design of structural elements: concrete, steelwork...

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Design of Structural Elements

Third Edition

Concrete, steelwork, masonry and timber designs toBritish Standards and Eurocodes

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Design ofStructuralElementsThird Edition

Concrete, steelwork, masonry andtimber designs to British Standardsand Eurocodes

Chanakya Arya

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First published 1994 by E & FN Spon

Second edition published 2003 by Spon Press

This edition published 2009by Taylor & Francis2 Park Square, Milton Park, Abingdon, Oxon OX14 4RN

Simultaneously published in the USA and Canadaby Taylor & Francis270 Madison Avenue, New York, NY 10016, USA

Taylor & Francis is an imprint of the Taylor & Francis Group, an informa business

© 1994, 2003, 2009 Chanakya Arya

All rights reserved. No part of this book may be reprinted or reproduced orutilised in any form or by any electronic, mechanical, or other means, nowknown or hereafter invented, including photocopying and recording, or inany information storage or retrieval system, without permission in writingfrom the publishers.

The publisher makes no representation, express or implied, with regardto the accuracy of the information contained in this book and cannot accept anylegal responsibility or liability for any errors 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 DataArya, Chanakya.Design of structural elements : concrete, steelwork, masonry, and timber designsto British standards and Eurocodes / Chanakya Arya. – 3rd ed.

p. cm.Includes bibliographical references and index.1. Structural design – Standards – Great Britain. 2. Structural design – Standards –Europe. I. Title. II. Title: Concrete, steelwork, masonry, and timber designto British standards and Eurocodes.TA658.A79 2009624.1′7–dc222008043080

ISBN10: 0-415-46719-5 (hbk)ISBN10: 0-415-46720-9 (pbk)ISBN10: 0-203-92650-1 (ebk)

ISBN13: 978-0-415-46719-3 (hbk)ISBN13: 978-0-415-46720-9 (pbk)ISBN13: 978-0-203-92650-5 (ebk)

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This edition published in the Taylor & Francis e-Library, 2009.

To purchase your own copy of this or any of Taylor & Francis or Routledge’scollection of thousands of eBooks please go to www.eBookstore.tandf.co.uk.

ISBN 0-203-92650-1 Master e-book ISBN

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Contents

Preface to the third edition viiPreface to the second edition ixPreface to the first edition xiAcknowledgements xiiiList of worked examples xv

PART ONE: INTRODUCTION TOSTRUCTURAL DESIGN

1 Philosophy of design 31.1 Introduction 31.2 Basis of design 41.3 Summary 8

Questions 8

2 Basic structural concepts andmaterial properties 9

2.1 Introduction 92.2 Design loads acting on structures 92.3 Design loads acting on elements 132.4 Structural analysis 172.5 Beam design 242.6 Column design 262.7 Summary 27

Questions 28

PART TWO: STRUCTURAL DESIGN TOBRITISH STANDARDS

3 Design in reinforced concreteto BS 8110 31

3.1 Introduction 313.2 Objectives and scope 313.3 Symbols 323.4 Basis of design 333.5 Material properties 333.6 Loading 353.7 Stress–strain curves 363.8 Durability and fire resistance 373.9 Beams 44

3.10 Slabs 933.11 Foundations 1153.12 Retaining walls 121

3.13 Design of short braced columns 1283.14 Summary 143

Questions 143

4 Design in structural steelworkto BS 5950 145

4.1 Introduction 1454.2 Iron and steel 1454.3 Structural steel and steel

sections 1464.4 Symbols 1484.5 General principles and design

methods 1494.6 Loading 1504.7 Design strengths 1514.8 Design of steel beams and joists 1514.9 Design of compression members 177

4.10 Floor systems for steel framedstructures 199

4.11 Design of connections 2184.12 Summary 236

Questions 237

5 Design in unreinforced masonryto BS 5628 239

5.1 Introduction 2395.2 Materials 2405.3 Masonry design 2455.4 Symbols 2455.5 Design of vertically loaded masonry

walls 2465.6 Design of laterally loaded wall

panels 2635.7 Summary 276

Questions 277

6 Design in timber to BS 5268 2796.1 Introduction 2796.2 Stress grading 2806.3 Grade stress and strength class 2806.4 Permissible stresses 282

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Contents

6.5 Timber design 2856.6 Symbols 2856.7 Flexural members 2876.8 Design of compression members 2986.9 Design of stud walls 303

6.10 Summary 305Questions 306

PART THREE: STRUCTURAL DESIGNTO THE EUROCODES

7 The structural Eurocodes:An introduction 309

7.1 Scope 3097.2 Benefits of Eurocodes 3097.3 Production of Eurocodes 3107.4 Format 3107.5 Problems associated with drafting

the Eurocodes 3107.6 Decimal point 3127.7 Implementation 3127.8 Maintenance 3127.9 Difference between national

standards and Eurocodes 312

8 Eurocode 2: Design of concretestructures 314

8.1 Introduction 3148.2 Structure of EC 2 3158.3 Symbols 3158.4 Material properties 3168.5 Actions 3178.6 Stress–strain diagrams 3238.7 Cover, fire, durability and bond 3248.8 Design of singly and doubly

reinforced rectangular beams 3278.9 Design of one-way solid slabs 350

8.10 Design of pad foundations 3578.11 Design of columns 361

9 Eurocode 3: Design of steelstructures 375

9.1 Introduction 3759.2 Structure of EC 3 3769.3 Principles and Application rules 3769.4 Nationally Determined

Parameters 3769.5 Symbols 3779.6 Member axes 3779.7 Basis of design 377

9.8 Actions 3789.9 Materials 378

9.10 Classification of cross-sections 3809.11 Design of beams 3809.12 Design of columns 4039.13 Connections 418

10 Eurocode 6: Design ofmasonry structures 434

10.1 Introduction 43410.2 Layout 43410.3 Principles/Application rules 43510.4 Nationally Determined

Parameters 43510.5 Symbols 43510.6 Basis of design 43610.7 Actions 43610.8 Design compressive strength 43710.9 Durability 441

10.10 Design of unreinforced masonrywalls subjected to verticalloading 441

10.11 Design of laterally loaded wallpanels 455

11 Eurocode 5: Design oftimber structures 458

11.1 Introduction 45811.2 Layout 45811.3 Principles/Application rules 45911.4 Nationally Determined

Parameters 45911.5 Symbols 45911.6 Basis of design 46011.7 Design of flexural members 46411.8 Design of columns 477

Appendix A Permissible stress and loadfactor design 481

Appendix B Dimensions and propertiesof steel universal beams andcolumns 485

Appendix C Buckling resistance ofunstiffened webs 489

Appendix D Second moment of area of acomposite beam 491

Appendix E References and furtherreading 493

Index 497

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Preface to thethird edition

Since publication of the second edition of Designof Structural Elements there have been two majordevelopments in the field of structural engineeringwhich have suggested this new edition.

The first and foremost of these is that theEurocodes for concrete, steel, masonry and timberdesign have now been converted to full EuroNorm(EN) status and, with the possible exception of thesteel code, all the associated UK National Annexeshave also been finalised and published. Therefore,these codes can now be used for structural design,although guidance on the timing and circumstancesunder which they must be used is still awaited.Thus, the content of Chapters 8 to 11 on, respec-tively, the design of concrete, steel, masonry andtimber structures has been completely revised tocomply with the EN versions of the Eurocodes forthese materials. The opportunity has been used toexpand Chapter 10 and include several workedexamples on the design of masonry walls subject toeither vertical or lateral loading or a combinationof both.

The second major development is that a numberof small but significant amendments have been

made to the 1997 edition of BS 8110: Part 1 onconcrete design, and new editions of BS 5628:Parts 1 and 3 on masonry design have recentlybeen published. These and other national stand-ards, e.g. BS 5950 for steel design and BS 5268for timber design, are still widely used in the UKand beyond. This situation is likely to persist forsome years, and therefore the decision was takento retain the chapters on British Standards andwhere necessary update the material to reflect latestdesign recommendations. This principally affectsthe material in Chapters 3 and 5 on concrete andmasonry design.

The chapters on Eurocodes are not self-containedbut include reference to relevant chapters on BritishStandards. This should not present any problemsto readers familiar with British Standards, but willmean that readers new to this subject will have torefer to two chapters from time to time to get themost from this book. This is not ideal, but shouldresult in the reader becoming familiar with bothBritish and European practices, which is probablynecessary during the transition phase from BritishStandards to Eurocodes.

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Preface to thesecond edition

The main motivation for preparing this newedition was to update the text in Chapters 4 and 6on steel and timber design to conform with thelatest editions of respectively BS 5950: Part 1 andBS 5268: Part 2. The opportunity has also beentaken to add new material to Chapters 3 and 4.Thus, Chapter 3 on concrete design now includesa new section and several new worked examples onthe analysis and design of continuous beams andslabs. Examples illustrating the analysis and designof two-way spanning slabs and columns subjectto axial load and bending have also been added.The section on concrete slabs has been updated. Adiscussion on flooring systems for steel framedstructures is featured in Chapter 4 together with asection and several worked examples on compositefloor design.

Work on converting Parts 1.1 of the Eurocodesfor concrete, steel, timber and masonry structures

to full EN status is still ongoing. Until such timethat these documents are approved the design rulesin pre-standard form, designated by ENV, remainvalid. The material in Chapters 8, 9 and 11 tothe ENV versions of EC2, EC3 and EC5 are stillcurrent. The first part of Eurocode 6 on masonrydesign was published in pre-standard form in1996, some three years after publication of the firstedition of this book. The material in Chapter 10has therefore been revised, so it now conforms tothe guidance given in the ENV.

I would like to thank the following who haveassisted with the preparation of this new edition: Pro-fessor Colin Baley for preparing Appendix C; FredLambert, Tony Threlfall, Charles Goodchild andPeter Watt for reviewing parts of the manuscript.

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Preface to thefirst edition

individual elements can be assessed,thereby enabling the designer to sizethe element.

Part Two contains four chapters covering thedesign and detailing of a number ofstructural elements, e.g. floors, beams,walls, columns, connections andfoundations to the latest British codesof practice for concrete, steelwork,masonry and timber design.

Part Three contains five chapters on the Euro-codes for these materials. The first ofthese describes the purpose, scope andproblems associated with drafting theEurocodes. The remaining chaptersdescribe the layout and contents ofEC2, EC3, EC5 and EC6 for designin concrete, steelwork, timber andmasonry respectively.

At the end of Chapters 1–6 a number of designproblems have been included for the student toattempt.

Although most of the tables and figures fromthe British Standards referred to in the text havebeen reproduced, it is expected that the readerwill have either the full Standard or the publica-tion Extracts from British Standards for Students ofStructural Design in order to gain the most fromthis book.

I would like to thank the following who haveassisted with the production of this book: PeterWright for co-authoring Chapters 1, 4 and 9; FredLambert, Tony Fewell, John Moran, David Smith,Tony Threlfall, Colin Taylor, Peter Watt and PeterSteer for reviewing various parts of the manuscript;Tony Fawcett for the drafting of the figures;and Associate Professor Noor Hana for help withproofreading.

C. AryaLondon

UK

Structural design is a key element of all degree anddiploma courses in civil and structural engineering.It involves the study of principles and procedurescontained in the latest codes of practice for struc-tural design for a range of materials, including con-crete, steel, masonry and timber.

Most textbooks on structural design consider onlyone construction material and, therefore, the studentmay end up buying several books on the subject.This is undesirable from the viewpoint of cost butalso because it makes it difficult for the studentto unify principles of structural design, because ofdiffering presentation approaches adopted by theauthors.

There are a number of combined textbooks whichinclude sections on several materials. However,these tend to concentrate on application of thecodes and give little explanation of the structuralprinciples involved or, indeed, an awareness ofmaterial properties and their design implications.Moreover, none of the books refer to the newEurocodes for structural design, which will eventu-ally replace British Standards.

The purpose of this book, then, is to describethe background to the principles and procedurescontained in the latest British Standards andEurocodes on the structural use of concrete, steel-work, masonry and timber. It is primarily aimed atstudents on civil and structural engineering degreeand diploma courses. Allied professionals such asarchitects, builders and surveyors will also find itappropriate. In so far as it includes five chapters onthe structural Eurocodes it will be of considerableinterest to practising engineers too.

The subject matter is divided into 11 chaptersand 3 parts:

Part One contains two chapters and explains theprinciples and philosophy of structuraldesign, focusing on the limit stateapproach. It also explains how theoverall loading on a structure and

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Acknowledgements

I am once again indebted to Tony Threlfall, for-merly of the British Cement Association and nowan independent consultant, for comprehensively re-viewing Chapter 8 and the material in Chapter 3on durability and fire resistance

I would also sincerely like to thank ProfessorR.S. Narayanan of the Clark Smith Partnershipfor reviewing Chapter 7, David Brown of theSteel Construction Institute for reviewing Chap-ter 9, Dr John Morton, an independent consultant,for reviewing Chapter 10, Dr Ali Arasteh of theBrick Development Association for reviewing Chap-ters 5 and 10, and Peter Steer, an independent

consultant, for reviewing Chapter 11. The contentsof these chapters are greatly improved due to theircomments.

A special thanks to John Aston for reading partsof the manuscript.

I am grateful to The Concrete Centre for per-mission to use extracts from their publications.Extracts from British Standards are reproduced withthe permission of BSI under licence number2008ET0037. Complete standards can be obtainedfrom BSI Customer Services, 389 Chiswick HighRoad, London W4 4AL.

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List of workedexamples

2.1 Self-weight of a reinforced concretebeam 10

2.2 Design loads on a floor beam 142.3 Design loads on floor beams and

columns 152.4 Design moments and shear forces in

beams using equilibrium equations 182.5 Design moments and shear forces in

beams using formulae 232.6 Elastic and plastic moments of

resistance of a beam section 262.7 Analysis of column section 27

3.1 Selection of minimum strength classand nominal concrete cover toreinforcement (BS 8110) 43

3.2 Design of bending reinforcement fora singly reinforced beam (BS 8110) 48

3.3 Design of shear reinforcement for abeam (BS 8110) 52

3.4 Sizing a concrete beam (BS 8110) 593.5 Design of a simply supported

concrete beam (BS 8110) 613.6 Analysis of a singly reinforced

concrete beam (BS 8110) 653.7 Design of bending reinforcement for

a doubly reinforced beam (BS 8110) 683.8 Analysis of a two-span continuous

beam using moment distribution 723.9 Analysis of a three span continuous

beam using moment distribution 763.10 Continuous beam design (BS 8110) 783.11 Design of a one-way spanning

concrete floor (BS 8110) 1003.12 Analysis of a one-way spanning

concrete floor (BS 8110) 1043.13 Continuous one-way spanning slab

design (BS 8110) 1063.14 Design of a two-way spanning

restrained slab (BS 8110) 1103.15 Design of a pad footing (BS8110) 117

3.16 Design of a cantilever retaining wall(BS 8110) 125

3.17 Classification of a concrete column(BS 8110) 131

3.18 Sizing a concrete column (BS 8110) 1333.19 Analysis of a column section

(BS 8110) 1343.20 Design of an axially loaded column

(BS 8110) 1393.21 Column supporting an approximately

symmetrical arrangement of beams(BS 8110) 140

3.22 Columns resisting an axial load andbending (BS 8110) 141

4.1 Selection of a beam section inS275 steel (BS 5950) 156

4.2 Selection of beam section inS460 steel (BS 5950) 158

4.3 Selection of a cantilever beamsection (BS 5950) 159

4.4 Deflection checks on steel beams(BS 5950) 161

4.5 Checks on web bearing and bucklingfor steel beams (BS 5950) 164

4.6 Design of a steel beam with webstiffeners (BS 5950) 164

4.7 Design of a laterally unrestrained steelbeam – simple method (BS 5950) 171

4.8 Design of a laterally unrestrainedbeam – rigorous method (BS 5950) 174

4.9 Checking for lateral instability in acantilever steel beam (BS 5950) 176

4.10 Design of an axially loaded column(BS 5950) 183

4.11 Column resisting an axial load andbending (BS 5950) 185

4.12 Design of a steel column in ‘simple’construction (BS 5950) 189

4.13 Encased steel column resisting an axialload (BS 5950) 193

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4.14 Encased steel column resisting anaxial load and bending (BS 5950) 195

4.15 Design of a steel column baseplate(BS 5950) 198

4.16 Advantages of compositeconstruction (BS 5950) 200

4.17 Moment capacity of a compositebeam (BS 5950) 209

4.18 Moment capacity of a compositebeam (BS 5950) 210

4.19 Design of a composite floor(BS 5950) 212

4.20 Design of a composite floorincorporating profiled metal decking(BS 5950) 215

4.21 Beam-to-column connection usingweb cleats (BS 5950) 224

4.22 Analysis of a bracket-to-columnconnection (BS 5950) 227

4.23 Analysis of a beam splice connection(BS 5950) 228

4.24 Analysis of a beam-to-columnconnection using an end plate(BS 5950) 232

4.25 Analysis of a welded beam-to-columnconnection (BS 5950) 235

5.1 Design of a load-bearing brick wall(BS 5628) 254

5.2 Design of a brick wall with ‘small’plan area (BS 5628) 255

5.3 Analysis of brick walls stiffened withpiers (BS 5628) 256

5.4 Design of single leaf brick and blockwalls (BS 5628) 258

5.5 Design of a cavity wall (BS 5628) 2615.6 Analysis of a one-way spanning wall

panel (BS 5628) 2715.7 Analysis of a two-way spanning panel

wall (BS 5628) 2725.8 Design of a two-way spanning

single-leaf panel wall (BS 5628) 2735.9 Analysis of a two-way spanning

cavity panel wall (BS 5628) 274

6.1 Design of a timber beam (BS 5268) 2916.2 Design of timber floor joists

(BS 5268) 2936.3 Design of a notched floor joist

(BS 5268) 2966.4 Analysis of a timber roof (BS 5268) 2966.5 Timber column resisting an axial

load (BS 5268) 301

6.6 Timber column resisting an axialload and moment (BS 5268) 302

6.7 Analysis of a stud wall (BS 5268) 304

8.1 Design actions for simply supportedbeam (EN 1990) 321

8.2 Bending reinforcement for a singlyreinforced beam (EC 2) 329

8.3 Bending reinforcement for a doublyreinforced beam (EC 2) 329

8.4 Design of shear reinforcement for abeam (EC 2) 334

8.5 Design of shear reinforcement atbeam support (EC 2) 335

8.6 Deflection check for concrete beams(EC 2) 338

8.7 Calculation of anchorage lengths(EC 2) 342

8.8 Design of a simply supportedbeam (EC 2) 345

8.9 Analysis of a singly reinforcedbeam (EC 2) 349

8.10 Design of a one-way spanningfloor (EC 2) 352

8.11 Analysis of one-way spanningfloor (EC 2) 355

8.12 Design of a pad foundation (EC 2) 3598.13 Column supporting an axial load

and uni-axial bending (EC 2) 3668.14 Classification of a column (EC 2) 3678.15 Classification of a column (EC 2) 3698.16 Column design (i) λ < λlim;

(ii) λ > λlim (EC 2) 3708.17 Column subjected to combined

axial load and biaxial bending(EC 2) 373

9.1 Analysis of a laterally restrainedbeam (EC 3) 384

9.2 Design of a laterally restrainedbeam (EC 3) 387

9.3 Design of a cantilever beam(EC 3) 391

9.4 Design of a beam with stiffeners(EC 3) 393

9.5 Analysis of a beam restrained at thesupports (EC 3) 401

9.6 Analysis of a beam restrained atmid-span and supports (EC 3) 402

9.7 Analysis of a column resisting anaxial load (EC 3) 408

9.8 Analysis of a column with a tie-beamat mid-height (EC 3) 410

List of worked examples

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9.9 Analysis of a column resisting anaxial load and moment (EC 3) 411

9.10 Analysis of a steel column in ‘simple’construction (EC 3) 415

9.11 Analysis of a column baseplate(EC 3) 417

9.12 Analysis of a tension splice connection(EC 3) 422

9.13 Shear resistance of a welded end plateto beam connection (EC 3) 424

9.14 Bolted beam-to-column connectionusing an end plate (EC 3) 426

9.15 Bolted beam-to-column connectionusing web cleats (EC 3) 429

10.1 Design of a loadbearing brickwall (EC 6) 444

10.2 Design of a brick wall with ‘small’plan area (EC 6) 446

10.3 Analysis of brick walls stiffened withpiers (EC 6) 447

10.4 Design of a cavity wall (EC 6) 45010.5 Block wall subject to axial load

and wind (EC 6) 45310.6 Analysis of a one-way spanning

wall panel (EC 6) 45610.7 Analysis of a two-way spanning

panel wall (EC 6) 457

11.1 Design of timber floor joists (EC 5) 46911.2 Design of a notched floor joist

(EC 5) 47511.3 Analysis of a solid timber beam

restrained at supports (EC 5) 47611.4 Analysis of a column resisting an

axial load (EC 5) 47811.5 Analysis of an eccentrically loaded

column (EC 5) 479

List of worked examples

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DedicationIn memory of Biji

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PART ONE

INTRODUCTIONTO STRUCTURALDESIGN

The primary aim of all structural design is toensure that the structure will perform satisfactorilyduring its design life. Specifically, the designer mustcheck that the structure is capable of carrying theloads safely and that it will not deform excessivelydue to the applied loads. This requires the de-signer to make realistic estimates of the strengths ofthe materials composing the structure and the load-ing to which it may be subject during its designlife. Furthermore, the designer will need a basicunderstanding of structural behaviour.

The work that follows has two objectives:

1. to describe the philosophy of structural design;2. to introduce various aspects of structural and

material behaviour.

Towards the first objective, Chapter 1 discusses thethree main philosophies of structural design, emphas-izing the limit state philosophy which forms the basesof design in many of the modern codes of practice.Chapter 2 then outlines a method of assessing thedesign loading acting on individual elements of astructure and how this information can be used, to-gether with the material properties, to size elements.

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Philosophy of design

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Chapter 1


This chapter is concerned with the philosophy of struc-tural design. The chapter describes the overall aims ofdesign and the many inputs into the design process.The primary aim of design is seen as the need to ensurethat at no point in the structure do the design loadsexceed the design strengths of the materials. This can beachieved by using the permissible stress or load factorphilosophies of design. However, both suffer from draw-backs and it is more common to design according tolimit state principles which involve considering all themechanisms by which a structure could become unfitfor its intended purpose during its design life.

1.1 IntroductionThe task of the structural engineer is to design astructure which satisfies the needs of the client andthe user. Specifically the structure should be safe,economical to build and maintain, and aesthetic-ally pleasing. But what does the design processinvolve?

Design is a word that means different things todifferent people. In dictionaries the word is de-scribed as a mental plan, preliminary sketch, pat-tern, construction, plot or invention. Even amongthose closely involved with the built environmentthere are considerable differences in interpretation.Architects, for example, may interpret design asbeing the production of drawings and models toshow what a new building will actually look like.To civil and structural engineers, however, design istaken to mean the entire planning process for a newbuilding structure, bridge, tunnel, road, etc., fromoutline concepts and feasibility studies throughmathematical calculations to working drawingswhich could show every last nut and bolt in theproject. Together with the drawings there will bebills of quantities, a specification and a contract,which will form the necessary legal and organiza-tional framework within which a contractor, under

the supervision of engineers and architects, can con-struct the scheme.

There are many inputs into the engineeringdesign process as illustrated by Fig. 1.1 including:

1. client brief2. experience3. imagination4. a site investigation5. model and laboratory tests6. economic factors7. environmental factors.

The starting-point for the designer is normallya conceptual brief from the client, who may be aprivate developer or perhaps a government body.The conceptual brief may simply consist of somesketches prepared by the client or perhaps a detailedset of architect’s drawings. Experience is cruciallyimportant, and a client will always demand thatthe firm he is employing to do the design has pre-vious experience designing similar structures.

Although imagination is thought by some tobe entirely the domain of the architect, this is notso. For engineers and technicians an imaginationof how elements of structure interrelate in threedimensions is essential, as is an appreciation ofthe loadings to which structures might be subjectin certain circumstances. In addition, imaginativesolutions to engineering problems are often requiredto save money, time, or to improve safety or quality.

A site investigation is essential to determine thestrength and other characteristics of the groundon which the structure will be founded. If the struc-ture is unusual in any way, or subject to abnormalloadings, model or laboratory tests may also be usedto help determine how the structure will behave.

In today’s economic climate a structural designermust be constantly aware of the cost implicationsof his or her design. On the one hand design shouldaim to achieve economy of materials in the struc-ture, but over-refinement can lead to an excessive

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number of different sizes and components in thestructure, and labour costs will rise. In additionthe actual cost of the designer’s time should not beexcessive, or this will undermine the employer’scompetitiveness. The idea is to produce a workabledesign achieving reasonable economy of materials,while keeping manufacturing and construction costsdown, and avoiding unnecessary design and researchexpenditure. Attention to detailing and buildabilityof structures cannot be overemphasized in design.Most failures are as a result of poor detailing ratherthan incorrect analysis.

Designers must also understand how the struc-ture will fit into the environment for which it isdesigned. Today many proposals for engineeringstructures stand or fall on this basis, so it is part ofthe designer’s job to try to anticipate and recon-cile the environmental priorities of the public andgovernment.

The engineering design process can often bedivided into two stages: (1) a feasibility study in-volving a comparison of the alternative forms ofstructure and selection of the most suitable type and(2) a detailed design of the chosen structure. Thesuccess of stage 1, the conceptual design, reliesto a large extent on engineering judgement andinstinct, both of which are the outcome of manyyears’ experience of designing structures. Stage 2,the detailed design, also requires these attributesbut is usually more dependent upon a thoroughunderstanding of the codes of practice for struc-tural design, e.g. BS 8110 and BS 5950. Thesedocuments are based on the amassed experience of

many generations of engineers, and the results ofresearch. They help to ensure safety and economyof construction, and that mistakes are not repeated.For instance, after the infamous disaster at theRonan Point block of flats in Newham, London,when a gas explosion caused a serious partial col-lapse, research work was carried out, and codes ofpractice were amended so that such structures couldsurvive a gas explosion, with damage being con-fined to one level.

The aim of this book is to look at the proceduresassociated with the detailed design of structuralelements such as beams, columns and slabs. Chap-ter 2 will help the reader to revise some basic the-ories of structural behaviour. Chapters 3–6 deal withdesign to British Standard (BS) codes of practicefor the structural use of concrete (BS 8110), struc-tural steelwork (BS 5950), masonry (BS 5628) andtimber (BS 5268). Chapter 7 introduces the newEurocodes (EC) for structural design and Chapters8–11 then describe the layout and design principlesin EC2, EC3, EC6 and EC5 for concrete, steel-work, masonry and timber respectively.

1.2 Basis of designTable 1.1 illustrates some risk factors that are asso-ciated with activities in which people engage. Itcan be seen that some degree of risk is associatedwith air and road travel. However, people normallyaccept that the benefits of mobility outweigh therisks. Staying in buildings, however, has always been

Fig. 1.1 Inputs into the design process.

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critical points, as stress due to loading exceeds thestrength of the material. In order for the structureto be safe the overlapping area must be kept to aminimum. The degree of overlap between the twocurves can be minimized by using one of three dis-tinct design philosophies, namely:

1. permissible stress design2. load factor method3. limit state design.

1.2.1 PERMISSIBLE STRESS DESIGNIn permissible stress design, sometimes referred toas modular ratio or elastic design, the stresses in thestructure at working loads are not allowed to exceeda certain proportion of the yield stress of the con-struction material, i.e. the stress levels are limitedto the elastic range. By assuming that the stress–strain relationship over this range is linear, it is pos-sible to calculate the actual stresses in the materialconcerned. Such an approach formed the basis of thedesign methods used in CP 114 (the forerunner ofBS 8110) and BS 449 (the forerunner of BS 5950).

However, although it modelled real building per-formance under actual conditions, this philosophyhad two major drawbacks. Firstly, permissible designmethods sometimes tended to overcomplicate thedesign process and also led to conservative solutions.Secondly, as the quality of materials increased andthe safety margins decreased, the assumption thatstress and strain are directly proportional becameunjustifiable for materials such as concrete, makingit impossible to estimate the true factors of safety.

1.2.2 LOAD FACTOR DESIGNLoad factor or plastic design was developed to takeaccount of the behaviour of the structure once theyield point of the construction material had beenreached. This approach involved calculating thecollapse load of the structure. The working load wasderived by dividing the collapse load by a load factor.This approach simplified methods of analysis andallowed actual factors of safety to be calculated.It was in fact permitted in CP 114 and BS 449but was slow in gaining acceptance and was even-tually superseded by the more comprehensive limitstate approach.

The reader is referred to Appendix A for an ex-ample illustrating the differences between the per-missible stress and load factor approaches to design.

1.2.3 LIMIT STATE DESIGNOriginally formulated in the former Soviet Unionin the 1930s and developed in Europe in the 1960s,

Table 1.1 Comparative death risk per 108

persons exposed

Mountaineering (international) 2700Air travel (international) 120Deep water trawling 59Car travel 56Coal mining 21Construction sites 8Manufacturing 2Accidents at home 2Fire at home 0.1Structural failures 0.002

Fig. 1.2 Relationship between stress and strength.

regarded as fairly safe. The risk of death or injurydue to structural failure is extremely low, but as wespend most of our life in buildings this is perhapsjust as well.

As far as the design of structures for safety isconcerned, it is seen as the process of ensuringthat stresses due to loading at all critical points in astructure have a very low chance of exceeding thestrength of materials used at these critical points.Figure 1.2 illustrates this in statistical terms.

In design there exist within the structure a numberof critical points (e.g. beam mid-spans) where thedesign process is concentrated. The normal distribu-tion curve on the left of Fig. 1.2 represents the actualmaximum material stresses at these critical pointsdue to the loading. Because loading varies accordingto occupancy and environmental conditions, andbecause design is an imperfect process, the materialstresses will vary about a modal value – the peak ofthe curve. Similarly the normal distribution curveon the right represents material strengths at thesecritical points, which are also not constant due tothe variability of manufacturing conditions.

The overlap between the two curves represents apossibility that failure may take place at one of the

Basis of design

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limit state design can perhaps be seen as a com-promise between the permissible and load factormethods. It is in fact a more comprehensive ap-proach which takes into account both methods inappropriate ways. Most modern structural codes ofpractice are now based on the limit state approach.BS 8110 for concrete, BS 5950 for structuralsteelwork, BS 5400 for bridges and BS 5628 formasonry are all limit state codes. The principalexceptions are the code of practice for design intimber, BS 5268, and the old (but still current)structural steelwork code, BS 449, both of whichare permissible stress codes. It should be noted, how-ever, that the Eurocode for timber (EC5), which isexpected to replace BS 5268 around 2010, is basedon limit state principles.

As limit state philosophy forms the basis of thedesign methods in most modern codes of practicefor structural design, it is essential that the designmethodology is fully understood. This then is thepurpose of the following subsections.

1.2.3.1 Ultimate and serviceabilitylimit states

The aim of limit state design is to achieve accept-able probabilities that a structure will not becomeunfit for its intended use during its design life, thatis, the structure will not reach a limit state. Thereare many ways in which a structure could becomeunfit for use, including excessive conditions of bend-ing, shear, compression, deflection and cracking(Fig. 1.3). Each of these mechanisms is a limit statewhose effect on the structure must be individuallyassessed.

Some of the above limit states, e.g. deflectionand cracking, principally affect the appearance ofthe structure. Others, e.g. bending, shear and com-pression, may lead to partial or complete collapseof the structure. Those limit states which can causefailure of the structure are termed ultimate limitstates. The others are categorized as serviceabilitylimit states. The ultimate limit states enable thedesigner to calculate the strength of the structure.Serviceability limit states model the behaviour of thestructure at working loads. In addition, there maybe other limit states which may adversely affectthe performance of the structure, e.g. durabilityand fire resistance, and which must therefore alsobe considered in design.

It is a matter of experience to be able to judgewhich limit states should be considered in thedesign of particular structures. Nevertheless, oncethis has been done, it is normal practice to base

the design on the most critical limit state and thencheck for the remaining limit states. For example,for reinforced concrete beams the ultimate limitstates of bending and shear are used to size thebeam. The design is then checked for the remain-ing limit states, e.g. deflection and cracking. Onthe other hand, the serviceability limit state ofdeflection is normally critical in the design of con-crete slabs. Again, once the designer has determineda suitable depth of slab, he/she must then makesure that the design satisfies the limit states of bend-ing, shear and cracking.

In assessing the effect of a particular limit stateon the structure, the designer will need to assumecertain values for the loading on the structure andthe strength of the materials composing the struc-ture. This requires an understanding of the con-cepts of characteristic and design values which arediscussed below.

1.2.3.2 Characteristic and design valuesAs stated at the outset, when checking whether aparticular member is safe, the designer cannot becertain about either the strength of the materialcomposing the member or, indeed, the load whichthe member must carry. The material strength maybe less than intended (a) because of its variablecomposition, and (b) because of the variability ofmanufacturing conditions during construction, andother effects such as corrosion. Similarly the loadin the member may be greater than anticipated (a)because of the variability of the occupancy or envir-onmental loading, and (b) because of unforeseencircumstances which may lead to an increase in thegeneral level of loading, errors in the analysis, errorsduring construction, etc.

In each case, item (a) is allowed for by using acharacteristic value. The characteristic strengthis the value below which the strength lies in only asmall number of cases. Similarly the characteristicload is the value above which the load lies in onlya small percentage of cases. In the case of strengththe characteristic value is determined from test re-sults using statistical principles, and is normallydefined as the value below which not more than5% of the test results fall. However, at this stagethere are insufficient data available to apply statist-ical principles to loads. Therefore the characteristicloads are normally taken to be the design loadsfrom other codes of practice, e.g. BS 648 and BS6399.

The overall effect of items under (b) is allowedfor using a partial safety factor: γm for strength

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Fig. 1.3 Typical modes of failure for beams and columns.

Basis of design

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and γ f for load. The design strength is obtained bydividing the characteristic strength by the partialsafety factor for strength:

Design strength

characteristic strength

m

=γ

(1.1)

The design load is obtained by multiplying thecharacteristic load by the partial safety factor forload:

Design load = characteristic load × γ f (1.2)

The value of γm will depend upon the propertiesof the actual construction material being used.Values for γ f depend on other factors which will bediscussed more fully in Chapter 2.

In general, once a preliminary assessment of thedesign loads has been made it is then possible tocalculate the maximum bending moments, shearforces and deflections in the structure (Chapter 2).The construction material must be capable ofwithstanding these forces otherwise failure of thestructure may occur, i.e.

Design strength ≥ design load (1.3)

Simplified procedures for calculating the moment,shear and axial load capacities of structural ele-ments together with acceptable deflection limitsare described in the appropriate codes of practice.

These allow the designer to rapidly assess the suit-ability of the proposed design. However, beforediscussing these procedures in detail, Chapter 2describes in general terms how the design loadsacting on the structure are estimated and used tosize individual elements of the structure.

1.3 SummaryThis chapter has examined the bases of threephilosophies of structural design: permissible stress,load factor and limit state. The chapter has con-centrated on limit state design since it forms thebasis of the design methods given in the codes ofpractice for concrete (BS 8110), structural steel-work (BS 5950) and masonry (BS 5628). The aimof limit state design is to ensure that a structurewill not become unfit for its intended use, that is,it will not reach a limit state during its design life.Two categories of limit states are examined indesign: ultimate and serviceability. The former isconcerned with overall stability and determiningthe collapse load of the structure; the latter exam-ines its behaviour under working loads. Structuraldesign principally involves ensuring that the loadsacting on the structure do not exceed its strengthand the first step in the design process then is toestimate the loads acting on the structure.

Questions1. Explain the difference between conceptual

design and detailed design.2. What is a code of practice and what is its

purpose in structural design?3. List the principal sources of uncertainty in

structural design and discuss how theseuncertainties are rationally allowed for indesign.

4. The characteristic strengths and designstrengths are related via the partial safetyfactor for materials. The partial safetyfactor for concrete is higher than for steelreinforcement. Discuss why this should be so.

5. Describe in general terms the ways inwhich a beam and column could becomeunfit for use.

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Chapter 2

Basic structuralconcepts andmaterial properties

This chapter is concerned with general methods ofsizing beams and columns in structures. The chapterdescribes how the characteristic and design loads actingon structures and on the individual elements are deter-mined. Methods of calculating the bending moments,shear forces and deflections in beams are outlined.Finally, the chapter describes general approaches tosizing beams according to elastic and plastic criteriaand sizing columns subject to axial loading.

2.1 IntroductionAll structures are composed of a number of inter-connected elements such as slabs, beams, columns,walls and foundations. Collectively, they enable theinternal and external loads acting on the structureto be safely transmitted down to the ground. Theactual way that this is achieved is difficult to modeland many simplifying, but conservative, assump-tions have to be made. For example, the degreeof fixity at column and beam ends is usually uncer-tain but, nevertheless, must be estimated as itsignificantly affects the internal forces in the element.Furthermore, it is usually assumed that the reactionfrom one element is a load on the next and thatthe sequence of load transfer between elementsoccurs in the order: ceiling/floor loads to beams tocolumns to foundations to ground (Fig. 2.1).

At the outset, the designer must make an assess-ment of the future likely level of loading, includingself-weight, to which the structure may be subjectduring its design life. Using computer methods orhand calculations the design loads acting on indi-vidual elements can then be evaluated. The designloads are used to calculate the bending moments,shear forces and deflections at critical points alongthe elements. Finally, suitable dimensions for theelement can be determined. This aspect requiresan understanding of the elementary theory ofbending and the behaviour of elements subject to

Fig. 2.1 Sequence of load transfer between elements of astructure.

compressive loading. These steps are summarizedin Fig. 2.2 and the following sections describe theprocedures associated with each step.

2.2 Design loads acting onstructures

The loads acting on a structure are divided intothree basic types: dead, imposed and wind. Foreach type of loading there will be characteristic anddesign values, as discussed in Chapter 1, which mustbe estimated. In addition, the designer will have todetermine the particular combination of loadingwhich is likely to produce the most adverse effecton the structure in terms of bending moments,shear forces and deflections.

2.2.1 DEAD LOADS, G k, gkDead loads are all the permanent loads acting onthe structure including self-weight, finishes, fixturesand partitions. The characteristic dead loads can be

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Basic structural concepts and material properties

10

Fig. 2.2 Design process.

Example 2.1 Self-weight of a reinforced concrete beamCalculate the self-weight of a reinforced concrete beam of breadth 300 mm, depth 600 mm and length 6000 mm.

From Table 2.1, unit mass of reinforced concrete is 2400 kg m−3. Assuming that the gravitational constant is10 m s−2 (strictly 9.807 m s−2), the unit weight of reinforced concrete, ρ, is

ρ = 2400 × 10 = 24 000 N m−3 = 24 kN m−3

Hence, the self-weight of beam, SW, is

SW = volume × unit weight

= (0.3 × 0.6 × 6)24 = 25.92 kN

estimated using the schedule of weights of buildingmaterials given in BS 648 (Table 2.1) or from manu-facturers’ literature. The symbols Gk and gk arenormally used to denote the total and uniformlydistributed characteristic dead loads respectively.

Estimation of the self-weight of an element tendsto be a cyclic process since its value can only beassessed once the element has been designed whichrequires prior knowledge of the self-weight of theelement. Generally, the self-weight of the elementis likely to be small in comparison with other deadand live loads and any error in estimation will tendto have a minimal effect on the overall design(Example 2.1).

2.2.2 IMPOSED LOADS Q k, qkImposed load, sometimes also referred to as liveload, represents the load due to the proposed oc-cupancy and includes the weights of the occupants,furniture and roof loads including snow. Sinceimposed loads tend to be much more variablethan dead loads they are more difficult to predict.

BS 6399: Part 1: 1984: Code of Practice for Dead andImposed Loads gives typical characteristic imposedfloor loads for different classes of structure, e.g.residential dwellings, educational institutions,hospitals, and parts of the same structure, e.g.balconies, corridors and toilet rooms (Table 2.2).

2.2.3 WIND LOADSWind pressure can either add to the other gravita-tional forces acting on the structure or, equallywell, exert suction or negative pressures on thestructure. Under particular situations, the latter maywell lead to critical conditions and must be con-sidered in design. The characteristic wind loadsacting on a structure can be assessed in accordancewith the recommendations given in CP 3: ChapterV: Part 2: 1972 Wind Loads or Part 2 of BS 6399:Code of Practice for Wind Loads.

Wind loading is important in the design of ma-sonry panel walls (Chapter 5). However beyond that,wind loading is not considered further since the em-phasis in this book is on the design of elements rather

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Table 2.1 Schedule of unit masses of building materials (based on BS 648)

AsphaltRoofing 2 layers, 19 mm thick 42 kg m−2

Damp-proofing, 19 mm thick 41 kg m−2

Roads and footpaths, 19 mm thick 44 kg m−2

Bitumen roofing feltsMineral surfaced bitumen 3.5 kg m−2

BlockworkSolid per 25 mm thick, stone 55 kg m−2

aggregateAerated per 25 mm thick 15 kg m−2

BoardBlockboard per 25 mm thick 12.5 kg m−2

BrickworkClay, solid per 25 mm thick 55 kg m−2

medium densityConcrete, solid per 25 mm thick 59 kg m−2

Cast stone 2250 kg m−3

ConcreteNatural aggregates 2400 kg m−3

Lightweight aggregates (structural) 1760 + 240/−160 kg m−3

FlagstonesConcrete, 50 mm thick 120 kg m−2

Glass fibreSlab, per 25 mm thick 2.0–5.0 kg m−2

Gypsum panels and partitionsBuilding panels 75 mm thick 44 kg m−2

LeadSheet, 2.5 mm thick 30 kg m−2

Linoleum3 mm thick 6 kg m−2

loads, γ f (Chapter 1). The value for γ f depends onseveral factors including the limit state underconsideration, i.e. ultimate or serviceability, theaccuracy of predicting the load and the particu-lar combination of loading which will produce theworst possible effect on the structure in terms ofbending moments, shear forces and deflections.

than structures, which generally involves investigat-ing the effects of dead and imposed loads only.

2.2.4 LOAD COMBINATIONS ANDDESIGN LOADS

The design loads are obtained by multiplying thecharacteristic loads by the partial safety factor for

PlasterTwo coats gypsum, 13 mm thick 22 kg m−2

Plastics sheeting (corrugated) 4.5 kg m−2

Plywoodper mm thick 0.7 kg m−2

Reinforced concrete 2400 kg m−3

RenderingCement: sand (1:3), 13 mm thick 30 kg m−2

ScreedingCement: sand (1:3), 13 mm thick 30 kg m−2

Slate tiles(depending upon thickness 24 –78 kg m−3

and source)

SteelSolid (mild) 7850 kg m−3

Corrugated roofing sheets, 10 kg m−2

per mm thick

Tarmacadam25 mm thick 60 kg m−2

Terrazzo25 mm thick 54 kg m−2

Tiling, roofClay 70 kg m−2

TimberSoftwood 590 kg m−3

Hardwood 1250 kg m−3

Water 1000 kg m−3

WoodwoolSlabs, 25 mm thick 15 kg m−2

Design loads acting on structures

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Table 2.2 Imposed loads for residential occupancy class

Floor area usage Intensity of distributed load Concentrated loadkN m−2 kN

Type 1. Self-contained dwelling unitsAll 1.5 1.4

Type 2. Apartment houses, boarding houses, lodginghouses, guest houses, hostels, residential clubs andcommunal areas in blocks of flatsBoiler rooms, motor rooms, fan rooms and the like 7.5 4.5

including the weight of machineryCommunal kitchens, laundries 3.0 4.5Dining rooms, lounges, billiard rooms 2.0 2.7Toilet rooms 2.0 –Bedrooms, dormitories 1.5 1.8Corridors, hallways, stairs, landings, footbridges, etc. 3.0 4.5Balconies Same as rooms to which 1.5 per metre run

they give access but with concentrated ata minimum of 3.0 the outer edge

Cat walks – 1.0 at 1 m centres

Type 3. Hotels and motelsBoiler rooms, motor rooms, fan rooms and the like, 7.5 4.5

including the weight of machineryAssembly areas without fixed seating, dance halls 5.0 3.6Bars 5.0 –Assembly areas with fixed seatinga 4.0 –Corridors, hallways, stairs, landings, footbridges, etc. 4.0 4.5Kitchens, laundries 3.0 4.5Dining rooms, lounges, billiard rooms 2.0 2.7Bedrooms 2.0 1.8Toilet rooms 2.0 –Balconies Same as rooms to which 1.5 per metre run

they give access but with concentrated at thea minimum of 4.0 outer edge

Cat walks – 1.0 at 1 m centres

Note. a Fixed seating is seating where its removal and the use of the space for other purposes are improbable.

In most of the simple structures which will beconsidered in this book, the worst possible com-bination will arise due to the maximum dead andmaximum imposed loads acting on the structuretogether. In such cases, the partial safety factors fordead and imposed loads are 1.4 and 1.6 respect-ively (Fig. 2.3) and hence the design load is given by

Fig. 2.3

Design load = 1.4Gk + 1.6Q k

However, it should be appreciated that theoret-ically the design dead loads can vary between thecharacteristic and ultimate values, i.e. 1.0Gk and1.4Gk. Similarly, the design imposed loads canvary between zero and the ultimate value, i.e. 0.0Qkand 1.6Qk. Thus for a simply supported beam withan overhang (Fig. 2.4(a)) the load cases shown inFigs 2.4(b)–(d) will need to be considered in orderto determine the design bending moments and shearforces in the beam.

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13

Fig. 2.4

Design loads acting on elements

Fig. 2.5 Typical beams and column support conditions.

commonly assumed support conditions at the endsof beams and columns respectively.

In design it is common to assume that all the jointsin the structure are pinned and that the sequence ofload transfer occurs in the order: ceiling/floor loads tobeams to columns to foundations to ground. Theseassumptions will considerably simplify calculationsand lead to conservative estimates of the designloads acting on individual elements of the struc-ture. The actual calculations to determine the forcesacting on the elements are best illustrated by anumber of worked examples as follows.

2.3 Design loads acting onelements

Once the design loads acting on the structure havebeen estimated it is then possible to calculate thedesign loads acting on individual elements. As waspointed out at the beginning of this chapter, thisusually requires the designer to make assumptionsregarding the support conditions and how the loadswill eventually be transmitted down to the ground.Figures 2.5(a) and (b) illustrate some of the more

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Example 2.2 Design loads on a floor beamA composite floor consisting of a 150 mm thick reinforced concrete slab supported on steel beams spanning 5 m andspaced at 3 m centres is to be designed to carry an imposed load of 3.5 kN m−2. Assuming that the unit mass of thesteel beams is 50 kg m−1 run, calculate the design loads on a typical internal beam.

UNIT WEIGHTS OF MATERIALSReinforced concreteFrom Table 2.1, unit mass of reinforced concrete is 2400 kg m−3. Assuming the gravitational constant is 10 m s−2, theunit weight of reinforced concrete is

2400 × 10 = 24 000 N m−3 = 24 kN m−3

Steel beamsUnit mass of beam = 50 kg m−1 runUnit weight of beam = 50 × 10 = 500 N m−1 run = 0.5 kN m−1 run

LOADINGSlab

Slab dead load ( g k) = self-weight = 0.15 × 24 = 3.6 kN m−2Slab imposed load (qk) = 3.5 kN m−2Slab ultimate load = 1.4g k + 1.6q k = 1.4 × 3.6 + 1.6 × 3.5

= 10.64 kN m−2

BeamBeam dead load ( gk) = self-weight = 0.5 kN m−1 runBeam ultimate load = 1.4g k = 1.4 × 0.5 = 0.7 kN m−1 run

DESIGN LOADEach internal beam supports a uniformly distributed load from a 3 m width of slab (hatched \\\\\\\) plus self-weight.Hence

Design load on beam = slab load + self-weight of beam

= 10.64 × 5 × 3 + 0.7 × 5

= 159.6 + 3.5 = 163.1 kN

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Design loads acting on elements

Example 2.3 Design loads on floor beams and columnsThe floor shown below with an overall depth of 225 mm is to be designed to carry an imposed load of 3 kN m−2 plusfloor finishes and ceiling loads of 1 kN m−2. Calculate the design loads acting on beams B1–C1, B2–C2 and B1–B3 andcolumns B1 and C1. Assume that all the column heights are 3 m and that the beam and column weights are 70 and60 kg m−1 run respectively.

UNIT WEIGHTS OF MATERIALSReinforced concreteFrom Table 2.1, unit mass of reinforced concrete is 2400 kg m−3. Assuming the gravitational constant is 10 m s−2, theunit weight of reinforced concrete is

2400 × 10 = 24 000 N m−3 = 24 kN m−3

Steel beamsUnit mass of beam = 70 kg m−1 runUnit weight of beam = 70 × 10 = 700 N m−1 run = 0.7 kN m−1 run

Steel columnsUnit mass of column = 60 kg m−1 runUnit weight of column = 60 × 10 = 600 N m−1 run = 0.6 kN m−1 run

LOADINGSlab

Slab dead load (g k) = self-weight + finishes= 0.225 × 24 + 1 = 6.4 kN m−2

Slab imposed load (q k) = 3 kN m−2Slab ultimate load = 1.4g k + 1.6q k

= 1.4 × 6.4 + 1.6 × 3 = 13.76 kN m−2

BeamBeam dead load (g k) = self-weight = 0.7 kN m−1 runBeam ultimate load = 1.4g k = 1.4 × 0.7 = 0.98 kN m−1 run

ColumnColumn dead load (g k) = 0.6 kN m−1 runColumn ultimate load = 1.4g k = 1.4 × 0.6 = 0.84 kN m−1 run

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DESIGN LOADS

Beam B1–C1Assuming that the slab is simply supported, beam B1–C1 supports a uniformly distributed load from a 1.5 m width ofslab (hatched //////) plus self-weight of beam. Hence

Design load on beam B1–C1 = slab load + self-weight of beam

= 13.76 × 6 × 1.5 + 0.98 × 6

= 123.84 + 5.88 = 129.72 kN

Since the beam is symmetrically loaded, RB2 and RC2 are the same and equal to 253.56/2 = 126.78 kN.

Beam B1–B3Assuming that the slab is simply supported, beam B1–B3 supports a uniformly distributed load from a 1.5 m width ofslab (shown cross-hatched) plus the self-weight of the beam and the reaction transmitted from beam B2–C2 whichacts as a point load at mid-span. Hence

Design load on beam B1–B3 = uniformly distributed load from slab plus self-weight of beam+ point load from reaction RB2

= (13.76 × 1.5 × 6 + 0.98 × 6) + 126.78

= 129.72 + 126.78 = 256.5 kN

Since the beam is symmetrically loaded,

RB1 = RC1 = 129.72/2 = 64.86 kN

Beam B2–C2Assuming that the slab is simply supported, beam B2–C2 supports a uniformly distributed load from a 3 m width ofslab (hatched \\\\\) plus its self-weight. Hence

Design load on beam B2–C2 = slab load + self-weight of beam

= 13.76 × 6 × 3 + 0.98 × 6

= 247.68 + 5.88 = 253.56 kN

Example 2.3 continued

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Structural analysis

Since the beam is symmetrically loaded,

RB1 = RB3 = 256.5/2 = 128.25 kN

Column B1

2.4 Structural analysisThe design axial loads can be used directly to size col-umns. Column design will be discussed more fullyin section 2.5. However, before flexural memberssuch as beams can be sized, the design bendingmoments and shear forces must be evaluated.Such calculations can be performed by a variety ofmethods as noted below, depending upon the com-plexity of the loading and support conditions:

1. equilibrium equations2. formulae3. computer methods.

Hand calculations are suitable for analysingstatically determinate structures such as simply sup-ported beams and slabs (section 2.4.1). For variousstandard load cases, formulae for calculating themaximum bending moments, shear forces anddeflections are available which can be used to rapidly

Column C1 supports the reactions from beams B1–C1 and C1–C3 and its self-weight. From the above, the reaction atC1 due to beam B1–C1 is 64.86 kN. Beam C1–C3 supports the reactions from B2–C2 (= 126.78 kN) and its self-weight (= 0.98 × 6) = 5.88 kN. Hence the reaction at C1 is (126.78 + 5.88)/2 = 66.33 kN. Since the column heightis 3 m, self-weight = 0.84 × 3 = 2.52 kN. Hence

Design load on column C1 = 64.86 + 66.33 + 2.52 = 133.71 kN

Column B1 supports the reactions from beams A1–B1, B1–C1 and B1–B3 and its self-weight. From the above, thereaction at B1 due to beam B1–C1 is 64.86 kN and from beam B1–B3 is 128.25 kN. Beam A1–B1 supports only itsself-weight = 0.98 × 3 = 2.94 kN. Hence reaction at B1 due to A1–B1 is 2.94/2 = 1.47 kN. Since the column height is3 m, self-weight of column = 0.84 × 3 = 2.52 kN. Hence

Design load on column B1 = 64.86 + 128.25 + 1.47 + 2.52

= 197.1 kN

Column C1


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analyse beams, as will be discussed in section 2.4.2.Alternatively, the designer may resort to using vari-ous commercially available computer packages, e.g.SAND. Their use is not considered in this book.

2.4.1 EQUILIBRIUM EQUATIONSIt can be demonstrated that if a body is in equilib-rium under the action of a system of external forces,all parts of the body must also be in equilibrium.

This principle can be used to determine the bend-ing moments and shear forces along a beam. Theactual procedure simply involves making fictitious‘cuts’ at intervals along the beam and applying theequilibrium equations given below to the cut por-tions of the beam.

Σ moments (M) = 0 (2.1)

Σ vertical forces (V ) = 0 (2.2)

From equation 2.1, taking moments about Z gives

126.78 × 1 − 42.26 × 1 × 0.5 − Mx=1 = 0

Hence

Mx=1 = 105.65 kN m

From equation 2.2, summing the vertical forces gives

126.78 − 42.26 × 1 − Vx=1 = 0

Example 2.4 Design moments and shear forces in beams usingequilibrium equations

Calculate the design bending moments and shear forces in beams B2–C2 and B1–B3 of Example 2.3.

BEAM B2–C2

Let the longitudinal centroidal axis of the beam be the x axis and x = 0 at support B2.

x ===== 0By inspection,

Moment at x = 0 (Mx=0) = 0

Shear force at x = 0 (Vx=0) = RB2 = 126.78 kN

x ===== 1Assuming that the beam is cut 1 m from support B2, i.e. x = 1 m, the moments and shear forces acting on the cutportion of the beam will be those shown in the free body diagram below:

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Structural analysis


126.78 × 2 − 42.26 × 2 × 1 − Mx=2 = 0

Hence

Mx=2 = 169.04 kN m

From equation 2.2, summing the vertical forces gives

126.78 − 42.26 × 2 − Vx=2 = 0

Hence

Vx=2 = 42.26 kN

If this process is repeated for values of x equal to 3, 4, 5 and 6 m, the following values of the moments and shearforces in the beam will result:

x(m) 0 1 2 3 4 5 6

M(kN m) 0 105.65 169.04 190.17 169.04 105.65 0V(kN ) 126.78 84.52 42.26 0 −42.26 −84.52 −126.78

This information is better presented diagrammatically as shown below:

Example 2.4 continuedHence

Vx=1 = 84.52 kN

x ===== 2The free body diagram for the beam, assuming that it has been cut 2 m from support B2, is shown below:

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20

Hence, the design moment (M) is 190.17 kN m. Note that this occurs at mid-span and coincides with the point ofzero shear force. The design shear force (V ) is 126.78 kN and occurs at the supports.

BEAM B1–B3

In (a) it is assumed that the beam is cut immediately to the left of the point load. In (b) the beam is cut immediatelyto the right of the point load. In (a), from equation 2.2, the shear force to the left of the cut, Vx=3,L, is given by

128.25 − 64.86 − Vx=3,L = 0

Hence

Vx=3,L = 63.39 kN


128.25 × 3 − 64.86 × 3/2 − Mx=3 = 0

Hence

Mx=3 = 287.46 kN m

In order to determine the shear force at x = 3 the following two load cases need to be considered:

Again, let the longitudinal centroidal axis of the beam be the x axis of the beam and set x = 0 at support B1.The steps outlined earlier can be used to determine the bending moments and shear forces at x = 0, 1 and 2 m andsince the beam is symmetrically loaded and supported, these values of bending moment and shear forces will applyat x = 4, 5 and 6 m respectively.

The bending moment at x = 3 m can be calculated by considering all the loading immediately to the left of thepoint load as shown below:


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21

Structural analysis

Example 2.4 continuedIn (b), from equation 2.2, the shear force to the right of the cut, Vx=3,R, is given by

128.25 − 64.86 − 126.78 − Vx=3,R = 0

Hence

Vx=3,R = −63.39 kN

Summarising the results in tabular and graphical form gives

x(m) 0 1 2 3 4 5 6

M(kN m) 0 117.44 213.26 287.46 213.26 117.44 0V(kN) 128.25 106.63 85.01 63.39|−63.39 −85.01 −106.63 −128.25

Hence, the design moment for beam B1–B3 is 287.46 kN m and occurs at mid-span and the design shear force is128.25 kN and occurs at the supports.

2.4.2 FORMULAEAn alternative method of determining the designbending moments and shear forces in beams in-volves using the formulae quoted in Table 2.3. Thetable also includes formulae for calculating themaximum deflections for each load case. The for-mulae can be derived using a variety of methods

for analysing statically indeterminate structures, e.g.slope deflection and virtual work, and the inter-ested reader is referred to any standard work onthis subject for background information. There willbe many instances in practical design where theuse of standard formulae is not convenient andequilibrium methods are preferable.

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22

Table 2.3 Bending moments, shear forces and deflections forvarious standard load cases

Loading Maximum Maximum Maximumbending shearing deflectionmoment force

WLEI

3

8

WL2

W

W2

WLEI

3

48

WL4

231296

3WLEI

WL6

W2

11768

3WLEI

W2

WL8

5384

3WLEI

W2

WL8

WL W

WLEI

3

3

WLEI

3

60

W2

WL6

WLEI

3

192(at supportsand at midspan)

WL8

W2

WL12

W2

WLEI

3

384at supports

WL24

at midspan

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23

Structural analysis

This load case can be solved using the principle of superposition which can be stated in general terms as follows: ‘Theeffect of several actions taking place simultaneously can be reproduced exactly by adding the effects of each caseseparately.’ Thus, the loading on beam B1–B3 can be considered to be the sum of a uniformly distributed load (Wudl)of 129.72 kN and a point load (Wpl) at mid-span of 126.78 kN.

By inspection, maximum moment and maximum shear force occur at beam mid-span and supports respectively. FromTable 2.3, design moment, M, is given by

M

W l

. = =

×8

253 56 68

= 190.17 kN m

and design shear force, V, is given by

V

W

. . = = =

2253 56

2126 78 kN

BEAM B1–B3

By inspection, the maximum bending moment and shear force for both load cases occur at beam mid-span andsupports respectively. Thus, the design moment, M, is given by

M

W l W l

.

. . = + =

×+

×=udl pl kN m

8 4129 72 6

8126 78 6

4287 46

and the design shear force, V, is given by

V

W W

.

. . = + = + =udl pl kN

2 2129 72

2126 78

2128 25

Example 2.5 Design moments and shear forces in beams using formulaeRepeat Example 2.4 using the formulae given in Table 2.3.

BEAM B2–C2

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24

2.5 Beam designHaving calculated the design bending moment andshear force, all that now remains to be done is toassess the size and strength of beam required.Generally, the ultimate limit state of bending willbe critical for medium-span beams which are mod-erately loaded and shear for short-span beamswhich are heavily loaded. For long-span beams theserviceability limit state of deflection may well becritical. Irrespective of the actual critical limitstate, once a preliminary assessment of the sizeand strength of beam needed has been made, it mustbe checked for the remaining limit states that mayinfluence its long-term integrity.

The processes involved in such a selection willdepend on whether the construction material be-haves (i) elastically or (ii) plastically. If the materialis elastic, it obeys Hooke’s law, that is, the stressin the material due to the applied load is directlyproportional to its strain (Fig. 2.6) where

Stress (σ) =

forcearea

Fig. 2.7 Strain and stress in an elastic beam.

and Strain (ε) =

change in lengthoriginal length

The slope of the graph of stress vs. strain (Fig. 2.6)is therefore constant, and this gradient is normallyreferred to as the elastic or Young’s modulus andis denoted by the letter E. It is given by

Young’s modulus (E) =

stressstrain

Note that strain is dimensionless but that bothstress and Young’s modulus are usually expressedin N mm−2.

A material is said to be plastic if it strains withouta change in stress. Plasticine and clay are plasticmaterials but so is steel beyond its yield point(Fig. 2.6). As will be seen in Chapter 3, reinforcedconcrete design also assumes that the materialbehaves plastically.

The structural implications of elastic and plasticbehaviour are best illustrated by considering howbending is resisted by the simplest of beams – arectangular section b wide and d deep.

2.5.1 ELASTIC CRITERIAWhen any beam is subject to load it bends as shownin Fig. 2.7(a). The top half of the beam is putinto compression and the bottom half into tension.In the middle, there is neither tension nor com-pression. This axis is normally termed the neutralaxis.

If the beam is elastic and stress and strain aredirectly proportional for the material, the variationin strain and stress from the top to middle to bot-tom is linear (Figs 2.7(b) and (c)). The maximumstress in compression and tension is σy. The aver-age stress in compression and tension is σy/2. Hencethe compressive force, Fc, and tensile force, Ft,acting on the section are equal and are given byFig. 2.6 Stress–strain plot for steel.

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25

atively if b and d are known, the required materialstrength can be evaluated.

Equation 2.3 is more usually written as

M = σZ (2.5)

or

M =

σIy

(2.6)

where Z is the elastic section modulus and is equalto bd 2/6 for a rectangular beam, I is the moment ofinertia or, more correctly, the second moment ofarea of the section and y the distance from theneutral axis (Fig. 2.7(a)). The elastic section modu-lus can be regarded as an index of the strength ofthe beam in bending. The second moment of areaabout an axis x–x in the plane of the section isdefined by

Ixx = ∫ y2dA (2.7)

Second moments of area of some common shapesare given in Table 2.4.

2.5.2 PLASTIC CRITERIAWhile the above approach would be suitable fordesign involving the use of materials which have alinear elastic behaviour, materials such as reinforcedconcrete and steel have a substantial plastic per-formance. In practice this means that on reachingan elastic yield point the material continues to de-form but with little or no change in maximum stress.

Figure 2.8 shows what this means in terms ofstresses in the beam. As the loading on the beamincreases extreme fibre stresses reach the yield point,σy, and remain constant as the beam continues tobend. A zone of plastic yielding begins to penetrateinto the interior of the beam, until a point is reachedimmediately prior to complete failure, when practic-ally all the cross-section has yielded plastically. Theaverage stress at failure is σy, rather than σy/2 as was

Fc = Ft = F = stress × area

= =

σ σy y2 2 4

bd bd

The tensile and compression forces are separ-ated by a distance s whose value is equal to 2d /3(Fig. 2.7(a)). Together they make up a couple, or mo-ment, which acts in the opposite sense to the designmoment. The value of this moment of resistance,Mr, is given by

M r = Fs =

bd d bdσ σy y4

23 6

2

= (2.3)

At equilibrium, the design moment in the beamwill equal the moment of resistance, i.e.

M = M r (2.4)

Provided that the yield strength of the material,i.e. σy is known, equations 2.3 and 2.4 can be usedto calculate suitable dimensions for the beamneeded to resist a particular design moment. Altern-

Fig. 2.8 Bending failure of a beam (a) below yield(elastic); ( b) at yield point (elastic); (c) beyond yield(partially plastic); (d) beyond yield ( fully plastic).

Beam design

I

DB dbyy =

− 3 3

12

Table 2.4 Second moments of area

Shape Second moment of area

Ixx = bd3/12

Iyy = db3/12

Ixx = bh3/36Iaa = bh3/12

Ixx = πR4/4Ipolar = 2Ixx

I

bdaa =

3

3

I

BD bdxx =

−

3 3

12

I

BD bdxx =

−

3 3

12

I

TB dtyy = +

212 12

3 3

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26

combination of buckling followed by crushing ofthe material(s) (Fig. 1.3). These columns will tendto have lower load-carrying capacities, a fact whichis taken into account by reducing the design stressesin the column. The design stresses are related tothe slenderness ratio of the column, which is foundto be a function of the following factors:

1. geometric properties of the column cross-section, e.g. lateral dimension of column, radiusof gyration;

2. length of column;3. support conditions (Fig. 2.5).

The radius of gyration, r, is a sectional propertywhich provides a measure of the column’s abilityto resist buckling. It is given by

r = (I/A)1/2 (2.11)

Generally, the higher the slenderness ratio, thegreater the tendency for buckling and hence thelower the load capacity of the column. Most prac-tical reinforced concrete columns are designed tofail by crushing and the design equations for thismedium are based on equation 2.10 (Chapter 3).Steel columns, on the other hand, are designed tofail by a combination of buckling and crushing.Empirical relations have been derived to predictthe design stress in steel columns in terms of theslenderness ratio (Chapter 4).

found to be the case when the material was assumedto have a linear-elastic behaviour. The moment ofresistance assuming plastic behaviour is given by

M Fs

bd d bdSr = = = =

σ σσy y y2 42

2

(2.8)

where S is the plastic section modulus and isequal to bd 2/4 for rectangular beams. By setting thedesign moment equal to the moment of resistanceof the beam its size and strength can be calculatedaccording to plastic criteria (Example 2.6).

2.6 Column designGenerally, column design is relatively straightfor-ward. The design process simply involves makingsure that the design load does not exceed the loadcapacity of the column, i.e.

Load capacity ≥ design axial load (2.9)

Where the column is required to resist a predomin-antly axial load, its load capacity is given by

Load capacity = design stress × area ofcolumn cross-section (2.10)

The design stress is related to the crushingstrength of the material(s) concerned. However, notall columns fail in this mode. Some fail due to a

Example 2.6 Elastic and plastic moments of resistance of abeam section

Calculate the moment of resistance of a beam 50 mm wide by 100 mm deep with σy = 20 N mm−2 according to (i)elastic criteria and (ii) plastic criteria.

ELASTIC CRITERIAFrom equation 2.3,

M

bdr,el y N mm

. = =× ×

= ×2 2

6

650 100 20

61 67 10σ

PLASTIC CRITERIAFrom equation 2.8,

M

bdr,pl

y

4N mm

. = =

× ×= ×

2 2650 100 20

42 5 10

σ

Hence it can be seen that the plastic moment of resistance of the section is greater than the maximum elasticmoment of resistance. This will always be the case but the actual difference between the two moments will dependupon the shape of the section.

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27

Summary

In most real situations, however, the loads willbe applied eccentric to the axes of the column.Columns will therefore be required to resist axialloads and bending moments about one or bothaxes. In some cases, the bending moments may besmall and can be accounted for in design simplyby reducing the allowable design stresses in the ma-terial(s). Where the moments are large, for instancecolumns along the outside edges of buildings,the analysis procedures become more complex andthe designer may have to resort to the use of designcharts. However, as different codes of practice treatthis in different ways, the details will be discussedseparately in the chapters which follow.

2.7 SummaryThis chapter has presented a simple but conserv-ative method for assessing the design loads actingon individual members of building structures. Thisassumes that all the joints in the structure arepinended and that the sequence of load transferoccurs in the order: ceiling /floor loads to beams tocolumns to foundations to ground. The design loadsare used to calculate the design bending moments,shear forces, axial loads and deflections experiencedby members. In combination with design strengthsand other properties of the construction medium,the sizes of the structural members are determined.

Example 2.7 Analysis of column sectionDetermine whether the reinforced concrete column cross-section shown below would be suitable to resist an axialload of 1500 kN. Assume that the design compressive strengths of the concrete and steel reinforcement are 14 and375 N mm−2 respectively.

Area of steel bars = 4 × (π202/4) = 1256 mm2

Net area of concrete = 300 × 300 − 1256 = 88 744 mm2

Load capacity of column = force in concrete + force in steel

= 14 × 88 744 + 375 × 1256

= 1 713 416 N = 1713.4 kN

Hence the column cross-section would be suitable to resist the design load of 1500 kN.

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28

D = 360 mm x x

t = 7 mm

T = 12 mmB = 170 mm

Questions1. For the beams shown, calculate and

sketch the bending moment and shearforce diagrams.

2. For the three load cases shown inFig. 2.4, sketch the bending moment andshear force diagrams and hence determinethe design bending moments and shearforces for the beam. Assume the mainspan is 6 m and the overhang is 2 m. Thecharacteristic dead and imposed loads arerespectively 20 kNm−1 and 10 kNm−1.

3. (a) Calculate the area and the major axismoment of inertia, elastic modulus,plastic modulus and the radius ofgyration of the steel I-section shownbelow.

according to (i) elastic and (ii) plasticcriteria. Use your results to determinethe working and collapse loads of abeam with this cross-section, 6 m longand simply supported at its ends.Assume the loading on the beam isuniformly distributed.

4. What are the most common ways inwhich columns can fail? List and discussthe factors that influence the load carryingcapacity of columns.

5. The water tank shown in Fig. 2.9 issubjected to the following characteristicdead (Gk), imposed (Q k) and wind loads(Wk) respectively(i) 200 kN, 100 kN and 50 kN (Load

case 1)(ii) 200 kN, 100 kN and 75 kN (Load

case 2).Assuming the support legs are pinned atthe base, determine the design axial forcesin both legs by considering the followingload combinations:(a) dead plus imposed(b) dead plus wind(c) dead plus imposed plus wind.Refer to Table 3.4 for relevant loadfactors.

(b) Assuming the design strength of steel,σy, is 275 Nmm−2, calculate themoment of resistance of the I-section Fig. 2.9

Gk,Qk1.2 m

5 m

Leg R

Leg L

2 m

Wk

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29

PART TWO

STRUCTURALDESIGN TO BRITISHSTANDARDS

Part One has discussed the principles of limit statedesign and outlined general approaches towardsassessing the sizes of beams and columns in build-ing structures. Since limit state design forms thebasis of the design methods in most modern codesof practice on structural design, there is consider-able overlap in the design procedures presented inthese codes.

The aim of this part of the book is to givedetailed guidance on the design of a number ofstructural elements in the four media: concrete,steel, masonry and timber to the current BritishStandards. The work has been divided into fourchapters as follows:

1. Chapter 3 discusses the design procedures inBS 8110: Structural use of concrete relating tobeams, slabs, pad foundations, retaining wallsand columns.

2. Chapter 4 discusses the design procedures in BS5950: Structural use of steelwork in buildings relat-ing to beams, columns, floors and connections.

3. Chapter 5 discusses the design procedures inBS 5628: Code of practice for use of masonry relat-ing to unreinforced loadbearing and panel walls.

4. Chapter 6 discusses the design procedures in BS5268: Structural use of timber relating to beams,columns and stud walling.

9780415467193_C03a 9/3/09, 12:41 PM29

Design in reinforced concrete to BS 8110

30

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31

Chapter 3

Design in reinforcedconcrete to BS 8110

the recommendations given in various documentsincluding BS 5400: Part 4: Code of practice fordesign of concrete bridges, BS 8007: Code of prac-tice for the design of concrete structures for retainingaqueous liquids and BS 8110: Structural use of con-crete. Since the primary aim of this book is to giveguidance on the design of structural elements, thisis best illustrated by considering the contents ofBS 8110.

BS 8110 is divided into the following three parts:

Part 1: Code of practice for design and construction.Part 2: Code of practice for special circumstances.Part 3: Design charts for singly reinforced beams, doubly

reinforced beams and rectangular columns.

Part 1 covers most of the material required foreveryday design. Since most of this chapter isconcerned with the contents of Part 1, it shouldbe assumed that all references to BS 8110 refer toPart 1 exclusively. Part 2 covers subjects such astorsional resistance, calculation of deflections andestimation of crack widths. These aspects of designare beyond the scope of this book and Part 2, there-fore, is not discussed here. Part 3 of BS 8110 con-tains charts for use in the design of singly reinforcedbeams, doubly reinforced beams and rectangularcolumns. A number of design examples illustratingthe use of these charts are included in the relevantsections of this chapter.

3.2 Objectives and scopeAll reinforced concrete building structures arecomposed of various categories of elements includ-ing slabs, beams, columns, walls and foundations(Fig. 3.1). Within each category is a range of ele-ment types. The aim of this chapter is to describethe element types and, for selected elements, togive guidance on their design.

This chapter is concerned with the detailed design ofreinforced concrete elements to British Standard 8110.A general discussion of the different types of commonlyoccurring beams, slabs, walls, foundations and col-umns is given together with a number of fully workedexamples covering the design of the following elements:singly and doubly reinforced beams, continuous beams,one-way and two-way spanning solid slabs, pad founda-tion, cantilever retaining wall and short braced columnssupporting axial loads and uni-axial or bi-axial bend-ing. The section which deals with singly reinforced beamsis, perhaps, the most important since it introduces thedesign procedures and equations which are common tothe design of the other elements mentioned above, withthe possible exception of columns.

3.1 IntroductionReinforced concrete is one of the principal materialsused in structural design. It is a composite material,consisting of steel reinforcing bars embedded inconcrete. These two materials have complementaryproperties. Concrete, on the one hand, has highcompressive strength but low tensile strength. Steelbars, on the other, can resist high tensile stressesbut will buckle when subjected to comparativelylow compressive stresses. Steel is much moreexpensive than concrete. By providing steel barspredominantly in those zones within a concretemember which will be subjected to tensile stresses,an economical structural material can be producedwhich is both strong in compression and strongin tension. In addition, the concrete provides cor-rosion protection and fire resistance to the morevulnerable embedded steel reinforcing bars.

Reinforced concrete is used in many civilengineering applications such as the constructionof structural frames, foundations, retaining walls,water retaining structures, highways and bridges.They are normally designed in accordance with

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Design in reinforced concrete to BS 8110

32

3. material properties4. loading5. stress–strain relationships6. durability and fire resistance.

The detailed design of beams, slabs, foundations,retaining walls and columns will be discussed insections 3.9, 3.10, 3.11, 3.12 and 3.13, respectively.

3.3 SymbolsFor the purpose of this book, the following sym-bols have been used. These have largely been takenfrom BS 8110. Note that in one or two cases thesame symbol is differently defined. Where thisoccurs the reader should use the definition mostappropriate to the element being designed.

Geometric properties:

b width of sectiond effective depth of the tension reinforcementh overall depth of sectionx depth to neutral axisz lever armd ′ depth to the compression reinforcementb effective spanc nominal cover to reinforcement

Bending:

Fk characteristic loadgk, Gk characteristic dead loadqk, Qk characteristic imposed loadwk, Wk characteristic wind loadfk characteristic strengthfcu characteristic compressive cube strength

of concretefy characteris

design of structural elements: concrete, steelwork...

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