions ofthe structure are considered in accordance with BS 8002Mobilisation factors are introduced into the calculation of thesoil strengths and the resulting pressures are used for bothserviceability and ultimate requirements For the subsequentdesign of the structure to BS 8110 the earth loads obtainedfrom BS 8002 are taken as characteristic values In designing tothe EC partial safety factors are applied to the soil propertiesfor the geotechnical aspects of the design and to the earthloads for the structural design

731 Types of retaining wall

Earth retention systems can be categorised into one of twogroups according to whether the earth is stabilised externally

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project a base backwards Under such conditions where thebase projection is entirely forwards the provision of a keybelow the base is necessary to prevent sliding by mobilising thepassive resistance of the soil in front of the base

For wall heights greater than about 8 m the stem thicknessof a cantilever wall becomes excessive In such cases a wallwith vertical counterforts can be used in which the slabspans horizontally between the counterforts For very highwalls in which the soil loading is considerable towardsthe bottom of the wall horizontal beams spanning betweenthe counterforts can be used By graduating the spacing of thebeams to suit the loading the vertical bending moments ineach span of the slab can be equalised and the slab thicknesskept the same

The factors affecting the design of a cantilever slab wall areusually considered per unit length of wall when the wall is ofconstant height but if the height varies a length of say 3 mcould be treated as a single unit For a wall with counterfortsthe length of a unit is taken as the distance between adjacentcounterforts The main factors to be considered in the design ofwalls on spread bases are stability against overturning groundbearing pressure resistance to sliding and internal resistance tobending moments and shearing forces Suitable dimensions forthe base to a cantilever wall can be estimated with the aid of thegraph given in Table 286

In BS 8002 for design purposes soil parameters are basedon representative shear strengths that have been reduced byapplying mobilisation factors Also for friction or adhesion ata soilndashstructure interface values not greater than 75 of thedesign shear strength are taken Allowance is made for a minimumsurcharge of 10 kNm2 applied to the surface of the retainedsoil and for a minimum depth of unplanned earth removal infront of the wall equal to 10 of the wall height but not lessthan 05 m

For overall equilibrium the effects of the disturbing forcesacting on the structure should not exceed the effects that canbe mobilised by the resisting forces No additional factors ofsafety are required with regard to overturning or sliding forwardsFor bases founded on clay soils both the short-term (usingundrained shear strength) and long-term (using drained shearstrength) conditions should be considered Checks on groundbearing are required for both the service and ultimate conditionswhere the design loading is the same for each but the bearingpressure distribution is different For the ultimate condition auniform distribution is considered with the centre of pressurecoincident with the centre of the applied force at the undersideof the base In general therefore the pressure diagram doesnot extend over the entire base In cases where resistance tosliding depends on base adhesion it is unclear as to whetherthe contact surface length should be based on the service or theultimate condition

The foregoing wall movements due to either overturning orsliding are independent of the general tendency of a bank or acutting to slip and carry the retaining wall along with it Thestrength and stability of the retaining wall have no bearingon such failures The precautions that must be taken to preventsuch failures are outside the scope of the design of a wall thatis constructed to retain the toe of the bank and are a problemin soil mechanics

Adequate drainage behind a retaining wall is important toreduce the water pressure on the wall For granular backfills of

high permeability no special drainage layer is needed but somemeans of draining away any water that has percolated throughthe backfill should be provided particularly where a wall isfounded on an impermeable material For cohesionless backfillsof medium to low permeability and for cohesive soils it is usualto provide a drainage layer behind the wall Various methodscan be used for instance (a) a blanket of rubble or coarseaggregate clean gravel or crushed stone (b) hand-placedpervious blocks as dry walling (c) graded filter drain wherethe back-filling consists of fine-grain material (d) a geotextilefilter used in combination with a permeable granular materialWater entering the drainage layer should drain into a drainagesystem which allows free exit of the water either by the provisionof weep-holes or by porous land drains and pipes laid at thebottom of the drainage layer and led to sumps or sewers viacatchpits Where weep-holes are being used they should be atleast 75 mm in diameter and at a spacing not more than 1 mhorizontally and 1ndash2 m vertically Puddled clay or concreteshould be placed directly below the weep-holes or pipes andin contact with the back of the wall to prevent water fromreaching the foundations

Vertical movement joints should be provided at intervalsdependent upon the expected temperature range the type ofthe structure and changes in the wall height or the nature of thefoundations Guidance on design options to accommodatemovement due to temperature and moisture change are given inBS 8007 and Highways Agency BD 2887

733 Embedded (or sheet) walls

Embedded walls are built of contiguous or interlocking pilesor diaphragm wall panels to form a continuous structure Thepiles may be of timber or concrete or steel and have lapped orV-shaped or tongued and grooved or interlocking jointsbetween adjacent piles Diaphragm wall panels are formed ofreinforced concrete using a bentonite or polymer suspension aspart of the construction process Excavation is carried out in thesuspension to a width equal to the thickness of the wallrequired The suspension is designed to maintain the stability ofthe slit trench during digging and until the diaphragm wall hasbeen concreted Wall panels are formed in predeterminedlengths with prefabricated reinforcement cages lowered into thetrench Concrete is cast in situ and placed by tremie it is vitalthat the wet concrete flows freely without segregation so as tosurround the reinforcement and displace the bentonite

Cantilever walls are suitable for only moderate height and itis preferable not to use cantilever walls when services orfoundations are located wholly or partly within the active soilzone since horizontal and vertical movement in the retainedmaterial can cause damage Anchored or propped walls canhave one or more levels of anchor or prop in the upper partof the wall They can be designed to have fixed or free earthsupport at the bottom as stability is derived mainly from theanchorages or props

Traditional methods of design although widely used allhave recognised shortcomings These methods are outlined inannex B of BS 8002 where comments are included on theapplicability and limitations of each method The design ofembedded walls is beyond the scope of this Handbook and forfurther information the reader should refer to BS 8002Highways Agency BD 4294 and ref 62

Foundations ground slabs retaining walls culverts and subways70

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74 CULVERTS AND SUBWAYS

Concrete culverts which can be either cast in situ or precastare usually of circular or rectangular cross section Box typestructures can also be used to form subways cattle creeps orbridges over minor roads

741 Pipe culverts

For conducting small streams or drains under embankmentsculverts can be built with precast reinforced concrete pipeswhich must be strong enough to resist vertical and horizontalpressures from the earth and other superimposed loads Thepipes should be laid on a bed of concrete and where passingunder a road should be surrounded with reinforced concrete atleast 150 mm thick The culvert should also be reinforced toresist longitudinal bending resulting from unequal vertical earthpressure and unequal settlement Due to the uncertaintyassociated with the magnitude and disposition of the earth pres-sures an accurate analysis of the bending moments is imprac-ticable A basic guide is to take the positive moments at the topand bottom of the pipe and the negative moments at the endsof a horizontal diameter as 00625qd 2 where d is the diameterof the circular pipe and q is the intensity of both the downwardpressure on the top and the upward pressure at the bottomassuming the pressure to be distributed uniformly on ahorizontal plane

742 Box culverts

The load on the top of a box culvert includes the weights of theearth covering and the top slab and the imposed load (if any)The weights of the walls and top slab (and any load that is onthem) produce an upward reaction from the ground Theweights of the bottom slab and water in the culvert are carrieddirectly on the ground below the slab and thus have no effectother than their contribution to the total bearing pressure Thehorizontal pressure due to the water in the culvert produces aninternal triangular load on the walls or a trapezoidal load if thesurface of the water outside the culvert is above the top whenthere will also be an upward pressure on the underside of thetop slab The magnitude and distribution of the earth pressureagainst the sides of the culvert can be calculated in accordancewith the information in section 91 consideration being givento the risk of the ground becoming waterlogged resulting inincreased pressure and the possibility of flotation Generallythere are only two load conditions to consider

1 Culvert empty maximum load on top slab weight of thewalls and maximum earth pressure on walls

2 Culvert full minimum load on top slab weight of the wallsminimum earth pressure and maximum internal hydrostaticpressure on walls (with possible upward pressure on top slab)

In some circumstances these conditions may not produce themaximum load effects at any particular section and the effectof every probable combination should be considered The crosssections should be designed for the combined effects of axialforce bending and shear as appropriate A simplistic analysiscan be used to determine the bending moments produced in amonolithic rectangular box by considering the four slabs asa continuous beam of four spans with equal moments at the end

supports However if the bending of the bottom slab tends toproduce a downward deflection the compressibility of theground and the consequent effect on the bending moments mustbe taken into account The loads can be conveniently dividedinto the following cases

1 A uniformly distributed load on the top slab and a uniformreaction from the ground under the bottom slab

2 A concentrated imposed load on the top slab and a uniformreaction from the ground under the bottom slab

3 Concentrated loads due to the weight of each wall and auniform reaction from the ground under the bottom slab

4 A triangular distributed horizontal pressure on each wall dueto the increase in earth pressure in the height of the wall

5 A uniformly distributed horizontal pressure on each walldue to pressure from the earth and any surcharge above thelevel of the top slab

6 Internal horizontal and possibly vertical pressures due towater in the culvert

Formulae for the bending moments at the corners of thebox due to each load case when the top and bottom slabsare the same thickness are given in Table 287 The limitingground conditions associated with the formulae should benoted

743 Subways

The design and construction of buried box type structureswhich could be complete boxes portal frames or structureswhere the walls are propped by the top slab are covered byrecommendations in Highways Agency standard BD 3187These recommendations do not apply to structures that areinstalled by methods such as thrust boring or pipe jacking

The nominal superimposed dead load consists of the weightof any road construction materials and the soil cover abovethe structure Due to negative arching of the fill material thestructure can be subjected to loads greater than the weight offill directly above it An allowance for this effect is made byconsidering a minimum load based on the weight of materialdirectly above the structure and a maximum load equal to theminimum load multiplied by 115 The nominal horizontalearth pressures on the walls of the box structure are based ona triangular distribution with the value of the earth pressurecoefficient taken as a maximum of 06 and a minimum of 02It is to be assumed that either the maximum or the minimumvalue can be applied to one wall irrespective of the value thatis applied to the other wall

Where the depth of cover measured from the finished roadsurface to the top of the structure is greater than 600 mm thenominal vertical live loads to be considered are the HA wheelload and the HB vehicle To determine the nominal vertical liveload pressure dispersion of the wheel loads may be taken tooccur from the contact area on the carriageway to the top ofthe structure at a slope of 2 vertically to 1 horizontally Forstructures where the depth of cover is in the range 200ndash600 mmfull highway loading is to be considered For HA load the KELmay be dispersed below the depth of 200 mm from the finishedroad surface Details of the nominal vertical live loads are givenin sections 248 and 249 and Table 25

Culverts and subways 71

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Part 2

Loads materials andstructures

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In this chapter unless otherwise stated all loads are given ascharacteristic or nominal (ie unfactored) values For designpurposes each value must be multiplied by the appropriatepartial safety factor for the particular load load combinationand limit-state being considered

Although unit weights of materials should be given strictlyin terms of mass per unit volume (eg kgm3) the designer isusually only concerned with the resulting gravitational forcesTo avoid the need for repetitive conversion unit weights aremore conveniently expressed in terms of force (eg kNm3)where 1 kN may be taken as 102 kilograms

81 DEAD LOAD

The data for the weights of construction materials given in thefollowing tables has been taken mainly from EC 1 Part 11 butalso from other sources such as BS 648

811 Concrete

The primary dead load is usually the weight of the concretestructure The weight of reinforced concrete varies with thedensity of the aggregate and the percentage of reinforcementIn UK practice a value of 24 kNm3 has traditionally beenused for normal weight concrete with normal percentages ofreinforcement but a value of 25 kNm3 is recommended inEC 1 Several typical weights for normal lightweight andheavyweight (as used for kentledge and nuclear-radiationshielding) concretes are given in Table 21 Weights are alsogiven for various forms and depths of concrete slabs

812 Other construction materials and finishes

Dead loads include such permanent weights as those of thefinishes and linings on walls floors stairs ceilings and roofsasphalt and other applied waterproofing layers partitionsdoors windows roof and pavement lights superstructures ofsteelwork masonry or timber concrete bases for machineryand tanks fillings of earth sand plain concrete or hardcorecork and other insulating materials rail tracks and ballastingrefractory linings and road surfacing In Table 21 the basicweights of various structural and other materials includingmetals stone timber and rail tracks are given

The average equivalent weights of various cladding types asgiven in Table 22 are useful in estimating the loads imposed

on a concrete substructure The weights of walls of variousconstructions are also given in Table 22 Where a concretelintel supports a brick wall it is generally not necessary toconsider the lintel as supporting the entire wall above it issufficient to allow only for the triangular areas indicated in thediagrams in Table 22

813 Partitions

The weight of a partition is determined by the material of whichit is made and the storey height When the position of thepartition is not known or the use of demountable partitions isenvisaged the equivalent uniformly distributed load given inTable 22 should be considered as an imposed load in the designof the supporting floor slabs

Weights of permanent partitions whose position is knownshould be included in the dead load Where the length of the par-tition is in the direction of span of the slab an equivalent UDLmay be used as given in Table 22 In the case of brick or similarlybonded partitions continuous over the slab supports some reliefof loading on the slab will occur due to the arching action of thepartition unless this is invalidated by the presence of doorways orother openings Where the partition is at right angles to the spanof the slab a concentrated line load should be applied at the appro-priate position The slab should then be designed for the combinedeffect of the distributed floor load and the concentrated load

82 IMPOSED LOADS

Imposed loads on structures include the weights of storedmaterials and the loads resulting from occupancy and trafficComprehensive data regarding the weights of stored materialsassociated with building industry and agriculture are given inEC 1 Part 11 Data for loads on floors due to livestock andagricultural vehicles are given in BS 5502 Part 22

821 Imposed loads on buildings

Data for the vertical loads on floors and horizontal loads onparapets barriers and balustrades are given in BS 6399 Part 1Loads are given in relation to the type of activityoccupancy forwhich the floor area will be used in service as follows

A Domestic and residential activitiesB Office and work areas not covered elsewhere

Chapter 8

Loads

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21Weights of construction materials and concrete floor slabs

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Weights of roofs and walls 22

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C Areas where people may congregateD Shopping areasE Areas susceptible to the accumulation of goodsFG Vehicle and traffic areas

Details of the imposed loads for categories A and B are givenin Table 23 Values are given for uniformly distributed andconcentrated loads These are not to be taken together butconsidered as two separate load cases The concentrated loadsnormally do not need to be considered for solid or other slabsthat are capable of effective lateral distribution When used forcalculating local effects such as bearing or the punching of thinflanges a square contact area of 50 mm side should be assumedin the absence of any other specific information

With certain exceptions the imposed loads on beams may bereduced according to the area of floor supported Loads oncolumns and foundations may be reduced according to eitherthe area of floor or the number of storeys supported Details ofthe reductions and the exceptions are given in Table 23

Data given in Table 24 for the load on flat or mono-pitchroofs has been taken from BS 6399 Part 3 The loads whichare additional to all surfacing materials include for snow andother incidental loads but exclude wind pressure For other roofshapes and the effects of local drifting of snow behind parapetsreference should be made to BS 6399 Part 3

For building structures designed to meet the requirementsof EC 2 Part 1 details of imposed and snow loads are given inEC 1 Parts 11 and 13 respectively

822 Imposed loads on highway bridges

The data given in Table 25 for the imposed load on highwaybridges have been taken from the Highways Agency documentBD 3701 Type HA loading consists of two parts a uniformload whose value varies with the lsquoloaded lengthrsquo and a singleKEL that is positioned so as to have the most severe effectThe loaded length is the length over which the application of theload increases the effect to be determined Influence lines maybe needed to determine critical loaded lengths for continuousspans and arches Loading is applied to one or more notionallanes and multiplied by appropriate lane factors The alternativeof a single wheel load also needs to be considered in certaincircumstances

Type HB is a unit loading represented by a 16ndashwheel vehicleof variable bogie spacing where one unit of loading is equivalentto 40 kN The number of units considered for a public highwayis normally between 30 and 45 according to the appropriateauthority The vehicle can be placed in any transverse positionon the carriageway displacing HA loading over a specified areasurrounding the vehicle

For further information on the application of combined HAand HB loading and details of other loads to be considered on

highway bridges reference should be made to BD 3701 andBD 6094 For information on loads to be considered for theassessment of existing highway bridges reference should bemade to BD 2101

823 Imposed loads on footbridges

The data given in Table 26 for the imposed load on bridges dueto pedestrian traffic have been taken from the Highways Agencydocument BD 3701 For further information on the pedestrianloading to be considered on elements of highway or railwaybridges that also support footways or cycle tracks and the ser-viceability vibration requirements of footbridges referenceshould be made to BD 3701

824 Imposed loads on railway bridges

The data given in Table 26 for the imposed load on railwaybridges has been taken from the Highways Agency documentBD 3701 Types RU and SW0 apply to main line railwaystype SW0 being considered as an additional and separate loadcase for continuous bridges For bridges with one or two tracksloads are to be applied to each track In other cases loads areto be applied as specified by the relevant authority

Type RL applies to passenger rapid transit railway systemswhere main line locomotives and rolling stock do not operateThe loading consists of a uniform load (or loads dependent onloaded length) combined with a single concentrated load posi-tioned so as to have the most severe effect The loading is to beapplied to each and every track An arrangement of two con-centrated loads is also to be considered for deck elementswhere this would have a more severe effect

For information on other loads to be considered on railwaybridges reference should be made BD 3701

83 WIND LOADS

The data given in Tables 27ndash29 for the wind loading onbuildings has been taken from the information given for thestandard method of design in BS 6399 Part 2 The effectivewind speed is determined from Table 27 Wind pressures andforces on rectangular buildings as defined in Table 28 aredetermined by using standard pressure coefficients given inTable 29 For data on other building shapes and different roofforms and details of the directional method of design referenceshould be made to BS 6399 Part 2

Details of the method used to assess wind loads on bridgestructures and the data to be used for effective wind speeds anddrag coefficients are given in BD 3701 For designs to EC 2wind loads are given in EC 1 Part 12

Loads78

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Imposed loads on floors of buildings 23

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Imposed loads on roofs of buildings 24

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Imposed loads on bridges ndash 1 25

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Imposed loads on bridges ndash 2 26

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Wind speeds (standard method of design) 27

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Wind pressures and forces (standard method of design) 28

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Pressure coefficients and size effect factors forrectangular buildings 29

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Chapter 9

Pressures due toretained materials

In this chapter unless otherwise stated all unit weights andother properties of materials are given as characteristic or rep-resentative (ie unfactored) values For design purposes eachvalue must be modified by appropriate partial safety or mobili-sation factors according to the basis of design and the code ofpractice employed

91 EARTH PRESSURES

The data given in Table 210 for the properties of soils has beentaken from BS 8002 Design values of earth pressure coeffi-cients are based on the design soil strength which is taken asthe lower of the peak soil strength reduced by a mobilisationfactor or the critical state strength

911 Pressures imposed by cohesionless soils

For the walls shown in Table 211 with a uniform normallyconsolidated soil a uniformly distributed surcharge and nowater pressure the pressure imposed on the wall increaseslinearly with depth and is given by

K(zq)

where is unit weight of soil z is depth below surface q issurcharge pressure (kNm2) K is at-rest active or passivecoefficient of earth pressure according to design conditions

A minimum live load surcharge of 10 kNm2 is specified inBS 8002 This may be reasonable for walls 5 m high and abovebut appears to be too large for low walls In this case valuessuch as 4 kNm2 for walls up to 2 m high 6 kNm2 for walls 3 mhigh and 8 kNm2 for walls 4 m high could be used InBD 3701 surcharge loads are given of 5 kNm2 for footpaths10 kNm2 for HA loading 12 kNm2 for 30 units of HB loading20 kNm2 for 45 units of HB loading and on areas occupiedby rail tracks 30 kNm2 for RL loading and 50 kNm2 forRU loading

If static ground water occurs at depth zw below the surfacethe total pressure imposed at z zw is given by

K [mz (s w)(zzw)q]w(zzw)

where m is moist bulk weight of soil s is saturated bulkweight of soil w is unit weight of water (981 kNm3)

912 At-rest pressures

For a level ground surface and a normally consolidated soilthat has not been subjected to removal of overburden thehorizontal earth pressure coefficient is given by

Ko 1sin

where is effective angle of shearing resistance of soilCompaction of the soil will result in earth pressures in the

upper layers of the soil mass that are higher than those givenby the above equation The diagram and equations given inTable 211 can be used to calculate the maximum horizontalpressure induced by the compaction of successive layers ofbackfill and determine the resultant earth pressure diagramThe effective line load for dead weight compaction rollers is theweight of the roller divided by its width For vibratory rollersthe dead weight of the roller plus the centrifugal force causedby the vibrating mechanism should be used The DOESpecification limits the mass of the roller to be used within 2 mof a wall to 1300 kgm

For a vertical wall retaining backfill with a ground surfacethat slopes upwards the horizontal earth pressure coefficientmay be taken as

Ko (1sin)(1sin)

where is slope angle The resultant pressure which acts in adirection parallel to the ground surface is given by

o Koz cos

913 Active pressures

Rankinersquos theory may be used to calculate the pressure on a ver-tical plane referred to as the lsquovirtual backrsquo of the wall For avertical wall and a level ground surface the Rankine horizontalearth pressure coefficient is given by

The solution applies particularly to the case of a smooth wall ora wall with no relative movement between the soil mass andthe back of the wall The charts given in Table 212 which arebased on the work of Caquot and Kerisel may be used generallyfor vertical walls with sloping ground or inclined walls with

Ka 1 sin

1 sin

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Properties of soils 210

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Earth pressure distributions on rigid walls 211

At-rest state for rigid wall Effects of soil compaction

Active state for rigid wall free to rotate about base or translate

Passive state for rigid wall free to rotate about base or translate

hc 1K

2Q1

zc K2Q1

c 2Q1

K earth pressure coefficientK o for unyielding structureK a for wall free to mobilise

fully active stateQ1 intensity of effective line load

imposed by compaction plant unit weight of soil maximum horizontal earth

pressure induced by compaction

Horizontal earth pressure distribution resulting from compaction

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Active earth pressure coefficients 212

Angle of Shearing Resistance f (degrees)

Coe

ffici

ent o

f Act

ive

Pre

ssur

e K

aC

oeffi

cien

t of A

ctiv

e P

ress

ure

Ka

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level ground The horizontal and vertical components of resultantpressure are given by

ah Kazcos() and av Kazsin()

where is wall inclination to vertical (positive or negative) isselected angle of wall friction (taken as positive)

914 Passive pressures

For a vertical wall and a level ground surface the Rankinehorizontal earth pressure coefficient is given by

The solution applies particularly to the case of a smooth wall ora wall with no relative movement between the soil mass and theback of the wall The charts given in Table 213 for verticalwalls with sloping ground and in Table 214 for inclined wallswith level ground are based on the work of Caquot and KeriselThe horizontal and vertical components of resultant pressureare given by

ph Kpzcos() and pv Kpzsin()

where is wall inclination to vertical (positive or negative) isselected angle of wall friction (taken as negative)

915 Cohesive soils

If a secant value (c 0) is selected the procedures given forcohesionless soils apply If tangent parameters (c ) are to beused the RankinendashBell equations may be used as follows

a Ka(zq)2cradicKa

p Kp(zq)2cradicKp

where c is effective cohesion The active earth pressure istheoretically negative to a depth given by

zo (2cradicKa q)

Where cracks which may form in the tension zone can becomefilled with water full hydrostatic pressure should be consideredover the depth zo If the surface is protected so that no surfacewater can accumulate in the tension cracks the earth pressureshould be taken as zero over the depth zo

916 Further considerations

For considerations such as earth pressures on embedded walls(with or without props) the effects of vertical concentratedloads and line loads and the effects of groundwater seepagereference should be made to BS 8002 For the pressures to beconsidered in the design of integral bridge abutments as aresult of thermal movements of the deck reference should bemade to Highways Agency document BA 4296

92 TANKS

The pressure imposed by a contained liquid is given by

wz

Kp 1 sin

1 sin

where w is unit weight of liquid (see EC 1 Part 11) and z isdepth below surface For a fully submerged granular materialthe total horizontal pressure on the walls is

K(w)(zzo)wz

where is unit weight of the material (including voids) zo isdepth to top of material K is material pressure coefficient If o

is unit weight of material (excluding voids) o(1 e)where e is ratio of volume of voids to volume of solids

The preceding equation applies to materials such as coal orbroken stone with an effective angle of shearing resistancewhen submerged of approximately 35o For submerged sand Kshould be taken as unity If the material floats (o w) thesimple hydrostatic pressure applies

93 SILOS

The data given in Tables 215 and 216 has been taken fromEurocode 1 Part 4 The pressures apply to silos of the formsshown in Table 215 subject to the following limitations

dimensions dc 50 m h 100 m hdc 10

eccentricities ei 025dc eo 025dc with no part of outlet ata distance greater than 03dc from centreline of silo

filling involves negligible inertia effects and impact loads

stored material is free-flowing (cohesion is less than 4 kPa fora sample pre-consolidated to 100 kPa) with a maximumparticle size not greater than 03dc

transition between vertical walled section and hopper is on asingle horizontal plane

Dimensions h ho h1 and z are measured from the equivalentsurface which is a level surface giving the same volume ofstored material as the actual surface at the maximum filling

Loads acting on a hopper are shown in Table 215 where thetensile force at the top of the hopper is required for the designof silo supports or a ring beam at the transition level The verticalcomponent of the force can be determined from force equilib-rium incorporating the vertical surcharge Cbpvo at the transitionlevel and the weight of the hopper contents The discharge loadon the hopper wall is affected by the flow pattern of the storedmaterial which may be mass flow or funnel flow according tothe characteristics of the hopper and the material The normalload due to pn is supplemented for mass flow silos only by akick load due to ps

Values of material properties and expressions to determineresulting pressures in the vertical walled and bottom sections ofa silo are given in Table 216 For squat silos (hdc 15) thehorizontal pressure ph may be reduced to zero at the level wherethe upper surface of the stored material meets the silo wallBelow this level a linear pressure variation may be assumedtaking K 10 until this pressure reaches the value appropriateto the depth z below the equivalent surface

Homogenizing silos and silos containing powders in whichthe speed of the rising surface of the material exceeds 10 mhshould be designed for both the fluidised and non-fluidised con-ditions For the fluidised condition the bulk unit weight of thematerial may be taken as 08 For information on test methodsto determine the properties of particulate materials referenceshould be made to EC 1 Part 4

Pressures due to retained materials90

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Passive earth pressure coefficents ndash 1 213

Angle of Shearing Resistance f (degrees)

Coe

ffici

ent o

f Pas

sive

Pre

ssur

e K

pR

educ

tion

Fac

tor

Rd

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214Passive earth pressure coefficents ndash 2

Angle of Shearing Resistance f (degrees)

Coe

ffici

ent o

f Pas

sive

Pre

ssur

e K

pR

educ

tion

Fac

tor

Rd

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215Silos ndash 1

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216Silos ndash 2

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101 CONSTITUENTS OF CONCRETE

1011 Cements and combinations

Manufactured cements are those made in a cement factoryWhere a mineral material is included it is generally added tothe cement clinker at the grinding stage The notation used forthese manufactured cements contains the prefix letters CEMWhen a concrete producer adds an addition such as pfa or ggbsto CEM I Portland cement in the mixer the resulting cement isknown as a mixer combination and is denoted by the prefixletter C Cements and combinations in general use are listed inTable 217 Further information on the different types and useof cements is given in section 311

1012 Aggregates

Overall grading limits for coarse and fine aggregates fromnatural sources in accordance with BS EN 12620 are given inTable 217 Further information is given in section 312

102 EARLY-AGE TEMPERATURES OF CONCRETE

The calculation of early thermal crack widths in a restrainedconcrete element requires knowledge of the temperature rise dueto the cement hydration Some typical early temperature histo-ries of various concrete walls and predicted temperature risesfor different cements are given in Table 218

The predicted temperature rise values for Portland cementconcretes in walls and slabs are taken from BS 8007 These aremaximum values selected from a range of values for Portlandcements obtained from different works (ref 11) The tempera-ture rises given for the other cements in concrete sections witha minimum dimension of 1 m should be taken as indicativeonly but could be used where other specific information isnot available

103 REINFORCEMENT

Reinforcement for concrete generally consists of steel bars orwelded steel mesh fabric that depend upon the provision of adurable concrete cover for protection against corrosion Theessential properties of bars to BS 4449 and wires to BS 4482both of which are in general conformity with BS EN 10080

are given in Table 219 For additional information on themanufacture and properties of steel reinforcement includingstainless steel refer to section 32

1031 Bars

Bars for normal use produced in the United Kingdom arehot-rolled to a characteristic strength of 500 MPa and achieveClass B or C ductility The bars are round in cross section withsets of parallel transverse ribs separated by longitudinal ribsThe nominal size is the diameter of a circle with an area equalto the effective cross-sectional area of the bar The maximumoverall size is approximately 15 greater than the nominal sizeValues of the total cross-sectional area provided in a concretesection according to the number or spacing of the bars fordifferent bar sizes are given in Table 220

The type and grade of reinforcement is designated as follows

Chapter 10

Concrete andreinforcement

Type of steel reinforcement Notation

Grade B500A B500B or B500C to BS 4449 HGrade B500A to BS 4449 AGrade B500B or B500C to BS 4449 BGrade B500C to BS 4449 CA specified grade and type of ribbed Sstainless steel to BS 6744

Reinforcement of a type not included above but Xwith material properties defined in the designor contract specification

Note In the description B500A and so on B indicates reinforcing steel

1032 Fabric

In the United Kingdom steel fabric reinforcement is generallyproduced to the requirements of BS 4483 using ribbed bar inaccordance with BS 4449 The exception is wrapping fabricwhere wire in accordance with BS 4482 may be used Fabric isproduced in a range of standard types or can be purpose-madeto the clientrsquos requirements Full details of the standard fabrictypes are given in Table 220

Type A is a square mesh with identical longitudinal barsand cross bars commonly used in ground slabs to provide aminimum amount of reinforcement in two directions Type B isa rectangular (structural) mesh that is particularly suitable for

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217Concrete cements and aggregate grading

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Concrete early-age temperatures 218

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219Reinforcement general properties

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Reinforcement cross-sectional areas of bars and fabric 220

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thin one-way spanning slabs where the longitudinal barsprovide the main reinforcement with the cross bars beingsufficient to meet the minimum requirements for secondaryreinforcement Type C is a rectangular (long) mesh where thecross bars are minimal which can be used in any solid slabincluding column bases by providing a separate sheet in eachdirection Type D is a rectangular (wrapping) mesh that is usedin the concrete encasement of structural steel sections

Standard fabric is normally supplied in stock sheet sizes of48 m (longitudinal bars) 24 m (cross bars) with end over-hangs equal to 05 the pitch of the perpendicular bar Tofacilitate fixing and to avoid a build-up of bar layers at the lapssheets with increased overhangs (flying ends) can be suppliedto order Sheets can also be supplied cut to size in lengths upto 12 m and prebent For guidance on the use of purpose-madefabrics reference should be made to BS 8666

1033 Cutting and bending tolerances

Bars are produced in stock lengths of 12 m and lengths up to18 m can be supplied to special order In most structures barsare required in shorter lengths and often need to be bent Thecutting and bending of reinforcement is generally specified tothe requirements of BS 8666 The tolerances on cutting andbending dimensions are as follows

For shape code 67 when the radius exceeds the value in thefollowing table straight bars will be supplied as the requiredcurvature can be obtained during fixing

Concrete and reinforcement100

Cutting and bending processes Tolerance (mm)

Cutting of straight lengths 25

Bending dimension (mm) 1000 5 1000 and 2000 5 10

2000 5 25

Length of wires in fabric L 5000 25L 5000 L 200

1034 Shape codes and bending dimensions

BS 8666 contains details of bar shapes designated by shapecodes as given in Tables 221 and 222 The information neededto cut and bend the bars to the required dimensions is enteredinto a bar schedule an example of which is shown in Table 223The standard shapes should be used wherever possible with therelevant dimensions entered into columns A to E of the barschedule All other shapes should be given a shape code 99 witha dimensioned sketch drawn over the columns A to E using twoparallel lines to indicate the bar thickness One of the bar dimen-sions should be indicated in parenthesis as a free dimensionDimensions should be given as a multiple of 5 mm and thetotal length determined in accordance with the equation givenin the table rounded up to a multiple of 25 mm To facilitatetransportation each bent bar should fit within an imaginaryrectangle the shorter side of which is not longer than 2750 mm

Most of the shape codes cater for bars bent to the minimumradius taken as 2d for d 16 and 35d for d 20 where d isthe bar size The minimum straight length needed beyond theend of the curved portion of a bend is 5d for a bob and 10d formost links For each bar size values of the minimum radius rand the minimum end projection P needed to form a bend aregiven in Table 219

Bars needing larger radius bends denoted by R except forshape codes 12 and 67 should be treated as a shape code 99

Maximum limit for which a preformed radius is required

Bar size (mm) 8 10 12 16

Radius (m) 275 35 425 75

Bar size (mm) 20 25 32 40

Radius (m) 14 30 43 58

For shape codes 12 13 22 and 33 the largest practical radiusfor producing a continuous curve is 200 mm and for a largerradius a series of short straight sections may be formed

1035 Deductions for variations

Cover to reinforcement is liable to variation due to the effect ofinevitable errors in the dimensions of formwork and in thecutting bending and fixing of the bars In cases where a bar isdetailed to fit between two concrete faces with no more thanthe nominal cover on each face (eg link in a beam) an appro-priate allowance for deviations should be applied The relevantdimension on the schedule should be determined as the nominaldimension of the concrete less the nominal cover on each faceless an allowance for deviations as follows

Total deductions to allow for permissible deviations onmember size and in cutting and bending of bars

Type of bar Distance between faces of Deductionconcrete member mm

Links and other Not more than 1 m 10bent bars Between 1 m and 2 m 15

Over 2 m 20

Straight bars Any length 40

The deductions recommended in the forgoing table are taken fromBS 8110 and allow for deviations on the member size of 5 mm fordimensions up to 2 m and 10 mm for dimensions over 2 m Wherethe permissible deviations on member size exceed these valueslarger deductions should be made or the cover increased

Example Determine the relevant bending dimensions for thebars shown in the following beam detail The completed barschedule for 6 beams thus is given in Table 223

Section A-A

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Reinforcement 101

Elevation of beam

Bar Shape Dimensions (mm) Length (mm) ndash see Tables 229 and 230mark code

01 00 L 8000 2 200 7600

02 00 L 8000 2 1250 5500

03 13 Bar requires radius of bend R 6d as a designrequirement This necessitates the use of shapecode 13 as dimension B would not provide 4dlength of straight between two bends L 1800 057 420 1300 16 25 3300

A 1800 C 1300 (design requirements)B 450 (2 10) 10 420

Note Dimension B is derived from dimension Aof bar mark 05 and includes a further deductionof 10 mm for tolerances on cutting and bending

04 00 L 8000 2 200 7600

05 51 A 500 (2 20) 10 450 L 2 (450 250 115) (25 16) (5 8) 1550B 300 (2 20) 10 250C D 115 (P in Table 219)r 16 (Table 219)Note Dimensions A and B include deductionsof 10 mm for permissible deviations

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221Reinforcement standard bar shapes and methodof measurement ndash 1

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222Reinforcement standard bar shapes and methodof measurement ndash 2

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Reinforcement typical bar schedule 223

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The formulae and coefficients in this chapter give values ofshearing forces bending moments slopes and deflections interms of the total load on the member For design purposes theload F must include the appropriate partial safety factors for thelimit-state being considered

111 SIMPLE BEAMS AND CANTILEVERS

The formulae for the reactions shearing forces and bendingmoments in freely supported beams (Tables 224 and 225) andsimple cantilevers (Tables 226 and 227) are obtained by therules of static equilibrium The slope and deflection formulaefor freely supported beams and simple cantilevers and all theformulae for propped cantilevers (Table 227) are for elasticbehaviour and members of constant cross section

112 BEAMS FIXED AT BOTH ENDS

The bending moments on a beam fixed at both ends can bederived from the principle that the area of the free-momentdiagram (ie the bending moment diagram due to the same loadimposed on a freely supported beam of equal span) is equal tothe area of the restraint-moment diagram Also the centresof area of the two diagrams are in the same vertical line Theshape of the free-moment diagram depends upon the particularcharacteristics of the imposed load but the restraint-momentdiagram is a trapezium For loads that are symmetrically disposedon the beam the centre of area of the free-moment diagram isat the mid-point of the span and thus the restraint-momentdiagram is a rectangle giving a restraint moment at eachsupport equal to the mean height of the free-moment diagram

The amount of shearing force in a beam with one or bothends fixed is calculated from the variation of the bendingmoment along the beam The shearing force resulting from therestraint moment alone is constant throughout the length of thebeam and equal to the difference between the two end-momentsdivided by the span (ie the rate of change of the restraintmoment) This shearing force is algebraically added to theshearing force due to the imposed load with the beam takenas freely supported Thus the support reaction is the sum (ordifference) of the restraint-moment shearing force and the free-moment shearing force For a beam that is symmetricallyloaded with both ends fixed the restraint moment at each endis the same and the shearing forces are identical to those for thesame beam freely supported The support reactions are bothequal to one-half of the total load on the span The formulae forthe reactions shearing forces bending moments slopes anddeflections for fully fixed spans (Table 225) are for elasticbehaviour and members of constant cross section

1121 Fixed-end moment coefficients

Fixed-end moment coefficients CAB and CBA are given inTable 228 for a variety of unsymmetrical and symmetricalimposed loadings on beams of constant cross section Morecomplex loading arrangements can generally be formed as acombination of the cases shown and the resulting fixed-endmoments found by superposition A full range of charts iscontained in Examples of the Design of Reinforced ConcreteBuildings for a member with a partial uniform or triangulardistribution of load placed anywhere within the span

Chapter 11

Cantilevers andsingle-span beams

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224Moments shears deflections general case for beams

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Moments shears deflections special cases for beams 225

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226Moments shears deflections general cases for cantilevers

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Moments shears deflections special cases for cantilevers 227

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228Fixed-end moment coefficients general data

The fixed-end moment coefficients CAB and CBA can be used as follows1 To obtain bending moments at supports of single-span beams fully fixed at both ends (Table 225)

MAB CABlAB and MBA CBAlAB (With symmetrical load MAB MBA)2 To obtain fixed-end moments for analysis of continuous beams by moment distribution methods (Table 236)

FEMAB CABlAB and FEMBA CBAlAB (With symmetrical load FEMAB FEMBA)3 To obtain loading factors for analysis of framed structures by slope-deflection methods (Table 260)

FAB CABlAB and FBA CBAlAB (With symmetrical load FAB FBA)4 To obtain loading factors for analysis of portal frames (Tables 263 and 264)

and z1 (With symmetrical loading CAB CBA and z1 05)=CAB 2CBA

2(CAB 2CBA)D

CAB CBA

2lAB

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The formulae and coefficients in this chapter give values ofshearing forces and bending moments in terms of the dead andlive loads on the member For design purposes these loads mustinclude the appropriate partial safety factors for the limit-stateconsidered and the Code of Practice employed

For the ULS the dead load factors are treated differently inBS 8110 and EC 2 For designs to BS 8110 values of either 14or 10 are applied separately to each span of the beam Fordesigns to EC 2 values of either 135 or 10 are applied to all thespans If the beam ends with a cantilever the effect of applyingvalues of either 135 or 115 separately to the cantilever and theadjacent span should also be considered Details of the designloads and of the effects of applying cantilever moments at oneor both ends of a continuous beam of two three four or fiveequal spans are given in Table 229

121 DETERMINATION OF MAXIMUM MOMENTS

1211 Incidence of live load

The values of the bending moments in the spans and at thesupports depend upon the incidence of the live load and forspans that are equal or approximately equal the dispositions oflive load shown in Table 229 give the maximum positivemoments in the spans and the maximum negative momentsat the supports Both BS 8110 and EC 2 consider a lesssevere incidence of live load when determining the maximumnegative moments at the supports According to BS 8110 theonly case that needs to be considered is when all the spans areloaded According to EC 2 all cases of two adjacent spansloaded should be considered but the loads shown as optionalin Table 229 may be ignored

Thus the maximum positive moments due to live load forthe system are obtained by considering two loading arrange-ments one with live load on all the odd-numbered spans andthe other with live load on all the even-numbered spans Fordesigns to BS 8110 the summation of the results for these twocases gives the maximum negative moments

It should be noted that for designs to EC 2 the UK NationalAnnex allows the BS 8110 loading arrangements to be used asan alternative to those recommended in the base document Inthis chapter the basic arrangements are used

1212 Shearing forces

The shearing forces in a continuous beam are determined byfirst considering each span as freely supported then addingalgebraically the rate of change of restraint moment for theparticular span Shearing forces for freely supported spans arereadily determined by the rules of static equilibrium The addi-tional shearing force which is constant throughout the span isequal to the difference in the support moments at each enddivided by the span

1213 Maximum positive moments

When the moments at the supports and the shearing forces havebeen determined the maximum positive moment in the spancan be obtained by first finding the position where the shearingforce is zero The maximum positive moment is then obtainedby subtracting the effect of the restraint moments which varieslinearly along the span from the freely supported moment atthis position

122 SOLUTIONS FOR EQUAL SPANS

1221 Coefficients for equal loads on equal spans

Approximate general solutions for the maximum bendingmoments and shearing forces in uniformly loaded beams ofthree or more spans are given in Table 229 Exact solutions forthe maximum bending moments in beams of two three fouror five equal spans are given in Tables 230 and 231 for eightdifferent load distributions The coefficients given for thesupport moments due to live load apply to the most onerousloading conditions For the less severe arrangements describedin section 1211 coefficients are shown in the square brackets [ ]for BS 8110 and the curved brackets ( ) for EC 2 The coefficientsin Table 232 enable the maximum shearing forces at thesupports to be determined

Example 1 Calculate the maximum ultimate moments inthe end and central spans and at the penultimate and interiorsupports for a beam continuous over five equal spans of 5 mwith characteristic dead and imposed loads of 20 kNm eachaccording to the requirements of BS 8110

Chapter 12

Continuous beams

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229Continuous beams general data

AKey B K A B J K

A B C H J KA B C

Adjustment to bending moment = M coefficient x applied bending momentAdjustment to shearing force = (V coefficent x applied bending moment)span

KJ

Optional

EC2 Consider live load on spans RS and ST only

Optional

R T U V

TS

S

To produce maximum positive moment in span ST

To produce maximum negative moment at support S

Simplifications BS8110 Consider live load on all spans

EC 2 EC 2

EC 2

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230Continuous beams moments from equal loadson equal spans ndash 1

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231Continuous beams moments from equal loadson equal spans ndash 2

Bending moment (coefficient) (total load on one span) (span)Bending moment coefficientsabove line apply to negative bending moment at supportsbelow line apply to positive bending moment in span

Coefficients apply when all spans are equal (may be used also whenshortest 85 longest) Loads on each loaded span are same

Second moment of area is same throughout all spansBending moment coefficients in square brackets (live load)apply if all spans are loaded (ie BS 8110 requirements)Bending moment coefficients in curved brackets (live load)apply if two adjacent spans are loaded (ie EC 2 requirements)

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232Continuous beams shears from equal loadson equal spans

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The design load consists of a dead load of 10gk 20 kNm anda live load of (04gk 16qk) 40 kNm Then from Table 230(using coefficients in square brackets for the live load) theultimate bending moments are as follows

Penultimate supportDead load 0105 20 52 525 kNm (negative)Imposed load 0105 40 52 1050 kNm (negative)Total 1575 kNm (negative)

Interior supportDead load 0079 20 52 395 kNm (negative)Imposed load 0079 40 52 790 kNm (negative)Total 1185 kNm (negative)

Near middle of end spanDead load 0078 20 52 390 kNm (positive)Imposed load 0100 40 52 1000 kNm (positive)Total 1390 kNm (positive)

Middle of central spanDead load 0046 20 52 230 kNm (positive)Imposed load 0086 40 52 860 kNm (positive)Total 1090 kNm (positive)

Example 2 Calculate the maximum ultimate moments forexample 1 according to the requirements of EC 2

The design load consists of a dead load of 135gk 27 kNmand a live load of 15qk 30 kNm Then from Table 230(using coefficients in curved brackets for the live load) theultimate bending moments are as follows

Penultimate supportDead load 0105 27 52 709 kNm (negative)Imposed load 0116 30 52 870 kNm (negative)Total 1579 kNm (negative)

Interior supportDead load 0079 27 52 533 kNm (negative)Imposed load 0106 30 52 795 kNm (negative)Total 1328 kNm (negative)

Near middle of end spanDead load 0078 27 52 527 kNm (positive)Imposed load 0100 30 52 750 kNm (positive)Total 1277 kNm (positive)

Middle of central spanDead load 0046 27 52 311 kNm (positive)Imposed load 0086 30 52 645 kNm (positive)Total 956 kNm (positive)

Example 3 Calculate the maximum ultimate moments forexample 1 according to the requirements of BS 8110 when a2 m long cantilever is provided at each end of the beamIncrease in moments due to dead and live loads on cantileversat critical positions is as follows

End supportDead load 05 20 22 400 kNm (negative)Imposed load 05 40 22 800 kNm (negative)Total 1200 kNm (negative)

Interior supportFrom Table 229 increase due to moments at end supports is0053 120 64 kNm (negative)

Decrease in moments due to dead load only on cantilevers atcritical positions is as follows

Penultimate supportFrom Table 229 decrease due to moments at end supports is0263 40 105 kNm (positive)

Middle of end spanFrom Table 229 decrease due to moments at end supports is[10 05(10 0263)] 40 147 (negative)

Middle of central spanFrom Table 229 decrease due to moments at end supports is0053 40 21 kNm (negative)

123 REDISTRIBUTION OF MOMENTS

As explained in section 422 for the ULS both BS 8110 andEC 2 permit the moments determined by a linear elastic analysisto be redistributed provided that the resulting distributionremains in equilibrium with the loads Although the conditionsaffecting the procedure are slightly different in the two codesthe general approach is to reduce the critical moments by achosen amount up to the maximum percentage permitted anddetermine the revised moments at other positions by equilibriumconsiderations

An important point to appreciate is that each particular loadcombination can be considered separately Thus if desired it ispossible to reduce the maximum moments in the spans and atthe supports For example the maximum support moments canbe reduced to values that are still greater than those that occurwith the maximum span moments The maximum spanmoments can then be reduced until the corresponding supportmoments are the same as the (reduced) maximum values

The principles of static equilibrium require that no changesshould be made to the moments in a cantilever or at a freelysupported end

1231 Code requirements

BS 8110 and EC 2 permit the maximum moments to be reducedby up to 30 provided that in the subsequent design of the rele-vant sections the depth of the neutral axis is limited accordingto the amount of redistribution (Note that there is no restrictionon the maximum percentage increase of moment) In EC 2 themaximum permitted reduction depends also on the ductilityof the reinforcement being 30 for reinforcement classes Band C but only 20 for class A

In BS 8110 it is stated that the ultimate resistance momentat a section should be at least 70 of the maximum momentat that section before redistribution In effect the process ofredistribution alters the positions of points of contra-flexureThe purpose of the code requirement is to ensure that at suchpoints on the diagram of redistributed moments (at which noflexural reinforcement is theoretically required) sufficientreinforcement is provided to cater for the moments that willoccur under service loading The redistribution procedure takesadvantage of the ability of continuous beams to develop plastichinges at critical sections prior to failure whilst also ensuringthat the response remains fully elastic under service loading Therequirements are discussed more fully in books on structuraldesign and in the Handbook to BS 8110

Continuous beams116

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1232 Redistribution procedure

The use of moment redistribution is illustrated in Table 233where a beam of three equal spans is examined in accordancewith the requirements of BS 8110 The uniformly distributeddead and live loads are each equal to 1200 units per span Themoment diagram for dead load on each span is shown in (a)Moment diagrams for the arrangements of live load that givethe maximum moments at the supports and in the spans areshown in (b) The moment envelope obtained by combiningthe diagrams for dead and live loads is shown in (c) (Notethat the vertical scale of diagrams (c)ndash(f) differs from thatof (a) and (b))

The redistribution procedure is normally used to reduce themaximum support moments One approach is to reduce these tothe values obtained when the span moments are greatest Thisis shown in (d) where the support moments have been reducedfrom 240 to 180 a reduction of 25 In this case no otheradjustment is needed to the moment envelope If the maximumsupport moments are reduced by 30 from 240 to 168 thespan moments must be increased as shown in (e) The 70requirement discussed in section 1132 determines the extentof the hogging region in the end span and the minimum valueof 21 in the middle span (For the load cases in EC2 themaximum support moments could be reduced by 30 from260 to 182 with no other adjustment needed to themoment envelope)

If the criterion is to reduce the maximum span moments inthe case of an up-stand beam say this may be achieved byincreasing the corresponding support moments This is shownin (f) where the maximum moment in the middle span has beenreduced by 30 from 120 to 84 by increasing the supportmoments from 180 to 216 (Note that the minimummoment in the middle span has increased from 30 to 66)The moment in the end span has also been reduced by 7 from217 to 202 The 70 requirement determines the extent of thesagging regions in both spans It is clear that any further reduc-tion of moment in the end span would result in a considerableincrease in the support moment For example a 30 reductionin the end span moment from 217 to 152 would increase thesupport moment to 346 and the minimum moment in themiddle span to 196

In view of the many factors involved it is difficult to give anygeneral rules as to whether to redistribute moments or by howmuch such decisions are basically matters of individual engi-neering judgement A useful approach is to first calculate theultimate resistance moments at the support sections provided bychosen arrangements of reinforcement and then redistribute themoment diagrams to suit The span sections can then be designedfor the resulting moments and a check made to ensure that allof the code requirements are satisfied Moment redistributionin general affects the shearing forces at the supports and it isrecommended that beam sections are designed for the greater ofthe shear forces calculated before and after redistribution

The use of moment distribution in systems where the beamsare analysed in conjunction with adjoining columns requiresfurther consideration In such cases it is important to ensure thatin any postulated collapse mechanism involving plastic hinges inthe columns these are the last hinges to form To this end it isrecommended that column sections should be designed for thegreater of the moments calculated before and after redistribution

1233 Bending moment diagrams

The moment diagrams and coefficients given in Tables 234and 235 cater for beams that are continuous over two threeand four or more equal spans They apply to cases where thesecond moment of area of the cross section is constant andthe loads on each loaded span are the same For conveniencecoefficients derived by elastic analysis before and after givenredistributions in accordance with the rules of both BS 8110and EC 2 are tabulated against the location points indicated inthe diagrams For example M12 is the coefficient correspondingto the maximum moment at the central support of a two-spanbeam while M13 is the coefficient that gives the moment at thissupport when the moment in the adjoining span is a maximumThus by means of the coefficients given the appropriate envelopeof maximum moments is obtained

Three load types are considered UDL throughout each spana central concentrated load and equal concentrated loadspositioned at the third-points of the span The span momentsdetermined by summing the individual maximum values givenseparately for dead and live loads in the case of uniformloading will be approximate but erring on the side of safetysince each maximum value occurs at a slightly different positionThe tabulated coefficients may also be used to determine thesupport moments resulting from combinations of the givenload types by summing the results for each type The corre-sponding span moments can then be determined as described insection 1213

Moment coefficients are given for redistribution values of10 and 30 respectively For the dead load all the supportmoments have been reduced by the full amount and the spanmoments increased to suit the adjusted values at the supportsFor the live loads all the support moments and for 10 redis-tribution all the span moments have been reduced by the fullamount For 30 redistribution each span moment has beenreduced to the minimum value required for equilibrium with thenew support moments As a result the BS 8110 span momentcoefficients are the same as those for the dead load Althoughthere is no particular merit in limiting redistribution to 10some BS 8110 design formulae for determining the ultimateresistance moment are related to this condition

For design purposes redistribution at a particular sectionrefers to the percentage change in the combined moment dueto the dead and live loads When using the tables the value forthe support moments will be either 10 or 30 but the valuesfor the span moments will need to be calculated for eachparticular case Consider for example a two-span beam sup-porting UDLs with gk qk and 30 redistribution accordingto the requirements of BS 8110

The design load consists of a dead load of 10gk and a liveload of (04gk 16qk) 20gk Then from Table 234 theultimate bending moments are as follows

Before redistributionM11 (0070gk 0096 2gk)l2 0262gkl2

After redistributionM11 (0085gk 0085 2gk)l2 0255gk l2

Redistribution 100 (0262 0255)0262 3

Thus a full 30 reduction of the maximum support moments isobtained with no increase in the maximum span moment

Redistribution of moments 117

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233Continuous beams moment redistribution

Uniformly distributed load (dead load = live load = 1200 units per span)

(b) Live load only Critical cases (for BS 8110)

(a) Dead load only

(c) Dead + live loads Critical cases

(d) Envelope obtained with some reduction of the support moments but with no increase of the span moments

(e) Envelope obtained with maximum reduction of the support moments and some increase of the span moments

(f) Envelope obtained with maximum reduction of moment in middle span and some reduction of moment in end spans

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234Continuous beams bending moment diagrams ndash 1

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235Continuous beams bending moment diagrams ndash 2

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124 ANALYSIS BY MOMENT DISTRIBUTION

The Hardy Cross moment distribution method of analysisin which support moments are derived by a step-by-stepprocess of successive approximations is described brieflyand shown by means of a worked example in Table 236 Themethod is able to accommodate span-to-span variationsin span length member size and loading arrangement Thelsquoprecise moment distributionrsquo method avoids the iterativeprocedure by using more complicated distribution and carry-over factors Span moments can be determined as describedin section 1213

For continuous beams of two three or four spans uniformcross-section and symmetrical loading the support momentsmay also be obtained by using the factors in Table 237

125 INFLUENCE LINES FOR CONTINUOUS BEAMS

The following procedure can be used to determine bendingmoments at chosen sections in a system of continuous beamsdue to a train of loads in any given position

1 Draw the beam system to a convenient scale

2 With the ordinates tabulated in the appropriate Table 238239 240 or 241 construct the influence line (for unit load)for the section being considered selecting a convenientscale for the bending moment

3 Plot on the influence line diagram the train of loads in whatis considered to be the most adverse position

4 Tabulate the value of (ordinate load) for each load

5 Add algebraically the values of (ordinate load) to obtainthe resultant bending moment at the section considered

6 Repeat for other positions of the load train to ensure that themost adverse position has been considered

The following example shows the direct use of the tabulatedinfluence lines for calculating the moments on a beam that iscontinuous over four spans with concentrated loads applied atspecified positions

Example Determine the bending moments at the penultimateleft-hand support of a system of four spans having a constantcross section and freely supported at the ends when loads of100 kN are applied at the mid-points of the first and third spansfrom the left-hand end The end spans are 8 m long and theinterior spans are 12 m long

The span ratio is 115151 and the ordinates are obtained fromTable 240 for penultimate support C

With load on first span (ordinate c)Bending moment (0082 100 8) 656 kNm

With load on third span (ordinate m)Bending moment (0035 100 8) 280 kNm

Net bending moment at penultimate support 376 kNm

Influence lines for continuous beams 121

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236Continuous beams moment distribution methods

HARDY CROSS MOMENT DISTRIBUTION

1 Consider each member to be fixed at ends calculate fixed-endmoments (FEMs) due to external loads on individual members bymeans of Table 228

2 Where members meet sum of bending moments must equalzero for equilibrium ie at B MBAMBC 0 Since FEM(ie FEMBA FEMBC) is unlikely to equal zero a balancingmoment ofFEM must be introduced at each support to achieveequilibrium

3 Distribute this balancing moment between members meeting ata joint in proportion to their relative stiffnesses K Il bymultiplying FEM by distribution factor D for each member(eg at B DBA KAB (KAB KBC) etc so that DBA DBC 1At a free end D 1 at a fully fixed end D0)

4 Applying a moment at one end of member induces momentof one-half of magnitude and of same sign at opposite endof member (termed carry-over) Thus distributed moment

FEM DBA at B of AB produces a moment of (12)FEM DBA at A and so on

5 These carried-over moments produce further unbalanced momentsat supports (eg moments carried over from A and C give rise tofurther moments at B) These must again be redistributed and thecarry-over process repeated

6 Repeat cycle of operations described in steps 2ndash5 until unbalancedmoments are negligible Then sum values obtained each sideof support

Various simplifications can be employed to shorten analysis Themost useful is that for dealing with a system that is freely supportedat the end If stiffness considered for end span when calculatingdistribution factors is taken as only three-quarters of actualstiffness and one-half of fixed-end moment at free support is addedto FEM at other end of span the span may then be treated as fixedand no further carrying over from free end back to penultimatesupport takes place

PRECISE MOMENT DISTRIBUTION

1 Calculate fixed-end moments (FEMs) as for Hardy Cross momentdistribution

2 Determine continuity factors for each span of system fromgeneral expression

where n is continuity factor for previous span and Kn and Kn1

are stiffnesses of two spans Work from left to right along system Ifleft-hand support (A in example below) is free take AB 0 for firstspan if A is fully fixed AB 05 (Intermediate fixity conditionsmay be assumed if desired by interpolation) Repeat the foregoingprocedure starting from right-hand end and working to left (to obtaincontinuity factor AB for span AB for example)

3 Calculate distribution factors (DFs) at junctions between spansfrom general expression

n1 12 Kn1

Kn(2n)

where AB and BA are continuity factors obtained in step 2Note that these distribution factors do not correspond to thoseused in Hardy Cross moment distribution Check that at eachsupport DF 1

4 Distribute the balancing moments FEM introduced at eachsupport to provide equilibrium for the unbalanced FEMs bymultiplying by the distribution factors obtained in step 3

5 Carry over the distributed balancing moments at the supportsby multiplying them by the continuity factors obtained instep 2 by working in opposite direction For example the momentcarried over from B to A is obtained by multiplying thedistributed moment at B by AB and so on This procedure isillustrated in example below Only a single carry-over operationin each direction is necessary

6 Sum values obtained to determine final moments

DFAB 12AB

1ABBA

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237Continuous beams unequal prismatic spans and loads

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238Continuous beams influence lines for two spans

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239Continuous beams influence lines for three spans

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240Continuous beams influence lines for four spans

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241Continuous beams influence lines for five or more spans

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Position Moment Shear

At outer support 0 040FNear middle of end span 008Fl mdashAt interior support 010Fl 060F

In monolithic building construction concrete floors can takevarious forms as shown in Table 242 Slabs can be solid orribbed and can span between beams in either one or two direc-tions or be supported directly by columns as a flat slab Slabelements occur also as decking in bridges and other forms ofplatform structures and as walling in rectangular tanks silosand other forms of retaining structures

131 ONE-WAY SLABS

For slabs carrying uniformly distributed load and continuousover three or more nearly equal spans approximate solutionsfor the ultimate bending moments and shearing forces for bothBS 8110 and EC 2 are given in Table 242 The supportmoments include an allowance for 20 redistribution in bothcases The differences in the values for the two codes occur asa result of the different load arrangements described in section441 However it should be noted that for designs to EC 2 theUK National Annex allows the use of the BS 8110 simplifiedload arrangement as an alternative to that recommended in thebase document For two equal spans the corresponding valuesfor both codes would be

serviceability requirements of cracking and deflection are metby compliance with simplified rules

1321 Uniformly loaded slabs (BS 8110 method)

For rectangular panels carrying uniformly distributed loadwhere the corners are prevented from lifting and adequateprovision is made for torsion the panel is consideredto be divided into middle and edge strips as shown inTable 242 The method may be used for continuous slabswhere the characteristic dead and imposed loads on adjacentpanels and the spans perpendicular to the lines of commonsupport are approximately the same as on the panel beingconsidered

The bending moments and shearing forces on the middlestrips for nine different panel types are given in Table 243Reinforcement meeting the minimum percentage requirementof the code should be provided in the edge strips At cornerswhere either one or both edges of the panel are discontinuoustorsion reinforcement is required This should consist of top andbottom reinforcement each with layers of bars placed parallelto the sides of the slab and extending from the edges a minimumdistance of one-fifth of the shorter span The area of reinforce-ment in each of the four layers as a proportion of that requiredfor the positive moment at mid-span should be three-quarterswhere both edges are discontinuous and three-eighths whereone edge is discontinuous At a discontinuous edge where theslab is monolithic with the support negative reinforcementequal to a half of that required for the positive moment atmid-span should be provided

Where because of differences between contiguous panelstwo different values are obtained for the negative moment at ashared continuous edge these values may be considered asfixed-end moments and moment distribution used to obtainequilibrium in the direction of span The revised negativemoments can then be used to adjust the positive moments atmid-span For each panel the sum of the mid-span moment andthe average of the support moments should be the same as theoriginal sum for that particular panel

When the long span exceeds twice the short span the slabshould be designed as spanning in the short direction In thelong direction the long span coefficient may still be used forthe negative moment at a continuous edge

Chapter 13

Slabs

For designs where elastic bending moments are required thecoefficients given for beams in Table 229 should be used

132 TWO-WAY SLABS

Various methods based on elastic or collapse considerationsare used to design slabs spanning in two directions Elasticmethods are appropriate if for example serviceability checkson crack widths are required as in the design of bridges andliquid-retaining structures Collapse methods are appropriatein cases such as floors in buildings and similar structureswhere the main criterion is the ultimate condition and the

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242Slabs general data

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243Two-way slabs uniformly loaded rectangular panels(BS 8110 method)

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Example Determine the bending moment coefficients in theshort span direction for the slab panel layout shown as follows

by finite element analysis with Poissonrsquos ratio taken as zerousing the following approximate relationships

Bending moments x x0 y0 y y0 x0

Torsion moments xy (1 )xy0

where is Poissonrsquos ratio and x0 y0 xy0 are coefficientscorresponding to 0 Thus if required the tabulated valuescan be readjusted to suit a Poissonrsquos ratio of zero as follows

Bending moments x0 104 (x02 02 y02)y0 104 (y02 02 x02)

Torsion moments xy0 125 xy02

For rectangular panels simply supported on four sides with noprovision to resist torsion at the corners or to prevent the cor-ners from lifting coefficients taken from BS 8110 are alsogiven in Table 244 The coefficients which are derived fromthe Grashof and Rankine formulae (see section 453) are givenby the following expressions

1323 Non-rectangular panels

For a non-rectangular panel supported along all of its edgesbending moments can be determined approximately from thedata given in Table 248 The information which is based onelastic analysis is applicable to panels that are trapezoidaltriangular polygonal or circular For guidance on using thisinformation including the arrangement of the reinforcementreference should be made to section 47

133 CONCENTRATED LOADS

1331 One-way slabs

For a slab simply supported along two opposite edges andcarrying a centrally placed load uniformly distributed over adefined area maximum elastic bending moments are given inTable 245 The coefficients which include for a Poissonrsquos ratioof 02 have been calculated from the data derived for arectangular panel infinitely long in one direction For designs toBS 8110 in which the ULS requirement is the main criteriona concentrated load placed in any position may be spread overa strip of effective width be as shown in Table 245 Parallel tothe supports a strip of width (x ay2) equally spaced eachside of the load has been considered

For slabs that are restrained at one or both edges maximumnegative and positive bending moments may be obtained bymultiplying the simply supported moment by the appropriatefactors given in Table 245 The factors which are given forboth fixed and continuous conditions are those appropriate toelastic beam behaviour

1332 Two-way slabs

For a rectangular panel freely supported along all four edges andcarrying a concentric load uniformly distributed over a definedarea maximum mid-span bending moments based on Pigeaudrsquostheory are given in Tables 246 and 247 Moment coefficients

my (ly lx)

2

8[1 (ly lx)4]

mx (ly lx)

4

8[1 (ly lx)4]

Concentrated loads 131

The upper half of the layout shows the coefficients obtainedfrom Table 243 with ly lx 9060 15 for panel types

AndashB (and CndashD) two adjacent edges discontinuousBndashC one short edge discontinuous

The lower half of the layout shows the coefficients obtained afterdistribution of the unbalanced support moments at B and CThe effective stiffnesses allowing for the effects of simple sup-ports at A and D and carry-over moments at B and C are

Span AndashB (and CndashD) 075Il Span BndashC 05Il

Distribution factors at B and C with no carry-overs are

BA (and CD) 075(075 050) 06BC (and CB) (10 06) 04

Moment coefficients at B and C after distribution are

0078 06 (0078 0058) 0066

Moment coefficients at mid-span after redistribution are

BA (and CD) 0059 05 (0078 0066) 0065BC 0043 (0066 0058) 0035

1322 Uniformly loaded slabs (elastic analysis)

For rectangular panels carrying uniformly distributed loadwhere the corners are prevented from lifting and adequateprovision is made for torsion maximum bending and torsionmoments are given in Table 244 for nine panel types Wherein continuous slabs the edge conditions in contiguous panelsresult in two different values being obtained for the negativemoment at a fixed edge the moment distribution procedureshown in section 1321 could be used but this would ignore theinter-dependence of the moments in the two directions A some-what complex procedure involving edge stiffness factors isderived and shown with fully worked examples in ref 21

The coefficients include for a Poissonrsquos ratio of 02 and havebeen calculated from data given in ref 21 which was derived

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244Two-way slabs uniformly loaded rectangular panels(elastic analysis)

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245One-way slabs concentrated loads

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246Two-way slabs rectangular panel with concentricconcentrated load ndash 1

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247Two-way slabs rectangular panel with concentricconcentrated load ndash 2

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248Two-way slabs non-rectangular panels (elastic analysis)

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given in the charts are used with an appropriate value of Poissonrsquosratio to calculate the bending moments The coefficients given atthe top right corner of each chart are for the limiting case whenthe load extends over the entire panel This case with Poissonrsquosratio taken as 02 is given also in Table 244

When using the chart for square panels (k 10) if ax ayx4 y4 and the resulting bending moment in each direction isgiven by F (1 )x4 In other cases coefficient x4 is based onthe direction chosen for ax and coefficient y4 is obtained byreversing ax and ay as shown in example 1 later

The maximum shearing forces V per unit length on a panelcarrying a concentrated load are given by Pigeaud as follows

ax ay at the centre of length ax V F(2ax ay)at the centre of length ay V F3ax

ay ax at the centre of length ax V F3ay

at the centre of length ay V F(2ay ax)

For panels that are restrained along all four edges Pigeaudrecommends that the mid-span moments be reduced by 20Alternatively the multipliers given for one-way slabs couldbe used in one or both directions as appropriate if the inter-dependence of the bending moments is ignored

Example 1 Consider a square panel freely supported alongall four edges carrying a concentric load with axlx 08 andayly 02 From Table 247 for k 1 the bending momentcoefficients are

For axlx 08 and ayly 02 x4 0072For axlx 02 and ayly 08 y4 0103

Maximum bending moments per unit width with 02 are

For span in direction of ax F(0072 02 0103) 0093FFor span in direction of ay F(0103 02 0072) 0118F

Example 2 A bridge deck is formed of a 200 mm thick slabsupported by longitudinal beams spaced at 2 m centres Theslab is covered with 100 mm thick surfacing Determine for theSLS the maximum positive bending moments in the slab due tothe local effects of live loading

The critical live load for serviceability is the HA wheel loadof 100 kN to which a partial load factor of 12 is applied Fora 100 mm 100 mm contact area and allowing for loaddispersal through the thickness of the surfacing and down tothe mid-depth of the slab (see section 249) the side of theresulting patch load is (300 100 200) 600 mm

The simply supported bending moment coefficients for acentrally placed load by interpolation from Table 245 are

For axlx aylx 6002000 03 mx 0206 my 0145

Allowing for continuity (interior span) in the direction of lx andapplying a partial load factor of 12 the positive bendingmoments per unit width are

In direction of lx

mx 064 0206 12 100 158 kNmm

At right angles to lx

my 0145 12 100 174 kNmm

For design purposes the bending moments determined earlierwill need to be combined with the moments due to the weightof the slab and surfacing and any transverse effects of theglobal deck analysis

Note Using the method given in BS 8110 with x 05lx theeffective width of the strip is

Allowing for continuity (interior span) in the direction of lx andapplying a partial load factor of 12 the positive bendingmoments per unit width are

In direction of lx

At right angles to lx

134 YIELD-LINE ANALYSIS

As stated in section 452 yield-line theory is too complex asubject to deal with adequately in the space available in thisHandbook The following notes are therefore intended merelyto introduce the designer to the basic concepts methods andproblems involved For further information see refs 23 to 28Application of yield-line theory to the design of rectangularslabs subjected to triangularly distributed loads is dealt with insection 1362

1341 Basic principles

When a reinforced concrete slab is loaded cracks form inthe regions where the applied moment exceeds the crackingresistance of the concrete As the load is increased beyond theservice value the concrete continues to crack eventually thereinforcement yields and the cracks extend to the corners ofthe slab dividing it into several areas separated by so-calledyield-lines as shown in diagrams (i)ndash(iii) on Table 249 Anyfurther increase in load will cause the slab to collapse Inthe design process the load corresponding to the formation ofthe entire system of yield lines is calculated and by applyingsuitable partial factors of safety the resistance moment thatmust be provided to avoid collapse is determined

For a slab of given shape it is usually possible to postulatedifferent modes of failure the critical mode depending on thesupport conditions the panel dimension and the relative propor-tions of reinforcement provided in each direction For exampleif the slab shown in diagram (i)(a) is reinforced sufficientlystrongly in the direction of the shorter span by comparison withthe longer span this mode of failure will be prevented and thatshown in diagram (i)(b) will occur instead Similarly if the slabwith one edge unsupported shown in diagrams (ii) is loadeduniformly pattern (b) will occur when the ratio of the longer toshorter side length (or longer to shorter lsquoreduced side lengthrsquo seesection 1346) exceeds radic2 otherwise pattern (a) will occur

All of these patterns may be modified by the formation ofcorner levers see section 1349

my 12 100 1842 10 06 1

0618 139kNmm

mx 064 12 100 10

2 18 1 06

2 20 181kNmm

be 06lx ay 06 20 06 18 m

Yield-line analysis 137

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249Two-way slabs yield-line theory general information

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1342 Rules for postulating yield-line patterns

Viable yield-line patterns must comply with the following rules

1 All yield lines must be straight

2 A yield line can only change direction at an intersection withanother yield-line

3 A yield line separating two elements of a slab must passthrough the intersection of their axes of rotation (Note thispoint may be at infinity)

4 All reinforcement intercepted by a yield line is assumed toyield at the line

1343 Methods of analysis

Two basic methods of analysis have been developed These arecommonly referred to as the lsquoworkrsquo or lsquovirtual workrsquo methodand the lsquoequilibriumrsquo method The former method involvesequating for the yield-line pattern postulated the work done bythe external loads on the various areas of the slab to obtain avirtual displacement to the work done by the internal forces informing the yield lines When the yield-line pattern is adjustedto its critical dimensions the ratio of the ultimate resistance tothe ultimate load reaches its maximum When analysing a slabalgebraically this situation can be ascertained by differentiatingthe expression representing the ratio and equating the differentialto zero in order to establish the critical dimensions Then byre-substituting these values into the original expression a formulagiving the required ultimate resistance for a slab of givendimensions and loading can be derived

The so-called equilibrium method is not a true equilibriummethod but a variant of the work method which also gives anupper-bound solution The method has the great advantagethat the resulting equations provide sufficient information ofthemselves to eliminate the unknown variables and thereforedifferentiation is unnecessary Although there are also otheradvantages the method is generally more limited in scope andis not described here for details see refs 23 and 27

1344 Virtual-work method

As explained earlier this method consists of equating thevirtual work done by the external loads in producing a givenvirtual displacement at some point on the slab to the work doneby the internal forces along the yield lines in rotating the slabelements To demonstrate the principles involved an analysiswill be given of the freely supported rectangular slab supportinga uniform load and reinforced to resist equal moments M eachway shown in diagram (i)(a) on Table 249

Clearly due to symmetry yield line OO will be midwaybetween AB and CD Similarly and thus only onedimension is unknown Consider first the external work done

The work done by an external load on an individual slabelement is equal to the area of the element times the displace-ment of its centroid times the unit load Thus for the triangularelement ADO with displacement at O

work done (12)(lx)(ly)(3)n lxlyn6

Similarly for the trapezoidal area ABOO with displacement at O and O

work done [(lx2)(ly)( 2) 2(12)(lx2)(ly)(23)]

(3 4)lxlyn12

Thus since the work done on BCO is the same as that doneon ADO and the work done on CDOO is the same as that onABOO for the entire slab

Total external work done 2[lxlyn6 (3 4)lxlyn12]

(3 2)lxlyn6

The internal work done in forming a yield line is equal to themoment along the yield line times the length of the line timesthe rotation A useful point to note is that where a yield line isformed at an angle to the direction of principal momentsinstead of considering its true length and rotation it is usuallysimpler to consider the components in the direction of theprincipal moments For example for yield line AO instead ofconsidering the actual length AO and the rotation at right anglesto AO consider length lx2 and the rotation about AB pluslength ly and the rotation about AD

Thus considering the component about AB of the yield linealong AOO B the length of the line is ly the moment is M andthe rotation is ( lx2) Hence

work done 2M(lylx)

Similarly for the yield line along DOO C the work done isagain 2M(lylx) (Length OO is considered twice because therotation between the elements separated by this length is dou-ble that occurring over the remaining length) Now consideringthe component about AD of the yield line AOD

work done M(lxly)

Since the work done on yield line BOC is similar for theentire slab

Total internal work done 2M(2lylx lxly)

Equating the external work done to the internal work done

(3 2)lxlyn6 2M(2lylx lxly)

or

To determine the critical value of ly the quotient in squarebrackets must be differentiated and equated to zero As Jones(ref 27) has pointed out to use the well-known relationship

simply as a means of maximising y uv it is convenient torearrange it in the form

Thus for the present example

This leads to a quadratic in the positive root of which is

12lx

ly4

3lx

ly2

lx

ly2

3 22

2 (lxly)2

3ndash 4

2

uv

dudxdvdx

dydx

v dudx

u dvdx v2 0

M n

12l 2

x 3 22

2 (lx ly)2

Yield-line analysis 139

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When substituted in the original equation this gives

1345 Concentrated and line loads

Concentrated and line loads are simpler to deal with thanuniform loads When considering the external work done thecontribution of a concentrated load is equal to the load timesthe relative deflection of the point at which it is applied In thecase of a line load the external work done over a given slab areais equal to the portion of the load carried on that area times therelative deflection at the centroid of the load

Yield lines tend to pass beneath heavy concentrated or lineloads since this maximises the external work done by suchloads When a concentrated load acts in isolation a so-calledcircular fan of yield lines tends to form this behaviour iscomplex and reference should be made to specialist textbooksfor details (refs 23 24 26)

1346 Affinity theorems

Section 1344 illustrates the work involved in analysing asimple freely supported slab with equal reinforcement in eachdirection (ie so-called isotropic reinforcement) If differentamounts of reinforcement are provided in each direction(ie so-called orthotropic reinforcement) or if continuity orfixity exists along one or more edges the formula needsmodifying accordingly

To avoid the need for a vast number of design formulae tocover all conceivable conditions it is possible to transformmost slabs with fixed or continuous edges and orthotropicreinforcement into their simpler freely supported isotropicequivalents by using the following affinity theorems skew slabscan be transformed similarly into rectangular forms

1 If an orthotropic slab is reinforced as shown in diagram (iv)(a)on Table 249 then it can be transformed into the simplerisotropic slab shown in diagram (iv)(b) All loads anddimensions in the direction of the principal co-ordinate axisremain unchanged but in the affine slab the distances in thedirection of the secondary co-ordinate axis are equal to theactual values divided by radic and the corresponding total loadsare equal to the original values divided by radic (The latterrequirement means that the intensity of a UDL per unit arearemains unchanged by the transformation since both the areaand the total load on that area are divided by radic)

Similarly a skew slab reinforced as shown in diagram(v)(a) can be transformed into the isotropic slab shown indiagram (v)(b) by dividing the original total load by sin(As before this requirement means that the intensity of aUDL per unit area remains unchanged by the transformation)

These rules can be combined when considering a skewslab with different reinforcement in each direction fordetails see ref 28 article 6

2 By using the reduced side lengths lxr and lyr an orthotropicslab that is continuous over one or more supports such as

M n

24lx

23 lx

ly2

lx

ly2

that in diagram (vi)(a) on Table 249 can be transformedinto the simpler freely supported isotropic slab shown indiagram (vi)(b)

For example for an orthotropic slab with fixed edges that isreinforced for positive moment M and negative moments i2Mand i4M in span direction lx and for positive moment M andnegative moments i1M and i3M in span direction ly it can beshown that

where

and

This is identical to the expression derived above for the freelysupported isotropic slab but with lxr and lyr substituted for lx

and ly Values of Mnlx2 corresponding to ratios of lylx can be

read directly from the scale on Table 249The validity of the analysis is based on the assumption that

lyr lxr If this is not the case the yield line pattern will be asshown in diagram (i)(b) on Table 249 and lxr and lyr should betransposed as shown in the following example

Example Design the slab in diagram (vii) on Table 249 tosupport an ultimate load n per unit area assuming that therelative moments of resistance are as shown

Since lx 4 m i2 32 and i4 0

Since ly 6 m 12 i1 1 and i3 0

Thus

1347 Superposition theorem

A problem that may arise when designing a slab to resist acombination of uniform concentrated and line loads some ofwhich may not always occur is that the critical pattern of yieldlines may well vary for different combinations of loads Also itis theoretically incorrect to sum the ultimate moments obtainedwhen considering the various loads individually since thesemoments may result from different yield line patterns HoweverJohansen has established the following superposition theorem

The sum of the ultimate moments for a series of loads is equalto or greater than those due to the sum of the loads

n

24 31023 310

7032

310703

2

0726 n

M n

24l 2xr3 lxr

lyr2

lxr

lyr2

ly r 2 6

(1 1 1)12 703 m

lxr 2 4

1 32 1 310 m

lyr 2ly

[1 i1 1 i3]lxr

2lx

1 i2 1 i4

M n

24l2xr3 lxr

lyr2

lxr

lyr2

Slabs140

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In other words if the ultimate moments corresponding to theyield-line patterns for each load considered separately areadded together the resulting value is equal to greater than thatof the system as a whole This theorem is demonstrated inref 28 article 4

1348 Empirical virtual-work analysis

An important advantage of collapse methods of design is thatthey can be readily applied to solve problems such as slabs thatare irregularly shaped or loaded or that contain large openingsThe analysis of such slabs using elastic methods is by comparisonextremely arduous

To solve such lsquoone-offrsquo problems it is clearly unrealistic todevelop standard algebraic design formulae The followingempirical trial-and-adjustment technique which involves adirect application of the virtual-work principles is easy tomaster and can be used to solve complex problems It is bestillustrated however by working through a simple problemsuch as the one considered in section 1346 There is of courseno need to employ the procedure in this case It is used hereonly to illustrate the method A more complicated example isgiven in ref 28 article 1 on which the description of themethod is based

In addition to the fundamental principles of virtual workdiscussed in section 1344 the present method depends alsoon the following principle If all yield lines (other than thosealong the supports) are positive and if none of them meets anunsupported edge except at right angles then no forces due toshear or torsion can occur at the yield lines Thus a separatevirtual-work balance for each slab area demarcated by the yieldlines can be taken

Example Consider the slab shown in diagram (vii) onTable 249 which is continuous over two adjacent edges freelysupported at the others and subjected to a uniform load n perunit area The ratios of the moments of resistance provided overthe continuous edges and in the secondary direction to thatin the principal direction are as shown

The step-by-step trial and adjustment process is as follows

1 Postulate a likely yield-line pattern

2 Give a virtual displacement of unity at some point andcalculate the relative displacement of any other yield-line

intersection points For the case considered if O is given adisplacement of unity the displacement at O will alsobe unity since OO is parallel to the axes of rotation of theadjoining slab areas

3 Choose reasonable arbitrary values for the dimensions thatmust be determined to define the yield-line pattern Thus inthe example initially ly is taken as 2 m ly as 1m and lx

as 25 m

4 Calculate the actual work done by the load n per unit areaand the internal work done by the moments of resistance Mfor each separate part of the slab and thus obtain ratios ofMn for each part

For example on area A the total load is 12 4 2 n 4nSince the centre of gravity moves through a distance of 13 thework done by the load on area A is 4n 13 4n3 Now sinceO is displaced by unity the rotation of area A about the supportis 1ly 12 The moments of M2 acting across both the pos-itive and negative yield lines each exert a total moment ofM2 4 2M Thus the total internal work done in rotatingarea A is (2M 2M) 12 2M Equating the internal andexternal work done on area A gives 2M 4n3 that isMn 23 Similarly for area C Mn 13

For convenience area B can be divided it into a rectangle (ofsize 3 m 25 m) and two triangles and the work done on eachpart calculated separately Since the centres of gravity movethrough a distance of 12 for the rectangle and 13 for eachtriangle the external work done is as follows

Rectangular area 3 25 12 n 375nTriangular areas 12 (2 1) 25 13 n 125nTotal 500n

Since the rotation is 125 the work done by the moments is(15M M) 6 125 6M Thus the virtual-work ratio isMn 56 0833 Likewise for area D Mn 34

5 Sum the separate values of internal and external workdone for the various slab areas and thus obtain a ratio ofMn for the entire slab This ratio will be lower than thecritical value unless the dimensions chosen arbitrarily instep 3 happen to be correct The calculations are best set outin tabular form as follows

Yield-line analysis 141

Area External work done Internal work done Mn

A 12 4 2 13 n 1333n [M2 M2] 4 12 2000M 0667B [3 25 12 12 3 25 13] n 5000n [3M2 M] 6 125 6000M 0833C 12 4 1 13 n 0667n [0 M2] 4 11 2000M 0333D [3 15 12 12 3 15 13] n 3000n [0 M] 6 115 4000M 0750

Total 10000n Total 14000M 0714

A 12 4 2071 13 n 1381n [M2 M2] 4 12071 1931M 0714B [2467 2424 12 12 3533 2424 13] n 4418n [3M2 M] 6 12424 6188M 0714C 12 4 1462 13 n 0975n [0 M2] 4 11462 1368M 0714D [2467 1576 12 12 3533 1576 13] n 2872n [0 M] 6 11576 3807M 0754

Total 9646n Total 13294M 0726

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6 By comparing the overall ratio obtained for Mn withthose due to each individual part it is possible to see howthe arbitrary dimensions should be adjusted so that the ratiosfor the individual parts become approximately equal to eachother and to that of the slab as a whole

The foregoing table shows calculations for initial values of Mnand also for a set of adjusted values An examination of theinitial values (for which the overall ratio is 0714) shows thatto obtain the same ratio for each area Mn needs to be increasedfor areas A and C and reduced for areas B and D For area Asince Mn is proportional to (ly)2 ly needs to be increased toradic(07140667) 2 2071 Similarly for area C ly needsto be increased to radic(07140333) 1 1462 If for area Bthe external work done is recalculated using the correctedvalues of ly and ly this gives

[2467 25 12 12 3533 25 13] n 4556n

The internal work done by the moments is unchanged and sothe revised value of Mn is 45566 0759 Thus since forarea B Mn is proportional to (lx)2 lx needs to be reducedto radic(07140759) 25 2424 For area D the external workdone is recalculated using the corrected values of all thevariable dimensions and the revised value of Mn obtained

7 Repeat steps 4 and 5 using the adjusted values for thearbitrary dimensions as shown earlier

8 Repeat this cyclic procedure until reasonable agreement isobtained between the values of Mn This ratio gives thevalue of M for which the required reinforcement must bedetermined for a given load n In the example the ratiosgiven by the second cycle are quite satisfactory Note thatalthough some of the dimensions originally guessed werenot particularly accurate the resulting error in the value ofMn obtained for the whole slab was only about 15 andthe required load-carrying capacity is not greatly affected bythe accuracy of the arbitrary dimensions

Concentrated loads and line loads occurring at boundariesbetween slab areas should be divided equally between the areasthat they adjoin and their contribution to the external workdone assessed as described in section 1345

As in all yield-line theory the above analysis is only valid ifthe yield-line pattern considered is the critical one Where thereis a reasonable alternative both patterns should be investigatedto determine which is critical

1349 Corner levers

Tests and elastic analyses of slabs show that the negativemoments along the edges reduce to zero near the corners andincrease rapidly away from these points Thus in slabs that arefixed or continuous at their edges negative yield lines tend toform across the corners and in conjunction with pairs of positiveyield lines result in the formation of additional triangular slabelements known as corner levers as shown in diagram (i)(a) onTable 250 If the slab is freely supported a similar mechanismis induced causing the corners to lift as shown in diagram(1)(b) If these mechanisms are substituted for the original yieldlines running into the corner of the slab the overall strength ofthe slab is correspondingly decreased by an amount dependingon the factors listed on Table 250 For a corner lever having anincluded angle of not less than 90o the strength reduction is not

likely to exceed 8ndash10 In such cases the main reinforcementcan be increased slightly and top reinforcement provided at thecorners of the slab to restrict cracking Recommendations takenfrom the Swedish Code of Practice are shown in diagram (ii) onTable 250

For acute-angled corners the decrease in strength is moresevere For a triangular slab ABC where no corner angle is lessthan 30o Johansen (ref 25) suggests that the calculated strengthwithout corner-lever action should be divided by a factor kgiven by the approximation

k (74 sin A sin B sin C)4

A mathematical determination of the true critical dimensions ofan individual corner lever involves much complex trial andadjustment However this is unnecessary as Jones and Wood(ref 23) have devised a direct design method that gives cornerlevers having dimensions such that the resulting adjustment instrength is similar to that due to the true mechanisms Thisdesign procedure is summarised on Table 250 and illustratedby the following example

The formulae derived by Jones and Wood and on which thegraphs in Table 250 are based are as follows

With fixed edges

where

and

With freely supported edges

where

and

Example Calculate the required resistance of the 5 m squareslab with fixed edges (i 1) shown on Table 250

For the transformed freely supported slab the reduced sidelength lr lradic(1 I) 5radic2 354 m Then for a square slabif the formation of corner levers is ignored the required resistancemoment M (124)nlr

2 0041 3542n 0521nNow from the appropriate graph on Table 250 with i 1

and 13 90o read off k1 11 and k2 42 Thus dimensionsa1 11radic0521 0794 m and a2 42radic0521 3032 m

By plotting these values on a diagram of the slab it is nowpossible to calculate the revised resistance moment required Ifthe deflection at the centre is unity the relative deflection at theapex of the corner levers is 30323536 0858 Thus therevised virtual-work equation is

[(13) 52 4 (12) 07942 (13) 0858] n

This reduces to 14036M 7973n so that M 0568n

4 2M3412 125 07942 0858

2471

cot [K2(1 i) (2 i)] tan (2)

K2 (4 i) 3 cot2 (2)

k2 k1

cos (2) cot sin (2)

k1 [K2 2(1 i) ]2 3 sec(2)

cot (K1 1) tan (2)K1 4 3 cot2 (2) 1

k2 k1

cos (2) cot sin (2)

k1 1K1 sin2(2)

16(1 i) sec(2)

Slabs142

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250Two-way slabs yield-line theory corner levers

Freely supportedslab edges

Corner leverFixed slabedges

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Note that this value of M is 9 greater than the uncorrectedvalue in other words the load supported by a square slab witha specified moment of resistance is actually 9 less than thatcalculated when corner levers are not taken into account

135 HILLERBORGrsquoS SIMPLE STRIP THEORY

1351 Moments in slabs

According to lower-bound (equilibrium) theory load acting ona slab is resisted by a combination of biaxial bending andtorsion In the simple strip method the torsion moment is takenas zero and load (or partial load) acting at any position on theslab is resisted by bending in one of two principal directionsThus in diagram (i) on Table 251 the load acting on theshaded areas is resisted by bending in direction ly and the loadacting on the remaining area is resisted by bending in direc-tion lx In principle there is an unlimited number of ways ofapportioning the load each of which will lead to a differentreinforcement layout while still meeting the collapse criteriaHowever the loading arrangement selected should also ensurethat the resulting design is simple economical and satisfactorywith regard to deflection and cracking under service loads

Some possible ways of apportioning the load on a freelysupported rectangular slab are shown in diagrams (i)ndash(iv) onTable 251 the notation adopted being given on the tablePerhaps the most immediately obvious arrangement is shown indiagram (i) for which Hillerborg originally suggested that 13could be taken as 45o where both adjacent edges are freelysupported However in ref 29 he recommends that 13 should bemade equal to tan1(lylx) as shown in diagram (i) The disad-vantage of the arrangement shown is that the bending moment(and thus the reinforcement theoretically required) variesacross strips 2 and 3 Since it is impractical to vary the rein-forcement continuously the usual approach is to calculatethe total moment acting on the strip divide by the width ofthe strip to obtain the average moment and provide a uniformdistribution of reinforcement to resist this moment To avoidhaving to integrate across the strip to obtain the total momentHillerborg recommends calculating the moment along thecentreline of the strip and then multiplying this value bythe correction factor

where lmax and lmin are the maximum and minimum loadedlengths of the strip Strictly speaking averaging the momentsas described violates the principles on which the method isbased and this device should only be used where the factor ofsafety will not be seriously impaired If the width of the stripover which the moments are to be averaged is large it is betterto sub-divide it and calculate the average moment for eachseparate part

An alternative arrangement that avoids the need to averagethe moments across the strips is shown in diagram (ii) This hasdisadvantages in that six different types of strip (and thus sixdifferent reinforcement layouts) must be considered and thatin strip 6 no moment theoretically occurs Such a strip mustnevertheless contain distribution reinforcement

1 (lmax lmin)

2

3(lmax lmin)2

So far the load on any one separate area has been carried inone direction only In diagram (iii) however the loads on thecorner areas are so divided that one-half is carried in eachdirection Hillerborg (ref 29) states that this very simple andpractical approach never requires more than 10 additionalreinforcement when compared to the theoretically more exactbut less practical solution shown in diagram (i) when lylx isbetween 11 and 4 An additional sophistication that can beintroduced is to apportion the load in each direction in thetwo-way spanning areas in such a way that the resulting rein-forcement across the shorter span corresponds to the minimumrequirement for secondary reinforcement Details of this andsimilar stratagems are given in ref 29

Diagram (iv) illustrates yet another arrangement that may beconsidered By dividing the corner areas into triangles andaveraging the moments over these widths as described earlierHillerborg shows that the moments in the side strips can bereduced to two-thirds of the values given by the arrangementused in diagram (iii)

1352 Loads on supporting beams

A particular feature of the strip method is that the boundariesbetween the different loaded areas also define the manner inwhich the loads are transferred to the supporting beamsFor example in diagram (i) the beams in direction lx supporttriangular areas of slab giving maximum loads of nlx

22ly attheir centres

136 BEAMS SUPPORTING RECTANGULAR PANELS

The loads on beams supporting uniformly loaded rectangularslab panels are distributed approximately as a triangular loadon the beams along the shorter edges lx and a trapezoidalload on the beams along the longer edges ly as shown in thediagrams on Table 252 For a beam supporting a single panelof slab that is either freely supported or subjected to the samedegree of restraint along all four edges where the beam span isequal to the length (or width) of the panel the equivalent UDLper unit length on the beam for the calculation of bendingmoments only is as follows

Short-span beam nlx3

Long-span beam (1 13k2) nlx2

where n is the total UDL per unit area on the slab appropriateto the limit-state being considered and k lylx For a beamsupporting two identical panels one on either side the fore-going equivalent loads are doubled If a beam supports morethan one panel in the direction of its length the distribution ofload is in the form of two or more triangles (or trapeziums)and the foregoing formulae are not applicable in such a casehowever it is sufficiently accurate if the total load on the beamis considered to be uniformly distributed

For slabs designed in accordance with the BS 8110 methodthe loads on the supporting beams may be determined from theslab shear forces given in Table 243 The loads are to be takenas uniformly distributed along the middle three-quarters ofthe beam length where the shear force for the short span of theslab is the load on the long-span beam and vice versa Theresulting beam fixed-end moments can be determined fromTable 228

Slabs144

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251Two-way slabs Hillerborgrsquos simple strip theory

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252Two-way slabs rectangular panels loads on beams(common values)

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Triangularly distributed loads 147

137 TRIANGULARLY DISTRIBUTED LOADS

In the design of rectangular tanks storage bunkers and someretaining structures conditions occur of wall panels spanningin two directions and subjected to distributions of pressurevarying linearly from zero at or near the top to a maximum atthe bottom For liquid-retaining structures with no provisionfor additional protection in the form of an internal lining orexternal tanking an elastic analysis is normally necessary as abasis for checking serviceability cracking In other cases ananalysis based on collapse methods may be justified

1371 Elastic analysis

The coefficients given in Table 253 enable the maximumvalues of bending moments and shearing forces on vertical andhorizontal strips of unit width to be determined for panels ofdifferent aspect ratios and edge conditions The latter are takenas fixed at the sides hinged or fixed at the bottom and hingedor free at the top The coefficients which are taken from ref 32were derived by a finite element analysis and include for aPoissonrsquos ratio of 02 For ratios less than 02 the momentscould be adjusted in the manner described in section 1322

The maximum negative bending moment at the bottom edgeand the maximum shear forces at the bottom and top edgesoccur halfway along the panel The other maximum momentsoccur at the positions indicated in the following table

From Table 253 for panel type 4 with lx lz 54 125 themaximum bending moments are as follows

Horizontal negative moment at corners

mx 0050 981 43 314 kNmm

Horizontal positive moment (at about 05lz 2 m above base)

mx 0022 981 43 138 kNmm

Vertical positive moment (at about 03lz 12 m above base)

mz 0021 981 43 132 kNmm

1372 Yield-line method

A feature of the collapse methods of designing two-way slabsis that the designer is free to choose the ratios between thetotal moments in each direction and between the positive andnegative moments In the case of liquid-retaining structureswhere it is important to ensure that the formation of cracksunder service load is minimised the ratios selected shouldcorrespond approximately to those given by elastic analysisThe following design procedure is thus suggested

1 Obtain maximum positive and negative service momentcoefficients from Table 253

2 Determine i1 ( i3) and i4 where mhposmzposi1 i3 mhnegmhpos and i4 mznegmzpos

3 Calculate lxr and lyr if the top edge is unsupported from

and

and if the top edge is freely supported from

and

4 Obtain M if the top edge is unsupported from the chart onTable 254 and if the top edge is supported from the scaleon Table 249 according to the values of f (or n f2) lx

(or lxr) lyr and i4 The basis of this approach is given below

Top edge of slab unsupported In ref 25 Johansen derivesthe following lsquoexactrsquo formulae according to the failure mode

For failure mode 1

where k is obtained by solving the quadratic equation

For failure mode 2

where is obtained by solving the following cubic equation

8lxr

lyr2

3 3ndash16lxr

lyr2 2 8 6 0

M f l 2

yr

96(6 8 3 2)

i241 i4 4 lx

lyr2 0

41 i4 lx

lyr2k2 4i4(1 i4) (6 4i4) lx

lyr2k

M f l 2

x

12k

lyr 2ly

[1 i1 1 i3 ]lxr

2lx

1 1 i4

lyr 2ly

[1 i1 1 i3 ]lxr

lx

1 i4Distance from bottom of panel to position ofmaximum horizontal moments (negativepositive)

Type of Heightlz for values of lx lzpanel 05 10 15 20 40

1 03 05 05 05 052 03 05 0710 0910 103 03 04 04 04 044 04 04 0506 0910 0910

Distance from bottom of panel to position ofmaximum vertical positive moment

Type of Heightlz for values of lxlzpanel 05 10 15 20 40

1 03 05 05 05 052 03 04 05 06 073 02 03 03 04 044 02 03 03 04 04

For a complete map of bending moment values at intervals ofone-tenth of the panel height and length see ref 32 Themoments obtained for an individual panel apply directly to asquare tank with hydrostatic loading For a rectangular tanka further distribution of the unequal negative moments at thecorners is needed (see Tables 275 and 276)

Example Determine due to internal hydrostatic loading themaximum service moments in the walls of a square tank thatcan be considered as free along the top edge and hinged alongthe bottom edge The tank is 5 m square 4 m deep and thewater level is to be taken to the top of the walls

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253Two-way slabs triangularly distributed load(elastic analysis)

Fixed

Fixed

Fixed

Panel 1 Panel 2 Panel 3 Panel 4

Hinged

Fixed

Fixed

Fixed

Free

Fixed

Fixed

Hinged

Hinged

Fixed

Hinged

Free

lx

lz

lx

Fixed

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254Two-way slabs triangularly distributed load(collapse method)

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The results of these calculations can be plotted graphically asshown in Table 254 from which coefficients of Mfl2

x can beread corresponding to given values of lyrlx and i4 The chainline on the chart indicates the values at which the failuremode changes

Top edge of slab supported In ref 25 Johansen showsthat the total moments resulting from yield-line analysisfor triangularly loaded slabs correspond to those obtainedwhen the same slab is loaded with a uniform load of one-halfthe maximum triangular load (ie n f 2) Hence the designexpressions in section 1346 and the scale on Table 249 foruniformly loaded slabs can again be used As before if theedges of the slab are restrained the reduced side lengths lxr andlyr should be calculated and substituted in the formula insteadof the actual side lengths lx and ly

Example Determine moments by yield-line analysis for thewalls of the square tank considered in the previous example insection 1371

Taking the service moment coefficients obtained previouslythe following values are the most suitable for

00220021 10 and i1 ( i3) 00500022 23

Hence

lyr 275 and lyrlx 27540 070

Then from the chart on Table 254 with i4 0

Mflx2 0018

Comparing this value with mx 0022 as obtained in theprevious example it can be seen that the elastic moments are12 times those determined by yield-line analysis Thus evenwith a partial load factor of 12 the yield-line moments are nogreater than the elastic service moments in this case

Note It can be seen from the chart on Table 254 that failuremode 2 applies for which

Hence

and yield-lines intersect at height lx 0335 40 134 mabove the base

138 FLAT SLABS (SIMPLIFIED METHOD)

The following notes and the data in Tables 255 and 256 arebased on the recommendations for the simplified method of flatslab design given in BS 8110 Alternatively and in cases wherethe following conditions are not met the structure can beanalysed by the equivalent frame method Other methods ofanalysis such as grillage finite-element and yield line may alsobe employed in which case the provisions given for the distri-bution of bending moments are a matter of judgement

32 8 6 0018 96 (40 275)2 366 0335

M fl 2

yr

96(6 8 3 2) 0018 fl2

x

2 521 2310

1381 Limitations of method

The system must comprise at least three rows of rectangularpanels in each direction The spans should be approximatelyequal in the direction being considered and the ratio of thelonger to the shorter sides of the panels should not exceed 2The conditions allowing the design to be based on the singleload case of all spans loaded with the maximum design load asexplained in section 441 and Table 242 must be met Alllateral forces must be resisted by shear walls or bracing

1382 Forms of construction

The slab can be of uniform thickness throughout or thickeneddrop panels can be introduced at the column positions Droppanels can extend to positions beyond which the slab can resistpunching shear without needing shear reinforcementAlternatively the panels can be further extended to positionswhere they may be considered to influence the distribution ofmoments within the slab In this case the smaller dimensionof the drop panel should be at least one-third of the smallerdimension of the surrounding slab panels

Columns can be of uniform cross section throughout or canbe provided with an enlarged head the effective dimensions ofwhich are limited according to the depth of the head as shownin Table 255 For a flared head the actual dimension is takento be the value at a depth 40 mm below the underside of the slabor drop

The effective diameter of a column or column head is thediameter of a circle whose area is equal to the cross-sectionalarea of the column or effective column head In no case is theeffective diameter hc to be taken greater than one-quarter ofthe shortest span framing into the column In cases where theedges of the slab are supported by walls hc can be taken asthe thickness of the wall

1383 Bending moments and shearing forces

The total design bending moments and shearing forces given inTable 255 are the same as those given for one-way slabs inTable 242 where the support moments include for 20 redis-tribution The requirements are applied independently in eachdirection Any negative moments greater than those at a dis-tance hc2 from the centreline of the column may be ignoredsubject to the following condition being met In each span thesum of the maximum positive moment and the average ofthe negative moments for the whole panel width where F is thetotal design load on the panel and l is the panel length betweencolumn centrelines must be not less than

(Fl8) (1 2hc3l)2

This condition is met in the case of designs that are based onthe single load case of all spans loaded with the maximumdesign load by taking the design negative moment as the valueat a distance hc3 from the centreline of the column Themoment at this position is obtained approximately by reducingthe value at the column centreline by 015Fhc

The total design moments on a panel should be apportionedbetween column and middle strips as shown in Table 255 Inthe following table the moment allocations given in BS 8110and EC 2 are shown for comparison

Slabs150

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255Flat slabs BS 8110 simplified method ndash 1

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256Flat slabs BS 8110 simplified method ndash 2

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At the edges of a slab the transfer of moments between the slaband an edge or corner column is limited by the effective breadthbe of the moment transfer strip given in Table 256 Themaximum design moment that can be transferred by this stripis given by the equation Mt max 015bed2fcu in BS 8110 andMt max 017bed2fck in EC 2

At internal columns two-thirds of the reinforcement neededto resist the negative moments in the column strips should beplaced in a width equal to half that of the column strip and cen-tral with the column Otherwise the reinforcement needed toresist the moment apportioned to a particular strip should bedistributed uniformly across the width of the strip

1384 Effective forces for punching shear

The critical consideration for shear in flat slab structures is thatof punching shear around the columns The design forceobtained by summing the shear forces on each side of the col-umn is multiplied by an enhancement factor to allow for theeffects of moment transfer as given in Table 256 The effectiveshear force is determined independently in each direction andthe design checked for the worse case

1385 Reservoir roofs

For reservoir roofs with simply supported ends where elasticmoments are required to check serviceability requirements thecoefficients given for beams in Table 230 could be used In thiscase the negative moments at the centrelines of the columnscould be reduced by 022Fhc to give approximately themoment at a distance hc2 from the centreline of the columnThis approach will still ensure that the minimum momentrequirement mentioned in section 1383 is met

Flat slabs (simplified method) 153

Design moment Column strip Middle strip

BS 8110 Negative 75 25Positive 55 45

EC 2 Negative 60ndash80 40ndash20Positive 50ndash70 50ndash30

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When using the formulae and coefficients in this chapter theloads must include the appropriate partial safety factors for thelimit-state being considered Design loads for the ULS inaccordance with the requirements of BS 8110 and EC 2 aregiven in Table 257

For many framed structures it is not necessary to carry out afull structural analysis of the complete frame as a single unitBS 8110 makes a distinction between frames supporting verticalloads only because lateral stability to the structure as a wholeis provided by other means such as shear walls and framessupporting both vertical and lateral loads Simplified models forthe purpose of analysis as described in section 491 are alsoshown in Table 257

The moment-distribution method used to analyse systems ofcontinuous beams as shown in Table 236 can be extended toapply to no-sway sub-frames (see section 492) as shown inTable 258 and single-bay sway frames (see section 493) asshown in Table 259

141 SLOPE-DEFLECTION METHOD OF ANALYSIS

Moment analysis of a restrained member by slope-deflection isbased on the following two principles The difference in slopebetween any two points in the length of the member is equalto the area of the MEI diagram between the two points Thedistance of any point on the member from a line drawn tangen-tially to the elastic curve at any other point the distance beingmeasured normal to the initial position of the member is equalto the moment (taken about the first point) of the MEI diagrambetween these two points In the foregoing M represents thebending moment E the modulus of elasticity of the materialand I the second moment of area of the member The bendingmoments that occur at the ends of a member subject to thedeformation and restraints shown in the moment diagram atthe top of Table 260 are given by the corresponding formulaeThe formulae which have been derived from a combinationof the basic principles are given in a general form and in thespecial form for members on non-elastic supports

The stiffness of a member is proportional to EIl but as E isassumed to be constant the term that varies in each member isthe stiffness factor K Il The terms FAB and FBA relate to theload on the member When there is no external load FAB andFBA are zero when the load is symmetrically disposed

FAB FBA Al Values of FAB FBA and Al for different loadcases are given in Table 228

The conventional signs for slope-deflection analyses are anexternal restraint moment acting clockwise is positive a slopeis positive if the rotation of the tangent to the elastic line isclockwise a deflection in the same direction as a positive slopeis positive

Example Establish the formulae for the bending moments ina column CAD into which is framed a beam AB The beam ishinged at B and the column is fixed at C and D (see diagramin Table 260) The beam only is loaded Assume there is nodisplacement of the joint A

From the general formulae given on Table 260

Therefore

Thus

For symmetrical loading

MAC 6KAC(AAB lAB)

3KAB 4KAC 4DAD

FAB FBA 2 15AAB lAB

MAB (MAC MAD)

MDA MAD 2

MAD MACKAD KAC

MCA 2EKAC13A MAC 2

MAC 4KAC(FAB FBA 2)

3KAB 4KAC 4KAD

E13A FAB FBA 2

3KAB 4KAC 4KAD

E13A(3KAB 4KAC 4KAC) (FAB FBA2) 0

MAB MAC MAD

MAC 4EKAC13A and MAD 4EKAD13A

MAB 3EKAB13A (FAB FBA 2)

Chapter 14

Framed structures

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257Frame analysis general data

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258Frame analysis moment-distribution method no sway

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259Frame analysis moment-distribution method with sway

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260Frame analysis slope-deflection data

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142 CONTINUOUS BEAMS AS FRAME MEMBERS

In many buildings the interaction of the columns and beamscan be considered with sufficient accuracy by applying one ofthe simplified models shown in Table 257 The simplified two-or three-span sub-frames in Table 261 are analysed on theassumption that the remote ends of the beams and columns arefixed Therefore for any internal span ST the ends of the beamsat R and U the ends of the lower columns at O and X and theends of the upper columns at P and Y are all assumed to befixed In addition the stiffness of the outer beams RS and TUis taken as one-half of the true value For the fixed-endmoments due to normal (ie downward-acting) loads positivenumerical values should be substituted into the tabulatedexpressions If the resulting sign of the support moment isnegative hogging with tension across the top face of the beamis indicated

1421 Internal spans

By slope-deflection methods it can be shown that

where

and is a factor representing the ratio of the assumed to theactual stiffness for the span concerned (ie here 12)

By eliminating 13ST and 13TS in the above and rearranging thefollowing basic formulae are obtained

These formulae which are lsquoexactrsquo within the limitationsof the fixity conditions of the sub-frame represent the case ofthree spans loaded and apply for example to the conditionof dead load For design to BS 8110 the maximum moments atsupports S and T occur when the live load also is applied to allthree spans For design to EC 2 the maximum moment at sup-port S occurs when the live load is applied to spans RS and STand the maximum moment at support T occurs when the live

(FST FSR) (4 DST)(FTU FTS)

MTS FTS DTS

4 DTS DTS2DST 1

DTS 1

(FTU FTS) (4 DTS)(FST FSR)

MST FST DST

4 DSTDTS2DTS 1

DST 1

KT KST KTX KTY KTU

KS KRS KSO KSP KST

13TS (KST 2)(FSR FST) KS(FTS FTU)

KSKT K2ST 4

13ST (KST 2)(FTS FTU) KT(FSR FST)

KS KT K2ST 4

MTS FTS KTS(13TS 13ST 2)

MST FST KST (13ST 13TS 2)

load is applied to spans ST and TU The maximum positivemoment in span ST is obtained when the live load is applied tothis span only for both Codes The appropriate formulae forthese conditions are also given in Table 261 For the ULS theBS 8110 loads are dead 10gk and live 04gk 16qk Thecorresponding loads in EC 2 are dead 135gk and live 15qk

In the foregoing dead and live loads are applied separatelyand the resulting moments are summed Alternatively bothdead and live loads can be applied in a single operation byevaluating the basic formulae with fixed-end moment valuescorresponding to (deadlive) load on the aforesaid spans anddead load only on the remaining spans To comply with EC 2for example in determining the maximum support moment at Sthe fixed-end moments FSR FST and FTS should be calculated fora load of 135gk 15qk while FTU should be evaluated for a loadof 135gk only This method is used in the following example

In accordance with both Codes the moments derived fromthese calculations may be redistributed if desired It should beemphasised that although the diagrams on Table 261 andin the following example are for uniform loads the methodand the formulae are applicable to any type of loadingprovided that the appropriate fixed-end moment coefficientsobtained from Table 228 are used

When the moments MST and MTS at the supports are knownthe positive and negative moments in the spans are obtained bycombining the diagram of free moments due to the design loadswith the diagram of corresponding support moments

1422 End spans

The formulae for any interior span ST are rewritten to applyto an end span AB by substituting A B C and so on for S TU etc (A is the end support and there is no span correspondingto RS) The modified stiffness and distribution factors are givenin Table 261 together with the moment formulae for bothspans loaded and load on span AB only The dead and liveloads should be evaluated and applied so as to obtain therequired support moments as described in section 1421

1423 Columns and adjoining spans

The outer members of the sub-frame have been taken as fullyfixed at their remote ends Thus for a member such as RS theslope-deflection equation is

MSR KRS 13SR

Since the rotation of all the members meeting at a joint is thesame 13SR 13ST Thus by eliminating 13SR and rearranging

Similarly

[2DST (FST FSR) 4(FTU FTS)]

MTU FTU DTU

4 DSTDTS

[2DTS (FTU FTS) 4(FST FSR)]

MSR FSR DSR

4 DSTDTS

Continuous beams as frame members 159

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261Frame analysis simplified sub-frames

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The expressions for the moments in the columns are similar tothe foregoing but FSR and FTU should be replaced by the initialfixed-end moment in the column concerned (normally zero)and the appropriate distribution factor for the column should besubstituted for DSR or DTU

Example Determine the critical ultimate bending moments inbeam ST of the system shown in the following figure belowwhich represents part of a multi-storey frame in accordancewith the requirements of BS 8110 and EC 2 respectively

Stiffness values (mm3)

For upper columns

For lower columns

Distribution factors (if stiffness values are divided by 106)

Support-moment equations

FTS 0132[0953(FST FSR) 3529(FTU FTS)] (4 0471)(FTU FTS)] [2 0471(10497 1)(FST FSR)

MTS FTS [0497 (4 0471 0497)]

FST 0125[1116(FTU FTS) 3503(FST FSR)] (4 0497)(FST FSR)] [2 0497(10471 1)(FTU FTS)

MST FST [0471 (4 0471 0497)]

DTS 273

273 05 273 053 086

273549

0497

DST 273

05 336 273 053 086

273580

0471

K 342 109

4 103 086 106

K 213 109

4 103 053 106

KST KTU 2185 109

8 103 273 106

KRS 2016 109

6 103 336 106

Continuous beams as frame members 161

BS 8110 requirements

Fixed-end moments

For dead load only 10gk 8 kNm

FRS FSR 8 6212 24 kNm

FST FTS FTU FUT 8 8212 427 kNm

For dead live load

14gk 16qk 14 8 16 10 272 kNm

FRS FSR 272 6212 816 kNm

FST FTS FTU FUT 272 8212 1451 kNm

Maximum moments on beam ST

At S (dead live load on all spans)

MST 1451 0125

[1116(1451 1451) 3503(1451 816)]

1451 278 1173 kNm

At T (dead live load on all spans)

MTS 1451 0132

[0953(1451 816) 3529(1451 1451)]

1451 80 1531 kNm

Mid-span (dead live load on ST dead load on RS and TU)

MST 1451 0125

[1116(427 1451) 3503(1451 240)]

1451 387 1064 kNm

MTS 1451 0132

[0953(1451 240) 3529(427 1451)]

1451 325 1126 kNm

wwwengbookspdfcom

Maximum positive span moment is then approximately

15FST 05(MST MTS) 15 1451 05(1064 1126)

2176 1095 1081 kNm

Maximum moments on column at S (dead live load on STdead load on RS and TU) Distribution factors for lower andupper columns respectively are

Bending moments for lower and upper columns respec-tively are

Alternatively using the method shown on Table 260 for aninterior column when the adjoining beams are analysed asa continuous system on knife-edge supports

KS 05KRS 05 KST KSO KSP

(168 137 053 086) 106 444 106 mm3

Maximum unbalanced fixed-end moment at S

(FSTFRS) 1451240 1211 kNm

MSO 0194 1211 235 kNm

MSP 0119 1211 144 kNm

It will be seen that in this example the results obtained by thetwo methods are almost identical

EC 2 requirements

Fixed-end moments For dead load only

135gk 135 8 108 kNm

FRS FSR 108 6212 324 kNm

FST FTS FTU FUT 108 8212 576 kNm

For dead live load

135gk 15qk 135 8 15 10 258 kNm

FRS FSR 258 6212 774 kNm

FST FTS FTU FUT 258 8212 1376 kNm

DSP 053444

0119DSO 086444

0194

142 kNm

[2 0497(1451ndash427) 4(1451ndash240)]

MSP [0091(4ndash0471 0497)] 231 kNm

[2 0497(1451ndash427) 4(1451ndash240)]

MSO [0148(4ndash0471 0497)]

DSP 053580

0091DSO 086580

0148

Maximum moments on beam ST At S (dead live load on RSand ST dead load on TU)

MST 1376 0125

[1116(576 1376) 3503(1376 774)]

1376 152 1224 kNm

At T (dead live load on ST and TU dead load on RS)

MTS 1376 0132

[0953(1376 324) 3529(1376 1376)]

1376 132 1508 kNm

Mid-span (dead live load on ST dead load on RS and TU)

MST 1376 0125

[1116(576 1376) 3503(1376324)]

1376 349 1027 kNm

MTS 1376 0132

[0953(1376324) 3529(576 1376)]

1376 240 1136 kNm

Maximum positive span moment is then approximately

15FST 05(MST MTS) 15 137605(1027 1136)

20641081 983 kNm

Maximum moments on column at S (dead live load on STdead load on RS and TU)

143 EFFECTS OF LATERAL LOADS

For many structures a close analysis of the bending momentsto which a frame is subjected due to wind or other horizontalloads is unwarranted In such cases the methods illustrated inTable 262 are sufficiently accurate Further information on theuse of these methods is given in section 411

144 PORTAL FRAMES

General formulae for the bending moments in single-storeysingle-bay rigid frames are given in Table 263 (rectangularframes) and Table 264 (gable frames) The formulae whichrelate to any vertical or horizontal load cater for frames withthe columns fixed or hinged at the base Formulae for specificload cases are given in Tables 265 and 266 Formulae for theforces and bending moments in frames with a hinge at the baseof each column and one at the ridge (ie three-hinged frames)are given in Table 267

121 kNm

[2 0497(1376 576) 4(1376 324)]

MSP [0091(4 0471 0497)] 197 kNm

[2 0497(1376 576) 4(1376 324)]

MSO [0148(4 0471 0497)]

Framed structures162

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262Frame analysis effects of lateral loads

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263Rectangular frames general cases

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264Gable frames general cases

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265Rectangular frames special cases

Not

efo

rmul

ae g

i ve

num

eric

al v

alue

s of

rea

ctio

ns a

nd b

endi

ng m

omen

tss

ee d

iagr

ams

for

dire

ctio

n of

act

ion

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266Gable frames special cases

Not

efo

rmul

ae g

ive

num

eric

al v

alue

s of

rea

ctio

ns a

nd b

endi

ng m

omen

tss

ee d

iagr

ams

for

dire

ctio

n of

act

ion

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267Three-hinged portal frames

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In modern multi-storey buildings lateral stability is providedby a system of frames or walls or a combination of both Noteson wall and frame systems are given in section 412 and differentstructural forms with information on typical building heightsand proportions given in ref 37 are shown in Table 268

151 LAYOUT OF SHEAR WALLS AND ALLOCATION OFLATERAL LOADS

Arrangements of shear walls and core units should be such thatthe building is stiff in both flexure and torsion Different planconfigurations with remarks as to their suitability are shown inTable 269 The lateral load transmitted to each wall is afunction of its stiffness and its position in relation to the shearcentre of the system The location of the shear centre can bereadily determined by calculating the moment of stiffness of thewalls about an arbitrary reference point as shown in Table 269The lateral load on each wall can then be determined from thegeneralised formulae given also in Table 269 For most wallswithout openings the dominant mode of deformation is bending(see section 4122) In this case K may be replaced with I inthe generalised formulae where I is the second moment of areaof the cross section

152 PIERCED SHEAR WALLS

In the case of walls pieced by openings the behaviour of theindividual wall sections is coupled to a variable degree Thedeflected shape of the pierced wall is a combination of frameand wall action The wall may be idealised as a plane frame oranalysed by elastic methods in which the flexibility of thebeams is represented as a continuous flexible medium

Solutions using the continuous connection model for a wallcontaining a single line of openings are given in Table 270 Thetotal lateral load F is considered in three different forms aconcentrated load applied at the top of the wall a uniformload distribution and a triangular load distribution with themaximum value at the top of the wall Formulae are given forthe maximum axial force at the base of each wall section themaximum shear force on a connecting beam (and the height ofthe beam above the base of the wall) and the maximumdeflection at the top of the wall Formulae whereby values atany level can be determined are given in ref 38

The main two parameters that define the performance of thewall are and which depend on the geometrical properties

of the wall and beam elements and on the number of lines ofopenings The formulae in Table 270 cater for a wall withdissimilar cross sections on either side of a single line ofopenings For identical cross sections the formulae become

where A1 and I1 refer to each portion of the wall For a wall withtwo symmetrical lines of openings the formulae for the para-meters become

where A1 and I1 refer to each outer portion and I2 refers tothe central portion of the wall Similarly the momentsbecome

There is no axial force in the central portion of the wall

153 INTERACTION OF SHEAR WALLS AND FRAMES

Although the interaction forces between solid walls piercedwalls and frames can vary considerably up the height of a build-ing the effect on the total lateral force resisted by each elementis less significant As a first approximation in order to deter-mine the forces at the bottom of each load-resisting element itis normally sufficient to consider the effect of a single interac-tion force at the top of the building This is equivalent to loadsharing in terms of relative stiffness The location of the shearcentre and the allocation of the lateral load can be determinedas indicated in Table 269 using the following formulae for thestiffness of each element

Solid wall

Pierced wall K 3EIw

H31 2l

21 3

(H)2

3(H)3

K 3EIw

H 3

M2 (M0 Nl )I2

2I1 I2M1

(M0 Nl )I1

2I1 I2

2 12lIe

(2I1 I2)hb3e

2 2

l 2l2 (2I1 I2)

A1

2 6lIe

I1hb3e

2 2

l l2 4I1

A1

Chapter 15

Shear wall structures

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268Structural forms for multi-storey buildings

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269Shear wall layout and lateral load allocation

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270Analysis of pierced shear walls

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Interaction of shear walls and frames 173

Frame

For the frame and are the sums of the second momentsof area of the columns and beams respectively for the lowestbay of height h and beam span lb The factor kc allows for areduction of Ic over the height of the building and is given bykc loge c(c 1) where c Ic top Ic bottom

Example 1 For the shear wall layout in the following figuredetermine the location of the shear centre and the lateral loadapplied to each wall using the formulae in Table 269 Alldimensions are in metres and the walls are 200 mm thick

IbIc

K 12EIc

h2Hkc lbIc

hIb

(0141 0309) F 045F

Example 2 The elevation of a pierced shear wall is shown inthe following figure Identical walls are provided at each end ofa 24 m long rectangular building The building is subjected to acharacteristic wind pressure of 125kNm2 acting on the broadface and the resistance provided by any other frames parallel tothe walls may be ignored The resulting forces and bendingmoments acting on the walls are to be determined

F3y F4y 36(44 0)(F 658)

60128 017F

F2z 20F1415

20(165 242)(F 658)

60128

Layout of shear walls

From symmetry the shear centre of the wall system must be onthe y-axis Similarly the shear centre of the two channelsections (taken together) is at the mid-point of the core unitThe second moment of area values (mm4) of walls 1 and 2 indirection z (discounting the stiffness of walls 3 and 4) are

The distance of the shear centre C from the reference point is

Eccentricity of total lateral force is e 90 242 658 mThe second moment of area values (mm4) of walls 3 and 4 indirection y are

Lateral forces on each wall in terms of total force F are

(0859 0309) F 055F1215F1415

1855F60128

1215(01 242)(F 658)

1215 2322 20 14082 2 36 442

F1z 1215F

1215 20

I3y I4y 02 603

12 36

yc 1215 01 200 165

1215 200 242m

I2z 18 303 14 263

12 20I1z

02 903

12 1215

Elevation of shear wall

If the total wind force acting on the building is shared equallybetween the two walls the horizontal force to be resisted byeach wall is F 125 12 60 900 kN

With reference to Table 270 the following values apply

A1 A2 0225 5 1125 m2

I1 I2 0225 5312 2344 m4

Ib 0225 06312 000405 m4

000333

be (2b 5d)3 (2 2 5 06)3 233

000157

0012860001577 72

2 2 23441125

2 2

l l2 A1 A2

A1A2(I1 I2)

12 7 0003332 2344 3 2333

2 12lIe

(I1 I2)hb3e

0004051 24 (062)2

Ie Ib

1 24(db)2

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Shear wall structures174

Hence 0113 00396 H 0113 60 68Maximum axial force at bottom of each wall element

N asymp

2470 kN

Moment at bottom of each wall element

M1 M2 (FH2 Nl)2 (900 602 2470 7)2 4855 kNm

900 60 0001572 001286 1

268

2

682

FH2

22 1 2

H

2(H)2

Maximum shear force in beams for H 68 occurs wherexH 028 (5th or 6th floor level) and Kv 057 Then

Vb 188 kN

Maximum bending moment in beam

Mb Vb b2 188 22 188 kNm

For subsequent design purposes the forces and moments due tothe characteristic wind load must be multiplied by a partialsafety factor appropriate to the load combination

900 3 000157 057001286

Fh2Kv

2

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In this chapter elastic analyses in terms of characteristic loadsand service stresses are indicated Where limit-state designmethods are employed care must be taken to include theappropriate partial safety factors for the load combination andlimit-state being considered Arch structures may be eitherthree-hinged or two-hinged or fixed-ended as shown in thediagrams at the top of Table 271

161 THREE-HINGED ARCH

For the general case of an unsymmetrical three-hinged archwith a load acting vertically horizontally or at an angle theexpressions for the horizontal and vertical components of thethrusts on the abutments are given in the lower part of Table271 For symmetrical arches the formulae for three-hingedportal frames given in Table 267 are applicable The bendingmoment at each hinge is zero and at any particular section thebending moment shearing force and axial thrust may be deter-mined by considering the loads and abutment reactions on oneside of the section Further information regarding the extent ofthe arch that should be loaded with imposed load in order toproduce the maximum effect at a particular section is given insection 4131

162 TWO-HINGED ARCH

For a symmetrical two-hinged arch the vertical component ofthrust on the abutments is the same as for a freely supportedbeam The horizontal component of thrust H is given by theformula in Table 271 where Mx is the bending moment on asection at distance x from the crown with the arch consideredas a freely supported beam Mx is given by the correspondingexpression in Table 271

The summations 13MxyaI and 13y2aI are taken over the wholelength of the arch In the formula for H which allows for theelastic shortening of the arch A is the average equivalent areaof the arch rib or slab and a is the length of a short segment ofthe axis of the arch where the coordinates of a are x and y asshown in Table 271 If I is the second moment of area of thearch at x then aI aI The bending moment at any section isgiven by Md MxHy

The procedure involves dividing the axis of the arch into aneven number of segments The calculations can be facilitatedby tabulating the successive steps The total bending moment is

required generally only at the crown (x 0 y yc) and thefirst quarter-point (x 025l) The moment Mc at the crown isthe bending moment for a freely supported beam minus HycFor the maximum positive moment at the crown the sum of thevalues of Mc for all elements of dead load is added to the valuesof Mc for only those values of imposed load that give positivevalues of Mc For the maximum negative moment at the crownthe sum of the values of Mc for all elements of dead load isadded to the values of Mc for only those values of imposed loadthat give negative values of Mc The moment at the first quarter-point is the bending moment for a freely supported beam minusHyq where yq is the vertical ordinate of the first quarter-pointThe bending moment is combined with the normal componentof H For an arch of large span it is worthwhile preparing theinfluence lines (F 1) for the bending moments at the crownand at the first quarter-point

163 FIXED ARCH

1631 Determination of thickness

The diagram at the top of Table 272 shows an approximatemethod of determining the section thickness at the crown andthe springing for a symmetrical arch with fixed ends Draw ahorizontal line through the crown C and find G the point ofintersection of the line with the vertical through the centre ofgravity of the total load on half the span of the arch Set outlength GT equal to the dead load on the half span drawn to aconvenient scale draw a horizontal line through T to intersectGS extended at R Draw lines RK perpendicular to GR and GKparallel to the tangent to the arch axis at S On the same scalethat was used to draw GT measure TR which equals Hc andGK which equals Hs If fcc is the maximum allowablecompressive stress in the concrete b the assumed breadth of thearch (1 m for a slab) hc the thickness at the crown and hs thethickness at the springing then approximately

hc 17Hcbfcc hs 2Hsbfcc

The method applies to spans from 12 to 60 m and spanriseratios from 4 to 8 The method does not depend on knowing theprofile of the arch (except for solid-spandrel earth-filled archeswhere the dead load is largely dependent on the shape of thearch) but the span and rise must be known With hc and hs thusdetermined the thrusts and bending moments at the crown

Chapter 16

Arches

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271Arches three-hinged and two-hinged arches

Three-hinged arch

Two-hinged arch

Unsymmetrical three-hinged arches General case

Types of archesInfluence line for section at x

Three-hinged arch

Two-hinged arch

Fixed arch

Note see section 162 for explanation of symbols in and notes on the formulae for two-hinged arches

Reaction formulae for general case

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272Arches fixed-ended arches

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Arches178

springing and quarter-points can be obtained and the stresses onthe assumed sections calculated If the stresses are shown to beunsuitable other dimensions must be tried and the calculationsreworked

1632 Determination of load effects

The following method is suitable for determining load effectsin any symmetrical fixed arch if the dimensions and shape ofthe arch are known or assumed and if the shape of the archmust conform to a particular profile Reference should be madeto section 4133 for general comments on this method

On half of the arch drawn to scale as in Table 272 plot thearch axis Divide the half-arch into k segments such that eachsegment has the same ratio aI aI where a is the length andI is the mean second moment of area of the segment based onthe thickness of the arch measured normal to the axis withallowance being made for the reinforcement

Coordinates x and y relative to the axis of the arch at thecrown are determined by measurement to the centre of lengthof each segment Calculate separately the dead and imposedloads on each segment Assume each load acts at the centre oflength of the segment In an open-spandrel arch the dead andimposed loads are concentrated on the arch at the columnpositions these should be taken as the centre of the segmentsbut it may not then be possible to maintain a constant value ofaI and the value of aI for each segment must be calculated thegeneral formulae in Table 272 are then applicable

For constant values of aI the forces and bending moment atthe crown are as follows

The summations are taken over one-half of the arch The termM1 is the moment at the centre of the segment of all the loadsfrom the centre of the segment to the crown Summations arealso made for all the loads on the other half of the arch forwhich Rc is negative

Due to the elastic shortening of the arch resulting from Hc

where A is the cross-sectional area of the segment calculated onthe same basis as I

Due to a rise (T) or a fall (T) in concrete temperature

where is the coefficient of linear expansion of the concrete Ec

is the short-term modulus of elasticity of the concrete and l1 is

Mc2 Hc2y k

Hc2 (T)kEcl1

2aIky2 y2

Mc1 Hc1ay k

Hc1 Hck(a A)

aIky2 y2

Mc M1 2Hcy

2k

Rc M1x 2x2

Hc kM1y yM1

2ky2 y2

the length of the arch axis Arch shortening due to Hc2 isneglected The effect of concrete shrinkage can be treated as atemperature fall in which T is replaced by (cs) where cs isthe shrinkage strain (allowing for reinforcement restraint) and is a reduction coefficient (typically taken as 043) to allow forthe effect of creep

The foregoing formulae are used to determine the effects ofthe various design loads in the following procedure Calculate(Hc Hc1) Rc and (Mc Mc1) for the dead load Calculate foreach load separately values of Hc2 and Mc2 for temperaturerise temperature fall and concrete shrinkage Calculate for theimposed load (represented as an equivalent uniform load)(Hc Hc1) and (Mc Mc1) To obtain the maximum positivebending moment at the crown (and the horizontal thrust) theimposed load should be applied to the middle third of the archmore or less (By considering the effect of the imposed load onone segment more and one segment less than those in the mid-dle third of the arch the number of segments that should beloaded to give the maximum positive bending moment at thecrown can be determined) With the imposed load applied onlyto those segments that are unloaded when calculating the max-imum positive moment at the crown the maximum negativemoment at the crown due to the imposed load is obtained Themaximum bending moments due to the imposed load are eachcombined with the bending moments due to dead load and archshortening and with the bending moments due to temperaturechange and concrete shrinkage in such a way that the mostadverse total values are obtained The corresponding thrusts arealso calculated and combined with the appropriate bendingmoments to check the design conditions at the crown

The bending moment at the springing due to load at a pointbetween the springing and the crown of the arch distant l fromthe springing (where l is the span of the arch) is

Ms (Mc Mc1) (Hc Hc1)yc Rcl2 Fl

where yc is the rise of the arch For the dead load the valuesdetermined for the crown are substituted in this expressionwith the term Fl replaced by F [(l2) x] To obtain themaximum negative bending moment at the springing theimposed load is applied to those parts of the arch extending 04of the span from the springing more or less (As before theeffect of applying the imposed load to one segment more andone segment less than this distance should be determined toensure that the most adverse loading disposition has beenconsidered) By applying load to only those segments that areunloaded when calculating the maximum negative bendingmoment the maximum positive bending moment is obtainedThese maximum bending moments are each combined with thebending moments due to dead load temperature change andconcrete shrinkage to obtain the most adverse total values Theconditions at the springing are then checked for the combinedeffects of the most adverse bending moments and the corre-sponding normal thrusts

The normal thrust at the springing is given by

Ns (Hc Hc1)cos13 Rssin13

In this expression the vertical component of thrust given byRs (total load on half-arch) Rc is calculated for the loadsused to determine (Hc Hc1)

The shearing force at the springing is given by

Vs (Hc Hc1)sin13 Rscos13

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Fixed parabolic arch 179

In this expression the maximum value is generally obtainedwhen the imposed load extends over the whole arch At thecrown the maximum shearing force is the maximum value ofRc due to any combination of dead and imposed load

The bending moment at a section with coordinates xq and yqdue to load at a point between the springing and the crown ofthe arch distant l from the springing is

Mq Mc Hcyq Rcxq F[xq (05 )l]

At the quarter-point xq l4 and the procedures to determinethe maximum bending moment normal thrust and shearingforce are similar to those described above The method given inTable 273 can be used to obtain data for influence lines

164 FIXED PARABOLIC ARCH

Formulae and guidance on using the data in Table 274 aregiven below See section 4134 for further comments

1641 Dead load and arch shortening

The horizontal thrust due to dead load alone is H k1gl2ycwhere g is the dead load per unit length at the crown l is thespan and yc is the rise of the arch axis The coefficient k1

depends on the dead load at the springing which varies with therisespan ratio and type of structure that is whether the arch isopen spandrel or solid spandrel or whether the dead load isuniform throughout the span

An elastic shortening results from the thrust along the archaxis (assuming rigid abutments) The counter-thrust H1 whileslightly reducing the thrust due to the dead load renders thisthrust eccentric and produces a positive bending moment at thecrown and a negative bending moment at each springing If h isthe thickness of the arch at the crown

in which the coefficient k2 depends on the relative thickness atthe crown and the springing

Due to dead load and elastic shortening the net thrusts Hc

and Hs at the crown and the springing respectively acting par-allel to the arch axis at these points are given by

where 13 is the angle between the horizontal and the tangent tothe arch axis at the springing with cos13 given in Table 274

The bending moments due to the eccentricities of Hc and Hs

are given by Mc k3ycH1 and Ms (k3 1) ycH1 respectively

1642 Temperature change

The additional horizontal thrust due to a rise in temperature orcorresponding counter-thrust due to a fall in temperature isgiven by

If T is the temperature change in oC with h and yc in metresthen H2 is in kN per metre width of arch The values of k4 inTable 274 are based on an elastic modulus Ec 20 kNmm2and a coefficient of linear expansion 12 micro-strainoC

H2 k4hyc

2

hT

Hs H

cos 13 H1 cos 13Hc H H1

H1 k2hyc

2

H

For any other values of Ec and k4 should be multiplied by avalue of (Ec20)(12) At the crown the increase or decrease innormal thrust due to a change in temperature is H2 and thebending moment is k3ycH2 account being taken of the sign ofH2 The normal thrust at the springing due to a change oftemperature is H2cos13 and whether this thrust increases ordecreases the thrust due to dead load depends on the sign of H2At the springing the bending moment is (1 k3)ycH2 the signbeing the same as that of H2

1643 Shrinkage of concrete

The effect of concrete shrinkage can be treated as a fall intemperature in which T is replaced by (cs)( 12) wherecs (micro-strain) is the shrinkage (allowing for reinforcementrestraint) and is a reduction coefficient (taken as 043) toallow for the effect of creep

1644 Imposed load

The maximum bending moments and corresponding thrustsand reactions are given by the following expressions whereq per unit length is the intensity of uniform load equivalent tothe specified imposed load

At the crown positive bending moment k5 ql2

horizontal thrust k6 ql2yc

At the springing negative bending moment k7 ql2

horizontal thrust k8 ql2yc

vertical reaction k9 ql

positive bending moment k10 ql2

horizontal thrust k11 ql2yc

vertical reaction k12 ql

The normal thrust at the springing where H is the horizontalthrust and R is the vertical reaction is given by

N Hcos13 Rradic(1 cos213)

1645 Dimensions of arch

The line of thrust and therefore the arch axis can be plotted asdescribed in section 4134 The thickness of the arch at thecrown and the springing having been determined the lines ofthe extrados and intrados can be plotted to give a parabolicvariation of thickness between the two extremes Thus thethickness normal to the axis of the arch at any point is given by[(hs h)2 h] where is the ratio of the distance of the pointfrom the crown measured along the axis of the arch to half thelength of the axis of the arch

Example Determine the bending moments and thrusts on afixed parabolic arch slab for an open-spandrel bridge

Specified values

Span 50 m measured horizontally between the intersection ofthe arch axis and the abutment Rise 75 m in the arch axisThickness 900 mm at springing 600 mm at crownDead load 12 kNm2 imposed load 15 kNm2Temperature range 24oC 12 micro-strain per oC

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273Arches computation chart for symmetrical fixed-ended arch

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274Arches fixed-ended parabolic arches

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Arches182

Ec 28 kNmm2 cs 200 micro-strain 043

Geometrical properties

yc

l

7550

015hs

h

900600

15

Inclination of arch axis at springing (Table 273) cos13 082A strip of slab 1 m wide is considered The coefficients takenfrom Table 274 are substituted in the expressions given insection 1644 Allowing for self-weight of arch slab

Total dead load at crown 12 06 24 264 kNm2

Loads and resulting bending moments and thrusts Moment ThrustkNmm kNm

Horizontal thrusts due to dead load and other actions

Dead load (k1 0140)H 0140 264 (50275) 1232

Arch shortening (k2 139)H1 139 (0675)2 1232 11

Temperature change (k4 422)H2 422 (2820) (0675)2 06 24 55

Concrete shrinkage (k4 422)H3 422 (2820) (0675)2 06 (20012) 043 16

Crown maximum positive bending moment and corresponding thrust

Dead load and arch shortening (k3 0243 thrusts H and H1 as above) 20Mc 0243 75 11Hc 1232 11 1221

Temperature fall (thrust H2 as above) 55Mc 0243 75 55 100

Shrinkage (thrust H3 as above) 16Mc 0243 75 16 29

Imposed load (k5 0005 k6 0064)Mc 0005 15 502 188Hc 0064 15 (50275) ____ 320

Totals 337 1470

Springing maximum negative bending moment and corresponding thrust

Dead load and arch shorteningMs (1 0243) 75 11 63Hs (1232082) 11 082 1494

Temperature fallMs (1 0243) 75 55 312Hs 55 082 45

ShrinkageMs (1 0243) 75 16 91Hs 16 082 13

Imposed load (k7 0020 k8 0038 k9 0352)Ms 0020 15 502 750H 0038 15 (50275) 190 kNm R 0352 15 50 264 kNmN 190 082 264radic(1 0822) 307

Totals 1216 1743

Springing maximum positive bending moment and corresponding thrust

Dead load and arch shortening (values as before) 63 1494Temperature rise (values for temperature fall) 312 45Shrinkage (neglect as partial in short term and beneficial in long term)Imposed load (k10 0023 k11 0089 k12 0151)

Ms 0023 15 502 863H 0089 15 (50275) 445 kNm R 0151 15 50 113 kNmN 445 082 113radic(1 0822) 430

Totals 1112 1969

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In the following containers are conveniently categorised astanks containing liquids and bunkers and silos containing drymaterials each category being subdivided into cylindrical andrectangular structures The intensity of pressure on the wallsof the structure is considered to be uniform at any level butvertically the pressure increases linearly from zero at the top toa maximum at the bottom

171 CYLINDRICAL TANKS

If the wall of a cylindrical tank has a sliding joint at the baseand is free at the top then when the tank is full no radial shearor vertical bending occurs The circumferential tension at depthz below the top is given by n ri z per unit height where ri

is the internal radius of the tank and is unit weight of theliquid If the wall is supported at the base in such a way that noradial movement can occur radial shear and vertical bendingresult and the circumferential tension is always zero at thebottom of the wall Values of circumferential tensions verticalmoments and radial shears according to values of the termheight2(2 mean radius thickness) can be obtained fromTables 275 and 276

The tables apply to walls with a free top and a bottom that iseither fixed or hinged The coefficients have been derivedby elastic analysis and allow for a Poissonrsquos ratio of 02 Forfurther information on the tables reference should be made topublications on cylindrical tanks such as refs 55 and 56 If anannular footing is provided at the base of the wall a hingeddetail can be formed although this is rarely done The footingnormally needs to be tied into the floor of the tank to preventradial movement Reliance solely on the frictional resistanceof the ground to the radial force on the footing is generally inad-equate and always uncertain If the joint between the wall andthe footing is continuous it is possible to develop a fixedcondition by widening the footing until a uniform distributionof bearing pressure is obtained In many cases the wall and thefloor are made continuous and it then becomes necessary toconsider the structural interaction of a cylindrical wall and aground supported circular slab Appropriate values for the stiff-ness of the member and the effect of edge loading can beobtained from Table 276 for walls and Table 277 for slabs

For slabs on an elastic foundation the values depend on theratio rrk where rk is the radius of relative stiffness defined insection 725 The value of rk is dependent on the modulus of

subgrade reaction for which data is given in section 724Taking rrk 0 which corresponds to a lsquoplasticrsquo soil state isappropriate for an empty tank liable to flotation

Example 1 Determine due to internal hydrostatic loadingmaximum service values for circumferential tension verticalmoment and radial shear in the wall of a cylindrical tank that isfree at the top edge and hinged at the bottom The wall is300 mm thick the tank is 6 m deep the mean radius is 10 mand the water level is taken to the top of the wall

From Table 275 for lz22rh 62(2 10 03) 6

n 0643 at zlz 07 m 0008 at zlz 08

v 0110 at zlz 10

n n lzr 0643 981 6 10 3785 kNm

m mlz3 0008 981 63 170 kNmm

v vlz2 0110 981 62 389 kNm

Example 2 Determine for the tank considered in example 1the corresponding values if the wall is fixed at the bottom

From Table 275 for lz22rh 62(2 10 03) 6

n 0514 at zlz 06 m 0005 at zlz 07

mb 0019 and v 0197 at zlz 10

n nlzr 0514 981 6 10 3025 kNm

m mlz3 0005 981 63 106 kNmm

m mlz3 0019 981 63 403 kNmm

v vlz2 0197 981 62 696 kNm

Consider a straight wall that is centrally placed on a footing ofwidth b (2a h) where a represents the distance from edgeof footing to face of wall The weight of liquid per unit lengthof wall on the inside of the footing is given by lza Assuminga uniform distribution of bearing pressure due to the liquid loadthe bending moment about the centre of the wall due to thepressure on the toe of the footing is

mlza (ab) (ah)2 giving a2(a h) 2(mlz)b

Substituting for b h and m lz 403(981 6) 0685 thefollowing cubic equation in a is obtained

a2 (a03) 137 (2a 03)

Chapter 17

Containmentstructures

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Cylindrical tanks elastic analysis ndash 1 275

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Cylindrical tanks elastic analysis ndash 2 276

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Cylindrical tanks elastic analysis ndash 3 277

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Cylindrical tanks 187

Solution of the equation by trial and error gives a 16 m andwidth b 2 16 03 35 m This width is correct for theannular footing provided the footing is positioned so that itscentre of area coincides with the centre of the wall

If c represents the distance from the outer edge of the footingto the centreline of the wall then for a unit length of wall thelengths of the trapezoidal area are (r c)r for the outer edgeand (r c b)r for the inner edge The usual formula for atrapezoidal area gives

[(r c)2(rcb)] br3[(rc) (rcb)] cr

Substituting for b and r and rearranging the terms gives

6c2 39c 805 0 from which c 165 m

Thus a fixed edge condition can be obtained by providing a35 m wide footing with the outer edge of the footing 15 mfrom the outer face of the wall

Example 3 Determine for the tank considered in example 1the corresponding values if the wall is continuous with the floorslab The slab is 400 mm thick and the soil on which the tank isto be built is described as well-compacted sand

Properties of cylindrical wall From Table 276 with lz22rh 6w 0783 and the wall stiffness is given by

Kw w Ec h3lz (0783 0336) Ec 00035 Ec

Also w1 00187 and the fixed edge moment when the tankis full is given by

Mw w1 lz3 00187 981 63 396 kNmm

Properties of circular slab on elastic subgrade From section724 for well-compacted sand a mean value of 75 MNm3 canbe taken for the modulus of subgrade reaction

From Table 277 for a slab on an elastic subgrade the radiusof relative stiffness with 02 is given by

rk [Ech31152 ks]025

Therefore taking Ec 33 kNmm2 and ks 75 MNm3

rk [33 109 043(1152 75 106)]025 125 m

With rrk 8 s 0468 and the slab stiffness is given by

Ks sEch3r (0468 04310) Ec 00030 Ec

The unit edge load on the slab due to the lsquoeffectiversquo weight ofthe wall is

Q 03 6 (24 981) 255 kNm

With rrk 8 s2 0088 and the corresponding fixed edgemoment is given by

Ms s2 Qr 0088 255 10 225 kNmm

Moment distribution at joint Since the calculated fixed edgemoments for the wall and the slab both act in the same directionthe joint will rotate when the notional restraint is removed Thiswill induce additional moments and change the circumferentialtensions in the wall At the joint the induced moments will be

proportional to the relative stiffness values of the two elementsaccording to the following distribution factors

wall slab

Element Wall SlabDistribution factor 054 046Fixed end moment 396 225Induced moment 335 286Final moment 61 61

It can be seen from the moments calculated for the wall thatthe joint rotation is close to that for a hinged base conditionThe use of a thinner slab or a lower value for the modulus ofsubgrade reaction will increase the rotation until a lsquoclosingrsquocorner moment develops and the circumferential tensions inthe wall exceed those obtained for a hinged base

Final forces and moments The final circumferential tensionsand vertical moments at various levels in the wall can beobtained by combining the results for load cases (1) and (3) inTable 275 where M is the induced moment The followingequations apply

n n1 lz r n3Mrlz2

(981 6 10) n1 (335 1062) n3

5886 n1 93 n3 kNm

m m1 lz3 m3M (981 63) m1 335 m3

2119 m1 335 m3

The resulting values for different values of zlz are shown in thefollowing tables

0003000035 00030

0460003500035 00030

054

Circumferential tensions in wall (kNm)

zlz

Load case (1) Load case (3) Final

n1 5886 n1 n3 93 n3force

05 0504 2967 334 311 327806 0514 3025 654 608 363307 0447 2631 103 958 358908 0301 1772 131 1218 299009 0112 659 114 1060 1719

Vertical moments in wall (kNmm)

zlz

Load case (1) Load case (3) Final

m1 2119 m1 m3 335 m3moment

06 00046 98 0037 12 8607 00051 108 0057 19 12708 00029 62 0252 84 14609 00041 87 0572 192 10510 00187 396 10 335 61

The final radial shear at the base of the wall is given by

v v1lz2 v3Mlz 0197 981 62 449 3356 445 kNm

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The new moment distribution factors are as follows

wall slab

Element Wall SlabDistribution factor 084 016Fixed end moment 22 1080Induced moment 926 176Final moment 904 904

It will be seen that in this example the final edge moment onthe slab is close to a fixed edge condition since the stiffnessof the wall is much greater than that of the slab Final values forthe moments and forces in the slab and the wall can now becalculated as shown in example 3

172 RECTANGULAR TANKS

The bending moments obtained by Table 253 for individualrectangular panels with triangularly distributed loads may beapplied without modification to continuous walls providingthere is no rotation about the vertical edges In a square tanktherefore moment coefficients can be taken directly fromTable 253 For a rectangular tank distribution of the unequalfixity moments at the wall junctions is needed and momentcoefficients for tanks of different span ratios are given in Tables278 and 279 The shearing forces given in Table 253 for theindividual panels may still be used

The tables give values for idealised edge conditions at the topand bottom of the wall The top of a wall should be taken as freefor an open tank and when a sliding joint is provided betweena roof and the top of the wall If a wall is continuous with theroof the joint will rotate when the tank is full and the conditionwill tend towards hinged When there is earth loading on thewall a closing-corner moment will develop when the tank isempty At the bottom of the wall a hinged condition may becreated by providing a narrow wall footing tied into the floorslab or by adopting a reinforced hinge detail Where the tankwall is continuous with the floor the deformation of the floorresulting from the wall loading is often complex and highlydependent on the assumed ground conditions The edge condi-tion at the bottom of the wall is generally uncertain but willtend towards hinged when the tank is full When there is earthloading on the wall the edge condition will tend towards fixedwhen the tank is empty If the floor is extended outwards toform a toe the condition at the base will be affected in the waydiscussed in section 171

In considering the horizontal spans the shear forces at thevertical edges of one wall result in axial forces in the adjacentwalls Thus for internal loading on a rectangular tank the shearforce at the end of the long wall is equal to the tensile force inthe short wall and vice versa In designing sections the com-bined effects of the bending moment the axial force and theshear force need to be considered

Example 5 Determine due to internal hydrostatic loadingthe maximum service moments and shear forces in the walls ofa rectangular tank that can be considered as free along the topedge and hinged along the bottom edge The tank is 6m long

0003000035 00007

0160003500035 00007

084

Containment structures188

Tangential moments in slab (kNmm)

rxrLoad case (1) Load case (2) Final

t1 286t1 t2 255t2moment

10 0378 108 0018 46 6208 0132 38 0009 23 6106 0020 06 0006 15 0904 0034 10 0 0 1002 0014 04 0002 05 090 0005 02 0002 05 07

Example 4 Determine for the tank considered in example 3the factor of safety against flotation and the moments in theslab if the worst credible groundwater level is 15 m above theunderside of the slab

The radius of the slab is 103 m and radius to the centre ofthe wall is 1015 m If the weight of the wall is spread uniformlyover the full area of the slab the total downward pressure dueto the weight of the slab and the wall is

[04 03 6 (2 10151032)] 24 178 kNm2

The upward pressure due to a 15 m depth of groundwater is147 kNm2 giving a safety factor of 178147 121 whichsatisfies the BS 8007 minimum requirement of 11 The unitedge load on the slab due to the weight of the wall is

Q 03 6 24 432 kNm

Taking rrk 0 (for a lsquoplasticrsquo soil condition) s 0104and s2 0250 giving the following values

Ks sEch3r (0104 04310)Ec 00007 Ec

Ms s2Qr ndash 0250 432 10 108 kNmm

The wall stiffness is the same as before and the effect of thegroundwater (11 m from the top of the slab) on the wall maybe taken as a simple cantilever giving values

Kw wEch3lz (0783 0336) Ec 00035 Ec

Mw w1lz3 981 1136 22 kNmm

Radial moments in slab (kNmm)

rxrLoad case (1) Load case (2) Final

r1 286r1 r2 255r2moment

10 10 286 0088 225 6108 0488 140 0013 33 17306 0028 08 0016 41 4904 0055 16 0003 08 0802 0023 07 0001 03 100 0005 02 0002 05 07

The final radial and tangential moments in the floor slab can beobtained by combining the results for load cases (1) and (2) inTable 277 The following equations apply

mr r1M r2 Qr and mt t1M t 2 Qr

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278Rectangular tanks triangularly distributed load(elastic analysis) ndash 1

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279Rectangular tanks triangularly distributed load(elastic analysis) ndash 2

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4m wide and 4m deep and the water level is to be taken to thetop of the walls

From Table 279 for lxlz 64 15 lylz 44 10 andtop edge free maximum bending moments are as follows

Horizontal negative moment at corners

mx my 0050 981 43 314 kNmm

Horizontal positive moments (at about 05lz 2m above base)

mx 0035 981 43 220 kNmm (long wall)

my 0011 981 43 69 kNmm (short wall)

Vertical positive moments (at about 03lz 12m above base)

mzx 0029 981 43 182 kNmm (long wall)

mzy 0013 981 43 82 kNmm (short wall)

From Table 253 for panel type 4 with lhlz 15 (long wall)and 10 (short wall) maximum shear forces are as follows

Horizontal shear forces at side edges

vhx 037 981 42 581 kNm (long wall)

vhy 031 981 42 487 kNm (short wall)

Vertical shear forces at bottom edge

vzx 026 981 42 408 kNm (long wall)

vzy 019 981 42 298 kNm (short wall)

From the above the horizontal tensile forces in the walls are

nhx 487 kNm (long wall) nhy 581 kNm (short wall)

173 SILOS

Notes on the properties of stored materials and the pressuresset up in silos of different forms and proportions are given insections 277 and 93 and Tables 215 and 216 Notes on thedesign of silo walls are given in section 641 For rectangularsilos that are divided into several compartments where thewalls span horizontally expressions for the bending momentsand reactions are given in Table 280

174 BOTTOMS OF ELEVATED TANKS AND SILOS

1741 Tanks

The figure here shows the floor of an elevated cylindrical tanksupported by beams in two different arrangements

area with the remainder of the floor load the weight of the walland the load from the roof being equally divided between theeight cantilever lengths An alternative to (b) is to place thecolumns almost under the wall in which case the cantilevers areunnecessary but secondary beams might be required

For a tank of large diameter a domed bottom of one of thetypes shown in Table 281 is more economical and althoughthe formwork is much more costly the saving in concrete andreinforcement compared to beam-and-slab construction can beconsiderable Ring beams A and C in the case of a simple domedbottom resist the horizontal component of thrust from the domeand the thickness of the dome is determined by the magnitudeof the thrust Expressions for the thrust and the vertical shearingforce around the edge of the dome and the resultant circumfer-ential tension in the ring beam are given in Table 281 Domesused to form the bottom or the roof can also be analysed by themethod given in Table 292

A bottom consisting of a central dome and an outer conicalpart as illustrated in Table 281 is economical for the largesttanks This form of construction is traditionally known as anIntze tank The outward thrust from the top of the conical partis resisted by the ring beam B and the difference between theinward thrust from the bottom of the conical part and the out-ward thrust from the domed part is resisted by the ring beam AExpressions for the forces are given in Table 281 and the pro-portions of the conical and domed parts can be chosen so thatthe resultant thrust on beam A is zero Suitable proportions forbottoms of this type are given in Table 281 and the volume ofa tank with these proportions is 0604 do

3 where do is the diam-eter of the tank The tank wall should be designed as describedpreviously account being taken of the vertical bending at thebase of the wall and the effect of this bending on the conicalpart The floor must be designed to resist in addition to theforces and bending moments already described any radialtension due to the vertical bending of the wall

Example 6 Determine for service loads the principal forcesin the bottom of a cylindrical tank of the Intze type where

do 10 m d 8 m 48o (giving cot 0900)

13 40o (giving cot 13 1192 and cosec13 155)

F1 2500 kN F2 2800 kN and F3 1300 kN

Then from Table 281 the following values are obtained

Unit vertical shearing force at periphery of dome

V1 F1(d) 2500(8) 100 kNm

Unit thrust at periphery of dome

N1 V1cosec13 100 155 155 kNm

(Note The required thickness of the dome at the springing isdetermined by the values of N1 and V1)

Unit outward horizontal thrust from dome on ring beam A

H1 V1cot 13 100 1192 1192 kNm

Unit vertical shearing force at inner edge of cone

V2 (F2 F3)(d) (2800 1300)(8) 163 kNm

Unit inward horizontal thrust from cone on ring beam A

H2 V2cot 163 0900 1468 kNm

Circumferential compression in ring beam A

NA 05d(H2 H1) 058 (14681192)1104 kN

Bottoms of elevated tanks and silos 191

(a) (b)

In (a) each beam spans between opposite columns and carriesone-quarter of the load on the floor The remaining half of thefloor load the weight of the wall and the load from the roof aretransferred to the columns through the wall In (b) each lengthof beam between the columns carries the load on the shaded

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Rectangular containers spanning horizontallymoments in walls 280

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Bottoms of elevated tanks and silos 281

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(Note If H1 exceeds H2 the circumferential force is tensile Theideal case occurs when H1 H2 and NA 0 see as follows)

Unit vertical shearing force at outer edge of cone

V3 F3(do) 1300(10) 414 kNm

Unit outward horizontal thrust from cone on ring beam B

H3 V3 cot 414 0900 372 kNm

Circumferential tension in ring beam B

NB 05do H3 05 10 372 1862 kN

The vertical wall must be reinforced for the circumferentialtension due to the horizontal pressure of the contained liquidgiven by n 05doz per unit depth The conical part of thetank must be reinforced to resist circumferential tension andthe reinforcement may be either distributed over the height ofthe conical portion or concentrated in the ring beams at the topand bottom In large-diameter Intze tanks the width of ringbeam A can be considerable and if this is so the weight ofwater immediately above the beam should not be taken to con-tribute to the forces on the dome and conical part With a widebeam F1 is taken as the weight of the contents over the net areaof the dome and d as the internal diameter of the ring beamF2 is taken as the weight of the contents over the net area of theconical part and d for use with F2 as the external diameterof the ring beam If this were to be done for a ring beam ofreasonable width in the forgoing example H1 and H2 would bemore nearly balanced

1742 Columns supporting elevated tanks

For a group of four columns on a square grid the thrusts andtensions in the columns due to wind loading on the tank can

be calculated as follows When the wind blows normal to theside of the group the thrust on each column on the leeward sideand the tension in each column on the windward side is Mw2dwhere d is the column spacing and Mw is the total moment dueto the wind When the wind blows normal to a diagonal of thegroup the thrust on a leeward corner column and tension in awindward corner column is Mwdradic2 For any other arrangementthe force on any column can be calculated from the equivalentsecond moment of area of the group

Consider the case when the wind blows normal to the X-Xaxis of the group of eight columns shown in the followingfigure The second moment of area of the group of columns aboutthe axis is 2 (d2)2 4 (0353 d)2 10 d2 The thruston the extreme leeward column is (05d10 d2) Mw Mw2dThe forces on each of the other columns in the group can bedetermined similarly by substituting the appropriate value forthe distance of the column from the axis

Containment structures194

1743 Silos

Notes on the pressures set up on hopper bottoms are given insections 277 and 93 and Tables 215 and 216 Notes on thedesign of hopper bottoms in the form of inverted truncatedpyramids are given in section 642 Expressions for bendingmoments and tensile forces are given in Table 281

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181 PAD FOUNDATIONS

Some general notes on the design of foundations are given insection 71 The size of a pad or spread foundation is usuallydetermined using service loads and allowable bearing valuesThe subsequent structural design is then determined by therequirements of the ULS Presumed allowable bearing valuesrecommended for preliminary design purposes in BS 8004 aregiven in Table 282

1811 Separate bases

An introduction to separate bases is given in section 716Diagrams of bearing pressure distributions and expressions forpressures and maximum bending moments in rectangular basessubjected to concentric and eccentric loading are given inTable 282

Example 1 The distribution of bearing pressure is requiredunder a base 3 m long 25 m wide and 600 mm thick when itsupports a concentrated load of 1000 kN at an eccentricity of300 mm in relation to its length

Weight of base Fb 30 25 06 24 108 kNTotal load Ftot Fb Fv 108 1000 1108 kN

Eccentricity of total load

etot Fve Ftot (1000 03)1108 027 m

Since etot l6 306 05 m the bearing pressure diagramis trapezoidal and the maximum and minimum pressures are

f (1 6etot l)Ftot bl (1 6 02730)[1108(25 30)] (154 and 046) 148 228 kNm2 and 68 kNm2

Note For the structural design of the base Fb and Fv shouldinclude safety factors appropriate to the ULS Bearing pressurescorresponding to these values of Fb and Fv reduced by Fbblshould then be used to determine bending moments and shearforces for the subsequent design

1812 Combined bases

When more than one column or load is carried on a single basethe centre of gravity of the total load should coincide if possiblewith the centre of area of the base Then assuming a rigid

base the resulting bearing pressure will be uniformly distributedThe base should be symmetrically disposed about the line of theloads and can be rectangular or trapezoidal on plan as shownin Table 283 Alternatively it could be made up as a series ofrectangles as shown at (b) in Table 284 In this last case eachrectangle should be proportioned so that the load upon it acts atthe centre of the area with the area of the rectangle being equalto the load divided by an allowable bearing pressure and thevalue of the pressure being the same for each rectangle

If it is not practical to proportion the combined base in thisway then the total load will be eccentric If the base is thickenough to be considered to act as a rigid member the groundbearing pressure will vary according to the diagram shown at(c) in Table 284 For a more flexible base the pressure will begreater immediately under the loads giving a distribution ofpressure as shown at (d) in Table 284

In the case of a uniform distribution or linear variation of pres-sure the longitudinal bending moment on the base at any sectionis the sum of the anti-clockwise moments of the loads to the leftof the section minus the clockwise moment due to the groundpressure between the section and the left-hand end of the baseThis method of analysis gives larger values for longitudinal bend-ing moments than if a non-linear variation is assumed Formulaefor combined bases carrying two loads are given in Table 283

Example 2 A strip base 15 m long and 15 m wide carries aline of five unequal concentrated loads arranged eccentricallyas shown in the following figure The bending moment is to bedetermined at the position of load F2 where the values of theloads and the distances from RH end are as follows

F1 500 kN F2 450 kN F3 400 kN F5 300 kN

z1 14 m z2 11 m z3 8 m z4 5 m z5 15 m

Chapter 18

Foundations andretaining walls

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282Foundations presumed allowable bearing values and separate bases

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283Foundations other bases and footings

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284Foundations inter-connected bases and rafts

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Pad foundations 199

The first step is to determine the eccentricity of the total loadand check that e (13Fz13F l2) l6 Hence

13Fz 500 14 450 11 400 8 350 5 300 15

17350 kNm

13F 500 450 400 350 300 2000 kN

e 173502000 152 1175 m l6 25 m

The maximum and minimum bearing pressures can now becalculated from the formula in Table 282 where

k (1 6el) (1 6 117515) 147 or 053

Hence with 13Fbl 2000(15 15) 889 kNm2

fmax 147 889 1307 kNm2

fmin 053 889 471 kNm2

Then for any section X-X at distance y from the left-hand endof the base the bearing pressure is

fx fmax (fmax fmin) yl

Considering the loads to the left of section X-X where x is thedistance of a load from the section the resultant bendingmoment on the base at section X-X is

M 13Fx (2fmax fx) by26

Thus at the position of load F2 where y 4 m and x1 3 m

fx 1307 (1307 471) 415 1084 kNm2

M 500 3 (2 1307 1084) 15 426 208 kNm

1813 Balanced bases

With reference to the diagrams on the upper right-hand side ofTable 284 (a) shows a system in which beam BC rests on a baseat A supports a column on the overhanging end C and iscounterbalanced at B The reaction at A which depends on therelative values of BC and BA can be provided by a basedesigned for a concentric load The counterbalance at B couldbe provided by load from another column as at (b) in which casethe dead load on this column needs to be sufficient to counter-balance the dead and imposed loads on the column at C and viceversa It is often possible to arrange for base A1 to be positioneddirectly under column B Formulae giving the values of thereactions at A and A1 are given in Table 283 where variouscombinations of values for F1 and F2 usually need to be consideredThus if F1 varies from F1 max to F1 min and F2 varies from F2 max toF2 min reaction R (see diagram in Table 283) will vary from

Rmax e F1max l to Rmin e F1min l

Hence R1 and R2 can have the following values

R1max F1max Fbase 1 Rmax Fbeam 2

R1min F1min Fbase 1 Rmin Fbeam 2

R2max F2max Fbase 2 Rmin Fbeam 2

R2min F2min Fbase 2 Rmax Fbeam 2

Therefore base 1 must be designed for a maximum load of R1max

and base 2 for R2max but R2min must always be positive From thereactions the shearing forces and bending moments on the beamcan be calculated In the absence of a convenient column loadbeing available at B a suitable anchorage must be provided by

other means such as a counterweight in mass concrete as at (c)or the provision of tension piles If the column to be supported isa corner column loading the foundation eccentrically in twodirections one parallel to each building line as at (d) it is some-times possible to use a diagonal balancing beam anchored by aninternal column D In other cases however the two wall beamsmeeting at the column can be designed as balancing beams toovercome the double eccentricity For beam EC the cantilevermoment is FE e1 where FE is the column load and the upwardforce on column C is FE e1 (l1 e1) For beam EF the corre-sponding values are FE e2 and FE e2 (l2 e2) respectively

1814 Rafts

The required thickness of a raft foundation is determined by theshearing forces and bending moments which depend on themagnitude and spacing of the loads If the thickness does notexceed 300 mm a solid slab as at (a) in the lower part ofTable 284 is generally the most convenient form If a slab atground level is required it is usually necessary to thicken theslab at the edge as at (b) to ensure that the edge of the raft isdeep enough to avoid weathering of the ground under the raft Ifa greater thickness is required beam-and-slab constructiondesigned as an inverted floor as at (c) is more efficient In caseswhere the total depth required exceeds 1 m or where a level topsurface is required a cellular construction consisting of a topand bottom slab with intermediate ribs as at (d) can be adopted

When the columns on a raft are not equally loaded or are notsymmetrically arranged the raft should be designed so that thecentre of area coincides with the centre of gravity of the loadsIn this case the pressure on the ground is uniform and therequired area is equal to the total load (including the weight ofthe raft) divided by the allowable bearing value If the coinci-dence of the centre of area of the raft and the centre of gravityof the loads is impractical due to the extent of the raft beinglimited on one or more sides the shape of the raft on planshould be so that the eccentricity ew of the total load Ftot is keptto a minimum as in the example shown at (f)

The maximum bearing pressure (which occurs at the cornershown at distance a1 from axis N-N on the plan and should notexceed the allowable bearing value) is given by

where Araft is the total area of the raft and Iraft is the secondmoment of area about the axis N-N which passes through thecentre of area and is normal to the line joining the centre of areaand the centre of gravity of the loads The pressure along axisN-N is FtotAraft and the minimum pressure (at the corner shownat distance a2 from axis N-N) is given by

When the three pressures have been determined the pressure atany other point or the mean pressure over any area can beassessed Having arranged for a rational system of beams orribs to divide the slab into suitable panels as suggested by thebroken lines at (f) the panels of slab and the beams can bedesigned for the bending moments and shear forces due to thenet upward pressures to which they are subjected

fmin Ftot

Araft

Ftot ewa2

Iraft

fmax Ftot

Araft

Ftot ew a1

Iraft

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182 OPEN-PILED STRUCTURES

Expressions from which the loads on groups of inclined andvertical piles supporting jetties and similar structures can beobtained are given in Table 285 For each probable conditionof load the forces acting on the superstructure are resolved intohorizontal and vertical components Fh and Fv the points ofapplication of which are also determined If the direction ofaction and position of the forces are opposite to those in thediagrams the signs in the formulae must be changed

Example 1 vertical piles only The adjacent figure showsa cross section through a jetty where the loads apply to one rowof piles (ie n 1 for each line and 13n 4) Since the groupis symmetrical x 05 90 45 m

M Fv e Fh h 800 075 100 48 120 kNm

The calculations for the loads on the piles are shown in thefollowing table from which the maximum load on any pile is212 kN For each pile the shear force is 1004 25 kN andthe bending moment is 25 482 60 kNm

Foundations and retaining walls200

Example 2 vertical and inclined piles The adjacent figureshows a cross section through a jetty where all the piles aredriven to the same depth Since A is the same for all the pilesif J Al is taken as unity for piles N1 and N4 then for piles N2

and N3 J 4radic(1 42) 097 Since the group is symmetrical132 0 133 not needed 134 0 k 131 135 xo 92 45 mand yo 0

M Fv e Fh h 800(525 45) 0 600 kNm

The calculation is then as shown in the following table fromwhich k 3826 0114 0436 The axial loads on the pilesare given by Nx kp (kw Fv kh Fh km M) hence

N1 10 (0261 800 0 0111 600) 1421 kN

N2 0941 (0261 800 219 100 0) 4026 kN

N3 0941 (0261 800 219 100 0) 96 kN

N4 10 (0261 800 0 0111 600) 2755 kN

Thus the maximum load on any pile is 4026 kN Note that theweight of the pile has to be added to the above values

kw km

Pile x x2 Axial loadno (m) (m2) (kwFv kmM)

N1 45 2025 025 45 0100 (025 X 800) (0100 X 120) 188kN45

N2 15 225 025 15 0033 200 (0033 X 120) 196kN45

N3 15 225 025 0033 200 4 204kN

N4 45 2025 025 0100 200 12 212kN

I nx2 4500 m4

xnI 1

n

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285Foundations loads on open-piled structures

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Foundations and retaining walls202

Example 3 inclined piles only The adjacent figure shows across section through a jetty where cot 13 5 Since the valuesof A and l are the same for each pile unity may be substitutedfor J Al For piles N1 and N3 B 5 and for piles N2 andN4 B 5 Since the group is symmetrical 132 0 133 notneeded 134 0 k 131 135 xo 3 m and yo 0

M Fv e Fh h 800(375 30) 0 600 kNm

The calculation is then as shown in the following table fromwhich k 3826 01538 05917 The axial loads on thepiles are given by Nx kp (kw Fv kh Fh km M) hence

N1 09806 (026 800 13 100 00867 600)

09806 (208 130 52) 2804 kN

N2 09806 (208 130 52) 255 kN

N3 09806 (208 130 52) 3824 kN

N4 09806 (208 130 52) 1275 kN

Thus the maximum load on any pile is 3824 kN to which theweight of the pile has to be added

Pile x Xno 1 5 (m) (m) 6 I kp kw kh km

N1 0114 450436 405

1 0 0 45 2025 1 0261 0 0111

N2 097 42 097

097 4

3826

1 + 42 1 + 42 radic(1 42) 4 0436

0913 0057 45 0 0 0941 0261 219 0

N3 0913 0057 45 0 0 0941 0261 219 0

N4 1 0 9 45 2025 1 0261 0 0111

Totals 3826 0114 mdash mdash 405 mdash mdash mdash mdash

Pile x Xno 1 5 (m) (m) 6I kp kw kh XI

N152 1 09615(3)2 5 01538 3846 3

152 152 radic(152) 05917 5 05917 3462

09615 00385 0 30 865 09806 026 13 00867

N2 3846

5 05917

09615 00385 0 30 865 09806 026 13 00867

N3 3

3462

09615 00385 60 30 865 09806 026 13 00867

N4 09615 00385 60 30 865 09806 026 13 00867

Total 3846 01538 mdash mdash 3462 mdash mdash mdash mdash

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chart given in Table 286 which is a modified version of a chartpublished in ref 63 The modified chart is applicable to designmethods in which the soil parameters incorporate either mobil-isation factors as in BS 8002 or partial factors of safety asin the Eurocodes

Stability against overturning is assured over the entire rangeof the chart and the maximum bearing pressure under serviceconditions can be investigated for all types of soil A uniformsurcharge that is small compared to the total forces acting onthe wall can be simply represented by an equivalent height ofsoil l can be replaced by le l q where q is the surchargepressure In more general cases le 3MhFh and 2FhKAle

2

can be used where Fh is the total horizontal force and Mh is thebending moment about underside of base due to Fh

The chart contains two curves showing where the bearingpressure is uniform and triangular respectively A uniformbearing condition is important when it is necessary to avoidtilting in order to minimise deflection at the top of the wallIt is generally advisable to maintain ground contact overthe full area of the base especially for clays where theoccurrence of ground water beneath the heel could soften theformation

Further information on the use of the chart for walls that aredesigned in accordance with BS 8002 and BS 8110 is given insection 284

184 BOX CULVERTS

Formulae for the bending moments in the corners of a boxculvert due to symmetrical load cases are given in Table 287Some notes on the different load cases and assumed groundconditions are given in section 742 Design requirements ofHighways Agency BD 3187 are summarised in section 743

Box culverts 203

Notes on design examples (1)ndash(3) In each case the pilegroup is symmetrical and is subjected to the same imposedloads Examples (2) and (3) are special cases of symmetricalgroups for which 134 0 and therefore yo 0 this conditionapplies only when the inclined piles are in symmetrical pairswith both pairs meeting at the same pile-cap

Example (3) requires the smallest pile but the differencein the maximum load is small between (2) and (3) Althoughthe maximum load on a pile is least in (1) the bendingmoment requires a pile of greater cross-sectional area to pro-vide the necessary resistance to combined bending and thrustIf the horizontal force is increased the superiority of (3) isgreater If Fh 200 kN (instead of 100 kN) the maximumpile loads are 236 kN (and a large bending moment of120 kNm) in (1) 609 kN in (2) and 510 kN in (3)Arrangement (2) is the most suitable when Fh is small IfFh 10 kN the maximum pile loads are 255 kN (and abending moment of 6 kNm) in (1) 217 kN in (2) and 268 kNin (3) Arrangement (1) is the most suitable when there is nohorizontal load

183 RETAINING WALLS

Various types of earth retention systems for which notes aregiven in section 731 are shown in Table 286 Information onthe pressures exerted by soils on retaining structures is given insection 91 and Tables 210 to 214

1831 Walls on spread bases

Several walls on spread bases for which notes are given insection 732 are shown in Table 286 Suitable dimensions fora base to a cantilever wall can be estimated with the aid of the

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Roker

Facing

Groutedbars

In-situ wall

Gravityelements(interlockingcribs)

Strul

Cantileverwall

Potential failure wedge

In-situwall

Trebocks

Anchors

Potentialfailurewedge

Potentialfailurewedge

Facingpanels

Strips or grids

286Retaining walls

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287Rectangular culverts

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191 STAIRS

Some general notes on stairs are given in section 614 Fordetails of characteristic imposed loads on stairs and landingsand horizontal loads on balustrades refer to BS 6399 Part 1Stairs forming monolithic structures are generally designed forthe uniformly distributed imposed load only Stairs that areformed of separate treads (usually precast) cantilevering from awall or central spine beam must be designed also for the alter-native concentrated load Some general information on stairtypes and dimensions is given in Table 288 and comprehensivedata is given in BS 5395

1911 Simple stairs

The term lsquosimplersquo is used here for a staircase that spans in thedirection of the stair flight between beams walls or landingslocated at the top and bottom of the flight The staircase mayinclude a section of landing spanning in the same direction andcontinuous with the stair flight For such staircases the followingstatements are contained in BS 8110

When staircases surrounding open wells include two spansthat intersect at right angles the load on the area common to bothspans may be divided equally between the spans When the stair-case is built at least 110 mm into a wall along part of all of itslength a 150 mm wide strip adjacent to the wall may be deductedfrom the loaded area When the staircase is built monolithicallyat its ends into structural members spanning at right angles to itsspan the effective span of the stairs should be taken as the hori-zontal distance between the centrelines of the supporting mem-bers where the width of each member is taken not more than18 m For a simply supported staircase the effective span shouldbe taken as the horizontal distance between the centrelines of thesupports or the clear distance between the faces of supports plusthe effective depth of the stair waist whichever is the lesser If astair flight occupies at least 60 of the effective span the per-missible spaneffective depth ratio calculated for a slab may beincreased by 15

1912 Free-standing stairs

In ref 64 Cusens and Kuang employ strainndashenergy principles todetermine expressions relating the horizontal restraint force Hand moment Mo at the mid-point of a free-standing stair

By solving the two resulting equations simultaneously thevalues of H and Mo obtained can then be substituted into generalexpressions to determine the forces and moments at any pointalong the structure

It is possible by neglecting subsidiary terms to simplify thebasic equations produced by Cusens and Kuang If this is donethe expressions given in Table 288 are obtained and these yieldH and Mo directly Comparisons of the solutions obtained bythese simplified expressions with those obtained using theexpressions presented by Cusens and Kuang show thatthe resulting variations are negligible for values in the rangeencountered in concrete design

The expressions given in Table 288 are based on a value ofGE 04 with C taken as half of the St Venant value for plainconcrete As assumed by Cusens and Kuang only half ofthe actual width of the landing is considered to determine its sec-ond moment of area

Example 1 Design a free-standing staircase assumed to befully fixed at the ends to support total ultimate loads per unitlength nf 169 kNm and nl 150 kNm The dimensions areas follows a 27 m b 14 m b1 18 m hf 100 mmhl 175 mm and 30o

From the expressions given in Table 288 H 8186 kNmand Mo 3587 kNmm If Cusens and Kuangrsquos more exactexpressions are used to analyse the structure H 8189 kNmand Mo 3567 kNmm Thus the errors involved by using theapproximate expressions are negligible for H and about 05for Mo If these values of H and Mo are substituted into the otherexpressions in Table 288 corresponding values of Mv Mh andT at any point in the structure can be found for various loadcombinations as shown in the table in page no 208

1913 Sawtooth stairs

In ref 65 Cusens shows that if axial shortening is neglected andthe strainndashenergy due to bending only is considered the mid-span moments for a so-called sawtooth stairs are given by thegeneral expression

Ms

where k (stiffness of tread)(stiffness of riser) and j is thenumber of treads

nl2(k11 kk12)

j2(k13 kk14)

Chapter 19

Miscellaneousstructures and details

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288Stairs general information

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If j is odd

k11 j216 j( j ndash 1)( j ndash 2)48

k12 ( j ndash 1)216 ( j ndash 1)( j ndash 2)( j ndash 3)48

k13 j2 k14 ( j ndash 1)2

If j is even

k11 j( j ndash 1)( j ndash 2)48

k12 ( j ndash 1)( j ndash 2)( j ndash 3)48

k13 ( j ndash 1)2 k14 ( j ndash 2)2

The chart on Table 289 gives coefficients to determine thesupport moments for various values of j and k Having found thesupport moment the maximum mid-span bending moment canbe determined by using the appropriate expression on the tableand deducting the support moment

Typical bending moment and shearing force diagrams for astair are also shown on Table 289 together with suggestedarrangements of reinforcement Because of the stair profilestress concentrations occur in the re-entrant corners and the realstresses to be resisted will be larger than those obtained from themoments To resist these increased stresses Cusens recom-mends providing twice the reinforcement theoretically requiredunless suitable fillets or haunches are incorporated at the junc-tions in which case the reinforcement need only be about 10more than is theoretically necessary The method of reinforcingshown in diagram (a) is very suitable but is generally only prac-tical if haunches are provided Otherwise the arrangementshown in diagram (b) should be adopted A further possibility isto arrange the bars as shown in diagram (a) on Table 363 forwall-to-wall corners

Example 2 A sawtooth stairs has seven treads each 300 mmwide with risers 180 mm high the thickness of both being100 mm The stairs which are 10 m wide are to be designed tosupport a characteristic imposed load of 30 kNm2 to therequirements of BS 8110

The self-weight of the treads and risers (assuming no finishesare required) is

gk 01 24 (03 018)03 384 kNm2

For design to BS 8110 total design ultimate load for a 10 mwide stair is given by

n 14 gk 16qk 14 384 16 30 1018 kNm

Since lt 300 mm lr 180 mm and ht hr 100 mm

k ht3lrhr

3lt 180300 06

From the chart on Table 289 for 7 treads and k 06 thesupport moment coefficient is ndash 0088 Thus

Ms ndash0088nl2 ndash0088 1018 (03 7)2 ndash395 kNm

Since j 7 is odd the free bending moment is given by

M (nl28)( j2 1)j2 1018 2128 5049 573 kNm

Hence the maximum moment at mid-span is

M0 M ndash Ms 573 ndash 395 178 kNm

1914 Helical stairs

By using strainndashenergy principles it is possible to formulate forsymmetrically loaded helical stairs fully fixed at the ends thefollowing two simultaneous equations in M0 and H

M0[K1(k5 sin213) k5 k7]

HR2[k4(k7 ndash K1)tan k5 sin cos (1 ndash K2)]

nR12[K1(k5 sin213ndash sin13) k5 k7 k6 k7R2R1] 0

M0[k4(k7 ndash K1) k5 (k7 ndash K2)]

HR2[12K1 tan ( 133 ndash 132sin213ndash 2k4)

12k7 tan ( 133 132sin213 2k4) 2k4 tan (k7 ndash K2)

k5 cos2 (tan K2 cot)] nR12[K1(k6 ndash k4) k4 k7

k7 (132sin 13 2k6)R2R1 (k7 ndash K2)(k5 k6R2R1)] 0

where

k4 13cos213ndash sin213 k5 13ndash sin213

k6 13cos13ndash sin13 k7 cos2 K2sin2

K1 GCEI1 K2 GCEI2 and 13 2

The equations can be solved on a programmable calculator orlarger machine to obtain coefficients k1 and k2 representing M0

and H respectively If the resulting values of M0 and H are thensubstituted into the equations given on Table 289 the bendingand torsional moments shearing forces and thrusts at any pointalong the stair can be easily calculated The critical quantity con-trolling helical stair design is usually the vertical moment Mvs atthe supports and a further coefficient can be derived to give thismoment directly

In ref 66 Santathadaporn and Cusens give 36 design chartsfor helical stairs covering ranges of of 60o to 720o of 20o

to 50o bh of 05 to 16 and R1R2 of 10 to 11 for GE 37The four design charts provided on Tables 290 and 291 havebeen recalculated for GE 04 with C taken as half of theSt Venant value for plain concrete These charts cover rangesof of 60o to 360o and of 20o to 40o with values for bh of5 and 10 and R1R2 of 10 and 11 being the ranges met most

14

12

18

14

12

13

12

13

12

12

Miscellaneous structures and details208

Application of Values of Mv (kNmm) Values of Mh (kNmm) Values of T (kNmm)imposed load

At O At B At D At A At B in OB Throughout AB At B in BC Throughout AB

Throughout 3587 1680 116 861 7367 8294 735 369Flights only 2481 922 160 1067 5036 5668 403 255Landing only 3132 1680 314 333 6486 7303 735 322

Note At point B the expressions give theoretical values for Mv (with imposed load applied throughout) that reduce abruptly from ndash 4195 kNmm in OB tondash368 kNmm in BC due to the intersection with flight AB Since the members in the actual structure are of finite width Cusens and Kuang recommendredistributing these moments across the intersection between the flight and the landing to give a value of (ndash4195 ndash 368)2 2282 kNmm

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289Stairs sawtooth and helical stairs

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290Design coefficients for helical stairs ndash 1

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291Design coefficients for helical stairs ndash 2

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frequently in helical stair design Interpolation between thevarious curves and charts will be sufficiently accurate forpreliminary design purposes

Example 3 A helical stairs having an angle of inclination tothe horizontal plane of 25o is to be designed to support acharacteristic imposed load of 30 kNm2 to the requirements ofBS 8110 The stairs are to be 12 m wide with a minimum slabthickness of 120 mm The radius to the inside of the stair isRi 900 mm and the angle turned through is 240o

Assuming the mean thickness on plan of the stairs (includingtreads and finishes) is 220 mm the self-weight of the stairs is022 24 53 kNm2 and the design ultimate load intensity

n 14 53 16 30 1222 kNm2

The radius of the centreline of the load is given by

R1 158 m

The radius of the centreline of the stairs

R2 09 05 12 15 m

Hence R1R2 15815 105 bh 1200120 10 andfrom the charts on Tables 290 and 291 interpolating asnecessary k1 ndash 012 k2 152 and k3 ndash 032 Thus fora 12 m wide stairs the total design values are

M0 12k1nR22 ndash012 1222 152 12 ndash396 kNm

H 12k2nR2 152 1222 15 12 334 kN

Mvs 12k3nR22 ndash032 1222 152 12 ndash1056 kNm

The slab should now be checked to ensure that the thicknessprovided is sufficient to resist Mvs Then assuming this is sothe foregoing values of M0 and H can be substituted into theequations for Mv Mn T N Vn and Vh given on Table 289 toobtain moments and forces at any point along the stairs Forexample where 13 60o Mv 111 kNm Mn ndash 4817 kNmT 005 kNm N ndash 365 kN Vn 968 kN and Vh 167 kNTypical distributions of moments and forces along the stair areshown on Table 289

192 NON-PLANAR ROOFS

1921 Prismatic structures

To design a simple prismatic roof or any structure comprising anumber of planar slabs for service load the resultant loadsQ acting at right angles to each slab and the unbalanced thrustsN acting in the plane of each slab are determined first taking intoaccount the thrust of one slab on another The slabs are thendesigned to resist the transverse bending moments due to theloads Q assuming continuity combined with the thrusts N Thelongitudinal forces F due to the slabs bending in their own planeunder the loads N are for any two adjacent slabs AB and BCcalculated from formula (2) in Table 292 where MAB and MBC

are found from formula (1) if the structure is freely supported atthe end of length L For each pair of slabs AB-BC BC-CD andso on there is an equation like formula (2) containing threeunknown forces F If there are n pairs there are (n ndash 1) equa-tions and (n 1) unknowns The conditions at the outer edgesa and z of the end slabs determine the forces F at these edgesfor example if the outer edges are unsupported Fa Fz 0

2(213 093)

3(212 092)

2(R3o R3

i )

3(R2o R2

i )

The simultaneous equations are solved for the remainingunknown forces FA FB FC and so on The longitudinal stress atany junction B is calculated from the formula (in Table 292) forfb Variation of the longitudinal stress from one function to thenext is rectilinear If fb is negative the stress is tensile and shouldbe resisted by reinforcement The shearing stresses are generallysmall

1922 Domes

A dome is designed for the total vertical load only that is forthe weights of the slab any covering on the slab any ceiling orother distributed load suspended from the slab and the imposedload The service load intensity w is the equivalent load per unitarea of surface of the dome Horizontal service loads due towind and the effects of shrinkage and changes in temperaturecan be allowed for by assuming an ample normal load or byinserting more reinforcement than that required for the normalload alone or by designing for stresses well below permissiblevalues or by combining any or all of these methods

Segmental domes Referring to the diagram and formulae inTable 292 for a unit strip at S the circumferential force actingin a horizontal plane is T and the corresponding force actingtangentially to the surface of the dome (the meridional thrust) isN At the plane where 13 is 51o48 that is the plane of ruptureT 0 Above this plane T is compressive reaching a maximumvalue of 05wr at the crown of the dome (13 0) Below thisplane T is tensile equalling 0167wr when 13 60o andwr 90o The meridional thrust N is 05wr at the crown0618wr at the plane of rupture 0667wr when 13 60o andwr when 13 90o that is N increases from the crown towardsthe support and has its greatest value at the support

For a concentrated load F applied at the crown of the domeT and N are given by the appropriate formulae in Table 292where T is tensile and N is compressive The load is assumed tobe concentrated on so small an area at the crown that it can betaken as a point load The theoretical stress at the crown is there-fore infinite but the practical impossibility of obtaining a pointload invalidates the use of the formulae when 13 0 or verynearly so For domes of varying thickness see ref 67

For a shallow dome approximate analysis only is sufficientand appropriate formulae are given in Table 292

Conical domes In a conical dome the circumferential forcesare compressive throughout and for any horizontal plane atdistance x from the apex are given by the expression for T inTable 292 the corresponding force in the direction of the slopebeing N The horizontal outward force per unit length of thecircumference at the bottom of the slope Tr needs to be resistedby the supports or a bottom ring beam

1923 Segmental shells

General notes on the design of cylindrical shell roofs and the useof Tables 293 and 294 are given in section 6 16

Membrane action Consideration of membrane action onlygives the following membrane forces per unit width of slab dueto the uniform load shown on Table 292 stresses are obtainedby dividing by the thickness of the shell h Negative values of Vindicate tension in the direction corresponding to an increase in

Miscellaneous structures and details212

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292Non-planar roofs general data

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293Shell roofs empirical design method ndash 1

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294Shell roofs empirical design method ndash 2

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x and a decrease in 13x In the case of F positive values indicatetension Reinforcement should be provided approximately inline with and to resist the principal tensile forces If the shell issupported along any edges the forces will be modified accord-ingly The values of the forces at any point are as follows

Tangential force

Fy ndash (g qcos13x) rcos13x

Longitudinal force

Fx ndash (1 ndash x)(xr)[gcos13x 15q(cos213x ndash sin213x)]

Shearing force

Vxy (2x ndash l)(g 15qcos13x)sin13x

Principal forces (due to membrane forces only)

Fp 05(Fx Fy) [(Fx ndash Fy)2 4V2xy]05

tan 2

At A (mid-point at edge 13x 13 x l2)

FyA ndash (g qcos13) rcos13

FxA ndash (l24r)[gcos13 15q(cos213ndash sin213)]

VxyA 0

At B (mid-point at crown 13x 0 x l2)

FyB ndash (g q)r

FxB ndash (l24r)(g 15q)

VxyB 0

At C (support at edge 13x 13 x 0)

FyC ndash (g qcos13) rcos13

FxC 0

VxyC ndash (g 15qcos13) lsin13

At D (support at crown 13x 0 x 0)

FyD ndash (g q) r

FxD 0 VxyD 0

Beam action If the ratio of the length of a cylindrical shell toits radius lr is not less than 25 the longitudinal forces canbe approximated with reasonable accuracy by calculating thesecond moment of area Ixx and the vertical distance from theneutral axis to the crown yndash from the approximate expressions

Ixx cong R2h (R ndash 3h2)(2 sincosndash 2sin2)

yndash cong R ndash [(R ndash 2h3)sin]

Then if n is the total uniform load per unit area acting on theshell (ie including self-weight etc) the maximum bendingmoment in a freely supported shell is given by

M (213rn)l28 13rnl24

Hence from the relationship MIxx = fy the horizontal forces atmid-span at the crown and springing of a shell of thickness h aregiven by the expressions

Fx(bottom) M[r(1 cos 13) y]h

IxxFx(top)

Myh

Ixx

2Vxy

Fx Fy

At the supports the total shearing force in the shell is

V (213rn)l2 13rnl

The shearing stress at any point is then given by v (VAx)(2hIxx)

where Axndash is the first moment of area about the neutral axis of thearea of cross section of shell above the point at which the shear-ing stress is being determined

The principal shortcoming of this approximate analysis is thatit does not indicate in any simple manner the magnitude of thetransverse moments that occur in the shell However a tabularmethod has been devised by which these moments can be eval-uated indirectly from the lateral components of the shearingstresses for details see ref 68

1924 Hyperbolic-paraboloidal shells

The simplest type of double-curved shell is generated by theintersection of two separate sets of inclined straight lines (parallelto axes XX and YY respectively as shown in the diagrams on Table292) Vertical sections through the shell at angles XX or YY areparabolic in shape while horizontal sections through the surfaceform hyperbolas hence the name hyperbolic paraboloid

Individual units such as those shown in diagrams (a) and (b)in Table 292 can be used separately being supported oncolumns or buttresses located at either the higher or the lowercorners Alternatively groups of units can be combined toachieve roofs having attractive and unusual shapes such as isshown in diagram (c) Some idea of the more unlikely forms thatcan be achieved may be obtained from ref 69

If the shell is shallow and the loading is uniform the shellbehaves as a membrane transferring uniform compressive andtensile forces of F2c (where F is the total load on the unit andc is the rise) acting parallel to the directions of principalcurvature to the edges of the shell The edge forces are thentransmitted back to the supports along beams at the shell edgesThe main problems that arise when designing these shells are theinteraction between the shell and the supporting edge membersthe design of the buttresses or ties needed to resist the horizon-tal component of the forces at the supports and the fact thatexcessive deflections at unsupported edges lead to stresses thatdiffer considerably from those predicted by simplified theories

Increasing the sharpness of curvature of the shell increases itsstability and reduces the forces and reactions within the shellbut to avoid the need for top forms the maximum slope shouldnot exceed 45o this corresponds to a value of 1radic2 for theratio ca or cb in the diagrams on Table 292 To ensure stabilityif a single unit is used the ratio should be not less than 15A useful introduction to the theory and design of hyperbolic-paraboloidal shells is given in ref 70

193 BEAMS CURVED ON PLAN

Some general notes on beams forming a circular arc on planare given in section 617 The following analyses apply only ifthe appropriate negative bending and torsional moments can bedeveloped at the end supports

1931 Concentrated loads

If a beam LR (see Table 295) curved on plan is subjected to aconcentrated load F such that the angle between the radius at

Miscellaneous structures and details216

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295Bow girders concentrated loads

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the point of application F of load F and the radius at themid-point O of the beam is 0 the following expressionsare applicable at any point X between F and L (ie 0)

M M0cos ndash T0sin V0rsin ndash Frsin ( ndash 0)

T M0sin T0cos V0r(1 ndash cos) ndash Fr[1 ndash cos ( ndash 0)]

V ndash V0 F

where M0 T0 and V0 are respectively the bending moment thetorsional moment and the shearing force at mid-span r isthe radius of curvature in plan and is the angle defining theposition of X as shown in the diagram on Table 295

If X is between F and O (ie X 0) terms containing Fare equal to zero If X is between O and R (ie X) signs of theterms containing sin should also be reversed Now by writingM0 K1Fr T0 K2Fr and V0 K3F

K1 K2 and K3

where k1 k2 k3 and so on are given by the following expressionsin which K EIGC (ie flexural rigiditytorsional rigidity)

k1 (K ndash 1)sin0(sin 213ndash sin 20)

(K ndash 1)cos0(sin213ndash sin20)

(K 1)(13ndash 0)sin0 ndash K(cos13ndash cos0)

k2 (K 1)13ndash 12(K ndash 1)sin213

k3 K(sin13ndash sin0) ndash 14 (K ndash 1)cos0(sin213ndash sin20)

(K ndash 1)sin0(sin213 ndash sin20)

(K 1)(13 ndash 0)cos0

k4 (K 1)13 (K ndash 1)sin213

k5 2Ksin13 ndash (K 1)13 ndash (K ndash 1)sin213

k6 (K ndash 1)cos0(sin213ndash sin20) K(13ndash 0)

(K 1)(13ndash 0)cos0

(K ndash 1)sin0(sin213ndash sin20)

K(1 cos0)(sin13ndash sin0)

Ksin0(cos13ndash cos0)

k7 k5

k8 (K ndash 1)sin213ndash 4Ksin13 (3K 1)13

For rectangular beams the graphs provided on Table 295 enablevalues of K1 K2 and K3 to be read directly for given values of 130 and hb (ie depthwidth) Values of G 04E and C J2as recommended in BS 8110 have been used

1932 Uniform load

For a curved beam with a UDL over the entire length owing tosymmetry the torsional moment (and also the shearing force) atmid-span is zero By integrating the foregoing formulae thebending and torsional moments at any point X along the beamare given by the expressions

M M0cos ndash nr2(1 ndash cos)

T M0sin ndash nr2( ndash sin)

If M0 K4nr2 where

K4 14[(1 K)sin13 K13 cos13]

213 (1 K) sin 213 (1 K)

12

12

12

14

12

12

12

12

12

12

14

k4k6 k3k7

k4k8 k5k7

k3k8 k5k6

k4k8 k5k7

k1

k2

these expressions can be rearranged to give the bending andtorsional moments at the supports and the maximum positivemoments in the span in terms of non-dimensional factors K4 K5K6 and K7 as shown on Tables 296 and 297 The factors can beread from the charts given on the tables

Example 1 A bow girder 450 mm deep and 450 mm widehas a radius of 4 m and subtends an angle of 90o The ends arerigidly fixed and the total UDL is 200 kN The maximummoments are to be determined

The distributed load per unit length

n 200(r2) 200( 42) 318 kNm

From the charts on Tables 296 and 297 (with hb 10 and13 45o) K4 0086 K5 ndash 023 K6 ndash 00175 K7 0023with 1 23o and 2 40o

Maximum positive bending moment (at midspan)

K4nr2 0086 318 42 438 kNm

Maximum negative bending moment (at supports)

K5nr2 ndash 023 318 42 ndash 117 kNm

Zero bending moment occurs at 1 23o

Maximum negative torsional moment (at supports)

K6nr2 ndash 00175 318 42 ndash 89 kNm

Maximum positive torsional moment (at 1 23o)

K7nr2 0023 318 42 117 kNm

Zero torsional moment occurs at 2 40o

Example 2 The beam in example 1 supports a concentratedload of 200 kN at a point where the radius subtends an angleof 15o from the left-hand end (ie 0 30o) Moments andshearing forces at midspan and the supports are required

From the charts on Table 295 (with hb 10 13 45o and0 30o) K1 0015 K2 ndash 00053 K3 0067

At midspan ( 0)

M0 0015 200 4 12 kNm

T0 ndash 00053 200 4 ndash 43 kNm

V0 0067 200 134 kN

At left-hand support ( 45o)

M M0cos 45o ndash T0sin 45o V0 rsin 45o ndash Frsin (45o ndash 30o) [12 ndash (ndash 43) 134 4) 0707 ndash 200 4 0259 ndash 158 kNm

T M0sin 45o T0cos 45o V0 r (1 ndash cos 45o) Fr (1 ndash cos 15o)

(12ndash43)070713440293ndash20040034 ndash 61 kNm

V V0 ndash F 134 ndash 200 1866 kN

At right-hand support ( 45o)

M M0cos 45o T0sin 45o ndash V0 rsin 45o

[12 ndash 43 ndash 134 4) 0707 ndash 324 kNm

T ndash M0sin 45o T0cos 45o V0 r (1 ndash cos 45o) (ndash 12 ndash 43) 0707 134 4 0293 42 kNm

V V0 134 kN

Miscellaneous structures and details218

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296Bow girders uniform loads ndash 1

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297Bow girders uniform loads ndash 2

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194 BEARINGS HINGES AND JOINTS

A comprehensive guide on bridge bearings and expansion jointsincluding a treatment of the design techniques used to accom-modate movements in bridges is contained in ref 71 Varioustypes of bridges for which notes are given in section 62 andtypical span ranges are shown in Table 298

1941 Hinges and bearings

A hinge is an element that can transmit thrust and transverseforce but permits rotation without restraint If it is vital for suchaction to be fully realised a steel hinge can be providedAlternatively hinges that are monolithic with the member canbe formed as indicated at (a) and (b) in Table 299 ThelsquoMesnagerrsquohinge shown at (a) has been used for example in theframes of large bunkers to isolate the container from the sub-structure or to provide a hinge at the base of the columns of ahinged frame bridge The hinge-bars lsquoarsquo resist the entirehorizontal shear force and the so-called throat of concrete at Dmust be sufficient to transfer the full compressive force from theupper to the lower part of the member The hinge-bars should bebound together by links lsquodrsquo and the main vertical bars lsquoersquo shouldterminate on each side of the slots B and C It may be advanta-geous during construction to provide bars extending across theslots and then cut these bars on completion of the frame Theslots should be filled with a bituminous material or lead or asimilar separating layer

The Freyssinet hinge shown at (b) has largely superseded theMesnager hinge In this case the large compressive stress acrossthe throat results in a high shearing resistance and the inclusionof bars crossing the throat can adversely affect the hinge Testshave shown that as a result of biaxial or triaxial restraint suchhinges can withstand compressive stresses of several times thecube strength without failure occurring The bursting tension oneach side of the throat normally governs the design of this typeof hinge

A number of other types of hinges and bearings that have beenused on various occasions are shown in Table 299

(c) a hinge formed by the convex end of a concrete memberbearing in a concave recess in the foundations

(d) a hinge suitable for the bearing of a girder where rotationbut not sliding is required

(e) a bearing for a girder where sliding is required

(f) a mechanical hinge suitable for the base of a large portalframe or the abutment of a large hinged arch bridge

(g) a hinge suitable for the crown of a three-hinged arch whenthe provision of a mechanical hinge is not justified

(h) a bearing suitable for the support of a freely suspended spanon a cantilever in an articulated bridge

Bridge bearings have to be able to accommodate the rotationsresulting from deflection of the deck under load They alsohave to be able to accommodate horizontal movements ofthe deck caused by prestress creep shrinkage and temperaturechange Some bearings allow horizontal movement in one direc-tion only and are restrained in the other direction whilst othertypes allow movement in any direction Elastomeric bearingsthat are formed of layers of steel plate embedded in rubbercan accommodate small horizontal shear movements

PTFE (polytetrafluoroethylene) bearings can give unlimitedfree sliding between low friction PTFE surfaces and steel platesPot bearings that incorporate rubber discs allow for small rota-tions while spherical bearings that move on a PTFE surface per-mit larger rotations Mechanical bearings such as rockers androllers can be used to provide either longitudinal fixity or resis-tance to lateral force Pot bearings special guide bearings or pinbearings are also used for this purpose Bearings need to beinspected regularly and may require maintenance or replace-ment during the lifetime of the bridge As this can be both diffi-cult and expensive it is very important that the structure isdesigned to make inspection maintenance and replacementpossible

1942 Movement joints

Movement joints are often required to allow free expansion andcontraction due to temperature changes and shrinkage in suchstructures as retaining walls reservoirs roads and long buildingsIn order to allow unrestrained deformation of the walls of cylin-drical containers sliding joints can be provided at the bottomand top of the wall Several types of joints for various purposesare shown in Table 2100 Figures (a)ndash(f) show some of the jointdetails recommended in BS 8007

Expansion joints at wide spacing may be desirable inlarge areas of walls and roofs that are not protected from solarheat gain or where a contained liquid is subjected to asubstantial temperature range Except for structures designed tobe fully continuous contraction joints of the type describedin section 2622 and at the maximum spacing specified inTable 345 should be provided The reinforcement should be cur-tailed to form a complete movement joint or made 50continuous to form a partial movement joint Waterstops arepositioned at the centre of wall sections and at the undersideof floor slabs that are supported on a smooth layer of blindingconcrete In basement walls waterstops are best positioned atthe external face where they are supported by the earth

The recommendations of BS 8007 with regard to the spacingof vertical joints may be applied also to earth-retaining wallsFor low walls with thin stems simple butt joints are generallyused However the effect of unequal deflection or tilting of onepart of a wall relative to the next will show at the joints For retain-ing walls higher than about 1 m a keyed joint can be usedAlternatively dowels passing through the joint with the ends onone side greased and sheathed can be used

Figure (g) shows alternative details at the joint betweenthe wall and floor of a cylindrical tank to minimise or eliminaterestraint at this position In the first case rubber or neoprenepads with known shear deformation characteristics are used Inthe second case action depends on a sliding membrane of PTFEor similar material These details are most commonly used forprestressed cylindrical tanks

Movement joints in buildings should divide the structure intoindividual sections passing through the whole structure aboveground level in one plane Joints at least 25 mm wide shouldbe provided at about 50 m centres both longitudinally andtransversely In the top storey and for open buildings andexposed slabs additional joints should be provided to give aspacing of about 25 m Joints also need to be incorporated in fin-ishes and cladding at movement joint locations Joints in wallsshould be made at column positions in which case a double

Bearings hinges and joints 221

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298Bridges

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299Hinges and bearings

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2100Movement joints

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column as shown at (h) can be provided The copper strip orother similar type of waterbar must be notched where the linksoccur the ends of the notched pieces being bent horizontally orcut off At joints in suspended floors and flat roofs a doublebeam can be provided Joints in floors should be sealed to pre-vent the accumulation of rubbish Roof joints should also beprovided with waterstops The provision of joints in large-areaindustrial ground floor slabs is considered in section 722 andrecommended details are given in ref 61

Expansion joints in bridges need to be either waterproof ordesigned to allow for drainage and should not badly disrupt the

riding quality of the deck Joints should also be designed torequire minimal maintenance during their lifetime and be able tobe replaced if necessary Compressible materials such as neopreneor rubber can accommodate small movements In this case jointscan be buried and covered by the surfacing This type of jointwhich consists of a small gap covered by a galvanised steel plateand a band of rubberised bitumen flexible binder to replace partof the surfacing is known as an asphaltic plug To accommodatelarger movements a flexible sealing element supported by steeledge beams is required Mechanical joints based on interlockingsets of steel toothed plates can be used for very large movements

Bearings hinges and joints 225

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The elastic analysis of a reinforced concrete section by themodular ratio method is applicable to the behaviour of thesection under service loads only The strength of the concrete intension is neglected and a linear stressndashstrain relationshipis assumed for both concrete and reinforcement The straindistribution across the section is also assumed to be linearThus the strain at any point on the section is proportional tothe distance of the point from the neutral axis and since thestressndashstrain relationship is linear the stress in the concrete isalso proportional to the distance from the neutral axis Thisgives a triangular distribution of stress ranging from zero atthe neutral axis to a maximum at the outermost point on thecompression face Assuming no slipping occurs between thereinforcement and the surrounding concrete the strain in bothmaterials is the same and the ratio of the stresses in the twomaterials depends on the ratio of the modulus of elasticity ofsteel and concrete known as the modular ratio e The value of Es

is taken as 200 kNmm2 but the value of E for concrete dependson several factors including the aggregate type the concretestrength and the load duration Commonly adopted values forsustained loads are 15 for normal-weight concrete and 30 forlightweight concrete The geometrical properties of reinforcedconcrete sections can be expressed in equivalent concrete unitsby multiplying the reinforcement area by e

201 PURE BENDING

Expressions for the properties of common reinforced concretesections are given in Tables 2102 and 2103 For sections thatare entirely in compression where the presence of the rein-forcement is ignored simplified expressions are given inTable 2101 The maximum stress in the concrete is given byfc for sections entirely in compression In other casesfc MK2 z unless expressed otherwise and the stress in theoutermost tension reinforcement is given by fs e fc (dx 1)

Expressions for the properties of rectangular and flangedsections are also given in Table 342 in connection with theserviceability calculation procedures contained in BS 8110 BS5400 and BS 8007 (see Chapter 26)

202 COMBINED BENDING AND AXIAL FORCE

The general analysis of any section subjected to direct thrustand uniaxial bending is considered in Table 2104 In the case

MxI

Chapter 20

Elastic analysis ofconcrete sections

of symmetrically reinforced rectangular columns the designcharts on Tables 2105 and 2106 apply The design charts onTable 2107 apply to annular sections such as hollow mastsInformation on uniaxial bending combined with direct tensionand biaxial bending and direct force is given in Tables 2108and 2109 respectively

2021 Symmetrically reinforced rectangularsection

For a symmetrically reinforced rectangular section subjected toaxial force N and bending moment M by equating forcesand taking moments about the mid-depth of the section thefollowing expressions are obtained

(1) For values of x h (ie entire section in compression)

where

A [1 (e 1) ]bh

(2) For values of x h (ie one face in tension)

where fc is maximum stress in concrete at compression faceand Ascbh is total reinforcement ratio

The stress in the tension reinforcement is then given by

fs e fc(dx 1)

For e 15 the charts given on Tables 2105 and 2106 canbe used directly For other values of e the curves for may beconsidered to represent values of [( e 1)14] in region (1)and (e 15) in region (2) For given values of M N and fc therequired value of can be readily determined For given values

dh

05 05edx 1d

h 05

Mbh2fc

05xh

05 x

3h 05(e1)1 h d

x

Nbhfc

05xh

05(e 1)1h d

x 05edh

1

I 112

(e 1)dh

12

bh3

Nbhfc

Abh

Ah2

21 Mbh2fc

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Geometric properties of uniform sections 2101

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Properties of reinforced concrete sections ndash 1 2102

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Properties of reinforced concrete sections ndash 2 2103

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Uniaxial bending and compression (modular ratio) 2104Equivalent area of strip

Equivalent area of transformed sectionDepth of centroid of transformed section

Moment of inertia of transformed section about centroid

Compressive stresses

fcr min Nd

Atr

Md(h x )Itr

fcr max Nd

Atr

MdxItr

Itr Atr(hs)2

12 (hc x )2

x Atr hcAtr

Atr Atr

Atr bshs (e 1)As

Assume a value of x

Depth to centre of tension where

If all bars are of the same size

Equivalent area of strip

Depth to centre of compression

Position of centroid of stressed area

Maximum stresses

Finally check the assumed value of x by substituting these stresses in

xd

1

1 fst(e fcr)

fst (d x)

S fcr

x (x hc)Atr Nd

fcr Nd x(e at x )

(at ac)(x hc)Atr

x eAs a Atr hc

eAs Atr

ac (x hc)hcAtr (x hc)Atr

Atr bshs (e 1)As

at a(a x)(a x)

S (a x)Asat SaS

Com

pres

siv e

str

esse

s on

lyC

ombi

ned

com

pres

sive

and

tens

ile s

tres

ses

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Symmetrically reinforced rectangular columns(modular ratio) ndash 1 2105

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2106Symmetrically reinforced rectangular columns(modular ratio) ndash 2

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2107Uniformly reinforced cylindrical columns(modular ratio)

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Uniaxial bending and tension (modular ratio) 2108

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Biaxial bending and compression (modular ratio) 2109

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Elastic analysis of concrete sections236

of M N and the value of fc is obtained by finding the point ofintersection of the curve and the eh line The eh line can beobtained by interpolation or can be drawn as follows on thegridline for Nbhfc 10 find the point where Mbh2fc is equalto the value of eh through this point draw a line from theorigin The value of xh can be obtained by interpolation or bysolving the equation

(xh)2 2[Nbhfc (e 05)](xh) (e dh 1) 0

Example 1 A 400 400 column reinforced with 4H32 barsis subjected to values of M 120 kNm and N 500 kN dueto service loads The maximum stresses in the concrete and thereinforcement are to be determined assuming e 15

Ascbh 32174002 002

eh MNh 120(500 04) 060

Allowing for 35 mm nominal cover to H8 links

d 400 (35 8 322) 340 mm dh 340400 085

From the chart for dh 340400 085 on Table 2105 at theintersection of 002 and eh 06

Nbhfc 025 xh 051

fc 500 103(025 4002) 125 Nmm2

x 051 400 204 mm

fs e fc (dx 1)

15 125 (340204 1) 125 Nmm2

2022 Uniformly reinforced annular section

The charts given on Table 2107 are based on the assumptionthat the bars may be represented with little loss of accuracy bya notional ring of reinforcement having the same total cross-sectional area and located at the centre of the sectionIf e 15 the charts can be used directly For other valuesof e the curves for and fs fc may be considered to representvalues of (e15) and (15e)( fsfc) respectively

Example 2 A cylindrical shaft with a mean radius of 1 m anda thickness of 100 mm is reinforced with 42H16 vertical barslocated at the centre of the section The shaft is subjected tovalues of M 2700 kNm and N 3600 kN due to serviceloads and the stresses in the concrete and the reinforcement areto be determined assuming e 15

Asc2rh 8446(2 1000 100) 00135

eh MNh 2400(3200 01) 75

A line corresponding to eh 75 can be drawn on the chart forhr 010 on Table 2107 from the origin to a point such asNrhfc 6 Mr2hfc 45 At the intersection of this line withan interpolated curve for 00135

Nrhfc 26 fsfc 6

fc 3200 103(26 1000 100) 123 Nmm2

fs 6 123 74 Nmm2

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Part 3

Design to British Codes

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In most British Codes the design requirements are set out inrelation to specified limit-state conditions Calculations todetermine the ability of a member (or assembly of members) tosatisfy a particular limit state are undertaken using design loadsand design strengths (or stresses) These design values aredetermined from characteristic loads and characteristicstrengths of materials (or stress limits) by the application ofpartial safety factors specified in the Code concerned

211 BUILDINGS

For buildings and other structures designed to BS 8110 thecharacteristic values of dead load Gk (Tables 21 and 22)imposed load Qk (Tables 23 and 24) and wind load Wk (Tables27ndash29) are specified in BS 6399 Parts 1 2 and 3

Design loads are given by

design load Fk f

where Fk is equal to Gk Qk or Wk as appropriate and f is thepartial safety factor appropriate to the load load combinationand limit-state being considered

The characteristic strength of a material fk means that valueof the cube strength of concrete fcu or the yield strength ofreinforcement fy below which 5 of all possible test resultswould be expected to fall In practice for concrete specified inaccordance with BS 8500 a dual classification is used forexample C2530 in which the characteristic strength of cylin-der test specimens is followed by the characteristic strength ofcube test specimens The characteristic strength of reinforcementis taken as the value specified in BS 4449 or BS 4482

Design strengths are given by

design strength fkm

where fk is either fcu or fy as appropriate and m is the partialsafety factor appropriate to the material and limit-state beingconsidered The appropriate factors are already incorporated inthe design equations provided in the Code

Details of the design requirements and partial safety factorsfor the ULS are given in Table 31 For the serviceability limitstates when calculations are required the partial safety factorsare taken as unity However for most designs explicit calcula-tions are unnecessary and the design requirement is met bycomplying with simple rules

For the ULS adverse and beneficial values are given fordead and imposed loads and these are to be applied separatelyto the loads on different parts of the structure to cause the mostsevere effects Thus for load combination 1 design loads mayvary from place to place with a maximum of (14Gk 16Qk)and a minimum of 10Gk (see also Table 230) For loadcombination 2 the maximum wind load of 14Wk with the min-imum dead load of 10Gk can result in a critical equilibriumcondition for tower structures For load combinations 2 and 3the design wind loads can sometimes be less than the minimumnotional horizontal load of 0015Gk associated with therequirement for robustness

The overall dimensions and stability of earth-retaining andfoundation structures are to be determined in accordance withthe appropriate codes for earth-retaining structures (BS 8002)and foundations (BS 8004) Design loads given in BS 8110 arethen used in to establish section sizes and reinforcement areasThe factor f should be applied to all earth and water pressuresexcept those that are derived from equilibrium with otherdesign loads such as bearing pressures below foundations

212 BRIDGES

For bridges and other structures designed to BS 5400 Part 4nominal and design values of dead load superimposed deadload wind load temperature live loads for highway footwayand railway bridges (Tables 25 and 26) and other loads aregiven in the Highways Agency Standard BD 3701

Design loads are given by

design load Fk fl

where Fk is the specified nominal load and fl is the partialsafety factor appropriate to the load load combination andlimit-state being considered A separate partial safety factor isthen used to allow for inaccuracies in the method of analysis

Design load effects are given by

design load effect effect of design load f3

where f3 is the partial safety factor appropriate to the methodof analysis and limit-state being considered If the method ofanalysis is linear elastic it is possible to replace fl by fl f3

in the calculation of the design loads

Chapter 21

Design requirementsand safety factors

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31Design requirements and partial safety factors (BS 8110)

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Liquid-retaining structures 241

The characteristic strength of a material fk means that valueof the cube strength of concrete fcu or the yield strength ofreinforcement fy below which 5 of all possible test resultswould be expected to fall In practice characteristic values arespecified in the manner described for design according toBS 8110 in section 211 The characteristic stress is the stressvalue at the assumed limit of linearity on the stressndashstrain curvefor the material

Design strengths are given by

design strength fkm

where fk is either fcu or fy as appropriate and m is the partialsafety factor appropriate to the material and limit-state beingconsidered The appropriate factors are already incorporated inthe design equations provided in the Code

Design stress limits (serviceability) are given by

design stress limit characteristic stressm

Details of the design requirements characteristic stresses andpartial safety factors are given in Tables 32 and 33

213 LIQUID-RETAINING STRUCTURES

For liquid-containing or liquid-excluding structures designed toBS 8007 the design basis is similar to that in BS 8110 but mod-ified for the limit-state of cracking Separate calculations ofcrack width are required for the effect of applied loads and theeffect of temperature and moisture change

Details of the design requirements and partial safety factors(updated according to BS 8110) are given in Table 34

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32Design requirements and partial safety factors (BS 5400) ndash 1

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33Design requirements and partial safety factors (BS 5400) ndash 2

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34Design requirements and partial safety factors (BS 8007)

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221 CONCRETE

2211 Strength and elastic properties

The characteristic strength of concrete means that value of the28-day cube strength below which 5 of all valid test results isexpected to fall In BS 8500 compressive strength classes areexpressed in terms of both characteristic cylinder strength andcharacteristic cube strength The recommended compressivestrength classes and mean values of the static modulus ofelasticity at 28 days are given in Table 35

In BS 8110 for normal-weight concrete the mean value ofthe static modulus of elasticity of concrete at 28 days is givenby the expression

Ec28 20 02fcu28 (kNmm2)

where fcu28 is the cube strength at 28 days The mean value ofthe modulus of elasticity at an age t 3 days can be estimatedfrom the expression

Ect Ec28 (04 06fcutfcu28)

where fcut is the cube strength at age t Where deflections are ofgreat importance and test data for concrete made with theaggregate to be used in the structure is not available a range ofvalues for Ec28 based on (mean value 6 kNmm2) should beconsidered

In BS 5400 slightly higher mean values of the modulus ofelasticity are used given by the expression

Ec 19 03fcu (kNmm2)

where fcu is the cube strength at the age considered

2212 Creep and shrinkage

The creep strain in concrete may be assumed to be directlyproportional to the applied stress for stresses not exceedingabout one-third of the cube strength at the age of loadingFor design to BS 8110 values of the creep coefficient (creepper unit of stress) according to the ambient relative humiditythe effective section thickness and the age of loading can beobtained from the figure in Table 35 Creep strain is partlyrecoverable if the stress is reduced The final recovery (after 1 year)

is approximately 03 (stress reduction)Eu where Eu is themodulus of elasticity at the age of unloading

For design to BS 8110 an estimate of the drying shrinkageof plain concrete according to the ambient relative humiditythe effective section thickness and the original water contentcan be obtained from the figure in Table 35 Aggregates witha high moisture movement such as some Scottish doleritesand whinstones and gravels containing these rocks produceconcrete with a high initial drying shrinkage (ref 12) Alsoaggregates with a low elastic modulus may result in higher thannormal concrete shrinkage

In BS 5400 values are obtained for the creep coefficient as aproduct of five partial coefficients and for the shrinkage strainas a product of four partial coefficients where the coefficientsare obtained from a series of figures

2213 Thermal properties

For design to BS 8110 values of the coefficient of thermalexpansion of concrete according to aggregate type are given inTable 35 In BS 5400 a value of 12 106 per oC is generallytaken except when limestone aggregates are used when a valueof 9 106 per oC is recommended

2214 Stressndashstrain curves

Typical short-term stressndashstrain curves for normal strengthconcretes in compression as described in section 316 andthe idealised curve given for design purposes in BS 8110and BS 5400 are shown in Table 36 In the expression givenfor the maximum design stress 067 takes account of the ratiobetween the cube strength and the strength in bending

222 REINFORCEMENT

2221 Strength and elastic properties

The characteristic strength of steel reinforcement made to therequirements of BS 4449 is 500 Nmm2 BS 4449 caters forround ribbed bars in three ductility classes grades B500AB500B and B500C Fabric reinforcement is manufacturedusing bars to BS 4449 except for wrapping fabric where wireto BS 4482 with a characteristic strength of 250 Nmm2 maybe used For more information on types properties and sizes of

Chapter 22

Properties of materials

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35Concrete (BS 8110) strength and deformation characteristics

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36Stress-strain curves (BS 8110 and BS 5400) concreteand reinforcement

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Properties of materials248

bar and fabric reinforcement reference should be made tosection 103 and Tables 219 and 220

2222 Stressndashstrain curves

Typical stressndashstrain curves for reinforcing steels in tension asdescribed in section 323 and the idealised curves given

for design purposes in BS 8110 and BS 5400 respectivelyfor reinforcement in tension or compression are shown inTable 36 For design purposes the modulus of elasticity of allreinforcing steels is taken as 200 kNmm2 The BS 5400stressndashstrain curve is the one that was used in CP110 prior tothat code being superseded by BS 8110

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In the following the term nominal cover is used to describe thedesign cover shown on the drawings It is the required coverto the first layer of bars including links The nominal covershould not be less than the values needed for durability andfire-resistance nor less than the nominal maximum size ofaggregate Also the nominal cover should be sufficient toensure that the cover to the main bars is not less than the barsize or for a group of bars in contact the equivalent bar sizeFor a group of bars the equivalent bar size is the diameter of acircle whose cross-sectional area is equal to the sum of the areasof the bars in the group

231 DURABILITY

2311 Exposure classes

Details of the classification system used in BS EN 206-1 andBS 8500-1 with informative examples applicable in the UKare given in Table 37 Often the concrete can be exposed tomore than one of the actions described in the table in whichcase a combination of the exposure classes will apply At thetime of drafting this Handbook the amendments necessitatedby the introduction of BS 8500 have not been incorporated inBS 5400 or BS 8007 (which is based on BS 81101985) Thesystem of exposure classes concrete grades and covers usedprior to BS 8500 is given in Table 39

2312 Concrete strength classes and covers

Concrete durability is dependent mainly on its constituents andlimitations on the maximum free watercement ratio and theminimum cement content are specified for each exposure classThese limitations result in minimum concrete strength classesfor particular cements For reinforced concrete the protection ofthe steel against corrosion depends on the cover The requiredthickness of cover is related to the exposure class the concretequality and the intended working life of the structure Details ofthe recommendations in BS 8500 are given in Table 38

The values given for the minimum cover apply for ordinarycarbon steel in concrete without special protection and for

structures with an intended working life of at least 50 yearsThe values given for the nominal cover include an allowance fortolerance of 10 mm which is recommended for buildings andis normally also sufficient for other types of structures

For uneven concrete surfaces (eg ribbed finish or exposedaggregate) the cover should be increased by at least 5 mm Ifconcrete is cast against an adequate blinding the nominal covershould generally be at least 40 mm For concrete cast directlyagainst the earth the nominal cover should generally be not lessthan 75 mm

232 FIRE-RESISTANCE

2321 Building regulations

The minimum period of fire-resistance required for elements ofthe structure according to the purpose group of a building andits height or for basements depth relative to the ground aregiven in Table 312 Building insurers may require longer fireperiods for storage facilities

2322 Nominal covers and minimum dimensions

The recommendations in BS 8110 regarding nominal cover fordifferent periods of fire-resistance are given in Table 310 Inthe table the cover applies to links for beams and columns butto main bars for floor slabs and ribs (even if links are provided)For two-way spanning solid slabs the cover may be taken asthe average for each direction For beams floors and ribs therequirements apply to the reinforcement in the bottom and sidefaces only The minimum thickness of floors includes anyconcrete screed on the top surface This is particularly impor-tant for ribbed slabs where the structural flange could be nomore than 75 mm thick The values given in the table apply tomembers whose dimensions comply with the minimum valuesgiven in Table 311

For cases where it is considered that special measures needto be taken to prevent the spalling of concrete a summary ofthe recommendations in BS 8110-2 is given in Table 310

Chapter 23

Durability andfire-resistance

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37Exposure classification (BS 8500)

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38Concrete quality and cover requirements fordurability (BS 8500)

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39Exposure conditions concrete and cover requirements(prior to BS 8500)

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310Fire resistance requirements (BS 8110) ndash 1

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311Fire resistance requirements (BS 8110) ndash 2

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312Building regulations minimum fire periods

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241 DESIGN ASSUMPTIONS

Basic assumptions regarding the design of cracked concretesections at the ULS are outlined in section 52 The tensilestrength of the concrete is neglected and strains are evaluatedon the assumption that plane sections before bending remainplane after bending Reinforcement stresses are then derivedfrom these strains on the basis of the design stressndashstraincurves shown on Table 36 The BS 5400 curve is the one thatwas used in CP 110 prior to that code being superseded byBS 8110 For the concrete stresses alternative assumptions arepermitted The design stressndashstrain curve for concrete shownon Table 36 gives a stress distribution that is a combination ofa parabola and rectangle The form of the data governing theshape of the curve causes the relative proportions of the twoparts to vary as the concrete strength changes Thus the totalcompressive force provided by the concrete is not linearlyrelated to fcu and the position of the centroid of the stress-block changes with fcu

Alternatively an equivalent rectangular stress distribution inthe concrete may be assumed as noted on Table 36 TheBS 5400 rectangular stress-block is the one that was usedin CP 110 prior to that code being superseded by BS 8110 InBS 5400 the use of the rectangular stress-block is prohibited inflanged ribbed and voided sections where the neutral axis liesoutside the flange although there appears to be no logicalreason for this restriction

For a rectangular area of width b and depth x the totalcompressive force is given by k1 fcubx and the distance of theforce from the compression face is given by k2x where values ofk1 (allowing for the term 067m)and k2 are given in thefollowing table according to the shape of the stress-block

Properties of concrete stress-blocks for rectangular area

Shape fcu Nmm2 k1 k2

Parabolic- 25 0405 0456rectangular 30 0401 0452(Table 36) 35 0397 0448

40 0394 044550 0388 0439

BS 8110 rectangular 0402 0450

BS 5400 rectangular 0400 0500

Using the rectangular stress-block is conservative in BS 5400but gives practically the same result as that obtained with theparabolic-rectangular form in BS 8110

242 BEAMS AND SLABS

Beams and slabs are generally subjected to bending only butsometimes are also required to resist an axial force for examplein a portal frame or in a floor acting as a prop between base-ment walls Axial thrusts not exceeding 01 fcu times the area ofthe cross section may be ignored in the analysis of the sectionsince the effect of the axial force is to increase the moment ofresistance Where a section is designed to resist bending onlythe value of the lever arm should not be taken greater than 095times the effective depth of the reinforcement

In cases where as a result of moment redistribution allowedin the analysis of the member the design moment is less than themaximum elastic moment at the section the neutral axis depthshould satisfy the requirement xd ( b 04) where b is theratio of design moment to maximum elastic moment

2421 Singly reinforced rectangular sections

Chapter 24

Bending and axialforce

The lever arm between the forces shown in the figure here isgiven by z (d k2x) from which x (d z)k2

Taking moments for the compressive force about the line ofaction of the tensile force gives

M k1 fcubxz k1 fcubz(d z)k2

The solution of the resulting quadratic equation in z gives

zd 05 where K Mbd 2fcu025(k2 k1)K 095

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Taking moments for the tensile force about the line of action ofthe compressive force gives

M As fsz from which As Mfsz

The strain in the reinforcement s 00035(1 xd)(xd) andfrom the BS 8110 design stressndashstrain curve the stress in thereinforcement is given by

fs sEs 700(1 xd)(xd) fy115

Thus for design to BS 8110 fs fy115 for values of

xd 805(805 fy) 0617 for fy 500 Nmm2

From the BS 5400 design stressndashstrain curve the stress in thereinforcement is given by

fs sEs 700(1 xd)(xd) 08fy115 or

08fy115 fs

Thus for design to BS 5400 fs 08fy115 for values of

xd 805(805 08fy) 0668 for fy 500 Nmm2

and fs fy115 for values of

xd 805(1265 fy) 0456 for fy 500 Nmm2

Design charts for fy 500 Nmm2 derived on the basis of theparabolic-rectangular stress-block for the concrete are given inTable 313 for BS 8110 and Table 323 for BS 5400 In the caseof BS 5400 unless the section is proportioned such thatfs fy115 the moment of resistance should provide at least115 times the applied design moment

Design tables based on the rectangular stress-blocks for theconcrete are given in Table 314 for BS 8110 and Table 324for BS 5400 The tables use non-dimensional parameters andare valid for fy 500 Nmm2 The formulae used to derive thetables and the limitations when redistribution of moment hasbeen allowed in the analysis of the member are also given

2422 Doubly reinforced rectangular sections

200fy

2300 fy35 45x d

x d fy 115

Thus for design to BS 8110 for values of

From the BS 5400 design stress-strain curve the stress in thereinforcement is given by

or

Thus for design to BS 5400 for values of

for fy500 Nmm2

and for values of

Equating the tensile and compressive forces gives

where the stress in the tension reinforcement is given by theexpressions used in section 2421

Design charts based on the rectangular stress-blocks forconcrete and for dd 01 and 015 respectively are given inTables 315 and 316 for BS 8110 and Tables 325 and 326 forBS 5400 The charts use non-dimensional parameters and weredetermined for fy 500 Nmm2 but may be safely used forfy 500 Nmm2 In determining the forces in the concrete noreduction has been made for the area of concrete displaced bythe compression reinforcement

2423 Design formulae for rectangular sections

Design formulae based on the rectangular stress-blocks forconcrete are given in BS 8110 and BS 5400 In both codes x islimited to 05d so that the formulae are automatically valid forredistribution of moment not exceeding 10

The stress in the tension reinforcement is taken as 087fy inboth codes although this is only strictly valid for xd 0456in BS 5400 The stress in the compression reinforcement istaken as 087fy in BS 8110 and 072fy in BS 5400 The codeequations which follow from the analyses in sections 2421and 2422 take different forms in BS 8110 and BS 5400

In BS 8110 the requirement for compression reinforcementdepends on the value of K Mbd2fcu compared to where

0156 for b 09

0402( b 04) 018( b 04)2 for 09 b 07

b is the ratio design moment to maximum elastic moment

For K compression reinforcement is not required and

As M087fyz

where

z d05 + lt 095d and x (d z)045

For K compression reinforcement is required and

(K ) bd2fcu087fy (d )

As bd2fcu087fyz

where

z d05 + and x (d z)045

For 0375 (for fy 500 Nmm2) should be replacedby 16(1 ) in the equations for and AsAsAsdx

Asdx

025 K09

KAs

dKAs

K

025K09

K

K

K

K

As fs k1 fcubx As f s

x d (7 3)(dd)f s 2000fy(2300 fy)

x d [805 (805 08fy)](dd) 2(dd)

f s 08fy 115

08fy 115 f s 200fy

2300 fy11535d

x 2000fy

2300 fy

f s sEs 700(1 dx) 08fy 115

x d [805(805 fy)](dd) 264(dd) for fy 500Nmm2

f s fy 115

Beams and slabs 257

The forces provided by the concrete and the reinforcement areshown in the figure here Taking moments for the two com-pressive forces about the line of action of the tensile force gives

M k1 fcubx(d k2x)

The strain in the reinforcement and fromthe BS 8110 design stressndashstrain curve the stress is given by

f s sEs 700(1 dx) fy 115

s 00035(1 dx)

As f s(d d)

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313BS 8110 Design chart for singly reinforced rectangularbeams

Sin

gly

rein

forc

ed b

eam

s (f

y=

500

Nm

m2 )

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314BS 8110 Design table for singly reinforced rectangularbeams

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315BS 8110 Design chart for doubly reinforced rectangularbeams ndash 1

Dou

bly

rein

forc

ed b

eam

s (f

y=

500

Nm

m2

d9d

=0

1)

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316BS 8110 Design chart for doubly reinforced rectangularbeams ndash 2

Dou

bly

rein

forc

ed b

eam

s (f

y=

500

Nm

m2

d9d

=0

15)

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In BS 5400 the moment of resistance of a section withoutcompression reinforcement is given by the equations

Mu As(087fy)z 015bd 2fcu

where

z d (1 11As fybdfcu) 095d

For a section with compression reinforcement the moment ofresistance is given by the equations

Mu 015bd2fcu (072fy)(d )

As(087fy) 02bdfcu (072fy)

The equations which are based on x 05d are consideredto be valid for values of dd 02 A variant on the equa-tions with x taken as a variable is included in HighwaysAgency BD4495 for assessment purposes These equationsshould not be used for values of x greater than 05d

2424 Flanged sections

In monolithic beam and slab construction where the web of thebeam projects below the slab the beam is considered as aflanged section for sagging moments The effective width of theflange may be taken as follows

T beam b bw 02lz actual flange widthL beam b bw 01lz actual flange width

lz is the distance between points of zero moment which for acontinuous beam may be taken as 07 times effective span

In most sections where the flange is in compression the depthof the neutral axis will be no greater than the thickness of theflange In this case the section can be considered to be rectangularwith b taken as the flange width The condition regarding theneutral axis depth can be confirmed initially by showing thatM k1 fcubhf (d k2hf) where hf is the thickness of the flangeAlternatively the section can be considered to be rectangularinitially and the neutral axis depth can be checked subsequently

As

dAs

Using the rectangular concrete stress-blocks in the foregoingequations gives k1 04 with k2 045 for BS 8110 and 05for BS 5400 This approach gives solutions that are lsquocorrectrsquowhen x hf but become slightly more conservative as(x hf) increases A different approach is used in BS 8110resulting in solutions that are lsquocorrectrsquo when x 05d butincreasingly conservative as (x hf) decreases As a result thesolution when x hf does not agree with that obtained byconsidering the section as rectangular with b taken as theflange width

2425 General analysis of sections

The analysis of a section of any shape with any arrangementof reinforcement involves a trial-and-error process Aninitial value is assumed for the neutral axis depth from whichthe concrete strains at the positions of the reinforcement canbe calculated The corresponding stresses in the reinforce-ment are determined and the resulting forces in thereinforcement and the concrete are obtained If the forces areout of balance the value of the neutral axis depth is changedand the process is repeated until equilibrium is achievedOnce the balanced condition has been found the resultantmoment of the forces about the neutral axis or any otherpoint is calculated

Example 1 The beam shown in the following figure is to bedesigned to the requirements of BS 8110 The design loads oneach span are as follows where Gk 160 kN andQk 120 kN

Fmax 14Gk 16Qk 416 kN Fmin 10Gk 160 kN

The section design is to be based on the following values

fcu 40 Nmm2 fy 500 Nmm2 cover to links 25 mm

For sagging moments effective width of flange

b bw 02lz 300 02(07 8000) 1420 mm

Allowing for 8 mm links and 32 mm main bars

d 500 (25 8 16) 450 mm say

Bending and axial force262

The figure here shows a flanged section where the neutral axisdepth is greater than the flange thickness The concrete forcecan be divided into two components and the required area oftension reinforcement is then given by

As As1 k1 fcu (b bw)hf 087fy

where

As1 area of reinforcement required to resist a moment M1

applied to a rectangular section of width bw and

M1 M k1 fcu (b bw)hf (d k2hf) 015bd2fcu

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In the calculations that follow solutions are obtained usingcharts and equations to demonstrate the use of each method

Maximum sagging moment For section to be designed asrectangular with b taken as the flange width bending momentshould satisfy the condition

M k1 fcubhf (d k2hf)

04 40 1420 150 (450 045 150) 106

1303 kNm (285 kNm)

Mbd2 285 106(1420 4502) 099 Nmm2

From chart in Table 313 100Asbd 024

As 00024 1420 450 1534 mm2

Alternatively K Mbd2fcu 09940 00248From Table 314 As fybdfcu 121K 00300

As 003 1420 450 40500 1534 mm2

Alternatively by calculation or from Table 314

Hence As M087fyz gives

As 285 106(087 500 095 450) 1533 mm2

Using 2H32 gives 1608 mm2

Maximum hogging moment

K Mbd2fcu 416 106(300 4502 40) 0171

From Table 314 As fybdfcu 0264

As 0264 300 450 40500 2851 mm2

Using 4H32 gives 3217 mm2

Although this is a valid solution it may be possible to reducethe area of tension reinforcement to a more suitable value byallowing for some compression reinforcement Consider theuse of 2H25 with 45 mm ( 01)

fybdfcu 982 500(300 450 40) 009

From the chart in Table 315 As fybdfcu 0225

As 0225 300 450 40500 2430 mm2

A solution can also be obtained using the design equations asfollows

With 2H25 for and assuming

K (087fy)(d )bd2fcu

0171982087500405(3004502 40)0100

Since is validdx 0375 f s 087fy

dx (dd)(x d) 010282 0355

x d (1 z d) 045 (1 0873) 045 0282

z d 05 025 0100 09 0873

dAsK

f s 087fyAs

As

ddd

z d 05 025 00248 09 0972 095

As bd2fcu 087fyz

982 0100 300 4502 40(087 500 0873 450)

2404 mm2 (compared to 2430 mm2 obtained from chart)

Using 3H32 gives 2413 mm2

Example 2 Suppose that in the previous example the maxi-mum hogging moment at B is reduced by 30 to 291 kNm

K Mbd2fcu 291 106(300 4502 40) 0120

b 291416 070 xd ( b 04) 030

From chart in Table 315 keeping to left of line for xd 03

fybdfcu 0025 As fybdfcu 0158

0025 300 450 40500 270 mm2

As 0158 300 450 40500 1706 mm2

A solution can also be obtained by using the design equationswith xd 03 as follows

0402(b 04) 018(b 04)2

0402 03 018 032 0104 (K 0120)

zd 05 0867

(K ) bd2fcu087fy (d ) 0016 300 4502 40 (087 500 405) 221 mm2

As bd2fcu087fyz 221 0104 300 4502 40(087 500 0867 450) 1710 mm2

Using 2H25 and 1H32 gives 1786 mm2

Since the reduced hogging moment for load case 1 is stillgreater than the elastic hogging moment for load case 2 thedesign sagging moment remains the same as in example 1

In the foregoing examples at the bottom of the beam 2H32bars would run the full length of each span with 2H25 splicebars at support B Other bars would be curtailed according tothe bending moment requirements and detailing rules

243 COLUMNS

In the Codes of Practice a column is a compression memberwhose greater overall cross-sectional dimension does notexceed four times its smaller dimension An effective heightand a slenderness ratio are determined in relation to major andminor axes of bending An effective height is a function of theclear height and depends upon the restraint conditions at theends of the column A slenderness ratio is defined as the effec-tive height divided by the depth of the cross section in the planeof bending The column is then considered to be either short orslender according to the slenderness ratios

Columns are subjected to combinations of bendingmoment and axial force and the cross section may need to bechecked for more than one combination of values In slendercolumns from an elastic analysis of the structure the initialmoment is increased by an additional moment induced bythe deflection of the column In BS 8110 this additional

KAs

dKAs

0250104 09

K

As

As

KAs

Columns 263

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moment contains a modification factor K the use of whichresults in an iteration process with K taken as 10 initiallyThe design charts in this chapter contain sets of K lines as anaid to the design process Details of the design procedures aregiven in Tables 321 and 322 for BS 8110 and Tables 331and 332 for BS 5400

2431 Rectangular columns

Design charts based on the rectangular stress-block for theconcrete and for values of dh 08 and 085 are given inTables 327 and 328 respectively On each curve a straight linehas been taken between the point where xh 08 and the pointwhere N Nuz The use of the rectangular stress-block resultsin Nuz being given by the equation

Nuzbhfcu 04 0714(Asc fybhfcu)

There are no K lines on the charts as no modification factor isused in the design of slender columns to BS 5400

2432 Circular columns

Bending and axial force264

The figure here shows a rectangular section in which thereinforcement is disposed equally on two opposite sides of ahorizontal axis through the mid-depth By resolving forces andtaking moments about the mid-depth of the section the followingequations are given for 0 xh 10

Nbhfcu k1(xh) 05(Asc fybhfcu)(ks1 ks2)

Mbh2fcu k1(xh)05 k2(xh) 05(Asc fybhfcu)(ks1 ks2) (dh 05)

For BS 8110 the stress factors ks1 and ks2 are given by

ks1 14(xh dh 1)(xh) 087

ks2 14(dh xh)(xh) 087

The maximum axial force Nuz is given by the equation

Nuzbhfcu 045 087(Asc fybhfcu)

Design charts based on the rectangular stress-block for theconcrete and for values of dh 08 and 085 are given inTables 317 and 318 respectively On each curve a straightline has been taken between the point where xh 10 andthe point where N Nuz The charts which were determinedfor fy 500 Nmm2 may be safely used for fy 500 Nmm2In determining the forces in the concrete no reduction hasbeen made for the area of concrete displaced by the com-pression reinforcement In the design of slender columns theK factor is used to modify the deflection corresponding to aload Nbal which for a symmetrically reinforced rectangularsection is given as 025bdfcu In the charts Nbal is taken as thevalue at which M is a maximum A line corresponding to Nbal

passes through a cusp on each curve For N Nbal K is takenas 10 For N Nbal K can be determined from the lines onthe chart

For BS 5400 the stress factors ks1 and ks2 are given by

ks1 14(xh dh 1)(xh) 07

07 ks1 025(33xh dh 1)(xh) 0714

ks2 14(dh xh)(xh) 07

07 ks2 025(dh 13xh)(xh) 087

The figure here shows a circular section containing six barsspaced equally around the circumference Solutions based on sixbars will be slightly conservative if more bars are used Thearrangement of the bars relative to the axis of bending affectsthe resistance of the section and the arrangement shown in thefigure is not the most critical in every case For some combina-tions of bending moment and axial force if the arrangementshown is rotated through 30o a slightly more critical conditionresults but the differences are small and may be reasonablyignored

The following analysis is based on a uniform stress-block forthe concrete of depth x and width hsin at the base (as shownin figure) Negative axial forces are included in order to cater formembers such as tension piles By resolving forces and takingmoments about the mid-depth of the section the followingequations are obtained where cos1 (1 2 xh) for0 x 10 and hs is the diameter of a circle through the centresof the bars

Nh2fcu kc (2 sin2)8 (12)(Asc fyAc fcu)(ks1 ks2 ks3)

Mh3fcu kc (3sin sin3)72 (277)(Asc fyAc fcu)(hsh)(ks1 ks3)

The minimum axial force Nmin is given by the equation

Nminh2fcu 087(4)(Asc fyAc fcu)

For BS 8110 kc 045 09 and the stress factors ks1 ks2

and ks3 are given by

087 ks1 14(0433hsh 05 xh)(xh) 087

087 ks2 14(05 xh)(xh) 087

087 ks3 14(05 0433hsh xh)(xh) 087

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The maximum axial force Nuz is given by the equation

Nuzh2fcu (4)045 087(Asc fyAc fcu)

Design charts for values of hsh 06 and 07 are given inTables 319 and 320 respectively The statements in section2431 on the derivation and use of the charts for rectangularsections apply also to those for circular sections

For BS 5400 kc 04 10 and the stress factors ks1 ks2

and ks3 are given by

087 ks1 025(0433hsh 05 13xh)(xh) 07

07 ks1 14(0433hsh 05 xh)(xh) 07

07 ks1 025(0433hsh 05 33xh)(xh) 0714

0714 ks2 025(05 33xh)(xh) 07

07 ks2 14(05 xh)(xh) 07

07 ks2 025(05 13 xh)(xh) 087

0714 ks3 025(0433hsh 05 33xh)(xh) 07

07 ks3 14(0433hsh 05 xh)(xh) 07

07 ks3 025(0433hsh 05 13xh)(xh) 087

The maximum axial force Nuz is given by the equation

Nuzh2fcu ( 4)04 072(Asc fyAc fcu)

Design charts for values of hsh 06 and 07 are given inTables 329 and 330 respectively The statements in section2431 on the derivation and use of the charts for rectangularsections apply also to those for circular sections

2433 Design formulae for short braced columns

Approximate formulae are given in BS 8110 for the design ofshort braced columns under specific conditions Where due tothe nature of the structure a column cannot be subjected tosignificant moments it may be considered adequate if thedesign ultimate axial load N 04fcuAc 075Asc fy

Columns supporting symmetrical arrangements of beamsthat are designed for uniformly distributed imposed load andhave spans that do not differ by more than 15 of the longermay be considered adequate if N 035fcuAc 067Asc fy

2434 General analysis of column sections

Any given cross section can be analysed by a trial-and-errorprocess For a section bent about one axis an initial value isassumed for the neutral axis depth from which the concretestrains at the positions of the reinforcement can be calculatedThe resulting stresses in the reinforcement are determined andthe forces in the reinforcement and concrete evaluated If theresultant force is not equal to the design axial force N the valueof the neutral axis depth is changed and the process repeateduntil equality is achieved The resultant moment of all theforces about the mid-depth of the section is then the moment ofresistance appropriate to N This approach is used to analyse arectangular section in example 6

Example 3 A 300 mm square braced column designed toBS 8110 for the following requirements

lo 375 m and 09 in both directions

Mx 54 kNm My 0 N 1800 kN

fcu 40 Nmm2 fy 500 Nmm2 cover to links 35 mm

Since leh 09 3750300 1125 15 the column isshort

Mmin N(005h) 1800 005 03 27 kNm (Mx)

Allowing for 8 mm links and 32 mm main bars

d 300 (35 8 16) 240 mm say

Mbh2fcu 54 106(300 3002 40) 005

Nbhfcu 1800 103(300 300 40) 05

From the design chart for dh 240300 08

Asc fybhfcu 022 (Table 317)

Asc 022 300 300 40500 1584 mm2

Using 4H25 gives 1963 mm2

Example 4 A 300 mm circular braced column designed toBS 8110 for the same requirements as example 3

Allowing for 8 mm links and 32 mm main bars

hs 300 2 (35 8 16) 180 mm say

Mh3fcu 54 106(3003 40) 005

Nh2fcu 1800 103(3002 40) 05

From the design chart for hsh 180300 06

Asc fyAc fcu 052 (Table 319)

Asc 052 (4) 3002 40500 2940 mm2

Using 6H25 gives 2945 mm2

Example 5 The column in example 3 but designed for biaxialbending with My 25 kNm and all other requirements as before

Since and Mx My the section may be designed for anincreased moment about the x-x axis (see Table 321)

1 (76)(Nbhfcu) 1 (76) 05 042

Mx My 54 042 25 645 kNm

Mbh2fcu 645 106(300 3002 40) 006

From the design chart for dh 240300 08

Asc fybhfcu 026 (Table 317)

Asc 026 300 300 40500 1872 mm2

Using 4H25 gives 1963 mm2

Example 6 A 300 mm square short column designed toBS 5400 for the following requirements

Mx 60 kNm My 40 kNm N 1800 kN

fcu 40 Nmm2 fy 500 Nmm2 d 240 mm

The section may be designed by assuming the reinforcement(4H32 say) and checking the condition (see Table 331)

Asc fybhfcu 3217 500(300 300 40) 045

Nbhfcu 1800 103(300 300 40) 05

Mx

bh

Columns 265

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317BS 8110 Design chart for rectangular columns ndash 1

Rec

tang

ular

col

umns

(f y

=50

0 N

mm

2 d

h=

08)

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318BS 8110 Design chart for rectangular columns ndash 2

Rec

tang

ular

col

umns

(f y

=50

0 N

mm

2 d

h=

085

)

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319BS 8110 Design chart for circular columns ndash 1

Circular columns (fy= 500 Nmm2 hs h = 06)

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320BS 8110 Design chart for circular columns ndash 2

Circular columns (fy= 500 Nmm2 hsh = 07)

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321BS 8110 Design procedure for columns ndash 1

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322BS 8110 Design procedure for columns ndash 2

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Bending and axial force272

From the design chart for dh 240300 08

Mubh2fcu 0075 Nuzbhfcu 072 (Table 327)

n 067 167(NNuz) 067 167(05072) 18

Since the column is square

Mux Muy 0075 300 3002 40 106 81 kNm

Since this value is less than 10 4H32 are sufficient

Example 7 The column in example 3 but taken as unbraced( 16) in the direction of Mx with all other requirements asbefore

Since lexh 16 3750300 20 10 the column is slender

The additional bending moment about the x-x axis is given by

Madd N(Kh2000)(lexb)2

With K 10 initially and since b h

Madd 1800 032000 202 108 kNm

M Mi Madd 54 108 162 kNm

Mbh2fcu 162 106(300 3002 40) 015

Nbhfcu 05 as before and from the design chart

Asc fybhfcu 063 (Table 317)

Asc 063 300 300 40500 4536 mm2

This requires 4T40 but it can be seen from the chart that withAsc fybhfcu 063 K is about 06 If we use 4H32

Asc fybhfcu 3217 500(300 300 40) 045

Nbhfcu 05 as before and from the design chart

Mbh2fcu 0107 and K 053

With K 053 corresponding to Asc fybhfcu 045

Madd 053 108 57 kNm M 54 57 111 kNm

Mbh2fcu 111 106(300 3002 40) 0103 0107

Thus 4H32 which gives 3217 mm2 is sufficient

Note that K can also be calculated from the equations given inBS 8110 as follows

Nuzbhfcu 045 087(Asc fybhfcu) 045 087 045 084

Nbalbhfcu 025(dh) 025 08 02

K (Nuz N)( Nuz Nbal) (084 05)(084 02) 053

Example 8 The following figure shows a rectangular sectionreinforced with 8H32 The ultimate moment of resistance of thesection about the major axis is to be determined in accordancewith the following requirements

N 2500 kN fcu 40 Nmm2 fy 500 Nmm2

Mx

Mux

n

My

Muy

n

6081

18

4081

18

086

Consider the bars in each half of the section to be replaced byan equivalent pair of bars Depth to the centre of area of the barsin one half of the section 60 2404 120 mm The sectioncan now be considered to be reinforced with four bars of areaAsc4 where d 600 120 480 mm

Asc fybhfcu 6434 500(300 600 40) 045

Nbhfcu 2500 103(300 600 40) 035

From the design chart for dh 480600 08

Mbh2fcu 014 (Table 317)

M 014 300 6002 40 106 605 kNm

The solution can be checked using a trial-and-error process toanalyse the original section as follows

N k1 fcubx (As1ks1 As2ks2 As3ks3)fy

where dh 540600 09 and ks1 ks2 and ks3 are givenby

ks1 14(xh dh 1)(xh) 087

ks2 14(05 xh)(xh) 087

ks3 14(dh xh)(xh) 087

With x 300 mm xh 05 ks1 087 ks2 0 and ks3 087

N 04 40 300 300 103 1440 kN (2500)

With x 360 mm xh 06 ks2 0233 ks3 07

N 04 40 300 360 103 (2413 087 1608 0233 2413 07) 500 103

1728 392 2120 kN (2500)

With x 390 mm xh 065 ks2 0323 ks3 0538

N 04 40 300 390 103 (2413 087 1608 0323 2413 0538) 500 103

1872 660 2532 kN (2500)

With x 387 mm xh 0645 ks2 0315 ks3 0553

N 04 40 300 387 103 (2413 087 1608 0315 2413 0553) 500 103

1858 636 2494 kN (asymp2500)

Taking moments about the mid-depth of the section gives

M k1 fcubx(05h k2x ) (As1ks1 As3ks3)(d 05h)fy

04 40 300 387 (300 045 387) 106

(241308724130553)(540300) 500106

233 412 645 kNm (605 obtained before)

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323BS 5400 Design chart for singly reinforced rectangularbeams

Sin

gly

rein

forc

ed b

eam

s (f

y=

500

Nm

m2 )

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324BS 5400 Design table for singly reinforced rectangularbeams

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325BS 5400 Design chart for doubly reinforced rectangularbeams ndash 1

Dou

bly

rein

forc

ed b

eam

s (f

y=

500

Nm

m2

d9d

=0

1)

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326BS 5400 Design chart for doubly reinforced rectangularbeams ndash 2

Dou

bly

rein

forc

ed b

eam

s (f

y=

500

Nm

m2

d9d

=0

15)

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327BS 5400 Design chart for rectangular columns ndash 1

Rec

tang

ular

col

umns

(f y

=50

0 N

mm

2 d

h=

08)

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328BS 5400 Design chart for rectangular columns ndash 2

Rec

tang

ular

col

umns

(f y

=50

0 N

mm

2 d

h=

085

)

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329BS 5400 Design chart for circular columns ndash 1

Rectangular columns (fy= 500 Nmm2 hsh = 06)

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330BS 5400 Design chart for circular columns ndash 2

Circular columns (fy= 500 Nmm2 hsh = 07)

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331BS 5400 Design procedure for columns ndash 1

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332BS 5400 Design procedure for columns ndash 2

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Chapter 25

Shear and torsion

251 SHEAR RESISTANCE

2511 Shear stress

In BS 8110 the design shear stress at any cross section in amember of uniform depth is calculated from

v Vbvd

where

V is the shear force due to ultimate loadsbv is the breadth of the section which for a flanged section

is taken as the average width of the web below the flanged is the effective depth to the tension reinforcement

For a member of varying depth the shear force is calculated asV (M tan13s)d where 13s is the angle between the tensionreinforcement and the compression face of the member Thenegative sign applies when moment and effective depth bothincrease in the same direction In no case should v exceed thelesser of 08radicfcu or 5 Nmm2 whatever the reinforcement InBS 5400 b is used in place of bv and the maximum value ofv is taken as the lesser of 075radicfcu or 475 Nmm2

2512 Concrete shear stress

In BS 8110 the design concrete shear stress vc is a function ofthe concrete grade the effective depth and the percentage ofeffective tension reinforcement at the section considered It isoften convenient to determine vc at the section where thereinforcement is least and use the same value throughout themember For sections at distance av 2d from the face of asupport or concentrated load vc may be multiplied by 2davproviding the tension reinforcement is adequately anchoredAlternatively as a simplification for beams carrying uniformload the section at distance d from the face of a support maybe designed without using this enhancement and the samereinforcement provided at sections closer to the support InBS 5400 the design concrete shear stress is obtained as svcwhere vc is the ultimate shear stress and s is a depth factor

2513 Shear reinforcement

Requirements for shear reinforcement depend on the value ofv in relation to vc (or svc) In slabs no shear reinforcement isrequired provided v does not exceed the concrete shear stress Inall beams of structural importance a minimum amount of shear

reinforcement in the form of links equivalent to a shearresistance Vsv 04 bvd is required The shear resistance can beincreased by introducing more links or bent-up bars can be usedto provide up to 50 of the total shear reinforcement Thecontribution of the shear reinforcement is determined on the basisof a truss analogy in which the bars act as tension members andinclined struts form within the concrete as shown in the figure here

System of bent-up bars used as shear reinforcement

In the figure the truss should be chosen so that both and are 45o and st 15d The design shear resistance providedby a system of bent-up bars is then given by

Vsb (Asbsb)(087fy)(stsin)

where st (d ndash d)(cot cot) 15d

For bars bent-up at 45o with fy 500 Nmm2

Vsb 0461Asb (dsb) 0307Asb kN

For bars bent-up at 60o with fy 500 Nmm2

Vsb 0594 Asb (d ndash d)sb 0376 Asb kN

In BS 8110 the shear resistance of members containing shearreinforcement is taken as the sum of the resistances providedseparately by the shear reinforcement and the concrete Thestrut analogy results in an additional longitudinal tensileforce that is effectively taken into account in the curtailmentrules for the longitudinal reinforcement In BS 5400 theminimum amount of shear reinforcement is required inaddition to that needed to cater for the difference between thedesign shear force and the concrete shear resistance It is alsonecessary to design the longitudinal reinforcement for theadditional force

Details of the design procedures for determining the shearresistances of members are given in Table 333 for BS 8110and Table 336 for BS 5400

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BS 8110 Shear resistance 333

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Design for torsion 285

2514 Shear under concentrated loads

The maximum shear stress at the edge of a concentrated loadshould not exceed the lesser of 08radicfcu or 5 Nmm2 Shear insolid slabs under concentrated loads can result in punching fail-ures on the inclined faces of truncated cones or pyramids Forcalculation purposes the shear perimeter is taken as the bound-ary of the smallest rectangle that nowhere comes closer to theedges of a loaded area than a specified distance The shearcapacity is checked first on a perimeter at distance 15d fromthe edge of the loaded area If the calculated shear stress is nogreater than vc no shear reinforcement is needed If the shearstress exceeds vc shear reinforcement is required within thefailure zone and further checks are needed on successiveperimeters at intervals of 075d until a perimeter is reachedwhere shear reinforcement is no longer required

Details of design procedures for shear under concentratedloads are given in Table 334 for BS 8110 and Tables 337 and338 for BS 5400

2515 Shear in bases

The shear strength of pad footings near concentrated loads isgoverned by the more severe of the following two conditions

(a) Shear along a vertical section extending for the full widthof the base In BS 8110 the concrete shear stress vc may bemultiplied by 2dav for all values of av 2d In BS 5400 thecritical section is taken at distance d from the face of the loadwith no enhancement of the concrete shear stress(b) Punching shear around the loaded area as described insection 2514 The reaction resulting from the soil bearingpressure within the shear perimeter may be deducted from thedesign load on the column when calculating the design shearforce acting on the section

The shear strength of pile caps is normally governed by theshear along a vertical section extending for the full width of thecap The critical section for shear is assumed to be located at20 of the pile diameter from the near face of the pile Thedesign shear force acting on this section is taken as the wholeof the reaction from the piles with centres lying outside thesection In BS 8110 the design concrete shear stress may bemultiplied by 2dav where av is the distance from the columnface to the critical section for strips of width up to three timesthe pile diameter centred on each pile In BS 5400 thisenhancement may be applied to strips of width equal to one pilediameter centred on each pile For pile caps designed by trussanalogy 80 of the tension reinforcement should be concen-trated in these strips

2516 Bottom loaded beams

Where load is applied near the bottom of a section sufficientvertical reinforcement to transmit the load to the top of thesection should be provided in addition to any reinforcementrequired to resist shear

252 DESIGN FOR TORSION

In normal beam-and-slab or framed construction calculationsfor torsion are not usually necessary adequate control of anytorsional cracking in beams being provided by the required

minimum shear reinforcement When it is judged necessaryto include torsional stiffness in the analysis of a structure ortorsional resistance is vital for static equilibrium membersshould be designed for the resulting torsional moment Thetorsional resistance of a section may be calculated on the basisof a thin-walled closed section in which equilibrium is satisfiedby a closed plastic shear flow Solid sections may be modelled asequivalent thin-walled sections Complex shapes may be dividedinto a series of sub-sections each of which is modelled as anequivalent thin-walled section and the total torsional resistancetaken as the sum of the resistances of the individual elementsWhen torsion reinforcement is required this should consist ofrectangular closed links together with longitudinal reinforce-ment Such reinforcement is additional to any requirements forshear and bending

Details of design procedures for torsion are given in Table 335for BS 8110 and Table 339 for BS 5400

Example 1 The beam shown in the following figure is to bedesigned for shear to the requirements of BS 8110 Details ofthe design loads and the bending requirements for which thetension reinforcement comprises 2H32 (bottom) and 3H32 (topat support B) are contained in example 1 of Chapter 24 Thedesign of the section is to be based on the following values

fcu 40 Nmm2 fy 500 Nmm2 d 450 mm

In the following calculations a simplified approach is used inwhich the critical section for shear is taken at distance d fromthe face of the support with no enhancement of the concreteshear stress In addition the value of vc is determined for thesection where the tension reinforcement is least and the samevalue of vc used throughout Since the beam will be providedwith shear reinforcement the value of vc may be taken as thatobtained for a section with d 400 mm

Based on 2H32 as effective tension reinforcement

100Asbvd 100 1608(300 450) 119

vc 078 Nmm2 (Table 333 for d 400 and fcu 40)

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BS 8110 Shear under concentrated loads 334

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BS 8110 Design for torsion 335

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BS 5400 Shear resistance 336

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BS 5400 Shear under concentrated loads ndash 1 337

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BS 5400 Shear under concentrated loads ndash 2 338

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BS 5400 Design for torsion 339

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Shear and torsion292

Consider H10 links and transverse spacing of legs at 200 mmto suit spacing of tension reinforcement that is 5 legs permetre Then required longitudinal spacing of links is given by

sv 5 78128 304 mm

Provide H10-300 with legs at 200 mm centres transversely

Example 3 A 280 mm thick flat slab is supported by 400 mmsquare columns arranged on a 72 m square grid Theslab which has been designed using the simplified method fordetermining moments contains as tension reinforcement inthe top of the slab at an interior support within a 18 m widestrip central with the column H16-180 in each directionThe slab is to be designed to the requirements of BS 8110(see Table 334) for a shear force resulting from the maximumdesign load applied to all panels adjacent to the column ofVt 954 kN

fcu 40 Nmm2 fy 500 Nmm2 d 240 mm (average)

For design using the simplified method the design effectiveshear force at an interior column Veff 115V 1098 kN

The maximum design shear stress at the column face

Veff uod 1098 103(4 400 240) 286 Nmm2 (50)

Based on H16-180 as effective tension reinforcement

100Asbvd 100 201(180 240) 046

vc 065 Nmm2 (Table 333 for d 240 and fcu 40)

The length of the first critical perimeter at 15d from the face ofthe column is 4 (3d 400) 4480 mm Thus the designshear stress at the first critical perimeter

v 1098 103(4480 240) 102 Nmm2 ( 157vc)

Since vc v 16vc and (v ndash vc) 04 Nmm2 the total areaof vertical links required within the failure zone is given by

Asv 04ud087fyv 04 4480 240(087 500)

989 mm2

20H8 will provide 1006 mm2 which should be arranged on twoperimeters at 05d 120 mm and 125d 300 mm from thecolumn face The inner perimeter should contain at least 40of the total that is 8H8 with 12H8 on the outer perimeter

The length of the second critical perimeter at 225d fromthe face of the column is 4 (45d 400) 5920 mm Thusthe design shear stress at the second critical perimeter

v 1098 103(5920 240) 077 Nmm2 (vc)

Asv 04 5920 240(087 500) 1307 mm2

12H8 are already provided by the outer perimeter of bars inthe first failure zone A further perimeter containing 16H8 togive a total of 28H8 will provide 1407 mm2 in the secondfailure zone The length of the third critical perimeter at 3d fromthe face of the column is 4 (6d 400) 7360 mm Thusthe design shear stress at the third critical perimeter

v 1098 103(7360 240) 062 Nmm2 (vc)

The reinforcement layout is shown in the figure followingwhere indicates the link positions and the spacing of thetension reinforcement has been adjusted so that the links can beanchored round the tension bars

Minimum link requirements are given by

Asv sv 04bv087fyv 04 300(087 500)

028 mm2mm

sv 075d 075 450 3375 mm

From Table 335 H8-300 provides 033 mm2mmShear resistance of section containing H8-300 is given by

Vu vcbvd (Asv sv)(087fyv)d

(078300033087500) 45010ndash3 170 kN

Based on a support of width 400 mm distance from centre ofsupport to critical section 200 450 650 mm The designload is 4168 52 kNm and the shear forces at the criticalsections are

End A V 172 ndash 065 52 138 kN 170 kN (H8-300)

End B V 260 ndash 065 52 226 kN 170 kN

v Vbvd 226 103(300 450) 168 Nmm2

Area of links required at end B is given by

Asvsv (v ndash vc)bv 087fyv (168 ndash 078) 300(087 500)

062 mm2mm

From Table 335 H8-150 provides 067 mm2mm

Note that if the concrete shear strength is taken as (2dav)vc forav 2d critical section is at av 2d and distance from centreof support to critical section 200 900 1100 mm HereV 203 kN v 150 Nmm2 Asvsv 050 mm2mm andfrom Table 335 H8-200 would be sufficient

Example 2 A 700 mm thick solid slab bridge deck is supportedat the end abutment on bearings spaced at 15 m centres Themaximum bearing reaction resulting from the worst arrange-ment of the design loads is 625 kN The tension reinforcement atthe end of the span is H25-200 The slab is to be designed forshear to the requirements of BS 5400 using the following values

fcu 40 Nmm2 fy 500 Nmm2 d 620 mm

If the critical section for punching is taken at 15d 930 mmfrom the edge of each bearing clearly the critical perimetersfrom adjacent bearings will overlap Therefore the slab will bedesigned for shear along a vertical section extending for thefull width of the slab Assuming that the bearing reaction canbe spread over a slab strip equal in width to the spacing ofthe bearings

v Vbd 625 103(1500 620) 067 Nmm2

Based on H25-200 as effective tension reinforcement

100Asbd 100 2454(1000 620) 040

vc 054 Nmm2 s 095 (Table 336 for fcu 40)

Area of links required at end of span is given by

Asvsv (v 04 ndash svc)b087fyv

(06704 ndash 095054)1000(087500)

128 mm2mm

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Design for torsion 293

Example 4 The following figure shows a channel sectionedge beam on the bottom flange of which bear 8 m long simplysupported contiguous floor units The edge beam which iscontinuous over several 14 m spans is prevented from lateralrotation at its supports The positions of the centroid and theshear centre of the section are shown in the figure and the beamis to be designed to the requirements of BS 8110

Characteristic loadsfloor units dead 35 kNm2 imposed 25 kNm2

edge beam dead 12 kNm

Design ultimate loadsfloor units (14 35 16 25) 82 356edge beam 14 12 168

524 kNm

fcu 40 Nmm2 fy 500 Nmm2 d 1440 mm

Bending moment shear force and torsional moment (aboutshear centre of section) at interior support (other than first)

M ndash 008 524 142 ndash 822 kNm (Table 229)

V 524 142 367 kN

T (356 0400 168 0192) 142 122 kNm

(Note In calculating V and T a coefficient of 05 rather than055 has been used since the dead load is dominant and thecritical section may be taken at the face of the support)

Considering beam as one large rectangle of size 250 1500and two small rectangles of size 200 300

13hmin3hmax 2503 1500 2 2003 300

(234 2 24) 109 282 109

Torsional moment to be considered on large rectangle

T1 122 234282 1012 kNm

Torsional moment to be considered on each small rectangle

T2 122 24282 104 kNm

Reinforcement required in large rectangle

Bending (see Table 314)

K Mbd2fcu 822 106(550 14402 40) 0018Since K 0043 As M087fyz where z 095d

As 822 106(087 500 095 1440) 1382 mm2

Shear (see Table 333)

v Vbvd 367 103(250 1440) 102 Nmm2

100Asbvd 100 1382(250 1440) 038vc 053 Nmm2 (for d 400 and fcu 40)

Asvsv bv(v ndash vc)087fyv 250 (102 ndash 053)(087 500)

028 mm2mm (total for all vertical legs)

Additional requirement for bottom loaded beam

Asvsv 356(087 500) 008 mm2mm (for inner leg)

Torsion (see Table 335)

vt 2T1[hmin2(hmax ndash hmin3)]

2 1012 106[2502 (1500 ndash 2503)] 229 Nmm2

(v vt) 102 229 331 Nmm2 (vtu 50)

Assuming 30 mm cover to links dimensions of links

x1 250 ndash 2 35 180 mm y1 1500 ndash 2 35 1430 mm

Asvsv T1[08x1y1(087fy)]

1012 106(08 180 1430 087 500)

113 mm2mm (total for 2 outer legs)

Total link requirement for shear torsion and bottom loadassuming single links with 2 legs

Asvsv 028 113 2 008 157 mm2mm

sv least of x1 180 mm y12 715 mm or 200 mm

From Table 335 H10-100 provides 157 mm2mm

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Shear and torsion294

Total area of longitudinal reinforcement for torsion

As (Asvsv)(fyvfy)(x1 y1)

113 10 (180 1430) 1819 mm2

Area of longitudinal reinforcement required in part of thesection between centrelines of flanges (1300 mm apart)

1819 1300(180 1430) 1469 mm2

From Table 228 14H12 provides 1583 mm2

Total area of tension reinforcement required at top of beam forbending and torsion

1382 05 (1819 ndash 1469) 1557 mm2

From Table 228 2H32 provides 1608 mm2

Reinforcement required in small rectangles

With link dimensions taken as x1 200 ndash 2 35 130 mmand y1 300 ndash 35 265 mm since y1 550 mm vt shouldnot exceed vtuy1550 50 265550 24 Nmm2

vt 2T2[hmin2(hmax ndash hmin3)]

2 104 106[2002 (300 ndash 2003)]

223 Nmm2 (24)

Area of link reinforcement required for torsion

Asvsv T2[08x1y1(087fy)]

104 106(08 130 265 087 500)

087 mm2mm (total for 2 outer legs)

sv least of x1 130 mm y12 132 mm or 200 mmThe lower rectangle should also be designed for the bendingand shear resulting from the load applied by the floor units

From Table 335 H8-100 provides 100 mm2mm

Area of longitudinal reinforcement for torsion

As (Asvsv)(fyvfy)(x1 y1)

087 10 (130 265) 344 mm2

From Table 228 4H12 provides 452 mm2

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Chapter 26

Deflection and cracking

261 DEFLECTION

Deflections of members under service load should not impairthe appearance or function of a structure For bridges there areno specific limits in BS 5400 but allowances are needed toensure that minimum clearances and satisfactory drainage areprovided Accurate predictions of deflections at different stagesof construction are also required

In BS 8110 the final deflection of members below thelevel of the supports after allowance for any pre-camber islimited to span250 A further limit to be taken as the lesserof span500 or 20 mm applies to the increase in deflectionthat occurs after the application of finishes cladding or par-titions in order to minimise any damage to such elementsThese requirements may be met by limiting the spaneffectivedepth ratio of the member to the values given in Table 340In this table the design service stress in the tensionreinforcement is shown as fs (58)(fyb)(As reqAs prov) InBS 8110 at the time of writing the term (58) which isapplicable to m 115 is given incorrectly as (23) whichis applicable to m 105

The spaneffective depth ratio limits take account ofnormal creep and shrinkage but if these are likely to beparticularly high (eg free shrinkage strain 000075 orcreep coefficient 3) the permissible spaneffective depthratio derived from the table should be reduced by up to 15The limiting ratios may be used also for designs where light-weight aggregate concrete is used except that for all beamsand slabs where the characteristic imposed load exceeds4 kNmm2 the values derived from Table 340 should bemultiplied by 085

In special circumstances when the calculation of deflectionis considered necessary the methods described in Tables 341and 342 can be used Careful consideration is needed in thecase of cantilevers where the usual formulae assume that thecantilever is rigidly fixed and remains horizontal at the rootWhere the cantilever forms the end of a continuous beam thedeflection at the end of the cantilever is likely to be eitherincreased or decreased by an amount l13 where l is the lengthof the cantilever measured to the centre of the supportand 13 is the rotation at the support Where a cantilever is con-nected to a substantially rigid structure some root rotationwill still occur and the effective length should be taken as thelength to the face of the support plus half the effective depth

262 CRACKING

2621 Buildings and bridges

Cracking of members under service load should not impair theappearance or durability of the structure In BS 8110 forbuildings the design surface crack width is generally limited to03 mm For members such as beams and slabs in which thenominal cover does not exceed 50 mm the crack widthrequirement may be met by limiting the gaps between tensionbars to specified values In circumstances where calculationis considered necessary crack width formulae are providedDetails of the bar spacing rules (see section 261 for com-ment on fs) and the crack width formulae are given inTable 343

In BS 5400 the design crack width limits apply only forload combination 1 where for highway bridges the live loadis generally taken as HA Crack widths are calculated for asurface taken at a distance from the outermost reinforcementequal to the nominal cover required for durability Thedesign crack width limits vary according to the exposureconditions as follows 025 mm (moderate or severe expo-sure) 015 mm (very severe exposure) and 010 mm(extreme exposure) In many cases these requirements arecritical and details of the crack width formulae are given inTable 343

In BS 8110 for cracking due to the effects of applied loadsthe modulus of elasticity of concrete is taken as Ec2 wherevalues of Ec are given in Table 35 In BS 5400 a value in therange Ec to Ec2 is taken according to the proportion of live topermanent load

Cracking due to restrained early thermal effects is consideredin BS 8110 Part 2 and Highways Agency BD2887 Inthese documents the restrained early thermal contraction isgiven by t 08RT where is a coefficient of expansion forthe mature concrete R is a restraint factor and T is a tempera-ture differential or fall The following values of R are given

Type of pour and restraint R

Base cast onto blinding 01ndash02Slab cast onto formwork 02ndash04Wall cast onto base slab 06ndash08Infill bays 08ndash10

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BS 8110 Deflection ndash 1 340

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BS 8110 Deflection ndash 2 341

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BS 8110 Deflection ndash 3 342

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BS 8110 (and BS 5400) Cracking 343

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Deflection and cracking300

For continuous support conditions and a flanged section withbwb 3001420 021 03 Table 340 gives

Basic spaneffective depth ratio 08 26 208

The estimated service stress in the reinforcement for a designwith no redistribution of the ultimate moment is given by

fs (58)fy (As reqAs prov) (58) 500 15341608 298 Nmm2

Modification factor for tension reinforcement

s 055 (477 fs)120(09 Mbd 2) 20 055 (477 298)[120 (09 099)] 134

Ignoring modification factor for compression reinforcement

Limiting spaneffective depth ratio 208 134 279

Actual spaneffective depth ratio 8000450 178

Allowing for H8 links with 25 mm cover the clear distancebetween the H32 bars in the bottom of the beam is given by

ab 300 2 (25 8 32) 170 mm

From Table 343 the limiting distance is given by

ab 47 000fs 47 000298 158 mm (170 mm)

The clear distance between the H32 bars could be reduced to150 mm by increasing the side cover to the links to 35 mmAlternatively the bars could be changed to 2H25 and 2H20

Example 2 A 280 mm thick flat slab supported by columnsarranged on a 72 m square grid is to be designed to the require-ments of BS 8110 Bending moments are to be determined usingthe simplified method where the total design ultimate load on apanel is 954 kN and the slab is to be checked for deflection

fcu 40 Nmm2 fy 500 Nmm2 cover to bars 25 mm

Allowing for the use of H12 bars in each direction and basedon the bars in the second layer of reinforcement

d 280 (25 12 6) 235 mm ld 7200235 306

In the following calculations s is based on the average valueof Mbd 2 determined for a full panel width From Table 262the design ultimate bending moment for an end span with acontinuous connection at the outer support is given by

M 0075Fl 0075 954 72 515 kNmMbd2 515 106(7200 2352) 130 Nmm2

From Table 340 for a continuous flat slab without drops thebasic spaneffective depth ratio 26 09 234

Modification factor for tension reinforcement if fs (58)fy

s 055 (477 fs)120(09 Mbd2) 20 055 (477 312)[120 (09 13)] 117

Assuming that no compression reinforcement is provided

Limiting spaneffective depth ratio 234 117 274

The limiting value of ld can be raised to 306 by increasing s

to 131 which reduces fs to 276 Nmm2 so that the area oftension reinforcement determined for the ULS should bemultiplied by the factor 312276 113

For an interior span where the bending moment coefficient is0063 compared to 0075 for the end span

Mbd 2 130 00630075 109 Nmm2

s 055 (477 312)[120 (09 109)] 124 (131)

2622 Liquid-retaining structures

In BS 8007 for structures where the retention or exclusion ofliquid is a prime consideration a design surface crack widthlimit of 02 mm generally applies In cases where the surfaceappearance is considered to be aesthetically critical the limit istaken as 01 mm Under liquid pressure cracks that extendthrough the entire thickness of a slab or a wall are expected toresult in some initial seepage but it is assumed that such crackswill heal autogeneously within 21 days for a 02 mm designcrack width and 7 days for a 01 mm design crack widthSeparate calculations using basically different crack width for-mulae are used for the effects of applied loads and the effectsof temperature and moisture change Details of the formulae foreach type of cracking are given in Table 344

In order to control any potential cracking due to the effectsof restrained thermal contraction and shrinkage three designoptions are given in which the reinforcement requirements arerelated to the incidence of any movement joints Details of thedesign options are given in Table 345 where the joint spacingrequirements refer to joints as being either complete or partialContraction joints can be formed by casting against stop endsor by incorporating crack-inducing waterstops The joints arecomplete if all of the reinforcement is discontinued at thejoints and partial if only 50 of the reinforcement is discon-tinued The joints need to incorporate waterstops and surfacesealants to ensure a liquid-tight structure

The reinforcement needed to control cracking in continuousor semi-continuous construction depends on the magnitude ofthe restrained contraction The restraint factor R (ie ratio ofrestrained to free contraction) may be taken as 05 generally butmore specific values for some common situations are also givenin Table 345 For particular sections and arrangements of rein-forcement limiting values for restrained contraction strain aregiven in Table 346

For cracking due to applied loading and concrete classes thatare typically either C2835 or C3240 the effective modularratio e 2EsEc may be taken as 15 In this case for singlyreinforced rectangular sections the elastic properties of thetransformed section in flexure are given in Table 347 For par-ticular sections and arrangements of reinforcement limitingvalues are given for service moments in Tables 348ndash350 anddirect tensile forces in Tables 351 and 352

Example 1 The beam shown in the following figure is to bechecked for deflection and cracking to the requirements ofBS 8110 The designs for bending and shear are contained inexample 1 of Chapters 24 and 25 respectively The tensionreinforcement comprising 3H32 (top at B) and 2H32 (bottom)is based on the following values

fcu 40 Nmm2 fy 500 Nmm2 cover to links 25 mm

For the bottom reinforcement with d 450 mm

b 1420 mm Mbd 2 099 Nmm2

As req 1534 mm2 As prov 1608 mm2

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BS 8007 Cracking 344

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BS 8007 Design options and restraint factors 345

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BS 8007 Design table for cracking due to temperature effects 346

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BS 8007 Elastic properties of cracked rectangularsections in flexure 347

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BS 8007 Design table for cracking due to flexure in slabs ndash 1 348

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BS 8007 Design table for cracking due to flexure in slabs ndash 2 349

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BS 8007 Design table for cracking due to flexure in slabs ndash 3 350

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351BS 8007 Design table for cracking due to directtension in walls ndash 1

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352BS 8007 Design table for cracking due to directtension in walls ndash 2

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The maximum design crack width is given by

wcr

3 735 000121[1 2 (735 45)(700 192)] 024 mm

From Table 32 the limiting design crack width for a bridgedeck soffit (severe exposure) is 025 mm

The service stress range in the reinforcement resulting from thelive load moment Mq is given by

(MqMs)fs (275600) 220 101 Nmm2

From Table 32 the limiting value for spans in the range5ndash200 m and bar sizes 16 mm is 120 Nmm2

Example 4 The wall of a cylindrical tank 75 m deep and15 m diameter is 300 mm thick The wall which is continuouswith the base slab is to be designed for temperature effects andthose due to internal hydrostatic pressure when the tank is fullof liquid

Design crack width 02 mm fcu 40 Nmm2

Cover to horizontal bars 52 mm fy 500 Nmm2

Effects of temperature change Allowing for concrete gradeC3240 with 350 kgm3 Portland cement at a placing temperatureof 20oC and a mean ambient temperature during construction of15oC the temperature rise for concrete placed within 18 mmplywood formwork

T1 25oC (Table 218)

As the wall is to be designed to resist hoop tension there willbe no vertical movement joints and allowance must be made fora fall in temperature due to seasonal variations Allowing forT2 15oC restraint factor R taken as 05 and coefficient ofthermal expansion taken as 12 106 per oC (Table 35)restrained total thermal contraction after the peak temperaturearising from hydration effects is given by

R (T1 T2) 05 12 106 (25 15) 240 106

From Table 346 for 02 mm crack width and a surface zonethickness of 3002 150 mm H16-200 (EF) will suffice

Effects of hydrostatic load Suppose that an elastic analysis ofthe tank assuming a floor 300 mm thick indicates a servicemaximum circumferential tension of 400 kNm This valueoccurs at a depth of 6 m and above this level the hoop ten-sions can be assumed to reduce approximately linearly tonear zero at the top of the wall

In BS 8110 for direct tension and fy 500 Nmm2 minimumreinforcement equal to 045 of the concrete cross section isrequired Hence for a wall 300 mm thick the minimum areaof reinforcement required on each face

As min 00045 1000 150 675 mm2m (H16-300)

From Table 351 for a 02 mm crack width and 40 mm coverthe following values are obtained for a 300 mm thick wall

H16-225 (EF) provides for N 408 kNmH16-300 (EF) provides for N 319 kNm

In order to cater for the effects of both temperature change andhydrostatic load H16-225 (EF) can be used for a height of

3acrm

1 2(acr cmin) (h dc)

Deflection and cracking310

In this case increasing s to 131 reduces fs to 295 Nmm2 sothat the area of tension reinforcement should be multiplied bythe factor 312295 106

Example 3 A simply supported 700 mm thick solid slabbridge deck is to be designed for bending and checked forcracking and fatigue to the requirements of BS 5400

fcu 40 Nmm2 fy 500 Nmm2 d 620 mm

The maximum longitudinal ultimate moment at mid-span forload combination 1 (375 units HB live load) is 950 kNmm

Mbd 2fcu 950 106(1000 6202 40) 0062As fybdfcu 0078 (Table 324)As 0078 1000 620 40500 3869 mm2m

Although H25-125 gives 3927 mm2m this is unlikely to besufficient with regard to cracking and fatigue which will bechecked on the basis of H25-100 (4909 mm2m)

The maximum longitudinal service moment at mid-span forload combination 1 (HA live load) is Ms 600 kNmm with

Mg 325 kNmm Mq 275 kNmm MqMg 085

In the following calculations the modulus of elasticity of theconcrete is taken as a value reflecting the relative proportions oflive load and permanent load Taking Ec for live load and Ec2for permanent load the effective value is given by

Eeff [1 05(1 MqMg)]Ec

From section 2211 Ec 19 03fcu 31 kNmm2

Eeff [1 05(1 085)] 31 226 kNmm2

From Table 342 the neutral axis depth x (or dc) is given by

xd where (EsEeff)(Asbd)

(200226) 4909(1000 620) 007

xd 007 031 x 192 mmStress in tension reinforcement is given by

fs MsAs(d x3) 600 106[4909 (620 1923)] 220 Nmm2

Strain in tension reinforcement

s fsEs 220(200 103) 00011

Strain at surface taken at distance cmin from outermost barsignoring stiffening effect of concrete

1 s (h x 10)(d x) 00011 (700 192 10)(620 192) 000128

Stiffening effect of concrete (with h)

109

38 1000 700 015 109(00011 4909) 000007

Strain at surface allowing for stiffening effect of concrete

m 1 2 000128 000007 000121

Distance from surface of bar where sb is bar spacing to pointmidway between bars on surface taken at distance cmin fromoutermost bars is given by

acr

125 735 mm(50)2 (70)2

(sb 2)2 (h d 10)2 2

2 38bthsAs

1 Mq

Mg

a

0072 2 007

2 2

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Cracking 311

30 m say and minimum reinforcement of H16-300 (EF) for theremaining height of 45 m

Example 5 A 200 mm thick roof slab to a reservoir is to bedesigned for serviceability cracking to the requirements ofBS 8007

Design crack width 02 mm Cover to bars 40 mm

(a) Sliding layer is provided between slab and perimeter wallMaximum service moment M 25 kNmm

From Table 348 H12-150 caters for M 252 kNmm withfs 243 Nmm2

(b) Slab is tied to perimeter wall Maximum service momentand direct tension M 25 kNmm and N 40 kNm

h 200 mm d 200 (40 122) 154 mm

Since MN (2540) 103 625 mm (d 05h) 54 mmone face will remain in compression and the section can bedesigned for a reduced moment M1 M N(d 05h) withthe tensile force acting at the level of the reinforcement

M1 25 40 0054 228 kNmm

From solution (a) above take fs 240 Nmm2 as a trialvalue

100M1fsbd2 100 228 106(240 1000 1542) 0400

From Table 347 by interpolation 100As1bd 0445

As1 000445 1000 154 685 mm2m

Total area of reinforcement required is given by

As As1 Nfs 685 40 103240 852 mm2m

Using H12-125 gives 905 mm2m This may not necessarily besufficient because the stiffening effect of the concrete inherentin solution (a) is reduced by the additional tension A crackwidth calculation is needed to confirm the solution

The stress in the reinforcement is given approximately by

fs 240 852905 226 Nmm2

100M1fsbd2 0400 240226 0425

From Table 347 by interpolation 100As1bd 0474

As1 000474 1000 154 730 mm2m

As As1 Nfs 730 40 103226 907 mm2m

This is near enough to the area given by H-125 no furtheriteration is needed and 100M1fsbd2 0425 may be assumed

From Table 347 by interpolation xd 0312 x 48 mm

Strain in the reinforcement

s fsEs 226(200 103) 000113

Strain at surface ignoring stiffening effect of concrete

1 s(h x)(d x) 000113 (200 48)(154 48) 000162

Stiffening effect of concrete at surface (with h)

2 bt(h x)23EsAs(d x)1000 (20048)2[3 200 103 905 (154 48)] 000040

Strain at surface allowing for stiffening effect of concrete

m 1 2 000162 000040 000122

Distance from surface of bar where sb is bar spacing topoint on surface midway between bars is given by

acr

6 716 mm

The maximum design crack width is given by

wcr

3 716 000122[1 2 (716 40)(200 48)] 019 mm

3acrm

1 2(acr cmin) (h x)

(625)2 (46)2

(sb 2)2 (h d)2 2

a

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Chapter 27

Considerations affecting design details

Codes of Practice contain numerous requirements that affectthe reinforcing details such as minimum and maximum areastying provisions anchorage and curtailment

Bars may be arranged individually in pairs or in bundles ofthree or four in contact In BS 8110 for the safe transmissionof bond forces the cover provided to the bars should be not lessthan the bar size or for a group of bars in contact the equivalentdiameter of a notional bar with the same total cross-sectionalarea as the group In BS 5400 the forgoing cover requirementis increased by 5 mm Requirements for cover with regard todurability and fire-resistance are given in Chapter 23 Gapsbetween bars (or groups of bars) generally should be not lessthan the greater of (hagg 5 mm) where hagg is the maximumsize of the coarse aggregate or the bar size (or the equivalentbar size for bars in groups) Details of reinforcement limits aregiven in Table 353 for BS 8110 and Table 359 for BS 5400

271 TIES IN STRUCTURES

For robustness the necessary interaction between elements isobtained by tying the structure together Where the structure isdivided into structurally independent sections each sectionshould have an appropriate tying system In the design of tiesthe reinforcement may be assumed to act at its characteristicstrength and only the specified tying forces need to be takeninto account Reinforcement provided for other purposes maybe considered to form part of or the whole of the ties Detailsof the tying requirements in BS 8110 are given in Table 354

272 ANCHORAGE AND LAP LENGTHS

At both sides of any cross section bars should be provided withan appropriate embedment length or other form of end anchorageIn BS 5400 it is also necessary to consider lsquolocal bondrsquo wherelarge changes of tensile force occur over short lengths ofreinforcement Critical sections for local bond are at simplysupported ends at points where tension bars stop and at pointsof contra-flexure However the last two points need not beconsidered if the anchorage bond stresses in the continuing barsdo not exceed 08 times the ultimate values

The radius of any bend in a reinforcing bar should conformto the minimum requirements of BS 8666 and should ensurethat the bearing stress at the mid-point of the curve does notexceed the maximum value given in BS 8110 or BS 5400 as

appropriate A link may be considered fully anchored if itpasses round another bar not less than its own size through anangle of 90o and continues beyond the end of the bend for aminimum length of eight diameters Details of anchoragelengths local bond stresses and bends in bars are given inTables 355 and 359 for BS 8110 and BS 5400 respectively

In BS 8007 for horizontal bars in direct tension the designultimate anchorage bond stress is taken as 70 of the valuegiven in BS 8110 Also for sections where bars are needed tocontrol cracking due to temperature and moisture effects therequired anchorage bond length lab 6 where crit isthe reinforcement ratio in the surface zone (Table 344) Thisvalue can exceed the anchorage bond length in BS 8110

Laps should be located if possible away from positions ofmaximum moment and should preferably be staggered Laps infabric should be layered or nested to keep the lapped wires orbars in one plane BS 8110 requires that at laps the sum of allthe reinforcement sizes in a particular layer should not exceed40 of the breadth of the section at that level When the sizeof both bars at a lap exceeds 20 mm and the cover is lessthan 15 times the size of the smaller bar links of size notless than one-quarter the size of the smaller bar and spacingnot greater then 200 mm should be provided throughout the laplength Details of lap lengths are given in Tables 355 and 359for BS 8110 and BS 5400 respectively

273 CURTAILMENT OF REINFORCEMENT

In flexural members it is generally advisable to stagger thecurtailment points of the tension reinforcement as allowedby the bending moment envelope Curtailed bars shouldextend beyond the points where in theory they are no longerneeded in accordance with certain conditions Details of thegeneral curtailment requirements in BS 8110 are given inTable 356 Simplified rules for beams and slabs are alsoshown in Tables 357 and 358 In BS 5400 the generalcurtailment procedure is the same as that in BS 8110 except forthe requirement at a simply supported end where condition (3)does not apply

Example 1 The design of the beam shown in the following fig-ure is given in example 1 of Chapters 24 (bending) and 25 (shear)The design ultimate loads on each span are Fmax 416 kN andFmin 160 kN The main reinforcement is as follows in the

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BS 8110 Reinforcement limits 353

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BS 8110 Provision of ties 354Internal ties

Peripheral tiesVertical ties

Column to wallties at each column(or wall)floorintersection

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BS 8110 Anchorage requirements 355

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spans 2H32 (bottom) at support B 3H32 (top) and 2H25(bottom) The width of each support is 400 mm and bars are tobe curtailed according to the requirements of BS 8110 In thefollowing calculations the use of the general curtailmentprocedure (Table 356) and the simplified curtailment rules(Table 357) are both examined

fcu 40 Nmm2 fy 500 Nmm2 d 450 mm

2 A point where the shear force is no more than half thedesign shear resistance at the section From the shear designcalculations in Chapter 25 with H8-300 links Vu 170 kNFor load case 1 distance x from B to the point where shearforce is 1702 85 kN is given by

x [1 (85 156)416]L 042 8 336 m

3 A point where the bending moment is no more than halfthe design resistance moment at the section As in thecalculations above but with M 2702 135 kNm05x2 3x 26 0 giving x 68 m and a distancefrom B of (80 68) 12 m

From the foregoing it can be seen that if the bar were curtailedat a distance of 12 m from B this would satisfy condition (3)and be more than 450 mm beyond the theoretical curtailmentpoint at 06 m from B However the bar should extend 13 mfrom B in order to provide a full tension anchorage of 11 mbeyond the face of the support It can be seen from the forego-ing calculations that checking conditions (2) and (3) is a tediousprocess and complying with condition (1) is a more practicalapproach even though it would mean curtailing the bar at(06 11) 17 m from B in this example

Suppose that the remaining 2H32 are continued to the point ofcontra-flexure in span BC for load case 2

The reaction at support C is given by

RC 05Fmin MBL 05 160 2888 44 kN

Distance from B to point of contra-flexure is given by

x L(1 2RCFmin) 8 (1 2 44160) 36 m

The bars need to extend beyond this point for a distance not lessthan d 12 450 mm Link support bars say 2H12 with alap of 300 mm could be used for the rest of the span

If the simplified curtailment rules are applied one bar out ofthree may be curtailed at 015L 45 145 m from the faceof the support that is at 165 m from B The other two bars maybe curtailed at 025L 20 m from the face of the support thatis at 22 m from B Beyond this point bars giving an area notless than 20 of the area required at B should be provided thatis 02 2413 483 mm2 (2H20 gives 628 mm2) Since thebars are in the top of a section as cast where the cover is lessthan 2 and the gap between adjacent laps is not less than6 l 14 (see Table 355) and the required lap length is

lbl 49 (AsreqAsprov) 49 20 483628 750 mm

Example 2 A typical floor to an 8-storey building consistsof a 280 mm thick flat slab supported by columns arrangedon a 72 m square grid The slab for which the characteristicloading is 80 kNm2 dead and 45 kNm2 imposed is to beprovided with ties to the requirements of BS 8110 Thedesign ultimate load on a panel is 954 kN and bendingmoments are to be determined by the simplified method (seesection 138)

fcu 40 Nmm2 fy 500 Nmm2 cover to bars 25 mm

Allowing for the use of H12 bars in each direction and basedon the bars in the second layer of reinforcement

d 280 (25 12 6) 235 mm say

Considerations affecting design details316

End anchorage At the bottom of each span the 2H32 will becontinued to the supports At the end support (simple) aneffective anchorage of 12 is required beyond the centre ofbearing This can be obtained by providing a 90o bend with aninternal radius of 35 provided the bend does not start beforethe centreline of the support Allowing for 50 mm end coverto the bars the distance from the outside face of the support tothe start of the bend is 50 45 194 mm This would besatisfactory since the width of the support is 400 mm If thesupport width were any less the 2H32 could be stopped atthe face of the support and lapped with 2H25 in which case itwould be necessary to reassess the shear design

Curtailment points for top bars The resistance momentprovided by 2H32 can be determined as follows

As fybdfcu 1608 500(300 450 40) 0149Mbd 2fcu 0111 (Table 314)M 0111 300 4502 40 106 270 kNm

For load case 1 reaction at A (or C) is given by

RA 05Fmax MBL 05 416 4168 156 kN

Distance x from A to point where M 270 kNm is given by

05(FmaxL) x2 RAx 05 (4168)x2 156x 270

Hence 05x2 3x 52 0 giving x 74 m Thus of the3H32 required at B one bar is no longer needed for flexure ata distance of (80 74) 06 m from B

The bar to be curtailed needs to extend beyond this point fora distance not less than d 12 450 mm to a positionwhere one of the following conditions is satisfied

1 A full tension anchorage length beyond the theoretical pointof curtailment that is 35 35 32 11 m

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From Table 255 the design ultimate sagging moment for aninterior panel is given by

M 0063Fl 0063 954 72 433 kNm

The total panel moment is to be apportioned between columnand middle strips where the width of each strip is 36 m Forthe column strip with 55 of the panel moment

M 055 433 238 kNmMbd2fcu 238 106(3600 2352 40) 0030

From Table 314 since Mbd 2fcu 0043 zd 095 Hence

As 238 106(087 500 095 235) 2504 mm2 (24H12-150 gives 2714 mm2)

For the middle strip with 45 of the panel moment

M 045 433 195 kNmMbd2fcu 195 106(3600 2352 40) 0025 ( 0043)

As 195 106(087 500 095 230) 2008 mm2 (18H12-200 gives 2036 mm2)

For the peripheral tie the tensile force is given by

Ft (20 4no) 60 kN (20 4 8) 52 kN

The required area of reinforcement acting at its characteristicstrength is given by

As Ftfy 52 103500 104 mm2 (1H12)

For the internal ties the tensile force is given by

Ft gt Ft kNm

52 = 125 kNm

If the internal ties are spread evenly in the slab the requiredarea of reinforcement acting at its characteristic strength

As 125 103500 250 mm2m (H12-400)

In this case alternate bars in both column and middle stripsneed to be made effectively continuous

If the internal ties are concentrated at the column lines the totalarea of reinforcement required in each group

As 250 72 1800 mm2 (16H12 gives 1810 mm2)

In this case the bars in the middle two-thirds of each columnstrip need to be made effectively continuous Since the bars arelocated at the bottom of the slab and the gap between each set

80 4575 72

5

Ftint gt qk

75 lr

5

Curtailment of reinforcement 317

of lapped bars exceeds 6 l 10 and a lap length of 35 issufficient (Table 355 for fcu 40 Nmm2)

Example 3 The following figure shows details of the rein-forcement at the junction between a 300 mm wide beam and a300 mm square column Bars 03 need to develop the maximumdesign stress at the column face and the required radius of bendis to be checked in accordance with the requirements of BS 8110

fcu 40 Nmm2 fy 500 Nmm2

The minimum radius of bend of the bars depends on the value ofab where ab is taken as either centre-to-centre distance betweenbars or (side cover plus bar size) whichever is less Henceab (300 75 2 25) 125 (75 25) 100 mm

From Table 355 for ab 10025 4 rmin 64 Thisvalue can be reduced slightly by considering the stress in the barat the start of the bend If r 6 distance from face of columnto start of bend 300 50 7 25 75 mm (ie 3) FromTable 355 the required anchorage length is 35 and rmin (1 335 ) 64 59 Thus r 6 is sufficient

Example 4 A 700 mm thick solid slab bridge deck is simplysupported at the end abutments The design of the slab to therequirements of BS 5400 is given in example 2 of Chapter 25(shear) and example 3 of Chapter 26 (bending cracking andfatigue) The tension reinforcement at the end of the span isH25-200 and the maximum design shear force is 625 kN act-ing on a 15 m wide strip of slab A local bond check is required

fcu 40 Nmm2 fy 500 Nmm2 d 620 mm

Bar perimeter provided by H25-200 in a 15 m strip of slab

13us (1500200) 25 589 mm

The local bond stress is given by the relationship

fbs V13usd 625 103(589 620) 171 Nmm2

From Table 359 ultimate local bond stress flbu 40 Nmm2

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BS 8110 Curtailment requirements 356

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BS 8110 Simplified curtailment rules for beams 357

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BS 8110 Simplified curtailment rules for slabs 358

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BS 5400 Considerations affecting design details 359

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Chapter 28

Miscellaneousmembers and details

281 LOAD-BEARING WALLS

In BS 8110 for the purpose of design a wall is defined as avertical load-bearing member whose length on plan exceedsfour times its thickness Otherwise the member is treated as acolumn A reinforced wall is one in which not less than therecommended minimum amount of vertical reinforcement isprovided and taken into account in the design Otherwise themember is treated as a plain concrete wall in which case thereinforcement is ignored for the purpose of design Limitingreinforcement requirements are given in Table 353 Bearingstresses under concentrated loads should not exceed 06fcu forconcrete strength classes C2025 Design requirements forreinforced and plain concrete walls are given in Table 360

282 PAD BASES

Notes on the distribution of pressure under pad foundations aregiven in section 181 and values for the structural design ofseparate bases are given in Table 282 Critical sections forbending are taken at the face a concrete column or the centreof a steel stanchion The design moment is taken as that due toall external loads and reactions to one side of the section

Generally tension reinforcement may be spread uniformlyacross the width of the base but the following requirementshould be satisfied where c is the column width and lc is thedistance from the centre of a column to the edge of the pad Iflc 075(c 3d) two-thirds of the reinforcement should beconcentrated within a zone that extends on either side for adistance no more than 15d from the face of the column Forbases with more than one column in the direction consideredlc should be taken as either half the column spacing or thedistance to the edge of the pad whichever is the greater

The pad should be examined for normal shear and punchingshear Normal shear is checked on vertical sections extendingacross the full width of the base Within any distance av 2dfrom the face of the column the shear strength may be taken as(2dav)vc For a concentric load the critical position occurs atav a2 2d where a is the distance from the column face tothe edge of the base For an eccentric load checks can be madeat av 05d d and so on to find the critical position

Punching shear is checked on a perimeter at a distance 15dfrom the face of the column The shear force at this position is that

due to the effective ground pressure acting on the area outsidethe perimeter For a concentric load with a 15d the checkfor punching shear is the critical shear condition If the mainreinforcement is taken into account in the determination of vcthe bars should extend a distance d beyond the shear perimeterIn this case the bars need to extend 25d beyond the face of thecolumn and will need to be bobbed at the end unless a 25din which case straight bars would suffice

The maximum clear spacing between the bars ab shouldsatisfy the following requirements for fy 500 Nmm2

100Asbd 03 03 04 05 06 075 10

ab (mm) 750 500 375 300 250 200 150

Example 1 A base is required to support a 600 mm squarecolumn subjected to vertical load only for which the valuesare 4600 kN (service) and 6800 kN (ultimate) The allowableground bearing value is 300 kNm2

fcu 35 Nmm2 fy 500 Nmm2 nominal cover 50 mm

Allowing 10 kNm2 for ground floor loading and extra over soildisplaced by concrete the net allowable bearing pressure canbe taken as 290 kNm2 Area of base required

Abase 4600290 1586 m2 Provide base 40 m square

Distance from face of column to edge of base a 1700 mm

Take depth of base 05a say h 850 mm

Allowing for 25 mm main bars average effective depth

d 850 ndash (50 25) 775 mm

Bearing pressure under base due to ultimate load on column

pu 680042 425 kNm2

Bending moment on base at face of column

M pu la22 425 4 1722 2456 kNm

K Mbd2 fcu 2456 106(4000 7752 35) 00292

From Table 314 since K 0043 As fybd fcu 121K and

As 121 00292 4000 775 35500 7670 mm2

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BS 8110 Load-bearing walls 360

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From Table 220

16H25-250 gives 7854 mm2 and

100Asbd 100 7854(4000 775) 025

Critical perimeter for punching shear occurs at 15d from faceof column where the length of side of the perimeter

l1 c 3d 600 3 775 2925 mm

Hence

V f (l 2 ndash l12) 425 (42 ndash 29252) 3164 kN

v V4l1d 3164 103(4 2925 775) 035 Nmm2

vc 0216

0216 (400775)14(025 35)13 038 Nmm2( v)

Critical position for shear on vertical section across full widthof base occurs at distance av a2 850 mm 2d from faceof column where

V 425 4 172 1445 kN

v Vbd 1445 103(4000 775) 047 Nmm2

vc(2dav) 038 (2 775850) 069 Nmm2 ( v)

283 PILE-CAPS

In BS 8110 (and BS 5400) a pile-cap may be designed by eitherbending theory or truss analogy In the latter case the truss isof a triangulated form with nodes at the centre of the loadedarea and at the intersections of the centrelines of the piles withthe tension reinforcement as shown for compact groups of twoto five piles in Table 361 Expressions for the tensile forcesare given taking into account the dimensions of the columnand also simplified expressions when the column dimensionsare ignored Bars to resist the tensile forces are to be locatedwithin zones extending not more than 15 times the pilediameter either side of the centre of the pile The bars are tobe provided with a tension anchorage beyond the centres of thepiles and the bearing stress on the concrete inside the bend inthe bars should be checked (see Table 355)

The design shear stress calculated at the perimeter of thecolumn should not exceed the lesser of 08radicfcu or 5 Nmm2Critical perimeters for checking shear resistance are shown inTable 361 where the whole of the shear force from piles withcentres lying outside the perimeter should be taken intoaccount The shear resistance is normally governed by shearalong a vertical section extending across the full width of thecap where the design concrete shear stress may be enhanced asshown in Table 361 If the pile spacing exceeds 3 times the pilediameter punching shear should also be considered

In BS 5400 shear enhancement applies to zones of widthequal to the pile diameter only and the concrete shear stress istaken as the average for the whole section Also the check forpunching shear is made at a distance of 15d from the face ofthe column with no enhancement of the shear strength

The following values are recommended for the thickness ofpile-caps where hp is the pile diameter

For hp 550 mm h (2hp 100) mm

For hp 550 mm h 8(hp ndash 100)3 mm

400d 14100As fcu

bd 13

Example 2 A pile-cap is required for a group of 4 450 mmdiameter piles arranged at 1350 mm centres on a square gridThe pile-cap supports a 450 mm square column subjected to anultimate design load of 4000 kN

fcu 35 Nmm2 fy 500 Nmm2

Allowing for an overhang of 150 mm beyond the face of the pilesize of pile-cap 1350 450 300 2100 mm square

Take depth of pile-cap as (2hp 100) 1000 mm

Assuming tension reinforcement to be 100 mm up from base ofpile-cap d 1000 ndash 100 900 mm

Using truss analogy with the apex of the truss at the centre ofthe column the tensile force between adjacent piles is

Ft 750 kN in each zone

As Ft087fy 750 103(087 500) 1724 mm2

Providing 4H25 gives 1963 mm2 and since the pile spacing isnot more than 3 times the pile diameter bars may be spreaduniformly across the pile-cap giving a total of 8H25-275 in eachdirection so that

100As bh 100 3926(2100 1000) 019 ( 013 min)

Critical position for shear on vertical section across full widthof pile-cap occurs at distance from face of column given by

av 05(l ndash c) ndash 03hp

05 (1350 ndash 450) ndash 03 450 315 mm

Portion of column load carried by two piles is 2000 kN thus

v Vbd 2000 103(2100 900) 106 Nmm2

vc 0216

0216 (400900)14 (019 35)13 033 Nmm2

vc (2dav) 033 (2 900315) 188 Nmm2 ( v)

Shear stress calculated at perimeter of column is

v Vud 4000 103(4 450 900) 247 Nmm2

Maximum shear strength 08radicfcu 473 Nmm2 ( v)

Taking ab as either centre-to-centre distance between bars or(side cover plus bar size) whichever is less

ab 275 (75 25) 100 mm ab 10025 4

From Table 355 minimum radius of bend rmin 7 say

284 RETAINING WALLS ON SPREAD BASES

General notes on the design of walls to BS 8002 are given insection 732 Design values of earth pressure coefficients arebased on a design soil strength which is taken as the lower ofeither the peak soil strength reduced by a mobilisation factor orthe critical state strength Design values of the soil strengthusing effective stress parameters are given by

design tan (tan max)12 tan crit

design c c12 tan crit

400d 14100As fcu

bd 13

4000 13508 900 Nl

8d

Miscellaneous members and details324

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BS 8110 Pile-caps 361

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Miscellaneous members and details326

where c max and crit are representative (ie conservative)values of effective cohesion peak effective angle of shearingresistance and critical state angle of shearing resistance for thesoil In the absence of reliable site investigation and soil testdata values may be derived from Table 210

Design values of friction and adhesion at the soil-structureinterface (wall or base) are given by

design tan (or b) 075 design tan

design cw (or cb) 075 cud 05 cu in which

cud cu 15 where cu is undrained shear strength

A minimum surcharge of 10 kNm2 applied to the surface of theretained soil and a minimum depth of unplanned earth removalin front of the wall equal to 10 of the wall height but not lessthan 05 m should be considered Wall friction should beignored in the determination of KA

Suitable dimensions for the base to a cantilever wall can beestimated with the aid of the chart given in Table 286 Forsliding the chart is valid for non-cohesive soils only Thusfor bases founded on clay soils the long-term condition can beinvestigated by using crit with c 0 For the short-termcondition the ratio does not enter into the calculations forsliding and taking the contact surface length as the full widthof the base is given by KAl2cb When has beendetermined from this equation the curve for radicKA on the chartcan be used to check the values of and that were obtainedfor the long-term condition

Example 3 A cantilever retaining wall on a spread base isrequired to support level ground and a footpath adjacent to aroad The existing ground may be excavated as necessary toconstruct the wall and the excavated ground behind the wall isto be reinstated by backfilling with a granular material Agraded drainage material will be provided behind the wall withan adequate drainage system at the bottom

Height of fill to be retained 40 m above top of baseSurcharge 100 mm surfacing plus 5 kNm2 live load

(design for minimum value of 10 kNm2)

Properties of retained soil (well graded sand and gravel)unit weight of soil 20 kNm3

max crit 35o design tanminus1 [(tan 35o)12] 30o

KA (1 ndash sin)(1 sin) 033

Properties of sub-base soil (medium sand)allowable bearing value pmax 200 kNm2

max 35o crit 32o design 30o (as fill)

design tan b 075 design tan 043

Take thickness of both wall (at bottom of stem) and base to beequal to (height of fill)10 400010 400 mm

Height of wall to underside of base l 40 04 44 m

Allowing for surcharge equivalent height of wall

le l q 44 1020 49 m

pmaxle 200(20 49) asymp 20

tan bradicKA 043radic033 075

From Table 286 radicKA 08 018 Hence

Width of base le (08radic033) 49 225 m

Toe projection (le) 018 225 04 m

Example 4 The sub-base for the wall described in example 3is a clay soil with properties as given here All other values areas specified in example 3

Properties of sub-base soil (firm clay)allowable bearing value pmax 100 kNm2 cu 50 kNm2

cud 5015 333 kNm2 design cb 502 25 kNm2

crit 25o (assumed plasticity index 30)

design tan b 075 design tan crit 033

For the long-term condition

pmaxle 100(20 49) asymp10

tan bradicKA 033radic033 057

From Table 286 radicKA 115 024 Hence

Width of base le (115radic033) 49 33 m say

Toe projection (le) 024 33 08 m

For the short-term condition

KA l2cb 033 20 49(2 25) 065

radicKA 065radic033 113 (115)

Since this value is less than that calculated for the long-termcondition the base dimensions are satisfactory

Example 5 The wall obtained in example 4 a cross sectionthrough which is shown here is to be designed to BS 8002

The vertical loads and bending moments about the front edgeof the base are

Load (kN) Moment (kNm)Surcharge 10 21 210 225 473Backfill 20 21 40 1680 225 3780Wall stem 24 04 40 384 10 384Wall base 24 04 33 317 165 523

Totals Fv 2591 5160

The horizontal loads and bending moments about the bottom ofthe base are

Load (kN) Moment (kNm)Surcharge 033 10 44 145 442 319Backfill 033 20 4422 639 443 937

Totals Fh 784 1256

Resultant moment Mnet 516 ndash 1256 3904 kNm

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Since the f values used for the horizontal and vertical loads arenot the same the bearing pressures must be recalculated

Fv 14 2591 3627 kN Mv 14 516 7224 kNm

Mnet 7224 ndash 12 1256 5717 kNm

a 57173627 1575 m e 165 ndash 1575 0075 m

pmax (362733)(1 6 007533) 125 kNm2

pmin (362733)(1 ndash 6 007533) 95 kNm2

Bearing pressure under base at inside face of wall

pwall 95 (125 ndash 95)(2133) 95 19 114 kNm2

Bending moment on base at inside face of wall

M 14 (10 20 4 24 04) 2122(95 2122 19 2126) 84 kNm

As 84 106[(087 500)(095 352)] 578 mm2m

Use H16-250 to fit in with vertical bars in wall

Shear force on base at inside face of wall

V 14 (10 20 4 24 04) 21(95 21 19 212) 734 kN

Vbd 734 103(1000 352) 021 Nmm2 ( vc)

For the base the bending moment and shear force have beencalculated for a bearing pressure diagram that varies linearly asindicated in BS 8110 If the pressure diagram assumed for theultimate bearing condition in example 5 is taken

pu Fv2a 3627(2 1575) 1151 kNm2

M 14 (10 20 4 24 04) 21221151 (315 ndash 12)22 886 kNm

V 14 (10 20 4 24 04) 211151 (315 ndash 12) 684 kN

285 RECOMMENDED DETAILS

The information given in Tables 362 and 363 has been takenfrom several sources including BS 8110 research undertakenby the Cement and Concrete Association (CampCA) and reportspublished by the Concrete Society

2851 Continuous nibs

The BS 8110 recommendations are shown in Table 362 As aresult of investigations by the CampCA various methods ofreinforcing continuous nibs were put forward Method (a) isefficient but it is difficult to incorporate the bars in shallow nibsif the bends in the bars are to meet the minimum code require-ments Method (b) is reasonably efficient and it is a simplematter to anchor the bars at the outer face of the nib The CampCArecommendations were based on an assumption of truss actionBS 8110 suggests that such nibs should be designed as shortcantilever slabs but both methods lead to similar amounts ofreinforcement

2852 Corbels

The information in Table 362 is based on the requirements ofBS 8110 and BS 5400 supplemented by recommendations

Recommended details 327

Distance from front edge of base to resultant vertical force

a MnetFv 39042591 150 m

Eccentricity of vertical force relative to centreline of base

e 332 ndash 15 015 m ( 336 055 m)

Maximum pressure at front of base

pmax (259133)(1 6 01533) 100 kNm2

Minimum pressure at back of base

pmin (259133)(1 ndash 6 01533) 57 kNm2

For the ultimate bearing condition a uniform distribution isconsidered of length lb 2a 2 15 30 m Pressure

pu Fvlb 259130 864 kNm2

The ultimate bearing resistance is given by the equation

qu (2 ) cud ic where ic 05[1 ]

ic 05[1 ] 073

qu (2 ) 333 073 125 kNm2 ( pu 864)

Resistance to sliding (long-term)

Fv tan b 2591 033 855 kN ( Fh 784)

Resistance to sliding (short-term)

(le) cb 33 25 825 kN ( Fh 784)

For resistance to sliding (short-term) the contact surface hasbeen taken as the full width of the base This is consideredreasonable since base adhesion is taken as only 075cud If thecontact surface is based on the pressure diagram assumed forthe ultimate bearing condition the resistance to sliding isreduced to lb cb 30 25 75 kN

Example 6 The structural design of the wall in example 5 isto be in accordance with the requirements of BS 8110

fcu 35 Nmm2 fy 500 Nmm2 nominal cover 40 mm

Allowing for H16 bars with 40 mm cover

d 400 ndash (40 8) 352 mm

For the ULS values of f are taken as 12 for the horizontalloads and 14 for all the vertical loads

The ultimate bending moment at the bottom of the wall stem

M 12 033 (10 422 20 436) 1162 kNmm

Mbd2fcu 1162 106(1000 3522 35) 00268

From Table 314 since K 0043 z 095d

As 1162 106[(087 500)(095 352)] 799 mm2m

From Table 220 H16-250 gives 804 mm2m

The ultimate shear force at the bottom of the wall stem

V 12 033 (10 4 20 422) 792 kNm

Vbd 792 103(1000 352) 023 Nmm2 ( vc)

From Table 343 the clear distance between bars should notexceed (47 000fs)(100Asbd) 750 mm Thus with

fs 087fyf 087 50012 362 Nmm2

ab (47 000362)[100 804(1000 352)] 568 mm

1 784(333 30)

1 Fh(cud lb)

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Recommended details nibs corbels and halving joints 362

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Recommended details intersections of members 363

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taken from CampCA research reports For the system of forcesshown in the figure the inclined force in the concrete

Fc Fvsin13 k1 fcub(x cos13) where tan13 (d ndash k2 x)av

In these expressions k1 and k2 are properties of the concretestress block as given in section 241 From these expressions aquadratic equation in xd (given in the table) can be derived InBS 8110 for values of xd (1 ndash 2avd)k2 a minimum valueof Ft 05Fv is taken for the tensile force

In the detail shown in the table the main bars are bent backto form a loop If the bars are welded to a crossbar they can becurtailed at the outer edge of the corbel In this case two addi-tional small diameter bars are needed to support the horizontalstirrups If the stirrups are required to pass outside the main barsin the column they should be detailed as two lapping U-barsfor ease of assembly

2853 Halving joints

The recommendations given in BS 5400 as a result of workcarried out by the CampCA are summarised in Table 362 Theinclined links must intersect the line of action of Fv If thiscannot be ensured (eg the inclined links could be displaced)or if horizontal forces can occur at the joint horizontal linksmust also be provided as shown

2854 Reinforcement details at frame corners

Research has shown that when frame corners are subjected tobending moments tending to close the corner the most likelycause of premature failure is due to bearing under the bend ofthe tension bars at the outside of the corner Provided that theradius of the bend is gradual and that sufficient anchorage isgiven for the lapping bars the use of simple details as shown in(a) or (b) on Table 363 is recommended

With lsquoopening cornersrsquo the problems are somewhatgreater and tests have shown that some details can fail at wellbelow their calculated strength In this case the detail shownin (d) is recommended If at all possible a concrete splayshould be formed within the corner and the diagonal rein-forcement Ass provided with appropriate cover If a splay isimpracticable the diagonal bars should be included withinthe corner itself

Detail (d) is suitable for reinforcement amounts up to about 1If more than this is required transverse links should be includedas shown in (e) The arrangement shown in (c) could be usedbut special attention needs to be paid to bending and fixingthe diagonal links which must be designed to resist all the forcein the main tension bars Care must also be taken to provideadequate cover to the bars at the inside of the corner

2855 Beam-column intersections

Research has shown that the forces in a joint between a beamand an end column are as shown in the sketch on Table 363Diagonal tensile forces occur at right angles to the theoreticalstrut that is shown To ensure that as a result diagonal cracksdo not form across the corner a design limit that is related tothe service condition is shown in the table To ensure that the

joint has sufficient ultimate strength an expression has alsobeen developed for a minimum amount of reinforcement that isneeded to extend from the top of the beam into the column atthe junction However for floor beams it has been shown thatthe U-bar detail shown in the table is satisfactory

While research indicates that unless it is carefully detailedas described the actual strength of the joint between a beamand an end column could be as little as half of the calculatedmoment capacity it seems that internal beam-column jointshave considerable reserves of strength Joints having a beam onone side of a column and a short cantilever on the other aremore prone to loss of strength and it is desirable in suchcircumstances to detail the joint as for an end column withthe beam reinforcement turned down into the column and thecantilever reinforcement extending into the beam

Example 7 A corbel is required to support a design ultimatevertical load of 500 kN at a distance of 200 mm from the faceof a column The load is applied through a bearing pad andthe 300 mm wide corbel is to be designed to BS 8110

fcu 30 Nmm2 fy 500 Nmm2 cover 40 mm

Minimum reinforcement based on Ft 05Fv and fyd 087fy

As 05 500 103(087 500) 575 mm2

Calculate shear resistance based on 2H20 (As 628 mm2) forvalues of d 400 mm where av 200 mm b 300 mm vc isobtained from Table 333 and Vc vc (2dav)bd

Miscellaneous members and details330

d vc (2dav)vc Vc ( 10ndash3)mm 100Asbd Nmm2 Nmm2 kN

500 042 050 250 375550 038 049 270 445600 035 047 282 508

Assuming that As will need to be increased as a result of thecorbel analysis with a corresponding increase in Vc considerh 600 mm with d 550 mm Hence

kv Fvbdfcu 500 103(300 550 30) 010

avd 200550 036 k1 040 k2 045 (section 241)

From Table 362 in the quadratic equation for xd

0452 040 045 03601 085

2 045 040 03601 234

1 0362 113 giving the equation

085(xd)2 ndash 234(xd ) 113 0 from which xd 0625

1 av

d 2

2k2 k1

kvav

d

k22

k1k2

kvav

d

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Hence x 0625 550 344 mm and the strain in the bars

s 00035(d ndash x)x 00035 206344 00021

fyd s Es 00021 200 103 420 Nmm2 ( 087fy)

Since (1 ndash 2avd)k2 (1 ndash 2 200550)045 0622 xd

Ft Fvav(d ndash k2x) Fv2 and As Fvav(d ndash k2x)fyd

As 500 103 200[(550 ndash 045 344) 420] 603 mm2

In this case 2H20 giving 628 mm2 is sufficient for the tensileforce but insufficient for the shear resistance Changing the

reinforcement to 3H20 gives 942 mm2 and increases the shearresistance as follows

100Asbd 100 942(300 550) 057

vc 056 Nmm2 Vc vc (2dav)bd 508 kN

Minimum area of horizontal links 05 942 471 mm2 tobe provided by 3H10 links (2 legs per link) which should belocated in the tension zone (ie extending over a depth of about200 mm below the main bars)

Recommended details 331

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Part 4

Design to EuropeanCodes

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Chapter 29

Design requirementsand safety factors

In the Eurocodes (ECs) design requirements are set out inrelation to specified limit-state conditions Calculations todetermine the ability of a member (or assembly of members)to satisfy a particular limit-state are undertaken using designactions (loads or deformations) and design strengths Thesedesign values are determined from either characteristic actionsor representative actions and from characteristic strengths ofmaterials by the application of partial safety factors

291 ACTIONS

Characteristic values of the actions to be used in the design ofbuildings and civil engineering structures are given in severalparts of EC 1 Actions on structures as follows

1991-1-1 Densities self-weight and imposed loads1991-1-2 Actions on structures exposed to fire1991-1-3 Snow loads1991-1-4 Wind loads1991-1-5 Thermal actions1991-1-6 Actions during execution1991-1-7 Accidental actions due to impact and explosions1991-2 Traffic loads on bridges1991-3 Actions induced by cranes and machinery1991-4 Actions on silos and tanks

A variable action (eg imposed load snow load wind loadthermal action) has the following representative values

characteristic value Qk

combination value 0Qk

frequent value 1Qk

quasi-permanent value 2Qk

The characteristic and combination values are used for theverification of the ultimate and irreversible serviceability limit-states The frequent and quasi-permanent values are usedfor the verification of ULSs involving accidental actions andreversible SLSs The quasi-permanent values are also used forthe calculation of long-term effects

Design actions (loads) are given by

design action (load) F Fk

where Fk is the specified characteristic value of the actionF is the value of the partial safety factor for the action (A for

accidental actions G for permanent actions Q for variableactions) and the limit state being considered and is either10 0 1 or 2 Recommended values of F and are given inEC 0 Basis of structural design

292 MATERIAL PROPERTIES

The characteristic strength of a material fk means that valueof either the cylinder strength fck or the cube strength fckcube ofconcrete or the yield strength fyk of reinforcement below whichnot more than 5 of all possible test results are expected tofall In practice the concrete strength is selected from a set ofstrength classes which in EC 2 are based on the characteristiccylinder strength The application rules in EC 2 are validfor reinforcement in accordance with BS EN 10080 whosespecified yield strength is in the range 400ndash600 MPa

Design strengths are given by

design strength fkM

where fk is either fck or fyk as appropriate and M is the value ofthe partial safety factor for the material ( C for concrete S forsteel reinforcement) and the limit-state being considered asspecified in EC 2

293 BUILDINGS

Details of the design requirements and partial safety factors aregiven in Table 41 Appropriate combinations of design actionsand values of are given on page 336

294 CONTAINMENT STRUCTURES

For structures containing liquids or granular solids the mainrepresentative value of the variable action resulting from theretained material should be taken as the characteristic value forall design situations Appropriate characteristic values are givenin EC 1 Part 4 Actions in silos and tanks

For the ULS where the maximum level of a retained liquidcan be clearly defined and the effective density of the liquid(allowing for any suspended solids) will not vary significantlythe value of Q applied to the resulting characteristic actionmay be taken as 12 Otherwise and for retained granularmaterials in silos Q 15 should be used

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Design requirements and safety factors336

For the SLS of cracking a classification of liquid-retainingstructures in relation to the required degree of protectionagainst leakage and the corresponding design requirementsgiven in EC 2 Part 3 are detailed here Silos containing dry

materials may generally be designed as Class 0 but where thestored material is particularly sensitive to moisture class 1 2or 3 may be appropriate

Limit-state and design consideration Combination of design actions

Ultimate (persistent and transient actions) 13Gj Gkj Q1 Qk1 13Qi 0i Qki ( j 1 i 1)

Ultimate (accidental action) (Ad 13Gkj (11 or 21) Qk1 132i Qki ( j 1 i 1)

Serviceability (function including damage to structural and 13Gkj Qk1 130i Qki ( j 1 i 1)non-structural elements eg partition walls etc)

Serviceability (comfort to user use of machinery avoiding 13Gkj 11 Qk1 132i Qki ( j 1 i 1)ponding of water etc)

Serviceability (appearance) 13Gkj 132i Qki ( j 1 i 1)

Note In the combination of design actions shown above Qk1 is the leading variable action and Qki are any accompanying variable actions Where necessary eachaction in turn should be considered as the leading variable action Serviceability design consideration and associated combination of design actions as specified in the UK National Annex

Values of for variable actions ( as specified in the UK National Annex) 0 1 2

Imposed loads (Category and type see EN 1991-1-1)A domestic residential area B office area 07 05 03C congregation area D shopping area 07 07 06E storage area 10 09 08F traffic area (vehicle weight 30 kN) 07 07 06G traffic area (30 kN vehicle weight 160 kN) 07 05 03H roof 07 0 0

Snow loads (see EN 1991-1-3)Sites located at altitude 1000 m above sea level 07 05 02Sites located at altitude 1000 m above sea level 05 02 0

Wind loads (see EN 1991-1-4) 05 02 0

Thermal actions (see EN 1991-1-5) 06 05 0

Class Leakage requirements Design provisions

0 Leakage acceptable or irrelevant The provisions in EN 1992-1-1 may be adopted (see Table 41)

1 Leakage limited to small amount The width of any cracks that can be expected to pass through the full thicknessSome surface staining or damp of the section should be limited to wk1 given bypatches acceptable 005 wk1 0225(1 hw 45h) 02 mm

where hwh is the hydraulic gradient (ie head of liquid divided by thicknessof section) at the depth under consideration Where the full thickness of the section is not cracked the provisions in EN 1992-1-1 apply (see Table 41)

2 Leakage minimal Appearance Cracks that might be expected to pass through the full thickness of the sectionnot to be impaired by staining should be avoided unless measures such as liners or water bars are included

3 No leakage permitted Special measures (eg liners or prestress) are required to ensure watertightness

Note In classes 1 and 2 to provide adequate assurance that cracks do not pass through the full width of a section the depth of the compression zone should beat least 02h 50 mm under quasi-permanent loading for all design conditions The depth should be calculated by linear elastic analysis assuming that concretein tension is neglected

Combinations of design actions on buildings

Values of for variable actions on buildings

Classification of liquid-retaining structures

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Design requirements and partial safety factors (EC 2 Part 1) 41

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Chapter 30

Properties of materials

301 CONCRETE

3011 Strength and elastic properties

The characteristic strength of concrete is defined as that level ofcompressive strength below which 5 of all valid test results isexpected to fall Strength classes are specified in terms of bothcylinder strength and equivalent cube strength Recommendedstrength classes with indicative values for the secant modulusof elasticity at 28 days are given in Table 42 The values givenfor normal-weight concrete are appropriate for concretesmade with quartzite aggregates For limestone and sandstoneaggregates these values should be reduced by 10 and 30respectively For basalt aggregates the values should be increasedby 20

Variation of the secant modulus of elasticity with time can beestimated by the expression

Ecm(t) [ fcm(t)fcm]03 Ecm

where Ecm(t) and fcm(t) are values at age t days and Ecm and fcm

are values determined at age 28 days

3012 Creep and shrinkage

The creep strain in concrete may be assumed to be directlyproportional to the applied stress for stresses not exceeding045fck(t0) where t0 is age of concrete at the time of loadingValues of the final creep coefficient (infin to) and the creepdevelopment coefficient c(t t0) according to the time underload can be obtained from Table 43 The procedure used todetermine the final creep coefficient is as follows

1 Determine point corresponding to age of loading to on theappropriate curve (N for normally hardening cement R forrapidly hardening cement S for slowly hardening cement)

2 Construct secant line from origin of curve to the pointcorresponding to to

3 Determine point corresponding to the notional size ofthe member ho on the appropriate curve for the concretestrength class

4 Cross horizontally from the point determined in 3 to intersectthe secant line determined in 2

5 Drop vertically from the intersection point determined in 4to obtain the required creep coefficient (infin to)

When the applied stress exceeds 045fck(t0) at time of loadingcreep non-linearity should be considered and (infin to) shouldbe increased as indicated in Table 42

The total shrinkage strain is composed of two componentsautogenous shrinkage strain and drying shrinkage strainThe autogenous shrinkage strain develops during hardening ofthe concrete a major part therefore develops in the early daysafter casting It should be considered specifically when newconcrete is cast against hardened concrete Drying shrinkagestrain develops slowly as it is a function of the migration ofthe water through the hardened concrete Final values of eachcomponent of shrinkage strain and development coefficientswith time are given in Table 42

3013 Thermal properties

Values of the coefficient of thermal expansion of concrete fornormal design purposes are given in Table 42

3014 Stressndashstrain curves

Idealised stressndashstrain curves for concrete in compression aregiven in Table 44 Curve A is part parabolic and part linear andcurve B is bi-linear A simplified rectangular diagram is alsogiven as a further option

302 REINFORCEMENT

3021 Strength and elastic properties

The characteristic yield strength of reinforcement according toEN 10080 is required to be in the range 400ndash600 MPa Thecharacteristic yield strength of reinforcement complying withBS 4449 is 500 MPa For further information on types proper-ties and preferred sizes of reinforcement reference should bemade to section 103 and Tables 219 and 220

3022 Stressndashstrain curves

Idealised bi-linear stressndashstrain curves for reinforcement intension or compression are shown in Table 44 Curve A has aninclined top branch up to a specified strain limit and curve Bhas a horizontal top branch with no need to check the strainlimit For design purposes the modulus of elasticity of steelreinforcement is taken as 200 GPa

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Concrete (EC 2) strength and deformation characteristics ndash 1 42

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Concrete (EC 2) strength and deformation characteristics ndash 2 43

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Stressndashstrain curves (EC 2) concrete and reinforcement 44

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Chapter 31

Durability andfire-resistance

In the following the concrete cover to the first layer of barsas shown on the drawings is described as the nominal coverIt is defined as a minimum cover plus an allowance in designfor deviation A minimum cover is required to ensure the safetransmission of bond forces the protection of steel againstcorrosion and an adequate fire-resistance In order to transmitbond forces safely and ensure adequate concrete compactionthe minimum cover should be not less than the bar diameter orfor bundled bars the equivalent diameter of a notional barhaving the same cross-sectional area as the bundle

311 DURABILITY

3111 Exposure classes

Details of the classification system used in BS EN 206ndash1 andBS 8500-1 with informative examples applicable in the UnitedKingdom are given in Table 45 Often the concrete can beexposed to more than one of the actions described in the tablein which case a combination of the exposure classes will apply

3112 Concrete strength classes and covers

Concrete durability is dependent mainly on its constituents andlimitations on the maximum free watercement ratio and theminimum cement content are specified for each exposure classThese limitations result in minimum concrete strength classesfor particular cements For reinforced concrete the protectionof the steel against corrosion depends on the cover The requiredthickness of cover is related to the exposure class the concretequality and the intended working life of the structure Details ofthe recommendations in BS 8500 are given in Table 46

The values given for the minimum cover apply for ordinarycarbon steel in concrete without special protection and forstructures with an intended working life of at least 50 years Thevalues given for the nominal cover include an allowance fortolerance of 10 mm which is recommended for buildings andis normally also sufficient for other types of structures

For uneven concrete surfaces (eg ribbed finish or exposedaggregate) the cover should be increased by at least 5 mm Ifin-situ concrete is placed against another concrete element(precast or in-situ) the minimum cover to the reinforcement atthe interface need be no more than that recommended foradequate bond provided the following conditions are metthe concrete strength class is at least C2530 the exposure time

of the concrete surface to an outdoor environment is no more than28 days and the interface has been roughened

Where concrete is cast against prepared ground (includingblinding) the nominal cover should be at least 50 mm Forconcrete cast directly against the earth the nominal covershould be at least 75 mm

312 FIRE-RESISTANCE

3121 Building regulations

The minimum period of fire-resistance required for elements ofthe structure according to the purpose group of a buildingand its height or for basements depth relative to the ground aregiven in Table 312 Building insurers may require longer fireperiods for storage facilities

3122 Design procedures

BS EN 1992-1-2 contains prescriptive rules in the form ofboth tabulated data and calculation models for standard fireexposure A procedure for a performance-based method usingfire-development models is also provided

The tabulated data tables give minimum dimensions for thesize of a member and the axis distance of the reinforcementThe axis distance is the nominal distance from the centre ofthe main reinforcement to the concrete surface as shown in thefollowing figure

Tabulated data is given for beams slabs and braced columnsfor which provision is made for the load level to be taken intoaccount In many cases for fire periods up to about two hoursthe cover required for other purposes will control For furtherinformation on all the design procedures reference should bemade to BS EN 1992-1-2

Section through member showing nominal axis distance

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Exposure classification (BS 8500) 45

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Concrete quality and cover requirements for durability (BS 8500) 46

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fck Maximum value of xd at section whereMPa moment redistribution is required

50 ( 04)

55 095 ( 04)60 092 ( 04)70 090 ( 04)

80 088 ( 04)

In the above expressions is the ratio of the design moment to themaximum elastic moment for values of

where class B and C reinforcement is usedwhere class A reinforcement is used

Based on values given in the UK National Annex10 0810 07

Chapter 32

Bending and axialforce

321 DESIGN ASSUMPTIONS

Basic assumptions regarding the design of cracked concretesections at the ULS are outlined in section 52 The tensilestrength of the concrete is neglected and strains are evaluatedon the assumption that plane sections before bending remainplane after bending Reinforcement stresses are then derivedfrom these strains on the basis of the design stress-strain curvesshown on Table 44 Two alternatives are given in which the topbranch of the curve is taken as either horizontal with no needto check the strain limit or inclined up to a specified strainlimit For the concrete stresses three alternative assumptionsare permitted as shown on Table 44 The design stressndashstraincurves give stress distributions that are a combination of eithera parabola and a rectangle or a triangle and a rectangleAlternatively a rectangular concrete stress distribution may beassumed Whichever alternative is used the proportions of thestress-block and the maximum strain are constant forfck 50 MPa but vary as the concrete strength changes forfck 50 MPa

For a rectangular area of width b and depth x the total com-pressive force is given by k1 fckbx and the distance of the forcefrom the compression face is given by k2x where values of k1

(including for the term ccc) and k2 according to the shape ofthe stress block are given in the table opposite

322 BEAMS AND SLABS

Beams and slabs are generally subjected to bending only butsometimes are also required to resist an axial force for examplein a portal frame or in a floor acting as a prop between basementwalls Axial thrusts not greater than 012fck times the area of thecross section may be ignored in the analysis of the sectionsince the effect of the axial force is to increase the moment ofresistance

In cases where as a result of moment redistributionallowed in the analysis of the member the design moment isless than the maximum elastic moment at the section the nec-essary ductility may be assumed without explicit verificationif the neutral axis depth satisfies the limits given in the tableopposite

Where plastic analysis is used the necessary ductility may beassumed without explicit verification if the neutral axis depth atany section satisfies the following requirement

xd 025 for concrete strength classes C5060xd 015 for concrete strength classes C5060

The following analyses and resulting design charts and tablesare applicable to concrete strength classes C5060

fck Shape of k1 k2 cuMPa stress-block

50 Parabola-rectangle 0459 0416Triangle-rectangle 0425 0389 00035Rectangular 0453 0400

55 Parabola-rectangle 0433 0400Triangle-rectangle 0402 0375 00031Rectangular 0435 0394

60 Parabola-rectangle 0417 0392Triangle-rectangle 0381 0363 00029Rectangular 0417 0388

70 Parabola-rectangle 0399 0383Triangle-rectangle 0357 0351 00027Rectangular 0382 0375

80 Parabola-rectangle 0385 0378Triangle-rectangle 0327 0340 00026Rectangular 0349 0363

90 Parabola-rectangle 0378 0375Triangle-rectangle 0316 0337 00026Rectangular 0317 0350

Properties of concrete stress-blocks for rectangular area

Limits of xd for moment redistribution

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Bending and axial force346

3221 Singly reinforced rectangular sectionsMbd2fck

As fykbdfck for ductility class

A B C

0010 00111 00108 001020012 00134 00130 001230014 00156 00152 001440016 00179 00174 001650018 00202 00197 00186

0020 00224 00219 002070022 00247 00241 002280024 00270 00264 002520026 00293 00286 002760028 00317 00309 00300

0030 00340 00331 003240032 00363 00354 003490034 00387 00379 003730036 00410 00403 003980038 00434 00428 00423

0040 00457 00453 004480042 00481 00478 004730044 00505 00503 004980046 00529 00528 005230048 00553 00554 00549

3222 Doubly reinforced rectangular sections

The lever arm between the forces shown in the figure here isgiven by z (d k2x) from which x (d z)k2

Taking moments for the compressive force about the line ofaction of the tensile force gives

M k1 fckbxz k1 fckbz(d z)k2

The solution of the resulting quadratic equation in z gives

where

Taking moments for the tensile force about the line of action ofthe compressive force gives

M As fs z from which As Mfs z

The strain in the reinforcement s 00035(1 xd)(xd) andfrom the design stressndashstrain curves the stress is given by

fs sEs 700(1 xd)(xd) ks fyk 115

If the top branch of the design stressndashstrain curve is taken ashorizontal (curve B) ks 10 and fs fyk 115 for values of

xd 805(805 fyk) 0617 for fyk 500 MPa

A design chart for fyk 500 MPa derived on the basis of theparabolic-rectangular stress-block for the concrete and curve Bfor the reinforcement is given in Table 47

If the top branch of the design stressndashstrain curve is taken asinclined (curve A) the value of ks depends on the strain and theductility class of the reinforcement The use of curve A isadvantageous in reducing the reinforcement requirement Themaximum reduction for a lightly reinforced section is close to5 8 and 15 for reinforcement ductility classes A B andC respectively For more heavily reinforced sections the reduc-tions are less and the particular ductility class makes very littledifference to the values obtained

A design table based on the rectangular stress-block for theconcrete and design curve A for the reinforcement is given inTable 48 The table uses non-dimensional parameters and isvalid for values of fck 50 MPa Values of ks determined forfyk 500 MPa were based on the most critical ductility classin each case namely class A for Mbd2fck 0046 and class Bfor Mbd2fck 0046 This can be seen from the followingtable which gives values of As fykbdfck for each ductility classin the range up to Mbd2fck 0048

M bd2fckz d 05 025 (k2 k1)

The forces provided by the concrete and the reinforcementare shown in the figure here Taking moments for the twocompressive forces about the line of action of the tensileforce gives

M k1 fckbx(d k2x)

The strain in the reinforcement and fromdesign stressndashstrain curve B the stress is given by

Thus fyk 115 for values of

for fyk 500 MPa

Equating the tensile and compressive forces gives

As fs k1 fckbx

where the stress in the tension reinforcement is given by theexpression derived in section 3221

As f s

x d [805 (805 fyk)](dd) 264(dd )

f s

f s s Es 700(1 dx) fyk 115

s 00035(1 dx)

As f s(dd)

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EC 2 Design chart for singly reinforced rectangular beams 47

Sin

gly

rein

forc

ed b

eam

s (f

yk=

500

MP

a)

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EC 2 Design table for singly reinforced rectangular beams 48

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Beams and slabs 349

Design charts based on the rectangular stress-block for theconcrete design curve B for the reinforcement and for valuesof and 015 respectively are given in Tables 49 and410 The charts which use non-dimensional parameters weredetermined for fyk 500 MPa but may be safely used forfyk 500 MPa In determining the forces in the concrete noreduction has been made for the area of concrete displaced bythe compression reinforcement

3223 Design formulae for rectangular sections

No design formulae are given in the code but the following arevalid for values of fck 50 MPa and fyk 500 MPa Theformulae are based on the rectangular stress-block for theconcrete and stresses of 087fyk in tension and compressionreinforcement The compression reinforcement requirementdepends on the value of Mbd2fck compared to where

0210 for 10

for

is the ratio design moment to maximum elastic momentwhere 07 for class B and C reinforcement and 08 forclass A reinforcement

For compression reinforcement is not required and

As M087fykz where

z d05 + and x (d z)04

For compression reinforcement is required and

where

z d05 + and x (d z)04

For (for fy 500 MPa) should be replaced by16(1 ) in the equations for and As

3224 Flanged sections

In monolithic beam and slab construction where the web of thebeam projects below the slab the beam is considered as aflanged section for sagging moments The effective width of theflange over which uniform conditions of stress can be assumedmay be taken as beff bw where

01(aw l0) 02l0 05aw for L beams 02(aw l0) 04l0 10aw for T beams

In the aforesaid expressions bw is the web width aw is the cleardistance between the webs of adjacent beams and l0 is the dis-tance between points of zero bending moment for the beam Ifleff is the effective span l0 may be taken as 085leff when thereis continuity at one end of the span only and 07leff when there iscontinuity at both ends of the span For up-stand beams whendesigning for hogging moments l0 may be taken as 03leff atinternal supports and 015leff at end supports

In most sections where the flange is in compression the depthof the neutral axis will be no greater than the thickness of theflange In this case the section can be considered to be rectangularwith b taken as the flange width The condition regarding theneutral axis depth can be confirmed initially by showing thatM k1 fckbhf (d k2hf) where hf is the thickness of the flange

bb

b

AsAsdxAsdx 0375

025 0882

As As bd2fck 087fykz

As ()bd2fck 087fyk(dd)

025 0882

10 0453( 04) 0181( 04)2

dd 01

Alternatively the section can be considered to be rectangularinitially and the neutral axis depth can be checked subsequently

The figure here shows a flanged section where the neutral axisdepth is greater than the flange thickness The concrete forcecan be divided into two components and the required area oftension reinforcement is then given by

As As1 k1 fck (b bw)hf 087fyk whereAs1 area of reinforcement required to resist a moment M1

applied to a rectangular section of width bw and

M1 M k1 fck (b bw)hf (d k2hf) bd2fck

Using the rectangular concrete stress-block in the forgoingequations gives k1 045 and k2 04 This approach givessolutions that are lsquocorrectrsquo when x hf but become slightlymore conservative as (x hf) increases

3225 General analysis of sections

The analysis of a section of any shape with any arrangement ofreinforcement involves a trial-and-error process An initialvalue is assumed for the neutral axis depth from which theconcrete strains at the positions of the reinforcement can becalculated The corresponding stresses in the reinforcement aredetermined and the resulting forces in the reinforcement andthe concrete are obtained If the forces are out of balance thevalue of the neutral axis depth is changed and the process isrepeated until equilibrium is achieved Once the balancedcondition has been found the resultant moment of the forcesabout the neutral axis or any convenient point is calculated

Example 1

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EC 2 Design chart for doubly reinforced rectangular beams ndash 1 49

Dou

bly

rein

forc

ed b

eam

s (f

yk=

500

MP

ad9

d=

01)

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410EC 2 Design chart for doubly reinforcedrectangular beams ndash 2

Dou

bly

rein

forc

ed b

eam

s (f

yk=

500

MP

ad9

d=

015

)

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Bending and axial force352

The beam shown in the figure here is part of a monolithic beamand slab floor in which the beams are spaced at 5 m centresThe maximum and minimum design loads on each span are asfollows where Gk 160 kN and Qk 120 kN

Fmax 135Gk 15Qk 396 kN Fmin 135 Gk 216 kN

The section design is to be based on the following values

fck 32 MPa fyk 500 MPa cover to links 35mm

For sagging moments and beff are given by

02(aw l0) 04 l0 aw

02(4700 085 8000) 2300 mmbeff bw 300 2300 2600 mm

Allowing for 8 mm links and 32 mm main bars

d 500 (35 8 16) 440 mm say

In the calculations that follow solutions are obtained usingcharts and equations to demonstrate the use of each method

Maximum sagging moment For section to be designed as rec-tangular with b taken as the flange width bending momentshould satisfy the condition

M k1 fckbhf (d k2hf) 045 32 2600 150 (440 04 150) 106

2134 kNm (258 kNm)

Mbd 2 258 106 (2600 4402) 051 MPa

From chart in Table 47 100Asbd 012

As 00012 2600 440 1373 mm2

Alternatively by calculation or from Table 48

Mbd2fck 05132 0016

Hence As M087fykz gives

As 258 106 (087 500 0985 440) 1369 mm2

Using 3H25 gives 1473 mm2

The above solutions are based on design stress-strain curve Bfor the reinforcement Solutions based on curve A also can beobtained from the table in section 3221 as follows

Mbd 2fck

As fykbdfck for ductility class

A B C

0016 00179 00174 00165As (mm2) 1310 1274 1208

Maximum hogging moment

Mbd2fck 396 106 (300 4402 32) 0213

From Table 48 this appears to be just beyond the range for asingly reinforced section From the chart in Table 49 a valueof As fykbdfck 034 can be obtained Although this is a validsolution it should be possible to reduce the area of tensionreinforcement to a more suitable value by allowing for somecompression reinforcement Consider the use of 2H25 whichgives 55 mm ( )dd 55 440 0125d

z d 05 025 0882 0016 0985

b

b

b

fykbdfck 982 500(300 440 32) 0116

Interpolating from charts in Tables 49 and 410 with

fykbdfck 01 As fykbdfck 0285 (for 0125)As 0285 300 440 32500 2408 mm2

Using 3H32 gives 2413 mm2

A solution can also be obtained using the design equations asfollows 087fy is valid for xd ( )0375 0333

With xd 0333 and zd (1 04xd) 0867

0453(xd)(zd) 0453 0333 0867 0131

( ) bd2fck 087fyk (d ) (0213 0131) 300 4402 32

(087 500 385) 910 mm2 (2H25 gives 982 mm2)

As bd 2fck 087fyk z 910 0131 300 4402 32

(087 500 0867 440) 2377 mm2 (3H32 gives 2413 mm2)

Example 2 Suppose that in the previous example the maxi-mum hogging moment at B is reduced by 20 to 317 kNmThis is valid for reinforcement of all ductility classes

Mbd2fck 317 106 (300 4402 32) 0171 317396 080 xd ( 04) 04

From chart in Table 410 keeping to left of line for xd 04

fykbdfck 005 Asfykbdfck 023 005 300 440 32500 423 mm2

Using 2H20 gives 628 mm2 (instead of 2H25 in example 1)

As 023 300 440 32500 1943 mm2

Using 4H25 gives 1963 mm2 (instead of 3H32 in example 1)

A solution can also be obtained by using the design equationsas in example 1 with xd 0333 zd 0867 0131

( )bd 2fck 087fyk (d ) 004 300 4402 32 (087 500 385) 444 mm2

As bd2fck 087fykz 444 0131 300 4402 32

(087 500 0867 440) 1912 mm2

Since the reduced hogging moment for load case 1 is stillgreater than the elastic hogging moment for load case 2 thedesign sagging moment remains the same as in example 1

In the foregoing examples at the bottom of the beam 2H25bars would run the full length of each span with 2H25 or 2H20splice bars at support B Other bars would be curtailed to suitthe bending moment requirements and detailing rules

323 COLUMNS

In the Code of Practice a column is taken to be a compressionmember whose greater overall cross sectional dimension is notmore than four times its smaller dimension An effective lengthand a slenderness ratio are determined in relation to its major

As

dAs

As

As

As

dAs

ddf s

ddAs

As

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Columns 353

and minor axes of bending The effective length is a function ofthe clear height and depends upon the restraint conditions at theends of the column A slenderness ratio is defined as the effec-tive length divided by the radius of gyration of the uncrackedconcrete section Columns should generally be designed for bothfirst order and second order effects but second order effects maybe ignored provided the slenderness ratio does not exceed a par-ticular limiting value This can vary considerably and has to bedetermined from an equation involving several factors Thesegenerally need to be calculated but default values are also given

Columns are subjected to combinations of bending momentand axial force and the cross section may need to be checkedfor more than one combination of values Several methods ofanalysis of varying complexity are available for determiningsecond order effects Many columns can be treated as isolatedmembers and a simplified method of design using equationsbased on an estimation of curvature is commonly used Theequations contain a modification factor Kr the use of whichresults in an iteration process with Kr taken as 10 initially Thedesign procedures are shown in Tables 415 and 416

In the code for sections subjected to pure axial load the concretestrain is limited to 0002 for values of fck 50 MPa In thiscase the design stress in the reinforcement should be limited to400 MPa However in other parts of the code the design stressin this condition is shown as fyd fyks 087fyk In thederivation of the charts in this chapter which apply for allvalues of fck 50 MPa and fyk 500 MPa the maximumcompressive stress in the reinforcement was taken as 087fykThe charts contain sets of Kr lines to aid the design process

3231 Rectangular columns

Design charts based on the rectangular stress-block for theconcrete and for values of dh 08 and 085 are given inTables 411 and 412 respectively On each curve a straight linehas been taken between the point where xh 10 and the pointwhere N Nu The charts which were determined for fyk 500MPa may be safely used for fy 500 MPa In determining theforces in the concrete no reduction has been made for the areaof concrete displaced by the compression reinforcement In thedesign of slender columns the Kr factor is used to modify thedeflection corresponding to a load Nbal at which the moment is amaximum A line corresponding to Nbal passes through a cusp oneach curve For N Nbal the Kr value is taken as 10 ForN Nbal Kr can be determined from the lines on the chart

3232 Circular columns

The following figure shows a circular section with six barsspaced equally around the circumference Solutions based on sixbars will be slightly conservative if more bars are used Thearrangement of the bars relative to the axis of bending affects theresistance of the section and the arrangement shown in the fig-ure is not the most critical in every case For some combinationsof bending moment and axial force if the arrangement shown isrotated through 30o a slightly more critical condition results butthe differences are small and may be reasonably ignored

The foregoing figure shows a rectangular section in which thereinforcement is disposed equally on two opposite sides of ahorizontal axis through the mid-depth By resolving forces andtaking moments about the mid-depth of the section the follow-ing equations are obtained for 0 xh 10

Nbhfck k1(xh) 05(As fykbhfck)(ks1 ks2)

Mbh2fck k1(xh)05 k2(xh) 05(As fykbhfck)(ks1 ks2)(dh 05)

The stress factors ks1 and ks2 are given by

ks1 14(xh dh 1)(xh) 087ks2 14(dh xh)(xh) 087

The maximum axial force Nu is given by the equation

Nubhfck 0567 087(As fykbhfck)

The following analysis is based on a uniform stress-block forthe concrete of depth x and width hsin at the base (as shownin figure) Negative axial forces are included in order to caterfor members such as tension piles By resolving forces andtaking moments about the mid-depth of the section the followingequations are obtained where cos1 (1 2 xh) for0 x 10 and hs is the diameter of a circle through thecentres of the bars

Nh2fck kc (2 sin2)8 (12)(As fykAc fck)(ks1 ks2 ks3)

Mh3fck kc (3sin sin3)72 ( 277)(As fykAc fck)(hsh)(ks1 ks3)

Because the width of the compression zone decreases in thedirection of the extreme compression fibre the design stressin the concrete has to be reduced by 10 Thus in the aboveequations kc 09 0567 051 and 08

The stress factors ks1 ks2 and ks3 are given by

087 ks1 14(0433hsh 05 xh)(xh) 087

087 ks2 14(05 xh)(xh) 087

087 ks3 14(05 0433hsh xh)(xh) 087

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EC 2 Design chart for rectangular columns ndash 1 411

Rec

tang

ular

col

umns

(f y

k=

500

MP

ad

h=

08)

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EC 2 Design chart for rectangular columns ndash 2 412

Rec

tang

ular

col

umns

(f y

k=

500

MP

ad

h=

085

)

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Bending and axial force356

To avoid irregularities in the charts the reduced design stress inthe concrete is used to determine the maximum axial force Nuwhich is given by the equation

Nuh2fck ( 4)051 087(As fykAc fck)

The minimum axial force Nmin is given by the equation

Nminh2fck 087( 4)(AsfykAcfck)

Design charts for values of hsh 06 and 07 are given inTables 413 and 414 respectively The statements in section3231 on the derivation and use of the charts for rectangularsections apply also to those for circular sections

3233 General analysis of column sections

Any given cross section can be analysed by a trial-and-errorprocess For a section bent about one axis an initial value isassumed for the neutral axis depth from which the concretestrains at the positions of the reinforcement can be calculatedThe resulting stresses in the reinforcement are determined andthe forces in the reinforcement and concrete evaluated If theresultant force is not equal to the design axial force N the valueof the neutral axis depth is changed and the process repeateduntil equality is achieved The resultant moment of all theforces about the mid-depth of the section is then the moment ofresistance appropriate to N This approach is used to analyse arectangular section in example 6

Example 3 A 300 mm square braced column designed forthe following requirements

l 50 m k 0675 at both joints in both directionsM02 40 kNm M01 20 kNm about x-x axisM0 negligible about y-y axis N 1720 kNfck 32 MPa fyk 500 MPa cover to links 35 mm

Allowing for 8 mm links and 32 mm main bars

d 300 (35 8 16) 240 mm say

From Table 415 effective length for a braced column wherethe joint stiffness is the same at both ends is given by

l0 [05(045 2k)(045 k)] l 08 50 40 m

Slenderness ratio l0 i 4000(300radic12) 462

From Table 415 with M01 05M02 C 22 and

Since lim second order moments need to be considered

Minimum design moment with e0 h30 30030 20 mm

Mmin Ne0 1720 002 34 kNm

Additional first order moment resulting from imperfectionswith 067 h 2radicl 2radic(50) 09 10

Mi N( hl0400) 1720 (09 40400) 16 kNm

Total first order moment for section at end 2 of the column

Mx M02 Mi 40 16 56 kNm (Mmin)

Mbh2fck 56 106(300 3002 32) 0065

Nbhfck 1720 103(300 300 32) 0600

lim 34 N Ac fck 34 1720 103 (3002 32) 44

From the design chart for dh 240300 08

As fykbhfck 025 (Table 411)As 025 300 300 32500 1440 mm2

Using 4H25 gives 1963 mm2

The section where the second order moment is greatest may bedesigned by first assuming the reinforcement (4H25 say)

As fykbhfck 1963 500(300 300 32) 034Nubhfck 086 (Table 411) and for Nbhfck 06Mubh2fck 009 Kr 04 (Table 411)Mux Muy 009 300 3002 32 106 78 kNm

Second order moment resulting from deflection with Kr 04and for fck 32 MPa and 46 K 13 (Table 416)

M2 N(KrKl02d)2000

1720 (04 13 402024)2000 30 kNm

Equivalent first order moment (near mid-height of column)

M0e 06M02 04M01 04M02 04 40 16 kNm

Total design moments (near mid-height of column)

Mx M0e Mi M2 16 16 30 62 kNm (Mmin)My M2 30 kNm

Design for biaxial bending may be ignored if the following twoconditions are satisfied (a) 05y x 2y and (b) for asquare column Mx is either 02My or 5My In this casesince condition (b) is not satisfied and a furthercheck is necessary as follows

For NNu 060086 07

n 092 083(NNu) 150

Hence

Since this value is less than 10 4H25 is sufficient

Example 4 A 350 mm circular braced column designed forthe same requirements as example 3 Thus l0 40 m as before

Slenderness ratio l0i 4000(3504) 457

Cross-sectional area Ac (4) 3502 962 103 mm2

Since second order moments need not be consideredAllowing for 8 mm links and 20 mm main bars

hs 350 2 (35 8 10) 244 mm

For the section at end 2 of the column

Mh3fck 56 106(3503 32) 0041Nh2fck 1720 103(3502 32) 044

From the design chart for hsh 244350 07

As fykAcfck 026 (Table 414)As 026 962 103 32500 1600 mm2

Using 6H20 gives 1885 mm2

lim

lim 34 N Ac fck 34 1720 (962 32) 455

Mx

Mux

n

My

Muy

n

6278

15

3078

15

095

Mx 2My

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EC 2 Design chart for circular columns ndash 1 413

Circular columns (fyk = 500 MPa hsh = 06)

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EC 2 Design chart for circular columns ndash 2 414

Circular columns (fyk = 500 MPa hsh = 07)

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EC 2 Design procedure for columns ndash 1 415

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EC 2 Design procedure for columns ndash 2 416

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Columns 361

Example 5 A 400 mm circular unbraced column designedfor the same requirements as example 3 Thus k 0675 asbefore

From Table 415 effective length for an unbraced columnwhere the joint stiffness is the same at both ends is given by

l0 [(1 2k)(1 k)] l 14 50 70 m

Slenderness ratio l0i 7000(4004) 70

Cross-sectional area Ac ( 4) 4002 1257 103 mm2

Since second order moments need to be consideredAllowing for 8 mm links and 32 mm main bars

hs 400 2 (35 8 16) 280 mm say

Additional first order moment resulting from imperfections

Mi N( hl0400) 1720 (09 70400) 27 kNm

Second order moment resulting from deflection with Kr 10(max) K 11 for fck 32 MPa and 70 (Table 416) andd h2 035hs 298 mm

M2 N(KrKl02d)2000

1720 (10 11 7020298)2000 155 kNm

Total design moments at end 2 of column

Mx M02 Mi M2 40 27 155 222 kNm (Mmin)My M2 155 kNm

Resultant uniaxial moment at end 2 of column

kNm

Mh3fck 270 106(4003 32) 013Nh2fck 1720 103(4002 32) 034

From the design chart for hsh 280400 07

As fykAcfck 084 (Table 414)As 084 1257 103 32500 6758 mm2

It can be seen from the chart that Kr 10 Using 8H32 gives

As fykAc fck 6434 500(1257 103 32) 080For Nh2fck 034 Mh3fck 0125 and Kr 075

Hence Mu 0125 4003 32 106 256 kNm

With Kr 075 modified M2 075 155 116 kNm

Total design moments at end 2 of column

Mx M02 Mi M2 40 27 116 183 kNmMy M2 116 kNm

Resultant uniaxial moment at end 2 of column

217 kNm

Since M Mu 8H32 is sufficient

Example 6 The column section in the following figure isreinforced with 8H32 arranged as shown The moment of resis-tance about the major axis is to be obtained for the followingrequirements

N 2300 kN fck 32 MPa fyk 500 MPa

M (1832 1162)

M (M2x M2

y) (2222 1552) 270

lim

lim 108 N Acfck 108 1720 (1257 32) 165

Consider the bars in each half of the section to be replaced byan equivalent pair of bars Depth to the centroid of the bars inone half of the section 60 2404 120 mm The section isnow considered to be reinforced with four equivalent barswhere d 600 120 480 mm

As fykbhfck 6434 500(300 600 32) 056Nbhfcu 2300 103(300 600 32) 040

From the design chart for dh 480600 08

Mubh2fck 018 (Table 411)Mu 018 300 6002 32 106 622 kNm

The solution can be checked using a trial-and-error process toanalyse the original section as follows

The axial load on the section is given by

N k1fckbx (As1ks1 As2ks2 As3ks3)fyk

where dh 540600 09 and ks1 ks2 and ks3 are given by

ks1 14(xh dh 1)(xh) 087ks2 14(05 xh)(xh) 087ks3 14(dh xh)(xh) 087

With x 300 mm xh 05 ks1 087 ks2 0 and ks3 087

N 045 32 300 300 103 1296 kN (2300)

With x 360 mm xh 06 ks2 0233 ks3 07

N 045 32 300 360 103 (2413 087 1608 0233 2413 07) 500 103

1555 392 1947 kN (2300)

With x 390 mm xh 065 ks2 0323 ks3 0538

N 045 32 300 390 103 (2413 087 1608 0323 2413 0538) 500 103

1685 660 2345 kN (2300)

With x 387 mm xh 0645 ks2 0315 ks3 0553

N 045 32 300 387 103 (2413 087 1608 0315 2413 0553) 500 103

1672 636 2308 kN

Since the internal and external forces are now sensibly equaltaking moments about the mid-depth of the section gives

Mu k1 fckbx(05h k2x ) (As1ks1 As3ks3)(d 05h)fy

045 32 300 387 (300 04 387) 106

(2413 087 2413 0553)(540 300) 500 106

243 412 655 kNm (622 obtained before)

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Chapter 33

Shear and torsion

331 SHEAR RESISTANCE

3311 Members without shear reinforcement

The design shear resistance at any cross section of a membernot requiring shear reinforcement can be calculated as

VRdc vRdcbwdwhere

bw is the minimum width of section in the tension zoned is the effective depth to the tension reinforcementvRdc is the design concrete shear stress

The design concrete shear stress is a function of the concretestrength the effective depth and the reinforcement percentageat the section considered To be effective this reinforcementshould extend a distance (lbd d) beyond the section wherelbd is the design anchorage length At a simple support for amember carrying predominantly uniform load the length lbd

may be taken from the face of the support The design shearresistance of members with and without axial load can bedetermined from the data given in Table 417

In the UK National Annex it is recommended that the shearstrength of concrete strength classes higher than C5060 isdetermined by tests unless there is evidence of satisfactorypast performance of the particular concrete mix including theaggregates used Alternatively the shear strength should belimited to that given for concrete strength class C5060

3312 Members with shear reinforcement

The design of members with shear reinforcement is based ona truss model in which the compression and tension chords

are spaced apart by a system of inclined concrete struts andvertical or inclined shear reinforcement Angle between thereinforcement and the axis of the member should be 45o

Angle 13 between the struts and the axis of the member maybe selected by the designer within the limits 10 cot13 25generally However for elements in which shear co-exists withexternally applied tension cot13 should be taken as 10 Theweb forces are Vsec13 in the struts and Vsec in the shearreinforcement over a panel length l z (cot cot13) wherez may normally be taken as 09d The width of each strutis equal to z (cot cot13) sin13 and the design value of themaximum shear force VRdmax is limited to the compressiveresistance provided by the struts which includes a strengthreduction factor for concrete cracked in shear The least shearreinforcement is required when cot13 is such that V VRdmaxThe truss model results in a force Ftd in the tension chord thatis additional to the force Mz due to bending but the sumFtd Mz need not be taken greater than Mmaxz where Mmax isthe maximum moment in the relevant hogging or sagging regionThe additional force Ftd can be taken into account by shiftingthe bending moment curve each side of any point of maximummoment by an amount al 05z(cot13 cot) For memberswithout shear reinforcement al d should be used The cur-tailment of the longitudinal reinforcement can then be based onthe modified bending moment diagram A design procedure todetermine the required area of shear reinforcement and detailsof the particular requirements for beams and slabs are given inTable 418

For most beams a minimum amount of shear reinforcementin the form of links is required irrespective of the magnitude ofthe shear force Thus there is no need to determine VRdc

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EC 2 Shear resistance ndash 1 417

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EC 2 Shear resistance ndash 2 418

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In members with inclined chords the shear components ofthe design forces in the chords may be added to the design shearresistance provided by the reinforcement In checking that thedesign shear force does not exceed VRdmax the same shearcomponents may be deducted from the shear force resultingfrom the design loads

3313 Shear under concentrated loads

In slabs and column bases the maximum shear stress at theperimeter of a concentrated load should not exceed vRdmaxShear in solid slabs under concentrated loads can result inpunching failures on the inclined faces of truncated conesor pyramids For design purposes a control perimeterforming the shortest boundary that nowhere comes closer tothe perimeter of the loaded area than a specified distanceshould be considered The basic control perimeter maygenerally be taken at a distance 2d from the perimeter ofthe loaded area

If the maximum shear stress here is no greater than vRdc noshear reinforcement is required Otherwise the position of thecontrol perimeter at which the maximum shear stress is equalto vRdc should be determined and shear reinforcement providedin the zone between this control perimeter and the perimeter ofthe loaded area

For flat slabs with enlarged column heads (or drop panels)where dH is the effective depth at the face of the column and thecolumn head (or drop) extends a distance lH 2dH beyond theface of the column a basic control perimeter at a distance 2dH

from the column face should be considered In addition a basiccontrol perimeter at a distance 2d from the column head (ordrop) should be considered

Control perimeters (in part or as a whole) at distances lessthan 2d should also be considered where a concentrated loadis applied close to a supported edge or is opposed by a highpressure (eg soil pressure on bases) In such cases values ofvRdc may be multiplied by 2da where a is the distance fromthe edge of the load to the control perimeter For column basesthe favourable action of the soil pressure may be taken intoaccount when determining the shear force acting at the controlperimeter

Details of design procedures for shear under concentratedloads are given in Table 419

3314 Bottom loaded beams

Where load is applied near the bottom of a section sufficientvertical reinforcement to transmit the load to the top of thesection should be provided in addition to any reinforcementrequired to resist shear

332 DESIGN FOR TORSION

In normal beam-and-slab or framed construction calculationsfor torsion are not usually necessary adequate control of anytorsional cracking in beams being provided by the requiredminimum shear reinforcement When it is judged necessaryto include torsional stiffness in the analysis of a structure ortorsional resistance is vital for static equilibrium membersshould be designed for the resulting torsional moment

Design for torsion 365

The torsional resistance may be calculated on the basis ofa thin-walled closed section in which equilibrium is satisfiedby a plastic shear flow A solid section may be modelled asan equivalent thin-walled section Complex shapes may bedivided into a series of sub-sections each of which is mod-elled as an equivalent thin-walled section and the totaltorsional resistance taken as the sum of the resistances of theindividual elements When torsion reinforcement isrequired this should consist of rectangular closed linkstogether with longitudinal reinforcement Such reinforcementis additional to the requirements for shear and bendingDetails of a suitable design procedure for torsion are givenin Table 420

Example 1 The beam shown in the following figure whichwas designed for bending in example 1 of Chapter 32 is to bedesigned for shear The maximum design load is 495 kNm andthe design is based on the following values

fck 32 MPa fywk 500 MPa d 440 mm

Since the load is uniformly distributed the critical sectionfor shear may be taken at distance d from the face of the sup-port Based on a support width of 400 mm distance fromcentre of support to critical section 200 440 640 mmAt end B

V 248 064 495 216 kNw V[bw z (1 fck250)fck]

216 103[300 09 440 (1 32250) 32] 0065

From Table 418 since w 0138 cot13 25 may be usedHence area of links required is given by

Asws Vfywd z cot13 216 103(087 500 09 440 25) 050 mm2mm

From Table 420 H8-200 provides 050 mm2mm

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EC 2 Shear under concentrated loads 419

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EC 2 Design for torsion 420

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Shear and torsion368

Minimum requirements for vertical links are given by

Asws (008radicfck) bw fyk (008radic32) 300500 027 mm2mm

s 075d 075 440 330 mm

From Table 420 H8-300 provides 033 mm2mm

VRds (Asws) fywd z cot13 033 087 500 09 440 25 103

142 kN

At end A V 160 064 495 128 kN ( VRds 142 kN)

Example 2 A 250 mm thick flat slab is supported by 400 mmsquare columns arranged on a 72 m square grid The slab con-tains as tension reinforcement in the top of the slab at an inte-rior support within a 18 m wide strip central with the columnH16-150 in each direction Lateral stability of the structuredoes not depend on frame action and the design shear forceresulting from the maximum design load applied to all panelsadjacent to the column is V 854 kN

fck 40 MPa fywk 500 MPa d 210 mm (average)

Since the lateral stability of the structure does not depend onframe action may be taken as 115 (Table 419)

Maximum shear stress adjacent to the column face

Vuod 115 854 103(4 400 210) 293 MPavRdmax 02 (1 fck250)fck

02 (1 40250) 40 672 MPa (293)

Based on H16-150 as effective tension reinforcement

100Aslbwd 100 201(150 210) 064vRdc 070 MPa (Table 417 for d 210 and fck 40)

The length of the first control perimeter at 2d from the faceof the column is 4 400 4 d 4239 mm Thus themaximum shear stress at the first control perimeter

Vu1d 115 854 103(4239 210) 110 MPa

Since v vRdc shear reinforcement is needed where effectivedesign strength fywdef 250 025d 300 MPa The areaneeded in one perimeter of vertical shear reinforcementat maximum radial spacing sr 075d 150 mm say isgiven by

Asw (v 075vRdc) u1 sr 15 fywdef

(110 075 070) 4239 150(15 300) 813 mm2

wmin u1 sr 15 (008radicfck) u1 sr 15fyk

(008radic40) 4239 150(15 500) 429 mm2

Using 12H10 gives 942 mm2

Length of control perimeter at which v vRdc is given by

u Vd vRdc 115 854 103(210 070) 6681 mm

Distance of this control perimeter from face of column is

a (6681 4 400)2 809 mm

The distance of the final perimeter of reinforcement from thecontrol perimeter where v vRdc should be 15d 315mm

Thus 4 perimeters of reinforcement with sr 150 mm and thefirst perimeter at 100 mm from the face of the column wouldbe suitable The reinforcement layout is shown in the followingfigure where indicates the link positions and the links canbe anchored round the tension bars

Example 3 The following figure shows a channel sectionedge beam on the bottom flange of which bear 8 m long simplysupported contiguous floor units The beam is continuous in14 m spans and is prevented from lateral rotation at thesupports The centroid and the shear centre of the sectionare shown

Characteristic loads

floor units dead 35 kNm2 imposed 25 kNm2

edge beam dead 12 kNm

Design ultimate loads

floor units (135 35 15 25) 82 338edge beam 135 12 162

500 kNm

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Design for torsion 369

fck 32 MPa fyk 500 MPa d 1440 mm

Bending moment shear force and torsional moment (aboutshear centre of section) at interior support (other than first)

M 009 50 142 882 kNm (Table 229)V 05 50 14 350 kNT 05 (338 0400 162 0192) 14 117 kNm

(Note In calculating V and T a coefficient of 05 rather than055 has been used since the dead load is dominant and thecritical section may be taken at the face of the support)

Considering beam as one large rectangle of size 250 1500and two small rectangles of size 200 300

13hmin3hmax 2503 1500 2 2003 300

(234 2 24) 109 282 109

Torsional moment to be considered on large rectangle

T1 117 234282 97 kNm

Torsional moment to be considered on each small rectangle

T2 117 24282 10 kNm

Reinforcement required in large rectangle

Shear and torsion (see Table 420) Assuming 30 mm cover toH10 links distance from surface of concrete to centre of H12longitudinal bars 46 mm

tefi Au 250 1500[2 (250 1500)] 107 mm ( 2 46 92 mm)

Ak (250 107) (1500 107) 1992 103 mm2

For values of (1 fck250)fck (1 32250) 32 279 MPaand z 1440 100 1340 mm (to centre of flange)

w [T12Ak tefi Vbw z](1 fck250)fck

[97(2 1992 107) 350(250 1340)] 103279 0119

Since w 0138 cot13 25 may be used (Table 418)For a system of closed links total area required in two legs fortorsion and shear is given by

Ass (T1Ak V z)fywd cot13 (971992 3501340) 103(087 500 25) 069 mm2mm

The inner legs of the links are also subjected to a vertical tensileforce resulting from the load of 338 kNm applied by the floorunits Additional area required in inner legs

Ass 338(087 500) 008 mm2mm

Total area required in two legs for torsion shear and theadditional vertical tensile force

Ass 069 2 008 085 mm2mm

The area of longitudinal reinforcement required for torsion isgiven by

Aslsl Tcot13 2Akfyd

97 103 25 (2 1992 087 500) 140 mm2mm

Different combinations of links and longitudinal bars can beobtained by changing the value of cot 13 as follows

Bars Links Longitudinal

Ass Size and Aslsl Size andcot 13 mm2mm spacing mm2mm spacing

25 085 H10-175 140 H12-15020 102 H10-150 112 H12-20016 124 H10-125 090 H12-250

s least of u8 35008 4375 mm 075d 1080 mmor hmin 250 mm sl 350 mm

Bending (see Table 48)

Mbd2fck 882 106(550 14402 32) 0024Asfykbdfck 0027 and xd 0054 (ie x 78 200 mm)

As 0027 550 1440 32500 1369 mm2

Total area of longitudinal bars required at top of beam forbending and torsion (equivalent to 2H12 say)

1369 226 1595 mm2

From Table 228 2H32 provides 1608 mm2

Reinforcement required in small rectangles

Torsion Assuming 30 mm cover to H8 links distance fromsurface of concrete to centre of H12 longitudinal bars 44 mm

tefi Au 200 300[2 (200 300)] 60 ( 2 44 88 mm)

Ak (200 88) (300 88) 237 103 mm2

w (T22Ak tefi)(1 fck250)fck

10 103(2 237 88 279) 0086

Since w 0138 cot 13 25 may be used For a system ofclosed links area required in two legs is given by

Asts T2Ak fywd cot13 10 103(237 087 500 25) 039 mm2mm

The lower rectangle is also subjected to bending resulting fromthe load of 338 kNm applied by the floor units The distanceof the load from the centre of the inner leg of the links in thelarge rectangle is 150 35 185 mm

M 338 0185 625 kNm

Taking the lever arm for the small rectangle as the distancebetween the centres of the top and bottom arms of the linksz 132 mm Additional area required in top arms of links

Ass Mfyd z 625 103(087 500 132) 011 mm2mm

Total area required in two arms for torsion and bending

Ass 039 2 011 061 mm2mm

The area of longitudinal reinforcement required for torsion isgiven by

Aslsl Tcot132Ak fyd

10 103 25(2 237 087 500) 121 mm2mm

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Shear and torsion370

Different combinations of links and longitudinal bars can beobtained by changing the value of cot13 as follows

Bars Links Longitudinal

Ass Size and Aslsl Size andcot 13 mm2mm spacing mm2mm spacing

25 061 H10-150 121 H12-17518 076 H10-125 088 H12-250

s lesser of u8 10008 125 mm or hmin 200 mm

The lower rectangle is also subjected to shear in the verticallongitudinal plane for which

Vbw d 338 103(1000 166) 021 MPa

From Table 417 vmin 056 MPa (fck 32 d 200)

From the foregoing calculations the reinforcement shown inthe figure opposite provides a practical arrangement in whichthe links comprise H10-125 for the large rectangle andH8-125 for the small rectangles The longitudinal bars are all

H12-250 apart from the 2H32 bars at the top of the largerectangle

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341 DEFLECTION

Deflections of members under service load should not impairthe function or appearance of a structure For buildings thedesign requirements and associated combinations of designactions to be considered are given in section 293 The finaldeflection of members below the level of the supports undercharacteristic loading after allowance for any pre-camber isgenerally limited to span250 A further limit of span500applies to the increase in deflection that occurs after theconstruction stage in order to minimise any damage to bothstructural and non-structural elements The requirements maybe met by complying with the limits on spaneffective depthratio given in Table 421

In special circumstances when the calculation of deflectionis considered necessary an adequate prediction can be madeusing the methods given in Table 422 Careful considerationis needed in the case of cantilevers where the usual formulaeassume that the cantilever is rigidly fixed and remains hori-zontal at the root Where the cantilever forms the end of acontinuous beam the deflection at the end of the cantilever islikely to be either increased or decreased by an amount l13where l is the length of the cantilever measured to the centreof the support and 13 is the rotation at the support Where acantilever is connected to a substantially rigid structure someroot rotation will still occur and the effective length shouldbe taken as the length to the face of the support plus half theeffective depth

342 CRACKING

3421 Building structures

Cracks in members under service load should not impair theappearance or durability of the structure For buildings thedesign requirements are given in Table 41 The calculatedcrack width under quasi-permanent loading or as a result ofrestrained deformations is generally limited to 03 mm For drysurfaces inside buildings where crack width has no effect ondurability a limit of 04 mm is recommended where there isa need to ensure an acceptable appearance However in theUK National Annex a limit of 03 mm is required in this situa-tion In the regions of concrete members where tension isexpected a calculated minimum amount of reinforcement isneeded in order to control cracking as given in Table 423

Where minimum reinforcement is provided the crack widthrequirements may be met by direct calculation or by limitingeither the bar size or the bar spacing as given in Table 424For the calculation of crack widths due to restrained imposeddeformation no guidance is given in Part 1 of the code but thefollowing equation is given in PD 6687 (see preface)

sm ndash cm 08Rimp

where R is a restraint factor (see section 2621) and imp is thefree strain due to temperature fall or drying shrinkage

3422 Liquid-retaining structures

For structures containing liquids design requirements related toleakage considerations are given in section 294 Where a smallamount of leakage with related surface staining or damppatches is acceptable for cracks that can be expected to passthrough the full thickness of the section the calculated crackwidth is limited to a value that varies according to the hydraulicgradient (ie head of liquid divided by thickness of section)The limits are 02 mm for hydraulic gradients 5 reducinguniformly to 005 mm for hydraulic gradients 35 Thus thelimits for a 300 mm thick wall to a 75 m deep tank would be02 mm at 15 m below the top 015 mm at 45 m below the topand 01 mm at 75 m below the top The limits apply under thequasi-permanent loading combination where the full charac-teristic value is taken for hydrostatic loading For members inaxial tension where at least the minimum reinforcement isprovided the crack width requirements may be met by directcalculation or by limiting either the bar size or the bar spacingas given in Table 425

In cases of bending with or without axial force where thefull thickness of the section is not cracked and not less than 02times the section thickness 50 mm is in compression thecrack width limit is 03 mm and Table 424 applies

For cracking due to restraint of imposed deformations suchas shrinkage and early thermal movements two distinct typesof restraint are considered For a concrete element restrained atits ends (eg an infill bay with construction joints between thenew section of concrete and the pre-existing sections) the crackformation is similar to that caused by external loading Anappropriate expression for the tensile strain contributing to thecrack width is given in Table 425 and for specified values of

Chapter 34

Deflection and cracking

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EC 2 Deflection ndash 1 421

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EC 2 Deflection ndash 2 422

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EC 2 Cracking ndash 1 423

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EC 2 Cracking ndash 2 424

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EC 2 Cracking ndash 3 425

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Cracking 377

cover section thickness and reinforcement content maximumvalues of fcteff are given in Table 426

For a concrete panel restrained along an edge (eg a wall castonto a pre-existing stiff base) the formation of the crack onlyinfluences the distribution of stresses locally and the crackwidth becomes a function of the restrained strain rather than thetensile strain capacity of the concrete In this case the tensilestrain contributing to the crack width is taken as Rax free wheretypical values of free can be estimated from the informationgiven in Table 425 The restraint factor Rax may be taken as 05generally or reference can be made to Table 345 where valuesare indicated for panels restrained along one two or three edgesrespectively For specified values of cover section thicknessand reinforcement content maximum values of Rax free aregiven in Table 427

It will be found that the calculated strain contributing to thecrack width for a panel restrained at its ends is normally morethan Rax free Thus the reinforcement required to limit a crackwidth to the required value is greater for a panel restrained atits ends than for a panel restrained along one or more edgesAlso for a specific crack width the reinforcement needed for apanel restrained along an edge is less than that in BS 8007since the design crack spacing is less than that in BS 8007

Example 1 The beam shown in the following figure is to bechecked for deflection and cracking The design for bendingand shear is shown in example 1 of Chapters 32 and 33 respec-tively The reinforcement in the bottom of each span 3H25(1473 mm2) is based on the following values

fck 32 MPa fyk 500 MPa beff 2600 mm d 440 mm

From Table 423 for fck 32 MPa and bending of a sectionwith h 500 mm by interpolation 100AsminAct 021

Asmin 00021 162 103 340 mm2 ( 1473 mm2)

026 ( f ctmfyk ) btd 00013 btd

026 (30500) 300 440 206 mm2

The design ultimate load is 396 kN and the quasi-permanentload where the value of 2 is obtained from section 293 is

Gk 2 Qk 160 03 120 196 kN

Hence the stress in the reinforcement under quasi-permanentloading is given approximately by

s (196396)(087fyk)(As req As prov)

Thus for the bars in the bottom of the beam

s (196396)(087 500)(13731473) 200 MPa

From Table 424 for wk 03 mm the crack width criterion ismet if

s 25 mm or the bar spacing 250 mm

The adjusted maximum bar size is given by

Depth of tension zone hcr h ndash x 500 ndash 128 372 mm

s s (fcteff 29)[kc hcr 2(h ndash d)]

25 (3029) [04 372(2 60)] 32 mm

Note It can be seen from the foregoing that all of the criteriaare comfortably satisfied and the checks for deflection andcracking are hardly necessary in this example

Example 2 A 250 mm thick flat slab is supported bycolumns which are arranged on a 72 m square grid Thecharacteristic loads are 72 kNm2 dead and 45 kNm2 imposedand the slab is to be checked for deflection and cracking

fck 32 MPa fyk 500 MPa cover to bars 25 mm

Allowing for the use of H12 bars in each direction and basedon the bars in the second layer of reinforcement

d 250 ndash (25 12 6) 205 mm ld 7200205 35

From Table 421 for a flat slab with spans 85 m

Basic spaneffective depth ratio 24

Total design ultimate load for a square panel is given by

F (135 72 15 45) 722 854 kN

From Table 262 the design ultimate bending moment for anend span with a continuous connection at the outer support is

M 0075Fl 0075 854 72 461 kNm

Mbd 2fck 461 106(7200 2052 32) 0048

As fykbdfck 0056 (Table 48)

100Asbd 100 0056 32500 036

From Table 421 for 100Asbd 01fck05 01 3205 057

s 055 00075fck(100Asbd)

0005fck05[ fck

05(100Asbd) ndash 10]15

055 00075 32036

0005 3205 (3205036 ndash 10)15 160

The actual spaneffective depth ratio 8000440 182

From Table 421 for the end span of a continuous beam and aflanged section with bbw 2600300 867 3

Basic spaneffective depth ratio 08 26 208

For members supporting partitions liable to be damaged byexcessive deflections the basic ratio should be multiplied by7span In this case the basic ratio 208 78 182

Since 100As reqbd 100 1373(2600 440) 012 is smallthe modification factor s is large ( 3) and the limiting ratiois more than three times the actual value

The neutral axis depth for the uncracked section ignoring theeffect of the reinforcement is given by

x

128 mm ( hf 150 mm)

Area of tension zone is given by

Act bw (h ndash hf) bf (hf ndash x) 300 350 2600 22 162 103 mm2

300 5002 2300 1502

2[300 500 2300 150]bwh2 (bf ndash bw)h2

f

2[bwh (bf ndash bw)hf]

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EC 2 Early thermal cracking in end restrained panels 426

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EC 2 Early thermal cracking in edge restrained panels 427

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Deflection and cracking380

Limiting spaneffective depth ratio 16 24 384 ( 35)

From Table 423 for fck 32 MPa and bending of a sectionwith h 300 mm 100AsminAct 024 With Act 05bh

100Asminbd 024 05 250205 015

026 (fctmfyk) 00013

026 30500 016 ( 036)

The design ultimate load is 854 kN and the quasi-permanentload where the value of 2 is obtained from section 293 is

Gk 2 Qk (72 03 45) 722 443 kN

Hence the stress in the reinforcement under quasi-permanentloading is given approximately by

s (443854)(087fyk) (443854)(087 500) 226 MPa

From Table 424 for wk 03 mm the crack width criterion ismet if

s 16 mm or the bar spacing 200 mmThe adjusted maximum bar size with hcr 05h is given by

s s (fcteff 29)[kc hcr 2(h ndash d)]

16 (3029) [04 125(2 45)] 9 mm

Area of reinforcement required to give 100Asbd 036 is

As 00036 1000 205 738 mm2m (H12-150)

Example 3 The wall of a cylindrical tank 75 m deep and15 m diameter is 300 mm thick The wall which is continuouswith the base slab is to be designed for temperature effects andthose due to internal hydrostatic pressure when the tank is fullof liquid

Design class 1 (see section 294) fyk 500 MPaCover to horizontal bars 40 mm fck 32 MPa

Effects of temperature change With fctm 03 fck(23) 30 MPa

and assuming that early thermal cracks will occur at a timewhen fcm(t) 24 MPa

fcteff [fcm(t)(fck 8)] fctm [24(32 8)] 30 18 MPa

The limiting crack width varies according to the hydraulicgradient (depth of liquid thickness of section) If the wall isdesigned to the recommendations for a panel restrained atits ends then suitable reinforcement details for 40 mmcover and fcteff 18 MPa selected from Table 426 aregiven here

Depth Hydraulic Design crack Reinforcement(m) gradient width (mm) required (EF)

15 5 02 H20-15045 15 015 H20-12575 25 01 H20-100

Note The table for wk 02 mm was used throughout bytaking effective values of fcteff 18(5wk) MPa

If the wall is designed to the recommendations for a panelrestrained along the edge the restrained tensile strain needs tobe estimated as follows

Allowing for concrete grade C3240 with 350 kgm3 Portlandcement at a placing temperature of 20oC and a mean ambient

temperature during construction of 15oC the temperature risefor concrete placed within 18 mm plywood formwork

T1 25oC (Table 219)

As the wall is to be designed to resist hoop tension there willbe no vertical movement joints and allowance must be made fora fall in temperature due to seasonal variations Allowing forT2 15oC restraint factor Rax taken as 05 and coefficient ofthermal expansion taken as 12 10ndash6 per oC (Table 35)restrained total thermal contraction after the peak temperaturearising from hydration effects is given by

Rax (T1 T2 ) 05 12 10ndash6 (25 15) 240 10ndash6

Hence suitable reinforcement details for 40 mm cover andRax 240 10ndash6 selected from Table 427 are given here

Depth Hydraulic Design crack Reinforcement(m) gradient width (mm) required (EF)

15 5 02 H20-25045 15 015 H20-15075 25 01 H20-100

Note The table for wk 02 mm was used throughout foreffective values of Rax [240(5wk)] 10ndash6 For H20 barsvalues for a 250 mm thick section apply for h 250 mm

From Table 423 the minimum reinforcement percentage forfcteff fctm in the case of a rectangular section in pure tensionwith h 300 mm and fck 32 MPa is 060 For the control ofearly thermal cracking the value is (1830) 06 036 andthe minimum area of reinforcement required on each face

Asmin 00036 150 1000 540 mm2m (H16-300)

Clearly the reinforcement needed for thermal crack controlgreatly exceeds this minimum requirement In the lower part ofthe wall the reinforcement provided is H20-100 (EF) and thecorresponding stress at a cracked section is

s fcteff (Act As) 18 150 10003142 86 MPa

This solution can be checked approximately by reference to thechart for maximum bar size on Table 425 as follows

For s 86 MPa and wk 01 mm s 55 mm say

s s

55 (1829) (01 30050) 20 mm

Effects of hydrostatic load Suppose that an elastic analysis ofthe tank assuming a floor 300 mm thick indicates a servicemaximum circumferential tension of 400 kNm This valueoccurs at a depth of 6 m where the design crack width is0125 mm Above this level the tensions can be assumed toreduce approximately linearly to near zero at the top of the wall

For a section reinforced with H20-100 (EF) the stress in thereinforcement s 400 1036284 64 MPa Since this isless than the stress due to fcteff the reinforcement needed forthermal crack control is also sufficient for the circumferentialtension This solution can be checked also by reference to thecharts on Table 425 which show that with s 64 MPa thebar size and the bar spacing are of no consequence

fcteff

2901h(hd)

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The code contains many requirements that affect the details ofthe reinforcement such as minimum and maximum areas tyingprovisions anchorage and curtailment

Bars may be set out individually or grouped in bundles oftwo or three in contact Bundles of four bars may also be usedfor vertical bars in compression and for bars in a lapped jointFor the safe transmission of bond forces the cover providedto the bars should be not less than the bar diameter or for abundle the equivalent diameter ( 55 mm) of a notional barwith the same sectional area as the bundle Requirements forcover with regard to durability are given in Chapter 31 Gapsbetween bars (or bundles) generally should be not less than thegreatest of (dg 5 mm) where dg is the maximum aggregatesize the bar diameter (or equivalent bar diameter for a bundle)or 20 mm Details of reinforcement limits are given in Table 428

At intermediate supports of continuous flanged beams thetotal area of tension reinforcement should be spread over theeffective width of the flange but a part of the reinforcementmay be concentrated over the web width

351 TIES IN STRUCTURES

Structures not specifically designed to withstand accidentalactions should be provided with a suitable tying system toprevent progressive collapse by providing alternative loadpaths after local damage Where the structure is divided intostructurally independent sections each section should have anappropriate tying system The reinforcement providing the tiesmay be assumed to act at its characteristic strength and onlythe specified tying forces need to be taken into accountReinforcement required for other purposes may be consideredto form part of or the whole of the ties Details of the tyingrequirements as specified in the UK National Annex are givenin Table 429

352 ANCHORAGE AND LAP LENGTHS

At both sides of any cross section bars should be providedwith an appropriate embedment length or other form of endanchorage For bent bars the basic tension anchorage length ismeasured along the centreline of the bar from the section inquestion to the end of the bar where

lbd 1 2 3 4 5 lbrqd lbmin

As a simplified alternative a tension anchorage for a standard bend hook or loop may be provided as an equivalentlength lbeq 1 lbrqd (see figure here) where 1 is taken as07 for covers perpendicular to the bend 3 Otherwise1 10

Chapter 35

Considerationsaffecting design details

Bends or hooks do not contribute to compression anchoragesDetails of anchorage lengths are given in Table 430

Laps should be located if possible away from positions ofmaximum moment and should generally be staggered Detailsof lap lengths are given in Table 431

The radius of any bend in a reinforcing bar should conformto the minimum requirements of BS 8666 and should ensurethat failure of the concrete inside the bend is prevented A linkmay be considered fully anchored if it passes round another barof not less than its own diameter through an angle of 90o andcontinues beyond the end of the bend for a minimum length of10 diameters 70 mm Details of bends in bars are given inTable 431 Additional rules for large diameter bars ( 40 mmaccording to the UK National Annex) and bundles are given inTable 432

353 CURTAILMENT OF REINFORCEMENT

In flexural members it is generally advisable to stagger thecurtailment points of the tension reinforcement as allowed bythe bending moment envelope Bars should be curtailed inaccordance with the rules set out in Table 432 and illustratedin the figure on page 387 Except at end supports every tensionbar should extend beyond the point at which in theory it is nolonger needed for flexural resistance for a distance not less thanal The bar should also extend beyond the point at which it isfully required to provide flexural resistance for a distance notless than al lbd At a simple end support the bars shouldextend for the anchorage length lbd necessary to develop theforce ∆Ftd

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EC 2 Reinforcement limits 428

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EC 2 Provision of ties 429

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EC 2 Anchorage requirements 430

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EC 2 Laps and bends in bars 431

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EC 2 Rules for curtailment large diameter bars and bundles 432

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Curtailment of reinforcement 387

Example 1 The beam shown in the following figure which wasdesigned in example 1 of Chapters 32 (bending) and 33 (shear)is to be checked for the reinforcement details The designultimate loads are Fmax 396 kN and Fmin 216 kN on eachspan The width of each support is 400 mm and the mainreinforcement is as follows spans 3H25 (bottom) support B3H32 (top) and 2H25 (bottom)

fck 32 MPa fyk 500 MPa cover to links 35 mm

For bars in the bottom of the section the bond condition isclassified as lsquogoodrsquo Thus from Table 430

lbrqd 35 (s435) 35 25 (172435) 346 mm

The design anchorage length is given by

lbd 1 2 3 4 5 lbrqd lbmin

Coefficients 1 and 2 depend on cd which is taken as either thecover to the main bar or half the gap between the main barswhichever is the lesser With 35 mm cover to H8 links coverto main bars is 45 mm and the gap between the bars is300 2 (45 25) 160 mm Hence cd 45 mm (or 18)

Since cd 3 1 10 for both bent bars and straight barsHence using the simplified approach (see section 352) forboth straight bars or standard bends lbeq 1 lbrqd 346 mm

For a 400 mm wide support allowing for 50 mm end cover tothe bars the anchorage length provided from the near face ofthe support is 350 mm ( lbeq)

For the basic approach the following values can be obtained

For straight bars 2 1 015(cd 1) 088

With no transverse reinforcement provided within the bearinglength 3 10 and 4 10

Transverse pressure due to reaction on support is given by

p V(bearing area) 150 103(400 300) 125 MPa

Hence with p 125 MPa 5 1 004p 095 and

lbd 25lbrqd 088 095 346 290 mm lbmin 10 250 mm ( 03lbrqd or 100 mm)

Curtailment points for bottom bars The resistance momentprovided by 2H25 at the bottom of the beam may be determinedas follows

As fykbdfck 982 500(2600 440 32) 00134Mbd2fcu 0012 (Table 48 or section 3221)M 0012 2600 4402 32 106 193 kNm

Illustration of the curtailment of longitudinal reinforcement taking account of resistance within anchorage lengths

End anchorage At the bottom of each span 2H25 ( 25 ofarea provided in the span) will be continued to the support Atthe end support the tensile force to be anchored isF 05Vcot13 in which 13 is the inclination of the concrete strutrequired for shear design In the shear design calculations inchapter 33 V 128 kN at the critical section and VRds 142 kNwhen cot13 25 Thus cot13 (VVRds) 25 225 could beused At the face of the support V 160 02 495 150 kNWith cot 13 225

F 05 150 225 169 kNs FAs 169 103982 172 MPa

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Considerations affecting design details388

Reaction at A for load case 2 is RA 160 kN

Distance x from A to point where M 193 kNm is given by

RAx 05(FmaxL)x2 160x2 05 (3968)x2 193

Hence 05x2 323x 39 0 giving x 16 m and 485 m

Thus of the 3H25 required in the span one bar is no longerneeded for flexure at 16 m and 485 m from the end support Atthese points V 80 kN and cot13 (80142) 25 14 issufficient Thus the bar needs to extend beyond these points fora distance al 045d cot13 045 440 14 280 mm

Curtailment points for top bars The resistance moment providedby 2H32 can be determined as follows

As fykbdfck 1608 500(300 440 32) 0190Mbd2fcu 0142 (Table 48)M 0142 300 4402 32 106 264 kNm

For load case 1 reaction at A (or C) is given by

RA 05Fmax MBL 05 396 3968 148 kN

Distance x from A to point where M 264 kNm is given by

05(FmaxL)x2 RAx 05 (3968) x2 148 x 264

Hence 05x2 3x 53 0 giving x 74 m Thus of the3H32 required at B one bar is no longer required for flexure atdistance (80 74) 06 m from B If this distance is lessthan lb rqd the point of curtailment will be determined by theneed to develop the full force in the bar at B Here cot13 25giving al 045dcot13 1125d As the bars are effectively ina slab of thickness 250 mm it seems reasonable to assumelsquogoodrsquo bond conditions giving lb rqd 35 (Table 430) Thusdistance from B (edge of support say) at which one bar may becurtailed is al lb rqd 1125 440 35 32 1615 mm

Suppose that the remaining bars are continued to the point ofcontra-flexure in span BC for load case 2

The reaction at support C is given by

RC 05Fmin MBL 05 216 3068 70 kN

Distance from B to point of contra-flexure is given by

x L(1 2RCFmin) 8 (1 2 70216) 28 m

Here V 70 kN and cot13 (70142) 25 125 is sufficientThus distance from B at which the remaining bars may becurtailed is 2800 045 440 125 3050 mm

Link support bars say 2H12 could be used for the remainderof the span

Example 2 A typical floor to an 8-storey building consists ofa 250 mm thick flat slab supported by columns arranged ona 72 m square grid The slab for which the characteristicloading is 72 kNm2 dead and 45 kNm2 imposed is to beprovided with ties to the requirements of the UK NationalAnnex The design panel load is 854 kN and bending momentsare to be determined by the simplified method (see section 138)

fck 32 MPa fyk 500 MPa cover to bars 25 mm

Allowing for the use of H12 bars in each direction and basedon the bars in the second layer of reinforcement

d 250 (25 12 6) 205 mm say

From Table 255 the design ultimate sagging moment for aninterior panel is given by

M 0063Fl 0063 854 72 388 kNm

The total panel moment is to be apportioned between columnand middle strips where the width of each strip is 36 m Forthe column strip with 60 of the panel moment

M 06 388 233 kNmMbd2fck 233 106(3600 2052 32) 0048As fykbdfck 0056 (Table 48)As 0056 3600 205 32500

2645 mm2 (24H12-150 gives 2714 mm2)

For the middle strip with 40 of the panel moment

M 04 388 155 kNmMbd 2fck 155 106(3600 2052 32) 0032As fykbdfck 0037 (Table 48)As 0037 3600 205 32500

1748 mm2 (16H12-225 gives 1810 mm2)

For the peripheral tie the tensile force is given by

Ftieper (20 4no) 60 kN (20 4 8) 52 kN

The required area of reinforcement acting at its characteristicstrength is given by

As Ftieper fyk 52 103500 104 mm2 (1H12)

For the internal ties the tensile force is given by

kNm

With Ft (20 4no) 60 kN (20 4 8) 52 kN

kNm

If the internal ties are spread evenly in the slab the requiredarea of reinforcement acting at its characteristic strength

As 117 103500 234 mm2m (H12-450)

In this case at least every third bar in the column strips andevery other bar in the middle strips need to be continuous

If the internal ties are concentrated at the column lines the totalarea of reinforcement required in each group

As 234 72 1685 mm2 (16H12 gives 1810 mm2)

In this case the bars in the middle two-thirds of each columnstrip need to be continuous For sections 250 mm deep thebond condition is lsquogoodrsquo and lbrqd 35 (Table 430) Laps inadjacent pairs of lapped bars should be staggered by 13l0where l0 is the design lap length (Table 431) With lbd lbrqd

and 6 14 for one in two bars lapped at the same section

l0 6 lbd 14 35 12 600 mm say

Example 3 The following figure shows details of thereinforcement at the junction between a 300 mm wide beamand a 300 mm square column Bars 03 need to develop themaximum design stress at the column face and the radius of

Ftieint 72 4575 72

5 52 117

Ftieint gk qk

75 lr

5Ft Ft

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Curtailment of reinforcement 389

bend necessary to avoid failure of the concrete inside the bendis to be determined

fck 32 MPa fyk 500 MPa

The minimum radius of bend of the bars depends on the valueof ab where ab is taken as half the centre-to-centre distancebetween adjacent bars or for bars adjacent to the face of amember the side cover plus half the bar size Thus in this caseab 75 252 875 mm

From Table 431 for ab 87525 35 rmin 74 Thisvalue can be reduced slightly by taking into account the stressreduction in the bar between the edge of the support and thestart of the bend If r 7 distance from face of column tostart of bend 300 50 8 25 50 mm (ie 2)

From Table 430 for lsquogoodrsquo bond conditions the requiredanchorage length is 35 and rmin (1 235) 74 7Thus r 7 is sufficient

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Chapter 36

Foundations andearth-retaining walls

361 GEOTECHNICAL DESIGN

Eurocode 7 provides in outline all the requirements for thedesign of geotechnical structures It classifies structures intothree categories according to their complexity and associatedrisk but concentrates on the design of conventional structureswith no exceptional risk These include spread raft and pilefoundations retaining structures bridge piers and abutmentsembankments and tunnels Limit-states of stability strengthand serviceability need to be considered The requirements ofthe ULS and SLS may be satisfied by the following methodsalone or in combination calculations prescriptive measurestesting observational procedures The calculation methodadopted in the United Kingdom for the ULS requires theconsideration of two combinations of partial factors for actionsand soil parameters as shown here

satisfactory if the ratio of design ultimate bearing capacityto service load is 3 This approach is not valid for soft claysand settlement calculations should always be carried out insuch cases

362 PAD BASES

Critical sections for bending are taken at the face of a columnor the centre of a steel stanchion The design moment is takenas that due to all external loads and reactions to one side of thesection Punching resistance should be verified at controlperimeters within 2d from the column periphery For baseswithout shear reinforcement the design shear resistance is

vRd vRdc (2dav) vmin (2dav)

where av ( 2d) is the distance from the column periphery tothe control perimeter being considered The net applied forceVred V V where V is the applied column load and V isthe resulting upward force within the control perimeter Forconcentric loading the punching shear stress is v Vredudwhere u is the length of the control perimeter For eccentricloading the maximum shear stress is Vredud where is amagnification factor determined from equations given in EC 2At the column perimeter the punching shear stress should notexceed vRdmax 02(1 fck250)fck

Normal shear resistance should also be verified on verticalsections at distance d from the column face extending across thefull width of the base where the design shear resistance isvRdc vmin Alternatively it would be reasonable to check atsections within 2d from the column face using the enhanceddesign shear resistance given for punching shear In this casefor concentric loading the critical position for normal shear andpunching shear occurs at av a2 2d where a is the distancefrom the column face to the edge of the base For eccentricloading checks can be made at av 05d d and so on to findthe critical position

If the tension reinforcement is included in the determinationof vRdc the bars should extend for at least (d lbd) beyondthe section considered (see also EC 2 section 9822) If thetension reinforcement is ignored in the shear calculationsstraight bars will usually suffice

Example 1 A base is required to support a 600 mm squarecolumn subjected to vertical load only for which the values

Partial safety factors for the ULS

Safety factor Safety factor for

Combination on actions F soil parameters M

G Q c cu

1 135 15 10 10 102 10 13 125 125 14

If the action is favourable to the effect being considered values of G 10and Q 0 should be used Subscripts refer to the following soil parameters

is angle of shearing resistance (in terms of effective stress) and factor

is applied to tanc is cohesion intercept (in terms of effective stress)cu is undrained shear strength

Generally combination 2 determines the size of the structurewith regard to overall stability bearing capacity sliding andsettlement and combination 1 governs the structural design ofthe members The required size of spread foundations may bedetermined by ULS calculations using soil parameter valuesderived from the geotechnical design report for the projectAlternatively the size may be determined by limiting the bearingpressure under the characteristic loads to a prescribed value ora calculated allowable bearing pressure may be used For theSLS the settlement of spread foundations should be checked bycalculation or may in the case of firm to stiff clays be taken as

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are 4250 kN (service) and 6000 kN (ultimate) The allowableground bearing value is 300 kNm2 (kPa)

fck 32 MPa fyk 500 MPa nominal cover 50 mm

Allowing 10 kNm2 for ground floor loading and extra over soildisplaced by concrete the net allowable bearing pressure can betaken as 290 kNm2 Area of base required

Abase 4250290 147 m2 Provide base 40 m square

Distance from face of column to edge of base a 1700 mm

Taking depth of base 05a say h 900 mm

Allowing for 25 mm main bars average effective depth

d 900 (50 25) 825 mm

Bearing pressure under base due to ultimate load on column

pu 600042 375 kNm2

Bending moment on base at face of column

M pula2 2 375 4 1722 2168 kNm

Mbd2 fck 2168 106 (4000 8252 32) 0025

From Table 48 As fykbdfck 00285 and

As 00285 4000 825 32500 6020 mm2

From Table 220

13H25-300 gives 6380 mm2 and

100Asbh 100 6380(4000 900) 018 ( 013 min)

For values of Gk 2500 kN Qk 1750 kN and 2 03 thequasi-permanent load is

Gk 2Qk 2500 03 1750 3025 kN

Hence the stress in the reinforcement under quasi-permanentloading is given approximately by

s (30256000)(087fyk)(As reqAs prov)

(30256000)(087 500)(60206380) 207 MPa

From Table 424 for wk 03 mm the crack width criterion ismet if

s 23 mm or the bar spacing 240 mm

The adjusted maximum bar size with hcr 05h is given by

s s ( fcteff 29)[kc hcr2(h d )]

23 (3029) [04 450(2 75)] 28 mm

In this case H25-300 satisfies the bar size criterion

For members without shear reinforcement distance from faceof column to critical position for punching shear (or normalshear) is given by av 05a 850 mm where perimeter

u 4c 2av 4 600 2 850 7740 mm

Area of base inside critical perimeter is

Au 4avc av2 4 085 06 0852 431 m2

Hence the net applied force and resulting shear stress are asfollows

Vred V V 375 (42 431) 4384 kN

v Vud 4384 103(7740 825) 069 MPa

From Table 417 for fck 32 MPa and 100Aslbd 020 thedesign concrete shear stress vRdc is determined by vmin

vmin 0035k32 fck12 where k 1 (200d)12 20

k 1 (200825)12 149

vmin 0035 14932 3212 036 MPa

Hence design shear resistance

vRd vmin (2dav) 036 (2 825850) 070 MPa (v)

At the column perimeter ignoring the small reduction V

v Vud 6000 103(4 600 825) 303 MPa

vRd max 02 (1 fck250)fck

02 (1 32250) 32 558 MPa (v)

363 PILE-CAPS

A pile-cap may be designed by either bending theory or trussanalogy (ie strut-and-tie) In the latter case the truss is of atriangulated form with nodes at the centre of the loaded areaand at the intersections of the centrelines of the piles with thetension reinforcement as shown for compact groups of two tofive piles in Table 361 Expressions for the tensile forces aregiven taking into account the dimensions of the column andalso simplified expressions when the column dimensions areignored Bars to resist the tensile forces are to be located withinzones extending not more than 15 times the pile diameter eitherside of the centre of the pile The bars are to be provided witha tension anchorage beyond the centres of the piles Thecompression caused by the pile reaction may be assumedto spread at 45o angles from the edge of the pile and taken intoaccount when calculating the anchorage length The bearingstress on the concrete inside the bend in the bars should bechecked (see Table 431)

Example 2 A pile-cap is required for a group of 4 450 mmdiameter piles arranged at 1350 mm centres on a square gridThe pile-cap supports a 450 mm square column subjected to anultimate design load of 4000 kN

fck 32 MPa fyk 500 MPa

Allowing for an overhang of 150 mm beyond the face of thepile size of pile-cap 1350 450 300 2100 mm square

Take depth of pile-cap as (2hp 100) 1000 mm

Assuming tension reinforcement to be 100 mm up from base ofpile-cap d 1000 100 900 mm

Using truss analogy with the apex of the truss at the centre ofthe column the tensile force between adjacent piles is

Ft 750 kN in each zone

As Ft 087fyk 750 103(087 500) 1724 mm2

For a pile spacing three times pile diameter the bars may bespread uniformly across the cap and a total for two ties of8H25-250 (giving 3926 mm2) in each direction can be used

100Asbd 100 3926(2100 900) 020 ( 013 min)

4000 13508 900

Nl8d

Pile-caps 391

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The bars should be provided with a tension anchorage beyondthe centre of the piles (see Table 430) For 25 mm diameterbars spaced at 250 mm centres half the gap between bars givescb 1125 mm (ie 45 ) So for bent bars and lsquogoodrsquo bondconditions lbd 1lbrqd 07 35 25

Taking ab as either half the centre-to-centre distance betweenbars or (side cover plus half bar size) whichever is less

ab 2502 175 125 mm ab 12525 5

From Table 431 minimum radius of bend rmin 7 say

Consider the critical section for shear to be located at 20 ofthe pile diameter inside the pile-cap Distance of this sectionfrom the column face

av 05(l c) 03hp

05 (1350 450) 03 450 315 mm

Length of corresponding square perimeter for punching shear

u 4 (l 06hp) 4 (1350 06 450) 4320 mm

Since length of perimeter of pile-cap 4 2100 8400 mmis less than 2u normal shear extending across the full width ofthe pile-cap is more critical than punching shear

The contribution of the column load to the shear force may bereduced by applying a factor av2d where 05d av 2dSince avd 315900 035 (05) 025

v Vbd 025 2000 103(2100 900) 027 MPa

From Table 417 for fck 32 MPa and 100Aslbd 020 thedesign concrete shear stress vRdc is determined by vmin

vmin 0035k32fck12 where k 1 (200d)12 20

k 1 (200900)12 147

vmin 0035 14732 3212 035 MPa (v)

Shear stress calculated at perimeter of column

v Vud 4000 103(4 450 900) 247 MPa

vRd max 02 (1 fck250) fck

02 (1 32250) 32 558 MPa (v)

364 RETAINING WALLS ON SPREAD BASES

General notes on walls on spread bases are given in section732 For design purposes the characteristic soil parameterwhich is defined as a cautious estimate of the value affecting theoccurrence of the limit-state is divided by a partial safety factor(see section 361) Design values of the soil strength at the ULS(combination 2) are given by

and

where and are characteristic values of cohesion interceptand angle of shearing resistance (in terms of effective stress)

Design values for shear resistance at the interface of the baseand sub-soil of friction (for drained conditions) and adhesion(for undrained conditions) are given by

tan tan for cast in-situ concrete

tan tan (23) for precast concrete

cud cu14 where cu is undrained shear strength

dd

dd

c

cd c 125tan d ( tan ) 125

Walls should be checked for ULS of overall stability bearingresistance and sliding The resistance of the ground should bedetermined for both long-term (ie drained) and short-term(ie undrained) conditions where appropriate

For eccentric loading the ground bearing is assumed to beuniformly distributed and coincident with the line of action ofthe resultant applied load The traditional practice of usingcharacteristic actions and allowable bearing pressures to limitground deformation and check bearing resistance may also beadopted by mutual agreement This approach assumes a linearvariation of bearing pressure for eccentric loading and it is stillnecessary to consider the ULS for the structural design and tocheck sliding

The partial safety factors for the SLS are given as unity butit is often prudent to use the ULS for the active force (as inBS 8002) In this case suitable dimensions for the wall base canbe estimated with the aid of the chart given in Table 286 Herethe value pmax applies for a linear pressure variation and if theground pressure is uniform and centred on the centre of gravityof the applied load the contact length is (l) where dependson whether the solution is (a) above or (b) below the curve forlsquozero pressure at heelrsquo shown on the chart as follows

(a) 4(1 )3 23 and p 075 pmax

(b) 43 3(1 ) 23 and p (1 )l

For sliding the chart applies directly to non-cohesive soilsThus for bases founded on clay the long-term condition canbe investigated by using with c 0 For the short-termcondition the ratio does not enter into the calculations forsliding and is given by KA l2cd When has beendetermined from this equation the curve for radicKA on the chartcan be used to check the values of and that were obtainedfor the long-term condition

Example 3 A cantilever retaining wall on a spread base isrequired to support level ground and a footpath adjacent to aroad The existing ground may be excavated as necessary toconstruct the wall and the excavated ground behind the wallis to be reinstated by backfilling with a granular material Agraded drainage material will be provided behind the wall withan adequate drainage system at the bottom

Height of fill to be retained 40 m above top of base

Surcharge 5 kNm2

Properties of retained soil (well graded sand and gravel)unit weight of soil 20 kNm3

35o tan1 [(tan 35o)125] 29o

KA (1 sin )(1 sin ) 035

Properties of sub-base soil (medium sand)allowable bearing value fmax 200 kNm2 (kPa)

35o 29o (as fill)tan d tan 055

Take thickness of both wall (at bottom of stem) and base to beequal to (height of fill)10 400010 400 mm

Height of wall to underside of base l 40 04 44 m

Allowing for surcharge equivalent height of wall

44 520 465 mle l q

d

d

dd

d

Foundations and earth-retaining walls392

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200(20 465) 215

radicKA 055radic035 093

From Table 286 radicKA 0745 038 Hence

Width of base le (0745radic035) 465 21 m say

Toe projection (le) 038 21 08 m

Since the solution lies above the curve for lsquozero pressure atheelrsquo on the chart for uniform bearing centred on the centre ofgravity of the applied load

4(1 )3 4 (1 038)(3 215) 0385 andp 075 pmax 075 200 150 kNm2 (kPa)

Example 4 The sub-base for the wall described in example 3is a clay soil with properties as given below All other values areas specified in example 3

Properties of sub-base soil (firm clay)

allowable bearing value pmax 100 kNm2 (kPa)cu 50 kNm2 cud cu14 5014 35 kNm2

25o tan1 [(tan 25o)125] 205o

tan d tan 037

For the long-term condition

pmaxle 100(20 465) 108

radicKA 037radic035 0625

From Table 286 radicKA 107 025 Hence

Width of base le (107radic035) 465 30 m say

Toe projection (le) 025 30 08 m say

Since the solution lies below the curve for lsquozero pressure atheelrsquo on the chart for uniform bearing centred on the centre ofgravity of the applied load

43 3(1 ) 43 1083(1 025) 085

p (1 )l (1 025) 20 465085 82 kNm2

For the short-term condition

KA l2cud 035 20 465(2 085 35) 055

radicKA 055radic035 093

Since this value is less than that obtained for the long-termcondition the base dimensions are satisfactory

Example 5 The wall obtained in example 4 a cross sectionthrough which is shown below is to be designed accordingto EC 7

tan d

d

d

tan d

pmax le The vertical loads and bending moments about the front edgeof the base are

Load (kN) Moment (kNm)Surcharge 5 18 90 21 189Backfill 20 18 40 1440 21 3024Wall stem 24 04 40 384 10 384Wall base 24 04 30 288 15 432

Totals Fv 2202 Mv 4029

The horizontal loads and bending moments about the bottom ofthe base are

Load (kN) Moment (kNm)Surcharge 035 5 44 77 442 170Backfill 035 20 4422 678 443 994

Totals Fh 755 Mh 1164

Resultant moment Mnet 4029 1164 2865 kNm

Distance from front edge of base to resultant vertical force

a Mnet Fv 28652202 130 m

Eccentricity of vertical force relative to centreline of base

e 302 130 020 m ( 306 05 m)

Maximum pressure at front of base

pmax (220230)(1 6 02030) 103 kNm2

Minimum pressure at back of base

pmin (220230)(1 6 02030) 44 kNm2

For the ultimate bearing condition a uniform distribution isconsidered of length lb 2a 2 13 26 m giving

pu Fvlb 220226 85 kNm2

The ultimate bearing resistance is given by the equation

qu (2 ) cud ic where ic 05[1 ]

ic 05[1 ] 070

qu (2 ) 35 070 126 kNm2 ( pu 85)

Resistance to sliding (long-term)

Fv tan b 2202 037 815 kN ( Fh 755)

Resistance to sliding (short-term)

cud lb 35 26 91 kN ( Fh 755)

Example 6 The structural design of the wall in example 5 isto be in accordance with the requirements of EC 2

fck 32 MPa fyk 500 MPa nominal cover 40 mm

Allowing for H16 bars with 40 mm cover

d 400 (40 8) 352 mm

For the ULS (combination 1) F 135 for all permanentactions and M 10 Thus

35o and KA (1 sin )(1 sin ) 027

The ultimate bending moment at the bottom of the wall stem

M 135 027 (5 422 20 436) 924 kNmm

ddd

1 755 (35 26)

1 Fh (cudlb)

Retaining walls on spread bases 393

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(Note that for combination 2 F KA 10 035 which is lessthan F KA 135 027 0365 for combination 1)

Mbd 2fck 924 106(1000 3522 32) 0023

From Table 48 As fykbd fck 0026 and

As 0026 1000 352 32500 586 mm2m

From Table 220 H12-150 gives 754 mm2m

The ultimate shear force at the bottom of the wall stem

V 135 027 (5 4 20 422) 656 kNm

From Table 417 vmin 0035k32fck12 where

k 1 (200d)12 1 (200352)12 175 ( 20)

vmin 0035 17532 3212 046 MPa

Vbd 656 103(1000 352) 019 MPa ( vmin)

Since the loads are permanent the stress in the reinforcementunder service loading is given approximately by

s (087fykF)(As reqAs prov)

(087 500135)(586754) 250 MPa

From Table 424 for wk 03 mm the crack width criterion ismet if

s 15 mm or the bar spacing 185 mm

The adjusted maximum bar size with hcr 05h is given by

s s ( fcteff 29)[kc hcr 2(h d)]

15 (3029) [04 200(2 48)] 13 mm

In this case H12-150 meets both requirements

For the wall base the loads and bending moments calculatedfor combination 2 (see example 4) can be modified to suit therevised parameters for combination 1 as follows

Fv 135 2202 2973 kN

Mv 135 4029 5439 kNm

Mh (0365035) 1164 1214 kNm

Resultant moment Mnet 5439 1214 4225 kNm

Distance from front edge of base to resultant vertical force

a Mnet Fv 42252973 142 m

Bearing contact length lb 2a 2 142 284 m

pu Fvlb 2973284 1047 kNm2

Note that values of both pu and qu are greater for combination 1than for combination 2 but 2 is still critical for bearing

Bending moment on base at inside face of wall

M 135 (5 20 4 24 04) 1822

1047 (284 12)22 661 kNm

Shear force on base at inside face of wall

V 135 (5 20 4 24 04) 18

1047 (284 12) 582 kNm

The bending moment and shear force are both less than thevalues at the bottom of the wall stem Thus H12-150 can beused to fit in with the vertical bars in the wall

Foundations and earth-retaining walls394

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Appendix

Mathematicalformulae and data

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Mathematical formulae and data396

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References and furtherreading

1 Institution of Structural EngineersConcrete Society Standardmethod of detailing structural concrete a manual for best practiceLondon The Institution of Structural Engineers 2006 p 188

2 CP2 Civil Engineering Code of Practice No2 Earth retainingstructures London The Institution of Structural Engineers 1951p 224

3 The Highways Agency BD 3701 Loads for highway bridgesDesign manual for roads and bridges London HMSO 2001 p 118

4 The Highways Agency BD 6494 The design of highway bridgesfor vehicle collision loads Design manual for roads and bridgesLondon HMSO 1994 p 13

5 The Highways Agency BD 2101 The assessment of highwaybridges and structures Design manual for roads and bridges LondonHMSO 2001 p 84

6 The Highways Agency BD 5293 The design of highway bridgeparapets Design manual for roads and bridges London HMSO1993 p 44

7 Department of Transport BD 3087 Backfilled retaining walls andbridge abutments London Department of Transport 1987 p 12

8 Caquot A and Kerisel J Tables for calculation of passive pressureactive pressure and bearing capacity of foundations (translated fromFrench by M A Bec London) Paris Gauthier-Villars 1948 p 121

9 The Highways Agency BA 4296 The design of integral bridgesDesign manual for roads and bridges London HMSO 1996 p 10

10 Blackledge G F and Binns R A Concrete practice CrowthorneBritish Cement Association Publication 48037 2002 p 71

11 CIRIA Report 91 Early-age thermal crack control in concreteLondon CIRIA 1981 p 160

12 BRE Digest 357 Shrinkage of natural aggregates in concreteWatford BRE 1991 p 4

13 BRE Special Digest 1 Concrete in aggressive ground Six partsWatford BRE 2005

14 Concrete Society Technical Report No 51 Guidance on the use ofstainless steel reinforcement Slough The Concrete Society 1998 p 55

15 Coates R C Coutie M G and Kong F K Structural analysisSunbury-on-Thames Nelson 1972 p 496

16 Rygol J Structural analysis by direct moment distributionLondon Crosby Lockwood 1968 p 407

17 Westergaard H M Computation of stresses in bridge slabs due towheel loads Public Roads Vol 2 No 1 March 1930 pp 1ndash23

18 Pucher A Influence surfaces for elastic plates Wien andNew York Springer Verlag 1964

19 Bares R Tables for the analysis of plates and beams based onelastic theory Berlin Bauverlag 1969

20 Timoshenko S P and Woinowsky-Krieger S Theory of plates andshells (second edition) New York McGraw-Hill 1959 p 580

21 Sarkar R K Slab design ndash elastic method (plates) Munich VerlagUNI-Druck 1975 p 191

22 Wang P C Numerical and matrix methods in structural mechanicsNew York Wiley 1966 p 426

23 Jones L L and Wood R H Yield-line analysis of slabs LondonThames and Hudson 1967 p 405

This book written by leading UK experts is the best English languagetext dealing with yield-line theory (essential for designers using themethod frequently and for more than lsquostandardrsquo solutions)

24 Johansen K W Yield-line theory London Cement and ConcreteAssociation 1962 p 181

This is an English translation of the original 1943 text on whichyield-line theory is founded

25 Johansen K W Yield-line formulae for slabs London Cementand Concrete Association 1972 p 106

Gives design formulae for virtually every lsquostandardrsquo slab shape andloading (essential for practical design purposes)

26 Wood R H Plastic and elastic design of slabs and platesLondon Thames and Hudson 1961 p 344

Relates collapse and elastic methods of slab analysis but mainlyfrom the viewpoint of research rather than practical design

27 Jones L L Ultimate load analysis of reinforced and prestressedconcrete structures London Chatto and Windus 1962 p 248

About half of this easily readable book deals with the yield-linemethod describing in detail the analysis of several lsquostandardrsquo slabs

28 Pannell F N Yield-line analysis Concrete and ConstructionalEngineering JunendashNov 1966

Basic application of virtual-work methods in slab design June1966 pp 209ndash216

Economical distribution of reinforcement in rectangular slabs July1966 pp 229ndash233

Edge conditions in flat plates Aug 1966 pp 290ndash294

General principle of superposition in the design of rigid-plasticplates Sept 1966 pp 323ndash326

Design of rectangular plates with banded orthotropic reinforcementOct 1966 pp 371ndash376

Non-rectangular slabs with orthotropic reinforcement Nov 1966pp 383ndash390

29 Hillerborg A Strip method of design London Viewpoint 1975p 225

This book is the English translation of the basic text on the stripmethod (both simple and advanced) by its originator It dealswith theory and gives appropriate design formulae for manyproblems

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References and further reading398

30 Fernando J S and Kemp K O A generalised strip deflexionmethod of reinforced concrete slab design Proceedings of theInstitution of Civil Engineers Part 2 Research and Theory March1978 pp 163ndash174

31 Taylor R Hayes B and Mohamedbhai G T G Coefficientsfor the design of slabs by the yield-line theory Concrete 3(5) 1969pp 171ndash172

32 Munshi J A Rectangular concrete tanks (revised fifth edition)Skokie Illinois Portland Cement Association 1998 p 188

This is the most detailed book on the subject with complete tablesgiving moments shears and deflections for plates and tanks withuseful worked examples

33 CIRIA Report 110 Design of reinforced concrete flat slabs toBS8110 London CIRIA 1985 p 48

34 Beeby A W The analysis of beams in plane frames according toCP110 London Cement and Concrete Association Publication44001 1978 p 34

35 Rygol J Structural analysis by direct moment distributionLondon Crosby Lockwood 1968 p 407

36 Naylor N Side-sway in symmetrical building frames TheStructural Engineer 28(4) 1950 pp 99ndash102

37 Orton A The way we build now form scale and techniqueLondon E amp FN Spon 1988 p 530

38 CIRIA Report 102 Design of shear wall buildings LondonCIRIA 1984 p 80

39 Eurocode 8 Design of structures for earthquake resistanceBrussels European Committee for Standardization 2004

40 Penelis G G and Kappos A J Earthquake-resistant concretestructures London E amp FN Spon 1997 p 572

41 Kruger H G Crack width calculation to BS 8007 for combinedflexure and direct tension The Structural Engineer 80(18) 2002pp 18ndash22

42 Kong F K Robins P J and Sharp G R The design of reinforcedconcrete deep beams in current practice The Structural Engineer53(4) 1975 pp 73ndash80

43 Ove Arup and Partners The design of deep beams in reinforcedconcrete CIRIA Guide No 2 1977 p 131

44 Concrete Society Technical Report No 42 Trough and wafflefloors Slough The Concrete Society 1992 p 34

45 Gibson J E The design of shell roofs (Third edition) LondonE amp FN Spon 1968 p 300

46 Chronowicz A The design of shells London Crosby Lockwood1959 p 202

47 Tottenham H A A simplified method of design for cylindricalshell roofs The Structural Engineer 32(6) 1954 pp 161ndash180

48 Bennett J D Empirical design of symmetrical cylindricalshells Proceedings of the colloquium on simplified calculationmethods Brussels 1961 Amsterdam North-Holland 1962pp 314ndash332

49 Salvadori and Levy Structural design in architecture EnglewoodCliffs Prentice-Hall 1967 p 457

50 Schulz M and Chedraui M Tables for circularly curved horizontalbeams with symmetrical uniform loads Journal of the AmericanConcrete Institute 28(11) 1957 pp 1033ndash1040

51 Spyropoulos P J Circularly curved beams transversely loadedJournal of the American Concrete Institute 60(10) 1963 pp 1457ndash1469

52 Concrete SocietyCBDG An introduction to concrete bridgesCamberley The Concrete Society 2006 p 32

53 Leonhardt Fritz Bridges Stuttgart Deutsche Verlags-Anstalt1982 p 308

54 Hambly E C Bridge deck behaviour (Second edition) LondonE amp FN Spon 1991 p 313

55 PCA Circular concrete tanks without prestressing SkokieIllinois Portland Cement Association p 54

56 Ghali A Circular storage tanks and silos London E amp FN Spon1979 p 210

57 CIRIA Reports 139 and 140 (Summary Report) Water-resistingbasement construction London CIRIA 1995 p 192 p 64

58 Irish K and Walker W P Foundations for reciprocating machinesLondon Cement and Concrete Association 1969 p 103

59 Barkan D D Dynamics of bases and foundations New YorkMcGraw Hill 1962 p 434

60 Tomlinson M J Pile design and construction practice LondonCement and Concrete Association 1977 p 413

61 Concrete Society Technical Report No 34 Concrete industrialground floors (Third edition) Crowthorne The Concrete Society2003 p 146

62 CIRIA Report 104 Design of retaining walls embedded in stiffclay London CIRIA 1984 p 146

63 Hairsine R C A design chart for determining the optimum baseproportions of free standing retaining walls Proceedings of theInstitution of Civil Engineers 51 (February) 1972 pp 295ndash318

64 Cusens A R and Kuang Jing-Gwo A simplified method ofanalysing free-standing stairs Concrete and ConstructionalEngineering 60(5) 1965 pp 167ndash172 and 194

65 Cusens A R Analysis of slabless stairs Concrete andConstructional Engineering 61(10) 1966 pp 359ndash364

66 Santathadaporn Sakda and Cusens A R Charts for the design ofhelical stairs Concrete and Constructional Engineering 61(2) 1966pp 46ndash54

67 Terrington J S and Turner L Design of non-planar roofsLondon Concrete Publications 1964 p 108

68 Krishna J and Jain O P The beam strength of reinforced concretecylindrical shells Civil Engineering and Public Works Review49(578) 1954 pp 838ndash840 and 49(579) 1954 pp 953ndash956

69 Faber C Candela the shell builder London The ArchitecturalPress 1963 p 240

70 Bennett J D Structural possibilities of hyperbolic paraboloidsLondon Reinforced Concrete Association February 1961 p 25

71 Lee D J Bridge bearings and expansion joints (Second edition)London E amp FN Spon 1994 p 212

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Numbers preceded by lsquotrsquo are Table Numbers

Actions see LoadsAdmixtures 18ndash19Affinity theorems 140Aggregates 16ndash17

size and grading 95 t217Anchorage bond see BondAnnular sections 236 t2107Arches

fixed 41ndash2 t272ndash4load effects in 178ndash9 t272

parabolic 42 179ndash82t274

thickness 175 t272three-hinged 41 175

t271two-hinged 41 175 t271see also Bridges

Areas 52ndash3 t2101

Barriers and balustrades 7Bars

anchorage 51 312 381t355 t359 t430 t432

bending schedules t223bends in 25 51 312 381

t219 t355 t359 t431considerations affecting

design details 51 312381 t353 t359 t428

curtailment 52 312 381t356ndash8 t432

cutting and bendingtolerances 100

lap lengths 51 312 381t355 t359 t431ndash2

shapes and dimensions25ndash6 100 t221ndash2

sizes 25 95 t220types 24ndash5 95see also Reinforcement

Basements 65Bases see FoundationsBeams

cantilevers see Cantileverscontinuous see Continuous

beamscurved 57 216 t295ndash7

concentrated load 216t295

uniform load 218 t296ndash7

deep 52

doubly reinforced sections257 346 t315ndash16t325ndash6 t49ndash10

fixed at both ends 105 t225t228

flanged see Flangedsections

imposed loads on 6 t23junctions with columns 330

t363single-span 29 105 t224ndash5singly reinforced sections

256 346 t313ndash14t323ndash4 t47ndash8

sizes and proportions 46supporting rectangular

panels 34 144 t252Bearings 62 221 t299

see also DetailsFoundations

Bending (alone or combinedwith axial force) 44ndash8

assumptions 44ndash5 t36 t44resistance of sections

beams 45ndash6columns 47ndash8slabs 46ndash7

see also individual membersBending moments

combined bases 195 t283continuous beams see

Continuous beamscylindrical tanks 60 183ndash8

t275ndash7flat slabs 35ndash6 150ndash3

t255ndash6rectangular tanks 60ndash1 188

t278ndash9silos 61ndash2 191 t280see also Beams Cantilevers

Structural analysisBiaxial bending see ColumnsBlinding layer 64Bond 51 312 381

anchorage lengths see Barsbends in bars see Barslap lengths see Bars

Bow girders see Beamscurved

Bridges 57ndash9deck 57ndash8 t298design considerations 59imposed loads

foot 8 78 t26railway 8ndash9 78 t26road 7ndash8 78 t25

integral 59partial safety factors 239

t32ndash3roofs see Roofsstairs see Stairssubstructures 58types 57ndash8 t298waterproofing 59wind loads 10see also Arches

Buildings 54ndash6dead loads 75 t22imposed loads

floors 6 75ndash8 t23roofs 7 78 t24

load-bearing walls see Wallsrobustness and provision of

ties 54ndash5 t354 t429wall and frame systems

40ndash1 t268wind loads 10 78 t27ndash9see also Floors

Foundations StairsWalls

Bunkers see Silos

Cantilevers 29 t226ndash7deflections 295 371

Cements and combinations14ndash16 95 t217

Characteristic loads see Loads

Characteristic strengthsconcrete t35 t42reinforcement t219

Columnsbiaxial bending t321 t331

t416circular 264 353 t319ndash20

t329ndash30 t413ndash14cylindrical (modular ratio)

236 t2107effective height t321 t331effective length t415elastic analysis of section

226 t2104 t2108ndash9imposed loads on 7 t23junctions with beams 330

t363loads and sizes 48rectangular 264 353

t317ndash18 t327ndash8t411ndash12

rectangular (modular ratio)t2105ndash6

short 47 265

slenderdesign procedure 263

352ndash3BS 8110 t321ndash2BS 5400 t331ndash2EC 2 t415ndash16

supporting elevated tanks61 194

see also Framed structuresConcentrated loads

analysis of memberscurved beams 216 t295solid slabs 31 34 131

t245ndash7bridges 7ndash9 78 t25ndash6dispersal of 9on floors 6 78 t23shear 49 285 365 t334

t337ndash8 t419Concrete 14ndash24 95

admixtures 18ndash19aggregates 16ndash17 95

t217alkali-silica attack 23carbonation 22ndash3cements 14ndash16 95

t217chemical attack 23compressive strength 21

245 338 t35 t42creep 21ndash2 245 338 t35

t43design strengths 239 335durability 22ndash4

cover to reinforcement 24t38ndash9 t46

exposure classes 23ndash4t37 t39 t45

early-age temperatures andcracking 20 95 t218

elastic properties 21 245338 t35 t42

fibre-reinforced 68fire resistance see Fire

resistancefreezethaw attack 23plastic cracking 19ndash20shrinkage 22 245 338 t35

t42specification 24stress-strain curves 22 t36

t44tensile strength 21 t344

t423thermal properties 22 245

338 t35 t42

Index

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Concrete (Continued)water 17weight 75 t21workability 19

Construction materialsweight 75 t21ndash2

Contained materials seeRetained materials

Containment structures 59ndash61see also Silos Tanks

Continuous beams 29ndash30arrangement of design loads

111 t229equal spans and loads 30

111 t229ndash32 t234ndash5influence lines for bending

moments 121 t238ndash41methods of analysis 29

moment distribution 29121 t236

moving loads 30redistribution of moments

see Moment redistributionsecond moment of area 29unequal spans and loads

t237Corbels 327 330ndash1 t362Cover to reinforcement 24

t38ndash9 t46Cracking 50ndash1 295 300 371

calculation procedures t343t424

crack width limits 295 371deemed-to-satisfy rules

t343 t424ndash7liquid-retaining structures

300 371 t344ndash52t425ndash7

minimum reinforcement t423

Cranes 7 13Creep see ConcreteCulverts 71

box culverts 71 t287pipe culverts 71subways 71

Curtailment see BarsCurvature see DeflectionCurved beams see BeamsCylindrical tanks see Tanks

Dead loads 6 75concrete 75 t21construction materials 75

t21ndash2partitions 75 t22

Deep beams 52Deep containers see SilosDeflection 49ndash50 295 371

calculation procedurest340ndash2 t421ndash2

cantilevers 295 371curvatures t341 t422deflection limits 295 371formulae for

beams t224ndash5cantilevers t226ndash7

spaneffective depth ratios295 371 t340 t421

Design of structural members44ndash53

see also individual members(eg Arches BeamsColumns Slabs StairsWalls)

Design principles and criteria 5 44

Design strengths seeConcrete Reinforcement

Detailscontinuous nibs 327 t362

corbels 327 t362corners and intersections

330 t363curtailment 52 312 381

t356 t432rules for beams t357rules for slabs t358

halving joints 330 t362Docks and dolphins see

Maritime structuresDomes see RoofsDrawings 4

Earth-retaining wallsembedded (or sheet) 70movement joints 221 t2100pressures behind 11ndash12 86

90 t210ndash14on spread bases 69ndash70

203 324 392 t286types 69 t286see also Retained materials

Earthquake-resistant structures 43

Economical structures 3Elastic analysis 52ndash3 226 236

biaxial bending andcompression t2109

design charts t2105ndash7properties of sections

t2101ndash3 t342uniaxial bending and

compression t2104uniaxial bending and tension

t2108Embedded walls see

Earth-retaining wallsEurocode loading standards 13Exposure classes 23ndash4 t37

t39 t45

Fabric 95 100 t220Fibre-reinforced concrete 68Fill materials 12Finite elements 38Fire resistance 27 249 342

cover to reinforcement 249t310ndash11

minimum fire periods t312Fixed-end moment coefficients

105 t228Flanged sections 46

effective flange width 262 349

elastic properties t342Flat slabs see SlabsFloors 55

forms of construction t242imposed loads 6 t23industrial ground see

Industrial ground floorsopenings in 55 t337weights of concrete t21

Footbridgesimposed loads 78 t26

Formwork 4Foundations 63ndash7

balanced and coupled bases64 199 t283ndash4

basements 65bearing pressures 63 t282blinding layer 64combined bases 64 195

t283eccentric loads 63imposed loads 7 t23for machines 66piers 65piled see Piled foundationsrafts 65 199 t284separate bases 64 t282

site inspection 63strip bases 65 195 t283types 64 t282ndash4wall footings 66 t283

Framed structures 36ndash8building code requirements

36 t257 t262columns in

non-sway frames 38ndash9t260ndash1

sway frames 39ndash40 t262continuous beams in 159effect of lateral loads 39ndash40

162 t262finite element method 38moment distribution method

no sway 37 t258with sway 37 t259

portal frames 38 162rigid joints t263ndash6hinged joints t267

properties of membersend conditions 42section properties 42ndash3

shear forces on members 37slope-deflection method

of analysis 37 154t260ndash2

see also Columns

Garages 6Geometric properties of

uniform sections t2101Ground water 86 90Gyration radius of 52 t2101

t415

Hillerborgrsquos strip method 33144 t251 t254

Hinges 62 221 t299Hoppers 12 62 90 194

t215ndash16 t281

Imposed loads 6ndash9 t23ndash6barriers and parapets 7bridges see Bridgesbuildings 6ndash7 75 t23floors 6 t23reduction on beams and

columns 7 78 t23roofs 7 78 t24structures subject to dynamic

loads 6structures supporting

cranes 7structures supporting

lifts 7underground tanks 60see also Eurocode loading

standardsIndustrial ground floors 67ndash9

construction methods 67ndash8methods of analysis 68ndash9modulus of subgrade

reaction 68reinforcement 68

Intersections 330 t363see also Joints

Janssenrsquos theory 12 90t215ndash16

Jetties see Maritime structures

Joints 52 330industrial ground

floors 67ndash8liquid-retaining structures

300 t345movement 62 221 t2100see also Bearings

Details

Lifts 7Limit state design 5

British codesbridges 239 241 t32ndash3buildings 239 t31liquid-retaining structures

241 t34loads 5ndash6 239properties of materials

concrete 245 t35ndash6reinforcement 245 t36

European codesactions 5ndash6 335buildings 335ndash6 t41containers 335ndash6properties of materials 335

concrete 338 t42ndash4reinforcement 338 t44

Liquid-retaining structures 241335ndash6 t34

see also Cracking JointsLoads 6ndash10

on bridges see Bridgesconcentrated see

Concentrated loadsdead see Dead loadsdynamic 6eccentric on foundations 63

t283imposed see Imposed loadson lintels t22moving loads on continuous

beams 30on piles see Piled

foundationswind see Wind loads

Maritime structures 10ndash11piled jetties 200 t285

Materials see AdmixturesAggregates Cementsand combinationsConcrete Reinforcement

Mathematical formulae 395ndash6Members see individual

members (eg ArchesBeams Columns SlabsStairs Walls)

Modular-ratio design seeElastic analysis

Modulus of subgrade reaction 68

Moment distribution 29continuous beams 121 t236framed structures 154

t258ndash9Moment redistribution 30 116

code requirements 116design procedure 117 t233moment diagrams for equal

spans t234ndash5

Neutral axis 44ndash5

Parapets 7Partial safety factors see

Safety factorsPartition loads 75 t22Passive pressures 90

t213ndash14Piers

bridges 58foundations 65see also Maritime structures

Piled foundations 66ndash7open-piled structures 67

200 t285pile-caps 66 324 t361piles in a group 67

Poissonrsquos ratio 131 137t246ndash7 t35 t42

Index400

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Precast concretebridge decks 58floors weights of t21

Pressures 11ndash13on earth-retaining walls 86

90 t211ndash14in silos 90 t215ndash16in tanks 90see also Wind loads

Properties of sections 42ndash3plain concrete t2101reinforced concrete t2102ndash3

Rafts 65 199 t284Railway bridges see BridgesReinforcement 24ndash7 95 100

bars 24 see also Barsfabric 25 see also Fabricfixing of 27mechanical and physical

properties t219prefabricated systems 26ndash7stainless steel 26stress-strain curves 25 t36

t44Reservoirs see TanksRetained materials

cohesionless soils 12 8690 t212ndash14

cohesive soils 12 90fill materials 12lateral pressures 11ndash12

t211liquids 11 86 90soil properties 11 t210see also Silos Tanks

Retaining walls seeEarth-retaining walls

Robustness 54ndash5ties 312 381 t354 t429

Roofsloads on 7 78 t24non-planar 56 212

cylindrical shells 56 212216 t292ndash4

domes 212 t292hyperbolic-paraboloidal

shells 216 t292prismatic 212 t292shell buckling 56ndash7

planar 56weights of t22

Safety factors 5British codes 239 t31ndash4

geotechnical design 324European codes 335 t41

geotechnical design 390Sea-walls see Maritime

structuresSection moduli t2101Serviceability limit states 5

295 300 336 371t31ndash4 t41

Shearin bases 285 322 390forces see individual

members (eg BeamsSlabs)

resistance with shearreinforcement 49 283362 t333 t336 t418

resistance without shearreinforcement 48 283362 t333 t336 t417

stress 283under concentrated loads 49

285 365 t334 t337ndash8t419

see also Structural analysisShear wall structures 40ndash1

arrangement of walls 40169 t269

interaction of walls andframes 41 169 173

walls containing openings40 169 t270

walls without openings 40169 t269

Sheet walls seeEarth-retaining walls

Shrinkagefixed parabolic arch 179 182see also Concrete

Silos 12ndash13 61ndash2 90hopper bottoms 62 t281stored material properties

and pressures 90t215ndash16

substructure 39 t262walls spanning horizontally

61ndash2 t280Slabs

flat 35ndash6reservoir roofs 153simplified method of

design 150 153t255ndash6

imposed loads 75 78 t23ndash4non-rectangular panels 34ndash5

131 t248one-way 31 128

concentrated loads 31131 t245

uniform load distribution31 128 t242

openings in 55 t337rectangular panels

concentrated loads 34131 137 t246ndash7

triangular load distribution33ndash4 147 t253ndash4

uniform load distribution33 128 131 t242ndash4

thickness of 46ndash7two-way 31ndash4 128 131

collapse methods 32ndash3elastic methods 32

strip 144 t251 t254yield-line 137 139ndash42

147 150 t249ndash50t254

types of 55 t242weights of t21

Slope-deflection method 37 154 t260

Soils see Retained materialsStairs 55ndash6 206 208 212

free-standing 206 t288helical 208 212 t289ndash91sawtooth 206 208 t289simple flights 206types and dimensions t288

Stored materials see SilosStress-strain curves

concrete 22 t36 t44reinforcement 25 t36 t44

Stresses see individual modes(eg Bond ShearTorsion)

Structural analysis 28ndash43properties of members

end conditions 42section properties 42ndash3

see also individual structures(eg Arches Continuousbeams Frames Shearwalls)

Structures 54ndash71earthquake-resistant 43economical 3see also individual

structures (eg BridgesBuildings FoundationsSilos Tanks)

Subways 71Superposition theorem 140Surcharge 86 203 326

Tankscylindrical 60 183 187ndash8

t275ndash7effects of temperature 61elevated 61 191 t281

substructure 39 t262joints 221 t2100octagonal 60pressure on walls 90rectangular 60ndash1 147 150

188 191 t253ndash4 t278ndash9

underground 60see also Cracking Liquid-

retaining structuresTemperature effects in

concrete at early-age 20 95 t218

fixed parabolic arch 179walls of tanks 61

Tensile strengthconcrete 21 t344 t423reinforcement 95 t219

Thermal properties of concrete22 245 338 t35 t42

Ties see RobustnessTorsion

design procedure 49 285365 t335 t339 t420

moments incurved beams 57 216

218 t295ndash7free-standing stairs 206

t288helical stairs 208 212

t289

Ultimate limit state 5 239 241335ndash6 t31ndash4 t41

Vehicle loads on bridges 7ndash978 t25ndash6

Vibrationfloors 6footbridges 8machine foundations 66

Virtual-work method 139ndash40

Walls 52 57load-bearing 57 322 t360weights of t22see also Earth-retaining

walls Shear walls SilosTanks

Water (for concrete) 17Water-tightness 59

basements 65Weights of

concrete 75 t21construction materials 75

t21ndash2partitions 75 t22roofs t22stored materials t216walls t22

Wharves see MaritimeStructures

Wheel loads 8 t25dispersal of 9

Wind loads 9ndash10 78on bridges 10on buildings 10effect of see Structural

analysiswind speed and pressure 10

t27ndash9

Yield-line analysis 32 137 139ndash42

affinity theorems 140basic concepts 137 139

t249concentrated loads 140corner levers 142 t250superposition theorem 140virtual work method 139

empirical analysis 141ndash2

Index 401

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ReynoldsrsquosReinforcedConcreteDesignerrsquosHandbook

Reynoldsrsquos Reinforced Concrete Designerrsquos Handbook has beencompletely rewritten and updated for this new edition to takeaccount of the numerous developments in design and practiceover the last 20 years These include significant revisions toBritish Standards and Codes of Practice and the introduction ofthe new Eurocodes The principal feature of the Handbook is thecollection of over 200 full-page tables and charts covering allaspects of structural analysis and reinforced concrete designThese together with extensive numerical examples will enableengineers to produce rapid and efficient designs for a large rangeof concrete structures conforming to the requirements of BS 5400BS 8007 BS 8110 and Eurocode 2

Design criteria safety factors loads and material propertiesare explained in the first part of the book Details are then givenof the analysis of structures ranging from single-span beamsand cantilevers to complex multi-bay frames shear walls

arches and containment structures Miscellaneous structuressuch as helical stairs shell roofs and bow girders are alsocovered

A large section of the Handbook presents detailed informationconcerning the design of various types of reinforced concreteelements according to current design methods and their use insuch structures as buildings bridges cylindrical and rectangulartanks silos foundations retaining walls culverts and subwaysAll of the design tables and charts in this section of the Handbookare completely new

This highly regarded work provides in one publication awealth of information presented in a practical and user-friendlyform It is a unique reference source for structural engineersspecialising in reinforced concrete design and will also be ofconsiderable interest to lecturers and students of structuralengineering

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Also available from Taylor amp Francis

Concrete Pavement Design GuidanceG Griffiths et al Hb ISBN 0ndash415ndash25451ndash5

Reinforced Concrete 3rd edP Bhatt et al Hb ISBN 0ndash415ndash30795ndash3

Pb ISBN 0ndash415ndash30796ndash1

Concrete BridgesP Mondorf Hb ISBN 0ndash415ndash39362ndash0

Reinforced amp Prestressed Concrete 4th edS Teng et al Hb ISBN 0ndash415ndash31627ndash8

Pb ISBN 0ndash415ndash31626ndashX

Concrete Mix Design Quality Control and Specification 3rd edK Day Hb ISBN 0ndash415ndash39313ndash2

Examples in Structural AnalysisW McKenzie Hb ISBN 0ndash415ndash37053ndash1

Pb ISBN 0ndash415ndash37054ndashX

Wind Loading of Structures 2nd edJ Holmes Hb ISBN 0ndash415ndash40946ndash2

Information and ordering details

For price availability and ordering visit our website wwwtandfcoukbuiltenvironment

Alternatively our books are available from all good bookshops

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ReynoldsrsquosReinforcedConcreteDesignerrsquosHandbookELEVENTH EDITION

Charles E ReynoldsBSc (Eng) CEng FICE

James C SteedmanBA CEng MICE MIStructE

and

Anthony J ThrelfallBEng DIC

wwwengbookspdfcom

First edition 1932 second edition 1939 third edition 1946 fourth edition 1948revised 1951 further revision 1954 fifth edition 1957 sixth edition 1961revised 1964 seventh edition 1971 revised 1972 eighth edition 1974 reprinted1976 ninth edition 1981 tenth edition 1988reprinted 1991 1994 (twice) 1995 1996 1997 1999 2002 2003

Eleventh edition published 2008by Taylor amp Francis2 Park Square Milton Park Abingdon Oxon OX14 4RN

Simultaneously published in the USA and Canadaby Taylor amp Francis270 Madison Ave New York NY 10016 USA

Taylor amp Francis is an imprint of the Taylor amp Francis Groupan informa business

copy 2008 Taylor and Francis

All rights reserved No part of this book may be reprinted or reproduced or utilised in any form or by any electronic mechanical or other meansnow known or hereafter invented including photocopying and recordingor in any information storage or retrieval system without permission in writing from the publishers

The publisher makes no representation express or implied with regard to the accuracy of the information contained in this book and cannot accept any legal responsibility or liability for any efforts or omissions that may be made

British Library Cataloguing in Publication DataA catalogue record for this book is available from the British Library

Library of Congress Cataloging-in-Publication DataReynolds Charles E (Charles Edward)

Reynoldsrsquos reinforced concrete designers handbook Charles E ReynoldsJames C Steedman and Anthony J Threlfall ndash 11th ed

p cmRev ed of Reinforced concrete designerrsquos handbook Charles E Reynolds

and James C Steedman 1988Includes bibliographical references and index1 Reinforced concrete construction ndash Handbooks manuals etc

I Steedman James C (James Cyril) II Threlfall A J III ReynoldsCharles E (Charles Edward) Reinforced concrete designerrsquos handbookIV Title

TA6832R48 200762418341ndashdc22 2006022625

ISBN10 0ndash419ndash25820ndash5 (hbk)ISBN10 0ndash419ndash25830ndash2 (pbk)ISBN10 0ndash203ndash08775ndash5 (ebk)

ISBN13 978ndash0ndash419ndash25820ndash9 (hbk)ISBN13 978ndash0ndash419ndash25830ndash8 (pbk)ISBN13 978ndash0ndash203ndash08775ndash6 (ebk)

This edition published in the Taylor amp Francis e-Library 2007

ldquoTo purchase your own copy of this or any of Taylor amp Francis or Routledgersquoscollection of thousands of eBooks please go to wwweBookstoretandfcoukrdquo

ISBN 0-203-08775-5 Master e-book ISBN

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List of tables viPreface to the eleventh edition ixThe authors xAcknowledgements xiSymbols and abbreviations xii

Part 1 ndash General information 11 Introduction 32 Design criteria safety factors and loads 53 Material properties 144 Structural analysis 285 Design of structural members 446 Buildings bridges and containment structures 547 Foundations ground slabs retaining walls

culverts and subways 63

Part 2 ndash Loads materials and structures 738 Loads 759 Pressures due to retained materials 86

10 Concrete and reinforcement 9511 Cantilevers and single-span beams 10512 Continuous beams 11113 Slabs 12814 Framed structures 15415 Shear wall structures 16916 Arches 17517 Containment structures 18318 Foundations and retaining walls 195

19 Miscellaneous structures and details 20620 Elastic analysis of concrete sections 226

Part 3 ndash Design to British Codes 23721 Design requirements and safety factors 23922 Properties of materials 24523 Durability and fire-resistance 24924 Bending and axial force 25625 Shear and torsion 28326 Deflection and cracking 29527 Considerations affecting design details 31228 Miscellaneous members and details 322

Part 4 ndash Design to European Codes 33329 Design requirements and safety factors 33530 Properties of materials 33831 Durability and fire-resistance 34232 Bending and axial force 34533 Shear and torsion 36234 Deflection and cracking 37135 Considerations affecting design details 38136 Foundations and earth-retaining walls 390

Appendix Mathematical formulae and data 395

References and further reading 397

Index 399

Contents

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21 Weights of construction materials and concrete

floor slabs22 Weights of roofs and walls23 Imposed loads on floors of buildings24 Imposed loads on roofs of buildings25 Imposed loads on bridges ndash 126 Imposed loads on bridges ndash 227 Wind speeds (standard method of design)28 Wind pressures and forces (standard method

of design)29 Pressure coefficients and size effect factors

for rectangular buildings210 Properties of soils211 Earth pressure distributions on rigid walls212 Active earth pressure coefficients213 Passive earth pressure coefficients ndash 1214 Passive earth pressure coefficients ndash 2215 Silos ndash 1216 Silos ndash 2217 Concrete cements and aggregate grading218 Concrete early-age temperatures219 Reinforcement general properties220 Reinforcement cross-sectional areas of bars

and fabric221 Reinforcement standard bar shapes and method of

measurement ndash 1222 Reinforcement standard bar shapes and method of

measurement ndash 2223 Reinforcement typical bar schedule224 Moments shears deflections general case for beams225 Moments shears deflections special cases for beams226 Moments shears deflections general cases for

cantilevers227 Moments shears deflections special cases for

cantilevers228 Fixed-end moment coefficients general data229 Continuous beams general data230 Continuous beams moments from equal loads on

equal spans ndash 1231 Continuous beams moments from equal loads on

equal spans ndash 2232 Continuous beams shears from equal loads on

equal spans233 Continuous beams moment redistribution234 Continuous beams bending moment diagrams ndash 1

235 Continuous beams bending moment diagrams ndash 2236 Continuous beams moment distribution methods237 Continuous beams unequal prismatic spans and loads238 Continuous beams influence lines for two spans239 Continuous beams influence lines for three spans240 Continuous beams influence lines for four spans241 Continuous beams influence lines for five or

more spans242 Slabs general data243 Two-way slabs uniformly loaded rectangular panels

(BS 8110 method)244 Two-way slabs uniformly loaded rectangular panels

(elastic analysis)245 One-way slabs concentrated loads246 Two-way slabs rectangular panel with concentric

concentrated load ndash 1247 Two-way slabs rectangular panel with concentric

concentrated load ndash 2248 Two-way slabs non-rectangular panels (elastic

analysis)249 Two-way slabs yield-line theory general information250 Two-way slabs yield-line theory corner levers251 Two-way slabs Hillerborgrsquos simple strip theory252 Two-way slabs rectangular panels loads on beams

(common values)253 Two-way slabs triangularly distributed load (elastic

analysis)254 Two-way slabs triangularly distributed load (collapse

method)255 Flat slabs BS 8110 simplified method ndash 1256 Flat slabs BS 8110 simplified method ndash 2257 Frame analysis general data258 Frame analysis moment-distribution method

no sway259 Frame analysis moment-distribution method

with sway260 Frame analysis slope-deflection data261 Frame analysis simplified sub-frames262 Frame analysis effects of lateral loads263 Rectangular frames general cases264 Gable frames general cases265 Rectangular frames special cases266 Gable frames special cases267 Three-hinged portal frames268 Structural forms for multi-storey buildings

List of tables

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List of tables vii

269 Shear wall layout and lateral load allocation270 Analysis of pierced shear walls271 Arches three-hinged and two-hinged arches272 Arches fixed-ended arches273 Arches computation chart for symmetrical

fixed-ended arch274 Arches fixed-ended parabolic arches275 Cylindrical tanks elastic analysis ndash 1276 Cylindrical tanks elastic analysis ndash 2277 Cylindrical tanks elastic analysis ndash 3278 Rectangular tanks triangularly distributed load

(elastic analysis) ndash 1279 Rectangular tanks triangularly distributed load

(elastic analysis) ndash 2280 Rectangular containers spanning horizontally

moments in walls281 Bottoms of elevated tanks and silos282 Foundations presumed allowable bearing values

and separate bases283 Foundations other bases and footings284 Foundations inter-connected bases and rafts285 Foundations loads on open-piled structures286 Retaining walls287 Rectangular culverts288 Stairs general information289 Stairs sawtooth and helical stairs290 Design coefficients for helical stairs ndash 1291 Design coefficients for helical stairs ndash 2292 Non-planar roofs general data293 Shell roofs empirical design method ndash 1294 Shell roofs empirical design method ndash 2295 Bow girders concentrated loads296 Bow girders uniform loads ndash 1297 Bow girders uniform loads ndash 2298 Bridges299 Hinges and bearings2100 Movement joints2101 Geometric properties of uniform sections2102 Properties of reinforced concrete sections ndash 12103 Properties of reinforced concrete sections ndash 22104 Uniaxial bending and compression (modular ratio)2105 Symmetrically reinforced rectangular columns

(modular ratio) ndash 12106 Symmetrically reinforced rectangular columns

(modular ratio) ndash 22107 Uniformly reinforced cylindrical columns

(modular ratio)2108 Uniaxial bending and tension (modular ratio)2109 Biaxial bending and compression (modular ratio)31 Design requirements and partial safety factors

(BS 8110)32 Design requirements and partial safety factors

(BS 5400) ndash 133 Design requirements and partial safety factors

(BS 5400) ndash 234 Design requirements and partial safety factors

(BS 8007)35 Concrete (BS 8110) strength and deformation

characteristics36 Stress-strain curves (BS 8110 and BS 5400) concrete

and reinforcement

37 Exposure classification (BS 8500)38 Concrete quality and cover requirements for durability

(BS 8500)39 Exposure conditions concrete and cover requirements

(prior to BS 8500)310 Fire resistance requirements (BS 8110) ndash 1311 Fire resistance requirements (BS 8110) ndash 2312 Building regulations minimum fire periods313 BS 8110 Design chart for singly reinforced

rectangular beams314 BS 8110 Design table for singly reinforced

rectangular beams315 BS 8110 Design chart for doubly reinforced

rectangular beams ndash 1316 BS 8110 Design chart for doubly reinforced

rectangular beams ndash 2317 BS 8110 Design chart for rectangular columns ndash 1318 BS 8110 Design chart for rectangular columns ndash 2319 BS 8110 Design chart for circular columns ndash 1320 BS 8110 Design chart for circular columns ndash 2321 BS 8110 Design procedure for columns ndash 1322 BS 8110 Design procedure for columns ndash 2323 BS 5400 Design chart for singly reinforced

rectangular beams324 BS 5400 Design table for singly reinforced

rectangular beams325 BS 5400 Design chart for doubly reinforced

rectangular beams ndash 1326 BS 5400 Design chart for doubly reinforced

rectangular beams ndash 2327 BS 5400 Design chart for rectangular columns ndash 1328 BS 5400 Design chart for rectangular columns ndash 2329 BS 5400 Design chart for circular columns ndash 1330 BS 5400 Design chart for circular columns ndash 2331 BS 5400 Design procedure for columns ndash 1332 BS 5400 Design procedure for columns ndash 2333 BS 8110 Shear resistance334 BS 8110 Shear under concentrated loads335 BS 8110 Design for torsion336 BS 5400 Shear resistance337 BS 5400 Shear under concentrated loads ndash 1338 BS 5400 Shear under concentrated loads ndash 2339 BS 5400 Design for torsion340 BS 8110 Deflection ndash 1341 BS 8110 Deflection ndash 2342 BS 8110 Deflection ndash 3343 BS 8110 (and BS 5400) Cracking344 BS 8007 Cracking345 BS 8007 Design options and restraint factors346 BS 8007 Design table for cracking due to temperature

effects347 BS 8007 Elastic properties of cracked rectangular

sections in flexure348 BS 8007 Design table for cracking due to flexure

in slabs ndash 1349 BS 8007 Design table for cracking due to flexure

in slabs ndash 2350 BS 8007 Design table for cracking due to flexure

in slabs ndash 3351 BS 8007 Design table for cracking due to direct

tension in walls ndash 1

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List of tablesviii

49 EC 2 Design chart for doubly reinforcedrectangular beams ndash 1

410 EC 2 Design chart for doubly reinforcedrectangular beams ndash 2

411 EC 2 Design chart for rectangular columns ndash 1412 EC 2 Design chart for rectangular columns ndash 2413 EC 2 Design chart for circular columns ndash 1414 EC 2 Design chart for circular columns ndash 2415 EC 2 Design procedure for columns ndash 1416 EC 2 Design procedure for columns ndash 2417 EC 2 Shear resistance ndash 1418 EC 2 Shear resistance ndash 2419 EC 2 Shear under concentrated loads420 EC 2 Design for torsion421 EC 2 Deflection ndash 1422 EC 2 Deflection ndash 2423 EC 2 Cracking ndash 1424 EC 2 Cracking ndash 2425 EC 2 Cracking ndash 3426 EC 2 Early thermal cracking in end restrained panels427 EC 2 Early thermal cracking in edge

restrained panels428 EC 2 Reinforcement limits429 EC 2 Provision of ties430 EC 2 Anchorage requirements431 EC 2 Laps and bends in bars432 EC 2 Rules for curtailment large diameter bars

and bundles

352 BS 8007 Design table for cracking due to directtension in walls ndash 2

353 BS 8110 Reinforcement limits354 BS 8110 Provision of ties355 BS 8110 Anchorage requirements356 BS 8110 Curtailment requirements357 BS 8110 Simplified curtailment rules for beams358 BS 8110 Simplified curtailment rules for slabs359 BS 5400 Considerations affecting design details360 BS 8110 Load-bearing walls361 BS 8110 Pile-caps362 Recommended details nibs corbels and halving joints363 Recommended details intersections of members41 Design requirements and partial safety factors

(EC 2 Part 1)42 Concrete (EC 2) strength and deformation

characteristics ndash 143 Concrete (EC 2) strength and deformation

characteristics ndash 244 Stressndashstrain curves (EC 2) concrete and

reinforcement45 Exposure classification (BS 8500)46 Concrete quality and cover requirements for durability

(BS 8500)47 EC 2 Design chart for singly reinforced

rectangular beams48 EC 2 Design table for singly reinforced

rectangular beams

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Since the last edition of Reynoldsrsquos Handbook considerabledevelopments in design and practice have occurred These includesignificant revisions to British standard specifications and codesof practice and the introduction of the Eurocodes Although cur-rent British codes are due to be withdrawn from 2008 onwardstheir use is likely to continue beyond that date at least in someEnglish-speaking countries outside the United Kingdom

One of the most significant changes has been in the systemfor classifying exposure conditions and selecting concretestrength and cover requirements for durability This is now dealtwith exclusively in BS 8500 which takes into account theparticular cementcombination type The notation used todefine concrete strength gives the cylinder strength as well asthe cube strength For structural design cube strength is usedin the British codes and cylinder strength in the Eurocodes

The characteristic yield strength of reinforcement has beenincreased to 500 Nmm2 (MPa) As a result new design aidshave become necessary and the Handbook includes tables andcharts for beams and columns (rectangular and circular)designed to both British and European codes Throughout theHandbook stress units are given as Nmm2 for British codesand MPa for European codes The decimal point is shown by afull stop (rather than a comma) in both cases

The basic layout of the Handbook is similar to the previousedition but the contents have been arranged in four separateparts for the convenience of the reader Also the opportunityhas been taken to omit a large amount of material that was nolonger relevant and to revise the entire text to reflect moderndesign and construction practice Part 1 is descriptive in formand covers design requirements loads materials structuralanalysis member design and forms of construction Frequentreference is made in Part 1 to the tables that are found in therest of the Handbook Although specific notes are attached tothese tables in Parts 2 3 and 4 much of the relevant text isembodied in Part 1 and the first part of the Handbook shouldalways be consulted

Part 2 has more detailed information on loads materialproperties and analysis in the form of tabulated data and chartsfor a large range of structural forms This material is largelyindependent of any specific code of practice Parts 3 and 4 cover

the design of members according to the requirements ofthe British and European codes respectively For each code thesame topics are covered in the same sequence so that the readercan move easily from one code to the other Each topic isillustrated by extensive numerical examples

In the Eurocodes some parameters are given recommendedvalues with the option of a national choice Choices also existwith regard to certain classes methods and procedures Thedecisions made by each country are given in a national annexPart 4 of the Handbook already incorporates the values given inthe UK national annex Further information concerning the useof Eurocode 2 is given in PD 6687 Background paper to theUK National Annex to BS EN 1992ndash1ndash1

The Handbook has been an invaluable source of reference forreinforced concrete engineers for over 70 years I madeextensive use of the sixth edition at the start of my professionalcareer 50 years ago This edition contains old and new infor-mation derived by many people and obtained from manysources past and present Although the selection inevitablyreflects the personal experience of the authors the informationhas been well tried and tested I owe a considerable debt ofgratitude to colleagues and mentors from whom I have learntmuch over the years and to the following organisations forpermission to include data for which they hold the copyright

British Cement AssociationBritish Standards InstitutionCabinet Office of Public Sector InformationConstruction Industry Research and Information AssociationPortland Cement AssociationThe Concrete Bridge Development GroupThe Concrete Society

Finally my sincere thanks go to Katy Low and all the staff atTaylor amp Francis Group and especially to my dear wife Joanwithout whose unstinting support this edition would never havebeen completed

Tony ThrelfallMarlow October 2006

Preface to theeleventh edition

wwwengbookspdfcom

Charles Edward Reynolds was born in 1900 and received hiseducation at Tiffin Boys School Kingston-on-Thames andBattersea Polytechnic After some years with Sir WilliamArroll BRC and Simon Carves he joined Leslie Turner andPartners and later C W Glover and Partners He was for someyears Technical Editor of Concrete Publications Ltd and thenbecame its Managing Editor combining this post with privatepractice In addition to the Reinforced Concrete DesignerrsquosHandbook of which almost 200000 copies have been soldsince it first appeared in 1932 Charles Reynolds was the authorof numerous other books papers and articles concerningconcrete and allied subjects Among his various professionalappointments he served on the council of the Junior Institutionof Engineers and was the Honorary Editor of its journal at hisdeath on Christmas Day 1971

James Cyril Steedman was educated at Varndean GrammarSchool and first was employed by British Rail whom he joinedin 1950 at the age of 16 In 1956 he began working for GKNReinforcements Ltd and later moved to Malcolm Glover andPartners His association with Charles Reynolds began whenafter the publication of numerous articles in the magazine

Concrete and Constructional Engineering he accepted anappointment as Technical Editor of Concrete Publications apost he held for seven years He then continued in privatepractice combining work for the Publications Division of theCement and Concrete Association with his own writing andother activities In 1981 he set up Jacys Computing Servicessubsequently devoting much of his time to the development ofmicro-computer software for reinforced concrete design He isthe joint author with Charles Reynolds of Examples of theDesign of Reinforced Concrete Buildings to BS 8110

Anthony John Threlfall was educated at Liverpool Institute forBoys after which he studied civil engineering at LiverpoolUniversity After eight years working for BRC Pierhead Ltdand IDC Ltd he took a diploma course in concrete structuresand technology at Imperial College For the next four years heworked for CEGB and Camus Ltd and then joined the Cementand Concrete Association in 1970 where he was engagedprimarily in education and training activities until 1993 Afterleaving the CampCA he has continued in private practice toprovide training in reinforced and prestressed concrete designand detailing

The authors

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The publishers would like to thank the following organisationsfor their kind permission to reproduce the following material

Permission to reproduce extracts from British Standards isgranted by BSI This applies to information in Tables 21 2324 27ndash210 215 216 219ndash223 242 243 245 255256 2100 31ndash311 321 322 331ndash345 353ndash361 41ndash46415ndash425 and 428ndash432 British Standards can be obtainedfrom BSI Customer Services 389 Chiswick High StreetLondon W4 4AL Tel 44 (0)20 8996 9001 emailcservicesbsi-globalcom

Information in section 31 and Tables 217ndash218 is reproducedwith permission from the British Cement Association andtaken from the publication Concrete Practice (ref 10)

Information in section 62 is reproduced with permissionfrom the Concrete Bridge Development Group and taken

from the publication An introduction to concrete bridges(ref 52)

Information in section 72 is reproduced with permissionfrom The Concrete Society and taken from TechnicalReport 34 Concrete industrial ground floors ndash A guide todesign and construction (ref 61) Technical Report 34 isavailable to purchase from The Concrete Bookshop wwwconcretebookshopcom Tel 0700 460 7777

Information in Chapter 15 and Table 270 is reproduced withpermission from CIRIA and taken from CIRIA Report 102Design of shear wall buildings London 1984 (ref 38)

Information in Tables 253 and 275ndash279 is reproduced withpermission from the Portland Cement Association (refs 32and 55)

Information in Tables 25 26 and 312 is reproduced withpermission from HMSO

Acknowledgements

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The symbols adopted in this book comply where appropriatewith those in the relevant codes of practice Although these arebased on an internationally agreed system for preparing nota-tions there are numerous differences between the British andthe European codes especially in the use of subscripts Whereadditional symbols are needed to represent properties not usedin the codes these have been selected in accordance with thebasic principles wherever possible

The amount and range of material contained in this bookmake it inevitable that the same symbols have to be used for

different purposes However care has been taken to ensure thatcode symbols are not duplicated except where this has beenfound unavoidable The notational principles adopted for con-crete design purposes are not necessarily best suited to otherbranches of engineering Consequently in those tables relatingto general structural analysis the notation employed in previ-ous editions of this book has generally been retained

Only the principal symbols that are common to all codes arelisted here all other symbols and abbreviations are defined inthe text and tables concerned

Symbols andabbreviations

Ac Area of concrete sectionAs Area of tension reinforcementAs Area of compression reinforcementAsc Area of longitudinal reinforcement in a columnC Torsional constantEc Static modulus of elasticity of concreteEs Modulus of elasticity of reinforcing steelF Action force or load (with appropriate

subscripts)G Shear modulus of concreteGk Characteristic permanent action or dead loadI Second moment of area of cross sectionK A constant (with appropriate subscripts)L Length spanM Bending momentN Axial forceQk Characteristic variable action or imposed loadR Reaction at supportS First moment of area of cross sectionT Torsional moment temperatureV Shear forceWk Characteristic wind load

a Dimension deflectionb Overall width of cross section or width of flanged Effective depth to tension reinforcementd Depth to compression reinforcementf Stress (with appropriate subscripts)fck Characteristic (cylinder) strength of concretefcu Characteristic (cube) strength of concretefyk Characteristic yield strength of reinforcementgk Characteristic dead load per unit areah Overall depth of cross section

i Radius of gyration of concrete sectionk A coefficient (with appropriate subscripts)l Length span (with appropriate subscripts)m Massqk Characteristic imposed load per unit arear Radius1r Curvaturet Thickness timeu Perimeter (with appropriate subscripts)v Shear stress (with appropriate subscripts)x Neutral axis depthz Lever arm of internal forces

Angle ratioe Modular ratio EsEc

Partial safety factor (with appropriate subscripts)c Compressive strain in concretes Strain in tension reinforcements Strain in compression reinforcement Diameter of reinforcing bar Creep coefficient (with appropriate subscripts) Slenderness ratio Poissonrsquos ratio Proportion of tension reinforcement Asbd Proportion of compression reinforcement As bd Stress (with appropriate subscripts) Factor defining representative value of action

BS British StandardEC EurocodeSLS Serviceability limit stateUDL Uniformly distributed loadULS Ultimate limit state

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Part 1

General information

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A structure is an assembly of members each of which under theaction of imposed loads and deformations is subjected tobending or direct force (either tensile or compressive) or to acombination of bending and direct force These effects may beaccompanied by shearing forces and sometimes by torsionImposed deformations occur as a result of concrete shrinkageand creep changes in temperature and differential settlementBehaviour of the structure in the event of fire or accidentaldamage resulting from impact or explosion may need to beexamined The conditions of exposure to environmental andchemical attack also need to be considered

Design includes selecting a suitable form of constructiondetermining the effects of imposed loads and deformationsand providing members of adequate stiffness and resistanceThe members should be arranged so as to combine efficientload transmission with ease of construction consistent withthe intended use of the structure and the nature of the siteExperience and sound judgement are often more important thanprecise calculations in achieving safe and economical structuresComplex mathematics should not be allowed to confuse a senseof good engineering The level of accuracy employed in thecalculations should be consistent throughout the designprocess wherever possible

Structural design is largely controlled by regulations or codesbut even within such bounds the designer needs to exercisejudgement in interpreting the requirements rather than designingto the minimum allowed by the letter of a clause In the UnitedKingdom for many years the design of reinforced concretestructures has been based on the recommendations of BritishStandards For buildings these include lsquoStructural use ofconcretersquo (BS 8110 Parts 1 2 and 3) and lsquoLoading on build-ingsrsquo (BS 6399 Parts 1 2 and 3) For other types of structureslsquoDesign of concrete bridgesrsquo (BS 5400 Part 4) and lsquoDesign ofconcrete structures for retaining aqueous liquidsrsquo (BS 8007)have been used Compliance with the particular requirements ofthe Building Regulations and the Highways Agency Standardsis also necessary in many cases

Since the last edition of this Handbook a comprehensiveset of harmonised Eurocodes (ECs) for the structural andgeotechnical design of buildings and civil engineering workshas been developed The Eurocodes were first introduced asEuronorme Voluntaire (ENV) standards intended for use inconjunction with a national application document (NAD) asan alternative to national codes for a limited number of years

These have now been largely replaced by Euronorme (EN)versions with each member state adding a National Annex(NA) containing nationally determined parameters in order toimplement the Eurocode as a national standard The relevantdocuments for concrete structures are EC 0 Basis of structuraldesign EC 1 Actions on structures and EC 2 Design of con-crete structures The last document is in four parts namely ndashPart 11 General rules and rules for buildings Part 12Structural fire design Part 2 Reinforced and prestressed con-crete bridges and Part 3 Liquid-retaining and containingstructures

The tables to be found in Parts 2 3 and 4 of this Handbookenable the designer to reduce the amount of arithmetical workinvolved in the analysis and design of members to the relevantstandards The use of such tables not only increases speed butalso eliminates inaccuracies provided the tables are thoroughlyunderstood and their applications and limitations are realisedIn the appropriate chapters of Part 1 and in the supplementaryinformation given on the pages preceding the tables the basisof the tabulated material is described Some general informa-tion is also provided The Appendix contains trigonometricaland other mathematical formulae and data

11 ECONOMICAL STRUCTURES

The cost of construction of a reinforced concrete structure isobviously affected by the prices of concrete reinforcementformwork and labour The most economical proportions ofmaterials and labour will depend on the current relationshipbetween the unit prices Economy in the use of formwork isgenerally achieved by uniformity of member size and the avoid-ance of complex shapes and intersections In particular casesthe use of available formwork of standard sizes may determinethe structural arrangement In the United Kingdom speed ofconstruction generally has a major impact on the overall costFast-track construction requires the repetitive use of a rapidformwork system and careful attention to both reinforcementdetails and concreting methods

There are also wider aspects of economy such as whetherthe anticipated life and use of a proposed structure warrant theuse of higher or lower factors of safety than usual or whetherthe use of a more expensive form of construction is warrantedby improvements in the integrity and appearance of the structureThe application of whole-life costing focuses attention on

Chapter 1

Introduction

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Introduction4

whether the initial cost of a construction of high qualitywith little or no subsequent maintenance is likely to be moreeconomical than a cheaper construction combined with theexpense of maintenance

The experience and method of working of the contractor theposition of the site and the nature of the available materials andeven the method of measuring the quantities together withnumerous other points all have their effect consciously or noton the designerrsquos attitude towards a contract So many andvaried are the factors involved that only experience and acontinuing study of design trends can give reliable guidanceAttempts to determine the most economical proportions for aparticular member based only on inclusive prices of concretereinforcement and formwork are likely to be misleading It isnevertheless possible to lay down certain principles

In broad terms the price of concrete increases with thecement content as does the durability and strength Concretegrades are often determined by durability requirements withdifferent grades used for foundations and superstructuresStrength is an important factor in the design of columns andbeams but rarely so in the case of slabs Nevertheless the samegrade is generally used for all parts of a superstructure exceptthat higher strength concrete may sometimes be used to reducethe size of heavily loaded columns

In the United Kingdom mild steel and high yield reinforce-ments have been used over the years but grade 500 is nowproduced as standard available in three ductility classes A B andC It is always uneconomical in material terms to use compressionreinforcement in beams and columns but the advantages gainedby being able to reduce member sizes and maintain the samecolumn size over several storeys generally offset the additionalmaterial costs For equal weights of reinforcement the combinedmaterial and fixing costs of small diameter bars are greater thanthose of large diameter bars It is generally sensible to use thelargest diameter bars consistent with the requirements for crackcontrol Fabric (welded mesh) is more expensive than barreinforcement in material terms but the saving in fixing time willoften result in an overall economy particularly in slabs and walls

Formwork is obviously cheaper if surfaces are plane and atright angles to each other and if there is repetition of use Thesimplest form of floor construction is a solid slab of constantthickness Beam and slab construction is more efficient struc-turally but less economical in formwork costs Two-way beamsystems complicate both formwork and reinforcement detailswith consequent delay in the construction programmeIncreased slab efficiency and economy over longer spans maybe obtained by using a ribbed form of construction Standardtypes of trough and waffle moulds are available in a range ofdepths Precasting usually reduces considerably the amountof formwork labour and erection time Individual mouldsare more expensive but can be used many more timesthan site formwork Structural connections are normally moreexpensive than with monolithic construction The economicaladvantage of precasting and the structural advantage of in situcasting may be combined in composite forms of construction

In many cases the most economical solution can only bedetermined by comparing the approximate costs of differentdesigns This may be necessary to decide say when a simplecantilever retaining wall ceases to be more economical thanone with counterforts or when a beam and slab bridge is moreeconomical than a voided slab The handbook Economic

Concrete Frame Elements published by the British CementAssociation on behalf of the Reinforced Concrete Councilenables designers to rapidly identify least-cost options for thesuperstructure of multi-storey buildings

12 DRAWINGS

In most drawing offices a practice has been developed to suitthe particular type of work done Computer aided drafting andreinforcement detailing is widely used The following observa-tions should be taken as general principles that accord with therecommendations in the manual Standard method of detailingstructural concrete published by the Institution of StructuralEngineers (ref 1)

It is important to ensure that on all drawings for a particularcontract the same conventions are adopted and uniformity ofsize and appearance are achieved In the preliminary stagesgeneral arrangement drawings of the whole structure are usuallyprepared to show the layout and sizes of beams columns slabswalls foundations and other members A scale of 1100 isrecommended although a larger scale may be necessary forcomplex structures Later these or similar drawings are devel-oped into working drawings and should show precisely suchparticulars as the setting-out of the structure in relation to anyadjacent buildings or other permanent works and the level ofsay the ground floor in relation to a fixed datum All principaldimensions such as distances between columns and walls andthe overall and intermediate heights should be shown Plansshould generally incorporate a gridline system with columnspositioned at the intersections Gridlines should be numbered 12 3 and so on in one direction and lettered A B C and soon in the other direction with the sequences starting at thelower left corner of the grid system The references canbe used to identify individual beams columns and othermembers on the reinforcement drawings

Outline drawings of the members are prepared to suitablescales such as 120 for beams and columns and 150 for slabsand walls with larger scales being used for cross sectionsReinforcement is shown and described in a standard way Theonly dimensions normally shown are those needed to positionthe bars It is generally preferable for the outline of the concreteto be indicated by a thin line and to show the reinforcement bybold lines The lines representing the bars should be shown inthe correct positions with due allowance for covers and thearrangement at intersections and laps so that the details onthe drawing represent as nearly as possible the appearanceof the reinforcement as fixed on site It is important to ensurethat the reinforcement does not interfere with the formation ofany holes or embedment of any other items in the concrete

A set of identical bars in a slab shown on plan might bedescribed as 20H16-03-150B1 This represents 20 numbergrade 500 bars of 16 mm nominal size bar mark 03 spaced at150 mm centres in the bottom outer layer The bar mark is anumber that uniquely identifies the bar on the drawing and thebar bending schedule Each different bar on a drawing is givena different bar mark Each set of bars is described only once onthe drawing The same bars on a cross section would be denotedsimply by the bar mark Bar bending schedules are prepared foreach drawing on separate forms according to recommendationsin BS 8666 Specification for scheduling dimensioning bendingand cutting of steel reinforcement for concrete

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There are two principal stages in the calculations requiredto design a reinforced concrete structure In the first stagecalculations are made to determine the effect on the structureof loads and imposed deformations in terms of appliedmoments and forces In the second stage calculations are madeto determine the capacity of the structure to withstand sucheffects in terms of resistance moments and forces

Factors of safety are introduced in order to allow for theuncertainties associated with the assumptions made and thevalues used at each stage For many years unfactored loadswere used in the first stage and total factors of safety wereincorporated in the material stresses used in the second stageThe stresses were intended to ensure both adequate safety andsatisfactory performance in service This simple approach waseventually replaced by a more refined method in which specificdesign criteria are set and partial factors of safety are incorpo-rated at each stage of the design process

21 DESIGN CRITERIA AND SAFETY FACTORS

A limit-state design concept is used in British and EuropeanCodes of Practice Ultimate (ULS) and serviceability (SLS)limit states need to be considered as well as durability and inthe case of buildings fire-resistance Partial safety factors areincorporated into loads (including imposed deformations) andmaterial strengths to ensure that the probability of failure (notsatisfying a design requirement) is acceptably low

In BS 8110 at the ULS a structure should be stable under allcombinations of dead imposed and wind load It should also berobust enough to withstand the effects of accidental loads dueto an unforeseen event such as a collision or explosion withoutdisproportionate collapse At the SLS the effects in normal useof deflection cracking and vibration should not cause thestructure to deteriorate or become unserviceable A deflectionlimit of span250 applies for the total sag of a beam or slabrelative to the level of the supports A further limit the lesser ofspan500 or 20 mm applies for the deflection that occurs afterthe application of finishes cladding and partitions so as to avoiddamage to these elements A limit of 03 mm generally appliesfor the width of a crack at any point on the concrete surface

In BS 5400 an additional partial safety factor is introducedThis is applied to the load effects and takes account of themethod of structural analysis that is used Also there are moreload types and combinations to be considered At the SLSthere are no specified deflection limits but the cracking limits

are more critical Crack width limits of 025 015 or 01 mmapply according to surface exposure conditions Compressivestress limits are also included but in many cases these do notneed to be checked Fatigue considerations require limitationson the reinforcement stress range for unwelded bars and morefundamental analysis if welding is involved Footbridges areto be analysed to ensure that either the fundamental naturalfrequency of vibration or the maximum vertical accelerationmeets specified requirements

In BS 8007 water-resistance is a primary design concernAny cracks that pass through the full thickness of a section arelikely to allow some seepage initially resulting in surfacestaining and damp patches Satisfactory performance dependsupon autogenous healing of such cracks taking place within afew weeks of first filling in the case of a containment vesselA crack width limit of 02 mm normally applies to all cracksirrespective of whether or not they pass completely through thesection Where the appearance of a structure is considered to beaesthetically critical a limit of 01 mm is recommended

There are significant differences between the structural andgeotechnical codes in British practice The approach to thedesign of foundations in BS 8004 is to use unfactored loadsand total factors of safety For the design of earth-retainingstructures CP2 (ref 2) used the same approach In 1994 CP2was replaced by BS 8002 in which mobilisation factors areintroduced into the calculation of soil strengths The resultingvalues are then used in BS 8002 for both serviceability andultimate requirements In BS 8110 the loads obtained fromBS 8002 are multiplied by a partial safety factor at the ULS

Although the design requirements are essentially the samein the British and European codes there are differences ofterminology and in the values of partial safety factors In theEurocodes loads are replaced by actions with dead loads as per-manent actions and all live loads as variable actions Each vari-able action is given several representative values to be used forparticular purposes The Eurocodes provide a more unifiedapproach to both structural and geotechnical design

Details of design requirements and partial safety factors to beapplied to loads and material strengths are given in Chapter 21for British Codes and Chapter 29 for Eurocodes

22 LOADS (ACTIONS)

The loads (actions) acting on a structure generally consist ofa combination of dead (permanent) and live (variable) loads

Chapter 2

Design criteria safetyfactors and loads

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Design criteria safety factors and loads6

In limit-state design a design load (action) is calculated bymultiplying the characteristic (or representative) value by anappropriate partial factor of safety The characteristic value isgenerally a value specified in a relevant standard or code Inparticular circumstances it may be a value given by a client ordetermined by a designer in consultation with the client

In BS 8110 characteristic dead imposed and wind loadsare taken as those defined in and calculated in accordancewith BS 6399 Parts 1 2 and 3 In BS 5400 characteristicdead and live loads are given in Part 2 but these have beensuperseded in practice by the loads in the appropriateHighways Agency standards These include BD 3701 andBD 6094 and for the assessment of existing bridgesBD 2101 (refs 3ndash5)

When EC 2 Part 11 was first introduced as an ENVdocument characteristic loads were taken as the values given inBS 6399 but with the specified wind load reduced by 10 Thiswas intended to compensate for the partial safety factor appliedto wind at the ULS being higher in the Eurocodes than in BS8110 Representative values were then obtained by multiplyingthe characteristic values by factors given in the NAD In theEN documents the characteristic values of all actions are givenin EC 1 and the factors to be used to determine representativevalues are given in EC 0

23 DEAD LOADS (PERMANENT ACTIONS)

Dead loads include the weights of the structure itself andall permanent fixtures finishes surfacing and so on Whenpermanent partitions are indicated they should be included asdead loads acting at the appropriate locations Where any doubtexists as to the permanency of the loads they should be treatedas imposed loads Dead loads can be calculated from the unitweights given in EC 1 Part 11 or from actual known weightsof the materials used Data for calculating dead loads are givenin Tables 21 and 22

24 LIVE LOADS (VARIABLE ACTIONS)

Live loads comprise any transient external loads imposed on thestructure in normal use due to gravitational dynamic andenvironmental effects They include loads due to occupancy(people furniture moveable equipment) traffic (road railpedestrian) retained material (earth liquids granular) snowwind temperature ground and water movement wave actionand so on Careful assessment of actual and probable loads is avery important factor in producing economical and efficientstructures Some imposed loads like those due to containedliquids can be determined precisely Other loads such as thoseon floors and bridges are very variable Snow and wind loadsare highly dependent on location Data for calculating loadsfrom stored materials are given in EC 1 Part 11

241 Floors

For most buildings the loads imposed on floors are specified inloading standards In BS 6399 Part 1 loads are specifiedaccording to the type of activity or occupancy involved Datafor residential buildings and for offices and particular workareas is given in Table 23 Imposed loads are given both as

a uniformly distributed load in kNm2 and a concentrated loadin kN The floor should be designed for the worst effects ofeither load The concentrated load needs to be considered forisolated short span members and for local effects such aspunching in a thin flange For this purpose a square contactarea with a 50 mm side may be assumed in the absence of anymore specific information Generally the concentrated loaddoes not need to be considered in slabs that are either solid orotherwise capable of effective lateral distribution Wherean allowance has to be made for non-permanent partitions auniformly distributed load equal to one-third of the load permetre run of the finished partitions may be used For offices theload used should not be less than 10 kNm2

The floors of garages are considered in two categoriesnamely those for cars and light vans and those for heaviervehicles In the lighter category the floor may be designedfor loads specified in the form described earlier In the heaviercategory the most adverse disposition of loads determinedfor the specific types of vehicle should be considered

The total imposed loads to be used for the design of beams maybe reduced by a percentage that increases with the area of floorsupported as given in Table 23 This does not apply to loads dueto storage vehicles plant or machinery For buildings designed tothe Eurocodes imposed loads are given in EC 1 Part 11

In all buildings it is advisable to affix a notice indicatingthe imposed load for which the floor is designed Floors ofindustrial buildings where plant and machinery are installedneed to be designed not only for the load when the plant is inrunning order but also for the probable load during erectionand testing which in some cases may be more severe Datafor loads imposed on the floors of agricultural buildings bylivestock and farm vehicles is given in BS 5502 Part 22

242 Structures subject to dynamic loads

The loads specified in BS 6399 Part 1 include allowancesfor small dynamic effects that should be sufficient for mostbuildings However the loading does not necessarily coverconditions resulting from rhythmical and synchronised crowdmovements or the operation of some types of machinery

Dynamic loads become significant when crowd movements(eg dancing jumping rhythmic stamping) are synchronisedIn practice this is usually associated with lively pop concertsor aerobics events where there is a strong musical beat Suchactivities can generate both horizontal and vertical loads Ifthe movement excites a natural frequency of the affected partof the structure resonance occurs which can greatly amplify theresponse Where such activities are likely to occur the structureshould be designed to either avoid any significant resonanceeffects or withstand the anticipated dynamic loads Somelimited guidance on dynamic loads caused by activities such asjumping and dancing is provided in BS 6399 Part 1 Annexe ATo avoid resonance effects the natural frequency of vibrationof the unloaded structure should be greater than 84 Hz for thevertical mode and greater than 40 Hz for the horizontal modes

Different types of machinery can give rise to a wide range ofdynamic loads and the potential resonant excitation of thesupporting structure should be considered Where necessaryspecialist advice should be sought

Footbridges are subject to particular requirements that willbe examined separately in the general context of bridges

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243 Parapets barriers and balustrades

Parapets barriers balustrades and other elements intended toretain stop or guide people should be designed for horizontalloads Values are given in BS 6399 Part 1 for a uniformlydistributed line load and for both uniformly distributed andconcentrated loads applied to the infill These are not takentogether but are applied as three separate load cases The lineload should be considered to act at a height of 11 m above adatum level taken as the finished level of the access platformor the pitch line drawn through the nosing of the stair treads

Vehicle barriers for car parking areas are also includedin BS 6399 Part 1 The horizontal force F as given in thefollowing equation is considered to act at bumper heightnormal to and uniformly distributed over any length of 15 m ofthe barrier By the fundamental laws of dynamics

F 05mv2(b c) (in kN)

m gross mass of vehicle (in kg)v speed of vehicle normal to barrier taken as 45 msecb deflection of barrier (in mm)c deformation of vehicle taken as 100 mm unless better

evidence is available

For car parks designed on the basis that the gross mass of thevehicles using it will not exceed 2500 kg (but taking as arepresentative value of the vehicle population m 1500 kg)and provided with rigid barriers (b 0) F is taken as 150 kNacting at a height of 375 mm above floor level It should benoted that bumper heights have been standardised at 445 mm

244 Roofs

The imposed loads given in Table 24 are additional to allsurfacing materials and include for snow and other incidentalloads but exclude wind pressure The snow load on the roofis determined by multiplying the estimated snow load on theground at the site location and altitude (the site snow load) byan appropriate snow load shape coefficient The main loadingconditions to be considered are

(a) a uniformly distributed snow load over the entire rooflikely to occur when snow falls with little or no wind

(b) a redistributed (or unevenly deposited) snow load likely tooccur in windy conditions

For flat or mono-pitch roofs it is sufficient to consider thesingle load case resulting from a uniform layer of snow asgiven in Table 24 For other roof shapes and for the effects oflocal drifting of snow behind parapets reference should bemade to BS 6399 Part 3 for further information

Minimum loads are given for roofs with no access (other thanthat necessary for cleaning and maintenance) and for roofswhere access is provided Roofs like floors should be designedfor the worst effects of either the distributed load or theconcentrated load For roofs with access the minimum loadwill exceed the snow load in most cases

If a flat roof is used for purposes such as a cafeacute playgroundor roof garden the appropriate imposed load for such a floorshould be allowed For buildings designed to the Eurocodessnow loads are given in EC 1 Part 13

245 Columns walls and foundations

Columns walls and foundations of buildings are designed forthe same loads as the slabs or beams that they support If theimposed loads on the beams are reduced according to the areaof floor supported the supporting members may be designedfor the same reduced loads Alternatively where two or morefloors are involved and the loads are not due to storage theimposed loads on columns or other supporting members maybe reduced by a percentage that increases with the number offloors supported as given in Table 23

246 Structures supporting cranes

Cranes and other hoisting equipment are often supported oncolumns in factories or similar buildings It is important that adimensioned diagram of the actual crane to be installed isobtained from the makers to ensure that the right clearances areprovided and the actual loads are taken into account For loadsdue to cranes reference should be made to BS 2573

For jib cranes running on rails on supporting gantries theload to which the structure is subjected depends on the actualdisposition of the weights of the crane The wheel loads aregenerally specified by the crane maker and should allow forthe static and dynamic effects of lifting discharging slewingtravelling and braking The maximum wheel load underpractical conditions may occur when the crane is stationary andhoisting the load at the maximum radius with the line of the jibdiagonally over one wheel

247 Structures supporting lifts

The effect of acceleration must be considered in addition to thestatic loads when calculating loads due to lifts and similarmachinery If a net static load F is subject to an accelerationa (ms2) the resulting load on the supporting structure isapproximately F (1 0098a) The average acceleration ofa passenger lift may be about 06 ms2 but the maximumacceleration will be considerably greater BS 2655 requiresthe supporting structure to be designed for twice the loadsuspended from the beams when the lift is at rest with anoverall factor of safety of 7 The deflection under the designload should not exceed span1500

248 Bridges

The analysis and design of bridges is now so complex thatit cannot be adequately treated in a book of this nature andreference should be made to specialist publications Howeverfor the guidance of designers the following notes regardingbridge loading are provided since they may also be applicableto ancillary construction and to structures having features incommon with bridges

Road bridges The loads to be considered in the design ofpublic road bridges in the United Kingdom are specified in theHighways Agency Standard BD 3701 Loads for HighwayBridges This is a revised version of BS 5400 Part 2 issuedby the Department of Transport rather than by BSI TheStandard includes a series of major amendments as agreed bythe BSI Technical Committee BD 3701 deals with both perma-nent loads (dead superimposed dead differential settlementearth pressure) and transient loads due to traffic use (vehicular

Live loads (variable actions) 7

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pedestrian) and environmental effects (wind temperature)The collision loads in BD 3701 may be applicable in certaincircumstances where agreed with the appropriate authority butin most cases the requirements of BD 6094 The design ofhighway bridges for vehicle collision loads will apply

Details of live loads due to traffic to be considered in thedesign of highway bridges are given in Table 25 Two typesof standard live loading are given in BD 3701 to representnormal traffic and abnormal vehicles respectively Loads areapplied to notional lanes of equal width The number ofnotional lanes is determined by the width of the carriagewaywhich includes traffic lanes hard shoulders and hard stripsand several typical examples are shown diagrammatically inBD 3701 Notional lanes are used rather than marked lanesin order to allow for changes of use and the introduction oftemporary contra-flow schemes

Type HA loading covers all the vehicles allowable under theRoad Vehicles (Construction and Use) and Road Vehicles(Authorised Weight) Regulations Values are given in terms ofa uniformly distributed load (UDL) and a single knife-edgeload (KEL) to be applied in combination to each notional laneThe specified intensity of the UDL (kNm) reduces as the loadedlength increases which allows for two effects At the shorterend it allows for loading in the vicinity of axles or bogies beinggreater than the average loading for the whole vehicle At thelonger end it takes account of the reducing percentage of heavygoods vehicles contained in the total vehicle population TheKEL of 120 kN is to be applied at any position within the UDLloaded length and spread over a length equal to the notional lanewidth In determining the loads consideration has been given tothe effects of impact vehicle overloading and unforeseen changesin traffic patterns The loading derived after application ofseparate factors for each of these effects was considered torepresent an ultimate load which was then divided by 15to obtain the specified nominal loads

The loads are multiplied by lane factors whose valuesdepend on the particular lane and the loaded length This isdefined as the length of the adverse area of the influence linethat is the length over which the load application increases themagnitude of the effect to be determined The lane factors takeaccount of the low probability of all lanes being fully loaded atthe same time They also for the shorter loaded lengths allowfor the effect of lateral bunching of vehicles As an alternativeto the combined loads a single wheel load of 100 kN appliedat any position is also to be considered

Type HB loading derives from the nature of exceptionalindustrial loads such as electrical transformers generatorspressure vessels and machine presses likely to use the roads inthe neighbouring area It is represented by a sixteen-wheelvehicle consisting of two bogies each one having two axleswith four wheels per axle Each axle represents one unit ofloading (equivalent to 10 kN) Bridges on public highwaysare designed for a specific number of units of HB loadingaccording to traffic use typically 45 units for trunk roads andmotorways 375 units for principal roads and 30 units for allother public roads Thus the maximum number of 45 unitscorresponds to a total vehicle load of 1800 kN with 450 kNper axle and 1125 kN per wheel The length of the vehicle isvariable according to the spacing of the bogies for which fivedifferent values are specified The HB vehicle can occupy anytransverse position on the carriageway and is considered to

displace HA loading over a specified area surrounding thevehicle Outside this area HA loading is applied as specifiedand shown by diagrams in BD 3701 The combined loadarrangement is normally critical for all but very long bridges

Road bridges may be subjected to forces other than those dueto dead load and traffic load These include forces due to windtemperature differential settlement and earth pressure Theeffects of centrifugal action and longitudinal actions due totraction braking and skidding must also be considered as wellas vehicle collision loads on supports and superstructure Fordetails of the loads to be considered on highway bridge parapetsreference should be made to BD 5293 (ref 6)

In the assessment of existing highway bridges traffic loadsare specified in the Highways Agency document BD 2101 TheAssessment of Highway Bridges and Structures In this casethe type HA loading is multiplied by a reduction factor thatvaries according to the road surface characteristics traffic flowconditions and vehicle weight restrictions Some of the contin-gency allowances incorporated in the design loading havealso been relaxed Vehicle weight categories of 40 38 25 1775 and 3 tonnes are considered as well as two groups offire engines For further information on reduction factors andspecific details of the axle weight and spacing values in eachcategory reference should be made BD 2101

Footbridges Details of live loads due to pedestrians to beconsidered in the design of footcycle track bridges are given inTable 26 A uniformly distributed load of 5 kNm2 is specified forloaded lengths up to 36 m Reduced loads may be used for bridgeswhere the loaded length exceeds 36 m except that specialconsideration is required in cases where exceptional crowds couldoccur For elements of highway bridges supporting footwayscycle tracks further reductions may be made in the pedestrian liveload where the width is greater than 2 m or the element alsosupports a carriageway When the footwaycycle track is notprotected from vehicular traffic by an effective barrier there is aseparate requirement to consider an accidental wheel loading

It is very important that consideration is given to vibrationthat could be induced in footcycle track bridges by resonancewith the movement of users or by deliberate excitation InBD 3701 the vibration requirement is deemed to be satisfiedin cases where the fundamental natural frequency of vibrationexceeds 5 Hz for the unloaded bridge in the vertical direction and15 Hz for the loaded bridge in the horizontal direction Whenthe fundamental natural frequency of vertical vibration fo doesnot exceed 5 Hz the maximum vertical acceleration shouldbe limited to 05radicfo ms2 Methods for determining the naturalfrequency of vibration and the maximum vertical accelerationare given in Appendix B of BD 3701 Where the fundamentalnatural frequency of horizontal vibration does not exceed15 Hz special consideration should be given to the possibilityof pedestrian excitation of lateral movements of unacceptablemagnitude Bridges possessing low mass and damping andexpected to be used by crowds of people are particularlysusceptible to such vibrations

Railway bridges Details of live loads to be considered in thedesign of railway bridges are given in Table 26 Two types ofstandard loading are given in BD 3701 type RU for mainline railways and type RL for passenger rapid transit systemsA further type SW0 is also included for main line railways

Design criteria safety factors and loads8

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The type RU loading was derived by a Committee of theInternational Union of Railways (UIC) to cover present andanticipated future loading on railways in Great Britain and on theContinent of Europe Nowadays motive power tends to be dieseland electric rather than steam and this produces axle loads andarrangements for locomotives that are similar to those for bogiefreight vehicles (these often being heavier than the locomotivesthat draw them) In addition to normal train loading whichcan be represented quite well by a uniformly distributed loadof 80 kNm railway bridges are occasionally subjected toexceptionally heavy abnormal loads For short loaded lengths itis necessary to introduce heavier concentrated loads to simulateindividual axles and to produce high shears at the ends Type RUloading consists of four concentrated loads of 250 kN precededand followed by a uniformly distributed load of 80 kNm For acontinuous bridge type SW0 loading is also to be considered asan additional and separate load case This loading consists oftwo uniformly distributed loads of 133 kNm each 15 m longseparated by a distance of 53 m Both types of loading whichare applied to each track or as specified by the relevant authoritywith half the track load acting on each rail are to be multipliedby appropriate dynamic factors to allow for impact lurchingoscillation and other dynamic effects The factors have beencalculated so that in combination with the specified loading theycover the effects of slow moving heavy and fast moving lightvehicles Exceptional vehicles are assumed to move at speeds notexceeding 80 kmh heavy wagons at speeds up to 120 kmh andpassenger trains at speeds up to 200 kmh

The type RL loading was derived by the London TransportExecutive to cover present and anticipated future loading onlines that carry only rapid transit passenger trains and lightengineersrsquo works trains Passenger trains include a variety ofstock of different ages loadings and gauges used on surfaceand tube lines Works trains include locomotives cranesand wagons used for maintenance purposes Locomotives areusually of the battery car type but diesel shunt varieties aresometimes used The rolling stock could include a 30t steamcrane 6t diesel cranes 20t hopper cranes and bolster wagonsThe heaviest train would comprise loaded hopper wagonshauled by battery cars Type RL loading consists of a singleconcentrated load of 200 kN coupled with a uniformly distrib-uted load of 50 kNm for loaded lengths up to 100 m Forloaded lengths in excess of 100 m the previous loading ispreceded and followed by a distributed load of 25 kNm Theloads are to be multiplied by appropriate dynamic factors Analternative bogie loading comprising two concentrated loadsone of 300 kN and the other of 150 kN spaced 24 m apart isalso to be considered on deck structures to check the ability ofthe deck to distribute the loads adequately

For full details of the locomotives and rolling stock coveredby each loading type and information on other loads to be con-sidered in the design of railway bridges due to the effects ofnosing centrifugal action traction and braking and in the eventof derailment reference should be made to BD 3701

249 Dispersal of wheel loads

A load from a wheel or similar concentrated load bearing on asmall but definite area of the supporting surface (called thecontact area) may be assumed to be further dispersed over anarea that depends on the combined thickness of any surfacing

material filling and underlying constructional material Thewidth of the contact area of a wheel on a slab is equal tothe width of the tyre The length of the contact area depends onthe type of tyre and the nature of the slab surface It is nearlyzero for steel tyres on steel plate or concrete The maximumcontact length is probably obtained with an iron wheel on loosemetalling or a pneumatic tyre on an asphalt surface

The wheel loads given in BD 3701 as part of the standardhighway loading are to be taken as uniformly distributed overa circular or square contact area assuming an effective pressureof 11 Nmm2 Thus for the HA single wheel load of 100 kNthe contact area becomes a 340 mm diameter circle or a squareof 300 mm side For the HB vehicle where 1 unit of loadingcorresponds to 25 kN per wheel the side of the square contactarea becomes approximately 260 mm for 30 units 290 mm for375 units and 320 mm for 45 units

Dispersal of the load beyond the contact area may be takenat a spread-to-depth ratio of 1 horizontally to 2 vertically forasphalt and similar surfacing so that the dimensions of thecontact area are increased by the thickness of the surfacingThe resulting boundary defines the loaded area to be used whenchecking for example the effects of punching shear on theunderlying structure

For a structural concrete slab 45o spread down to the levelof the neutral axis may be taken Since for the purpose ofstructural analysis the position of the neutral axis is usuallytaken at the mid-depth of the section the dimensions of thecontact area are further increased by the total thickness of theslab The resulting boundary defines the area of the patch loadto be used in the analysis

The concentrated loads specified in BD 3701 as part of therailway loading will be distributed both longitudinally bythe continuous rails to more than one sleeper and transverselyover a certain area of deck by the sleeper and ballast It maybe assumed that two-thirds of a concentrated load applied toone sleeper will be transmitted to the deck by that sleeper andthe remainder will be transmitted equally to the adjacent sleeperon either side Where the depth of ballast is at least 200 mmthe distribution may be assumed to be half to the sleeper lyingunder the load and half equally to the adjacent sleeper oneither side The load acting on the sleeper from each rail maybe distributed uniformly over the ballast at the level of theunderside of the sleeper for a distance taken symmetricallyabout the centreline of the rail of 800 mm or twice the distancefrom the centreline of the rail to the nearer end of the sleeperwhichever is the lesser Dispersal of the loads applied to theballast may be taken at an angle of 5o to the vertical down tothe supporting structure The distribution of concentrated loadsapplied to a track without ballast will depend on the relativestiffness of the rail the rail support and the bridge deck itself

25 WIND LOADS

All structures built above ground level are affected by thewind to a greater or lesser extent Wind comprises a randomfluctuating velocity component (turbulence or lsquogustinessrsquo)superimposed on a steady mean component The turbulenceincreases with the roughness of the terrain due to frictionaleffects between the wind and features on the ground such asbuildings and vegetation On the other hand the frictionaleffects also reduce the mean wind velocity

Wind loads 9

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Wind loads are dynamic and fluctuate continuously in bothmagnitude and position Some relatively flexible structuressuch as tall slender masts towers and chimneys suspensionbridges and other cable-stayed structures may be susceptible todynamic excitation in which case lateral deflections will be animportant consideration However the vast majority of build-ings are sufficiently stiff for the deflections to be small inwhich case the structure may be designed as if it was static

251 Wind speed and pressure

The local wind climate at any site in the United Kingdom can bepredicted reliably using statistical methods in conjunction withboundary-layer wind flow models However the complexity offlow around structures is not sufficiently well understood toallow wind pressures or distributions to be determined directlyFor this reason the procedure used in most modern wind codesis to treat the calculation of wind speed in a fully probabilisticmanner whilst continuing to use deterministic values of pressurecoefficients This is the approach adopted in BS 6399 Part 2which offers a choice of two methods for calculating wind loadsas follows

standard method uses a simplified procedure to obtain aneffective wind speed which is used with standard pressurecoefficients for orthogonal load cases

directional method provides a more precise assessment ofeffective wind speeds for particular wind directions which isused with directional pressure coefficients for load cases ofany orientation

The starting point for both methods is the basic hourly-meanwind speed at a height of 10 m in standard lsquocountryrsquo terrainhaving an annual risk (probability) of being exceeded of 002(ie a mean recurrence interval of 50 years) A map of basicwind speeds covering Great Britain and Ireland is provided

The basic hourly-mean wind speed is corrected according tothe site altitude and if required the wind direction seasonand probability to obtain an effective site wind speed This isfurther modified by a site terrain and building height factorto obtain an effective gust wind speed Ve ms which is used tocalculate an appropriate dynamic pressure q 0613Ve

2 Nm2Topographic effects are incorporated in the altitude factor for

the standard method and in the terrain and building factor for thedirectional method The standard method can be used in hand-based calculations and gives a generally conservative resultwithin its range of applicability The directional method is lessconservative and is not limited to orthogonal design cases Theloading is assessed in more detail but with the penalty ofincreased complexity and computational effort For furtherdetails of the directional method reference should be made toBS 6399 Part 2

252 Buildings

The standard method of BS 6399 Part 2 is the source of theinformation in Tables 27ndash29 The basic wind speed andthe correction factors are given in Table 27 The altitudefactor depends on the location of the structure in relation tothe local topography In terrain with upwind slopes exceeding005 the effects of topography are taken to be significant for

certain designated zones of the upwind and downwind slopesIn this case further reference should be made to BS 6399Part 2 When the orientation of the building is known the windspeed may be adjusted according to the direction under consid-eration Where the building height is greater than the crosswindbreadth for the direction being considered a reduction in thelateral load may be obtained by dividing the building into anumber of parts For buildings in town terrain the effectiveheight may be reduced as a result of the shelter afforded bystructures upwind of the site For details of the adjustmentsbased on wind direction division of buildings into parts and theinfluence of shelter on effective height reference should bemade to BS 6399 Part 2

When the wind acts on a building the windward faces aresubjected to direct positive pressure the magnitude of whichcannot exceed the available kinetic energy of the wind Asthe wind is deflected around the sides and over the roof of thebuilding it is accelerated lowering the pressure locally onthe building surface especially just downwind of the eavesridge and corners These local areas where the acceleration ofthe flow is greatest can experience very large wind suctionsThe surfaces of enclosed buildings are also subjected to internalpressures Values for both external and internal pressures areobtained by multiplying the dynamic pressure by appropriatepressure coefficients and size effect factors The overall force ona rectangular building is determined from the normal forces onthe windward-facing and leeward-facing surfaces the frictionaldrag forces on surfaces parallel to the direction of the wind anda dynamic augmentation factor that depends on the buildingheight and type

Details of the dimensions used to define surface pressuresand forces and values for dynamic augmentation factors andfrictional drag coefficients are given in Table 28 Size effectfactors and external and internal pressure coefficients for thewalls of rectangular buildings are given in Table 29 Furtherinformation including pressure coefficients for various roofforms free-standing walls and cylindrical structures such assilos tanks and chimneys and procedures for more-complexbuilding shapes are given in BS 6399 Part 2 For buildingsdesigned to the Eurocodes data for wind loading is given inEC 1 Part 12

253 Bridges

The approach used for calculating wind loads in BD 3701 is ahybrid mix of the methods given in BS 6399 Part 2 The direc-tional method is used to calculate the effective wind speed asthis gives a better estimate of wind speeds in towns and for sitesaffected by topography In determining the wind speed theprobability factor is taken as 105 appropriate to a return periodof 120 years Directional effective wind speeds are derived fororthogonal load cases and used with standard drag coefficientsto obtain wind loads on different elements of the structure suchas decks parapets and piers For details of the proceduresreference must be made to BD 3701

26 MARITIME STRUCTURES

The forces acting upon sea walls dolphins wharves jettiespiers docks and similar maritime structures include those dueto winds and waves blows and pulls from vessels the loads

Design criteria safety factors and loads10

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from cranes roads railways and stored goods imposed on thedeck and the pressures of earth retained behind the structure

For wharves or jetties of solid construction the energy ofimpact due to blows from vessels berthing is absorbed by themass of the structure usually without damage to the structureor vessel if fendering is provided With open constructionconsisting of braced piles or piers supporting the deck in whichthe mass of the structure is comparatively small the forcesresulting from impact must be considered The forces dependon the weight and speed of approach of the vessel on theamount of fendering and on the flexibility of the structureIn general a large vessel will approach at a low speed and asmall vessel at a higher speed Some typical examples are a1000 tonne vessel at 03 ms a 10 000 tonne vessel at 02 msand a 100 000 tonne vessel at 015 ms The kinetic energy of avessel displacing F tonnes approaching at a speed V ms isequal to 0514FV 2 kNm Hence the kinetic energy of a2000 tonne vessel at 03 ms and a 5000 tonne vessel at 02 msis about 100 kNm in each case If the direction of approachof a vessel is normal to the face of a jetty the whole of thisenergy must be absorbed on impact More commonly a vesselapproaches at an angle with the face of the jetty and touchesfirst at one point about which the vessel swings The energythen to be absorbed is 0514F[(Vsin13)2 ()2] with 13 theangle of approach of the vessel with the face of the jetty theradius of gyration (m) of the vessel about the point of impactand the angular velocity (radianss) of the vessel about the pointof impact The numerical values of the terms in the expressionare difficult to assess accurately and can vary considerablyunder different conditions of tide and wind and with differentvessels and methods of berthing

The kinetic energy of approach is absorbed partly by theresistance of the water but mainly by the fendering elasticdeformation of the structure and the vessel movement of theground and also by energy lsquolostrsquo upon impact The relativecontributions are difficult to assess but only about half ofthe total kinetic energy of the vessel may be imparted to thestructure and the fendering The force to which the structure issubjected is calculated by equating the product of the force andhalf the elastic horizontal displacement of the structure to thekinetic energy imparted Ordinary timber fenders appliedto reinforced concrete jetties cushion the blow but may notsubstantially reduce the force on the structure Spring fendersor suspended fenders can however absorb a large proportion ofthe kinetic energy Timber fenders independent of the jetty aresometimes provided to protect the structure from impact

The combined action of wind waves currents and tides on avessel moored to a jetty is usually transmitted by the vesselpressing directly against the side of the structure or by pullson mooring ropes secured to bollards The pulls on bollardsdue to the foregoing causes or during berthing vary with thesize of the vessel For vessels of up to 20 000 tonnes loadeddisplacement bollards are required at intervals of 15ndash30 m withload capacities according to the vessel displacement of 100 kNup to 2000 tonnes 300 kN up to 10 000 tonnes and 600 kN upto 20 000 tonnes

The effects of wind and waves acting on a marine structureare much reduced if an open construction is adopted and ifprovision is made for the relief of pressures due to water and airtrapped below the deck The force is not however relateddirectly to the proportion of solid vertical face presented to

the action of the wind and waves The pressures imposed areimpossible to assess with accuracy except for sea walls andsimilar structures where the depth of water at the face of the wallis such that breaking waves do not occur In this case the forceis due to simple hydrostatic pressure and can be evaluated forthe highest anticipated wave level with appropriate allowancefor wind surge In the Thames estuary for example the lattercan raise the high-tide level to 15 m above normal

A wave breaking against a sea wall causes a shock pressureadditional to the hydrostatic pressure which reaches its peakvalue at about mean water level and diminishes rapidly belowthis level and more slowly above it The shock pressure can beas much as 10 times the hydrostatic value and pressures up to650 kNm2 are possible with waves 45ndash6 m high The shape ofthe face of the wall the slope of the foreshore and the depthof water at the wall affect the maximum pressure and thedistribution of the pressure For information on the loads to beconsidered in the design of all types of maritime structuresreference should be made to BS 6349 Parts 1 to 7

27 RETAINED AND CONTAINED MATERIALS

The pressures imposed by materials on retaining structures orcontainment vessels are uncertain except when the retainedor contained material is a liquid In this case at any depth zbelow the free surface of the liquid the intensity of pressurenormal to the contact surface is equal to the vertical pressuregiven by the simple hydrostatic expression z wz where w

is unit weight of liquid (eg 981 kNm3 for water) For soilsand stored granular materials the pressures are considerablyinfluenced by the effective shear strength of the material

271 Properties of soils

For simplicity of analysis it is conventional to express the shearstrength of a soil by the equation

c n tan

where cis effective cohesion of soil is effective angle ofshearing resistance of soil n is effective normal pressure

Values of cand are not intrinsic soil properties and canonly be assumed constant within the stress range for which theyhave been evaluated For recommended fill materials it isgenerally sufficient to adopt a soil model with c 0 Such amodel gives a conservative estimate of the shear strength of thesoil and is analytically simple to apply in design Data takenfrom BS 8002 is given in Table 210 for unit weights of soilsand effective angles of shearing resistance

272 Lateral soil pressures

The lateral pressure exerted by a soil on a retaining structuredepends on the initial state of stress and the subsequent strainwithin the soil Where there has been no lateral strain eitherbecause the soil has not been disturbed during construction or thesoil has been prevented from lateral movement during placementan at-rest state of equilibrium exists Additional lateral strain isneeded to change the initial stress conditions Depending on themagnitude of the strain involved the final state of stress in the soilmass can be anywhere between the two failure conditions knownas the active and passive states of plastic equilibrium

Retained and contained materials 11

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The problem of determining lateral pressures at the limitingequilibrium conditions has been approached in different waysby different investigators In Coulomb theory the force actingon a retaining wall is determined by considering the limitingequilibrium of a soil wedge bounded by the rear face of thewall the ground surface and a planar failure surface Shearingresistance is assumed to have been mobilised both on the backof the wall and on the failure surface Rankine theory givesthe complete state of stress in a cohesionless soil mass whichis assumed to have expanded or compressed to a state of plasticequilibrium The stress conditions require that the earthpressure on a vertical plane should act in a direction parallel tothe ground surface Caquot and Kerisel produced tables ofearth pressure coefficients derived by a method that directlyintegrates the equilibrium equations along combined planar andlogarithmic spiral failure surfaces

273 Fill materials

A wide range of fill materials may be used behind retainingwalls All materials should be properly investigated and classi-fied Industrial chemical and domestic waste shale mudstoneand steel slag peaty or highly organic soil should not be usedas fill Selected cohesionless granular materials placed in acontrolled manner such as well-graded small rock-fills gravelsand sands are particularly suitable The use of cohesive soilscan result in significant economies by avoiding the need toimport granular materials but may also involve additionalproblems during design and construction The cohesive soilshould be within a range suitable for adequate compactionThe placement moisture content should be close to the finalequilibrium value to avoid either the swelling of clays placedtoo dry or the consolidation of clays placed too wet Suchproblems will be minimised if the fill is limited to clays with aliquid limit not exceeding 45 and a plasticity index notexceeding 25 Chalk with a saturation moisture content notexceeding 20 is acceptable as fill and may be compacted asfor a well-graded granular material Conditioned pulverizedfuel ash (PFA) from a single source may also be used it shouldbe supplied at a moisture-content of 80ndash100 of the optimumvalue For further guidance on the suitability of fill materialsreference should be made to relevant Transport ResearchLaboratory publications DoT Standard BD 3087 (ref 7)and BS 8002

274 Pressures imposed by cohesionless soils

Earth pressure distributions on unyielding walls and on rigidwalls free to translate or rotate about the base are shown inTable 211 For a normally consolidated soil the pressure on thewall increases linearly with depth Compaction results in higherearth pressures in the upper layers of the soil mass

Expressions for the pressures imposed in the at-rest activeand passive states including the effects of uniform surchargeand static ground water are given in sections 911ndash914Charts of earth pressure coefficients based on the work ofCaquot and Kerisel (ref 8) are given in Tables 212ndash214These may be used generally for vertical walls with slopingground or inclined walls with level ground

275 Cohesive soils

Clays in the long term behave as granular soils exhibitingfriction and dilation If a secant value (c 0) is used theprocedures for cohesionless soils apply If tangent parameters(c ) are used the RankinendashBell equations apply as given insection 915 In the short term if a clay soil is subjected torapid shearing a total stress analysis should be undertakenusing the undrained shear strength (see BS 8002)

276 Further considerations

For considerations such as earth pressures on embedded walls(with or without props) the effects of vertical concentrated loadsand line loads and the effects of groundwater seepage referenceshould be made to specialist books and BS 8002 For the pres-sures to be considered in the design of integral bridge abutmentsas a result of thermal movements of the deck reference shouldbe made to the Highways Agency document BA 4296 (ref 9)

277 Silos

Silos which may also be referred to as bunkers or bins aredeep containers used to store particulate materials In a deepcontainer the linear increase of pressure with depth found inshallow containers is modified When a deep container is filleda slight settlement of the fill activates the frictional resistancebetween the stored material and the wall This induces verticalload in the silo wall but reduces the vertical pressure in thematerial and the lateral pressures on the wall Janssen devel-oped a theory by which expressions have been derived for thepressures on the walls of a silo containing a granular materialhaving uniform properties The ratio of horizontal to verticalpressure in the fill is assumed constant and a Rankine coeffi-cient is generally used Eccentric filling (or discharge) tends toproduce variations in lateral pressure round the silo wall Anallowance is made by considering additional patch loads takento act on any part of the wall

Unloading a silo disturbs the equilibrium of the containedmass If the silo is unloaded from the top the frictional load onthe wall may be reversed as the mass re-expands but the lateralpressures remain similar to those during filling With a free-flowing material unloading at the bottom of the silo from thecentre of a hopper two different flow patterns are possibledepending on the characteristics of the hopper and the materialThese patterns are termed funnel flow (or core flow) and massflow respectively In the former a channel of flowing materialdevelops within a confined zone above the outlet the materialadjacent to the wall near the outlet remaining stationary Theflow channel can intersect the vertical walled section of the siloor extend to the surface of the stored material In mass flowwhich occurs particularly in steep-sided hoppers all the storedmaterial is mobilised during discharge Such flow can developat varying levels within the mass of material contained in anytall silo owing to the formation of a lsquoself-hopperrsquo with highlocal pressures arising where parallel flow starts to diverge fromthe walls Both flow patterns give rise to increases in lateralpressure from the stable filled condition Mass flow results alsoin a substantial local kick load at the intersection of the hopperand the vertical walled section

Design criteria safety factors and loads12

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When calculating pressures care should be taken to allow forthe inherent variability of the material properties In generalconcrete silo design is not sensitive to vertical wall load sovalues of maximum unit weight in conjunction with maximumor minimum consistent coefficients of friction should be usedData taken from EC 1 Part 4 for the properties of storedmaterials and the pressures on the walls and bottoms of silosare given in Tables 215 and 216

Fine powders like cement and flour can become fluidisedin silos either owing to rapid filling or through aeration tofacilitate discharge In such cases the design should allow forboth non-fluidised and fluidised conditions

28 EUROCODE LOADING STANDARDS

Eurocode 1 Actions on Structures is one of nine internationalunified codes of practice that have been published by the

European Committee for Standardization (CEN) The codewhich contains comprehensive information on all the actions(loads) normally necessary for consideration in the design ofbuilding and civil engineering structures consists of ten partsas follows

1991-1-1 Densities self-weight and imposed loads1991-1-2 Actions on structures exposed to fire1991-1-3 Snow loads1991-1-4 Wind loads1991-1-5 Thermal actions1991-1-6 Actions during execution1991-1-7 Accidental actions due to impact and explosions1991-2 Traffic loads on bridges1991-3 Actions induced by cranes and machinery1991-4 Actions on silos and tanks

Eurocode loading standards 13

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The requirements of concrete and its constituent materialsand of reinforcement are specified in Regulations Standardsand Codes of Practice Only those properties that concern thedesigner directly because they influence the behaviour anddurability of the structure are dealt with in this chapter

31 CONCRETE

Concrete is a structural material composed of crushed rockor gravel and sand bound together with a hardened paste ofcement and water A large range of cements and aggregateschemical admixtures and additions can be used to producea range of concretes having the required properties in boththe fresh and hardened states for many different structuralapplications The following information is taken mainly fromref 10 where a fuller treatment of the subject will be found

311 Cements and combinations

Portland cements are made from limestone and clay or otherchemically similar suitable raw materials which are burnedtogether in a rotary kiln to form a clinker rich in calciumsilicates This clinker is ground to a fine powder with a smallproportion of gypsum (calcium sulphate) which regulates therate of setting when the cement is mixed with water Overthe years several types of Portland cement have been developed

As well as cement for general use (which used to be knownas ordinary Portland cement) cements for rapid hardeningfor protection against attack by freezing and thawing or bychemicals and white cement for architectural finishes are alsomade The cements contain the same active compounds but indifferent proportions By incorporating other materials duringmanufacture an even wider range of cements is made includingair-entraining cement and combinations of Portland cementwith mineral additions Materials other than those in Portlandcements are used in cements for special purposes for examplecalcium aluminate cement is used for refractory concrete

The setting and hardening process that occurs when cementis mixed with water results from a chemical reaction known ashydration The process produces heat and is irreversible Settingis the gradual stiffening whereby the cement paste changesfrom a workable to a hardened state Subsequently the strengthof the hardened mass increases rapidly at first but slowinggradually This gain of strength continues as long as moisture ispresent to maintain the chemical reaction

Portland cements can be either inter-ground or blendedwith mineral materials at the cement factory or combined withadditions in the concrete mixer The most frequently used ofthese additional materials in the United Kingdom and therelevant British Standards are pulverized-fuel ash (pfa) toBS 3892 fly ash to BS EN 450 ground granulated blastfurnaceslag (ggbs) to BS 6699 and limestone fines to BS 7979 Otheradditions include condensed silica fume and metakaolin Theseare intended for specialised uses of concrete beyond the scopeof this book

The inclusion of pfa fly ash and ggbs has been particularlyuseful in massive concrete sections where they have been usedprimarily to reduce the temperature rise of the concrete withcorresponding reductions in temperature differentials and peaktemperatures The risk of early thermal contraction cracking isthereby also reduced The use of these additional materialsis also one of the options available for minimising the risk ofdamage due to alkalindashsilica reaction which can occur withsome aggregates and for increasing the resistance of concreteto sulfate attack Most additions react slowly at early stagesunder normal temperatures and at low temperature the reac-tion particularly in the case of ggbs can become considerablyretarded and make little contribution to the early strength ofconcrete However provided the concrete is not allowed to dryout the use of such additions can increase the long-termstrength and impermeability of the concrete

When the terms lsquowater-cement ratiorsquo and lsquocement contentrsquoare used in British Standards these are understood to includecombinations The word lsquobinderrsquo which is sometimes used isinterchangeable with the word lsquocementrsquo or lsquocombinationrsquo

The two methods of incorporating mineral additions makelittle or no difference to the properties of the concrete but therecently introduced notation system includes a unique code thatidentifies both composition and production method The typesof cement and combinations in most common usage are shownwith their notation in Table 217

Portland cement The most commonly used cement wasknown formerly as OPC in British Standards By grinding thecement clinker more finely cement with a more rapid earlystrength development is produced known formerly as RHPCBoth types are now designated as

Portland cement CEM I conforming to BS EN 197-1

Chapter 3

Material properties

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Cements are now classified in terms of both their standardstrength derived from their performance at 28 days and at anearly age normally two days using a specific laboratory testbased on a standard mortar prism This is termed the strengthclass for example CEM I 425N where 425 (Nmm2) is thestandard strength and N indicates a normal early strength

The most common standard strength classes for cements are425 and 525 These can take either N (normal) or R (rapid)identifiers depending on the early strength characteristics ofthe product CEM I in bags is generally a 425N cementwhereas CEM I for bulk supply tends to be 425R or 525 NCement corresponding to the former RHPC is now producedin the United Kingdom within the 525 strength class Thesecements are often used to advantage by precast concretemanufacturers to achieve a more rapid turn round of mouldsand on site when it is required to reduce the time for which theformwork must remain in position The cements which gener-ates more early heat than CEM I 425N can also be useful incold weather conditions

It is worth noting that the specified setting times of cementpastes relate to the performance of a cement paste of standardconsistence in a particular test made under closely controlledconditions of temperature and humidity the stiffening andsetting of concrete on site are not directly related to thesestandard setting regimes and are more dependent on factorssuch as workability cement content use of admixtures thetemperature of the concrete and the ambient conditions

Sulfate-resisting Portland cement SRPC This is a Portlandcement with a low tricalcium aluminate (C3A) content forwhich the British Standard is BS 4027 When concrete madewith CEM I cement is exposed to the sulfate solutions that arefound in some soils and groundwaters a reaction can occurbetween the sulfate and the hydrates from the C3A in thecement causing deterioration of the concrete By limitingthe C3A content in SRPC cement with a superior resistanceto sulfate attack is obtained SRPC normally has a low-alkalicontent but otherwise it is similar to other Portland cements inbeing non-resistant to strong acids The strength properties ofSRPC are similar to those of CEM I 425N but slightly lessearly heat is normally produced This can be an advantage inmassive concrete and in thick sections SRPC is not normallyused in combination with pfa or ggbs

Blastfurnace slag cements These are cements incorporatingggbs which is a by-product of iron smelting obtained byquenching selected molten slag to form granules The slag canbe inter-ground or blended with Portland cement clinker atcertain cement works to produce

Portland-slag cement CEM IIA-S with a slag content of6ndash35 conformimg to BS EN 197-1 or more commonly

Blastfurnace cement CEM IIIA with a slag content of36ndash65 conforming to BS EN 197-1

Alternatively the granules may be ground down separately to awhite powder with a fineness similar to that of cement and thencombined in the concrete mixer with CEM I cement to producea blastfurnace cement Typical mixer combinations of 40ndash50ggbs with CEM I cement have a notation CIIIA and at this levelof addition 28-day strengths are similar to those obtained withCEM I 425N

As ggbs has little hydraulic activity of its own it is referredto as lsquoa latent hydraulic binderrsquo Cements incorporating ggbsgenerate less heat and gain strength more slowly with lowerearly age strengths than those obtained with CEM I cementThe aforementioned blastfurnace cements can be used insteadof CEM I cement but because the early strength developmentis slower particularly in cold weather it may not be suitablewhere early removal of formwork is required They are amoderately low-heat cement and can therefore be used toadvantage to reduce early heat of hydration in thick sectionsWhen the proportion of ggbs is 66ndash80 CEM IIIA and CIIIAbecome CEM IIIB and CIIIB respectively These were knownformerly as high-slag blastfurnace cements and are specifiedbecause of their lower heat characteristics or to impart resis-tance to sulfate attack

Because the reaction between ggbs and lime released by thePortland cement is dependent on the availability of moistureextra care has to be taken in curing concrete containing thesecements or combinations to prevent premature drying out andto permit the development of strength

Pulverized-fuel ash and fly ash cements The ash resultingfrom the burning of pulverized coal in power station furnaces isknown in the concrete sector as pfa or fly ash The ash whichis fine enough to be carried away in the flue gases is removedfrom the gases by electrostatic precipitators to prevent atmos-pheric pollution The resulting material is a fine powder ofglassy spheres that can have pozzolanic properties that iswhen mixed into concrete it can react chemically with thelime that is released during the hydration of Portland cementThe products of this reaction are cementitious and in certaincircumstances pfa or fly ash can be used as a replacement forpart of the Portland cement provided in the concrete

The required properties of ash to be used as a cementitiouscomponent in concrete are specified in BS EN 450 withadditional UK provisions for pfa made in BS 3892 Part 1 Flyash in the context of BS EN 450 means lsquocoal fly ashrsquo ratherthan ash produced from other combustible materials and fly ashconforming to BS EN 450 can be coarser than that conformingto BS 3892 Part 1

Substitution of these types of cement for Portland cement isnot a straightforward replacement of like for like and thefollowing points have to be borne in mind when consideringthe use of pfa concrete

Pfa reacts more slowly than Portland cement At early ageand particularly at low temperatures pfa contributes lessstrength in order to achieve the same 28-day compressivestrength the amount of cementitious material may need to beincreased typically by about 10 The potential strengthafter three months is likely to be greater than CEM I providedthe concrete is kept in a moist environment for example inunderwater structures or concrete in the ground

The water demand of pfa for equal consistence may beless than that of Portland cement

The density of pfa is about three-quarters that of Portlandcement

The reactivity of pfa and its effect on water demand and hencestrength depend on the particular pfa and Portland cementwith which it is used A change in the source of either materialmay result in a change in the replacement level required

Concrete 15

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When pfa is to be air-entrained the admixture dosage ratemay have to be increased or a different formulation thatproduces a more stable air bubble structure used

Portland-fly ash cement comprises in effect a mixture ofCEM I and pfa When the ash is inter-ground or blended withPortland cement clinker at an addition rate of 20ndash35 themanufactured cement is known as Portland-fly ash cementCEM IIB-V conforming to BS EN 197-1 When this combina-tion is produced in a concrete mixer it has the notation CIIB-Vconforming to BS 8500 Part 2

Typical ash proportions are 25ndash30 and these cements canbe used in concrete for most purposes They are likely to havea slower rate of strength development compared with CEM IWhen the cement contains 25ndash40 ash it may be used toimpart resistance to sulfate attack and can also be beneficial inreducing the harmful effects of alkalindashsilica reaction Wherehigher replacement levels of ash are used for improved low-heatcharacteristics the resulting product is pozzolanic (pfa) cementwith the notation if manufactured CEM IVB-V conforming toBS EN 197-1 or if combined in the concrete mixer CIVB-Vconforming to BS 8500 Part 2

Because the pozzolanic reaction between pfa or fly ash andfree lime is dependent on the availability of moisture extra carehas to be taken in curing concrete containing mineral additionsto prevent premature drying out and to permit the developmentof strength

Portland-limestone cement Portland cement incorporating6ndash35 of carefully selected fine limestone powder is knownas Portland-limestone cement conforming to BS EN 197-1When a 425N product is manufactured the typical limestoneproportion is 10ndash20 and the notation is CEM IIA-L or CEMIIA-LL It is most popular in continental Europe but its usageis growing in the United Kingdom Decorative precast andreconstituted stone concretes benefit from its lighter colouringand it is also used for general-purpose concrete in non-aggressiveand moderately aggressive environments

312 Aggregates

The term lsquoaggregatersquo is used to describe the gravels crushedrocks and sands that are mixed with cement and water to pro-duce concrete As aggregates form the bulk of the volume ofconcrete and can significantly affect its performance the selec-tion of suitable material is extremely important Fine aggregatesinclude natural sand crushed rock or crushed gravel that is fineenough to pass through a sieve with 4 mm apertures (formerly5 mm as specified in BS 882) Coarse aggregates compriselarger particles of gravel crushed gravel or crushed rock Mostconcrete is produced from natural aggregates that are specifiedto conform to the requirements of BS EN 12620 together withthe UK Guidance Document PD 6682-1 Manufactured light-weight aggregates are also sometimes used

Aggregates should be hard and should not contain materialsthat are likely to decompose or undergo volumetric changeswhen exposed to the weather Some examples of undesirablematerials are lignite coal pyrite and lumps of clay Coal andlignite may swell and decompose leaving small holes on thesurface of the concrete lumps of clay may soften and formweak pockets and pyrite may decompose causing iron oxide

stains to appear on the concrete surface When exposed tooxygen pyrite has been known to contribute to sulfate attackHigh-strength concretes may call for special propertiesThe mechanical properties of aggregates for heavy-dutyconcrete floors and for pavement wearing surfaces may have tobe specially selected Most producers of aggregate are ableto provide information about these properties and referencewhen necessary should be made to BS EN 12620

There are no simple tests for aggregate durability or theirresistance to freezethaw exposure conditions and assessmentof particular aggregates is best based on experience of theproperties of concrete made with the type of aggregate andknowledge of its source Some flint gravels with a white porouscortex may be frost-susceptible because of the high waterabsorption of the cortex resulting in pop-outs on the surface ofthe concrete when subjected to freezethaw cycles

Aggregates must be clean and free from organic impuritiesThe particles should be free from coatings of dust or clay asthese prevent proper bonding of the material An excessiveamount of fine dust or stone lsquoflourrsquo can prevent the particles ofstone from being properly coated with cement and lower thestrength of the concrete Gravels and sands are usually washedby the suppliers to remove excess fines (eg clay and silt) andother impurities which otherwise could result in a poor-qualityconcrete However too much washing can also remove allfine material passing the 025 mm sieve This may result in aconcrete mix lacking in cohesion and in particular one that isunsuitable for placing by pump Sands deficient in fines alsotend to increase the bleeding characteristics of the concreteleading to poor vertical finishes due to water scour

Where the colour of a concrete surface finish is importantsupplies of aggregate should be obtained from the one sourcethroughout the job whenever practicable This is particularlyimportant for the sand ndash and for the coarse aggregate when anexposed-aggregate finish is required

Size and grading The maximum size of coarse aggregateto be used is dependent on the type of work to be done Forreinforced concrete it should be such that the concrete can beplaced without difficulty surrounding all the reinforcementthoroughly and filling the corners of the formwork In theUnited Kingdom it is usual for the coarse aggregate to havea maximum size of 20 mm Smaller aggregate usually with amaximum size of 10 mm may be needed for concrete that is tobe placed through congested reinforcement and in thin sectionswith small covers In this case the cement content may haveto be increased by 10ndash20 to achieve the same strength andworkability as that obtained with a 20 mm maximum-sizedaggregate because both sand and water contents usually haveto be increased to produce a cohesive mix Larger aggregatewith a maximum size of 40 mm can be used for foundationsand mass concrete where there are no restrictions to the flowof the concrete It should be noted however that this sort ofconcrete is not always available from ready-mixed concreteproducers The use of a larger aggregate results in a slightlyreduced water demand and hence a slightly reduced cementcontent for a given strength and workability

The proportions of the different sizes of particles makingup the aggregate which are found by sieving are known asthe aggregate lsquogradingrsquo The grading is given in terms of thepercentage by mass passing the various sieves Continuously

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graded aggregates for concrete contain particles ranging in sizefrom the largest to the smallest in gap-graded aggregatessome of the intermediate sizes are absent Gap grading may benecessary to achieve certain surface finishes Sieves used formaking a sieve analysis should conform to BS EN 933-2Recommended sieve sizes typically range from 80 to 2 mm forcoarse aggregates and from 8 to 025 mm for fine aggregatesTests should be carried out in accordance with the proceduregiven in BS EN 933-1

An aggregate containing a high proportion of large particlesis referred to as being lsquocoarselyrsquo graded and one containing ahigh proportion of small particles as lsquofinelyrsquo graded Overallgrading limits for coarse fine and lsquoall-inrsquo aggregates arecontained in BS EN 12620 and PD 6682-1 All-in aggregatescomprising both coarse and fine materials should not be usedfor structural reinforced concrete work because the gradingwill vary considerably from time to time and hence from batchto batch thus resulting in excessive variation in the consistenceand the strength of the concrete To ensure that the properamount of sand is present the separate delivery storage andbatching of coarse and fine materials is essential Graded coarseaggregates that have been produced by layer loading (ie fillinga lorry with say two grabs of material size 10ndash20 mm andone grab of material size 4ndash10 mm) are seldom satisfactorybecause the unmixed materials will not be uniformly gradedThe producer should ensure that such aggregates are effectivelymixed before loading into lorries

For a high degree of control over concrete production andparticularly if high-quality surface finishes are required it isnecessary for the coarse aggregate to be delivered stored andbatched using separate single sizes

The overall grading limits for coarse and fine aggregates asrecommended in BS EN 12620 are given in Table 217 Thelimits vary according to the aggregate size indicated as dD inmillimetres where d is the lower limiting sieve size and D isthe upper limiting sieve size for example 420 Additionally thecoarsenessfineness of the fine aggregate is assessed againstthe percentage passing the 05 mm sieve to give a CP MPFP grading This compares with the C (coarse) M (medium)F (fine) grading used formerly in BS 882 Good concrete canbe made using sand within the overall limits but there may beoccasions such as where a high degree of control is requiredor a high-quality surface finish is to be achieved when it isnecessary to specify the grading to even closer limits On theother hand sand whose grading falls outside the overall limitsmay still produce perfectly satisfactory concrete Maintaining areasonably uniform grading is generally more important thanthe grading limits themselves

Marine-dredged aggregates Large quantities of aggregatesobtained by dredging marine deposits have been widely andsatisfactorily used for making concrete for many years Ifpresent in sufficient quantities hollow andor flat shells canaffect the properties of both fresh and hardened concrete andtwo categories for shell content are given in BS EN 12620 Inorder to reduce the corrosion risk of embedded metal limitsfor the chloride content of concrete are given in BS EN 206-1and BS 8500 To conform to these limits it is necessary formarine-dredged aggregates to be carefully and efficientlywashed in fresh water that is frequently changed in order toreduce the salt content Chloride contents should be checked

frequently throughout aggregate production in accordance withthe method given in BS EN 1744-1

Some sea-dredged sands tend to have a preponderance ofone size of particle and a deficiency in the amount passing the025 mm sieve This can lead to mixes prone to bleeding unlessmix proportions are adjusted to overcome the problemIncreasing the cement content by 5ndash10 can often offset thelack of fine particles in the sand Beach sands are generallyunsuitable for good-quality concrete since they are likely tohave high concentrations of chloride due to the accumulation ofsalt crystals above the high-tide mark They are also oftensingle-sized which can make the mix design difficult

Lightweight aggregates In addition to natural gravels andcrushed rocks a number of manufactured aggregates are alsoavailable for use in concrete Aggregates such as sintered pfaare required to conform to BS EN 13055-1 and PD 6682-4

Lightweight aggregate has been used in concrete for manyyears ndash the Romans used pumice in some of their constructionwork Small quantities of pumice are imported and still used inthe United Kingdom mainly in lightweight concrete blocksbut most lightweight aggregate concrete uses manufacturedaggregate

All lightweight materials are relatively weak because of theirhigher porosity which gives them reduced weight The resultinglimitation on aggregate strength is not normally a problemsince the concrete strength that can be obtained still exceedsmost structural requirements Lightweight aggregates are usedto reduce the weight of structural elements and to giveimproved thermal insulation and fire resistance

313 Water

The water used for mixing concrete should be free fromimpurities that could adversely affect the process of hydrationand consequently the properties of concrete For examplesome organic matter can cause retardation whilst chloridesmay not only accelerate the stiffening process but also causeembedded steel such as reinforcement to corrode Otherchemicals like sulfate solutions and acids can have harmfullong-term effects by dissolving the cement paste in concreteIt is important therefore to be sure of the quality of water If itcomes from an unknown source such as a pond or boreholeit needs to be tested BS EN 1008 specifies requirements forthe quality of the water and gives procedures for checking itssuitability for use in concrete

Drinking water is suitable of course and it is usual simplyto obtain a supply from the local water utility Some recycledwater is being increasingly used in the interests of reducing theenvironmental impact of concrete production Seawater hasalso been used successfully in mass concrete with no embeddedsteel Recycled water systems are usually found at large-scalepermanent mixing plants such as precast concrete factories andready-mixed concrete depots where water that has been usedfor cleaning the plant and washing out mixers can be collectedfiltered and stored for re-use Some systems are able to reclaimup to a half of the mixing water in this way Large volumesettlement tanks are normally required The tanks do not needto be particularly deep but should have a large surface area andideally the water should be made to pass through a series ofsuch tanks becoming progressively cleaner at each stage

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314 Admixtures

An admixture is a material usually a liquid which is added toa batch of concrete during mixing to modify the properties ofthe fresh or the hardened concrete in some way Most admix-tures benefit concrete by reducing the amount of free waterneeded for a given level of consistence often in addition tosome other specific improvement Permeability is therebyreduced and durability increased There are occasions when theuse of an admixture is not only desirable but also essentialBecause admixtures are added to concrete mixes in smallquantities they should be used only when a high degree ofcontrol can be exercised Incorrect dosage of an admixture canadversely affect strength and other properties of the concreteRequirements for the following main types of admixture arespecified in BS EN 934-2

Normal water-reducing admixtures Commonly knownas plasticisers or workability aids these act by reducing theinter-particle attraction within the cement to produce a moreuniform dispersion of the cement grains The cement paste isbetter lsquolubricatedrsquo and hence the amount of water needed toobtain a given consistency can be reduced The use of theseadmixtures can be beneficial in one of three ways

When added to a normal concrete at normal dosage theyproduce an increase in slump of about 50 mm This can beuseful in high-strength concrete rich in cement which wouldotherwise be too stiff to place

The water content can be reduced while maintaining the samecement content and consistence class the reduction in watercement ratio (about 10) results in increased strength andimproved durability This can also be useful for reducingbleeding in concrete prone to this problem and for increasingthe cohesion and thereby reducing segregation in concrete ofhigh consistence or in harsh mixes that sometimes arise withangular aggregates or low sand contents or when the sand isdeficient in fines

The cement content can be reduced while maintaining thesame strength and consistence class The watercement ratiois kept constant and the water and cement contents arereduced accordingly This approach should never be used ifthereby the cement content would be reduced below theminimum specified amount

Too big a dosage may result in retardation andor a degree ofair-entrainment without necessarily increasing workabilityand therefore may be of no benefit in the fresh concrete

Accelerating water-reducing admixtures Acceleratorsact by increasing the initial rate of chemical reaction betweenthe cement and the water so that the concrete stiffens hardensand develops strength more quickly They have a negligibleeffect on consistence and the 28-day strengths are seldomaffected Accelerating admixtures have been used mainlyduring cold weather when the slowing down of the chemicalreaction between cement and water at low temperature couldbe offset by the increased speed of reaction resulting fromthe accelerator The most widely used accelerator used to becalcium chloride but because the presence of chlorides even insmall amounts increases the risk of corrosion modern standardsprohibit the use of admixtures containing chlorides in all concrete

containing embedded metal Accelerators are sometimesmarketed under other names such as hardeners or anti-freezersbut no accelerator is a true anti-freeze and the use of anaccelerator does not avoid the need to protect the concrete incold weather by keeping it warm (with insulation) after ithas been placed

Retarding water-reducing admixtures These slow downthe initial reaction between cement and water by reducingthe rate of water penetration to the cement By slowing down thegrowth of the hydration products the concrete stays workablelonger than it otherwise would The length of time during whichconcrete remains workable depends on its temperature consis-tence class and watercement ratio and on the amount of retarderused Although the occasions justifying the use of retarders in theUnited Kingdom are limited these admixtures can be helpfulwhen one or more of the following conditions apply

In warm weather when the ambient temperature is higherthan about 20oC to prevent early stiffening (lsquogoing-offrsquo) andloss of workability which would otherwise make the placingand finishing of the concrete difficult

When a large concrete pour which will take several hours tocomplete must be constructed so that concrete already placeddoes not harden before the subsequent concrete can bemerged with it (ie without a cold joint)

When the complexity of a slip-forming operation requires aslow rate of rise

When there is a delay of more than 30 minutes betweenmixing and placing ndash for example when ready-mixed concreteis being used over long-haul distances or there are risks oftraffic delays This can be seriously aggravated during hotweather especially if the cement content is high

The retardation can be varied by altering the dosage a delayof 4ndash6 hours is usual but longer delays can be obtained forspecial purposes While the reduction in early strength ofconcrete may affect formwork-striking times the 7-day and28-day strengths are not likely to be significantly affectedRetarded concrete needs careful proportioning to minimisebleeding due to the longer period during which the concreteremains fresh

Air-entraining admixtures These may be organic resinsor synthetic surfactants that entrain a controlled amount of airin concrete in the form of small air bubbles The bubbles needto be about 50 microns in diameter and well dispersed Themain reason for using an air-entraining admixture is that thepresence of tiny bubbles in the hardened concrete increases itsresistance to the action of freezing and thawing especiallywhen this is aggravated by the application of de-icing saltsand fluids Saturated concrete ndash as most external paving willbe ndash can be seriously affected by the freezing of water inthe capillary voids which will expand and try to burst it If theconcrete is air-entrained the air bubbles which intersect thecapillaries stay unfilled with water even when the concrete issaturated Thus the bubbles act as pressure relief valves andcushion the expansive effect by providing voids into which thewater can expand as it freezes without disrupting the concreteWhen the ice melts surface tension effects draw the water backout of the bubbles

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Air-entrained concrete should be specified and used for allforms of external paving from major roads and airfieldrunways down to garage drives and footpaths which are likelyto be subjected to severe freezing and to de-icing salts The saltsmay be applied directly or come from the spray of passingtraffic or by dripping from the underside of vehicles

Air-entrainment also affects the properties of the freshconcrete The minute air bubbles act like ball bearings and havea plasticising effect resulting in a higher consistence Concretethat is lacking in cohesion or harsh or which tends to bleedexcessively is greatly improved by air-entrainment The riskof plastic settlement and plastic-shrinkage cracking is alsoreduced There is also evidence that colour uniformity isimproved and surface blemishes reduced One factor that has tobe taken into account when using air-entrainment is that thestrength of the concrete is reduced by about 5 for every 1 ofair entrained However the plasticising effect of the admixturemeans that the water content of the concrete can be reducedwhich will offset most of the strength loss that would otherwiseoccur but even so some increase in the cement content is likelyto be required

High-range water-reducing admixtures Commonlyknown as superplasticizers these have a considerable plasticizingeffect on concrete They are used for one of two reasons

To greatly increase the consistence of a concrete mix so thata lsquoflowingrsquo concrete is produced that is easy both to placeand to compact some such concretes are completely self-compacting and free from segregation

To produce high-strength concrete by reducing the watercontent to a much greater extent than can be achieved byusing a normal plasticizer (water-reducing admixture)

A flowing concrete is usually obtained by first producing aconcrete whose slump is in the range 50ndash90 mm and thenadding the superplasticizer which increases the slump to over200 mm This high consistence lasts for only a limited periodof time stiffening and hardening of the concrete then proceednormally Because of this time limitation when ready-mixedconcrete is being used it is usual for the superplasticizer to beadded to the concrete on site rather than at the batching ormixing plant Flowing concrete can be more susceptible tosegregation and bleeding so it is essential for the mix designand proportions to allow for the use of a superplasticizer As ageneral guide a conventionally designed mix needs to bemodified by increasing the sand content by about 5 A highdegree of control over the batching of all the constituents isessential especially the water because if the consistence of theconcrete is not correct at the time of adding the superplasticizerexcessive flow and segregation will occur

The use of flowing concrete is likely to be limited to workwhere the advantages in ease and speed of placing offset theincreased cost of the concrete ndash considerably more than withother admixtures Typical examples are where reinforcement isparticularly congested making both placing and vibrationdifficult and where large areas such as slabs would benefitfrom a flowing easily placed concrete The fluidity of flowingconcrete increases the pressures on formwork which should bedesigned to resist full hydrostatic pressure

When used to produce high-strength concrete reductions inwater content of as much as 30 can be obtained by using

superplasticizers compared to 10 with normal plasticizers asa result 1-day and 28-day strengths can be increased by as muchas 50 Such high-strength water-reduced concrete is used bothfor high-performance in situ concrete construction and for themanufacture of precast units where the increased early strengthallows earlier demoulding

315 Properties of fresh and hardening concrete

Workability It is vital that the workability of concrete ismatched to the requirements of the construction process Theease or difficulty of placing concrete in sections of varioussizes and shapes the type of compaction equipment neededthe complexity of the reinforcement the size and skills of theworkforce are amongst the items to be considered In generalthe more difficult it is to work the concrete the higher shouldbe the level of workability But the concrete must also havesufficient cohesiveness in order to resist segregation andbleeding Concrete needs to be particularly cohesive if it is tobe pumped or allowed to fall from a considerable height

The workability of fresh concrete is increasingly referred toin British and European standards as consistence The slumptest is the best-known method for testing consistence and theslump classes given in BS EN 206-1 are S1 (10ndash40 mm)S2 (50ndash90 mm) S3 (100ndash150 mm) S4 (160ndash210 mm) Threeother test methods recognised in BS EN 206-1 all with theirown unique consistency classes are namely Vebe timedegree of compactability and flow diameter

Plastic cracking There are two basic types of plastic cracksplastic settlement cracks which can develop in deep sectionsand often follow the pattern of the reinforcement and plasticshrinkage cracks which are most likely to develop in slabsBoth types form while the concrete is still in its plastic statebefore it has set or hardened and depending on the weatherconditions within about one to six hours after the concrete hasbeen placed and compacted They are often not noticed until thefollowing day Both types of crack are related to the extent towhich the fresh concrete bleeds

Fresh concrete is a suspension of solids in water and after ithas been compacted there is a tendency for the solids (bothaggregates and cement) to settle The sedimentation processdisplaces water which is pushed upwards and if excessiveappears as a layer on the surface This bleed water may notalways be seen since it can evaporate on hot or windy daysfaster than it rises to the surface Bleeding can generally bereduced by increasing the cohesiveness of the concrete This isusually achieved by one or more of the following meansincreasing the cement content increasing the sand contentusing a finer sand using less water air-entrainment using arounded natural sand rather than an angular crushed one Therate of bleeding will be influenced by the drying conditionsespecially wind and bleeding will take place for longer on colddays Similarly concrete containing a retarder tends to bleed fora longer period of time due to the slower stiffening rate ofthe concrete and the use of retarders will in general increasethe risk of plastic cracking

Plastic settlement cracks caused by differential settlementare directly related to the amount of bleeding They tend tooccur in deep sections particularly deep beams but they may

Concrete 19

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also develop in columns and walls This is because the deeperthe section the greater the sedimentation or settlement thatcan occur However cracks will form only where somethingprevents the concrete lsquosolidsrsquo from settling freely The mostcommon cause of this is the reinforcement fixed at the top ofdeep sections the concrete will be seen to lsquohang-uprsquo over thebars and the pattern of cracks will directly reflect the layout ofthe reinforcement below Plastic settlement cracks can alsooccur in trough and waffle slabs or at any section where thereis a significant change in the depth of concrete If alterationsto the concrete for example the use of an air-entraining orwater-reducing admixture cannot be made due to contractualor economic reasons the most effective way of eliminatingplastic settlement cracking is to re-vibrate the concrete afterthe cracks have formed Such re-vibration is acceptable whenthe concrete is still plastic enough to be capable of beinglsquofluidizedrsquo by a poker but not so stiff that a hole is left when thepoker is withdrawn The prevailing weather conditions willdetermine the timing of the operation

Plastic shrinkage cracks occur in horizontal slabs such asfloors and pavements They usually take the form of one ormore diagonal cracks at 05ndash2 m centres that do not extendto the slab edges or they form a very large pattern of mapcracking Such cracks are most common in concrete placed onhot or windy days because they are caused by the rate ofevaporation of moisture from the surface exceeding the rateof bleeding Clearly plastic shrinkage cracks can be reducedby preventing the loss of moisture from the concrete surface inthe critical first few hours While sprayed-on resin-based curingcompounds are very efficient at curing concrete that has alreadyhardened they cannot be used on fresh concrete until the freebleed water has evaporated This is too late to prevent plasticshrinkage cracking and so the only alternative is to protect theconcrete for the first few hours with polythene sheeting Thisneeds to be supported clear of the concrete by means of blocksor timber but with all the edges held down to prevent a wind-tunnel effect It has been found that plastic shrinkage crackingis virtually non-existent when air-entrainment is used

The main danger from plastic cracking is the possibility ofmoisture ingress leading to corrosion of any reinforcement Ifthe affected surface is to be covered subsequently by eithermore concrete or a screed no treatment is usually necessaryIn other cases often the best repair is to brush dry cement(dampened down later) or wet grout into the cracks the day afterthey form and while they are still clean this encourages naturalor autogenous healing

Early thermal cracking The reaction of cement with wateror hydration is a chemical reaction that produces heat If thisheat development exceeds the rate of heat loss the concretetemperature will rise Subsequently the concrete will cool andcontract Typical temperature histories of different concretesections are shown in the figure on Table 218

If the contraction of the concrete were unrestrained therewould be no cracking at this stage However in practice thereis nearly always some form of restraint inducing tension andhence a risk of cracks forming The restraint can occur due toboth external and internal influences Concrete is externallyrestrained when for example it is cast onto a previously castbase such as a wall kicker or between two already hardenedsections such as in infill bay in a wall or slab without the

provision of a contraction joint Internal restraint occurs forexample because the surfaces of an element will cool fasterthan the core producing a temperature differential When thisdifferential is large such as in thick sections surface cracksmay form at an early stage Subsequently as the core of thesection cools these surface cracks will tend to close in theabsence of any external restraints Otherwise the cracks willpenetrate into the core and link up to form continuous cracksthrough the whole section

The main factors affecting the temperature rise in concreteare the dimensions of the section the cement content andtype the initial temperature of the concrete and the ambienttemperature the type of formwork and the use of admixturesThicker sections retain more heat giving rise to higher peaktemperatures and cool down more slowly Within the coreof very thick sections adiabatic conditions obtain and abovea thickness of about 15 m there is little further increasein the temperature of the concrete The heat generated isdirectly related to the cement content For Portland cementconcretes in sections of thickness 1 m and more the temper-ature rise in the core is likely to be about 14oC for every100 kgm3 of cement Thinner sections will exhibit lowertemperature rises

Different cement types generate heat at different rates Thepeak temperature and the total amount of heat produced byhydration depend upon both the fineness and the chemistry ofthe cement As a guide the cements whose strength developsmost rapidly tend to produce the most heat Sulfate-resistingcement generally gives off less heat than CEM I and cementsthat are inter-ground or combined with mineral additions suchas pfa or ggbs are often chosen for massive constructionbecause of their low heat of hydration

A higher initial temperature results in a greater temperaturerise for example concrete in a 500 mm thick section placedat 10oC could have a temperature rise of 30oC but the sameconcrete placed at 20oC may have a temperature rise of 40oCSteel and GRP formwork will allow the heat generated to bedissipated more quickly than will timber formwork resultingin lower temperature rises especially in thinner sectionsTimber formwork andor additional insulation will reduce thetemperature differential between the core and the surface ofthe section but this differential could increase significantlywhen the formwork is struck Retarding water-reducers willdelay the onset of hydration but do not reduce the total heatgenerated Accelerating water-reducers will increase the rate ofheat evolution and the temperature rise

The problem of early thermal cracking is usually confined toslabs and walls Walls are particularly susceptible becausethey are often lightly reinforced in the horizontal directionand the timber formwork tends to act as a thermal insulatorencouraging a larger temperature rise The problem could bereduced by lowering the cement content and using cementwith a lower heat of hydration or one containing ggbs or pfaHowever there are practical and economic limits to thesemeasures often dictated by the specification requirements forthe strength and durability of the concrete itself In practicecracking due to external restraint is generally dealt with byproviding crack control reinforcement and contraction jointsWith very thick sections and very little external restraint ifthe temperature differential can be controlled by insulating theconcrete surfaces for a few days cracking can be avoided

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Typical values of the temperature rise in walls and slabs forPortland cement concretes as well as comparative values forconcrete using other cements are given in Table 218 Furtherdata on predicted temperature rises is given in ref 11

316 Properties of hardened concrete

Compressive strength The strength of concrete is specifiedas a strength class or grade namely the 28-day characteristiccompressive strength of specimens made from fresh concreteunder standardised conditions The results of strength tests areused routinely for control of production and contractual confor-mity purposes The characteristic strength is defined as that levelof strength below which 5 of all valid test results is expectedto fall Test cubes either 100 mm or 150 mm are the specimensnormally used in the United Kingdom and most other Europeancountries but cylinders are used elsewhere Because theirbasic shapes (ratio of height to cross-sectional dimension) aredifferent the strength test results are also different cylindersbeing weaker than cubes For normal-weight aggregates theconcrete cylinder strength is about 80 of the correspondingcube strength For lightweight aggregates cylinder strengths areabout 90 of the corresponding cube strengths

In British Codes of Practice like BS 8110 strength gradesused to be specified in terms of cube strength (eg C30) asshown in Table 39 Nowadays strength classes are specified interms of both cylinder strength and equivalent cube strength(eg C2530) as shown in Tables 35 and 42

In principle compressive strengths can be determined fromcores cut from the hardened concrete Core tests are normallymade only when there is some doubt about the quality ofconcrete placed (eg if the cube results are unsatisfactory) orto assist in determining the strength and quality of an existingstructure for which records are not available Great care isnecessary in the interpretation of the results of core tests andsamples drilled from in situ concrete are expected to be lowerin strength than cubes made cured and tested under standardlaboratory conditions The standard reference for core testingis BS EN 12504-1

Tensile strength The direct tensile strength of concrete asa proportion of the cube strength varies from about one-tenthfor low-strength concretes to one-twentieth for high-strengthconcretes The proportion is affected by the aggregate used andthe compressive strength is therefore only a very general guideto the tensile strength For specific design purposes in regard tocracking and shear strength analytical relationships betweenthe tensile strength and the specified cylindercube strength areprovided in codes of practice

The indirect tensile strength (or cylinder splitting strength) isseldom specified nowadays Flexural testing of specimens maybe used on some airfield runway contracts where the methodof design is based on the modulus of rupture and for someprecast concrete products such as flags and kerbs

Elastic properties The initial behaviour of concrete underservice load is almost elastic but under sustained loading thestrain increases with time Stressndashstrain tests cannot be carriedout instantaneously and there is always a degree of non-linearityand a residual strain upon unloading For practical purpose theinitial deformation is considered to be elastic (recoverable

upon unloading) and the subsequent increase in strain undersustained stress is defined as creep The elastic modulus onloading defined in this way is a secant modulus related to aspecific stress level The value of the modulus of elasticity ofconcrete is influenced mainly by the aggregate used With aparticular aggregate the value increases with the strength of theconcrete and the age at loading In special circumstances Forexample where deflection calculations are of great importanceload tests should be carried out on concrete made with theaggregate to be used in the actual structure For most designpurposes specific values of the mean elastic modulus at28 days and of Poissonrsquos ratio are given in Table 35 forBS 8110 and Table 42 for EC 2

Creep The increase in strain beyond the initial elastic valuethat occurs in concrete under a sustained constant stress aftertaking into account other time-dependent deformations notassociated with stress is defined as creep If the stress isremoved after some time the strain decreases immediately byan amount that is less than the original elastic value becauseof the increase in the modulus of elasticity with age This isfollowed by a further gradual decrease in strain The creeprecovery is always less than the preceding creep so that thereis always a residual deformation

The creep source in normal-weight concrete is the hardenedcement paste The aggregate restrains the creep in the paste sothat the stiffer the aggregate and the higher its volumetricproportion the lower is the creep of the concrete Creep isalso affected by the watercement ratio as is the porosity andstrength of the concrete For constant cement paste contentcreep is reduced by a decrease in the watercement ratio

The most important external factor influencing creep is therelative humidity of the air surrounding the concrete For aspecimen that is cured at a relative humidity of 100 thenloaded and exposed to different environments the lower therelative humidity the higher is the creep The values are muchreduced in the case of specimens that have been allowed todry prior to the application of load The influence of relativehumidity on creep is dependent on the size of the memberWhen drying occurs at constant relative humidity the largerthe specimen the smaller is the creep This size effect isexpressed in terms of the volumesurface area ratio of themember If no drying occurs as in mass concrete the creep isindependent of size

Creep is inversely proportional to concrete strength at the ageof loading over a wide range of concrete mixes Thus for agiven type of cement the creep decreases as the age andconsequently the strength of the concrete at application of theload increases The type of cement temperature and curingconditions all influence the development of strength with age

The influence of temperature on creep is important in the useof concrete for nuclear pressure vessels and containers forstoring liquefied gases The time at which the temperature ofconcrete rises relative to the time at which load is appliedaffects the creepndashtemperature relation If saturated concrete isheated and loaded at the same time the creep is greater thanwhen the concrete is heated during the curing period prior to theapplication of load At low temperatures creep behaviour isaffected by the formation of ice As the temperature falls creepdecreases until the formation of ice causes an increase in creepbut below the ice point creep again decreases

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Creep is normally assumed to be directly proportional toapplied stress within the service range and the term specificcreep is used for creep per unit of stress At stresses aboveabout one-third of the cube strength (45 cylinder strength)the formation of micro-cracks causes the creepndashstress relationto become non-linear creep increasing at an increasing rate

The effect of creep is unfavourable in some circumstances(eg increased deflection) and favourable in others (eg reliefof stress due to restraint of imposed deformations such asdifferential settlement seasonal temperature change)

For normal exposure conditions (inside and outside) creepcoefficients according to ambient relative humidity effectivesection thickness (notional size) and age of loading are givenin Table 35 for BS 8110 and Table 43 for EC 2

Shrinkage Withdrawal of water from hardened concretekept in unsaturated air causes drying shrinkage If concretethat has been left to dry in air of a given relative humidity issubsequently placed in water (or a higher relative humidity)it will swell due to absorption of water by the cement pasteHowever not all of the initial drying shrinkage is recoveredeven after prolonged storage in water For the usual rangeof concretes the reversible moisture movement representsabout 40ndash70 of the drying shrinkage A pattern of alternatewetting and drying will occur in normal outdoor conditionsThe magnitude of the cyclic movement clearly depends uponthe duration of the wetting and drying periods but drying ismuch slower than wetting The consequence of prolonged dryweather can be reversed by a short period of rain More stableconditions exist indoors (dry) and in the ground or in contactwith water (eg reservoirs and tanks)

Shrinkage of hardened concrete under drying conditions isinfluenced by several factors in a similar manner to creep Theintrinsic shrinkage of the cement paste increases with thewatercement ratio so that for a given aggregate proportionconcrete shrinkage is also a function of watercement ratio

The relative humidity of the air surrounding the membergreatly affects the magnitude of concrete shrinkage accordingto the volumesurface area ratio of the member The lowershrinkage value of large members is due to the fact that dryingis restricted to the outer parts of the concrete the shrinkage ofwhich is restrained by the non-shrinking core Clearly shrink-able aggregates present special problems and can greatlyincrease concrete shrinkage (ref 12)

For normal exposure conditions (inside and outside) valuesof drying shrinkage according to ambient relative humidity andeffective section thickness (notional size) are given in Table 35for BS 8110 and Table 42 for EC 2

Thermal properties The coefficient of thermal expansionof concrete depends on both the composition of the concreteand its moisture condition at the time of the temperaturechange The thermal coefficient of the cement paste is higherthan that of the aggregate which exerts a restraining influenceon the movement of the cement paste The coefficient of thermalexpansion of a normally cured paste varies from the lowestvalues when the paste is either totally dry or saturated to amaximum at a relative humidity of about 70 Values for theaggregate are related to their mineralogical composition

A value for the coefficient of thermal expansion of concreteis needed in the design of structures such as chimneys tanks

containing hot liquids bridges and other elevated structuresexposed to significant solar effects and for large expanses ofconcrete where provision must be made to accommodate theeffects of temperature change in controlled cracking or byproviding movement joints For normal design purposes valuesof the coefficient of thermal expansion of concrete accordingto the type of aggregate are given in Table 35 for BS 8110 andTable 42 for EC 2

Short-term stressndashstrain curves For normal low to mediumstrength unconfined concrete the stressndashstrain relationship incompression is approximately linear up to about one-third ofthe cube strength (40 of cylinder strength) With increasingstress the strain increases at an increasing rate and a peakstress (cylinder strength) is reached at a strain of about 0002With increasing strain the stress reduces until failure occurs ata strain of about 00035 For higher strength concretes the peakstress occurs at strains 0002 and the failure occurs atstrains 00035 the failure being progressively more brittle asthe concrete strength increases

For design purposes the short-term stressndashstrain curve isgenerally idealised to a form in which the initial portion isparabolic or linear and the remainder is at a uniform stress Afurther simplification in the form of an equivalent rectangularstress block may be made subsequently Typical stressndashstraincurves and those recommended for design purposes are givenin Table 36 for BS 8110 and Table 44 for EC 2

317 Durability of concrete

Concrete has to be durable in natural environments rangingfrom mild to extremely aggressive and resistant to factors suchas weathering freezethaw attack chemical attack and abrasionIn addition for concrete containing reinforcement the surfaceconcrete must provide adequate protection against the ingressof moisture and air which would eventually cause corrosion ofthe embedded steel

Strength alone is not necessarily a reliable guide to concretedurability many other factors have to be taken into accountthe most important being the degree of impermeability This isdependent mainly on the constituents of the concrete in partic-ular the free watercement ratio and in the provision of fullcompaction to eliminate air voids and effective curing toensure continuing hydration

Concrete has a tendency to be permeable as a result ofthe capillary voids in the cement paste matrix In order for theconcrete to be sufficiently workable it is common to use farmore water than is actually necessary for the hydration of thecement When the concrete dries out the space previouslyoccupied by the excess water forms capillary voids Providedthe concrete has been fully compacted and properly cured thevoids are extremely small the number and the size of the voidsdecreasing as the free watercement ratio is reduced The moreopen the structure of the cement paste the easier it is for airmoisture and harmful chemicals to penetrate

Carbonation Steel reinforcement that is embedded in goodconcrete with an adequate depth of cover is protectedagainst corrosion by the highly alkaline pore water in thehardened cement paste Loss of alkalinity of the concrete canbe caused by the carbon dioxide in the air reacting with and

Material properties22

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neutralising the free lime If this reaction which is calledcarbonation reaches the reinforcement then corrosion willoccur in moist environments Carbonation is a slow processthat progresses from the surface and is dependent on thepermeability of the concrete and the humidity of the environ-ment Provided the depth of cover and quality of concreterecommended for the anticipated exposure conditions areachieved corrosion due to carbonation should not occur duringthe intended lifetime of the structure

Freezethaw attack The resistance of concrete to freezingand thawing depends on its impermeability and the degreeof saturation on being exposed to frost the higher the degree ofsaturation the more liable the concrete is to damage The useof salt for de-icing roads and pavements greatly increases therisk of freezethaw damage

The benefits of air-entrained concrete have been referred toin section 314 where it was recommended that all exposedhorizontal paved areas from roads and runways to footpathsand garage drives and marine structures should be made ofair-entrained concrete Similarly parts of structures adjacent tohighways and in car parks which could be splashed or comeinto contact with salt solutions used for de-icing should alsouse air-entrained concrete Alternatively the cube strength ofthe concrete should be 50 Nmm2 or more Whilst C4050concrete is suitable for many situations it does not have thesame freezethaw resistance as air-entrained concrete

Chemical attack Portland cement concrete is liable toattack by acids and acid fumes including the organic acidsoften produced when foodstuffs are being processed Vinegarfruit juices silage effluent sour milk and sugar solutions can allattack concrete Concrete made with Portland cement is notrecommended for use in acidic conditions where the pH valueis 55 or less without careful consideration of the exposurecondition and the intended construction Alkalis have little effecton concrete

For concrete that is exposed to made-up ground includingcontaminated and industrial material specialist advice shouldbe sought in determining the design chemical class so that asuitable concrete can be specified The most common formof chemical attack that concretes have to resist is the effect ofsolutions of sulfates present in some soils and groundwaters

In all cases of chemical attack concrete resistance is related tofree watercement ratio cement content type of cement and thedegree of compaction Well-compacted concrete will always bemore resistant to sulfate attack than one less well compactedregardless of cement type Recommendations for concreteexposed to sulfate-containing groundwater and for chemicallycontaminated brownfield sites are incorporated in BS 8500-1

Alkalindashsilica reaction ASR is a reaction that can occur inconcrete between certain siliceous constituents present in theaggregate and the alkalis ndash sodium and potassium hydroxide ndashthat are released during cement hydration A gelatinous productis formed which imbibes pore fluid and in so doing expandsinducing an internal stress within the concrete The reactionwill cause damage to the concrete only when the followingthree conditions occur simultaneously

A reactive form of silica is present in the aggregate in criticalquantities

The pore solution contains ions of sodium potassium andhydroxyl and is of a sufficiently high alkalinity

A continuing supply of water is available

If any one of these factors is absent then damage from ASRwill not occur and no precautions are necessary It is possiblefor the reaction to take place in the concrete without inducingexpansion Damage may not occur even when the reactionproduct is present throughout the concrete as the gel may fillcracks induced by some other mechanism Recommendationsare available for minimising the risk of damage from ASR innew concrete construction based on ensuring that at least oneof the three aforementioned conditions is absent

Exposure classes For design and specification purposes theenvironment to which concrete will be exposed during itsintended life is classified into various levels of severity Foreach category minimum requirements regarding the qualityof the concrete and the cover to the reinforcement are givenin Codes of Practice In British Codes for many years theexposure conditions were mild moderate severe very severeand most severe (or in BS 5400 extreme) with abrasive as afurther category Details of the classification system that wasused in BS 8110 and BS 5400 are given in Table 39

In BS EN 206-1 BS 8500-1 and EC 2 the conditionsare classified in terms of exposure to particular actions withvarious levels of severity in each category The followingcategories are considered

1 No risk of corrosion or attack

2 Corrosion induced by carbonation

3 Corrosion induced by chlorides other than from seawater

4 Corrosion induced by chlorides from seawater

5 Freezethaw attack

6 Chemical attack

If the concrete is exposed to more than one of these actions theenvironmental conditions are expressed as a combination ofexposure classes Details of each class in categories 1ndash5 withdescriptions and informative examples applicable in the UnitedKingdom are given in Tables 37 and 45 For concrete exposedto chemical attack the exposure classes given in BS EN 206-1cover only natural ground with static water which represents alimited proportion of the aggressive ground conditions found inthe United Kingdom In the complementary British StandardBS 8500-1 more comprehensive recommendations are providedbased on the approach used in ref 13

On this basis an ACEC (aggressive chemical environmentfor concrete) class is determined according to the chemicals inthe ground the type of soil and the mobility and acidity of thegroundwater The chemicals in the ground are expressed as adesign sulfate class (DS) in which the measured sulfate contentis increased to take account of materials that may oxidise intosulfate for example pyrite and other aggressive species suchas hydrochloric or nitric acid Magnesium ion content is alsoincluded in this classification Soil is classified as natural orfor sites that may contain chemical residues from previousindustrial use or imported wastes as brownfield Water in theground is classified as either static or mobile and according toits pH value

Concrete 23

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Based on the ACEC classification and according to the sizeof the section and the selected structural performance level therequired concrete quality expressed as a design chemicalclass (DC) and any necessary additional protective measures(APMs) can be determined The structural performance level isclassified as low normal or high in relation to the intendedservice life the vulnerability of the structural details and thesecurity of structures retaining hazardous materials

Concrete quality and cover to reinforcement Concretedurability is dependent mainly on its constituents particularlythe free watercement ratio The ratio can be reduced and thedurability of the concrete enhanced by increasing the cementcontent andor using admixtures to reduce the amount of freewater needed for a particular level of consistence subject tospecified minimum requirements being met for the cementcontent By limiting the maximum free watercement ratio andthe minimum cement content a minimum strength class can beobtained for particular cements and combinations

Where concrete containing reinforcement is exposed to airand moisture or is subject to contact with chlorides from anysource the protection of the steel against corrosion depends onthe concrete cover The required thickness is related to theexposure class the concrete quality and the intended workinglife of the structure Recommended values for an intendedworking life of at least 50 years are given in Tables 38 and 46(BS 8500) and 39 (prior to BS 8500)

Codes of Practice also specify values for the covers neededto ensure the safe transmission of bond forces and provide anadequate fire-resistance for the reinforced concrete member Inaddition allowance may need to be made for abrasion or forsurface treatments such as bush hammering In BS 8110 valuesused to be given for a nominal cover to be provided to all rein-forcement including links on the basis that the actual covershould not be less than the nominal cover minus 5 mm In BS8500 values are given for a minimum cover to which anallowance for tolerance (normally 10 mm) is then added

Concrete specification Details of how to specify con-crete and what to specify are given in BS 8500-1 Threetypes ndash designed prescribed and standardised prescribedconcretes ndash are recognised by BS EN 206-1 but BS 8500adds two more ndash designated and proprietary concretes

Designed concretes are ones where the concrete producer isresponsible for selecting the mix proportions to provide theperformance defined by the specifier Conformity of designedconcretes is usually judged by strength testing of 100 mm or150 mm cubes which in BS 8500 is the responsibility ofthe concrete producer Prescribed concretes are ones where thespecification states the mix proportions in order to satisfyparticular performance requirements in terms of the mass ofeach constituent Such concretes are seldom necessary butmight be used where particular properties or special surfacefinishes are required Standardised prescribed concretes that areintended for site production using basic equipment and controlare given in BS 8500-2 Whilst conformity does not depend onstrength testing assumed characteristic strengths are given forthe purposes of design Designated concretes are a wide-ranginggroup of concretes that provide for most types of concreteconstruction The producer must operate a recognized accreditedthird party certification system and is responsible for ensuring

that the concrete conforms to the specification given inBS 8500-2 Proprietary concretes are intended to provide forinstances when a concrete producer would give assurance of theperformance of concrete without being required to declare itscomposition

For conditions where corrosion induced by chlorides doesnot apply structural concretes should generally be specifiedas either designated concretes or designed concretes Whereexposure to corrosion due to chlorides is applicable only thedesigned concrete method of specifying is appropriate Anexception to this situation is where an exposed aggregate ortooled finish that removes the concrete surface is required Inthese cases in order to get an acceptable finish a special mixdesign is needed Initial testing including trial panels shouldbe undertaken and from the results of these tests a prescribedconcrete can be specified For housing applications both adesignated concrete and a standardised prescribed concrete canbe specified as acceptable alternatives This would allow aconcrete producer with accredited certification to quote forsupplying a designated concrete and the site contractor or aconcrete producer without accredited certification to quote forsupplying a standardised prescribed concrete

32 REINFORCEMENT

Reinforcement for concrete generally consists of deformedsteel bars or welded steel mesh fabric Normal reinforcementrelies entirely upon the alkaline environment provided by adurable concrete cover for its protection against corrosion Inspecial circumstances galvanised epoxy-coated or stainlesssteel can be used Fibre-reinforced polymer materials havealso been developed So far in the United Kingdom thesematerials have been used mainly for external strengthening anddamage repair applications

321 Bar reinforcement

In the United Kingdom reinforcing bars are generally specifiedordered and delivered to the requirements of BS 4449 Thiscaters for steel bars with a yield strength of 500 MPa in threeductility classes grades B500A B500B and B500C Bars areround in cross section having two or more rows of uniformlyspaced transverse ribs with or without longitudinal ribs Thepattern of transverse ribs varies with the grade and can beused as a means of identification Information with regardto the basic properties of reinforcing bars to BS 4449 whichis in general conformity with BS EN 10080 is given inTable 219

All reinforcing bars are produced by a hot-rolling process inwhich a cast steel billet is reheated to 1100ndash1200oC andthen rolled in a mill to reduce its cross section and impart therib pattern There are two common methods for achievingthe required mechanical properties in hot-rolled bars in-lineheat treatment and micro-alloying In the former method whichis sometimes referred to as the quench-and-self-temper (QST)process high-pressure water sprays quench the bar surface as itexits the rolling mill producing a bar with a hard temperedouter layer and a softer more ductile core Most reinforcingbars in the United Kingdom are of this type and achieve class Bor class C ductility In the micro-alloying method strengthis achieved by adding small amounts of alloying elements

Material properties24

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during the steel-making process Micro-alloy steels normallyachieve class C ductility Another method that can be usedto produce high-yield bars involves a cold-twisting process toform bars that are identified by spiralling longitudinal ribs Thisprocess has been obsolete in the United Kingdom for sometime but round ribbed twisted bars can be found in someexisting structures

In addition to bars being produced in cut straight lengthsbillets are also rolled into coil for diameters up to 16 mm Inthis form the product is ideal for automated processes suchas link bending QST micro-alloying and cold deformationprocesses are all used for high-yield coil Cold deformation isapplied by continuous stretching which is less detrimental toductility than the cold-twisting process mentioned previouslyCoil products have to be de-coiled before use and automaticlink bending machines incorporate straightening rolls Largerde-coiling machines are also used to produce straight lengths

322 Fabric reinforcement

Steel fabric reinforcement is an arrangement of longitudinal barsand cross bars welded together at their intersections in a shearresistant manner In the United Kingdom fabric is producedunder a closely controlled factory-based manufacturing processto the requirements of BS 4483 In fabric for structural purposesribbed bars complying with BS 4449 are used For wrappingfabric as described later wire complying with BS 4482 may beused Wire can be produced from hot-rolled rod by eitherdrawing the rod through a die to produce plain wire or coldrolling the rod to form indented or ribbed wires In BS 4482provision is made for plain round wire with a yield strength of250 MPa and plain indented or ribbed wires with a yieldstrength of 500 MPa

In BS 4483 provision is made for fabric reinforcement tobe either of a standard type or purpose made to the clientrsquosrequirements The standard fabric types have regular mesharrangements and bar sizes and are defined by identifiablereference numbers Type A is a square mesh with identical longbars and cross bars commonly used in ground slabs Type B isa rectangular (structural) mesh that is particularly suitable foruse in thin one-way spanning slabs Type C is a rectangular(long) mesh that can be used in pavements and in two-wayspanning slabs by providing separate sheets in each directionType D is a rectangular (wrapping) mesh that is used in theconcrete encasement of structural steel sections The stock sizeof standard fabric sheets is 48 m 24 m and merchantsize sheets are also available in a 36 m 20 m size Fulldetails of the preferred range of standard fabric types are givenin Table 220

Purpose-made fabrics specified by the customer can haveany combination of wire size and spacing in either direction Inpractice manufacturers may sub-divide purpose-made fabricsinto two categories special (also called scheduled) and bespoke(also called detailed) Special fabrics consist of the standardwire size combinations but with non-standard overhangs andsheet dimensions up to 12 m 33 m Sheets with so-calledflying ends are used to facilitate the lapping of adjacent sheets

Bespoke fabrics involve a more complex arrangement inwhich the wire size spacing and length can be varied within thesheet These products are made to order for each contract as areplacement for conventional loose bar assemblies The use of

bespoke fabrics is appropriate on contracts with a large amountof repeatability and generally manufacturers would require aminimum tonnage order for commercial viability

323 Stressndashstrain curves

For hot-rolled reinforcement the stressndashstrain relationship intension is linear up to yield when there is a pronounced increaseof strain at constant stress (yield strength) Further smallincreases of stress resulting in work hardening are accompaniedby considerable elongation A maximum stress (tensile strength)is reached beyond which further elongation is accompanied bya stress reduction to failure Micro-alloy bars are characterisedby high ductility (high level of uniform elongation and high ratioof tensile strengthyield strength) For QST bars the stressndashstraincurve is of similar shape but with slightly less ductility

Cold-processed reinforcing steels show continuous yieldingbehaviour with no defined yield point The work-hardeningcapacity is lower than for the hot-rolled reinforcement withthe uniform elongation level being particularly reduced Thecharacteristic strength is defined as the 02 proof stress(ie a stress which on unloading would result in a residualstrain of 02) and the initial part of the stressndashstrain curve islinear to beyond 80 of this value

For design purposes the yield or 02 proof condition isnormally critical and the stressndashstrain curves are idealised to abi-linear or sometimes tri-linear form Typical stressndashstraincurves and those recommended for design purposes are givenin Table 36 for BS 8110 and Table 44 for EC 2

324 Bar sizes and bends

The nominal size of a bar is the diameter of a circle with an areaequal to the effective cross-sectional area of the bar The rangeof nominal sizes (millimetres) is from 6 to 50 with preferredsizes of 8 10 12 16 20 25 32 and 40 Values of the totalcross-sectional area provided in a concrete section according tothe number or spacing of the bars for different bar sizes aregiven in Table 220

Bends in bars should be formed around standard mandrels onbar-bending machines In BS 8666 the minimum radius ofbend r is standardised as 2d for d 16 and 35d for d 20where d is the bar size Values of r for each different bar sizeand values of the minimum end projection P needed to formthe bend are given in Table 219 In some cases (eg wherebars are highly stressed) the bars need to be bent to a radiuslarger than the minimum value in order to satisfy the designrequirements and the required radius R is then specified on thebar-bending schedule

Reinforcement should not be bent or straightened on site ina way that could damage or fracture the bars All bars shouldpreferably be bent at ambient temperature but when the steeltemperature is below 5oC special precautions may be neededsuch as reducing the speed of bending or with the engineerrsquosapproval increasing the radius of bending Alternatively thebars may be warmed to a temperature not exceeding 100oC

325 Bar shapes and bending dimensions

Bars are produced in stock lengths of 12 m and lengths upto 18 m can be supplied to special order In most structures

Reinforcement 25

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bars are required in shorter lengths and often need to be bentThe cutting and bending of reinforcement is generally specifiedto the requirements of BS 8666 This contains recommendedbar shapes designated by shape code numbers which areshown in Tables 221 and 222 The information needed to cutand bend the bars to the required dimensions is entered into abar schedule an example of which is shown in Table 223 Eachschedule is related to a member on a particular drawing bymeans of the bar schedule reference number

In cases where a bar is detailed to fit between two concretefaces with no more than the nominal cover on each face (eg linksin beams) an allowance for deviations is required This is tocater for variation due to the effect of inevitable errors in thedimensions of the formwork and the cutting bending andfixing of the bars Details of the deductions to be made to allowfor these deviations and calculations to determine the bendingdimensions in a typical example are given in section 1035with the completed bar schedule in Table 223

326 Stainless steel reinforcement

The type of reinforcement to be used in a structure is usuallyselected on the basis of initial costs This normally results inthe use of carbon steel reinforcement which is around 15of the cost of stainless steel For some structures however theselective use of stainless steel reinforcement ndash on exposedsurfaces for example ndash can be justified In Highways Agencydocument BA 8402 it is recommended that stainless steelreinforcement should be used in splash zones abutmentsparapet edges and soffits and where the chances of chlorideattack are greatest It is generally considered that where theconcrete is saturated and oxygen movement limited stainlesssteel is not required Adherence to these guidelines can meanthat the use of stainless steel reinforcement only marginallyincreases construction costs while significantly reducing thewhole-life costs of the structure and increasing its usable life

Stainless steels are produced by adding elements to iron toachieve the required compositional balance The additionalelements besides chromium can include nickel manganesemolybdenum and titanium with the level of carbon beingcontrolled during processing These alloying elements affectthe steelrsquos microstructure as well as its mechanical propertiesand corrosion resistance Four ranges of stainless steel areproduced two of which are recommended for reinforcement toconcrete because of their high resistance to corrosionAustenitic stainless steels for which chromium and nickel arethe main alloying elements have good general propertiesincluding corrosion resistance and are normally suitable formost applications Duplex stainless steels which have highchromium and low nickel contents provide greater corrosionresistance for the most demanding environments

In the United Kingdom austenitic stainless steel reinforcementhas been produced to the requirements of BS 6744 which isbroadly aligned to conventional reinforcement practice Thusplain and ribbed bars are available in the same characteristicstrengths and range of preferred sizes as normal carbon steelreinforcement Traditionally stainless steel reinforcement hasonly been stocked in maximum lengths of 6 m for all sizesBars are currently available in lengths up to 12 m for sizes upto 16 mm For larger sizes bars can be supplied to order in

lengths up to 8 m Comprehensive data and recommendationson the use of stainless steel reinforcement are given in ref 14

327 Prefabricated reinforcement systems

In order to speed construction by reducing the time needed tofix reinforcement it is important to be able to pre-assemblemuch of the reinforcement This can be achieved on site givenadequate space and a ready supply of skilled personnel Inmany cases with careful planning and collaboration at an earlystage the use of reinforcement assemblies prefabricated by thesupplier can provide considerable benefits

A common application is the use of fabric reinforcement asdescribed in sections 323 and 1032 The preferred range ofdesignated fabrics can be routinely used in slabs and walls Incases involving large areas with long spans and considerablerepetition made-to-order fabrics can be specially designed tosuit specific projects Provision for small holes and openingscan be made by cutting the fabric on site after placing thesheets and adding loose trimming bars as necessary Whilesheets of fabric can be readily handled normally they areawkward to lift over column starter bars In such cases it isgenerally advisable to provide the reinforcement local to thecolumn as loose bars fixed in the conventional manner

A more recent development is the use of slab reinforcementrolls that can be unrolled directly into place on site Each made-to-order roll consists of reinforcement of the required size andspacing in one direction welded to thin metal bands and rolledaround hoops that are later discarded Rolls can be produced upto a maximum bar length of 15 m and a weight of 5 tonnes Thewidth of the sheet when fully rolled out could be more than50 m depending upon the bar size and spacing The full rangeof preferred bar sizes can be used and the bar spacing andlength can be varied within the same roll For each area of slaband for each surface to be reinforced two rolls are requiredThese are delivered to site craned into position and unrolled oncontinuous bar supports Each roll provides the bars in onedirection with those in the lower layer resting on conventionalspacers or chairs

The need to provide punching shear reinforcement in solidflat slabs in the vicinity of the columns has resulted in severalproprietary reinforcement systems Vertical reinforcement isrequired in potential shear failure zones around the columnsuntil a position is reached at which the slab can withstand theshear stresses without reinforcement Conventional links aredifficult and time-consuming to set out and fix Single-leggedlinks are provided with a hook at the top and a 90o bend at thebottom Each link has to be hooked over a top bar in the slaband the 90o bend pushed under a bottom bar and tied in place

Shear ladders can be used in which a row of single-leggedlinks are connected by three straight anchor bars welded toform a robust single unit The ladders provide the required shearreinforcement and act as chairs to support the top bars The sizespacing and height of the links can be varied to suit the designrequirements Shear hoops consist of U-shaped links welded toupper and lower hoops to form a three-dimensional unit Byusing hoops of increasing size shear reinforcement can beprovided on successive perimeters

Shear band strips with a castellated profile are made from25 mm wide high-tensile steel strip in a variety of gauges to

Material properties26

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cater for different shear capacities The strip has perforatedholes along the length to help with anchorage and fixing Thepeaks and troughs of the profile are spaced to coincide withthe spacing of the main reinforcement Stud rails consist of arow of steel studs welded to a flat steel strip or a pair of rodsThe studs are fabricated from plain or deformed reinforcingbars with an enlarged head welded to one or both ends Thesize spacing and height of the studs can be varied to suit theshear requirements and the slab depth

The use of reinforcement continuity strips is a simple andeffective means of providing reinforcement continuity acrossconstruction joints A typical application occurs at a junctionbetween a wall and a slab that is to be cast at a later stageThe strips comprise a set of special pre-bent bars housed in agalvanised indented steel casing that is fabricated off-site ina factory-controlled environment On site the entire unit is castinto the front face of the wall After the formwork is struck thelid of the casing is removed to reveal the legs of the barscontained within the casing The legs are then straightenedoutwards by the contractor ready for lapping with the mainreinforcement in the slab The casing remains embedded in thewall creating a rebate into which the slab concrete flows andeliminating the need for traditional joint preparation

328 Fixing of reinforcement

Reinforcing bars need to be tied together to prevent their beingdisplaced and provide a rigid system Bar assemblies and fabricreinforcement need to be supported by spacers and chairs toensure that the required cover is achieved and kept duringthe subsequent placing and compaction of concrete Spacersshould be fixed to the links bars or fabric wires that are nearestto the concrete surface to which the cover is specifiedRecommendations for the specification and use of spacers andchairs and the tying of reinforcement are given in BS 7973Parts 1 and 2 These include details of the number and positionof spacers and the frequency of tying

33 FIRE-RESISTANCE

Building structures need to conform in the event of fire toperformance requirements stated in the Building RegulationsFor stability the elements of the structure need to providea specified minimum period of fire-resistance in relation to astandard test The required fire period depends on the purposegroup of the building and the height or for basements depth ofthe building relative to the ground as given in Table 312Building insurers may require longer fire periods for storagefacilities where the value of the contents and the costs ofreinstatement of the structure are particularly important

In BS 8110 design for fire-resistance is considered at twolevels Part 1 contains simple recommendations suitable formost purposes Part 2 contains a more detailed treatment with

a choice of three methods involving tabulated data furnacetests or fire engineering calculations The tabulated data is inthe form of minimum specified values of member size andconcrete cover The cover is given to the main reinforcementand in the case of beams and ribs can vary in relation to theactual width of the section The recommendations in Part 1 arebased on the same data but the presentation is different intwo respects values are given for the nominal cover to all rein-forcement (this includes an allowance for links in the case ofbeams and columns) and the values do not vary in relation tothe width of the section The required nominal covers to allreinforcement and minimum dimensions for various membersare given in Tables 310 and 311 respectively

In the event of a fire in a building the vulnerable elementsare the floor construction above the fire and any supportingcolumns or walls The fire-resistance of the floor members(beams ribs and slabs) depends upon the protection providedto the bottom reinforcement The steel begins to lose strengthat a temperature of 300oC losses of 50 and 75 occurringat temperatures of about 560oC and 700oC respectively Theconcrete cover needs to be sufficient to delay the time takento reach a temperature likely to result in structural failure Adistinction is made between simply supported spans wherea 50 loss of strength in the bottom reinforcement could becritical and continuous spans where a greater loss is allowedbecause the top reinforcement will retain its full capacity

If the cover becomes excessive there is a risk of prematurespalling of the concrete in the event of fire Concretes madewith aggregates containing a high proportion of silica are the mostsusceptible In cases where the nominal cover needs to exceed40 mm additional measures should be considered and severalpossible courses of action are described in Part 2 of BS 8110The preferred approach is to reduce the cover by providingadditional protection in the form of an applied finish or a falseceiling or by using lightweight aggregates or sacrificial steelThe last measure refers to the provision of more steel than isnecessary for normal purposes so that a greater loss of strengthcan be allowed in the event of fire If the nominal cover doesexceed 40 mm then supplementary reinforcement in the formof welded steel fabric should be placed within the thicknessof the cover at 20 mm from the concrete surface There areconsiderable practical difficulties with this approach and it mayconflict with the requirements for durability in some cases

For concrete made with lightweight aggregate the nominalcover requirements are all reduced and the risk of prematurespalling only needs to be considered when the cover exceeds50 mm The detailed requirements for lightweight aggregateconcrete and guidance on the additional protection provided byselected applied finishes are given in Table 310

EC 2 contains a more flexible approach to fire safety designbased on the concept of lsquoload ratiorsquo which is the ratio of theload applied at the fire limit-state to the capacity of the elementat ambient temperature

Fire-resistance 27

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Torsion-less beams are designed as linear elements subjectedto bending moments and shear forces The values for freelysupported beams and cantilevers are readily determinedby the simple rules of static equilibrium but the analysis ofcontinuous beams and statically indeterminate frames is morecomplex Historically various analytical techniques have beendeveloped and used as self-contained methods to solve partic-ular problems In time it was realised that the methodscould be divided into two basic categories flexibility methods(otherwise known as action methods compatibility methods orforce methods) and displacement methods (otherwise known asstiffness methods or equilibrium methods) The behaviour ofthe structure is considered in terms of unknown forces in thefirst category and unknown displacements in the secondcategory For each method a particular solution obtained bymodifying the structure to make it statically determinate iscombined with a complementary solution in which the effectof each modification is determined Consider the case of acontinuous beam For the flexibility methods the particularsolution involves removing redundant actions (ie the continuitybetween the individual members) to leave a series of discon-nected spans For the displacement methods the particularsolution involves restricting the rotations andor displacementsthat would otherwise occur at the joints

To clarify further the main differences between the methodsin the two categories consider a propped cantilever With theflexibility approach the first step is to remove the prop andcalculate the deflection at the position of the prop due to theaction of the applied loads this gives the particular solutionThe next step is to calculate the concentrated load needed at theposition of the prop to restore the deflection to zero this givesthe complementary solution The calculated load is the reactionin the prop knowing this enables the moments and forces in thepropped cantilever to be simply determined If the displacementapproach is used the first step is to consider the span as fullyfixed at both ends and calculate the moment at the propped enddue to the applied loads this gives the particular solution Thenext step is to release the restraint at the propped end and applyan equal and opposite moment to restore the rotation to zero thisgives the complementary solution By combining the momentdiagrams the resulting moments and forces can be determined

In general there are several unknowns and irrespectiveof the method of analysis used the preparation and solution ofa set of simultaneous equations is required The resulting

relationship between forces and displacements embodies a seriesof coefficients that can be set out concisely in matrix formIf flexibility methods are used the resulting matrix is built up offlexibility coefficients each of which represents a displacementproduced by a unit action Similarly if stiffness methods areused the resulting matrix is formed of stiffness coefficients eachof which represents an action produced by a unit displacementThe solution of matrix equations either by matrix inversionor by a systematic elimination process is ideally suited tocomputer technology To this end methods have been devised(the so-called matrix stiffness and matrix flexibility methods)for which the computer both sets up and solves the simultaneousequations (ref 15)

Here it is worthwhile to summarise the basic purpose ofthe analysis Calculating the bending moments on individualfreely supported spans ensures that equilibrium is maintainedThe analytical procedure that is undertaken involves linearlytransforming these free-moment diagrams in a manner that iscompatible with the allowable deformations of the structureUnder ultimate load conditions deformations at the criticalsections must remain within the limits that the sections canwithstand and under service load conditions deformationsmust not result in excessive deflection or cracking or both Ifthe analysis is able to ensure that these requirements are met itwill be entirely satisfactory for its purpose endeavouring toobtain painstakingly precise results by over-complex methodsis unjustified in view of the many uncertainties involved

To determine at any section the effects of the applied loadsand support reactions the basic relationships are as follows

Shear force (forces on one side of section) rate of change of bending moment

Bending moment (moments of forces on one side of section) (shear force) area of shear force diagram

Slope (curvature) area of curvature diagram

Deflection (slope) area of slope diagram

For elastic behaviour curvature MEI where M is bendingmoment E is modulus of elasticity of concrete I is secondmoment of area of section For the purposes of structural

Chapter 4

Structural analysis

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analysis to determine bending moments due to applied loadsI values may normally be based on the gross concrete sectionIn determining deflections however due allowance needs to bemade for the effects of cracking and in the long term for theeffects of concrete creep and shrinkage

41 SINGLE-SPAN BEAMS AND CANTILEVERS

Formulae to determine the shearing forces bending momentsand deflections produced by various general loads on beamsfreely supported at the ends are given in Table 224 Similarexpressions for some particular load arrangements commonlyencountered on beams either freely supported or fully fixedat both ends with details of the maximum values are given inTable 225 The same information but relating to simple andpropped cantilevers is given in Tables 226 and 227 respectivelyCombinations of loads can be considered by summing theresults obtained for each individual load

In Tables 224ndash227 expressions are also given for the slopesat the beam supports and the free (or propped) end of a cantileverInformation regarding the slope at other points is seldomrequired If needed it is usually a simple matter to obtain theslope by differentiating the deflection formula with respect to xIf the resulting expression is equated to zero and solved toobtain x the point of maximum deflection will have been foundThis value of x can then be substituted into the original formulato obtain the maximum deflection

Coefficients to determine the fixed-end moments producedby various symmetrical and unsymmetrical loads on beamsfully fixed at both ends are given in Table 228 Loadings notshown can usually be considered by using the tabulated casesin combination For the general case of a partial uniform ortriangular distribution of load placed anywhere on a membera full range of charts is contained in Examples of the Design ofReinforced Concrete Buildings The charts give deflection andmoment coefficients for beams (freely supported or fully fixedat both ends) and cantilevers (simple or propped)

42 CONTINUOUS BEAMS

Historically various methods of structural analysis have beendeveloped for determining the bending moments and shearingforces on beams continuous over two or more spans Most ofthese have been stiffness methods which are generally bettersuited than flexibility methods to hand computation Some ofthese approaches such as the theorem of three-moments and themethods of fixed points and characteristic points were includedin the previous edition of this Handbook If beams having twothree or four spans are of uniform cross section and supportloads that are symmetrical on each individual span formulaeand coefficients can be derived that enable the support momentsto be determined by direct calculation Such a method is givenin Table 237 More generally in order to avoid the need to solvelarge sets of simultaneous equations methods involving succes-sive approximations have been devised Despite the general useof computers hand methods can still be very useful in dealingwith routine problems The ability to use hand methods alsodevelops in the engineer an appreciation of analysis that isinvaluable in applying output from the computer

When bending moments are calculated with the spans takenas the distances between the centres of supports the critical

negative moment in monolithic forms of construction can beconsidered as that occurring at the edge of the support Whenthe supports are of considerable width the span can be taken asthe clear distance between the supports plus the effective depthof the beam or an additional span can be introduced thatis equal to the width of the support minus the effective depth ofthe beam The load on this additional span should be takenas the support reaction spread uniformly over the width of thesupport If a beam is constructed monolithically with a verywide and massive support the effect of continuity with the spanor spans beyond the support may be negligible in which casethe beam should be treated as fixed at the support

The second moment of area of a reinforced concrete beamof uniform depth may still vary throughout its length due tovariations in the amount of reinforcement and also becausewhen acting with an adjoining slab a down-stand beam maybe considered as a flanged section at mid-span but a simplerectangular section at the supports It is common practicehowever to neglect these variations for beams of uniformdepth and use the value of I for the plain rectangular section Itis often assumed that a continuous beam is freely supportedat the ends even when beam and support are constructedmonolithically Some provision should still be made for theeffects of end restraint

421 Analysis by moment distribution

Probably the best-known and simplest system for analysingcontinuous beams by hand is that of moment distributionas devised by Hardy Cross in 1929 The method whichderives from slope-deflection principles is described briefly inTable 236 It employs a system of successive approximationsthat may be terminated as soon as the required degree ofaccuracy has been reached A particular advantage of this andsimilar methods is that even after only one distribution cycleit is often clear whether or not the final values will be acceptableIf not the analysis can be discontinued and unnecessary workavoided The method is simple to remember and apply andthe step-by-step procedure gives the engineer a lsquofeelrsquo for thebehaviour of the system It can be applied albeit less easily tothe analysis of systems containing non-prismatic members andto frames Hardy Cross moment distribution is described inmany textbooks dealing with structural analysis

Over the years the Hardy Cross method of analysis begotvarious offspring One of these is known as precise momentdistribution (also called the coefficient of restraint method ordirect moment distribution) The procedure is very similar tonormal moment distribution but the distribution and carryoverfactors are so adjusted that an exact solution is obtainedafter one distribution in each direction The method thus hasthe advantage of removing the necessity to decide when toterminate the analysis Brief details are given in Table 236 andthe method is described in more detail in Examples of theDesign of Reinforced Concrete Buildings (see also ref 16)

It should be noted that the load arrangements that producethe greatest negative bending moments at the supports are notnecessarily those that produce the greatest positive bendingmoments in the spans The design loads to be considered inBS 8110 and EC 2 and the arrangements of live load that givethe greatest theoretical bending moments as well as the lessonerous code requirements are given in Table 229 Some live

Continuous beams 29

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load arrangements can result in negative bending momentsthroughout adjacent unloaded spans

422 Redistribution of bending moments

For the ULS the bending moments obtained by linear elasticanalysis may be adjusted on the basis that some redistributionof moments can occur prior to collapse This enables the effectsof both service and ultimate loadings to be assessed without theneed to undertake a separate analysis using plastic-hinge tech-niques for the ultimate condition The theoretical justificationfor moment redistribution is clearly explained in the Handbookto BS 8110 Since the reduction of moment at a section assumesthe formation of a plastic hinge at that position prior to theultimate condition being reached it is necessary to limit thereduction in order to restrict the amount of plastic-hinge rotationand control the cracking that occurs under serviceabilityconditions For these reasons the maximum ratio of neutralaxis depth to effective depth and the maximum distancebetween tension bars are each limited according to the requiredamount of redistribution

Such adjustments are useful in reducing the inequalitiesbetween negative and positive moments and minimising theamount of reinforcement that must be provided at a particularsection such as the intersection between beam and columnwhere concreting may otherwise be more difficult due to thecongestion of reinforcement Both BS 8110 and EC 2 allowthe use of moment redistribution the procedure which may beapplied to any system that has been analysed by the so-calledexact methods is described in section 123 with an illustratedexample provided in Table 233

423 Coefficients for equal loads onequal spans

For beams that are continuous over a number of equal spanswith equal loads on each loaded span the maximum bendingmoments and shearing forces can be tabulated In Tables 230and 231 maximum bending moment coefficients are given foreach span and at each support for two three four and five equalspans with identical loads on each span which is the usualdisposition of the dead load on a beam Coefficients are alsogiven for the most adverse incidence of live loads and in thecase of the support moments for the arrangements of live loadrequired by BS 8110 (values in square brackets) and by EC 2(values in curved brackets) It should be noted that the maximumbending moments due to live load do not occur at all thesections simultaneously The types of load considered are auniformly distributed load a central point load two equal loadsapplied at the third-points of the span and trapezoidal loads ofvarious proportions In Table 232 coefficients are given for themaximum shearing forces for each type of load with identicalloads on each span and due to the most adverse incidence oflive loads

424 Bending moment diagrams for equal spans

In Tables 234 and 235 bending moment coefficients forvarious arrangements of dead and live loads with sketches

of the resulting moment envelopes are given for beams oftwo and three spans and for a theoretically infinite systemThis information enables appropriate bending moment diagramsto be plotted quickly and accurately The load types consideredare a uniformly distributed load a central point load and twoequal loads at the third points of the span Values are givenfor identical loads on each span (for example dead load) andfor the arrangements of live load required by BS 8110 andEC 2 As the coefficients have been calculated by exactmethods moment redistribution is allowed at the ultimate statein accordance with the requirements of BS 8110 and EC 2 Inaddition to the coefficients obtained by linear elastic analysisvalues are given for conditions in which the maximum supportmoments are reduced by either 10 or 30 as described insection 1233 Coefficients are also given for the positivesupport moments and negative span moments that occur undersome arrangements of live load

425 Solutions for routine design

A precise determination of theoretical bending moments andshearing forces on continuous beams is not always necessary Itshould also be appreciated that the general assumptions ofunyielding knife-edge supports uniform sectional propertiesand uniform distributions of live load are hardly realistic Theindeterminate nature of these factors often leads in practice tothe adoption of values based on approximate coefficients InTable 229 values in accordance with the recommendationsof BS 8110 and EC 2 are given for bending moments andshearing forces on uniformly loaded beams of three or morespans The values are applicable when the characteristicimposed load is not greater than the characteristic dead loadand the variations in span do not exceed 15 of the longestspan The same coefficients may be used with service loads orultimate loads and the resulting bending moments may beconsidered to be without redistribution

43 MOVING LOADS ON CONTINUOUS BEAMS

Bending moments caused by moving loads such as those due tovehicles traversing a series of continuous spans are most easilycalculated with the aid of influence lines An influence line is acurve with the span of the beam taken as the base the ordinateof the curve at any point being the value of the bending momentproduced at a particular section when a unit load acts at thepoint The data given in Tables 238ndash241 enable the influencelines for the critical sections of beams continuous over twothree four and five or more spans to be drawn By plotting theposition of the load on the beam (to scale) the bending momentsat the section being considered can be derived as explained inthe example given in Chapter 12 The curves given for equalspans can be used directly but the corresponding curves forunequal spans need to be plotted from the data tabulated

The bending moment due to a load at any point is equal tothe ordinate of the influence line at the point multiplied by theproduct of the load and the span the length of the shortest spanbeing used when the spans are unequal The influence lines inthe tables are drawn for a symmetrical inequality of spans Thesymbols on each curve indicate the section of the beam andthe ratio of span lengths to which the curve applies

Structural analysis30

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44 ONE-WAY SLABS

In monolithic building construction the column layout oftenforms a rectangular grid Continuous beams may be provided inone direction or two orthogonal directions to support slabs thatmay be solid or ribbed in cross section Alternatively the slabsmay be supported directly on the columns as a flat slab Severaldifferent forms of slab construction are shown in Table 242These are considered in more detail in the general context ofbuilding structures in Chapter 6

Where beams are provided in one direction only the slab isa one-way slab Where beams are provided in two orthogonaldirections the slab is a two-way slab However if the longerside of a slab panel exceeds twice the shorter side the slab isgenerally designed as a one-way slab A flat slab is designedas a one-way slab in each direction Bending moments andshearing forces are usually determined on strips of unit widthfor solid slabs and strips of width equal to the spacing of theribs for ribbed slabs

The comments in section 425 and the coefficients for theroutine design of beams given in Table 229 apply equally toone-way spanning slabs This is particularly true when elasticmoments due to service loads are required However lightlyreinforced slabs are highly ductile members and allowanceis generally made for redistribution of elastic moments atthe ULS

441 Uniformly distributed load

For slabs carrying uniformly distributed loads and continuousover three or more nearly equal spans approximate solutionsfor ultimate bending moments and shearing forces accordingto BS 8110 and EC 2 are given in Table 242 In both cases thesupport moments include an allowance for 20 redistributionbut the situation regarding the span moments is somewhatdifferent in the two codes

In BS 8110 a simplified arrangement of the design loadsis permitted where the characteristic imposed load doesnot exceed 125 the characteristic dead load or 5 kNm2excluding partitions and the area of each bay exceeds 30 m2Design for a single load case of maximum design load on allspans is considered sufficient providing the support momentsare reduced by 20 and the span moments are increasedto maintain equilibrium Although the resulting moments arecompatible with yield-line theory the span moments are lessthan those that would occur in the case of alternate spans beingloaded with maximum load and minimum load The implicitredistribution of the span moments the effect of which on thereinforcement stress under service loads would be detrimentalto the deflection of the beam is ignored in the subsequentdesign In EC 2 this simplification is not included and thevalues given for the span moments are the same as those forbeams in Table 229

Provision is made in Table 242 for conditions where aslab is continuous with the end support The restrainingelement may vary from a substantial wall to a small edgebeam and allowance has been made for both eventualitiesThe support moment is given as 004Fl but the reducedspan moment is based on the support moment being no morethan 002Fl

442 Concentrated loads

When a slab supported on two opposite sides carries a loadconcentrated on a limited area of the slab such as a wheelload on the deck of a bridge conventional elastic methods ofanalysis based on isotropic plate theory are often used Thesemay be in the form of equations as derived by Westergaard(ref 17) or influence surfaces as derived by Pucher (ref 18)Another approach is to extend to one-way spanning slabs thetheory applied to slabs spanning in two directions For examplethe curves given in Table 247 for a slab infinitely long in thedirection ly can be used to evaluate directly the bendingmoments in the direction of and at right angles to the spanof a one-way slab carrying a concentrated load this methodhas been used to produce the data for elastic analysis givenin Table 245

For designs in which the ULS requirement is the maincriterion a much simpler approach is to assume that a certainwidth of slab carries the entire load In BS 8110 for examplethe effective width for solid slabs is taken as the load widthplus 24x(1 xl) x being the distance from the nearer supportto the section under consideration and l the span Thus themaximum width at mid-span is equal to the load width plus06l Where the concentrated load is near an unsupported edgeof a slab the effective width should not exceed 12x(1 xl)plus the distance of the slab edge from the further edge of theload Expressions for the resulting bending moments are givenin Table 245 For ribbed slabs the effective width will dependon the ratio of the transverse and longitudinal flexural rigiditiesof the slab but need not be taken less than the load width plus4xl(1 x l) metres

The solutions referred to so far are for single-span slabs thatare simply supported at each end The effects of end-fixity orcontinuity may be allowed for approximately by multiplyingthe moment for the simply supported case by an appropriatefactor The factors given in Table 245 are derived by elasticbeam analysis

45 TWO-WAY SLABS

When a slab is supported other than on two opposite sides onlythe precise amount and distribution of the load taken by eachsupport and consequently the magnitude of the bendingmoments on the slab are not easily calculated if assumptionsresembling real conditions are made Therefore approximateanalyses are generally used The method applicable in anyparticular case depends on the shape of the slab panel theconditions of restraint at the supports and the type of load

Two basic methods are commonly used to analyse slabsthat span in two directions The theory of plates which isbased on elastic analysis is particularly appropriate to thebehaviour under service loads Yield-line theory considersthe behaviour of the slab as a collapse condition approachesHillerborgrsquos strip method is a less well-known alternative tothe use of yield-line in this case In some circumstances itis convenient to use coefficients derived by an elastic analysiswith loads that are factored to represent ULS conditions Thisapproach is used in BS 8110 for the case of a simply supportedslab with corners that are not held down or reinforced fortorsion It is also normal practice to use elastic analysis for

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both service and ULS conditions in the design of bridge decksand liquid-retaining structures For elastic analyses a Poissonrsquosratio of 02 is recommended in BS 8110 and BS 5400 Part 4In EC 2 the values given are 02 for uncracked concrete and 0for cracked concrete

The analysis must take account of the support conditionswhich are often idealised as being free or hinged or fixed andwhether or not the corners of the panels are held down A freecondition refers to an unsupported edge as for example the topof a wall of an uncovered rectangular tank The condition ofbeing freely or simply supported with the corners not helddown may occur when a slab is not continuous and the edgesbear directly on masonry walls or structural steelwork If theedge of the slab is built into a substantial masonry wall or isconstructed monolithically with a reinforced concrete beam orwall a condition of partial restraint exists Such restraint maybe allowed for when computing the bending moments on theslab but the support must be able to resist the torsion andorbending effects and the slab must be reinforced to resist thenegative bending moment A slab can be considered as fixedalong an edge if there is no change in the slope of the slab atthe support irrespective of the incidence of the load A fixedcondition could be assumed if the polar second moment of areaof the beam or other support is very large Continuity over asupport generally implies a condition of restraint less rigid thanfixity that is the slope of the slab at the support depends uponthe incidence of load not only on the panel under considerationbut also on adjacent panels

451 Elastic methods

The so-called exact theory of the elastic bending of platesspanning in two directions derives from work by Lagrangewho produced the governing differential equation for platebending in 1811 and Navier who in 1820 described the useof a double trigonometric series to analyse freely supportedrectangular plates Pigeaud and others later developed theanalysis of panels freely supported along all four edges

Many standard elastic solutions have been produced butalmost all of these are restricted to square rectangular andcircular slabs (see for example refs 19 20 and 21) Exactanalysis of a slab having an arbitrary shape and supportconditions with a general arrangement of loading would beextremely complex To deal with such problems numericaltechniques such as finite differences and finite elementshave been devised Some notes on finite elements are givenin section 497 Finite-difference methods are considered inref 15 (useful introduction) and ref 22 (detailed treatment)The methods are suited particularly to computer-based analysisand continuing software developments have led to the techniquesbeing readily available for routine office use

452 Collapse methods

Unlike in frame design where the converse is generally trueit is normally easier to analyse slabs by collapse methods thanby elastic methods The most-widely known methods ofplastic analysis of slabs are the yield-line method developedby K W Johansen and the so-called strip method devised byArne Hillerborg

It is generally impossible to calculate the precise ultimateresistance of a slab by collapse theory since such elements are

highly indeterminate Instead two separate solutions can befound ndash one being upper bound and the other lower boundWith solutions of the first type a collapse mechanism is firstpostulated Then if the slab is deformed the energy absorbedin inducing ultimate moments along the yield lines is equal tothe work done on the slab by the applied load in producing thisdeformation Thus the load determined is the maximum thatthe slab will support before failure occurs However since suchmethods do not investigate conditions between the postulatedyield lines to ensure that the moments in these areas do notexceed the ultimate resistance of the slab there is no guaranteethat the minimum possible collapse load has been found Thisis an inevitable shortcoming of upper-bound solutions such asthose given by Johansenrsquos theory

Conversely lower-bound solutions will generally result in thedetermination of collapse loads that are less than the maximumthat the slab can actually carry The procedure here is to choosea distribution of ultimate moments that ensures that equilibriumis satisfied throughout and that nowhere is the resistance of theslab exceeded

Most of the literature dealing with the methods of Johansenand Hillerborg assumes that any continuous supports at the slabedges are rigid and unyielding This assumption is also madethroughout the material given in Part 2 of this book Howeverif the slab is supported on beams of finite strength it is possiblefor collapse mechanisms to form in which the yield lines passthrough the supporting beams These beams would then becomepart of the mechanism considered and such a possibility shouldbe taken into account when using collapse methods to analysebeam-and-slab construction

Yield-line analysis Johansenrsquos method requires the designerto first postulate an appropriate collapse mechanism for the slabbeing considered according to the rules given in section 1342Variable dimensions (such as ly on diagram (iv)(a) in Table 249)may then be adjusted to obtain the maximum ultimate resistancefor a given load (ie the maximum ratio of MF) This maximumvalue can be found in various ways for example by tabulatingthe work equation as shown in section 1348 using actualnumerical values and employing a trial-and-adjustment processAlternatively the work equation may be expressed algebraicallyand by substituting various values for the maximum ratio ofMF may be read from a graph relating to MF Anothermethod is to use calculus to differentiate the equation and thenby setting this equal to zero determine the critical value of This method cannot always be used however (see ref 23)

As already explained although such processes enable themaximum resistance for a given mode of failure to be foundthey do not indicate whether the yield-line pattern considered isthe critical one A further disadvantage of such a method is thatunlike Hillerborgrsquos method it gives no direct indication of theresulting distribution of load on the supports Although it seemspossible to use the yield-line pattern as a basis for apportioningthe loaded areas of slab to particular supports there is no realjustification for this assumption (see ref 23) In spite of theseshortcomings yield-line theory is extremely useful A consid-erable advantage is that it can be applied relatively easily tosolve problems that are almost intractable by other means

Yield-line theory is too complex to deal with adequately in thisHandbook indeed several textbooks are completely or almostcompletely devoted to the subject (refs 23ndash28) In section 134

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and Tables 249 and 250 notes and examples are given on therules for choosing yield-line patterns for analysis on theoreticaland empirical methods of analysis on simplifications that canbe made by using so-called affinity theorems and on the effectsof corner levers

Strip method Hillerborg devised his strip method in orderto obtain a lower-bound solution for the collapse load whileachieving a good economical arrangement of reinforcement Aslong as the reinforcement provided is sufficient to cater for thecalculated moments the strip method enables such a lower-boundsolution to be obtained (Hillerborg and others sometimes referto the strip method as the equilibrium theory this should nothowever be confused with the equilibrium method of yield-lineanalysis) In Hillerborgrsquos original theory (now known as thesimple strip method) it is assumed that at failure no load isresisted by torsion and thus all load is carried by flexure ineither of two principal directions The theory results in simplesolutions giving full information regarding the moments overthe whole slab to resist a unique collapse load the reinforcementbeing placed economically in bands Brief notes on the use ofsimple strip theory to design rectangular slabs supportinguniform loads are given in section 135 and Table 251

However the simple strip theory is unable to deal withconcentrated loads andor supports and leads to difficultieswith free edges To overcome such problems Hillerborg laterdeveloped his advanced strip method which involves the use ofcomplex moment fields Although this development extendsthe scope of the simple strip method it somewhat spoils thesimplicity and directness of the original concept A full treat-ment of both the simple and advanced strip theories is givenin ref 29

A further disadvantage of both Hillerborgrsquos and Johansenrsquosmethods is that being based on conditions at failure onlythey permit unwary designers to adopt load distributions thatmay differ widely from those that would occur under serviceloads with the risk of unforeseen cracking A development thateliminates this problem as well as overcoming the limitationsarising from simple strip theory is the so-called strip-deflectionmethod due to Fernando and Kemp (ref 30) With this methodthe distribution of load in either principal direction is notselected arbitrarily by the designer (as in the Hillerborg methodor by choosing the ratio of reinforcement provided in eachdirection as in the yield-line method) but is calculated so as toensure compatibility of deflection in mutually orthogonal stripsThe method results in sets of simultaneous equations (usuallyeight) the solution of which requires computer assistance

453 Rectangular panel with uniformlydistributed load

The bending moments in rectangular panels depend on thesupport conditions and the ratio of the lengths of the sides ofthe panel The ultimate bending moment coefficients given inBS 8110 are derived from a yield-line analysis in which thevalues of the coefficients have been adjusted to suit the divisionof the panel into middle and edge strips as shown in Table 242Reinforcement to resist the bending moments calculated fromthe data given in Table 243 is required only within the middlestrips which are of width equal to three-quarters of the panelwidth in each direction The ratio of the negative moment at

a continuous edge to the positive moment at mid-span has beenchosen as 43 to conform approximately to the serviceabilityrequirements For further details on the derivation of the coef-ficients see ref 31 Nine types of panel are considered inorder to cater for all possible combinations of edge conditionsWhere two different values are obtained for the negativemoment at a continuous edge because of differences betweenthe contiguous panels the values may be treated as fixed-endmoments and distributed elastically in the direction of spanThe procedure is illustrated by means of a worked example insection 1321 Minimum reinforcement as given in BS 8110is to be provided in the edge strips Torsion reinforcement isrequired at corners where either one or both edges of the panelare discontinuous Values for the shearing forces at the ends ofthe middle strips are also given in Table 243

Elastic bending moment coefficients for the same types ofpanel (except that the edge conditions are now defined as fixedor hinged rather than continuous or discontinuous) are givenin Table 244 The information has been prepared from datagiven in ref 21 which was derived by finite element analysisand includes for a Poissonrsquos ratio of 02 For ratios less than 02the positive moments at mid-span are reduced slightly and thetorsion moments at the corners are increased The coefficientsmay be adjusted to suit a Poissonrsquos ratio of zero as explainedin section 1322

The simplified analysis due to Grashof and Rankine can beused for a rectangular panel simply supported on four sideswhen no provision is made to resist torsion at the corners orto prevent the corners from lifting A solution is obtained byconsidering uniform distributions of load along orthogonalstrips in each direction and equating the elastic deflections atthe middle of the strips The proportions of load carried by eachstrip are then obtained as a function of the ratio of the spansand the resulting mid-span moments are calculated Bendingmoment coefficients for this case are also provided in Table 244and basic formulae are given in section 1322

454 Rectangular panel with triangularlydistributed load

In the design of rectangular tanks storage bunkers and someretaining structures cases occur of wall panels spanning in twodirections and subjected to triangular distributions of pressureThe intensity of pressure is uniform at any level but verticallythe pressure increases linearly from zero at the top to a maxi-mum at the bottom Elastic bending moment and shear forcecoefficients are given for four different types of panel to caterfor the most common combinations of edge conditions inTable 253 The information has been prepared from data givenin ref 32 which was derived by finite element analysis andincludes for a Poissonrsquos ratio of 02 For ratios less than 02 thebending moments would be affected in the manner discussed insection 453

The bending moments given for individual panels fixed atthe sides may be applied without modification to continuouswalls provided there is no rotation about the vertical edges Ina square tank therefore moment coefficients can be takendirectly from Table 253 For a rectangular tank distribution ofthe unequal negative moments at the corners is needed

An alternative method of designing the panels would be touse yield-line theory If the resulting structure is to be used

Two-way slabs 33

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to store liquids however extreme care must be taken to ensurethat the adopted proportions of span to support moment andvertical to horizontal moment conform closely to those givenby elastic analyses Otherwise the predicted service momentsand calculated crack widths will be invalid and the structuremay be unsuitable for its intended purpose In the case of struc-tures with non-fluid contents such considerations may be lessimportant This matter is discussed in section 1362

Johansen has shown (ref 24) for a panel fixed or freelysupported along the top edge that the total ultimate momentacting on the panel is identical to that on a similar panel withthe same total load uniformly distributed Furthermore as in thecase of the uniformly loaded slab considered in section 1346a restrained slab may be analysed as if it were freely supportedby employing so-called reduced side lengths to represent theeffects of continuity or fixity Of course unlike the uniformlyloaded slab along the bottom edge of the panel where the load-ing is greatest a higher ratio of support to span moment shouldbe adopted than at the top edge of the panel If the panel isunsupported along the top edge its behaviour is controlledby different collapse mechanisms The relevant expressionsdeveloped by Johansen (ref 24) are represented graphically inTable 254 Triangularly loaded panels can also be designed bymeans of Hillerborgrsquos strip method (ref 29) shown also inTable 254

455 Rectangular panels with concentratedloads

Elastic methods can be used to analyse rectangular panelscarrying concentrated loads The curves in Tables 246 and 247based on Pigeaudrsquos theory give bending moments on a panelfreely supported along all four edges with restrained corners andcarrying a load uniformly distributed over a defined area sym-metrically disposed upon the panel Wheel loads and similarlyhighly concentrated loads are considered to be dispersedthrough the thickness of any surfacing down to the top of theslab or farther down to the mid-depth of the slab as describedin section 249 The dimensions ax and ay of the resultingboundary are used to determine axlx and ayly for which thebending moment factors x4 and y4 are read off the curvesaccording to the ratio of spans k lylx

For a total load F acting on the area ax by ay the positivebending moments per unit width of slab are given by theexpressions in Tables 246 and 247 in which the value ofPoissonrsquos ratio is normally taken as 02 The curves are drawnfor k values of 10 125 radic2 ( 141 approx) 167 20 25 andinfinity For intermediate values of k the values of x4 and y4

can be interpolated from the values above and below the givenvalue of k The use of the curves for k 10 which apply to asquare panel is explained in section 1332

The curves for k infin apply to panels where ly is very muchgreater than lx and can be used to determine the transverse andlongitudinal bending moments for a long narrow panel sup-ported on the two long edges only This chart has been used toproduce the elastic data for one-way slabs given in Table 245as mentioned in section 442

For panels that are restrained along all four edges Pigeaudrecommends that the mid-span moments be reduced by 20Alternatively the multipliers given for one-way slabs could beused if the inter-dependence of the bending moments in the

two directions is ignored Pigeaudrsquos recommendations for themaximum shearing forces are given in section 1332

To determine the load on the supporting beams the rulesin section 46 for a load distributed over the entire panel aresufficiently accurate for a load concentrated at the centre ofthe panel This is not always the critical case for live loads suchas a load imposed by a wheel on a bridge deck since themaximum load on the beam occurs when the wheel is passingover the beam in which case the beam carries the whole load

Johansenrsquos yield-line theory and Hillerborgrsquos strip methodcan also be used to analyse slabs carrying concentrated loadsAppropriate yield-line formulae are given in ref 24 or themethod described in section 1348 may be used For detailsof the analysis involved if the advanced strip method is usedsee ref 29

46 BEAMS SUPPORTING RECTANGULAR PANELS

When designing beams supporting a uniformly loaded panelthat is freely supported along all four edges or with the samedegree of fixity along all four edges it is generally accepted thateach of the beams along the shorter edges of the panel carriesload on an area in the shape of a 45o isosceles triangle whosebase is equal to the length of the shorter side for example eachbeam carries a triangularly distributed load Each beam alongthe longer edges of the panel carries the load on a trapezoidalarea The amount of load carried by each beam is given bythe diagram and expressions in the top left-hand corner ofTable 252 In the case of a square panel each beam carries atriangularly distributed load equal to one-quarter of the totalload on the panel For beams with triangular and trapezoidaldistributions of loading fixed-end moments and moments forcontinuous beams are given in Tables 228 230 and 231

When a panel is fixed or continuous along one two orthree supports and freely supported on the remaining edges thesub-division of the total load to the various supporting beamscan be determined from the diagrams and expressions on theleft-hand side of Table 252 If the panel is unsupported alongone edge or two adjacent edges the loads on the supportingbeams at the remaining edges are as given on the right-handside of Table 252 The expressions which are given in terms ofa service load w may be applied also to an ultimate load n

For slabs designed in accordance with the BS 8110 methodthe loads on the supporting beams may be determined from theshear forces given in Table 243 The relevant loads are takenas uniformly distributed along the middle three-quarters of thebeam length and the resulting fixed-end moments can bedetermined from Table 228

47 NON-RECTANGULAR PANELS

When a panel that is not rectangular is supported along all itsedges and is of such proportions that main reinforcement intwo directions seems desirable the bending moments can bedetermined approximately from the data given in Table 248The information derived from elastic analyses is applicable toa trapezoidal panel approximately symmetrical about one axisto a panel that in plan is an isosceles triangle (or nearly so) andto panels that are regular polygons or circular The case of atriangular panel continuous or partially restrained along threeedges occurs in pyramidal hopper bottoms For this case

Structural analysis34

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reinforcement determined for the positive moments shouldextend over the entire area of the panel and provision must bemade for the negative moments and for the direct tensions thatact simultaneously with the bending moments

If the shape of a panel is approximately square the bendingmoments for a square slab of the same area should be usedA slab having the shape of a regular polygon with five or moresides can be treated as a circular slab with the diameter takenas the mean of the diameters obtained for the inscribed andcircumscribed circles for regular hexagons and octagons themean diameters are given in Table 248

For a panel circular in plan that is freely supported or fullyfixed along the circumference and carries a load concentratedsymmetrically about the centre on a circular area the totalbending moment to be considered acting across each of twomutually perpendicular diameters is given by the appropriateexpressions in Table 248 These are based on the expressionsderived by Timoshenko and Woinowski-Krieger (ref 20) Ingeneral the radial and tangential moments vary according to theposition being considered A circular panel can therefore bedesigned by one of the following elastic methods

1 Design for the maximum positive bending moment at thecentre of the panel and reduce the amount of reinforcementor the thickness of the slab towards the circumference If thepanel is not truly freely supported at the edge provide forthe appropriate negative bending moment

2 Design for the average positive bending moment across adiameter and retain the same thickness of slab and amountof reinforcement throughout the entire area of the panel Ifthe panel is not truly freely supported at the edge providefor the appropriate negative bending moment

The reinforcement required for the positive bending momentsin each of the preceding methods must be provided in twodirections mutually at right angles the reinforcement for thenegative bending moment should be provided by radial barsnormal to and equally spaced around the circumference or bysome equivalent arrangement

Both circular and other non-rectangular shapes of slab mayconveniently be designed for ULS conditions by using yield-line theory the method of obtaining solutions for slabs ofvarious shapes is described in detail in ref 24

48 FLAT SLABS

The design of flat slabs that is beamless slabs supporteddirectly on columns has often been based on empirical rulesModern codes place much greater emphasis on the analysis ofsuch structures as a series of continuous frames Other methodssuch as grillage finite element and yield-line analysis may beemployed The principles described hereafter and summarisedin section 138 and Table 255 are in accordance with thesimplified method given in BS 8110 This type of slab can beof uniform thickness throughout or can incorporate thickeneddrop panels at the column positions The columns may be ofuniform cross section throughout or may be provided with anenlarged head as indicated in Table 255

The simplified method may be used for slabs consisting ofrectangular panels with at least three spans of approximatelyequal length in each direction where the ratio of the longer tothe shorter side of each panel does not exceed 2 Each panel is

divided into column and middle strips where the width of acolumn strip is taken as one-half of the shorter dimension of thepanel and bending moments determined for a full panel widthare then distributed between column and middle strips as shownin Table 255 If drops of dimensions not less than one-third ofthe shorter dimension of the panel are provided the width of thecolumn strip can be taken as the width of the drop In this casethe apportionment of the bending moments between columnand middle strips is modified accordingly

The slab thickness must be sufficient to satisfy appropriatedeflection criteria with a minimum thickness of 125 mm andprovide resistance to shearing forces and bending momentsPunching shear around the columns is a critical considerationfor which shear reinforcement can be provided in slabs not lessthan 200 mm thick The need for shear reinforcement can beavoided if drop panels or column heads of sufficient size areprovided Holes of limited dimensions may be formed in certainareas of the slab according to recommendations given in BS8110 Larger openings should be appropriately framed withbeams designed to carry the slab loads to the columns

481 Bending moments

The total bending moments for a full panel width at principalsections in each direction of span are given in Table 255 Panelwidths are taken between the centrelines of adjacent bays andpanel lengths between the centrelines of columns Momentscalculated at the centrelines of the supports may be reduced asexplained in section 1383 The slab is effectively designedas one-way spanning in each direction and the commentscontained in section 441 also apply here

At the edges of a flat slab the transfer of moments betweenthe slab and an edge or corner column may be limited by theeffective breadth of the moment transfer strip as shown inTable 256 The structural arrangement should be chosen toensure that the moment capacity of the transfer strip is at least50 of the outer support moment given in Table 255

482 Shearing forces

For punching shear calculations the design force obtained bysumming the shear forces on two opposite sides of a column ismultiplied by a shear enhancement factor to allow for theeffects of moment transfer as shown in Table 256 Criticalperimeters for punching shear occur at distances of 15d fromthe faces of columns column heads and drops where d is theeffective depth of the slab or drop as shown in Table 255

483 Reinforcement

At internal columns two-thirds of the reinforcement neededto resist the negative moments in the column strips should beplaced in a width equal to half that of the column strip andcentral with the column Otherwise the reinforcement neededto resist the moment apportioned to a particular strip should bedistributed uniformly across the full width of the strip

484 Alternative analysis

A more general equivalent frame method for the analysis offlat slabs is described in BS 8110 The bending moments and

Flat slabs 35

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shearing forces are calculated by considering the structure asa series of continuous frames transversely and longitudinallyThe method is described in detail in Examples of the design ofreinforced concrete buildings For further information on bothequivalent frame and grillage methods of analysis of flat slabstructures see ref 33

49 FRAMED STRUCTURES

A structure is statically determinate if the forces and bendingmoments can be determined by the direct application of theprinciples of equilibrium Some examples include cantilevers(whether a simple bracket or a roof of a grandstand) a freelysupported beam a truss with pin-joints and a three-hinged archor frame A statically indeterminate structure is one in whichthere is a redundancy of members or supports or both andwhich can be analysed only by considering the elastic defor-mations under load Typical examples of such structures includerestrained beams continuous beams portal frames and othernon-triangulated structures with rigid joints and two-hinged andfixed-end arches The general notes relating to the analysis ofstatically determinate and indeterminate beam systems given insections 41 and 42 are equally valid when analysing framesProviding a frame can be represented sufficiently accurately byan idealised two-dimensional line structure it can be analysedby any of the methods mentioned earlier (and various othersof course)

The analysis of a two-dimensional frame is somewhat morecomplex than that of a beam system If the configuration ofthe frame or the applied loading (or both) is unsymmetricalside-sway will almost invariably occur making the requiredanalysis considerably longer Many more combinations of load(vertical and horizontal) may need to be considered to obtainthe critical moments Different partial safety factors may applyto different load combinations The critical design conditionsfor some columns may not necessarily be those correspondingto the maximum moment loading producing a reduced momenttogether with an increased axial thrust may be more criticalHowever to combat such complexities it is often possible tosimplify the calculations by introducing a degree of approxi-mation For instance when considering wind loads acting onregular multi-bay frames points of contra-flexure may beassumed to occur at the centres of all the beams and columns(see Table 262) thus rendering the frame statically determinateIn the case of frames that are not required to provide lateralstability the beams at each level acting with the columns aboveand below that level may be considered to form a separatesub-frame for analysis

Beeby (ref 34) has shown that if the many uncertaintiesinvolved in frame analysis are considered there is little tochoose as far as accuracy is concerned between analysing aframe as a single complete structure as a set of sub-frames oras a series of continuous beams with attached columns Ifthe effect of the columns is not included in the analysis of thebeams some of the calculated moments in the beams will begreater than those actually likely to occur

It may not always be possible to represent the true frame asan idealised two-dimensional line structure and analysis as afully three-dimensional space frame may be necessary If thestructure consists of large solid areas such as walls it may notbe possible to represent it adequately by a skeletal frame

The finite-element method of analysis is particularly suited tosolve such problems and is summarised briefly later

In the following pages the analysis of primary frames by themethods of slope deflection and various forms of momentdistribution is described Rigorous analysis of complex rigidframes generally requires an amount of calculation out ofall proportion to the real accuracy of the results and someapproximate solutions are therefore given for common casesof building frames and similar structures When a suitablepreliminary design has been justified by using approximatemethods an exhaustive exact analysis may be undertaken byemploying an established computer program

491 Building code requirements

For most framed structures it is not necessary to carry out afull structural analysis of the complete frame as a single unitand various simplifications are shown in Table 257 BS 8110distinguishes between frames subjected to vertical loads onlybecause overall lateral stability to the structure is provided byother means such as shear walls and frames that are requiredto support both vertical and lateral loads Load combinationsconsisting of (1) dead and imposed (2) dead and wind and(3) dead imposed and wind are also given in Table 257

For frames that are not required to provide lateral stabilitythe construction at each floor may be considered as a separatesub-frame formed from the beams at that level together withthe columns above and below The columns should be taken asfixed in position and direction at their remote ends unless theassumption of a pinned end would be more reasonable (eg ifa foundation detail is considered unable to develop momentrestraint) The sub-frame should then be analysed for therequired arrangements of dead and live loads

As a further simplification each individual beam span maybe considered separately by analysing a sub-frame consisting ofthe span in question together with at each end the upper andlower columns and the adjacent span These members areregarded as fixed at their remote ends with the stiffness of theouter spans taken as only one-half of their true value This sim-plified sub-frame should then be analysed for the loadingrequirements previously mentioned Formulae giving bendingmoments due to various loading arrangements acting on thesimplified sub-frame obtained by slope-deflection methods asdescribed in section 1421 are given in Table 261 Since themethod is lsquoexactrsquo the calculated bending moments may beredistributed within the limits permitted by the Codes Themethod is dealt with in more detail in Examples of the designof reinforced concrete buildings

BS 8110 also allows analysis of the beams at each floor as acontinuous system neglecting the restraint provided by thecolumns entirely so that the continuous beam is assumed to beresting on knife-edge supports Column moments are thenobtained by considering at each joint a sub-frame consistingof the upper and lower columns together with the adjacentbeams regarded as fixed at their remote ends and with theirstiffness taken as one-half of the true value

For frames that are required to provide lateral stability to thestructure as a whole load combinations 1 and 3 both need to beconsidered For combination 3 the following two-stage methodof analysis is allowed for frames of three or more approxi-mately equal bays First each floor is considered as a separate

Structural analysis36

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sub-frame for the effect of vertical loading as describedpreviously Next the complete structural frame is consideredfor the effect of lateral loading assuming that a positionof contra-flexure (ie zero bending moment) occurs at themid-point of each member This analysis corresponds to thatdescribed for building frames in section 4113 and the methodset out in diagram (c) of Table 262 may thus be used Themoments obtained from each of these analyses should thenbe summed and compared with those resulting from loadcombination 1 For tall narrow buildings and other cantileverstructures such as masts pylons and towers load combination2 should also be considered

492 Moment-distribution method no sway

In some circumstances a framed structure may not be subjectto side-sway for example if the frame is braced by other stiffelements within the structure or if both the configuration andthe loading are symmetrical Similarly if a vertically loadedframe is being analysed as a set of sub-frames as permitted inBS 8110 the effects of any side-sway may be ignored In suchcases Hardy Cross moment distribution may be used to evaluatethe moments in the beam and column system The procedurewhich is outlined in Table 258 is similar to the one used toanalyse systems of continuous beams

Precise moment distribution may also be used to solvesuch systems Here the method which is also summarised inTable 258 is slightly more complex to apply than in theequivalent continuous beam case Each time a moment iscarried over the unbalanced moment in the member must bedistributed between the remaining members meeting at the jointin proportion to the relative restraint that each provides Alsothe expression for the continuity factors is more difficultto evaluate Nevertheless the method is a valid alternative tothe conventional moment-distribution method It is describedin more detail in Examples of the design of reinforcedconcrete buildings

493 Moment-distribution method with sway

If sway occurs analysis by moment distribution increases incomplexity since in addition to the influence of the originalloading with no sway it is necessary to consider the effect ofeach degree of sway freedom separately in terms of unknownsway forces The separate results are then combined to obtainthe unknown sway values and hence the final moments Theprocedure is outlined in Table 259

The advantages of precise moment distribution are largelynullified if sway occurs but details of the procedure in suchcases are given in ref 35

To determine the moments in single-bay frames subjected toside sway Naylor (ref 36) devised an ingenious variant ofmoment distribution details of which are given in Table 259The method can also be used to analyse Vierendeel girders

494 Slope-deflection method

The principles of the slope-deflection method of analysing arestrained member are given in Table 260 and section 141together with basic formulae and formulae for the bending

moments in special cases When there is no deflection of oneend of the member relative to the other (eg when the supportsare not elastic as assumed) when the ends of the memberare either hinged or fixed and when the load on the member issymmetrically disposed the general expressions are simplifiedand the resulting formulae for some common cases of restrainedmembers are also given in Table 260

The bending moments on a framed structure are determinedby applying the formulae to each member successively Thealgebraic sum of the bending moments at any joint must equalzero When it is assumed that there is no deflection (or settle-ment) a of one support relative to the other there are as manyformulae for the end moments as there are unknowns andtherefore the restraint moments and the slopes at the endsof the members can be evaluated For symmetrical frameson unyielding foundations and carrying symmetrical verticalloads it is common to neglect the change in the position of thejoints due to the small elastic contractions of the members andthe assumption of a 0 is reasonably correct If the founda-tions or other supports settle unequally under the load thisassumption is not justified and the term a must be assigned avalue for the members affected

If a symmetrical or unsymmetrical frame is subjected to ahorizontal force the resulting sway causes lateral movementof the joints It is common in this case to assume that there isno elastic shortening of the members Sufficient formulae toenable the additional unknowns to be evaluated are obtainedby equating the reaction normal to the member that is theshear force on the member to the rate of change of bendingmoment Sway occurs also in unsymmetrical frames subjectto vertical loads and in any frame on which the load is notsymmetrically disposed

Slope-deflection methods have been used to derive bendingmoment formulae for the simplified sub-frames illustratedon Table 260 These simplified sub-frames correspond tothose referred to in BS 8110 as a basis for determiningthe bending moments in the individual members of a framesubjected to vertical loads only The method is describedin section 142

An example of applying the slope-deflection formulae to asimple problem of a beam hinged at one end and framed intoa column at the other end is given in section 141

495 Shearing forces on members of a frame

The shearing forces on any member forming part of a frame canbe simply determined once the bending moments have beenfound by considering the rate of change of the bendingmoment The uniform shearing force on a member AB due toend restraint only is (MAB MBA)lAB account being taken ofthe signs of the bending moment Thus if both of the restraintmoments are clockwise the shearing force is the numerical sumof the moments divided by the length of the member If onerestraint moment acts in a direction contrary to the other theshearing force is the numerical difference in the momentsdivided by the length of the member For a member with end Bhinged the shearing force due to the restraint moment at A isMABlAB The variable shearing forces caused by the loadson the member should be algebraically added to the uniformshearing force due to the restraint moments as indicated fora continuous beam in section 1112

Framed structures 37

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496 Portal frames

A common type of frame used in single-storey buildings is theportal frame with either a horizontal top member or twoinclined top members meeting at the ridge In Tables 263 and264 general formulae for the moments at both ends of thecolumns and at the ridge where appropriate are given togetherwith expressions for the forces at the bases of the columnsThe formulae relate to any vertical or horizontal load and toframes fixed or hinged at the bases In Tables 265 and 266corresponding formulae for special conditions of loading onframes of one bay are given

Frames of the foregoing types are statically indeterminatebut frames with a hinge at the base of each column and one atthe ridge that is a three-hinged frame can be readily analysedFormulae for the forces and bending moments are given inTable 267 for three-hinged frames Approximate expressionsare also given for certain modified forms of these frames such aswhen the ends of the columns are embedded in the foundationsand when a tie-rod is provided at eaves level

497 Finite elements

In conventional structural analysis numerous approximationsare introduced and the engineer is normally content to acceptthe resulting simplification Actual elements are considered asidealised one-dimensional linear members deformations due toaxial force and shear are assumed to be sufficiently small to beneglected and so on

In general such assumptions are valid and the results of theanalysis are sufficiently close to the values that would occurin the actual structure to be acceptable However when themember sizes become large in relation to the structure theyform the system of skeletal simplification breaks down Thisoccurs for example with the design of such elements as deepbeams shear walls and slabs of various types

One of the methods developed to deal with such so-calledcontinuum structures is that known as finite elements Thestructure is subdivided arbitrarily into a set of individualelements (usually triangular or rectangular in shape) which arethen considered to be inter-connected only at their corners(nodes) Although the resulting reduction in continuity mightseem to indicate that the substitute system would be muchmore flexible than the original structure this is not the case ifthe substitution is undertaken carefully since the adjoiningedges of the elements tend not to separate and thus simulatecontinuity A stiffness matrix for the substitute structure cannow be prepared and analysed using a computer in a similarway to that already described

Theoretically the pattern of elements chosen might bethought to have a marked effect on the validity of the resultsHowever although the use of a smaller mesh consisting ofa larger number of elements can often increase the accuracyof the analysis it is normal for surprisingly good results to beobtained by experienced analysts when using a rather coarsegrid consisting of only a few large elements

410 COLUMNS IN NON-SWAY FRAMES

In monolithic beam-and-column construction subjected tovertical loads only provision is still needed for the bending

moments produced on the columns due to the rigidity of thejoints The external columns of a building are subjected togreater moments than the internal columns (other conditionsbeing equal) The magnitude of the moment depends on therelative stiffness and the end conditions of the members

The two principal cases for beamndashcolumn connections areat intermediate points on the column (eg floor beams) and atthe top of the column (eg roof beam) Since each member canbe hinged fully fixed or partially restrained at its remote endthere are many possible combinations

In the first case the maximum restraint moment at the jointbetween a beam and an external column occurs when theremote end of the beam is hinged and the remote ends of thecolumn are fixed as indicated in Table 260 The minimumrestraint moment at the joint occurs when the remote end ofthe beam is fixed and the remote ends of the column are bothhinged as also indicated in Table 260 Real conditions inpractice generally lie between these extremes and with anycondition of fixity of the remote ends of the column themoment at the joint decreases as the degree of fixity at theremote end of the beam increases With any degree of fixity atthe remote end of the beam the moment at the joint increasesvery slightly as the degree of fixity at the remote ends of thecolumn increase

Formulae for maximum and minimum bending moments aregiven in Table 260 for a number of single-bay frames Themoment on the beam at the joint is divided between the upperand lower columns in the ratio of their stiffness factors K whenthe conditions at the ends of the two columns are identicalWhen one column is hinged at the end and the other is fixedthe solution given for two columns with fixed ends can still beused by taking the effective stiffness factor of the column withthe hinged end as 075K

For cases where the beamndashcolumn connection is at the top ofthe column the formulae given in Table 260 may be used bytaking the stiffness factors for the upper columns as zero

4101 Internal columns

For the frames of ordinary buildings the bending moments onthe upper and lower internal columns can be computed from theexpressions given at the bottom of Table 260 these formulaeconform to the method to be used when the beams are analysedas a continuous system on knife-edge supports as describedin clause 32125 of BS 8110 When the spans are unequal thegreatest bending moments on the column are when the value ofMes (see Table 260) is greatest which is generally when thelonger beam is loaded with (dead live) load while the shorterbeam carries dead load only

Another method of determining moments in the columnsaccording to the Code requirements is to use the simplifiedsub-frame formulae given on Table 261 Then consideringcolumn SO for example the column moment is given by

where DSO DST and DTS are distribution factors FS and FT arefixed-end moments at S and T respectively (see Table 261)This moment is additional to any initial fixed-end momentacting on SO

DSO2DTSFT 4FS

4DSTDTS

Structural analysis38

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To determine the maximum moment in the column it may benecessary to examine two separate simplified sub-frames inwhich each column is embodied at each floor level (ie thecolumn at joint S say is part of two sub-frames comprisingbeams QR to ST and RS to TU respectively) However themaximum moments usually occur when the central beam ofthe sub-frame is the longer of the two beams adjoining thecolumn being investigated as specified in the Code

4102 End columns

The bending moments due to continuity between the beams andthe columns vary more for end columns than for internalcolumns The lack of uniformity in the end conditions affectsthe moments determined by the simplified method describedearlier more significantly than for internal columns Howevereven though the values obtained by the simplified methodsare more approximate than for internal columns they are stillsufficiently accurate for ordinary buildings The simplifiedformulae given on Table 260 conform to clause 32125 ofBS 8110 while the alternative simplified sub-frame methoddescribed for internal columns may also be used

4103 Corner columns

Corner columns are generally subjected to bending momentsfrom beams in two directions at right angles These momentscan be independently calculated by considering two frames(also at right angles) but practical methods of column designdepend on both the relative magnitudes of the moments andthe direct load and the relevant limit-state condition Thesemethods are described in later sections of the Handbook

4104 Use of approximate methods

The methods hitherto described for evaluating the columnmoments in beam-and-column construction with rigid jointsinvolve significant calculation including the second momentof area of the members Often in practice and especially inthe preparation of preliminary schemes approximate methodsare very useful The final design should be checked by moreaccurate methods

The column can be designed provisionally for a direct loadincreased to allow for the effects of bending In determiningthe total column load at any particular level the load from thefloor immediately above that level should be multiplied by thefollowing factors internal columns 125 end columns 15 andcorner column 20

411 COLUMNS IN SWAY FRAMES

In exposed structures such as water towers bunkers and silosand in frames that are required to provide lateral stability to abuilding the columns must be designed to resist the effects ofwind When conditions do not warrant a close analysis of thebending moments to which a frame is subjected due to wind orother lateral forces the methods described in the following andshown in Table 262 are sufficiently accurate

4111 Open braced towers

For columns (of identical cross section) with braced cornersforming an open tower such as that supporting an elevated

water tank the expressions at (a) in Table 262 give bendingmoments and shearing forces on the columns and braces dueto the effect of a horizontal force at the head of the columns

In general the bending moment on the column is the shearforce on the column multiplied by half the distance between thebraces If a column is not continuous or is insufficiently bracedat one end as at an isolated foundation the bending moment atthe other end is twice this value

The bending moment on the brace at an external column isthe sum of the bending moments on the column at the points ofintersection with the brace The shearing force on the brace isequal to the change of bending moment from one end of thebrace to the other end divided by the length of the braceThese shearing forces and bending moments are additional tothose caused by the dead weight of the brace and any externalloads to which it may be subjected

The overturning moment on the frame causes an additionaldirect load on the leeward column and a corresponding relief ofload on the windward column The maximum value of thisdirect load is equal to the overturning moment at the footof the columns divided by the distance between the centres ofthe columns

The expressions in Table 262 for the bending moments andforces on the columns and braces apply for columns that arevertical or near vertical If the columns are inclined then theshearing force on a brace is 2Mb divided by the length ofthe brace being considered

4112 Columns supporting massivesuperstructures

The case illustrated at (b) in Table 262 is common in silos andbunkers where a superstructure of considerable rigidity iscarried on comparatively short columns If the columns arefixed at the base the bending moment on a single column isFh2J where J is the number of columns if they are all of thesame size the significance of the other symbols is indicated inTable 262

If the columns are of different sizes the total shearing forceon any one line of columns should be divided between them inproportion to the second moment of area of each column sincethey are all deflected by the same amount If J1 is the numberof columns with second moment of area I1 J2 is the number ofcolumns with second moment of area I2 and so on the totalsecond moment of area I J1I1 J2I2 and so on Then onany column having a second moment of area Ij the bendingmoment is FhIj2I as given in diagram (b) in Table 262Alternatively the total horizontal force can be divided amongthe columns in proportion to their cross-sectional areas (thusgiving uniform shear stress) in which case the formula for thebending moment on any column with cross-sectional area Aj isFhAj2A where A is the sum of the cross-sectional areas ofall the columns resisting the total shearing force F

4113 Building frames

In the frame of a multi-storey multi-bay building the effect ofthe wind may be small compared to that of other loads andin this case it is sufficiently accurate to divide the horizontalshearing force between the columns on the basis that an endcolumn resists half the amount on an internal column If in the

Columns in sway frames 39

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plane of the lateral force F Jt is the total number of columns inone frame the effective number of columns for the purpose ofcalculating the bending moment on an internal column is Jt 1the two end columns being equivalent to one internal columnsee diagram (c) in Table 262 In a building frame subjectedto wind pressure the forces on each panel (or storey height)F1 F2 F3 and so on are generally divided into equal shearingforces at the head and base of each storey height of columnsThe shearing force at the bottom of any internal column istoreys from the top is (F Fi2)( Jt 1) where F F1 F2 F3 Fi 1 The bending moment is then the shearingforce multiplied by half the storey height

A bending moment and a corresponding shearing force arecaused on the floor beams in the same way as on the braces ofan open braced tower At an internal column the sum of thebending moments on the two adjacent beams is equal to the sumof the moments at the base of the upper column and the head ofthe lower column

The above method of analysis for determining the effects oflateral loading corresponds to that described in section 491and recommended in BS 8110 for a frame of three or moreapproximately equal bays

412 WALL AND FRAME SYSTEMS

In all forms of construction the effects of wind force increasein significance as the height of the structure increases Oneway of reducing lateral sway and improving stability is byincreasing the sectional size of the component members ofsway frames However this will have a direct consequenceof increasing storey height and building cost

Often a better way is to provide a suitable arrangement ofwalls linked to flexible frames The walls can be external orinternal be placed around lift shafts and stairwells to form corestructures or be a combination of types Sometimes core wallsare constructed in advance of the rest of the structure to avoidsubsequent delays The lateral stiffness of systems with acentral core can be increased by providing deep cantilevermembers at the top of the core structure to which the exteriorcolumns are connected Another approach is to increase theload on the central core by replacing the exterior columns byhangers suspended from the cantilever members at the top ofthe building This also avoids the need for exterior columns atground level and their attendant foundations As buildings gettaller the lateral stability requirements are of paramount impor-tance The structural efficiency can be increased by replacingthe building facade by a rigidly jointed framework so that theouter shell acts effectively as a closed-box

Some different structural forms consisting of assemblies ofmulti-storey frames shear walls and cores with an indicationof typical heights and proportions taken from ref 37 areshown in Table 268

4121 Shear wall structures

The lateral stability of low- to medium-rise buildings is oftenobtained by providing a suitable system of stiff shear walls Thearrangement of the walls should be such that the building is stiffin both flexure and torsion In rectangular buildings externalshear walls in the short direction can be used to resist lateralloads acting on the wide faces with rigid frames or infill panels

in the long direction In buildings of square plan form a strongcentral service core surrounded by flexible external framescan be used If strong points are placed at both ends of a longbuilding the restraint provided to the subsequent shrinkageand thermal movements of floors and roof should be carefullyconsidered

In all cases the floors and roof are considered to act as stiffplates so that at each level the horizontal displacements of allwalls and columns are taken to be the same provided the totallateral load acts through the shear centre of the system If thetotal lateral load acts eccentrically then the additional effectof the resulting torsion moment needs to be considered Theanalysis and design of shear wall buildings is covered in ref 38from which much of the following treatment is based Severaldifferent plan configurations of shear walls and core units withnotes on their suitability are shown in Table 269

4122 Walls without openings

The lateral load transmitted to an individual wall is a functionof its position and its relative stiffness The total deflection of acantilever wall under lateral load is a combination of bendingand shear deformations However for a uniformly distributedload the shear deformation is less than 10 of the total forHD 3 in the case of plane walls and HD 5 in the case offlanged walls with BD 05 (where B is width of flange D isdepth of web and H is height of wall) Thus for most shearwalls without openings the dominant mode of deformation isbending and the stiffness of the wall can be related directly to thesecond moment of area of the cross section I Then for a totallateral load F applied at the shear centre of a system of parallelwalls the shearing force on an individual wall j is FIjIj

The position of the shear centre along a given axis y can bereadily determined by calculating the moment of stiffness ofeach wall about an arbitrary reference point on the axis Thedistance from the point to the shear centre yc Ijyj Ij

If the total lateral load acts at distance yo along the axis theresulting horizontal moment is F(yondashyc) Then if the torsionstiffness of individual walls is neglected the total shearingforce on wall j is

Fj FIj Ij F(yo yc)IjyjIj (yj yc)2

More generalised formulae in which a wall system is related totwo perpendicular axes are given in Table 269 The aboveanalysis takes no account of rotation at the base of the walls

4123 Walls containing openings

In the case of walls pierced by openings the behaviour ofthe individual wall sections is coupled to a variable degree Theconnections between the individual sections are provided eitherby beams that form part of the wall or by floor slabs or by acombination of both The pierced wall may be analysed byelastic methods in which the flexibility of the coupling elementsis represented as a continuous flexible medium Alternativelythe pierced wall may be idealised as an equivalent plane frameusing a lsquowide columnrsquo analogy

The basis of the continuous connection model is described insection 152 and analytical solutions for a wall containing asingle line of openings are given in Table 270

Structural analysis40

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4124 Interaction of shear walls and frames

The interaction forces between solid walls pierced walls andframes can vary significantly up the height of a building asa result of the differences in the free deflected shapes ofeach structural form The deformation of solid walls is mainlyflexural whereas pierced walls deform in a shearndashflexure modeand frames deform in an almost pure shear manner As a resulttowards the bottom of a building solid walls attract load whilstframes and to a lesser extent pierced walls shed load Thebehaviour is reversed towards the top of a building Thusalthough the distribution of load intensity between the differentelements is far from uniform up the building the total lateralforce resisted by each varies by a smaller amount

As a first approximation the shearing force at the bottom ofeach load-resisting element can be determined by considering asingle interaction force at the top of the building Formulae bywhich the effective stiffness of pierced walls and frames can bedetermined are given in section 153

413 ARCHES

Arch construction in reinforced concrete occurs sometimes inroofs but mainly in bridges An arch may be three-hingedtwo-hinged or fixed-ended (see diagrams in Table 271) andmay be symmetrical or unsymmetrical right or skew singleor one of a series of arches mutually dependent upon eachother The following consideration is limited to symmetrical andunsymmetrical three-hinged arches and to symmetrical two-hinged and fixed-end arches reference should be made to otherpublications for information on more complex types

Arch construction may comprise an arch slab (or vault) or aseries of parallel arch ribs The deck of an arch bridge may besupported by columns or transverse walls carried on an archslab or ribs when the structure may have open spandrels or thedeck may be below the crown of the arch either at the level ofthe springing (as in a bowstring girder) or at some intermediatelevel A bowstring girder is generally regarded as a two-hingedarch with the horizontal component of thrust resisted by a tiewhich normally forms part of the deck If earth or other filling isprovided to support the deck an arch slab and spandrel walls arerequired and the bridge is a closed or solid-spandrel structure

4131 Three-hinged arch

An arch with a hinge at each springing and at the crown isstatically determinate The thrusts on the abutments and thebending moments and shearing forces on the arch itself arenot affected by a small movement of one abutment relative tothe other This type of arch is therefore used when there is apossibility of unequal settlement of the abutments

For any load in any position the thrust on the abutmentscan be determined by the equations of static equilibrium Forthe general case of an unsymmetrical arch with a load actingvertically horizontally or at an angle the expressions for thehorizontal and vertical components of the thrusts are givenin the lower part of Table 271 For symmetrical arches the for-mulae given in Table 267 for the thrusts on three-hinged framesapply or similar formulae can be obtained from the generalexpressions in Table 271 The vertical component is the same asthe vertical reaction for a freely supported beam The bending

moment at any cross section of the arch is the algebraic sum ofthe moments of the loads and reactions on one side of thesection There is no bending moment at a hinge The shearingforce is likewise the algebraic sum of the loads and reactionsresolved at right angles to the arch axis at the section and actingon one side of the section The thrust at any section is the sumof the loads and reactions resolved parallel to the axis of thearch at the section and acting on one side of the section

The extent of the arch that should be loaded with imposedload to give the maximum bending moment or shearing forceor thrust at a particular cross section can be determined byconstructing a series of influence lines A typical influence linefor a three-hinged arch and the formulae necessary to constructan influence line for unit load in any position are given in theupper part of Table 271

4132 Two-hinged arch

The hinges of a two-hinged arch are placed at the abutmentsso that as in a three-hinged arch only thrusts are transmitted tothe abutments and there is no bending moment on the archat the springing The vertical component of the thrust from asymmetrical two-hinged arch is the same as the reaction fora freely supported beam Formulae for the thrusts and bendingmoments are given in Table 271 and notes in section 162

4133 Fixed arch

An arch with fixed ends exerts in addition to the vertical andhorizontal thrusts a bending moment on the abutments Like atwo-hinged arch and unlike a three-hinged arch a fixed-endarch is statically indeterminate and the stresses are affected bychanges of temperature and shrinkage of the concrete As it isassumed in the general theory that the abutments cannot moveor rotate the arch can only be used in such conditions

A cross section of a fixed-arch rib or slab is subjected to abending moment and a thrust the magnitudes of which have tobe determined The design of a fixed arch is a matter of trial andadjustment since both the dimensions and the shape of the archaffect the calculations but it is possible to select preliminarysizes that reduce the repetition of arithmetic work to a minimumA suggested method of determining possible sections at thecrown and springing as given in Table 272 and explained insection 1631 is based on first treating the fixed arch as ahinged arch and then estimating the size of the cross sectionsby greatly reducing the maximum stresses

The general formulae for thrusts and bending moments on asymmetrical fixed arch of any profile are given in Table 272and notes on the application and modification of the formulaeare given in section 163 The calculations necessary to solvethe general and modified formulae are tedious but are easedsomewhat by preparing them in tabular form The form givenin Table 272 is particularly suitable for open-spandrel archbridges because the appropriate formulae do not assume a con-stant value of aI the ratio of the length of a segment of the archto the mean second moment of area of the segment

For large span arches calculations are made much easier andmore accurate by preparing and using influence lines for thebending moment and thrust at the crown the springing and thequarter points of the arch Typical influence lines are given inTable 272 and such diagrams can be constructed by considering

Arches 41

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the passage over the arch of a single concentrated unit load andapplying the formulae for this condition The effect of the deadload and of the most adverse disposition of imposed load canbe readily calculated from these diagrams If the specifiedimposed load includes a moving concentrated load such as aKEL the influence lines are almost essential for determiningthe most adverse position The case of the positive bendingmoment at the crown is an exception when the most adverseposition of the load is at the crown A method of determiningthe data to establish the ordinates of the influence lines is givenin Table 273

4134 Fixed parabolic arches

In Table 274 and in section 164 consideration is given tosymmetrical fixed arches that can have either open or solidspandrels and be either arch ribs or arch slabs The method isbased on that of Strassner as developed by H Carpenter andthe principal assumption is that the axis of the arch is made tocoincide with the line of thrust due to the dead load This resultsin an economical structure and a simple calculation methodThe shape of the axis of the arch is approximately that of aparabola and this method can therefore be used only when thedesigner is free to select the profile of the arch The parabolicform may not be the most economic for large spans althoughit is almost so and a profile that produces an arch axis coinci-dent with the line of thrust for the dead load plus one-half of theimposed load may be more satisfactory If the increase in thethickness of the arch from crown to springing is of a parabolicform only the bending moments and thrusts at the crownand the springing need to be investigated The necessaryformulae are given in section 164 where these include a seriesof coefficients values of which are given in Table 274 Theapplication of the method is also illustrated by an examplegiven in section 164 The component forces and momentsare considered in the following treatment

The thrusts due to the dead load are relieved somewhat by theeffect of the compression causing elastic shortening of the archFor arches with small ratios of rise to span and arches that arethick in comparison with the span the stresses due to archshortening may be excessive This can be overcome by intro-ducing temporary hinges at the crown and the springing whicheliminate all bending stresses due to dead load The hinges arefilled with concrete after arch shortening and much of theshrinkage of the concrete have taken place

An additional horizontal thrust due to a temperature rise ora corresponding counter-thrust due to a temperature fall willaffect the stresses in the arch and careful consideration mustbe given to the likely temperature range The shrinkage of theconcrete that occurs after completion of the arch produces acounter-thrust the magnitude of which is modified by creep

The extent of the imposed load on an arch necessary toproduce the maximum stresses in the critical sections can bedetermined from influence lines and the following values areapproximately correct for parabolic arches The maximumpositive moment at the crown occurs when the middle third of thearch is loaded the maximum negative moment at a springingoccurs when four-tenths of the span adjacent to the springing isloaded the maximum positive moment at the springing occurswhen six-tenths of the span furthest away from the springing

is loaded In the expressions given in section 1644 the imposedload is expressed in terms of an equivalent UDL

When the normal thrusts and bending moments on the mainsections have been determined the areas of reinforcement andstresses at the crown and springing can be calculated Allthat now remains is to consider the intermediate sections anddetermine the profile of the axis of the arch If the dead loadis uniform throughout (or practically so) the axis will be aparabola but if the dead load is not uniform the axis must beshaped to coincide with the resulting line of thrust This canbe obtained graphically by plotting force-and-link polygonsthe necessary data being the magnitudes of the dead load thehorizontal thrust due to dead load and the vertical reaction(equal to the dead load on half the span) of the springing Theline of thrust and therefore the axis of the arch having beenestablished and the thickness of the arch at the crown and thespringing having been determined the lines of the extradosand the intrados can be plotted to give a parabolic variation ofthickness between the two extremes

414 PROPERTIES OF MEMBERS

4141 End conditions

Since the results given by the more precise methods of elasticanalysis vary considerably with the conditions of restraint atthe ends of the members it is important that the assumedconditions are reasonably obtained in the actual constructionAbsolute fixity is difficult to attain unless a beam or column isembedded monolithically in a comparatively large mass ofconcrete Embedment of a beam in a masonry wall representsmore nearly the condition of a hinge and should normally beconsidered as such A continuous beam supported internallyon a beam or column is only partly restrained and where thesupport at the outer end of an end span is a beam a hinge shouldbe assumed With the ordinary type of pad foundation designedsimply for a uniform ground bearing pressure under the directload on a column the condition at the foot of the column shouldalso be considered as a hinge A column built on a pile-capsupported by two three or four piles is not absolutely fixed buta bending moment can be developed if the resulting verticalreaction (upwards and downwards) and the horizontal thrust canbe resisted by the piles The foot of a column can be consideredas fixed if it is monolithic with a substantial raft foundation

In two-hinged and three-hinged arches hinged frames andsome bridge types where the assumption of a hinged joint mustbe fully realised it is necessary to form a definite hinge in theconstruction This can be done by inserting a steel hinge (orsimilar) or by forming a hinge within the frame

4142 Section properties

For the elastic analysis of continuous structures the sectionproperties need to be known Three bases for calculating thesecond moment of area of a reinforced concrete section are gen-erally recognised in codes of practice as follows

1 The concrete section the entire concrete area but ignoringthe reinforcement

Structural analysis42

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2 The gross section the entire concrete area together with thereinforcement on the basis of a modular ratio (ie ratio ofmodulus of elasticity values of steel and concrete)

3 The transformed section the concrete area in compressiontogether with the reinforcement on the basis of modular ratio

For methods 2 and 3 the modular ratio should be based on aneffective modulus of elasticity of concrete taking account ofthe creep effects of long-term loading In BS 8110 a modularratio of 15 is recommended unless a more accurate figure can bedetermined However until the reinforcement has been deter-mined or assumed calculation of the section properties in thisway cannot be made with any precision Moreover the sectionproperties vary considerably along the length of the member asthe distribution of reinforcement and for method 3 the depthof concrete in compression change The extent and effect ofcracking on the section properties is particularly difficult toassess for a continuous beam in beam-and-slab construction inwhich the beam behaves as a flanged section in the spans wherethe bending moments are positive but is designed as a rectan-gular section towards the supports where the bending momentsare negative

Method 1 is the simplest one to apply and the only practicalapproach when beginning a new design but one of the othermethods could be used when checking the ability of existingstructures to carry revised loadings and for new structureswhen a separate analysis for the SLSs is required In all cases itis important that the method used to assess the section propertiesis the same for all the members involved in the calculationWhere a single stiffness value is to be used to characterise amember method 1 (or 2) is likely to provide the most accurateoverall solution Method 3 will only be appropriate where thevariations in section properties over the length of membersare properly taken into account

415 EARTHQUAKE-RESISTANT STRUCTURES

Earthquakes are ground vibrations that are caused mainly byfracture of the earthrsquos crust or by sudden movement along analready existing fault During a seismic excitation structuresare caused to oscillate in response to the forced motion of thefoundations The affected structure needs to be able to resistthe resulting horizontal load and also dissipate the impartedkinetic energy over successive deformation cycles It would beuneconomical to design the structure to withstand a majorearthquake elastically and the normal approach is to provide itwith sufficient strength and ductility to withstand such an eventby responding inelastically provided that the critical regions

and the connections between members are designed speciallyto ensure adequate ductility

Significant advances have been made in the seismic designof structures in recent years and very sophisticated codes ofpractice have been introduced (ref 39) A design horizontalseismic load is recommended that depends on the importanceof the structure the seismic zone the ground conditionsthe natural period of vibration of the structure and the availableductility of the structure Design load effects in the structureare determined either by linear-elastic structural analysis forthe equivalent static loading or by dynamic analysis When alinear-elastic method is used the design and detailing of themembers needs to ensure that in the event of a more severeearthquake the post-elastic deformation of the structure willbe adequately ductile For example in a multi-storey framesufficient flexural and shear strength should be provided in thecolumns to ensure that plastic hinges form in the beams inorder to avoid a column side-sway mechanism The properdetailing of the reinforcement is also a very important aspectin ensuring ductile behaviour At the plastic hinge regions ofmoment resisting frames in addition to longitudinal tensionreinforcement it is essential to provide adequate compressionreinforcement Transverse reinforcement is also necessary toact as shear reinforcement to prevent premature buckling ofthe longitudinal compression reinforcement and to confine thecompressed concrete

Buildings should be regular in plan and elevation withoutre-entrant angles and discontinuities in transferring verticalloads to the ground Unsymmetrical layouts resulting in largetorsion effects flat slab floor systems without any beams andlarge discontinuities in infill systems (such as open groundstoreys) should be avoided Footings should be founded at thesame level and should be interconnected by a mat foundationor by a grid of foundation beams Only one foundation typeshould in general be used for the same structure unless thestructure is formed of dynamically independent units

An alternative to the conventional ductile design approach isto use a seismic isolation scheme In this case the structure issupported on flexible bearings so that the period of vibration ofthe combined structure and supporting system is long enoughfor the structure to be isolated from the predominant earthquakeground motion frequencies In addition extra damping isintroduced into the system by mechanical energy dissipatingdevices in order to reduce the response of the structure to theearthquake and keep the deflections of the flexible systemwithin acceptable limits

A detailed treatment of the design of earthquake-resistingconcrete structures is contained in ref 40

Earthquake-resistant structures 43

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51 PRINCIPLES AND REQUIREMENTS

In modern Codes of Practice a limit-state design concept isused Ultimate (ULS) and serviceability (SLS) limit-states areconsidered as well as durability and in the case of buildingsfire-resistance Partial safety factors are incorporated in bothloads and material strengths to ensure that the probability offailure (ie not satisfying a design requirement) is acceptablylow For British Codes (BS 8110 BS 5400 BS 8007) detailsare given of design requirements and partial safety factors inChapter 21 material properties in Chapter 22 durability andfire-resistance in Chapter 23 For EC 2 corresponding data aregiven in Chapters 29 30 and 31 respectively

Members are first designed to satisfy the most critical limit-state and then checked to ensure that the other limit-statesare not reached For most members the critical condition to beconsidered is the ULS on which the required resistances of themember in bending shear and torsion are based The require-ments of the various SLSs such as deflection and crackingare considered later However since the selection of an adequatespan to effective depth ratio to prevent excessive deflection andthe choice of a suitable bar spacing to avoid excessive crackingcan also be affected by the reinforcement stress the designprocess is generally interactive Nevertheless it is normal tostart with the requirements of the ULS

52 RESISTANCE TO BENDING AND AXIAL FORCE

Typically beams and slabs are members subjected to bendingwhile columns are subjected to a combination of bending andaxial force In this context a beam is defined as a member inBS 8110 with a clear span not less than twice the effectivedepth and in EC 2 as a member with a span not less than threetimes the overall depth Otherwise the member is treated as adeep beam for which different design methods are applicableA column is defined as a member in which the greater overallcross-sectional dimension does not exceed four times thesmaller dimension Otherwise the member is considered as awall for which a different design approach is adopted Somebeams for example in portal frames and slabs for example inretaining walls are subjected to bending and axial force Insuch cases small axial forces that are beneficial in providingresistance to bending are generally ignored in design

521 Basic assumptions

For the analysis of sections in bending or combined bendingand axial force at the ULS the following basic assumptionsare made

The resistance of the concrete in tension is ignored

The distribution of strain across the section is linear that issections that are plane before bending remain plane afterbending the strain at a point being proportional to its distancefrom the axis of zero strain (neutral axis) In columns ifthe axial force is dominant the neutral axis can lie outsidethe section

Stressndashstrain relationships for concrete in compression andfor reinforcement in tension and compression are thoseshown in the diagrams on Table 36 for BS 8110 andBS 5400 and Table 44 for EC 2

The maximum strain in the concrete in compression is 00035except for EC 2 where the strains shown in the followingdiagram and described in the following paragraph apply

h

0

0

c2

c2

cu

(37)h

For sections subjected to pure axial compression the strain islimited to c2 For sections partly in tension the compressivestrain is limited to cu For intermediate conditions the straindiagram is obtained by taking the compressive strain as c2 at alevel equal to 37 of the section depth from the more highlycompressed face For concrete strength classes C5060 thelimiting strains are c2 0002 and cu 00035 For higherstrength concretes other values are given in Table 44

Strain distribution at ULS in EC 2

Chapter 5

Design of structuralmembers

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In all codes for sections partly in tension the shape of thebasic concrete stress-block is a combination of a parabolaand a rectangle In EC 2 a form consisting of a triangle anda rectangle is also given In all codes a simplified rectangularstress distribution may also be used If the compression zoneis rectangular the compressive force and the distance of theforce from the compression face can be readily determined foreach stress-block and the resulting properties are given insection 241 for BS 8110 and section 321 for EC 2

The stresses in the reinforcement depend on the strains in theadjacent concrete which depend in turn on the depth of theneutral axis and the position of the reinforcement in relationto the concrete surfaces The effect of these factors will beexamined separately for beams and columns

522 Beams

Depth of neutral axis This is significant because the valueof xd where x is the neutral axis depth and d is the effectivedepth of the tension reinforcement not only affects the stress inthe reinforcement but also limits the amount of moment redis-tribution allowed at a given section In BS 8110 where becauseof moment redistribution allowed in the analysis of a memberthe design moment is less than the maximum elastic momentthe requirement xd (b 04) should be satisfied whereb is the ratio of design moment to maximum elastic momentThus for reductions in moment of 10 20 and 30 xdmust not exceed 05 04 and 03 respectively In EC 2 asmodified by the UK National Annex similar restrictions applyfor concrete strength classes C5060

and xd 0456 and dx 043 for BS 5400 For design toEC 2 considerations similar to those in BS 8110 apply

Effect of axial force The following figure shows a sectionthat is subjected to a bending moment M and an axial force Nin which a simplified rectangular stress distribution has beenassumed for the compression in the concrete The stress blockis shown divided into two parts of depths dc and (h 2dc)providing resistance to the bending moment M and the axialforce N respectively where 0 dc 05h

Resistance to bending and axial force 45

Section Strain diagram

Section Forces

The figure here shows a typical strain diagram for a sectioncontaining both tension and compression reinforcement Forthe bi-linear stressndashstrain curve in BS 8110 the maximumdesign stresses in the reinforcement are fy115 for values of s

and s fy115Es From the strain diagram this gives

and

In BS 5400 the reinforcement stressndashstrain curve is tri-linear withmaximum design stresses of fy115 in tension and 2000fy (2300 fy) in compression These stresses apply for values ofs 0002 fy115Es and s 0002 giving

With cu 00035 fy 500 Nmm2 and Es 200 kNmm2 thecritical values are xd 0617 and for BS 8110dx 038

dx (cu0002) cu

x d cu (cu 0002 fy 115Es) and

dx (cu fy 115Es) cux d cu (cu fy 115Es)

The depth dc (and the force in the tension reinforcement) aredetermined by the bending moment given by

M bdc(d 05dc) fcd

Thus for analysis of the section axial forces may be ignoredfor values satisfying the condition

Nb(h 2dc) fcd

Combining the two requirements gives

Nbhfcd 2M(d05dc)

In the limit when dc 05h this gives

Nbhfcd 2M(d025h)congbhfcd 3Mh

For BS 8110 the condition becomes N 045bhfcu 3Mhwhich being simplified to N 01bhfcu is reasonably valid forMbh2fcu 012 For EC 2 the same condition becomesN 0567bhfck 3Mh which may be reasonably simplified toN 012bhfck for Mbh2fck 015

Analysis of section Any given section can be analysed by atrial-and-error process An initial value is assumed for theneutral axis depth from which the concrete strains at the rein-forcement positions can be calculated The correspondingstresses in the reinforcement are determined and the resultingforces in the reinforcement and the concrete are obtained If theforces are out of balance the value of the neutral axis depth ischanged and the process is repeated until equilibrium isachieved Once the balanced condition has been found theresultant moment of all the forces about the neutral axis or anyother suitable point is calculated

Singly reinforced rectangular sections For a section thatis reinforced in tension only and subjected to a moment M aquadratic equation in x can be obtained by taking moments forthe compressive force in the concrete about the line of actionof the tension reinforcement The resulting value of x can beused to determine the strain diagram from which the strain in

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the reinforcement and hence the stress can be calculated Therequired area of reinforcement can then be determined fromthe tensile force whose magnitude is equal to the compressiveforce in the concrete If the calculated value of x exceeds thelimit required for any redistribution of moment then a doublyreinforced section will be necessary

In designs to BS 8110 and BS 5400 the lever arm betweenthe tensile and compressive forces is to be taken not greater than095d Furthermore it is a requirement in BS 5400 that if xexceeds the limiting value for using the maximum designstress then the resistance moment should be at least 115MAnalyses are included in section 2421 for both BS 8110 andBS 5400 and in section 3221 for EC 2 Design charts basedon the parabolic-rectangular stress-block for concrete withfy 500 Nmm2 are given in Tables 313 323 and 47 forBS 8110 BS 5400 and EC 2 respectively Design tables basedon the rectangular stress-blocks for concrete are given inTables 314 324 and 48 for BS 8110 BS 5400 and EC 2respectively These tables use non-dimensional parameters andare applicable for values of fy 500 Nmm2

Doubly reinforced rectangular sections A sectionneeding both tension and compression reinforcement andsubjected to a moment M can be designed by first selecting asuitable value for x such as the limiting value for using themaximum design stress in the tension reinforcement or satisfy-ing the condition necessary for moment redistribution Therequired force to be provided by the compression reinforcementcan be derived by taking moments for the compressive forcesin the concrete and the reinforcement about the line of actionof the tensile reinforcement The force to be provided by thetension reinforcement is equal to the sum of the compressiveforces The reinforcement areas can now be determined takingdue account of the strains appropriate to the value of x selected

Analyses are included in section 2422 for both BS 8110and BS 5400 and in section 3222 for EC 2 Design chartsbased on the rectangular stress-blocks for concrete are given inTables 315 and 316 for BS 8110 Tables 325 and 326 forBS 5400 and Tables 49 and 410 for EC 2

Design formulae for rectangular sections Designformulae based on the rectangular stress-blocks for concreteare given in BS 8110 and BS 5400 In both codes x is limitedto 05d so that the formulae are automatically valid for redistri-bution of moment not greater than 10 The design stress intension reinforcement is taken 087fy although this is onlystrictly valid for xd 0456 in BS 5400 The design stresses inany compression reinforcement are taken as 087fy in BS 8110and 072fy in BS 5400 Design formulae are given in section2423 for BS 8110 and BS 5400 Although not included inEC 2 appropriate formulae are given in section 3223

Flanged sections In monolithic beam and slab constructionwhere the web of the beam projects below the slab the beam isconsidered as a flanged section for sagging moments Theeffective width of the flange over which uniform conditionsof stress can be assumed is limited to values stipulated in thecodes In most sections where the flange is in compressionthe depth of the neutral axis will be no greater than the flangethickness In such cases the section can be considered to berectangular with b taken as the flange width If the depth of

the neutral axis does exceed the thickness of the flange thesection can be designed by dividing the compression zoneinto portions comprising the web and the outlying flangesDetails of the flange widths and design procedures are given insections 2424 for BS 8110 and 3224 for EC 2

Beam sizes The dimensions of beams are mainly determinedby the need to provide resistance to moment and shear In thecase of beams supporting items such as cladding partitions orsensitive equipment service deflections can also be criticalOther factors such as clearances below beams dimensions ofbrick and block courses widths of supporting members andsuitable sizes of formwork also need to be taken into accountFor initial design purposes typical spaneffective depth ratiosfor beams in buildings are given in the following table

Design of structural members46

Spaneffective depth ratios for initial design of beams

Span conditionsUltimate design load

25 kNm 50 kNm 100 kNm

Cantilever 9 7 5Simply supported 18 14 10Continuous 22 17 12

The effective span of a continuous beam is generally taken asthe distance between centres of supports At a simple supportor at an encastre end the centre of action may be taken at adistance not greater than half of the effective depth fromthe face of the support Beam widths are often taken as half theoverall depth of the beam with a minimum of 300 mm If amuch wider band beam is used the spaneffective depth ratiocan be increased significantly to the limit necessitated bydeflection considerations

In BS 8110 and BS 5400 to ensure lateral stability simplysupported and continuous beams should be so proportioned thatthe clear distance between lateral restraints is not greater than60bc or 250bc

2d whichever is the lesser For cantilevers inwhich lateral restraint is provided only at the support the cleardistance from the end of the cantilever to the face of the sup-port should not exceed 25bc or 100bc

2d whichever is the lesserone In the foregoing bc is the breadth of the compression faceof the beam (measured midway between restraints) orcantilever In EC 2 second order effects in relation to lateralstability may be ignored if the distance between lateralrestraints is not greater than 50bc(hbc)13 and h 25bc

523 Slabs

Solid slabs are generally designed as rectangular strips of unitwidth and singly reinforced sections are normally sufficientRibbed slabs are designed as flanged sections of width equalto the rib spacing for sagging moments Continuous ribbedslabs are often made solid in support regions so as to developsufficient resistance to hogging moments and shear forcesAlternatively in BS 8110 ribbed slabs may be designed as aseries of simply supported spans with a minimum amountof reinforcement provided in the hogging regions to controlthe cracking The amount of reinforcement recommended is

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25 of that in the middle of the adjoining spans extending intothe spans for at least 15 of the span length

The thickness of slabs is normally determined by deflectionconsiderations which sometimes result in the use of reducedreinforcement stresses to meet code requirements Typicalspaneffective depth ratios for slabs designed to BS 8110 aregiven in the following table

contain a modification factor the use of which necessitates aniteration process with the factor taken as 10 initially Details ofthe design procedures are given in Tables 321 and 322 forBS 8110 Tables 331 and 332 for BS 5400 and Tables 415 and416 for EC 2

Analysis of section Any given section can be analysed by atrial-and-error process For a section bent about one axis aninitial value is assumed for the neutral axis depth from whichthe concrete strains at the positions of the reinforcement can becalculated The resulting stresses in the reinforcement aredetermined and the forces in the reinforcement and concreteare evaluated If the resultant force is not equal to the designaxial force N the value of the neutral axis depth is changed andthe process repeated until equality is achieved The sum of themoments of all the forces about the mid-depth of the section isthen the moment of resistance appropriate to N For a section inbiaxial bending initial values have to be assumed for the depthand the inclination of the neutral axis and the design processwould be extremely tedious without the aid of an interactivecomputer program

For design purposes charts for symmetrically reinforcedrectangular and circular sections bent about one axis can bereadily derived For biaxial bending conditions approximatedesign methods have been developed that utilise the solutionsobtained for uniaxial bending

Rectangular sections The figure here shows a rectangularsection with reinforcement in the faces parallel to the axisof bending

Resistance to bending and axial force 47

Spaneffective depth ratios for initial design of solid slabs

Span conditionsCharacteristic imposed load

5 kNm2 10 kNm2

Cantilever 11 10Simply supported

One-way span 27 24Two-way span 30 27

ContinuousOne-way span 34 30Two-way span 44 40

Flat slab (no drops) 30 27

In the table here the characteristic imposed load should includefor all finishes partitions and services For two-way spans theratios given apply to square panels For rectangular panelswhere the length is twice the breadth the ratios given for one-wayspans should be used For other cases ratios may be obtainedby interpolation The ratios apply to the shorter span for two-wayslabs and the longer span for flat slabs For ribbed slabs exceptfor cantilevers the ratios given in the table should be reducedby 20

524 Columns

The second order effects associated with lateral stability are animportant consideration in column design An effective height(or length in EC 2) and a slenderness ratio are determined inrelation to major and minor axes of bending An effective heightor length is a function of the clear height and depends upon theconditions of restraint at the ends of the column A clear distinc-tion exists between a braced column with effective height clear height and an unbraced column with effective height clearheight A braced column is one that is fully retrained in positionat the ends as in a structure where resistance to all the lateralforces in a particular plane is provided by stiff walls or bracingAn unbraced column is one that is considered to contribute tothe lateral stability of the structure as in a sway frame

In BS 8110 and BS 5400 a slenderness ratio is defined asthe effective height divided by the depth of the cross section inthe plane of bending A column is then considered as eithershort or slender according to the slenderness ratios Bracedcolumns are often short in which case second order effects maybe ignored In EC 2 the slenderness ratio is defined as theeffective length divided by the radius of gyration of the crosssection

Columns are subjected to combinations of bending momentand axial force and the cross section may need to be checked formore than one combination of values In slender columns theinitial moments obtained from an elastic analysis of the structureare increased by additional moments induced by the deflectionof the column In BS 8110 and EC 2 these additional moments

Resolving forces and taking moments about the mid-depth ofthe section gives the following equations for 0 x h

N k1bxfc As1 fs1 As2 fs2

M k1bxfc (05h k2x) As1 fs1 (05h d) As2 fs2 (d 05h)

where fs1 and fs2 are determined by the stressndashstrain curves for thereinforcement and depend on the value of x Values of k1 and k2

are determined by the concrete stress-block and fc is equal to fcu

in BS 8110 and BS 5400 and fck in EC 2For symmetrically reinforced sections As1 As2 Asc2

and d h d Design charts based on a rectangular stress-blockfor the concrete with values of fy 500 Nmm2 and dh 08and 085 respectively are given in Tables 317 and 318 forBS 8110 Tables 327 and 328 for BS 5400 and Tables 411 and412 for EC 2 Approximate design methods for biaxial bendingare given in Tables 321 331 and 416 for design to BS 8110BS 5400 and EC 2 respectively

Section Forces

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Circular sections The figure here shows a circular sectionwith six bars spaced equally around the circumference Six is theminimum number of bars recommended in the codes and solu-tions based on six bars will be slightly conservative if more barsare used The arrangement of bars relative to the axis of bendingaffects the resistance of the section and it can be shown that thearrangement in the figure is not the most critical in every casebut the variations are small and may be reasonably ignored

braced structures are typically square in cross section withsizes being determined mainly by the magnitude of the axialloads In multi-storey buildings column sizes are often keptconstant over several storeys with the reinforcement changingin relation to the axial load For initial design purposes typicalload capacities for short braced square columns in buildings aregiven in the following table

Design of structural members48

The following analysis is based on a uniform stress-block forthe concrete of depth x and width hsin at the base (as shownin the figure) Resolving forces and taking moments about themid-depth of the section where hs is the diameter of a circlethrough the centres of the bars gives the following equationsfor 0 x h

N [(2 sin2)8]h2fcd (Asc3)( fs1 fs2 fs3)M [(3sin sin3)72]h3fcd 0433(Asc3)( fs1 fs3)hs

where fs1 fs2 and fs3 are determined by the stressndashstrain curvesfor the reinforcement and depend on the value of x Values of fcd

and respectively are taken as 045fcu and 09 in BS 811004fcu and 10 in BS 5400 and 051fck and 08 in EC 2

Design charts derived for values of fy 500 Nmm2 andhsh 06 and 07 respectively are given in Tables 319and 320 for BS 8110 Tables 329 and 330 for BS 5400 andTables 413 and 414 for EC 2 Sections subjected to biaxialmoments Mx and My can be designed for the resultant moment

Design formulae In BS 8110 two approximate formulae aregiven for the design of short braced columns under specificconditions Columns which due to the nature of the structurecannot be subjected to significant moments for example columnsthat provide support to very stiff beams or beams on bearingsmay be considered adequate if N 040fcuAc 067Asc fy

Columns supporting symmetrical arrangements of beamsthat are designed for uniformly distributed imposed load andhave spans that do not differ by more than 15 of the longermay be considered adequate if N 035fcuAc 060Asc fy

BS 5400 contains general formulae for rectangular sectionsin the form of a trial-and-error procedure and two simplifiedformulae for specific applications details of which are given inTable 332

Column sizes Columns in unbraced structures are likely tobe rectangular in cross section due to the dominant effect ofbending moments in the plane of the structure Columns in

M (M2x M2

y)

Concrete Column Reinforcement percentageclass size

1 2 3 4

C2530 300 300 1370 1660 1950 2240350 350 1860 2260 2650 3050400 400 2430 2950 3470 3980450 450 3080 3730 4390 5040500 500 3800 4610 5420 6230

C3240 300 300 1720 2010 2300 2580350 350 2350 2740 3130 3520400 400 3070 3580 4090 4600450 450 3880 4530 5170 5820500 500 4790 5590 6390 7190

C4050 300 300 2080 2360 2650 2930350 350 2830 3220 3600 3990400 400 3700 4200 4710 5210450 450 4680 5320 5960 6600500 500 5780 6570 7360 8150

Ultimate design loads (kN) for short braced columns

In the foregoing table the loads were derived from the BS 8110equation for columns that are not subjected to significantmoments with fy 500 Nmm2 In determining the columnloads the ultimate load from the floor directly above the levelbeing considered should be multiplied by the following factorsto compensate for the effects of bending internal column 125edge column 15 corner column 20 The total imposed loadsmay be reduced according to the number of floors supportedThe reductions for 2 3 4 5ndash10 and over 10 floors are 1020 30 40 and 50 respectively

53 RESISTANCE TO SHEAR

Much research by many investigators has been undertaken in aneffort to develop a better understanding of the behaviour ofreinforced concrete subjected to shear As a result of thisresearch various theories have been proposed to explain themechanism of shear transfer in cracked sections and provide asatisfactory basis for designing shear reinforcement In theevent of overloading sudden failure can occur at the onset ofshear cracking in members without shear reinforcement As aconsequence a minimum amount of shear reinforcement in theform of links is required in nearly all beams Resistance to shearcan be increased by adding more shear reinforcement but even-tually the resistance is limited by the capacity of the inclinedstruts that form within the web of the section

531 Members without shear reinforcement

In an uncracked section shear results in a system of mutuallyorthogonal diagonal tension and compression stresses Whenthe diagonal tension stress reaches the tensile strength of the

Section Forces

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concrete a diagonal crack occurs This simple concept rarelyapplies to reinforced concrete since members such as beamsand slabs are generally cracked in flexure In current practiceit is more useful to refer to the nominal shear stress v Vbdwhere b is the breadth of the section in the tension zone Thisstress can then be related to empirical limiting values derivedfrom test data The limiting value vc depends on the concretestrength the effective depth and the reinforcement percentageat the section considered To be effective this reinforcementshould continue beyond the section for a specified minimumdistance as given in Codes of Practice For values of v vc noshear reinforcement is required in slabs but for most beams aspecified minimum amount in the form of links is required

At sections close to supports the shear strength is enhancedand for members carrying generally uniform load the criticalsection may be taken at d from the face of the support Whereconcentrated loads are applied close to supports in memberssuch as corbels and pile-caps some of the load is transmittedby direct strut action This effect is taken into account in theCodes of Practice by either enhancing the shear strength of thesection or reducing the design load In members subjected tobending and axial load the shear strength is increased due tocompression and reduced due to tension

Details of design procedures in Codes of Practice are givenin Table 333 for BS 8110 Table 336 for BS 5400 andTable 417 for EC 2

532 Members with shear reinforcement

The design of members with shear reinforcement is based on atruss model in which the tension and compression chords arespaced apart by a system of inclined concrete struts and uprightor inclined shear reinforcement Most reinforcement is in theform of upright links but bent-up bars may be used for upto 50 of the total shear reinforcement in beams The trussmodel results in a force in the tension chord additional to thatdue to bending This can be taken into account directly in thedesign of the tension reinforcement or indirectly by shiftingthe bending moment curve each side of any point of maximumbending moment

In BS 8110 shear reinforcement is required to cater for thedifference between the shear force and the shear resistance ofthe section without shear reinforcement Equations are givenfor upright links based on concrete struts inclined at about 45oand for bent-up bars where the inclination of the concrete strutsmay be varied between specified limits In BS 5400 a specifiedminimum amount of link reinforcement is required in additionto that needed to cater for the difference between the shear forceand the shear resistance of the section without shear reinforce-ment The forces in the inclined concrete struts are restrictedindirectly by limiting the maximum value of the nominal shearstress to specified values

In EC 2 shear reinforcement is required to cater for the entireshear force and the strength of the inclined concrete struts ischecked explicitly The inclination of the struts may be variedbetween specified limits for links as well as bent-up bars Incases where upright links are combined with bent-up bars thestrut inclination needs to be the same for both

Details of design procedures in Codes of Practice are givenin Table 333 for BS 8110 Table 336 for BS 5400 andTable 418 for EC 2

533 Shear under concentrated loads

Suspended slabs and foundations are often subjected to largeloads or reactions acting on small areas Shear in solid slabsunder concentrated loads can result in punching failures on theinclined faces of truncated cones or pyramids For designpurposes shear stresses are checked on given perimeters atspecified distances from the edges of the loaded area Where aload or reaction is eccentric with regard to a shear perimeter(eg at the edges of a slab and in cases of moment transferbetween a slab and a column) an allowance is made for theeffect of the eccentricity In cases where v exceeds vc linksbent-up bars or other proprietary products may be provided inslabs not less than 200 mm deep

Details of design procedures in Codes of Practice are givenin Table 334 for BS 8110 Tables 337 and 338 for BS 5400and Table 419 for EC 2

54 RESISTANCE TO TORSION

In normal beam-and-slab or framed construction calculationsfor torsion are not usually necessary adequate control of anytorsional cracking in beams being provided by the requiredminimum shear reinforcement When it is judged necessary toinclude torsional stiffness in the analysis of a structure ortorsional resistance is vital for static equilibrium membersshould be designed for the resulting torsional moment Thetorsional resistance of a section may be calculated on the basisof a thin-walled closed section in which equilibrium is satisfiedby a closed plastic shear flow Solid sections may be modelled asequivalent thin-walled sections Complex shapes may be dividedinto a series of sub-sections each of which is modelled as anequivalent thin-walled section and the total torsional resistancetaken as the sum of the resistances of the individual elementsWhen torsion reinforcement is required this should consist ofrectangular closed links together with longitudinal reinforce-ment Such reinforcement is additional to any requirements forshear and bending

Details of design procedures in Codes of Practice are givenin Table 335 for BS 8110 Table 339 for BS 5400 andTable 420 for EC 2

55 DEFLECTION

The deflections of members under service loading should notimpair the appearance or function of a structure An accurateprediction of deflections at different stages of construction mayalso be necessary in bridges for example For buildings thefinal deflection of members below the support level afterallowance for any pre-camber is limited to span250 In orderto minimise any damage to non-structural elements such asfinishes cladding or partitions that part of the deflection thatoccurs after the construction stage is also limited to span500In BS 8110 this limit is taken as 20 mm for spans 10 m

The behaviour of a reinforced concrete beam under serviceloading can be divided into two basic phases before and aftercracking During the uncracked phase the member behaveselastically as a homogeneous material This phase is ended bythe load at which the first flexural crack forms The cracks resultin a gradual reduction in stiffness with increasing load duringthe cracked phase The concrete between the cracks continues

Deflection 49

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to provide some tensile resistance though less on average thanthe tensile strength of the concrete Thus the member is stifferthan the value calculated on the assumption that the concretecarries no tension This additional stiffness known as lsquotensionstiffeningrsquo is highly significant in lightly reinforced memberssuch as slabs but has only a relatively minor effect on thedeflection of heavily reinforced members These concepts areillustrated in the following figure

assumptions made in their derivation provide a useful basisfor estimating long-term deflections of members in buildingsas follows

Details of spaneffective depth ratios and explicit calculationprocedures are given in Tables 340 to 342 for BS 8110 andTables 421 and 422 for EC 2

56 CRACKING

Cracks in members under service loading should not impairthe appearance durability or water-tightness of a structure InBS 8110 for buildings the design crack width is generallylimited to 03 mm In BS 5400 for bridges the limit variesbetween 025 mm and 010 mm depending on the exposureconditions In BS 8007 for structures to retain liquids a limitof 02 mm usually applies Under liquid pressure continuouscracks that extend through the full thickness of a slab or wallare likely to result in some initial seepage but such cracks areexpected to self-heal within a few weeks If the appearance ofa liquid-retaining structure is considered aesthetically critical acrack width limit of 01 mm applies

In EC 2 for most buildings the design crack width is generallylimited to 03 mm but for internal dry surfaces a limitof 04 mm is considered sufficient For liquid-retainingstructures a classification system according to the degree ofprotection required against leakage is introduced Where asmall amount of leakage is acceptable for cracks that passthrough the full thickness of the section the crack width limitvaries according to the hydraulic gradient (ie head of liquiddivided by thickness of section) The limits are 02 mm forhydraulic gradients 5 reducing uniformly to 005 mmfor hydraulic gradients 35

In order to control cracking in the regions where tension isexpected it is necessary to ensure that the tensile capacity ofthe reinforcement at yielding is not less than the tensile force inthe concrete just before cracking Thus a minimum amount ofreinforcement is required according to the strength of thereinforcing steel and the tensile strength of the concrete atthe time when cracks may first be expected to occur Cracks dueto restrained early thermal effects in continuous walls and someslabs may occur within a few days of the concrete being placedIn other members it may be several weeks before the appliedload reaches a level at which cracking occurs

Crack widths are influenced by several factors including thecover bar size bar spacing and stress in the reinforcement Thestress may need to be reduced in order to meet the crack widthlimit Design formulae are given in Codes of Practice in whichstrain calculated on the basis of no tension in the concrete isreduced by a value that decreases with increasing amounts oftension reinforcement For cracks that are caused by appliedloading the same formulae are used in BS 8110 BS 5400 andBS 8007 For cracks that are caused by restraint to temperatureeffects and shrinkage fundamentally different formulae areincluded in BS 8007 Here it is assumed that bond slip occursat each crack and the crack width increases in direct proportionto the contraction of the concrete

Deflection actual spaneffective depth ratio

limiting spaneffective depth ratiospan250

Design of structural members50

Schematic load-deflection response

In BS 8110 for the purpose of analysis lsquotension stiffeningrsquo isrepresented by a triangular stress distribution in the concreteincreasing from zero at the neutral axis to a maximum value atthe tension face At the level of the tension reinforcement theconcrete stress is taken as 1 Nmm2 for short-term loads and055 Nmm2 for long-term loads irrespective of the strain in thetension reinforcement In EC 2 a more general approach isadopted in which the deformation of a section which could bea curvature or in the case of pure tension an extension or acombination of these is calculated first for a homogeneousuncracked section 1 and second for a cracked section ignor-ing tension in the concrete 2 The deformation of the sectionunder the design loading is then obtained as

2 (1 )1

where is a distribution coefficient that takes account of thedegree of cracking according to the nature and duration ofthe loading and the stress in the tension reinforcement underthe load causing first cracking in relation to the stress under thedesign service load

When assessing long-term deflections allowances need to bemade for the effect of concrete creep and shrinkage Creep canbe taken into account by using an effective modulus of elasticityEceff Ec(1 ) where Ec is the short-term value and is acreep coefficient Shrinkage deformations can be calculatedseparately and added to those due to loading

Generally explicit calculation of deflections is unnecessaryto satisfy code requirements and simple rules in the form oflimiting spaneffective depth ratios are provided in BS 8110 andEC 2 These are considered adequate for avoiding deflectionproblems in normal circumstances and subject to the particular

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Generally for design to BS 8110 and EC 2 there is no needto calculate crack widths explicitly and simple rules that limiteither bar size or bar spacing according to the stress in thereinforcement are provided Details of both rules and crackwidth formulae are given in Table 343 for BS 8110 and BS 5400Tables 344 and 345 for BS 8007 and Tables 423ndash425 forEC 2 Additional design aids derived from the crack widthformulae are provided in Tables 346ndash352 for BS 8007 andTables 426 and 427 for EC 2

57 REINFORCEMENT CONSIDERATIONS

Codes of Practice contain many requirements affecting thereinforcement details such as minimum and maximum areasanchorage and lap lengths bends in bars and curtailment Thereinforcement may be curtailed in relation to the bendingmoment diagram provided there is always enough anchorageto develop the necessary design force in each bar at every crosssection Particular requirements apply at the positions wherebars are curtailed and at simple supports

Bars may be set out individually in pairs or in bundles ofthree or four in contact For the safe transmission of bondforces the cover provided to the bars should be not less thanthe bar size or for a group of bars in contact the equivalentdiameter of a notional bar with the same cross-sectional area asthe group Gaps between bars (or groups of bars) should benot less than the greater of (aggregate size plus 5 mm) or thebar size (or equivalent bar diameter for a group) Details ofreinforcement limits and requirements for containing bars incompression are given in Table 353 for BS 8110 Table 359for BS 5400 and Table 428 for EC 2

571 Anchorage lengths

At both sides of any cross section the reinforcement should beprovided with an appropriate embedment length or other formof end anchorage In earlier codes it was also necessary to con-sider lsquolocal bondrsquo at sections where large changes of tensileforce occur over short lengths of reinforcement and thisrequirement remains in BS 5400

Assuming a uniform bond stress between concrete and thesurface of a bar the required anchorage length is given by

lbreq (design force in bar)(bond stress perimeter of bar) fsd (24)fbd () ( fsdfbd)(4)

where fsd is the design stress in the bar at the position fromwhich the anchorage is measured The design bond stress fbd

depends on the strength of the concrete the type of bar and inEC 2 the location of the bar within the concrete section duringconcreting For example the bond condition is classified aslsquogoodrsquo in the bottom 250 mm of any section and in the top300 mm of a section 600 mm deep In other locations thebond condition is classified as lsquopoorrsquo Also in EC 2 the basicanchorage length in tension can be multiplied by severalcoefficients that take account of factors such as the bar shapethe cover and the effect of transverse reinforcement or pressureFor bars of diameter 40 mm and bars grouped in pairs orbundles additional considerations apply Details of design

anchorage lengths in tension and compression are given inTable 355 for BS 8110 Table 359 for BS 5400 and Tables 430and 432 for EC 2

572 Lap lengths

Forces can be transferred between reinforcement by lappingwelding or joining bars with mechanical devices (couplers)Connections should be placed whenever possible away frompositions of high stress and should preferably be staggeredIn Codes of Practice the necessary lap length is obtained bymultiplying the required anchorage length by a coefficient

In BS 8110 for bars in compression the coefficient is 125For bars in tension the coefficient is 10 14 or 20 accordingto the cover the gap between adjacent laps in the same layerand the location of the bar in the section In slabs where thecover is not less than twice the bar size and the gap betweenadjacent laps is not less than six times the bar size or 75 mm afactor of 10 applies Larger factors are frequently necessary incolumns typically 14 and beams typically 14 for bottom barsand 20 for top bars The sum of all the reinforcement sizes ina particular layer should not exceed 40 of the width of thesection at that level When the size of both bars at a lap exceeds20 mm and the cover is less than 15 times the size of thesmaller bar links at a maximum spacing of 200 mm arerequired throughout the lap length

In EC 2 for bars in tension or compression the lap coefficientvaries from 10 to 15 according to the percentage of lapped barsrelative to the total area of bars at the section considered andtransverse reinforcement is required at each end of the lap zoneDetails of lap lengths are given in Table 355 for BS 8110Table 359 for BS 5400 and Tables 431 and 432 for EC 2

573 Bends in bars

The radius of any bend in a reinforcing bar should conform tothe minimum requirements of BS 8666 and should ensure thatfailure of the concrete inside the bend is prevented For barsbent to the minimum radius according to BS 8666 it is notnecessary to check for concrete failure if the anchorage of thebar does not require a length more than 5 beyond the end ofthe bend (see Table 227) It is also not necessary to check forconcrete failure where the plane of the bend is not close to aconcrete face and there is a transverse bar not less than its ownsize inside the bend This applies in particular to a link whichmay be considered fully anchored if it passes round anotherbar not less than its own size through an angle of 90o andcontinues beyond the end of the bend for a length not less than8 in BS 8110 and 10 in EC 2

In cases when a bend occurs at a position where the bar ishighly stressed the bearing stress inside the bend needs to bechecked and the radius of bend will need to be more than theminimum given in BS 8666 This situation occurs typically atmonolithic connections between members for example junc-tion of beam and end column and in short members such ascorbels and pile caps The design bearing stress is limitedaccording to the concrete strength and the confinementperpendicular to the plane of the bend Details of bends in barsare given in Table 355 for BS 8110 Table 359 for BS 5400and Table 431 for EC 2

Reinforcement considerations 51

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574 Curtailment of reinforcement

In flexural members it is generally advisable to stagger thecurtailment points of the tension reinforcement as allowed bythe bending moment envelope Bars to be curtailed need toextend beyond the points where in theory they are no longerneeded for flexural resistance for a number of reasons butmainly to ensure that the shear resistance of the section is notreduced locally Clearly of course no reinforcement shouldbe curtailed at a point less than a full anchorage length from asection where it is required to be fully stressed

In BS 8110 and BS 5400 except at end supports every barshould extend beyond the point at which in theory it is no longerrequired for a distance not less than the greater of the effectivedepth of the member or 12 times the bar size In addition barscurtailed in a tension zone should satisfy at least one of threealternative conditions one requires a full anchorage length onerequires the designer to determine the position where the shearresistance is twice the shear force and the other requires thedesigner to determine the position where the bending resistanceis twice the bending moment The simplest approach is to complywith the first option by providing a full anchorage lengthbeyond the point where in theory the bar is no longer requiredeven if this requires a longer extension than is absolutelynecessary in some cases Details of the requirements are givenin Table 356

In BS 8110 simplified rules are also given for beams andslabs where the loads are mainly uniformly distributed and inthe case of continuous members the spans are approximatelyequal Details of the rules are given in Tables 357 and 358

At simple end supports the tension bars should extend foran effective anchorage length of 12 times the bar size beyondthe centre of the support but no bend should begin before thecentre of the support In cases where the width of the supportexceeds the effective depth of the member the centre ofthe support may be assumed at half the effective depth from theface of the support In BS 8110 for slabs in cases where thedesign shear force is less than half the shear resistance anchor-age can be obtained by extending the bars beyond the centre ofthe support for a distance equal to one third of the supportwidth 30 mm

In EC 2 the extension al of a tension bar beyond the pointwhere in theory it is no longer required for flexural resistance isdirectly related to the shear force at the section For memberswith upright shear links al 05zcot13 where z is the lever armand 13 is the inclination of the concrete struts (see section 3512)Taking z 09d al 045dcot13 where cot13 is selected by thedesigner in the range 10 cot13 25 If the value of cot 13 usedin the shear design calculations is unknown al 1125d can beassumed For members with no shear reinforcement al d isused At simple end supports bottom bars should extend for ananchorage length beyond the face of the support The tensileforce to be anchored is given by F 05Vcot 13 and F 125Vcan be conservatively taken in all cases Details of the curtailmentrequirements are given in Table 432

58 DEEP BEAMS

In designing normal (shallow) beams of the proportions morecommonly used in construction plane sections are assumed toremain plane after loading This assumption is not strictly true

but the errors resulting from it only become significant whenthe depth of the beam becomes equal to or more than abouthalf the span The beam is then classed as a deep beam anddifferent methods of analysis and design need to be usedThese methods take into account not only the overall appliedmoments and shears but also the stress patterns and internaldeformations within the beam

For a single-span deep beam after the concrete in tensionhas cracked the structural behaviour is similar to a tied archThe centre of the compression force in the arch rises from thesupport to a height at the crown equal to about half the span ofthe beam The tension force in the tie is roughly constant alongits length since the bending moment and the lever arm undergosimilar variations along the length of the beam For a continuousdeep beam the structural behaviour is analogous to a separatetied arch system for each span combined with a suspensionsystem centred over each internal support

In BS 8110 for the design of beams of clear span less thantwice the effective depth the designer is referred to specialistliterature In EC 2 a deep beam is classified as a beam whoseeffective span is less than three times its overall depth Briefdetails of suitable methods of design based on the result ofextensive experimental work by various investigators are givenin ref 42 and a comprehensive well-produced design guide iscontained in ref 43

59 WALLS

Information concerning the design of load-bearing walls inaccordance with BS 8110 is given in section 618 Retainingwalls and other similar elements that are subjected mainly totransverse bending where the design vertical load is less than01fcu times the area of the cross section are treated as slabs

510 DETAILS

It has long been realised that the calculated strength of areinforced concrete member cannot be attained if the details ofthe required reinforcement are unsatisfactory Research by theformer Cement and Concrete Association and others has shownthat this applies particularly at joints and intersections Thedetails commonly used in wall-to-base and wall-to-walljunctions in retaining structures and containment vessels wherethe action of the applied load is to lsquoopenrsquo the corner are notalways effective

On Tables 362 and 363 are shown recommended details thathave emerged from the results of reported research The designinformation given in BS 8110 and BS 5400 for nibs corbels andhalving joints is included and supplemented by informationgiven elsewhere In general however details that are primarilyintended for precast concrete construction have not beenincluded as they fall outside the scope of this book

511 ELASTIC ANALYSIS OF CONCRETE SECTIONS

The geometrical properties of various figures the shapes ofwhich conform to the cross sections of common reinforcedconcrete members are given in Table 2101 The data includeexpressions for the area section modulus second moment ofarea and radius of gyration The values that are derived fromthese expressions are applicable in cases when the amount of

Design of structural members52

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reinforcement provided need not be taken into account in theanalysis of the structure (see section 141)

The data given in Tables 2102 and 2103 are applicable toreinforced concrete members with rectilinear and polygonalcross sections when the reinforcement provided is taken intoaccount on the basis of the modular ratio Two conditions areconsidered (1) when the entire section is subjected to stressand (2) when for members subjected to bending the concretein tension is not taken into account The data given for the

former condition are the effective area the centre and secondmoment of area the modulus and radius of gyration For thecondition when a member is subjected to bending and theconcrete in tension is assumed to be ineffective data giveninclude the position of the neutral axis the lever-arm and theresistance moment

Design procedures for sections subjected to bending andaxial force with design charts for rectangular and cylindricalcolumns are given in Tables 2104ndash2109

Elastic analysis of concrete sections 53

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The loads and consequent bending moments and forces onthe principal types of structural components and the designresistances of such components have been dealt with in thepreceding chapters In this chapter some complete structurescomprising assemblies or special cases of such componentsand their foundations are considered

61 BUILDINGS

Buildings may be constructed entirely of reinforced concreteor one or more elements of the roof floors walls stairs andfoundations may be of reinforced concrete in conjunction witha steel frame Alternatively the building may consist of interiorand exterior walls of cast in situ reinforced concrete supportingthe floors and roof with the columns and beams being formedin the thickness of the walls Again the entire structure or partsthereof may be built of precast concrete elements connectedtogether during construction

The design of the various parts of a building is the subjectof Examples of the Design of Buildings That book includesillustrative calculations and drawings for a typical six-storeymultipurpose building This section provides a brief guide tocomponent design

611 Robustness and provision of ties

The progressive collapse of one corner of a London tower blockin 1968 as a result of an explosion caused by a gas leak in adomestic appliance on the eighteenth floor led to recommen-dations to consider such accidental actions in the design of allbuildings Regulations require a building to be designed andconstructed so that in the event of an accident the buildingwill not collapse to an extent disproportionate to the causeBuildings are divided into classes depending on the type andoccupancy including the likelihood of accidents and thenumber of occupants that may be affected with a statementof the design measures to be taken in each of the classes TheBS 8110 normal requirements for lsquorobustnessrsquo automaticallysatisfy the regulations for all buildings except those wherespecific account is to be taken of likely hazards

The layout and form of the structure should be checked toensure that it is inherently stable and robust In some cases itmay be necessary to protect certain elements from vehicularimpact by providing bollards or earth banks All structures

should be able to resist a notional ultimate horizontal forceequal to 15 of the characteristic dead load of the structureThis force effectively replaces the design wind load in caseswhere the exposed surface area of the building is small

Wherever possible continuous horizontal and vertical tiesshould be provided throughout the building to resist specifiedforces The magnitude of the force increases with the numberof storeys for buildings of less than 10 storeys but remainsconstant thereafter The requirements may be met by usingreinforcement that is necessary for normal design purposes inbeams slabs columns and walls Only the tying forces needto be considered and the full characteristic strength of thereinforcement may be taken into account Horizontal ties arerequired in floors and roofs at the periphery and internally intwo perpendicular directions The internal ties which may bespread uniformly over the entire building or concentrated atbeam and column positions are to be properly anchored atthe peripheral tie Vertical ties are required in all columns andload-bearing walls from top to bottom and all external columnsand walls are to be tied into each floor and roof For regulatorypurposes some buildings are exempt from the vertical tyingrequirement Details of the tying requirements are given inTable 354

For in situ construction proper attention to reinforcementdetailing is all that is normally necessary to meet the tyingrequirements Precast forms of construction generally requiremore care and recommended details to obtain continuity ofhorizontal ties are given in the code of practice If ties cannotbe provided other strategies should be adopted as described inPart 2 of the code These strategies are presented in the contextof residential buildings of five or more storeys where eachelement that cannot be tied is to be considered as notionallyremoved one at a time in each storey in turn The requirementis that any resulting collapse should be limited in extent withthe remaining structure being able to bridge the gap causedby the removal of the element If this requirement cannot besatisfied then the element in question is considered as a keyelement In this case the element and its connections need tobe able to resist a design ultimate load of 34 kNm2 consideredto act from any direction BS 8110 is vague with regard to theextent of collapse associated with this approach but a moreclearly defined statement is given in the building regulationsHere a key element is any untied member whose removalwould put at risk of collapse within the storey in question

Chapter 6

Buildings bridges andcontainmentstructures

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and the immediately adjacent storeys more than 15 of thearea of the storey (or 70 m2 if less)

In EC 2 similar principles apply in that structures notspecifically designed to withstand accidental actions should beprovided with a suitable tying system to prevent progressivecollapse by providing alternative load paths after local damageThe UK National Annex specifies compliance with the BS 8110requirements as given in Table 429

612 Floors

Suspended concrete floors can be of monolithic constructionin the form of beam-and-slab (solid or ribbed) or flat slab(solid or waffle) or can consist of precast concrete slab unitssupported on concrete or steel beams or comprise one ofseveral other hybrid forms Examples of monolithic forms ofconstruction are shown in the figure on Table 242

Two-way beam and solid slab systems can involve a layoutof long span secondary beams supported by usually shorterspan main beams The resulting slab panels may be designed astwo-way spanning if the longer side is less than twice theshorter side However such two-way beam systems tend tocomplicate both formwork and reinforcement details with aconsequent delay in the construction programme A one-waybeam and solid slab system is best suited to a rectangular gridof columns with long span beams and shorter span slabs If aribbed slab is used a system of long span slabs supported byshorter span beams is preferable If wide beams are used thebeam can be incorporated within the depth of the ribbed slab

In BS 8110 ribbed slabs include construction in which ribsare cast in situ between rows of blocks that remain part of thecompleted floor This type of construction is no longer used inthe United Kingdom although blocks are incorporated in someprecast and composite construction The formers for ribbedslabs can be of steel glassfibre or polypropylene Standardmoulds are available that provide tapered ribs with a minimumwidth of 125 mm spaced at 600 mm (troughs) and 900 mm(waffles) The ribs are connected by a structural concrete toppingwith a minimum thickness of 50 mm for trough moulds and75 mm for waffle moulds In most structures to obtain thenecessary fire-resistance either the thickness of topping has toexceed these minimum values or a non-structural screed addedat a later stage of construction The spacing of the ribs may beincreased to a maximum of 1500 mm by using purpose-madeformers Comprehensive details of trough and waffle floorsare contained in ref 44

BS 8110 and EC 2 contain recommendations for both solidand ribbed slabs spanning between beams or supported directlyby columns (flat slabs) Ribs in waffle slabs and ribs reinforcedwith a single bar in trough slabs do not require links unlessneeded for shear or fire-resistance Ribs in trough slabs whichare reinforced with more than one bar should be provided withsome links to help maintain the correct cover The spacing ofthese links may be in the range 10ndash15 m according to the sizeof the main bars Structural toppings are normally reinforcedwith a welded steel fabric

Information on the weight of concrete floor slabs is given inTable 21 and details of imposed loads on floors are givenin Table 23 Detailed guidance on the analysis of slabs isgiven in Chapters 4 and 13 More general guidance includinginitial sizing suggestions is given in section 523

613 Openings in floors

Large openings (eg stairwells) should generally be providedwith beams around the opening Holes for pipes ducts andother services should generally be formed when the slabis constructed and the cutting of such holes should not bepermitted afterwards unless done under the supervision of acompetent engineer Small isolated holes may generally beignored structurally with the reinforcement needed for a slabwithout holes simply displaced locally to avoid the hole

In other cases the area of slab around an opening or groupof closely spaced holes needs to be strengthened with extrareinforcement The cross-sectional area of additional bars to beplaced parallel to the principal reinforcement should be at leastequal to the area of principal reinforcement interrupted by theopening Also for openings of dimensions exceeding 500 mmadditional bars should be placed diagonally across the cornersof the opening Openings with dimensions greater than 1000 mmshould be regarded as structurally significant and the area ofslab around the opening designed accordingly

The effect of an opening in the proximity of a concentratedload or supporting column on the shearing resistance of theslab is shown in Table 337

614 Stairs

Structural stairs may be tucked away out of sight within a fireenclosure or they may form a principal architectural feature Inthe former case the stairs can be designed and constructed assimply and cheaply as possible but in the latter case much moretime and trouble is likely to be expended on the design

Several stair types are illustrated on Table 288 Variousprocedures for analysing the more common types of stairhave been developed and some of these are described onTables 288ndash291 These theoretical procedures are based onthe concept of an idealised line structure and when detailingthe reinforcement for the resulting stairs additional bars shouldbe included to limit the formation of cracks at the points ofhigh stress concentration that inevitably occur The lsquothree-dimensionalrsquo nature of the actual structure and the stiffeningeffect of the triangular tread areas both of which are usuallyignored when analysing the structure will result in actual stressdistributions that differ from those calculated and this mustbe remembered when detailing The stair types illustrated onTable 288 and others can also be investigated by finite-elementmethods and similar procedures suitable for computer analysisWith such methods it is often possible to take account of thethree-dimensional nature of the stair

Simple straight flights of stairs can span either transversely(ie across the flight) or longitudinally (ie along the flight)When spanning transversely supports must be provided on bothsides of the flight by either walls or stringer beams In this casethe waist or thinnest part of the stair construction need be nomore than 60 mm thick say the effective lever arm for resistingthe bending moment being about half of the maximumthickness from the nose to the soffit measured at right anglesto the soffit When the stair spans longitudinally deflectionconsiderations can determine the waist thickness

In principle the design requirements for beams and slabsapply also to staircases but designers cannot be expected todetermine the deflections likely to occur in the more complex

Buildings 55

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stair types BS 8110 deals only with simple types and allows amodified spaneffective depth ratio to be used The bendingmoments should be calculated from the ultimate load due to thetotal weight of the stairs and imposed load measured on plancombined with the horizontal span Stresses produced bythe longitudinal thrust are small and generally neglected in thedesign of simple systems Unless circumstances otherwisedictate suitable step dimensions for a semi-public stairs are 165mm rise and 275 mm going which with a 25 mm nosing orundercut gives a tread of 300 mm Private stairs may be steeperand those in public buildings should be less steep In eachcase optimum proportions are given by the relationship(2 rise going) 600 mm Different forms of constructionand further details on stair dimensions are given in BS 5395

Finally it should be remembered that the prime purpose of astair is to provide safe pedestrian access between the floors itconnects As such it is of vital importance in the event of a fireand a principal design consideration must be to provide adequatefire-resistance

615 Planar roofs

The design and construction of a flat reinforced concrete roofare essentially the same as for a floor A water-tight coveringsuch as asphalt or bituminous felt is generally necessary andwith a solid slab some form of thermal insulation is normallyrequired For ordinary buildings the slab is generally built leveland a drainage slope of the order of 1 in 120 is formed byadding a mortar topping The topping is laid directly onto theconcrete and below the water-tight covering and can formthe thermal insulation if it is made of a sufficient thickness oflightweight concrete or other material having low thermalconductivity

Planar slabs with a continuous steep slope are not commonin reinforced concrete except for mansard roofs The roofcovering is generally of metal or asbestos-cement sheeting orsome lightweight material Such coverings and roof glazingrequire purlins for their support and although these are often ofsteel precast concrete purlins are also used especially if theroof structure is of reinforced concrete

616 Non-planar roofs

Roofs that are not planar other than the simple pitched roofsconsidered in the foregoing can be constructed as a series ofplanar slabs (prismatic or hipped-plate construction) or assingle- or double-curved shells Single-curved roofs such assegmental or cylindrical shells are classified as developablesurfaces Such surfaces are not as stiff as double-curved roofsor their prismatic counterparts which cannot be lsquoopened uprsquointo plates without some shrinking or stretching taking place

If the curvature of a double-curved shell is similar in alldirections the surface is known as synclastic A typical case isa dome where the curvature is identical in all directions Ifthe shell curves in opposite directions over certain areas thesurface is termed anticlastic (saddle shaped) The hyperbolic-paraboloidal shell is a well-known example and is the specialcase where such a double-curved surface is generated by twosets of straight lines An elementary analysis of some of thesestructural forms is dealt with in section 192 and Table 292but reference should be made to specialist publications for

more comprehensive analyses and more complex structuresSolutions for many particular shell types have been producedand in addition general methods have been developed foranalysing shell forms of any shape by means of a computerShells like all statically indeterminate structures are affectedby such secondary effects as shrinkage temperature change andsettlement and a designer must always bear in mind the factthat the stresses arising from these effects can modify quiteconsiderably those due to normal dead and imposed load InTable 281 simple expressions are given for the forces indomed slabs such as are used for the bottoms and roofs of somecylindrical tanks In a building a domed roof generally hasa much larger rise to span ratio and where the dome is partof a spherical surface and has an approximately uniform thick-ness overall the analysis given in Table 292 applies Shallowsegmental domes and truncated cones are also dealt with inTable 292

Cylindrical shells Segmental or cylindrical roofs are usuallydesigned as shell structures Thin curved slabs that behave asshells are assumed to offer no resistance to bending nor todeform under applied distributed loads Except near edge andend stiffeners the shell is subjected only to membrane forcesnamely a direct force acting longitudinally in the plane of theslab a direct force acting tangentially to the curve of the slaband a shearing force Formulae for these membrane forces aregiven in section 1923 In practice the boundary conditionsdue to either the presence or absence of edge or valley beamsend diaphragms continuity and so on affect the displacementsand forces that would otherwise occur as a result of membraneaction Thus as when analysing any indeterminate structure(such as a continuous beam system) the effects due to theseboundary restraints need to be combined with the staticallydeterminate stresses arising from the membrane action

Shell roofs can be arbitrarily subdivided into lsquoshortrsquo (wherethe ratio of length l to radius r is less than about 05) lsquolongrsquo(where l r exceeds 25) and lsquointermediatersquo For short shellsthe influence of the edge forces is slight in comparison withmembrane action and the stresses can be reasonably taken asthose due to the latter only If the shell is long the membraneaction is relatively insignificant and an approximate solutioncan be obtained by considering the shell to act as a beam withcurved flanges as described in section 1923

For the initial analysis of intermediate shells no equivalentshort-cut method has yet been devised The standard method ofsolution is described in various textbooks (eg refs 45 and 46)Such methods involve the solution of eight simultaneousequations if the shell or the loading is unsymmetrical or four ifsymmetry is present by matrix inversion or other means Bymaking certain simplifying assumptions and providing tables ofcoefficients Tottenham (ref 47) developed a popular simplifieddesign method which is rapid and requires the solution of threesimultaneous equations only J D Bennett also developed amethod of designing long and intermediate shells based on ananalysis of actual designs of more than 250 roofs The methodwhich involves the use of simple formulae incorporatingempirical coefficients is summarised on Tables 293 and 294For further details see ref 48

Buckling of shells A major concern in the design of anyshell is the possibility of buckling since the loads at which

Buildings bridges and containment structures56

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buckling occurs as established by tests often differ from thevalues predicted by theory Ref 49 indicates that for domessubtending angles of about 90o the critical external pressure atwhich buckling occurs according to both theory and tests isgiven by p 03E(hr)2 where E is the elastic modulus ofconcrete and h is the thickness and r the radius of the domeFor a shallow dome with spanrise cong 10 p 015E(hr)2 Afactor of safety against buckling of 2 to 3 should be adoptedSynclastic shells having a radius ranging from r1 to r2 may beconsidered as an equivalent dome with a radius of r radic(r1 r2)

For a cylindrical shell buckling is unlikely if the shell isshort In the case of long shells p 06E(hr)2

Anticlastic surfaces are more rigid than single-curved shellsand the buckling pressure for a saddle-shaped shell supportedon edge stiffeners safely exceeds that of a cylinder having acurvature equal to that of the anticlastic shell at the stiffenerFor a hyperbolic-paraboloidal shell with straight boundariesthe buckling load obtained from tests is slightly more than thevalue given by n E(ch)22ab where a and b are the lengthsof the sides of the shell c is the rise and h the thickness this isonly half of the value predicted theoretically

617 Curved beams

When bow girders and beams that are not rectilinear in planare subjected to vertical loading torsional moments occur inaddition to the normal bending moments and shearing forcesBeams forming a circular arc in plan may comprise part of acomplete circular system with equally spaced supports andequal loads on each span such systems occur in silos towersand similar cylindrical structures Equivalent conditions canalso occur in beams where the circle is incomplete provided theappropriate negative bending and torsional moments can bedeveloped at the end supports This type of circular beam canoccur in structures such as balconies

On Tables 295ndash297 charts are given that enable a rapidevaluation of the bending moments torsional moments andshear forces occurring in curved beams due to uniform andconcentrated loads The formulae on which the charts arebased are given in section 193 and on the tables concernedThe expressions have been developed from those in ref 50 foruniform loads and ref 51 for concentrated loads In both casesthe results have been recalculated to take into account values ofG 04E and C J2

618 Load-bearing walls

In building codes for design purposes a wall is defined as avertical load-bearing member whose length on plan exceedsfour times its thickness Otherwise the member is treated as acolumn in which case the effects of slenderness in relation toboth major and minor axes of bending need to be considered(section 524) A reinforced wall is one in which not lessthan the recommended minimum amount of reinforcement isprovided and taken into account in the design Otherwise themember is treated as a plain concrete wall in which case thereinforcement is ignored for design purposes

A single planar wall in general can be subjected to verticaland horizontal in-plane forces acting together with in-planeand transverse moments The in-plane forces and moment canbe combined to obtain at any particular level a longitudinal

shear force and a linear distribution of vertical force If thein-plane eccentricity of the vertical force exceeds one-sixth ofthe length of the wall reinforcement can be provided to resistthe tension that develops at one end of the wall In a plain wallsince the tensile strength of the concrete is ignored the distrib-ution of vertical load is similar to that for the bearing pressuredue to an eccentric load on a footing Flanged walls and coreshapes can be treated in a similar way to obtain the resultingdistribution of vertical force Any unit length of the wall cannow be designed as a column subjected to vertical loadcombined with bending about the minor axis due to anytransverse moment

In BS 8110 the effective height of a wall in relation to itsthickness depends upon the effect of any lateral supports andwhether the wall is braced or unbraced A braced wall is onethat is supported laterally by floors andor other walls able totransmit lateral forces from the wall to the principal structuralbracing or to the foundations The principal structural bracingcomprise strong points shear walls or other suitable elementsgiving lateral stability to a structure as a whole An unbracedwall provides its own lateral stability and the overall stabilityof multi-storey buildings should not in any direction dependon such walls alone The slenderness ratio of a wall is definedas the effective height divided by the thickness and the wall isconsidered lsquostockyrsquo if the slenderness ratio does not exceed15 for a braced wall or 10 for an unbraced wall Otherwise awall is considered slender in which case it must be designed foran additional transverse moment

The design of plain concrete walls in BS 8110 is similar tothat of unreinforced masonry walls in BS 5328 Equations aregiven for the maximum design ultimate axial load taking intoaccount the transverse eccentricity of the load including anadditional eccentricity in the case of slender walls The basicrequirements for the design of reinforced and plain concretewalls are summarised in Table 360

62 BRIDGES

As stated in section 248 the analysis and design of bridges isnow so complex that it cannot be adequately covered in a bookof this type and reference should be made to specialist publi-cations However for the guidance of designers who may haveto deal with structures having features in common with bridgesbrief notes on some aspects of their design and constructionare provided Most of the following information is taken fromref 52 which also contains other references for further reading

621 Types of bridges

For short spans the simplest and most cost-effective form ofdeck construction is a cast in situ reinforced concrete solid slabSingle span slabs are often connected monolithically to theabutments to form a portal frame A precast box-shaped rein-forced concrete culvert can be used as a simple form of framedbridge and is particularly economical for short span (up toabout 6 m) bridges that have to be built on relatively poorground obviating the need for piled foundations

As the span increases the high self-weight of a solid slabbecomes a major disadvantage The weight can be reduced byproviding voids within the slab using polystyrene formersThese are usually of circular section enabling the concrete to

Bridges 57

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flow freely under them to the deck soffit Reinforced concretevoided slabs are economical for spans up to about 25 m Theintroduction of prestressing enables such construction to beeconomical over longer spans and prestressed voided slabswith internal bonded tendons can be used for spans up toabout 50 m If a bridge location does not suit cast in situ slabconstruction precast concrete beams can be used Severaldifferent types of high quality factory-made components thatcan be rapidly erected on site are manufactured Precast beamconstruction is particularly useful for bridging over live roadsrailways and waterways where any interruptions to trafficmust be minimised Pre-tensioned inverted T-beams placedside-by-side and then infilled with concrete provide a viablealternative to a reinforced concrete solid slab for spans upto about 18 m Composite forms of construction consistingtypically of a 200 mm thick cast in situ slab supported onpre-tensioned beams spaced at about 15 m centres can be usedfor spans in the range 12ndash40 m

For very long spans prestressed concrete box girders are theusual form for bridge decks ndash the details of the design beingdictated by the method of construction The span-by-spanmethod is used in multi-span viaducts with individual spans ofup to 60 m A span plus a cantilever of about one quarter thenext span is first constructed This is then prestressed and thefalsework moved forward after which a full span length isformed and stressed back to the previous cantilever In situ con-struction is used for smaller spans but as spans increase so alsodoes the cost of the falsework To minimise the cost the weightof the concrete to be supported at any one time is reduced bydividing each span into a series of transverse segments Thesesegments which can be cast in situ or precast are normallyerected on either side of each pier to form balanced cantileversand then stressed together Further segments are then addedextending the cantilevers to mid-span where an in situ concreteclosure is formed to make the spans continuous During erectionthe leading segments are supported from gantries erected on thepiers or completed parts of the deck and work can advancesimultaneously on several fronts When the segments are precasteach unit is match-cast against the previous one and thenseparated for transportation and erection Finally an epoxyresin is applied to the matching faces before the units arestressed together

Straight or curved bridges of single radius and of constantcross section can also be built in short lengths from one orboth ends The bridge is then pushed out in stages from theabutments a system known as incremental launching Archbridges in spans up to 250 m and beyond can be constructedeither in situ or using precast segments which are prestressedtogether and held on stays until the whole arch is complete

For spans in excess of 250 m the decks of suspensionand cable-stayed bridges can be of in situ concrete ndash constructedusing travelling formwork ndash or of precast segments stressedtogether For a comprehensive treatment of the aestheticsand design of bridges by one of the worldrsquos most eminentbridge engineers see ref 53 Brief information on typicalstructural forms and span ranges is given in Table 298

622 Substructures

A bridge is supported at the ends on abutments and may haveintermediate piers where the positions of the supports and the

lengths of the spans are determined by the topography of theground and the need to ensure unimpeded traffic under thebridge The overall appearance of the bridge structure is verydependent on the relative proportions of the deck and itssupports The abutments are usually constructed of reinforcedconcrete but in some circumstances mass concrete withoutreinforcement can provide a simple and durable solution

Contiguous bored piles or diaphragm walling can be used toform an abutment wall in cases where the wall is to be formedbefore the main excavation is carried out Although the cost ofthis type of construction is high it can be offset against savingsin the amount of land required the cost of temporary works andconstruction time A facing of in situ or precast concrete orblockwork will normally be required after excavation Reinforcedearth construction can be used where there is an embankmentbehind the abutment in which case a precast facing is oftenapplied The selection of appropriate ties and fittings is partic-ularly important since replacement of the ties during the life ofthe structure is very difficult

Where a bridge is constructed over a cutting it is usuallypossible to form a bank-seat abutment on firm undisturbedground Alternatively bank seats can be constructed on piledfoundations However where bridges over motorways aredesigned to allow for future widening of the carriageway theabutment is likely to be taken down to full depth so that it canbe exposed at a later date when the widening is carried out

The design of wing walls is determined by the topography ofthe site and can have a major effect on the appearance of thebridge Wing walls are often taken back at an angle from theface of the abutment for both economy and appearance Castin situ concrete is normally used but precast concrete retainingwall units are also available from manufacturers Concrete cribwalling is also used and its appearance makes it particularlysuitable for rural situations Filling material must be carefullyselected to ensure that it does not flow out and the fill mustbe properly drained It is important to limit the differentialsettlement that could occur between an abutment and its wingwalls The problem can be avoided if the wing walls cantileverfrom the abutment and the whole structure is supported onone foundation

The simplest and most economic form of pier is a verticalmember or group of members of uniform cross section Thismight be square rectangular circular or elliptical Shaping ofpiers can be aesthetically beneficial but complex shapes willsignificantly increase the cost unless considerable reuse of theforms is possible Raking piers and abutments can help toreduce spans for high bridges but they also require expensivepropping and support structures This in turn complicates theconstruction process and considerably increases costs

The choice of foundation to abutments and piers is usuallybetween spread footings and piling Where ground conditionspermit a spread footing will provide a simple and economicsolution Piling will be needed where the ground conditionsare poor and cannot be improved the bridge is over a river orestuary the water table is high or site restrictions prevent theconstruction of a spread footing It is sometimes possible toimprove the ground by consolidating grouting or applying asurcharge by constructing the embankments well in advance ofthe bridge structure Differential settlement of foundations willbe affected by the construction sequence and needs to becontrolled In the early stages of construction the abutments

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are likely to settle more than the piers but the piers will settlelater when the deck is constructed

623 Integral bridges

For road bridges in the United Kingdom experience has shownthat with all forms of construction continuous structures aregenerally more durable than structures with discontinuous spansThis is mainly because joints between spans have often allowedsalty water to leak through to piers and abutments HighwaysAgency standard BD 5701 says that in principle all bridgesshould be designed as continuous over intermediate supportsunless special circumstances exist The connections betweenspans may be made to provide full structural continuity or inbeam and slab construction continuity of the deck slab only

Bridges with lengths up to 60 m and skews up to 30o shouldalso be designed as integral bridges in which the abutmentsare connected directly to the deck and no movement joints areprovided to allow for expansion or contraction When the designerconsiders that an integral bridge is inappropriate the agreementof the overseeing organisation must be obtained HighwaysAgency document BA 5701 has figures indicating a variety ofcontinuity and abutment details

624 Design considerations

Whether the bridge is carrying a road railway waterway or justpedestrians it will be subject to various types of load

Self-weight and loads from surfacing parapets and so on

Environmental (eg wind snow temperature effects)

Traffic

Accidental loads (eg impact)

Temporary loads (during construction and maintenance)

Bridges in the United Kingdom are generally designed to therequirements of BS 5400 and several related Highways Agencystandards Details of the traffic loads to be considered forroad railway and footbridges are given in section 248 andTables 25 and 26 Details of structural design requirementsincluding the load combinations to be considered are given insection 212 and Tables 32 and 33

The application of traffic load to any one area of a bridgedeck causes the deck to bend transversely and twist therebyspreading load to either side The assessment of how much ofthe load is shared in this way and the extent to which it isspread across the deck depends on the bending torsion andshear stiffness of the deck in the longitudinal and transversedirections Computer methods are generally used to analysethe structure for load effects the most versatile method beinggrillage analysis which treats the deck as a two-dimensionalseries of beam elements in both directions This method canbe used for solid slab beam and slab and voided slabs wherethe cross-sectional area of the voids does not exceed 60 ofthe area of the deck Box girders are now generally formed asone or two cells without any transverse diaphragms These areusually quite stiff in torsion but can distort under load givingrise to warping stresses in the walls and slabs of the box It isthen necessary to use three-dimensional analytical methodssuch as 3D space frame folded plate (for decks of uniformcross section) or the generalised 3D finite element method

An excellent treatment of the behaviour and analysis of bridgedecks is provided in ref 54

It is usual to assume that movement of abutments and wingwalls will occur and to take these into account in the designof the deck and the substructure Normally the backfill used isa free-draining material and satisfactory drainage facilities areprovided If these conditions do not apply then higher designpressures must be considered Due allowance must be madealso for the compaction of the fill during construction and thesubsequent effects of traffic loading The Highways Agencydocument BA 4296 shows several forms of integral abutmentwith guidance on their behaviour Abutments to frame bridges areconsidered to rock bodily under the effect of deck movementsEmbedded abutments such as piled and diaphragm wallsare considered to flex and pad foundations to bank seats areconsidered to slide Notional earth pressure distributionsresulting from deck expansion are also given for frame andembedded abutments

Creep shrinkage and temperature movements in bridgedecks can all affect the forces applied to the abutments Piersand to a lesser extent abutments are vulnerable to impact loadsfrom vehicles or shipping and must be designed to resistimpact or be protected from it Substructures of bridges overrivers and estuaries are also subjected to scouring and lateralforces due to water flow unless properly protected

625 Waterproofing of bridge decks

Over the years mastic asphalt has been extensively used forwaterproofing bridge decks but good weather conditions arerequired if it is to be laid satisfactorily Preformed bituminoussheeting is less sensitive to laying conditions but moisturetrapped below the sheeting can cause subsequent lifting Theuse of hot-bonded heavy-duty reinforced sheet membranes ifproperly laid can provide a completely water-tight layer Thesheets which are 3ndash4 mm in thickness have good punctureresistance and it is not necessary to protect the membrane fromasphalt laid on top Sprayed acrylic and polyurethane water-proofing membranes are also used These bond well to theconcrete deck surface with little or no risk of blowing or liftingA tack coat must be applied over the membrane and a protec-tive asphalt layer is placed before the final surfacing is carriedout Some bridges have depended upon the use of a dense highquality concrete to resist the penetration of water without anapplied waterproofing layer In such cases it can be advanta-geous to include silica fume or some similar very fine powderedaddition in the concrete

63 CONTAINMENT STRUCTURES

Weights of stored materials are given in EC 1 Part 11 and thecalculation of horizontal pressures due to liquids and granularmaterials contained in tanks reservoirs bunkers and silosis explained in sections 92 and 93 in conjunction withTables 215 and 216 This section deals with the design ofcontainment structures and the calculation of the forces andbending moments produced by the pressure of the containedmaterials Where containers are required to be watertight thestructural design should follow the recommendations givenin either BS 8007 or EC 2 Part 3 as indicated in sections 213and 294 respectively In the following notes containers are

Containment structures 59

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conveniently classified as either tanks containing liquids orbunkers and silos containing dry materials

631 Underground tanks

Underground storage tanks are subjected to external pressuresdue to the surrounding earth in addition to internal waterpressure The empty structure should also be investigated forpossible flotation if the earth can become waterlogged Earthpressure at-rest conditions should generally be assumed fordesign purposes but for reservoirs where the earth is banked upagainst the walls it would be more reasonable to assume activeconditions Storage tanks are normally filled to check for water-tightness before any backfill material is placed and there isalways a risk that such material could be excavated in the futureTherefore no reduction to the internal hydrostatic pressure byreason of the external earth pressure should be made when atank is full

The earth covering on the roof of a reservoir in its final stateacts uniformly over the entire area but it is usually sensible totreat it as an imposed load This is to cater for non-uniformconditions that can occur when the earth is being placed inposition and if it becomes necessary to remove the earth formaintenance purposes Problems can arise in partially buriedreservoirs due to solar radiation causing thermal expansion ofthe roof The effect of such movement on a perimeter wall willbe minimised if no connection is made between the roof and thewall until reflective gravel or some other protective materialhas been placed on the roof Alternatively restraint to thedeflection of the wall can be minimised by providing a durablecompressible material between the wall and the soil Thisprevents the build-up of large passive earth pressures in theupper portion of the soil and allows the wall to deflect as a longflexible cantilever

632 Cylindrical tanks

The wall of a cylindrical tank is primarily designed to resist ringtensions due to the horizontal pressures of the contained liquidIf the wall is free at the top and free-to-slide at the bottom thenwhen the tank is full the ring tension at depth z is given bynzr where is the unit weight of liquid and r is the internalradius of the tank In this condition when the tank is full novertical bending or radial shear exists

If the wall is connected to the floor in such a way that noradial movement occurs at the base the ring tension will be zeroat the bottom of the wall The ring tensions are affectedthroughout the lower part of the wall and significant verticalbending and radial shear occurs Elastic analysis can be usedto derive equations involving trigonometric and hyperbolicfunctions and solutions expressed in the form of tables areincluded in publications (eg refs 55 and 56) Coefficients todetermine values of circumferential tensions vertical bendingmoments and radial shears for particular values of the termheight2(2 mean radius thickness) are given in Tables 275and 276

The tables apply to idealised boundary conditions in whichthe bottom of the wall is either hinged or fixed It is possible todevelop these conditions if an annular footing is provided at thebottom of the wall The footing should be tied into the floor ofthe tank to prevent radial movement If the footing is narrow

there will be little resistance to rotation and a hinged conditioncould be reasonably assumed It is also possible to form a hingeby providing horizontal grooves at each side of the wall so thatthe contact between the wall and the footing is reduced to anarrow throat The vertical bars are then bent to cross over atthe centre of the wall but this detail is rarely used At the otherextreme if the wall footing is made wide enough it is possibleto get a uniform distribution of bearing pressure In this casethere will be no rotation and a fixed condition can be assumedIn many cases the wall and the floor slab are made continuousand it is necessary to consider the interaction between the twoelements Appropriate values for the stiffness of the memberand the effect of edge loading can be obtained from Tables 276and 277

For slabs on an elastic foundation the values depend on theratio rrk where rk is the radius of relative stiffness defined insection 725 The value of rk is dependent on the modulus ofsubgrade reaction for which data is given in section 724Taking rrk 0 which corresponds to a lsquoplasticrsquo soil state isappropriate for an empty tank liable to flotation

633 Octagonal tanks

If the wall of a tank forms in plan a series of straight sidesinstead of being circular the formwork may be less costly butextra reinforcement and possibly an increased thickness ofconcrete is needed to resist the horizontal bending momentsthat are produced in addition to the ring tension If the tankforms a regular octagon the bending moments in each side areq l212 at the corners and q l 224 at the centre where l is thelength of the side and q is the lsquoeffectiversquo lateral pressure at depthz If the wall is free at the top and free-to-slide at the bottomq z In other cases q nr where n is the ring tension atdepth z and r is the lsquoeffectiversquo radius (ie half the distancebetween opposite sides) If the tank does not form a regularoctagon but the length and thickness of the sides are alternatelyl1 h1 and l2 h2 the horizontal bending moment at the junctionof any two sides is

634 Rectangular tanks

The walls of large rectangular reservoirs are sometimes built indiscontinuous lengths in order to minimise restraints to theeffects of early thermal contraction and shrinkage If the wallbase is discontinuous with the main floor slab each wall unit isdesigned to be independently stable and no slip membrane isprovided between the wall base and the blinding concreteAlternatively the base to each wall unit can be tied into theadjacent panel of floor slab Roof slabs can be connected to theperimeter walls or simply supported with a sliding jointbetween the top of the wall and the underside of the slab Insuch forms of construction except for the effect of any cornerjunctions the walls span vertically either as a cantilever orwith ends that are simply supported or restrained depending onthe particular details

A cantilever wall is statically determinate and if supportinga roof is also isolated from the effect of roof movement Thedeflection at the top of the wall is an important consideration

q12l1

3 l23h1

h23 l1 l2h1

h23

Buildings bridges and containment structures60

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and the base needs to be carefully proportioned in order tominimise the effect of base tilting The problem of excessivedeflection can be overcome and the wall thickness reduced ifthe wall is tied into the roof If the wall is also provided with anarrow footing tied into the floor it can be designed as simplysupported although considerable reliance is being put in theability of the joint to accept continual rotation If the wallfooting is made wide enough it is possible to obtain a uniformdistribution of bearing pressure in which case there will be norotation and a fixed condition can be assumed In cases wherethe wall and floor slab are made continuous the interactionbetween the two elements should be considered

Smaller rectangular tanks are generally constructed withoutmovement joints so that structural continuity is obtained inboth horizontal and vertical planes Bending moments andshear forces in individual rectangular panels with idealisededge conditions when subjected to hydrostatic loading aregiven in Table 253 For a rectangular tank distribution of theunequal fixity moments obtained at the wall junctions isneeded and moment coefficients for tanks of different spanratios are given in Tables 278 and 279 The shearing forcesgiven in Table 253 for individual panels may still be used

The tables give values for tanks where the top of the wall iseither hinged or free and the bottom is either hinged or fixedThe edge conditions are generally uncertain and tend to varywith the loading conditions as discussed in section 172 Forthe horizontal spans the shear forces at the vertical edges ofone wall result in axial forces in the adjacent walls Thus forinternal loading the shear force at the end of a long wall isequal to the tensile force in the short wall and vice versa Indesigning sections the combined effects of bending momentaxial force and shear force need to be considered

635 Elevated tanks

The type of bottom provided to an elevated cylindrical tankdepends on the diameter of the tank and the depth of water Forsmall tanks a flat beamless slab is satisfactory but beams arenecessary for tanks exceeding about 3 m diameter Someappropriate examples which include bottoms with beams anddomed bottoms are included in section 174 and Table 281

It is important that there should be no unequal settlement ofthe foundations of columns supporting an elevated tank and araft should be provided in cases where such problems couldoccur In addition to the bending moments and shear forces dueto the wind pressure on the tank as described in sections 25and 83 the wind force causes a thrust on the columns on theleeward side and tension in the columns on the windward sideThe values of the thrusts and tensions can be calculated fromthe expressions given for columns supporting elevated tanks insection 1742

636 Effects of temperature

If the walls of a tank are subjected to significant temperatureeffects due to solar radiation or the storage of warm liquids theresulting moments and forces need to be determined by anappropriate analysis The structure can usually be analysedseparately for temperature change (expansion or contraction)and temperature differential (gradient through section) For awall with all of the edges notionally clamped the temperature

differential results in bending moments causing compressionon the warm face and tension on the cold face given by

where E is the modulus of elasticity of concrete I is secondmoment of area of the section h is thickness of wall is thecoefficient of thermal expansion of concrete 13 is temperaturedifference between the two surfaces is Poissonrsquos ratio Forcracked sections may be taken as zero but the value of I shouldallow for the tension stiffening effect of the concrete The effectof releasing the notional restraints at edges that are free orhinged modifies the moment field and in cylindrical tankscauses additional ring tensions For further information onthermal effects in cylindrical tanks reference can be made toeither the Australian or the New Zealand standard Code ofPractice for liquid-retaining concrete structures

64 SILOS

Silos which may also be referred to as bunkers or bins aredeep containers used to store particulate materials In a deepcontainer the linear increase of pressure with depth found inshallow containers is modified Allowances are made for theeffects of filling and unloading as described in section 277The properties of materials commonly stored in silos andexpressions for the pressures set up in silos of different formsand proportions are given in Tables 215 and 216

641 Walls

Silo walls are designed to resist the bending moments andtensions caused by the pressure of the contained material If thewall spans horizontally it is designed for the combined effectsIf the wall spans vertically horizontal reinforcement is neededto resist the axial tension and vertical reinforcement to resist thebending In this case the effect of the horizontal bendingmoments due to continuity at the corners should also beconsidered For walls spanning horizontally the bendingmoments and forces depend on the number and arrangement ofthe compartments Where there are several compartmentsthe intermediate walls act as ties between the outer walls Forvarious arrangements of intermediate walls expressions forthe negative bending moments on the outer walls of the silosare given in Table 280 Corresponding expressions for thereactions which are a measure of the axial tensions in thewalls are also given The positive bending moments can bereadily calculated when the negative bending moments atthe wall corners are known An external wall is subjected tothe maximum combined effects when the adjacent compartmentis full An internal cross-wall is subjected to the maximumbending moments when the compartment on one side of thewall is full and to maximum axial tension (but zero bending)when the compartments on both sides are full In small silosthe proportions of the wall panels may be such that they spanboth horizontally and vertically in which case Table 253 canbe used to calculate the bending moments

In the case of an elevated silo the whole load is generallytransferred to the columns by the walls and when the clear spanis greater than twice the depth the wall can be designed as ashallow beam Otherwise the recommendations for deep beamsshould be followed (see section 58 and ref 43) The effect of

M EI13(1v)h

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wind loads on large structures should be calculated The effectof both the tensile force in the windward walls of the empty siloand the compressive force in the leeward walls of the full siloare important In the latter condition the effect of the eccentricforce on the inside face of the wall due to the proportion of theweight of the contents supported by friction must be combinedwith the force due to the wind At the base and the top ofthe wall there are additional bending effects due to continuityof the wall with the bottom and the covers or roof over thecompartments

642 Hopper bottoms

The design of sloping hopper bottoms in the form of invertedtruncated pyramids consists of finding for each sloping sidethe centre of pressure the intensity of pressure normal to theslope at this point and the mean span The bending momentsat the centre and edge of each sloping side are calculated Thehorizontal tensile force is computed and combined with thebending moment to determine the horizontal reinforcementrequired The tensile force acting along the slope at the centreof pressure is combined with the bending moment at this pointto find the inclined reinforcement needed in the bottom of theslab At the top of the slope the bending moment and theinclined component of the hanging-up force are combined todetermine the reinforcement needed in the top of the slab

For each sloping side the centre of pressure and the mean spancan be obtained by inscribing on a normal plan a circle thattouches three of the sides The diameter of this circle is the meanspan and its centre is the centre of pressure The total intensityof load normal to the slope at this point is the sum of the normalcomponents of the vertical and horizontal pressures and the deadweight of the slab Expressions for determining the pressures onthe slab are given in Table 216 Expressions for determiningthe bending moments and tensile forces acting along the slopeand horizontally are given in Table 281 When using thismethod it should be noted that although the horizontal span ofthe slab reduces considerably towards the outlet the amount

of reinforcement should not be reduced below that calculatedfor the centre of pressure This is because in determining thebending moment based on the mean span adequate transversesupport from reinforcement towards the base is assumed

The hanging-up force along the slope has both vertical andhorizontal components the former being resisted by the wallsacting as beams The horizontal component acting inwardstends to produce horizontal bending moments on the beam atthe top of the slope but this is opposed by a correspondingoutward force due to the pressure of the contained material Thelsquohip-beamrsquo at the top of the slope needs to be designed both toresist the inward pull from the hopper bottom when the hopperis full and the silo above is only partly filled and also for thecase when the arching of the fill concentrates the outward forcesdue to the peak lateral pressure on the beam during unloadingThis is especially important in the case of mass-flow silos(see section 277)

65 BEARINGS HINGES AND JOINTS

In the construction of frames and arches hinges are needed atpoints where it is assumed that there is no bending moment Inbridges bearings are often required at abutments and piers totransfer loads from the deck to the supports Various types ofbearings and hinges for different purposes are illustrated inTable 299 with associated notes in section 1941

Movement joints are often required in concrete structures toallow free expansion and contraction Fluctuating movementsoccur due to diurnal solar effects and seasonal changes ofhumidity and temperature Progressive movements occur due toconcrete creep drying shrinkage and ground settlementMovement joints may also be provided in structures wherebecause of abrupt changes of loading or ground conditionspronounced changes occur in the size or type of foundationVarious types of joints for different purposes are illustrated inTable 2100 with associated notes on their construction andapplication in section 1942

Buildings bridges and containment structures62

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71 FOUNDATIONS

The design of the foundations for a structure comprises threestages The first is to determine from an inspection of the sitetogether with field data on soil profiles and laboratory testing ofsoil samples the nature of the ground The second stage is toselect the stratum on which to impose the load the bearingcapacity and the type of foundation These decisions dependnot only on the nature of the ground but also on the type ofstructure and different solutions may need to be consideredReference should be made to BS 8004 Code of Practice forfoundations The third stage is to design the foundation totransfer and distribute load from the structure to the ground

711 Site inspection

The objective of a site inspection is to determine the nature ofthe top stratum and the underlying strata in order to detect anyweak strata that may impair the bearing capacity of the stratumselected for the foundation Generally the depth to whichknowledge of the strata is obtained should be not less than oneand a half times the width of an isolated foundation or thewidth of a structure with closely spaced footings

The nature of the ground can be determined by digging trialholes by sinking bores or by driving piles A trial hole can betaken down to only moderate depths but the undisturbed soilcan be examined and the difficulties of excavation with theneed or otherwise of timbering and groundwater pumping canbe determined Bores can be taken very much deeper than trialholes and stratum samples at different depths obtained forlaboratory testing A test pile does not indicate the type ofsoil it has been driven through but it is useful in showing thethickness of the top crust and the depth below poorer soil atwhich a firm stratum is found A sufficient number of any ofthese tests should be taken to enable the engineer to ascertainthe nature of the ground under all parts of the foundationsReference should be made to BS 5930 Code of practice for siteinvestigations and BS 1377 Methods of test for soils for civilengineering purposes

712 Bearing pressures

The pressure that can be safely imposed on a thick stratum ofsoil commonly encountered is in some districts stipulated in

local by-laws The pressures recommended for preliminarydesign purposes in BS 8004 are given in Table 282 but thesevalues should be used with caution since several factors cannecessitate the use of lower values Allowable pressures maygenerally be exceeded by the weight of soil excavated down tothe foundation level but if this increase is allowed any fillmaterial applied on top of the foundation must be included inthe total load If the resistance of the soil is uncertain a studyof local records for existing buildings on the same soil can beuseful as may the results of a ground-bearing test

Failure of a foundation can occur due to consolidation of theground causing settlement or rupture of the ground due toshearing The shape of the surface along which shear failureoccurs under a strip footing is an almost circular arc startingfrom one edge of the footing passing under the footing andthen continuing as a tangent to the arc to intersect the groundsurface at an angle depending on the angle of internal frictionof the soil Thus the average shear resistance depends onthe angle of internal resistance of the soil and on the depthof the footing below the ground surface In a cohesionless soilthe bearing resistance not only increases as the depth increasesbut is proportional to the width of the footing In a cohesive soilthe bearing resistance also increases with the width of footingbut the increase is less than for a non-cohesive soil

Except when bearing directly on rock foundations for all butsingle-storey buildings or other light structures should betaken down at least 1 m below the ground surface in order toobtain undisturbed soil that is sufficiently consolidated In claysoils a depth of at least 15 m is needed in the United Kingdomto ensure protection of the bearing stratum from weathering

713 Eccentric loads

When a rigid foundation is subjected to concentric loadingthat is when the centre of gravity of the loads coincides withthe centre of area of the foundation the bearing pressure on theground is uniform and equal to the total applied load divided bythe total area When a load is eccentrically placed on a base ora concentric load and a bending moment are applied to a basethe bearing pressure is not uniform For a load that is eccentricabout one axis of a rectangular base the bearing pressure variesfrom a maximum at the side nearer the centre of gravity of theload to a minimum at the opposite side or to zero at some inter-mediate position The pressure variation is usually assumed to

Chapter 7

Foundations groundslabs retaining wallsculverts and subways

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Foundations ground slabs retaining walls culverts and subways64

be linear in which case the maximum and minimum pressuresare given by the formulae in Table 282 For large eccentricitiesthere may be a part of the foundation where there is no bearingpressure Although this state may be satisfactory for transientconditions (such as those due to wind) it is preferable for thefoundation to be designed so that contact with the ground existsover the whole area under normal service conditions

714 Blinding layer

For reinforced concrete footings or other construction wherethere is no underlying mass concrete forming an integral part ofthe foundation the bottom of the excavation should be coveredwith a layer of lean concrete to protect the soil and provide aclean surface on which to place the reinforcement The thicknessof this blinding layer is typically 50ndash75 mm depending on thesurface condition of the excavation

715 Foundation types

The most suitable type of foundation depends primarily on thedepth at which the bearing stratum lies and the allowable bearingpressure which determines the foundation area Data relatingto some common types of separate and combined pad founda-tions suitable for sites where the bearing stratum is found closeto the surface are given in Tables 282 and 283 Several typesof inter-connected bases and rafts are given in Table 284 Inchoosing a foundation suitable for a particular purpose thenature of the structure should also be considered Sometimes itmay be decided to accept the risk of settlement in preference toproviding a more expensive foundation For silos and fixed-endarches the risk of unequal settlement of the foundations mustbe avoided at all costs but for gantries and the bases of largesteel tanks a simple foundation can be provided and probablesettlement allowed for in the design of the superstructure Inmining districts where it is reasonable to expect some subsidencea rigid raft foundation should be provided for small structuresto allow the structure to move as a whole For large structuresa raft may not be economical and the structure should bedesigned either to be flexible or as several separate elementson independent raft foundations

716 Separate bases

The simplest form of foundation for an individual column orstanchion is a reinforced concrete pad Such bases are widelyused on ground that is strong and on weaker grounds wherethe structure and the cladding are light and flexible For basesthat are small in area or founded on rock a block of plain ornominally reinforced concrete can be used The thickness ofthe block is made sufficient for the load to be transferred to theground under the base at an angle of dispersion through theblock of not less than 45o to the horizontal

To reduce the risk of unequal settlement the column basesizes for a building founded on a compressible soil should be inproportion to the dead load carried by each column Bases for thecolumns of a storage structure should be in proportion to the totalload excluding the effects of wind In all cases the pressure onthe ground under any base due to combined dead and imposedload including wind load and any bending at the base of thecolumn should not exceed the allowable bearing value

In the design of a separate base the area of a concentricallyloaded base is determined by dividing the maximum serviceload by the allowable bearing pressure The subsequentstructural design is then governed by the requirements of theultimate limit state The base thickness is usually determined byshear considerations governed by the more severe of two con-ditions ndash either shear along a vertical section extending acrossthe full width of the base or punching shear around the loadedarea ndash where the second condition is normally critical Thecritical section for the bending moment at a vertical sectionextending across the full width of the base is taken at the faceof the column for a reinforced concrete column and at the cen-tre of the base for a steel stanchion The tension reinforcementis usually spread uniformly over the full width of the base butin some cases it may need to be arranged so that there is aconcentration of reinforcement beneath the column Outsidethis central zone the remaining reinforcement must still con-form to minimum requirements It is also necessary for tensionreinforcement to comply with the bar spacing limitations forcrack control

If the base cannot be placed centrally under the column thebearing pressure varies linearly The base is then preferablyrectangular and modified formulae for bearing pressures andbending moments are given in Table 282 A base supportingfor example a column of a portal frame may be subjected to anapplied moment and horizontal shear force in addition to avertical load Such a base can be made equivalent to a base witha concentric load by placing the base under the column with aneccentricity that offsets the effect of the moment and horizontalforce This procedure is impractical if the direction of theapplied moment and horizontal force is reversible for exampledue to wind In this case the base should be placed centrallyunder the column and designed as eccentrically loaded for thetwo different conditions

717 Combined bases

If the size of the bases required for adjacent columns is suchthat independent bases would overlap two or more columnscan be provided with a common foundation Suitable typesfor two columns are shown in Table 283 for concentrically andeccentrically loaded cases Reinforcement is required top andbottom and the critical condition for shear is along a verticalsection extending across the full width of the base For someconditions of loading on the columns the total load on the basemay be concentric while for other conditions the total load iseccentric and both cases have to be considered Some notes oncombined bases are given in section 1812

718 Balanced and coupled bases

When it is not possible to place an adequate base centrallyunder a column owing to restrictions of the site and when forsuch conditions the eccentricity would result in inadmissibleground pressures a balanced foundation as shown in Tables 283and 284 and described in section 1813 is provided A beamis introduced and the effect of the cantilever moment causedby the offset column load is counterbalanced by load from anadjacent column This situation occurs frequently for externalcolumns of buildings on sites in built-up areas

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Sometimes as in the case of bases under the towers of atrestle or gantry pairs of bases are subjected to moments andhorizontal forces acting in the same direction on each base Insuch conditions the bases can be connected by a stiff beam thatconverts the effects of the moments and horizontal forces intoequal and opposite vertical reactions then each base can bedesigned as concentrically loaded Such a pair of coupled basesis shown in Table 283 which also gives formulae for thereactions and the bending moments on the beam

719 Strip bases and rafts

When the columns or other supports of a structure are closelyspaced in one direction it is common to provide a continuousbase similar to a footing for a wall Particulars of the designof strip bases are given in Table 283 Some notes on thesebases in relation to the diagrams in Table 284 together with anexample are given in section 1812

When the columns or other supports are closely spaced intwo directions or when the column loads are so high and theallowable bearing pressure is so low that a group of separatebases would totally cover the space between the columns asingle raft foundation of one of the types shown at (a)ndash(d) inTable 284 should be provided Notes on these designs are givenin section 1814

The analysis of a raft foundation supporting a set of equalloads that are symmetrically arranged is usually based on theassumption of uniformly distributed pressure on the groundThe design is similar to that for an inverted floor upon whichthe load is that portion of the ground pressure that is due to theconcentrated loads only Notes on the design of a raft for whichthe columns are not symmetrically disposed are also includedin section 1814 An example of the design of a raft foundationis given in Examples of the Design of Buildings

7110 Basements

The floor of a basement for which a typical cross section isshown at (e) in the lower part of Table 284 is typically a raftsince the weights of the ground floor over the basementthe walls and other structure above the ground floor and thebasement itself are carried on the ground under the floor ofthe basement For water-tightness it is common to construct thewall and the floor of the basement monolithically In mostcases although the average ground pressure is low the spansare large resulting in high bending moments and a thick floorif the total load is taken as uniform over the whole area Sincethe greater part of the load is transmitted through the wallsand any internal columns it is more rational and economicalto transfer the load on strips and pads placed immediatelyunder the walls and columns The resulting cantilever actiondetermines the required thickness of these portions and theremainder of the floor can generally be made thinner

Where basements are in water-bearing soils the effect ofhydrostatic pressure must be taken into account The upwardwater pressure is uniform below the whole area of the floorwhich must be capable of resisting the total pressure lessthe weight of the floor The walls must be designed to resist thehorizontal pressures due to the waterlogged ground and thebasement must be prevented from floating Two conditions needto be considered Upon completion the total weight of the

basement and superimposed dead load must exceed the worstcredible upward force due to the water by a substantial marginDuring construction there must always be an excess ofdownward load If these conditions cannot be satisfied oneof the following steps should be taken

1 The level of the groundwater near the basement should becontrolled by pumping or other means

2 Temporary vents should be formed in the basement floor orat the base of the walls to enable water to freely enter thebasement thereby equalising the external and internalpressures The vents should be sealed when sufficient deadload from the superstructure has been obtained

3 The basement should be temporarily flooded to a depth suchthat the weight of water in the basement together with thedead load exceeds the total upward force on the structure

While the basement is under construction method 1 normallyhas to be used but once the basement is complete method 3 hasthe merit of simplicity Basements are generally designedand constructed in accordance with the recommendations ofBS 8102 supplemented by the guidance provided in reportsproduced by CIRIA (ref 57) BS 8102 defines four gradesof internal environment each grade requiring a different levelof protection against water and moisture ingress Three types ofconstruction are described to provide either A tanked or Bintegral or C drained protection

Type A refers to concrete or masonry construction whereadded protection is provided by a continuous barrier system Anexternal tanking is generally preferred so that any externalwater pressure will force the membrane against the structureThis is normally only practicable where the construction is byconventional methods in excavation that is open or supportedby temporary sheet piling The structure should be monolithicthroughout and special care should be taken when a structureis supported on piles to avoid rupture of the membrane due tosettlement of the fill supported by the basement wall

Type B refers to concrete construction where the structureitself is expected to be sufficient without added protection Astructure designed to the requirements of BS 8007 is expectedto inhibit the ingress of water to the level required for a utilitygrade basement It is considered that this standard can also beachieved in basements constructed by using diaphragm wallssecant pile walls and permanent sheet piling If necessary theperformance can be improved by internal ventilation and theaddition of a vapour-proof barrier

Type C refers to concrete or masonry construction whereadded protection is provided by an internal ventilated drainedcavity This method is applicable to all types of constructionand can provide a high level of protection It is particularlyuseful for deep basements using diaphragm walls secant pilewalls contiguous piles or steel sheet piling

7111 Foundation piers

When a satisfactory bearing stratum is found at a depth of15ndash5 m below the natural ground level piers can be formedfrom the bearing stratum up to ground level The constructionof columns or other supporting members can then begin on thetop of the piers at ground level Such piers are generally squarein cross section and most economically constructed in plain

Foundations 65

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concrete When piers are impractical by reason of the depth atwhich a firm stratum occurs or due to the nature of the groundshort bored piles can be used

7112 Wall footings

When the load on a strip footing is distributed uniformly overthe whole length as in the general case of a wall footing theprincipal effects are due to the transverse cantilever action ofthe projecting portion of the footing If the wall is of concreteand built monolithically with the footing the critical bendingmoment is at the face of the wall If the wall is of masonry themaximum bending moment is at the centre of the footingExpressions for these moments are given in Table 283 If theprojection is less than the thickness of the base the transversebending moment may be ignored but the thickness should besuch that the shear strength is not exceeded Whether or not awall footing is designed for transverse bending longitudinalreinforcement is generally included to give some resistance tomoments due to unequal settlement and non-uniformity ofbearing In cases where a deep narrow trench is excavated downto a firm stratum plain concrete fill is normally used

7113 Foundations for machines

The area of a concrete base supporting a machine or enginemust be sufficient to spread the load onto the ground withoutexceeding the allowable bearing value It is advantageous if thecentre of area of the base coincides with the centre of gravityof the loads when the machine is working as this reducesthe risk of unequal settlement If vibration from the machine istransmitted to the ground the bearing pressure should beconsiderably lower than normally taken especially if the groundis clay or contains a large proportion of clay It is often importantthat the vibration of a machine should not be transmitted toadjacent structures either directly or via the ground In suchcases a layer of insulating material should be placed betweenthe concrete base carrying the machine and the groundSometimes the base is enclosed in a pit lined with insulatingmaterial In exceptional cases a machine base may stand onsprings or more elaborate damping devices may be installed Inall cases the base should be separated from any surroundingarea of concrete ground floor

With light machines the ground bearing pressure may not bethe factor that determines the size of the concrete base as thearea occupied by the machine and its frame may require a baseof larger area The position of the holding-down bolts generallydetermines the length and width of the base which shouldextend 150 mm or more beyond the outer edges of the holes leftfor the bolts The depth of the base must be such that the bottomis on a satisfactory bearing stratum and there is enough thick-ness to accommodate the holding-down bolts If the machineexerts an uplift force on any part of the base the dimensions ofthe base must be such that the part that is subjected to uplift hasenough weight to resist the uplift force with a suitable marginof safety All the supports of any one machine should be carriedon a single base and any sudden changes in the depth and widthof the base should be avoided This reduces the risk of fracturesthat might result in unequal settlements which could throw themachine out of alignment Reinforcement should be providedto resist all tensile forces

Advice on the design of reinforced concrete foundations tosupport vibrating machinery is given in ref 58 which givespractical solutions for the design of raft piled and massivefoundations Comprehensive information on the dynamics ofmachine foundations is included in ref 59

7114 Piled foundations

Where the upper soil strata is compressible or too weak tosupport the loads transmitted by a structure piles can be usedto transmit the load to underlying bedrock or a stronger soillayer using end-bearing piles Where bedrock is not located ata reasonable depth piles can be used to gradually transmit thestructural loads to the soil using friction piles

Horizontal forces due to wind loading on tall structures orearth pressure on retaining structures can be resisted by pilesacting in bending or by using raking piles Foundations forsome structures such as transmission towers and the roofs tosports stadiums are subjected to upward forces that can beresisted by tension piles Bridge abutments and piers adjacentto water can be constructed with piled foundations to counterthe possible detrimental effects of erosion

There are two basic categories of piles Displacement pilesare driven into the ground in the form of either a preformedsolid concrete pile or a hollow tube Alternatively a void canbe formed in the ground by driving a closed-ended tube thebottom of which is plugged with concrete or aggregate Thisallows the tube to be withdrawn and the void to be filled withconcrete It also allows the base of the pile to be enlarged inorder to increase the bearing capacity Non-displacement orlsquocast-in-placersquo piles are formed by boring or excavating theground to create a void into which steel reinforcement andconcrete can be placed In some soils the excavation needs tobe supported to stop the sides from falling in this is achievedeither with casings or by the use of drilling mud (bentonite)For further information on piles including aspects such aspile driving load testing and assessment of bearing capacityreference should be made to specialist textbooks (ref 60)

7115 Pile-caps

Rarely does a foundation element consist of a single pile Inmost cases piles are arranged in groups or rows with the topsof the piles connected by caps or beams Generally concrete ispoured directly onto the ground and encases the tops of the pilesto a depth of about 75 mm The thickness of the cap must besufficient to ensure that the imposed load is spread equallybetween the piles For typical arrangements of two to five pilesforming a compact group load can be transmitted by dispersionthrough the cap Inclined struts extending from the load to thetop of each pile are held together by tension reinforcement inthe bottom of the cap to form a space frame The struts areusually taken to intersect at the top of the cap at the centreof the loaded area but expressions have also been developedthat take into account the dimensions of the loaded areaInformation regarding the design of such pile-caps andstandardised arrangements and dimensions for groups of twoto five piles are given in Table 361

The thickness of a pile-cap designed by dispersion theory isusually determined by shear considerations along a verticalsection extending across the full width of the cap If the pile

Foundations ground slabs retaining walls culverts and subways66

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spacing exceeds three pile diameters it is also necessary todesign for punching shear In all cases the shear stress at theperimeter of the loaded area should not exceed the maximumdesign value related to the compressive strength of the strutsThe reinforcement in the bottom of the pile-cap should beprovided at each end with a full tension anchorage measuredfrom the centre of the pile Pile-caps can also be designed bybending theory but this is generally more appropriate where alarge number of piles are involved In such cases punchingshear is likely to be a critical consideration

7116 Loads on piles in a group

If a group of n piles is connected by a rigid pile-cap and thecentres of gravity of the load Fv and the piles are coincident eachpile will be equally loaded and will be subjected to a load FvnIf the centre of gravity of the load is displaced a distance e fromthe centre of gravity of the piles the load on any one pile is

where is the sum of the squares of the distance of each pilemeasured from an axis that passes through the centre of gravityof the group of piles and is at right angles to the line joining thiscentre of gravity and the centre of gravity of the load and a1 isthe distance of the pile considered from this axis (positive if onthe same side of the axis as the centre of gravity of the load andnegative if on the opposite side) If the structure supported onthe group of piles is subjected to a bending moment M whichis transmitted to the foundations the expression given for theload on any pile can be used by substituting e MFv

The total load that can be carried on a group of piles is notnecessarily the safe load calculated for one pile multipliedby the number of piles Some allowance has to be made for theoverlapping of the zones of stress in the soil supporting thepiles The reduction due to this effect is greatest for piles thatare supported mainly by friction For piles supported entirely oralmost entirely by end bearing the maximum safe load on agroup cannot greatly exceed the safe bearing load on the areaof bearing stratum covered by the group

7117 Loads on open-piled structures

The loads and forces to which wharves jetties and similarmaritime structures are subjected are dealt with in section 26Such structures can be solid walls made of plain or reinforcedconcrete as are most dock walls A quay or similar watersidewall is more often a sheet pile-wall as described in section 733or it can be an open-piled structure similar to a jetty The loadson groups of inclined and vertical piles for such structures areconsidered in Table 285

For each probable condition of load the external forces areresolved into horizontal and vertical components Fh and Fvthe points of application of which are also determined If thedirection of action and position are opposite to those shown inthe diagrams the signs in the formulae must be changed It isassumed that the piles are surmounted by a rigid pile-cap orsuperstructure The effects on each pile when all the piles arevertical are based on a simple but approximate method ofanalysis Since a pile offers very little resistance to bending

a2

Fv1n

ea1

a2

structures with vertical piles only are not suitable when Fh isdominant In a group containing inclined piles Fh can beresisted by a system of axial forces and the bending momentsand shear forces in the piles are negligible The analysis used inTable 285 is based on the assumption that each pile is hingedat the head and toe Although this assumption is not accuratethe analysis predicts the behaviour reasonably well Three designsof the same typical jetty using different pile arrangements aregiven in section 182

72 INDUSTRIAL GROUND FLOORS

Most forms of activity in buildings ndash from manufacturingstorage and distribution to retail and recreation ndash need a firmplatform on which to operate Concrete ground floors arealmost invariably used for such purposes Although in manyparts of the world conventional manufacturing activity hasdeclined in recent years there has been a steady growth indistribution warehousing and retail operations to serve theneeds of industry and society The scale of such facilities andthe speed with which they are constructed has also increasedwith higher and heavier racking and storage equipment beingused These all make greater demands on concrete floors Thefollowing information is taken mainly from ref 61 where acomprehensive treatment of the subject will be found

721 Floor uses

In warehouses materials handling equipment is used in twodistinct areas according to whether the movement of traffic isfree or defined In free-movement areas vehicles can travelrandomly in any direction This typically occurs in factoriesretail outlets low-level storage and food distribution centresIn defined-movement areas vehicles use fixed paths in verynarrow aisles This usually occurs where high-level storageracking is being employed and distribution and warehousefacilities often combine areas of free movement for low-levelactivities such as unloading and packing alongside areas ofdefined movement for high-level storage The two floor usesrequire different tolerances on surface regularity

722 Construction methods

A ground-supported industrial floor slab is made up of layersof materials comprising a sub-base a slip membranemethanebarrier and a concrete slab of appropriate thickness providinga suitable wearing surface Various construction methods can beused to form the concrete slab

Large areas of floor up to several thousand square metres inextent can be laid in a continuous operation Fixed forms areused up to 50 m apart at the edges of the area only Concrete isdischarged into the area and spread either manually or bymachine Surface levels are controlled either manually using atarget staff in conjunction with a laser level transmitter or bydirect control of a laser-guided spreading machine After thefloor has been laid and finished the area is sub-divided intopanels typically on a 6 m grid in both directions This is achievedby making saw cuts in the top surface for a depth of at leastone-quarter of the depth of the slab creating a line of weaknessin the slab that induces a crack below the saw cut As a result ofconcrete shrinkage each sawn joint will open by a small amount

Industrial ground floors 67

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With such large-area construction there are limitations on theaccuracy of level and surface regularity that can be achievedand the construction is most commonly used for free-movementfloor areas

The large-area construction method can also be employedwithout sub-dividing the area into small panels In this case nosawn joints are made but steel fibres are incorporated in theconcrete mix to control the distribution and width of the cracksthat occur as a result of shrinkage The formed joints at theedges of the area will typically open by about 20 mm

Floors can also be formed as a series of long strips typically4ndash6 m wide with forms along each side Strips can be laidalternately with infill strips laid later or consecutively orbetween lsquoleave-in-placersquo screed rails Concrete is poured in acontinuous operation in each strip after which transverse sawcuts are made about 6 m apart to accommodate longitudinalshrinkage As formwork can be set to tight tolerances and thedistance between the forms is relatively small the long-stripmethod lends itself to the construction of very flat floors and isparticularly suitable for defined-movement floor areas

723 Reinforcement

Steel fibres usually manufactured from cold-drawn wire arecommonly used in ground-supported slabs The fibres vary inlength up to about 60 mm with aspect ratios (lengthnominaldiameter) from 20 to 100 and a variety of cross sections Inorder to increase pull-out resistance the fibres have enlargedflattened or hooked ends roughened surface textures or wavyprofiles The composite concrete slab can have considerableductility dependent on fibre type dosage tensile strength andanchorage mechanism The ductility is commonly measuredusing the Japanese Standard test method which uses beams ina third-point loading arrangement Load-deflection curves areplotted as the load increases until the first crack and thendecreases with increasing deflection The ductility value isexpressed as the average load to a deflection of 3 mm dividedby the load to first crack This measure is commonly known asthe equivalent flexural strength ratio In large-area floors withshrinkage joints at the edges only fibre dosages in the order of35ndash45 kgm3 are used to control the distribution and width ofcracks In floors with additional sawn joints fibre dosages in therange 20ndash30 kgm3 are typically used

In large area floors with additional sawn joints steel fabricreinforcement (type A) can be placed in the bottom of the slabwith typically 50 mm of cover The proportion of reinforcementused is typically 01ndash0125 of the effective cross section bdwhich is small enough to ensure that the reinforcement willyield at the sawn joints as the concrete shrinks and alsosufficient to provide the slab with adequate rotational capacityafter cracking

724 Modulus of subgrade reaction

For design purposes the subgrade is assumed to be an elasticmedium characterised by a modulus of subgrade reaction ksdefined as the load per unit area causing unit deflection It canbe shown that errors of up to 50 in the value of ks have onlya small effect on the slab thickness required for flexural designHowever deflections are more sensitive to ks values and long-term settlement due to soil consolidation under load can be

Foundations ground slabs retaining walls culverts and subways68

Soil typeValues of ks (MNm3)

Lower Upper

Fine or slightly compacted sand 15 30Well compacted sand 50 100Very well compacted sand 100 150Loam or clay (moist) 30 60Loam or clay (dry) 80 100Clay with sand 80 100Crushed stone with sand 100 150Coarse crushed stone 200 250Well-compacted crushed stone 200 300

much larger than the elastic deflections calculated as part of theslab design

In principle the value of ks used in design should be relatedto the range of influence of the load but it is normal practice tobase ks on a loaded area of diameter 750 mm To this endit is strongly recommended that the value of ks is determinedfrom a BS plate-loading test using a 750 mm diameter plateand a fixed settlement of 125 mm If a smaller plate is used ora value of ks appropriate to a particular area is required thefollowing approximate relationship may be assumed

ks 05(103D)2k075

where D is the diameter of the loaded area and k075 is a valuefor D 075 m This gives values of ksk075 as follows

D (m) 03 045 075 12 infin

ksk075 20 14 10 08 05

In the absence of more accurate information typical values ofks according to the soil type are given in the following table

725 Methods of analysis

Traditionally ground-supported slabs have been designed byelastic methods using equations developed by Westergaard inthe 1920s Such slabs are relatively thick and an assessment ofdeflections and other in-service requirements has generallybeen unnecessary Using plastic methods of analysis thinnerslabs can be designed and the need to investigate in-servicerequirements and load-transfer across joints has become moreimportant The use of plastic analysis assumes that the slab hasadequate ductility after cracking that is it contains sufficientfibres or reinforcement as described in section 723 to give anequivalent flexural strength ratio in the range 03ndash05 Plainconcrete slabs and slabs with less than the minimum recom-mended amounts of fibres or reinforcement should still bedesigned by elastic methods

Westergaard assumed that a ground-supported concrete slabis a homogeneous isotropic elastic solid in equilibrium withthe subgrade reactions being vertical only and proportional tothe deflections of the slab He also introduced the concept of theradius of relative stiffness rk given by the relationship

where Ec is the short-term modulus of elasticity of concreteh is the slab thickness ks is the modulus of subgrade reaction

rk [Ech312(1v2)ks]

025

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and is Poissonrsquos ratio The physical significance of rk isillustrated in the following figure showing the approximatedistribution of elastic bending moments for a single internalconcentrated load The bending moment is positive (tensionat the bottom of the slab) with a maximum value at the loadposition Along radial lines it remains positive reducing to zero atrk from the load It then becomes negative reaching a maximumat 2rk from the load with the maximum negative moment (tensionat the top of the slab) significantly less than the maximumpositive moment The moment approaches zero at 3rk fromthe load

or internally as shown in Table 286 An externally stabilisedsystem uses an external structural wall to mobilise stabilisingforces An internally stabilised system utilises reinforcementsinstalled within the soil and extending beyond the potentialfailure zone

Traditional retaining walls can be considered as externallystabilised systems one of the most common forms being thereinforced concrete cantilever wall Retaining walls on spreadfoundations together with gravity structures support the soilby weight and stiffness to resist forward sliding overturningand excessive soil movements The equilibrium of cantileverwalls can also be obtained by embedment of the lower part ofthe wall Anchored or propped walls obtain their equilibriumpartly by embedment of the lower part of the wall and partlyfrom an anchorage or prop system that provides support to theupper part of the wall

Internally stabilised walls built above ground are known asreinforced soil structures By placing reinforcement within thesoil a composite material can be produced that is strong intension as well as compression A key aspect of reinforced soilwalls is its incremental form of construction being built up alayer at a time starting from a small plain concrete strip footingIn this way construction is always at ground level the structureis always stable and progress can be very rapid The result ofthe incremental construction is that the soil is partitioned witheach layer receiving support from a locally inserted reinforcingelement The process is the opposite of what occurs in aconventional wall where pressures exerted by the backfill areintegrated to produce an overall force to be resisted by the struc-ture The materials used in a reinforced soil structure comprisea facing (usually reinforced concrete) soil reinforcement (inthe form of flat strips anchors or grids made from eithergalvanised steel or synthetic material) and soil (usually a well-graded cohesionless material) Reinforced soil structuresare more economic than equivalent structures using externallystabilised methods

Internal soil stabilisation used in the formation of cuttingsor excavations is known as soil nailing The process is againincremental with each stage of excavation limited in depth sothat the soil is able to support itself The exposed soil face isprotected usually by a covering of light mesh reinforcementand spray applied concrete Holes are drilled into the soil andreinforcement in the form of steel bars installed and groutedWith both reinforced soil and soil nailing great care is takento make sure that the reinforcing members do not corrode ordeteriorate Hybrid systems combining elements of internallyand externally stabilised soils are also used

732 Walls on spread bases

Various walls on spread bases are shown in Table 286 Acantilever wall is suitable for walls of moderate height If thesoil to be retained can be excavated during construction ofthe wall or the wall is required to retain an embankment thebase can project backwards This is always advantageous asthe earth supported on the base assists in counterbalancing theoverturning effect due to the horizontal pressures exerted bythe soil However a base that projects mainly backwards butpartly forwards is usually necessary in order to limit the bearingpressure at the toe to an allowable value Sometimes due tothe proximity of adjacent property it may be impossible to

Retaining walls 69

Approximate distribution of elastic bending moments for an internal concentrated load on a ground-supported slab

As the load is increased the tensile stresses at the bottom ofthe slab under the load will reach the flexural strength of theconcrete Radial tension cracks will form at the bottom ofthe slab and provided there is sufficient ductility the slab willyield Redistribution of moments will occur with a reduction inthe positive moment at the load position and a substantialincrease in the negative moments some distance away Withfurther increases in load the positive moment at the loadposition will remain constant and the negative moments willincrease until the tensile stresses at the top of the slab reach theflexural strength of the concrete at which stage failure isassumed For further information on the analysis and designmethod with fully worked examples see ref 61

73 RETAINING WALLS

Information on soil properties and the pressures exerted bysoils on retaining structures is given in section 91 andTables 210ndash214 This section deals with the design of wallsto retain soils and materials with similar engineering propertiesIn designing to British Codes of Practice the geotechnicalaspects of the design which govern the size and proportions ofthe structure are considered in accordance with BS 8002Mobilisation factors are introduced into the calculation of thesoil strengths and the resulting pressures are used for bothserviceability and ultimate requirements For the subsequentdesign of the structure to BS 8110 the earth loads obtainedfrom BS 8002 are taken as characteristic values In designing tothe EC partial safety factors are applied to the soil propertiesfor the geotechnical aspects of the design and to the earthloads for the structural design

731 Types of retaining wall

Earth retention systems can be categorised into one of twogroups according to whether the earth is stabilised externally

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project a base backwards Under such conditions where thebase projection is entirely forwards the provision of a keybelow the base is necessary to prevent sliding by mobilising thepassive resistance of the soil in front of the base

For wall heights greater than about 8 m the stem thicknessof a cantilever wall becomes excessive In such cases a wallwith vertical counterforts can be used in which the slabspans horizontally between the counterforts For very highwalls in which the soil loading is considerable towardsthe bottom of the wall horizontal beams spanning betweenthe counterforts can be used By graduating the spacing of thebeams to suit the loading the vertical bending moments ineach span of the slab can be equalised and the slab thicknesskept the same

The factors affecting the design of a cantilever slab wall areusually considered per unit length of wall when the wall is ofconstant height but if the height varies a length of say 3 mcould be treated as a single unit For a wall with counterfortsthe length of a unit is taken as the distance between adjacentcounterforts The main factors to be considered in the design ofwalls on spread bases are stability against overturning groundbearing pressure resistance to sliding and internal resistance tobending moments and shearing forces Suitable dimensions forthe base to a cantilever wall can be estimated with the aid of thegraph given in Table 286

In BS 8002 for design purposes soil parameters are basedon representative shear strengths that have been reduced byapplying mobilisation factors Also for friction or adhesion ata soilndashstructure interface values not greater than 75 of thedesign shear strength are taken Allowance is made for a minimumsurcharge of 10 kNm2 applied to the surface of the retainedsoil and for a minimum depth of unplanned earth removal infront of the wall equal to 10 of the wall height but not lessthan 05 m

For overall equilibrium the effects of the disturbing forcesacting on the structure should not exceed the effects that canbe mobilised by the resisting forces No additional factors ofsafety are required with regard to overturning or sliding forwardsFor bases founded on clay soils both the short-term (usingundrained shear strength) and long-term (using drained shearstrength) conditions should be considered Checks on groundbearing are required for both the service and ultimate conditionswhere the design loading is the same for each but the bearingpressure distribution is different For the ultimate condition auniform distribution is considered with the centre of pressurecoincident with the centre of the applied force at the undersideof the base In general therefore the pressure diagram doesnot extend over the entire base In cases where resistance tosliding depends on base adhesion it is unclear as to whetherthe contact surface length should be based on the service or theultimate condition

The foregoing wall movements due to either overturning orsliding are independent of the general tendency of a bank or acutting to slip and carry the retaining wall along with it Thestrength and stability of the retaining wall have no bearingon such failures The precautions that must be taken to preventsuch failures are outside the scope of the design of a wall thatis constructed to retain the toe of the bank and are a problemin soil mechanics

Adequate drainage behind a retaining wall is important toreduce the water pressure on the wall For granular backfills of

high permeability no special drainage layer is needed but somemeans of draining away any water that has percolated throughthe backfill should be provided particularly where a wall isfounded on an impermeable material For cohesionless backfillsof medium to low permeability and for cohesive soils it is usualto provide a drainage layer behind the wall Various methodscan be used for instance (a) a blanket of rubble or coarseaggregate clean gravel or crushed stone (b) hand-placedpervious blocks as dry walling (c) graded filter drain wherethe back-filling consists of fine-grain material (d) a geotextilefilter used in combination with a permeable granular materialWater entering the drainage layer should drain into a drainagesystem which allows free exit of the water either by the provisionof weep-holes or by porous land drains and pipes laid at thebottom of the drainage layer and led to sumps or sewers viacatchpits Where weep-holes are being used they should be atleast 75 mm in diameter and at a spacing not more than 1 mhorizontally and 1ndash2 m vertically Puddled clay or concreteshould be placed directly below the weep-holes or pipes andin contact with the back of the wall to prevent water fromreaching the foundations

Vertical movement joints should be provided at intervalsdependent upon the expected temperature range the type ofthe structure and changes in the wall height or the nature of thefoundations Guidance on design options to accommodatemovement due to temperature and moisture change are given inBS 8007 and Highways Agency BD 2887

733 Embedded (or sheet) walls

Embedded walls are built of contiguous or interlocking pilesor diaphragm wall panels to form a continuous structure Thepiles may be of timber or concrete or steel and have lapped orV-shaped or tongued and grooved or interlocking jointsbetween adjacent piles Diaphragm wall panels are formed ofreinforced concrete using a bentonite or polymer suspension aspart of the construction process Excavation is carried out in thesuspension to a width equal to the thickness of the wallrequired The suspension is designed to maintain the stability ofthe slit trench during digging and until the diaphragm wall hasbeen concreted Wall panels are formed in predeterminedlengths with prefabricated reinforcement cages lowered into thetrench Concrete is cast in situ and placed by tremie it is vitalthat the wet concrete flows freely without segregation so as tosurround the reinforcement and displace the bentonite

Cantilever walls are suitable for only moderate height and itis preferable not to use cantilever walls when services orfoundations are located wholly or partly within the active soilzone since horizontal and vertical movement in the retainedmaterial can cause damage Anchored or propped walls canhave one or more levels of anchor or prop in the upper partof the wall They can be designed to have fixed or free earthsupport at the bottom as stability is derived mainly from theanchorages or props

Traditional methods of design although widely used allhave recognised shortcomings These methods are outlined inannex B of BS 8002 where comments are included on theapplicability and limitations of each method The design ofembedded walls is beyond the scope of this Handbook and forfurther information the reader should refer to BS 8002Highways Agency BD 4294 and ref 62

Foundations ground slabs retaining walls culverts and subways70

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74 CULVERTS AND SUBWAYS

Concrete culverts which can be either cast in situ or precastare usually of circular or rectangular cross section Box typestructures can also be used to form subways cattle creeps orbridges over minor roads

741 Pipe culverts

For conducting small streams or drains under embankmentsculverts can be built with precast reinforced concrete pipeswhich must be strong enough to resist vertical and horizontalpressures from the earth and other superimposed loads Thepipes should be laid on a bed of concrete and where passingunder a road should be surrounded with reinforced concrete atleast 150 mm thick The culvert should also be reinforced toresist longitudinal bending resulting from unequal vertical earthpressure and unequal settlement Due to the uncertaintyassociated with the magnitude and disposition of the earth pres-sures an accurate analysis of the bending moments is imprac-ticable A basic guide is to take the positive moments at the topand bottom of the pipe and the negative moments at the endsof a horizontal diameter as 00625qd 2 where d is the diameterof the circular pipe and q is the intensity of both the downwardpressure on the top and the upward pressure at the bottomassuming the pressure to be distributed uniformly on ahorizontal plane

742 Box culverts

The load on the top of a box culvert includes the weights of theearth covering and the top slab and the imposed load (if any)The weights of the walls and top slab (and any load that is onthem) produce an upward reaction from the ground Theweights of the bottom slab and water in the culvert are carrieddirectly on the ground below the slab and thus have no effectother than their contribution to the total bearing pressure Thehorizontal pressure due to the water in the culvert produces aninternal triangular load on the walls or a trapezoidal load if thesurface of the water outside the culvert is above the top whenthere will also be an upward pressure on the underside of thetop slab The magnitude and distribution of the earth pressureagainst the sides of the culvert can be calculated in accordancewith the information in section 91 consideration being givento the risk of the ground becoming waterlogged resulting inincreased pressure and the possibility of flotation Generallythere are only two load conditions to consider

1 Culvert empty maximum load on top slab weight of thewalls and maximum earth pressure on walls

2 Culvert full minimum load on top slab weight of the wallsminimum earth pressure and maximum internal hydrostaticpressure on walls (with possible upward pressure on top slab)

In some circumstances these conditions may not produce themaximum load effects at any particular section and the effectof every probable combination should be considered The crosssections should be designed for the combined effects of axialforce bending and shear as appropriate A simplistic analysiscan be used to determine the bending moments produced in amonolithic rectangular box by considering the four slabs asa continuous beam of four spans with equal moments at the end

supports However if the bending of the bottom slab tends toproduce a downward deflection the compressibility of theground and the consequent effect on the bending moments mustbe taken into account The loads can be conveniently dividedinto the following cases

1 A uniformly distributed load on the top slab and a uniformreaction from the ground under the bottom slab

2 A concentrated imposed load on the top slab and a uniformreaction from the ground under the bottom slab

3 Concentrated loads due to the weight of each wall and auniform reaction from the ground under the bottom slab

4 A triangular distributed horizontal pressure on each wall dueto the increase in earth pressure in the height of the wall

5 A uniformly distributed horizontal pressure on each walldue to pressure from the earth and any surcharge above thelevel of the top slab

6 Internal horizontal and possibly vertical pressures due towater in the culvert

Formulae for the bending moments at the corners of thebox due to each load case when the top and bottom slabsare the same thickness are given in Table 287 The limitingground conditions associated with the formulae should benoted

743 Subways

The design and construction of buried box type structureswhich could be complete boxes portal frames or structureswhere the walls are propped by the top slab are covered byrecommendations in Highways Agency standard BD 3187These recommendations do not apply to structures that areinstalled by methods such as thrust boring or pipe jacking

The nominal superimposed dead load consists of the weightof any road construction materials and the soil cover abovethe structure Due to negative arching of the fill material thestructure can be subjected to loads greater than the weight offill directly above it An allowance for this effect is made byconsidering a minimum load based on the weight of materialdirectly above the structure and a maximum load equal to theminimum load multiplied by 115 The nominal horizontalearth pressures on the walls of the box structure are based ona triangular distribution with the value of the earth pressurecoefficient taken as a maximum of 06 and a minimum of 02It is to be assumed that either the maximum or the minimumvalue can be applied to one wall irrespective of the value thatis applied to the other wall

Where the depth of cover measured from the finished roadsurface to the top of the structure is greater than 600 mm thenominal vertical live loads to be considered are the HA wheelload and the HB vehicle To determine the nominal vertical liveload pressure dispersion of the wheel loads may be taken tooccur from the contact area on the carriageway to the top ofthe structure at a slope of 2 vertically to 1 horizontally Forstructures where the depth of cover is in the range 200ndash600 mmfull highway loading is to be considered For HA load the KELmay be dispersed below the depth of 200 mm from the finishedroad surface Details of the nominal vertical live loads are givenin sections 248 and 249 and Table 25

Culverts and subways 71

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Part 2

Loads materials andstructures

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In this chapter unless otherwise stated all loads are given ascharacteristic or nominal (ie unfactored) values For designpurposes each value must be multiplied by the appropriatepartial safety factor for the particular load load combinationand limit-state being considered

Although unit weights of materials should be given strictlyin terms of mass per unit volume (eg kgm3) the designer isusually only concerned with the resulting gravitational forcesTo avoid the need for repetitive conversion unit weights aremore conveniently expressed in terms of force (eg kNm3)where 1 kN may be taken as 102 kilograms

81 DEAD LOAD

The data for the weights of construction materials given in thefollowing tables has been taken mainly from EC 1 Part 11 butalso from other sources such as BS 648

811 Concrete

The primary dead load is usually the weight of the concretestructure The weight of reinforced concrete varies with thedensity of the aggregate and the percentage of reinforcementIn UK practice a value of 24 kNm3 has traditionally beenused for normal weight concrete with normal percentages ofreinforcement but a value of 25 kNm3 is recommended inEC 1 Several typical weights for normal lightweight andheavyweight (as used for kentledge and nuclear-radiationshielding) concretes are given in Table 21 Weights are alsogiven for various forms and depths of concrete slabs

812 Other construction materials and finishes

Dead loads include such permanent weights as those of thefinishes and linings on walls floors stairs ceilings and roofsasphalt and other applied waterproofing layers partitionsdoors windows roof and pavement lights superstructures ofsteelwork masonry or timber concrete bases for machineryand tanks fillings of earth sand plain concrete or hardcorecork and other insulating materials rail tracks and ballastingrefractory linings and road surfacing In Table 21 the basicweights of various structural and other materials includingmetals stone timber and rail tracks are given

The average equivalent weights of various cladding types asgiven in Table 22 are useful in estimating the loads imposed

on a concrete substructure The weights of walls of variousconstructions are also given in Table 22 Where a concretelintel supports a brick wall it is generally not necessary toconsider the lintel as supporting the entire wall above it issufficient to allow only for the triangular areas indicated in thediagrams in Table 22

813 Partitions

The weight of a partition is determined by the material of whichit is made and the storey height When the position of thepartition is not known or the use of demountable partitions isenvisaged the equivalent uniformly distributed load given inTable 22 should be considered as an imposed load in the designof the supporting floor slabs

Weights of permanent partitions whose position is knownshould be included in the dead load Where the length of the par-tition is in the direction of span of the slab an equivalent UDLmay be used as given in Table 22 In the case of brick or similarlybonded partitions continuous over the slab supports some reliefof loading on the slab will occur due to the arching action of thepartition unless this is invalidated by the presence of doorways orother openings Where the partition is at right angles to the spanof the slab a concentrated line load should be applied at the appro-priate position The slab should then be designed for the combinedeffect of the distributed floor load and the concentrated load

82 IMPOSED LOADS

Imposed loads on structures include the weights of storedmaterials and the loads resulting from occupancy and trafficComprehensive data regarding the weights of stored materialsassociated with building industry and agriculture are given inEC 1 Part 11 Data for loads on floors due to livestock andagricultural vehicles are given in BS 5502 Part 22

821 Imposed loads on buildings

Data for the vertical loads on floors and horizontal loads onparapets barriers and balustrades are given in BS 6399 Part 1Loads are given in relation to the type of activityoccupancy forwhich the floor area will be used in service as follows

A Domestic and residential activitiesB Office and work areas not covered elsewhere

Chapter 8

Loads

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21Weights of construction materials and concrete floor slabs

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Weights of roofs and walls 22

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C Areas where people may congregateD Shopping areasE Areas susceptible to the accumulation of goodsFG Vehicle and traffic areas

Details of the imposed loads for categories A and B are givenin Table 23 Values are given for uniformly distributed andconcentrated loads These are not to be taken together butconsidered as two separate load cases The concentrated loadsnormally do not need to be considered for solid or other slabsthat are capable of effective lateral distribution When used forcalculating local effects such as bearing or the punching of thinflanges a square contact area of 50 mm side should be assumedin the absence of any other specific information

With certain exceptions the imposed loads on beams may bereduced according to the area of floor supported Loads oncolumns and foundations may be reduced according to eitherthe area of floor or the number of storeys supported Details ofthe reductions and the exceptions are given in Table 23

Data given in Table 24 for the load on flat or mono-pitchroofs has been taken from BS 6399 Part 3 The loads whichare additional to all surfacing materials include for snow andother incidental loads but exclude wind pressure For other roofshapes and the effects of local drifting of snow behind parapetsreference should be made to BS 6399 Part 3

For building structures designed to meet the requirementsof EC 2 Part 1 details of imposed and snow loads are given inEC 1 Parts 11 and 13 respectively

822 Imposed loads on highway bridges

The data given in Table 25 for the imposed load on highwaybridges have been taken from the Highways Agency documentBD 3701 Type HA loading consists of two parts a uniformload whose value varies with the lsquoloaded lengthrsquo and a singleKEL that is positioned so as to have the most severe effectThe loaded length is the length over which the application of theload increases the effect to be determined Influence lines maybe needed to determine critical loaded lengths for continuousspans and arches Loading is applied to one or more notionallanes and multiplied by appropriate lane factors The alternativeof a single wheel load also needs to be considered in certaincircumstances

Type HB is a unit loading represented by a 16ndashwheel vehicleof variable bogie spacing where one unit of loading is equivalentto 40 kN The number of units considered for a public highwayis normally between 30 and 45 according to the appropriateauthority The vehicle can be placed in any transverse positionon the carriageway displacing HA loading over a specified areasurrounding the vehicle

For further information on the application of combined HAand HB loading and details of other loads to be considered on

highway bridges reference should be made to BD 3701 andBD 6094 For information on loads to be considered for theassessment of existing highway bridges reference should bemade to BD 2101

823 Imposed loads on footbridges

The data given in Table 26 for the imposed load on bridges dueto pedestrian traffic have been taken from the Highways Agencydocument BD 3701 For further information on the pedestrianloading to be considered on elements of highway or railwaybridges that also support footways or cycle tracks and the ser-viceability vibration requirements of footbridges referenceshould be made to BD 3701

824 Imposed loads on railway bridges

The data given in Table 26 for the imposed load on railwaybridges has been taken from the Highways Agency documentBD 3701 Types RU and SW0 apply to main line railwaystype SW0 being considered as an additional and separate loadcase for continuous bridges For bridges with one or two tracksloads are to be applied to each track In other cases loads areto be applied as specified by the relevant authority

Type RL applies to passenger rapid transit railway systemswhere main line locomotives and rolling stock do not operateThe loading consists of a uniform load (or loads dependent onloaded length) combined with a single concentrated load posi-tioned so as to have the most severe effect The loading is to beapplied to each and every track An arrangement of two con-centrated loads is also to be considered for deck elementswhere this would have a more severe effect

For information on other loads to be considered on railwaybridges reference should be made BD 3701

83 WIND LOADS

The data given in Tables 27ndash29 for the wind loading onbuildings has been taken from the information given for thestandard method of design in BS 6399 Part 2 The effectivewind speed is determined from Table 27 Wind pressures andforces on rectangular buildings as defined in Table 28 aredetermined by using standard pressure coefficients given inTable 29 For data on other building shapes and different roofforms and details of the directional method of design referenceshould be made to BS 6399 Part 2

Details of the method used to assess wind loads on bridgestructures and the data to be used for effective wind speeds anddrag coefficients are given in BD 3701 For designs to EC 2wind loads are given in EC 1 Part 12

Loads78

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Imposed loads on floors of buildings 23

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Imposed loads on roofs of buildings 24

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Imposed loads on bridges ndash 1 25

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Imposed loads on bridges ndash 2 26

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Wind speeds (standard method of design) 27

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Wind pressures and forces (standard method of design) 28

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Pressure coefficients and size effect factors forrectangular buildings 29

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Chapter 9

Pressures due toretained materials

In this chapter unless otherwise stated all unit weights andother properties of materials are given as characteristic or rep-resentative (ie unfactored) values For design purposes eachvalue must be modified by appropriate partial safety or mobili-sation factors according to the basis of design and the code ofpractice employed

91 EARTH PRESSURES

The data given in Table 210 for the properties of soils has beentaken from BS 8002 Design values of earth pressure coeffi-cients are based on the design soil strength which is taken asthe lower of the peak soil strength reduced by a mobilisationfactor or the critical state strength

911 Pressures imposed by cohesionless soils

For the walls shown in Table 211 with a uniform normallyconsolidated soil a uniformly distributed surcharge and nowater pressure the pressure imposed on the wall increaseslinearly with depth and is given by

K(zq)

where is unit weight of soil z is depth below surface q issurcharge pressure (kNm2) K is at-rest active or passivecoefficient of earth pressure according to design conditions

A minimum live load surcharge of 10 kNm2 is specified inBS 8002 This may be reasonable for walls 5 m high and abovebut appears to be too large for low walls In this case valuessuch as 4 kNm2 for walls up to 2 m high 6 kNm2 for walls 3 mhigh and 8 kNm2 for walls 4 m high could be used InBD 3701 surcharge loads are given of 5 kNm2 for footpaths10 kNm2 for HA loading 12 kNm2 for 30 units of HB loading20 kNm2 for 45 units of HB loading and on areas occupiedby rail tracks 30 kNm2 for RL loading and 50 kNm2 forRU loading

If static ground water occurs at depth zw below the surfacethe total pressure imposed at z zw is given by

K [mz (s w)(zzw)q]w(zzw)

where m is moist bulk weight of soil s is saturated bulkweight of soil w is unit weight of water (981 kNm3)

912 At-rest pressures

For a level ground surface and a normally consolidated soilthat has not been subjected to removal of overburden thehorizontal earth pressure coefficient is given by

Ko 1sin

where is effective angle of shearing resistance of soilCompaction of the soil will result in earth pressures in the

upper layers of the soil mass that are higher than those givenby the above equation The diagram and equations given inTable 211 can be used to calculate the maximum horizontalpressure induced by the compaction of successive layers ofbackfill and determine the resultant earth pressure diagramThe effective line load for dead weight compaction rollers is theweight of the roller divided by its width For vibratory rollersthe dead weight of the roller plus the centrifugal force causedby the vibrating mechanism should be used The DOESpecification limits the mass of the roller to be used within 2 mof a wall to 1300 kgm

For a vertical wall retaining backfill with a ground surfacethat slopes upwards the horizontal earth pressure coefficientmay be taken as

Ko (1sin)(1sin)

where is slope angle The resultant pressure which acts in adirection parallel to the ground surface is given by

o Koz cos

913 Active pressures

Rankinersquos theory may be used to calculate the pressure on a ver-tical plane referred to as the lsquovirtual backrsquo of the wall For avertical wall and a level ground surface the Rankine horizontalearth pressure coefficient is given by

The solution applies particularly to the case of a smooth wall ora wall with no relative movement between the soil mass andthe back of the wall The charts given in Table 212 which arebased on the work of Caquot and Kerisel may be used generallyfor vertical walls with sloping ground or inclined walls with

Ka 1 sin

1 sin

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Properties of soils 210

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Earth pressure distributions on rigid walls 211

At-rest state for rigid wall Effects of soil compaction

Active state for rigid wall free to rotate about base or translate

Passive state for rigid wall free to rotate about base or translate

hc 1K

2Q1

zc K2Q1

c 2Q1

K earth pressure coefficientK o for unyielding structureK a for wall free to mobilise

fully active stateQ1 intensity of effective line load

imposed by compaction plant unit weight of soil maximum horizontal earth

pressure induced by compaction

Horizontal earth pressure distribution resulting from compaction

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Active earth pressure coefficients 212

Angle of Shearing Resistance f (degrees)

Coe

ffici

ent o

f Act

ive

Pre

ssur

e K

aC

oeffi

cien

t of A

ctiv

e P

ress

ure

Ka

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level ground The horizontal and vertical components of resultantpressure are given by

ah Kazcos() and av Kazsin()

where is wall inclination to vertical (positive or negative) isselected angle of wall friction (taken as positive)

914 Passive pressures

For a vertical wall and a level ground surface the Rankinehorizontal earth pressure coefficient is given by

The solution applies particularly to the case of a smooth wall ora wall with no relative movement between the soil mass and theback of the wall The charts given in Table 213 for verticalwalls with sloping ground and in Table 214 for inclined wallswith level ground are based on the work of Caquot and KeriselThe horizontal and vertical components of resultant pressureare given by

ph Kpzcos() and pv Kpzsin()

where is wall inclination to vertical (positive or negative) isselected angle of wall friction (taken as negative)

915 Cohesive soils

If a secant value (c 0) is selected the procedures given forcohesionless soils apply If tangent parameters (c ) are to beused the RankinendashBell equations may be used as follows

a Ka(zq)2cradicKa

p Kp(zq)2cradicKp

where c is effective cohesion The active earth pressure istheoretically negative to a depth given by

zo (2cradicKa q)

Where cracks which may form in the tension zone can becomefilled with water full hydrostatic pressure should be consideredover the depth zo If the surface is protected so that no surfacewater can accumulate in the tension cracks the earth pressureshould be taken as zero over the depth zo

916 Further considerations

For considerations such as earth pressures on embedded walls(with or without props) the effects of vertical concentratedloads and line loads and the effects of groundwater seepagereference should be made to BS 8002 For the pressures to beconsidered in the design of integral bridge abutments as aresult of thermal movements of the deck reference should bemade to Highways Agency document BA 4296

92 TANKS

The pressure imposed by a contained liquid is given by

wz

Kp 1 sin

1 sin

where w is unit weight of liquid (see EC 1 Part 11) and z isdepth below surface For a fully submerged granular materialthe total horizontal pressure on the walls is

K(w)(zzo)wz

where is unit weight of the material (including voids) zo isdepth to top of material K is material pressure coefficient If o

is unit weight of material (excluding voids) o(1 e)where e is ratio of volume of voids to volume of solids

The preceding equation applies to materials such as coal orbroken stone with an effective angle of shearing resistancewhen submerged of approximately 35o For submerged sand Kshould be taken as unity If the material floats (o w) thesimple hydrostatic pressure applies

93 SILOS

The data given in Tables 215 and 216 has been taken fromEurocode 1 Part 4 The pressures apply to silos of the formsshown in Table 215 subject to the following limitations

dimensions dc 50 m h 100 m hdc 10

eccentricities ei 025dc eo 025dc with no part of outlet ata distance greater than 03dc from centreline of silo

filling involves negligible inertia effects and impact loads

stored material is free-flowing (cohesion is less than 4 kPa fora sample pre-consolidated to 100 kPa) with a maximumparticle size not greater than 03dc

transition between vertical walled section and hopper is on asingle horizontal plane

Dimensions h ho h1 and z are measured from the equivalentsurface which is a level surface giving the same volume ofstored material as the actual surface at the maximum filling

Loads acting on a hopper are shown in Table 215 where thetensile force at the top of the hopper is required for the designof silo supports or a ring beam at the transition level The verticalcomponent of the force can be determined from force equilib-rium incorporating the vertical surcharge Cbpvo at the transitionlevel and the weight of the hopper contents The discharge loadon the hopper wall is affected by the flow pattern of the storedmaterial which may be mass flow or funnel flow according tothe characteristics of the hopper and the material The normalload due to pn is supplemented for mass flow silos only by akick load due to ps

Values of material properties and expressions to determineresulting pressures in the vertical walled and bottom sections ofa silo are given in Table 216 For squat silos (hdc 15) thehorizontal pressure ph may be reduced to zero at the level wherethe upper surface of the stored material meets the silo wallBelow this level a linear pressure variation may be assumedtaking K 10 until this pressure reaches the value appropriateto the depth z below the equivalent surface

Homogenizing silos and silos containing powders in whichthe speed of the rising surface of the material exceeds 10 mhshould be designed for both the fluidised and non-fluidised con-ditions For the fluidised condition the bulk unit weight of thematerial may be taken as 08 For information on test methodsto determine the properties of particulate materials referenceshould be made to EC 1 Part 4

Pressures due to retained materials90

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Passive earth pressure coefficents ndash 1 213

Angle of Shearing Resistance f (degrees)

Coe

ffici

ent o

f Pas

sive

Pre

ssur

e K

pR

educ

tion

Fac

tor

Rd

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214Passive earth pressure coefficents ndash 2

Angle of Shearing Resistance f (degrees)

Coe

ffici

ent o

f Pas

sive

Pre

ssur

e K

pR

educ

tion

Fac

tor

Rd

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215Silos ndash 1

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216Silos ndash 2

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101 CONSTITUENTS OF CONCRETE

1011 Cements and combinations

Manufactured cements are those made in a cement factoryWhere a mineral material is included it is generally added tothe cement clinker at the grinding stage The notation used forthese manufactured cements contains the prefix letters CEMWhen a concrete producer adds an addition such as pfa or ggbsto CEM I Portland cement in the mixer the resulting cement isknown as a mixer combination and is denoted by the prefixletter C Cements and combinations in general use are listed inTable 217 Further information on the different types and useof cements is given in section 311

1012 Aggregates

Overall grading limits for coarse and fine aggregates fromnatural sources in accordance with BS EN 12620 are given inTable 217 Further information is given in section 312

102 EARLY-AGE TEMPERATURES OF CONCRETE

The calculation of early thermal crack widths in a restrainedconcrete element requires knowledge of the temperature rise dueto the cement hydration Some typical early temperature histo-ries of various concrete walls and predicted temperature risesfor different cements are given in Table 218

The predicted temperature rise values for Portland cementconcretes in walls and slabs are taken from BS 8007 These aremaximum values selected from a range of values for Portlandcements obtained from different works (ref 11) The tempera-ture rises given for the other cements in concrete sections witha minimum dimension of 1 m should be taken as indicativeonly but could be used where other specific information isnot available

103 REINFORCEMENT

Reinforcement for concrete generally consists of steel bars orwelded steel mesh fabric that depend upon the provision of adurable concrete cover for protection against corrosion Theessential properties of bars to BS 4449 and wires to BS 4482both of which are in general conformity with BS EN 10080

are given in Table 219 For additional information on themanufacture and properties of steel reinforcement includingstainless steel refer to section 32

1031 Bars

Bars for normal use produced in the United Kingdom arehot-rolled to a characteristic strength of 500 MPa and achieveClass B or C ductility The bars are round in cross section withsets of parallel transverse ribs separated by longitudinal ribsThe nominal size is the diameter of a circle with an area equalto the effective cross-sectional area of the bar The maximumoverall size is approximately 15 greater than the nominal sizeValues of the total cross-sectional area provided in a concretesection according to the number or spacing of the bars fordifferent bar sizes are given in Table 220

The type and grade of reinforcement is designated as follows

Chapter 10

Concrete andreinforcement

Type of steel reinforcement Notation

Grade B500A B500B or B500C to BS 4449 HGrade B500A to BS 4449 AGrade B500B or B500C to BS 4449 BGrade B500C to BS 4449 CA specified grade and type of ribbed Sstainless steel to BS 6744

Reinforcement of a type not included above but Xwith material properties defined in the designor contract specification

Note In the description B500A and so on B indicates reinforcing steel

1032 Fabric

In the United Kingdom steel fabric reinforcement is generallyproduced to the requirements of BS 4483 using ribbed bar inaccordance with BS 4449 The exception is wrapping fabricwhere wire in accordance with BS 4482 may be used Fabric isproduced in a range of standard types or can be purpose-madeto the clientrsquos requirements Full details of the standard fabrictypes are given in Table 220

Type A is a square mesh with identical longitudinal barsand cross bars commonly used in ground slabs to provide aminimum amount of reinforcement in two directions Type B isa rectangular (structural) mesh that is particularly suitable for

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217Concrete cements and aggregate grading

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Concrete early-age temperatures 218

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219Reinforcement general properties

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Reinforcement cross-sectional areas of bars and fabric 220

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thin one-way spanning slabs where the longitudinal barsprovide the main reinforcement with the cross bars beingsufficient to meet the minimum requirements for secondaryreinforcement Type C is a rectangular (long) mesh where thecross bars are minimal which can be used in any solid slabincluding column bases by providing a separate sheet in eachdirection Type D is a rectangular (wrapping) mesh that is usedin the concrete encasement of structural steel sections

Standard fabric is normally supplied in stock sheet sizes of48 m (longitudinal bars) 24 m (cross bars) with end over-hangs equal to 05 the pitch of the perpendicular bar Tofacilitate fixing and to avoid a build-up of bar layers at the lapssheets with increased overhangs (flying ends) can be suppliedto order Sheets can also be supplied cut to size in lengths upto 12 m and prebent For guidance on the use of purpose-madefabrics reference should be made to BS 8666

1033 Cutting and bending tolerances

Bars are produced in stock lengths of 12 m and lengths up to18 m can be supplied to special order In most structures barsare required in shorter lengths and often need to be bent Thecutting and bending of reinforcement is generally specified tothe requirements of BS 8666 The tolerances on cutting andbending dimensions are as follows

For shape code 67 when the radius exceeds the value in thefollowing table straight bars will be supplied as the requiredcurvature can be obtained during fixing

Concrete and reinforcement100

Cutting and bending processes Tolerance (mm)

Cutting of straight lengths 25

Bending dimension (mm) 1000 5 1000 and 2000 5 10

2000 5 25

Length of wires in fabric L 5000 25L 5000 L 200

1034 Shape codes and bending dimensions

BS 8666 contains details of bar shapes designated by shapecodes as given in Tables 221 and 222 The information neededto cut and bend the bars to the required dimensions is enteredinto a bar schedule an example of which is shown in Table 223The standard shapes should be used wherever possible with therelevant dimensions entered into columns A to E of the barschedule All other shapes should be given a shape code 99 witha dimensioned sketch drawn over the columns A to E using twoparallel lines to indicate the bar thickness One of the bar dimen-sions should be indicated in parenthesis as a free dimensionDimensions should be given as a multiple of 5 mm and thetotal length determined in accordance with the equation givenin the table rounded up to a multiple of 25 mm To facilitatetransportation each bent bar should fit within an imaginaryrectangle the shorter side of which is not longer than 2750 mm

Most of the shape codes cater for bars bent to the minimumradius taken as 2d for d 16 and 35d for d 20 where d isthe bar size The minimum straight length needed beyond theend of the curved portion of a bend is 5d for a bob and 10d formost links For each bar size values of the minimum radius rand the minimum end projection P needed to form a bend aregiven in Table 219

Bars needing larger radius bends denoted by R except forshape codes 12 and 67 should be treated as a shape code 99

Maximum limit for which a preformed radius is required

Bar size (mm) 8 10 12 16

Radius (m) 275 35 425 75

Bar size (mm) 20 25 32 40

Radius (m) 14 30 43 58

For shape codes 12 13 22 and 33 the largest practical radiusfor producing a continuous curve is 200 mm and for a largerradius a series of short straight sections may be formed

1035 Deductions for variations

Cover to reinforcement is liable to variation due to the effect ofinevitable errors in the dimensions of formwork and in thecutting bending and fixing of the bars In cases where a bar isdetailed to fit between two concrete faces with no more thanthe nominal cover on each face (eg link in a beam) an appro-priate allowance for deviations should be applied The relevantdimension on the schedule should be determined as the nominaldimension of the concrete less the nominal cover on each faceless an allowance for deviations as follows

Total deductions to allow for permissible deviations onmember size and in cutting and bending of bars

Type of bar Distance between faces of Deductionconcrete member mm

Links and other Not more than 1 m 10bent bars Between 1 m and 2 m 15

Over 2 m 20

Straight bars Any length 40

The deductions recommended in the forgoing table are taken fromBS 8110 and allow for deviations on the member size of 5 mm fordimensions up to 2 m and 10 mm for dimensions over 2 m Wherethe permissible deviations on member size exceed these valueslarger deductions should be made or the cover increased

Example Determine the relevant bending dimensions for thebars shown in the following beam detail The completed barschedule for 6 beams thus is given in Table 223

Section A-A

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Reinforcement 101

Elevation of beam

Bar Shape Dimensions (mm) Length (mm) ndash see Tables 229 and 230mark code

01 00 L 8000 2 200 7600

02 00 L 8000 2 1250 5500

03 13 Bar requires radius of bend R 6d as a designrequirement This necessitates the use of shapecode 13 as dimension B would not provide 4dlength of straight between two bends L 1800 057 420 1300 16 25 3300

A 1800 C 1300 (design requirements)B 450 (2 10) 10 420

Note Dimension B is derived from dimension Aof bar mark 05 and includes a further deductionof 10 mm for tolerances on cutting and bending

04 00 L 8000 2 200 7600

05 51 A 500 (2 20) 10 450 L 2 (450 250 115) (25 16) (5 8) 1550B 300 (2 20) 10 250C D 115 (P in Table 219)r 16 (Table 219)Note Dimensions A and B include deductionsof 10 mm for permissible deviations

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221Reinforcement standard bar shapes and methodof measurement ndash 1

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222Reinforcement standard bar shapes and methodof measurement ndash 2

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Reinforcement typical bar schedule 223

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The formulae and coefficients in this chapter give values ofshearing forces bending moments slopes and deflections interms of the total load on the member For design purposes theload F must include the appropriate partial safety factors for thelimit-state being considered

111 SIMPLE BEAMS AND CANTILEVERS

The formulae for the reactions shearing forces and bendingmoments in freely supported beams (Tables 224 and 225) andsimple cantilevers (Tables 226 and 227) are obtained by therules of static equilibrium The slope and deflection formulaefor freely supported beams and simple cantilevers and all theformulae for propped cantilevers (Table 227) are for elasticbehaviour and members of constant cross section

112 BEAMS FIXED AT BOTH ENDS

The bending moments on a beam fixed at both ends can bederived from the principle that the area of the free-momentdiagram (ie the bending moment diagram due to the same loadimposed on a freely supported beam of equal span) is equal tothe area of the restraint-moment diagram Also the centresof area of the two diagrams are in the same vertical line Theshape of the free-moment diagram depends upon the particularcharacteristics of the imposed load but the restraint-momentdiagram is a trapezium For loads that are symmetrically disposedon the beam the centre of area of the free-moment diagram isat the mid-point of the span and thus the restraint-momentdiagram is a rectangle giving a restraint moment at eachsupport equal to the mean height of the free-moment diagram

The amount of shearing force in a beam with one or bothends fixed is calculated from the variation of the bendingmoment along the beam The shearing force resulting from therestraint moment alone is constant throughout the length of thebeam and equal to the difference between the two end-momentsdivided by the span (ie the rate of change of the restraintmoment) This shearing force is algebraically added to theshearing force due to the imposed load with the beam takenas freely supported Thus the support reaction is the sum (ordifference) of the restraint-moment shearing force and the free-moment shearing force For a beam that is symmetricallyloaded with both ends fixed the restraint moment at each endis the same and the shearing forces are identical to those for thesame beam freely supported The support reactions are bothequal to one-half of the total load on the span The formulae forthe reactions shearing forces bending moments slopes anddeflections for fully fixed spans (Table 225) are for elasticbehaviour and members of constant cross section

1121 Fixed-end moment coefficients

Fixed-end moment coefficients CAB and CBA are given inTable 228 for a variety of unsymmetrical and symmetricalimposed loadings on beams of constant cross section Morecomplex loading arrangements can generally be formed as acombination of the cases shown and the resulting fixed-endmoments found by superposition A full range of charts iscontained in Examples of the Design of Reinforced ConcreteBuildings for a member with a partial uniform or triangulardistribution of load placed anywhere within the span

Chapter 11

Cantilevers andsingle-span beams

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224Moments shears deflections general case for beams

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Moments shears deflections special cases for beams 225

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226Moments shears deflections general cases for cantilevers

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Moments shears deflections special cases for cantilevers 227

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228Fixed-end moment coefficients general data

The fixed-end moment coefficients CAB and CBA can be used as follows1 To obtain bending moments at supports of single-span beams fully fixed at both ends (Table 225)

MAB CABlAB and MBA CBAlAB (With symmetrical load MAB MBA)2 To obtain fixed-end moments for analysis of continuous beams by moment distribution methods (Table 236)

FEMAB CABlAB and FEMBA CBAlAB (With symmetrical load FEMAB FEMBA)3 To obtain loading factors for analysis of framed structures by slope-deflection methods (Table 260)

FAB CABlAB and FBA CBAlAB (With symmetrical load FAB FBA)4 To obtain loading factors for analysis of portal frames (Tables 263 and 264)

and z1 (With symmetrical loading CAB CBA and z1 05)=CAB 2CBA

2(CAB 2CBA)D

CAB CBA

2lAB

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The formulae and coefficients in this chapter give values ofshearing forces and bending moments in terms of the dead andlive loads on the member For design purposes these loads mustinclude the appropriate partial safety factors for the limit-stateconsidered and the Code of Practice employed

For the ULS the dead load factors are treated differently inBS 8110 and EC 2 For designs to BS 8110 values of either 14or 10 are applied separately to each span of the beam Fordesigns to EC 2 values of either 135 or 10 are applied to all thespans If the beam ends with a cantilever the effect of applyingvalues of either 135 or 115 separately to the cantilever and theadjacent span should also be considered Details of the designloads and of the effects of applying cantilever moments at oneor both ends of a continuous beam of two three four or fiveequal spans are given in Table 229

121 DETERMINATION OF MAXIMUM MOMENTS

1211 Incidence of live load

The values of the bending moments in the spans and at thesupports depend upon the incidence of the live load and forspans that are equal or approximately equal the dispositions oflive load shown in Table 229 give the maximum positivemoments in the spans and the maximum negative momentsat the supports Both BS 8110 and EC 2 consider a lesssevere incidence of live load when determining the maximumnegative moments at the supports According to BS 8110 theonly case that needs to be considered is when all the spans areloaded According to EC 2 all cases of two adjacent spansloaded should be considered but the loads shown as optionalin Table 229 may be ignored

Thus the maximum positive moments due to live load forthe system are obtained by considering two loading arrange-ments one with live load on all the odd-numbered spans andthe other with live load on all the even-numbered spans Fordesigns to BS 8110 the summation of the results for these twocases gives the maximum negative moments

It should be noted that for designs to EC 2 the UK NationalAnnex allows the BS 8110 loading arrangements to be used asan alternative to those recommended in the base document Inthis chapter the basic arrangements are used

1212 Shearing forces

The shearing forces in a continuous beam are determined byfirst considering each span as freely supported then addingalgebraically the rate of change of restraint moment for theparticular span Shearing forces for freely supported spans arereadily determined by the rules of static equilibrium The addi-tional shearing force which is constant throughout the span isequal to the difference in the support moments at each enddivided by the span

1213 Maximum positive moments

When the moments at the supports and the shearing forces havebeen determined the maximum positive moment in the spancan be obtained by first finding the position where the shearingforce is zero The maximum positive moment is then obtainedby subtracting the effect of the restraint moments which varieslinearly along the span from the freely supported moment atthis position

122 SOLUTIONS FOR EQUAL SPANS

1221 Coefficients for equal loads on equal spans

Approximate general solutions for the maximum bendingmoments and shearing forces in uniformly loaded beams ofthree or more spans are given in Table 229 Exact solutions forthe maximum bending moments in beams of two three fouror five equal spans are given in Tables 230 and 231 for eightdifferent load distributions The coefficients given for thesupport moments due to live load apply to the most onerousloading conditions For the less severe arrangements describedin section 1211 coefficients are shown in the square brackets [ ]for BS 8110 and the curved brackets ( ) for EC 2 The coefficientsin Table 232 enable the maximum shearing forces at thesupports to be determined

Example 1 Calculate the maximum ultimate moments inthe end and central spans and at the penultimate and interiorsupports for a beam continuous over five equal spans of 5 mwith characteristic dead and imposed loads of 20 kNm eachaccording to the requirements of BS 8110

Chapter 12

Continuous beams

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229Continuous beams general data

AKey B K A B J K

A B C H J KA B C

Adjustment to bending moment = M coefficient x applied bending momentAdjustment to shearing force = (V coefficent x applied bending moment)span

KJ

Optional

EC2 Consider live load on spans RS and ST only

Optional

R T U V

TS

S

To produce maximum positive moment in span ST

To produce maximum negative moment at support S

Simplifications BS8110 Consider live load on all spans

EC 2 EC 2

EC 2

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230Continuous beams moments from equal loadson equal spans ndash 1

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231Continuous beams moments from equal loadson equal spans ndash 2

Bending moment (coefficient) (total load on one span) (span)Bending moment coefficientsabove line apply to negative bending moment at supportsbelow line apply to positive bending moment in span

Coefficients apply when all spans are equal (may be used also whenshortest 85 longest) Loads on each loaded span are same

Second moment of area is same throughout all spansBending moment coefficients in square brackets (live load)apply if all spans are loaded (ie BS 8110 requirements)Bending moment coefficients in curved brackets (live load)apply if two adjacent spans are loaded (ie EC 2 requirements)

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232Continuous beams shears from equal loadson equal spans

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The design load consists of a dead load of 10gk 20 kNm anda live load of (04gk 16qk) 40 kNm Then from Table 230(using coefficients in square brackets for the live load) theultimate bending moments are as follows

Penultimate supportDead load 0105 20 52 525 kNm (negative)Imposed load 0105 40 52 1050 kNm (negative)Total 1575 kNm (negative)

Interior supportDead load 0079 20 52 395 kNm (negative)Imposed load 0079 40 52 790 kNm (negative)Total 1185 kNm (negative)

Near middle of end spanDead load 0078 20 52 390 kNm (positive)Imposed load 0100 40 52 1000 kNm (positive)Total 1390 kNm (positive)

Middle of central spanDead load 0046 20 52 230 kNm (positive)Imposed load 0086 40 52 860 kNm (positive)Total 1090 kNm (positive)

Example 2 Calculate the maximum ultimate moments forexample 1 according to the requirements of EC 2

The design load consists of a dead load of 135gk 27 kNmand a live load of 15qk 30 kNm Then from Table 230(using coefficients in curved brackets for the live load) theultimate bending moments are as follows

Penultimate supportDead load 0105 27 52 709 kNm (negative)Imposed load 0116 30 52 870 kNm (negative)Total 1579 kNm (negative)

Interior supportDead load 0079 27 52 533 kNm (negative)Imposed load 0106 30 52 795 kNm (negative)Total 1328 kNm (negative)

Near middle of end spanDead load 0078 27 52 527 kNm (positive)Imposed load 0100 30 52 750 kNm (positive)Total 1277 kNm (positive)

Middle of central spanDead load 0046 27 52 311 kNm (positive)Imposed load 0086 30 52 645 kNm (positive)Total 956 kNm (positive)

Example 3 Calculate the maximum ultimate moments forexample 1 according to the requirements of BS 8110 when a2 m long cantilever is provided at each end of the beamIncrease in moments due to dead and live loads on cantileversat critical positions is as follows

End supportDead load 05 20 22 400 kNm (negative)Imposed load 05 40 22 800 kNm (negative)Total 1200 kNm (negative)

Interior supportFrom Table 229 increase due to moments at end supports is0053 120 64 kNm (negative)

Decrease in moments due to dead load only on cantilevers atcritical positions is as follows

Penultimate supportFrom Table 229 decrease due to moments at end supports is0263 40 105 kNm (positive)

Middle of end spanFrom Table 229 decrease due to moments at end supports is[10 05(10 0263)] 40 147 (negative)

Middle of central spanFrom Table 229 decrease due to moments at end supports is0053 40 21 kNm (negative)

123 REDISTRIBUTION OF MOMENTS

As explained in section 422 for the ULS both BS 8110 andEC 2 permit the moments determined by a linear elastic analysisto be redistributed provided that the resulting distributionremains in equilibrium with the loads Although the conditionsaffecting the procedure are slightly different in the two codesthe general approach is to reduce the critical moments by achosen amount up to the maximum percentage permitted anddetermine the revised moments at other positions by equilibriumconsiderations

An important point to appreciate is that each particular loadcombination can be considered separately Thus if desired it ispossible to reduce the maximum moments in the spans and atthe supports For example the maximum support moments canbe reduced to values that are still greater than those that occurwith the maximum span moments The maximum spanmoments can then be reduced until the corresponding supportmoments are the same as the (reduced) maximum values

The principles of static equilibrium require that no changesshould be made to the moments in a cantilever or at a freelysupported end

1231 Code requirements

BS 8110 and EC 2 permit the maximum moments to be reducedby up to 30 provided that in the subsequent design of the rele-vant sections the depth of the neutral axis is limited accordingto the amount of redistribution (Note that there is no restrictionon the maximum percentage increase of moment) In EC 2 themaximum permitted reduction depends also on the ductilityof the reinforcement being 30 for reinforcement classes Band C but only 20 for class A

In BS 8110 it is stated that the ultimate resistance momentat a section should be at least 70 of the maximum momentat that section before redistribution In effect the process ofredistribution alters the positions of points of contra-flexureThe purpose of the code requirement is to ensure that at suchpoints on the diagram of redistributed moments (at which noflexural reinforcement is theoretically required) sufficientreinforcement is provided to cater for the moments that willoccur under service loading The redistribution procedure takesadvantage of the ability of continuous beams to develop plastichinges at critical sections prior to failure whilst also ensuringthat the response remains fully elastic under service loading Therequirements are discussed more fully in books on structuraldesign and in the Handbook to BS 8110

Continuous beams116

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1232 Redistribution procedure

The use of moment redistribution is illustrated in Table 233where a beam of three equal spans is examined in accordancewith the requirements of BS 8110 The uniformly distributeddead and live loads are each equal to 1200 units per span Themoment diagram for dead load on each span is shown in (a)Moment diagrams for the arrangements of live load that givethe maximum moments at the supports and in the spans areshown in (b) The moment envelope obtained by combiningthe diagrams for dead and live loads is shown in (c) (Notethat the vertical scale of diagrams (c)ndash(f) differs from thatof (a) and (b))

The redistribution procedure is normally used to reduce themaximum support moments One approach is to reduce these tothe values obtained when the span moments are greatest Thisis shown in (d) where the support moments have been reducedfrom 240 to 180 a reduction of 25 In this case no otheradjustment is needed to the moment envelope If the maximumsupport moments are reduced by 30 from 240 to 168 thespan moments must be increased as shown in (e) The 70requirement discussed in section 1132 determines the extentof the hogging region in the end span and the minimum valueof 21 in the middle span (For the load cases in EC2 themaximum support moments could be reduced by 30 from260 to 182 with no other adjustment needed to themoment envelope)

If the criterion is to reduce the maximum span moments inthe case of an up-stand beam say this may be achieved byincreasing the corresponding support moments This is shownin (f) where the maximum moment in the middle span has beenreduced by 30 from 120 to 84 by increasing the supportmoments from 180 to 216 (Note that the minimummoment in the middle span has increased from 30 to 66)The moment in the end span has also been reduced by 7 from217 to 202 The 70 requirement determines the extent of thesagging regions in both spans It is clear that any further reduc-tion of moment in the end span would result in a considerableincrease in the support moment For example a 30 reductionin the end span moment from 217 to 152 would increase thesupport moment to 346 and the minimum moment in themiddle span to 196

In view of the many factors involved it is difficult to give anygeneral rules as to whether to redistribute moments or by howmuch such decisions are basically matters of individual engi-neering judgement A useful approach is to first calculate theultimate resistance moments at the support sections provided bychosen arrangements of reinforcement and then redistribute themoment diagrams to suit The span sections can then be designedfor the resulting moments and a check made to ensure that allof the code requirements are satisfied Moment redistributionin general affects the shearing forces at the supports and it isrecommended that beam sections are designed for the greater ofthe shear forces calculated before and after redistribution

The use of moment distribution in systems where the beamsare analysed in conjunction with adjoining columns requiresfurther consideration In such cases it is important to ensure thatin any postulated collapse mechanism involving plastic hinges inthe columns these are the last hinges to form To this end it isrecommended that column sections should be designed for thegreater of the moments calculated before and after redistribution

1233 Bending moment diagrams

The moment diagrams and coefficients given in Tables 234and 235 cater for beams that are continuous over two threeand four or more equal spans They apply to cases where thesecond moment of area of the cross section is constant andthe loads on each loaded span are the same For conveniencecoefficients derived by elastic analysis before and after givenredistributions in accordance with the rules of both BS 8110and EC 2 are tabulated against the location points indicated inthe diagrams For example M12 is the coefficient correspondingto the maximum moment at the central support of a two-spanbeam while M13 is the coefficient that gives the moment at thissupport when the moment in the adjoining span is a maximumThus by means of the coefficients given the appropriate envelopeof maximum moments is obtained

Three load types are considered UDL throughout each spana central concentrated load and equal concentrated loadspositioned at the third-points of the span The span momentsdetermined by summing the individual maximum values givenseparately for dead and live loads in the case of uniformloading will be approximate but erring on the side of safetysince each maximum value occurs at a slightly different positionThe tabulated coefficients may also be used to determine thesupport moments resulting from combinations of the givenload types by summing the results for each type The corre-sponding span moments can then be determined as described insection 1213

Moment coefficients are given for redistribution values of10 and 30 respectively For the dead load all the supportmoments have been reduced by the full amount and the spanmoments increased to suit the adjusted values at the supportsFor the live loads all the support moments and for 10 redis-tribution all the span moments have been reduced by the fullamount For 30 redistribution each span moment has beenreduced to the minimum value required for equilibrium with thenew support moments As a result the BS 8110 span momentcoefficients are the same as those for the dead load Althoughthere is no particular merit in limiting redistribution to 10some BS 8110 design formulae for determining the ultimateresistance moment are related to this condition

For design purposes redistribution at a particular sectionrefers to the percentage change in the combined moment dueto the dead and live loads When using the tables the value forthe support moments will be either 10 or 30 but the valuesfor the span moments will need to be calculated for eachparticular case Consider for example a two-span beam sup-porting UDLs with gk qk and 30 redistribution accordingto the requirements of BS 8110

The design load consists of a dead load of 10gk and a liveload of (04gk 16qk) 20gk Then from Table 234 theultimate bending moments are as follows

Before redistributionM11 (0070gk 0096 2gk)l2 0262gkl2

After redistributionM11 (0085gk 0085 2gk)l2 0255gk l2

Redistribution 100 (0262 0255)0262 3

Thus a full 30 reduction of the maximum support moments isobtained with no increase in the maximum span moment

Redistribution of moments 117

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233Continuous beams moment redistribution

Uniformly distributed load (dead load = live load = 1200 units per span)

(b) Live load only Critical cases (for BS 8110)

(a) Dead load only

(c) Dead + live loads Critical cases

(d) Envelope obtained with some reduction of the support moments but with no increase of the span moments

(e) Envelope obtained with maximum reduction of the support moments and some increase of the span moments

(f) Envelope obtained with maximum reduction of moment in middle span and some reduction of moment in end spans

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234Continuous beams bending moment diagrams ndash 1

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235Continuous beams bending moment diagrams ndash 2

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124 ANALYSIS BY MOMENT DISTRIBUTION

The Hardy Cross moment distribution method of analysisin which support moments are derived by a step-by-stepprocess of successive approximations is described brieflyand shown by means of a worked example in Table 236 Themethod is able to accommodate span-to-span variationsin span length member size and loading arrangement Thelsquoprecise moment distributionrsquo method avoids the iterativeprocedure by using more complicated distribution and carry-over factors Span moments can be determined as describedin section 1213

For continuous beams of two three or four spans uniformcross-section and symmetrical loading the support momentsmay also be obtained by using the factors in Table 237

125 INFLUENCE LINES FOR CONTINUOUS BEAMS

The following procedure can be used to determine bendingmoments at chosen sections in a system of continuous beamsdue to a train of loads in any given position

1 Draw the beam system to a convenient scale

2 With the ordinates tabulated in the appropriate Table 238239 240 or 241 construct the influence line (for unit load)for the section being considered selecting a convenientscale for the bending moment

3 Plot on the influence line diagram the train of loads in whatis considered to be the most adverse position

4 Tabulate the value of (ordinate load) for each load

5 Add algebraically the values of (ordinate load) to obtainthe resultant bending moment at the section considered

6 Repeat for other positions of the load train to ensure that themost adverse position has been considered

The following example shows the direct use of the tabulatedinfluence lines for calculating the moments on a beam that iscontinuous over four spans with concentrated loads applied atspecified positions

Example Determine the bending moments at the penultimateleft-hand support of a system of four spans having a constantcross section and freely supported at the ends when loads of100 kN are applied at the mid-points of the first and third spansfrom the left-hand end The end spans are 8 m long and theinterior spans are 12 m long

The span ratio is 115151 and the ordinates are obtained fromTable 240 for penultimate support C

With load on first span (ordinate c)Bending moment (0082 100 8) 656 kNm

With load on third span (ordinate m)Bending moment (0035 100 8) 280 kNm

Net bending moment at penultimate support 376 kNm

Influence lines for continuous beams 121

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236Continuous beams moment distribution methods

HARDY CROSS MOMENT DISTRIBUTION

1 Consider each member to be fixed at ends calculate fixed-endmoments (FEMs) due to external loads on individual members bymeans of Table 228

2 Where members meet sum of bending moments must equalzero for equilibrium ie at B MBAMBC 0 Since FEM(ie FEMBA FEMBC) is unlikely to equal zero a balancingmoment ofFEM must be introduced at each support to achieveequilibrium

3 Distribute this balancing moment between members meeting ata joint in proportion to their relative stiffnesses K Il bymultiplying FEM by distribution factor D for each member(eg at B DBA KAB (KAB KBC) etc so that DBA DBC 1At a free end D 1 at a fully fixed end D0)

4 Applying a moment at one end of member induces momentof one-half of magnitude and of same sign at opposite endof member (termed carry-over) Thus distributed moment

FEM DBA at B of AB produces a moment of (12)FEM DBA at A and so on

5 These carried-over moments produce further unbalanced momentsat supports (eg moments carried over from A and C give rise tofurther moments at B) These must again be redistributed and thecarry-over process repeated

6 Repeat cycle of operations described in steps 2ndash5 until unbalancedmoments are negligible Then sum values obtained each sideof support

Various simplifications can be employed to shorten analysis Themost useful is that for dealing with a system that is freely supportedat the end If stiffness considered for end span when calculatingdistribution factors is taken as only three-quarters of actualstiffness and one-half of fixed-end moment at free support is addedto FEM at other end of span the span may then be treated as fixedand no further carrying over from free end back to penultimatesupport takes place

PRECISE MOMENT DISTRIBUTION

1 Calculate fixed-end moments (FEMs) as for Hardy Cross momentdistribution

2 Determine continuity factors for each span of system fromgeneral expression

where n is continuity factor for previous span and Kn and Kn1

are stiffnesses of two spans Work from left to right along system Ifleft-hand support (A in example below) is free take AB 0 for firstspan if A is fully fixed AB 05 (Intermediate fixity conditionsmay be assumed if desired by interpolation) Repeat the foregoingprocedure starting from right-hand end and working to left (to obtaincontinuity factor AB for span AB for example)

3 Calculate distribution factors (DFs) at junctions between spansfrom general expression

n1 12 Kn1

Kn(2n)

where AB and BA are continuity factors obtained in step 2Note that these distribution factors do not correspond to thoseused in Hardy Cross moment distribution Check that at eachsupport DF 1

4 Distribute the balancing moments FEM introduced at eachsupport to provide equilibrium for the unbalanced FEMs bymultiplying by the distribution factors obtained in step 3

5 Carry over the distributed balancing moments at the supportsby multiplying them by the continuity factors obtained instep 2 by working in opposite direction For example the momentcarried over from B to A is obtained by multiplying thedistributed moment at B by AB and so on This procedure isillustrated in example below Only a single carry-over operationin each direction is necessary

6 Sum values obtained to determine final moments

DFAB 12AB

1ABBA

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237Continuous beams unequal prismatic spans and loads

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238Continuous beams influence lines for two spans

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239Continuous beams influence lines for three spans

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240Continuous beams influence lines for four spans

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241Continuous beams influence lines for five or more spans

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Position Moment Shear

At outer support 0 040FNear middle of end span 008Fl mdashAt interior support 010Fl 060F

In monolithic building construction concrete floors can takevarious forms as shown in Table 242 Slabs can be solid orribbed and can span between beams in either one or two direc-tions or be supported directly by columns as a flat slab Slabelements occur also as decking in bridges and other forms ofplatform structures and as walling in rectangular tanks silosand other forms of retaining structures

131 ONE-WAY SLABS

For slabs carrying uniformly distributed load and continuousover three or more nearly equal spans approximate solutionsfor the ultimate bending moments and shearing forces for bothBS 8110 and EC 2 are given in Table 242 The supportmoments include an allowance for 20 redistribution in bothcases The differences in the values for the two codes occur asa result of the different load arrangements described in section441 However it should be noted that for designs to EC 2 theUK National Annex allows the use of the BS 8110 simplifiedload arrangement as an alternative to that recommended in thebase document For two equal spans the corresponding valuesfor both codes would be

serviceability requirements of cracking and deflection are metby compliance with simplified rules

1321 Uniformly loaded slabs (BS 8110 method)

For rectangular panels carrying uniformly distributed loadwhere the corners are prevented from lifting and adequateprovision is made for torsion the panel is consideredto be divided into middle and edge strips as shown inTable 242 The method may be used for continuous slabswhere the characteristic dead and imposed loads on adjacentpanels and the spans perpendicular to the lines of commonsupport are approximately the same as on the panel beingconsidered

The bending moments and shearing forces on the middlestrips for nine different panel types are given in Table 243Reinforcement meeting the minimum percentage requirementof the code should be provided in the edge strips At cornerswhere either one or both edges of the panel are discontinuoustorsion reinforcement is required This should consist of top andbottom reinforcement each with layers of bars placed parallelto the sides of the slab and extending from the edges a minimumdistance of one-fifth of the shorter span The area of reinforce-ment in each of the four layers as a proportion of that requiredfor the positive moment at mid-span should be three-quarterswhere both edges are discontinuous and three-eighths whereone edge is discontinuous At a discontinuous edge where theslab is monolithic with the support negative reinforcementequal to a half of that required for the positive moment atmid-span should be provided

Where because of differences between contiguous panelstwo different values are obtained for the negative moment at ashared continuous edge these values may be considered asfixed-end moments and moment distribution used to obtainequilibrium in the direction of span The revised negativemoments can then be used to adjust the positive moments atmid-span For each panel the sum of the mid-span moment andthe average of the support moments should be the same as theoriginal sum for that particular panel

When the long span exceeds twice the short span the slabshould be designed as spanning in the short direction In thelong direction the long span coefficient may still be used forthe negative moment at a continuous edge

Chapter 13

Slabs

For designs where elastic bending moments are required thecoefficients given for beams in Table 229 should be used

132 TWO-WAY SLABS

Various methods based on elastic or collapse considerationsare used to design slabs spanning in two directions Elasticmethods are appropriate if for example serviceability checkson crack widths are required as in the design of bridges andliquid-retaining structures Collapse methods are appropriatein cases such as floors in buildings and similar structureswhere the main criterion is the ultimate condition and the

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242Slabs general data

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243Two-way slabs uniformly loaded rectangular panels(BS 8110 method)

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Example Determine the bending moment coefficients in theshort span direction for the slab panel layout shown as follows

by finite element analysis with Poissonrsquos ratio taken as zerousing the following approximate relationships

Bending moments x x0 y0 y y0 x0

Torsion moments xy (1 )xy0

where is Poissonrsquos ratio and x0 y0 xy0 are coefficientscorresponding to 0 Thus if required the tabulated valuescan be readjusted to suit a Poissonrsquos ratio of zero as follows

Bending moments x0 104 (x02 02 y02)y0 104 (y02 02 x02)

Torsion moments xy0 125 xy02

For rectangular panels simply supported on four sides with noprovision to resist torsion at the corners or to prevent the cor-ners from lifting coefficients taken from BS 8110 are alsogiven in Table 244 The coefficients which are derived fromthe Grashof and Rankine formulae (see section 453) are givenby the following expressions

1323 Non-rectangular panels

For a non-rectangular panel supported along all of its edgesbending moments can be determined approximately from thedata given in Table 248 The information which is based onelastic analysis is applicable to panels that are trapezoidaltriangular polygonal or circular For guidance on using thisinformation including the arrangement of the reinforcementreference should be made to section 47

133 CONCENTRATED LOADS

1331 One-way slabs

For a slab simply supported along two opposite edges andcarrying a centrally placed load uniformly distributed over adefined area maximum elastic bending moments are given inTable 245 The coefficients which include for a Poissonrsquos ratioof 02 have been calculated from the data derived for arectangular panel infinitely long in one direction For designs toBS 8110 in which the ULS requirement is the main criteriona concentrated load placed in any position may be spread overa strip of effective width be as shown in Table 245 Parallel tothe supports a strip of width (x ay2) equally spaced eachside of the load has been considered

For slabs that are restrained at one or both edges maximumnegative and positive bending moments may be obtained bymultiplying the simply supported moment by the appropriatefactors given in Table 245 The factors which are given forboth fixed and continuous conditions are those appropriate toelastic beam behaviour

1332 Two-way slabs

For a rectangular panel freely supported along all four edges andcarrying a concentric load uniformly distributed over a definedarea maximum mid-span bending moments based on Pigeaudrsquostheory are given in Tables 246 and 247 Moment coefficients

my (ly lx)

2

8[1 (ly lx)4]

mx (ly lx)

4

8[1 (ly lx)4]

Concentrated loads 131

The upper half of the layout shows the coefficients obtainedfrom Table 243 with ly lx 9060 15 for panel types

AndashB (and CndashD) two adjacent edges discontinuousBndashC one short edge discontinuous

The lower half of the layout shows the coefficients obtained afterdistribution of the unbalanced support moments at B and CThe effective stiffnesses allowing for the effects of simple sup-ports at A and D and carry-over moments at B and C are

Span AndashB (and CndashD) 075Il Span BndashC 05Il

Distribution factors at B and C with no carry-overs are

BA (and CD) 075(075 050) 06BC (and CB) (10 06) 04

Moment coefficients at B and C after distribution are

0078 06 (0078 0058) 0066

Moment coefficients at mid-span after redistribution are

BA (and CD) 0059 05 (0078 0066) 0065BC 0043 (0066 0058) 0035

1322 Uniformly loaded slabs (elastic analysis)

For rectangular panels carrying uniformly distributed loadwhere the corners are prevented from lifting and adequateprovision is made for torsion maximum bending and torsionmoments are given in Table 244 for nine panel types Wherein continuous slabs the edge conditions in contiguous panelsresult in two different values being obtained for the negativemoment at a fixed edge the moment distribution procedureshown in section 1321 could be used but this would ignore theinter-dependence of the moments in the two directions A some-what complex procedure involving edge stiffness factors isderived and shown with fully worked examples in ref 21

The coefficients include for a Poissonrsquos ratio of 02 and havebeen calculated from data given in ref 21 which was derived

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244Two-way slabs uniformly loaded rectangular panels(elastic analysis)

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245One-way slabs concentrated loads

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246Two-way slabs rectangular panel with concentricconcentrated load ndash 1

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247Two-way slabs rectangular panel with concentricconcentrated load ndash 2

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248Two-way slabs non-rectangular panels (elastic analysis)

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given in the charts are used with an appropriate value of Poissonrsquosratio to calculate the bending moments The coefficients given atthe top right corner of each chart are for the limiting case whenthe load extends over the entire panel This case with Poissonrsquosratio taken as 02 is given also in Table 244

When using the chart for square panels (k 10) if ax ayx4 y4 and the resulting bending moment in each direction isgiven by F (1 )x4 In other cases coefficient x4 is based onthe direction chosen for ax and coefficient y4 is obtained byreversing ax and ay as shown in example 1 later

The maximum shearing forces V per unit length on a panelcarrying a concentrated load are given by Pigeaud as follows

ax ay at the centre of length ax V F(2ax ay)at the centre of length ay V F3ax

ay ax at the centre of length ax V F3ay

at the centre of length ay V F(2ay ax)

For panels that are restrained along all four edges Pigeaudrecommends that the mid-span moments be reduced by 20Alternatively the multipliers given for one-way slabs couldbe used in one or both directions as appropriate if the inter-dependence of the bending moments is ignored

Example 1 Consider a square panel freely supported alongall four edges carrying a concentric load with axlx 08 andayly 02 From Table 247 for k 1 the bending momentcoefficients are

For axlx 08 and ayly 02 x4 0072For axlx 02 and ayly 08 y4 0103

Maximum bending moments per unit width with 02 are

For span in direction of ax F(0072 02 0103) 0093FFor span in direction of ay F(0103 02 0072) 0118F

Example 2 A bridge deck is formed of a 200 mm thick slabsupported by longitudinal beams spaced at 2 m centres Theslab is covered with 100 mm thick surfacing Determine for theSLS the maximum positive bending moments in the slab due tothe local effects of live loading

The critical live load for serviceability is the HA wheel loadof 100 kN to which a partial load factor of 12 is applied Fora 100 mm 100 mm contact area and allowing for loaddispersal through the thickness of the surfacing and down tothe mid-depth of the slab (see section 249) the side of theresulting patch load is (300 100 200) 600 mm

The simply supported bending moment coefficients for acentrally placed load by interpolation from Table 245 are

For axlx aylx 6002000 03 mx 0206 my 0145

Allowing for continuity (interior span) in the direction of lx andapplying a partial load factor of 12 the positive bendingmoments per unit width are

In direction of lx

mx 064 0206 12 100 158 kNmm

At right angles to lx

my 0145 12 100 174 kNmm

For design purposes the bending moments determined earlierwill need to be combined with the moments due to the weightof the slab and surfacing and any transverse effects of theglobal deck analysis

Note Using the method given in BS 8110 with x 05lx theeffective width of the strip is

Allowing for continuity (interior span) in the direction of lx andapplying a partial load factor of 12 the positive bendingmoments per unit width are

In direction of lx

At right angles to lx

134 YIELD-LINE ANALYSIS

As stated in section 452 yield-line theory is too complex asubject to deal with adequately in the space available in thisHandbook The following notes are therefore intended merelyto introduce the designer to the basic concepts methods andproblems involved For further information see refs 23 to 28Application of yield-line theory to the design of rectangularslabs subjected to triangularly distributed loads is dealt with insection 1362

1341 Basic principles

When a reinforced concrete slab is loaded cracks form inthe regions where the applied moment exceeds the crackingresistance of the concrete As the load is increased beyond theservice value the concrete continues to crack eventually thereinforcement yields and the cracks extend to the corners ofthe slab dividing it into several areas separated by so-calledyield-lines as shown in diagrams (i)ndash(iii) on Table 249 Anyfurther increase in load will cause the slab to collapse Inthe design process the load corresponding to the formation ofthe entire system of yield lines is calculated and by applyingsuitable partial factors of safety the resistance moment thatmust be provided to avoid collapse is determined

For a slab of given shape it is usually possible to postulatedifferent modes of failure the critical mode depending on thesupport conditions the panel dimension and the relative propor-tions of reinforcement provided in each direction For exampleif the slab shown in diagram (i)(a) is reinforced sufficientlystrongly in the direction of the shorter span by comparison withthe longer span this mode of failure will be prevented and thatshown in diagram (i)(b) will occur instead Similarly if the slabwith one edge unsupported shown in diagrams (ii) is loadeduniformly pattern (b) will occur when the ratio of the longer toshorter side length (or longer to shorter lsquoreduced side lengthrsquo seesection 1346) exceeds radic2 otherwise pattern (a) will occur

All of these patterns may be modified by the formation ofcorner levers see section 1349

my 12 100 1842 10 06 1

0618 139kNmm

mx 064 12 100 10

2 18 1 06

2 20 181kNmm

be 06lx ay 06 20 06 18 m

Yield-line analysis 137

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249Two-way slabs yield-line theory general information

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1342 Rules for postulating yield-line patterns

Viable yield-line patterns must comply with the following rules

1 All yield lines must be straight

2 A yield line can only change direction at an intersection withanother yield-line

3 A yield line separating two elements of a slab must passthrough the intersection of their axes of rotation (Note thispoint may be at infinity)

4 All reinforcement intercepted by a yield line is assumed toyield at the line

1343 Methods of analysis

Two basic methods of analysis have been developed These arecommonly referred to as the lsquoworkrsquo or lsquovirtual workrsquo methodand the lsquoequilibriumrsquo method The former method involvesequating for the yield-line pattern postulated the work done bythe external loads on the various areas of the slab to obtain avirtual displacement to the work done by the internal forces informing the yield lines When the yield-line pattern is adjustedto its critical dimensions the ratio of the ultimate resistance tothe ultimate load reaches its maximum When analysing a slabalgebraically this situation can be ascertained by differentiatingthe expression representing the ratio and equating the differentialto zero in order to establish the critical dimensions Then byre-substituting these values into the original expression a formulagiving the required ultimate resistance for a slab of givendimensions and loading can be derived

The so-called equilibrium method is not a true equilibriummethod but a variant of the work method which also gives anupper-bound solution The method has the great advantagethat the resulting equations provide sufficient information ofthemselves to eliminate the unknown variables and thereforedifferentiation is unnecessary Although there are also otheradvantages the method is generally more limited in scope andis not described here for details see refs 23 and 27

1344 Virtual-work method

As explained earlier this method consists of equating thevirtual work done by the external loads in producing a givenvirtual displacement at some point on the slab to the work doneby the internal forces along the yield lines in rotating the slabelements To demonstrate the principles involved an analysiswill be given of the freely supported rectangular slab supportinga uniform load and reinforced to resist equal moments M eachway shown in diagram (i)(a) on Table 249

Clearly due to symmetry yield line OO will be midwaybetween AB and CD Similarly and thus only onedimension is unknown Consider first the external work done

The work done by an external load on an individual slabelement is equal to the area of the element times the displace-ment of its centroid times the unit load Thus for the triangularelement ADO with displacement at O

work done (12)(lx)(ly)(3)n lxlyn6

Similarly for the trapezoidal area ABOO with displacement at O and O

work done [(lx2)(ly)( 2) 2(12)(lx2)(ly)(23)]

(3 4)lxlyn12

Thus since the work done on BCO is the same as that doneon ADO and the work done on CDOO is the same as that onABOO for the entire slab

Total external work done 2[lxlyn6 (3 4)lxlyn12]

(3 2)lxlyn6

The internal work done in forming a yield line is equal to themoment along the yield line times the length of the line timesthe rotation A useful point to note is that where a yield line isformed at an angle to the direction of principal momentsinstead of considering its true length and rotation it is usuallysimpler to consider the components in the direction of theprincipal moments For example for yield line AO instead ofconsidering the actual length AO and the rotation at right anglesto AO consider length lx2 and the rotation about AB pluslength ly and the rotation about AD

Thus considering the component about AB of the yield linealong AOO B the length of the line is ly the moment is M andthe rotation is ( lx2) Hence

work done 2M(lylx)

Similarly for the yield line along DOO C the work done isagain 2M(lylx) (Length OO is considered twice because therotation between the elements separated by this length is dou-ble that occurring over the remaining length) Now consideringthe component about AD of the yield line AOD

work done M(lxly)

Since the work done on yield line BOC is similar for theentire slab

Total internal work done 2M(2lylx lxly)

Equating the external work done to the internal work done

(3 2)lxlyn6 2M(2lylx lxly)

or

To determine the critical value of ly the quotient in squarebrackets must be differentiated and equated to zero As Jones(ref 27) has pointed out to use the well-known relationship

simply as a means of maximising y uv it is convenient torearrange it in the form

Thus for the present example

This leads to a quadratic in the positive root of which is

12lx

ly4

3lx

ly2

lx

ly2

3 22

2 (lxly)2

3ndash 4

2

uv

dudxdvdx

dydx

v dudx

u dvdx v2 0

M n

12l 2

x 3 22

2 (lx ly)2

Yield-line analysis 139

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When substituted in the original equation this gives

1345 Concentrated and line loads

Concentrated and line loads are simpler to deal with thanuniform loads When considering the external work done thecontribution of a concentrated load is equal to the load timesthe relative deflection of the point at which it is applied In thecase of a line load the external work done over a given slab areais equal to the portion of the load carried on that area times therelative deflection at the centroid of the load

Yield lines tend to pass beneath heavy concentrated or lineloads since this maximises the external work done by suchloads When a concentrated load acts in isolation a so-calledcircular fan of yield lines tends to form this behaviour iscomplex and reference should be made to specialist textbooksfor details (refs 23 24 26)

1346 Affinity theorems

Section 1344 illustrates the work involved in analysing asimple freely supported slab with equal reinforcement in eachdirection (ie so-called isotropic reinforcement) If differentamounts of reinforcement are provided in each direction(ie so-called orthotropic reinforcement) or if continuity orfixity exists along one or more edges the formula needsmodifying accordingly

To avoid the need for a vast number of design formulae tocover all conceivable conditions it is possible to transformmost slabs with fixed or continuous edges and orthotropicreinforcement into their simpler freely supported isotropicequivalents by using the following affinity theorems skew slabscan be transformed similarly into rectangular forms

1 If an orthotropic slab is reinforced as shown in diagram (iv)(a)on Table 249 then it can be transformed into the simplerisotropic slab shown in diagram (iv)(b) All loads anddimensions in the direction of the principal co-ordinate axisremain unchanged but in the affine slab the distances in thedirection of the secondary co-ordinate axis are equal to theactual values divided by radic and the corresponding total loadsare equal to the original values divided by radic (The latterrequirement means that the intensity of a UDL per unit arearemains unchanged by the transformation since both the areaand the total load on that area are divided by radic)

Similarly a skew slab reinforced as shown in diagram(v)(a) can be transformed into the isotropic slab shown indiagram (v)(b) by dividing the original total load by sin(As before this requirement means that the intensity of aUDL per unit area remains unchanged by the transformation)

These rules can be combined when considering a skewslab with different reinforcement in each direction fordetails see ref 28 article 6

2 By using the reduced side lengths lxr and lyr an orthotropicslab that is continuous over one or more supports such as

M n

24lx

23 lx

ly2

lx

ly2

that in diagram (vi)(a) on Table 249 can be transformedinto the simpler freely supported isotropic slab shown indiagram (vi)(b)

For example for an orthotropic slab with fixed edges that isreinforced for positive moment M and negative moments i2Mand i4M in span direction lx and for positive moment M andnegative moments i1M and i3M in span direction ly it can beshown that

where

and

This is identical to the expression derived above for the freelysupported isotropic slab but with lxr and lyr substituted for lx

and ly Values of Mnlx2 corresponding to ratios of lylx can be

read directly from the scale on Table 249The validity of the analysis is based on the assumption that

lyr lxr If this is not the case the yield line pattern will be asshown in diagram (i)(b) on Table 249 and lxr and lyr should betransposed as shown in the following example

Example Design the slab in diagram (vii) on Table 249 tosupport an ultimate load n per unit area assuming that therelative moments of resistance are as shown

Since lx 4 m i2 32 and i4 0

Since ly 6 m 12 i1 1 and i3 0

Thus

1347 Superposition theorem

A problem that may arise when designing a slab to resist acombination of uniform concentrated and line loads some ofwhich may not always occur is that the critical pattern of yieldlines may well vary for different combinations of loads Also itis theoretically incorrect to sum the ultimate moments obtainedwhen considering the various loads individually since thesemoments may result from different yield line patterns HoweverJohansen has established the following superposition theorem

The sum of the ultimate moments for a series of loads is equalto or greater than those due to the sum of the loads

n

24 31023 310

7032

310703

2

0726 n

M n

24l 2xr3 lxr

lyr2

lxr

lyr2

ly r 2 6

(1 1 1)12 703 m

lxr 2 4

1 32 1 310 m

lyr 2ly

[1 i1 1 i3]lxr

2lx

1 i2 1 i4

M n

24l2xr3 lxr

lyr2

lxr

lyr2

Slabs140

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In other words if the ultimate moments corresponding to theyield-line patterns for each load considered separately areadded together the resulting value is equal to greater than thatof the system as a whole This theorem is demonstrated inref 28 article 4

1348 Empirical virtual-work analysis

An important advantage of collapse methods of design is thatthey can be readily applied to solve problems such as slabs thatare irregularly shaped or loaded or that contain large openingsThe analysis of such slabs using elastic methods is by comparisonextremely arduous

To solve such lsquoone-offrsquo problems it is clearly unrealistic todevelop standard algebraic design formulae The followingempirical trial-and-adjustment technique which involves adirect application of the virtual-work principles is easy tomaster and can be used to solve complex problems It is bestillustrated however by working through a simple problemsuch as the one considered in section 1346 There is of courseno need to employ the procedure in this case It is used hereonly to illustrate the method A more complicated example isgiven in ref 28 article 1 on which the description of themethod is based

In addition to the fundamental principles of virtual workdiscussed in section 1344 the present method depends alsoon the following principle If all yield lines (other than thosealong the supports) are positive and if none of them meets anunsupported edge except at right angles then no forces due toshear or torsion can occur at the yield lines Thus a separatevirtual-work balance for each slab area demarcated by the yieldlines can be taken

Example Consider the slab shown in diagram (vii) onTable 249 which is continuous over two adjacent edges freelysupported at the others and subjected to a uniform load n perunit area The ratios of the moments of resistance provided overthe continuous edges and in the secondary direction to thatin the principal direction are as shown

The step-by-step trial and adjustment process is as follows

1 Postulate a likely yield-line pattern

2 Give a virtual displacement of unity at some point andcalculate the relative displacement of any other yield-line

intersection points For the case considered if O is given adisplacement of unity the displacement at O will alsobe unity since OO is parallel to the axes of rotation of theadjoining slab areas

3 Choose reasonable arbitrary values for the dimensions thatmust be determined to define the yield-line pattern Thus inthe example initially ly is taken as 2 m ly as 1m and lx

as 25 m

4 Calculate the actual work done by the load n per unit areaand the internal work done by the moments of resistance Mfor each separate part of the slab and thus obtain ratios ofMn for each part

For example on area A the total load is 12 4 2 n 4nSince the centre of gravity moves through a distance of 13 thework done by the load on area A is 4n 13 4n3 Now sinceO is displaced by unity the rotation of area A about the supportis 1ly 12 The moments of M2 acting across both the pos-itive and negative yield lines each exert a total moment ofM2 4 2M Thus the total internal work done in rotatingarea A is (2M 2M) 12 2M Equating the internal andexternal work done on area A gives 2M 4n3 that isMn 23 Similarly for area C Mn 13

For convenience area B can be divided it into a rectangle (ofsize 3 m 25 m) and two triangles and the work done on eachpart calculated separately Since the centres of gravity movethrough a distance of 12 for the rectangle and 13 for eachtriangle the external work done is as follows

Rectangular area 3 25 12 n 375nTriangular areas 12 (2 1) 25 13 n 125nTotal 500n

Since the rotation is 125 the work done by the moments is(15M M) 6 125 6M Thus the virtual-work ratio isMn 56 0833 Likewise for area D Mn 34

5 Sum the separate values of internal and external workdone for the various slab areas and thus obtain a ratio ofMn for the entire slab This ratio will be lower than thecritical value unless the dimensions chosen arbitrarily instep 3 happen to be correct The calculations are best set outin tabular form as follows

Yield-line analysis 141

Area External work done Internal work done Mn

A 12 4 2 13 n 1333n [M2 M2] 4 12 2000M 0667B [3 25 12 12 3 25 13] n 5000n [3M2 M] 6 125 6000M 0833C 12 4 1 13 n 0667n [0 M2] 4 11 2000M 0333D [3 15 12 12 3 15 13] n 3000n [0 M] 6 115 4000M 0750

Total 10000n Total 14000M 0714

A 12 4 2071 13 n 1381n [M2 M2] 4 12071 1931M 0714B [2467 2424 12 12 3533 2424 13] n 4418n [3M2 M] 6 12424 6188M 0714C 12 4 1462 13 n 0975n [0 M2] 4 11462 1368M 0714D [2467 1576 12 12 3533 1576 13] n 2872n [0 M] 6 11576 3807M 0754

Total 9646n Total 13294M 0726

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6 By comparing the overall ratio obtained for Mn withthose due to each individual part it is possible to see howthe arbitrary dimensions should be adjusted so that the ratiosfor the individual parts become approximately equal to eachother and to that of the slab as a whole

The foregoing table shows calculations for initial values of Mnand also for a set of adjusted values An examination of theinitial values (for which the overall ratio is 0714) shows thatto obtain the same ratio for each area Mn needs to be increasedfor areas A and C and reduced for areas B and D For area Asince Mn is proportional to (ly)2 ly needs to be increased toradic(07140667) 2 2071 Similarly for area C ly needsto be increased to radic(07140333) 1 1462 If for area Bthe external work done is recalculated using the correctedvalues of ly and ly this gives

[2467 25 12 12 3533 25 13] n 4556n

The internal work done by the moments is unchanged and sothe revised value of Mn is 45566 0759 Thus since forarea B Mn is proportional to (lx)2 lx needs to be reducedto radic(07140759) 25 2424 For area D the external workdone is recalculated using the corrected values of all thevariable dimensions and the revised value of Mn obtained

7 Repeat steps 4 and 5 using the adjusted values for thearbitrary dimensions as shown earlier

8 Repeat this cyclic procedure until reasonable agreement isobtained between the values of Mn This ratio gives thevalue of M for which the required reinforcement must bedetermined for a given load n In the example the ratiosgiven by the second cycle are quite satisfactory Note thatalthough some of the dimensions originally guessed werenot particularly accurate the resulting error in the value ofMn obtained for the whole slab was only about 15 andthe required load-carrying capacity is not greatly affected bythe accuracy of the arbitrary dimensions

Concentrated loads and line loads occurring at boundariesbetween slab areas should be divided equally between the areasthat they adjoin and their contribution to the external workdone assessed as described in section 1345

As in all yield-line theory the above analysis is only valid ifthe yield-line pattern considered is the critical one Where thereis a reasonable alternative both patterns should be investigatedto determine which is critical

1349 Corner levers

Tests and elastic analyses of slabs show that the negativemoments along the edges reduce to zero near the corners andincrease rapidly away from these points Thus in slabs that arefixed or continuous at their edges negative yield lines tend toform across the corners and in conjunction with pairs of positiveyield lines result in the formation of additional triangular slabelements known as corner levers as shown in diagram (i)(a) onTable 250 If the slab is freely supported a similar mechanismis induced causing the corners to lift as shown in diagram(1)(b) If these mechanisms are substituted for the original yieldlines running into the corner of the slab the overall strength ofthe slab is correspondingly decreased by an amount dependingon the factors listed on Table 250 For a corner lever having anincluded angle of not less than 90o the strength reduction is not

likely to exceed 8ndash10 In such cases the main reinforcementcan be increased slightly and top reinforcement provided at thecorners of the slab to restrict cracking Recommendations takenfrom the Swedish Code of Practice are shown in diagram (ii) onTable 250

For acute-angled corners the decrease in strength is moresevere For a triangular slab ABC where no corner angle is lessthan 30o Johansen (ref 25) suggests that the calculated strengthwithout corner-lever action should be divided by a factor kgiven by the approximation

k (74 sin A sin B sin C)4

A mathematical determination of the true critical dimensions ofan individual corner lever involves much complex trial andadjustment However this is unnecessary as Jones and Wood(ref 23) have devised a direct design method that gives cornerlevers having dimensions such that the resulting adjustment instrength is similar to that due to the true mechanisms Thisdesign procedure is summarised on Table 250 and illustratedby the following example

The formulae derived by Jones and Wood and on which thegraphs in Table 250 are based are as follows

With fixed edges

where

and

With freely supported edges

where

and

Example Calculate the required resistance of the 5 m squareslab with fixed edges (i 1) shown on Table 250

For the transformed freely supported slab the reduced sidelength lr lradic(1 I) 5radic2 354 m Then for a square slabif the formation of corner levers is ignored the required resistancemoment M (124)nlr

2 0041 3542n 0521nNow from the appropriate graph on Table 250 with i 1

and 13 90o read off k1 11 and k2 42 Thus dimensionsa1 11radic0521 0794 m and a2 42radic0521 3032 m

By plotting these values on a diagram of the slab it is nowpossible to calculate the revised resistance moment required Ifthe deflection at the centre is unity the relative deflection at theapex of the corner levers is 30323536 0858 Thus therevised virtual-work equation is

[(13) 52 4 (12) 07942 (13) 0858] n

This reduces to 14036M 7973n so that M 0568n

4 2M3412 125 07942 0858

2471

cot [K2(1 i) (2 i)] tan (2)

K2 (4 i) 3 cot2 (2)

k2 k1

cos (2) cot sin (2)

k1 [K2 2(1 i) ]2 3 sec(2)

cot (K1 1) tan (2)K1 4 3 cot2 (2) 1

k2 k1

cos (2) cot sin (2)

k1 1K1 sin2(2)

16(1 i) sec(2)

Slabs142

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250Two-way slabs yield-line theory corner levers

Freely supportedslab edges

Corner leverFixed slabedges

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Note that this value of M is 9 greater than the uncorrectedvalue in other words the load supported by a square slab witha specified moment of resistance is actually 9 less than thatcalculated when corner levers are not taken into account

135 HILLERBORGrsquoS SIMPLE STRIP THEORY

1351 Moments in slabs

According to lower-bound (equilibrium) theory load acting ona slab is resisted by a combination of biaxial bending andtorsion In the simple strip method the torsion moment is takenas zero and load (or partial load) acting at any position on theslab is resisted by bending in one of two principal directionsThus in diagram (i) on Table 251 the load acting on theshaded areas is resisted by bending in direction ly and the loadacting on the remaining area is resisted by bending in direc-tion lx In principle there is an unlimited number of ways ofapportioning the load each of which will lead to a differentreinforcement layout while still meeting the collapse criteriaHowever the loading arrangement selected should also ensurethat the resulting design is simple economical and satisfactorywith regard to deflection and cracking under service loads

Some possible ways of apportioning the load on a freelysupported rectangular slab are shown in diagrams (i)ndash(iv) onTable 251 the notation adopted being given on the tablePerhaps the most immediately obvious arrangement is shown indiagram (i) for which Hillerborg originally suggested that 13could be taken as 45o where both adjacent edges are freelysupported However in ref 29 he recommends that 13 should bemade equal to tan1(lylx) as shown in diagram (i) The disad-vantage of the arrangement shown is that the bending moment(and thus the reinforcement theoretically required) variesacross strips 2 and 3 Since it is impractical to vary the rein-forcement continuously the usual approach is to calculatethe total moment acting on the strip divide by the width ofthe strip to obtain the average moment and provide a uniformdistribution of reinforcement to resist this moment To avoidhaving to integrate across the strip to obtain the total momentHillerborg recommends calculating the moment along thecentreline of the strip and then multiplying this value bythe correction factor

where lmax and lmin are the maximum and minimum loadedlengths of the strip Strictly speaking averaging the momentsas described violates the principles on which the method isbased and this device should only be used where the factor ofsafety will not be seriously impaired If the width of the stripover which the moments are to be averaged is large it is betterto sub-divide it and calculate the average moment for eachseparate part

An alternative arrangement that avoids the need to averagethe moments across the strips is shown in diagram (ii) This hasdisadvantages in that six different types of strip (and thus sixdifferent reinforcement layouts) must be considered and thatin strip 6 no moment theoretically occurs Such a strip mustnevertheless contain distribution reinforcement

1 (lmax lmin)

2

3(lmax lmin)2

So far the load on any one separate area has been carried inone direction only In diagram (iii) however the loads on thecorner areas are so divided that one-half is carried in eachdirection Hillerborg (ref 29) states that this very simple andpractical approach never requires more than 10 additionalreinforcement when compared to the theoretically more exactbut less practical solution shown in diagram (i) when lylx isbetween 11 and 4 An additional sophistication that can beintroduced is to apportion the load in each direction in thetwo-way spanning areas in such a way that the resulting rein-forcement across the shorter span corresponds to the minimumrequirement for secondary reinforcement Details of this andsimilar stratagems are given in ref 29

Diagram (iv) illustrates yet another arrangement that may beconsidered By dividing the corner areas into triangles andaveraging the moments over these widths as described earlierHillerborg shows that the moments in the side strips can bereduced to two-thirds of the values given by the arrangementused in diagram (iii)

1352 Loads on supporting beams

A particular feature of the strip method is that the boundariesbetween the different loaded areas also define the manner inwhich the loads are transferred to the supporting beamsFor example in diagram (i) the beams in direction lx supporttriangular areas of slab giving maximum loads of nlx

22ly attheir centres

136 BEAMS SUPPORTING RECTANGULAR PANELS

The loads on beams supporting uniformly loaded rectangularslab panels are distributed approximately as a triangular loadon the beams along the shorter edges lx and a trapezoidalload on the beams along the longer edges ly as shown in thediagrams on Table 252 For a beam supporting a single panelof slab that is either freely supported or subjected to the samedegree of restraint along all four edges where the beam span isequal to the length (or width) of the panel the equivalent UDLper unit length on the beam for the calculation of bendingmoments only is as follows

Short-span beam nlx3

Long-span beam (1 13k2) nlx2

where n is the total UDL per unit area on the slab appropriateto the limit-state being considered and k lylx For a beamsupporting two identical panels one on either side the fore-going equivalent loads are doubled If a beam supports morethan one panel in the direction of its length the distribution ofload is in the form of two or more triangles (or trapeziums)and the foregoing formulae are not applicable in such a casehowever it is sufficiently accurate if the total load on the beamis considered to be uniformly distributed

For slabs designed in accordance with the BS 8110 methodthe loads on the supporting beams may be determined from theslab shear forces given in Table 243 The loads are to be takenas uniformly distributed along the middle three-quarters ofthe beam length where the shear force for the short span of theslab is the load on the long-span beam and vice versa Theresulting beam fixed-end moments can be determined fromTable 228

Slabs144

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251Two-way slabs Hillerborgrsquos simple strip theory

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252Two-way slabs rectangular panels loads on beams(common values)

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Triangularly distributed loads 147

137 TRIANGULARLY DISTRIBUTED LOADS

In the design of rectangular tanks storage bunkers and someretaining structures conditions occur of wall panels spanningin two directions and subjected to distributions of pressurevarying linearly from zero at or near the top to a maximum atthe bottom For liquid-retaining structures with no provisionfor additional protection in the form of an internal lining orexternal tanking an elastic analysis is normally necessary as abasis for checking serviceability cracking In other cases ananalysis based on collapse methods may be justified

1371 Elastic analysis

The coefficients given in Table 253 enable the maximumvalues of bending moments and shearing forces on vertical andhorizontal strips of unit width to be determined for panels ofdifferent aspect ratios and edge conditions The latter are takenas fixed at the sides hinged or fixed at the bottom and hingedor free at the top The coefficients which are taken from ref 32were derived by a finite element analysis and include for aPoissonrsquos ratio of 02 For ratios less than 02 the momentscould be adjusted in the manner described in section 1322

The maximum negative bending moment at the bottom edgeand the maximum shear forces at the bottom and top edgesoccur halfway along the panel The other maximum momentsoccur at the positions indicated in the following table

From Table 253 for panel type 4 with lx lz 54 125 themaximum bending moments are as follows

Horizontal negative moment at corners

mx 0050 981 43 314 kNmm

Horizontal positive moment (at about 05lz 2 m above base)

mx 0022 981 43 138 kNmm

Vertical positive moment (at about 03lz 12 m above base)

mz 0021 981 43 132 kNmm

1372 Yield-line method

A feature of the collapse methods of designing two-way slabsis that the designer is free to choose the ratios between thetotal moments in each direction and between the positive andnegative moments In the case of liquid-retaining structureswhere it is important to ensure that the formation of cracksunder service load is minimised the ratios selected shouldcorrespond approximately to those given by elastic analysisThe following design procedure is thus suggested

1 Obtain maximum positive and negative service momentcoefficients from Table 253

2 Determine i1 ( i3) and i4 where mhposmzposi1 i3 mhnegmhpos and i4 mznegmzpos

3 Calculate lxr and lyr if the top edge is unsupported from

and

and if the top edge is freely supported from

and

4 Obtain M if the top edge is unsupported from the chart onTable 254 and if the top edge is supported from the scaleon Table 249 according to the values of f (or n f2) lx

(or lxr) lyr and i4 The basis of this approach is given below

Top edge of slab unsupported In ref 25 Johansen derivesthe following lsquoexactrsquo formulae according to the failure mode

For failure mode 1

where k is obtained by solving the quadratic equation

For failure mode 2

where is obtained by solving the following cubic equation

8lxr

lyr2

3 3ndash16lxr

lyr2 2 8 6 0

M f l 2

yr

96(6 8 3 2)

i241 i4 4 lx

lyr2 0

41 i4 lx

lyr2k2 4i4(1 i4) (6 4i4) lx

lyr2k

M f l 2

x

12k

lyr 2ly

[1 i1 1 i3 ]lxr

2lx

1 1 i4

lyr 2ly

[1 i1 1 i3 ]lxr

lx

1 i4Distance from bottom of panel to position ofmaximum horizontal moments (negativepositive)

Type of Heightlz for values of lx lzpanel 05 10 15 20 40

1 03 05 05 05 052 03 05 0710 0910 103 03 04 04 04 044 04 04 0506 0910 0910

Distance from bottom of panel to position ofmaximum vertical positive moment

Type of Heightlz for values of lxlzpanel 05 10 15 20 40

1 03 05 05 05 052 03 04 05 06 073 02 03 03 04 044 02 03 03 04 04

For a complete map of bending moment values at intervals ofone-tenth of the panel height and length see ref 32 Themoments obtained for an individual panel apply directly to asquare tank with hydrostatic loading For a rectangular tanka further distribution of the unequal negative moments at thecorners is needed (see Tables 275 and 276)

Example Determine due to internal hydrostatic loading themaximum service moments in the walls of a square tank thatcan be considered as free along the top edge and hinged alongthe bottom edge The tank is 5 m square 4 m deep and thewater level is to be taken to the top of the walls

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253Two-way slabs triangularly distributed load(elastic analysis)

Fixed

Fixed

Fixed

Panel 1 Panel 2 Panel 3 Panel 4

Hinged

Fixed

Fixed

Fixed

Free

Fixed

Fixed

Hinged

Hinged

Fixed

Hinged

Free

lx

lz

lx

Fixed

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254Two-way slabs triangularly distributed load(collapse method)

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The results of these calculations can be plotted graphically asshown in Table 254 from which coefficients of Mfl2

x can beread corresponding to given values of lyrlx and i4 The chainline on the chart indicates the values at which the failuremode changes

Top edge of slab supported In ref 25 Johansen showsthat the total moments resulting from yield-line analysisfor triangularly loaded slabs correspond to those obtainedwhen the same slab is loaded with a uniform load of one-halfthe maximum triangular load (ie n f 2) Hence the designexpressions in section 1346 and the scale on Table 249 foruniformly loaded slabs can again be used As before if theedges of the slab are restrained the reduced side lengths lxr andlyr should be calculated and substituted in the formula insteadof the actual side lengths lx and ly

Example Determine moments by yield-line analysis for thewalls of the square tank considered in the previous example insection 1371

Taking the service moment coefficients obtained previouslythe following values are the most suitable for

00220021 10 and i1 ( i3) 00500022 23

Hence

lyr 275 and lyrlx 27540 070

Then from the chart on Table 254 with i4 0

Mflx2 0018

Comparing this value with mx 0022 as obtained in theprevious example it can be seen that the elastic moments are12 times those determined by yield-line analysis Thus evenwith a partial load factor of 12 the yield-line moments are nogreater than the elastic service moments in this case

Note It can be seen from the chart on Table 254 that failuremode 2 applies for which

Hence

and yield-lines intersect at height lx 0335 40 134 mabove the base

138 FLAT SLABS (SIMPLIFIED METHOD)

The following notes and the data in Tables 255 and 256 arebased on the recommendations for the simplified method of flatslab design given in BS 8110 Alternatively and in cases wherethe following conditions are not met the structure can beanalysed by the equivalent frame method Other methods ofanalysis such as grillage finite-element and yield line may alsobe employed in which case the provisions given for the distri-bution of bending moments are a matter of judgement

32 8 6 0018 96 (40 275)2 366 0335

M fl 2

yr

96(6 8 3 2) 0018 fl2

x

2 521 2310

1381 Limitations of method

The system must comprise at least three rows of rectangularpanels in each direction The spans should be approximatelyequal in the direction being considered and the ratio of thelonger to the shorter sides of the panels should not exceed 2The conditions allowing the design to be based on the singleload case of all spans loaded with the maximum design load asexplained in section 441 and Table 242 must be met Alllateral forces must be resisted by shear walls or bracing

1382 Forms of construction

The slab can be of uniform thickness throughout or thickeneddrop panels can be introduced at the column positions Droppanels can extend to positions beyond which the slab can resistpunching shear without needing shear reinforcementAlternatively the panels can be further extended to positionswhere they may be considered to influence the distribution ofmoments within the slab In this case the smaller dimensionof the drop panel should be at least one-third of the smallerdimension of the surrounding slab panels

Columns can be of uniform cross section throughout or canbe provided with an enlarged head the effective dimensions ofwhich are limited according to the depth of the head as shownin Table 255 For a flared head the actual dimension is takento be the value at a depth 40 mm below the underside of the slabor drop

The effective diameter of a column or column head is thediameter of a circle whose area is equal to the cross-sectionalarea of the column or effective column head In no case is theeffective diameter hc to be taken greater than one-quarter ofthe shortest span framing into the column In cases where theedges of the slab are supported by walls hc can be taken asthe thickness of the wall

1383 Bending moments and shearing forces

The total design bending moments and shearing forces given inTable 255 are the same as those given for one-way slabs inTable 242 where the support moments include for 20 redis-tribution The requirements are applied independently in eachdirection Any negative moments greater than those at a dis-tance hc2 from the centreline of the column may be ignoredsubject to the following condition being met In each span thesum of the maximum positive moment and the average ofthe negative moments for the whole panel width where F is thetotal design load on the panel and l is the panel length betweencolumn centrelines must be not less than

(Fl8) (1 2hc3l)2

This condition is met in the case of designs that are based onthe single load case of all spans loaded with the maximumdesign load by taking the design negative moment as the valueat a distance hc3 from the centreline of the column Themoment at this position is obtained approximately by reducingthe value at the column centreline by 015Fhc

The total design moments on a panel should be apportionedbetween column and middle strips as shown in Table 255 Inthe following table the moment allocations given in BS 8110and EC 2 are shown for comparison

Slabs150

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255Flat slabs BS 8110 simplified method ndash 1

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256Flat slabs BS 8110 simplified method ndash 2

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At the edges of a slab the transfer of moments between the slaband an edge or corner column is limited by the effective breadthbe of the moment transfer strip given in Table 256 Themaximum design moment that can be transferred by this stripis given by the equation Mt max 015bed2fcu in BS 8110 andMt max 017bed2fck in EC 2

At internal columns two-thirds of the reinforcement neededto resist the negative moments in the column strips should beplaced in a width equal to half that of the column strip and cen-tral with the column Otherwise the reinforcement needed toresist the moment apportioned to a particular strip should bedistributed uniformly across the width of the strip

1384 Effective forces for punching shear

The critical consideration for shear in flat slab structures is thatof punching shear around the columns The design forceobtained by summing the shear forces on each side of the col-umn is multiplied by an enhancement factor to allow for theeffects of moment transfer as given in Table 256 The effectiveshear force is determined independently in each direction andthe design checked for the worse case

1385 Reservoir roofs

For reservoir roofs with simply supported ends where elasticmoments are required to check serviceability requirements thecoefficients given for beams in Table 230 could be used In thiscase the negative moments at the centrelines of the columnscould be reduced by 022Fhc to give approximately themoment at a distance hc2 from the centreline of the columnThis approach will still ensure that the minimum momentrequirement mentioned in section 1383 is met

Flat slabs (simplified method) 153

Design moment Column strip Middle strip

BS 8110 Negative 75 25Positive 55 45

EC 2 Negative 60ndash80 40ndash20Positive 50ndash70 50ndash30

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When using the formulae and coefficients in this chapter theloads must include the appropriate partial safety factors for thelimit-state being considered Design loads for the ULS inaccordance with the requirements of BS 8110 and EC 2 aregiven in Table 257

For many framed structures it is not necessary to carry out afull structural analysis of the complete frame as a single unitBS 8110 makes a distinction between frames supporting verticalloads only because lateral stability to the structure as a wholeis provided by other means such as shear walls and framessupporting both vertical and lateral loads Simplified models forthe purpose of analysis as described in section 491 are alsoshown in Table 257

The moment-distribution method used to analyse systems ofcontinuous beams as shown in Table 236 can be extended toapply to no-sway sub-frames (see section 492) as shown inTable 258 and single-bay sway frames (see section 493) asshown in Table 259

141 SLOPE-DEFLECTION METHOD OF ANALYSIS

Moment analysis of a restrained member by slope-deflection isbased on the following two principles The difference in slopebetween any two points in the length of the member is equalto the area of the MEI diagram between the two points Thedistance of any point on the member from a line drawn tangen-tially to the elastic curve at any other point the distance beingmeasured normal to the initial position of the member is equalto the moment (taken about the first point) of the MEI diagrambetween these two points In the foregoing M represents thebending moment E the modulus of elasticity of the materialand I the second moment of area of the member The bendingmoments that occur at the ends of a member subject to thedeformation and restraints shown in the moment diagram atthe top of Table 260 are given by the corresponding formulaeThe formulae which have been derived from a combinationof the basic principles are given in a general form and in thespecial form for members on non-elastic supports

The stiffness of a member is proportional to EIl but as E isassumed to be constant the term that varies in each member isthe stiffness factor K Il The terms FAB and FBA relate to theload on the member When there is no external load FAB andFBA are zero when the load is symmetrically disposed

FAB FBA Al Values of FAB FBA and Al for different loadcases are given in Table 228

The conventional signs for slope-deflection analyses are anexternal restraint moment acting clockwise is positive a slopeis positive if the rotation of the tangent to the elastic line isclockwise a deflection in the same direction as a positive slopeis positive

Example Establish the formulae for the bending moments ina column CAD into which is framed a beam AB The beam ishinged at B and the column is fixed at C and D (see diagramin Table 260) The beam only is loaded Assume there is nodisplacement of the joint A

From the general formulae given on Table 260

Therefore

Thus

For symmetrical loading

MAC 6KAC(AAB lAB)

3KAB 4KAC 4DAD

FAB FBA 2 15AAB lAB

MAB (MAC MAD)

MDA MAD 2

MAD MACKAD KAC

MCA 2EKAC13A MAC 2

MAC 4KAC(FAB FBA 2)

3KAB 4KAC 4KAD

E13A FAB FBA 2

3KAB 4KAC 4KAD

E13A(3KAB 4KAC 4KAC) (FAB FBA2) 0

MAB MAC MAD

MAC 4EKAC13A and MAD 4EKAD13A

MAB 3EKAB13A (FAB FBA 2)

Chapter 14

Framed structures

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257Frame analysis general data

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258Frame analysis moment-distribution method no sway

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259Frame analysis moment-distribution method with sway

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260Frame analysis slope-deflection data

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142 CONTINUOUS BEAMS AS FRAME MEMBERS

In many buildings the interaction of the columns and beamscan be considered with sufficient accuracy by applying one ofthe simplified models shown in Table 257 The simplified two-or three-span sub-frames in Table 261 are analysed on theassumption that the remote ends of the beams and columns arefixed Therefore for any internal span ST the ends of the beamsat R and U the ends of the lower columns at O and X and theends of the upper columns at P and Y are all assumed to befixed In addition the stiffness of the outer beams RS and TUis taken as one-half of the true value For the fixed-endmoments due to normal (ie downward-acting) loads positivenumerical values should be substituted into the tabulatedexpressions If the resulting sign of the support moment isnegative hogging with tension across the top face of the beamis indicated

1421 Internal spans

By slope-deflection methods it can be shown that

where

and is a factor representing the ratio of the assumed to theactual stiffness for the span concerned (ie here 12)

By eliminating 13ST and 13TS in the above and rearranging thefollowing basic formulae are obtained

These formulae which are lsquoexactrsquo within the limitationsof the fixity conditions of the sub-frame represent the case ofthree spans loaded and apply for example to the conditionof dead load For design to BS 8110 the maximum moments atsupports S and T occur when the live load also is applied to allthree spans For design to EC 2 the maximum moment at sup-port S occurs when the live load is applied to spans RS and STand the maximum moment at support T occurs when the live

(FST FSR) (4 DST)(FTU FTS)

MTS FTS DTS

4 DTS DTS2DST 1

DTS 1

(FTU FTS) (4 DTS)(FST FSR)

MST FST DST

4 DSTDTS2DTS 1

DST 1

KT KST KTX KTY KTU

KS KRS KSO KSP KST

13TS (KST 2)(FSR FST) KS(FTS FTU)

KSKT K2ST 4

13ST (KST 2)(FTS FTU) KT(FSR FST)

KS KT K2ST 4

MTS FTS KTS(13TS 13ST 2)

MST FST KST (13ST 13TS 2)

load is applied to spans ST and TU The maximum positivemoment in span ST is obtained when the live load is applied tothis span only for both Codes The appropriate formulae forthese conditions are also given in Table 261 For the ULS theBS 8110 loads are dead 10gk and live 04gk 16qk Thecorresponding loads in EC 2 are dead 135gk and live 15qk

In the foregoing dead and live loads are applied separatelyand the resulting moments are summed Alternatively bothdead and live loads can be applied in a single operation byevaluating the basic formulae with fixed-end moment valuescorresponding to (deadlive) load on the aforesaid spans anddead load only on the remaining spans To comply with EC 2for example in determining the maximum support moment at Sthe fixed-end moments FSR FST and FTS should be calculated fora load of 135gk 15qk while FTU should be evaluated for a loadof 135gk only This method is used in the following example

In accordance with both Codes the moments derived fromthese calculations may be redistributed if desired It should beemphasised that although the diagrams on Table 261 andin the following example are for uniform loads the methodand the formulae are applicable to any type of loadingprovided that the appropriate fixed-end moment coefficientsobtained from Table 228 are used

When the moments MST and MTS at the supports are knownthe positive and negative moments in the spans are obtained bycombining the diagram of free moments due to the design loadswith the diagram of corresponding support moments

1422 End spans

The formulae for any interior span ST are rewritten to applyto an end span AB by substituting A B C and so on for S TU etc (A is the end support and there is no span correspondingto RS) The modified stiffness and distribution factors are givenin Table 261 together with the moment formulae for bothspans loaded and load on span AB only The dead and liveloads should be evaluated and applied so as to obtain therequired support moments as described in section 1421

1423 Columns and adjoining spans

The outer members of the sub-frame have been taken as fullyfixed at their remote ends Thus for a member such as RS theslope-deflection equation is

MSR KRS 13SR

Since the rotation of all the members meeting at a joint is thesame 13SR 13ST Thus by eliminating 13SR and rearranging

Similarly

[2DST (FST FSR) 4(FTU FTS)]

MTU FTU DTU

4 DSTDTS

[2DTS (FTU FTS) 4(FST FSR)]

MSR FSR DSR

4 DSTDTS

Continuous beams as frame members 159

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261Frame analysis simplified sub-frames

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The expressions for the moments in the columns are similar tothe foregoing but FSR and FTU should be replaced by the initialfixed-end moment in the column concerned (normally zero)and the appropriate distribution factor for the column should besubstituted for DSR or DTU

Example Determine the critical ultimate bending moments inbeam ST of the system shown in the following figure belowwhich represents part of a multi-storey frame in accordancewith the requirements of BS 8110 and EC 2 respectively

Stiffness values (mm3)

For upper columns

For lower columns

Distribution factors (if stiffness values are divided by 106)

Support-moment equations

FTS 0132[0953(FST FSR) 3529(FTU FTS)] (4 0471)(FTU FTS)] [2 0471(10497 1)(FST FSR)

MTS FTS [0497 (4 0471 0497)]

FST 0125[1116(FTU FTS) 3503(FST FSR)] (4 0497)(FST FSR)] [2 0497(10471 1)(FTU FTS)

MST FST [0471 (4 0471 0497)]

DTS 273

273 05 273 053 086

273549

0497

DST 273

05 336 273 053 086

273580

0471

K 342 109

4 103 086 106

K 213 109

4 103 053 106

KST KTU 2185 109

8 103 273 106

KRS 2016 109

6 103 336 106

Continuous beams as frame members 161

BS 8110 requirements

Fixed-end moments

For dead load only 10gk 8 kNm

FRS FSR 8 6212 24 kNm

FST FTS FTU FUT 8 8212 427 kNm

For dead live load

14gk 16qk 14 8 16 10 272 kNm

FRS FSR 272 6212 816 kNm

FST FTS FTU FUT 272 8212 1451 kNm

Maximum moments on beam ST

At S (dead live load on all spans)

MST 1451 0125

[1116(1451 1451) 3503(1451 816)]

1451 278 1173 kNm

At T (dead live load on all spans)

MTS 1451 0132

[0953(1451 816) 3529(1451 1451)]

1451 80 1531 kNm

Mid-span (dead live load on ST dead load on RS and TU)

MST 1451 0125

[1116(427 1451) 3503(1451 240)]

1451 387 1064 kNm

MTS 1451 0132

[0953(1451 240) 3529(427 1451)]

1451 325 1126 kNm

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Maximum positive span moment is then approximately

15FST 05(MST MTS) 15 1451 05(1064 1126)

2176 1095 1081 kNm

Maximum moments on column at S (dead live load on STdead load on RS and TU) Distribution factors for lower andupper columns respectively are

Bending moments for lower and upper columns respec-tively are

Alternatively using the method shown on Table 260 for aninterior column when the adjoining beams are analysed asa continuous system on knife-edge supports

KS 05KRS 05 KST KSO KSP

(168 137 053 086) 106 444 106 mm3

Maximum unbalanced fixed-end moment at S

(FSTFRS) 1451240 1211 kNm

MSO 0194 1211 235 kNm

MSP 0119 1211 144 kNm

It will be seen that in this example the results obtained by thetwo methods are almost identical

EC 2 requirements

Fixed-end moments For dead load only

135gk 135 8 108 kNm

FRS FSR 108 6212 324 kNm

FST FTS FTU FUT 108 8212 576 kNm

For dead live load

135gk 15qk 135 8 15 10 258 kNm

FRS FSR 258 6212 774 kNm

FST FTS FTU FUT 258 8212 1376 kNm

DSP 053444

0119DSO 086444

0194

142 kNm

[2 0497(1451ndash427) 4(1451ndash240)]

MSP [0091(4ndash0471 0497)] 231 kNm

[2 0497(1451ndash427) 4(1451ndash240)]

MSO [0148(4ndash0471 0497)]

DSP 053580

0091DSO 086580

0148

Maximum moments on beam ST At S (dead live load on RSand ST dead load on TU)

MST 1376 0125

[1116(576 1376) 3503(1376 774)]

1376 152 1224 kNm

At T (dead live load on ST and TU dead load on RS)

MTS 1376 0132

[0953(1376 324) 3529(1376 1376)]

1376 132 1508 kNm

Mid-span (dead live load on ST dead load on RS and TU)

MST 1376 0125

[1116(576 1376) 3503(1376324)]

1376 349 1027 kNm

MTS 1376 0132

[0953(1376324) 3529(576 1376)]

1376 240 1136 kNm

Maximum positive span moment is then approximately

15FST 05(MST MTS) 15 137605(1027 1136)

20641081 983 kNm

Maximum moments on column at S (dead live load on STdead load on RS and TU)

143 EFFECTS OF LATERAL LOADS

For many structures a close analysis of the bending momentsto which a frame is subjected due to wind or other horizontalloads is unwarranted In such cases the methods illustrated inTable 262 are sufficiently accurate Further information on theuse of these methods is given in section 411

144 PORTAL FRAMES

General formulae for the bending moments in single-storeysingle-bay rigid frames are given in Table 263 (rectangularframes) and Table 264 (gable frames) The formulae whichrelate to any vertical or horizontal load cater for frames withthe columns fixed or hinged at the base Formulae for specificload cases are given in Tables 265 and 266 Formulae for theforces and bending moments in frames with a hinge at the baseof each column and one at the ridge (ie three-hinged frames)are given in Table 267

121 kNm

[2 0497(1376 576) 4(1376 324)]

MSP [0091(4 0471 0497)] 197 kNm

[2 0497(1376 576) 4(1376 324)]

MSO [0148(4 0471 0497)]

Framed structures162

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262Frame analysis effects of lateral loads

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263Rectangular frames general cases

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264Gable frames general cases

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265Rectangular frames special cases

Not

efo

rmul

ae g

i ve

num

eric

al v

alue

s of

rea

ctio

ns a

nd b

endi

ng m

omen

tss

ee d

iagr

ams

for

dire

ctio

n of

act

ion

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266Gable frames special cases

Not

efo

rmul

ae g

ive

num

eric

al v

alue

s of

rea

ctio

ns a

nd b

endi

ng m

omen

tss

ee d

iagr

ams

for

dire

ctio

n of

act

ion

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267Three-hinged portal frames

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In modern multi-storey buildings lateral stability is providedby a system of frames or walls or a combination of both Noteson wall and frame systems are given in section 412 and differentstructural forms with information on typical building heightsand proportions given in ref 37 are shown in Table 268

151 LAYOUT OF SHEAR WALLS AND ALLOCATION OFLATERAL LOADS

Arrangements of shear walls and core units should be such thatthe building is stiff in both flexure and torsion Different planconfigurations with remarks as to their suitability are shown inTable 269 The lateral load transmitted to each wall is afunction of its stiffness and its position in relation to the shearcentre of the system The location of the shear centre can bereadily determined by calculating the moment of stiffness of thewalls about an arbitrary reference point as shown in Table 269The lateral load on each wall can then be determined from thegeneralised formulae given also in Table 269 For most wallswithout openings the dominant mode of deformation is bending(see section 4122) In this case K may be replaced with I inthe generalised formulae where I is the second moment of areaof the cross section

152 PIERCED SHEAR WALLS

In the case of walls pieced by openings the behaviour of theindividual wall sections is coupled to a variable degree Thedeflected shape of the pierced wall is a combination of frameand wall action The wall may be idealised as a plane frame oranalysed by elastic methods in which the flexibility of thebeams is represented as a continuous flexible medium

Solutions using the continuous connection model for a wallcontaining a single line of openings are given in Table 270 Thetotal lateral load F is considered in three different forms aconcentrated load applied at the top of the wall a uniformload distribution and a triangular load distribution with themaximum value at the top of the wall Formulae are given forthe maximum axial force at the base of each wall section themaximum shear force on a connecting beam (and the height ofthe beam above the base of the wall) and the maximumdeflection at the top of the wall Formulae whereby values atany level can be determined are given in ref 38

The main two parameters that define the performance of thewall are and which depend on the geometrical properties

of the wall and beam elements and on the number of lines ofopenings The formulae in Table 270 cater for a wall withdissimilar cross sections on either side of a single line ofopenings For identical cross sections the formulae become

where A1 and I1 refer to each portion of the wall For a wall withtwo symmetrical lines of openings the formulae for the para-meters become

where A1 and I1 refer to each outer portion and I2 refers tothe central portion of the wall Similarly the momentsbecome

There is no axial force in the central portion of the wall

153 INTERACTION OF SHEAR WALLS AND FRAMES

Although the interaction forces between solid walls piercedwalls and frames can vary considerably up the height of a build-ing the effect on the total lateral force resisted by each elementis less significant As a first approximation in order to deter-mine the forces at the bottom of each load-resisting element itis normally sufficient to consider the effect of a single interac-tion force at the top of the building This is equivalent to loadsharing in terms of relative stiffness The location of the shearcentre and the allocation of the lateral load can be determinedas indicated in Table 269 using the following formulae for thestiffness of each element

Solid wall

Pierced wall K 3EIw

H31 2l

21 3

(H)2

3(H)3

K 3EIw

H 3

M2 (M0 Nl )I2

2I1 I2M1

(M0 Nl )I1

2I1 I2

2 12lIe

(2I1 I2)hb3e

2 2

l 2l2 (2I1 I2)

A1

2 6lIe

I1hb3e

2 2

l l2 4I1

A1

Chapter 15

Shear wall structures

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268Structural forms for multi-storey buildings

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269Shear wall layout and lateral load allocation

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270Analysis of pierced shear walls

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Interaction of shear walls and frames 173

Frame

For the frame and are the sums of the second momentsof area of the columns and beams respectively for the lowestbay of height h and beam span lb The factor kc allows for areduction of Ic over the height of the building and is given bykc loge c(c 1) where c Ic top Ic bottom

Example 1 For the shear wall layout in the following figuredetermine the location of the shear centre and the lateral loadapplied to each wall using the formulae in Table 269 Alldimensions are in metres and the walls are 200 mm thick

IbIc

K 12EIc

h2Hkc lbIc

hIb

(0141 0309) F 045F

Example 2 The elevation of a pierced shear wall is shown inthe following figure Identical walls are provided at each end ofa 24 m long rectangular building The building is subjected to acharacteristic wind pressure of 125kNm2 acting on the broadface and the resistance provided by any other frames parallel tothe walls may be ignored The resulting forces and bendingmoments acting on the walls are to be determined

F3y F4y 36(44 0)(F 658)

60128 017F

F2z 20F1415

20(165 242)(F 658)

60128

Layout of shear walls

From symmetry the shear centre of the wall system must be onthe y-axis Similarly the shear centre of the two channelsections (taken together) is at the mid-point of the core unitThe second moment of area values (mm4) of walls 1 and 2 indirection z (discounting the stiffness of walls 3 and 4) are

The distance of the shear centre C from the reference point is

Eccentricity of total lateral force is e 90 242 658 mThe second moment of area values (mm4) of walls 3 and 4 indirection y are

Lateral forces on each wall in terms of total force F are

(0859 0309) F 055F1215F1415

1855F60128

1215(01 242)(F 658)

1215 2322 20 14082 2 36 442

F1z 1215F

1215 20

I3y I4y 02 603

12 36

yc 1215 01 200 165

1215 200 242m

I2z 18 303 14 263

12 20I1z

02 903

12 1215

Elevation of shear wall

If the total wind force acting on the building is shared equallybetween the two walls the horizontal force to be resisted byeach wall is F 125 12 60 900 kN

With reference to Table 270 the following values apply

A1 A2 0225 5 1125 m2

I1 I2 0225 5312 2344 m4

Ib 0225 06312 000405 m4

000333

be (2b 5d)3 (2 2 5 06)3 233

000157

0012860001577 72

2 2 23441125

2 2

l l2 A1 A2

A1A2(I1 I2)

12 7 0003332 2344 3 2333

2 12lIe

(I1 I2)hb3e

0004051 24 (062)2

Ie Ib

1 24(db)2

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Shear wall structures174

Hence 0113 00396 H 0113 60 68Maximum axial force at bottom of each wall element

N asymp

2470 kN

Moment at bottom of each wall element

M1 M2 (FH2 Nl)2 (900 602 2470 7)2 4855 kNm

900 60 0001572 001286 1

268

2

682

FH2

22 1 2

H

2(H)2

Maximum shear force in beams for H 68 occurs wherexH 028 (5th or 6th floor level) and Kv 057 Then

Vb 188 kN

Maximum bending moment in beam

Mb Vb b2 188 22 188 kNm

For subsequent design purposes the forces and moments due tothe characteristic wind load must be multiplied by a partialsafety factor appropriate to the load combination

900 3 000157 057001286

Fh2Kv

2

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In this chapter elastic analyses in terms of characteristic loadsand service stresses are indicated Where limit-state designmethods are employed care must be taken to include theappropriate partial safety factors for the load combination andlimit-state being considered Arch structures may be eitherthree-hinged or two-hinged or fixed-ended as shown in thediagrams at the top of Table 271

161 THREE-HINGED ARCH

For the general case of an unsymmetrical three-hinged archwith a load acting vertically horizontally or at an angle theexpressions for the horizontal and vertical components of thethrusts on the abutments are given in the lower part of Table271 For symmetrical arches the formulae for three-hingedportal frames given in Table 267 are applicable The bendingmoment at each hinge is zero and at any particular section thebending moment shearing force and axial thrust may be deter-mined by considering the loads and abutment reactions on oneside of the section Further information regarding the extent ofthe arch that should be loaded with imposed load in order toproduce the maximum effect at a particular section is given insection 4131

162 TWO-HINGED ARCH

For a symmetrical two-hinged arch the vertical component ofthrust on the abutments is the same as for a freely supportedbeam The horizontal component of thrust H is given by theformula in Table 271 where Mx is the bending moment on asection at distance x from the crown with the arch consideredas a freely supported beam Mx is given by the correspondingexpression in Table 271

The summations 13MxyaI and 13y2aI are taken over the wholelength of the arch In the formula for H which allows for theelastic shortening of the arch A is the average equivalent areaof the arch rib or slab and a is the length of a short segment ofthe axis of the arch where the coordinates of a are x and y asshown in Table 271 If I is the second moment of area of thearch at x then aI aI The bending moment at any section isgiven by Md MxHy

The procedure involves dividing the axis of the arch into aneven number of segments The calculations can be facilitatedby tabulating the successive steps The total bending moment is

required generally only at the crown (x 0 y yc) and thefirst quarter-point (x 025l) The moment Mc at the crown isthe bending moment for a freely supported beam minus HycFor the maximum positive moment at the crown the sum of thevalues of Mc for all elements of dead load is added to the valuesof Mc for only those values of imposed load that give positivevalues of Mc For the maximum negative moment at the crownthe sum of the values of Mc for all elements of dead load isadded to the values of Mc for only those values of imposed loadthat give negative values of Mc The moment at the first quarter-point is the bending moment for a freely supported beam minusHyq where yq is the vertical ordinate of the first quarter-pointThe bending moment is combined with the normal componentof H For an arch of large span it is worthwhile preparing theinfluence lines (F 1) for the bending moments at the crownand at the first quarter-point

163 FIXED ARCH

1631 Determination of thickness

The diagram at the top of Table 272 shows an approximatemethod of determining the section thickness at the crown andthe springing for a symmetrical arch with fixed ends Draw ahorizontal line through the crown C and find G the point ofintersection of the line with the vertical through the centre ofgravity of the total load on half the span of the arch Set outlength GT equal to the dead load on the half span drawn to aconvenient scale draw a horizontal line through T to intersectGS extended at R Draw lines RK perpendicular to GR and GKparallel to the tangent to the arch axis at S On the same scalethat was used to draw GT measure TR which equals Hc andGK which equals Hs If fcc is the maximum allowablecompressive stress in the concrete b the assumed breadth of thearch (1 m for a slab) hc the thickness at the crown and hs thethickness at the springing then approximately

hc 17Hcbfcc hs 2Hsbfcc

The method applies to spans from 12 to 60 m and spanriseratios from 4 to 8 The method does not depend on knowing theprofile of the arch (except for solid-spandrel earth-filled archeswhere the dead load is largely dependent on the shape of thearch) but the span and rise must be known With hc and hs thusdetermined the thrusts and bending moments at the crown

Chapter 16

Arches

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271Arches three-hinged and two-hinged arches

Three-hinged arch

Two-hinged arch

Unsymmetrical three-hinged arches General case

Types of archesInfluence line for section at x

Three-hinged arch

Two-hinged arch

Fixed arch

Note see section 162 for explanation of symbols in and notes on the formulae for two-hinged arches

Reaction formulae for general case

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272Arches fixed-ended arches

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Arches178

springing and quarter-points can be obtained and the stresses onthe assumed sections calculated If the stresses are shown to beunsuitable other dimensions must be tried and the calculationsreworked

1632 Determination of load effects

The following method is suitable for determining load effectsin any symmetrical fixed arch if the dimensions and shape ofthe arch are known or assumed and if the shape of the archmust conform to a particular profile Reference should be madeto section 4133 for general comments on this method

On half of the arch drawn to scale as in Table 272 plot thearch axis Divide the half-arch into k segments such that eachsegment has the same ratio aI aI where a is the length andI is the mean second moment of area of the segment based onthe thickness of the arch measured normal to the axis withallowance being made for the reinforcement

Coordinates x and y relative to the axis of the arch at thecrown are determined by measurement to the centre of lengthof each segment Calculate separately the dead and imposedloads on each segment Assume each load acts at the centre oflength of the segment In an open-spandrel arch the dead andimposed loads are concentrated on the arch at the columnpositions these should be taken as the centre of the segmentsbut it may not then be possible to maintain a constant value ofaI and the value of aI for each segment must be calculated thegeneral formulae in Table 272 are then applicable

For constant values of aI the forces and bending moment atthe crown are as follows

The summations are taken over one-half of the arch The termM1 is the moment at the centre of the segment of all the loadsfrom the centre of the segment to the crown Summations arealso made for all the loads on the other half of the arch forwhich Rc is negative

Due to the elastic shortening of the arch resulting from Hc

where A is the cross-sectional area of the segment calculated onthe same basis as I

Due to a rise (T) or a fall (T) in concrete temperature

where is the coefficient of linear expansion of the concrete Ec

is the short-term modulus of elasticity of the concrete and l1 is

Mc2 Hc2y k

Hc2 (T)kEcl1

2aIky2 y2

Mc1 Hc1ay k

Hc1 Hck(a A)

aIky2 y2

Mc M1 2Hcy

2k

Rc M1x 2x2

Hc kM1y yM1

2ky2 y2

the length of the arch axis Arch shortening due to Hc2 isneglected The effect of concrete shrinkage can be treated as atemperature fall in which T is replaced by (cs) where cs isthe shrinkage strain (allowing for reinforcement restraint) and is a reduction coefficient (typically taken as 043) to allow forthe effect of creep

The foregoing formulae are used to determine the effects ofthe various design loads in the following procedure Calculate(Hc Hc1) Rc and (Mc Mc1) for the dead load Calculate foreach load separately values of Hc2 and Mc2 for temperaturerise temperature fall and concrete shrinkage Calculate for theimposed load (represented as an equivalent uniform load)(Hc Hc1) and (Mc Mc1) To obtain the maximum positivebending moment at the crown (and the horizontal thrust) theimposed load should be applied to the middle third of the archmore or less (By considering the effect of the imposed load onone segment more and one segment less than those in the mid-dle third of the arch the number of segments that should beloaded to give the maximum positive bending moment at thecrown can be determined) With the imposed load applied onlyto those segments that are unloaded when calculating the max-imum positive moment at the crown the maximum negativemoment at the crown due to the imposed load is obtained Themaximum bending moments due to the imposed load are eachcombined with the bending moments due to dead load and archshortening and with the bending moments due to temperaturechange and concrete shrinkage in such a way that the mostadverse total values are obtained The corresponding thrusts arealso calculated and combined with the appropriate bendingmoments to check the design conditions at the crown

The bending moment at the springing due to load at a pointbetween the springing and the crown of the arch distant l fromthe springing (where l is the span of the arch) is

Ms (Mc Mc1) (Hc Hc1)yc Rcl2 Fl

where yc is the rise of the arch For the dead load the valuesdetermined for the crown are substituted in this expressionwith the term Fl replaced by F [(l2) x] To obtain themaximum negative bending moment at the springing theimposed load is applied to those parts of the arch extending 04of the span from the springing more or less (As before theeffect of applying the imposed load to one segment more andone segment less than this distance should be determined toensure that the most adverse loading disposition has beenconsidered) By applying load to only those segments that areunloaded when calculating the maximum negative bendingmoment the maximum positive bending moment is obtainedThese maximum bending moments are each combined with thebending moments due to dead load temperature change andconcrete shrinkage to obtain the most adverse total values Theconditions at the springing are then checked for the combinedeffects of the most adverse bending moments and the corre-sponding normal thrusts

The normal thrust at the springing is given by

Ns (Hc Hc1)cos13 Rssin13

In this expression the vertical component of thrust given byRs (total load on half-arch) Rc is calculated for the loadsused to determine (Hc Hc1)

The shearing force at the springing is given by

Vs (Hc Hc1)sin13 Rscos13

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Fixed parabolic arch 179

In this expression the maximum value is generally obtainedwhen the imposed load extends over the whole arch At thecrown the maximum shearing force is the maximum value ofRc due to any combination of dead and imposed load

The bending moment at a section with coordinates xq and yqdue to load at a point between the springing and the crown ofthe arch distant l from the springing is

Mq Mc Hcyq Rcxq F[xq (05 )l]

At the quarter-point xq l4 and the procedures to determinethe maximum bending moment normal thrust and shearingforce are similar to those described above The method given inTable 273 can be used to obtain data for influence lines

164 FIXED PARABOLIC ARCH

Formulae and guidance on using the data in Table 274 aregiven below See section 4134 for further comments

1641 Dead load and arch shortening

The horizontal thrust due to dead load alone is H k1gl2ycwhere g is the dead load per unit length at the crown l is thespan and yc is the rise of the arch axis The coefficient k1

depends on the dead load at the springing which varies with therisespan ratio and type of structure that is whether the arch isopen spandrel or solid spandrel or whether the dead load isuniform throughout the span

An elastic shortening results from the thrust along the archaxis (assuming rigid abutments) The counter-thrust H1 whileslightly reducing the thrust due to the dead load renders thisthrust eccentric and produces a positive bending moment at thecrown and a negative bending moment at each springing If h isthe thickness of the arch at the crown

in which the coefficient k2 depends on the relative thickness atthe crown and the springing

Due to dead load and elastic shortening the net thrusts Hc

and Hs at the crown and the springing respectively acting par-allel to the arch axis at these points are given by

where 13 is the angle between the horizontal and the tangent tothe arch axis at the springing with cos13 given in Table 274

The bending moments due to the eccentricities of Hc and Hs

are given by Mc k3ycH1 and Ms (k3 1) ycH1 respectively

1642 Temperature change

The additional horizontal thrust due to a rise in temperature orcorresponding counter-thrust due to a fall in temperature isgiven by

If T is the temperature change in oC with h and yc in metresthen H2 is in kN per metre width of arch The values of k4 inTable 274 are based on an elastic modulus Ec 20 kNmm2and a coefficient of linear expansion 12 micro-strainoC

H2 k4hyc

2

hT

Hs H

cos 13 H1 cos 13Hc H H1

H1 k2hyc

2

H

For any other values of Ec and k4 should be multiplied by avalue of (Ec20)(12) At the crown the increase or decrease innormal thrust due to a change in temperature is H2 and thebending moment is k3ycH2 account being taken of the sign ofH2 The normal thrust at the springing due to a change oftemperature is H2cos13 and whether this thrust increases ordecreases the thrust due to dead load depends on the sign of H2At the springing the bending moment is (1 k3)ycH2 the signbeing the same as that of H2

1643 Shrinkage of concrete

The effect of concrete shrinkage can be treated as a fall intemperature in which T is replaced by (cs)( 12) wherecs (micro-strain) is the shrinkage (allowing for reinforcementrestraint) and is a reduction coefficient (taken as 043) toallow for the effect of creep

1644 Imposed load

The maximum bending moments and corresponding thrustsand reactions are given by the following expressions whereq per unit length is the intensity of uniform load equivalent tothe specified imposed load

At the crown positive bending moment k5 ql2

horizontal thrust k6 ql2yc

At the springing negative bending moment k7 ql2

horizontal thrust k8 ql2yc

vertical reaction k9 ql

positive bending moment k10 ql2

horizontal thrust k11 ql2yc

vertical reaction k12 ql

The normal thrust at the springing where H is the horizontalthrust and R is the vertical reaction is given by

N Hcos13 Rradic(1 cos213)

1645 Dimensions of arch

The line of thrust and therefore the arch axis can be plotted asdescribed in section 4134 The thickness of the arch at thecrown and the springing having been determined the lines ofthe extrados and intrados can be plotted to give a parabolicvariation of thickness between the two extremes Thus thethickness normal to the axis of the arch at any point is given by[(hs h)2 h] where is the ratio of the distance of the pointfrom the crown measured along the axis of the arch to half thelength of the axis of the arch

Example Determine the bending moments and thrusts on afixed parabolic arch slab for an open-spandrel bridge

Specified values

Span 50 m measured horizontally between the intersection ofthe arch axis and the abutment Rise 75 m in the arch axisThickness 900 mm at springing 600 mm at crownDead load 12 kNm2 imposed load 15 kNm2Temperature range 24oC 12 micro-strain per oC

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273Arches computation chart for symmetrical fixed-ended arch

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274Arches fixed-ended parabolic arches

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Arches182

Ec 28 kNmm2 cs 200 micro-strain 043

Geometrical properties

yc

l

7550

015hs

h

900600

15

Inclination of arch axis at springing (Table 273) cos13 082A strip of slab 1 m wide is considered The coefficients takenfrom Table 274 are substituted in the expressions given insection 1644 Allowing for self-weight of arch slab

Total dead load at crown 12 06 24 264 kNm2

Loads and resulting bending moments and thrusts Moment ThrustkNmm kNm

Horizontal thrusts due to dead load and other actions

Dead load (k1 0140)H 0140 264 (50275) 1232

Arch shortening (k2 139)H1 139 (0675)2 1232 11

Temperature change (k4 422)H2 422 (2820) (0675)2 06 24 55

Concrete shrinkage (k4 422)H3 422 (2820) (0675)2 06 (20012) 043 16

Crown maximum positive bending moment and corresponding thrust

Dead load and arch shortening (k3 0243 thrusts H and H1 as above) 20Mc 0243 75 11Hc 1232 11 1221

Temperature fall (thrust H2 as above) 55Mc 0243 75 55 100

Shrinkage (thrust H3 as above) 16Mc 0243 75 16 29

Imposed load (k5 0005 k6 0064)Mc 0005 15 502 188Hc 0064 15 (50275) ____ 320

Totals 337 1470

Springing maximum negative bending moment and corresponding thrust

Dead load and arch shorteningMs (1 0243) 75 11 63Hs (1232082) 11 082 1494

Temperature fallMs (1 0243) 75 55 312Hs 55 082 45

ShrinkageMs (1 0243) 75 16 91Hs 16 082 13

Imposed load (k7 0020 k8 0038 k9 0352)Ms 0020 15 502 750H 0038 15 (50275) 190 kNm R 0352 15 50 264 kNmN 190 082 264radic(1 0822) 307

Totals 1216 1743

Springing maximum positive bending moment and corresponding thrust

Dead load and arch shortening (values as before) 63 1494Temperature rise (values for temperature fall) 312 45Shrinkage (neglect as partial in short term and beneficial in long term)Imposed load (k10 0023 k11 0089 k12 0151)

Ms 0023 15 502 863H 0089 15 (50275) 445 kNm R 0151 15 50 113 kNmN 445 082 113radic(1 0822) 430

Totals 1112 1969

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In the following containers are conveniently categorised astanks containing liquids and bunkers and silos containing drymaterials each category being subdivided into cylindrical andrectangular structures The intensity of pressure on the wallsof the structure is considered to be uniform at any level butvertically the pressure increases linearly from zero at the top toa maximum at the bottom

171 CYLINDRICAL TANKS

If the wall of a cylindrical tank has a sliding joint at the baseand is free at the top then when the tank is full no radial shearor vertical bending occurs The circumferential tension at depthz below the top is given by n ri z per unit height where ri

is the internal radius of the tank and is unit weight of theliquid If the wall is supported at the base in such a way that noradial movement can occur radial shear and vertical bendingresult and the circumferential tension is always zero at thebottom of the wall Values of circumferential tensions verticalmoments and radial shears according to values of the termheight2(2 mean radius thickness) can be obtained fromTables 275 and 276

The tables apply to walls with a free top and a bottom that iseither fixed or hinged The coefficients have been derivedby elastic analysis and allow for a Poissonrsquos ratio of 02 Forfurther information on the tables reference should be made topublications on cylindrical tanks such as refs 55 and 56 If anannular footing is provided at the base of the wall a hingeddetail can be formed although this is rarely done The footingnormally needs to be tied into the floor of the tank to preventradial movement Reliance solely on the frictional resistanceof the ground to the radial force on the footing is generally inad-equate and always uncertain If the joint between the wall andthe footing is continuous it is possible to develop a fixedcondition by widening the footing until a uniform distributionof bearing pressure is obtained In many cases the wall and thefloor are made continuous and it then becomes necessary toconsider the structural interaction of a cylindrical wall and aground supported circular slab Appropriate values for the stiff-ness of the member and the effect of edge loading can beobtained from Table 276 for walls and Table 277 for slabs

For slabs on an elastic foundation the values depend on theratio rrk where rk is the radius of relative stiffness defined insection 725 The value of rk is dependent on the modulus of

subgrade reaction for which data is given in section 724Taking rrk 0 which corresponds to a lsquoplasticrsquo soil state isappropriate for an empty tank liable to flotation

Example 1 Determine due to internal hydrostatic loadingmaximum service values for circumferential tension verticalmoment and radial shear in the wall of a cylindrical tank that isfree at the top edge and hinged at the bottom The wall is300 mm thick the tank is 6 m deep the mean radius is 10 mand the water level is taken to the top of the wall

From Table 275 for lz22rh 62(2 10 03) 6

n 0643 at zlz 07 m 0008 at zlz 08

v 0110 at zlz 10

n n lzr 0643 981 6 10 3785 kNm

m mlz3 0008 981 63 170 kNmm

v vlz2 0110 981 62 389 kNm

Example 2 Determine for the tank considered in example 1the corresponding values if the wall is fixed at the bottom

From Table 275 for lz22rh 62(2 10 03) 6

n 0514 at zlz 06 m 0005 at zlz 07

mb 0019 and v 0197 at zlz 10

n nlzr 0514 981 6 10 3025 kNm

m mlz3 0005 981 63 106 kNmm

m mlz3 0019 981 63 403 kNmm

v vlz2 0197 981 62 696 kNm

Consider a straight wall that is centrally placed on a footing ofwidth b (2a h) where a represents the distance from edgeof footing to face of wall The weight of liquid per unit lengthof wall on the inside of the footing is given by lza Assuminga uniform distribution of bearing pressure due to the liquid loadthe bending moment about the centre of the wall due to thepressure on the toe of the footing is

mlza (ab) (ah)2 giving a2(a h) 2(mlz)b

Substituting for b h and m lz 403(981 6) 0685 thefollowing cubic equation in a is obtained

a2 (a03) 137 (2a 03)

Chapter 17

Containmentstructures

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Cylindrical tanks elastic analysis ndash 1 275

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Cylindrical tanks elastic analysis ndash 2 276

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Cylindrical tanks elastic analysis ndash 3 277

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Cylindrical tanks 187

Solution of the equation by trial and error gives a 16 m andwidth b 2 16 03 35 m This width is correct for theannular footing provided the footing is positioned so that itscentre of area coincides with the centre of the wall

If c represents the distance from the outer edge of the footingto the centreline of the wall then for a unit length of wall thelengths of the trapezoidal area are (r c)r for the outer edgeand (r c b)r for the inner edge The usual formula for atrapezoidal area gives

[(r c)2(rcb)] br3[(rc) (rcb)] cr

Substituting for b and r and rearranging the terms gives

6c2 39c 805 0 from which c 165 m

Thus a fixed edge condition can be obtained by providing a35 m wide footing with the outer edge of the footing 15 mfrom the outer face of the wall

Example 3 Determine for the tank considered in example 1the corresponding values if the wall is continuous with the floorslab The slab is 400 mm thick and the soil on which the tank isto be built is described as well-compacted sand

Properties of cylindrical wall From Table 276 with lz22rh 6w 0783 and the wall stiffness is given by

Kw w Ec h3lz (0783 0336) Ec 00035 Ec

Also w1 00187 and the fixed edge moment when the tankis full is given by

Mw w1 lz3 00187 981 63 396 kNmm

Properties of circular slab on elastic subgrade From section724 for well-compacted sand a mean value of 75 MNm3 canbe taken for the modulus of subgrade reaction

From Table 277 for a slab on an elastic subgrade the radiusof relative stiffness with 02 is given by

rk [Ech31152 ks]025

Therefore taking Ec 33 kNmm2 and ks 75 MNm3

rk [33 109 043(1152 75 106)]025 125 m

With rrk 8 s 0468 and the slab stiffness is given by

Ks sEch3r (0468 04310) Ec 00030 Ec

The unit edge load on the slab due to the lsquoeffectiversquo weight ofthe wall is

Q 03 6 (24 981) 255 kNm

With rrk 8 s2 0088 and the corresponding fixed edgemoment is given by

Ms s2 Qr 0088 255 10 225 kNmm

Moment distribution at joint Since the calculated fixed edgemoments for the wall and the slab both act in the same directionthe joint will rotate when the notional restraint is removed Thiswill induce additional moments and change the circumferentialtensions in the wall At the joint the induced moments will be

proportional to the relative stiffness values of the two elementsaccording to the following distribution factors

wall slab

Element Wall SlabDistribution factor 054 046Fixed end moment 396 225Induced moment 335 286Final moment 61 61

It can be seen from the moments calculated for the wall thatthe joint rotation is close to that for a hinged base conditionThe use of a thinner slab or a lower value for the modulus ofsubgrade reaction will increase the rotation until a lsquoclosingrsquocorner moment develops and the circumferential tensions inthe wall exceed those obtained for a hinged base

Final forces and moments The final circumferential tensionsand vertical moments at various levels in the wall can beobtained by combining the results for load cases (1) and (3) inTable 275 where M is the induced moment The followingequations apply

n n1 lz r n3Mrlz2

(981 6 10) n1 (335 1062) n3

5886 n1 93 n3 kNm

m m1 lz3 m3M (981 63) m1 335 m3

2119 m1 335 m3

The resulting values for different values of zlz are shown in thefollowing tables

0003000035 00030

0460003500035 00030

054

Circumferential tensions in wall (kNm)

zlz

Load case (1) Load case (3) Final

n1 5886 n1 n3 93 n3force

05 0504 2967 334 311 327806 0514 3025 654 608 363307 0447 2631 103 958 358908 0301 1772 131 1218 299009 0112 659 114 1060 1719

Vertical moments in wall (kNmm)

zlz

Load case (1) Load case (3) Final

m1 2119 m1 m3 335 m3moment

06 00046 98 0037 12 8607 00051 108 0057 19 12708 00029 62 0252 84 14609 00041 87 0572 192 10510 00187 396 10 335 61

The final radial shear at the base of the wall is given by

v v1lz2 v3Mlz 0197 981 62 449 3356 445 kNm

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The new moment distribution factors are as follows

wall slab

Element Wall SlabDistribution factor 084 016Fixed end moment 22 1080Induced moment 926 176Final moment 904 904

It will be seen that in this example the final edge moment onthe slab is close to a fixed edge condition since the stiffnessof the wall is much greater than that of the slab Final values forthe moments and forces in the slab and the wall can now becalculated as shown in example 3

172 RECTANGULAR TANKS

The bending moments obtained by Table 253 for individualrectangular panels with triangularly distributed loads may beapplied without modification to continuous walls providingthere is no rotation about the vertical edges In a square tanktherefore moment coefficients can be taken directly fromTable 253 For a rectangular tank distribution of the unequalfixity moments at the wall junctions is needed and momentcoefficients for tanks of different span ratios are given in Tables278 and 279 The shearing forces given in Table 253 for theindividual panels may still be used

The tables give values for idealised edge conditions at the topand bottom of the wall The top of a wall should be taken as freefor an open tank and when a sliding joint is provided betweena roof and the top of the wall If a wall is continuous with theroof the joint will rotate when the tank is full and the conditionwill tend towards hinged When there is earth loading on thewall a closing-corner moment will develop when the tank isempty At the bottom of the wall a hinged condition may becreated by providing a narrow wall footing tied into the floorslab or by adopting a reinforced hinge detail Where the tankwall is continuous with the floor the deformation of the floorresulting from the wall loading is often complex and highlydependent on the assumed ground conditions The edge condi-tion at the bottom of the wall is generally uncertain but willtend towards hinged when the tank is full When there is earthloading on the wall the edge condition will tend towards fixedwhen the tank is empty If the floor is extended outwards toform a toe the condition at the base will be affected in the waydiscussed in section 171

In considering the horizontal spans the shear forces at thevertical edges of one wall result in axial forces in the adjacentwalls Thus for internal loading on a rectangular tank the shearforce at the end of the long wall is equal to the tensile force inthe short wall and vice versa In designing sections the com-bined effects of the bending moment the axial force and theshear force need to be considered

Example 5 Determine due to internal hydrostatic loadingthe maximum service moments and shear forces in the walls ofa rectangular tank that can be considered as free along the topedge and hinged along the bottom edge The tank is 6m long

0003000035 00007

0160003500035 00007

084

Containment structures188

Tangential moments in slab (kNmm)

rxrLoad case (1) Load case (2) Final

t1 286t1 t2 255t2moment

10 0378 108 0018 46 6208 0132 38 0009 23 6106 0020 06 0006 15 0904 0034 10 0 0 1002 0014 04 0002 05 090 0005 02 0002 05 07

Example 4 Determine for the tank considered in example 3the factor of safety against flotation and the moments in theslab if the worst credible groundwater level is 15 m above theunderside of the slab

The radius of the slab is 103 m and radius to the centre ofthe wall is 1015 m If the weight of the wall is spread uniformlyover the full area of the slab the total downward pressure dueto the weight of the slab and the wall is

[04 03 6 (2 10151032)] 24 178 kNm2

The upward pressure due to a 15 m depth of groundwater is147 kNm2 giving a safety factor of 178147 121 whichsatisfies the BS 8007 minimum requirement of 11 The unitedge load on the slab due to the weight of the wall is

Q 03 6 24 432 kNm

Taking rrk 0 (for a lsquoplasticrsquo soil condition) s 0104and s2 0250 giving the following values

Ks sEch3r (0104 04310)Ec 00007 Ec

Ms s2Qr ndash 0250 432 10 108 kNmm

The wall stiffness is the same as before and the effect of thegroundwater (11 m from the top of the slab) on the wall maybe taken as a simple cantilever giving values

Kw wEch3lz (0783 0336) Ec 00035 Ec

Mw w1lz3 981 1136 22 kNmm

Radial moments in slab (kNmm)

rxrLoad case (1) Load case (2) Final

r1 286r1 r2 255r2moment

10 10 286 0088 225 6108 0488 140 0013 33 17306 0028 08 0016 41 4904 0055 16 0003 08 0802 0023 07 0001 03 100 0005 02 0002 05 07

The final radial and tangential moments in the floor slab can beobtained by combining the results for load cases (1) and (2) inTable 277 The following equations apply

mr r1M r2 Qr and mt t1M t 2 Qr

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278Rectangular tanks triangularly distributed load(elastic analysis) ndash 1

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279Rectangular tanks triangularly distributed load(elastic analysis) ndash 2

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4m wide and 4m deep and the water level is to be taken to thetop of the walls

From Table 279 for lxlz 64 15 lylz 44 10 andtop edge free maximum bending moments are as follows

Horizontal negative moment at corners

mx my 0050 981 43 314 kNmm

Horizontal positive moments (at about 05lz 2m above base)

mx 0035 981 43 220 kNmm (long wall)

my 0011 981 43 69 kNmm (short wall)

Vertical positive moments (at about 03lz 12m above base)

mzx 0029 981 43 182 kNmm (long wall)

mzy 0013 981 43 82 kNmm (short wall)

From Table 253 for panel type 4 with lhlz 15 (long wall)and 10 (short wall) maximum shear forces are as follows

Horizontal shear forces at side edges

vhx 037 981 42 581 kNm (long wall)

vhy 031 981 42 487 kNm (short wall)

Vertical shear forces at bottom edge

vzx 026 981 42 408 kNm (long wall)

vzy 019 981 42 298 kNm (short wall)

From the above the horizontal tensile forces in the walls are

nhx 487 kNm (long wall) nhy 581 kNm (short wall)

173 SILOS

Notes on the properties of stored materials and the pressuresset up in silos of different forms and proportions are given insections 277 and 93 and Tables 215 and 216 Notes on thedesign of silo walls are given in section 641 For rectangularsilos that are divided into several compartments where thewalls span horizontally expressions for the bending momentsand reactions are given in Table 280

174 BOTTOMS OF ELEVATED TANKS AND SILOS

1741 Tanks

The figure here shows the floor of an elevated cylindrical tanksupported by beams in two different arrangements

area with the remainder of the floor load the weight of the walland the load from the roof being equally divided between theeight cantilever lengths An alternative to (b) is to place thecolumns almost under the wall in which case the cantilevers areunnecessary but secondary beams might be required

For a tank of large diameter a domed bottom of one of thetypes shown in Table 281 is more economical and althoughthe formwork is much more costly the saving in concrete andreinforcement compared to beam-and-slab construction can beconsiderable Ring beams A and C in the case of a simple domedbottom resist the horizontal component of thrust from the domeand the thickness of the dome is determined by the magnitudeof the thrust Expressions for the thrust and the vertical shearingforce around the edge of the dome and the resultant circumfer-ential tension in the ring beam are given in Table 281 Domesused to form the bottom or the roof can also be analysed by themethod given in Table 292

A bottom consisting of a central dome and an outer conicalpart as illustrated in Table 281 is economical for the largesttanks This form of construction is traditionally known as anIntze tank The outward thrust from the top of the conical partis resisted by the ring beam B and the difference between theinward thrust from the bottom of the conical part and the out-ward thrust from the domed part is resisted by the ring beam AExpressions for the forces are given in Table 281 and the pro-portions of the conical and domed parts can be chosen so thatthe resultant thrust on beam A is zero Suitable proportions forbottoms of this type are given in Table 281 and the volume ofa tank with these proportions is 0604 do

3 where do is the diam-eter of the tank The tank wall should be designed as describedpreviously account being taken of the vertical bending at thebase of the wall and the effect of this bending on the conicalpart The floor must be designed to resist in addition to theforces and bending moments already described any radialtension due to the vertical bending of the wall

Example 6 Determine for service loads the principal forcesin the bottom of a cylindrical tank of the Intze type where

do 10 m d 8 m 48o (giving cot 0900)

13 40o (giving cot 13 1192 and cosec13 155)

F1 2500 kN F2 2800 kN and F3 1300 kN

Then from Table 281 the following values are obtained

Unit vertical shearing force at periphery of dome

V1 F1(d) 2500(8) 100 kNm

Unit thrust at periphery of dome

N1 V1cosec13 100 155 155 kNm

(Note The required thickness of the dome at the springing isdetermined by the values of N1 and V1)

Unit outward horizontal thrust from dome on ring beam A

H1 V1cot 13 100 1192 1192 kNm

Unit vertical shearing force at inner edge of cone

V2 (F2 F3)(d) (2800 1300)(8) 163 kNm

Unit inward horizontal thrust from cone on ring beam A

H2 V2cot 163 0900 1468 kNm

Circumferential compression in ring beam A

NA 05d(H2 H1) 058 (14681192)1104 kN

Bottoms of elevated tanks and silos 191

(a) (b)

In (a) each beam spans between opposite columns and carriesone-quarter of the load on the floor The remaining half of thefloor load the weight of the wall and the load from the roof aretransferred to the columns through the wall In (b) each lengthof beam between the columns carries the load on the shaded

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Rectangular containers spanning horizontallymoments in walls 280

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Bottoms of elevated tanks and silos 281

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(Note If H1 exceeds H2 the circumferential force is tensile Theideal case occurs when H1 H2 and NA 0 see as follows)

Unit vertical shearing force at outer edge of cone

V3 F3(do) 1300(10) 414 kNm

Unit outward horizontal thrust from cone on ring beam B

H3 V3 cot 414 0900 372 kNm

Circumferential tension in ring beam B

NB 05do H3 05 10 372 1862 kN

The vertical wall must be reinforced for the circumferentialtension due to the horizontal pressure of the contained liquidgiven by n 05doz per unit depth The conical part of thetank must be reinforced to resist circumferential tension andthe reinforcement may be either distributed over the height ofthe conical portion or concentrated in the ring beams at the topand bottom In large-diameter Intze tanks the width of ringbeam A can be considerable and if this is so the weight ofwater immediately above the beam should not be taken to con-tribute to the forces on the dome and conical part With a widebeam F1 is taken as the weight of the contents over the net areaof the dome and d as the internal diameter of the ring beamF2 is taken as the weight of the contents over the net area of theconical part and d for use with F2 as the external diameterof the ring beam If this were to be done for a ring beam ofreasonable width in the forgoing example H1 and H2 would bemore nearly balanced

1742 Columns supporting elevated tanks

For a group of four columns on a square grid the thrusts andtensions in the columns due to wind loading on the tank can

be calculated as follows When the wind blows normal to theside of the group the thrust on each column on the leeward sideand the tension in each column on the windward side is Mw2dwhere d is the column spacing and Mw is the total moment dueto the wind When the wind blows normal to a diagonal of thegroup the thrust on a leeward corner column and tension in awindward corner column is Mwdradic2 For any other arrangementthe force on any column can be calculated from the equivalentsecond moment of area of the group

Consider the case when the wind blows normal to the X-Xaxis of the group of eight columns shown in the followingfigure The second moment of area of the group of columns aboutthe axis is 2 (d2)2 4 (0353 d)2 10 d2 The thruston the extreme leeward column is (05d10 d2) Mw Mw2dThe forces on each of the other columns in the group can bedetermined similarly by substituting the appropriate value forthe distance of the column from the axis

Containment structures194

1743 Silos

Notes on the pressures set up on hopper bottoms are given insections 277 and 93 and Tables 215 and 216 Notes on thedesign of hopper bottoms in the form of inverted truncatedpyramids are given in section 642 Expressions for bendingmoments and tensile forces are given in Table 281

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181 PAD FOUNDATIONS

Some general notes on the design of foundations are given insection 71 The size of a pad or spread foundation is usuallydetermined using service loads and allowable bearing valuesThe subsequent structural design is then determined by therequirements of the ULS Presumed allowable bearing valuesrecommended for preliminary design purposes in BS 8004 aregiven in Table 282

1811 Separate bases

An introduction to separate bases is given in section 716Diagrams of bearing pressure distributions and expressions forpressures and maximum bending moments in rectangular basessubjected to concentric and eccentric loading are given inTable 282

Example 1 The distribution of bearing pressure is requiredunder a base 3 m long 25 m wide and 600 mm thick when itsupports a concentrated load of 1000 kN at an eccentricity of300 mm in relation to its length

Weight of base Fb 30 25 06 24 108 kNTotal load Ftot Fb Fv 108 1000 1108 kN

Eccentricity of total load

etot Fve Ftot (1000 03)1108 027 m

Since etot l6 306 05 m the bearing pressure diagramis trapezoidal and the maximum and minimum pressures are

f (1 6etot l)Ftot bl (1 6 02730)[1108(25 30)] (154 and 046) 148 228 kNm2 and 68 kNm2

Note For the structural design of the base Fb and Fv shouldinclude safety factors appropriate to the ULS Bearing pressurescorresponding to these values of Fb and Fv reduced by Fbblshould then be used to determine bending moments and shearforces for the subsequent design

1812 Combined bases

When more than one column or load is carried on a single basethe centre of gravity of the total load should coincide if possiblewith the centre of area of the base Then assuming a rigid

base the resulting bearing pressure will be uniformly distributedThe base should be symmetrically disposed about the line of theloads and can be rectangular or trapezoidal on plan as shownin Table 283 Alternatively it could be made up as a series ofrectangles as shown at (b) in Table 284 In this last case eachrectangle should be proportioned so that the load upon it acts atthe centre of the area with the area of the rectangle being equalto the load divided by an allowable bearing pressure and thevalue of the pressure being the same for each rectangle

If it is not practical to proportion the combined base in thisway then the total load will be eccentric If the base is thickenough to be considered to act as a rigid member the groundbearing pressure will vary according to the diagram shown at(c) in Table 284 For a more flexible base the pressure will begreater immediately under the loads giving a distribution ofpressure as shown at (d) in Table 284

In the case of a uniform distribution or linear variation of pres-sure the longitudinal bending moment on the base at any sectionis the sum of the anti-clockwise moments of the loads to the leftof the section minus the clockwise moment due to the groundpressure between the section and the left-hand end of the baseThis method of analysis gives larger values for longitudinal bend-ing moments than if a non-linear variation is assumed Formulaefor combined bases carrying two loads are given in Table 283

Example 2 A strip base 15 m long and 15 m wide carries aline of five unequal concentrated loads arranged eccentricallyas shown in the following figure The bending moment is to bedetermined at the position of load F2 where the values of theloads and the distances from RH end are as follows

F1 500 kN F2 450 kN F3 400 kN F5 300 kN

z1 14 m z2 11 m z3 8 m z4 5 m z5 15 m

Chapter 18

Foundations andretaining walls

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282Foundations presumed allowable bearing values and separate bases

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283Foundations other bases and footings

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284Foundations inter-connected bases and rafts

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Pad foundations 199

The first step is to determine the eccentricity of the total loadand check that e (13Fz13F l2) l6 Hence

13Fz 500 14 450 11 400 8 350 5 300 15

17350 kNm

13F 500 450 400 350 300 2000 kN

e 173502000 152 1175 m l6 25 m

The maximum and minimum bearing pressures can now becalculated from the formula in Table 282 where

k (1 6el) (1 6 117515) 147 or 053

Hence with 13Fbl 2000(15 15) 889 kNm2

fmax 147 889 1307 kNm2

fmin 053 889 471 kNm2

Then for any section X-X at distance y from the left-hand endof the base the bearing pressure is

fx fmax (fmax fmin) yl

Considering the loads to the left of section X-X where x is thedistance of a load from the section the resultant bendingmoment on the base at section X-X is

M 13Fx (2fmax fx) by26

Thus at the position of load F2 where y 4 m and x1 3 m

fx 1307 (1307 471) 415 1084 kNm2

M 500 3 (2 1307 1084) 15 426 208 kNm

1813 Balanced bases

With reference to the diagrams on the upper right-hand side ofTable 284 (a) shows a system in which beam BC rests on a baseat A supports a column on the overhanging end C and iscounterbalanced at B The reaction at A which depends on therelative values of BC and BA can be provided by a basedesigned for a concentric load The counterbalance at B couldbe provided by load from another column as at (b) in which casethe dead load on this column needs to be sufficient to counter-balance the dead and imposed loads on the column at C and viceversa It is often possible to arrange for base A1 to be positioneddirectly under column B Formulae giving the values of thereactions at A and A1 are given in Table 283 where variouscombinations of values for F1 and F2 usually need to be consideredThus if F1 varies from F1 max to F1 min and F2 varies from F2 max toF2 min reaction R (see diagram in Table 283) will vary from

Rmax e F1max l to Rmin e F1min l

Hence R1 and R2 can have the following values

R1max F1max Fbase 1 Rmax Fbeam 2

R1min F1min Fbase 1 Rmin Fbeam 2

R2max F2max Fbase 2 Rmin Fbeam 2

R2min F2min Fbase 2 Rmax Fbeam 2

Therefore base 1 must be designed for a maximum load of R1max

and base 2 for R2max but R2min must always be positive From thereactions the shearing forces and bending moments on the beamcan be calculated In the absence of a convenient column loadbeing available at B a suitable anchorage must be provided by

other means such as a counterweight in mass concrete as at (c)or the provision of tension piles If the column to be supported isa corner column loading the foundation eccentrically in twodirections one parallel to each building line as at (d) it is some-times possible to use a diagonal balancing beam anchored by aninternal column D In other cases however the two wall beamsmeeting at the column can be designed as balancing beams toovercome the double eccentricity For beam EC the cantilevermoment is FE e1 where FE is the column load and the upwardforce on column C is FE e1 (l1 e1) For beam EF the corre-sponding values are FE e2 and FE e2 (l2 e2) respectively

1814 Rafts

The required thickness of a raft foundation is determined by theshearing forces and bending moments which depend on themagnitude and spacing of the loads If the thickness does notexceed 300 mm a solid slab as at (a) in the lower part ofTable 284 is generally the most convenient form If a slab atground level is required it is usually necessary to thicken theslab at the edge as at (b) to ensure that the edge of the raft isdeep enough to avoid weathering of the ground under the raft Ifa greater thickness is required beam-and-slab constructiondesigned as an inverted floor as at (c) is more efficient In caseswhere the total depth required exceeds 1 m or where a level topsurface is required a cellular construction consisting of a topand bottom slab with intermediate ribs as at (d) can be adopted

When the columns on a raft are not equally loaded or are notsymmetrically arranged the raft should be designed so that thecentre of area coincides with the centre of gravity of the loadsIn this case the pressure on the ground is uniform and therequired area is equal to the total load (including the weight ofthe raft) divided by the allowable bearing value If the coinci-dence of the centre of area of the raft and the centre of gravityof the loads is impractical due to the extent of the raft beinglimited on one or more sides the shape of the raft on planshould be so that the eccentricity ew of the total load Ftot is keptto a minimum as in the example shown at (f)

The maximum bearing pressure (which occurs at the cornershown at distance a1 from axis N-N on the plan and should notexceed the allowable bearing value) is given by

where Araft is the total area of the raft and Iraft is the secondmoment of area about the axis N-N which passes through thecentre of area and is normal to the line joining the centre of areaand the centre of gravity of the loads The pressure along axisN-N is FtotAraft and the minimum pressure (at the corner shownat distance a2 from axis N-N) is given by

When the three pressures have been determined the pressure atany other point or the mean pressure over any area can beassessed Having arranged for a rational system of beams orribs to divide the slab into suitable panels as suggested by thebroken lines at (f) the panels of slab and the beams can bedesigned for the bending moments and shear forces due to thenet upward pressures to which they are subjected

fmin Ftot

Araft

Ftot ewa2

Iraft

fmax Ftot

Araft

Ftot ew a1

Iraft

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182 OPEN-PILED STRUCTURES

Expressions from which the loads on groups of inclined andvertical piles supporting jetties and similar structures can beobtained are given in Table 285 For each probable conditionof load the forces acting on the superstructure are resolved intohorizontal and vertical components Fh and Fv the points ofapplication of which are also determined If the direction ofaction and position of the forces are opposite to those in thediagrams the signs in the formulae must be changed

Example 1 vertical piles only The adjacent figure showsa cross section through a jetty where the loads apply to one rowof piles (ie n 1 for each line and 13n 4) Since the groupis symmetrical x 05 90 45 m

M Fv e Fh h 800 075 100 48 120 kNm

The calculations for the loads on the piles are shown in thefollowing table from which the maximum load on any pile is212 kN For each pile the shear force is 1004 25 kN andthe bending moment is 25 482 60 kNm

Foundations and retaining walls200

Example 2 vertical and inclined piles The adjacent figureshows a cross section through a jetty where all the piles aredriven to the same depth Since A is the same for all the pilesif J Al is taken as unity for piles N1 and N4 then for piles N2

and N3 J 4radic(1 42) 097 Since the group is symmetrical132 0 133 not needed 134 0 k 131 135 xo 92 45 mand yo 0

M Fv e Fh h 800(525 45) 0 600 kNm

The calculation is then as shown in the following table fromwhich k 3826 0114 0436 The axial loads on the pilesare given by Nx kp (kw Fv kh Fh km M) hence

N1 10 (0261 800 0 0111 600) 1421 kN

N2 0941 (0261 800 219 100 0) 4026 kN

N3 0941 (0261 800 219 100 0) 96 kN

N4 10 (0261 800 0 0111 600) 2755 kN

Thus the maximum load on any pile is 4026 kN Note that theweight of the pile has to be added to the above values

kw km

Pile x x2 Axial loadno (m) (m2) (kwFv kmM)

N1 45 2025 025 45 0100 (025 X 800) (0100 X 120) 188kN45

N2 15 225 025 15 0033 200 (0033 X 120) 196kN45

N3 15 225 025 0033 200 4 204kN

N4 45 2025 025 0100 200 12 212kN

I nx2 4500 m4

xnI 1

n

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285Foundations loads on open-piled structures

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Foundations and retaining walls202

Example 3 inclined piles only The adjacent figure shows across section through a jetty where cot 13 5 Since the valuesof A and l are the same for each pile unity may be substitutedfor J Al For piles N1 and N3 B 5 and for piles N2 andN4 B 5 Since the group is symmetrical 132 0 133 notneeded 134 0 k 131 135 xo 3 m and yo 0

M Fv e Fh h 800(375 30) 0 600 kNm

The calculation is then as shown in the following table fromwhich k 3826 01538 05917 The axial loads on thepiles are given by Nx kp (kw Fv kh Fh km M) hence

N1 09806 (026 800 13 100 00867 600)

09806 (208 130 52) 2804 kN

N2 09806 (208 130 52) 255 kN

N3 09806 (208 130 52) 3824 kN

N4 09806 (208 130 52) 1275 kN

Thus the maximum load on any pile is 3824 kN to which theweight of the pile has to be added

Pile x Xno 1 5 (m) (m) 6 I kp kw kh km

N1 0114 450436 405

1 0 0 45 2025 1 0261 0 0111

N2 097 42 097

097 4

3826

1 + 42 1 + 42 radic(1 42) 4 0436

0913 0057 45 0 0 0941 0261 219 0

N3 0913 0057 45 0 0 0941 0261 219 0

N4 1 0 9 45 2025 1 0261 0 0111

Totals 3826 0114 mdash mdash 405 mdash mdash mdash mdash

Pile x Xno 1 5 (m) (m) 6I kp kw kh XI

N152 1 09615(3)2 5 01538 3846 3

152 152 radic(152) 05917 5 05917 3462

09615 00385 0 30 865 09806 026 13 00867

N2 3846

5 05917

09615 00385 0 30 865 09806 026 13 00867

N3 3

3462

09615 00385 60 30 865 09806 026 13 00867

N4 09615 00385 60 30 865 09806 026 13 00867

Total 3846 01538 mdash mdash 3462 mdash mdash mdash mdash

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chart given in Table 286 which is a modified version of a chartpublished in ref 63 The modified chart is applicable to designmethods in which the soil parameters incorporate either mobil-isation factors as in BS 8002 or partial factors of safety asin the Eurocodes

Stability against overturning is assured over the entire rangeof the chart and the maximum bearing pressure under serviceconditions can be investigated for all types of soil A uniformsurcharge that is small compared to the total forces acting onthe wall can be simply represented by an equivalent height ofsoil l can be replaced by le l q where q is the surchargepressure In more general cases le 3MhFh and 2FhKAle

2

can be used where Fh is the total horizontal force and Mh is thebending moment about underside of base due to Fh

The chart contains two curves showing where the bearingpressure is uniform and triangular respectively A uniformbearing condition is important when it is necessary to avoidtilting in order to minimise deflection at the top of the wallIt is generally advisable to maintain ground contact overthe full area of the base especially for clays where theoccurrence of ground water beneath the heel could soften theformation

Further information on the use of the chart for walls that aredesigned in accordance with BS 8002 and BS 8110 is given insection 284

184 BOX CULVERTS

Formulae for the bending moments in the corners of a boxculvert due to symmetrical load cases are given in Table 287Some notes on the different load cases and assumed groundconditions are given in section 742 Design requirements ofHighways Agency BD 3187 are summarised in section 743

Box culverts 203

Notes on design examples (1)ndash(3) In each case the pilegroup is symmetrical and is subjected to the same imposedloads Examples (2) and (3) are special cases of symmetricalgroups for which 134 0 and therefore yo 0 this conditionapplies only when the inclined piles are in symmetrical pairswith both pairs meeting at the same pile-cap

Example (3) requires the smallest pile but the differencein the maximum load is small between (2) and (3) Althoughthe maximum load on a pile is least in (1) the bendingmoment requires a pile of greater cross-sectional area to pro-vide the necessary resistance to combined bending and thrustIf the horizontal force is increased the superiority of (3) isgreater If Fh 200 kN (instead of 100 kN) the maximumpile loads are 236 kN (and a large bending moment of120 kNm) in (1) 609 kN in (2) and 510 kN in (3)Arrangement (2) is the most suitable when Fh is small IfFh 10 kN the maximum pile loads are 255 kN (and abending moment of 6 kNm) in (1) 217 kN in (2) and 268 kNin (3) Arrangement (1) is the most suitable when there is nohorizontal load

183 RETAINING WALLS

Various types of earth retention systems for which notes aregiven in section 731 are shown in Table 286 Information onthe pressures exerted by soils on retaining structures is given insection 91 and Tables 210 to 214

1831 Walls on spread bases

Several walls on spread bases for which notes are given insection 732 are shown in Table 286 Suitable dimensions fora base to a cantilever wall can be estimated with the aid of the

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Roker

Facing

Groutedbars

In-situ wall

Gravityelements(interlockingcribs)

Strul

Cantileverwall

Potential failure wedge

In-situwall

Trebocks

Anchors

Potentialfailurewedge

Potentialfailurewedge

Facingpanels

Strips or grids

286Retaining walls

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287Rectangular culverts

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191 STAIRS

Some general notes on stairs are given in section 614 Fordetails of characteristic imposed loads on stairs and landingsand horizontal loads on balustrades refer to BS 6399 Part 1Stairs forming monolithic structures are generally designed forthe uniformly distributed imposed load only Stairs that areformed of separate treads (usually precast) cantilevering from awall or central spine beam must be designed also for the alter-native concentrated load Some general information on stairtypes and dimensions is given in Table 288 and comprehensivedata is given in BS 5395

1911 Simple stairs

The term lsquosimplersquo is used here for a staircase that spans in thedirection of the stair flight between beams walls or landingslocated at the top and bottom of the flight The staircase mayinclude a section of landing spanning in the same direction andcontinuous with the stair flight For such staircases the followingstatements are contained in BS 8110

When staircases surrounding open wells include two spansthat intersect at right angles the load on the area common to bothspans may be divided equally between the spans When the stair-case is built at least 110 mm into a wall along part of all of itslength a 150 mm wide strip adjacent to the wall may be deductedfrom the loaded area When the staircase is built monolithicallyat its ends into structural members spanning at right angles to itsspan the effective span of the stairs should be taken as the hori-zontal distance between the centrelines of the supporting mem-bers where the width of each member is taken not more than18 m For a simply supported staircase the effective span shouldbe taken as the horizontal distance between the centrelines of thesupports or the clear distance between the faces of supports plusthe effective depth of the stair waist whichever is the lesser If astair flight occupies at least 60 of the effective span the per-missible spaneffective depth ratio calculated for a slab may beincreased by 15

1912 Free-standing stairs

In ref 64 Cusens and Kuang employ strainndashenergy principles todetermine expressions relating the horizontal restraint force Hand moment Mo at the mid-point of a free-standing stair

By solving the two resulting equations simultaneously thevalues of H and Mo obtained can then be substituted into generalexpressions to determine the forces and moments at any pointalong the structure

It is possible by neglecting subsidiary terms to simplify thebasic equations produced by Cusens and Kuang If this is donethe expressions given in Table 288 are obtained and these yieldH and Mo directly Comparisons of the solutions obtained bythese simplified expressions with those obtained using theexpressions presented by Cusens and Kuang show thatthe resulting variations are negligible for values in the rangeencountered in concrete design

The expressions given in Table 288 are based on a value ofGE 04 with C taken as half of the St Venant value for plainconcrete As assumed by Cusens and Kuang only half ofthe actual width of the landing is considered to determine its sec-ond moment of area

Example 1 Design a free-standing staircase assumed to befully fixed at the ends to support total ultimate loads per unitlength nf 169 kNm and nl 150 kNm The dimensions areas follows a 27 m b 14 m b1 18 m hf 100 mmhl 175 mm and 30o

From the expressions given in Table 288 H 8186 kNmand Mo 3587 kNmm If Cusens and Kuangrsquos more exactexpressions are used to analyse the structure H 8189 kNmand Mo 3567 kNmm Thus the errors involved by using theapproximate expressions are negligible for H and about 05for Mo If these values of H and Mo are substituted into the otherexpressions in Table 288 corresponding values of Mv Mh andT at any point in the structure can be found for various loadcombinations as shown in the table in page no 208

1913 Sawtooth stairs

In ref 65 Cusens shows that if axial shortening is neglected andthe strainndashenergy due to bending only is considered the mid-span moments for a so-called sawtooth stairs are given by thegeneral expression

Ms

where k (stiffness of tread)(stiffness of riser) and j is thenumber of treads

nl2(k11 kk12)

j2(k13 kk14)

Chapter 19

Miscellaneousstructures and details

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288Stairs general information

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If j is odd

k11 j216 j( j ndash 1)( j ndash 2)48

k12 ( j ndash 1)216 ( j ndash 1)( j ndash 2)( j ndash 3)48

k13 j2 k14 ( j ndash 1)2

If j is even

k11 j( j ndash 1)( j ndash 2)48

k12 ( j ndash 1)( j ndash 2)( j ndash 3)48

k13 ( j ndash 1)2 k14 ( j ndash 2)2

The chart on Table 289 gives coefficients to determine thesupport moments for various values of j and k Having found thesupport moment the maximum mid-span bending moment canbe determined by using the appropriate expression on the tableand deducting the support moment

Typical bending moment and shearing force diagrams for astair are also shown on Table 289 together with suggestedarrangements of reinforcement Because of the stair profilestress concentrations occur in the re-entrant corners and the realstresses to be resisted will be larger than those obtained from themoments To resist these increased stresses Cusens recom-mends providing twice the reinforcement theoretically requiredunless suitable fillets or haunches are incorporated at the junc-tions in which case the reinforcement need only be about 10more than is theoretically necessary The method of reinforcingshown in diagram (a) is very suitable but is generally only prac-tical if haunches are provided Otherwise the arrangementshown in diagram (b) should be adopted A further possibility isto arrange the bars as shown in diagram (a) on Table 363 forwall-to-wall corners

Example 2 A sawtooth stairs has seven treads each 300 mmwide with risers 180 mm high the thickness of both being100 mm The stairs which are 10 m wide are to be designed tosupport a characteristic imposed load of 30 kNm2 to therequirements of BS 8110

The self-weight of the treads and risers (assuming no finishesare required) is

gk 01 24 (03 018)03 384 kNm2

For design to BS 8110 total design ultimate load for a 10 mwide stair is given by

n 14 gk 16qk 14 384 16 30 1018 kNm

Since lt 300 mm lr 180 mm and ht hr 100 mm

k ht3lrhr

3lt 180300 06

From the chart on Table 289 for 7 treads and k 06 thesupport moment coefficient is ndash 0088 Thus

Ms ndash0088nl2 ndash0088 1018 (03 7)2 ndash395 kNm

Since j 7 is odd the free bending moment is given by

M (nl28)( j2 1)j2 1018 2128 5049 573 kNm

Hence the maximum moment at mid-span is

M0 M ndash Ms 573 ndash 395 178 kNm

1914 Helical stairs

By using strainndashenergy principles it is possible to formulate forsymmetrically loaded helical stairs fully fixed at the ends thefollowing two simultaneous equations in M0 and H

M0[K1(k5 sin213) k5 k7]

HR2[k4(k7 ndash K1)tan k5 sin cos (1 ndash K2)]

nR12[K1(k5 sin213ndash sin13) k5 k7 k6 k7R2R1] 0

M0[k4(k7 ndash K1) k5 (k7 ndash K2)]

HR2[12K1 tan ( 133 ndash 132sin213ndash 2k4)

12k7 tan ( 133 132sin213 2k4) 2k4 tan (k7 ndash K2)

k5 cos2 (tan K2 cot)] nR12[K1(k6 ndash k4) k4 k7

k7 (132sin 13 2k6)R2R1 (k7 ndash K2)(k5 k6R2R1)] 0

where

k4 13cos213ndash sin213 k5 13ndash sin213

k6 13cos13ndash sin13 k7 cos2 K2sin2

K1 GCEI1 K2 GCEI2 and 13 2

The equations can be solved on a programmable calculator orlarger machine to obtain coefficients k1 and k2 representing M0

and H respectively If the resulting values of M0 and H are thensubstituted into the equations given on Table 289 the bendingand torsional moments shearing forces and thrusts at any pointalong the stair can be easily calculated The critical quantity con-trolling helical stair design is usually the vertical moment Mvs atthe supports and a further coefficient can be derived to give thismoment directly

In ref 66 Santathadaporn and Cusens give 36 design chartsfor helical stairs covering ranges of of 60o to 720o of 20o

to 50o bh of 05 to 16 and R1R2 of 10 to 11 for GE 37The four design charts provided on Tables 290 and 291 havebeen recalculated for GE 04 with C taken as half of theSt Venant value for plain concrete These charts cover rangesof of 60o to 360o and of 20o to 40o with values for bh of5 and 10 and R1R2 of 10 and 11 being the ranges met most

14

12

18

14

12

13

12

13

12

12

Miscellaneous structures and details208

Application of Values of Mv (kNmm) Values of Mh (kNmm) Values of T (kNmm)imposed load

At O At B At D At A At B in OB Throughout AB At B in BC Throughout AB

Throughout 3587 1680 116 861 7367 8294 735 369Flights only 2481 922 160 1067 5036 5668 403 255Landing only 3132 1680 314 333 6486 7303 735 322

Note At point B the expressions give theoretical values for Mv (with imposed load applied throughout) that reduce abruptly from ndash 4195 kNmm in OB tondash368 kNmm in BC due to the intersection with flight AB Since the members in the actual structure are of finite width Cusens and Kuang recommendredistributing these moments across the intersection between the flight and the landing to give a value of (ndash4195 ndash 368)2 2282 kNmm

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289Stairs sawtooth and helical stairs

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290Design coefficients for helical stairs ndash 1

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291Design coefficients for helical stairs ndash 2

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frequently in helical stair design Interpolation between thevarious curves and charts will be sufficiently accurate forpreliminary design purposes

Example 3 A helical stairs having an angle of inclination tothe horizontal plane of 25o is to be designed to support acharacteristic imposed load of 30 kNm2 to the requirements ofBS 8110 The stairs are to be 12 m wide with a minimum slabthickness of 120 mm The radius to the inside of the stair isRi 900 mm and the angle turned through is 240o

Assuming the mean thickness on plan of the stairs (includingtreads and finishes) is 220 mm the self-weight of the stairs is022 24 53 kNm2 and the design ultimate load intensity

n 14 53 16 30 1222 kNm2

The radius of the centreline of the load is given by

R1 158 m

The radius of the centreline of the stairs

R2 09 05 12 15 m

Hence R1R2 15815 105 bh 1200120 10 andfrom the charts on Tables 290 and 291 interpolating asnecessary k1 ndash 012 k2 152 and k3 ndash 032 Thus fora 12 m wide stairs the total design values are

M0 12k1nR22 ndash012 1222 152 12 ndash396 kNm

H 12k2nR2 152 1222 15 12 334 kN

Mvs 12k3nR22 ndash032 1222 152 12 ndash1056 kNm

The slab should now be checked to ensure that the thicknessprovided is sufficient to resist Mvs Then assuming this is sothe foregoing values of M0 and H can be substituted into theequations for Mv Mn T N Vn and Vh given on Table 289 toobtain moments and forces at any point along the stairs Forexample where 13 60o Mv 111 kNm Mn ndash 4817 kNmT 005 kNm N ndash 365 kN Vn 968 kN and Vh 167 kNTypical distributions of moments and forces along the stair areshown on Table 289

192 NON-PLANAR ROOFS

1921 Prismatic structures

To design a simple prismatic roof or any structure comprising anumber of planar slabs for service load the resultant loadsQ acting at right angles to each slab and the unbalanced thrustsN acting in the plane of each slab are determined first taking intoaccount the thrust of one slab on another The slabs are thendesigned to resist the transverse bending moments due to theloads Q assuming continuity combined with the thrusts N Thelongitudinal forces F due to the slabs bending in their own planeunder the loads N are for any two adjacent slabs AB and BCcalculated from formula (2) in Table 292 where MAB and MBC

are found from formula (1) if the structure is freely supported atthe end of length L For each pair of slabs AB-BC BC-CD andso on there is an equation like formula (2) containing threeunknown forces F If there are n pairs there are (n ndash 1) equa-tions and (n 1) unknowns The conditions at the outer edgesa and z of the end slabs determine the forces F at these edgesfor example if the outer edges are unsupported Fa Fz 0

2(213 093)

3(212 092)

2(R3o R3

i )

3(R2o R2

i )

The simultaneous equations are solved for the remainingunknown forces FA FB FC and so on The longitudinal stress atany junction B is calculated from the formula (in Table 292) forfb Variation of the longitudinal stress from one function to thenext is rectilinear If fb is negative the stress is tensile and shouldbe resisted by reinforcement The shearing stresses are generallysmall

1922 Domes

A dome is designed for the total vertical load only that is forthe weights of the slab any covering on the slab any ceiling orother distributed load suspended from the slab and the imposedload The service load intensity w is the equivalent load per unitarea of surface of the dome Horizontal service loads due towind and the effects of shrinkage and changes in temperaturecan be allowed for by assuming an ample normal load or byinserting more reinforcement than that required for the normalload alone or by designing for stresses well below permissiblevalues or by combining any or all of these methods

Segmental domes Referring to the diagram and formulae inTable 292 for a unit strip at S the circumferential force actingin a horizontal plane is T and the corresponding force actingtangentially to the surface of the dome (the meridional thrust) isN At the plane where 13 is 51o48 that is the plane of ruptureT 0 Above this plane T is compressive reaching a maximumvalue of 05wr at the crown of the dome (13 0) Below thisplane T is tensile equalling 0167wr when 13 60o andwr 90o The meridional thrust N is 05wr at the crown0618wr at the plane of rupture 0667wr when 13 60o andwr when 13 90o that is N increases from the crown towardsthe support and has its greatest value at the support

For a concentrated load F applied at the crown of the domeT and N are given by the appropriate formulae in Table 292where T is tensile and N is compressive The load is assumed tobe concentrated on so small an area at the crown that it can betaken as a point load The theoretical stress at the crown is there-fore infinite but the practical impossibility of obtaining a pointload invalidates the use of the formulae when 13 0 or verynearly so For domes of varying thickness see ref 67

For a shallow dome approximate analysis only is sufficientand appropriate formulae are given in Table 292

Conical domes In a conical dome the circumferential forcesare compressive throughout and for any horizontal plane atdistance x from the apex are given by the expression for T inTable 292 the corresponding force in the direction of the slopebeing N The horizontal outward force per unit length of thecircumference at the bottom of the slope Tr needs to be resistedby the supports or a bottom ring beam

1923 Segmental shells

General notes on the design of cylindrical shell roofs and the useof Tables 293 and 294 are given in section 6 16

Membrane action Consideration of membrane action onlygives the following membrane forces per unit width of slab dueto the uniform load shown on Table 292 stresses are obtainedby dividing by the thickness of the shell h Negative values of Vindicate tension in the direction corresponding to an increase in

Miscellaneous structures and details212

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292Non-planar roofs general data

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293Shell roofs empirical design method ndash 1

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294Shell roofs empirical design method ndash 2

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x and a decrease in 13x In the case of F positive values indicatetension Reinforcement should be provided approximately inline with and to resist the principal tensile forces If the shell issupported along any edges the forces will be modified accord-ingly The values of the forces at any point are as follows

Tangential force

Fy ndash (g qcos13x) rcos13x

Longitudinal force

Fx ndash (1 ndash x)(xr)[gcos13x 15q(cos213x ndash sin213x)]

Shearing force

Vxy (2x ndash l)(g 15qcos13x)sin13x

Principal forces (due to membrane forces only)

Fp 05(Fx Fy) [(Fx ndash Fy)2 4V2xy]05

tan 2

At A (mid-point at edge 13x 13 x l2)

FyA ndash (g qcos13) rcos13

FxA ndash (l24r)[gcos13 15q(cos213ndash sin213)]

VxyA 0

At B (mid-point at crown 13x 0 x l2)

FyB ndash (g q)r

FxB ndash (l24r)(g 15q)

VxyB 0

At C (support at edge 13x 13 x 0)

FyC ndash (g qcos13) rcos13

FxC 0

VxyC ndash (g 15qcos13) lsin13

At D (support at crown 13x 0 x 0)

FyD ndash (g q) r

FxD 0 VxyD 0

Beam action If the ratio of the length of a cylindrical shell toits radius lr is not less than 25 the longitudinal forces canbe approximated with reasonable accuracy by calculating thesecond moment of area Ixx and the vertical distance from theneutral axis to the crown yndash from the approximate expressions

Ixx cong R2h (R ndash 3h2)(2 sincosndash 2sin2)

yndash cong R ndash [(R ndash 2h3)sin]

Then if n is the total uniform load per unit area acting on theshell (ie including self-weight etc) the maximum bendingmoment in a freely supported shell is given by

M (213rn)l28 13rnl24

Hence from the relationship MIxx = fy the horizontal forces atmid-span at the crown and springing of a shell of thickness h aregiven by the expressions

Fx(bottom) M[r(1 cos 13) y]h

IxxFx(top)

Myh

Ixx

2Vxy

Fx Fy

At the supports the total shearing force in the shell is

V (213rn)l2 13rnl

The shearing stress at any point is then given by v (VAx)(2hIxx)

where Axndash is the first moment of area about the neutral axis of thearea of cross section of shell above the point at which the shear-ing stress is being determined

The principal shortcoming of this approximate analysis is thatit does not indicate in any simple manner the magnitude of thetransverse moments that occur in the shell However a tabularmethod has been devised by which these moments can be eval-uated indirectly from the lateral components of the shearingstresses for details see ref 68

1924 Hyperbolic-paraboloidal shells

The simplest type of double-curved shell is generated by theintersection of two separate sets of inclined straight lines (parallelto axes XX and YY respectively as shown in the diagrams on Table292) Vertical sections through the shell at angles XX or YY areparabolic in shape while horizontal sections through the surfaceform hyperbolas hence the name hyperbolic paraboloid

Individual units such as those shown in diagrams (a) and (b)in Table 292 can be used separately being supported oncolumns or buttresses located at either the higher or the lowercorners Alternatively groups of units can be combined toachieve roofs having attractive and unusual shapes such as isshown in diagram (c) Some idea of the more unlikely forms thatcan be achieved may be obtained from ref 69

If the shell is shallow and the loading is uniform the shellbehaves as a membrane transferring uniform compressive andtensile forces of F2c (where F is the total load on the unit andc is the rise) acting parallel to the directions of principalcurvature to the edges of the shell The edge forces are thentransmitted back to the supports along beams at the shell edgesThe main problems that arise when designing these shells are theinteraction between the shell and the supporting edge membersthe design of the buttresses or ties needed to resist the horizon-tal component of the forces at the supports and the fact thatexcessive deflections at unsupported edges lead to stresses thatdiffer considerably from those predicted by simplified theories

Increasing the sharpness of curvature of the shell increases itsstability and reduces the forces and reactions within the shellbut to avoid the need for top forms the maximum slope shouldnot exceed 45o this corresponds to a value of 1radic2 for theratio ca or cb in the diagrams on Table 292 To ensure stabilityif a single unit is used the ratio should be not less than 15A useful introduction to the theory and design of hyperbolic-paraboloidal shells is given in ref 70

193 BEAMS CURVED ON PLAN

Some general notes on beams forming a circular arc on planare given in section 617 The following analyses apply only ifthe appropriate negative bending and torsional moments can bedeveloped at the end supports

1931 Concentrated loads

If a beam LR (see Table 295) curved on plan is subjected to aconcentrated load F such that the angle between the radius at

Miscellaneous structures and details216

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295Bow girders concentrated loads

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the point of application F of load F and the radius at themid-point O of the beam is 0 the following expressionsare applicable at any point X between F and L (ie 0)

M M0cos ndash T0sin V0rsin ndash Frsin ( ndash 0)

T M0sin T0cos V0r(1 ndash cos) ndash Fr[1 ndash cos ( ndash 0)]

V ndash V0 F

where M0 T0 and V0 are respectively the bending moment thetorsional moment and the shearing force at mid-span r isthe radius of curvature in plan and is the angle defining theposition of X as shown in the diagram on Table 295

If X is between F and O (ie X 0) terms containing Fare equal to zero If X is between O and R (ie X) signs of theterms containing sin should also be reversed Now by writingM0 K1Fr T0 K2Fr and V0 K3F

K1 K2 and K3

where k1 k2 k3 and so on are given by the following expressionsin which K EIGC (ie flexural rigiditytorsional rigidity)

k1 (K ndash 1)sin0(sin 213ndash sin 20)

(K ndash 1)cos0(sin213ndash sin20)

(K 1)(13ndash 0)sin0 ndash K(cos13ndash cos0)

k2 (K 1)13ndash 12(K ndash 1)sin213

k3 K(sin13ndash sin0) ndash 14 (K ndash 1)cos0(sin213ndash sin20)

(K ndash 1)sin0(sin213 ndash sin20)

(K 1)(13 ndash 0)cos0

k4 (K 1)13 (K ndash 1)sin213

k5 2Ksin13 ndash (K 1)13 ndash (K ndash 1)sin213

k6 (K ndash 1)cos0(sin213ndash sin20) K(13ndash 0)

(K 1)(13ndash 0)cos0

(K ndash 1)sin0(sin213ndash sin20)

K(1 cos0)(sin13ndash sin0)

Ksin0(cos13ndash cos0)

k7 k5

k8 (K ndash 1)sin213ndash 4Ksin13 (3K 1)13

For rectangular beams the graphs provided on Table 295 enablevalues of K1 K2 and K3 to be read directly for given values of 130 and hb (ie depthwidth) Values of G 04E and C J2as recommended in BS 8110 have been used

1932 Uniform load

For a curved beam with a UDL over the entire length owing tosymmetry the torsional moment (and also the shearing force) atmid-span is zero By integrating the foregoing formulae thebending and torsional moments at any point X along the beamare given by the expressions

M M0cos ndash nr2(1 ndash cos)

T M0sin ndash nr2( ndash sin)

If M0 K4nr2 where

K4 14[(1 K)sin13 K13 cos13]

213 (1 K) sin 213 (1 K)

12

12

12

14

12

12

12

12

12

12

14

k4k6 k3k7

k4k8 k5k7

k3k8 k5k6

k4k8 k5k7

k1

k2

these expressions can be rearranged to give the bending andtorsional moments at the supports and the maximum positivemoments in the span in terms of non-dimensional factors K4 K5K6 and K7 as shown on Tables 296 and 297 The factors can beread from the charts given on the tables

Example 1 A bow girder 450 mm deep and 450 mm widehas a radius of 4 m and subtends an angle of 90o The ends arerigidly fixed and the total UDL is 200 kN The maximummoments are to be determined

The distributed load per unit length

n 200(r2) 200( 42) 318 kNm

From the charts on Tables 296 and 297 (with hb 10 and13 45o) K4 0086 K5 ndash 023 K6 ndash 00175 K7 0023with 1 23o and 2 40o

Maximum positive bending moment (at midspan)

K4nr2 0086 318 42 438 kNm

Maximum negative bending moment (at supports)

K5nr2 ndash 023 318 42 ndash 117 kNm

Zero bending moment occurs at 1 23o

Maximum negative torsional moment (at supports)

K6nr2 ndash 00175 318 42 ndash 89 kNm

Maximum positive torsional moment (at 1 23o)

K7nr2 0023 318 42 117 kNm

Zero torsional moment occurs at 2 40o

Example 2 The beam in example 1 supports a concentratedload of 200 kN at a point where the radius subtends an angleof 15o from the left-hand end (ie 0 30o) Moments andshearing forces at midspan and the supports are required

From the charts on Table 295 (with hb 10 13 45o and0 30o) K1 0015 K2 ndash 00053 K3 0067

At midspan ( 0)

M0 0015 200 4 12 kNm

T0 ndash 00053 200 4 ndash 43 kNm

V0 0067 200 134 kN

At left-hand support ( 45o)

M M0cos 45o ndash T0sin 45o V0 rsin 45o ndash Frsin (45o ndash 30o) [12 ndash (ndash 43) 134 4) 0707 ndash 200 4 0259 ndash 158 kNm

T M0sin 45o T0cos 45o V0 r (1 ndash cos 45o) Fr (1 ndash cos 15o)

(12ndash43)070713440293ndash20040034 ndash 61 kNm

V V0 ndash F 134 ndash 200 1866 kN

At right-hand support ( 45o)

M M0cos 45o T0sin 45o ndash V0 rsin 45o

[12 ndash 43 ndash 134 4) 0707 ndash 324 kNm

T ndash M0sin 45o T0cos 45o V0 r (1 ndash cos 45o) (ndash 12 ndash 43) 0707 134 4 0293 42 kNm

V V0 134 kN

Miscellaneous structures and details218

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296Bow girders uniform loads ndash 1

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297Bow girders uniform loads ndash 2

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194 BEARINGS HINGES AND JOINTS

A comprehensive guide on bridge bearings and expansion jointsincluding a treatment of the design techniques used to accom-modate movements in bridges is contained in ref 71 Varioustypes of bridges for which notes are given in section 62 andtypical span ranges are shown in Table 298

1941 Hinges and bearings

A hinge is an element that can transmit thrust and transverseforce but permits rotation without restraint If it is vital for suchaction to be fully realised a steel hinge can be providedAlternatively hinges that are monolithic with the member canbe formed as indicated at (a) and (b) in Table 299 ThelsquoMesnagerrsquohinge shown at (a) has been used for example in theframes of large bunkers to isolate the container from the sub-structure or to provide a hinge at the base of the columns of ahinged frame bridge The hinge-bars lsquoarsquo resist the entirehorizontal shear force and the so-called throat of concrete at Dmust be sufficient to transfer the full compressive force from theupper to the lower part of the member The hinge-bars should bebound together by links lsquodrsquo and the main vertical bars lsquoersquo shouldterminate on each side of the slots B and C It may be advanta-geous during construction to provide bars extending across theslots and then cut these bars on completion of the frame Theslots should be filled with a bituminous material or lead or asimilar separating layer

The Freyssinet hinge shown at (b) has largely superseded theMesnager hinge In this case the large compressive stress acrossthe throat results in a high shearing resistance and the inclusionof bars crossing the throat can adversely affect the hinge Testshave shown that as a result of biaxial or triaxial restraint suchhinges can withstand compressive stresses of several times thecube strength without failure occurring The bursting tension oneach side of the throat normally governs the design of this typeof hinge

A number of other types of hinges and bearings that have beenused on various occasions are shown in Table 299

(c) a hinge formed by the convex end of a concrete memberbearing in a concave recess in the foundations

(d) a hinge suitable for the bearing of a girder where rotationbut not sliding is required

(e) a bearing for a girder where sliding is required

(f) a mechanical hinge suitable for the base of a large portalframe or the abutment of a large hinged arch bridge

(g) a hinge suitable for the crown of a three-hinged arch whenthe provision of a mechanical hinge is not justified

(h) a bearing suitable for the support of a freely suspended spanon a cantilever in an articulated bridge

Bridge bearings have to be able to accommodate the rotationsresulting from deflection of the deck under load They alsohave to be able to accommodate horizontal movements ofthe deck caused by prestress creep shrinkage and temperaturechange Some bearings allow horizontal movement in one direc-tion only and are restrained in the other direction whilst othertypes allow movement in any direction Elastomeric bearingsthat are formed of layers of steel plate embedded in rubbercan accommodate small horizontal shear movements

PTFE (polytetrafluoroethylene) bearings can give unlimitedfree sliding between low friction PTFE surfaces and steel platesPot bearings that incorporate rubber discs allow for small rota-tions while spherical bearings that move on a PTFE surface per-mit larger rotations Mechanical bearings such as rockers androllers can be used to provide either longitudinal fixity or resis-tance to lateral force Pot bearings special guide bearings or pinbearings are also used for this purpose Bearings need to beinspected regularly and may require maintenance or replace-ment during the lifetime of the bridge As this can be both diffi-cult and expensive it is very important that the structure isdesigned to make inspection maintenance and replacementpossible

1942 Movement joints

Movement joints are often required to allow free expansion andcontraction due to temperature changes and shrinkage in suchstructures as retaining walls reservoirs roads and long buildingsIn order to allow unrestrained deformation of the walls of cylin-drical containers sliding joints can be provided at the bottomand top of the wall Several types of joints for various purposesare shown in Table 2100 Figures (a)ndash(f) show some of the jointdetails recommended in BS 8007

Expansion joints at wide spacing may be desirable inlarge areas of walls and roofs that are not protected from solarheat gain or where a contained liquid is subjected to asubstantial temperature range Except for structures designed tobe fully continuous contraction joints of the type describedin section 2622 and at the maximum spacing specified inTable 345 should be provided The reinforcement should be cur-tailed to form a complete movement joint or made 50continuous to form a partial movement joint Waterstops arepositioned at the centre of wall sections and at the undersideof floor slabs that are supported on a smooth layer of blindingconcrete In basement walls waterstops are best positioned atthe external face where they are supported by the earth

The recommendations of BS 8007 with regard to the spacingof vertical joints may be applied also to earth-retaining wallsFor low walls with thin stems simple butt joints are generallyused However the effect of unequal deflection or tilting of onepart of a wall relative to the next will show at the joints For retain-ing walls higher than about 1 m a keyed joint can be usedAlternatively dowels passing through the joint with the ends onone side greased and sheathed can be used

Figure (g) shows alternative details at the joint betweenthe wall and floor of a cylindrical tank to minimise or eliminaterestraint at this position In the first case rubber or neoprenepads with known shear deformation characteristics are used Inthe second case action depends on a sliding membrane of PTFEor similar material These details are most commonly used forprestressed cylindrical tanks

Movement joints in buildings should divide the structure intoindividual sections passing through the whole structure aboveground level in one plane Joints at least 25 mm wide shouldbe provided at about 50 m centres both longitudinally andtransversely In the top storey and for open buildings andexposed slabs additional joints should be provided to give aspacing of about 25 m Joints also need to be incorporated in fin-ishes and cladding at movement joint locations Joints in wallsshould be made at column positions in which case a double

Bearings hinges and joints 221

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298Bridges

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299Hinges and bearings

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2100Movement joints

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column as shown at (h) can be provided The copper strip orother similar type of waterbar must be notched where the linksoccur the ends of the notched pieces being bent horizontally orcut off At joints in suspended floors and flat roofs a doublebeam can be provided Joints in floors should be sealed to pre-vent the accumulation of rubbish Roof joints should also beprovided with waterstops The provision of joints in large-areaindustrial ground floor slabs is considered in section 722 andrecommended details are given in ref 61

Expansion joints in bridges need to be either waterproof ordesigned to allow for drainage and should not badly disrupt the

riding quality of the deck Joints should also be designed torequire minimal maintenance during their lifetime and be able tobe replaced if necessary Compressible materials such as neopreneor rubber can accommodate small movements In this case jointscan be buried and covered by the surfacing This type of jointwhich consists of a small gap covered by a galvanised steel plateand a band of rubberised bitumen flexible binder to replace partof the surfacing is known as an asphaltic plug To accommodatelarger movements a flexible sealing element supported by steeledge beams is required Mechanical joints based on interlockingsets of steel toothed plates can be used for very large movements

Bearings hinges and joints 225

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The elastic analysis of a reinforced concrete section by themodular ratio method is applicable to the behaviour of thesection under service loads only The strength of the concrete intension is neglected and a linear stressndashstrain relationshipis assumed for both concrete and reinforcement The straindistribution across the section is also assumed to be linearThus the strain at any point on the section is proportional tothe distance of the point from the neutral axis and since thestressndashstrain relationship is linear the stress in the concrete isalso proportional to the distance from the neutral axis Thisgives a triangular distribution of stress ranging from zero atthe neutral axis to a maximum at the outermost point on thecompression face Assuming no slipping occurs between thereinforcement and the surrounding concrete the strain in bothmaterials is the same and the ratio of the stresses in the twomaterials depends on the ratio of the modulus of elasticity ofsteel and concrete known as the modular ratio e The value of Es

is taken as 200 kNmm2 but the value of E for concrete dependson several factors including the aggregate type the concretestrength and the load duration Commonly adopted values forsustained loads are 15 for normal-weight concrete and 30 forlightweight concrete The geometrical properties of reinforcedconcrete sections can be expressed in equivalent concrete unitsby multiplying the reinforcement area by e

201 PURE BENDING

Expressions for the properties of common reinforced concretesections are given in Tables 2102 and 2103 For sections thatare entirely in compression where the presence of the rein-forcement is ignored simplified expressions are given inTable 2101 The maximum stress in the concrete is given byfc for sections entirely in compression In other casesfc MK2 z unless expressed otherwise and the stress in theoutermost tension reinforcement is given by fs e fc (dx 1)

Expressions for the properties of rectangular and flangedsections are also given in Table 342 in connection with theserviceability calculation procedures contained in BS 8110 BS5400 and BS 8007 (see Chapter 26)

202 COMBINED BENDING AND AXIAL FORCE

The general analysis of any section subjected to direct thrustand uniaxial bending is considered in Table 2104 In the case

MxI

Chapter 20

Elastic analysis ofconcrete sections

of symmetrically reinforced rectangular columns the designcharts on Tables 2105 and 2106 apply The design charts onTable 2107 apply to annular sections such as hollow mastsInformation on uniaxial bending combined with direct tensionand biaxial bending and direct force is given in Tables 2108and 2109 respectively

2021 Symmetrically reinforced rectangularsection

For a symmetrically reinforced rectangular section subjected toaxial force N and bending moment M by equating forcesand taking moments about the mid-depth of the section thefollowing expressions are obtained

(1) For values of x h (ie entire section in compression)

where

A [1 (e 1) ]bh

(2) For values of x h (ie one face in tension)

where fc is maximum stress in concrete at compression faceand Ascbh is total reinforcement ratio

The stress in the tension reinforcement is then given by

fs e fc(dx 1)

For e 15 the charts given on Tables 2105 and 2106 canbe used directly For other values of e the curves for may beconsidered to represent values of [( e 1)14] in region (1)and (e 15) in region (2) For given values of M N and fc therequired value of can be readily determined For given values

dh

05 05edx 1d

h 05

Mbh2fc

05xh

05 x

3h 05(e1)1 h d

x

Nbhfc

05xh

05(e 1)1h d

x 05edh

1

I 112

(e 1)dh

12

bh3

Nbhfc

Abh

Ah2

21 Mbh2fc

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Geometric properties of uniform sections 2101

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Properties of reinforced concrete sections ndash 1 2102

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Properties of reinforced concrete sections ndash 2 2103

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Uniaxial bending and compression (modular ratio) 2104Equivalent area of strip

Equivalent area of transformed sectionDepth of centroid of transformed section

Moment of inertia of transformed section about centroid

Compressive stresses

fcr min Nd

Atr

Md(h x )Itr

fcr max Nd

Atr

MdxItr

Itr Atr(hs)2

12 (hc x )2

x Atr hcAtr

Atr Atr

Atr bshs (e 1)As

Assume a value of x

Depth to centre of tension where

If all bars are of the same size

Equivalent area of strip

Depth to centre of compression

Position of centroid of stressed area

Maximum stresses

Finally check the assumed value of x by substituting these stresses in

xd

1

1 fst(e fcr)

fst (d x)

S fcr

x (x hc)Atr Nd

fcr Nd x(e at x )

(at ac)(x hc)Atr

x eAs a Atr hc

eAs Atr

ac (x hc)hcAtr (x hc)Atr

Atr bshs (e 1)As

at a(a x)(a x)

S (a x)Asat SaS

Com

pres

siv e

str

esse

s on

lyC

ombi

ned

com

pres

sive

and

tens

ile s

tres

ses

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Symmetrically reinforced rectangular columns(modular ratio) ndash 1 2105

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2106Symmetrically reinforced rectangular columns(modular ratio) ndash 2

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2107Uniformly reinforced cylindrical columns(modular ratio)

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Uniaxial bending and tension (modular ratio) 2108

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Biaxial bending and compression (modular ratio) 2109

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Elastic analysis of concrete sections236

of M N and the value of fc is obtained by finding the point ofintersection of the curve and the eh line The eh line can beobtained by interpolation or can be drawn as follows on thegridline for Nbhfc 10 find the point where Mbh2fc is equalto the value of eh through this point draw a line from theorigin The value of xh can be obtained by interpolation or bysolving the equation

(xh)2 2[Nbhfc (e 05)](xh) (e dh 1) 0

Example 1 A 400 400 column reinforced with 4H32 barsis subjected to values of M 120 kNm and N 500 kN dueto service loads The maximum stresses in the concrete and thereinforcement are to be determined assuming e 15

Ascbh 32174002 002

eh MNh 120(500 04) 060

Allowing for 35 mm nominal cover to H8 links

d 400 (35 8 322) 340 mm dh 340400 085

From the chart for dh 340400 085 on Table 2105 at theintersection of 002 and eh 06

Nbhfc 025 xh 051

fc 500 103(025 4002) 125 Nmm2

x 051 400 204 mm

fs e fc (dx 1)

15 125 (340204 1) 125 Nmm2

2022 Uniformly reinforced annular section

The charts given on Table 2107 are based on the assumptionthat the bars may be represented with little loss of accuracy bya notional ring of reinforcement having the same total cross-sectional area and located at the centre of the sectionIf e 15 the charts can be used directly For other valuesof e the curves for and fs fc may be considered to representvalues of (e15) and (15e)( fsfc) respectively

Example 2 A cylindrical shaft with a mean radius of 1 m anda thickness of 100 mm is reinforced with 42H16 vertical barslocated at the centre of the section The shaft is subjected tovalues of M 2700 kNm and N 3600 kN due to serviceloads and the stresses in the concrete and the reinforcement areto be determined assuming e 15

Asc2rh 8446(2 1000 100) 00135

eh MNh 2400(3200 01) 75

A line corresponding to eh 75 can be drawn on the chart forhr 010 on Table 2107 from the origin to a point such asNrhfc 6 Mr2hfc 45 At the intersection of this line withan interpolated curve for 00135

Nrhfc 26 fsfc 6

fc 3200 103(26 1000 100) 123 Nmm2

fs 6 123 74 Nmm2

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Part 3

Design to British Codes

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In most British Codes the design requirements are set out inrelation to specified limit-state conditions Calculations todetermine the ability of a member (or assembly of members) tosatisfy a particular limit state are undertaken using design loadsand design strengths (or stresses) These design values aredetermined from characteristic loads and characteristicstrengths of materials (or stress limits) by the application ofpartial safety factors specified in the Code concerned

211 BUILDINGS

For buildings and other structures designed to BS 8110 thecharacteristic values of dead load Gk (Tables 21 and 22)imposed load Qk (Tables 23 and 24) and wind load Wk (Tables27ndash29) are specified in BS 6399 Parts 1 2 and 3

Design loads are given by

design load Fk f

where Fk is equal to Gk Qk or Wk as appropriate and f is thepartial safety factor appropriate to the load load combinationand limit-state being considered

The characteristic strength of a material fk means that valueof the cube strength of concrete fcu or the yield strength ofreinforcement fy below which 5 of all possible test resultswould be expected to fall In practice for concrete specified inaccordance with BS 8500 a dual classification is used forexample C2530 in which the characteristic strength of cylin-der test specimens is followed by the characteristic strength ofcube test specimens The characteristic strength of reinforcementis taken as the value specified in BS 4449 or BS 4482

Design strengths are given by

design strength fkm

where fk is either fcu or fy as appropriate and m is the partialsafety factor appropriate to the material and limit-state beingconsidered The appropriate factors are already incorporated inthe design equations provided in the Code

Details of the design requirements and partial safety factorsfor the ULS are given in Table 31 For the serviceability limitstates when calculations are required the partial safety factorsare taken as unity However for most designs explicit calcula-tions are unnecessary and the design requirement is met bycomplying with simple rules

For the ULS adverse and beneficial values are given fordead and imposed loads and these are to be applied separatelyto the loads on different parts of the structure to cause the mostsevere effects Thus for load combination 1 design loads mayvary from place to place with a maximum of (14Gk 16Qk)and a minimum of 10Gk (see also Table 230) For loadcombination 2 the maximum wind load of 14Wk with the min-imum dead load of 10Gk can result in a critical equilibriumcondition for tower structures For load combinations 2 and 3the design wind loads can sometimes be less than the minimumnotional horizontal load of 0015Gk associated with therequirement for robustness

The overall dimensions and stability of earth-retaining andfoundation structures are to be determined in accordance withthe appropriate codes for earth-retaining structures (BS 8002)and foundations (BS 8004) Design loads given in BS 8110 arethen used in to establish section sizes and reinforcement areasThe factor f should be applied to all earth and water pressuresexcept those that are derived from equilibrium with otherdesign loads such as bearing pressures below foundations

212 BRIDGES

For bridges and other structures designed to BS 5400 Part 4nominal and design values of dead load superimposed deadload wind load temperature live loads for highway footwayand railway bridges (Tables 25 and 26) and other loads aregiven in the Highways Agency Standard BD 3701

Design loads are given by

design load Fk fl

where Fk is the specified nominal load and fl is the partialsafety factor appropriate to the load load combination andlimit-state being considered A separate partial safety factor isthen used to allow for inaccuracies in the method of analysis

Design load effects are given by

design load effect effect of design load f3

where f3 is the partial safety factor appropriate to the methodof analysis and limit-state being considered If the method ofanalysis is linear elastic it is possible to replace fl by fl f3

in the calculation of the design loads

Chapter 21

Design requirementsand safety factors

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31Design requirements and partial safety factors (BS 8110)

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Liquid-retaining structures 241

The characteristic strength of a material fk means that valueof the cube strength of concrete fcu or the yield strength ofreinforcement fy below which 5 of all possible test resultswould be expected to fall In practice characteristic values arespecified in the manner described for design according toBS 8110 in section 211 The characteristic stress is the stressvalue at the assumed limit of linearity on the stressndashstrain curvefor the material

Design strengths are given by

design strength fkm

where fk is either fcu or fy as appropriate and m is the partialsafety factor appropriate to the material and limit-state beingconsidered The appropriate factors are already incorporated inthe design equations provided in the Code

Design stress limits (serviceability) are given by

design stress limit characteristic stressm

Details of the design requirements characteristic stresses andpartial safety factors are given in Tables 32 and 33

213 LIQUID-RETAINING STRUCTURES

For liquid-containing or liquid-excluding structures designed toBS 8007 the design basis is similar to that in BS 8110 but mod-ified for the limit-state of cracking Separate calculations ofcrack width are required for the effect of applied loads and theeffect of temperature and moisture change

Details of the design requirements and partial safety factors(updated according to BS 8110) are given in Table 34

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32Design requirements and partial safety factors (BS 5400) ndash 1

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33Design requirements and partial safety factors (BS 5400) ndash 2

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34Design requirements and partial safety factors (BS 8007)

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221 CONCRETE

2211 Strength and elastic properties

The characteristic strength of concrete means that value of the28-day cube strength below which 5 of all valid test results isexpected to fall In BS 8500 compressive strength classes areexpressed in terms of both characteristic cylinder strength andcharacteristic cube strength The recommended compressivestrength classes and mean values of the static modulus ofelasticity at 28 days are given in Table 35

In BS 8110 for normal-weight concrete the mean value ofthe static modulus of elasticity of concrete at 28 days is givenby the expression

Ec28 20 02fcu28 (kNmm2)

where fcu28 is the cube strength at 28 days The mean value ofthe modulus of elasticity at an age t 3 days can be estimatedfrom the expression

Ect Ec28 (04 06fcutfcu28)

where fcut is the cube strength at age t Where deflections are ofgreat importance and test data for concrete made with theaggregate to be used in the structure is not available a range ofvalues for Ec28 based on (mean value 6 kNmm2) should beconsidered

In BS 5400 slightly higher mean values of the modulus ofelasticity are used given by the expression

Ec 19 03fcu (kNmm2)

where fcu is the cube strength at the age considered

2212 Creep and shrinkage

The creep strain in concrete may be assumed to be directlyproportional to the applied stress for stresses not exceedingabout one-third of the cube strength at the age of loadingFor design to BS 8110 values of the creep coefficient (creepper unit of stress) according to the ambient relative humiditythe effective section thickness and the age of loading can beobtained from the figure in Table 35 Creep strain is partlyrecoverable if the stress is reduced The final recovery (after 1 year)

is approximately 03 (stress reduction)Eu where Eu is themodulus of elasticity at the age of unloading

For design to BS 8110 an estimate of the drying shrinkageof plain concrete according to the ambient relative humiditythe effective section thickness and the original water contentcan be obtained from the figure in Table 35 Aggregates witha high moisture movement such as some Scottish doleritesand whinstones and gravels containing these rocks produceconcrete with a high initial drying shrinkage (ref 12) Alsoaggregates with a low elastic modulus may result in higher thannormal concrete shrinkage

In BS 5400 values are obtained for the creep coefficient as aproduct of five partial coefficients and for the shrinkage strainas a product of four partial coefficients where the coefficientsare obtained from a series of figures

2213 Thermal properties

For design to BS 8110 values of the coefficient of thermalexpansion of concrete according to aggregate type are given inTable 35 In BS 5400 a value of 12 106 per oC is generallytaken except when limestone aggregates are used when a valueof 9 106 per oC is recommended

2214 Stressndashstrain curves

Typical short-term stressndashstrain curves for normal strengthconcretes in compression as described in section 316 andthe idealised curve given for design purposes in BS 8110and BS 5400 are shown in Table 36 In the expression givenfor the maximum design stress 067 takes account of the ratiobetween the cube strength and the strength in bending

222 REINFORCEMENT

2221 Strength and elastic properties

The characteristic strength of steel reinforcement made to therequirements of BS 4449 is 500 Nmm2 BS 4449 caters forround ribbed bars in three ductility classes grades B500AB500B and B500C Fabric reinforcement is manufacturedusing bars to BS 4449 except for wrapping fabric where wireto BS 4482 with a characteristic strength of 250 Nmm2 maybe used For more information on types properties and sizes of

Chapter 22

Properties of materials

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35Concrete (BS 8110) strength and deformation characteristics

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36Stress-strain curves (BS 8110 and BS 5400) concreteand reinforcement

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Properties of materials248

bar and fabric reinforcement reference should be made tosection 103 and Tables 219 and 220

2222 Stressndashstrain curves

Typical stressndashstrain curves for reinforcing steels in tension asdescribed in section 323 and the idealised curves given

for design purposes in BS 8110 and BS 5400 respectivelyfor reinforcement in tension or compression are shown inTable 36 For design purposes the modulus of elasticity of allreinforcing steels is taken as 200 kNmm2 The BS 5400stressndashstrain curve is the one that was used in CP110 prior tothat code being superseded by BS 8110

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In the following the term nominal cover is used to describe thedesign cover shown on the drawings It is the required coverto the first layer of bars including links The nominal covershould not be less than the values needed for durability andfire-resistance nor less than the nominal maximum size ofaggregate Also the nominal cover should be sufficient toensure that the cover to the main bars is not less than the barsize or for a group of bars in contact the equivalent bar sizeFor a group of bars the equivalent bar size is the diameter of acircle whose cross-sectional area is equal to the sum of the areasof the bars in the group

231 DURABILITY

2311 Exposure classes

Details of the classification system used in BS EN 206-1 andBS 8500-1 with informative examples applicable in the UKare given in Table 37 Often the concrete can be exposed tomore than one of the actions described in the table in whichcase a combination of the exposure classes will apply At thetime of drafting this Handbook the amendments necessitatedby the introduction of BS 8500 have not been incorporated inBS 5400 or BS 8007 (which is based on BS 81101985) Thesystem of exposure classes concrete grades and covers usedprior to BS 8500 is given in Table 39

2312 Concrete strength classes and covers

Concrete durability is dependent mainly on its constituents andlimitations on the maximum free watercement ratio and theminimum cement content are specified for each exposure classThese limitations result in minimum concrete strength classesfor particular cements For reinforced concrete the protection ofthe steel against corrosion depends on the cover The requiredthickness of cover is related to the exposure class the concretequality and the intended working life of the structure Details ofthe recommendations in BS 8500 are given in Table 38

The values given for the minimum cover apply for ordinarycarbon steel in concrete without special protection and for

structures with an intended working life of at least 50 yearsThe values given for the nominal cover include an allowance fortolerance of 10 mm which is recommended for buildings andis normally also sufficient for other types of structures

For uneven concrete surfaces (eg ribbed finish or exposedaggregate) the cover should be increased by at least 5 mm Ifconcrete is cast against an adequate blinding the nominal covershould generally be at least 40 mm For concrete cast directlyagainst the earth the nominal cover should generally be not lessthan 75 mm

232 FIRE-RESISTANCE

2321 Building regulations

The minimum period of fire-resistance required for elements ofthe structure according to the purpose group of a building andits height or for basements depth relative to the ground aregiven in Table 312 Building insurers may require longer fireperiods for storage facilities

2322 Nominal covers and minimum dimensions

The recommendations in BS 8110 regarding nominal cover fordifferent periods of fire-resistance are given in Table 310 Inthe table the cover applies to links for beams and columns butto main bars for floor slabs and ribs (even if links are provided)For two-way spanning solid slabs the cover may be taken asthe average for each direction For beams floors and ribs therequirements apply to the reinforcement in the bottom and sidefaces only The minimum thickness of floors includes anyconcrete screed on the top surface This is particularly impor-tant for ribbed slabs where the structural flange could be nomore than 75 mm thick The values given in the table apply tomembers whose dimensions comply with the minimum valuesgiven in Table 311

For cases where it is considered that special measures needto be taken to prevent the spalling of concrete a summary ofthe recommendations in BS 8110-2 is given in Table 310

Chapter 23

Durability andfire-resistance

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37Exposure classification (BS 8500)

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38Concrete quality and cover requirements fordurability (BS 8500)

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39Exposure conditions concrete and cover requirements(prior to BS 8500)

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310Fire resistance requirements (BS 8110) ndash 1

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311Fire resistance requirements (BS 8110) ndash 2

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312Building regulations minimum fire periods

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241 DESIGN ASSUMPTIONS

Basic assumptions regarding the design of cracked concretesections at the ULS are outlined in section 52 The tensilestrength of the concrete is neglected and strains are evaluatedon the assumption that plane sections before bending remainplane after bending Reinforcement stresses are then derivedfrom these strains on the basis of the design stressndashstraincurves shown on Table 36 The BS 5400 curve is the one thatwas used in CP 110 prior to that code being superseded byBS 8110 For the concrete stresses alternative assumptions arepermitted The design stressndashstrain curve for concrete shownon Table 36 gives a stress distribution that is a combination ofa parabola and rectangle The form of the data governing theshape of the curve causes the relative proportions of the twoparts to vary as the concrete strength changes Thus the totalcompressive force provided by the concrete is not linearlyrelated to fcu and the position of the centroid of the stress-block changes with fcu

Alternatively an equivalent rectangular stress distribution inthe concrete may be assumed as noted on Table 36 TheBS 5400 rectangular stress-block is the one that was usedin CP 110 prior to that code being superseded by BS 8110 InBS 5400 the use of the rectangular stress-block is prohibited inflanged ribbed and voided sections where the neutral axis liesoutside the flange although there appears to be no logicalreason for this restriction

For a rectangular area of width b and depth x the totalcompressive force is given by k1 fcubx and the distance of theforce from the compression face is given by k2x where values ofk1 (allowing for the term 067m)and k2 are given in thefollowing table according to the shape of the stress-block

Properties of concrete stress-blocks for rectangular area

Shape fcu Nmm2 k1 k2

Parabolic- 25 0405 0456rectangular 30 0401 0452(Table 36) 35 0397 0448

40 0394 044550 0388 0439

BS 8110 rectangular 0402 0450

BS 5400 rectangular 0400 0500

Using the rectangular stress-block is conservative in BS 5400but gives practically the same result as that obtained with theparabolic-rectangular form in BS 8110

242 BEAMS AND SLABS

Beams and slabs are generally subjected to bending only butsometimes are also required to resist an axial force for examplein a portal frame or in a floor acting as a prop between base-ment walls Axial thrusts not exceeding 01 fcu times the area ofthe cross section may be ignored in the analysis of the sectionsince the effect of the axial force is to increase the moment ofresistance Where a section is designed to resist bending onlythe value of the lever arm should not be taken greater than 095times the effective depth of the reinforcement

In cases where as a result of moment redistribution allowedin the analysis of the member the design moment is less than themaximum elastic moment at the section the neutral axis depthshould satisfy the requirement xd ( b 04) where b is theratio of design moment to maximum elastic moment

2421 Singly reinforced rectangular sections

Chapter 24

Bending and axialforce

The lever arm between the forces shown in the figure here isgiven by z (d k2x) from which x (d z)k2

Taking moments for the compressive force about the line ofaction of the tensile force gives

M k1 fcubxz k1 fcubz(d z)k2

The solution of the resulting quadratic equation in z gives

zd 05 where K Mbd 2fcu025(k2 k1)K 095

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Taking moments for the tensile force about the line of action ofthe compressive force gives

M As fsz from which As Mfsz

The strain in the reinforcement s 00035(1 xd)(xd) andfrom the BS 8110 design stressndashstrain curve the stress in thereinforcement is given by

fs sEs 700(1 xd)(xd) fy115

Thus for design to BS 8110 fs fy115 for values of

xd 805(805 fy) 0617 for fy 500 Nmm2

From the BS 5400 design stressndashstrain curve the stress in thereinforcement is given by

fs sEs 700(1 xd)(xd) 08fy115 or

08fy115 fs

Thus for design to BS 5400 fs 08fy115 for values of

xd 805(805 08fy) 0668 for fy 500 Nmm2

and fs fy115 for values of

xd 805(1265 fy) 0456 for fy 500 Nmm2

Design charts for fy 500 Nmm2 derived on the basis of theparabolic-rectangular stress-block for the concrete are given inTable 313 for BS 8110 and Table 323 for BS 5400 In the caseof BS 5400 unless the section is proportioned such thatfs fy115 the moment of resistance should provide at least115 times the applied design moment

Design tables based on the rectangular stress-blocks for theconcrete are given in Table 314 for BS 8110 and Table 324for BS 5400 The tables use non-dimensional parameters andare valid for fy 500 Nmm2 The formulae used to derive thetables and the limitations when redistribution of moment hasbeen allowed in the analysis of the member are also given

2422 Doubly reinforced rectangular sections

200fy

2300 fy35 45x d

x d fy 115

Thus for design to BS 8110 for values of

From the BS 5400 design stress-strain curve the stress in thereinforcement is given by

or

Thus for design to BS 5400 for values of

for fy500 Nmm2

and for values of

Equating the tensile and compressive forces gives

where the stress in the tension reinforcement is given by theexpressions used in section 2421

Design charts based on the rectangular stress-blocks forconcrete and for dd 01 and 015 respectively are given inTables 315 and 316 for BS 8110 and Tables 325 and 326 forBS 5400 The charts use non-dimensional parameters and weredetermined for fy 500 Nmm2 but may be safely used forfy 500 Nmm2 In determining the forces in the concrete noreduction has been made for the area of concrete displaced bythe compression reinforcement

2423 Design formulae for rectangular sections

Design formulae based on the rectangular stress-blocks forconcrete are given in BS 8110 and BS 5400 In both codes x islimited to 05d so that the formulae are automatically valid forredistribution of moment not exceeding 10

The stress in the tension reinforcement is taken as 087fy inboth codes although this is only strictly valid for xd 0456in BS 5400 The stress in the compression reinforcement istaken as 087fy in BS 8110 and 072fy in BS 5400 The codeequations which follow from the analyses in sections 2421and 2422 take different forms in BS 8110 and BS 5400

In BS 8110 the requirement for compression reinforcementdepends on the value of K Mbd2fcu compared to where

0156 for b 09

0402( b 04) 018( b 04)2 for 09 b 07

b is the ratio design moment to maximum elastic moment

For K compression reinforcement is not required and

As M087fyz

where

z d05 + lt 095d and x (d z)045

For K compression reinforcement is required and

(K ) bd2fcu087fy (d )

As bd2fcu087fyz

where

z d05 + and x (d z)045

For 0375 (for fy 500 Nmm2) should be replacedby 16(1 ) in the equations for and AsAsAsdx

Asdx

025 K09

KAs

dKAs

K

025K09

K

K

K

K

As fs k1 fcubx As f s

x d (7 3)(dd)f s 2000fy(2300 fy)

x d [805 (805 08fy)](dd) 2(dd)

f s 08fy 115

08fy 115 f s 200fy

2300 fy11535d

x 2000fy

2300 fy

f s sEs 700(1 dx) 08fy 115

x d [805(805 fy)](dd) 264(dd) for fy 500Nmm2

f s fy 115

Beams and slabs 257

The forces provided by the concrete and the reinforcement areshown in the figure here Taking moments for the two com-pressive forces about the line of action of the tensile force gives

M k1 fcubx(d k2x)

The strain in the reinforcement and fromthe BS 8110 design stressndashstrain curve the stress is given by

f s sEs 700(1 dx) fy 115

s 00035(1 dx)

As f s(d d)

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313BS 8110 Design chart for singly reinforced rectangularbeams

Sin

gly

rein

forc

ed b

eam

s (f

y=

500

Nm

m2 )

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314BS 8110 Design table for singly reinforced rectangularbeams

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315BS 8110 Design chart for doubly reinforced rectangularbeams ndash 1

Dou

bly

rein

forc

ed b

eam

s (f

y=

500

Nm

m2

d9d

=0

1)

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316BS 8110 Design chart for doubly reinforced rectangularbeams ndash 2

Dou

bly

rein

forc

ed b

eam

s (f

y=

500

Nm

m2

d9d

=0

15)

wwwengbookspdfcom

In BS 5400 the moment of resistance of a section withoutcompression reinforcement is given by the equations

Mu As(087fy)z 015bd 2fcu

where

z d (1 11As fybdfcu) 095d

For a section with compression reinforcement the moment ofresistance is given by the equations

Mu 015bd2fcu (072fy)(d )

As(087fy) 02bdfcu (072fy)

The equations which are based on x 05d are consideredto be valid for values of dd 02 A variant on the equa-tions with x taken as a variable is included in HighwaysAgency BD4495 for assessment purposes These equationsshould not be used for values of x greater than 05d

2424 Flanged sections

In monolithic beam and slab construction where the web of thebeam projects below the slab the beam is considered as aflanged section for sagging moments The effective width of theflange may be taken as follows

T beam b bw 02lz actual flange widthL beam b bw 01lz actual flange width

lz is the distance between points of zero moment which for acontinuous beam may be taken as 07 times effective span

In most sections where the flange is in compression the depthof the neutral axis will be no greater than the thickness of theflange In this case the section can be considered to be rectangularwith b taken as the flange width The condition regarding theneutral axis depth can be confirmed initially by showing thatM k1 fcubhf (d k2hf) where hf is the thickness of the flangeAlternatively the section can be considered to be rectangularinitially and the neutral axis depth can be checked subsequently

As

dAs

Using the rectangular concrete stress-blocks in the foregoingequations gives k1 04 with k2 045 for BS 8110 and 05for BS 5400 This approach gives solutions that are lsquocorrectrsquowhen x hf but become slightly more conservative as(x hf) increases A different approach is used in BS 8110resulting in solutions that are lsquocorrectrsquo when x 05d butincreasingly conservative as (x hf) decreases As a result thesolution when x hf does not agree with that obtained byconsidering the section as rectangular with b taken as theflange width

2425 General analysis of sections

The analysis of a section of any shape with any arrangementof reinforcement involves a trial-and-error process Aninitial value is assumed for the neutral axis depth from whichthe concrete strains at the positions of the reinforcement canbe calculated The corresponding stresses in the reinforce-ment are determined and the resulting forces in thereinforcement and the concrete are obtained If the forces areout of balance the value of the neutral axis depth is changedand the process is repeated until equilibrium is achievedOnce the balanced condition has been found the resultantmoment of the forces about the neutral axis or any otherpoint is calculated

Example 1 The beam shown in the following figure is to bedesigned to the requirements of BS 8110 The design loads oneach span are as follows where Gk 160 kN andQk 120 kN

Fmax 14Gk 16Qk 416 kN Fmin 10Gk 160 kN

The section design is to be based on the following values

fcu 40 Nmm2 fy 500 Nmm2 cover to links 25 mm

For sagging moments effective width of flange

b bw 02lz 300 02(07 8000) 1420 mm

Allowing for 8 mm links and 32 mm main bars

d 500 (25 8 16) 450 mm say

Bending and axial force262

The figure here shows a flanged section where the neutral axisdepth is greater than the flange thickness The concrete forcecan be divided into two components and the required area oftension reinforcement is then given by

As As1 k1 fcu (b bw)hf 087fy

where

As1 area of reinforcement required to resist a moment M1

applied to a rectangular section of width bw and

M1 M k1 fcu (b bw)hf (d k2hf) 015bd2fcu

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In the calculations that follow solutions are obtained usingcharts and equations to demonstrate the use of each method

Maximum sagging moment For section to be designed asrectangular with b taken as the flange width bending momentshould satisfy the condition

M k1 fcubhf (d k2hf)

04 40 1420 150 (450 045 150) 106

1303 kNm (285 kNm)

Mbd2 285 106(1420 4502) 099 Nmm2

From chart in Table 313 100Asbd 024

As 00024 1420 450 1534 mm2

Alternatively K Mbd2fcu 09940 00248From Table 314 As fybdfcu 121K 00300

As 003 1420 450 40500 1534 mm2

Alternatively by calculation or from Table 314

Hence As M087fyz gives

As 285 106(087 500 095 450) 1533 mm2

Using 2H32 gives 1608 mm2

Maximum hogging moment

K Mbd2fcu 416 106(300 4502 40) 0171

From Table 314 As fybdfcu 0264

As 0264 300 450 40500 2851 mm2

Using 4H32 gives 3217 mm2

Although this is a valid solution it may be possible to reducethe area of tension reinforcement to a more suitable value byallowing for some compression reinforcement Consider theuse of 2H25 with 45 mm ( 01)

fybdfcu 982 500(300 450 40) 009

From the chart in Table 315 As fybdfcu 0225

As 0225 300 450 40500 2430 mm2

A solution can also be obtained using the design equations asfollows

With 2H25 for and assuming

K (087fy)(d )bd2fcu

0171982087500405(3004502 40)0100

Since is validdx 0375 f s 087fy

dx (dd)(x d) 010282 0355

x d (1 z d) 045 (1 0873) 045 0282

z d 05 025 0100 09 0873

dAsK

f s 087fyAs

As

ddd

z d 05 025 00248 09 0972 095

As bd2fcu 087fyz

982 0100 300 4502 40(087 500 0873 450)

2404 mm2 (compared to 2430 mm2 obtained from chart)

Using 3H32 gives 2413 mm2

Example 2 Suppose that in the previous example the maxi-mum hogging moment at B is reduced by 30 to 291 kNm

K Mbd2fcu 291 106(300 4502 40) 0120

b 291416 070 xd ( b 04) 030

From chart in Table 315 keeping to left of line for xd 03

fybdfcu 0025 As fybdfcu 0158

0025 300 450 40500 270 mm2

As 0158 300 450 40500 1706 mm2

A solution can also be obtained by using the design equationswith xd 03 as follows

0402(b 04) 018(b 04)2

0402 03 018 032 0104 (K 0120)

zd 05 0867

(K ) bd2fcu087fy (d ) 0016 300 4502 40 (087 500 405) 221 mm2

As bd2fcu087fyz 221 0104 300 4502 40(087 500 0867 450) 1710 mm2

Using 2H25 and 1H32 gives 1786 mm2

Since the reduced hogging moment for load case 1 is stillgreater than the elastic hogging moment for load case 2 thedesign sagging moment remains the same as in example 1

In the foregoing examples at the bottom of the beam 2H32bars would run the full length of each span with 2H25 splicebars at support B Other bars would be curtailed according tothe bending moment requirements and detailing rules

243 COLUMNS

In the Codes of Practice a column is a compression memberwhose greater overall cross-sectional dimension does notexceed four times its smaller dimension An effective heightand a slenderness ratio are determined in relation to major andminor axes of bending An effective height is a function of theclear height and depends upon the restraint conditions at theends of the column A slenderness ratio is defined as the effec-tive height divided by the depth of the cross section in the planeof bending The column is then considered to be either short orslender according to the slenderness ratios

Columns are subjected to combinations of bendingmoment and axial force and the cross section may need to bechecked for more than one combination of values In slendercolumns from an elastic analysis of the structure the initialmoment is increased by an additional moment induced bythe deflection of the column In BS 8110 this additional

KAs

dKAs

0250104 09

K

As

As

KAs

Columns 263

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moment contains a modification factor K the use of whichresults in an iteration process with K taken as 10 initiallyThe design charts in this chapter contain sets of K lines as anaid to the design process Details of the design procedures aregiven in Tables 321 and 322 for BS 8110 and Tables 331and 332 for BS 5400

2431 Rectangular columns

Design charts based on the rectangular stress-block for theconcrete and for values of dh 08 and 085 are given inTables 327 and 328 respectively On each curve a straight linehas been taken between the point where xh 08 and the pointwhere N Nuz The use of the rectangular stress-block resultsin Nuz being given by the equation

Nuzbhfcu 04 0714(Asc fybhfcu)

There are no K lines on the charts as no modification factor isused in the design of slender columns to BS 5400

2432 Circular columns

Bending and axial force264

The figure here shows a rectangular section in which thereinforcement is disposed equally on two opposite sides of ahorizontal axis through the mid-depth By resolving forces andtaking moments about the mid-depth of the section the followingequations are given for 0 xh 10

Nbhfcu k1(xh) 05(Asc fybhfcu)(ks1 ks2)

Mbh2fcu k1(xh)05 k2(xh) 05(Asc fybhfcu)(ks1 ks2) (dh 05)

For BS 8110 the stress factors ks1 and ks2 are given by

ks1 14(xh dh 1)(xh) 087

ks2 14(dh xh)(xh) 087

The maximum axial force Nuz is given by the equation

Nuzbhfcu 045 087(Asc fybhfcu)

Design charts based on the rectangular stress-block for theconcrete and for values of dh 08 and 085 are given inTables 317 and 318 respectively On each curve a straightline has been taken between the point where xh 10 andthe point where N Nuz The charts which were determinedfor fy 500 Nmm2 may be safely used for fy 500 Nmm2In determining the forces in the concrete no reduction hasbeen made for the area of concrete displaced by the com-pression reinforcement In the design of slender columns theK factor is used to modify the deflection corresponding to aload Nbal which for a symmetrically reinforced rectangularsection is given as 025bdfcu In the charts Nbal is taken as thevalue at which M is a maximum A line corresponding to Nbal

passes through a cusp on each curve For N Nbal K is takenas 10 For N Nbal K can be determined from the lines onthe chart

For BS 5400 the stress factors ks1 and ks2 are given by

ks1 14(xh dh 1)(xh) 07

07 ks1 025(33xh dh 1)(xh) 0714

ks2 14(dh xh)(xh) 07

07 ks2 025(dh 13xh)(xh) 087

The figure here shows a circular section containing six barsspaced equally around the circumference Solutions based on sixbars will be slightly conservative if more bars are used Thearrangement of the bars relative to the axis of bending affectsthe resistance of the section and the arrangement shown in thefigure is not the most critical in every case For some combina-tions of bending moment and axial force if the arrangementshown is rotated through 30o a slightly more critical conditionresults but the differences are small and may be reasonablyignored

The following analysis is based on a uniform stress-block forthe concrete of depth x and width hsin at the base (as shownin figure) Negative axial forces are included in order to cater formembers such as tension piles By resolving forces and takingmoments about the mid-depth of the section the followingequations are obtained where cos1 (1 2 xh) for0 x 10 and hs is the diameter of a circle through the centresof the bars

Nh2fcu kc (2 sin2)8 (12)(Asc fyAc fcu)(ks1 ks2 ks3)

Mh3fcu kc (3sin sin3)72 (277)(Asc fyAc fcu)(hsh)(ks1 ks3)

The minimum axial force Nmin is given by the equation

Nminh2fcu 087(4)(Asc fyAc fcu)

For BS 8110 kc 045 09 and the stress factors ks1 ks2

and ks3 are given by

087 ks1 14(0433hsh 05 xh)(xh) 087

087 ks2 14(05 xh)(xh) 087

087 ks3 14(05 0433hsh xh)(xh) 087

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The maximum axial force Nuz is given by the equation

Nuzh2fcu (4)045 087(Asc fyAc fcu)

Design charts for values of hsh 06 and 07 are given inTables 319 and 320 respectively The statements in section2431 on the derivation and use of the charts for rectangularsections apply also to those for circular sections

For BS 5400 kc 04 10 and the stress factors ks1 ks2

and ks3 are given by

087 ks1 025(0433hsh 05 13xh)(xh) 07

07 ks1 14(0433hsh 05 xh)(xh) 07

07 ks1 025(0433hsh 05 33xh)(xh) 0714

0714 ks2 025(05 33xh)(xh) 07

07 ks2 14(05 xh)(xh) 07

07 ks2 025(05 13 xh)(xh) 087

0714 ks3 025(0433hsh 05 33xh)(xh) 07

07 ks3 14(0433hsh 05 xh)(xh) 07

07 ks3 025(0433hsh 05 13xh)(xh) 087

The maximum axial force Nuz is given by the equation

Nuzh2fcu ( 4)04 072(Asc fyAc fcu)

Design charts for values of hsh 06 and 07 are given inTables 329 and 330 respectively The statements in section2431 on the derivation and use of the charts for rectangularsections apply also to those for circular sections

2433 Design formulae for short braced columns

Approximate formulae are given in BS 8110 for the design ofshort braced columns under specific conditions Where due tothe nature of the structure a column cannot be subjected tosignificant moments it may be considered adequate if thedesign ultimate axial load N 04fcuAc 075Asc fy

Columns supporting symmetrical arrangements of beamsthat are designed for uniformly distributed imposed load andhave spans that do not differ by more than 15 of the longermay be considered adequate if N 035fcuAc 067Asc fy

2434 General analysis of column sections

Any given cross section can be analysed by a trial-and-errorprocess For a section bent about one axis an initial value isassumed for the neutral axis depth from which the concretestrains at the positions of the reinforcement can be calculatedThe resulting stresses in the reinforcement are determined andthe forces in the reinforcement and concrete evaluated If theresultant force is not equal to the design axial force N the valueof the neutral axis depth is changed and the process repeateduntil equality is achieved The resultant moment of all theforces about the mid-depth of the section is then the moment ofresistance appropriate to N This approach is used to analyse arectangular section in example 6

Example 3 A 300 mm square braced column designed toBS 8110 for the following requirements

lo 375 m and 09 in both directions

Mx 54 kNm My 0 N 1800 kN

fcu 40 Nmm2 fy 500 Nmm2 cover to links 35 mm

Since leh 09 3750300 1125 15 the column isshort

Mmin N(005h) 1800 005 03 27 kNm (Mx)

Allowing for 8 mm links and 32 mm main bars

d 300 (35 8 16) 240 mm say

Mbh2fcu 54 106(300 3002 40) 005

Nbhfcu 1800 103(300 300 40) 05

From the design chart for dh 240300 08

Asc fybhfcu 022 (Table 317)

Asc 022 300 300 40500 1584 mm2

Using 4H25 gives 1963 mm2

Example 4 A 300 mm circular braced column designed toBS 8110 for the same requirements as example 3

Allowing for 8 mm links and 32 mm main bars

hs 300 2 (35 8 16) 180 mm say

Mh3fcu 54 106(3003 40) 005

Nh2fcu 1800 103(3002 40) 05

From the design chart for hsh 180300 06

Asc fyAc fcu 052 (Table 319)

Asc 052 (4) 3002 40500 2940 mm2

Using 6H25 gives 2945 mm2

Example 5 The column in example 3 but designed for biaxialbending with My 25 kNm and all other requirements as before

Since and Mx My the section may be designed for anincreased moment about the x-x axis (see Table 321)

1 (76)(Nbhfcu) 1 (76) 05 042

Mx My 54 042 25 645 kNm

Mbh2fcu 645 106(300 3002 40) 006

From the design chart for dh 240300 08

Asc fybhfcu 026 (Table 317)

Asc 026 300 300 40500 1872 mm2

Using 4H25 gives 1963 mm2

Example 6 A 300 mm square short column designed toBS 5400 for the following requirements

Mx 60 kNm My 40 kNm N 1800 kN

fcu 40 Nmm2 fy 500 Nmm2 d 240 mm

The section may be designed by assuming the reinforcement(4H32 say) and checking the condition (see Table 331)

Asc fybhfcu 3217 500(300 300 40) 045

Nbhfcu 1800 103(300 300 40) 05

Mx

bh

Columns 265

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317BS 8110 Design chart for rectangular columns ndash 1

Rec

tang

ular

col

umns

(f y

=50

0 N

mm

2 d

h=

08)

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318BS 8110 Design chart for rectangular columns ndash 2

Rec

tang

ular

col

umns

(f y

=50

0 N

mm

2 d

h=

085

)

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319BS 8110 Design chart for circular columns ndash 1

Circular columns (fy= 500 Nmm2 hs h = 06)

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320BS 8110 Design chart for circular columns ndash 2

Circular columns (fy= 500 Nmm2 hsh = 07)

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321BS 8110 Design procedure for columns ndash 1

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322BS 8110 Design procedure for columns ndash 2

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Bending and axial force272

From the design chart for dh 240300 08

Mubh2fcu 0075 Nuzbhfcu 072 (Table 327)

n 067 167(NNuz) 067 167(05072) 18

Since the column is square

Mux Muy 0075 300 3002 40 106 81 kNm

Since this value is less than 10 4H32 are sufficient

Example 7 The column in example 3 but taken as unbraced( 16) in the direction of Mx with all other requirements asbefore

Since lexh 16 3750300 20 10 the column is slender

The additional bending moment about the x-x axis is given by

Madd N(Kh2000)(lexb)2

With K 10 initially and since b h

Madd 1800 032000 202 108 kNm

M Mi Madd 54 108 162 kNm

Mbh2fcu 162 106(300 3002 40) 015

Nbhfcu 05 as before and from the design chart

Asc fybhfcu 063 (Table 317)

Asc 063 300 300 40500 4536 mm2

This requires 4T40 but it can be seen from the chart that withAsc fybhfcu 063 K is about 06 If we use 4H32

Asc fybhfcu 3217 500(300 300 40) 045

Nbhfcu 05 as before and from the design chart

Mbh2fcu 0107 and K 053

With K 053 corresponding to Asc fybhfcu 045

Madd 053 108 57 kNm M 54 57 111 kNm

Mbh2fcu 111 106(300 3002 40) 0103 0107

Thus 4H32 which gives 3217 mm2 is sufficient

Note that K can also be calculated from the equations given inBS 8110 as follows

Nuzbhfcu 045 087(Asc fybhfcu) 045 087 045 084

Nbalbhfcu 025(dh) 025 08 02

K (Nuz N)( Nuz Nbal) (084 05)(084 02) 053

Example 8 The following figure shows a rectangular sectionreinforced with 8H32 The ultimate moment of resistance of thesection about the major axis is to be determined in accordancewith the following requirements

N 2500 kN fcu 40 Nmm2 fy 500 Nmm2

Mx

Mux

n

My

Muy

n

6081

18

4081

18

086

Consider the bars in each half of the section to be replaced byan equivalent pair of bars Depth to the centre of area of the barsin one half of the section 60 2404 120 mm The sectioncan now be considered to be reinforced with four bars of areaAsc4 where d 600 120 480 mm

Asc fybhfcu 6434 500(300 600 40) 045

Nbhfcu 2500 103(300 600 40) 035

From the design chart for dh 480600 08

Mbh2fcu 014 (Table 317)

M 014 300 6002 40 106 605 kNm

The solution can be checked using a trial-and-error process toanalyse the original section as follows

N k1 fcubx (As1ks1 As2ks2 As3ks3)fy

where dh 540600 09 and ks1 ks2 and ks3 are givenby

ks1 14(xh dh 1)(xh) 087

ks2 14(05 xh)(xh) 087

ks3 14(dh xh)(xh) 087

With x 300 mm xh 05 ks1 087 ks2 0 and ks3 087

N 04 40 300 300 103 1440 kN (2500)

With x 360 mm xh 06 ks2 0233 ks3 07

N 04 40 300 360 103 (2413 087 1608 0233 2413 07) 500 103

1728 392 2120 kN (2500)

With x 390 mm xh 065 ks2 0323 ks3 0538

N 04 40 300 390 103 (2413 087 1608 0323 2413 0538) 500 103

1872 660 2532 kN (2500)

With x 387 mm xh 0645 ks2 0315 ks3 0553

N 04 40 300 387 103 (2413 087 1608 0315 2413 0553) 500 103

1858 636 2494 kN (asymp2500)

Taking moments about the mid-depth of the section gives

M k1 fcubx(05h k2x ) (As1ks1 As3ks3)(d 05h)fy

04 40 300 387 (300 045 387) 106

(241308724130553)(540300) 500106

233 412 645 kNm (605 obtained before)

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323BS 5400 Design chart for singly reinforced rectangularbeams

Sin

gly

rein

forc

ed b

eam

s (f

y=

500

Nm

m2 )

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324BS 5400 Design table for singly reinforced rectangularbeams

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325BS 5400 Design chart for doubly reinforced rectangularbeams ndash 1

Dou

bly

rein

forc

ed b

eam

s (f

y=

500

Nm

m2

d9d

=0

1)

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326BS 5400 Design chart for doubly reinforced rectangularbeams ndash 2

Dou

bly

rein

forc

ed b

eam

s (f

y=

500

Nm

m2

d9d

=0

15)

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327BS 5400 Design chart for rectangular columns ndash 1

Rec

tang

ular

col

umns

(f y

=50

0 N

mm

2 d

h=

08)

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328BS 5400 Design chart for rectangular columns ndash 2

Rec

tang

ular

col

umns

(f y

=50

0 N

mm

2 d

h=

085

)

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329BS 5400 Design chart for circular columns ndash 1

Rectangular columns (fy= 500 Nmm2 hsh = 06)

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330BS 5400 Design chart for circular columns ndash 2

Circular columns (fy= 500 Nmm2 hsh = 07)

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331BS 5400 Design procedure for columns ndash 1

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332BS 5400 Design procedure for columns ndash 2

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Chapter 25

Shear and torsion

251 SHEAR RESISTANCE

2511 Shear stress

In BS 8110 the design shear stress at any cross section in amember of uniform depth is calculated from

v Vbvd

where

V is the shear force due to ultimate loadsbv is the breadth of the section which for a flanged section

is taken as the average width of the web below the flanged is the effective depth to the tension reinforcement

For a member of varying depth the shear force is calculated asV (M tan13s)d where 13s is the angle between the tensionreinforcement and the compression face of the member Thenegative sign applies when moment and effective depth bothincrease in the same direction In no case should v exceed thelesser of 08radicfcu or 5 Nmm2 whatever the reinforcement InBS 5400 b is used in place of bv and the maximum value ofv is taken as the lesser of 075radicfcu or 475 Nmm2

2512 Concrete shear stress

In BS 8110 the design concrete shear stress vc is a function ofthe concrete grade the effective depth and the percentage ofeffective tension reinforcement at the section considered It isoften convenient to determine vc at the section where thereinforcement is least and use the same value throughout themember For sections at distance av 2d from the face of asupport or concentrated load vc may be multiplied by 2davproviding the tension reinforcement is adequately anchoredAlternatively as a simplification for beams carrying uniformload the section at distance d from the face of a support maybe designed without using this enhancement and the samereinforcement provided at sections closer to the support InBS 5400 the design concrete shear stress is obtained as svcwhere vc is the ultimate shear stress and s is a depth factor

2513 Shear reinforcement

Requirements for shear reinforcement depend on the value ofv in relation to vc (or svc) In slabs no shear reinforcement isrequired provided v does not exceed the concrete shear stress Inall beams of structural importance a minimum amount of shear

reinforcement in the form of links equivalent to a shearresistance Vsv 04 bvd is required The shear resistance can beincreased by introducing more links or bent-up bars can be usedto provide up to 50 of the total shear reinforcement Thecontribution of the shear reinforcement is determined on the basisof a truss analogy in which the bars act as tension members andinclined struts form within the concrete as shown in the figure here

System of bent-up bars used as shear reinforcement

In the figure the truss should be chosen so that both and are 45o and st 15d The design shear resistance providedby a system of bent-up bars is then given by

Vsb (Asbsb)(087fy)(stsin)

where st (d ndash d)(cot cot) 15d

For bars bent-up at 45o with fy 500 Nmm2

Vsb 0461Asb (dsb) 0307Asb kN

For bars bent-up at 60o with fy 500 Nmm2

Vsb 0594 Asb (d ndash d)sb 0376 Asb kN

In BS 8110 the shear resistance of members containing shearreinforcement is taken as the sum of the resistances providedseparately by the shear reinforcement and the concrete Thestrut analogy results in an additional longitudinal tensileforce that is effectively taken into account in the curtailmentrules for the longitudinal reinforcement In BS 5400 theminimum amount of shear reinforcement is required inaddition to that needed to cater for the difference between thedesign shear force and the concrete shear resistance It is alsonecessary to design the longitudinal reinforcement for theadditional force

Details of the design procedures for determining the shearresistances of members are given in Table 333 for BS 8110and Table 336 for BS 5400

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BS 8110 Shear resistance 333

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Design for torsion 285

2514 Shear under concentrated loads

The maximum shear stress at the edge of a concentrated loadshould not exceed the lesser of 08radicfcu or 5 Nmm2 Shear insolid slabs under concentrated loads can result in punching fail-ures on the inclined faces of truncated cones or pyramids Forcalculation purposes the shear perimeter is taken as the bound-ary of the smallest rectangle that nowhere comes closer to theedges of a loaded area than a specified distance The shearcapacity is checked first on a perimeter at distance 15d fromthe edge of the loaded area If the calculated shear stress is nogreater than vc no shear reinforcement is needed If the shearstress exceeds vc shear reinforcement is required within thefailure zone and further checks are needed on successiveperimeters at intervals of 075d until a perimeter is reachedwhere shear reinforcement is no longer required

Details of design procedures for shear under concentratedloads are given in Table 334 for BS 8110 and Tables 337 and338 for BS 5400

2515 Shear in bases

The shear strength of pad footings near concentrated loads isgoverned by the more severe of the following two conditions

(a) Shear along a vertical section extending for the full widthof the base In BS 8110 the concrete shear stress vc may bemultiplied by 2dav for all values of av 2d In BS 5400 thecritical section is taken at distance d from the face of the loadwith no enhancement of the concrete shear stress(b) Punching shear around the loaded area as described insection 2514 The reaction resulting from the soil bearingpressure within the shear perimeter may be deducted from thedesign load on the column when calculating the design shearforce acting on the section

The shear strength of pile caps is normally governed by theshear along a vertical section extending for the full width of thecap The critical section for shear is assumed to be located at20 of the pile diameter from the near face of the pile Thedesign shear force acting on this section is taken as the wholeof the reaction from the piles with centres lying outside thesection In BS 8110 the design concrete shear stress may bemultiplied by 2dav where av is the distance from the columnface to the critical section for strips of width up to three timesthe pile diameter centred on each pile In BS 5400 thisenhancement may be applied to strips of width equal to one pilediameter centred on each pile For pile caps designed by trussanalogy 80 of the tension reinforcement should be concen-trated in these strips

2516 Bottom loaded beams

Where load is applied near the bottom of a section sufficientvertical reinforcement to transmit the load to the top of thesection should be provided in addition to any reinforcementrequired to resist shear

252 DESIGN FOR TORSION

In normal beam-and-slab or framed construction calculationsfor torsion are not usually necessary adequate control of anytorsional cracking in beams being provided by the required

minimum shear reinforcement When it is judged necessaryto include torsional stiffness in the analysis of a structure ortorsional resistance is vital for static equilibrium membersshould be designed for the resulting torsional moment Thetorsional resistance of a section may be calculated on the basisof a thin-walled closed section in which equilibrium is satisfiedby a closed plastic shear flow Solid sections may be modelled asequivalent thin-walled sections Complex shapes may be dividedinto a series of sub-sections each of which is modelled as anequivalent thin-walled section and the total torsional resistancetaken as the sum of the resistances of the individual elementsWhen torsion reinforcement is required this should consist ofrectangular closed links together with longitudinal reinforce-ment Such reinforcement is additional to any requirements forshear and bending

Details of design procedures for torsion are given in Table 335for BS 8110 and Table 339 for BS 5400

Example 1 The beam shown in the following figure is to bedesigned for shear to the requirements of BS 8110 Details ofthe design loads and the bending requirements for which thetension reinforcement comprises 2H32 (bottom) and 3H32 (topat support B) are contained in example 1 of Chapter 24 Thedesign of the section is to be based on the following values

fcu 40 Nmm2 fy 500 Nmm2 d 450 mm

In the following calculations a simplified approach is used inwhich the critical section for shear is taken at distance d fromthe face of the support with no enhancement of the concreteshear stress In addition the value of vc is determined for thesection where the tension reinforcement is least and the samevalue of vc used throughout Since the beam will be providedwith shear reinforcement the value of vc may be taken as thatobtained for a section with d 400 mm

Based on 2H32 as effective tension reinforcement

100Asbvd 100 1608(300 450) 119

vc 078 Nmm2 (Table 333 for d 400 and fcu 40)

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BS 8110 Shear under concentrated loads 334

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BS 8110 Design for torsion 335

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BS 5400 Shear resistance 336

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BS 5400 Shear under concentrated loads ndash 1 337

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BS 5400 Shear under concentrated loads ndash 2 338

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BS 5400 Design for torsion 339

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Shear and torsion292

Consider H10 links and transverse spacing of legs at 200 mmto suit spacing of tension reinforcement that is 5 legs permetre Then required longitudinal spacing of links is given by

sv 5 78128 304 mm

Provide H10-300 with legs at 200 mm centres transversely

Example 3 A 280 mm thick flat slab is supported by 400 mmsquare columns arranged on a 72 m square grid Theslab which has been designed using the simplified method fordetermining moments contains as tension reinforcement inthe top of the slab at an interior support within a 18 m widestrip central with the column H16-180 in each directionThe slab is to be designed to the requirements of BS 8110(see Table 334) for a shear force resulting from the maximumdesign load applied to all panels adjacent to the column ofVt 954 kN

fcu 40 Nmm2 fy 500 Nmm2 d 240 mm (average)

For design using the simplified method the design effectiveshear force at an interior column Veff 115V 1098 kN

The maximum design shear stress at the column face

Veff uod 1098 103(4 400 240) 286 Nmm2 (50)

Based on H16-180 as effective tension reinforcement

100Asbvd 100 201(180 240) 046

vc 065 Nmm2 (Table 333 for d 240 and fcu 40)

The length of the first critical perimeter at 15d from the face ofthe column is 4 (3d 400) 4480 mm Thus the designshear stress at the first critical perimeter

v 1098 103(4480 240) 102 Nmm2 ( 157vc)

Since vc v 16vc and (v ndash vc) 04 Nmm2 the total areaof vertical links required within the failure zone is given by

Asv 04ud087fyv 04 4480 240(087 500)

989 mm2

20H8 will provide 1006 mm2 which should be arranged on twoperimeters at 05d 120 mm and 125d 300 mm from thecolumn face The inner perimeter should contain at least 40of the total that is 8H8 with 12H8 on the outer perimeter

The length of the second critical perimeter at 225d fromthe face of the column is 4 (45d 400) 5920 mm Thusthe design shear stress at the second critical perimeter

v 1098 103(5920 240) 077 Nmm2 (vc)

Asv 04 5920 240(087 500) 1307 mm2

12H8 are already provided by the outer perimeter of bars inthe first failure zone A further perimeter containing 16H8 togive a total of 28H8 will provide 1407 mm2 in the secondfailure zone The length of the third critical perimeter at 3d fromthe face of the column is 4 (6d 400) 7360 mm Thusthe design shear stress at the third critical perimeter

v 1098 103(7360 240) 062 Nmm2 (vc)

The reinforcement layout is shown in the figure followingwhere indicates the link positions and the spacing of thetension reinforcement has been adjusted so that the links can beanchored round the tension bars

Minimum link requirements are given by

Asv sv 04bv087fyv 04 300(087 500)

028 mm2mm

sv 075d 075 450 3375 mm

From Table 335 H8-300 provides 033 mm2mmShear resistance of section containing H8-300 is given by

Vu vcbvd (Asv sv)(087fyv)d

(078300033087500) 45010ndash3 170 kN

Based on a support of width 400 mm distance from centre ofsupport to critical section 200 450 650 mm The designload is 4168 52 kNm and the shear forces at the criticalsections are

End A V 172 ndash 065 52 138 kN 170 kN (H8-300)

End B V 260 ndash 065 52 226 kN 170 kN

v Vbvd 226 103(300 450) 168 Nmm2

Area of links required at end B is given by

Asvsv (v ndash vc)bv 087fyv (168 ndash 078) 300(087 500)

062 mm2mm

From Table 335 H8-150 provides 067 mm2mm

Note that if the concrete shear strength is taken as (2dav)vc forav 2d critical section is at av 2d and distance from centreof support to critical section 200 900 1100 mm HereV 203 kN v 150 Nmm2 Asvsv 050 mm2mm andfrom Table 335 H8-200 would be sufficient

Example 2 A 700 mm thick solid slab bridge deck is supportedat the end abutment on bearings spaced at 15 m centres Themaximum bearing reaction resulting from the worst arrange-ment of the design loads is 625 kN The tension reinforcement atthe end of the span is H25-200 The slab is to be designed forshear to the requirements of BS 5400 using the following values

fcu 40 Nmm2 fy 500 Nmm2 d 620 mm

If the critical section for punching is taken at 15d 930 mmfrom the edge of each bearing clearly the critical perimetersfrom adjacent bearings will overlap Therefore the slab will bedesigned for shear along a vertical section extending for thefull width of the slab Assuming that the bearing reaction canbe spread over a slab strip equal in width to the spacing ofthe bearings

v Vbd 625 103(1500 620) 067 Nmm2

Based on H25-200 as effective tension reinforcement

100Asbd 100 2454(1000 620) 040

vc 054 Nmm2 s 095 (Table 336 for fcu 40)

Area of links required at end of span is given by

Asvsv (v 04 ndash svc)b087fyv

(06704 ndash 095054)1000(087500)

128 mm2mm

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Design for torsion 293

Example 4 The following figure shows a channel sectionedge beam on the bottom flange of which bear 8 m long simplysupported contiguous floor units The edge beam which iscontinuous over several 14 m spans is prevented from lateralrotation at its supports The positions of the centroid and theshear centre of the section are shown in the figure and the beamis to be designed to the requirements of BS 8110

Characteristic loadsfloor units dead 35 kNm2 imposed 25 kNm2

edge beam dead 12 kNm

Design ultimate loadsfloor units (14 35 16 25) 82 356edge beam 14 12 168

524 kNm

fcu 40 Nmm2 fy 500 Nmm2 d 1440 mm

Bending moment shear force and torsional moment (aboutshear centre of section) at interior support (other than first)

M ndash 008 524 142 ndash 822 kNm (Table 229)

V 524 142 367 kN

T (356 0400 168 0192) 142 122 kNm

(Note In calculating V and T a coefficient of 05 rather than055 has been used since the dead load is dominant and thecritical section may be taken at the face of the support)

Considering beam as one large rectangle of size 250 1500and two small rectangles of size 200 300

13hmin3hmax 2503 1500 2 2003 300

(234 2 24) 109 282 109

Torsional moment to be considered on large rectangle

T1 122 234282 1012 kNm

Torsional moment to be considered on each small rectangle

T2 122 24282 104 kNm

Reinforcement required in large rectangle

Bending (see Table 314)

K Mbd2fcu 822 106(550 14402 40) 0018Since K 0043 As M087fyz where z 095d

As 822 106(087 500 095 1440) 1382 mm2

Shear (see Table 333)

v Vbvd 367 103(250 1440) 102 Nmm2

100Asbvd 100 1382(250 1440) 038vc 053 Nmm2 (for d 400 and fcu 40)

Asvsv bv(v ndash vc)087fyv 250 (102 ndash 053)(087 500)

028 mm2mm (total for all vertical legs)

Additional requirement for bottom loaded beam

Asvsv 356(087 500) 008 mm2mm (for inner leg)

Torsion (see Table 335)

vt 2T1[hmin2(hmax ndash hmin3)]

2 1012 106[2502 (1500 ndash 2503)] 229 Nmm2

(v vt) 102 229 331 Nmm2 (vtu 50)

Assuming 30 mm cover to links dimensions of links

x1 250 ndash 2 35 180 mm y1 1500 ndash 2 35 1430 mm

Asvsv T1[08x1y1(087fy)]

1012 106(08 180 1430 087 500)

113 mm2mm (total for 2 outer legs)

Total link requirement for shear torsion and bottom loadassuming single links with 2 legs

Asvsv 028 113 2 008 157 mm2mm

sv least of x1 180 mm y12 715 mm or 200 mm

From Table 335 H10-100 provides 157 mm2mm

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Shear and torsion294

Total area of longitudinal reinforcement for torsion

As (Asvsv)(fyvfy)(x1 y1)

113 10 (180 1430) 1819 mm2

Area of longitudinal reinforcement required in part of thesection between centrelines of flanges (1300 mm apart)

1819 1300(180 1430) 1469 mm2

From Table 228 14H12 provides 1583 mm2

Total area of tension reinforcement required at top of beam forbending and torsion

1382 05 (1819 ndash 1469) 1557 mm2

From Table 228 2H32 provides 1608 mm2

Reinforcement required in small rectangles

With link dimensions taken as x1 200 ndash 2 35 130 mmand y1 300 ndash 35 265 mm since y1 550 mm vt shouldnot exceed vtuy1550 50 265550 24 Nmm2

vt 2T2[hmin2(hmax ndash hmin3)]

2 104 106[2002 (300 ndash 2003)]

223 Nmm2 (24)

Area of link reinforcement required for torsion

Asvsv T2[08x1y1(087fy)]

104 106(08 130 265 087 500)

087 mm2mm (total for 2 outer legs)

sv least of x1 130 mm y12 132 mm or 200 mmThe lower rectangle should also be designed for the bendingand shear resulting from the load applied by the floor units

From Table 335 H8-100 provides 100 mm2mm

Area of longitudinal reinforcement for torsion

As (Asvsv)(fyvfy)(x1 y1)

087 10 (130 265) 344 mm2

From Table 228 4H12 provides 452 mm2

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Chapter 26

Deflection and cracking

261 DEFLECTION

Deflections of members under service load should not impairthe appearance or function of a structure For bridges there areno specific limits in BS 5400 but allowances are needed toensure that minimum clearances and satisfactory drainage areprovided Accurate predictions of deflections at different stagesof construction are also required

In BS 8110 the final deflection of members below thelevel of the supports after allowance for any pre-camber islimited to span250 A further limit to be taken as the lesserof span500 or 20 mm applies to the increase in deflectionthat occurs after the application of finishes cladding or par-titions in order to minimise any damage to such elementsThese requirements may be met by limiting the spaneffectivedepth ratio of the member to the values given in Table 340In this table the design service stress in the tensionreinforcement is shown as fs (58)(fyb)(As reqAs prov) InBS 8110 at the time of writing the term (58) which isapplicable to m 115 is given incorrectly as (23) whichis applicable to m 105

The spaneffective depth ratio limits take account ofnormal creep and shrinkage but if these are likely to beparticularly high (eg free shrinkage strain 000075 orcreep coefficient 3) the permissible spaneffective depthratio derived from the table should be reduced by up to 15The limiting ratios may be used also for designs where light-weight aggregate concrete is used except that for all beamsand slabs where the characteristic imposed load exceeds4 kNmm2 the values derived from Table 340 should bemultiplied by 085

In special circumstances when the calculation of deflectionis considered necessary the methods described in Tables 341and 342 can be used Careful consideration is needed in thecase of cantilevers where the usual formulae assume that thecantilever is rigidly fixed and remains horizontal at the rootWhere the cantilever forms the end of a continuous beam thedeflection at the end of the cantilever is likely to be eitherincreased or decreased by an amount l13 where l is the lengthof the cantilever measured to the centre of the supportand 13 is the rotation at the support Where a cantilever is con-nected to a substantially rigid structure some root rotationwill still occur and the effective length should be taken as thelength to the face of the support plus half the effective depth

262 CRACKING

2621 Buildings and bridges

Cracking of members under service load should not impair theappearance or durability of the structure In BS 8110 forbuildings the design surface crack width is generally limited to03 mm For members such as beams and slabs in which thenominal cover does not exceed 50 mm the crack widthrequirement may be met by limiting the gaps between tensionbars to specified values In circumstances where calculationis considered necessary crack width formulae are providedDetails of the bar spacing rules (see section 261 for com-ment on fs) and the crack width formulae are given inTable 343

In BS 5400 the design crack width limits apply only forload combination 1 where for highway bridges the live loadis generally taken as HA Crack widths are calculated for asurface taken at a distance from the outermost reinforcementequal to the nominal cover required for durability Thedesign crack width limits vary according to the exposureconditions as follows 025 mm (moderate or severe expo-sure) 015 mm (very severe exposure) and 010 mm(extreme exposure) In many cases these requirements arecritical and details of the crack width formulae are given inTable 343

In BS 8110 for cracking due to the effects of applied loadsthe modulus of elasticity of concrete is taken as Ec2 wherevalues of Ec are given in Table 35 In BS 5400 a value in therange Ec to Ec2 is taken according to the proportion of live topermanent load

Cracking due to restrained early thermal effects is consideredin BS 8110 Part 2 and Highways Agency BD2887 Inthese documents the restrained early thermal contraction isgiven by t 08RT where is a coefficient of expansion forthe mature concrete R is a restraint factor and T is a tempera-ture differential or fall The following values of R are given

Type of pour and restraint R

Base cast onto blinding 01ndash02Slab cast onto formwork 02ndash04Wall cast onto base slab 06ndash08Infill bays 08ndash10

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BS 8110 Deflection ndash 1 340

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BS 8110 Deflection ndash 2 341

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BS 8110 Deflection ndash 3 342

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BS 8110 (and BS 5400) Cracking 343

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Deflection and cracking300

For continuous support conditions and a flanged section withbwb 3001420 021 03 Table 340 gives

Basic spaneffective depth ratio 08 26 208

The estimated service stress in the reinforcement for a designwith no redistribution of the ultimate moment is given by

fs (58)fy (As reqAs prov) (58) 500 15341608 298 Nmm2

Modification factor for tension reinforcement

s 055 (477 fs)120(09 Mbd 2) 20 055 (477 298)[120 (09 099)] 134

Ignoring modification factor for compression reinforcement

Limiting spaneffective depth ratio 208 134 279

Actual spaneffective depth ratio 8000450 178

Allowing for H8 links with 25 mm cover the clear distancebetween the H32 bars in the bottom of the beam is given by

ab 300 2 (25 8 32) 170 mm

From Table 343 the limiting distance is given by

ab 47 000fs 47 000298 158 mm (170 mm)

The clear distance between the H32 bars could be reduced to150 mm by increasing the side cover to the links to 35 mmAlternatively the bars could be changed to 2H25 and 2H20

Example 2 A 280 mm thick flat slab supported by columnsarranged on a 72 m square grid is to be designed to the require-ments of BS 8110 Bending moments are to be determined usingthe simplified method where the total design ultimate load on apanel is 954 kN and the slab is to be checked for deflection

fcu 40 Nmm2 fy 500 Nmm2 cover to bars 25 mm

Allowing for the use of H12 bars in each direction and basedon the bars in the second layer of reinforcement

d 280 (25 12 6) 235 mm ld 7200235 306

In the following calculations s is based on the average valueof Mbd 2 determined for a full panel width From Table 262the design ultimate bending moment for an end span with acontinuous connection at the outer support is given by

M 0075Fl 0075 954 72 515 kNmMbd2 515 106(7200 2352) 130 Nmm2

From Table 340 for a continuous flat slab without drops thebasic spaneffective depth ratio 26 09 234

Modification factor for tension reinforcement if fs (58)fy

s 055 (477 fs)120(09 Mbd2) 20 055 (477 312)[120 (09 13)] 117

Assuming that no compression reinforcement is provided

Limiting spaneffective depth ratio 234 117 274

The limiting value of ld can be raised to 306 by increasing s

to 131 which reduces fs to 276 Nmm2 so that the area oftension reinforcement determined for the ULS should bemultiplied by the factor 312276 113

For an interior span where the bending moment coefficient is0063 compared to 0075 for the end span

Mbd 2 130 00630075 109 Nmm2

s 055 (477 312)[120 (09 109)] 124 (131)

2622 Liquid-retaining structures

In BS 8007 for structures where the retention or exclusion ofliquid is a prime consideration a design surface crack widthlimit of 02 mm generally applies In cases where the surfaceappearance is considered to be aesthetically critical the limit istaken as 01 mm Under liquid pressure cracks that extendthrough the entire thickness of a slab or a wall are expected toresult in some initial seepage but it is assumed that such crackswill heal autogeneously within 21 days for a 02 mm designcrack width and 7 days for a 01 mm design crack widthSeparate calculations using basically different crack width for-mulae are used for the effects of applied loads and the effectsof temperature and moisture change Details of the formulae foreach type of cracking are given in Table 344

In order to control any potential cracking due to the effectsof restrained thermal contraction and shrinkage three designoptions are given in which the reinforcement requirements arerelated to the incidence of any movement joints Details of thedesign options are given in Table 345 where the joint spacingrequirements refer to joints as being either complete or partialContraction joints can be formed by casting against stop endsor by incorporating crack-inducing waterstops The joints arecomplete if all of the reinforcement is discontinued at thejoints and partial if only 50 of the reinforcement is discon-tinued The joints need to incorporate waterstops and surfacesealants to ensure a liquid-tight structure

The reinforcement needed to control cracking in continuousor semi-continuous construction depends on the magnitude ofthe restrained contraction The restraint factor R (ie ratio ofrestrained to free contraction) may be taken as 05 generally butmore specific values for some common situations are also givenin Table 345 For particular sections and arrangements of rein-forcement limiting values for restrained contraction strain aregiven in Table 346

For cracking due to applied loading and concrete classes thatare typically either C2835 or C3240 the effective modularratio e 2EsEc may be taken as 15 In this case for singlyreinforced rectangular sections the elastic properties of thetransformed section in flexure are given in Table 347 For par-ticular sections and arrangements of reinforcement limitingvalues are given for service moments in Tables 348ndash350 anddirect tensile forces in Tables 351 and 352

Example 1 The beam shown in the following figure is to bechecked for deflection and cracking to the requirements ofBS 8110 The designs for bending and shear are contained inexample 1 of Chapters 24 and 25 respectively The tensionreinforcement comprising 3H32 (top at B) and 2H32 (bottom)is based on the following values

fcu 40 Nmm2 fy 500 Nmm2 cover to links 25 mm

For the bottom reinforcement with d 450 mm

b 1420 mm Mbd 2 099 Nmm2

As req 1534 mm2 As prov 1608 mm2

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BS 8007 Cracking 344

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BS 8007 Design options and restraint factors 345

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BS 8007 Design table for cracking due to temperature effects 346

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BS 8007 Elastic properties of cracked rectangularsections in flexure 347

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BS 8007 Design table for cracking due to flexure in slabs ndash 1 348

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BS 8007 Design table for cracking due to flexure in slabs ndash 2 349

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BS 8007 Design table for cracking due to flexure in slabs ndash 3 350

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351BS 8007 Design table for cracking due to directtension in walls ndash 1

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352BS 8007 Design table for cracking due to directtension in walls ndash 2

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The maximum design crack width is given by

wcr

3 735 000121[1 2 (735 45)(700 192)] 024 mm

From Table 32 the limiting design crack width for a bridgedeck soffit (severe exposure) is 025 mm

The service stress range in the reinforcement resulting from thelive load moment Mq is given by

(MqMs)fs (275600) 220 101 Nmm2

From Table 32 the limiting value for spans in the range5ndash200 m and bar sizes 16 mm is 120 Nmm2

Example 4 The wall of a cylindrical tank 75 m deep and15 m diameter is 300 mm thick The wall which is continuouswith the base slab is to be designed for temperature effects andthose due to internal hydrostatic pressure when the tank is fullof liquid

Design crack width 02 mm fcu 40 Nmm2

Cover to horizontal bars 52 mm fy 500 Nmm2

Effects of temperature change Allowing for concrete gradeC3240 with 350 kgm3 Portland cement at a placing temperatureof 20oC and a mean ambient temperature during construction of15oC the temperature rise for concrete placed within 18 mmplywood formwork

T1 25oC (Table 218)

As the wall is to be designed to resist hoop tension there willbe no vertical movement joints and allowance must be made fora fall in temperature due to seasonal variations Allowing forT2 15oC restraint factor R taken as 05 and coefficient ofthermal expansion taken as 12 106 per oC (Table 35)restrained total thermal contraction after the peak temperaturearising from hydration effects is given by

R (T1 T2) 05 12 106 (25 15) 240 106

From Table 346 for 02 mm crack width and a surface zonethickness of 3002 150 mm H16-200 (EF) will suffice

Effects of hydrostatic load Suppose that an elastic analysis ofthe tank assuming a floor 300 mm thick indicates a servicemaximum circumferential tension of 400 kNm This valueoccurs at a depth of 6 m and above this level the hoop ten-sions can be assumed to reduce approximately linearly tonear zero at the top of the wall

In BS 8110 for direct tension and fy 500 Nmm2 minimumreinforcement equal to 045 of the concrete cross section isrequired Hence for a wall 300 mm thick the minimum areaof reinforcement required on each face

As min 00045 1000 150 675 mm2m (H16-300)

From Table 351 for a 02 mm crack width and 40 mm coverthe following values are obtained for a 300 mm thick wall

H16-225 (EF) provides for N 408 kNmH16-300 (EF) provides for N 319 kNm

In order to cater for the effects of both temperature change andhydrostatic load H16-225 (EF) can be used for a height of

3acrm

1 2(acr cmin) (h dc)

Deflection and cracking310

In this case increasing s to 131 reduces fs to 295 Nmm2 sothat the area of tension reinforcement should be multiplied bythe factor 312295 106

Example 3 A simply supported 700 mm thick solid slabbridge deck is to be designed for bending and checked forcracking and fatigue to the requirements of BS 5400

fcu 40 Nmm2 fy 500 Nmm2 d 620 mm

The maximum longitudinal ultimate moment at mid-span forload combination 1 (375 units HB live load) is 950 kNmm

Mbd 2fcu 950 106(1000 6202 40) 0062As fybdfcu 0078 (Table 324)As 0078 1000 620 40500 3869 mm2m

Although H25-125 gives 3927 mm2m this is unlikely to besufficient with regard to cracking and fatigue which will bechecked on the basis of H25-100 (4909 mm2m)

The maximum longitudinal service moment at mid-span forload combination 1 (HA live load) is Ms 600 kNmm with

Mg 325 kNmm Mq 275 kNmm MqMg 085

In the following calculations the modulus of elasticity of theconcrete is taken as a value reflecting the relative proportions oflive load and permanent load Taking Ec for live load and Ec2for permanent load the effective value is given by

Eeff [1 05(1 MqMg)]Ec

From section 2211 Ec 19 03fcu 31 kNmm2

Eeff [1 05(1 085)] 31 226 kNmm2

From Table 342 the neutral axis depth x (or dc) is given by

xd where (EsEeff)(Asbd)

(200226) 4909(1000 620) 007

xd 007 031 x 192 mmStress in tension reinforcement is given by

fs MsAs(d x3) 600 106[4909 (620 1923)] 220 Nmm2

Strain in tension reinforcement

s fsEs 220(200 103) 00011

Strain at surface taken at distance cmin from outermost barsignoring stiffening effect of concrete

1 s (h x 10)(d x) 00011 (700 192 10)(620 192) 000128

Stiffening effect of concrete (with h)

109

38 1000 700 015 109(00011 4909) 000007

Strain at surface allowing for stiffening effect of concrete

m 1 2 000128 000007 000121

Distance from surface of bar where sb is bar spacing to pointmidway between bars on surface taken at distance cmin fromoutermost bars is given by

acr

125 735 mm(50)2 (70)2

(sb 2)2 (h d 10)2 2

2 38bthsAs

1 Mq

Mg

a

0072 2 007

2 2

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Cracking 311

30 m say and minimum reinforcement of H16-300 (EF) for theremaining height of 45 m

Example 5 A 200 mm thick roof slab to a reservoir is to bedesigned for serviceability cracking to the requirements ofBS 8007

Design crack width 02 mm Cover to bars 40 mm

(a) Sliding layer is provided between slab and perimeter wallMaximum service moment M 25 kNmm

From Table 348 H12-150 caters for M 252 kNmm withfs 243 Nmm2

(b) Slab is tied to perimeter wall Maximum service momentand direct tension M 25 kNmm and N 40 kNm

h 200 mm d 200 (40 122) 154 mm

Since MN (2540) 103 625 mm (d 05h) 54 mmone face will remain in compression and the section can bedesigned for a reduced moment M1 M N(d 05h) withthe tensile force acting at the level of the reinforcement

M1 25 40 0054 228 kNmm

From solution (a) above take fs 240 Nmm2 as a trialvalue

100M1fsbd2 100 228 106(240 1000 1542) 0400

From Table 347 by interpolation 100As1bd 0445

As1 000445 1000 154 685 mm2m

Total area of reinforcement required is given by

As As1 Nfs 685 40 103240 852 mm2m

Using H12-125 gives 905 mm2m This may not necessarily besufficient because the stiffening effect of the concrete inherentin solution (a) is reduced by the additional tension A crackwidth calculation is needed to confirm the solution

The stress in the reinforcement is given approximately by

fs 240 852905 226 Nmm2

100M1fsbd2 0400 240226 0425

From Table 347 by interpolation 100As1bd 0474

As1 000474 1000 154 730 mm2m

As As1 Nfs 730 40 103226 907 mm2m

This is near enough to the area given by H-125 no furtheriteration is needed and 100M1fsbd2 0425 may be assumed

From Table 347 by interpolation xd 0312 x 48 mm

Strain in the reinforcement

s fsEs 226(200 103) 000113

Strain at surface ignoring stiffening effect of concrete

1 s(h x)(d x) 000113 (200 48)(154 48) 000162

Stiffening effect of concrete at surface (with h)

2 bt(h x)23EsAs(d x)1000 (20048)2[3 200 103 905 (154 48)] 000040

Strain at surface allowing for stiffening effect of concrete

m 1 2 000162 000040 000122

Distance from surface of bar where sb is bar spacing topoint on surface midway between bars is given by

acr

6 716 mm

The maximum design crack width is given by

wcr

3 716 000122[1 2 (716 40)(200 48)] 019 mm

3acrm

1 2(acr cmin) (h x)

(625)2 (46)2

(sb 2)2 (h d)2 2

a

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Chapter 27

Considerations affecting design details

Codes of Practice contain numerous requirements that affectthe reinforcing details such as minimum and maximum areastying provisions anchorage and curtailment

Bars may be arranged individually in pairs or in bundles ofthree or four in contact In BS 8110 for the safe transmissionof bond forces the cover provided to the bars should be not lessthan the bar size or for a group of bars in contact the equivalentdiameter of a notional bar with the same total cross-sectionalarea as the group In BS 5400 the forgoing cover requirementis increased by 5 mm Requirements for cover with regard todurability and fire-resistance are given in Chapter 23 Gapsbetween bars (or groups of bars) generally should be not lessthan the greater of (hagg 5 mm) where hagg is the maximumsize of the coarse aggregate or the bar size (or the equivalentbar size for bars in groups) Details of reinforcement limits aregiven in Table 353 for BS 8110 and Table 359 for BS 5400

271 TIES IN STRUCTURES

For robustness the necessary interaction between elements isobtained by tying the structure together Where the structure isdivided into structurally independent sections each sectionshould have an appropriate tying system In the design of tiesthe reinforcement may be assumed to act at its characteristicstrength and only the specified tying forces need to be takeninto account Reinforcement provided for other purposes maybe considered to form part of or the whole of the ties Detailsof the tying requirements in BS 8110 are given in Table 354

272 ANCHORAGE AND LAP LENGTHS

At both sides of any cross section bars should be provided withan appropriate embedment length or other form of end anchorageIn BS 5400 it is also necessary to consider lsquolocal bondrsquo wherelarge changes of tensile force occur over short lengths ofreinforcement Critical sections for local bond are at simplysupported ends at points where tension bars stop and at pointsof contra-flexure However the last two points need not beconsidered if the anchorage bond stresses in the continuing barsdo not exceed 08 times the ultimate values

The radius of any bend in a reinforcing bar should conformto the minimum requirements of BS 8666 and should ensurethat the bearing stress at the mid-point of the curve does notexceed the maximum value given in BS 8110 or BS 5400 as

appropriate A link may be considered fully anchored if itpasses round another bar not less than its own size through anangle of 90o and continues beyond the end of the bend for aminimum length of eight diameters Details of anchoragelengths local bond stresses and bends in bars are given inTables 355 and 359 for BS 8110 and BS 5400 respectively

In BS 8007 for horizontal bars in direct tension the designultimate anchorage bond stress is taken as 70 of the valuegiven in BS 8110 Also for sections where bars are needed tocontrol cracking due to temperature and moisture effects therequired anchorage bond length lab 6 where crit isthe reinforcement ratio in the surface zone (Table 344) Thisvalue can exceed the anchorage bond length in BS 8110

Laps should be located if possible away from positions ofmaximum moment and should preferably be staggered Laps infabric should be layered or nested to keep the lapped wires orbars in one plane BS 8110 requires that at laps the sum of allthe reinforcement sizes in a particular layer should not exceed40 of the breadth of the section at that level When the sizeof both bars at a lap exceeds 20 mm and the cover is lessthan 15 times the size of the smaller bar links of size notless than one-quarter the size of the smaller bar and spacingnot greater then 200 mm should be provided throughout the laplength Details of lap lengths are given in Tables 355 and 359for BS 8110 and BS 5400 respectively

273 CURTAILMENT OF REINFORCEMENT

In flexural members it is generally advisable to stagger thecurtailment points of the tension reinforcement as allowedby the bending moment envelope Curtailed bars shouldextend beyond the points where in theory they are no longerneeded in accordance with certain conditions Details of thegeneral curtailment requirements in BS 8110 are given inTable 356 Simplified rules for beams and slabs are alsoshown in Tables 357 and 358 In BS 5400 the generalcurtailment procedure is the same as that in BS 8110 except forthe requirement at a simply supported end where condition (3)does not apply

Example 1 The design of the beam shown in the following fig-ure is given in example 1 of Chapters 24 (bending) and 25 (shear)The design ultimate loads on each span are Fmax 416 kN andFmin 160 kN The main reinforcement is as follows in the

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BS 8110 Reinforcement limits 353

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BS 8110 Provision of ties 354Internal ties

Peripheral tiesVertical ties

Column to wallties at each column(or wall)floorintersection

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BS 8110 Anchorage requirements 355

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spans 2H32 (bottom) at support B 3H32 (top) and 2H25(bottom) The width of each support is 400 mm and bars are tobe curtailed according to the requirements of BS 8110 In thefollowing calculations the use of the general curtailmentprocedure (Table 356) and the simplified curtailment rules(Table 357) are both examined

fcu 40 Nmm2 fy 500 Nmm2 d 450 mm

2 A point where the shear force is no more than half thedesign shear resistance at the section From the shear designcalculations in Chapter 25 with H8-300 links Vu 170 kNFor load case 1 distance x from B to the point where shearforce is 1702 85 kN is given by

x [1 (85 156)416]L 042 8 336 m

3 A point where the bending moment is no more than halfthe design resistance moment at the section As in thecalculations above but with M 2702 135 kNm05x2 3x 26 0 giving x 68 m and a distancefrom B of (80 68) 12 m

From the foregoing it can be seen that if the bar were curtailedat a distance of 12 m from B this would satisfy condition (3)and be more than 450 mm beyond the theoretical curtailmentpoint at 06 m from B However the bar should extend 13 mfrom B in order to provide a full tension anchorage of 11 mbeyond the face of the support It can be seen from the forego-ing calculations that checking conditions (2) and (3) is a tediousprocess and complying with condition (1) is a more practicalapproach even though it would mean curtailing the bar at(06 11) 17 m from B in this example

Suppose that the remaining 2H32 are continued to the point ofcontra-flexure in span BC for load case 2

The reaction at support C is given by

RC 05Fmin MBL 05 160 2888 44 kN

Distance from B to point of contra-flexure is given by

x L(1 2RCFmin) 8 (1 2 44160) 36 m

The bars need to extend beyond this point for a distance not lessthan d 12 450 mm Link support bars say 2H12 with alap of 300 mm could be used for the rest of the span

If the simplified curtailment rules are applied one bar out ofthree may be curtailed at 015L 45 145 m from the faceof the support that is at 165 m from B The other two bars maybe curtailed at 025L 20 m from the face of the support thatis at 22 m from B Beyond this point bars giving an area notless than 20 of the area required at B should be provided thatis 02 2413 483 mm2 (2H20 gives 628 mm2) Since thebars are in the top of a section as cast where the cover is lessthan 2 and the gap between adjacent laps is not less than6 l 14 (see Table 355) and the required lap length is

lbl 49 (AsreqAsprov) 49 20 483628 750 mm

Example 2 A typical floor to an 8-storey building consistsof a 280 mm thick flat slab supported by columns arrangedon a 72 m square grid The slab for which the characteristicloading is 80 kNm2 dead and 45 kNm2 imposed is to beprovided with ties to the requirements of BS 8110 Thedesign ultimate load on a panel is 954 kN and bendingmoments are to be determined by the simplified method (seesection 138)

fcu 40 Nmm2 fy 500 Nmm2 cover to bars 25 mm

Allowing for the use of H12 bars in each direction and basedon the bars in the second layer of reinforcement

d 280 (25 12 6) 235 mm say

Considerations affecting design details316

End anchorage At the bottom of each span the 2H32 will becontinued to the supports At the end support (simple) aneffective anchorage of 12 is required beyond the centre ofbearing This can be obtained by providing a 90o bend with aninternal radius of 35 provided the bend does not start beforethe centreline of the support Allowing for 50 mm end coverto the bars the distance from the outside face of the support tothe start of the bend is 50 45 194 mm This would besatisfactory since the width of the support is 400 mm If thesupport width were any less the 2H32 could be stopped atthe face of the support and lapped with 2H25 in which case itwould be necessary to reassess the shear design

Curtailment points for top bars The resistance momentprovided by 2H32 can be determined as follows

As fybdfcu 1608 500(300 450 40) 0149Mbd 2fcu 0111 (Table 314)M 0111 300 4502 40 106 270 kNm

For load case 1 reaction at A (or C) is given by

RA 05Fmax MBL 05 416 4168 156 kN

Distance x from A to point where M 270 kNm is given by

05(FmaxL) x2 RAx 05 (4168)x2 156x 270

Hence 05x2 3x 52 0 giving x 74 m Thus of the3H32 required at B one bar is no longer needed for flexure ata distance of (80 74) 06 m from B

The bar to be curtailed needs to extend beyond this point fora distance not less than d 12 450 mm to a positionwhere one of the following conditions is satisfied

1 A full tension anchorage length beyond the theoretical pointof curtailment that is 35 35 32 11 m

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From Table 255 the design ultimate sagging moment for aninterior panel is given by

M 0063Fl 0063 954 72 433 kNm

The total panel moment is to be apportioned between columnand middle strips where the width of each strip is 36 m Forthe column strip with 55 of the panel moment

M 055 433 238 kNmMbd2fcu 238 106(3600 2352 40) 0030

From Table 314 since Mbd 2fcu 0043 zd 095 Hence

As 238 106(087 500 095 235) 2504 mm2 (24H12-150 gives 2714 mm2)

For the middle strip with 45 of the panel moment

M 045 433 195 kNmMbd2fcu 195 106(3600 2352 40) 0025 ( 0043)

As 195 106(087 500 095 230) 2008 mm2 (18H12-200 gives 2036 mm2)

For the peripheral tie the tensile force is given by

Ft (20 4no) 60 kN (20 4 8) 52 kN

The required area of reinforcement acting at its characteristicstrength is given by

As Ftfy 52 103500 104 mm2 (1H12)

For the internal ties the tensile force is given by

Ft gt Ft kNm

52 = 125 kNm

If the internal ties are spread evenly in the slab the requiredarea of reinforcement acting at its characteristic strength

As 125 103500 250 mm2m (H12-400)

In this case alternate bars in both column and middle stripsneed to be made effectively continuous

If the internal ties are concentrated at the column lines the totalarea of reinforcement required in each group

As 250 72 1800 mm2 (16H12 gives 1810 mm2)

In this case the bars in the middle two-thirds of each columnstrip need to be made effectively continuous Since the bars arelocated at the bottom of the slab and the gap between each set

80 4575 72

5

Ftint gt qk

75 lr

5

Curtailment of reinforcement 317

of lapped bars exceeds 6 l 10 and a lap length of 35 issufficient (Table 355 for fcu 40 Nmm2)

Example 3 The following figure shows details of the rein-forcement at the junction between a 300 mm wide beam and a300 mm square column Bars 03 need to develop the maximumdesign stress at the column face and the required radius of bendis to be checked in accordance with the requirements of BS 8110

fcu 40 Nmm2 fy 500 Nmm2

The minimum radius of bend of the bars depends on the value ofab where ab is taken as either centre-to-centre distance betweenbars or (side cover plus bar size) whichever is less Henceab (300 75 2 25) 125 (75 25) 100 mm

From Table 355 for ab 10025 4 rmin 64 Thisvalue can be reduced slightly by considering the stress in the barat the start of the bend If r 6 distance from face of columnto start of bend 300 50 7 25 75 mm (ie 3) FromTable 355 the required anchorage length is 35 and rmin (1 335 ) 64 59 Thus r 6 is sufficient

Example 4 A 700 mm thick solid slab bridge deck is simplysupported at the end abutments The design of the slab to therequirements of BS 5400 is given in example 2 of Chapter 25(shear) and example 3 of Chapter 26 (bending cracking andfatigue) The tension reinforcement at the end of the span isH25-200 and the maximum design shear force is 625 kN act-ing on a 15 m wide strip of slab A local bond check is required

fcu 40 Nmm2 fy 500 Nmm2 d 620 mm

Bar perimeter provided by H25-200 in a 15 m strip of slab

13us (1500200) 25 589 mm

The local bond stress is given by the relationship

fbs V13usd 625 103(589 620) 171 Nmm2

From Table 359 ultimate local bond stress flbu 40 Nmm2

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BS 8110 Curtailment requirements 356

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BS 8110 Simplified curtailment rules for beams 357

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BS 8110 Simplified curtailment rules for slabs 358

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BS 5400 Considerations affecting design details 359

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Chapter 28

Miscellaneousmembers and details

281 LOAD-BEARING WALLS

In BS 8110 for the purpose of design a wall is defined as avertical load-bearing member whose length on plan exceedsfour times its thickness Otherwise the member is treated as acolumn A reinforced wall is one in which not less than therecommended minimum amount of vertical reinforcement isprovided and taken into account in the design Otherwise themember is treated as a plain concrete wall in which case thereinforcement is ignored for the purpose of design Limitingreinforcement requirements are given in Table 353 Bearingstresses under concentrated loads should not exceed 06fcu forconcrete strength classes C2025 Design requirements forreinforced and plain concrete walls are given in Table 360

282 PAD BASES

Notes on the distribution of pressure under pad foundations aregiven in section 181 and values for the structural design ofseparate bases are given in Table 282 Critical sections forbending are taken at the face a concrete column or the centreof a steel stanchion The design moment is taken as that due toall external loads and reactions to one side of the section

Generally tension reinforcement may be spread uniformlyacross the width of the base but the following requirementshould be satisfied where c is the column width and lc is thedistance from the centre of a column to the edge of the pad Iflc 075(c 3d) two-thirds of the reinforcement should beconcentrated within a zone that extends on either side for adistance no more than 15d from the face of the column Forbases with more than one column in the direction consideredlc should be taken as either half the column spacing or thedistance to the edge of the pad whichever is the greater

The pad should be examined for normal shear and punchingshear Normal shear is checked on vertical sections extendingacross the full width of the base Within any distance av 2dfrom the face of the column the shear strength may be taken as(2dav)vc For a concentric load the critical position occurs atav a2 2d where a is the distance from the column face tothe edge of the base For an eccentric load checks can be madeat av 05d d and so on to find the critical position

Punching shear is checked on a perimeter at a distance 15dfrom the face of the column The shear force at this position is that

due to the effective ground pressure acting on the area outsidethe perimeter For a concentric load with a 15d the checkfor punching shear is the critical shear condition If the mainreinforcement is taken into account in the determination of vcthe bars should extend a distance d beyond the shear perimeterIn this case the bars need to extend 25d beyond the face of thecolumn and will need to be bobbed at the end unless a 25din which case straight bars would suffice

The maximum clear spacing between the bars ab shouldsatisfy the following requirements for fy 500 Nmm2

100Asbd 03 03 04 05 06 075 10

ab (mm) 750 500 375 300 250 200 150

Example 1 A base is required to support a 600 mm squarecolumn subjected to vertical load only for which the valuesare 4600 kN (service) and 6800 kN (ultimate) The allowableground bearing value is 300 kNm2

fcu 35 Nmm2 fy 500 Nmm2 nominal cover 50 mm

Allowing 10 kNm2 for ground floor loading and extra over soildisplaced by concrete the net allowable bearing pressure canbe taken as 290 kNm2 Area of base required

Abase 4600290 1586 m2 Provide base 40 m square

Distance from face of column to edge of base a 1700 mm

Take depth of base 05a say h 850 mm

Allowing for 25 mm main bars average effective depth

d 850 ndash (50 25) 775 mm

Bearing pressure under base due to ultimate load on column

pu 680042 425 kNm2

Bending moment on base at face of column

M pu la22 425 4 1722 2456 kNm

K Mbd2 fcu 2456 106(4000 7752 35) 00292

From Table 314 since K 0043 As fybd fcu 121K and

As 121 00292 4000 775 35500 7670 mm2

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BS 8110 Load-bearing walls 360

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From Table 220

16H25-250 gives 7854 mm2 and

100Asbd 100 7854(4000 775) 025

Critical perimeter for punching shear occurs at 15d from faceof column where the length of side of the perimeter

l1 c 3d 600 3 775 2925 mm

Hence

V f (l 2 ndash l12) 425 (42 ndash 29252) 3164 kN

v V4l1d 3164 103(4 2925 775) 035 Nmm2

vc 0216

0216 (400775)14(025 35)13 038 Nmm2( v)

Critical position for shear on vertical section across full widthof base occurs at distance av a2 850 mm 2d from faceof column where

V 425 4 172 1445 kN

v Vbd 1445 103(4000 775) 047 Nmm2

vc(2dav) 038 (2 775850) 069 Nmm2 ( v)

283 PILE-CAPS

In BS 8110 (and BS 5400) a pile-cap may be designed by eitherbending theory or truss analogy In the latter case the truss isof a triangulated form with nodes at the centre of the loadedarea and at the intersections of the centrelines of the piles withthe tension reinforcement as shown for compact groups of twoto five piles in Table 361 Expressions for the tensile forcesare given taking into account the dimensions of the columnand also simplified expressions when the column dimensionsare ignored Bars to resist the tensile forces are to be locatedwithin zones extending not more than 15 times the pilediameter either side of the centre of the pile The bars are tobe provided with a tension anchorage beyond the centres of thepiles and the bearing stress on the concrete inside the bend inthe bars should be checked (see Table 355)

The design shear stress calculated at the perimeter of thecolumn should not exceed the lesser of 08radicfcu or 5 Nmm2Critical perimeters for checking shear resistance are shown inTable 361 where the whole of the shear force from piles withcentres lying outside the perimeter should be taken intoaccount The shear resistance is normally governed by shearalong a vertical section extending across the full width of thecap where the design concrete shear stress may be enhanced asshown in Table 361 If the pile spacing exceeds 3 times the pilediameter punching shear should also be considered

In BS 5400 shear enhancement applies to zones of widthequal to the pile diameter only and the concrete shear stress istaken as the average for the whole section Also the check forpunching shear is made at a distance of 15d from the face ofthe column with no enhancement of the shear strength

The following values are recommended for the thickness ofpile-caps where hp is the pile diameter

For hp 550 mm h (2hp 100) mm

For hp 550 mm h 8(hp ndash 100)3 mm

400d 14100As fcu

bd 13

Example 2 A pile-cap is required for a group of 4 450 mmdiameter piles arranged at 1350 mm centres on a square gridThe pile-cap supports a 450 mm square column subjected to anultimate design load of 4000 kN

fcu 35 Nmm2 fy 500 Nmm2

Allowing for an overhang of 150 mm beyond the face of the pilesize of pile-cap 1350 450 300 2100 mm square

Take depth of pile-cap as (2hp 100) 1000 mm

Assuming tension reinforcement to be 100 mm up from base ofpile-cap d 1000 ndash 100 900 mm

Using truss analogy with the apex of the truss at the centre ofthe column the tensile force between adjacent piles is

Ft 750 kN in each zone

As Ft087fy 750 103(087 500) 1724 mm2

Providing 4H25 gives 1963 mm2 and since the pile spacing isnot more than 3 times the pile diameter bars may be spreaduniformly across the pile-cap giving a total of 8H25-275 in eachdirection so that

100As bh 100 3926(2100 1000) 019 ( 013 min)

Critical position for shear on vertical section across full widthof pile-cap occurs at distance from face of column given by

av 05(l ndash c) ndash 03hp

05 (1350 ndash 450) ndash 03 450 315 mm

Portion of column load carried by two piles is 2000 kN thus

v Vbd 2000 103(2100 900) 106 Nmm2

vc 0216

0216 (400900)14 (019 35)13 033 Nmm2

vc (2dav) 033 (2 900315) 188 Nmm2 ( v)

Shear stress calculated at perimeter of column is

v Vud 4000 103(4 450 900) 247 Nmm2

Maximum shear strength 08radicfcu 473 Nmm2 ( v)

Taking ab as either centre-to-centre distance between bars or(side cover plus bar size) whichever is less

ab 275 (75 25) 100 mm ab 10025 4

From Table 355 minimum radius of bend rmin 7 say

284 RETAINING WALLS ON SPREAD BASES

General notes on the design of walls to BS 8002 are given insection 732 Design values of earth pressure coefficients arebased on a design soil strength which is taken as the lower ofeither the peak soil strength reduced by a mobilisation factor orthe critical state strength Design values of the soil strengthusing effective stress parameters are given by

design tan (tan max)12 tan crit

design c c12 tan crit

400d 14100As fcu

bd 13

4000 13508 900 Nl

8d

Miscellaneous members and details324

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BS 8110 Pile-caps 361

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Miscellaneous members and details326

where c max and crit are representative (ie conservative)values of effective cohesion peak effective angle of shearingresistance and critical state angle of shearing resistance for thesoil In the absence of reliable site investigation and soil testdata values may be derived from Table 210

Design values of friction and adhesion at the soil-structureinterface (wall or base) are given by

design tan (or b) 075 design tan

design cw (or cb) 075 cud 05 cu in which

cud cu 15 where cu is undrained shear strength

A minimum surcharge of 10 kNm2 applied to the surface of theretained soil and a minimum depth of unplanned earth removalin front of the wall equal to 10 of the wall height but not lessthan 05 m should be considered Wall friction should beignored in the determination of KA

Suitable dimensions for the base to a cantilever wall can beestimated with the aid of the chart given in Table 286 Forsliding the chart is valid for non-cohesive soils only Thusfor bases founded on clay soils the long-term condition can beinvestigated by using crit with c 0 For the short-termcondition the ratio does not enter into the calculations forsliding and taking the contact surface length as the full widthof the base is given by KAl2cb When has beendetermined from this equation the curve for radicKA on the chartcan be used to check the values of and that were obtainedfor the long-term condition

Example 3 A cantilever retaining wall on a spread base isrequired to support level ground and a footpath adjacent to aroad The existing ground may be excavated as necessary toconstruct the wall and the excavated ground behind the wall isto be reinstated by backfilling with a granular material Agraded drainage material will be provided behind the wall withan adequate drainage system at the bottom

Height of fill to be retained 40 m above top of baseSurcharge 100 mm surfacing plus 5 kNm2 live load

(design for minimum value of 10 kNm2)

Properties of retained soil (well graded sand and gravel)unit weight of soil 20 kNm3

max crit 35o design tanminus1 [(tan 35o)12] 30o

KA (1 ndash sin)(1 sin) 033

Properties of sub-base soil (medium sand)allowable bearing value pmax 200 kNm2

max 35o crit 32o design 30o (as fill)

design tan b 075 design tan 043

Take thickness of both wall (at bottom of stem) and base to beequal to (height of fill)10 400010 400 mm

Height of wall to underside of base l 40 04 44 m

Allowing for surcharge equivalent height of wall

le l q 44 1020 49 m

pmaxle 200(20 49) asymp 20

tan bradicKA 043radic033 075

From Table 286 radicKA 08 018 Hence

Width of base le (08radic033) 49 225 m

Toe projection (le) 018 225 04 m

Example 4 The sub-base for the wall described in example 3is a clay soil with properties as given here All other values areas specified in example 3

Properties of sub-base soil (firm clay)allowable bearing value pmax 100 kNm2 cu 50 kNm2

cud 5015 333 kNm2 design cb 502 25 kNm2

crit 25o (assumed plasticity index 30)

design tan b 075 design tan crit 033

For the long-term condition

pmaxle 100(20 49) asymp10

tan bradicKA 033radic033 057

From Table 286 radicKA 115 024 Hence

Width of base le (115radic033) 49 33 m say

Toe projection (le) 024 33 08 m

For the short-term condition

KA l2cb 033 20 49(2 25) 065

radicKA 065radic033 113 (115)

Since this value is less than that calculated for the long-termcondition the base dimensions are satisfactory

Example 5 The wall obtained in example 4 a cross sectionthrough which is shown here is to be designed to BS 8002

The vertical loads and bending moments about the front edgeof the base are

Load (kN) Moment (kNm)Surcharge 10 21 210 225 473Backfill 20 21 40 1680 225 3780Wall stem 24 04 40 384 10 384Wall base 24 04 33 317 165 523

Totals Fv 2591 5160

The horizontal loads and bending moments about the bottom ofthe base are

Load (kN) Moment (kNm)Surcharge 033 10 44 145 442 319Backfill 033 20 4422 639 443 937

Totals Fh 784 1256

Resultant moment Mnet 516 ndash 1256 3904 kNm

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Since the f values used for the horizontal and vertical loads arenot the same the bearing pressures must be recalculated

Fv 14 2591 3627 kN Mv 14 516 7224 kNm

Mnet 7224 ndash 12 1256 5717 kNm

a 57173627 1575 m e 165 ndash 1575 0075 m

pmax (362733)(1 6 007533) 125 kNm2

pmin (362733)(1 ndash 6 007533) 95 kNm2

Bearing pressure under base at inside face of wall

pwall 95 (125 ndash 95)(2133) 95 19 114 kNm2

Bending moment on base at inside face of wall

M 14 (10 20 4 24 04) 2122(95 2122 19 2126) 84 kNm

As 84 106[(087 500)(095 352)] 578 mm2m

Use H16-250 to fit in with vertical bars in wall

Shear force on base at inside face of wall

V 14 (10 20 4 24 04) 21(95 21 19 212) 734 kN

Vbd 734 103(1000 352) 021 Nmm2 ( vc)

For the base the bending moment and shear force have beencalculated for a bearing pressure diagram that varies linearly asindicated in BS 8110 If the pressure diagram assumed for theultimate bearing condition in example 5 is taken

pu Fv2a 3627(2 1575) 1151 kNm2

M 14 (10 20 4 24 04) 21221151 (315 ndash 12)22 886 kNm

V 14 (10 20 4 24 04) 211151 (315 ndash 12) 684 kN

285 RECOMMENDED DETAILS

The information given in Tables 362 and 363 has been takenfrom several sources including BS 8110 research undertakenby the Cement and Concrete Association (CampCA) and reportspublished by the Concrete Society

2851 Continuous nibs

The BS 8110 recommendations are shown in Table 362 As aresult of investigations by the CampCA various methods ofreinforcing continuous nibs were put forward Method (a) isefficient but it is difficult to incorporate the bars in shallow nibsif the bends in the bars are to meet the minimum code require-ments Method (b) is reasonably efficient and it is a simplematter to anchor the bars at the outer face of the nib The CampCArecommendations were based on an assumption of truss actionBS 8110 suggests that such nibs should be designed as shortcantilever slabs but both methods lead to similar amounts ofreinforcement

2852 Corbels

The information in Table 362 is based on the requirements ofBS 8110 and BS 5400 supplemented by recommendations

Recommended details 327

Distance from front edge of base to resultant vertical force

a MnetFv 39042591 150 m

Eccentricity of vertical force relative to centreline of base

e 332 ndash 15 015 m ( 336 055 m)

Maximum pressure at front of base

pmax (259133)(1 6 01533) 100 kNm2

Minimum pressure at back of base

pmin (259133)(1 ndash 6 01533) 57 kNm2

For the ultimate bearing condition a uniform distribution isconsidered of length lb 2a 2 15 30 m Pressure

pu Fvlb 259130 864 kNm2

The ultimate bearing resistance is given by the equation

qu (2 ) cud ic where ic 05[1 ]

ic 05[1 ] 073

qu (2 ) 333 073 125 kNm2 ( pu 864)

Resistance to sliding (long-term)

Fv tan b 2591 033 855 kN ( Fh 784)

Resistance to sliding (short-term)

(le) cb 33 25 825 kN ( Fh 784)

For resistance to sliding (short-term) the contact surface hasbeen taken as the full width of the base This is consideredreasonable since base adhesion is taken as only 075cud If thecontact surface is based on the pressure diagram assumed forthe ultimate bearing condition the resistance to sliding isreduced to lb cb 30 25 75 kN

Example 6 The structural design of the wall in example 5 isto be in accordance with the requirements of BS 8110

fcu 35 Nmm2 fy 500 Nmm2 nominal cover 40 mm

Allowing for H16 bars with 40 mm cover

d 400 ndash (40 8) 352 mm

For the ULS values of f are taken as 12 for the horizontalloads and 14 for all the vertical loads

The ultimate bending moment at the bottom of the wall stem

M 12 033 (10 422 20 436) 1162 kNmm

Mbd2fcu 1162 106(1000 3522 35) 00268

From Table 314 since K 0043 z 095d

As 1162 106[(087 500)(095 352)] 799 mm2m

From Table 220 H16-250 gives 804 mm2m

The ultimate shear force at the bottom of the wall stem

V 12 033 (10 4 20 422) 792 kNm

Vbd 792 103(1000 352) 023 Nmm2 ( vc)

From Table 343 the clear distance between bars should notexceed (47 000fs)(100Asbd) 750 mm Thus with

fs 087fyf 087 50012 362 Nmm2

ab (47 000362)[100 804(1000 352)] 568 mm

1 784(333 30)

1 Fh(cud lb)

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Recommended details nibs corbels and halving joints 362

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Recommended details intersections of members 363

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taken from CampCA research reports For the system of forcesshown in the figure the inclined force in the concrete

Fc Fvsin13 k1 fcub(x cos13) where tan13 (d ndash k2 x)av

In these expressions k1 and k2 are properties of the concretestress block as given in section 241 From these expressions aquadratic equation in xd (given in the table) can be derived InBS 8110 for values of xd (1 ndash 2avd)k2 a minimum valueof Ft 05Fv is taken for the tensile force

In the detail shown in the table the main bars are bent backto form a loop If the bars are welded to a crossbar they can becurtailed at the outer edge of the corbel In this case two addi-tional small diameter bars are needed to support the horizontalstirrups If the stirrups are required to pass outside the main barsin the column they should be detailed as two lapping U-barsfor ease of assembly

2853 Halving joints

The recommendations given in BS 5400 as a result of workcarried out by the CampCA are summarised in Table 362 Theinclined links must intersect the line of action of Fv If thiscannot be ensured (eg the inclined links could be displaced)or if horizontal forces can occur at the joint horizontal linksmust also be provided as shown

2854 Reinforcement details at frame corners

Research has shown that when frame corners are subjected tobending moments tending to close the corner the most likelycause of premature failure is due to bearing under the bend ofthe tension bars at the outside of the corner Provided that theradius of the bend is gradual and that sufficient anchorage isgiven for the lapping bars the use of simple details as shown in(a) or (b) on Table 363 is recommended

With lsquoopening cornersrsquo the problems are somewhatgreater and tests have shown that some details can fail at wellbelow their calculated strength In this case the detail shownin (d) is recommended If at all possible a concrete splayshould be formed within the corner and the diagonal rein-forcement Ass provided with appropriate cover If a splay isimpracticable the diagonal bars should be included withinthe corner itself

Detail (d) is suitable for reinforcement amounts up to about 1If more than this is required transverse links should be includedas shown in (e) The arrangement shown in (c) could be usedbut special attention needs to be paid to bending and fixingthe diagonal links which must be designed to resist all the forcein the main tension bars Care must also be taken to provideadequate cover to the bars at the inside of the corner

2855 Beam-column intersections

Research has shown that the forces in a joint between a beamand an end column are as shown in the sketch on Table 363Diagonal tensile forces occur at right angles to the theoreticalstrut that is shown To ensure that as a result diagonal cracksdo not form across the corner a design limit that is related tothe service condition is shown in the table To ensure that the

joint has sufficient ultimate strength an expression has alsobeen developed for a minimum amount of reinforcement that isneeded to extend from the top of the beam into the column atthe junction However for floor beams it has been shown thatthe U-bar detail shown in the table is satisfactory

While research indicates that unless it is carefully detailedas described the actual strength of the joint between a beamand an end column could be as little as half of the calculatedmoment capacity it seems that internal beam-column jointshave considerable reserves of strength Joints having a beam onone side of a column and a short cantilever on the other aremore prone to loss of strength and it is desirable in suchcircumstances to detail the joint as for an end column withthe beam reinforcement turned down into the column and thecantilever reinforcement extending into the beam

Example 7 A corbel is required to support a design ultimatevertical load of 500 kN at a distance of 200 mm from the faceof a column The load is applied through a bearing pad andthe 300 mm wide corbel is to be designed to BS 8110

fcu 30 Nmm2 fy 500 Nmm2 cover 40 mm

Minimum reinforcement based on Ft 05Fv and fyd 087fy

As 05 500 103(087 500) 575 mm2

Calculate shear resistance based on 2H20 (As 628 mm2) forvalues of d 400 mm where av 200 mm b 300 mm vc isobtained from Table 333 and Vc vc (2dav)bd

Miscellaneous members and details330

d vc (2dav)vc Vc ( 10ndash3)mm 100Asbd Nmm2 Nmm2 kN

500 042 050 250 375550 038 049 270 445600 035 047 282 508

Assuming that As will need to be increased as a result of thecorbel analysis with a corresponding increase in Vc considerh 600 mm with d 550 mm Hence

kv Fvbdfcu 500 103(300 550 30) 010

avd 200550 036 k1 040 k2 045 (section 241)

From Table 362 in the quadratic equation for xd

0452 040 045 03601 085

2 045 040 03601 234

1 0362 113 giving the equation

085(xd)2 ndash 234(xd ) 113 0 from which xd 0625

1 av

d 2

2k2 k1

kvav

d

k22

k1k2

kvav

d

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Hence x 0625 550 344 mm and the strain in the bars

s 00035(d ndash x)x 00035 206344 00021

fyd s Es 00021 200 103 420 Nmm2 ( 087fy)

Since (1 ndash 2avd)k2 (1 ndash 2 200550)045 0622 xd

Ft Fvav(d ndash k2x) Fv2 and As Fvav(d ndash k2x)fyd

As 500 103 200[(550 ndash 045 344) 420] 603 mm2

In this case 2H20 giving 628 mm2 is sufficient for the tensileforce but insufficient for the shear resistance Changing the

reinforcement to 3H20 gives 942 mm2 and increases the shearresistance as follows

100Asbd 100 942(300 550) 057

vc 056 Nmm2 Vc vc (2dav)bd 508 kN

Minimum area of horizontal links 05 942 471 mm2 tobe provided by 3H10 links (2 legs per link) which should belocated in the tension zone (ie extending over a depth of about200 mm below the main bars)

Recommended details 331

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Part 4

Design to EuropeanCodes

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Chapter 29

Design requirementsand safety factors

In the Eurocodes (ECs) design requirements are set out inrelation to specified limit-state conditions Calculations todetermine the ability of a member (or assembly of members)to satisfy a particular limit-state are undertaken using designactions (loads or deformations) and design strengths Thesedesign values are determined from either characteristic actionsor representative actions and from characteristic strengths ofmaterials by the application of partial safety factors

291 ACTIONS

Characteristic values of the actions to be used in the design ofbuildings and civil engineering structures are given in severalparts of EC 1 Actions on structures as follows

1991-1-1 Densities self-weight and imposed loads1991-1-2 Actions on structures exposed to fire1991-1-3 Snow loads1991-1-4 Wind loads1991-1-5 Thermal actions1991-1-6 Actions during execution1991-1-7 Accidental actions due to impact and explosions1991-2 Traffic loads on bridges1991-3 Actions induced by cranes and machinery1991-4 Actions on silos and tanks

A variable action (eg imposed load snow load wind loadthermal action) has the following representative values

characteristic value Qk

combination value 0Qk

frequent value 1Qk

quasi-permanent value 2Qk

The characteristic and combination values are used for theverification of the ultimate and irreversible serviceability limit-states The frequent and quasi-permanent values are usedfor the verification of ULSs involving accidental actions andreversible SLSs The quasi-permanent values are also used forthe calculation of long-term effects

Design actions (loads) are given by

design action (load) F Fk

where Fk is the specified characteristic value of the actionF is the value of the partial safety factor for the action (A for

accidental actions G for permanent actions Q for variableactions) and the limit state being considered and is either10 0 1 or 2 Recommended values of F and are given inEC 0 Basis of structural design

292 MATERIAL PROPERTIES

The characteristic strength of a material fk means that valueof either the cylinder strength fck or the cube strength fckcube ofconcrete or the yield strength fyk of reinforcement below whichnot more than 5 of all possible test results are expected tofall In practice the concrete strength is selected from a set ofstrength classes which in EC 2 are based on the characteristiccylinder strength The application rules in EC 2 are validfor reinforcement in accordance with BS EN 10080 whosespecified yield strength is in the range 400ndash600 MPa

Design strengths are given by

design strength fkM

where fk is either fck or fyk as appropriate and M is the value ofthe partial safety factor for the material ( C for concrete S forsteel reinforcement) and the limit-state being considered asspecified in EC 2

293 BUILDINGS

Details of the design requirements and partial safety factors aregiven in Table 41 Appropriate combinations of design actionsand values of are given on page 336

294 CONTAINMENT STRUCTURES

For structures containing liquids or granular solids the mainrepresentative value of the variable action resulting from theretained material should be taken as the characteristic value forall design situations Appropriate characteristic values are givenin EC 1 Part 4 Actions in silos and tanks

For the ULS where the maximum level of a retained liquidcan be clearly defined and the effective density of the liquid(allowing for any suspended solids) will not vary significantlythe value of Q applied to the resulting characteristic actionmay be taken as 12 Otherwise and for retained granularmaterials in silos Q 15 should be used

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Design requirements and safety factors336

For the SLS of cracking a classification of liquid-retainingstructures in relation to the required degree of protectionagainst leakage and the corresponding design requirementsgiven in EC 2 Part 3 are detailed here Silos containing dry

materials may generally be designed as Class 0 but where thestored material is particularly sensitive to moisture class 1 2or 3 may be appropriate

Limit-state and design consideration Combination of design actions

Ultimate (persistent and transient actions) 13Gj Gkj Q1 Qk1 13Qi 0i Qki ( j 1 i 1)

Ultimate (accidental action) (Ad 13Gkj (11 or 21) Qk1 132i Qki ( j 1 i 1)

Serviceability (function including damage to structural and 13Gkj Qk1 130i Qki ( j 1 i 1)non-structural elements eg partition walls etc)

Serviceability (comfort to user use of machinery avoiding 13Gkj 11 Qk1 132i Qki ( j 1 i 1)ponding of water etc)

Serviceability (appearance) 13Gkj 132i Qki ( j 1 i 1)

Note In the combination of design actions shown above Qk1 is the leading variable action and Qki are any accompanying variable actions Where necessary eachaction in turn should be considered as the leading variable action Serviceability design consideration and associated combination of design actions as specified in the UK National Annex

Values of for variable actions ( as specified in the UK National Annex) 0 1 2

Imposed loads (Category and type see EN 1991-1-1)A domestic residential area B office area 07 05 03C congregation area D shopping area 07 07 06E storage area 10 09 08F traffic area (vehicle weight 30 kN) 07 07 06G traffic area (30 kN vehicle weight 160 kN) 07 05 03H roof 07 0 0

Snow loads (see EN 1991-1-3)Sites located at altitude 1000 m above sea level 07 05 02Sites located at altitude 1000 m above sea level 05 02 0

Wind loads (see EN 1991-1-4) 05 02 0

Thermal actions (see EN 1991-1-5) 06 05 0

Class Leakage requirements Design provisions

0 Leakage acceptable or irrelevant The provisions in EN 1992-1-1 may be adopted (see Table 41)

1 Leakage limited to small amount The width of any cracks that can be expected to pass through the full thicknessSome surface staining or damp of the section should be limited to wk1 given bypatches acceptable 005 wk1 0225(1 hw 45h) 02 mm

where hwh is the hydraulic gradient (ie head of liquid divided by thicknessof section) at the depth under consideration Where the full thickness of the section is not cracked the provisions in EN 1992-1-1 apply (see Table 41)

2 Leakage minimal Appearance Cracks that might be expected to pass through the full thickness of the sectionnot to be impaired by staining should be avoided unless measures such as liners or water bars are included

3 No leakage permitted Special measures (eg liners or prestress) are required to ensure watertightness

Note In classes 1 and 2 to provide adequate assurance that cracks do not pass through the full width of a section the depth of the compression zone should beat least 02h 50 mm under quasi-permanent loading for all design conditions The depth should be calculated by linear elastic analysis assuming that concretein tension is neglected

Combinations of design actions on buildings

Values of for variable actions on buildings

Classification of liquid-retaining structures

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Design requirements and partial safety factors (EC 2 Part 1) 41

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Chapter 30

Properties of materials

301 CONCRETE

3011 Strength and elastic properties

The characteristic strength of concrete is defined as that level ofcompressive strength below which 5 of all valid test results isexpected to fall Strength classes are specified in terms of bothcylinder strength and equivalent cube strength Recommendedstrength classes with indicative values for the secant modulusof elasticity at 28 days are given in Table 42 The values givenfor normal-weight concrete are appropriate for concretesmade with quartzite aggregates For limestone and sandstoneaggregates these values should be reduced by 10 and 30respectively For basalt aggregates the values should be increasedby 20

Variation of the secant modulus of elasticity with time can beestimated by the expression

Ecm(t) [ fcm(t)fcm]03 Ecm

where Ecm(t) and fcm(t) are values at age t days and Ecm and fcm

are values determined at age 28 days

3012 Creep and shrinkage

The creep strain in concrete may be assumed to be directlyproportional to the applied stress for stresses not exceeding045fck(t0) where t0 is age of concrete at the time of loadingValues of the final creep coefficient (infin to) and the creepdevelopment coefficient c(t t0) according to the time underload can be obtained from Table 43 The procedure used todetermine the final creep coefficient is as follows

1 Determine point corresponding to age of loading to on theappropriate curve (N for normally hardening cement R forrapidly hardening cement S for slowly hardening cement)

2 Construct secant line from origin of curve to the pointcorresponding to to

3 Determine point corresponding to the notional size ofthe member ho on the appropriate curve for the concretestrength class

4 Cross horizontally from the point determined in 3 to intersectthe secant line determined in 2

5 Drop vertically from the intersection point determined in 4to obtain the required creep coefficient (infin to)

When the applied stress exceeds 045fck(t0) at time of loadingcreep non-linearity should be considered and (infin to) shouldbe increased as indicated in Table 42

The total shrinkage strain is composed of two componentsautogenous shrinkage strain and drying shrinkage strainThe autogenous shrinkage strain develops during hardening ofthe concrete a major part therefore develops in the early daysafter casting It should be considered specifically when newconcrete is cast against hardened concrete Drying shrinkagestrain develops slowly as it is a function of the migration ofthe water through the hardened concrete Final values of eachcomponent of shrinkage strain and development coefficientswith time are given in Table 42

3013 Thermal properties

Values of the coefficient of thermal expansion of concrete fornormal design purposes are given in Table 42

3014 Stressndashstrain curves

Idealised stressndashstrain curves for concrete in compression aregiven in Table 44 Curve A is part parabolic and part linear andcurve B is bi-linear A simplified rectangular diagram is alsogiven as a further option

302 REINFORCEMENT

3021 Strength and elastic properties

The characteristic yield strength of reinforcement according toEN 10080 is required to be in the range 400ndash600 MPa Thecharacteristic yield strength of reinforcement complying withBS 4449 is 500 MPa For further information on types proper-ties and preferred sizes of reinforcement reference should bemade to section 103 and Tables 219 and 220

3022 Stressndashstrain curves

Idealised bi-linear stressndashstrain curves for reinforcement intension or compression are shown in Table 44 Curve A has aninclined top branch up to a specified strain limit and curve Bhas a horizontal top branch with no need to check the strainlimit For design purposes the modulus of elasticity of steelreinforcement is taken as 200 GPa

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Concrete (EC 2) strength and deformation characteristics ndash 1 42

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Concrete (EC 2) strength and deformation characteristics ndash 2 43

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Stressndashstrain curves (EC 2) concrete and reinforcement 44

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Chapter 31

Durability andfire-resistance

In the following the concrete cover to the first layer of barsas shown on the drawings is described as the nominal coverIt is defined as a minimum cover plus an allowance in designfor deviation A minimum cover is required to ensure the safetransmission of bond forces the protection of steel againstcorrosion and an adequate fire-resistance In order to transmitbond forces safely and ensure adequate concrete compactionthe minimum cover should be not less than the bar diameter orfor bundled bars the equivalent diameter of a notional barhaving the same cross-sectional area as the bundle

311 DURABILITY

3111 Exposure classes

Details of the classification system used in BS EN 206ndash1 andBS 8500-1 with informative examples applicable in the UnitedKingdom are given in Table 45 Often the concrete can beexposed to more than one of the actions described in the tablein which case a combination of the exposure classes will apply

3112 Concrete strength classes and covers

Concrete durability is dependent mainly on its constituents andlimitations on the maximum free watercement ratio and theminimum cement content are specified for each exposure classThese limitations result in minimum concrete strength classesfor particular cements For reinforced concrete the protectionof the steel against corrosion depends on the cover The requiredthickness of cover is related to the exposure class the concretequality and the intended working life of the structure Details ofthe recommendations in BS 8500 are given in Table 46

The values given for the minimum cover apply for ordinarycarbon steel in concrete without special protection and forstructures with an intended working life of at least 50 years Thevalues given for the nominal cover include an allowance fortolerance of 10 mm which is recommended for buildings andis normally also sufficient for other types of structures

For uneven concrete surfaces (eg ribbed finish or exposedaggregate) the cover should be increased by at least 5 mm Ifin-situ concrete is placed against another concrete element(precast or in-situ) the minimum cover to the reinforcement atthe interface need be no more than that recommended foradequate bond provided the following conditions are metthe concrete strength class is at least C2530 the exposure time

of the concrete surface to an outdoor environment is no more than28 days and the interface has been roughened

Where concrete is cast against prepared ground (includingblinding) the nominal cover should be at least 50 mm Forconcrete cast directly against the earth the nominal covershould be at least 75 mm

312 FIRE-RESISTANCE

3121 Building regulations

The minimum period of fire-resistance required for elements ofthe structure according to the purpose group of a buildingand its height or for basements depth relative to the ground aregiven in Table 312 Building insurers may require longer fireperiods for storage facilities

3122 Design procedures

BS EN 1992-1-2 contains prescriptive rules in the form ofboth tabulated data and calculation models for standard fireexposure A procedure for a performance-based method usingfire-development models is also provided

The tabulated data tables give minimum dimensions for thesize of a member and the axis distance of the reinforcementThe axis distance is the nominal distance from the centre ofthe main reinforcement to the concrete surface as shown in thefollowing figure

Tabulated data is given for beams slabs and braced columnsfor which provision is made for the load level to be taken intoaccount In many cases for fire periods up to about two hoursthe cover required for other purposes will control For furtherinformation on all the design procedures reference should bemade to BS EN 1992-1-2

Section through member showing nominal axis distance

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Exposure classification (BS 8500) 45

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Concrete quality and cover requirements for durability (BS 8500) 46

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fck Maximum value of xd at section whereMPa moment redistribution is required

50 ( 04)

55 095 ( 04)60 092 ( 04)70 090 ( 04)

80 088 ( 04)

In the above expressions is the ratio of the design moment to themaximum elastic moment for values of

where class B and C reinforcement is usedwhere class A reinforcement is used

Based on values given in the UK National Annex10 0810 07

Chapter 32

Bending and axialforce

321 DESIGN ASSUMPTIONS

Basic assumptions regarding the design of cracked concretesections at the ULS are outlined in section 52 The tensilestrength of the concrete is neglected and strains are evaluatedon the assumption that plane sections before bending remainplane after bending Reinforcement stresses are then derivedfrom these strains on the basis of the design stress-strain curvesshown on Table 44 Two alternatives are given in which the topbranch of the curve is taken as either horizontal with no needto check the strain limit or inclined up to a specified strainlimit For the concrete stresses three alternative assumptionsare permitted as shown on Table 44 The design stressndashstraincurves give stress distributions that are a combination of eithera parabola and a rectangle or a triangle and a rectangleAlternatively a rectangular concrete stress distribution may beassumed Whichever alternative is used the proportions of thestress-block and the maximum strain are constant forfck 50 MPa but vary as the concrete strength changes forfck 50 MPa

For a rectangular area of width b and depth x the total com-pressive force is given by k1 fckbx and the distance of the forcefrom the compression face is given by k2x where values of k1

(including for the term ccc) and k2 according to the shape ofthe stress block are given in the table opposite

322 BEAMS AND SLABS

Beams and slabs are generally subjected to bending only butsometimes are also required to resist an axial force for examplein a portal frame or in a floor acting as a prop between basementwalls Axial thrusts not greater than 012fck times the area of thecross section may be ignored in the analysis of the sectionsince the effect of the axial force is to increase the moment ofresistance

In cases where as a result of moment redistributionallowed in the analysis of the member the design moment isless than the maximum elastic moment at the section the nec-essary ductility may be assumed without explicit verificationif the neutral axis depth satisfies the limits given in the tableopposite

Where plastic analysis is used the necessary ductility may beassumed without explicit verification if the neutral axis depth atany section satisfies the following requirement

xd 025 for concrete strength classes C5060xd 015 for concrete strength classes C5060

The following analyses and resulting design charts and tablesare applicable to concrete strength classes C5060

fck Shape of k1 k2 cuMPa stress-block

50 Parabola-rectangle 0459 0416Triangle-rectangle 0425 0389 00035Rectangular 0453 0400

55 Parabola-rectangle 0433 0400Triangle-rectangle 0402 0375 00031Rectangular 0435 0394

60 Parabola-rectangle 0417 0392Triangle-rectangle 0381 0363 00029Rectangular 0417 0388

70 Parabola-rectangle 0399 0383Triangle-rectangle 0357 0351 00027Rectangular 0382 0375

80 Parabola-rectangle 0385 0378Triangle-rectangle 0327 0340 00026Rectangular 0349 0363

90 Parabola-rectangle 0378 0375Triangle-rectangle 0316 0337 00026Rectangular 0317 0350

Properties of concrete stress-blocks for rectangular area

Limits of xd for moment redistribution

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Bending and axial force346

3221 Singly reinforced rectangular sectionsMbd2fck

As fykbdfck for ductility class

A B C

0010 00111 00108 001020012 00134 00130 001230014 00156 00152 001440016 00179 00174 001650018 00202 00197 00186

0020 00224 00219 002070022 00247 00241 002280024 00270 00264 002520026 00293 00286 002760028 00317 00309 00300

0030 00340 00331 003240032 00363 00354 003490034 00387 00379 003730036 00410 00403 003980038 00434 00428 00423

0040 00457 00453 004480042 00481 00478 004730044 00505 00503 004980046 00529 00528 005230048 00553 00554 00549

3222 Doubly reinforced rectangular sections

The lever arm between the forces shown in the figure here isgiven by z (d k2x) from which x (d z)k2

Taking moments for the compressive force about the line ofaction of the tensile force gives

M k1 fckbxz k1 fckbz(d z)k2

The solution of the resulting quadratic equation in z gives

where

Taking moments for the tensile force about the line of action ofthe compressive force gives

M As fs z from which As Mfs z

The strain in the reinforcement s 00035(1 xd)(xd) andfrom the design stressndashstrain curves the stress is given by

fs sEs 700(1 xd)(xd) ks fyk 115

If the top branch of the design stressndashstrain curve is taken ashorizontal (curve B) ks 10 and fs fyk 115 for values of

xd 805(805 fyk) 0617 for fyk 500 MPa

A design chart for fyk 500 MPa derived on the basis of theparabolic-rectangular stress-block for the concrete and curve Bfor the reinforcement is given in Table 47

If the top branch of the design stressndashstrain curve is taken asinclined (curve A) the value of ks depends on the strain and theductility class of the reinforcement The use of curve A isadvantageous in reducing the reinforcement requirement Themaximum reduction for a lightly reinforced section is close to5 8 and 15 for reinforcement ductility classes A B andC respectively For more heavily reinforced sections the reduc-tions are less and the particular ductility class makes very littledifference to the values obtained

A design table based on the rectangular stress-block for theconcrete and design curve A for the reinforcement is given inTable 48 The table uses non-dimensional parameters and isvalid for values of fck 50 MPa Values of ks determined forfyk 500 MPa were based on the most critical ductility classin each case namely class A for Mbd2fck 0046 and class Bfor Mbd2fck 0046 This can be seen from the followingtable which gives values of As fykbdfck for each ductility classin the range up to Mbd2fck 0048

M bd2fckz d 05 025 (k2 k1)

The forces provided by the concrete and the reinforcementare shown in the figure here Taking moments for the twocompressive forces about the line of action of the tensileforce gives

M k1 fckbx(d k2x)

The strain in the reinforcement and fromdesign stressndashstrain curve B the stress is given by

Thus fyk 115 for values of

for fyk 500 MPa

Equating the tensile and compressive forces gives

As fs k1 fckbx

where the stress in the tension reinforcement is given by theexpression derived in section 3221

As f s

x d [805 (805 fyk)](dd) 264(dd )

f s

f s s Es 700(1 dx) fyk 115

s 00035(1 dx)

As f s(dd)

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EC 2 Design chart for singly reinforced rectangular beams 47

Sin

gly

rein

forc

ed b

eam

s (f

yk=

500

MP

a)

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EC 2 Design table for singly reinforced rectangular beams 48

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Beams and slabs 349

Design charts based on the rectangular stress-block for theconcrete design curve B for the reinforcement and for valuesof and 015 respectively are given in Tables 49 and410 The charts which use non-dimensional parameters weredetermined for fyk 500 MPa but may be safely used forfyk 500 MPa In determining the forces in the concrete noreduction has been made for the area of concrete displaced bythe compression reinforcement

3223 Design formulae for rectangular sections

No design formulae are given in the code but the following arevalid for values of fck 50 MPa and fyk 500 MPa Theformulae are based on the rectangular stress-block for theconcrete and stresses of 087fyk in tension and compressionreinforcement The compression reinforcement requirementdepends on the value of Mbd2fck compared to where

0210 for 10

for

is the ratio design moment to maximum elastic momentwhere 07 for class B and C reinforcement and 08 forclass A reinforcement

For compression reinforcement is not required and

As M087fykz where

z d05 + and x (d z)04

For compression reinforcement is required and

where

z d05 + and x (d z)04

For (for fy 500 MPa) should be replaced by16(1 ) in the equations for and As

3224 Flanged sections

In monolithic beam and slab construction where the web of thebeam projects below the slab the beam is considered as aflanged section for sagging moments The effective width of theflange over which uniform conditions of stress can be assumedmay be taken as beff bw where

01(aw l0) 02l0 05aw for L beams 02(aw l0) 04l0 10aw for T beams

In the aforesaid expressions bw is the web width aw is the cleardistance between the webs of adjacent beams and l0 is the dis-tance between points of zero bending moment for the beam Ifleff is the effective span l0 may be taken as 085leff when thereis continuity at one end of the span only and 07leff when there iscontinuity at both ends of the span For up-stand beams whendesigning for hogging moments l0 may be taken as 03leff atinternal supports and 015leff at end supports

In most sections where the flange is in compression the depthof the neutral axis will be no greater than the thickness of theflange In this case the section can be considered to be rectangularwith b taken as the flange width The condition regarding theneutral axis depth can be confirmed initially by showing thatM k1 fckbhf (d k2hf) where hf is the thickness of the flange

bb

b

AsAsdxAsdx 0375

025 0882

As As bd2fck 087fykz

As ()bd2fck 087fyk(dd)

025 0882

10 0453( 04) 0181( 04)2

dd 01

Alternatively the section can be considered to be rectangularinitially and the neutral axis depth can be checked subsequently

The figure here shows a flanged section where the neutral axisdepth is greater than the flange thickness The concrete forcecan be divided into two components and the required area oftension reinforcement is then given by

As As1 k1 fck (b bw)hf 087fyk whereAs1 area of reinforcement required to resist a moment M1

applied to a rectangular section of width bw and

M1 M k1 fck (b bw)hf (d k2hf) bd2fck

Using the rectangular concrete stress-block in the forgoingequations gives k1 045 and k2 04 This approach givessolutions that are lsquocorrectrsquo when x hf but become slightlymore conservative as (x hf) increases

3225 General analysis of sections

The analysis of a section of any shape with any arrangement ofreinforcement involves a trial-and-error process An initialvalue is assumed for the neutral axis depth from which theconcrete strains at the positions of the reinforcement can becalculated The corresponding stresses in the reinforcement aredetermined and the resulting forces in the reinforcement andthe concrete are obtained If the forces are out of balance thevalue of the neutral axis depth is changed and the process isrepeated until equilibrium is achieved Once the balancedcondition has been found the resultant moment of the forcesabout the neutral axis or any convenient point is calculated

Example 1

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EC 2 Design chart for doubly reinforced rectangular beams ndash 1 49

Dou

bly

rein

forc

ed b

eam

s (f

yk=

500

MP

ad9

d=

01)

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410EC 2 Design chart for doubly reinforcedrectangular beams ndash 2

Dou

bly

rein

forc

ed b

eam

s (f

yk=

500

MP

ad9

d=

015

)

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Bending and axial force352

The beam shown in the figure here is part of a monolithic beamand slab floor in which the beams are spaced at 5 m centresThe maximum and minimum design loads on each span are asfollows where Gk 160 kN and Qk 120 kN

Fmax 135Gk 15Qk 396 kN Fmin 135 Gk 216 kN

The section design is to be based on the following values

fck 32 MPa fyk 500 MPa cover to links 35mm

For sagging moments and beff are given by

02(aw l0) 04 l0 aw

02(4700 085 8000) 2300 mmbeff bw 300 2300 2600 mm

Allowing for 8 mm links and 32 mm main bars

d 500 (35 8 16) 440 mm say

In the calculations that follow solutions are obtained usingcharts and equations to demonstrate the use of each method

Maximum sagging moment For section to be designed as rec-tangular with b taken as the flange width bending momentshould satisfy the condition

M k1 fckbhf (d k2hf) 045 32 2600 150 (440 04 150) 106

2134 kNm (258 kNm)

Mbd 2 258 106 (2600 4402) 051 MPa

From chart in Table 47 100Asbd 012

As 00012 2600 440 1373 mm2

Alternatively by calculation or from Table 48

Mbd2fck 05132 0016

Hence As M087fykz gives

As 258 106 (087 500 0985 440) 1369 mm2

Using 3H25 gives 1473 mm2

The above solutions are based on design stress-strain curve Bfor the reinforcement Solutions based on curve A also can beobtained from the table in section 3221 as follows

Mbd 2fck

As fykbdfck for ductility class

A B C

0016 00179 00174 00165As (mm2) 1310 1274 1208

Maximum hogging moment

Mbd2fck 396 106 (300 4402 32) 0213

From Table 48 this appears to be just beyond the range for asingly reinforced section From the chart in Table 49 a valueof As fykbdfck 034 can be obtained Although this is a validsolution it should be possible to reduce the area of tensionreinforcement to a more suitable value by allowing for somecompression reinforcement Consider the use of 2H25 whichgives 55 mm ( )dd 55 440 0125d

z d 05 025 0882 0016 0985

b

b

b

fykbdfck 982 500(300 440 32) 0116

Interpolating from charts in Tables 49 and 410 with

fykbdfck 01 As fykbdfck 0285 (for 0125)As 0285 300 440 32500 2408 mm2

Using 3H32 gives 2413 mm2

A solution can also be obtained using the design equations asfollows 087fy is valid for xd ( )0375 0333

With xd 0333 and zd (1 04xd) 0867

0453(xd)(zd) 0453 0333 0867 0131

( ) bd2fck 087fyk (d ) (0213 0131) 300 4402 32

(087 500 385) 910 mm2 (2H25 gives 982 mm2)

As bd 2fck 087fyk z 910 0131 300 4402 32

(087 500 0867 440) 2377 mm2 (3H32 gives 2413 mm2)

Example 2 Suppose that in the previous example the maxi-mum hogging moment at B is reduced by 20 to 317 kNmThis is valid for reinforcement of all ductility classes

Mbd2fck 317 106 (300 4402 32) 0171 317396 080 xd ( 04) 04

From chart in Table 410 keeping to left of line for xd 04

fykbdfck 005 Asfykbdfck 023 005 300 440 32500 423 mm2

Using 2H20 gives 628 mm2 (instead of 2H25 in example 1)

As 023 300 440 32500 1943 mm2

Using 4H25 gives 1963 mm2 (instead of 3H32 in example 1)

A solution can also be obtained by using the design equationsas in example 1 with xd 0333 zd 0867 0131

( )bd 2fck 087fyk (d ) 004 300 4402 32 (087 500 385) 444 mm2

As bd2fck 087fykz 444 0131 300 4402 32

(087 500 0867 440) 1912 mm2

Since the reduced hogging moment for load case 1 is stillgreater than the elastic hogging moment for load case 2 thedesign sagging moment remains the same as in example 1

In the foregoing examples at the bottom of the beam 2H25bars would run the full length of each span with 2H25 or 2H20splice bars at support B Other bars would be curtailed to suitthe bending moment requirements and detailing rules

323 COLUMNS

In the Code of Practice a column is taken to be a compressionmember whose greater overall cross sectional dimension is notmore than four times its smaller dimension An effective lengthand a slenderness ratio are determined in relation to its major

As

dAs

As

As

As

dAs

ddf s

ddAs

As

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Columns 353

and minor axes of bending The effective length is a function ofthe clear height and depends upon the restraint conditions at theends of the column A slenderness ratio is defined as the effec-tive length divided by the radius of gyration of the uncrackedconcrete section Columns should generally be designed for bothfirst order and second order effects but second order effects maybe ignored provided the slenderness ratio does not exceed a par-ticular limiting value This can vary considerably and has to bedetermined from an equation involving several factors Thesegenerally need to be calculated but default values are also given

Columns are subjected to combinations of bending momentand axial force and the cross section may need to be checkedfor more than one combination of values Several methods ofanalysis of varying complexity are available for determiningsecond order effects Many columns can be treated as isolatedmembers and a simplified method of design using equationsbased on an estimation of curvature is commonly used Theequations contain a modification factor Kr the use of whichresults in an iteration process with Kr taken as 10 initially Thedesign procedures are shown in Tables 415 and 416

In the code for sections subjected to pure axial load the concretestrain is limited to 0002 for values of fck 50 MPa In thiscase the design stress in the reinforcement should be limited to400 MPa However in other parts of the code the design stressin this condition is shown as fyd fyks 087fyk In thederivation of the charts in this chapter which apply for allvalues of fck 50 MPa and fyk 500 MPa the maximumcompressive stress in the reinforcement was taken as 087fykThe charts contain sets of Kr lines to aid the design process

3231 Rectangular columns

Design charts based on the rectangular stress-block for theconcrete and for values of dh 08 and 085 are given inTables 411 and 412 respectively On each curve a straight linehas been taken between the point where xh 10 and the pointwhere N Nu The charts which were determined for fyk 500MPa may be safely used for fy 500 MPa In determining theforces in the concrete no reduction has been made for the areaof concrete displaced by the compression reinforcement In thedesign of slender columns the Kr factor is used to modify thedeflection corresponding to a load Nbal at which the moment is amaximum A line corresponding to Nbal passes through a cusp oneach curve For N Nbal the Kr value is taken as 10 ForN Nbal Kr can be determined from the lines on the chart

3232 Circular columns

The following figure shows a circular section with six barsspaced equally around the circumference Solutions based on sixbars will be slightly conservative if more bars are used Thearrangement of the bars relative to the axis of bending affects theresistance of the section and the arrangement shown in the fig-ure is not the most critical in every case For some combinationsof bending moment and axial force if the arrangement shown isrotated through 30o a slightly more critical condition results butthe differences are small and may be reasonably ignored

The foregoing figure shows a rectangular section in which thereinforcement is disposed equally on two opposite sides of ahorizontal axis through the mid-depth By resolving forces andtaking moments about the mid-depth of the section the follow-ing equations are obtained for 0 xh 10

Nbhfck k1(xh) 05(As fykbhfck)(ks1 ks2)

Mbh2fck k1(xh)05 k2(xh) 05(As fykbhfck)(ks1 ks2)(dh 05)

The stress factors ks1 and ks2 are given by

ks1 14(xh dh 1)(xh) 087ks2 14(dh xh)(xh) 087

The maximum axial force Nu is given by the equation

Nubhfck 0567 087(As fykbhfck)

The following analysis is based on a uniform stress-block forthe concrete of depth x and width hsin at the base (as shownin figure) Negative axial forces are included in order to caterfor members such as tension piles By resolving forces andtaking moments about the mid-depth of the section the followingequations are obtained where cos1 (1 2 xh) for0 x 10 and hs is the diameter of a circle through thecentres of the bars

Nh2fck kc (2 sin2)8 (12)(As fykAc fck)(ks1 ks2 ks3)

Mh3fck kc (3sin sin3)72 ( 277)(As fykAc fck)(hsh)(ks1 ks3)

Because the width of the compression zone decreases in thedirection of the extreme compression fibre the design stressin the concrete has to be reduced by 10 Thus in the aboveequations kc 09 0567 051 and 08

The stress factors ks1 ks2 and ks3 are given by

087 ks1 14(0433hsh 05 xh)(xh) 087

087 ks2 14(05 xh)(xh) 087

087 ks3 14(05 0433hsh xh)(xh) 087

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EC 2 Design chart for rectangular columns ndash 1 411

Rec

tang

ular

col

umns

(f y

k=

500

MP

ad

h=

08)

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EC 2 Design chart for rectangular columns ndash 2 412

Rec

tang

ular

col

umns

(f y

k=

500

MP

ad

h=

085

)

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Bending and axial force356

To avoid irregularities in the charts the reduced design stress inthe concrete is used to determine the maximum axial force Nuwhich is given by the equation

Nuh2fck ( 4)051 087(As fykAc fck)

The minimum axial force Nmin is given by the equation

Nminh2fck 087( 4)(AsfykAcfck)

Design charts for values of hsh 06 and 07 are given inTables 413 and 414 respectively The statements in section3231 on the derivation and use of the charts for rectangularsections apply also to those for circular sections

3233 General analysis of column sections

Any given cross section can be analysed by a trial-and-errorprocess For a section bent about one axis an initial value isassumed for the neutral axis depth from which the concretestrains at the positions of the reinforcement can be calculatedThe resulting stresses in the reinforcement are determined andthe forces in the reinforcement and concrete evaluated If theresultant force is not equal to the design axial force N the valueof the neutral axis depth is changed and the process repeateduntil equality is achieved The resultant moment of all theforces about the mid-depth of the section is then the moment ofresistance appropriate to N This approach is used to analyse arectangular section in example 6

Example 3 A 300 mm square braced column designed forthe following requirements

l 50 m k 0675 at both joints in both directionsM02 40 kNm M01 20 kNm about x-x axisM0 negligible about y-y axis N 1720 kNfck 32 MPa fyk 500 MPa cover to links 35 mm

Allowing for 8 mm links and 32 mm main bars

d 300 (35 8 16) 240 mm say

From Table 415 effective length for a braced column wherethe joint stiffness is the same at both ends is given by

l0 [05(045 2k)(045 k)] l 08 50 40 m

Slenderness ratio l0 i 4000(300radic12) 462

From Table 415 with M01 05M02 C 22 and

Since lim second order moments need to be considered

Minimum design moment with e0 h30 30030 20 mm

Mmin Ne0 1720 002 34 kNm

Additional first order moment resulting from imperfectionswith 067 h 2radicl 2radic(50) 09 10

Mi N( hl0400) 1720 (09 40400) 16 kNm

Total first order moment for section at end 2 of the column

Mx M02 Mi 40 16 56 kNm (Mmin)

Mbh2fck 56 106(300 3002 32) 0065

Nbhfck 1720 103(300 300 32) 0600

lim 34 N Ac fck 34 1720 103 (3002 32) 44

From the design chart for dh 240300 08

As fykbhfck 025 (Table 411)As 025 300 300 32500 1440 mm2

Using 4H25 gives 1963 mm2

The section where the second order moment is greatest may bedesigned by first assuming the reinforcement (4H25 say)

As fykbhfck 1963 500(300 300 32) 034Nubhfck 086 (Table 411) and for Nbhfck 06Mubh2fck 009 Kr 04 (Table 411)Mux Muy 009 300 3002 32 106 78 kNm

Second order moment resulting from deflection with Kr 04and for fck 32 MPa and 46 K 13 (Table 416)

M2 N(KrKl02d)2000

1720 (04 13 402024)2000 30 kNm

Equivalent first order moment (near mid-height of column)

M0e 06M02 04M01 04M02 04 40 16 kNm

Total design moments (near mid-height of column)

Mx M0e Mi M2 16 16 30 62 kNm (Mmin)My M2 30 kNm

Design for biaxial bending may be ignored if the following twoconditions are satisfied (a) 05y x 2y and (b) for asquare column Mx is either 02My or 5My In this casesince condition (b) is not satisfied and a furthercheck is necessary as follows

For NNu 060086 07

n 092 083(NNu) 150

Hence

Since this value is less than 10 4H25 is sufficient

Example 4 A 350 mm circular braced column designed forthe same requirements as example 3 Thus l0 40 m as before

Slenderness ratio l0i 4000(3504) 457

Cross-sectional area Ac (4) 3502 962 103 mm2

Since second order moments need not be consideredAllowing for 8 mm links and 20 mm main bars

hs 350 2 (35 8 10) 244 mm

For the section at end 2 of the column

Mh3fck 56 106(3503 32) 0041Nh2fck 1720 103(3502 32) 044

From the design chart for hsh 244350 07

As fykAcfck 026 (Table 414)As 026 962 103 32500 1600 mm2

Using 6H20 gives 1885 mm2

lim

lim 34 N Ac fck 34 1720 (962 32) 455

Mx

Mux

n

My

Muy

n

6278

15

3078

15

095

Mx 2My

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EC 2 Design chart for circular columns ndash 1 413

Circular columns (fyk = 500 MPa hsh = 06)

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EC 2 Design chart for circular columns ndash 2 414

Circular columns (fyk = 500 MPa hsh = 07)

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EC 2 Design procedure for columns ndash 1 415

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EC 2 Design procedure for columns ndash 2 416

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Columns 361

Example 5 A 400 mm circular unbraced column designedfor the same requirements as example 3 Thus k 0675 asbefore

From Table 415 effective length for an unbraced columnwhere the joint stiffness is the same at both ends is given by

l0 [(1 2k)(1 k)] l 14 50 70 m

Slenderness ratio l0i 7000(4004) 70

Cross-sectional area Ac ( 4) 4002 1257 103 mm2

Since second order moments need to be consideredAllowing for 8 mm links and 32 mm main bars

hs 400 2 (35 8 16) 280 mm say

Additional first order moment resulting from imperfections

Mi N( hl0400) 1720 (09 70400) 27 kNm

Second order moment resulting from deflection with Kr 10(max) K 11 for fck 32 MPa and 70 (Table 416) andd h2 035hs 298 mm

M2 N(KrKl02d)2000

1720 (10 11 7020298)2000 155 kNm

Total design moments at end 2 of column

Mx M02 Mi M2 40 27 155 222 kNm (Mmin)My M2 155 kNm

Resultant uniaxial moment at end 2 of column

kNm

Mh3fck 270 106(4003 32) 013Nh2fck 1720 103(4002 32) 034

From the design chart for hsh 280400 07

As fykAcfck 084 (Table 414)As 084 1257 103 32500 6758 mm2

It can be seen from the chart that Kr 10 Using 8H32 gives

As fykAc fck 6434 500(1257 103 32) 080For Nh2fck 034 Mh3fck 0125 and Kr 075

Hence Mu 0125 4003 32 106 256 kNm

With Kr 075 modified M2 075 155 116 kNm

Total design moments at end 2 of column

Mx M02 Mi M2 40 27 116 183 kNmMy M2 116 kNm

Resultant uniaxial moment at end 2 of column

217 kNm

Since M Mu 8H32 is sufficient

Example 6 The column section in the following figure isreinforced with 8H32 arranged as shown The moment of resis-tance about the major axis is to be obtained for the followingrequirements

N 2300 kN fck 32 MPa fyk 500 MPa

M (1832 1162)

M (M2x M2

y) (2222 1552) 270

lim

lim 108 N Acfck 108 1720 (1257 32) 165

Consider the bars in each half of the section to be replaced byan equivalent pair of bars Depth to the centroid of the bars inone half of the section 60 2404 120 mm The section isnow considered to be reinforced with four equivalent barswhere d 600 120 480 mm

As fykbhfck 6434 500(300 600 32) 056Nbhfcu 2300 103(300 600 32) 040

From the design chart for dh 480600 08

Mubh2fck 018 (Table 411)Mu 018 300 6002 32 106 622 kNm

The solution can be checked using a trial-and-error process toanalyse the original section as follows

The axial load on the section is given by

N k1fckbx (As1ks1 As2ks2 As3ks3)fyk

where dh 540600 09 and ks1 ks2 and ks3 are given by

ks1 14(xh dh 1)(xh) 087ks2 14(05 xh)(xh) 087ks3 14(dh xh)(xh) 087

With x 300 mm xh 05 ks1 087 ks2 0 and ks3 087

N 045 32 300 300 103 1296 kN (2300)

With x 360 mm xh 06 ks2 0233 ks3 07

N 045 32 300 360 103 (2413 087 1608 0233 2413 07) 500 103

1555 392 1947 kN (2300)

With x 390 mm xh 065 ks2 0323 ks3 0538

N 045 32 300 390 103 (2413 087 1608 0323 2413 0538) 500 103

1685 660 2345 kN (2300)

With x 387 mm xh 0645 ks2 0315 ks3 0553

N 045 32 300 387 103 (2413 087 1608 0315 2413 0553) 500 103

1672 636 2308 kN

Since the internal and external forces are now sensibly equaltaking moments about the mid-depth of the section gives

Mu k1 fckbx(05h k2x ) (As1ks1 As3ks3)(d 05h)fy

045 32 300 387 (300 04 387) 106

(2413 087 2413 0553)(540 300) 500 106

243 412 655 kNm (622 obtained before)

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Chapter 33

Shear and torsion

331 SHEAR RESISTANCE

3311 Members without shear reinforcement

The design shear resistance at any cross section of a membernot requiring shear reinforcement can be calculated as

VRdc vRdcbwdwhere

bw is the minimum width of section in the tension zoned is the effective depth to the tension reinforcementvRdc is the design concrete shear stress

The design concrete shear stress is a function of the concretestrength the effective depth and the reinforcement percentageat the section considered To be effective this reinforcementshould extend a distance (lbd d) beyond the section wherelbd is the design anchorage length At a simple support for amember carrying predominantly uniform load the length lbd

may be taken from the face of the support The design shearresistance of members with and without axial load can bedetermined from the data given in Table 417

In the UK National Annex it is recommended that the shearstrength of concrete strength classes higher than C5060 isdetermined by tests unless there is evidence of satisfactorypast performance of the particular concrete mix including theaggregates used Alternatively the shear strength should belimited to that given for concrete strength class C5060

3312 Members with shear reinforcement

The design of members with shear reinforcement is based ona truss model in which the compression and tension chords

are spaced apart by a system of inclined concrete struts andvertical or inclined shear reinforcement Angle between thereinforcement and the axis of the member should be 45o

Angle 13 between the struts and the axis of the member maybe selected by the designer within the limits 10 cot13 25generally However for elements in which shear co-exists withexternally applied tension cot13 should be taken as 10 Theweb forces are Vsec13 in the struts and Vsec in the shearreinforcement over a panel length l z (cot cot13) wherez may normally be taken as 09d The width of each strutis equal to z (cot cot13) sin13 and the design value of themaximum shear force VRdmax is limited to the compressiveresistance provided by the struts which includes a strengthreduction factor for concrete cracked in shear The least shearreinforcement is required when cot13 is such that V VRdmaxThe truss model results in a force Ftd in the tension chord thatis additional to the force Mz due to bending but the sumFtd Mz need not be taken greater than Mmaxz where Mmax isthe maximum moment in the relevant hogging or sagging regionThe additional force Ftd can be taken into account by shiftingthe bending moment curve each side of any point of maximummoment by an amount al 05z(cot13 cot) For memberswithout shear reinforcement al d should be used The cur-tailment of the longitudinal reinforcement can then be based onthe modified bending moment diagram A design procedure todetermine the required area of shear reinforcement and detailsof the particular requirements for beams and slabs are given inTable 418

For most beams a minimum amount of shear reinforcementin the form of links is required irrespective of the magnitude ofthe shear force Thus there is no need to determine VRdc

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EC 2 Shear resistance ndash 1 417

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EC 2 Shear resistance ndash 2 418

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In members with inclined chords the shear components ofthe design forces in the chords may be added to the design shearresistance provided by the reinforcement In checking that thedesign shear force does not exceed VRdmax the same shearcomponents may be deducted from the shear force resultingfrom the design loads

3313 Shear under concentrated loads

In slabs and column bases the maximum shear stress at theperimeter of a concentrated load should not exceed vRdmaxShear in solid slabs under concentrated loads can result inpunching failures on the inclined faces of truncated conesor pyramids For design purposes a control perimeterforming the shortest boundary that nowhere comes closer tothe perimeter of the loaded area than a specified distanceshould be considered The basic control perimeter maygenerally be taken at a distance 2d from the perimeter ofthe loaded area

If the maximum shear stress here is no greater than vRdc noshear reinforcement is required Otherwise the position of thecontrol perimeter at which the maximum shear stress is equalto vRdc should be determined and shear reinforcement providedin the zone between this control perimeter and the perimeter ofthe loaded area

For flat slabs with enlarged column heads (or drop panels)where dH is the effective depth at the face of the column and thecolumn head (or drop) extends a distance lH 2dH beyond theface of the column a basic control perimeter at a distance 2dH

from the column face should be considered In addition a basiccontrol perimeter at a distance 2d from the column head (ordrop) should be considered

Control perimeters (in part or as a whole) at distances lessthan 2d should also be considered where a concentrated loadis applied close to a supported edge or is opposed by a highpressure (eg soil pressure on bases) In such cases values ofvRdc may be multiplied by 2da where a is the distance fromthe edge of the load to the control perimeter For column basesthe favourable action of the soil pressure may be taken intoaccount when determining the shear force acting at the controlperimeter

Details of design procedures for shear under concentratedloads are given in Table 419

3314 Bottom loaded beams

Where load is applied near the bottom of a section sufficientvertical reinforcement to transmit the load to the top of thesection should be provided in addition to any reinforcementrequired to resist shear

332 DESIGN FOR TORSION

In normal beam-and-slab or framed construction calculationsfor torsion are not usually necessary adequate control of anytorsional cracking in beams being provided by the requiredminimum shear reinforcement When it is judged necessaryto include torsional stiffness in the analysis of a structure ortorsional resistance is vital for static equilibrium membersshould be designed for the resulting torsional moment

Design for torsion 365

The torsional resistance may be calculated on the basis ofa thin-walled closed section in which equilibrium is satisfiedby a plastic shear flow A solid section may be modelled asan equivalent thin-walled section Complex shapes may bedivided into a series of sub-sections each of which is mod-elled as an equivalent thin-walled section and the totaltorsional resistance taken as the sum of the resistances of theindividual elements When torsion reinforcement isrequired this should consist of rectangular closed linkstogether with longitudinal reinforcement Such reinforcementis additional to the requirements for shear and bendingDetails of a suitable design procedure for torsion are givenin Table 420

Example 1 The beam shown in the following figure whichwas designed for bending in example 1 of Chapter 32 is to bedesigned for shear The maximum design load is 495 kNm andthe design is based on the following values

fck 32 MPa fywk 500 MPa d 440 mm

Since the load is uniformly distributed the critical sectionfor shear may be taken at distance d from the face of the sup-port Based on a support width of 400 mm distance fromcentre of support to critical section 200 440 640 mmAt end B

V 248 064 495 216 kNw V[bw z (1 fck250)fck]

216 103[300 09 440 (1 32250) 32] 0065

From Table 418 since w 0138 cot13 25 may be usedHence area of links required is given by

Asws Vfywd z cot13 216 103(087 500 09 440 25) 050 mm2mm

From Table 420 H8-200 provides 050 mm2mm

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EC 2 Shear under concentrated loads 419

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EC 2 Design for torsion 420

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Shear and torsion368

Minimum requirements for vertical links are given by

Asws (008radicfck) bw fyk (008radic32) 300500 027 mm2mm

s 075d 075 440 330 mm

From Table 420 H8-300 provides 033 mm2mm

VRds (Asws) fywd z cot13 033 087 500 09 440 25 103

142 kN

At end A V 160 064 495 128 kN ( VRds 142 kN)

Example 2 A 250 mm thick flat slab is supported by 400 mmsquare columns arranged on a 72 m square grid The slab con-tains as tension reinforcement in the top of the slab at an inte-rior support within a 18 m wide strip central with the columnH16-150 in each direction Lateral stability of the structuredoes not depend on frame action and the design shear forceresulting from the maximum design load applied to all panelsadjacent to the column is V 854 kN

fck 40 MPa fywk 500 MPa d 210 mm (average)

Since the lateral stability of the structure does not depend onframe action may be taken as 115 (Table 419)

Maximum shear stress adjacent to the column face

Vuod 115 854 103(4 400 210) 293 MPavRdmax 02 (1 fck250)fck

02 (1 40250) 40 672 MPa (293)

Based on H16-150 as effective tension reinforcement

100Aslbwd 100 201(150 210) 064vRdc 070 MPa (Table 417 for d 210 and fck 40)

The length of the first control perimeter at 2d from the faceof the column is 4 400 4 d 4239 mm Thus themaximum shear stress at the first control perimeter

Vu1d 115 854 103(4239 210) 110 MPa

Since v vRdc shear reinforcement is needed where effectivedesign strength fywdef 250 025d 300 MPa The areaneeded in one perimeter of vertical shear reinforcementat maximum radial spacing sr 075d 150 mm say isgiven by

Asw (v 075vRdc) u1 sr 15 fywdef

(110 075 070) 4239 150(15 300) 813 mm2

wmin u1 sr 15 (008radicfck) u1 sr 15fyk

(008radic40) 4239 150(15 500) 429 mm2

Using 12H10 gives 942 mm2

Length of control perimeter at which v vRdc is given by

u Vd vRdc 115 854 103(210 070) 6681 mm

Distance of this control perimeter from face of column is

a (6681 4 400)2 809 mm

The distance of the final perimeter of reinforcement from thecontrol perimeter where v vRdc should be 15d 315mm

Thus 4 perimeters of reinforcement with sr 150 mm and thefirst perimeter at 100 mm from the face of the column wouldbe suitable The reinforcement layout is shown in the followingfigure where indicates the link positions and the links canbe anchored round the tension bars

Example 3 The following figure shows a channel sectionedge beam on the bottom flange of which bear 8 m long simplysupported contiguous floor units The beam is continuous in14 m spans and is prevented from lateral rotation at thesupports The centroid and the shear centre of the sectionare shown

Characteristic loads

floor units dead 35 kNm2 imposed 25 kNm2

edge beam dead 12 kNm

Design ultimate loads

floor units (135 35 15 25) 82 338edge beam 135 12 162

500 kNm

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Design for torsion 369

fck 32 MPa fyk 500 MPa d 1440 mm

Bending moment shear force and torsional moment (aboutshear centre of section) at interior support (other than first)

M 009 50 142 882 kNm (Table 229)V 05 50 14 350 kNT 05 (338 0400 162 0192) 14 117 kNm

(Note In calculating V and T a coefficient of 05 rather than055 has been used since the dead load is dominant and thecritical section may be taken at the face of the support)

Considering beam as one large rectangle of size 250 1500and two small rectangles of size 200 300

13hmin3hmax 2503 1500 2 2003 300

(234 2 24) 109 282 109

Torsional moment to be considered on large rectangle

T1 117 234282 97 kNm

Torsional moment to be considered on each small rectangle

T2 117 24282 10 kNm

Reinforcement required in large rectangle

Shear and torsion (see Table 420) Assuming 30 mm cover toH10 links distance from surface of concrete to centre of H12longitudinal bars 46 mm

tefi Au 250 1500[2 (250 1500)] 107 mm ( 2 46 92 mm)

Ak (250 107) (1500 107) 1992 103 mm2

For values of (1 fck250)fck (1 32250) 32 279 MPaand z 1440 100 1340 mm (to centre of flange)

w [T12Ak tefi Vbw z](1 fck250)fck

[97(2 1992 107) 350(250 1340)] 103279 0119

Since w 0138 cot13 25 may be used (Table 418)For a system of closed links total area required in two legs fortorsion and shear is given by

Ass (T1Ak V z)fywd cot13 (971992 3501340) 103(087 500 25) 069 mm2mm

The inner legs of the links are also subjected to a vertical tensileforce resulting from the load of 338 kNm applied by the floorunits Additional area required in inner legs

Ass 338(087 500) 008 mm2mm

Total area required in two legs for torsion shear and theadditional vertical tensile force

Ass 069 2 008 085 mm2mm

The area of longitudinal reinforcement required for torsion isgiven by

Aslsl Tcot13 2Akfyd

97 103 25 (2 1992 087 500) 140 mm2mm

Different combinations of links and longitudinal bars can beobtained by changing the value of cot 13 as follows

Bars Links Longitudinal

Ass Size and Aslsl Size andcot 13 mm2mm spacing mm2mm spacing

25 085 H10-175 140 H12-15020 102 H10-150 112 H12-20016 124 H10-125 090 H12-250

s least of u8 35008 4375 mm 075d 1080 mmor hmin 250 mm sl 350 mm

Bending (see Table 48)

Mbd2fck 882 106(550 14402 32) 0024Asfykbdfck 0027 and xd 0054 (ie x 78 200 mm)

As 0027 550 1440 32500 1369 mm2

Total area of longitudinal bars required at top of beam forbending and torsion (equivalent to 2H12 say)

1369 226 1595 mm2

From Table 228 2H32 provides 1608 mm2

Reinforcement required in small rectangles

Torsion Assuming 30 mm cover to H8 links distance fromsurface of concrete to centre of H12 longitudinal bars 44 mm

tefi Au 200 300[2 (200 300)] 60 ( 2 44 88 mm)

Ak (200 88) (300 88) 237 103 mm2

w (T22Ak tefi)(1 fck250)fck

10 103(2 237 88 279) 0086

Since w 0138 cot 13 25 may be used For a system ofclosed links area required in two legs is given by

Asts T2Ak fywd cot13 10 103(237 087 500 25) 039 mm2mm

The lower rectangle is also subjected to bending resulting fromthe load of 338 kNm applied by the floor units The distanceof the load from the centre of the inner leg of the links in thelarge rectangle is 150 35 185 mm

M 338 0185 625 kNm

Taking the lever arm for the small rectangle as the distancebetween the centres of the top and bottom arms of the linksz 132 mm Additional area required in top arms of links

Ass Mfyd z 625 103(087 500 132) 011 mm2mm

Total area required in two arms for torsion and bending

Ass 039 2 011 061 mm2mm

The area of longitudinal reinforcement required for torsion isgiven by

Aslsl Tcot132Ak fyd

10 103 25(2 237 087 500) 121 mm2mm

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Shear and torsion370

Different combinations of links and longitudinal bars can beobtained by changing the value of cot13 as follows

Bars Links Longitudinal

Ass Size and Aslsl Size andcot 13 mm2mm spacing mm2mm spacing

25 061 H10-150 121 H12-17518 076 H10-125 088 H12-250

s lesser of u8 10008 125 mm or hmin 200 mm

The lower rectangle is also subjected to shear in the verticallongitudinal plane for which

Vbw d 338 103(1000 166) 021 MPa

From Table 417 vmin 056 MPa (fck 32 d 200)

From the foregoing calculations the reinforcement shown inthe figure opposite provides a practical arrangement in whichthe links comprise H10-125 for the large rectangle andH8-125 for the small rectangles The longitudinal bars are all

H12-250 apart from the 2H32 bars at the top of the largerectangle

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341 DEFLECTION

Deflections of members under service load should not impairthe function or appearance of a structure For buildings thedesign requirements and associated combinations of designactions to be considered are given in section 293 The finaldeflection of members below the level of the supports undercharacteristic loading after allowance for any pre-camber isgenerally limited to span250 A further limit of span500applies to the increase in deflection that occurs after theconstruction stage in order to minimise any damage to bothstructural and non-structural elements The requirements maybe met by complying with the limits on spaneffective depthratio given in Table 421

In special circumstances when the calculation of deflectionis considered necessary an adequate prediction can be madeusing the methods given in Table 422 Careful considerationis needed in the case of cantilevers where the usual formulaeassume that the cantilever is rigidly fixed and remains hori-zontal at the root Where the cantilever forms the end of acontinuous beam the deflection at the end of the cantilever islikely to be either increased or decreased by an amount l13where l is the length of the cantilever measured to the centreof the support and 13 is the rotation at the support Where acantilever is connected to a substantially rigid structure someroot rotation will still occur and the effective length shouldbe taken as the length to the face of the support plus half theeffective depth

342 CRACKING

3421 Building structures

Cracks in members under service load should not impair theappearance or durability of the structure For buildings thedesign requirements are given in Table 41 The calculatedcrack width under quasi-permanent loading or as a result ofrestrained deformations is generally limited to 03 mm For drysurfaces inside buildings where crack width has no effect ondurability a limit of 04 mm is recommended where there isa need to ensure an acceptable appearance However in theUK National Annex a limit of 03 mm is required in this situa-tion In the regions of concrete members where tension isexpected a calculated minimum amount of reinforcement isneeded in order to control cracking as given in Table 423

Where minimum reinforcement is provided the crack widthrequirements may be met by direct calculation or by limitingeither the bar size or the bar spacing as given in Table 424For the calculation of crack widths due to restrained imposeddeformation no guidance is given in Part 1 of the code but thefollowing equation is given in PD 6687 (see preface)

sm ndash cm 08Rimp

where R is a restraint factor (see section 2621) and imp is thefree strain due to temperature fall or drying shrinkage

3422 Liquid-retaining structures

For structures containing liquids design requirements related toleakage considerations are given in section 294 Where a smallamount of leakage with related surface staining or damppatches is acceptable for cracks that can be expected to passthrough the full thickness of the section the calculated crackwidth is limited to a value that varies according to the hydraulicgradient (ie head of liquid divided by thickness of section)The limits are 02 mm for hydraulic gradients 5 reducinguniformly to 005 mm for hydraulic gradients 35 Thus thelimits for a 300 mm thick wall to a 75 m deep tank would be02 mm at 15 m below the top 015 mm at 45 m below the topand 01 mm at 75 m below the top The limits apply under thequasi-permanent loading combination where the full charac-teristic value is taken for hydrostatic loading For members inaxial tension where at least the minimum reinforcement isprovided the crack width requirements may be met by directcalculation or by limiting either the bar size or the bar spacingas given in Table 425

In cases of bending with or without axial force where thefull thickness of the section is not cracked and not less than 02times the section thickness 50 mm is in compression thecrack width limit is 03 mm and Table 424 applies

For cracking due to restraint of imposed deformations suchas shrinkage and early thermal movements two distinct typesof restraint are considered For a concrete element restrained atits ends (eg an infill bay with construction joints between thenew section of concrete and the pre-existing sections) the crackformation is similar to that caused by external loading Anappropriate expression for the tensile strain contributing to thecrack width is given in Table 425 and for specified values of

Chapter 34

Deflection and cracking

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EC 2 Deflection ndash 1 421

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EC 2 Deflection ndash 2 422

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EC 2 Cracking ndash 1 423

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EC 2 Cracking ndash 2 424

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EC 2 Cracking ndash 3 425

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Cracking 377

cover section thickness and reinforcement content maximumvalues of fcteff are given in Table 426

For a concrete panel restrained along an edge (eg a wall castonto a pre-existing stiff base) the formation of the crack onlyinfluences the distribution of stresses locally and the crackwidth becomes a function of the restrained strain rather than thetensile strain capacity of the concrete In this case the tensilestrain contributing to the crack width is taken as Rax free wheretypical values of free can be estimated from the informationgiven in Table 425 The restraint factor Rax may be taken as 05generally or reference can be made to Table 345 where valuesare indicated for panels restrained along one two or three edgesrespectively For specified values of cover section thicknessand reinforcement content maximum values of Rax free aregiven in Table 427

It will be found that the calculated strain contributing to thecrack width for a panel restrained at its ends is normally morethan Rax free Thus the reinforcement required to limit a crackwidth to the required value is greater for a panel restrained atits ends than for a panel restrained along one or more edgesAlso for a specific crack width the reinforcement needed for apanel restrained along an edge is less than that in BS 8007since the design crack spacing is less than that in BS 8007

Example 1 The beam shown in the following figure is to bechecked for deflection and cracking The design for bendingand shear is shown in example 1 of Chapters 32 and 33 respec-tively The reinforcement in the bottom of each span 3H25(1473 mm2) is based on the following values

fck 32 MPa fyk 500 MPa beff 2600 mm d 440 mm

From Table 423 for fck 32 MPa and bending of a sectionwith h 500 mm by interpolation 100AsminAct 021

Asmin 00021 162 103 340 mm2 ( 1473 mm2)

026 ( f ctmfyk ) btd 00013 btd

026 (30500) 300 440 206 mm2

The design ultimate load is 396 kN and the quasi-permanentload where the value of 2 is obtained from section 293 is

Gk 2 Qk 160 03 120 196 kN

Hence the stress in the reinforcement under quasi-permanentloading is given approximately by

s (196396)(087fyk)(As req As prov)

Thus for the bars in the bottom of the beam

s (196396)(087 500)(13731473) 200 MPa

From Table 424 for wk 03 mm the crack width criterion ismet if

s 25 mm or the bar spacing 250 mm

The adjusted maximum bar size is given by

Depth of tension zone hcr h ndash x 500 ndash 128 372 mm

s s (fcteff 29)[kc hcr 2(h ndash d)]

25 (3029) [04 372(2 60)] 32 mm

Note It can be seen from the foregoing that all of the criteriaare comfortably satisfied and the checks for deflection andcracking are hardly necessary in this example

Example 2 A 250 mm thick flat slab is supported bycolumns which are arranged on a 72 m square grid Thecharacteristic loads are 72 kNm2 dead and 45 kNm2 imposedand the slab is to be checked for deflection and cracking

fck 32 MPa fyk 500 MPa cover to bars 25 mm

Allowing for the use of H12 bars in each direction and basedon the bars in the second layer of reinforcement

d 250 ndash (25 12 6) 205 mm ld 7200205 35

From Table 421 for a flat slab with spans 85 m

Basic spaneffective depth ratio 24

Total design ultimate load for a square panel is given by

F (135 72 15 45) 722 854 kN

From Table 262 the design ultimate bending moment for anend span with a continuous connection at the outer support is

M 0075Fl 0075 854 72 461 kNm

Mbd 2fck 461 106(7200 2052 32) 0048

As fykbdfck 0056 (Table 48)

100Asbd 100 0056 32500 036

From Table 421 for 100Asbd 01fck05 01 3205 057

s 055 00075fck(100Asbd)

0005fck05[ fck

05(100Asbd) ndash 10]15

055 00075 32036

0005 3205 (3205036 ndash 10)15 160

The actual spaneffective depth ratio 8000440 182

From Table 421 for the end span of a continuous beam and aflanged section with bbw 2600300 867 3

Basic spaneffective depth ratio 08 26 208

For members supporting partitions liable to be damaged byexcessive deflections the basic ratio should be multiplied by7span In this case the basic ratio 208 78 182

Since 100As reqbd 100 1373(2600 440) 012 is smallthe modification factor s is large ( 3) and the limiting ratiois more than three times the actual value

The neutral axis depth for the uncracked section ignoring theeffect of the reinforcement is given by

x

128 mm ( hf 150 mm)

Area of tension zone is given by

Act bw (h ndash hf) bf (hf ndash x) 300 350 2600 22 162 103 mm2

300 5002 2300 1502

2[300 500 2300 150]bwh2 (bf ndash bw)h2

f

2[bwh (bf ndash bw)hf]

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EC 2 Early thermal cracking in end restrained panels 426

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EC 2 Early thermal cracking in edge restrained panels 427

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Deflection and cracking380

Limiting spaneffective depth ratio 16 24 384 ( 35)

From Table 423 for fck 32 MPa and bending of a sectionwith h 300 mm 100AsminAct 024 With Act 05bh

100Asminbd 024 05 250205 015

026 (fctmfyk) 00013

026 30500 016 ( 036)

The design ultimate load is 854 kN and the quasi-permanentload where the value of 2 is obtained from section 293 is

Gk 2 Qk (72 03 45) 722 443 kN

Hence the stress in the reinforcement under quasi-permanentloading is given approximately by

s (443854)(087fyk) (443854)(087 500) 226 MPa

From Table 424 for wk 03 mm the crack width criterion ismet if

s 16 mm or the bar spacing 200 mmThe adjusted maximum bar size with hcr 05h is given by

s s (fcteff 29)[kc hcr 2(h ndash d)]

16 (3029) [04 125(2 45)] 9 mm

Area of reinforcement required to give 100Asbd 036 is

As 00036 1000 205 738 mm2m (H12-150)

Example 3 The wall of a cylindrical tank 75 m deep and15 m diameter is 300 mm thick The wall which is continuouswith the base slab is to be designed for temperature effects andthose due to internal hydrostatic pressure when the tank is fullof liquid

Design class 1 (see section 294) fyk 500 MPaCover to horizontal bars 40 mm fck 32 MPa

Effects of temperature change With fctm 03 fck(23) 30 MPa

and assuming that early thermal cracks will occur at a timewhen fcm(t) 24 MPa

fcteff [fcm(t)(fck 8)] fctm [24(32 8)] 30 18 MPa

The limiting crack width varies according to the hydraulicgradient (depth of liquid thickness of section) If the wall isdesigned to the recommendations for a panel restrained atits ends then suitable reinforcement details for 40 mmcover and fcteff 18 MPa selected from Table 426 aregiven here

Depth Hydraulic Design crack Reinforcement(m) gradient width (mm) required (EF)

15 5 02 H20-15045 15 015 H20-12575 25 01 H20-100

Note The table for wk 02 mm was used throughout bytaking effective values of fcteff 18(5wk) MPa

If the wall is designed to the recommendations for a panelrestrained along the edge the restrained tensile strain needs tobe estimated as follows

Allowing for concrete grade C3240 with 350 kgm3 Portlandcement at a placing temperature of 20oC and a mean ambient

temperature during construction of 15oC the temperature risefor concrete placed within 18 mm plywood formwork

T1 25oC (Table 219)

As the wall is to be designed to resist hoop tension there willbe no vertical movement joints and allowance must be made fora fall in temperature due to seasonal variations Allowing forT2 15oC restraint factor Rax taken as 05 and coefficient ofthermal expansion taken as 12 10ndash6 per oC (Table 35)restrained total thermal contraction after the peak temperaturearising from hydration effects is given by

Rax (T1 T2 ) 05 12 10ndash6 (25 15) 240 10ndash6

Hence suitable reinforcement details for 40 mm cover andRax 240 10ndash6 selected from Table 427 are given here

Depth Hydraulic Design crack Reinforcement(m) gradient width (mm) required (EF)

15 5 02 H20-25045 15 015 H20-15075 25 01 H20-100

Note The table for wk 02 mm was used throughout foreffective values of Rax [240(5wk)] 10ndash6 For H20 barsvalues for a 250 mm thick section apply for h 250 mm

From Table 423 the minimum reinforcement percentage forfcteff fctm in the case of a rectangular section in pure tensionwith h 300 mm and fck 32 MPa is 060 For the control ofearly thermal cracking the value is (1830) 06 036 andthe minimum area of reinforcement required on each face

Asmin 00036 150 1000 540 mm2m (H16-300)

Clearly the reinforcement needed for thermal crack controlgreatly exceeds this minimum requirement In the lower part ofthe wall the reinforcement provided is H20-100 (EF) and thecorresponding stress at a cracked section is

s fcteff (Act As) 18 150 10003142 86 MPa

This solution can be checked approximately by reference to thechart for maximum bar size on Table 425 as follows

For s 86 MPa and wk 01 mm s 55 mm say

s s

55 (1829) (01 30050) 20 mm

Effects of hydrostatic load Suppose that an elastic analysis ofthe tank assuming a floor 300 mm thick indicates a servicemaximum circumferential tension of 400 kNm This valueoccurs at a depth of 6 m where the design crack width is0125 mm Above this level the tensions can be assumed toreduce approximately linearly to near zero at the top of the wall

For a section reinforced with H20-100 (EF) the stress in thereinforcement s 400 1036284 64 MPa Since this isless than the stress due to fcteff the reinforcement needed forthermal crack control is also sufficient for the circumferentialtension This solution can be checked also by reference to thecharts on Table 425 which show that with s 64 MPa thebar size and the bar spacing are of no consequence

fcteff

2901h(hd)

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The code contains many requirements that affect the details ofthe reinforcement such as minimum and maximum areas tyingprovisions anchorage and curtailment

Bars may be set out individually or grouped in bundles oftwo or three in contact Bundles of four bars may also be usedfor vertical bars in compression and for bars in a lapped jointFor the safe transmission of bond forces the cover providedto the bars should be not less than the bar diameter or for abundle the equivalent diameter ( 55 mm) of a notional barwith the same sectional area as the bundle Requirements forcover with regard to durability are given in Chapter 31 Gapsbetween bars (or bundles) generally should be not less than thegreatest of (dg 5 mm) where dg is the maximum aggregatesize the bar diameter (or equivalent bar diameter for a bundle)or 20 mm Details of reinforcement limits are given in Table 428

At intermediate supports of continuous flanged beams thetotal area of tension reinforcement should be spread over theeffective width of the flange but a part of the reinforcementmay be concentrated over the web width

351 TIES IN STRUCTURES

Structures not specifically designed to withstand accidentalactions should be provided with a suitable tying system toprevent progressive collapse by providing alternative loadpaths after local damage Where the structure is divided intostructurally independent sections each section should have anappropriate tying system The reinforcement providing the tiesmay be assumed to act at its characteristic strength and onlythe specified tying forces need to be taken into accountReinforcement required for other purposes may be consideredto form part of or the whole of the ties Details of the tyingrequirements as specified in the UK National Annex are givenin Table 429

352 ANCHORAGE AND LAP LENGTHS

At both sides of any cross section bars should be providedwith an appropriate embedment length or other form of endanchorage For bent bars the basic tension anchorage length ismeasured along the centreline of the bar from the section inquestion to the end of the bar where

lbd 1 2 3 4 5 lbrqd lbmin

As a simplified alternative a tension anchorage for a standard bend hook or loop may be provided as an equivalentlength lbeq 1 lbrqd (see figure here) where 1 is taken as07 for covers perpendicular to the bend 3 Otherwise1 10

Chapter 35

Considerationsaffecting design details

Bends or hooks do not contribute to compression anchoragesDetails of anchorage lengths are given in Table 430

Laps should be located if possible away from positions ofmaximum moment and should generally be staggered Detailsof lap lengths are given in Table 431

The radius of any bend in a reinforcing bar should conformto the minimum requirements of BS 8666 and should ensurethat failure of the concrete inside the bend is prevented A linkmay be considered fully anchored if it passes round another barof not less than its own diameter through an angle of 90o andcontinues beyond the end of the bend for a minimum length of10 diameters 70 mm Details of bends in bars are given inTable 431 Additional rules for large diameter bars ( 40 mmaccording to the UK National Annex) and bundles are given inTable 432

353 CURTAILMENT OF REINFORCEMENT

In flexural members it is generally advisable to stagger thecurtailment points of the tension reinforcement as allowed bythe bending moment envelope Bars should be curtailed inaccordance with the rules set out in Table 432 and illustratedin the figure on page 387 Except at end supports every tensionbar should extend beyond the point at which in theory it is nolonger needed for flexural resistance for a distance not less thanal The bar should also extend beyond the point at which it isfully required to provide flexural resistance for a distance notless than al lbd At a simple end support the bars shouldextend for the anchorage length lbd necessary to develop theforce ∆Ftd

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EC 2 Reinforcement limits 428

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EC 2 Provision of ties 429

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EC 2 Anchorage requirements 430

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EC 2 Laps and bends in bars 431

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EC 2 Rules for curtailment large diameter bars and bundles 432

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Curtailment of reinforcement 387

Example 1 The beam shown in the following figure which wasdesigned in example 1 of Chapters 32 (bending) and 33 (shear)is to be checked for the reinforcement details The designultimate loads are Fmax 396 kN and Fmin 216 kN on eachspan The width of each support is 400 mm and the mainreinforcement is as follows spans 3H25 (bottom) support B3H32 (top) and 2H25 (bottom)

fck 32 MPa fyk 500 MPa cover to links 35 mm

For bars in the bottom of the section the bond condition isclassified as lsquogoodrsquo Thus from Table 430

lbrqd 35 (s435) 35 25 (172435) 346 mm

The design anchorage length is given by

lbd 1 2 3 4 5 lbrqd lbmin

Coefficients 1 and 2 depend on cd which is taken as either thecover to the main bar or half the gap between the main barswhichever is the lesser With 35 mm cover to H8 links coverto main bars is 45 mm and the gap between the bars is300 2 (45 25) 160 mm Hence cd 45 mm (or 18)

Since cd 3 1 10 for both bent bars and straight barsHence using the simplified approach (see section 352) forboth straight bars or standard bends lbeq 1 lbrqd 346 mm

For a 400 mm wide support allowing for 50 mm end cover tothe bars the anchorage length provided from the near face ofthe support is 350 mm ( lbeq)

For the basic approach the following values can be obtained

For straight bars 2 1 015(cd 1) 088

With no transverse reinforcement provided within the bearinglength 3 10 and 4 10

Transverse pressure due to reaction on support is given by

p V(bearing area) 150 103(400 300) 125 MPa

Hence with p 125 MPa 5 1 004p 095 and

lbd 25lbrqd 088 095 346 290 mm lbmin 10 250 mm ( 03lbrqd or 100 mm)

Curtailment points for bottom bars The resistance momentprovided by 2H25 at the bottom of the beam may be determinedas follows

As fykbdfck 982 500(2600 440 32) 00134Mbd2fcu 0012 (Table 48 or section 3221)M 0012 2600 4402 32 106 193 kNm

Illustration of the curtailment of longitudinal reinforcement taking account of resistance within anchorage lengths

End anchorage At the bottom of each span 2H25 ( 25 ofarea provided in the span) will be continued to the support Atthe end support the tensile force to be anchored isF 05Vcot13 in which 13 is the inclination of the concrete strutrequired for shear design In the shear design calculations inchapter 33 V 128 kN at the critical section and VRds 142 kNwhen cot13 25 Thus cot13 (VVRds) 25 225 could beused At the face of the support V 160 02 495 150 kNWith cot 13 225

F 05 150 225 169 kNs FAs 169 103982 172 MPa

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Considerations affecting design details388

Reaction at A for load case 2 is RA 160 kN

Distance x from A to point where M 193 kNm is given by

RAx 05(FmaxL)x2 160x2 05 (3968)x2 193

Hence 05x2 323x 39 0 giving x 16 m and 485 m

Thus of the 3H25 required in the span one bar is no longerneeded for flexure at 16 m and 485 m from the end support Atthese points V 80 kN and cot13 (80142) 25 14 issufficient Thus the bar needs to extend beyond these points fora distance al 045d cot13 045 440 14 280 mm

Curtailment points for top bars The resistance moment providedby 2H32 can be determined as follows

As fykbdfck 1608 500(300 440 32) 0190Mbd2fcu 0142 (Table 48)M 0142 300 4402 32 106 264 kNm

For load case 1 reaction at A (or C) is given by

RA 05Fmax MBL 05 396 3968 148 kN

Distance x from A to point where M 264 kNm is given by

05(FmaxL)x2 RAx 05 (3968) x2 148 x 264

Hence 05x2 3x 53 0 giving x 74 m Thus of the3H32 required at B one bar is no longer required for flexure atdistance (80 74) 06 m from B If this distance is lessthan lb rqd the point of curtailment will be determined by theneed to develop the full force in the bar at B Here cot13 25giving al 045dcot13 1125d As the bars are effectively ina slab of thickness 250 mm it seems reasonable to assumelsquogoodrsquo bond conditions giving lb rqd 35 (Table 430) Thusdistance from B (edge of support say) at which one bar may becurtailed is al lb rqd 1125 440 35 32 1615 mm

Suppose that the remaining bars are continued to the point ofcontra-flexure in span BC for load case 2

The reaction at support C is given by

RC 05Fmin MBL 05 216 3068 70 kN

Distance from B to point of contra-flexure is given by

x L(1 2RCFmin) 8 (1 2 70216) 28 m

Here V 70 kN and cot13 (70142) 25 125 is sufficientThus distance from B at which the remaining bars may becurtailed is 2800 045 440 125 3050 mm

Link support bars say 2H12 could be used for the remainderof the span

Example 2 A typical floor to an 8-storey building consists ofa 250 mm thick flat slab supported by columns arranged ona 72 m square grid The slab for which the characteristicloading is 72 kNm2 dead and 45 kNm2 imposed is to beprovided with ties to the requirements of the UK NationalAnnex The design panel load is 854 kN and bending momentsare to be determined by the simplified method (see section 138)

fck 32 MPa fyk 500 MPa cover to bars 25 mm

Allowing for the use of H12 bars in each direction and basedon the bars in the second layer of reinforcement

d 250 (25 12 6) 205 mm say

From Table 255 the design ultimate sagging moment for aninterior panel is given by

M 0063Fl 0063 854 72 388 kNm

The total panel moment is to be apportioned between columnand middle strips where the width of each strip is 36 m Forthe column strip with 60 of the panel moment

M 06 388 233 kNmMbd2fck 233 106(3600 2052 32) 0048As fykbdfck 0056 (Table 48)As 0056 3600 205 32500

2645 mm2 (24H12-150 gives 2714 mm2)

For the middle strip with 40 of the panel moment

M 04 388 155 kNmMbd 2fck 155 106(3600 2052 32) 0032As fykbdfck 0037 (Table 48)As 0037 3600 205 32500

1748 mm2 (16H12-225 gives 1810 mm2)

For the peripheral tie the tensile force is given by

Ftieper (20 4no) 60 kN (20 4 8) 52 kN

The required area of reinforcement acting at its characteristicstrength is given by

As Ftieper fyk 52 103500 104 mm2 (1H12)

For the internal ties the tensile force is given by

kNm

With Ft (20 4no) 60 kN (20 4 8) 52 kN

kNm

If the internal ties are spread evenly in the slab the requiredarea of reinforcement acting at its characteristic strength

As 117 103500 234 mm2m (H12-450)

In this case at least every third bar in the column strips andevery other bar in the middle strips need to be continuous

If the internal ties are concentrated at the column lines the totalarea of reinforcement required in each group

As 234 72 1685 mm2 (16H12 gives 1810 mm2)

In this case the bars in the middle two-thirds of each columnstrip need to be continuous For sections 250 mm deep thebond condition is lsquogoodrsquo and lbrqd 35 (Table 430) Laps inadjacent pairs of lapped bars should be staggered by 13l0where l0 is the design lap length (Table 431) With lbd lbrqd

and 6 14 for one in two bars lapped at the same section

l0 6 lbd 14 35 12 600 mm say

Example 3 The following figure shows details of thereinforcement at the junction between a 300 mm wide beamand a 300 mm square column Bars 03 need to develop themaximum design stress at the column face and the radius of

Ftieint 72 4575 72

5 52 117

Ftieint gk qk

75 lr

5Ft Ft

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Curtailment of reinforcement 389

bend necessary to avoid failure of the concrete inside the bendis to be determined

fck 32 MPa fyk 500 MPa

The minimum radius of bend of the bars depends on the valueof ab where ab is taken as half the centre-to-centre distancebetween adjacent bars or for bars adjacent to the face of amember the side cover plus half the bar size Thus in this caseab 75 252 875 mm

From Table 431 for ab 87525 35 rmin 74 Thisvalue can be reduced slightly by taking into account the stressreduction in the bar between the edge of the support and thestart of the bend If r 7 distance from face of column tostart of bend 300 50 8 25 50 mm (ie 2)

From Table 430 for lsquogoodrsquo bond conditions the requiredanchorage length is 35 and rmin (1 235) 74 7Thus r 7 is sufficient

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Chapter 36

Foundations andearth-retaining walls

361 GEOTECHNICAL DESIGN

Eurocode 7 provides in outline all the requirements for thedesign of geotechnical structures It classifies structures intothree categories according to their complexity and associatedrisk but concentrates on the design of conventional structureswith no exceptional risk These include spread raft and pilefoundations retaining structures bridge piers and abutmentsembankments and tunnels Limit-states of stability strengthand serviceability need to be considered The requirements ofthe ULS and SLS may be satisfied by the following methodsalone or in combination calculations prescriptive measurestesting observational procedures The calculation methodadopted in the United Kingdom for the ULS requires theconsideration of two combinations of partial factors for actionsand soil parameters as shown here

satisfactory if the ratio of design ultimate bearing capacityto service load is 3 This approach is not valid for soft claysand settlement calculations should always be carried out insuch cases

362 PAD BASES

Critical sections for bending are taken at the face of a columnor the centre of a steel stanchion The design moment is takenas that due to all external loads and reactions to one side of thesection Punching resistance should be verified at controlperimeters within 2d from the column periphery For baseswithout shear reinforcement the design shear resistance is

vRd vRdc (2dav) vmin (2dav)

where av ( 2d) is the distance from the column periphery tothe control perimeter being considered The net applied forceVred V V where V is the applied column load and V isthe resulting upward force within the control perimeter Forconcentric loading the punching shear stress is v Vredudwhere u is the length of the control perimeter For eccentricloading the maximum shear stress is Vredud where is amagnification factor determined from equations given in EC 2At the column perimeter the punching shear stress should notexceed vRdmax 02(1 fck250)fck

Normal shear resistance should also be verified on verticalsections at distance d from the column face extending across thefull width of the base where the design shear resistance isvRdc vmin Alternatively it would be reasonable to check atsections within 2d from the column face using the enhanceddesign shear resistance given for punching shear In this casefor concentric loading the critical position for normal shear andpunching shear occurs at av a2 2d where a is the distancefrom the column face to the edge of the base For eccentricloading checks can be made at av 05d d and so on to findthe critical position

If the tension reinforcement is included in the determinationof vRdc the bars should extend for at least (d lbd) beyondthe section considered (see also EC 2 section 9822) If thetension reinforcement is ignored in the shear calculationsstraight bars will usually suffice

Example 1 A base is required to support a 600 mm squarecolumn subjected to vertical load only for which the values

Partial safety factors for the ULS

Safety factor Safety factor for

Combination on actions F soil parameters M

G Q c cu

1 135 15 10 10 102 10 13 125 125 14

If the action is favourable to the effect being considered values of G 10and Q 0 should be used Subscripts refer to the following soil parameters

is angle of shearing resistance (in terms of effective stress) and factor

is applied to tanc is cohesion intercept (in terms of effective stress)cu is undrained shear strength

Generally combination 2 determines the size of the structurewith regard to overall stability bearing capacity sliding andsettlement and combination 1 governs the structural design ofthe members The required size of spread foundations may bedetermined by ULS calculations using soil parameter valuesderived from the geotechnical design report for the projectAlternatively the size may be determined by limiting the bearingpressure under the characteristic loads to a prescribed value ora calculated allowable bearing pressure may be used For theSLS the settlement of spread foundations should be checked bycalculation or may in the case of firm to stiff clays be taken as

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are 4250 kN (service) and 6000 kN (ultimate) The allowableground bearing value is 300 kNm2 (kPa)

fck 32 MPa fyk 500 MPa nominal cover 50 mm

Allowing 10 kNm2 for ground floor loading and extra over soildisplaced by concrete the net allowable bearing pressure can betaken as 290 kNm2 Area of base required

Abase 4250290 147 m2 Provide base 40 m square

Distance from face of column to edge of base a 1700 mm

Taking depth of base 05a say h 900 mm

Allowing for 25 mm main bars average effective depth

d 900 (50 25) 825 mm

Bearing pressure under base due to ultimate load on column

pu 600042 375 kNm2

Bending moment on base at face of column

M pula2 2 375 4 1722 2168 kNm

Mbd2 fck 2168 106 (4000 8252 32) 0025

From Table 48 As fykbdfck 00285 and

As 00285 4000 825 32500 6020 mm2

From Table 220

13H25-300 gives 6380 mm2 and

100Asbh 100 6380(4000 900) 018 ( 013 min)

For values of Gk 2500 kN Qk 1750 kN and 2 03 thequasi-permanent load is

Gk 2Qk 2500 03 1750 3025 kN

Hence the stress in the reinforcement under quasi-permanentloading is given approximately by

s (30256000)(087fyk)(As reqAs prov)

(30256000)(087 500)(60206380) 207 MPa

From Table 424 for wk 03 mm the crack width criterion ismet if

s 23 mm or the bar spacing 240 mm

The adjusted maximum bar size with hcr 05h is given by

s s ( fcteff 29)[kc hcr2(h d )]

23 (3029) [04 450(2 75)] 28 mm

In this case H25-300 satisfies the bar size criterion

For members without shear reinforcement distance from faceof column to critical position for punching shear (or normalshear) is given by av 05a 850 mm where perimeter

u 4c 2av 4 600 2 850 7740 mm

Area of base inside critical perimeter is

Au 4avc av2 4 085 06 0852 431 m2

Hence the net applied force and resulting shear stress are asfollows

Vred V V 375 (42 431) 4384 kN

v Vud 4384 103(7740 825) 069 MPa

From Table 417 for fck 32 MPa and 100Aslbd 020 thedesign concrete shear stress vRdc is determined by vmin

vmin 0035k32 fck12 where k 1 (200d)12 20

k 1 (200825)12 149

vmin 0035 14932 3212 036 MPa

Hence design shear resistance

vRd vmin (2dav) 036 (2 825850) 070 MPa (v)

At the column perimeter ignoring the small reduction V

v Vud 6000 103(4 600 825) 303 MPa

vRd max 02 (1 fck250)fck

02 (1 32250) 32 558 MPa (v)

363 PILE-CAPS

A pile-cap may be designed by either bending theory or trussanalogy (ie strut-and-tie) In the latter case the truss is of atriangulated form with nodes at the centre of the loaded areaand at the intersections of the centrelines of the piles with thetension reinforcement as shown for compact groups of two tofive piles in Table 361 Expressions for the tensile forces aregiven taking into account the dimensions of the column andalso simplified expressions when the column dimensions areignored Bars to resist the tensile forces are to be located withinzones extending not more than 15 times the pile diameter eitherside of the centre of the pile The bars are to be provided witha tension anchorage beyond the centres of the piles Thecompression caused by the pile reaction may be assumedto spread at 45o angles from the edge of the pile and taken intoaccount when calculating the anchorage length The bearingstress on the concrete inside the bend in the bars should bechecked (see Table 431)

Example 2 A pile-cap is required for a group of 4 450 mmdiameter piles arranged at 1350 mm centres on a square gridThe pile-cap supports a 450 mm square column subjected to anultimate design load of 4000 kN

fck 32 MPa fyk 500 MPa

Allowing for an overhang of 150 mm beyond the face of thepile size of pile-cap 1350 450 300 2100 mm square

Take depth of pile-cap as (2hp 100) 1000 mm

Assuming tension reinforcement to be 100 mm up from base ofpile-cap d 1000 100 900 mm

Using truss analogy with the apex of the truss at the centre ofthe column the tensile force between adjacent piles is

Ft 750 kN in each zone

As Ft 087fyk 750 103(087 500) 1724 mm2

For a pile spacing three times pile diameter the bars may bespread uniformly across the cap and a total for two ties of8H25-250 (giving 3926 mm2) in each direction can be used

100Asbd 100 3926(2100 900) 020 ( 013 min)

4000 13508 900

Nl8d

Pile-caps 391

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The bars should be provided with a tension anchorage beyondthe centre of the piles (see Table 430) For 25 mm diameterbars spaced at 250 mm centres half the gap between bars givescb 1125 mm (ie 45 ) So for bent bars and lsquogoodrsquo bondconditions lbd 1lbrqd 07 35 25

Taking ab as either half the centre-to-centre distance betweenbars or (side cover plus half bar size) whichever is less

ab 2502 175 125 mm ab 12525 5

From Table 431 minimum radius of bend rmin 7 say

Consider the critical section for shear to be located at 20 ofthe pile diameter inside the pile-cap Distance of this sectionfrom the column face

av 05(l c) 03hp

05 (1350 450) 03 450 315 mm

Length of corresponding square perimeter for punching shear

u 4 (l 06hp) 4 (1350 06 450) 4320 mm

Since length of perimeter of pile-cap 4 2100 8400 mmis less than 2u normal shear extending across the full width ofthe pile-cap is more critical than punching shear

The contribution of the column load to the shear force may bereduced by applying a factor av2d where 05d av 2dSince avd 315900 035 (05) 025

v Vbd 025 2000 103(2100 900) 027 MPa

From Table 417 for fck 32 MPa and 100Aslbd 020 thedesign concrete shear stress vRdc is determined by vmin

vmin 0035k32fck12 where k 1 (200d)12 20

k 1 (200900)12 147

vmin 0035 14732 3212 035 MPa (v)

Shear stress calculated at perimeter of column

v Vud 4000 103(4 450 900) 247 MPa

vRd max 02 (1 fck250) fck

02 (1 32250) 32 558 MPa (v)

364 RETAINING WALLS ON SPREAD BASES

General notes on walls on spread bases are given in section732 For design purposes the characteristic soil parameterwhich is defined as a cautious estimate of the value affecting theoccurrence of the limit-state is divided by a partial safety factor(see section 361) Design values of the soil strength at the ULS(combination 2) are given by

and

where and are characteristic values of cohesion interceptand angle of shearing resistance (in terms of effective stress)

Design values for shear resistance at the interface of the baseand sub-soil of friction (for drained conditions) and adhesion(for undrained conditions) are given by

tan tan for cast in-situ concrete

tan tan (23) for precast concrete

cud cu14 where cu is undrained shear strength

dd

dd

c

cd c 125tan d ( tan ) 125

Walls should be checked for ULS of overall stability bearingresistance and sliding The resistance of the ground should bedetermined for both long-term (ie drained) and short-term(ie undrained) conditions where appropriate

For eccentric loading the ground bearing is assumed to beuniformly distributed and coincident with the line of action ofthe resultant applied load The traditional practice of usingcharacteristic actions and allowable bearing pressures to limitground deformation and check bearing resistance may also beadopted by mutual agreement This approach assumes a linearvariation of bearing pressure for eccentric loading and it is stillnecessary to consider the ULS for the structural design and tocheck sliding

The partial safety factors for the SLS are given as unity butit is often prudent to use the ULS for the active force (as inBS 8002) In this case suitable dimensions for the wall base canbe estimated with the aid of the chart given in Table 286 Herethe value pmax applies for a linear pressure variation and if theground pressure is uniform and centred on the centre of gravityof the applied load the contact length is (l) where dependson whether the solution is (a) above or (b) below the curve forlsquozero pressure at heelrsquo shown on the chart as follows

(a) 4(1 )3 23 and p 075 pmax

(b) 43 3(1 ) 23 and p (1 )l

For sliding the chart applies directly to non-cohesive soilsThus for bases founded on clay the long-term condition canbe investigated by using with c 0 For the short-termcondition the ratio does not enter into the calculations forsliding and is given by KA l2cd When has beendetermined from this equation the curve for radicKA on the chartcan be used to check the values of and that were obtainedfor the long-term condition

Example 3 A cantilever retaining wall on a spread base isrequired to support level ground and a footpath adjacent to aroad The existing ground may be excavated as necessary toconstruct the wall and the excavated ground behind the wallis to be reinstated by backfilling with a granular material Agraded drainage material will be provided behind the wall withan adequate drainage system at the bottom

Height of fill to be retained 40 m above top of base

Surcharge 5 kNm2

Properties of retained soil (well graded sand and gravel)unit weight of soil 20 kNm3

35o tan1 [(tan 35o)125] 29o

KA (1 sin )(1 sin ) 035

Properties of sub-base soil (medium sand)allowable bearing value fmax 200 kNm2 (kPa)

35o 29o (as fill)tan d tan 055

Take thickness of both wall (at bottom of stem) and base to beequal to (height of fill)10 400010 400 mm

Height of wall to underside of base l 40 04 44 m

Allowing for surcharge equivalent height of wall

44 520 465 mle l q

d

d

dd

d

Foundations and earth-retaining walls392

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200(20 465) 215

radicKA 055radic035 093

From Table 286 radicKA 0745 038 Hence

Width of base le (0745radic035) 465 21 m say

Toe projection (le) 038 21 08 m

Since the solution lies above the curve for lsquozero pressure atheelrsquo on the chart for uniform bearing centred on the centre ofgravity of the applied load

4(1 )3 4 (1 038)(3 215) 0385 andp 075 pmax 075 200 150 kNm2 (kPa)

Example 4 The sub-base for the wall described in example 3is a clay soil with properties as given below All other values areas specified in example 3

Properties of sub-base soil (firm clay)

allowable bearing value pmax 100 kNm2 (kPa)cu 50 kNm2 cud cu14 5014 35 kNm2

25o tan1 [(tan 25o)125] 205o

tan d tan 037

For the long-term condition

pmaxle 100(20 465) 108

radicKA 037radic035 0625

From Table 286 radicKA 107 025 Hence

Width of base le (107radic035) 465 30 m say

Toe projection (le) 025 30 08 m say

Since the solution lies below the curve for lsquozero pressure atheelrsquo on the chart for uniform bearing centred on the centre ofgravity of the applied load

43 3(1 ) 43 1083(1 025) 085

p (1 )l (1 025) 20 465085 82 kNm2

For the short-term condition

KA l2cud 035 20 465(2 085 35) 055

radicKA 055radic035 093

Since this value is less than that obtained for the long-termcondition the base dimensions are satisfactory

Example 5 The wall obtained in example 4 a cross sectionthrough which is shown below is to be designed accordingto EC 7

tan d

d

d

tan d

pmax le The vertical loads and bending moments about the front edgeof the base are

Load (kN) Moment (kNm)Surcharge 5 18 90 21 189Backfill 20 18 40 1440 21 3024Wall stem 24 04 40 384 10 384Wall base 24 04 30 288 15 432

Totals Fv 2202 Mv 4029

The horizontal loads and bending moments about the bottom ofthe base are

Load (kN) Moment (kNm)Surcharge 035 5 44 77 442 170Backfill 035 20 4422 678 443 994

Totals Fh 755 Mh 1164

Resultant moment Mnet 4029 1164 2865 kNm

Distance from front edge of base to resultant vertical force

a Mnet Fv 28652202 130 m

Eccentricity of vertical force relative to centreline of base

e 302 130 020 m ( 306 05 m)

Maximum pressure at front of base

pmax (220230)(1 6 02030) 103 kNm2

Minimum pressure at back of base

pmin (220230)(1 6 02030) 44 kNm2

For the ultimate bearing condition a uniform distribution isconsidered of length lb 2a 2 13 26 m giving

pu Fvlb 220226 85 kNm2

The ultimate bearing resistance is given by the equation

qu (2 ) cud ic where ic 05[1 ]

ic 05[1 ] 070

qu (2 ) 35 070 126 kNm2 ( pu 85)

Resistance to sliding (long-term)

Fv tan b 2202 037 815 kN ( Fh 755)

Resistance to sliding (short-term)

cud lb 35 26 91 kN ( Fh 755)

Example 6 The structural design of the wall in example 5 isto be in accordance with the requirements of EC 2

fck 32 MPa fyk 500 MPa nominal cover 40 mm

Allowing for H16 bars with 40 mm cover

d 400 (40 8) 352 mm

For the ULS (combination 1) F 135 for all permanentactions and M 10 Thus

35o and KA (1 sin )(1 sin ) 027

The ultimate bending moment at the bottom of the wall stem

M 135 027 (5 422 20 436) 924 kNmm

ddd

1 755 (35 26)

1 Fh (cudlb)

Retaining walls on spread bases 393

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(Note that for combination 2 F KA 10 035 which is lessthan F KA 135 027 0365 for combination 1)

Mbd 2fck 924 106(1000 3522 32) 0023

From Table 48 As fykbd fck 0026 and

As 0026 1000 352 32500 586 mm2m

From Table 220 H12-150 gives 754 mm2m

The ultimate shear force at the bottom of the wall stem

V 135 027 (5 4 20 422) 656 kNm

From Table 417 vmin 0035k32fck12 where

k 1 (200d)12 1 (200352)12 175 ( 20)

vmin 0035 17532 3212 046 MPa

Vbd 656 103(1000 352) 019 MPa ( vmin)

Since the loads are permanent the stress in the reinforcementunder service loading is given approximately by

s (087fykF)(As reqAs prov)

(087 500135)(586754) 250 MPa

From Table 424 for wk 03 mm the crack width criterion ismet if

s 15 mm or the bar spacing 185 mm

The adjusted maximum bar size with hcr 05h is given by

s s ( fcteff 29)[kc hcr 2(h d)]

15 (3029) [04 200(2 48)] 13 mm

In this case H12-150 meets both requirements

For the wall base the loads and bending moments calculatedfor combination 2 (see example 4) can be modified to suit therevised parameters for combination 1 as follows

Fv 135 2202 2973 kN

Mv 135 4029 5439 kNm

Mh (0365035) 1164 1214 kNm

Resultant moment Mnet 5439 1214 4225 kNm

Distance from front edge of base to resultant vertical force

a Mnet Fv 42252973 142 m

Bearing contact length lb 2a 2 142 284 m

pu Fvlb 2973284 1047 kNm2

Note that values of both pu and qu are greater for combination 1than for combination 2 but 2 is still critical for bearing

Bending moment on base at inside face of wall

M 135 (5 20 4 24 04) 1822

1047 (284 12)22 661 kNm

Shear force on base at inside face of wall

V 135 (5 20 4 24 04) 18

1047 (284 12) 582 kNm

The bending moment and shear force are both less than thevalues at the bottom of the wall stem Thus H12-150 can beused to fit in with the vertical bars in the wall

Foundations and earth-retaining walls394

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Appendix

Mathematicalformulae and data

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Mathematical formulae and data396

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References and furtherreading

1 Institution of Structural EngineersConcrete Society Standardmethod of detailing structural concrete a manual for best practiceLondon The Institution of Structural Engineers 2006 p 188

2 CP2 Civil Engineering Code of Practice No2 Earth retainingstructures London The Institution of Structural Engineers 1951p 224

3 The Highways Agency BD 3701 Loads for highway bridgesDesign manual for roads and bridges London HMSO 2001 p 118

4 The Highways Agency BD 6494 The design of highway bridgesfor vehicle collision loads Design manual for roads and bridgesLondon HMSO 1994 p 13

5 The Highways Agency BD 2101 The assessment of highwaybridges and structures Design manual for roads and bridges LondonHMSO 2001 p 84

6 The Highways Agency BD 5293 The design of highway bridgeparapets Design manual for roads and bridges London HMSO1993 p 44

7 Department of Transport BD 3087 Backfilled retaining walls andbridge abutments London Department of Transport 1987 p 12

8 Caquot A and Kerisel J Tables for calculation of passive pressureactive pressure and bearing capacity of foundations (translated fromFrench by M A Bec London) Paris Gauthier-Villars 1948 p 121

9 The Highways Agency BA 4296 The design of integral bridgesDesign manual for roads and bridges London HMSO 1996 p 10

10 Blackledge G F and Binns R A Concrete practice CrowthorneBritish Cement Association Publication 48037 2002 p 71

11 CIRIA Report 91 Early-age thermal crack control in concreteLondon CIRIA 1981 p 160

12 BRE Digest 357 Shrinkage of natural aggregates in concreteWatford BRE 1991 p 4

13 BRE Special Digest 1 Concrete in aggressive ground Six partsWatford BRE 2005

14 Concrete Society Technical Report No 51 Guidance on the use ofstainless steel reinforcement Slough The Concrete Society 1998 p 55

15 Coates R C Coutie M G and Kong F K Structural analysisSunbury-on-Thames Nelson 1972 p 496

16 Rygol J Structural analysis by direct moment distributionLondon Crosby Lockwood 1968 p 407

17 Westergaard H M Computation of stresses in bridge slabs due towheel loads Public Roads Vol 2 No 1 March 1930 pp 1ndash23

18 Pucher A Influence surfaces for elastic plates Wien andNew York Springer Verlag 1964

19 Bares R Tables for the analysis of plates and beams based onelastic theory Berlin Bauverlag 1969

20 Timoshenko S P and Woinowsky-Krieger S Theory of plates andshells (second edition) New York McGraw-Hill 1959 p 580

21 Sarkar R K Slab design ndash elastic method (plates) Munich VerlagUNI-Druck 1975 p 191

22 Wang P C Numerical and matrix methods in structural mechanicsNew York Wiley 1966 p 426

23 Jones L L and Wood R H Yield-line analysis of slabs LondonThames and Hudson 1967 p 405

This book written by leading UK experts is the best English languagetext dealing with yield-line theory (essential for designers using themethod frequently and for more than lsquostandardrsquo solutions)

24 Johansen K W Yield-line theory London Cement and ConcreteAssociation 1962 p 181

This is an English translation of the original 1943 text on whichyield-line theory is founded

25 Johansen K W Yield-line formulae for slabs London Cementand Concrete Association 1972 p 106

Gives design formulae for virtually every lsquostandardrsquo slab shape andloading (essential for practical design purposes)

26 Wood R H Plastic and elastic design of slabs and platesLondon Thames and Hudson 1961 p 344

Relates collapse and elastic methods of slab analysis but mainlyfrom the viewpoint of research rather than practical design

27 Jones L L Ultimate load analysis of reinforced and prestressedconcrete structures London Chatto and Windus 1962 p 248

About half of this easily readable book deals with the yield-linemethod describing in detail the analysis of several lsquostandardrsquo slabs

28 Pannell F N Yield-line analysis Concrete and ConstructionalEngineering JunendashNov 1966

Basic application of virtual-work methods in slab design June1966 pp 209ndash216

Economical distribution of reinforcement in rectangular slabs July1966 pp 229ndash233

Edge conditions in flat plates Aug 1966 pp 290ndash294

General principle of superposition in the design of rigid-plasticplates Sept 1966 pp 323ndash326

Design of rectangular plates with banded orthotropic reinforcementOct 1966 pp 371ndash376

Non-rectangular slabs with orthotropic reinforcement Nov 1966pp 383ndash390

29 Hillerborg A Strip method of design London Viewpoint 1975p 225

This book is the English translation of the basic text on the stripmethod (both simple and advanced) by its originator It dealswith theory and gives appropriate design formulae for manyproblems

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References and further reading398

30 Fernando J S and Kemp K O A generalised strip deflexionmethod of reinforced concrete slab design Proceedings of theInstitution of Civil Engineers Part 2 Research and Theory March1978 pp 163ndash174

31 Taylor R Hayes B and Mohamedbhai G T G Coefficientsfor the design of slabs by the yield-line theory Concrete 3(5) 1969pp 171ndash172

32 Munshi J A Rectangular concrete tanks (revised fifth edition)Skokie Illinois Portland Cement Association 1998 p 188

This is the most detailed book on the subject with complete tablesgiving moments shears and deflections for plates and tanks withuseful worked examples

33 CIRIA Report 110 Design of reinforced concrete flat slabs toBS8110 London CIRIA 1985 p 48

34 Beeby A W The analysis of beams in plane frames according toCP110 London Cement and Concrete Association Publication44001 1978 p 34

35 Rygol J Structural analysis by direct moment distributionLondon Crosby Lockwood 1968 p 407

36 Naylor N Side-sway in symmetrical building frames TheStructural Engineer 28(4) 1950 pp 99ndash102

37 Orton A The way we build now form scale and techniqueLondon E amp FN Spon 1988 p 530

38 CIRIA Report 102 Design of shear wall buildings LondonCIRIA 1984 p 80

39 Eurocode 8 Design of structures for earthquake resistanceBrussels European Committee for Standardization 2004

40 Penelis G G and Kappos A J Earthquake-resistant concretestructures London E amp FN Spon 1997 p 572

41 Kruger H G Crack width calculation to BS 8007 for combinedflexure and direct tension The Structural Engineer 80(18) 2002pp 18ndash22

42 Kong F K Robins P J and Sharp G R The design of reinforcedconcrete deep beams in current practice The Structural Engineer53(4) 1975 pp 73ndash80

43 Ove Arup and Partners The design of deep beams in reinforcedconcrete CIRIA Guide No 2 1977 p 131

44 Concrete Society Technical Report No 42 Trough and wafflefloors Slough The Concrete Society 1992 p 34

45 Gibson J E The design of shell roofs (Third edition) LondonE amp FN Spon 1968 p 300

46 Chronowicz A The design of shells London Crosby Lockwood1959 p 202

47 Tottenham H A A simplified method of design for cylindricalshell roofs The Structural Engineer 32(6) 1954 pp 161ndash180

48 Bennett J D Empirical design of symmetrical cylindricalshells Proceedings of the colloquium on simplified calculationmethods Brussels 1961 Amsterdam North-Holland 1962pp 314ndash332

49 Salvadori and Levy Structural design in architecture EnglewoodCliffs Prentice-Hall 1967 p 457

50 Schulz M and Chedraui M Tables for circularly curved horizontalbeams with symmetrical uniform loads Journal of the AmericanConcrete Institute 28(11) 1957 pp 1033ndash1040

51 Spyropoulos P J Circularly curved beams transversely loadedJournal of the American Concrete Institute 60(10) 1963 pp 1457ndash1469

52 Concrete SocietyCBDG An introduction to concrete bridgesCamberley The Concrete Society 2006 p 32

53 Leonhardt Fritz Bridges Stuttgart Deutsche Verlags-Anstalt1982 p 308

54 Hambly E C Bridge deck behaviour (Second edition) LondonE amp FN Spon 1991 p 313

55 PCA Circular concrete tanks without prestressing SkokieIllinois Portland Cement Association p 54

56 Ghali A Circular storage tanks and silos London E amp FN Spon1979 p 210

57 CIRIA Reports 139 and 140 (Summary Report) Water-resistingbasement construction London CIRIA 1995 p 192 p 64

58 Irish K and Walker W P Foundations for reciprocating machinesLondon Cement and Concrete Association 1969 p 103

59 Barkan D D Dynamics of bases and foundations New YorkMcGraw Hill 1962 p 434

60 Tomlinson M J Pile design and construction practice LondonCement and Concrete Association 1977 p 413

61 Concrete Society Technical Report No 34 Concrete industrialground floors (Third edition) Crowthorne The Concrete Society2003 p 146

62 CIRIA Report 104 Design of retaining walls embedded in stiffclay London CIRIA 1984 p 146

63 Hairsine R C A design chart for determining the optimum baseproportions of free standing retaining walls Proceedings of theInstitution of Civil Engineers 51 (February) 1972 pp 295ndash318

64 Cusens A R and Kuang Jing-Gwo A simplified method ofanalysing free-standing stairs Concrete and ConstructionalEngineering 60(5) 1965 pp 167ndash172 and 194

65 Cusens A R Analysis of slabless stairs Concrete andConstructional Engineering 61(10) 1966 pp 359ndash364

66 Santathadaporn Sakda and Cusens A R Charts for the design ofhelical stairs Concrete and Constructional Engineering 61(2) 1966pp 46ndash54

67 Terrington J S and Turner L Design of non-planar roofsLondon Concrete Publications 1964 p 108

68 Krishna J and Jain O P The beam strength of reinforced concretecylindrical shells Civil Engineering and Public Works Review49(578) 1954 pp 838ndash840 and 49(579) 1954 pp 953ndash956

69 Faber C Candela the shell builder London The ArchitecturalPress 1963 p 240

70 Bennett J D Structural possibilities of hyperbolic paraboloidsLondon Reinforced Concrete Association February 1961 p 25

71 Lee D J Bridge bearings and expansion joints (Second edition)London E amp FN Spon 1994 p 212

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Numbers preceded by lsquotrsquo are Table Numbers

Actions see LoadsAdmixtures 18ndash19Affinity theorems 140Aggregates 16ndash17

size and grading 95 t217Anchorage bond see BondAnnular sections 236 t2107Arches

fixed 41ndash2 t272ndash4load effects in 178ndash9 t272

parabolic 42 179ndash82t274

thickness 175 t272three-hinged 41 175

t271two-hinged 41 175 t271see also Bridges

Areas 52ndash3 t2101

Barriers and balustrades 7Bars

anchorage 51 312 381t355 t359 t430 t432

bending schedules t223bends in 25 51 312 381

t219 t355 t359 t431considerations affecting

design details 51 312381 t353 t359 t428

curtailment 52 312 381t356ndash8 t432

cutting and bendingtolerances 100

lap lengths 51 312 381t355 t359 t431ndash2

shapes and dimensions25ndash6 100 t221ndash2

sizes 25 95 t220types 24ndash5 95see also Reinforcement

Basements 65Bases see FoundationsBeams

cantilevers see Cantileverscontinuous see Continuous

beamscurved 57 216 t295ndash7

concentrated load 216t295

uniform load 218 t296ndash7

deep 52

doubly reinforced sections257 346 t315ndash16t325ndash6 t49ndash10

fixed at both ends 105 t225t228

flanged see Flangedsections

imposed loads on 6 t23junctions with columns 330

t363single-span 29 105 t224ndash5singly reinforced sections

256 346 t313ndash14t323ndash4 t47ndash8

sizes and proportions 46supporting rectangular

panels 34 144 t252Bearings 62 221 t299

see also DetailsFoundations

Bending (alone or combinedwith axial force) 44ndash8

assumptions 44ndash5 t36 t44resistance of sections

beams 45ndash6columns 47ndash8slabs 46ndash7

see also individual membersBending moments

combined bases 195 t283continuous beams see

Continuous beamscylindrical tanks 60 183ndash8

t275ndash7flat slabs 35ndash6 150ndash3

t255ndash6rectangular tanks 60ndash1 188

t278ndash9silos 61ndash2 191 t280see also Beams Cantilevers

Structural analysisBiaxial bending see ColumnsBlinding layer 64Bond 51 312 381

anchorage lengths see Barsbends in bars see Barslap lengths see Bars

Bow girders see Beamscurved

Bridges 57ndash9deck 57ndash8 t298design considerations 59imposed loads

foot 8 78 t26railway 8ndash9 78 t26road 7ndash8 78 t25

integral 59partial safety factors 239

t32ndash3roofs see Roofsstairs see Stairssubstructures 58types 57ndash8 t298waterproofing 59wind loads 10see also Arches

Buildings 54ndash6dead loads 75 t22imposed loads

floors 6 75ndash8 t23roofs 7 78 t24

load-bearing walls see Wallsrobustness and provision of

ties 54ndash5 t354 t429wall and frame systems

40ndash1 t268wind loads 10 78 t27ndash9see also Floors

Foundations StairsWalls

Bunkers see Silos

Cantilevers 29 t226ndash7deflections 295 371

Cements and combinations14ndash16 95 t217

Characteristic loads see Loads

Characteristic strengthsconcrete t35 t42reinforcement t219

Columnsbiaxial bending t321 t331

t416circular 264 353 t319ndash20

t329ndash30 t413ndash14cylindrical (modular ratio)

236 t2107effective height t321 t331effective length t415elastic analysis of section

226 t2104 t2108ndash9imposed loads on 7 t23junctions with beams 330

t363loads and sizes 48rectangular 264 353

t317ndash18 t327ndash8t411ndash12

rectangular (modular ratio)t2105ndash6

short 47 265

slenderdesign procedure 263

352ndash3BS 8110 t321ndash2BS 5400 t331ndash2EC 2 t415ndash16

supporting elevated tanks61 194

see also Framed structuresConcentrated loads

analysis of memberscurved beams 216 t295solid slabs 31 34 131

t245ndash7bridges 7ndash9 78 t25ndash6dispersal of 9on floors 6 78 t23shear 49 285 365 t334

t337ndash8 t419Concrete 14ndash24 95

admixtures 18ndash19aggregates 16ndash17 95

t217alkali-silica attack 23carbonation 22ndash3cements 14ndash16 95

t217chemical attack 23compressive strength 21

245 338 t35 t42creep 21ndash2 245 338 t35

t43design strengths 239 335durability 22ndash4

cover to reinforcement 24t38ndash9 t46

exposure classes 23ndash4t37 t39 t45

early-age temperatures andcracking 20 95 t218

elastic properties 21 245338 t35 t42

fibre-reinforced 68fire resistance see Fire

resistancefreezethaw attack 23plastic cracking 19ndash20shrinkage 22 245 338 t35

t42specification 24stress-strain curves 22 t36

t44tensile strength 21 t344

t423thermal properties 22 245

338 t35 t42

Index

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Concrete (Continued)water 17weight 75 t21workability 19

Construction materialsweight 75 t21ndash2

Contained materials seeRetained materials

Containment structures 59ndash61see also Silos Tanks

Continuous beams 29ndash30arrangement of design loads

111 t229equal spans and loads 30

111 t229ndash32 t234ndash5influence lines for bending

moments 121 t238ndash41methods of analysis 29

moment distribution 29121 t236

moving loads 30redistribution of moments

see Moment redistributionsecond moment of area 29unequal spans and loads

t237Corbels 327 330ndash1 t362Cover to reinforcement 24

t38ndash9 t46Cracking 50ndash1 295 300 371

calculation procedures t343t424

crack width limits 295 371deemed-to-satisfy rules

t343 t424ndash7liquid-retaining structures

300 371 t344ndash52t425ndash7

minimum reinforcement t423

Cranes 7 13Creep see ConcreteCulverts 71

box culverts 71 t287pipe culverts 71subways 71

Curtailment see BarsCurvature see DeflectionCurved beams see BeamsCylindrical tanks see Tanks

Dead loads 6 75concrete 75 t21construction materials 75

t21ndash2partitions 75 t22

Deep beams 52Deep containers see SilosDeflection 49ndash50 295 371

calculation procedurest340ndash2 t421ndash2

cantilevers 295 371curvatures t341 t422deflection limits 295 371formulae for

beams t224ndash5cantilevers t226ndash7

spaneffective depth ratios295 371 t340 t421

Design of structural members44ndash53

see also individual members(eg Arches BeamsColumns Slabs StairsWalls)

Design principles and criteria 5 44

Design strengths seeConcrete Reinforcement

Detailscontinuous nibs 327 t362

corbels 327 t362corners and intersections

330 t363curtailment 52 312 381

t356 t432rules for beams t357rules for slabs t358

halving joints 330 t362Docks and dolphins see

Maritime structuresDomes see RoofsDrawings 4

Earth-retaining wallsembedded (or sheet) 70movement joints 221 t2100pressures behind 11ndash12 86

90 t210ndash14on spread bases 69ndash70

203 324 392 t286types 69 t286see also Retained materials

Earthquake-resistant structures 43

Economical structures 3Elastic analysis 52ndash3 226 236

biaxial bending andcompression t2109

design charts t2105ndash7properties of sections

t2101ndash3 t342uniaxial bending and

compression t2104uniaxial bending and tension

t2108Embedded walls see

Earth-retaining wallsEurocode loading standards 13Exposure classes 23ndash4 t37

t39 t45

Fabric 95 100 t220Fibre-reinforced concrete 68Fill materials 12Finite elements 38Fire resistance 27 249 342

cover to reinforcement 249t310ndash11

minimum fire periods t312Fixed-end moment coefficients

105 t228Flanged sections 46

effective flange width 262 349

elastic properties t342Flat slabs see SlabsFloors 55

forms of construction t242imposed loads 6 t23industrial ground see

Industrial ground floorsopenings in 55 t337weights of concrete t21

Footbridgesimposed loads 78 t26

Formwork 4Foundations 63ndash7

balanced and coupled bases64 199 t283ndash4

basements 65bearing pressures 63 t282blinding layer 64combined bases 64 195

t283eccentric loads 63imposed loads 7 t23for machines 66piers 65piled see Piled foundationsrafts 65 199 t284separate bases 64 t282

site inspection 63strip bases 65 195 t283types 64 t282ndash4wall footings 66 t283

Framed structures 36ndash8building code requirements

36 t257 t262columns in

non-sway frames 38ndash9t260ndash1

sway frames 39ndash40 t262continuous beams in 159effect of lateral loads 39ndash40

162 t262finite element method 38moment distribution method

no sway 37 t258with sway 37 t259

portal frames 38 162rigid joints t263ndash6hinged joints t267

properties of membersend conditions 42section properties 42ndash3

shear forces on members 37slope-deflection method

of analysis 37 154t260ndash2

see also Columns

Garages 6Geometric properties of

uniform sections t2101Ground water 86 90Gyration radius of 52 t2101

t415

Hillerborgrsquos strip method 33144 t251 t254

Hinges 62 221 t299Hoppers 12 62 90 194

t215ndash16 t281

Imposed loads 6ndash9 t23ndash6barriers and parapets 7bridges see Bridgesbuildings 6ndash7 75 t23floors 6 t23reduction on beams and

columns 7 78 t23roofs 7 78 t24structures subject to dynamic

loads 6structures supporting

cranes 7structures supporting

lifts 7underground tanks 60see also Eurocode loading

standardsIndustrial ground floors 67ndash9

construction methods 67ndash8methods of analysis 68ndash9modulus of subgrade

reaction 68reinforcement 68

Intersections 330 t363see also Joints

Janssenrsquos theory 12 90t215ndash16

Jetties see Maritime structures

Joints 52 330industrial ground

floors 67ndash8liquid-retaining structures

300 t345movement 62 221 t2100see also Bearings

Details

Lifts 7Limit state design 5

British codesbridges 239 241 t32ndash3buildings 239 t31liquid-retaining structures

241 t34loads 5ndash6 239properties of materials

concrete 245 t35ndash6reinforcement 245 t36

European codesactions 5ndash6 335buildings 335ndash6 t41containers 335ndash6properties of materials 335

concrete 338 t42ndash4reinforcement 338 t44

Liquid-retaining structures 241335ndash6 t34

see also Cracking JointsLoads 6ndash10

on bridges see Bridgesconcentrated see

Concentrated loadsdead see Dead loadsdynamic 6eccentric on foundations 63

t283imposed see Imposed loadson lintels t22moving loads on continuous

beams 30on piles see Piled

foundationswind see Wind loads

Maritime structures 10ndash11piled jetties 200 t285

Materials see AdmixturesAggregates Cementsand combinationsConcrete Reinforcement

Mathematical formulae 395ndash6Members see individual

members (eg ArchesBeams Columns SlabsStairs Walls)

Modular-ratio design seeElastic analysis

Modulus of subgrade reaction 68

Moment distribution 29continuous beams 121 t236framed structures 154

t258ndash9Moment redistribution 30 116

code requirements 116design procedure 117 t233moment diagrams for equal

spans t234ndash5

Neutral axis 44ndash5

Parapets 7Partial safety factors see

Safety factorsPartition loads 75 t22Passive pressures 90

t213ndash14Piers

bridges 58foundations 65see also Maritime structures

Piled foundations 66ndash7open-piled structures 67

200 t285pile-caps 66 324 t361piles in a group 67

Poissonrsquos ratio 131 137t246ndash7 t35 t42

Index400

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Precast concretebridge decks 58floors weights of t21

Pressures 11ndash13on earth-retaining walls 86

90 t211ndash14in silos 90 t215ndash16in tanks 90see also Wind loads

Properties of sections 42ndash3plain concrete t2101reinforced concrete t2102ndash3

Rafts 65 199 t284Railway bridges see BridgesReinforcement 24ndash7 95 100

bars 24 see also Barsfabric 25 see also Fabricfixing of 27mechanical and physical

properties t219prefabricated systems 26ndash7stainless steel 26stress-strain curves 25 t36

t44Reservoirs see TanksRetained materials

cohesionless soils 12 8690 t212ndash14

cohesive soils 12 90fill materials 12lateral pressures 11ndash12

t211liquids 11 86 90soil properties 11 t210see also Silos Tanks

Retaining walls seeEarth-retaining walls

Robustness 54ndash5ties 312 381 t354 t429

Roofsloads on 7 78 t24non-planar 56 212

cylindrical shells 56 212216 t292ndash4

domes 212 t292hyperbolic-paraboloidal

shells 216 t292prismatic 212 t292shell buckling 56ndash7

planar 56weights of t22

Safety factors 5British codes 239 t31ndash4

geotechnical design 324European codes 335 t41

geotechnical design 390Sea-walls see Maritime

structuresSection moduli t2101Serviceability limit states 5

295 300 336 371t31ndash4 t41

Shearin bases 285 322 390forces see individual

members (eg BeamsSlabs)

resistance with shearreinforcement 49 283362 t333 t336 t418

resistance without shearreinforcement 48 283362 t333 t336 t417

stress 283under concentrated loads 49

285 365 t334 t337ndash8t419

see also Structural analysisShear wall structures 40ndash1

arrangement of walls 40169 t269

interaction of walls andframes 41 169 173

walls containing openings40 169 t270

walls without openings 40169 t269

Sheet walls seeEarth-retaining walls

Shrinkagefixed parabolic arch 179 182see also Concrete

Silos 12ndash13 61ndash2 90hopper bottoms 62 t281stored material properties

and pressures 90t215ndash16

substructure 39 t262walls spanning horizontally

61ndash2 t280Slabs

flat 35ndash6reservoir roofs 153simplified method of

design 150 153t255ndash6

imposed loads 75 78 t23ndash4non-rectangular panels 34ndash5

131 t248one-way 31 128

concentrated loads 31131 t245

uniform load distribution31 128 t242

openings in 55 t337rectangular panels

concentrated loads 34131 137 t246ndash7

triangular load distribution33ndash4 147 t253ndash4

uniform load distribution33 128 131 t242ndash4

thickness of 46ndash7two-way 31ndash4 128 131

collapse methods 32ndash3elastic methods 32

strip 144 t251 t254yield-line 137 139ndash42

147 150 t249ndash50t254

types of 55 t242weights of t21

Slope-deflection method 37 154 t260

Soils see Retained materialsStairs 55ndash6 206 208 212

free-standing 206 t288helical 208 212 t289ndash91sawtooth 206 208 t289simple flights 206types and dimensions t288

Stored materials see SilosStress-strain curves

concrete 22 t36 t44reinforcement 25 t36 t44

Stresses see individual modes(eg Bond ShearTorsion)

Structural analysis 28ndash43properties of members

end conditions 42section properties 42ndash3

see also individual structures(eg Arches Continuousbeams Frames Shearwalls)

Structures 54ndash71earthquake-resistant 43economical 3see also individual

structures (eg BridgesBuildings FoundationsSilos Tanks)

Subways 71Superposition theorem 140Surcharge 86 203 326

Tankscylindrical 60 183 187ndash8

t275ndash7effects of temperature 61elevated 61 191 t281

substructure 39 t262joints 221 t2100octagonal 60pressure on walls 90rectangular 60ndash1 147 150

188 191 t253ndash4 t278ndash9

underground 60see also Cracking Liquid-

retaining structuresTemperature effects in

concrete at early-age 20 95 t218

fixed parabolic arch 179walls of tanks 61

Tensile strengthconcrete 21 t344 t423reinforcement 95 t219

Thermal properties of concrete22 245 338 t35 t42

Ties see RobustnessTorsion

design procedure 49 285365 t335 t339 t420

moments incurved beams 57 216

218 t295ndash7free-standing stairs 206

t288helical stairs 208 212

t289

Ultimate limit state 5 239 241335ndash6 t31ndash4 t41

Vehicle loads on bridges 7ndash978 t25ndash6

Vibrationfloors 6footbridges 8machine foundations 66

Virtual-work method 139ndash40

Walls 52 57load-bearing 57 322 t360weights of t22see also Earth-retaining

walls Shear walls SilosTanks

Water (for concrete) 17Water-tightness 59

basements 65Weights of

concrete 75 t21construction materials 75

t21ndash2partitions 75 t22roofs t22stored materials t216walls t22

Wharves see MaritimeStructures

Wheel loads 8 t25dispersal of 9

Wind loads 9ndash10 78on bridges 10on buildings 10effect of see Structural

analysiswind speed and pressure 10

t27ndash9

Yield-line analysis 32 137 139ndash42

affinity theorems 140basic concepts 137 139

t249concentrated loads 140corner levers 142 t250superposition theorem 140virtual work method 139

empirical analysis 141ndash2

Index 401

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Also available from Taylor amp Francis

Concrete Pavement Design GuidanceG Griffiths et al Hb ISBN 0ndash415ndash25451ndash5

Reinforced Concrete 3rd edP Bhatt et al Hb ISBN 0ndash415ndash30795ndash3

Pb ISBN 0ndash415ndash30796ndash1

Concrete BridgesP Mondorf Hb ISBN 0ndash415ndash39362ndash0

Reinforced amp Prestressed Concrete 4th edS Teng et al Hb ISBN 0ndash415ndash31627ndash8

Pb ISBN 0ndash415ndash31626ndashX

Concrete Mix Design Quality Control and Specification 3rd edK Day Hb ISBN 0ndash415ndash39313ndash2

Examples in Structural AnalysisW McKenzie Hb ISBN 0ndash415ndash37053ndash1

Pb ISBN 0ndash415ndash37054ndashX

Wind Loading of Structures 2nd edJ Holmes Hb ISBN 0ndash415ndash40946ndash2

Information and ordering details

For price availability and ordering visit our website wwwtandfcoukbuiltenvironment

Alternatively our books are available from all good bookshops

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ReynoldsrsquosReinforcedConcreteDesignerrsquosHandbookELEVENTH EDITION

Charles E ReynoldsBSc (Eng) CEng FICE

James C SteedmanBA CEng MICE MIStructE

and

Anthony J ThrelfallBEng DIC

wwwengbookspdfcom

First edition 1932 second edition 1939 third edition 1946 fourth edition 1948revised 1951 further revision 1954 fifth edition 1957 sixth edition 1961revised 1964 seventh edition 1971 revised 1972 eighth edition 1974 reprinted1976 ninth edition 1981 tenth edition 1988reprinted 1991 1994 (twice) 1995 1996 1997 1999 2002 2003

Eleventh edition published 2008by Taylor amp Francis2 Park Square Milton Park Abingdon Oxon OX14 4RN

Simultaneously published in the USA and Canadaby Taylor amp Francis270 Madison Ave New York NY 10016 USA

Taylor amp Francis is an imprint of the Taylor amp Francis Groupan informa business

copy 2008 Taylor and Francis

All rights reserved No part of this book may be reprinted or reproduced or utilised in any form or by any electronic mechanical or other meansnow known or hereafter invented including photocopying and recordingor in any information storage or retrieval system without permission in writing from the publishers

The publisher makes no representation express or implied with regard to the accuracy of the information contained in this book and cannot accept any legal responsibility or liability for any efforts or omissions that may be made

British Library Cataloguing in Publication DataA catalogue record for this book is available from the British Library

Library of Congress Cataloging-in-Publication DataReynolds Charles E (Charles Edward)

Reynoldsrsquos reinforced concrete designers handbook Charles E ReynoldsJames C Steedman and Anthony J Threlfall ndash 11th ed

p cmRev ed of Reinforced concrete designerrsquos handbook Charles E Reynolds

and James C Steedman 1988Includes bibliographical references and index1 Reinforced concrete construction ndash Handbooks manuals etc

I Steedman James C (James Cyril) II Threlfall A J III ReynoldsCharles E (Charles Edward) Reinforced concrete designerrsquos handbookIV Title

TA6832R48 200762418341ndashdc22 2006022625

ISBN10 0ndash419ndash25820ndash5 (hbk)ISBN10 0ndash419ndash25830ndash2 (pbk)ISBN10 0ndash203ndash08775ndash5 (ebk)

ISBN13 978ndash0ndash419ndash25820ndash9 (hbk)ISBN13 978ndash0ndash419ndash25830ndash8 (pbk)ISBN13 978ndash0ndash203ndash08775ndash6 (ebk)

This edition published in the Taylor amp Francis e-Library 2007

ldquoTo purchase your own copy of this or any of Taylor amp Francis or Routledgersquoscollection of thousands of eBooks please go to wwweBookstoretandfcoukrdquo

ISBN 0-203-08775-5 Master e-book ISBN

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List of tables viPreface to the eleventh edition ixThe authors xAcknowledgements xiSymbols and abbreviations xii

Part 1 ndash General information 11 Introduction 32 Design criteria safety factors and loads 53 Material properties 144 Structural analysis 285 Design of structural members 446 Buildings bridges and containment structures 547 Foundations ground slabs retaining walls

culverts and subways 63

Part 2 ndash Loads materials and structures 738 Loads 759 Pressures due to retained materials 86

10 Concrete and reinforcement 9511 Cantilevers and single-span beams 10512 Continuous beams 11113 Slabs 12814 Framed structures 15415 Shear wall structures 16916 Arches 17517 Containment structures 18318 Foundations and retaining walls 195

19 Miscellaneous structures and details 20620 Elastic analysis of concrete sections 226

Part 3 ndash Design to British Codes 23721 Design requirements and safety factors 23922 Properties of materials 24523 Durability and fire-resistance 24924 Bending and axial force 25625 Shear and torsion 28326 Deflection and cracking 29527 Considerations affecting design details 31228 Miscellaneous members and details 322

Part 4 ndash Design to European Codes 33329 Design requirements and safety factors 33530 Properties of materials 33831 Durability and fire-resistance 34232 Bending and axial force 34533 Shear and torsion 36234 Deflection and cracking 37135 Considerations affecting design details 38136 Foundations and earth-retaining walls 390

Appendix Mathematical formulae and data 395

References and further reading 397

Index 399

Contents

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21 Weights of construction materials and concrete

floor slabs22 Weights of roofs and walls23 Imposed loads on floors of buildings24 Imposed loads on roofs of buildings25 Imposed loads on bridges ndash 126 Imposed loads on bridges ndash 227 Wind speeds (standard method of design)28 Wind pressures and forces (standard method

of design)29 Pressure coefficients and size effect factors

for rectangular buildings210 Properties of soils211 Earth pressure distributions on rigid walls212 Active earth pressure coefficients213 Passive earth pressure coefficients ndash 1214 Passive earth pressure coefficients ndash 2215 Silos ndash 1216 Silos ndash 2217 Concrete cements and aggregate grading218 Concrete early-age temperatures219 Reinforcement general properties220 Reinforcement cross-sectional areas of bars

and fabric221 Reinforcement standard bar shapes and method of

measurement ndash 1222 Reinforcement standard bar shapes and method of

measurement ndash 2223 Reinforcement typical bar schedule224 Moments shears deflections general case for beams225 Moments shears deflections special cases for beams226 Moments shears deflections general cases for

cantilevers227 Moments shears deflections special cases for

cantilevers228 Fixed-end moment coefficients general data229 Continuous beams general data230 Continuous beams moments from equal loads on

equal spans ndash 1231 Continuous beams moments from equal loads on

equal spans ndash 2232 Continuous beams shears from equal loads on

equal spans233 Continuous beams moment redistribution234 Continuous beams bending moment diagrams ndash 1

235 Continuous beams bending moment diagrams ndash 2236 Continuous beams moment distribution methods237 Continuous beams unequal prismatic spans and loads238 Continuous beams influence lines for two spans239 Continuous beams influence lines for three spans240 Continuous beams influence lines for four spans241 Continuous beams influence lines for five or

more spans242 Slabs general data243 Two-way slabs uniformly loaded rectangular panels

(BS 8110 method)244 Two-way slabs uniformly loaded rectangular panels

(elastic analysis)245 One-way slabs concentrated loads246 Two-way slabs rectangular panel with concentric

concentrated load ndash 1247 Two-way slabs rectangular panel with concentric

concentrated load ndash 2248 Two-way slabs non-rectangular panels (elastic

analysis)249 Two-way slabs yield-line theory general information250 Two-way slabs yield-line theory corner levers251 Two-way slabs Hillerborgrsquos simple strip theory252 Two-way slabs rectangular panels loads on beams

(common values)253 Two-way slabs triangularly distributed load (elastic

analysis)254 Two-way slabs triangularly distributed load (collapse

method)255 Flat slabs BS 8110 simplified method ndash 1256 Flat slabs BS 8110 simplified method ndash 2257 Frame analysis general data258 Frame analysis moment-distribution method

no sway259 Frame analysis moment-distribution method

with sway260 Frame analysis slope-deflection data261 Frame analysis simplified sub-frames262 Frame analysis effects of lateral loads263 Rectangular frames general cases264 Gable frames general cases265 Rectangular frames special cases266 Gable frames special cases267 Three-hinged portal frames268 Structural forms for multi-storey buildings

List of tables

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List of tables vii

269 Shear wall layout and lateral load allocation270 Analysis of pierced shear walls271 Arches three-hinged and two-hinged arches272 Arches fixed-ended arches273 Arches computation chart for symmetrical

fixed-ended arch274 Arches fixed-ended parabolic arches275 Cylindrical tanks elastic analysis ndash 1276 Cylindrical tanks elastic analysis ndash 2277 Cylindrical tanks elastic analysis ndash 3278 Rectangular tanks triangularly distributed load

(elastic analysis) ndash 1279 Rectangular tanks triangularly distributed load

(elastic analysis) ndash 2280 Rectangular containers spanning horizontally

moments in walls281 Bottoms of elevated tanks and silos282 Foundations presumed allowable bearing values

and separate bases283 Foundations other bases and footings284 Foundations inter-connected bases and rafts285 Foundations loads on open-piled structures286 Retaining walls287 Rectangular culverts288 Stairs general information289 Stairs sawtooth and helical stairs290 Design coefficients for helical stairs ndash 1291 Design coefficients for helical stairs ndash 2292 Non-planar roofs general data293 Shell roofs empirical design method ndash 1294 Shell roofs empirical design method ndash 2295 Bow girders concentrated loads296 Bow girders uniform loads ndash 1297 Bow girders uniform loads ndash 2298 Bridges299 Hinges and bearings2100 Movement joints2101 Geometric properties of uniform sections2102 Properties of reinforced concrete sections ndash 12103 Properties of reinforced concrete sections ndash 22104 Uniaxial bending and compression (modular ratio)2105 Symmetrically reinforced rectangular columns

(modular ratio) ndash 12106 Symmetrically reinforced rectangular columns

(modular ratio) ndash 22107 Uniformly reinforced cylindrical columns

(modular ratio)2108 Uniaxial bending and tension (modular ratio)2109 Biaxial bending and compression (modular ratio)31 Design requirements and partial safety factors

(BS 8110)32 Design requirements and partial safety factors

(BS 5400) ndash 133 Design requirements and partial safety factors

(BS 5400) ndash 234 Design requirements and partial safety factors

(BS 8007)35 Concrete (BS 8110) strength and deformation

characteristics36 Stress-strain curves (BS 8110 and BS 5400) concrete

and reinforcement

37 Exposure classification (BS 8500)38 Concrete quality and cover requirements for durability

(BS 8500)39 Exposure conditions concrete and cover requirements

(prior to BS 8500)310 Fire resistance requirements (BS 8110) ndash 1311 Fire resistance requirements (BS 8110) ndash 2312 Building regulations minimum fire periods313 BS 8110 Design chart for singly reinforced

rectangular beams314 BS 8110 Design table for singly reinforced

rectangular beams315 BS 8110 Design chart for doubly reinforced

rectangular beams ndash 1316 BS 8110 Design chart for doubly reinforced

rectangular beams ndash 2317 BS 8110 Design chart for rectangular columns ndash 1318 BS 8110 Design chart for rectangular columns ndash 2319 BS 8110 Design chart for circular columns ndash 1320 BS 8110 Design chart for circular columns ndash 2321 BS 8110 Design procedure for columns ndash 1322 BS 8110 Design procedure for columns ndash 2323 BS 5400 Design chart for singly reinforced

rectangular beams324 BS 5400 Design table for singly reinforced

rectangular beams325 BS 5400 Design chart for doubly reinforced

rectangular beams ndash 1326 BS 5400 Design chart for doubly reinforced

rectangular beams ndash 2327 BS 5400 Design chart for rectangular columns ndash 1328 BS 5400 Design chart for rectangular columns ndash 2329 BS 5400 Design chart for circular columns ndash 1330 BS 5400 Design chart for circular columns ndash 2331 BS 5400 Design procedure for columns ndash 1332 BS 5400 Design procedure for columns ndash 2333 BS 8110 Shear resistance334 BS 8110 Shear under concentrated loads335 BS 8110 Design for torsion336 BS 5400 Shear resistance337 BS 5400 Shear under concentrated loads ndash 1338 BS 5400 Shear under concentrated loads ndash 2339 BS 5400 Design for torsion340 BS 8110 Deflection ndash 1341 BS 8110 Deflection ndash 2342 BS 8110 Deflection ndash 3343 BS 8110 (and BS 5400) Cracking344 BS 8007 Cracking345 BS 8007 Design options and restraint factors346 BS 8007 Design table for cracking due to temperature

effects347 BS 8007 Elastic properties of cracked rectangular

sections in flexure348 BS 8007 Design table for cracking due to flexure

in slabs ndash 1349 BS 8007 Design table for cracking due to flexure

in slabs ndash 2350 BS 8007 Design table for cracking due to flexure

in slabs ndash 3351 BS 8007 Design table for cracking due to direct

tension in walls ndash 1

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List of tablesviii

49 EC 2 Design chart for doubly reinforcedrectangular beams ndash 1

410 EC 2 Design chart for doubly reinforcedrectangular beams ndash 2

411 EC 2 Design chart for rectangular columns ndash 1412 EC 2 Design chart for rectangular columns ndash 2413 EC 2 Design chart for circular columns ndash 1414 EC 2 Design chart for circular columns ndash 2415 EC 2 Design procedure for columns ndash 1416 EC 2 Design procedure for columns ndash 2417 EC 2 Shear resistance ndash 1418 EC 2 Shear resistance ndash 2419 EC 2 Shear under concentrated loads420 EC 2 Design for torsion421 EC 2 Deflection ndash 1422 EC 2 Deflection ndash 2423 EC 2 Cracking ndash 1424 EC 2 Cracking ndash 2425 EC 2 Cracking ndash 3426 EC 2 Early thermal cracking in end restrained panels427 EC 2 Early thermal cracking in edge

restrained panels428 EC 2 Reinforcement limits429 EC 2 Provision of ties430 EC 2 Anchorage requirements431 EC 2 Laps and bends in bars432 EC 2 Rules for curtailment large diameter bars

and bundles

352 BS 8007 Design table for cracking due to directtension in walls ndash 2

353 BS 8110 Reinforcement limits354 BS 8110 Provision of ties355 BS 8110 Anchorage requirements356 BS 8110 Curtailment requirements357 BS 8110 Simplified curtailment rules for beams358 BS 8110 Simplified curtailment rules for slabs359 BS 5400 Considerations affecting design details360 BS 8110 Load-bearing walls361 BS 8110 Pile-caps362 Recommended details nibs corbels and halving joints363 Recommended details intersections of members41 Design requirements and partial safety factors

(EC 2 Part 1)42 Concrete (EC 2) strength and deformation

characteristics ndash 143 Concrete (EC 2) strength and deformation

characteristics ndash 244 Stressndashstrain curves (EC 2) concrete and

reinforcement45 Exposure classification (BS 8500)46 Concrete quality and cover requirements for durability

(BS 8500)47 EC 2 Design chart for singly reinforced

rectangular beams48 EC 2 Design table for singly reinforced

rectangular beams

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Since the last edition of Reynoldsrsquos Handbook considerabledevelopments in design and practice have occurred These includesignificant revisions to British standard specifications and codesof practice and the introduction of the Eurocodes Although cur-rent British codes are due to be withdrawn from 2008 onwardstheir use is likely to continue beyond that date at least in someEnglish-speaking countries outside the United Kingdom

One of the most significant changes has been in the systemfor classifying exposure conditions and selecting concretestrength and cover requirements for durability This is now dealtwith exclusively in BS 8500 which takes into account theparticular cementcombination type The notation used todefine concrete strength gives the cylinder strength as well asthe cube strength For structural design cube strength is usedin the British codes and cylinder strength in the Eurocodes

The characteristic yield strength of reinforcement has beenincreased to 500 Nmm2 (MPa) As a result new design aidshave become necessary and the Handbook includes tables andcharts for beams and columns (rectangular and circular)designed to both British and European codes Throughout theHandbook stress units are given as Nmm2 for British codesand MPa for European codes The decimal point is shown by afull stop (rather than a comma) in both cases

The basic layout of the Handbook is similar to the previousedition but the contents have been arranged in four separateparts for the convenience of the reader Also the opportunityhas been taken to omit a large amount of material that was nolonger relevant and to revise the entire text to reflect moderndesign and construction practice Part 1 is descriptive in formand covers design requirements loads materials structuralanalysis member design and forms of construction Frequentreference is made in Part 1 to the tables that are found in therest of the Handbook Although specific notes are attached tothese tables in Parts 2 3 and 4 much of the relevant text isembodied in Part 1 and the first part of the Handbook shouldalways be consulted

Part 2 has more detailed information on loads materialproperties and analysis in the form of tabulated data and chartsfor a large range of structural forms This material is largelyindependent of any specific code of practice Parts 3 and 4 cover

the design of members according to the requirements ofthe British and European codes respectively For each code thesame topics are covered in the same sequence so that the readercan move easily from one code to the other Each topic isillustrated by extensive numerical examples

In the Eurocodes some parameters are given recommendedvalues with the option of a national choice Choices also existwith regard to certain classes methods and procedures Thedecisions made by each country are given in a national annexPart 4 of the Handbook already incorporates the values given inthe UK national annex Further information concerning the useof Eurocode 2 is given in PD 6687 Background paper to theUK National Annex to BS EN 1992ndash1ndash1

The Handbook has been an invaluable source of reference forreinforced concrete engineers for over 70 years I madeextensive use of the sixth edition at the start of my professionalcareer 50 years ago This edition contains old and new infor-mation derived by many people and obtained from manysources past and present Although the selection inevitablyreflects the personal experience of the authors the informationhas been well tried and tested I owe a considerable debt ofgratitude to colleagues and mentors from whom I have learntmuch over the years and to the following organisations forpermission to include data for which they hold the copyright

British Cement AssociationBritish Standards InstitutionCabinet Office of Public Sector InformationConstruction Industry Research and Information AssociationPortland Cement AssociationThe Concrete Bridge Development GroupThe Concrete Society

Finally my sincere thanks go to Katy Low and all the staff atTaylor amp Francis Group and especially to my dear wife Joanwithout whose unstinting support this edition would never havebeen completed

Tony ThrelfallMarlow October 2006

Preface to theeleventh edition

wwwengbookspdfcom

Charles Edward Reynolds was born in 1900 and received hiseducation at Tiffin Boys School Kingston-on-Thames andBattersea Polytechnic After some years with Sir WilliamArroll BRC and Simon Carves he joined Leslie Turner andPartners and later C W Glover and Partners He was for someyears Technical Editor of Concrete Publications Ltd and thenbecame its Managing Editor combining this post with privatepractice In addition to the Reinforced Concrete DesignerrsquosHandbook of which almost 200000 copies have been soldsince it first appeared in 1932 Charles Reynolds was the authorof numerous other books papers and articles concerningconcrete and allied subjects Among his various professionalappointments he served on the council of the Junior Institutionof Engineers and was the Honorary Editor of its journal at hisdeath on Christmas Day 1971

James Cyril Steedman was educated at Varndean GrammarSchool and first was employed by British Rail whom he joinedin 1950 at the age of 16 In 1956 he began working for GKNReinforcements Ltd and later moved to Malcolm Glover andPartners His association with Charles Reynolds began whenafter the publication of numerous articles in the magazine

Concrete and Constructional Engineering he accepted anappointment as Technical Editor of Concrete Publications apost he held for seven years He then continued in privatepractice combining work for the Publications Division of theCement and Concrete Association with his own writing andother activities In 1981 he set up Jacys Computing Servicessubsequently devoting much of his time to the development ofmicro-computer software for reinforced concrete design He isthe joint author with Charles Reynolds of Examples of theDesign of Reinforced Concrete Buildings to BS 8110

Anthony John Threlfall was educated at Liverpool Institute forBoys after which he studied civil engineering at LiverpoolUniversity After eight years working for BRC Pierhead Ltdand IDC Ltd he took a diploma course in concrete structuresand technology at Imperial College For the next four years heworked for CEGB and Camus Ltd and then joined the Cementand Concrete Association in 1970 where he was engagedprimarily in education and training activities until 1993 Afterleaving the CampCA he has continued in private practice toprovide training in reinforced and prestressed concrete designand detailing

The authors

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The publishers would like to thank the following organisationsfor their kind permission to reproduce the following material

Permission to reproduce extracts from British Standards isgranted by BSI This applies to information in Tables 21 2324 27ndash210 215 216 219ndash223 242 243 245 255256 2100 31ndash311 321 322 331ndash345 353ndash361 41ndash46415ndash425 and 428ndash432 British Standards can be obtainedfrom BSI Customer Services 389 Chiswick High StreetLondon W4 4AL Tel 44 (0)20 8996 9001 emailcservicesbsi-globalcom

Information in section 31 and Tables 217ndash218 is reproducedwith permission from the British Cement Association andtaken from the publication Concrete Practice (ref 10)

Information in section 62 is reproduced with permissionfrom the Concrete Bridge Development Group and taken

from the publication An introduction to concrete bridges(ref 52)

Information in section 72 is reproduced with permissionfrom The Concrete Society and taken from TechnicalReport 34 Concrete industrial ground floors ndash A guide todesign and construction (ref 61) Technical Report 34 isavailable to purchase from The Concrete Bookshop wwwconcretebookshopcom Tel 0700 460 7777

Information in Chapter 15 and Table 270 is reproduced withpermission from CIRIA and taken from CIRIA Report 102Design of shear wall buildings London 1984 (ref 38)

Information in Tables 253 and 275ndash279 is reproduced withpermission from the Portland Cement Association (refs 32and 55)

Information in Tables 25 26 and 312 is reproduced withpermission from HMSO

Acknowledgements

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The symbols adopted in this book comply where appropriatewith those in the relevant codes of practice Although these arebased on an internationally agreed system for preparing nota-tions there are numerous differences between the British andthe European codes especially in the use of subscripts Whereadditional symbols are needed to represent properties not usedin the codes these have been selected in accordance with thebasic principles wherever possible

The amount and range of material contained in this bookmake it inevitable that the same symbols have to be used for

different purposes However care has been taken to ensure thatcode symbols are not duplicated except where this has beenfound unavoidable The notational principles adopted for con-crete design purposes are not necessarily best suited to otherbranches of engineering Consequently in those tables relatingto general structural analysis the notation employed in previ-ous editions of this book has generally been retained

Only the principal symbols that are common to all codes arelisted here all other symbols and abbreviations are defined inthe text and tables concerned

Symbols andabbreviations

Ac Area of concrete sectionAs Area of tension reinforcementAs Area of compression reinforcementAsc Area of longitudinal reinforcement in a columnC Torsional constantEc Static modulus of elasticity of concreteEs Modulus of elasticity of reinforcing steelF Action force or load (with appropriate

subscripts)G Shear modulus of concreteGk Characteristic permanent action or dead loadI Second moment of area of cross sectionK A constant (with appropriate subscripts)L Length spanM Bending momentN Axial forceQk Characteristic variable action or imposed loadR Reaction at supportS First moment of area of cross sectionT Torsional moment temperatureV Shear forceWk Characteristic wind load

a Dimension deflectionb Overall width of cross section or width of flanged Effective depth to tension reinforcementd Depth to compression reinforcementf Stress (with appropriate subscripts)fck Characteristic (cylinder) strength of concretefcu Characteristic (cube) strength of concretefyk Characteristic yield strength of reinforcementgk Characteristic dead load per unit areah Overall depth of cross section

i Radius of gyration of concrete sectionk A coefficient (with appropriate subscripts)l Length span (with appropriate subscripts)m Massqk Characteristic imposed load per unit arear Radius1r Curvaturet Thickness timeu Perimeter (with appropriate subscripts)v Shear stress (with appropriate subscripts)x Neutral axis depthz Lever arm of internal forces

Angle ratioe Modular ratio EsEc

Partial safety factor (with appropriate subscripts)c Compressive strain in concretes Strain in tension reinforcements Strain in compression reinforcement Diameter of reinforcing bar Creep coefficient (with appropriate subscripts) Slenderness ratio Poissonrsquos ratio Proportion of tension reinforcement Asbd Proportion of compression reinforcement As bd Stress (with appropriate subscripts) Factor defining representative value of action

BS British StandardEC EurocodeSLS Serviceability limit stateUDL Uniformly distributed loadULS Ultimate limit state

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Part 1

General information

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A structure is an assembly of members each of which under theaction of imposed loads and deformations is subjected tobending or direct force (either tensile or compressive) or to acombination of bending and direct force These effects may beaccompanied by shearing forces and sometimes by torsionImposed deformations occur as a result of concrete shrinkageand creep changes in temperature and differential settlementBehaviour of the structure in the event of fire or accidentaldamage resulting from impact or explosion may need to beexamined The conditions of exposure to environmental andchemical attack also need to be considered

Design includes selecting a suitable form of constructiondetermining the effects of imposed loads and deformationsand providing members of adequate stiffness and resistanceThe members should be arranged so as to combine efficientload transmission with ease of construction consistent withthe intended use of the structure and the nature of the siteExperience and sound judgement are often more important thanprecise calculations in achieving safe and economical structuresComplex mathematics should not be allowed to confuse a senseof good engineering The level of accuracy employed in thecalculations should be consistent throughout the designprocess wherever possible

Structural design is largely controlled by regulations or codesbut even within such bounds the designer needs to exercisejudgement in interpreting the requirements rather than designingto the minimum allowed by the letter of a clause In the UnitedKingdom for many years the design of reinforced concretestructures has been based on the recommendations of BritishStandards For buildings these include lsquoStructural use ofconcretersquo (BS 8110 Parts 1 2 and 3) and lsquoLoading on build-ingsrsquo (BS 6399 Parts 1 2 and 3) For other types of structureslsquoDesign of concrete bridgesrsquo (BS 5400 Part 4) and lsquoDesign ofconcrete structures for retaining aqueous liquidsrsquo (BS 8007)have been used Compliance with the particular requirements ofthe Building Regulations and the Highways Agency Standardsis also necessary in many cases

Since the last edition of this Handbook a comprehensiveset of harmonised Eurocodes (ECs) for the structural andgeotechnical design of buildings and civil engineering workshas been developed The Eurocodes were first introduced asEuronorme Voluntaire (ENV) standards intended for use inconjunction with a national application document (NAD) asan alternative to national codes for a limited number of years

These have now been largely replaced by Euronorme (EN)versions with each member state adding a National Annex(NA) containing nationally determined parameters in order toimplement the Eurocode as a national standard The relevantdocuments for concrete structures are EC 0 Basis of structuraldesign EC 1 Actions on structures and EC 2 Design of con-crete structures The last document is in four parts namely ndashPart 11 General rules and rules for buildings Part 12Structural fire design Part 2 Reinforced and prestressed con-crete bridges and Part 3 Liquid-retaining and containingstructures

The tables to be found in Parts 2 3 and 4 of this Handbookenable the designer to reduce the amount of arithmetical workinvolved in the analysis and design of members to the relevantstandards The use of such tables not only increases speed butalso eliminates inaccuracies provided the tables are thoroughlyunderstood and their applications and limitations are realisedIn the appropriate chapters of Part 1 and in the supplementaryinformation given on the pages preceding the tables the basisof the tabulated material is described Some general informa-tion is also provided The Appendix contains trigonometricaland other mathematical formulae and data

11 ECONOMICAL STRUCTURES

The cost of construction of a reinforced concrete structure isobviously affected by the prices of concrete reinforcementformwork and labour The most economical proportions ofmaterials and labour will depend on the current relationshipbetween the unit prices Economy in the use of formwork isgenerally achieved by uniformity of member size and the avoid-ance of complex shapes and intersections In particular casesthe use of available formwork of standard sizes may determinethe structural arrangement In the United Kingdom speed ofconstruction generally has a major impact on the overall costFast-track construction requires the repetitive use of a rapidformwork system and careful attention to both reinforcementdetails and concreting methods

There are also wider aspects of economy such as whetherthe anticipated life and use of a proposed structure warrant theuse of higher or lower factors of safety than usual or whetherthe use of a more expensive form of construction is warrantedby improvements in the integrity and appearance of the structureThe application of whole-life costing focuses attention on

Chapter 1

Introduction

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Introduction4

whether the initial cost of a construction of high qualitywith little or no subsequent maintenance is likely to be moreeconomical than a cheaper construction combined with theexpense of maintenance

The experience and method of working of the contractor theposition of the site and the nature of the available materials andeven the method of measuring the quantities together withnumerous other points all have their effect consciously or noton the designerrsquos attitude towards a contract So many andvaried are the factors involved that only experience and acontinuing study of design trends can give reliable guidanceAttempts to determine the most economical proportions for aparticular member based only on inclusive prices of concretereinforcement and formwork are likely to be misleading It isnevertheless possible to lay down certain principles

In broad terms the price of concrete increases with thecement content as does the durability and strength Concretegrades are often determined by durability requirements withdifferent grades used for foundations and superstructuresStrength is an important factor in the design of columns andbeams but rarely so in the case of slabs Nevertheless the samegrade is generally used for all parts of a superstructure exceptthat higher strength concrete may sometimes be used to reducethe size of heavily loaded columns

In the United Kingdom mild steel and high yield reinforce-ments have been used over the years but grade 500 is nowproduced as standard available in three ductility classes A B andC It is always uneconomical in material terms to use compressionreinforcement in beams and columns but the advantages gainedby being able to reduce member sizes and maintain the samecolumn size over several storeys generally offset the additionalmaterial costs For equal weights of reinforcement the combinedmaterial and fixing costs of small diameter bars are greater thanthose of large diameter bars It is generally sensible to use thelargest diameter bars consistent with the requirements for crackcontrol Fabric (welded mesh) is more expensive than barreinforcement in material terms but the saving in fixing time willoften result in an overall economy particularly in slabs and walls

Formwork is obviously cheaper if surfaces are plane and atright angles to each other and if there is repetition of use Thesimplest form of floor construction is a solid slab of constantthickness Beam and slab construction is more efficient struc-turally but less economical in formwork costs Two-way beamsystems complicate both formwork and reinforcement detailswith consequent delay in the construction programmeIncreased slab efficiency and economy over longer spans maybe obtained by using a ribbed form of construction Standardtypes of trough and waffle moulds are available in a range ofdepths Precasting usually reduces considerably the amountof formwork labour and erection time Individual mouldsare more expensive but can be used many more timesthan site formwork Structural connections are normally moreexpensive than with monolithic construction The economicaladvantage of precasting and the structural advantage of in situcasting may be combined in composite forms of construction

In many cases the most economical solution can only bedetermined by comparing the approximate costs of differentdesigns This may be necessary to decide say when a simplecantilever retaining wall ceases to be more economical thanone with counterforts or when a beam and slab bridge is moreeconomical than a voided slab The handbook Economic

Concrete Frame Elements published by the British CementAssociation on behalf of the Reinforced Concrete Councilenables designers to rapidly identify least-cost options for thesuperstructure of multi-storey buildings

12 DRAWINGS

In most drawing offices a practice has been developed to suitthe particular type of work done Computer aided drafting andreinforcement detailing is widely used The following observa-tions should be taken as general principles that accord with therecommendations in the manual Standard method of detailingstructural concrete published by the Institution of StructuralEngineers (ref 1)

It is important to ensure that on all drawings for a particularcontract the same conventions are adopted and uniformity ofsize and appearance are achieved In the preliminary stagesgeneral arrangement drawings of the whole structure are usuallyprepared to show the layout and sizes of beams columns slabswalls foundations and other members A scale of 1100 isrecommended although a larger scale may be necessary forcomplex structures Later these or similar drawings are devel-oped into working drawings and should show precisely suchparticulars as the setting-out of the structure in relation to anyadjacent buildings or other permanent works and the level ofsay the ground floor in relation to a fixed datum All principaldimensions such as distances between columns and walls andthe overall and intermediate heights should be shown Plansshould generally incorporate a gridline system with columnspositioned at the intersections Gridlines should be numbered 12 3 and so on in one direction and lettered A B C and soon in the other direction with the sequences starting at thelower left corner of the grid system The references canbe used to identify individual beams columns and othermembers on the reinforcement drawings

Outline drawings of the members are prepared to suitablescales such as 120 for beams and columns and 150 for slabsand walls with larger scales being used for cross sectionsReinforcement is shown and described in a standard way Theonly dimensions normally shown are those needed to positionthe bars It is generally preferable for the outline of the concreteto be indicated by a thin line and to show the reinforcement bybold lines The lines representing the bars should be shown inthe correct positions with due allowance for covers and thearrangement at intersections and laps so that the details onthe drawing represent as nearly as possible the appearanceof the reinforcement as fixed on site It is important to ensurethat the reinforcement does not interfere with the formation ofany holes or embedment of any other items in the concrete

A set of identical bars in a slab shown on plan might bedescribed as 20H16-03-150B1 This represents 20 numbergrade 500 bars of 16 mm nominal size bar mark 03 spaced at150 mm centres in the bottom outer layer The bar mark is anumber that uniquely identifies the bar on the drawing and thebar bending schedule Each different bar on a drawing is givena different bar mark Each set of bars is described only once onthe drawing The same bars on a cross section would be denotedsimply by the bar mark Bar bending schedules are prepared foreach drawing on separate forms according to recommendationsin BS 8666 Specification for scheduling dimensioning bendingand cutting of steel reinforcement for concrete

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There are two principal stages in the calculations requiredto design a reinforced concrete structure In the first stagecalculations are made to determine the effect on the structureof loads and imposed deformations in terms of appliedmoments and forces In the second stage calculations are madeto determine the capacity of the structure to withstand sucheffects in terms of resistance moments and forces

Factors of safety are introduced in order to allow for theuncertainties associated with the assumptions made and thevalues used at each stage For many years unfactored loadswere used in the first stage and total factors of safety wereincorporated in the material stresses used in the second stageThe stresses were intended to ensure both adequate safety andsatisfactory performance in service This simple approach waseventually replaced by a more refined method in which specificdesign criteria are set and partial factors of safety are incorpo-rated at each stage of the design process

21 DESIGN CRITERIA AND SAFETY FACTORS

A limit-state design concept is used in British and EuropeanCodes of Practice Ultimate (ULS) and serviceability (SLS)limit states need to be considered as well as durability and inthe case of buildings fire-resistance Partial safety factors areincorporated into loads (including imposed deformations) andmaterial strengths to ensure that the probability of failure (notsatisfying a design requirement) is acceptably low

In BS 8110 at the ULS a structure should be stable under allcombinations of dead imposed and wind load It should also berobust enough to withstand the effects of accidental loads dueto an unforeseen event such as a collision or explosion withoutdisproportionate collapse At the SLS the effects in normal useof deflection cracking and vibration should not cause thestructure to deteriorate or become unserviceable A deflectionlimit of span250 applies for the total sag of a beam or slabrelative to the level of the supports A further limit the lesser ofspan500 or 20 mm applies for the deflection that occurs afterthe application of finishes cladding and partitions so as to avoiddamage to these elements A limit of 03 mm generally appliesfor the width of a crack at any point on the concrete surface

In BS 5400 an additional partial safety factor is introducedThis is applied to the load effects and takes account of themethod of structural analysis that is used Also there are moreload types and combinations to be considered At the SLSthere are no specified deflection limits but the cracking limits

are more critical Crack width limits of 025 015 or 01 mmapply according to surface exposure conditions Compressivestress limits are also included but in many cases these do notneed to be checked Fatigue considerations require limitationson the reinforcement stress range for unwelded bars and morefundamental analysis if welding is involved Footbridges areto be analysed to ensure that either the fundamental naturalfrequency of vibration or the maximum vertical accelerationmeets specified requirements

In BS 8007 water-resistance is a primary design concernAny cracks that pass through the full thickness of a section arelikely to allow some seepage initially resulting in surfacestaining and damp patches Satisfactory performance dependsupon autogenous healing of such cracks taking place within afew weeks of first filling in the case of a containment vesselA crack width limit of 02 mm normally applies to all cracksirrespective of whether or not they pass completely through thesection Where the appearance of a structure is considered to beaesthetically critical a limit of 01 mm is recommended

There are significant differences between the structural andgeotechnical codes in British practice The approach to thedesign of foundations in BS 8004 is to use unfactored loadsand total factors of safety For the design of earth-retainingstructures CP2 (ref 2) used the same approach In 1994 CP2was replaced by BS 8002 in which mobilisation factors areintroduced into the calculation of soil strengths The resultingvalues are then used in BS 8002 for both serviceability andultimate requirements In BS 8110 the loads obtained fromBS 8002 are multiplied by a partial safety factor at the ULS

Although the design requirements are essentially the samein the British and European codes there are differences ofterminology and in the values of partial safety factors In theEurocodes loads are replaced by actions with dead loads as per-manent actions and all live loads as variable actions Each vari-able action is given several representative values to be used forparticular purposes The Eurocodes provide a more unifiedapproach to both structural and geotechnical design

Details of design requirements and partial safety factors to beapplied to loads and material strengths are given in Chapter 21for British Codes and Chapter 29 for Eurocodes

22 LOADS (ACTIONS)

The loads (actions) acting on a structure generally consist ofa combination of dead (permanent) and live (variable) loads

Chapter 2

Design criteria safetyfactors and loads

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Design criteria safety factors and loads6

In limit-state design a design load (action) is calculated bymultiplying the characteristic (or representative) value by anappropriate partial factor of safety The characteristic value isgenerally a value specified in a relevant standard or code Inparticular circumstances it may be a value given by a client ordetermined by a designer in consultation with the client

In BS 8110 characteristic dead imposed and wind loadsare taken as those defined in and calculated in accordancewith BS 6399 Parts 1 2 and 3 In BS 5400 characteristicdead and live loads are given in Part 2 but these have beensuperseded in practice by the loads in the appropriateHighways Agency standards These include BD 3701 andBD 6094 and for the assessment of existing bridgesBD 2101 (refs 3ndash5)

When EC 2 Part 11 was first introduced as an ENVdocument characteristic loads were taken as the values given inBS 6399 but with the specified wind load reduced by 10 Thiswas intended to compensate for the partial safety factor appliedto wind at the ULS being higher in the Eurocodes than in BS8110 Representative values were then obtained by multiplyingthe characteristic values by factors given in the NAD In theEN documents the characteristic values of all actions are givenin EC 1 and the factors to be used to determine representativevalues are given in EC 0

23 DEAD LOADS (PERMANENT ACTIONS)

Dead loads include the weights of the structure itself andall permanent fixtures finishes surfacing and so on Whenpermanent partitions are indicated they should be included asdead loads acting at the appropriate locations Where any doubtexists as to the permanency of the loads they should be treatedas imposed loads Dead loads can be calculated from the unitweights given in EC 1 Part 11 or from actual known weightsof the materials used Data for calculating dead loads are givenin Tables 21 and 22

24 LIVE LOADS (VARIABLE ACTIONS)

Live loads comprise any transient external loads imposed on thestructure in normal use due to gravitational dynamic andenvironmental effects They include loads due to occupancy(people furniture moveable equipment) traffic (road railpedestrian) retained material (earth liquids granular) snowwind temperature ground and water movement wave actionand so on Careful assessment of actual and probable loads is avery important factor in producing economical and efficientstructures Some imposed loads like those due to containedliquids can be determined precisely Other loads such as thoseon floors and bridges are very variable Snow and wind loadsare highly dependent on location Data for calculating loadsfrom stored materials are given in EC 1 Part 11

241 Floors

For most buildings the loads imposed on floors are specified inloading standards In BS 6399 Part 1 loads are specifiedaccording to the type of activity or occupancy involved Datafor residential buildings and for offices and particular workareas is given in Table 23 Imposed loads are given both as

a uniformly distributed load in kNm2 and a concentrated loadin kN The floor should be designed for the worst effects ofeither load The concentrated load needs to be considered forisolated short span members and for local effects such aspunching in a thin flange For this purpose a square contactarea with a 50 mm side may be assumed in the absence of anymore specific information Generally the concentrated loaddoes not need to be considered in slabs that are either solid orotherwise capable of effective lateral distribution Wherean allowance has to be made for non-permanent partitions auniformly distributed load equal to one-third of the load permetre run of the finished partitions may be used For offices theload used should not be less than 10 kNm2

The floors of garages are considered in two categoriesnamely those for cars and light vans and those for heaviervehicles In the lighter category the floor may be designedfor loads specified in the form described earlier In the heaviercategory the most adverse disposition of loads determinedfor the specific types of vehicle should be considered

The total imposed loads to be used for the design of beams maybe reduced by a percentage that increases with the area of floorsupported as given in Table 23 This does not apply to loads dueto storage vehicles plant or machinery For buildings designed tothe Eurocodes imposed loads are given in EC 1 Part 11

In all buildings it is advisable to affix a notice indicatingthe imposed load for which the floor is designed Floors ofindustrial buildings where plant and machinery are installedneed to be designed not only for the load when the plant is inrunning order but also for the probable load during erectionand testing which in some cases may be more severe Datafor loads imposed on the floors of agricultural buildings bylivestock and farm vehicles is given in BS 5502 Part 22

242 Structures subject to dynamic loads

The loads specified in BS 6399 Part 1 include allowancesfor small dynamic effects that should be sufficient for mostbuildings However the loading does not necessarily coverconditions resulting from rhythmical and synchronised crowdmovements or the operation of some types of machinery

Dynamic loads become significant when crowd movements(eg dancing jumping rhythmic stamping) are synchronisedIn practice this is usually associated with lively pop concertsor aerobics events where there is a strong musical beat Suchactivities can generate both horizontal and vertical loads Ifthe movement excites a natural frequency of the affected partof the structure resonance occurs which can greatly amplify theresponse Where such activities are likely to occur the structureshould be designed to either avoid any significant resonanceeffects or withstand the anticipated dynamic loads Somelimited guidance on dynamic loads caused by activities such asjumping and dancing is provided in BS 6399 Part 1 Annexe ATo avoid resonance effects the natural frequency of vibrationof the unloaded structure should be greater than 84 Hz for thevertical mode and greater than 40 Hz for the horizontal modes

Different types of machinery can give rise to a wide range ofdynamic loads and the potential resonant excitation of thesupporting structure should be considered Where necessaryspecialist advice should be sought

Footbridges are subject to particular requirements that willbe examined separately in the general context of bridges

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243 Parapets barriers and balustrades

Parapets barriers balustrades and other elements intended toretain stop or guide people should be designed for horizontalloads Values are given in BS 6399 Part 1 for a uniformlydistributed line load and for both uniformly distributed andconcentrated loads applied to the infill These are not takentogether but are applied as three separate load cases The lineload should be considered to act at a height of 11 m above adatum level taken as the finished level of the access platformor the pitch line drawn through the nosing of the stair treads

Vehicle barriers for car parking areas are also includedin BS 6399 Part 1 The horizontal force F as given in thefollowing equation is considered to act at bumper heightnormal to and uniformly distributed over any length of 15 m ofthe barrier By the fundamental laws of dynamics

F 05mv2(b c) (in kN)

m gross mass of vehicle (in kg)v speed of vehicle normal to barrier taken as 45 msecb deflection of barrier (in mm)c deformation of vehicle taken as 100 mm unless better

evidence is available

For car parks designed on the basis that the gross mass of thevehicles using it will not exceed 2500 kg (but taking as arepresentative value of the vehicle population m 1500 kg)and provided with rigid barriers (b 0) F is taken as 150 kNacting at a height of 375 mm above floor level It should benoted that bumper heights have been standardised at 445 mm

244 Roofs

The imposed loads given in Table 24 are additional to allsurfacing materials and include for snow and other incidentalloads but exclude wind pressure The snow load on the roofis determined by multiplying the estimated snow load on theground at the site location and altitude (the site snow load) byan appropriate snow load shape coefficient The main loadingconditions to be considered are

(a) a uniformly distributed snow load over the entire rooflikely to occur when snow falls with little or no wind

(b) a redistributed (or unevenly deposited) snow load likely tooccur in windy conditions

For flat or mono-pitch roofs it is sufficient to consider thesingle load case resulting from a uniform layer of snow asgiven in Table 24 For other roof shapes and for the effects oflocal drifting of snow behind parapets reference should bemade to BS 6399 Part 3 for further information

Minimum loads are given for roofs with no access (other thanthat necessary for cleaning and maintenance) and for roofswhere access is provided Roofs like floors should be designedfor the worst effects of either the distributed load or theconcentrated load For roofs with access the minimum loadwill exceed the snow load in most cases

If a flat roof is used for purposes such as a cafeacute playgroundor roof garden the appropriate imposed load for such a floorshould be allowed For buildings designed to the Eurocodessnow loads are given in EC 1 Part 13

245 Columns walls and foundations

Columns walls and foundations of buildings are designed forthe same loads as the slabs or beams that they support If theimposed loads on the beams are reduced according to the areaof floor supported the supporting members may be designedfor the same reduced loads Alternatively where two or morefloors are involved and the loads are not due to storage theimposed loads on columns or other supporting members maybe reduced by a percentage that increases with the number offloors supported as given in Table 23

246 Structures supporting cranes

Cranes and other hoisting equipment are often supported oncolumns in factories or similar buildings It is important that adimensioned diagram of the actual crane to be installed isobtained from the makers to ensure that the right clearances areprovided and the actual loads are taken into account For loadsdue to cranes reference should be made to BS 2573

For jib cranes running on rails on supporting gantries theload to which the structure is subjected depends on the actualdisposition of the weights of the crane The wheel loads aregenerally specified by the crane maker and should allow forthe static and dynamic effects of lifting discharging slewingtravelling and braking The maximum wheel load underpractical conditions may occur when the crane is stationary andhoisting the load at the maximum radius with the line of the jibdiagonally over one wheel

247 Structures supporting lifts

The effect of acceleration must be considered in addition to thestatic loads when calculating loads due to lifts and similarmachinery If a net static load F is subject to an accelerationa (ms2) the resulting load on the supporting structure isapproximately F (1 0098a) The average acceleration ofa passenger lift may be about 06 ms2 but the maximumacceleration will be considerably greater BS 2655 requiresthe supporting structure to be designed for twice the loadsuspended from the beams when the lift is at rest with anoverall factor of safety of 7 The deflection under the designload should not exceed span1500

248 Bridges

The analysis and design of bridges is now so complex thatit cannot be adequately treated in a book of this nature andreference should be made to specialist publications Howeverfor the guidance of designers the following notes regardingbridge loading are provided since they may also be applicableto ancillary construction and to structures having features incommon with bridges

Road bridges The loads to be considered in the design ofpublic road bridges in the United Kingdom are specified in theHighways Agency Standard BD 3701 Loads for HighwayBridges This is a revised version of BS 5400 Part 2 issuedby the Department of Transport rather than by BSI TheStandard includes a series of major amendments as agreed bythe BSI Technical Committee BD 3701 deals with both perma-nent loads (dead superimposed dead differential settlementearth pressure) and transient loads due to traffic use (vehicular

Live loads (variable actions) 7

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pedestrian) and environmental effects (wind temperature)The collision loads in BD 3701 may be applicable in certaincircumstances where agreed with the appropriate authority butin most cases the requirements of BD 6094 The design ofhighway bridges for vehicle collision loads will apply

Details of live loads due to traffic to be considered in thedesign of highway bridges are given in Table 25 Two typesof standard live loading are given in BD 3701 to representnormal traffic and abnormal vehicles respectively Loads areapplied to notional lanes of equal width The number ofnotional lanes is determined by the width of the carriagewaywhich includes traffic lanes hard shoulders and hard stripsand several typical examples are shown diagrammatically inBD 3701 Notional lanes are used rather than marked lanesin order to allow for changes of use and the introduction oftemporary contra-flow schemes

Type HA loading covers all the vehicles allowable under theRoad Vehicles (Construction and Use) and Road Vehicles(Authorised Weight) Regulations Values are given in terms ofa uniformly distributed load (UDL) and a single knife-edgeload (KEL) to be applied in combination to each notional laneThe specified intensity of the UDL (kNm) reduces as the loadedlength increases which allows for two effects At the shorterend it allows for loading in the vicinity of axles or bogies beinggreater than the average loading for the whole vehicle At thelonger end it takes account of the reducing percentage of heavygoods vehicles contained in the total vehicle population TheKEL of 120 kN is to be applied at any position within the UDLloaded length and spread over a length equal to the notional lanewidth In determining the loads consideration has been given tothe effects of impact vehicle overloading and unforeseen changesin traffic patterns The loading derived after application ofseparate factors for each of these effects was considered torepresent an ultimate load which was then divided by 15to obtain the specified nominal loads

The loads are multiplied by lane factors whose valuesdepend on the particular lane and the loaded length This isdefined as the length of the adverse area of the influence linethat is the length over which the load application increases themagnitude of the effect to be determined The lane factors takeaccount of the low probability of all lanes being fully loaded atthe same time They also for the shorter loaded lengths allowfor the effect of lateral bunching of vehicles As an alternativeto the combined loads a single wheel load of 100 kN appliedat any position is also to be considered

Type HB loading derives from the nature of exceptionalindustrial loads such as electrical transformers generatorspressure vessels and machine presses likely to use the roads inthe neighbouring area It is represented by a sixteen-wheelvehicle consisting of two bogies each one having two axleswith four wheels per axle Each axle represents one unit ofloading (equivalent to 10 kN) Bridges on public highwaysare designed for a specific number of units of HB loadingaccording to traffic use typically 45 units for trunk roads andmotorways 375 units for principal roads and 30 units for allother public roads Thus the maximum number of 45 unitscorresponds to a total vehicle load of 1800 kN with 450 kNper axle and 1125 kN per wheel The length of the vehicle isvariable according to the spacing of the bogies for which fivedifferent values are specified The HB vehicle can occupy anytransverse position on the carriageway and is considered to

displace HA loading over a specified area surrounding thevehicle Outside this area HA loading is applied as specifiedand shown by diagrams in BD 3701 The combined loadarrangement is normally critical for all but very long bridges

Road bridges may be subjected to forces other than those dueto dead load and traffic load These include forces due to windtemperature differential settlement and earth pressure Theeffects of centrifugal action and longitudinal actions due totraction braking and skidding must also be considered as wellas vehicle collision loads on supports and superstructure Fordetails of the loads to be considered on highway bridge parapetsreference should be made to BD 5293 (ref 6)

In the assessment of existing highway bridges traffic loadsare specified in the Highways Agency document BD 2101 TheAssessment of Highway Bridges and Structures In this casethe type HA loading is multiplied by a reduction factor thatvaries according to the road surface characteristics traffic flowconditions and vehicle weight restrictions Some of the contin-gency allowances incorporated in the design loading havealso been relaxed Vehicle weight categories of 40 38 25 1775 and 3 tonnes are considered as well as two groups offire engines For further information on reduction factors andspecific details of the axle weight and spacing values in eachcategory reference should be made BD 2101

Footbridges Details of live loads due to pedestrians to beconsidered in the design of footcycle track bridges are given inTable 26 A uniformly distributed load of 5 kNm2 is specified forloaded lengths up to 36 m Reduced loads may be used for bridgeswhere the loaded length exceeds 36 m except that specialconsideration is required in cases where exceptional crowds couldoccur For elements of highway bridges supporting footwayscycle tracks further reductions may be made in the pedestrian liveload where the width is greater than 2 m or the element alsosupports a carriageway When the footwaycycle track is notprotected from vehicular traffic by an effective barrier there is aseparate requirement to consider an accidental wheel loading

It is very important that consideration is given to vibrationthat could be induced in footcycle track bridges by resonancewith the movement of users or by deliberate excitation InBD 3701 the vibration requirement is deemed to be satisfiedin cases where the fundamental natural frequency of vibrationexceeds 5 Hz for the unloaded bridge in the vertical direction and15 Hz for the loaded bridge in the horizontal direction Whenthe fundamental natural frequency of vertical vibration fo doesnot exceed 5 Hz the maximum vertical acceleration shouldbe limited to 05radicfo ms2 Methods for determining the naturalfrequency of vibration and the maximum vertical accelerationare given in Appendix B of BD 3701 Where the fundamentalnatural frequency of horizontal vibration does not exceed15 Hz special consideration should be given to the possibilityof pedestrian excitation of lateral movements of unacceptablemagnitude Bridges possessing low mass and damping andexpected to be used by crowds of people are particularlysusceptible to such vibrations

Railway bridges Details of live loads to be considered in thedesign of railway bridges are given in Table 26 Two types ofstandard loading are given in BD 3701 type RU for mainline railways and type RL for passenger rapid transit systemsA further type SW0 is also included for main line railways

Design criteria safety factors and loads8

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The type RU loading was derived by a Committee of theInternational Union of Railways (UIC) to cover present andanticipated future loading on railways in Great Britain and on theContinent of Europe Nowadays motive power tends to be dieseland electric rather than steam and this produces axle loads andarrangements for locomotives that are similar to those for bogiefreight vehicles (these often being heavier than the locomotivesthat draw them) In addition to normal train loading whichcan be represented quite well by a uniformly distributed loadof 80 kNm railway bridges are occasionally subjected toexceptionally heavy abnormal loads For short loaded lengths itis necessary to introduce heavier concentrated loads to simulateindividual axles and to produce high shears at the ends Type RUloading consists of four concentrated loads of 250 kN precededand followed by a uniformly distributed load of 80 kNm For acontinuous bridge type SW0 loading is also to be considered asan additional and separate load case This loading consists oftwo uniformly distributed loads of 133 kNm each 15 m longseparated by a distance of 53 m Both types of loading whichare applied to each track or as specified by the relevant authoritywith half the track load acting on each rail are to be multipliedby appropriate dynamic factors to allow for impact lurchingoscillation and other dynamic effects The factors have beencalculated so that in combination with the specified loading theycover the effects of slow moving heavy and fast moving lightvehicles Exceptional vehicles are assumed to move at speeds notexceeding 80 kmh heavy wagons at speeds up to 120 kmh andpassenger trains at speeds up to 200 kmh

The type RL loading was derived by the London TransportExecutive to cover present and anticipated future loading onlines that carry only rapid transit passenger trains and lightengineersrsquo works trains Passenger trains include a variety ofstock of different ages loadings and gauges used on surfaceand tube lines Works trains include locomotives cranesand wagons used for maintenance purposes Locomotives areusually of the battery car type but diesel shunt varieties aresometimes used The rolling stock could include a 30t steamcrane 6t diesel cranes 20t hopper cranes and bolster wagonsThe heaviest train would comprise loaded hopper wagonshauled by battery cars Type RL loading consists of a singleconcentrated load of 200 kN coupled with a uniformly distrib-uted load of 50 kNm for loaded lengths up to 100 m Forloaded lengths in excess of 100 m the previous loading ispreceded and followed by a distributed load of 25 kNm Theloads are to be multiplied by appropriate dynamic factors Analternative bogie loading comprising two concentrated loadsone of 300 kN and the other of 150 kN spaced 24 m apart isalso to be considered on deck structures to check the ability ofthe deck to distribute the loads adequately

For full details of the locomotives and rolling stock coveredby each loading type and information on other loads to be con-sidered in the design of railway bridges due to the effects ofnosing centrifugal action traction and braking and in the eventof derailment reference should be made to BD 3701

249 Dispersal of wheel loads

A load from a wheel or similar concentrated load bearing on asmall but definite area of the supporting surface (called thecontact area) may be assumed to be further dispersed over anarea that depends on the combined thickness of any surfacing

material filling and underlying constructional material Thewidth of the contact area of a wheel on a slab is equal tothe width of the tyre The length of the contact area depends onthe type of tyre and the nature of the slab surface It is nearlyzero for steel tyres on steel plate or concrete The maximumcontact length is probably obtained with an iron wheel on loosemetalling or a pneumatic tyre on an asphalt surface

The wheel loads given in BD 3701 as part of the standardhighway loading are to be taken as uniformly distributed overa circular or square contact area assuming an effective pressureof 11 Nmm2 Thus for the HA single wheel load of 100 kNthe contact area becomes a 340 mm diameter circle or a squareof 300 mm side For the HB vehicle where 1 unit of loadingcorresponds to 25 kN per wheel the side of the square contactarea becomes approximately 260 mm for 30 units 290 mm for375 units and 320 mm for 45 units

Dispersal of the load beyond the contact area may be takenat a spread-to-depth ratio of 1 horizontally to 2 vertically forasphalt and similar surfacing so that the dimensions of thecontact area are increased by the thickness of the surfacingThe resulting boundary defines the loaded area to be used whenchecking for example the effects of punching shear on theunderlying structure

For a structural concrete slab 45o spread down to the levelof the neutral axis may be taken Since for the purpose ofstructural analysis the position of the neutral axis is usuallytaken at the mid-depth of the section the dimensions of thecontact area are further increased by the total thickness of theslab The resulting boundary defines the area of the patch loadto be used in the analysis

The concentrated loads specified in BD 3701 as part of therailway loading will be distributed both longitudinally bythe continuous rails to more than one sleeper and transverselyover a certain area of deck by the sleeper and ballast It maybe assumed that two-thirds of a concentrated load applied toone sleeper will be transmitted to the deck by that sleeper andthe remainder will be transmitted equally to the adjacent sleeperon either side Where the depth of ballast is at least 200 mmthe distribution may be assumed to be half to the sleeper lyingunder the load and half equally to the adjacent sleeper oneither side The load acting on the sleeper from each rail maybe distributed uniformly over the ballast at the level of theunderside of the sleeper for a distance taken symmetricallyabout the centreline of the rail of 800 mm or twice the distancefrom the centreline of the rail to the nearer end of the sleeperwhichever is the lesser Dispersal of the loads applied to theballast may be taken at an angle of 5o to the vertical down tothe supporting structure The distribution of concentrated loadsapplied to a track without ballast will depend on the relativestiffness of the rail the rail support and the bridge deck itself

25 WIND LOADS

All structures built above ground level are affected by thewind to a greater or lesser extent Wind comprises a randomfluctuating velocity component (turbulence or lsquogustinessrsquo)superimposed on a steady mean component The turbulenceincreases with the roughness of the terrain due to frictionaleffects between the wind and features on the ground such asbuildings and vegetation On the other hand the frictionaleffects also reduce the mean wind velocity

Wind loads 9

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Wind loads are dynamic and fluctuate continuously in bothmagnitude and position Some relatively flexible structuressuch as tall slender masts towers and chimneys suspensionbridges and other cable-stayed structures may be susceptible todynamic excitation in which case lateral deflections will be animportant consideration However the vast majority of build-ings are sufficiently stiff for the deflections to be small inwhich case the structure may be designed as if it was static

251 Wind speed and pressure

The local wind climate at any site in the United Kingdom can bepredicted reliably using statistical methods in conjunction withboundary-layer wind flow models However the complexity offlow around structures is not sufficiently well understood toallow wind pressures or distributions to be determined directlyFor this reason the procedure used in most modern wind codesis to treat the calculation of wind speed in a fully probabilisticmanner whilst continuing to use deterministic values of pressurecoefficients This is the approach adopted in BS 6399 Part 2which offers a choice of two methods for calculating wind loadsas follows

standard method uses a simplified procedure to obtain aneffective wind speed which is used with standard pressurecoefficients for orthogonal load cases

directional method provides a more precise assessment ofeffective wind speeds for particular wind directions which isused with directional pressure coefficients for load cases ofany orientation

The starting point for both methods is the basic hourly-meanwind speed at a height of 10 m in standard lsquocountryrsquo terrainhaving an annual risk (probability) of being exceeded of 002(ie a mean recurrence interval of 50 years) A map of basicwind speeds covering Great Britain and Ireland is provided

The basic hourly-mean wind speed is corrected according tothe site altitude and if required the wind direction seasonand probability to obtain an effective site wind speed This isfurther modified by a site terrain and building height factorto obtain an effective gust wind speed Ve ms which is used tocalculate an appropriate dynamic pressure q 0613Ve

2 Nm2Topographic effects are incorporated in the altitude factor for

the standard method and in the terrain and building factor for thedirectional method The standard method can be used in hand-based calculations and gives a generally conservative resultwithin its range of applicability The directional method is lessconservative and is not limited to orthogonal design cases Theloading is assessed in more detail but with the penalty ofincreased complexity and computational effort For furtherdetails of the directional method reference should be made toBS 6399 Part 2

252 Buildings

The standard method of BS 6399 Part 2 is the source of theinformation in Tables 27ndash29 The basic wind speed andthe correction factors are given in Table 27 The altitudefactor depends on the location of the structure in relation tothe local topography In terrain with upwind slopes exceeding005 the effects of topography are taken to be significant for

certain designated zones of the upwind and downwind slopesIn this case further reference should be made to BS 6399Part 2 When the orientation of the building is known the windspeed may be adjusted according to the direction under consid-eration Where the building height is greater than the crosswindbreadth for the direction being considered a reduction in thelateral load may be obtained by dividing the building into anumber of parts For buildings in town terrain the effectiveheight may be reduced as a result of the shelter afforded bystructures upwind of the site For details of the adjustmentsbased on wind direction division of buildings into parts and theinfluence of shelter on effective height reference should bemade to BS 6399 Part 2

When the wind acts on a building the windward faces aresubjected to direct positive pressure the magnitude of whichcannot exceed the available kinetic energy of the wind Asthe wind is deflected around the sides and over the roof of thebuilding it is accelerated lowering the pressure locally onthe building surface especially just downwind of the eavesridge and corners These local areas where the acceleration ofthe flow is greatest can experience very large wind suctionsThe surfaces of enclosed buildings are also subjected to internalpressures Values for both external and internal pressures areobtained by multiplying the dynamic pressure by appropriatepressure coefficients and size effect factors The overall force ona rectangular building is determined from the normal forces onthe windward-facing and leeward-facing surfaces the frictionaldrag forces on surfaces parallel to the direction of the wind anda dynamic augmentation factor that depends on the buildingheight and type

Details of the dimensions used to define surface pressuresand forces and values for dynamic augmentation factors andfrictional drag coefficients are given in Table 28 Size effectfactors and external and internal pressure coefficients for thewalls of rectangular buildings are given in Table 29 Furtherinformation including pressure coefficients for various roofforms free-standing walls and cylindrical structures such assilos tanks and chimneys and procedures for more-complexbuilding shapes are given in BS 6399 Part 2 For buildingsdesigned to the Eurocodes data for wind loading is given inEC 1 Part 12

253 Bridges

The approach used for calculating wind loads in BD 3701 is ahybrid mix of the methods given in BS 6399 Part 2 The direc-tional method is used to calculate the effective wind speed asthis gives a better estimate of wind speeds in towns and for sitesaffected by topography In determining the wind speed theprobability factor is taken as 105 appropriate to a return periodof 120 years Directional effective wind speeds are derived fororthogonal load cases and used with standard drag coefficientsto obtain wind loads on different elements of the structure suchas decks parapets and piers For details of the proceduresreference must be made to BD 3701

26 MARITIME STRUCTURES

The forces acting upon sea walls dolphins wharves jettiespiers docks and similar maritime structures include those dueto winds and waves blows and pulls from vessels the loads

Design criteria safety factors and loads10

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from cranes roads railways and stored goods imposed on thedeck and the pressures of earth retained behind the structure

For wharves or jetties of solid construction the energy ofimpact due to blows from vessels berthing is absorbed by themass of the structure usually without damage to the structureor vessel if fendering is provided With open constructionconsisting of braced piles or piers supporting the deck in whichthe mass of the structure is comparatively small the forcesresulting from impact must be considered The forces dependon the weight and speed of approach of the vessel on theamount of fendering and on the flexibility of the structureIn general a large vessel will approach at a low speed and asmall vessel at a higher speed Some typical examples are a1000 tonne vessel at 03 ms a 10 000 tonne vessel at 02 msand a 100 000 tonne vessel at 015 ms The kinetic energy of avessel displacing F tonnes approaching at a speed V ms isequal to 0514FV 2 kNm Hence the kinetic energy of a2000 tonne vessel at 03 ms and a 5000 tonne vessel at 02 msis about 100 kNm in each case If the direction of approachof a vessel is normal to the face of a jetty the whole of thisenergy must be absorbed on impact More commonly a vesselapproaches at an angle with the face of the jetty and touchesfirst at one point about which the vessel swings The energythen to be absorbed is 0514F[(Vsin13)2 ()2] with 13 theangle of approach of the vessel with the face of the jetty theradius of gyration (m) of the vessel about the point of impactand the angular velocity (radianss) of the vessel about the pointof impact The numerical values of the terms in the expressionare difficult to assess accurately and can vary considerablyunder different conditions of tide and wind and with differentvessels and methods of berthing

The kinetic energy of approach is absorbed partly by theresistance of the water but mainly by the fendering elasticdeformation of the structure and the vessel movement of theground and also by energy lsquolostrsquo upon impact The relativecontributions are difficult to assess but only about half ofthe total kinetic energy of the vessel may be imparted to thestructure and the fendering The force to which the structure issubjected is calculated by equating the product of the force andhalf the elastic horizontal displacement of the structure to thekinetic energy imparted Ordinary timber fenders appliedto reinforced concrete jetties cushion the blow but may notsubstantially reduce the force on the structure Spring fendersor suspended fenders can however absorb a large proportion ofthe kinetic energy Timber fenders independent of the jetty aresometimes provided to protect the structure from impact

The combined action of wind waves currents and tides on avessel moored to a jetty is usually transmitted by the vesselpressing directly against the side of the structure or by pullson mooring ropes secured to bollards The pulls on bollardsdue to the foregoing causes or during berthing vary with thesize of the vessel For vessels of up to 20 000 tonnes loadeddisplacement bollards are required at intervals of 15ndash30 m withload capacities according to the vessel displacement of 100 kNup to 2000 tonnes 300 kN up to 10 000 tonnes and 600 kN upto 20 000 tonnes

The effects of wind and waves acting on a marine structureare much reduced if an open construction is adopted and ifprovision is made for the relief of pressures due to water and airtrapped below the deck The force is not however relateddirectly to the proportion of solid vertical face presented to

the action of the wind and waves The pressures imposed areimpossible to assess with accuracy except for sea walls andsimilar structures where the depth of water at the face of the wallis such that breaking waves do not occur In this case the forceis due to simple hydrostatic pressure and can be evaluated forthe highest anticipated wave level with appropriate allowancefor wind surge In the Thames estuary for example the lattercan raise the high-tide level to 15 m above normal

A wave breaking against a sea wall causes a shock pressureadditional to the hydrostatic pressure which reaches its peakvalue at about mean water level and diminishes rapidly belowthis level and more slowly above it The shock pressure can beas much as 10 times the hydrostatic value and pressures up to650 kNm2 are possible with waves 45ndash6 m high The shape ofthe face of the wall the slope of the foreshore and the depthof water at the wall affect the maximum pressure and thedistribution of the pressure For information on the loads to beconsidered in the design of all types of maritime structuresreference should be made to BS 6349 Parts 1 to 7

27 RETAINED AND CONTAINED MATERIALS

The pressures imposed by materials on retaining structures orcontainment vessels are uncertain except when the retainedor contained material is a liquid In this case at any depth zbelow the free surface of the liquid the intensity of pressurenormal to the contact surface is equal to the vertical pressuregiven by the simple hydrostatic expression z wz where w

is unit weight of liquid (eg 981 kNm3 for water) For soilsand stored granular materials the pressures are considerablyinfluenced by the effective shear strength of the material

271 Properties of soils

For simplicity of analysis it is conventional to express the shearstrength of a soil by the equation

c n tan

where cis effective cohesion of soil is effective angle ofshearing resistance of soil n is effective normal pressure

Values of cand are not intrinsic soil properties and canonly be assumed constant within the stress range for which theyhave been evaluated For recommended fill materials it isgenerally sufficient to adopt a soil model with c 0 Such amodel gives a conservative estimate of the shear strength of thesoil and is analytically simple to apply in design Data takenfrom BS 8002 is given in Table 210 for unit weights of soilsand effective angles of shearing resistance

272 Lateral soil pressures

The lateral pressure exerted by a soil on a retaining structuredepends on the initial state of stress and the subsequent strainwithin the soil Where there has been no lateral strain eitherbecause the soil has not been disturbed during construction or thesoil has been prevented from lateral movement during placementan at-rest state of equilibrium exists Additional lateral strain isneeded to change the initial stress conditions Depending on themagnitude of the strain involved the final state of stress in the soilmass can be anywhere between the two failure conditions knownas the active and passive states of plastic equilibrium

Retained and contained materials 11

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The problem of determining lateral pressures at the limitingequilibrium conditions has been approached in different waysby different investigators In Coulomb theory the force actingon a retaining wall is determined by considering the limitingequilibrium of a soil wedge bounded by the rear face of thewall the ground surface and a planar failure surface Shearingresistance is assumed to have been mobilised both on the backof the wall and on the failure surface Rankine theory givesthe complete state of stress in a cohesionless soil mass whichis assumed to have expanded or compressed to a state of plasticequilibrium The stress conditions require that the earthpressure on a vertical plane should act in a direction parallel tothe ground surface Caquot and Kerisel produced tables ofearth pressure coefficients derived by a method that directlyintegrates the equilibrium equations along combined planar andlogarithmic spiral failure surfaces

273 Fill materials

A wide range of fill materials may be used behind retainingwalls All materials should be properly investigated and classi-fied Industrial chemical and domestic waste shale mudstoneand steel slag peaty or highly organic soil should not be usedas fill Selected cohesionless granular materials placed in acontrolled manner such as well-graded small rock-fills gravelsand sands are particularly suitable The use of cohesive soilscan result in significant economies by avoiding the need toimport granular materials but may also involve additionalproblems during design and construction The cohesive soilshould be within a range suitable for adequate compactionThe placement moisture content should be close to the finalequilibrium value to avoid either the swelling of clays placedtoo dry or the consolidation of clays placed too wet Suchproblems will be minimised if the fill is limited to clays with aliquid limit not exceeding 45 and a plasticity index notexceeding 25 Chalk with a saturation moisture content notexceeding 20 is acceptable as fill and may be compacted asfor a well-graded granular material Conditioned pulverizedfuel ash (PFA) from a single source may also be used it shouldbe supplied at a moisture-content of 80ndash100 of the optimumvalue For further guidance on the suitability of fill materialsreference should be made to relevant Transport ResearchLaboratory publications DoT Standard BD 3087 (ref 7)and BS 8002

274 Pressures imposed by cohesionless soils

Earth pressure distributions on unyielding walls and on rigidwalls free to translate or rotate about the base are shown inTable 211 For a normally consolidated soil the pressure on thewall increases linearly with depth Compaction results in higherearth pressures in the upper layers of the soil mass

Expressions for the pressures imposed in the at-rest activeand passive states including the effects of uniform surchargeand static ground water are given in sections 911ndash914Charts of earth pressure coefficients based on the work ofCaquot and Kerisel (ref 8) are given in Tables 212ndash214These may be used generally for vertical walls with slopingground or inclined walls with level ground

275 Cohesive soils

Clays in the long term behave as granular soils exhibitingfriction and dilation If a secant value (c 0) is used theprocedures for cohesionless soils apply If tangent parameters(c ) are used the RankinendashBell equations apply as given insection 915 In the short term if a clay soil is subjected torapid shearing a total stress analysis should be undertakenusing the undrained shear strength (see BS 8002)

276 Further considerations

For considerations such as earth pressures on embedded walls(with or without props) the effects of vertical concentrated loadsand line loads and the effects of groundwater seepage referenceshould be made to specialist books and BS 8002 For the pres-sures to be considered in the design of integral bridge abutmentsas a result of thermal movements of the deck reference shouldbe made to the Highways Agency document BA 4296 (ref 9)

277 Silos

Silos which may also be referred to as bunkers or bins aredeep containers used to store particulate materials In a deepcontainer the linear increase of pressure with depth found inshallow containers is modified When a deep container is filleda slight settlement of the fill activates the frictional resistancebetween the stored material and the wall This induces verticalload in the silo wall but reduces the vertical pressure in thematerial and the lateral pressures on the wall Janssen devel-oped a theory by which expressions have been derived for thepressures on the walls of a silo containing a granular materialhaving uniform properties The ratio of horizontal to verticalpressure in the fill is assumed constant and a Rankine coeffi-cient is generally used Eccentric filling (or discharge) tends toproduce variations in lateral pressure round the silo wall Anallowance is made by considering additional patch loads takento act on any part of the wall

Unloading a silo disturbs the equilibrium of the containedmass If the silo is unloaded from the top the frictional load onthe wall may be reversed as the mass re-expands but the lateralpressures remain similar to those during filling With a free-flowing material unloading at the bottom of the silo from thecentre of a hopper two different flow patterns are possibledepending on the characteristics of the hopper and the materialThese patterns are termed funnel flow (or core flow) and massflow respectively In the former a channel of flowing materialdevelops within a confined zone above the outlet the materialadjacent to the wall near the outlet remaining stationary Theflow channel can intersect the vertical walled section of the siloor extend to the surface of the stored material In mass flowwhich occurs particularly in steep-sided hoppers all the storedmaterial is mobilised during discharge Such flow can developat varying levels within the mass of material contained in anytall silo owing to the formation of a lsquoself-hopperrsquo with highlocal pressures arising where parallel flow starts to diverge fromthe walls Both flow patterns give rise to increases in lateralpressure from the stable filled condition Mass flow results alsoin a substantial local kick load at the intersection of the hopperand the vertical walled section

Design criteria safety factors and loads12

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When calculating pressures care should be taken to allow forthe inherent variability of the material properties In generalconcrete silo design is not sensitive to vertical wall load sovalues of maximum unit weight in conjunction with maximumor minimum consistent coefficients of friction should be usedData taken from EC 1 Part 4 for the properties of storedmaterials and the pressures on the walls and bottoms of silosare given in Tables 215 and 216

Fine powders like cement and flour can become fluidisedin silos either owing to rapid filling or through aeration tofacilitate discharge In such cases the design should allow forboth non-fluidised and fluidised conditions

28 EUROCODE LOADING STANDARDS

Eurocode 1 Actions on Structures is one of nine internationalunified codes of practice that have been published by the

European Committee for Standardization (CEN) The codewhich contains comprehensive information on all the actions(loads) normally necessary for consideration in the design ofbuilding and civil engineering structures consists of ten partsas follows

1991-1-1 Densities self-weight and imposed loads1991-1-2 Actions on structures exposed to fire1991-1-3 Snow loads1991-1-4 Wind loads1991-1-5 Thermal actions1991-1-6 Actions during execution1991-1-7 Accidental actions due to impact and explosions1991-2 Traffic loads on bridges1991-3 Actions induced by cranes and machinery1991-4 Actions on silos and tanks

Eurocode loading standards 13

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The requirements of concrete and its constituent materialsand of reinforcement are specified in Regulations Standardsand Codes of Practice Only those properties that concern thedesigner directly because they influence the behaviour anddurability of the structure are dealt with in this chapter

31 CONCRETE

Concrete is a structural material composed of crushed rockor gravel and sand bound together with a hardened paste ofcement and water A large range of cements and aggregateschemical admixtures and additions can be used to producea range of concretes having the required properties in boththe fresh and hardened states for many different structuralapplications The following information is taken mainly fromref 10 where a fuller treatment of the subject will be found

311 Cements and combinations

Portland cements are made from limestone and clay or otherchemically similar suitable raw materials which are burnedtogether in a rotary kiln to form a clinker rich in calciumsilicates This clinker is ground to a fine powder with a smallproportion of gypsum (calcium sulphate) which regulates therate of setting when the cement is mixed with water Overthe years several types of Portland cement have been developed

As well as cement for general use (which used to be knownas ordinary Portland cement) cements for rapid hardeningfor protection against attack by freezing and thawing or bychemicals and white cement for architectural finishes are alsomade The cements contain the same active compounds but indifferent proportions By incorporating other materials duringmanufacture an even wider range of cements is made includingair-entraining cement and combinations of Portland cementwith mineral additions Materials other than those in Portlandcements are used in cements for special purposes for examplecalcium aluminate cement is used for refractory concrete

The setting and hardening process that occurs when cementis mixed with water results from a chemical reaction known ashydration The process produces heat and is irreversible Settingis the gradual stiffening whereby the cement paste changesfrom a workable to a hardened state Subsequently the strengthof the hardened mass increases rapidly at first but slowinggradually This gain of strength continues as long as moisture ispresent to maintain the chemical reaction

Portland cements can be either inter-ground or blendedwith mineral materials at the cement factory or combined withadditions in the concrete mixer The most frequently used ofthese additional materials in the United Kingdom and therelevant British Standards are pulverized-fuel ash (pfa) toBS 3892 fly ash to BS EN 450 ground granulated blastfurnaceslag (ggbs) to BS 6699 and limestone fines to BS 7979 Otheradditions include condensed silica fume and metakaolin Theseare intended for specialised uses of concrete beyond the scopeof this book

The inclusion of pfa fly ash and ggbs has been particularlyuseful in massive concrete sections where they have been usedprimarily to reduce the temperature rise of the concrete withcorresponding reductions in temperature differentials and peaktemperatures The risk of early thermal contraction cracking isthereby also reduced The use of these additional materialsis also one of the options available for minimising the risk ofdamage due to alkalindashsilica reaction which can occur withsome aggregates and for increasing the resistance of concreteto sulfate attack Most additions react slowly at early stagesunder normal temperatures and at low temperature the reac-tion particularly in the case of ggbs can become considerablyretarded and make little contribution to the early strength ofconcrete However provided the concrete is not allowed to dryout the use of such additions can increase the long-termstrength and impermeability of the concrete

When the terms lsquowater-cement ratiorsquo and lsquocement contentrsquoare used in British Standards these are understood to includecombinations The word lsquobinderrsquo which is sometimes used isinterchangeable with the word lsquocementrsquo or lsquocombinationrsquo

The two methods of incorporating mineral additions makelittle or no difference to the properties of the concrete but therecently introduced notation system includes a unique code thatidentifies both composition and production method The typesof cement and combinations in most common usage are shownwith their notation in Table 217

Portland cement The most commonly used cement wasknown formerly as OPC in British Standards By grinding thecement clinker more finely cement with a more rapid earlystrength development is produced known formerly as RHPCBoth types are now designated as

Portland cement CEM I conforming to BS EN 197-1

Chapter 3

Material properties

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Cements are now classified in terms of both their standardstrength derived from their performance at 28 days and at anearly age normally two days using a specific laboratory testbased on a standard mortar prism This is termed the strengthclass for example CEM I 425N where 425 (Nmm2) is thestandard strength and N indicates a normal early strength

The most common standard strength classes for cements are425 and 525 These can take either N (normal) or R (rapid)identifiers depending on the early strength characteristics ofthe product CEM I in bags is generally a 425N cementwhereas CEM I for bulk supply tends to be 425R or 525 NCement corresponding to the former RHPC is now producedin the United Kingdom within the 525 strength class Thesecements are often used to advantage by precast concretemanufacturers to achieve a more rapid turn round of mouldsand on site when it is required to reduce the time for which theformwork must remain in position The cements which gener-ates more early heat than CEM I 425N can also be useful incold weather conditions

It is worth noting that the specified setting times of cementpastes relate to the performance of a cement paste of standardconsistence in a particular test made under closely controlledconditions of temperature and humidity the stiffening andsetting of concrete on site are not directly related to thesestandard setting regimes and are more dependent on factorssuch as workability cement content use of admixtures thetemperature of the concrete and the ambient conditions

Sulfate-resisting Portland cement SRPC This is a Portlandcement with a low tricalcium aluminate (C3A) content forwhich the British Standard is BS 4027 When concrete madewith CEM I cement is exposed to the sulfate solutions that arefound in some soils and groundwaters a reaction can occurbetween the sulfate and the hydrates from the C3A in thecement causing deterioration of the concrete By limitingthe C3A content in SRPC cement with a superior resistanceto sulfate attack is obtained SRPC normally has a low-alkalicontent but otherwise it is similar to other Portland cements inbeing non-resistant to strong acids The strength properties ofSRPC are similar to those of CEM I 425N but slightly lessearly heat is normally produced This can be an advantage inmassive concrete and in thick sections SRPC is not normallyused in combination with pfa or ggbs

Blastfurnace slag cements These are cements incorporatingggbs which is a by-product of iron smelting obtained byquenching selected molten slag to form granules The slag canbe inter-ground or blended with Portland cement clinker atcertain cement works to produce

Portland-slag cement CEM IIA-S with a slag content of6ndash35 conformimg to BS EN 197-1 or more commonly

Blastfurnace cement CEM IIIA with a slag content of36ndash65 conforming to BS EN 197-1

Alternatively the granules may be ground down separately to awhite powder with a fineness similar to that of cement and thencombined in the concrete mixer with CEM I cement to producea blastfurnace cement Typical mixer combinations of 40ndash50ggbs with CEM I cement have a notation CIIIA and at this levelof addition 28-day strengths are similar to those obtained withCEM I 425N

As ggbs has little hydraulic activity of its own it is referredto as lsquoa latent hydraulic binderrsquo Cements incorporating ggbsgenerate less heat and gain strength more slowly with lowerearly age strengths than those obtained with CEM I cementThe aforementioned blastfurnace cements can be used insteadof CEM I cement but because the early strength developmentis slower particularly in cold weather it may not be suitablewhere early removal of formwork is required They are amoderately low-heat cement and can therefore be used toadvantage to reduce early heat of hydration in thick sectionsWhen the proportion of ggbs is 66ndash80 CEM IIIA and CIIIAbecome CEM IIIB and CIIIB respectively These were knownformerly as high-slag blastfurnace cements and are specifiedbecause of their lower heat characteristics or to impart resis-tance to sulfate attack

Because the reaction between ggbs and lime released by thePortland cement is dependent on the availability of moistureextra care has to be taken in curing concrete containing thesecements or combinations to prevent premature drying out andto permit the development of strength

Pulverized-fuel ash and fly ash cements The ash resultingfrom the burning of pulverized coal in power station furnaces isknown in the concrete sector as pfa or fly ash The ash whichis fine enough to be carried away in the flue gases is removedfrom the gases by electrostatic precipitators to prevent atmos-pheric pollution The resulting material is a fine powder ofglassy spheres that can have pozzolanic properties that iswhen mixed into concrete it can react chemically with thelime that is released during the hydration of Portland cementThe products of this reaction are cementitious and in certaincircumstances pfa or fly ash can be used as a replacement forpart of the Portland cement provided in the concrete

The required properties of ash to be used as a cementitiouscomponent in concrete are specified in BS EN 450 withadditional UK provisions for pfa made in BS 3892 Part 1 Flyash in the context of BS EN 450 means lsquocoal fly ashrsquo ratherthan ash produced from other combustible materials and fly ashconforming to BS EN 450 can be coarser than that conformingto BS 3892 Part 1

Substitution of these types of cement for Portland cement isnot a straightforward replacement of like for like and thefollowing points have to be borne in mind when consideringthe use of pfa concrete

Pfa reacts more slowly than Portland cement At early ageand particularly at low temperatures pfa contributes lessstrength in order to achieve the same 28-day compressivestrength the amount of cementitious material may need to beincreased typically by about 10 The potential strengthafter three months is likely to be greater than CEM I providedthe concrete is kept in a moist environment for example inunderwater structures or concrete in the ground

The water demand of pfa for equal consistence may beless than that of Portland cement

The density of pfa is about three-quarters that of Portlandcement

The reactivity of pfa and its effect on water demand and hencestrength depend on the particular pfa and Portland cementwith which it is used A change in the source of either materialmay result in a change in the replacement level required

Concrete 15

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When pfa is to be air-entrained the admixture dosage ratemay have to be increased or a different formulation thatproduces a more stable air bubble structure used

Portland-fly ash cement comprises in effect a mixture ofCEM I and pfa When the ash is inter-ground or blended withPortland cement clinker at an addition rate of 20ndash35 themanufactured cement is known as Portland-fly ash cementCEM IIB-V conforming to BS EN 197-1 When this combina-tion is produced in a concrete mixer it has the notation CIIB-Vconforming to BS 8500 Part 2

Typical ash proportions are 25ndash30 and these cements canbe used in concrete for most purposes They are likely to havea slower rate of strength development compared with CEM IWhen the cement contains 25ndash40 ash it may be used toimpart resistance to sulfate attack and can also be beneficial inreducing the harmful effects of alkalindashsilica reaction Wherehigher replacement levels of ash are used for improved low-heatcharacteristics the resulting product is pozzolanic (pfa) cementwith the notation if manufactured CEM IVB-V conforming toBS EN 197-1 or if combined in the concrete mixer CIVB-Vconforming to BS 8500 Part 2

Because the pozzolanic reaction between pfa or fly ash andfree lime is dependent on the availability of moisture extra carehas to be taken in curing concrete containing mineral additionsto prevent premature drying out and to permit the developmentof strength

Portland-limestone cement Portland cement incorporating6ndash35 of carefully selected fine limestone powder is knownas Portland-limestone cement conforming to BS EN 197-1When a 425N product is manufactured the typical limestoneproportion is 10ndash20 and the notation is CEM IIA-L or CEMIIA-LL It is most popular in continental Europe but its usageis growing in the United Kingdom Decorative precast andreconstituted stone concretes benefit from its lighter colouringand it is also used for general-purpose concrete in non-aggressiveand moderately aggressive environments

312 Aggregates

The term lsquoaggregatersquo is used to describe the gravels crushedrocks and sands that are mixed with cement and water to pro-duce concrete As aggregates form the bulk of the volume ofconcrete and can significantly affect its performance the selec-tion of suitable material is extremely important Fine aggregatesinclude natural sand crushed rock or crushed gravel that is fineenough to pass through a sieve with 4 mm apertures (formerly5 mm as specified in BS 882) Coarse aggregates compriselarger particles of gravel crushed gravel or crushed rock Mostconcrete is produced from natural aggregates that are specifiedto conform to the requirements of BS EN 12620 together withthe UK Guidance Document PD 6682-1 Manufactured light-weight aggregates are also sometimes used

Aggregates should be hard and should not contain materialsthat are likely to decompose or undergo volumetric changeswhen exposed to the weather Some examples of undesirablematerials are lignite coal pyrite and lumps of clay Coal andlignite may swell and decompose leaving small holes on thesurface of the concrete lumps of clay may soften and formweak pockets and pyrite may decompose causing iron oxide

stains to appear on the concrete surface When exposed tooxygen pyrite has been known to contribute to sulfate attackHigh-strength concretes may call for special propertiesThe mechanical properties of aggregates for heavy-dutyconcrete floors and for pavement wearing surfaces may have tobe specially selected Most producers of aggregate are ableto provide information about these properties and referencewhen necessary should be made to BS EN 12620

There are no simple tests for aggregate durability or theirresistance to freezethaw exposure conditions and assessmentof particular aggregates is best based on experience of theproperties of concrete made with the type of aggregate andknowledge of its source Some flint gravels with a white porouscortex may be frost-susceptible because of the high waterabsorption of the cortex resulting in pop-outs on the surface ofthe concrete when subjected to freezethaw cycles

Aggregates must be clean and free from organic impuritiesThe particles should be free from coatings of dust or clay asthese prevent proper bonding of the material An excessiveamount of fine dust or stone lsquoflourrsquo can prevent the particles ofstone from being properly coated with cement and lower thestrength of the concrete Gravels and sands are usually washedby the suppliers to remove excess fines (eg clay and silt) andother impurities which otherwise could result in a poor-qualityconcrete However too much washing can also remove allfine material passing the 025 mm sieve This may result in aconcrete mix lacking in cohesion and in particular one that isunsuitable for placing by pump Sands deficient in fines alsotend to increase the bleeding characteristics of the concreteleading to poor vertical finishes due to water scour

Where the colour of a concrete surface finish is importantsupplies of aggregate should be obtained from the one sourcethroughout the job whenever practicable This is particularlyimportant for the sand ndash and for the coarse aggregate when anexposed-aggregate finish is required

Size and grading The maximum size of coarse aggregateto be used is dependent on the type of work to be done Forreinforced concrete it should be such that the concrete can beplaced without difficulty surrounding all the reinforcementthoroughly and filling the corners of the formwork In theUnited Kingdom it is usual for the coarse aggregate to havea maximum size of 20 mm Smaller aggregate usually with amaximum size of 10 mm may be needed for concrete that is tobe placed through congested reinforcement and in thin sectionswith small covers In this case the cement content may haveto be increased by 10ndash20 to achieve the same strength andworkability as that obtained with a 20 mm maximum-sizedaggregate because both sand and water contents usually haveto be increased to produce a cohesive mix Larger aggregatewith a maximum size of 40 mm can be used for foundationsand mass concrete where there are no restrictions to the flowof the concrete It should be noted however that this sort ofconcrete is not always available from ready-mixed concreteproducers The use of a larger aggregate results in a slightlyreduced water demand and hence a slightly reduced cementcontent for a given strength and workability

The proportions of the different sizes of particles makingup the aggregate which are found by sieving are known asthe aggregate lsquogradingrsquo The grading is given in terms of thepercentage by mass passing the various sieves Continuously

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graded aggregates for concrete contain particles ranging in sizefrom the largest to the smallest in gap-graded aggregatessome of the intermediate sizes are absent Gap grading may benecessary to achieve certain surface finishes Sieves used formaking a sieve analysis should conform to BS EN 933-2Recommended sieve sizes typically range from 80 to 2 mm forcoarse aggregates and from 8 to 025 mm for fine aggregatesTests should be carried out in accordance with the proceduregiven in BS EN 933-1

An aggregate containing a high proportion of large particlesis referred to as being lsquocoarselyrsquo graded and one containing ahigh proportion of small particles as lsquofinelyrsquo graded Overallgrading limits for coarse fine and lsquoall-inrsquo aggregates arecontained in BS EN 12620 and PD 6682-1 All-in aggregatescomprising both coarse and fine materials should not be usedfor structural reinforced concrete work because the gradingwill vary considerably from time to time and hence from batchto batch thus resulting in excessive variation in the consistenceand the strength of the concrete To ensure that the properamount of sand is present the separate delivery storage andbatching of coarse and fine materials is essential Graded coarseaggregates that have been produced by layer loading (ie fillinga lorry with say two grabs of material size 10ndash20 mm andone grab of material size 4ndash10 mm) are seldom satisfactorybecause the unmixed materials will not be uniformly gradedThe producer should ensure that such aggregates are effectivelymixed before loading into lorries

For a high degree of control over concrete production andparticularly if high-quality surface finishes are required it isnecessary for the coarse aggregate to be delivered stored andbatched using separate single sizes

The overall grading limits for coarse and fine aggregates asrecommended in BS EN 12620 are given in Table 217 Thelimits vary according to the aggregate size indicated as dD inmillimetres where d is the lower limiting sieve size and D isthe upper limiting sieve size for example 420 Additionally thecoarsenessfineness of the fine aggregate is assessed againstthe percentage passing the 05 mm sieve to give a CP MPFP grading This compares with the C (coarse) M (medium)F (fine) grading used formerly in BS 882 Good concrete canbe made using sand within the overall limits but there may beoccasions such as where a high degree of control is requiredor a high-quality surface finish is to be achieved when it isnecessary to specify the grading to even closer limits On theother hand sand whose grading falls outside the overall limitsmay still produce perfectly satisfactory concrete Maintaining areasonably uniform grading is generally more important thanthe grading limits themselves

Marine-dredged aggregates Large quantities of aggregatesobtained by dredging marine deposits have been widely andsatisfactorily used for making concrete for many years Ifpresent in sufficient quantities hollow andor flat shells canaffect the properties of both fresh and hardened concrete andtwo categories for shell content are given in BS EN 12620 Inorder to reduce the corrosion risk of embedded metal limitsfor the chloride content of concrete are given in BS EN 206-1and BS 8500 To conform to these limits it is necessary formarine-dredged aggregates to be carefully and efficientlywashed in fresh water that is frequently changed in order toreduce the salt content Chloride contents should be checked

frequently throughout aggregate production in accordance withthe method given in BS EN 1744-1

Some sea-dredged sands tend to have a preponderance ofone size of particle and a deficiency in the amount passing the025 mm sieve This can lead to mixes prone to bleeding unlessmix proportions are adjusted to overcome the problemIncreasing the cement content by 5ndash10 can often offset thelack of fine particles in the sand Beach sands are generallyunsuitable for good-quality concrete since they are likely tohave high concentrations of chloride due to the accumulation ofsalt crystals above the high-tide mark They are also oftensingle-sized which can make the mix design difficult

Lightweight aggregates In addition to natural gravels andcrushed rocks a number of manufactured aggregates are alsoavailable for use in concrete Aggregates such as sintered pfaare required to conform to BS EN 13055-1 and PD 6682-4

Lightweight aggregate has been used in concrete for manyyears ndash the Romans used pumice in some of their constructionwork Small quantities of pumice are imported and still used inthe United Kingdom mainly in lightweight concrete blocksbut most lightweight aggregate concrete uses manufacturedaggregate

All lightweight materials are relatively weak because of theirhigher porosity which gives them reduced weight The resultinglimitation on aggregate strength is not normally a problemsince the concrete strength that can be obtained still exceedsmost structural requirements Lightweight aggregates are usedto reduce the weight of structural elements and to giveimproved thermal insulation and fire resistance

313 Water

The water used for mixing concrete should be free fromimpurities that could adversely affect the process of hydrationand consequently the properties of concrete For examplesome organic matter can cause retardation whilst chloridesmay not only accelerate the stiffening process but also causeembedded steel such as reinforcement to corrode Otherchemicals like sulfate solutions and acids can have harmfullong-term effects by dissolving the cement paste in concreteIt is important therefore to be sure of the quality of water If itcomes from an unknown source such as a pond or boreholeit needs to be tested BS EN 1008 specifies requirements forthe quality of the water and gives procedures for checking itssuitability for use in concrete

Drinking water is suitable of course and it is usual simplyto obtain a supply from the local water utility Some recycledwater is being increasingly used in the interests of reducing theenvironmental impact of concrete production Seawater hasalso been used successfully in mass concrete with no embeddedsteel Recycled water systems are usually found at large-scalepermanent mixing plants such as precast concrete factories andready-mixed concrete depots where water that has been usedfor cleaning the plant and washing out mixers can be collectedfiltered and stored for re-use Some systems are able to reclaimup to a half of the mixing water in this way Large volumesettlement tanks are normally required The tanks do not needto be particularly deep but should have a large surface area andideally the water should be made to pass through a series ofsuch tanks becoming progressively cleaner at each stage

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314 Admixtures

An admixture is a material usually a liquid which is added toa batch of concrete during mixing to modify the properties ofthe fresh or the hardened concrete in some way Most admix-tures benefit concrete by reducing the amount of free waterneeded for a given level of consistence often in addition tosome other specific improvement Permeability is therebyreduced and durability increased There are occasions when theuse of an admixture is not only desirable but also essentialBecause admixtures are added to concrete mixes in smallquantities they should be used only when a high degree ofcontrol can be exercised Incorrect dosage of an admixture canadversely affect strength and other properties of the concreteRequirements for the following main types of admixture arespecified in BS EN 934-2

Normal water-reducing admixtures Commonly knownas plasticisers or workability aids these act by reducing theinter-particle attraction within the cement to produce a moreuniform dispersion of the cement grains The cement paste isbetter lsquolubricatedrsquo and hence the amount of water needed toobtain a given consistency can be reduced The use of theseadmixtures can be beneficial in one of three ways

When added to a normal concrete at normal dosage theyproduce an increase in slump of about 50 mm This can beuseful in high-strength concrete rich in cement which wouldotherwise be too stiff to place

The water content can be reduced while maintaining the samecement content and consistence class the reduction in watercement ratio (about 10) results in increased strength andimproved durability This can also be useful for reducingbleeding in concrete prone to this problem and for increasingthe cohesion and thereby reducing segregation in concrete ofhigh consistence or in harsh mixes that sometimes arise withangular aggregates or low sand contents or when the sand isdeficient in fines

The cement content can be reduced while maintaining thesame strength and consistence class The watercement ratiois kept constant and the water and cement contents arereduced accordingly This approach should never be used ifthereby the cement content would be reduced below theminimum specified amount

Too big a dosage may result in retardation andor a degree ofair-entrainment without necessarily increasing workabilityand therefore may be of no benefit in the fresh concrete

Accelerating water-reducing admixtures Acceleratorsact by increasing the initial rate of chemical reaction betweenthe cement and the water so that the concrete stiffens hardensand develops strength more quickly They have a negligibleeffect on consistence and the 28-day strengths are seldomaffected Accelerating admixtures have been used mainlyduring cold weather when the slowing down of the chemicalreaction between cement and water at low temperature couldbe offset by the increased speed of reaction resulting fromthe accelerator The most widely used accelerator used to becalcium chloride but because the presence of chlorides even insmall amounts increases the risk of corrosion modern standardsprohibit the use of admixtures containing chlorides in all concrete

containing embedded metal Accelerators are sometimesmarketed under other names such as hardeners or anti-freezersbut no accelerator is a true anti-freeze and the use of anaccelerator does not avoid the need to protect the concrete incold weather by keeping it warm (with insulation) after ithas been placed

Retarding water-reducing admixtures These slow downthe initial reaction between cement and water by reducingthe rate of water penetration to the cement By slowing down thegrowth of the hydration products the concrete stays workablelonger than it otherwise would The length of time during whichconcrete remains workable depends on its temperature consis-tence class and watercement ratio and on the amount of retarderused Although the occasions justifying the use of retarders in theUnited Kingdom are limited these admixtures can be helpfulwhen one or more of the following conditions apply

In warm weather when the ambient temperature is higherthan about 20oC to prevent early stiffening (lsquogoing-offrsquo) andloss of workability which would otherwise make the placingand finishing of the concrete difficult

When a large concrete pour which will take several hours tocomplete must be constructed so that concrete already placeddoes not harden before the subsequent concrete can bemerged with it (ie without a cold joint)

When the complexity of a slip-forming operation requires aslow rate of rise

When there is a delay of more than 30 minutes betweenmixing and placing ndash for example when ready-mixed concreteis being used over long-haul distances or there are risks oftraffic delays This can be seriously aggravated during hotweather especially if the cement content is high

The retardation can be varied by altering the dosage a delayof 4ndash6 hours is usual but longer delays can be obtained forspecial purposes While the reduction in early strength ofconcrete may affect formwork-striking times the 7-day and28-day strengths are not likely to be significantly affectedRetarded concrete needs careful proportioning to minimisebleeding due to the longer period during which the concreteremains fresh

Air-entraining admixtures These may be organic resinsor synthetic surfactants that entrain a controlled amount of airin concrete in the form of small air bubbles The bubbles needto be about 50 microns in diameter and well dispersed Themain reason for using an air-entraining admixture is that thepresence of tiny bubbles in the hardened concrete increases itsresistance to the action of freezing and thawing especiallywhen this is aggravated by the application of de-icing saltsand fluids Saturated concrete ndash as most external paving willbe ndash can be seriously affected by the freezing of water inthe capillary voids which will expand and try to burst it If theconcrete is air-entrained the air bubbles which intersect thecapillaries stay unfilled with water even when the concrete issaturated Thus the bubbles act as pressure relief valves andcushion the expansive effect by providing voids into which thewater can expand as it freezes without disrupting the concreteWhen the ice melts surface tension effects draw the water backout of the bubbles

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Air-entrained concrete should be specified and used for allforms of external paving from major roads and airfieldrunways down to garage drives and footpaths which are likelyto be subjected to severe freezing and to de-icing salts The saltsmay be applied directly or come from the spray of passingtraffic or by dripping from the underside of vehicles

Air-entrainment also affects the properties of the freshconcrete The minute air bubbles act like ball bearings and havea plasticising effect resulting in a higher consistence Concretethat is lacking in cohesion or harsh or which tends to bleedexcessively is greatly improved by air-entrainment The riskof plastic settlement and plastic-shrinkage cracking is alsoreduced There is also evidence that colour uniformity isimproved and surface blemishes reduced One factor that has tobe taken into account when using air-entrainment is that thestrength of the concrete is reduced by about 5 for every 1 ofair entrained However the plasticising effect of the admixturemeans that the water content of the concrete can be reducedwhich will offset most of the strength loss that would otherwiseoccur but even so some increase in the cement content is likelyto be required

High-range water-reducing admixtures Commonlyknown as superplasticizers these have a considerable plasticizingeffect on concrete They are used for one of two reasons

To greatly increase the consistence of a concrete mix so thata lsquoflowingrsquo concrete is produced that is easy both to placeand to compact some such concretes are completely self-compacting and free from segregation

To produce high-strength concrete by reducing the watercontent to a much greater extent than can be achieved byusing a normal plasticizer (water-reducing admixture)

A flowing concrete is usually obtained by first producing aconcrete whose slump is in the range 50ndash90 mm and thenadding the superplasticizer which increases the slump to over200 mm This high consistence lasts for only a limited periodof time stiffening and hardening of the concrete then proceednormally Because of this time limitation when ready-mixedconcrete is being used it is usual for the superplasticizer to beadded to the concrete on site rather than at the batching ormixing plant Flowing concrete can be more susceptible tosegregation and bleeding so it is essential for the mix designand proportions to allow for the use of a superplasticizer As ageneral guide a conventionally designed mix needs to bemodified by increasing the sand content by about 5 A highdegree of control over the batching of all the constituents isessential especially the water because if the consistence of theconcrete is not correct at the time of adding the superplasticizerexcessive flow and segregation will occur

The use of flowing concrete is likely to be limited to workwhere the advantages in ease and speed of placing offset theincreased cost of the concrete ndash considerably more than withother admixtures Typical examples are where reinforcement isparticularly congested making both placing and vibrationdifficult and where large areas such as slabs would benefitfrom a flowing easily placed concrete The fluidity of flowingconcrete increases the pressures on formwork which should bedesigned to resist full hydrostatic pressure

When used to produce high-strength concrete reductions inwater content of as much as 30 can be obtained by using

superplasticizers compared to 10 with normal plasticizers asa result 1-day and 28-day strengths can be increased by as muchas 50 Such high-strength water-reduced concrete is used bothfor high-performance in situ concrete construction and for themanufacture of precast units where the increased early strengthallows earlier demoulding

315 Properties of fresh and hardening concrete

Workability It is vital that the workability of concrete ismatched to the requirements of the construction process Theease or difficulty of placing concrete in sections of varioussizes and shapes the type of compaction equipment neededthe complexity of the reinforcement the size and skills of theworkforce are amongst the items to be considered In generalthe more difficult it is to work the concrete the higher shouldbe the level of workability But the concrete must also havesufficient cohesiveness in order to resist segregation andbleeding Concrete needs to be particularly cohesive if it is tobe pumped or allowed to fall from a considerable height

The workability of fresh concrete is increasingly referred toin British and European standards as consistence The slumptest is the best-known method for testing consistence and theslump classes given in BS EN 206-1 are S1 (10ndash40 mm)S2 (50ndash90 mm) S3 (100ndash150 mm) S4 (160ndash210 mm) Threeother test methods recognised in BS EN 206-1 all with theirown unique consistency classes are namely Vebe timedegree of compactability and flow diameter

Plastic cracking There are two basic types of plastic cracksplastic settlement cracks which can develop in deep sectionsand often follow the pattern of the reinforcement and plasticshrinkage cracks which are most likely to develop in slabsBoth types form while the concrete is still in its plastic statebefore it has set or hardened and depending on the weatherconditions within about one to six hours after the concrete hasbeen placed and compacted They are often not noticed until thefollowing day Both types of crack are related to the extent towhich the fresh concrete bleeds

Fresh concrete is a suspension of solids in water and after ithas been compacted there is a tendency for the solids (bothaggregates and cement) to settle The sedimentation processdisplaces water which is pushed upwards and if excessiveappears as a layer on the surface This bleed water may notalways be seen since it can evaporate on hot or windy daysfaster than it rises to the surface Bleeding can generally bereduced by increasing the cohesiveness of the concrete This isusually achieved by one or more of the following meansincreasing the cement content increasing the sand contentusing a finer sand using less water air-entrainment using arounded natural sand rather than an angular crushed one Therate of bleeding will be influenced by the drying conditionsespecially wind and bleeding will take place for longer on colddays Similarly concrete containing a retarder tends to bleed fora longer period of time due to the slower stiffening rate ofthe concrete and the use of retarders will in general increasethe risk of plastic cracking

Plastic settlement cracks caused by differential settlementare directly related to the amount of bleeding They tend tooccur in deep sections particularly deep beams but they may

Concrete 19

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also develop in columns and walls This is because the deeperthe section the greater the sedimentation or settlement thatcan occur However cracks will form only where somethingprevents the concrete lsquosolidsrsquo from settling freely The mostcommon cause of this is the reinforcement fixed at the top ofdeep sections the concrete will be seen to lsquohang-uprsquo over thebars and the pattern of cracks will directly reflect the layout ofthe reinforcement below Plastic settlement cracks can alsooccur in trough and waffle slabs or at any section where thereis a significant change in the depth of concrete If alterationsto the concrete for example the use of an air-entraining orwater-reducing admixture cannot be made due to contractualor economic reasons the most effective way of eliminatingplastic settlement cracking is to re-vibrate the concrete afterthe cracks have formed Such re-vibration is acceptable whenthe concrete is still plastic enough to be capable of beinglsquofluidizedrsquo by a poker but not so stiff that a hole is left when thepoker is withdrawn The prevailing weather conditions willdetermine the timing of the operation

Plastic shrinkage cracks occur in horizontal slabs such asfloors and pavements They usually take the form of one ormore diagonal cracks at 05ndash2 m centres that do not extendto the slab edges or they form a very large pattern of mapcracking Such cracks are most common in concrete placed onhot or windy days because they are caused by the rate ofevaporation of moisture from the surface exceeding the rateof bleeding Clearly plastic shrinkage cracks can be reducedby preventing the loss of moisture from the concrete surface inthe critical first few hours While sprayed-on resin-based curingcompounds are very efficient at curing concrete that has alreadyhardened they cannot be used on fresh concrete until the freebleed water has evaporated This is too late to prevent plasticshrinkage cracking and so the only alternative is to protect theconcrete for the first few hours with polythene sheeting Thisneeds to be supported clear of the concrete by means of blocksor timber but with all the edges held down to prevent a wind-tunnel effect It has been found that plastic shrinkage crackingis virtually non-existent when air-entrainment is used

The main danger from plastic cracking is the possibility ofmoisture ingress leading to corrosion of any reinforcement Ifthe affected surface is to be covered subsequently by eithermore concrete or a screed no treatment is usually necessaryIn other cases often the best repair is to brush dry cement(dampened down later) or wet grout into the cracks the day afterthey form and while they are still clean this encourages naturalor autogenous healing

Early thermal cracking The reaction of cement with wateror hydration is a chemical reaction that produces heat If thisheat development exceeds the rate of heat loss the concretetemperature will rise Subsequently the concrete will cool andcontract Typical temperature histories of different concretesections are shown in the figure on Table 218

If the contraction of the concrete were unrestrained therewould be no cracking at this stage However in practice thereis nearly always some form of restraint inducing tension andhence a risk of cracks forming The restraint can occur due toboth external and internal influences Concrete is externallyrestrained when for example it is cast onto a previously castbase such as a wall kicker or between two already hardenedsections such as in infill bay in a wall or slab without the

provision of a contraction joint Internal restraint occurs forexample because the surfaces of an element will cool fasterthan the core producing a temperature differential When thisdifferential is large such as in thick sections surface cracksmay form at an early stage Subsequently as the core of thesection cools these surface cracks will tend to close in theabsence of any external restraints Otherwise the cracks willpenetrate into the core and link up to form continuous cracksthrough the whole section

The main factors affecting the temperature rise in concreteare the dimensions of the section the cement content andtype the initial temperature of the concrete and the ambienttemperature the type of formwork and the use of admixturesThicker sections retain more heat giving rise to higher peaktemperatures and cool down more slowly Within the coreof very thick sections adiabatic conditions obtain and abovea thickness of about 15 m there is little further increasein the temperature of the concrete The heat generated isdirectly related to the cement content For Portland cementconcretes in sections of thickness 1 m and more the temper-ature rise in the core is likely to be about 14oC for every100 kgm3 of cement Thinner sections will exhibit lowertemperature rises

Different cement types generate heat at different rates Thepeak temperature and the total amount of heat produced byhydration depend upon both the fineness and the chemistry ofthe cement As a guide the cements whose strength developsmost rapidly tend to produce the most heat Sulfate-resistingcement generally gives off less heat than CEM I and cementsthat are inter-ground or combined with mineral additions suchas pfa or ggbs are often chosen for massive constructionbecause of their low heat of hydration

A higher initial temperature results in a greater temperaturerise for example concrete in a 500 mm thick section placedat 10oC could have a temperature rise of 30oC but the sameconcrete placed at 20oC may have a temperature rise of 40oCSteel and GRP formwork will allow the heat generated to bedissipated more quickly than will timber formwork resultingin lower temperature rises especially in thinner sectionsTimber formwork andor additional insulation will reduce thetemperature differential between the core and the surface ofthe section but this differential could increase significantlywhen the formwork is struck Retarding water-reducers willdelay the onset of hydration but do not reduce the total heatgenerated Accelerating water-reducers will increase the rate ofheat evolution and the temperature rise

The problem of early thermal cracking is usually confined toslabs and walls Walls are particularly susceptible becausethey are often lightly reinforced in the horizontal directionand the timber formwork tends to act as a thermal insulatorencouraging a larger temperature rise The problem could bereduced by lowering the cement content and using cementwith a lower heat of hydration or one containing ggbs or pfaHowever there are practical and economic limits to thesemeasures often dictated by the specification requirements forthe strength and durability of the concrete itself In practicecracking due to external restraint is generally dealt with byproviding crack control reinforcement and contraction jointsWith very thick sections and very little external restraint ifthe temperature differential can be controlled by insulating theconcrete surfaces for a few days cracking can be avoided

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Typical values of the temperature rise in walls and slabs forPortland cement concretes as well as comparative values forconcrete using other cements are given in Table 218 Furtherdata on predicted temperature rises is given in ref 11

316 Properties of hardened concrete

Compressive strength The strength of concrete is specifiedas a strength class or grade namely the 28-day characteristiccompressive strength of specimens made from fresh concreteunder standardised conditions The results of strength tests areused routinely for control of production and contractual confor-mity purposes The characteristic strength is defined as that levelof strength below which 5 of all valid test results is expectedto fall Test cubes either 100 mm or 150 mm are the specimensnormally used in the United Kingdom and most other Europeancountries but cylinders are used elsewhere Because theirbasic shapes (ratio of height to cross-sectional dimension) aredifferent the strength test results are also different cylindersbeing weaker than cubes For normal-weight aggregates theconcrete cylinder strength is about 80 of the correspondingcube strength For lightweight aggregates cylinder strengths areabout 90 of the corresponding cube strengths

In British Codes of Practice like BS 8110 strength gradesused to be specified in terms of cube strength (eg C30) asshown in Table 39 Nowadays strength classes are specified interms of both cylinder strength and equivalent cube strength(eg C2530) as shown in Tables 35 and 42

In principle compressive strengths can be determined fromcores cut from the hardened concrete Core tests are normallymade only when there is some doubt about the quality ofconcrete placed (eg if the cube results are unsatisfactory) orto assist in determining the strength and quality of an existingstructure for which records are not available Great care isnecessary in the interpretation of the results of core tests andsamples drilled from in situ concrete are expected to be lowerin strength than cubes made cured and tested under standardlaboratory conditions The standard reference for core testingis BS EN 12504-1

Tensile strength The direct tensile strength of concrete asa proportion of the cube strength varies from about one-tenthfor low-strength concretes to one-twentieth for high-strengthconcretes The proportion is affected by the aggregate used andthe compressive strength is therefore only a very general guideto the tensile strength For specific design purposes in regard tocracking and shear strength analytical relationships betweenthe tensile strength and the specified cylindercube strength areprovided in codes of practice

The indirect tensile strength (or cylinder splitting strength) isseldom specified nowadays Flexural testing of specimens maybe used on some airfield runway contracts where the methodof design is based on the modulus of rupture and for someprecast concrete products such as flags and kerbs

Elastic properties The initial behaviour of concrete underservice load is almost elastic but under sustained loading thestrain increases with time Stressndashstrain tests cannot be carriedout instantaneously and there is always a degree of non-linearityand a residual strain upon unloading For practical purpose theinitial deformation is considered to be elastic (recoverable

upon unloading) and the subsequent increase in strain undersustained stress is defined as creep The elastic modulus onloading defined in this way is a secant modulus related to aspecific stress level The value of the modulus of elasticity ofconcrete is influenced mainly by the aggregate used With aparticular aggregate the value increases with the strength of theconcrete and the age at loading In special circumstances Forexample where deflection calculations are of great importanceload tests should be carried out on concrete made with theaggregate to be used in the actual structure For most designpurposes specific values of the mean elastic modulus at28 days and of Poissonrsquos ratio are given in Table 35 forBS 8110 and Table 42 for EC 2

Creep The increase in strain beyond the initial elastic valuethat occurs in concrete under a sustained constant stress aftertaking into account other time-dependent deformations notassociated with stress is defined as creep If the stress isremoved after some time the strain decreases immediately byan amount that is less than the original elastic value becauseof the increase in the modulus of elasticity with age This isfollowed by a further gradual decrease in strain The creeprecovery is always less than the preceding creep so that thereis always a residual deformation

The creep source in normal-weight concrete is the hardenedcement paste The aggregate restrains the creep in the paste sothat the stiffer the aggregate and the higher its volumetricproportion the lower is the creep of the concrete Creep isalso affected by the watercement ratio as is the porosity andstrength of the concrete For constant cement paste contentcreep is reduced by a decrease in the watercement ratio

The most important external factor influencing creep is therelative humidity of the air surrounding the concrete For aspecimen that is cured at a relative humidity of 100 thenloaded and exposed to different environments the lower therelative humidity the higher is the creep The values are muchreduced in the case of specimens that have been allowed todry prior to the application of load The influence of relativehumidity on creep is dependent on the size of the memberWhen drying occurs at constant relative humidity the largerthe specimen the smaller is the creep This size effect isexpressed in terms of the volumesurface area ratio of themember If no drying occurs as in mass concrete the creep isindependent of size

Creep is inversely proportional to concrete strength at the ageof loading over a wide range of concrete mixes Thus for agiven type of cement the creep decreases as the age andconsequently the strength of the concrete at application of theload increases The type of cement temperature and curingconditions all influence the development of strength with age

The influence of temperature on creep is important in the useof concrete for nuclear pressure vessels and containers forstoring liquefied gases The time at which the temperature ofconcrete rises relative to the time at which load is appliedaffects the creepndashtemperature relation If saturated concrete isheated and loaded at the same time the creep is greater thanwhen the concrete is heated during the curing period prior to theapplication of load At low temperatures creep behaviour isaffected by the formation of ice As the temperature falls creepdecreases until the formation of ice causes an increase in creepbut below the ice point creep again decreases

Concrete 21

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Creep is normally assumed to be directly proportional toapplied stress within the service range and the term specificcreep is used for creep per unit of stress At stresses aboveabout one-third of the cube strength (45 cylinder strength)the formation of micro-cracks causes the creepndashstress relationto become non-linear creep increasing at an increasing rate

The effect of creep is unfavourable in some circumstances(eg increased deflection) and favourable in others (eg reliefof stress due to restraint of imposed deformations such asdifferential settlement seasonal temperature change)

For normal exposure conditions (inside and outside) creepcoefficients according to ambient relative humidity effectivesection thickness (notional size) and age of loading are givenin Table 35 for BS 8110 and Table 43 for EC 2

Shrinkage Withdrawal of water from hardened concretekept in unsaturated air causes drying shrinkage If concretethat has been left to dry in air of a given relative humidity issubsequently placed in water (or a higher relative humidity)it will swell due to absorption of water by the cement pasteHowever not all of the initial drying shrinkage is recoveredeven after prolonged storage in water For the usual rangeof concretes the reversible moisture movement representsabout 40ndash70 of the drying shrinkage A pattern of alternatewetting and drying will occur in normal outdoor conditionsThe magnitude of the cyclic movement clearly depends uponthe duration of the wetting and drying periods but drying ismuch slower than wetting The consequence of prolonged dryweather can be reversed by a short period of rain More stableconditions exist indoors (dry) and in the ground or in contactwith water (eg reservoirs and tanks)

Shrinkage of hardened concrete under drying conditions isinfluenced by several factors in a similar manner to creep Theintrinsic shrinkage of the cement paste increases with thewatercement ratio so that for a given aggregate proportionconcrete shrinkage is also a function of watercement ratio

The relative humidity of the air surrounding the membergreatly affects the magnitude of concrete shrinkage accordingto the volumesurface area ratio of the member The lowershrinkage value of large members is due to the fact that dryingis restricted to the outer parts of the concrete the shrinkage ofwhich is restrained by the non-shrinking core Clearly shrink-able aggregates present special problems and can greatlyincrease concrete shrinkage (ref 12)

For normal exposure conditions (inside and outside) valuesof drying shrinkage according to ambient relative humidity andeffective section thickness (notional size) are given in Table 35for BS 8110 and Table 42 for EC 2

Thermal properties The coefficient of thermal expansionof concrete depends on both the composition of the concreteand its moisture condition at the time of the temperaturechange The thermal coefficient of the cement paste is higherthan that of the aggregate which exerts a restraining influenceon the movement of the cement paste The coefficient of thermalexpansion of a normally cured paste varies from the lowestvalues when the paste is either totally dry or saturated to amaximum at a relative humidity of about 70 Values for theaggregate are related to their mineralogical composition

A value for the coefficient of thermal expansion of concreteis needed in the design of structures such as chimneys tanks

containing hot liquids bridges and other elevated structuresexposed to significant solar effects and for large expanses ofconcrete where provision must be made to accommodate theeffects of temperature change in controlled cracking or byproviding movement joints For normal design purposes valuesof the coefficient of thermal expansion of concrete accordingto the type of aggregate are given in Table 35 for BS 8110 andTable 42 for EC 2

Short-term stressndashstrain curves For normal low to mediumstrength unconfined concrete the stressndashstrain relationship incompression is approximately linear up to about one-third ofthe cube strength (40 of cylinder strength) With increasingstress the strain increases at an increasing rate and a peakstress (cylinder strength) is reached at a strain of about 0002With increasing strain the stress reduces until failure occurs ata strain of about 00035 For higher strength concretes the peakstress occurs at strains 0002 and the failure occurs atstrains 00035 the failure being progressively more brittle asthe concrete strength increases

For design purposes the short-term stressndashstrain curve isgenerally idealised to a form in which the initial portion isparabolic or linear and the remainder is at a uniform stress Afurther simplification in the form of an equivalent rectangularstress block may be made subsequently Typical stressndashstraincurves and those recommended for design purposes are givenin Table 36 for BS 8110 and Table 44 for EC 2

317 Durability of concrete

Concrete has to be durable in natural environments rangingfrom mild to extremely aggressive and resistant to factors suchas weathering freezethaw attack chemical attack and abrasionIn addition for concrete containing reinforcement the surfaceconcrete must provide adequate protection against the ingressof moisture and air which would eventually cause corrosion ofthe embedded steel

Strength alone is not necessarily a reliable guide to concretedurability many other factors have to be taken into accountthe most important being the degree of impermeability This isdependent mainly on the constituents of the concrete in partic-ular the free watercement ratio and in the provision of fullcompaction to eliminate air voids and effective curing toensure continuing hydration

Concrete has a tendency to be permeable as a result ofthe capillary voids in the cement paste matrix In order for theconcrete to be sufficiently workable it is common to use farmore water than is actually necessary for the hydration of thecement When the concrete dries out the space previouslyoccupied by the excess water forms capillary voids Providedthe concrete has been fully compacted and properly cured thevoids are extremely small the number and the size of the voidsdecreasing as the free watercement ratio is reduced The moreopen the structure of the cement paste the easier it is for airmoisture and harmful chemicals to penetrate

Carbonation Steel reinforcement that is embedded in goodconcrete with an adequate depth of cover is protectedagainst corrosion by the highly alkaline pore water in thehardened cement paste Loss of alkalinity of the concrete canbe caused by the carbon dioxide in the air reacting with and

Material properties22

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neutralising the free lime If this reaction which is calledcarbonation reaches the reinforcement then corrosion willoccur in moist environments Carbonation is a slow processthat progresses from the surface and is dependent on thepermeability of the concrete and the humidity of the environ-ment Provided the depth of cover and quality of concreterecommended for the anticipated exposure conditions areachieved corrosion due to carbonation should not occur duringthe intended lifetime of the structure

Freezethaw attack The resistance of concrete to freezingand thawing depends on its impermeability and the degreeof saturation on being exposed to frost the higher the degree ofsaturation the more liable the concrete is to damage The useof salt for de-icing roads and pavements greatly increases therisk of freezethaw damage

The benefits of air-entrained concrete have been referred toin section 314 where it was recommended that all exposedhorizontal paved areas from roads and runways to footpathsand garage drives and marine structures should be made ofair-entrained concrete Similarly parts of structures adjacent tohighways and in car parks which could be splashed or comeinto contact with salt solutions used for de-icing should alsouse air-entrained concrete Alternatively the cube strength ofthe concrete should be 50 Nmm2 or more Whilst C4050concrete is suitable for many situations it does not have thesame freezethaw resistance as air-entrained concrete

Chemical attack Portland cement concrete is liable toattack by acids and acid fumes including the organic acidsoften produced when foodstuffs are being processed Vinegarfruit juices silage effluent sour milk and sugar solutions can allattack concrete Concrete made with Portland cement is notrecommended for use in acidic conditions where the pH valueis 55 or less without careful consideration of the exposurecondition and the intended construction Alkalis have little effecton concrete

For concrete that is exposed to made-up ground includingcontaminated and industrial material specialist advice shouldbe sought in determining the design chemical class so that asuitable concrete can be specified The most common formof chemical attack that concretes have to resist is the effect ofsolutions of sulfates present in some soils and groundwaters

In all cases of chemical attack concrete resistance is related tofree watercement ratio cement content type of cement and thedegree of compaction Well-compacted concrete will always bemore resistant to sulfate attack than one less well compactedregardless of cement type Recommendations for concreteexposed to sulfate-containing groundwater and for chemicallycontaminated brownfield sites are incorporated in BS 8500-1

Alkalindashsilica reaction ASR is a reaction that can occur inconcrete between certain siliceous constituents present in theaggregate and the alkalis ndash sodium and potassium hydroxide ndashthat are released during cement hydration A gelatinous productis formed which imbibes pore fluid and in so doing expandsinducing an internal stress within the concrete The reactionwill cause damage to the concrete only when the followingthree conditions occur simultaneously

A reactive form of silica is present in the aggregate in criticalquantities

The pore solution contains ions of sodium potassium andhydroxyl and is of a sufficiently high alkalinity

A continuing supply of water is available

If any one of these factors is absent then damage from ASRwill not occur and no precautions are necessary It is possiblefor the reaction to take place in the concrete without inducingexpansion Damage may not occur even when the reactionproduct is present throughout the concrete as the gel may fillcracks induced by some other mechanism Recommendationsare available for minimising the risk of damage from ASR innew concrete construction based on ensuring that at least oneof the three aforementioned conditions is absent

Exposure classes For design and specification purposes theenvironment to which concrete will be exposed during itsintended life is classified into various levels of severity Foreach category minimum requirements regarding the qualityof the concrete and the cover to the reinforcement are givenin Codes of Practice In British Codes for many years theexposure conditions were mild moderate severe very severeand most severe (or in BS 5400 extreme) with abrasive as afurther category Details of the classification system that wasused in BS 8110 and BS 5400 are given in Table 39

In BS EN 206-1 BS 8500-1 and EC 2 the conditionsare classified in terms of exposure to particular actions withvarious levels of severity in each category The followingcategories are considered

1 No risk of corrosion or attack

2 Corrosion induced by carbonation

3 Corrosion induced by chlorides other than from seawater

4 Corrosion induced by chlorides from seawater

5 Freezethaw attack

6 Chemical attack

If the concrete is exposed to more than one of these actions theenvironmental conditions are expressed as a combination ofexposure classes Details of each class in categories 1ndash5 withdescriptions and informative examples applicable in the UnitedKingdom are given in Tables 37 and 45 For concrete exposedto chemical attack the exposure classes given in BS EN 206-1cover only natural ground with static water which represents alimited proportion of the aggressive ground conditions found inthe United Kingdom In the complementary British StandardBS 8500-1 more comprehensive recommendations are providedbased on the approach used in ref 13

On this basis an ACEC (aggressive chemical environmentfor concrete) class is determined according to the chemicals inthe ground the type of soil and the mobility and acidity of thegroundwater The chemicals in the ground are expressed as adesign sulfate class (DS) in which the measured sulfate contentis increased to take account of materials that may oxidise intosulfate for example pyrite and other aggressive species suchas hydrochloric or nitric acid Magnesium ion content is alsoincluded in this classification Soil is classified as natural orfor sites that may contain chemical residues from previousindustrial use or imported wastes as brownfield Water in theground is classified as either static or mobile and according toits pH value

Concrete 23

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Based on the ACEC classification and according to the sizeof the section and the selected structural performance level therequired concrete quality expressed as a design chemicalclass (DC) and any necessary additional protective measures(APMs) can be determined The structural performance level isclassified as low normal or high in relation to the intendedservice life the vulnerability of the structural details and thesecurity of structures retaining hazardous materials

Concrete quality and cover to reinforcement Concretedurability is dependent mainly on its constituents particularlythe free watercement ratio The ratio can be reduced and thedurability of the concrete enhanced by increasing the cementcontent andor using admixtures to reduce the amount of freewater needed for a particular level of consistence subject tospecified minimum requirements being met for the cementcontent By limiting the maximum free watercement ratio andthe minimum cement content a minimum strength class can beobtained for particular cements and combinations

Where concrete containing reinforcement is exposed to airand moisture or is subject to contact with chlorides from anysource the protection of the steel against corrosion depends onthe concrete cover The required thickness is related to theexposure class the concrete quality and the intended workinglife of the structure Recommended values for an intendedworking life of at least 50 years are given in Tables 38 and 46(BS 8500) and 39 (prior to BS 8500)

Codes of Practice also specify values for the covers neededto ensure the safe transmission of bond forces and provide anadequate fire-resistance for the reinforced concrete member Inaddition allowance may need to be made for abrasion or forsurface treatments such as bush hammering In BS 8110 valuesused to be given for a nominal cover to be provided to all rein-forcement including links on the basis that the actual covershould not be less than the nominal cover minus 5 mm In BS8500 values are given for a minimum cover to which anallowance for tolerance (normally 10 mm) is then added

Concrete specification Details of how to specify con-crete and what to specify are given in BS 8500-1 Threetypes ndash designed prescribed and standardised prescribedconcretes ndash are recognised by BS EN 206-1 but BS 8500adds two more ndash designated and proprietary concretes

Designed concretes are ones where the concrete producer isresponsible for selecting the mix proportions to provide theperformance defined by the specifier Conformity of designedconcretes is usually judged by strength testing of 100 mm or150 mm cubes which in BS 8500 is the responsibility ofthe concrete producer Prescribed concretes are ones where thespecification states the mix proportions in order to satisfyparticular performance requirements in terms of the mass ofeach constituent Such concretes are seldom necessary butmight be used where particular properties or special surfacefinishes are required Standardised prescribed concretes that areintended for site production using basic equipment and controlare given in BS 8500-2 Whilst conformity does not depend onstrength testing assumed characteristic strengths are given forthe purposes of design Designated concretes are a wide-ranginggroup of concretes that provide for most types of concreteconstruction The producer must operate a recognized accreditedthird party certification system and is responsible for ensuring

that the concrete conforms to the specification given inBS 8500-2 Proprietary concretes are intended to provide forinstances when a concrete producer would give assurance of theperformance of concrete without being required to declare itscomposition

For conditions where corrosion induced by chlorides doesnot apply structural concretes should generally be specifiedas either designated concretes or designed concretes Whereexposure to corrosion due to chlorides is applicable only thedesigned concrete method of specifying is appropriate Anexception to this situation is where an exposed aggregate ortooled finish that removes the concrete surface is required Inthese cases in order to get an acceptable finish a special mixdesign is needed Initial testing including trial panels shouldbe undertaken and from the results of these tests a prescribedconcrete can be specified For housing applications both adesignated concrete and a standardised prescribed concrete canbe specified as acceptable alternatives This would allow aconcrete producer with accredited certification to quote forsupplying a designated concrete and the site contractor or aconcrete producer without accredited certification to quote forsupplying a standardised prescribed concrete

32 REINFORCEMENT

Reinforcement for concrete generally consists of deformedsteel bars or welded steel mesh fabric Normal reinforcementrelies entirely upon the alkaline environment provided by adurable concrete cover for its protection against corrosion Inspecial circumstances galvanised epoxy-coated or stainlesssteel can be used Fibre-reinforced polymer materials havealso been developed So far in the United Kingdom thesematerials have been used mainly for external strengthening anddamage repair applications

321 Bar reinforcement

In the United Kingdom reinforcing bars are generally specifiedordered and delivered to the requirements of BS 4449 Thiscaters for steel bars with a yield strength of 500 MPa in threeductility classes grades B500A B500B and B500C Bars areround in cross section having two or more rows of uniformlyspaced transverse ribs with or without longitudinal ribs Thepattern of transverse ribs varies with the grade and can beused as a means of identification Information with regardto the basic properties of reinforcing bars to BS 4449 whichis in general conformity with BS EN 10080 is given inTable 219

All reinforcing bars are produced by a hot-rolling process inwhich a cast steel billet is reheated to 1100ndash1200oC andthen rolled in a mill to reduce its cross section and impart therib pattern There are two common methods for achievingthe required mechanical properties in hot-rolled bars in-lineheat treatment and micro-alloying In the former method whichis sometimes referred to as the quench-and-self-temper (QST)process high-pressure water sprays quench the bar surface as itexits the rolling mill producing a bar with a hard temperedouter layer and a softer more ductile core Most reinforcingbars in the United Kingdom are of this type and achieve class Bor class C ductility In the micro-alloying method strengthis achieved by adding small amounts of alloying elements

Material properties24

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during the steel-making process Micro-alloy steels normallyachieve class C ductility Another method that can be usedto produce high-yield bars involves a cold-twisting process toform bars that are identified by spiralling longitudinal ribs Thisprocess has been obsolete in the United Kingdom for sometime but round ribbed twisted bars can be found in someexisting structures

In addition to bars being produced in cut straight lengthsbillets are also rolled into coil for diameters up to 16 mm Inthis form the product is ideal for automated processes suchas link bending QST micro-alloying and cold deformationprocesses are all used for high-yield coil Cold deformation isapplied by continuous stretching which is less detrimental toductility than the cold-twisting process mentioned previouslyCoil products have to be de-coiled before use and automaticlink bending machines incorporate straightening rolls Largerde-coiling machines are also used to produce straight lengths

322 Fabric reinforcement

Steel fabric reinforcement is an arrangement of longitudinal barsand cross bars welded together at their intersections in a shearresistant manner In the United Kingdom fabric is producedunder a closely controlled factory-based manufacturing processto the requirements of BS 4483 In fabric for structural purposesribbed bars complying with BS 4449 are used For wrappingfabric as described later wire complying with BS 4482 may beused Wire can be produced from hot-rolled rod by eitherdrawing the rod through a die to produce plain wire or coldrolling the rod to form indented or ribbed wires In BS 4482provision is made for plain round wire with a yield strength of250 MPa and plain indented or ribbed wires with a yieldstrength of 500 MPa

In BS 4483 provision is made for fabric reinforcement tobe either of a standard type or purpose made to the clientrsquosrequirements The standard fabric types have regular mesharrangements and bar sizes and are defined by identifiablereference numbers Type A is a square mesh with identical longbars and cross bars commonly used in ground slabs Type B isa rectangular (structural) mesh that is particularly suitable foruse in thin one-way spanning slabs Type C is a rectangular(long) mesh that can be used in pavements and in two-wayspanning slabs by providing separate sheets in each directionType D is a rectangular (wrapping) mesh that is used in theconcrete encasement of structural steel sections The stock sizeof standard fabric sheets is 48 m 24 m and merchantsize sheets are also available in a 36 m 20 m size Fulldetails of the preferred range of standard fabric types are givenin Table 220

Purpose-made fabrics specified by the customer can haveany combination of wire size and spacing in either direction Inpractice manufacturers may sub-divide purpose-made fabricsinto two categories special (also called scheduled) and bespoke(also called detailed) Special fabrics consist of the standardwire size combinations but with non-standard overhangs andsheet dimensions up to 12 m 33 m Sheets with so-calledflying ends are used to facilitate the lapping of adjacent sheets

Bespoke fabrics involve a more complex arrangement inwhich the wire size spacing and length can be varied within thesheet These products are made to order for each contract as areplacement for conventional loose bar assemblies The use of

bespoke fabrics is appropriate on contracts with a large amountof repeatability and generally manufacturers would require aminimum tonnage order for commercial viability

323 Stressndashstrain curves

For hot-rolled reinforcement the stressndashstrain relationship intension is linear up to yield when there is a pronounced increaseof strain at constant stress (yield strength) Further smallincreases of stress resulting in work hardening are accompaniedby considerable elongation A maximum stress (tensile strength)is reached beyond which further elongation is accompanied bya stress reduction to failure Micro-alloy bars are characterisedby high ductility (high level of uniform elongation and high ratioof tensile strengthyield strength) For QST bars the stressndashstraincurve is of similar shape but with slightly less ductility

Cold-processed reinforcing steels show continuous yieldingbehaviour with no defined yield point The work-hardeningcapacity is lower than for the hot-rolled reinforcement withthe uniform elongation level being particularly reduced Thecharacteristic strength is defined as the 02 proof stress(ie a stress which on unloading would result in a residualstrain of 02) and the initial part of the stressndashstrain curve islinear to beyond 80 of this value

For design purposes the yield or 02 proof condition isnormally critical and the stressndashstrain curves are idealised to abi-linear or sometimes tri-linear form Typical stressndashstraincurves and those recommended for design purposes are givenin Table 36 for BS 8110 and Table 44 for EC 2

324 Bar sizes and bends

The nominal size of a bar is the diameter of a circle with an areaequal to the effective cross-sectional area of the bar The rangeof nominal sizes (millimetres) is from 6 to 50 with preferredsizes of 8 10 12 16 20 25 32 and 40 Values of the totalcross-sectional area provided in a concrete section according tothe number or spacing of the bars for different bar sizes aregiven in Table 220

Bends in bars should be formed around standard mandrels onbar-bending machines In BS 8666 the minimum radius ofbend r is standardised as 2d for d 16 and 35d for d 20where d is the bar size Values of r for each different bar sizeand values of the minimum end projection P needed to formthe bend are given in Table 219 In some cases (eg wherebars are highly stressed) the bars need to be bent to a radiuslarger than the minimum value in order to satisfy the designrequirements and the required radius R is then specified on thebar-bending schedule

Reinforcement should not be bent or straightened on site ina way that could damage or fracture the bars All bars shouldpreferably be bent at ambient temperature but when the steeltemperature is below 5oC special precautions may be neededsuch as reducing the speed of bending or with the engineerrsquosapproval increasing the radius of bending Alternatively thebars may be warmed to a temperature not exceeding 100oC

325 Bar shapes and bending dimensions

Bars are produced in stock lengths of 12 m and lengths upto 18 m can be supplied to special order In most structures

Reinforcement 25

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bars are required in shorter lengths and often need to be bentThe cutting and bending of reinforcement is generally specifiedto the requirements of BS 8666 This contains recommendedbar shapes designated by shape code numbers which areshown in Tables 221 and 222 The information needed to cutand bend the bars to the required dimensions is entered into abar schedule an example of which is shown in Table 223 Eachschedule is related to a member on a particular drawing bymeans of the bar schedule reference number

In cases where a bar is detailed to fit between two concretefaces with no more than the nominal cover on each face (eg linksin beams) an allowance for deviations is required This is tocater for variation due to the effect of inevitable errors in thedimensions of the formwork and the cutting bending andfixing of the bars Details of the deductions to be made to allowfor these deviations and calculations to determine the bendingdimensions in a typical example are given in section 1035with the completed bar schedule in Table 223

326 Stainless steel reinforcement

The type of reinforcement to be used in a structure is usuallyselected on the basis of initial costs This normally results inthe use of carbon steel reinforcement which is around 15of the cost of stainless steel For some structures however theselective use of stainless steel reinforcement ndash on exposedsurfaces for example ndash can be justified In Highways Agencydocument BA 8402 it is recommended that stainless steelreinforcement should be used in splash zones abutmentsparapet edges and soffits and where the chances of chlorideattack are greatest It is generally considered that where theconcrete is saturated and oxygen movement limited stainlesssteel is not required Adherence to these guidelines can meanthat the use of stainless steel reinforcement only marginallyincreases construction costs while significantly reducing thewhole-life costs of the structure and increasing its usable life

Stainless steels are produced by adding elements to iron toachieve the required compositional balance The additionalelements besides chromium can include nickel manganesemolybdenum and titanium with the level of carbon beingcontrolled during processing These alloying elements affectthe steelrsquos microstructure as well as its mechanical propertiesand corrosion resistance Four ranges of stainless steel areproduced two of which are recommended for reinforcement toconcrete because of their high resistance to corrosionAustenitic stainless steels for which chromium and nickel arethe main alloying elements have good general propertiesincluding corrosion resistance and are normally suitable formost applications Duplex stainless steels which have highchromium and low nickel contents provide greater corrosionresistance for the most demanding environments

In the United Kingdom austenitic stainless steel reinforcementhas been produced to the requirements of BS 6744 which isbroadly aligned to conventional reinforcement practice Thusplain and ribbed bars are available in the same characteristicstrengths and range of preferred sizes as normal carbon steelreinforcement Traditionally stainless steel reinforcement hasonly been stocked in maximum lengths of 6 m for all sizesBars are currently available in lengths up to 12 m for sizes upto 16 mm For larger sizes bars can be supplied to order in

lengths up to 8 m Comprehensive data and recommendationson the use of stainless steel reinforcement are given in ref 14

327 Prefabricated reinforcement systems

In order to speed construction by reducing the time needed tofix reinforcement it is important to be able to pre-assemblemuch of the reinforcement This can be achieved on site givenadequate space and a ready supply of skilled personnel Inmany cases with careful planning and collaboration at an earlystage the use of reinforcement assemblies prefabricated by thesupplier can provide considerable benefits

A common application is the use of fabric reinforcement asdescribed in sections 323 and 1032 The preferred range ofdesignated fabrics can be routinely used in slabs and walls Incases involving large areas with long spans and considerablerepetition made-to-order fabrics can be specially designed tosuit specific projects Provision for small holes and openingscan be made by cutting the fabric on site after placing thesheets and adding loose trimming bars as necessary Whilesheets of fabric can be readily handled normally they areawkward to lift over column starter bars In such cases it isgenerally advisable to provide the reinforcement local to thecolumn as loose bars fixed in the conventional manner

A more recent development is the use of slab reinforcementrolls that can be unrolled directly into place on site Each made-to-order roll consists of reinforcement of the required size andspacing in one direction welded to thin metal bands and rolledaround hoops that are later discarded Rolls can be produced upto a maximum bar length of 15 m and a weight of 5 tonnes Thewidth of the sheet when fully rolled out could be more than50 m depending upon the bar size and spacing The full rangeof preferred bar sizes can be used and the bar spacing andlength can be varied within the same roll For each area of slaband for each surface to be reinforced two rolls are requiredThese are delivered to site craned into position and unrolled oncontinuous bar supports Each roll provides the bars in onedirection with those in the lower layer resting on conventionalspacers or chairs

The need to provide punching shear reinforcement in solidflat slabs in the vicinity of the columns has resulted in severalproprietary reinforcement systems Vertical reinforcement isrequired in potential shear failure zones around the columnsuntil a position is reached at which the slab can withstand theshear stresses without reinforcement Conventional links aredifficult and time-consuming to set out and fix Single-leggedlinks are provided with a hook at the top and a 90o bend at thebottom Each link has to be hooked over a top bar in the slaband the 90o bend pushed under a bottom bar and tied in place

Shear ladders can be used in which a row of single-leggedlinks are connected by three straight anchor bars welded toform a robust single unit The ladders provide the required shearreinforcement and act as chairs to support the top bars The sizespacing and height of the links can be varied to suit the designrequirements Shear hoops consist of U-shaped links welded toupper and lower hoops to form a three-dimensional unit Byusing hoops of increasing size shear reinforcement can beprovided on successive perimeters

Shear band strips with a castellated profile are made from25 mm wide high-tensile steel strip in a variety of gauges to

Material properties26

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cater for different shear capacities The strip has perforatedholes along the length to help with anchorage and fixing Thepeaks and troughs of the profile are spaced to coincide withthe spacing of the main reinforcement Stud rails consist of arow of steel studs welded to a flat steel strip or a pair of rodsThe studs are fabricated from plain or deformed reinforcingbars with an enlarged head welded to one or both ends Thesize spacing and height of the studs can be varied to suit theshear requirements and the slab depth

The use of reinforcement continuity strips is a simple andeffective means of providing reinforcement continuity acrossconstruction joints A typical application occurs at a junctionbetween a wall and a slab that is to be cast at a later stageThe strips comprise a set of special pre-bent bars housed in agalvanised indented steel casing that is fabricated off-site ina factory-controlled environment On site the entire unit is castinto the front face of the wall After the formwork is struck thelid of the casing is removed to reveal the legs of the barscontained within the casing The legs are then straightenedoutwards by the contractor ready for lapping with the mainreinforcement in the slab The casing remains embedded in thewall creating a rebate into which the slab concrete flows andeliminating the need for traditional joint preparation

328 Fixing of reinforcement

Reinforcing bars need to be tied together to prevent their beingdisplaced and provide a rigid system Bar assemblies and fabricreinforcement need to be supported by spacers and chairs toensure that the required cover is achieved and kept duringthe subsequent placing and compaction of concrete Spacersshould be fixed to the links bars or fabric wires that are nearestto the concrete surface to which the cover is specifiedRecommendations for the specification and use of spacers andchairs and the tying of reinforcement are given in BS 7973Parts 1 and 2 These include details of the number and positionof spacers and the frequency of tying

33 FIRE-RESISTANCE

Building structures need to conform in the event of fire toperformance requirements stated in the Building RegulationsFor stability the elements of the structure need to providea specified minimum period of fire-resistance in relation to astandard test The required fire period depends on the purposegroup of the building and the height or for basements depth ofthe building relative to the ground as given in Table 312Building insurers may require longer fire periods for storagefacilities where the value of the contents and the costs ofreinstatement of the structure are particularly important

In BS 8110 design for fire-resistance is considered at twolevels Part 1 contains simple recommendations suitable formost purposes Part 2 contains a more detailed treatment with

a choice of three methods involving tabulated data furnacetests or fire engineering calculations The tabulated data is inthe form of minimum specified values of member size andconcrete cover The cover is given to the main reinforcementand in the case of beams and ribs can vary in relation to theactual width of the section The recommendations in Part 1 arebased on the same data but the presentation is different intwo respects values are given for the nominal cover to all rein-forcement (this includes an allowance for links in the case ofbeams and columns) and the values do not vary in relation tothe width of the section The required nominal covers to allreinforcement and minimum dimensions for various membersare given in Tables 310 and 311 respectively

In the event of a fire in a building the vulnerable elementsare the floor construction above the fire and any supportingcolumns or walls The fire-resistance of the floor members(beams ribs and slabs) depends upon the protection providedto the bottom reinforcement The steel begins to lose strengthat a temperature of 300oC losses of 50 and 75 occurringat temperatures of about 560oC and 700oC respectively Theconcrete cover needs to be sufficient to delay the time takento reach a temperature likely to result in structural failure Adistinction is made between simply supported spans wherea 50 loss of strength in the bottom reinforcement could becritical and continuous spans where a greater loss is allowedbecause the top reinforcement will retain its full capacity

If the cover becomes excessive there is a risk of prematurespalling of the concrete in the event of fire Concretes madewith aggregates containing a high proportion of silica are the mostsusceptible In cases where the nominal cover needs to exceed40 mm additional measures should be considered and severalpossible courses of action are described in Part 2 of BS 8110The preferred approach is to reduce the cover by providingadditional protection in the form of an applied finish or a falseceiling or by using lightweight aggregates or sacrificial steelThe last measure refers to the provision of more steel than isnecessary for normal purposes so that a greater loss of strengthcan be allowed in the event of fire If the nominal cover doesexceed 40 mm then supplementary reinforcement in the formof welded steel fabric should be placed within the thicknessof the cover at 20 mm from the concrete surface There areconsiderable practical difficulties with this approach and it mayconflict with the requirements for durability in some cases

For concrete made with lightweight aggregate the nominalcover requirements are all reduced and the risk of prematurespalling only needs to be considered when the cover exceeds50 mm The detailed requirements for lightweight aggregateconcrete and guidance on the additional protection provided byselected applied finishes are given in Table 310

EC 2 contains a more flexible approach to fire safety designbased on the concept of lsquoload ratiorsquo which is the ratio of theload applied at the fire limit-state to the capacity of the elementat ambient temperature

Fire-resistance 27

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Torsion-less beams are designed as linear elements subjectedto bending moments and shear forces The values for freelysupported beams and cantilevers are readily determinedby the simple rules of static equilibrium but the analysis ofcontinuous beams and statically indeterminate frames is morecomplex Historically various analytical techniques have beendeveloped and used as self-contained methods to solve partic-ular problems In time it was realised that the methodscould be divided into two basic categories flexibility methods(otherwise known as action methods compatibility methods orforce methods) and displacement methods (otherwise known asstiffness methods or equilibrium methods) The behaviour ofthe structure is considered in terms of unknown forces in thefirst category and unknown displacements in the secondcategory For each method a particular solution obtained bymodifying the structure to make it statically determinate iscombined with a complementary solution in which the effectof each modification is determined Consider the case of acontinuous beam For the flexibility methods the particularsolution involves removing redundant actions (ie the continuitybetween the individual members) to leave a series of discon-nected spans For the displacement methods the particularsolution involves restricting the rotations andor displacementsthat would otherwise occur at the joints

To clarify further the main differences between the methodsin the two categories consider a propped cantilever With theflexibility approach the first step is to remove the prop andcalculate the deflection at the position of the prop due to theaction of the applied loads this gives the particular solutionThe next step is to calculate the concentrated load needed at theposition of the prop to restore the deflection to zero this givesthe complementary solution The calculated load is the reactionin the prop knowing this enables the moments and forces in thepropped cantilever to be simply determined If the displacementapproach is used the first step is to consider the span as fullyfixed at both ends and calculate the moment at the propped enddue to the applied loads this gives the particular solution Thenext step is to release the restraint at the propped end and applyan equal and opposite moment to restore the rotation to zero thisgives the complementary solution By combining the momentdiagrams the resulting moments and forces can be determined

In general there are several unknowns and irrespectiveof the method of analysis used the preparation and solution ofa set of simultaneous equations is required The resulting

relationship between forces and displacements embodies a seriesof coefficients that can be set out concisely in matrix formIf flexibility methods are used the resulting matrix is built up offlexibility coefficients each of which represents a displacementproduced by a unit action Similarly if stiffness methods areused the resulting matrix is formed of stiffness coefficients eachof which represents an action produced by a unit displacementThe solution of matrix equations either by matrix inversionor by a systematic elimination process is ideally suited tocomputer technology To this end methods have been devised(the so-called matrix stiffness and matrix flexibility methods)for which the computer both sets up and solves the simultaneousequations (ref 15)

Here it is worthwhile to summarise the basic purpose ofthe analysis Calculating the bending moments on individualfreely supported spans ensures that equilibrium is maintainedThe analytical procedure that is undertaken involves linearlytransforming these free-moment diagrams in a manner that iscompatible with the allowable deformations of the structureUnder ultimate load conditions deformations at the criticalsections must remain within the limits that the sections canwithstand and under service load conditions deformationsmust not result in excessive deflection or cracking or both Ifthe analysis is able to ensure that these requirements are met itwill be entirely satisfactory for its purpose endeavouring toobtain painstakingly precise results by over-complex methodsis unjustified in view of the many uncertainties involved

To determine at any section the effects of the applied loadsand support reactions the basic relationships are as follows

Shear force (forces on one side of section) rate of change of bending moment

Bending moment (moments of forces on one side of section) (shear force) area of shear force diagram

Slope (curvature) area of curvature diagram

Deflection (slope) area of slope diagram

For elastic behaviour curvature MEI where M is bendingmoment E is modulus of elasticity of concrete I is secondmoment of area of section For the purposes of structural

Chapter 4

Structural analysis

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analysis to determine bending moments due to applied loadsI values may normally be based on the gross concrete sectionIn determining deflections however due allowance needs to bemade for the effects of cracking and in the long term for theeffects of concrete creep and shrinkage

41 SINGLE-SPAN BEAMS AND CANTILEVERS

Formulae to determine the shearing forces bending momentsand deflections produced by various general loads on beamsfreely supported at the ends are given in Table 224 Similarexpressions for some particular load arrangements commonlyencountered on beams either freely supported or fully fixedat both ends with details of the maximum values are given inTable 225 The same information but relating to simple andpropped cantilevers is given in Tables 226 and 227 respectivelyCombinations of loads can be considered by summing theresults obtained for each individual load

In Tables 224ndash227 expressions are also given for the slopesat the beam supports and the free (or propped) end of a cantileverInformation regarding the slope at other points is seldomrequired If needed it is usually a simple matter to obtain theslope by differentiating the deflection formula with respect to xIf the resulting expression is equated to zero and solved toobtain x the point of maximum deflection will have been foundThis value of x can then be substituted into the original formulato obtain the maximum deflection

Coefficients to determine the fixed-end moments producedby various symmetrical and unsymmetrical loads on beamsfully fixed at both ends are given in Table 228 Loadings notshown can usually be considered by using the tabulated casesin combination For the general case of a partial uniform ortriangular distribution of load placed anywhere on a membera full range of charts is contained in Examples of the Design ofReinforced Concrete Buildings The charts give deflection andmoment coefficients for beams (freely supported or fully fixedat both ends) and cantilevers (simple or propped)

42 CONTINUOUS BEAMS

Historically various methods of structural analysis have beendeveloped for determining the bending moments and shearingforces on beams continuous over two or more spans Most ofthese have been stiffness methods which are generally bettersuited than flexibility methods to hand computation Some ofthese approaches such as the theorem of three-moments and themethods of fixed points and characteristic points were includedin the previous edition of this Handbook If beams having twothree or four spans are of uniform cross section and supportloads that are symmetrical on each individual span formulaeand coefficients can be derived that enable the support momentsto be determined by direct calculation Such a method is givenin Table 237 More generally in order to avoid the need to solvelarge sets of simultaneous equations methods involving succes-sive approximations have been devised Despite the general useof computers hand methods can still be very useful in dealingwith routine problems The ability to use hand methods alsodevelops in the engineer an appreciation of analysis that isinvaluable in applying output from the computer

When bending moments are calculated with the spans takenas the distances between the centres of supports the critical

negative moment in monolithic forms of construction can beconsidered as that occurring at the edge of the support Whenthe supports are of considerable width the span can be taken asthe clear distance between the supports plus the effective depthof the beam or an additional span can be introduced thatis equal to the width of the support minus the effective depth ofthe beam The load on this additional span should be takenas the support reaction spread uniformly over the width of thesupport If a beam is constructed monolithically with a verywide and massive support the effect of continuity with the spanor spans beyond the support may be negligible in which casethe beam should be treated as fixed at the support

The second moment of area of a reinforced concrete beamof uniform depth may still vary throughout its length due tovariations in the amount of reinforcement and also becausewhen acting with an adjoining slab a down-stand beam maybe considered as a flanged section at mid-span but a simplerectangular section at the supports It is common practicehowever to neglect these variations for beams of uniformdepth and use the value of I for the plain rectangular section Itis often assumed that a continuous beam is freely supportedat the ends even when beam and support are constructedmonolithically Some provision should still be made for theeffects of end restraint

421 Analysis by moment distribution

Probably the best-known and simplest system for analysingcontinuous beams by hand is that of moment distributionas devised by Hardy Cross in 1929 The method whichderives from slope-deflection principles is described briefly inTable 236 It employs a system of successive approximationsthat may be terminated as soon as the required degree ofaccuracy has been reached A particular advantage of this andsimilar methods is that even after only one distribution cycleit is often clear whether or not the final values will be acceptableIf not the analysis can be discontinued and unnecessary workavoided The method is simple to remember and apply andthe step-by-step procedure gives the engineer a lsquofeelrsquo for thebehaviour of the system It can be applied albeit less easily tothe analysis of systems containing non-prismatic members andto frames Hardy Cross moment distribution is described inmany textbooks dealing with structural analysis

Over the years the Hardy Cross method of analysis begotvarious offspring One of these is known as precise momentdistribution (also called the coefficient of restraint method ordirect moment distribution) The procedure is very similar tonormal moment distribution but the distribution and carryoverfactors are so adjusted that an exact solution is obtainedafter one distribution in each direction The method thus hasthe advantage of removing the necessity to decide when toterminate the analysis Brief details are given in Table 236 andthe method is described in more detail in Examples of theDesign of Reinforced Concrete Buildings (see also ref 16)

It should be noted that the load arrangements that producethe greatest negative bending moments at the supports are notnecessarily those that produce the greatest positive bendingmoments in the spans The design loads to be considered inBS 8110 and EC 2 and the arrangements of live load that givethe greatest theoretical bending moments as well as the lessonerous code requirements are given in Table 229 Some live

Continuous beams 29

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load arrangements can result in negative bending momentsthroughout adjacent unloaded spans

422 Redistribution of bending moments

For the ULS the bending moments obtained by linear elasticanalysis may be adjusted on the basis that some redistributionof moments can occur prior to collapse This enables the effectsof both service and ultimate loadings to be assessed without theneed to undertake a separate analysis using plastic-hinge tech-niques for the ultimate condition The theoretical justificationfor moment redistribution is clearly explained in the Handbookto BS 8110 Since the reduction of moment at a section assumesthe formation of a plastic hinge at that position prior to theultimate condition being reached it is necessary to limit thereduction in order to restrict the amount of plastic-hinge rotationand control the cracking that occurs under serviceabilityconditions For these reasons the maximum ratio of neutralaxis depth to effective depth and the maximum distancebetween tension bars are each limited according to the requiredamount of redistribution

Such adjustments are useful in reducing the inequalitiesbetween negative and positive moments and minimising theamount of reinforcement that must be provided at a particularsection such as the intersection between beam and columnwhere concreting may otherwise be more difficult due to thecongestion of reinforcement Both BS 8110 and EC 2 allowthe use of moment redistribution the procedure which may beapplied to any system that has been analysed by the so-calledexact methods is described in section 123 with an illustratedexample provided in Table 233

423 Coefficients for equal loads onequal spans

For beams that are continuous over a number of equal spanswith equal loads on each loaded span the maximum bendingmoments and shearing forces can be tabulated In Tables 230and 231 maximum bending moment coefficients are given foreach span and at each support for two three four and five equalspans with identical loads on each span which is the usualdisposition of the dead load on a beam Coefficients are alsogiven for the most adverse incidence of live loads and in thecase of the support moments for the arrangements of live loadrequired by BS 8110 (values in square brackets) and by EC 2(values in curved brackets) It should be noted that the maximumbending moments due to live load do not occur at all thesections simultaneously The types of load considered are auniformly distributed load a central point load two equal loadsapplied at the third-points of the span and trapezoidal loads ofvarious proportions In Table 232 coefficients are given for themaximum shearing forces for each type of load with identicalloads on each span and due to the most adverse incidence oflive loads

424 Bending moment diagrams for equal spans

In Tables 234 and 235 bending moment coefficients forvarious arrangements of dead and live loads with sketches

of the resulting moment envelopes are given for beams oftwo and three spans and for a theoretically infinite systemThis information enables appropriate bending moment diagramsto be plotted quickly and accurately The load types consideredare a uniformly distributed load a central point load and twoequal loads at the third points of the span Values are givenfor identical loads on each span (for example dead load) andfor the arrangements of live load required by BS 8110 andEC 2 As the coefficients have been calculated by exactmethods moment redistribution is allowed at the ultimate statein accordance with the requirements of BS 8110 and EC 2 Inaddition to the coefficients obtained by linear elastic analysisvalues are given for conditions in which the maximum supportmoments are reduced by either 10 or 30 as described insection 1233 Coefficients are also given for the positivesupport moments and negative span moments that occur undersome arrangements of live load

425 Solutions for routine design

A precise determination of theoretical bending moments andshearing forces on continuous beams is not always necessary Itshould also be appreciated that the general assumptions ofunyielding knife-edge supports uniform sectional propertiesand uniform distributions of live load are hardly realistic Theindeterminate nature of these factors often leads in practice tothe adoption of values based on approximate coefficients InTable 229 values in accordance with the recommendationsof BS 8110 and EC 2 are given for bending moments andshearing forces on uniformly loaded beams of three or morespans The values are applicable when the characteristicimposed load is not greater than the characteristic dead loadand the variations in span do not exceed 15 of the longestspan The same coefficients may be used with service loads orultimate loads and the resulting bending moments may beconsidered to be without redistribution

43 MOVING LOADS ON CONTINUOUS BEAMS

Bending moments caused by moving loads such as those due tovehicles traversing a series of continuous spans are most easilycalculated with the aid of influence lines An influence line is acurve with the span of the beam taken as the base the ordinateof the curve at any point being the value of the bending momentproduced at a particular section when a unit load acts at thepoint The data given in Tables 238ndash241 enable the influencelines for the critical sections of beams continuous over twothree four and five or more spans to be drawn By plotting theposition of the load on the beam (to scale) the bending momentsat the section being considered can be derived as explained inthe example given in Chapter 12 The curves given for equalspans can be used directly but the corresponding curves forunequal spans need to be plotted from the data tabulated

The bending moment due to a load at any point is equal tothe ordinate of the influence line at the point multiplied by theproduct of the load and the span the length of the shortest spanbeing used when the spans are unequal The influence lines inthe tables are drawn for a symmetrical inequality of spans Thesymbols on each curve indicate the section of the beam andthe ratio of span lengths to which the curve applies

Structural analysis30

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44 ONE-WAY SLABS

In monolithic building construction the column layout oftenforms a rectangular grid Continuous beams may be provided inone direction or two orthogonal directions to support slabs thatmay be solid or ribbed in cross section Alternatively the slabsmay be supported directly on the columns as a flat slab Severaldifferent forms of slab construction are shown in Table 242These are considered in more detail in the general context ofbuilding structures in Chapter 6

Where beams are provided in one direction only the slab isa one-way slab Where beams are provided in two orthogonaldirections the slab is a two-way slab However if the longerside of a slab panel exceeds twice the shorter side the slab isgenerally designed as a one-way slab A flat slab is designedas a one-way slab in each direction Bending moments andshearing forces are usually determined on strips of unit widthfor solid slabs and strips of width equal to the spacing of theribs for ribbed slabs

The comments in section 425 and the coefficients for theroutine design of beams given in Table 229 apply equally toone-way spanning slabs This is particularly true when elasticmoments due to service loads are required However lightlyreinforced slabs are highly ductile members and allowanceis generally made for redistribution of elastic moments atthe ULS

441 Uniformly distributed load

For slabs carrying uniformly distributed loads and continuousover three or more nearly equal spans approximate solutionsfor ultimate bending moments and shearing forces accordingto BS 8110 and EC 2 are given in Table 242 In both cases thesupport moments include an allowance for 20 redistributionbut the situation regarding the span moments is somewhatdifferent in the two codes

In BS 8110 a simplified arrangement of the design loadsis permitted where the characteristic imposed load doesnot exceed 125 the characteristic dead load or 5 kNm2excluding partitions and the area of each bay exceeds 30 m2Design for a single load case of maximum design load on allspans is considered sufficient providing the support momentsare reduced by 20 and the span moments are increasedto maintain equilibrium Although the resulting moments arecompatible with yield-line theory the span moments are lessthan those that would occur in the case of alternate spans beingloaded with maximum load and minimum load The implicitredistribution of the span moments the effect of which on thereinforcement stress under service loads would be detrimentalto the deflection of the beam is ignored in the subsequentdesign In EC 2 this simplification is not included and thevalues given for the span moments are the same as those forbeams in Table 229

Provision is made in Table 242 for conditions where aslab is continuous with the end support The restrainingelement may vary from a substantial wall to a small edgebeam and allowance has been made for both eventualitiesThe support moment is given as 004Fl but the reducedspan moment is based on the support moment being no morethan 002Fl

442 Concentrated loads

When a slab supported on two opposite sides carries a loadconcentrated on a limited area of the slab such as a wheelload on the deck of a bridge conventional elastic methods ofanalysis based on isotropic plate theory are often used Thesemay be in the form of equations as derived by Westergaard(ref 17) or influence surfaces as derived by Pucher (ref 18)Another approach is to extend to one-way spanning slabs thetheory applied to slabs spanning in two directions For examplethe curves given in Table 247 for a slab infinitely long in thedirection ly can be used to evaluate directly the bendingmoments in the direction of and at right angles to the spanof a one-way slab carrying a concentrated load this methodhas been used to produce the data for elastic analysis givenin Table 245

For designs in which the ULS requirement is the maincriterion a much simpler approach is to assume that a certainwidth of slab carries the entire load In BS 8110 for examplethe effective width for solid slabs is taken as the load widthplus 24x(1 xl) x being the distance from the nearer supportto the section under consideration and l the span Thus themaximum width at mid-span is equal to the load width plus06l Where the concentrated load is near an unsupported edgeof a slab the effective width should not exceed 12x(1 xl)plus the distance of the slab edge from the further edge of theload Expressions for the resulting bending moments are givenin Table 245 For ribbed slabs the effective width will dependon the ratio of the transverse and longitudinal flexural rigiditiesof the slab but need not be taken less than the load width plus4xl(1 x l) metres

The solutions referred to so far are for single-span slabs thatare simply supported at each end The effects of end-fixity orcontinuity may be allowed for approximately by multiplyingthe moment for the simply supported case by an appropriatefactor The factors given in Table 245 are derived by elasticbeam analysis

45 TWO-WAY SLABS

When a slab is supported other than on two opposite sides onlythe precise amount and distribution of the load taken by eachsupport and consequently the magnitude of the bendingmoments on the slab are not easily calculated if assumptionsresembling real conditions are made Therefore approximateanalyses are generally used The method applicable in anyparticular case depends on the shape of the slab panel theconditions of restraint at the supports and the type of load

Two basic methods are commonly used to analyse slabsthat span in two directions The theory of plates which isbased on elastic analysis is particularly appropriate to thebehaviour under service loads Yield-line theory considersthe behaviour of the slab as a collapse condition approachesHillerborgrsquos strip method is a less well-known alternative tothe use of yield-line in this case In some circumstances itis convenient to use coefficients derived by an elastic analysiswith loads that are factored to represent ULS conditions Thisapproach is used in BS 8110 for the case of a simply supportedslab with corners that are not held down or reinforced fortorsion It is also normal practice to use elastic analysis for

Two-way slabs 31

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both service and ULS conditions in the design of bridge decksand liquid-retaining structures For elastic analyses a Poissonrsquosratio of 02 is recommended in BS 8110 and BS 5400 Part 4In EC 2 the values given are 02 for uncracked concrete and 0for cracked concrete

The analysis must take account of the support conditionswhich are often idealised as being free or hinged or fixed andwhether or not the corners of the panels are held down A freecondition refers to an unsupported edge as for example the topof a wall of an uncovered rectangular tank The condition ofbeing freely or simply supported with the corners not helddown may occur when a slab is not continuous and the edgesbear directly on masonry walls or structural steelwork If theedge of the slab is built into a substantial masonry wall or isconstructed monolithically with a reinforced concrete beam orwall a condition of partial restraint exists Such restraint maybe allowed for when computing the bending moments on theslab but the support must be able to resist the torsion andorbending effects and the slab must be reinforced to resist thenegative bending moment A slab can be considered as fixedalong an edge if there is no change in the slope of the slab atthe support irrespective of the incidence of the load A fixedcondition could be assumed if the polar second moment of areaof the beam or other support is very large Continuity over asupport generally implies a condition of restraint less rigid thanfixity that is the slope of the slab at the support depends uponthe incidence of load not only on the panel under considerationbut also on adjacent panels

451 Elastic methods

The so-called exact theory of the elastic bending of platesspanning in two directions derives from work by Lagrangewho produced the governing differential equation for platebending in 1811 and Navier who in 1820 described the useof a double trigonometric series to analyse freely supportedrectangular plates Pigeaud and others later developed theanalysis of panels freely supported along all four edges

Many standard elastic solutions have been produced butalmost all of these are restricted to square rectangular andcircular slabs (see for example refs 19 20 and 21) Exactanalysis of a slab having an arbitrary shape and supportconditions with a general arrangement of loading would beextremely complex To deal with such problems numericaltechniques such as finite differences and finite elementshave been devised Some notes on finite elements are givenin section 497 Finite-difference methods are considered inref 15 (useful introduction) and ref 22 (detailed treatment)The methods are suited particularly to computer-based analysisand continuing software developments have led to the techniquesbeing readily available for routine office use

452 Collapse methods

Unlike in frame design where the converse is generally trueit is normally easier to analyse slabs by collapse methods thanby elastic methods The most-widely known methods ofplastic analysis of slabs are the yield-line method developedby K W Johansen and the so-called strip method devised byArne Hillerborg

It is generally impossible to calculate the precise ultimateresistance of a slab by collapse theory since such elements are

highly indeterminate Instead two separate solutions can befound ndash one being upper bound and the other lower boundWith solutions of the first type a collapse mechanism is firstpostulated Then if the slab is deformed the energy absorbedin inducing ultimate moments along the yield lines is equal tothe work done on the slab by the applied load in producing thisdeformation Thus the load determined is the maximum thatthe slab will support before failure occurs However since suchmethods do not investigate conditions between the postulatedyield lines to ensure that the moments in these areas do notexceed the ultimate resistance of the slab there is no guaranteethat the minimum possible collapse load has been found Thisis an inevitable shortcoming of upper-bound solutions such asthose given by Johansenrsquos theory

Conversely lower-bound solutions will generally result in thedetermination of collapse loads that are less than the maximumthat the slab can actually carry The procedure here is to choosea distribution of ultimate moments that ensures that equilibriumis satisfied throughout and that nowhere is the resistance of theslab exceeded

Most of the literature dealing with the methods of Johansenand Hillerborg assumes that any continuous supports at the slabedges are rigid and unyielding This assumption is also madethroughout the material given in Part 2 of this book Howeverif the slab is supported on beams of finite strength it is possiblefor collapse mechanisms to form in which the yield lines passthrough the supporting beams These beams would then becomepart of the mechanism considered and such a possibility shouldbe taken into account when using collapse methods to analysebeam-and-slab construction

Yield-line analysis Johansenrsquos method requires the designerto first postulate an appropriate collapse mechanism for the slabbeing considered according to the rules given in section 1342Variable dimensions (such as ly on diagram (iv)(a) in Table 249)may then be adjusted to obtain the maximum ultimate resistancefor a given load (ie the maximum ratio of MF) This maximumvalue can be found in various ways for example by tabulatingthe work equation as shown in section 1348 using actualnumerical values and employing a trial-and-adjustment processAlternatively the work equation may be expressed algebraicallyand by substituting various values for the maximum ratio ofMF may be read from a graph relating to MF Anothermethod is to use calculus to differentiate the equation and thenby setting this equal to zero determine the critical value of This method cannot always be used however (see ref 23)

As already explained although such processes enable themaximum resistance for a given mode of failure to be foundthey do not indicate whether the yield-line pattern considered isthe critical one A further disadvantage of such a method is thatunlike Hillerborgrsquos method it gives no direct indication of theresulting distribution of load on the supports Although it seemspossible to use the yield-line pattern as a basis for apportioningthe loaded areas of slab to particular supports there is no realjustification for this assumption (see ref 23) In spite of theseshortcomings yield-line theory is extremely useful A consid-erable advantage is that it can be applied relatively easily tosolve problems that are almost intractable by other means

Yield-line theory is too complex to deal with adequately in thisHandbook indeed several textbooks are completely or almostcompletely devoted to the subject (refs 23ndash28) In section 134

Structural analysis32

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and Tables 249 and 250 notes and examples are given on therules for choosing yield-line patterns for analysis on theoreticaland empirical methods of analysis on simplifications that canbe made by using so-called affinity theorems and on the effectsof corner levers

Strip method Hillerborg devised his strip method in orderto obtain a lower-bound solution for the collapse load whileachieving a good economical arrangement of reinforcement Aslong as the reinforcement provided is sufficient to cater for thecalculated moments the strip method enables such a lower-boundsolution to be obtained (Hillerborg and others sometimes referto the strip method as the equilibrium theory this should nothowever be confused with the equilibrium method of yield-lineanalysis) In Hillerborgrsquos original theory (now known as thesimple strip method) it is assumed that at failure no load isresisted by torsion and thus all load is carried by flexure ineither of two principal directions The theory results in simplesolutions giving full information regarding the moments overthe whole slab to resist a unique collapse load the reinforcementbeing placed economically in bands Brief notes on the use ofsimple strip theory to design rectangular slabs supportinguniform loads are given in section 135 and Table 251

However the simple strip theory is unable to deal withconcentrated loads andor supports and leads to difficultieswith free edges To overcome such problems Hillerborg laterdeveloped his advanced strip method which involves the use ofcomplex moment fields Although this development extendsthe scope of the simple strip method it somewhat spoils thesimplicity and directness of the original concept A full treat-ment of both the simple and advanced strip theories is givenin ref 29

A further disadvantage of both Hillerborgrsquos and Johansenrsquosmethods is that being based on conditions at failure onlythey permit unwary designers to adopt load distributions thatmay differ widely from those that would occur under serviceloads with the risk of unforeseen cracking A development thateliminates this problem as well as overcoming the limitationsarising from simple strip theory is the so-called strip-deflectionmethod due to Fernando and Kemp (ref 30) With this methodthe distribution of load in either principal direction is notselected arbitrarily by the designer (as in the Hillerborg methodor by choosing the ratio of reinforcement provided in eachdirection as in the yield-line method) but is calculated so as toensure compatibility of deflection in mutually orthogonal stripsThe method results in sets of simultaneous equations (usuallyeight) the solution of which requires computer assistance

453 Rectangular panel with uniformlydistributed load

The bending moments in rectangular panels depend on thesupport conditions and the ratio of the lengths of the sides ofthe panel The ultimate bending moment coefficients given inBS 8110 are derived from a yield-line analysis in which thevalues of the coefficients have been adjusted to suit the divisionof the panel into middle and edge strips as shown in Table 242Reinforcement to resist the bending moments calculated fromthe data given in Table 243 is required only within the middlestrips which are of width equal to three-quarters of the panelwidth in each direction The ratio of the negative moment at

a continuous edge to the positive moment at mid-span has beenchosen as 43 to conform approximately to the serviceabilityrequirements For further details on the derivation of the coef-ficients see ref 31 Nine types of panel are considered inorder to cater for all possible combinations of edge conditionsWhere two different values are obtained for the negativemoment at a continuous edge because of differences betweenthe contiguous panels the values may be treated as fixed-endmoments and distributed elastically in the direction of spanThe procedure is illustrated by means of a worked example insection 1321 Minimum reinforcement as given in BS 8110is to be provided in the edge strips Torsion reinforcement isrequired at corners where either one or both edges of the panelare discontinuous Values for the shearing forces at the ends ofthe middle strips are also given in Table 243

Elastic bending moment coefficients for the same types ofpanel (except that the edge conditions are now defined as fixedor hinged rather than continuous or discontinuous) are givenin Table 244 The information has been prepared from datagiven in ref 21 which was derived by finite element analysisand includes for a Poissonrsquos ratio of 02 For ratios less than 02the positive moments at mid-span are reduced slightly and thetorsion moments at the corners are increased The coefficientsmay be adjusted to suit a Poissonrsquos ratio of zero as explainedin section 1322

The simplified analysis due to Grashof and Rankine can beused for a rectangular panel simply supported on four sideswhen no provision is made to resist torsion at the corners orto prevent the corners from lifting A solution is obtained byconsidering uniform distributions of load along orthogonalstrips in each direction and equating the elastic deflections atthe middle of the strips The proportions of load carried by eachstrip are then obtained as a function of the ratio of the spansand the resulting mid-span moments are calculated Bendingmoment coefficients for this case are also provided in Table 244and basic formulae are given in section 1322

454 Rectangular panel with triangularlydistributed load

In the design of rectangular tanks storage bunkers and someretaining structures cases occur of wall panels spanning in twodirections and subjected to triangular distributions of pressureThe intensity of pressure is uniform at any level but verticallythe pressure increases linearly from zero at the top to a maxi-mum at the bottom Elastic bending moment and shear forcecoefficients are given for four different types of panel to caterfor the most common combinations of edge conditions inTable 253 The information has been prepared from data givenin ref 32 which was derived by finite element analysis andincludes for a Poissonrsquos ratio of 02 For ratios less than 02 thebending moments would be affected in the manner discussed insection 453

The bending moments given for individual panels fixed atthe sides may be applied without modification to continuouswalls provided there is no rotation about the vertical edges Ina square tank therefore moment coefficients can be takendirectly from Table 253 For a rectangular tank distribution ofthe unequal negative moments at the corners is needed

An alternative method of designing the panels would be touse yield-line theory If the resulting structure is to be used

Two-way slabs 33

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to store liquids however extreme care must be taken to ensurethat the adopted proportions of span to support moment andvertical to horizontal moment conform closely to those givenby elastic analyses Otherwise the predicted service momentsand calculated crack widths will be invalid and the structuremay be unsuitable for its intended purpose In the case of struc-tures with non-fluid contents such considerations may be lessimportant This matter is discussed in section 1362

Johansen has shown (ref 24) for a panel fixed or freelysupported along the top edge that the total ultimate momentacting on the panel is identical to that on a similar panel withthe same total load uniformly distributed Furthermore as in thecase of the uniformly loaded slab considered in section 1346a restrained slab may be analysed as if it were freely supportedby employing so-called reduced side lengths to represent theeffects of continuity or fixity Of course unlike the uniformlyloaded slab along the bottom edge of the panel where the load-ing is greatest a higher ratio of support to span moment shouldbe adopted than at the top edge of the panel If the panel isunsupported along the top edge its behaviour is controlledby different collapse mechanisms The relevant expressionsdeveloped by Johansen (ref 24) are represented graphically inTable 254 Triangularly loaded panels can also be designed bymeans of Hillerborgrsquos strip method (ref 29) shown also inTable 254

455 Rectangular panels with concentratedloads

Elastic methods can be used to analyse rectangular panelscarrying concentrated loads The curves in Tables 246 and 247based on Pigeaudrsquos theory give bending moments on a panelfreely supported along all four edges with restrained corners andcarrying a load uniformly distributed over a defined area sym-metrically disposed upon the panel Wheel loads and similarlyhighly concentrated loads are considered to be dispersedthrough the thickness of any surfacing down to the top of theslab or farther down to the mid-depth of the slab as describedin section 249 The dimensions ax and ay of the resultingboundary are used to determine axlx and ayly for which thebending moment factors x4 and y4 are read off the curvesaccording to the ratio of spans k lylx

For a total load F acting on the area ax by ay the positivebending moments per unit width of slab are given by theexpressions in Tables 246 and 247 in which the value ofPoissonrsquos ratio is normally taken as 02 The curves are drawnfor k values of 10 125 radic2 ( 141 approx) 167 20 25 andinfinity For intermediate values of k the values of x4 and y4

can be interpolated from the values above and below the givenvalue of k The use of the curves for k 10 which apply to asquare panel is explained in section 1332

The curves for k infin apply to panels where ly is very muchgreater than lx and can be used to determine the transverse andlongitudinal bending moments for a long narrow panel sup-ported on the two long edges only This chart has been used toproduce the elastic data for one-way slabs given in Table 245as mentioned in section 442

For panels that are restrained along all four edges Pigeaudrecommends that the mid-span moments be reduced by 20Alternatively the multipliers given for one-way slabs could beused if the inter-dependence of the bending moments in the

two directions is ignored Pigeaudrsquos recommendations for themaximum shearing forces are given in section 1332

To determine the load on the supporting beams the rulesin section 46 for a load distributed over the entire panel aresufficiently accurate for a load concentrated at the centre ofthe panel This is not always the critical case for live loads suchas a load imposed by a wheel on a bridge deck since themaximum load on the beam occurs when the wheel is passingover the beam in which case the beam carries the whole load

Johansenrsquos yield-line theory and Hillerborgrsquos strip methodcan also be used to analyse slabs carrying concentrated loadsAppropriate yield-line formulae are given in ref 24 or themethod described in section 1348 may be used For detailsof the analysis involved if the advanced strip method is usedsee ref 29

46 BEAMS SUPPORTING RECTANGULAR PANELS

When designing beams supporting a uniformly loaded panelthat is freely supported along all four edges or with the samedegree of fixity along all four edges it is generally accepted thateach of the beams along the shorter edges of the panel carriesload on an area in the shape of a 45o isosceles triangle whosebase is equal to the length of the shorter side for example eachbeam carries a triangularly distributed load Each beam alongthe longer edges of the panel carries the load on a trapezoidalarea The amount of load carried by each beam is given bythe diagram and expressions in the top left-hand corner ofTable 252 In the case of a square panel each beam carries atriangularly distributed load equal to one-quarter of the totalload on the panel For beams with triangular and trapezoidaldistributions of loading fixed-end moments and moments forcontinuous beams are given in Tables 228 230 and 231

When a panel is fixed or continuous along one two orthree supports and freely supported on the remaining edges thesub-division of the total load to the various supporting beamscan be determined from the diagrams and expressions on theleft-hand side of Table 252 If the panel is unsupported alongone edge or two adjacent edges the loads on the supportingbeams at the remaining edges are as given on the right-handside of Table 252 The expressions which are given in terms ofa service load w may be applied also to an ultimate load n

For slabs designed in accordance with the BS 8110 methodthe loads on the supporting beams may be determined from theshear forces given in Table 243 The relevant loads are takenas uniformly distributed along the middle three-quarters of thebeam length and the resulting fixed-end moments can bedetermined from Table 228

47 NON-RECTANGULAR PANELS

When a panel that is not rectangular is supported along all itsedges and is of such proportions that main reinforcement intwo directions seems desirable the bending moments can bedetermined approximately from the data given in Table 248The information derived from elastic analyses is applicable toa trapezoidal panel approximately symmetrical about one axisto a panel that in plan is an isosceles triangle (or nearly so) andto panels that are regular polygons or circular The case of atriangular panel continuous or partially restrained along threeedges occurs in pyramidal hopper bottoms For this case

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reinforcement determined for the positive moments shouldextend over the entire area of the panel and provision must bemade for the negative moments and for the direct tensions thatact simultaneously with the bending moments

If the shape of a panel is approximately square the bendingmoments for a square slab of the same area should be usedA slab having the shape of a regular polygon with five or moresides can be treated as a circular slab with the diameter takenas the mean of the diameters obtained for the inscribed andcircumscribed circles for regular hexagons and octagons themean diameters are given in Table 248

For a panel circular in plan that is freely supported or fullyfixed along the circumference and carries a load concentratedsymmetrically about the centre on a circular area the totalbending moment to be considered acting across each of twomutually perpendicular diameters is given by the appropriateexpressions in Table 248 These are based on the expressionsderived by Timoshenko and Woinowski-Krieger (ref 20) Ingeneral the radial and tangential moments vary according to theposition being considered A circular panel can therefore bedesigned by one of the following elastic methods

1 Design for the maximum positive bending moment at thecentre of the panel and reduce the amount of reinforcementor the thickness of the slab towards the circumference If thepanel is not truly freely supported at the edge provide forthe appropriate negative bending moment

2 Design for the average positive bending moment across adiameter and retain the same thickness of slab and amountof reinforcement throughout the entire area of the panel Ifthe panel is not truly freely supported at the edge providefor the appropriate negative bending moment

The reinforcement required for the positive bending momentsin each of the preceding methods must be provided in twodirections mutually at right angles the reinforcement for thenegative bending moment should be provided by radial barsnormal to and equally spaced around the circumference or bysome equivalent arrangement

Both circular and other non-rectangular shapes of slab mayconveniently be designed for ULS conditions by using yield-line theory the method of obtaining solutions for slabs ofvarious shapes is described in detail in ref 24

48 FLAT SLABS

The design of flat slabs that is beamless slabs supporteddirectly on columns has often been based on empirical rulesModern codes place much greater emphasis on the analysis ofsuch structures as a series of continuous frames Other methodssuch as grillage finite element and yield-line analysis may beemployed The principles described hereafter and summarisedin section 138 and Table 255 are in accordance with thesimplified method given in BS 8110 This type of slab can beof uniform thickness throughout or can incorporate thickeneddrop panels at the column positions The columns may be ofuniform cross section throughout or may be provided with anenlarged head as indicated in Table 255

The simplified method may be used for slabs consisting ofrectangular panels with at least three spans of approximatelyequal length in each direction where the ratio of the longer tothe shorter side of each panel does not exceed 2 Each panel is

divided into column and middle strips where the width of acolumn strip is taken as one-half of the shorter dimension of thepanel and bending moments determined for a full panel widthare then distributed between column and middle strips as shownin Table 255 If drops of dimensions not less than one-third ofthe shorter dimension of the panel are provided the width of thecolumn strip can be taken as the width of the drop In this casethe apportionment of the bending moments between columnand middle strips is modified accordingly

The slab thickness must be sufficient to satisfy appropriatedeflection criteria with a minimum thickness of 125 mm andprovide resistance to shearing forces and bending momentsPunching shear around the columns is a critical considerationfor which shear reinforcement can be provided in slabs not lessthan 200 mm thick The need for shear reinforcement can beavoided if drop panels or column heads of sufficient size areprovided Holes of limited dimensions may be formed in certainareas of the slab according to recommendations given in BS8110 Larger openings should be appropriately framed withbeams designed to carry the slab loads to the columns

481 Bending moments

The total bending moments for a full panel width at principalsections in each direction of span are given in Table 255 Panelwidths are taken between the centrelines of adjacent bays andpanel lengths between the centrelines of columns Momentscalculated at the centrelines of the supports may be reduced asexplained in section 1383 The slab is effectively designedas one-way spanning in each direction and the commentscontained in section 441 also apply here

At the edges of a flat slab the transfer of moments betweenthe slab and an edge or corner column may be limited by theeffective breadth of the moment transfer strip as shown inTable 256 The structural arrangement should be chosen toensure that the moment capacity of the transfer strip is at least50 of the outer support moment given in Table 255

482 Shearing forces

For punching shear calculations the design force obtained bysumming the shear forces on two opposite sides of a column ismultiplied by a shear enhancement factor to allow for theeffects of moment transfer as shown in Table 256 Criticalperimeters for punching shear occur at distances of 15d fromthe faces of columns column heads and drops where d is theeffective depth of the slab or drop as shown in Table 255

483 Reinforcement

At internal columns two-thirds of the reinforcement neededto resist the negative moments in the column strips should beplaced in a width equal to half that of the column strip andcentral with the column Otherwise the reinforcement neededto resist the moment apportioned to a particular strip should bedistributed uniformly across the full width of the strip

484 Alternative analysis

A more general equivalent frame method for the analysis offlat slabs is described in BS 8110 The bending moments and

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shearing forces are calculated by considering the structure asa series of continuous frames transversely and longitudinallyThe method is described in detail in Examples of the design ofreinforced concrete buildings For further information on bothequivalent frame and grillage methods of analysis of flat slabstructures see ref 33

49 FRAMED STRUCTURES

A structure is statically determinate if the forces and bendingmoments can be determined by the direct application of theprinciples of equilibrium Some examples include cantilevers(whether a simple bracket or a roof of a grandstand) a freelysupported beam a truss with pin-joints and a three-hinged archor frame A statically indeterminate structure is one in whichthere is a redundancy of members or supports or both andwhich can be analysed only by considering the elastic defor-mations under load Typical examples of such structures includerestrained beams continuous beams portal frames and othernon-triangulated structures with rigid joints and two-hinged andfixed-end arches The general notes relating to the analysis ofstatically determinate and indeterminate beam systems given insections 41 and 42 are equally valid when analysing framesProviding a frame can be represented sufficiently accurately byan idealised two-dimensional line structure it can be analysedby any of the methods mentioned earlier (and various othersof course)

The analysis of a two-dimensional frame is somewhat morecomplex than that of a beam system If the configuration ofthe frame or the applied loading (or both) is unsymmetricalside-sway will almost invariably occur making the requiredanalysis considerably longer Many more combinations of load(vertical and horizontal) may need to be considered to obtainthe critical moments Different partial safety factors may applyto different load combinations The critical design conditionsfor some columns may not necessarily be those correspondingto the maximum moment loading producing a reduced momenttogether with an increased axial thrust may be more criticalHowever to combat such complexities it is often possible tosimplify the calculations by introducing a degree of approxi-mation For instance when considering wind loads acting onregular multi-bay frames points of contra-flexure may beassumed to occur at the centres of all the beams and columns(see Table 262) thus rendering the frame statically determinateIn the case of frames that are not required to provide lateralstability the beams at each level acting with the columns aboveand below that level may be considered to form a separatesub-frame for analysis

Beeby (ref 34) has shown that if the many uncertaintiesinvolved in frame analysis are considered there is little tochoose as far as accuracy is concerned between analysing aframe as a single complete structure as a set of sub-frames oras a series of continuous beams with attached columns Ifthe effect of the columns is not included in the analysis of thebeams some of the calculated moments in the beams will begreater than those actually likely to occur

It may not always be possible to represent the true frame asan idealised two-dimensional line structure and analysis as afully three-dimensional space frame may be necessary If thestructure consists of large solid areas such as walls it may notbe possible to represent it adequately by a skeletal frame

The finite-element method of analysis is particularly suited tosolve such problems and is summarised briefly later

In the following pages the analysis of primary frames by themethods of slope deflection and various forms of momentdistribution is described Rigorous analysis of complex rigidframes generally requires an amount of calculation out ofall proportion to the real accuracy of the results and someapproximate solutions are therefore given for common casesof building frames and similar structures When a suitablepreliminary design has been justified by using approximatemethods an exhaustive exact analysis may be undertaken byemploying an established computer program

491 Building code requirements

For most framed structures it is not necessary to carry out afull structural analysis of the complete frame as a single unitand various simplifications are shown in Table 257 BS 8110distinguishes between frames subjected to vertical loads onlybecause overall lateral stability to the structure is provided byother means such as shear walls and frames that are requiredto support both vertical and lateral loads Load combinationsconsisting of (1) dead and imposed (2) dead and wind and(3) dead imposed and wind are also given in Table 257

For frames that are not required to provide lateral stabilitythe construction at each floor may be considered as a separatesub-frame formed from the beams at that level together withthe columns above and below The columns should be taken asfixed in position and direction at their remote ends unless theassumption of a pinned end would be more reasonable (eg ifa foundation detail is considered unable to develop momentrestraint) The sub-frame should then be analysed for therequired arrangements of dead and live loads

As a further simplification each individual beam span maybe considered separately by analysing a sub-frame consisting ofthe span in question together with at each end the upper andlower columns and the adjacent span These members areregarded as fixed at their remote ends with the stiffness of theouter spans taken as only one-half of their true value This sim-plified sub-frame should then be analysed for the loadingrequirements previously mentioned Formulae giving bendingmoments due to various loading arrangements acting on thesimplified sub-frame obtained by slope-deflection methods asdescribed in section 1421 are given in Table 261 Since themethod is lsquoexactrsquo the calculated bending moments may beredistributed within the limits permitted by the Codes Themethod is dealt with in more detail in Examples of the designof reinforced concrete buildings

BS 8110 also allows analysis of the beams at each floor as acontinuous system neglecting the restraint provided by thecolumns entirely so that the continuous beam is assumed to beresting on knife-edge supports Column moments are thenobtained by considering at each joint a sub-frame consistingof the upper and lower columns together with the adjacentbeams regarded as fixed at their remote ends and with theirstiffness taken as one-half of the true value

For frames that are required to provide lateral stability to thestructure as a whole load combinations 1 and 3 both need to beconsidered For combination 3 the following two-stage methodof analysis is allowed for frames of three or more approxi-mately equal bays First each floor is considered as a separate

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sub-frame for the effect of vertical loading as describedpreviously Next the complete structural frame is consideredfor the effect of lateral loading assuming that a positionof contra-flexure (ie zero bending moment) occurs at themid-point of each member This analysis corresponds to thatdescribed for building frames in section 4113 and the methodset out in diagram (c) of Table 262 may thus be used Themoments obtained from each of these analyses should thenbe summed and compared with those resulting from loadcombination 1 For tall narrow buildings and other cantileverstructures such as masts pylons and towers load combination2 should also be considered

492 Moment-distribution method no sway

In some circumstances a framed structure may not be subjectto side-sway for example if the frame is braced by other stiffelements within the structure or if both the configuration andthe loading are symmetrical Similarly if a vertically loadedframe is being analysed as a set of sub-frames as permitted inBS 8110 the effects of any side-sway may be ignored In suchcases Hardy Cross moment distribution may be used to evaluatethe moments in the beam and column system The procedurewhich is outlined in Table 258 is similar to the one used toanalyse systems of continuous beams

Precise moment distribution may also be used to solvesuch systems Here the method which is also summarised inTable 258 is slightly more complex to apply than in theequivalent continuous beam case Each time a moment iscarried over the unbalanced moment in the member must bedistributed between the remaining members meeting at the jointin proportion to the relative restraint that each provides Alsothe expression for the continuity factors is more difficultto evaluate Nevertheless the method is a valid alternative tothe conventional moment-distribution method It is describedin more detail in Examples of the design of reinforcedconcrete buildings

493 Moment-distribution method with sway

If sway occurs analysis by moment distribution increases incomplexity since in addition to the influence of the originalloading with no sway it is necessary to consider the effect ofeach degree of sway freedom separately in terms of unknownsway forces The separate results are then combined to obtainthe unknown sway values and hence the final moments Theprocedure is outlined in Table 259

The advantages of precise moment distribution are largelynullified if sway occurs but details of the procedure in suchcases are given in ref 35

To determine the moments in single-bay frames subjected toside sway Naylor (ref 36) devised an ingenious variant ofmoment distribution details of which are given in Table 259The method can also be used to analyse Vierendeel girders

494 Slope-deflection method

The principles of the slope-deflection method of analysing arestrained member are given in Table 260 and section 141together with basic formulae and formulae for the bending

moments in special cases When there is no deflection of oneend of the member relative to the other (eg when the supportsare not elastic as assumed) when the ends of the memberare either hinged or fixed and when the load on the member issymmetrically disposed the general expressions are simplifiedand the resulting formulae for some common cases of restrainedmembers are also given in Table 260

The bending moments on a framed structure are determinedby applying the formulae to each member successively Thealgebraic sum of the bending moments at any joint must equalzero When it is assumed that there is no deflection (or settle-ment) a of one support relative to the other there are as manyformulae for the end moments as there are unknowns andtherefore the restraint moments and the slopes at the endsof the members can be evaluated For symmetrical frameson unyielding foundations and carrying symmetrical verticalloads it is common to neglect the change in the position of thejoints due to the small elastic contractions of the members andthe assumption of a 0 is reasonably correct If the founda-tions or other supports settle unequally under the load thisassumption is not justified and the term a must be assigned avalue for the members affected

If a symmetrical or unsymmetrical frame is subjected to ahorizontal force the resulting sway causes lateral movementof the joints It is common in this case to assume that there isno elastic shortening of the members Sufficient formulae toenable the additional unknowns to be evaluated are obtainedby equating the reaction normal to the member that is theshear force on the member to the rate of change of bendingmoment Sway occurs also in unsymmetrical frames subjectto vertical loads and in any frame on which the load is notsymmetrically disposed

Slope-deflection methods have been used to derive bendingmoment formulae for the simplified sub-frames illustratedon Table 260 These simplified sub-frames correspond tothose referred to in BS 8110 as a basis for determiningthe bending moments in the individual members of a framesubjected to vertical loads only The method is describedin section 142

An example of applying the slope-deflection formulae to asimple problem of a beam hinged at one end and framed intoa column at the other end is given in section 141

495 Shearing forces on members of a frame

The shearing forces on any member forming part of a frame canbe simply determined once the bending moments have beenfound by considering the rate of change of the bendingmoment The uniform shearing force on a member AB due toend restraint only is (MAB MBA)lAB account being taken ofthe signs of the bending moment Thus if both of the restraintmoments are clockwise the shearing force is the numerical sumof the moments divided by the length of the member If onerestraint moment acts in a direction contrary to the other theshearing force is the numerical difference in the momentsdivided by the length of the member For a member with end Bhinged the shearing force due to the restraint moment at A isMABlAB The variable shearing forces caused by the loadson the member should be algebraically added to the uniformshearing force due to the restraint moments as indicated fora continuous beam in section 1112

Framed structures 37

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496 Portal frames

A common type of frame used in single-storey buildings is theportal frame with either a horizontal top member or twoinclined top members meeting at the ridge In Tables 263 and264 general formulae for the moments at both ends of thecolumns and at the ridge where appropriate are given togetherwith expressions for the forces at the bases of the columnsThe formulae relate to any vertical or horizontal load and toframes fixed or hinged at the bases In Tables 265 and 266corresponding formulae for special conditions of loading onframes of one bay are given

Frames of the foregoing types are statically indeterminatebut frames with a hinge at the base of each column and one atthe ridge that is a three-hinged frame can be readily analysedFormulae for the forces and bending moments are given inTable 267 for three-hinged frames Approximate expressionsare also given for certain modified forms of these frames such aswhen the ends of the columns are embedded in the foundationsand when a tie-rod is provided at eaves level

497 Finite elements

In conventional structural analysis numerous approximationsare introduced and the engineer is normally content to acceptthe resulting simplification Actual elements are considered asidealised one-dimensional linear members deformations due toaxial force and shear are assumed to be sufficiently small to beneglected and so on

In general such assumptions are valid and the results of theanalysis are sufficiently close to the values that would occurin the actual structure to be acceptable However when themember sizes become large in relation to the structure theyform the system of skeletal simplification breaks down Thisoccurs for example with the design of such elements as deepbeams shear walls and slabs of various types

One of the methods developed to deal with such so-calledcontinuum structures is that known as finite elements Thestructure is subdivided arbitrarily into a set of individualelements (usually triangular or rectangular in shape) which arethen considered to be inter-connected only at their corners(nodes) Although the resulting reduction in continuity mightseem to indicate that the substitute system would be muchmore flexible than the original structure this is not the case ifthe substitution is undertaken carefully since the adjoiningedges of the elements tend not to separate and thus simulatecontinuity A stiffness matrix for the substitute structure cannow be prepared and analysed using a computer in a similarway to that already described

Theoretically the pattern of elements chosen might bethought to have a marked effect on the validity of the resultsHowever although the use of a smaller mesh consisting ofa larger number of elements can often increase the accuracyof the analysis it is normal for surprisingly good results to beobtained by experienced analysts when using a rather coarsegrid consisting of only a few large elements

410 COLUMNS IN NON-SWAY FRAMES

In monolithic beam-and-column construction subjected tovertical loads only provision is still needed for the bending

moments produced on the columns due to the rigidity of thejoints The external columns of a building are subjected togreater moments than the internal columns (other conditionsbeing equal) The magnitude of the moment depends on therelative stiffness and the end conditions of the members

The two principal cases for beamndashcolumn connections areat intermediate points on the column (eg floor beams) and atthe top of the column (eg roof beam) Since each member canbe hinged fully fixed or partially restrained at its remote endthere are many possible combinations

In the first case the maximum restraint moment at the jointbetween a beam and an external column occurs when theremote end of the beam is hinged and the remote ends of thecolumn are fixed as indicated in Table 260 The minimumrestraint moment at the joint occurs when the remote end ofthe beam is fixed and the remote ends of the column are bothhinged as also indicated in Table 260 Real conditions inpractice generally lie between these extremes and with anycondition of fixity of the remote ends of the column themoment at the joint decreases as the degree of fixity at theremote end of the beam increases With any degree of fixity atthe remote end of the beam the moment at the joint increasesvery slightly as the degree of fixity at the remote ends of thecolumn increase

Formulae for maximum and minimum bending moments aregiven in Table 260 for a number of single-bay frames Themoment on the beam at the joint is divided between the upperand lower columns in the ratio of their stiffness factors K whenthe conditions at the ends of the two columns are identicalWhen one column is hinged at the end and the other is fixedthe solution given for two columns with fixed ends can still beused by taking the effective stiffness factor of the column withthe hinged end as 075K

For cases where the beamndashcolumn connection is at the top ofthe column the formulae given in Table 260 may be used bytaking the stiffness factors for the upper columns as zero

4101 Internal columns

For the frames of ordinary buildings the bending moments onthe upper and lower internal columns can be computed from theexpressions given at the bottom of Table 260 these formulaeconform to the method to be used when the beams are analysedas a continuous system on knife-edge supports as describedin clause 32125 of BS 8110 When the spans are unequal thegreatest bending moments on the column are when the value ofMes (see Table 260) is greatest which is generally when thelonger beam is loaded with (dead live) load while the shorterbeam carries dead load only

Another method of determining moments in the columnsaccording to the Code requirements is to use the simplifiedsub-frame formulae given on Table 261 Then consideringcolumn SO for example the column moment is given by

where DSO DST and DTS are distribution factors FS and FT arefixed-end moments at S and T respectively (see Table 261)This moment is additional to any initial fixed-end momentacting on SO

DSO2DTSFT 4FS

4DSTDTS

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To determine the maximum moment in the column it may benecessary to examine two separate simplified sub-frames inwhich each column is embodied at each floor level (ie thecolumn at joint S say is part of two sub-frames comprisingbeams QR to ST and RS to TU respectively) However themaximum moments usually occur when the central beam ofthe sub-frame is the longer of the two beams adjoining thecolumn being investigated as specified in the Code

4102 End columns

The bending moments due to continuity between the beams andthe columns vary more for end columns than for internalcolumns The lack of uniformity in the end conditions affectsthe moments determined by the simplified method describedearlier more significantly than for internal columns Howevereven though the values obtained by the simplified methodsare more approximate than for internal columns they are stillsufficiently accurate for ordinary buildings The simplifiedformulae given on Table 260 conform to clause 32125 ofBS 8110 while the alternative simplified sub-frame methoddescribed for internal columns may also be used

4103 Corner columns

Corner columns are generally subjected to bending momentsfrom beams in two directions at right angles These momentscan be independently calculated by considering two frames(also at right angles) but practical methods of column designdepend on both the relative magnitudes of the moments andthe direct load and the relevant limit-state condition Thesemethods are described in later sections of the Handbook

4104 Use of approximate methods

The methods hitherto described for evaluating the columnmoments in beam-and-column construction with rigid jointsinvolve significant calculation including the second momentof area of the members Often in practice and especially inthe preparation of preliminary schemes approximate methodsare very useful The final design should be checked by moreaccurate methods

The column can be designed provisionally for a direct loadincreased to allow for the effects of bending In determiningthe total column load at any particular level the load from thefloor immediately above that level should be multiplied by thefollowing factors internal columns 125 end columns 15 andcorner column 20

411 COLUMNS IN SWAY FRAMES

In exposed structures such as water towers bunkers and silosand in frames that are required to provide lateral stability to abuilding the columns must be designed to resist the effects ofwind When conditions do not warrant a close analysis of thebending moments to which a frame is subjected due to wind orother lateral forces the methods described in the following andshown in Table 262 are sufficiently accurate

4111 Open braced towers

For columns (of identical cross section) with braced cornersforming an open tower such as that supporting an elevated

water tank the expressions at (a) in Table 262 give bendingmoments and shearing forces on the columns and braces dueto the effect of a horizontal force at the head of the columns

In general the bending moment on the column is the shearforce on the column multiplied by half the distance between thebraces If a column is not continuous or is insufficiently bracedat one end as at an isolated foundation the bending moment atthe other end is twice this value

The bending moment on the brace at an external column isthe sum of the bending moments on the column at the points ofintersection with the brace The shearing force on the brace isequal to the change of bending moment from one end of thebrace to the other end divided by the length of the braceThese shearing forces and bending moments are additional tothose caused by the dead weight of the brace and any externalloads to which it may be subjected

The overturning moment on the frame causes an additionaldirect load on the leeward column and a corresponding relief ofload on the windward column The maximum value of thisdirect load is equal to the overturning moment at the footof the columns divided by the distance between the centres ofthe columns

The expressions in Table 262 for the bending moments andforces on the columns and braces apply for columns that arevertical or near vertical If the columns are inclined then theshearing force on a brace is 2Mb divided by the length ofthe brace being considered

4112 Columns supporting massivesuperstructures

The case illustrated at (b) in Table 262 is common in silos andbunkers where a superstructure of considerable rigidity iscarried on comparatively short columns If the columns arefixed at the base the bending moment on a single column isFh2J where J is the number of columns if they are all of thesame size the significance of the other symbols is indicated inTable 262

If the columns are of different sizes the total shearing forceon any one line of columns should be divided between them inproportion to the second moment of area of each column sincethey are all deflected by the same amount If J1 is the numberof columns with second moment of area I1 J2 is the number ofcolumns with second moment of area I2 and so on the totalsecond moment of area I J1I1 J2I2 and so on Then onany column having a second moment of area Ij the bendingmoment is FhIj2I as given in diagram (b) in Table 262Alternatively the total horizontal force can be divided amongthe columns in proportion to their cross-sectional areas (thusgiving uniform shear stress) in which case the formula for thebending moment on any column with cross-sectional area Aj isFhAj2A where A is the sum of the cross-sectional areas ofall the columns resisting the total shearing force F

4113 Building frames

In the frame of a multi-storey multi-bay building the effect ofthe wind may be small compared to that of other loads andin this case it is sufficiently accurate to divide the horizontalshearing force between the columns on the basis that an endcolumn resists half the amount on an internal column If in the

Columns in sway frames 39

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plane of the lateral force F Jt is the total number of columns inone frame the effective number of columns for the purpose ofcalculating the bending moment on an internal column is Jt 1the two end columns being equivalent to one internal columnsee diagram (c) in Table 262 In a building frame subjectedto wind pressure the forces on each panel (or storey height)F1 F2 F3 and so on are generally divided into equal shearingforces at the head and base of each storey height of columnsThe shearing force at the bottom of any internal column istoreys from the top is (F Fi2)( Jt 1) where F F1 F2 F3 Fi 1 The bending moment is then the shearingforce multiplied by half the storey height

A bending moment and a corresponding shearing force arecaused on the floor beams in the same way as on the braces ofan open braced tower At an internal column the sum of thebending moments on the two adjacent beams is equal to the sumof the moments at the base of the upper column and the head ofthe lower column

The above method of analysis for determining the effects oflateral loading corresponds to that described in section 491and recommended in BS 8110 for a frame of three or moreapproximately equal bays

412 WALL AND FRAME SYSTEMS

In all forms of construction the effects of wind force increasein significance as the height of the structure increases Oneway of reducing lateral sway and improving stability is byincreasing the sectional size of the component members ofsway frames However this will have a direct consequenceof increasing storey height and building cost

Often a better way is to provide a suitable arrangement ofwalls linked to flexible frames The walls can be external orinternal be placed around lift shafts and stairwells to form corestructures or be a combination of types Sometimes core wallsare constructed in advance of the rest of the structure to avoidsubsequent delays The lateral stiffness of systems with acentral core can be increased by providing deep cantilevermembers at the top of the core structure to which the exteriorcolumns are connected Another approach is to increase theload on the central core by replacing the exterior columns byhangers suspended from the cantilever members at the top ofthe building This also avoids the need for exterior columns atground level and their attendant foundations As buildings gettaller the lateral stability requirements are of paramount impor-tance The structural efficiency can be increased by replacingthe building facade by a rigidly jointed framework so that theouter shell acts effectively as a closed-box

Some different structural forms consisting of assemblies ofmulti-storey frames shear walls and cores with an indicationof typical heights and proportions taken from ref 37 areshown in Table 268

4121 Shear wall structures

The lateral stability of low- to medium-rise buildings is oftenobtained by providing a suitable system of stiff shear walls Thearrangement of the walls should be such that the building is stiffin both flexure and torsion In rectangular buildings externalshear walls in the short direction can be used to resist lateralloads acting on the wide faces with rigid frames or infill panels

in the long direction In buildings of square plan form a strongcentral service core surrounded by flexible external framescan be used If strong points are placed at both ends of a longbuilding the restraint provided to the subsequent shrinkageand thermal movements of floors and roof should be carefullyconsidered

In all cases the floors and roof are considered to act as stiffplates so that at each level the horizontal displacements of allwalls and columns are taken to be the same provided the totallateral load acts through the shear centre of the system If thetotal lateral load acts eccentrically then the additional effectof the resulting torsion moment needs to be considered Theanalysis and design of shear wall buildings is covered in ref 38from which much of the following treatment is based Severaldifferent plan configurations of shear walls and core units withnotes on their suitability are shown in Table 269

4122 Walls without openings

The lateral load transmitted to an individual wall is a functionof its position and its relative stiffness The total deflection of acantilever wall under lateral load is a combination of bendingand shear deformations However for a uniformly distributedload the shear deformation is less than 10 of the total forHD 3 in the case of plane walls and HD 5 in the case offlanged walls with BD 05 (where B is width of flange D isdepth of web and H is height of wall) Thus for most shearwalls without openings the dominant mode of deformation isbending and the stiffness of the wall can be related directly to thesecond moment of area of the cross section I Then for a totallateral load F applied at the shear centre of a system of parallelwalls the shearing force on an individual wall j is FIjIj

The position of the shear centre along a given axis y can bereadily determined by calculating the moment of stiffness ofeach wall about an arbitrary reference point on the axis Thedistance from the point to the shear centre yc Ijyj Ij

If the total lateral load acts at distance yo along the axis theresulting horizontal moment is F(yondashyc) Then if the torsionstiffness of individual walls is neglected the total shearingforce on wall j is

Fj FIj Ij F(yo yc)IjyjIj (yj yc)2

More generalised formulae in which a wall system is related totwo perpendicular axes are given in Table 269 The aboveanalysis takes no account of rotation at the base of the walls

4123 Walls containing openings

In the case of walls pierced by openings the behaviour ofthe individual wall sections is coupled to a variable degree Theconnections between the individual sections are provided eitherby beams that form part of the wall or by floor slabs or by acombination of both The pierced wall may be analysed byelastic methods in which the flexibility of the coupling elementsis represented as a continuous flexible medium Alternativelythe pierced wall may be idealised as an equivalent plane frameusing a lsquowide columnrsquo analogy

The basis of the continuous connection model is described insection 152 and analytical solutions for a wall containing asingle line of openings are given in Table 270

Structural analysis40

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4124 Interaction of shear walls and frames

The interaction forces between solid walls pierced walls andframes can vary significantly up the height of a building asa result of the differences in the free deflected shapes ofeach structural form The deformation of solid walls is mainlyflexural whereas pierced walls deform in a shearndashflexure modeand frames deform in an almost pure shear manner As a resulttowards the bottom of a building solid walls attract load whilstframes and to a lesser extent pierced walls shed load Thebehaviour is reversed towards the top of a building Thusalthough the distribution of load intensity between the differentelements is far from uniform up the building the total lateralforce resisted by each varies by a smaller amount

As a first approximation the shearing force at the bottom ofeach load-resisting element can be determined by considering asingle interaction force at the top of the building Formulae bywhich the effective stiffness of pierced walls and frames can bedetermined are given in section 153

413 ARCHES

Arch construction in reinforced concrete occurs sometimes inroofs but mainly in bridges An arch may be three-hingedtwo-hinged or fixed-ended (see diagrams in Table 271) andmay be symmetrical or unsymmetrical right or skew singleor one of a series of arches mutually dependent upon eachother The following consideration is limited to symmetrical andunsymmetrical three-hinged arches and to symmetrical two-hinged and fixed-end arches reference should be made to otherpublications for information on more complex types

Arch construction may comprise an arch slab (or vault) or aseries of parallel arch ribs The deck of an arch bridge may besupported by columns or transverse walls carried on an archslab or ribs when the structure may have open spandrels or thedeck may be below the crown of the arch either at the level ofthe springing (as in a bowstring girder) or at some intermediatelevel A bowstring girder is generally regarded as a two-hingedarch with the horizontal component of thrust resisted by a tiewhich normally forms part of the deck If earth or other filling isprovided to support the deck an arch slab and spandrel walls arerequired and the bridge is a closed or solid-spandrel structure

4131 Three-hinged arch

An arch with a hinge at each springing and at the crown isstatically determinate The thrusts on the abutments and thebending moments and shearing forces on the arch itself arenot affected by a small movement of one abutment relative tothe other This type of arch is therefore used when there is apossibility of unequal settlement of the abutments

For any load in any position the thrust on the abutmentscan be determined by the equations of static equilibrium Forthe general case of an unsymmetrical arch with a load actingvertically horizontally or at an angle the expressions for thehorizontal and vertical components of the thrusts are givenin the lower part of Table 271 For symmetrical arches the for-mulae given in Table 267 for the thrusts on three-hinged framesapply or similar formulae can be obtained from the generalexpressions in Table 271 The vertical component is the same asthe vertical reaction for a freely supported beam The bending

moment at any cross section of the arch is the algebraic sum ofthe moments of the loads and reactions on one side of thesection There is no bending moment at a hinge The shearingforce is likewise the algebraic sum of the loads and reactionsresolved at right angles to the arch axis at the section and actingon one side of the section The thrust at any section is the sumof the loads and reactions resolved parallel to the axis of thearch at the section and acting on one side of the section

The extent of the arch that should be loaded with imposedload to give the maximum bending moment or shearing forceor thrust at a particular cross section can be determined byconstructing a series of influence lines A typical influence linefor a three-hinged arch and the formulae necessary to constructan influence line for unit load in any position are given in theupper part of Table 271

4132 Two-hinged arch

The hinges of a two-hinged arch are placed at the abutmentsso that as in a three-hinged arch only thrusts are transmitted tothe abutments and there is no bending moment on the archat the springing The vertical component of the thrust from asymmetrical two-hinged arch is the same as the reaction fora freely supported beam Formulae for the thrusts and bendingmoments are given in Table 271 and notes in section 162

4133 Fixed arch

An arch with fixed ends exerts in addition to the vertical andhorizontal thrusts a bending moment on the abutments Like atwo-hinged arch and unlike a three-hinged arch a fixed-endarch is statically indeterminate and the stresses are affected bychanges of temperature and shrinkage of the concrete As it isassumed in the general theory that the abutments cannot moveor rotate the arch can only be used in such conditions

A cross section of a fixed-arch rib or slab is subjected to abending moment and a thrust the magnitudes of which have tobe determined The design of a fixed arch is a matter of trial andadjustment since both the dimensions and the shape of the archaffect the calculations but it is possible to select preliminarysizes that reduce the repetition of arithmetic work to a minimumA suggested method of determining possible sections at thecrown and springing as given in Table 272 and explained insection 1631 is based on first treating the fixed arch as ahinged arch and then estimating the size of the cross sectionsby greatly reducing the maximum stresses

The general formulae for thrusts and bending moments on asymmetrical fixed arch of any profile are given in Table 272and notes on the application and modification of the formulaeare given in section 163 The calculations necessary to solvethe general and modified formulae are tedious but are easedsomewhat by preparing them in tabular form The form givenin Table 272 is particularly suitable for open-spandrel archbridges because the appropriate formulae do not assume a con-stant value of aI the ratio of the length of a segment of the archto the mean second moment of area of the segment

For large span arches calculations are made much easier andmore accurate by preparing and using influence lines for thebending moment and thrust at the crown the springing and thequarter points of the arch Typical influence lines are given inTable 272 and such diagrams can be constructed by considering

Arches 41

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the passage over the arch of a single concentrated unit load andapplying the formulae for this condition The effect of the deadload and of the most adverse disposition of imposed load canbe readily calculated from these diagrams If the specifiedimposed load includes a moving concentrated load such as aKEL the influence lines are almost essential for determiningthe most adverse position The case of the positive bendingmoment at the crown is an exception when the most adverseposition of the load is at the crown A method of determiningthe data to establish the ordinates of the influence lines is givenin Table 273

4134 Fixed parabolic arches

In Table 274 and in section 164 consideration is given tosymmetrical fixed arches that can have either open or solidspandrels and be either arch ribs or arch slabs The method isbased on that of Strassner as developed by H Carpenter andthe principal assumption is that the axis of the arch is made tocoincide with the line of thrust due to the dead load This resultsin an economical structure and a simple calculation methodThe shape of the axis of the arch is approximately that of aparabola and this method can therefore be used only when thedesigner is free to select the profile of the arch The parabolicform may not be the most economic for large spans althoughit is almost so and a profile that produces an arch axis coinci-dent with the line of thrust for the dead load plus one-half of theimposed load may be more satisfactory If the increase in thethickness of the arch from crown to springing is of a parabolicform only the bending moments and thrusts at the crownand the springing need to be investigated The necessaryformulae are given in section 164 where these include a seriesof coefficients values of which are given in Table 274 Theapplication of the method is also illustrated by an examplegiven in section 164 The component forces and momentsare considered in the following treatment

The thrusts due to the dead load are relieved somewhat by theeffect of the compression causing elastic shortening of the archFor arches with small ratios of rise to span and arches that arethick in comparison with the span the stresses due to archshortening may be excessive This can be overcome by intro-ducing temporary hinges at the crown and the springing whicheliminate all bending stresses due to dead load The hinges arefilled with concrete after arch shortening and much of theshrinkage of the concrete have taken place

An additional horizontal thrust due to a temperature rise ora corresponding counter-thrust due to a temperature fall willaffect the stresses in the arch and careful consideration mustbe given to the likely temperature range The shrinkage of theconcrete that occurs after completion of the arch produces acounter-thrust the magnitude of which is modified by creep

The extent of the imposed load on an arch necessary toproduce the maximum stresses in the critical sections can bedetermined from influence lines and the following values areapproximately correct for parabolic arches The maximumpositive moment at the crown occurs when the middle third of thearch is loaded the maximum negative moment at a springingoccurs when four-tenths of the span adjacent to the springing isloaded the maximum positive moment at the springing occurswhen six-tenths of the span furthest away from the springing

is loaded In the expressions given in section 1644 the imposedload is expressed in terms of an equivalent UDL

When the normal thrusts and bending moments on the mainsections have been determined the areas of reinforcement andstresses at the crown and springing can be calculated Allthat now remains is to consider the intermediate sections anddetermine the profile of the axis of the arch If the dead loadis uniform throughout (or practically so) the axis will be aparabola but if the dead load is not uniform the axis must beshaped to coincide with the resulting line of thrust This canbe obtained graphically by plotting force-and-link polygonsthe necessary data being the magnitudes of the dead load thehorizontal thrust due to dead load and the vertical reaction(equal to the dead load on half the span) of the springing Theline of thrust and therefore the axis of the arch having beenestablished and the thickness of the arch at the crown and thespringing having been determined the lines of the extradosand the intrados can be plotted to give a parabolic variation ofthickness between the two extremes

414 PROPERTIES OF MEMBERS

4141 End conditions

Since the results given by the more precise methods of elasticanalysis vary considerably with the conditions of restraint atthe ends of the members it is important that the assumedconditions are reasonably obtained in the actual constructionAbsolute fixity is difficult to attain unless a beam or column isembedded monolithically in a comparatively large mass ofconcrete Embedment of a beam in a masonry wall representsmore nearly the condition of a hinge and should normally beconsidered as such A continuous beam supported internallyon a beam or column is only partly restrained and where thesupport at the outer end of an end span is a beam a hinge shouldbe assumed With the ordinary type of pad foundation designedsimply for a uniform ground bearing pressure under the directload on a column the condition at the foot of the column shouldalso be considered as a hinge A column built on a pile-capsupported by two three or four piles is not absolutely fixed buta bending moment can be developed if the resulting verticalreaction (upwards and downwards) and the horizontal thrust canbe resisted by the piles The foot of a column can be consideredas fixed if it is monolithic with a substantial raft foundation

In two-hinged and three-hinged arches hinged frames andsome bridge types where the assumption of a hinged joint mustbe fully realised it is necessary to form a definite hinge in theconstruction This can be done by inserting a steel hinge (orsimilar) or by forming a hinge within the frame

4142 Section properties

For the elastic analysis of continuous structures the sectionproperties need to be known Three bases for calculating thesecond moment of area of a reinforced concrete section are gen-erally recognised in codes of practice as follows

1 The concrete section the entire concrete area but ignoringthe reinforcement

Structural analysis42

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2 The gross section the entire concrete area together with thereinforcement on the basis of a modular ratio (ie ratio ofmodulus of elasticity values of steel and concrete)

3 The transformed section the concrete area in compressiontogether with the reinforcement on the basis of modular ratio

For methods 2 and 3 the modular ratio should be based on aneffective modulus of elasticity of concrete taking account ofthe creep effects of long-term loading In BS 8110 a modularratio of 15 is recommended unless a more accurate figure can bedetermined However until the reinforcement has been deter-mined or assumed calculation of the section properties in thisway cannot be made with any precision Moreover the sectionproperties vary considerably along the length of the member asthe distribution of reinforcement and for method 3 the depthof concrete in compression change The extent and effect ofcracking on the section properties is particularly difficult toassess for a continuous beam in beam-and-slab construction inwhich the beam behaves as a flanged section in the spans wherethe bending moments are positive but is designed as a rectan-gular section towards the supports where the bending momentsare negative

Method 1 is the simplest one to apply and the only practicalapproach when beginning a new design but one of the othermethods could be used when checking the ability of existingstructures to carry revised loadings and for new structureswhen a separate analysis for the SLSs is required In all cases itis important that the method used to assess the section propertiesis the same for all the members involved in the calculationWhere a single stiffness value is to be used to characterise amember method 1 (or 2) is likely to provide the most accurateoverall solution Method 3 will only be appropriate where thevariations in section properties over the length of membersare properly taken into account

415 EARTHQUAKE-RESISTANT STRUCTURES

Earthquakes are ground vibrations that are caused mainly byfracture of the earthrsquos crust or by sudden movement along analready existing fault During a seismic excitation structuresare caused to oscillate in response to the forced motion of thefoundations The affected structure needs to be able to resistthe resulting horizontal load and also dissipate the impartedkinetic energy over successive deformation cycles It would beuneconomical to design the structure to withstand a majorearthquake elastically and the normal approach is to provide itwith sufficient strength and ductility to withstand such an eventby responding inelastically provided that the critical regions

and the connections between members are designed speciallyto ensure adequate ductility

Significant advances have been made in the seismic designof structures in recent years and very sophisticated codes ofpractice have been introduced (ref 39) A design horizontalseismic load is recommended that depends on the importanceof the structure the seismic zone the ground conditionsthe natural period of vibration of the structure and the availableductility of the structure Design load effects in the structureare determined either by linear-elastic structural analysis forthe equivalent static loading or by dynamic analysis When alinear-elastic method is used the design and detailing of themembers needs to ensure that in the event of a more severeearthquake the post-elastic deformation of the structure willbe adequately ductile For example in a multi-storey framesufficient flexural and shear strength should be provided in thecolumns to ensure that plastic hinges form in the beams inorder to avoid a column side-sway mechanism The properdetailing of the reinforcement is also a very important aspectin ensuring ductile behaviour At the plastic hinge regions ofmoment resisting frames in addition to longitudinal tensionreinforcement it is essential to provide adequate compressionreinforcement Transverse reinforcement is also necessary toact as shear reinforcement to prevent premature buckling ofthe longitudinal compression reinforcement and to confine thecompressed concrete

Buildings should be regular in plan and elevation withoutre-entrant angles and discontinuities in transferring verticalloads to the ground Unsymmetrical layouts resulting in largetorsion effects flat slab floor systems without any beams andlarge discontinuities in infill systems (such as open groundstoreys) should be avoided Footings should be founded at thesame level and should be interconnected by a mat foundationor by a grid of foundation beams Only one foundation typeshould in general be used for the same structure unless thestructure is formed of dynamically independent units

An alternative to the conventional ductile design approach isto use a seismic isolation scheme In this case the structure issupported on flexible bearings so that the period of vibration ofthe combined structure and supporting system is long enoughfor the structure to be isolated from the predominant earthquakeground motion frequencies In addition extra damping isintroduced into the system by mechanical energy dissipatingdevices in order to reduce the response of the structure to theearthquake and keep the deflections of the flexible systemwithin acceptable limits

A detailed treatment of the design of earthquake-resistingconcrete structures is contained in ref 40

Earthquake-resistant structures 43

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51 PRINCIPLES AND REQUIREMENTS

In modern Codes of Practice a limit-state design concept isused Ultimate (ULS) and serviceability (SLS) limit-states areconsidered as well as durability and in the case of buildingsfire-resistance Partial safety factors are incorporated in bothloads and material strengths to ensure that the probability offailure (ie not satisfying a design requirement) is acceptablylow For British Codes (BS 8110 BS 5400 BS 8007) detailsare given of design requirements and partial safety factors inChapter 21 material properties in Chapter 22 durability andfire-resistance in Chapter 23 For EC 2 corresponding data aregiven in Chapters 29 30 and 31 respectively

Members are first designed to satisfy the most critical limit-state and then checked to ensure that the other limit-statesare not reached For most members the critical condition to beconsidered is the ULS on which the required resistances of themember in bending shear and torsion are based The require-ments of the various SLSs such as deflection and crackingare considered later However since the selection of an adequatespan to effective depth ratio to prevent excessive deflection andthe choice of a suitable bar spacing to avoid excessive crackingcan also be affected by the reinforcement stress the designprocess is generally interactive Nevertheless it is normal tostart with the requirements of the ULS

52 RESISTANCE TO BENDING AND AXIAL FORCE

Typically beams and slabs are members subjected to bendingwhile columns are subjected to a combination of bending andaxial force In this context a beam is defined as a member inBS 8110 with a clear span not less than twice the effectivedepth and in EC 2 as a member with a span not less than threetimes the overall depth Otherwise the member is treated as adeep beam for which different design methods are applicableA column is defined as a member in which the greater overallcross-sectional dimension does not exceed four times thesmaller dimension Otherwise the member is considered as awall for which a different design approach is adopted Somebeams for example in portal frames and slabs for example inretaining walls are subjected to bending and axial force Insuch cases small axial forces that are beneficial in providingresistance to bending are generally ignored in design

521 Basic assumptions

For the analysis of sections in bending or combined bendingand axial force at the ULS the following basic assumptionsare made

The resistance of the concrete in tension is ignored

The distribution of strain across the section is linear that issections that are plane before bending remain plane afterbending the strain at a point being proportional to its distancefrom the axis of zero strain (neutral axis) In columns ifthe axial force is dominant the neutral axis can lie outsidethe section

Stressndashstrain relationships for concrete in compression andfor reinforcement in tension and compression are thoseshown in the diagrams on Table 36 for BS 8110 andBS 5400 and Table 44 for EC 2

The maximum strain in the concrete in compression is 00035except for EC 2 where the strains shown in the followingdiagram and described in the following paragraph apply

h

0

0

c2

c2

cu

(37)h

For sections subjected to pure axial compression the strain islimited to c2 For sections partly in tension the compressivestrain is limited to cu For intermediate conditions the straindiagram is obtained by taking the compressive strain as c2 at alevel equal to 37 of the section depth from the more highlycompressed face For concrete strength classes C5060 thelimiting strains are c2 0002 and cu 00035 For higherstrength concretes other values are given in Table 44

Strain distribution at ULS in EC 2

Chapter 5

Design of structuralmembers

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In all codes for sections partly in tension the shape of thebasic concrete stress-block is a combination of a parabolaand a rectangle In EC 2 a form consisting of a triangle anda rectangle is also given In all codes a simplified rectangularstress distribution may also be used If the compression zoneis rectangular the compressive force and the distance of theforce from the compression face can be readily determined foreach stress-block and the resulting properties are given insection 241 for BS 8110 and section 321 for EC 2

The stresses in the reinforcement depend on the strains in theadjacent concrete which depend in turn on the depth of theneutral axis and the position of the reinforcement in relationto the concrete surfaces The effect of these factors will beexamined separately for beams and columns

522 Beams

Depth of neutral axis This is significant because the valueof xd where x is the neutral axis depth and d is the effectivedepth of the tension reinforcement not only affects the stress inthe reinforcement but also limits the amount of moment redis-tribution allowed at a given section In BS 8110 where becauseof moment redistribution allowed in the analysis of a memberthe design moment is less than the maximum elastic momentthe requirement xd (b 04) should be satisfied whereb is the ratio of design moment to maximum elastic momentThus for reductions in moment of 10 20 and 30 xdmust not exceed 05 04 and 03 respectively In EC 2 asmodified by the UK National Annex similar restrictions applyfor concrete strength classes C5060

and xd 0456 and dx 043 for BS 5400 For design toEC 2 considerations similar to those in BS 8110 apply

Effect of axial force The following figure shows a sectionthat is subjected to a bending moment M and an axial force Nin which a simplified rectangular stress distribution has beenassumed for the compression in the concrete The stress blockis shown divided into two parts of depths dc and (h 2dc)providing resistance to the bending moment M and the axialforce N respectively where 0 dc 05h

Resistance to bending and axial force 45

Section Strain diagram

Section Forces

The figure here shows a typical strain diagram for a sectioncontaining both tension and compression reinforcement Forthe bi-linear stressndashstrain curve in BS 8110 the maximumdesign stresses in the reinforcement are fy115 for values of s

and s fy115Es From the strain diagram this gives

and

In BS 5400 the reinforcement stressndashstrain curve is tri-linear withmaximum design stresses of fy115 in tension and 2000fy (2300 fy) in compression These stresses apply for values ofs 0002 fy115Es and s 0002 giving

With cu 00035 fy 500 Nmm2 and Es 200 kNmm2 thecritical values are xd 0617 and for BS 8110dx 038

dx (cu0002) cu

x d cu (cu 0002 fy 115Es) and

dx (cu fy 115Es) cux d cu (cu fy 115Es)

The depth dc (and the force in the tension reinforcement) aredetermined by the bending moment given by

M bdc(d 05dc) fcd

Thus for analysis of the section axial forces may be ignoredfor values satisfying the condition

Nb(h 2dc) fcd

Combining the two requirements gives

Nbhfcd 2M(d05dc)

In the limit when dc 05h this gives

Nbhfcd 2M(d025h)congbhfcd 3Mh

For BS 8110 the condition becomes N 045bhfcu 3Mhwhich being simplified to N 01bhfcu is reasonably valid forMbh2fcu 012 For EC 2 the same condition becomesN 0567bhfck 3Mh which may be reasonably simplified toN 012bhfck for Mbh2fck 015

Analysis of section Any given section can be analysed by atrial-and-error process An initial value is assumed for theneutral axis depth from which the concrete strains at the rein-forcement positions can be calculated The correspondingstresses in the reinforcement are determined and the resultingforces in the reinforcement and the concrete are obtained If theforces are out of balance the value of the neutral axis depth ischanged and the process is repeated until equilibrium isachieved Once the balanced condition has been found theresultant moment of all the forces about the neutral axis or anyother suitable point is calculated

Singly reinforced rectangular sections For a section thatis reinforced in tension only and subjected to a moment M aquadratic equation in x can be obtained by taking moments forthe compressive force in the concrete about the line of actionof the tension reinforcement The resulting value of x can beused to determine the strain diagram from which the strain in

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the reinforcement and hence the stress can be calculated Therequired area of reinforcement can then be determined fromthe tensile force whose magnitude is equal to the compressiveforce in the concrete If the calculated value of x exceeds thelimit required for any redistribution of moment then a doublyreinforced section will be necessary

In designs to BS 8110 and BS 5400 the lever arm betweenthe tensile and compressive forces is to be taken not greater than095d Furthermore it is a requirement in BS 5400 that if xexceeds the limiting value for using the maximum designstress then the resistance moment should be at least 115MAnalyses are included in section 2421 for both BS 8110 andBS 5400 and in section 3221 for EC 2 Design charts basedon the parabolic-rectangular stress-block for concrete withfy 500 Nmm2 are given in Tables 313 323 and 47 forBS 8110 BS 5400 and EC 2 respectively Design tables basedon the rectangular stress-blocks for concrete are given inTables 314 324 and 48 for BS 8110 BS 5400 and EC 2respectively These tables use non-dimensional parameters andare applicable for values of fy 500 Nmm2

Doubly reinforced rectangular sections A sectionneeding both tension and compression reinforcement andsubjected to a moment M can be designed by first selecting asuitable value for x such as the limiting value for using themaximum design stress in the tension reinforcement or satisfy-ing the condition necessary for moment redistribution Therequired force to be provided by the compression reinforcementcan be derived by taking moments for the compressive forcesin the concrete and the reinforcement about the line of actionof the tensile reinforcement The force to be provided by thetension reinforcement is equal to the sum of the compressiveforces The reinforcement areas can now be determined takingdue account of the strains appropriate to the value of x selected

Analyses are included in section 2422 for both BS 8110and BS 5400 and in section 3222 for EC 2 Design chartsbased on the rectangular stress-blocks for concrete are given inTables 315 and 316 for BS 8110 Tables 325 and 326 forBS 5400 and Tables 49 and 410 for EC 2

Design formulae for rectangular sections Designformulae based on the rectangular stress-blocks for concreteare given in BS 8110 and BS 5400 In both codes x is limitedto 05d so that the formulae are automatically valid for redistri-bution of moment not greater than 10 The design stress intension reinforcement is taken 087fy although this is onlystrictly valid for xd 0456 in BS 5400 The design stresses inany compression reinforcement are taken as 087fy in BS 8110and 072fy in BS 5400 Design formulae are given in section2423 for BS 8110 and BS 5400 Although not included inEC 2 appropriate formulae are given in section 3223

Flanged sections In monolithic beam and slab constructionwhere the web of the beam projects below the slab the beam isconsidered as a flanged section for sagging moments Theeffective width of the flange over which uniform conditionsof stress can be assumed is limited to values stipulated in thecodes In most sections where the flange is in compressionthe depth of the neutral axis will be no greater than the flangethickness In such cases the section can be considered to berectangular with b taken as the flange width If the depth of

the neutral axis does exceed the thickness of the flange thesection can be designed by dividing the compression zoneinto portions comprising the web and the outlying flangesDetails of the flange widths and design procedures are given insections 2424 for BS 8110 and 3224 for EC 2

Beam sizes The dimensions of beams are mainly determinedby the need to provide resistance to moment and shear In thecase of beams supporting items such as cladding partitions orsensitive equipment service deflections can also be criticalOther factors such as clearances below beams dimensions ofbrick and block courses widths of supporting members andsuitable sizes of formwork also need to be taken into accountFor initial design purposes typical spaneffective depth ratiosfor beams in buildings are given in the following table

Design of structural members46

Spaneffective depth ratios for initial design of beams

Span conditionsUltimate design load

25 kNm 50 kNm 100 kNm

Cantilever 9 7 5Simply supported 18 14 10Continuous 22 17 12

The effective span of a continuous beam is generally taken asthe distance between centres of supports At a simple supportor at an encastre end the centre of action may be taken at adistance not greater than half of the effective depth fromthe face of the support Beam widths are often taken as half theoverall depth of the beam with a minimum of 300 mm If amuch wider band beam is used the spaneffective depth ratiocan be increased significantly to the limit necessitated bydeflection considerations

In BS 8110 and BS 5400 to ensure lateral stability simplysupported and continuous beams should be so proportioned thatthe clear distance between lateral restraints is not greater than60bc or 250bc

2d whichever is the lesser For cantilevers inwhich lateral restraint is provided only at the support the cleardistance from the end of the cantilever to the face of the sup-port should not exceed 25bc or 100bc

2d whichever is the lesserone In the foregoing bc is the breadth of the compression faceof the beam (measured midway between restraints) orcantilever In EC 2 second order effects in relation to lateralstability may be ignored if the distance between lateralrestraints is not greater than 50bc(hbc)13 and h 25bc

523 Slabs

Solid slabs are generally designed as rectangular strips of unitwidth and singly reinforced sections are normally sufficientRibbed slabs are designed as flanged sections of width equalto the rib spacing for sagging moments Continuous ribbedslabs are often made solid in support regions so as to developsufficient resistance to hogging moments and shear forcesAlternatively in BS 8110 ribbed slabs may be designed as aseries of simply supported spans with a minimum amountof reinforcement provided in the hogging regions to controlthe cracking The amount of reinforcement recommended is

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25 of that in the middle of the adjoining spans extending intothe spans for at least 15 of the span length

The thickness of slabs is normally determined by deflectionconsiderations which sometimes result in the use of reducedreinforcement stresses to meet code requirements Typicalspaneffective depth ratios for slabs designed to BS 8110 aregiven in the following table

contain a modification factor the use of which necessitates aniteration process with the factor taken as 10 initially Details ofthe design procedures are given in Tables 321 and 322 forBS 8110 Tables 331 and 332 for BS 5400 and Tables 415 and416 for EC 2

Analysis of section Any given section can be analysed by atrial-and-error process For a section bent about one axis aninitial value is assumed for the neutral axis depth from whichthe concrete strains at the positions of the reinforcement can becalculated The resulting stresses in the reinforcement aredetermined and the forces in the reinforcement and concreteare evaluated If the resultant force is not equal to the designaxial force N the value of the neutral axis depth is changed andthe process repeated until equality is achieved The sum of themoments of all the forces about the mid-depth of the section isthen the moment of resistance appropriate to N For a section inbiaxial bending initial values have to be assumed for the depthand the inclination of the neutral axis and the design processwould be extremely tedious without the aid of an interactivecomputer program

For design purposes charts for symmetrically reinforcedrectangular and circular sections bent about one axis can bereadily derived For biaxial bending conditions approximatedesign methods have been developed that utilise the solutionsobtained for uniaxial bending

Rectangular sections The figure here shows a rectangularsection with reinforcement in the faces parallel to the axisof bending

Resistance to bending and axial force 47

Spaneffective depth ratios for initial design of solid slabs

Span conditionsCharacteristic imposed load

5 kNm2 10 kNm2

Cantilever 11 10Simply supported

One-way span 27 24Two-way span 30 27

ContinuousOne-way span 34 30Two-way span 44 40

Flat slab (no drops) 30 27

In the table here the characteristic imposed load should includefor all finishes partitions and services For two-way spans theratios given apply to square panels For rectangular panelswhere the length is twice the breadth the ratios given for one-wayspans should be used For other cases ratios may be obtainedby interpolation The ratios apply to the shorter span for two-wayslabs and the longer span for flat slabs For ribbed slabs exceptfor cantilevers the ratios given in the table should be reducedby 20

524 Columns

The second order effects associated with lateral stability are animportant consideration in column design An effective height(or length in EC 2) and a slenderness ratio are determined inrelation to major and minor axes of bending An effective heightor length is a function of the clear height and depends upon theconditions of restraint at the ends of the column A clear distinc-tion exists between a braced column with effective height clear height and an unbraced column with effective height clearheight A braced column is one that is fully retrained in positionat the ends as in a structure where resistance to all the lateralforces in a particular plane is provided by stiff walls or bracingAn unbraced column is one that is considered to contribute tothe lateral stability of the structure as in a sway frame

In BS 8110 and BS 5400 a slenderness ratio is defined asthe effective height divided by the depth of the cross section inthe plane of bending A column is then considered as eithershort or slender according to the slenderness ratios Bracedcolumns are often short in which case second order effects maybe ignored In EC 2 the slenderness ratio is defined as theeffective length divided by the radius of gyration of the crosssection

Columns are subjected to combinations of bending momentand axial force and the cross section may need to be checked formore than one combination of values In slender columns theinitial moments obtained from an elastic analysis of the structureare increased by additional moments induced by the deflectionof the column In BS 8110 and EC 2 these additional moments

Resolving forces and taking moments about the mid-depth ofthe section gives the following equations for 0 x h

N k1bxfc As1 fs1 As2 fs2

M k1bxfc (05h k2x) As1 fs1 (05h d) As2 fs2 (d 05h)

where fs1 and fs2 are determined by the stressndashstrain curves for thereinforcement and depend on the value of x Values of k1 and k2

are determined by the concrete stress-block and fc is equal to fcu

in BS 8110 and BS 5400 and fck in EC 2For symmetrically reinforced sections As1 As2 Asc2

and d h d Design charts based on a rectangular stress-blockfor the concrete with values of fy 500 Nmm2 and dh 08and 085 respectively are given in Tables 317 and 318 forBS 8110 Tables 327 and 328 for BS 5400 and Tables 411 and412 for EC 2 Approximate design methods for biaxial bendingare given in Tables 321 331 and 416 for design to BS 8110BS 5400 and EC 2 respectively

Section Forces

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Circular sections The figure here shows a circular sectionwith six bars spaced equally around the circumference Six is theminimum number of bars recommended in the codes and solu-tions based on six bars will be slightly conservative if more barsare used The arrangement of bars relative to the axis of bendingaffects the resistance of the section and it can be shown that thearrangement in the figure is not the most critical in every casebut the variations are small and may be reasonably ignored

braced structures are typically square in cross section withsizes being determined mainly by the magnitude of the axialloads In multi-storey buildings column sizes are often keptconstant over several storeys with the reinforcement changingin relation to the axial load For initial design purposes typicalload capacities for short braced square columns in buildings aregiven in the following table

Design of structural members48

The following analysis is based on a uniform stress-block forthe concrete of depth x and width hsin at the base (as shownin the figure) Resolving forces and taking moments about themid-depth of the section where hs is the diameter of a circlethrough the centres of the bars gives the following equationsfor 0 x h

N [(2 sin2)8]h2fcd (Asc3)( fs1 fs2 fs3)M [(3sin sin3)72]h3fcd 0433(Asc3)( fs1 fs3)hs

where fs1 fs2 and fs3 are determined by the stressndashstrain curvesfor the reinforcement and depend on the value of x Values of fcd

and respectively are taken as 045fcu and 09 in BS 811004fcu and 10 in BS 5400 and 051fck and 08 in EC 2

Design charts derived for values of fy 500 Nmm2 andhsh 06 and 07 respectively are given in Tables 319and 320 for BS 8110 Tables 329 and 330 for BS 5400 andTables 413 and 414 for EC 2 Sections subjected to biaxialmoments Mx and My can be designed for the resultant moment

Design formulae In BS 8110 two approximate formulae aregiven for the design of short braced columns under specificconditions Columns which due to the nature of the structurecannot be subjected to significant moments for example columnsthat provide support to very stiff beams or beams on bearingsmay be considered adequate if N 040fcuAc 067Asc fy

Columns supporting symmetrical arrangements of beamsthat are designed for uniformly distributed imposed load andhave spans that do not differ by more than 15 of the longermay be considered adequate if N 035fcuAc 060Asc fy

BS 5400 contains general formulae for rectangular sectionsin the form of a trial-and-error procedure and two simplifiedformulae for specific applications details of which are given inTable 332

Column sizes Columns in unbraced structures are likely tobe rectangular in cross section due to the dominant effect ofbending moments in the plane of the structure Columns in

M (M2x M2

y)

Concrete Column Reinforcement percentageclass size

1 2 3 4

C2530 300 300 1370 1660 1950 2240350 350 1860 2260 2650 3050400 400 2430 2950 3470 3980450 450 3080 3730 4390 5040500 500 3800 4610 5420 6230

C3240 300 300 1720 2010 2300 2580350 350 2350 2740 3130 3520400 400 3070 3580 4090 4600450 450 3880 4530 5170 5820500 500 4790 5590 6390 7190

C4050 300 300 2080 2360 2650 2930350 350 2830 3220 3600 3990400 400 3700 4200 4710 5210450 450 4680 5320 5960 6600500 500 5780 6570 7360 8150

Ultimate design loads (kN) for short braced columns

In the foregoing table the loads were derived from the BS 8110equation for columns that are not subjected to significantmoments with fy 500 Nmm2 In determining the columnloads the ultimate load from the floor directly above the levelbeing considered should be multiplied by the following factorsto compensate for the effects of bending internal column 125edge column 15 corner column 20 The total imposed loadsmay be reduced according to the number of floors supportedThe reductions for 2 3 4 5ndash10 and over 10 floors are 1020 30 40 and 50 respectively

53 RESISTANCE TO SHEAR

Much research by many investigators has been undertaken in aneffort to develop a better understanding of the behaviour ofreinforced concrete subjected to shear As a result of thisresearch various theories have been proposed to explain themechanism of shear transfer in cracked sections and provide asatisfactory basis for designing shear reinforcement In theevent of overloading sudden failure can occur at the onset ofshear cracking in members without shear reinforcement As aconsequence a minimum amount of shear reinforcement in theform of links is required in nearly all beams Resistance to shearcan be increased by adding more shear reinforcement but even-tually the resistance is limited by the capacity of the inclinedstruts that form within the web of the section

531 Members without shear reinforcement

In an uncracked section shear results in a system of mutuallyorthogonal diagonal tension and compression stresses Whenthe diagonal tension stress reaches the tensile strength of the

Section Forces

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concrete a diagonal crack occurs This simple concept rarelyapplies to reinforced concrete since members such as beamsand slabs are generally cracked in flexure In current practiceit is more useful to refer to the nominal shear stress v Vbdwhere b is the breadth of the section in the tension zone Thisstress can then be related to empirical limiting values derivedfrom test data The limiting value vc depends on the concretestrength the effective depth and the reinforcement percentageat the section considered To be effective this reinforcementshould continue beyond the section for a specified minimumdistance as given in Codes of Practice For values of v vc noshear reinforcement is required in slabs but for most beams aspecified minimum amount in the form of links is required

At sections close to supports the shear strength is enhancedand for members carrying generally uniform load the criticalsection may be taken at d from the face of the support Whereconcentrated loads are applied close to supports in memberssuch as corbels and pile-caps some of the load is transmittedby direct strut action This effect is taken into account in theCodes of Practice by either enhancing the shear strength of thesection or reducing the design load In members subjected tobending and axial load the shear strength is increased due tocompression and reduced due to tension

Details of design procedures in Codes of Practice are givenin Table 333 for BS 8110 Table 336 for BS 5400 andTable 417 for EC 2

532 Members with shear reinforcement

The design of members with shear reinforcement is based on atruss model in which the tension and compression chords arespaced apart by a system of inclined concrete struts and uprightor inclined shear reinforcement Most reinforcement is in theform of upright links but bent-up bars may be used for upto 50 of the total shear reinforcement in beams The trussmodel results in a force in the tension chord additional to thatdue to bending This can be taken into account directly in thedesign of the tension reinforcement or indirectly by shiftingthe bending moment curve each side of any point of maximumbending moment

In BS 8110 shear reinforcement is required to cater for thedifference between the shear force and the shear resistance ofthe section without shear reinforcement Equations are givenfor upright links based on concrete struts inclined at about 45oand for bent-up bars where the inclination of the concrete strutsmay be varied between specified limits In BS 5400 a specifiedminimum amount of link reinforcement is required in additionto that needed to cater for the difference between the shear forceand the shear resistance of the section without shear reinforce-ment The forces in the inclined concrete struts are restrictedindirectly by limiting the maximum value of the nominal shearstress to specified values

In EC 2 shear reinforcement is required to cater for the entireshear force and the strength of the inclined concrete struts ischecked explicitly The inclination of the struts may be variedbetween specified limits for links as well as bent-up bars Incases where upright links are combined with bent-up bars thestrut inclination needs to be the same for both

Details of design procedures in Codes of Practice are givenin Table 333 for BS 8110 Table 336 for BS 5400 andTable 418 for EC 2

533 Shear under concentrated loads

Suspended slabs and foundations are often subjected to largeloads or reactions acting on small areas Shear in solid slabsunder concentrated loads can result in punching failures on theinclined faces of truncated cones or pyramids For designpurposes shear stresses are checked on given perimeters atspecified distances from the edges of the loaded area Where aload or reaction is eccentric with regard to a shear perimeter(eg at the edges of a slab and in cases of moment transferbetween a slab and a column) an allowance is made for theeffect of the eccentricity In cases where v exceeds vc linksbent-up bars or other proprietary products may be provided inslabs not less than 200 mm deep

Details of design procedures in Codes of Practice are givenin Table 334 for BS 8110 Tables 337 and 338 for BS 5400and Table 419 for EC 2

54 RESISTANCE TO TORSION

In normal beam-and-slab or framed construction calculationsfor torsion are not usually necessary adequate control of anytorsional cracking in beams being provided by the requiredminimum shear reinforcement When it is judged necessary toinclude torsional stiffness in the analysis of a structure ortorsional resistance is vital for static equilibrium membersshould be designed for the resulting torsional moment Thetorsional resistance of a section may be calculated on the basisof a thin-walled closed section in which equilibrium is satisfiedby a closed plastic shear flow Solid sections may be modelled asequivalent thin-walled sections Complex shapes may be dividedinto a series of sub-sections each of which is modelled as anequivalent thin-walled section and the total torsional resistancetaken as the sum of the resistances of the individual elementsWhen torsion reinforcement is required this should consist ofrectangular closed links together with longitudinal reinforce-ment Such reinforcement is additional to any requirements forshear and bending

Details of design procedures in Codes of Practice are givenin Table 335 for BS 8110 Table 339 for BS 5400 andTable 420 for EC 2

55 DEFLECTION

The deflections of members under service loading should notimpair the appearance or function of a structure An accurateprediction of deflections at different stages of construction mayalso be necessary in bridges for example For buildings thefinal deflection of members below the support level afterallowance for any pre-camber is limited to span250 In orderto minimise any damage to non-structural elements such asfinishes cladding or partitions that part of the deflection thatoccurs after the construction stage is also limited to span500In BS 8110 this limit is taken as 20 mm for spans 10 m

The behaviour of a reinforced concrete beam under serviceloading can be divided into two basic phases before and aftercracking During the uncracked phase the member behaveselastically as a homogeneous material This phase is ended bythe load at which the first flexural crack forms The cracks resultin a gradual reduction in stiffness with increasing load duringthe cracked phase The concrete between the cracks continues

Deflection 49

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to provide some tensile resistance though less on average thanthe tensile strength of the concrete Thus the member is stifferthan the value calculated on the assumption that the concretecarries no tension This additional stiffness known as lsquotensionstiffeningrsquo is highly significant in lightly reinforced memberssuch as slabs but has only a relatively minor effect on thedeflection of heavily reinforced members These concepts areillustrated in the following figure

assumptions made in their derivation provide a useful basisfor estimating long-term deflections of members in buildingsas follows

Details of spaneffective depth ratios and explicit calculationprocedures are given in Tables 340 to 342 for BS 8110 andTables 421 and 422 for EC 2

56 CRACKING

Cracks in members under service loading should not impairthe appearance durability or water-tightness of a structure InBS 8110 for buildings the design crack width is generallylimited to 03 mm In BS 5400 for bridges the limit variesbetween 025 mm and 010 mm depending on the exposureconditions In BS 8007 for structures to retain liquids a limitof 02 mm usually applies Under liquid pressure continuouscracks that extend through the full thickness of a slab or wallare likely to result in some initial seepage but such cracks areexpected to self-heal within a few weeks If the appearance ofa liquid-retaining structure is considered aesthetically critical acrack width limit of 01 mm applies

In EC 2 for most buildings the design crack width is generallylimited to 03 mm but for internal dry surfaces a limitof 04 mm is considered sufficient For liquid-retainingstructures a classification system according to the degree ofprotection required against leakage is introduced Where asmall amount of leakage is acceptable for cracks that passthrough the full thickness of the section the crack width limitvaries according to the hydraulic gradient (ie head of liquiddivided by thickness of section) The limits are 02 mm forhydraulic gradients 5 reducing uniformly to 005 mmfor hydraulic gradients 35

In order to control cracking in the regions where tension isexpected it is necessary to ensure that the tensile capacity ofthe reinforcement at yielding is not less than the tensile force inthe concrete just before cracking Thus a minimum amount ofreinforcement is required according to the strength of thereinforcing steel and the tensile strength of the concrete atthe time when cracks may first be expected to occur Cracks dueto restrained early thermal effects in continuous walls and someslabs may occur within a few days of the concrete being placedIn other members it may be several weeks before the appliedload reaches a level at which cracking occurs

Crack widths are influenced by several factors including thecover bar size bar spacing and stress in the reinforcement Thestress may need to be reduced in order to meet the crack widthlimit Design formulae are given in Codes of Practice in whichstrain calculated on the basis of no tension in the concrete isreduced by a value that decreases with increasing amounts oftension reinforcement For cracks that are caused by appliedloading the same formulae are used in BS 8110 BS 5400 andBS 8007 For cracks that are caused by restraint to temperatureeffects and shrinkage fundamentally different formulae areincluded in BS 8007 Here it is assumed that bond slip occursat each crack and the crack width increases in direct proportionto the contraction of the concrete

Deflection actual spaneffective depth ratio

limiting spaneffective depth ratiospan250

Design of structural members50

Schematic load-deflection response

In BS 8110 for the purpose of analysis lsquotension stiffeningrsquo isrepresented by a triangular stress distribution in the concreteincreasing from zero at the neutral axis to a maximum value atthe tension face At the level of the tension reinforcement theconcrete stress is taken as 1 Nmm2 for short-term loads and055 Nmm2 for long-term loads irrespective of the strain in thetension reinforcement In EC 2 a more general approach isadopted in which the deformation of a section which could bea curvature or in the case of pure tension an extension or acombination of these is calculated first for a homogeneousuncracked section 1 and second for a cracked section ignor-ing tension in the concrete 2 The deformation of the sectionunder the design loading is then obtained as

2 (1 )1

where is a distribution coefficient that takes account of thedegree of cracking according to the nature and duration ofthe loading and the stress in the tension reinforcement underthe load causing first cracking in relation to the stress under thedesign service load

When assessing long-term deflections allowances need to bemade for the effect of concrete creep and shrinkage Creep canbe taken into account by using an effective modulus of elasticityEceff Ec(1 ) where Ec is the short-term value and is acreep coefficient Shrinkage deformations can be calculatedseparately and added to those due to loading

Generally explicit calculation of deflections is unnecessaryto satisfy code requirements and simple rules in the form oflimiting spaneffective depth ratios are provided in BS 8110 andEC 2 These are considered adequate for avoiding deflectionproblems in normal circumstances and subject to the particular

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Generally for design to BS 8110 and EC 2 there is no needto calculate crack widths explicitly and simple rules that limiteither bar size or bar spacing according to the stress in thereinforcement are provided Details of both rules and crackwidth formulae are given in Table 343 for BS 8110 and BS 5400Tables 344 and 345 for BS 8007 and Tables 423ndash425 forEC 2 Additional design aids derived from the crack widthformulae are provided in Tables 346ndash352 for BS 8007 andTables 426 and 427 for EC 2

57 REINFORCEMENT CONSIDERATIONS

Codes of Practice contain many requirements affecting thereinforcement details such as minimum and maximum areasanchorage and lap lengths bends in bars and curtailment Thereinforcement may be curtailed in relation to the bendingmoment diagram provided there is always enough anchorageto develop the necessary design force in each bar at every crosssection Particular requirements apply at the positions wherebars are curtailed and at simple supports

Bars may be set out individually in pairs or in bundles ofthree or four in contact For the safe transmission of bondforces the cover provided to the bars should be not less thanthe bar size or for a group of bars in contact the equivalentdiameter of a notional bar with the same cross-sectional area asthe group Gaps between bars (or groups of bars) should benot less than the greater of (aggregate size plus 5 mm) or thebar size (or equivalent bar diameter for a group) Details ofreinforcement limits and requirements for containing bars incompression are given in Table 353 for BS 8110 Table 359for BS 5400 and Table 428 for EC 2

571 Anchorage lengths

At both sides of any cross section the reinforcement should beprovided with an appropriate embedment length or other formof end anchorage In earlier codes it was also necessary to con-sider lsquolocal bondrsquo at sections where large changes of tensileforce occur over short lengths of reinforcement and thisrequirement remains in BS 5400

Assuming a uniform bond stress between concrete and thesurface of a bar the required anchorage length is given by

lbreq (design force in bar)(bond stress perimeter of bar) fsd (24)fbd () ( fsdfbd)(4)

where fsd is the design stress in the bar at the position fromwhich the anchorage is measured The design bond stress fbd

depends on the strength of the concrete the type of bar and inEC 2 the location of the bar within the concrete section duringconcreting For example the bond condition is classified aslsquogoodrsquo in the bottom 250 mm of any section and in the top300 mm of a section 600 mm deep In other locations thebond condition is classified as lsquopoorrsquo Also in EC 2 the basicanchorage length in tension can be multiplied by severalcoefficients that take account of factors such as the bar shapethe cover and the effect of transverse reinforcement or pressureFor bars of diameter 40 mm and bars grouped in pairs orbundles additional considerations apply Details of design

anchorage lengths in tension and compression are given inTable 355 for BS 8110 Table 359 for BS 5400 and Tables 430and 432 for EC 2

572 Lap lengths

Forces can be transferred between reinforcement by lappingwelding or joining bars with mechanical devices (couplers)Connections should be placed whenever possible away frompositions of high stress and should preferably be staggeredIn Codes of Practice the necessary lap length is obtained bymultiplying the required anchorage length by a coefficient

In BS 8110 for bars in compression the coefficient is 125For bars in tension the coefficient is 10 14 or 20 accordingto the cover the gap between adjacent laps in the same layerand the location of the bar in the section In slabs where thecover is not less than twice the bar size and the gap betweenadjacent laps is not less than six times the bar size or 75 mm afactor of 10 applies Larger factors are frequently necessary incolumns typically 14 and beams typically 14 for bottom barsand 20 for top bars The sum of all the reinforcement sizes ina particular layer should not exceed 40 of the width of thesection at that level When the size of both bars at a lap exceeds20 mm and the cover is less than 15 times the size of thesmaller bar links at a maximum spacing of 200 mm arerequired throughout the lap length

In EC 2 for bars in tension or compression the lap coefficientvaries from 10 to 15 according to the percentage of lapped barsrelative to the total area of bars at the section considered andtransverse reinforcement is required at each end of the lap zoneDetails of lap lengths are given in Table 355 for BS 8110Table 359 for BS 5400 and Tables 431 and 432 for EC 2

573 Bends in bars

The radius of any bend in a reinforcing bar should conform tothe minimum requirements of BS 8666 and should ensure thatfailure of the concrete inside the bend is prevented For barsbent to the minimum radius according to BS 8666 it is notnecessary to check for concrete failure if the anchorage of thebar does not require a length more than 5 beyond the end ofthe bend (see Table 227) It is also not necessary to check forconcrete failure where the plane of the bend is not close to aconcrete face and there is a transverse bar not less than its ownsize inside the bend This applies in particular to a link whichmay be considered fully anchored if it passes round anotherbar not less than its own size through an angle of 90o andcontinues beyond the end of the bend for a length not less than8 in BS 8110 and 10 in EC 2

In cases when a bend occurs at a position where the bar ishighly stressed the bearing stress inside the bend needs to bechecked and the radius of bend will need to be more than theminimum given in BS 8666 This situation occurs typically atmonolithic connections between members for example junc-tion of beam and end column and in short members such ascorbels and pile caps The design bearing stress is limitedaccording to the concrete strength and the confinementperpendicular to the plane of the bend Details of bends in barsare given in Table 355 for BS 8110 Table 359 for BS 5400and Table 431 for EC 2

Reinforcement considerations 51

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574 Curtailment of reinforcement

In flexural members it is generally advisable to stagger thecurtailment points of the tension reinforcement as allowed bythe bending moment envelope Bars to be curtailed need toextend beyond the points where in theory they are no longerneeded for flexural resistance for a number of reasons butmainly to ensure that the shear resistance of the section is notreduced locally Clearly of course no reinforcement shouldbe curtailed at a point less than a full anchorage length from asection where it is required to be fully stressed

In BS 8110 and BS 5400 except at end supports every barshould extend beyond the point at which in theory it is no longerrequired for a distance not less than the greater of the effectivedepth of the member or 12 times the bar size In addition barscurtailed in a tension zone should satisfy at least one of threealternative conditions one requires a full anchorage length onerequires the designer to determine the position where the shearresistance is twice the shear force and the other requires thedesigner to determine the position where the bending resistanceis twice the bending moment The simplest approach is to complywith the first option by providing a full anchorage lengthbeyond the point where in theory the bar is no longer requiredeven if this requires a longer extension than is absolutelynecessary in some cases Details of the requirements are givenin Table 356

In BS 8110 simplified rules are also given for beams andslabs where the loads are mainly uniformly distributed and inthe case of continuous members the spans are approximatelyequal Details of the rules are given in Tables 357 and 358

At simple end supports the tension bars should extend foran effective anchorage length of 12 times the bar size beyondthe centre of the support but no bend should begin before thecentre of the support In cases where the width of the supportexceeds the effective depth of the member the centre ofthe support may be assumed at half the effective depth from theface of the support In BS 8110 for slabs in cases where thedesign shear force is less than half the shear resistance anchor-age can be obtained by extending the bars beyond the centre ofthe support for a distance equal to one third of the supportwidth 30 mm

In EC 2 the extension al of a tension bar beyond the pointwhere in theory it is no longer required for flexural resistance isdirectly related to the shear force at the section For memberswith upright shear links al 05zcot13 where z is the lever armand 13 is the inclination of the concrete struts (see section 3512)Taking z 09d al 045dcot13 where cot13 is selected by thedesigner in the range 10 cot13 25 If the value of cot 13 usedin the shear design calculations is unknown al 1125d can beassumed For members with no shear reinforcement al d isused At simple end supports bottom bars should extend for ananchorage length beyond the face of the support The tensileforce to be anchored is given by F 05Vcot 13 and F 125Vcan be conservatively taken in all cases Details of the curtailmentrequirements are given in Table 432

58 DEEP BEAMS

In designing normal (shallow) beams of the proportions morecommonly used in construction plane sections are assumed toremain plane after loading This assumption is not strictly true

but the errors resulting from it only become significant whenthe depth of the beam becomes equal to or more than abouthalf the span The beam is then classed as a deep beam anddifferent methods of analysis and design need to be usedThese methods take into account not only the overall appliedmoments and shears but also the stress patterns and internaldeformations within the beam

For a single-span deep beam after the concrete in tensionhas cracked the structural behaviour is similar to a tied archThe centre of the compression force in the arch rises from thesupport to a height at the crown equal to about half the span ofthe beam The tension force in the tie is roughly constant alongits length since the bending moment and the lever arm undergosimilar variations along the length of the beam For a continuousdeep beam the structural behaviour is analogous to a separatetied arch system for each span combined with a suspensionsystem centred over each internal support

In BS 8110 for the design of beams of clear span less thantwice the effective depth the designer is referred to specialistliterature In EC 2 a deep beam is classified as a beam whoseeffective span is less than three times its overall depth Briefdetails of suitable methods of design based on the result ofextensive experimental work by various investigators are givenin ref 42 and a comprehensive well-produced design guide iscontained in ref 43

59 WALLS

Information concerning the design of load-bearing walls inaccordance with BS 8110 is given in section 618 Retainingwalls and other similar elements that are subjected mainly totransverse bending where the design vertical load is less than01fcu times the area of the cross section are treated as slabs

510 DETAILS

It has long been realised that the calculated strength of areinforced concrete member cannot be attained if the details ofthe required reinforcement are unsatisfactory Research by theformer Cement and Concrete Association and others has shownthat this applies particularly at joints and intersections Thedetails commonly used in wall-to-base and wall-to-walljunctions in retaining structures and containment vessels wherethe action of the applied load is to lsquoopenrsquo the corner are notalways effective

On Tables 362 and 363 are shown recommended details thathave emerged from the results of reported research The designinformation given in BS 8110 and BS 5400 for nibs corbels andhalving joints is included and supplemented by informationgiven elsewhere In general however details that are primarilyintended for precast concrete construction have not beenincluded as they fall outside the scope of this book

511 ELASTIC ANALYSIS OF CONCRETE SECTIONS

The geometrical properties of various figures the shapes ofwhich conform to the cross sections of common reinforcedconcrete members are given in Table 2101 The data includeexpressions for the area section modulus second moment ofarea and radius of gyration The values that are derived fromthese expressions are applicable in cases when the amount of

Design of structural members52

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reinforcement provided need not be taken into account in theanalysis of the structure (see section 141)

The data given in Tables 2102 and 2103 are applicable toreinforced concrete members with rectilinear and polygonalcross sections when the reinforcement provided is taken intoaccount on the basis of the modular ratio Two conditions areconsidered (1) when the entire section is subjected to stressand (2) when for members subjected to bending the concretein tension is not taken into account The data given for the

former condition are the effective area the centre and secondmoment of area the modulus and radius of gyration For thecondition when a member is subjected to bending and theconcrete in tension is assumed to be ineffective data giveninclude the position of the neutral axis the lever-arm and theresistance moment

Design procedures for sections subjected to bending andaxial force with design charts for rectangular and cylindricalcolumns are given in Tables 2104ndash2109

Elastic analysis of concrete sections 53

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The loads and consequent bending moments and forces onthe principal types of structural components and the designresistances of such components have been dealt with in thepreceding chapters In this chapter some complete structurescomprising assemblies or special cases of such componentsand their foundations are considered

61 BUILDINGS

Buildings may be constructed entirely of reinforced concreteor one or more elements of the roof floors walls stairs andfoundations may be of reinforced concrete in conjunction witha steel frame Alternatively the building may consist of interiorand exterior walls of cast in situ reinforced concrete supportingthe floors and roof with the columns and beams being formedin the thickness of the walls Again the entire structure or partsthereof may be built of precast concrete elements connectedtogether during construction

The design of the various parts of a building is the subjectof Examples of the Design of Buildings That book includesillustrative calculations and drawings for a typical six-storeymultipurpose building This section provides a brief guide tocomponent design

611 Robustness and provision of ties

The progressive collapse of one corner of a London tower blockin 1968 as a result of an explosion caused by a gas leak in adomestic appliance on the eighteenth floor led to recommen-dations to consider such accidental actions in the design of allbuildings Regulations require a building to be designed andconstructed so that in the event of an accident the buildingwill not collapse to an extent disproportionate to the causeBuildings are divided into classes depending on the type andoccupancy including the likelihood of accidents and thenumber of occupants that may be affected with a statementof the design measures to be taken in each of the classes TheBS 8110 normal requirements for lsquorobustnessrsquo automaticallysatisfy the regulations for all buildings except those wherespecific account is to be taken of likely hazards

The layout and form of the structure should be checked toensure that it is inherently stable and robust In some cases itmay be necessary to protect certain elements from vehicularimpact by providing bollards or earth banks All structures

should be able to resist a notional ultimate horizontal forceequal to 15 of the characteristic dead load of the structureThis force effectively replaces the design wind load in caseswhere the exposed surface area of the building is small

Wherever possible continuous horizontal and vertical tiesshould be provided throughout the building to resist specifiedforces The magnitude of the force increases with the numberof storeys for buildings of less than 10 storeys but remainsconstant thereafter The requirements may be met by usingreinforcement that is necessary for normal design purposes inbeams slabs columns and walls Only the tying forces needto be considered and the full characteristic strength of thereinforcement may be taken into account Horizontal ties arerequired in floors and roofs at the periphery and internally intwo perpendicular directions The internal ties which may bespread uniformly over the entire building or concentrated atbeam and column positions are to be properly anchored atthe peripheral tie Vertical ties are required in all columns andload-bearing walls from top to bottom and all external columnsand walls are to be tied into each floor and roof For regulatorypurposes some buildings are exempt from the vertical tyingrequirement Details of the tying requirements are given inTable 354

For in situ construction proper attention to reinforcementdetailing is all that is normally necessary to meet the tyingrequirements Precast forms of construction generally requiremore care and recommended details to obtain continuity ofhorizontal ties are given in the code of practice If ties cannotbe provided other strategies should be adopted as described inPart 2 of the code These strategies are presented in the contextof residential buildings of five or more storeys where eachelement that cannot be tied is to be considered as notionallyremoved one at a time in each storey in turn The requirementis that any resulting collapse should be limited in extent withthe remaining structure being able to bridge the gap causedby the removal of the element If this requirement cannot besatisfied then the element in question is considered as a keyelement In this case the element and its connections need tobe able to resist a design ultimate load of 34 kNm2 consideredto act from any direction BS 8110 is vague with regard to theextent of collapse associated with this approach but a moreclearly defined statement is given in the building regulationsHere a key element is any untied member whose removalwould put at risk of collapse within the storey in question

Chapter 6

Buildings bridges andcontainmentstructures

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and the immediately adjacent storeys more than 15 of thearea of the storey (or 70 m2 if less)

In EC 2 similar principles apply in that structures notspecifically designed to withstand accidental actions should beprovided with a suitable tying system to prevent progressivecollapse by providing alternative load paths after local damageThe UK National Annex specifies compliance with the BS 8110requirements as given in Table 429

612 Floors

Suspended concrete floors can be of monolithic constructionin the form of beam-and-slab (solid or ribbed) or flat slab(solid or waffle) or can consist of precast concrete slab unitssupported on concrete or steel beams or comprise one ofseveral other hybrid forms Examples of monolithic forms ofconstruction are shown in the figure on Table 242

Two-way beam and solid slab systems can involve a layoutof long span secondary beams supported by usually shorterspan main beams The resulting slab panels may be designed astwo-way spanning if the longer side is less than twice theshorter side However such two-way beam systems tend tocomplicate both formwork and reinforcement details with aconsequent delay in the construction programme A one-waybeam and solid slab system is best suited to a rectangular gridof columns with long span beams and shorter span slabs If aribbed slab is used a system of long span slabs supported byshorter span beams is preferable If wide beams are used thebeam can be incorporated within the depth of the ribbed slab

In BS 8110 ribbed slabs include construction in which ribsare cast in situ between rows of blocks that remain part of thecompleted floor This type of construction is no longer used inthe United Kingdom although blocks are incorporated in someprecast and composite construction The formers for ribbedslabs can be of steel glassfibre or polypropylene Standardmoulds are available that provide tapered ribs with a minimumwidth of 125 mm spaced at 600 mm (troughs) and 900 mm(waffles) The ribs are connected by a structural concrete toppingwith a minimum thickness of 50 mm for trough moulds and75 mm for waffle moulds In most structures to obtain thenecessary fire-resistance either the thickness of topping has toexceed these minimum values or a non-structural screed addedat a later stage of construction The spacing of the ribs may beincreased to a maximum of 1500 mm by using purpose-madeformers Comprehensive details of trough and waffle floorsare contained in ref 44

BS 8110 and EC 2 contain recommendations for both solidand ribbed slabs spanning between beams or supported directlyby columns (flat slabs) Ribs in waffle slabs and ribs reinforcedwith a single bar in trough slabs do not require links unlessneeded for shear or fire-resistance Ribs in trough slabs whichare reinforced with more than one bar should be provided withsome links to help maintain the correct cover The spacing ofthese links may be in the range 10ndash15 m according to the sizeof the main bars Structural toppings are normally reinforcedwith a welded steel fabric

Information on the weight of concrete floor slabs is given inTable 21 and details of imposed loads on floors are givenin Table 23 Detailed guidance on the analysis of slabs isgiven in Chapters 4 and 13 More general guidance includinginitial sizing suggestions is given in section 523

613 Openings in floors

Large openings (eg stairwells) should generally be providedwith beams around the opening Holes for pipes ducts andother services should generally be formed when the slabis constructed and the cutting of such holes should not bepermitted afterwards unless done under the supervision of acompetent engineer Small isolated holes may generally beignored structurally with the reinforcement needed for a slabwithout holes simply displaced locally to avoid the hole

In other cases the area of slab around an opening or groupof closely spaced holes needs to be strengthened with extrareinforcement The cross-sectional area of additional bars to beplaced parallel to the principal reinforcement should be at leastequal to the area of principal reinforcement interrupted by theopening Also for openings of dimensions exceeding 500 mmadditional bars should be placed diagonally across the cornersof the opening Openings with dimensions greater than 1000 mmshould be regarded as structurally significant and the area ofslab around the opening designed accordingly

The effect of an opening in the proximity of a concentratedload or supporting column on the shearing resistance of theslab is shown in Table 337

614 Stairs

Structural stairs may be tucked away out of sight within a fireenclosure or they may form a principal architectural feature Inthe former case the stairs can be designed and constructed assimply and cheaply as possible but in the latter case much moretime and trouble is likely to be expended on the design

Several stair types are illustrated on Table 288 Variousprocedures for analysing the more common types of stairhave been developed and some of these are described onTables 288ndash291 These theoretical procedures are based onthe concept of an idealised line structure and when detailingthe reinforcement for the resulting stairs additional bars shouldbe included to limit the formation of cracks at the points ofhigh stress concentration that inevitably occur The lsquothree-dimensionalrsquo nature of the actual structure and the stiffeningeffect of the triangular tread areas both of which are usuallyignored when analysing the structure will result in actual stressdistributions that differ from those calculated and this mustbe remembered when detailing The stair types illustrated onTable 288 and others can also be investigated by finite-elementmethods and similar procedures suitable for computer analysisWith such methods it is often possible to take account of thethree-dimensional nature of the stair

Simple straight flights of stairs can span either transversely(ie across the flight) or longitudinally (ie along the flight)When spanning transversely supports must be provided on bothsides of the flight by either walls or stringer beams In this casethe waist or thinnest part of the stair construction need be nomore than 60 mm thick say the effective lever arm for resistingthe bending moment being about half of the maximumthickness from the nose to the soffit measured at right anglesto the soffit When the stair spans longitudinally deflectionconsiderations can determine the waist thickness

In principle the design requirements for beams and slabsapply also to staircases but designers cannot be expected todetermine the deflections likely to occur in the more complex

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stair types BS 8110 deals only with simple types and allows amodified spaneffective depth ratio to be used The bendingmoments should be calculated from the ultimate load due to thetotal weight of the stairs and imposed load measured on plancombined with the horizontal span Stresses produced bythe longitudinal thrust are small and generally neglected in thedesign of simple systems Unless circumstances otherwisedictate suitable step dimensions for a semi-public stairs are 165mm rise and 275 mm going which with a 25 mm nosing orundercut gives a tread of 300 mm Private stairs may be steeperand those in public buildings should be less steep In eachcase optimum proportions are given by the relationship(2 rise going) 600 mm Different forms of constructionand further details on stair dimensions are given in BS 5395

Finally it should be remembered that the prime purpose of astair is to provide safe pedestrian access between the floors itconnects As such it is of vital importance in the event of a fireand a principal design consideration must be to provide adequatefire-resistance

615 Planar roofs

The design and construction of a flat reinforced concrete roofare essentially the same as for a floor A water-tight coveringsuch as asphalt or bituminous felt is generally necessary andwith a solid slab some form of thermal insulation is normallyrequired For ordinary buildings the slab is generally built leveland a drainage slope of the order of 1 in 120 is formed byadding a mortar topping The topping is laid directly onto theconcrete and below the water-tight covering and can formthe thermal insulation if it is made of a sufficient thickness oflightweight concrete or other material having low thermalconductivity

Planar slabs with a continuous steep slope are not commonin reinforced concrete except for mansard roofs The roofcovering is generally of metal or asbestos-cement sheeting orsome lightweight material Such coverings and roof glazingrequire purlins for their support and although these are often ofsteel precast concrete purlins are also used especially if theroof structure is of reinforced concrete

616 Non-planar roofs

Roofs that are not planar other than the simple pitched roofsconsidered in the foregoing can be constructed as a series ofplanar slabs (prismatic or hipped-plate construction) or assingle- or double-curved shells Single-curved roofs such assegmental or cylindrical shells are classified as developablesurfaces Such surfaces are not as stiff as double-curved roofsor their prismatic counterparts which cannot be lsquoopened uprsquointo plates without some shrinking or stretching taking place

If the curvature of a double-curved shell is similar in alldirections the surface is known as synclastic A typical case isa dome where the curvature is identical in all directions Ifthe shell curves in opposite directions over certain areas thesurface is termed anticlastic (saddle shaped) The hyperbolic-paraboloidal shell is a well-known example and is the specialcase where such a double-curved surface is generated by twosets of straight lines An elementary analysis of some of thesestructural forms is dealt with in section 192 and Table 292but reference should be made to specialist publications for

more comprehensive analyses and more complex structuresSolutions for many particular shell types have been producedand in addition general methods have been developed foranalysing shell forms of any shape by means of a computerShells like all statically indeterminate structures are affectedby such secondary effects as shrinkage temperature change andsettlement and a designer must always bear in mind the factthat the stresses arising from these effects can modify quiteconsiderably those due to normal dead and imposed load InTable 281 simple expressions are given for the forces indomed slabs such as are used for the bottoms and roofs of somecylindrical tanks In a building a domed roof generally hasa much larger rise to span ratio and where the dome is partof a spherical surface and has an approximately uniform thick-ness overall the analysis given in Table 292 applies Shallowsegmental domes and truncated cones are also dealt with inTable 292

Cylindrical shells Segmental or cylindrical roofs are usuallydesigned as shell structures Thin curved slabs that behave asshells are assumed to offer no resistance to bending nor todeform under applied distributed loads Except near edge andend stiffeners the shell is subjected only to membrane forcesnamely a direct force acting longitudinally in the plane of theslab a direct force acting tangentially to the curve of the slaband a shearing force Formulae for these membrane forces aregiven in section 1923 In practice the boundary conditionsdue to either the presence or absence of edge or valley beamsend diaphragms continuity and so on affect the displacementsand forces that would otherwise occur as a result of membraneaction Thus as when analysing any indeterminate structure(such as a continuous beam system) the effects due to theseboundary restraints need to be combined with the staticallydeterminate stresses arising from the membrane action

Shell roofs can be arbitrarily subdivided into lsquoshortrsquo (wherethe ratio of length l to radius r is less than about 05) lsquolongrsquo(where l r exceeds 25) and lsquointermediatersquo For short shellsthe influence of the edge forces is slight in comparison withmembrane action and the stresses can be reasonably taken asthose due to the latter only If the shell is long the membraneaction is relatively insignificant and an approximate solutioncan be obtained by considering the shell to act as a beam withcurved flanges as described in section 1923

For the initial analysis of intermediate shells no equivalentshort-cut method has yet been devised The standard method ofsolution is described in various textbooks (eg refs 45 and 46)Such methods involve the solution of eight simultaneousequations if the shell or the loading is unsymmetrical or four ifsymmetry is present by matrix inversion or other means Bymaking certain simplifying assumptions and providing tables ofcoefficients Tottenham (ref 47) developed a popular simplifieddesign method which is rapid and requires the solution of threesimultaneous equations only J D Bennett also developed amethod of designing long and intermediate shells based on ananalysis of actual designs of more than 250 roofs The methodwhich involves the use of simple formulae incorporatingempirical coefficients is summarised on Tables 293 and 294For further details see ref 48

Buckling of shells A major concern in the design of anyshell is the possibility of buckling since the loads at which

Buildings bridges and containment structures56

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buckling occurs as established by tests often differ from thevalues predicted by theory Ref 49 indicates that for domessubtending angles of about 90o the critical external pressure atwhich buckling occurs according to both theory and tests isgiven by p 03E(hr)2 where E is the elastic modulus ofconcrete and h is the thickness and r the radius of the domeFor a shallow dome with spanrise cong 10 p 015E(hr)2 Afactor of safety against buckling of 2 to 3 should be adoptedSynclastic shells having a radius ranging from r1 to r2 may beconsidered as an equivalent dome with a radius of r radic(r1 r2)

For a cylindrical shell buckling is unlikely if the shell isshort In the case of long shells p 06E(hr)2

Anticlastic surfaces are more rigid than single-curved shellsand the buckling pressure for a saddle-shaped shell supportedon edge stiffeners safely exceeds that of a cylinder having acurvature equal to that of the anticlastic shell at the stiffenerFor a hyperbolic-paraboloidal shell with straight boundariesthe buckling load obtained from tests is slightly more than thevalue given by n E(ch)22ab where a and b are the lengthsof the sides of the shell c is the rise and h the thickness this isonly half of the value predicted theoretically

617 Curved beams

When bow girders and beams that are not rectilinear in planare subjected to vertical loading torsional moments occur inaddition to the normal bending moments and shearing forcesBeams forming a circular arc in plan may comprise part of acomplete circular system with equally spaced supports andequal loads on each span such systems occur in silos towersand similar cylindrical structures Equivalent conditions canalso occur in beams where the circle is incomplete provided theappropriate negative bending and torsional moments can bedeveloped at the end supports This type of circular beam canoccur in structures such as balconies

On Tables 295ndash297 charts are given that enable a rapidevaluation of the bending moments torsional moments andshear forces occurring in curved beams due to uniform andconcentrated loads The formulae on which the charts arebased are given in section 193 and on the tables concernedThe expressions have been developed from those in ref 50 foruniform loads and ref 51 for concentrated loads In both casesthe results have been recalculated to take into account values ofG 04E and C J2

618 Load-bearing walls

In building codes for design purposes a wall is defined as avertical load-bearing member whose length on plan exceedsfour times its thickness Otherwise the member is treated as acolumn in which case the effects of slenderness in relation toboth major and minor axes of bending need to be considered(section 524) A reinforced wall is one in which not lessthan the recommended minimum amount of reinforcement isprovided and taken into account in the design Otherwise themember is treated as a plain concrete wall in which case thereinforcement is ignored for design purposes

A single planar wall in general can be subjected to verticaland horizontal in-plane forces acting together with in-planeand transverse moments The in-plane forces and moment canbe combined to obtain at any particular level a longitudinal

shear force and a linear distribution of vertical force If thein-plane eccentricity of the vertical force exceeds one-sixth ofthe length of the wall reinforcement can be provided to resistthe tension that develops at one end of the wall In a plain wallsince the tensile strength of the concrete is ignored the distrib-ution of vertical load is similar to that for the bearing pressuredue to an eccentric load on a footing Flanged walls and coreshapes can be treated in a similar way to obtain the resultingdistribution of vertical force Any unit length of the wall cannow be designed as a column subjected to vertical loadcombined with bending about the minor axis due to anytransverse moment

In BS 8110 the effective height of a wall in relation to itsthickness depends upon the effect of any lateral supports andwhether the wall is braced or unbraced A braced wall is onethat is supported laterally by floors andor other walls able totransmit lateral forces from the wall to the principal structuralbracing or to the foundations The principal structural bracingcomprise strong points shear walls or other suitable elementsgiving lateral stability to a structure as a whole An unbracedwall provides its own lateral stability and the overall stabilityof multi-storey buildings should not in any direction dependon such walls alone The slenderness ratio of a wall is definedas the effective height divided by the thickness and the wall isconsidered lsquostockyrsquo if the slenderness ratio does not exceed15 for a braced wall or 10 for an unbraced wall Otherwise awall is considered slender in which case it must be designed foran additional transverse moment

The design of plain concrete walls in BS 8110 is similar tothat of unreinforced masonry walls in BS 5328 Equations aregiven for the maximum design ultimate axial load taking intoaccount the transverse eccentricity of the load including anadditional eccentricity in the case of slender walls The basicrequirements for the design of reinforced and plain concretewalls are summarised in Table 360

62 BRIDGES

As stated in section 248 the analysis and design of bridges isnow so complex that it cannot be adequately covered in a bookof this type and reference should be made to specialist publi-cations However for the guidance of designers who may haveto deal with structures having features in common with bridgesbrief notes on some aspects of their design and constructionare provided Most of the following information is taken fromref 52 which also contains other references for further reading

621 Types of bridges

For short spans the simplest and most cost-effective form ofdeck construction is a cast in situ reinforced concrete solid slabSingle span slabs are often connected monolithically to theabutments to form a portal frame A precast box-shaped rein-forced concrete culvert can be used as a simple form of framedbridge and is particularly economical for short span (up toabout 6 m) bridges that have to be built on relatively poorground obviating the need for piled foundations

As the span increases the high self-weight of a solid slabbecomes a major disadvantage The weight can be reduced byproviding voids within the slab using polystyrene formersThese are usually of circular section enabling the concrete to

Bridges 57

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flow freely under them to the deck soffit Reinforced concretevoided slabs are economical for spans up to about 25 m Theintroduction of prestressing enables such construction to beeconomical over longer spans and prestressed voided slabswith internal bonded tendons can be used for spans up toabout 50 m If a bridge location does not suit cast in situ slabconstruction precast concrete beams can be used Severaldifferent types of high quality factory-made components thatcan be rapidly erected on site are manufactured Precast beamconstruction is particularly useful for bridging over live roadsrailways and waterways where any interruptions to trafficmust be minimised Pre-tensioned inverted T-beams placedside-by-side and then infilled with concrete provide a viablealternative to a reinforced concrete solid slab for spans upto about 18 m Composite forms of construction consistingtypically of a 200 mm thick cast in situ slab supported onpre-tensioned beams spaced at about 15 m centres can be usedfor spans in the range 12ndash40 m

For very long spans prestressed concrete box girders are theusual form for bridge decks ndash the details of the design beingdictated by the method of construction The span-by-spanmethod is used in multi-span viaducts with individual spans ofup to 60 m A span plus a cantilever of about one quarter thenext span is first constructed This is then prestressed and thefalsework moved forward after which a full span length isformed and stressed back to the previous cantilever In situ con-struction is used for smaller spans but as spans increase so alsodoes the cost of the falsework To minimise the cost the weightof the concrete to be supported at any one time is reduced bydividing each span into a series of transverse segments Thesesegments which can be cast in situ or precast are normallyerected on either side of each pier to form balanced cantileversand then stressed together Further segments are then addedextending the cantilevers to mid-span where an in situ concreteclosure is formed to make the spans continuous During erectionthe leading segments are supported from gantries erected on thepiers or completed parts of the deck and work can advancesimultaneously on several fronts When the segments are precasteach unit is match-cast against the previous one and thenseparated for transportation and erection Finally an epoxyresin is applied to the matching faces before the units arestressed together

Straight or curved bridges of single radius and of constantcross section can also be built in short lengths from one orboth ends The bridge is then pushed out in stages from theabutments a system known as incremental launching Archbridges in spans up to 250 m and beyond can be constructedeither in situ or using precast segments which are prestressedtogether and held on stays until the whole arch is complete

For spans in excess of 250 m the decks of suspensionand cable-stayed bridges can be of in situ concrete ndash constructedusing travelling formwork ndash or of precast segments stressedtogether For a comprehensive treatment of the aestheticsand design of bridges by one of the worldrsquos most eminentbridge engineers see ref 53 Brief information on typicalstructural forms and span ranges is given in Table 298

622 Substructures

A bridge is supported at the ends on abutments and may haveintermediate piers where the positions of the supports and the

lengths of the spans are determined by the topography of theground and the need to ensure unimpeded traffic under thebridge The overall appearance of the bridge structure is verydependent on the relative proportions of the deck and itssupports The abutments are usually constructed of reinforcedconcrete but in some circumstances mass concrete withoutreinforcement can provide a simple and durable solution

Contiguous bored piles or diaphragm walling can be used toform an abutment wall in cases where the wall is to be formedbefore the main excavation is carried out Although the cost ofthis type of construction is high it can be offset against savingsin the amount of land required the cost of temporary works andconstruction time A facing of in situ or precast concrete orblockwork will normally be required after excavation Reinforcedearth construction can be used where there is an embankmentbehind the abutment in which case a precast facing is oftenapplied The selection of appropriate ties and fittings is partic-ularly important since replacement of the ties during the life ofthe structure is very difficult

Where a bridge is constructed over a cutting it is usuallypossible to form a bank-seat abutment on firm undisturbedground Alternatively bank seats can be constructed on piledfoundations However where bridges over motorways aredesigned to allow for future widening of the carriageway theabutment is likely to be taken down to full depth so that it canbe exposed at a later date when the widening is carried out

The design of wing walls is determined by the topography ofthe site and can have a major effect on the appearance of thebridge Wing walls are often taken back at an angle from theface of the abutment for both economy and appearance Castin situ concrete is normally used but precast concrete retainingwall units are also available from manufacturers Concrete cribwalling is also used and its appearance makes it particularlysuitable for rural situations Filling material must be carefullyselected to ensure that it does not flow out and the fill mustbe properly drained It is important to limit the differentialsettlement that could occur between an abutment and its wingwalls The problem can be avoided if the wing walls cantileverfrom the abutment and the whole structure is supported onone foundation

The simplest and most economic form of pier is a verticalmember or group of members of uniform cross section Thismight be square rectangular circular or elliptical Shaping ofpiers can be aesthetically beneficial but complex shapes willsignificantly increase the cost unless considerable reuse of theforms is possible Raking piers and abutments can help toreduce spans for high bridges but they also require expensivepropping and support structures This in turn complicates theconstruction process and considerably increases costs

The choice of foundation to abutments and piers is usuallybetween spread footings and piling Where ground conditionspermit a spread footing will provide a simple and economicsolution Piling will be needed where the ground conditionsare poor and cannot be improved the bridge is over a river orestuary the water table is high or site restrictions prevent theconstruction of a spread footing It is sometimes possible toimprove the ground by consolidating grouting or applying asurcharge by constructing the embankments well in advance ofthe bridge structure Differential settlement of foundations willbe affected by the construction sequence and needs to becontrolled In the early stages of construction the abutments

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are likely to settle more than the piers but the piers will settlelater when the deck is constructed

623 Integral bridges

For road bridges in the United Kingdom experience has shownthat with all forms of construction continuous structures aregenerally more durable than structures with discontinuous spansThis is mainly because joints between spans have often allowedsalty water to leak through to piers and abutments HighwaysAgency standard BD 5701 says that in principle all bridgesshould be designed as continuous over intermediate supportsunless special circumstances exist The connections betweenspans may be made to provide full structural continuity or inbeam and slab construction continuity of the deck slab only

Bridges with lengths up to 60 m and skews up to 30o shouldalso be designed as integral bridges in which the abutmentsare connected directly to the deck and no movement joints areprovided to allow for expansion or contraction When the designerconsiders that an integral bridge is inappropriate the agreementof the overseeing organisation must be obtained HighwaysAgency document BA 5701 has figures indicating a variety ofcontinuity and abutment details

624 Design considerations

Whether the bridge is carrying a road railway waterway or justpedestrians it will be subject to various types of load

Self-weight and loads from surfacing parapets and so on

Environmental (eg wind snow temperature effects)

Traffic

Accidental loads (eg impact)

Temporary loads (during construction and maintenance)

Bridges in the United Kingdom are generally designed to therequirements of BS 5400 and several related Highways Agencystandards Details of the traffic loads to be considered forroad railway and footbridges are given in section 248 andTables 25 and 26 Details of structural design requirementsincluding the load combinations to be considered are given insection 212 and Tables 32 and 33

The application of traffic load to any one area of a bridgedeck causes the deck to bend transversely and twist therebyspreading load to either side The assessment of how much ofthe load is shared in this way and the extent to which it isspread across the deck depends on the bending torsion andshear stiffness of the deck in the longitudinal and transversedirections Computer methods are generally used to analysethe structure for load effects the most versatile method beinggrillage analysis which treats the deck as a two-dimensionalseries of beam elements in both directions This method canbe used for solid slab beam and slab and voided slabs wherethe cross-sectional area of the voids does not exceed 60 ofthe area of the deck Box girders are now generally formed asone or two cells without any transverse diaphragms These areusually quite stiff in torsion but can distort under load givingrise to warping stresses in the walls and slabs of the box It isthen necessary to use three-dimensional analytical methodssuch as 3D space frame folded plate (for decks of uniformcross section) or the generalised 3D finite element method

An excellent treatment of the behaviour and analysis of bridgedecks is provided in ref 54

It is usual to assume that movement of abutments and wingwalls will occur and to take these into account in the designof the deck and the substructure Normally the backfill used isa free-draining material and satisfactory drainage facilities areprovided If these conditions do not apply then higher designpressures must be considered Due allowance must be madealso for the compaction of the fill during construction and thesubsequent effects of traffic loading The Highways Agencydocument BA 4296 shows several forms of integral abutmentwith guidance on their behaviour Abutments to frame bridges areconsidered to rock bodily under the effect of deck movementsEmbedded abutments such as piled and diaphragm wallsare considered to flex and pad foundations to bank seats areconsidered to slide Notional earth pressure distributionsresulting from deck expansion are also given for frame andembedded abutments

Creep shrinkage and temperature movements in bridgedecks can all affect the forces applied to the abutments Piersand to a lesser extent abutments are vulnerable to impact loadsfrom vehicles or shipping and must be designed to resistimpact or be protected from it Substructures of bridges overrivers and estuaries are also subjected to scouring and lateralforces due to water flow unless properly protected

625 Waterproofing of bridge decks

Over the years mastic asphalt has been extensively used forwaterproofing bridge decks but good weather conditions arerequired if it is to be laid satisfactorily Preformed bituminoussheeting is less sensitive to laying conditions but moisturetrapped below the sheeting can cause subsequent lifting Theuse of hot-bonded heavy-duty reinforced sheet membranes ifproperly laid can provide a completely water-tight layer Thesheets which are 3ndash4 mm in thickness have good punctureresistance and it is not necessary to protect the membrane fromasphalt laid on top Sprayed acrylic and polyurethane water-proofing membranes are also used These bond well to theconcrete deck surface with little or no risk of blowing or liftingA tack coat must be applied over the membrane and a protec-tive asphalt layer is placed before the final surfacing is carriedout Some bridges have depended upon the use of a dense highquality concrete to resist the penetration of water without anapplied waterproofing layer In such cases it can be advanta-geous to include silica fume or some similar very fine powderedaddition in the concrete

63 CONTAINMENT STRUCTURES

Weights of stored materials are given in EC 1 Part 11 and thecalculation of horizontal pressures due to liquids and granularmaterials contained in tanks reservoirs bunkers and silosis explained in sections 92 and 93 in conjunction withTables 215 and 216 This section deals with the design ofcontainment structures and the calculation of the forces andbending moments produced by the pressure of the containedmaterials Where containers are required to be watertight thestructural design should follow the recommendations givenin either BS 8007 or EC 2 Part 3 as indicated in sections 213and 294 respectively In the following notes containers are

Containment structures 59

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conveniently classified as either tanks containing liquids orbunkers and silos containing dry materials

631 Underground tanks

Underground storage tanks are subjected to external pressuresdue to the surrounding earth in addition to internal waterpressure The empty structure should also be investigated forpossible flotation if the earth can become waterlogged Earthpressure at-rest conditions should generally be assumed fordesign purposes but for reservoirs where the earth is banked upagainst the walls it would be more reasonable to assume activeconditions Storage tanks are normally filled to check for water-tightness before any backfill material is placed and there isalways a risk that such material could be excavated in the futureTherefore no reduction to the internal hydrostatic pressure byreason of the external earth pressure should be made when atank is full

The earth covering on the roof of a reservoir in its final stateacts uniformly over the entire area but it is usually sensible totreat it as an imposed load This is to cater for non-uniformconditions that can occur when the earth is being placed inposition and if it becomes necessary to remove the earth formaintenance purposes Problems can arise in partially buriedreservoirs due to solar radiation causing thermal expansion ofthe roof The effect of such movement on a perimeter wall willbe minimised if no connection is made between the roof and thewall until reflective gravel or some other protective materialhas been placed on the roof Alternatively restraint to thedeflection of the wall can be minimised by providing a durablecompressible material between the wall and the soil Thisprevents the build-up of large passive earth pressures in theupper portion of the soil and allows the wall to deflect as a longflexible cantilever

632 Cylindrical tanks

The wall of a cylindrical tank is primarily designed to resist ringtensions due to the horizontal pressures of the contained liquidIf the wall is free at the top and free-to-slide at the bottom thenwhen the tank is full the ring tension at depth z is given bynzr where is the unit weight of liquid and r is the internalradius of the tank In this condition when the tank is full novertical bending or radial shear exists

If the wall is connected to the floor in such a way that noradial movement occurs at the base the ring tension will be zeroat the bottom of the wall The ring tensions are affectedthroughout the lower part of the wall and significant verticalbending and radial shear occurs Elastic analysis can be usedto derive equations involving trigonometric and hyperbolicfunctions and solutions expressed in the form of tables areincluded in publications (eg refs 55 and 56) Coefficients todetermine values of circumferential tensions vertical bendingmoments and radial shears for particular values of the termheight2(2 mean radius thickness) are given in Tables 275and 276

The tables apply to idealised boundary conditions in whichthe bottom of the wall is either hinged or fixed It is possible todevelop these conditions if an annular footing is provided at thebottom of the wall The footing should be tied into the floor ofthe tank to prevent radial movement If the footing is narrow

there will be little resistance to rotation and a hinged conditioncould be reasonably assumed It is also possible to form a hingeby providing horizontal grooves at each side of the wall so thatthe contact between the wall and the footing is reduced to anarrow throat The vertical bars are then bent to cross over atthe centre of the wall but this detail is rarely used At the otherextreme if the wall footing is made wide enough it is possibleto get a uniform distribution of bearing pressure In this casethere will be no rotation and a fixed condition can be assumedIn many cases the wall and the floor slab are made continuousand it is necessary to consider the interaction between the twoelements Appropriate values for the stiffness of the memberand the effect of edge loading can be obtained from Tables 276and 277

For slabs on an elastic foundation the values depend on theratio rrk where rk is the radius of relative stiffness defined insection 725 The value of rk is dependent on the modulus ofsubgrade reaction for which data is given in section 724Taking rrk 0 which corresponds to a lsquoplasticrsquo soil state isappropriate for an empty tank liable to flotation

633 Octagonal tanks

If the wall of a tank forms in plan a series of straight sidesinstead of being circular the formwork may be less costly butextra reinforcement and possibly an increased thickness ofconcrete is needed to resist the horizontal bending momentsthat are produced in addition to the ring tension If the tankforms a regular octagon the bending moments in each side areq l212 at the corners and q l 224 at the centre where l is thelength of the side and q is the lsquoeffectiversquo lateral pressure at depthz If the wall is free at the top and free-to-slide at the bottomq z In other cases q nr where n is the ring tension atdepth z and r is the lsquoeffectiversquo radius (ie half the distancebetween opposite sides) If the tank does not form a regularoctagon but the length and thickness of the sides are alternatelyl1 h1 and l2 h2 the horizontal bending moment at the junctionof any two sides is

634 Rectangular tanks

The walls of large rectangular reservoirs are sometimes built indiscontinuous lengths in order to minimise restraints to theeffects of early thermal contraction and shrinkage If the wallbase is discontinuous with the main floor slab each wall unit isdesigned to be independently stable and no slip membrane isprovided between the wall base and the blinding concreteAlternatively the base to each wall unit can be tied into theadjacent panel of floor slab Roof slabs can be connected to theperimeter walls or simply supported with a sliding jointbetween the top of the wall and the underside of the slab Insuch forms of construction except for the effect of any cornerjunctions the walls span vertically either as a cantilever orwith ends that are simply supported or restrained depending onthe particular details

A cantilever wall is statically determinate and if supportinga roof is also isolated from the effect of roof movement Thedeflection at the top of the wall is an important consideration

q12l1

3 l23h1

h23 l1 l2h1

h23

Buildings bridges and containment structures60

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and the base needs to be carefully proportioned in order tominimise the effect of base tilting The problem of excessivedeflection can be overcome and the wall thickness reduced ifthe wall is tied into the roof If the wall is also provided with anarrow footing tied into the floor it can be designed as simplysupported although considerable reliance is being put in theability of the joint to accept continual rotation If the wallfooting is made wide enough it is possible to obtain a uniformdistribution of bearing pressure in which case there will be norotation and a fixed condition can be assumed In cases wherethe wall and floor slab are made continuous the interactionbetween the two elements should be considered

Smaller rectangular tanks are generally constructed withoutmovement joints so that structural continuity is obtained inboth horizontal and vertical planes Bending moments andshear forces in individual rectangular panels with idealisededge conditions when subjected to hydrostatic loading aregiven in Table 253 For a rectangular tank distribution of theunequal fixity moments obtained at the wall junctions isneeded and moment coefficients for tanks of different spanratios are given in Tables 278 and 279 The shearing forcesgiven in Table 253 for individual panels may still be used

The tables give values for tanks where the top of the wall iseither hinged or free and the bottom is either hinged or fixedThe edge conditions are generally uncertain and tend to varywith the loading conditions as discussed in section 172 Forthe horizontal spans the shear forces at the vertical edges ofone wall result in axial forces in the adjacent walls Thus forinternal loading the shear force at the end of a long wall isequal to the tensile force in the short wall and vice versa Indesigning sections the combined effects of bending momentaxial force and shear force need to be considered

635 Elevated tanks

The type of bottom provided to an elevated cylindrical tankdepends on the diameter of the tank and the depth of water Forsmall tanks a flat beamless slab is satisfactory but beams arenecessary for tanks exceeding about 3 m diameter Someappropriate examples which include bottoms with beams anddomed bottoms are included in section 174 and Table 281

It is important that there should be no unequal settlement ofthe foundations of columns supporting an elevated tank and araft should be provided in cases where such problems couldoccur In addition to the bending moments and shear forces dueto the wind pressure on the tank as described in sections 25and 83 the wind force causes a thrust on the columns on theleeward side and tension in the columns on the windward sideThe values of the thrusts and tensions can be calculated fromthe expressions given for columns supporting elevated tanks insection 1742

636 Effects of temperature

If the walls of a tank are subjected to significant temperatureeffects due to solar radiation or the storage of warm liquids theresulting moments and forces need to be determined by anappropriate analysis The structure can usually be analysedseparately for temperature change (expansion or contraction)and temperature differential (gradient through section) For awall with all of the edges notionally clamped the temperature

differential results in bending moments causing compressionon the warm face and tension on the cold face given by

where E is the modulus of elasticity of concrete I is secondmoment of area of the section h is thickness of wall is thecoefficient of thermal expansion of concrete 13 is temperaturedifference between the two surfaces is Poissonrsquos ratio Forcracked sections may be taken as zero but the value of I shouldallow for the tension stiffening effect of the concrete The effectof releasing the notional restraints at edges that are free orhinged modifies the moment field and in cylindrical tankscauses additional ring tensions For further information onthermal effects in cylindrical tanks reference can be made toeither the Australian or the New Zealand standard Code ofPractice for liquid-retaining concrete structures

64 SILOS

Silos which may also be referred to as bunkers or bins aredeep containers used to store particulate materials In a deepcontainer the linear increase of pressure with depth found inshallow containers is modified Allowances are made for theeffects of filling and unloading as described in section 277The properties of materials commonly stored in silos andexpressions for the pressures set up in silos of different formsand proportions are given in Tables 215 and 216

641 Walls

Silo walls are designed to resist the bending moments andtensions caused by the pressure of the contained material If thewall spans horizontally it is designed for the combined effectsIf the wall spans vertically horizontal reinforcement is neededto resist the axial tension and vertical reinforcement to resist thebending In this case the effect of the horizontal bendingmoments due to continuity at the corners should also beconsidered For walls spanning horizontally the bendingmoments and forces depend on the number and arrangement ofthe compartments Where there are several compartmentsthe intermediate walls act as ties between the outer walls Forvarious arrangements of intermediate walls expressions forthe negative bending moments on the outer walls of the silosare given in Table 280 Corresponding expressions for thereactions which are a measure of the axial tensions in thewalls are also given The positive bending moments can bereadily calculated when the negative bending moments atthe wall corners are known An external wall is subjected tothe maximum combined effects when the adjacent compartmentis full An internal cross-wall is subjected to the maximumbending moments when the compartment on one side of thewall is full and to maximum axial tension (but zero bending)when the compartments on both sides are full In small silosthe proportions of the wall panels may be such that they spanboth horizontally and vertically in which case Table 253 canbe used to calculate the bending moments

In the case of an elevated silo the whole load is generallytransferred to the columns by the walls and when the clear spanis greater than twice the depth the wall can be designed as ashallow beam Otherwise the recommendations for deep beamsshould be followed (see section 58 and ref 43) The effect of

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wind loads on large structures should be calculated The effectof both the tensile force in the windward walls of the empty siloand the compressive force in the leeward walls of the full siloare important In the latter condition the effect of the eccentricforce on the inside face of the wall due to the proportion of theweight of the contents supported by friction must be combinedwith the force due to the wind At the base and the top ofthe wall there are additional bending effects due to continuityof the wall with the bottom and the covers or roof over thecompartments

642 Hopper bottoms

The design of sloping hopper bottoms in the form of invertedtruncated pyramids consists of finding for each sloping sidethe centre of pressure the intensity of pressure normal to theslope at this point and the mean span The bending momentsat the centre and edge of each sloping side are calculated Thehorizontal tensile force is computed and combined with thebending moment to determine the horizontal reinforcementrequired The tensile force acting along the slope at the centreof pressure is combined with the bending moment at this pointto find the inclined reinforcement needed in the bottom of theslab At the top of the slope the bending moment and theinclined component of the hanging-up force are combined todetermine the reinforcement needed in the top of the slab

For each sloping side the centre of pressure and the mean spancan be obtained by inscribing on a normal plan a circle thattouches three of the sides The diameter of this circle is the meanspan and its centre is the centre of pressure The total intensityof load normal to the slope at this point is the sum of the normalcomponents of the vertical and horizontal pressures and the deadweight of the slab Expressions for determining the pressures onthe slab are given in Table 216 Expressions for determiningthe bending moments and tensile forces acting along the slopeand horizontally are given in Table 281 When using thismethod it should be noted that although the horizontal span ofthe slab reduces considerably towards the outlet the amount

of reinforcement should not be reduced below that calculatedfor the centre of pressure This is because in determining thebending moment based on the mean span adequate transversesupport from reinforcement towards the base is assumed

The hanging-up force along the slope has both vertical andhorizontal components the former being resisted by the wallsacting as beams The horizontal component acting inwardstends to produce horizontal bending moments on the beam atthe top of the slope but this is opposed by a correspondingoutward force due to the pressure of the contained material Thelsquohip-beamrsquo at the top of the slope needs to be designed both toresist the inward pull from the hopper bottom when the hopperis full and the silo above is only partly filled and also for thecase when the arching of the fill concentrates the outward forcesdue to the peak lateral pressure on the beam during unloadingThis is especially important in the case of mass-flow silos(see section 277)

65 BEARINGS HINGES AND JOINTS

In the construction of frames and arches hinges are needed atpoints where it is assumed that there is no bending moment Inbridges bearings are often required at abutments and piers totransfer loads from the deck to the supports Various types ofbearings and hinges for different purposes are illustrated inTable 299 with associated notes in section 1941

Movement joints are often required in concrete structures toallow free expansion and contraction Fluctuating movementsoccur due to diurnal solar effects and seasonal changes ofhumidity and temperature Progressive movements occur due toconcrete creep drying shrinkage and ground settlementMovement joints may also be provided in structures wherebecause of abrupt changes of loading or ground conditionspronounced changes occur in the size or type of foundationVarious types of joints for different purposes are illustrated inTable 2100 with associated notes on their construction andapplication in section 1942

Buildings bridges and containment structures62

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71 FOUNDATIONS

The design of the foundations for a structure comprises threestages The first is to determine from an inspection of the sitetogether with field data on soil profiles and laboratory testing ofsoil samples the nature of the ground The second stage is toselect the stratum on which to impose the load the bearingcapacity and the type of foundation These decisions dependnot only on the nature of the ground but also on the type ofstructure and different solutions may need to be consideredReference should be made to BS 8004 Code of Practice forfoundations The third stage is to design the foundation totransfer and distribute load from the structure to the ground

711 Site inspection

The objective of a site inspection is to determine the nature ofthe top stratum and the underlying strata in order to detect anyweak strata that may impair the bearing capacity of the stratumselected for the foundation Generally the depth to whichknowledge of the strata is obtained should be not less than oneand a half times the width of an isolated foundation or thewidth of a structure with closely spaced footings

The nature of the ground can be determined by digging trialholes by sinking bores or by driving piles A trial hole can betaken down to only moderate depths but the undisturbed soilcan be examined and the difficulties of excavation with theneed or otherwise of timbering and groundwater pumping canbe determined Bores can be taken very much deeper than trialholes and stratum samples at different depths obtained forlaboratory testing A test pile does not indicate the type ofsoil it has been driven through but it is useful in showing thethickness of the top crust and the depth below poorer soil atwhich a firm stratum is found A sufficient number of any ofthese tests should be taken to enable the engineer to ascertainthe nature of the ground under all parts of the foundationsReference should be made to BS 5930 Code of practice for siteinvestigations and BS 1377 Methods of test for soils for civilengineering purposes

712 Bearing pressures

The pressure that can be safely imposed on a thick stratum ofsoil commonly encountered is in some districts stipulated in

local by-laws The pressures recommended for preliminarydesign purposes in BS 8004 are given in Table 282 but thesevalues should be used with caution since several factors cannecessitate the use of lower values Allowable pressures maygenerally be exceeded by the weight of soil excavated down tothe foundation level but if this increase is allowed any fillmaterial applied on top of the foundation must be included inthe total load If the resistance of the soil is uncertain a studyof local records for existing buildings on the same soil can beuseful as may the results of a ground-bearing test

Failure of a foundation can occur due to consolidation of theground causing settlement or rupture of the ground due toshearing The shape of the surface along which shear failureoccurs under a strip footing is an almost circular arc startingfrom one edge of the footing passing under the footing andthen continuing as a tangent to the arc to intersect the groundsurface at an angle depending on the angle of internal frictionof the soil Thus the average shear resistance depends onthe angle of internal resistance of the soil and on the depthof the footing below the ground surface In a cohesionless soilthe bearing resistance not only increases as the depth increasesbut is proportional to the width of the footing In a cohesive soilthe bearing resistance also increases with the width of footingbut the increase is less than for a non-cohesive soil

Except when bearing directly on rock foundations for all butsingle-storey buildings or other light structures should betaken down at least 1 m below the ground surface in order toobtain undisturbed soil that is sufficiently consolidated In claysoils a depth of at least 15 m is needed in the United Kingdomto ensure protection of the bearing stratum from weathering

713 Eccentric loads

When a rigid foundation is subjected to concentric loadingthat is when the centre of gravity of the loads coincides withthe centre of area of the foundation the bearing pressure on theground is uniform and equal to the total applied load divided bythe total area When a load is eccentrically placed on a base ora concentric load and a bending moment are applied to a basethe bearing pressure is not uniform For a load that is eccentricabout one axis of a rectangular base the bearing pressure variesfrom a maximum at the side nearer the centre of gravity of theload to a minimum at the opposite side or to zero at some inter-mediate position The pressure variation is usually assumed to

Chapter 7

Foundations groundslabs retaining wallsculverts and subways

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Foundations ground slabs retaining walls culverts and subways64

be linear in which case the maximum and minimum pressuresare given by the formulae in Table 282 For large eccentricitiesthere may be a part of the foundation where there is no bearingpressure Although this state may be satisfactory for transientconditions (such as those due to wind) it is preferable for thefoundation to be designed so that contact with the ground existsover the whole area under normal service conditions

714 Blinding layer

For reinforced concrete footings or other construction wherethere is no underlying mass concrete forming an integral part ofthe foundation the bottom of the excavation should be coveredwith a layer of lean concrete to protect the soil and provide aclean surface on which to place the reinforcement The thicknessof this blinding layer is typically 50ndash75 mm depending on thesurface condition of the excavation

715 Foundation types

The most suitable type of foundation depends primarily on thedepth at which the bearing stratum lies and the allowable bearingpressure which determines the foundation area Data relatingto some common types of separate and combined pad founda-tions suitable for sites where the bearing stratum is found closeto the surface are given in Tables 282 and 283 Several typesof inter-connected bases and rafts are given in Table 284 Inchoosing a foundation suitable for a particular purpose thenature of the structure should also be considered Sometimes itmay be decided to accept the risk of settlement in preference toproviding a more expensive foundation For silos and fixed-endarches the risk of unequal settlement of the foundations mustbe avoided at all costs but for gantries and the bases of largesteel tanks a simple foundation can be provided and probablesettlement allowed for in the design of the superstructure Inmining districts where it is reasonable to expect some subsidencea rigid raft foundation should be provided for small structuresto allow the structure to move as a whole For large structuresa raft may not be economical and the structure should bedesigned either to be flexible or as several separate elementson independent raft foundations

716 Separate bases

The simplest form of foundation for an individual column orstanchion is a reinforced concrete pad Such bases are widelyused on ground that is strong and on weaker grounds wherethe structure and the cladding are light and flexible For basesthat are small in area or founded on rock a block of plain ornominally reinforced concrete can be used The thickness ofthe block is made sufficient for the load to be transferred to theground under the base at an angle of dispersion through theblock of not less than 45o to the horizontal

To reduce the risk of unequal settlement the column basesizes for a building founded on a compressible soil should be inproportion to the dead load carried by each column Bases for thecolumns of a storage structure should be in proportion to the totalload excluding the effects of wind In all cases the pressure onthe ground under any base due to combined dead and imposedload including wind load and any bending at the base of thecolumn should not exceed the allowable bearing value

In the design of a separate base the area of a concentricallyloaded base is determined by dividing the maximum serviceload by the allowable bearing pressure The subsequentstructural design is then governed by the requirements of theultimate limit state The base thickness is usually determined byshear considerations governed by the more severe of two con-ditions ndash either shear along a vertical section extending acrossthe full width of the base or punching shear around the loadedarea ndash where the second condition is normally critical Thecritical section for the bending moment at a vertical sectionextending across the full width of the base is taken at the faceof the column for a reinforced concrete column and at the cen-tre of the base for a steel stanchion The tension reinforcementis usually spread uniformly over the full width of the base butin some cases it may need to be arranged so that there is aconcentration of reinforcement beneath the column Outsidethis central zone the remaining reinforcement must still con-form to minimum requirements It is also necessary for tensionreinforcement to comply with the bar spacing limitations forcrack control

If the base cannot be placed centrally under the column thebearing pressure varies linearly The base is then preferablyrectangular and modified formulae for bearing pressures andbending moments are given in Table 282 A base supportingfor example a column of a portal frame may be subjected to anapplied moment and horizontal shear force in addition to avertical load Such a base can be made equivalent to a base witha concentric load by placing the base under the column with aneccentricity that offsets the effect of the moment and horizontalforce This procedure is impractical if the direction of theapplied moment and horizontal force is reversible for exampledue to wind In this case the base should be placed centrallyunder the column and designed as eccentrically loaded for thetwo different conditions

717 Combined bases

If the size of the bases required for adjacent columns is suchthat independent bases would overlap two or more columnscan be provided with a common foundation Suitable typesfor two columns are shown in Table 283 for concentrically andeccentrically loaded cases Reinforcement is required top andbottom and the critical condition for shear is along a verticalsection extending across the full width of the base For someconditions of loading on the columns the total load on the basemay be concentric while for other conditions the total load iseccentric and both cases have to be considered Some notes oncombined bases are given in section 1812

718 Balanced and coupled bases

When it is not possible to place an adequate base centrallyunder a column owing to restrictions of the site and when forsuch conditions the eccentricity would result in inadmissibleground pressures a balanced foundation as shown in Tables 283and 284 and described in section 1813 is provided A beamis introduced and the effect of the cantilever moment causedby the offset column load is counterbalanced by load from anadjacent column This situation occurs frequently for externalcolumns of buildings on sites in built-up areas

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Sometimes as in the case of bases under the towers of atrestle or gantry pairs of bases are subjected to moments andhorizontal forces acting in the same direction on each base Insuch conditions the bases can be connected by a stiff beam thatconverts the effects of the moments and horizontal forces intoequal and opposite vertical reactions then each base can bedesigned as concentrically loaded Such a pair of coupled basesis shown in Table 283 which also gives formulae for thereactions and the bending moments on the beam

719 Strip bases and rafts

When the columns or other supports of a structure are closelyspaced in one direction it is common to provide a continuousbase similar to a footing for a wall Particulars of the designof strip bases are given in Table 283 Some notes on thesebases in relation to the diagrams in Table 284 together with anexample are given in section 1812

When the columns or other supports are closely spaced intwo directions or when the column loads are so high and theallowable bearing pressure is so low that a group of separatebases would totally cover the space between the columns asingle raft foundation of one of the types shown at (a)ndash(d) inTable 284 should be provided Notes on these designs are givenin section 1814

The analysis of a raft foundation supporting a set of equalloads that are symmetrically arranged is usually based on theassumption of uniformly distributed pressure on the groundThe design is similar to that for an inverted floor upon whichthe load is that portion of the ground pressure that is due to theconcentrated loads only Notes on the design of a raft for whichthe columns are not symmetrically disposed are also includedin section 1814 An example of the design of a raft foundationis given in Examples of the Design of Buildings

7110 Basements

The floor of a basement for which a typical cross section isshown at (e) in the lower part of Table 284 is typically a raftsince the weights of the ground floor over the basementthe walls and other structure above the ground floor and thebasement itself are carried on the ground under the floor ofthe basement For water-tightness it is common to construct thewall and the floor of the basement monolithically In mostcases although the average ground pressure is low the spansare large resulting in high bending moments and a thick floorif the total load is taken as uniform over the whole area Sincethe greater part of the load is transmitted through the wallsand any internal columns it is more rational and economicalto transfer the load on strips and pads placed immediatelyunder the walls and columns The resulting cantilever actiondetermines the required thickness of these portions and theremainder of the floor can generally be made thinner

Where basements are in water-bearing soils the effect ofhydrostatic pressure must be taken into account The upwardwater pressure is uniform below the whole area of the floorwhich must be capable of resisting the total pressure lessthe weight of the floor The walls must be designed to resist thehorizontal pressures due to the waterlogged ground and thebasement must be prevented from floating Two conditions needto be considered Upon completion the total weight of the

basement and superimposed dead load must exceed the worstcredible upward force due to the water by a substantial marginDuring construction there must always be an excess ofdownward load If these conditions cannot be satisfied oneof the following steps should be taken

1 The level of the groundwater near the basement should becontrolled by pumping or other means

2 Temporary vents should be formed in the basement floor orat the base of the walls to enable water to freely enter thebasement thereby equalising the external and internalpressures The vents should be sealed when sufficient deadload from the superstructure has been obtained

3 The basement should be temporarily flooded to a depth suchthat the weight of water in the basement together with thedead load exceeds the total upward force on the structure

While the basement is under construction method 1 normallyhas to be used but once the basement is complete method 3 hasthe merit of simplicity Basements are generally designedand constructed in accordance with the recommendations ofBS 8102 supplemented by the guidance provided in reportsproduced by CIRIA (ref 57) BS 8102 defines four gradesof internal environment each grade requiring a different levelof protection against water and moisture ingress Three types ofconstruction are described to provide either A tanked or Bintegral or C drained protection

Type A refers to concrete or masonry construction whereadded protection is provided by a continuous barrier system Anexternal tanking is generally preferred so that any externalwater pressure will force the membrane against the structureThis is normally only practicable where the construction is byconventional methods in excavation that is open or supportedby temporary sheet piling The structure should be monolithicthroughout and special care should be taken when a structureis supported on piles to avoid rupture of the membrane due tosettlement of the fill supported by the basement wall

Type B refers to concrete construction where the structureitself is expected to be sufficient without added protection Astructure designed to the requirements of BS 8007 is expectedto inhibit the ingress of water to the level required for a utilitygrade basement It is considered that this standard can also beachieved in basements constructed by using diaphragm wallssecant pile walls and permanent sheet piling If necessary theperformance can be improved by internal ventilation and theaddition of a vapour-proof barrier

Type C refers to concrete or masonry construction whereadded protection is provided by an internal ventilated drainedcavity This method is applicable to all types of constructionand can provide a high level of protection It is particularlyuseful for deep basements using diaphragm walls secant pilewalls contiguous piles or steel sheet piling

7111 Foundation piers

When a satisfactory bearing stratum is found at a depth of15ndash5 m below the natural ground level piers can be formedfrom the bearing stratum up to ground level The constructionof columns or other supporting members can then begin on thetop of the piers at ground level Such piers are generally squarein cross section and most economically constructed in plain

Foundations 65

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concrete When piers are impractical by reason of the depth atwhich a firm stratum occurs or due to the nature of the groundshort bored piles can be used

7112 Wall footings

When the load on a strip footing is distributed uniformly overthe whole length as in the general case of a wall footing theprincipal effects are due to the transverse cantilever action ofthe projecting portion of the footing If the wall is of concreteand built monolithically with the footing the critical bendingmoment is at the face of the wall If the wall is of masonry themaximum bending moment is at the centre of the footingExpressions for these moments are given in Table 283 If theprojection is less than the thickness of the base the transversebending moment may be ignored but the thickness should besuch that the shear strength is not exceeded Whether or not awall footing is designed for transverse bending longitudinalreinforcement is generally included to give some resistance tomoments due to unequal settlement and non-uniformity ofbearing In cases where a deep narrow trench is excavated downto a firm stratum plain concrete fill is normally used

7113 Foundations for machines

The area of a concrete base supporting a machine or enginemust be sufficient to spread the load onto the ground withoutexceeding the allowable bearing value It is advantageous if thecentre of area of the base coincides with the centre of gravityof the loads when the machine is working as this reducesthe risk of unequal settlement If vibration from the machine istransmitted to the ground the bearing pressure should beconsiderably lower than normally taken especially if the groundis clay or contains a large proportion of clay It is often importantthat the vibration of a machine should not be transmitted toadjacent structures either directly or via the ground In suchcases a layer of insulating material should be placed betweenthe concrete base carrying the machine and the groundSometimes the base is enclosed in a pit lined with insulatingmaterial In exceptional cases a machine base may stand onsprings or more elaborate damping devices may be installed Inall cases the base should be separated from any surroundingarea of concrete ground floor

With light machines the ground bearing pressure may not bethe factor that determines the size of the concrete base as thearea occupied by the machine and its frame may require a baseof larger area The position of the holding-down bolts generallydetermines the length and width of the base which shouldextend 150 mm or more beyond the outer edges of the holes leftfor the bolts The depth of the base must be such that the bottomis on a satisfactory bearing stratum and there is enough thick-ness to accommodate the holding-down bolts If the machineexerts an uplift force on any part of the base the dimensions ofthe base must be such that the part that is subjected to uplift hasenough weight to resist the uplift force with a suitable marginof safety All the supports of any one machine should be carriedon a single base and any sudden changes in the depth and widthof the base should be avoided This reduces the risk of fracturesthat might result in unequal settlements which could throw themachine out of alignment Reinforcement should be providedto resist all tensile forces

Advice on the design of reinforced concrete foundations tosupport vibrating machinery is given in ref 58 which givespractical solutions for the design of raft piled and massivefoundations Comprehensive information on the dynamics ofmachine foundations is included in ref 59

7114 Piled foundations

Where the upper soil strata is compressible or too weak tosupport the loads transmitted by a structure piles can be usedto transmit the load to underlying bedrock or a stronger soillayer using end-bearing piles Where bedrock is not located ata reasonable depth piles can be used to gradually transmit thestructural loads to the soil using friction piles

Horizontal forces due to wind loading on tall structures orearth pressure on retaining structures can be resisted by pilesacting in bending or by using raking piles Foundations forsome structures such as transmission towers and the roofs tosports stadiums are subjected to upward forces that can beresisted by tension piles Bridge abutments and piers adjacentto water can be constructed with piled foundations to counterthe possible detrimental effects of erosion

There are two basic categories of piles Displacement pilesare driven into the ground in the form of either a preformedsolid concrete pile or a hollow tube Alternatively a void canbe formed in the ground by driving a closed-ended tube thebottom of which is plugged with concrete or aggregate Thisallows the tube to be withdrawn and the void to be filled withconcrete It also allows the base of the pile to be enlarged inorder to increase the bearing capacity Non-displacement orlsquocast-in-placersquo piles are formed by boring or excavating theground to create a void into which steel reinforcement andconcrete can be placed In some soils the excavation needs tobe supported to stop the sides from falling in this is achievedeither with casings or by the use of drilling mud (bentonite)For further information on piles including aspects such aspile driving load testing and assessment of bearing capacityreference should be made to specialist textbooks (ref 60)

7115 Pile-caps

Rarely does a foundation element consist of a single pile Inmost cases piles are arranged in groups or rows with the topsof the piles connected by caps or beams Generally concrete ispoured directly onto the ground and encases the tops of the pilesto a depth of about 75 mm The thickness of the cap must besufficient to ensure that the imposed load is spread equallybetween the piles For typical arrangements of two to five pilesforming a compact group load can be transmitted by dispersionthrough the cap Inclined struts extending from the load to thetop of each pile are held together by tension reinforcement inthe bottom of the cap to form a space frame The struts areusually taken to intersect at the top of the cap at the centreof the loaded area but expressions have also been developedthat take into account the dimensions of the loaded areaInformation regarding the design of such pile-caps andstandardised arrangements and dimensions for groups of twoto five piles are given in Table 361

The thickness of a pile-cap designed by dispersion theory isusually determined by shear considerations along a verticalsection extending across the full width of the cap If the pile

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spacing exceeds three pile diameters it is also necessary todesign for punching shear In all cases the shear stress at theperimeter of the loaded area should not exceed the maximumdesign value related to the compressive strength of the strutsThe reinforcement in the bottom of the pile-cap should beprovided at each end with a full tension anchorage measuredfrom the centre of the pile Pile-caps can also be designed bybending theory but this is generally more appropriate where alarge number of piles are involved In such cases punchingshear is likely to be a critical consideration

7116 Loads on piles in a group

If a group of n piles is connected by a rigid pile-cap and thecentres of gravity of the load Fv and the piles are coincident eachpile will be equally loaded and will be subjected to a load FvnIf the centre of gravity of the load is displaced a distance e fromthe centre of gravity of the piles the load on any one pile is

where is the sum of the squares of the distance of each pilemeasured from an axis that passes through the centre of gravityof the group of piles and is at right angles to the line joining thiscentre of gravity and the centre of gravity of the load and a1 isthe distance of the pile considered from this axis (positive if onthe same side of the axis as the centre of gravity of the load andnegative if on the opposite side) If the structure supported onthe group of piles is subjected to a bending moment M whichis transmitted to the foundations the expression given for theload on any pile can be used by substituting e MFv

The total load that can be carried on a group of piles is notnecessarily the safe load calculated for one pile multipliedby the number of piles Some allowance has to be made for theoverlapping of the zones of stress in the soil supporting thepiles The reduction due to this effect is greatest for piles thatare supported mainly by friction For piles supported entirely oralmost entirely by end bearing the maximum safe load on agroup cannot greatly exceed the safe bearing load on the areaof bearing stratum covered by the group

7117 Loads on open-piled structures

The loads and forces to which wharves jetties and similarmaritime structures are subjected are dealt with in section 26Such structures can be solid walls made of plain or reinforcedconcrete as are most dock walls A quay or similar watersidewall is more often a sheet pile-wall as described in section 733or it can be an open-piled structure similar to a jetty The loadson groups of inclined and vertical piles for such structures areconsidered in Table 285

For each probable condition of load the external forces areresolved into horizontal and vertical components Fh and Fvthe points of application of which are also determined If thedirection of action and position are opposite to those shown inthe diagrams the signs in the formulae must be changed It isassumed that the piles are surmounted by a rigid pile-cap orsuperstructure The effects on each pile when all the piles arevertical are based on a simple but approximate method ofanalysis Since a pile offers very little resistance to bending

a2

Fv1n

ea1

a2

structures with vertical piles only are not suitable when Fh isdominant In a group containing inclined piles Fh can beresisted by a system of axial forces and the bending momentsand shear forces in the piles are negligible The analysis used inTable 285 is based on the assumption that each pile is hingedat the head and toe Although this assumption is not accuratethe analysis predicts the behaviour reasonably well Three designsof the same typical jetty using different pile arrangements aregiven in section 182

72 INDUSTRIAL GROUND FLOORS

Most forms of activity in buildings ndash from manufacturingstorage and distribution to retail and recreation ndash need a firmplatform on which to operate Concrete ground floors arealmost invariably used for such purposes Although in manyparts of the world conventional manufacturing activity hasdeclined in recent years there has been a steady growth indistribution warehousing and retail operations to serve theneeds of industry and society The scale of such facilities andthe speed with which they are constructed has also increasedwith higher and heavier racking and storage equipment beingused These all make greater demands on concrete floors Thefollowing information is taken mainly from ref 61 where acomprehensive treatment of the subject will be found

721 Floor uses

In warehouses materials handling equipment is used in twodistinct areas according to whether the movement of traffic isfree or defined In free-movement areas vehicles can travelrandomly in any direction This typically occurs in factoriesretail outlets low-level storage and food distribution centresIn defined-movement areas vehicles use fixed paths in verynarrow aisles This usually occurs where high-level storageracking is being employed and distribution and warehousefacilities often combine areas of free movement for low-levelactivities such as unloading and packing alongside areas ofdefined movement for high-level storage The two floor usesrequire different tolerances on surface regularity

722 Construction methods

A ground-supported industrial floor slab is made up of layersof materials comprising a sub-base a slip membranemethanebarrier and a concrete slab of appropriate thickness providinga suitable wearing surface Various construction methods can beused to form the concrete slab

Large areas of floor up to several thousand square metres inextent can be laid in a continuous operation Fixed forms areused up to 50 m apart at the edges of the area only Concrete isdischarged into the area and spread either manually or bymachine Surface levels are controlled either manually using atarget staff in conjunction with a laser level transmitter or bydirect control of a laser-guided spreading machine After thefloor has been laid and finished the area is sub-divided intopanels typically on a 6 m grid in both directions This is achievedby making saw cuts in the top surface for a depth of at leastone-quarter of the depth of the slab creating a line of weaknessin the slab that induces a crack below the saw cut As a result ofconcrete shrinkage each sawn joint will open by a small amount

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With such large-area construction there are limitations on theaccuracy of level and surface regularity that can be achievedand the construction is most commonly used for free-movementfloor areas

The large-area construction method can also be employedwithout sub-dividing the area into small panels In this case nosawn joints are made but steel fibres are incorporated in theconcrete mix to control the distribution and width of the cracksthat occur as a result of shrinkage The formed joints at theedges of the area will typically open by about 20 mm

Floors can also be formed as a series of long strips typically4ndash6 m wide with forms along each side Strips can be laidalternately with infill strips laid later or consecutively orbetween lsquoleave-in-placersquo screed rails Concrete is poured in acontinuous operation in each strip after which transverse sawcuts are made about 6 m apart to accommodate longitudinalshrinkage As formwork can be set to tight tolerances and thedistance between the forms is relatively small the long-stripmethod lends itself to the construction of very flat floors and isparticularly suitable for defined-movement floor areas

723 Reinforcement

Steel fibres usually manufactured from cold-drawn wire arecommonly used in ground-supported slabs The fibres vary inlength up to about 60 mm with aspect ratios (lengthnominaldiameter) from 20 to 100 and a variety of cross sections Inorder to increase pull-out resistance the fibres have enlargedflattened or hooked ends roughened surface textures or wavyprofiles The composite concrete slab can have considerableductility dependent on fibre type dosage tensile strength andanchorage mechanism The ductility is commonly measuredusing the Japanese Standard test method which uses beams ina third-point loading arrangement Load-deflection curves areplotted as the load increases until the first crack and thendecreases with increasing deflection The ductility value isexpressed as the average load to a deflection of 3 mm dividedby the load to first crack This measure is commonly known asthe equivalent flexural strength ratio In large-area floors withshrinkage joints at the edges only fibre dosages in the order of35ndash45 kgm3 are used to control the distribution and width ofcracks In floors with additional sawn joints fibre dosages in therange 20ndash30 kgm3 are typically used

In large area floors with additional sawn joints steel fabricreinforcement (type A) can be placed in the bottom of the slabwith typically 50 mm of cover The proportion of reinforcementused is typically 01ndash0125 of the effective cross section bdwhich is small enough to ensure that the reinforcement willyield at the sawn joints as the concrete shrinks and alsosufficient to provide the slab with adequate rotational capacityafter cracking

724 Modulus of subgrade reaction

For design purposes the subgrade is assumed to be an elasticmedium characterised by a modulus of subgrade reaction ksdefined as the load per unit area causing unit deflection It canbe shown that errors of up to 50 in the value of ks have onlya small effect on the slab thickness required for flexural designHowever deflections are more sensitive to ks values and long-term settlement due to soil consolidation under load can be

Foundations ground slabs retaining walls culverts and subways68

Soil typeValues of ks (MNm3)

Lower Upper

Fine or slightly compacted sand 15 30Well compacted sand 50 100Very well compacted sand 100 150Loam or clay (moist) 30 60Loam or clay (dry) 80 100Clay with sand 80 100Crushed stone with sand 100 150Coarse crushed stone 200 250Well-compacted crushed stone 200 300

much larger than the elastic deflections calculated as part of theslab design

In principle the value of ks used in design should be relatedto the range of influence of the load but it is normal practice tobase ks on a loaded area of diameter 750 mm To this endit is strongly recommended that the value of ks is determinedfrom a BS plate-loading test using a 750 mm diameter plateand a fixed settlement of 125 mm If a smaller plate is used ora value of ks appropriate to a particular area is required thefollowing approximate relationship may be assumed

ks 05(103D)2k075

where D is the diameter of the loaded area and k075 is a valuefor D 075 m This gives values of ksk075 as follows

D (m) 03 045 075 12 infin

ksk075 20 14 10 08 05

In the absence of more accurate information typical values ofks according to the soil type are given in the following table

725 Methods of analysis

Traditionally ground-supported slabs have been designed byelastic methods using equations developed by Westergaard inthe 1920s Such slabs are relatively thick and an assessment ofdeflections and other in-service requirements has generallybeen unnecessary Using plastic methods of analysis thinnerslabs can be designed and the need to investigate in-servicerequirements and load-transfer across joints has become moreimportant The use of plastic analysis assumes that the slab hasadequate ductility after cracking that is it contains sufficientfibres or reinforcement as described in section 723 to give anequivalent flexural strength ratio in the range 03ndash05 Plainconcrete slabs and slabs with less than the minimum recom-mended amounts of fibres or reinforcement should still bedesigned by elastic methods

Westergaard assumed that a ground-supported concrete slabis a homogeneous isotropic elastic solid in equilibrium withthe subgrade reactions being vertical only and proportional tothe deflections of the slab He also introduced the concept of theradius of relative stiffness rk given by the relationship

where Ec is the short-term modulus of elasticity of concreteh is the slab thickness ks is the modulus of subgrade reaction

rk [Ech312(1v2)ks]

025

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and is Poissonrsquos ratio The physical significance of rk isillustrated in the following figure showing the approximatedistribution of elastic bending moments for a single internalconcentrated load The bending moment is positive (tensionat the bottom of the slab) with a maximum value at the loadposition Along radial lines it remains positive reducing to zero atrk from the load It then becomes negative reaching a maximumat 2rk from the load with the maximum negative moment (tensionat the top of the slab) significantly less than the maximumpositive moment The moment approaches zero at 3rk fromthe load

or internally as shown in Table 286 An externally stabilisedsystem uses an external structural wall to mobilise stabilisingforces An internally stabilised system utilises reinforcementsinstalled within the soil and extending beyond the potentialfailure zone

Traditional retaining walls can be considered as externallystabilised systems one of the most common forms being thereinforced concrete cantilever wall Retaining walls on spreadfoundations together with gravity structures support the soilby weight and stiffness to resist forward sliding overturningand excessive soil movements The equilibrium of cantileverwalls can also be obtained by embedment of the lower part ofthe wall Anchored or propped walls obtain their equilibriumpartly by embedment of the lower part of the wall and partlyfrom an anchorage or prop system that provides support to theupper part of the wall

Internally stabilised walls built above ground are known asreinforced soil structures By placing reinforcement within thesoil a composite material can be produced that is strong intension as well as compression A key aspect of reinforced soilwalls is its incremental form of construction being built up alayer at a time starting from a small plain concrete strip footingIn this way construction is always at ground level the structureis always stable and progress can be very rapid The result ofthe incremental construction is that the soil is partitioned witheach layer receiving support from a locally inserted reinforcingelement The process is the opposite of what occurs in aconventional wall where pressures exerted by the backfill areintegrated to produce an overall force to be resisted by the struc-ture The materials used in a reinforced soil structure comprisea facing (usually reinforced concrete) soil reinforcement (inthe form of flat strips anchors or grids made from eithergalvanised steel or synthetic material) and soil (usually a well-graded cohesionless material) Reinforced soil structuresare more economic than equivalent structures using externallystabilised methods

Internal soil stabilisation used in the formation of cuttingsor excavations is known as soil nailing The process is againincremental with each stage of excavation limited in depth sothat the soil is able to support itself The exposed soil face isprotected usually by a covering of light mesh reinforcementand spray applied concrete Holes are drilled into the soil andreinforcement in the form of steel bars installed and groutedWith both reinforced soil and soil nailing great care is takento make sure that the reinforcing members do not corrode ordeteriorate Hybrid systems combining elements of internallyand externally stabilised soils are also used

732 Walls on spread bases

Various walls on spread bases are shown in Table 286 Acantilever wall is suitable for walls of moderate height If thesoil to be retained can be excavated during construction ofthe wall or the wall is required to retain an embankment thebase can project backwards This is always advantageous asthe earth supported on the base assists in counterbalancing theoverturning effect due to the horizontal pressures exerted bythe soil However a base that projects mainly backwards butpartly forwards is usually necessary in order to limit the bearingpressure at the toe to an allowable value Sometimes due tothe proximity of adjacent property it may be impossible to

Retaining walls 69

Approximate distribution of elastic bending moments for an internal concentrated load on a ground-supported slab

As the load is increased the tensile stresses at the bottom ofthe slab under the load will reach the flexural strength of theconcrete Radial tension cracks will form at the bottom ofthe slab and provided there is sufficient ductility the slab willyield Redistribution of moments will occur with a reduction inthe positive moment at the load position and a substantialincrease in the negative moments some distance away Withfurther increases in load the positive moment at the loadposition will remain constant and the negative moments willincrease until the tensile stresses at the top of the slab reach theflexural strength of the concrete at which stage failure isassumed For further information on the analysis and designmethod with fully worked examples see ref 61

73 RETAINING WALLS

Information on soil properties and the pressures exerted bysoils on retaining structures is given in section 91 andTables 210ndash214 This section deals with the design of wallsto retain soils and materials with similar engineering propertiesIn designing to British Codes of Practice the geotechnicalaspects of the design which govern the size and proportions ofthe structure are considered in accordance with BS 8002Mobilisation factors are introduced into the calculation of thesoil strengths and the resulting pressures are used for bothserviceability and ultimate requirements For the subsequentdesign of the structure to BS 8110 the earth loads obtainedfrom BS 8002 are taken as characteristic values In designing tothe EC partial safety factors are applied to the soil propertiesfor the geotechnical aspects of the design and to the earthloads for the structural design

731 Types of retaining wall

Earth retention systems can be categorised into one of twogroups according to whether the earth is stabilised externally

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project a base backwards Under such conditions where thebase projection is entirely forwards the provision of a keybelow the base is necessary to prevent sliding by mobilising thepassive resistance of the soil in front of the base

For wall heights greater than about 8 m the stem thicknessof a cantilever wall becomes excessive In such cases a wallwith vertical counterforts can be used in which the slabspans horizontally between the counterforts For very highwalls in which the soil loading is considerable towardsthe bottom of the wall horizontal beams spanning betweenthe counterforts can be used By graduating the spacing of thebeams to suit the loading the vertical bending moments ineach span of the slab can be equalised and the slab thicknesskept the same

The factors affecting the design of a cantilever slab wall areusually considered per unit length of wall when the wall is ofconstant height but if the height varies a length of say 3 mcould be treated as a single unit For a wall with counterfortsthe length of a unit is taken as the distance between adjacentcounterforts The main factors to be considered in the design ofwalls on spread bases are stability against overturning groundbearing pressure resistance to sliding and internal resistance tobending moments and shearing forces Suitable dimensions forthe base to a cantilever wall can be estimated with the aid of thegraph given in Table 286

In BS 8002 for design purposes soil parameters are basedon representative shear strengths that have been reduced byapplying mobilisation factors Also for friction or adhesion ata soilndashstructure interface values not greater than 75 of thedesign shear strength are taken Allowance is made for a minimumsurcharge of 10 kNm2 applied to the surface of the retainedsoil and for a minimum depth of unplanned earth removal infront of the wall equal to 10 of the wall height but not lessthan 05 m

For overall equilibrium the effects of the disturbing forcesacting on the structure should not exceed the effects that canbe mobilised by the resisting forces No additional factors ofsafety are required with regard to overturning or sliding forwardsFor bases founded on clay soils both the short-term (usingundrained shear strength) and long-term (using drained shearstrength) conditions should be considered Checks on groundbearing are required for both the service and ultimate conditionswhere the design loading is the same for each but the bearingpressure distribution is different For the ultimate condition auniform distribution is considered with the centre of pressurecoincident with the centre of the applied force at the undersideof the base In general therefore the pressure diagram doesnot extend over the entire base In cases where resistance tosliding depends on base adhesion it is unclear as to whetherthe contact surface length should be based on the service or theultimate condition

The foregoing wall movements due to either overturning orsliding are independent of the general tendency of a bank or acutting to slip and carry the retaining wall along with it Thestrength and stability of the retaining wall have no bearingon such failures The precautions that must be taken to preventsuch failures are outside the scope of the design of a wall thatis constructed to retain the toe of the bank and are a problemin soil mechanics

Adequate drainage behind a retaining wall is important toreduce the water pressure on the wall For granular backfills of

high permeability no special drainage layer is needed but somemeans of draining away any water that has percolated throughthe backfill should be provided particularly where a wall isfounded on an impermeable material For cohesionless backfillsof medium to low permeability and for cohesive soils it is usualto provide a drainage layer behind the wall Various methodscan be used for instance (a) a blanket of rubble or coarseaggregate clean gravel or crushed stone (b) hand-placedpervious blocks as dry walling (c) graded filter drain wherethe back-filling consists of fine-grain material (d) a geotextilefilter used in combination with a permeable granular materialWater entering the drainage layer should drain into a drainagesystem which allows free exit of the water either by the provisionof weep-holes or by porous land drains and pipes laid at thebottom of the drainage layer and led to sumps or sewers viacatchpits Where weep-holes are being used they should be atleast 75 mm in diameter and at a spacing not more than 1 mhorizontally and 1ndash2 m vertically Puddled clay or concreteshould be placed directly below the weep-holes or pipes andin contact with the back of the wall to prevent water fromreaching the foundations

Vertical movement joints should be provided at intervalsdependent upon the expected temperature range the type ofthe structure and changes in the wall height or the nature of thefoundations Guidance on design options to accommodatemovement due to temperature and moisture change are given inBS 8007 and Highways Agency BD 2887

733 Embedded (or sheet) walls

Embedded walls are built of contiguous or interlocking pilesor diaphragm wall panels to form a continuous structure Thepiles may be of timber or concrete or steel and have lapped orV-shaped or tongued and grooved or interlocking jointsbetween adjacent piles Diaphragm wall panels are formed ofreinforced concrete using a bentonite or polymer suspension aspart of the construction process Excavation is carried out in thesuspension to a width equal to the thickness of the wallrequired The suspension is designed to maintain the stability ofthe slit trench during digging and until the diaphragm wall hasbeen concreted Wall panels are formed in predeterminedlengths with prefabricated reinforcement cages lowered into thetrench Concrete is cast in situ and placed by tremie it is vitalthat the wet concrete flows freely without segregation so as tosurround the reinforcement and displace the bentonite

Cantilever walls are suitable for only moderate height and itis preferable not to use cantilever walls when services orfoundations are located wholly or partly within the active soilzone since horizontal and vertical movement in the retainedmaterial can cause damage Anchored or propped walls canhave one or more levels of anchor or prop in the upper partof the wall They can be designed to have fixed or free earthsupport at the bottom as stability is derived mainly from theanchorages or props

Traditional methods of design although widely used allhave recognised shortcomings These methods are outlined inannex B of BS 8002 where comments are included on theapplicability and limitations of each method The design ofembedded walls is beyond the scope of this Handbook and forfurther information the reader should refer to BS 8002Highways Agency BD 4294 and ref 62

Foundations ground slabs retaining walls culverts and subways70

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74 CULVERTS AND SUBWAYS

Concrete culverts which can be either cast in situ or precastare usually of circular or rectangular cross section Box typestructures can also be used to form subways cattle creeps orbridges over minor roads

741 Pipe culverts

For conducting small streams or drains under embankmentsculverts can be built with precast reinforced concrete pipeswhich must be strong enough to resist vertical and horizontalpressures from the earth and other superimposed loads Thepipes should be laid on a bed of concrete and where passingunder a road should be surrounded with reinforced concrete atleast 150 mm thick The culvert should also be reinforced toresist longitudinal bending resulting from unequal vertical earthpressure and unequal settlement Due to the uncertaintyassociated with the magnitude and disposition of the earth pres-sures an accurate analysis of the bending moments is imprac-ticable A basic guide is to take the positive moments at the topand bottom of the pipe and the negative moments at the endsof a horizontal diameter as 00625qd 2 where d is the diameterof the circular pipe and q is the intensity of both the downwardpressure on the top and the upward pressure at the bottomassuming the pressure to be distributed uniformly on ahorizontal plane

742 Box culverts

The load on the top of a box culvert includes the weights of theearth covering and the top slab and the imposed load (if any)The weights of the walls and top slab (and any load that is onthem) produce an upward reaction from the ground Theweights of the bottom slab and water in the culvert are carrieddirectly on the ground below the slab and thus have no effectother than their contribution to the total bearing pressure Thehorizontal pressure due to the water in the culvert produces aninternal triangular load on the walls or a trapezoidal load if thesurface of the water outside the culvert is above the top whenthere will also be an upward pressure on the underside of thetop slab The magnitude and distribution of the earth pressureagainst the sides of the culvert can be calculated in accordancewith the information in section 91 consideration being givento the risk of the ground becoming waterlogged resulting inincreased pressure and the possibility of flotation Generallythere are only two load conditions to consider

1 Culvert empty maximum load on top slab weight of thewalls and maximum earth pressure on walls

2 Culvert full minimum load on top slab weight of the wallsminimum earth pressure and maximum internal hydrostaticpressure on walls (with possible upward pressure on top slab)

In some circumstances these conditions may not produce themaximum load effects at any particular section and the effectof every probable combination should be considered The crosssections should be designed for the combined effects of axialforce bending and shear as appropriate A simplistic analysiscan be used to determine the bending moments produced in amonolithic rectangular box by considering the four slabs asa continuous beam of four spans with equal moments at the end

supports However if the bending of the bottom slab tends toproduce a downward deflection the compressibility of theground and the consequent effect on the bending moments mustbe taken into account The loads can be conveniently dividedinto the following cases

1 A uniformly distributed load on the top slab and a uniformreaction from the ground under the bottom slab

2 A concentrated imposed load on the top slab and a uniformreaction from the ground under the bottom slab

3 Concentrated loads due to the weight of each wall and auniform reaction from the ground under the bottom slab

4 A triangular distributed horizontal pressure on each wall dueto the increase in earth pressure in the height of the wall

5 A uniformly distributed horizontal pressure on each walldue to pressure from the earth and any surcharge above thelevel of the top slab

6 Internal horizontal and possibly vertical pressures due towater in the culvert

Formulae for the bending moments at the corners of thebox due to each load case when the top and bottom slabsare the same thickness are given in Table 287 The limitingground conditions associated with the formulae should benoted

743 Subways

The design and construction of buried box type structureswhich could be complete boxes portal frames or structureswhere the walls are propped by the top slab are covered byrecommendations in Highways Agency standard BD 3187These recommendations do not apply to structures that areinstalled by methods such as thrust boring or pipe jacking

The nominal superimposed dead load consists of the weightof any road construction materials and the soil cover abovethe structure Due to negative arching of the fill material thestructure can be subjected to loads greater than the weight offill directly above it An allowance for this effect is made byconsidering a minimum load based on the weight of materialdirectly above the structure and a maximum load equal to theminimum load multiplied by 115 The nominal horizontalearth pressures on the walls of the box structure are based ona triangular distribution with the value of the earth pressurecoefficient taken as a maximum of 06 and a minimum of 02It is to be assumed that either the maximum or the minimumvalue can be applied to one wall irrespective of the value thatis applied to the other wall

Where the depth of cover measured from the finished roadsurface to the top of the structure is greater than 600 mm thenominal vertical live loads to be considered are the HA wheelload and the HB vehicle To determine the nominal vertical liveload pressure dispersion of the wheel loads may be taken tooccur from the contact area on the carriageway to the top ofthe structure at a slope of 2 vertically to 1 horizontally Forstructures where the depth of cover is in the range 200ndash600 mmfull highway loading is to be considered For HA load the KELmay be dispersed below the depth of 200 mm from the finishedroad surface Details of the nominal vertical live loads are givenin sections 248 and 249 and Table 25

Culverts and subways 71

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Part 2

Loads materials andstructures

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In this chapter unless otherwise stated all loads are given ascharacteristic or nominal (ie unfactored) values For designpurposes each value must be multiplied by the appropriatepartial safety factor for the particular load load combinationand limit-state being considered

Although unit weights of materials should be given strictlyin terms of mass per unit volume (eg kgm3) the designer isusually only concerned with the resulting gravitational forcesTo avoid the need for repetitive conversion unit weights aremore conveniently expressed in terms of force (eg kNm3)where 1 kN may be taken as 102 kilograms

81 DEAD LOAD

The data for the weights of construction materials given in thefollowing tables has been taken mainly from EC 1 Part 11 butalso from other sources such as BS 648

811 Concrete

The primary dead load is usually the weight of the concretestructure The weight of reinforced concrete varies with thedensity of the aggregate and the percentage of reinforcementIn UK practice a value of 24 kNm3 has traditionally beenused for normal weight concrete with normal percentages ofreinforcement but a value of 25 kNm3 is recommended inEC 1 Several typical weights for normal lightweight andheavyweight (as used for kentledge and nuclear-radiationshielding) concretes are given in Table 21 Weights are alsogiven for various forms and depths of concrete slabs

812 Other construction materials and finishes

Dead loads include such permanent weights as those of thefinishes and linings on walls floors stairs ceilings and roofsasphalt and other applied waterproofing layers partitionsdoors windows roof and pavement lights superstructures ofsteelwork masonry or timber concrete bases for machineryand tanks fillings of earth sand plain concrete or hardcorecork and other insulating materials rail tracks and ballastingrefractory linings and road surfacing In Table 21 the basicweights of various structural and other materials includingmetals stone timber and rail tracks are given

The average equivalent weights of various cladding types asgiven in Table 22 are useful in estimating the loads imposed

on a concrete substructure The weights of walls of variousconstructions are also given in Table 22 Where a concretelintel supports a brick wall it is generally not necessary toconsider the lintel as supporting the entire wall above it issufficient to allow only for the triangular areas indicated in thediagrams in Table 22

813 Partitions

The weight of a partition is determined by the material of whichit is made and the storey height When the position of thepartition is not known or the use of demountable partitions isenvisaged the equivalent uniformly distributed load given inTable 22 should be considered as an imposed load in the designof the supporting floor slabs

Weights of permanent partitions whose position is knownshould be included in the dead load Where the length of the par-tition is in the direction of span of the slab an equivalent UDLmay be used as given in Table 22 In the case of brick or similarlybonded partitions continuous over the slab supports some reliefof loading on the slab will occur due to the arching action of thepartition unless this is invalidated by the presence of doorways orother openings Where the partition is at right angles to the spanof the slab a concentrated line load should be applied at the appro-priate position The slab should then be designed for the combinedeffect of the distributed floor load and the concentrated load

82 IMPOSED LOADS

Imposed loads on structures include the weights of storedmaterials and the loads resulting from occupancy and trafficComprehensive data regarding the weights of stored materialsassociated with building industry and agriculture are given inEC 1 Part 11 Data for loads on floors due to livestock andagricultural vehicles are given in BS 5502 Part 22

821 Imposed loads on buildings

Data for the vertical loads on floors and horizontal loads onparapets barriers and balustrades are given in BS 6399 Part 1Loads are given in relation to the type of activityoccupancy forwhich the floor area will be used in service as follows

A Domestic and residential activitiesB Office and work areas not covered elsewhere

Chapter 8

Loads

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21Weights of construction materials and concrete floor slabs

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Weights of roofs and walls 22

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C Areas where people may congregateD Shopping areasE Areas susceptible to the accumulation of goodsFG Vehicle and traffic areas

Details of the imposed loads for categories A and B are givenin Table 23 Values are given for uniformly distributed andconcentrated loads These are not to be taken together butconsidered as two separate load cases The concentrated loadsnormally do not need to be considered for solid or other slabsthat are capable of effective lateral distribution When used forcalculating local effects such as bearing or the punching of thinflanges a square contact area of 50 mm side should be assumedin the absence of any other specific information

With certain exceptions the imposed loads on beams may bereduced according to the area of floor supported Loads oncolumns and foundations may be reduced according to eitherthe area of floor or the number of storeys supported Details ofthe reductions and the exceptions are given in Table 23

Data given in Table 24 for the load on flat or mono-pitchroofs has been taken from BS 6399 Part 3 The loads whichare additional to all surfacing materials include for snow andother incidental loads but exclude wind pressure For other roofshapes and the effects of local drifting of snow behind parapetsreference should be made to BS 6399 Part 3

For building structures designed to meet the requirementsof EC 2 Part 1 details of imposed and snow loads are given inEC 1 Parts 11 and 13 respectively

822 Imposed loads on highway bridges

The data given in Table 25 for the imposed load on highwaybridges have been taken from the Highways Agency documentBD 3701 Type HA loading consists of two parts a uniformload whose value varies with the lsquoloaded lengthrsquo and a singleKEL that is positioned so as to have the most severe effectThe loaded length is the length over which the application of theload increases the effect to be determined Influence lines maybe needed to determine critical loaded lengths for continuousspans and arches Loading is applied to one or more notionallanes and multiplied by appropriate lane factors The alternativeof a single wheel load also needs to be considered in certaincircumstances

Type HB is a unit loading represented by a 16ndashwheel vehicleof variable bogie spacing where one unit of loading is equivalentto 40 kN The number of units considered for a public highwayis normally between 30 and 45 according to the appropriateauthority The vehicle can be placed in any transverse positionon the carriageway displacing HA loading over a specified areasurrounding the vehicle

For further information on the application of combined HAand HB loading and details of other loads to be considered on

highway bridges reference should be made to BD 3701 andBD 6094 For information on loads to be considered for theassessment of existing highway bridges reference should bemade to BD 2101

823 Imposed loads on footbridges

The data given in Table 26 for the imposed load on bridges dueto pedestrian traffic have been taken from the Highways Agencydocument BD 3701 For further information on the pedestrianloading to be considered on elements of highway or railwaybridges that also support footways or cycle tracks and the ser-viceability vibration requirements of footbridges referenceshould be made to BD 3701

824 Imposed loads on railway bridges

The data given in Table 26 for the imposed load on railwaybridges has been taken from the Highways Agency documentBD 3701 Types RU and SW0 apply to main line railwaystype SW0 being considered as an additional and separate loadcase for continuous bridges For bridges with one or two tracksloads are to be applied to each track In other cases loads areto be applied as specified by the relevant authority

Type RL applies to passenger rapid transit railway systemswhere main line locomotives and rolling stock do not operateThe loading consists of a uniform load (or loads dependent onloaded length) combined with a single concentrated load posi-tioned so as to have the most severe effect The loading is to beapplied to each and every track An arrangement of two con-centrated loads is also to be considered for deck elementswhere this would have a more severe effect

For information on other loads to be considered on railwaybridges reference should be made BD 3701

83 WIND LOADS

The data given in Tables 27ndash29 for the wind loading onbuildings has been taken from the information given for thestandard method of design in BS 6399 Part 2 The effectivewind speed is determined from Table 27 Wind pressures andforces on rectangular buildings as defined in Table 28 aredetermined by using standard pressure coefficients given inTable 29 For data on other building shapes and different roofforms and details of the directional method of design referenceshould be made to BS 6399 Part 2

Details of the method used to assess wind loads on bridgestructures and the data to be used for effective wind speeds anddrag coefficients are given in BD 3701 For designs to EC 2wind loads are given in EC 1 Part 12

Loads78

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Imposed loads on floors of buildings 23

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Imposed loads on roofs of buildings 24

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Imposed loads on bridges ndash 1 25

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Imposed loads on bridges ndash 2 26

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Wind speeds (standard method of design) 27

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Wind pressures and forces (standard method of design) 28

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Pressure coefficients and size effect factors forrectangular buildings 29

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Chapter 9

Pressures due toretained materials

In this chapter unless otherwise stated all unit weights andother properties of materials are given as characteristic or rep-resentative (ie unfactored) values For design purposes eachvalue must be modified by appropriate partial safety or mobili-sation factors according to the basis of design and the code ofpractice employed

91 EARTH PRESSURES

The data given in Table 210 for the properties of soils has beentaken from BS 8002 Design values of earth pressure coeffi-cients are based on the design soil strength which is taken asthe lower of the peak soil strength reduced by a mobilisationfactor or the critical state strength

911 Pressures imposed by cohesionless soils

For the walls shown in Table 211 with a uniform normallyconsolidated soil a uniformly distributed surcharge and nowater pressure the pressure imposed on the wall increaseslinearly with depth and is given by

K(zq)

where is unit weight of soil z is depth below surface q issurcharge pressure (kNm2) K is at-rest active or passivecoefficient of earth pressure according to design conditions

A minimum live load surcharge of 10 kNm2 is specified inBS 8002 This may be reasonable for walls 5 m high and abovebut appears to be too large for low walls In this case valuessuch as 4 kNm2 for walls up to 2 m high 6 kNm2 for walls 3 mhigh and 8 kNm2 for walls 4 m high could be used InBD 3701 surcharge loads are given of 5 kNm2 for footpaths10 kNm2 for HA loading 12 kNm2 for 30 units of HB loading20 kNm2 for 45 units of HB loading and on areas occupiedby rail tracks 30 kNm2 for RL loading and 50 kNm2 forRU loading

If static ground water occurs at depth zw below the surfacethe total pressure imposed at z zw is given by

K [mz (s w)(zzw)q]w(zzw)

where m is moist bulk weight of soil s is saturated bulkweight of soil w is unit weight of water (981 kNm3)

912 At-rest pressures

For a level ground surface and a normally consolidated soilthat has not been subjected to removal of overburden thehorizontal earth pressure coefficient is given by

Ko 1sin

where is effective angle of shearing resistance of soilCompaction of the soil will result in earth pressures in the

upper layers of the soil mass that are higher than those givenby the above equation The diagram and equations given inTable 211 can be used to calculate the maximum horizontalpressure induced by the compaction of successive layers ofbackfill and determine the resultant earth pressure diagramThe effective line load for dead weight compaction rollers is theweight of the roller divided by its width For vibratory rollersthe dead weight of the roller plus the centrifugal force causedby the vibrating mechanism should be used The DOESpecification limits the mass of the roller to be used within 2 mof a wall to 1300 kgm

For a vertical wall retaining backfill with a ground surfacethat slopes upwards the horizontal earth pressure coefficientmay be taken as

Ko (1sin)(1sin)

where is slope angle The resultant pressure which acts in adirection parallel to the ground surface is given by

o Koz cos

913 Active pressures

Rankinersquos theory may be used to calculate the pressure on a ver-tical plane referred to as the lsquovirtual backrsquo of the wall For avertical wall and a level ground surface the Rankine horizontalearth pressure coefficient is given by

The solution applies particularly to the case of a smooth wall ora wall with no relative movement between the soil mass andthe back of the wall The charts given in Table 212 which arebased on the work of Caquot and Kerisel may be used generallyfor vertical walls with sloping ground or inclined walls with

Ka 1 sin

1 sin

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Properties of soils 210

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Earth pressure distributions on rigid walls 211

At-rest state for rigid wall Effects of soil compaction

Active state for rigid wall free to rotate about base or translate

Passive state for rigid wall free to rotate about base or translate

hc 1K

2Q1

zc K2Q1

c 2Q1

K earth pressure coefficientK o for unyielding structureK a for wall free to mobilise

fully active stateQ1 intensity of effective line load

imposed by compaction plant unit weight of soil maximum horizontal earth

pressure induced by compaction

Horizontal earth pressure distribution resulting from compaction

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Active earth pressure coefficients 212

Angle of Shearing Resistance f (degrees)

Coe

ffici

ent o

f Act

ive

Pre

ssur

e K

aC

oeffi

cien

t of A

ctiv

e P

ress

ure

Ka

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level ground The horizontal and vertical components of resultantpressure are given by

ah Kazcos() and av Kazsin()

where is wall inclination to vertical (positive or negative) isselected angle of wall friction (taken as positive)

914 Passive pressures

For a vertical wall and a level ground surface the Rankinehorizontal earth pressure coefficient is given by

The solution applies particularly to the case of a smooth wall ora wall with no relative movement between the soil mass and theback of the wall The charts given in Table 213 for verticalwalls with sloping ground and in Table 214 for inclined wallswith level ground are based on the work of Caquot and KeriselThe horizontal and vertical components of resultant pressureare given by

ph Kpzcos() and pv Kpzsin()

where is wall inclination to vertical (positive or negative) isselected angle of wall friction (taken as negative)

915 Cohesive soils

If a secant value (c 0) is selected the procedures given forcohesionless soils apply If tangent parameters (c ) are to beused the RankinendashBell equations may be used as follows

a Ka(zq)2cradicKa

p Kp(zq)2cradicKp

where c is effective cohesion The active earth pressure istheoretically negative to a depth given by

zo (2cradicKa q)

Where cracks which may form in the tension zone can becomefilled with water full hydrostatic pressure should be consideredover the depth zo If the surface is protected so that no surfacewater can accumulate in the tension cracks the earth pressureshould be taken as zero over the depth zo

916 Further considerations

For considerations such as earth pressures on embedded walls(with or without props) the effects of vertical concentratedloads and line loads and the effects of groundwater seepagereference should be made to BS 8002 For the pressures to beconsidered in the design of integral bridge abutments as aresult of thermal movements of the deck reference should bemade to Highways Agency document BA 4296

92 TANKS

The pressure imposed by a contained liquid is given by

wz

Kp 1 sin

1 sin

where w is unit weight of liquid (see EC 1 Part 11) and z isdepth below surface For a fully submerged granular materialthe total horizontal pressure on the walls is

K(w)(zzo)wz

where is unit weight of the material (including voids) zo isdepth to top of material K is material pressure coefficient If o

is unit weight of material (excluding voids) o(1 e)where e is ratio of volume of voids to volume of solids

The preceding equation applies to materials such as coal orbroken stone with an effective angle of shearing resistancewhen submerged of approximately 35o For submerged sand Kshould be taken as unity If the material floats (o w) thesimple hydrostatic pressure applies

93 SILOS

The data given in Tables 215 and 216 has been taken fromEurocode 1 Part 4 The pressures apply to silos of the formsshown in Table 215 subject to the following limitations

dimensions dc 50 m h 100 m hdc 10

eccentricities ei 025dc eo 025dc with no part of outlet ata distance greater than 03dc from centreline of silo

filling involves negligible inertia effects and impact loads

stored material is free-flowing (cohesion is less than 4 kPa fora sample pre-consolidated to 100 kPa) with a maximumparticle size not greater than 03dc

transition between vertical walled section and hopper is on asingle horizontal plane

Dimensions h ho h1 and z are measured from the equivalentsurface which is a level surface giving the same volume ofstored material as the actual surface at the maximum filling

Loads acting on a hopper are shown in Table 215 where thetensile force at the top of the hopper is required for the designof silo supports or a ring beam at the transition level The verticalcomponent of the force can be determined from force equilib-rium incorporating the vertical surcharge Cbpvo at the transitionlevel and the weight of the hopper contents The discharge loadon the hopper wall is affected by the flow pattern of the storedmaterial which may be mass flow or funnel flow according tothe characteristics of the hopper and the material The normalload due to pn is supplemented for mass flow silos only by akick load due to ps

Values of material properties and expressions to determineresulting pressures in the vertical walled and bottom sections ofa silo are given in Table 216 For squat silos (hdc 15) thehorizontal pressure ph may be reduced to zero at the level wherethe upper surface of the stored material meets the silo wallBelow this level a linear pressure variation may be assumedtaking K 10 until this pressure reaches the value appropriateto the depth z below the equivalent surface

Homogenizing silos and silos containing powders in whichthe speed of the rising surface of the material exceeds 10 mhshould be designed for both the fluidised and non-fluidised con-ditions For the fluidised condition the bulk unit weight of thematerial may be taken as 08 For information on test methodsto determine the properties of particulate materials referenceshould be made to EC 1 Part 4

Pressures due to retained materials90

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Passive earth pressure coefficents ndash 1 213

Angle of Shearing Resistance f (degrees)

Coe

ffici

ent o

f Pas

sive

Pre

ssur

e K

pR

educ

tion

Fac

tor

Rd

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214Passive earth pressure coefficents ndash 2

Angle of Shearing Resistance f (degrees)

Coe

ffici

ent o

f Pas

sive

Pre

ssur

e K

pR

educ

tion

Fac

tor

Rd

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215Silos ndash 1

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216Silos ndash 2

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101 CONSTITUENTS OF CONCRETE

1011 Cements and combinations

Manufactured cements are those made in a cement factoryWhere a mineral material is included it is generally added tothe cement clinker at the grinding stage The notation used forthese manufactured cements contains the prefix letters CEMWhen a concrete producer adds an addition such as pfa or ggbsto CEM I Portland cement in the mixer the resulting cement isknown as a mixer combination and is denoted by the prefixletter C Cements and combinations in general use are listed inTable 217 Further information on the different types and useof cements is given in section 311

1012 Aggregates

Overall grading limits for coarse and fine aggregates fromnatural sources in accordance with BS EN 12620 are given inTable 217 Further information is given in section 312

102 EARLY-AGE TEMPERATURES OF CONCRETE

The calculation of early thermal crack widths in a restrainedconcrete element requires knowledge of the temperature rise dueto the cement hydration Some typical early temperature histo-ries of various concrete walls and predicted temperature risesfor different cements are given in Table 218

The predicted temperature rise values for Portland cementconcretes in walls and slabs are taken from BS 8007 These aremaximum values selected from a range of values for Portlandcements obtained from different works (ref 11) The tempera-ture rises given for the other cements in concrete sections witha minimum dimension of 1 m should be taken as indicativeonly but could be used where other specific information isnot available

103 REINFORCEMENT

Reinforcement for concrete generally consists of steel bars orwelded steel mesh fabric that depend upon the provision of adurable concrete cover for protection against corrosion Theessential properties of bars to BS 4449 and wires to BS 4482both of which are in general conformity with BS EN 10080

are given in Table 219 For additional information on themanufacture and properties of steel reinforcement includingstainless steel refer to section 32

1031 Bars

Bars for normal use produced in the United Kingdom arehot-rolled to a characteristic strength of 500 MPa and achieveClass B or C ductility The bars are round in cross section withsets of parallel transverse ribs separated by longitudinal ribsThe nominal size is the diameter of a circle with an area equalto the effective cross-sectional area of the bar The maximumoverall size is approximately 15 greater than the nominal sizeValues of the total cross-sectional area provided in a concretesection according to the number or spacing of the bars fordifferent bar sizes are given in Table 220

The type and grade of reinforcement is designated as follows

Chapter 10

Concrete andreinforcement

Type of steel reinforcement Notation

Grade B500A B500B or B500C to BS 4449 HGrade B500A to BS 4449 AGrade B500B or B500C to BS 4449 BGrade B500C to BS 4449 CA specified grade and type of ribbed Sstainless steel to BS 6744

Reinforcement of a type not included above but Xwith material properties defined in the designor contract specification

Note In the description B500A and so on B indicates reinforcing steel

1032 Fabric

In the United Kingdom steel fabric reinforcement is generallyproduced to the requirements of BS 4483 using ribbed bar inaccordance with BS 4449 The exception is wrapping fabricwhere wire in accordance with BS 4482 may be used Fabric isproduced in a range of standard types or can be purpose-madeto the clientrsquos requirements Full details of the standard fabrictypes are given in Table 220

Type A is a square mesh with identical longitudinal barsand cross bars commonly used in ground slabs to provide aminimum amount of reinforcement in two directions Type B isa rectangular (structural) mesh that is particularly suitable for

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217Concrete cements and aggregate grading

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Concrete early-age temperatures 218

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219Reinforcement general properties

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Reinforcement cross-sectional areas of bars and fabric 220

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thin one-way spanning slabs where the longitudinal barsprovide the main reinforcement with the cross bars beingsufficient to meet the minimum requirements for secondaryreinforcement Type C is a rectangular (long) mesh where thecross bars are minimal which can be used in any solid slabincluding column bases by providing a separate sheet in eachdirection Type D is a rectangular (wrapping) mesh that is usedin the concrete encasement of structural steel sections

Standard fabric is normally supplied in stock sheet sizes of48 m (longitudinal bars) 24 m (cross bars) with end over-hangs equal to 05 the pitch of the perpendicular bar Tofacilitate fixing and to avoid a build-up of bar layers at the lapssheets with increased overhangs (flying ends) can be suppliedto order Sheets can also be supplied cut to size in lengths upto 12 m and prebent For guidance on the use of purpose-madefabrics reference should be made to BS 8666

1033 Cutting and bending tolerances

Bars are produced in stock lengths of 12 m and lengths up to18 m can be supplied to special order In most structures barsare required in shorter lengths and often need to be bent Thecutting and bending of reinforcement is generally specified tothe requirements of BS 8666 The tolerances on cutting andbending dimensions are as follows

For shape code 67 when the radius exceeds the value in thefollowing table straight bars will be supplied as the requiredcurvature can be obtained during fixing

Concrete and reinforcement100

Cutting and bending processes Tolerance (mm)

Cutting of straight lengths 25

Bending dimension (mm) 1000 5 1000 and 2000 5 10

2000 5 25

Length of wires in fabric L 5000 25L 5000 L 200

1034 Shape codes and bending dimensions

BS 8666 contains details of bar shapes designated by shapecodes as given in Tables 221 and 222 The information neededto cut and bend the bars to the required dimensions is enteredinto a bar schedule an example of which is shown in Table 223The standard shapes should be used wherever possible with therelevant dimensions entered into columns A to E of the barschedule All other shapes should be given a shape code 99 witha dimensioned sketch drawn over the columns A to E using twoparallel lines to indicate the bar thickness One of the bar dimen-sions should be indicated in parenthesis as a free dimensionDimensions should be given as a multiple of 5 mm and thetotal length determined in accordance with the equation givenin the table rounded up to a multiple of 25 mm To facilitatetransportation each bent bar should fit within an imaginaryrectangle the shorter side of which is not longer than 2750 mm

Most of the shape codes cater for bars bent to the minimumradius taken as 2d for d 16 and 35d for d 20 where d isthe bar size The minimum straight length needed beyond theend of the curved portion of a bend is 5d for a bob and 10d formost links For each bar size values of the minimum radius rand the minimum end projection P needed to form a bend aregiven in Table 219

Bars needing larger radius bends denoted by R except forshape codes 12 and 67 should be treated as a shape code 99

Maximum limit for which a preformed radius is required

Bar size (mm) 8 10 12 16

Radius (m) 275 35 425 75

Bar size (mm) 20 25 32 40

Radius (m) 14 30 43 58

For shape codes 12 13 22 and 33 the largest practical radiusfor producing a continuous curve is 200 mm and for a largerradius a series of short straight sections may be formed

1035 Deductions for variations

Cover to reinforcement is liable to variation due to the effect ofinevitable errors in the dimensions of formwork and in thecutting bending and fixing of the bars In cases where a bar isdetailed to fit between two concrete faces with no more thanthe nominal cover on each face (eg link in a beam) an appro-priate allowance for deviations should be applied The relevantdimension on the schedule should be determined as the nominaldimension of the concrete less the nominal cover on each faceless an allowance for deviations as follows

Total deductions to allow for permissible deviations onmember size and in cutting and bending of bars

Type of bar Distance between faces of Deductionconcrete member mm

Links and other Not more than 1 m 10bent bars Between 1 m and 2 m 15

Over 2 m 20

Straight bars Any length 40

The deductions recommended in the forgoing table are taken fromBS 8110 and allow for deviations on the member size of 5 mm fordimensions up to 2 m and 10 mm for dimensions over 2 m Wherethe permissible deviations on member size exceed these valueslarger deductions should be made or the cover increased

Example Determine the relevant bending dimensions for thebars shown in the following beam detail The completed barschedule for 6 beams thus is given in Table 223

Section A-A

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Reinforcement 101

Elevation of beam

Bar Shape Dimensions (mm) Length (mm) ndash see Tables 229 and 230mark code

01 00 L 8000 2 200 7600

02 00 L 8000 2 1250 5500

03 13 Bar requires radius of bend R 6d as a designrequirement This necessitates the use of shapecode 13 as dimension B would not provide 4dlength of straight between two bends L 1800 057 420 1300 16 25 3300

A 1800 C 1300 (design requirements)B 450 (2 10) 10 420

Note Dimension B is derived from dimension Aof bar mark 05 and includes a further deductionof 10 mm for tolerances on cutting and bending

04 00 L 8000 2 200 7600

05 51 A 500 (2 20) 10 450 L 2 (450 250 115) (25 16) (5 8) 1550B 300 (2 20) 10 250C D 115 (P in Table 219)r 16 (Table 219)Note Dimensions A and B include deductionsof 10 mm for permissible deviations

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221Reinforcement standard bar shapes and methodof measurement ndash 1

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222Reinforcement standard bar shapes and methodof measurement ndash 2

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Reinforcement typical bar schedule 223

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The formulae and coefficients in this chapter give values ofshearing forces bending moments slopes and deflections interms of the total load on the member For design purposes theload F must include the appropriate partial safety factors for thelimit-state being considered

111 SIMPLE BEAMS AND CANTILEVERS

The formulae for the reactions shearing forces and bendingmoments in freely supported beams (Tables 224 and 225) andsimple cantilevers (Tables 226 and 227) are obtained by therules of static equilibrium The slope and deflection formulaefor freely supported beams and simple cantilevers and all theformulae for propped cantilevers (Table 227) are for elasticbehaviour and members of constant cross section

112 BEAMS FIXED AT BOTH ENDS

The bending moments on a beam fixed at both ends can bederived from the principle that the area of the free-momentdiagram (ie the bending moment diagram due to the same loadimposed on a freely supported beam of equal span) is equal tothe area of the restraint-moment diagram Also the centresof area of the two diagrams are in the same vertical line Theshape of the free-moment diagram depends upon the particularcharacteristics of the imposed load but the restraint-momentdiagram is a trapezium For loads that are symmetrically disposedon the beam the centre of area of the free-moment diagram isat the mid-point of the span and thus the restraint-momentdiagram is a rectangle giving a restraint moment at eachsupport equal to the mean height of the free-moment diagram

The amount of shearing force in a beam with one or bothends fixed is calculated from the variation of the bendingmoment along the beam The shearing force resulting from therestraint moment alone is constant throughout the length of thebeam and equal to the difference between the two end-momentsdivided by the span (ie the rate of change of the restraintmoment) This shearing force is algebraically added to theshearing force due to the imposed load with the beam takenas freely supported Thus the support reaction is the sum (ordifference) of the restraint-moment shearing force and the free-moment shearing force For a beam that is symmetricallyloaded with both ends fixed the restraint moment at each endis the same and the shearing forces are identical to those for thesame beam freely supported The support reactions are bothequal to one-half of the total load on the span The formulae forthe reactions shearing forces bending moments slopes anddeflections for fully fixed spans (Table 225) are for elasticbehaviour and members of constant cross section

1121 Fixed-end moment coefficients

Fixed-end moment coefficients CAB and CBA are given inTable 228 for a variety of unsymmetrical and symmetricalimposed loadings on beams of constant cross section Morecomplex loading arrangements can generally be formed as acombination of the cases shown and the resulting fixed-endmoments found by superposition A full range of charts iscontained in Examples of the Design of Reinforced ConcreteBuildings for a member with a partial uniform or triangulardistribution of load placed anywhere within the span

Chapter 11

Cantilevers andsingle-span beams

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224Moments shears deflections general case for beams

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Moments shears deflections special cases for beams 225

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226Moments shears deflections general cases for cantilevers

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Moments shears deflections special cases for cantilevers 227

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228Fixed-end moment coefficients general data

The fixed-end moment coefficients CAB and CBA can be used as follows1 To obtain bending moments at supports of single-span beams fully fixed at both ends (Table 225)

MAB CABlAB and MBA CBAlAB (With symmetrical load MAB MBA)2 To obtain fixed-end moments for analysis of continuous beams by moment distribution methods (Table 236)

FEMAB CABlAB and FEMBA CBAlAB (With symmetrical load FEMAB FEMBA)3 To obtain loading factors for analysis of framed structures by slope-deflection methods (Table 260)

FAB CABlAB and FBA CBAlAB (With symmetrical load FAB FBA)4 To obtain loading factors for analysis of portal frames (Tables 263 and 264)

and z1 (With symmetrical loading CAB CBA and z1 05)=CAB 2CBA

2(CAB 2CBA)D

CAB CBA

2lAB

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The formulae and coefficients in this chapter give values ofshearing forces and bending moments in terms of the dead andlive loads on the member For design purposes these loads mustinclude the appropriate partial safety factors for the limit-stateconsidered and the Code of Practice employed

For the ULS the dead load factors are treated differently inBS 8110 and EC 2 For designs to BS 8110 values of either 14or 10 are applied separately to each span of the beam Fordesigns to EC 2 values of either 135 or 10 are applied to all thespans If the beam ends with a cantilever the effect of applyingvalues of either 135 or 115 separately to the cantilever and theadjacent span should also be considered Details of the designloads and of the effects of applying cantilever moments at oneor both ends of a continuous beam of two three four or fiveequal spans are given in Table 229

121 DETERMINATION OF MAXIMUM MOMENTS

1211 Incidence of live load

The values of the bending moments in the spans and at thesupports depend upon the incidence of the live load and forspans that are equal or approximately equal the dispositions oflive load shown in Table 229 give the maximum positivemoments in the spans and the maximum negative momentsat the supports Both BS 8110 and EC 2 consider a lesssevere incidence of live load when determining the maximumnegative moments at the supports According to BS 8110 theonly case that needs to be considered is when all the spans areloaded According to EC 2 all cases of two adjacent spansloaded should be considered but the loads shown as optionalin Table 229 may be ignored

Thus the maximum positive moments due to live load forthe system are obtained by considering two loading arrange-ments one with live load on all the odd-numbered spans andthe other with live load on all the even-numbered spans Fordesigns to BS 8110 the summation of the results for these twocases gives the maximum negative moments

It should be noted that for designs to EC 2 the UK NationalAnnex allows the BS 8110 loading arrangements to be used asan alternative to those recommended in the base document Inthis chapter the basic arrangements are used

1212 Shearing forces

The shearing forces in a continuous beam are determined byfirst considering each span as freely supported then addingalgebraically the rate of change of restraint moment for theparticular span Shearing forces for freely supported spans arereadily determined by the rules of static equilibrium The addi-tional shearing force which is constant throughout the span isequal to the difference in the support moments at each enddivided by the span

1213 Maximum positive moments

When the moments at the supports and the shearing forces havebeen determined the maximum positive moment in the spancan be obtained by first finding the position where the shearingforce is zero The maximum positive moment is then obtainedby subtracting the effect of the restraint moments which varieslinearly along the span from the freely supported moment atthis position

122 SOLUTIONS FOR EQUAL SPANS

1221 Coefficients for equal loads on equal spans

Approximate general solutions for the maximum bendingmoments and shearing forces in uniformly loaded beams ofthree or more spans are given in Table 229 Exact solutions forthe maximum bending moments in beams of two three fouror five equal spans are given in Tables 230 and 231 for eightdifferent load distributions The coefficients given for thesupport moments due to live load apply to the most onerousloading conditions For the less severe arrangements describedin section 1211 coefficients are shown in the square brackets [ ]for BS 8110 and the curved brackets ( ) for EC 2 The coefficientsin Table 232 enable the maximum shearing forces at thesupports to be determined

Example 1 Calculate the maximum ultimate moments inthe end and central spans and at the penultimate and interiorsupports for a beam continuous over five equal spans of 5 mwith characteristic dead and imposed loads of 20 kNm eachaccording to the requirements of BS 8110

Chapter 12

Continuous beams

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229Continuous beams general data

AKey B K A B J K

A B C H J KA B C

Adjustment to bending moment = M coefficient x applied bending momentAdjustment to shearing force = (V coefficent x applied bending moment)span

KJ

Optional

EC2 Consider live load on spans RS and ST only

Optional

R T U V

TS

S

To produce maximum positive moment in span ST

To produce maximum negative moment at support S

Simplifications BS8110 Consider live load on all spans

EC 2 EC 2

EC 2

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230Continuous beams moments from equal loadson equal spans ndash 1

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231Continuous beams moments from equal loadson equal spans ndash 2

Bending moment (coefficient) (total load on one span) (span)Bending moment coefficientsabove line apply to negative bending moment at supportsbelow line apply to positive bending moment in span

Coefficients apply when all spans are equal (may be used also whenshortest 85 longest) Loads on each loaded span are same

Second moment of area is same throughout all spansBending moment coefficients in square brackets (live load)apply if all spans are loaded (ie BS 8110 requirements)Bending moment coefficients in curved brackets (live load)apply if two adjacent spans are loaded (ie EC 2 requirements)

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232Continuous beams shears from equal loadson equal spans

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The design load consists of a dead load of 10gk 20 kNm anda live load of (04gk 16qk) 40 kNm Then from Table 230(using coefficients in square brackets for the live load) theultimate bending moments are as follows

Penultimate supportDead load 0105 20 52 525 kNm (negative)Imposed load 0105 40 52 1050 kNm (negative)Total 1575 kNm (negative)

Interior supportDead load 0079 20 52 395 kNm (negative)Imposed load 0079 40 52 790 kNm (negative)Total 1185 kNm (negative)

Near middle of end spanDead load 0078 20 52 390 kNm (positive)Imposed load 0100 40 52 1000 kNm (positive)Total 1390 kNm (positive)

Middle of central spanDead load 0046 20 52 230 kNm (positive)Imposed load 0086 40 52 860 kNm (positive)Total 1090 kNm (positive)

Example 2 Calculate the maximum ultimate moments forexample 1 according to the requirements of EC 2

The design load consists of a dead load of 135gk 27 kNmand a live load of 15qk 30 kNm Then from Table 230(using coefficients in curved brackets for the live load) theultimate bending moments are as follows

Penultimate supportDead load 0105 27 52 709 kNm (negative)Imposed load 0116 30 52 870 kNm (negative)Total 1579 kNm (negative)

Interior supportDead load 0079 27 52 533 kNm (negative)Imposed load 0106 30 52 795 kNm (negative)Total 1328 kNm (negative)

Near middle of end spanDead load 0078 27 52 527 kNm (positive)Imposed load 0100 30 52 750 kNm (positive)Total 1277 kNm (positive)

Middle of central spanDead load 0046 27 52 311 kNm (positive)Imposed load 0086 30 52 645 kNm (positive)Total 956 kNm (positive)

Example 3 Calculate the maximum ultimate moments forexample 1 according to the requirements of BS 8110 when a2 m long cantilever is provided at each end of the beamIncrease in moments due to dead and live loads on cantileversat critical positions is as follows

End supportDead load 05 20 22 400 kNm (negative)Imposed load 05 40 22 800 kNm (negative)Total 1200 kNm (negative)

Interior supportFrom Table 229 increase due to moments at end supports is0053 120 64 kNm (negative)

Decrease in moments due to dead load only on cantilevers atcritical positions is as follows

Penultimate supportFrom Table 229 decrease due to moments at end supports is0263 40 105 kNm (positive)

Middle of end spanFrom Table 229 decrease due to moments at end supports is[10 05(10 0263)] 40 147 (negative)

Middle of central spanFrom Table 229 decrease due to moments at end supports is0053 40 21 kNm (negative)

123 REDISTRIBUTION OF MOMENTS

As explained in section 422 for the ULS both BS 8110 andEC 2 permit the moments determined by a linear elastic analysisto be redistributed provided that the resulting distributionremains in equilibrium with the loads Although the conditionsaffecting the procedure are slightly different in the two codesthe general approach is to reduce the critical moments by achosen amount up to the maximum percentage permitted anddetermine the revised moments at other positions by equilibriumconsiderations

An important point to appreciate is that each particular loadcombination can be considered separately Thus if desired it ispossible to reduce the maximum moments in the spans and atthe supports For example the maximum support moments canbe reduced to values that are still greater than those that occurwith the maximum span moments The maximum spanmoments can then be reduced until the corresponding supportmoments are the same as the (reduced) maximum values

The principles of static equilibrium require that no changesshould be made to the moments in a cantilever or at a freelysupported end

1231 Code requirements

BS 8110 and EC 2 permit the maximum moments to be reducedby up to 30 provided that in the subsequent design of the rele-vant sections the depth of the neutral axis is limited accordingto the amount of redistribution (Note that there is no restrictionon the maximum percentage increase of moment) In EC 2 themaximum permitted reduction depends also on the ductilityof the reinforcement being 30 for reinforcement classes Band C but only 20 for class A

In BS 8110 it is stated that the ultimate resistance momentat a section should be at least 70 of the maximum momentat that section before redistribution In effect the process ofredistribution alters the positions of points of contra-flexureThe purpose of the code requirement is to ensure that at suchpoints on the diagram of redistributed moments (at which noflexural reinforcement is theoretically required) sufficientreinforcement is provided to cater for the moments that willoccur under service loading The redistribution procedure takesadvantage of the ability of continuous beams to develop plastichinges at critical sections prior to failure whilst also ensuringthat the response remains fully elastic under service loading Therequirements are discussed more fully in books on structuraldesign and in the Handbook to BS 8110

Continuous beams116

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1232 Redistribution procedure

The use of moment redistribution is illustrated in Table 233where a beam of three equal spans is examined in accordancewith the requirements of BS 8110 The uniformly distributeddead and live loads are each equal to 1200 units per span Themoment diagram for dead load on each span is shown in (a)Moment diagrams for the arrangements of live load that givethe maximum moments at the supports and in the spans areshown in (b) The moment envelope obtained by combiningthe diagrams for dead and live loads is shown in (c) (Notethat the vertical scale of diagrams (c)ndash(f) differs from thatof (a) and (b))

The redistribution procedure is normally used to reduce themaximum support moments One approach is to reduce these tothe values obtained when the span moments are greatest Thisis shown in (d) where the support moments have been reducedfrom 240 to 180 a reduction of 25 In this case no otheradjustment is needed to the moment envelope If the maximumsupport moments are reduced by 30 from 240 to 168 thespan moments must be increased as shown in (e) The 70requirement discussed in section 1132 determines the extentof the hogging region in the end span and the minimum valueof 21 in the middle span (For the load cases in EC2 themaximum support moments could be reduced by 30 from260 to 182 with no other adjustment needed to themoment envelope)

If the criterion is to reduce the maximum span moments inthe case of an up-stand beam say this may be achieved byincreasing the corresponding support moments This is shownin (f) where the maximum moment in the middle span has beenreduced by 30 from 120 to 84 by increasing the supportmoments from 180 to 216 (Note that the minimummoment in the middle span has increased from 30 to 66)The moment in the end span has also been reduced by 7 from217 to 202 The 70 requirement determines the extent of thesagging regions in both spans It is clear that any further reduc-tion of moment in the end span would result in a considerableincrease in the support moment For example a 30 reductionin the end span moment from 217 to 152 would increase thesupport moment to 346 and the minimum moment in themiddle span to 196

In view of the many factors involved it is difficult to give anygeneral rules as to whether to redistribute moments or by howmuch such decisions are basically matters of individual engi-neering judgement A useful approach is to first calculate theultimate resistance moments at the support sections provided bychosen arrangements of reinforcement and then redistribute themoment diagrams to suit The span sections can then be designedfor the resulting moments and a check made to ensure that allof the code requirements are satisfied Moment redistributionin general affects the shearing forces at the supports and it isrecommended that beam sections are designed for the greater ofthe shear forces calculated before and after redistribution

The use of moment distribution in systems where the beamsare analysed in conjunction with adjoining columns requiresfurther consideration In such cases it is important to ensure thatin any postulated collapse mechanism involving plastic hinges inthe columns these are the last hinges to form To this end it isrecommended that column sections should be designed for thegreater of the moments calculated before and after redistribution

1233 Bending moment diagrams

The moment diagrams and coefficients given in Tables 234and 235 cater for beams that are continuous over two threeand four or more equal spans They apply to cases where thesecond moment of area of the cross section is constant andthe loads on each loaded span are the same For conveniencecoefficients derived by elastic analysis before and after givenredistributions in accordance with the rules of both BS 8110and EC 2 are tabulated against the location points indicated inthe diagrams For example M12 is the coefficient correspondingto the maximum moment at the central support of a two-spanbeam while M13 is the coefficient that gives the moment at thissupport when the moment in the adjoining span is a maximumThus by means of the coefficients given the appropriate envelopeof maximum moments is obtained

Three load types are considered UDL throughout each spana central concentrated load and equal concentrated loadspositioned at the third-points of the span The span momentsdetermined by summing the individual maximum values givenseparately for dead and live loads in the case of uniformloading will be approximate but erring on the side of safetysince each maximum value occurs at a slightly different positionThe tabulated coefficients may also be used to determine thesupport moments resulting from combinations of the givenload types by summing the results for each type The corre-sponding span moments can then be determined as described insection 1213

Moment coefficients are given for redistribution values of10 and 30 respectively For the dead load all the supportmoments have been reduced by the full amount and the spanmoments increased to suit the adjusted values at the supportsFor the live loads all the support moments and for 10 redis-tribution all the span moments have been reduced by the fullamount For 30 redistribution each span moment has beenreduced to the minimum value required for equilibrium with thenew support moments As a result the BS 8110 span momentcoefficients are the same as those for the dead load Althoughthere is no particular merit in limiting redistribution to 10some BS 8110 design formulae for determining the ultimateresistance moment are related to this condition

For design purposes redistribution at a particular sectionrefers to the percentage change in the combined moment dueto the dead and live loads When using the tables the value forthe support moments will be either 10 or 30 but the valuesfor the span moments will need to be calculated for eachparticular case Consider for example a two-span beam sup-porting UDLs with gk qk and 30 redistribution accordingto the requirements of BS 8110

The design load consists of a dead load of 10gk and a liveload of (04gk 16qk) 20gk Then from Table 234 theultimate bending moments are as follows

Before redistributionM11 (0070gk 0096 2gk)l2 0262gkl2

After redistributionM11 (0085gk 0085 2gk)l2 0255gk l2

Redistribution 100 (0262 0255)0262 3

Thus a full 30 reduction of the maximum support moments isobtained with no increase in the maximum span moment

Redistribution of moments 117

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233Continuous beams moment redistribution

Uniformly distributed load (dead load = live load = 1200 units per span)

(b) Live load only Critical cases (for BS 8110)

(a) Dead load only

(c) Dead + live loads Critical cases

(d) Envelope obtained with some reduction of the support moments but with no increase of the span moments

(e) Envelope obtained with maximum reduction of the support moments and some increase of the span moments

(f) Envelope obtained with maximum reduction of moment in middle span and some reduction of moment in end spans

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234Continuous beams bending moment diagrams ndash 1

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235Continuous beams bending moment diagrams ndash 2

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124 ANALYSIS BY MOMENT DISTRIBUTION

The Hardy Cross moment distribution method of analysisin which support moments are derived by a step-by-stepprocess of successive approximations is described brieflyand shown by means of a worked example in Table 236 Themethod is able to accommodate span-to-span variationsin span length member size and loading arrangement Thelsquoprecise moment distributionrsquo method avoids the iterativeprocedure by using more complicated distribution and carry-over factors Span moments can be determined as describedin section 1213

For continuous beams of two three or four spans uniformcross-section and symmetrical loading the support momentsmay also be obtained by using the factors in Table 237

125 INFLUENCE LINES FOR CONTINUOUS BEAMS

The following procedure can be used to determine bendingmoments at chosen sections in a system of continuous beamsdue to a train of loads in any given position

1 Draw the beam system to a convenient scale

2 With the ordinates tabulated in the appropriate Table 238239 240 or 241 construct the influence line (for unit load)for the section being considered selecting a convenientscale for the bending moment

3 Plot on the influence line diagram the train of loads in whatis considered to be the most adverse position

4 Tabulate the value of (ordinate load) for each load

5 Add algebraically the values of (ordinate load) to obtainthe resultant bending moment at the section considered

6 Repeat for other positions of the load train to ensure that themost adverse position has been considered

The following example shows the direct use of the tabulatedinfluence lines for calculating the moments on a beam that iscontinuous over four spans with concentrated loads applied atspecified positions

Example Determine the bending moments at the penultimateleft-hand support of a system of four spans having a constantcross section and freely supported at the ends when loads of100 kN are applied at the mid-points of the first and third spansfrom the left-hand end The end spans are 8 m long and theinterior spans are 12 m long

The span ratio is 115151 and the ordinates are obtained fromTable 240 for penultimate support C

With load on first span (ordinate c)Bending moment (0082 100 8) 656 kNm

With load on third span (ordinate m)Bending moment (0035 100 8) 280 kNm

Net bending moment at penultimate support 376 kNm

Influence lines for continuous beams 121

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236Continuous beams moment distribution methods

HARDY CROSS MOMENT DISTRIBUTION

1 Consider each member to be fixed at ends calculate fixed-endmoments (FEMs) due to external loads on individual members bymeans of Table 228

2 Where members meet sum of bending moments must equalzero for equilibrium ie at B MBAMBC 0 Since FEM(ie FEMBA FEMBC) is unlikely to equal zero a balancingmoment ofFEM must be introduced at each support to achieveequilibrium

3 Distribute this balancing moment between members meeting ata joint in proportion to their relative stiffnesses K Il bymultiplying FEM by distribution factor D for each member(eg at B DBA KAB (KAB KBC) etc so that DBA DBC 1At a free end D 1 at a fully fixed end D0)

4 Applying a moment at one end of member induces momentof one-half of magnitude and of same sign at opposite endof member (termed carry-over) Thus distributed moment

FEM DBA at B of AB produces a moment of (12)FEM DBA at A and so on

5 These carried-over moments produce further unbalanced momentsat supports (eg moments carried over from A and C give rise tofurther moments at B) These must again be redistributed and thecarry-over process repeated

6 Repeat cycle of operations described in steps 2ndash5 until unbalancedmoments are negligible Then sum values obtained each sideof support

Various simplifications can be employed to shorten analysis Themost useful is that for dealing with a system that is freely supportedat the end If stiffness considered for end span when calculatingdistribution factors is taken as only three-quarters of actualstiffness and one-half of fixed-end moment at free support is addedto FEM at other end of span the span may then be treated as fixedand no further carrying over from free end back to penultimatesupport takes place

PRECISE MOMENT DISTRIBUTION

1 Calculate fixed-end moments (FEMs) as for Hardy Cross momentdistribution

2 Determine continuity factors for each span of system fromgeneral expression

where n is continuity factor for previous span and Kn and Kn1

are stiffnesses of two spans Work from left to right along system Ifleft-hand support (A in example below) is free take AB 0 for firstspan if A is fully fixed AB 05 (Intermediate fixity conditionsmay be assumed if desired by interpolation) Repeat the foregoingprocedure starting from right-hand end and working to left (to obtaincontinuity factor AB for span AB for example)

3 Calculate distribution factors (DFs) at junctions between spansfrom general expression

n1 12 Kn1

Kn(2n)

where AB and BA are continuity factors obtained in step 2Note that these distribution factors do not correspond to thoseused in Hardy Cross moment distribution Check that at eachsupport DF 1

4 Distribute the balancing moments FEM introduced at eachsupport to provide equilibrium for the unbalanced FEMs bymultiplying by the distribution factors obtained in step 3

5 Carry over the distributed balancing moments at the supportsby multiplying them by the continuity factors obtained instep 2 by working in opposite direction For example the momentcarried over from B to A is obtained by multiplying thedistributed moment at B by AB and so on This procedure isillustrated in example below Only a single carry-over operationin each direction is necessary

6 Sum values obtained to determine final moments

DFAB 12AB

1ABBA

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237Continuous beams unequal prismatic spans and loads

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238Continuous beams influence lines for two spans

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239Continuous beams influence lines for three spans

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240Continuous beams influence lines for four spans

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241Continuous beams influence lines for five or more spans

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Position Moment Shear

At outer support 0 040FNear middle of end span 008Fl mdashAt interior support 010Fl 060F

In monolithic building construction concrete floors can takevarious forms as shown in Table 242 Slabs can be solid orribbed and can span between beams in either one or two direc-tions or be supported directly by columns as a flat slab Slabelements occur also as decking in bridges and other forms ofplatform structures and as walling in rectangular tanks silosand other forms of retaining structures

131 ONE-WAY SLABS

For slabs carrying uniformly distributed load and continuousover three or more nearly equal spans approximate solutionsfor the ultimate bending moments and shearing forces for bothBS 8110 and EC 2 are given in Table 242 The supportmoments include an allowance for 20 redistribution in bothcases The differences in the values for the two codes occur asa result of the different load arrangements described in section441 However it should be noted that for designs to EC 2 theUK National Annex allows the use of the BS 8110 simplifiedload arrangement as an alternative to that recommended in thebase document For two equal spans the corresponding valuesfor both codes would be

serviceability requirements of cracking and deflection are metby compliance with simplified rules

1321 Uniformly loaded slabs (BS 8110 method)

For rectangular panels carrying uniformly distributed loadwhere the corners are prevented from lifting and adequateprovision is made for torsion the panel is consideredto be divided into middle and edge strips as shown inTable 242 The method may be used for continuous slabswhere the characteristic dead and imposed loads on adjacentpanels and the spans perpendicular to the lines of commonsupport are approximately the same as on the panel beingconsidered

The bending moments and shearing forces on the middlestrips for nine different panel types are given in Table 243Reinforcement meeting the minimum percentage requirementof the code should be provided in the edge strips At cornerswhere either one or both edges of the panel are discontinuoustorsion reinforcement is required This should consist of top andbottom reinforcement each with layers of bars placed parallelto the sides of the slab and extending from the edges a minimumdistance of one-fifth of the shorter span The area of reinforce-ment in each of the four layers as a proportion of that requiredfor the positive moment at mid-span should be three-quarterswhere both edges are discontinuous and three-eighths whereone edge is discontinuous At a discontinuous edge where theslab is monolithic with the support negative reinforcementequal to a half of that required for the positive moment atmid-span should be provided

Where because of differences between contiguous panelstwo different values are obtained for the negative moment at ashared continuous edge these values may be considered asfixed-end moments and moment distribution used to obtainequilibrium in the direction of span The revised negativemoments can then be used to adjust the positive moments atmid-span For each panel the sum of the mid-span moment andthe average of the support moments should be the same as theoriginal sum for that particular panel

When the long span exceeds twice the short span the slabshould be designed as spanning in the short direction In thelong direction the long span coefficient may still be used forthe negative moment at a continuous edge

Chapter 13

Slabs

For designs where elastic bending moments are required thecoefficients given for beams in Table 229 should be used

132 TWO-WAY SLABS

Various methods based on elastic or collapse considerationsare used to design slabs spanning in two directions Elasticmethods are appropriate if for example serviceability checkson crack widths are required as in the design of bridges andliquid-retaining structures Collapse methods are appropriatein cases such as floors in buildings and similar structureswhere the main criterion is the ultimate condition and the

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242Slabs general data

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243Two-way slabs uniformly loaded rectangular panels(BS 8110 method)

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Example Determine the bending moment coefficients in theshort span direction for the slab panel layout shown as follows

by finite element analysis with Poissonrsquos ratio taken as zerousing the following approximate relationships

Bending moments x x0 y0 y y0 x0

Torsion moments xy (1 )xy0

where is Poissonrsquos ratio and x0 y0 xy0 are coefficientscorresponding to 0 Thus if required the tabulated valuescan be readjusted to suit a Poissonrsquos ratio of zero as follows

Bending moments x0 104 (x02 02 y02)y0 104 (y02 02 x02)

Torsion moments xy0 125 xy02

For rectangular panels simply supported on four sides with noprovision to resist torsion at the corners or to prevent the cor-ners from lifting coefficients taken from BS 8110 are alsogiven in Table 244 The coefficients which are derived fromthe Grashof and Rankine formulae (see section 453) are givenby the following expressions

1323 Non-rectangular panels

For a non-rectangular panel supported along all of its edgesbending moments can be determined approximately from thedata given in Table 248 The information which is based onelastic analysis is applicable to panels that are trapezoidaltriangular polygonal or circular For guidance on using thisinformation including the arrangement of the reinforcementreference should be made to section 47

133 CONCENTRATED LOADS

1331 One-way slabs

For a slab simply supported along two opposite edges andcarrying a centrally placed load uniformly distributed over adefined area maximum elastic bending moments are given inTable 245 The coefficients which include for a Poissonrsquos ratioof 02 have been calculated from the data derived for arectangular panel infinitely long in one direction For designs toBS 8110 in which the ULS requirement is the main criteriona concentrated load placed in any position may be spread overa strip of effective width be as shown in Table 245 Parallel tothe supports a strip of width (x ay2) equally spaced eachside of the load has been considered

For slabs that are restrained at one or both edges maximumnegative and positive bending moments may be obtained bymultiplying the simply supported moment by the appropriatefactors given in Table 245 The factors which are given forboth fixed and continuous conditions are those appropriate toelastic beam behaviour

1332 Two-way slabs

For a rectangular panel freely supported along all four edges andcarrying a concentric load uniformly distributed over a definedarea maximum mid-span bending moments based on Pigeaudrsquostheory are given in Tables 246 and 247 Moment coefficients

my (ly lx)

2

8[1 (ly lx)4]

mx (ly lx)

4

8[1 (ly lx)4]

Concentrated loads 131

The upper half of the layout shows the coefficients obtainedfrom Table 243 with ly lx 9060 15 for panel types

AndashB (and CndashD) two adjacent edges discontinuousBndashC one short edge discontinuous

The lower half of the layout shows the coefficients obtained afterdistribution of the unbalanced support moments at B and CThe effective stiffnesses allowing for the effects of simple sup-ports at A and D and carry-over moments at B and C are

Span AndashB (and CndashD) 075Il Span BndashC 05Il

Distribution factors at B and C with no carry-overs are

BA (and CD) 075(075 050) 06BC (and CB) (10 06) 04

Moment coefficients at B and C after distribution are

0078 06 (0078 0058) 0066

Moment coefficients at mid-span after redistribution are

BA (and CD) 0059 05 (0078 0066) 0065BC 0043 (0066 0058) 0035

1322 Uniformly loaded slabs (elastic analysis)

For rectangular panels carrying uniformly distributed loadwhere the corners are prevented from lifting and adequateprovision is made for torsion maximum bending and torsionmoments are given in Table 244 for nine panel types Wherein continuous slabs the edge conditions in contiguous panelsresult in two different values being obtained for the negativemoment at a fixed edge the moment distribution procedureshown in section 1321 could be used but this would ignore theinter-dependence of the moments in the two directions A some-what complex procedure involving edge stiffness factors isderived and shown with fully worked examples in ref 21

The coefficients include for a Poissonrsquos ratio of 02 and havebeen calculated from data given in ref 21 which was derived

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244Two-way slabs uniformly loaded rectangular panels(elastic analysis)

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245One-way slabs concentrated loads

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246Two-way slabs rectangular panel with concentricconcentrated load ndash 1

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247Two-way slabs rectangular panel with concentricconcentrated load ndash 2

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248Two-way slabs non-rectangular panels (elastic analysis)

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given in the charts are used with an appropriate value of Poissonrsquosratio to calculate the bending moments The coefficients given atthe top right corner of each chart are for the limiting case whenthe load extends over the entire panel This case with Poissonrsquosratio taken as 02 is given also in Table 244

When using the chart for square panels (k 10) if ax ayx4 y4 and the resulting bending moment in each direction isgiven by F (1 )x4 In other cases coefficient x4 is based onthe direction chosen for ax and coefficient y4 is obtained byreversing ax and ay as shown in example 1 later

The maximum shearing forces V per unit length on a panelcarrying a concentrated load are given by Pigeaud as follows

ax ay at the centre of length ax V F(2ax ay)at the centre of length ay V F3ax

ay ax at the centre of length ax V F3ay

at the centre of length ay V F(2ay ax)

For panels that are restrained along all four edges Pigeaudrecommends that the mid-span moments be reduced by 20Alternatively the multipliers given for one-way slabs couldbe used in one or both directions as appropriate if the inter-dependence of the bending moments is ignored

Example 1 Consider a square panel freely supported alongall four edges carrying a concentric load with axlx 08 andayly 02 From Table 247 for k 1 the bending momentcoefficients are

For axlx 08 and ayly 02 x4 0072For axlx 02 and ayly 08 y4 0103

Maximum bending moments per unit width with 02 are

For span in direction of ax F(0072 02 0103) 0093FFor span in direction of ay F(0103 02 0072) 0118F

Example 2 A bridge deck is formed of a 200 mm thick slabsupported by longitudinal beams spaced at 2 m centres Theslab is covered with 100 mm thick surfacing Determine for theSLS the maximum positive bending moments in the slab due tothe local effects of live loading

The critical live load for serviceability is the HA wheel loadof 100 kN to which a partial load factor of 12 is applied Fora 100 mm 100 mm contact area and allowing for loaddispersal through the thickness of the surfacing and down tothe mid-depth of the slab (see section 249) the side of theresulting patch load is (300 100 200) 600 mm

The simply supported bending moment coefficients for acentrally placed load by interpolation from Table 245 are

For axlx aylx 6002000 03 mx 0206 my 0145

Allowing for continuity (interior span) in the direction of lx andapplying a partial load factor of 12 the positive bendingmoments per unit width are

In direction of lx

mx 064 0206 12 100 158 kNmm

At right angles to lx

my 0145 12 100 174 kNmm

For design purposes the bending moments determined earlierwill need to be combined with the moments due to the weightof the slab and surfacing and any transverse effects of theglobal deck analysis

Note Using the method given in BS 8110 with x 05lx theeffective width of the strip is

Allowing for continuity (interior span) in the direction of lx andapplying a partial load factor of 12 the positive bendingmoments per unit width are

In direction of lx

At right angles to lx

134 YIELD-LINE ANALYSIS

As stated in section 452 yield-line theory is too complex asubject to deal with adequately in the space available in thisHandbook The following notes are therefore intended merelyto introduce the designer to the basic concepts methods andproblems involved For further information see refs 23 to 28Application of yield-line theory to the design of rectangularslabs subjected to triangularly distributed loads is dealt with insection 1362

1341 Basic principles

When a reinforced concrete slab is loaded cracks form inthe regions where the applied moment exceeds the crackingresistance of the concrete As the load is increased beyond theservice value the concrete continues to crack eventually thereinforcement yields and the cracks extend to the corners ofthe slab dividing it into several areas separated by so-calledyield-lines as shown in diagrams (i)ndash(iii) on Table 249 Anyfurther increase in load will cause the slab to collapse Inthe design process the load corresponding to the formation ofthe entire system of yield lines is calculated and by applyingsuitable partial factors of safety the resistance moment thatmust be provided to avoid collapse is determined

For a slab of given shape it is usually possible to postulatedifferent modes of failure the critical mode depending on thesupport conditions the panel dimension and the relative propor-tions of reinforcement provided in each direction For exampleif the slab shown in diagram (i)(a) is reinforced sufficientlystrongly in the direction of the shorter span by comparison withthe longer span this mode of failure will be prevented and thatshown in diagram (i)(b) will occur instead Similarly if the slabwith one edge unsupported shown in diagrams (ii) is loadeduniformly pattern (b) will occur when the ratio of the longer toshorter side length (or longer to shorter lsquoreduced side lengthrsquo seesection 1346) exceeds radic2 otherwise pattern (a) will occur

All of these patterns may be modified by the formation ofcorner levers see section 1349

my 12 100 1842 10 06 1

0618 139kNmm

mx 064 12 100 10

2 18 1 06

2 20 181kNmm

be 06lx ay 06 20 06 18 m

Yield-line analysis 137

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249Two-way slabs yield-line theory general information

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1342 Rules for postulating yield-line patterns

Viable yield-line patterns must comply with the following rules

1 All yield lines must be straight

2 A yield line can only change direction at an intersection withanother yield-line

3 A yield line separating two elements of a slab must passthrough the intersection of their axes of rotation (Note thispoint may be at infinity)

4 All reinforcement intercepted by a yield line is assumed toyield at the line

1343 Methods of analysis

Two basic methods of analysis have been developed These arecommonly referred to as the lsquoworkrsquo or lsquovirtual workrsquo methodand the lsquoequilibriumrsquo method The former method involvesequating for the yield-line pattern postulated the work done bythe external loads on the various areas of the slab to obtain avirtual displacement to the work done by the internal forces informing the yield lines When the yield-line pattern is adjustedto its critical dimensions the ratio of the ultimate resistance tothe ultimate load reaches its maximum When analysing a slabalgebraically this situation can be ascertained by differentiatingthe expression representing the ratio and equating the differentialto zero in order to establish the critical dimensions Then byre-substituting these values into the original expression a formulagiving the required ultimate resistance for a slab of givendimensions and loading can be derived

The so-called equilibrium method is not a true equilibriummethod but a variant of the work method which also gives anupper-bound solution The method has the great advantagethat the resulting equations provide sufficient information ofthemselves to eliminate the unknown variables and thereforedifferentiation is unnecessary Although there are also otheradvantages the method is generally more limited in scope andis not described here for details see refs 23 and 27

1344 Virtual-work method

As explained earlier this method consists of equating thevirtual work done by the external loads in producing a givenvirtual displacement at some point on the slab to the work doneby the internal forces along the yield lines in rotating the slabelements To demonstrate the principles involved an analysiswill be given of the freely supported rectangular slab supportinga uniform load and reinforced to resist equal moments M eachway shown in diagram (i)(a) on Table 249

Clearly due to symmetry yield line OO will be midwaybetween AB and CD Similarly and thus only onedimension is unknown Consider first the external work done

The work done by an external load on an individual slabelement is equal to the area of the element times the displace-ment of its centroid times the unit load Thus for the triangularelement ADO with displacement at O

work done (12)(lx)(ly)(3)n lxlyn6

Similarly for the trapezoidal area ABOO with displacement at O and O

work done [(lx2)(ly)( 2) 2(12)(lx2)(ly)(23)]

(3 4)lxlyn12

Thus since the work done on BCO is the same as that doneon ADO and the work done on CDOO is the same as that onABOO for the entire slab

Total external work done 2[lxlyn6 (3 4)lxlyn12]

(3 2)lxlyn6

The internal work done in forming a yield line is equal to themoment along the yield line times the length of the line timesthe rotation A useful point to note is that where a yield line isformed at an angle to the direction of principal momentsinstead of considering its true length and rotation it is usuallysimpler to consider the components in the direction of theprincipal moments For example for yield line AO instead ofconsidering the actual length AO and the rotation at right anglesto AO consider length lx2 and the rotation about AB pluslength ly and the rotation about AD

Thus considering the component about AB of the yield linealong AOO B the length of the line is ly the moment is M andthe rotation is ( lx2) Hence

work done 2M(lylx)

Similarly for the yield line along DOO C the work done isagain 2M(lylx) (Length OO is considered twice because therotation between the elements separated by this length is dou-ble that occurring over the remaining length) Now consideringthe component about AD of the yield line AOD

work done M(lxly)

Since the work done on yield line BOC is similar for theentire slab

Total internal work done 2M(2lylx lxly)

Equating the external work done to the internal work done

(3 2)lxlyn6 2M(2lylx lxly)

or

To determine the critical value of ly the quotient in squarebrackets must be differentiated and equated to zero As Jones(ref 27) has pointed out to use the well-known relationship

simply as a means of maximising y uv it is convenient torearrange it in the form

Thus for the present example

This leads to a quadratic in the positive root of which is

12lx

ly4

3lx

ly2

lx

ly2

3 22

2 (lxly)2

3ndash 4

2

uv

dudxdvdx

dydx

v dudx

u dvdx v2 0

M n

12l 2

x 3 22

2 (lx ly)2

Yield-line analysis 139

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When substituted in the original equation this gives

1345 Concentrated and line loads

Concentrated and line loads are simpler to deal with thanuniform loads When considering the external work done thecontribution of a concentrated load is equal to the load timesthe relative deflection of the point at which it is applied In thecase of a line load the external work done over a given slab areais equal to the portion of the load carried on that area times therelative deflection at the centroid of the load

Yield lines tend to pass beneath heavy concentrated or lineloads since this maximises the external work done by suchloads When a concentrated load acts in isolation a so-calledcircular fan of yield lines tends to form this behaviour iscomplex and reference should be made to specialist textbooksfor details (refs 23 24 26)

1346 Affinity theorems

Section 1344 illustrates the work involved in analysing asimple freely supported slab with equal reinforcement in eachdirection (ie so-called isotropic reinforcement) If differentamounts of reinforcement are provided in each direction(ie so-called orthotropic reinforcement) or if continuity orfixity exists along one or more edges the formula needsmodifying accordingly

To avoid the need for a vast number of design formulae tocover all conceivable conditions it is possible to transformmost slabs with fixed or continuous edges and orthotropicreinforcement into their simpler freely supported isotropicequivalents by using the following affinity theorems skew slabscan be transformed similarly into rectangular forms

1 If an orthotropic slab is reinforced as shown in diagram (iv)(a)on Table 249 then it can be transformed into the simplerisotropic slab shown in diagram (iv)(b) All loads anddimensions in the direction of the principal co-ordinate axisremain unchanged but in the affine slab the distances in thedirection of the secondary co-ordinate axis are equal to theactual values divided by radic and the corresponding total loadsare equal to the original values divided by radic (The latterrequirement means that the intensity of a UDL per unit arearemains unchanged by the transformation since both the areaand the total load on that area are divided by radic)

Similarly a skew slab reinforced as shown in diagram(v)(a) can be transformed into the isotropic slab shown indiagram (v)(b) by dividing the original total load by sin(As before this requirement means that the intensity of aUDL per unit area remains unchanged by the transformation)

These rules can be combined when considering a skewslab with different reinforcement in each direction fordetails see ref 28 article 6

2 By using the reduced side lengths lxr and lyr an orthotropicslab that is continuous over one or more supports such as

M n

24lx

23 lx

ly2

lx

ly2

that in diagram (vi)(a) on Table 249 can be transformedinto the simpler freely supported isotropic slab shown indiagram (vi)(b)

For example for an orthotropic slab with fixed edges that isreinforced for positive moment M and negative moments i2Mand i4M in span direction lx and for positive moment M andnegative moments i1M and i3M in span direction ly it can beshown that

where

and

This is identical to the expression derived above for the freelysupported isotropic slab but with lxr and lyr substituted for lx

and ly Values of Mnlx2 corresponding to ratios of lylx can be

read directly from the scale on Table 249The validity of the analysis is based on the assumption that

lyr lxr If this is not the case the yield line pattern will be asshown in diagram (i)(b) on Table 249 and lxr and lyr should betransposed as shown in the following example

Example Design the slab in diagram (vii) on Table 249 tosupport an ultimate load n per unit area assuming that therelative moments of resistance are as shown

Since lx 4 m i2 32 and i4 0

Since ly 6 m 12 i1 1 and i3 0

Thus

1347 Superposition theorem

A problem that may arise when designing a slab to resist acombination of uniform concentrated and line loads some ofwhich may not always occur is that the critical pattern of yieldlines may well vary for different combinations of loads Also itis theoretically incorrect to sum the ultimate moments obtainedwhen considering the various loads individually since thesemoments may result from different yield line patterns HoweverJohansen has established the following superposition theorem

The sum of the ultimate moments for a series of loads is equalto or greater than those due to the sum of the loads

n

24 31023 310

7032

310703

2

0726 n

M n

24l 2xr3 lxr

lyr2

lxr

lyr2

ly r 2 6

(1 1 1)12 703 m

lxr 2 4

1 32 1 310 m

lyr 2ly

[1 i1 1 i3]lxr

2lx

1 i2 1 i4

M n

24l2xr3 lxr

lyr2

lxr

lyr2

Slabs140

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In other words if the ultimate moments corresponding to theyield-line patterns for each load considered separately areadded together the resulting value is equal to greater than thatof the system as a whole This theorem is demonstrated inref 28 article 4

1348 Empirical virtual-work analysis

An important advantage of collapse methods of design is thatthey can be readily applied to solve problems such as slabs thatare irregularly shaped or loaded or that contain large openingsThe analysis of such slabs using elastic methods is by comparisonextremely arduous

To solve such lsquoone-offrsquo problems it is clearly unrealistic todevelop standard algebraic design formulae The followingempirical trial-and-adjustment technique which involves adirect application of the virtual-work principles is easy tomaster and can be used to solve complex problems It is bestillustrated however by working through a simple problemsuch as the one considered in section 1346 There is of courseno need to employ the procedure in this case It is used hereonly to illustrate the method A more complicated example isgiven in ref 28 article 1 on which the description of themethod is based

In addition to the fundamental principles of virtual workdiscussed in section 1344 the present method depends alsoon the following principle If all yield lines (other than thosealong the supports) are positive and if none of them meets anunsupported edge except at right angles then no forces due toshear or torsion can occur at the yield lines Thus a separatevirtual-work balance for each slab area demarcated by the yieldlines can be taken

Example Consider the slab shown in diagram (vii) onTable 249 which is continuous over two adjacent edges freelysupported at the others and subjected to a uniform load n perunit area The ratios of the moments of resistance provided overthe continuous edges and in the secondary direction to thatin the principal direction are as shown

The step-by-step trial and adjustment process is as follows

1 Postulate a likely yield-line pattern

2 Give a virtual displacement of unity at some point andcalculate the relative displacement of any other yield-line

intersection points For the case considered if O is given adisplacement of unity the displacement at O will alsobe unity since OO is parallel to the axes of rotation of theadjoining slab areas

3 Choose reasonable arbitrary values for the dimensions thatmust be determined to define the yield-line pattern Thus inthe example initially ly is taken as 2 m ly as 1m and lx

as 25 m

4 Calculate the actual work done by the load n per unit areaand the internal work done by the moments of resistance Mfor each separate part of the slab and thus obtain ratios ofMn for each part

For example on area A the total load is 12 4 2 n 4nSince the centre of gravity moves through a distance of 13 thework done by the load on area A is 4n 13 4n3 Now sinceO is displaced by unity the rotation of area A about the supportis 1ly 12 The moments of M2 acting across both the pos-itive and negative yield lines each exert a total moment ofM2 4 2M Thus the total internal work done in rotatingarea A is (2M 2M) 12 2M Equating the internal andexternal work done on area A gives 2M 4n3 that isMn 23 Similarly for area C Mn 13

For convenience area B can be divided it into a rectangle (ofsize 3 m 25 m) and two triangles and the work done on eachpart calculated separately Since the centres of gravity movethrough a distance of 12 for the rectangle and 13 for eachtriangle the external work done is as follows

Rectangular area 3 25 12 n 375nTriangular areas 12 (2 1) 25 13 n 125nTotal 500n

Since the rotation is 125 the work done by the moments is(15M M) 6 125 6M Thus the virtual-work ratio isMn 56 0833 Likewise for area D Mn 34

5 Sum the separate values of internal and external workdone for the various slab areas and thus obtain a ratio ofMn for the entire slab This ratio will be lower than thecritical value unless the dimensions chosen arbitrarily instep 3 happen to be correct The calculations are best set outin tabular form as follows

Yield-line analysis 141

Area External work done Internal work done Mn

A 12 4 2 13 n 1333n [M2 M2] 4 12 2000M 0667B [3 25 12 12 3 25 13] n 5000n [3M2 M] 6 125 6000M 0833C 12 4 1 13 n 0667n [0 M2] 4 11 2000M 0333D [3 15 12 12 3 15 13] n 3000n [0 M] 6 115 4000M 0750

Total 10000n Total 14000M 0714

A 12 4 2071 13 n 1381n [M2 M2] 4 12071 1931M 0714B [2467 2424 12 12 3533 2424 13] n 4418n [3M2 M] 6 12424 6188M 0714C 12 4 1462 13 n 0975n [0 M2] 4 11462 1368M 0714D [2467 1576 12 12 3533 1576 13] n 2872n [0 M] 6 11576 3807M 0754

Total 9646n Total 13294M 0726

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6 By comparing the overall ratio obtained for Mn withthose due to each individual part it is possible to see howthe arbitrary dimensions should be adjusted so that the ratiosfor the individual parts become approximately equal to eachother and to that of the slab as a whole

The foregoing table shows calculations for initial values of Mnand also for a set of adjusted values An examination of theinitial values (for which the overall ratio is 0714) shows thatto obtain the same ratio for each area Mn needs to be increasedfor areas A and C and reduced for areas B and D For area Asince Mn is proportional to (ly)2 ly needs to be increased toradic(07140667) 2 2071 Similarly for area C ly needsto be increased to radic(07140333) 1 1462 If for area Bthe external work done is recalculated using the correctedvalues of ly and ly this gives

[2467 25 12 12 3533 25 13] n 4556n

The internal work done by the moments is unchanged and sothe revised value of Mn is 45566 0759 Thus since forarea B Mn is proportional to (lx)2 lx needs to be reducedto radic(07140759) 25 2424 For area D the external workdone is recalculated using the corrected values of all thevariable dimensions and the revised value of Mn obtained

7 Repeat steps 4 and 5 using the adjusted values for thearbitrary dimensions as shown earlier

8 Repeat this cyclic procedure until reasonable agreement isobtained between the values of Mn This ratio gives thevalue of M for which the required reinforcement must bedetermined for a given load n In the example the ratiosgiven by the second cycle are quite satisfactory Note thatalthough some of the dimensions originally guessed werenot particularly accurate the resulting error in the value ofMn obtained for the whole slab was only about 15 andthe required load-carrying capacity is not greatly affected bythe accuracy of the arbitrary dimensions

Concentrated loads and line loads occurring at boundariesbetween slab areas should be divided equally between the areasthat they adjoin and their contribution to the external workdone assessed as described in section 1345

As in all yield-line theory the above analysis is only valid ifthe yield-line pattern considered is the critical one Where thereis a reasonable alternative both patterns should be investigatedto determine which is critical

1349 Corner levers

Tests and elastic analyses of slabs show that the negativemoments along the edges reduce to zero near the corners andincrease rapidly away from these points Thus in slabs that arefixed or continuous at their edges negative yield lines tend toform across the corners and in conjunction with pairs of positiveyield lines result in the formation of additional triangular slabelements known as corner levers as shown in diagram (i)(a) onTable 250 If the slab is freely supported a similar mechanismis induced causing the corners to lift as shown in diagram(1)(b) If these mechanisms are substituted for the original yieldlines running into the corner of the slab the overall strength ofthe slab is correspondingly decreased by an amount dependingon the factors listed on Table 250 For a corner lever having anincluded angle of not less than 90o the strength reduction is not

likely to exceed 8ndash10 In such cases the main reinforcementcan be increased slightly and top reinforcement provided at thecorners of the slab to restrict cracking Recommendations takenfrom the Swedish Code of Practice are shown in diagram (ii) onTable 250

For acute-angled corners the decrease in strength is moresevere For a triangular slab ABC where no corner angle is lessthan 30o Johansen (ref 25) suggests that the calculated strengthwithout corner-lever action should be divided by a factor kgiven by the approximation

k (74 sin A sin B sin C)4

A mathematical determination of the true critical dimensions ofan individual corner lever involves much complex trial andadjustment However this is unnecessary as Jones and Wood(ref 23) have devised a direct design method that gives cornerlevers having dimensions such that the resulting adjustment instrength is similar to that due to the true mechanisms Thisdesign procedure is summarised on Table 250 and illustratedby the following example

The formulae derived by Jones and Wood and on which thegraphs in Table 250 are based are as follows

With fixed edges

where

and

With freely supported edges

where

and

Example Calculate the required resistance of the 5 m squareslab with fixed edges (i 1) shown on Table 250

For the transformed freely supported slab the reduced sidelength lr lradic(1 I) 5radic2 354 m Then for a square slabif the formation of corner levers is ignored the required resistancemoment M (124)nlr

2 0041 3542n 0521nNow from the appropriate graph on Table 250 with i 1

and 13 90o read off k1 11 and k2 42 Thus dimensionsa1 11radic0521 0794 m and a2 42radic0521 3032 m

By plotting these values on a diagram of the slab it is nowpossible to calculate the revised resistance moment required Ifthe deflection at the centre is unity the relative deflection at theapex of the corner levers is 30323536 0858 Thus therevised virtual-work equation is

[(13) 52 4 (12) 07942 (13) 0858] n

This reduces to 14036M 7973n so that M 0568n

4 2M3412 125 07942 0858

2471

cot [K2(1 i) (2 i)] tan (2)

K2 (4 i) 3 cot2 (2)

k2 k1

cos (2) cot sin (2)

k1 [K2 2(1 i) ]2 3 sec(2)

cot (K1 1) tan (2)K1 4 3 cot2 (2) 1

k2 k1

cos (2) cot sin (2)

k1 1K1 sin2(2)

16(1 i) sec(2)

Slabs142

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250Two-way slabs yield-line theory corner levers

Freely supportedslab edges

Corner leverFixed slabedges

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Note that this value of M is 9 greater than the uncorrectedvalue in other words the load supported by a square slab witha specified moment of resistance is actually 9 less than thatcalculated when corner levers are not taken into account

135 HILLERBORGrsquoS SIMPLE STRIP THEORY

1351 Moments in slabs

According to lower-bound (equilibrium) theory load acting ona slab is resisted by a combination of biaxial bending andtorsion In the simple strip method the torsion moment is takenas zero and load (or partial load) acting at any position on theslab is resisted by bending in one of two principal directionsThus in diagram (i) on Table 251 the load acting on theshaded areas is resisted by bending in direction ly and the loadacting on the remaining area is resisted by bending in direc-tion lx In principle there is an unlimited number of ways ofapportioning the load each of which will lead to a differentreinforcement layout while still meeting the collapse criteriaHowever the loading arrangement selected should also ensurethat the resulting design is simple economical and satisfactorywith regard to deflection and cracking under service loads

Some possible ways of apportioning the load on a freelysupported rectangular slab are shown in diagrams (i)ndash(iv) onTable 251 the notation adopted being given on the tablePerhaps the most immediately obvious arrangement is shown indiagram (i) for which Hillerborg originally suggested that 13could be taken as 45o where both adjacent edges are freelysupported However in ref 29 he recommends that 13 should bemade equal to tan1(lylx) as shown in diagram (i) The disad-vantage of the arrangement shown is that the bending moment(and thus the reinforcement theoretically required) variesacross strips 2 and 3 Since it is impractical to vary the rein-forcement continuously the usual approach is to calculatethe total moment acting on the strip divide by the width ofthe strip to obtain the average moment and provide a uniformdistribution of reinforcement to resist this moment To avoidhaving to integrate across the strip to obtain the total momentHillerborg recommends calculating the moment along thecentreline of the strip and then multiplying this value bythe correction factor

where lmax and lmin are the maximum and minimum loadedlengths of the strip Strictly speaking averaging the momentsas described violates the principles on which the method isbased and this device should only be used where the factor ofsafety will not be seriously impaired If the width of the stripover which the moments are to be averaged is large it is betterto sub-divide it and calculate the average moment for eachseparate part

An alternative arrangement that avoids the need to averagethe moments across the strips is shown in diagram (ii) This hasdisadvantages in that six different types of strip (and thus sixdifferent reinforcement layouts) must be considered and thatin strip 6 no moment theoretically occurs Such a strip mustnevertheless contain distribution reinforcement

1 (lmax lmin)

2

3(lmax lmin)2

So far the load on any one separate area has been carried inone direction only In diagram (iii) however the loads on thecorner areas are so divided that one-half is carried in eachdirection Hillerborg (ref 29) states that this very simple andpractical approach never requires more than 10 additionalreinforcement when compared to the theoretically more exactbut less practical solution shown in diagram (i) when lylx isbetween 11 and 4 An additional sophistication that can beintroduced is to apportion the load in each direction in thetwo-way spanning areas in such a way that the resulting rein-forcement across the shorter span corresponds to the minimumrequirement for secondary reinforcement Details of this andsimilar stratagems are given in ref 29

Diagram (iv) illustrates yet another arrangement that may beconsidered By dividing the corner areas into triangles andaveraging the moments over these widths as described earlierHillerborg shows that the moments in the side strips can bereduced to two-thirds of the values given by the arrangementused in diagram (iii)

1352 Loads on supporting beams

A particular feature of the strip method is that the boundariesbetween the different loaded areas also define the manner inwhich the loads are transferred to the supporting beamsFor example in diagram (i) the beams in direction lx supporttriangular areas of slab giving maximum loads of nlx

22ly attheir centres

136 BEAMS SUPPORTING RECTANGULAR PANELS

The loads on beams supporting uniformly loaded rectangularslab panels are distributed approximately as a triangular loadon the beams along the shorter edges lx and a trapezoidalload on the beams along the longer edges ly as shown in thediagrams on Table 252 For a beam supporting a single panelof slab that is either freely supported or subjected to the samedegree of restraint along all four edges where the beam span isequal to the length (or width) of the panel the equivalent UDLper unit length on the beam for the calculation of bendingmoments only is as follows

Short-span beam nlx3

Long-span beam (1 13k2) nlx2

where n is the total UDL per unit area on the slab appropriateto the limit-state being considered and k lylx For a beamsupporting two identical panels one on either side the fore-going equivalent loads are doubled If a beam supports morethan one panel in the direction of its length the distribution ofload is in the form of two or more triangles (or trapeziums)and the foregoing formulae are not applicable in such a casehowever it is sufficiently accurate if the total load on the beamis considered to be uniformly distributed

For slabs designed in accordance with the BS 8110 methodthe loads on the supporting beams may be determined from theslab shear forces given in Table 243 The loads are to be takenas uniformly distributed along the middle three-quarters ofthe beam length where the shear force for the short span of theslab is the load on the long-span beam and vice versa Theresulting beam fixed-end moments can be determined fromTable 228

Slabs144

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251Two-way slabs Hillerborgrsquos simple strip theory

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252Two-way slabs rectangular panels loads on beams(common values)

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Triangularly distributed loads 147

137 TRIANGULARLY DISTRIBUTED LOADS

In the design of rectangular tanks storage bunkers and someretaining structures conditions occur of wall panels spanningin two directions and subjected to distributions of pressurevarying linearly from zero at or near the top to a maximum atthe bottom For liquid-retaining structures with no provisionfor additional protection in the form of an internal lining orexternal tanking an elastic analysis is normally necessary as abasis for checking serviceability cracking In other cases ananalysis based on collapse methods may be justified

1371 Elastic analysis

The coefficients given in Table 253 enable the maximumvalues of bending moments and shearing forces on vertical andhorizontal strips of unit width to be determined for panels ofdifferent aspect ratios and edge conditions The latter are takenas fixed at the sides hinged or fixed at the bottom and hingedor free at the top The coefficients which are taken from ref 32were derived by a finite element analysis and include for aPoissonrsquos ratio of 02 For ratios less than 02 the momentscould be adjusted in the manner described in section 1322

The maximum negative bending moment at the bottom edgeand the maximum shear forces at the bottom and top edgesoccur halfway along the panel The other maximum momentsoccur at the positions indicated in the following table

From Table 253 for panel type 4 with lx lz 54 125 themaximum bending moments are as follows

Horizontal negative moment at corners

mx 0050 981 43 314 kNmm

Horizontal positive moment (at about 05lz 2 m above base)

mx 0022 981 43 138 kNmm

Vertical positive moment (at about 03lz 12 m above base)

mz 0021 981 43 132 kNmm

1372 Yield-line method

A feature of the collapse methods of designing two-way slabsis that the designer is free to choose the ratios between thetotal moments in each direction and between the positive andnegative moments In the case of liquid-retaining structureswhere it is important to ensure that the formation of cracksunder service load is minimised the ratios selected shouldcorrespond approximately to those given by elastic analysisThe following design procedure is thus suggested

1 Obtain maximum positive and negative service momentcoefficients from Table 253

2 Determine i1 ( i3) and i4 where mhposmzposi1 i3 mhnegmhpos and i4 mznegmzpos

3 Calculate lxr and lyr if the top edge is unsupported from

and

and if the top edge is freely supported from

and

4 Obtain M if the top edge is unsupported from the chart onTable 254 and if the top edge is supported from the scaleon Table 249 according to the values of f (or n f2) lx

(or lxr) lyr and i4 The basis of this approach is given below

Top edge of slab unsupported In ref 25 Johansen derivesthe following lsquoexactrsquo formulae according to the failure mode

For failure mode 1

where k is obtained by solving the quadratic equation

For failure mode 2

where is obtained by solving the following cubic equation

8lxr

lyr2

3 3ndash16lxr

lyr2 2 8 6 0

M f l 2

yr

96(6 8 3 2)

i241 i4 4 lx

lyr2 0

41 i4 lx

lyr2k2 4i4(1 i4) (6 4i4) lx

lyr2k

M f l 2

x

12k

lyr 2ly

[1 i1 1 i3 ]lxr

2lx

1 1 i4

lyr 2ly

[1 i1 1 i3 ]lxr

lx

1 i4Distance from bottom of panel to position ofmaximum horizontal moments (negativepositive)

Type of Heightlz for values of lx lzpanel 05 10 15 20 40

1 03 05 05 05 052 03 05 0710 0910 103 03 04 04 04 044 04 04 0506 0910 0910

Distance from bottom of panel to position ofmaximum vertical positive moment

Type of Heightlz for values of lxlzpanel 05 10 15 20 40

1 03 05 05 05 052 03 04 05 06 073 02 03 03 04 044 02 03 03 04 04

For a complete map of bending moment values at intervals ofone-tenth of the panel height and length see ref 32 Themoments obtained for an individual panel apply directly to asquare tank with hydrostatic loading For a rectangular tanka further distribution of the unequal negative moments at thecorners is needed (see Tables 275 and 276)

Example Determine due to internal hydrostatic loading themaximum service moments in the walls of a square tank thatcan be considered as free along the top edge and hinged alongthe bottom edge The tank is 5 m square 4 m deep and thewater level is to be taken to the top of the walls

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253Two-way slabs triangularly distributed load(elastic analysis)

Fixed

Fixed

Fixed

Panel 1 Panel 2 Panel 3 Panel 4

Hinged

Fixed

Fixed

Fixed

Free

Fixed

Fixed

Hinged

Hinged

Fixed

Hinged

Free

lx

lz

lx

Fixed

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254Two-way slabs triangularly distributed load(collapse method)

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The results of these calculations can be plotted graphically asshown in Table 254 from which coefficients of Mfl2

x can beread corresponding to given values of lyrlx and i4 The chainline on the chart indicates the values at which the failuremode changes

Top edge of slab supported In ref 25 Johansen showsthat the total moments resulting from yield-line analysisfor triangularly loaded slabs correspond to those obtainedwhen the same slab is loaded with a uniform load of one-halfthe maximum triangular load (ie n f 2) Hence the designexpressions in section 1346 and the scale on Table 249 foruniformly loaded slabs can again be used As before if theedges of the slab are restrained the reduced side lengths lxr andlyr should be calculated and substituted in the formula insteadof the actual side lengths lx and ly

Example Determine moments by yield-line analysis for thewalls of the square tank considered in the previous example insection 1371

Taking the service moment coefficients obtained previouslythe following values are the most suitable for

00220021 10 and i1 ( i3) 00500022 23

Hence

lyr 275 and lyrlx 27540 070

Then from the chart on Table 254 with i4 0

Mflx2 0018

Comparing this value with mx 0022 as obtained in theprevious example it can be seen that the elastic moments are12 times those determined by yield-line analysis Thus evenwith a partial load factor of 12 the yield-line moments are nogreater than the elastic service moments in this case

Note It can be seen from the chart on Table 254 that failuremode 2 applies for which

Hence

and yield-lines intersect at height lx 0335 40 134 mabove the base

138 FLAT SLABS (SIMPLIFIED METHOD)

The following notes and the data in Tables 255 and 256 arebased on the recommendations for the simplified method of flatslab design given in BS 8110 Alternatively and in cases wherethe following conditions are not met the structure can beanalysed by the equivalent frame method Other methods ofanalysis such as grillage finite-element and yield line may alsobe employed in which case the provisions given for the distri-bution of bending moments are a matter of judgement

32 8 6 0018 96 (40 275)2 366 0335

M fl 2

yr

96(6 8 3 2) 0018 fl2

x

2 521 2310

1381 Limitations of method

The system must comprise at least three rows of rectangularpanels in each direction The spans should be approximatelyequal in the direction being considered and the ratio of thelonger to the shorter sides of the panels should not exceed 2The conditions allowing the design to be based on the singleload case of all spans loaded with the maximum design load asexplained in section 441 and Table 242 must be met Alllateral forces must be resisted by shear walls or bracing

1382 Forms of construction

The slab can be of uniform thickness throughout or thickeneddrop panels can be introduced at the column positions Droppanels can extend to positions beyond which the slab can resistpunching shear without needing shear reinforcementAlternatively the panels can be further extended to positionswhere they may be considered to influence the distribution ofmoments within the slab In this case the smaller dimensionof the drop panel should be at least one-third of the smallerdimension of the surrounding slab panels

Columns can be of uniform cross section throughout or canbe provided with an enlarged head the effective dimensions ofwhich are limited according to the depth of the head as shownin Table 255 For a flared head the actual dimension is takento be the value at a depth 40 mm below the underside of the slabor drop

The effective diameter of a column or column head is thediameter of a circle whose area is equal to the cross-sectionalarea of the column or effective column head In no case is theeffective diameter hc to be taken greater than one-quarter ofthe shortest span framing into the column In cases where theedges of the slab are supported by walls hc can be taken asthe thickness of the wall

1383 Bending moments and shearing forces

The total design bending moments and shearing forces given inTable 255 are the same as those given for one-way slabs inTable 242 where the support moments include for 20 redis-tribution The requirements are applied independently in eachdirection Any negative moments greater than those at a dis-tance hc2 from the centreline of the column may be ignoredsubject to the following condition being met In each span thesum of the maximum positive moment and the average ofthe negative moments for the whole panel width where F is thetotal design load on the panel and l is the panel length betweencolumn centrelines must be not less than

(Fl8) (1 2hc3l)2

This condition is met in the case of designs that are based onthe single load case of all spans loaded with the maximumdesign load by taking the design negative moment as the valueat a distance hc3 from the centreline of the column Themoment at this position is obtained approximately by reducingthe value at the column centreline by 015Fhc

The total design moments on a panel should be apportionedbetween column and middle strips as shown in Table 255 Inthe following table the moment allocations given in BS 8110and EC 2 are shown for comparison

Slabs150

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255Flat slabs BS 8110 simplified method ndash 1

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256Flat slabs BS 8110 simplified method ndash 2

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At the edges of a slab the transfer of moments between the slaband an edge or corner column is limited by the effective breadthbe of the moment transfer strip given in Table 256 Themaximum design moment that can be transferred by this stripis given by the equation Mt max 015bed2fcu in BS 8110 andMt max 017bed2fck in EC 2

At internal columns two-thirds of the reinforcement neededto resist the negative moments in the column strips should beplaced in a width equal to half that of the column strip and cen-tral with the column Otherwise the reinforcement needed toresist the moment apportioned to a particular strip should bedistributed uniformly across the width of the strip

1384 Effective forces for punching shear

The critical consideration for shear in flat slab structures is thatof punching shear around the columns The design forceobtained by summing the shear forces on each side of the col-umn is multiplied by an enhancement factor to allow for theeffects of moment transfer as given in Table 256 The effectiveshear force is determined independently in each direction andthe design checked for the worse case

1385 Reservoir roofs

For reservoir roofs with simply supported ends where elasticmoments are required to check serviceability requirements thecoefficients given for beams in Table 230 could be used In thiscase the negative moments at the centrelines of the columnscould be reduced by 022Fhc to give approximately themoment at a distance hc2 from the centreline of the columnThis approach will still ensure that the minimum momentrequirement mentioned in section 1383 is met

Flat slabs (simplified method) 153

Design moment Column strip Middle strip

BS 8110 Negative 75 25Positive 55 45

EC 2 Negative 60ndash80 40ndash20Positive 50ndash70 50ndash30

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When using the formulae and coefficients in this chapter theloads must include the appropriate partial safety factors for thelimit-state being considered Design loads for the ULS inaccordance with the requirements of BS 8110 and EC 2 aregiven in Table 257

For many framed structures it is not necessary to carry out afull structural analysis of the complete frame as a single unitBS 8110 makes a distinction between frames supporting verticalloads only because lateral stability to the structure as a wholeis provided by other means such as shear walls and framessupporting both vertical and lateral loads Simplified models forthe purpose of analysis as described in section 491 are alsoshown in Table 257

The moment-distribution method used to analyse systems ofcontinuous beams as shown in Table 236 can be extended toapply to no-sway sub-frames (see section 492) as shown inTable 258 and single-bay sway frames (see section 493) asshown in Table 259

141 SLOPE-DEFLECTION METHOD OF ANALYSIS

Moment analysis of a restrained member by slope-deflection isbased on the following two principles The difference in slopebetween any two points in the length of the member is equalto the area of the MEI diagram between the two points Thedistance of any point on the member from a line drawn tangen-tially to the elastic curve at any other point the distance beingmeasured normal to the initial position of the member is equalto the moment (taken about the first point) of the MEI diagrambetween these two points In the foregoing M represents thebending moment E the modulus of elasticity of the materialand I the second moment of area of the member The bendingmoments that occur at the ends of a member subject to thedeformation and restraints shown in the moment diagram atthe top of Table 260 are given by the corresponding formulaeThe formulae which have been derived from a combinationof the basic principles are given in a general form and in thespecial form for members on non-elastic supports

The stiffness of a member is proportional to EIl but as E isassumed to be constant the term that varies in each member isthe stiffness factor K Il The terms FAB and FBA relate to theload on the member When there is no external load FAB andFBA are zero when the load is symmetrically disposed

FAB FBA Al Values of FAB FBA and Al for different loadcases are given in Table 228

The conventional signs for slope-deflection analyses are anexternal restraint moment acting clockwise is positive a slopeis positive if the rotation of the tangent to the elastic line isclockwise a deflection in the same direction as a positive slopeis positive

Example Establish the formulae for the bending moments ina column CAD into which is framed a beam AB The beam ishinged at B and the column is fixed at C and D (see diagramin Table 260) The beam only is loaded Assume there is nodisplacement of the joint A

From the general formulae given on Table 260

Therefore

Thus

For symmetrical loading

MAC 6KAC(AAB lAB)

3KAB 4KAC 4DAD

FAB FBA 2 15AAB lAB

MAB (MAC MAD)

MDA MAD 2

MAD MACKAD KAC

MCA 2EKAC13A MAC 2

MAC 4KAC(FAB FBA 2)

3KAB 4KAC 4KAD

E13A FAB FBA 2

3KAB 4KAC 4KAD

E13A(3KAB 4KAC 4KAC) (FAB FBA2) 0

MAB MAC MAD

MAC 4EKAC13A and MAD 4EKAD13A

MAB 3EKAB13A (FAB FBA 2)

Chapter 14

Framed structures

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257Frame analysis general data

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258Frame analysis moment-distribution method no sway

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259Frame analysis moment-distribution method with sway

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260Frame analysis slope-deflection data

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142 CONTINUOUS BEAMS AS FRAME MEMBERS

In many buildings the interaction of the columns and beamscan be considered with sufficient accuracy by applying one ofthe simplified models shown in Table 257 The simplified two-or three-span sub-frames in Table 261 are analysed on theassumption that the remote ends of the beams and columns arefixed Therefore for any internal span ST the ends of the beamsat R and U the ends of the lower columns at O and X and theends of the upper columns at P and Y are all assumed to befixed In addition the stiffness of the outer beams RS and TUis taken as one-half of the true value For the fixed-endmoments due to normal (ie downward-acting) loads positivenumerical values should be substituted into the tabulatedexpressions If the resulting sign of the support moment isnegative hogging with tension across the top face of the beamis indicated

1421 Internal spans

By slope-deflection methods it can be shown that

where

and is a factor representing the ratio of the assumed to theactual stiffness for the span concerned (ie here 12)

By eliminating 13ST and 13TS in the above and rearranging thefollowing basic formulae are obtained

These formulae which are lsquoexactrsquo within the limitationsof the fixity conditions of the sub-frame represent the case ofthree spans loaded and apply for example to the conditionof dead load For design to BS 8110 the maximum moments atsupports S and T occur when the live load also is applied to allthree spans For design to EC 2 the maximum moment at sup-port S occurs when the live load is applied to spans RS and STand the maximum moment at support T occurs when the live

(FST FSR) (4 DST)(FTU FTS)

MTS FTS DTS

4 DTS DTS2DST 1

DTS 1

(FTU FTS) (4 DTS)(FST FSR)

MST FST DST

4 DSTDTS2DTS 1

DST 1

KT KST KTX KTY KTU

KS KRS KSO KSP KST

13TS (KST 2)(FSR FST) KS(FTS FTU)

KSKT K2ST 4

13ST (KST 2)(FTS FTU) KT(FSR FST)

KS KT K2ST 4

MTS FTS KTS(13TS 13ST 2)

MST FST KST (13ST 13TS 2)

load is applied to spans ST and TU The maximum positivemoment in span ST is obtained when the live load is applied tothis span only for both Codes The appropriate formulae forthese conditions are also given in Table 261 For the ULS theBS 8110 loads are dead 10gk and live 04gk 16qk Thecorresponding loads in EC 2 are dead 135gk and live 15qk

In the foregoing dead and live loads are applied separatelyand the resulting moments are summed Alternatively bothdead and live loads can be applied in a single operation byevaluating the basic formulae with fixed-end moment valuescorresponding to (deadlive) load on the aforesaid spans anddead load only on the remaining spans To comply with EC 2for example in determining the maximum support moment at Sthe fixed-end moments FSR FST and FTS should be calculated fora load of 135gk 15qk while FTU should be evaluated for a loadof 135gk only This method is used in the following example

In accordance with both Codes the moments derived fromthese calculations may be redistributed if desired It should beemphasised that although the diagrams on Table 261 andin the following example are for uniform loads the methodand the formulae are applicable to any type of loadingprovided that the appropriate fixed-end moment coefficientsobtained from Table 228 are used

When the moments MST and MTS at the supports are knownthe positive and negative moments in the spans are obtained bycombining the diagram of free moments due to the design loadswith the diagram of corresponding support moments

1422 End spans

The formulae for any interior span ST are rewritten to applyto an end span AB by substituting A B C and so on for S TU etc (A is the end support and there is no span correspondingto RS) The modified stiffness and distribution factors are givenin Table 261 together with the moment formulae for bothspans loaded and load on span AB only The dead and liveloads should be evaluated and applied so as to obtain therequired support moments as described in section 1421

1423 Columns and adjoining spans

The outer members of the sub-frame have been taken as fullyfixed at their remote ends Thus for a member such as RS theslope-deflection equation is

MSR KRS 13SR

Since the rotation of all the members meeting at a joint is thesame 13SR 13ST Thus by eliminating 13SR and rearranging

Similarly

[2DST (FST FSR) 4(FTU FTS)]

MTU FTU DTU

4 DSTDTS

[2DTS (FTU FTS) 4(FST FSR)]

MSR FSR DSR

4 DSTDTS

Continuous beams as frame members 159

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261Frame analysis simplified sub-frames

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The expressions for the moments in the columns are similar tothe foregoing but FSR and FTU should be replaced by the initialfixed-end moment in the column concerned (normally zero)and the appropriate distribution factor for the column should besubstituted for DSR or DTU

Example Determine the critical ultimate bending moments inbeam ST of the system shown in the following figure belowwhich represents part of a multi-storey frame in accordancewith the requirements of BS 8110 and EC 2 respectively

Stiffness values (mm3)

For upper columns

For lower columns

Distribution factors (if stiffness values are divided by 106)

Support-moment equations

FTS 0132[0953(FST FSR) 3529(FTU FTS)] (4 0471)(FTU FTS)] [2 0471(10497 1)(FST FSR)

MTS FTS [0497 (4 0471 0497)]

FST 0125[1116(FTU FTS) 3503(FST FSR)] (4 0497)(FST FSR)] [2 0497(10471 1)(FTU FTS)

MST FST [0471 (4 0471 0497)]

DTS 273

273 05 273 053 086

273549

0497

DST 273

05 336 273 053 086

273580

0471

K 342 109

4 103 086 106

K 213 109

4 103 053 106

KST KTU 2185 109

8 103 273 106

KRS 2016 109

6 103 336 106

Continuous beams as frame members 161

BS 8110 requirements

Fixed-end moments

For dead load only 10gk 8 kNm

FRS FSR 8 6212 24 kNm

FST FTS FTU FUT 8 8212 427 kNm

For dead live load

14gk 16qk 14 8 16 10 272 kNm

FRS FSR 272 6212 816 kNm

FST FTS FTU FUT 272 8212 1451 kNm

Maximum moments on beam ST

At S (dead live load on all spans)

MST 1451 0125

[1116(1451 1451) 3503(1451 816)]

1451 278 1173 kNm

At T (dead live load on all spans)

MTS 1451 0132

[0953(1451 816) 3529(1451 1451)]

1451 80 1531 kNm

Mid-span (dead live load on ST dead load on RS and TU)

MST 1451 0125

[1116(427 1451) 3503(1451 240)]

1451 387 1064 kNm

MTS 1451 0132

[0953(1451 240) 3529(427 1451)]

1451 325 1126 kNm

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Maximum positive span moment is then approximately

15FST 05(MST MTS) 15 1451 05(1064 1126)

2176 1095 1081 kNm

Maximum moments on column at S (dead live load on STdead load on RS and TU) Distribution factors for lower andupper columns respectively are

Bending moments for lower and upper columns respec-tively are

Alternatively using the method shown on Table 260 for aninterior column when the adjoining beams are analysed asa continuous system on knife-edge supports

KS 05KRS 05 KST KSO KSP

(168 137 053 086) 106 444 106 mm3

Maximum unbalanced fixed-end moment at S

(FSTFRS) 1451240 1211 kNm

MSO 0194 1211 235 kNm

MSP 0119 1211 144 kNm

It will be seen that in this example the results obtained by thetwo methods are almost identical

EC 2 requirements

Fixed-end moments For dead load only

135gk 135 8 108 kNm

FRS FSR 108 6212 324 kNm

FST FTS FTU FUT 108 8212 576 kNm

For dead live load

135gk 15qk 135 8 15 10 258 kNm

FRS FSR 258 6212 774 kNm

FST FTS FTU FUT 258 8212 1376 kNm

DSP 053444

0119DSO 086444

0194

142 kNm

[2 0497(1451ndash427) 4(1451ndash240)]

MSP [0091(4ndash0471 0497)] 231 kNm

[2 0497(1451ndash427) 4(1451ndash240)]

MSO [0148(4ndash0471 0497)]

DSP 053580

0091DSO 086580

0148

Maximum moments on beam ST At S (dead live load on RSand ST dead load on TU)

MST 1376 0125

[1116(576 1376) 3503(1376 774)]

1376 152 1224 kNm

At T (dead live load on ST and TU dead load on RS)

MTS 1376 0132

[0953(1376 324) 3529(1376 1376)]

1376 132 1508 kNm

Mid-span (dead live load on ST dead load on RS and TU)

MST 1376 0125

[1116(576 1376) 3503(1376324)]

1376 349 1027 kNm

MTS 1376 0132

[0953(1376324) 3529(576 1376)]

1376 240 1136 kNm

Maximum positive span moment is then approximately

15FST 05(MST MTS) 15 137605(1027 1136)

20641081 983 kNm

Maximum moments on column at S (dead live load on STdead load on RS and TU)

143 EFFECTS OF LATERAL LOADS

For many structures a close analysis of the bending momentsto which a frame is subjected due to wind or other horizontalloads is unwarranted In such cases the methods illustrated inTable 262 are sufficiently accurate Further information on theuse of these methods is given in section 411

144 PORTAL FRAMES

General formulae for the bending moments in single-storeysingle-bay rigid frames are given in Table 263 (rectangularframes) and Table 264 (gable frames) The formulae whichrelate to any vertical or horizontal load cater for frames withthe columns fixed or hinged at the base Formulae for specificload cases are given in Tables 265 and 266 Formulae for theforces and bending moments in frames with a hinge at the baseof each column and one at the ridge (ie three-hinged frames)are given in Table 267

121 kNm

[2 0497(1376 576) 4(1376 324)]

MSP [0091(4 0471 0497)] 197 kNm

[2 0497(1376 576) 4(1376 324)]

MSO [0148(4 0471 0497)]

Framed structures162

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262Frame analysis effects of lateral loads

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263Rectangular frames general cases

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264Gable frames general cases

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265Rectangular frames special cases

Not

efo

rmul

ae g

i ve

num

eric

al v

alue

s of

rea

ctio

ns a

nd b

endi

ng m

omen

tss

ee d

iagr

ams

for

dire

ctio

n of

act

ion

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266Gable frames special cases

Not

efo

rmul

ae g

ive

num

eric

al v

alue

s of

rea

ctio

ns a

nd b

endi

ng m

omen

tss

ee d

iagr

ams

for

dire

ctio

n of

act

ion

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267Three-hinged portal frames

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In modern multi-storey buildings lateral stability is providedby a system of frames or walls or a combination of both Noteson wall and frame systems are given in section 412 and differentstructural forms with information on typical building heightsand proportions given in ref 37 are shown in Table 268

151 LAYOUT OF SHEAR WALLS AND ALLOCATION OFLATERAL LOADS

Arrangements of shear walls and core units should be such thatthe building is stiff in both flexure and torsion Different planconfigurations with remarks as to their suitability are shown inTable 269 The lateral load transmitted to each wall is afunction of its stiffness and its position in relation to the shearcentre of the system The location of the shear centre can bereadily determined by calculating the moment of stiffness of thewalls about an arbitrary reference point as shown in Table 269The lateral load on each wall can then be determined from thegeneralised formulae given also in Table 269 For most wallswithout openings the dominant mode of deformation is bending(see section 4122) In this case K may be replaced with I inthe generalised formulae where I is the second moment of areaof the cross section

152 PIERCED SHEAR WALLS

In the case of walls pieced by openings the behaviour of theindividual wall sections is coupled to a variable degree Thedeflected shape of the pierced wall is a combination of frameand wall action The wall may be idealised as a plane frame oranalysed by elastic methods in which the flexibility of thebeams is represented as a continuous flexible medium

Solutions using the continuous connection model for a wallcontaining a single line of openings are given in Table 270 Thetotal lateral load F is considered in three different forms aconcentrated load applied at the top of the wall a uniformload distribution and a triangular load distribution with themaximum value at the top of the wall Formulae are given forthe maximum axial force at the base of each wall section themaximum shear force on a connecting beam (and the height ofthe beam above the base of the wall) and the maximumdeflection at the top of the wall Formulae whereby values atany level can be determined are given in ref 38

The main two parameters that define the performance of thewall are and which depend on the geometrical properties

of the wall and beam elements and on the number of lines ofopenings The formulae in Table 270 cater for a wall withdissimilar cross sections on either side of a single line ofopenings For identical cross sections the formulae become

where A1 and I1 refer to each portion of the wall For a wall withtwo symmetrical lines of openings the formulae for the para-meters become

where A1 and I1 refer to each outer portion and I2 refers tothe central portion of the wall Similarly the momentsbecome

There is no axial force in the central portion of the wall

153 INTERACTION OF SHEAR WALLS AND FRAMES

Although the interaction forces between solid walls piercedwalls and frames can vary considerably up the height of a build-ing the effect on the total lateral force resisted by each elementis less significant As a first approximation in order to deter-mine the forces at the bottom of each load-resisting element itis normally sufficient to consider the effect of a single interac-tion force at the top of the building This is equivalent to loadsharing in terms of relative stiffness The location of the shearcentre and the allocation of the lateral load can be determinedas indicated in Table 269 using the following formulae for thestiffness of each element

Solid wall

Pierced wall K 3EIw

H31 2l

21 3

(H)2

3(H)3

K 3EIw

H 3

M2 (M0 Nl )I2

2I1 I2M1

(M0 Nl )I1

2I1 I2

2 12lIe

(2I1 I2)hb3e

2 2

l 2l2 (2I1 I2)

A1

2 6lIe

I1hb3e

2 2

l l2 4I1

A1

Chapter 15

Shear wall structures

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268Structural forms for multi-storey buildings

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269Shear wall layout and lateral load allocation

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270Analysis of pierced shear walls

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Interaction of shear walls and frames 173

Frame

For the frame and are the sums of the second momentsof area of the columns and beams respectively for the lowestbay of height h and beam span lb The factor kc allows for areduction of Ic over the height of the building and is given bykc loge c(c 1) where c Ic top Ic bottom

Example 1 For the shear wall layout in the following figuredetermine the location of the shear centre and the lateral loadapplied to each wall using the formulae in Table 269 Alldimensions are in metres and the walls are 200 mm thick

IbIc

K 12EIc

h2Hkc lbIc

hIb

(0141 0309) F 045F

Example 2 The elevation of a pierced shear wall is shown inthe following figure Identical walls are provided at each end ofa 24 m long rectangular building The building is subjected to acharacteristic wind pressure of 125kNm2 acting on the broadface and the resistance provided by any other frames parallel tothe walls may be ignored The resulting forces and bendingmoments acting on the walls are to be determined

F3y F4y 36(44 0)(F 658)

60128 017F

F2z 20F1415

20(165 242)(F 658)

60128

Layout of shear walls

From symmetry the shear centre of the wall system must be onthe y-axis Similarly the shear centre of the two channelsections (taken together) is at the mid-point of the core unitThe second moment of area values (mm4) of walls 1 and 2 indirection z (discounting the stiffness of walls 3 and 4) are

The distance of the shear centre C from the reference point is

Eccentricity of total lateral force is e 90 242 658 mThe second moment of area values (mm4) of walls 3 and 4 indirection y are

Lateral forces on each wall in terms of total force F are

(0859 0309) F 055F1215F1415

1855F60128

1215(01 242)(F 658)

1215 2322 20 14082 2 36 442

F1z 1215F

1215 20

I3y I4y 02 603

12 36

yc 1215 01 200 165

1215 200 242m

I2z 18 303 14 263

12 20I1z

02 903

12 1215

Elevation of shear wall

If the total wind force acting on the building is shared equallybetween the two walls the horizontal force to be resisted byeach wall is F 125 12 60 900 kN

With reference to Table 270 the following values apply

A1 A2 0225 5 1125 m2

I1 I2 0225 5312 2344 m4

Ib 0225 06312 000405 m4

000333

be (2b 5d)3 (2 2 5 06)3 233

000157

0012860001577 72

2 2 23441125

2 2

l l2 A1 A2

A1A2(I1 I2)

12 7 0003332 2344 3 2333

2 12lIe

(I1 I2)hb3e

0004051 24 (062)2

Ie Ib

1 24(db)2

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Shear wall structures174

Hence 0113 00396 H 0113 60 68Maximum axial force at bottom of each wall element

N asymp

2470 kN

Moment at bottom of each wall element

M1 M2 (FH2 Nl)2 (900 602 2470 7)2 4855 kNm

900 60 0001572 001286 1

268

2

682

FH2

22 1 2

H

2(H)2

Maximum shear force in beams for H 68 occurs wherexH 028 (5th or 6th floor level) and Kv 057 Then

Vb 188 kN

Maximum bending moment in beam

Mb Vb b2 188 22 188 kNm

For subsequent design purposes the forces and moments due tothe characteristic wind load must be multiplied by a partialsafety factor appropriate to the load combination

900 3 000157 057001286

Fh2Kv

2

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In this chapter elastic analyses in terms of characteristic loadsand service stresses are indicated Where limit-state designmethods are employed care must be taken to include theappropriate partial safety factors for the load combination andlimit-state being considered Arch structures may be eitherthree-hinged or two-hinged or fixed-ended as shown in thediagrams at the top of Table 271

161 THREE-HINGED ARCH

For the general case of an unsymmetrical three-hinged archwith a load acting vertically horizontally or at an angle theexpressions for the horizontal and vertical components of thethrusts on the abutments are given in the lower part of Table271 For symmetrical arches the formulae for three-hingedportal frames given in Table 267 are applicable The bendingmoment at each hinge is zero and at any particular section thebending moment shearing force and axial thrust may be deter-mined by considering the loads and abutment reactions on oneside of the section Further information regarding the extent ofthe arch that should be loaded with imposed load in order toproduce the maximum effect at a particular section is given insection 4131

162 TWO-HINGED ARCH

For a symmetrical two-hinged arch the vertical component ofthrust on the abutments is the same as for a freely supportedbeam The horizontal component of thrust H is given by theformula in Table 271 where Mx is the bending moment on asection at distance x from the crown with the arch consideredas a freely supported beam Mx is given by the correspondingexpression in Table 271

The summations 13MxyaI and 13y2aI are taken over the wholelength of the arch In the formula for H which allows for theelastic shortening of the arch A is the average equivalent areaof the arch rib or slab and a is the length of a short segment ofthe axis of the arch where the coordinates of a are x and y asshown in Table 271 If I is the second moment of area of thearch at x then aI aI The bending moment at any section isgiven by Md MxHy

The procedure involves dividing the axis of the arch into aneven number of segments The calculations can be facilitatedby tabulating the successive steps The total bending moment is

required generally only at the crown (x 0 y yc) and thefirst quarter-point (x 025l) The moment Mc at the crown isthe bending moment for a freely supported beam minus HycFor the maximum positive moment at the crown the sum of thevalues of Mc for all elements of dead load is added to the valuesof Mc for only those values of imposed load that give positivevalues of Mc For the maximum negative moment at the crownthe sum of the values of Mc for all elements of dead load isadded to the values of Mc for only those values of imposed loadthat give negative values of Mc The moment at the first quarter-point is the bending moment for a freely supported beam minusHyq where yq is the vertical ordinate of the first quarter-pointThe bending moment is combined with the normal componentof H For an arch of large span it is worthwhile preparing theinfluence lines (F 1) for the bending moments at the crownand at the first quarter-point

163 FIXED ARCH

1631 Determination of thickness

The diagram at the top of Table 272 shows an approximatemethod of determining the section thickness at the crown andthe springing for a symmetrical arch with fixed ends Draw ahorizontal line through the crown C and find G the point ofintersection of the line with the vertical through the centre ofgravity of the total load on half the span of the arch Set outlength GT equal to the dead load on the half span drawn to aconvenient scale draw a horizontal line through T to intersectGS extended at R Draw lines RK perpendicular to GR and GKparallel to the tangent to the arch axis at S On the same scalethat was used to draw GT measure TR which equals Hc andGK which equals Hs If fcc is the maximum allowablecompressive stress in the concrete b the assumed breadth of thearch (1 m for a slab) hc the thickness at the crown and hs thethickness at the springing then approximately

hc 17Hcbfcc hs 2Hsbfcc

The method applies to spans from 12 to 60 m and spanriseratios from 4 to 8 The method does not depend on knowing theprofile of the arch (except for solid-spandrel earth-filled archeswhere the dead load is largely dependent on the shape of thearch) but the span and rise must be known With hc and hs thusdetermined the thrusts and bending moments at the crown

Chapter 16

Arches

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271Arches three-hinged and two-hinged arches

Three-hinged arch

Two-hinged arch

Unsymmetrical three-hinged arches General case

Types of archesInfluence line for section at x

Three-hinged arch

Two-hinged arch

Fixed arch

Note see section 162 for explanation of symbols in and notes on the formulae for two-hinged arches

Reaction formulae for general case

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272Arches fixed-ended arches

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Arches178

springing and quarter-points can be obtained and the stresses onthe assumed sections calculated If the stresses are shown to beunsuitable other dimensions must be tried and the calculationsreworked

1632 Determination of load effects

The following method is suitable for determining load effectsin any symmetrical fixed arch if the dimensions and shape ofthe arch are known or assumed and if the shape of the archmust conform to a particular profile Reference should be madeto section 4133 for general comments on this method

On half of the arch drawn to scale as in Table 272 plot thearch axis Divide the half-arch into k segments such that eachsegment has the same ratio aI aI where a is the length andI is the mean second moment of area of the segment based onthe thickness of the arch measured normal to the axis withallowance being made for the reinforcement

Coordinates x and y relative to the axis of the arch at thecrown are determined by measurement to the centre of lengthof each segment Calculate separately the dead and imposedloads on each segment Assume each load acts at the centre oflength of the segment In an open-spandrel arch the dead andimposed loads are concentrated on the arch at the columnpositions these should be taken as the centre of the segmentsbut it may not then be possible to maintain a constant value ofaI and the value of aI for each segment must be calculated thegeneral formulae in Table 272 are then applicable

For constant values of aI the forces and bending moment atthe crown are as follows

The summations are taken over one-half of the arch The termM1 is the moment at the centre of the segment of all the loadsfrom the centre of the segment to the crown Summations arealso made for all the loads on the other half of the arch forwhich Rc is negative

Due to the elastic shortening of the arch resulting from Hc

where A is the cross-sectional area of the segment calculated onthe same basis as I

Due to a rise (T) or a fall (T) in concrete temperature

where is the coefficient of linear expansion of the concrete Ec

is the short-term modulus of elasticity of the concrete and l1 is

Mc2 Hc2y k

Hc2 (T)kEcl1

2aIky2 y2

Mc1 Hc1ay k

Hc1 Hck(a A)

aIky2 y2

Mc M1 2Hcy

2k

Rc M1x 2x2

Hc kM1y yM1

2ky2 y2

the length of the arch axis Arch shortening due to Hc2 isneglected The effect of concrete shrinkage can be treated as atemperature fall in which T is replaced by (cs) where cs isthe shrinkage strain (allowing for reinforcement restraint) and is a reduction coefficient (typically taken as 043) to allow forthe effect of creep

The foregoing formulae are used to determine the effects ofthe various design loads in the following procedure Calculate(Hc Hc1) Rc and (Mc Mc1) for the dead load Calculate foreach load separately values of Hc2 and Mc2 for temperaturerise temperature fall and concrete shrinkage Calculate for theimposed load (represented as an equivalent uniform load)(Hc Hc1) and (Mc Mc1) To obtain the maximum positivebending moment at the crown (and the horizontal thrust) theimposed load should be applied to the middle third of the archmore or less (By considering the effect of the imposed load onone segment more and one segment less than those in the mid-dle third of the arch the number of segments that should beloaded to give the maximum positive bending moment at thecrown can be determined) With the imposed load applied onlyto those segments that are unloaded when calculating the max-imum positive moment at the crown the maximum negativemoment at the crown due to the imposed load is obtained Themaximum bending moments due to the imposed load are eachcombined with the bending moments due to dead load and archshortening and with the bending moments due to temperaturechange and concrete shrinkage in such a way that the mostadverse total values are obtained The corresponding thrusts arealso calculated and combined with the appropriate bendingmoments to check the design conditions at the crown

The bending moment at the springing due to load at a pointbetween the springing and the crown of the arch distant l fromthe springing (where l is the span of the arch) is

Ms (Mc Mc1) (Hc Hc1)yc Rcl2 Fl

where yc is the rise of the arch For the dead load the valuesdetermined for the crown are substituted in this expressionwith the term Fl replaced by F [(l2) x] To obtain themaximum negative bending moment at the springing theimposed load is applied to those parts of the arch extending 04of the span from the springing more or less (As before theeffect of applying the imposed load to one segment more andone segment less than this distance should be determined toensure that the most adverse loading disposition has beenconsidered) By applying load to only those segments that areunloaded when calculating the maximum negative bendingmoment the maximum positive bending moment is obtainedThese maximum bending moments are each combined with thebending moments due to dead load temperature change andconcrete shrinkage to obtain the most adverse total values Theconditions at the springing are then checked for the combinedeffects of the most adverse bending moments and the corre-sponding normal thrusts

The normal thrust at the springing is given by

Ns (Hc Hc1)cos13 Rssin13

In this expression the vertical component of thrust given byRs (total load on half-arch) Rc is calculated for the loadsused to determine (Hc Hc1)

The shearing force at the springing is given by

Vs (Hc Hc1)sin13 Rscos13

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Fixed parabolic arch 179

In this expression the maximum value is generally obtainedwhen the imposed load extends over the whole arch At thecrown the maximum shearing force is the maximum value ofRc due to any combination of dead and imposed load

The bending moment at a section with coordinates xq and yqdue to load at a point between the springing and the crown ofthe arch distant l from the springing is

Mq Mc Hcyq Rcxq F[xq (05 )l]

At the quarter-point xq l4 and the procedures to determinethe maximum bending moment normal thrust and shearingforce are similar to those described above The method given inTable 273 can be used to obtain data for influence lines

164 FIXED PARABOLIC ARCH

Formulae and guidance on using the data in Table 274 aregiven below See section 4134 for further comments

1641 Dead load and arch shortening

The horizontal thrust due to dead load alone is H k1gl2ycwhere g is the dead load per unit length at the crown l is thespan and yc is the rise of the arch axis The coefficient k1

depends on the dead load at the springing which varies with therisespan ratio and type of structure that is whether the arch isopen spandrel or solid spandrel or whether the dead load isuniform throughout the span

An elastic shortening results from the thrust along the archaxis (assuming rigid abutments) The counter-thrust H1 whileslightly reducing the thrust due to the dead load renders thisthrust eccentric and produces a positive bending moment at thecrown and a negative bending moment at each springing If h isthe thickness of the arch at the crown

in which the coefficient k2 depends on the relative thickness atthe crown and the springing

Due to dead load and elastic shortening the net thrusts Hc

and Hs at the crown and the springing respectively acting par-allel to the arch axis at these points are given by

where 13 is the angle between the horizontal and the tangent tothe arch axis at the springing with cos13 given in Table 274

The bending moments due to the eccentricities of Hc and Hs

are given by Mc k3ycH1 and Ms (k3 1) ycH1 respectively

1642 Temperature change

The additional horizontal thrust due to a rise in temperature orcorresponding counter-thrust due to a fall in temperature isgiven by

If T is the temperature change in oC with h and yc in metresthen H2 is in kN per metre width of arch The values of k4 inTable 274 are based on an elastic modulus Ec 20 kNmm2and a coefficient of linear expansion 12 micro-strainoC

H2 k4hyc

2

hT

Hs H

cos 13 H1 cos 13Hc H H1

H1 k2hyc

2

H

For any other values of Ec and k4 should be multiplied by avalue of (Ec20)(12) At the crown the increase or decrease innormal thrust due to a change in temperature is H2 and thebending moment is k3ycH2 account being taken of the sign ofH2 The normal thrust at the springing due to a change oftemperature is H2cos13 and whether this thrust increases ordecreases the thrust due to dead load depends on the sign of H2At the springing the bending moment is (1 k3)ycH2 the signbeing the same as that of H2

1643 Shrinkage of concrete

The effect of concrete shrinkage can be treated as a fall intemperature in which T is replaced by (cs)( 12) wherecs (micro-strain) is the shrinkage (allowing for reinforcementrestraint) and is a reduction coefficient (taken as 043) toallow for the effect of creep

1644 Imposed load

The maximum bending moments and corresponding thrustsand reactions are given by the following expressions whereq per unit length is the intensity of uniform load equivalent tothe specified imposed load

At the crown positive bending moment k5 ql2

horizontal thrust k6 ql2yc

At the springing negative bending moment k7 ql2

horizontal thrust k8 ql2yc

vertical reaction k9 ql

positive bending moment k10 ql2

horizontal thrust k11 ql2yc

vertical reaction k12 ql

The normal thrust at the springing where H is the horizontalthrust and R is the vertical reaction is given by

N Hcos13 Rradic(1 cos213)

1645 Dimensions of arch

The line of thrust and therefore the arch axis can be plotted asdescribed in section 4134 The thickness of the arch at thecrown and the springing having been determined the lines ofthe extrados and intrados can be plotted to give a parabolicvariation of thickness between the two extremes Thus thethickness normal to the axis of the arch at any point is given by[(hs h)2 h] where is the ratio of the distance of the pointfrom the crown measured along the axis of the arch to half thelength of the axis of the arch

Example Determine the bending moments and thrusts on afixed parabolic arch slab for an open-spandrel bridge

Specified values

Span 50 m measured horizontally between the intersection ofthe arch axis and the abutment Rise 75 m in the arch axisThickness 900 mm at springing 600 mm at crownDead load 12 kNm2 imposed load 15 kNm2Temperature range 24oC 12 micro-strain per oC

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273Arches computation chart for symmetrical fixed-ended arch

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274Arches fixed-ended parabolic arches

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Arches182

Ec 28 kNmm2 cs 200 micro-strain 043

Geometrical properties

yc

l

7550

015hs

h

900600

15

Inclination of arch axis at springing (Table 273) cos13 082A strip of slab 1 m wide is considered The coefficients takenfrom Table 274 are substituted in the expressions given insection 1644 Allowing for self-weight of arch slab

Total dead load at crown 12 06 24 264 kNm2

Loads and resulting bending moments and thrusts Moment ThrustkNmm kNm

Horizontal thrusts due to dead load and other actions

Dead load (k1 0140)H 0140 264 (50275) 1232

Arch shortening (k2 139)H1 139 (0675)2 1232 11

Temperature change (k4 422)H2 422 (2820) (0675)2 06 24 55

Concrete shrinkage (k4 422)H3 422 (2820) (0675)2 06 (20012) 043 16

Crown maximum positive bending moment and corresponding thrust

Dead load and arch shortening (k3 0243 thrusts H and H1 as above) 20Mc 0243 75 11Hc 1232 11 1221

Temperature fall (thrust H2 as above) 55Mc 0243 75 55 100

Shrinkage (thrust H3 as above) 16Mc 0243 75 16 29

Imposed load (k5 0005 k6 0064)Mc 0005 15 502 188Hc 0064 15 (50275) ____ 320

Totals 337 1470

Springing maximum negative bending moment and corresponding thrust

Dead load and arch shorteningMs (1 0243) 75 11 63Hs (1232082) 11 082 1494

Temperature fallMs (1 0243) 75 55 312Hs 55 082 45

ShrinkageMs (1 0243) 75 16 91Hs 16 082 13

Imposed load (k7 0020 k8 0038 k9 0352)Ms 0020 15 502 750H 0038 15 (50275) 190 kNm R 0352 15 50 264 kNmN 190 082 264radic(1 0822) 307

Totals 1216 1743

Springing maximum positive bending moment and corresponding thrust

Dead load and arch shortening (values as before) 63 1494Temperature rise (values for temperature fall) 312 45Shrinkage (neglect as partial in short term and beneficial in long term)Imposed load (k10 0023 k11 0089 k12 0151)

Ms 0023 15 502 863H 0089 15 (50275) 445 kNm R 0151 15 50 113 kNmN 445 082 113radic(1 0822) 430

Totals 1112 1969

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In the following containers are conveniently categorised astanks containing liquids and bunkers and silos containing drymaterials each category being subdivided into cylindrical andrectangular structures The intensity of pressure on the wallsof the structure is considered to be uniform at any level butvertically the pressure increases linearly from zero at the top toa maximum at the bottom

171 CYLINDRICAL TANKS

If the wall of a cylindrical tank has a sliding joint at the baseand is free at the top then when the tank is full no radial shearor vertical bending occurs The circumferential tension at depthz below the top is given by n ri z per unit height where ri

is the internal radius of the tank and is unit weight of theliquid If the wall is supported at the base in such a way that noradial movement can occur radial shear and vertical bendingresult and the circumferential tension is always zero at thebottom of the wall Values of circumferential tensions verticalmoments and radial shears according to values of the termheight2(2 mean radius thickness) can be obtained fromTables 275 and 276

The tables apply to walls with a free top and a bottom that iseither fixed or hinged The coefficients have been derivedby elastic analysis and allow for a Poissonrsquos ratio of 02 Forfurther information on the tables reference should be made topublications on cylindrical tanks such as refs 55 and 56 If anannular footing is provided at the base of the wall a hingeddetail can be formed although this is rarely done The footingnormally needs to be tied into the floor of the tank to preventradial movement Reliance solely on the frictional resistanceof the ground to the radial force on the footing is generally inad-equate and always uncertain If the joint between the wall andthe footing is continuous it is possible to develop a fixedcondition by widening the footing until a uniform distributionof bearing pressure is obtained In many cases the wall and thefloor are made continuous and it then becomes necessary toconsider the structural interaction of a cylindrical wall and aground supported circular slab Appropriate values for the stiff-ness of the member and the effect of edge loading can beobtained from Table 276 for walls and Table 277 for slabs

For slabs on an elastic foundation the values depend on theratio rrk where rk is the radius of relative stiffness defined insection 725 The value of rk is dependent on the modulus of

subgrade reaction for which data is given in section 724Taking rrk 0 which corresponds to a lsquoplasticrsquo soil state isappropriate for an empty tank liable to flotation

Example 1 Determine due to internal hydrostatic loadingmaximum service values for circumferential tension verticalmoment and radial shear in the wall of a cylindrical tank that isfree at the top edge and hinged at the bottom The wall is300 mm thick the tank is 6 m deep the mean radius is 10 mand the water level is taken to the top of the wall

From Table 275 for lz22rh 62(2 10 03) 6

n 0643 at zlz 07 m 0008 at zlz 08

v 0110 at zlz 10

n n lzr 0643 981 6 10 3785 kNm

m mlz3 0008 981 63 170 kNmm

v vlz2 0110 981 62 389 kNm

Example 2 Determine for the tank considered in example 1the corresponding values if the wall is fixed at the bottom

From Table 275 for lz22rh 62(2 10 03) 6

n 0514 at zlz 06 m 0005 at zlz 07

mb 0019 and v 0197 at zlz 10

n nlzr 0514 981 6 10 3025 kNm

m mlz3 0005 981 63 106 kNmm

m mlz3 0019 981 63 403 kNmm

v vlz2 0197 981 62 696 kNm

Consider a straight wall that is centrally placed on a footing ofwidth b (2a h) where a represents the distance from edgeof footing to face of wall The weight of liquid per unit lengthof wall on the inside of the footing is given by lza Assuminga uniform distribution of bearing pressure due to the liquid loadthe bending moment about the centre of the wall due to thepressure on the toe of the footing is

mlza (ab) (ah)2 giving a2(a h) 2(mlz)b

Substituting for b h and m lz 403(981 6) 0685 thefollowing cubic equation in a is obtained

a2 (a03) 137 (2a 03)

Chapter 17

Containmentstructures

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Cylindrical tanks elastic analysis ndash 1 275

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Cylindrical tanks elastic analysis ndash 2 276

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Cylindrical tanks elastic analysis ndash 3 277

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Cylindrical tanks 187

Solution of the equation by trial and error gives a 16 m andwidth b 2 16 03 35 m This width is correct for theannular footing provided the footing is positioned so that itscentre of area coincides with the centre of the wall

If c represents the distance from the outer edge of the footingto the centreline of the wall then for a unit length of wall thelengths of the trapezoidal area are (r c)r for the outer edgeand (r c b)r for the inner edge The usual formula for atrapezoidal area gives

[(r c)2(rcb)] br3[(rc) (rcb)] cr

Substituting for b and r and rearranging the terms gives

6c2 39c 805 0 from which c 165 m

Thus a fixed edge condition can be obtained by providing a35 m wide footing with the outer edge of the footing 15 mfrom the outer face of the wall

Example 3 Determine for the tank considered in example 1the corresponding values if the wall is continuous with the floorslab The slab is 400 mm thick and the soil on which the tank isto be built is described as well-compacted sand

Properties of cylindrical wall From Table 276 with lz22rh 6w 0783 and the wall stiffness is given by

Kw w Ec h3lz (0783 0336) Ec 00035 Ec

Also w1 00187 and the fixed edge moment when the tankis full is given by

Mw w1 lz3 00187 981 63 396 kNmm

Properties of circular slab on elastic subgrade From section724 for well-compacted sand a mean value of 75 MNm3 canbe taken for the modulus of subgrade reaction

From Table 277 for a slab on an elastic subgrade the radiusof relative stiffness with 02 is given by

rk [Ech31152 ks]025

Therefore taking Ec 33 kNmm2 and ks 75 MNm3

rk [33 109 043(1152 75 106)]025 125 m

With rrk 8 s 0468 and the slab stiffness is given by

Ks sEch3r (0468 04310) Ec 00030 Ec

The unit edge load on the slab due to the lsquoeffectiversquo weight ofthe wall is

Q 03 6 (24 981) 255 kNm

With rrk 8 s2 0088 and the corresponding fixed edgemoment is given by

Ms s2 Qr 0088 255 10 225 kNmm

Moment distribution at joint Since the calculated fixed edgemoments for the wall and the slab both act in the same directionthe joint will rotate when the notional restraint is removed Thiswill induce additional moments and change the circumferentialtensions in the wall At the joint the induced moments will be

proportional to the relative stiffness values of the two elementsaccording to the following distribution factors

wall slab

Element Wall SlabDistribution factor 054 046Fixed end moment 396 225Induced moment 335 286Final moment 61 61

It can be seen from the moments calculated for the wall thatthe joint rotation is close to that for a hinged base conditionThe use of a thinner slab or a lower value for the modulus ofsubgrade reaction will increase the rotation until a lsquoclosingrsquocorner moment develops and the circumferential tensions inthe wall exceed those obtained for a hinged base

Final forces and moments The final circumferential tensionsand vertical moments at various levels in the wall can beobtained by combining the results for load cases (1) and (3) inTable 275 where M is the induced moment The followingequations apply

n n1 lz r n3Mrlz2

(981 6 10) n1 (335 1062) n3

5886 n1 93 n3 kNm

m m1 lz3 m3M (981 63) m1 335 m3

2119 m1 335 m3

The resulting values for different values of zlz are shown in thefollowing tables

0003000035 00030

0460003500035 00030

054

Circumferential tensions in wall (kNm)

zlz

Load case (1) Load case (3) Final

n1 5886 n1 n3 93 n3force

05 0504 2967 334 311 327806 0514 3025 654 608 363307 0447 2631 103 958 358908 0301 1772 131 1218 299009 0112 659 114 1060 1719

Vertical moments in wall (kNmm)

zlz

Load case (1) Load case (3) Final

m1 2119 m1 m3 335 m3moment

06 00046 98 0037 12 8607 00051 108 0057 19 12708 00029 62 0252 84 14609 00041 87 0572 192 10510 00187 396 10 335 61

The final radial shear at the base of the wall is given by

v v1lz2 v3Mlz 0197 981 62 449 3356 445 kNm

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The new moment distribution factors are as follows

wall slab

Element Wall SlabDistribution factor 084 016Fixed end moment 22 1080Induced moment 926 176Final moment 904 904

It will be seen that in this example the final edge moment onthe slab is close to a fixed edge condition since the stiffnessof the wall is much greater than that of the slab Final values forthe moments and forces in the slab and the wall can now becalculated as shown in example 3

172 RECTANGULAR TANKS

The bending moments obtained by Table 253 for individualrectangular panels with triangularly distributed loads may beapplied without modification to continuous walls providingthere is no rotation about the vertical edges In a square tanktherefore moment coefficients can be taken directly fromTable 253 For a rectangular tank distribution of the unequalfixity moments at the wall junctions is needed and momentcoefficients for tanks of different span ratios are given in Tables278 and 279 The shearing forces given in Table 253 for theindividual panels may still be used

The tables give values for idealised edge conditions at the topand bottom of the wall The top of a wall should be taken as freefor an open tank and when a sliding joint is provided betweena roof and the top of the wall If a wall is continuous with theroof the joint will rotate when the tank is full and the conditionwill tend towards hinged When there is earth loading on thewall a closing-corner moment will develop when the tank isempty At the bottom of the wall a hinged condition may becreated by providing a narrow wall footing tied into the floorslab or by adopting a reinforced hinge detail Where the tankwall is continuous with the floor the deformation of the floorresulting from the wall loading is often complex and highlydependent on the assumed ground conditions The edge condi-tion at the bottom of the wall is generally uncertain but willtend towards hinged when the tank is full When there is earthloading on the wall the edge condition will tend towards fixedwhen the tank is empty If the floor is extended outwards toform a toe the condition at the base will be affected in the waydiscussed in section 171

In considering the horizontal spans the shear forces at thevertical edges of one wall result in axial forces in the adjacentwalls Thus for internal loading on a rectangular tank the shearforce at the end of the long wall is equal to the tensile force inthe short wall and vice versa In designing sections the com-bined effects of the bending moment the axial force and theshear force need to be considered

Example 5 Determine due to internal hydrostatic loadingthe maximum service moments and shear forces in the walls ofa rectangular tank that can be considered as free along the topedge and hinged along the bottom edge The tank is 6m long

0003000035 00007

0160003500035 00007

084

Containment structures188

Tangential moments in slab (kNmm)

rxrLoad case (1) Load case (2) Final

t1 286t1 t2 255t2moment

10 0378 108 0018 46 6208 0132 38 0009 23 6106 0020 06 0006 15 0904 0034 10 0 0 1002 0014 04 0002 05 090 0005 02 0002 05 07

Example 4 Determine for the tank considered in example 3the factor of safety against flotation and the moments in theslab if the worst credible groundwater level is 15 m above theunderside of the slab

The radius of the slab is 103 m and radius to the centre ofthe wall is 1015 m If the weight of the wall is spread uniformlyover the full area of the slab the total downward pressure dueto the weight of the slab and the wall is

[04 03 6 (2 10151032)] 24 178 kNm2

The upward pressure due to a 15 m depth of groundwater is147 kNm2 giving a safety factor of 178147 121 whichsatisfies the BS 8007 minimum requirement of 11 The unitedge load on the slab due to the weight of the wall is

Q 03 6 24 432 kNm

Taking rrk 0 (for a lsquoplasticrsquo soil condition) s 0104and s2 0250 giving the following values

Ks sEch3r (0104 04310)Ec 00007 Ec

Ms s2Qr ndash 0250 432 10 108 kNmm

The wall stiffness is the same as before and the effect of thegroundwater (11 m from the top of the slab) on the wall maybe taken as a simple cantilever giving values

Kw wEch3lz (0783 0336) Ec 00035 Ec

Mw w1lz3 981 1136 22 kNmm

Radial moments in slab (kNmm)

rxrLoad case (1) Load case (2) Final

r1 286r1 r2 255r2moment

10 10 286 0088 225 6108 0488 140 0013 33 17306 0028 08 0016 41 4904 0055 16 0003 08 0802 0023 07 0001 03 100 0005 02 0002 05 07

The final radial and tangential moments in the floor slab can beobtained by combining the results for load cases (1) and (2) inTable 277 The following equations apply

mr r1M r2 Qr and mt t1M t 2 Qr

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278Rectangular tanks triangularly distributed load(elastic analysis) ndash 1

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279Rectangular tanks triangularly distributed load(elastic analysis) ndash 2

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4m wide and 4m deep and the water level is to be taken to thetop of the walls

From Table 279 for lxlz 64 15 lylz 44 10 andtop edge free maximum bending moments are as follows

Horizontal negative moment at corners

mx my 0050 981 43 314 kNmm

Horizontal positive moments (at about 05lz 2m above base)

mx 0035 981 43 220 kNmm (long wall)

my 0011 981 43 69 kNmm (short wall)

Vertical positive moments (at about 03lz 12m above base)

mzx 0029 981 43 182 kNmm (long wall)

mzy 0013 981 43 82 kNmm (short wall)

From Table 253 for panel type 4 with lhlz 15 (long wall)and 10 (short wall) maximum shear forces are as follows

Horizontal shear forces at side edges

vhx 037 981 42 581 kNm (long wall)

vhy 031 981 42 487 kNm (short wall)

Vertical shear forces at bottom edge

vzx 026 981 42 408 kNm (long wall)

vzy 019 981 42 298 kNm (short wall)

From the above the horizontal tensile forces in the walls are

nhx 487 kNm (long wall) nhy 581 kNm (short wall)

173 SILOS

Notes on the properties of stored materials and the pressuresset up in silos of different forms and proportions are given insections 277 and 93 and Tables 215 and 216 Notes on thedesign of silo walls are given in section 641 For rectangularsilos that are divided into several compartments where thewalls span horizontally expressions for the bending momentsand reactions are given in Table 280

174 BOTTOMS OF ELEVATED TANKS AND SILOS

1741 Tanks

The figure here shows the floor of an elevated cylindrical tanksupported by beams in two different arrangements

area with the remainder of the floor load the weight of the walland the load from the roof being equally divided between theeight cantilever lengths An alternative to (b) is to place thecolumns almost under the wall in which case the cantilevers areunnecessary but secondary beams might be required

For a tank of large diameter a domed bottom of one of thetypes shown in Table 281 is more economical and althoughthe formwork is much more costly the saving in concrete andreinforcement compared to beam-and-slab construction can beconsiderable Ring beams A and C in the case of a simple domedbottom resist the horizontal component of thrust from the domeand the thickness of the dome is determined by the magnitudeof the thrust Expressions for the thrust and the vertical shearingforce around the edge of the dome and the resultant circumfer-ential tension in the ring beam are given in Table 281 Domesused to form the bottom or the roof can also be analysed by themethod given in Table 292

A bottom consisting of a central dome and an outer conicalpart as illustrated in Table 281 is economical for the largesttanks This form of construction is traditionally known as anIntze tank The outward thrust from the top of the conical partis resisted by the ring beam B and the difference between theinward thrust from the bottom of the conical part and the out-ward thrust from the domed part is resisted by the ring beam AExpressions for the forces are given in Table 281 and the pro-portions of the conical and domed parts can be chosen so thatthe resultant thrust on beam A is zero Suitable proportions forbottoms of this type are given in Table 281 and the volume ofa tank with these proportions is 0604 do

3 where do is the diam-eter of the tank The tank wall should be designed as describedpreviously account being taken of the vertical bending at thebase of the wall and the effect of this bending on the conicalpart The floor must be designed to resist in addition to theforces and bending moments already described any radialtension due to the vertical bending of the wall

Example 6 Determine for service loads the principal forcesin the bottom of a cylindrical tank of the Intze type where

do 10 m d 8 m 48o (giving cot 0900)

13 40o (giving cot 13 1192 and cosec13 155)

F1 2500 kN F2 2800 kN and F3 1300 kN

Then from Table 281 the following values are obtained

Unit vertical shearing force at periphery of dome

V1 F1(d) 2500(8) 100 kNm

Unit thrust at periphery of dome

N1 V1cosec13 100 155 155 kNm

(Note The required thickness of the dome at the springing isdetermined by the values of N1 and V1)

Unit outward horizontal thrust from dome on ring beam A

H1 V1cot 13 100 1192 1192 kNm

Unit vertical shearing force at inner edge of cone

V2 (F2 F3)(d) (2800 1300)(8) 163 kNm

Unit inward horizontal thrust from cone on ring beam A

H2 V2cot 163 0900 1468 kNm

Circumferential compression in ring beam A

NA 05d(H2 H1) 058 (14681192)1104 kN

Bottoms of elevated tanks and silos 191

(a) (b)

In (a) each beam spans between opposite columns and carriesone-quarter of the load on the floor The remaining half of thefloor load the weight of the wall and the load from the roof aretransferred to the columns through the wall In (b) each lengthof beam between the columns carries the load on the shaded

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Rectangular containers spanning horizontallymoments in walls 280

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Bottoms of elevated tanks and silos 281

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(Note If H1 exceeds H2 the circumferential force is tensile Theideal case occurs when H1 H2 and NA 0 see as follows)

Unit vertical shearing force at outer edge of cone

V3 F3(do) 1300(10) 414 kNm

Unit outward horizontal thrust from cone on ring beam B

H3 V3 cot 414 0900 372 kNm

Circumferential tension in ring beam B

NB 05do H3 05 10 372 1862 kN

The vertical wall must be reinforced for the circumferentialtension due to the horizontal pressure of the contained liquidgiven by n 05doz per unit depth The conical part of thetank must be reinforced to resist circumferential tension andthe reinforcement may be either distributed over the height ofthe conical portion or concentrated in the ring beams at the topand bottom In large-diameter Intze tanks the width of ringbeam A can be considerable and if this is so the weight ofwater immediately above the beam should not be taken to con-tribute to the forces on the dome and conical part With a widebeam F1 is taken as the weight of the contents over the net areaof the dome and d as the internal diameter of the ring beamF2 is taken as the weight of the contents over the net area of theconical part and d for use with F2 as the external diameterof the ring beam If this were to be done for a ring beam ofreasonable width in the forgoing example H1 and H2 would bemore nearly balanced

1742 Columns supporting elevated tanks

For a group of four columns on a square grid the thrusts andtensions in the columns due to wind loading on the tank can

be calculated as follows When the wind blows normal to theside of the group the thrust on each column on the leeward sideand the tension in each column on the windward side is Mw2dwhere d is the column spacing and Mw is the total moment dueto the wind When the wind blows normal to a diagonal of thegroup the thrust on a leeward corner column and tension in awindward corner column is Mwdradic2 For any other arrangementthe force on any column can be calculated from the equivalentsecond moment of area of the group

Consider the case when the wind blows normal to the X-Xaxis of the group of eight columns shown in the followingfigure The second moment of area of the group of columns aboutthe axis is 2 (d2)2 4 (0353 d)2 10 d2 The thruston the extreme leeward column is (05d10 d2) Mw Mw2dThe forces on each of the other columns in the group can bedetermined similarly by substituting the appropriate value forthe distance of the column from the axis

Containment structures194

1743 Silos

Notes on the pressures set up on hopper bottoms are given insections 277 and 93 and Tables 215 and 216 Notes on thedesign of hopper bottoms in the form of inverted truncatedpyramids are given in section 642 Expressions for bendingmoments and tensile forces are given in Table 281

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181 PAD FOUNDATIONS

Some general notes on the design of foundations are given insection 71 The size of a pad or spread foundation is usuallydetermined using service loads and allowable bearing valuesThe subsequent structural design is then determined by therequirements of the ULS Presumed allowable bearing valuesrecommended for preliminary design purposes in BS 8004 aregiven in Table 282

1811 Separate bases

An introduction to separate bases is given in section 716Diagrams of bearing pressure distributions and expressions forpressures and maximum bending moments in rectangular basessubjected to concentric and eccentric loading are given inTable 282

Example 1 The distribution of bearing pressure is requiredunder a base 3 m long 25 m wide and 600 mm thick when itsupports a concentrated load of 1000 kN at an eccentricity of300 mm in relation to its length

Weight of base Fb 30 25 06 24 108 kNTotal load Ftot Fb Fv 108 1000 1108 kN

Eccentricity of total load

etot Fve Ftot (1000 03)1108 027 m

Since etot l6 306 05 m the bearing pressure diagramis trapezoidal and the maximum and minimum pressures are

f (1 6etot l)Ftot bl (1 6 02730)[1108(25 30)] (154 and 046) 148 228 kNm2 and 68 kNm2

Note For the structural design of the base Fb and Fv shouldinclude safety factors appropriate to the ULS Bearing pressurescorresponding to these values of Fb and Fv reduced by Fbblshould then be used to determine bending moments and shearforces for the subsequent design

1812 Combined bases

When more than one column or load is carried on a single basethe centre of gravity of the total load should coincide if possiblewith the centre of area of the base Then assuming a rigid

base the resulting bearing pressure will be uniformly distributedThe base should be symmetrically disposed about the line of theloads and can be rectangular or trapezoidal on plan as shownin Table 283 Alternatively it could be made up as a series ofrectangles as shown at (b) in Table 284 In this last case eachrectangle should be proportioned so that the load upon it acts atthe centre of the area with the area of the rectangle being equalto the load divided by an allowable bearing pressure and thevalue of the pressure being the same for each rectangle

If it is not practical to proportion the combined base in thisway then the total load will be eccentric If the base is thickenough to be considered to act as a rigid member the groundbearing pressure will vary according to the diagram shown at(c) in Table 284 For a more flexible base the pressure will begreater immediately under the loads giving a distribution ofpressure as shown at (d) in Table 284

In the case of a uniform distribution or linear variation of pres-sure the longitudinal bending moment on the base at any sectionis the sum of the anti-clockwise moments of the loads to the leftof the section minus the clockwise moment due to the groundpressure between the section and the left-hand end of the baseThis method of analysis gives larger values for longitudinal bend-ing moments than if a non-linear variation is assumed Formulaefor combined bases carrying two loads are given in Table 283

Example 2 A strip base 15 m long and 15 m wide carries aline of five unequal concentrated loads arranged eccentricallyas shown in the following figure The bending moment is to bedetermined at the position of load F2 where the values of theloads and the distances from RH end are as follows

F1 500 kN F2 450 kN F3 400 kN F5 300 kN

z1 14 m z2 11 m z3 8 m z4 5 m z5 15 m

Chapter 18

Foundations andretaining walls

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282Foundations presumed allowable bearing values and separate bases

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283Foundations other bases and footings

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284Foundations inter-connected bases and rafts

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Pad foundations 199

The first step is to determine the eccentricity of the total loadand check that e (13Fz13F l2) l6 Hence

13Fz 500 14 450 11 400 8 350 5 300 15

17350 kNm

13F 500 450 400 350 300 2000 kN

e 173502000 152 1175 m l6 25 m

The maximum and minimum bearing pressures can now becalculated from the formula in Table 282 where

k (1 6el) (1 6 117515) 147 or 053

Hence with 13Fbl 2000(15 15) 889 kNm2

fmax 147 889 1307 kNm2

fmin 053 889 471 kNm2

Then for any section X-X at distance y from the left-hand endof the base the bearing pressure is

fx fmax (fmax fmin) yl

Considering the loads to the left of section X-X where x is thedistance of a load from the section the resultant bendingmoment on the base at section X-X is

M 13Fx (2fmax fx) by26

Thus at the position of load F2 where y 4 m and x1 3 m

fx 1307 (1307 471) 415 1084 kNm2

M 500 3 (2 1307 1084) 15 426 208 kNm

1813 Balanced bases

With reference to the diagrams on the upper right-hand side ofTable 284 (a) shows a system in which beam BC rests on a baseat A supports a column on the overhanging end C and iscounterbalanced at B The reaction at A which depends on therelative values of BC and BA can be provided by a basedesigned for a concentric load The counterbalance at B couldbe provided by load from another column as at (b) in which casethe dead load on this column needs to be sufficient to counter-balance the dead and imposed loads on the column at C and viceversa It is often possible to arrange for base A1 to be positioneddirectly under column B Formulae giving the values of thereactions at A and A1 are given in Table 283 where variouscombinations of values for F1 and F2 usually need to be consideredThus if F1 varies from F1 max to F1 min and F2 varies from F2 max toF2 min reaction R (see diagram in Table 283) will vary from

Rmax e F1max l to Rmin e F1min l

Hence R1 and R2 can have the following values

R1max F1max Fbase 1 Rmax Fbeam 2

R1min F1min Fbase 1 Rmin Fbeam 2

R2max F2max Fbase 2 Rmin Fbeam 2

R2min F2min Fbase 2 Rmax Fbeam 2

Therefore base 1 must be designed for a maximum load of R1max

and base 2 for R2max but R2min must always be positive From thereactions the shearing forces and bending moments on the beamcan be calculated In the absence of a convenient column loadbeing available at B a suitable anchorage must be provided by

other means such as a counterweight in mass concrete as at (c)or the provision of tension piles If the column to be supported isa corner column loading the foundation eccentrically in twodirections one parallel to each building line as at (d) it is some-times possible to use a diagonal balancing beam anchored by aninternal column D In other cases however the two wall beamsmeeting at the column can be designed as balancing beams toovercome the double eccentricity For beam EC the cantilevermoment is FE e1 where FE is the column load and the upwardforce on column C is FE e1 (l1 e1) For beam EF the corre-sponding values are FE e2 and FE e2 (l2 e2) respectively

1814 Rafts

The required thickness of a raft foundation is determined by theshearing forces and bending moments which depend on themagnitude and spacing of the loads If the thickness does notexceed 300 mm a solid slab as at (a) in the lower part ofTable 284 is generally the most convenient form If a slab atground level is required it is usually necessary to thicken theslab at the edge as at (b) to ensure that the edge of the raft isdeep enough to avoid weathering of the ground under the raft Ifa greater thickness is required beam-and-slab constructiondesigned as an inverted floor as at (c) is more efficient In caseswhere the total depth required exceeds 1 m or where a level topsurface is required a cellular construction consisting of a topand bottom slab with intermediate ribs as at (d) can be adopted

When the columns on a raft are not equally loaded or are notsymmetrically arranged the raft should be designed so that thecentre of area coincides with the centre of gravity of the loadsIn this case the pressure on the ground is uniform and therequired area is equal to the total load (including the weight ofthe raft) divided by the allowable bearing value If the coinci-dence of the centre of area of the raft and the centre of gravityof the loads is impractical due to the extent of the raft beinglimited on one or more sides the shape of the raft on planshould be so that the eccentricity ew of the total load Ftot is keptto a minimum as in the example shown at (f)

The maximum bearing pressure (which occurs at the cornershown at distance a1 from axis N-N on the plan and should notexceed the allowable bearing value) is given by

where Araft is the total area of the raft and Iraft is the secondmoment of area about the axis N-N which passes through thecentre of area and is normal to the line joining the centre of areaand the centre of gravity of the loads The pressure along axisN-N is FtotAraft and the minimum pressure (at the corner shownat distance a2 from axis N-N) is given by

When the three pressures have been determined the pressure atany other point or the mean pressure over any area can beassessed Having arranged for a rational system of beams orribs to divide the slab into suitable panels as suggested by thebroken lines at (f) the panels of slab and the beams can bedesigned for the bending moments and shear forces due to thenet upward pressures to which they are subjected

fmin Ftot

Araft

Ftot ewa2

Iraft

fmax Ftot

Araft

Ftot ew a1

Iraft

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182 OPEN-PILED STRUCTURES

Expressions from which the loads on groups of inclined andvertical piles supporting jetties and similar structures can beobtained are given in Table 285 For each probable conditionof load the forces acting on the superstructure are resolved intohorizontal and vertical components Fh and Fv the points ofapplication of which are also determined If the direction ofaction and position of the forces are opposite to those in thediagrams the signs in the formulae must be changed

Example 1 vertical piles only The adjacent figure showsa cross section through a jetty where the loads apply to one rowof piles (ie n 1 for each line and 13n 4) Since the groupis symmetrical x 05 90 45 m

M Fv e Fh h 800 075 100 48 120 kNm

The calculations for the loads on the piles are shown in thefollowing table from which the maximum load on any pile is212 kN For each pile the shear force is 1004 25 kN andthe bending moment is 25 482 60 kNm

Foundations and retaining walls200

Example 2 vertical and inclined piles The adjacent figureshows a cross section through a jetty where all the piles aredriven to the same depth Since A is the same for all the pilesif J Al is taken as unity for piles N1 and N4 then for piles N2

and N3 J 4radic(1 42) 097 Since the group is symmetrical132 0 133 not needed 134 0 k 131 135 xo 92 45 mand yo 0

M Fv e Fh h 800(525 45) 0 600 kNm

The calculation is then as shown in the following table fromwhich k 3826 0114 0436 The axial loads on the pilesare given by Nx kp (kw Fv kh Fh km M) hence

N1 10 (0261 800 0 0111 600) 1421 kN

N2 0941 (0261 800 219 100 0) 4026 kN

N3 0941 (0261 800 219 100 0) 96 kN

N4 10 (0261 800 0 0111 600) 2755 kN

Thus the maximum load on any pile is 4026 kN Note that theweight of the pile has to be added to the above values

kw km

Pile x x2 Axial loadno (m) (m2) (kwFv kmM)

N1 45 2025 025 45 0100 (025 X 800) (0100 X 120) 188kN45

N2 15 225 025 15 0033 200 (0033 X 120) 196kN45

N3 15 225 025 0033 200 4 204kN

N4 45 2025 025 0100 200 12 212kN

I nx2 4500 m4

xnI 1

n

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285Foundations loads on open-piled structures

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Foundations and retaining walls202

Example 3 inclined piles only The adjacent figure shows across section through a jetty where cot 13 5 Since the valuesof A and l are the same for each pile unity may be substitutedfor J Al For piles N1 and N3 B 5 and for piles N2 andN4 B 5 Since the group is symmetrical 132 0 133 notneeded 134 0 k 131 135 xo 3 m and yo 0

M Fv e Fh h 800(375 30) 0 600 kNm

The calculation is then as shown in the following table fromwhich k 3826 01538 05917 The axial loads on thepiles are given by Nx kp (kw Fv kh Fh km M) hence

N1 09806 (026 800 13 100 00867 600)

09806 (208 130 52) 2804 kN

N2 09806 (208 130 52) 255 kN

N3 09806 (208 130 52) 3824 kN

N4 09806 (208 130 52) 1275 kN

Thus the maximum load on any pile is 3824 kN to which theweight of the pile has to be added

Pile x Xno 1 5 (m) (m) 6 I kp kw kh km

N1 0114 450436 405

1 0 0 45 2025 1 0261 0 0111

N2 097 42 097

097 4

3826

1 + 42 1 + 42 radic(1 42) 4 0436

0913 0057 45 0 0 0941 0261 219 0

N3 0913 0057 45 0 0 0941 0261 219 0

N4 1 0 9 45 2025 1 0261 0 0111

Totals 3826 0114 mdash mdash 405 mdash mdash mdash mdash

Pile x Xno 1 5 (m) (m) 6I kp kw kh XI

N152 1 09615(3)2 5 01538 3846 3

152 152 radic(152) 05917 5 05917 3462

09615 00385 0 30 865 09806 026 13 00867

N2 3846

5 05917

09615 00385 0 30 865 09806 026 13 00867

N3 3

3462

09615 00385 60 30 865 09806 026 13 00867

N4 09615 00385 60 30 865 09806 026 13 00867

Total 3846 01538 mdash mdash 3462 mdash mdash mdash mdash

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chart given in Table 286 which is a modified version of a chartpublished in ref 63 The modified chart is applicable to designmethods in which the soil parameters incorporate either mobil-isation factors as in BS 8002 or partial factors of safety asin the Eurocodes

Stability against overturning is assured over the entire rangeof the chart and the maximum bearing pressure under serviceconditions can be investigated for all types of soil A uniformsurcharge that is small compared to the total forces acting onthe wall can be simply represented by an equivalent height ofsoil l can be replaced by le l q where q is the surchargepressure In more general cases le 3MhFh and 2FhKAle

2

can be used where Fh is the total horizontal force and Mh is thebending moment about underside of base due to Fh

The chart contains two curves showing where the bearingpressure is uniform and triangular respectively A uniformbearing condition is important when it is necessary to avoidtilting in order to minimise deflection at the top of the wallIt is generally advisable to maintain ground contact overthe full area of the base especially for clays where theoccurrence of ground water beneath the heel could soften theformation

Further information on the use of the chart for walls that aredesigned in accordance with BS 8002 and BS 8110 is given insection 284

184 BOX CULVERTS

Formulae for the bending moments in the corners of a boxculvert due to symmetrical load cases are given in Table 287Some notes on the different load cases and assumed groundconditions are given in section 742 Design requirements ofHighways Agency BD 3187 are summarised in section 743

Box culverts 203

Notes on design examples (1)ndash(3) In each case the pilegroup is symmetrical and is subjected to the same imposedloads Examples (2) and (3) are special cases of symmetricalgroups for which 134 0 and therefore yo 0 this conditionapplies only when the inclined piles are in symmetrical pairswith both pairs meeting at the same pile-cap

Example (3) requires the smallest pile but the differencein the maximum load is small between (2) and (3) Althoughthe maximum load on a pile is least in (1) the bendingmoment requires a pile of greater cross-sectional area to pro-vide the necessary resistance to combined bending and thrustIf the horizontal force is increased the superiority of (3) isgreater If Fh 200 kN (instead of 100 kN) the maximumpile loads are 236 kN (and a large bending moment of120 kNm) in (1) 609 kN in (2) and 510 kN in (3)Arrangement (2) is the most suitable when Fh is small IfFh 10 kN the maximum pile loads are 255 kN (and abending moment of 6 kNm) in (1) 217 kN in (2) and 268 kNin (3) Arrangement (1) is the most suitable when there is nohorizontal load

183 RETAINING WALLS

Various types of earth retention systems for which notes aregiven in section 731 are shown in Table 286 Information onthe pressures exerted by soils on retaining structures is given insection 91 and Tables 210 to 214

1831 Walls on spread bases

Several walls on spread bases for which notes are given insection 732 are shown in Table 286 Suitable dimensions fora base to a cantilever wall can be estimated with the aid of the

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Roker

Facing

Groutedbars

In-situ wall

Gravityelements(interlockingcribs)

Strul

Cantileverwall

Potential failure wedge

In-situwall

Trebocks

Anchors

Potentialfailurewedge

Potentialfailurewedge

Facingpanels

Strips or grids

286Retaining walls

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287Rectangular culverts

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191 STAIRS

Some general notes on stairs are given in section 614 Fordetails of characteristic imposed loads on stairs and landingsand horizontal loads on balustrades refer to BS 6399 Part 1Stairs forming monolithic structures are generally designed forthe uniformly distributed imposed load only Stairs that areformed of separate treads (usually precast) cantilevering from awall or central spine beam must be designed also for the alter-native concentrated load Some general information on stairtypes and dimensions is given in Table 288 and comprehensivedata is given in BS 5395

1911 Simple stairs

The term lsquosimplersquo is used here for a staircase that spans in thedirection of the stair flight between beams walls or landingslocated at the top and bottom of the flight The staircase mayinclude a section of landing spanning in the same direction andcontinuous with the stair flight For such staircases the followingstatements are contained in BS 8110

When staircases surrounding open wells include two spansthat intersect at right angles the load on the area common to bothspans may be divided equally between the spans When the stair-case is built at least 110 mm into a wall along part of all of itslength a 150 mm wide strip adjacent to the wall may be deductedfrom the loaded area When the staircase is built monolithicallyat its ends into structural members spanning at right angles to itsspan the effective span of the stairs should be taken as the hori-zontal distance between the centrelines of the supporting mem-bers where the width of each member is taken not more than18 m For a simply supported staircase the effective span shouldbe taken as the horizontal distance between the centrelines of thesupports or the clear distance between the faces of supports plusthe effective depth of the stair waist whichever is the lesser If astair flight occupies at least 60 of the effective span the per-missible spaneffective depth ratio calculated for a slab may beincreased by 15

1912 Free-standing stairs

In ref 64 Cusens and Kuang employ strainndashenergy principles todetermine expressions relating the horizontal restraint force Hand moment Mo at the mid-point of a free-standing stair

By solving the two resulting equations simultaneously thevalues of H and Mo obtained can then be substituted into generalexpressions to determine the forces and moments at any pointalong the structure

It is possible by neglecting subsidiary terms to simplify thebasic equations produced by Cusens and Kuang If this is donethe expressions given in Table 288 are obtained and these yieldH and Mo directly Comparisons of the solutions obtained bythese simplified expressions with those obtained using theexpressions presented by Cusens and Kuang show thatthe resulting variations are negligible for values in the rangeencountered in concrete design

The expressions given in Table 288 are based on a value ofGE 04 with C taken as half of the St Venant value for plainconcrete As assumed by Cusens and Kuang only half ofthe actual width of the landing is considered to determine its sec-ond moment of area

Example 1 Design a free-standing staircase assumed to befully fixed at the ends to support total ultimate loads per unitlength nf 169 kNm and nl 150 kNm The dimensions areas follows a 27 m b 14 m b1 18 m hf 100 mmhl 175 mm and 30o

From the expressions given in Table 288 H 8186 kNmand Mo 3587 kNmm If Cusens and Kuangrsquos more exactexpressions are used to analyse the structure H 8189 kNmand Mo 3567 kNmm Thus the errors involved by using theapproximate expressions are negligible for H and about 05for Mo If these values of H and Mo are substituted into the otherexpressions in Table 288 corresponding values of Mv Mh andT at any point in the structure can be found for various loadcombinations as shown in the table in page no 208

1913 Sawtooth stairs

In ref 65 Cusens shows that if axial shortening is neglected andthe strainndashenergy due to bending only is considered the mid-span moments for a so-called sawtooth stairs are given by thegeneral expression

Ms

where k (stiffness of tread)(stiffness of riser) and j is thenumber of treads

nl2(k11 kk12)

j2(k13 kk14)

Chapter 19

Miscellaneousstructures and details

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288Stairs general information

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If j is odd

k11 j216 j( j ndash 1)( j ndash 2)48

k12 ( j ndash 1)216 ( j ndash 1)( j ndash 2)( j ndash 3)48

k13 j2 k14 ( j ndash 1)2

If j is even

k11 j( j ndash 1)( j ndash 2)48

k12 ( j ndash 1)( j ndash 2)( j ndash 3)48

k13 ( j ndash 1)2 k14 ( j ndash 2)2

The chart on Table 289 gives coefficients to determine thesupport moments for various values of j and k Having found thesupport moment the maximum mid-span bending moment canbe determined by using the appropriate expression on the tableand deducting the support moment

Typical bending moment and shearing force diagrams for astair are also shown on Table 289 together with suggestedarrangements of reinforcement Because of the stair profilestress concentrations occur in the re-entrant corners and the realstresses to be resisted will be larger than those obtained from themoments To resist these increased stresses Cusens recom-mends providing twice the reinforcement theoretically requiredunless suitable fillets or haunches are incorporated at the junc-tions in which case the reinforcement need only be about 10more than is theoretically necessary The method of reinforcingshown in diagram (a) is very suitable but is generally only prac-tical if haunches are provided Otherwise the arrangementshown in diagram (b) should be adopted A further possibility isto arrange the bars as shown in diagram (a) on Table 363 forwall-to-wall corners

Example 2 A sawtooth stairs has seven treads each 300 mmwide with risers 180 mm high the thickness of both being100 mm The stairs which are 10 m wide are to be designed tosupport a characteristic imposed load of 30 kNm2 to therequirements of BS 8110

The self-weight of the treads and risers (assuming no finishesare required) is

gk 01 24 (03 018)03 384 kNm2

For design to BS 8110 total design ultimate load for a 10 mwide stair is given by

n 14 gk 16qk 14 384 16 30 1018 kNm

Since lt 300 mm lr 180 mm and ht hr 100 mm

k ht3lrhr

3lt 180300 06

From the chart on Table 289 for 7 treads and k 06 thesupport moment coefficient is ndash 0088 Thus

Ms ndash0088nl2 ndash0088 1018 (03 7)2 ndash395 kNm

Since j 7 is odd the free bending moment is given by

M (nl28)( j2 1)j2 1018 2128 5049 573 kNm

Hence the maximum moment at mid-span is

M0 M ndash Ms 573 ndash 395 178 kNm

1914 Helical stairs

By using strainndashenergy principles it is possible to formulate forsymmetrically loaded helical stairs fully fixed at the ends thefollowing two simultaneous equations in M0 and H

M0[K1(k5 sin213) k5 k7]

HR2[k4(k7 ndash K1)tan k5 sin cos (1 ndash K2)]

nR12[K1(k5 sin213ndash sin13) k5 k7 k6 k7R2R1] 0

M0[k4(k7 ndash K1) k5 (k7 ndash K2)]

HR2[12K1 tan ( 133 ndash 132sin213ndash 2k4)

12k7 tan ( 133 132sin213 2k4) 2k4 tan (k7 ndash K2)

k5 cos2 (tan K2 cot)] nR12[K1(k6 ndash k4) k4 k7

k7 (132sin 13 2k6)R2R1 (k7 ndash K2)(k5 k6R2R1)] 0

where

k4 13cos213ndash sin213 k5 13ndash sin213

k6 13cos13ndash sin13 k7 cos2 K2sin2

K1 GCEI1 K2 GCEI2 and 13 2

The equations can be solved on a programmable calculator orlarger machine to obtain coefficients k1 and k2 representing M0

and H respectively If the resulting values of M0 and H are thensubstituted into the equations given on Table 289 the bendingand torsional moments shearing forces and thrusts at any pointalong the stair can be easily calculated The critical quantity con-trolling helical stair design is usually the vertical moment Mvs atthe supports and a further coefficient can be derived to give thismoment directly

In ref 66 Santathadaporn and Cusens give 36 design chartsfor helical stairs covering ranges of of 60o to 720o of 20o

to 50o bh of 05 to 16 and R1R2 of 10 to 11 for GE 37The four design charts provided on Tables 290 and 291 havebeen recalculated for GE 04 with C taken as half of theSt Venant value for plain concrete These charts cover rangesof of 60o to 360o and of 20o to 40o with values for bh of5 and 10 and R1R2 of 10 and 11 being the ranges met most

14

12

18

14

12

13

12

13

12

12

Miscellaneous structures and details208

Application of Values of Mv (kNmm) Values of Mh (kNmm) Values of T (kNmm)imposed load

At O At B At D At A At B in OB Throughout AB At B in BC Throughout AB

Throughout 3587 1680 116 861 7367 8294 735 369Flights only 2481 922 160 1067 5036 5668 403 255Landing only 3132 1680 314 333 6486 7303 735 322

Note At point B the expressions give theoretical values for Mv (with imposed load applied throughout) that reduce abruptly from ndash 4195 kNmm in OB tondash368 kNmm in BC due to the intersection with flight AB Since the members in the actual structure are of finite width Cusens and Kuang recommendredistributing these moments across the intersection between the flight and the landing to give a value of (ndash4195 ndash 368)2 2282 kNmm

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289Stairs sawtooth and helical stairs

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290Design coefficients for helical stairs ndash 1

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291Design coefficients for helical stairs ndash 2

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frequently in helical stair design Interpolation between thevarious curves and charts will be sufficiently accurate forpreliminary design purposes

Example 3 A helical stairs having an angle of inclination tothe horizontal plane of 25o is to be designed to support acharacteristic imposed load of 30 kNm2 to the requirements ofBS 8110 The stairs are to be 12 m wide with a minimum slabthickness of 120 mm The radius to the inside of the stair isRi 900 mm and the angle turned through is 240o

Assuming the mean thickness on plan of the stairs (includingtreads and finishes) is 220 mm the self-weight of the stairs is022 24 53 kNm2 and the design ultimate load intensity

n 14 53 16 30 1222 kNm2

The radius of the centreline of the load is given by

R1 158 m

The radius of the centreline of the stairs

R2 09 05 12 15 m

Hence R1R2 15815 105 bh 1200120 10 andfrom the charts on Tables 290 and 291 interpolating asnecessary k1 ndash 012 k2 152 and k3 ndash 032 Thus fora 12 m wide stairs the total design values are

M0 12k1nR22 ndash012 1222 152 12 ndash396 kNm

H 12k2nR2 152 1222 15 12 334 kN

Mvs 12k3nR22 ndash032 1222 152 12 ndash1056 kNm

The slab should now be checked to ensure that the thicknessprovided is sufficient to resist Mvs Then assuming this is sothe foregoing values of M0 and H can be substituted into theequations for Mv Mn T N Vn and Vh given on Table 289 toobtain moments and forces at any point along the stairs Forexample where 13 60o Mv 111 kNm Mn ndash 4817 kNmT 005 kNm N ndash 365 kN Vn 968 kN and Vh 167 kNTypical distributions of moments and forces along the stair areshown on Table 289

192 NON-PLANAR ROOFS

1921 Prismatic structures

To design a simple prismatic roof or any structure comprising anumber of planar slabs for service load the resultant loadsQ acting at right angles to each slab and the unbalanced thrustsN acting in the plane of each slab are determined first taking intoaccount the thrust of one slab on another The slabs are thendesigned to resist the transverse bending moments due to theloads Q assuming continuity combined with the thrusts N Thelongitudinal forces F due to the slabs bending in their own planeunder the loads N are for any two adjacent slabs AB and BCcalculated from formula (2) in Table 292 where MAB and MBC

are found from formula (1) if the structure is freely supported atthe end of length L For each pair of slabs AB-BC BC-CD andso on there is an equation like formula (2) containing threeunknown forces F If there are n pairs there are (n ndash 1) equa-tions and (n 1) unknowns The conditions at the outer edgesa and z of the end slabs determine the forces F at these edgesfor example if the outer edges are unsupported Fa Fz 0

2(213 093)

3(212 092)

2(R3o R3

i )

3(R2o R2

i )

The simultaneous equations are solved for the remainingunknown forces FA FB FC and so on The longitudinal stress atany junction B is calculated from the formula (in Table 292) forfb Variation of the longitudinal stress from one function to thenext is rectilinear If fb is negative the stress is tensile and shouldbe resisted by reinforcement The shearing stresses are generallysmall

1922 Domes

A dome is designed for the total vertical load only that is forthe weights of the slab any covering on the slab any ceiling orother distributed load suspended from the slab and the imposedload The service load intensity w is the equivalent load per unitarea of surface of the dome Horizontal service loads due towind and the effects of shrinkage and changes in temperaturecan be allowed for by assuming an ample normal load or byinserting more reinforcement than that required for the normalload alone or by designing for stresses well below permissiblevalues or by combining any or all of these methods

Segmental domes Referring to the diagram and formulae inTable 292 for a unit strip at S the circumferential force actingin a horizontal plane is T and the corresponding force actingtangentially to the surface of the dome (the meridional thrust) isN At the plane where 13 is 51o48 that is the plane of ruptureT 0 Above this plane T is compressive reaching a maximumvalue of 05wr at the crown of the dome (13 0) Below thisplane T is tensile equalling 0167wr when 13 60o andwr 90o The meridional thrust N is 05wr at the crown0618wr at the plane of rupture 0667wr when 13 60o andwr when 13 90o that is N increases from the crown towardsthe support and has its greatest value at the support

For a concentrated load F applied at the crown of the domeT and N are given by the appropriate formulae in Table 292where T is tensile and N is compressive The load is assumed tobe concentrated on so small an area at the crown that it can betaken as a point load The theoretical stress at the crown is there-fore infinite but the practical impossibility of obtaining a pointload invalidates the use of the formulae when 13 0 or verynearly so For domes of varying thickness see ref 67

For a shallow dome approximate analysis only is sufficientand appropriate formulae are given in Table 292

Conical domes In a conical dome the circumferential forcesare compressive throughout and for any horizontal plane atdistance x from the apex are given by the expression for T inTable 292 the corresponding force in the direction of the slopebeing N The horizontal outward force per unit length of thecircumference at the bottom of the slope Tr needs to be resistedby the supports or a bottom ring beam

1923 Segmental shells

General notes on the design of cylindrical shell roofs and the useof Tables 293 and 294 are given in section 6 16

Membrane action Consideration of membrane action onlygives the following membrane forces per unit width of slab dueto the uniform load shown on Table 292 stresses are obtainedby dividing by the thickness of the shell h Negative values of Vindicate tension in the direction corresponding to an increase in

Miscellaneous structures and details212

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292Non-planar roofs general data

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293Shell roofs empirical design method ndash 1

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294Shell roofs empirical design method ndash 2

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x and a decrease in 13x In the case of F positive values indicatetension Reinforcement should be provided approximately inline with and to resist the principal tensile forces If the shell issupported along any edges the forces will be modified accord-ingly The values of the forces at any point are as follows

Tangential force

Fy ndash (g qcos13x) rcos13x

Longitudinal force

Fx ndash (1 ndash x)(xr)[gcos13x 15q(cos213x ndash sin213x)]

Shearing force

Vxy (2x ndash l)(g 15qcos13x)sin13x

Principal forces (due to membrane forces only)

Fp 05(Fx Fy) [(Fx ndash Fy)2 4V2xy]05

tan 2

At A (mid-point at edge 13x 13 x l2)

FyA ndash (g qcos13) rcos13

FxA ndash (l24r)[gcos13 15q(cos213ndash sin213)]

VxyA 0

At B (mid-point at crown 13x 0 x l2)

FyB ndash (g q)r

FxB ndash (l24r)(g 15q)

VxyB 0

At C (support at edge 13x 13 x 0)

FyC ndash (g qcos13) rcos13

FxC 0

VxyC ndash (g 15qcos13) lsin13

At D (support at crown 13x 0 x 0)

FyD ndash (g q) r

FxD 0 VxyD 0

Beam action If the ratio of the length of a cylindrical shell toits radius lr is not less than 25 the longitudinal forces canbe approximated with reasonable accuracy by calculating thesecond moment of area Ixx and the vertical distance from theneutral axis to the crown yndash from the approximate expressions

Ixx cong R2h (R ndash 3h2)(2 sincosndash 2sin2)

yndash cong R ndash [(R ndash 2h3)sin]

Then if n is the total uniform load per unit area acting on theshell (ie including self-weight etc) the maximum bendingmoment in a freely supported shell is given by

M (213rn)l28 13rnl24

Hence from the relationship MIxx = fy the horizontal forces atmid-span at the crown and springing of a shell of thickness h aregiven by the expressions

Fx(bottom) M[r(1 cos 13) y]h

IxxFx(top)

Myh

Ixx

2Vxy

Fx Fy

At the supports the total shearing force in the shell is

V (213rn)l2 13rnl

The shearing stress at any point is then given by v (VAx)(2hIxx)

where Axndash is the first moment of area about the neutral axis of thearea of cross section of shell above the point at which the shear-ing stress is being determined

The principal shortcoming of this approximate analysis is thatit does not indicate in any simple manner the magnitude of thetransverse moments that occur in the shell However a tabularmethod has been devised by which these moments can be eval-uated indirectly from the lateral components of the shearingstresses for details see ref 68

1924 Hyperbolic-paraboloidal shells

The simplest type of double-curved shell is generated by theintersection of two separate sets of inclined straight lines (parallelto axes XX and YY respectively as shown in the diagrams on Table292) Vertical sections through the shell at angles XX or YY areparabolic in shape while horizontal sections through the surfaceform hyperbolas hence the name hyperbolic paraboloid

Individual units such as those shown in diagrams (a) and (b)in Table 292 can be used separately being supported oncolumns or buttresses located at either the higher or the lowercorners Alternatively groups of units can be combined toachieve roofs having attractive and unusual shapes such as isshown in diagram (c) Some idea of the more unlikely forms thatcan be achieved may be obtained from ref 69

If the shell is shallow and the loading is uniform the shellbehaves as a membrane transferring uniform compressive andtensile forces of F2c (where F is the total load on the unit andc is the rise) acting parallel to the directions of principalcurvature to the edges of the shell The edge forces are thentransmitted back to the supports along beams at the shell edgesThe main problems that arise when designing these shells are theinteraction between the shell and the supporting edge membersthe design of the buttresses or ties needed to resist the horizon-tal component of the forces at the supports and the fact thatexcessive deflections at unsupported edges lead to stresses thatdiffer considerably from those predicted by simplified theories

Increasing the sharpness of curvature of the shell increases itsstability and reduces the forces and reactions within the shellbut to avoid the need for top forms the maximum slope shouldnot exceed 45o this corresponds to a value of 1radic2 for theratio ca or cb in the diagrams on Table 292 To ensure stabilityif a single unit is used the ratio should be not less than 15A useful introduction to the theory and design of hyperbolic-paraboloidal shells is given in ref 70

193 BEAMS CURVED ON PLAN

Some general notes on beams forming a circular arc on planare given in section 617 The following analyses apply only ifthe appropriate negative bending and torsional moments can bedeveloped at the end supports

1931 Concentrated loads

If a beam LR (see Table 295) curved on plan is subjected to aconcentrated load F such that the angle between the radius at

Miscellaneous structures and details216

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295Bow girders concentrated loads

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the point of application F of load F and the radius at themid-point O of the beam is 0 the following expressionsare applicable at any point X between F and L (ie 0)

M M0cos ndash T0sin V0rsin ndash Frsin ( ndash 0)

T M0sin T0cos V0r(1 ndash cos) ndash Fr[1 ndash cos ( ndash 0)]

V ndash V0 F

where M0 T0 and V0 are respectively the bending moment thetorsional moment and the shearing force at mid-span r isthe radius of curvature in plan and is the angle defining theposition of X as shown in the diagram on Table 295

If X is between F and O (ie X 0) terms containing Fare equal to zero If X is between O and R (ie X) signs of theterms containing sin should also be reversed Now by writingM0 K1Fr T0 K2Fr and V0 K3F

K1 K2 and K3

where k1 k2 k3 and so on are given by the following expressionsin which K EIGC (ie flexural rigiditytorsional rigidity)

k1 (K ndash 1)sin0(sin 213ndash sin 20)

(K ndash 1)cos0(sin213ndash sin20)

(K 1)(13ndash 0)sin0 ndash K(cos13ndash cos0)

k2 (K 1)13ndash 12(K ndash 1)sin213

k3 K(sin13ndash sin0) ndash 14 (K ndash 1)cos0(sin213ndash sin20)

(K ndash 1)sin0(sin213 ndash sin20)

(K 1)(13 ndash 0)cos0

k4 (K 1)13 (K ndash 1)sin213

k5 2Ksin13 ndash (K 1)13 ndash (K ndash 1)sin213

k6 (K ndash 1)cos0(sin213ndash sin20) K(13ndash 0)

(K 1)(13ndash 0)cos0

(K ndash 1)sin0(sin213ndash sin20)

K(1 cos0)(sin13ndash sin0)

Ksin0(cos13ndash cos0)

k7 k5

k8 (K ndash 1)sin213ndash 4Ksin13 (3K 1)13

For rectangular beams the graphs provided on Table 295 enablevalues of K1 K2 and K3 to be read directly for given values of 130 and hb (ie depthwidth) Values of G 04E and C J2as recommended in BS 8110 have been used

1932 Uniform load

For a curved beam with a UDL over the entire length owing tosymmetry the torsional moment (and also the shearing force) atmid-span is zero By integrating the foregoing formulae thebending and torsional moments at any point X along the beamare given by the expressions

M M0cos ndash nr2(1 ndash cos)

T M0sin ndash nr2( ndash sin)

If M0 K4nr2 where

K4 14[(1 K)sin13 K13 cos13]

213 (1 K) sin 213 (1 K)

12

12

12

14

12

12

12

12

12

12

14

k4k6 k3k7

k4k8 k5k7

k3k8 k5k6

k4k8 k5k7

k1

k2

these expressions can be rearranged to give the bending andtorsional moments at the supports and the maximum positivemoments in the span in terms of non-dimensional factors K4 K5K6 and K7 as shown on Tables 296 and 297 The factors can beread from the charts given on the tables

Example 1 A bow girder 450 mm deep and 450 mm widehas a radius of 4 m and subtends an angle of 90o The ends arerigidly fixed and the total UDL is 200 kN The maximummoments are to be determined

The distributed load per unit length

n 200(r2) 200( 42) 318 kNm

From the charts on Tables 296 and 297 (with hb 10 and13 45o) K4 0086 K5 ndash 023 K6 ndash 00175 K7 0023with 1 23o and 2 40o

Maximum positive bending moment (at midspan)

K4nr2 0086 318 42 438 kNm

Maximum negative bending moment (at supports)

K5nr2 ndash 023 318 42 ndash 117 kNm

Zero bending moment occurs at 1 23o

Maximum negative torsional moment (at supports)

K6nr2 ndash 00175 318 42 ndash 89 kNm

Maximum positive torsional moment (at 1 23o)

K7nr2 0023 318 42 117 kNm

Zero torsional moment occurs at 2 40o

Example 2 The beam in example 1 supports a concentratedload of 200 kN at a point where the radius subtends an angleof 15o from the left-hand end (ie 0 30o) Moments andshearing forces at midspan and the supports are required

From the charts on Table 295 (with hb 10 13 45o and0 30o) K1 0015 K2 ndash 00053 K3 0067

At midspan ( 0)

M0 0015 200 4 12 kNm

T0 ndash 00053 200 4 ndash 43 kNm

V0 0067 200 134 kN

At left-hand support ( 45o)

M M0cos 45o ndash T0sin 45o V0 rsin 45o ndash Frsin (45o ndash 30o) [12 ndash (ndash 43) 134 4) 0707 ndash 200 4 0259 ndash 158 kNm

T M0sin 45o T0cos 45o V0 r (1 ndash cos 45o) Fr (1 ndash cos 15o)

(12ndash43)070713440293ndash20040034 ndash 61 kNm

V V0 ndash F 134 ndash 200 1866 kN

At right-hand support ( 45o)

M M0cos 45o T0sin 45o ndash V0 rsin 45o

[12 ndash 43 ndash 134 4) 0707 ndash 324 kNm

T ndash M0sin 45o T0cos 45o V0 r (1 ndash cos 45o) (ndash 12 ndash 43) 0707 134 4 0293 42 kNm

V V0 134 kN

Miscellaneous structures and details218

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296Bow girders uniform loads ndash 1

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297Bow girders uniform loads ndash 2

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194 BEARINGS HINGES AND JOINTS

A comprehensive guide on bridge bearings and expansion jointsincluding a treatment of the design techniques used to accom-modate movements in bridges is contained in ref 71 Varioustypes of bridges for which notes are given in section 62 andtypical span ranges are shown in Table 298

1941 Hinges and bearings

A hinge is an element that can transmit thrust and transverseforce but permits rotation without restraint If it is vital for suchaction to be fully realised a steel hinge can be providedAlternatively hinges that are monolithic with the member canbe formed as indicated at (a) and (b) in Table 299 ThelsquoMesnagerrsquohinge shown at (a) has been used for example in theframes of large bunkers to isolate the container from the sub-structure or to provide a hinge at the base of the columns of ahinged frame bridge The hinge-bars lsquoarsquo resist the entirehorizontal shear force and the so-called throat of concrete at Dmust be sufficient to transfer the full compressive force from theupper to the lower part of the member The hinge-bars should bebound together by links lsquodrsquo and the main vertical bars lsquoersquo shouldterminate on each side of the slots B and C It may be advanta-geous during construction to provide bars extending across theslots and then cut these bars on completion of the frame Theslots should be filled with a bituminous material or lead or asimilar separating layer

The Freyssinet hinge shown at (b) has largely superseded theMesnager hinge In this case the large compressive stress acrossthe throat results in a high shearing resistance and the inclusionof bars crossing the throat can adversely affect the hinge Testshave shown that as a result of biaxial or triaxial restraint suchhinges can withstand compressive stresses of several times thecube strength without failure occurring The bursting tension oneach side of the throat normally governs the design of this typeof hinge

A number of other types of hinges and bearings that have beenused on various occasions are shown in Table 299

(c) a hinge formed by the convex end of a concrete memberbearing in a concave recess in the foundations

(d) a hinge suitable for the bearing of a girder where rotationbut not sliding is required

(e) a bearing for a girder where sliding is required

(f) a mechanical hinge suitable for the base of a large portalframe or the abutment of a large hinged arch bridge

(g) a hinge suitable for the crown of a three-hinged arch whenthe provision of a mechanical hinge is not justified

(h) a bearing suitable for the support of a freely suspended spanon a cantilever in an articulated bridge

Bridge bearings have to be able to accommodate the rotationsresulting from deflection of the deck under load They alsohave to be able to accommodate horizontal movements ofthe deck caused by prestress creep shrinkage and temperaturechange Some bearings allow horizontal movement in one direc-tion only and are restrained in the other direction whilst othertypes allow movement in any direction Elastomeric bearingsthat are formed of layers of steel plate embedded in rubbercan accommodate small horizontal shear movements

PTFE (polytetrafluoroethylene) bearings can give unlimitedfree sliding between low friction PTFE surfaces and steel platesPot bearings that incorporate rubber discs allow for small rota-tions while spherical bearings that move on a PTFE surface per-mit larger rotations Mechanical bearings such as rockers androllers can be used to provide either longitudinal fixity or resis-tance to lateral force Pot bearings special guide bearings or pinbearings are also used for this purpose Bearings need to beinspected regularly and may require maintenance or replace-ment during the lifetime of the bridge As this can be both diffi-cult and expensive it is very important that the structure isdesigned to make inspection maintenance and replacementpossible

1942 Movement joints

Movement joints are often required to allow free expansion andcontraction due to temperature changes and shrinkage in suchstructures as retaining walls reservoirs roads and long buildingsIn order to allow unrestrained deformation of the walls of cylin-drical containers sliding joints can be provided at the bottomand top of the wall Several types of joints for various purposesare shown in Table 2100 Figures (a)ndash(f) show some of the jointdetails recommended in BS 8007

Expansion joints at wide spacing may be desirable inlarge areas of walls and roofs that are not protected from solarheat gain or where a contained liquid is subjected to asubstantial temperature range Except for structures designed tobe fully continuous contraction joints of the type describedin section 2622 and at the maximum spacing specified inTable 345 should be provided The reinforcement should be cur-tailed to form a complete movement joint or made 50continuous to form a partial movement joint Waterstops arepositioned at the centre of wall sections and at the undersideof floor slabs that are supported on a smooth layer of blindingconcrete In basement walls waterstops are best positioned atthe external face where they are supported by the earth

The recommendations of BS 8007 with regard to the spacingof vertical joints may be applied also to earth-retaining wallsFor low walls with thin stems simple butt joints are generallyused However the effect of unequal deflection or tilting of onepart of a wall relative to the next will show at the joints For retain-ing walls higher than about 1 m a keyed joint can be usedAlternatively dowels passing through the joint with the ends onone side greased and sheathed can be used

Figure (g) shows alternative details at the joint betweenthe wall and floor of a cylindrical tank to minimise or eliminaterestraint at this position In the first case rubber or neoprenepads with known shear deformation characteristics are used Inthe second case action depends on a sliding membrane of PTFEor similar material These details are most commonly used forprestressed cylindrical tanks

Movement joints in buildings should divide the structure intoindividual sections passing through the whole structure aboveground level in one plane Joints at least 25 mm wide shouldbe provided at about 50 m centres both longitudinally andtransversely In the top storey and for open buildings andexposed slabs additional joints should be provided to give aspacing of about 25 m Joints also need to be incorporated in fin-ishes and cladding at movement joint locations Joints in wallsshould be made at column positions in which case a double

Bearings hinges and joints 221

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298Bridges

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299Hinges and bearings

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2100Movement joints

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column as shown at (h) can be provided The copper strip orother similar type of waterbar must be notched where the linksoccur the ends of the notched pieces being bent horizontally orcut off At joints in suspended floors and flat roofs a doublebeam can be provided Joints in floors should be sealed to pre-vent the accumulation of rubbish Roof joints should also beprovided with waterstops The provision of joints in large-areaindustrial ground floor slabs is considered in section 722 andrecommended details are given in ref 61

Expansion joints in bridges need to be either waterproof ordesigned to allow for drainage and should not badly disrupt the

riding quality of the deck Joints should also be designed torequire minimal maintenance during their lifetime and be able tobe replaced if necessary Compressible materials such as neopreneor rubber can accommodate small movements In this case jointscan be buried and covered by the surfacing This type of jointwhich consists of a small gap covered by a galvanised steel plateand a band of rubberised bitumen flexible binder to replace partof the surfacing is known as an asphaltic plug To accommodatelarger movements a flexible sealing element supported by steeledge beams is required Mechanical joints based on interlockingsets of steel toothed plates can be used for very large movements

Bearings hinges and joints 225

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The elastic analysis of a reinforced concrete section by themodular ratio method is applicable to the behaviour of thesection under service loads only The strength of the concrete intension is neglected and a linear stressndashstrain relationshipis assumed for both concrete and reinforcement The straindistribution across the section is also assumed to be linearThus the strain at any point on the section is proportional tothe distance of the point from the neutral axis and since thestressndashstrain relationship is linear the stress in the concrete isalso proportional to the distance from the neutral axis Thisgives a triangular distribution of stress ranging from zero atthe neutral axis to a maximum at the outermost point on thecompression face Assuming no slipping occurs between thereinforcement and the surrounding concrete the strain in bothmaterials is the same and the ratio of the stresses in the twomaterials depends on the ratio of the modulus of elasticity ofsteel and concrete known as the modular ratio e The value of Es

is taken as 200 kNmm2 but the value of E for concrete dependson several factors including the aggregate type the concretestrength and the load duration Commonly adopted values forsustained loads are 15 for normal-weight concrete and 30 forlightweight concrete The geometrical properties of reinforcedconcrete sections can be expressed in equivalent concrete unitsby multiplying the reinforcement area by e

201 PURE BENDING

Expressions for the properties of common reinforced concretesections are given in Tables 2102 and 2103 For sections thatare entirely in compression where the presence of the rein-forcement is ignored simplified expressions are given inTable 2101 The maximum stress in the concrete is given byfc for sections entirely in compression In other casesfc MK2 z unless expressed otherwise and the stress in theoutermost tension reinforcement is given by fs e fc (dx 1)

Expressions for the properties of rectangular and flangedsections are also given in Table 342 in connection with theserviceability calculation procedures contained in BS 8110 BS5400 and BS 8007 (see Chapter 26)

202 COMBINED BENDING AND AXIAL FORCE

The general analysis of any section subjected to direct thrustand uniaxial bending is considered in Table 2104 In the case

MxI

Chapter 20

Elastic analysis ofconcrete sections

of symmetrically reinforced rectangular columns the designcharts on Tables 2105 and 2106 apply The design charts onTable 2107 apply to annular sections such as hollow mastsInformation on uniaxial bending combined with direct tensionand biaxial bending and direct force is given in Tables 2108and 2109 respectively

2021 Symmetrically reinforced rectangularsection

For a symmetrically reinforced rectangular section subjected toaxial force N and bending moment M by equating forcesand taking moments about the mid-depth of the section thefollowing expressions are obtained

(1) For values of x h (ie entire section in compression)

where

A [1 (e 1) ]bh

(2) For values of x h (ie one face in tension)

where fc is maximum stress in concrete at compression faceand Ascbh is total reinforcement ratio

The stress in the tension reinforcement is then given by

fs e fc(dx 1)

For e 15 the charts given on Tables 2105 and 2106 canbe used directly For other values of e the curves for may beconsidered to represent values of [( e 1)14] in region (1)and (e 15) in region (2) For given values of M N and fc therequired value of can be readily determined For given values

dh

05 05edx 1d

h 05

Mbh2fc

05xh

05 x

3h 05(e1)1 h d

x

Nbhfc

05xh

05(e 1)1h d

x 05edh

1

I 112

(e 1)dh

12

bh3

Nbhfc

Abh

Ah2

21 Mbh2fc

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Geometric properties of uniform sections 2101

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Properties of reinforced concrete sections ndash 1 2102

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Properties of reinforced concrete sections ndash 2 2103

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Uniaxial bending and compression (modular ratio) 2104Equivalent area of strip

Equivalent area of transformed sectionDepth of centroid of transformed section

Moment of inertia of transformed section about centroid

Compressive stresses

fcr min Nd

Atr

Md(h x )Itr

fcr max Nd

Atr

MdxItr

Itr Atr(hs)2

12 (hc x )2

x Atr hcAtr

Atr Atr

Atr bshs (e 1)As

Assume a value of x

Depth to centre of tension where

If all bars are of the same size

Equivalent area of strip

Depth to centre of compression

Position of centroid of stressed area

Maximum stresses

Finally check the assumed value of x by substituting these stresses in

xd

1

1 fst(e fcr)

fst (d x)

S fcr

x (x hc)Atr Nd

fcr Nd x(e at x )

(at ac)(x hc)Atr

x eAs a Atr hc

eAs Atr

ac (x hc)hcAtr (x hc)Atr

Atr bshs (e 1)As

at a(a x)(a x)

S (a x)Asat SaS

Com

pres

siv e

str

esse

s on

lyC

ombi

ned

com

pres

sive

and

tens

ile s

tres

ses

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Symmetrically reinforced rectangular columns(modular ratio) ndash 1 2105

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2106Symmetrically reinforced rectangular columns(modular ratio) ndash 2

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2107Uniformly reinforced cylindrical columns(modular ratio)

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Uniaxial bending and tension (modular ratio) 2108

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Biaxial bending and compression (modular ratio) 2109

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Elastic analysis of concrete sections236

of M N and the value of fc is obtained by finding the point ofintersection of the curve and the eh line The eh line can beobtained by interpolation or can be drawn as follows on thegridline for Nbhfc 10 find the point where Mbh2fc is equalto the value of eh through this point draw a line from theorigin The value of xh can be obtained by interpolation or bysolving the equation

(xh)2 2[Nbhfc (e 05)](xh) (e dh 1) 0

Example 1 A 400 400 column reinforced with 4H32 barsis subjected to values of M 120 kNm and N 500 kN dueto service loads The maximum stresses in the concrete and thereinforcement are to be determined assuming e 15

Ascbh 32174002 002

eh MNh 120(500 04) 060

Allowing for 35 mm nominal cover to H8 links

d 400 (35 8 322) 340 mm dh 340400 085

From the chart for dh 340400 085 on Table 2105 at theintersection of 002 and eh 06

Nbhfc 025 xh 051

fc 500 103(025 4002) 125 Nmm2

x 051 400 204 mm

fs e fc (dx 1)

15 125 (340204 1) 125 Nmm2

2022 Uniformly reinforced annular section

The charts given on Table 2107 are based on the assumptionthat the bars may be represented with little loss of accuracy bya notional ring of reinforcement having the same total cross-sectional area and located at the centre of the sectionIf e 15 the charts can be used directly For other valuesof e the curves for and fs fc may be considered to representvalues of (e15) and (15e)( fsfc) respectively

Example 2 A cylindrical shaft with a mean radius of 1 m anda thickness of 100 mm is reinforced with 42H16 vertical barslocated at the centre of the section The shaft is subjected tovalues of M 2700 kNm and N 3600 kN due to serviceloads and the stresses in the concrete and the reinforcement areto be determined assuming e 15

Asc2rh 8446(2 1000 100) 00135

eh MNh 2400(3200 01) 75

A line corresponding to eh 75 can be drawn on the chart forhr 010 on Table 2107 from the origin to a point such asNrhfc 6 Mr2hfc 45 At the intersection of this line withan interpolated curve for 00135

Nrhfc 26 fsfc 6

fc 3200 103(26 1000 100) 123 Nmm2

fs 6 123 74 Nmm2

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Part 3

Design to British Codes

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In most British Codes the design requirements are set out inrelation to specified limit-state conditions Calculations todetermine the ability of a member (or assembly of members) tosatisfy a particular limit state are undertaken using design loadsand design strengths (or stresses) These design values aredetermined from characteristic loads and characteristicstrengths of materials (or stress limits) by the application ofpartial safety factors specified in the Code concerned

211 BUILDINGS

For buildings and other structures designed to BS 8110 thecharacteristic values of dead load Gk (Tables 21 and 22)imposed load Qk (Tables 23 and 24) and wind load Wk (Tables27ndash29) are specified in BS 6399 Parts 1 2 and 3

Design loads are given by

design load Fk f

where Fk is equal to Gk Qk or Wk as appropriate and f is thepartial safety factor appropriate to the load load combinationand limit-state being considered

The characteristic strength of a material fk means that valueof the cube strength of concrete fcu or the yield strength ofreinforcement fy below which 5 of all possible test resultswould be expected to fall In practice for concrete specified inaccordance with BS 8500 a dual classification is used forexample C2530 in which the characteristic strength of cylin-der test specimens is followed by the characteristic strength ofcube test specimens The characteristic strength of reinforcementis taken as the value specified in BS 4449 or BS 4482

Design strengths are given by

design strength fkm

where fk is either fcu or fy as appropriate and m is the partialsafety factor appropriate to the material and limit-state beingconsidered The appropriate factors are already incorporated inthe design equations provided in the Code

Details of the design requirements and partial safety factorsfor the ULS are given in Table 31 For the serviceability limitstates when calculations are required the partial safety factorsare taken as unity However for most designs explicit calcula-tions are unnecessary and the design requirement is met bycomplying with simple rules

For the ULS adverse and beneficial values are given fordead and imposed loads and these are to be applied separatelyto the loads on different parts of the structure to cause the mostsevere effects Thus for load combination 1 design loads mayvary from place to place with a maximum of (14Gk 16Qk)and a minimum of 10Gk (see also Table 230) For loadcombination 2 the maximum wind load of 14Wk with the min-imum dead load of 10Gk can result in a critical equilibriumcondition for tower structures For load combinations 2 and 3the design wind loads can sometimes be less than the minimumnotional horizontal load of 0015Gk associated with therequirement for robustness

The overall dimensions and stability of earth-retaining andfoundation structures are to be determined in accordance withthe appropriate codes for earth-retaining structures (BS 8002)and foundations (BS 8004) Design loads given in BS 8110 arethen used in to establish section sizes and reinforcement areasThe factor f should be applied to all earth and water pressuresexcept those that are derived from equilibrium with otherdesign loads such as bearing pressures below foundations

212 BRIDGES

For bridges and other structures designed to BS 5400 Part 4nominal and design values of dead load superimposed deadload wind load temperature live loads for highway footwayand railway bridges (Tables 25 and 26) and other loads aregiven in the Highways Agency Standard BD 3701

Design loads are given by

design load Fk fl

where Fk is the specified nominal load and fl is the partialsafety factor appropriate to the load load combination andlimit-state being considered A separate partial safety factor isthen used to allow for inaccuracies in the method of analysis

Design load effects are given by

design load effect effect of design load f3

where f3 is the partial safety factor appropriate to the methodof analysis and limit-state being considered If the method ofanalysis is linear elastic it is possible to replace fl by fl f3

in the calculation of the design loads

Chapter 21

Design requirementsand safety factors

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31Design requirements and partial safety factors (BS 8110)

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Liquid-retaining structures 241

The characteristic strength of a material fk means that valueof the cube strength of concrete fcu or the yield strength ofreinforcement fy below which 5 of all possible test resultswould be expected to fall In practice characteristic values arespecified in the manner described for design according toBS 8110 in section 211 The characteristic stress is the stressvalue at the assumed limit of linearity on the stressndashstrain curvefor the material

Design strengths are given by

design strength fkm

where fk is either fcu or fy as appropriate and m is the partialsafety factor appropriate to the material and limit-state beingconsidered The appropriate factors are already incorporated inthe design equations provided in the Code

Design stress limits (serviceability) are given by

design stress limit characteristic stressm

Details of the design requirements characteristic stresses andpartial safety factors are given in Tables 32 and 33

213 LIQUID-RETAINING STRUCTURES

For liquid-containing or liquid-excluding structures designed toBS 8007 the design basis is similar to that in BS 8110 but mod-ified for the limit-state of cracking Separate calculations ofcrack width are required for the effect of applied loads and theeffect of temperature and moisture change

Details of the design requirements and partial safety factors(updated according to BS 8110) are given in Table 34

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32Design requirements and partial safety factors (BS 5400) ndash 1

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33Design requirements and partial safety factors (BS 5400) ndash 2

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34Design requirements and partial safety factors (BS 8007)

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221 CONCRETE

2211 Strength and elastic properties

The characteristic strength of concrete means that value of the28-day cube strength below which 5 of all valid test results isexpected to fall In BS 8500 compressive strength classes areexpressed in terms of both characteristic cylinder strength andcharacteristic cube strength The recommended compressivestrength classes and mean values of the static modulus ofelasticity at 28 days are given in Table 35

In BS 8110 for normal-weight concrete the mean value ofthe static modulus of elasticity of concrete at 28 days is givenby the expression

Ec28 20 02fcu28 (kNmm2)

where fcu28 is the cube strength at 28 days The mean value ofthe modulus of elasticity at an age t 3 days can be estimatedfrom the expression

Ect Ec28 (04 06fcutfcu28)

where fcut is the cube strength at age t Where deflections are ofgreat importance and test data for concrete made with theaggregate to be used in the structure is not available a range ofvalues for Ec28 based on (mean value 6 kNmm2) should beconsidered

In BS 5400 slightly higher mean values of the modulus ofelasticity are used given by the expression

Ec 19 03fcu (kNmm2)

where fcu is the cube strength at the age considered

2212 Creep and shrinkage

The creep strain in concrete may be assumed to be directlyproportional to the applied stress for stresses not exceedingabout one-third of the cube strength at the age of loadingFor design to BS 8110 values of the creep coefficient (creepper unit of stress) according to the ambient relative humiditythe effective section thickness and the age of loading can beobtained from the figure in Table 35 Creep strain is partlyrecoverable if the stress is reduced The final recovery (after 1 year)

is approximately 03 (stress reduction)Eu where Eu is themodulus of elasticity at the age of unloading

For design to BS 8110 an estimate of the drying shrinkageof plain concrete according to the ambient relative humiditythe effective section thickness and the original water contentcan be obtained from the figure in Table 35 Aggregates witha high moisture movement such as some Scottish doleritesand whinstones and gravels containing these rocks produceconcrete with a high initial drying shrinkage (ref 12) Alsoaggregates with a low elastic modulus may result in higher thannormal concrete shrinkage

In BS 5400 values are obtained for the creep coefficient as aproduct of five partial coefficients and for the shrinkage strainas a product of four partial coefficients where the coefficientsare obtained from a series of figures

2213 Thermal properties

For design to BS 8110 values of the coefficient of thermalexpansion of concrete according to aggregate type are given inTable 35 In BS 5400 a value of 12 106 per oC is generallytaken except when limestone aggregates are used when a valueof 9 106 per oC is recommended

2214 Stressndashstrain curves

Typical short-term stressndashstrain curves for normal strengthconcretes in compression as described in section 316 andthe idealised curve given for design purposes in BS 8110and BS 5400 are shown in Table 36 In the expression givenfor the maximum design stress 067 takes account of the ratiobetween the cube strength and the strength in bending

222 REINFORCEMENT

2221 Strength and elastic properties

The characteristic strength of steel reinforcement made to therequirements of BS 4449 is 500 Nmm2 BS 4449 caters forround ribbed bars in three ductility classes grades B500AB500B and B500C Fabric reinforcement is manufacturedusing bars to BS 4449 except for wrapping fabric where wireto BS 4482 with a characteristic strength of 250 Nmm2 maybe used For more information on types properties and sizes of

Chapter 22

Properties of materials

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35Concrete (BS 8110) strength and deformation characteristics

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36Stress-strain curves (BS 8110 and BS 5400) concreteand reinforcement

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Properties of materials248

bar and fabric reinforcement reference should be made tosection 103 and Tables 219 and 220

2222 Stressndashstrain curves

Typical stressndashstrain curves for reinforcing steels in tension asdescribed in section 323 and the idealised curves given

for design purposes in BS 8110 and BS 5400 respectivelyfor reinforcement in tension or compression are shown inTable 36 For design purposes the modulus of elasticity of allreinforcing steels is taken as 200 kNmm2 The BS 5400stressndashstrain curve is the one that was used in CP110 prior tothat code being superseded by BS 8110

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In the following the term nominal cover is used to describe thedesign cover shown on the drawings It is the required coverto the first layer of bars including links The nominal covershould not be less than the values needed for durability andfire-resistance nor less than the nominal maximum size ofaggregate Also the nominal cover should be sufficient toensure that the cover to the main bars is not less than the barsize or for a group of bars in contact the equivalent bar sizeFor a group of bars the equivalent bar size is the diameter of acircle whose cross-sectional area is equal to the sum of the areasof the bars in the group

231 DURABILITY

2311 Exposure classes

Details of the classification system used in BS EN 206-1 andBS 8500-1 with informative examples applicable in the UKare given in Table 37 Often the concrete can be exposed tomore than one of the actions described in the table in whichcase a combination of the exposure classes will apply At thetime of drafting this Handbook the amendments necessitatedby the introduction of BS 8500 have not been incorporated inBS 5400 or BS 8007 (which is based on BS 81101985) Thesystem of exposure classes concrete grades and covers usedprior to BS 8500 is given in Table 39

2312 Concrete strength classes and covers

Concrete durability is dependent mainly on its constituents andlimitations on the maximum free watercement ratio and theminimum cement content are specified for each exposure classThese limitations result in minimum concrete strength classesfor particular cements For reinforced concrete the protection ofthe steel against corrosion depends on the cover The requiredthickness of cover is related to the exposure class the concretequality and the intended working life of the structure Details ofthe recommendations in BS 8500 are given in Table 38

The values given for the minimum cover apply for ordinarycarbon steel in concrete without special protection and for

structures with an intended working life of at least 50 yearsThe values given for the nominal cover include an allowance fortolerance of 10 mm which is recommended for buildings andis normally also sufficient for other types of structures

For uneven concrete surfaces (eg ribbed finish or exposedaggregate) the cover should be increased by at least 5 mm Ifconcrete is cast against an adequate blinding the nominal covershould generally be at least 40 mm For concrete cast directlyagainst the earth the nominal cover should generally be not lessthan 75 mm

232 FIRE-RESISTANCE

2321 Building regulations

The minimum period of fire-resistance required for elements ofthe structure according to the purpose group of a building andits height or for basements depth relative to the ground aregiven in Table 312 Building insurers may require longer fireperiods for storage facilities

2322 Nominal covers and minimum dimensions

The recommendations in BS 8110 regarding nominal cover fordifferent periods of fire-resistance are given in Table 310 Inthe table the cover applies to links for beams and columns butto main bars for floor slabs and ribs (even if links are provided)For two-way spanning solid slabs the cover may be taken asthe average for each direction For beams floors and ribs therequirements apply to the reinforcement in the bottom and sidefaces only The minimum thickness of floors includes anyconcrete screed on the top surface This is particularly impor-tant for ribbed slabs where the structural flange could be nomore than 75 mm thick The values given in the table apply tomembers whose dimensions comply with the minimum valuesgiven in Table 311

For cases where it is considered that special measures needto be taken to prevent the spalling of concrete a summary ofthe recommendations in BS 8110-2 is given in Table 310

Chapter 23

Durability andfire-resistance

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37Exposure classification (BS 8500)

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38Concrete quality and cover requirements fordurability (BS 8500)

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39Exposure conditions concrete and cover requirements(prior to BS 8500)

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310Fire resistance requirements (BS 8110) ndash 1

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311Fire resistance requirements (BS 8110) ndash 2

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312Building regulations minimum fire periods

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241 DESIGN ASSUMPTIONS

Basic assumptions regarding the design of cracked concretesections at the ULS are outlined in section 52 The tensilestrength of the concrete is neglected and strains are evaluatedon the assumption that plane sections before bending remainplane after bending Reinforcement stresses are then derivedfrom these strains on the basis of the design stressndashstraincurves shown on Table 36 The BS 5400 curve is the one thatwas used in CP 110 prior to that code being superseded byBS 8110 For the concrete stresses alternative assumptions arepermitted The design stressndashstrain curve for concrete shownon Table 36 gives a stress distribution that is a combination ofa parabola and rectangle The form of the data governing theshape of the curve causes the relative proportions of the twoparts to vary as the concrete strength changes Thus the totalcompressive force provided by the concrete is not linearlyrelated to fcu and the position of the centroid of the stress-block changes with fcu

Alternatively an equivalent rectangular stress distribution inthe concrete may be assumed as noted on Table 36 TheBS 5400 rectangular stress-block is the one that was usedin CP 110 prior to that code being superseded by BS 8110 InBS 5400 the use of the rectangular stress-block is prohibited inflanged ribbed and voided sections where the neutral axis liesoutside the flange although there appears to be no logicalreason for this restriction

For a rectangular area of width b and depth x the totalcompressive force is given by k1 fcubx and the distance of theforce from the compression face is given by k2x where values ofk1 (allowing for the term 067m)and k2 are given in thefollowing table according to the shape of the stress-block

Properties of concrete stress-blocks for rectangular area

Shape fcu Nmm2 k1 k2

Parabolic- 25 0405 0456rectangular 30 0401 0452(Table 36) 35 0397 0448

40 0394 044550 0388 0439

BS 8110 rectangular 0402 0450

BS 5400 rectangular 0400 0500

Using the rectangular stress-block is conservative in BS 5400but gives practically the same result as that obtained with theparabolic-rectangular form in BS 8110

242 BEAMS AND SLABS

Beams and slabs are generally subjected to bending only butsometimes are also required to resist an axial force for examplein a portal frame or in a floor acting as a prop between base-ment walls Axial thrusts not exceeding 01 fcu times the area ofthe cross section may be ignored in the analysis of the sectionsince the effect of the axial force is to increase the moment ofresistance Where a section is designed to resist bending onlythe value of the lever arm should not be taken greater than 095times the effective depth of the reinforcement

In cases where as a result of moment redistribution allowedin the analysis of the member the design moment is less than themaximum elastic moment at the section the neutral axis depthshould satisfy the requirement xd ( b 04) where b is theratio of design moment to maximum elastic moment

2421 Singly reinforced rectangular sections

Chapter 24

Bending and axialforce

The lever arm between the forces shown in the figure here isgiven by z (d k2x) from which x (d z)k2

Taking moments for the compressive force about the line ofaction of the tensile force gives

M k1 fcubxz k1 fcubz(d z)k2

The solution of the resulting quadratic equation in z gives

zd 05 where K Mbd 2fcu025(k2 k1)K 095

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Taking moments for the tensile force about the line of action ofthe compressive force gives

M As fsz from which As Mfsz

The strain in the reinforcement s 00035(1 xd)(xd) andfrom the BS 8110 design stressndashstrain curve the stress in thereinforcement is given by

fs sEs 700(1 xd)(xd) fy115

Thus for design to BS 8110 fs fy115 for values of

xd 805(805 fy) 0617 for fy 500 Nmm2

From the BS 5400 design stressndashstrain curve the stress in thereinforcement is given by

fs sEs 700(1 xd)(xd) 08fy115 or

08fy115 fs

Thus for design to BS 5400 fs 08fy115 for values of

xd 805(805 08fy) 0668 for fy 500 Nmm2

and fs fy115 for values of

xd 805(1265 fy) 0456 for fy 500 Nmm2

Design charts for fy 500 Nmm2 derived on the basis of theparabolic-rectangular stress-block for the concrete are given inTable 313 for BS 8110 and Table 323 for BS 5400 In the caseof BS 5400 unless the section is proportioned such thatfs fy115 the moment of resistance should provide at least115 times the applied design moment

Design tables based on the rectangular stress-blocks for theconcrete are given in Table 314 for BS 8110 and Table 324for BS 5400 The tables use non-dimensional parameters andare valid for fy 500 Nmm2 The formulae used to derive thetables and the limitations when redistribution of moment hasbeen allowed in the analysis of the member are also given

2422 Doubly reinforced rectangular sections

200fy

2300 fy35 45x d

x d fy 115

Thus for design to BS 8110 for values of

From the BS 5400 design stress-strain curve the stress in thereinforcement is given by

or

Thus for design to BS 5400 for values of

for fy500 Nmm2

and for values of

Equating the tensile and compressive forces gives

where the stress in the tension reinforcement is given by theexpressions used in section 2421

Design charts based on the rectangular stress-blocks forconcrete and for dd 01 and 015 respectively are given inTables 315 and 316 for BS 8110 and Tables 325 and 326 forBS 5400 The charts use non-dimensional parameters and weredetermined for fy 500 Nmm2 but may be safely used forfy 500 Nmm2 In determining the forces in the concrete noreduction has been made for the area of concrete displaced bythe compression reinforcement

2423 Design formulae for rectangular sections

Design formulae based on the rectangular stress-blocks forconcrete are given in BS 8110 and BS 5400 In both codes x islimited to 05d so that the formulae are automatically valid forredistribution of moment not exceeding 10

The stress in the tension reinforcement is taken as 087fy inboth codes although this is only strictly valid for xd 0456in BS 5400 The stress in the compression reinforcement istaken as 087fy in BS 8110 and 072fy in BS 5400 The codeequations which follow from the analyses in sections 2421and 2422 take different forms in BS 8110 and BS 5400

In BS 8110 the requirement for compression reinforcementdepends on the value of K Mbd2fcu compared to where

0156 for b 09

0402( b 04) 018( b 04)2 for 09 b 07

b is the ratio design moment to maximum elastic moment

For K compression reinforcement is not required and

As M087fyz

where

z d05 + lt 095d and x (d z)045

For K compression reinforcement is required and

(K ) bd2fcu087fy (d )

As bd2fcu087fyz

where

z d05 + and x (d z)045

For 0375 (for fy 500 Nmm2) should be replacedby 16(1 ) in the equations for and AsAsAsdx

Asdx

025 K09

KAs

dKAs

K

025K09

K

K

K

K

As fs k1 fcubx As f s

x d (7 3)(dd)f s 2000fy(2300 fy)

x d [805 (805 08fy)](dd) 2(dd)

f s 08fy 115

08fy 115 f s 200fy

2300 fy11535d

x 2000fy

2300 fy

f s sEs 700(1 dx) 08fy 115

x d [805(805 fy)](dd) 264(dd) for fy 500Nmm2

f s fy 115

Beams and slabs 257

The forces provided by the concrete and the reinforcement areshown in the figure here Taking moments for the two com-pressive forces about the line of action of the tensile force gives

M k1 fcubx(d k2x)

The strain in the reinforcement and fromthe BS 8110 design stressndashstrain curve the stress is given by

f s sEs 700(1 dx) fy 115

s 00035(1 dx)

As f s(d d)

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313BS 8110 Design chart for singly reinforced rectangularbeams

Sin

gly

rein

forc

ed b

eam

s (f

y=

500

Nm

m2 )

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314BS 8110 Design table for singly reinforced rectangularbeams

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315BS 8110 Design chart for doubly reinforced rectangularbeams ndash 1

Dou

bly

rein

forc

ed b

eam

s (f

y=

500

Nm

m2

d9d

=0

1)

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316BS 8110 Design chart for doubly reinforced rectangularbeams ndash 2

Dou

bly

rein

forc

ed b

eam

s (f

y=

500

Nm

m2

d9d

=0

15)

wwwengbookspdfcom

In BS 5400 the moment of resistance of a section withoutcompression reinforcement is given by the equations

Mu As(087fy)z 015bd 2fcu

where

z d (1 11As fybdfcu) 095d

For a section with compression reinforcement the moment ofresistance is given by the equations

Mu 015bd2fcu (072fy)(d )

As(087fy) 02bdfcu (072fy)

The equations which are based on x 05d are consideredto be valid for values of dd 02 A variant on the equa-tions with x taken as a variable is included in HighwaysAgency BD4495 for assessment purposes These equationsshould not be used for values of x greater than 05d

2424 Flanged sections

In monolithic beam and slab construction where the web of thebeam projects below the slab the beam is considered as aflanged section for sagging moments The effective width of theflange may be taken as follows

T beam b bw 02lz actual flange widthL beam b bw 01lz actual flange width

lz is the distance between points of zero moment which for acontinuous beam may be taken as 07 times effective span

In most sections where the flange is in compression the depthof the neutral axis will be no greater than the thickness of theflange In this case the section can be considered to be rectangularwith b taken as the flange width The condition regarding theneutral axis depth can be confirmed initially by showing thatM k1 fcubhf (d k2hf) where hf is the thickness of the flangeAlternatively the section can be considered to be rectangularinitially and the neutral axis depth can be checked subsequently

As

dAs

Using the rectangular concrete stress-blocks in the foregoingequations gives k1 04 with k2 045 for BS 8110 and 05for BS 5400 This approach gives solutions that are lsquocorrectrsquowhen x hf but become slightly more conservative as(x hf) increases A different approach is used in BS 8110resulting in solutions that are lsquocorrectrsquo when x 05d butincreasingly conservative as (x hf) decreases As a result thesolution when x hf does not agree with that obtained byconsidering the section as rectangular with b taken as theflange width

2425 General analysis of sections

The analysis of a section of any shape with any arrangementof reinforcement involves a trial-and-error process Aninitial value is assumed for the neutral axis depth from whichthe concrete strains at the positions of the reinforcement canbe calculated The corresponding stresses in the reinforce-ment are determined and the resulting forces in thereinforcement and the concrete are obtained If the forces areout of balance the value of the neutral axis depth is changedand the process is repeated until equilibrium is achievedOnce the balanced condition has been found the resultantmoment of the forces about the neutral axis or any otherpoint is calculated

Example 1 The beam shown in the following figure is to bedesigned to the requirements of BS 8110 The design loads oneach span are as follows where Gk 160 kN andQk 120 kN

Fmax 14Gk 16Qk 416 kN Fmin 10Gk 160 kN

The section design is to be based on the following values

fcu 40 Nmm2 fy 500 Nmm2 cover to links 25 mm

For sagging moments effective width of flange

b bw 02lz 300 02(07 8000) 1420 mm

Allowing for 8 mm links and 32 mm main bars

d 500 (25 8 16) 450 mm say

Bending and axial force262

The figure here shows a flanged section where the neutral axisdepth is greater than the flange thickness The concrete forcecan be divided into two components and the required area oftension reinforcement is then given by

As As1 k1 fcu (b bw)hf 087fy

where

As1 area of reinforcement required to resist a moment M1

applied to a rectangular section of width bw and

M1 M k1 fcu (b bw)hf (d k2hf) 015bd2fcu

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In the calculations that follow solutions are obtained usingcharts and equations to demonstrate the use of each method

Maximum sagging moment For section to be designed asrectangular with b taken as the flange width bending momentshould satisfy the condition

M k1 fcubhf (d k2hf)

04 40 1420 150 (450 045 150) 106

1303 kNm (285 kNm)

Mbd2 285 106(1420 4502) 099 Nmm2

From chart in Table 313 100Asbd 024

As 00024 1420 450 1534 mm2

Alternatively K Mbd2fcu 09940 00248From Table 314 As fybdfcu 121K 00300

As 003 1420 450 40500 1534 mm2

Alternatively by calculation or from Table 314

Hence As M087fyz gives

As 285 106(087 500 095 450) 1533 mm2

Using 2H32 gives 1608 mm2

Maximum hogging moment

K Mbd2fcu 416 106(300 4502 40) 0171

From Table 314 As fybdfcu 0264

As 0264 300 450 40500 2851 mm2

Using 4H32 gives 3217 mm2

Although this is a valid solution it may be possible to reducethe area of tension reinforcement to a more suitable value byallowing for some compression reinforcement Consider theuse of 2H25 with 45 mm ( 01)

fybdfcu 982 500(300 450 40) 009

From the chart in Table 315 As fybdfcu 0225

As 0225 300 450 40500 2430 mm2

A solution can also be obtained using the design equations asfollows

With 2H25 for and assuming

K (087fy)(d )bd2fcu

0171982087500405(3004502 40)0100

Since is validdx 0375 f s 087fy

dx (dd)(x d) 010282 0355

x d (1 z d) 045 (1 0873) 045 0282

z d 05 025 0100 09 0873

dAsK

f s 087fyAs

As

ddd

z d 05 025 00248 09 0972 095

As bd2fcu 087fyz

982 0100 300 4502 40(087 500 0873 450)

2404 mm2 (compared to 2430 mm2 obtained from chart)

Using 3H32 gives 2413 mm2

Example 2 Suppose that in the previous example the maxi-mum hogging moment at B is reduced by 30 to 291 kNm

K Mbd2fcu 291 106(300 4502 40) 0120

b 291416 070 xd ( b 04) 030

From chart in Table 315 keeping to left of line for xd 03

fybdfcu 0025 As fybdfcu 0158

0025 300 450 40500 270 mm2

As 0158 300 450 40500 1706 mm2

A solution can also be obtained by using the design equationswith xd 03 as follows

0402(b 04) 018(b 04)2

0402 03 018 032 0104 (K 0120)

zd 05 0867

(K ) bd2fcu087fy (d ) 0016 300 4502 40 (087 500 405) 221 mm2

As bd2fcu087fyz 221 0104 300 4502 40(087 500 0867 450) 1710 mm2

Using 2H25 and 1H32 gives 1786 mm2

Since the reduced hogging moment for load case 1 is stillgreater than the elastic hogging moment for load case 2 thedesign sagging moment remains the same as in example 1

In the foregoing examples at the bottom of the beam 2H32bars would run the full length of each span with 2H25 splicebars at support B Other bars would be curtailed according tothe bending moment requirements and detailing rules

243 COLUMNS

In the Codes of Practice a column is a compression memberwhose greater overall cross-sectional dimension does notexceed four times its smaller dimension An effective heightand a slenderness ratio are determined in relation to major andminor axes of bending An effective height is a function of theclear height and depends upon the restraint conditions at theends of the column A slenderness ratio is defined as the effec-tive height divided by the depth of the cross section in the planeof bending The column is then considered to be either short orslender according to the slenderness ratios

Columns are subjected to combinations of bendingmoment and axial force and the cross section may need to bechecked for more than one combination of values In slendercolumns from an elastic analysis of the structure the initialmoment is increased by an additional moment induced bythe deflection of the column In BS 8110 this additional

KAs

dKAs

0250104 09

K

As

As

KAs

Columns 263

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moment contains a modification factor K the use of whichresults in an iteration process with K taken as 10 initiallyThe design charts in this chapter contain sets of K lines as anaid to the design process Details of the design procedures aregiven in Tables 321 and 322 for BS 8110 and Tables 331and 332 for BS 5400

2431 Rectangular columns

Design charts based on the rectangular stress-block for theconcrete and for values of dh 08 and 085 are given inTables 327 and 328 respectively On each curve a straight linehas been taken between the point where xh 08 and the pointwhere N Nuz The use of the rectangular stress-block resultsin Nuz being given by the equation

Nuzbhfcu 04 0714(Asc fybhfcu)

There are no K lines on the charts as no modification factor isused in the design of slender columns to BS 5400

2432 Circular columns

Bending and axial force264

The figure here shows a rectangular section in which thereinforcement is disposed equally on two opposite sides of ahorizontal axis through the mid-depth By resolving forces andtaking moments about the mid-depth of the section the followingequations are given for 0 xh 10

Nbhfcu k1(xh) 05(Asc fybhfcu)(ks1 ks2)

Mbh2fcu k1(xh)05 k2(xh) 05(Asc fybhfcu)(ks1 ks2) (dh 05)

For BS 8110 the stress factors ks1 and ks2 are given by

ks1 14(xh dh 1)(xh) 087

ks2 14(dh xh)(xh) 087

The maximum axial force Nuz is given by the equation

Nuzbhfcu 045 087(Asc fybhfcu)

Design charts based on the rectangular stress-block for theconcrete and for values of dh 08 and 085 are given inTables 317 and 318 respectively On each curve a straightline has been taken between the point where xh 10 andthe point where N Nuz The charts which were determinedfor fy 500 Nmm2 may be safely used for fy 500 Nmm2In determining the forces in the concrete no reduction hasbeen made for the area of concrete displaced by the com-pression reinforcement In the design of slender columns theK factor is used to modify the deflection corresponding to aload Nbal which for a symmetrically reinforced rectangularsection is given as 025bdfcu In the charts Nbal is taken as thevalue at which M is a maximum A line corresponding to Nbal

passes through a cusp on each curve For N Nbal K is takenas 10 For N Nbal K can be determined from the lines onthe chart

For BS 5400 the stress factors ks1 and ks2 are given by

ks1 14(xh dh 1)(xh) 07

07 ks1 025(33xh dh 1)(xh) 0714

ks2 14(dh xh)(xh) 07

07 ks2 025(dh 13xh)(xh) 087

The figure here shows a circular section containing six barsspaced equally around the circumference Solutions based on sixbars will be slightly conservative if more bars are used Thearrangement of the bars relative to the axis of bending affectsthe resistance of the section and the arrangement shown in thefigure is not the most critical in every case For some combina-tions of bending moment and axial force if the arrangementshown is rotated through 30o a slightly more critical conditionresults but the differences are small and may be reasonablyignored

The following analysis is based on a uniform stress-block forthe concrete of depth x and width hsin at the base (as shownin figure) Negative axial forces are included in order to cater formembers such as tension piles By resolving forces and takingmoments about the mid-depth of the section the followingequations are obtained where cos1 (1 2 xh) for0 x 10 and hs is the diameter of a circle through the centresof the bars

Nh2fcu kc (2 sin2)8 (12)(Asc fyAc fcu)(ks1 ks2 ks3)

Mh3fcu kc (3sin sin3)72 (277)(Asc fyAc fcu)(hsh)(ks1 ks3)

The minimum axial force Nmin is given by the equation

Nminh2fcu 087(4)(Asc fyAc fcu)

For BS 8110 kc 045 09 and the stress factors ks1 ks2

and ks3 are given by

087 ks1 14(0433hsh 05 xh)(xh) 087

087 ks2 14(05 xh)(xh) 087

087 ks3 14(05 0433hsh xh)(xh) 087

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The maximum axial force Nuz is given by the equation

Nuzh2fcu (4)045 087(Asc fyAc fcu)

Design charts for values of hsh 06 and 07 are given inTables 319 and 320 respectively The statements in section2431 on the derivation and use of the charts for rectangularsections apply also to those for circular sections

For BS 5400 kc 04 10 and the stress factors ks1 ks2

and ks3 are given by

087 ks1 025(0433hsh 05 13xh)(xh) 07

07 ks1 14(0433hsh 05 xh)(xh) 07

07 ks1 025(0433hsh 05 33xh)(xh) 0714

0714 ks2 025(05 33xh)(xh) 07

07 ks2 14(05 xh)(xh) 07

07 ks2 025(05 13 xh)(xh) 087

0714 ks3 025(0433hsh 05 33xh)(xh) 07

07 ks3 14(0433hsh 05 xh)(xh) 07

07 ks3 025(0433hsh 05 13xh)(xh) 087

The maximum axial force Nuz is given by the equation

Nuzh2fcu ( 4)04 072(Asc fyAc fcu)

Design charts for values of hsh 06 and 07 are given inTables 329 and 330 respectively The statements in section2431 on the derivation and use of the charts for rectangularsections apply also to those for circular sections

2433 Design formulae for short braced columns

Approximate formulae are given in BS 8110 for the design ofshort braced columns under specific conditions Where due tothe nature of the structure a column cannot be subjected tosignificant moments it may be considered adequate if thedesign ultimate axial load N 04fcuAc 075Asc fy

Columns supporting symmetrical arrangements of beamsthat are designed for uniformly distributed imposed load andhave spans that do not differ by more than 15 of the longermay be considered adequate if N 035fcuAc 067Asc fy

2434 General analysis of column sections

Any given cross section can be analysed by a trial-and-errorprocess For a section bent about one axis an initial value isassumed for the neutral axis depth from which the concretestrains at the positions of the reinforcement can be calculatedThe resulting stresses in the reinforcement are determined andthe forces in the reinforcement and concrete evaluated If theresultant force is not equal to the design axial force N the valueof the neutral axis depth is changed and the process repeateduntil equality is achieved The resultant moment of all theforces about the mid-depth of the section is then the moment ofresistance appropriate to N This approach is used to analyse arectangular section in example 6

Example 3 A 300 mm square braced column designed toBS 8110 for the following requirements

lo 375 m and 09 in both directions

Mx 54 kNm My 0 N 1800 kN

fcu 40 Nmm2 fy 500 Nmm2 cover to links 35 mm

Since leh 09 3750300 1125 15 the column isshort

Mmin N(005h) 1800 005 03 27 kNm (Mx)

Allowing for 8 mm links and 32 mm main bars

d 300 (35 8 16) 240 mm say

Mbh2fcu 54 106(300 3002 40) 005

Nbhfcu 1800 103(300 300 40) 05

From the design chart for dh 240300 08

Asc fybhfcu 022 (Table 317)

Asc 022 300 300 40500 1584 mm2

Using 4H25 gives 1963 mm2

Example 4 A 300 mm circular braced column designed toBS 8110 for the same requirements as example 3

Allowing for 8 mm links and 32 mm main bars

hs 300 2 (35 8 16) 180 mm say

Mh3fcu 54 106(3003 40) 005

Nh2fcu 1800 103(3002 40) 05

From the design chart for hsh 180300 06

Asc fyAc fcu 052 (Table 319)

Asc 052 (4) 3002 40500 2940 mm2

Using 6H25 gives 2945 mm2

Example 5 The column in example 3 but designed for biaxialbending with My 25 kNm and all other requirements as before

Since and Mx My the section may be designed for anincreased moment about the x-x axis (see Table 321)

1 (76)(Nbhfcu) 1 (76) 05 042

Mx My 54 042 25 645 kNm

Mbh2fcu 645 106(300 3002 40) 006

From the design chart for dh 240300 08

Asc fybhfcu 026 (Table 317)

Asc 026 300 300 40500 1872 mm2

Using 4H25 gives 1963 mm2

Example 6 A 300 mm square short column designed toBS 5400 for the following requirements

Mx 60 kNm My 40 kNm N 1800 kN

fcu 40 Nmm2 fy 500 Nmm2 d 240 mm

The section may be designed by assuming the reinforcement(4H32 say) and checking the condition (see Table 331)

Asc fybhfcu 3217 500(300 300 40) 045

Nbhfcu 1800 103(300 300 40) 05

Mx

bh

Columns 265

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317BS 8110 Design chart for rectangular columns ndash 1

Rec

tang

ular

col

umns

(f y

=50

0 N

mm

2 d

h=

08)

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318BS 8110 Design chart for rectangular columns ndash 2

Rec

tang

ular

col

umns

(f y

=50

0 N

mm

2 d

h=

085

)

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319BS 8110 Design chart for circular columns ndash 1

Circular columns (fy= 500 Nmm2 hs h = 06)

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320BS 8110 Design chart for circular columns ndash 2

Circular columns (fy= 500 Nmm2 hsh = 07)

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321BS 8110 Design procedure for columns ndash 1

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322BS 8110 Design procedure for columns ndash 2

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Bending and axial force272

From the design chart for dh 240300 08

Mubh2fcu 0075 Nuzbhfcu 072 (Table 327)

n 067 167(NNuz) 067 167(05072) 18

Since the column is square

Mux Muy 0075 300 3002 40 106 81 kNm

Since this value is less than 10 4H32 are sufficient

Example 7 The column in example 3 but taken as unbraced( 16) in the direction of Mx with all other requirements asbefore

Since lexh 16 3750300 20 10 the column is slender

The additional bending moment about the x-x axis is given by

Madd N(Kh2000)(lexb)2

With K 10 initially and since b h

Madd 1800 032000 202 108 kNm

M Mi Madd 54 108 162 kNm

Mbh2fcu 162 106(300 3002 40) 015

Nbhfcu 05 as before and from the design chart

Asc fybhfcu 063 (Table 317)

Asc 063 300 300 40500 4536 mm2

This requires 4T40 but it can be seen from the chart that withAsc fybhfcu 063 K is about 06 If we use 4H32

Asc fybhfcu 3217 500(300 300 40) 045

Nbhfcu 05 as before and from the design chart

Mbh2fcu 0107 and K 053

With K 053 corresponding to Asc fybhfcu 045

Madd 053 108 57 kNm M 54 57 111 kNm

Mbh2fcu 111 106(300 3002 40) 0103 0107

Thus 4H32 which gives 3217 mm2 is sufficient

Note that K can also be calculated from the equations given inBS 8110 as follows

Nuzbhfcu 045 087(Asc fybhfcu) 045 087 045 084

Nbalbhfcu 025(dh) 025 08 02

K (Nuz N)( Nuz Nbal) (084 05)(084 02) 053

Example 8 The following figure shows a rectangular sectionreinforced with 8H32 The ultimate moment of resistance of thesection about the major axis is to be determined in accordancewith the following requirements

N 2500 kN fcu 40 Nmm2 fy 500 Nmm2

Mx

Mux

n

My

Muy

n

6081

18

4081

18

086

Consider the bars in each half of the section to be replaced byan equivalent pair of bars Depth to the centre of area of the barsin one half of the section 60 2404 120 mm The sectioncan now be considered to be reinforced with four bars of areaAsc4 where d 600 120 480 mm

Asc fybhfcu 6434 500(300 600 40) 045

Nbhfcu 2500 103(300 600 40) 035

From the design chart for dh 480600 08

Mbh2fcu 014 (Table 317)

M 014 300 6002 40 106 605 kNm

The solution can be checked using a trial-and-error process toanalyse the original section as follows

N k1 fcubx (As1ks1 As2ks2 As3ks3)fy

where dh 540600 09 and ks1 ks2 and ks3 are givenby

ks1 14(xh dh 1)(xh) 087

ks2 14(05 xh)(xh) 087

ks3 14(dh xh)(xh) 087

With x 300 mm xh 05 ks1 087 ks2 0 and ks3 087

N 04 40 300 300 103 1440 kN (2500)

With x 360 mm xh 06 ks2 0233 ks3 07

N 04 40 300 360 103 (2413 087 1608 0233 2413 07) 500 103

1728 392 2120 kN (2500)

With x 390 mm xh 065 ks2 0323 ks3 0538

N 04 40 300 390 103 (2413 087 1608 0323 2413 0538) 500 103

1872 660 2532 kN (2500)

With x 387 mm xh 0645 ks2 0315 ks3 0553

N 04 40 300 387 103 (2413 087 1608 0315 2413 0553) 500 103

1858 636 2494 kN (asymp2500)

Taking moments about the mid-depth of the section gives

M k1 fcubx(05h k2x ) (As1ks1 As3ks3)(d 05h)fy

04 40 300 387 (300 045 387) 106

(241308724130553)(540300) 500106

233 412 645 kNm (605 obtained before)

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323BS 5400 Design chart for singly reinforced rectangularbeams

Sin

gly

rein

forc

ed b

eam

s (f

y=

500

Nm

m2 )

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324BS 5400 Design table for singly reinforced rectangularbeams

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325BS 5400 Design chart for doubly reinforced rectangularbeams ndash 1

Dou

bly

rein

forc

ed b

eam

s (f

y=

500

Nm

m2

d9d

=0

1)

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326BS 5400 Design chart for doubly reinforced rectangularbeams ndash 2

Dou

bly

rein

forc

ed b

eam

s (f

y=

500

Nm

m2

d9d

=0

15)

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327BS 5400 Design chart for rectangular columns ndash 1

Rec

tang

ular

col

umns

(f y

=50

0 N

mm

2 d

h=

08)

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328BS 5400 Design chart for rectangular columns ndash 2

Rec

tang

ular

col

umns

(f y

=50

0 N

mm

2 d

h=

085

)

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329BS 5400 Design chart for circular columns ndash 1

Rectangular columns (fy= 500 Nmm2 hsh = 06)

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330BS 5400 Design chart for circular columns ndash 2

Circular columns (fy= 500 Nmm2 hsh = 07)

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331BS 5400 Design procedure for columns ndash 1

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332BS 5400 Design procedure for columns ndash 2

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Chapter 25

Shear and torsion

251 SHEAR RESISTANCE

2511 Shear stress

In BS 8110 the design shear stress at any cross section in amember of uniform depth is calculated from

v Vbvd

where

V is the shear force due to ultimate loadsbv is the breadth of the section which for a flanged section

is taken as the average width of the web below the flanged is the effective depth to the tension reinforcement

For a member of varying depth the shear force is calculated asV (M tan13s)d where 13s is the angle between the tensionreinforcement and the compression face of the member Thenegative sign applies when moment and effective depth bothincrease in the same direction In no case should v exceed thelesser of 08radicfcu or 5 Nmm2 whatever the reinforcement InBS 5400 b is used in place of bv and the maximum value ofv is taken as the lesser of 075radicfcu or 475 Nmm2

2512 Concrete shear stress

In BS 8110 the design concrete shear stress vc is a function ofthe concrete grade the effective depth and the percentage ofeffective tension reinforcement at the section considered It isoften convenient to determine vc at the section where thereinforcement is least and use the same value throughout themember For sections at distance av 2d from the face of asupport or concentrated load vc may be multiplied by 2davproviding the tension reinforcement is adequately anchoredAlternatively as a simplification for beams carrying uniformload the section at distance d from the face of a support maybe designed without using this enhancement and the samereinforcement provided at sections closer to the support InBS 5400 the design concrete shear stress is obtained as svcwhere vc is the ultimate shear stress and s is a depth factor

2513 Shear reinforcement

Requirements for shear reinforcement depend on the value ofv in relation to vc (or svc) In slabs no shear reinforcement isrequired provided v does not exceed the concrete shear stress Inall beams of structural importance a minimum amount of shear

reinforcement in the form of links equivalent to a shearresistance Vsv 04 bvd is required The shear resistance can beincreased by introducing more links or bent-up bars can be usedto provide up to 50 of the total shear reinforcement Thecontribution of the shear reinforcement is determined on the basisof a truss analogy in which the bars act as tension members andinclined struts form within the concrete as shown in the figure here

System of bent-up bars used as shear reinforcement

In the figure the truss should be chosen so that both and are 45o and st 15d The design shear resistance providedby a system of bent-up bars is then given by

Vsb (Asbsb)(087fy)(stsin)

where st (d ndash d)(cot cot) 15d

For bars bent-up at 45o with fy 500 Nmm2

Vsb 0461Asb (dsb) 0307Asb kN

For bars bent-up at 60o with fy 500 Nmm2

Vsb 0594 Asb (d ndash d)sb 0376 Asb kN

In BS 8110 the shear resistance of members containing shearreinforcement is taken as the sum of the resistances providedseparately by the shear reinforcement and the concrete Thestrut analogy results in an additional longitudinal tensileforce that is effectively taken into account in the curtailmentrules for the longitudinal reinforcement In BS 5400 theminimum amount of shear reinforcement is required inaddition to that needed to cater for the difference between thedesign shear force and the concrete shear resistance It is alsonecessary to design the longitudinal reinforcement for theadditional force

Details of the design procedures for determining the shearresistances of members are given in Table 333 for BS 8110and Table 336 for BS 5400

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BS 8110 Shear resistance 333

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Design for torsion 285

2514 Shear under concentrated loads

The maximum shear stress at the edge of a concentrated loadshould not exceed the lesser of 08radicfcu or 5 Nmm2 Shear insolid slabs under concentrated loads can result in punching fail-ures on the inclined faces of truncated cones or pyramids Forcalculation purposes the shear perimeter is taken as the bound-ary of the smallest rectangle that nowhere comes closer to theedges of a loaded area than a specified distance The shearcapacity is checked first on a perimeter at distance 15d fromthe edge of the loaded area If the calculated shear stress is nogreater than vc no shear reinforcement is needed If the shearstress exceeds vc shear reinforcement is required within thefailure zone and further checks are needed on successiveperimeters at intervals of 075d until a perimeter is reachedwhere shear reinforcement is no longer required

Details of design procedures for shear under concentratedloads are given in Table 334 for BS 8110 and Tables 337 and338 for BS 5400

2515 Shear in bases

The shear strength of pad footings near concentrated loads isgoverned by the more severe of the following two conditions

(a) Shear along a vertical section extending for the full widthof the base In BS 8110 the concrete shear stress vc may bemultiplied by 2dav for all values of av 2d In BS 5400 thecritical section is taken at distance d from the face of the loadwith no enhancement of the concrete shear stress(b) Punching shear around the loaded area as described insection 2514 The reaction resulting from the soil bearingpressure within the shear perimeter may be deducted from thedesign load on the column when calculating the design shearforce acting on the section

The shear strength of pile caps is normally governed by theshear along a vertical section extending for the full width of thecap The critical section for shear is assumed to be located at20 of the pile diameter from the near face of the pile Thedesign shear force acting on this section is taken as the wholeof the reaction from the piles with centres lying outside thesection In BS 8110 the design concrete shear stress may bemultiplied by 2dav where av is the distance from the columnface to the critical section for strips of width up to three timesthe pile diameter centred on each pile In BS 5400 thisenhancement may be applied to strips of width equal to one pilediameter centred on each pile For pile caps designed by trussanalogy 80 of the tension reinforcement should be concen-trated in these strips

2516 Bottom loaded beams

Where load is applied near the bottom of a section sufficientvertical reinforcement to transmit the load to the top of thesection should be provided in addition to any reinforcementrequired to resist shear

252 DESIGN FOR TORSION

In normal beam-and-slab or framed construction calculationsfor torsion are not usually necessary adequate control of anytorsional cracking in beams being provided by the required

minimum shear reinforcement When it is judged necessaryto include torsional stiffness in the analysis of a structure ortorsional resistance is vital for static equilibrium membersshould be designed for the resulting torsional moment Thetorsional resistance of a section may be calculated on the basisof a thin-walled closed section in which equilibrium is satisfiedby a closed plastic shear flow Solid sections may be modelled asequivalent thin-walled sections Complex shapes may be dividedinto a series of sub-sections each of which is modelled as anequivalent thin-walled section and the total torsional resistancetaken as the sum of the resistances of the individual elementsWhen torsion reinforcement is required this should consist ofrectangular closed links together with longitudinal reinforce-ment Such reinforcement is additional to any requirements forshear and bending

Details of design procedures for torsion are given in Table 335for BS 8110 and Table 339 for BS 5400

Example 1 The beam shown in the following figure is to bedesigned for shear to the requirements of BS 8110 Details ofthe design loads and the bending requirements for which thetension reinforcement comprises 2H32 (bottom) and 3H32 (topat support B) are contained in example 1 of Chapter 24 Thedesign of the section is to be based on the following values

fcu 40 Nmm2 fy 500 Nmm2 d 450 mm

In the following calculations a simplified approach is used inwhich the critical section for shear is taken at distance d fromthe face of the support with no enhancement of the concreteshear stress In addition the value of vc is determined for thesection where the tension reinforcement is least and the samevalue of vc used throughout Since the beam will be providedwith shear reinforcement the value of vc may be taken as thatobtained for a section with d 400 mm

Based on 2H32 as effective tension reinforcement

100Asbvd 100 1608(300 450) 119

vc 078 Nmm2 (Table 333 for d 400 and fcu 40)

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BS 8110 Shear under concentrated loads 334

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BS 8110 Design for torsion 335

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BS 5400 Shear resistance 336

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BS 5400 Shear under concentrated loads ndash 1 337

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BS 5400 Shear under concentrated loads ndash 2 338

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BS 5400 Design for torsion 339

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Shear and torsion292

Consider H10 links and transverse spacing of legs at 200 mmto suit spacing of tension reinforcement that is 5 legs permetre Then required longitudinal spacing of links is given by

sv 5 78128 304 mm

Provide H10-300 with legs at 200 mm centres transversely

Example 3 A 280 mm thick flat slab is supported by 400 mmsquare columns arranged on a 72 m square grid Theslab which has been designed using the simplified method fordetermining moments contains as tension reinforcement inthe top of the slab at an interior support within a 18 m widestrip central with the column H16-180 in each directionThe slab is to be designed to the requirements of BS 8110(see Table 334) for a shear force resulting from the maximumdesign load applied to all panels adjacent to the column ofVt 954 kN

fcu 40 Nmm2 fy 500 Nmm2 d 240 mm (average)

For design using the simplified method the design effectiveshear force at an interior column Veff 115V 1098 kN

The maximum design shear stress at the column face

Veff uod 1098 103(4 400 240) 286 Nmm2 (50)

Based on H16-180 as effective tension reinforcement

100Asbvd 100 201(180 240) 046

vc 065 Nmm2 (Table 333 for d 240 and fcu 40)

The length of the first critical perimeter at 15d from the face ofthe column is 4 (3d 400) 4480 mm Thus the designshear stress at the first critical perimeter

v 1098 103(4480 240) 102 Nmm2 ( 157vc)

Since vc v 16vc and (v ndash vc) 04 Nmm2 the total areaof vertical links required within the failure zone is given by

Asv 04ud087fyv 04 4480 240(087 500)

989 mm2

20H8 will provide 1006 mm2 which should be arranged on twoperimeters at 05d 120 mm and 125d 300 mm from thecolumn face The inner perimeter should contain at least 40of the total that is 8H8 with 12H8 on the outer perimeter

The length of the second critical perimeter at 225d fromthe face of the column is 4 (45d 400) 5920 mm Thusthe design shear stress at the second critical perimeter

v 1098 103(5920 240) 077 Nmm2 (vc)

Asv 04 5920 240(087 500) 1307 mm2

12H8 are already provided by the outer perimeter of bars inthe first failure zone A further perimeter containing 16H8 togive a total of 28H8 will provide 1407 mm2 in the secondfailure zone The length of the third critical perimeter at 3d fromthe face of the column is 4 (6d 400) 7360 mm Thusthe design shear stress at the third critical perimeter

v 1098 103(7360 240) 062 Nmm2 (vc)

The reinforcement layout is shown in the figure followingwhere indicates the link positions and the spacing of thetension reinforcement has been adjusted so that the links can beanchored round the tension bars

Minimum link requirements are given by

Asv sv 04bv087fyv 04 300(087 500)

028 mm2mm

sv 075d 075 450 3375 mm

From Table 335 H8-300 provides 033 mm2mmShear resistance of section containing H8-300 is given by

Vu vcbvd (Asv sv)(087fyv)d

(078300033087500) 45010ndash3 170 kN

Based on a support of width 400 mm distance from centre ofsupport to critical section 200 450 650 mm The designload is 4168 52 kNm and the shear forces at the criticalsections are

End A V 172 ndash 065 52 138 kN 170 kN (H8-300)

End B V 260 ndash 065 52 226 kN 170 kN

v Vbvd 226 103(300 450) 168 Nmm2

Area of links required at end B is given by

Asvsv (v ndash vc)bv 087fyv (168 ndash 078) 300(087 500)

062 mm2mm

From Table 335 H8-150 provides 067 mm2mm

Note that if the concrete shear strength is taken as (2dav)vc forav 2d critical section is at av 2d and distance from centreof support to critical section 200 900 1100 mm HereV 203 kN v 150 Nmm2 Asvsv 050 mm2mm andfrom Table 335 H8-200 would be sufficient

Example 2 A 700 mm thick solid slab bridge deck is supportedat the end abutment on bearings spaced at 15 m centres Themaximum bearing reaction resulting from the worst arrange-ment of the design loads is 625 kN The tension reinforcement atthe end of the span is H25-200 The slab is to be designed forshear to the requirements of BS 5400 using the following values

fcu 40 Nmm2 fy 500 Nmm2 d 620 mm

If the critical section for punching is taken at 15d 930 mmfrom the edge of each bearing clearly the critical perimetersfrom adjacent bearings will overlap Therefore the slab will bedesigned for shear along a vertical section extending for thefull width of the slab Assuming that the bearing reaction canbe spread over a slab strip equal in width to the spacing ofthe bearings

v Vbd 625 103(1500 620) 067 Nmm2

Based on H25-200 as effective tension reinforcement

100Asbd 100 2454(1000 620) 040

vc 054 Nmm2 s 095 (Table 336 for fcu 40)

Area of links required at end of span is given by

Asvsv (v 04 ndash svc)b087fyv

(06704 ndash 095054)1000(087500)

128 mm2mm

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Design for torsion 293

Example 4 The following figure shows a channel sectionedge beam on the bottom flange of which bear 8 m long simplysupported contiguous floor units The edge beam which iscontinuous over several 14 m spans is prevented from lateralrotation at its supports The positions of the centroid and theshear centre of the section are shown in the figure and the beamis to be designed to the requirements of BS 8110

Characteristic loadsfloor units dead 35 kNm2 imposed 25 kNm2

edge beam dead 12 kNm

Design ultimate loadsfloor units (14 35 16 25) 82 356edge beam 14 12 168

524 kNm

fcu 40 Nmm2 fy 500 Nmm2 d 1440 mm

Bending moment shear force and torsional moment (aboutshear centre of section) at interior support (other than first)

M ndash 008 524 142 ndash 822 kNm (Table 229)

V 524 142 367 kN

T (356 0400 168 0192) 142 122 kNm

(Note In calculating V and T a coefficient of 05 rather than055 has been used since the dead load is dominant and thecritical section may be taken at the face of the support)

Considering beam as one large rectangle of size 250 1500and two small rectangles of size 200 300

13hmin3hmax 2503 1500 2 2003 300

(234 2 24) 109 282 109

Torsional moment to be considered on large rectangle

T1 122 234282 1012 kNm

Torsional moment to be considered on each small rectangle

T2 122 24282 104 kNm

Reinforcement required in large rectangle

Bending (see Table 314)

K Mbd2fcu 822 106(550 14402 40) 0018Since K 0043 As M087fyz where z 095d

As 822 106(087 500 095 1440) 1382 mm2

Shear (see Table 333)

v Vbvd 367 103(250 1440) 102 Nmm2

100Asbvd 100 1382(250 1440) 038vc 053 Nmm2 (for d 400 and fcu 40)

Asvsv bv(v ndash vc)087fyv 250 (102 ndash 053)(087 500)

028 mm2mm (total for all vertical legs)

Additional requirement for bottom loaded beam

Asvsv 356(087 500) 008 mm2mm (for inner leg)

Torsion (see Table 335)

vt 2T1[hmin2(hmax ndash hmin3)]

2 1012 106[2502 (1500 ndash 2503)] 229 Nmm2

(v vt) 102 229 331 Nmm2 (vtu 50)

Assuming 30 mm cover to links dimensions of links

x1 250 ndash 2 35 180 mm y1 1500 ndash 2 35 1430 mm

Asvsv T1[08x1y1(087fy)]

1012 106(08 180 1430 087 500)

113 mm2mm (total for 2 outer legs)

Total link requirement for shear torsion and bottom loadassuming single links with 2 legs

Asvsv 028 113 2 008 157 mm2mm

sv least of x1 180 mm y12 715 mm or 200 mm

From Table 335 H10-100 provides 157 mm2mm

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Shear and torsion294

Total area of longitudinal reinforcement for torsion

As (Asvsv)(fyvfy)(x1 y1)

113 10 (180 1430) 1819 mm2

Area of longitudinal reinforcement required in part of thesection between centrelines of flanges (1300 mm apart)

1819 1300(180 1430) 1469 mm2

From Table 228 14H12 provides 1583 mm2

Total area of tension reinforcement required at top of beam forbending and torsion

1382 05 (1819 ndash 1469) 1557 mm2

From Table 228 2H32 provides 1608 mm2

Reinforcement required in small rectangles

With link dimensions taken as x1 200 ndash 2 35 130 mmand y1 300 ndash 35 265 mm since y1 550 mm vt shouldnot exceed vtuy1550 50 265550 24 Nmm2

vt 2T2[hmin2(hmax ndash hmin3)]

2 104 106[2002 (300 ndash 2003)]

223 Nmm2 (24)

Area of link reinforcement required for torsion

Asvsv T2[08x1y1(087fy)]

104 106(08 130 265 087 500)

087 mm2mm (total for 2 outer legs)

sv least of x1 130 mm y12 132 mm or 200 mmThe lower rectangle should also be designed for the bendingand shear resulting from the load applied by the floor units

From Table 335 H8-100 provides 100 mm2mm

Area of longitudinal reinforcement for torsion

As (Asvsv)(fyvfy)(x1 y1)

087 10 (130 265) 344 mm2

From Table 228 4H12 provides 452 mm2

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Chapter 26

Deflection and cracking

261 DEFLECTION

Deflections of members under service load should not impairthe appearance or function of a structure For bridges there areno specific limits in BS 5400 but allowances are needed toensure that minimum clearances and satisfactory drainage areprovided Accurate predictions of deflections at different stagesof construction are also required

In BS 8110 the final deflection of members below thelevel of the supports after allowance for any pre-camber islimited to span250 A further limit to be taken as the lesserof span500 or 20 mm applies to the increase in deflectionthat occurs after the application of finishes cladding or par-titions in order to minimise any damage to such elementsThese requirements may be met by limiting the spaneffectivedepth ratio of the member to the values given in Table 340In this table the design service stress in the tensionreinforcement is shown as fs (58)(fyb)(As reqAs prov) InBS 8110 at the time of writing the term (58) which isapplicable to m 115 is given incorrectly as (23) whichis applicable to m 105

The spaneffective depth ratio limits take account ofnormal creep and shrinkage but if these are likely to beparticularly high (eg free shrinkage strain 000075 orcreep coefficient 3) the permissible spaneffective depthratio derived from the table should be reduced by up to 15The limiting ratios may be used also for designs where light-weight aggregate concrete is used except that for all beamsand slabs where the characteristic imposed load exceeds4 kNmm2 the values derived from Table 340 should bemultiplied by 085

In special circumstances when the calculation of deflectionis considered necessary the methods described in Tables 341and 342 can be used Careful consideration is needed in thecase of cantilevers where the usual formulae assume that thecantilever is rigidly fixed and remains horizontal at the rootWhere the cantilever forms the end of a continuous beam thedeflection at the end of the cantilever is likely to be eitherincreased or decreased by an amount l13 where l is the lengthof the cantilever measured to the centre of the supportand 13 is the rotation at the support Where a cantilever is con-nected to a substantially rigid structure some root rotationwill still occur and the effective length should be taken as thelength to the face of the support plus half the effective depth

262 CRACKING

2621 Buildings and bridges

Cracking of members under service load should not impair theappearance or durability of the structure In BS 8110 forbuildings the design surface crack width is generally limited to03 mm For members such as beams and slabs in which thenominal cover does not exceed 50 mm the crack widthrequirement may be met by limiting the gaps between tensionbars to specified values In circumstances where calculationis considered necessary crack width formulae are providedDetails of the bar spacing rules (see section 261 for com-ment on fs) and the crack width formulae are given inTable 343

In BS 5400 the design crack width limits apply only forload combination 1 where for highway bridges the live loadis generally taken as HA Crack widths are calculated for asurface taken at a distance from the outermost reinforcementequal to the nominal cover required for durability Thedesign crack width limits vary according to the exposureconditions as follows 025 mm (moderate or severe expo-sure) 015 mm (very severe exposure) and 010 mm(extreme exposure) In many cases these requirements arecritical and details of the crack width formulae are given inTable 343

In BS 8110 for cracking due to the effects of applied loadsthe modulus of elasticity of concrete is taken as Ec2 wherevalues of Ec are given in Table 35 In BS 5400 a value in therange Ec to Ec2 is taken according to the proportion of live topermanent load

Cracking due to restrained early thermal effects is consideredin BS 8110 Part 2 and Highways Agency BD2887 Inthese documents the restrained early thermal contraction isgiven by t 08RT where is a coefficient of expansion forthe mature concrete R is a restraint factor and T is a tempera-ture differential or fall The following values of R are given

Type of pour and restraint R

Base cast onto blinding 01ndash02Slab cast onto formwork 02ndash04Wall cast onto base slab 06ndash08Infill bays 08ndash10

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BS 8110 Deflection ndash 1 340

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BS 8110 Deflection ndash 2 341

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BS 8110 Deflection ndash 3 342

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BS 8110 (and BS 5400) Cracking 343

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Deflection and cracking300

For continuous support conditions and a flanged section withbwb 3001420 021 03 Table 340 gives

Basic spaneffective depth ratio 08 26 208

The estimated service stress in the reinforcement for a designwith no redistribution of the ultimate moment is given by

fs (58)fy (As reqAs prov) (58) 500 15341608 298 Nmm2

Modification factor for tension reinforcement

s 055 (477 fs)120(09 Mbd 2) 20 055 (477 298)[120 (09 099)] 134

Ignoring modification factor for compression reinforcement

Limiting spaneffective depth ratio 208 134 279

Actual spaneffective depth ratio 8000450 178

Allowing for H8 links with 25 mm cover the clear distancebetween the H32 bars in the bottom of the beam is given by

ab 300 2 (25 8 32) 170 mm

From Table 343 the limiting distance is given by

ab 47 000fs 47 000298 158 mm (170 mm)

The clear distance between the H32 bars could be reduced to150 mm by increasing the side cover to the links to 35 mmAlternatively the bars could be changed to 2H25 and 2H20

Example 2 A 280 mm thick flat slab supported by columnsarranged on a 72 m square grid is to be designed to the require-ments of BS 8110 Bending moments are to be determined usingthe simplified method where the total design ultimate load on apanel is 954 kN and the slab is to be checked for deflection

fcu 40 Nmm2 fy 500 Nmm2 cover to bars 25 mm

Allowing for the use of H12 bars in each direction and basedon the bars in the second layer of reinforcement

d 280 (25 12 6) 235 mm ld 7200235 306

In the following calculations s is based on the average valueof Mbd 2 determined for a full panel width From Table 262the design ultimate bending moment for an end span with acontinuous connection at the outer support is given by

M 0075Fl 0075 954 72 515 kNmMbd2 515 106(7200 2352) 130 Nmm2

From Table 340 for a continuous flat slab without drops thebasic spaneffective depth ratio 26 09 234

Modification factor for tension reinforcement if fs (58)fy

s 055 (477 fs)120(09 Mbd2) 20 055 (477 312)[120 (09 13)] 117

Assuming that no compression reinforcement is provided

Limiting spaneffective depth ratio 234 117 274

The limiting value of ld can be raised to 306 by increasing s

to 131 which reduces fs to 276 Nmm2 so that the area oftension reinforcement determined for the ULS should bemultiplied by the factor 312276 113

For an interior span where the bending moment coefficient is0063 compared to 0075 for the end span

Mbd 2 130 00630075 109 Nmm2

s 055 (477 312)[120 (09 109)] 124 (131)

2622 Liquid-retaining structures

In BS 8007 for structures where the retention or exclusion ofliquid is a prime consideration a design surface crack widthlimit of 02 mm generally applies In cases where the surfaceappearance is considered to be aesthetically critical the limit istaken as 01 mm Under liquid pressure cracks that extendthrough the entire thickness of a slab or a wall are expected toresult in some initial seepage but it is assumed that such crackswill heal autogeneously within 21 days for a 02 mm designcrack width and 7 days for a 01 mm design crack widthSeparate calculations using basically different crack width for-mulae are used for the effects of applied loads and the effectsof temperature and moisture change Details of the formulae foreach type of cracking are given in Table 344

In order to control any potential cracking due to the effectsof restrained thermal contraction and shrinkage three designoptions are given in which the reinforcement requirements arerelated to the incidence of any movement joints Details of thedesign options are given in Table 345 where the joint spacingrequirements refer to joints as being either complete or partialContraction joints can be formed by casting against stop endsor by incorporating crack-inducing waterstops The joints arecomplete if all of the reinforcement is discontinued at thejoints and partial if only 50 of the reinforcement is discon-tinued The joints need to incorporate waterstops and surfacesealants to ensure a liquid-tight structure

The reinforcement needed to control cracking in continuousor semi-continuous construction depends on the magnitude ofthe restrained contraction The restraint factor R (ie ratio ofrestrained to free contraction) may be taken as 05 generally butmore specific values for some common situations are also givenin Table 345 For particular sections and arrangements of rein-forcement limiting values for restrained contraction strain aregiven in Table 346

For cracking due to applied loading and concrete classes thatare typically either C2835 or C3240 the effective modularratio e 2EsEc may be taken as 15 In this case for singlyreinforced rectangular sections the elastic properties of thetransformed section in flexure are given in Table 347 For par-ticular sections and arrangements of reinforcement limitingvalues are given for service moments in Tables 348ndash350 anddirect tensile forces in Tables 351 and 352

Example 1 The beam shown in the following figure is to bechecked for deflection and cracking to the requirements ofBS 8110 The designs for bending and shear are contained inexample 1 of Chapters 24 and 25 respectively The tensionreinforcement comprising 3H32 (top at B) and 2H32 (bottom)is based on the following values

fcu 40 Nmm2 fy 500 Nmm2 cover to links 25 mm

For the bottom reinforcement with d 450 mm

b 1420 mm Mbd 2 099 Nmm2

As req 1534 mm2 As prov 1608 mm2

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BS 8007 Cracking 344

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BS 8007 Design options and restraint factors 345

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BS 8007 Design table for cracking due to temperature effects 346

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BS 8007 Elastic properties of cracked rectangularsections in flexure 347

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BS 8007 Design table for cracking due to flexure in slabs ndash 1 348

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BS 8007 Design table for cracking due to flexure in slabs ndash 2 349

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BS 8007 Design table for cracking due to flexure in slabs ndash 3 350

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351BS 8007 Design table for cracking due to directtension in walls ndash 1

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352BS 8007 Design table for cracking due to directtension in walls ndash 2

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The maximum design crack width is given by

wcr

3 735 000121[1 2 (735 45)(700 192)] 024 mm

From Table 32 the limiting design crack width for a bridgedeck soffit (severe exposure) is 025 mm

The service stress range in the reinforcement resulting from thelive load moment Mq is given by

(MqMs)fs (275600) 220 101 Nmm2

From Table 32 the limiting value for spans in the range5ndash200 m and bar sizes 16 mm is 120 Nmm2

Example 4 The wall of a cylindrical tank 75 m deep and15 m diameter is 300 mm thick The wall which is continuouswith the base slab is to be designed for temperature effects andthose due to internal hydrostatic pressure when the tank is fullof liquid

Design crack width 02 mm fcu 40 Nmm2

Cover to horizontal bars 52 mm fy 500 Nmm2

Effects of temperature change Allowing for concrete gradeC3240 with 350 kgm3 Portland cement at a placing temperatureof 20oC and a mean ambient temperature during construction of15oC the temperature rise for concrete placed within 18 mmplywood formwork

T1 25oC (Table 218)

As the wall is to be designed to resist hoop tension there willbe no vertical movement joints and allowance must be made fora fall in temperature due to seasonal variations Allowing forT2 15oC restraint factor R taken as 05 and coefficient ofthermal expansion taken as 12 106 per oC (Table 35)restrained total thermal contraction after the peak temperaturearising from hydration effects is given by

R (T1 T2) 05 12 106 (25 15) 240 106

From Table 346 for 02 mm crack width and a surface zonethickness of 3002 150 mm H16-200 (EF) will suffice

Effects of hydrostatic load Suppose that an elastic analysis ofthe tank assuming a floor 300 mm thick indicates a servicemaximum circumferential tension of 400 kNm This valueoccurs at a depth of 6 m and above this level the hoop ten-sions can be assumed to reduce approximately linearly tonear zero at the top of the wall

In BS 8110 for direct tension and fy 500 Nmm2 minimumreinforcement equal to 045 of the concrete cross section isrequired Hence for a wall 300 mm thick the minimum areaof reinforcement required on each face

As min 00045 1000 150 675 mm2m (H16-300)

From Table 351 for a 02 mm crack width and 40 mm coverthe following values are obtained for a 300 mm thick wall

H16-225 (EF) provides for N 408 kNmH16-300 (EF) provides for N 319 kNm

In order to cater for the effects of both temperature change andhydrostatic load H16-225 (EF) can be used for a height of

3acrm

1 2(acr cmin) (h dc)

Deflection and cracking310

In this case increasing s to 131 reduces fs to 295 Nmm2 sothat the area of tension reinforcement should be multiplied bythe factor 312295 106

Example 3 A simply supported 700 mm thick solid slabbridge deck is to be designed for bending and checked forcracking and fatigue to the requirements of BS 5400

fcu 40 Nmm2 fy 500 Nmm2 d 620 mm

The maximum longitudinal ultimate moment at mid-span forload combination 1 (375 units HB live load) is 950 kNmm

Mbd 2fcu 950 106(1000 6202 40) 0062As fybdfcu 0078 (Table 324)As 0078 1000 620 40500 3869 mm2m

Although H25-125 gives 3927 mm2m this is unlikely to besufficient with regard to cracking and fatigue which will bechecked on the basis of H25-100 (4909 mm2m)

The maximum longitudinal service moment at mid-span forload combination 1 (HA live load) is Ms 600 kNmm with

Mg 325 kNmm Mq 275 kNmm MqMg 085

In the following calculations the modulus of elasticity of theconcrete is taken as a value reflecting the relative proportions oflive load and permanent load Taking Ec for live load and Ec2for permanent load the effective value is given by

Eeff [1 05(1 MqMg)]Ec

From section 2211 Ec 19 03fcu 31 kNmm2

Eeff [1 05(1 085)] 31 226 kNmm2

From Table 342 the neutral axis depth x (or dc) is given by

xd where (EsEeff)(Asbd)

(200226) 4909(1000 620) 007

xd 007 031 x 192 mmStress in tension reinforcement is given by

fs MsAs(d x3) 600 106[4909 (620 1923)] 220 Nmm2

Strain in tension reinforcement

s fsEs 220(200 103) 00011

Strain at surface taken at distance cmin from outermost barsignoring stiffening effect of concrete

1 s (h x 10)(d x) 00011 (700 192 10)(620 192) 000128

Stiffening effect of concrete (with h)

109

38 1000 700 015 109(00011 4909) 000007

Strain at surface allowing for stiffening effect of concrete

m 1 2 000128 000007 000121

Distance from surface of bar where sb is bar spacing to pointmidway between bars on surface taken at distance cmin fromoutermost bars is given by

acr

125 735 mm(50)2 (70)2

(sb 2)2 (h d 10)2 2

2 38bthsAs

1 Mq

Mg

a

0072 2 007

2 2

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Cracking 311

30 m say and minimum reinforcement of H16-300 (EF) for theremaining height of 45 m

Example 5 A 200 mm thick roof slab to a reservoir is to bedesigned for serviceability cracking to the requirements ofBS 8007

Design crack width 02 mm Cover to bars 40 mm

(a) Sliding layer is provided between slab and perimeter wallMaximum service moment M 25 kNmm

From Table 348 H12-150 caters for M 252 kNmm withfs 243 Nmm2

(b) Slab is tied to perimeter wall Maximum service momentand direct tension M 25 kNmm and N 40 kNm

h 200 mm d 200 (40 122) 154 mm

Since MN (2540) 103 625 mm (d 05h) 54 mmone face will remain in compression and the section can bedesigned for a reduced moment M1 M N(d 05h) withthe tensile force acting at the level of the reinforcement

M1 25 40 0054 228 kNmm

From solution (a) above take fs 240 Nmm2 as a trialvalue

100M1fsbd2 100 228 106(240 1000 1542) 0400

From Table 347 by interpolation 100As1bd 0445

As1 000445 1000 154 685 mm2m

Total area of reinforcement required is given by

As As1 Nfs 685 40 103240 852 mm2m

Using H12-125 gives 905 mm2m This may not necessarily besufficient because the stiffening effect of the concrete inherentin solution (a) is reduced by the additional tension A crackwidth calculation is needed to confirm the solution

The stress in the reinforcement is given approximately by

fs 240 852905 226 Nmm2

100M1fsbd2 0400 240226 0425

From Table 347 by interpolation 100As1bd 0474

As1 000474 1000 154 730 mm2m

As As1 Nfs 730 40 103226 907 mm2m

This is near enough to the area given by H-125 no furtheriteration is needed and 100M1fsbd2 0425 may be assumed

From Table 347 by interpolation xd 0312 x 48 mm

Strain in the reinforcement

s fsEs 226(200 103) 000113

Strain at surface ignoring stiffening effect of concrete

1 s(h x)(d x) 000113 (200 48)(154 48) 000162

Stiffening effect of concrete at surface (with h)

2 bt(h x)23EsAs(d x)1000 (20048)2[3 200 103 905 (154 48)] 000040

Strain at surface allowing for stiffening effect of concrete

m 1 2 000162 000040 000122

Distance from surface of bar where sb is bar spacing topoint on surface midway between bars is given by

acr

6 716 mm

The maximum design crack width is given by

wcr

3 716 000122[1 2 (716 40)(200 48)] 019 mm

3acrm

1 2(acr cmin) (h x)

(625)2 (46)2

(sb 2)2 (h d)2 2

a

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Chapter 27

Considerations affecting design details

Codes of Practice contain numerous requirements that affectthe reinforcing details such as minimum and maximum areastying provisions anchorage and curtailment

Bars may be arranged individually in pairs or in bundles ofthree or four in contact In BS 8110 for the safe transmissionof bond forces the cover provided to the bars should be not lessthan the bar size or for a group of bars in contact the equivalentdiameter of a notional bar with the same total cross-sectionalarea as the group In BS 5400 the forgoing cover requirementis increased by 5 mm Requirements for cover with regard todurability and fire-resistance are given in Chapter 23 Gapsbetween bars (or groups of bars) generally should be not lessthan the greater of (hagg 5 mm) where hagg is the maximumsize of the coarse aggregate or the bar size (or the equivalentbar size for bars in groups) Details of reinforcement limits aregiven in Table 353 for BS 8110 and Table 359 for BS 5400

271 TIES IN STRUCTURES

For robustness the necessary interaction between elements isobtained by tying the structure together Where the structure isdivided into structurally independent sections each sectionshould have an appropriate tying system In the design of tiesthe reinforcement may be assumed to act at its characteristicstrength and only the specified tying forces need to be takeninto account Reinforcement provided for other purposes maybe considered to form part of or the whole of the ties Detailsof the tying requirements in BS 8110 are given in Table 354

272 ANCHORAGE AND LAP LENGTHS

At both sides of any cross section bars should be provided withan appropriate embedment length or other form of end anchorageIn BS 5400 it is also necessary to consider lsquolocal bondrsquo wherelarge changes of tensile force occur over short lengths ofreinforcement Critical sections for local bond are at simplysupported ends at points where tension bars stop and at pointsof contra-flexure However the last two points need not beconsidered if the anchorage bond stresses in the continuing barsdo not exceed 08 times the ultimate values

The radius of any bend in a reinforcing bar should conformto the minimum requirements of BS 8666 and should ensurethat the bearing stress at the mid-point of the curve does notexceed the maximum value given in BS 8110 or BS 5400 as

appropriate A link may be considered fully anchored if itpasses round another bar not less than its own size through anangle of 90o and continues beyond the end of the bend for aminimum length of eight diameters Details of anchoragelengths local bond stresses and bends in bars are given inTables 355 and 359 for BS 8110 and BS 5400 respectively

In BS 8007 for horizontal bars in direct tension the designultimate anchorage bond stress is taken as 70 of the valuegiven in BS 8110 Also for sections where bars are needed tocontrol cracking due to temperature and moisture effects therequired anchorage bond length lab 6 where crit isthe reinforcement ratio in the surface zone (Table 344) Thisvalue can exceed the anchorage bond length in BS 8110

Laps should be located if possible away from positions ofmaximum moment and should preferably be staggered Laps infabric should be layered or nested to keep the lapped wires orbars in one plane BS 8110 requires that at laps the sum of allthe reinforcement sizes in a particular layer should not exceed40 of the breadth of the section at that level When the sizeof both bars at a lap exceeds 20 mm and the cover is lessthan 15 times the size of the smaller bar links of size notless than one-quarter the size of the smaller bar and spacingnot greater then 200 mm should be provided throughout the laplength Details of lap lengths are given in Tables 355 and 359for BS 8110 and BS 5400 respectively

273 CURTAILMENT OF REINFORCEMENT

In flexural members it is generally advisable to stagger thecurtailment points of the tension reinforcement as allowedby the bending moment envelope Curtailed bars shouldextend beyond the points where in theory they are no longerneeded in accordance with certain conditions Details of thegeneral curtailment requirements in BS 8110 are given inTable 356 Simplified rules for beams and slabs are alsoshown in Tables 357 and 358 In BS 5400 the generalcurtailment procedure is the same as that in BS 8110 except forthe requirement at a simply supported end where condition (3)does not apply

Example 1 The design of the beam shown in the following fig-ure is given in example 1 of Chapters 24 (bending) and 25 (shear)The design ultimate loads on each span are Fmax 416 kN andFmin 160 kN The main reinforcement is as follows in the

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BS 8110 Reinforcement limits 353

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BS 8110 Provision of ties 354Internal ties

Peripheral tiesVertical ties

Column to wallties at each column(or wall)floorintersection

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BS 8110 Anchorage requirements 355

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spans 2H32 (bottom) at support B 3H32 (top) and 2H25(bottom) The width of each support is 400 mm and bars are tobe curtailed according to the requirements of BS 8110 In thefollowing calculations the use of the general curtailmentprocedure (Table 356) and the simplified curtailment rules(Table 357) are both examined

fcu 40 Nmm2 fy 500 Nmm2 d 450 mm

2 A point where the shear force is no more than half thedesign shear resistance at the section From the shear designcalculations in Chapter 25 with H8-300 links Vu 170 kNFor load case 1 distance x from B to the point where shearforce is 1702 85 kN is given by

x [1 (85 156)416]L 042 8 336 m

3 A point where the bending moment is no more than halfthe design resistance moment at the section As in thecalculations above but with M 2702 135 kNm05x2 3x 26 0 giving x 68 m and a distancefrom B of (80 68) 12 m

From the foregoing it can be seen that if the bar were curtailedat a distance of 12 m from B this would satisfy condition (3)and be more than 450 mm beyond the theoretical curtailmentpoint at 06 m from B However the bar should extend 13 mfrom B in order to provide a full tension anchorage of 11 mbeyond the face of the support It can be seen from the forego-ing calculations that checking conditions (2) and (3) is a tediousprocess and complying with condition (1) is a more practicalapproach even though it would mean curtailing the bar at(06 11) 17 m from B in this example

Suppose that the remaining 2H32 are continued to the point ofcontra-flexure in span BC for load case 2

The reaction at support C is given by

RC 05Fmin MBL 05 160 2888 44 kN

Distance from B to point of contra-flexure is given by

x L(1 2RCFmin) 8 (1 2 44160) 36 m

The bars need to extend beyond this point for a distance not lessthan d 12 450 mm Link support bars say 2H12 with alap of 300 mm could be used for the rest of the span

If the simplified curtailment rules are applied one bar out ofthree may be curtailed at 015L 45 145 m from the faceof the support that is at 165 m from B The other two bars maybe curtailed at 025L 20 m from the face of the support thatis at 22 m from B Beyond this point bars giving an area notless than 20 of the area required at B should be provided thatis 02 2413 483 mm2 (2H20 gives 628 mm2) Since thebars are in the top of a section as cast where the cover is lessthan 2 and the gap between adjacent laps is not less than6 l 14 (see Table 355) and the required lap length is

lbl 49 (AsreqAsprov) 49 20 483628 750 mm

Example 2 A typical floor to an 8-storey building consistsof a 280 mm thick flat slab supported by columns arrangedon a 72 m square grid The slab for which the characteristicloading is 80 kNm2 dead and 45 kNm2 imposed is to beprovided with ties to the requirements of BS 8110 Thedesign ultimate load on a panel is 954 kN and bendingmoments are to be determined by the simplified method (seesection 138)

fcu 40 Nmm2 fy 500 Nmm2 cover to bars 25 mm

Allowing for the use of H12 bars in each direction and basedon the bars in the second layer of reinforcement

d 280 (25 12 6) 235 mm say

Considerations affecting design details316

End anchorage At the bottom of each span the 2H32 will becontinued to the supports At the end support (simple) aneffective anchorage of 12 is required beyond the centre ofbearing This can be obtained by providing a 90o bend with aninternal radius of 35 provided the bend does not start beforethe centreline of the support Allowing for 50 mm end coverto the bars the distance from the outside face of the support tothe start of the bend is 50 45 194 mm This would besatisfactory since the width of the support is 400 mm If thesupport width were any less the 2H32 could be stopped atthe face of the support and lapped with 2H25 in which case itwould be necessary to reassess the shear design

Curtailment points for top bars The resistance momentprovided by 2H32 can be determined as follows

As fybdfcu 1608 500(300 450 40) 0149Mbd 2fcu 0111 (Table 314)M 0111 300 4502 40 106 270 kNm

For load case 1 reaction at A (or C) is given by

RA 05Fmax MBL 05 416 4168 156 kN

Distance x from A to point where M 270 kNm is given by

05(FmaxL) x2 RAx 05 (4168)x2 156x 270

Hence 05x2 3x 52 0 giving x 74 m Thus of the3H32 required at B one bar is no longer needed for flexure ata distance of (80 74) 06 m from B

The bar to be curtailed needs to extend beyond this point fora distance not less than d 12 450 mm to a positionwhere one of the following conditions is satisfied

1 A full tension anchorage length beyond the theoretical pointof curtailment that is 35 35 32 11 m

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From Table 255 the design ultimate sagging moment for aninterior panel is given by

M 0063Fl 0063 954 72 433 kNm

The total panel moment is to be apportioned between columnand middle strips where the width of each strip is 36 m Forthe column strip with 55 of the panel moment

M 055 433 238 kNmMbd2fcu 238 106(3600 2352 40) 0030

From Table 314 since Mbd 2fcu 0043 zd 095 Hence

As 238 106(087 500 095 235) 2504 mm2 (24H12-150 gives 2714 mm2)

For the middle strip with 45 of the panel moment

M 045 433 195 kNmMbd2fcu 195 106(3600 2352 40) 0025 ( 0043)

As 195 106(087 500 095 230) 2008 mm2 (18H12-200 gives 2036 mm2)

For the peripheral tie the tensile force is given by

Ft (20 4no) 60 kN (20 4 8) 52 kN

The required area of reinforcement acting at its characteristicstrength is given by

As Ftfy 52 103500 104 mm2 (1H12)

For the internal ties the tensile force is given by

Ft gt Ft kNm

52 = 125 kNm

If the internal ties are spread evenly in the slab the requiredarea of reinforcement acting at its characteristic strength

As 125 103500 250 mm2m (H12-400)

In this case alternate bars in both column and middle stripsneed to be made effectively continuous

If the internal ties are concentrated at the column lines the totalarea of reinforcement required in each group

As 250 72 1800 mm2 (16H12 gives 1810 mm2)

In this case the bars in the middle two-thirds of each columnstrip need to be made effectively continuous Since the bars arelocated at the bottom of the slab and the gap between each set

80 4575 72

5

Ftint gt qk

75 lr

5

Curtailment of reinforcement 317

of lapped bars exceeds 6 l 10 and a lap length of 35 issufficient (Table 355 for fcu 40 Nmm2)

Example 3 The following figure shows details of the rein-forcement at the junction between a 300 mm wide beam and a300 mm square column Bars 03 need to develop the maximumdesign stress at the column face and the required radius of bendis to be checked in accordance with the requirements of BS 8110

fcu 40 Nmm2 fy 500 Nmm2

The minimum radius of bend of the bars depends on the value ofab where ab is taken as either centre-to-centre distance betweenbars or (side cover plus bar size) whichever is less Henceab (300 75 2 25) 125 (75 25) 100 mm

From Table 355 for ab 10025 4 rmin 64 Thisvalue can be reduced slightly by considering the stress in the barat the start of the bend If r 6 distance from face of columnto start of bend 300 50 7 25 75 mm (ie 3) FromTable 355 the required anchorage length is 35 and rmin (1 335 ) 64 59 Thus r 6 is sufficient

Example 4 A 700 mm thick solid slab bridge deck is simplysupported at the end abutments The design of the slab to therequirements of BS 5400 is given in example 2 of Chapter 25(shear) and example 3 of Chapter 26 (bending cracking andfatigue) The tension reinforcement at the end of the span isH25-200 and the maximum design shear force is 625 kN act-ing on a 15 m wide strip of slab A local bond check is required

fcu 40 Nmm2 fy 500 Nmm2 d 620 mm

Bar perimeter provided by H25-200 in a 15 m strip of slab

13us (1500200) 25 589 mm

The local bond stress is given by the relationship

fbs V13usd 625 103(589 620) 171 Nmm2

From Table 359 ultimate local bond stress flbu 40 Nmm2

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BS 8110 Curtailment requirements 356

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BS 8110 Simplified curtailment rules for beams 357

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BS 8110 Simplified curtailment rules for slabs 358

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BS 5400 Considerations affecting design details 359

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Chapter 28

Miscellaneousmembers and details

281 LOAD-BEARING WALLS

In BS 8110 for the purpose of design a wall is defined as avertical load-bearing member whose length on plan exceedsfour times its thickness Otherwise the member is treated as acolumn A reinforced wall is one in which not less than therecommended minimum amount of vertical reinforcement isprovided and taken into account in the design Otherwise themember is treated as a plain concrete wall in which case thereinforcement is ignored for the purpose of design Limitingreinforcement requirements are given in Table 353 Bearingstresses under concentrated loads should not exceed 06fcu forconcrete strength classes C2025 Design requirements forreinforced and plain concrete walls are given in Table 360

282 PAD BASES

Notes on the distribution of pressure under pad foundations aregiven in section 181 and values for the structural design ofseparate bases are given in Table 282 Critical sections forbending are taken at the face a concrete column or the centreof a steel stanchion The design moment is taken as that due toall external loads and reactions to one side of the section

Generally tension reinforcement may be spread uniformlyacross the width of the base but the following requirementshould be satisfied where c is the column width and lc is thedistance from the centre of a column to the edge of the pad Iflc 075(c 3d) two-thirds of the reinforcement should beconcentrated within a zone that extends on either side for adistance no more than 15d from the face of the column Forbases with more than one column in the direction consideredlc should be taken as either half the column spacing or thedistance to the edge of the pad whichever is the greater

The pad should be examined for normal shear and punchingshear Normal shear is checked on vertical sections extendingacross the full width of the base Within any distance av 2dfrom the face of the column the shear strength may be taken as(2dav)vc For a concentric load the critical position occurs atav a2 2d where a is the distance from the column face tothe edge of the base For an eccentric load checks can be madeat av 05d d and so on to find the critical position

Punching shear is checked on a perimeter at a distance 15dfrom the face of the column The shear force at this position is that

due to the effective ground pressure acting on the area outsidethe perimeter For a concentric load with a 15d the checkfor punching shear is the critical shear condition If the mainreinforcement is taken into account in the determination of vcthe bars should extend a distance d beyond the shear perimeterIn this case the bars need to extend 25d beyond the face of thecolumn and will need to be bobbed at the end unless a 25din which case straight bars would suffice

The maximum clear spacing between the bars ab shouldsatisfy the following requirements for fy 500 Nmm2

100Asbd 03 03 04 05 06 075 10

ab (mm) 750 500 375 300 250 200 150

Example 1 A base is required to support a 600 mm squarecolumn subjected to vertical load only for which the valuesare 4600 kN (service) and 6800 kN (ultimate) The allowableground bearing value is 300 kNm2

fcu 35 Nmm2 fy 500 Nmm2 nominal cover 50 mm

Allowing 10 kNm2 for ground floor loading and extra over soildisplaced by concrete the net allowable bearing pressure canbe taken as 290 kNm2 Area of base required

Abase 4600290 1586 m2 Provide base 40 m square

Distance from face of column to edge of base a 1700 mm

Take depth of base 05a say h 850 mm

Allowing for 25 mm main bars average effective depth

d 850 ndash (50 25) 775 mm

Bearing pressure under base due to ultimate load on column

pu 680042 425 kNm2

Bending moment on base at face of column

M pu la22 425 4 1722 2456 kNm

K Mbd2 fcu 2456 106(4000 7752 35) 00292

From Table 314 since K 0043 As fybd fcu 121K and

As 121 00292 4000 775 35500 7670 mm2

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BS 8110 Load-bearing walls 360

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From Table 220

16H25-250 gives 7854 mm2 and

100Asbd 100 7854(4000 775) 025

Critical perimeter for punching shear occurs at 15d from faceof column where the length of side of the perimeter

l1 c 3d 600 3 775 2925 mm

Hence

V f (l 2 ndash l12) 425 (42 ndash 29252) 3164 kN

v V4l1d 3164 103(4 2925 775) 035 Nmm2

vc 0216

0216 (400775)14(025 35)13 038 Nmm2( v)

Critical position for shear on vertical section across full widthof base occurs at distance av a2 850 mm 2d from faceof column where

V 425 4 172 1445 kN

v Vbd 1445 103(4000 775) 047 Nmm2

vc(2dav) 038 (2 775850) 069 Nmm2 ( v)

283 PILE-CAPS

In BS 8110 (and BS 5400) a pile-cap may be designed by eitherbending theory or truss analogy In the latter case the truss isof a triangulated form with nodes at the centre of the loadedarea and at the intersections of the centrelines of the piles withthe tension reinforcement as shown for compact groups of twoto five piles in Table 361 Expressions for the tensile forcesare given taking into account the dimensions of the columnand also simplified expressions when the column dimensionsare ignored Bars to resist the tensile forces are to be locatedwithin zones extending not more than 15 times the pilediameter either side of the centre of the pile The bars are tobe provided with a tension anchorage beyond the centres of thepiles and the bearing stress on the concrete inside the bend inthe bars should be checked (see Table 355)

The design shear stress calculated at the perimeter of thecolumn should not exceed the lesser of 08radicfcu or 5 Nmm2Critical perimeters for checking shear resistance are shown inTable 361 where the whole of the shear force from piles withcentres lying outside the perimeter should be taken intoaccount The shear resistance is normally governed by shearalong a vertical section extending across the full width of thecap where the design concrete shear stress may be enhanced asshown in Table 361 If the pile spacing exceeds 3 times the pilediameter punching shear should also be considered

In BS 5400 shear enhancement applies to zones of widthequal to the pile diameter only and the concrete shear stress istaken as the average for the whole section Also the check forpunching shear is made at a distance of 15d from the face ofthe column with no enhancement of the shear strength

The following values are recommended for the thickness ofpile-caps where hp is the pile diameter

For hp 550 mm h (2hp 100) mm

For hp 550 mm h 8(hp ndash 100)3 mm

400d 14100As fcu

bd 13

Example 2 A pile-cap is required for a group of 4 450 mmdiameter piles arranged at 1350 mm centres on a square gridThe pile-cap supports a 450 mm square column subjected to anultimate design load of 4000 kN

fcu 35 Nmm2 fy 500 Nmm2

Allowing for an overhang of 150 mm beyond the face of the pilesize of pile-cap 1350 450 300 2100 mm square

Take depth of pile-cap as (2hp 100) 1000 mm

Assuming tension reinforcement to be 100 mm up from base ofpile-cap d 1000 ndash 100 900 mm

Using truss analogy with the apex of the truss at the centre ofthe column the tensile force between adjacent piles is

Ft 750 kN in each zone

As Ft087fy 750 103(087 500) 1724 mm2

Providing 4H25 gives 1963 mm2 and since the pile spacing isnot more than 3 times the pile diameter bars may be spreaduniformly across the pile-cap giving a total of 8H25-275 in eachdirection so that

100As bh 100 3926(2100 1000) 019 ( 013 min)

Critical position for shear on vertical section across full widthof pile-cap occurs at distance from face of column given by

av 05(l ndash c) ndash 03hp

05 (1350 ndash 450) ndash 03 450 315 mm

Portion of column load carried by two piles is 2000 kN thus

v Vbd 2000 103(2100 900) 106 Nmm2

vc 0216

0216 (400900)14 (019 35)13 033 Nmm2

vc (2dav) 033 (2 900315) 188 Nmm2 ( v)

Shear stress calculated at perimeter of column is

v Vud 4000 103(4 450 900) 247 Nmm2

Maximum shear strength 08radicfcu 473 Nmm2 ( v)

Taking ab as either centre-to-centre distance between bars or(side cover plus bar size) whichever is less

ab 275 (75 25) 100 mm ab 10025 4

From Table 355 minimum radius of bend rmin 7 say

284 RETAINING WALLS ON SPREAD BASES

General notes on the design of walls to BS 8002 are given insection 732 Design values of earth pressure coefficients arebased on a design soil strength which is taken as the lower ofeither the peak soil strength reduced by a mobilisation factor orthe critical state strength Design values of the soil strengthusing effective stress parameters are given by

design tan (tan max)12 tan crit

design c c12 tan crit

400d 14100As fcu

bd 13

4000 13508 900 Nl

8d

Miscellaneous members and details324

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BS 8110 Pile-caps 361

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Miscellaneous members and details326

where c max and crit are representative (ie conservative)values of effective cohesion peak effective angle of shearingresistance and critical state angle of shearing resistance for thesoil In the absence of reliable site investigation and soil testdata values may be derived from Table 210

Design values of friction and adhesion at the soil-structureinterface (wall or base) are given by

design tan (or b) 075 design tan

design cw (or cb) 075 cud 05 cu in which

cud cu 15 where cu is undrained shear strength

A minimum surcharge of 10 kNm2 applied to the surface of theretained soil and a minimum depth of unplanned earth removalin front of the wall equal to 10 of the wall height but not lessthan 05 m should be considered Wall friction should beignored in the determination of KA

Suitable dimensions for the base to a cantilever wall can beestimated with the aid of the chart given in Table 286 Forsliding the chart is valid for non-cohesive soils only Thusfor bases founded on clay soils the long-term condition can beinvestigated by using crit with c 0 For the short-termcondition the ratio does not enter into the calculations forsliding and taking the contact surface length as the full widthof the base is given by KAl2cb When has beendetermined from this equation the curve for radicKA on the chartcan be used to check the values of and that were obtainedfor the long-term condition

Example 3 A cantilever retaining wall on a spread base isrequired to support level ground and a footpath adjacent to aroad The existing ground may be excavated as necessary toconstruct the wall and the excavated ground behind the wall isto be reinstated by backfilling with a granular material Agraded drainage material will be provided behind the wall withan adequate drainage system at the bottom

Height of fill to be retained 40 m above top of baseSurcharge 100 mm surfacing plus 5 kNm2 live load

(design for minimum value of 10 kNm2)

Properties of retained soil (well graded sand and gravel)unit weight of soil 20 kNm3

max crit 35o design tanminus1 [(tan 35o)12] 30o

KA (1 ndash sin)(1 sin) 033

Properties of sub-base soil (medium sand)allowable bearing value pmax 200 kNm2

max 35o crit 32o design 30o (as fill)

design tan b 075 design tan 043

Take thickness of both wall (at bottom of stem) and base to beequal to (height of fill)10 400010 400 mm

Height of wall to underside of base l 40 04 44 m

Allowing for surcharge equivalent height of wall

le l q 44 1020 49 m

pmaxle 200(20 49) asymp 20

tan bradicKA 043radic033 075

From Table 286 radicKA 08 018 Hence

Width of base le (08radic033) 49 225 m

Toe projection (le) 018 225 04 m

Example 4 The sub-base for the wall described in example 3is a clay soil with properties as given here All other values areas specified in example 3

Properties of sub-base soil (firm clay)allowable bearing value pmax 100 kNm2 cu 50 kNm2

cud 5015 333 kNm2 design cb 502 25 kNm2

crit 25o (assumed plasticity index 30)

design tan b 075 design tan crit 033

For the long-term condition

pmaxle 100(20 49) asymp10

tan bradicKA 033radic033 057

From Table 286 radicKA 115 024 Hence

Width of base le (115radic033) 49 33 m say

Toe projection (le) 024 33 08 m

For the short-term condition

KA l2cb 033 20 49(2 25) 065

radicKA 065radic033 113 (115)

Since this value is less than that calculated for the long-termcondition the base dimensions are satisfactory

Example 5 The wall obtained in example 4 a cross sectionthrough which is shown here is to be designed to BS 8002

The vertical loads and bending moments about the front edgeof the base are

Load (kN) Moment (kNm)Surcharge 10 21 210 225 473Backfill 20 21 40 1680 225 3780Wall stem 24 04 40 384 10 384Wall base 24 04 33 317 165 523

Totals Fv 2591 5160

The horizontal loads and bending moments about the bottom ofthe base are

Load (kN) Moment (kNm)Surcharge 033 10 44 145 442 319Backfill 033 20 4422 639 443 937

Totals Fh 784 1256

Resultant moment Mnet 516 ndash 1256 3904 kNm

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Since the f values used for the horizontal and vertical loads arenot the same the bearing pressures must be recalculated

Fv 14 2591 3627 kN Mv 14 516 7224 kNm

Mnet 7224 ndash 12 1256 5717 kNm

a 57173627 1575 m e 165 ndash 1575 0075 m

pmax (362733)(1 6 007533) 125 kNm2

pmin (362733)(1 ndash 6 007533) 95 kNm2

Bearing pressure under base at inside face of wall

pwall 95 (125 ndash 95)(2133) 95 19 114 kNm2

Bending moment on base at inside face of wall

M 14 (10 20 4 24 04) 2122(95 2122 19 2126) 84 kNm

As 84 106[(087 500)(095 352)] 578 mm2m

Use H16-250 to fit in with vertical bars in wall

Shear force on base at inside face of wall

V 14 (10 20 4 24 04) 21(95 21 19 212) 734 kN

Vbd 734 103(1000 352) 021 Nmm2 ( vc)

For the base the bending moment and shear force have beencalculated for a bearing pressure diagram that varies linearly asindicated in BS 8110 If the pressure diagram assumed for theultimate bearing condition in example 5 is taken

pu Fv2a 3627(2 1575) 1151 kNm2

M 14 (10 20 4 24 04) 21221151 (315 ndash 12)22 886 kNm

V 14 (10 20 4 24 04) 211151 (315 ndash 12) 684 kN

285 RECOMMENDED DETAILS

The information given in Tables 362 and 363 has been takenfrom several sources including BS 8110 research undertakenby the Cement and Concrete Association (CampCA) and reportspublished by the Concrete Society

2851 Continuous nibs

The BS 8110 recommendations are shown in Table 362 As aresult of investigations by the CampCA various methods ofreinforcing continuous nibs were put forward Method (a) isefficient but it is difficult to incorporate the bars in shallow nibsif the bends in the bars are to meet the minimum code require-ments Method (b) is reasonably efficient and it is a simplematter to anchor the bars at the outer face of the nib The CampCArecommendations were based on an assumption of truss actionBS 8110 suggests that such nibs should be designed as shortcantilever slabs but both methods lead to similar amounts ofreinforcement

2852 Corbels

The information in Table 362 is based on the requirements ofBS 8110 and BS 5400 supplemented by recommendations

Recommended details 327

Distance from front edge of base to resultant vertical force

a MnetFv 39042591 150 m

Eccentricity of vertical force relative to centreline of base

e 332 ndash 15 015 m ( 336 055 m)

Maximum pressure at front of base

pmax (259133)(1 6 01533) 100 kNm2

Minimum pressure at back of base

pmin (259133)(1 ndash 6 01533) 57 kNm2

For the ultimate bearing condition a uniform distribution isconsidered of length lb 2a 2 15 30 m Pressure

pu Fvlb 259130 864 kNm2

The ultimate bearing resistance is given by the equation

qu (2 ) cud ic where ic 05[1 ]

ic 05[1 ] 073

qu (2 ) 333 073 125 kNm2 ( pu 864)

Resistance to sliding (long-term)

Fv tan b 2591 033 855 kN ( Fh 784)

Resistance to sliding (short-term)

(le) cb 33 25 825 kN ( Fh 784)

For resistance to sliding (short-term) the contact surface hasbeen taken as the full width of the base This is consideredreasonable since base adhesion is taken as only 075cud If thecontact surface is based on the pressure diagram assumed forthe ultimate bearing condition the resistance to sliding isreduced to lb cb 30 25 75 kN

Example 6 The structural design of the wall in example 5 isto be in accordance with the requirements of BS 8110

fcu 35 Nmm2 fy 500 Nmm2 nominal cover 40 mm

Allowing for H16 bars with 40 mm cover

d 400 ndash (40 8) 352 mm

For the ULS values of f are taken as 12 for the horizontalloads and 14 for all the vertical loads

The ultimate bending moment at the bottom of the wall stem

M 12 033 (10 422 20 436) 1162 kNmm

Mbd2fcu 1162 106(1000 3522 35) 00268

From Table 314 since K 0043 z 095d

As 1162 106[(087 500)(095 352)] 799 mm2m

From Table 220 H16-250 gives 804 mm2m

The ultimate shear force at the bottom of the wall stem

V 12 033 (10 4 20 422) 792 kNm

Vbd 792 103(1000 352) 023 Nmm2 ( vc)

From Table 343 the clear distance between bars should notexceed (47 000fs)(100Asbd) 750 mm Thus with

fs 087fyf 087 50012 362 Nmm2

ab (47 000362)[100 804(1000 352)] 568 mm

1 784(333 30)

1 Fh(cud lb)

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Recommended details nibs corbels and halving joints 362

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Recommended details intersections of members 363

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taken from CampCA research reports For the system of forcesshown in the figure the inclined force in the concrete

Fc Fvsin13 k1 fcub(x cos13) where tan13 (d ndash k2 x)av

In these expressions k1 and k2 are properties of the concretestress block as given in section 241 From these expressions aquadratic equation in xd (given in the table) can be derived InBS 8110 for values of xd (1 ndash 2avd)k2 a minimum valueof Ft 05Fv is taken for the tensile force

In the detail shown in the table the main bars are bent backto form a loop If the bars are welded to a crossbar they can becurtailed at the outer edge of the corbel In this case two addi-tional small diameter bars are needed to support the horizontalstirrups If the stirrups are required to pass outside the main barsin the column they should be detailed as two lapping U-barsfor ease of assembly

2853 Halving joints

The recommendations given in BS 5400 as a result of workcarried out by the CampCA are summarised in Table 362 Theinclined links must intersect the line of action of Fv If thiscannot be ensured (eg the inclined links could be displaced)or if horizontal forces can occur at the joint horizontal linksmust also be provided as shown

2854 Reinforcement details at frame corners

Research has shown that when frame corners are subjected tobending moments tending to close the corner the most likelycause of premature failure is due to bearing under the bend ofthe tension bars at the outside of the corner Provided that theradius of the bend is gradual and that sufficient anchorage isgiven for the lapping bars the use of simple details as shown in(a) or (b) on Table 363 is recommended

With lsquoopening cornersrsquo the problems are somewhatgreater and tests have shown that some details can fail at wellbelow their calculated strength In this case the detail shownin (d) is recommended If at all possible a concrete splayshould be formed within the corner and the diagonal rein-forcement Ass provided with appropriate cover If a splay isimpracticable the diagonal bars should be included withinthe corner itself

Detail (d) is suitable for reinforcement amounts up to about 1If more than this is required transverse links should be includedas shown in (e) The arrangement shown in (c) could be usedbut special attention needs to be paid to bending and fixingthe diagonal links which must be designed to resist all the forcein the main tension bars Care must also be taken to provideadequate cover to the bars at the inside of the corner

2855 Beam-column intersections

Research has shown that the forces in a joint between a beamand an end column are as shown in the sketch on Table 363Diagonal tensile forces occur at right angles to the theoreticalstrut that is shown To ensure that as a result diagonal cracksdo not form across the corner a design limit that is related tothe service condition is shown in the table To ensure that the

joint has sufficient ultimate strength an expression has alsobeen developed for a minimum amount of reinforcement that isneeded to extend from the top of the beam into the column atthe junction However for floor beams it has been shown thatthe U-bar detail shown in the table is satisfactory

While research indicates that unless it is carefully detailedas described the actual strength of the joint between a beamand an end column could be as little as half of the calculatedmoment capacity it seems that internal beam-column jointshave considerable reserves of strength Joints having a beam onone side of a column and a short cantilever on the other aremore prone to loss of strength and it is desirable in suchcircumstances to detail the joint as for an end column withthe beam reinforcement turned down into the column and thecantilever reinforcement extending into the beam

Example 7 A corbel is required to support a design ultimatevertical load of 500 kN at a distance of 200 mm from the faceof a column The load is applied through a bearing pad andthe 300 mm wide corbel is to be designed to BS 8110

fcu 30 Nmm2 fy 500 Nmm2 cover 40 mm

Minimum reinforcement based on Ft 05Fv and fyd 087fy

As 05 500 103(087 500) 575 mm2

Calculate shear resistance based on 2H20 (As 628 mm2) forvalues of d 400 mm where av 200 mm b 300 mm vc isobtained from Table 333 and Vc vc (2dav)bd

Miscellaneous members and details330

d vc (2dav)vc Vc ( 10ndash3)mm 100Asbd Nmm2 Nmm2 kN

500 042 050 250 375550 038 049 270 445600 035 047 282 508

Assuming that As will need to be increased as a result of thecorbel analysis with a corresponding increase in Vc considerh 600 mm with d 550 mm Hence

kv Fvbdfcu 500 103(300 550 30) 010

avd 200550 036 k1 040 k2 045 (section 241)

From Table 362 in the quadratic equation for xd

0452 040 045 03601 085

2 045 040 03601 234

1 0362 113 giving the equation

085(xd)2 ndash 234(xd ) 113 0 from which xd 0625

1 av

d 2

2k2 k1

kvav

d

k22

k1k2

kvav

d

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Hence x 0625 550 344 mm and the strain in the bars

s 00035(d ndash x)x 00035 206344 00021

fyd s Es 00021 200 103 420 Nmm2 ( 087fy)

Since (1 ndash 2avd)k2 (1 ndash 2 200550)045 0622 xd

Ft Fvav(d ndash k2x) Fv2 and As Fvav(d ndash k2x)fyd

As 500 103 200[(550 ndash 045 344) 420] 603 mm2

In this case 2H20 giving 628 mm2 is sufficient for the tensileforce but insufficient for the shear resistance Changing the

reinforcement to 3H20 gives 942 mm2 and increases the shearresistance as follows

100Asbd 100 942(300 550) 057

vc 056 Nmm2 Vc vc (2dav)bd 508 kN

Minimum area of horizontal links 05 942 471 mm2 tobe provided by 3H10 links (2 legs per link) which should belocated in the tension zone (ie extending over a depth of about200 mm below the main bars)

Recommended details 331

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Part 4

Design to EuropeanCodes

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Chapter 29

Design requirementsand safety factors

In the Eurocodes (ECs) design requirements are set out inrelation to specified limit-state conditions Calculations todetermine the ability of a member (or assembly of members)to satisfy a particular limit-state are undertaken using designactions (loads or deformations) and design strengths Thesedesign values are determined from either characteristic actionsor representative actions and from characteristic strengths ofmaterials by the application of partial safety factors

291 ACTIONS

Characteristic values of the actions to be used in the design ofbuildings and civil engineering structures are given in severalparts of EC 1 Actions on structures as follows

1991-1-1 Densities self-weight and imposed loads1991-1-2 Actions on structures exposed to fire1991-1-3 Snow loads1991-1-4 Wind loads1991-1-5 Thermal actions1991-1-6 Actions during execution1991-1-7 Accidental actions due to impact and explosions1991-2 Traffic loads on bridges1991-3 Actions induced by cranes and machinery1991-4 Actions on silos and tanks

A variable action (eg imposed load snow load wind loadthermal action) has the following representative values

characteristic value Qk

combination value 0Qk

frequent value 1Qk

quasi-permanent value 2Qk

The characteristic and combination values are used for theverification of the ultimate and irreversible serviceability limit-states The frequent and quasi-permanent values are usedfor the verification of ULSs involving accidental actions andreversible SLSs The quasi-permanent values are also used forthe calculation of long-term effects

Design actions (loads) are given by

design action (load) F Fk

where Fk is the specified characteristic value of the actionF is the value of the partial safety factor for the action (A for

accidental actions G for permanent actions Q for variableactions) and the limit state being considered and is either10 0 1 or 2 Recommended values of F and are given inEC 0 Basis of structural design

292 MATERIAL PROPERTIES

The characteristic strength of a material fk means that valueof either the cylinder strength fck or the cube strength fckcube ofconcrete or the yield strength fyk of reinforcement below whichnot more than 5 of all possible test results are expected tofall In practice the concrete strength is selected from a set ofstrength classes which in EC 2 are based on the characteristiccylinder strength The application rules in EC 2 are validfor reinforcement in accordance with BS EN 10080 whosespecified yield strength is in the range 400ndash600 MPa

Design strengths are given by

design strength fkM

where fk is either fck or fyk as appropriate and M is the value ofthe partial safety factor for the material ( C for concrete S forsteel reinforcement) and the limit-state being considered asspecified in EC 2

293 BUILDINGS

Details of the design requirements and partial safety factors aregiven in Table 41 Appropriate combinations of design actionsand values of are given on page 336

294 CONTAINMENT STRUCTURES

For structures containing liquids or granular solids the mainrepresentative value of the variable action resulting from theretained material should be taken as the characteristic value forall design situations Appropriate characteristic values are givenin EC 1 Part 4 Actions in silos and tanks

For the ULS where the maximum level of a retained liquidcan be clearly defined and the effective density of the liquid(allowing for any suspended solids) will not vary significantlythe value of Q applied to the resulting characteristic actionmay be taken as 12 Otherwise and for retained granularmaterials in silos Q 15 should be used

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Design requirements and safety factors336

For the SLS of cracking a classification of liquid-retainingstructures in relation to the required degree of protectionagainst leakage and the corresponding design requirementsgiven in EC 2 Part 3 are detailed here Silos containing dry

materials may generally be designed as Class 0 but where thestored material is particularly sensitive to moisture class 1 2or 3 may be appropriate

Limit-state and design consideration Combination of design actions

Ultimate (persistent and transient actions) 13Gj Gkj Q1 Qk1 13Qi 0i Qki ( j 1 i 1)

Ultimate (accidental action) (Ad 13Gkj (11 or 21) Qk1 132i Qki ( j 1 i 1)

Serviceability (function including damage to structural and 13Gkj Qk1 130i Qki ( j 1 i 1)non-structural elements eg partition walls etc)

Serviceability (comfort to user use of machinery avoiding 13Gkj 11 Qk1 132i Qki ( j 1 i 1)ponding of water etc)

Serviceability (appearance) 13Gkj 132i Qki ( j 1 i 1)

Note In the combination of design actions shown above Qk1 is the leading variable action and Qki are any accompanying variable actions Where necessary eachaction in turn should be considered as the leading variable action Serviceability design consideration and associated combination of design actions as specified in the UK National Annex

Values of for variable actions ( as specified in the UK National Annex) 0 1 2

Imposed loads (Category and type see EN 1991-1-1)A domestic residential area B office area 07 05 03C congregation area D shopping area 07 07 06E storage area 10 09 08F traffic area (vehicle weight 30 kN) 07 07 06G traffic area (30 kN vehicle weight 160 kN) 07 05 03H roof 07 0 0

Snow loads (see EN 1991-1-3)Sites located at altitude 1000 m above sea level 07 05 02Sites located at altitude 1000 m above sea level 05 02 0

Wind loads (see EN 1991-1-4) 05 02 0

Thermal actions (see EN 1991-1-5) 06 05 0

Class Leakage requirements Design provisions

0 Leakage acceptable or irrelevant The provisions in EN 1992-1-1 may be adopted (see Table 41)

1 Leakage limited to small amount The width of any cracks that can be expected to pass through the full thicknessSome surface staining or damp of the section should be limited to wk1 given bypatches acceptable 005 wk1 0225(1 hw 45h) 02 mm

where hwh is the hydraulic gradient (ie head of liquid divided by thicknessof section) at the depth under consideration Where the full thickness of the section is not cracked the provisions in EN 1992-1-1 apply (see Table 41)

2 Leakage minimal Appearance Cracks that might be expected to pass through the full thickness of the sectionnot to be impaired by staining should be avoided unless measures such as liners or water bars are included

3 No leakage permitted Special measures (eg liners or prestress) are required to ensure watertightness

Note In classes 1 and 2 to provide adequate assurance that cracks do not pass through the full width of a section the depth of the compression zone should beat least 02h 50 mm under quasi-permanent loading for all design conditions The depth should be calculated by linear elastic analysis assuming that concretein tension is neglected

Combinations of design actions on buildings

Values of for variable actions on buildings

Classification of liquid-retaining structures

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Design requirements and partial safety factors (EC 2 Part 1) 41

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Chapter 30

Properties of materials

301 CONCRETE

3011 Strength and elastic properties

The characteristic strength of concrete is defined as that level ofcompressive strength below which 5 of all valid test results isexpected to fall Strength classes are specified in terms of bothcylinder strength and equivalent cube strength Recommendedstrength classes with indicative values for the secant modulusof elasticity at 28 days are given in Table 42 The values givenfor normal-weight concrete are appropriate for concretesmade with quartzite aggregates For limestone and sandstoneaggregates these values should be reduced by 10 and 30respectively For basalt aggregates the values should be increasedby 20

Variation of the secant modulus of elasticity with time can beestimated by the expression

Ecm(t) [ fcm(t)fcm]03 Ecm

where Ecm(t) and fcm(t) are values at age t days and Ecm and fcm

are values determined at age 28 days

3012 Creep and shrinkage

The creep strain in concrete may be assumed to be directlyproportional to the applied stress for stresses not exceeding045fck(t0) where t0 is age of concrete at the time of loadingValues of the final creep coefficient (infin to) and the creepdevelopment coefficient c(t t0) according to the time underload can be obtained from Table 43 The procedure used todetermine the final creep coefficient is as follows

1 Determine point corresponding to age of loading to on theappropriate curve (N for normally hardening cement R forrapidly hardening cement S for slowly hardening cement)

2 Construct secant line from origin of curve to the pointcorresponding to to

3 Determine point corresponding to the notional size ofthe member ho on the appropriate curve for the concretestrength class

4 Cross horizontally from the point determined in 3 to intersectthe secant line determined in 2

5 Drop vertically from the intersection point determined in 4to obtain the required creep coefficient (infin to)

When the applied stress exceeds 045fck(t0) at time of loadingcreep non-linearity should be considered and (infin to) shouldbe increased as indicated in Table 42

The total shrinkage strain is composed of two componentsautogenous shrinkage strain and drying shrinkage strainThe autogenous shrinkage strain develops during hardening ofthe concrete a major part therefore develops in the early daysafter casting It should be considered specifically when newconcrete is cast against hardened concrete Drying shrinkagestrain develops slowly as it is a function of the migration ofthe water through the hardened concrete Final values of eachcomponent of shrinkage strain and development coefficientswith time are given in Table 42

3013 Thermal properties

Values of the coefficient of thermal expansion of concrete fornormal design purposes are given in Table 42

3014 Stressndashstrain curves

Idealised stressndashstrain curves for concrete in compression aregiven in Table 44 Curve A is part parabolic and part linear andcurve B is bi-linear A simplified rectangular diagram is alsogiven as a further option

302 REINFORCEMENT

3021 Strength and elastic properties

The characteristic yield strength of reinforcement according toEN 10080 is required to be in the range 400ndash600 MPa Thecharacteristic yield strength of reinforcement complying withBS 4449 is 500 MPa For further information on types proper-ties and preferred sizes of reinforcement reference should bemade to section 103 and Tables 219 and 220

3022 Stressndashstrain curves

Idealised bi-linear stressndashstrain curves for reinforcement intension or compression are shown in Table 44 Curve A has aninclined top branch up to a specified strain limit and curve Bhas a horizontal top branch with no need to check the strainlimit For design purposes the modulus of elasticity of steelreinforcement is taken as 200 GPa

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Concrete (EC 2) strength and deformation characteristics ndash 1 42

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Concrete (EC 2) strength and deformation characteristics ndash 2 43

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Stressndashstrain curves (EC 2) concrete and reinforcement 44

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Chapter 31

Durability andfire-resistance

In the following the concrete cover to the first layer of barsas shown on the drawings is described as the nominal coverIt is defined as a minimum cover plus an allowance in designfor deviation A minimum cover is required to ensure the safetransmission of bond forces the protection of steel againstcorrosion and an adequate fire-resistance In order to transmitbond forces safely and ensure adequate concrete compactionthe minimum cover should be not less than the bar diameter orfor bundled bars the equivalent diameter of a notional barhaving the same cross-sectional area as the bundle

311 DURABILITY

3111 Exposure classes

Details of the classification system used in BS EN 206ndash1 andBS 8500-1 with informative examples applicable in the UnitedKingdom are given in Table 45 Often the concrete can beexposed to more than one of the actions described in the tablein which case a combination of the exposure classes will apply

3112 Concrete strength classes and covers

Concrete durability is dependent mainly on its constituents andlimitations on the maximum free watercement ratio and theminimum cement content are specified for each exposure classThese limitations result in minimum concrete strength classesfor particular cements For reinforced concrete the protectionof the steel against corrosion depends on the cover The requiredthickness of cover is related to the exposure class the concretequality and the intended working life of the structure Details ofthe recommendations in BS 8500 are given in Table 46

The values given for the minimum cover apply for ordinarycarbon steel in concrete without special protection and forstructures with an intended working life of at least 50 years Thevalues given for the nominal cover include an allowance fortolerance of 10 mm which is recommended for buildings andis normally also sufficient for other types of structures

For uneven concrete surfaces (eg ribbed finish or exposedaggregate) the cover should be increased by at least 5 mm Ifin-situ concrete is placed against another concrete element(precast or in-situ) the minimum cover to the reinforcement atthe interface need be no more than that recommended foradequate bond provided the following conditions are metthe concrete strength class is at least C2530 the exposure time

of the concrete surface to an outdoor environment is no more than28 days and the interface has been roughened

Where concrete is cast against prepared ground (includingblinding) the nominal cover should be at least 50 mm Forconcrete cast directly against the earth the nominal covershould be at least 75 mm

312 FIRE-RESISTANCE

3121 Building regulations

The minimum period of fire-resistance required for elements ofthe structure according to the purpose group of a buildingand its height or for basements depth relative to the ground aregiven in Table 312 Building insurers may require longer fireperiods for storage facilities

3122 Design procedures

BS EN 1992-1-2 contains prescriptive rules in the form ofboth tabulated data and calculation models for standard fireexposure A procedure for a performance-based method usingfire-development models is also provided

The tabulated data tables give minimum dimensions for thesize of a member and the axis distance of the reinforcementThe axis distance is the nominal distance from the centre ofthe main reinforcement to the concrete surface as shown in thefollowing figure

Tabulated data is given for beams slabs and braced columnsfor which provision is made for the load level to be taken intoaccount In many cases for fire periods up to about two hoursthe cover required for other purposes will control For furtherinformation on all the design procedures reference should bemade to BS EN 1992-1-2

Section through member showing nominal axis distance

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Exposure classification (BS 8500) 45

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Concrete quality and cover requirements for durability (BS 8500) 46

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fck Maximum value of xd at section whereMPa moment redistribution is required

50 ( 04)

55 095 ( 04)60 092 ( 04)70 090 ( 04)

80 088 ( 04)

In the above expressions is the ratio of the design moment to themaximum elastic moment for values of

where class B and C reinforcement is usedwhere class A reinforcement is used

Based on values given in the UK National Annex10 0810 07

Chapter 32

Bending and axialforce

321 DESIGN ASSUMPTIONS

Basic assumptions regarding the design of cracked concretesections at the ULS are outlined in section 52 The tensilestrength of the concrete is neglected and strains are evaluatedon the assumption that plane sections before bending remainplane after bending Reinforcement stresses are then derivedfrom these strains on the basis of the design stress-strain curvesshown on Table 44 Two alternatives are given in which the topbranch of the curve is taken as either horizontal with no needto check the strain limit or inclined up to a specified strainlimit For the concrete stresses three alternative assumptionsare permitted as shown on Table 44 The design stressndashstraincurves give stress distributions that are a combination of eithera parabola and a rectangle or a triangle and a rectangleAlternatively a rectangular concrete stress distribution may beassumed Whichever alternative is used the proportions of thestress-block and the maximum strain are constant forfck 50 MPa but vary as the concrete strength changes forfck 50 MPa

For a rectangular area of width b and depth x the total com-pressive force is given by k1 fckbx and the distance of the forcefrom the compression face is given by k2x where values of k1

(including for the term ccc) and k2 according to the shape ofthe stress block are given in the table opposite

322 BEAMS AND SLABS

Beams and slabs are generally subjected to bending only butsometimes are also required to resist an axial force for examplein a portal frame or in a floor acting as a prop between basementwalls Axial thrusts not greater than 012fck times the area of thecross section may be ignored in the analysis of the sectionsince the effect of the axial force is to increase the moment ofresistance

In cases where as a result of moment redistributionallowed in the analysis of the member the design moment isless than the maximum elastic moment at the section the nec-essary ductility may be assumed without explicit verificationif the neutral axis depth satisfies the limits given in the tableopposite

Where plastic analysis is used the necessary ductility may beassumed without explicit verification if the neutral axis depth atany section satisfies the following requirement

xd 025 for concrete strength classes C5060xd 015 for concrete strength classes C5060

The following analyses and resulting design charts and tablesare applicable to concrete strength classes C5060

fck Shape of k1 k2 cuMPa stress-block

50 Parabola-rectangle 0459 0416Triangle-rectangle 0425 0389 00035Rectangular 0453 0400

55 Parabola-rectangle 0433 0400Triangle-rectangle 0402 0375 00031Rectangular 0435 0394

60 Parabola-rectangle 0417 0392Triangle-rectangle 0381 0363 00029Rectangular 0417 0388

70 Parabola-rectangle 0399 0383Triangle-rectangle 0357 0351 00027Rectangular 0382 0375

80 Parabola-rectangle 0385 0378Triangle-rectangle 0327 0340 00026Rectangular 0349 0363

90 Parabola-rectangle 0378 0375Triangle-rectangle 0316 0337 00026Rectangular 0317 0350

Properties of concrete stress-blocks for rectangular area

Limits of xd for moment redistribution

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Bending and axial force346

3221 Singly reinforced rectangular sectionsMbd2fck

As fykbdfck for ductility class

A B C

0010 00111 00108 001020012 00134 00130 001230014 00156 00152 001440016 00179 00174 001650018 00202 00197 00186

0020 00224 00219 002070022 00247 00241 002280024 00270 00264 002520026 00293 00286 002760028 00317 00309 00300

0030 00340 00331 003240032 00363 00354 003490034 00387 00379 003730036 00410 00403 003980038 00434 00428 00423

0040 00457 00453 004480042 00481 00478 004730044 00505 00503 004980046 00529 00528 005230048 00553 00554 00549

3222 Doubly reinforced rectangular sections

The lever arm between the forces shown in the figure here isgiven by z (d k2x) from which x (d z)k2

Taking moments for the compressive force about the line ofaction of the tensile force gives

M k1 fckbxz k1 fckbz(d z)k2

The solution of the resulting quadratic equation in z gives

where

Taking moments for the tensile force about the line of action ofthe compressive force gives

M As fs z from which As Mfs z

The strain in the reinforcement s 00035(1 xd)(xd) andfrom the design stressndashstrain curves the stress is given by

fs sEs 700(1 xd)(xd) ks fyk 115

If the top branch of the design stressndashstrain curve is taken ashorizontal (curve B) ks 10 and fs fyk 115 for values of

xd 805(805 fyk) 0617 for fyk 500 MPa

A design chart for fyk 500 MPa derived on the basis of theparabolic-rectangular stress-block for the concrete and curve Bfor the reinforcement is given in Table 47

If the top branch of the design stressndashstrain curve is taken asinclined (curve A) the value of ks depends on the strain and theductility class of the reinforcement The use of curve A isadvantageous in reducing the reinforcement requirement Themaximum reduction for a lightly reinforced section is close to5 8 and 15 for reinforcement ductility classes A B andC respectively For more heavily reinforced sections the reduc-tions are less and the particular ductility class makes very littledifference to the values obtained

A design table based on the rectangular stress-block for theconcrete and design curve A for the reinforcement is given inTable 48 The table uses non-dimensional parameters and isvalid for values of fck 50 MPa Values of ks determined forfyk 500 MPa were based on the most critical ductility classin each case namely class A for Mbd2fck 0046 and class Bfor Mbd2fck 0046 This can be seen from the followingtable which gives values of As fykbdfck for each ductility classin the range up to Mbd2fck 0048

M bd2fckz d 05 025 (k2 k1)

The forces provided by the concrete and the reinforcementare shown in the figure here Taking moments for the twocompressive forces about the line of action of the tensileforce gives

M k1 fckbx(d k2x)

The strain in the reinforcement and fromdesign stressndashstrain curve B the stress is given by

Thus fyk 115 for values of

for fyk 500 MPa

Equating the tensile and compressive forces gives

As fs k1 fckbx

where the stress in the tension reinforcement is given by theexpression derived in section 3221

As f s

x d [805 (805 fyk)](dd) 264(dd )

f s

f s s Es 700(1 dx) fyk 115

s 00035(1 dx)

As f s(dd)

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EC 2 Design chart for singly reinforced rectangular beams 47

Sin

gly

rein

forc

ed b

eam

s (f

yk=

500

MP

a)

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EC 2 Design table for singly reinforced rectangular beams 48

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Beams and slabs 349

Design charts based on the rectangular stress-block for theconcrete design curve B for the reinforcement and for valuesof and 015 respectively are given in Tables 49 and410 The charts which use non-dimensional parameters weredetermined for fyk 500 MPa but may be safely used forfyk 500 MPa In determining the forces in the concrete noreduction has been made for the area of concrete displaced bythe compression reinforcement

3223 Design formulae for rectangular sections

No design formulae are given in the code but the following arevalid for values of fck 50 MPa and fyk 500 MPa Theformulae are based on the rectangular stress-block for theconcrete and stresses of 087fyk in tension and compressionreinforcement The compression reinforcement requirementdepends on the value of Mbd2fck compared to where

0210 for 10

for

is the ratio design moment to maximum elastic momentwhere 07 for class B and C reinforcement and 08 forclass A reinforcement

For compression reinforcement is not required and

As M087fykz where

z d05 + and x (d z)04

For compression reinforcement is required and

where

z d05 + and x (d z)04

For (for fy 500 MPa) should be replaced by16(1 ) in the equations for and As

3224 Flanged sections

In monolithic beam and slab construction where the web of thebeam projects below the slab the beam is considered as aflanged section for sagging moments The effective width of theflange over which uniform conditions of stress can be assumedmay be taken as beff bw where

01(aw l0) 02l0 05aw for L beams 02(aw l0) 04l0 10aw for T beams

In the aforesaid expressions bw is the web width aw is the cleardistance between the webs of adjacent beams and l0 is the dis-tance between points of zero bending moment for the beam Ifleff is the effective span l0 may be taken as 085leff when thereis continuity at one end of the span only and 07leff when there iscontinuity at both ends of the span For up-stand beams whendesigning for hogging moments l0 may be taken as 03leff atinternal supports and 015leff at end supports

In most sections where the flange is in compression the depthof the neutral axis will be no greater than the thickness of theflange In this case the section can be considered to be rectangularwith b taken as the flange width The condition regarding theneutral axis depth can be confirmed initially by showing thatM k1 fckbhf (d k2hf) where hf is the thickness of the flange

bb

b

AsAsdxAsdx 0375

025 0882

As As bd2fck 087fykz

As ()bd2fck 087fyk(dd)

025 0882

10 0453( 04) 0181( 04)2

dd 01

Alternatively the section can be considered to be rectangularinitially and the neutral axis depth can be checked subsequently

The figure here shows a flanged section where the neutral axisdepth is greater than the flange thickness The concrete forcecan be divided into two components and the required area oftension reinforcement is then given by

As As1 k1 fck (b bw)hf 087fyk whereAs1 area of reinforcement required to resist a moment M1

applied to a rectangular section of width bw and

M1 M k1 fck (b bw)hf (d k2hf) bd2fck

Using the rectangular concrete stress-block in the forgoingequations gives k1 045 and k2 04 This approach givessolutions that are lsquocorrectrsquo when x hf but become slightlymore conservative as (x hf) increases

3225 General analysis of sections

The analysis of a section of any shape with any arrangement ofreinforcement involves a trial-and-error process An initialvalue is assumed for the neutral axis depth from which theconcrete strains at the positions of the reinforcement can becalculated The corresponding stresses in the reinforcement aredetermined and the resulting forces in the reinforcement andthe concrete are obtained If the forces are out of balance thevalue of the neutral axis depth is changed and the process isrepeated until equilibrium is achieved Once the balancedcondition has been found the resultant moment of the forcesabout the neutral axis or any convenient point is calculated

Example 1

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EC 2 Design chart for doubly reinforced rectangular beams ndash 1 49

Dou

bly

rein

forc

ed b

eam

s (f

yk=

500

MP

ad9

d=

01)

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410EC 2 Design chart for doubly reinforcedrectangular beams ndash 2

Dou

bly

rein

forc

ed b

eam

s (f

yk=

500

MP

ad9

d=

015

)

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Bending and axial force352

The beam shown in the figure here is part of a monolithic beamand slab floor in which the beams are spaced at 5 m centresThe maximum and minimum design loads on each span are asfollows where Gk 160 kN and Qk 120 kN

Fmax 135Gk 15Qk 396 kN Fmin 135 Gk 216 kN

The section design is to be based on the following values

fck 32 MPa fyk 500 MPa cover to links 35mm

For sagging moments and beff are given by

02(aw l0) 04 l0 aw

02(4700 085 8000) 2300 mmbeff bw 300 2300 2600 mm

Allowing for 8 mm links and 32 mm main bars

d 500 (35 8 16) 440 mm say

In the calculations that follow solutions are obtained usingcharts and equations to demonstrate the use of each method

Maximum sagging moment For section to be designed as rec-tangular with b taken as the flange width bending momentshould satisfy the condition

M k1 fckbhf (d k2hf) 045 32 2600 150 (440 04 150) 106

2134 kNm (258 kNm)

Mbd 2 258 106 (2600 4402) 051 MPa

From chart in Table 47 100Asbd 012

As 00012 2600 440 1373 mm2

Alternatively by calculation or from Table 48

Mbd2fck 05132 0016

Hence As M087fykz gives

As 258 106 (087 500 0985 440) 1369 mm2

Using 3H25 gives 1473 mm2

The above solutions are based on design stress-strain curve Bfor the reinforcement Solutions based on curve A also can beobtained from the table in section 3221 as follows

Mbd 2fck

As fykbdfck for ductility class

A B C

0016 00179 00174 00165As (mm2) 1310 1274 1208

Maximum hogging moment

Mbd2fck 396 106 (300 4402 32) 0213

From Table 48 this appears to be just beyond the range for asingly reinforced section From the chart in Table 49 a valueof As fykbdfck 034 can be obtained Although this is a validsolution it should be possible to reduce the area of tensionreinforcement to a more suitable value by allowing for somecompression reinforcement Consider the use of 2H25 whichgives 55 mm ( )dd 55 440 0125d

z d 05 025 0882 0016 0985

b

b

b

fykbdfck 982 500(300 440 32) 0116

Interpolating from charts in Tables 49 and 410 with

fykbdfck 01 As fykbdfck 0285 (for 0125)As 0285 300 440 32500 2408 mm2

Using 3H32 gives 2413 mm2

A solution can also be obtained using the design equations asfollows 087fy is valid for xd ( )0375 0333

With xd 0333 and zd (1 04xd) 0867

0453(xd)(zd) 0453 0333 0867 0131

( ) bd2fck 087fyk (d ) (0213 0131) 300 4402 32

(087 500 385) 910 mm2 (2H25 gives 982 mm2)

As bd 2fck 087fyk z 910 0131 300 4402 32

(087 500 0867 440) 2377 mm2 (3H32 gives 2413 mm2)

Example 2 Suppose that in the previous example the maxi-mum hogging moment at B is reduced by 20 to 317 kNmThis is valid for reinforcement of all ductility classes

Mbd2fck 317 106 (300 4402 32) 0171 317396 080 xd ( 04) 04

From chart in Table 410 keeping to left of line for xd 04

fykbdfck 005 Asfykbdfck 023 005 300 440 32500 423 mm2

Using 2H20 gives 628 mm2 (instead of 2H25 in example 1)

As 023 300 440 32500 1943 mm2

Using 4H25 gives 1963 mm2 (instead of 3H32 in example 1)

A solution can also be obtained by using the design equationsas in example 1 with xd 0333 zd 0867 0131

( )bd 2fck 087fyk (d ) 004 300 4402 32 (087 500 385) 444 mm2

As bd2fck 087fykz 444 0131 300 4402 32

(087 500 0867 440) 1912 mm2

Since the reduced hogging moment for load case 1 is stillgreater than the elastic hogging moment for load case 2 thedesign sagging moment remains the same as in example 1

In the foregoing examples at the bottom of the beam 2H25bars would run the full length of each span with 2H25 or 2H20splice bars at support B Other bars would be curtailed to suitthe bending moment requirements and detailing rules

323 COLUMNS

In the Code of Practice a column is taken to be a compressionmember whose greater overall cross sectional dimension is notmore than four times its smaller dimension An effective lengthand a slenderness ratio are determined in relation to its major

As

dAs

As

As

As

dAs

ddf s

ddAs

As

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Columns 353

and minor axes of bending The effective length is a function ofthe clear height and depends upon the restraint conditions at theends of the column A slenderness ratio is defined as the effec-tive length divided by the radius of gyration of the uncrackedconcrete section Columns should generally be designed for bothfirst order and second order effects but second order effects maybe ignored provided the slenderness ratio does not exceed a par-ticular limiting value This can vary considerably and has to bedetermined from an equation involving several factors Thesegenerally need to be calculated but default values are also given

Columns are subjected to combinations of bending momentand axial force and the cross section may need to be checkedfor more than one combination of values Several methods ofanalysis of varying complexity are available for determiningsecond order effects Many columns can be treated as isolatedmembers and a simplified method of design using equationsbased on an estimation of curvature is commonly used Theequations contain a modification factor Kr the use of whichresults in an iteration process with Kr taken as 10 initially Thedesign procedures are shown in Tables 415 and 416

In the code for sections subjected to pure axial load the concretestrain is limited to 0002 for values of fck 50 MPa In thiscase the design stress in the reinforcement should be limited to400 MPa However in other parts of the code the design stressin this condition is shown as fyd fyks 087fyk In thederivation of the charts in this chapter which apply for allvalues of fck 50 MPa and fyk 500 MPa the maximumcompressive stress in the reinforcement was taken as 087fykThe charts contain sets of Kr lines to aid the design process

3231 Rectangular columns

Design charts based on the rectangular stress-block for theconcrete and for values of dh 08 and 085 are given inTables 411 and 412 respectively On each curve a straight linehas been taken between the point where xh 10 and the pointwhere N Nu The charts which were determined for fyk 500MPa may be safely used for fy 500 MPa In determining theforces in the concrete no reduction has been made for the areaof concrete displaced by the compression reinforcement In thedesign of slender columns the Kr factor is used to modify thedeflection corresponding to a load Nbal at which the moment is amaximum A line corresponding to Nbal passes through a cusp oneach curve For N Nbal the Kr value is taken as 10 ForN Nbal Kr can be determined from the lines on the chart

3232 Circular columns

The following figure shows a circular section with six barsspaced equally around the circumference Solutions based on sixbars will be slightly conservative if more bars are used Thearrangement of the bars relative to the axis of bending affects theresistance of the section and the arrangement shown in the fig-ure is not the most critical in every case For some combinationsof bending moment and axial force if the arrangement shown isrotated through 30o a slightly more critical condition results butthe differences are small and may be reasonably ignored

The foregoing figure shows a rectangular section in which thereinforcement is disposed equally on two opposite sides of ahorizontal axis through the mid-depth By resolving forces andtaking moments about the mid-depth of the section the follow-ing equations are obtained for 0 xh 10

Nbhfck k1(xh) 05(As fykbhfck)(ks1 ks2)

Mbh2fck k1(xh)05 k2(xh) 05(As fykbhfck)(ks1 ks2)(dh 05)

The stress factors ks1 and ks2 are given by

ks1 14(xh dh 1)(xh) 087ks2 14(dh xh)(xh) 087

The maximum axial force Nu is given by the equation

Nubhfck 0567 087(As fykbhfck)

The following analysis is based on a uniform stress-block forthe concrete of depth x and width hsin at the base (as shownin figure) Negative axial forces are included in order to caterfor members such as tension piles By resolving forces andtaking moments about the mid-depth of the section the followingequations are obtained where cos1 (1 2 xh) for0 x 10 and hs is the diameter of a circle through thecentres of the bars

Nh2fck kc (2 sin2)8 (12)(As fykAc fck)(ks1 ks2 ks3)

Mh3fck kc (3sin sin3)72 ( 277)(As fykAc fck)(hsh)(ks1 ks3)

Because the width of the compression zone decreases in thedirection of the extreme compression fibre the design stressin the concrete has to be reduced by 10 Thus in the aboveequations kc 09 0567 051 and 08

The stress factors ks1 ks2 and ks3 are given by

087 ks1 14(0433hsh 05 xh)(xh) 087

087 ks2 14(05 xh)(xh) 087

087 ks3 14(05 0433hsh xh)(xh) 087

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EC 2 Design chart for rectangular columns ndash 1 411

Rec

tang

ular

col

umns

(f y

k=

500

MP

ad

h=

08)

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EC 2 Design chart for rectangular columns ndash 2 412

Rec

tang

ular

col

umns

(f y

k=

500

MP

ad

h=

085

)

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Bending and axial force356

To avoid irregularities in the charts the reduced design stress inthe concrete is used to determine the maximum axial force Nuwhich is given by the equation

Nuh2fck ( 4)051 087(As fykAc fck)

The minimum axial force Nmin is given by the equation

Nminh2fck 087( 4)(AsfykAcfck)

Design charts for values of hsh 06 and 07 are given inTables 413 and 414 respectively The statements in section3231 on the derivation and use of the charts for rectangularsections apply also to those for circular sections

3233 General analysis of column sections

Any given cross section can be analysed by a trial-and-errorprocess For a section bent about one axis an initial value isassumed for the neutral axis depth from which the concretestrains at the positions of the reinforcement can be calculatedThe resulting stresses in the reinforcement are determined andthe forces in the reinforcement and concrete evaluated If theresultant force is not equal to the design axial force N the valueof the neutral axis depth is changed and the process repeateduntil equality is achieved The resultant moment of all theforces about the mid-depth of the section is then the moment ofresistance appropriate to N This approach is used to analyse arectangular section in example 6

Example 3 A 300 mm square braced column designed forthe following requirements

l 50 m k 0675 at both joints in both directionsM02 40 kNm M01 20 kNm about x-x axisM0 negligible about y-y axis N 1720 kNfck 32 MPa fyk 500 MPa cover to links 35 mm

Allowing for 8 mm links and 32 mm main bars

d 300 (35 8 16) 240 mm say

From Table 415 effective length for a braced column wherethe joint stiffness is the same at both ends is given by

l0 [05(045 2k)(045 k)] l 08 50 40 m

Slenderness ratio l0 i 4000(300radic12) 462

From Table 415 with M01 05M02 C 22 and

Since lim second order moments need to be considered

Minimum design moment with e0 h30 30030 20 mm

Mmin Ne0 1720 002 34 kNm

Additional first order moment resulting from imperfectionswith 067 h 2radicl 2radic(50) 09 10

Mi N( hl0400) 1720 (09 40400) 16 kNm

Total first order moment for section at end 2 of the column

Mx M02 Mi 40 16 56 kNm (Mmin)

Mbh2fck 56 106(300 3002 32) 0065

Nbhfck 1720 103(300 300 32) 0600

lim 34 N Ac fck 34 1720 103 (3002 32) 44

From the design chart for dh 240300 08

As fykbhfck 025 (Table 411)As 025 300 300 32500 1440 mm2

Using 4H25 gives 1963 mm2

The section where the second order moment is greatest may bedesigned by first assuming the reinforcement (4H25 say)

As fykbhfck 1963 500(300 300 32) 034Nubhfck 086 (Table 411) and for Nbhfck 06Mubh2fck 009 Kr 04 (Table 411)Mux Muy 009 300 3002 32 106 78 kNm

Second order moment resulting from deflection with Kr 04and for fck 32 MPa and 46 K 13 (Table 416)

M2 N(KrKl02d)2000

1720 (04 13 402024)2000 30 kNm

Equivalent first order moment (near mid-height of column)

M0e 06M02 04M01 04M02 04 40 16 kNm

Total design moments (near mid-height of column)

Mx M0e Mi M2 16 16 30 62 kNm (Mmin)My M2 30 kNm

Design for biaxial bending may be ignored if the following twoconditions are satisfied (a) 05y x 2y and (b) for asquare column Mx is either 02My or 5My In this casesince condition (b) is not satisfied and a furthercheck is necessary as follows

For NNu 060086 07

n 092 083(NNu) 150

Hence

Since this value is less than 10 4H25 is sufficient

Example 4 A 350 mm circular braced column designed forthe same requirements as example 3 Thus l0 40 m as before

Slenderness ratio l0i 4000(3504) 457

Cross-sectional area Ac (4) 3502 962 103 mm2

Since second order moments need not be consideredAllowing for 8 mm links and 20 mm main bars

hs 350 2 (35 8 10) 244 mm

For the section at end 2 of the column

Mh3fck 56 106(3503 32) 0041Nh2fck 1720 103(3502 32) 044

From the design chart for hsh 244350 07

As fykAcfck 026 (Table 414)As 026 962 103 32500 1600 mm2

Using 6H20 gives 1885 mm2

lim

lim 34 N Ac fck 34 1720 (962 32) 455

Mx

Mux

n

My

Muy

n

6278

15

3078

15

095

Mx 2My

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EC 2 Design chart for circular columns ndash 1 413

Circular columns (fyk = 500 MPa hsh = 06)

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EC 2 Design chart for circular columns ndash 2 414

Circular columns (fyk = 500 MPa hsh = 07)

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EC 2 Design procedure for columns ndash 1 415

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EC 2 Design procedure for columns ndash 2 416

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Columns 361

Example 5 A 400 mm circular unbraced column designedfor the same requirements as example 3 Thus k 0675 asbefore

From Table 415 effective length for an unbraced columnwhere the joint stiffness is the same at both ends is given by

l0 [(1 2k)(1 k)] l 14 50 70 m

Slenderness ratio l0i 7000(4004) 70

Cross-sectional area Ac ( 4) 4002 1257 103 mm2

Since second order moments need to be consideredAllowing for 8 mm links and 32 mm main bars

hs 400 2 (35 8 16) 280 mm say

Additional first order moment resulting from imperfections

Mi N( hl0400) 1720 (09 70400) 27 kNm

Second order moment resulting from deflection with Kr 10(max) K 11 for fck 32 MPa and 70 (Table 416) andd h2 035hs 298 mm

M2 N(KrKl02d)2000

1720 (10 11 7020298)2000 155 kNm

Total design moments at end 2 of column

Mx M02 Mi M2 40 27 155 222 kNm (Mmin)My M2 155 kNm

Resultant uniaxial moment at end 2 of column

kNm

Mh3fck 270 106(4003 32) 013Nh2fck 1720 103(4002 32) 034

From the design chart for hsh 280400 07

As fykAcfck 084 (Table 414)As 084 1257 103 32500 6758 mm2

It can be seen from the chart that Kr 10 Using 8H32 gives

As fykAc fck 6434 500(1257 103 32) 080For Nh2fck 034 Mh3fck 0125 and Kr 075

Hence Mu 0125 4003 32 106 256 kNm

With Kr 075 modified M2 075 155 116 kNm

Total design moments at end 2 of column

Mx M02 Mi M2 40 27 116 183 kNmMy M2 116 kNm

Resultant uniaxial moment at end 2 of column

217 kNm

Since M Mu 8H32 is sufficient

Example 6 The column section in the following figure isreinforced with 8H32 arranged as shown The moment of resis-tance about the major axis is to be obtained for the followingrequirements

N 2300 kN fck 32 MPa fyk 500 MPa

M (1832 1162)

M (M2x M2

y) (2222 1552) 270

lim

lim 108 N Acfck 108 1720 (1257 32) 165

Consider the bars in each half of the section to be replaced byan equivalent pair of bars Depth to the centroid of the bars inone half of the section 60 2404 120 mm The section isnow considered to be reinforced with four equivalent barswhere d 600 120 480 mm

As fykbhfck 6434 500(300 600 32) 056Nbhfcu 2300 103(300 600 32) 040

From the design chart for dh 480600 08

Mubh2fck 018 (Table 411)Mu 018 300 6002 32 106 622 kNm

The solution can be checked using a trial-and-error process toanalyse the original section as follows

The axial load on the section is given by

N k1fckbx (As1ks1 As2ks2 As3ks3)fyk

where dh 540600 09 and ks1 ks2 and ks3 are given by

ks1 14(xh dh 1)(xh) 087ks2 14(05 xh)(xh) 087ks3 14(dh xh)(xh) 087

With x 300 mm xh 05 ks1 087 ks2 0 and ks3 087

N 045 32 300 300 103 1296 kN (2300)

With x 360 mm xh 06 ks2 0233 ks3 07

N 045 32 300 360 103 (2413 087 1608 0233 2413 07) 500 103

1555 392 1947 kN (2300)

With x 390 mm xh 065 ks2 0323 ks3 0538

N 045 32 300 390 103 (2413 087 1608 0323 2413 0538) 500 103

1685 660 2345 kN (2300)

With x 387 mm xh 0645 ks2 0315 ks3 0553

N 045 32 300 387 103 (2413 087 1608 0315 2413 0553) 500 103

1672 636 2308 kN

Since the internal and external forces are now sensibly equaltaking moments about the mid-depth of the section gives

Mu k1 fckbx(05h k2x ) (As1ks1 As3ks3)(d 05h)fy

045 32 300 387 (300 04 387) 106

(2413 087 2413 0553)(540 300) 500 106

243 412 655 kNm (622 obtained before)

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Chapter 33

Shear and torsion

331 SHEAR RESISTANCE

3311 Members without shear reinforcement

The design shear resistance at any cross section of a membernot requiring shear reinforcement can be calculated as

VRdc vRdcbwdwhere

bw is the minimum width of section in the tension zoned is the effective depth to the tension reinforcementvRdc is the design concrete shear stress

The design concrete shear stress is a function of the concretestrength the effective depth and the reinforcement percentageat the section considered To be effective this reinforcementshould extend a distance (lbd d) beyond the section wherelbd is the design anchorage length At a simple support for amember carrying predominantly uniform load the length lbd

may be taken from the face of the support The design shearresistance of members with and without axial load can bedetermined from the data given in Table 417

In the UK National Annex it is recommended that the shearstrength of concrete strength classes higher than C5060 isdetermined by tests unless there is evidence of satisfactorypast performance of the particular concrete mix including theaggregates used Alternatively the shear strength should belimited to that given for concrete strength class C5060

3312 Members with shear reinforcement

The design of members with shear reinforcement is based ona truss model in which the compression and tension chords

are spaced apart by a system of inclined concrete struts andvertical or inclined shear reinforcement Angle between thereinforcement and the axis of the member should be 45o

Angle 13 between the struts and the axis of the member maybe selected by the designer within the limits 10 cot13 25generally However for elements in which shear co-exists withexternally applied tension cot13 should be taken as 10 Theweb forces are Vsec13 in the struts and Vsec in the shearreinforcement over a panel length l z (cot cot13) wherez may normally be taken as 09d The width of each strutis equal to z (cot cot13) sin13 and the design value of themaximum shear force VRdmax is limited to the compressiveresistance provided by the struts which includes a strengthreduction factor for concrete cracked in shear The least shearreinforcement is required when cot13 is such that V VRdmaxThe truss model results in a force Ftd in the tension chord thatis additional to the force Mz due to bending but the sumFtd Mz need not be taken greater than Mmaxz where Mmax isthe maximum moment in the relevant hogging or sagging regionThe additional force Ftd can be taken into account by shiftingthe bending moment curve each side of any point of maximummoment by an amount al 05z(cot13 cot) For memberswithout shear reinforcement al d should be used The cur-tailment of the longitudinal reinforcement can then be based onthe modified bending moment diagram A design procedure todetermine the required area of shear reinforcement and detailsof the particular requirements for beams and slabs are given inTable 418

For most beams a minimum amount of shear reinforcementin the form of links is required irrespective of the magnitude ofthe shear force Thus there is no need to determine VRdc

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EC 2 Shear resistance ndash 1 417

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EC 2 Shear resistance ndash 2 418

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In members with inclined chords the shear components ofthe design forces in the chords may be added to the design shearresistance provided by the reinforcement In checking that thedesign shear force does not exceed VRdmax the same shearcomponents may be deducted from the shear force resultingfrom the design loads

3313 Shear under concentrated loads

In slabs and column bases the maximum shear stress at theperimeter of a concentrated load should not exceed vRdmaxShear in solid slabs under concentrated loads can result inpunching failures on the inclined faces of truncated conesor pyramids For design purposes a control perimeterforming the shortest boundary that nowhere comes closer tothe perimeter of the loaded area than a specified distanceshould be considered The basic control perimeter maygenerally be taken at a distance 2d from the perimeter ofthe loaded area

If the maximum shear stress here is no greater than vRdc noshear reinforcement is required Otherwise the position of thecontrol perimeter at which the maximum shear stress is equalto vRdc should be determined and shear reinforcement providedin the zone between this control perimeter and the perimeter ofthe loaded area

For flat slabs with enlarged column heads (or drop panels)where dH is the effective depth at the face of the column and thecolumn head (or drop) extends a distance lH 2dH beyond theface of the column a basic control perimeter at a distance 2dH

from the column face should be considered In addition a basiccontrol perimeter at a distance 2d from the column head (ordrop) should be considered

Control perimeters (in part or as a whole) at distances lessthan 2d should also be considered where a concentrated loadis applied close to a supported edge or is opposed by a highpressure (eg soil pressure on bases) In such cases values ofvRdc may be multiplied by 2da where a is the distance fromthe edge of the load to the control perimeter For column basesthe favourable action of the soil pressure may be taken intoaccount when determining the shear force acting at the controlperimeter

Details of design procedures for shear under concentratedloads are given in Table 419

3314 Bottom loaded beams

Where load is applied near the bottom of a section sufficientvertical reinforcement to transmit the load to the top of thesection should be provided in addition to any reinforcementrequired to resist shear

332 DESIGN FOR TORSION

In normal beam-and-slab or framed construction calculationsfor torsion are not usually necessary adequate control of anytorsional cracking in beams being provided by the requiredminimum shear reinforcement When it is judged necessaryto include torsional stiffness in the analysis of a structure ortorsional resistance is vital for static equilibrium membersshould be designed for the resulting torsional moment

Design for torsion 365

The torsional resistance may be calculated on the basis ofa thin-walled closed section in which equilibrium is satisfiedby a plastic shear flow A solid section may be modelled asan equivalent thin-walled section Complex shapes may bedivided into a series of sub-sections each of which is mod-elled as an equivalent thin-walled section and the totaltorsional resistance taken as the sum of the resistances of theindividual elements When torsion reinforcement isrequired this should consist of rectangular closed linkstogether with longitudinal reinforcement Such reinforcementis additional to the requirements for shear and bendingDetails of a suitable design procedure for torsion are givenin Table 420

Example 1 The beam shown in the following figure whichwas designed for bending in example 1 of Chapter 32 is to bedesigned for shear The maximum design load is 495 kNm andthe design is based on the following values

fck 32 MPa fywk 500 MPa d 440 mm

Since the load is uniformly distributed the critical sectionfor shear may be taken at distance d from the face of the sup-port Based on a support width of 400 mm distance fromcentre of support to critical section 200 440 640 mmAt end B

V 248 064 495 216 kNw V[bw z (1 fck250)fck]

216 103[300 09 440 (1 32250) 32] 0065

From Table 418 since w 0138 cot13 25 may be usedHence area of links required is given by

Asws Vfywd z cot13 216 103(087 500 09 440 25) 050 mm2mm

From Table 420 H8-200 provides 050 mm2mm

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EC 2 Shear under concentrated loads 419

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EC 2 Design for torsion 420

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Shear and torsion368

Minimum requirements for vertical links are given by

Asws (008radicfck) bw fyk (008radic32) 300500 027 mm2mm

s 075d 075 440 330 mm

From Table 420 H8-300 provides 033 mm2mm

VRds (Asws) fywd z cot13 033 087 500 09 440 25 103

142 kN

At end A V 160 064 495 128 kN ( VRds 142 kN)

Example 2 A 250 mm thick flat slab is supported by 400 mmsquare columns arranged on a 72 m square grid The slab con-tains as tension reinforcement in the top of the slab at an inte-rior support within a 18 m wide strip central with the columnH16-150 in each direction Lateral stability of the structuredoes not depend on frame action and the design shear forceresulting from the maximum design load applied to all panelsadjacent to the column is V 854 kN

fck 40 MPa fywk 500 MPa d 210 mm (average)

Since the lateral stability of the structure does not depend onframe action may be taken as 115 (Table 419)

Maximum shear stress adjacent to the column face

Vuod 115 854 103(4 400 210) 293 MPavRdmax 02 (1 fck250)fck

02 (1 40250) 40 672 MPa (293)

Based on H16-150 as effective tension reinforcement

100Aslbwd 100 201(150 210) 064vRdc 070 MPa (Table 417 for d 210 and fck 40)

The length of the first control perimeter at 2d from the faceof the column is 4 400 4 d 4239 mm Thus themaximum shear stress at the first control perimeter

Vu1d 115 854 103(4239 210) 110 MPa

Since v vRdc shear reinforcement is needed where effectivedesign strength fywdef 250 025d 300 MPa The areaneeded in one perimeter of vertical shear reinforcementat maximum radial spacing sr 075d 150 mm say isgiven by

Asw (v 075vRdc) u1 sr 15 fywdef

(110 075 070) 4239 150(15 300) 813 mm2

wmin u1 sr 15 (008radicfck) u1 sr 15fyk

(008radic40) 4239 150(15 500) 429 mm2

Using 12H10 gives 942 mm2

Length of control perimeter at which v vRdc is given by

u Vd vRdc 115 854 103(210 070) 6681 mm

Distance of this control perimeter from face of column is

a (6681 4 400)2 809 mm

The distance of the final perimeter of reinforcement from thecontrol perimeter where v vRdc should be 15d 315mm

Thus 4 perimeters of reinforcement with sr 150 mm and thefirst perimeter at 100 mm from the face of the column wouldbe suitable The reinforcement layout is shown in the followingfigure where indicates the link positions and the links canbe anchored round the tension bars

Example 3 The following figure shows a channel sectionedge beam on the bottom flange of which bear 8 m long simplysupported contiguous floor units The beam is continuous in14 m spans and is prevented from lateral rotation at thesupports The centroid and the shear centre of the sectionare shown

Characteristic loads

floor units dead 35 kNm2 imposed 25 kNm2

edge beam dead 12 kNm

Design ultimate loads

floor units (135 35 15 25) 82 338edge beam 135 12 162

500 kNm

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Design for torsion 369

fck 32 MPa fyk 500 MPa d 1440 mm

Bending moment shear force and torsional moment (aboutshear centre of section) at interior support (other than first)

M 009 50 142 882 kNm (Table 229)V 05 50 14 350 kNT 05 (338 0400 162 0192) 14 117 kNm

(Note In calculating V and T a coefficient of 05 rather than055 has been used since the dead load is dominant and thecritical section may be taken at the face of the support)

Considering beam as one large rectangle of size 250 1500and two small rectangles of size 200 300

13hmin3hmax 2503 1500 2 2003 300

(234 2 24) 109 282 109

Torsional moment to be considered on large rectangle

T1 117 234282 97 kNm

Torsional moment to be considered on each small rectangle

T2 117 24282 10 kNm

Reinforcement required in large rectangle

Shear and torsion (see Table 420) Assuming 30 mm cover toH10 links distance from surface of concrete to centre of H12longitudinal bars 46 mm

tefi Au 250 1500[2 (250 1500)] 107 mm ( 2 46 92 mm)

Ak (250 107) (1500 107) 1992 103 mm2

For values of (1 fck250)fck (1 32250) 32 279 MPaand z 1440 100 1340 mm (to centre of flange)

w [T12Ak tefi Vbw z](1 fck250)fck

[97(2 1992 107) 350(250 1340)] 103279 0119

Since w 0138 cot13 25 may be used (Table 418)For a system of closed links total area required in two legs fortorsion and shear is given by

Ass (T1Ak V z)fywd cot13 (971992 3501340) 103(087 500 25) 069 mm2mm

The inner legs of the links are also subjected to a vertical tensileforce resulting from the load of 338 kNm applied by the floorunits Additional area required in inner legs

Ass 338(087 500) 008 mm2mm

Total area required in two legs for torsion shear and theadditional vertical tensile force

Ass 069 2 008 085 mm2mm

The area of longitudinal reinforcement required for torsion isgiven by

Aslsl Tcot13 2Akfyd

97 103 25 (2 1992 087 500) 140 mm2mm

Different combinations of links and longitudinal bars can beobtained by changing the value of cot 13 as follows

Bars Links Longitudinal

Ass Size and Aslsl Size andcot 13 mm2mm spacing mm2mm spacing

25 085 H10-175 140 H12-15020 102 H10-150 112 H12-20016 124 H10-125 090 H12-250

s least of u8 35008 4375 mm 075d 1080 mmor hmin 250 mm sl 350 mm

Bending (see Table 48)

Mbd2fck 882 106(550 14402 32) 0024Asfykbdfck 0027 and xd 0054 (ie x 78 200 mm)

As 0027 550 1440 32500 1369 mm2

Total area of longitudinal bars required at top of beam forbending and torsion (equivalent to 2H12 say)

1369 226 1595 mm2

From Table 228 2H32 provides 1608 mm2

Reinforcement required in small rectangles

Torsion Assuming 30 mm cover to H8 links distance fromsurface of concrete to centre of H12 longitudinal bars 44 mm

tefi Au 200 300[2 (200 300)] 60 ( 2 44 88 mm)

Ak (200 88) (300 88) 237 103 mm2

w (T22Ak tefi)(1 fck250)fck

10 103(2 237 88 279) 0086

Since w 0138 cot 13 25 may be used For a system ofclosed links area required in two legs is given by

Asts T2Ak fywd cot13 10 103(237 087 500 25) 039 mm2mm

The lower rectangle is also subjected to bending resulting fromthe load of 338 kNm applied by the floor units The distanceof the load from the centre of the inner leg of the links in thelarge rectangle is 150 35 185 mm

M 338 0185 625 kNm

Taking the lever arm for the small rectangle as the distancebetween the centres of the top and bottom arms of the linksz 132 mm Additional area required in top arms of links

Ass Mfyd z 625 103(087 500 132) 011 mm2mm

Total area required in two arms for torsion and bending

Ass 039 2 011 061 mm2mm

The area of longitudinal reinforcement required for torsion isgiven by

Aslsl Tcot132Ak fyd

10 103 25(2 237 087 500) 121 mm2mm

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Shear and torsion370

Different combinations of links and longitudinal bars can beobtained by changing the value of cot13 as follows

Bars Links Longitudinal

Ass Size and Aslsl Size andcot 13 mm2mm spacing mm2mm spacing

25 061 H10-150 121 H12-17518 076 H10-125 088 H12-250

s lesser of u8 10008 125 mm or hmin 200 mm

The lower rectangle is also subjected to shear in the verticallongitudinal plane for which

Vbw d 338 103(1000 166) 021 MPa

From Table 417 vmin 056 MPa (fck 32 d 200)

From the foregoing calculations the reinforcement shown inthe figure opposite provides a practical arrangement in whichthe links comprise H10-125 for the large rectangle andH8-125 for the small rectangles The longitudinal bars are all

H12-250 apart from the 2H32 bars at the top of the largerectangle

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341 DEFLECTION

Deflections of members under service load should not impairthe function or appearance of a structure For buildings thedesign requirements and associated combinations of designactions to be considered are given in section 293 The finaldeflection of members below the level of the supports undercharacteristic loading after allowance for any pre-camber isgenerally limited to span250 A further limit of span500applies to the increase in deflection that occurs after theconstruction stage in order to minimise any damage to bothstructural and non-structural elements The requirements maybe met by complying with the limits on spaneffective depthratio given in Table 421

In special circumstances when the calculation of deflectionis considered necessary an adequate prediction can be madeusing the methods given in Table 422 Careful considerationis needed in the case of cantilevers where the usual formulaeassume that the cantilever is rigidly fixed and remains hori-zontal at the root Where the cantilever forms the end of acontinuous beam the deflection at the end of the cantilever islikely to be either increased or decreased by an amount l13where l is the length of the cantilever measured to the centreof the support and 13 is the rotation at the support Where acantilever is connected to a substantially rigid structure someroot rotation will still occur and the effective length shouldbe taken as the length to the face of the support plus half theeffective depth

342 CRACKING

3421 Building structures

Cracks in members under service load should not impair theappearance or durability of the structure For buildings thedesign requirements are given in Table 41 The calculatedcrack width under quasi-permanent loading or as a result ofrestrained deformations is generally limited to 03 mm For drysurfaces inside buildings where crack width has no effect ondurability a limit of 04 mm is recommended where there isa need to ensure an acceptable appearance However in theUK National Annex a limit of 03 mm is required in this situa-tion In the regions of concrete members where tension isexpected a calculated minimum amount of reinforcement isneeded in order to control cracking as given in Table 423

Where minimum reinforcement is provided the crack widthrequirements may be met by direct calculation or by limitingeither the bar size or the bar spacing as given in Table 424For the calculation of crack widths due to restrained imposeddeformation no guidance is given in Part 1 of the code but thefollowing equation is given in PD 6687 (see preface)

sm ndash cm 08Rimp

where R is a restraint factor (see section 2621) and imp is thefree strain due to temperature fall or drying shrinkage

3422 Liquid-retaining structures

For structures containing liquids design requirements related toleakage considerations are given in section 294 Where a smallamount of leakage with related surface staining or damppatches is acceptable for cracks that can be expected to passthrough the full thickness of the section the calculated crackwidth is limited to a value that varies according to the hydraulicgradient (ie head of liquid divided by thickness of section)The limits are 02 mm for hydraulic gradients 5 reducinguniformly to 005 mm for hydraulic gradients 35 Thus thelimits for a 300 mm thick wall to a 75 m deep tank would be02 mm at 15 m below the top 015 mm at 45 m below the topand 01 mm at 75 m below the top The limits apply under thequasi-permanent loading combination where the full charac-teristic value is taken for hydrostatic loading For members inaxial tension where at least the minimum reinforcement isprovided the crack width requirements may be met by directcalculation or by limiting either the bar size or the bar spacingas given in Table 425

In cases of bending with or without axial force where thefull thickness of the section is not cracked and not less than 02times the section thickness 50 mm is in compression thecrack width limit is 03 mm and Table 424 applies

For cracking due to restraint of imposed deformations suchas shrinkage and early thermal movements two distinct typesof restraint are considered For a concrete element restrained atits ends (eg an infill bay with construction joints between thenew section of concrete and the pre-existing sections) the crackformation is similar to that caused by external loading Anappropriate expression for the tensile strain contributing to thecrack width is given in Table 425 and for specified values of

Chapter 34

Deflection and cracking

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EC 2 Deflection ndash 1 421

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EC 2 Deflection ndash 2 422

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EC 2 Cracking ndash 1 423

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EC 2 Cracking ndash 2 424

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EC 2 Cracking ndash 3 425

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Cracking 377

cover section thickness and reinforcement content maximumvalues of fcteff are given in Table 426

For a concrete panel restrained along an edge (eg a wall castonto a pre-existing stiff base) the formation of the crack onlyinfluences the distribution of stresses locally and the crackwidth becomes a function of the restrained strain rather than thetensile strain capacity of the concrete In this case the tensilestrain contributing to the crack width is taken as Rax free wheretypical values of free can be estimated from the informationgiven in Table 425 The restraint factor Rax may be taken as 05generally or reference can be made to Table 345 where valuesare indicated for panels restrained along one two or three edgesrespectively For specified values of cover section thicknessand reinforcement content maximum values of Rax free aregiven in Table 427

It will be found that the calculated strain contributing to thecrack width for a panel restrained at its ends is normally morethan Rax free Thus the reinforcement required to limit a crackwidth to the required value is greater for a panel restrained atits ends than for a panel restrained along one or more edgesAlso for a specific crack width the reinforcement needed for apanel restrained along an edge is less than that in BS 8007since the design crack spacing is less than that in BS 8007

Example 1 The beam shown in the following figure is to bechecked for deflection and cracking The design for bendingand shear is shown in example 1 of Chapters 32 and 33 respec-tively The reinforcement in the bottom of each span 3H25(1473 mm2) is based on the following values

fck 32 MPa fyk 500 MPa beff 2600 mm d 440 mm

From Table 423 for fck 32 MPa and bending of a sectionwith h 500 mm by interpolation 100AsminAct 021

Asmin 00021 162 103 340 mm2 ( 1473 mm2)

026 ( f ctmfyk ) btd 00013 btd

026 (30500) 300 440 206 mm2

The design ultimate load is 396 kN and the quasi-permanentload where the value of 2 is obtained from section 293 is

Gk 2 Qk 160 03 120 196 kN

Hence the stress in the reinforcement under quasi-permanentloading is given approximately by

s (196396)(087fyk)(As req As prov)

Thus for the bars in the bottom of the beam

s (196396)(087 500)(13731473) 200 MPa

From Table 424 for wk 03 mm the crack width criterion ismet if

s 25 mm or the bar spacing 250 mm

The adjusted maximum bar size is given by

Depth of tension zone hcr h ndash x 500 ndash 128 372 mm

s s (fcteff 29)[kc hcr 2(h ndash d)]

25 (3029) [04 372(2 60)] 32 mm

Note It can be seen from the foregoing that all of the criteriaare comfortably satisfied and the checks for deflection andcracking are hardly necessary in this example

Example 2 A 250 mm thick flat slab is supported bycolumns which are arranged on a 72 m square grid Thecharacteristic loads are 72 kNm2 dead and 45 kNm2 imposedand the slab is to be checked for deflection and cracking

fck 32 MPa fyk 500 MPa cover to bars 25 mm

Allowing for the use of H12 bars in each direction and basedon the bars in the second layer of reinforcement

d 250 ndash (25 12 6) 205 mm ld 7200205 35

From Table 421 for a flat slab with spans 85 m

Basic spaneffective depth ratio 24

Total design ultimate load for a square panel is given by

F (135 72 15 45) 722 854 kN

From Table 262 the design ultimate bending moment for anend span with a continuous connection at the outer support is

M 0075Fl 0075 854 72 461 kNm

Mbd 2fck 461 106(7200 2052 32) 0048

As fykbdfck 0056 (Table 48)

100Asbd 100 0056 32500 036

From Table 421 for 100Asbd 01fck05 01 3205 057

s 055 00075fck(100Asbd)

0005fck05[ fck

05(100Asbd) ndash 10]15

055 00075 32036

0005 3205 (3205036 ndash 10)15 160

The actual spaneffective depth ratio 8000440 182

From Table 421 for the end span of a continuous beam and aflanged section with bbw 2600300 867 3

Basic spaneffective depth ratio 08 26 208

For members supporting partitions liable to be damaged byexcessive deflections the basic ratio should be multiplied by7span In this case the basic ratio 208 78 182

Since 100As reqbd 100 1373(2600 440) 012 is smallthe modification factor s is large ( 3) and the limiting ratiois more than three times the actual value

The neutral axis depth for the uncracked section ignoring theeffect of the reinforcement is given by

x

128 mm ( hf 150 mm)

Area of tension zone is given by

Act bw (h ndash hf) bf (hf ndash x) 300 350 2600 22 162 103 mm2

300 5002 2300 1502

2[300 500 2300 150]bwh2 (bf ndash bw)h2

f

2[bwh (bf ndash bw)hf]

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EC 2 Early thermal cracking in end restrained panels 426

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EC 2 Early thermal cracking in edge restrained panels 427

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Deflection and cracking380

Limiting spaneffective depth ratio 16 24 384 ( 35)

From Table 423 for fck 32 MPa and bending of a sectionwith h 300 mm 100AsminAct 024 With Act 05bh

100Asminbd 024 05 250205 015

026 (fctmfyk) 00013

026 30500 016 ( 036)

The design ultimate load is 854 kN and the quasi-permanentload where the value of 2 is obtained from section 293 is

Gk 2 Qk (72 03 45) 722 443 kN

Hence the stress in the reinforcement under quasi-permanentloading is given approximately by

s (443854)(087fyk) (443854)(087 500) 226 MPa

From Table 424 for wk 03 mm the crack width criterion ismet if

s 16 mm or the bar spacing 200 mmThe adjusted maximum bar size with hcr 05h is given by

s s (fcteff 29)[kc hcr 2(h ndash d)]

16 (3029) [04 125(2 45)] 9 mm

Area of reinforcement required to give 100Asbd 036 is

As 00036 1000 205 738 mm2m (H12-150)

Example 3 The wall of a cylindrical tank 75 m deep and15 m diameter is 300 mm thick The wall which is continuouswith the base slab is to be designed for temperature effects andthose due to internal hydrostatic pressure when the tank is fullof liquid

Design class 1 (see section 294) fyk 500 MPaCover to horizontal bars 40 mm fck 32 MPa

Effects of temperature change With fctm 03 fck(23) 30 MPa

and assuming that early thermal cracks will occur at a timewhen fcm(t) 24 MPa

fcteff [fcm(t)(fck 8)] fctm [24(32 8)] 30 18 MPa

The limiting crack width varies according to the hydraulicgradient (depth of liquid thickness of section) If the wall isdesigned to the recommendations for a panel restrained atits ends then suitable reinforcement details for 40 mmcover and fcteff 18 MPa selected from Table 426 aregiven here

Depth Hydraulic Design crack Reinforcement(m) gradient width (mm) required (EF)

15 5 02 H20-15045 15 015 H20-12575 25 01 H20-100

Note The table for wk 02 mm was used throughout bytaking effective values of fcteff 18(5wk) MPa

If the wall is designed to the recommendations for a panelrestrained along the edge the restrained tensile strain needs tobe estimated as follows

Allowing for concrete grade C3240 with 350 kgm3 Portlandcement at a placing temperature of 20oC and a mean ambient

temperature during construction of 15oC the temperature risefor concrete placed within 18 mm plywood formwork

T1 25oC (Table 219)

As the wall is to be designed to resist hoop tension there willbe no vertical movement joints and allowance must be made fora fall in temperature due to seasonal variations Allowing forT2 15oC restraint factor Rax taken as 05 and coefficient ofthermal expansion taken as 12 10ndash6 per oC (Table 35)restrained total thermal contraction after the peak temperaturearising from hydration effects is given by

Rax (T1 T2 ) 05 12 10ndash6 (25 15) 240 10ndash6

Hence suitable reinforcement details for 40 mm cover andRax 240 10ndash6 selected from Table 427 are given here

Depth Hydraulic Design crack Reinforcement(m) gradient width (mm) required (EF)

15 5 02 H20-25045 15 015 H20-15075 25 01 H20-100

Note The table for wk 02 mm was used throughout foreffective values of Rax [240(5wk)] 10ndash6 For H20 barsvalues for a 250 mm thick section apply for h 250 mm

From Table 423 the minimum reinforcement percentage forfcteff fctm in the case of a rectangular section in pure tensionwith h 300 mm and fck 32 MPa is 060 For the control ofearly thermal cracking the value is (1830) 06 036 andthe minimum area of reinforcement required on each face

Asmin 00036 150 1000 540 mm2m (H16-300)

Clearly the reinforcement needed for thermal crack controlgreatly exceeds this minimum requirement In the lower part ofthe wall the reinforcement provided is H20-100 (EF) and thecorresponding stress at a cracked section is

s fcteff (Act As) 18 150 10003142 86 MPa

This solution can be checked approximately by reference to thechart for maximum bar size on Table 425 as follows

For s 86 MPa and wk 01 mm s 55 mm say

s s

55 (1829) (01 30050) 20 mm

Effects of hydrostatic load Suppose that an elastic analysis ofthe tank assuming a floor 300 mm thick indicates a servicemaximum circumferential tension of 400 kNm This valueoccurs at a depth of 6 m where the design crack width is0125 mm Above this level the tensions can be assumed toreduce approximately linearly to near zero at the top of the wall

For a section reinforced with H20-100 (EF) the stress in thereinforcement s 400 1036284 64 MPa Since this isless than the stress due to fcteff the reinforcement needed forthermal crack control is also sufficient for the circumferentialtension This solution can be checked also by reference to thecharts on Table 425 which show that with s 64 MPa thebar size and the bar spacing are of no consequence

fcteff

2901h(hd)

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The code contains many requirements that affect the details ofthe reinforcement such as minimum and maximum areas tyingprovisions anchorage and curtailment

Bars may be set out individually or grouped in bundles oftwo or three in contact Bundles of four bars may also be usedfor vertical bars in compression and for bars in a lapped jointFor the safe transmission of bond forces the cover providedto the bars should be not less than the bar diameter or for abundle the equivalent diameter ( 55 mm) of a notional barwith the same sectional area as the bundle Requirements forcover with regard to durability are given in Chapter 31 Gapsbetween bars (or bundles) generally should be not less than thegreatest of (dg 5 mm) where dg is the maximum aggregatesize the bar diameter (or equivalent bar diameter for a bundle)or 20 mm Details of reinforcement limits are given in Table 428

At intermediate supports of continuous flanged beams thetotal area of tension reinforcement should be spread over theeffective width of the flange but a part of the reinforcementmay be concentrated over the web width

351 TIES IN STRUCTURES

Structures not specifically designed to withstand accidentalactions should be provided with a suitable tying system toprevent progressive collapse by providing alternative loadpaths after local damage Where the structure is divided intostructurally independent sections each section should have anappropriate tying system The reinforcement providing the tiesmay be assumed to act at its characteristic strength and onlythe specified tying forces need to be taken into accountReinforcement required for other purposes may be consideredto form part of or the whole of the ties Details of the tyingrequirements as specified in the UK National Annex are givenin Table 429

352 ANCHORAGE AND LAP LENGTHS

At both sides of any cross section bars should be providedwith an appropriate embedment length or other form of endanchorage For bent bars the basic tension anchorage length ismeasured along the centreline of the bar from the section inquestion to the end of the bar where

lbd 1 2 3 4 5 lbrqd lbmin

As a simplified alternative a tension anchorage for a standard bend hook or loop may be provided as an equivalentlength lbeq 1 lbrqd (see figure here) where 1 is taken as07 for covers perpendicular to the bend 3 Otherwise1 10

Chapter 35

Considerationsaffecting design details

Bends or hooks do not contribute to compression anchoragesDetails of anchorage lengths are given in Table 430

Laps should be located if possible away from positions ofmaximum moment and should generally be staggered Detailsof lap lengths are given in Table 431

The radius of any bend in a reinforcing bar should conformto the minimum requirements of BS 8666 and should ensurethat failure of the concrete inside the bend is prevented A linkmay be considered fully anchored if it passes round another barof not less than its own diameter through an angle of 90o andcontinues beyond the end of the bend for a minimum length of10 diameters 70 mm Details of bends in bars are given inTable 431 Additional rules for large diameter bars ( 40 mmaccording to the UK National Annex) and bundles are given inTable 432

353 CURTAILMENT OF REINFORCEMENT

In flexural members it is generally advisable to stagger thecurtailment points of the tension reinforcement as allowed bythe bending moment envelope Bars should be curtailed inaccordance with the rules set out in Table 432 and illustratedin the figure on page 387 Except at end supports every tensionbar should extend beyond the point at which in theory it is nolonger needed for flexural resistance for a distance not less thanal The bar should also extend beyond the point at which it isfully required to provide flexural resistance for a distance notless than al lbd At a simple end support the bars shouldextend for the anchorage length lbd necessary to develop theforce ∆Ftd

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EC 2 Reinforcement limits 428

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EC 2 Provision of ties 429

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EC 2 Anchorage requirements 430

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EC 2 Laps and bends in bars 431

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EC 2 Rules for curtailment large diameter bars and bundles 432

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Curtailment of reinforcement 387

Example 1 The beam shown in the following figure which wasdesigned in example 1 of Chapters 32 (bending) and 33 (shear)is to be checked for the reinforcement details The designultimate loads are Fmax 396 kN and Fmin 216 kN on eachspan The width of each support is 400 mm and the mainreinforcement is as follows spans 3H25 (bottom) support B3H32 (top) and 2H25 (bottom)

fck 32 MPa fyk 500 MPa cover to links 35 mm

For bars in the bottom of the section the bond condition isclassified as lsquogoodrsquo Thus from Table 430

lbrqd 35 (s435) 35 25 (172435) 346 mm

The design anchorage length is given by

lbd 1 2 3 4 5 lbrqd lbmin

Coefficients 1 and 2 depend on cd which is taken as either thecover to the main bar or half the gap between the main barswhichever is the lesser With 35 mm cover to H8 links coverto main bars is 45 mm and the gap between the bars is300 2 (45 25) 160 mm Hence cd 45 mm (or 18)

Since cd 3 1 10 for both bent bars and straight barsHence using the simplified approach (see section 352) forboth straight bars or standard bends lbeq 1 lbrqd 346 mm

For a 400 mm wide support allowing for 50 mm end cover tothe bars the anchorage length provided from the near face ofthe support is 350 mm ( lbeq)

For the basic approach the following values can be obtained

For straight bars 2 1 015(cd 1) 088

With no transverse reinforcement provided within the bearinglength 3 10 and 4 10

Transverse pressure due to reaction on support is given by

p V(bearing area) 150 103(400 300) 125 MPa

Hence with p 125 MPa 5 1 004p 095 and

lbd 25lbrqd 088 095 346 290 mm lbmin 10 250 mm ( 03lbrqd or 100 mm)

Curtailment points for bottom bars The resistance momentprovided by 2H25 at the bottom of the beam may be determinedas follows

As fykbdfck 982 500(2600 440 32) 00134Mbd2fcu 0012 (Table 48 or section 3221)M 0012 2600 4402 32 106 193 kNm

Illustration of the curtailment of longitudinal reinforcement taking account of resistance within anchorage lengths

End anchorage At the bottom of each span 2H25 ( 25 ofarea provided in the span) will be continued to the support Atthe end support the tensile force to be anchored isF 05Vcot13 in which 13 is the inclination of the concrete strutrequired for shear design In the shear design calculations inchapter 33 V 128 kN at the critical section and VRds 142 kNwhen cot13 25 Thus cot13 (VVRds) 25 225 could beused At the face of the support V 160 02 495 150 kNWith cot 13 225

F 05 150 225 169 kNs FAs 169 103982 172 MPa

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Considerations affecting design details388

Reaction at A for load case 2 is RA 160 kN

Distance x from A to point where M 193 kNm is given by

RAx 05(FmaxL)x2 160x2 05 (3968)x2 193

Hence 05x2 323x 39 0 giving x 16 m and 485 m

Thus of the 3H25 required in the span one bar is no longerneeded for flexure at 16 m and 485 m from the end support Atthese points V 80 kN and cot13 (80142) 25 14 issufficient Thus the bar needs to extend beyond these points fora distance al 045d cot13 045 440 14 280 mm

Curtailment points for top bars The resistance moment providedby 2H32 can be determined as follows

As fykbdfck 1608 500(300 440 32) 0190Mbd2fcu 0142 (Table 48)M 0142 300 4402 32 106 264 kNm

For load case 1 reaction at A (or C) is given by

RA 05Fmax MBL 05 396 3968 148 kN

Distance x from A to point where M 264 kNm is given by

05(FmaxL)x2 RAx 05 (3968) x2 148 x 264

Hence 05x2 3x 53 0 giving x 74 m Thus of the3H32 required at B one bar is no longer required for flexure atdistance (80 74) 06 m from B If this distance is lessthan lb rqd the point of curtailment will be determined by theneed to develop the full force in the bar at B Here cot13 25giving al 045dcot13 1125d As the bars are effectively ina slab of thickness 250 mm it seems reasonable to assumelsquogoodrsquo bond conditions giving lb rqd 35 (Table 430) Thusdistance from B (edge of support say) at which one bar may becurtailed is al lb rqd 1125 440 35 32 1615 mm

Suppose that the remaining bars are continued to the point ofcontra-flexure in span BC for load case 2

The reaction at support C is given by

RC 05Fmin MBL 05 216 3068 70 kN

Distance from B to point of contra-flexure is given by

x L(1 2RCFmin) 8 (1 2 70216) 28 m

Here V 70 kN and cot13 (70142) 25 125 is sufficientThus distance from B at which the remaining bars may becurtailed is 2800 045 440 125 3050 mm

Link support bars say 2H12 could be used for the remainderof the span

Example 2 A typical floor to an 8-storey building consists ofa 250 mm thick flat slab supported by columns arranged ona 72 m square grid The slab for which the characteristicloading is 72 kNm2 dead and 45 kNm2 imposed is to beprovided with ties to the requirements of the UK NationalAnnex The design panel load is 854 kN and bending momentsare to be determined by the simplified method (see section 138)

fck 32 MPa fyk 500 MPa cover to bars 25 mm

Allowing for the use of H12 bars in each direction and basedon the bars in the second layer of reinforcement

d 250 (25 12 6) 205 mm say

From Table 255 the design ultimate sagging moment for aninterior panel is given by

M 0063Fl 0063 854 72 388 kNm

The total panel moment is to be apportioned between columnand middle strips where the width of each strip is 36 m Forthe column strip with 60 of the panel moment

M 06 388 233 kNmMbd2fck 233 106(3600 2052 32) 0048As fykbdfck 0056 (Table 48)As 0056 3600 205 32500

2645 mm2 (24H12-150 gives 2714 mm2)

For the middle strip with 40 of the panel moment

M 04 388 155 kNmMbd 2fck 155 106(3600 2052 32) 0032As fykbdfck 0037 (Table 48)As 0037 3600 205 32500

1748 mm2 (16H12-225 gives 1810 mm2)

For the peripheral tie the tensile force is given by

Ftieper (20 4no) 60 kN (20 4 8) 52 kN

The required area of reinforcement acting at its characteristicstrength is given by

As Ftieper fyk 52 103500 104 mm2 (1H12)

For the internal ties the tensile force is given by

kNm

With Ft (20 4no) 60 kN (20 4 8) 52 kN

kNm

If the internal ties are spread evenly in the slab the requiredarea of reinforcement acting at its characteristic strength

As 117 103500 234 mm2m (H12-450)

In this case at least every third bar in the column strips andevery other bar in the middle strips need to be continuous

If the internal ties are concentrated at the column lines the totalarea of reinforcement required in each group

As 234 72 1685 mm2 (16H12 gives 1810 mm2)

In this case the bars in the middle two-thirds of each columnstrip need to be continuous For sections 250 mm deep thebond condition is lsquogoodrsquo and lbrqd 35 (Table 430) Laps inadjacent pairs of lapped bars should be staggered by 13l0where l0 is the design lap length (Table 431) With lbd lbrqd

and 6 14 for one in two bars lapped at the same section

l0 6 lbd 14 35 12 600 mm say

Example 3 The following figure shows details of thereinforcement at the junction between a 300 mm wide beamand a 300 mm square column Bars 03 need to develop themaximum design stress at the column face and the radius of

Ftieint 72 4575 72

5 52 117

Ftieint gk qk

75 lr

5Ft Ft

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Curtailment of reinforcement 389

bend necessary to avoid failure of the concrete inside the bendis to be determined

fck 32 MPa fyk 500 MPa

The minimum radius of bend of the bars depends on the valueof ab where ab is taken as half the centre-to-centre distancebetween adjacent bars or for bars adjacent to the face of amember the side cover plus half the bar size Thus in this caseab 75 252 875 mm

From Table 431 for ab 87525 35 rmin 74 Thisvalue can be reduced slightly by taking into account the stressreduction in the bar between the edge of the support and thestart of the bend If r 7 distance from face of column tostart of bend 300 50 8 25 50 mm (ie 2)

From Table 430 for lsquogoodrsquo bond conditions the requiredanchorage length is 35 and rmin (1 235) 74 7Thus r 7 is sufficient

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Chapter 36

Foundations andearth-retaining walls

361 GEOTECHNICAL DESIGN

Eurocode 7 provides in outline all the requirements for thedesign of geotechnical structures It classifies structures intothree categories according to their complexity and associatedrisk but concentrates on the design of conventional structureswith no exceptional risk These include spread raft and pilefoundations retaining structures bridge piers and abutmentsembankments and tunnels Limit-states of stability strengthand serviceability need to be considered The requirements ofthe ULS and SLS may be satisfied by the following methodsalone or in combination calculations prescriptive measurestesting observational procedures The calculation methodadopted in the United Kingdom for the ULS requires theconsideration of two combinations of partial factors for actionsand soil parameters as shown here

satisfactory if the ratio of design ultimate bearing capacityto service load is 3 This approach is not valid for soft claysand settlement calculations should always be carried out insuch cases

362 PAD BASES

Critical sections for bending are taken at the face of a columnor the centre of a steel stanchion The design moment is takenas that due to all external loads and reactions to one side of thesection Punching resistance should be verified at controlperimeters within 2d from the column periphery For baseswithout shear reinforcement the design shear resistance is

vRd vRdc (2dav) vmin (2dav)

where av ( 2d) is the distance from the column periphery tothe control perimeter being considered The net applied forceVred V V where V is the applied column load and V isthe resulting upward force within the control perimeter Forconcentric loading the punching shear stress is v Vredudwhere u is the length of the control perimeter For eccentricloading the maximum shear stress is Vredud where is amagnification factor determined from equations given in EC 2At the column perimeter the punching shear stress should notexceed vRdmax 02(1 fck250)fck

Normal shear resistance should also be verified on verticalsections at distance d from the column face extending across thefull width of the base where the design shear resistance isvRdc vmin Alternatively it would be reasonable to check atsections within 2d from the column face using the enhanceddesign shear resistance given for punching shear In this casefor concentric loading the critical position for normal shear andpunching shear occurs at av a2 2d where a is the distancefrom the column face to the edge of the base For eccentricloading checks can be made at av 05d d and so on to findthe critical position

If the tension reinforcement is included in the determinationof vRdc the bars should extend for at least (d lbd) beyondthe section considered (see also EC 2 section 9822) If thetension reinforcement is ignored in the shear calculationsstraight bars will usually suffice

Example 1 A base is required to support a 600 mm squarecolumn subjected to vertical load only for which the values

Partial safety factors for the ULS

Safety factor Safety factor for

Combination on actions F soil parameters M

G Q c cu

1 135 15 10 10 102 10 13 125 125 14

If the action is favourable to the effect being considered values of G 10and Q 0 should be used Subscripts refer to the following soil parameters

is angle of shearing resistance (in terms of effective stress) and factor

is applied to tanc is cohesion intercept (in terms of effective stress)cu is undrained shear strength

Generally combination 2 determines the size of the structurewith regard to overall stability bearing capacity sliding andsettlement and combination 1 governs the structural design ofthe members The required size of spread foundations may bedetermined by ULS calculations using soil parameter valuesderived from the geotechnical design report for the projectAlternatively the size may be determined by limiting the bearingpressure under the characteristic loads to a prescribed value ora calculated allowable bearing pressure may be used For theSLS the settlement of spread foundations should be checked bycalculation or may in the case of firm to stiff clays be taken as

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are 4250 kN (service) and 6000 kN (ultimate) The allowableground bearing value is 300 kNm2 (kPa)

fck 32 MPa fyk 500 MPa nominal cover 50 mm

Allowing 10 kNm2 for ground floor loading and extra over soildisplaced by concrete the net allowable bearing pressure can betaken as 290 kNm2 Area of base required

Abase 4250290 147 m2 Provide base 40 m square

Distance from face of column to edge of base a 1700 mm

Taking depth of base 05a say h 900 mm

Allowing for 25 mm main bars average effective depth

d 900 (50 25) 825 mm

Bearing pressure under base due to ultimate load on column

pu 600042 375 kNm2

Bending moment on base at face of column

M pula2 2 375 4 1722 2168 kNm

Mbd2 fck 2168 106 (4000 8252 32) 0025

From Table 48 As fykbdfck 00285 and

As 00285 4000 825 32500 6020 mm2

From Table 220

13H25-300 gives 6380 mm2 and

100Asbh 100 6380(4000 900) 018 ( 013 min)

For values of Gk 2500 kN Qk 1750 kN and 2 03 thequasi-permanent load is

Gk 2Qk 2500 03 1750 3025 kN

Hence the stress in the reinforcement under quasi-permanentloading is given approximately by

s (30256000)(087fyk)(As reqAs prov)

(30256000)(087 500)(60206380) 207 MPa

From Table 424 for wk 03 mm the crack width criterion ismet if

s 23 mm or the bar spacing 240 mm

The adjusted maximum bar size with hcr 05h is given by

s s ( fcteff 29)[kc hcr2(h d )]

23 (3029) [04 450(2 75)] 28 mm

In this case H25-300 satisfies the bar size criterion

For members without shear reinforcement distance from faceof column to critical position for punching shear (or normalshear) is given by av 05a 850 mm where perimeter

u 4c 2av 4 600 2 850 7740 mm

Area of base inside critical perimeter is

Au 4avc av2 4 085 06 0852 431 m2

Hence the net applied force and resulting shear stress are asfollows

Vred V V 375 (42 431) 4384 kN

v Vud 4384 103(7740 825) 069 MPa

From Table 417 for fck 32 MPa and 100Aslbd 020 thedesign concrete shear stress vRdc is determined by vmin

vmin 0035k32 fck12 where k 1 (200d)12 20

k 1 (200825)12 149

vmin 0035 14932 3212 036 MPa

Hence design shear resistance

vRd vmin (2dav) 036 (2 825850) 070 MPa (v)

At the column perimeter ignoring the small reduction V

v Vud 6000 103(4 600 825) 303 MPa

vRd max 02 (1 fck250)fck

02 (1 32250) 32 558 MPa (v)

363 PILE-CAPS

A pile-cap may be designed by either bending theory or trussanalogy (ie strut-and-tie) In the latter case the truss is of atriangulated form with nodes at the centre of the loaded areaand at the intersections of the centrelines of the piles with thetension reinforcement as shown for compact groups of two tofive piles in Table 361 Expressions for the tensile forces aregiven taking into account the dimensions of the column andalso simplified expressions when the column dimensions areignored Bars to resist the tensile forces are to be located withinzones extending not more than 15 times the pile diameter eitherside of the centre of the pile The bars are to be provided witha tension anchorage beyond the centres of the piles Thecompression caused by the pile reaction may be assumedto spread at 45o angles from the edge of the pile and taken intoaccount when calculating the anchorage length The bearingstress on the concrete inside the bend in the bars should bechecked (see Table 431)

Example 2 A pile-cap is required for a group of 4 450 mmdiameter piles arranged at 1350 mm centres on a square gridThe pile-cap supports a 450 mm square column subjected to anultimate design load of 4000 kN

fck 32 MPa fyk 500 MPa

Allowing for an overhang of 150 mm beyond the face of thepile size of pile-cap 1350 450 300 2100 mm square

Take depth of pile-cap as (2hp 100) 1000 mm

Assuming tension reinforcement to be 100 mm up from base ofpile-cap d 1000 100 900 mm

Using truss analogy with the apex of the truss at the centre ofthe column the tensile force between adjacent piles is

Ft 750 kN in each zone

As Ft 087fyk 750 103(087 500) 1724 mm2

For a pile spacing three times pile diameter the bars may bespread uniformly across the cap and a total for two ties of8H25-250 (giving 3926 mm2) in each direction can be used

100Asbd 100 3926(2100 900) 020 ( 013 min)

4000 13508 900

Nl8d

Pile-caps 391

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The bars should be provided with a tension anchorage beyondthe centre of the piles (see Table 430) For 25 mm diameterbars spaced at 250 mm centres half the gap between bars givescb 1125 mm (ie 45 ) So for bent bars and lsquogoodrsquo bondconditions lbd 1lbrqd 07 35 25

Taking ab as either half the centre-to-centre distance betweenbars or (side cover plus half bar size) whichever is less

ab 2502 175 125 mm ab 12525 5

From Table 431 minimum radius of bend rmin 7 say

Consider the critical section for shear to be located at 20 ofthe pile diameter inside the pile-cap Distance of this sectionfrom the column face

av 05(l c) 03hp

05 (1350 450) 03 450 315 mm

Length of corresponding square perimeter for punching shear

u 4 (l 06hp) 4 (1350 06 450) 4320 mm

Since length of perimeter of pile-cap 4 2100 8400 mmis less than 2u normal shear extending across the full width ofthe pile-cap is more critical than punching shear

The contribution of the column load to the shear force may bereduced by applying a factor av2d where 05d av 2dSince avd 315900 035 (05) 025

v Vbd 025 2000 103(2100 900) 027 MPa

From Table 417 for fck 32 MPa and 100Aslbd 020 thedesign concrete shear stress vRdc is determined by vmin

vmin 0035k32fck12 where k 1 (200d)12 20

k 1 (200900)12 147

vmin 0035 14732 3212 035 MPa (v)

Shear stress calculated at perimeter of column

v Vud 4000 103(4 450 900) 247 MPa

vRd max 02 (1 fck250) fck

02 (1 32250) 32 558 MPa (v)

364 RETAINING WALLS ON SPREAD BASES

General notes on walls on spread bases are given in section732 For design purposes the characteristic soil parameterwhich is defined as a cautious estimate of the value affecting theoccurrence of the limit-state is divided by a partial safety factor(see section 361) Design values of the soil strength at the ULS(combination 2) are given by

and

where and are characteristic values of cohesion interceptand angle of shearing resistance (in terms of effective stress)

Design values for shear resistance at the interface of the baseand sub-soil of friction (for drained conditions) and adhesion(for undrained conditions) are given by

tan tan for cast in-situ concrete

tan tan (23) for precast concrete

cud cu14 where cu is undrained shear strength

dd

dd

c

cd c 125tan d ( tan ) 125

Walls should be checked for ULS of overall stability bearingresistance and sliding The resistance of the ground should bedetermined for both long-term (ie drained) and short-term(ie undrained) conditions where appropriate

For eccentric loading the ground bearing is assumed to beuniformly distributed and coincident with the line of action ofthe resultant applied load The traditional practice of usingcharacteristic actions and allowable bearing pressures to limitground deformation and check bearing resistance may also beadopted by mutual agreement This approach assumes a linearvariation of bearing pressure for eccentric loading and it is stillnecessary to consider the ULS for the structural design and tocheck sliding

The partial safety factors for the SLS are given as unity butit is often prudent to use the ULS for the active force (as inBS 8002) In this case suitable dimensions for the wall base canbe estimated with the aid of the chart given in Table 286 Herethe value pmax applies for a linear pressure variation and if theground pressure is uniform and centred on the centre of gravityof the applied load the contact length is (l) where dependson whether the solution is (a) above or (b) below the curve forlsquozero pressure at heelrsquo shown on the chart as follows

(a) 4(1 )3 23 and p 075 pmax

(b) 43 3(1 ) 23 and p (1 )l

For sliding the chart applies directly to non-cohesive soilsThus for bases founded on clay the long-term condition canbe investigated by using with c 0 For the short-termcondition the ratio does not enter into the calculations forsliding and is given by KA l2cd When has beendetermined from this equation the curve for radicKA on the chartcan be used to check the values of and that were obtainedfor the long-term condition

Example 3 A cantilever retaining wall on a spread base isrequired to support level ground and a footpath adjacent to aroad The existing ground may be excavated as necessary toconstruct the wall and the excavated ground behind the wallis to be reinstated by backfilling with a granular material Agraded drainage material will be provided behind the wall withan adequate drainage system at the bottom

Height of fill to be retained 40 m above top of base

Surcharge 5 kNm2

Properties of retained soil (well graded sand and gravel)unit weight of soil 20 kNm3

35o tan1 [(tan 35o)125] 29o

KA (1 sin )(1 sin ) 035

Properties of sub-base soil (medium sand)allowable bearing value fmax 200 kNm2 (kPa)

35o 29o (as fill)tan d tan 055

Take thickness of both wall (at bottom of stem) and base to beequal to (height of fill)10 400010 400 mm

Height of wall to underside of base l 40 04 44 m

Allowing for surcharge equivalent height of wall

44 520 465 mle l q

d

d

dd

d

Foundations and earth-retaining walls392

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200(20 465) 215

radicKA 055radic035 093

From Table 286 radicKA 0745 038 Hence

Width of base le (0745radic035) 465 21 m say

Toe projection (le) 038 21 08 m

Since the solution lies above the curve for lsquozero pressure atheelrsquo on the chart for uniform bearing centred on the centre ofgravity of the applied load

4(1 )3 4 (1 038)(3 215) 0385 andp 075 pmax 075 200 150 kNm2 (kPa)

Example 4 The sub-base for the wall described in example 3is a clay soil with properties as given below All other values areas specified in example 3

Properties of sub-base soil (firm clay)

allowable bearing value pmax 100 kNm2 (kPa)cu 50 kNm2 cud cu14 5014 35 kNm2

25o tan1 [(tan 25o)125] 205o

tan d tan 037

For the long-term condition

pmaxle 100(20 465) 108

radicKA 037radic035 0625

From Table 286 radicKA 107 025 Hence

Width of base le (107radic035) 465 30 m say

Toe projection (le) 025 30 08 m say

Since the solution lies below the curve for lsquozero pressure atheelrsquo on the chart for uniform bearing centred on the centre ofgravity of the applied load

43 3(1 ) 43 1083(1 025) 085

p (1 )l (1 025) 20 465085 82 kNm2

For the short-term condition

KA l2cud 035 20 465(2 085 35) 055

radicKA 055radic035 093

Since this value is less than that obtained for the long-termcondition the base dimensions are satisfactory

Example 5 The wall obtained in example 4 a cross sectionthrough which is shown below is to be designed accordingto EC 7

tan d

d

d

tan d

pmax le The vertical loads and bending moments about the front edgeof the base are

Load (kN) Moment (kNm)Surcharge 5 18 90 21 189Backfill 20 18 40 1440 21 3024Wall stem 24 04 40 384 10 384Wall base 24 04 30 288 15 432

Totals Fv 2202 Mv 4029

The horizontal loads and bending moments about the bottom ofthe base are

Load (kN) Moment (kNm)Surcharge 035 5 44 77 442 170Backfill 035 20 4422 678 443 994

Totals Fh 755 Mh 1164

Resultant moment Mnet 4029 1164 2865 kNm

Distance from front edge of base to resultant vertical force

a Mnet Fv 28652202 130 m

Eccentricity of vertical force relative to centreline of base

e 302 130 020 m ( 306 05 m)

Maximum pressure at front of base

pmax (220230)(1 6 02030) 103 kNm2

Minimum pressure at back of base

pmin (220230)(1 6 02030) 44 kNm2

For the ultimate bearing condition a uniform distribution isconsidered of length lb 2a 2 13 26 m giving

pu Fvlb 220226 85 kNm2

The ultimate bearing resistance is given by the equation

qu (2 ) cud ic where ic 05[1 ]

ic 05[1 ] 070

qu (2 ) 35 070 126 kNm2 ( pu 85)

Resistance to sliding (long-term)

Fv tan b 2202 037 815 kN ( Fh 755)

Resistance to sliding (short-term)

cud lb 35 26 91 kN ( Fh 755)

Example 6 The structural design of the wall in example 5 isto be in accordance with the requirements of EC 2

fck 32 MPa fyk 500 MPa nominal cover 40 mm

Allowing for H16 bars with 40 mm cover

d 400 (40 8) 352 mm

For the ULS (combination 1) F 135 for all permanentactions and M 10 Thus

35o and KA (1 sin )(1 sin ) 027

The ultimate bending moment at the bottom of the wall stem

M 135 027 (5 422 20 436) 924 kNmm

ddd

1 755 (35 26)

1 Fh (cudlb)

Retaining walls on spread bases 393

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(Note that for combination 2 F KA 10 035 which is lessthan F KA 135 027 0365 for combination 1)

Mbd 2fck 924 106(1000 3522 32) 0023

From Table 48 As fykbd fck 0026 and

As 0026 1000 352 32500 586 mm2m

From Table 220 H12-150 gives 754 mm2m

The ultimate shear force at the bottom of the wall stem

V 135 027 (5 4 20 422) 656 kNm

From Table 417 vmin 0035k32fck12 where

k 1 (200d)12 1 (200352)12 175 ( 20)

vmin 0035 17532 3212 046 MPa

Vbd 656 103(1000 352) 019 MPa ( vmin)

Since the loads are permanent the stress in the reinforcementunder service loading is given approximately by

s (087fykF)(As reqAs prov)

(087 500135)(586754) 250 MPa

From Table 424 for wk 03 mm the crack width criterion ismet if

s 15 mm or the bar spacing 185 mm

The adjusted maximum bar size with hcr 05h is given by

s s ( fcteff 29)[kc hcr 2(h d)]

15 (3029) [04 200(2 48)] 13 mm

In this case H12-150 meets both requirements

For the wall base the loads and bending moments calculatedfor combination 2 (see example 4) can be modified to suit therevised parameters for combination 1 as follows

Fv 135 2202 2973 kN

Mv 135 4029 5439 kNm

Mh (0365035) 1164 1214 kNm

Resultant moment Mnet 5439 1214 4225 kNm

Distance from front edge of base to resultant vertical force

a Mnet Fv 42252973 142 m

Bearing contact length lb 2a 2 142 284 m

pu Fvlb 2973284 1047 kNm2

Note that values of both pu and qu are greater for combination 1than for combination 2 but 2 is still critical for bearing

Bending moment on base at inside face of wall

M 135 (5 20 4 24 04) 1822

1047 (284 12)22 661 kNm

Shear force on base at inside face of wall

V 135 (5 20 4 24 04) 18

1047 (284 12) 582 kNm

The bending moment and shear force are both less than thevalues at the bottom of the wall stem Thus H12-150 can beused to fit in with the vertical bars in the wall

Foundations and earth-retaining walls394

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Appendix

Mathematicalformulae and data

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Mathematical formulae and data396

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References and furtherreading

1 Institution of Structural EngineersConcrete Society Standardmethod of detailing structural concrete a manual for best practiceLondon The Institution of Structural Engineers 2006 p 188

2 CP2 Civil Engineering Code of Practice No2 Earth retainingstructures London The Institution of Structural Engineers 1951p 224

3 The Highways Agency BD 3701 Loads for highway bridgesDesign manual for roads and bridges London HMSO 2001 p 118

4 The Highways Agency BD 6494 The design of highway bridgesfor vehicle collision loads Design manual for roads and bridgesLondon HMSO 1994 p 13

5 The Highways Agency BD 2101 The assessment of highwaybridges and structures Design manual for roads and bridges LondonHMSO 2001 p 84

6 The Highways Agency BD 5293 The design of highway bridgeparapets Design manual for roads and bridges London HMSO1993 p 44

7 Department of Transport BD 3087 Backfilled retaining walls andbridge abutments London Department of Transport 1987 p 12

8 Caquot A and Kerisel J Tables for calculation of passive pressureactive pressure and bearing capacity of foundations (translated fromFrench by M A Bec London) Paris Gauthier-Villars 1948 p 121

9 The Highways Agency BA 4296 The design of integral bridgesDesign manual for roads and bridges London HMSO 1996 p 10

10 Blackledge G F and Binns R A Concrete practice CrowthorneBritish Cement Association Publication 48037 2002 p 71

11 CIRIA Report 91 Early-age thermal crack control in concreteLondon CIRIA 1981 p 160

12 BRE Digest 357 Shrinkage of natural aggregates in concreteWatford BRE 1991 p 4

13 BRE Special Digest 1 Concrete in aggressive ground Six partsWatford BRE 2005

14 Concrete Society Technical Report No 51 Guidance on the use ofstainless steel reinforcement Slough The Concrete Society 1998 p 55

15 Coates R C Coutie M G and Kong F K Structural analysisSunbury-on-Thames Nelson 1972 p 496

16 Rygol J Structural analysis by direct moment distributionLondon Crosby Lockwood 1968 p 407

17 Westergaard H M Computation of stresses in bridge slabs due towheel loads Public Roads Vol 2 No 1 March 1930 pp 1ndash23

18 Pucher A Influence surfaces for elastic plates Wien andNew York Springer Verlag 1964

19 Bares R Tables for the analysis of plates and beams based onelastic theory Berlin Bauverlag 1969

20 Timoshenko S P and Woinowsky-Krieger S Theory of plates andshells (second edition) New York McGraw-Hill 1959 p 580

21 Sarkar R K Slab design ndash elastic method (plates) Munich VerlagUNI-Druck 1975 p 191

22 Wang P C Numerical and matrix methods in structural mechanicsNew York Wiley 1966 p 426

23 Jones L L and Wood R H Yield-line analysis of slabs LondonThames and Hudson 1967 p 405

This book written by leading UK experts is the best English languagetext dealing with yield-line theory (essential for designers using themethod frequently and for more than lsquostandardrsquo solutions)

24 Johansen K W Yield-line theory London Cement and ConcreteAssociation 1962 p 181

This is an English translation of the original 1943 text on whichyield-line theory is founded

25 Johansen K W Yield-line formulae for slabs London Cementand Concrete Association 1972 p 106

Gives design formulae for virtually every lsquostandardrsquo slab shape andloading (essential for practical design purposes)

26 Wood R H Plastic and elastic design of slabs and platesLondon Thames and Hudson 1961 p 344

Relates collapse and elastic methods of slab analysis but mainlyfrom the viewpoint of research rather than practical design

27 Jones L L Ultimate load analysis of reinforced and prestressedconcrete structures London Chatto and Windus 1962 p 248

About half of this easily readable book deals with the yield-linemethod describing in detail the analysis of several lsquostandardrsquo slabs

28 Pannell F N Yield-line analysis Concrete and ConstructionalEngineering JunendashNov 1966

Basic application of virtual-work methods in slab design June1966 pp 209ndash216

Economical distribution of reinforcement in rectangular slabs July1966 pp 229ndash233

Edge conditions in flat plates Aug 1966 pp 290ndash294

General principle of superposition in the design of rigid-plasticplates Sept 1966 pp 323ndash326

Design of rectangular plates with banded orthotropic reinforcementOct 1966 pp 371ndash376

Non-rectangular slabs with orthotropic reinforcement Nov 1966pp 383ndash390

29 Hillerborg A Strip method of design London Viewpoint 1975p 225

This book is the English translation of the basic text on the stripmethod (both simple and advanced) by its originator It dealswith theory and gives appropriate design formulae for manyproblems

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References and further reading398

30 Fernando J S and Kemp K O A generalised strip deflexionmethod of reinforced concrete slab design Proceedings of theInstitution of Civil Engineers Part 2 Research and Theory March1978 pp 163ndash174

31 Taylor R Hayes B and Mohamedbhai G T G Coefficientsfor the design of slabs by the yield-line theory Concrete 3(5) 1969pp 171ndash172

32 Munshi J A Rectangular concrete tanks (revised fifth edition)Skokie Illinois Portland Cement Association 1998 p 188

This is the most detailed book on the subject with complete tablesgiving moments shears and deflections for plates and tanks withuseful worked examples

33 CIRIA Report 110 Design of reinforced concrete flat slabs toBS8110 London CIRIA 1985 p 48

34 Beeby A W The analysis of beams in plane frames according toCP110 London Cement and Concrete Association Publication44001 1978 p 34

35 Rygol J Structural analysis by direct moment distributionLondon Crosby Lockwood 1968 p 407

36 Naylor N Side-sway in symmetrical building frames TheStructural Engineer 28(4) 1950 pp 99ndash102

37 Orton A The way we build now form scale and techniqueLondon E amp FN Spon 1988 p 530

38 CIRIA Report 102 Design of shear wall buildings LondonCIRIA 1984 p 80

39 Eurocode 8 Design of structures for earthquake resistanceBrussels European Committee for Standardization 2004

40 Penelis G G and Kappos A J Earthquake-resistant concretestructures London E amp FN Spon 1997 p 572

41 Kruger H G Crack width calculation to BS 8007 for combinedflexure and direct tension The Structural Engineer 80(18) 2002pp 18ndash22

42 Kong F K Robins P J and Sharp G R The design of reinforcedconcrete deep beams in current practice The Structural Engineer53(4) 1975 pp 73ndash80

43 Ove Arup and Partners The design of deep beams in reinforcedconcrete CIRIA Guide No 2 1977 p 131

44 Concrete Society Technical Report No 42 Trough and wafflefloors Slough The Concrete Society 1992 p 34

45 Gibson J E The design of shell roofs (Third edition) LondonE amp FN Spon 1968 p 300

46 Chronowicz A The design of shells London Crosby Lockwood1959 p 202

47 Tottenham H A A simplified method of design for cylindricalshell roofs The Structural Engineer 32(6) 1954 pp 161ndash180

48 Bennett J D Empirical design of symmetrical cylindricalshells Proceedings of the colloquium on simplified calculationmethods Brussels 1961 Amsterdam North-Holland 1962pp 314ndash332

49 Salvadori and Levy Structural design in architecture EnglewoodCliffs Prentice-Hall 1967 p 457

50 Schulz M and Chedraui M Tables for circularly curved horizontalbeams with symmetrical uniform loads Journal of the AmericanConcrete Institute 28(11) 1957 pp 1033ndash1040

51 Spyropoulos P J Circularly curved beams transversely loadedJournal of the American Concrete Institute 60(10) 1963 pp 1457ndash1469

52 Concrete SocietyCBDG An introduction to concrete bridgesCamberley The Concrete Society 2006 p 32

53 Leonhardt Fritz Bridges Stuttgart Deutsche Verlags-Anstalt1982 p 308

54 Hambly E C Bridge deck behaviour (Second edition) LondonE amp FN Spon 1991 p 313

55 PCA Circular concrete tanks without prestressing SkokieIllinois Portland Cement Association p 54

56 Ghali A Circular storage tanks and silos London E amp FN Spon1979 p 210

57 CIRIA Reports 139 and 140 (Summary Report) Water-resistingbasement construction London CIRIA 1995 p 192 p 64

58 Irish K and Walker W P Foundations for reciprocating machinesLondon Cement and Concrete Association 1969 p 103

59 Barkan D D Dynamics of bases and foundations New YorkMcGraw Hill 1962 p 434

60 Tomlinson M J Pile design and construction practice LondonCement and Concrete Association 1977 p 413

61 Concrete Society Technical Report No 34 Concrete industrialground floors (Third edition) Crowthorne The Concrete Society2003 p 146

62 CIRIA Report 104 Design of retaining walls embedded in stiffclay London CIRIA 1984 p 146

63 Hairsine R C A design chart for determining the optimum baseproportions of free standing retaining walls Proceedings of theInstitution of Civil Engineers 51 (February) 1972 pp 295ndash318

64 Cusens A R and Kuang Jing-Gwo A simplified method ofanalysing free-standing stairs Concrete and ConstructionalEngineering 60(5) 1965 pp 167ndash172 and 194

65 Cusens A R Analysis of slabless stairs Concrete andConstructional Engineering 61(10) 1966 pp 359ndash364

66 Santathadaporn Sakda and Cusens A R Charts for the design ofhelical stairs Concrete and Constructional Engineering 61(2) 1966pp 46ndash54

67 Terrington J S and Turner L Design of non-planar roofsLondon Concrete Publications 1964 p 108

68 Krishna J and Jain O P The beam strength of reinforced concretecylindrical shells Civil Engineering and Public Works Review49(578) 1954 pp 838ndash840 and 49(579) 1954 pp 953ndash956

69 Faber C Candela the shell builder London The ArchitecturalPress 1963 p 240

70 Bennett J D Structural possibilities of hyperbolic paraboloidsLondon Reinforced Concrete Association February 1961 p 25

71 Lee D J Bridge bearings and expansion joints (Second edition)London E amp FN Spon 1994 p 212

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Numbers preceded by lsquotrsquo are Table Numbers

Actions see LoadsAdmixtures 18ndash19Affinity theorems 140Aggregates 16ndash17

size and grading 95 t217Anchorage bond see BondAnnular sections 236 t2107Arches

fixed 41ndash2 t272ndash4load effects in 178ndash9 t272

parabolic 42 179ndash82t274

thickness 175 t272three-hinged 41 175

t271two-hinged 41 175 t271see also Bridges

Areas 52ndash3 t2101

Barriers and balustrades 7Bars

anchorage 51 312 381t355 t359 t430 t432

bending schedules t223bends in 25 51 312 381

t219 t355 t359 t431considerations affecting

design details 51 312381 t353 t359 t428

curtailment 52 312 381t356ndash8 t432

cutting and bendingtolerances 100

lap lengths 51 312 381t355 t359 t431ndash2

shapes and dimensions25ndash6 100 t221ndash2

sizes 25 95 t220types 24ndash5 95see also Reinforcement

Basements 65Bases see FoundationsBeams

cantilevers see Cantileverscontinuous see Continuous

beamscurved 57 216 t295ndash7

concentrated load 216t295

uniform load 218 t296ndash7

deep 52

doubly reinforced sections257 346 t315ndash16t325ndash6 t49ndash10

fixed at both ends 105 t225t228

flanged see Flangedsections

imposed loads on 6 t23junctions with columns 330

t363single-span 29 105 t224ndash5singly reinforced sections

256 346 t313ndash14t323ndash4 t47ndash8

sizes and proportions 46supporting rectangular

panels 34 144 t252Bearings 62 221 t299

see also DetailsFoundations

Bending (alone or combinedwith axial force) 44ndash8

assumptions 44ndash5 t36 t44resistance of sections

beams 45ndash6columns 47ndash8slabs 46ndash7

see also individual membersBending moments

combined bases 195 t283continuous beams see

Continuous beamscylindrical tanks 60 183ndash8

t275ndash7flat slabs 35ndash6 150ndash3

t255ndash6rectangular tanks 60ndash1 188

t278ndash9silos 61ndash2 191 t280see also Beams Cantilevers

Structural analysisBiaxial bending see ColumnsBlinding layer 64Bond 51 312 381

anchorage lengths see Barsbends in bars see Barslap lengths see Bars

Bow girders see Beamscurved

Bridges 57ndash9deck 57ndash8 t298design considerations 59imposed loads

foot 8 78 t26railway 8ndash9 78 t26road 7ndash8 78 t25

integral 59partial safety factors 239

t32ndash3roofs see Roofsstairs see Stairssubstructures 58types 57ndash8 t298waterproofing 59wind loads 10see also Arches

Buildings 54ndash6dead loads 75 t22imposed loads

floors 6 75ndash8 t23roofs 7 78 t24

load-bearing walls see Wallsrobustness and provision of

ties 54ndash5 t354 t429wall and frame systems

40ndash1 t268wind loads 10 78 t27ndash9see also Floors

Foundations StairsWalls

Bunkers see Silos

Cantilevers 29 t226ndash7deflections 295 371

Cements and combinations14ndash16 95 t217

Characteristic loads see Loads

Characteristic strengthsconcrete t35 t42reinforcement t219

Columnsbiaxial bending t321 t331

t416circular 264 353 t319ndash20

t329ndash30 t413ndash14cylindrical (modular ratio)

236 t2107effective height t321 t331effective length t415elastic analysis of section

226 t2104 t2108ndash9imposed loads on 7 t23junctions with beams 330

t363loads and sizes 48rectangular 264 353

t317ndash18 t327ndash8t411ndash12

rectangular (modular ratio)t2105ndash6

short 47 265

slenderdesign procedure 263

352ndash3BS 8110 t321ndash2BS 5400 t331ndash2EC 2 t415ndash16

supporting elevated tanks61 194

see also Framed structuresConcentrated loads

analysis of memberscurved beams 216 t295solid slabs 31 34 131

t245ndash7bridges 7ndash9 78 t25ndash6dispersal of 9on floors 6 78 t23shear 49 285 365 t334

t337ndash8 t419Concrete 14ndash24 95

admixtures 18ndash19aggregates 16ndash17 95

t217alkali-silica attack 23carbonation 22ndash3cements 14ndash16 95

t217chemical attack 23compressive strength 21

245 338 t35 t42creep 21ndash2 245 338 t35

t43design strengths 239 335durability 22ndash4

cover to reinforcement 24t38ndash9 t46

exposure classes 23ndash4t37 t39 t45

early-age temperatures andcracking 20 95 t218

elastic properties 21 245338 t35 t42

fibre-reinforced 68fire resistance see Fire

resistancefreezethaw attack 23plastic cracking 19ndash20shrinkage 22 245 338 t35

t42specification 24stress-strain curves 22 t36

t44tensile strength 21 t344

t423thermal properties 22 245

338 t35 t42

Index

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Concrete (Continued)water 17weight 75 t21workability 19

Construction materialsweight 75 t21ndash2

Contained materials seeRetained materials

Containment structures 59ndash61see also Silos Tanks

Continuous beams 29ndash30arrangement of design loads

111 t229equal spans and loads 30

111 t229ndash32 t234ndash5influence lines for bending

moments 121 t238ndash41methods of analysis 29

moment distribution 29121 t236

moving loads 30redistribution of moments

see Moment redistributionsecond moment of area 29unequal spans and loads

t237Corbels 327 330ndash1 t362Cover to reinforcement 24

t38ndash9 t46Cracking 50ndash1 295 300 371

calculation procedures t343t424

crack width limits 295 371deemed-to-satisfy rules

t343 t424ndash7liquid-retaining structures

300 371 t344ndash52t425ndash7

minimum reinforcement t423

Cranes 7 13Creep see ConcreteCulverts 71

box culverts 71 t287pipe culverts 71subways 71

Curtailment see BarsCurvature see DeflectionCurved beams see BeamsCylindrical tanks see Tanks

Dead loads 6 75concrete 75 t21construction materials 75

t21ndash2partitions 75 t22

Deep beams 52Deep containers see SilosDeflection 49ndash50 295 371

calculation procedurest340ndash2 t421ndash2

cantilevers 295 371curvatures t341 t422deflection limits 295 371formulae for

beams t224ndash5cantilevers t226ndash7

spaneffective depth ratios295 371 t340 t421

Design of structural members44ndash53

see also individual members(eg Arches BeamsColumns Slabs StairsWalls)

Design principles and criteria 5 44

Design strengths seeConcrete Reinforcement

Detailscontinuous nibs 327 t362

corbels 327 t362corners and intersections

330 t363curtailment 52 312 381

t356 t432rules for beams t357rules for slabs t358

halving joints 330 t362Docks and dolphins see

Maritime structuresDomes see RoofsDrawings 4

Earth-retaining wallsembedded (or sheet) 70movement joints 221 t2100pressures behind 11ndash12 86

90 t210ndash14on spread bases 69ndash70

203 324 392 t286types 69 t286see also Retained materials

Earthquake-resistant structures 43

Economical structures 3Elastic analysis 52ndash3 226 236

biaxial bending andcompression t2109

design charts t2105ndash7properties of sections

t2101ndash3 t342uniaxial bending and

compression t2104uniaxial bending and tension

t2108Embedded walls see

Earth-retaining wallsEurocode loading standards 13Exposure classes 23ndash4 t37

t39 t45

Fabric 95 100 t220Fibre-reinforced concrete 68Fill materials 12Finite elements 38Fire resistance 27 249 342

cover to reinforcement 249t310ndash11

minimum fire periods t312Fixed-end moment coefficients

105 t228Flanged sections 46

effective flange width 262 349

elastic properties t342Flat slabs see SlabsFloors 55

forms of construction t242imposed loads 6 t23industrial ground see

Industrial ground floorsopenings in 55 t337weights of concrete t21

Footbridgesimposed loads 78 t26

Formwork 4Foundations 63ndash7

balanced and coupled bases64 199 t283ndash4

basements 65bearing pressures 63 t282blinding layer 64combined bases 64 195

t283eccentric loads 63imposed loads 7 t23for machines 66piers 65piled see Piled foundationsrafts 65 199 t284separate bases 64 t282

site inspection 63strip bases 65 195 t283types 64 t282ndash4wall footings 66 t283

Framed structures 36ndash8building code requirements

36 t257 t262columns in

non-sway frames 38ndash9t260ndash1

sway frames 39ndash40 t262continuous beams in 159effect of lateral loads 39ndash40

162 t262finite element method 38moment distribution method

no sway 37 t258with sway 37 t259

portal frames 38 162rigid joints t263ndash6hinged joints t267

properties of membersend conditions 42section properties 42ndash3

shear forces on members 37slope-deflection method

of analysis 37 154t260ndash2

see also Columns

Garages 6Geometric properties of

uniform sections t2101Ground water 86 90Gyration radius of 52 t2101

t415

Hillerborgrsquos strip method 33144 t251 t254

Hinges 62 221 t299Hoppers 12 62 90 194

t215ndash16 t281

Imposed loads 6ndash9 t23ndash6barriers and parapets 7bridges see Bridgesbuildings 6ndash7 75 t23floors 6 t23reduction on beams and

columns 7 78 t23roofs 7 78 t24structures subject to dynamic

loads 6structures supporting

cranes 7structures supporting

lifts 7underground tanks 60see also Eurocode loading

standardsIndustrial ground floors 67ndash9

construction methods 67ndash8methods of analysis 68ndash9modulus of subgrade

reaction 68reinforcement 68

Intersections 330 t363see also Joints

Janssenrsquos theory 12 90t215ndash16

Jetties see Maritime structures

Joints 52 330industrial ground

floors 67ndash8liquid-retaining structures

300 t345movement 62 221 t2100see also Bearings

Details

Lifts 7Limit state design 5

British codesbridges 239 241 t32ndash3buildings 239 t31liquid-retaining structures

241 t34loads 5ndash6 239properties of materials

concrete 245 t35ndash6reinforcement 245 t36

European codesactions 5ndash6 335buildings 335ndash6 t41containers 335ndash6properties of materials 335

concrete 338 t42ndash4reinforcement 338 t44

Liquid-retaining structures 241335ndash6 t34

see also Cracking JointsLoads 6ndash10

on bridges see Bridgesconcentrated see

Concentrated loadsdead see Dead loadsdynamic 6eccentric on foundations 63

t283imposed see Imposed loadson lintels t22moving loads on continuous

beams 30on piles see Piled

foundationswind see Wind loads

Maritime structures 10ndash11piled jetties 200 t285

Materials see AdmixturesAggregates Cementsand combinationsConcrete Reinforcement

Mathematical formulae 395ndash6Members see individual

members (eg ArchesBeams Columns SlabsStairs Walls)

Modular-ratio design seeElastic analysis

Modulus of subgrade reaction 68

Moment distribution 29continuous beams 121 t236framed structures 154

t258ndash9Moment redistribution 30 116

code requirements 116design procedure 117 t233moment diagrams for equal

spans t234ndash5

Neutral axis 44ndash5

Parapets 7Partial safety factors see

Safety factorsPartition loads 75 t22Passive pressures 90

t213ndash14Piers

bridges 58foundations 65see also Maritime structures

Piled foundations 66ndash7open-piled structures 67

200 t285pile-caps 66 324 t361piles in a group 67

Poissonrsquos ratio 131 137t246ndash7 t35 t42

Index400

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Precast concretebridge decks 58floors weights of t21

Pressures 11ndash13on earth-retaining walls 86

90 t211ndash14in silos 90 t215ndash16in tanks 90see also Wind loads

Properties of sections 42ndash3plain concrete t2101reinforced concrete t2102ndash3

Rafts 65 199 t284Railway bridges see BridgesReinforcement 24ndash7 95 100

bars 24 see also Barsfabric 25 see also Fabricfixing of 27mechanical and physical

properties t219prefabricated systems 26ndash7stainless steel 26stress-strain curves 25 t36

t44Reservoirs see TanksRetained materials

cohesionless soils 12 8690 t212ndash14

cohesive soils 12 90fill materials 12lateral pressures 11ndash12

t211liquids 11 86 90soil properties 11 t210see also Silos Tanks

Retaining walls seeEarth-retaining walls

Robustness 54ndash5ties 312 381 t354 t429

Roofsloads on 7 78 t24non-planar 56 212

cylindrical shells 56 212216 t292ndash4

domes 212 t292hyperbolic-paraboloidal

shells 216 t292prismatic 212 t292shell buckling 56ndash7

planar 56weights of t22

Safety factors 5British codes 239 t31ndash4

geotechnical design 324European codes 335 t41

geotechnical design 390Sea-walls see Maritime

structuresSection moduli t2101Serviceability limit states 5

295 300 336 371t31ndash4 t41

Shearin bases 285 322 390forces see individual

members (eg BeamsSlabs)

resistance with shearreinforcement 49 283362 t333 t336 t418

resistance without shearreinforcement 48 283362 t333 t336 t417

stress 283under concentrated loads 49

285 365 t334 t337ndash8t419

see also Structural analysisShear wall structures 40ndash1

arrangement of walls 40169 t269

interaction of walls andframes 41 169 173

walls containing openings40 169 t270

walls without openings 40169 t269

Sheet walls seeEarth-retaining walls

Shrinkagefixed parabolic arch 179 182see also Concrete

Silos 12ndash13 61ndash2 90hopper bottoms 62 t281stored material properties

and pressures 90t215ndash16

substructure 39 t262walls spanning horizontally

61ndash2 t280Slabs

flat 35ndash6reservoir roofs 153simplified method of

design 150 153t255ndash6

imposed loads 75 78 t23ndash4non-rectangular panels 34ndash5

131 t248one-way 31 128

concentrated loads 31131 t245

uniform load distribution31 128 t242

openings in 55 t337rectangular panels

concentrated loads 34131 137 t246ndash7

triangular load distribution33ndash4 147 t253ndash4

uniform load distribution33 128 131 t242ndash4

thickness of 46ndash7two-way 31ndash4 128 131

collapse methods 32ndash3elastic methods 32

strip 144 t251 t254yield-line 137 139ndash42

147 150 t249ndash50t254

types of 55 t242weights of t21

Slope-deflection method 37 154 t260

Soils see Retained materialsStairs 55ndash6 206 208 212

free-standing 206 t288helical 208 212 t289ndash91sawtooth 206 208 t289simple flights 206types and dimensions t288

Stored materials see SilosStress-strain curves

concrete 22 t36 t44reinforcement 25 t36 t44

Stresses see individual modes(eg Bond ShearTorsion)

Structural analysis 28ndash43properties of members

end conditions 42section properties 42ndash3

see also individual structures(eg Arches Continuousbeams Frames Shearwalls)

Structures 54ndash71earthquake-resistant 43economical 3see also individual

structures (eg BridgesBuildings FoundationsSilos Tanks)

Subways 71Superposition theorem 140Surcharge 86 203 326

Tankscylindrical 60 183 187ndash8

t275ndash7effects of temperature 61elevated 61 191 t281

substructure 39 t262joints 221 t2100octagonal 60pressure on walls 90rectangular 60ndash1 147 150

188 191 t253ndash4 t278ndash9

underground 60see also Cracking Liquid-

retaining structuresTemperature effects in

concrete at early-age 20 95 t218

fixed parabolic arch 179walls of tanks 61

Tensile strengthconcrete 21 t344 t423reinforcement 95 t219

Thermal properties of concrete22 245 338 t35 t42

Ties see RobustnessTorsion

design procedure 49 285365 t335 t339 t420

moments incurved beams 57 216

218 t295ndash7free-standing stairs 206

t288helical stairs 208 212

t289

Ultimate limit state 5 239 241335ndash6 t31ndash4 t41

Vehicle loads on bridges 7ndash978 t25ndash6

Vibrationfloors 6footbridges 8machine foundations 66

Virtual-work method 139ndash40

Walls 52 57load-bearing 57 322 t360weights of t22see also Earth-retaining

walls Shear walls SilosTanks

Water (for concrete) 17Water-tightness 59

basements 65Weights of

concrete 75 t21construction materials 75

t21ndash2partitions 75 t22roofs t22stored materials t216walls t22

Wharves see MaritimeStructures

Wheel loads 8 t25dispersal of 9

Wind loads 9ndash10 78on bridges 10on buildings 10effect of see Structural

analysiswind speed and pressure 10

t27ndash9

Yield-line analysis 32 137 139ndash42

affinity theorems 140basic concepts 137 139

t249concentrated loads 140corner levers 142 t250superposition theorem 140virtual work method 139

empirical analysis 141ndash2

Index 401

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