concrete structures: stresses and deformations, third edition

Administrator
20008d9bcoverv05bjpg

Concrete StructuresStresses and Deformations

Third Edition

Also available from Spon Press

Abnormal Loading on StructuresExperimental and Numerical ModellingF K Garas K S Virdi R Matthews ampJ L Clarke

Autogenous Shrinkage ofConcreteEdited by E Tazawa

Bridge Deck Behaviour3rd EditionE C Hambly

Bridge LoadsAn International PerspectiveC OrsquoConnor amp P Shaw

Circular Storage Tanks and Silos2nd EditionA Ghali

Concrete Ground FloorsN Williamson

Concrete Masonry DesignerrsquosHandbook2nd EditionJ J Roberts A K Tovey amp A Fried

Design Aids for Eurocode 2Design of Concrete StructuresEdited by The Concrete Societies of TheUK The Netherlands and Germany

Design of Offshore ConcreteStructuresI Holand E Jersin amp O T Gudmestad

Dynamic Loading and Design ofStructuresA J Kappos

Earthquake Resistant ConcreteStructuresG G Penelis amp A J Kappos

Global Structural Analysis ofBuildingsK A Zalka

Introduction to Eurocode 2Design of Concrete StructuresD Beckett amp A Alexandrou

Monitoring and Assessment ofStructuresG Armer

Structural AnalysisA Unified Classical and MatrixApproachA Ghali amp A M Neville

Structural Defects ReferenceManual for Low-rise BuildingsM F Atkinson

Wind Loading of StructuresJ D Holmes


Third Edition

A GhaliProfessor The University of CalgaryCanada

R FavreProfessor Swiss Federal Institute of Technology (EPFL)Lausanne Switzerland

M ElbadryAssociate Professor The University of CalgaryCanada

London and New York

First published 1986 by E amp FN Spon

Second edition first published 1994

Third edition first published 2002by Spon Press11 New Fetter Lane London EC4P 4EE

Simultaneously published in the USA and Canadaby Spon Press29 West 35th Street New York NY 10001

Spon Press is an imprint of the Taylor amp Francis Group

copy 1986 1994 A Ghali and R Favrecopy 2002 A Ghali R Favre and M Elbadry

The right of A Ghali R Favre and M Elbadry to be identified as theAuthors of this Work has been asserted by them in accordance withthe Copyright Designs and Patents Act 1988

All rights reserved No part of this book may be reprinted orreproduced or utilised in any form or by any electronicmechanical or other means now known or hereafterinvented including photocopying and recording or in anyinformation storage or retrieval system without permission inwriting from the publishers

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

Library of Congress Cataloging in Publication DataA catalog record for this book has been requested

ISBN 0ndash415ndash24721ndash7

This edition published in the Taylor amp Francis e-Library 2006

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

ISBN 0-203-98752-7 Master e-book ISBN

(Print Edition)

Contents

Preface to the third edition xivAcknowledgements xviNote xviiThe SI system of units and British equivalents xviiiNotation xx

1 Creep and shrinkage of concrete and relaxation of steel 1

11 Introduction 112 Creep of concrete 213 Shrinkage of concrete 414 Relaxation of prestressed steel 515 Reduced relaxation 716 Creep superposition 817 The aging coefficient χ definition 1018 Equation for the aging coefficient χ 1119 Relaxation of concrete 12110 Step-by-step calculation of the relaxation function for concrete 14111 Age-adjusted elasticity modulus 17

1111 Transformed section 171112 Age-adjusted flexibility and stiffness 18

112 General 18

2 Stress and strain of uncracked sections 20

21 Introduction 2022 Sign convention 2223 Strain stress and curvature in composite and

homogeneous cross-sections 22231 Basic equations 25

24 Strain and stress due to non-linear temperature variation 27Example 21 Rectangular section with parabolic

temperature variation 2925 Time-dependent stress and strain in a composite section 30

251 Instantaneous stress and strain at age t0 31252 Changes in stress and strain during the period t0 to t 33

Example 22 Post-tensioned section 37Example 23 Pre-tensioned section 43Example 24 Composite section steel and post-

tensioned concrete 44Example 25 Composite section pre-tensioned and

cast-in-situ parts 4926 Summary of analysis of time-dependent strain and stress 5727 Examples worked out in British units 61

Example 26 Stresses and strains in a pre-tensioned section 61Example 27 Bridge section steel box and post-tensioned

slab 6428 General 68

3 Special cases of uncracked sections and calculationof displacements 69

31 Introduction 7032 Prestress loss in a section with one layer of reinforcement 70

321 Changes in strain in curvature and in stress due tocreep shrinkage and relaxation 74

Example 31 Post-tensioned section without non-prestressed steel 75

33 Effects of presence of non-prestressed steel 7834 Reinforced concrete section without prestress effects of

creep and shrinkage 79Example 32 Section subjected to uniform shrinkage 81Example 33 Section subjected to normal force and

moment 8335 Approximate equations for axial strain and curvature

due to creep 8536 Graphs for rectangular sections 8537 Multi-stage prestressing 8738 Calculation of displacements 88

381 Unit load theory 89382 Method of elastic weights 89

vi Contents

Example 34 Simple beam derivation of equations fordisplacements 92

Example 35 Simplified calculations of displacements 9339 Example worked out in British units 95

Example 36 Parametric study 95310 General 98

4 Time-dependent internal forces in uncracked structuresanalysis by the force method 100

41 Introduction 10142 The force method 10343 Analysis of time-dependent changes of internal forces

by the force method 105Example 41 Shrinkage effect on a portal frame 108Example 42 Continuous prestressed beam constructed

in two stages 109Example 43 Three-span continuous beam composed of

precast elements 113Example 44 Post-tensioned continuous beam 116

44 Movement of supports of continuous structures 121Example 45 Two-span continuous beam settlement of

central support 12545 Accounting for the reinforcement 128

Example 46 Three-span precast post-tensioned bridge 12846 Step-by-step analysis by the force method 13647 Example worked out in British units 141

Example 47 Two-span bridge steel box andpost-tensioned deck 141

48 General 144

5 Time-dependent internal forces in uncracked structuresanalysis by the displacement method 146

51 Introduction 14652 The displacement method 14753 Time-dependent changes in fixed-end forces in a

homogeneous member 149Example 51 Cantilever restraint of creep displacements 152

54 Analysis of time-dependent changes in internal forces incontinuous structures 153

Contents vii

55 Continuous composite structures 15456 Time-dependent changes in the fixed-end forces in a

composite member 15657 Artificial restraining forces 158

Example 52 Steel bridge frame with concrete deckeffects of shrinkage 160

Example 53 Composite frame effects of creep 16458 Step-by-step analysis by the displacement method 17259 General 175

6 Analysis of time-dependent internal forces with conventionalcomputer programs 176

61 Introduction 17762 Assumptions and limitations 17763 Problem statement 17964 Computer programs 17965 Two computer runs 18466 Equivalent temperature parameters 18667 Multi-stage loading 18868 Examples 188

Example 61 Propped cantilever 188Example 62 Cantilever construction method 192Example 63 Cable-stayed shed 193Example 64 Composite space truss 197Example 65 Prestressed portal frame 201

69 General 205

7 Stress and strain of cracked sections 207

71 Introduction 20872 Basic assumptions 20973 Sign convention 20974 Instantaneous stress and strain 210

741 Remarks on determination of neutral axis position 213742 Neutral axis position in a T or rectangular fully

cracked section 214743 Graphs and tables for the properties of transformed

fully cracked rectangular and T sections 216Example 71 Cracked T section subjected to bending moment 234Example 72 Cracked T section subjected to M and N 236

viii Contents

75 Effects of creep and shrinkage on a reinforced concretesection without prestress 237751 Approximate equation for the change in curvature

due to creep in a reinforced concrete sectionsubjected to bending 243

Example 73 Cracked T section creep and shrinkageeffects 243

76 Partial prestressed sections 246Example 74 Pre-tensioned tie before and after cracking 250Example 75 Pre-tensioned section in flexure live-load

cracking 25477 Flow chart 24978 Example worked out in British units 260

Example 76 The section of Example 26 live-loadcracking 260

79 General 262

8 Displacements of cracked members 264

81 Introduction 26582 Basic assumptions 26683 Strain due to axial tension 266

Example 81 Mean axial strain in a tie 27184 Curvature due to bending 271

841 Provisions of codes 274Example 82 Rectangular section subjected to bending

moment 27585 Curvature due to a bending moment combined with an

axial force 276Example 83 Rectangular section subjected to M and N 278

851 Effect of load history 28086 Summary and idealized model for calculation of

deformations of cracked members subjected to N andor M 281861 Note on crack width calculation 284

87 Time-dependent deformations of cracked members 284Example 84 Non-prestressed simple beam variation of

curvature over span 285Example 85 Pre-tensioned simple beam variation of

curvature over span 29088 Shear deformations 293

Contents ix

89 Angle of twist due to torsion 293891 Twisting of an uncracked member 294892 Twisting of a fully cracked member 295

810 Examples worked out in British units 298Example 86 Live-load deflection of a cracked

pre-tensioned beam 298Example 87 Parametric study 299

811 General 301

9 Simplified prediction of deflections 303

91 Introduction 30392 Curvature coefficients κ 30493 Deflection prediction by interpolation between

uncracked and cracked states 306931 Instantaneous and creep deflections 308932 Deflection of beams due to uniform shrinkage 309933 Total deflection 313

94 Interpolation procedure the lsquobilinear methodrsquo 31495 Effective moment of inertia 315

Example 91 Use of curvature coefficients member inflexure 315

96 Simplified procedure for calculation of curvature at asection subjected to M and N 318

97 Deflections by the bilinear method members subjectedto M and N 320Example 92 Use of curvature coefficients member

subjected to M and N 32398 Estimation of probable deflection method of lsquoglobal

coefficientsrsquo 325981 Instantaneous plus creep deflection 325982 Shrinkage deflection 327Example 93 Non-prestressed beam use of global

coefficients 330Example 94 Prestressed beam use of global coefficients 330

99 Deflection of two-way slab systems 332991 Geometric relation 333992 Curvature-bending moment relations 335993 Effects of cracking and creep 336Example 95 Interior panel 338Example 96 Edge panel 341

x Contents

994 Deflection of two-way slabs due to uniform shrinkage 345Example 97 Edge panel 345

910 General 348

10 Effects of temperature 349

101 Introduction 350102 Sources of heat in concrete structures 351103 Shape of temperature distribution in bridge cross-sections 352104 Heat transfer equation 354105 Material properties 357106 Stresses in the transverse direction in a bridge cross-section 357107 Self-equilibrating stresses 360108 Continuity stresses 361

Example 101 Continuous bridge girder 363109 Typical temperature distributions in bridge sections 3661010 Effect of creep on thermal response 368

Example 102 Wall stresses developed by heat ofhydration 371

1011 Effect of cracking on thermal response 3741012 General 378

11 Control of cracking 380

111 Introduction 380112 Variation of tensile strength of concrete 381113 Force-induced and displacement-induced cracking 382

1131 Example of a member subjected to bending 3841132 Example of a member subjected to axial force

(worked out in British units) 387114 Advantage of partial prestressing 391115 Minimum reinforcement to avoid yielding of steel 391116 Early thermal cracking 393117 Amount of reinforcement to limit crack width 394

1171 Fatigue of steel 3951172 Graph for the change in steel stress in a

rectangular cracked section 395Example 111 Non-prestressed section crack width

calculation 397118 Considerations in crack control 399119 Cracking of high-strength concrete 401

Contents xi

1110 Examples worked out in British units 402Example 112 Prestressed section crack width calculation 402Example 113 Overhanging slab reinforcement to

control thermal cracking 4031111 General 406

12 Design for serviceability of prestressed concrete 407

121 Introduction 407122 Permanent state 408123 Balanced deflection factor 408124 Design of prestressing level 409125 Examples of design of prestress level in bridges 413

Example 121 Bridges continuous over three spans 413Example 122 Simply-supported bridges 415Example 123 Effects of variation of span to thickness

ratio on βD 416126 Transient stresses 419127 Residual opening of cracks 419128 Water-tightness 421129 Control of residual crack opening 4221210 Recommended longitudinal non-prestressed steel in

closed-box bridge sections 4221211 Residual curvature 4221212 General 426

13 Non-linear analysis of plane frames 428

131 Introduction 428132 Reference axis 429133 Idealization of plane frames 429134 Tangent stiffness matrix of a member 431135 Examples of stiffness matrices 434

Example 131 Stiffness matrix of an uncrackedprismatic cantilever 434

Example 132 Tangent stiffness matrix of a cracked cantilever 437136 Fixed-end forces 439137 Fixed-end forces due to temperature 440138 Numerical integration 442139 Iterative analysis 4431310 Convergence criteria 445

xii Contents

1311 Incremental method 4461312 Examples of statically indeterminate structures 447

Example 133 Demonstration of the iterative analysis 447Example 134 Deflection of a non-prestressed concrete slab 452Example 135 Prestressed continuous beam analysed

by the incremental method 4541313 General 456

14 Serviceability of members reinforced with fibre-reinforced polymers 457

141 Introduction 457142 Properties of FRP reinforcements for concrete 458143 Strain in reinforcement and width of cracks 459144 Design of cross-sectional area of FRP for

non-prestressed flexural members 460145 Curvature and deflections of flexural members 463146 Relationship between deflection mean curvature and

strain in reinforcement 464147 Ratio of span to minimum thickness 466

1471 Minimum thickness comparison betweenmembers reinforced with steel and with FRP 467

1472 Empirical equation for ratio of length tominimum thickness 468

148 Design examples for deflection control 469Example 141 A simple beam 469Example 142 Verification of the ratio of span to deflection 470

149 Deformability of sections in flexure 4711410 Prestressing with FRP 4721411 General 473

Appendix A Time functions for modulus of elasticity creepshrinkage and aging coefficient of concrete 474A1 CEB-FIP Model Code 1990 (MC-90) 474

A11 Parameters affecting creep 475A12 Effect of temperature on maturity 475A13 Modulus of elasticity 476A14 Development of strength and modulus of

elasticity with time 476A15 Tensile strength 477A16 Creep under stress not exceeding 40 per cent of

mean compressive strength 477

Contents xiii

A17 Effect of type of cement on creep 479A18 Creep under high stress 479A19 Shrinkage 479

A2 Eurocode 2ndash1991 (EC2ndash91) 480A3 ACI Committee 209 481

A31 Creep 482A32 Shrinkage 483

A4 British Standard BS 8110 485A41 Modulus of elasticity of concrete 485A42 Tensile strength of concrete 485A43 Creep 486A44 Shrinkage 486

A5 Computer code for creep and aging coefficients 486A6 Graphs for creep and aging coefficients 488A7 Approximate equation for aging coefficient 489

Appendix B Relaxation reduction coefficient χχr 534

Appendix C Elongation end rotation and central deflection of abeam in terms of the values of axial strain and curvature at anumber of sections 538

Appendix D Depth of compression zone in a fully cracked T section 542

Appendix E Crack width and crack spacing 544E1 Introduction 544E2 Crack spacing 546E3 Eurocode 2ndash1991(EC2ndash91) 547E4 CEB-FIP 1990(MC-90) 548E5 ACI318-89 and ACI318-99 550E6 British Standard BS 8110 552

Appendix F Values of curvature coefficients κs κφ and κcs 555

Appendix G Description of computer programs provided atwwwsponpresscomconcretestructures 568G1 Introduction 568G2 Computer program CREEP 569

G21 Input and output of CREEP 569G22 FORTRAN code 569G23 Example input file for CREEP 570

xiv Contents

G3 Computer program SCS (Stresses in Cracked Sections) 570G31 Input and output of SCS 570G32 Units and sign convention 571G33 Example input file for SCS 571

G4 Computer program TDA (Time-Dependent Analysis) 571G41 Input data 572G42 Units and sign convention 573G43 Prestressing duct 573G44 Example input file for TDA 573

Further reading 575Index 577

Contents xv

Preface to the third edition

Concrete structures must have adequate safety factor against failure and mustalso exhibit satisfactory performance in service This book is concerned withthe checks on stresses and deformations that can be done in design to ensuresatisfactory serviceability of reinforced concrete structures with or withoutprestressing The following are qualities which are essential for a satisfactoryperformance

1 No excessive deflection should occur under the combined effect of pre-stressing the self-weight of the structures and the superimposed deadload

2 Deflections and crack width should not be excessive under the abovementioned loads combined with live and other transitory loads settle-ment of support and temperature variations This makes it necessary tocontrol stress in the reinforcement which is one of the main parametersaffecting width of cracks Durability of concrete structures is closelylinked to the extent of cracking

Because of creep and shrinkage of concrete and relaxation of prestressedreinforcement the stresses in the concrete and in the reinforcement vary withtime In addition when the structure is statically indeterminate the reactionsand the internal forces are also time dependent The strains and consequentlythe displacement change considerably with time due to the same effects andalso due to cracking The purpose of this text is to present the most effectivemethods for prediction of the true stresses and deformations during the lifeof the structure

The mechanical properties that enter in calculation of stress and strain arethe modulus of elasticity creep and shrinkage of concrete and modulus ofelasticity of reinforcements These properties differ from project to projectand from one country to another The methods of analysis presented in thetext allow the designer to account for the effects of variance in these param-eters Appendix A based on the latest two European codes British Standardsand American Concrete Institute practice gives guidance on the choice of

values of these parameters for use in design Appendix E also based on thesame sources deals with crack width and crack spacing

The methods of analysis of stresses and deformations presented in thechapters of the text are applicable in design of concrete structures regardlessof codes Thus future code revisions as well as codes of other countries maybe employed

Some of the examples in the text are dimensionless Some examples areworked out in the SI units and others in the so-called British units customaryto engineers in the USA the input data and the main results are given inboth SI and British Units It is hoped that the use of both systems of unitswill make the text equally accessible to readers in all countries Working outdifferent examples in the two systems of units is considered more useful thanthe simpler task of working each example in both units

In the second edition a chapter discussing control of cracking was addedFour new chapters are added in the third edition The new Chapter 6 explainshow linear computer programs routinely used by almost all structural engin-eers can be employed for analysis of the time-dependent effects of creepshrinkage and relaxation Chapter 12 discusses the choice of amount anddistribution of prestressed and non-prestressed reinforcements to achievebest serviceability Fibre-reinforced polymer (FRP) bars and strands aresometimes used as reinforcement of concrete in lieu of steel Chapter 14 isconcerned with serviceability of concrete structures reinforced with thesematerials The effect of cracking on the reactions and the internal forces ofstatically indeterminate reinforced concrete structures requires non-linearanalysis discussed in Chapter 13

The analysis procedures presented in the text can in part be executed usingcomputer programs provided on wwwsponpresscomconcretestructures foruse as an optional companion to this book The new Appendix G describesthe programs on the website and how they can be used

Mr S Youakim doctoral candidate and Mr R Gayed MSc student atthe University of Calgary prepared the figures and checked the revisions inthe third edition Mrs K Knoll-Williams typed the new material We aregrateful to them as well as to those who have helped in the earlier editions

A GhaliR Favre

M ElbadryCalgary Canada

Lausanne SwitzerlandJanuary 2002

Preface to the third edition xvii

Acknowledgements

This book was produced through the collaboration of A Ghali with R Favreand his research group mainly during sabbatical leaves spent at the SwissFederal Institute of Technology Lausanne For completion of the work onthe first edition A Ghali was granted a Killam Resident Fellowship at theUniversity of Calgary for which he is very grateful

The authors would like to thank those who helped in the preparationof the first edition of the book In Lausanne Dr M Koprna ResearchAssociate reviewed parts of the text and collaborated in writing Chapter 8and Appendix A Mr J Trevino Research Assistant made a considerablecontribution by providing solutions or checking the numerical examplesand preparing the manuscript for the publisher Mr B-F Gardel preparedthe figures In Calgary Mr M Elbadry and Mr A Mokhtar graduatestudents checked parts of the text Mr B Unterberger prepared by computerthe graphs of Appendix F Miss C Larkin produced an excellent typescript

The authors deeply appreciate the work of Dr S El-Gabalawy of theDepartment of English at the University of Calgary who revised themanuscript

Figures A1 and A2 are reproduced with permission of BSI under licencenumber 2001SK0331 Complete standards can be obtained from BSICustomer Services 389 Chiswick High Road London W4 4AL (tel 0208996 9001)

Note

It has been assumed that the design and assessment of structures areentrusted to experienced civil engineers and that calculations are carried outunder the direction of appropriately experienced and qualified supervisorsUsers of this book are expected to draw upon other works on the subjectincluding national and international codes of practice and are expected toverify the appropriateness and content of information they draw from thisbook

The SI system of units andBritish equivalents

Lengthmetre (m) 1 m = 3937 in

1 m = 3281 ft

Areasquare metre (m2) 1 m2 = 1550 in2

1m2 = 1076 ft2

Volumecubic metre (m3) 1m3 = 3532 ft3

Moment of inertiametre to the power four (m4) 1m4 = 2403 times 103 in4

Forcenewton (N) 1N = 02248 lb

Load intensitynewton per metre (Nm) 1Nm = 006852 lbftnewton per square metre (Nm2) 1Nm2 = 2088 times 10minus3 lbft2

Momentnewton metre (N-m) 1N-m = 8851 lb-in

1N-m = 07376 times 10minus3 kip-ft1kN-m = 8851kip-in

Stressnewton per square metre (pascal) 1Pa = 1450 times 10minus6 lbin2

1MPa = 01450ksi

Curvature(metre) minus 1 1mminus1 = 00254 inminus1

Temperature changedegree Celsius (degC) 1 degC = (59) degFahrenheit

Energy and powerjoule (J) = 1N-m 1J = 07376 lb-ftwatt (W) = 1Js 1W = 07376 lb-fts

1W = 3416 Btuh

Nomenclature for decimal multiples in the SI system109 giga (G)106 mega (M)103 kilo (k)10minus3 milli (m)

The SI system of units and British equivalents xxi

Notation

The following is a list of symbols which are common in various chapters ofthe book All symbols are defined in the text when they first appear and againwhen they are used in equations which are expected to be frequently appliedThe sign convention adopted throughout the text is also indicated whereapplicable

A Cross-sectional areaA Vector of actions (internal forces or reactions)A B and I Area first moment of area and moment of inertia of the

age-adjusted transformed section composed of area ofconcrete plus α times area of reinforcement

B First moment of area For B see Ab Breadth of a rectangular section or width of the flange of a

T-sectionc Depth of compression zone in a fully cracked sectionD Displacementd Distance between extreme compressive fibre to the bottom

reinforcement layerE Modulus of elasticityEc = Ec(t0)[1 + χφ(t t0)] = age-adjusted elasticity modulus of

concretee EccentricityF Forcef Stress related to strength of concrete or steel[ f ] Flexibility matrixfct Tensile strength of concreteh Height of a cross-sectionI Moment of inertia For I see Ai j m n Integersl Length of a memberM Bending moment In a horizontal beam a positive moment

produces tension at the bottom fibre

Mr andor Nr Values of the bending moment andor the axial force whichare just sufficient to produce cracking

N Normal force positive when tensileP Forcer Radius of gyrationr(t t0) Relaxation function = concrete stress at time t due to a unit

strain imposed at time t0 and sustained to time t[S] Stiffness matrixsr Spacing between cracksT Temperaturet Time or age (generally in days)W Section modulus (length3)y Coordinate defining location of a fibre or a reinforcement

layer y is measured in the downward direction from a speci-fied reference point

α = EsEc(t0) = ratio of elasticity modulus of steel to elasticitymodulus of concrete at age t0

α = α[1 + χφ(t t0)] = EsEc = ratio of elasticity modulus of steelto the age-adjusted elasticity modulus of concrete

αt Coefficient of thermal expansion (degreeminus1)ε Normal strain positive for elongationζ Coefficient of interpolation between strain curvature

and deflection values for non-cracked and fully crackedconditions (states 1 and 2 respectively)

η Dimensionless multiplier for calculation of time-dependentchange in axial strain

κ Dimensionless multiplier for calculation time-dependentchange of curvature

ν Poissonrsquos ratioξ Dimensionless shape functionρ ρprime Ratio of tension and of compression reinforcement to the

area (bd) ρ = Asbd ρprime = Aprimesbdσ Normal stress positive when tensileτ Instant of timeφ(t t0) Creep coefficient of concrete = ratio of creep to the

instantaneous strain due to a stress applied at time t0 andsustained to time t

χ(t t0) Aging coefficient of concrete (generally between 06 and 09see Section 17 and Figs A6ndash45)

χφ(t t0) = χ(t t0) φ(t t0) = aging coefficient times creep coefficientχr Relaxation reduction coefficient for prestressed steelψ Curvature (lengthminus1) Positive curvature corresponds to

positive bending moment Braces indicate a vector ie a matrix of one column

Notation xxiii

[ ] A rectangular or a square matrixrarr Single-headed arrows indicate a displacement (translation

or rotation) or a force (a concentrated load or a couple)rarrrarr Double-headed arrow indicates a couple or a rotation

its direction is that of the rotation of a right-hand screwprogressing in the direction of the arrow

Subscriptsc Concretecs Shrinkagem Meanns Non-prestressed steelO Reference point0 Initial or instantaneouspr Relaxation in prestressed steelps Prestressed steels Steelst Total steel prestressed and non-prestressedu Unit force effect unit displacement effectφ Creep effect12 Uncracked or cracked state

xxiv Notation

Creep and shrinkage ofconcrete and relaxation of steel

11 Introduction

The stress and strain in a reinforced or prestressed concrete structure aresubject to change for a long period of time during which creep and shrinkageof concrete and relaxation of the steel used for prestressing develop grad-ually For analysis of the time-dependent stresses and deformations it isnecessary to employ time functions for strain or stress in the materialsinvolved In this chapter the basic equations necessary for the analysis arepresented The important parameters that affect the stresses or the strains are

The lsquoSaddledomersquo Olympic Ice Stadium Calgary Canada (Courtesy Genestar StructuresLtd and J Bobrowski and Partners Ltd)

Chapter 1

included in the equations but it is beyond the scope of this book to examinehow these parameters vary with the variations of the material properties

The modulus of elasticity of concrete increases with its age A stressapplied on concrete produces instantaneous strain if the stress is sustainedthe strain will progressively increase with time due to creep Thus the magni-tude of the instantaneous strain and creep depends upon the age of concreteat loading and the length of the period after loading Other parametersaffecting the magnitude of creep as well as shrinkage are related to the qualityof concrete and the environment in which it is kept Creep and shrinkage arealso affected by the shape of the concrete member considered

Steel subjected to stress higher than 50 per cent of its strength exhibitssome creep In practice steel used for prestressing may be subjected in serviceconditions to a stress 05 to 08 its strength If a tendon is stretched betweentwo fixed points constant strain is sustained but the stress will decrease pro-gressively due to creep This relaxation in tension is of concern in calculationof the time-dependent prestress loss and the associated deformations ofprestressed concrete members

Several equations are available to express the modulus of elasticity of con-crete creep shrinkage and relaxation of steel as functions of time Examplesof such expressions that are considered most convenient for practical applica-tions are given in Appendix A However the equations and the procedures ofanalysis presented in the chapters of this book do not depend upon the choiceof these time functions

In this chapter the effect of cracking is not included Combining theeffects of creep shrinkage and relaxation of steel with the effect of crackingon the deformations of concrete structures will be discussed in Chapters 7 89 and 13

12 Creep of concrete

A typical stressndashstrain curve for concrete is shown in Fig 11 It is commonpractice to assume that the stress in concrete is proportional to strain inservice conditions The strain occurring during the application of the stress(or within seconds thereafter) is referred to as the instantaneous strain and isexpressed as follows

εc(t0) =σc(t0)

Ec(t0)(11)

where σc(t0) is the concrete stress and Ec(t0) is the modulus of elasticity ofconcrete at age t0 the time of application of the stress The value of Ec thesecant modulus defined in Fig 11 depends upon the magnitude of the stressbut this dependence is ignored in practical applications The value Ec is gen-erally assumed to be proportional to the square or cubic root of concrete

2 Concrete Structures

strength which depends on the age of concrete at loading1 Expressions for Ec

in terms of the strength and age of concrete are given in Appendix AUnder sustained stress the strain increases with time due to creep and the

total strain ndash instantaneous plus creep ndash at time t (see Fig 12) is

εc(t) =σc(t0)

Ec(t0) [1 + φ(t t0)] (12)

where φ(t t0) is a dimensionless coefficient and is a function of the age atloading t0 and the age t for which the strain is calculated The coefficient φrepresents the ratio of creep to the instantaneous strain its value increaseswith the decrease of age at loading t0 and the increase of the length of theperiod (t minus t0) during which the stress is sustained When for example t0 isone month and t infinity the creep coefficient may be between 2 and 4depending on the quality of concrete the ambient temperature and humidityas well as the dimensions of the element considered2 Appendix A givesexpressions and graphs for the creep coefficient according to MC-90 ACICommittee 209 and British Standard BS 81103

Figure 11 Stressndashstrain curve for concrete Ec(t0) = secant modulus of elasticityt0 = age of concrete at loading

Creep and shrinkage of concrete and relaxation of steel 3

13 Shrinkage of concrete

Drying of concrete in air results in shrinkage while concrete kept under waterswells When the change in volume by shrinkage or by swelling is restrainedstresses develop In reinforced concrete structures the restraint may be causedby the reinforcing steel by the supports or by the difference in volume changeof various parts of the structure We are concerned here with the stressescaused by shrinkage which is generally larger in absolute value than swellingand occurs more frequently However there is no difference in the treatmentexcept in the sign of the term representing the amount of volume change Thesymbol εcs will be used for the free (unrestrained) strain due to shrinkage orswelling In order to comply with the sign convention for other causes ofstrain εcs is considered positive when it represents elongation Thus shrinkageof concrete εcs is a negative quantity

Stresses caused by shrinkage are generally reduced by the effect of creep ofconcrete Thus the effects of these two simultaneous phenomena must beconsidered in stress analysis For this purpose the amount of free shrinkageand an expression for its variation with time are needed Shrinkage starts todevelop at time ts when moist curing stops The strain that develops dueto free shrinkage between ts and a later instant t may be expressed as follows

εcs(t ts) = εcs0 βs(t minus ts) (13)

where εcs0 is the total shrinkage that occurs after concrete hardening up to

Figure 12 Creep of concrete under the effect of sustained stress


time infinity The value of εcs0 depends upon the quality of concrete and theambient air humidity The function βs(t minus ts) adopted by MC-90 dependsupon the size and shape of the element considered (see Appendix A)

The free shrinkage εcs(t2 t1) occurring between any two instants t1 and t2

can be determined as the difference between the two values obtained byEquation (13) substituting t2 and t1 for t

14 Relaxation of prestressed steel

The effect of creep on prestressing steel is commonly evaluated by a relaxa-tion test in which a tendon is stretched and maintained at a constant lengthand temperature and the loss in tension is measured over a long period Therelaxation under constant strain as in a constant-length test is referred to asintrinsic relaxation ∆σpr An equation widely used in the US and Canada forthe intrinsic relaxation at any time τ of stress-relieved wires or strands is4

∆σpr

σp0

= minuslog(τ minus t0)

10 σp0

fpy

minus 055 (14)

where fpy is the lsquoyieldrsquo stress defined as the stress at a strain of 001 The ratiofpy to the characteristic tensile stress fptk varies between 08 and 090 with thelower value for prestressing bars and the higher value for low-relaxationstrands ( (τ minus t0) is the period in hours for which the tendon is stretched)

The amount of intrinsic relaxation depends on the quality of steel TheMC-905 refers to three classes of relaxation and represents the relaxation as afraction of the initial stress σp0 Steels of the first class include cold-drawnwires and strands the second class includes quenched and tempered wiresand cold-drawn wires and strands which are treated (stabilized) to achievelow relaxation The third class of intermediate relaxation is for bars

For a given steel and duration of relaxation test the intrinsic relaxationincreases quickly as the initial stress in steel approaches its strength In theabsence of reliable relaxation tests MC-90 suggests the intrinsic relaxationvalues shown in Fig 13 for duration of 1000 hours and assumes that therelaxation after 50 years and more is three times these values

The Eurocode 2-916 (EC2ndash91) allows use of relaxation values differingslightly from MC-90 The values of EC2ndash91 are given between brackets in thegraphs of Fig 13

The following equation may be employed to give the ratio of the ultimateintrinsic relaxation to the initial stress

∆σprinfin

σp0

= minus η(λ minus 04)2 (15)

where


λ =σp0

fptk

(16)

∆σprinfin is the value of intrinsic relaxation of stress in prestressed steel at infin-ity The symbol ∆ is used throughout this book to indicate an increment Therelaxation represents a reduction in tension hence it is a negative quantity σp0

is the initial stress in prestressed steel fptk the characteristic tensile strengthand η the dimensionless coefficient depending on the quality of prestressedsteel

Equation (15) is applicable only when λ ge 04 below this value theintrinsic relaxation is negligible

When the value of the ultimate intrinsic relaxation is known for a particu-lar initial stress Equation (15) can be solved for the value of η Subsequentuse of the same equation gives the variation of relaxation with the changeof σp0

Intrinsic relaxation tests are often reported for time equals 1000h How-ever for analysis of the effects of relaxation of steel on stresses and deform-ations in prestressed concrete structures it is often necessary to employexpressions that give development of the intrinsic relaxation with time Suchexpressions are included in Appendix B

Relaxation increases rapidly with temperature The values suggested in

Figure 13 Intrinsic relaxation of prestressing steel according to MC-90 Thesymbols | pr 1000 | and | prinfin | represent respectively absolute values ofintrinsic relaxation after 1000 hours and after 50 years or more p0 = initialstress fptk = characteristic tensile strength The values indicated betweenbrackets are for 1000 hours relaxation according to EC2ndash91


Fig 13 are for normal temperatures (20 degC) With higher temperaturescaused for example by steam curing larger relaxation loss is to beexpected

15 Reduced relaxation

The magnitude of the intrinsic relaxation is heavily dependent on the value ofthe initial stress Compare two tendons with the same initial stress one in aconstant-length relaxation test and the other in a prestressed concrete mem-ber The force in the latter tendon decreases more rapidly because of theeffects of shrinkage and creep The reduction in tension caused by these twofactors has the same effect on the relaxation as if the initial stress weresmaller Thus the relaxation value to be used in prediction of the lossof prestress in a concrete structure should be smaller than the intrinsicrelaxation obtained from a constant-length test

The reduced relaxation value to be used in the calculation of loss ofprestress in concrete structures can be expressed as follows

∆σpr = χr∆σpr (17)

where ∆σpr is the intrinsic relaxation as would occur in a constant-lengthrelaxation test χr is a dimensionless coefficient smaller than unity The valueof χr can be obtained from Table 11 or Fig 14 The graph gives the value ofχr as a function of λ the ratio of the initial tensile stress to the characteristictensile strength of the prestress steel (Equation (16) ) and

Ω = minus ∆σps minus ∆σpr

σp0 (18)

where σp0 is the initial tensile stress in prestress steel ∆σps is the change instress in the prestressed steel due to the combined effect of creep shrinkage

Table 11 Relaxation reduction coefficient r

055 060 065 070 075 080

00 1000 1000 1000 1000 1000 100001 06492 06978 07282 07490 07642 0775702 04168 04820 05259 05573 05806 0598703 02824 03393 03832 04166 04425 0463004 02118 02546 02897 03188 03429 0362705 01694 02037 02318 02551 02748 02917


and relaxation and ∆σpr is the intrinsic relaxation as would occur in aconstant-length relaxation test

The value of the total loss is generally not known a priori because itdepends upon the reduced relaxation Iteration is here required the total lossis calculated using an estimated value of the reduction factor χr (for example07) which is later adjusted if necessary (see Example 31)

Appendix B gives the derivation of the relaxation reduction coefficientvalues in Table 11 and the graphs in Fig 14 The values given in the tableand the graphs may be approximated by Equation (B11)

16 Creep superposition

Equation (12) implies the assumption that the total strain instantaneousplus creep is proportional to the applied stress This linear relationship whichis generally true within the range of stresses in service conditions allowssuperposition of the strain due to stress increments or decrements and due toshrinkage Thus when the magnitude of the applied stress changes with timethe total strain of concrete due to the applied stress and shrinkage is given by(Fig 15)

Figure 14 Relaxation reduction coefficient r


εc(t) = σc(t0) 1 + φ(t t0)

Ec(t0)+

∆σc(t)

0

1 + φ(t τ)

Ec(τ) dσc(τ) + εcs(t t0) (19)

where

Equation (19) implies the assumption that a unit stress increment ordecrement introduced at the same age and maintained for the same timeproduces the same absolute value of creep This equation is the basis of the

Figure 15 Stress versus time and strain versus time for a concrete member subjected touniaxial stress of magnitude varying with time

t0 and t = ages of concrete when the initial stress is applied and when thestrain is considered

τ = an intermediate age between t0 and tσc(t0) = initial stress applied at age t0

dσc(τ) = an elemental stress (increment or decrement) applied at age τEc(τ) = modulus of elasticity of concrete at the age τ

φ(t τ) = coefficient of creep at time t for loading at age τεcs(t t0) = free shrinkage occurring between the ages t0 and t


methods presented in this book for analysis of the time-dependent stressesand deformations of concrete structures

17 The aging coefficient definition

The integral in Equation (19) represents the instantaneous strain plus creepdue to an increment in concrete stress of magnitude ∆σc (Fig 15) Thisincrement is gradually introduced during the period t0 to t A stress intro-duced gradually in this manner produces creep of smaller magnitude com-pared to a stress of the same magnitude applied at age t0 and sustained duringthe period (t minus t0) In the following equation the stress increment ∆σc(t) istreated as if it were introduced with its full magnitude at age t0 and sustainedto age t but the creep coefficient φ(t t0) is replaced by a reduced valuewhich equals χφ(t t0) where χ = χ(t t0) is a dimensionless multiplier (smallerthan 1) which is referred to as the aging coefficient With this importantsimplification the integral in Equation (19) can be eliminated as follows

εc(t) = σc(t0) 1 + φ(t t0)

Ec(t0)+ ∆σc(t)

1 + χφ(t t0)

Ec(t0)+ εcs(t t0) (110)

Equation (110) gives the strain which occurs during a period t0 to t due tothe combined effect of free shrinkage and a stress which varies in magnitudeduring the same period The first term on the right-hand side of Equation(110) is the instantaneous strain plus creep due to a stress of magnitude σc(t0)introduced at time t0 and sustained without change in magnitude until time tThe second term is the instantaneous strain plus creep due to a stress incre-ment (or decrement) of a magnitude changing gradually from zero at t0 to avalue ∆σc(t) at time t The last term is simply the free shrinkage occurringduring the considered period

For a practical example in which the stress on concrete varies with time asdescribed above consider a prestressed concrete cross-section At time t0 theprestressing is introduced causing compression on the concrete which grad-ually changes with time due to the losses caused by the combined effects ofcreep shrinkage and relaxation of the prestressed steel

Use of the aging coefficient χ as in Equation (110) greatly simplifies theanalysis of strain caused by a gradually introduced stress increment ∆σc orinversely the magnitude of the stress increment ∆σc can be expressed in termsof the strain it produces The aging coefficient will be extensively used in thistext for the analysis of the time-dependent stresses and strains in prestressedand reinforced concrete members

In practical computations the aging coefficient can be taken from a tableor a graph (see Appendix A) or simply assumed its value generally variesbetween 06 and 09 The method of calculating the aging coefficient will bediscussed in Section 18 but this may not be of prime concern in practical


design However it is important that the reader understands at this stage themeaning of the aging coefficient and how it is used in Equation (110)

18 Equation for the aging coefficient

The stress variation between t0 and t (Fig 15) may be expressed as

ξ1 =σc(τ) minus σc(t0)

∆σc(t)(111)

where ξ1 is a dimensionless time function defining the shape of the stressndashtimecurve the value of ξ1 at any time τ is equal to the ratio of the stress changebetween t0 and τ to the total change during the period (t minus t0) The value ofthe shape function ξ1 varies between 0 and 1 as τ changes from t0 to t

Differentiation of Equation (111) with respect to time gives

dσc(τ)

dτ= ∆σc(t)

dξ1

dτ(112)

Substitution of Equation (112) into (19) gives

εc(t) = σc(t0)1 + φ(t t0)

Ec(t0)+ ∆σc(t)

t

t0

1 + φ(t τ)

Ec(τ) dξ1

dτ dτ

+ εcs (t t0) (113)

Comparison of Equation (113) with (110) gives the following expressionfor the aging coefficient

χ(t t0) =Ec(t0)

φ(t t0)

t

t0

1 + φ(t τ)

Ec(τ) dξ1

dτ dτ minus1

φ(t t0)(114)

Three functions of time are included in Equation (114) ξ1 Ec(τ) andφ(t τ) of which the last two depend upon the quality of concrete and theambient air Examples of expressions that can be used for the variables Ec andφ are given in Appendix A

In practical applications the actual shape of variation of stress σc(τ) is oftenunknown and the function ξ1 defining this shape must be assumed In pre-paration of the graphs and the table for the aging coefficient χ presentedin Appendix A the time function ξ1 is assumed to have the same shape asthat of the timendashrelaxation curve for concrete which will be discussed inSection 19

As mentioned in the preceding section the aging coefficient χ is intended


for use in the calculation of strain due to stress which varies with time as forexample in the cross-section of a prestressed member made from one ormore types of concrete (composite section) Shrinkage creep and relaxationresult in gradual change in stresses in the concrete and the steel The use ofprecalculated values of the coefficient χ in the analysis of strain or stress byEquation (110) in such examples implies an assumption of the shape ofvariation of the stress during the period (t minus t0) The margin of error causedby this approximation is generally small

We have seen from the equations of this section that χ and φ are functionsof t0 and t the ages of the concrete at loading and at the time the strain isconsidered The product χφ often occurs in the equations of this book tosimplify the notation we will use the symbol χφ to mean

χφ(t t0) equiv χ(t t0) φ(t t0)

19 Relaxation of concrete

In the discussion presented in this section we exclude the effect of shrinkageand consider only the effect of creep

When a concrete member is subjected at age t0 to an imposed strain εc theinstantaneous stress will be

σc(t0) = εcEc(t0) (115)

where Ec(t0) is the modulus of elasticity of concrete at age t0 If subsequentlythe length of the member is maintained constant the strain εc will not changebut the stress will gradually decrease because of creep (Fig 16) The value ofstress at any time t gt t0 may be expressed as follows

σc(t) = εcr(t t0) (116)

where r(t t0) is the relaxation function to be determined in the followingsection The value r(t t0) is the stress at age t due to a unit strain introduced atage t0 and sustained constant during the period (t minus t0)

At any instant τ between t0 and t the magnitude of the relaxed stress ∆σc(τ)may be expressed as follows

∆σc(τ) = ξ[∆σc(t)] (117)

where ∆σc(τ) is the stress increment (the stress relaxed) during the period t0 to τ

∆σc(τ) = σc(τ) minus σc(t0) (118)

Similarly the stress increment during the period t0 to t is


∆σc(t) = σc(t) minus σc(t0) (119)

The symbol ξ is a dimensionless shape function representing for any valueτ the ratio of the stress relaxed during the period (τ minus t0) to the stress relaxedduring the whole period (t minus t0) Thus

ξ =∆σc(τ)

∆σc(t)(120)

The value of ξ is 0 and 1 when τ = t0 and t respectively The shape function ξhas the same significance as ξ1 adopted in the preceding section (Equation(111) )

Referring to Fig 16 the strain value εc which exists at time t may beconsidered as being the result of (a) an initial stress σc(t0) introduced at aget0 and maintained constant up to age t and (b) a stress increment ∆σc(t)introduced gradually during the period (t minus t0) Thus using Equation (110)

εc = σc(t0) 1 + φ (t t0)

Ec(t0)+ ∆σc(t)

1 + χφ(t t0)

Ec(t0)(121)

Substitution of Equations (115) (116) and (119) and (121) gives

Figure 16 Variation of stress with time due to a strain c imposed at age t0 and maintainedconstant thereafter (phenomenon of relaxation)


εc = εc[1 + φ(t t0)] + εc[r(t t0) minus Ec(t0)] 1 + χφ(t t0)

Ec(t0)(122)

We recall that the symbol χφ(t t0) indicates the product of two functionsχ and φ of the time variables t and t0 The constant strain value εc inEquation (122) cancels out and by algebraic manipulation of the remainingterms we can express the aging coefficient χ in terms of Ec(t0) r(t t0) andφ(t t0)

χ(t t0) =1

1 minus r(t t0)Ec(t0)minus

1

φ(t t0)(123)

A step-by-step numerical procedure will be discussed in the following sec-tion for the derivation of the relaxation curve in Fig 16 The relaxationfunction r(t t0) obtained in this way can be used to calculate the agingcoefficient χ(t t0) by Equation (123)

110 Step-by-step calculation of the relaxationfunction for concrete

The step-by-step numerical procedure introduced in this section can be usedfor the calculation of the time-dependent stresses and deformations in con-crete structures It is intended for computer use and is particularly suitable forstructures built or loaded in several stages as for example in the segmentalconstruction method of prestressed structures In this section a step-by-stepmethod will be used to derive the relaxation function r(τ t0) Furtherdevelopment of the method is deferred to Sections 46 and 58

The value of the relaxation function r(t t0) is defined as the stress at time tdue to a unit strain introduced at time t0 and sustained without change duringthe period (t minus t0) (see Equation (116) )

Consider a concrete member subjected to uniaxial stress and assume thatthe magnitude of stress varies with time as shown in Fig 17(b) At age t0 aninitial stress value σc(t0) is introduced and subsequently increased graduallyor step-wise during the period t0 to t When the variation of stress with time isknown the step-by-step analysis to be described can be used to find the strainat any time τ between t0 and t Alternatively if the strain is known themethod can be used to determine the time variation of stress

Divide the period (t minus t0) into intervals (Fig 17(a) ) and assume that thestress is introduced in increments at the middle of the intervals Thus (∆σc)i isintroduced at the middle of the ith interval For a sudden increase in stressconsider an increment introduced at an interval of zero length (for example(∆σc)l and (∆σc)k) in Fig 17(b) ) The symbols tj minus 1

2 tj and tj + 1

2 are used to refer

to the instant (or the age of concrete) at the start the middle and the end of


the jth interval respectively The strain at the end of the ith interval can becalculated by Equation (19) replacing the first two terms by a summation asfollows

εc(ti + 12) =

i

j = 1(∆σc)j

1 + φ(ti + 12 tj)

Ec(tj) + εcs(ti + 1

2 t0) (124)

The summation represents the superposition of strain caused by stress incre-ments When the magnitude of the increments is known the sum gives thestrain In the case when the strain is known the stress increments can bedetermined in steps The stress at the end of the ith interval is

σc(ti + 12) =

i

j = 1

(∆σc)j (125)

Consider now the case when a strain εc is imposed at the time t0 andsustained constant up to time t The corresponding stress introduced at t0 is

Figure 17 Division of (a) time into intervals and (b) stress into increments for step-by-step analysis


εcEc(t0) and its value will gradually drop following the relaxation functionaccording to Equation (116) Assume that the time after t0 is divided intointervals as in Fig 17(a) and apply Equation (116) at the end of the ithinterval

σc(ti + 12) = εcr(ti + 1

2 t0) (126)

Substitution of Equation (125) into (126) gives the value of the relaxationfunction at the end of the ith interval

r(ti + 12 t0) =

1

εc

i

j = 1

(∆σc)j (127)

Rewrite Equation (124) separating the last term of the summation

εc(ti + 12) = (∆σc)i

1 + φ(ti + 12 ti)

Ec(ti)+

i minus 1

j = 1

(∆σc)j 1 + φ(ti + 1

2 tj)

Ec(tj)

+ εcs(ti + 12 t0) (128)

Consider that the strain εc(ti + 12) is known at the end of all intervals and it is

required to find the stress increments Values of the modulus of elasticity ofconcrete creep coefficients and free shrinkage are also assumed to be knownfor all intervals as needed in Equation (128) In the step-by-step analysis thestress increment for any interval is determined after the increments of allthe preceding intervals have been determined Thus Equation (128) can besolved for the only unknown stress increment (∆σc)i

(∆σc)i =Ec(ti)

1 + φ(ti + 12 ti)

εc(ti + 12) minus εcs(ti + 1

2 t0)

minus i minus 1

j = 1

(∆σc)j 1 + φ(ti + 1

2 tj)

Ec(tj) (129)

Successive application of this equation with i = 1 2 gives the stressincrements Equations (129) and (127) can be employed in this manner todetermine the relaxation function r(t t0) For this purpose εcs(ti + 1

2 t0) = 0 and

εc(ti + 12) = εc = constant for all i values εc may be conveniently chosen equal to

unity This procedure is employed to calculate r(t t0) which is subsequentlysubstituted in Equation (123) to determine the aging coefficient χ(t t0) inpreparation of the graphs in part (b) of each of Figs A6 to A45 and TableA3 in Appendix A The same appendix also includes an example plot bycomputer of the relaxation function (see Fig A5)


The aging coefficient χ(t t0) calculated by the above procedure dependsmainly upon t0 and t other factors affecting χ are the time functions φ(t τ)and Ec(τ) The graphs and table presented for χ in Appendix A are based ontime functions for φ and Ec in accordance with MC-90 and the ACI Com-mittee 209 report7 respectively Choice of other functions results in smallchange in the value of χ but this change may be ignored in practice Since χ isalways used as a multiplier to φ which is rarely accurately determinedhigh accuracy in the derivation of χ is hardly justified Appendix Gincludes information about computer programs that perform the step-by-stepcalculations discussed in this section The programs can be executed onmicro-computers using the software provided on the Internet as optionalcompanion of this book (See web address in Appendix G)

111 Age-adjusted elasticity modulus

The three terms in Equation (110) represent the strain in concrete at age tdue to a stress σc(t0) introduced at age t0 and sustained during the period(t minus t0) a stress increment of magnitude zero at t0 increasing gradually to afinal value ∆σc(t) at age t and the free shrinkage occurring during the period(t minus t0) This equation may be rewritten as follows

εc(t) = σc(t0) 1 + φ(t t0)

Ec(t0)+

∆σc(t)

Ec(t t0)+ εcs(t t0) (130)

where

Ec(t t0) =Ec(t0)

1 + χφ(t t0)(131)

Ec(t t0) is the age-adjusted elasticity modulus to be used in the calculation ofthe total strain increment instantaneous plus creep due to a stress incre-ment of magnitude developing gradually from zero to a value ∆σc(t) Thusthe strain increment in the period (t minus t0) caused by the stress ∆σc(t) is givenby

∆εc(t) =∆σc(t)

Ec(t t0) (132)

1111 Transformed section

In various chapters of this book the term transformed section is employed tomean a cross-section of a reinforced concrete member for which the actualarea is replaced by a transformed area equal to the area of concrete plus αtimes the area of steel where


α(t0) =Es(Eps or Ens)

Ec(t0)(133)

where Es is the modulus of elasticity of the reinforcement When prestressedor non-prestressed steel are involved the subscripts ps or ns are employed torefer to the two types of reinforcement Ec(t0) is the modulus of elasticity ofconcrete at age t0 It thus follows that α is also a function of t0

In the analysis of stresses due to forces gradually developed during aperiod t0 to t we will use in Chapter 2 the term age-adjusted transformedsection to mean a transformed section for which the actual area is replaced bya transformed area composed of the area of concrete plus α times the area ofsteel where

α(t t0) =Es(Eps or Ens)

Ec(t t0) (134)

1112 Age-adjusted flexibility and stiffness

Similarly when the age-adjusted modulus of elasticity of concrete is used inthe calculation of a flexibility or stiffness of a structure the result is referredto as an age-adjusted flexibility or age-adjusted stiffness

112 General

Creep and shrinkage of concrete and relaxation of steel result in deform-ations and in stresses that vary with time This chapter presents the basicequations for two methods for the analysis of time-dependent stresses anddeformations in reinforced and prestressed concrete structures The first issuitable for hand computation and requires knowledge of an aging coefficientχ (generally between 06 and 09) which may be taken from a graph or atable (see Appendix A) The second is a step-by-step numerical procedureintended for computer use In Chapters 2 to 9 the first method is extensivelyemployed for the analysis of changes of stress and internal forces caused bycreep shrinkage and relaxation of steel in statically determinate andindeterminate structures The second method namely the step-by-step pro-cedure is employed for the same purpose in Sections 46 and 58 AppendixA gives equations and graphs for the material parameters discussed in thischapter based upon requirements of codes and technical committeerecommendations

Notes1 See Neville AM (1997) Properties of Concrete 4th edn Wiley New York2 See Neville AM Dilger WH and Brooks JJ (1983) Creep of Plain and

Structural Concrete Construction Press London


3 Comiteacute Euro-International du Beacuteton (CEB) ndash Feacutedeacuteration Internationale de laPreacutecontrainte (FIP) (1990) Model Code for Concrete Structures (MC-90) CEBThomas Telford London 1993 American Concrete Institute (ACI) Committee209 (1992) Prediction of Creep Shrinkage and Temperature Effects in ConcreteStructures 209R-92 ACI Detroit Michigan 47 pp British Standard BS 8110 Part1 1997 and Part 2 1985 Structural Use of Concrete British Standards Institute 2Park Street London W1A 2BS Part I is reproduced by Deco 15210 Stagg StreetVan Nuys Ca 91405ndash1092 USA

4 Based on Magura D Sozen MA and Siess CP (1964) A study of stressrelaxation in prestressing reinforcement PCI Journal 9 (2) 13ndash57

5 See reference mentioned in note 3 above6 Eurocode 2 (1991) Design of Concrete Structures Part 1 General Rules and Rules

for Buildings European Prestandard ENV 1992ndash1 1991E European Committeefor Standardization rue de Stassart 36 B-1050 Brussels Belgium

7 See reference mentioned in note 3 above


Stress and strain ofuncracked sections

21 Introduction

Cross-sections of concrete frames or beams are often composed of threetypes of material concrete prestressed steel and non-prestressed reinforce-ment In some cases concrete of more than one type is employed in onecross-section as for example in T-sections where the web is precast and theflanges are cast in situ Concrete exhibits the properties of creep and shrink-age and prestressed steel loses part of its tension due to relaxation Thus thecomponents forming one section tend to have different strains Howeverbecause of the bond the difference in strain is restrained Thus the stresses in

Pre-tensioned element of double tee cross-section at time of cutting of prestressedstrands (Courtesy Prestressed Concrete Institute Chicago)

Chapter 2

concrete and the two types of reinforcement change with time as creepshrinkage and relaxation develop

This chapter is concerned with the calculation of the time-dependentstresses and the associated strain and curvature in individual cross-sectionsof reinforced prestressed or composite members Cross-sections composed ofconcrete and structural steel sections are treated in the same way as reinforcedconcrete members with the only difference that the steel section has a flexuralrigidity which is not ignored

The cross-sections considered are assumed to have one axis of symmetryand to be subjected to a bending and an axial force caused by prestressing orby other loading Perfect bond is assumed between concrete and steel thus atany fibre the strains in concrete and steel are equal Plane cross-sections areassumed to remain plane after deformation No cracking is assumed in theanalysis procedures presented in this chapter analysis of cracked sections istreated in Chapter 7

Prestressing is generally applied in one of two ways pre-tensioning or post-tensioning With pre-tensioning a tendon is stretched in the form in whichthe concrete member is cast After the concrete has attained sufficientstrength the tendon is cut Because of bond with concrete the tendon cannotregain its original length and thus a compressive force is transferred to theconcrete causing shortening of the member accompanied by an instant-aneous loss of a part of the prestress in the tendon We here assume that thechange in strain in steel that occurs during transfer is compatible with theconcrete strain at the same fibre The slip that usually occurs at the extremitiesof the member is ignored

With post-tensioning the tendon passes through a duct which is placed inthe concrete before casting After attaining a specified strength tension isapplied on the tendon and it is anchored to the concrete at the two ends andlater the duct is grouted with cement mortar During tensioning of the ten-don before its anchorage the strain in steel and concrete are not compatibleconcrete shortens without causing instantaneous loss of the prestress forceAfter transfer perfect bond is assumed between the tendon the grout theduct and the concrete outside the duct This assumption is not justified when thetendon is left unbonded However in most practical calculations the incom-patibility in strain which may develop after prestress transfer between thestrain in an unbonded tendon and the adjacent concrete is generally ignored

In this chapter we are concerned with the stress strain and deformations ofa member for which the elongations or end rotation are not restrained by thesupports or by continuity with other members Creep shrinkage and relaxa-tion of steel change the distribution of stress and strain in the section but donot change the reactions and the induced stress resultants (values of the axialforce or bending moment acting on the section) Analysis of the time-dependent effects on continuous beams and other statically indeterminatestructures are discussed in Chapters 4 and 5

Stress and strain of uncracked sections 21

Creep and shrinkage of concrete and relaxation of prestressed steel resultin prestress loss and thus in time-dependent change of the internal forces (theresultant of stresses) on the concrete cross-section Generally in a prestressedsection non-prestressed reinforcement is also present The time effects ofcreep shrinkage and relaxation usually produce a reduction of tension inthe prestressed steel and of compression in the concrete and an increase ofcompression in the non-prestressed steel

At the time of prestressing or at a later date external loads are oftenintroduced as for example the self-weight The internal forces due to suchloading and the time of their application are assumed to be known Theinitial prestressing is assumed to be known but the changes in the stress inthe prestressed and non-prestressed steels and the concrete are determinedby the analysis

22 Sign convention

The following sign convention will be adopted in all chapters Axial force N ispositive when tensile In a horizontal beam a bending moment M that pro-duces tension at the bottom fibre and the corresponding curvature ψ arepositive Tensile stress σ and the corresponding strain ε are positive thus thevalue of shrinkage of concrete εcs is generally a negative quantity The sym-bol P indicates the absolute value of the prestress force ∆ represents anincrement or decrement when positive or negative respectively Thus the lossof tension in the prestressed steel due to creep shrinkage and relaxation isgenerally a negative quantity

23 Strain stress and curvature in composite andhomogeneous cross-sections

Fig 21(a) is the cross-section of a member composed of different materialsand having an axis of symmetry For the analysis of stresses due to normalforce or moment on the section we replace the actual section by a trans-formed section for which the actual area of any part i is replaced by atransformed area given by (EiEref)Ai where Eref is an arbitrarily chosen valueof a reference modulus of elasticity Ei is the modulus of elasticity of part iof the section The member is thus considered to have a modulus of elasticityEref and cross-section properties for example area and moment of inertiaequal to those of the transformed section

In reinforced and prestressed concrete cross-sections the reference modu-lus is taken to be equal to Ec the modulus of elasticity of concrete of one ofthe parts and the reinforcement area prestressed and non-prestressed isreplaced by α times the actual area where α is the ratio of the modulus ofelasticity of the reinforcement to the modulus of elasticity of concrete (seeEquation (133) )


Assume that the cross-section in Fig 21(a) is subjected to a force N nor-mal to the section situated at any point on the symmetry axis Such a force isstatically equivalent to a system composed of a normal force N at a referencepoint O and a bending moment M as shown in Fig 21(a) The equationsmost commonly used in calculations of stress strain and curvature at thecross-section are generally based on the assumption that O is the centroid ofthe transformed section

When considering the effects of creep we shall use for the analysis of thesame cross-section different elasticity moduli for concrete and superpose thestresses from several analyses (see Section 25) Changing the value of Ec willresult in a change of the centroid of the transformed section To avoid thisdifficulty we derive the equations below for the strain curvature and thestress distribution of a cross-section without requiring that the referencepoint O be the centroid of the cross-section Thus O is an arbitrarily chosenreference point on the axis of symmetry

The strain distribution is assumed to be linear as shown in Fig 21(b)in other words a plane cross-section is assumed to remain plane afterdeformation At any fibre at a distance y from the reference point O thestrain is

ε = εO + ψy (21)

where εO is the strain at the reference point and ψ is the curvature Thedistance y is positive when the point considered is below the reference point

Figure 21 Analysis of strain distribution in a composite cross-section by Equation (215)(a) positive M N and y (b) strain distribution


When the fibre considered is in the ith part of the composite section thestress at the fibre is

σ = Ei(εO + ψy) (22)

Integration of the stress over the area of the cross-section and taking themoment about an axis through O gives

N = σdA (23)

M = σydA (24)

The integral is to be performed for all parts of the cross-sectionSubstitution of Equation (22) into (23) and (24) gives

N = εO m

i = 1

Ei dA + ψ m

i = 1

Ei ydA (25)

M = εO m

i = 1

Ei y dA + ψ m

i = 1

Ei y2dA (26)

Thus summations in Equations (25) and (26) are to be performed fromi = 1 to m where m is the number of parts in the cross-section Equations (25)and (26) may be rewritten

N = Eref(AεO + Bψ) (27)

M = Eref(BεO + Iψ) (28)

where A B and I are the transformed cross-section area and its first andsecond moment about an axis through O

For a composite section A B and I are derived by summing up thecontribution of the parts

A = m

i = 1

Ei

Eref

Ai (29)

B = m

i = 1

Ei

Eref

Bi (210)

I = m

i = 1

Ei

Eref

Ii (211)


where Ai Bi and Ii are respectively the area of the ith part and its first andsecond moment about an axis through O A reinforcement layer may betreated as one part

Equations (27) and (28) may be rewritten in the matrix form

NM = Eref ABB

I εO

ψ (212)

This equation may be used to find N and M when εO and ψ are known orwhen N and M are known the equation may be solved for the axial strain andcurvature

εO

ψ =1

ErefAB

B

I minus1

NM (213)

The inverse of the 2 times 2 matrix in this equation is

ABB

I minus1

=1

(AI minus B2) I

minusB

minusB

A (214)

Substitution in Equation (213) gives the axial strain at O and the curvature

εO

ψ =1

Eref(AI minus B2) I

minusB

minusB

A NM (215)

When the reference point O is chosen at the centroid of the transformedsection B = 0 and Equation (215) takes the more familiar form

εO

ψ =1

Eref

NA

MI (216)

231 Basic equations

The equations derived above give the stresses and the strains in a cross-section subjected to a normal force and a bending moment (Fig 21)Extensive use of these equations will be made throughout this book inanalysis of reinforced composite or non-composite cross-sections Because ofthis the basic equations are summarized below and the symbols defined foreasy reference

ε = εO + ψy σ = E(εO + ψy) (217)

N = E(AεO + Bψ) M = E(BεO + Iψ) (218)


εO =IN minus BM

E(AI minus B2)ψ =

minusBN + AM

E(AI minus B2)(219)

σO =IN minus BM

AI minus B2γ =

minusBN + AM

AI minus B2(220)

where

When the section is composed of more than one material (eg concrete partsof different age prestressed non-prestressed steel structural steel) E inEquation (217) is the modulus of elasticity of the material for which thestress is calculated A B and I are properties of a transformed section com-posed of the cross-section areas of the individual materials each multipliedby its modulus of elasticity divided by a reference modulus whose value is tobe used in Equations (218) and (219)

Figure 22 Cross-section of a member subjected to a rise of temperature which variesnon-linearly over the depth

A B and I = cross-sectional area and its first and second moment about ahorizontal axis through reference point O respectively

E = modulus of elasticityy = coordinate of any fibre with respect to a horizontal axis

through reference point O y is measured downward (Fig 22)N = normal forceM = bending moment about a horizontal axis through reference

point Oε and σ = strain and stress at any fibreεO and σO = strain and stress at reference point Oψ and γ = dεdy (the curvature) and dσdy respectively


24 Strain and stress due to non-lineartemperature variation

Analysis of the change in stresses due to creep shrinkage of concrete andrelaxation of prestressed steel in concrete structures can be done in the sameway as the analysis of stresses due to temperature (as will be shown in Sec-tions 25 54 to 56 and 107) For this reason we shall consider here thestrain and stress in a cross-section subjected to a temperature rise of magni-tude T(y) which varies over the depth of the section in an arbitrary fashion(Fig 22)

In a statically determinate frame uniform or linearly varying temperatureover the depth of the cross-section of a member produces no stresses Whenthe temperature variation is non-linear (Fig 22) stresses are producedbecause each fibre being attached to adjacent fibres is not free to acquire thefull expansion due to temperature The stresses produced in this way in anindividual cross-section must be self-equilibrating in other words the tem-perature stress in a statically determinate structure corresponds to no changein the stress resultants (the internal forces) We shall discuss below the analy-sis of the stresses produced by a rise of temperature which varies non-linearlyover the depth of a member of a statically determinate framed structure

The self-equilibrating stresses caused by non-linear temperature variationover the cross-sections of statically determinate frame are sometimes referredto as the eigenstresses If the structure is statically indeterminate the elonga-tions andor the rotations of the joints of the members are restrained orprevented resulting in a statically indeterminate set of reactions which arealso self-equilibrating but these will produce statically indeterminate internalforces and corresponding stresses Statically indeterminate forces producedby temperature will be discussed in Section 108 The present section is con-cerned with the axial strain the curvature and the self-equilibrating stressesin a cross-section of a statically determinate structure subjected to a rise oftemperature which varies non-linearly over the depth of the section (Fig 22)

The hypothetical strain that would occur at any fibre if it were free is

εf = αtT (221)

where T = T(y) the temperature rise at any fibre at a distance y below areference point O and αt = coefficient of thermal expansion

If this strain is artificially prevented the stress in the restrained conditionwill be

σrestrained = minusEεf (222)

where E is the modulus of elasticity which is considered for simplicity to beconstant over the whole depth of the section


The resultant of this stress may be represented by an axial force ∆N at areference point O and a bending moment ∆M given by

∆N = σrestrained dA (223)

∆M = σrestrainedy dA (224)

Substitution of Equation (222) into (223) and (224) gives

∆N = minus Eεf dA (225)

∆M = minus Eεfy dA (226)

The artificial restraint is now released by the application of a force minus ∆N atO and a bending moment minus ∆M the resulting axial strain and curvature areobtained by Equation (219) and the corresponding stress by Equation (217)

∆εO

∆ψ =1

E(AI minus B2) I

minusB

minusB

A minus∆N

minus∆M (227)

∆σ = E[∆εO + (∆ψ)y] (228)

where A B and I are the area and its first and second moment about an axisthrough the reference point O When O is at the centroid of the section B = 0and Equation (227) becomes

∆εO

∆ψ =1

E minus∆NA

minus∆MI (229)

The actual stress due to temperature is the sum of σrestrained and ∆σ thus(Equations (222) and (228) )

σ = E [minus εf + ∆εO + (∆ψ)y] (230)

The equations of the present section are applicable for composite cross-sections having more than one material in this case A B and I are propertiesof a transformed section with modulus of elasticity of E = Eref The trans-formed section is composed of parts of cross-section area α times the actualareas of individual parts where α is the ratio of the modulus of elasticity ofthe part considered to Eref (see Equations (133) and (29)ndash(211) ) When thechange in temperature occurs at age t0 and takes a short time to develop suchthat creep may be ignored Eref = Ec(t0) where Ec(t0) is the modulus of elas-ticity at age t0 of one of the concrete parts chosen as reference When thechange in temperature develops gradually during a period t0 to t α is replacedby α and Ec(t0) by the age-adjusted modulus of elasticity Ec(t t0) as discussed


in Section 1111 (see Equation (134) ) The analysis in this way accounts forthe fact that creep of concrete alleviates the stresses due to temperature

Example 21 Rectangular section with parabolic temperaturevariation

Calculate the axial strain the curvature and the stress distribution in amember of a rectangular section subjected to a rise of temperaturewhich varies over the depth in the form of a parabola of the mth degree(Figs 23(a) and (b) ) The elongation and rotation at the member

Figure 23 Temperature stresses in a statically determinate member (Example 21)(a) cross-section (b) variation of the magnitude of rise of temperatureover depth (c) strain (d) stress (self-equilibrating)


ends are assumed to occur freely (structure statically determinateexternally)

Choose the reference point at the middle of the depth Equations(225) and (226) give

∆N

∆M = αtTtopE

minusbh

m + 1

bh2m

2(m + 1)(m + 2)

With A = bh I = bh312 Equation (229) gives

εO =αtTtop

m + 1

ψ = minusαtTtop

h

6m

(m + 1)(m + 2)

The variation of strain over the cross-section is shown in Fig 23(c)The corresponding stress calculated by Equation (230) is shown in Fig23(d) The values given in Figs 23(c) and (d) are calculated assumingthe temperature rise to vary over the depth as a parabola of fifth degree(m = 5) and other data as follows b = 1m h = 1m αt = 10minus5 per degC andTtop = 30 degC E = 25GPa Or in British units b = 40 in h = 40 in αt = 56times 10minus6 per degF and Ttop = 54 degF E = 3600ksi

25 Time-dependent stress and strain in acomposite section

The equations derived in Sections 23 and 24 will be employed here to findthe strain and the stress in a composite or reinforced concrete section whichmay have prestressed and non-prestressed steel Examples of the sectionsconsidered are shown in Fig 24

Consider a section (Fig 25(a)) subjected at age t0 to a prestressing force Pan axial force N at an arbitrarily chosen reference point O and a bendingmoment M It is required to find the strain the curvature and the stress inconcrete and steel at age t0 immediately after prestressing at age t where t isgreater than t0 Assumed to be known are the cross-section dimensions andthe reinforcement areas the magnitudes of P N and M the modulus ofelasticity of concrete Ec(t0) at age t0 the shrinkage εcs(t t0) that would occur at


any fibre if it were free the creep coefficient φ(t t0) and the aging coefficientχ(t t0)

The intrinsic relaxation ∆σpr that occurs during the period (t minus t0) is alsoassumed to be known A reduced relaxation value ∆σpr = χr(∆σpr) will be usedin the analysis The reduction factor χr must be assumed at the start of theanalysis and adjusted later if necessary as will be further discussed in Section32 here we assume that the reduced relaxation value ∆σpr is known

251 Instantaneous stress and strain at age t0

Before we can apply Equation (219) we must combine N and M with theprestressing forces into an equivalent normal force at O and a moment

Figure 24 Examples of cross-sections treated in Section 25

Figure 25 Analysis of time-dependent stress and strain in a composite section Allvariables are shown in their positive directions (a) cross-section (b) strain att0 (c) change in strain during the period t minus t0


Nequivalent

Mequivalent = N minus ΣPi

M minus ΣPiypsi (231)

where the subscript i refers to the ith prestressed steel layer and ypsi is itsdistance below the reference point O The summation in this equation is to beperformed for the prestressed steel layers Here we assume that the prestress isintroduced in one stage multi-stage prestressing will be discussed in Section37 P is the absolute value of the prestressing force

The instantaneous axial strain and curvature immediately after pre-stressing (Equation 215) are given by

εO(t0)

ψ(t0) =

1

Eref(AI minus B2) I

minusB

minusB

A NMequivalent

(232)

where A B and I are respectively the area and its first and second moment ofthe transformed section at time t0 (see Section 1111) the modulus of elas-ticity of concrete to be used here is Ec(t0) for the individual parts of thesection Eref is a reference modulus of elasticity which may be chosen equal toEc1(t0) the modulus at age t0 for concrete of part 1 (see Equations (29) to(211) )

When the reference point O is at the centroid of the transformed section attime t0 B = 0 and Equation (232) becomes

εO(t0)

ψ(t0) =

1

Eref

Nequivalent

A

Mequivalent

I

(233)

With post-tensioning the area of prestressed duct should be deductedfrom the area of concrete and the area of the prestressed steel excluded whencalculating the properties of the transformed section to be used in Equation(232) or (233)

The instantaneous strain and stress in concrete at any fibre (Equation(217) ) are

εc(t0) = εO(t0) + ψ(t0)y (234)

σc(t0) = [Ec(t0)]i[εO(t0) + ψ(t0)y] (235)

where y is the distance below the reference point O of the layer consideredand the subscript i refers to the number of the concrete part of the fibreconsidered

The instantaneous stress in the non-prestressed steel is


σns(t0) = Ens[εO(t0) + ψ(t0)yns] (236)

In the case of pretensioning the stress in the prestressed steel immediatelyafter transfer is

σps(t0) = (σps)initial + Eps[εO(t0) + ψ(t0)yps] (237)

where (σps)initial is the stress in prestressed steel before transfer The secondterm in this equation represents the instantaneous change in stress (generallya loss of tension due to shortening of concrete) Thus the instantaneousprestress change (the loss) in pretensioned tendon at the time of transfer is

(∆σps)inst = Eps[εO(t0) + ψ(t0)y] (238)

With post-tensioning compatibility of strain in the tendon does not takeplace at this stage and thus no instantaneous loss occurs1 The stresses inpost-tensioned tendon immediately before and after transfer are the same

σps(t0) = (σps)initial (239)

where (σps)initial is the initial stress in prestressed steel given by the prestressedforce P divided by the cross-section area of prestressed steel

252 Changes in stress and strain during the period t0 to t

In this step of the analysis we deal with a cross-section for which the initialstress and strain are known Creep shrinkage and relaxation of steel result instress redistribution between the various materials involved The analysis tobe presented here gives the stress changes in each material occurring during aspecified period of time In some cases the cross-section of the member ischanged at the beginning of the period for example by the addition of a partcast in situ to a precast section (see Fig 24) In such a case the initial stress inthe added part is known to be zero Assuming perfect bond the two partsbehave as one cross-section thus creep shrinkage and relaxation of any partwill affect both parts

The change in strain during the period t0 to t (Fig 25(c) ) is defined by theincrements ∆εO and ∆ψ in the axial strain and curvature To determine thesewe follow a similar procedure to that in Section 24 The change of straindue to creep and shrinkage of concrete and relaxation of prestressed steel isfirst artificially restrained by application of an axial force ∆N at O and abending moment ∆M Subsequently these restraining forces are removed bythe application of equal and opposite forces on the composite sectionresulting in the following changes in axial strain and in curvature (Equation(219) )


∆εO

∆ψ =1

Ec(AI minus B2) I

minusB

minusB

A minus∆N

minus∆M (240)

where A B and I are respectively the area of the age-adjusted transformedsection and its first and second moment about an axis through the referencepoint O (see Section 1111) Ec = Eref = Ec(t t0) is the age-adjusted elasticitymodulus of one of the concrete types chosen as reference material (Equation(131) ) The restraining forces are calculated as a sum of three terms

∆N

∆M = ∆N

∆Mcreep

+ ∆N

∆Mshrinkage

+ ∆N

∆Mrelaxation

(241)

If creep were free to occur the axial strain and curvature would increaseduring the period t0 to t by the amounts φ(t t0) ε(t0) and φ(t t0) ψ(t0) Theforces necessary to prevent these deformations may be determined byEquation (218)

∆N

∆Mcreep

= minusm

i = 1

EcφAc

Bc

Bc

Ic εO(t0)

ψ(t0)

i

(242)

The subscript i refers to the ith part of the section and m is the totalnumber of concrete parts Aci Bci and Ici are respectively the area of con-crete of the ith part and its first and second moment about an axisthrough the reference point O Eci = [Ec(t t0)]i and φi = [φ(t t0)]i are the age-adjusted modulus of elasticity and creep coefficient for concrete in the ithpart

When applying Equation (242) it should be noted that [εO(t0)]i and [ψ(t0)]i

are two quantities defining a straight line of the strain distribution on the ithpart and the value [εO(t0)]i is the strain at the reference point O (which may notbe situated in the ith part (see Example 24) )

In Equation (242) it is assumed that all loads are applied at age t0 in casethere are other loads introduced at an earlier age the vector φεO ψ must bereplaced by a vector of two values equal to the total axial strain and curva-ture due to creep if it were free This is equal to the sum of products ofinstantaneous strains and curvatures by appropriate creep coefficients (seepart (d) of solution of Example 25)

The forces required to prevent shrinkage are

∆N

∆Mshrinkage

= minus m

i = 1

EcεcsAc

Bc

i

(243)

where εcs = εcs(t t0) is the free shrinkage for the period t0 to t


The age-adjusted moduli of elasticity are used in Equations (240) (242)and (243) (indicated by a bar as superscript) because the forces ∆N and ∆Mare gradually developed between the instants t0 and t

The forces necessary to prevent the strain due to relaxation of prestressedsteel are

∆N

∆Mrelaxation

= Aps∆σ-pr

Apsyps∆σ-pr

i

(244)

The subscript i in this equation refers to a prestressed steel layer Aps is itscross-section area and yps its distance below the reference point O and ∆σpr isthe reduced relaxation during the period t0 to t

The stress in concrete required to prevent creep and shrinkage at anyfibre is

σrestrained = minusEc(t t0)[φ(t t0)εc(t0) + εcs] (245)

where εc(t0) is the instantaneous strain determined earlier (Equation (234) )In Equation (245) we assume that all loads are applied at t0 in the case whenother loads are introduced earlier the quantity (φεc) must be replaced by thesum of products of instantaneous strains by appropriate creep coefficient (seepart (d) of solution of Example 25)

The stress increments that develop during the period (t minus t0) are as followsIn concrete at any fibre in the ith part

∆σc = σrestrained + Ec(t t0)(∆εO + y∆ψ) (246)

in non-prestressed steel

∆σns = Ens(∆εO + yns∆ψ) (247)

and in prestressed steel

∆σps = ∆σpr + Eps(∆εO + yps∆ψ) (248)

The last equation gives the change in prestress due to creep shrinkage andrelaxation Multiplication of ∆σps by Aps gives the loss of tension in theprestressed steel

The procedure of analysis presented in this section is demonstrated by thefollowing examples The input data and the main results are given in allexamples throughout this book in both SI and British units however theexamples are worked out either in SI units or in British units


Example 22 Post-tensioned section

A prestress force P = 1400 times 103N (315kip) and a bending moment M =390 times 103 N-m (3450kip-in) are applied at age t0 on the rectangularpost-tensioned concrete section shown in Fig 26(a) Calculate thestresses the axial strain and curvature at age t0 and t given the followingdata Ec(t0) = 300GPa (4350ksi) Ens = Eps = 200GPa (29 times 103 ksi)uniform free shrinkage value εcs(t t0) = minus240 times 10minus6 φ(t t0) = 3 χ = 08reduced relaxation ∆σpr = minus80MPa (minus12ksi) The dimensions of thesection and cross-section areas of the reinforcement and the prestressduct are indicated in Fig 26(a)

(a) Stress and strain at age t0

Calculation of the properties of the transformed section at time t0 isdone in Table 21 The reference modulus of elasticity Eref = Ec(t0) =300GPa The forces introduced at age t0 are equivalent to Equation(231) is

NMequivalent

= minus1400 times 103

390 times 103 minus 1400 times 103 times 045 = minus1400 times 103 N

minus240 times 103 N-m

The instantaneous axial strain at O and curvature (Equation (232) ) is

εO(t0)

ψ(t0) =

1

30 times 109[03712 times 4688 times 10minus3 minus (0208 times 10minus3)2]

times 4688 times 10minus3

minus0208 times 10minus3


03712 minus1400 times 103

minus240 times 103

= 10minus6 minus126

minus170mminus1

The concrete stress at top and bottom fibres (Equation (235) ) is

(σc(t0) )top = 30 times 109[minus126 + (minus06)(minus170)] 10minus6

= minus0706MPa (minus0102ksi)

(σc(t0) )bot = 30 times 109[minus126 + 06(minus170)] 10minus6


The stress distribution is shown in Fig 26(b)


Figure 26 Analysis of stress and strain in the cross-section of a post-tensionedmember (Example 22) (a) cross-section dimensions (b) condition at aget0 immediately after prestress (c) changes caused by creep shrinkage andrelaxation


Tabl

e 2

1C

alcu

latio

n of

A B

and

I of

tra

nsfo

rmed

sec

tion

at t

ime

t 0

Prop

ertie

s of

are

aPr

oper

ties

of t

rans

form

ed a

rea

AB

IAE

Ere

fBE

Ere

fIE

Ere

f

(m2)

(m3)

(m4)

(m2)

(m3)

(m4)

Conc

rete

035

45minus1

625

times 1

0minus341

84

times 1

0minus30

3545

minus16

25 times

10minus3

418

4 times

10minus3

Non

-pre

stre

ssed

ste

el25

00 times

10minus6

027

5 times

10minus3

075

6 times

10minus3

001

671

833

times 1

0minus35

04 times

10minus3

Pres

tres

sed

stee

lmdash

mdashmdash

mdashmdash

mdash

Prop

ertie

s of

037

120

208

times 1

0minus346

88

times 1

0minus3

trans

form

ed s

ectio

nA

BI

(b) Changes in stress and strain due to creep shrinkageand relaxation

The age-adjusted elasticity modulus of concrete (Equation (131) ) is

Ec(t t0) =30 times 109

1 + 08 times 3= 882GPa

The stress in concrete at the top and bottom fibres when the straindue to creep and shrinkage is artificially restrained (Equations (234)and (245) ) is

(σc restrained)top = minus882 times 109[3 times 10minus6(minus126 + 170 times 06) minus240 times 10minus6]

= 2741MPa (0398ksi)

(σc restrained)bot = minus882 times 109[3 times 10minus6(minus126 minus 170 times 06) minus240 times 10minus6]

= 8145MPa (1181ksi)

The restraining forces (Equations (241) to (244) ) are

∆N

∆Mcreep

= minus882 times 109 times 3 03545



4184 times 10minus3

times minus126

minus170 10minus6 = 106 1175N

01828N-m

∆N

∆Mshrinkage

= minus882 times 109(minus 240 times 10minus6) 03545


= 106 0750N

minus00034N-m

∆N

∆Mrelaxation

= 1120 times 10minus6(minus80 times 106)

1120 times 10minus6 times 045 (minus80 times 106)

= 106minus0090N

minus00403N-m

∆N

∆M = 106 1175 + 0750 minus 0090

01828 minus 00034 minus 00403 =1061835 N

0139 N-m

Calculation of the properties of the age-adjusted transformed section isperformed in Table 22 using Eref = Ec(t t0) = 882GPa and α(t t0) =2268 (Equation (131) )


Tabl

e 2

2C

alcu

latio

n of

A B

and

I of

the

age

-adj

uste

d tr

ansf

orm

ed s

ectio

n

Prop

ertie

s of

are

aPr

oper

ties

of tr

ansf

orm

ed a

rea

AB

IAE

Ere

fBE

Ere

fIE

Ere

f

(m2)

(m3)

(m4)

(m2)

(m3)

(m4)

Conc

rete

035

45minus1

625

times 1

0minus341

84

times 1

0minus30

3545

minus16

25 times

10minus3

418

4 times

10minus3

Non

-pre

stre

ssed

ste

el25

00 times

10minus6

027

5 times

10minus3

075

6 times

10minus3

005

676

236

times 1

0minus317

24

times 1

0minus3

Pres

tres

sed

stee

l11

20 times

10minus6

050

4 times

10minus3

022

7 times

10minus3

002

5411

429

times 1

0minus35

15 times

10minus3

Prop

ertie

s of

age

-adj

uste

d0

4366

160

40 times

10minus3

641

2 times

10minus3

trans

form

ed s

ectio

nA

BI

The prestress duct is usually grouted shortly after the prestresshence its area may be included in Table 22 but this is ignored here

∆εO

∆ψ =

1





04366 minus1835

minus0139 106

= 10minus6minus470

minus128mminus1

Increments of stress that will develop during the period (t minus t0) inconcrete non-prestressed steel and prestressed steel are (Equations(246ndash48) )

(∆σc)top = 2741 times 106 + 882[minus471 + (minus06)(minus128)]103


(∆σc)bot = 8145 times 106 + 882[minus471 + 06(minus128)]103

= 3313MPa (0481ksi)

∆σns2 = 200[minus471 + (minus055)(minus128)]103

= minus 801MPa (minus116ksi)

∆σns1 = 200[minus471 + 055(minus128)]103


∆σps = minus80 times 106 + 200[minus471 + 045(minus128)]103


The last value is the loss of prestress in the tendon Fig 26(b) showsthe distributions of stress and strain on the concrete and the resultantsof forces on concrete and steel at age t0 The changes in these variablescaused by creep shrinkage and relaxation are shown in Fig 26(c)From these figures it is seen that the loss in tension in the prestressedsteel due to these effects is 208kN or 15 of the original tension(1400kN) The resultant compressive force on the concrete at age t0 is1329kN and the difference (1400 minus 1329 = 71kN) represents the com-pression in the non-prestressed steel The loss in compression in con-crete due to creep shrinkage and relaxation amounts to 451kN which is


32 of the initial compression in the concrete (1329kN) Thehigher percentage is caused by the compression picked up by thenon-prestressed steel as creep and shrinkage develop

The results of this example may be checked by verifying that the sumof the changes of the resultants of stress in concrete and steel is zeroThus the system of forces shown in Fig 26(c) is in equilibrium

A check on compatibility can be made by verifying that the change instrain in prestress steel caused by (∆σps minus ∆σpr) is equal to the change instrain in concrete at the prestressed steel level

In Fig 27 we assumed that the cross-section analysed in this

Figure 27 Axial strain curvature and prestress loss in a post-tensioned span (beamof Example 22)


example is at the centre of span of a simply supported beam Theabsolute value P of the prestressing force at time t0 is assumed constantat all sections while the dead load bending moment M is assumed tovary as a parabola The profile of the prestress tendon is assumed aparabola as shown The graphs in this figure show the variation overthe span of εO(t0) ψ(t0) ∆εO ∆ψ ∆σps which are respectively the axialstrain and curvature at t0 and the changes during the period (t minus t0) inaxial strain in curvature and in tension in prestress steel due tothe combined effects of creep shrinkage and relaxation The valuesof (εO + ∆εO) and (ψ + ∆ψ) will be used in Example 35 to calculatedisplacement values at time t

Example 23 Pre-tensioned section

Solve the same problem as in Example 22 assuming that pre-tensioningis employed (the duct shown in Fig 26(a) is eliminated)


The prestressed steel must now be included in the calculation of theproperties of the transformed section at t0 With this modification andconsidering that there is no prestress duct in this case calculation of thearea properties of the transformed section in the same way as in Table21 gives A = 03805m2 B = 4413 times 10minus3 m3 I = 4877 times 10minus3 m4

The forces applied on the section at t0 are the same as in Example 22Equation (232) gives the strain and the curvature at the reference pointimmediately after prestress transfer

εO(t0) = minus120 times 10minus6 ψ(t0) = minus153 times 10minus6 mminus1

The change in stress in the prestressed steel at transfer (Equation(238) ) are

(∆σps)inst = 200[minus120 + 045(minus153)]103 = minus378MPa

Multiplying this value by the area of the prestressed steel gives theinstantaneous prestress loss (minus43kN)

The stresses and strain introduced at transfer and the correspondingresultants of stresses are shown in Fig 28(a)


Using these results and following the same procedure as in Example22 the time-dependent changes in stress and strain are calculated andthe results shown in Fig 28(b)

Figure 28 Stress and strain distribution in the section of Fig 25(a) assuming thatpre-tensioning is used (Example 23) (a) condition at age t0 immediatelyafter prestress transfer (b) changes caused by creep shrinkage andrelaxation

Example 24 Composite section steel and post-tensioned concrete

Figure 29(a) shows the cross-section of a composite simply supportedbeam made of steel plate girder and a prestressed post-tensioned con-crete slab The plate girder is placed first in position and shored Then


the deck slab is cast in situ but its connection to the steel girder isdelayed by means of pockets left out around the anchor studs Thepockets are cast only after the application of the prestress Assume thatat age t0 the prestress is applied shortly after the anchorage of the deckto the steel girder is realized and the shoring removed It is required tofind the stress and strain distribution occurring immediately afterremoval of the shores (age t0) and the changes in these values at time

Figure 29 Analysis of stress and strain in a composite cross-section (Example 24)(a) cross-section properties (b) stress and strain immediately afterremoval of shores (c) changes caused by creep shrinkage and relaxation


t due to creep shrinkage and relaxation using the following data pre-stressing force P = 4500 times 103 N(1010kip) bending moment introducedat age t0 M = 2800 times 103 N-m (24800kip-in) φ(t t0) = 25 χ = 075εcs(t t0) = minus350 times 10minus6 reduced relaxation of the prestressed steel∆σpr = minus90MPa (minus13ksi) Ec(t0) = 30GPa (4350ksi) The moduli ofelasticity of the plate girder the prestressed and non-prestressed steelare equal Es = Ens = Eps = 200GPa (29000ksi) The dimensions andproperties of the cross-section area of concrete prestressed and non-prestressed steel are given in Fig 29(a) The centroid of the steel girderits cross-section area and moment of inertia about an axis through itscentroid are also given in the same figure

(a) Stress and strain at age t0 before connection of slab tosteel girder

Immediately after prestress the steel girder has no stress and the stressand strain need to be calculated only in the concrete slab Because thecentroid of the reinforcement coincides with the centroid of concretethe prestress produces no curvature and the strain is uniform over thedepth of the slab of magnitude = minus110 times 10minus6 and the correspondingconcrete stress = minus3305MPa Here the difference between the cross-section area of prestressed steel and that of prestress ducts is ignored

(b) Stress and strain immediately after removal of shores (age t0)The reference point O is chosen at the centroid of the steel girder Theproperties of the transformed section are calculated in Table 23 Eref ischosen equal to Ec(t0) = 30GPa

Table 23 Properties of the transformed section used in calculation of stress attime t0

Properties of areas Properties of transformed area

A B I AEEref BEEref IEEref

(m2) (m3) (m4) (m2) (m3) (m4)

Concrete 13081 minus15828 19205 13081 minus15828 19205Non-prestressed

steel 8000 times 10minus6 minus00097 00117 00533 minus00645 00781Prestressed steel 3900 times 10minus6 minus00047 00057 00260 minus00315 00381Steel girder 39000 times 10minus6 0 00150 02600 0 01000

Properties of 16474 minus16788 21367transformed section A B I


Axial force at O and bending moment introduced at removal ofshores is

NM = 02800 times 103 N-m

The axial strain at O and the curvature caused by these forces(Equation (232) ) is

εO(t0)

ψ(t0) =

1

30 times 109(16474 times 21367 minus 167882)

times 21367

16788

16788

16474 0

2800 times 103

= 10minus6 223

219mminus1

The values of εO(t0) and ψ(t0) are used to find the strain at any fibreand hence the corresponding stress Superposition of these stresses andstrains and of the values determined in (a) above gives the stress andstrain distributions shown in Fig 29(b)

(c) Changes in stress and strain due to creep shrinkageand relaxation

Age-adjusted elasticity modulus is


1 + 075 times 25= 10435GPa

In the restrained condition stress in concrete is (Equation (245) )

(σc restrained)top = minus10435 times 109[25(minus176 times 10minus6) minus350 times 10minus6]

= 824MPa

(σc restrained)bot = minus10435 times 109[25(minus128 times 106) minus350 times 10minus6]

= 699MPa

To calculate the axial force at O and the bending moment necessaryto prevent creep by Equation (242) we must find (εO)1 and ψ1 defining


the straight-line distribution of strain in part 1 the deck slab (Fig29(b) ) These values are (εO)1 = 113 times 10minus6 ψ1 = 219 times 10minus6 mminus1

∆N

∆Mcreep

= minus10435 times 109 times 25

times 13081

minus15828

minus15828

19205 113

219 10minus6

= 106 5187 N

minus6306 N-mThe forces required to prevent strain due to shrinkage and relaxation

(Equation (243) and (244) ) are

∆N

∆Mshrinkage

= minus10435 times 109(minus350 times 10minus6) 13081

minus15828

= 106 4777 N

minus5781 N-m

∆N

∆Mrelaxation


3900 times 10minus6)(minus121)(minus90 times 106)

= 106 minus0351 N

0425 N-mThe total restraining forces are

∆N

∆M = 106 5187 + 4777 minus 0351

minus6306 minus 5781 + 0425 = 106 9613 N

minus11662 N-mProperties of the age-adjusted transformed section are calculated in a

similar way as in Table 23 giving

A = 2284m2 B = minus1859m3 I = 2542m4

Eref used in the calculation of the above values is

Eref = Ec(t t0) = 10435GPa

Increments in axial strain and curvature when the restraining forcesare removed (Equation (240) ) are


∆εO

∆ψ =

106

10435 times 109(2284 times 2542 minus 18592) 2542

1859

1859

2284minus 9613

11662

= 10minus6 minus112

357mminus1

The corresponding stress and strain distributions are shown inFig 29(c) The stresses are calculated by Equation (246)

Example 25 Composite section pre-tensioned and cast-in-situ parts

The cross-section shown in Fig 210 is composed of a precast pre-tensioned beam (part 1) and a slab cast in situ (part 2) It is required tofind the stress and strain distribution in the section immediately afterprestressing and the changes in these values occurring between pre-stressing and casting of the deck slab and after a long period usingthe following data

Ages of precast beam at the time of prestress t1 = 3 days and at the

Figure 210 Analysis of stress and strain in a cross-section composed of precast andcast in situ parts (Example 25)


time of casting of the deck slab t2 = 60 days the final stress and strainare required at age t3 = infin The prestress force P = 4100 times 103 N(920kip) the bending moment due to self-weight of the prestress beam(which is introduced at the same time as the prestress) M1 = 1400 times 103

N-m (12400kip-in) additional bending moment introduced at age t2

(representing the effect of the weight of the slab plus superimposeddead load) M2 = 1850 times 103 N-m (16400kip-in) The modulus of elas-ticity of concrete of the precast beam Ec1(3) = 25GPa (3600ksi) andEc1(60) = 37GPa (5400ksi)

Soon after hardening of the concrete the composite action starts todevelop gradually Here we will ignore the small composite actionoccurring during the first three days Consider that age t2 = 60 days forthe precast beam corresponds to age = 3 days of the deck at which timethe modulus of elasticity of the deck Ec2(3) = 23GPa (3300ksi)

Creep and aging coefficients and the free shrinkage values to be usedare

Concrete part 1[φ(60 3)]1 = 120 [φ(infin 3)]1 = 230 [φ(infin 60)]1 = 227[χ(60 3)]1 = 086 [χ(infin 60)]1 = 080[εcs(60 3)]1 = minus57 times 10minus6 [εcs(infin 60)]1 = minus 205 times 10minus6

Concrete part 2[φ(infin 3)]2 = 240 [χ(infin 3)]2 = 078[εcs(infin 3)]2 = minus269 times 10minus6

Reduced relaxation ∆σpr = minus85MPa (12ksi) of which minus15MPa(22ksi) in the first 57 days Modulus of elasticity of the prestressed andnon-prestressed steels = 200GPa

The dimensions and properties of areas of the concrete and steel inthe two parts are given in Fig 210

(a) Stress and strain immediately after prestressing of theprecast beam

The geometric properties of the precast beam are calculated in Table24 with the reference point O chosen at the centroid of concrete cross-section and Eref = Ec1(3) = 25GPa

The prestress force and the bending moment introduced at t1 areequivalent to an axial force at O plus a bending moment given byEquation (231)


NMequivalent




Instantaneous axial strain and curvature at t1 = 3 days (Equation (232) )are

εO (t1)

ψ(t1) =

1

25 times 109 (05583 times 01264 minus 001172)01264

minus00117

minus00117

05583

times minus4100 times 103

minus773 times 103 = 10minus6 minus289

minus218mminus1

The above values of εO and ψ are used to calculate the strain at anyfibre and the corresponding stress (Fig 211(a) ) The strain at the levelof prestress tendon is minus405 times 10minus6 The instantaneous prestress lossis minus256kN (62 of the initial force)

(b) Change in stress and strain occurring between t = 3 days andt = 60 days


Ec(60 3) =25 times 109

1 + 086 times 120= 1230GPa (1780ksi)

The stress in concrete required to artificially restrain creep andshrinkage (Equation (245) ) is

Table 24 Properties of the precast section employed in calculation of stress andstrain at time t1 = 3 days

Properties of areaProperties of transformedarea


(m2) (m3) (m4) (m2) (m3) (m4)

Concrete 05090 00 01090 05090 00 01090Non-prestressed

steel 3000 times 10minus6 minus210 times 10minus6 1282 times 10minus6 00240 minus00017 00103Prestressed steel 3160 times 10minus6 1675 times 10minus6 888 times 10minus6 00253 00134 00071

Properties of 05583 00117 01264transformed section A B I


(σc restrained)top = minus1230 times 109[12(minus 121 times 10minus6) minus57 times 10minus6] = 2487MPa

(σc restrained)bot = minus1230 times 109[12(minus 426 times 10minus6) minus57 times 10minus6] = 6989MPa

Strain due to creep shrinkage and relaxation can be restrained by thefollowing forces (Equations (242) (243) and (244) )

Figure 211 Stress and strain in the precast beam of Example 25 (a) conditions atage t1 = 3 days (b) changes caused by creep shrinkage and relaxationoccurring between t1 = 3 days and t2 = 60 days (c) instantaneous changesat t2 caused by introduction of moment M2 = 1850 times 103 N-m


∆N

∆Mcreep


0

0

01090 minus289

minus218 10minus6

= 106 2171 N

0351 N-m

∆N

∆Mshrinkage


0 = 106 0357 N

0

∆N

∆Mrelaxation

= 3160 times 10minus6(minus15 times 106

3160 times 10minus6 times 053(minus15 times 106) = 106minus0047 N

minus0025 N-mThe total restraining forces are

∆N

∆M = 106 2171 + 0357 minus 0047

00351 + 0 minus 0025 = 106 2481 N

0326 N-mWith Eref = Ec(60 3) the properties of the age-adjusted transformed

section are calculated in the same way as in Table 24 giving A =06092m2 B = 00238m3 I = 01443m4

Removal of the restraining forces results in the following increments ofaxial strain and curvature during the period t1 to t2 (Equation (240) )

∆εO(t2 t1)

∆ψ(t2 t1) = 10minus6 minus326

minus130mminus1The corresponding incremental stress and strain distributions are

shown in Fig 211(b) (The stresses are calculated by Equation (246))The stress at t2 = 60 days may be obtained by superposition of the

diagrams in Fig 211(a) and (b)

(c) Instantaneous increments of stress and strain at t2 = 60 daysThe bending moment M = 1850 times 103 N-m is resisted only by theprestressed beam The properties of the transformed section are calcu-lated in the same way in Table 24 using Eref = Ec(60) = 37GPa givingA = 05423m2 B = 00079m3 I = 01207m4 Substitution in Equation(232) gives the instantaneous increments in axial strain and curvatureoccurring at t2

∆εO(t2)

∆ψ(t2) = 10minus6 minus6

415mminus1


The corresponding stress and strain distributions are shown inFig 211(c)

(d) Changes in stress and strain due to creep shrinkage andrelaxation during the period t2 = 60 days to t3 = infin

The age-adjusted moduli of elasticity for the precast beam and the deckslab are

Ec1 (infin 60) =37 times 109

1 + 08 times 227= 1314GPa (1900ksi)


1 + 078 times 240= 801GPa (1160ksi)

The stresses shown in Figs 211(a) (b) and (c) are introduced atvarious ages and thus have different coefficients for creep occurringduring the period considered In the following the stresses in Figs211(a) and (b) are combined and treated as if the combined stress wereintroduced when the age of the precast beam is 3 days thus the creepcoefficient to be used is φ(infin 3) minus φ(60 3) = 230 minus 120 = 110 Thestress in Fig 211(c) is introduced when the precast beam is 60 daysold thus the coefficient of creep for the period considered is φ(infin 60) =227

For more accuracy the stress in Fig 211(b) which is gradually intro-duced between the age 3 days and 60 days may be treated as if it wereintroduced at some intermediate time t such that

1

Ec(t)[1 + φ(60 t)] =

1

Ec(3) [1 + χ(60 3) φ(60 3)]

Using this approach would result in a slightly smaller coefficient than110 adopted above

The stresses in the precast beam necessary to artificially restrain creepand shrinkage (Equation (245) ) are

(∆σc restrained)top = minus1314 times 109[110 minus121 times 10minus6 minus029 times 106

25 times 109 + 227 (minus325 times 10minus6) + (minus205 times 10minus6)]

= 14304MPa


(∆σc restrained)bot = minus1314 times 109[110minus426 times 10minus6 +197 times 106

25 times 109 + 227(255 times 10minus6) + (minus205 times 10minus6)]

= 0106MPa

The stress in the restrained condition in the deck slab is constant overits thickness and is equal to Equation (245)

σc restrained = minus 801 times 109 (minus269 times 10minus6) = 2155MPa

The properties of the age-adjusted transformed section for the periodt2 to t3 are calculated in Table 25 using Eref = Ec1 (infin 60) = 1314GPa

The forces necessary to restrain creep shrinkage and relaxationduring the period t2 to t3 are (Equations (241) to (244) )

∆N

∆Mcreep


0

0

0109

times

110minus289 +095 times 1012

25 times 109 + 227(minus6)

110minus218 +161 times 1012

25 times 109 + 227(415)

10minus6

= 106 1937 N

minus1108 N-m

The term between the curly brackets represents the changes in axialstrain and curvature that would occur due to creep if it wereunrestrained The deck slab is not included in this equation because nostress is applied on the slab before the period considered

∆N

∆Mshrinkage


0

minus801 times 109(minus269 times 10minus6) 0495

minus04307

= 106 2437 N

minus0928 N-m


∆N

∆Mrelaxation

= 3160 times 10minus6 (minus70 times 106)

3160 times 10minus6 times 053 (minus70 times 106) = 106 minus0221 N

minus0117N-m

Figure 212 Changes in stress and strain in the composite section of Example 25due to creep shrinkage and relaxation occurring between casting of thedeck slab t2 = 60 days and t3 = infin

Table 25 Properties of the composite age-adjusted transformed section used incalculation of the changes of stress and strain between t2 = 60 daysand t3 = infin

Properties of area Properties of transformedarea


(m2) (m3) (m4) (m2) (m3) (m4)

Concrete ofdeck slab 0495 minus04307 03763 03017 minus02625 02294

Non-prestressedsteel in deckslab 5000 times 10minus6 minus4350 times 10minus6 3785 times 10minus6 00761 minus00662 005763

Concrete inbeam 05090 00 01090 05090 00 01090

Non-prestressedsteel in beam 3000 times 10minus6 minus210 times 10minus6 1282 times 10minus6 00457 minus00032 00195

Prestressed steel 3160 times 10minus6 1675 times 10minus6 8876 times 10minus6 00481 00255 00135

Properties of the age-adjusted 09806 minus03064 04290transformed section A B I


The total restraining forces

∆N

∆M = 106 1937 + 2437 minus 0221

minus1108 minus 0928 minus 0117 = 106 4153 N

minus2153 N-m

The increments of axial strain and curvature during the period t2 to t3

are obtained by substitution in Equation (240) and are plotted inFig 212

∆εO(t3 t2)


195mminus1

The corresponding change in stress is calculated by Equation (246) andplotted also in Fig 212

26 Summary of analysis of time-dependent strainand stress

The procedure of analysis given in this chapter can be performed in foursteps Figure 213 outlines the four steps to determine the instantaneous andthe time-dependent changes in strain and stress in a non-cracked prestressedsection For quick reference the symbols used are defined again below andthe four steps summarized

Notation

A areaB first moment of areaE modulus of elasticityEc age-adjusted elasticity modulus of concreteI second moment of areaM bending moment about an axis through ON normal force at OP absolute value of prestressingt time or age of concretey coordinate (Fig 25)σ stressα ratio of modulus of steel to that of concrete at time t0

α ratio of modulus of elasticity of steel to Ec

χ aging coefficient of concrete∆ increment


∆σpr reduced relaxation of prestressed steelε strainφ creep coefficientψ curvature (slope of strain diagram = dεdy)γ slope of stress diagram (= dσdy)

Subscripts

c concretecs shrinkagens non-prestressed steelO arbitrary reference point0 time of prestressingps prestressed steel

Four analysis steps

Step 1 Apply the initial prestressing force and the dead load or other bend-ing moment which becomes effective at the time of prestressing t0 on atransformed section composed of Ac plus (αpsAps + Ansαns) Here the trans-formed section includes only the prestressed and the non-prestressed steel

Figure 213 Steps of analysis of time-dependent strain and stress


bonded to the concrete at the prestress transfer Thus Aps should be includedwhen pre-tentioning is used but when all the prestressing is post-tensionedin one stage Aps and the area of the duct should be excluded

When the structure is statistically indeterminate the indeterminate normalforce and moment should be included in the forces on the section Determinethe resultants N and M of all forces on the section

Apply Equation (219) to determine εO(t0) and ψ(t0) which define distribu-tion of the instantaneous strain Multiplication by Ec(t0) or application ofEquation (220) gives σO(t0) and γ(t0) which define the instantaneous stressdistribution

Step 2 Determine the hypothetical change in the period t0 to t in straindistribution due to creep and shrinkage if they were free to occur The strainchange at O is equal to [φ(t t0) εO(t0) + εcs] and the change in curvature is [φ(tt0) ψ(t0)]

Step 3 Calculate artificial stress which when gradually introduced on theconcrete during the period t0 to t will prevent occurrence of the strain deter-mined in step 2 The restraining stress at any fibre y is (Equations (234) and(245) )

∆σrestrained = minusEc φ(t t0)[εO(t0) + ψ(t0)y] + εcs (249)

Step 4 Determine by Equation (218) a force at O and a moment which arethe resultants of ∆σrestrained The change in concrete strain due to relaxation ofthe prestressed steel can be artificially prevented by the application at thelevel of the prestressed steel of a force equal to Aps ∆σpr substitute this forceby a force of the same magnitude at O plus a couple Summing up gives∆Nrestrained and ∆Mrestrained the restraining normal force and the couple requiredto prevent artificially the strain change due to creep shrinkage andrelaxation

To eliminate the artificial restraint apply ∆N ∆Mrestrained in reverseddirections on an age-adjusted transformed section composed of Ac plus(αAps + αAns) calculate the corresponding changes in strains and stresses byEquations (219) and (217)

The strain distribution at time t is the sum of the strains determined insteps 1 and 4 while the corresponding stress is the sum of the stresses at t0

calculated in step 1 and the time-dependent changes calculated in steps 3 and4 (Equations (246)ndash(248) )

Commentary

1 The flow chart in Fig 214 shows how the four steps of analysis canbe applied in a general case to determine the instantaneous and


time-dependent increments of stress and strain due to the application attime t0 of a normal force N and a moment M on a section for which theinitial values of stresses and strains are known

If after step 1 the stress at the extreme fibre exceeds the tensile strengthof concrete the calculation in step 1 must be repeated using A B and I ofa cracked section in which the concrete in tension is ignored The depthof the compression zone c must be determined prior to applying steps 2 3and 4 to a cracked section (see Chapter 7) The flow chart also outlinesthe analysis in a less common case in which cracking occurs during theperiod t0 to t and thus will be detected only at the end of step 4

Figure 214 Flow chart for calculating stress and strain increments in a section due to anormal force N and a bending moment M introduced at time t0 and sustainedto time t


2 Superposition of strains or stresses in the various steps can be done bysumming up the increments ∆εO ∆ψ ∆σO or ∆γ This is possible becauseof the use of the same reference point O in all steps

3 The four steps give the stress and strain at time t without preceding theanalysis by an estimate of the loss in tension in the prestressed steel Noempirical equation is involved for loss calculation

4 The analysis satisfies the requirements of compatibility and equilibriumthe strain changes in concrete and steel are equal at all reinforcementlayers the time-dependent effect changes the distribution of stressesbetween the concrete and the reinforcements but does not change thestress resultants

5 The same four-step analysis applies to reinforced concrete sections with-out prestressing simply by setting Aps = 0 the effects of cracking will bediscussed in Chapter 7

6 The same procedure can be used for analysis of composite sections madeof more than one type of concrete cast or prestressed in stages or madeof concrete and structural steel

27 Examples worked out in British units

Example 26 Stresses and strains in a pre-tensioned section

The pretensioned cross-section shown in Fig 215 is subjected at time t0

to a prestressing force 600kip (2700kN) and a bending moment10560kip-in (1193kN-m) Find the extreme fibre stresses at time t0

immediately after prestressing and at time t after occurrence of creepshrinkage and relaxation The following data are given Ec(t0) = 3600ksi(25GPa) Ens = 29000ksi (200GPa) Eps = 27000ksi (190GPa) φ(t t0) =3 χ = 08 ∆σpr(t t0) = minus13ksi (90MPa) εcs(t t0) = minus300 times 10minus6

Step 1 The reference point O is chosen at the top fibre The presentingand the bending moment introduced at t0 are equivalent to ∆N at O anda moment ∆M about an axis through O calculated by Equation (231)

∆N = minus600kip ∆M = minus9840kip-in

The ratios of the elasticity moduli Ens and Eps to Ec (t0) are (Equation(133) )

αns = 806 αps = 750

Use of these values to calculate the area properties of the trans-formed section at time t0 gives


A = 1158 in2 B = 19819 in3 I = 547200 in4

Substitution in Equations (219) and (217) gives (Fig 215(b) )

∆εO(t0) = minus154 times 10minus6 ∆ψ(t0) = 0562 times 10minus6 inminus1

∆σc(t0)top bot = minus0553 minus0472 ksi

Figure 215 Analysis of strain and stress in a pre-tensioned section (Example 26)(a) cross-section dimensions (b) strain and stress at t0 (c) strain andstress at t


Step 2 Hypothetical changes in strain at O and in curvature if creepand shrinkage were free to occur are

(∆εO)free = 3(minus154 times 10minus6) minus300 times 106 = minus762 times 10minus6

(∆ψ)free = 3(0562 times 10minus6) = 169 times 10minus6 inminus1

Step 3 The age-adjusted elasticity modulus (Equation (131) )

Ec(t t0) = 1059ksi

The area properties of concrete cross-section are

Ac = 1023 in2 Bc = 16000 in3 Ic = 410800 in4

Artificial stress to prevent strain changes due to creep and shrinkage(Equation (245) )

∆σrestrainedtop bot = 0807 0734 ksi

Step 4 Substitution of Ac Bc Ic Ec (minus∆εO)free and (minusψ)free in Equation(218) gives the forces necessary to restrain creep and shrinkage

∆Ncreep + shrinkage = 795kip ∆Mcreep + shrinkage = 12151kip-in

Forces necessary to prevent strain change due to relaxation ofprestressed steel are (Equation (244) )

∆Nrelaxation = minus 39kip ∆Mrelaxation = minus1326kip-in

Summing (Equation (241) )

∆Nrestrained = 756kip ∆Mrestrained = 10825kip-in

The ratios of the elasticity moduli Ens and Eps to Ec (t t0) are(Equation (134) )

αns = 2739 αps = 2550

The area properties of the age-adjusted section are


A = 1483 in2 B = 28950 in3 I = 874600 in4

Substitution of these three values Ec and minus∆N minus∆Mrestrained inEquation (219) gives the changes in strain in the period t0 to t

∆εO(t t0) = minus716 times 10minus6 ∆ψ(t t0) = 1203 times 10minus6 inminus1

Substitution of these two values in Equation (217) and addition of∆σrestrained gives the changes in stress in the period t0 to t

∆σ(t t0)top bot = 0047 0485 ksi

Addition of these two stress values to the stresses determined in step1 gives the stresses at time t

∆σ(t)top bot = minus0506 0013 ksi

The strain and stress distributions at t0 and t are shown in Fig 215

Example 27 Bridge section steel box and post-tensioned slab

Figure 216 shows the cross-section of a simply supported bridge ofspan 144 ft (439m) The deck is made out of precast rectangular seg-ments assembled in their final position above a structural steel U-shaped section by straight longitudinal post-tensioned tendons Eachprecast segment covers the full width of the bridge In the longitudinaldirection each segments covers a fraction of the span At completionof installation of the precast elements the structural steel sectioncarries without shoring a uniform load = 54kipft (79kNm) repre-senting the weight of concrete and structural steel Shortly afterprestressing the bridge section is made composite by connecting thedeck slab to the structural steel section This is achieved by the castingof concrete to fill holes in the precast deck at the locations of pro-truding steel studs welded to the top flanges of the structural steelsection Determine the strain and stress distributions in concrete andstructural steel at the mid-span section at time t0 shortly after prestress-ing and at time t after occurrence of creep shrinkage and relaxation

Consider that the post-tensioning and the connection of concrete to


structural steel occur at the same time t0 Assume that during prestress-ing the deck slides freely over the structural steel section The followingdata are given Initial total prestressing force excluding loss by frictionand anchor set = 2200kip (9800kN) creep coefficient φ(t t0) = 22aging coefficient χ(t t0) = 08 free shrinkage εcs(t t0) = minus220 times 10minus6reduced relaxation ∆σpr = minus60ksi (minus48MPa) modulus of elasticity ofconcrete Ec(t0) = 4300ksi (30GPa) modulus of elasticity of prestressedsteel non-prestressed steel or structural steel = 28000ksi (190GPa)

Figure 216 Composite cross-section of a bridge (Example 27) (a) cross-sectiondimensions (b) strain and stress at t0 (c) strain and stress at t


Strain and stress at time t0

At completion of the installation of the precast elements the concreteand steel act as separate sections the concrete section is subjected to theprestressing force 2200kip at the centroid and the steel section is sub-jected to a bending moment = 168000kip-in The area of the trans-formed concrete section composed of Ac and αns (excluding the area ofducts) = A = 4535 in2 with αns = 651 The strain and stress distributionsat this stage are shown in Fig 216(b)

Strain and stress at time tAfter connection of concrete and steel at time t0 the section becomescomposite Select the reference point O at the centroid of the structuralsteel and follow the four analysis steps outlined in Section 26

Step 1 The instantaneous strain and stress at t0 have been determinedin Fig 216(b)

Step 2 If creep and shrinkage were free to occur the change in strainbetween t0 and t would be

(∆εO)free = minus220 times 10minus6 minus 22(1128 times 10minus6)= minus4682 times 10minus6 (∆ψ)free = 0

Step 3 The age-adjusted elasticity modulus of concrete (Equation(131) ) Ec = 1558ksi

The stress necessary to restrain creep and shrinkage (Equation(245) ) is

(σc)restraint = minus1558 (minus4682 times 106) = 0729ksi

Step 4 The area properties of concrete are

Ac = 4471 in2 Bc = minus216800 in3 Ic = 1057 times 106 in4

The forces necessary to restrain creep and shrinkage (Equation(218) ) are

(∆N)creep + shrinkage = 3261kip (∆M)creep + shrinkage

= minus1582 times 103 kip-in


Forces necessary to restrain relaxation (Equation 244) are

(∆N)relaxation = minus84kip (∆M)relaxation = 4074kip-in

The total restraining forces are

∆N = 3177kip ∆M = minus1541 times 103 kip-in

Properties of the age-adjusted transformed section are

A = 10170 in2 B = minus2421 times 103 in3 I = 1629 times 106 in4

Apply minus∆N and minus∆M on the age-adjusted section and use Equation(219) to calculate the changes in strain between t0 and t


Adding the change in strain to the initial strain in Fig 216(b) givesthe total strain at time t shown in Fig 216(c)

The time-dependent change in stress in concrete is calculated byEquation (246)

[∆σc(t t0)]top = 0729 + 1558[minus867 times 10minus6

+ 4784 times 10minus6 (minus56)] = 0177 ksi

[∆σc(t t0)]bot = 0729 + 1558[minus867 times 10minus6

+ 4784 times 10minus6(minus40)] = 0296ksi

Adding these stresses to the initial stress (Fig 216(b) ) gives thetotal stress at time t shown in Fig 216(c) It is interesting to note thechange in the resultant force on the concrete (the area of the concretecross-section multiplied by the stress at its centroid) The values of theresultants are minus2180 and minus1130kip at time t0 and t respectively Thesubstantial drop in compressive force is due to the fact that the time-dependent shortening of the concrete is restrained after its attachmentto a relatively stiff structural steel section


28 General

A general procedure is presented in Section 25 which gives the stress andstrain distribution at any time in a composite cross-section accounting for theeffects of creep shrinkage and relaxation of prestress The analysis employsthe aging coefficient χ to calculate the instantaneous strain and creep due to astress increment which is gradually introduced in the same way as if it wereintroduced all at once The analysis employs equations which can be easilyprogrammed on desk calculators or small computers

We have seen that the axial strain and curvature and the correspondingstrain are calculated in two steps with single stage prestress or in more stepswhen the prestress is applied in more than one step If a computer isemployed more accuracy can be achieved if the time is divided into incre-ments and a step-by-step calculation is performed to determine the timedevelopment of stress and strain (see Sections 46 and 58) In this casethe aging coefficient χ is not needed and the approximation involved in theassumption used for its derivation is eliminated (see Sections 17 and 110)

When the equations of Section 25 are used the loss of prestress due tocreep shrinkage and relaxation is accounted for and the effects of the loss onthe strain and stress distributions are directly obtained Among the time-dependent variables obtained by the analysis is the loss of tension in theprestressed steel (Equation 248) ) However of more interest in design is theloss of compression in the concrete because it is this value which governsthe possibility of cracking when the strength of concrete in tension isapproached The loss of tension in the prestressed steel is equal in absolutevalue to the loss of compression on the concrete only in a concrete sectionwithout non-prestressed reinforcement In general the loss in compression isgreater in absolute value than the loss in tension in prestressed steel Thedifference represents the compression picked up gradually by the non-prestressed steel as creep shrinkage and relaxation develop This will be fur-ther discussed in the following chapter where the time-dependent effects willbe considered for sections without non-prestressed steel or with one or morelayers of this reinforcement

Note

1 The loss due to friction or anchor setting are excluded in this discussion theprestress force P is the force in the tendon excluding the losses due to these effects


Special cases of uncrackedsections and calculationof displacements

Bow River Calgary Canada Continuous bridge over 430m (1410 ft) Cantilever slabs arecast on forms moving on box girder cast in earlier stages (Courtesy KVN HeavyConstruction Division of the Foundation Co of Canada Ltd and Stanley AssociatesEngineering Ltd Calgary)

Chapter 3

31 Introduction

In the preceding chapter we presented a method of analysis of the time-dependent stresses and strains in composite sections composed of more thanone type of concrete or of concrete and structural steel sections with orwithout prestressed or non-prestressed reinforcement In the special casewhen the section is composed of one type of concrete and the prestressed andnon-prestressed steel are situated (or approximately considered to be) in onelayer the analysis leads to simplified equations which are presented in thischapter Another special case which is also examined in this chapter is across-section which has reinforcement without prestressing and we willconsider the effects of creep and shrinkage However discussion of the effectsof cracking is excluded from the present chapter and deferred to Chapters 7and 8

The procedures of analysis presented in Chapter 2 and in the present chap-ter give the values of the axial strain and the curvature at any section of aframed structure at any time after loading These can be used to calculate thedisplacements (the translation and the rotation) at any section or at a jointThis is a geometry problem generally treated in books of structural analysisIn Section 38 two methods which will be employed in the chapters to followare reviewed the unit load theory based on the principle of virtual work andthe method of elastic weights The two methods are applicable for cracked oruncracked structures

32 Prestress loss in a section with one layerof reinforcement

The method of analysis in the preceding chapter gives the loss of prestressamong other values of stress and strain in composite cross-sections with anumber of layers of reinforcements When the total reinforcement pre-stressed and non-prestressed are closely located such that it is possibleto assume that the total reinforcement is concentrated at one fibre it maybe expedient to calculate the loss of prestress by an equation ndash to be givenbelow ndash then find the time-dependent strain and curvature by superposing theeffect of the initial forces and the prestress loss

Consider a prestressed concrete member with a cross-section shown in Fig31 The section has a total reinforcement area

Ast = Ans + Aps (31)

where Ans and Aps are the areas of the non-prestressed reinforcement and theprestress steel respectively A reference point O is chosen at the centroid of theconcrete section The total reinforcement Ast is assumed to be concentrated inone fibre at coordinate yst The moduli of elasticity of the two types of


reinforcement are assumed to be the same thus one symbol Est is used for themodulus of elasticity of the total steel

Est = Ens = Eps (32)

The prestress force P is applied at age t0 at the same time as a bendingmoment and an axial force It is required to calculate the prestress loss andcalculate the changes in axial strain and curvature and in stresses in steel andconcrete due to creep shrinkage and relaxation

Creep shrinkage and relaxation cause changes in the distribution of stressin concrete and the two steel types but any time the sum of the total changesin the forces in the three materials must be zero thus

∆Pc = minus∆Pns minus ∆Pps (33)

where ∆Pc is the change in the resultant force on the concrete ∆Pps is thechange in the force in the prestress tendon and ∆Pns is the change in the forcein the non-prestressed reinforcement

We recall according to our sign convention (see Section 22) that a posi-tive ∆P means an increase in tension Thus generally ∆Pc is a positive valuewhile ∆Pps is negative

The loss in tension in the tendon is equal to the loss of the compressiveforce on concrete (∆Pc = minus∆Pps) only in the absence of non-prestressedreinforcement

The change of the resultant force on concrete due to creep shrinkage and

Figure 31 Definition of symbols used in Equations (31) to (314)

Special cases of uncracked sections and calculation of displacements 71

relaxation can be calculated by the following equation which is applicable forpost-tensioned and pre-tensioned members

∆Pc = minusφ(t t0)σcst(t0)Ast[EstEc(t0)] + εcs(t t0)EstAst + ∆σprAps

1 +Ast

Ac

Est

Ec(t t0)1 +

y2st

r2c

(34)

This equation can of course be used when the section has only one type ofsteel (substituting Ans or Aps = 0 in Equations (31) and (34) ) When used fora reinforced concrete section without prestressing Equation (34) gives thechange in the resultant force in concrete due to creep and shrinkage We arehere assuming that no cracking occurs

The symbols used in Equation (34) are defined below

Post-tensioned and pre-tensioned members differ only in the calculationof σcst With post-tensioning the area of the cross-section to be used in thecalculation of σcst includes the cross-section areas of the non-prestressed steeland of concrete excluding the area of prestress duct With pre-tensioningthe cross-section to be employed is composed of the areas of concrete andprestressed and non-prestressed steel (see Examples 22 and 23)

The procedure of analysis adopted in Section 25 can be employed to

Ac = cross-section of concreter2

c = IcAc where Ic is the moment of inertia of concrete section aboutan axis through its centroid

Ast Est = the total cross-section area of reinforcement and its modulus ofelasticity one modulus of elasticity is assumed for the two types ofsteel

yst = the y-coordinate of a fibre at which the total reinforcement isassumed to be concentrated y is measured downwards from pointO the centroid of concrete area

Ec(t0) = modulus of elasticity of concrete at age t0

Ec(t t0) = age-adjusted elasticity modulus of concrete given by Equation(131)

φ(t t0) = creep coefficient at time t for age at loading t0εcs(t t0) = the shrinkage that would occur during the period (t minus t0) in free

(unrestrained) concrete∆σpr = the intrinsic relaxation of the prestressed steel multiplied by the

reduction coefficient χr (see Fig 14 or Table 11)σcst(t0) = stress of concrete at age t0 at the same fibre as the centroid of

the total steel reinforcement This is the instantaneous stressexisting immediately after application of prestress (if any) andother simultaneous loading for example the member self-weight


calculate ∆Pc and the same result as by Equation (34) should be obtainedHowever in the special case considered here Equation (34) can be derivedmore easily as follows

During the period (t minus t0) the changes of the forces in the prestressed steeland the non-prestressed reinforcement are

∆Pps = Aps∆σps (35)

∆Pns = Ans∆σns (36)

The change of resultant force on the concrete during the same period(Equation (33) ) is

∆Pc = minus (Aps∆σps + Ans∆σns) (37)

For compatibility the strain in the non-prestressed steel in the prestressedsteel and in the concrete at the fibre with y = yst must be equal Thus

∆σns

Est

=∆σps minus ∆σpr

Est

=σcst(t0)φ(t t0)

Ec(t0)+ εcs(t t0) +

1

Ec(t t0)∆Pc

Ac

+∆Pc y2

st

Ic (38)

The relaxation is deducted from the total change in prestress steel because therelaxation represents a change in stress without associated strain The firstand second terms on the second line are the strains in concrete due to creepand shrinkage The last term is the instantaneous strain plus creep due to aforce ∆Pc This term represents the strain recovery associated with prestressloss Solution of Equations (35ndash8) for ∆Pc gives Equation (34)

The last term in Equation (38) can be presented in this form only when yst

is measured from point O the centroid of Ac and Ic is the moment of inertia ofthe area of concrete about an axis through its centroid Thus in determin-ation of the values yst and r2

c to be substituted in Equation (34) point O mustbe chosen at the centroid of Ac

The reduced relaxation ∆σpr to be used in Equation (34) is given byEquation (17) which is repeated here


where ∆σpr is the intrinsic relaxation that would develop during a period (t minust0) in a tendon stretched between two fixed points χr is a reduction factor (seeSection 15 and Appendix B)

The relaxation reduction factor χr may be taken from Table 11 or Fig 14


The value χr depends upon the magnitude of the total loss ∆σps which isgenerally not known Thus for calculation of the total loss due to creepshrinkage and relaxation an assumed value of ∆σpr is substituted in Equation(34) to give a first estimate of ∆Pc This answer is used to obtain an improvedreduced relaxation value and Equation (34) is used again to calculate a betterestimate of ∆Pc In most cases a first estimate of χr = 07 followed by oneiteration gives sufficient accuracy

321 Changes in strain in curvature and in stress due tocreep shrinkage and relaxation

The changes in axial strain at O or in curvature during the period (t ndash t0) maybe expressed as the sum of the free shrinkage the creep due to the prestressand external applied loads plus the instantaneous strain (or curvature) pluscreep produced by the force ∆Pc which acts at a distance yst below O Thus

∆εO = εcs(t t0) + φ(t t0) εO (t0) +∆Pc

Ec(t t0)Ac

(310)

∆ψ = φ(t t0)ψ(t0) +∆Pc yst

Ec(t t0)Ic

(311)

The change in concrete stress at any fibre due to creep shrinkage andrelaxation is

∆σc =∆Pc

Ac

+∆Pc yst

Ic

y

where y is the coordinate of the fibre considered y is measured downwardsfrom the centroid of concrete area Substitution of Ic = Acr

2c in the last

equation gives

∆σc =∆Pc

Ac

1 +yyst

r2c (312)

The changes in stress in the prestressed steel and in the non-prestressedreinforcement caused by creep shrinkage and relaxation are

∆σns = Est(∆εO + yns∆ψ) (313)

∆σps = Est(∆εO + yps∆ψ) + ∆σpr (314)

where yns and yps are the y coordinates of a non-prestressed and prestressedsteel layer respectively


Equation (34) may be used to calculate ∆Pc also in the case when the cross-section has more than one layer of reinforcement and when the centroid ofthe prestressed and the non-prestressed steels do not coincide In this case∆Pc must be considered to act at the centroid of the total steel area and theequations of this section may be used to calculate the changes in strain incurvature and in stress The solution in this way involves approximationacceptable in most practical calculations the compatibility relations(Equation 38) are not satisfied exactly at all layers of reinforcement

Example 31 Post-tensioned section without non-prestressed steel

Calculate the prestress loss due to creep shrinkage and relaxation in thepost-tensioned cross-section of Example 22 (Fig 26(a) ) ignoring thenon-prestressed steel Assume that the intrinsic relaxation ∆σprinfin =minus115MPa (minus167ksi) The reduced relaxation value is to be calculatedemploying the graph in Fig 14 assuming that the characteristic tensilestrength of the prestressed steel fptk = 1770MPa (257ksi) Use thecalculated prestress loss to find the axial strain and curvature at t = infin

In this example Ans = 0 Ast = Aps and yst = yps Because of the prestressduct the centroid of concrete section is slightly shifted upwards fromcentre (Fig 32(a) ) With this shift we have y2

st = 02059m2 Ic = 42588 times10minus3 m4 Ac = 0357m2 r2

c = 01193m2For calculation of the stress and the strain at age t0 immediately after

prestressing consider a plain concrete section subjected to a compres-sive force P = 1400kN at eccentricity yst = 0454m plus a bendingmoment M = 390kN-m Calculation of stress and strain distributions ina conventional way using a modulus of elasticity equal to Ec(t0) =30GPa gives the results shown in Fig 32(b) From this figure the stressin concrete at age t0 at the level of the prestress steel is

σcps(t0) = minus6533MPa (minus09475ksi)

As first estimate assume the relaxation reduction factor χr = 07 thusthe reduced relaxation is

∆σprinfin = 07(minus115) = minus8051 MPa (minus116ksi)

The age-adjusted elasticity modulus of concrete is calculated byEquation (131) giving Ec(t t0) = 8824GPa Substitution in Equation(34) gives


Figu

re3

2A

naly

sis

of s

tres

s an

d st

rain

in a

sec

tion

with

one

laye

r of

pre

stre

ssed

ste

el (E

xam

ple

31)

(a)

cro

ss-s

ectio

n (b

) con

ditio

nsim

med

iate

ly a

fter

pre

stre

ss (

c) c

hang

es d

ue t

o cr

eep

shr

inka

ge a

nd r

elax

atio

n

∆Pc = minus [3(minus6533 times 106)(20030)1120 times 10minus6 + (minus240 times 10minus6)200 times 109

times 1120 times 10minus6 + (minus80 times 106)1120 times 10minus6]

times 1 +1120 times 10minus6

0357

200

8824 1 +

02059

01193minus1

= 2427kN (5456kip)

In the absence of non-prestressed steel ∆Pc = minus∆Pps Thus thechange in stress in the tendon is

∆σps =∆Pps

Aps

= minus∆Pc

Aps


1120 times 10minus6 Pa = minus2167MPa (minus3143ksi)

We can now find an improved estimate of χr The initial stress in thetendon is

σp0 =1400 times 103

1120 times 10minus6 Pa = 1250MPa (181ksi)

The coefficients λ and Ω are (see Equation (17) )

λ =σp0

fptk

=1250

1770= 0706 Ω =

2167 minus 115

1250= 0081

Entering these values in the graph of Fig 14 we obtain χr = 080An improved estimate of the reduced relaxation is

∆σprinfin = 080(minus115) = minus92MPa (minus133ksi)

Equation (35) may be used again to obtain a more accurate valueof the prestress loss (∆σps = minus2259MPa) Further iteration is hardlynecessary in practical calculations

Application of Equations (312) (310) and (311)2 gives the changesin concrete stress the axial strain and curvature due to creep shrinkageand relaxation as follows (Fig 32(c) )

(∆σc)top =2427 times 103

0357 1 +

(minus0596)0454

01193 Pa



(∆σc)bot =2427 times 103

0357 1 +

0604 times 0454

01193 Pa = 2241MPa (0325ksi)

∆εO = minus240 times 10minus6 + 3(minus131 times 10minus6) +2427 times 103

8824 times 109 times 0357

= minus556 times 10minus6

∆ψ = 3(minus192 times 10minus6) +2427 times 103 times 0454

8824 times 109 times 42588 times 10minus3

= minus283 times 10minus6 mminus1(minus719 times 10minus6 inminus1)

Solution of the above problem employing the equations of Section25 would give identical results

33 Effects of presence of non-prestressed steel

Presence of non-prestressed reinforcement in a prestressed member reducesthe loss in tension in the prestressed steel A part of the compressive forceintroduced by prestressing will be taken by the non-prestressed steel at thetime of prestressing and the magnitude of the compressive force in thisreinforcement substantially increases with time As a result at t = infin theremaining compressive force in the concrete in a member with non-prestressed steel is much smaller compared with the compressive force on amember without such reinforcement

The loss of tension in the prestressed steel is equal to the loss of compres-sion in the concrete only when there is no non-prestressed reinforcementComparing absolute values the loss in compression in concrete is generallylarger than the reduction in tension in prestressed steel the difference is thecompression picked up by the non-prestressed steel during the period of loss

The axial strain and curvature are also much affected Presence of non-prestressed steel substantially decreases the axial strain and curvature at t =infin Thus the non-prestressed reinforcement should be accounted for in calcu-lations to predict the displacements as will be further discussed in Chapter 8

A comparison is made in Table 31 of the results of Examples 22 and31 in which two sections are analysed The data for the two examples areidentical with the only difference being the absence of non-prestressedreinforcement in Example 31 (see Figs 26(a) and 32(a) )


34 Reinforced concrete section without prestresseffects of creep and shrinkage

The procedure of analysis of Section 25 when applied to a reinforced con-crete section without prestress can be simplified as shown below

Consider a reinforced concrete section with several layers of reinforcement(Fig 33) subjected to a normal force N and a bending moment M thatproduce no cracking The equations presented in this section give the changesdue to creep and shrinkage in axial strain in curvature and in stress in con-crete and steel during a period (t minus t0) where t gt t0 and t0 is the age of concreteat the time of application of N and M The force N is assumed to act atreference point O chosen at the centroid of the age-adjusted transformed sec-tion of area Ac plus [α(t t0)As] where α(t t0) is a ratio of elasticity moduligiven by Equation (134)

Following the procedure of analysis in Section 25 two equations may bederived for the changes in axial strain and in curvature during the period t0 tot (the derivation is given at the end of this section)

∆εO = ηφ(t t0)[εO(t0) + ψ(t0)yc] + εcs(t t0) (315)

∆ψ = κφ(t t0)ψ(t0) + εO(t0)yc

r2c

+ εcs(t t0)yc

r2c (316)

where εO(t0) and ψ(t0) are instantaneous axial strain at O and curvature at age

Table 31 Comparison of strains curvatures and losses of prestress in two identical cross-sections with and without non-prestressed reinforcement (Examples 22 and31)

Symbolused

Withoutnon-prestressedreinforcement

Withnon-prestressedreinforcement

Axial strain immediately after prestress O minus131 times 10minus6 minus126 times 10minus6

Curvature immediately after prestress minus192 times 10minus6 mminus1 minus170 times 10minus6 mminus1

Change in axial strain due to creepshrinkage and relaxation O minus556 times 10minus6 minus470 times 10minus6

Change in curvature due to creepshrinkage and relaxation minus283 times 10minus6 mminus1 minus128 times 10minus6 mminus1

Axial strain at time t = infin O + O minus687 times 10minus6 minus596 times 10minus6

Curvature at time t = infin + minus475 times 10minus6 mminus1 minus298 times 10minus6 mminus1

Change in force in prestressed steel(the loss) Apsps minus243kN minus208kN

Axial force on concrete immediatelyafter prestress intc(t0)dAc minus1400kN minus1329kN

Axial force on concrete at t = infin intc(t)dAc minus1157kN minus878kNChange in force on concrete Pc int[c(t0) minus 243kN 451kN

c(t)]dAc


t0 η and κ are the ratios of the area and moment of inertia of the concretesection to the area and moment of inertia of the age-adjusted transformedsection (see Section 1111) thus

η = AcA (317)

κ = IcI (318)

Ac and A are areas of the concrete section and of the age-adjusted trans-formed section Ic and I are moments of inertia of the concrete area and ofthe age-adjusted transformed section about an axis through O the centroid ofthe age-adjusted transformed section

The values η and κ smaller than unity represent the effect of thereinforcement in reducing the absolute value of the change in axial strain andin curvature due to creep and shrinkage or applied forces For this reason ηand κ will be referred to as axial strain and curvature reduction coefficients

r2c = IcAc is the radius of gyration of the concrete area yc is the y-

coordinate of the centroid c of the concrete area y is measured in the down-ward direction from the reference point O thus in Fig 33 yc is a negativevalue

The change in stress in concrete at any fibre during the period t0 to t (seeEquations (245) and (246) ) is

∆σc = Ec(t t0) minus [εO(t0) + ψ(t0)y]φ(t t0) minus εcs(t t0) + ∆εO + ∆ψy (319)

where Ec is the age-adjusted modulus of elasticity of concrete (see Equation(131) )

The change in steel stress may be calculated by Equation (247)

Figure 33 Definition of symbols in Equations (315) to (323) for analysis of effects ofcreep and shrinkage in a reinforced concrete uncracked section


For the derivation of Equations (315) (316) and (319) apply Equations(242) and (243) to calculate the forces necessary to artificially preventdeformations due to creep and shrinkage

∆N

∆Mcreep

= minusEc(t t0)φ(t t0) Ac

Ac yc

Ac yc

Ic εO(t0)

ψ(t0) (320)

∆N

∆Mshrinkage

= minusEc(t t0)εcs(t t0) Ac

Ac yc (321)

The sum of Equations (320) and (321) gives the forces necessary torestrain creep and shrinkage

∆N

∆M = minusEc(t t0)

φ(t t0)[εO(t0) + ψ(t0)yc] + εcs(t t0)Ac

times φ(t t0)ψ(t0) + εO(t0)yc

r2c + εcs(t t0)

yc

r2cAcr

2c (322)

The artificial restraint may now be eliminated by application of minus∆N andminus∆M on the age-adjusted transformed section With the reference point Ochosen at the centroid of this section the first moment of area B must bezero The axial strain and curvature due to minus∆N and minus∆M can be calculatedby Equation (229) giving

∆εO

∆ψ =1

Ec(t t0) minus∆NA

minus∆MI (323)

Substitution of Equation (322) into (323) gives Equations (315) and (316)Equation (319) can be obtained by substitution of Equation (245) into(246)

Example 32 Section subjected to uniform shrinkage

Find the stress and strain distribution in the cross-section in Fig 34(a)due to uniform free shrinkage εcs(t t0) = minus300 times 10minus6 using the followingdata Ec(t0) = 30GPa (4350ksi) Es = 200GPa (29000ksi) φ(t t0) = 3χ = 08 The section dimensions and reinforcement areas are given inFig 34(a)

The age-adjusted modulus of elasticity of concrete and thecorresponding modular ratio are (Equations (131) and (134) )



1 + 08 times 3= 8824GPa (1280ksi)

α(t t0) =200

8824= 22665

The age-adjusted transformed section composed of Ac plus αAs has acentroid O at a distance 0551m below the top fibre The centroid of theconcrete area is at 0497m below top thus yc = minus0054 The area andmoment of inertia of concrete section about an axis through O are

Figure 34 Analysis of changes in stress and in strain due to shrinkage and creep in areinforced concrete section (Examples 32 and 33) (a) cross-sectiondimensions (b) changes in stress and strain due to shrinkage (c) stressand strain at age t0 due to axial force of minus1300kN (minus292kip) at mid-height and a bending moment of 350kN-m (3100kip-in) (d) changesin stress and strain due to creep


Ac = 02963m2 Ic = 2526 times 10minus3 m4 r2c = IcAc = 8475 times 10minus3 m2

The area and moment of inertia of the age-adjusted transformedsection about an axis through O are

A = 03811m2 I = 3750 times 10minus3 m4

The axial strain and curvature reduction coefficient (Equations (317)and (318) ) are

η =02963

03811= 0777 κ =

2526

3750= 0674

Substitution in Equations (315) and (316) gives the changes in axialstrain and in curvature due to shrinkage

∆εO = 0777(minus300 times 10minus6) = minus233 times 10minus6

∆ψ = 0674 (minus300 times 10minus6)minus0054

8445 times 10minus3

= 129 times 10minus6 mminus1 (323 inminus1)

The changes in concrete stress due to shrinkage (Equation (319) )are

(∆σc)top = 8824 times 109[minus (minus300) + (minus233) + 129(minus0551)]10minus6 Pa


(∆σc)bot = 8824 times 109[minus (minus300) + (minus233) + 129(0449)]10minus6 Pa

= 1102MPa (0159ksi)

The changes in stress and strain distributions caused by shrinkage areshown in Fig 34(b)

Example 33 Section subjected to normal force and moment

The same cross-section of Example 32 (Fig 34(a) ) is subjected at aget0 to an axial force = minus1300kN at mid-height and a bending moment of


350kN-m It is required to find the changes during the period (t minus t0) inaxial strain curvature and in concrete stress due to creep Use the samedata as in Example 32 but do not consider shrinkage Assume nocracking

The applied forces are a bending moment of 350kN-m and an axialforce of minus 1300kN at mid-height Replacing these by equivalent coupleand axial force at the reference point O gives (see Fig 34(a) )

N = minus1300kN (minus292kip) M = 350 + 1300(0051)

= 4163kN-m (3685kip-in)

These two values are substituted in Equation (232) to give theinstantaneous axial strain and curvature

εO(t0) = minus120 times 10minus6 ψ(t0) = 428 times 10minus6 mminus1 (109 inminus1)

The stress and strain distributions at age t0 are shown in Fig 34(c)The modulus of elasticity of concrete used for calculating the values ofthis figure is Ec(t0) = 30GPa

The values Ec(t t0) α(t t0) η and κ are the same as in Example 32Substitution in Equations (315) and (316) gives the changes in axial

strain and curvature due to creep (Fig 34(d) )

∆εO = 07773[minus120 + 428(minus0054)]10minus6 = minus334 times 10minus6

∆ψ = 0674 3428 + (minus120)minus 0054

8445 times 10minus310minus6= 1021 times 10minus6 mminus1 (2592 times 10minus6 inminus1)

The corresponding changes in concrete stress (Equation (319) ) are

(∆σc)top = 8824 times 109minus [minus120 + 428(minus0551)]3

+ (minus334) + 1021(minus0551)

= 1508MPa (0219ksi)

(∆σc)bot = 8824 times 109minus [minus120 + 428(0449)]3 + (minus334)

+ 1021(0449) = minus0813MPa (minus0118ksi)


35 Approximate equations for axial strain andcurvature due to creep

The changes in axial strain and curvature due to creep and shrinkage in areinforced concrete section without prestressing subjected to a normalforce and a bending moment are given by Equations (315) and (316) Whenconsidering only the effect of creep the equations become

∆εO = ηφ(t t0)[εO(t0) + ψ(t0)yc]) (324)

∆ψ = κφ(t t0) ψ(t0) + εO(t0) yc

r2c (325)

where εO = εO(t0) is the instantaneous strain at the reference point chosen atthe centroid of the age-adjusted transformed section ψ(t0) is the instantaneouscurvature yc is the y-coordinate of point c the centroid of concrete area r2

c =Ic Ac with Ac and Ic being the area of concrete and its moment of inertiaabout an axis through O φ(t t0) are creep coefficients η and κ are axial strainand curvature reduction coefficients (see Equations (317) and (318) )

When the section is subjected only to an axial force at O or to abending moment without axial force Equations (324) and (325) may beapproximated as follows

(a) Creep due to axial force The change in axial strain due to creep in areinforced section subjected to axial force

∆εO ηεO(t0)φ(t t0) (326)

(b) Creep due to bending moment The change in curvature due to creep in areinforced concrete section subjected to bending moment

∆ψ κψ(t0)φ(t t0) (327)

Equation (326) is derived from Equation (324) ignoring the term [ψ(t0)yc]because it is small compared to εO (t0) Similarly ignoring the term [εO(t0)ycr

2c]

in Equation (325) leads to Equation (327)If the section is without reinforcement ∆εO and ∆ψ due to creep would

simply be equal to φ times the instantaneous values In a section withreinforcement we need to multiply further by the reduction coefficients η

and κ

36 Graphs for rectangular sections

The graphs in Fig 35 for rectangular non-cracked sections can be employedto determine the position of the centroid O and moment of inertia I (or I )


Figure 35 Position of the centroid and moment of inertia I (or I) of transformed (orage-adjusted transformed) non-cracked rectangular section about an axisthrough the centroid (dprime = 01 h d = 09 h)


about an axis through O of the transformed (or age-adjusted transformed)section The transformed section is composed of the area of concrete Ac bhplus αAs (or αAs) where (see Equations (131) and (134) )

α = α(t0) = EsEc(t0) (328)

α = α(t t0) = Es[1 + χφ(t t0)]Ec(t0) (329)

b and h are breadth and height of the sectionThe values of I (or I ) may be used in the calculations for the instantaneous

curvature by Equation (216) or the change in curvature due to creep andshrinkage by Equation (240) (setting B = 0)

The top graph in Fig 35 gives the coordinate yc of the centroid ofthe concrete area (mid-height of the section) with respect to point O It isto be noted that in the common case when As is larger than Asprime yc has anegative value As and Asprime are the cross-section areas of the bottom and topreinforcement (Fig 35)

The bottom graph in Fig 35 gives the curvature reduction coefficient

κ =Ic

I (or I )(330)

where Ic is the moment of inertia of the concrete area Ac about an axisthrough O Ic is given by

Ic bh3

12 1 + 12y2

c

h2 (331)

In this equation the area Ac is considered equal to bh and its centroid atmid-height In other words the space occupied by the reinforcement isignored The graphs in Fig 35 are calculated assuming that the distancebetween the centroid of the top or bottom reinforcement and the nearbyextreme fibre is equal to 01 h A small error results when the graphs are usedwith this distance between 005 h and 015 h

37 Multi-stage prestressing

Consider a cross-section with a number of prestress tendons which are pre-stressed at different stages of construction This is often used in bridge con-struction where ducts are left in the concrete for the prestress cables to beinserted and prestressed in stages to suit the development of forces due tothe structure self-weight as the construction proceeds

In the procedure presented in Section 25 with one-stage prestressingthe axial strain and curvature were calculated in two steps one for the


instantaneous values occurring at time t0 and the other for the incrementsdeveloping during the period t0 to t due to creep shrinkage and relaxationWith multi-stage prestress the two steps are repeated for each prestress stage

Assume that the prestress is applied at age t0 and t1 and we are interested inthe stress and strain at these two ages and at a later age t2 The analysis is to bedone in four steps to calculate the following

1 εO(t0) and ψ(t0) are the instantaneous strain at reference point O and thecurvature immediately after application of the first prestress

2 ∆εO(t1 t0) and ∆ψ(t1 t0) are the changes in strain at reference point O andin curvature during the period t0 to t1

3 ∆εO(t1) and ∆ψ(t1) are the additional instantaneous strain at referencepoint O and curvature immediately after second prestress

4 ∆εO(t2 t1) and ∆ψ(t2 t1) are the additional change in strain at referencepoint O and curvature during the period t1 to t2

In each of the four steps appropriate values must be used for the propertiesof the cross-section the modulus of elasticity of concrete the shrinkage andcreep coefficients and the relaxation all these values vary according to the ageor the ages considered in each step

It is to be noted that when the prestress is introduced in stage 2 instan-taneous prestress loss occurs in the tendons prestressed in stage 1 This isaccounted for in the increments calculated in step 3

38 Calculation of displacements

In various sections of Chapters 2 and 3 equations are given for calculation ofthe axial strain and the curvature and the changes in these values caused bytemperature creep and shrinkage of concrete and relaxation of prestressedsteel

The present section is concerned with the methods of calculation of dis-placements in a framed structure for which the axial strain and curvature areknown at various sections The term lsquodisplacementrsquo is used throughout thisbook to mean a translation or a rotation at a coordinate A coordinate issimply an arrow drawn at a section or a joint of a structure to indicate thelocation and the positive direction of a displacement

Once the axial strain and curvature are known calculation of the dis-placement is a problem of geometry and thus the methods of calculation arethe same whether the material of the structure is linear or non-linear andwhether cracking has occurred or not


381 Unit load theory

The most effective method to find the displacement at a coordinate j is theunit load theory based on the principle of virtual work3 For this purpose afictitious virtual system of forces in equilibrium is related to the actual dis-placements and strains in the structure The virtual system of forces is com-posed of a single force Fj = 1 and the corresponding reactions at the supportswhere the displacements in actual structure are known to be zero When sheardeformations are ignored the displacement at any coordinate j on a planeframe is given by

Dj = εONuj dl + ψMuj dl (332)

where εO and ψ are the axial strain at a reference point O and the curvature ψin any cross-section of the frame Nuj and Muj are the axial normal force andbending moment at any section due to unit virtual force at coordinate j Thecross-section is assumed to have a principal axis in the plane of the frame andthe reference point O is arbitrarily chosen on this principal axis The axialforce Nuj acts at O and Muj is a bending moment about an axis through O Theintegral in Equation (332) is to be performed over the length of all membersof the frame

The principle of virtual work relates the deformations of the actual struc-ture to any virtual system of forces in equilibrium Thus in a staticallyindeterminate structure the unit virtual load may be applied on a releasedstatically determinate structure obtained by removal of redundants Thisresults in an important simplification of the calculation of Nuj and Muj and inthe evaluation of the integrals in Equation (332) For example consider thetransverse deflection at a section of a continuous beam of several spans Theunit virtual load may be applied at the section considered on a releasedstructure composed of simple beams Thus Muj will be zero for all spansexcept one while Nuj is zero everywhere

Only the second integral in Equation (332) needs to be evaluated and thevalue of the integral is zero for all spans except one

382 Method of elastic weights

The rotation and the deflection in a beam may be calculated respectively asthe shearing force and the bending moment in a conjugate beam subjected to atransverse load of intensity numerically equal to the curvature ψ for theactual beam This load is referred to as elastic load (Fig 36)

The method of elastic weights is applicable for continuous beams Theconjugate beam is of the same length as the actual beam but the conditionsof the supports are changed4 whereas for a simple beam the conjugate andactual beam are the same (Fig 36(a) and (b) )


The ψ-diagram in the actual beam is treated as the load on the conjugatebeam as shown in Fig 36(b) Positive curvature is positive (downward) loadIt can be shown that the shear V and the moment M in the conjugate beamare equal respectively to the rotation θ and the deflection D at the correspond-ing point of the actual beam The calculation of the reactions and bending

Figure 36 Actual and conjugate beams (a) deflection of actual beam (b) elastic load onconjugate beam

Figure 37 Equivalent concentrated loads which produce the same bending moment at thenodes and reactions at the supports of a statically determinate beam subjectedto variable load (a) variable load intensity (b) equivalent concentrated load at i


moments in a beam due to irregular elastic loading may be simplified by theuse of equivalent concentrated elastic loads applied at chosen node points(Fig 37(a) (b) ) At any node i the equivalent concentrated load Qi is equaland opposite to the sum of the reactions at i of two simply supported beamsbetween i minus 1 i and i + 1 carrying the same elastic load as the conjugate beambetween the nodes considered (Fig 37(b) )

The equivalent concentrated loads for straight-line and second-degreeparabolic variations are given in Fig 38(a) and (b) The formulae for theparabolic variation can of course be used to approximate other curves

The method of elastic weights and the equivalent concentrated loads are

Figure 38 Equivalent concentrated load for (a) straight-line and (b) parabolic varying load


used to derive a set of equations presented in Appendix C for the elongationend rotations and central deflection of a beam in terms of the values of axialstrain and curvature at a number of equally spaced sections of a simple beam(Fig C1) The same equations are applicable to a general member of a planeframe but the equations in this case give deflections and rotations measuredfrom a straight line joining the two displaced ends of the member (Line A lsquoBrsquoin Fig C2) Appendix C also includes equations for the displacements of acantilever

Example 34 Simple beam derivation of equations fordisplacements

Express the displacements D1 to D4 in the simple beam in Fig C1(a) interms of the axial strain ε and the curvature ψ at three sections(Fig C1(b) )

Assume parabolic variation of εO and ψ between the three sectionsEquivalent concentrated elastic loads at the three nodes are (see

Fig 38(b) )

Q =

7l48

l24

minusl48

l8

5l12

l8

minusl48

l24

7l48

ψ (a)

The first column of the 3 times 3 matrix represents the equivalent con-centrated forces at the three nodes when ψ1 = 1 while ψ2 = ψ3 = 0 Theother two columns are derived in a similar way

The displacements may be expressed in terms of Q

D2 = [0 minus 05 minus 1]Q (b)

D3 = [1 05 0]Q (c)

D4 = l [0 025 0] Q (d)

The first element in each of the 1 times 3 matrices in the last threeequations represents the shear at one of the two ends or the bendingmoment at the centre of the conjugate simple beam subjected to Q1 = 1while Q2 = Q3 = 0 The other two elements of each matrix are derived ina similar way

Substitution of Equation (a) in each of equations (b) (c) and (d)respectively gives Equations (C6) (C7) and (C8)


The sum of the elements of the first column in the 3 times 3 matrix inEquation (a) is l6 this is equal to the total elastic load on the beamwhich is the integral int ψ dl when ψ1 = 1 while ψ2 = ψ3 = 0 The sum ofthe elements in the second and third column of the matrix is equal tosimilar integrals

The displacement at coordinate 1 in Fig C1(a) is equal to the changein length of the beam thus

D1 = εO dl (e)

This integral is to be evaluated over the length of the beam for thevariable εO when (εO)1 = 1 while (εO)2 = (εO)3 = 0 and this procedure has tobe repeated two more times each time setting one of the εo values equalto unity and the others zero Thus summing the elements in each col-umn of the matrix in Equation (a) and changing the variable ψ to εO wecan express the displacement D1 in terms of the axial strain at the threenodes

D1 = l

6

2l

3

l

6 εO

Appendix C lists a series of expressions derived by the same pro-cedure as Example 34 The variation of εO and ψ is assumed eitherlinear or parabolic and the number of nodes either 3 or 5

Example 35 Simplified calculation of displacements

Use the values of the axial strain and curvature at mid-span and at theends of the post-tensioned simple beam in Fig 27 to calculate thevertical deflection at point C the centre of the span and the horizontalmovement of the roller at B at time t after occurrence of creep shrink-age and relaxation Assume parabolic variation of the axial strain andcurvature between the three sections

We prepare the vectors εO and ψ to be used in the equations ofAppendix C (see table p 94)

The deflection at the centre (Equation (C8) ) is given by

(186)2

96 [1 10 1]

64minus298

64 10minus6 = minus00103m = minus103mm

The minus sign indicates upward deflection


The change of length at the level of the reference axis (Equation(C5) ) is

186

6 [1 4 1]

minus605minus596minus605

10minus6 = minus00111m = minus111mm

(Here Equation (C1) could have been used assuming straight-linevariation between the section at mid-span and the two ends but theanswer will not change within the significant figures employed)

Rotation at the left end A (Equation (C7) ) is

186

6 [1 2 0]

64minus298

64 10minus6 = minus165 times 10minus3 radian

The minus sign means an anticlockwise rotationThe same rotation but opposite sign occurs at B The change in

length of AB (on the bottom fibre) is

minus00111 + 2 times 06(minus165 times 10minus3) = minus00131m

Horizontal movement of the roller at B is minus00131m = minus131mm Theminus indicates shortening of AB and hence B moves to the left

Left end Mid-span Right end

Axial strain at t0 O(t0)Change in axial strain O

minus126 times10minus6

minus479 times10minus6minus126 times10minus6



Total axial strain at time t minus605 times10minus6 minus596 times10minus6 minus605 times10minus6

Curvature at t0 (t0)Change in curvature

4 times10minus6

60 times10minus6minus170 times10minus6 mminus1

minus128 times10minus6 mminus14 times10minus6

60 times10minus6

Total curvature at time t 64 times10minus6 minus298 times10minus6 mminus1 64 times10minus6


39 Example worked out in British units

Example 36 Parametric study

The structure shown in Fig 39(a) represents a 1 ft wide (305mm) stripof a post-tensioned simply supported solid slab At time t0 the structureis subjected to dead load q = 040kipft (58kNm) and an initial pre-stressing force P = 290kip (1300kN) which is assumed constant overthe length The objectives of this example are to study the effects of thepresence of the non-prestressed steel on the stress distributions betweenconcrete and the reinforcement and on the mid-span deflection at time tafter occurrence of creep shrinkage and relaxation Non-prestressedsteel of equal cross-section area Ans is provided at top and bottom Thesteel ratio ρns = Ansbh is considered variable between zero and 1 per cent

The modulus of elasticity of concrete Ec(t0) = 4350ksi (30GPa) thechange in Ec with time is ignored The modulus of elasticity of theprestressed and the non-prestressed steel Es = 29000ksi (200GPa)Other data are

φ(t t0) = 30 εcs(t t0) = minus300 times 10minus6∆σpr = minus93ksi (minus64MPa)

The effects of varying the values of φ and εcs on the results will also bediscussed

The dead load q produces a bending moment at mid-span =1500kip-in (169kN-m)

Only the results of the analyses are given and discussed below Forease in verifying the results the simplest cross-section is selected Alsothe variation of the initial prestressing force P because of friction isignored and the difference in the cross-section area of the tendon andthe area of the prestressing duct is neglected

Table 32 gives the concrete stresses at midspan at time t after occur-rence of creep shrinkage and relaxation It can be seen that the stress atthe bottom fibre varies between minus1026 and minus502psi (minus708 andminus346MPa) as the non-prestressed steel ratio ρns is increased from zeroto 1

In other words ignoring the non-prestressed steel substantially over-estimates the compressive stress provided by prestressing to prevent orto control cracking by subsequent live load the overestimation is of thesame order of magnitude as the tensile strength of concrete The


compressive stress reserve commonly intended to counteract the ten-sion due to live load is substantially eroded as a result of the presenceof the non-prestressed steel On the other hand the non-prestressedsteel is beneficial in controlling the width of cracks (see Example 76)

Figure 39 Post-tensioned slab (Example 36) (a) slab dimensions and materialparameters (b) relative time-dependent change in forces in concreteprestressed steel and non-prestressed steel at mid-span cross-section


Table 32 also gives the force changes ∆Pc ∆Pns and ∆Pps in theconcrete the non-prestressed and the prestressed steel due to creepshrinkage and relaxation The sign of ∆Pc is positive indicating adecrease of the initial compressive force in concrete The negative ∆Pns

indicates an increase in compression Also the negative ∆Pps indicatesloss of tension in the prestressing tendon

The non-dimensional graphs in Fig 39(b) represent the variationof ρns versus ( | ∆Pps | ∆ref) or (∆Pc∆ref) where ∆ref is a referenceforce equal to | ∆Pps | when ρns = 0 in which case | ∆Pps | = ∆Pc Thedifference between the ordinates of the two curves in Fig 39(b) repre-sents the relative increase in compressive force in the non-prestressedsteel

Unless ρns = 0 the absolute value of the tension loss in prestressingsteel | ∆Pps | should not be considered as equal to the compression lossin concrete because this will overestimate the compression remaining inconcrete after losses

Table 32 also gives the deflection at the centre of span with varying

Table 32 Stress and deflection at mid-span in non-cracked slab Example 36

Non-prestressed steel ratio ns (percent)

0 02 04 06 08 10 04 withreduced amp cs

Concrete stresses at top minus302 minus276 minus246 minus215 minus184 minus155 minus250time t (psi) bot minus1026 minus879 minus759 minus659 minus574 minus502 minus969

Change of Concrete Pc 52 76 97 114 128 140 59force in the three Non-

Pns 0 minus28 minus52 minus72 minus88 minus102 minus28materials prestressed

between steel

t0 and t Prestressed Pps minus52 minus48 minus45 minus42 minus40 minus38 minus30(kips) steel

Deflection at time t minus923 minus794 minus696 minus621 minus560 minus510 minus553before application of the live load (10minus3 in)

Ratio of deflection at 256 232 213 200 188 178 169time t before applicationof live load to the instantaneous deflection

Steel stresses at time tbefore live load

ns

(bot)minus36 minus33 minus30 minus28 minus26 minus24 minus20

application (ksi) ps 159 161 163 165 167 168 173


ρns The negative sign indicates camber It is clear that the camber willbe overestimated if non-prestressed steel is ignored Also it can be seenthat the deflection after creep shrinkage and relaxation cannot beaccurately predicted by multiplying the instantaneous deflection by aconstant number because such a number must vary with ρns and withthe creep shrinkage and relaxation parameters

Effects of varying creep and shrinkage parametersIt is sometimes argued that the effort required for an accurate analysisof the strain and the stress is not justified because accurate values of thecreep coefficient φ and the free shrinkage εcs are not commonly availableA more rational approach for important structures is to perform accur-ate analyses using upper and lower bounds of the parameters φ and εcs

The analysis is repeated in the above example for the case ρns = 04with φ = 15 and εcs = minus150 times 10minus6 (instead of 30 and minus300 times 10minus6) Theresults shown in the last column of Table 32 indicate that reducing φand εcs by a factor of 2 has some effect but the effect is not as importantas the effect of ignoring the non-prestressed steel

310 General

The loss in tension in prestressed steel ∆Pps caused by creep shrinkage andrelaxation is equal in absolute value to the loss in compression on the con-crete ∆Pc only in a cross-section without non-prestressed reinforcement Ingeneral the value of ∆Pc is greater in absolute value than ∆Pps the differencedepends on several variables one of which of course is the amount of non-prestressed reinforcement (in Example 22 ∆Pps = minus208kN and ∆Pc =451kN see Fig 26) The presence of non-prestressed reinforcement maysubstantially reduce the instantaneous strains and to a greater extent thetime-dependent strains Thus the non-prestressed steel must be taken intoconsideration for accurate prediction of deformations of prestressedstructures

Equation (34) gives the value of ∆Pc when the prestressed steel and thenon-prestressed reinforcement are at one level and the force ∆Pc is situated atthis level Once ∆Pc is known it may be used to calculate the changes instresses and in strain variation over the section The same procedure may alsobe employed involving approximation when the section has more than onelayer of reinforcement

The methods discussed in Section 38 can be used to determine the dis-placements when the axial strain εO and the curvature ψ are known at all


sections (or at a number of chosen sections) Here the calculation represents asolution of a problem of geometry and thus the same methods are equallyapplicable in structures with or without cracking

Notes

1 The value of the reduced relaxation = 80MPa is used below in order to compare theresults with those of Example 22

2 The value ∆σps = minus2167MPa (not the slightly improved value obtained after iter-ation) is substituted in these equations in order to be able to compare the resultswith Example 22 where the reduced relaxation was minus80MPa (see section 34)

3 See Ghali A and Neville AM (1997) Structural Analysis A Unified Classical andMatrix Approach 4th edn E amp FN Spon London (Sections 65 66 72 and 73)

4 See p 187 of the reference mentioned in note 3 above


Time-dependent internalforces in uncracked structuresanalysis by the force method

Pasco-Kennewick Intercity Bridge Wa USA Segmentally assembled concrete cable-stayedbridge (Courtesy A Grant and Associates Olympia)

Chapter 4

41 Introduction

The preceding two chapters were concerned with the analysis of stress andstrain in an uncracked reinforced or prestressed concrete section subjected tointernal forces for which the magnitude and the time of application areknown Creep and shrinkage of concrete and relaxation of steel were con-sidered to affect the distribution of stress and strain and the magnitude of theprestressing force in a prestressed member but it was assumed that theelongation or curvature occur without restraint by the supports or bycontinuity with other members which is the case in a statically determinatestructure The present chapter is concerned with the analysis of changes ininternal forces due to creep shrinkage and relaxation of steel in staticallyindeterminate structures

Consider the effect of creep on a statically indeterminate structure madeup of concrete as a homogeneous material neglecting the presence ofreinforcement A sustained load of given magnitude produces strains anddisplacements that increase with time but this is not accompanied by anychange in the internal forces or in the reactions at the supports Creep effecton displacements in such a case can be accounted for simply by using ndash for themodulus of elasticity of the material ndash a reduced value equal to E(1 + φ)where φ is the creep coefficient

On the other hand if the structure is composed of parts that have differentcreep coefficients or if its boundary conditions change the internal forceswill of course be affected by creep Concrete structures are generally con-structed in stages thus made up of concrete of different ages and hencedifferent creep coefficients Precast parts are often made continuous withother members by casting joints or by prestressing and hence the boundaryconditions for the members change during construction For all these casesstatically indeterminate forces gradually develop with time

Change in the length of members due to shrinkage when restrained pro-duces internal forces But because shrinkage develops gradually with timeshrinkage is always accompanied by creep and thus the internal forces due toshrinkage are well below the values that would develop if the shrinkage wereto occur alone

Similarly the internal forces that develop due to gradual differentialsettlements of the supports in a continuous structure are greatly reduced bythe effect of creep that occurs simultaneously with the settlement

In the present chapter and in Chapter 5 we shall consider the analysis ofthe changes in internal forces in a statically indeterminate structure due tocreep shrinkage and differential settlement of supports The well-knownforce or displacement method of structural analysis may be employed Ineither method two types of forces (or displacements) are to be considered (a)external applied forces (or imposed displacements) introduced at their fullvalues at instant t0 and sustained without change in magnitude up to a later

Time-dependent internal forces in uncracked structures 101

time t and (b) forces (or displacements) developed gradually between zerovalue at time t0 to their full values at time t The first type of forces causeinstantaneous displacement which is subsequently increased by the ratio φwhere φ = φ(t t0) coefficient for creep at time t for age at loading t0 Thesecond type of forces produce at time t a total displacement instantaneousplus creep (1 + χφ) times the instantaneous displacement that would occur ifthe full value of the force is introduced at t0 where χ = χ(t t0) the agingcoefficient (see Section 17) This implies that the internal forces (or the dis-placements) develop with time at the same rate as relaxation of concrete (seeSection 19)

Use of the coefficients φ or χφ to calculate the increase in displacement dueto creep ndash in the same way as done with strain ndash is strictly correct only whenthe structure considered is made of homogeneous material In the precedingtwo chapters we have seen that in a statically determinate structure thepresence of reinforcement reduces the axial strain and curvature caused bycreep and hence reduces the associated displacements (see Section 33 and34) The presence of reinforcement has a similar effect on the displacementin a statically indeterminate structure but has a smaller effect on the staticallyindeterminate forces Thus the reinforcement is often ignored when thechanges in the statically indeterminate forces due to creep or shrinkage areconsidered The prestress loss due to creep shrinkage and relaxation ispredicted separately and is substituted by a set of external applied forcesHowever the presence of prestressed or non-prestressed reinforcementshould not be ignored when prediction of the displacement is the objectiveof the analysis or when more accuracy is desired Also the forces caused bythe movements of the supports will be underestimated if the presence of thereinforcement is ignored a correction to offset this error is suggested inSection 44

Section 42 serves as a review of the general force method of structuralanalysis and introduces the symbols and terminology adopted The analysisby the force method involves calculations of displacements due to knownexternal forces applied on a statically determinate released structure It alsoinvolves calculations of displacements of the released structure due to unitvalues of the statically indeterminate redundants In Sections 43 and 44 theforce method is applied to calculate the time-dependent changes in internalforces caused by creep shrinkage of concrete relaxation of steel and move-ment of supports in statically indeterminate structures In these two sectionswe shall ignore the presence of the reinforcement when calculating the dis-placements involved in the analysis by the force method However a correc-tion is suggested in Section 44 to account for the reinforcement and avoidunderestimation of the statically indeterminate forces caused by movementsof supports

An alternative solution which also employs the force method is presentedin Section 45 The presence of all reinforcement is accounted for and the


effect of prestress loss is automatically included The general procedure ofSection 25 is applied in a number of sections to calculate the axial strain andthe curvature in a statically determinate released structure The displacementsinvolved in the analysis by the force method are calculated by numericalintegration of the curvature andor axial strain (see Section 38) Naturallyaccounting for the reinforcement involves more computation (see Section 45)

42 The force method1

Consider for example the continuous beam shown in Fig 41(a) subjectedto vertical loads as shown Here we shall consider the simple case when allthe loads are applied at the same time and the beam is made of homo-geneous material The purpose of the analysis may be to find thereactions the internal forces or the displacements the term lsquoactionrsquo will beused here to mean any of these The analysis by the force method involves fivesteps

Step 1 Select a number of releases n by the removal of internal or externalforces (redundants) Removal of the redundant forces F leaves a staticallydeterminate structure for example the continuous beam in Fig 41(a) isreleased in Fig 41(b) A system of n coordinates on the released structureindicates the chosen positive directions for the released forces and thecorresponding displacements

Step 2 With the given external loads applied on the released structure cal-culate the displacements D at the n coordinates These representinconsistencies to be eliminated by the redundant forces The values As ofthe actions are also determined at the desired positions of the released struc-ture In the example considered D represent the angular discontinuities atthe intermediate supports (Fig 41(c) )

Step 3 The released structures are subjected to a force F1 = 1 and the dis-placements f11 f21 fn1 at the n coordinates are determined (see Fig41(d) ) The process is repeated for unit values of the forces at each of the ncoordinates respectively Thus a set of flexibility coefficients is generatedwhich forms the flexibility matrix [ f ]n times n a general element fij is the displace-ment of the released structure at coordinate i due to a unit force Fj = 1 Thevalues of the actions [Au] are also determined due to unit values of the redun-dants any column j of the matrix [Au] is composed of the actions due to aforce Fj = 1 on the released structure

Step 4 The values of the redundant forces necessary to eliminate the incon-sistencies in the displacements are determined by solving the compatibilityequation


[ f ] F = minusD (41)

The compatibility Equation (41) ensures that the forces F are of sucha magnitude that the displacements of the released structure becomecompatible with the actual structure

Figure 41 Analysis of a continuous beam by the force method (a) statically indeterminatestructure (b) released structure and coordinate system (c) external forcesapplied on released structure (d) generation of flexibility matrix [f ]


Step 5 The values A of the actions in the actual statically indeterminatestructure are obtained by adding the values As in the released structurecalculated in step 2 to the values caused by the redundants This may beexpressed by the following superposition equation

A = As + [Au] F (42)

43 Analysis of time-dependent changes of internalforces by the force method

Forces applied at time t0 on a structure made up of homogeneous materialproduce instantaneous strain which will increase due to creep If the magni-tude of the forces is maintained constant strain at time t will be φ times theinstantaneous strain where φ = φ(t t0) is the creep coefficient at time t whenthe age at loading is t0 Because the material is homogeneous the increase ofstrain by the ratio φ at all points results in the same increase in the displace-ments Thus the creep coefficient φ used for strain can be applied directly todisplacements

In the example considered in Section 42 (Fig 41) assume that theexternal loads are applied at time t0 and the structure is made up of homo-geneous material At any time t greater than t0 creep increases the values ofD and [ f ] to (1 + φ) times the values at t0 This results in no change in thestatically indeterminate forces and in the internal forces The change in theactual statically indeterminate structure is only in the displacements whichare magnified by the ratio (1 + φ) The same conclusion can be reached byconsidering that the modulus of elasticity of the structure is Ec(t0)(1 + φ) andperforming a conventional elastic analysis where Ec(t0) is the modulus ofelasticity at age t0

Now let us consider a case in which creep affects the internal forcesAssume for example that the beam in Fig 41(a) is made of three precastsimple beams which are prestressed and placed in position at age t0 and madecontinuous shortly after The instantaneous deflections which occur at t0 dueto the self-weight of the beam are those of simple beams with modulus ofelasticity Ec(t0) Further deflection due to creep occurs after the beams havebecome continuous The angular rotation of the beam ends at B and C mustbe compatible This will result in the gradual development of the redundants∆F which represent in this case the changes in the bending moments atcoordinates 1 and 2 caused by creep

To find the changes in the reactions the internal forces or the displace-ments at any section occurring during a time interval t0 to t the analysisfollows the five steps of the force method as outlined in Section 42 with themodifications discussed below The time-dependent changes considered heremay be caused by creep as in the above-mentioned example or by shrinkageor support settlement or a combination of these


In step 2 of the force method calculate ∆D the changes in the displace-ment of the released structure at the coordinates that occur between t0 and tThe displacement ∆D may be expressed as a sum of four terms

∆D = ∆Dloads + ∆Dprestress loss + ∆Dshrinkage + ∆Dsettlement (43)

∆Dloads represents the displacements due to creep under the effect ofprestress and other loads introduced at time t0 and sustained at their fullvalues up to time t eg the structure self-weight For calculation of theelements of this vector multiply the instantaneous displacement at t0 by thecreep coefficient φ(t t0) If the loads are applied at t0 and the continuity isintroduced at t1 and we are concerned with the changes in the displacementbetween t1 and a later time t2 the creep coefficient would be [φ(t2 t0) minusφ(t1 t0)]

∆Dprestress loss represents the displacements due to creep under the effect ofprestress loss during the period t0 to t The loss of prestress should not beignored in practice when the dead load and the load balanced by the prestressare of the same order of magnitude and of opposite signs Thus the accuracyof the analysis may be sensitive to the accuracy in calculating and accountingfor the effect of prestress loss Prestress loss may be represented by a set offorces in the opposite direction to the prestress forces The prestress lossdevelops gradually between time t0 and t thus displacement due to the pre-stress loss is equal to [1 + χφ(t t0)] times the instantaneous displacement dueto the same forces if they were applied at time t0

∆Dshrinkage and ∆Dsettlement are displacements occurring in the releasedstatically determinate structure thus in this step of analysis no forces areinvolved These displacements are determined by geometry using the shrink-age or settlement values which would occur without restraint during theperiod (t minus t0)

In the same step (2) also calculate ∆As the changes in the values of therequired actions in the released structure occurring during the same period

In step 3 generate an age-adjusted flexibility matrix [ f ] composed of thedisplacements of the released structure at the coordinates due to unit valuesof the redundants These unit forces are assumed to be introduced graduallyfrom zero at t0 to unity at t Any element f ij represents the instantaneous pluscreep displacements at coordinate i due to a unit force gradually introducedat coordinate j [ f ] is generated in the same way as [ f ] using for the calcula-tion of the displacements the age-adjusted modulus of elasticity given byEquation (131) which is repeated here

Ec =Ec

1 + χφ(44)

where Ec = Ec(t t0) Ec = Ec(t0) is the modulus of elasticity of concrete at age t0


The matrix [∆Au] which is composed of the changes in the values of theactions due to unit change in the values of the redundant is the same as [Au]discussed in Section 42 Only when one of the actions is a displacementshould the corresponding Au value be magnified by the appropriate (1 + χφ)as explained above for the flexibility coefficients

In step 4 of the analysis we find the changes in the redundants occurringbetween t0 and t by solving the compatibility equations

[ f ] ∆F = minus∆D (45)

In step 5 the changes in the actions caused by creep are determined bysubstitution in the equation

∆A = ∆As + [∆Au] ∆F (46)

The value of the aging coefficient χ to be used in the above analysis may betaken from the graphs or the table in Appendix A This implies that theprestress loss and the statically indeterminate forces develop with time at thesame rate as the relaxation of concrete (see Section 18)

It is to be noted that the analysis discussed in the present section is con-cerned only with the changes ∆A in the values of the actions developingduring a given period of time Addition of ∆A to A the values of theactions at the beginning of the period gives the final values of the actionsCalculation of A requires a separate analysis and may require use of theforce method (Equation (42) ) As an example consider a staticallyindeterminate structure made up of parts of different creep properties andsubjected at time t0 to an external applied load or sudden settlement To findthe values of any actions at a later time the analysis is to be performed in twostages and the force method may be used for each In stage 1 determine Athe values of the actions at age t0 immediately after application of the load orthe settlement The moduli of elasticity to be used in this analysis are theappropriate values for individual parts of the structure for instantaneousloading at time t0 In the second stage the time-dependent changes ∆A aredetermined using the procedure described in the present section

In the method of analysis suggested here the presence of the reinforcementis to be consistently ignored in the calculation of the displacement vector∆D and the age-adjusted flexibility matrix [ f ] this results in general in anoverestimation of the elements of the two matrices However the consistencytends to reduce the error in calculation of the statically indeterminate forces∆F by solution of Equation (45) For the same reason the forces due toprestress loss ndash to be used in the calculation of ∆Dprestress loss ndash should beevaluated ignoring the presence of the non-prestressed steel The error result-ing from this approximation is generally acceptable in practice for calculationof the internal forces in statically indeterminate structures The internal


forces calculated in this way may subsequently be employed to predict deflec-tions but the presence of reinforcement should not be ignored in the calcula-tion of axial strains and curvature from which the displacements can bedetermined as discussed in Chapters 2 and 3

An alternative procedure of analysis using Equation (45) is discussed inSection 45 in which the presence of the reinforcement is accounted for in thecalculation of [ f ] and ∆D

Example 41 Shrinkage effect on a portal frame

Find the bending moment diagram in the concrete frame in Fig 42(a)due to shrinkage that gradually develops between a period t0 to t Theframe has a constant cross-section the moment of inertia of theconcrete area is Ic Ignore deformations due to axial forces

The analysis for this problem is the same as that for a drop of tem-perature that produces a free strain equal to εcs(t t0) The only difference

Figure 42 Analysis of internal forces caused by shrinkage in a plane frame (a) framedimensions (b) bending moment diagram


is in the modulus of elasticity to be used in the analysis With shrinkageuse the age-adjusted elasticity modulus

Ec(t t0) =Ec(t0)

1 + χφ(t t0)

The bending moment diagram for this frame and the reactions arederived by a conventional elastic analysis eg by use of the generalforce method or by moment distribution2 the results are given inFig 42(b) Note that the shrinkage εcs is a negative value after applica-tion of the multiplier the ordinates in Fig 42(b) will have reversedsigns

To calculate the stress in concrete at any fibre we should use thevalues of the internal forces as calculated by this analysis andthe section properties Ac and Ic of the concrete excluding thereinforcement

Example 42 Continuous beam constructed in two stages

The continuous prestressed beam ABC (Fig 43(a) ) is cast in twostages AB is cast first and at age 7 days it is prestressed and its formsremoved span BC is cast in a second stage and its prestressing andremoval of forms are performed when the ages of AB and BC are 60and 7 days respectively Find the bending moment diagram at timeinfinity due to the self-weight of the beam only using the followingcreep and aging coefficients

φ(infin 7) = 27 χ(infin 7) = 074 φ(60 7) = 11

φ(infin 60) = 23 χ(infin 60) = 078

Ratio of elasticity moduli for concrete at ages 60 and 7 days are

Ec(60)Ec(7) = 126

Let t be the time measured from day of casting of AB A staticallydeterminate released structure and a system of one coordinate areshown in Fig 43(b) At t = 60 uniform load q is applied on span BC ofthe continuous beam ABC which has moduli of elasticity Ec(60) for AB


and Ec(7) for BC We use here the force method Displacement of thereleased structure is

D1 = (D1)AB + (D1)BC

= 0 +ql3

24Ec(7)Ic

Figure 43 Analysis of internal forces in a continuous beam with different creepcoefficients and different ages at loading of spans (Example 42) (a)continuous beam stripped in two stages (b) statically determinatereleased structure (c) bending moment diagram at t = infin


Flexibility coefficient is

f11 = ( f11)AB + ( f11)BC

=l

3Ec(60)Ic

+l

3Ec(7)Ic

The statically indeterminate bending moment at B at t = 60 is

F1 = minus D1 f11

Substitution of Ec(60) = 126Ec(7) in the above equations gives

F1 = minus00697 ql 2

The broken line (a) in Fig 43(c) represents the bending momentdiagram immediately after removal of the formwork of BC If after thisevent the beam is released again creep will produce between t = 60and infin the following change in displacement

∆D1 = (∆D1)AB + (∆D1)BC

The first term on the right-hand side of this equation representseffects of creep on span AB due to load q introduced at t = 7 and thestatically indeterminate force F1 introduced at t = 60 thus

(∆D1)AB =ql 3

24Ec(7)Ic

[φ(infin 7) minus φ(60 7)] minus00697 ql 3

3Ec(60)Ic

φ(infin 60)

On BC the distributed load q and the force F are introduced at t = 7creep produces a change in slope at B

(∆D1)BC =ql 3

24Ec(7)Ic

φ(infin 7) minus00697 ql 3

3Ec(7)Ic

φ(infin 7)

Substitution of the values of φ and Ec(60) = 126 Ec(7) in the aboveequations gives

∆D1 = 00720 ql 3

Ec(7)Ic


The age-adjusted flexibility coefficient f11 is the sum of the rotationsat the ends of the two spans due to a redundant force F1 graduallyintroduced between t = 60 and infin

f11 = ( f11)AB + ( f11)BC

( f11)AB =l

3Ec (infin 60)Ic

( f11)BC =l

3Ec(infin 7)Ic

The age-adjusted moduli (Equation (131) ) are

Ec(infin 60) =Ec(60)

1 + χφ(infin 60)= 1

1 + 078 times 23 Ec(60) = 045Ec(7)

Ec(infin 7) =Ec(7)

1 + χφ(infin 7)= 1

1 + 074 times 27 Ec(7) = 034Ec(7)

Thus

f11 =l

3 times 045Ec(7)Ic

+l

3 times 034Ec(7)Ic

= 1724 l

Ec(7)Ic

Solution of the compatibility Equation (45) gives the staticallyindeterminate moment at support B developing gradually betweent = 60 and infin

∆F1 = minus f minus111 ∆D1 = minus00418ql 2

The statically indeterminate bending moment at B at t = infin is

minus00697ql 2 minus00418ql 2 = minus01115ql 2

The bending moment diagram is shown in Fig 43(c) The twobroken lines in the same figure indicate the bending moment diagramwhen (a) the two construction stages are considered but creep isignored and (b) the beam is cast prestressed and the forms removed inthe two spans simultaneously


Example 43 Three-span continuous beam composed of precastelements

Three precast prestressed simple beams are prestressed and made con-tinuous at age t0 by a reinforced concrete joint cast in situ (Fig 44(a) )It is required to find the bending moment diagram at a later age t Theprestress tendon profile for each beam is as shown in Fig 44(b) Thefollowing data are given The initial prestress at age t0 creates auniformly distributed upward load of intensity (23)q thus

2

3q =

8Pa

l 2

where P is the absolute value of the prestress force a and l are defined inFig 44(b) q is the weight per unit length of beam Prestress loss is to beassumed uniform and equal to 15 per cent of the initial prestress Creepcoefficient φ(t t0) = 25 aging coefficient χ(t t0) = 08 Ignore crackingat the joint

Two statical systems need to be analysed (a) Simple beams withmodulus of elasticity Ec(t0) subjected to the self-weight qunit lengthdownwards plus a set of self-equilibrating forces representing the initialprestress (Fig 44(c) ) the bending moment for this system is shown inthe same figure (b) A continuous beam subjected to a set of self-equilibrating forces representing the prestress loss and redundant con-necting moments caused by creep the modulus of elasticity to be usedwith this loading is the age-adjusted modulus Ec(t t0) The analysisfor the statically indeterminate bending moment due to loadings iscalculated below

A statically determinate released structure is shown in Fig 44(d)Because of symmetry the two coordinates representing the connectingmoments at B and C are given the same number 1

Change in displacement in the released structure during the period t0

to t (Equation (43) ) are

∆D1 = (∆D1)load + (∆D1)prestress loss

(∆D1)load = D1(t0)φ(t t0)

where D1(t0) is the instantaneous displacement of the released structuredue to the loading in Fig 44(c) Using Equation (C6) Appendix C


Figure 44 Bending moment developed by creep in precast simple beams madecontinuous by casting joints (Example 43) (a) three simple beams madecontinuous at age t0 by a cast in situ joint (b) typical prestress tendonprofile for all beams (c) loads and diagram of the bending momentsintroduced at age t0 (d) statically determinate released structure andcoordinate system (e) forces and bending moment due to prestress lossin one span of released structure (f) statically indeterminate bendingmoments (g) bending moment diagram at time t


D1(t0) =10minus3ql 2

Ec(t0)Ic

2 timesl

6 (2 times 695 + 1 times 278) = 556 times 10minus3

ql 3

Ec(t0)Ic

(∆D1)load = 556 times 10minus3 ql 3

Ec(t0)Ic

25 = 1390 times 10minus3 ql 3

Ec(t0)Ic

The age-adjusted modulus of elasticity of concrete (Equation (131) )is

Ec(t t0) = 1

1 + 08 times 25 Ec(t0) =1

3Ec(t0)

A set of self-equilibrating forces3 representing the prestress loss andthe corresponding bending moment diagram for a typical span of thereleased structure is shown in Fig 44(e) The displacement due to theseforces using a modulus of elasticity Ec = Ec(t0)3 (see Equation (C6) ) is

(∆D1)presstress loss =10minus3ql 2

[Ec(t0)3]Ic

2 timesl

6 (2 times 83 minus 1 times 42)

= 125 times 10minus3 ql 3

Ec(t0)Ic

(∆D1) = (1390 + 125)10minus3 ql 3

Ec(t0)Ic


Ec(t0)Ic

Age-adjusted flexibility coefficient is

f11 =l

[Ec(t0)3]Ic

13 +1

2 = 25 l

Ec(t0)Ic

Substituting in Equation (45) and solving gives

∆F1 = minus1515

25 10minus3ql 2 = minus606 times 10minus3ql 2

The statically indeterminate bending moment developed by creep


and prestress loss is shown in Fig 44(f) The diagrams of the bendingmoment at time t (Fig 44(g) ) are obtained by the superposition of thediagrams in Fig 44(c) (e) and (f) Note that the negative bendingmoment at the joints B and C (= minus00606 ql 2) is higher in absolute valuethan the bending moment on the adjacent sections the higher value isplotted over a short length representing the length of the cast in situjoint

Example 44

A two-span continuous beam ABC (Fig 45(a) ) is built in two stagesPart AD is cast first and its scaffolding removed at time t0 immediatelyafter prestressing Shortly after part DC is cast and at time t1 pre-stressed and its scaffolding removed Find the bending moment dia-gram for the beam at a much later time t2 due to prestressing plus theself-weight of the beam q per unit length The initial prestress createsan upward load of intensity of 075 q and the prestress loss is 15 percent of the initial value

Assume that the time is measured from the day of casting of part ADand that the prestress for DC is applied at time t1 when the age of DC ist0 The following material properties are assumed to be known (datacorresponds to t0 = 7 days t1 = 60 days and t2 = infin)

φ(t1 t0) = 11 χ(t1 t0) = 079 φ(t2 t0) = 27 χ(t2 t0) = 074

φ(t2 t1) = 23 χ(t2 t1) = 078 Ec(t1)Ec(t0) = 126

The prestress loss starts to develop immediately after prestressingHowever for simplicity of presentation we assume here that the loss is15 per cent of the initial forces and the total amount of the loss occursduring the period t1 to t2

Three statical systems need to be analysed

(a) A simple beam with an overhang (Fig 45(b) ) subjected at t0 to adownward load q and a system of self-equilibrating forces representingthe initial prestress forces on AD The bending moment diagram forthis system is shown in Fig 45(c)

(b) A continuous beam subjected at time t1 to the self-weight of partDC and the forces due to the prestress of stage 2 (Fig 45(d) ) The


moduli of elasticity to be used are Ec(t1) for AD and Ec(t0) for DC Theinstantaneous bending moment diagram corresponding to this loadingis shown in Fig 45(e)

(c) A continuous beam subjected to a set of self-equilibrating forcesrepresenting the prestress loss and redundant forces caused by creepWith this system use the age-adjusted elasticity moduli

(Ec)AD = Ec(t2 t1) =Ec(t1)

1 + 078 times 23= 036Ec(t1) = 045 Ec(t0)

(Ec)DC = Ec(t2 t0) =Ec(t0)

1 + 074 times 27= 034Ec(t0)

The released structure and the coordinate system shown in Fig 45(f)will be used below to calculate the redundant force F1 due to creep andprestress loss

The term (∆D1)loads is the displacement in the released structurecaused by creep Using virtual work (Equation (332) )

(∆D1)loads =1

Ec(t0)Ic

D

AMcMu1 dl [φ(t2 t0) minus φ(t1 t0)]

+1

Ec(t1)Ic

D

AMeMu1 dl φ(t2 t1)

+1

Ec(t0)Ic

C

DMeMu1 dl φ(t2 t0)

where Mc and Me are the bending moments shown in parts (c) and (e) ofFig 45 and Mu1 is the bending moment due to a unit value of theredundant at coordinate 1 Fig 45(g) The values of the three integrals4

are indicated separately in the following equation

(∆D1)loads = 247 times 10minus3 ql 3

Ec(t0)Ic

minus 215 times 10minus3

timesql 3

Ec(t1)Ic

+ 201 times 10minus3 ql 3

Ec(t0)Ic


Ec(t0)Ic



Figure 45 Analysis of the instantaneous and time-dependent bending moment in acontinuous beam built and prestressed in two stages (Example 44) (a) acontinuous beam cast and prestressed in two stages (b) loads introducedat time t0 (c) bending moment for the beam and loads in (b) (d) loadsintroduced at time t1 on a continuous beam (e) bending moment for thebeam and loads in (d) (f) statically determinate released structure andcoordinate system (g) bending moment due to the unit value of theredundant F1 (h) loads representing the prestress loss (i) bendingmoment in the released structure due to prestress loss (j) final bendingmoments at time t2


A system of forces representing the prestress loss is applied on thereleased structure in Fig 45(h) and the corresponding bendingmoment is shown in Fig 45(i) The displacement at coordinate 1 due toprestress loss is

(∆D1)prestress loss =1

(Ec)ADIc

D

AMiMu1 dl +

1

(Ec)DCIc

C

DMiMu1 dl

where Mi is the bending moment shown in part (i) of Fig 45The values of the two integrals in this equation are separately

indicated in the following


045Ec(t0)minus006 times 10minus3

ql 3

Ic +

1

034Ec(t0)

24 times 10minus3 ql 3

Ic = 70 times 10minus3

ql 3

Ec(t0)Ic

∆D1 = (277 + 70)10minus3ql 3

Ec(t0)Ic


Ec(t0)Ic

The age-adjusted flexibility coefficient

f 11 =1

(Ec)ADIc

D

AM2

u1 dl +1

(Ec)DCIc

C

DM 2

u1 dl =251l

Ec(t0)Ic

Substitution in Equation (45) and solving for the redundant value

∆F1 = minus347

251times 10minus3ql 2 = minus138 times 10minus3ql 2

The bending moment diagram at time t2 shown in Fig 45(j) isobtained by superposition of ∆F1 times Mu1 Mc Me and Mi The twobroken curves shown in Fig 45(j) are approximate bending momentdiagrams obtained as follows (a) considering the construction stagesbut ignoring creep (b) ignoring the construction stages and creep thusapplying the dead load and 085 the prestress forces directly on acontinuous beam

Figure 45(j) indicates that the bending moment diagram for astructure built in stages is gradually modified by creep to approach the


bending moment which would occur if the structure were built in onestage

It should be noted that at some sections the bending moment duringthe construction stages is higher than in the final stage

44 Movement of supports of continuousstructures

Sudden movement of a support in a statically indeterminate concrete struc-ture produces instantaneous changes in the reactions and in the internalforces subsequently these forces decrease gradually with time due to theeffect of creep (ie relaxation occurs) In actual structures the movement ofsupports such as the settlement due to soil consolidation develops graduallyover a period of time creep also occurs during the same period and maycontinue to develop after the maximum settlement is reached Thus thechanges in internal forces start from zero at the beginning of settlementreaching maximum values at or near the end of the period of settlement andsubsequent creep results in relaxation (reduction in values) of the inducedforces

This is illustrated by considering the reaction F at B caused by a downwardsettlement δ of the central support of the continuous beam in Fig 46(a)When δ is sudden a force of magnitude Fsudden is instantaneously induced atB If subsequently δ is maintained constant creep of concrete causes relax-ation of the reaction as shown by curve A in Fig 46(b) Curve B in the samefigure represents the variation of the force F when the magnitude of thesettlement is changed from zero to δ over a period of time The force Fincreases from zero to a maximum value Fmax ndash which is generally muchsmaller than Fsudden ndash and then decreases gradually

Consider a continuous homogeneous structure subjected to supportmovements that develop gradually from 0 at age t0 to final values δ at age t1The values at t1 and at subsequent time t2 of a reaction or internal forceinduced by the movement of supports may be calculated by the equations (forwhich the proof is given later in this section)

F(t1) = Fsudden

1

1 + χ φ(t1 t0)(47)

F(t2) = F(t1) 1 minusEc(t1)

Ec(te)

φ(t2 te) minus φ(t1 te)

1 + χφ(t2 t1) (48)

where Fsudden is the value of the reaction or internal force when δ occurs


suddenly The value Fsudden is obtained by conventional elastic analysis inwhich the value of the modulus of elasticity of concrete Ec = Ec(t0) and thecross-section properties of the members are those of plain concrete sections φand χ are creep and aging coefficients which are functions of the time whencreep is considered and the age at loading (see Sections 12 and 17) te is anage between t0 and t1 The value te can be determined by trial from the graphsor equations of Appendix A such that

1

Ec(te)[1 + φ(t1 te)] =

1

Ec(t0)[1 + χφ (t1t0)] (49)

Figure 46 Time-dependent forces caused by support settlement in a continuous beam(a) continuous beam (b) reaction at central support versus time A suddensettlement B progressive settlement


In other words a stress increment introduced at the effective time te andsustained without change in value to t1 produces at t1 a total strain of thesame magnitude as would occur when the value of the stress increment isintroduced gradually from zero at t0 to full value at t1

If the movement of supports is introduced suddenly at age t0 Equation(48) can be used to find the induced forces at any time t after t0 by substitu-tion of t1 = t0 = te and t2 = t thus

F(t) = Fsudden 1 minusφ(t t0)

1 + χφ(t t0) (410)

The term between large parentheses in Equation (410) is equal to therelaxation function r(t t0) divided by Ec(t0) see Equation (123)

The presence of the reinforcement may be accounted for as follows Incalculation of Fsudden use the cross-section properties of a transformed sectioncomposed of the area of concrete plus α times the area of steel where α = EsEc(t0) also replace each creep coefficient φ(ti tj) in Equations (47) and (48)by [κφ(ti tj)] where

κ = Ic I (411)

κ is the curvature reduction factor (see Section 34) I = I(ti tj) is the momentof inertia of an age-adjusted transformed section for which Eref = Ec(ti tj) (seeEquation (131) and Section 1111) Ic is the moment of inertia of concreteBoth Ic and I are moments of inertia about an axis through the centroidof the age-adjusted transformed section The above treatment is basedapproximately on Equation (327) which gives the change in curvature due tocreep as the product (κφ) times the instantaneous curvature No distinction ismade between the effects of the reinforcement on axial strain and oncurvature

For proof of Equations (47) and (48) consider as an example the struc-ture in Fig 46(a) The instantaneous reaction at B due to a suddensettlement δ

Fsudden = 6

l 3Ec(t0)Icδ (412)

where Ic is the moment of inertia of a concrete cross-section about an axisthrough its centroid The term in the large parentheses represents thestiffness that is the force when δ is unity

Now consider that the settlement is introduced gradually from zero at t0 upto δ at t1 the reaction at B will also develop gradually from zero to a valueF(t1) during the same period The displacement δ may be expressed in termsof F(t1)


δ = l 3

6Ec(t1 t0)Ic F(t1) (413)

The term in the large parentheses is the age-adjusted flexibility or the dis-placement due to a unit increment of force introduced gradually Ec(t1 t0) isthe age-adjusted modulus of elasticity of concrete (see Equation (131) )

Ec(t1 t0) =Ec(t0)

1 + χφ(t1 t0)(414)

Equation (413) implies that the force F and hence δ are developed withtime at the same rate as relaxation of concrete (see Section 18)

Substitution of Equations (414) and (412) into (413) gives Equation(47)

Under the effect of the force F(t1) free creep would increase the deflectionby the hypothetical increment

∆δ = l 3

6Ec(te)Ic F(t1) [φ(t2 te) minus φ(t1 te)] (415)

In this equation F(t1) is treated as if it were applied in its entire value at theeffective time te

Because the support settlement does not change during the period t1 to t2an increment of force ∆F must develop such that

∆δ + l 3

6Ec(t2 t1)Ic ∆F = 0 (416)

where

Ec(t2 t1) =Ec(t1)

1 + χφ(t2 t1)(417)

The force at B at time t2 is

F(t2) = F(t1) + ∆F (418)

Solving for ∆F in Equation (416) and substitution of (415) and (417) intoEquation (418) gives Equation (48)

The ascending part of curve B in Fig 46(b) represents simultaneous grad-ual increase in force and in settlement while the descending part representsthe relaxation due to creep Thus one would expect curve B to be broken at t1

as shown by the broken line In practice movement of supports such as that


caused by consolidation of clays occurs over an infinite period of time How-ever it is reasonable to consider for the analysis of forces that the full settle-ment occurs between ages t0 and t1 with the period (t1 t0) representing thetime necessary for the major part (say 95 per cent) of the consolidation tooccur With settlement due to consolidation of soil the transition betweenthe ascending and descending parts of curve B Fig 46(b) would be smoothas shown by the continuous line

Example 45 Two-span continuous beam settlement of centralsupport

The continuous concrete beam shown in Fig 46(a) is subjected to adownwards settlement at B Find the time variation of the force F andthe reaction at the central support Express F in terms of Fsudden thevalue of the instantaneous reaction when the settlement δ is suddenlyintroduced Consider two cases

(a) δ introduced suddenly at t0 = 14 days and maintained constant to t2

= 10 000 days(b) Settlement introduced gradually from zero at t0 = 14 days to a value

δ at t1 = 104 days maintained constant thereafter up to t2 = 10 000days

Use the following creep and aging coefficients

The value te = 23 days is obtained by trial such that Equation (49) issatisfied The ratio Ec(t1)Ec(te) = 1077

Use of Equation (410) with t0 = 14 and t = 104 500 2000 and 10 000gives the values of F(t) which are plotted in Fig 47 curve A

ti tj (tj ti ) (tj ti )

1414141423232323

104104104

104500

200010000

104500

200010000

5002000

10000

114179226257101172220255117170196

079076076076

081080079


When the settlement is gradually introduced the starting value of Fis zero at t0 = 14 days Substitution in Equation (47) with t0 = 14 and t1 =104 days gives the value of F(t1) at the end of the period in which thesettlement is introduced Use of Equation (48) with te = 23 t1 = 104and t2 = 500 2000 and 10 000 gives the values of F(t2) plotted on curveB Fig 47

In practice interest is in the maximum value of F this is approxi-mately equal to the value F(t1) with t1 being the end of the period inwhich the settlement occurs Although Equations (47) and (48) givethe maximum value of F and its variation after the maximum isreached the two equations do not give the values of F between t0 and t1

(the ascending part of curve B Fig 47)The above example is solved by a step-by-step procedure (see Section

46) assuming that the variation of settlement with time follows theequation

δ(t)

δ(infin)= 1 minus exp minus 3(t minus t0)

t095 minus t0 (419)

where δ(t) and δ(infin) are the settlement at any time t and the ultimate

Figure 47 Values of the reaction at the central support versus time in a continuousbeam subjected to settlement of a support (Example 45) A period ofsettlement (t1 minus t0) = 0 B (t1 minus t0) = 90 days


settlement at time infinity t0 is the time at which the settlement startst095 is the time at which 95 per cent of the ultimate settlement occursEquation (419) closely approximates the standard-time consolidationcurve for clays given by Terzaghi and Peck (in the form of a table)5

The results of the step-by-step analysis (employing Equation (431) )are shown in Fig 48 in which the period (t095 minus t0) ndash the time duringwhich 95 per cent of settlement occurs ndash is considered equal to 0 10 3090 365 days or 5 years The graphs show the variation of F with timethe values of F are expressed in terms of Fsudden which is the instant-aneous reaction at B if the full settlement occurs suddenly at t0 = 14days The broken curve represents the case when (t095 minus t0) = 5 yearswith creep ignored The curves in Fig 48 show clearly the pronouncedeffect of creep on the forces induced by slow settlement of a support

When the settlement is sudden the curve for F versus time has thesame shape as the relaxation function r(t t0) which represents the stressvariation with time due to a strain imposed at age t0 and sustainedconstant to age t (see Fig A3 Appendix A) The sudden drop AB offorce at age t0 (Fig 48) is caused by the creep which develops in thefirst few days but is considered as if it occurs at time t0

Figure 48 Time versus reaction by slow settlement of support occurring in a periodof 0 10 30 365 days or 5 years (Example 45)


45 Accounting for the reinforcement

Analysis of the time-dependent changes in the internal forces in a staticallyindeterminate structure by Equation (45) involves calculation of the dis-placements of a statically determinate released structure to generate itsage-adjusted flexibility matrix [ f ] and the vector ∆D of the changes indisplacements occurring between two specified instants t0 and t By theprocedure of analysis presented in Section 25 we can determine the changes∆εO and ∆ψ in the axial strain and curvature in a section of a staticallydeterminate structure taking into account the presence of the reinforcementThe analysis gives the effects of creep shrinkage and relaxation of steel onthe stress and strain distribution and thus the prestress loss in a prestressedsection is automatically accounted for

Once ∆εO and ∆ψ are determined the changes ∆D in the displacementsat the coordinates may be calculated by virtual work or by numerical integra-tion (see Section 38) The equations given in Appendix C may be used forthis purpose This procedure of analysis described above is employed inExample 46

Example 46 Three-span precast post-tensioned bridge

A three-span bridge (Fig 49(a) ) is made up of precast post-tensionedsimple beams for which the cross-section at mid-spans is shown in Fig49(b) The beams are prestressed at age t placed in position and madecontinuous at age t0 by casting concrete at the joints and by continuousprestress tendons as shown in Fig 49(c) It is required to find thebending moment diagram at time t later than t0 Assume no cracks areproduced at the casting joint and that the joint results in perfect con-tinuity Also calculate the deflection at time t0 at the centre of AB andthe change in this value during the period t0 to t

To simplify the presentation we shall assume that the differencebetween t and t0 is small and consider that the prestressing placing thebeams in positions and casting of the joints all occur at age t0 We shallalso ignore the area of the cast in situ concrete (hatched area in Fig49(b) ) Other data are area of concrete section for one beam Ac =078m2 (1200 in2) moment of inertia about an axis through the centroidof the concrete area Ic = 0159m4 (382 times 103 in4) dead load of theprecast and cast in situ concrete (assumed to come into effect at age t0) =91kNm2 of area of deck or the dead load per beam = 1957kNm(1344kipft) A superimposed dead load of 50kNm2 (1075kNm perbeam (0737kipft) ) is applied shortly after the structure is made con-


tinuous Again for the sake of simplicity we shall consider that thesuperimposed load is applied at t0 on the continuous structure

The prestress in each beam is achieved by straight tendons A andparabolic tendons B and C The prestressing of A and B is applied tosimple beams while C is inserted after placing the beams in positionand the cable runs continuous over the whole length of the bridgeFurther we shall consider that cables B and C have identical profiles(Fig 49(d) ) The cross-section areas of prestress steel Aps are 430 1000and 1000mm2 (067 155 155 in2) for tendons A B and C respectivelythe initial prestress forces are 500 1160 and 1160kN (112 260 and260kip) Consider that these forces exclude friction loss and that theprestress force is constant over the full length of a tendon

Non-prestressed steel of total area Ans = 3750mm2 (581 in2) is dis-tributed over all surfaces of the cross-section thus we here assume thatAns has the same centroid as Ac (point O in Fig 49(b) ) and that the

Figure 49 Continuous precast bridge of Example 46 (a) three-span bridge (b)cross-section of one beam at mid-span (c) joint of precast beams atsupports B and C (d) typical prestress tendon profiles in precast beams


moment of inertia of the area Ans about an axis through the samecentroid is Ic(AnsAc) = 0764 times 10minus3 m4 this is equivalent to consideringthat the radius of gyration for Ans is the same as that of Ac

The material properties are modulus of elasticity for all reinforce-ment Eps = Ens = 200GPa (29000ksi) modulus of elasticity of concreteat age t0 Ec(t0) = 28GPa (4100ksi) creep coefficient φ(t t0) = 26 agingcoefficient χ(t t0) = 08 free shrinkage during the period (t minus t0) = εcs(tt0) = minus240 times 10minus6 reduced relaxation during the same period ∆σpr =minus90MPa (minus13ksi)

At t0 the self-weight and the prestress of tendons A and B are appliedon simple beams while tendon C and the superimposed dead load areapplied on a continuous beam The bending moments for the simpleand the continuous beams are calculated separately and then super-posed the result is shown in Fig 410(a) Two values of the bendingmoment are indicated at B with the larger value being the bendingmoment in the joint cast in situ

With the axial force and bending moment known at time t0 theinstantaneous axial strain at the reference point O εO(t0) and the curva-ture ψ(t0) are calculated (by Equation (232) ) at a number of sectionsand given in Table 41 The reference point O is chosen at the centroidof the concrete and the reference modulus of elasticity used in thecalculation of area properties is Eref = Ec(t0)

The properties of the transformed section at age t0 in Table 41 arecalculated for a section composed of Ac plus (α(t0)Ans) The area ofprestress steel should have been accounted for in the calculation of thedeformations due to the superimposed dead load but this is ignoredhere

The changes in axial strain and in curvature ∆εo and ∆ψ during theperiod t0 to t are calculated by Equation (240) and the results are givenin Table 42 These calculations involve the properties of the age-adjusted transformed section which are included in Table 41 using asreference modulus Eref = Ec(t t0) = 909GPa (1320ksi) (Equation(131) )

The released structure and the coordinate system are shown in Fig410(b) Because of symmetry the change in displacement ∆D1 needs tobe calculated only at coordinate 1 and can be calculated from the curva-ture increments ∆ψ in Table 42 The increment in displacement ∆D1 isequal to the sum of the changes in rotation at B of members BA andBC treated as simple beams Employing Equations (C6) and (C7) gives


Figure 410 Analysis of the statically indeterminate forces and bending momentdiagrams at t0 and t for the continuous bridge of Fig 49 (a) bendingmoment at time t0 (b) released structure and coordinate system(c) bending moment diagrams due to F1 = 1 and F2 = 1 (d) staticallyindeterminate bending moment developed during the period t0 to t(e) bending moment due to prestress loss (f) final bending momentat time t


Tabl

e 4

1C

ross

-sec

tion

prop

ertie

s1 and

cal

cula

tion

of in

stan

tane

ous

axia

l str

ain

and

curv

atur

e fo

r a

cont

inuo

us b

ridg

e (E

xam

ple

46)

Forc

es a

pplie

d at

time

t 0 e

quiva

lent

sof

pre

stre

ss fo

rce

Inst

anta

neou

sSe

ctio

n nu

mbe

rTr

ansf

orm

ed s

ectio

nAg

e-ad

just

ed tr

ansf

orm

edan

d de

ad-lo

adax

ial s

train

and

(see

Co

ncre

te s

ectio

npr

oper

ties

at ti

me

t 0se

ctio

n pr

oper

ties

bend

ing

mom

ent

curv

atur

eFi

g 4

10(b

))pr

oper

ties

E ref

=E c

(t0)

=28

GPa

E ref

=E c

(t t

0)=

909

GPa

(Equ

atio

n (2

31)

)(E

quat

ion(

232

))

A cB c

I cA

BI

AB

IN

M O

(t0)

(t

0)

10

780

015

90

8068

00

1645

091

60minus0

003

60

1835

minus28

20

189

minus125

410

20

780

015

90

8068

00

1645

091

600

0393

020

47minus2

82

024

5minus1

2553

23

078

00

159

080

680

016

450

9160

minus00

036

018

35minus2

82

008

8minus1

2519

14

078

00

159

080

680

016

450

9160

003

930

2047

minus28

20

195

minus125

423

Mul

tiplie

rsm

2m

3m

4m

2m

3m

4m

2m

3m

410

6 N10

6 N-m

10minus6

10minus6

mminus1

1R

efer

ence

poi

nt O

is c

hose

n at

the

com

mon

cen

troi

d of

Ac o

r A n

s

Tabl

e 4

2C

hang

es in

axi

al s

trai

n an

d in

cur

vatu

re o

f the

rel

ease

d st

ruct

ure

duri

ng t

he p

erio

d t 0

to

t in

Exam

ple

46

Calcu

latio

n of

rest

rain

ing

forc

es

Tota

lCh

ange

s in

axi

al s

train

Sect

ion

num

ber

Cree

pSh

rinka

geRe

laxa

tion

rest

rain

ing

forc

esan

d in

cur

vatu

re(s

ee F

ig 4

10(

b))

(Equ

atio

n (2

42)

)(E

quat

ion

(24

3))

(Equ

atio

n (2

44)

)(E

quat

ion

(24

1))

(Equ

atio

n (2

40)

)

N

M

N

M

N

M

N

M

O

12

304

minus01

541

170

20

minus02

190

0148

378

7minus0

139

3minus4

5574

62

230

4minus0

199

91

702

0minus0

219

minus01

607

378

7minus0

360

6minus4

6728

34

32

304

minus00

718

170

20

minus02

190

0148

378

7minus0

057

0minus4

5525

34

230

4minus0

159

01

702

0minus0

219

minus01

607

378

7minus0

319

7minus4

6626

13

Mul

tiplie

rs10

6 N10

6 N-m

106 N

106 N

-m10

6 N10

6 N-m

106 N

106 N

-m10

minus610

minus6 m

minus1

the change in displacement of the released structure during the time t0

to t

∆D1 =25

6 [0 2 1]

7462834253

10minus6 +25

6 [1 2 0]

2532613253

10minus6 = 4750 times 10minus6 radian

Use of Equation (C8) and the curvature values ψ(t0) from Table 41gives the instantaneous deflection at middle of span AB as

(25)2

96[1 10 1]

410532191

10minus6 = 385 times 10minus3 m = 385mm (0152 in)

The change in deflection of the released structure during the period t0

to t (using ∆ψ values from Table 42 and Equation (C8) ) is

(25)2

96 [1 10 1]

7462834253

10minus6 = 1910 times 10minus3 m

= 191mm (0752 in)

For calculation of the age-adjusted flexibility coefficient apply F1 = 1at coordinate 1 the diagram of the corresponding bending moment Mu1

is shown in Fig 410(c) Division of the ordinates of this diagram byErefIcentroid at sections 1 to 4 gives the curvatures due to F1 = 1 Icentroid isthe moment of inertia of the age-adjusted transformed section about anaxis through the centroid

Icentroid = I minusB2

A

The values of the curvatures due to F1 = 1 calculated in this fashionat the four sections considered are

ψu1 = 10minus9 0 02710 05995 02710mminus1N-m


The value f 11 is the sum of the rotations just to the left and to the rightof section 3 caused by F1 = 1 These rotations can be calculated fromthe above curvatures using Equations (C6) and (C7) giving

f 11 =25

6(2 times 02710 + 1 times 05995)2 times 10minus9 = 9513 times 10minus9 (N-m)minus1

The age-adjusted flexibility coefficient f 12 is the rotation at coordinate1 due to F2 = 1 Using a similar procedure as above gives

f 12 =25

6(2 times 02710) 10minus9 = 2258 times 10minus9 (N-m)minus1

The deflection at the centre of AB due to F1 = 1 (by Equation (C8) )

(25)2

96(10 times 02710 + 05995) 10minus9 = 2155 times 10minus9 mN-m

The force F2 = 1 produces no deflection at the centre of ABBecause of symmetry the two redundants are equal and can be

determined by solving one equation

( f 11 + f 12)∆F1 = minus ∆D1

Thus

∆F1 = ∆F2 =minus4750 times 10minus6

(9513 + 2258)10minus9= minus0404 times 106 N-m

The statically indeterminate bending moment diagram developedduring the period t0 to t is shown in Fig 410(d)

When considering the bending moment due to prestressing it is acommon practice to consider the effect of the forces of the tendon onthe concrete structure or on the concrete plus the non-prestressed steelwhen this steel is present To determine the bending moment at time twe calculate ∆σps (the prestress loss) in each tendon by Equation (248)The summation Σ(minusAps∆σps yps) performed for all the tendons at anysection gives the change in the bending moment of the released struc-ture due to the prestress loss where Aps is the cross-section area of a


tendon and yps is its distance below point O This is calculated forvarious sections and plotted in Fig 410(e) The final bending momentat time t is the superposition of the diagrams in Fig 410(a) (d) and (e)and the result is given in Fig 410(f)

The change in deflection of the actual structure can now becalculated by the superposition Equation (46) which is repeated here

∆A = ∆As + [∆Au] ∆F

where ∆As is the change in deflection of the released structure [∆Au]are the changes in deflection due to F1 = 1 and due to F2 = 1 ∆F arethe time-dependent redundant forces Substitution of the values calcu-lated above gives the change in deflection at the centre of AB during theperiod t0 to t

1910 times 10minus3 + 10minus9[2155 0] minus0404minus0404 106 = 1039 times 10minus3 m

= 1039mm (0409 in)

46 Step-by-step analysis by the force method

A step-by-step numerical procedure is presented in Section 110 for calcula-tion of the strain of concrete caused by stress which is introduced graduallyor step-wise in an arbitrary fashion The procedure is also used to calculatethe stress caused by imposed strain which is either constant with time (relax-ation problem) or varying in arbitrary fashion

In this and in Section 58 we shall use a similar procedure to calculate theinternal forces in statically indeterminate structures caused by creep shrink-age and settlement of supports In the present section the force method isemployed for structures in which individual cross-sections are composed ofhomogeneous material (presence of reinforcement ignored) In Section 58the step-by-step analysis is applied with the displacement method in concretestructures with composite cross-sections taking into account the effect of thereinforcement

The advantages of the step-by-step analysis are (a) the time variation offorces or imposed displacement can be of any form (not necessarily affine tothe time-relaxation curve as implied when the aging coefficient is used) (b)the method is applicable with any time functions chosen for creep shrinkageor relaxation of steel or modulus of elasticity of concrete (c) the changes incross-section properties eg due to cracking or modification of support con-


ditions can be accounted for in any time interval The step-by-step analysishowever involves a relatively large number of repetitive computations whichmakes it particularly suitable when a computer is used

In the step-by-step analysis the time is divided into intervals the internalforces the stresses or the displacements at the end of a time interval arecalculated in terms of the forces or stresses applied in the first interval and theincrements which have occurred in the preceding intervals Increments offorces or stresses are introduced at the middle of the intervals (Fig 411)Instantaneous applied loads such as prestressing are assumed for the sakeof consistency to occur at the middle of an interval of length zero (egintervals 1 and k in Fig 411) Accurate results can be obtained with a smallnumber of intervals (5 or 6) the length of the intervals should be relativelyshort in the early stages when the rates of change of modulus of elasticitycreep and shrinkage of concrete and often settlement of supports aregreatest

The general force method of structural analysis involves solution of thecompatibility equation (see Section 42)

[ f ] F = minusD (420)

where [ f ] is the flexibility matrix D are displacements of the released

Figure 411 Division of (a) time into intervals (b) stresses into increments


structure F are the redundant forces The displacements D representinconsistencies in the released structure (with respect to the actualstructure) The redundants F must therefore be applied to eliminate theinconsistencies

Any element of the flexibility matrix fmn is equal to the displacement atcoordinate m due to unit load applied at coordinate n Because of creep ofconcrete the value of any element of the matrix [ f ] depends upon the timefor which the displacement is considered and the age of concrete at the timeof the introduction of the unit load Thus we use here the symbol [ f (ti + 1

2 tj)]

to represent the matrix of flexibility at time ti + 12 when the age at loading is tj

The subscripts i minus 12 i and i + 1

2 respectively refer to the beginning the middleand the end of interval i

The forces Fi + 12 and the displacement Di + 1

2 at the end of any interval i

may be expressed as the sum of incremental forces ∆Fj and displacements∆Dj occurring at the middle of the intervals j = 1 2 i Thus

Fi + 12=

i

j = 1

∆Fj (421)

Di + 12=

i

j = 1

∆Dj (422)

The compatibility Equation (420) applied at the end of the ith intervalmay be written in the form

i

j = 1

[ f (ti + 12 tj)]∆Fj = minus

i

j = 1

∆Dj (423)

The analysis for ∆Fi can be done in steps in each step a new increment iscalculated In the ith step the values ∆F1 ∆F2 ∆Fi minus 1 are knownfrom the preceding steps and Equation (423) can be used to determine ∆FiEquation (423) may be rewritten by separating the last term of the summa-tion on the left-hand side and the substitution of Equation (422)

[ f (ti + 12 ti)] ∆Fi = minus Di + 1

2minus

i minus 1

j = 1

[f(ti + 12 tj)]∆Fj (424)

This recurrent equation can be solved successively with i = 1 2 todetermine the values of the vector ∆F1 ∆F2 and so on

The flexibility matrices involved in the analysis differ only in the modulusof elasticity and the creep coefficient to be employed in the calculation


The vector Di + 12 represents the total displacement of the released struc-

ture caused by applied loads shrinkage or supports settlement The dis-placement due to the applied load generally includes the instantaneous pluscreep but instantaneous displacements should be excluded if the loading isapplied prior to the start of the period for which the changes of the internalforces are required

The use of the recurrent Equation (424) is demonstrated below for astructure with one degree of indeterminacy

Application The two-span continuous concrete beam in Fig 412(a) is sub-jected to a settlement of the central support the magnitude of which δ(t)varies with time in an arbitrary form Equation (424) will be used to find thedownward reaction F at the central support

A statically determinate released structure with one coordinate is shown inFig 412(b) The instantaneous displacement at coordinate 1 due to a unitforce at the same coordinate

f instantaneous =l 3

6EcIc

(425)

Figure 412 Reaction due to settlement of support of a continuous beam by a step-by-stepprocedure employing Equation (424) (a) continuous beam (b) staticallydeterminate released structure


where Ic is the moment of inertia of the section Ec is the modulus of elasticityof concrete at the time of application of the load We have only one coordin-ate thus we use F to mean F1 and f for f11 If the unit load is applied at tj itwill produce at time ti + 1

2 a displacement

f(ti + 12 tj) = C

1

Ec(tj) [1 + φ(ti + 1

2 tj)] (426)

where C is a constant independent of time related to the geometry of thestructure

C =l 3

6Ic

(427)

At the end of any interval i

Di + 12= minusδ(ti + 1

2) (428)

The minus sign is included in this equation because it represents a dis-placement caused by the redundant force F (rather than eliminated by it)

Substitution of Equations (426) and (428) into Equation (424) gives

f (ti + 12 ti)(∆F)i = δ(ti + 1

2) minus C

i minus 1

j = 11 + φ(ti + 1

2 tj)

Ec(tj) (∆F)j (429)

The magnitude of the reaction at the central support at the end of the ithinterval is

F(ti + 12) = F(ti minus 1

2) + (∆F)i (430)

Solving Equation (429) for (∆F)i and substitution in Equation (430) gives


2)

+ [ f (ti + 12 ti)]

minus1δ(ti + 12) minus C

i minus 1

j = 11 + φ(ti + 1

2 tj)

Ec(tj) (∆F)j (431)

Equation (431) has been used to derive the graphs in Fig 48 (seeExample 45)



Example 47 Two-span bridge steel box and post-tensioned deck

The same bridge cross-section and method of construction of Example27 are used for a continuous bridge of two equal spans each = l = 144 ft(439m) what will be the stress distribution at the section over thecentral support at time t Again assume that at completion of installa-tion of the precast elements the structural steel section alone acting ascontinuous beam carries the weight of concrete and structural steel

The bending moment over the central support at time t0 immediatelyafter installation of the precast deck = minusql 28 = minus54(144)28 =minus14000kip-ft = minus168000kip-in The distribution of stress on the struc-tural steel due to this bending moment is shown in Fig 413(a) thesame figure shows the stress distribution in the concrete deck cross-section due to the axial prestressing force introduced at time t0

The five steps of the force method (Sections 42 and 43) are followedto determine the time-dependent change in stresses in the section abovethe central support

Step 1 A statically determinate released structure is shown in Fig413(b) The stress values required are

∆A = ∆σc top ∆σc bot ∆σs top ∆σs bot

These represent the stress changes in the period t0 to t at top andbottom fibres of the concrete and the structural steel

Step 2 In Example 27 we determined the time-dependent change incurvature at mid-span section as

∆ψ(t t0) = 4784 times 10minus6 inminus1

The same change in curvature occurs at any other section of thereleased structure Thus the change in displacement in the releasedstructure at coordinate 1 is

∆D1(t t0) = ∆ψl = 4784 times 10minus6 (144 times 12) = 8266 times 10minus6

∆As in the present problem represents the stress changes in the


Figure 413 Analysis of stress distribution over the central support of a two-spancontinuous bridge (Example 47) (for bridge cross-section see Fig216(a) ) (a) stress at time t0 (b) released structure and coordinatesystem (c) stress distribution at time t


released structure during the period t0 to t These are calculated inExample 27 and are constant over the span (the stress values in Fig216(c) minus the values in Fig 216(b) )

∆As =

01770296

minus77851324

ksi


Select the reference point O as shown in Fig 216(a) properties ofthe age-adjusted section are


Due to F1 = 1kip-in the changes in strain at B are (Equation (219) )

(∆εOB)due to F1 = 1 = 1451 times 10minus9 (kip-in)minus1

(∆ψB)due to F1 = 1 = 6097 times 10minus12 inminus1(kip-in)minus1

This change in curvature varies linearly between the above value at Band zero at the two ends The corresponding change in displacement atcoordinate 1 is

f 11 = ∆ψB 2l

3= 6097 times 10minus12 2 times 144 times 12

3 = 7024 times 10minus9(kip-in)minus1

The stress changes at B in the four fibres considered due to F1 = 1kip-in are Equations (219) and (217)

[∆Au] = 10minus6

minus3059minus1539

minus2765988428

ksi

kip-in

Step 4 The time-dependent change in the statically indeterminateforce (Equation (45) )


∆F1(t t0) = (7024 times 10minus9)minus1(minus8266 times 10minus6) = minus117700kip-in

Step 5 The stress changes in the period t0 to t are (Equation (46) )

∆A =

01770296

minus77851324

+ 10minus6

minus3059minus1539

minus2765988428

(minus117700) =

05370427

minus4530minus9084

ksi

Addition of these stress changes to the stress values at time t0 givesthe total stress distribution at time t at the section above the centralsupport (Fig 413(c) ) It is interesting to compare the initial stress(minus0485ksi) introduced by prestressing at time t0 to the remaining com-pression at time t in the present example and in Example 27 (Fig216(c) ) In the present example the remaining compression in concreteat time t dropped to almost zero

48 General

The stresses in all reinforced or prestressed concrete structures staticallydeterminate or indeterminate change with time due to the effects of creepshrinkage of concrete and relaxation of prestress steel In a staticallydeterminate structure the distribution of stresses over the area of concreteand reinforcement in any section varies with time but the resultant of stressesin the two components combined remains unchanged This is not the casewith statically indeterminate structures where statically indeterminate reac-tions are produced causing gradual changes in the stress resultants in thesections

The force method employed in this chapter to analyse the time-dependentinternal forces is intended for computations by hand or using small deskcalculators It is of course possible to prepare computer programs to doparts of the computations or all the computations for a certain type of struc-ture (for example continuous beams) However for a more general computerprogram it is more convenient to use the displacement method which is thesubject of the following chapter

Notes

1 For more detailed presentation and examples see reference mentioned in note 3of Chapter 3

2 See reference mentioned in note 3 of Chapter 3


3 See Section 511 of the reference mentioned in note 3 of Chapter 34 A simple method for the evaluation of integrals for the calculation of displacements

by virtual work can be found in Section 64 of the reference mentioned in note 3 ofChapter 3

5 Terzaghi K and Peck RB (1966) Soil Mechanics in Engineering Practice WileyNew York page 240


Time-dependent internalforces in uncracked structuresanalysis by the displacementmethod

51 Introduction

The force method is employed in Chapter 4 to calculate the time-dependentforces in a statically indeterminate structure caused by shrinkage and creep ofconcrete relaxation of prestressed steel and movement of the supports Thegeneral displacement method of structural analysis can be used for the samepurpose Computer programs for the elastic analysis of frames are nowwidely used by engineers and they are usually based on the displacementmethod In Section 52 we shall review the general displacement methodand in Section 53 indicate how a conventional computer program forthe analysis of an elastic framed structure can be used to determine thetime-dependent changes in internal forces

Cast in situ segmental construction of lsquoPont de la Feacutegirersquo near Lausanne Switzerland

Chapter 5

A step-by-step procedure suitable for computer use is presented in Sec-tion 58 It accounts for the effects of creep shrinkage of concrete andrelaxation of steel in statically determinate or indeterminate structures Thecross-section may be made up of one concrete type or composite and thestructure may be composed of members of different ages and the presenceof non-prestressed steel is accounted for in the analysis The loading pre-stressing forces or prescribed support displacements may be introducedgradually at an arbitrary rate or in stages and the boundary conditions maybe changed in any stage The method is suitable when precast segments areassembled and made continuous by prestressing or by cast in situ concreteor both The lsquosegmentalrsquo (or lsquocantileverrsquo) method of construction mainlyused for bridges is an example of a case in which the step-by-step analysisis most fitting

52 The displacement method

This section serves as a review of the general displacement method of analy-sis of framed structures while the following two sections will indicate howthis method can be used for the analysis of time-dependent changes ininternal forces

To explain the method consider for example the plane frame shown in Fig51(a) subjected to external applied loads (not shown in the figure) Assumethat it is required to find m actions A representing reaction componentsinternal forces or displacements at chosen sections The analysis by thedisplacement method involves five steps

Step 1 A coordinate system is established to identify the locations and thepositive directions of the joint displacements (Fig 51(b) ) The number ofcoordinates n is equal to the number of possible independent joint displace-ments (degrees of freedom) There are generally two translations and a rota-tion at a free (unsupported) joint of a plane frame The number of unknowndisplacements may be reduced by ignoring the axial deformations Forexample by considering that the length of the members of the frame in Fig51(b) remains unchanged the degrees of freedom are reduced to coordinates1 3 6 and 9

Step 2 Restraining forces F are introduced at the n coordinates to preventthe joint displacements The forces F are calculated by summing the fixed-end forces for the members meeting at the joints Also determine Ar valuesof the actions with the joints in the restrained position

Step 3 The structure is now assumed to be deformed such that the dis-placement at coordinate j Dj = 1 with the displacements prevented at all theother coordinates The forces S1j S2j Snj required to hold the frame in


this configuration are determined at the n coordinates This process isrepeated for unit values of displacement at each of the coordinates respect-ively Thus a set of n times n stiffness coefficients is calculated which forms thestiffness matrix [S]n times n of the structure a general element Sij is the forcerequired at coordinate i due to a unit displacement at coordinate j The valuesof the actions [Au] are also determined due to unit values of the displace-ments any column j of the matrix [Au] is composed of the values of theactions at the desired locations due to Dj = 1

Step 4 The displacement D in the actual (unrestrained) structure isobtained by solving the equilibrium equation

[S] D = minus F (51)

The equilibrium Equation (51) indicates that the displacements D mustbe of such a magnitude that the artificial restraining forces F areeliminated

Step 5 Finally the required values A of the actions in the actual structure

Figure 51 Example of a coordinate system (b) employed for the analysis of a plane frame(a) by the displacement method


are obtained by adding the values Ar in the restrained structure (calculatedin step 2) to the values caused by the joint displacements This is expressed bythe superposition equation

Am times 1 = Arm times 1 + [Au]m times n Dn times 1 (52)

53 Time-dependent changes in fixed-end forces ina homogeneous member

In the analysis of statically indeterminate structures by the displacementmethod the internal forces and the forces at the ends of individual memberswith fixed ends must be known in advance In this section we shall considerfor a homogeneous beam with totally fixed ends the changes in the fixed-endforces caused by creep shrinkage and settlement of supports

The totally fixed beam in Fig 52(a) is made up of homogeneous materialand subjected at age t0 to a set of external applied loads such as gravity loadsor prestress forces Consider the changes in the forces at the ends of the beamand hence the internal forces at any section that will occur during a laterperiod t1 to t2 due to creep gradual settlement of supports shrinkage andprestress loss We here assume that the amount of prestress loss is a knownvalue not affected by the internal forces resulting from the loss We alsoconsider the case when the support conditions change at time t1

Forces applied on the beam and sustained without change in magnitude oralteration on the boundary conditions produce no changes in the internalforces due to creep (see Section 43) However under the same loads but withchanges in support conditions creep results in changes in the internal forcesas will be further discussed below

The forces at the beam ends induced by shrinkage or gradual settlement ofthe support may be determined through conventional analysis by the forcemethod in which the modulus of elasticity is the age-adjusted modulus (seeExample 41)

Prestress loss of a known magnitude may be represented as a set of self-equilibrating forces1 in the same way as the prestress itself but generally witha reversed sign and smaller magnitude The prestress loss is represented by asystem of forces at the anchors and at the sections where the cable changesdirection The prestress loss develops gradually with time and so do the stat-ically indeterminate forces it induces Thus the changes in the internal forcesdue to prestress loss are independent of the value of modulus of elasticity tobe used in the analysis

In Section 25 we have seen that creep shrinkage and relaxation produce ina statically determinate structure changes in the stress distribution but thestress resultants N and M remain unchanged N and M are the resultants ofnormal stress on the entire section composed of its three components con-crete non-prestressed reinforcement and prestressed steel However it is a


common practice to calculate the internal forces due to prestressing by con-sidering the forces exerted by the prestress tendons on the remainder of thestructure the concrete and non-prestressed reinforcement This is the mean-ing adopted here where reference is made to the internal forces caused byprestress loss

Now consider that the beam in Fig 52(a) is constructed in three differentways At age t0 we assume that the external loads are applied on one of thestatically determinate systems in Fig 52(b) (c) or (d) Subsequently at age t1

the beam is made totally fixed as shown in Fig 52(a) Time-dependentchanges in the forces at the end of the member will gradually develop theequations derived below can be used to calculate the member-end forces atany time t2 later than t1

A system of three coordinates 1 minus 3 is defined in each of Fig 52(e) (f)

Figure 52 Analysis of the time-dependent changes in the end forces of a member causedby fixity introduced after loading (a) totally fixed beam subjected at time t0 to asystem of forces (b) (c) (d) statically determinate beams loaded at time t0statical system changed to totally fixed beam at time t1 (e) (f) (g) coordinatesystems


and (g) If the statically determinate system in Fig 52(b) (c) or (d) is leftunchanged during the period t1 to t2 creep will change the displacement atthe coordinates by the amount

∆D = D(t0)[φ(t2 t0) minus φ(t1 t0)] (53)

where D(t0) are the instantaneous displacement at t0 due to the externalloads on the statically determinate system φ(ti tj) is the coefficient for creep atti when the age at loading is tj

The age-adjusted flexibility matrix is

[ f ] = [ f ][1 + χφ(t2 t1)] (54)

where χ = χ(t2 t1) is the aging coefficient (see Section 17) [ f ] is the flexibilitymatrix of a statically determinate beam (Fig 52(b) (c) or (d) ) The modulusof elasticity to be used in the calculation of the elements of [ f ] is Ec(t1)

The compatibility Equation (45) can now be applied which is repeatedhere

[ f ]∆F = minus∆D (55)

Substitution of Equations (53) and (54) in Equation (55) and solutiongives the changes in the three end forces developed during the period t1 to t2

∆F = φ(t2 t0) minus φ(t1 t0)

1 + χφ(t2 t1) F (56)

where

F = [ f ]minus1 minusD(t0) (57)

The three values F in Equation (57) are equal to the three fixed-endforces at the same coordinates when calculated in a conventional way ie foran elastic beam subjected to the external loads in Fig 52(a) with no creep orchange in support conditions

Equation (56) gives the changes occurring during the period t1 to t2 inthree of the six end forces The changes in the other three are the staticalequilibrants of the first three As an example see the three forces indicated bybroken arrows at the left end of the beam in Fig 52(f) It can be seen that thefinal fixed-end forces at time t2 will not be the same in the three beamsconsidered above


Example 51 Cantilever restraint of creep displacements

The cantilever in Fig 53(a) is subjected at age t0 to a uniformly distrib-uted load qunit length At age t1 end B is made totally fixed Find theforces at the two ends at a later time t2 Use the following creep andaging coefficients φ(t1 t0) = 09 φ(t2 t0) = 26 χ(t2 t1) = 08 φ(t2 t1) =245

If the beam were totally fixed at the two ends with no creep or

Figure 53 Analysis of time-dependent forces in a cantilever transformed into atotally fixed beam after loading (Example 51) (a) forces acting at time t0(b) changes in end forces between t1 and t2 (c) total forces at t2


change in support the end forces at B caused by the load q wouldbe

F = minusql2

ql 212Forces developed at end B of the cantilever during the period t1 to t2

(Equation (56) ) are

∆F = 26 minus 09

1 + 08 times 245 minusql2

ql 212 = minus02872ql

00479ql 2 These two forces and their equilibrants at end A are shown in Fig

53(b) Superposition of the forces at the member ends in Fig 53(a)and (b) gives the end forces at time t2 shown in Fig 53(c)

54 Analysis of time-dependent changes in internalforces in continuous structures

The method of analysis is explained using as an example the plane frameshown in Fig 54 This bridge structure is made up of three precast pre-stressed beams AB CD and EF At age t0 prestress is applied in the factoryat which time each of the three members acted as a simple beam subjected toits self-weight and to the prestress Precast elements in the form of a T areused for the inclined columns GH and IJ Assume that the casting of theelements in the factory is done at the same time for all the elements Theprecast elements are erected at age t1 with provisional supports at B C D andE and shortly after the structure is made continuous by casting joints at BC D and E and post-tensioned cables inserted through ducts along the deckA to F At the same time the shores at B C D and E are removed The

Figure 54 A frame composed of precast parts made continuous by cast in situ joints andpost-tensioning


analysis described below is concerned with the changes in the internal forcesoccurring between t1 and a later time t2

We assume that a computer program is available for the analysis of elasticplane frames and will indicate here how such a program can be used for thisproblem The axis of the frame is usually taken through the centroid ofthe cross-section and three degrees of freedom assumed at each joint For theframe considered here the joints are at the supports the corners and at B CD and E

The analysis is to be done in two stages employing the same computerprogram in each to analyse a continuous frame The presence of thereinforcement is ignored here hence the moment of inertia of any cross-section is that of a plain concrete section In the first stage calculate thedisplacements and the internal forces occurring instantaneously at t1 after thecontinuity prestressing and removal of the shores The modulus of elasticityto be used for the members is Ec(t1) and the loads to be applied are downwardconcentrated loads at B C D and E which are equal and opposite to thereaction on the shores due to the self-weight of the precast elements beforecontinuity In addition apply a set of self-equilibrating forces representingthe effect of prestressing introduced at age t1

In the second stage consider the effect of the forces developing graduallybetween t1 and t2 The modulus of elasticity to be used is the age-adjustedmodulus Ec(t2 t1) (see Equation (131) ) The forces to be applied form asystem of self-equilibrating forces ndash ∆F where ∆F are the changes in thefixed-end forces due to creep shrinkage and prestress loss Here each memberis treated as a separate beam with fixed ends and the changes in the six forcesat the member ends calculated according to the procedure of Section 53 (seeFig 52) The six self-equilibrating forces calculated for each beam maybe reversed and applied directly to the frame at the appropriate jointsAlternatively the three forces to be applied at each joint are calculated byassemblage of forces at the ends of the members meeting at the joint

The displacements and internal forces obtained in the analysis in the twostages mentioned above when superimposed on the corresponding valuesexisting prior to t1 give the final values existing at time t2 Use of con-ventional linear computer programs to perform this analysis is discussed indetail with examples in Chapter 6

55 Continuous composite structures

In this section we consider the time-dependent changes of internal forces in astatically indeterminate structure which has composite cross-sections Con-sider the frame in Fig 55(a) which has a composite cross-section for the partAD The composite section is made up either of steel and concrete (Fig55(c) ) or prestressed precast beam and cast in situ deck (Fig 55(b) ) Due toshrinkage creep and prestress loss internal forces develop and the changes


for a specified period may be determined by application of the displacementmethod to the continuous frame in two stages as discussed in the precedingsection The first stage is concerned with the joint displacements and themember-end forces produced at time t0 immediately after application ofloads The joints are artificially locked in this position causing time-dependent fixed-end forces to develop gradually during a specified period t0

to t In the second stage of analysis the artificial restraining forces areremoved producing changes in joint displacements and member-end forcescalculated by a second application of the displacement method The follow-ing are additional remarks to be considered in the second stage of analysiswhen calculating the cross-section properties and the changes in fixed-endforces in composite members

For any of the composite sections in Fig 55 the cross-section to be usedin the second stage of analysis should be the age-adjusted transformed sec-tion (see Section 1111) The fixed-end forces to be used in the same stage areto be determined at the centroid of the age-adjusted transformed section

The age-adjusted modulus of elasticity of concrete depends upon t0 and tthe ages of concrete at the beginning and the end of the period consideredThus the centroid of the transformed section will be changing when analys-ing for different time periods or when considering the instantaneous effects of

Figure 55 Example of a continuous composite structure (a) statically indeterminate frame(b) (c) alternative composite cross-sections for part AD of the frame in (a)


applied loads This difficulty may be avoided by assuming that the axis of theframe passes through an arbitrary reference point in the cross-section butthis will result in coupling the effects of the axial forces and bending on theaxial strain and curvature (see Section 23 and Equation (213) ) Some com-puter programs allow the reference axis of the frame to be different from thecentroidal axis but in general this facility is not available Hence it may benecessary to determine the position of the centroid of the transformed sec-tion and calculate the fixed-end forces with respect to the centroid at the endsections of each member Determining the correct position of the centroid isparticularly important when considering the effect of the shrinkage of thedeck slab

Use of the displacement method for the analysis of a framed structureinvolves the assumption that the internal forces and the forces at the ends of amember with fixed ends are known a priori Due to creep and shrinkage thestress distribution in a composite statically determinate member changes withtime (see Section 25) and if the member is statically indeterminate thereactions and the stress resultants are also time-dependent The staticallyindeterminate changes in internal forces in a composite member with fixedends are discussed in the following section

56 Time-dependent changes in the fixed-endforces in a composite member

Consider a member of a continuous structure subjected at time t0 to externalapplied forces including prestressing Assume that the axial force N and thebending moment M are known at all sections at time t0 (determined byconventional analysis) Immediately after application of the loads the jointsare totally fixed as shown in Fig 56(a) for a typical member which isassumed to have a composite cross-section The time-dependent changes inthe fixed-end forces due to creep and shrinkage of concrete and relaxation ofprestressed steel are here analysed by the force method

A system of three coordinates is chosen on a statically determinate releasedstructure in Fig 56(b) The analysis involves the solution of the followingequation (see Equation (45) )

[ f ] ∆F = minus ∆D (58)

where [ f ] is the age-adjusted flexibility matrix of the released structurecorresponding to the three coordinates ∆F are the changes in theredundants during the period t0 to t ∆D are the changes during the sameperiod in the displacements of the released structure

Coordinate 1 in Fig 56(b) is assumed to be at the centroid of theage-adjusted transformed section (see Section 1111)


Solution of Equation (58) gives

∆F = [ f ]minus1 minus∆D (59)

where [ f ]minus1 is the age-adjusted stiffness corresponding to the coordinatesystem in Fig 56(b) For a member with constant cross-section2

[ f ]minus1 = Ec

l

A

0

0

0

4I

2I

0

2I

4I

(510)

where l is the length of member A and I are the area and moment of inertiaabout an axis through the centroid of the age-adjusted transformed sectionfor which Eref = Ec(t t0) is the age-adjusted elasticity modulus Substitution ofEquation (510) into (59) gives

Figure 56 Analysis of changes of internal forces due to creep shrinkage and relaxationin a composite member with fixed ends (a) composite member beam endsfixed at t0 after application of external loads (b) statically determinate releasedstructure and coordinate system


∆F = Ec

l

A

0

0

0

4I

2I

0

2I

4I

minus∆D (511)

The changes ∆D in the displacements of the released structure may bedetermined by numerical integration or by virtual work using the equation(see Section 38)

∆D =int(∆εO)

int(∆ψ)

int(∆ψ)

Nul

Mu2

Mu3

dl

dl

dl

(512)

where ∆εO and ∆ψ are the changes during the period considered in the strainat the reference point O and in the curvature in any section Nu1 Mu2 and Mu3

are axial force and bending moments due to unit force at the threecoordinates

∆εO and ∆ψ may be calculated by the method presented in Section 25using Equation (240) which is rewritten here

∆εO

∆ψ = 1

Ec(AI minus B2) I

minusB

minusB

A minus∆N

minus∆M (513)

where ∆N ∆M are a normal force at O and a bending moment required toartificially prevent the change in strain in the section during the period t0 to tB is the first moment of area of the age-adjusted transformed section aboutan axis through the reference point O

Because the reference point O is chosen at the centroid of A the value B iszero and Equation (513) is simplified to

∆εO

∆ψ = minus1

Ec

∆NA

∆MI (514)

The value ∆N ∆M is obtained by summing up the forces required toprevent creep shrinkage and relaxation (see Equations (241) to (244) )

In Examples 52 and 53 composite frames are analysed for the effects ofcreep and shrinkage using the procedure discussed in Sections 55 and 56

57 Artificial restraining forces

In Sections 55 and 56 a method is suggested for the analysis of the forcesdeveloped by creep shrinkage of concrete and relaxation of prestress steel ina continuous structure The procedure presented in Section 25 is employedin which the strain due to creep shrinkage and relaxation is first restrained by


the introduction of the internal forces ∆N and ∆M (Equation (241) ) whichare subsequently released while the member ends are allowed to displacefreely as in a simple beam Then the member ends are restrained by theintroduction of the fixed-end forces This artificial restraint is also to beremoved by the application of a set of equal and opposite forces at the jointson the continuous structure (see Example 52 to follow) An alternative pro-cedure is to determine a set of external applied forces preventing the straindue to creep shrinkage and relaxation at all sections and then remove thisartificial restraint in one step by applying a set of equal and opposite forceson the continuous composite structure The same method will be employed inSection 106 for the analysis of the effect of temperature on the continuousstructure in which the cross-section andor the temperature distribution isnon-uniform

The artificial restraining internal forces ∆N and ∆M can be introduced bythe application of external forces at the ends of members as well as tangentialand transverse forces as shown in Fig 57 The intensities p and q of thetangential and transverse artificial restraining load are given by

p = minusd

dx(∆N) (515)

q = minusd2

dx2(∆M) (516)

Two additional shear forces at the ends are necessary for equilibrium Theset of self-equilibrating forces shown in Fig 57 is to be reversed and appliedon the continuous composite structure

When a computer is used each member may be subdivided into parts forwhich the axial force ∆N may be considered constant while ∆M varies as astraight line In this way Equations (515) and (516) will give p = 0 and q = 0and hence the restraining forces need to be applied only at the nodes

Figure 57 A set of self-equilibrating forces applied on a member to artificially prevent thestrain due to creep shrinkage and relaxation


Example 52 Steel bridge frame with concrete deck effects ofshrinkage

The bridge frame in Fig 58(a) has a composite section for part AD(Fig 58(b) ) and a steel section for the columns BE and CF It isrequired to find the changes in the reactions and in the stress distribu-tion in the cross-section at G due to uniform shrinkage of deck slaboccurring during a period t0 to t1

The cross-section properties of members are for columns BE andCF area = 20000mm2 (31 in2) and moment of inertia about an axisthrough centroid = 0012m4 (29000 in4) for part AD the steel

Figure 58 Analysis of statically indeterminate forces caused by creep and shrinkagein a composite frame (Examples 52 and 53) (a) frame dimensions(b) cross-section properties for part AD (c) location of centroid ofage-adjusted transformed section composed of area of concrete plus times area of steel


cross-section area = 39000mm2 (60 in2) and moment of inertia about itscentroid = 0015m4 (36000 in4)

The material properties are

Ec(t0) = 30GPa (4350ksi) Es = 200GPa (29000ksi)

φ(t1 t0) = 25 χ(t1 t0) = 08 εcs(t1 t0) = minus270 times 10minus6

The following cross-section properties for part AD are needed in theanalysis

Age-adjusted transformed section

Ec(t1 t0) = 30 times 109

1 + 08 times 25 = 10GPa α(t1 t0) =

200

10 = 20

The age-adjusted transformed section is composed of Ac = 132m2

plus αAs = 20 times 0039 = 0780m2 A reference point O is chosen at thecentroid of the age-adjusted transformed section at 1361m above bot-tom fibre (Fig 58(c) ) Using Eref = Ec = 10GPa the properties of theage-adjusted transformed section are

A = 210m2 B = 0 I = 10232m4

Transformed section at t0

Ec(t0) = 30GPa α(t0) = 200

30 = 6667 Eref = Ec(t0)

Area and its first and second moment about an axis through thereference point O

A = 158m2 B = minus03947m3 I = 05221m4

The centroid of this transformed section is 1611m above the bottomfibre and moment of inertia about an axis through the centroid is04234m4

Concrete deck slab Area first and second moment of the concretedeck slab alone about an axis though the reference point O

Ac = 132m2 Bc = minus05927m3 Ic = 02714m4


The resultant of stresses if shrinkage were restrained at all sections ofAD (Equation (243) )

∆N

∆M = minus10 times 109(minus270 times10minus6) 132

minus05927 = 3564 times 106 N


Because ∆N and ∆M are constant in all sections of members AB BCand CD shrinkage can be prevented at all sections by the application ofexternal forces only at the ends of the members as shown in Fig 59(a)The stress distribution in the restrained condition is the same for allsections of AD and is shown in Fig 59(b)

The restraining forces at the member ends are assembled at the jointsand applied in a reversed direction on the continuous frame (Fig59(c) ) The forces at the end of members at each of joints B and Ccancel out leaving only forces at A and D Now the continuous frame inFig 59(c) is to be analysed in a conventional way by computer or byhand giving the internal forces shown in Fig 59(d) The properties ofthe cross-sections of the members to be used in the analysis are the age-adjusted transformed section properties using the same Eref = 10GPafor AD as well as for the columns Line AD in Fig 59(c) is at the levelof the centroid of the age-adjusted transformed section (1361m abovethe soffit of the section in Fig 58(c) ) In the analysis of the continuousframe the upper 1361m of each of members BE and CF in Fig 59(c)is considered rigid

The forces in Fig 59(d) represent the internal forces which willeliminate the artificial restraint introduced in Fig 59(a)

The statically indeterminate reactions caused by shrinkage are equalto the superposition of the reactions in Fig 59(a) and (c) But since theforces in Fig 59(a) produce no reactions the reactions shown in Fig59(d) represent the total statically indeterminate values The internalforces in Fig 59(d) represent resultants of stresses in concrete and steelof the age-adjusted transformed sections caused by elimination of theartificial restraint

To find the stress distribution at any section we have to superpose thestress distribution in the restrained condition (Fig 59(b) ) to the stressdistribution caused by the internal forces in Fig 59(d) applied on theage-adjusted transformed section The superposition is performed inFig 510 for the cross-section at G


Figure 59 Analysis of internal forces caused by shrinkage in the compositecontinuous frame of Example 52 (a) resultants of stresses to restrainshrinkage of concrete at the ends of members AB BC or CD (b) stressdistribution in any section of AD at time t1 if shrinkage were fullyrestrained (c) forces in (a) assembled and applied in a reversed directionon the continuous frame (the reactions corresponding to the appliedforces are included in the figure) (d) bending moment and axial forcediagrams for the frame in (c)


Figure 510 Analysis of stresses at section G due to shrinkage in a compositecontinuous frame of Example 52 (a) stress distribution due to N =minus3469MN at O and M = minus0056MN-m applied on age-adjustedtransformed section (b) total stress due to shrinkage (superpositionof Figs 59(b) and 510(a) )

Example 53 Composite frame effects of creep

The frame in Fig 511(a) has a composite cross-section for part BC anda steel section for the columns BE and CF The dimensions of the cross-sections and the properties of the materials are the same as for memberBC in Example 52 see Fig 58 The properties of the cross-sections ofthe columns BE and CF are given in Fig 511(a) At time t0 a uniformlydistributed downward load of intensity q = 40kNm is applied on BCand sustained to a later time t1 It is required to find the change in thebending moment due to creep during the period t0 to t1 Use the samecreep and aging coefficients as in Example 52 Also find the stressdistribution and the deflection at section G at time t1

The properties of the cross-section for member BC are the same asfor part AD of the frame of Example 52 and thus this part of thecalculation is not discussed here


A conventional elastic analysis is performed for a continuous framesubjected to the load q giving the bending moment at time t0 shown inFig 511(b) The moments of inertia of the cross-sections used in theanalysis are 04234m4 for BC and 0080m4 for the columns These arethe centroidal moments of inertia of transformed sections using Eref =Ec(t0) = 30GPa for all members The centroid of the transformed sec-tion at age t0 for member BC is 1611m above the bottom fibre hence

Figure 511 Composite frame of Example 53 (a) frame dimension (for cross-section of member BC see Fig 58(b)) (b) bending moment diagramat t0 (c) released structure for analysis of changes of fixed-end forcesin BC


the length of the columns used in the analysis is 11611m The axialforce in member BC at time t0 is minus02431MN

If immediately after application of the load at time t0 joints B and Cwere locked preventing displacements creep would produce change inthe forces at the ends of member BC For calculation of these changesrelease the member as a simple beam as shown in Fig 511(c) Thechanges ∆εO and ∆ψ in the axial strain and curvature due to creep in thereleased structure are calculated at various sections by successive appli-cations of Equations (232) (242) and (240) and the results are givenin Tables 51 and 52

In the preparation of the two tables the reference point O at whichthe axial strain is calculated is considered at the centroid of the age-adjusted transformed section The values of the axial force and bendingmoments in member BC of the frame in Fig 511(b) are transformed totheir statical equivalents before listing in Table 51 (The centroidal axisis moved downwards 0250m the value 0250 times 02431 = 0062MN-mis added to the bending moment ordinates shown for part BC inFig 511(b)

The changes in the displacements D at the three coordinates of Fig511(c) are calculated assuming parabolic variation of ∆εO and ∆ψ overthe length BC and employing Equations (C5ndash7) The values obtainedare listed in Table 52

The forces necessary to prevent the displacements at the threecoordinates are (Equation (511)

∆F = 10 times 109

33

210

0

0

0

4(10232)

2(10232)

0

2(10232)

4(10232)

1691

807

minus807

10minus6

=

10761MN

05004MN-m

minus05004MN-m

The three forces ∆F together with their three equilibrants areshown at the member ends in Fig 512(a) This set of self-equilibratingforces is reversed in direction and applied on the frame in Fig 512(b)Analysis of this frame by a conventional method gives the member-end forces shown in Fig 512(c) The properties of the cross-sectionfor member BC used in the analysis are those of the age-adjusted


Tabl

e 5

1In

stan

tane

ous

axia

l str

ain

and

curv

atur

e at

t 0 i

mm

edia

tely

aft

er a

pplic

atio

n of

the

load

q (E

xam

ple

53

Fig

51

1)

Prop

ertie

s of

Defl

ectio

ntra

nsfo

rmed

1 sec

tion

atAx

ial s

train

and

at G

age

t 0In

tern

al fo

rces

curv

atur

e at

t 0Pr

oper

ties

of(E

quat

ion

(Ere

f=

E c(t 0

)=30

GPa

)in

trodu

ced

at t 0

(Equ

atio

n (2

32)

)co

ncre

te a

rea

(C8

))

Mem

ber

Sect

ion

AB

IN

M O

(t 0)

(t 0

)A c

B cI c

D(t

0)

B1

58minus0

394

70

5221

minus02

431

minus18

21minus4

21

minus148

11

32minus0

592

70

2714

BCG

158

minus03

947

052

21minus0

243

13

624

649

280

51

32minus0

592

70

2714

284

6C

158

minus03

947

052

21minus0

243

1minus1

821

minus42

1minus1

481

132

minus05

927

027

14

Mul

tiplie

rm

2m

3m

410

6 N10

6 N-m

10minus6

10minus6

mminus1

m2

m3

m4

10minus3

m

1T

he r

efer

ence

poi

nt O

is a

t th

e ce

ntro

id o

f age

-adj

uste

d tr

ansf

orm

ed s

ectio

n (F

ig 5

8(c

))

Tabl

e 5

2C

hang

es in

axi

al s

trai

n an

d in

cur

vatu

re a

nd c

orre

spon

ding

elo

ngat

ion

and

end

rota

tions

of t

he r

elea

sed

stru

ctur

e in

Fig

51

1(c)

Chan

ges

inCh

ange

inCh

ange

s in

axi

aldi

spla

cem

ents

at t

hede

flect

ion

Inte

rnal

forc

es to

Prop

ertie

s of

age

-adj

uste

dst

rain

and

inco

ordi

nate

s in

Fig

at

Gre

stra

in c

reep

trans

form

ed s

ectio

ncu

rvat

ure

511

(c) (

Equa

tions

(Equ

atio

n(E

quat

ion

(24

2))

(Ere

f=

E c(t 1

t0)

=10

GPa

)(E

quat

ion

(24

0))

(C5

ndash7))

(C8

))

Mem

ber

Sect

ion

N

M

AB

I

O

D1

D

2

D3

D

Bminus0

805

20

3810

210

01

0232

383

minus37

2BC

G2

015

minus09

415

210

01

0232

minus96

092

0minus1

691

minus807

807

959

Cminus0

805

20

3810

210

01

0232

383

minus37

2

Mul

tiplie

rs10

6 N10

6 N-m

m2

m3

m4

10minus6

10minus6

mminus1

10minus6

m10

minus6 1

0minus610

minus3 m

radi

anra

dian

Figu

re5

12A

naly

sis

of s

tatic

ally

inde

term

inat

e fo

rces

cau

sed

by c

reep

in a

com

posi

te fr

ame

(Exa

mpl

e 5

3) (

a) fi

xed-

end

forc

es d

ue t

ocr

eep

Ar

(b) a

ssem

blag

e of

fixe

d-en

d fo

rces

and

rev

ersa

l of d

irec

tion

minusF

(c)

mem

ber-

end

forc

es d

ue to

join

t dis

plac

emen

t[A

u]

D

(d) t

otal

mem

ber-

end

forc

es d

ue t

o cr

eep

= su

m (a

) and

(c)

(e) b

endi

ng m

omen

t di

agra

m a

t t 1

transformed section Superposition of the forces in Fig 512(a) and (c)gives the member-end forces caused by creep (Fig 512(d) ) Followingthe notations used with the displacement method in Section 52 theforces in Fig 512(a) (b) and (c) represent respectively Ar minusF and[Au] D

The bending moment at end B of member BC = minus1821 minus 0289 =minus2110MN-m which is the sum of the bending moment at time t0 (seeTable 51) and the change due to creep The bending moments at vari-ous sections are calculated in a similar way and plotted in Fig 512(e)

The stress distribution at time t1 is determined in Table 53 bysuperposition of

(a) Stress at time t0 calculated for N = minus02431 MN and M =3624MN-m applied on the transformed section at t0 The corres-ponding strain distribution is defined by εO(t0) = 649 times 10minus6 andψ(t0) = 2805 times 10minus6 mminus1 (Table 51) The stress values are calculatedby multiplication of the strain by Es = 200GPa for the steel or Ec(t0)= 30GPa for concrete

(b) Stress required to restrain creep which is equal to the product of[minus φ(t1 t0)Ec(t1 t0)Ec(t0)] and the stress in concrete calculated in (a)

(c) Stress due to minus∆N = minus2015MN and minus∆M = 09415MN-m appliedon the age-adjusted transformed section The corresponding straindistribution is defined by ∆εO = minus960 times 10minus6 and ∆ψ = 920 times10minus6 mminus1 (Table 52)

(d) Stress due to the statically indeterminate forces produced by creepaxial force = minus0038MN and moment = minus0289MN-m applied onthe age-adjusted transformed section

The stress values for the above four stages and their superposition arelisted in Table 53 at the top and bottom fibres of concrete and steel

The deflection at G is calculated by superposition of

(a) The deflection at time t0 calculated from the curvature ψ(t0) usingEquation (C8) which gives D(t0) = 2846 times 10minus3 m (Table 51)

(b) The deflection due to creep in the released system calculated fromthe curvatures ∆ψ giving ∆D = 959 times 10minus3 m (Table 52)

(c) The deflection due to a statically indeterminate moment due tocreep = minus0289MN-m constant over BC This gives a deflection ofminus384 times 10minus3 m Hence the total deflection at time t1 is


Tabl

e 5

3St

ress

dis

trib

utio

n at

sec

tion

G (E

xam

ple

53)

Stre

ss in

sta

ges

(MPa

)Cr

eep

effe

ct =

Stre

ss a

t tim

e t 1

=At

tim

e t 0

Cree

p ef

fect

(b)+

(c)+

(d)

(a)+

(b)+

(c)+

(d)

(a)

(b)

(c)

(d)

MPa

MPa

ksi

Top

of c

oncr

ete

minus27

562

297

minus14

740

140

096

31

793

minus02

60Bo

ttom

of c

oncr

ete

minus09

060

755

minus12

720

078

minus04

39minus1

345

minus01

95To

p of

ste

elminus6

04

0minus2

544

155

minus23

89minus2

993

minus43

4Bo

ttom

of s

teel

893

30

584

minus80

4minus2

20

871

312

64

(2846 + 959 minus 384)10minus3 = 3421 times 10minus3 m

= 3421mm (1347 in)

We can see by comparing the bending-moment diagrams in Figs511(b) and 512(e) that creep increases the absolute values of the bend-ing moment in the columns Creep reduces the effective modulus ofelasticity of concrete thus the flexural rigidity of BC is reduced whilethe rigidity of the steel column is unchanged The change in relativerigidity is the reason for the increase in bending moment in the columnsIt follows from this discussion that if the same composite cross-sectionis used in all members creep will not result in any changes in internalforces or reactions However this is a hypothetical situation in practicethe shrinkage which occurs at the same time as creep will result in achange in the bending moments

58 Step-by-step analysis by the displacementmethod

Modern concrete structures are often composed of precast or cast in situelements assembled by prestressing Bridges built by the segmental methodare examples The basis of a step-by-step numerical procedure similar to thatpresented in Section 46 but using the general displacement method of analy-sis will be presented here The time is divided into intervals and the changesin stresses or internal forces are considered to occur at the middle of theintervals (Fig 411)

Three different materials are generally involved concrete prestressed steeland non-prestressed reinforcement In the three materials the strainsdeveloped between t0 the beginning of the first interval and ti + 1

2 the end of

the ith interval are given by (see Equation (124) )

εc(ti + 12) =

i

j = 11 + φ(ti + 1

2 tj)

Ec(tj) (∆σc)j + εcs(ti + 1

2 t0) (517)

εps(ti + 12) =

1

Eps

i

j = 1

[(∆σps)j minus (∆σpr)j] (518)

εns(ti + 12) =

1

Ens

i

j = 1

(∆σns)j (519)


where σ and ε are the stress and strain with subscripts c ps and ns referring toconcrete prestressed and non-prestressed steel respectively t is the age withsubscript i (or j) indicating the middle of the ith (or jth) interval t0 is the ageat the beginning of the period for which the analysis is considered εcs(ti + 1

2 t0)

is the shrinkage that would occur in concrete if it were free during the periodt0 to ti + 1

2 ∆σpr is the reduced relaxation of prestressed steel (see Section 15)

(∆σ)j is the change of stress at the middle of the jth intervalThe change in strain in the ith interval can be separated by taking the

difference between the strain values calculated by each of the last threeequations at the ends of the intervals i minus l and i

(∆εc)i = 1 + φ(ti + 1

2 ti)

Ec(ti) (∆σc)i

+ i minus 1

j = 1

(∆σc)j

Ec(tj)[φ(ti + 1

2 tj) minus φ(ti minus 1

2 tj)] + (∆εcs)i (520)

(∆εps)i = (∆σps)i

Eps

minus (∆σpr)i

Eps

(521)

(∆εns)i = (∆σns)i

Ens

(522)

The last equation is a linear relationship between stress and strain in thenon-prestressed steel Equations (520) and (521) may be rewritten inpseudolinear forms

(∆εc)i = (∆σc)i

(Ece)i

+ (∆εc)i (523)


Eps

+ (∆εps)i (524)

where (Ece)i is an effective modulus of elasticity of concrete to be used in anelastic analysis for the ith interval

(Ece)i = Ec(ti)

1 + φ(ti + 12 ti)

(525)

(∆εc)i is equal to the sum of the second and third terms on the right-hand sideof Equation (520) Similarly (∆εps)i is equal to the last term in Equation(521) The terms (∆ε)i in Equations (523) and (524) represent an lsquoinitialrsquodeformation independent of the stress increment in the ith interval Thus


(∆ε)i can be determined if the stress increments in the preceding incrementsare known

In the step-by-step analysis a complete analysis of the structure is per-formed for each time interval Thus when the analysis is done for the ithinterval the stress increments in the preceding intervals have been previouslydetermined In this way the initial strains (∆ε)i are known values which canbe treated as if they were produced by a change in temperature of knownmagnitude

In the analysis of a plane frame by the displacement method three nodaldisplacements are determined at each joint translations in two orthogonaldirections and a rotation With the usual assumption that a plane cross-section remains plane during deformation the strain and hence the stress atany fibre in a cross-section of a member can be calculated from the nodaldisplacements at its ends

In the step-by-step procedure a linear elastic analysis is executed for eachtime interval by the conventional displacement method The cross-sectionproperties to be used in this analysis are those of a transformed sectioncomposed of the area of concrete plus αi times the area of steel where αi is aratio varying with the interval and for the ith interval

αi = Es

(Ece)i

(526)

where Es is the respective modulus of elasticity of prestressed or non-prestressed steel

In any interval i the three materials are considered as if they were sub-jected to a change of temperature producing the free strain (∆ε)i of knownmagnitude The corresponding stress (∆σ)i in the three materials areunknowns to be determined by the analysis for the ith interval the values(∆σ)i represent the stress due to external loading (if any) applied at the middleof the ith interval plus the stress due to the fictitious change in temperaturementioned above

Analysis of stress due to arbitrary temperature distribution involves thefollowing steps First the strain due to temperature ( (∆ε)i in our case)is artificially restrained by internal forces ∆N and ∆M in each section (seeEquations (225) and (226) ) This is equivalent to the application of a setof self-equilibrating forces (see Fig 57 and Equations (515) and (516) )The artificial restraint is then removed by application of a set of equal andopposite forces

An example of analysis by this method and a listing of a computer pro-gram which performs the analysis can be found in the references mentioned inNote 3


59 General

Chapters 2 to 5 are concerned with the analysis of stresses and deformationsin uncracked reinforced or prestressed concrete structures accounting for theeffects of the applied load including prestressing creep and shrinkage ofconcrete and relaxation of prestressed steel Creep is assumed to be pro-portional to stress and thus instantaneous strain and creep have a linearrelationship with stress Shrinkage and relaxation result in changes in con-crete stress and must therefore also produce creep In spite of this inter-dependence the analysis is linear which means that superposition of stressesstrains or displacements applies and the stresses or deformations due toapplied loads or due to shrinkage or due to relaxation are proportional to thecause Because of the linearity conventional linear computer programscan be employed for the time-dependent analysis This is demonstrated byexamples in Chapter 6

Creep shrinkage and relaxation change stresses in concrete and steel Instatically determinate structures the change is in the partitioning of theinternal forces between concrete prestressed and non-prestressed steel butthe resultants in the three components combined remain unchanged In stat-ically indeterminate structures the reactions and the internal forces generallychange with time

Chapters 7 8 9 and 13 are concerned with the analysis of stresses anddeformations when the tensile strength of concrete is exceeded at somesections of a structure producing cracking The behavior is no longer linear

Notes

1 See Section 145 of the reference mentioned in note 3 of Chapter 32 See Appendix D of the reference mentioned in note 3 of Chapter 33 A computer program in FORTRAN for analysis of the time-dependent displace-

ments internal forces and stresses reinforced and prestressed concrete structuresincluding the effects of cracking is available See Elbadry M and Ghali AManual of Computer Program CPF Cracked Plane Frames in Prestressed ConcreteResearch Report No CE85-2 revised 1993 Department of Civil Engineering TheUniversity of Calgary Calgary Alberta Canada


Analysis of time-dependentinternal forces withconventional computerprograms

The Confederation Bridge connecting Prince Edward Island and New Brunswick CanadaFloating crane installing a 190m long segment on a pier

Chapter 6

61 Introduction

Computers are routinely used in practice to analyse structures particularlywhen linear stressndashstrain relationship of the material is acceptable and whendisplacements are small These assumptions are commonly accepted in theanalysis of structures in service Thus many of the available computer pro-grams perform linear analysis in which the strain is proportional to the stressand superposition of displacements strains stresses and internal forces isallowed The present chapter demonstrates how conventional linearcomputer programs can be employed for approximate analysis of thetime-dependent effects of creep and shrinkage of concrete and relaxation ofprestressed steel Only framed structures are considered here These can beidealized as assemblages of beams (bars) Thus the computer programs ofconcern are those for plane or space frames plane or space trusses or planegrids1

The procedure discussed in this chapter can be used to solve time-dependent problems of common occurrence in practice As an example con-sider the effects of shortening due to creep and shrinkage of a prestressedfloor supported on columns constructed in an earlier stage Analysis of theeffect of differential shortening of columns in a high-rise building providesanother example the compressive stress and the change in length due to creepare commonly greater in interior than exterior columns Bridge structures arefrequently composed of members (segments) precast or cast-in-situ made ofconcrete of different ages or of concrete and steel (eg cable stays) Theprecast members are erected with or without the use of temporary supportsand made continuous with cast-in-situ joints or with post-tensioned tendonsIn all these cases the time-dependent analysis can be done by the applicationand the superposition of the results of conventional linear computerprograms

62 Assumptions and limitations

Immediate strain and creep of concrete are proportional to the stress (com-pressive or tensile) and the effect of cracking is ignored Structures are ideal-ized as prismatic bars (members) connected at nodes The cross-sectional areaproperties of any bar are those of a homogeneous section Thus the presenceof the reinforcing bars or the tendons in a cross-section is ignored in calcula-tion of the cross-sectional area properties Alternatively a tendon or areinforcing bar can be treated as a separate member connected to the nodes byrigid arms (Fig 61(a) ) The axes of members coincide with their centroidalaxes Because the cross-section of an individual member is considered homo-geneous no transformed cross-sectional area properties are required and thevariation of the location of the centroids of transformed sections due tocreep of concrete does not need to be considered A composite member

Analysis of time-dependent internal forces 177

whose cross section consists of a precast part and a cast-in-situ part or ofconcrete and steel is treated as two homogeneous members connected byrigid arms joining the centroids of the two parts (Fig 61(b) )

With the idealization using short rigid arms as shown in Figs 61(a) and(b) the actual member should be divided into a number of short members(say 10 see Example 65) The internal forces obtained by analysis should beconsidered representatives of the actual structure only at mid-length of theshort members If the external loads are applied only at the nodes the bend-ing moment at mid-length of a member is the average of the two bendingmoment values at the two ends and the shearing force and the axial force areconstants

Figure 61 Idealization of members (a) prestressed member idealized as two bars (b)composite member idealized as two bars of different material properties


63 Problem statement

Consider a framed structure composed of members cast prestressed orloaded in stages each of these is treated as an event occurring at a specificinstant Introduction or removal of a support is considered an event Thesubscript j is used to refer to the effects of the event occurring at the instant tjIt is required to determine the changes in displacements internal forces andreactions that occur between tj and a later instant tk due to creep and shrink-age of concrete and relaxation of the prestressed reinforcement When thechanges in internal forces are known the corresponding changes in strainsand stresses can be determined by basic equations (eg Equations (219) and(220) ) Section 64 describes two computer runs to solve this problem using alinear computer program

As discussed in Chapter 5 in a statically determinate structure the time-dependent phenomena affect only the displacements while the reactions andthe internal forces remain constants The stress and stress resultants on a partof a composite cross-section can change with time but when the structure isstatically determinate the stress resultants in any cross section as a whole donot change with time In other words only the repartition of forces betweenthe parts of a cross-section varies with time without change in the resultantsof stresses in all the parts combined

64 Computer programs

This section describes the input and the output of typical linear computerprograms for the analysis of framed structures based on the displacementmethod (Section 52) Global axes must be defined by the user The positionof the nodes is specified by their coordinates (x y) or (x y z) for plane orspace structures respectively Figure 62 shows global axes the nodal dis-placements (the degrees of freedom) and the order of numbering of thecoordinates representing displacements or forces at a typical node of the fivetypes of framed structures plane truss space truss plane frame space frameand grid The analysis gives the nodal displacements D and the forces onthe supported nodes in the global directions It also gives a member endforces A for individual members in the directions of their local axes Figure63 shows local coordinates and their numbering and the positive directionsof the member end forces for each of the five types of framed structures Anasterisk is used here in reference to local axes and local coordinates ofmembers

Input data description The input data must give the nodal coordinates thenode numbers at the two ends of each member and its cross-sectional area Inaddition for a cross-section of a member of a plane frame the input mustinclude the second moment of area I about a centroidal principal axis


perpendicular to the plane of the frame for a space frame member the inputmust give Iy Iz and J the second moment of area about centroidal principalaxes y and z and the torsion constant for a member of a grid the inputmust include the second moment of area Iz about centroidal principal axisz and the torsion constant J All members are assumed to have constantcross-sections

Images of an input data file are shown in each of Figs 64 and 65 The firstinput file is for computer program SPACET2 for the analysis of a space truss(Fig 610) to be discussed in Example 64 Section 68 The second input fileis for computer program PLANEF3 for the analysis of a plane frame (Fig611) to be discussed in Example 65 Section 68 The integers and the realvalues given on the left-hand sides of Figs 64 and 65 are the input data to beused by the computer the words and the symbols on the right-hand side ofthe figures indicate to the user the contents of each data line

Notation The symbols employed in Figs 64 and 65 are defined below

NJ NM NSJ and NLC number of joints number of members number ofsupported joints and number of load casesrespectively

Figure 62 Global axes degrees of freedom and the order of numbering of the coordinatesat typical nodes


JS and JE the joint numbers at the start and at the end of amember

a I and ar cross-sectional area second moment of area andthe reduced cross-sectional area for shear deform-ation (A large value is entered in Fig 65 becauseshear deformation is ignored this is also done inother examples of this chapter where PLANEF isemployed)

Figure 63 Local coordinates for typical members


Fx Fy Fz Mz forces at a joint applied in directions of the nodalcoordinates defined in Fig 62

Ar fixed-end forces these are the forces produced atfixed member ends due to external load tempera-ture variation shrinkage creep or relaxation

Support conditions A restraint indicator integer 1 or 0 in the input datasignifies a free or a prescribed displacement in direction of one of the global

Figure 64 Image of input file (abbreviated) for computer program SPACET (SpaceTrusses) see note 1 p 206) The input data are for the space truss ofExample 64 Fig 610


axes The integer 0 denotes that the displacement has a prescribed real valueincluded in the input when a support prevents the displacement the pre-scribed value should be 00 When the restraint indicator is 1 it signifies thatthe displacement is free an arbitrary (dummy) real value should be entered inthe space of prescribed displacement

Load data These are given in two sets of lines each set is terminated by alsquodummyrsquo line which starts by an integer gtNLC The first set is for forcesapplied at the nodes The second set gives the fixed-end forces Ar for indi-vidual members two forces and six forces must be entered respectively for amember of a space truss and a member of a plane frame The fixed-end forcesare included in the data only for members subjected to forces away from

Figure 65 Image of input file (abbreviated) for computer program PLANEF (PlaneFrames) see note 1 p 206 The input data are for the plane frame ofExample 65 Fig 611


nodes or for members subjected to temperature variation The values of thefixed-end forces are to be calculated by well-known equations given in manytexts Some computer programs calculate Ar from input data describing theloads on the members with such programs Ar is not part of the input data

Member end forces In the displacement method of analysis which is thebasis of all computer programs the member end forces for a member aredetermined by the superposition equation (see step 5 in Section 52)

A = Ar + [Au] D (61)

where D is a vector of the displacements at the two ends of the memberafter they are transformed from the directions of the global axes to the direc-tions of the local axes of the member (Fig 63) [Au] which has the samemeaning as the memberrsquos stiffness matrix consists of the member end forcesdue to separate unit values of the displacements D1 D2 It is to benoted that for a concrete member [Au] is directly proportionate to the modu-lus of elasticity of concrete at the age considered For the presentation thatfollows define the symbol

AD = [Au] D (62)

AD = A minus Ar (63)

AD which is equal to the second term on the right-hand side of Equation(61) is a vector of self-equilibrating forces that would be produced at themember ends by the introduction of the displacements D at its two nodes

65 Two computer runs

The problem stated in Section 63 can be solved by two computer runs usingan appropriate linear computer program such as the ones described in Sec-tion 64 For simplicity of presentation we consider the case of the structuresubjected to a single event at time tj that can be the application of externalloads andor prestressing or temperature change The analysis is for the time-dependent effects of creep and shrinkage of concrete and relaxation of pre-stressed steel between tj and a later instant tk Two computer runs arerequired

Computer run 1 First the structure is analysed for the instantaneous forcesintroduced at tj The modulus of elasticity of concrete members is Ec(tj) Theresults give the instantaneous displacements D(tj) the reactions and themember end forces A(tj)

The effect of prestressing introduced at tj can be included in this run bytreating the forces exerted by the tendons on the concrete as any other


external force Alternatively when a tendon is idealized as a member (Fig61(a)) two axial restraining forces are to be entered for this member

Ar(tj)prestress =

plusmn Apsσp (tj) (64)

where Aps and σp (tj) are the cross-sectional area of the tendon and its stress attime tj respectively The minus and the plus sign are respectively for the forceat the first and second ends of the member (Fig 63)

Computer run 2 In this run the structure is idealized with the modulus ofelasticity of concrete being the age-adjusted modulus Ec (tk tj) given byEquation (131) which is repeated here

Ec(tk tj) = Ec(tj)

1 + χφ(tk tj)(65)

where φ (tk tj) is creep coefficient at time tk for loading at time tj χ (equiv χ (tk tj) )is the aging coefficient Ec(tj) is the modulus of elasticity of concrete at time tjThe vector of fixed-end forces Ar(tk tj) is to be entered as loading datawhere Ar(tk tj) is a vector of hypothetical forces that can be introducedgradually in the period tj to tk to prevent the changes in nodal displacementsat member ends The elements of the vector Ar(tk tj) for any member com-prise a set of forces in equilibrium Calculation of the elements of the vectorAr(tk tj) is discussed below considering the separate effect of each of creepshrinkage and relaxation

Member fixed-end forces due to creep The member end forces that restrainnodal displacements due to creep are

Ar(tk tj)creep = minusEc(tk tj)

Ec (tj)φ(tk tj) AD (tj) (66)

The vector AD(tj) is given by Equation (63) using the results and theinput data of Computer run 1 For the derivation of Equation (66) considerthe hypothetical displacements change [φ(tk tj) D(tk)] as if they wereunrestrained Premultiplication of this vector by [minusAu] and substitution ofEquation (62) give the values of the restraining forces for a member whoseelasticity modulus is Ec(tj) Multiplication of the ratio [Ec (tk tj)Ec(tj)] toaccount for the fact that the restraining forces are gradually introduced givesEquation (66)

Member end forces due to shrinkage The change of length of a concretemember subjected to shrinkage εcs(tk tj) can be prevented by the gradualintroduction of axial member-end forces


Ar(tk tj)axial shrinkage = plusmn [Ec(tk tj)Ac]εcs (67)

where Ac is the cross-sectional area of concrete member the plus and theminus signs are respectively for the forces at the first and the second node of amember (see Fig 63) Note that for shrinkage εcs is a negative value

Member end forces due to relaxation When the effect of prestressing is repre-sented in Computer run 1 by external forces exerted by the tendons on theconcrete it is only necessary in Computer run 2 to use an estimated prestressloss due to creep shrinkage and relaxation combined to calculate externalforces on the concrete in the same way as for the prestress in Computer run 1(with reversed signs and reduced magnitudes) When a tendon is idealized asan individual member the relaxation effect can be represented in Computerrun 2 by two axial restraining forces

Ar (tk tj)axial relaxation = Aps∆σpr (tk tj) (68)

where ∆σpr (tk tj) is the reduced relaxation the negative and the positive signsin this equation are respectively for the force at the first and the second nodeof the member (Fig 63) In verifying or in applying Equation (68) note that∆σpr (tk tj) is commonly a negative value When the tendons are idealized asseparate members and Equation (68) is used no estimated value of loss ofprestress due to creep shrinkage and relaxation is needed the analysis willmore accurately give the combined effect of creep shrinkage and relaxationand the time-dependent changes in the internal forces

66 Equivalent temperature parameters

In the preceding section two computer runs are proposed to analyse thetime-dependent effects of creep shrinkage and relaxation In Computerrun 2 the values of self-equilibrating forces Ar(tk tj) are entered as inputdata for individual members It will be shown below that fictitious tem-perature parameters to be calculated by Equations (69) and (610) can beemployed as thermal data for computer programs that do not accept Aras input

As example consider a plane frame member AB having six end forces Ar(Fig 66(a) ) The six forces represent a system in equilibrium Figure 66(b)represents a conjugate beam of the same length and cross section as the beamin Fig 66(a) but subdivided by a mid-length node The conjugate beam issubjected to a rise of temperature TO for its two parts and temperature gradi-ents T prime1 and T prime2 for parts AC and CB respectively where T prime = dTdy with ybeing the coordinate of any fibre measured downward from the centroidalaxis It can be verified that the conjugate beam with ends A and B fixed hasthe same forces at the ends A and B as the actual member when


TO

T prime1T prime2

= 1

Ecα

1A00

0l(6I)5l(6I)

0minus1I

minus1I

Ar1

Ar2

Ar3

(69)

where A and I is the cross-sectional area and its second moment about cen-troidal axis l is length of member Ec (equivEc(tk tj) ) is the age-adjusted modulusand α is an arbitrary thermal expansion coefficient The same values of Ec

and α used in Equation (69) must be entered as input in Computer run 2 Thefictitious temperature parameters T0 T prime1 and T prime2 can be expressed (by anequation similar to Equation (69) ) in terms of the fixed-end forces at end Binstead of end A to give the same result The subdivision of members intotwo parts should not be done in Computer run 1 Also the subdivision is notnecessary in Computer run 2 when the structure is a plane or a space truss Inthis case the input in Computer run 2 is a uniform rise of temperature T0where

TO = Ar (tk tj)

αEc (tk tj)A(610)

where Ar(tk tj) is an axial force at the first-end of the member (given by

Figure 66 Equivalent temperature parameters (a) actual member of a plane frame (b)conjugate beam subjected to rise of temperature producing the same forces atends A and B as in the actual beam


Equation (64) (66) or (67) The first and the second nodes of members andthe positive sign convention for member-end forces are defined in Fig 63

67 Multi-stage loading

The problem stated in Section 63 can be solved when the analysis for thetime-dependent changes between time tj and time tk are required for the effectof events 1 to j with the last event j occurring at tj with events 1 to (jminus1)occurring at earlier instants t1 t2 tjminus1 We recall the term lsquoeventrsquo refersto the application of forces the introduction of prestressing the casting a newmember or the removal or the introduction of a support The two computerruns as discussed in Section 65 are to be applied differing only in the calcula-tion of the fixed-end forces Ar(tk tj) to be included in the input of Com-puter run 2 These forces are to be determined by a summation to replaceEquation (66) The summation is to superimpose the effect of creep due tothe forces introduced at t1 t2 tj as well as due to the gradual changes ininternal forces in the intervals (t2 minus t1) (t3 minus t2) (tj minus tjminus1) As exampleEquation (611) gives contribution to Ar(tk tj)creep of the loads introduced attime ti where ti lt tj lt tk

Ar(tk tj)creep load introduced at ti

= minusEc (tk tj)

Ec (ti)[φ(tk ti) minus φ(tj ti)]AD (ti) (611)

The vector AD(ti) is to be determined by Equation (63) using the resultsof a computer run having an input that includes the modulus of elasticityEc(ti) and the loading introduced at ti

When the structure is subjected to more than one or two events severalcomputer runs are required In this case it is more practical to apply the step-by-step procedure discussed in Section 58 employing a specialized computerprogram (see eg note 3 page 175)

68 Examples

The following are analysis examples of structures subjected to a single or twoevents and it is required to determine the change(s) in displacements andorinternal forces or stresses between time tj and a later time tk

Example 61 Propped cantilever

The cantilever AB in Fig 67(a) is subjected at time t0 to a uniform loadq At time t1 a simple support is introduced at B thus preventing the


increase in deflection at B due to creep Determine the change in the endforces at A and B between time t1 and a later time t2 Given data φ(t1 t0)= 09 φ(t2 t0) = 26 φ(t2 t1) = 245 χ(t2 t1) = 08 Ignore the differencebetween Ec(t0) and Ec(t1)

The computer program PLANEF is used here but the results can bechecked by hand computation as discussed in Chapter 5 Table 61shows the input data and the results of Computer run 1 analyzing theimmediate effect of the load introduced at time t0 Each of Ec(t0) q and lare considered equal to unity the support conditions are those of acantilever encastreacute at A and free at B (Fig 67(a) ) the end forces for atotally fixed beam subjected to uniform load are entered as the loadinput

Ar(t0) = 0 minus05 ql minus00833 ql 2 0 minus05 ql 00833 ql 2

Figure 67 Propped cantilever Example 61 (a) cantilever loaded at time t0 (b)member end forces developed between time t1 and t2 due to theintroduction of support B at t1


The result of this computer run includes the member end forcesimmediately after load application

A(t0) = 0 minusql minus05ql 2 0 0 0

As expected these are the forces at the ends of a cantilever ApplyEquation (63) to obtain

AD(t0) = 0 minus05 ql minus04167ql 2 0 05 ql minus00833 ql 2

These are the changes in end forces produced by varying the nodaldisplacements form null when the nodal displacements are preventedto the values D included in the results of Computer run 1 Creepfreely increases these displacements in the period t0 to t1 The hypo-thetical end forces that can prevent further increase in the period t1 to t2

are (Equation (611) )

Ar(t2 t1) = minusEc(t2 t1)

Ec(t0)[φ(t2 t0) minus φ(t1 t0)] AD(t0)

The age-adjusted elasticity modulus is (Equation (65) )

Ec (t2 t1) = Ec(t1)

1 + χφ(t2 t1) =

Ec(t1)

1 + 08(245) = 03378 Ec(t0)

Substitution in Equation (611) gives a set of self-equilibrating endforces to be used as load input data in Computer run 2

Table 61 Input and results of Computer run 1 with program PLANEF Example 61

Analysis results load case No 1

Nodal displacementsNode

12

u00000E+0000000E+00

v41668Eminus0712500E+00

10417Eminus0616667E+00

Forces at the supported nodesNode

1Fx

00000E+00Fy

minus10000E+01Mz

minus50000E+00

Member end forcesMember

1F1

00000E+00F2

minus10000E+01F3

minus50000E+00F4

00000E+00F5

minus11102Eminus15F6

55511Eminus15


Ar(t2 t1) = minus03378(26 minus 09) AD(t0)

Ar(t2 t1) = 0 02872 ql 02393 ql 2 0 minus02872 ql 00479 ql 2

The same forces are obtained in Example 51 (Fig 53(b) ) Table 62includes the input data and the analysis results of Computer run 2 Wenote that the age-adjusted elasticity modulus is used and the supportconditions are those of end encastreacute at A and simply supported at BThe required changes in member end forces between time t1 and t2 are aset of self-equilibrating forces (Fig 67(b) ) which are copied here

A(t2 t1) = 0 02155 ql 02154 ql 2 0 minus02155 ql 0

Table 62 Input (abbreviated) and results of Computer run 2 using program PLANEFExample 61

Input dataNumber of joints = 2 Number of members = 1 Number of load cases = 1Number of joints with prescribed displacements = 2 Elasticity modulus = 03378

Nodal coordinates and element informationSame as in Table 61

Support conditionsNode

12

Restraint indicatorsu v 0 0 01 0 1

Prescribed displacementsu v

00000E+00 00000E+00 00000E+00

00000E+00 00000E+00 00000E+00

Forces applied at the nodesSame as in Table 61

Member end forces with nodal displacement restrainedLd case Member

1 1Ar1

0000E+00Ar2

2872E+00Ar3

2393E+00Ar4

0000E+00Ar5

minus2872E+00Ar6

4786Eminus01



12

u00000E+0000000E+00

v17688Eminus07

minus17688Eminus07

17688Eminus07

minus35377Eminus01

Forces applied at the supported nodesNode

12

Fx

00000E+00

00000E+00

Fy

21550E+00minus21550E+00

Mz

21540E+00

00000E+00


1F1

00000E+00F2

21550E+00F3

21540E+00F4

00000E+00F5

minus21550E+00F6

00000E+00


Example 62 Cantilever construction method

The girder ABC (Fig 68(a) ) is constructed as two separate cantileverssubjected at time t0 to a uniform load qunit length representing theself weight At time t1 the two cantilevers are made continuous at Bby a cast-in-situ joint Determine the changes in member end forcesfor AB between t1 and a later time t2 Use the same creep and agingcoefficients as in Example 1 and ignore the difference between Ec(t0)and Ec(t1)

Because of symmetry the computer analysis needs to be done for halfthe structure only (say part AB) Computer run 1 and calculation ofAr(t2 t1) for use as load input in Computer run 2 are the same as inExample 1 (Table 61) In the current problem the support conditions atend B in Computer run 2 must be as indicated below with the remaininginput data as in Table 62

Figure 68 Cantilever construction Example 62 (a) girder ABC constructed as twoseparate cantilevers subjected to uniform load at time t0 (b) member endforces Ar(t2 t1) calculated in Example 61 (c) changes in member end forcesin Computer run 2 (d) superposition of member end forces in Fig 67(a) and Fig 68 (c) to give A(t2) (e) bending moment diagram at time t2


Parts of the input and the results of Computer run 2 are presented inFig 68 rather than in a table Figure 68(b) shows Ar(t2 t1) these arethe forces that can artificially prevent the changes due to creep in thedisplacements at ends A and B of the cantilever AB Figure 68(c) showsthe results of Computer run 2 the computer applies the forcesAr(t2t1) in a reversed direction and determines the correspondingchanges in member end forces and superposes them on Ar(t2 t1)Figure 68(d) shows the sum of the forces in Fig 68(c) and Fig 67(a)this gives the forces on member AB at time t2 The bending momentdiagram at time t2 is shown in Fig 68(e)

Node Restraint indicators Restraint displacements

u v θ u v θ2 0 1 0 00 00 00

Example 63 Cable-stayed shed

The line AB in Fig 69 represents the centroidal axis of a concretecantilever At time t1 the cantilever is subjected to its own weight q =25kN-m and a prestressing force P(t1) = 200kN introduced by the steelcable AC Calculate the changes in deflection at the tip of the cantileverand in the force in the cable in the period t1 to a later time t2 caused by

Figure 69 A cable-stayed shed Example 63


creep and shrinkage of concrete and relaxation of prestressed steelIgnore cracking and presence of reinforcement in AB Given data Ec(t1)= 25GPa φ(t2 t1) = 2 χ = 08 εcs(t2 t1) = minus300 times 10minus6 ∆σpr = minus50MPaCross-sectional area properties for AB Ac = 10m2 I = 01m4 For thecable As = 250mm2 Es = 200GPa

Table 63 gives the input and the results of Computer run 1 using theprogram PLANEF During the tensioning the change in cable lengthcan occur independently from the deformation of concrete thus thetranslation at the tip of the cantilever is not compatible with the elonga-tion of the cable For this reason the analysis in Table 63 is for a

Table 63 Input data and results of Computer run 1 using program PLANEF Example63 Fig 69

Input dataNumber of joints = 2 Number of members = 1 Number of load cases = 1Number of joints with prescribed displacements = 1 Elasticity modulus = 250E+09

Nodal coordinatesNode

12

x y00 00

100 00

Element informationElement

11st node

12nd node

2a

10000E+01I

10000E+00

Support conditions Node

2

Restraint indicatorsu v 0 0 0


00000E+00 00000E+00 00000E+00

Forces applied at the nodesLoad case

1Node

1Fx

17890E+06Fy

minus89440E+06Mz

00000E+00

Member end forces with nodal displacement restrainedLdcase

1Member

1Ar1

0000E+00Ar2

minus1250E+06Ar3

minus2083E+06Ar4

0000E+00Ar5

minus1250E+06Ar6

2083E+06

Analysis results load case No 1Nodal displacements

Node12

u71560Eminus0471560Eminus10

v57534Eminus0311853Eminus08

12200Eminus03

minus14730Eminus09


2Fx

minus17890E+06Fy

minus16056E+06Mz

35560E+06

Member end forces Member

1F1

17890E+06F2

minus89440E+05F3


minus17890E+06F5

minus16056E+06F6

35560E+06


cantilever (without the cable) subjected to a uniform load q = 25kNcombined with Fx = 1789 and Fy = minus894kN at node A where Fx and Fy

are the forces exerted by the cable on the concrete member at time t1The modulus of elasticity in Computer run 1 is equal to Ec(t1) = 25GPaThe results of Computer run 1 include the member end forces of AB attime t1

A(t1)AB = 1789kN minus894kN 0 minus1789kN minus1606kN3556kN-m

The changes in the forces at end of member AB from the fixed-endstatus are (Equation (63) )

AD (t1)AB = 1789kN 356kN 2083kN-m minus1789kN minus356kN1473kN-m

In Computer run 2 (Table 64) the structure is composed of the twomembers AB and AC and the modulus of elasticity used is (Equation(65) )

Ec(t2 t1) = 25GPa

1 + 08(20) = 9615GPa

A transformed cross-sectional area equal to AsEsEc is used for thecable a negligible value is entered for I The end forces that can arti-ficially prevent the time-dependent changes in displacements due tocreep at the two nodes of member AB are (Equation (66) )

Ar(t2 t1)creep AB = minus 9615

250 (20)AD(t1)AB

Ar(t2 t1)creep AB = minus1376kN minus274kN minus1602kN-m 1376kN274kN minus1133kN-m

The axial force that can artificially prevent the change in length dueto shrinkage of AB (Equation (67) )

Ar(t2 t1)axial shrinkage AB = plusmn[9615GPa (minus300 times10minus6) (10m2) = 28845kN


The restraining forces for creep and shrinkage are entered on separatelines as load data (for member 1) in Computer run 2 (Table 64)

The relaxation in cable AC is a loss of tension presented in Computerrun 2 by an axial compressive force in the member thus the memberend forces to be used in the input (Equation (68) )

Table 64 Input data and results of Computer run 2 using program PLANEFExample 63 Fig 69



123

x y00 00

100 00100 minus50


12

1st node11

2nd node23

a10000E+0152000Eminus02

I10000E+0010000Eminus06


23



00000E+00 00000E+00 00000E+00

00000E+00 00000E+00 00000E+00


1Node

1Fx

00000E+00Fy

00000E+00Mz

00000E+00


111

Member111

Ar1

minus1376E+06minus2885E+071250E+05

Ar2

minus2740E+050000E+000000E+00

Ar3

minus1602E+060000E+000000E+00

Ar4

1376E+06

2885E+07minus1250E+05

Ar5

2740E+05

0000E+00

0000E+00

Ar6

minus1133E+050000E+000000E+00



123

u31270Eminus0231270Eminus0812014Eminus08


minus24022Eminus08

minus16116Eminus03minus49711Eminus09minus56920Eminus09


23

Fx

minus15478E+05minus15478E+05

Fy

minus77388E+0477388E+04

Mz

77890E+05

19580E+01


12

F1minus15478E+0517305E+05

F277388E+0432547E+00

F3minus16808E+0116808E+01

F415478E+05

minus17305E+05

F5minus77388E+04minus32547E+00

F677890E+0519580E+01


Ar(t2 t1)axial relaxation AC = (250 times 10minus6 m2)(minus50MPa) = plusmn 125kN

The forces Ar due to relaxation are entered on a separate line (formember 2) in the input data in Table 64

The results of Computer run 2 (Table 64) include the deflectionincrease at the tip of the cantilever v = 39mm and the changes in theend forces in member Ac representing a drop of 173kN in the tensileforce in the cable

Example 64 Composite space truss

Figure 610(a) depicts a cross-section of a concrete floor slab supportedby structural steel members The structure is idealized as a space trussshown in pictorial view elevation and top views in Figs 610(b) (c) and(d) The truss has a span of 360m but for symmetry half the span isanalysed Consider that the half truss is subjected at time t1 to down-ward forces P at each of nodes 1 2 10 and 11 and 2P at each of nodes4 5 7 and 8 where P = 40kN Find the deflection at mid-span at time t1

and the change in deflection at the same location occurring betweentime t1 and a later time t2 due to creep and shrinkage of concrete Givendata for concrete Ec(t1) = 25GPa φ(t2 t1) = 225 χ(t2 t1) = 08 εcs =minus400 times 10minus6 for structural steel Es = 200GPa The material for members1 to 6 is concrete all other members are structural steel The crosssectional areas of members are

For each of members 1 to 6 the cross-sectional area = 04m2

For each of members 11 to 13 the cross-sectional area = 9100mm2

For each of member 14 of 25 the cross-sectional area = 2300mm2

For each remaining members the cross-sectional area = 1200mm2

Light steel members running along lines 1ndash10 and 2ndash11 may be neces-sary during construction these are here ignored

The computer program SPACET (space trusses) is used in two runsIn Computer run 1 the modulus of elasticity is Ec(t1) = 25GPa a trans-formed cross-sectional area = AsEsEc(t1) is entered for the steel mem-bers of the truss An image of the input file (abbreviated) is shown inFig 64 Table 65 shows the results which include the deflection atmid-span (nodes 10 or 11) at time t1 = 558mm



Ec (t2 t1) = 25GPa

1 + 08(225) = 8929GPa

This modulus is used in Computer run 2 and a transformed cross-sectional area = AsEsEc is entered for the steel members The load dataare the two axial end forces Ar(t2 t1)creep calculated by Equation (66)for each of the concrete members (1 to 6)

Figure 610 Concrete floor slab on structural steel members idealized as a spacetruss (Example 64) (a) cross-section (b) pictorial view with the diagonalmembers in the x-y plane omitted for clarity (c) elevation (d) top view


Ar(t2 t1)creep = minusEc (t2 t1)

Ec (t1) φ(t2 t1) AD(t1)

The values of AD(t1) are calculated by Equation (63) using the resultsof Computer run 1 and noting that Ar(t1) = 0 for all members Theartificial restraining forces are calculated below for member 1 asexample

AD(t1)member 1 = 19209 minus19209kN

Table 65 Abbreviated results of Computer run 1 Example 64 Space trussimmediate displacements and forces at time t1

Nodal displacementsNode u v w

123456789

101112

83225Eminus03

83225Eminus03minus92308Eminus0271699Eminus0371699Eminus03

minus65934Eminus0240567Eminus0340567Eminus03


minus23736Eminus08

60029Eminus03minus60029Eminus0310672Eminus24



minus27428Eminus0327428Eminus0300000E+00

44022Eminus08

44022Eminus08

14765Eminus01

29254Eminus01

29254Eminus01

39067Eminus01

48840Eminus01

48840Eminus01

52367Eminus01

55758Eminus01

55758Eminus01

00000E+00

Forces at the supported nodesNode Fx Fy Mz

129

101112

00000E+00

00000E+00

00000E+00minus72000E+06minus72000E+0614400E+07

00000E+00

00000E+00minus45097Eminus0900000E+0000000E+0000000E+00

minus24000E+06minus24000E+0600000E+0000000E+0000000E+0000000E+00

Member end forcesMember F1 F2

1234567

31

19209E+06

51887E+06

67612E+06

19209E+06

51887E+06

67612E+06

96046E+05

43425E+04

minus19209E+06minus51887E+06minus67612E+06minus19209E+06minus51887E+06minus67612E+06minus96046E+05

minus43425E+04


Table 66 Abbreviated input and results of Computer run 2 Example 64 Spacetruss Analysis of changes in displacements and internal forces betweentime t1 and t2

Elasticity modulus = 89286E+10Member end forces with nodal displacement restrained

Ld case Member Ar1 Ar2

111111111111

123456123456

minus1544E+06minus4170E+06minus5433E+06minus1544E+06minus4170E+06minus5433E+06minus1429E+07minus1429E+07minus1429E+07minus1429E+07minus1429E+07minus1429E+07

1544E+06

4170E+06

5433E+06

1544E+06

4170E+06

5433E+06

1429E+07

1429E+07

1429E+07

1429E+07

1429E+07

1429E+07

Analysis resultsNodal displacements

Node u v w123456789

101112

87131Eminus02




minus28656Eminus23

minus20020Eminus0320020Eminus03

minus14889Eminus22minus42660Eminus0342660Eminus0349631Eminus23


minus12550Eminus45minus48412Eminus0348412Eminus0300000E+00

28363Eminus22minus13222Eminus2286130Eminus0214988Eminus0114988Eminus0120936Eminus0124353Eminus0124353Eminus0127301Eminus0127543Eminus0127543Eminus0100000E+00


123456789

10 262728293031

minus64064E+05minus72448E+05minus77460E+05minus64064E+05minus72448E+05minus77460E+05minus32032E+05minus68256E+05minus74954E+05minus38730E+05

71626E+05

71626E+05

80999E+05

80999E+05

86603E+05

86603E+05

64064E+05

72448E+05

77460E+05

64064E+05

72448E+05

77460E+05

32032E+05

68256E+05

74954E+05

38730E+05

minus71626E+05minus71626E+05minus80999E+05minus80999E+05minus86603E+05minus86603E+05


Ar(t2 t1)creep member 1 = minus8929

250 (225) 19209 minus19209

= minus1544 1544kNThe restraining forces for shrinkage are the same for any of the

concrete members (1 to 6) Equation (67) gives

Ar(t2 t1)shrinkage members 1 to 6 = 8929GPa (minus400 times 10minus6) 04 1 minus1

= minus14286 14286kN

Table 66 gives abbreviated input and results of Computer run 2Because this structure is statically determinate externally creep andshrinkage do not affect the reactions (omitted in Table 66) Thechanges in displacements due to creep and shrinkage in the period t1 tot2 are given in Table 66 including the change of mid-span deflection of275mm (node 10 or 11) The changes in member end forces are given inTable 66 only for the members where the change is non zero

Example 65 Prestressed portal frame

Figure 611(b) represents a portal frame idealization Member BC has apost-tensioned T-section shown in Fig 611(a) Member AC is non-prestressed The prestressing steel tendon having parabolic profile isidealized as straight steel members connected by rigid arms to nodes onthe x-axis through the centroid of the gross concrete section of memberBC At time t1 member BC is subjected to a uniform gravity loadq = 26kN-m (representing self weight and superimposed dead load)combined with a prestressing force P = 2640kN assumed constantover the length of the tendon Find the changes in the force in thetendon and the deflection at mid-span due to creep and shrinkage ofconcrete and relaxation of prestressed steel occurring between t1 and alater time t2 Ignore presence of non-prestressed reinforcement andcracking Given data modulus of elasticity of concrete at time t1Ec(t1) = 25GPa creep and aging coefficients φ(t2 t1) = 20 and χ(t2 t1)= 08 free shrinkage εcs(t2 t1) = minus300 times 10minus6 reduced relaxation (Sec-tion 15) ∆σpr = minus60MPa modulus of elasticity of prestressed steel =200GPa The cross-sections of the members have the following areaproperties


Member AB cross-sectional area = 016m2 second moment of area =213 times 10minus3 m4Member BC cross-sectional area = 0936m2 second moment of area =5586 times 10minus3 m4Tendon cross-sectional area Aps = 2200mm2 and negligible second-moment of area

The problem is solved by two computer runs using programPLANEF (Plane Frames note 1 p 206) The input file for Computerrun 1 is shown in Fig 65 The modulus of elasticity used is Ec(t1) =25GPa While the prestressing is being introduced the tendon canelongate independently from the concrete Thus in Computer run 1

Figure 611 Prestressed portal frame of Example 65 (a) cross-section of memberBC (c) idealization of half the structure


negligible cross-sectional areas is entered for members 7 to 11 whichrepresent the tendon in this way the tendon does not contribute to thestiffness of the frame Two axial forces are entered as Ar to representinitial tension = 2640kN in each of members 7 to 11 Large values areentered for the cross-sectional area properties to represent rigid mem-bers 12 to 17 connecting the nodes of the tendon to nodes on thecentroid of BC Table 67 gives abbreviated results of Computer run 1

The age-adjusted elasticity modulus of concrete is used in Computerrun 2 (Equation (65) )

Ec(t2 t1) = 25GPa

1 + 08(20) = 9615GPa

Table 67 Abbreviated results of Computer run 1 for the portal frame of Example65 using program PLANEF


123456789

10111213

uminus48997Eminus0813645Eminus0210922Eminus0281954Eminus0354654Eminus0327332Eminus0313086Eminus1315449Eminus0276927Eminus0325185Eminus03

minus55401Eminus05minus58490Eminus0442656Eminus20

v39000Eminus0939000Eminus0361245Eminus0211027Eminus0114764Eminus0117100Eminus0117894Eminus0139001Eminus0361245Eminus0211027Eminus0114764Eminus0117100Eminus0117894Eminus01

wminus84349Eminus0925057Eminus0222427Eminus0218195Eminus0212780Eminus0265836Eminus0320542Eminus1325057Eminus0222427Eminus0218195Eminus0212780Eminus0265836Eminus0327272Eminus19


17

13

Fx

25047E+05minus26649E+0726399E+07

Fy

minus31200E+0600000E+0000000E+00

Mz

35933E+05minus38885E+06minus51617E+00

Member end forces

Member F1 F2 F3 F4 F5 F6123456789

1011

31200E+06

26544E+07

26586E+07

26618E+07

26639E+07

26649E+07minus26400E+07minus26400E+07minus26400E+07minus26400E+07minus26400E+07

25047E+05minus44157E+05minus34051E+05minus24165E+05minus14436E+05minus48013E+04minus12826Eminus02minus86862Eminus03minus50763Eminus03minus24704Eminus03minus71642Eminus04

35933E+05

10001E+06

20538E+06

28612E+06

34321E+06

37732E+06

11827Eminus02

33526Eminus02

50243Eminus02

61547Eminus02

67716Eminus02

minus31200E+06minus26544E+07minus26586E+07minus26618E+07minus26639E+07minus26649E+0726400E+0726400E+0726400E+0726400E+0726400E+07

minus25047E+0544157E+0534051E+0524165E+0514436E+0548013E+0412826Eminus0286862Eminus0350763Eminus0324704Eminus0371642Eminus04



Table 68 Abbreviated input data and results of Computer run 2 for the portal frameof Example 65 using program PLANEF

Member end forces with nodal displacement restrained

Ldcase Member Ar1 Ar2 Ar3 Ar4 Ar5 Ar6

11111111111111111

123456123456789

1011

minus2400E+06minus2042E+07minus2045E+07minus2048E+07minus2049E+07minus2050E+07minus4615E+06minus2700E+07minus2700E+07minus2700E+07minus2700E+07minus2700E+071320E+061320E+061320E+061320E+061320E+06

minus1927E+053397E+052619E+051859E+051111E+053693E+040000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+00

minus2764E+05minus7693E+05minus1580E+06minus2201E+06minus2640E+06minus2903E+060000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+00

2400E+06

2042E+07

2045E+07

2048E+07

2049E+07

2050E+07

4615E+06

2700E+07

2700E+07

2700E+07

2700E+07


1927E+05minus3397E+05minus2619E+05minus1859E+05minus1111E+05minus3693E+040000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+00



Nodal displacements

Node123456789

10111213

uminus10241Eminus0759852Eminus0248107Eminus0236251Eminus0224258Eminus0212157Eminus0258205Eminus1364525Eminus0239544Eminus0221176Eminus0297055Eminus0334761Eminus0376495Eminus15


minus14497Eminus0864905Eminus0259470Eminus0248319Eminus0233686Eminus0217224Eminus0253870Eminus1364905Eminus0259470Eminus0248319Eminus0233686Eminus0217224Eminus0213720Eminus18

Forces at the supported nodes

Node17

13

Fx

86860E+03

19104E+06minus19190E+06

Fy

26096Eminus04

00000E+00

00000E+00

Mz

minus38882E+04minus93086E+05minus99877E+00

Member end forces

Member F1 F2 F3 F4 F5 F6123456789

1011

minus26096Eminus04minus33753E+06minus29902E+06minus24999E+06minus21136E+06minus19103E+0633977E+0630063E+0625117E+0621233E+0619191E+06



minus22704Eminus0328021Eminus0248621Eminus0260855Eminus0267470Eminus02

26096Eminus04

33753E+05

29902E+06

24999E+06

21136E+06





Table 68 gives abbreviated input and results of Computer run 2 Atransformed cross-sectional area = ApsEsEc (t2 t1) = 00022(2009615) =004576m2 is entered for each of members 7 to 11 The forces Ar thatcan restrain the nodal displacements due to creep and shrinkage ofmembers 1 to 6 are entered separately for each of the two causes Alsothe forces Arrelaxation are entered for members 7 to 11 These forces arecalculated using Equations (66) (67) and (68) As example we showbelow the calculation for member 2

Ar(t2 t1)creep member 2 = minusEc (t2 t1)

Ec (t1) φ(t2 t1) AD(t1)member 2

= minus9615

25 (20) 26544kN minus442kN 1000kNm minus26544kN

442kN minus2060kNm

= minus20420kN 3397kN minus769kNm 20420kN minus3397kN1585kNm

Ar(t2 t1)axial shrinkage member 2 = plusmn [Ec(t2 t1) εcs (t2 t1) Ac member 2]

= plusmn[9615 times 109 (minus300 times 10minus6) (0936)]

= 2700kN

Ar(t2 t1)axial relaxation member 2 = Aps member 2 ∆σpr (t2 t1)

= (2200 times 10minus6) (minus60 times 106) = plusmn 1320kN

The results in Table 68 include the increase in deflection at mid-spanin the period t1 to t2 (node 7) = 485mm They also include the change intension in the tendon member 11 = minus1919kN this represents a drop intension at a section halfway between nodes 6 and 7 The changein force in the tendon is the combined effect of creep shrinkage andrelaxation and the accompanying variation of internal forces

69 General

Conventional linear computer programs for framed structures are employedin this chapter to calculate approximately the time-dependent effects of creepand shrinkage of concrete and relaxation of prestressed steel in various struc-tures A number of computer runs (at least two) depending on the numberof load stages is required for the analysis The approach can be useful in the


absence of specialized computer programs that can perform the analysismore accurately in a single computer run for structures constructed andorloaded in stages Cracking requires non-linear analysis that cannot be con-sidered in the procedure presented in this chapter The non-linear analysis thatconsiders cracking is discussed in Chapter 13

Notes

1 See for example the computer programs described in an appendix of Ghali A andNeville A M Structural Analysis A Unified and Classical Approach 4th ednE amp FN Spon London 1997 831 pp This set of programs is available from LilianeGhali 3911 Vincent Drive NW Calgary Canada T3A 0G9 The set includes theprograms PLANEF (Plane Frames) and SPACET (Space Trusses) used to solveexamples in this chapter

2 See Note 1 above3 See Note 1 above


Stress and strain ofcracked sections

Western Canadian Place Calgary Partial prestressing used in all floors (Courtesy CohosEvamy amp Partners Calgary)

Chapter 7

71 Introduction

Cracks occur in reinforced and partially prestressed members when thestresses exceed the tensile strength of concrete After cracking the stresses inconcrete normal to the plane of the crack cannot be tensile Thus the internalforces in a section at the crack location must be resisted by the reinforcementand the uncracked part of the concrete cross-section The part of the concretecross-section area which continues to be effective in resisting the internalforces is subjected mainly to compression and some tension not exceeding thetensile strength of concrete At sections away from cracks concrete in tensionalso contributes in resisting the internal forces and hence to the stiffness ofthe member

Two extreme states are to be considered in the calculation of displacementsin a cracked member as will be further discussed in Chapter 8 In state 1 thefull area of the concrete cross-section is considered effective and the strains inthe concrete and the reinforcement are assumed to be compatible In state 2concrete in tension is ignored thus the cross-section is assumed to be com-posed of the reinforcement and concrete in compression The cross-section instate 2 is said to be fully cracked

The actual elongation or curvature of a cracked member can be calculatedby interpolation between the two extreme states 1 and 2

In Chapters 2 and 3 we analysed the stresses axial strain and curvature inan uncracked section including the effects of creep shrinkage and relaxationof prestressed steel The section was assumed to be subjected to an axial forceandor a bending moment The values and the time of application of theseforces were assumed to be known With prestressing the initial prestress forcewas assumed to be known but the changes in the stresses in the prestressedand non-prestressed steel due to creep shrinkage and relaxation were deter-mined by the analysis The full concrete cross-section area was considered tobe effective whether the stresses were tensile or compressive

In the present chapter fully cracked reinforced concrete sections withoutprestressing are analysed The section is assumed to be subjected to an axialforce N and a bending moment M of known magnitudes With the con-crete in tension ignored these forces are resisted by the concrete in compres-sion and by the reinforcement The analysis will give axial strain curvatureand corresponding stresses immediately after application of N and M andafter a period of time in which creep and shrinkage occur

Analysis of a partially prestressed section is also included in this chapterThe section is assumed to remain in state 1 (uncracked) under the effect ofprestress and loads of long duration such as the dead load After a givenperiod of time during which creep shrinkage and relaxation have occurredlive load is assumed to be applied producing cracking With this assumptionthe equations of Chapter 2 can be used to determine the stress and strainin concrete and the reinforcement at the time of prestressing and after a


specified period during which creep shrinkage and relaxation have occurredThe additional internal forces produced by the live load are assumed to pro-duce instantaneous changes in stress and strain and also cause crackingwhich reduces the effective area of the section The instantaneous changes instress and strain are calculated but no time-dependent effects are consideredIt is believed that these assumptions are not too restrictive and they representmost practical situations Other assumptions adopted in the analysis arestated in the following section

If the load which produces cracking is sustained the effects of creep andshrinkage which occur after cracking are the same as for a reinforced concretesection without prestressing

72 Basic assumptions

Concrete in the tension zone is assumed to be ineffective in resisting internalforces acting on a cracked cross-section The effective area of the cross-section is composed of the area of the compressive zone and the area ofreinforcement

Plane cross-sections are assumed to remain plane after the deformationand strains in concrete and steel are assumed to be compatible These twoassumptions are satisfied by using in the analysis the area properties of atransformed fully cracked section composed of Ac the area of the compres-sion zone and αAs where α = EsEc Es is the modulus of elasticity of thereinforcement Ec is the modulus of elasticity of concrete at the time ofapplication of the load when the analysis is concerned with instantaneousstress and strain When creep and shrinkage are considered Ec is theage-adjusted modulus (see Section 111)

Due to creep and shrinkage the depth of the compression zone changesthus Ac is time-dependent In the analysis of stress and strain changes due tocreep and shrinkage during a time interval Ac is considered a constant equalto the area of the compression zone at the beginning of the time interval Thisassumption greatly simplifies the analysis but involves negligible error

73 Sign convention

A positive bending moment M produces compression at the top fibre (Fig71(a) ) The axial force N is positive when tensile N acts at an arbitrarilychosen reference point O The eccentricity e = MN and the coordinatey of any fibre are measured downward from O Tensile stress and thecorresponding strain are positive Positive M produces positive curvature ψ

The above is a review of some of the conventions adopted throughout thisbook (see Section 22)

Stress and strain of cracked sections 209

74 Instantaneous stress and strain

Consider a concrete section reinforced by a number of layers of steel andsubjected to a bending moment M and a normal force N at an arbitrarilychosen reference point O (Fig 71(a) ) The values of M and N are suchthat the top fibre is in compression and the bottom is in tension producingcracking at the bottom face

The equations graphs and tables presented in this section and subsections741 and 742 are based on the assumption that the top fibre is in compres-sion and the bottom part of the section is cracked due to tension When thebottom part of the section is in compression and the tension zone and crack-ing are at the top the equations apply if the direction of the y-axis is reversedand all reference to the top fibre will be considered to mean the bottom fibreIn this case the flange of a T section will be at the tension zone the graphsand tables for a rectangular section (of width equal to the width of the web)will apply as long as the compression zone is a rectangle

Not included here are the situations when the stresses over all the sectionare of the same sign When all the stresses are compressive the equations foruncracked sections presented in Chapter 2 apply When all the stresses aretensile the concrete is assumed to be ineffective in the fully cracked state 2and the internal forces are resisted only by the steel In this case creep andshrinkage have no effect on the stress and strain distribution over the section

The stress and strain distributions shown in Fig 71(b) and (c) are assumedto be produced by the combined effect of M and N as shown in Fig 71(a)The resultant of M and N is located at eccentricity e given by

e = MN (71)

Figure 71 Stress (c) and strain (b) distributions in a fully cracked reinforced concretesection (a) (state 2) subjected to M and N Convention for positive M N y yn

and ys


Positive e means that the resultant is situated below the reference point OThe location of the neutral axis depends on the value of e not on theseparate values of M and N This is true in an uncracked or a fully crackedstate 1 and 2 but the depth of the compression zone is of course not thesame in the two states In the analysis presented below the area consideredeffective in resisting the internal forces is composed of the area of the com-pression zone plus the area of the reinforcement The equations given belowenable determination of the depth c of the compressive zone without timeeffect and when c is known the properties of the transformed area can bedetermined and the stress and strain calculated in the same way as for anuncracked section

The strain at any fibre (Fig 71(b) ) is

ε = εO + yψ (72)

The y-coordinate of the neutral axis is

yn = minusεOψ (73)

The stress in concrete at any fibre is

σc = Ec1 minus y

ynεO y lt yn (74)

0 y yn (75)

It may be noted that in Fig 71(b) εO is a negative quantity since O ischosen in the compression zone The stress in any steel layer at coordinateys is

σs = Es1 minus ys

ynεO (76)

Integrating the stresses over the area and taking moment about an axisthrough O gives

εOEcyn

yt1 minus

y

yn dA + Es Σ As1 minus

ys

yn = N (77)

εOEcyn

yt

y1 minus y

yn dA + Es Σ As ys1 minus

ys

yn = M (78)

where


The summations in Equations (77) and (78) are for all steel layersWhen the section is subjected to bending moment only N can be set equal

to zero in Equation (77) giving the following equation which can be solvedfor the coordinate yn defining the position of the neutral axis

yn

yt

(yn minus y)dA + α Σ [As(yn minus ys)] = 0 (79)

where α = EsEcEquation (79) indicates that when N = 0 the first moment of the trans-

formed area of the fully cracked cross-section about the neutral axis is zeroThus the neutral axis is at the centroid of the transformed fully crackedsection (the area of concrete in compression plus α times the area ofreinforcement)

When N ne 0 the neutral axis does not coincide with the centroid of thetransformed area The equation to be solved for yn is obtained by division ofEquation (78) by (77)

yn

yt

y(yn minus y)dA + α Σ [As ys(yn minus ys)]

yn

yt

(yn minus y)dA + α Σ [As(yn minus ys)]

minus e = 0 (710)

For an arbitrary cross-section the value yn that satisfies Equation (79) or(710) may be determined by trial In subsection 741 Equations (79) and(710) are applied for a cross-section in the form of a T or a rectangle

Once the position of the neutral axis is determined the properties of thetransformed fully cracked section are determined in the conventional waygiving A the area B the first moment and I the moment of inertia about anaxis through the reference point O Now the general equations of Section 23may be applied to determine εO ψ and the stress at any fibre

dA = an elemental area of concrete in compressionAs and ys = the area of steel in one layer of reinforcement and its coordinate

measured downwards from the reference point Oyt = the y-coordinate at the top fibreyn = the y-coordinate of the neutral axis

Es and Ec = the moduli of elasticity of steel and concrete


741 Remarks on determination of neutral axis position

Equations (79) or (710) can be used to determine the position of the neutralaxis and thus the depth c of compression zone for any section having avertical axis of symmetry Equation (79) applies when the section is subjectedto a moment M without a normal force Equation (710) applies when M iscombined with a normal force N

For a section of arbitrary shape a trial value of the coordinate yn of theneutral axis is assumed the integral in Equation (79) or the two integrals inEquation (710) are evaluated ignoring concrete in tension By iteration avalue yn between yt and yb is determined to satisfy one or the other of thetwo equations where yt and yb are the y coordinates of the top and bottomfibres respectively

Both Equations (79) and (710) are based on the assumption that theextreme top and bottom fibres are in compression and in tension respectivelyThus the equations apply when

σt1 0 while σb1 0 (711)

where σ is stress at concrete fibre the subscripts t and b refer to top andbottom fibres and the subscript 1 refers to state 1 in which cracking isignored When the extreme top and bottom fibres are in tension and com-pression respectively Equation (79) or (710) applies when the directionof the y-axis is reversed to point upwards and the symbol yt in the equa-tions is treated as coordinate of bottom fibre It is here assumed that atleast one of σt1 and σb1 exceeds the tensile strength of concrete causingcracking

When a section is subjected to a moment without a normal force solutionof Equation (79) gives the position of the neutral axis at the centroid of thetransformed section with concrete in tension ignored In this case the equa-tion has a solution yn between yt and the y-coordinate of the extreme tensionreinforcement However when a section is subjected to a normal force Ncombined with a moment M the neutral axis can be not within the height ofthe section in which case Equation (710) has no solution for yn that isbetween yt and yb The following are limitations on the use of Equation(710) depending upon the values of M and N It is here assumed that thecompression zone is at top fibre

(1) When N is compressive both σt1 and σb1 are compressive when

I1 minus ytB1

B1 minus ytA1

M

N

I1 minus ybB1

B1 minus ybA1

(712)

where A1 B1 and I1 are area of transformed uncracked section (state 1)


and its first moment and second moment about an axis through thereference point O (Fig 71) In this case the section is uncracked and useof Equation (710) is not needed

(2) When the section is made of plain concrete without reinforcementEquation (710) applies only when resultant force is compressive andsituated within the height of the section that is when

yt M

N yb (713)

(3) When the section has two or more reinforcement layers and the normalforce N is tensile Equation (710) applies only when

ΣIs minus yt ΣBs

ΣBs minus yt ΣAs

M

N

ΣIs minus yb ΣBs

ΣBs minus yb ΣAs

(714)

where ΣAs ΣBs and ΣIs are sum of cross-sectional areas of reinforcementlayers and their first and second moments about an axis through thereference point O (Fig 71) This inequality gives lower and upper limitsof a range of (MN) within which Equation (710) does not apply Thelower and the upper limits of the range are respectively equal to the thirdand the first terms in Equation (714) When (MN) is equal to the lowerlimit or to the upper limit the neutral axis coincides with the bottom ortop fibres respectively In other words when (MN) is within this rangeEquation (710) has no solution for yn that lies between yt and yb In thiscase the resultant tensile force is resisted entirely by the reinforcementthe strain and stress in any reinforcement layer can be determined byEquations (219) and (220) substituting ΣAs ΣBs and ΣIs for A B and Irespectively

(4) When the section has only one reinforcement layer and the normal forceN is tensile the compression zone is at top fibre and Equation (710)applies when (MN) ys where ys is the y-coordinate of the reinforce-ment layer But when (MN) lt ys the compression one is at the bottomthe direction of the y axis must be reversed the coordinate of thereinforcement layer becomes (minusys) before Equation (710) can be applied

742 Neutral axis position in a T or rectangular fullycracked section

The equations of the preceding section are applied below for a T sectionreinforced by steel layers Ans and Aprimens near the bottom and top fibres (Fig72) The section is also assumed to have one layer of prestress steel Aps

situated anywhere in the tension zone Presence of Aps simply adds an area


αpsAps to the transformed effective area where αps = EpsEc with Eps and Ec

being the moduli of elasticity of prestressed steel and concrete Theequations presented below are applicable for a rectangular section by settingbw = b

Consider the case when the section in Fig 72 is subjected to a positivebending moment without an axial force Application of Equation (79) givesthe following quadratic equation from which the depth c of the compressionzone can be determined

12bwc2 + [hf (b minus bw) + αnsAns + αpsAps + (αns minus 1) Aprimens]c

minus [12 (b minus bw)h2

f + αnsAnsdns + αpsApsdps

+ (αns minus 1) Aprimensd primens] = 0 when c hf (715)

where dns dps and d primens are distances from the extreme compression fibre to thereinforcements Ans Aps and Aprimens respectively b and bw are widths of the flangeand of the web respectively and hf is the thickness of the flange αns = EnsEcwith Ens being the modulus of elasticity of non-prestressed steel

Solution of the quadratic Equation (715) gives the depth of the compres-sion zone in a T section subjected to bending moment

c = minusa2 + radic(a2

2 minus 4a1a3)

2a1

(716)

where

a1 = bw2 (717)

Figure 72 Definition of symbols employed in Section 742


a2 = hf(b minus bw) + αnsAns + αpsAps + (αns minus 1)Aprimens (718)

a3 = minus12h

2f(b minus bw) minus αnsAnsdns minus αpsApsdps minus (αns minus 1)Aprimensdprimens (719)

When the section is subjected to a bending moment M and a normalforce N in any position the two actions may be replaced by a resultantnormal force N at the appropriate eccentricity Let es be the eccentricity ofthe resultant measured downwards from the bottom reinforcement Thus es

is a negative quantity when the resultant normal force is situated above Ans

(Fig 72) The depth of the compression zone c can be determined bysolving the following cubic equation which is derived from Equation(710)

bw(12c2)(dns minus 13c)

+ (b minus bw)hf[c(dns minus 12hf) minus 12hf(dns minus 23hf)]

+ (αns minus 1)Aprimens(c minus dprimens)(dns minus dprimens) minus αpsAps(dps minus c)(dns minus dps)

+ es[bw(12c2) + (b minus bw)hf(c minus 12hf) + (αns minus 1)Aprimens(c minus dprimens)

minus αpsAps(dps minus c) minus αnsAns(dns minus c)] = 0 when c hf (720)

Equation (720) may be conveniently solved by trial employing aprogrammable calculator A direct solution is also possible (see Appendix D)

In the derivations of Equations (715) and (720) the height c of the com-pression zone is assumed to be greater or equal to hf (Fig 72) If c lt hf thearea for the fully cracked T section in Fig 72 will be the same as that fora rectangular section of width b Equation (715) or (720) applies for arectangular section simply by setting bw = b

It should be noted that Equation (720) applies when the top fibre of the Tsection is in compression while the bottom fibre is in tension This occursonly when the normal force is tensile situated below the centroid of thetensile reinforcement (Ans plus Aps) or when the normal force is compressivesituated above approximately 07 the depth of the section

743 Graphs and tables for the properties of transformedfully cracked rectangular and T sections

Figure 73 shows a T section subjected to a bending moment or to a bendingmoment combined with an axial force that produces cracking The section isprovided with only one layer of reinforcement As in the tension zone Thegraphs and tables presented below give the depth c of the compression zonethe distance y between the extreme compression fibre and the centroid of thetransformed fully cracked section and its moment of inertia I about an axisthrough the centroid Each of c y and I depends on the dimensions of the


section and the product αAs where α = EsEc the ratio of elasticity moduli ofsteel and concrete The computer programs described in Appendix G can beused in lieu of the graphs and the tables

For the use of the graphs or the tables with a section having more than onelayer of steel in the tension zone with different elasticity moduli (as forexample in the section in Fig 72) the value αAs to be used in the graphs ortables is

αAs = ΣαiAsi (721)

and the area αAs is to be considered situated at distance d from the top edgegiven by

d = ΣαiAsidi

ΣαiAsi

(722)

When the section is subjected to bending without axial force the height cof the compression zone depends on αAs and the dimensions d hf b and bw

(Fig 73) When the section is subjected to a moment M and a normal forceN the height c is a function of the same parameters plus es where es is theeccentricity of the resultant of M and N measured downwards from thetension reinforcement (Fig 73)

The graphs in Fig 74 give the value of c for a fully cracked rectangularsection subjected to a moment and a normal force This pair of forces must be

Figure 73 Definition of symbols used in the graphs on Figs 74 to 76 and Tables 71 to 74


Figu

re7

4D

epth

of t

he c

ompr

essi

on z

one

in a

fully

cra

cked

rec

tang

ular

sec

tion

subj

ecte

d to

ecc

entr

ic n

orm

al fo

rce

replaced by statical equivalents M and N with N located at the same level asAs The resultant of the pair is thus a force N situated at a distance

es = MN (723)

where es is an eccentricity of the resultant measured downwards fromAs

The use of the graphs in Fig 74 is limited to a rectangular cracked sectionwith the compression zone at the top part of the section This occurs onlywhen N is tension and esd has a value greater than zero or when N is com-pression and esd has a value smaller than minus07 The limiting values 0 and minus07are approximate quantities which depend upon the reinforcement ratios ρ andρprime where

ρ = Asbd (724)

and

ρprime = Aprimesbd (725)

where b is the breadth of the section and d is the distance between the bottomreinforcement As and the extreme compression fibre Aprimes is the area of anadditional layer near the top situated at a distance dprime from the extreme com-pression fibre

The case of a section subjected to a positive moment with no axial force isthe same as for es = infin with N a small tensile force or es = minusinfin with N a smallcompressive force In each of the graphs in Fig 74 the curve labelled es = plusmninfinis to be used when the section is subjected to a moment without axial forceOther curves are usable when N is tension or compression

Figs 75 and 76 give the position of the centroid and the moment of inertiaof a transformed fully cracked rectangular section for which the depth c ofthe compression zone is predetermined

Tables 71 and 72 can be used for the same purpose as Fig 74 when thesection is in the form of a T In order to reduce the number of variables thetables are limited to T sections without steel in the compression zone or whenthis reinforcement is ignored The two tables naturally give the identicalresults as Fig 74 in the special case when ρprime = 0 and bwb = 1 where b and bw

are widths of flange and web respectivelyOnce the depth c of the compression zone of a fully cracked T section is

determined Tables 73 and 74 can be used to determine the centroid and themoment of inertia about an axis through the centroid of the transformedsection


Figu

re7

5D

ista

nce

from

the

top

fibr

e to

the

cen

troi

d of

the

tra

nsfo

rmed

are

a of

a fu

lly c

rack

ed r

ecta

ngul

ar s

ectio

n

Figu

re7

6M

omen

t of

iner

tia a

bout

an

axis

thr

ough

the

cen

troi

d of

the

tra

nsfo

rmed

are

a of

a fu

lly c

rack

ed r

ecta

ngul

ar s

ectio

n

Tabl

e 7

1D

epth

of c

ompr

essi

on z

one

in a

fully

cra

cked

T s

ectio

n su

bjec

ted

to e

ccen

tric

com

pres

sive

forc

e c

= (c

oeffi

cien

t fr

om t

able

times 1

0minus3)d

b wb

= 0

05

e sd

= minus

09

e sd

= minus

10

e sd

= minus

15

e sd

= minus

20

e sd

= minus

50

e sd

= minus

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

000

582

090

982

958

632

024

615

013

713

713

752

5252

5252

4343

4343

4334

3434

3434

3131

3131

310

010

828

912

836

613

339

333

214

173

170

170

7772

7272

7261

6060

6060

4848

4848

4843

4343

4343

003

085

392

385

968

340

349

037

228

324

523

916

212

312

012

012

012

410

210

210

210

291

8282

8282

7974

7474

740

050

870

931

875

725

453

567

461

361

304

278

227

164

151

151

151

177

133

129

129

129

129

105

105

105

105

111

9595

9595

007

588

693

989

176

250

462

953

443

336

331

429

220

918

418

118

123

216

815

515

515

517

113

012

712

712

714

711

611

511

511

50

100

898

946

902

788

544

671

585

488

412

345

345

249

214

204

204

278

200

178

176

176

208

153

145

145

145

180

136

131

131

131

012

590

895

191

180

957

870

362

553

145

237

338

928

424

122

622

431

722

920

019

419

424

117

516

116

016

020

915

414

614

614

60

150

916

955

919

825

606

729

656

566

487

398

426

316

266

246

242

352

256

220

210

210

271

196

176

174

174

236

172

159

158

158

020

092

896

193

185

065

276

770

362

154

344

348

637

131

228

327

141

030

425

824

023

732

323

420

519

819

828

420

518

318

018

0

b wb

= 0

1

e sd

= minus

09

e sd

= minus

10

e sd

= minus

15

e sd

= minus

20

e sd

= minus

50

e sd

= minus

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

001

064

770

065

352

433

829

220

817

317

017

077

7272

7272

6160

6060

6048

4848

4848

4343

4343

430

020

670

717

676

562

371

373

290

231

212

211

120

100

100

100

100

9384

8484

8470

6868

6868

6261

6161

610

060

731

766

734

651

467

519

452

378

325

294

242

181

165

164

164

192

147

140

140

140

143

115

114

114

114

124

103

103

103

103

010

076

879

777

070

153

159

153

446

440

334

432

224

521

320

420

426

319

817

817

617

620

015

314

514

514

517

413

513

113

113

10

150

799

824

801

742

588

649

600

535

473

397

394

308

264

246

242

329

251

220

210

210

257

194

176

174

174

226

171

159

158

158

020

082

284

382

477

263

068

964

558

652

544

044

835

930

828

227

138

229

725

624

023

730

423

120

419

819

826

920

318

318

018

00

250

840

859

841

794

663

720

679

624

566

477

491

401

346

315

296

424

335

289

268

260

343

263

230

219

218

306

232

206

199

199

030

085

487

185

581

269

074

470

765

560

050

952

743

737

934

431

846

036

931

929

328

037

729

225

423

923

633

825

922

721

721

60

400

875

890

876

840

731

780

748

702

651

560

583

496

435

395

358

518

426

370

338

315

434

343

297

275

267

392

306

266

249

245

Tabl

e 7

1co

ntin

ued

b wb

= 0

2

e sd

= minus

09

e sd

= minus

10

e sd

= minus

15

e sd

= minus

20

e sd

= minus

50

e sd

= minus

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

002

054

256

954

648

536

831

826

922

721

121

111

610

010

010

010

092

8484

8484

7068

6868

6862

6161

6161

004

058

260

458

453

341

739

535

130

427

226

017

714

213

713

713

714

111

711

611

611

610

794

9494

9493

8585

8585

012

066

968

467

063

453

453

450

245

941

736

531

826

123

222

122

126

621

519

519

119

120

916

815

815

715

718

414

914

314

314

30

200

717

730

718

688

599

604

577

539

499

436

400

340

302

281

271

343

285

253

240

237

277

225

203

198

198

247

199

183

180

180

030

075

776

875

873

265

465

963

660

356

650

047

041

036

834

031

841

235

031

229

128

034

128

225

123

923

630

725

122

521

721

60

400

785

794

786

764

693

698

678

648

614

548

521

463

419

388

357

464

402

360

334

315

390

328

292

274

267

354

294

263

248

245

050

080

781

580

878

772

372

870

968

265

058

656

150

546

142

839

150

544

340

037

134

543

036

732

830

629

339

433

129

527

727

00

600

824

831

825

806

746

751

734

709

679

618

593

540

496

462

421

538

478

434

404

372

464

400

359

334

316

427

363

324

303

291

080

085

085

685

083

478

278

777

275

072

366

664

459

455

251

847

259

253

449

045

841

951

945

541

238

335

748

141

637

434

932

8

b wb

= 0

5

e sd

= minus

09

e sd

= minus

10

e sd

= minus

15

e sd

= minus

20

e sd

= minus

50

e sd

= minus

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

005

047

248

047

345

841

833

532

130

328

827

817

415

715

115

115

114

413

112

912

912

911

410

510

510

510

510

195

9595

950

100

525

531

526

513

477

408

396

381

364

342

242

221

209

204

204

205

185

177

176

176

165

149

145

145

145

147

133

131

131

131

030

063

263

663

262

359

554

353

552

350

848

038

336

134

333

031

833

731

429

728

628

028

225

924

523

823

625

623

422

121

721

60

500

687

690

688

680

656

611

604

594

581

555

460

440

422

407

388

413

390

372

358

344

353

329

311

300

293

324

299

283

274

270

075

073

273

573

372

670

566

666

065

164

061

552

650

648

947

445

247

945

643

842

340

441

639

237

336

034

638

536

034

233

031

91

000

764

766

764

758

740

704

699

691

681

658

573

555

538

523

500

527

506

488

472

451

464

440

421

406

389

431

407

388

374

360

125

078

879

078

878

376

673

372

872

171

269

161

059

357

756

253

956

554

552

751

249

050

247

946

044

542

646

944

542

641

139

41

500

807

808

807

802

786

756

752

745

737

717

639

623

608

594

571

596

577

560

545

522

534

511

492

477

457

501

477

458

443

424

200

083

583

683

583

181

779

178

778

177

475

668

567

165

764

462

264

462

761

159

757

458

456

354

553

050

855

252

951

049

547

4

Tabl

e 7

1co

ntin

ued b w

b =

10

e sd

minus09

minus10

minus15

minus20

minus50

minusinfin

010

044

834

020

417

614

513

10

200

507

412

271

237

198

180

060

062

054

640

836

731

629

11

000

678

614

483

441

385

358

150

072

466

854

650

444

741

72

000

757

706

592

550

493

463

250

078

173

562

758

753

050

03

000

801

758

656

617

560

530

400

083

079

270

066

360

957

9

Tabl

e 7

2D

epth

of t

he c

ompr

essi

on z

one

in a

fully

cra

cked

T s

ectio

n su

bjec

ted

to e

ccen

tric

ten

sile

forc

e c

= (c

oeffi

cien

t fr

om t

able

times 1

0minus3)d

b wb

= 0

05

e sd

= 1

e sd

= 5

e sd

= 1

0e s

d =

20

e sd

= 5

0e s

d =

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

000

59

99

99

1818

1818

1822

2222

2222

2525

2525

2528

2828

2828

3131

3131

310

010

1313

1313

1325

2525

2525

3131

3131

3135

3535

3535

4040

4040

4043

4343

4343

003

023

2323

2323

4343

4343

4353

5353

5353

6261

6161

6171

6868

6868

7974

7474

740

050

2929

2929

2956

5656

5656

7168

6868

6885

7878

7878

9987

8787

8711

195

9595

950

075

3636

3636

3671

6868

6868

9283

8383

8311

295

9595

9513

010

610

610

610

614

711

611

511

511

50

100

4242

4242

4286

7979

7979

112

9595

9595

137

110

109

109

109

159

123

121

121

121

180

136

131

131

131

012

546

4646

4646

9988

8888

8813

110

610

610

610

616

012

312

112

112

118

613

913

413

413

420

915

414

614

614

60

150

5151

5151

5111

396

9696

9614

911

711

611

611

618

113

713

213

213

221

015

514

614

614

623

617

215

915

815

80

200

5959

5959

5913

811

011

011

011

018

213

713

313

313

322

016

215

115

115

125

418

516

816

716

728

420

518

318

018

0

b wb

= 0

1

e sd

= 1

e sd

= 5

e sd

= 1

0e s

d =

20

e sd

= 5

0e s

d =

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

001

013

1313

1313

2525

2525

2531

3131

3131

3535

3535

3540

4040

4040

4343

4343

430

020

1818

1818

1835

3535

3535

4343

4343

4350

5050

5050

5656

5656

5662

6161

6161

006

032

3232

3232

6261

6161

6179

7474

7474

9585

8585

8511

095

9595

9512

410

310

310

310

30

100

4242

4242

4285

7979

7979

111

9595

9595

134

110

109

109

109

155

123

121

121

121

174

135

131

131

131

015

051

5151

5151

111

9696

9696

145

117

116

116

116

175

136

132

132

132

202

154

146

146

146

226

171

159

158

158

020

059

5959

5959

135

110

110

110

110

176

137

133

133

133

211

161

151

151

151

242

183

168

167

167

269

203

183

180

180

025

067

6565

6565

156

124

122

122

122

203

156

147

147

147

243

184

168

168

168

276

209

188

185

185

306

232

206

199

199

030

075

7171

7171

176

137

133

133

133

228

174

161

160

160

271

205

185

182

182

307

234

207

201

201

338

259

227

217

216

040

090

8282

8282

213

162

152

152

152

272

206

186

183

183

320

244

215

208

207

359

277

242

229

228

392

306

266

249

245

Tabl

e 7

2co

ntin

ued

b wb

= 0

2

e sd

= 1

e sd

= 5

e sd

= 2

0e s

d =

10

e sd

= 5

0e s

d =

infin

h fd

h fd

h fd

h fd

h f d

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

002

018

1818

1818

3535

3535

3543

4343

4343

5050

5050

5056

5656

5656

6261

6161

610

040

2626

2626

2650

5050

5050

6261

6161

6173

7070

7070

8378

7878

7893

8585

8585

012

045

4545

4545

9486

8686

8612

110

410

410

410

414

512

011

911

911

916

513

513

213

213

218

414

914

314

314

30

200

5959

5959

5913

011

011

011

011

016

713

613

313

313

319

716

015

115

115

122

418

116

816

716

724

719

918

318

018

00

300

7571

7171

7116

713

713

313

313

321

217

116

116

016

024

920

118

518

218

228

122

820

620

120

130

725

122

521

721

60

400

8982

8282

8219

916

115

215

215

225

120

218

518

318

329

223

721

420

820

732

626

824

022

922

835

429

426

324

824

50

500

102

9292

9292

227

183

170

169

169

283

230

208

203

203

328

269

241

230

229

364

302

270

255

251

394

331

295

277

270

060

011

510

010

010

010

025

220

318

618

418

431

225

522

922

122

035

929

726

625

224

839

633

329

727

927

142

736

332

430

329

10

800

138

116

115

115

115

295

240

217

210

210

361

299

268

253

250

410

346

309

290

281

449

384

344

322

306

481

416

374

349

328

b wb

= 0

5

e sd

= 1

e sd

= 5

e sd

= 1

0e s

d =

20

e sd

= 5

0e s

d =

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

005

029

2929

2929

5656

5656

5670

6868

6868

8178

7878

7892

8787

8787

101

9595

9595

010

042

4242

4242

8279

7979

7910

295

9595

9511

910

910

910

910

913

412

212

112

112

114

713

313

113

113

10

300

7371

7171

7114

913

513

313

313

318

416

616

116

016

021

319

218

418

218

223

621

520

420

120

125

623

422

121

721

60

500

9792

9292

9219

617

717

016

916

923

921

720

620

320

327

324

923

623

022

930

027

626

125

325

132

429

928

327

427

00

750

122

112

111

111

111

240

218

207

204

204

290

266

251

244

243

328

303

287

278

273

359

334

317

306

298

385

360

342

330

319

100

014

312

912

812

812

827

625

223

923

323

233

130

528

928

027

537

234

632

931

730

840

537

936

134

833

643

140

738

837

436

01

250

161

145

142

142

142

306

282

267

259

256

365

339

321

310

302

408

382

364

351

339

442

417

398

384

369

469

445

426

411

394

150

017

716

015

515

515

533

330

829

128

227

739

436

834

933

732

743

941

339

438

036

647

344

842

941

539

850

147

745

844

342

42

000

206

186

178

177

177

378

352

334

322

313

442

416

397

383

369

489

464

444

429

412

524

500

481

466

446

552

529

510

495

474

Tabl

e 7

2co

ntin

ued b w

b =

10

e sd

15

10

20

50

infin

010

042

7995

109

121

131

020

059

110

133

151

167

180

060

010

018

422

024

827

129

11

000

128

232

275

308

335

358

150

015

527

732

636

439

341

72

000

177

313

367

406

438

463

250

019

734

440

044

247

450

03

000

214

370

429

472

504

530

400

024

341

447

652

055

357

9

Tabl

e 7

3D

ista

nce

from

top

fibr

e to

cen

troi

d of

tra

nsfo

rmed

1 fully

cra

cked

T s

ectio

n y

= (c

oeffi

cien

t fr

om t

able

times 1

0minus3)d

b wb

= 0

05

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

000

536

5454

5454

4659

7910

210

262

6785

105

151

108

9510

311

915

918

414

513

914

717

727

220

718

718

520

30

010

4559

5959

5954

6382

104

104

6972

8710

815

211

499

106

121

160

188

148

142

149

178

276

210

189

186

204

003

079

7777

7777

8581

9411

311

398

8899

116

158

138

113

116

129

166

207

160

151

156

183

290

221

197

193

209

005

011

195

9595

9511

497

105

121

121

125

104

110

125

163

160

127

127

137

171

224

173

160

163

188

304

231

206

199

213

007

514

811

611

611

611

614

911

711

913

213

215

612

312

313

517

018

614

513

914

717

724

518

717

117

219

432

024

421

520

821

90

100

182

136

136

136

136

180

136

133

142

142

185

141

136

145

177

211

161

151

156

184

265

202

182

181

200

336

256

225

215

225

012

521

415

515

515

515

520

915

514

615

215

221

215

914

915

518

423

417

716

316

519

028

421

519

318

920

635

126

823

522

323

00

150

243

173

173

173

173

237

172

159

162

162

237

176

161

164

190

256

192

174

174

196

302

229

203

198

212

366

279

244

231

236

020

029

520

820

820

820

828

620

518

318

118

128

420

718

518

320

329

622

119

719

220

933

525

422

321

422

439

330

126

224

624

7

b wb

= 0

1

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

001

046

5959

5959

6267

8410

410

487

8294

111

152

152

127

127

136

168

250

202

186

184

201

357

290

259

245

246

002

063

6868

6868

7675

8910

810

899

9099

115

155

161

133

131

140

170

256

207

189

187

203

361

294

262

248

248

006

012

510

310

310

310

312

810

711

212

612

614

311

912

013

116

619

415

714

915

418

027

922

520

419

921

237

830

827

425

825

60

100

178

136

136

136

136

174

137

134

142

142

183

146

140

147

177

224

179

167

168

190

301

243

219

211

221

394

322

286

268

264

015

023

717

317

317

317

322

617

115

916

216

222

917

716

416

619

026

020

618

818

520

232

726

423

622

623

241

333

930

028

127

40

200

288

208

208

208

208

272

203

183

181

181

269

207

187

184

203

292

231

207

201

214

351

284

253

240

242

431

354

314

293

283

025

033

224

024

024

024

031

223

320

619

919

930

623

420

820

221

532

125

422

621

722

637

430

326

925

425

344

736

932

730

429

20

300

372

269

269

269

269

348

260

227

217

217

339

259

228

218

227

348

276

244

232

237

394

321

284

267

262

463

384

340

316

301

040

043

832

132

132

132

141

030

926

725

025

039

630

626

625

025

039

731

627

826

125

843

235

431

229

128

249

241

036

433

731

8

1T

he s

ame

tabl

e m

ay b

e us

ed fo

r ag

e-ad

just

ed t

rans

form

ed s

ectio

n re

plac

ing

b

y ndash

= E

s[1 +

(t

t0)

]E c

(t0)

Tabl

e 7

3co

ntin

ued

b wb

= 0

2

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

002

064

6868

6868

8581

9210

810

811

710

510

912

115

519

717

016

216

418

430

826

824

723

724

042

537

534

432

531

20

040

9386

8686

8610

796

103

117

117

134

118

119

129

161

208

179

169

170

188

315

274

252

242

244

430

380

348

329

315

012

019

415

115

115

115

118

415

114

515

015

019

616

415

615

918

225

021

319

819

420

734

229

827

326

125

944

839

736

434

432

70

200

274

208

208

208

208

250

200

183

181

181

250

206

189

187

203

287

244

224

217

224

367

320

293

279

273

465

413

379

357

339

030

035

526

926

926

926

931

825

322

621

721

730

725

222

821

922

732

928

025

524

424

539

634

731

730

029

148

543

239

737

435

30

400

419

321

321

321

321

375

299

264

250

250

357

294

263

250

250

366

313

284

269

265

422

371

339

320

307

503

449

413

389

366

050

047

236

636

636

636

642

334

129

927

927

939

933

129

527

727

139

934

331

129

328

444

639

336

033

932

352

046

642

940

437

90

600

516

406

406

406

406

464

377

331

307

307

437

365

324

303

291

429

370

335

315

302

468

414

379

357

338

536

482

444

419

391

080

058

547

247

247

247

253

143

938

735

735

749

942

237

634

932

848

141

938

035

533

550

845

241

439

036

656

551

147

244

541

5

b wb

= 0

5

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h f d

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

005

010

195

9595

9512

011

211

412

112

115

614

614

414

716

324

322

922

121

822

236

134

433

232

431

748

246

344

943

842

30

100

154

136

136

136

136

152

140

138

142

142

179

166

162

163

177

256

241

233

229

231

368

352

339

331

323

487

468

454

442

428

030

031

526

926

926

926

926

223

622

221

721

725

923

922

722

322

730

528

727

626

926

739

838

036

735

734

750

548

747

246

044

40

500

424

366

366

366

366

346

312

291

279

279

324

299

284

274

271

348

328

314

306

299

424

406

392

382

370

522

504

489

476

460

075

052

045

745

745

745

742

838

836

234

534

539

236

334

333

032

039

437

335

734

733

645

543

642

141

039

654

252

450

849

647

91

000

589

524

524

524

524

491

449

420

400

400

447

416

394

378

362

434

412

395

383

369

482

463

447

435

420

560

542

526

514

496

125

064

057

755

755

757

754

250

046

844

644

649

346

143

741

939

947

044

742

941

539

950

748

747

145

944

357

755

954

353

151

21

500

680

620

620

620

620

584

541

509

485

485

532

500

474

456

433

501

477

459

444

427

529

509

493

480

463

593

574

559

546

528

200

073

868

368

368

368

364

860

757

454

954

959

456

253

651

649

055

352

951

049

547

456

854

853

251

950

062

160

358

757

455

5

Tabl

e 7

3co

ntin

ued b w

b =

10

cd

12

35

75

10

010

013

614

217

726

438

350

40

200

208

181

203

278

391

509

060

040

630

729

133

042

152

81

000

524

400

362

375

448

545

150

061

948

543

342

347

956

52

000

683

549

489

464

506

583

250

072

859

953

650

053

159

93

000

762

640

574

531

553

615

400

080

969

963

558

359

264

2

Tabl

e 7

4M

omen

t of

iner

tia o

f a t

rans

form

ed fu

lly c

rack

ed T

sec

tion

abou

t ce

ntro

idal

axi

s I =

(coe

ffici

ent

from

tab

le times

10minus4

)bd3

b wb

= 0

05

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

000

54

55

55

55

710

107

78

1126

1718

1820

3246

5050

5159

9911

111

411

411

80

010

99

99

99

1011

1414

1111

1215

2921

2222

2435

5054

5454

6210

211

511

711

712

10

030

2727

2727

2727

2728

3030

2828

2931

4336

3838

3949

6268

6869

7511

212

713

013

113

40

050

4343

4343

4343

4344

4646

4445

4547

5851

5353

5463

7582

8383

8912

213

914

314

314

60

075

6263

6363

6362

6364

6565

6264

6465

7568

7272

7280

8999

100

100

105

133

153

159

159

161

010

079

8282

8282

7982

8383

8379

8383

8492

8490

9090

9710

311

517

711

712

114

516

717

417

517

60

125

9510

110

110

110

195

101

101

101

101

9610

110

210

210

999

107

108

108

114

116

130

133

134

137

155

181

189

190

191

015

011

011

811

811

811

811

011

811

911

911

911

111

811

912

012

511

312

312

512

513

012

914

514

915

015

316

619

420

320

520

60

200

137

151

151

151

151

138

151

153

153

153

138

151

154

154

157

139

155

158

159

162

152

174

180

181

183

185

219

231

234

235

b wb

= 0

1

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

001

09

99

99

1010

1114

1413

1314

1629

2931

3132

4175

8587

8791

157

185

195

197

198

002

018

1818

1818

1919

2022

2221

2222

2436

3639

3940

4880

9194

9497

161

190

200

203

204

006

051

5151

5151

5152

5253

5352

5454

5564

6368

6869

7510

211

612

012

012

217

720

922

222

522

60

100

7982

8282

8280

8283

8383

8084

8484

9288

9697

9710

212

213

914

414

514

719

222

824

224

724

80

150

111

118

118

118

118

111

118

119

119

119

112

119

120

120

125

117

128

131

131

134

145

167

174

176

177

210

250

267

273

275

020

013

815

115

115

115

114

015

115

315

315

314

015

115

415

415

714

315

916

316

316

516

719

320

320

520

622

727

229

129

930

10

250

162

181

181

181

181

165

182

186

186

186

165

182

186

186

189

167

187

193

194

196

188

218

230

233

234

242

292

314

323

326

030

018

320

920

920

920

918

721

021

621

721

718

821

021

721

821

918

921

422

222

422

520

624

225

626

126

225

731

133

634

735

10

400

218

258

258

258

258

225

261

273

276

276

228

261

273

276

277

229

264

277

281

282

241

286

305

313

315

284

348

378

392

398

Tabl

e 7

4co

ntin

ued

b wb

= 0

2

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h f d

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

002

018

1818

1818

1919

2022

2224

2425

2636

4853

5354

5911

813

614

314

514

624

829

031

332

332

70

040

3535

3535

3536

3636

3838

3940

4041

5161

6667

6772

128

147

155

157

158

255

298

321

332

337

012

093

9797

9797

9497

9798

9895

9910

010

010

510

811

812

112

112

316

418

819

820

220

228

032

835

436

737

40

200

140

151

151

151

151

143

151

153

153

153

143

152

154

154

157

151

165

170

171

173

197

226

239

244

245

303

356

386

401

409

030

018

720

920

920

920

919

421

121

721

721

719

521

121

721

821

919

922

022

823

023

123

527

028

729

529

733

139

042

344

145

20

400

224

258

258

258

258

236

263

273

276

276

239

264

274

276

277

241

269

281

285

286

270

311

332

342

346

356

421

458

479

493

050

025

530

130

130

130

127

231

032

533

033

027

831

132

533

033

227

931

433

033

733

930

234

937

538

739

338

045

049

251

553

20

600

280

339

339

339

339

303

351

372

380

380

312

354

373

381

383

313

355

376

385

389

331

385

414

429

438

402

478

524

550

570

080

032

040

140

140

140

135

442

045

346

946

936

842

745

747

147

837

242

945

947

448

138

344

948

750

852

144

352

858

261

464

1

b wb

= 0

5

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

005

043

4343

4343

4445

4546

4651

5252

5358

9299

101

102

103

225

244

255

261

264

483

523

551

569

585

010

081

8282

8282

8283

8383

8385

8888

8892

120

128

131

132

133

245

265

278

284

287

497

537

566

584

602

030

019

720

920

920

920

920

721

421

721

721

720

721

521

821

821

922

423

624

224

524

532

134

636

137

037

654

759

262

364

566

50

500

276

301

301

301

301

303

319

327

330

330

307

321

328

331

332

314

332

341

346

348

390

419

438

449

458

595

643

677

701

725

075

034

438

738

738

738

739

642

444

044

844

841

043

344

645

245

541

343

745

145

946

446

850

352

654

055

364

970

274

076

779

61

000

394

452

452

452

452

469

508

532

546

546

493

525

545

556

564

498

529

548

560

568

539

578

606

624

640

699

756

798

828

861

125

043

150

250

250

250

252

757

760

962

962

956

360

463

064

665

957

361

063

565

066

360

364

767

970

072

074

680

785

288

592

21

500

459

542

542

542

542

575

634

674

700

700

622

671

704

725

744

639

682

712

731

749

661

710

746

770

795

789

853

902

938

980

200

050

160

260

260

260

264

872

477

981

681

671

778

182

685

788

975

080

584

487

189

976

282

186

489

592

986

693

899

310

3410

85

Tabl

e 7

4co

ntin

ued

b wb

= 1

0

cd

12

35

75

10

010

082

8392

159

390

858

020

015

115

315

721

242

788

20

600

339

380

383

405

568

974

100

045

254

656

457

269

610

601

500

542

700

744

753

839

1159

200

060

281

688

990

796

812

492

500

645

906

1007

1041

1083

1333

300

067

797

811

0611

5811

8814

104

000

722

1086

1261

1354

1370

1547

Example 71 Cracked T section subjected to bending

The T section shown in Fig 77(a) is subjected to a bending moment of1000kN-m (8850kip-in) It is required to find the stress and strain dis-tributions ignoring the concrete in tension Effects of creep and shrink-age are not considered in this example The cross-section dimensionsare indicated in Fig 77(a) Ec = 30 GPa (4350 ksi) Es = 200 GPa(29000ksi)

Figure 77 Strain and stress distributions in a fully cracked reinforced concretesection (Examples 71 and 72) (a) cross-section dimensions (b) effectivearea strain and stress due to bending (Example 71) (c) effective areastrain and stress due to bending and normal force (Example 72)


α = Es

Ec

= 200

30 = 6667

In the absence of prestress steel Aps = 0 and the symbols Ans and Aprimens

have the same meaning as As and AprimesSubstitution in Equations (717ndash19) gives

a1 = 015m a2 = 17407 times 10minus3 m2 a3 = minus40812 times 10minus3 m3

Equation (716) gives the depth of the compression zone

c = minus17407 times 10minus3 +radic[(17407 times 10minus3)2 + 4(015)(40812 times 10minus3)]

2 times 015

= 0200m (79 in)

The moment of inertia of the transformed section about the centroi-dal axis (which is the same as the neutral axis)

I = 0302003

3 + (15 minus 03)0120122

12 + 0142

+ 6667(0004)10002 + 5667(00006)0152

= 3054 times 10minus3 m4 (353 ft4)

Alternatively if Aprimes is ignored Tables 71 and 74 can be used giving c= y = 0202m and I = 3046 times 10minus3 m4 The curvature

ψ = 1000 times 103

30 times 109 times 3054 times 10minus3 = 1091 times 10minus6 mminus1 (28 times 10minus6 inminus1)

Stress at the top fibre = 30 times 109 times 1091 times 10minus6(minus0200)

= minus655 MPa (minus0950ksi)

Stress in steel = 200 times 109 times 1091 times 10minus6(1000)

= 2182 MPa (3165ksi)

Strain and stress distributions are shown in Fig 77(b)


Example 72 Cracked T section subjected to M and N

Solve Example 71 assuming that the section is subjected to a bendingmoment of 1000kN-m (8850kip-in) and a normal force of minus800kN(minus180kip) at a point 10m (40 in) below the top edge of the section Thecross-section dimensions and moduli of elasticity of steel and concreteare the same as in Example 71 (Fig 77(a) )

The resultant force on the section is a normal force of minus800kN at adistance 025m above the top edge Thus es = minus(025 + 120) = minus145mSubstituting in Equation (720) and solving for c the height of thecompression zone gives

c = 0444m (175 in)

The effective area is shown in Fig 77(c) The transformed section iscomposed of the area of concrete in compression plus α(As + Aprimes) with α= 20030 = 6667 The distance between point O the centroid of thetransformed section and the top edge is calculated to be y = 0229m(Fig 77(c) ) The area and moment of inertia of the transformedsection about an axis through its centroid

A = 03073m2 I = 3173 times 10minus3 m4

If Aprimes is ignored Tables 71 73 and 74 may be used giving

c = 046m y = 024m I = 30 times 10minus3 m4

Transform the given bending moment and normal force into anequivalent system of a normal force N at the centroid of the trans-formed section combined with a bending moment M

N = minus800kN

M = 1000 times 103 minus 800 times 103(1000 minus 0229)

= 3832kN m (3400kip in)

The strain at O and the curvature (Equation (216) )

εO = 1

30 times 109 minus800 times 103

03073 = minus87 times 10minus6


ψ = 1

30 times 109 3832 times 103

3173 times 10minus3

= 403 times 10minus6 mminus1 (102 times 10minus6 inminus1)

Stress at the top fibre = 30 times 109[minus87 + 403(minus0229)]10minus6 = minus538 MPaStress in bottom steel = 200 times 109[minus87 + 403 times 0971]10minus6 = 608 MPa

The strain and stress distributions are shown in Fig 77(c)

75 Effects of creep and shrinkage on a reinforcedconcrete section without prestress

Consider a cross-section cracked due to the application of a positive bendingmoment M and an axial tensile (or compressive) force N at an arbitrarilychosen point O (Fig 71(a) ) The internal forces M and N are assumed tohave been introduced at age t0 The instantaneous strain and stress distribu-tions immediately after application of M and N are assumed to be available(see Section 74) It is required to find the changes in strain and in stress dueto creep and shrinkage occurring between t0 and t where t gt t0

In a fully cracked section only the part of the concrete area subjected tocompression is considered effective in resisting the internal forces Creep andshrinkage generally result in a shift of the neutral axis towards the bottom ofthe section Thus to be strictly consistent the effective area of the cross-section must be modified according to the new position of the neutral axisHowever this would hamper the validity of the superposition involved in theanalysis To avoid this difficulty the effective area of the cracked section isassumed to be unchanged by creep or shrinkage The error resulting from thisassumption can be assessed at the end of the analysis and corrected by iter-ation procedure But because the error is usually small the iteration is hardlyjustified

With the above simplification the analysis for the changes in axial strainand in curvature and the corresponding stresses can be done by the proceduregiven in Section 252 The resulting equations are given in Section 34 andrepeated here

A reference point O is chosen at the centroid of the age-adjusted trans-formed section composed of the area of the compression zone plus (t t0)times the area of steel (Figs 78 and 79(a) ) where (t t0) = EsEc(t t0) withEc(t t0) the age-adjusted modulus of elasticity of concrete (see Equation(131) ) Creep and shrinkage produce the following changes in axial strain atO in curvature and in stresses

∆εO = η[φ(t t0)(εO + ψyc) + εcs(t t0)] (726)


∆ψ = κφ(t t0) ψ + εO

yc

r2c + εcs(t t0)

yc

r2c (727)

∆σc = Ec(t t0)[minusφ(t t0)(εO + ψy) minus εcs(t t0) + ∆εO + ∆ψy] (728)

∆σs = Es(∆εO + ∆ψys) (729)

where

r2c = IcAc (730)

with Ac and Ic being the area of the compression zone and its moment ofinertia about an axis through O

Figure 78 Curvature reduction κ for a fully cracked rectangular section

εO ψ = the axial strain at O and the curvature at time t0 immediately afterapplication of M and N (Fig 79(b) )

φ(t t0) = coefficient for creep at time t for age at loading t0

εcs(t t0) = the shrinkage that would occur in concrete if it were free duringthe period (t minus t0)

yc = the y-coordinate of the centroid of the concrete area in compres-sion (based on the stress distribution at age t0) yc is measureddownwards from O


Figu

re7

9A

naly

sis

of c

hang

es in

str

ain

and

stre

ss d

ue t

o cr

eep

and

shri

nkag

e of

a fu

lly c

rack

ed r

einf

orce

d co

ncre

te s

ectio

n (E

xam

ple

73)

(a

) effe

ctiv

e ar

ea (

b) s

trai

n an

d st

ress

dis

trib

utio

ns a

t t0

(c) c

hang

es in

str

ain

and

stre

ss d

ue to

cre

ep a

nd s

hrin

kage

(d)

str

ain

and

stre

ss a

t t

Tabl

e 7

5C

urva

ture

red

uctio

n fa

ctor

κ fo

r a

fully

cra

cked

T s

ectio

n (fo

r us

e in

Equ

atio

n (6

23)

) κ =

coe

ffici

ent

from

tab

le times

10minus3

b wb

= 0

05

cd

= 1

0c

d=

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

ndash

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

000

543

160

160

160

160

170

236

420

623

623

408

426

522

661

862

774

773

781

809

891

928

927

927

929

942

973

971

971

971

973

001

036

9494

9494

105

140

270

455

455

266

276

356

496

758

636

632

643

681

804

868

865

864

868

892

948

945

944

943

947

003

059

5858

5858

8075

126

229

229

146

135

171

258

516

394

381

387

425

582

700

691

686

691

736

865

857

852

851

860

005

090

6565

6565

9874

9816

516

513

611

112

818

539

731

228

928

931

846

160

058

457

658

063

080

178

878

077

778

90

075

127

8181

8181

128

8692

134

134

149

110

112

149

315

274

240

233

252

372

524

500

488

489

539

741

721

710

705

717

010

016

110

010

010

010

015

710

196

123

123

170

118

110

134

267

262

219

206

217

318

479

447

431

429

474

697

670

655

649

660

012

519

311

811

811

811

818

611

810

312

012

019

213

011

412

923

726

321

119

319

828

245

141

239

238

842

766

363

161

360

461

40

150

223

136

136

136

136

213

134

112

121

121

215

143

121

128

218

270

210

187

187

257

435

389

366

358

392

637

600

579

568

576

020

027

617

117

117

117

126

216

713

313

013

025

817

113

813

519

629

221

918

718

022

842

036

333

332

134

460

255

652

951

551

8

b wb

= 0

1

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

ndash

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

001

041

9494

9494

160

179

286

455

455

387

387

430

530

758

758

759

760

771

834

925

925

924

924

930

973

972

971

971

971

002

047

6363

6363

107

111

175

300

300

256

250

281

366

613

618

617

616

631

717

863

863

860

860

870

949

947

945

944

944

006

010

271

7171

7111

184

9814

914

916

413

914

418

335

839

037

737

038

046

969

669

168

468

169

786

986

385

785

485

30

100

156

100

100

100

100

149

103

9712

312

317

413

412

514

526

732

430

028

629

036

060

259

057

957

358

980

979

879

078

478

20

150

215

136

136

136

136

198

133

113

121

121

204

150

130

135

218

301

265

245

241

292

535

516

499

490

501

754

738

725

717

712

020

026

717

117

117

117

124

316

413

313

013

023

917

214

313

919

630

325

723

022

125

649

947

145

143

844

471

469

367

766

665

90

250

313

203

203

203

203

284

193

154

142

142

273

196

160

148

186

314

260

228

214

237

479

445

420

405

405

686

660

640

626

616

030

035

323

323

323

323

332

122

117

415

715

730

521

917

716

018

433

026

823

221

422

646

942

940

138

337

866

463

461

259

658

30

400

421

288

288

288

288

385

271

214

186

186

362

265

213

186

189

365

292

249

224

220

466

417

382

359

345

638

601

573

553

534

Tabl

e 7

5co

ntin

ued

b wb

= 0

2

cd

= 1

0c

d =

20

cd

= 3

0c

d =

75

cd

= 5

0c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

ndash

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

002

049

6363

6363

150

148

194

300

300

356

355

366

417

613

733

741

739

741

775

919

921

921

920

921

973

973

972

971

971

004

071

6161

6161

114

103

124

189

189

241

233

238

274

448

590

596

593

594

637

853

857

856

854

855

949

948

947

945

944

012

016

911

411

411

411

415

311

510

512

012

018

515

814

615

424

237

837

336

335

939

168

468

668

167

667

586

986

786

385

985

40

200

251

171

171

171

171

214

157

132

130

130

215

174

152

147

196

329

313

296

287

305

595

592

583

575

571

811

807

800

794

786

030

033

423

323

323

323

328

221

017

215

715

726

320

917

816

418

432

229

527

325

826

353

652

751

450

249

376

075

274

273

372

20

400

400

288

288

288

288

340

257

211

186

186

310

247

208

187

189

336

301

273

254

248

507

492

475

461

446

723

712

700

689

674

050

045

533

533

533

533

539

030

024

621

621

635

328

323

821

220

035

631

528

326

024

649

347

445

443

741

769

868

466

965

663

80

600

500

376

376

376

376

433

338

280

244

244

391

316

268

236

215

378

332

297

271

250

489

466

443

423

400

680

663

647

632

611

080

057

144

544

544

544

550

340

333

829

529

545

737

632

128

324

742

437

133

029

926

749

646

643

841

538

465

963

861

759

957

3

b wb

= 0

5

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

ndash

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

005

074

6565

6565

135

130

135

165

165

305

309

307

316

397

691

701

702

701

707

909

912

912

911

972

972

972

972

971

010

012

310

010

010

010

012

411

411

012

312

321

521

320

921

226

754

255

255

355

155

683

784

284

384

284

194

794

794

794

694

50

300

287

233

233

233

233

211

184

166

157

157

206

192

180

172

184

354

356

352

347

344

661

667

667

665

660

866

866

865

864

861

050

040

133

533

533

533

529

626

023

421

621

625

923

822

020

720

032

432

131

330

529

657

658

057

957

556

780

880

880

780

479

90

750

500

429

429

429

429

382

340

307

283

283

325

298

276

258

239

335

326

315

304

289

525

526

523

517

506

758

758

755

752

744

100

057

250

050

050

050

044

940

536

934

134

138

235

232

730

627

936

034

733

432

130

250

350

249

649

047

672

472

371

971

570

51

250

625

556

556

556

556

504

458

421

391

391

431

400

372

349

318

388

374

358

344

321

496

493

486

478

462

701

699

694

689

678

150

066

760

060

060

060

054

950

346

543

443

447

444

141

338

835

341

740

038

436

834

349

749

248

447

545

868

568

267

767

165

92

000

727

667

667

667

667

618

574

536

504

504

542

509

480

454

415

470

451

433

415

387

512

504

494

484

463

669

664

658

650

636

Tabl

e 7

5co

ntin

ued

b wb

= 1

0

cd

ndash

12

35

75

10

010

010

012

326

766

090

297

10

200

171

130

196

510

826

945

060

037

624

421

533

664

686

31

000

500

341

279

318

563

805

150

060

043

435

333

751

575

52

000

667

504

415

367

497

722

250

071

455

846

639

949

369

93

000

750

602

510

431

496

685

400

080

066

857

948

751

567

0

η and κ are axial strain and curvature reduction factors given by

η = AcA (731)

κ = IcI (732)

where A and I are the area and moment of inertia about an axis through O ofan age-adjusted transformed section composed of Ac plus α(t t0)As Thecoefficient η and κ represent the restraining effect of the reinforcement on theaxial strain and curvature due to creep and shrinkage

Figure 78 and Table 75 give the values of κ for fully cracked rectangularand T sections respectively Use of Fig 78 and Table 75 must be precededby determination of c (from Fig 74 and Table 71 or 72) Location of Othe centroid of the age-adjusted transformed fully cracked section may bedetermined by the graphs of Fig 75 or Table 73 replacing α by α

Provided that the depth c is known calculation of the axial strain reduc-tion factor η by Equation (727) involves simple calculation thus no tables orgraphs are provided here for η

751 Approximate equation for the change in curvature due tocreep in a reinforced concrete section subjected to bending

An approximation Equation (327) is suggested in Section 35 for the curva-ture due to creep in a reinforced concrete section subjected to bending with-out axial force Extension of use of this approximation for a cracked sectionwould result in a relatively larger margin of error This is so because the termεOycr

2c for the cracked section is not negligible enough compared to ψ to

justify ignoring the first of these two quantities when using Equation (727)

Example 73 Cracked T section creep and shrinkage effects

Find the changes in strain and stress distributions due to creep andshrinkage in the cross-section of Example 72 (Fig 77(a) ) Considerthat the result of Example 72 represents the stress and strain at age t0

and use the following data

φ(t t0) = 25 χ(t t0) = 075 εcs(t t0) = minus300 times 10minus6

The effective area of the section is considered unchangeable withtime Thus using the result of Example 72 the depth of the effectivepart of the section c = 0444 and the stress distribution at time t0 is asshown in Fig 77(c)

The area of the effective part of concrete Ac = 02766m2 Thedistance of the centroid of Ac from top yc = 0138m (Fig 79(a) )


The age-adjusted modulus of elasticity of concrete (Equation(131) )

Ec(t t0) = 30 times 109

1 + 075 times 25 = 1043GPa (1500ksi)

α(t t0) = 200

1043 = 1917

The area of a transformed section composed of Ac plus α(As + Aprimes) is

A = 03648m2 (560 in2)

For use of Equations (726ndash31) a reference point O must be chosenat the centroid of the transformed effective area This centroid iscalculated and is found to be at y = 0358m below the top edge

The moment of inertia of Ac about an axis through O is

Ic = 1756 times 10minus3 m4 r2c = IcAc = 00635m2

The moment of inertia of the transformed section is

I = 7301 times 10minus3 m4

The axial strain and curvature reduction factors (Equations (730)and (731) ) are

η = 02766

03648 = 07582 κ =

1756

7301 = 02404

If the area Aprimes is ignored Tables 73 to 75 can be used to calculate yI and κ

The y-coordinate of the centroid of Ac (see Fig 79(a) ) is

yc = minus(0358 minus 0138) = minus0220m

The strain and stress distributions at time t0 are shown in Fig 79(b)(copied from the result of Example 72 Fig 77(c) )

εO = minus35 times 10minus6 ψ = 403 times 10minus6 mminus1


(Note that the reference point O is lower in Fig 79(b) compared toFig 77(c))

Changes in strain at O and in curvature due to creep and shrinkage(Equations (726) and (727) ) are

∆εO = 0758225[minus35 + 403(minus022)]10minus6 minus300 times 10minus6 = minus462 times 10minus6

∆ψ = 0240425403 minus 35 (minus022)

0063510minus6 minus300 times 10minus6 (minus022)

00635= 565 times 10minus6 mminus1

Changes in concrete stresses due to creep and shrinkage (Equation(728) ) are at the top edge

(∆σc)top = 1043 times 109 minus25[minus35 + 403(minus0358)]

+ 300 minus 462 + 565(minus 0358) 10minus6

= 0876MPa (0127ksi)

at the lower edge of the effective area

(∆σc)at 0444m below top edge = 1043 times 109(300 minus 462 + 565 times 0086)10minus6


Changes in stress in steel due to creep and shrinkage (Equation (729) )are

(∆σs)bot = 200 times 109 (minus462 + 565 times 0842)10minus6 = 28MPa (041ksi)

(∆σs)top = 200 times 109 (minus462 minus 565 times 0308)10minus6


The changes in strain and stress distributions due to creep andshrinkage are shown in Fig 79(c) The final strain and stress distribu-tions at time t are obtained by summing up the values in Figs 79(b)and (c) the results are shown in Fig 79(d)

From the stress distribution in Fig 79(d) it is seen that the neutralaxis has moved downwards due to the effects of creep and shrinkageThus according to the above solution a part of the area of the web of


height 0159m above the neutral axis at time t is ignored although itwould have been subjected to compressive stress Because the ignoredarea is close to the neutral axis the error involved is small

76 Partial prestressed sections

Consider a prestressed concrete section which is also reinforced by non-prestressed steel The prestress is applied at age t0 at which time a part of thedead load is also introduced and shortly after a superimposed dead load isapplied At a much later date t the live load comes into effect and producescracking What is the procedure of analysis to determine the strain and stressdistributions at age t after cracking The term partial prestressing is usedthroughout this book to refer to the case when the prestressing forces are notsufficient to prevent cracking at all load stages

We shall assume here that all the time-dependent changes due to creep andshrinkage of concrete and relaxation of prestressed steel take place prior toage t and that no cracking occurs up to this date Thus the method ofanalysis presented in Section 25 for uncracked sections can be applied todetermine the strain and stress distributions at age t just before application ofthe live load The problem that needs to be discussed in the present sectionmay be stated as follows Given the stress distribution in an uncracked sectionreinforced by prestressed and non-prestressed reinforcement what are theinstantaneous changes in stress and strain caused by the application of anadditional bending moment and axial force causing cracking

Figure 710(a) shows a cross-section with several layers of prestressed andnon-prestressed reinforcement At time t the distribution of stress on thesection is assumed to be known and σc(t) the concrete stress is assumed tovary linearly over the depth without producing cracking This stress distribu-tion may be completely defined by the stress value σO(t) at an arbitrary refer-ence point O and stress diagram slope γ(t) = dσdy The additional bendingmoment M and axial force N at O are applied producing cracking of thesection It is required to find the changes in strain and in stress due to M andN

Partition each of M and N in two parts such that (see Fig 710(c) and (e) )

M = M1 + M2 (733)

N = N1 + N2 (734)

M1 and N1 represent the part of the internal forces that will bring the stressesin the concrete to zero and M2 and N2 represent the remainder of the internalforces With M1 and N1 the section is in state 1 (uncracked) Cracking is


Figure 710 Analysis of strain and stress in a partially prestressed section (a) cross-sectiondimensions (b) concrete stress σc(t) immediately before application of M andN (c) decompression forces M1 and N1 on uncracked section (d) strain andstress changes due to application of M1 and N1 (e) M2 and N2 on a fully crackedsection (f) strain and stress changes due to application of M2 and N2


produced only by the part M2 combined with N2 Thus for the analysis twoloading stages need to be considered

(1) M1 and N1 applied on uncracked section(2) M2 and N2 applied on a fully cracked section

The strain changes in the two stages are given by Fig 710(d) and (f)

(∆ε)1 = (∆εO)1 + (∆ψ)1y (735)

(∆ε)2 = (∆εO)2 + (∆ψ)2 y (736)

The total instantaneous change in strain due to M and N is

∆ε = (∆ε)1 + (∆ε)2 (737)

The stress produced in stage 1 is simply equal to the stress in Fig 710(b)reversed in sign as shown in Fig 710(d) The corresponding strain in stage 1is obtained by division of stress values by Ec(t) the strain distribution in stage1 is also shown in Fig 710(d) Thus the stress in concrete is zero afterapplication of M1 and N1 The final stress in concrete is given by the analysisof the effects of M2 and N2 only (Fig 710(f) ) It should however be notedthat M1 and N1 bring to zero the stress in concrete but not in steel

The values of M1 and N1 are equal and opposite to the resultants ofstresses σc(t) on the concrete and α times this stress on steel with σc(t) beingthe stress existing before application of M and N (Fig 710(b) ) M1 and N1

are sometimes referred1 to as decompression forces because σc(t) is generallycompressive (In all the stress and strain diagrams in Fig 710 the variablesεO ψ σO and γ are plotted as positive quantities) The decompression forcesare given by

N1 = minus σdA (738)

M1 = minus σydA (739)

When the stress varies over the full height of the section as one straight linethe integrals in Equations (738) and (739) may be eliminated (see Equations(22ndash8) )

N1 = minus(AσO + Bγ) (740)

M1 = minus(BσO + Iγ) (741)

where A is the area of a transformed section composed of the full con-crete area plus α times the area of steel prestressed and non-prestressed


α = EsEc(t) with Es and Ec(t) being the moduli of elasticity of steel and ofconcrete at the time of application of M and N B and I are the first andsecond moments of the same transformed area about an axis through thereference point O σO = σO(t) is the stress in concrete at the reference point Oat time t immediately before application of the live load γ = γ(t) is the slope ofthe stress diagram

γ = γ(t) = d

dyσ(t) (742)

If O is chosen at the centroid of the above-mentioned transformed areaB = 0 and Equations (740) and (741) become

N1 = minusAσO (743)

M1 = minusIγ (744)

The changes in axial strain and curvature due to M1 and N1 simply are

(∆εO)1 = minus1

Ec

σO (745)

(∆ψ)1 = minus1

Ec

γ (746)

The strain and stress distributions due to M2 and N2 require more elaboratecalculation following the procedure for a cracked section presented inSection 74

In a composite section made of more than one type of concrete the distri-bution of stress σ(t) is generally represented by one straight line for each partof the section and thus Equations (740) and (741) must be adjusted If weassume that cracking occurs only in one part say part i of the cross-sectionthe values σO and γ in Equations (740) and (741) are to be substituted by σi

and γi which define one straight line of distribution of stress σ(t) over part iother non-cracked parts are to be treated in the same way as the non-prestressed steel but using appropriate moduli of elasticity Ec(t) The forcesN1 and M1 calculated in this way represent the decompression forces whichwill bring the stress in concrete to zero in part i of the section the stress inother parts will change but will not necessarily become zero

77 Flow chart

The steps of analysis of the strain and stress presented in Chapter 2 andthe present chapter apply to the whole range from reinforced concrete


without prestressing to fully prestressed concrete where no cracking isallowed

The flow chart in Fig 214 shows how the procedures discussed in the twochapters can be applied in a general case to determine the instantaneous andtime-dependent changes in strain and stress due to the application at time t0

of a normal force N and a bending moment M on a section for which theinitial strain and stress are known

Example 74 Pre-tensioned tie before and after cracking

Fig 711 shows a square cross-section of a precast pretensioned tieImmediately before transfer the force in the tendon is 1100kN(247kip) the age of concrete t0 and no dead load is simultaneouslyapplied with the prestress At a much older age t a normal tensile force1200kN (270kip) is applied at the centre of the section It is required tofind the axial strain and stress in the concrete and steel immediatelyafter prestressing and just before and after application of the 1200kNforce The following data are given the moduli of elasticity of concreteand steel Ec(t0) = 24GPa (3480ksi) Ec(t) = 35GPa (5076ksi) Es =200GPa (29000ksi) (for prestressed and non-prestressed reinforce-ments) creep coefficient φ(t t0) = 24 aging coefficient χ(t t0) = 080during the period (t minus t0) the reduced relaxation ∆σpr = minus90MPa(minus13ksi) and the free shrinkage εcs(t t0) = minus270 times 10minus6

(a) Strain and stress immediately after transferThe area of the transformed section is composed of Ac + α(Aps + Ans)where α = EsEc(t0)

Figure 711 Cross-section of a partially prestressed tie analysed for strain and stressin Example 74


Ac = 030 times 030 minus (930 + 1000)10minus6 = 00881m2

α = 20024 = 833

A = 00881 + 833(930 + 1000)10minus6 = 01042m2

The axial strain at transfer (Equation (233) ) is

ε(t0) = minus1100 times 103

24 times 109 times 01042 = minus440 times 10minus6

The stress in concrete (Equation (235) ) is

σ(t0) = 24 times 109(minus440 times 10minus6) = minus10559MPa (minus1532ksi)

The stress in non-prestressed and in prestressed steel is

σns = 200 times 109(minus440 times 10minus6) = minus880MPa (minus128ksi)

σps = 1100 times 103

930 times 10minus6 + 200 times 109(minus440 times 10minus6)

= 10948MPa (1588ksi)

(b) Changes in strain and in stress due to creep shrinkage andrelaxation

The transformed section to be used here is composed of Ac + α(Aps +Ans) where α = EsEc(t t0)

Using Equation (131)

Ec = 24 times 109

1 + 24 times 08 = 8215GPa (1192ksi)

α = 200

8215 = 2433

The transformed area

A = 00881 + 2433(930 + 1000)10minus6 = 01351m2

The artificial force that would be necessary to prevent strain due tocreep shrinkage and relaxation (Equations (241ndash44) ) is


∆N = minus8215 times 109 times 24 times 00881 (minus440 times 10minus6)

minus8215 times 109(minus270 times 10minus6)00881 + 930 times 10minus6(minus90 times 106)

= 08759 times 106 N (1969kip)

The change in axial strain in concrete when the restraint is removed(Equation (240) ) is

∆ε = minus 08758 times 106


The change in concrete stress (Equations (245) and (246) ) is

∆σ = minus8215 times 109[24(minus440 times 10minus6) minus270 times 10minus6]

+ 8215 times 109(minus789 times 10minus6)

= 4407MPa (06392ksi)

Changes in stress in non-prestressed and prestressed steels (Equations(247) and (248) ) are

∆σns = 200 times 109(minus789 times 10minus6) = minus1579MPa (minus2290ksi)

∆σps = minus90 times 106 + 200 times 109(minus789 times 10minus6)


The stress in concrete after creep shrinkage and relaxation is

σ(t) = minus10559 + 4407 = minus6152MPa (minus08923ksi)

(c) Changes in strain and stress in the decompression stageThe transformed area to be used here is composed of Ac + α(Aps +Ans) where α = EsEc(t)

α = 20035 = 571

The transformed area is

A = 00881 + 571(930 + 1000)10minus6 = 00991m2


The decompression force (Equation (743) ) is

N1 = minus00991(minus6152 times 106) = 6098kN (1371kip)

The change in strain due to N1 (Equation (745) ) is

(∆ε)1 = 6152 times 106

35 times 109 = 176 times 10minus6

The change in stress in the two types of reinforcement is

(∆σns)1 = (∆σps)1 = 200 times 109 times 176 times 10minus6 = 352MPa (511ksi)

(d) Changes in strain and stress in the cracking stageAll the concrete area will be in tension thus the transformed area iscomposed of α(Aps + Ans) with α the same as in (c) above

Transformed area is

A = 571(930 + 1000)10minus6 = 00110m2

Force producing cracking (Equation (734) ) is

N2 = 1200 minus 6098 = 5902kN (113kip)


(∆ε)2 = 5902 times 103

35 times 109 times 00110 = 1530 times 10minus6

The change in stress in any of the two types of reinforcement is

(∆σns)2 = (∆σps)2 = 200 times 109 times 1530 times 10minus6

= 3060MPa (444ksi)

In this example the entire concrete area is ineffective in stage 2 andN2 is resisted only by the steel with cross-section Aps + Ans = 1930 times10minus6 m2 With Es = 200GPa for the two types of steel the strainincrement (∆ε)2 can also be calculated as follows


(∆ε)2 = 5902 times 103

200 times 109 times 1930 times 10minus6 = 1530 times 10minus6

The stress in non-prestressed steel is

minus880 minus 1579 + 352 + 3060 = 953MPa (138ksi)

The stress in prestressed steel is

10948 minus 2479 + 352 + 3060 = 11881MPa (1723ksi)

The strain in the non-prestressed steel immediately before cracking isthe sum of strain values calculated in steps (a) (b) and (c) = (minus440 minus 789+ 176)10minus6 = minus1053times10minus6 At cracking the change in strain in the pre-stressed or non-prestressed steel is (∆ε)2 = 1530 times 10minus6 Thus the strainin the non-prestressed steel after cracking is 477 times 10minus6 At this stage theconcrete is not participating in resisting any force The strains inconcrete and steel are no more compatible and slip must occur in thevicinity of cracks This will be discussed further in Chapter 8

Example 75 Pre-tensioned section in flexure live-load cracking

Fig 712(a) shows the cross-section of a pre-tensioned partially pre-stressed beam A 700kN-m (6200kip-in) bending moment due to adead load is introduced at age t0 at the same time as the prestress trans-fer This bending moment includes the effect of the superimposed deadload introduced shortly after transfer but is considered here as if it wereapplied simultaneously with the prestress transfer At time t long aftert0 a live load is applied producing a bending moment of 400kN-m(3540kip-in) Find the strain and stress distributions immediately afterapplication of the live load bending moment given the following data

Tension in prestressed tendon before transfer = 1250kN (281kip)moduli of elasticity of concrete at ages t0 and t Ec(t0) = 24GPa(3480ksi) and Ec(t) = 30GPa (4350ksi) Es = 200GPa (29000ksi) for allreinforcements φ(t t0) = 20 χ(t t0) = 08 reduced relaxation for theperiod (t minus t0) = minus90MPa (minus13ksi) shrinkage during the same periodεcs(t t0) = minus300 times 10minus6

As in Example 74 the analysis may be done in five parts


Figure 712 Analysis of instantaneous and time-dependent strain and stress in apre-tensioned cross-section before cracking (Example 75) (a) cross-section dimensions (b) strain and stress immediately after prestresstransfer (c) changes in stress and strain due to creep shrinkage andrelaxation


(a) Strain and stress immediately after transferThe calculations in this part follow the procedure presented in Section23 and applied in Example 22 Thus here only the results of the calcu-lations are presented (Fig 712(b) ) The stress in the bottom non-prestressed reinforcement σns = minus56MPa and in the prestressed steelσps = 10305MPa

(b) Changes in strain and in stress due to creep shrinkageand relaxation

The analysis for this part follows the method discussed in Section 25and applied in Example 22 The results are shown in Fig 712(c) Thechanges in stress in the bottom non-prestressed steel ∆σns = minus168MPaand in the prestress steel ∆σps = minus1245MPa

After occurrence of the time-dependent changes the distribution ofstress σ(t) becomes as shown in Fig 713(b)

(c) Changes in strain and stress in the decompression stageThe transformed area to be used here is composed of Ac plus α times thearea of all reinforcements where α = EsEc(t) Ac = area of concretesection = 02768m2

α = 20030 = 6667

Choose reference point O at the centroid of Ac at 0303m below thetop edge (Fig 713(a) ) The moment of inertia of Ac about an axisthrough O = 2178 times 10minus3 m4 Ac = 02768m2

The area of the transformed section its first and second momentsabout an axis through O are

A = 02768 + 6667(1600 + 1200 + 400)10minus6 = 02981m2

B = 6667(1600 times 0547 + 1200 times 0447 minus 400 times 0253)10minus6

= 8734 times 10minus3 m3

I = 2178 times 10minus3 + 6667(1600 times 05472

+ 1200 times 04472 + 400 times 02532)10minus6


The stress distribution in Fig 713(b) may be defined by the value ofstress at O and the slope of diagram


Figure 713 Changes in strain and stress in the cross-section of Fig 712 due to abending moment producing cracking (Example 75) (a) effective cross-section area before cracking (b) stress at time t immediately beforeapplication of bending moment due to live load (Fig 712(b) (c) )(c) strain and stress changes in the decompression stage (d) effectivecross-section area after cracking (e) changes in strain and stress atcracking


σO(t) = minus3490MPa γ(t) = 9737MPam

The decompression forces (Equations (740) and (741) ) are

N1 = minus [02981(minus3490 times 106) + 8734 times 10minus3 times 9737 times 106]

= 0955 times 106N

M1 = minus [8734 times 10minus3(minus3490 times 106) + 2674 times 10minus3 times 9737 times 106]

= minus2299 times 103 N-m

The changes in strain at O and in curvature (Equations (745) and(746) ) are

(∆εO)1 = 3490 times 106


(∆ψ)1 = minus9737 times 106

30 times 109 = minus325 times 10minus6 mminus1

The changes in stress in the bottom reinforcement and in theprestressed steel are

(∆σns)1 = 200 times 109(116 minus 325 times 0547)10minus6 = minus123MPa

(∆σps)1 = 200 times 109(116 minus 325 times 0447)10minus6 = minus58MPa

The changes in strain and in stress distributions in the decompressionstage are shown in Fig 713(c)

(d) Changes in strain and stress in the cracking stageInternal forces producing cracking (Equations (733) and (734) ) are

M2 = 400 times 103 minus (minus2299 times 103) = 6299 times 103 N-m

N2 = 0 minus 0955 times 106 = minus0955 times 106 N

Eccentricity of the resultant of M2 and N2 measured from the bottomreinforcement

es = 6299 times 103

minus0955 times 106minus 0547 = minus1206m


Substitution in Equation (720) and solution by trial or use of Table71 gives the depth of the compression zone (Fig 713(d) )

c = 0263m

The transformed section to be used here is composed of the area ofconcrete in compression plus α times the area of all reinforcementswhere α = EsEc(t) = 20030 = 6667

The transformed area its first and second moments about an axisthrough the reference point O are (Tables 73 and 4 may be used for thispurpose)

A = 01736m2 B = minus25484 times 10minus3 m3 I = 13270 times 10minus3 m4

Changes in axial strain and in curvature produced by M2 and N2

(Equation (215) with Eref = 30GPa) are

(∆εO)2 = 68 times 10minus6 (∆ψ)2 = 1714 times 10minus6 mminus1

The distributions of strain and stress changes are shown in Fig713(e)

The changes in stress in the bottom reinforcement and in theprestress steel are

(∆σns)2 = 200 times 109(68 + 1714 times 0547)10minus6 = 2011MPa (2917ksi)

(∆σps)2 = 200 times 109(68 + 1714 times 0447)10minus6 = 1668MPa (2419ksi)

(e) Strain and stress immediately after crackingThe stress diagram in Fig 713(e) obtained by multiplying the straindiagram in the same figure by the value Ec(t) = 30GPa represents thefinal stress in concrete after cracking The final stress in the reinforce-ment may be obtained by summing up the stress values calculated abovein steps (a) to (d) Thus the stress in the bottom non-prestressed steel is

minus56 minus 168 minus 123 + 2011 = 1664MPa (2413ksi)

The stress in the prestressed steel is

10305 minus 1245 minus 58 + 1668 = 10670MPa (155ksi)


Similarly summing up the strains (Fig 712(b) and (c) and Figs713(c) and (e) ) gives the strain at the reference point O

(minus181 minus 569 + 116 + 68)10minus6 = minus566 times 10minus6

and curvature

(280 + 887 minus 325 + 1714)10minus6 = 2556 times 10minus6 mminus1 (6492 times 10minus6 inminus1)


Example 76 The section of Example 26 live-load cracking

The cross-section of Example 26 (Fig 215) is subjected at time t to anadditional bending moment = 9600kip-in (1080kN-m) representingthe effect of live load Determine the stress and strain distributionsimmediately after application of the live load moment Consider Ec(t) =4000ksi (28GPa) other data are the same as in Example 26 Assumethat cracking occurs due to the live load moment

The strain and stress distributions existing at time t before applica-tion of the live load have been determined in Example 26 (Fig215(c) ) The stress parameters are

σo(t) = minus0506ksi γ(t) = 00130ksiin

(a) The decompression stageProperties of the transformed section at time t are (assuming thereference point O at top fibre)

A = 1145 in2 B = 1943 times 103 in3 I = 5336 times 103 in4

The decompression forces are (Equations (740) and (741) )

N1 = minus [1145(minus0506) + 1943 times 103(00130)] = 327kip

M1 = minus [1943 times 103(minus0506) + 5336 times 103(00130)] = 2908kip-in

The strain changes by decompression are (Equations (745) and(746) )


(∆εO)1 = minus1

4000(minus0506) = 127 times 10minus6

(∆ψ)1 = minus1

4000(00130) = minus324 times 10minus6 inminus1

(b) The cracking stateThe forces on the section due to live load are N = 0 M = 9600kip-inThe forces to be applied on a fully cracked section are (Equations (733)and (734) )

N2 = 0 minus 327 = minus327kip M2 = 9600 minus 2908 = 6692kip-in

This combination of forces is equivalent to a compressive forceof minus327kip at 205 in above the top edge (es = minus575 in see Fig714(a) )

The depth of the compressive zone c = 139 in (by solving Equation(720) ) The properties of the fully cracked section are

A = 635 in2 B = 5824 in3 I = 1418 times 103 in4

The changes in strain and stress due to N2 and M2 are (Equations(219) and (220) )

(∆εO)2 = minus380 times 10minus6 (∆ψ)2 = 2741 times 10minus6 inminus1

(∆σO)2 = minus1520ksi (∆γ)2 = 01096ksiin

The corresponding change in stress in the bottom non-prestressedsteel layer is (Equation (217) )

(∆σns)2 = 29000 [minus380 times 10minus6 + 2741 times 10minus6(37)] = 1839ksi

Summing up the strain changes calculated above to the strain existingbefore the live load application (Fig 215(c) ) gives the total strain aftercracking (Fig 714(b) )

εO = minus870 times 10minus6 + 127 times 10minus6 minus 380 times 10minus6 = minus1123 times 10minus6

ψ = 1259 times 10minus6 minus 324 times 10minus6 + 2741 times 10minus6 = 3676 times 10minus6inminus1


The total stress in concrete after the live load application is simplyequal to the stress due to N2 and M2 (Fig 714(b) )

Figure 714 Analysis of strain and stress distributions in a partially prestressedcracked section (Example 76)

79 General

The methods of analysis presented in this chapter give the axial strain andcurvature in a fully cracked section referred to as state 2 In this state theconcrete in tension is fully ignored It is well established that concretealthough weak in tension participates in the rigidity of cracked members aswill be discussed in Chapter 8 The axial strain and curvature calculated for afully cracked section do not represent actual values from which displacementscan be calculated They only represent upper bounds of strain and curvature


Lower bounds of strain and curvature can be obtained by assuming thatthere is no cracking and using the full area of the concrete section regardlessof stress value or its sign The values of axial strain and curvature to be usedin calculation of displacements are derived by interpolation between thesetwo bounds How this is done is the subject of the next chapter

The procedures of analysis presented in this chapter can be easily pro-grammed for modern desk calculators or microcomputers and indeed thisshould be the route to follow if such computations are to be done repeatedlyThe analysis can also be done by the computer programs SCS and TDAdescribed in Appendix G The programs can be executed on microcomputersusing the software provided on the Internet The program SCS Stresses inCracked Sections determines the neutral axis position and calculates the stressand the strain distributions ignoring concrete in tension The program TDATime Dependent Analysis accounts for the effects of creep and shrinkage ofconcrete and relaxation of prestressed reinforcement

Note

1 See Tadros MK (1982) Expedient serviceability analysis of cracked prestressedconcrete beams Prestressed Concrete Inst J 27 No 6 67ndash86


Displacements ofcracked members

Launching girder for construction of a cast in situ 2043km long bridge at Lake GruyegravereSwitzerland

Chapter 8

81 Introduction

Cracks are generally expected to occur in reinforced concrete structures with-out or with partial prestress when the tensile stresses exceed the strength ofconcrete in tension Reduction in stiffness of members due to cracking mustbe considered in the calculation of displacements in reinforced concretestructures This chapter presents a method to predict the elongation andcurvature of a reinforced concrete cracked member subjected to axial forceandor bending moment

The weakest section in a cracked member is obviously at the location of acrack Away from a crack the concrete in the tension zone is capable ofresisting some tensile stress and thus contributing to the stiffness (the rigidity)of the member Thus the stiffness of a cracked member varies from a min-imum value at the location of the crack to a maximum value midway betweencracks For calculation of displacements a mean value of the flexibility of themember is employed

Two extreme states are considered the uncracked condition in whichconcrete and steel are assumed to behave elastically and exhibit compatibledeformations and the fully cracked condition with the concrete in tensionignored The elongations or the curvatures are calculated with these twoassumptions and the actual deformations in a cracked member are pre-dicted by interpolation between these two extreme conditions A dimension-less coefficient ζ is used for the interpolation it represents the extent ofcracking At the start of cracking ζ = 0 and its value approaches unity withthe increase of the values of the applied axial force andor bendingmoment

The same coefficient ζ can be used to predict the width of a crack Thespacings between the cracks depend on several factors other than the magni-tude of the applied loads There exist several empirical equations for theprediction of the spacing between cracks and no doubt more equations willevolve from research A chosen procedure for the prediction of crack spacingis presented in Appendix E In the numerical examples of the present chapterthe spacing between the cracks will simply be assumed when the width ofcracks is calculated

The interpolation procedure described above gives mean values of axialstrain and curvature at various sections of a structure which can be sub-sequently employed for calculation of displacements (see Section 38) Theinternal forces are assumed to be known and supposed to be of a magnitudewell below the ultimate strength of the sections (service conditions) Thechange in internal forces due to cracking in a statically indeterminatestructure is briefly discussed in Section 811

The reduction in stiffness due to cracking associated with shear stress ismore difficult to evaluate The truss model often employed in the calculationof the ultimate strength of members subjected to shearing force or twisting

Displacements of cracked members 265

moment is sometimes used to assess an upper limit of the displacements aftercracking This is briefly discussed at the end of this chapter


Consider a reinforced concrete member subjected to an axial force or a bend-ing moment When the stress in concrete has never exceeded its tensilestrength the member is free from cracks The reinforcement and concreteundergo compatible strains This condition is referred to as state 1

When the tensile strength in concrete is exceeded cracks occur At thelocation of a crack the tensile stress is assumed to be resisted completely bythe reinforcement The tensile zone is assumed to be fully cracked and thiscondition is referred to as state 2

In both conditions states 1 and 2 Bernoullirsquos assumption is adoptednamely plane cross-sections remain plane after deformation or crackingAnalysis of strain and stress in states 1 and 2 in accordance with theseassumptions is covered in Chapters 2 3 and 7

In a section situated between two cracks bond between the concrete andthe reinforcing bars restrains the elongation of the steel and thus a part ofthe tensile force in the reinforcement at a crack is transmitted to the concretesituated between the cracks The stress and strain in the section will be in anintermediate condition between states 1 and 2 Thus the strain in areinforcing bar varies from a maximum value at the cracks to a minimumvalue midway between the cracks The rigidity varies between consecutivecracks in a similar way Therefore an effective or mean value of the memberstiffness must be considered in the calculation of the elongation or curvatureof the member1 The contribution of the concrete in the tension zone tothe rigidity of the member is sometimes referred to as tension stiffeningIgnoring the effect of tension stiffening generally results in overestimationof deflection or crack width To account for tension-stiffening effectsadditional assumptions are required which will be discussed in the followingsections

83 Strain due to axial tension

A reinforced concrete member subjected to axial tension N (Fig 81(a) ) willbe free from cracks when the value of N is lower than

Nr = fct(Ac + αAs) = fctA1 (81)

where fct is the strength of concrete in tension Nr is the value of the axialforce that produces first cracking Ac and As are the cross-section areas ofconcrete and steel and α = EsEc with Es being the modulus of elasticity ofsteel and Ec the secant modulus of elasticity of concrete for a loading of short


duration (The effect of creep is not considered in this section) A1 is the areaof a transformed section in state 1 composed of Ac plus αAs

Just before cracking the section is in state 1 the stress in concrete is fct andthe stress in steel is αfct Immediately after cracking the section at a crack is instate 2 the stress in steel

σsr = NrAs (82)

When Nr is reached the first crack occurs At the crack the tensile stress inconcrete drops to zero and the total tension is resisted by the steel reinforce-ment (state 2) The sudden increase in stress in steel produces strain in steel

Figure 81 Stresses in a reinforced concrete member cracked due to axial force (a)cracking of a tie (b) stress in reinforcement (c) bond stress (d) stress inconcrete (c fct)


that is incompatible with the strain in the adjacent concrete and results inwidening of the crack

Away from the crack concrete bonded to the reinforcement tends torestrain its elongation and the bond stress τ transmits a part of the tensileforce from the bar to the surrounding concrete At a certain distance s fromthe first crack strain compatibility is recovered (state 1) and the tensilestrength in concrete is again reached causing a second crack (Fig 81(a) )

Figure 81(b) (c) and (d) shows the variation of steel stress bond stress andconcrete stress over the length of a cracked member subjected to an axialforce N gt Nr

At a crack the section is in state 2 the concrete stress is zero and the steelstress and strain when N gt Nr

σs2 = NAs (83)

εs2 = NEsAs (84)

Midway between consecutive cracks the tensile stress in concrete has someunknown value smaller than fct and the steel stress has value smaller than σs2Thus the strain in the reinforcement varies along the length of the member amean value of the steel strain is

εsm = ∆ll (85)

where l is the original length of the member and ∆l is the member extensionThe symbol εsm represents an overall mean strain value for the cracked mem-ber Obviously εsm is smaller than εs2 which is the steel strain at the crackedsection Let

εsm = εs2 minus ∆εs (86)

where ∆εs is a reduction in steel strain caused by the participation of concretein carrying the tensile stress between the cracks Fig 82 shows the variationof the mean strain εsm with the applied load N it follows a curve situatedbetween the two straight lines representing εs1 and εs2 Here εs1 is a hypo-thetical strain in the reinforcement assuming that state 1 continues to applywhen N gt Nr Thus

εs1 = εc1 = N

Ec(Ac + αAs) =

N

EcA1

(87)

where A1 is the area of the transformed section in state 1The value ∆εs represents the difference between the mean steel strain εsm

and the steel strain in a fully cracked section This difference has a maximum


value ∆εs max at the start of cracking when N = Nr Based on experimentalevidence it is assumed that ∆εs has hyperbolic variation with σs2 as follows

∆εs = ∆εs max σsr

σs2

(88)

From the geometry of the graph in Fig 82

∆εs max = (εs2 minus εs1)σsr

σs2

(89)

Substitution of Equations (88) and (89) into Equation (86) gives for acracked member an overall strain value which is also the mean strain in steel

εsm = (1 minus ζ)εs1 + ζεs2 (810)

where ζ is a dimensionless coefficient between 0 and 1 representing theextent of cracking ζ = 0 for an uncracked section (N lt Nr) and 0 lt ζ lt 1 for acracked section The value of ζ is given by

ζ = 1 minus σsr

σs2

2

(with σs2 gt σsr) (811)

Figure 82 Axial force versus mean strain in a member subjected to axial tension


or

ζ = 1 minus Nr

N2

(with N gt Nr) (812)

In Equation (810) the mean strain in steel is determined by interpolationbetween the steel strains εs1 and εs2 in states 1 and 2 The interpolation co-efficient ζ depends upon the ratio of the steel stresses σsr and σs2 in a fullycracked section when the applied forces are Nr and N respectively The use ofthis equation will be extended in the following sections to be applied formembers subjected to bending

In order to take into account the bond properties of the reinforcing barsand the influence of duration of the application or the repetition of loadingthe Eurocode 2ndash19912 (EC2ndash91) introduces the coefficients β1 and β2 intoEquation (811) as follows

ζ = 1 minus β1β2σsr

σs2

2

(with σs2 σsr) (813)

where β1 = 1 and 05 for high bond bars and for plain bars respectively β2 = 1and 05 respectively for first loading and for loads applied in a sustainedmanner or for a large number of load cycles

With this modification the graph of εsm (Fig 82) will have a horizontalplateau at cracking level as shown in Fig 83 (line AC)

The second term in Equation (810) (ζεs2) represents the supplementarystrain of steel compared with the strain of concrete3 Thus the average widthof a crack is

wm = srmζεs2 (814)

Figure 83 Mean strain in the reinforcement of a cracked member (according to EC2ndash91)


where srm is the average spacing between cracks it depends upon factorsincluding the bond properties the amount of cover of the reinforcement andthe shape of distribution of tensile stress over the section Empirical equa-tions based on experiments are generally used to predict value of srm This isfurther discussed in Appendix E

Example 81 Mean axial strain in a tie

Find the mean strain excluding the effect of creep in a reinforced con-crete member (Fig 81(a) ) having a square cross-section 020 times 020m2

(62 in2) subjected to an axial tensile force N = 200kN (45kip) given thefollowing data As = 804mm2 (125 in2) Es = 200GPa (29000ksi) Ec =30GPa (4350ksi) fct = 20MPa (290psi) β1 = 1 and β2 = 05 What is thewidth of a crack assuming srm = 200mm (8 in)

Equation (81) gives Nr = 891kN (200kip) The stresses in steelassuming state 2 prevails (Equations (82) and (83) )

σsr = Nr

As

= 111MPa σs2 = 249MPa

Substitution in Equation (813) gives ζ = 090 The strains in steel due toN calculated with the assumption that the section is in states 1 and 2are (Equations (87) and (84) )

εs1 = 150 times 10minus6 εs2 = 1244 times 10minus6

The mean strain for the member (Equation (810) ) is

εsm = 150 times 10minus6(1 minus 090) + 1244 times 10minus6 times 090 = 1134 times 10minus6

The width of a crack (Equation (814) ) is

wm = 200 times 090 times 1244 times 10minus6 = 022mm(88 times 10minus3 in)

84 Curvature due to bending

A reinforced concrete member subjected to a bending moment (Fig 84) willbe free from cracks when the bending moment is less than

Mr = W1 fct (815)


where Mr is the value of the bending moment that produces first cracking W1

is the section modulus in state 1 Thus W1 is calculated for the cross-sectionarea of concrete plus α times the cross-section area of steel fct is the tensilestrength of concrete in flexure (modulus of rupture)

For a bending moment M gt Mr cracking occurs and the steel stress alongthe reinforcement varies from a maximum value at the crack location to aminimum value at the middle of the spacing between the cracks Assumingthat the concrete between the cracks has the same effect on the mean strain insteel as in the case of axial force Equation (810) can be adopted Thus


where

ζ = 1 minus β1β2 σsr

σs2

2

= 1 minus β1β2Mr

M2

(817)

Here σsr and σs2 are the steel stresses calculated for Mr and M withassumption that the section is fully cracked

For spacing between cracks srm the width of one crack can be calculated byEquation (814) which is repeated here


The curvature at an uncracked or a cracked section can be expressed interms of the bending moment and flexural rigidity or in terms of strains asfollows

ψ = M

EI(819)

Figure 84 A reinforced concrete member in flexure


or

ψ = εs minus (εc)top

d(820)

where ψ is the curvature E is the modulus of elasticity I is the moment ofinertia of the section εs is the strain in steel reinforcement and (εc)top is thestrain at the extreme fibre of the compression zone and d is the distancebetween steel in tension and the extreme compression fibre (Fig 84) Assumethat cracking has an effect on curvature similar to its effect on the strain inaxial tension Thus the mean curvature is expressed in this form

ψm = (1 minus ζ)ψ1 + ζψ2 (821)

where ψ1 and ψ2 are the curvatures corresponding to a bending moment Mwith the assumptions that the section is in states 1 and 2 respectively

Thus the coefficient ζ is employed to interpolate between the curvatures instates 1 and 2 to obtain the mean curvature This is illustrated in the momentndashcurvature graph in Fig 85 The cracked member has a mean flexural rigiditygiven by

(EI)m = M

ψm

(822)

The curvatures ψ1 and ψ2 are given by

Figure 85 Moment versus curvature in a reinforced concrete member in flexure


ψ1 = M

EcI1

(823)

ψ2 = M

EcI2

(824)

where I1 and I2 are the moments of inertia of a transformed uncracked andfully cracked section about an axis through their respective centroids Ec = Eref

is the modulus of elasticity of concrete the value used as a reference elasticitymodulus in the calculation of I1 and I2 The use of Equation (821) is demon-strated in Example 82

841 Provisions of codes

The interpolation between states 1 and 2 to calculate the mean curvature asdone in Equation (821) is adopted in MC-90 and EC2ndash914 The EC2ndash91allows use of the same coefficient ζ to calculate by the same equation meanvalues for deformation parameters such as curvature strain rotation ordeflection

The MC-90 considers that the M-ψ relation shown by the lines ABCD inFig 85 is most representative of actual practice with the exception of thepart EBC This part is replaced by the dashed line which is an extension ofthe curve CD (Equation (821) ) until it intersects AB at point E Thus forpractical application the Mminusψ relation follows the straight line AE when 0 M (Mr radicβ ) where (Mrradicβ ) represents a reduced value of the crackingmoment β = β1β2

When (Mr radicβ ) M My the Mminusψ relation is the non-linear part EDfollowing hyperbolic Equation (821) where My is the moment which pro-duces yielding of the reinforcement If the concrete is in a virgin state and theloading is of short-term character the Mminusψ relation is more closely pre-sented by the lines ABCD Replacement of the part EBC by EC takes intoconsideration the behaviour of a member which has been cracked due toloads shrinkage and temperature variations during construction

The MC-90 also differs in the value of the coefficient β2 which isconsidered equal to 08 (instead of 10) for first loading

The deflection of members can be calculated most accurately by numericalintegration of the curvatures at various sections (see Appendix C) TheEC2ndash91 allows for simplicity to calculate the deflection twice assuming thewhole member to be in uncracked and fully cracked condition in turn (states1 and 2) and then to employ Equation (821) substituting the deflectionvalues for the curvatures

ACI318-015 also allows a similar interpolation between the moment ofinertia of a gross concrete section neglecting the reinforcement and themoment of inertia of a transformed fully cracked section to calculate an


lsquoeffective moment of inertiarsquo Ie to be used in the deflection calculationThis is based on an empirical equation by Branson discussed further inSection 95

Example 82 Rectangular section subjected to bending moment

Calculate the mean curvature in a reinforced concrete member of arectangular cross-section (Fig 84) due to a bending moment M =250kN-m (221kip-ft) excluding creep effect and employing the follow-ing data b = 400mm (16 in) h = 800mm (32 in) d = 750mm (30 in) dprime =50mm (2 in) As = 2120mm2 (329 in2) Aprimes = 760mm2 (118 in2) Es =200GPa (29000ksi) Ec = 30GPa (4350ksi) fct = 25MPa (360psi)β1 = 1 and β2 = 05

Assuming the spacing between cracks srm = 300mm (12 in) find thewidth of a crack

The moment of inertia and the section modulus of transformeduncracked section are (graphs of Fig 35 may be employed)

I1 = 00191m4 W1 = 00488m3

Equation (815) gives Mr = 122kN-m (900kip-ft) Substitution inEquation (817) gives ζ = 088

Depth of compression zone in state 2 (by Equation (716) or thegraphs of Fig 84)

c = 0191m (752 in)

The centroid of the transformed fully cracked section coincides withthe neutral axis The moment of inertia (calculated from first principlesor by use of graphs of Fig 76) is

I2 = 000543m4

The curvatures due to M = 250kN-m assuming the section to be instates 1 and 2 (Equations (823) and (824) ) are

ψ1 = 437 times 10minus6 mminus1 ψ2 = 1530 times 10minus6 mminus1

The mean curvature (Equation (821) ) is

ψm = [(1 minus 088)437 + 088 times 1530]10minus6 = 1400 times 10minus6 mminus1


The strain in steel in state 2 is

εs2 = ψ2 ys = 1530 times 10minus6(075 minus 0191) = 856 times 10minus6


wm = 300 times 088 times 856 times 10minus6 = 023mm (00091 in)

85 Curvature due to a bending moment combinedwith an axial force

Fig 86 shows a reinforced concrete member subjected to a bending momentM and an axial force N at the centroid of the transformed uncracked sectionThe values of M and N are assumed to be large enough to produce crackingat the bottom fibre

The use of the equations of Section 84 will be extended to calculate themean steel strain and the mean curvature in a cracked member subjected to Nand M

The eccentricity of the axial force is

e = MN (825)

Our sign convention is as follows N is positive when tensile and M ispositive when it produces tension at the bottom fibre It thus follows that eis positive when the resultant of M and N is situated below the centroid ofthe transformed uncracked section (Fig 86)

Without change in eccentricity we can find the values of Nr and the corre-sponding Mr that produce at the bottom fibre a tensile stress fct the strengthof concrete in tension

Nr = fct 1

A +

e

Wbot

minus1

1(826)

Mr = eNr (827)

where A1 and W1 are the area and section modulus with respect to the bottomfibre of the transformed uncracked section

Equation (826) of course does not apply when the bottom fibre is incompression This occurs when the resultant normal force on the section istensile acting at a point above the top edge of the core of the transformeduncracked section (e minus (WbotA)1) This occurs also when the resultantnormal force is compressive acting within the core ( (WtopA)1 e minus (Wbot


A)1) where Wbot and Wtop are section moduli with respect to the bottomand the top fibres the subscript 1 refers to the transformed uncrackedsection

When N gt Nr and M gt Mr cracking occurs and the mean strain in thereinforcement can be calculated by

εm = (1 minus ζ)εs1 + ζεs2 (828)

where


σs2

2

(829)

or

ζ = 1 minus β1β2Mr

M2

(830)

or

Figure 86 Curvature due to an eccentric normal force on a reinforced concrete section instates 1 and 2


ζ = 1 minus β1β2Nr

N2

(831)

The symbols in Equations (828) and (829) are defined below

It is to be noted that in a fully cracked section the position of the neutralaxis depends on the eccentricity e = MN not on the separate values of M andN Because e is assumed to be unchanged (MN) = (MrNr) and

σsr

σs2

= Mr

M =

Nr

N(832)

Assuming that the cracks are spaced at a distance srm the width of a crack


The mean curvature in the cracked member

ψm = (1 minus ζ)ψ1 + ζψ2 (834)

where ψ1 and ψ2 are the curvatures corresponding to a bending moment Mand an axial force N with the assumptions that the section is in states 1 and 2respectively

εs1 and εs2 = strain in the bottom steel due to M combined with N on a sectionin states 1 and 2 respectively

σs2 = stress in the bottom steel due to M and N on a section in state 2σsr = stress in the bottom steel due to Mr and Nr on a section in state 2

Example 83 Rectangular section subjected to M and N

Calculate the mean curvature for the reinforced concrete section ofExample 82 subjected to M = 250kN-m (184kip-ft) combined with anaxial force N = minus200kN (minus45kip) at mid-height All other data are thesame as in Example 82 Assuming spacing between cracks srm =300mm find the width of a crack

The area of the transformed section in state 1

A1 = 0336m2

The centroid of A1 is very close to mid-height the eccentricity isconsidered to be measured from mid-height


e = 250

minus200 = minus125m

Substitution in Equations (826) and (827) (with fct = 25MPa W1 =00488m3 see Example 82) gives

Mr = 138kN-m

Substitution in Equation (830) gives

ζ = 085

The presence of N does not change the curvature in state 1 from what iscalculated in Example 82 Thus

ψ1 = 437 times 10minus6 mminus1

Solution of Equation (720) or use of graphs in Fig 74 gives the depthof the compression zone

c = 0241m (949 in)

Distance from the top fibre to the centroid of the transformed section instate 2 (Fig 75) is

y = 0195m (768 in)

The area and the moment of inertia of the transformed section in state2 about an axis through its centroid (Fig 76) are

A2 = 0115m2 I2 = 000544m4

The applied forces N = minus200kN at mid-height combined with M =250kN-m may be replaced by an equivalent system of Nprime = minus200kN atthe centroid of the transformed section in state 2 combined with Mprime =209kN-m

The curvature in state 2 is

ψ2 = 209 times 103

30 times 109 times 000544 = 1280 times 10minus6 mminus1



ψm = [(1 minus 085)437 + 085 times 1280]10minus6


The axial strain at the centroid of the fully cracked section is

εO2 = minus200 times 103


The strain in the bottom steel in state 2 is

εs2 = 10minus6[minus580 + 1280(075 minus 0195)] = 652 times 10minus6

Crack width (Equation (833) ) is


851 Effect of load history

Calculation of Nr and Mr by Equations (826) and (827) implies that Mand N are increased simultaneously from zero until cracking occurs with-out change in the eccentricity e = MN This represents the case when Mand N are caused by an external applied load of a gradually increasingmagnitude

If N is introduced first and maintained at a constant value andsubsequently M is gradually increased cracking will occur when

Mr = fct minus N

A1W1 (835)

This means that the values of Mr and the coefficient ζ representing theextent of cracking depend upon the history of loading

An important case in practice is when the axial force N is a compressiveforce due to partial prestressing The axial force N is generally introducedwith its full value before the cracking bending moment Mr Thus use ofEquation (835) is more appropriate and the first cracking occurs due to thecombination N and Mr

In a fully cracked section the position of the neutral axis depends upon theeccentricity e = MN Thus the combination of Mr and N has a differentneutral axis from the combination of M and N With the two combinations


the ratio σsrσs2 is not equal to MrM It is therefore necessary to calculate σsr

and σs2 separately for a fully cracked section once due to Mr and N andanother time with M and N The ratio (σsrσs2) can then be used to determine ζby Equation (829) rather than the ratio (MrM) with Equation (830) Thiswould result in slightly different values for the mean curvature and crackwidth In Example 83 this modification would give Mr = 151kN-m ζ = 088ψm = 1170 times 10minus6 mminus1 and wm = 017mm compared with ψm = 1150 times 10minus6 mminus1

and wm = 017mm previously calculated Because the difference is small it issuggested that Equations (826) (827) and (830) (or (831) ) be employed inall cases regardless of loading history

86 Summary and idealized model for calculationof deformations of cracked members subjectedto N andor M

In the preceding sections equations were presented for calculation of aninterpolation coefficient ζ for calculation of the mean strain in a reinforcedconcrete member subjected to axial tension (Equation (812) ) and the curva-ture due to a bending moment without or combined with an axial force(Equations (817) and (830) respectively) These equations are repeated hereand the symbols are defined again for easy reference

Axial tension (Fig 81)The mean axial strain

εOm = (1 minus ζ)εO1 + ζεO2 (836)

where


N2

(837)

Nr = f ct(Ac + αAs) (838)

α = EsEc (839)

εO1 and εO2 = axial strain values due to N calculated with the assumptionsthat the section is uncracked and fully cracked (states 1 and 2)respectively

ζ = dimensionless coefficient employed for interpolation betweenthe steel strain values in states 1 and 2

Nr = value of the axial force that produces tension in the concreteequal to its strength fct The value Nr is given by


β1 = 1 or 05 for high bond or for plain bars respectivelyβ2 = 1 or 05 The value 1 is to be used for first loading and 05 is for the case

when the load is applied in a sustained manner or with a large number ofload cycles

Bending moment (Fig 84)The mean curvature

ψm = (1 minus ζ)ψ1 + ζψ2 (840)

where


M2

(841)

Mr = fctW1 (842)

where W1 is the modulus of the transformed section Other symbols are thesame as defined earlier in this section

Bending moment combined with axial force (Fig 86)The mean axial strain and curvature


ψm = (1 minus ζ)ψ1 + ζψ2 (844)

where


M2

= 1 minus β1β2Nr

N2

(845)

The pairs εO1 with ψ1 and εO2 with ψ2 are values of axial strain at a referencepoint O and the curvatures calculated with the assumptions that the section isrespectively uncracked and fully cracked (states 1 and 2) Mr and Nr are thevalues of the bending moment and the corresponding axial force that pro-duces tensile stress fct at the extreme fibre The eccentricity e is assumed to beunchanged thus

ψ1 and ψ2 = values of the curvature due to M calculated with the assump-tions that the section is respectively uncracked and fully cracked(states 1 and 2)

Mr = the value of the bending moment that produces tensile stress fct

at the extreme fibre


e = M

N =

Mr

Nr

(846)

The value of Mr is given by

Mr = efct 1

A1

+ e

W1

minus1

(847)

Other symbols have the same meaning as defined earlier in this sectionThe mean crack width due to any of the above internal forces is given by


where srm is the mean crack spacing εs2 is the steel stress due to N andor Mon a fully cracked section

When the section is subjected to N or M or N and M combined theinterpolation coefficient ζ may be expressed in terms of concrete stresses

ζ = 1 minus β1β2 fct

σ1 max

2

(849)

where σ1 max is the value of the tensile stress at the extreme fibre which wouldoccur due to the applied N andor M with the assumption of no cracking(state 1) fct is the concrete strength in tension If the stress is caused mainly byflexure (see Section 84) fct will represent the tensile strength in flexure whichis sometimes called the modulus of rupture and considered somewhat largerthan the value for axial tension

Equation (849) gives the same result as Equation (837) because the samelinear relationship between fct and Nr applies between σ1 max and N SimilarlyEquation (849) gives the same result as Equation (841) or (845)

Fig 87 shows a physical model which idealizes the behaviour of a crackedmember in accordance with the equations of this section An element of unitlength is considered to be composed of two parts a part of length (1 minus ζ) instate 1 (uncracked) and a part of length ζ in state 2 (fully cracked)

The axial deformation of this idealized member and the angular rotationper unit length (the curvature) are the same as in the actual cracked member

Equations (843) to (849) are applicable for partially prestressed sectionsbut it must be noted in this case that M and N represent the part of thebending moment and of the normal force after deduction of the decompres-sion forces (ie use the values of M2 and N2 obtained by Equations (733) and(734) This is further explained by Example 85


861 Note on crack width calculation

The value of εs2 to be used in the crack width calculation by Equation (848)is the steel strain due to N and M on a fully cracked section ignoring theconcrete in tension Here it is assumed that the stress on concrete is zero priorto the application of N and M If the section is subjected to initial stress duefor example to the effect of shrinkage occurring prior to the application of Nand M the forces N and M should be replaced by N2 = N minus N1 and M2 = M minusM1 where N1 and M1 are two forces just sufficient to eliminate the initialstress The values of N1 and M1 may be calculated by Equations (740) and(741) which are used to calculate the decompression forces in partiallyprestressed sections

87 Time-dependent deformations ofcracked members

Partially prestressed members are often designed in such a way that crackingdoes not occur under the effect of the dead load Thus cracking due to thelive load is of a transient nature hence the effects of creep shrinkage

Figure 87 Representation of an element of unit length of a cracked member by a modelcomposed of uncracked and fully cracked parts such that the extension orcurvature is the same as in the actual member


and relaxation of prestress steel need be considered only for uncrackedsections

If this is not the case or in the case of a reinforced concrete member wherecracking occurs for a load of long duration the time-dependent effects maybe accounted for in the calculation of the axial strain and curvature in states 1and 2 as covered in Chapters 2 3 and 7 Interpolation between the two statesmay be done to find the deformations in the cracked member accounting forthe tension stiffening The equations presented in Section 86 for the inter-polation coefficient ζ are applicable (noting that with a loading of longduration β2 = 05)

Example 84 Non-prestressed simple beam variation of curvatureover span

The reinforced concrete simple beam of the constant cross-sectionshown in Fig 88(a) has bottom and top steel area ratios ρ = 06 percent and ρprime = 015 per cent At time t0 uniform load q = 170kNm(117kipft) is applied It is required to find the curvatures at t0 and at alater time t and to draw sketches of the variations of the curvature overthe span The following data are given

Es = 200GPa (29000ksi) Ec(t0) = 300GPa (4350ksi) fct = 25MPa(036ksi) β1 = 10 β2 = 10 for calculation of instantaneous curvatureand 05 for long-term curvature creep coefficient φ(t t0) = 25 agingcoefficient χ(t t0) = 08 free shrinkage εcs(t t0) = minus250 times 10minus6

What is the deflection at mid-span at time t

(a) Curvature at time t0

The following sectionsrsquo properties will be used in the analysis ofcurvatures at t0

Transformed uncracked section (state 1) Area A1 = 02027m2 centroidO1 is at 0331m below top edge moment of inertia about an axisthrough O1 I1 = 7436 times 10minus3 m4 section modulus W1 = 2333 times 10minus3 m3

Transformed cracked section (state 2) Depth of compression zone(Equation (716) ) c = 0145m centroid O2 lies on neutral axis momentof inertia about an axis through O2 I2 = 1809 times 10minus3 m4

The bending moment at mid-span = 17 times 828 = 136kN-m Thebending moment which produces cracking (Equation (815) )

Mr = 2333 times 10minus3 times 25 times 106 = 583kN-m


The interpolation coefficient for instantaneous curvature (Equation(841) ) is

ζ = 1 minus 10 times 10rsquo 583

136 2

= 082

The interpolation coefficient for long-term curvature (Equation (841) ) is

Figure 88 Curvature in a reinforced concrete beam (Example 84) (a) span load andcross-section dimensions (b) curvature at time t0 (c) curvature at time t


ζ = 1 minus 10 times 05 583

136 2

= 091

The curvature at t0 assuming states 1 and 2 (Equations (823) and(824) )

State 1

ψ1(t0) = 136 times 103

30 times 109 times 7436 times 10minus3 = 610 times 10minus6 mminus1

State 2

ψ2(t0) = 136 times 103


InterpolationMean curvature at time t0 (Equation (840) )

ψ(t0) = (1 minus 082)610 times 10minus6 + 082 times 2506 times 10minus6 = 2157 times 10minus6 mminus1

With parabolic variation of the bending moment over the span thevalue Mr = 583kN-m is reached at distance 098m from the supportThus cracking occurs over the central 605m (198 ft) of the span

Fig 88(b) shows the variation of the curvatures at time t0 with theassumptions of states 1 and 2 the mean curvature is also shown withthe broken curve

(b) Curvatures at time tThe age-adjusted modulus of elasticity of concrete (Equation (131) )

Ec(t t0) = 30 times 109

1 + 08 times 25 = 10GPa

α = Es

Ec(t t0) =

200

10 = 20

The following sectionsrsquo properties are required for the age-adjustedtransformed sections in states 1 and 2


Age-adjusted transformed section in state 1 A1 = 02207m2 centroid O1

is at 0344m below top edge Moment of inertia about an axis throughO1 I1 = 8724 times 10minus3 m4 y = coordinate of the centroid of the concretearea (measured downwards from O1) yc = minus0020m area of concreteAc = 01937m2 moment of inertia of Ac about an axis through O1 Ic =6937 times 10minus3 m4 r2

c = IcAc = 3534 times 10minus3 m2The curvature reduction factor (Equation (318) ) is

κ1 = 6937 times 10minus3

8724 times 10minus3 = 0795

Age-adjusted transformed section in state 2 A2 = 701 times 10minus3m2 cen-troid O2 is at 0233m below top edge moment of inertia about an axisthrough O2 I2 = 4277 times 10minus3 m4 y-coordinate of centroid of concretearea in compression (measured downwards from O2) yc = minus0161marea of the compression zone Ac = 00431m2 moment of inertia of Ac

about an axis through O2 Ic = 1190 times 10minus3m4 r2c = IcAc = 2762 times

10minus3 m2The curvature reduction factor (Equation (731) ) is

κ2 = 1190

4277 = 0278

Changes in curvature due to creep and shrinkage

State 1The curvature at t0 = 610 times 10minus6 mminus1 the corresponding axial strain at O1

= 610 times 10minus6 (0344 minus 0331) = 8 times 10minus6The change in curvature during the period t0 to t (Equation (316) )

∆ψ = 079525610 times 10minus6 + 8 times 10minus6 minus0020

3534 times 10minus3

+ (minus250 times 10minus6)minus0020

3534 times 10minus3= 1299 times 10minus6 mminus1

The curvature at time t (state 1)

ψ1(t) = (610 + 1299)10minus6 = 1909 times 10minus6 mminus1


State 2The curvature at t0 = 2506 times 10minus6 mminus1 the corresponding axial strain atO2 = 2506 times 10minus6 (0233 minus 0145) = 222 times 10minus6

The change in curvature during the period t0 to t (Equation (727) )


2762 times 10minus3





InterpolationMean curvature at time t (Equation (840) )

ψ(t) = (1 minus 091)1909 times 10minus6 + 091 times 3754 times 10minus6

= 3584 times 10minus6 mminus1

= 9113 times 10minus6 inminus1

The curvature at the end section is caused only by shrinkage and maybe calculated by Equation (316) However if we ignore this value andcalculate the deflection by assuming parabolic variation of curvaturewith zero at ends and maximum at the centre we obtain (Equation(C8) )

Deflection at centre = 3584 times 10minus6 82

96

= 00239m

= 239mm (0948 in)

By numerical integration a more accurate value of the deflection atthe centre is 235mm (0925 in)

It can be seen in Fig 88(b) and (c)6 that once Mr is exceeded the line


representing the mean curvature starts to deviate from the curve forstate 1 and quickly becomes closer to the curve for state 2 Thusprediction of deflection in design may start by considering state 2 whichgives the upper bound for deflection and the designer may find thiscomputation sufficient when the upper bound is not excessive

Solution of this example is done using no graphs in order to demon-strate the computation for a general case with any cross-section How-ever with a rectangular section the graphs in Figs 35 and 74ndash76 canbe used to determine the section properties involved in the calculationAdditional graphs are presented in Chapter 9 which further simplifythe prediction of deflection when the cross-section is a rectangle

Example 85 Pre-tensioned simple beam variation of curvatureover span

Find the mean curvature at a section at mid-span of a partiallyprestressed beam shown in Fig 89(a) after application of a live

Figure 89 Curvature of a partially prestressed beam (Example 85) (a) tendonprofile (b) curvature after creep and shrinkage and application of liveload For beam cross-section see Fig 712(a)


load producing cracking Also sketch the corresponding variationof curvature over the span and calculate the deflection at thecentre

Fig 712(a) shows the cross-section at mid-span The section is con-stant over the span with the exception of the location of the prestressedsteel The beam is pretensioned with a tendon depressed at points B andC resulting in the profile shown in Fig 89(a) The beam carries uni-form dead and live loads of intensities 140 and 80kNm respectively(096 and 055kipft) resulting in bending moments at mid-span of 700and 400kN-m (6200 and 3540kip-in) Assume a high-bond quality ofreinforcement and tensile strength of concrete fct = 25MPa Other dataare the same as in Example 75

The stress and strain in the section at mid-span have been analysed inExample 75 The curvature in state 2 is obtained by summing up thevalues of curvatures shown in Fig 712(b) and (c) and 713(c) and (e)This gives the following value of curvature in state 2


Cracking is produced at time t only after application of a live loadImmediately before application of the live load after occurrence ofprestress loss the curvature at mid-span is 1167 times 10minus6 mminus1 (sum ofcurvature values indicated in Fig 712(b) and (c) ) Assuming no crack-ing (state 1) the live load would produce additional curvature of 499 times10minus6 mminus1 This is calculated by dividing the live-load moment by[Ec(t)I1(t)] where Ec(t) = 30GPa is the modulus of elasticity of concreteat time t and I1(t) = 2674 times 10minus3 m4 is the centroidal moment of inertiaof transformed uncracked section at time t Thus after live-loadapplication the total curvature in state 1 is

ψ1 = (1167 + 499)10minus6 = 1670 times 10minus6 mminus1

The stress at the bottom fibre due to the live-load moment on theuncracked section is 8580MPa Addition of this value to the stress of2323MPa existing before application of the live load (Fig 713(b) )gives the stress at the bottom fibre after the live-load application withthe assumptions of state 1

σ1 max = 2323 + 8580 = 10903MPa


The interpolation coefficient between states 1 and 2 (Equation (849) ) is


σ1 max

2

= 1 minus 10 times 10 25

109032

= 095

β1 = 10 because of the high-bond quality of the reinforcement and β2

= 10 assuming that the deflection is calculated for non-repetitiveloading

The mean curvature at mid-span (Equation (844) ) is

ψm = (1 minus 095)1670 times 10minus6 + 095 times 2556 times 10minus6

= 2510 times 10minus6 mminus1(638 times 10minus6 inminus1)

The curvature variation over the span is shown in Fig 89(b)7 Thelength of the zone where cracking occurs is 148m Over this zone threelines are plotted for curvatures in states 1 and 2 and mean curvature

If we assume parabolic variation and use the values of the meancurvature at the ends and the centre we obtain by Equation (C8)

Deflection at the centre = 202

96 [2(minus402) + 10 times 2510]10minus6

= 1012mm (399 in)

Using five sections instead of three and employing Equation (C16)gives a more accurate value for the central deflection after applicationof live load of 862mm (339 in) The dead-load deflection includingeffects of creep shrinkage and relaxation is 384mm (151 in)

In the design of a partially prestressed cross-section the amount ofnon-prestressed steel may be decreased and the prestressed steelincreased such that the ultimate strength in flexure is unchanged Theamount of deflection is one criterion for the decision on the amounts ofprestressed steel and non-prestressed reinforcement The calculateddeflection in this example may be considered excessive Assuming thatthe yield stresses of the non-prestressed reinforcement and the pre-stressed steel are 400 and 1600MPa (58 and 230ksi) the area of the


bottom non-prestressed reinforcement may be reduced from 1600 to400mm2 with the addition of prestressed steel of area 300mm2 at thesame level without substantial change in the flexural strength of thesection If the stress before transfer is the same in all prestressed steel asin the original design the tension in the added prestressed steel beforetransfer is 3125kN

With the second design the curvatures in states 1 and 2 at mid-spanafter application of the live load will respectively be 1109 times 10minus6 and1976 times 10minus6 mminus1 and the corresponding mean curvature will be 1897 times10minus6 mminus1 The deflection just before and after the application of the liveload will respectively be 60 and 431mm (024 and 170 in) and thelength of the cracked zone after the live-load application will be 125m(408 ft)

88 Shear deformations

Reinforced concrete members are often designed in such a way that inclinedcracks due to shear are expected to occur even at service load After thedevelopment of such cracks shear deformations can be large To predict theultimate shear strength the behaviour of a beam cracked by shear is oftenidealized as that of a truss model in which compression is resisted by concreteand tension by stirrups and flexural reinforcement The same model has beenemployed for evaluation of the deflection in a member cracked by shearHowever the computation involves several assumptions and relies on empir-ical rules The mean shear deformations are somewhere between those givenby an uncracked member and those given by a truss model8

89 Angle of twist due to torsion

Cracks due to twisting moments in reinforced concrete members result inreduction of the torsional rigidity The reduction in rigidity due to crackingby torsion is much more important than the corresponding reduction in caseof flexure In the following the angles of twist per unit length θ1 and θ2 arederived for uncracked or fully cracked conditions (states 1 and 2) This giveslower and upper bounds of the angle of twist When the value of the twistingmoment exceeds the value Tr that produces first cracking the angle of twistper unit length θm will be some value between θ1 and θ2 but it is difficult tofind an expression that can reliably predict the value of θm For this reasononly expressions for θ1 and θ2 will be derived below

In many structures for example grids or curved beams the drastic


reduction in stiffness due to cracking by torsion results in a redistribution ofstresses and the twisting moment drops at the expense of an increase of thebending moment in another section without excessive deformation of thestructure When such redistribution cannot occur excessive deformations dueto torsion must be avoided for example by the introduction of appropriateprestressing

891 Twisting of an uncracked member

According to the theory of elasticity the angle of twist per unit length is

θ1 = T

GcJ1

(850)

where T is the twisting moment Gc is the shear modulus of concrete and J1 isthe torsion constant For a rectangular section

J1 = cb313 minus 021 b

c 1 minus b4

12c4 (851)

where c and b are the two sides of the rectangle with b c The maximumshear stress is at the middle of the longer side c and its value

τmax = T

microbc2(852)

where micro is a dimensionless coefficient which varies with the aspect ratio cb asfollows9

For a closed hollow section

J1 = 4A20 [int(dst)]minus1 (853)

where t is the wall thickness A0 is the area enclosed by a line throughthe centre of the thickness and the integral is carried out over thecircumference

The shear flow (the shearing force per unit length of the circumference) isgiven by

cb 10 15 175 20 25 30 40 60 80 100 infin

0208 0231 0239 0246 0258 0267 0282 0299 0307 0313 0333


τt = T2A0 (854)

where τ is the shear stress

892 Twisting of a fully cracked member

The discussion here is limited to a hollow box section (Fig 810(a) ) The trussmodel usually adopted in the calculation of strength of reinforced concretemembers in shear or torsion is used here for the calculation of the angle oftwist in state 2 After cracking the sides of the hollow section are assumed toact as a truss in which the compression is resisted by concrete inclinedmembers and the tension is resisted by the stirrups and by the longitudinal

Figure 810 Torsion in a box girder (a) cross-section (b) free body diagram of a wall of acracked box girder


reinforcement assumed to be lumped at the four corners Fig 810(b) is afree-body diagram showing the forces acting on a part of the wall of the boxFig 811(a) is a truss idealization of a cracked box girder The members in thehidden faces of the box are not shown for clarity The external applied twist-ing moment is replaced by the forces τth and τtb as shown where τt is theshear flow (Equation (854) )

τt = T2hb (855)

where h and b are height and breadth of the truss model (see Fig 811(a) )

Figure 811 Truss idealization of a box girder cracked by twisting (see Fig 810) (a) spacetruss model (b) a typical panel of truss model


A typical panel of the space truss is shown in Fig 811(b) By staticsthe forces in the twelve members of the panel due to a unit twisting momentare

forces resisted by stirrups

F1 = F3 = 1

2hF2 = F4 =

1

2b (856)

forces in the longitudinal bars

F5 = F6 = F7 = F8 = x

2hb (857)

forces in the diagonal members

F9 = F11 = minus 1

2b sin α1

F10 = F12 = minus 1

2h sin α2

(858)

where α1 and α2 are angles defined in Fig 811(a) It is suggested that thedistance x in Fig 811(b) be selected such that the angles α1 and α2 are close to45 degrees

The angle of twist per unit length of the cracked member is considered thesame as the relative rotation of the two cross-sections of the panel in Fig811(b) divided by the distance x between them Considering virtual work theangle of twist per unit length is

θ = T

x 12

i = 1F

2l

AEi

(859)

where Fi is the force in the ith member due to a unit twisting moment li is itslength Ei = Es for the members in tension (the stirrups and the longitudinalbars) and Ei = Ec for the diagonal members in compression Ai is thecross-section area For members representing the stirrups (i = 1 to 4)

Ai = Av

x

s(860)

where Av is the cross-sectional area of a stirrup and s is the spacing betweenstirrups For longitudinal bars Ai is the area of the longitudinal reinforcementlumped at one corner The area of the diagonal compression member isusually considered equal to


A9 = A11 = th cos α1 (861)

and

A10 = A12 = tb cos α2 (862)


Example 86 Live-load deflection of a cracked pre-tensioned beam

Consider that the analyses conducted in Examples 26 and 76 are forthe cross-section at the centre of a simply supported beam of span 80 ft(24m) What is the deflection at mid-span after application of the liveload Assume fct = 050ksi (34MPa) β1 = 10 β2 = 05 Other data arethe same as in Examples 26 and 76 Assume parabolic variation ofcurvature over the span and ignore the curvature at the two ends

The curvature at mid-span after application of the live loadcalculated at a cracked section in Example 76 is

ψ2 = 3676 times 10minus6 inminus1

Properties of the transformed uncracked section at time t are (Ec(t) =4000ksi and reference point O at top fibre)


The curvature change due to M = 9600kip-in applied on anuncracked section is (Equation (219) )

(∆εO1)live load = minus200 times 10minus6 (∆ψ1)live load = 1177 times 10minus6inminus1

The corresponding stress change at bottom fibre is

(∆σbot)live load = 1083ksi

Add the change in stress to the stress value existing before applicationof the live load to obtain the total stress ignoring cracking

(σbot)non-cracked = 0013 + 1083 = 1096ksi

Similarly the total curvature ignoring cracking


ψ1 = 1259 times 10minus6 + 1177 times 10minus6 = 2436 times 10minus6 inminus1

The interpolation coefficient (Equation (849) )

ζ = 1 minus 10(05) 0500

10962

= 0896

The mean curvature (Equation (844) )

ψm = (1 minus 0896)2436 times 10minus6 + 0896(3676 times 10minus6) = 3547 times10minus6 inminus1

The deflection at mid-span after application of the live load(Equation (C8) )

Dmid-span = (80 times 12)2

96 [0 + 10(3547 times 10minus6) + 0] = 341 in


At time t after occurrence of creep shrinkage and relaxation the struc-ture of Example 36 (Fig 39(a) ) is subjected to a uniform live load p =100 kipft (146kN-m) The intensity p is sufficient to produce crackingat mid-span The tensile strength of concrete at time t is fct = 0360ksi(250MPa) The modulus of elasticity of concrete Ec(t) = 4350ksi(300GPa) Other data are the same as in Example 36 The objective ofthe analysis is to determine the stresses crack width and mid-spandeflection immediately after application of p and to study the influenceof varying the non-prestressed steel ratio ρns = Ansbh on the resultswhere Ans is the non-prestressed bottom steel the same amount of non-prestressed steel is also provided at the top The effects of varying thecreep and shrinkage parameters will also be discussed

The live-load bending moment at mid-span is 3750kip-in Table 81which gives the results of the analysis includes the load intensity ρcr andthe corresponding mid-span bending moment Mcr when cracking firstoccurs The table also gives the stress changes ∆σps and ∆σns in the stressin the prestressed and non-prestressed steels due to p = 1kipft The lastcolumn of the table gives the results for the case ρns = 04 per cent and


reduced creep and shrinkage parameters φ(t t0) = 15 and εcs = minus150 times10minus6 (from φ(t t0) = 30 and εcs = minus300 times 10minus6)

Based on the results in Table 81 the following remarks can be made

(a) Presence of the non-prestressed steel reduces the deflection inother words the deflection is overestimated if the presence of thenon-prestressed steel is ignored

(b) The level of loading at which cracking occurs drops because of thepresence of the non-prestressed steel thus for certain load inten-sity ignoring the non-prestressed steel may indicate that crackingdoes not occur contrary to reality

(c) The steel stress increments ∆σps and ∆σns decrease with the increasein ρns Thus presence of the non-prestressed steel increases safetyagainst fatigue

(d) The width of cracks can be controlled by the appropriate choice ofρns The ratio of the mean crack width wm to the mean crack

Table 81 Stresses mid-span deflection and crack width after live-load application ofthe structure of Example 87 (Fig 39(a))

04 withNon-prestressed steel ratio reducedns (per cent) 0 02 04 06 08 10 13 amp cs

Live-load bending momentat which cracking occurs(kip-in) 2600 2400 2300 2200 2100 2000 2700

Ratio of uniform loadintensity pcr at which crackingoccurs to p (p = 1kipft) 069 065 061 058 056 054 073

Deflection after application ofp (10minus3 in) 1250 1229 1182 1128 1074 1022 976

Steel stresses after ns(bot) minus7 minus6 minus5 minus5 minus4 minus4 minus2application of p (ksi) ps 180 181 182 183 183 183 187

Stress changes ns(bot) 29 27 25 23 21 20 18in steel caused byp (ksi) ps 22 20 19 17 16 15 14

Ratio of crack width to crackspacing (10minus3) 072 068 065 062 058 055 039

Conversion factors 1kipft = 146kNm 1ksi = 69MPa


spacing srm is given in the table rather than the value of wm This isso because wm is proportional to srm (see Equation (848) ) and thevalue srm depends on ρns and on how the non-prestressed steel isarranged in the section In general Srm decreases with the increase inρns Thus wm decreases faster than the ratio wmsrm as ρns isincreased

(e) It is interesting to note that the stress in the bottom non-prestressedsteel is compressive in spite of cracking

811 General

Strain in cracked sections is determined by two analyses ignoring cracking(state 1) and assuming that the concrete cannot carry any stress in tension(state 2) Values of the axial strain and curvature are determined in the twostates and the values in the actual condition are obtained by interpolationbetween the two analyses using an empirical coefficient ζ In this wayaccount is made of the additional stiffness which concrete in tension providesto a section in state 2

Branson10 accounts for the tension stiffening by interpolation betweenmoments of inertia of the cross-section in states 1 and 2 about axes throughtheir respective centroids using an empirical interpolation Equation (926)to calculate an lsquoeffectiversquo moment of inertia to be used in calculation ofdeflection More important than the type of empirical procedure to be usedfor the interpolation is the correct analysis of the two limiting states 1 and 2

It should be noted that when the section changes from state 1 to state 2 thecentroid is shifted towards the compression zone Thus in the case of asection subjected to an eccentricity normal force eg prestressing a substan-tial change in eccentricity is associated with cracking The moment about acentroidal axis changes and so does the moment of inertia of the sectionThis is automatically accounted for when the equations used to calculate theaxial strain and curvature employ cross-section properties (A B and I ) withrespect to a reference point O used for both states 1 and 2

Cracking changes cross-section properties and thus is associated withalteration in the reactions and internal forces when the structure is staticallyindeterminate Analysis of these statically indeterminate forces has to bemade by iterative methods which are treated in references on structuralanalysis (see also Chapter 13) The equations presented in this chapter whichgive the axial strain and curvature due to specified values of M and N can beincorporated in an iterative analysis to determine the statically indeterminateforces in cracked reinforced or prestressed concrete structures11


Notes

1 Favre R Beeby AW Falkner H Koprna M and Schiessl P (1985) Crackingand Deformations CEB Manual Printed and distributed by the Swiss FederalInstitute of Technology Lausanne Switzerland

2 See the reference mentioned in Note 5 page 193 See reference mentioned in Note 1 above4 See references mentioned in Notes 2 and 5 on page 19 respectively5 ACI Committee 318 Building Code Requirement for Structural Concrete 2001

American Concrete Institute Farmington Hills Michigan 48333-90946 The graphs in Fig 88(b) and (c) are prepared using the computer program RPM

lsquoReinforced and Prestressed Membersrsquo Elbadry M and Ghali A AmericanConcrete Institute PO Box 9094 Farmington Hills MI 48333-9094 USA RPManalyses strain stress change in length end rotation and deflection of a reinforcedmember with or without prestressing The member can have variable depth andcan be a simple beam a cantilever or can be part of a continuous beam or a frameCracking tension stiffening creep and shrinkage of concrete and relaxation ofprestressing reinforcement are accounted for

7 This figure was prepared using the computer program RPM see Note 6 above8 See reference in Note 1 above9 Timoshenko S and Young D (1962) Elements of Strength of Materials 4th edn

Van Nostrand Princeton New Jersey pp 91ndash210 See reference mentioned in Note 2 page 348 See also Branson DE and

Trost H (1982) Application of the I-effective method in calculation deflectionsof partially prestressed members Prestressed Concrete Institute Journal ChicagoIllinois K27 No 5 SeptndashOct pp 62ndash77

11 A computer program in FORTRAN for analysis of the time-dependent internalforces stresses and displacements in cracked reinforced and prestressed concretestructures is available See Elbadry M and Ghali A Manual of ComputerProgram CPF Cracked Plane Frames in Prestressed Concrete Research Report NoCE85-2 revised 1993 Department of Civil Engineering University of CalgaryCalgary Alberta Canada


Simplified predictionof deflections

91 Introduction

In many practical situations designers are interested in prediction of prob-able maximum deflections of reinforced concrete members Accuracy in pre-diction is often of little or no concern For this purpose two methods arepresented in this chapter for prediction of maximum deflections of reinforcedconcrete members accounting for long-term effects of creep and shrinkageWe are here concerned only with the transverse deflection associated withcurvature ψ in simple or continuous members subjected to bending momentswith or without axial forces Prestressed beams are treated as reinforced con-crete members for which the magnitude of the axial force and the bending

Long-term deflection and cracking of a reinforced concrete slab A test series conducted atthe Swiss Federal Institute of Technology Lausanne

Chapter 9

moment are known thus prestress losses must be determined by a separateestimate

In earlier chapters we discussed how to obtain the axial strain εo andcurvature ψ at sections of reinforced concrete frames in uncracked andcracked conditions (states 1 and 2) Ignoring concrete in tension in state 2underestimates the rigidity of the sections To account for the stiffening effectof concrete in the tension zone mean values of axial strain and curvature arecalculated by empirical interpolation between values of εO and ψ in states 1and 2 The mean values can be employed to calculate displacements at anysection by conventional methods which generally require knowledge of thevariation of the mean values over the entire length of all members (seeSection 38 or Appendix C)

The interpolation mentioned above is done by using a coefficient ζ whichdepends upon the value of the internal forces Thus any member even with aconstant cross-section behaves in general as a beam of variable rigidity Thesimplified methods presented in this chapter avoid this difficulty by calcula-tion of limiting deflection values for states 1 and 2 considering the memberto have a constant section in each state thus well-known expressions fordeflection of members of constant rigidity can be applied The interpolationis then done for the two limiting deflection values rather than for axial strainor curvature

Through conventional linear analysis a lsquobasicrsquo deflection value is calcu-lated assuming that the member is made of homogeneous elastic materialwithout cracking The basic value is then multiplied by coefficients whichaccount for the stiffening effect of reinforcement cracking and creep Thedeflection due to shrinkage is determined by a simple expression which alsoincludes a coefficient depending on the amount of reinforcement and itsposition in the section The coefficients needed in these calculations whenthe section is rectangular are presented in graphs in this chapter and inAppendix F This appendix also includes expressions for the coefficients forcross-sections of any shape

Section 99 is concerned with the deflection of flat slabs by simplifiedprocedures similar to the methods suggested for beams

92 Curvature coefficients

Consider a reinforced concrete cross-section without prestressing (Fig 91)subjected to a bending moment M introduced at time t0 The followingexpressions give the instantaneous curvature and the changes in curvaturecaused by creep and shrinkage between t0 and a later time t

ψ(t0) = κsψc (91)

(∆ψ)φ = ψ(t0)φκφ (92)


(∆ψ)cs = minus εcs

d κcs (93)

where ψc is the instantaneous curvature at a hypothetical uncracked concretesection without reinforcement

ψc = M

Ec(t0)Ig

(94)

Following the sign convention adopted through this book positivestrain represents elongation hence shrinkage of concrete is a negativequantity

A positive bending moment produces positive curvature (Fig 91) In across-section with top and bottom reinforcements shrinkage is restrained bythe reinforcement and the result is smaller shrinkage at the face of the sectionwith heavier reinforcement In a simple beam subjected to gravity load theheavier reinforcement is generally at the bottom Thus the curvature due toshrinkage is of the same sign as the curvature due to the positive bendingmoment due to load For the same reason in a cantilever with heavierreinforcement at the top curvatures due to shrinkage and due to gravity loadare cumulative

κs κ and κcs are dimensionless coefficients depending on the geometri-cal properties of the cross-section the ratio α(t0) = EsEc(t0) and the product

Figure 91 Curvature at a reinforced concrete cross-section subjected to bending moment

Ig = moment of inertia of gross concrete section about an axisthrough its centroid

Ec(t0) = modulus of elasticity of concrete at time t0

ψ(t0) = instantaneous curvature(∆ψ)φ and (∆ψ)cs = curvature changes caused by creep and by shrinkage

εcs = εcs(t t0) = value of free shrinkage during the period considered

Simplified prediction of deflections 305

χφ(t t0) where Es is the modulus of elasticity of steel Ec(t0) is the modulus ofelasticity of concrete at time t0 φ and χ are creep and aging coefficientsfunctions of the ages t0 and t (see Section 17)

Equations (91)ndash(94) are applicable to uncracked sections in state 1 or fullycracked sections in state 2 employing coefficients κs1 κφ1 and κcs1 for state 1and κs2 κφ2 and κcs2 for state 2

For state 2 cracking is assumed to occur at t0 due to the bending momentM The concrete in tension is ignored thus the cross-section in state 2 iscomposed of the area of concrete in compression plus the area of thereinforcement For T or rectangular cross-sections the depth of the compres-sion zone may be determined by Equation (716) The geometrical propertiesof the cracked section are assumed to undergo no further changes during theperiod of creep and shrinkage

The graphs in Figs F1 to F10 of Appendix F give the values of the κ-coefficients in the two states for rectangular cross-sections For easy referencethe variables in these graphs are listed in Table 91 Expressions for the coef-ficients for a general cross-section are also given in Appendix F These arederived from Equations (216) and (316)

93 Deflection prediction by interpolationbetween uncracked and cracked states

In a simplified procedure suggested in Section 94 the probable maximumdeflection in reinforced concrete members including the effects of creep andshrinkage is predicted by empirical interpolation between lower and upperbounds D1 and D2 The values of D1 and D2 are determined assuming themember to have a constant cross-section in states 1 and 2 respectively Anempirical coefficient ζ is employed to determine the probable deflectionbetween the two limits D1 and D2 The difference between this simplifiedprocedure and the method discussed in Chapter 8 is that the interpolation isperformed on the deflection at one section to be defined below rather than onthe curvature at various sections of the member

The interpolation coefficient used in Chapter 8 depends on the value of thebending moment and the cracking moment at the section considered (seeEquation (841) ) Here the interpolation coefficient for deflection is based onthe bending moment at one section which is referred to as the lsquodeterminantrsquosection Similarly the properties of the cross-section in states 1 and 2 will bebased on the reinforcement at the determinant section

If we apply any of Equations (C4) (C8) (C12) or (C16) to calculate thedeflection in a simple beam in terms of curvature at various sections itbecomes evident that the maximum deflection is largely dependent upon thecurvature at mid-span This is so because the largest curvature is at thissection and this value is multiplied by the largest coefficient in each equationThus for a simple beam the determinant section is considered at mid-span


Table 91 Graphs for curvature coefficients s and cs for rectangular reinforced con-crete sections (Figs F1 to F10)

ParametersFigure number

Coefficient dh dprimeh in Appendix F

10s1 09 0 to 02 ndash F1

08

10s2 09 0 to 02 ndash F2

08

1010 0 to 02 20 F3

3040

101 09 0 to 02 20 F4

3040

1020

08 0 to 02 30 F540

100 20 F6

3040

1020

2 08 to 10 01 30 F740

1002 20 F8

3040

10 anycs1 08 0 to 02 value F9

0cs2 08 to 10 20 F10

02

Note The value of some parameters is indicated by a range for which the graph may be employed Forpreparation of the graphs the value at the middle of the range is used in the calculations


Similarly for a cantilever the determinant section is at or near its fixed end(see Equations (C17) (C19) and (C21) )

Equations (C4) (C8) (C12) and (C16) are also applicable to members ofcontinuous structures Because the coefficient of curvature is largest for thevalue at mid-span this section may again be considered determinant But inthis case the curvatures at the end sections are generally not small and mayhave a comparatively larger influence on the calculated deflection It shouldbe recognized here that we are dealing with an approximation the choice ofthe determinant section is a matter of judgement

931 Instantaneous and creep deflections

Consider a simple beam (Fig 92) subjected at age t0 to a uniform load ofintensity q Variation of the instantaneous deflection D(t0) at mid-span withthe load intensity is as shown For any load intensity q the deflection D(t0) issome value between lower and upper bounds D1(t0) and D2(t0) where

D1(t0) = instantaneous deflection in state 1 all sections are assumed to beuncracked

D2(t0) = instantaneous deflection in state 2 all sections are assumed to befully cracked Contribution of concrete in tension to the stiffness ofthe member is ignored

A basic deflection value Dc is calculated by conventional analysis assum-ing the load q applied on a member of linear homogeneous materialof modulus of elasticity Ec(t0) and having a constant cross-section witha moment of inertia equal to that of the gross concrete section withoutconsidering reinforcement

Figure 92 Instantaneous deflection at mid-span of a reinforced concrete simple beamversus intensity of load


The instantaneous deflections at mid-span in states 1 and 2 are determinedby the equations

D1(t0) = κs1Dc (95)

D2(t0) = κs2Dc (96)

κs1 and κs2 are curvature coefficients calculated at a determinant sectionwhich in this case is the section at mid-span Equations (95) and (96) followfrom Equation (91) if we consider that the reinforcement effect on theflexural rigidity is the same in all sections as in the determinant section

Similarly the changes in deflection due to creep in states 1 and 2 (seeEquation (92) ) are

(∆D1)φ = D1(t0)φκφ1 (97)

(∆D2)φ = D2(t0)φκφ2 (98)

κφ1 and κφ2 are curvature coefficients related to creep to be calculated for thedeterminant section

932 Deflection of beams due to uniform shrinkage

Consider a non-prestressed reinforced concrete simple beam without cracks(Fig 93(a) ) Assume that the cross-section is constant with area of the bot-tom reinforcement As larger than the top reinforcement Aprimes Uniform shrink-age of concrete occurring during a specified period produces at all sections acurvature of magnitude given by

(∆ψ)cs = εcs Ac yc

I (99)

or

(∆ψ)cs = minusεcs κcs

d(910)

where εcs is the value of the free shrinkage (generally a negative quantity) Ac

is the cross-section area of concrete I is the moment of inertia of the age-adjusted transformed section about its centroid yc is the y-coordinate of thecentroid of Ac y is measured downwards from point O the centroid of theage-adjusted transformed section Note that in the cross-section consideredin Fig 93(a) yc is a negative value The age-adjusted transformed section iscomposed of Ac plus α times area of the reinforcement where α = EsEc Es is


the modulus of elasticity of steel Ec = Ec(1 + χφ) is the age-adjusted modulusof elasticity of concrete φ and χ are creep and aging coefficients and Ec is themodulus of elasticity of concrete at the start of the period considered κcs isthe curvature coefficient defined in Section 92 and given in graphs in Figs F9and F10 Equations (99) and (910) may be derived from Equation (316)

For a simple beam of span l the deflection at mid-span due to shrinkage is

(∆D)cs = (∆ψ)cs

l2

8(911)

Figure 93 Deflection and stresses produced by shrinkage of reinforced concrete simple(a) and continuous (b) beams


The stress at any fibre due to shrinkage (see Equations (315) and (319) ) is

(∆σ)cs = Ec(∆ψ)cs y minus εcs 1 minus Ac

A (912)

where A is the area of the age-adjusted section y is the coordinate of the fibreconsidered At the bottom fibre y = yb the stress is the largest tensile

(∆σ)cs bot = Ec (∆ψ)cs yb minus εcs 1 minus Ac

A (913)

The stress distribution shown in Fig 93(a) is calculated for a rectangularcross-section with ρ = 1 per cent dh = 09 Ec = 25GPa (3600ksi) and χφ = 2For a free shrinkage εcs = minus300 times 10minus6 the tensile stress at the bottom fibre is106MPa (0154ksi) Presence of this tensile stress allows cracking to occur atsmaller external applied loads (smaller value of Nr or Mr see Sections 84 and85)

Thus it can be concluded that uniform shrinkage can affect the deflectionin direct and indirect ways First it produces curvature which increases thedeflection in a simple beam Second it produces tension at the bottom fibreenhancing cracking and causing further increase in deflection

In a cantilever heavier reinforcement is commonly at the top and thecurvature due to shrinkage will be given by Equation (99)

The corresponding downward deflection at the free end is

(∆D)cs = minus(∆ψ)cs

l2

2(914)

Note that (∆ψ)cs in this case is a negative valueEquation (99) or (910) may also be used to calculate the curvature due to

shrinkage at a fully cracked section (compare Equations (316) and (727) )Here cracking is assumed to have occurred due to loads applied prior toshrinkage For the fully cracked section concrete in the tension zone isignored and the geometric properties of the cracked section assumedunchanged by the effect of shrinkage

For a cracked simple beam of length l the deflection due to uniformshrinkage may be determined by interpolation between the limiting values

(∆D1)cs = minusεcsκcs1 l2

8d(915)


8d(916)


κcs = minusAcd

Iyc (917)

The subscripts 1 and 2 are employed with κcs to refer to uncracked and fullycracked states

Equations (915ndash16) are derived by combining Equations (910) and (911)and (917) by comparing Equation (910) with Equations (316) and (727)equations for a cantilever can be derived in a similar way The curvaturecoefficients κcs1 and κcs2 are to be calculated for the lsquodeterminantrsquo sectionwhich is at mid-span for a simple beam and at the fixed end for a cantilever(see Section 93)

In statically indeterminate structures hyperstatic forces develop which tendto reduce the deflection due to shrinkage Consider as an example the interiorspan of a continuous beam of equal spans (Fig 93(b) ) Assume that thespan shown is sufficiently far from the end spans such that the rotations at Aand B are zero Use the force method (see Section 42) to calculate the static-ally indeterminate connecting moments This gives for a beam of constantcross-section M = minusEcI(∆ψ)cs where (∆ψ)cs represents the curvature if thebeam were simply supported The curvature due to the connecting momentsis of constant value equal to minus (∆ψ)cs Thus the statically indeterminate beamhas no curvature and no deflection due to shrinkage and the concrete stress isuniform tensile of magnitude

(∆σ)cs = minusεcsEc 1 minus Ac

A (918)

Note that the stress in this case depends only on the sum of the reinforcementareas (As + Aprimes) not on their locations in the cross-section For a rectangularsection with 1 per cent reinforcement εcs = minus300 times 10minus6 and χφ = 2 (∆σ)cs =045MPa (0065ksi) (Fig 93(b) )

The statically indeterminate reactions and bending moments caused byuniform shrinkage in continuous beams of constant cross-section having twoto five equal spans are given in Fig 107 This figure intended for the effect oftemperature is also usable for the effect of shrinkage the only difference isthat the multiplier (∆ψ) used for the values of the figure represents the changein curvature due to uniform shrinkage of a simple beam (Equation (99) or(910) ) The deflection is largest for the end span and its value at the middleof the span may be expressed as follows

Deflection at the centre of a continuous span= reduction coefficient times deflection of a simple beam (919)

The reduction coefficient for an end span is respectively 025 040 036 and037 when the number of spans is 2 3 4 and 5 The values of the reduction


coefficient given here apply only when the cross-section and the reinforce-ment are constant within the span other values for the coefficient aresuggested later in this subsection for the more common case when As andAprimes vary within the span When As and Aprimes are constant the tensile stress atbottom fibre in a section at the middle of an end span may be approximatedby the average of the values calculated by Equations (913) and (918)

(∆σ)cs bot = Ec (∆ψ)cs

2yb minus εcs 1 minus

Ac

A (920)

Note that (∆ψ)cs is the value of curvature which would occur in a simplysupported beam (Equation (99) or (910) )

The curvature (∆ψ)cs due to shrinkage depends mainly on (As minus Aprimes) Inactual continuous beams the bottom reinforcement is larger than the topreinforcement at mid-span but the reverse is true at the supports The curva-ture (∆ψ)cs of any span when released as a simple beam (Fig 93(b) ) will bepositive at mid-span and negative at the supports This has the effect ofreducing the absolute value of the statically indeterminate connectingmoment |M| It can be shown that in the interior span of a continuous beamof rectangular cross-section (Fig 93(b) ) the statically indeterminate con-necting moments M = 0 when the absolute value |As minus Aprimes| is constant withthe heavier steel at the bottom for only the middle half of the span and at thetop for the remainder of the span It can also be shown that the deflection inthis case is half the value for a simple beam (Equation (911) ) For a moregeneral case accurate calculation of the value of the connecting moment andthe deflection due to shrinkage must account for the values of As and Aprimes atvarious sections of the span

As approximation the change of location of the heavier reinforcementbetween top and bottom in a common case may be accounted for by the useof Equation (919) with the reduction coefficient 05 for an interior span and07 for an end span This coefficient is to be multiplied by the deflection of asimple beam of a constant cross-section based on the reinforcement at mid-span The tensile stress at bottom fibre at the same section may be approxi-mated by Equation (913) this implies ignoring the effect of the staticallyindeterminate connecting moment

933 Total deflection

The deflections due to applied load including the effects of creep and shrink-age for states 1 and 2 are (by superposition)

D1 = D1(t0) + (∆D1)φ + (∆D1)cs (921)

D2 = D2(t0) + (∆D2)φ + (∆D2)cs (922)


94 Interpolation procedure the lsquobilinear methodrsquo

We consider in this section the maximum deflection of a member in flexurewithout axial force The case of a member subjected to axial force combinedwith bending will be discussed in Section 97

The probable maximum deflection for the member considered in the pre-ceding section (Fig 92) is determined by interpolation between the lower andupper bounds D1 and D2 Thus the deflection due to load including creepand shrinkage is

D = (1 minus ζ)D1 + ζD2 (923)

where ζ is the interpolation coefficient for which the following empiricalequation is suggested

ζ = 1 minus β1β2 Mr

M(924)

where M is the bending moment at the determinant section due to load-ing Mr is the value of bending moment which produces in state 1 a tensilestress fct at extreme fibre fct is the modulus of rupture (tensile strengthof concrete in flexure) Mr is given by Equation (846) which is repeatedhere

Mr = fctW1 (925)

where W1 is the section modulus of the transformed uncracked section attime t0 As an approximation Wg the section modulus of the gross concretearea may be employed in Equation (925) in lieu of W1

The coefficient β1 = 10 or 05 for high-bond reinforcements or plain barsrespectively β2 represents the influence of the duration of application andrepetition of loading β2 = 1 at first loading and 05 for loads applied in asustained manner or for a large number of load cycles

Equation (845) for the interpolation coefficient used for curvature differsfrom the interpolation coefficient for deflection (Equation (924) ) only in theterm (MrM) which is raised here to the power 1 instead of 2 The two equa-tions are merely empirical and the difference between the two is only justifiedby a better correlation with test results or with more accurate computationmethods

With the assumptions involved in the calculation of the deflection in states1 and 2 due to applied load the two values D1 and D2 vary linearly with theapplied load or with the value of the free shrinkage εcs Thus Equation (923)


interpolates between two straight lines For this reason the procedure isreferred to as the lsquobilinear methodrsquo1

95 Effective moment of inertia

An analogous approach for estimation of the instantaneous deflection due toload on a cracked reinforced concrete member is based on calculation of anlsquoeffective moment of inertiarsquo Ie to be assumed constant over the full length ofthe member Several empirical expressions have been suggested The bestknown is by Branson2

Ie = Mr

Mm

Ig + 1 minus Mr

Mm

I2 with M Mr (926)

where

The power m = 3 Branson uses the same equation with m = 4 when Ie isintended for calculation of curvature in an individual section

Ie = an effective moment of inertiaIg = moment of inertia of gross concrete area about its centroidal axis neg-

lecting reinforcementI2 = moment of inertia of transformed fully cracked section (state 2) about

its centroidal axisM = maximum moment in the member at the stage for which the deflection is

computedMr = The moment which produces cracking

Example 91 Use of curvature coefficients member in flexure

Figure 94 shows a reinforced concrete simple beam of a rectangularcross-section A uniform load q = 17kNm is applied at time t0Calculate the deflection at time t at mid-span including effects of creepand shrinkage The ratio ρ and ρprime for the bottom and top reinforcementsare

ρ = As

bd = 06 per cent ρprime =

Aprimesbd

= 015 per cent

Other data are Es = 200GPa (29 times 103 ksi) Ec(t0) = 300GPa (4350ksi)φ(t t0) = 25 χ(t t0) = 08 εcs(t t0) = minus250 times 10minus6 fct = 25MPa (036ksi)


Ig = bh3

12 =

03 times (065)3

12 = 6866 times 10minus3 m4

Basic deflection

Dc = 5

384

ql4

Ec(t0)Ig

= 5

384


30 times 109 times 6866 times 10minus3 = 440mm

The following curvature coefficients can be read from the graphs inFigs F1 F2 F4 F7 F9 and F10 (or by Equations (F1ndash3) )

κs1 = 092 κφ1 = 079 κcs1 = 027

κs2 = 380 κφ2 = 014 κcs2 = 097

Instantaneous deflections in states 1 and 2 (Equations (95) and (96) )

D1(t0) = 092 times 440 = 405mm

D2(t0) = 380 times 440 = 1672mm

Changes in deflections in the two states due to creep (Equations (97)and (98) ) are

(∆D1)φ = 405 times 25 times 079 = 800mm

(∆D2)φ = 1672 times 25 times 014 = 585mm

Figure 94 Beam of Examples 91 92 and 93


Changes in deflections due to shrinkage (Equations (915) and (916) )are

(∆D1)cs = 250 times 10minus6 times 027 82

8 times 06 = 090mm


8 times 06 = 323mm

Lower and upper bounds on deflection at time t (Equations (921) and(922) ) are

D1 = 405 + 800 + 090 = 1295mm

D2 = 1672 + 585 + 323 = 258mm

Value of bending moment which produces cracking in state 1 at mid-span (Equation (925) ) is

Mr bh2

6fct =

03 times (065)2

6 25 times 106 = 528kN-m

(The reinforcement could be included in calculations of section modu-lus but this is ignored here) Actual bending moment at mid-span is

M = 17 times 82

8 = 1360kN-m

Interpolation coefficient using β1 = 1 assuming high-bond reinforce-ment and β2 = 05 for sustained loading (Equation (924) ) is


1360 = 081

Probable deflection at time t (Equation 923) is

(1 minus 081)1295 + 081 times 258 = 234mm (0920 in)

The deflection for the same beam is calculated by a more accurateprocedure involving numerical integration in Example 84 The answersare almost identical


96 Simplified procedure for calculation ofcurvature at a section subjected to M and N

Favre et al3 suggest the following approximation for the mean curvature at acracked section subjected to a moment and a normal force

Consider a reinforced concrete section subjected at time t0 to a moment Mand a normal force N located at the centroid of the gross concrete section(Fig 95) The force N in this figure is assumed to be compressive but thediscussion applies also when the normal force is tensile The graph repre-sents the variation of instantaneous curvature excluding creep when N iskept constant and M increased gradually from zero The straight line ABrepresents the curvature ψ1(t0) in state 1 In the case when the section hasheavier reinforcement at the bottom than at the top the centroid of thetransformed uncracked section at time t0 is slightly lower than the centroid

Figure 95 Moment versus curvature in presence of constant normal force (a) actual graph(b) idealized graph


of the gross concrete section thus the compressive force N produces apositive moment about an axis through the centroid of the transformedsection This is why the line AB is slightly shifted from the origin inFig 95(a)

When M is zero the neutral axis is outside the section indicating that allthe stress are of one sign (compression in the case considered here) When Mreaches a certain level tensile stress is produced at the bottom fibre this isrepresented by point C in Fig 95(a) If M is further increased and concretein tension ignored the curvature will follow the broken curve shown Thenon-linear behaviour is caused by change in position of the neutral axisaltering the size of the compression zone Thus the geometrical properties ofthe cracked cross-section vary as M changes However as M increases thebroken curve in Fig 95(a) gradually approaches the straight line OD paral-lel to the line AF of Fig 95(b) which represents the curvature when the cross-section properties of the cracked cross-section are those of a section in state 2subjected to a bending moment without a normal force As an approxima-tion we accept the two straight lines AO and OD for the curvature in state 2in lieu of the broken curve

When a section is subjected to M without N the neutral axis in state 2coincides with the centroid of the transformed fully cracked section (its pos-ition can be determined by Equation (716) We further assume that the partof concrete considered effective does not change with time With theseassumptions the curvature coefficients κs κφ and κcs can be employed to findthe instantaneous and time-dependent curvatures in states 1 and 2 account-ing for the effects of creep and shrinkage as discussed in Sections 931 to933

The momentndashcurvature relation in Fig 95(a) is further simplified in Fig95(b) by ignoring the small curvature when M = 0 thus the line AB is movedparallel to itself bringing A to the origin The line AF in this figure representsthe curvature in state 2 when the section is subjected to M without N Thusthe presence of N has resulted simply in translation of AF without a changein the slope to OD

The mean curvature can now be obtained by empirical interpolationbetween the two straight lines OB and OD

Fig 95(b) is an idealized representation of M versus the instantaneouscurvature ψ(t0) A graph of M versus the change in curvature due to creep(∆ψ)φ or M versus the instantaneous plus creep curvature [ψ(t0) + (∆ψ)φ]would be of the same form as in Fig 95(a) differing only in the slopes of thelines ED and AB Let us now consider that Fig 95(b) represents the instant-aneous plus creep curvature and write expressions for parameters related tothe geometry of the figure

(slope)AB = Ec(t0)Ig

κs1(1 + κφ1)(927)


(slope)OD = Ec(t0)Ig

κs2(1 + κφ2)(928)

The value MO at the intersection of AB and CD (Fig 95(b) ) does not varywith time (in the usual range of variation of χφ) Favre et al employ thefollowing expression for the value of MO

MO minusN |y12| 1

1 minus (κs1κs2)(929)

where |y12| the absolute value is the distance between the centroid of thetransformed section at time t0 in state 1 and the centroid of the transformedsection at the same time in state 2 subjected to M without axial force

The length EA in Fig 95(b) represents a hypothetical curvature ψ2N is thevalue of curvature due to the normal force N on a cracked section (state 2)with M = 0 From geometry

ψ2N = minusMO 1

(slope)OD

minus 1

(slope)AB (930)

Equation (929) is applicable when N is tension or compression Accordingto the sign convention followed throughout this book N is positive whentensile

97 Deflections by the bilinear method memberssubjected to M and N

This section is concerned with the maximum deflection of a reinforced con-crete member subjected to a moment M which may vary over the length ofthe member combined with a constant axial force

Fig 96 represents a simple beam subjected to a normal force N at thecentroid of the gross concrete section combined with a uniform load q Acompressive normal force is indicated in the figure but the discussion appliesalso when N is tensile The idealized M-ψ relationship in Fig 95(b) will beused to extend the use of the bilinear method for calculation of the probablemaximum deflection in the member shown in Fig 96

The graphs in Fig 96 represent the variation of M at the determinantsection (caused by the variation of q) with the corresponding instantaneousdeflection at mid-span Line AB represents the deflection D1 in state 1 LineAF represents deflection in state 2 in the absence of the normal force LineED represents the deflection D2 in state 2 due to M and N the length EArepresents the deflection due to a (negative) bending moment equal to N |y12|where y12 is the upward shift of the centroid of the transformed section asstate 1 is changed to state 2


If the deflection due to creep is included the MndashD diagram in Fig 96 willnot differ in form but lines AB and ED will have smaller slopes Inclusion ofdeflection due to shrinkage will cause the two lines to translate (to the right)without change in slope

In the bilinear method suggested by Favre et al4 two deflection values arecalculated

D1 = maximum deflection assuming that the member has a constantuncracked cross-section (state 1)

D2 = maximum deflection assuming that the member is subjected to bend-ing with no axial force and has a constant fully cracked cross-section(state 2)

The probable maximum deflection is determined by interpolation betweenthese two values using the equation

D = (1 minus ζ)D1 + ζD2 (931)

where ζ is the interpolation coefficient given empirically by one of thefollowing four equations

When (β1β2Mr) MO (Fig 97(b) )

ζ = 1 minus β1β2

Mr

M(932)

0 for M lt Mr (933)

When (β1β2Mr) lt MO

Figure 96 Maximum deflection versus bending moment at the determinant section in asimple beam


ζ = 1 minusMO

M(934)

0 for M lt MO (935)

Mo is given by Equation (929) Mr is the value of the bending momentwhich produces cracking in the presence of the axial force Mr is given byEquation (835) which is repeated here

Mr = fct minus N

A1W1 (936)

where fct is the strength of concrete in tension A1 and W1 are respectively thearea and section modulus of the transformed uncracked section at time t0 Asan approximation the area Ag and section modulus Wg of the gross concretesection may be used instead of A1 and W1

The coefficient β1 = 1 for high-bond reinforcement and 05 for plain barsβ2 = 1 for first loading and 05 for loads applied in a sustained manner or ina large number of load cycles

Comparing the equations of this section with Section 94 it can be seen

Figure 97 Summary of the bilinear method for prediction of maximum deflection ofreinforced concrete members (a) member subjected to bending momentwithout axial force (see Fig 92) (b) member subjected to bending momentcombined with axial force (see Fig 96)


that they differ only in the equations for Mr (Equations (925) and (936) ) andfor the interpolation coefficient in the case when (β1β2Mr) lt MO (Equations(934) and (935) )

Figure 97 gives a concise presentation of the bilinear method for predic-tion of probable maximum deflections in reinforced concrete members sub-jected to a bending moment or a bending moment combined with an axialforce It should be noted that the chosen interpolation equations result in aprobable deflection D which varies linearly with the bending moment M at thedeterminant section The horizontal distance between the parallel lines D andD2 in Fig 97(a) represents the stiffening effect of concrete in tension How-ever the distance between the lines D and D2 represents the tension stiffeningcombined with the effect of an additional bending moment resulting from theshift of centroid of the transformed section when cracking occurs

Example 92 Use of curvature coefficients member subjected toM and N

Consider the same beam of Example 91 (Fig 94) subjected at time t0

to a uniform downward load q = 17kNm (117kipft) combined withan axial compressive force N = minus400kN (899kip) at mid-height of thesection It is required to find the maximum deflection at time t gt t0

including the effect of creep but without shrinkage using the bilinearmethod Other data are the same as in Example 91

The calculations are identical to the case of simple bending withoutaxial force (Example 91) except for the cracking moment Mr and theinterpolation coefficient ζ We give here some values calculated inExample 91

Basic deflection = 440mmCurvature coefficients

κs1 = 092 κφ1 = 079 κs2 = 38 κφ2 = 014

Instantaneous deflections in states 1 and 2

D1(t0) = 405mm D2(t0) = 1672mm

Changes in deflections in the two states due to creep

(∆D1)φ = 800mm (∆D2)φ = 585mm

Lower bound on deflection


D1 = 405 + 800 = 1205mm

Upper bound on deflection assuming no axial force

D2 = 1672 + 585 = 2257mm

Bending moment at the determinant section (mid-span)

M = 1360kN-m

Cracking bending moment (Equation (936) )

Mr = 25 times 106 minus minus400 times 103

03 times 065 03(065)2

6 = 961kN-m

The centroid of the transformed uncracked section at time t0 is at331mm below top edge In the cracked stage when the section is sub-jected to bending without axial force the depth of the compressionzone c = 145mm (Equation (716) ) This is also equal to the distancebetween the top edge and the centroid of the transformed crackedsection Thus the shift of the centroid as the section changes from state1 to 2 is

|y12| = 331 minus 145 = 186mm

The value MO by Equation (929) is

MO = minus (minus400 times 103)0186 1

1 minus (09238) = 981kN-m

β1 and β2 = 10 and 05 the same as in Example 91

β1β2Mr = 05 times 961 = 481kN-m lt MO

Interpolation coefficient (Equation (934) ) is

ζ = 1 minus 981

1360 = 028

The probable deflection (Equation (931) ) is


D = (1 minus 028)1205 + 028 times 2257 = 150mm (059 in)

If the deflection due to shrinkage is excluded in Example 91 theprobable deflection will be 206mm (0810 in) Thus the compressiveforce N reduces the deflection in this example by 27 per cent

98 Estimation of probable deflection method oflsquoglobal coefficientsrsquo

In the majority of cases in practical design particularly in preliminary stud-ies the engineer is only interested in an estimate of the probable deflectionTo this effect Favre et al5 have prepared graphs based on the bilinearmethod permitting a simple and rapid estimation (within plusmn30 per cent) oflong-term deflections due to sustained loads and shrinkage

981 Instantaneous plus creep deflection

Equations (95ndash8) (921ndash23) can be combined in one equation for the deflec-tion due to a sustained load including the effect of creep (but not shrinkage)

D = Dc[(1 minus ζ)κs1(1 + φκφ1) + ζκs2(1 + φκφ2)] (937)

where ζ is the interpolation coefficient (Equation (924) )Based on parametric study this equation may be replaced by the following

approximation

D Dcκth

d 3

(1 minus 20 ρprime) (938)

This equation was derived for rectangular sections h is total height d is thedistance between tension reinforcement and extreme compressive fibre ρprime =Aprimes bd b is the breadth of section and Aprimes is the cross-section area of compres-sion reinforcement

κt is a global correction coefficient which depends on the level of loadingexpressed by the ratio (MrM) at the determinant section creep coefficient φand the product αρ with α = EsEc(t0) and ρ = Asbd As is the cross-sectionarea of tension reinforcement

The graphs in Fig 98 give the global correction coefficient κt These wereprepared by calculating a value κt such that the terms between the squarebrackets in Equations (937) and (938) are equal The following parametersare assumed constants dh = 09 dprimeh = 01 α = EsEc(t0) = 7 χ = 08 β1 = 1 and


Figu

re9

8G

loba

l coe

ffici

ent

κt f

or c

alcu

latio

n of

inst

anta

neou

s pl

us c

reep

def

lect

ions

of u

ncra

cked

or

crac

ked

mem

bers

by

Equa

tion

(93

8)

β2 = 05 (assuming use of high-bond reinforcement and sustained load) Thecompression steel reduces the long-term deflection by approximately 5 to 10per cent In preparation of the graphs of Fig 98 ρprime = Aprimesbd is consideredzero but the term (1 minus 20ρprime) approximately accounts for the effect of thecompression reinforcement

Equation (938) is applicable for cracked or uncracked members When thebending moment at the determinant section does not exceed cracking (M Mr) ζ = 0 and the corresponding graph in Fig 98 may be employed todetermine κt Comparison of the values of κt for uncracked and crackedmembers shows that when M is close to Mr it is important to determinewhether cracking occurred or not because the value κt and hence thedeflection can increase by a factor of 1 to 3 once cracking occurs

The approximate Equation (938) may be employed for members havingcross-sections other than rectangular but with less accuracy For this pur-pose when calculating ρ and ρprime the section is transformed into a rectangle ofthe same height and with a width calculated such that the moment of inertiaof the gross area is the same Calculation of Mr should be based on sectionmodulus of the actual section

The tensile reinforcement has a great influence on deflection in the crackedstate (M Mr) on the other hand its influence is small in the uncracked stateThe amount of the tensile reinforcement is accounted for in κt and itsposition is included in Equation (938) by the ratio hd

The value Mr of the cracking moment at the determinant section andconsequently the tensile strength of concrete fct (see Equations (925) and(936) ) play an important role particularly when the bending moment in thevicinity of the determinant section is close to Mr because the deflection maythen vary greatly On the other hand the influence of fct diminishes in thecracked stage

The method of global coefficients was designed for members subjected toflexure without axial force If bending is combined with axial compressionproduced for example by prestressing the method may be used but again withless accuracy The effect of the axial force will be limited to increasing thevalue Mr (Equation (936) )

982 Shrinkage deflection

Equations (915 16) (921ndash23) may be combined in one equation for thedeflection at mid-span of a cracked reinforced concrete simple beam due toshrinkage

(∆D)cs = minusεcs l 2

8d [(1 minus ζ) κcs1 + ζκcs2] (939)

where εcs is the value of free shrinkage of concrete (generally a negative


quantity) εcs is assumed uniform κcs1 and κcs2 are coefficients for the calcula-tion of curvature at the determinant section assumed uncracked andfully cracked respectively (Equation (910) ) Values of κcs1 and κcs2 may bedetermined by Equation (917) or the graphs of Figs F9 and F10 ζ is aninterpolation coefficient given by Equation (924) which is repeated below


M(940)

Mr is the value of the bending moment which produces cracking (Equation(925) ) M is the bending moment at the determinant section (at mid-span)M is assumed to have been applied before the occurrence of shrinkage

The term inside the square brackets in Equation (939) may be combined inone global coefficient for shrinkage deflection

κtcs = (1 minus ζ) κcs1 + ζκcs2 (941)

The deflection due to shrinkage in a simple beam is

(∆D)cs = minusεcs

l 2

8dκtcs (942)

Shrinkage deflection in continuous beams can be predicted by multiplica-tion of the simple-beam deflection calculated by Equation (942) by areduction factor (see Section 932 near its end)

In a similar way an equation may be derived for the shrinkage deflection atthe free end of a cantilever


l 2

2dκtcs (942a)

The determinant section in this case is at the fixed end where the bendingmoment produces cracking at the top thus when the graphs in Figs F9 andF10 are used the pairs (As with d) and (Aprimes with d prime) must refer to the topand bottom reinforcements respectively Equations (942) and (942a) areapplicable to uncracked and cracked members

In the common case when β1β2 = 05 the interpolation coefficient ζ for acracked member is a value between 05 and 10 The graphs in Fig 99 givethe values of the global coefficient for shrinkage deflection κtcs calculated forrectangular sections with the assumptions ζ = 05 075 and 10 dh = 09 andd primeh = 01 and χφ = 20 The graphs may be used to calculate approximatevalues of κtcs for sections other than rectangles or when the values dh d primeh orχφ are different


Figure 99 Global coefficient κtcs for calculation of shrinkage deflection of crackedmembers by Equation (942) or (942a)


When the member is uncracked ζ = 0 and κtcs = κcs1 (Equation (917) orFig F9)

Example 93 Non-prestressed beam use of global coefficients

Estimate the deflection at mid-span for the beam of Example 91 (Fig94) by the method of global coefficients

The following values calculated in Example 91 are required hereBasic deflection Dc = 440mm

Mr = 528kN-m M = 1360kN-m ζ = 081

αρ = 200

30 times

06

100 = 004

Mr

M =

528

1360 = 039

Entering the last two values in the graph for φ = 25 in Fig 98 givesκt = 38 The probable instantaneous plus creep deflection (Equation(938) ) is

44 times 38 065

06 3

1 minus 20 times 015

100 = 206mm

Entering the graph of Fig 99 with ζ = 081 αρ = 004 and ρprimeρ = 025gives κtcs = 085 Thus the deflection due to shrinkage (Equation (942) )is

(∆D)cs = minus (minus250 times 10minus6) 82(085)

8(06) = 28mm

Estimated value of deflection including effects of creep and shrinkage is

D = 206 + 28 = 234mm (094 in)

Example 94 Prestressed beam use of global coefficients

Estimate the deflection at mid-span of the prestressed beam in Fig 910due to the effects of a sustained load q = 20kNm (14kipft) combined


with prestressing Assume that the effective prestress after loss bal-ances 40 per cent of the dead load Use the method of global coef-ficients The beam has a rectangular cross-section as shown the area ofnon-prestressed steel at the bottom is 500mm2 (078 in2) and at the top200mm2 (031 in2) the area of prestressed steel is 200mm2 (031 in2)Other data are Es = 200GPa (29000ksi) Ec(t0) = 30GPa (4350ksi) fct =25MPa (036ksi) φ = 25 The prestress duct is grouted aftertensioning

Prestress force necessary to balance 40 per cent of q (at time t afterloss) is

P(t) = 04 times 20 times 103 times 82

8 times 0275 = 2327kN

Bending moment at mid-span due to dead load and effective prestressforce is

M = 20 times 60

100 times

82

8 = 960kN-m

Value of bending moment producing cracking (Equation (936) ) is


03 times 065 03(065)2

6 = 780kN-m

Mr

M =

78

96 = 081

Figure 910 Prestressed beam of Example 94


Total steel ratio

ρ = (500 + 200)10minus6


α = 200

30 = 667 αρ = 0026

Graph for φ = 25 in Fig 98 gives κt = 44 The basic deflection due tounbalanced load (Equation (C8) ) is

960 times 103

30 times 109(03 times 065312)

82

96 = 311mm

ρprime = 200 times 10minus6

03 times 060 = 00011

The probable deflection at mid-span (Equation (938) ) is

D 311 times 44065

0603

(1 minus 20 times 00011) = 171mm (0671 in)

Included in the data given in this example is the value of the effectiveprestress after loss due to creep shrinkage and relaxation However inpractice the initial prestress is known and the effective prestress must becalculated by an estimation of the amount of loss In this example theprestress balances only 40 per cent of the dead load but when theupward load produced by prestressing is of almost the same magnitudeas the downward gravity load the long-term deflection is mainly due toprestress loss Hence in such a case the estimate of deflection is largelyaffected by accuracy in the calculation of prestress loss

99 Deflection of two-way slab systems

This section is concerned with prediction of the maximum deflection inreinforced concrete floor systems taking into account the effects of creepshrinkage and cracking The method presented is applicable to slab systemswith or without beams between supports The supports are either columns orwalls arranged in a rectangular pattern

Calculation of the bending moments in two-way slab systems is extensively


covered in codes and books on structural design6 Tables and other designaids7 are available for this purpose In this section we assume that the bend-ing moment values at the supports and at mid-span are available ndash in the twodirections ndash at the centre lines of columns and at the centre lines of panelsAlso we assume that the reinforcement has been chosen and it is required todetermine the long-term deflection at the centre of the panels

991 Geometric relation

The deflection at the centre of a straight member relative to its ends can becalculated from the curvatures at three sections using the equation

δ = l 2

96 (ψ1 + 10ψ2 + ψ3) (943)

where

Equation (943) is based on the assumption that the variation of curvaturefollows a second degree parabola defined by the three ψ-values employedThis geometric relation which can be proved by double integration is validfor simply supported and for continuous members It is of course applicableto a strip of a slab

In most practical applications the main loading is the member self-weightproducing curvature which varies as second-degree parabola when the flex-ural rigidity is constant When cracking occurs the flexural rigidity is nolonger constant and the ψ-values will be changed Use of Equation (943) forcalculation of deflection of a member of variable cross-section or for acracked member results in tolerable error acceptable in practice as long asthe three ψ-values employed are determined with appropriate account of theflexural rigidity at the respective sections

Figure 911(a) is the top view of a two-way slab with rectangular panelsThe deflected shape of a typical panel is shown in Fig 911(b) The deflec-tion D at the centre of the panel can be expressed as the sum of deflectionsof a strip joining two columns and a strip running along a centre line ofthe panel One of the two following equations may be used (see Fig911(c) )

δ = deflection at centre measured perpendicular to the memberfrom the straight line joining the two ends (see Equation(C8) and Fig C2)

l = length of memberψ1 ψ2 and ψ3 = curvatures at the left end the centre and the right end of the

member


D = δEF + 1

2(δAB + δDC) (944)

or

D = δHI + 1

2(δAD + δBC) (945)

where D is the deflection at the centre of the panel δ represents the deflectionat the centre of a column or a middle strip with respect to its ends

Figure 911 Deflection at the centre of a panel of a two-way slab system (a) top view ofsystem (b) deflection of a typical panel (c) definition of symbols employed inEquations (944) and (945)


The values of δ required in any of the equations may be calculated byEquation (943)

Application of Equation (944) or (945) should theoretically give the sameanswer for the deflection at the centre this is in fact a check on compatibilityHowever practical application of the two equations results in differentanswers and it is here suggested that the deflection be considered equal to theaverage of the two answers The two answers may differ for the followingreasons (a) the true curvature variation is not parabolic (b) the curvaturevalues cannot be accurately determined The curvature is usually calculatedusing bending moment values based on elastic analysis which does notaccount for cracking or may account for cracking in an empirical way

992 Curvature-bending moment relations

In the elastic state without consideration of the effects of the presence of thereinforcement creep shrinkage or cracking the curvatures in x and y direc-tions at any point of a slab can be calculated from the bending moments dueto the applied load as follows

Mx = EcIg slab (ψx + ν ψy) My = EcIg slab (νψx + ψy) (946)

ψx = 12

Ech3 (Mx minus ν My) ψy =

12

Ech3 (minus ν Mx + My) (947)

where Ec is the modulus of elasticity of concrete and ν its Poissonrsquos ratio(normally close to 02) Mx and My are bending moment values in a strip ofunit width running in the x and y directions Ig slab is an effective moment ofinertia of the gross concrete area of the strip

Ig slab = h3

12(1 minus ν2)(948)

where h is slab thicknessWhen the floor system has beams the curvature of a beam is

ψbeam = M

EcIg beam

(949)

where Ig beam is the moment of inertia of the gross concrete cross-section Ifthe beam is monolithic with the slab the beam cross-section includes a por-tion of the slab on each side of the beam of width equal to the projection ofthe beam above or below the slab (This width may also be determined byother empirical rules)

The equations presented in this subsection give curvature values to be


substituted in Equation (943) to determine δ-values for column and middlestrips followed by Equation (944) or (945) to obtain the deflection at thecentre of two-way slab panels This gives a basic deflection value Dbasic whichdoes not account for the reinforcement creep or cracking The true deflectioncan be much higher than the value Dbasic (five to eight times) as will bediscussed in the following subsection

Table 928 may be employed to find the basic deflection value Dbasic at thecentre of an interior panel and at the centre of column strips for two-way slabsystems with or without beams It should be noted that the deflection valuesgiven in this table are based on elastic analysis of an interior panel Exteriorpanels usually have larger deflection Other limitations of Table 92 arementioned with the table

993 Effects of cracking and creep

In this subsection an approximate procedure is presented to account for theeffect of the reinforcement ratio cracking and creep on the deflection at thecentre of panels of two-way slab systems The effect of shrinkage will bediscussed in the following subsection

The deflection at the centre of a panel can be considered as the sum ofdeflection values δ for column and middle strips (see Equation (944) or(945) and Fig 911(c) ) The symbol δ represents the deflection at the middleof a strip relative to its ends First basic δ-values are determined using Equa-tion (943) and curvature values based on gross concrete sections withoutconsideration of cracking or creep The basic δ-values may also be extractedfrom Table 92 or from alternative sources

The deflection δ of a strip is largely influenced by the curvature at its mid-span Thus the reinforcement ratio ρ at mid-span section of the strip is usedto determine coefficients κs1 and κφ1 and also κs2 and κφ2 when crackingoccurs These coefficients are employed as multipliers to the basic δ-values toapproximately account for the effects of creep and cracking in the same wayas discussed for beams in Section 94

The suggested procedure is more clearly explained in steps given belowThe steps are to be followed after the bending moments and the reinforce-ments in the middle and column strips have been determined

(1) Calculate curvatures at ends and at mid-span for a column strip runningin the x or y direction and for a middle strip running in the perpendiculardirection In this step use the moment values corresponding to the serviceload for which the deflection is required Ignore the reinforcement creepshrinkage and cracking (Equations (946ndash49) )

(2) Use the curvatures to determine basic deflection δbasic of the two stripsrelative to their ends (by Equation (943) )

Alternatively steps 1 and 2 may be replaced by a design aid which givesδ-values such as Table 92


Table 92 Basic deflections for interior panels of two-way slab systems

Deflection at A B or C = (coefficient times 10minus3)ql 4

(EI) slab

(EI) beam cl = 00 cl = 01 cl = 02

sl s(EI) slab A B C A B C A B C

10 00 581 435 441 304 289 17302 438 299 same 340 207 same 240 122 same05 331 198 as 271 141 as 205 085 as10 260 130 B 222 092 B 179 056 B20 206 077 184 054 158 03340 174 043 159 030 144 018

08 00 420 378 230 301 262 131 189 155 05702 316 271 149 237 192 088 159 116 04005 146 195 099 191 138 059 136 085 02810 191 134 063 154 095 038 117 058 01820 147 083 036 124 058 022 100 036 01140 116 048 019 103 033 012 089 020 006

06 00 327 321 099 234 228 040 143 137 00802 256 246 063 189 178 027 119 108 00605 201 187 040 150 134 017 098 098 00410 153 135 025 116 096 011 079 059 00320 110 087 013 085 061 006 061 037 00240 077 051 007 063 035 003 048 022 001

04 00 284 284 031 204 204 00402 231 230 020 166 165 00305 183 181 012 131 128 00210 137 134 007 098 094 00120 093 088 004 066 061 00140 059 053 002 042 036 000

(El)slab = Ec

h3

12(1 minus ν 2)q = load intensity h = slab thickness ν = Poissonrsquos ratioBasic deflection values do not account for the effects of the reinforcement creep or cracking


(3) For the cross-section at the middle of the two strips determine thecurvature coefficients κs1 and κφ1 for a non-cracked section using graphs ofFigs F1 F3ndash5 or Equations (F1) and (F2) For each of the two stripscalculate the instantaneous plus creep value of δ in the uncracked state 1

δ1 = δbasicκs1(1 + φκφ1) (950)

Calculate the value Mr of the bending moment which produces cracking Ifthe bending moment at the centre of a strip is less than or equal to Mr nocracking occurs and δ = δ1 where δ is the deflection at the middle of the striprelative to its ends a value to be used in step 5

(4) When cracking occurs at the mid-span section of any of the two stripsdetermine the curvature coefficients κs2 and κφ2 for a fully cracked sectionusing graphs of Figs F2 F6 F7 and F8 or Equations (F1) and (F2) andcalculate the instantaneous plus creep deflection for a fully cracked strip


Calculate the interpolation coefficient using Equation (924) and determinethe δ-value including effects of creep and cracking

δ = (1 minus ζ)δ1 + ζδ2 (952)

(5) Add the δ-values of a column and a middle strip according to Equation(944) or (945) to obtain the deflection at the centre of the panel For a morereliable answer two possible patterns of strips may be used and the probabledeflection considered equal to the average of the answers from the twopatterns

When the column strips running in one direction have different δ-values anaverage value is to be used in Equations (944) and (945) as shown in Fig911(c)

Example 95 Interior panel

Figure 912 is a top view of an interior square panel of a two-way slabsupported directly on columns It is required to calculate the long-termdeflection due to a uniform load 840kNm2 (175 lbft2) which repre-sents the dead load plus a part of the live load The bending moments9

due to this load are indicated in Fig 912(b) for a section at mid-span ofa column and a middle strip The reinforcement cross-section areas10 atthese two locations are given in Fig 912(a) Other data are slab thick-ness h = 020m (8 in) average distance from top of slab to centroid of


bottom reinforcements in x and y directions d = 016m (63 in) Modu-lus of elasticity of concrete at time t0 when the load is applied Ec (t0) =25GPa (3600ksi) creep coefficient φ = 25 aging coefficient χ = 08tensile strength of concrete in flexure (modulus of rupture) fct =20MPa (290psi) modulus of elasticity of the reinforcement = 200GPa(29000ksi)

Effective moment of inertia of the gross concrete section of a strip ofunit width (Equation (948) ) is

Ig slab = (02)3

12(1 minus 022) = 694 times 10minus6 m4m

Poissonrsquos ratio is assumed equal to 02In this example the basic deflection can be calculated using coef-

ficients from Table 92 which gives

deflection at centre of panel

000482ql 4

EcIg slab

= 560mm (0221 in)

Deflection at mid-span of column strip

000342ql 4

EcIg slab

= 397mm (0157 in)

Figure 912 Calculation for deflection at the centre of an interior panel of a two-wayslab accounting for creep and cracking (Example 95) (a) top view of aninterior panel (b) bending moments at mid-span of column and middlestrips due to service load 84kNm2 (175 lbft2)


where q (load intensity) = 84kNm2 and l the span measured centre tocentre of columns = 700m

The basic deflections of the middle section of column and middlestrips relative to their ends are

δABbasic = 397mm δEFbasic = 163mm

The effects of creep and cracking are calculated separately below foreach of the two strips

Column stripThe following parameters are determined for a section of unit width atthe middle of the strip b = 100m d = 016m ρ = Asbd = 406 times 10minus3χφ = 20 α = 80 Curvature coefficients for the section in uncrackedstate 1 (from graphs of Figs F1 and F5 or Equations (F1) and (F2) )

κs1 = 098 κφ1 = 093

In the fully cracked state 2 the depth of compression zone is 0036mand the curvature coefficients are (Figs F2 and F6 or Equations (F1)and (F2) )

κs2 = 70 κφ2 = 014

Lower and upper bounds of deflection of the strip corresponding tostates 1 and 2 (Equations (950) and (951) ) are

δ1 = 397(098)[1 + 25(093)] = 1294mm

δ2 = 397(70)[1 + 25(014)] = 3752mm

Value of the bending moment which produces cracking (Equation(925) ) is

Mr = ƒctW1 = 20 times 106 10 times 022

6 = 133kN-m

(the reinforcement is ignored in calculation of W1) The interpolationcoefficient (Equation (924) ) is


1 minus β1β2

Mr

M = 1 minus 10(05)

133

186 = 064

The deflection at mid-span relative to the ends of the column stripincluding effects of cracking and creep (Equation (952) ) is

δAB = (1 minus 064)1294 + 064(3752) = 2867mm

Middle stripThe value of the bending moment at mid-span does not exceed Mr thusno cracking occurs The cross-section has the following parameters

b = 100m d = 016m p = Asbd = 281 times 10minus3 χφ = 20

Curvature coefficients for the section in the uncracked state 1

κsl = 098 κφ1 = 095

The deflection at mid-span relative to the ends of a middle strip includ-ing effect of creep (Equation (950) ) is

δEF = 163(098)[1 + 25(095)] = 538mm

Deflection at centre of panel including effects of creep and cracking(Equation (944) is

D = 2867 + 538 = 3405mm (1341 in)

Example 96 Edge panel

Figure 913(a) is a typical bay of a two-way slab system of equal spans700m in the x and y directions The slab is provided by edge beamsrunning in the y direction It is required to find the deflection at thecentre of an edge panel ABCD due to load 840kN-m2 (175 lbft2) Thecorresponding bending moments11 in column and in middle strips andin the edge beam are indicated in Fig 913(b) and the reinforcements atmid-span sections in Fig 913(c) Other data are the same as inExample 95


The basic deflection cannot be calculated by the use of a design aidsuch as Table 92 because it does not apply Thus we employ theequations of subsections 991 and 992

The effective moment of inertia of the gross concrete section of astrip of unit width (Equation (948) ) is 694 times 10minus6 m4m

Figure 913 Two-way slab of Example 96 (a) top view of an interior bay of system(b) bending moment in slab and in edge beam of panel ABCD 1kN-mm = 225 lb-ftft (c) reinforcement cross-section areas at mid-spanof column and middle strips 1mm2m = 0472 times 10minus3 in2ft


Consider a column strip along AB the curvature in the x direction atthe two ends is

ψ1 = minus234 times 103

(25 times 109)(694 times 10minus6) = minus1349 times 10minus6 mminus1



Curvature at middle of the strip (Equation 946) is

ψ2 = 12[281 minus 02 (minus144)] 103

(25 times 109) (02)3 = 1859 times 10minus6 mminus1

Basic deflection at mid-span of the strip relative to its ends (Equation(943) ) is

δABbasic = (655)2

96 (minus1349 + 10 times 1859 minus 2726)10minus6 = 649mm

For a middle strip along EF the basic curvatures at the ends and at themiddle are (Equation (946) )

ψ1 = ψ3 = 12[minus144 minus 02(281)]103

(25 times 109) (02)3 = minus1201 times 10minus6 mminus1

ψ2 = 12[101 minus 02(187)] 103


Basic deflection at mid-span relative to ends (Equation (943) ) is

δEFbasic = (700)2

96 (minus1201 + 10 times 382 minus 1201) 10minus6 = 072mm

Basic deflection at centre of panel (point G Equation (944) ) is

Dbasic = 649 + 072 = 721mm (0284 in)

A second pattern of strips may be used to calculate Dbasic as theaverage δ-value for column strips BC and AD plus δ-value for the


Tabl

e 9

3D

eflec

tion

at c

entr

e of

pan

el c

alcu

late

d fr

om t

wo

stri

p pa

tter

ns E

xam

ple

96

Strip

Bend

ing

Curv

atur

eCu

rvat

ure

Inte

r-D

eflec

tion

mom

ent a

tG

eom

etric

alco

effic

ient

s an

dco

effic

ient

s an

dpo

latio

nof

Basic

mid

-spa

nCr

acki

ngpr

oper

ties

of s

ectio

nde

flect

ion

inde

flect

ion

inco

effic

ient

strip

defle

ctio

nsp

an

mom

ent

at m

id-s

pan

uncr

acke

d st

ate

fully

-cra

cked

sta

te

basic

MM

r

A sd

κ

s1κ

1

1

κs2

κ

2

2

(m

m)

(kN

-m)

(kN

-m)

(mm

2 )(m

)(1

0minus3)

(mm

)(m

m)

(mm

)

AB

649

281

133

900

017

05

290

960

8819

91

468

016

424

60

7637

05

EF0

7210

113

335

00

155

226

099

097

244

1ndash

ndashndash

244

Firs

t es

timat

e of

defl

ectio

n at

cen

tre

of p

anel

D =

A

B +

EF

= 3

705

+ 2

44

= 39

49

mm

BC3

5518

613

36

500

155

419

098

093

115

77

480

1435

85

064

271

12

003

004

089

AD

21

4641

220

44

top

092

079

400

821

009

146

80

7512

03

kN-m

kN-m

500

045

222

bot

tom

HI

451

187

133

600

017

03

530

970

9114

33

655

013

391

40

6430

21

Seco

nd e

stim

ate

of d

eflec

tion

at c

entr

e of

pan

el D

= ndashsup1 sup2

(BC

+

AD)+

H

I=

ndashsup1 sup2(2

711

+12

03)

+30

21

= 49

78

Defl

ectio

n at

cen

tre

of p

anel

incl

udin

g ef

fect

s of

cre

ep a

nd c

rack

ing

=39

4+

497

8

2=

446

mm

(17

6in

)

1W

hen

M

Mr c

rack

ing

does

not

occ

ur

=

1 and

the

col

umns

for

κs2 κ

2

2 a

nd

are

left

bla

nk

2T

he e

dge

beam

is t

reat

ed a

s a

T-s

ectio

n w

ith fl

ange

wid

th =

05

0m

3T

his

line

give

s Aprime

s dprime

and

prime f

or t

he t

op r

einf

orce

men

t of

str

ip A

D

middle strip along HI This gives Dbasic = 702mm (0276 in) (see Table93)

The effects of the presence of the reinforcement and cracking on theinitial deflection and the effect of creep on the long-term deflection arecalculated using the κ-coefficients as in Example 95 A summary of thecomputations is given in Table 93

994 Deflection of two-way slabs due to uniform shrinkage

The reinforcement in a slab restrains shrinkage resulting in curvature andstresses which tend to increase the deflection The deflection due to shrinkage ina two-way slab is of course dependent upon the amount of reinforcement in twodirections and on the extent of cracking in the two directions As an approxima-tion it is here suggested to calculate the deflection at the centre of a slab panelas the sum of shrinkage deflections at mid-spans of column and middlestrips treated as beams using the equations of Section 932 (see Fig 93)

As for beams the deflection is affected by shrinkage in two ways The firstis a direct effect shrinkage produces curvature which increases deflectionThe second effect is indirect shrinkage produces tensile stresses at bottomfibre at mid-span of the strips and hence enhances cracking This may beapproximately accounted for by an appropriate reduction of the tensilestrength of concrete in flexure (the modulus of rupture) ƒct when calculatingthe value Mr of the bending moment which produces cracking (Equation(925) ) as will be demonstrated in the following example


Calculate the deflection in the two-way slab panel of Example 95 dueto shrinkage εcs = minus300 times 10minus6

Column strip AB (Fig 912)For a section at mid-span As = 900mm2m d = 017m h = 020m dh =085 α = EsEc = 8 αρ = 00424 χφ = 2 Equation (F3) or Figs F9 andF10 give for the uncracked and the fully cracked states κcs1 = 030 andκcs2 = 112

Deflection due to shrinkage in non-cracked and fully cracked stripswhen simply supported (Equations (915) and (916) ) is

(∆D1)cs = 300 times 10minus6(030)(655)2

8(017) = 288mm


(∆D2)cs = 300 times 10minus6(112)(655)2

8(017) = 1061mm

Interpolation between these two values using ζ = 076 (see Table 93)gives for the simply supported strip

(∆D)cs = (1 minus 076)288 + 076(1061) = 875mm

The slab is continuous over three equal spans in the direction of stripAB multiply the simple-beam deflection by 07 according to Equation(919) to obtain the deflection at mid-span of strips AB or DC relativeto their ends

δAB = δDC = 07(875) = 613mm

Middle strip EF (Fig 912)For a section at mid-span As = 350mm2m d = 0155m h = 020m dh= 078 α = EsEc = 8 αρ = 00181 χφ = 2 Equation (F3) or Fig F9 givesfor the uncracked state κcs1 = 010

Deflection due to shrinkage in the uncracked state for a simplysupported strip is

(∆D1)cs = 300 times 10minus6(010)(700)2

8(0155) = 119mm

The bending moment due to applied load is not sufficient to producecracking at mid-span the fully cracked state need not be consideredThe deflection of the strip if it were simply supported is

(∆D)cs = 119mm

For an interior span of a continuous strip multiply the simple-beamdeflection by 05 according to Equation (919) thus

δEF = 05(119) = 060mm

The deflection at the centre of the slab panel due to shrinkage(Equation (944) with δDC = δAB) is

(∆D)cs = 613 + 060 = 673mm


Consider an alternative strip pattern deflection at centre of panel

= 1

2(δAD + δBC) + δHI Similar calculations as above give δAD = 140mm δBC

= 395mm δHI = 592mm The deflection at the centre of the panel dueto shrinkage is 860mm

The average of the two values obtained by the strip patterns con-sidered is the probable deflection at the centre of the panel due toshrinkage and is equal to 77mm (030 in) Addition of this value to thedeflection value 446mm (176 in) calculated in Table 93 gives the totaldeflection including the effects of creep shrinkage and cracking

For the indirect effect of shrinkage we determine the tensile stress atbottom fibre at mid-span of strips In this problem the indirect effect ofshrinkage is small and will be calculated for strip AB only At mid-spanwe have Ec = Ec(1 + χφ) = 833GPa α = EsEc = 24 Ac = 01991m2 A =02207m2 distance between centroid of A (the age-adjusted trans-formed section) and the bottom fibre yb = 0093m

Curvature due to shrinkage if the strip were simply supported(Equation (910) with κcs1 = 030) is

(∆ψ)cs = 300 times 10minus6 030

017 = 536 times 10minus6 mminus1

Tensile stress at bottom fibre caused by shrinkage (Equation (913) ) is

∆σbot = 833 times 109(536 times 10minus6)0093 minus (minus300 times 10minus6) 1 minus01991

02207= 0660MPa

The value of tensile strength of concrete in flexure fct = 20MPa maynow be reduced to 1340MPa Cracking occurs at a reduced bendingmoment (Equation (925) ) Mr = 893kN-mm The correspondinginterpolation coefficient (Equation (924) ) ζ = 084 which is largerthan the value ζ = 076 calculated in Table 93 This means that theindirect effect of shrinkage is to bring the deflection closer to the fullycracked state and gives δAB = 3885mm instead of 3705mm calculatedin Table 93

From the above it can be seen that the indirect effect of shrinkage ismore conveniently accounted for by estimating a reduced value of ƒct

and then using it in the calculations for Table 93


910 General

The simplified procedures of deflection calculation presented in this chapterare justified by extensive studies12 comparing the results with more accuratemethods using a wide range of the parameters involved

Codes of various countries specify limits to the maximum deflection whichare not discussed here but only a brief discussion is made below of problemswhich may result from excessive deflection

Visibly large deflections are a cause of anxiety for owners and occupants ofstructures However the human eye is not generally speaking very sensitive todeflections and relatively large values can be tolerated An exception is when theeye can be situated at the same level as the bottom of the member If appearanceis the only concern one should avoid deflections greater than the span250

Excessive deflections can produce cracking in partitions or cause damageto other non-structural elements eg glass panels A limit on acceptabledeflections in such cases is often suggested to be the smaller of span500 or10mm However unacceptable damages have been reported with deflectionsas small as span1000

The age of concrete when various dead loads are applied is generally notknown at the time of design In prediction of deflection all dead loads maybe assumed to be introduced simultaneously at a chosen average age

Notes

1 See reference mentioned in Note 1 page 3022 Branson DE (1977) Deformation of Concrete Structures McGraw-Hill New

York3 See reference mentioned in Note 1 page 3024 See reference mentioned in Note 1 page 3025 See reference mentioned in Note 1 page 3026 See for example the following references

ACI 318ndash01 Building Code Requirements for Structural Concrete and Commen-tary American Concrete Institute Farmington Hills Michigan 48333ndash9094Park R and Gamble WL (1980) Reinforced Concrete Slabs Wiley New York

7 See for example the following referencesTimoshenko S and Woinowsky-Krieger S (1959) Theory of Plates and ShellsMcGraw-Hill New YorkSzilard R (1974) Theory of Analysis of Plates Classical and Numerical MethodsPrentice Hall Englewood Cliffs New Jersey

8 Table 92 is extracted from Vanderbilt MD Sozen MA and Siess CP (1965)Deflections of multiple-panel reinforced concrete floor slabs J Struct Div AmSoc Civil Engrs 91 No ST4 August 77ndash101

9 The bending moment values are determined by the lsquoDirect Design Methodrsquo of thefirst two references mentioned in Note 6 above

10 The reinforcement cross-section area approximately corresponds to an ultimatestrength design with ultimate load 148kNm2 and yield strength of reinforcement= 400MPa (58ksi)

11 See Notes 9 and 10 above12 See reference mentioned in Note 1 above


Effects of temperature

Post-tensioned precast segmental bridge erected by means of a launching trussKishwaukee River Illinois (Courtesy Prestressed Concrete Institute Chicago)

Chapter 10

101 Introduction

It is well known that changes in temperature can produce stresses in concretestructures of the same order of magnitude as the dead or live loads Howeverthe stresses due to temperature are produced only when the thermal expan-sion or contraction is restrained High tensile stresses due to temperatureoften result in cracking of concrete once this occurs the restraint to thermalexpansion or contraction of concrete is gradually removed and its stressesreduced

Most design codes require that temperature effects be considered althoughin many cases very little guidance is given on how this can be done Thermalstresses can be substantially reduced and the risk of damage caused by tem-perature eliminated by provision of expansion joints and sufficient well-distributed reinforcements For this reason and because of the complexity ofthe problem many structures are designed by following empirical rules fordetails (eg Equation (E1) ) with virtually no calculation of the effects oftemperature However for important structures exposed to large temperaturevariations eg structures with members of relatively large depth exposed tothe weather it is appropriate to have assessment of the magnitude of tem-perature variations and the corresponding stresses This chapter attempts to

Precast concrete liquid storage tank Gold Beach Oregon (Courtesy PrestressedConcrete Institute Chicago)


solve some of the problems involved Particular attention is given to bridgesuperstructures

Bridges are usually provided with expansion joints which allow thelongitudinal movement due to temperature expansion in the direction ofthe bridge axis Even with these joints important stresses can develop par-ticularly when the structure is statically indeterminate The stresses in thelongitudinal direction in a bridge cross-section will be analysed here treatingthe structure as a beam

The first part of this chapter is concerned with temperature distribution inbridge cross-sections other sections focus on analysis of the correspondingstresses Effects of creep and cracking on the response of concrete structuresto temperature variations will be briefly discussed

102 Sources of heat in concrete structures

The chemical reaction of hydration of cement generates heat over the curingperiod1 A significant rise of temperature may occur in thick members whenthe dissipation of heat by conduction and convection from the surfaces is at asmaller rate than the liberated heat of hydration Because the conductivity ofconcrete is relatively low steep temperature gradients can occur between theinterior of a large concrete mass and the surfaces so that the resulting stressesmay produce cracking A temperature rise of 30 to 50 degC (54 to 90 degF) can beexpected in members thicker than 05m (16) ft2

The stresses due to heat of hydration occur at an early age and are thusconsiderably relieved by creep Prediction of temperature distribution and thecorresponding stresses and deformations due to heat of hydration creep andshrinkage is a complex problem which has been treated only in simplifiedcases3

Exposed concrete structures eg bridges continuously lose and gain heatfrom solar radiation convection and re-radiation to or from the surround-ing air Analysis of heat flow in a body is generally a three-dimensionalproblem For a concrete slab or wall or for a bridge cross-section it issufficient to treat it as a one- or two-dimensional problem The major partof this chapter is concerned with temperature distribution and the corres-ponding stresses in bridge cross-sections4 The temperature at any instant isassumed constant over the bridge length but variable over the cross-section

The temperature distribution over a bridge cross-section varies with timeand depends upon several variables

1 geometry of the cross-section2 thermal conductivity specific heat and density of the material3 nature and colour of the exposed surfaces expressed in terms of solar

radiation absorptivity emissivity and convection coefficients

Effects of temperature 351

4 orientation of the bridge axis latitude and altitude of the location5 time of the day and the season6 diurnal variations of ambient air temperature and wind speed7 degree of cloudiness and turbidity of the atmosphere

In daytime especially in summer heat gain is greater than heat loss result-ing in a rise of temperature The reverse occurs in winter nights and thetemperature of the structure drops Figure 101 is a schematic representationof heat flow for a bridge deck during daytime in summer Incident solarradiation is partly absorbed and the rest is reflected The absorbed energyheats the surface and produces a temperature gradient through the deck Theamount of absorbed radiation depends upon the nature and colour of thesurface the absorptivity is higher in a dark rough surface compared to asmooth surface of light colour Some of the absorbed heat of radiation is lostto the air by convection and re-radiation from the surface The amount ofconvection depends upon wind velocity and the temperatures of the air andthe surface

103 Shape of temperature distribution in bridgecross-sections

Bridges are generally provided with bearings which allow free longitudinaltranslation of the superstructure A change in temperature which varies lin-early over the cross-section of a simply supported bridge produces nostresses When the temperature variation is non-linear the same bridge will besubjected to stresses because any fibre being attached to other fibres cannot

Figure 101 Heat transfer processes for a bridge deck in daytime in summer


exhibit free temperature expansion Thermal stresses in the cross-section of astatically determinate structure will be referred to as the self-equilibratingstresses

Figure 102 shows the strain and stress distribution and the deflection ofa simple beam subjected to a rise of temperature which varies linearly ornon-linearly over the depth of the section Two lines are shown for thestrain distribution in the case of non-linear temperature variations Thebroken line represents the hypothetical strain which would occur if eachfibre were free to expand But because plane cross-sections tend to remainplane the actual strain distribution is linear as shown The differencebetween the ordinates of the broken line and of the straight line representsexpansion or contraction which is restrained by the self-equilibratingstresses Calculation of the actual strain and the self-equilibrating stressin a statically determinate structure was discussed in Section 24 andExample 21

In a continuous bridge a temperature rise varying linearly or non-linearlyover the cross-section produces statically indeterminate reactions andinternal forces The stresses due to these forces are referred to as continuitystresses

A change in temperature which is uniform over a bridge cross-section willresult in a longitudinal free translation at the bearings without change instresses or in transverse deflections Thus for the purpose of calculation ofstresses or deflections the temperature variation over the cross-section maybe measured from an arbitrary datum Fig 103 represents the distribution

Figure 102 Deflection strain and stress distribution in a simple beam due to a rise oftemperature which varies linearly or non-linearly over the depth


over the cross-section of a bridge of a temperature rise which may be con-sidered for design of box- and T-girders The distribution is a combination ofstraight lines and a fifth-degree parabola and is based on finite differenceanalyses by Priestley5 The temperature ordinates shown in Fig 103 aremeasured from a datum representing the temperature over the zone in thevicinity of mid-height of the web

The temperature distribution in Fig 103 represents the conditions in theearly afternoon of a hot summer day A temperature distribution of the sameform but reversed in sign (with smaller ordinates) is often suggested fordesign to consider the conditions in winter during the night or early in themorning

104 Heat transfer equation

With the assumption that the temperature distribution over a bridge cross-section is constant over the bridge length no heat flow occurs in the longi-tudinal direction and the following well-known equation applies for the heatflow in the plane of the cross-section

kpart2T

partx2 + k

part2T

party2 + Q = pc

partT

partt(101)

where

Figure 103 Distribution of a rise of temperature suggested by Priestley for design ofbridges or T or box sections


Heat flow at any instant at any point on the cross-section boundariesfollows the equation

kpartT

partxnx + k

partT

partyny + q = 0 (102)

where nx and ny are direction cosines of an outwards vector normal to theboundary q is the amount of heat transfer per unit time per unit area of theboundary in units of Wm2 (or Btuh ft2)

The value of q which varies with time and with position of the point onthe boundary is the sum of three components

q = qs minus qc minus qr (103)

where

The solar radiation can be expressed

qs = αaIs (104)

where αa is a dimensionless solar radiation absorptivity coefficient less than10 Is is the total heat from sun rays reaching the surface per unit area perunit time

The solar energy incident upon a surface normal to rays of the sun at apoint on the outer edge of the earthrsquos atmosphere is almost constant andequal to 1350Wm2 (428Btu(h ft2) ) However seasonal variation of the dis-tance between the sun and the earth produces variation in radiation in therange of plusmn 3 per cent

T = the temperature at any point (x y) at any instant tk = thermal conductivity which is the rate of heat flow by conduction per unit

area per unit temperature gradient The units of k are W(m degC) (or Btu(h ft degF) )

Q = amount of heat generated within the body (eg by hydration of cement)per unit time per unit volume Wm3 (or Btu(h ft3) )

ρ = density kgm3 (or lbft3)c = specific heat capacity that is the quantity of heat required to increase

the temperature of the unit mass of the material by one degree J(kg degC)(or Btu(lb degF) )

qs = the solar radiation that is the heat gain due to sun raysqc = the convection due to temperature difference between surface and airqr = the re-radiation from the surface to the surrounding air


Only a portion of this solar radiation reaches the earthrsquos surface becauseof the atmosphere which acts like a filter The amount of radiation whichreaches the surface of the earth depends upon the length of the path of thesunrsquos rays through the atmosphere hence on the latitude and altitude It alsodepends upon the air pollution The angle of incidence of the sunrsquos rays onthe surface also affects the amount of solar radiation6

The maximum solar radiation in summer on a horizontal surface in Europeand North America (around latitude 50) is in the order of 800ndash900Wm2

(250ndash300Btu(h ft2) )The amount of heat transfer by convection is given by Newtonrsquos law of

cooling

qc = hc(T minus Ta) (105)

where T and Ta are temperatures of the surface and of the surrounding airrespectively hc is the convection heat transfer coefficient (W(m2 degC) or Btu(hft2 degF) ) The value hc depends mainly upon wind speed and to a small degreeon the orientation and configuration of the surface and type of material7

The amount of re-radiation from the surface to the air is given by theStefan-Boltzmann law which may be written in the form

qr = hr(T minus Ta) (106)

where hr is the radiation heat transfer coefficient given by

hr = Csαe [(T + T)2 + (Ta + T)2](T + Ta + 2T) (107)

where Cs = Stefan-Boltzmann constant = 567 times 10minus8 (W(m2K4) or (0171 times10minus8 Btu(h ft2 degR4) ) T = constant = 273 used to convert temperature fromdegrees Celsius (degC) to degrees Kelvin (K) (or = 460 to convert degrees Fahr-enheit (degF) to degrees Rankin (degR) αe is a dimensionless coefficient of emissiv-ity of the surface and takes a value between 0 and 1 The latter value is for anideal radiator the black body

Equation (107) indicates that hr can be calculated only when the tempera-ture T of the surface is known However because hr is only slightly affectedby T in a time-incremental solution of Equations (101) and (102) anapproximate value of hr can be employed based on earlier values of T

The convection and re-radiation coefficients hc and hr may be combined inone overall heat transfer coefficient for the surface

h = hc + hr (108)

and the heat flow by convection and re-radiation qcr can be expressed by oneequation combining Equations (105) and (106)


qcr = h(T minus Ta) (109)

where

qcr = qc + qr (1010)

For analysis of temperature distribution over the thickness of a slab ora wall it is sufficient to employ a simplified one-dimensional form ofEquations (101) and (102) by dropping out the term involving x (or y)

Numerical solution of the differential Equation (101) subject to theboundary condition expressed by Equation (102) gives the temperature dis-tributions at various time intervals Finite difference or finite elements8

methods may be employed

105 Material properties

From the preceding sections it is seen that a number of values related tothermal properties of the material are involved in heat transfer analyses Forconcrete the material properties vary over wide ranges depending mainly oncomposition and moisture content

Table 101 gives several material properties which may be employed foranalysis of temperature distribution and the corresponding stresses in bridgecross-sections

The following values may be employed for the convection heat transfercoefficient hc (W(m2 degC))(or Btu(h ft2 degF) ) based on a wind speed of 1ms(3 fts) for all surfaces of a box-section bridge except for the inner surfaces ofthe box where the wind speed is considered zero

106 Stresses in the transverse direction in abridge cross-section

In Section 103 we discussed analysis of self-equilibrating and continuitythermal stresses in the direction of the axis of a bridge Equally importantstresses occur in the transverse direction in a closed box cross-section

W(m2 degC) Btu(h ft2 degF)

Top surface of concrete deckAsphalt coverBottom surface of a cantileverInner surfaces of boxOutside box surface

8588603575

1516110613


Figure 104(a) represents the cross-section of a box-girder bridge sub-jected to a rise of temperature which is assumed to be constant over thelength of the bridge but varies arbitrarily over the cross-section Stresses inthe transverse direction may be calculated by considering a closed-planeframe made up of a strip between two cross-sections of the box a unitdistance apart (Fig 104(b) ) The method of analysis discussed below isapplicable to any plane frame of a variable cross-section subjected to atemperature rise which varies non-linearly over any individual cross-section(Fig 104(b) ) and varies from section to section The material is assumed tobe homogeneous and elastic

If temperature expansion is artificially restrained the restraining stress atany fibre will be

σrestraint = minusEαt

(1 minus ν2)T(y) (1011)

where T(y) is the temperature rise at the fibre considered αt is the coefficientof thermal expansion E is the modulus of elasticity and ν is Poissonrsquos ratioThe term (1 minus ν2) is included in this equation because the expansion of thestrip (of unit width) considered here is restrained by the presence of adjacentidentical strips When considering an isolated plane frame or when expansionin the direction of the bridge axis is free to occur which is the common casethe term (1 minus ν2) should be dropped

At any section 1 minus 1 the restraining stresses have the following resultants

Table 101 Material properties

Concrete Steel Asphalt

Thermal conductivity k 15ndash25 (087ndash15) 45 (26) 10 (060)W(m degC)or[Btu(h ft degF)]

Specific heat c J(kg degC) 840ndash1200 (020ndash029) 460 (011) 920 (022)or [Btu(lb degF)]

Density kgm3 or 2400 (150) 7800 (490) 2100 (130)(lbft3)

Solar radiationabsorptivity

065ndash080 07(rusted)

09

coefficient a

(dimensionless)Radiation emissivity

coefficient c

09 08 09

(dimensionless)Coefficient of thermal

expansion t per degC80 times 10minus6 (44 times 10minus6) 12 times 10minus6 (67 times 10minus6) ndash

(or per degF)


∆N = thickness

σrestraintdy (1012)

∆M = thickness

yσrestraintdy (1013)

∆N is a normal force at the centroid of the section Both ∆N and ∆M mayvary with s as a result of the variation of temperature or the thickness wheres is a distance measured on the frame from an arbitrary origin as shown inFig 104 The element ds in Fig 104(a) is isolated as a free body in Fig104(c) Considering equilibrium we can see that tangential and transverseforces of magnitudes per unit length equal to p and q must exist where

p = minusd(∆N)

ds(1014)

q = minus d2(∆M)

ds2(1015)

In other words the loads p and q must be applied in order to restrainartificially the thermal expansion The set of restraining forces for a typicalmember represents a system in equilibrium (Fig 104(d) )

Figure 104 Forces necessary for the artificial restraint of the transverse expansion due totemperature in a box-girder bridge (a) cross-section of a bridge treated as aplane frame ABCD (b) section 1ndash1 (c) free-body diagram (d) a set of self-equilibrating restraining forces for a typical member of a frame


The artificial restraint must now be eliminated by the application ndash to allmembers ndash of forces equal and opposite to those shown in Fig 104(d) Theinternal forces and stresses due to this loading on the continuous frame are tobe analysed by a conventional method (eg the displacement method seeSection 52) Let the stress at any fibre obtained from such an analysis be ∆σThe actual stress due to temperature is given by superposition

σ = σrestraint + ∆σ (1016)

107 Self-equilibrating stresses

Analysis of the stresses in a direction parallel to the axis of a bridge due totemperature is discussed here and in the following section As mentioned inSection 103 longitudinal stress (referred to as self-equilibrating stresses)occurs in a statically determinate bridge only when the temperature distribu-tion is non-linear Equation (230) can be used to calculate the self-equilibrating stress in a bridge cross-section when the temperature is assumedto vary in the vertical (y) direction The equations given below are usablewhen the temperature T varies in both the horizontal and vertical directions

Consider a statically determinate bridge having the cross-section shown inFig 105 In general the temperature distribution varies in both the x and ydirections Consider the stress strain and curvature caused by a temperaturerise with a given temperature distribution T(x y) If the temperatureexpansion is artificially restrained a normal stress will be produced and itsmagnitude at any fibre will be

σrestraint = minus EαtT(x y) (1017)

and the stress resultants on any section are

Figure 105 Cross-section of a statically determinate bridge symbols and sign conventionsused in Equations (1017ndash25)


∆N = σrestraint dx dy (1018)

∆Mx = σrestraint ydx dy (1019)

∆My = σrestraint xdx dy (1020)

where ∆N is a normal force at the centroidTo remove the artificial restraint we apply on the cross-section the forces

minus∆N minus∆Mx and minus∆My producing at any point the stress

∆σ = minus ∆N

A +

∆Mx

Ix

y + ∆My

Iy

x (1021)

where A is the area of the cross-section Ix and Iy are moments of inertiaabout centroidal axes x and y

The self-equilibrating stresses are given by superposition


The normal strain at the centroid O and the curvatures in the yz and xzplanes respectively are

∆εO = minus∆N

EA(1023)

∆ψx = minus ∆Mx

EIx

(1024)

∆ψy = minus ∆My

EIy

(1025)

Substitution of Equations (1017ndash20) in the last three equations wouldshow that the values ∆εO ∆ψx and ∆ψy are independent of the value of E

108 Continuity stresses

Equations (1023ndash25) give the axial strain and the curvatures at any cross-section of a statically determinate beam These can be used to calculate thedisplacements at member ends If these displacements are not free to occur asfor example in a continuous structure statically indeterminate forces developproducing continuity stresses which must be added to the self-equilibratingstresses to produce the total stresses at any section Analysis of the staticallyindeterminate forces can be performed by the general force or displacementmethods (see Sections 42 and 52)


Consider the simple beam shown in Fig 106(a) which has a constantcross-section subjected to a rise of temperature varying over the depth asshown in Fig 106(b) The displacements at the coordinates shown at themember ends are given by

D1 = minusD2 = ∆ψl

2(1026)

D3 = ∆εOl (1027)

where ∆εO and ∆ψ are respectively the normal strain at the centroid and thecurvature at any section caused by the temperature change and are given bysubstitution of Equations (1017ndash20) into (1023) and (1024) (dropping theintegral with respect to x)

∆εO = αt

A T b dy (1028)

∆ψ = αt

I T b ydy (1029)

where b = b(y) is the breadth of the cross-section at any fibre A and I are thecross-section area and moment of inertia about the centroidal axis

If the same beam AB is made continuous with an identical span BC asshown for the first beam in Fig 107 the rotation at B cannot occur and astatically indeterminate connecting moment must be produced at B The

Figure 106 Displacements at the ends of a simple beam due to temperature(a) coordinate system (b) temperature distribution


value of the connecting moment may be calculated by the force method (seeExample 101)

Figure 107 gives the statically indeterminate bending moment diagramsand the reactions in continuous beams of constant cross-section having twoto five equal spans subjected to a rise of temperature which varies overthe depth of the section in arbitrary shape (Fig 106(b) ) The staticallyindeterminate values are expressed in terms of the quantity ∆ψ the curvatureat any section when the static indeterminacy is released The numerical valueof ∆ψ is obtained by evaluation of the integral in Equation (1029) (seeExample 101)

Figure 107 Statically indeterminate forces due to temperature rise in continuous beams ofequal spans = curvature in a statically determinate beam = (tI) T bydy(see Fig 106(b) )

Example 101 Continuous bridge girder

Find the stress distribution at support B of the continuous bridgeshown in Figs 108(a) and (b) due to a rise of temperature whose dis-tribution varies over the depth of the cross-section as suggested byPriestley (Fig 103) Consider E = 300GPa (4350ksi) and αt = 10 times10minus5 per degC (56 times 10minus6 per degF) Ignore rigidity of the pavement

In accordance with the rules in Fig 103 the temperature rise to beconsidered in the analysis varies over the top 12m (4 ft) as a fifth-degreeparabola (Fig 108(b) )


The cross-section area A is 0877m2 the centroid O is at 0969mabove the soffit the moment of inertia I about a horizontal axisthrough the centroid is 01615m4

Hypothetical strain that would occur at any fibre if it were free toexpand (Equation (221) ) is

εf = 10 times 10minus522 y

125

where y is a distance in metres measured upwards from a point 02mabove the soffit (see Fig 108(b) )

The stress necessary to prevent this expansion (Equation (222) ) is

σrestraint = minus(30 times 109)10minus5 times 22 y

125

= minus(2652 times 106)(y)5 N-m2

The resultants of this stress are (Equations (223) and (224) )

Figure 108 Analysis of stress distribution due to temperature in a bridge girder(Example 101) (a) elevation and cross-section (b) rise of temperature(c) self-equilibrating stresses in any section (d) released structure (e)continuity stresses at B (f) total temperature stresses at B


∆N = minus(2652 times 106) 035102

0y5 dy + 25

120

102y5 dy

= minus2229 times 106 N

∆M = minus(2652 times 106) 035 102

0 (0769 minus y)y5 dy

+2512

102 (0769 minus y)y5dy = 07438 times 106 N-m

Release the artificial restraint by application of (minus∆N) and (minus∆M) onthe cross-section the resulting axial strain and curvature are (Equation(229) )

∆εo

∆ψ = 1

30 times 109 times

2229 times 106

0877

minus07438 times 106

01615

= 8473 times 10minus6

minus1535 times 10minus6 mminus1

The stress in a statically determinate beam (the self-equilibratingstresses Equation (230) ) is

σself-equilibrating = [2542 minus 4606y minus 2652(0769 minus y)5] MPa

for y = minus0431 to 0769m

or

σself-equilibrating = (2542 minus 4606y) MPa

for y = 0769 to 0969m

where y is the distance in metres measured downwards from thecentroid O The distribution of the self-equilibrating stress is shown inFig 108(c)

We use the force method for the analysis of the statically indeter-minate forces The structure is released by the introduction of hingesat B and C as shown in Fig 108(d) The displacements of thereleased structure at the two coordinates indicated are (Equation(1026) )


D1 = D2 = minus1535 times 10minus6 18

2 +

24


The displacements at the two coordinates due to F1 = 1 at coordinate 1(the flexibility coefficients)9

f11 = l

3EIAB

+ l

3EIBC

= 1

3 times 30 times 109 times 01615 (18 + 24) = 2890 times 10minus9

f21 = l

6EIBC


Because of symmetry one compatibility equation only is necessary tosolve for the two redundant forces (F1 = F2)

(f11 + f12)F1 = minusD1

F1 = minusD1

f11 + f12

= 8675 times 103 N-m

Thus the statically indeterminate bending moment at B or C is8675kN-m The stress distribution due to this bending moment (thecontinuity stress) is shown in Fig 108(e) The total stress distributionat B due to temperature is the sum of the values in Fig 108(c) and (e)the result is shown in Fig 108(f)

In a prestressed concrete bridge the prestress force near the top fibreover an interior support is often relatively high resulting in small or nocompressive stress at the bottom fibre in service conditions The ther-mal tensile stress in this area (see Fig 108(e) ) may cause vertical cracksnear the soffit in the vicinity of the support The consequence of thiscracking may be alleviated by provision of reinforcement and reductionof the prestress force (partial prestressing)

109 Typical temperature distributions inbridge sections

Concrete bridges of the same depth but with different cross-section shapeshave almost the same temperature distribution However the temperature


distribution and the resulting stresses vary considerably with the cross-sectiondepth With greater depth higher temperature stresses occur

Figure 101010 shows the temperature distributions and the correspondingself-equilibrating stresses in three bridges of the same depth but with thecross-section configurations shown in Fig 109 Fig 1011 shows thedistributions of temperature and self-equilibrating stresses in five cross-sections varying in depth from 025 to 225m (10ndash89 in)

Figure 109 Bridge cross-sections analysed by Elbadry and Ghali to study effects of sectionshape and depth on temperature distribution (a) solid slab (b) cellular slab(c) box girder

Figure 1010 Temperature and self-equilibrating stresses in three bridge cross-sectionswith the same depth (Fig 109) (summer conditions in Calgary Canada) fullcurve solid slab dotted curve cellular slab broken curve box girder


Composite cross-sections may exhibit a high temperature differencebetween concrete and steel (45 degC (81 degF) ) when the vertical sides of the websare exposed to the sun

1010 Effect of creep on thermal response

In Sections 107 and 108 we considered the stresses produced by a tempera-ture rise which varies in an arbitrary manner over the cross-section of aconcrete beam Now we shall consider that the rise of temperature developsgradually with time during a period t0 to t where t0 and t are the ages ofconcrete at the start and at the end respectively Assuming that the tempera-ture expansion is artificially prevented the normal stress which will beproduced at any fibre

σrestraint = minusEαT(x y) (1030)

where E = Ec(t t0) is the age-adjusted elasticity modulus of concrete asdefined by Equation (131) which is repeated here for the sake ofconvenience

Ec(t t0) = Ec(t0)

1 + χφ(t t0)(1031)

where

For values of χ and φ see Appendix A The aging coefficient χ = 10 when thestress is introduced in its entire value at time t0 and maintained constant totime t The value of χ is less than 10 when the stress is introduced gradually(see graphs in Figs A6 to A45)

The equations derived in Sections 107 and 108 for the self-equilibratingand the continuity stresses are applicable in the case considered here with Ereplaced by E The change in temperature due to weather conditions occursover a period of time (several hours or days) during which some creepoccurs Thus it may be more appropriate to employ E rather than theinstantaneous elasticity modulus of concrete This will generally result in asmaller absolute value of the calculated stresses due to temperature

Heat of hydration of cement causes a rise of temperature which maydevelop gradually to a peak over a period of time for example one week the


φ(t t0) = the ratio of creep during the period (t minus t0) to the instantaneousstrain due to a stress introduced at age t0

χ = χ(t t0) = the aging coefficient


Figure 1011 Distribution of temperature and self-equilibrating stress in bridge cross-sections of different depths and of shapes shown in Fig 109 (summerconditions in Calgary Canada) (a) solid slab (b) cellular slab (c) box girder


temperature rise subsequently vanishes slowly over a much longer periodThe stresses due to this temperature change may be analysed in steps bydividing the time into intervals and considering that increments of tempera-ture or stresses occur suddenly at the middle of the intervals For each inter-val an appropriate creep coefficient and modulus of elasticity is employed(see Section 58) Considering creep in this fashion will result in substantiallydifferent stresses from a calculation in which creep and change in modulus ofelasticity are ignored

In fact considering these time-dependent effects may indicate that thestresses developed at peak temperature reverse signs after a long time whenthe heat of hydration is completely lost11 This can be seen in Example 102which treats the problem using a step-by-step numerical analysis

A general procedure for a step-by-step procedure of stress analysis ofconcrete structures is discussed in Section 58 Consider here the applicationof the method for analysis of the self-equilibrating stresses in a cross-section of a concrete member due to a rise of temperature which varies withtime Divide the time during which the temperature change occurs into anumber of intervals The symbols timinus 1

2 ti and ti+ 1

2 represent the age of

concrete at the beginning middle and end of the ith interval At the endof any interval i the strain due to free temperature expansion is thesummation

αt i

j = 1

(∆T)j (1032)

This strain is prevented artificially by the introduction of stress (∆σrestraint)j

at the middle of the intervals The combined strain caused by temperatureand these stress increments is zero For the end of the ith interval we canwrite

αt i

j = 1

(∆T)j + i

j = 1

(∆σrestraint)j

Ec(tj) [1 + φ(ti + 1

2 tj)] = 0 (1033)

where Ec(tj) is the modulus of elasticity of concrete at the middle of the jthinterval φ(ti + 1

2 tj) is the ratio of creep occurring between the middle of the jth

interval and the end of the ith interval to the instantaneous strain when astress is introduced at tj The summation in the second term of the equationrepresents the instantaneous strain plus creep caused by the stress incrementsduring the intervals 1 2 i

In a step-by-step analysis when Equation (933) is applied at any interval ithe stress increments are known for the earlier intervals Thus the equationcan be solved for the stress increment in the ith interval giving


(∆σrestraint)i = minus Ec(ti)

1 + φ(ti + 12 ti)

αti

j = 1

(∆T )j

+ i minus 1

j = 1

(∆σrestraint)j

Ec(tj) [1 + φ (ti + 1

2 tj)] (1034)

The resultant of the stress increment (∆σrestraint)i for the ith interval is to beintegrated over the area of the cross-section to determine the correspondingstress resultants (∆N)i (∆Mx)i and (∆My)i Equal and opposite forces areapplied on the cross-section to remove the artificial restraint the correspond-ing stress strain or curvature are derived using Equations (1021ndash25) employ-ing a modulus of elasticity E = Ec(ti) The analysis in this way gives thechanges in the self-equilibrating stresses in the individual increments andthese may be summed up to find the stress at any time

Example 102 Wall stress developed by heat of hydration

Figure 1012(a) represents a typical distribution of the temperature risedeveloped by the heat of hydration in a concrete wall of thickness b LetTc be the difference of temperature rise between the middle surface ofthe wall and its faces assume the distribution to be parabolic at alltimes The value of the temperature rise Tc is assumed to be zero at theage of casting reaches a peak value Tc max at the age of 6 days and dropsto 006 Tc max at the age of 50 days It is required to determine the self-equilibrating stress distributions at ages 6 and 50 days Assume thattemperature expansion of the wall is free to occur

Use three time intervals for which the interval limits are 0 2 6 and50 days and assume the temperature increments at the wall centre linein the three intervals Tc max053 047 minus094 Assume the moduli ofelasticity of concrete at the middle of the three intervals Ec(28) 044073 100 where Ec(28) is the modulus of elasticity at age 28 days (seeEquation (A37) )

The following creep coefficients are required in the step-by-stepanalysis

φ(2 1) = 056 φ(6 1) = 060 φ(50 1) = 082

φ(6 4) = 055 φ(50 4) = 076 φ(50 28) = 048

Successive application of Equation (1034) with i = 1 2 and 3 gives


Figure 1012 Self-equilibrating stresses caused by heat of hydration of cement in athick concrete wall (Example 102) (a) assumed distribution oftemperature at any time (b) self-equilibrating stress at age 6 days (timeat which Tc max occurs) (c) self-equilibrating stress at age 50 days (timewhen the temperature rise is almost lost)


the following values of the stress increments at wall centre line if thetemperature expansion is artificially prevented

(∆σrestraint)1 = αtTc maxEc(28) minus 044

1 + 056 (053) = minus0150[αtTc maxEc(28)]


1 + 055 (053 + 047)

minus 0150

044 (1 + 060)

= minus0214[αtTc maxEc(28)]


1 + 048

times (053 + 047 minus 094) minus 0150

044 (1 + 082) minus

0214

073(1 + 076)

= 0727 [αtTc maxEc(28)]

Summation of the increments gives the following values of stresses atages 6 and 50 days

∆σrestraint(6) = minus0364[αtTc maxEc(28)]

∆σrestraint(50) = 0363[αtTc maxEc(28)]

The increments of self-equilibrating stress may be calculated separ-ately for each interval and then the increments added to give the stressdistributions shown in Figs 1012(b) and (c) at ages 6 and 50 daysrespectively The same results will be reached if the change in tempera-ture is considered instantaneous and the modulus of elasticity E =Ec(28) and the temperature distributions parabolic with values at thewall centre line of 0364 Tc max and minus0363 Tc max for the stresses at 6 and50 days respectively

For any symmetrical temperature distribution T as considered inthis example the self-equilibrating stress for an elastic material may becalculated by the equation

σ = minusEαtT + (Eαt) Taverage (1035)

where Taverage is the average temperature


Figures 1012(b) and (c) indicate that the stresses at the outer fibresof the wall are tensile at age 6 days but they become compressive at age50 days To have an idea about the magnitude of the self-equilibratingstress at the surfaces assume the following values Tc max = 30 degC (54 degF)wall thickness b = 10m (33 ft) (Ec(28) = 300GPa (4350ksi) αt = 1 times10minus5 per degC (06 times 10minus5 per degF) This gives the following stresses at theouter surface 219MPa (0317ksi) at age 6 days and minus218MPa(minus317ksi) at age 50 days Note that the dimension b does not directlyaffect the stress but of course it affects the value Tc max and the creepcoefficients

The stress reversal at the older age may be explained as follows Astress introduced at an early age causes a relatively large strain becauseof a smaller modulus of elasticity and larger creep A rise of tempera-ture at an early age can be restrained by a stress smaller in absolutevalue than the corresponding stress for a drop of temperature of thesame magnitude but occurring at an older age Thus the self-equilibrating stress developed while the temperature is rising is morethan offset by the self-equilibrating stress produced by the subsequentcooling

In this example we assumed that the wall is free to expand and wecalculated the self-equilibrating stresses Much larger stress wouldoccur if the wall edges were not free this can be seen by comparing themagnitude of σrestrained with the self-equilibrating stress in Fig 1012(c)

It can be seen from this example that for the analysis of the stressesdue to heat of hydration it is necessary to know the temperature distri-bution and its history as well as the mechanical properties of concreteat various ages starting from the time of hardening The informationrequired for the analysis is not usually easy to obtain

It is well known that creep of concrete is of a larger magnitude whenthe temperature is higher This is of importance when concrete is sub-jected to elevated temperature eg in power plants The step-by-stepanalysis discussed above and in Section 58 may be applied using valuesof creep coefficients that are functions of the temperature

1011 Effect of cracking on thermal response

In general the absolute values of stresses caused by temperature in a crackedreinforced concrete cross-section are smaller than in an uncracked section


Calculation of stresses caused by temperature on cracked structures is com-plex Simplifying assumptions are necessary to make the calculations reason-ably simple In the following we shall consider that cracking is produced byloads other than temperature and assume that the depth of the compressionzone remains unchanged by the effect of temperature With these assump-tions the analysis of the self-equilibrating stresses in a statically determinatestructure or the continuity stresses in an indeterminate structure may beperformed in the same way as discussed in Sections 107 and 108 But theactual cross-section of the members must be replaced by a transformedsection composed of the area of concrete plus α times the area of steel whereα = EsEc is the ratio of the modulus of elasticity of steel to that of concrete

For qualitative assessment of the effect of cracking we consider the cross-section in Fig 1013(a) of a statically determinate structure and calculatethe self-equilibrating stresses and the strain due to a temperature rise which

Figure 1013 Distributions of strain and self-equilibrating stresses due to temperature in ahomogeneous elastic cross-section (a) concrete cross-section (b)distribution of temperature rise (c) strain (d) stress


varies over the depth as shown in Fig 1013(b) Figures 1013(c) and (d) showthe distributions of strain and stress if the section is considered of homo-geneous elastic material The values shown may be checked by the methodpresented in Section 107 with the following data

Ec = 300GPa (4350ksi) αt = 1 times 10minus5 per degC (06 times 10minus5 per degF)

The same cross-section is considered cracked at the bottom or at the topand provided with 1 per cent reinforcement (6000mm2 (930 in2) ) at thecracked face (Figs 1014(a) and (d) ) Concrete is ignored over a cracked zoneof depth 0467m (184 in) The distributions of strain and the self-equilibrating stresses due to the temperature rise in Fig 1013(b) are shown inFigs 1014(b) and (c) when the cracking is at the bottom and in Figs 1014(e)and (f) when the cracking is at the top The values shown in these figures areobtained by application of Equations (221) (222) (229) and (230) andemploying the following properties of the transformed section α = EsEc =667 area A = 0223m2 (346 in2) moment of inertia about centroidal axis I =900 times 10minus3 m4 (21600 in4)

Comparison of the stress values in Figs 1013(a) 1014(c) and 1014(f)indicates that the self-equilibrating stresses caused by temperature are gener-ally smaller in the cracked section However the corresponding strain valuesand particularly the curvatures ∆ψ are not much different (Figs 1013(c)1014(b) and 1014(e) ) It follows that the strains and hence the displacements

Figure 1014 Distributions of strain and self-equilibrating stresses in a cracked reinforcedconcrete section due to a temperature rise shown in Fig 1013(b)(a) cracking at bottom face (b) strain (c) stress (d) cracking at top face(e) strain (f) stress


(eg elongations or rotations at member ends) due to temperature in a static-ally determinate cracked structure may be approximated by considering thecross-sections to be homogeneous elastic and with no cracks

It also follows that calculation of the statically indeterminate forces pro-duced by temperature using the force method (see Section 42) can be sim-plified by calculation of the displacements D of a statically determinateuncracked structure and considering cracking only when calculating theflexibility matrix [ f ] If for example the continuous beam in Fig 108(a)(Example 101) was cracked the flexibility coefficients would be (see Equation(332) )

f11 = M 2

u1 dl

EcIf12 =

Mu1Mu2dl

EcIf22 =

M 2u2 dl

EcI

where Mu1 and Mu2 are bending moments due to unit couples applied atcoordinates 1 and 2 on a released structure (Fig 108(d) ) I is the moment ofinertia of a transformed cracked section (or uncracked where no crackingoccurs) The displacements D calculated in Example 101 may be employedwithout change in the case of a cracked continuous beam The result of suchcalculation will generally give smaller statically indeterminate forces Fwhen cracking is considered

It should be mentioned that when the concrete in the cracked zone iscompletely ignored as suggested above the flexibility coefficients fij are gen-erally overestimated hence the calculated statically indeterminate forces willbe somewhat lower than the true values

The above discussion indicates that the stresses or the internal forces pro-duced by temperature depend upon the extent of cracking caused by otherloading Thus at service conditions when no or little cracking occurs thestresses or internal forces induced by a temperature increment are large com-pared to the effect of the same increment when introduced at a higher levelclose to the ultimate strength of the structure This may be seen by consider-ing the identical moment-curvature diagrams shown in Fig 1015 which aretypical for a reinforced concrete cross-section An increment (∆ψ) in thecurvature due to temperature is introduced in Fig 1015(a) in the servicecondition near the linear part of the graph The same increment introducednear the ultimate strength of the section (Fig 1015(b) ) produces a smallerincrement in moment compared to the increment in moment in Fig 1015(a)Thus the effect of temperature is of less significance at the ultimate load thanat service conditions The effects of temperature must be considered in designfor service conditions and sufficient reinforcement provided to ensure that thecracks are closely distributed and the crack width within acceptable limits asopposed to wide cracks far apart


1012 General

Design of concrete structures for the effects of temperature is complex Thetemperature distribution is not easy to predict and is variable with time Thecombination of the effect of temperature and that of other loadings is notclearly specified in design codes The stresses developed due to temperatureare affected by the age of concrete and by creep A further complicationresults from cracking which limits the validity of superposition This chapterby no means provides a complete solution to all these problems

The stresses due to temperature are generally smaller in a cracked structurecompared to a structure without cracks This favours the use of limited pre-stressing and provision of non-prestressed steel to furnish the necessarystrength with allowance for cracking as opposed to a design in which thetotal strength is provided for by prestressed reinforcement without cracking

Figure 1015 Comparison of bending-moment increments corresponding to an incrementof curvature introduced at service conditions (a) or near ultimate limitstate (b)


It is believed that with the first design the structure is less vulnerableto damages at peak temperatures when sufficient and well-distributednon-prestressed reinforcement is provided the cracks will be of small width

Notes

1 For information on the amount and rate of heat generated and additional refer-ences see Neville AM (1997) Properties of Concrete 4th ed Wiley New York

2 See Leonhardt F (1982) Prevention of Damages in Bridges Proceedings of theNinth International Congress of the FIP Stockholm Commission Reports Vol 1June

3 See Thurston SJ Priestley N and Cooke N (1980) Thermal analysis of thickconcrete sections American Concrete Institute Journal No 77ndash38 SeptndashOct347ndash57

4 For more details and references see Elbadry MM and Ghali A (1983) Nonlin-ear temperature distribution and its effects on bridges International Association ofBridge and Structural Engineering Proceedings pp 6683 Periodica 31983

5 Priestley MJN (1976) Design of thermal gradients for concrete bridges NewZealand Engineering 31 No 9 September 213ndash19

6 Extensive discussion and equations for evaluation of the solar radiation can befound in Duffie JA and Beckmann WA (1974) Solar Energy Thermal Pro-cesses Wiley New York

7 See Kreith F (1983) Principles of Heat Transfer 3rd edn Intext EducationalPublishers New York

8 See Elbadry MM and Ghali A (1983) Temperature variations in concretebridges Proc Am Soc Civil Engrs J Structural Div 109 No 10 2355ndash74

Also see Elbadry MM and Ghali A (1982) User manual and computerprogram FETAB finite element thermal analysis of bridges Research Report NoCE82ndash10 Department of Civil Engineering The University of Calgary CanadaOctober

9 See Appendix B of the reference mentioned in Note 3 page 9910 Figures 109ndash1011 are taken (with permission) from the last reference mentioned

in Note 4 above The values shown in the last two figures are based on heattransfer analyses with climatic conditions representing a summer day in CalgaryCanada (air temperature extremes in 24 hours 10 and 30 degC (50 and 86 degF) ) Thetemperature and stress graphs represent the variations when maximum absolutevalues occur (in the afternoon)

11 See Zienkiewicz OC (1961) Analysis of visco-elastic behaviour of concretestructures with particular reference to thermal stresses J Am Concrete InstProc 58 No 4 October 383ndash94


Control of cracking

111 Introduction

Cracking occurs in a concrete member when the stress in concrete reaches thetensile strength fct The value of fct depends upon several parameters Choiceof the appropriate value of fct is the first difficulty in the analysis of crackprediction to be discussed in this chapter

When a statically determinate member is subjected to external applied loadof sufficient magnitude to produce cracking the member stiffness drops andan increase in displacement occurs As long as the external applied load issustained there will be no change in the internal forces On the other handwhen cracking of a statistically indeterminate structure is due to temperature

Viaduc de Sylans-France (Courtesy Bouygues Contractor France)

Chapter 11

variation volumetric change or settlement of supports a reduction of stiff-ness occurs and the magnitude of the internal forces drops from the valuesexisting before cracking Cracking in the first and second cases will bereferred to as force-induced and displacement-induced cracking and analysisof the two types of cracking will be discussed

Provision of bonded reinforcement of sufficient magnitude and appropri-ate detailing can effectively limit the mean crack width to any specified valueThe amount of reinforcement required for crack control is discussed belowThe equations presented include two parameters which have to be predictedby empirical expressions The two parameters are the mean crack spacing srm

and the coefficient ζ used to account for the additional stiffness which con-crete in tension provides to the state of full cracking where concrete in tensionis ignored (see Equation 848) The empirical expressions used here for ζ andsrm are adopted from codes (see Equation 845 and Appendix E) the accuracyof the predicted crack width depends upon these empirical expressions

Concrete of very high strength reaching 80ndash100MPa (12000ndash15000psi) isincreasingly used in practice The increase in fct can prevent cracking But ifcracking takes place the crack width will generally be smaller because of theimproved bond between the concrete and the reinforcing bars Cracking ofhigh-strength concrete members will be discussed

112 Variation of tensile strength of concrete

The value of the normal force Nr andor the bending moment Mr at whichcracking of a section occurs is directly proportional to the tensile strength ofconcrete fct It is important to use an appropriate value of fct to predictwhether or not cracking will occur and to account for the effect of cracking inthe calculations of the probable deflections The minimum reinforcementrequired for control of cracking also depends upon the value fct which will befurther discussed in Section 115

The values of fct determined in tests for a given concrete composition candiffer from the average value fctm by plus or minus 30 per cent The value of fct

in a member varies from section to section as a result cracks do not all format the same load level Furthermore in a structure the value of fct is generallysmaller than the value measured by the testing of cylinders made out of thesame concrete the difference between the two values is larger in members oflarger size This can be attributed to microcracking and to surface shrinkagecracking resulting from the rapid loss of moisture in freshly placed concrete

Appendix A includes equations for fct according to codes and technicalcommittee reports

Control of cracking 381

113 Force-induced and displacement-induced cracking

Figures 111(a) and (b) show the variation of displacement D with the axialforce N in two experiments in which a reinforced concrete member is sub-jected to an axial force or imposed end displacement The two graphs areidentical in the range 0 N Nr where Nr is the force producing the firstcrack After cracking the behaviour depends upon the way the experimentis conducted Figure 111(a) represents the case when the force N is con-trolled during the experiment specified increments of N are applied and thecorresponding D is measured In Figure 111(b) the displacement D is con-trolled by imposing specified increments and measuring the correspondingvalue of N

In the force-controlled test the occurrence of a crack is accompanied by asudden increase in D without change in N In the displacement-controlledtest formation of a crack is accompanied by a sudden drop in the value of N

The cracks in both tests (Figs 111(a) and 111(b) ) correspond to the samevalues of N Nr1 Nr2 Nrn The first crack occurs at the weakest sectionwhen the stress in concrete reaches its tensile strength fct1 (the correspondingstrain εc 0001) The second crack occurs at the second weakest sectionwhen the stress reaches a value fct2 slightly greater than fct1 The distancebetween the two cracks cannot be smaller than the crack spacing sr At acrack the stress in the concrete is zero the distance sr is necessary for trans-mission by bond of the force from the reinforcement to the concrete untilthe stress σc again reaches the tensile strength This is discussed in AppendixE which mentions the parameters that affect the crack spacing and givesempirical equations for estimation of the mean crack spacing srm The max-imum number of cracks that can occur n is equal to the integer part of thequotient (lsrm) a subsequent increase of N or D causes a widening of theexisting cracks

The formation of each crack is accompanied by a reduction of the memberstiffness (increase in flexibility) this is demonstrated in Fig 111(b) by areduction in the slope of the NndashD diagram

The force-controlled test represents the effects of external applied forces ona statically determinate structure the cracking in this case is referred to asforce-induced cracking The behaviour of a statically indeterminate structuresubjected to external applied forces is more complex because of the changesin the statically indeterminate internal forces due to cracking The behaviourin the displacement-controlled test can occur in statically indeterminatestructures due to the effects of temperature variation shrinkage of concreteor settlement of supports

With force-induced cracking the stabilized cracking is reached when N gtNrn Because Nrn is not substantially larger than the force Nr1 we can expectstabilized cracking in most cases when the cracking force is exceeded


Figu

re11

1A

rei

nfor

ced

conc

rete

mem

ber

subj

ecte

d to

(a)

axi

al fo

rce

N (

b) im

pose

d en

d di

spla

cem

ent

D

On the other hand stabilized cracking rarely occurs with displacement-induced cracking If the first crack is formed at a displacement ∆D1 sub-sequent cracks require increments ∆D2 ∆Dn the magnitude of eachincrement is greater than the preceding one This is further explained in thefollowing example of a beam subjected to a bending moment produced by atemperature gradient

1131 Example of a member subjected to bending1

A member representing an interior span of a continuous beam of infinitenumber of equal spans is shown in Fig 112(a) The member has a rect-angular cross-section and is subjected to a rise of temperature which varieslinearly over the depth with a difference of ∆T degrees between the top andbottom fibres It is required to study the variation of the bending moment M

Figure 112 Development of cracks and statically indeterminate moment due totemperature gradient (a) an interior span of a continuous beam (b) variationof M with ∆T (Elbadry (1988) see Note 1 page 406)


with ∆T as ∆T increases from zero to a value ∆Tn causing stabilized crackformation Assume the average crack spacing srm = 06m (24 in) Consider thetensile stress of concrete at which the successive cracks form to vary as2

fcti = fct1[1 + 350αt(∆Ti minus ∆T1)] (111)

where fcti is the tensile strength of concrete at the ith crack ∆Ti is the value of∆T at which the ith crack is formed αt is the coefficient of thermal expansion

The cross-section geometrical data are given in Fig 112(a) Other data arefct1 = 21MPa (030ksi) αt = 10 times 10minus6 per degC (56 times 10minus6 per degF) moduli ofelasticity of concrete and steel are Ec = 25GPa (3600ksi) and Es = 200GPa(29000ksi)

Assume that yielding of the reinforcement does not occur when the lastcrack occurs the nth crack with n = 8

The value of M at which the ith crack occurs is

Mri = fctiW1 (112)

where W1 is the section modulus in the non-cracked state 1For simplicity we assume that the beam does not lose symmetry about the

centre of the span as a result of crack formation This assumption makes thestructure statically indeterminate to the first degree with the indeterminateforce being a bending moment M whose magnitude is constant over thelength of the span Before cracking M is given by (line OA in Fig 112(b) )

M = αtEcI1 ∆T

h (0 le ∆T le ∆T1) (113)

where I1 is the second moment of area about the centroidal axis of thetransformed non-cracked section (state 1)

Setting M = Mr1 and ∆T = ∆T1 in Equation (113) and solving gives thevalue of ∆T at which the first crack occurs

∆T1 = Mr1h

αtEcI1

(114)

After occurrence of the ith crack and before formation of the next crackthe span l can be considered to be composed of a non-cracked part of length(l minus i srm) with flexural rigidity EcI1 and a cracked part of length i srm with amean flexural rigidity EcIrm given by the equation

1

EcIrm

= (1 minus ζ) 1

EcI1

+ ζ 1

EcI2 (115)


where ζ is the coefficient of interpolation between the curvatures in states 1and 2 (Equation (845) ) I2 is the second moment of area about the centroidalaxis of the transformed fully cracked section (state 2) Equation (115) isderived from Equation (840) by substitution for ψm ψ1 and ψ2 by 1(EcIrm) 1(Ec I1) and 1(Ec I2) respectively each of these quantities represents curvaturedue to a unit moment


Irm = I1I2

ζI1 + (1 minus ζ)I2

(116)

If we assume that high-bond deformed bars are used the interpolationcoefficient just after cracking is ζ = 05 (by Equation (841) substituting M =Mr and assuming β1 = 1 β2 = 05)

After formation of the ith crack and before occurrence of the next crackthe value of M is represented by (line CD in Fig 112(b) )

M = αt∆T

h EcI1 1 +

i srm

l I1

Irm

minus 1minus1

(with ∆Ti ∆T ∆Ti + 1) (117)

This equation can be derived by the force method (Section 42)Equation (117) can be solved for the value Mri at which the ith crack is

formed (by substituting ∆Ti for ∆T and i minus 1 for i)

Mri = αt∆Ti

h EcI11 +

(i minus 1)srm

l I1

Irm

minus1minus1

(118)

The values of Mri calculated by Equations (112) and (118) depend upon∆Ti whose value can be determined by elimination of Mri and solving the twoequations

The mean crack width may be calculated by Equation (848)


where srm is the mean crack spacing (see Appendix E) εs2 the steel strain in afully cracked section (state 2) may be calculated by

εs2 = M

yCTAsEs

(1110)

where yCT is the distance between the tension steel and the resultant of com-pression on the section As and Es are the cross-section area and modulus ofelasticity of the tension steel


The value of M varies in the stage of crack formation as shown in Fig112(b) Just before formation of the second crack M = Mr2 = 101 kN-msubstituting this value in Equations (119) and (1110) gives (yCT 09d =0225m)

εs2 = 1496 times 10minus6 wm = 045mm (0018 in)

The value of wm can be reduced to any specified limit by increasing the steelarea As On the other hand the mean crack width can become much larger ifAs is reduced below a minimum at which σs2 = Esεs2 = fy where fy is the yieldstrength of the steel The minimum value of the steel ratio required to avoidthis situation is discussed in Section 116

This procedure can be employed to determine ∆Ti for i = 2 3 nSubstitution of the value of ∆Ti in Equation (118) gives the larger of two M-ordinates corresponding to ∆Ti required to construct the graph in Fig112(b) Equation (117) gives the lesser ordinate The values of ∆T1 andthe two M-ordinates corresponding to the first crack can be determined byEquations (112) (114) and (117)

The results of the above analysis are plotted in Fig 112(b) for com-parison the dashed line OB is included to represent the case when concrete intension is ignored The values of ∆Ti and the corresponding ordinates arelisted below

1132 Example of a member subjected to axial force(worked out in British units)

It is required to study the variation of N versus ε(= Dl) for a member oflength l subjected to an imposed end displacement D (Fig 113) in the range 0le ε le εs where εs = Dsl with Ds being the displacement at which stabilizedcracking occurs Assume that yielding of the reinforcement does not occur inthis range Consider average crack spacing srm = 12 in (300mm) the value ofthe tensile strength of concrete at which successive cracks form is3

fcti = fct1[1 + 350(εi minus ε1)] (1111)

where fcti is the tensile strength of concrete at the location of the ith crack εi =Dil with Di being the imposed displacement at which the ith crack is formed

The cross-section geometrical data are given in Fig 113 Other data arefct1 = 035ksi (24MPa) Ec = 4150ksi (286GPa) Es = 29000ksi (200GPa)

The equations derived in Section 1131 apply for a member subjected to an

Crack numberMri (larger ordinate kN-m)M (lesser ordinate kN-m)

19758

210172

310582

411090

511698

6122106

7129114

8136122


imposed axial end displacement by changing some of the parameters asgiven below

Nri = fctiA1 (1112)

N = EcA1ε (0 ε ε1) (1113)

ε1 = Nr1

EcA1

(1114)

1

EcArm

= (1 minus ζ) 1

EcA1

+ ζ 1

EcA2 (1115)

Arm = A1A2

ζA1 + (1 minus ζ)A2

(1116)

Figure 113 Axial force N versus ε (= Dl ) in a member subjected to imposed axial enddisplacement


N = εEcA11 + i srm

l A1

Arm

minus 1 minus1

(with εi le ε le εi + 1) (1117)

Nri = εiEcA11 + (i minus 1)srm

l A1

Arm

minus 1 minus1

(1118)

where

The transformed section area A1 = 1126 in2 A2 = 461 in2 Using ζ = 05Equation (1116) gives Arm = 886 in2

The number of cracks at crack stabilization n = lsrm = 4 cracksEquations (1112) and (1114) give Nr1 = 394kip ε1 = 84 times 10minus6 Substitut-

ing the value of ε1 in Equation (1117) gives N = 100kip this is the lowerordinate plotted for ε = ε1

Setting i = 2 3 and 4 and solving Equations (1111) (1112) and (1118) forεi and substitution of this value in Equations (1117) and (1118) give all thevalues required for plotting the graph in Fig 113 The following is a list ofthe values of εi and the corresponding ordinates for i = 1 2 4

Discussion of resultsIf the same example is analysed with a reduced value of the steel area As thevertical drops in the N-value at each crack formation will be larger and thedegradation of the slope of the Nndashε graph will be faster with the successivecrack formations Furthermore the value given for srm should be increasedbecause of the reduction in As (see Appendix E) As a result the number ofcracks will be smaller and the cracks will be wider

When the steel ratio ρ = AsAc is reduced below a limiting value ρmin y the

A1 and A2 = areas of transformed sections in non-cracked and in fullycracked states A1 = Ac(1 + αρ) A2 = Acαρ α = EsEc Ac = area ofconcrete ρ = AsAc

Arm = mean transformed cross-section areaN = axial normal force

Nri = value of N just before formation of the ith cracksrm = mean crack spacing

ε = Dl where D is imposed displacementεi = Di l where Di is the imposed displacement at which the ith crack

is formed

Crack number ii

Nri(kip)N (lesser ordinate kip)

184 times 10minus6

394100

2362 times 10minus6

430246

3700 times 10minus6

476334

41119 times 10minus6

534411


Nndashε graph (Fig 113) will exhibit a large drop in the value of N at the forma-tion of the first crack (at ε = ε1 and N = Nr1) Subsequent increase in N willoccur at a relatively low rate to reach a limiting value equal to Ny = As fy ltNr1 where fy is the yield strength At this point the mean crack width wm =srmζ fy Es Any further increase of the imposed displacement D will be accom-panied by an increase of the same magnitude in the crack width while thevalue of N remains constant equal to Ny and no further cracks develop Thuswhen the steel ratio ρ le ρmin y a single usually excessively wide crack occursEquations will be derived in Section 115 for ρmin y in reinforced concretesections with or without prestressing

Experimental verificationJaccoud4 conducted experiments on reinforced concrete prisms subjected toan imposed axial end displacement (Fig 114) The geometrical and materialdata are given in the figure all parameters have values approximately equal tothe values employed in the above example with the exception of As which isreduced but is still sufficient to avoid yielding Fig 114 compares the graphsof σc (= NA1) versus ε obtained by experiment and by analysis setting fcti =constant (as observed for this specimen) and using Equations (1112) to(1118)

Figure 114 Comparison of analysis with experimental results of Jaccoud (see Note 2page 406) A prism subjected to imposed end displacement


114 Advantage of partial prestressing

Cracking results in considerable reduction of the statically indeterminateforces caused by an imposed displacement or restraint of volumetric changesdue to temperature or shrinkage Without cracking the graph in Fig 112(b)would be the straight line OA extended upwards This would be the case inprestressed structures in which cracking is not allowed However a designwhich does not allow tensile stresses requires a high prestressing level Inaddition to being costly the higher compression due to prestressing increasesthe losses due to creep and can produce excessive deformations Thus it isbeneficial to use partial prestressing allowing cracking to occur while con-trolling the width of cracks by provision of adequate non-prestressed steelThe amount of steel required for crack control is discussed in the followingsections

115 Minimum reinforcement to avoid yieldingof steel

If the reinforcement in a cross-section of a member is below a minimumratio ρmin y yielding of the reinforcement occurs at the formation of thefirst crack such a crack will be excessively wide and formation of severalcracks with limited width does not take place This is true when cracking isinduced by applied forces or imposed displacements The minimumreinforcement cross-section area As min y and the corresponding steel ratioρmin y to ensure that wide isolated cracks do not occur due to yielding aredetermined below

Consider a section subjected to axial tension N The value of N justsufficient to produce cracking is

Nr = fctA1 (1119)

where fct is the tensile strength of concrete and A1 = Ac(1 + αρ) is the area ofthe transformed non-cracked section (in state 1) α = EsEc where Ec is themodulus of elasticity of concrete at the time considered and Es the modulusof elasticity of the reinforcement ρ = AsAc

Immediately after cracking the force Nr is resisted entirely by thereinforcement thus σs = NrAs Setting in this equation σs = fy and As = As min y

gives the minimum steel ratio to ensure non-yielding at cracking

ρmin y = fct

fy 1

1 minus α(fctfy) (1120)

where ρmin y = As min y AcThe term inside the square brackets in Equation (1120) is approximately


equal to unity thus the equation is frequently written in the simpler formρmin y = fctfy

Derivation of Equation (1120) implies that the normal force is equal to Nr

just before and just after formation of the first crack However in the case ofdisplacement-induced cracking (defined in Section 113) a sudden drop ofthe value of the normal force takes place once the crack is formed Sub-sequent increase in the imposed displacement will increase the normal forceto a value N le Nr

When the section is prestressed the first crack occurs when N2 = Nr whereN2 = N minus N1 with N1 being the decompression force (Equation 740) Equation(1120) can be used substituting for fy the yield stress of the non-prestressedsteel the resulting value ρmin y will be equal to (Aps + As min y)Ac where Aps isthe area of the prestressed steel and As min y is the minimum area of the non-prestressed steel required to ensure no yielding Use of Equation (1120) inthis way implies the assumption that α is the same for all reinforcements

The case when cracking is produced by a normal force N applied at areference point O and a bending moment M about an axis through O isconsidered below Assuming that the pair N and M are just sufficient toproduce cracking we can write

fct = N

A1

+ M

W1

(1121)

where A1 and W1 are respectively the transformed non-cracked cross-sectionarea and section modulus (in state 1) Equation (1121) applies only when thereference point O is at the centroid of the transformed non-cracked section ifthis is not the case the statical equivalents of the normal force and themoment must be determined to be used in the equation The stress in steel atthe crack can be calculated and equated to fy to give ρmin y

ρmin y = (NesyCT) + N

fybd(1122)

where ρmin y = (As min y bd) yCT is an absolute value equal to the distancebetween resultant tension and resultant compression when the concrete intension is ignored (Fig 115) b and d are defined in Fig 115 es is theeccentricity of the resultant of M and N measured downward from the cen-troid of the tension steel es = (MN) minus ys with ys being the y coordinate of thecentroid of the tension steel

Calculation of yCT will involve determination of the depth c of thecompression zone by solving Equation (710) or (79)

When the cross-section is prestressed Equations (1121) and (1122) can beapplied by substituting N2 and M2 for N and M where N2 = N minus N1 and M2 =M minus M1 with N1 and M1 being the decompression forces (Equations (740)


and (741) ) Again the value of fy should be equal to the yield strength of thenon-prestressed steel and the resulting value ρmin y = (Aps + As min y)(bd) withAs min y being the minimum non-prestressed steel area

In the case of a reinforced rectangular section without prestressing sub-jected to bending moment without axial force Equations (1121) and (1122)give

ρmin y = As min y

bd 024

fct

fy

(1123)

which is derived by assuming d = 09h and yCT = 09dIt is to be noted that in Equations (1121) and (1122) the values of N and

M are just sufficient to produce the first crack The two equations apply toreinforced concrete sections with or without prestressing subjected to anycombination of N and M satisfying Equation (1121) Thus for the case ofaxial tension M = 0 and es = 0 Equations (1121) and (1122) give the sameresults as Equations (1119) and (1120)

116 Early thermal cracking

In many cases cracking of concrete structures occurs at an early age due toheat of hydration of cement When the heat of hydration is generated at ahigher rate than heat dissipation rise of temperature and expansion of theconcrete occurs The expansion is followed by contraction as the concretecools down to the ambient temperature These volumetric changes are inmost cases partially restrained and stresses result the magnitude of stressmay be assumed proportionate to the modulus of elasticity of concrete Ec

Figure 115 Stress distribution in a cracked prestressed section Positive sign conventionfor N M and y


The value of Ec when the contraction occurs is large compared with thecorresponding value at earlier stages of hydration when the expansionoccurs As a result the tensile stress during contraction exceeds the compres-sive stress which has occurred during expansion The difference is frequentlysufficient to exceed the tensile strength fct of the young concrete

In a member without reinforcement one wide crack is induced Provisionof reinforcement controls the crack so that the member remains serviceableThe amount of reinforcement required for this purpose may be calculated bythe equations in Section 115 substituting an appropriate (relatively low)value5 of fct representing the tensile strength of concrete at an early age (seeAppendix A)

117 Amount of reinforcement to limit crack width

The width of force-induced or displacement-induced cracks can be limited toany specified value by provision of sufficient area of bonded reinforcement inthe tension zone The same objective can be achieved by limiting the steelstrain at a cracked section The following equations can be used for thispurpose for a reinforced concrete section with or without prestressing It isassumed that the section is subjected to normal force N at a reference point Oand a bending moment about a horizontal axis through the reference point(Fig 115)


es = M

N minus ys (with N ne 0) (1125)

εs2 = (NesyCT) + N

EsAs

(1126)

σs2 = Esεs2 (1127)

As = (NesyCT) + N

σs2

(1128)

where

As = sum of the cross-section areas of prestressed and non-prestressedsteel resisting the tension

Es = modulus of elasticity of steel assumed the same for the pre-stressed and the non-prestressed steel

es = eccentricity of the resultant of N and M measured downwardfrom the centroid of As Equation (1126) implies the assumptionthat the resultant tension is at the centroid of As εs2 for any


The product (Nes) in Equation (1126) is to be replaced by the value of Mwhen N = 0 Calculation of yCT is commonly preceded by determination ofthe depth c of the compression zone (Equation (710) or (79) )

1171 Fatigue of steel

Fatigue can occur in the non-prestressed steel or in the prestressed steel whencyclic change in steel stress σs2 is relatively large Equations (1126) and(1127) can be used to calculate the magnitude of σs2 when the values of Nand M producing the cyclic change in stress are known For given values of Nand M the change in steel stress σs2 can be critical for fatigue when the totalsteel ratio prestressed and non-prestressed is small This can be the case inpartially prestressed structures where a smaller area of steel is used com-pared to the case of no prestressing

The allowable limits of σs2 to avoid fatigue failure are given in variouscodes Approximate values are 125MPa (18ksi) for non-prestressed deformedbars and 10ndash12 per cent of the ultimate strength for prestressed steel

1172 Graph for the change in steel stress in a rectangularcracked section

Equations (1126) and (1127) are used to derive the graph in Fig 116 forrectangular sections subjected to a normal force N at O at mid-height of thesection and a bending moment M about a horizontal axis through O To use

reinforcement layer can be more accurately calculated by theprocedure given in Chapter 8

N and M = normal force at reference point O and bending moment about ahorizontal axis through O the sign conventions for N and M aredefined in Fig 115 It is assumed that the stress in concrete σc = 0prior to the introduction of N and M If this is not the casesubstitute N and M in Equations (1125) (1126) and (1128)by N2 = N minus N1 and M2 = M minus M1 where N1 and M1 are thelsquodecompression forcesrsquo (Equations (740) and (741) )

srm = crack spacing (see Appendix E)wm = mean crack widthyCT = absolute value equal to the distance between resultant tension

and compression forcesεs2 = strain increment in steel (prestressed or non-prestressed) due to N

and M with concrete in tension ignoredσs2 = stress increment in steel (prestressed or non-prestressed) due to N

and M with concrete in tension ignoredζ = coefficient of interpolation between state 1 where cracking is dis-

regarded and state 2 where concrete in tension is ignored


Figu

re11

6C

hang

e in

ste

el s

tres

s

s2 in

a r

ecta

ngul

ar c

rack

ed s

ectio

n du

e to

a n

orm

al fo

rce

N a

nd a

ben

ding

mom

ent

M T

he v

alue

of

s2 is

equ

al t

o th

e va

lue

read

from

the

gra

ph m

ultip

lied

by

c m

ax w

here

c

max

is t

he e

xtre

me

fibre

str

ess

igno

ring

cra

ckin

g Po

sitiv

e si

gn c

onve

ntio

n fo

r N

and

M a

rein

dica

ted

the graph enter the dimensionless parameters ρ(= Asbd) and (MNd) andread the ordinate value representing the ratio σs2σc max where σs2 is stressincrement in steel (prestressed and non-prestressed) due to N and M withconcrete in tension ignored σc max is the extreme fibre stress due to N and Mwith cracking ignored The value of σc max may be calculated by Equations(217) and (220) using properties of the transformed non-cracked section

When the graph is used for a prestressed section As represents the sum ofthe cross-section area of the prestressed and the non-prestressed steel thisimplies an approximation by the assumption that the resultant tension is atthe centroid of As In all cases it should be noted that N and M represent thevalues of the normal force and bending moment after deduction of thedecompression forces (see definition of N and M given below Equation(1128) )

Assumed parameters which have small influence used in preparing thegraph are α = EsEc = 70 Aprimes = As d = 09h dprime = 01h (Fig 116)

For given values of N and M the graph in Fig 116 may be used tocalculate the steel ratio ρ(= As(bd) ) required to limit the mean crack widthwm to a specified value For this purpose determine the extreme fibre stressσc max in the non-cracked section and calculate ζ by Equation (849) Using anassumed crack spacing srm (Appendix E) determine εs2 and σs2 by Equations(1126) and (1127) Enter the graphs with the values of (σs2σc max) and [M(Nd)] and read the value of ρ

Example 111

What is the change in steel stress σs2 and the mean crack width due to abending moment = 40kN-m (350kip-in) and an axial force = 0 appliedat time t on the section shown in Fig 117 The free shrinkage occurringprior to age t is εcs(t t0) = minus400 times 10minus6 where t0 is the age of concretewhen curing is stopped Use the following data Es = 200GPa(29000ksi) α = EsEc(t) = 7 creep coefficient φ = 25 aging coefficient χ= 08 Ec(t0) = Ec(t) mean crack spacing srm = 400mm (16 in) inter-polation coefficient ζ = 08

Determine the bottom steel area required to limit the mean crackwidth to 02mm

The age-adjusted modulus of elasticity of concrete Equation (131)Ec(t t0) = 952GPa Properties of the age-adjusted transformed sectionare A = 03720m2 B = 0 I = 3287 times 10minus3 m4 The stress in concreteat time t due to shrinkage is constant over the section and its value isσc(t) = 0774MPa (Equations (315) and (319) )

Properties of the transformed non-cracked section at time t are A =


03216m2 B = 0 I = 2561 times 10minus3 m4 The decompression forces are M1 =0 and by Equation (743)

N1 = minus 03216(0774 times 106) = minus249kN

Forces producing cracking after deducting the decompression forcesare

N = 0 minus (minus249) = 249kN M = 40kN-m

Stress at the extreme fibre ignoring cracking is

σc max = 249

03216 +

40(015)

2561 times 10minus3 = 312MPa

The steel ratio ρ = 1800(1000 times 270) = 067 per cent M(Nd) = 40(249 times 027) = 060 Entering the graph in Fig 116 with the values of ρand M(Nd) gives (σs2σc max) = 51 thus

σs2 = 51(312) = 160MPa (232ksi)

The mean crack width (Equations (1124) and (1127) )

Figure 117 Cross-section subjected to shrinkage and a bending moment analysedto determine crack width (Example 111)


wm = 400(080) 160

200 times 103 = 026mm (0010 in)

To limit the mean crack width to 02mm σs2 is to be reduced to σs2 =125MPa (181ksi) Enter the graph in Fig 116 with ordinate = 125312 = 40 and M(Nd) = 06 to read ρ 104 per cent The increase insteel ratio will change σc(t) and M(Nd) values hence iteration isrequired Repetition of the analysis will give a more accurate value ofρ = 10 per cent or As = 2700mm2 (419 in2)

118 Considerations in crack control

This section discusses the motives and the most important measures for con-trol of cracking

There are three motives for crack control durability by reducing risk ofcorrosion of reinforcement aesthetic appearance and functional require-ments such as gas or liquid tightness or hygiene (cracks can be the focusof development of pathogenic microbes) The three motives are discussedseparately below

Corrosion of reinforcementThere is no general agreement on the influence of width of cracks on corro-sion of reinforcing steel Some research6 indicates that intensity of observedcorrosion is not dependent upon width of cracks w as long as wmaximum islimited to 03ndash05mm (001ndash02 in) the lower limit is for cracks running par-allel to reinforcing bars producing a higher risk of corrosion Only the lengthof time (a few years) before initiation of corrosion is influenced bycrack width but this period is relatively short and has little influence on thelongevity of structures

On the other hand there is agreement that thickness and porosity of con-crete covering the reinforcement are important parameters influencing corro-sion Improving the quality of concrete (mainly by limiting the waterndashcementratio) and at the same time controlling the crack width are at present con-sidered important to control cracking Thus it is prudent to specify the crackwidth dependent upon the aggressiveness of the environment7 Also stricterrequirements should be applied to prestressed structures because the pre-stressed steel is more susceptible to corrosion than ordinary reinforcing bars[wmaximum 02mm (0008 in) at the exposed concrete surface and w = 0 at thelevel of the prestressed steel]


Aesthetic appearanceCracks of width smaller than 03mm (001 in) generally are not of muchconcern to the public However owners and users of structures are in generalsensitive to the aesthetic damage of appearance when wide cracks developObviously the crack width of tolerable appearance is subjective and depend-ent on many factors such as the distance between the crack and the observerthe lighting and the roughness of the surface

Gas or liquid tightnessThe need for tightness depends upon the nature of gas or liquid to be con-tained in the structure It is theoretically possible to specify and expect astructure with no cracks It is more realistic however to specify a limit forcrack width Research and experience have shown that water-retaining struc-tures can be water tight with crack width w = 01 to 02mm (0004ndash0009 in)Such a crack even when it traverses the full thickness of a wall may allowmoisture to penetrate after first occurrence of the crack but lsquohealingrsquo andstopping of leakage occurs within a few days

Measures for control of crackingControlled cracks of width w = 010 to 030mm (0004 to 0012 in) generallydo not undermine the use the durability or the appearance of concrete struc-tures On the other hand uncontrolled wide cracks (w gt 05mm) must beavoided This objective may be achieved by

1 reducing the risk of cracking by measures such as use of appropriate mixcuring casting sequence and construction joints of concrete provisionof temporary or permanent expansion joints prestressing etc

2 provision of minimum bonded reinforcement in all parts of reinforcedconcrete or prestressed structures when cracking is probable during con-struction or during use of the structure Thus designers should considerthe combinations of direct and indirect loads (settlement of supportstemperature and volumetric changes etc) which can produce tensilestress close to or exceeding the tensile strength of concrete

3 limiting the steel stress calculated with concrete in tension ignored Thisdesign check is commonly done considering only the quasi-permanentloads and allowing wider partly reversible cracks to occur due toadditional transient loads

4 provision of prestressing even at a low level can be effective in reducingcrack width This is particularly the case when cracking is caused byflexure When cracking is caused by a normal force provision ofprestressing is effective only when the element considered is free toshorten


119 Cracking of high-strength concrete

The effects of use of high-strength concrete (HSC) on cracking of reinforcedconcrete structures with or without prestressing are briefly discussed hereFor this purpose consider a reinforced concrete member subjected to an axialforce N as shown in Fig 111(a) If concrete of higher strength is used forthe member cracks will occur at higher N values because of the increase ofthe tensile strength (see the dashed lines in Fig 111(a) ) Also because of theimproved bond slip between the concrete and the reinforcing bars occurs athigher bond stress this decreases the crack spacing (see Appendix E) and alsoincreases the effect of the tension stiffening thus reducing the width ofcracks Therefore use of HSC may prevent the cracking or when N is greaterthan the cracking value the crack width will be smaller compared with anidentical member with the same reinforcement but lower concrete strength

On the other hand when N is caused by an imposed displacement (Fig111(b) ) the first crack will be formed at a higher N value and higher steelstress will occur at the crack This means that a larger steel ratio is necessaryto avoid yielding of steel at cracking (see Equation (1120) ) In spite of thehigher steel stress at the crack the crack width will increase only slightlybecause of the increase of tension stiffening effect due to improved bond

The effects of use of HSC on cracking of members subjected to bendingis not different from what is discussed above This is evident in Fig 118which summarizes the results of long-term tests8 on simply supported slabs

Figure 118 Mean crack width wm and mean crack spacing srm observed in tests onreinforced concrete slabs of varying concrete strength


subjected to two symmetrically located equal forces The results are given fortwo load levels represented by the ratio σs2fy where σs2 is steel stress at acracked section in the central part of the span (zone of constant bending) andfy is the yield strength of the steel (460MPa)

The empirical equations given in some codes to predict crack spacing or toaccount for tension stiffening do not accurately represent structures made ofHSC This status will no doubt change because concrete strength higher than50MPa (7000psi) and reaching up to 80 or 100MPa (12 000 or 15 000psi) isincreasingly used in modern structures


Example 112 Prestressed section crack width calculation

Figure 119 shows the stress distribution in a prestressed concrete sec-tion at time t after occurrence of creep shrinkage and relaxation (thesame cross-section was analysed in Example 22 Fig 26) Find thecrack width after application of live-load bending M = 7000kip-in(790kN-m) about an axis through reference point O Use the follow-ing data σcO(t) = minus0360ksi (minus248MPa) γ(t) = minus638 times 10minus3 ksiin(minus173MPam) Es = 29 000ksi (200GPa) for all reinforcements α(t) =Es Ec(t) = 7 mean crack spacing srm = 16 in (400mm) interpolationcoefficient ζ = 09

Properties of the transformed non-cracked section at time t

Figure 119 Prestressed cross-section analysed to determine crack width afterapplication of live-load bending moment (Example 112)


A = 5983 in2 B = 2704 in3 I = 118 500 in4


N1 = 2132kip M1 = 8354kip-in

Forces producing cracking after deducting the decompressionforces

N2 = minus2132kip M2 = 7000 minus 8354 = 61646kip-in

Stress at the extreme fibre ignoring cracking (Equations (219) and(217) ) σc max = 0868ksi

The tension is resisted by the prestressed and the bottom non-prestressed steel the total steel area resisting tension = 174 + 233 =407 in2 and its centroid is at depth d = 436 in The steel ratio ρ = 407(12 times 436) = 078 per cent The value M(Nd) = 61646(minus2132 times 436)= minus0663 Entering the graph in Fig 116 with the values of ρ andM(Nd) gives (σs2σc max) = 213 Thus

σs2 = 213(0868) = 185ksi

The mean crack width (Equations (1124) and (1127) ) is

wm = 16(09) 185

29 000 = 00092 in (023mm)

A more accurate analysis using the equation of Chapter 8 gives c =203 in σs2 in the bottom non-prestressed steel = 178ksi mean crackwidth = 00088 in (022mm)

Example 113 Overhanging slab reinforcement to controlthermal cracking

Figure 1110 represents top view and section in a reinforced concreteslab extending as a cantilever from the floor of a building Transversecracks can occur in the cantilever due to temperature differencebetween the outside air and the interior heated building It is required tocalculate the concrete stress which would occur ignoring cracking due


to a temperature drop of 55 degF (31 degC) of the cantilever below thetemperature of the interior part of the floor Determine the reinforce-ment required to avoid yielding of the steel Consider the coefficient ofthermal expansion αt = 44 times 10minus6 per degree Fahrenheit (80 times 10minus6 perdegree Celsius) Ec = 2900ksi (20GPa) fy = 60ksi (410MPa) Es = 29000ksi (200GPa)

Strain if thermal contraction were free to occur

εfree = 44 times 10minus6 (minus55) = minus242 times 10minus6

Assuming the strain is completely prevented the stress in concretewill be

σc = minusEcεfree = minus2900(minus242 times 10minus6) = 702psi (484MPa)

Figure 1110 Minimum reinforcement requirement to control cracking due totemperature in an overhanging reinforced concrete slab in a building(Example 113) (a) top view (b) section AndashA


Cracking will no doubt occur before this hypothetical high stressvalue is reached The minimum reinforcement ratio necessary to avoidyielding (Eq (1120) ) is

ρmin y = 360

60 times 103

1

1 minus29000

2900 360

60 times 103= 00063

This steel ratio is approximately provided by using 12 in bars (Asbar

= 020 in2 (130mm2) ) with spacing s = 8 in at top and bottom Thecorresponding steel ratio is

ρ = Asbar

sts2 =

020

(8 times 8)2 = 00063

where ts = slab thicknessThe ACI 318ndash899 assumes an effective tension area to be used in

crack analysis (Fig E 3(b) ) Accordingly the effective tension area perbar is 24 in2 if this value is used to replace the quantity (sts 2) the aboveequation gives Asbar = 00063(24) = 015 in2 (98mm2)

The MC-90 and the EC-91 assume an effective tension area definedin Fig E2(c) accordingly the effective tension area will be 30 in2 andthe required cross-section area per bar will be 00063 (30) = 019 in2

(120mm2)

CommentaryIn the above example the stress σx ignoring cracking is based on the assump-tion that the volumetric change in the x direction is fully restrained Thissimple analysis is sufficient to give an approximate value of the stress σx andto conclude that fct will be exceeded thus cracking will occur Analysis basedon elastic theory shows that the stress σx at section AB varies between theapproximate value calculated above (assuming complete restraint) and asmaller value at point A the tip of the cantilever When lb = 2 4 or 8 thevalue of stress at A is respectively equal to 9 55 or 96 per cent of the stresscalculated assuming complete restraint In this example lb = 8 thus the stressσx is approximately constant over the section AndashB

Typical crack pattern is shown in Fig 1110(a) the width and spacings ofcracks depend upon the reinforcement provided mainly in the x directionThe type of stress distribution and crack pattern described above occurs inpractice in walls where the volumetric change of the wall due to temperature


or shrinkage is restrained by the wall footings Another example where simi-lar crack patterns can occur is in bridge superstructures10 where parts of thecross-section are cast separately using longitudinal casting joints (forexample parapets or overhanging parts of the deck cast separately to themain deck)

1111 General

This chapter discusses the parameters that influence crack width and givesequations to determine the steel area of bonded reinforcement required tolimit the width of cracks to a specified value The analysis is approximatebecause it includes empirical parameters srm and ζ to predict crack spacingand to account for tension stiffening The empirical coefficients for srm and ζare to be determined from codes (some code expression are given in AppendixE and Equation (845)

Notes

1 This example is based on Elbadry MM (1988) Serviceability of ReinforcedConcrete Structures PhD thesis Department of Civil Engineering University ofCalgary Calgary Canada 294 pp

2 Equation (111) is derived from experiments reported in Jaccoud JP (1987)Armature minimale pour le controcircle de la fissuration des structures en beacuteton PhDthesis Deacutepartement de Geacutenie Civil Eacutecole Polytechnique Feacutedeacuterale de LausanneLausanne Switzerland 195 pp

3 Equation (1111) is derived from experiments reported in Jaccoud (1987) seereference mentioned in Note 2 above

4 See reference mentioned in Note 2 above5 A suggested equation can also be found in Department of Transport Highways

and Traffic (1987) Department Advice Note BA2487 Early Thermal Cracking ofConcrete 16 pp DOEDTp Publication Sales Unit Bldg 1 Victoria Rd 5Ruislip Middlesex HA4 0NZ UK

6 Schiessl P (1985) Mindestbewehrung zur Vermeidung klaffender Risse Institut furBetonstahl und Stahlbetonbau Munich Bericht 284

7 Beeby AW (1983) lsquoCracking cover and corrosion of reinforcementrsquo ConcreteInternational 28 (2) Feb pp 35ndash40

8 See Jaccoud J-P Charif H and Farra B (1993) Cracking Behaviour of HSCStructures and Practical Considerations for Design Publication No 139 IBAPSwiss Federal Institute of Technology Lausanne Switzerland

9 ACI 318ndash89 Building Code Requirements for Reinforced Concrete InstituteFarmington Hills Michigan 48333ndash9094 The clause referred to here has beendropped out in subsequent issues of the Code

10 See reference mentioned in Note 5 above


Design for serviceability ofprestressed concrete

121 Introduction

Prestressed concrete structures commonly contain non-prestressedreinforcement to control cracks that develop before introduction of theprestressing or when the structure is in service The appropriate design ofthe non-prestressed reinforcement and the prestressing forces can controldeflections and limit the opening of cracks This chapter1 discusses the choiceof the level of prestressing forces and the amount of non-prestressedreinforcement to achieve these objectives

Viaduct at Gruyegravere Switzerland

Chapter 12

122 Permanent state

Durability of concrete structures is closely linked to their serviceability in thepermanent state This is defined here as the state of the structures subjected tosustained loads such as the prestressing the self-weight and the superimposeddead loads and the quasi-permanent live loads A prestressed structure maybe designed such that cracks occur only under the effect of exceptional liveload combined with temperature variation Such cracks open and close ineach cycle that the loads are applied However cyclic loading of crackedstructures produces residual opening of cracks and residual deflections whichare discussed in this chapter Some bridges exhibit increasing deflections afterseveral decades in service This is attributed partly to the irreversible curva-ture which adds to the deflections due to the effects of creep shrinkage andrelaxation

The approach adopted in this chapter to achieve satisfactory serviceabilityis to limit the tensile stress in the permanent state to a specified value Thiscan be achieved by designing the prestressing such that its effect combinedwith the sustained loads produce stress in concrete not exceeding the specifiedvalue In the permanent state the structure has no or only limited cracks thusany elastic analysis ignoring cracking is considered adequate to calculate thestress in concrete due to the sustained loads and the prestressing combinedIn addition to the prestressing the structure should have non-prestressedreinforcement the design of which is discussed below

123 Balanced deflection factor

The balanced deflection factor βD is defined as the ratio between theelastic deflections at mid-span due to prestressing and due to sustainedquasi-permanent loads

βD = minusD(Pm)

D(q)(121)

where D(Pm) is the deflection at mid-span due to prestressing In the calcula-tion of this deflection the prestressing force in a tendon is taken equal to themean of the initial prestressing force excluding friction loss and the forceremaining after losses due to creep and shrinkage of concrete and relaxationof prestressed steel D(q) is the elastic (immediate) deflection at mid-span dueto permanent and quasi-permanent load

A parabolic tendon having constant prestressing force exerts on a prismaticconcrete member a uniform upward load If in addition the permanent loadis uniform downward the balanced deflection factor βD is the same as thewell-known balancing load factor which is equal to minus the ratio ofthe intensities of the upward and the downward loads However use of the


balanced deflection factor is preferred here because it applies with any tendonprofile and with members having variable depth

The significance of the balanced deflection factor is explained by Fig 121which depicts the strain distribution in a section of members having βD = 0and 1 In the former the strain at the centroid εO = 0 and the curvature ψ ne 0in the latter εO ne 0 and ψ = 0 We recognize that with βD = 0 the member isnon-prestressed and with βD = 1 the prestressing is just sufficient to eliminatethe deflection

In determining βD by Equation (121) the deflection is calculated using thecross-sectional area properties of gross concrete sections and an estimatedreduction of the prestressing forces to account for the time-dependent lossesdue to creep shrinkage and relaxation Because the analysis is concerned withthe behaviour of the structure during its service life it is suggested that theprestressing force used in calculating βD be the average of the values beforeand after the time-dependent losses

124 Design of prestressing level

In the design of a prestressed structure the level of prestressing expressed bythe balanced deflection factor βD can be a means of controlling cracks inservice condition For this purpose the structure can be designed such that thestress at a specified fibre due to prestressing combined with sustained quasi-permanent loads σperm satisfy the condition

σperm σallowable (122)

where σallowable is an allowable stress value depending upon the width of cracksthat can be tolerated and the amount of non-prestressed reinforcement that is

Figure 121 Strain distribution in a cross-section of members having the balanceddeflection factors D = 0 and 1

Design for serviceability of prestressed concrete 409

provided For example at the extreme top fibre of a bridge deck exceptionallive load combined with temperature variation can produce transient stress∆σ and cause cracking when

∆σ + σperm fct (123)

where fct is the tensile strength of concrete Thus when fct = 3MPa andσperm is equal to an allowable value say σallowable = minus2MPa cracking occurswhen ∆σ 5MPa The cracks partially close when the live and the thermalloads are removed With the repetition of these loads the cracks will havesome residual opening and the structure will exhibit residual deformations(see Sections 127 and 1211)

Equation (128) is derived below to give for a continuous or a simple beamsubjected to uniform permanent downward load qunit length a design valueof the balanced deflection factor βD This value represents the prestress forcePm that satisfies Equation (122) at any fibre of a specified section The bridgecross-sections shown in Fig 122 will be used as examples for the applicationof Equation (128) and to study the sensitivity of the design to the choiceof βD

Figure 123 shows a typical parabolic tendon profile for a simple or acontinuous span The tendon exerts on the concrete a uniform load whoseintensity is

qprestress = minus8 Pm f0l2 (124)

where Pm is the absolute value of the mean prestress force an average of thevalues before and after the loss due to creep shrinkage and relaxation l isthe span f0 is the distance between the chord joining the tendon ends over thesupports and the tendon profile at mid-span this distance is measured in thedirection of the normal to the centroidal axis of the beam The value of Pm isassumed to be constant within each span at a simply supported end thetendon has zero eccentricity the value (qprestressq) is assumed to be the same inall spans thus when the spans are unequal f0 is assumed to vary such thatqprestress is the same for all spans (Equation (124) ) The intensity q of the per-manent or quasi-permanent load is assumed to be constant and equal in allspans With these assumptions the balanced deflection factor βD is the sameas the balanced load factor thus

βD = minusqprestress

q(125)

Mprestress = minusβDMq (126)

where Mprestress and Mq are bending moments at any section due to the


prestressing force Pm and due to the sustained load q respectively Thepermanent stress at any fibre due to Pm and q combined is

σperm = Mq y

I (1 minus βD) minus

Pm

A(127)

where A and I are the area and the second moment of area about centroidalaxis of the gross concrete section y is the coordinate of the fibre consideredmeasured downward from the centroidal axis (Fig 122) Substituting Equa-tion (125) to (127) in Equation (122) and solving for the balanced deflectionfactor gives

Figure 122 Bridge cross-sections considered in Examples 121 to 123 (a) closed section(b) open section


βD 1 minus σallowable σq

1 + I(8αqA f0 y)(128)

where σq is a hypothetical value of the stress that would occur at the fibreconsidered if q were applied without prestressing and the section ishomogeneous noncracked

σq = Mq y

I(129)

αq is a dimensionless coefficient defined as

αq = Mq (ql 2) (1210)

The mean value of the prestressing force required is (by Equations (124) and(125) )

Pm = βD ql 2

8f0

(1211)

For a simple span Equation (128) is to be applied at the section at mid-span to give βD and the result substituted in Equation (1211) to give Pm In acontinuous beam the two equations should be applied for critical sectionsover the interior supports and at (or close to) mid-spans The largest Pm thusobtained should be adopted in design

Figure 123 Typical cable profile in a span of a simple or a continuous beam Assumptionused in derivation of Equation 128 When end A or B is simply supported thetendon eccentricity eA or eB is zero


For a chosen value of βD the permanent stress is (Equations (127) and(1211) )

σperm = σq[1 minus βD (1 + I (8αqAf0y)] (1212)

125 Examples of design of prestress levelin bridges

Figures 122(a) and 122(b) represent cross-sections of bridges that will beused in design examples and in parametric studies The thickness h willbe varied as well as the span to thickness ratio lh

Example 121 Bridges continuous over three spans

Consider a bridge deck having a constant cross-section shown in Fig122(a) continuous over three spans 07l l and 07l with l = 60m andh = 3m The cross-sectional area properties are A = 725m2 I =951m4 the y coordinate of the top-fibre is y = ndash1168m In addition toits self-weight (24kN-m3) the deck carries a sustained dead load of325kN-m Thus the total permanent load is q = 325 + 24 (725) =2065kN-m Assume a parabolic tendon profile (Fig 123) with f0 =h minus 01m Determine the balanced deflection factor βD and the requiredmean prestressing force Pm such that the permanent stress at top fibreover the two interior supports equal the allowable stress σallowable =minus2MPa

For a continuous beam of the specified spans the bending momentover the two interior supports is (by elastic analysis)

Mq = minus00763 ql 2 thus αq = minus00763

The value αq can be more accurately calculated by considering thefact that over a short distance over the supports the actual profile ofthe tendon should be convex to avoid sudden change in direction

The hypothetical stress value at top fibre if q were applied withoutprestressing (Equation (129) )

σq = minus00763 (2065 times 103) (60)2 (minus1168)

951 = 697MPa

The balanced deflection factor is (Equation (128) )


βD 1 minus (minus20)697

1 + 951[8(minus00763)(725)(29)(minus1168)] = 079

The absolute value of mean prestressing force should exceed(Equation (1211) )

Pm = 079 (2065 times 103)(60)2

8(29) = 25 200kN

Table 121 gives the variation of βD for the same structure when theallowable stress σallowable is varied between +24 and minus44MPa The sametable indicates the sensitivity of the permanent stress σperm to the vari-ation of the prestressing force The table shows the typical result thatthe absolute value of the sustained compressive stress drops rapidlywith the decrease of βD In this example a change of βD from 10 to 06varies σperm from minus44 to 01MPa We recall that βD = 10 corresponds tozero curvature (Fig 121) or zero deflection With βD = 06 the curva-ture the deflection and the cracking can be critical particularly whenthe transient stress due to live load and temperature ∆σ is high Table121 also demonstrates the typical result that the reserve compressivestress σperm can be substantially eroded when the actual time-dependentprestress loss is greater than estimated (eg when Pm actual = 09 Pm) Thiscan cause σperm to become small compressive or even tensile causing theresidual cracking and the residual deflection to be critical after cyclicapplication of exceptional live load and temperature variation

The values on the last line of Table 121 are calculated for the opencross-section shown in Fig 122(b) with h = 3m and area properties A= 600m2 I = 510m4 yt = minus0813m q = 1765kN-m other data are the

Table 121 Variation of the permanent stress perm with the balanced deflectionfactor D The table also gives variation of the required D for a givenallowable stress allowable Three-span bridge of Example 121 Fig 122(a)and (b) with h = 3m

Balanced deflection factor D 04 05 06 07 08 09 10

Top fibre stress perm or allowable

(MPa) closed box section Fig 122(a) 241 127 013 minus100 minus214 minus328 minus442


(MPa) open section Fig 122(b) 281 158 035 minus088 minus211 minus333 minus456


same as above The same remarks made above about the sensitivity ofσperm to the choice of βD apply to the open cross-section

For comparison we give below the results when the above calcula-tions are repeated for the open cross-section in Fig 122(b) with q =1765kN-m and σallowable = minus2 MPa (unchanged)

σq = 773MPa

βD 079

Pm = 21 600kN

Example 122 Simply supported bridges

Determine the balanced deflection factor βD and the required meanforce Pm using the same data as in Example 121 but for a simplysupported span l = 60m and h = 3m σallowable = minus2MPa at the bottomextreme fibre of mid-span section

(a) Closed cross-section (Fig 122a) q = 2065kN-m y = 1832m A =725m2 I = 951m4

Mq = 0125 ql 2 αq = 0125

The hypothetical stress value at bottom fibre (y = 1832m) if q wereapplied without prestress

σq = 125(2065 times 103) (60)2 1832

951 = 1790MPa


βD 1minus (minus20)1790

1 + 951[8(0125)(725)(29)(1832)] = 089

Pm = 089 (2065 times 103)(60)2

8(29) = 28 500kN

(b) Open cross-section (Fig 122(b) ) q = 1765kN-m y = 2187m


A = 600m2 I = 510m4 Repetition of the above calculation using thisdata gives

σq = 3405MPa

βD = 085

Pm = 23 300kN

Example 123 Effects of variation of span to thickness ratio on βD

For the closed bridge cross-section shown in Fig 122(a) determine thevalue of βD required for an allowable stress above the interior supportsσallowable = minus2MPa Assume that the bridge deck is continuous over threespans of lengths 07l l and 07l Consider l = 30 60 and 90m and lh= 20 25 and 30 In all cases use f0 = h minus01m and q = self-weight plus325kNm specific weight of concrete 24kN-m3

Calculations similar to Example 121 give the results in Table 122which indicate that βD varies between 094 and 069 the lower value isapproached with the increase in l or in frasll h The values of the meanprestress force Pm for each case are also given in the same table

Table 122 Variation of the required balanced deflection factor D with the span land span to thickness ratio lh Bridge deck with spans 07l l and 07lclosed cross-section (Fig 122a) sustained load q = self-weight plus325kN-m allowable permanent stress at top fibre above interior sup-ports allowable = minus2 MPa

lh l = 30 l = 60m l = 90m

D Pm (kN) D Pm (kN) D Pm (kN)

20 094 12 800 079 25 400 075 41 60025 087 14 500 073 27 500 071 45 50030 083 16 400 072 31 100 069 49 700

126 Transient stresses

The transient stress in concrete ∆σ caused by variable actions such as liveload and temperature variation when added to the sustained stress σperm

can produce irreversible opening of cracks and irreversible deformationsThe values of ∆σ and σperm will be used below (Section 129) to give the


non-prestressed reinforcement ratio required for control of residual crackopening Again homogenous uncracked sections are assumed in calculating∆σ and σperm

The magnitude of the transient stress ∆σ is different from structure tostructure depending upon the type of live load and the climate As exampleconsider a bridge continuous over three spans 07l l and 07l and having abox or open cross-section shown in Fig 122(a) or (b) A temperature risevarying over the depth as shown in Fig 124 produces at the section at themiddle of the interior span the stress distributions2 shown in Fig 125 Herethe span is varied between l = 30 and 90m maintaining the span to thicknessratio frasll h =20 It can be seen that in this case the maximum tensile stress dueto temperature rise is of the order 2 MPa (03ksi) and occurs near the bottomfibre The temperature distribution in Fig 124 may be representative of thecondition in the afternoon of a summer day in moderate climate A distribu-tion of the same shape but with half the temperature values and the signreversed (representing drop of temperature) may appear in the night or in theearly morning in winter This drop in temperature produces at the sectionover the interior supports the stress distributions shown in Fig 125 with thestress values multiplied by minus05 Here the maximum tensile stress is againclose to 2 MPa (03ksi) occurring at the top fibre It is to be noted thatbecause of the roller supports the constant part of the temperature riseshown in Fig 124 produces no stress In calculation of the stresses presentedin Fig 125 the coefficient of thermal expansion is taken equal to 10 times 10minus6

per degree Celsuis (56 times 10minus6 per degree Fahrenheit) and the modulus ofelasticity is considered equal to 30 GPa (4400ksi) The stresses presented arethe sum of self-equilibrating stresses and stresses due to staticallyindeterminate moment (See Example 101)

In the same bridges the maximum stress ∆σtraffic due to exceptionally heavyload3 (convoy weighing 4000kN (900kip) ) is between 18 and 47MPa (026

Figure 124 Distribution of temperature rise over the height of a bridge cross-section usedin the analyses whose results are shown in Fig 125


and 068ksi) in the bottom fibre at the section in the middle of the interiorspan The lower stress value is for interior span l = 30m and closed cross-section (Fig 122(a) ) the upper value is for l = 90m and open cross-section(Fig 122(b) ) At the section over the interior support the maximum value is∆σtraffic = 15 to 2MPa (02 to 03ksi) with the lower value being for the closedsection

The thermal stresses given above are calculated for bridges continuous oversupports that allow free axial elongation Thus the distribution of tempera-ture rise is divided into a constant part and a variable part (hatched) in Fig124 only the latter part produces thermal stresses Cracking is ignored in thecalculation of the thermal stresses presented in Fig 125 The staticallyindeterminate bending moments due to temperature variation are pro-portionate to the flexural rigidity EI In a cracked zone the flexural rigidity issmaller than in the uncracked zone if this variation in flexural rigidity istaken into account (as discussed in Chapter 13) the calculated staticallyindeterminate bending moments can be much smaller than what is obtainedwith linear analysis This fact makes it advantageous to allow cracking bynot choosing a too high value of the balanced deflection factor βD particu-larly in climates with temperature extremes Details of the linear analysis thatgive the thermal stress distributions in Fig 125 are given in Section 59 inGhali amp Neville4

Figure 125 Distribution of stress due to temperature rise (Fig 124) in a bridge continuousover three spans Cross-section at the middle of the central span Constantcross-section with height h = l20 l = variable The cross-section is eitherclosed or open (Fig 122) 1MPa = 0145ksi 1m = 328 ft 1 degC = 18 degF


127 Residual opening of cracks

Consider a prestressed concrete member containing non-prestressed re-inforcement While the sustained stress σperm at extreme fibre of a cross-section is compressive assume that the member is subjected to cyclic loadsproducing cyclic stress change at the same fibre Assume that the maximumvalue ∆σ represents in practice the transient effect of live load andor tem-perature variation It is assumed here that the value ∆σ is calculated for anoncracked section Experiments and observations have shown that in eachload cycle the crack opens and closes and the width varies between wmax andwres where the former is the maximum width and the latter is a residual crackwidth caused by a permanent damage of the bond between the concrete andthe reinforcement

Experiments5 on a prestressed member subjected to axial tension producedby a displacement controlled actuator show that the residual crack widthwres is highly dependent upon the sustained compressive stress σperm Figure126 shows the results of a series of tests in which σperm is varied between 0and minus35MPa It can be seen that as the absolute value of σperm is varied fromzero (reinforced non-prestressed member) to 35MPa wres varies from012mm to almost zero The ordinates plotted in the graph are crack widthsmeasured after 9000 cycles the widths measured after one cycle are not sub-stantially different The experiments repeated with different non-prestressedsteel ratio ρns (= Ansgross concrete cross-sectional area) show that wres isslightly influenced by ρns This is contrary to what is generally observed forreinforced non-prestressed elements

The specified concrete strength and the diameter of the non-prestressedbars (for a given ρns value) have an important effect on the maximum width ofcrack at the peak of the transient stress however the same two parametershave negligible influence on the residual crack width wres

The residual crack width may be considered an important parameter indesign because it is closely related to the durability of the structure Charac-teristic limit values wk res = 02 to 005mm are recommended or required bysome codes in the design of prestressed concrete structure the lower value isfor the case when water-tightness is required as for example the deck slab ofa bridge The characteristic value wk res is assumed to be equal to 15 times themean value obtained in experiments or by analysis

128 Water-tightness

Avoiding or limiting leakage is one of the reasons of controlling cracking instructural members that are permanently or occasionally in contact withwater Examples of such structures are water tanks tunnels parking floorsand bridge decks The repeated flow of contaminated water often withde-icing salts has negative impact on the structuresrsquo durability


For a specified liquid the rate of flow through a crack in a floor or a wall(mass per unit time per unit length) is proportional to the gradient of liquidpressure and inversely proportional to the thickness of the structural elementbut more importantly to the cube of the crack width6 The crack width is thusthe governing factor for liquid-tightness

When water is permanently in contact with a cracked concrete surface itwill cause swelling that tends to close cracks having widths less than 02mmHowever this self-sealing process is hampered when the crack is repeatedlywidened and the increase in width exceeds 01mm

Figure 126 Variation of residual crack with wres with the sustained prestress sus Cyclicimposed elongation ∆l l varying between 200 times 10minus6 and 400 times 10minus6Specified concrete strength f primec 30MPa (4400psi) Reference is mentionedin Note 1 of this chapter


129 Control of residual crack opening

The experimental work on prestressed concrete members subjected to axialtension may be considered representative of the upper or the lower slabs of aprestressed box cross-section subjected to bending moment that producecracking The results of the experiments have been used to verify an ana-lytical model that calculates the width of residual cracks wres Involved in theanalysis are equations for the bond stress and the tensile stress in the vicinityof a crack under the effect of cyclic loading The analytical model is thenused in parametric studies and in the development of the design charts7

presented in Figs 127 (a) to (e)For a specified characteristic residual crack width wres the solid lines in

the charts give the non-prestressed steel ratio ρns as functions of the perman-ent stress σperm and the maximum transient stress ∆σ The dashed straightlines in the graphs give a non-prestressed steel ratio that must be exceeded toavoid yielding of this reinforcement when the maximum transient loading isapplied The specified yield stress of the non-prestressed steel is assumedequal to 400 500 or 800 MPa The residual crack width wres is slightlyinfluenced by the specified concrete strength or by the diameter of the non-prestressed steel bar thus these two parameters are absent from the designcharts

The charts give the recommended value for ρns from the input parametersσperm ∆σ wk res and σs max(equiv fy) The symbols are redefined below and someremarks are given on the calculation of the parameters involved

fy = specified yield stress of non-prestressed reinforcementwk res = specified characteristic residual crack width a value equal to 15

times the mean residual crack width∆σ = maximum change in tensile stress in concrete due to transient actions

(eg live load and temperature variation) In calculation of ∆σcracking is ignored

ρns = AnsAcef where Ans is cross-sectional area of non-prestressedreinforcement within an effective concrete area Acef The charts arederived for sections subjected to axial tension Their use is recom-mended without adjustment for a slab whose thickness is less than03m in the tension face of a box section in flexure in this case Acef isequal to the cross-sectional area of the slab it is also empiricallyrecommended that σperm and ∆σ be calculated at a distance 01maway from the extreme fibre Use of the same charts is extended toother cases by empirically taking Acef as specified in Fig E2 (seeAppendix E)

σperm = permanent stress at a fibre due to dead loads and other quasi-permanent loads combined with prestress The value of σperm is calcu-lated for a gross concrete section (uncracked) at a fibre that will


1210 Recommended8 longitudinalnon-prestressed steel in closed-boxbridge sections

Figure 128 represents top view of the top and bottom slabs respectively of abridge having the closed cross-section in Fig 122(a) The two slabs are div-ided into zones labelled 1 2 3 and 4 Use of the charts in Figs 127(a) to (e)is recommended to give ρns but this ratio should be greater or equal to 04per cent Near the tip of the cantilevers over a width equal to one third of thewidth of the overhang bc and a length equal the full length of the deck ρns isempirically recommended to be not less than 06 per cent This is because theslab edges are vulnerable to cracking caused by shrinkage of the tip which isoften cast subsequent to the casting of the deck The edges are also subjectedto stresses due to temperature gradient in the horizontal direction often notconsidered in calculation of the transient stress ∆σ In the longitudinal direc-tion for a length equal l5 on either side of an interior support (zones 1 and4) ρns is to be controlled by σperm and ∆σ at the section over the support forthe remainder of the length of the deck (zones 2 and 3) ρns is to be controlledby σperm and ∆σ at the section at mid-span

Commonly the thickness of the top and the bottom slabs are variable in thetransverse direction (as opposed to the simplified bridge cross-sections in Fig122) It is suggested that the cross-sectional area of the non-prestressedreinforcement per unit width of the slab be equal to the value of ρns asrecommended above for each zone multiplied by the minimum slab thicknessin the zone in this way the diameter of the bars and their spacing can beconstant in each zone

1211 Residual curvature

The graph in Fig 85 represents the relationship between the bendingmoment M and the curvature ψ for a reinforced non-prestressed sectionsubjected to no normal force The same figure is shown in Fig 129(a) forcomparison with the M-ψ graph when the section is subjected to a normalforce N lt 0 caused by prestressing Here the coefficient β = β1β2 = 05representing the case of repetitive loading and the use of high bond barsThe graph AFG in Fig 129(b) represents M versus ψ1 or M versus meancurvature ψm after cracking for a section subjected to a constant normal

subsequently be in tension when the transient load is applied Thefibre considered may be empirically taken at 01m away from theextreme fibre of a slab of a box section alternatively at the centroidof Acef The prestressing forces to be used in calculating σperm is anaverage of the values of the prestressing forces before and after thetime-dependent losses


force N lt 0 (compressive) and variable M The graph AED in the samefigure represents the case when N = 0 (same as AED in Fig 129(a)) It canbe seen that the effect of prestressing is to move ED upward to FG thus thevalue Mr radic05 the upper limit of the noncracked state is increased also aftercracking ψm is reduced We recall that Mr is the value of the moment justsufficient to cause a virgin section to crack Mr can be determined from theequation

Figure 127 Choice of the non-prestressed steel ratio ns to limit characteristic crackwidth wk res for variable maximum transient stress ∆ The charts can alsogive the mean residual crack width (= wk res15) for given values of ns and∆ (a) to (e) ∆ = 2 3 4 5 and 6 MPa respectively 1 MPa = 0145 ksi


fct = N

A1

+ M

W1

(1213)

where fct is the tensile strength of concrete A1 and W1 are the area and thesection modulus of the transformed uncracked section respectively

Figure 128 Division of top and bottom of slabs of a closed-box section into zones in whichthe recommended non-prestressed steel ratio ρns is constant


Figure 129 Moment curvature relationship (a) N = 0 (b) comparison of the meancurvatures ψm with N = 0 and with N lt 0 (compressive) (c) residualcurvature in a prestressed section subjected to transient momentincrement ∆M producing cracking


Figure 129(c) represents a prestressed section for which the moment Mperm

due to prestressing and sustained load is below Mr A transient momentincrement ∆M brings the moment level to Mmax and the curvature to ψmaxRemoval of ∆M does not fully recover the curvature and the section is leftwith a residual curvature ψres as the moment level returns to Mperm The des-cending branch HI of the M-ψ curve is defined by the straight line HIJ whereJ is a point whose ordinate is minusMmax and situated on the straight line FAJThis graphical construction that gives ψres is based on extensive experiments9

The residual curvature can be calculated by the equation

ψres = Mmax + Mperm

2Mmax

ψm Mmax minus

Mmax

EI1 (1214)

where ψm Mmax is the mean curvature corresponding to Mmax the value of the

mean curvature is given by Equation (821)The residual curvature can cause an increase in deflection which is

additional to the deformation caused by the time-dependent effects of creepshrinkage and relaxation Studies of the bridge cross-sections shown in Fig122 and discussed in Section 125 show that the residual curvature increasesas the balanced deflection factor βD is decreased Significant residual curva-ture and deflection that should be avoided occurs when βD 06 and l 60m For structures expected to have long service life such as bridges it isrecommended to adopt βD between 08 and 10 It is to be noted that for thesame value of βD the permanent stress σperm is a compressive stress of smallerabsolute value for smaller spans This can be seen by setting a value of βD inEquation (128) and solving for the allowable stress Thus cracking occursat a smaller moment increment ∆M when the span is smaller For this reasonβD should be closer to 10 for smaller spans

1212 General

This chapter is concerned with prestressed structures in which crackingoccurs under transient live load andor temperature variation These repeti-tious actions produce residual opening of cracks and residual curvatureassociated with residual deflections that are additional to the time-dependenteffects of creep shrinkage and concrete Structures should be designed suchthat in the permanent state the opening of cracks and the deflections are notexcessive

A balanced deflection factor βD is defined as equal to the absolute value ofthe permanent deflection due to prestressing divided by the deflection due topermanent load Choice of the parameters βD and the corresponding amountof prestressing are important factors in limiting the residual opening ofcracks and the residual deflections

The residual width of cracks also depends upon the amount of the


non-prestressed steel The charts presented can be used to give the non-prestressed steel ratios that limit the residual crack widths to specified valuesChoice of low value for βD requires higher non-prestressed steel ratio

The value of βD may be selected between 07 and 10 depending upon thetype of the structure The selected value of βD can be 10 or even more instructures exposed to large variable loads such as railway bridges Forbridges built by incremental launching βD between 05 and 06 is possiblebecause of the favourable effect of high axial prestressing force needed forlaunching (Such an axial force does not produce deflection thus is notaccounted for by the parameter βD) In prestressed concrete floor slabs βD canbe as low as 05

Notes

1 For further reference on bridge design relating to concepts in this chapter seeLaurencet P Rotilio J-D Jaccoud J-P and Favre R Influence des actionsvariables sur lrsquoeacutetat permanent des ponts en beacuteton preacutecontraint Swiss Federal Instituteof Technology IBAP Lausanne Switzerland May 1999 168 pp

2 Slightly different stress values are presented in a figure by Rotilio J-D Con-tributions des actions variables aux deacuteformations a long terme des ponts en beacutetonDoctorate thesis No1870 Swiss Federal Institute of Technology IBAP LausanneSwitzerland 1997

3 See the reference in Note 2 above4 See the reference in Note 3 page 995 See Note 1 above6 Mivelaz P Jaccoud J-P and Favre R lsquoExperimental Study of Air and Water Flow

through Cracked Concrete Tension Membersrsquo 4th International Symposium onUtilization of High-StrengthHigh-Performance Concrete Paris 1996 PublicationIBAP No 173 Swiss Federal Institute of Technology Lausanne SwitzerlandJanuary 1996

7 The charts are taken from the reference mentioned in Note 1 above8 The recommendations are taken from the reference in Note 1 above9 Details of the experiments and their results are given in the reference in Note 1

above


Non-linear analysis ofplane frames

131 Introduction

In statically indeterminate structures the reduction in member stiffness dueto cracking is accompanied by changes in the reactions and internal forcesThe changes can result in an increase in deflection An example of this situ-ation is the so-called lsquoredistributionrsquo of moments associated with cracking incontinuous beams and slabs The relatively large negative moments overinterior supports produce cracking accompanied by a drop in the absolutevalue of the negative moment and an increase in the positive moment and inthe deflection Cracking also causes considerable reduction in the stresses andinternal forces induced by temperature variations support movements or anyother type of imposed deformations (see Chapter 11)

Arch bridge in Switzerland

Chapter 13

The present chapter is concerned with the analysis of reinforced concreteplane frames with or without prestressing accounting for the effects ofcracking The general displacement method of analysis (Section 52) is usedfor this purpose The structures are here considered in service condition inwhich the stressndashstrain relation for concrete can be assumed linear Crackinghowever causes non-linearity therefore the analysis requires iterative compu-tations using a computer Analysis beyond service condition up to failurerequires consideration of the non-linear stressndashstrain relationship for the con-crete and the reinforcement this is beyond the scope of this book Sheardeformations are ignored in the analysis presented below

When the structure is prestressed it is assumed that the stresses in thestructure are known after accounting for the effects of creep and shrinkage ofconcrete and relaxation of the prestressing reinforcement Furthermore it isassumed that no cracking has occurred at this stage and the analysis isrequired to find the effects of additional loading (eg live loads ortemperature variations) that may produce cracking

132 Reference axis

In the analysis which will follow we will use A1 B1 and I1 which are theproperties of transformed uncracked sections composed of the area of con-crete plus the area of the reinforcement Ans or Aps multiplied by αns or αpswhere αns = (EnsEc) or αps = (EpsEc) with Ens and Eps being the moduli ofelasticity of non-prestressed and prestressed reinforcement respectively Wewill also use A2 B2 and I2 the properties of transformed cracked sectioncalculated in the same manner but the concrete in tension is ignored Thesymbols A B and I represent respectively the cross-sectional area its firstand second moment about an axis through a reference point Obviously theuncracked and the cracked transformed sections do not have the same cen-troid Thus it is more suitable for frame analysis to place the nodes on a fixednon-centroidal axis of members The axis is through reference point O whichmay be chosen at any fibre for example at the top fibre However choice ofthe reference axes at or close to the centroid of the gross concrete sectionmay be advantageous in some cases thus for practice it is recommended toselect the reference axis at ndash or close to ndash this centroid in all cases

133 Idealization of plane frames

Figure 131(a) shows a typical reinforced concrete plane frame with or with-out prestressing The frame is idealized as an assemblage of straight non-prismatic beam elements connected at the joints (nodes) see Fig 131(b)Each node has three displacement components two translations and arotation in direction of arbitrarily chosen global axes x y and z

For each member a local system of axes x y and z is defined in Fig

Non-linear analysis of plane frames 429

132(a) The axis x coincides with the reference axis of the member and isdirected from end node O1 to end node O2 The y- and z-axes are mutuallyperpendicular to the x-axis with y lying in the plane of the frame Thepurpose of the analysis is to determine the changes in nodal displacements inglobal directions and for each member the internal forces at its ends alongthe six local coordinates 1 to 6 Define a number of cross-sections arbitrar-ily spaced the first section is at node O1 and the last section is at node O2 (Fig133(a) ) For each cross-section the geometry (including the areas Ans andorAps of the reinforcement layers and their depths) and the parameters σO γindefining the distribution of the initial stress (Fig 133(b) ) are part of thegiven data External applied loads on an individual member are given atthe same sections in directions of the local axes Also external forces may begiven as nodal forces in the directions of the global axes Figure 133(b) showsthe positive sign convention for σO and γ and the normal force N and thebending moment M

Figure 131 Idealization of a plane frame (a) a typical plane frame (b) idealized structure


134 Tangent stiffness matrix of a member

The tangent stiffness matrix of a typical member (Fig 132(a) ) relates theforces and the displacements in local coordinates

[S] D = F (131)

where F and D represent small increments of nodal forces and nodal

Figure 132 Coordinate system for plane frame analysis (a) local coordinates representingdisplacements D or forces F at ends of a typical member (b) system offorces in equilibrium caused by a displacement introduced at coordinate 1 2or 3 while end O2 is totally fixed


displacements A typical element Sij of the tangent stiffness matrix is equal tothe force increment at coordinate i due to a small unit displacement atcoordinate j The cross-sectional area properties A B and I are assumedvariable where A B and I are the area its first moment and its secondmoment about an axis through the reference point O (Fig 133(a) ) Thevariation of the cross-sectional area properties can be caused by variation ofgeometry or by cracking

The tangent stiffness matrix [S] can be partitioned into 3 times 3 submatrices

[S] = [S11]

[S21]

[S12]

[S22] (132)

Figure 133 Input data defining the geometry and initial stresses (a) a number of sectionsdefined on a typical member (b) cross-section geometry and initial stressdistribution at a typical section also positive sign convention for σO γ N and M


The submatrices in the first row contain forces at coordinates 1 2 and 3 atnode O1 The equilibrants of these forces at coordinates 4 5 and 6 atnode O2 form the elements of the submatrices in the second row Thus thethree elements of any column of [S11] and the three elements of the samecolumn of [S21] represent a system of forces in equilibrium (Fig 132(b) )This equilibrium relationship may be expressed as

[S21] = [R] [S11] (133)

where

[R] =minus1

0

0

0

minus1

l

0

0

minus1

(134)

with l being the memberrsquos length

Because of this equilibrium relationship and symmetry of [S] Equation(132) may be rewritten as

[S] = [S11]

[R][S11]

[S11] [R]T

[R] [S11] [R]T (135)

The matrix [S11] can be determined by

[S11] = [f ]minus1 (136)

where [f ] is the flexibility matrix of the member when it is treated as acantilever fixed at node O2 (Fig 132(b) ) Any element f ij of the flexibilitymatrix is equal to the change in displacement at coordinate i due to a smallunit increment of force at coordinate j Using virtual work (unit load theory)a typical element of [f ] is given by

f ij = l

0Nui ε Oujdx +

l

0Muiψujdx (137)

where Nui and Mui are the normal force and bending moment at any section ata distance x from end O1 due to a virtual unit force at coordinate i with i = 12 or 3 εOuj and ψuj are the changes in the strain in the same section atreference point O and in the curvature produced by a small unit force appliedat coordinate j with j = 1 2 or 3 For F1 = 1 Nu1 = minus1 and Mu1 = 0 for F2 = 1Nu2 = 0 and Mu2 = minusx and for F3 = 1 Nu3 = 0 while Mu3 = 1 Substitution inEquation (137) gives


f 1j = minusl

0εOujdx f 2j = minus

l

0ψujx dx f 3j =

l

0ψujdx with j = 1 2 3 (138)

The integrals in this equation are evaluated numerically (Section 138) usingvalues of εO and ψ determined by Equation (219) at a number of sections forwhich the geometry and cross-sectional area of reinforcement are known Forcracked sections εO and ψ represent mean values determined by Equations(843) and (844) This requires that the depth c of the compression zone andthe interpolation coefficient ζ (defined in Section 83) be known The twoparameters depend upon the stresses existing before introducing the incre-ments in the forces at the ends Thus the tangent stiffness depends upon thestress level and the state of cracking of the member

In order to generate the tangent stiffness matrix for the structure the tan-gent stiffness matrices [S] of individual members must be transformed fromthe local coordinate systems to the global system

[Smember] = [T]T [S] [T] (139)

where [Smember] is the member stiffness matrix in global coordinates [T] is atransformation matrix given by

[T] = [t][0]

[0]

[t] [t] =c

minuss

0

s

c

0

0

0

1

(1310)

where c = cos α and s = sin α with α being the angle between the global x-direction and the local x-axis (Fig 132(a) ) The matrix [T] can be used fortransformation of member end forces and displacements from local to globalor vice versa

D = [T]D Fglobal = [T]T F (1311)

135 Examples of stiffness matrices

Example 131 Stiffness matrix of an uncrackedprismatic cantilever

Derive the stiffness matrix with respect to non-centroidal coordinatesshown in Fig 134(a) for an uncracked cantilever having a constantcross-section with properties A B and I What are the displacements atthe three coordinates due to a downward force P applied at the free end


Assume that the member has a rectangular cross-section of width b andheight h

The normal strain εOuj and curvature ψuj at any section are obtainedby the application of a unit force at each of the three coordinates at theend O1 and the use of Equation (219)

For F1 = 1 Nu1 = minus1 and Mu1 = 0

εOu1 = minusI

E(AI minus B2)and ψu1 =

B

E(AI minus B2)(1312)

For F2 = 1 Nu2 = 0 and Mu2 = minusx

εOu2 = Bx

E(AI minus B2)and ψu2 = minus

Ax

E (AI minus B2)(1313)

For F3 = 1 Nu3 = 0 and Mu3 = 1

εOu3 = minusB


A


Substitution of the above expressions for εOu and ψu in Equation (138)gives

I minusBl

2B

[f] = l

E(AI minus B2)minus

Bl

2

Al 2

3minus

Al

2(1315)

B minusAl

2A

Application of Equations (136) (134) and (135) gives the stiffnessmatrix corresponding to the six local coordinates in Fig 132(b)


When O is chosen at the centroid of the cross-section B will be equal tozero and the matrix [S] in Equation (1316) will reduce to the con-ventional form of the stiffness matrix for a plane frame member1 The 3times 3 submatrix at the top left-hand corner of this matrix is the stiffnessmatrix of the cantilever in Fig 134(a)

For the cantilever with a rectangular section the area propertieswith the reference point O at the top fibre are A = b h B =b h2

2 I =

b h3

3 Substitution in Equation (1316) gives

1 0 minush

2

[S]cantilever = Ebh

l0

h

l 2

h 2

2l

minush

2

h2

2l

7h2

12(1317)

The displacements at the free end due to the applied force P are

D = [S]minus1

0

P

0

= l

Ebh3

4h2

minus3hl

6h

minus3hl

4l 2

minus6l

6h

minus6l

12

0

P

0

= Pl 2

Ebh3

minus3h

4l

minus6

(1318)

If the procedure followed in this example is redone with the referenceaxis chosen through the cross-sectional centroid D1 would be zerowith the other two displacements unchanged The top fibre at the tip of

A

l

012(AI minus B2)

Al 3

symmetrical

minusB

l

6(AI minus B2)

Al 2

4AI minus 3B2

Al

brvbarbrvbarbrvbarbrvbarbrvbarbrvbarbrvbarbrvbar

[S] = E ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash (1316)

minusA

l

0 B

l

brvbarbrvbarbrvbarbrvbar

A

l

0 minus12(AI minus B2)

Al 3minus6(AI minus B2)

Al 2


012(AI minus B2)

Al 3

B

l

6(AI minus B 2)

Al 2

2AI minus 3B 2

Albrvbarbrvbar minus

B

lminus6(AI minus B 2)

AI 2

4AI minus 3B 2

Al


the cantilever would move horizontally outwards at a distance equal toh2 multiplied by D3 which is the same answer as obtained above

Example 132 Tangent stiffness matrix of a cracked cantilever

Find the tangent stiffness matrix corresponding to the coordinates inFig 134(a) for the cantilever of Example 131 assuming that it has aconstant concrete cross-section reinforced with non-prestressed steel ofareas Ans1 = Ans2 = 001 bh (Fig 134(b) ) Also find the three displace-ments at the three coordinates due to a downward force P applied at thefree end Assume that initially the cantilever has been cracked due to anegative bending moment having the same value at all sections suchthat c and ζ are constant Given data The elasticity moduli of steel andconcrete are Es = 200GPa and Ec = 25GPa c = 0275 h and ζ = 08 thegiven c value is determined by Equation (716) the compression zoneis at the bottom of the section the transformed cross-sectional areaproperties are A1 = 06840 h2 B1 = 03420 h3 I1 = 02344 h4 A2 = 02549h2 B2 = 01849 h3 I2 = 01582 h4 where the subscripts 1 and 2 refer to theuncracked and the fully cracked states respectively Give the answer interms of P and Ec Assume that P is small such that it causes negligiblechange in the value of ζ

At any section of the cantilever the strain parameters εO and ψ arecalculated assuming that the sections are uncracked (using A1 B1 andI1) and again assuming that all sections are fully cracked (using A2 B2and I2) Then the mean strain parameters are determined using ζ = 08

Figure 134 The cantilever considered in Examples 131 and 132 (a) coordinatesystem (b) cracked reinforced section considered in Example 132


in Equations (843) and (844) The results of these calculations arepresented in Table 131 We give below as example the calculations forF1 = 1

For F1 = 1 Nu1 = minus1 and Mu1 = 0 at any section Apply Equation(1312) for uncracked section

εO = minus02344

Ec h2 [06840(02344) minus (03420)2] = minus5405 times 10minus3 (Ec h

2)minus1

ψ = 03420

Ec h3[06840(02344) minus (03420)2] = 7886 times 10minus3(Ec h

3)minus1

Apply the same equation for a fully cracked section

εO = minus01582

Ec h2[02549(01582) minus (01849)2] = minus25780 times 10minus3(Ec h

2)minus1

ψ = 01849


3)minus1

Mean parameters (with ζ = 08)

εO

ψ

mean

= (1 minus ζ)εO

ψ

uncracked

+ ζ εO

ψ

fully-cracked

= minus21700 h

25680 (Ec h3)minus1

The flexibility coefficients are determined by Equation (138) with theintegrals evaluated explicitly giving

Table 131 Strain parameters at any section of the cantilever of Example 132 due tounit force F1 F2 or F3 applied at the free end

Force Strain Uncracked Fully cracked Mean Multiplierapplied parameters

F1 = 1 Ou1 minus5405 minus25780 minus21700 10minus3(h2Ec)minus1

u1 7886 30130 25680 10minus3 (h3Ec)minus1

F2 = 1 Ou2 7886x 30130x 25680x 10minus3(h3Ec)minus1

u2 minus15770x minus41530x minus36380x 10minus3(h4Ec)minus1


u3 15770 41530 36380 10minus3(h4Ec)minus1


21700 minus25680

h l

225680

h

[f ] = 10minus3l

Ech2

minus25680

h l

236380

h2 l2

3 minus36380

h2 l

225680

hminus

36380

h2 l

236380

h2

The stiffness matrix is

02799 0 minus01976h

[S] = [f ]minus1 =Ec h2

l0 03295

h2

l201647

h2

l

minus01976h 01647h2

l02493 h2

The displacements at the free end are

D = [S]minus1 0

P

0 =

Pl2

Ec h4

minus1284h

1213l

minus1819

136 Fixed-end forces

When the external forces are applied at intermediate sections away from thenodes (Fig 132(a) ) the analysis by the displacement method involves calcu-lation of the actions Ar These are the values of the end forces due to theapplied loads with the member ends totally fixed The vector Ar may bepartitioned into two 3 times 1 vectors

Ar = ArO1

ArO2 (1319)

Consider the case when a nonprismatic member is subjected to the three forcecomponents Q shown in Fig 132(a) at one section located at a distance xQ

from the member end O1 The values of ArO1 can be determined by the forcemethod (Section 42) using as the released structure the cantilever in Fig132(b) This gives


ArO1 = minus[f ]minus1 Ds (1320)

where [f ] is the flexibility matrix derived by Equation (138) Ds is vectorof the displacements of the released structure these are given by virtualwork

D1s = minusl

0εO dx D2s = minus

l

0ψ x dx D3s =

l

0ψ dx (1321)

where εO and ψ are the strain at the reference point O and the curvatureproduced in the released structure at any section at a distance x from O1 Hereagain Equation (219) is to be used to determine εO and ψ at different sectionsand the integrals are evaluated numerically (Section 138)

The forces at end O2 can be determined by equilibrium

ArO2 = [R] ArO1 minus

Q1

Q2

Q3 minus Q2(l minus xQ)

(1322)

where [R] is the matrix defined in Equation (134)Distributed loads in the transverse or the tangential direction of a member

can be replaced by statical equivalent concentrated forces in the samedirection The following equation can be used for this purpose

QA

QB =

s

6 2

1

1

2 qA

qB (1323)

where qA and qB are intensities of the distributed applied load at two sectionsA and B spaced at a distance s QA and QB are statical equivalent concentratedforces assuming that q varies linearly over the spacing s Replacement of thedistributed load by concentrated forces (rather than concentrated forcescombined with moments) in this way involves an error which is consideredhere negligible when s is small compared to the member length (say s l8)The fixed-end forces due to the external forces applied at more than onesection can be obtained by summation using Equations (1319) to (1322) foreach section where forces are applied

137 Fixed-end forces due to temperature

Consider a nonprismatic uncracked member subjected to temperature risewhich varies over the length of the member and also over the height of itscross-section (Fig 135(c) ) Such temperature distribution can occur in prac-tice on a summer day in bridges of variable cross-section We will consider


here the fixed-end forces Ar at the six coordinates shown in Fig 132(a)caused by the temperature rise

Apply the force method (Section 42) using as a released structure a canti-lever fixed at the right-hand end O2 (Fig 135(b) ) Equations (1319) to(1321) and Equation (1322) (with Q1 to Q3 set equal to zero) can be used togive the required fixed-end forces the values of εO and ψ at any section to besubstituted in Equation (1321) can be determined by

Figure 135 Analysis of fixed-end forces and numerical integration (a) typical nonprismaticmember (b) released structure (c) temperature rise that varies nonlinearlyover the height of a typical cross-section (d) deflected shape


εO = εcentroid + yOψ (1324)

εcentroid = αt

Al

T dA1 ψ = αt

I1 centroidalT y dA1 (1325)

yO = minusB1

A1

I1 centroidal = I1 minusB2

1

A1

(1326)

where αt is the thermal expansion coefficient assumed the same for the con-crete and the reinforcement T is the temperature rise at any fibre (Fig135(c) ) y is the coordinate of any fibre measured downward from the cen-troid of the transformed section yO is the y-coordinate of the reference pointO A1 B1 and I1 are the area of the transformed uncracked section and its firstand second moments about an axis through the reference point O I1 centroidal isthe second moment of area of the uncracked transformed section about anaxis through its centroid

A temperature rise that varies over the depth as a straight line defined byTO = εOαt and dTdy = ψαt produces the same fixed-end forces as the nonlin-ear distribution having the same values of εO and ψ determined by Equations(1324) and (1325) However only when the variation of temperature is non-linear self-equilibrating stresses given by Equation (230) occur The nonlin-ear analysis presented in this chapter does not explicitly account for the self-equilibrating stresses Presence of these stresses can cause cracking at a sec-tion to occur at lower load level As approximation the value of fct may bereduced by an estimated value of the tensile stress at the extreme fibre esti-mated by Equation (230)

138 Numerical integration

Consider a typical member (Fig 135(a) ) and define a number of its crosssections (say ge 9) In evaluating the integrals involving εO and ψ at the sectionsit will be considered sufficiently accurate to assume that the parameters varylinearly over a typical spacing AB Thus the well-known trapezoidal rule canbe employed to determine the areas below the broken lines representing thevariation of εO and ψ over the member length this gives the value of the firstand the last integrals in each of Equations (138) and (1321) The remainingintegral in each of the two equations can be evaluated over the spacing s oftwo consecutive sections by (Fig 135(b) )

sψ x dx =

s

6(2ψA xA + ψA xB + ψB xA + 2ψB xB) (1327)

where the subscripts A and B refer to values of ψ and x at the ends of thespacing s This equation can be employed for each spacing the sum of theresults gives the value of the integral over the member length


When the curvatures ψ have been determined at a number of specifiedsections on a member Equations (1328) and (1329) can be used to deter-mine the deflection DC where DC is the transverse distance between the chordand the deflected member at any section C (Fig 135(c) ) The chord is thestraight line joining the nodes O1 and O2 in their displaced position Againthe contribution of the curvature over a typical spacing s can be calculatedseparately and the contributions of all spacings can be summed up togive DC

Contribution of spacing AB to DC (Fig 135(c) )

l minus xC

l

s ψ x dx =

l minus xC

l s

6(2ψA xA + ψA xB + ψB xA + 2ψB xB)

when xB xC (1328)

xC

l

s ψ(l minus x)dx =

xC

l l s(ψA + ψB)

2 minus

s

6(2ψA xA + ψA xB

+ ψB xA + 2ψB xB) when xA xC (1329)

where xC is the distance between node O1 and the point considered

139 Iterative analysis

The analysis described below applies the displacement method in iterativecycles Each cycle starts with known values of the parameters σO γ c and ζ ateach section of individual members where σO is the stress at reference pointO γ is the slope of the stress diagram (Fig 133(b) ) c here means depthof the part of the section in which the concrete is not ignored thus for acracked section c is the depth of the compression zone but for an uncrackedsection c = h with h being the full height of the section ζ is the interpolationcoefficient In each iteration these values are updated For the analysis of anon-prestressed reinforced concrete frame the initial stresses are assumednull and the sections are assumed uncracked thus at all sections σO = 0 γ = 0c = h ζ = 0 For a prestressed frame the initial stresses are defined by the givenparameters σOin and γin and again the sections are assumed uncracked thusc = h and ζ = 0 The cycles of analysis are repeated until the residual vectorFresidual becomes approximately equal to 0 generation of the vectorFresidual is explained below The analysis cycle is completed in three steps

Step 1 Determine by conventional linear analysis the nodal displacementsand the member end forces This involves generation of stiffness matrices[S] of individual members transformation of [S] from local member direc-tions to global directions assemblage of the transformed matrices ([T]T [S][T]) to obtain the stiffness matrix [S] of the structure adjustment of the


stiffness matrix according to the support conditions of the structure andsolution of the equilibrium equations

[S] D = minusF (1330)

where F is a vector of forces that can restrain the displacements at thenodes F is generated by summing up the forces applied at the nodes withreversed sign to the transformed fixed-end forces ([T]TAr) for each memberwhere [T] is a transformation matrix defined by Equation (1310) Solution ofEquation (1330) gives the nodal displacements D in the global directionsThese are used to find the member end forces for individual members

A = Ar + [S] D (1331)

D6times1 is a vector generated by transformation of three displacement com-ponents at each of the two nodes at the member ends (Equations (1311) )

In the first cycle Ar is determined by the force method ignoring cracking(see Sections 136 and 137) In other cycles Ar is given by Equation(1334) In each cycle the memberrsquos tangent stiffness matrix [S] is calculatedconsidering the updated c and ζ values for the sections (values existing at theend of the preceding cycle)

Step 2 Update the nodal displacements D and the member end forcesA by adding the values determined in step 1 of this cycle to the valuesexisting at end of the preceding cycle For each member the forces at the endsand the forces at intermediate sections represent a system in equilibriumDetermine the values of normal force and the bending moment in all sec-tions Use these to update c ζ σO and γ and calculate the strain parameters εO

and ψ Apply Equation (1321) to determine Ds a vector of the threedisplacements at node O1 relative to node O2 Calculate for each member theerror in nodal displacements of node O1 relative to node O2 by

Derror = [H] D minus Ds (1332)

where

[H] =1

0

0

0

1

0

0

0

1

minus1

0

0

0

minus1

0

0

l

minus1

(1333)

D are the displacements in local directions at the two ends of the mem-ber The elements of D are obtained by transformation using Equation(1311) of the displacements of the two nodes associated with the memberthese are part of the updated global displacements D


Step 3 Calculate the residual member end forces by the equation

Aresidual = [S11] Derror

[R] [S11] Derror (1334)

where [R] is the matrix defined by Equation (134) [S11] is a 3 times 3 matrix thetop left-hand part of the partitioned matrix in Equation (132) Generationof [S11] is to be based on the updated c and ζ determined in step 2 Trans-form Aresidual to global directions (Equation (1311) ) and sum up for allmembers to generate a vector of residual nodal forces Fresidual If Fresidual issmaller than a specified tolerance (Section 1310) terminate the analysisotherwise set F = Fresidual and go to step 1 to start a new cycle

1310 Convergence criteria

The iteration cycles discussed above may be terminated when

FT F12residual le αtolerance FT F12

cycle 1 (1335)

where αtolerance is a small specified value say between 001 and 0001 Thiscriterion ensures that the residual forces are small compared to the nodalforces applied on the structure

When the analysis is for the effect of prescribed displacements F can benull while the nonzero forces are the support reactions R In this case theconvergence criterion may be

FT F12residual le αtolerance RT R12

cycle 1 (1336)

where Rcycle is a vector of the reaction components determined in step 1 ofcycle 1

The elements of F or R have either the unit of force or force-lengthThe elements with the latter units may be derived by an arbitrary lengthbefore application of Equation (1335) or (1336) The arbitrary length maybe chosen equal to the larger overall dimension of the frame in the global xand y direction

The objective of the iteration cycles discussed in Section 139 is to bring toa tolerable level Derror (the membersrsquo displacements calculated by Equation(1332) ) It is possible that the error in one of the three displacements changessign in two consecutive iterations with no convergence This can occur whenat one section of the member a new crack is developed due to a small changein the internal forces thus causing a sudden change in the mean strainparameters This iteration problem can be avoided by adopting Equations


(843) and (844) in the calculation of εOm and ψm for a section when the stressat the extreme fibre exceeds radicβ fct instead of fct where β = β1β2 defined inSection 83 When the section is subjected to M without N this change willmake the moment-curvature graph in Fig 85 follow the curve AED insteadof ABCD (see Section 841)

1311 Incremental method

In this section we discuss the simpler technique known as the incrementalmethod The load vector F generated as described in Section 139 is div-ided into m increments δi F with i = 1 to m where δ is a dimensionlessincrement multiplier The load increments are applied one at a time and anelastic analysis is performed For each increment the following equilibriumequation is solved

([S]∆D)i = δi F (1337)

The tangent stiffness matrix [S]i depends upon the state of cracking (c andζ) of all sections reached in the preceding increment (i minus 1) The incrementsof displacements and the stresses calculated for each load increment areused to update these parameters for each node and for each section Thesolution of the problem is achieved at the end of calculations in the lastincrement m

A typical plot of the displacement at any coordinate versus the correspond-ing nodal force is shown in Fig 136 from which it is seen that the error forany increment exceeds the preceding ones However the accuracy can beimproved by using smaller increments thus increasing the computation thisis not a hindrance with most computers and most structures The advantagesof the incremental method are its simplicity and robustness (no convergencecriterion has to be satisfied)

Because in the analysis considered in this chapter the stressndashstrain rela-tionships of the concrete and the reinforcement are assumed linear the struc-ture behaves linearly until cracking occurs at one section Thus the first loadincrement multiplier δ1 can represent the estimated load level that producesfirst cracking

For each load increment a new analysis is performed this allows forchanges in the support conditions if necessary (eg to represent a construc-tion stage) It should be noted that a section cracked in the ith incrementremains cracked in a later increment j even when N combined with Mbecome insufficient to produce cracking In such a case a part or all of thetension zone becomes in compression and the crack closes If in a subsequentincrement N combined with M produce compression at the extreme fibrethe crack closes further change of N andor M will cause the crack to reopenwhen the stress at the same extreme fibre is greater than zero and not


necessarily greater than fct This should be observed in the storage and in theuse of the parameters representing the state of cracking of the sections

1312 Examples of statically indeterminatestructures

Figure 136 Variation of displacement with force at a typical coordinate Typical result ofthe incremental method

Example 133 Demonstration of the iterative analysis

Perform two iteration cycles to determine the bending moment MB atthe interior support of the reinforced concrete beam shown in Fig 137due to concentrated load Q = 600kN (135kip) at mid-span Considerfct = 25MPa (036ksi) Ec = 300GPa (4350ksi) and Es = 200GPa(29000ksi) β1 = 1 β2 = 05 (definitions of β1 and β2 are given in Section83)

The reference axis is chosen at mid-height of cross-section The struc-ture has only one member and only three unknown displacements atcoordinates 1 2 and 3 at node A need to be considered (of which D2 =0) The local coordinates 1 2and 3of member AB are the same as


the global coordinates The cross-sectional area properties before crack-ing are

A1 = 2213 times 10minus3 m2 B1 = 0 I1 = 9595 times 10minus3 m4

Eleven sections equally spaced at 01l are considered in the analysis Thepositive or negative moment that is just sufficient to produce crackingis

Mr = plusmn fctW1 = plusmn685kN-m

The units used in all calculations given below are Newton and metre

Iteration cycle 1In this cycle the structure is assumed uncracked A linear analysis gives

MA cycle 1 = 0 MBcycle 1 = minus135 times 103 MC cycle1 = 1125 times 103 (a)

The corresponding displacements at the three coordinates at end A(Fig 137(b) ) are

Figure 137 Reinforced concrete beam analysed in Example 133 (a) beamdimensions (b) coordinate system for member AC with end C fixed


Dcycle 1 = Dcycle 1 = 10minus6

0

0

9380

(b)

Treat AB as a cantilever fixed at B and subjected to a bending momentdiagram composed of two straight lines joining the above three M-values The magnitudes of these moments indicate that cracking occursat several sections in the negative and in the positive moment zonesCalculate εO and ψ at the eleven sections and apply Equation (1321) toobtain

Dscycle 1 = 10minus6

minus1119

6922

1530

(c)

For the member considered end B is fixed thus the displacements ofend A with respect to end B is the same as D From Equation (1332)determine Derror

Derror cycle 1 = DminusDs = 10minus6

1119

minus6922

minus5922

(d)

The flexibility matrix of the cracked cantilever is (Equation(138) )

fcycle 1 = 10minus9

5569

minus16790

minus1685

minus16790

4839

minus5692

minus1685

minus5692

8207

(e)

Inversion of [f] gives the stiffness matrix of the cracked member

[S]cycle 1 = [f]minus1cycle 1 = 106

2162

6908

5235

6908

1343

9454

5235

9454

7883

(f )

The residual member end forces (Equation (1334) )


AO1residual cycle 1 = [S] Derror = 103

1631

minus7165

minus5357

AO2residual cycle 1 = 103

minus1631

7165

minus3242

(g)

Thus the residual forces are

Fresidual cycle 1 = AO 1residual cycle 1 = 103

1631

minus7165

minus5357

(h)

Iteration cycle 2The nodal forces to be used in this cycle are F = Fresidual Because thisstructure has only one member [S] to be used in this cycle is the same as[S] in the preceding cycle however with the support conditions of theactual structure the stiffness matrix becomes

[S]cycle 1 = 106

2162

6908

5235

6908

1343times106

9454

5235

9454

7883

(i)

This is the same as in Equation (f) above but with S22 multiplied by alarge number (106) to cause the displacement at coordinate 2 to be zero

The equilibrium equations and their solutions are

[S]Dcycle 2 = minusFresidual cycle 1 = minus103

1631

minus7165

minus5357

Dcycle 2 = 10minus6

minus1095

0

1406

Dupdated cycle 2 = 10minus6

minus1095

0

2344

The member end forces (Equation (1331) )

Acycle 2 = Arcycle 2 + [S]cycle 1 Dcycle 2 Arcycle 2 = Aresidual cycle 1


AO1cycle 2 = 103

1631

minus7165

minus5357

+ [S]

minus1095

0

1406

10minus6 = 103

0

minus1430

0

AO2cycle 2 = 103

0

1430

minus1717

Update the member end forces by adding the end forces to the endforces in cycle 1 This gives MB = (minus135 + 1717) 103 = minus1178kN-m Thecorresponding bending moments at A and C are MA = 0 and MC =1211kN-m

Again the bending moment varies linearly between A and C andbetween C and B Calculation of εO and ψ at the eleven sections andintegration using Equation (1321) gives


minus1402

minus9574

3904

Apply Equation (1322) and note that D and D are the same

Derror cycle 2 = 10minus6

minus1095

0

2344

minus

minus1402

minus9574

3904

= 10minus6

3074

9574

minus1559

The computations are proceeded similar to cycle 1 giving at the end ofcycle 2 the residual nodal forces for use in the next cycle

Fresidual cycle 2 = 103

3117

0340

minus1145

Performing more iterations will give a more accurate value MB =minus1263kN-m


Example 134 Deflection of a non-prestressed concrete slab

Figure 138(a) represents a concrete slab continuous over two equalspans l = 7m The cross-section of a strip of width 1m is shown in thefigure The bottom reinforcement covers the full length of the spanswhile the top reinforcement covers a distance of 21m on both sides ofsupport B It is required to calculate the immediate deflection (withoutthe effects of creep or shrinkage of concrete or temperature variation)due to uniform load q = 75kNm treating the strip as a continuousbeam Given data Ec = 20GPa (2900ksi) Es = 200GPa (29000ksi) fct =25MPa (036ksi) β1 = 1 and β2 = 05

The moment that is just sufficient to produce cracking at top fibre atB or at bottom fibre near mid-span is

(Mr)top or bot = fctWtop or bot = minus286 or 276kN-m

where W is the section modulus the subscripts top and bot refer to theextreme top and bottom fibres respectively The first cracking occurs atthe top fibre at a load intensity qr = 47kNm The correspondingdeflection at mid-span Dr = 21mm Ignoring cracking in calculation ofthe statically indeterminate moment would give MB = minus0125ql 2 =minus459kN-m and the maximum positive moment in the spans = 25kN-m Because the latter value is lower than Mr one would concludebased on this calculation that cracking occurs only in the negativemoment zone The corresponding deflection at mid-span is equal to32mm

An analysis using a computer program that accounts for cracking(with sections placed at one tenth of the span) gives MB = minus388kN-mand indicates that the values of Mr are exceeded at B and also in thevicinity of the point of maximum moment in the span The correspond-ing deflection = 65mm which indicates that ignoring cracking in calcu-lation of MB leads to underestimation of deflection

The value of the dimensionless coefficient [minusMB(ql2)] is 0125 beforecracking When cracking develops in the negative moment zone andbefore occurrence of cracking in the positive moment zone the value ofthis coefficient drops For the load intensity considered above q =75kNm = 16 qr the dimensionless coefficient is equal to [minus(minus388)(75times 72)] = 0106 It is to be noted that the value of this coefficient dependsupon the relative flexural rigidity of the negative and positive momentzones As cracking develops in the positive zone by the increase of the


Figure 138 Slab continuous over two spans analysed in Example 134 (a) slabgeometry (b) variation of the deflection at mid-span with theload intensity


load intensity q in this example the dimensionless coefficient [minusMB (ql 2)]tends to approach again the value 0125

Figure 138(b) is a plot of (DDr) versus (qqr) where D is the deflec-tion at mid-span q is the load intensity and the subscript r refers to thestate of the first cracking The value of (qqr) is varied from zero to 22in steps of 03 before cracking (0 le (qqr) le 10) the graph is linear Aftercracking the values plotted in Fig 138(b) are based on a nonlinearanalysis of MB giving the following values for the coefficient [minusMB(ql2)] = 10minus3 125 98 105 122 118 for qqr = 10 13 16 19 22respectively In the analyses reported here the sections considered ineach span are spaced at l10

Example 135 Prestressed continuous beam analysed by theincremental method

Plot the deflection D at mid-span versus the intensity q of a uniformlydistributed live load covering two equal spans of the post-tensionedbeam shown in Figs 139(a) and (b) The parameters σOin and γin

defining the distribution of initial stresses at various sections existingbefore application of the live load are given in Table 132 The vari-ations of the parameters between sections 1 and 16 and between sec-tions 17 and 22 are parabolic Given data fct = 25MPa (036ksi) Ec =300GPa (4350ksi) Ens = Eps = 200GPa (29000ksi) β1 = 10 and β2 = 05definitions of β1 and β2 are given in Section 83 The prestressing forceand the self weight of the beam producing σO in and γin given in Table132 are respectively P = 2200kN (4946kip) and 180kNm (123kipft) The given value of P accounts for the time-dependent losses thus

Table 132 Initial stress parameters at selected sections of the post-tensioned beamof Example 135

Section Distance from Initial stress at Initial slope ofnumber support A (m) reference point O stress diagram

Oin (MPa) γin (MPam)

1 000 minus2939 01509 1000 minus2939 0843

16 1875 minus2939 091817 1875 minus2855 minus015220 2250 minus2857 minus083222 2500 minus2856 minus1414


Figure 139 Prestressed continuous beam analysed in Example 135 (a) beamdimensions (b) cross-section (c) variation of the deflection at mid-spanwith the load intensity


these are not to be considered here The initial deflection before theapplication of the live load is Din = 11mm (0043 in)

Because of symmetry one span is considered Twenty-two equally-spaced sections are used of which two sections (16 and 17) at zerodistance apart are assumed at C The reference axis is chosen at 0368m(145 in) below the top fibre First cracking occurs at qr = 134kNm(0921kipft) the corresponding deflection at mid-span Dr = 128mm(05 in) After cracking the intensity of the live load is increased up to2qr in steps of 01qr Figure 139(c) is a plot of (DDr) versus (qqr)

1313 General

At service load the stress in concrete is sufficiently low such that linearstressndashstrain relationship can be used for uncracked concrete thus the non-linearity of the analysis is required only after cracking In spite of this simpli-fication the amount of computations is large making it necessary to use acomputer The incremental or the iterative methods presented in this chaptercan be the basis of a computer program for the analysis of reinforced con-crete frames with or without prestressing accounting for cracking

For simplicity of presentation a single loading stage is considered in thischapter The same analysis with minor adjustment can be used for multi-stage loading For each stage the analysis can be applied and the results usedto update the stress the strain and the state of cracking at all sections beforestarting the analysis for a new loading stage

Note

1 See the reference mentioned in footnote 3 of Chapter 3


Serviceability of membersreinforced with fibre-reinforced polymers

141 Introduction

Corrosion of steel reinforcement increases its volume and causes spallingand deterioration of concrete To avoid corrosion stainless steel can beused as reinforcement for concrete Fibre-reinforced polymers (FRP) prod-ucts do not corrode thus they are used in lieu of steel Several FRP prod-ucts in the shapes of bars cables or grids are in use They have high tensilestrength which can exceed that of steel But many FRP bars have smallermodulus of elasticity compared to that of steel This makes membersreinforced with FRP more vulnerable to excessive deflection and widecracks when the members are non-prestressed This chapter1 discussesdesign of the amount of FRP reinforcement and choice of span to

Fibre-reinforced polymer bars

Chapter 14

thickness ratio of concrete members to avoid excessive deflections and widecracks

142 Properties of FRP reinforcementsfor concrete

FRP bars are made of continuous fibres bonded by impregnating them withmatrices such as epoxy resins and vinyl ester resin Three main types of fibresare used to produce FRP bars carbon aramid and glass Carbon fibres aremade from petroleum or coal pitch and polyacrilic nitril The stressndashstrainrelationships for FRP bars in tension is linear up to failure by rupture Valuesof the tensile strengths ffu and the moduli of elasticity Ef of carbon aramidand glass FRP bars are compared in Table 141 with the nominal yield stressand modulus of elasticity of steel reinforcing bars The values given in thetable for FRP are approximate the mechanical properties depend uponthe constituents as well as the manufacturing process the manufacturerscommonly provide the mechanical properties of their products

FRP bars have almost no adhesion to concrete The force in FRP barsembedded in concrete is developed by interlocking with deformations on thesurface of the bars (similar to deformations on steel bars) Sand-coated barsbraided bars and strands are FRP products intended to have when embeddedin concrete resistance to slipping comparable to that of steel reinforcements

Thermal expansion coefficients of steel and concrete are 12 times 10minus6 and 8 times10minus6 per degree Celcius (67 times 10minus6 and 44 times 10minus6 per degree Fahrenheit)respectively The difference between the coefficients of thermal expansion ofconcrete and FRP products is generally greater than the difference betweenconcrete and steel Furthermore FRP bars have substantially larger coef-ficient of thermal expansion in the radial direction than that of concrete (21times 10minus6 to 23 times 10minus6 per degree Celsius for glass and carbon FRP and 60 times 10minus6

to 80 times 10minus6 per degree Celsius for aramid FRP) In the longitudinal direc-tion the thermal expansion coefficient is 6 times 10minus6 to 8 times 10minus6 for glass FRPwhile carbon and aramid FRP have zero or small negative thermal expansioncoefficient The incompatibility of thermal expansion coefficients of concreteand FRP products has some adverse effect The high thermal expansion

Table 141 Properties of types of reinforcements for concrete structures

Reinforcementtype

Modulus of elasticity Tensile strength of FRP or yield stress for steel

GPa 103 ksi MPa ksi

Glass FRP 40 6 550 80Aramid FRP 80 12 1200 170Carbon FRP 150 22 2000 300Steel 200 29 400 and 500 60 and 73


coefficient in radial direction of FRP bars can cause hoop tensile stress in theconcrete adjacent to the bar and produce radial hair cracks when the tem-perature rises Once cracking occurs the tensile stresses are relieved thus theradial hair cracks do not extend to the surface of concrete member when thecover is not overly small It should be noted that the thermal expansioncoefficients of FRP vary with the method of production the type of fibre andthe resin matrix The values mentioned above are only approximate Againthe manufacturer commonly provides information on thermal properties

The compressive strength of FRP bars is relatively low and their contribu-tion to ultimate strength as compression reinforcement in concrete sections isoften not considered Also because the modulus of elasticity of FRP isrelatively low particularly in compression the contribution of FRP barssituated in the compression zone to the flexural rigidity of cracked membersis ignored

Carbon FRP has tensile strength that exceeds the tensile strength of steelused for prestressing To avoid very wide cracks the high strength of FRPcannot be fully used in non-prestressed concrete members When used forprestressing information about relaxation of carbon FRP is required thisdata should take into account the temperature and the ratio of the initialtensile stress to the tensile strength At 20 degree Celsius and initial stress 70to 80 percent of the tensile strength the relaxation of FRP in 05 millionhours (57yrs) is approximately 15 percent regardless of the type of fibre Themanufacturers of FRP for prestressing should provide relaxation data

Certain FRP products are vulnerable to rupture when they are subjected tosustained tensile stress This phenomenon referred to as creep ruptureoccurs in a shorter time when the ratio of the sustained stress to the tensilestrength is larger To control width of cracks in non-prestressed members thepermissible strain in FRP in service should be relatively low compared to thetensile strength The permissible strain in service proposed in Section 143 fornon-prestressed FRP is below the strain that can produce creep ruptureHowever when FRP is used for prestressing the ratio of stress at transfer tothe tensile strength should be small compared to the permissible ratio forprestressing steel

The basic assumptions in analysis of stresses strains and displacements ofsteel-reinforced concrete structures in service are also adopted when FRP isemployed Thus concrete and reinforcement are assumed to have linearstressndashstrain relationships Sections that are plane before deformation remainplane after deformation Concrete in tension in a cracked section is ignoredthe tension stiffening effect is accounted for empirically by interpolationbetween the uncracked state and the state of full cracking The analysis pro-cedures and equations presented in the remainder of the book for structuresreinforced or prestressed with steel can be applied with FRP using theappropriate characteristic material properties However because of some ofthe differences of properties of FRP and steel particularly in the moduli of

Serviceability of members reinforced with fibre-reinforced polymers 459

elasticity the design of sections with FRP may be governed by serviceabilityrequirements (control of deflection and crack width) rather than by ultimatestrength

143 Strain in reinforcement and width of cracks

Widths of cracks in flexural members depend upon crack spacing quality ofbond between concrete and reinforcing bars and above all upon strain inreinforcement For steel-reinforced sections codes explicitly or implicitly limitthe stain in steel in service to approximately 1200 times 10minus6 The correspondingmean crack width to be expected is 04mm (001 or 002 in) Corrosion ofsteel is one of the reasons to control crack width Because FRP bars do notcorrode wider cracks are commonly tolerated In this chapter a permissiblestrain εfservice in service in FRP bars is taken equal to 2000 times 10minus6 Theanticipated mean crack width where this strain is reached is (20001200) timesthe width when steel is employed that is 07mm (002 or 003 in) The corres-ponding permissible stress in the FRP in service is σfservice = 80 160 and300MPa (12 24 and 44ksi) for glass aramid and carbon FRP respectivelyWhen these values are treated as the permissible service stresses the design ofthe required cross-sectional area Af of the FRP will be governed by thisserviceability requirement rather than by ultimate strength This is sobecause the three values of σfservice mentioned above do not exceed 15 percentof the tensile strengths of the three types of FRP (see Table 141)

144 Design of cross-sectional area of FRP for non-prestressed flexural members

The equations presented below give the cross-sectional area Af of FRPrequired for a non-prestressed section (Fig 141) subjected in service to amoment Mservice or a moment Mservice combined with a normal force Nserviceacting at the centroid of the tension reinforcement The cross-sectional areaAf is calculated such that the strain in FRP in service be equal to a specifiedvalue εfservice It is assumed that concrete dimensions of the section have beenselected Choice of the overall height hf of the sections is discussed in Section147

For a specified value of the permissible strain εfservice in FRP in service thestress is σfservice = Ef εfservice where Ef is modulus of elasticity of FRP Ignoringconcrete in tension the cross-sectional area of FRP required when the sec-tion is subjected to a bending moment without normal force is (Fig 141(a) )

Af = Mservice(σfserviceyCT) (141)

where Mservice is the moment in service yCT is the distance between theresultants of tensile and compressive stresses in the section The depth of


compression zone c can be calculated by Equations (612) or (616) here thereinforcement in the compression zone is ignored and the section has noprestressed reinforcement For T- or rectangular sections having one layer oftension reinforcement and subjected to Mservice only without Nservice

(Fig 141(a) ) Equation 612 reduces to

c = 1

2 minusa1 + radica21 + 4a2 (when Nservice = 0) (142)

a1 = 2hflange

bw

(b minus bw) + 2α Af

bw

(with bw = b when c hflange) (143)

a2 = h2

flange

bw

(b minus bw) + 2α Afd

bw


where α = EfEc with Ec being the modulus of elasticity of concrete b bw d

Figure 141 Strain and stress distributions in a section with FRP (a) bending momentMservice without normal force (b) bending moment Mservice combined withnormal force Nservice


and hflange are defined in Fig 141(a) When c le hflange or when the section isrectangular set b = bw in Equations (143) and (144)

The distance yCT between the resultants of tensile and compressive stressesis given by Equation 145 or 146 which apply when the section is subjectedto Mservice only or when Mservice is combined with Nservice

yCT = d minus c

3with c hf (145)

When c is substantially greater than hflange use Equation 146 for yCT

yCT = d minus c

3 bc2 minus (b minus bw)(c minus hflange)

2 (c + 2hflange)c

bc2 minus (b minus bw)(c minus hflange)2 (with c hflange) (146)

For a given reinforcement ratio ρf (=Af bd ) and specified value for εf servicethe curvature at a cracked section due to Mservice or due to Mservice combinedwith Nservice can be calculated by (see Equation (1128))

ψ2service = εf service

d minus c(147)

When the section is subjected in service to normal force Nservice combinedwith moment Mservice the required cross-sectional area of FRP is given by(Fig 141(b) )

Af = MserviceyCT + Nservice

σf service

(148)

The depth c of the compression zone can be determined by Equation(610) or (616) then Equations (145) to (147) can be applied to give yCT

and ψ2 service In Equation (148) Mservice = efNservice where ef is the eccentricityof the normal force measured downwards from the centroid of Af

To calculate Af by Equation (141) or by Equation (148) yCT must beknown But yCT depends upon Af Thus iteration is required first a value ofyCT is assumed to obtain an approximate value of Af In the second iteration(which is often sufficient) the approximate value of Af is used to calculate yCT

and Equation (141) or (148) is reusedThe above equations are based on the assumption that the amount of

tension reinforcement required is governed by the allowable strain in thetension reinforcement εf service (= σf serviceEf) in service It will be shown belowthat deflection of a member is also governed by the value εf service at mid-lengthsection


145 Curvature and deflection of flexural members

In this section we consider the momentndashmean curvature relationship formembers reinforced with FRP the mean curvatures calculated at varioussections can be used to give deflections in service (eg by the equations inAppendix C) For simplicity the subscript lsquoservicersquo is omitted in Sections 145to 148 which are concerned with deflections or curvatures in service condi-tion Section 74 presents equations that give the mean curvature ψm of amember reinforced with non-prestressed steel bars subjected to bendingmoment M gt Mr with Mr being the moment that produces first crackingThe equations are repeated here with the symbols adjusted for use when FRPis employed

ψm = (1 minus ζ) ψ1 + ζ ψ2 (1410)

ψ1 = M

EcI1

ψ2 = M

EcI2

(1411)

ζ = 1 minus β Mr

M2

(1412)

or

ζ = 1 minus β fct

σ1 max

2

(1413)

where I1 is the second moment of areas of a transformed area consisting of Ac

plus αAf with Ac and Af being the cross-sectional areas of concrete and FRP(ignoring the bars in compression) and α = EfEc where Ef and Ec are themoduli of elasticity of FRP and concrete respectively fct is tensile strength ofconcrete σ1 max is stress at extreme fibre in state 1 where cracking is ignoredThe coefficient β replaces the product of β1 and β2 which account for the bondquality and the effect of repetitious loading with FRP bars a value of β =05 is recommended This recommendation is based on comparisons of pub-lished experimental deflections of numerous beams reinforced with differentFRP types with the values of deflections calculated from curvatures using theabove equations2

Equation (1410) can be rewritten in the form

ψm = M

EIem

(1414)

where Iem is an effective second moment of area for use in calculation of meancurvature in members subjected to bending moment without axial force


Iem = I1I2

I2 + 1 minus 05 Mr

M2

(I1 minus I2)

(1415)

For a cracked member subjected to a normal force N combined withmoment M the mean axial strain εom and the mean curvature ψm can becalculated by Equations (736) and (740) Validity of these equations againwith β1 β2 replaced by β = 05 for prestressed members using carbon FRP hasbeen verified3 by comparison with published experimental data of severalbeams

146 Relationship between deflection meancurvature and strain in reinforcement

Consider a straight non-prestressed concrete member having steel or FRPreinforcement The deflection Dcentre at mid-length section varies almost lin-early with the strain εs or εf in the reinforcement at the same section Thesymbol Dcentre represents the transverse deflection measured from the chordjoining the two ends of a continuous or simply-supported member The linearrelationship of Dcentre versus εf is demonstrated below for a simply supportedbeam reinforced with FRP and will be used in the following section indeveloping an equation for recommended span to thickness ratio

Using virtual work the deflection Dcentre can be expressed as

Dcentre = 1

2 l2

0ψx dx +

1

2

l

l2ψ(l minus x)dx (1416)

This is a geometric relationship applicable for any straight member Herethe symbol ψ stands for ψm in the cracked part of the length and for ψ1 in theuncracked part l is length of member When ψ is assumed to vary as asecond-degree parabola between ψend1 ψcentre and ψend 2 (curvature at the twoends and at mid-length) Equation (1416) gives

Dcentre = l 2

96 (ψend 1 + 10ψcentre + ψend 2) (1417)

Equation (1416) is employed to calculate Dcentre for a simple beamreinforced with glass FRP and having the cross-section shown in Fig 142(a)The integrals in the equation are evaluated numerically assuming ψ to varylinearly over short segments of length 0025 l Figure 142(b) is a plot of thedeflection Dcentre versus the curvature ψm centre at the centre of a simple beamsubjected to uniform load whose intensity q is varied between zero and 4qrwhere qr is the value just sufficient to produce cracking at mid-span section


Figure 142 Simple beam example showing that Dcentre is almost proportionate to m andεf at mid-length section (a) cross-section of beam (b) deflection versuscurvature at mid-length (c) deflection versus strain in FRP at mid-lengthsection


The following data are adopted fct = 2MPa Ec = 25GPa Ef = 40GPa and l =7m where fct and Ec are tensile stress and modulus of elasticity of concreteEf is modulus of elasticity of glass FRP The dashed line in Fig 142(b) isplotted by the equation

Dcentre = 5

48ψm centre l

2 (1418)

This is the same as Equation (1417) applied to a simple beam having zerocurvatures at the ends The difference between the ordinates of the solid lineand the dashed line indicates the error in deflection calculation when Equa-tion (1418) is used this equation overestimates the deflection because itignores the fact that the parts of the simple beam adjacent to the ends areuncracked It can be seen that this simplification overestimates the deflectionby a small margin particularly when the load intensity q is higher but notclose to qr This example shows that the deflection at the centre of a simplebeam is almost linearly dependent upon the mean curvature at the samesection

When Equation (1417) is applied to a continuous member Dcentre willdepend on the curvatures ψend 1 and ψend 2 at the ends but will mainly dependupon the curvature ψcentre at mid-length (because this value is multiplied by10) In Section 147 we will adopt the assumption that the deflection atthe centre of reinforced concrete cracked members continuous or simplysupported is proportionate to the mean curvature ψm centre at mid-length

Figure 142(c) shows the variation of Dcentre with the strain εf in the tensionreinforcement at mid-span section of the same simple beam consideredabove The value εf is calculated ignoring the concrete in tension as the loadintensity is increased from qr to 4qr The dashed straight line in Fig 142(c)connects the origin to the ordinate corresponding to the load intensity 4qrFrom this example it is also concluded that Dcentre is almost linearly dependentupon εf Again in Section 147 we will adopt the assumption that the deflec-tion at the centre of reinforced concrete cracked members is proportionate tothe strain in the tension reinforcement at a cracked section at mid-length Thetwo assumptions based on the above study will be used below to develop anempirical equation for the ratio span to minimum thickness of membersreinforced with FRP

147 Ratio of span to minimum thickness

In design of concrete members the overall depths (here referred to as thick-ness) of members eg thicknesses of slabs are often selected based on codesor guidelines which give a ratio of span to minimum thickness (lh) thatnormally avoids excessive deflection The codes and the guidelines areintended for steel-reinforced members an adjustment to the ratio


recommended by the codes and the guidelines will be derived below for usefor members with FRP The subscripts f and s are employed to refer to FRPand steel respectively

The deflection Dcentre at centre of a member is considered here (based onSection 146) to be almost proportional to the strain ε calculated in thetension reinforcement at a cracked section at mid-length Thus for a giventhickness when the strain εf in service in FRP is allowed to be greater thanthe permissible strain εs in service in steel Dcentre will be greater when FRP isemployed This can be avoided by adopting a minimum thickness hf gt hs suchthat Dcentre become the same when FRP or steel are used where hf and hs arethe minimum thickness of member reinforced with FRP and steel respect-ively Equation 1425 presented below gives a recommended value for (lh)f

in terms of the values specified in codes or guidelines for (lh)s and thepermissible strain in service εf in FRP When the equation is used in designthe ratio of length to deflection lDcentre will be the same regardless of thereinforcement material

Different types of FRP have different elasticity moduli Thus identicalmembers with different FRP types have different deflections However whenthe thickness is determined by Equation (1425) and the amount ofreinforcement is designed such that εf at mid-length section in service is equalto a specified value the deflection will be the same with all FRP types

1471 Minimum thickness comparison between membersreinforced with steel and with FRP

Figures 141(a) and (b) show the strain variation over the depth of a crackedreinforced concrete section With any reinforcement material the curvaturemay be expressed by the equation

ψ2 = εs or εf

βh(1419)

where

β = d minus c

h(1420)

where c is depth of compression zone (Equation (142) ) The mean curvature(Equation (1410) ) may be expressed as

ψm = ηψ2 (1421)

where


η = 1 minus 05 fct

σ1 max

2

1 minus I2

I1 (1422)

Use of Equations (1418) (1419) and (1421) gives the deflections at mid-length of a cracked simply-supported concrete member as function of strainεs or f in the reinforcement at mid-length section

Dcentre = 5

48 l 2

h ηβs or f

εs or f (1423)

Application of this equation to a member reinforced with FRP and to aconjugate member reinforced with steel and equating (lDcentre) for the twobeams gives

l

hf = l

hs (ηβ)s

(ηβ)f εs

εf (1424)

This equation can be used in design to select thickness hf for a memberreinforced with FRP guided by (lh)s for steel-reinforced members for thispurpose substitute εs = 1200 times 10minus6 (or a not substantially different value)representing the commonly allowable value of strain in steel in service Thevalue of εf may be taken equal to 2000 times 10minus6 or a different value dependingupon the acceptable crack width (see Section 143) The term between squarebrackets in Equation (1424) is a dimensionless parameter which can beexpressed empirically as a function of (εsεf) as discussed in Section 1472

1472 Empirical equation for ratio of length to minimumthickness

The empirical Equation (1425) given below (as approximation of Equation(1424) ) is based on a parametric study4 of T and rectangular sectionsreinforced with FRP varying the thickness hf hflange (bbw) and εf where hf andhflange are the overall height and the flange thickness b and bw are width offlange and web and εf is the permissible strain in FRP in service The ratio (l h)f length to minimum thickness of a member reinforced with FRP is

l

hf

= l

hs

εs

εf

αd

(1425)

where αd = 05 for rectangular sections for T-sections αd is

αd = 05 + 3b

100bw

minus b

80hs

(1426)


When (l h)f for FRP-reinforced member satisfies Equation (1425) its ratio l Dcentre will be approximately equal to that of a conjugate steel-reinforcedmember having its length to thickness ratio equal to (l h)s The upper limit ofthe difference between (l Dcentre)f and (l Dcentre)s will be approximately 11per cent of the latter ratio Equation (1425) is intended to give the minimumthickness for FRP-reinforced concrete members using codes or guidelinesthat recommend (l h)s values for steel-reinforced members Equation (1425)can be employed with any type of FRP The permissible strain in serviceεf used in Equation (1425) is to be also used in calculating the requiredcross-sectional area of FRP (Section 144)

The design of members reinforced with FRP avoiding excessive deflectionand crack opening can be done by the steps outlined below Given memberlength l cross-sectional dimensions except hf and internal forces in serviceMservice with or without Nservice

1 Select the minimum thickness hs for a conjugate steel-reinforced memberusing code or guideline the corresponding permissible strain in steel inservice εs = 1200 times 10minus6 or a value not substantially different explicitlygiven or implied by the same code or guideline

2 Apply Equation (1425) to obtain minimum thickness hf of the memberwith FRP the permissible strain in FRP in service may be taken εf = 2000times 10minus6 or different value depending upon the tolerable crack width(Section 143)

3 Apply Equation (141) or (148) to calculate the required cross-sectionalarea of FRP

148 Design examples for deflection control

In the following examples a simple beam is designed following the steps of thepreceding section then the ratio l Dcentre for the beam with FRP is comparedwith the same ratio for a conjugate steel-reinforced beam

Example 141 A simple beam

Determine the minimum thickness hf and the cross-sectional area ofglass FRP required for a simple beam of span 8m to carry a uniformservice load q = 16kN-m Given data Ec = 25GPa Ef = 40GPa assumea T-section with b = 2 m bw = 04m hflange = 015m and d = 085hf thepermissible strain in FRP in service εf = 2000 times 10minus6 For the conjugatesteel-reinforced beam take the permissible strain in steel in serviceεs = 1200 times 10minus6 and (l h)s = 16

If the beam is reinforced with steel the thickness would be


hs = 8m

16 = 05m

Application of Equations (1426) and (1425) gives

αd = 05 + 3(2)

100(04) minus

2

80(05) = 060

l

hf

= 16 1200times10minus6

2000times10minus606

= 118

Thus the minimum thickness for beam with FRP is

hf = 8

118 = 068m take hf = 07 m d = 06m

The bending moment at mid-span in service

M = (16kNm) (8m)2

8 = 1280kNm

The permissible stress in FRP in service is σf = Ef εf = 40GPa (2000 times10minus6) = 80MPa Substitution of this value in Equation (141) with anestimated value yCT = 09d = 054m gives

Af 128kN minus m

80MPa (054) = 296times10minus3 m2 = 2960mm2

Application of Equations (142) and (145) gives c = 51 mm and yCT =0583m Substitution of this value in Equation (141) gives a moreaccurate value Af = 2740mm2

Example 142 Verification of the ratio of span to deflection

Compare l Dcentre for the beam designed in Example 141 with that of aconjugate steel-reinforced beam carrying the same load intensity andhaving the same thickness hs = hf = 07m and d = 06m but (l h)s = 16


Additional data fct = 2MPa Es = 200GPa εs = 1200 times 10minus6 Other dataare the same as in Example 141

(a) Deflection of beam with FRP The following is calculated at mid-spare section by Equations (1411) (1413) (1410) and (1418) (with Af

= 2740mm2 and M = 128kN-m)

I1 = 219times10minus3 m4 c = 0049 m I2 = 141times10minus3 m4

ψ1 = 234times 10minus6 mminus1 ψ2 = 3633 times 10minus6 mminus1

σ1 max = 278MPa ζ = 0742

ψm = 2754 times 10minus6 mminus1

Dcentre = 184mm

(l Dcentre)f = 80m

00184m = 435

(b) Beam with steel With steel the beam has a longer span ls = 16(07)= 112m The bending moment at mid-span

M = (16kN-m)(112)2

8 = 2509kN-m

As = 1830mm2 εs = 1200 times 106

I1 = 234 times 10minus3 m4 c = 0087m I2 = 429 times 10minus3 m4

ψ1 = 428 times 10minus6 ψ2 = 2338

σ1 max = 501MPa ζ = 0920


Dcentre = 286mm

l Dcentres = 112m

00286m = 392

149 Deformability of sections in flexure

The discussion in this section is limited to non-prestressed sections Failure ofsteel-reinforced sections by flexure is ductile exhibiting large curvature before


the ultimate moment is reached Unlike steel FRP continues to exhibit linearstressndashstrain relationship up to rupture without yielding or strain hardeningFor this reason FRP-reinforced sections in flexure should have sufficientlylarge curvature before failure by rupture of the FRP this requirement is hereconsidered satisfied when the section has a deformability factor ge 4 Thedeformability factor is defined as the ratio of the products of moment multi-plied by curvature at ultimate and at service This factor is an approximateindicator of the ratio of strain energy values per unit length of the flexuralmember at ultimate and at service Parametric studies5 show that steel-reinforced sections have deformability factor greater than 4 except when thesteel ratio ρs = As(bd) is greater than the balanced ratio in which case thedeformability factor is slightly below 40 The parametric studies also showthat design of cross-sectional areas Af in flexural sections with FRP based ona permissible strain in service εf service (as discussed in Section 144) will nor-mally result in sections having deformability factors greater than 4 Thus Af

is governed by the serviceability requirement and there is no need to check thedeformability except in the unusual case when the FRP ratio ρf = Af (bd ) isgreater than 015 f primecσf service where b is width of section at extreme compressivefibre f primec is specified concrete strength σf service (= Ef εf service) is permissible stressin FRP in service In the parametric studies referred to here the strain in FRPin service is assumed εf service = 2000 times 10minus6 and the modulus of elasticity of theFRP Ef = 40GPa to 150GPa

1410 Prestressing with FRP

In non-prestressed sections the stress in FRP reinforcement in service is arelatively small fraction of the tensile strength because of control of width ofcracks The high strength of FRP particularly with carbon fibres can bemore efficiently utilized when the FRP is employed for prestressingAppropriate permissible stresses at jacking and at transfer should be adoptedaccounting for the vulnerability of FRP to creep rupture (eg 70 and 60per cent of the tensile strength at jacking and at transfers respectively)Fatigue rupture should also be considered in setting the permissible stressesThe deformability should also be considered noting that the discussion in thepreceding section does not apply

With FRP types that have low moduli of elasticity compared to steel theloss of prestress force in the tendons due to creep and shrinkage of concrete isrelatively small An appropriate value of the intrinsic relaxation dependingupon the type of the FRP should be used The procedure in Appendix B forcalculating the relaxation reduction coefficient χr can be used with FRP butnot the graph and the empirical equation that give χr for prestressing steel


1411 General

Properties of FRP for use as reinforcement in concrete vary with the type offibres the resin and the manufacturing process For the design using thesematerials their properties should be established with certainty The differenceof modulus of elasticity of FRP from that of steel and its influence on thedesign for serviceability are presented in this chapter The basic assumptionsand equations adopted in calculation of stresses strains and displacements ofsteel-reinforced concrete structures apply when FRP is used

Notes

1 For further reading on properties of FRP and its design for concrete members seeJapan Society of Civil Engineers (1993) State-of-the-Art Report on ContinuousFiber Reinforcing Materials ed A Machida Concrete Engg Series 3 See alsoISIS Canada (2001) Reinforcing New Concrete Structures with Fibre ReinforcedPolymers Design Manual No 3 ISIS Canada Intelligent Sensing for InnovativeStructures A Canadian Network of Centres of Excellence 227 Engineering Build-ing University of Manitoba American Concrete Institute Committee (2001)report ACI 4401R-01 Guide for the Design and Construction of ConcreteReinforced with FRP Bars 41 pp

2 Hall Tara S (2000) Deflections of Concrete Members Reinforced with FibreReinforced Polymer (FRP) Bars MSc Thesis Department of Civil EngineeringUniversity of Calgary Calgary Canada

3 Ariyawardena N (2000) Prestressed Concrete with Internal or External TendonsBehaviour and Analysis PhD Thesis Department of Civil Engineering Universityof Calgary Calgary Canada

4 Ghali A Hall T and Bobey W (2001) lsquoMinimum Thickness of Concrete Mem-bers Reinforced with Fibre Reinforced Polymer (FRP) Barsrsquo Canadian J of CivilEngg 28 No 4 pp 583ndash592

5 Newhook J Ghali A and Tadros G (2002) ldquoConcrete Flexural MembersReinforced with FRP Design for Cracking and Deformabilityrdquo Canadian J ofCivil Engg 29 No 1


Time functions for modulus ofelasticity creep shrinkage andaging coefficient of concrete

The equations and graphs presented below are based on the requirements ofthe CEB-FIP Model Code for Concrete Structures 1990 (MC-90) and ACICommittee 209 Prediction of Creep Shrinkage and Temperature Effectsin Concrete Structures 19921 It is expected that the requirements of thecodes will change in future editions and it is for this reason that this materialis presented in an appendix rather than in the body of the text Thus allequations and methods of analysis included in the body of the text areindependent of the time functions to be used for Ec φ and εcs the modulus ofelasticity creep coefficient and free shrinkage of concrete Requirementsof Eurocode 2ndash19912 (EC2ndash91) and British Standard BS8110ndash19973 are alsodiscussed

A1 CEB-FIP Model Code 1990 (MC-90)

In this code the symbol φ is used differently from the way it is used in this bookfor this reason we adopt the symbol φCEB for the creep coefficient employed inMC-90 Equation (12) expresses the total strain at time t instantaneous pluscreep due to a constant stress σc(t0) introduced at time t0 as follows

εc(t) = σc(t0)

Ec(t0) [1 + φ(t t0)] (A1)

where Ec(t0) is the modulus of elasticity at age t0 φ(t t0) is the ratio of creep tothe instantaneous strain

The strain εc(t) is expressed in MC-90 by the equation

εc(t) = σc(t0)

Ec(t0) 1 +

Ec(t0)

Ec(28) φCEB(t t0) (A2)

where Ec(28) is the modulus of elasticity at age 28 days Comparison ofEquations (A1) and (A2) indicates that

Appendix A

φ = φCEB Ec(t0)

Ec(28)(A3)

Thus the numerical values of the creep coefficient φCEB calculated accord-ing to MC-90 must be multiplied by the ratio Ec(t0)Ec(28) to obtain the valueof the creep coefficient for use in the equations of this book The graphs andequations for the creep coefficient presented in this appendix include thisadjustment thus they can be used directly in the equations of the bookwithout further adjustment

A11 Parameters affecting creep

Creep depends upon the age at loading t0 and the length of the period t0 to twhere t is the instant at which the value of creep is considered In the equa-tions which will follow t0 and t are in days Creep also depends upon therelative humidity RH (per cent) and the notional size h0 (mm) defined by

h0 = 2Ac

u(A4)

where Ac and u are the area and perimeter in contact with the atmosphere ofthe cross-section of the considered member

The value of creep coefficient is inversely proportional to radicfcm where fcm

(MPa) is the mean compressive strength of concrete at age 28 days The valuefcm may be estimated by

fcm = fck + 8MPa (A5)

fck (MPa) is characteristic compressive strength of cylinders 150mm in diam-eter and 300mm in height stored in water at 20 plusmn 2 degC and tested at the age of28 days The value fck is the strength below which 5 per cent of all possiblestrength measurements may be expected to fall

The graphs presented in this appendix give creep coefficient φ(t t0) andaging coefficient χ(t t0) for selected combinations of the parameters fck RHand h0 The graphs are based on the code equations given below which arevalid for mean temperature of 20 degC taking into account seasonal variationsbetween minus20 degC and + 40 degC

A12 Effect of temperature on maturity

When prevailing temperature is higher or lower than 20 degrees Celsius theeffect of temperature on the maturity of concrete may be accounted for bythe use of adjusted age tT in lieu of t0 or t in the equations or graphs presentedbelow The adjusted age is given by

Time functions 475

tT = n

i = 1

∆tiexp 1365 minus 4000

273 + T(∆ti) (A6)

where tT is the concrete age adjusted for temperature ∆ti is the number ofdays in which a temperature T(∆ti) degree Celsius prevails For applicationof (A6) the age t0 or t is to be divided into n intervals and a prevailingtemperature is to be assumed for each

A13 Modulus of elasticity

The modulus of elasticity of concrete Ec(28) (MPa) at age 28 days fornormal-weight concrete can be estimated by

Ec(28) = 21 500 ( fcmfcm0)13 (A7)

where fcmo = 10MPaWhen the mean compressive strength fcm MPa is not known Ec(28) may beestimated from the characteristic compressive strength fck (for MPa) at 28days by the equation

Ec(28) = 21 500[( fck + ∆f )fcm0]13 (A8)

where ∆f = 8MPaEquations (A7) and (A8) apply when quartzitic aggregates are used For

other aggregates multiply Ec(28) by a factor varying between 07 and 12Equations (A7) and (A8) give the tangent modulus of elasticity which is

equal to the slope of the stressndashstrain diagram at the origin This modulus isthe value to be employed with the creep coefficient given by Equation (A16)and the graphs presented in this section to calculate the strain at any time (seeEquation (A1) )

When the modulus of elasticity is for use in an elastic analysis withoutconsidering creep the value Ec(28) should be reduced by a factor of 085 toaccount for the quasi-instantaneous strain which occurs shortly (within oneday) after loading

A14 Development of strength and modulus of elasticitywith time

The mean concrete strength fcm(t) at age t (days) may be estimated from thestrength fcm at 28 days by

fcm(t) = βcc(t) fcm (A9)

476 Appendix A

where

βcc(t) = exp[s(1 minus radic28t)] (A10)

with s being a coefficient depending on type of cement s is equal to 02025 and 038 respectively for rapidly hardening high-strength cementsfor normal and rapidly hardening cements and for slowly hardeningcements

The modulus of elasticity of concrete at age t may be estimated by

Ec(t) = βE(t)Ec(28) (A11)

with

βE(t) = radicβcc(t) (A12)

A15 Tensile strength

The tensile strength of concrete may be subject to large variation by environ-mental effects Upper and lower values of the characteristic axial tensilestrength fctk (MPa) may be estimated by

fctk min = 095( fckfck0)23 (A13)

fctk max = 185( fckfck0)23 (A14)

where fck0 = 10MPaCaution should be taken when the tensile strength of concrete is used in

analysis of displacements The value of the tensile strength assumed in suchanalysis will indicate whether cracking occurred or not Cracking can sub-stantially increase displacements Thus when the displacements are criticaltheir analysis should be based on the minimum value of tensile strength(Equation (A13) )

A16 Creep under stress not exceeding 40 per cent of meancompressive strength

We recall the definition of the creep coefficient φ(t t0) as the ratio of creep attime t to the instantaneous strain due to a constant stress introduced at timet0 MC-90 gives a coefficient φCEB(t t0) which is equal to φ(t t0) divided byβE(t0) (see Equations (A3) and (A11) ) where

βE(t0) = Ec(t0)

Ec(28)(A15)

Time functions 477

The equation given below for φ(t t0) is valid for compressive stress notexceeding 040 fcm(t0) relative humidity RH = 40 to 100 per cent mean tem-perature 5 to 30 degrees Celsius and fck between 12 and 80MPa The sameequation applies when the stress is tensile

The equation given by MC-90 for φCEB is adjusted below to give the creepcoefficient φ(t t0) as defined above

φ(t t0) = φ0βc(t minus t0)βE(t0) (A16)

where βc is a coefficient describing development of creep with time afterloading φ0 is a notional creep coefficient given by

φ0 = φRHβ(fcm)β(t0) (A17)

φRH = 1 + 1 minus (RH100)

046(h0href)13

(A18)

where href = 100mm

β(fcm) = 53

radicfcmfcmo

(A19)

where fcm0 = 10MPa

β(t0) = 1

01 + t002

(A20)

The symbol h0 (mm) is the notional size of member defined by Equation(A4) Development of creep with time is expressed by

βc(t minus t0) = t minus t0

βH + t minus t0

03

(A21)

βH (mm) is a function of the notional size h0 (mm) and the relative humidityRH (per cent)

βH = 150h0

href

[1 + (0012 RH)18] + 250 le 1500mm (A22)

where href = 100mm

478 Appendix A

A17 Effect of type of cement on creep

Creep of concrete depends on the degree of hydration needed at the age ofloading t0 and thus on the type of cement This effect can be accounted for bymodifying t0 using equation

t0 = t0 T 9

2 + (t0 T)12+ 1

a

ge 05 (A23)

where t0 T (days) is the age of concrete at loading adjusted by Equation (A6)for substantial deviation of prevailing temperature from 20 degrees Celsiusα is coefficient equal to minus10 0 or 10 respectively for slowly hardeningcement for normal or rapidly hardening cement and for rapidly hardeninghigh-strength cements

A18 Creep under high stress

Creep under compressive stress in the range (04 to 06) fcm(t0) can be calcu-lated by Equation (A16) replacing φ0 by φ0k given by

φ0k = φ0 exp[15(k minus 04)] (A24)

where k is the applied stress divided by fcm(t0)

A19 Shrinkage

Shrinkage starts at time ts (days) when curing is stopped On the other handconcrete immersed in water at time ts starts to swell The shrinkage or theswelling at any time t (days) may be estimated by

εcs(t ts) = εcs0βs(t minus ts) (A25)

where βs(t minus ts) is a function describing the development of shrinkage orswelling with time given by

βs(t minus ts) = t minus ts

350(h0href)2 + t minus ts

05

(A26)

where h0(mm) is the notional size defined by Equation (A4) and href =100mmεcs0 is the notional shrinkage given by

εcs0 = εs( fcm)βRH (A27)

Time functions 479

where

εs(fcm) = 10minus6[160 + 10βsc(9 minus fcmfcm0)] (A28)

with βsc equalling 4 5 or 8 respectively for slowly hardening cementsfor normal or rapidly hardening cements and for rapidly hardeninghigh-strength cements fcm0 = 10MPa

βRH = minus1551 minus RH

1003

for 40 le RH lt 99 (A29)

βRH = +025 for RH ge 99 (immersed in water) (A30)

Positive βRH indicates swelling RH (per cent) is relative humidity

A2 Eurocode 2ndash1991 (EC2ndash91)

The values of the parameters discussed in the preceding section for MC-90will not be much different when estimated in accordance with EC2ndash91 Someof the differences between the two codes are summarized below

EC2ndash91 gives Equation (A31) for estimation of secant modulus of elas-ticity (MPa) for normal weight concrete at age t0 days The secant modulus isdefined as stress divided by strain at a stress level = σc(t0) = 04 fck(t0) (see Fig11)

Ec(t0) = 095 21 500[(fck(t0) + 8)10]13 (A31)

where fck(t0) (MPa) is characteristic compressive stress at age t0EC2ndash91 contains a table of creep coefficients for normal-weight concrete at

t = infin due to compressive stress not exceeding 045 fck(t0) introduced at age t0the value t0 = 1 7 28 90 and 365 days The EC2ndash91 values are here adjustedby multiplication by βE(t0) (given by Equations (A12) and (A10) and theresulting coefficients are presented in Table A1 The values φ(infin t0) given inTable A1 can be used with the secant modulus of elasticity (Equation (A31)and Equation (A1) ) to calculate the total strain instantaneous plus ultimatecreep after a very long time

Table A2 from EC2ndash91 gives final shrinkage values of normal-weightconcrete (εcs(infin ts) ) where ts is time when curing is stopped

The values given in Tables A1 and A2 apply for a range of mean temper-atures between 10 and 20 degrees Celsius (taking into account seasonalvariations between minus20 and +40 degrees)

As a complement to Tables A1 and A2 EC2ndash91 gives for use when moreaccuracy is required the same equations as MC-90 for creep coefficient andshrinkage which are presented above in Section A1

480 Appendix A

The quantity inside the curly brackets in Equation (A31) is the tangentmodulus of elasticity according to MC-90 (slope of stressndashstrain diagram atthe origin see Equation (A8) ) Thus the EC2ndash91 allows calculation of thestrain ε(t) by Equation (A1) using the creep coefficient φ(t t0) given byEquation (A16) and the tangent modulus of elasticity The graphs for creepand aging coefficients presented in Section A6 are based on Equation(A16) thus they are in accordance with EC2ndash91 At age 28 days EC2ndash91considers that the tangent modulus of elasticity is equal to 105 the secantmodulus

A3 ACI Committee 209

A large number of variables affect the magnitude of creep and shrinkagewhich is discussed in some detail in the report of the American ConcreteInstitute Committee 2094 The following equations are considered applicablein lsquostandard conditionsrsquo The term lsquostandard conditionsrsquo is defined in thereport by ranges for a number of variables related to the material proper-ties the climate and the sizes of members Reference must be made to thementioned report when the conditions are different from what is specifiedbelow

Table A1 Final creep coefficients (infin t0) of normal-weight concrete subjected to com-pressive stress not exceeding 045 fck(t0) (based on EC2ndash91)

Age atloading

Notional size h0 (mm) defined by Equation (A4)

t0 (days) 50 150 600 50 150 600

Dry atmosphere (inside) Humid atmosphere (outside)(RH = 50 per cent) (RH = 80 per cent)

1 32 26 21 20 18 157 34 28 22 22 19 17

28 32 25 20 19 17 1590 28 23 17 17 15 13

365 22 18 13 13 11 11

Table A2 Final shrinkage strain cs of normal-weight concrete (based on EC2ndash91)

Location of member Relative humidity Notional size h0 (mm) defined byper cent Equation (A4)

le150 600

Inside 50 minus600 times 10minus6 minus500 times 10minus6

Outside 80 minus330 times 10minus6 minus280 times 10minus6

Time functions 481

A31 Creep

The coefficient for creep at time t for age at loading t0 is given by

φ(t t0) = (t minus t0)

06

10 + (t minus t0)06

φu (A32)

where

φu = φ(tinfin t0) (A33)

φu is the ultimate creep after a very long time (10000 days) for age at loadingt0 The value φu is given by

φu = 235 γc (A34)

where γc is a correction factor the product of several multipliers dependingupon ambient relative humidity average thickness of the member or itsvolume-to-surface ratio and on the temperature For relative humidity of 40per cent average thickness 6 in (015m) or volume-to-surface ratio of 15 inand temperature 70 degF (21 degC) all the multipliers are equal to unity In thiscase γc may be calculated as a function of the age at loading t0

γc = 125 tminus01180 (A35)

or

γc = 1113 tminus00940 (A36)

Equations (A35) and (A36) are respectively applicable for moist-cured con-crete and for 1ndash3 days steam-cured concrete The two equations give γc 10when t0 = 7 and 3 days respectively

The ratio of the modulus of elasticity at any age t0 days to the value at age28 days

Ec(t0)

Ec(28) = t0

α + βt0

12 (A37)

The coefficients α and β are constants depending upon the type of cement andcuring used For cement Type I α = 4 and β = 085

The value of Ec to be employed with the equations presented in this sectionmay be estimated by the ACI318(1989) Code5 equation

Ec = w15c 33 radicf primec (A38)

482 Appendix A

where Ec (psi) and f primec (psi) are the modulus of elasticity of concrete and itsspecified compressive strength wc (lb per cu ft) is the unit weight of concreteFor normal-weight concrete Ec (psi) may be taken as 57000 radicf primec Equation(A38) may be rewritten using SI units


with Ec (MPa) and f primec (MPa) and wc (kgm3) the corresponding value of Ec

(MPa) for normal-weight concrete is 4730 radicf primec (MPa)Equation (A38) or (A39) gives the secant modulus of elasticity which is

the slope of the secant drawn from the origin to a point corresponding to 040f primec on the instantaneous (1ndash5 minutes) stressndashstrain curve

Use of Equation (A38) or (A39) will overestimate Ec when f primec is higherthan 6000 psi (40MPa) in which case the following equation is suggested6 fornormal-weight concrete

Ec = 40000 radicf primec + 106 psi (A40)

Ec = 3300 radicf primec + 7000MPa (A41)

From Equations (A32ndash34) the ratio of the creep coefficient φ(tinfin t0) toφ(tinfin 7) for moist-cured concrete may be expressed as

φ(tinfin t0)

φ(tinfin 7) = 125tminus0118

0 (A42)

Bazant7 employs Equations (A31) (A32) (A36) and (A41) to calculate φ(tt0) and uses a numerical procedure similar to the method in Section 110 tocalculate the values of the aging coefficient χ(t t0) given in Table A3

A32 Shrinkage

For moist-cured concrete the free shrinkage which occurs between t0 = 7 daysand any time t

εcs(t t0) = t minus t0

35 + (t minus t0)(εcs)u with t0 = 7 (A43)

and for steam-cured concrete the shrinkage between t0 = 1 to 3 days and anytime t

εcs(t t0 = 1 to 3) = t minus (1 to 3)

55 + (t minus 1 to 3) (εcs)u (A44)

Time functions 483

where (εcs)u is the ultimate free shrinkage corresponding to tinfin (say at 10000days) The ultimate free shrinkage is given by

(εcs)u = minus780 times 106γcs (A45)

where γcs is a correction factor the product of a number of multipliers whichdepends upon the same factors mentioned above for γc The correction factorγcs = 10 when the period of initial moist curing is 7 days the relative humidityof the ambient air is 40 per cent the average thickness is 6 in (015m) or thevolume-to-surface ratio is 15 in

The free shrinkage between any two ages t0 and t can be calculated as thedifference of shrinkage for the periods (t minus 7) and (t0 minus 7)

εcs(t t0) = εcs(t 7) minus εcs(t0 7) (A46)

Table A3 Aging coefficient (t t0) calculated by Bazant

Value of

(t minus t0) (tinfin 7) t0 = 10 t0 = 102 t0 = 103 t0 = 104 (t t0)

(tinfin t0)

10 days 05 0525 0804 0811 080915 0720 0826 0825 0820 027325 0774 0842 0837 083035 0806 0856 0848 0839

102 05 0505 0888 0916 0915days 15 0739 0919 0932 0928 0608

25 0804 0935 0943 093835 0839 0946 0951 0946

103 05 0511 0912 0973 0981days 15 0732 0943 0981 0985 0857

25 0795 0956 0985 098835 0830 0964 0987 0990

104 05 0461 0887 0956 0965days 15 0702 0924 0966 0972 0954

25 0770 0940 0972 097635 0808 0950 0977 0980

(tinfin t0)

(tinfin 7)

0960 0731 0558 0425

Ec(t0)

Ec(28)

0895 1060 1083 1089

484 Appendix A

Equation (A43) is applicable for each of the two terms in Equation (A46) Ina similar way Equation (A44) can be employed to calculate εcs(t t0) forsteam-cured concrete

A4 British Standard BS 81108

Part 2 of BS 8110 gives equations for modulus of elasticity creep and shrink-age of concrete that result in level of accuracy greater than that in BS 8110Part 1 The equations presented below are taken from Part 2

A41 Modulus of elasticity of concrete

A mean value of the elasticity modulus of normal-weight concrete is givenby

Ec(28) = K0 + 02 fcu(28) (A47)

where 28 is the age of concrete in days Ec is the static modulus of elasticityfcu is the characteristic cube strength below which 5 per cent of all possible testresults would be expected to fall K0 = 20GPa a constant related to themodulus of elasticity of the aggregate For lightweight aggregate concretethe value of elasticity modulus given by Equation A47 should be multipliedby (w2400)2 where w is the density of concrete in kgm3 When Ec is forcalculation of deflections that are of great importance BS 8110 states thattests should be carried out on concrete made with the aggregate to be used inthe structure With unknown aggregate at the design stage the standardadvises to consider a range of Ec(28) based on K0 = 14 to 26GPa

Variation of the elasticity modulus with the age of concrete t is expressedby

Ec(t) = Ec(28) 04 + 06 fcu (t)

fcu (28) with t ge 3 days (A48)

The value of fcu(t) to be used in this equation is given in Table A4 Forlightweight concrete having density w multiply the values in Table A4 by[w(kgm3)2400]2

A42 Tensile strength of concrete

The British Standard BS 8110 does not specify tensile strength of concreteHowever non-prestressed sections subjected to bending moment can be con-sidered uncracked when the stress in concrete at the level of the tensionreinforcement is less than 1MPa When the section is considered cracked thestress in concrete in tension is assumed to vary linearly over the tension zone

Time functions 485

the value of the tensile stress is zero at the neutral axis and at the level of thetension reinforcement the concrete stress is equal to 1MPa or 055MPa inshort term and in long term respectively (see Fig E4 Appendix E)

For prestressed sections class 2 flexural tensile stress is permitted withoutvisible cracks the allowable tensile stress is 045 radicfcu or 036radicfcu for pre-tensioned and post-tensioned members respectively The allowable tensilestress may be increased by up to 17MPa in certain conditions

A43 Creep

Final creep is assumed to occur in 30 years The creep coefficient φ (30yrs t0)is given by the graph in Fig A1 where t0 is age of concrete at loadingin days In this figure the effective thickness is twice the cross-sectional areaof a member divided by the exposed perimeter When concrete is exposedto constant relative humidity 40 60 and 80 per cent the final creep isassumed to develop in the first month at six months and 30 monthsrespectively

A44 Shrinkage

British Standard BS 8110 gives the graph in Fig A2 for an estimate of dryingshrinkage of plain normal-weight concrete as function of the relative humid-ity and the effective thickness (defined the same as in the preceding section)The graph is for concrete made without water-reducing admixtures (watercontent about 190 litrem3) Shrinkage is considered proportional to watercontent in the range 150 to 230 litrem3

A5 Computer code for creep and agingcoefficients

The computer code in FORTRAN described below employs the step-by-stepprocedure given in Section 110 and Equations (123) (125) (127) and (129)

Table A4 Variation of cube strength in MPa with age of concrete according to BritishStandard BS 8110

Grade Characteristic Cube strength at an age ofstrength fcu(28)

7 days 2 months 3 months 6 months 1 year

20 200 135 220 230 240 25025 250 165 275 290 300 31030 300 200 330 350 360 37040 400 280 440 455 475 50050 500 360 540 555 575 600

486 Appendix A

to calculate the relaxation function r(t t0) and the aging coefficient χ(t t0)The values of Ec(t) and φ(t t0) required in the analysis are based on theequations of MC-90 (see Section A2)

Figure A3 is a listing of subroutine named Chicoef for which the inputdata are fck h0 RH t and t0 where fck (MPa) is characteristic compressivestrength (at 28 days) h0 (mm) is notional size (Equation (A4) ) RH (per cent)is relative humidity t0 and t are ages of concrete at the start and at the end ofthe loading (or relaxation) period The output gives the relaxation functionr(τ t0) varying τ between t0 and t and the aging coefficient χ(t t0)

The subroutine Chicoef employs a subroutine named Phicoef (see FigA4) which calculates φ(t2 t1) as a function of fck h0 RH t1 and t2 where t1 isthe age of concrete at loading and t2 is the age at the end of a period in whichthe load is sustained The computer programs provided on the Internet forthis book include FORTRAN files for the subroutines Chicoef and Phicoefthus they can be revised as may become necessary The present subroutinesare employed to produce an executable file also included on the disc that canperform the calculations on a micro-computer (see Appendix G)

Figure A5 is an example plot of the relaxation function r(t t0) prepared bythe above computer codes with t0 = 3 days and 120 days fck = 30MPa h0 =400mm and RH = 50 per cent The broken line on the same graph representsthe variation of Ec with time

Figure A1 Creep coefficient φ (30 yrs t0) where t0 is age of loading Reproduced from BS8110 Part 2 1985 with kind permission of BSI

Time functions 487

A6 Graphs for creep and aging coefficients

The graphs in Figs A6 to A45 based on MC-90 give the values of the creepcoefficient φ(t t0) and the aging coefficients for selected sets of characteristiccompressive stress fck notional size h0 (Equation (A4) ) and relative humidityRH The coefficient φ(t t0) is the ratio of creep in the period t0 to t divided bythe instantaneous strain εc(t0) The value of Ec(t0) to be used in calculatingεc(t0) is the tangent elasticity modulus given by Equations (A8) and (A11)As mentioned earlier the graphs are in accordance with EC2ndash91

Table A4 lists the values of fck (MPa) h0 (mm) and RH (per cent) selectedfor the graphs

Figure A2 Free shrinkage of normal-weight concrete Reproduced from BS 8110 Part 21985 with kind permission of BSI

488 Appendix A

A7 Approximate equation for aging coefficient

It can be seen from any of the aging coefficient graphs in Figs A6 to A45that for a specified age at start of loading t0 the value of χ(t t0) is almostconstant when (t minus t0) ge 1 year In this case the aging coefficient can beapproximated by the empirical equation9

χ(t t0) χ(30 times 103 t0) radict0

1 + radict0

(A49)

The error in this equation is less than plusmn 10 per cent when t0 gt 28 days and can

This figure is continued on next page

Time functions 489

reach plusmn 20 per cent when t0 = 3 days The equation underestimates the valueof χ when creep is high that is when fck RH and h0 are relatively small Moreaccuracy can be achieved by replacing the constant 10 in the denominator onthe right-hand side of Equation (A49) by a variable9 05 to 20 which is afunction of fck RH and h0

Figure A3 Computer code in FORTRAN for relaxation function r(t t0) and aging coefficientχ(t t0)

490 Appendix A

Figure A4 Computer code in FORTRAN for calculation of creep coefficient (t2 t1)according to MC-90

Figure A5 Example of relaxation function r(t t0) and variation of Ec with time fck = 30MPa(4500 psi) RH = 50 per cent h0 = 400mm (16 in)

Time functions 491

Table A5 List of graphs for (t t0) and χ (t t0) in Figs A6 to A45

Characteristic Relative humidity Notional size FigureCompressive fck RH (per cent) h0 (Eq A4) number

MPa psi mm in

100 4 A650 200 8 A7

400 16 A820 3000 1000 40 A9

100 4 A1080 200 8 A11

400 16 A121000 40 A13

100 4 A1450 200 8 A15

400 16 A1630 4500 1000 40 A17

100 4 A1880 200 8 A19

400 16 A201000 40 A21

100 4 A2250 200 8 A23

400 16 A2440 6000 1000 40 A25

100 4 A2680 200 8 A27

400 16 A281000 40 A29

100 4 A3050 200 8 A31

400 16 A3260 9000 1000 40 A33

100 4 A3480 200 8 A35

400 16 A361000 40 A37

100 4 A3850 200 8 A39

400 16 A4080 12000 1000 40 A41

100 4 A4280 200 8 A43

400 16 A441000 40 A45

492 Appendix A

Figure A6 Creep and aging coefficients of concrete when fck = 20MPa (3000psi) RH =50 per cent h0 = 100mm (4 in)

Time functions 493


494 Appendix A


Time functions 495


496 Appendix A


Time functions 497


498 Appendix A


Time functions 499


500 Appendix A


Time functions 501


502 Appendix A


Time functions 503


504 Appendix A


Time functions 505


506 Appendix A


Time functions 507


508 Appendix A


Time functions 509


510 Appendix A


Time functions 511


512 Appendix A


Time functions 513


514 Appendix A


Time functions 515


516 Appendix A


Time functions 517


518 Appendix A


Time functions 519


520 Appendix A


Time functions 521


522 Appendix A


Time functions 523


524 Appendix A

Figure A38 Creep and aging coefficients of concrete when fck = 80MPa (12000psi) RH= 50 per cent h0 = 100mm (4 in)

Time functions 525


526 Appendix A


Time functions 527


528 Appendix A


Time functions 529


530 Appendix A


Time functions 531


532 Appendix A

Notes

1 See references mentioned in Note 2 page 192 See reference mentioned in Note 5 page 193 British Standard BS 8110 Part 1 1997 and Part 2 1985 Structural Use of Con-

crete British Standards Institute 2 Park Street London W1A 2BS Part 1 is repro-duced by Deco 15210 Stagg St Van Nuys Ca 91405-1092 USA

4 See reference in Note 2 page 195 ACI 318(2001) Building Code Requirements for Reinforcements for Reinforced

Concrete American Concrete Institute Detroit Michigan 482196 Carrasquillo RL Nilson AH and Slate FO (1981) Properties of high-

strength concrete subject to short-term loads ACI Journal 78 (3) 171ndash87 Bazant ZP (1972) Prediction of concrete creep effects using age-adjusted effect-

ive modulus method J Proc Amer Concrete Inst 69 (4) 212ndash178 See Note 3 above9 Chiorino MA and G Lacidogna (1991) Approximate values of the aging coef-

ficient for the age-adjusted effective modulus method in linear analysis of creepstructures Report No 31 Department of Structural Engineering Politecnico diTorino Turin Italy See also Report No 35 (1992) by the second author for themore accurate version of Equation (A49)

Time functions 533

Relaxation reduction coefficient r

In a concrete structure the relaxation of prestressed steel is commonlysmaller than the intrinsic relaxation that would occur in a constant-lengthtest with the same initial stress The coefficient χr is a multiplier to be appliedto the intrinsic relaxation to obtain a reduced relaxation value to be employedin calculation of the loss of prestress in a prestressed concrete cross-section(see Sections 252 and 32) Values of χr are given by the graphs in Fig 14and Table 11 or by Equation (B11) The equations used for preparation ofthe graphs and the table are here derived

Consider a tendon stretched at time t0 and its length kept constant up totime t Let ∆σpr(t) be the intrinsic relaxation during the period (t minus t0) itsvalue depends on the quality of steel and the initial tension σp0 (see Equation(15) )

∆σpr(t) = minusηtσp0(λ minus 04)2

0

λ ge 04

λ lt 04(B1)

where

λ = σp0

fptk

(B2)

and fptk is the characteristic tensile strength ηt is a dimensionless coefficientdepending upon the steel quality and length of the period (t minus t0) (when (t minus t0)is infinity ηt becomes equal to η where η is defined in Section 14) The minussign in Equation (B1) indicates that the relaxation is a reduction of stresshence a negative increment

When the tendon is employed to prestress a concrete member with a stressat time t0 equal to σp0 a commonly smaller amount of relaxation occursduring the period (t minus t0) given by

∆σpr(t) = χr∆σpr(t) (B3)

Appendix B

where ∆σpr(t) is the reduced relaxation value χr is the relaxation reduc-tion coefficient It will be shown below that the relaxation reductioncoefficient

χr = 1

0 (1 minus Ωξ) λ(1 minus Ωξ) minus 04)

λ minus 04 2

dξ (B4)

where Ω is the ratio of (total change in prestressndashintrinsic relaxation) to theinitial prestress that is

Ω = minus∆σps(t) minus ∆σpr(t)

σp0

(B5)

where ∆σps(t) is the change in stress in the prestressed steel during the period(t minus t0) due to the combined effect of creep shrinkage and relaxation ∆σpr(t)is the intrinsic relaxation in the same period

ξ is a dimensionless time function defining the shape of the stress-timecurve (Fig B1) The value ξ varies between 0 and 1 as τ varies between t0 and

Figure B1 Stress versus strain in a constant-length relaxation test Definition of the shapefunction ξ

Relaxation reduction coefficient 535

t where τ represents any intermediate instant Thus the intrinsic relaxation atthe instant τ is

∆σpr(τ) = [∆σpr(t)]ξ (B6)

The prestress loss ∆σps(τ) due to the combined effects of creep shrinkageand relaxation is assumed to vary during the period t0 to t following the sameshape function thus

∆σps(τ) = [∆σps(t)]ξ (B7)

At any instant τ the absolute value |∆σps(τ) minus ∆σpr(τ)| represents a reductionin tension caused by shortening of the tendon Thus the elemental change inrelaxation at the instant τ is the same as in a tendon with a reduced initialtension of value

σp0(τ) = σp0(1 minus Ωξ) (B8)

Employing Equations (B6) and (B1) a differential value of the intrinsicrelaxation is expressed as (see Fig B1)

d[σpr(τ)] = [minusηtσp0(λ minus 04)2]dξ (B9)

Similarly a differential value of the reduced relaxation

d[σpr(τ)] = minus ηtσp0(1 minus Ωξ)[λ(1 minus Ωξ) minus04]2dξ

where λ(1 minus Ωξ) ge 04 (B10)

The value between curly brackets obtained by substitution of Equation (B8)in (B1) represents the intrinsic relaxation at time t for a tendon with areduced initial tension σp0(τ) The differential of the reduced relaxation iszero when λ(1 minus Ωξ) lt 04

Integration of each of Equations (B10) and (B9) and then division givesEquation (B4)

The graphs in Fig 14 and Table 11 for the relaxation reduction coefficientχr are determined by expressing the integral in Equation (B4) in a closed formand substitution of chosen values of λ and Ω noting the above-mentionedrestriction to Equation (B10) The restriction is tantamount to replacing theupper limit of the integral in Equation (B4) by the smaller of the two values 1and [(λ minus 04)(λΩ)]

The closed form expression for χr based on Equation (B4) is ratherlengthy Instead the following expression obtained by curve fitting fromFig 14 may be employed as an approximation to the relaxation reductioncoefficient

536 Appendix B

χr = exp[(minus67 + 53 λ)Ω] (B11)

In most cases Ω is positive and χr lt 1 In exceptional situations Ω is negativeand χr gt 1

The value of the intrinsic relaxation for any type of steel is commonlyreported from tests in which a tendon is stretched between two fixed pointsfor a period (τ minus t0) = 1000 hours The value of the intrinsic relaxation may beexpressed as a function of the period (τ minus t0)

∆σprinfin 1

16 ln τ minus t0

10 + 10 le (τ minus t0) le 1000 (B12)

∆σpr (τ minus t0) = ∆σprinfin τ minus t0

05 times 10602

1000 lt (τ minus t0) le 05 times 106 (B13)

∆σprinfin (τ minus t0) le 05 times 106 (B14)

The quantity (τ minus (t0) is the period in hours for which the tendon is stretched∆σpr infin is the ultimate intrinsic relaxation Equations (B12ndash14) closely followthe requirements of MC901 and FIP report on prestressing steel2

Notes

1 See reference mentioned in Note 2 page 192 Feacutedeacuteration Internationale de la Preacutecontrainte (1976) Report on Prestressing Steel

Part 1 Types and Properties FIP53 Aug published by Cement and ConcreteAssociation Wexham Springs Slough S13 6PL England


Elongation end rotation and centraldeflection of a beam in terms of thevalues of axial strain and curvatureat a number of sections

A number of geometry relations are given below to express the elongationthe end rotations and the deflection at the middle of a beam in terms of theaxial strain εO at the centroid and the curvature ψ at a number of equallyspaced sections Fig C1(a) defines four coordinates at which the displace-ments are considered εO and ψ are assumed to be known at three or fivesections as shown in Fig C1(b) and (c) The variation of εO and ψ is assumedto be linear between each two consecutive sections Parabolic variation is alsoconsidered for the three- and five-section systems of Fig C1(b) and (c)

Figure C1 Coordinates and station points referred to in Equations (C1ndash16) (a) coordinatesystem (b) three sections (Equations (C1ndash8) ) (c) five sections (Equations(C9ndash12) or (C13ndash16) )

Appendix C

The equations presented in parts (a) to (e) of this appendix are not limitedto the simple beam shown in Fig C1(a) they are applicable to any memberof a framed structure Fig C2 shows a straight member AB and its deflected

Figure C2 Original and deformed shape of a member of a framed structure

Figure C3 Coordinates and station points referred to in Equations (C17ndash22) (a)coordinate system (b) two sections (Equations (C17) (C18) ) (c) threesections (Equations (C19) (C20) ) (d) five sections (Equations (C21) (C22) )

Elongation end rotation and central deflection 539

shape AprimeBprime The displacement D1 in this case represents the elongation of themember while D2 to D4 are as indicated in the figure

In parts (e) to (g) of this appendix equations are given for deflection androtation at the free end of a cantilever in terms of curvature at a number ofsections (Fig C3)

Positive εO means elongation and positive curvature corresponds toelongation and shortening at the bottom and top fibres respectively

(a) Three sections straight-line variation (Fig C1)

D1 = l

4 [1 2 1] εO (C1)

D2 = minusl

24 [1 6 5] ψ (C2)

D3 = l

24 [5 6 1] ψ (C3)

D4 = l 2

48 [1 4 1] ψ (C4)

(b) Three sections parabolic variation (Fig C1)

D1 = l

6 [1 4 1]εO (C5)

D2 = minusl

6 [0 2 1]ψ (C6)

D3 = l

6 [1 2 0]ψ (C7)

D4 = l 2

96 [1 10 1]ψ (C8)

(c) Five sections straight-line variation (Fig C1)

D1 = l

8 [1 2 2 2 1]εO (C9)

D2 = minusl

96 [1 6 12 18 11]ψ (C10)

D3 = l

96 [11 18 12 6 1]ψ (C11)

540 Appendix C

D4 = l 2

192 [1 6 10 6 1]ψ (C12)

(d) Five sections parabolic variation between sections 1 2 and 3 andsections 3 4 and 5

D1 = l

12 [1 4 2 4 1]εO (C13)

D2 = minusl

12 [0 1 1 3 1]ψ (C14)

D3 = l

12 [1 3 1 1 0]ψ (C15)

D4 = l 2

24 [0 1 1 1 0]ψ (C16)

(e) Cantilever two sections straight-line variation (Fig C3)

D1 = minusl 2

6 [1 2]ψ (C17)

D2 = l

2 [1 1]ψ (C18)

(f) Cantilever three sections straight-line variation (Fig C3)

D1 = minusl 2

24 [1 6 5]ψ (C19)

D2 = l

4 [1 2 1]ψ (C20)

(g) Cantilever five sections parabolic variation between sections 1 2 and3 and sections 3 4 and 5

D1 = minusl 2

12 [0 1 1 3 1]ψ (C21)

D2 = l

12 [1 4 2 4 1]ψ (C22)


Depth of compression zone in afully cracked T section

Equation (720) can be solved to give the depth c of the compression zone ofa T section (Fig 72) subjected to an eccentric normal force N which pro-duces compression and tension at the top and bottom fibres respectively Theequation may be rewritten as a cubic polynomial

c3 + a1c2 + a2c + a3 = 0 (D1)

where

a1 = minus3(dns + es) (D2)

a2 = minus6

bw

[hf (b minus bw)(dns + es minus 12hf) + Aprimens(αns minus 1)(dns + es minus d primens)

+ αpsAps(dns + es minus dps) + αnsAnses] (D3)

a3 = 6

bw

[12h

2f(b minus bw)(dns + es minus 23hf) + Aprimensd primens(αns minus 1)(dns + es minus d primens)

+ αpsApsdps(dns + es minus dps) + αnsAnsdnses] (D4)

The symbols related to the geometry of the cross-section and position ofthe normal force are defined in Fig 72 αns = EnsEc and αps = EpsEc with Ens

and Eps being moduli of elasticity of the non-prestressed and the prestressedsteel and Ec is the modulus of elasticity of concrete at the time of applicationof the normal force The limitations of Equation (720) mentioned in Section742 also apply to Equations (D1) to (D4)

If the section has an additional steel layer of cross-section area Ansi at adistance dnsi below the top edge additional terms [αnsAnsi(dns + es minus dnsi)] and[αnsAnsidnsi(dns + es minus dnsi)] should be included inside the square brackets inEquations (D3) and (D4) respectively When the added layer is situatedin the compression zone αns should be substituted by (αns minus 1)

Solution of the cubic equation (D1) is given by substitution in thefollowing equations1

Appendix D

a4 = a2 minus a2

1

3(D5)

a5 = 2 a1

3 3

minus a1a2

3 + a3 (D6)

a6 = a4

3 3

+ a5

2 2

(D7)

When a6 is positive the cubic equation has only one real solution

c = minus a5

2 + radica6

12

+ minus a5

2 minus radica6

12

minus a1

3(D8)

When any of the quantities between brackets in the first two terms onthe right-hand side of this equation is negative the term should be replacedby [minus (absolute value of the quantity)

13 ]

When a6 is zero or negative the cubic equation has three real solutions butonly one is meaningful (with c between zero and dns) The three solutions aregiven by

cos θ = minusa5

2[minus(a43)3]1ne2(D9)

c1 = 2 minus a4

3 12

cosθ3 minus a1

3(D10)

c2 3 = minus2 minus a4

3 12 cosθ3 plusmn 60 minus

a1

3 (D11)

Note

1 Korn GA and Korn TM (1968) Mathematical Handbook for Scientists andEngineers 2nd edn McGraw-Hill New York see page 23

Depth of compression zone in a fully cracked T section 543

Crack width and crack spacing

E1 Introduction

Cracks in reinforced and partially prestressed concrete structures areexpected to occur but with adequate and well detailed reinforcement it ispossible to limit the width of cracks to a small value such that appearance orperformance of the structure are not hampered

Accurate prediction of crack width is not possible Many equations andmethods have been suggested but most are merely empirical rules resultingfrom observations or testing Furthermore there is no agreement on whatcrack width should be permitted for different types of structures Thisappendix discusses the main parameters which affect crack width and giveequations which may be used in common situations

External load applied on a concrete structure produces cracking when thetensile strength of concrete is exceeded When the reinforcement is designedto provide ultimate strength in accordance with any of the existing codesload-induced cracks rarely exceed a width of 05mm (002 in) Cracks oflarger width occur only when the structure is subjected to loads larger thanwhat it is designed for or when there is a misconception of the staticalbehaviour of the structure which results in yielding of the reinforcementunder service loads

Internal forces and stresses develop due to temperature shrinkage andsettlements of supports only when the movement due to these effects isrestrained The magnitude of the forces produced by the restraint dependupon the stiffness of the members and hence the forces are much smaller in acracked structure compared to a structure without cracks When adequatereinforcement is provided cracks caused by restraint are generally of smallwidth and the number of cracks increases with the increase in the restrainedmovement There is no generally accepted procedure for design of reinforce-ment necessary to control cracking caused by restraint One approach is toprovide reinforcement at all tension zones of a minimum ratio (See Section116)

Appendix E

ρmin = fct

fsy

(E1)

where fct is the tensile strength of concrete fsy is the yield strength of steelThis equation is based on the assumption that the tensile force carried by theconcrete immediately before cracking is transmitted to the reinforcementcausing stress which does not exceed its yield strength With fct = 2MPa andfsy = 400MPa (03 and 60ksi) Equation (E1) gives ρmin = 00051

Cracking can also occur due to causes other than what is discussed aboveMuch wider cracks can occur during the first few hours after placing ofconcrete while it is in a plastic state These are caused by shrinkage or bysettlement of the plastic concrete in the forms Cracks occur when movementof concrete is restrained by the reinforcement or by the formwork Plasticcracking cannot be controlled by provision of reinforcement it can only beachieved by attention to mix design and avoidance of conditions which pro-duce rapid drying during the first hour after placing This type of crack is notdiscussed any further below

Permissible crack width varies with design codes Acceptable values varybetween 01 and 04mm The smaller value may be suitable for water-retaining structures and the larger value for structures in dry air or withprotective membrane

The width of cracks depends mainly on stress in steel after cracking Otherfactors affecting crack width are thickness of concrete cover to reinforce-ment diameter of bars their spacing and the way they are arranged in thecross-section bond properties of the bars concrete strength and the shape ofstrain distribution Load-induced cracks unlike displacement-induced cracks(Section 113) increase in width with the duration of loading

In Section 83ndash6 the following expression was derived for the average widthof cracks which run in a direction perpendicular to the main reinforcement inmembers subjected to an axial force bending moment or both (Equation(848)

wm = srmζεs2 (E2)

where srm is the spacing between cracks this will be discussed in the followingsection εs2 is the steel strain calculated for a transformed section in which theconcrete in tension is ignored (state 2) ζ is a dimensionless coefficientbetween 0 and 1 representing the effect of the participation of concrete in thetension zone to the stiffness of the member the so-called tension stiffeningeffect The product (ζεs2) represents average excess in strain in the reinforce-ment relative to the surrounding concrete (Further explanation of the mean-ing of the symbol εs2 and its calculation are given in Section 861)

The value of the coefficient ζ depends upon the ratio (NrN) or (MrM)where N and M are the values of the axial normal force or bending moment

Crack width and crack spacing 545

on the section the subscript r refers to the value of N and M which producestensile stress fct at the extreme fibre

E2 Crack spacing

A semi-empirical equation is presented below for prediction of spacingbetween transverse cracks in members subjected to axial force or bending2

Fig E1 shows a reinforced concrete member subjected to an axial forceof magnitude just sufficient to produce the first crack At the cracked sec-tion the stress in concrete is zero (state 2) and the axial force is carriedentirely by steel At some distance sr0 from the crack the cross-section is inuncracked state 1 and the stress in concrete is fct the strength of concrete intension the force in steel at this section is only a fraction of the axial forceThe remaining part of the force is transmitted to the concrete by bondstress over the length sr0 Assuming fbm is the average value of the bondstress we can write

Ac fct = sr0 fbm 4As

db (E3)

where Ac and As are the cross-section areas of concrete and steel the quantity(4Asdb) is the sum of bar perimeters assuming that the bars have equaldiameter db

For a given type of reinforcement the bond stress fbm may be consideredproportional to fct thus

κ1 = fct

fbm

(E4)

where κ1 is a dimensionless coefficient depending upon bond properties of thereinforcing bars

Substitution of Equation (E4) into (E3) gives

Figure E1 Stress in concrete after first crack in a member subjected to axial force

546 Appendix E

sr0 = κ1 db

4ρ(E5)

The symbol sr0 represents the distance between the first crack and thecross-section at which the concrete stress reaches fct Subsequent smallincrease in applied force causes second and third cracks to occur at adistance sr0 on either side of the first crack and so on until a so-called stabil-ized crack pattern is obtained Further increase in load does not produce newcracks

Restraint which occurs when a member with fixed ends attempts toshorten due to shrinkage or temperature drop may produce only few cracksso that a stabilized state of cracking does not usually occur This is becausecracking is associated with a reduction in stiffness and hence alleviation ofrestraining forces

Experiments indicate that crack spacing is affected by other parameters notincluded in Equation (E5) namely the concrete cover and the spacing ofbars For this reason Equation (E5) is empirically modified in practice

EC2ndash91 and MC-903 give equations for the characteristic maximum crackwidth wk The two codes consider the value wk = 030mm (0012 in) underquasi-permanent loading as satisfactory for reinforced concrete members(without prestressing) This limit may be relaxed when the exposure condi-tions are such that crack width has no influence on durability (for examplethe interior of buildings for habitation or offices) A lower limit for wk shouldbe specified in accordance with the client when de-icing agents are expected tobe used on top of tensioned zones

For prestressed concrete members the two codes limit in general the valueof the characteristic crack width to wk = 02mm (0008 in) furthermore forcertain exposure conditions it is required that under frequent load combin-ations the prestressed tendons lie at least 25mm (1 in) within concrete incompression or no tension is allowed within the section

The two codes differ in the equation to be used in calculation of wk as givenbelow Provisions of ACI318-89 code are also discussed

E3 Eurocode 2ndash1991 (EC2ndash91)

The EC2ndash91 employs Equation (E2) to calculate the average crack width wmbut the code defines the design or characteristic maximum crack width wkas

wk = βwm (E6)

For load-induced cracking the value of the coefficient β to be used inEquation (E6) is β = 17 or 13 respectively for sections whose minimumdimension exceeds 800mm (30 in) or is smaller than 300mm (12 in)


According to EC2ndash91 the average crack spacing srm (mm) to be used inEquation (E2) is

srm = 50 + κ1κ2 db

4ρr

(E7)

where

κ2 = ε1 + ε2

2ε1

(E8)

where ε1 is the greater and ε2 the lesser tensile strain values (assessed on thebasis of fully cracked section) at upper and lower boundaries of the effectivetension area Acef defined in Fig E2 The steel ratio ρr is defined as

ρr = As

Acef

(E9)

The effective tension area is generally equal to 25 times the distance fromthe tension face of the section to the centroid of As (see Fig E2) but theheight of the effective area should not be greater than (h minus c)3 where h isthe height of the section and c is the depth of the compression zone

E4 CEB-FIP 1990 (MC-90)

MC-90 gives the following equation for calculation of the design crackwidth

wk = ls max(εs2 minus βεsr2 minus εcs) (E10)

where

db = bar diameter (mm)κ1 = coefficient depending upon bond quality κ1 = 08 for high bond bars and

16 for plain bars When cracking is due to restraint of intrinsic imposeddeformations (for example restraint of shrinkage) the coefficient κ1 is tobe replaced by 08 κ1 The multiplier 08 should generally be used but forrectangular sections of height h the multiplier should be equal to 08 forh 03m (12 in) and equal to 05 for h 08m (30 in)

κ2 = coefficient depending upon the shape of the strain diagram κ2 = 05 inthe case of bending without axial force κ2 = 10 in the case of axialtension In the case of eccentric tension

548 Appendix E

εsr2 = fctm(t)

ρrEs

(1 + αρr) (E11)

α = EsEc(t) (E12)

ρr and Acef are defined in Equation (E9) and Fig E2

Figure E2 Effective area Acef for use in Equations (E9 and E13) (a) beam (b) slab(c) member in tension (reference MC-90 or ECndash91)

εcs = the free shrinkage of concrete generally a negative valueεsr2 = the steel strain at the crack under a force causing stress equal to fctm(t)

within Acef

fctm(t) = the mean value of the tensile strength at time t at which the crackoccurred (Equations (A13) and (A14) )

εs2 = steel strain at the crackls max = the length over which slip between steel and concrete occurs This

length is given by Equation (E13) or Equation (E14) for stabilizedcracking and for single crack formation respectively


ls max = db

36 ρr

(E13)

ls max = σs2db

2τbk(1 + αρr)(E14)

db = diameter of reinforcing bar

E5 ACI318-89 and ACI318-99

The American Concrete Institute Building Code Requirements for Rein-forced Concrete ACI318-894 controls flexural cracking by limiting the stressin steel at a cracked section due to service load to 60 per cent of the specifiedyield strength Alternatively the parameter z defined by Equation (E15)must not exceed 175 or 145kipin (306 times 106 or 254 times 106 Nm) for interiorand exterior exposure respectively The parameter z is defined as

z = fs 3radicdcA (forcelength) (E15)

where

σs2 = steel stress at the crackτbk = bond stress given in Table E1 (assuming deformed bars are used)

β = empirical coefficient to assess the average strain within ls max The valueof β is given in Table E1 (assuming that deformed bars are used)

fs = stress in reinforcement at service load this is to be calculated for a fullycracked section (state 2)

dc = thickness of concrete cover measured from extreme tension fibre tocentre of bar located closest thereto (Fig E3(a) )

A = effective tension area of concrete surrounding the flexural tensionreinforcement and having the same centroid as that reinforcementdivided by the number of bars (Fig E3(a) ) When the flexural reinforce-ment consists of different bar sizes the number of bars is to be com-puted as the total area of reinforcement divided by the area of the largestbar or wire used

Table E1 Value of and τbk for use in Equations (E10) and (E14)

Single-crack formation Stabilized cracking

τbk τbk

Short-term or instantaneous 06 18 fctm(t) 06 18 fctm(t)loading

Long-term or repeated loading 06 135 fctm(t) 038 18 fctm(t)

550 Appendix E

Equation (E 15) is based on the GergelyndashLutz5 expression for maximumcrack width wmax (Equation (E16) ) corresponding to limiting crack widthsof 0016 and 0013 in (040 and 033mm) The GergelyndashLutz equationpredicts the maximum crack width as

wmax = (76 times 10minus6) βz (kip-in units) (E16)

wmax = (11 times 10minus12) βz (N-m units) (E17)

where β is the ratio of the distances from the neutral axis to the extremetension fibre and to the centroid of the main reinforcement The limitingvalues for z and wmax given above are based on an average value β = 12 whichapplies for beams For slabs the average value of β 135 thus for consist-ency the maximum values for z are to be reduced by the ratio 12135

Derivation of Equation (E16) involves the assumption that the maximumcrack spacing is4

Srm = 4te (E18)

where te is an increased effective cover defined as

te = dc 1 + s

4dc

2

(E19)

where s is the bar spacing (Fig E3(b) )

Figure E3 Definitions of symbols A dc and s for use in Equations (E15) and (E19) (a) beamfor 5 bars (b) slab (references ACI 318-89 and ACI 224-86)


The American Concrete Institute code ACI 318-996 replaces the require-ment for the parameter z presented above by setting a limit to the spacing sbetween bars in the zone of maximum tension in a cross-section as the smallerof

s(in) = 540

σs (ksi) minus25 (cover (in) ) (E20)

s(mm) = 95

σs (MPa) minus25 (cover (mm)) (E21)

and

s(in) = 12(36)

σs (ksi)(E22)

s(mm) = 76

σs (MPa)(E23)

where σs is stress in the reinforcement at service computed as the unfactoredmoment divided by the product of the steel area and the internal momentarm The code permits to take the stress in steel as 60 percent of specifiedyield strength

The parameter z in earlier ACI codes was based on empirical equationsusing a calculated crack width of 04mm (0016 in) The ACI 318R-99 codecommentary recognizes that crack widths are highly variable and adoptsthe Equations E20 to E23 that intend to control surface cracks to a widththat is generally acceptable in practice but may vary widely in a given struc-ture At the same time the ACI 318-99 code states that the bar spacingrequirement is not sufficient and requires special investigations and precau-tions for structures subject to very aggressive exposure or designed to bewatertight

Similar to the commentary of the earlier code ACI 318R-99 states thatcontrol of cracking is particularly important when reinforcement with a yieldstress in excess of 40ksi (300MPa) is used The commentary lists referencesto laboratory tests involving deformed bars that confirm that crack width atservice load is proportional to steel stress

E6 British Standard BS 8110

British Standard BS 8110 Part 2 19857 gives Equation (E24) for lsquodesignwidthrsquo of flexural crack at a particular point on the surface of a memberThe equation gives the design width of crack with acceptably small chanceof being exceeded actual cracks occasionally exceeding this width are

552 Appendix E

considered acceptable Provided the strain in the tension reinforcement is lessor equal to 08 fy Es the design crack width at surface may be calculated by

wat service = 3aεsurface

1 + 2a minus cover

h minus c (E24)

In assessing εsurface assume that the stress in the tension zone is as shown inFig E4 and take the modulus of elasticity of concrete Ec half the instant-aneous value Where shrinkage is abnormally high (gt 600 times 10minus6) εsurface

should be increased by 50 per cent of the expected shrinkage otherwiseshrinkage may be ignored This approach makes a notional allowance forlong-term effects

Notes

1 This steel ratio which may be adopted for water-retaining or structures exposed toweather is relatively high compared to the value of 0002 or 00018 required by ACI318-89 Code for shrinkage and temperature reinforcement at right angles to the

Figure E4 Assumptions in calculation of strain and stress distributions according to BritishStandard BS 8110

fy = specified characteristic strength of reinforcementEs = modulus of elasticity of reinforcementa = distance from point considered to the surface of the nearest longi-

tudinal barh = height of sectionc = depth of compression zone

εsurface = strain at the tension face (Fig E4)


principal reinforcement in structural slabs See Building Code Requirements forReinforced Concrete ACI 318-89 American Concrete Institute Detroit Michigan48219 (Section 712)

2 See reference mentioned in Note 2 page 193 See references mentioned in Notes 2 and 3 page 194 See reference mentioned in Note 9 page 4065 See ACI Committee 224 Cracking for Concrete Members in Direct Tension

ACI2242R-86 American Concrete Institute Farmington Hills Michigan48333-9094 USA

6 ACI 318 (1999) Building Code Requirements for Structural Concrete (318-99) andCommentary (318R-99) American Concrete Institute Farmington Hills Michigan48333-9094

7 See Note 3 page 533

554 Appendix E

Values of curvature coefficientsκs κ and κcs

Equations (91) to (94) give the instantaneous curvature at time t0 and thechanges in curvature during a period t0 to t caused by creep and shrinkage ata reinforced concrete section without prestress subjected to a bendingmoment M applied at t0 The equations include curvature coefficients κs κφ

and κcs which are evaluated in this appendixFigs F1 to F10 give values of κs κφ and κcs for rectangular sections the

additional subscript 1 or 2 is used to refer to the two states of no cracking andfull cracking respectively

For a general cross-section the curvature coefficients may be calculated bythe following expressions which can be derived from comparison of Equa-tions (91) to (94) with Equations (216) and (316)

κs = Ig

I(F1)

κφ = Ic + Ac yc∆y

I(F2)

κcs = minusA

cycd

I(F3)

The above equations are applicable to uncracked and cracked sections (seeEquation (627) ) for this reason the subscripts 1 and 2 are omitted Thesymbols in the equations are defined below

κs = coefficient smaller than unity which represents the stiffening effect ofthe presence of reinforcement on instantaneous curvature (Equation(91) )

κφ = coefficient smaller than unity representing the restraining effect ofreinforcement on creep curvature (Equation (92) )

κcs = coefficient to be used in Equation (93) for the curvature due toshrinkage

Appendix F

Figure F1 Curvature coefficient s1 for rectangular uncracked sections

556 Appendix F

Figure F2 Curvature coefficient s2 for rectangular cracked sections

Values of curvature coefficients 557

Figure F3 Curvature coefficient φ1 for rectangular uncracked sections dh = 10

558 Appendix F




560 Appendix F

Figure F6 Curvature coefficient φ2 for rectangular cracked sections dprimeh = 0



562 Appendix F



Figure F9 Curvature coefficient cs1 for rectangular uncracked sections

564 Appendix F

Figure F10 Curvature coefficient cs2 for rectangular cracked sections


Other symbols in Equations (F1) to (F3) are geometrical properties of thetransformed section at time t0 and the age-adjusted transformed section (FigF11(a) and (b) ) The first is composed of the concrete area plus α times thearea of steel where α = α(t0) = EsEc(t0) Es is the modulus of elasticity of steeland Ec(t0) is the modulus of elasticity of concrete at t0 The age-adjustedtransformed section is composed of the area of concrete plus α times thearea of reinforcement α = α(t t0) = EsEc(t t0) Ec(t t0) is the age-adjustedmodulus of elasticity of concrete (Equation (131) ) The geometrical sectionproperties included in Equation F1 to F3 are

Figure F11 Transformed section at t0 and age-adjusted transformed section in states 1 and2 (a) uncracked section (b) cracked section concrete in tension ignored

I = moment of inertia of the transformed section at time t0 about an axisthrough its centroid

Ig = moment of inertia of the gross concrete area about an axis through itscentroid

I = moment of inertia of the age-adjusted transformed section about anaxis through its centroid

566 Appendix F

In the usual case when the tension steel As near the bottom fibre is largerthan the compression steel and α is larger than α the values of ∆y and yc arerespectively positive and negative as shown in Fig F11(a) and (b)

When Equations F1 to F3 are used for a fully cracked section the symbolsrefer to properties of transformed cross-sections for which the concrete intension is ignored The depth of the compression zone is determined byEquation (716) which is applicable for the case when the section is subjectedto a bending moment without normal force

Ic = moment of inertia of concrete area Ac about an axis through centroid ofage-adjusted transformed section

Ac = area of concrete considered effective = entire concrete area in state 1but only area of compression zone in state 2

yc = the y-coordinate of the centroid of Ac measured downwards from thecentroid of the age-adjusted transformed section

∆y = the y-coordinate of the centroid of the age-adjusted transformed sec-tion measured downwards from the centroid of the transformedsection at t0

d = distance between extreme compression fibre and centroid of tensionsteel


Description of computer programsprovided atwwwsponpresscomconcretestructuresThe following password will be required to access the siteCONCRETE(NB the password needs to be in capital letters)

G1 Introduction

At the above web site three computer programs are provided as optionalcompanions of this book The programs are for use on IBM personal com-puters or compatibles One program gives creep and aging coefficient andrelaxation function for concrete the other two are for analyses of stress andstrains in individual sections A more comprehensive computer program thatperforms these anlyses for a number of sections and calculates deflectionsand rotations is

RPM ldquoReinforced and Prestressed membersrdquo Elbadry M and Ghali AAmerican Concrete Institute PO Box 9094 Farmington Hills MI 48333ndash9094 USA

The names of the three programs in the above-mentioned web site are

bull CREEPbull SCS (Stresses in Cracked Sections)bull TDA (Time-Dependent Analysis)

The files are listed below separately for each program

The files in the web address should be copied in a directory of arbitrary nameThe files CREEPIN SCSIN and TDAIN are input files of example prob-lems To generate an input file for a new problem edit the relevant file withthe ending lsquoINrsquo replacing the problem title the integers and the real values bythe data of the problem to be solved Before editing any input file exampleit should be copied ndash for future reference ndash in a file of arbitrary name To

MANUALCRP MANUALSCS MANUALTDACREEPIN SCSIN TDAINCREEPEXE SCSEXE TDAEXECREEPFOR

Appendix G

run a program type its name while the computer is set on DOS prompt andpress lsquoEnterrsquo The results will be written by the computer in a file namedCREEPOUT SCSOUT or TDAOUT Sections G2 to G4 describe thethree programs

The following are DOS commands that may be used with the computer onDOS prompt After typing each command press lsquoEnterrsquo

G2 Computer program CREEP

The program CREEP calculates the creep coefficient for concrete the relax-ation function and the aging coefficient in accordance with CEB-FIP ModelCode 19901 (See Section A1) Use of the program CREEP gives results notmuch different from the answers calculated in accordance with Eurocode2 ndash19912

G21 Input and output of CREEP

The input data file named CREEPIN has three lines of data

bull Title of problem (less than 76 characters)bull Values of fck ho and RH in MPa mm and per cent respectivelybull Concrete ages t0 and t in days

The output file CREEPOUT includes r(τ t0) χ (τ t0) Ec(t0) φ (t t0) χ (t t0)r (t t0) where

Command What the command achievesmd JOE Make a new directory named JOEcd JOE Change directory by opening JOEcopy a Copy all files in drive Acopy CREEPIN CREEPINBAK Copy an existing file in a new fileedit CREEPIN Open a file to read it or edit itCREEP Run the program named CREEPedit CREEPOUT Open a file to read it or edit it

fck (MPa) = characteristic compressive strength of cylinders 150mm indiameter and 300mm in height stored in water at 20 plusmn 2 degCand tested at the age of 28 days

ho (mm) = notional size = 2Acu with Ac and u being the area and theperimeter in contact with the atmosphere of the cross-sectionof the considered member

RH (per cent) = relative humidityt0 (days) = age of concrete at loadingt (days) = age of concrete at the end of a period in which the load is

sustained

Description of computer programs provided 569

G22 FORTRAN code

The file CREEPFOR presents a listing of FORTRAN statements whichincludes subroutine named Phicoef to calculate φ (t t0) using equations ofCEB-FIP Model Code 1990 (see Section A1) This subroutine can bechanged when use of other equations is required A manual for quickreference is included in the web address in the file MANUALCRP

G23 Example input file for CREEP

The file CREEPIN can generate the data to plot one of the two relaxationfunctions in Fig A3 The three lines of data for this problem are

Title Relaxation function Fig A3 for t0 = 3 days300 4000 500 fck (MPa) ho (mm) RH (per cent)30 300000 t0 t (days)

G3 Computer program SCS (Stresses inCracked Sections)

The program SCS calculates stresses and strains in a reinforced concretesection subjected to a bending moment M with or without a normal force NThe section can be composed of any number of trapezoidal concrete layersand any number of reinforcement layers The layers can have different elas-ticity moduli Ec and Es Prestressed and non-prestressed reinforcement aretreated in the same way First the stresses are calculated for uncracked sec-tion If stress in concrete at an extreme fibre exceeds the tensile strength fctthe analysis is redone ignoring concrete in tension

G31 Input and output of SCS

The input and output files are named SCSIN and SCSOUT Running theprogram must be preceded by preparation of the input file in which the dataare presented as follows

bull Title of problem (less than 76 characters)bull Number of concrete and reinforcement layers NCL and NRL

respectively

τ (days) = a time varying between t0 and tEc (t0) = modulus of elasticity of concrete at age t0

φ (t t0) = ratio of creep to instantaneous strainχ (t t0) = aging coefficient of concreter (t t0) = relaxation function = concrete stress at time t due to a unit

strain imposed at time t0 and sustained to time t

570 Appendix G

bull A set of NCL lines each line describes consecutively a trapezoidal con-crete layer starting by the top layerLayer number widths at top and at bottom height and elasticitymodulus Ec

bull A set of NRL lines each line describes a reinforcement layerLayer number cross-sectional area depth ds below top fibre and elas-ticity modulus Es When NRL = 0 skip this set of lines

bull Values of M N and fct

The computer writes the results in file SCSOUT which includes the strainand stress parameters that define their distributions and area properties ofthe cross-section When cracking occurs the output includes depth c of thecompression zone

G32 Units and sign convention

The basic units used are force unit and length unit Any units for these twomust be consistently used As example when Newton and metre are used forforce and length respectively M must be in Newton-metre and Ec Es and fct

in Newton per metre squaredThe reference point O is at top fibre When the resultant force on the

section is a normal force N at any position on vertical symmetry axis it mustbe substituted by statical equivalent normal force N at O combined with amoment M The y-coordinate of any fibre and the depth ds of anyreinforcement area are measured downward from the top fibre A tensilestress and the associated strain are positive A positive moment M producestensile stress at bottom fibre and induces positive curvature

Prestressing duct When it is required to deduct a cavity such as a prestress-ing duct from concrete area enter it as a reinforcement layer having a nega-tive cross-sectional area a dummy real value say a zero should be entered forthe modulus of elasticity

G34 Example input file for SCS

The following is file SCSIN for analysis of the section in Example 76 in thecracking stage

T-section Example 76 cracking stage N2 = minus327 kip M2 = 6692 kip in2 3 Number of concrete layers number of reinforcement layers1 80 80 4 4000 Layer no widths at top amp bot ht modulus Ec

2 20 20 36 40001 4 2 29000 Reinft layer no area depth ds modulus Es


2 3 34 270003 10 37 290006692 minus327 00 Moment M Normal force N and fct

G4 Computer program TDA(Time-Dependent Analysis)

A section composed of any number of trapezoidal layers and any number ofnon-prestressed reinforcement layers is considered All concrete layers havethe same elasticity modulus The section may have a single prestressedreinforcement layer which can be pretensioned or post-tensioned The pre-stressing is introduced simultaneously with a normal force N at top fibre andmoment M about an axis at top fibre After a period during which creep andshrinkage of concrete and relaxation of prestressed steel occur additionalnormal force and moment are introduced representing effect of live loadThe purpose of this program is to calculate the strain and the stress immedi-ately after prestressing after occurrence of creep shrinkage and relaxationand after application of the live load

G41 Input data for TDA

The input and output files have the names TDAIN and TDAOUT Runningthe program must be preceded by preparation of the input file with the datapresented as follows

bull Title of problem (less than 76 characters)bull Numbers of concrete and reinforcement layers NCL and NRL

respectivelybull A set of NCL lines each line describes a consecutive trapezoidal concrete

layer starting by the top layer layer number widths at top and bottomheight and elasticity modulus Ec at the time of prestressing (first loadingstage) The same value of Ec must be entered for all layers

bull A set of NRL lines each line describes a reinforcement layer layernumber cross-sectional area depth ds below top fibre and elasticitymodulus Es When NRL = 0 skip this set of lines

bull Values of M N and fct Value of fct is tensile strength at time of first stageof loading M and N are values of moment and axial force introduced atfirst stage No prestressing is included in values of M and N

bull Iprestress Ilayer prestress force Itda where Iprestress = 0 1 or 2 mean-ing no prestress pretensioning or post-tensioning respectively Ilayer isthe number of the layer that is prestressed Itdata = 0 or 1 meaning thetime-dependent analysis is not required or required respectively WhenIprestress = 0 enter 0 and 00 for the layer number and the prestressingforce respectively

572 Appendix G

bull Creep coefficient aging coefficient free shrinkage and reduced relax-ation Omit this line when Itdata = 0

bull Values of M N fct Ec Enter here magnitudes of moment and normalforce introduced after the time-dependent changes give also fct and Ec atthis instant Omit this line when Itdata = 0


The references point O is at top fibre A normal force N at any position onvertical symmetry axis is substituted by statical equivalent normal force N atO combined with a moment M The y-coordinate of any fibre and depth ds

of any reinforcement layer are measured downwards from the top fibre Atensile stress and the associated strain are positive A positive moment Mproduces a tensile stress at bottom fibre and induces a positive curvature Thefree shrinkage is commonly a negative value indicating shortening thereduced relaxation is also negative indicating loss of tension Any basic unitsof force and length can be adopted all parameters must be entered using thesame basic units

G43 Prestressing duct

When it is required to deduct a cavity such as a prestressing duct fromconcrete area it should be entered as a reinforcement layer having a negativecross-sectional area a dummy real value say a zero should be entered for themodulus of elasticity The prestressed steel in the duct must be entered on aseparate line

G44 Example input file for TDA

The input file presented below is for solution of Examples 26 and 76 TheT-section of a pretensioned beam (Fig 215(a) ) is to be analyzed for the time-dependent effects occurring between the time of prestress and a later instantAt this instant a bending moment is applied representing effect of live loadThe immediate strain and stress due to live load are also required The pre-stress transfer is accompanied by a given bending moment due to the self-weight In this problem basic units used for force and length are kip and inrespectively The input data file is

T-section of Examples 26 and 76 (Fig 215)2 3 No of concrete layers no of reinforcement layers1 80 80 4 3600 Layer no widths at top amp bot ht Ec

2 20 20 36 36001 4 2 29000 Reinft layer no area depth ds Es

2 3 34 27000


3 10 37 2900010560 0 00 M N and fct1 2 600 1 Iprestress Ilayer prestress force Itda3 8 minus300 e-6 minus13 Creep coef aging coef fr shrge red relaxn9600 0 0 4000 M N fct Ec

Notes

1 See reference mentioned in Note 2 p 192 See reference mentioned in Note 5 p 19

574 Appendix G

Further reading

The following are selected relevant books Extensive lists of references can be found ineach of them

Branson DE (1977) Deformation of Concrete Structures McGraw-Hill New YorkFavre R Beeby AW Falkner H Koprna M and Schiessl P (1985) Cracking and

Deformation Comiteacute Euro-International de Beacuteton (CEB) Federal Institute ofTechnology Lausanne Switzerland

Favre R Koprna M and Radojicic A (1980) Effects differeacutes Fissuration etDeacuteformations des Structures en Beacuteton Georgi Saint-Saphorin VD Switzerland

Favre R Jaccoud J-P Koprna M and Radojicic A (1990) Dimensionnement desstructures en beacuteton volume 8 of traiteacute de Geacutenie Civil Presses polytechniques etuniversitaires romandes Lausanne Switzerland

Gilbert RI (1988) Time Effects in Concrete Structures Elsevier AmsterdamNeville AM Dilger WH and Brooks JJ (1983) Creep of Plain and Structural

Concrete Construction Press London

Index

ACI see American Concrete InstituteAge-adjusted

elasticity modulus of concrete 17flexibility 18 106 151stiffness 18transformed section 18 40

Ageing coefficient of concretecomputer code for 489ndash490definition 10equation for 11 489factors affecting 17graphs and table 484 493ndash532

American Concrete Institute 3 19 274302 474 481 545 550ndash552554

Branson equation for effective moment ofinertia 315

ldquoBilinearrdquo method for deflectionprediction 313 320

Bridgescomposite see Composite structuresconstruction see Segmental

constructionprestressed see Prestressingthermal effects on see Temperature

British Standard 3 19 485 533 552British units

examples worked out in 61 95 141260 298 402 403

Cantilever method of construction seeSegmental construction

CEB see Comiteacute Euro-International duBeacuteton

Coefficient of thermal expansion 358Comiteacute Euro-International du Beacuteton

474 548

Composite structurespartially prestressed 249stress and strain in sections 22 25

30ndash35examples of calculations 44ndash49

49ndash57 64ndash67time-dependent changes

in fixed-end forces 156in internal forces 154 160 163

Computer programscompanion of this book

availability on the Internet 568address of web site on the Internet

568description 568conventional linear for framed

structures 177 206description 568

CPF computer program (CrackedPlane Frames) 175 302

CREEP computer program 569code in FORTRAN 569example input file for CREEP 570input and output of 569

linear for framed structures 177 206availability 206description of 179ndash184multi-stage loading 188use for time-dependent analysis

176ndash206cable-stayed shed example 193cantilever construction example

192composite space truss 201equivalent temperature

parameters 186ndash187prestressed portal frame example

205

propped cantilever example 188PLANEF computer program (Plane

Frames) linear analysis 177 180181 183 189 191 194 196 202203 204 206

availability 206RPM computer program (Reinforced

and Prestressed Members) 302568

SCS computer program (Stresses inCracked Sections) 570

example input file for SCS 571input and output files for 570units and sign convention 571

SPACET computer program (SpaceTrusses) linear analysis 177 180182 197 206

availability 206TDA computer program (Time

Dependent Analysis) 571ndash574example input file for DA 573input of 572units and sign convention 573

Conductivity see TemperatureConjugate beam see Elastic weightsConstruction stages see Multi-stage

constructionContinuous structures see Statically

indeterminate structuresCracking

aesthetic appearance 400changes in stress andor strain at 246

255 256 262 391 395control of 380

minimum reinforcement for controlof 391

corrosion of reinforcement effect on399

creep and shrinkage effects after237

deformations of cracked membersequations and calculationssummary 281

examples of calculations 271 275278 285 290 298 299

displacement induced 382example analysis 387

force-induced 382example analysis member subjected

to axial force 387example analysis member subjected

to bending 384

fully-cracked sectionsdefinition 208rectangle properties 216 218ndash221stress and strain 210stress and strain calculation

examples 234 236 243 250 254260

thermal 393T-shape properties 215 222ndash233

gas or liquid tightness effect on 399400

idealization model 283interpolation between uncracked

and cracked states 264ndash294mean curvature due to bending 273mean curvature due to bending

combined with axial force 277mean strain due to axial tension 269mean strain and curvature with

partial prestressing 283 290variation of curvature over the

length 285 290heat of hydration due to 393high-strength concrete of 401plastic 545of prestressed sections 208 246reduction of stiffness due to 218 245

547reduction of temperature stresses after

350 374spacing of cracks 544 546ndash552stabilized crack pattern 547temperature due to example

overhanging slab 403width of cracks mean value 265 270

544ndash554amount of reinforcement to limit

crack width 394permissible 545

yielding of steel at a cracked section391

Creep of concretecement type effect on 479coefficient of computer code for 486

definition 2ndash3equations and graphs for 477 479

480 481 488ndash532deflection change due to 308 313 315

323deflection of slabs due to 336effects on

composite sections 44 54 55

578 Index

cracked sections 237cracked sections with prestressing

209internal forces analysis by

conventional computer programs177ndash206

internal forces calculation examples109 113 152

internal forces in staticallyindeterminate structures 101 121146 149 175

internal forces in structures built instages 105 109 113

internal forces in structures withcomposite members 141 154156 160 164

prestressed sections 35 44 49 6064 74

reinforced concrete section withoutprestressing 79 85 86 237

high stress due to 479parameters affecting 2 475relative humidity effect on 478restraining effect of the reinforcement

on 238 239step-by-step analysis 14ndash7 127 136

142 172under sustained stress 3ndash4temperature effect on 474 475thickness of member effect on 475 492time functions for 474 478 479 481under varying stress 9ndash11 17

Creep of steel see Relaxation of steelCurvature

bending moment relationship in slabs335

coefficients non-prestressed sectionssubjected to bending 303 556

of cracked membersexample of calculation 275mean value due to bending 271 303

combined with axial force 276318

due to temperature 376 377examples of calculations at a fully

cracked section 254 260variation over the length of 285

290creep and shrinkage effects on

sections without prestressing 237reduction factor to account for the

reinforcement 238 240

deflection expression in terms ofcurvatures at a number ofsections 333 538ndash541

equation 25as intensity of elastic load 89ndash90non-prestressed steel effect on 78variation over the length of uncracked

beam 42variation with time 30 33 74 79 133

240examples of uncracked sections 35

41 437 43 44 49 61 64 75 8183

Decompression forces see Partialprestressing

Deflectioncalculation from curvature at a

number of sections 333 538ndash541of cracked members 285 290determinant section for calculation of

309of floors 332geometric relationship with curvature

333 538ndash541interpolation between uncracked and

cracked states 306limitations 348prediction by simplified calculations

ldquobilinearrdquo method 313 318examples 315 323 330 333ldquoglobal coefficientsrdquo method

325ndash327instantaneous-plus-creep deflection

due to bending 308shrinkage deflection 309

see also DisplacementDensity of materials 358Depth of compression zone in a fully

cracked section see Neutral axisposition

Design for serviceability of prestressedconcrete 407ndash427

balanced deflection factor 408balancing load factor 408non-prestressed steel recommended

ratio in box-girder bridges 422permanent state 408prestressing level 409ndash413residual crack opening 419

control of 421residual curvature 422

Index 579

transient stresses 416ndash419distribution of thermal stresses over

bridges section 418water tightness 419

Determinant section see DeflectionDisplacement see DeflectionDisplacement calculation

from axial strain and curvatures at anumber of sections 538ndash541

for cracked members 264 266by elastic weight 70 89by unit-load theory 89by virtual work 70 89 119 120

Displacement method of analysis effectsof temperature by the 175

review 146step-by-step 147 172time-dependent internal forces by the

146

Effective moment of inertia 315Elastic weights method of deflection

calculation 70 89Emissivity of a surface see TemperatureEquivalent concentrated load 90 91Eurocode 5 19 270 474 480 547

569

Fatigue of steel 395Feacutedeacuteration Internationale de la

Preacutecontrainte 4 5 6 19 474 537548 569 570

Fibre-reinforced polymersadhesion to concrete 458aramid 458 460carbon 458 460compressive strength 458creep rupture 459glass 458 460modulus of elasticity 460properties of 458ndash459relaxation 459serviceability of members reinforced

with 458ndash473curvature and deflection of flexural

members 463deformability of sections in flexure

471deflection control

design example for 469ndash470verification of the ratio of span to

deflection 470ndash471

design of cross-sectional area ofFRP for non-prestressed flexuralmembers 460ndash462

prestressing with FRP 472ratio of span to minimum thickness

466ndash469empirical equation for 468ndash469

relationship between deflectionmean curvature and strain inreinforcement 464ndash466

strain in reinforcement and width ofcracks 459ndash460

tensile strength 460thermal expansion coefficient 458

FIP see Feacutedeacuteration Internationale de laPreacutecontrainte

Fixed-end forcestime-dependent changes 149

examples of calculation 152Flexibility increase due to cracking 265

295mean flexibility 265

Flexibility matrixage-adjusted 18 106 152definition 103

Floors see Two-way slab systemsForces artificial restraining 158Force method of analysis of statically

indeterminate structureseffect of temperature 363review 103step-by-step 136time-dependent changes in internal

force by the 105FRP see Fibre-reinforced polymers

Heat see TemperatureHeat of hydration of cement 351

368ndash373stress due to example of calculation 371

Heat transfer equation 354High-strength concrete

cracking of 401creep of 477shrinkage of 479

Indeterminate structures see Staticallyindeterminate

Interpolationcoefficient for between uncracked and

fully cracked states 265 266ndash271301ndash302 304 306

580 Index

procedure for deflection prediction(the ldquobilinearrdquo method) 313

Loss of prestress see Prestress loss

Maturity of concrete 475MC-90 see CEB FIPMean curvature due to bending on a

cracked member 271Mean strain due to axial tension on a

cracked member 266Modulus of elasticity of concrete 476

age-adjusted 17secant 3time variation 476

Multi-stage construction see Time-dependent changes

Multi-stage prestressing 87step-by-step analysis 136

Neutral axis position in a fully crackedsection 210 213 542ndash543

remarks on determination of 213Non-linear analysis of plane frames

428ndash456convergence criteria 445ndash446examples of statically indeterminate

structures 447ndash456demonstration of the iterative

analysis 447ndash451deflection of non-prestressed

concrete slab 452ndash454prestressed continuous beam

454ndash456fixed-end forces 439ndash440

due to temperature 440ndash442idealization of plane frames 429ndash431incremental method 446ndash447

example 454ndash456iterative analysis 443ndash445non-linearity due to cracking 429numerical integration 442ndash443reference axis of a member 429tangent stiffness matrix of a member

429uncracked member example

431ndash437cracked member example 437ndash439

Partial prestressingdecompression forces 248

composite section 249

definition 208 246effects of creep and shrinkage 209

246example of deflection calculation of

cracked members with 286 290298 299

examples of stress and straincalculations in a cross sectionwith 250 254 260 290 298 299

mean strain and curvature in memberswith 283

reduction of deflection by 292 298temperature effects in structures with

377ndash378time-dependent deformations with 283variation of curvature along a beam

with 290Post-tensioning

accounting for cross-sectional area ofducts 32

continuity of precast elements by 116128 153

definition 21examples 37 43 44 64 75 95 116

128 141 299 330Precast elements made continuous

by cast-in-situ joints 64 95 116 128141

by prestressing 116 128 153Prestressing

methods of 21in multi-stages 61 68 87ndash88partial see Partial prestressingself-equilibrating forces due to 114

149 152Pre-tensioning


298instantaneous loss in 33instantaneous stress and strain due to

33 43 72

Radiation see TemperatureRelative humidity effect on creep 478

effect on shrinkage 479Relaxation of concrete 12ndash17 127 491Relaxation function 12ndash17Relaxation of prestressed steel changes in

stress and strain in a prestressedsection due to 21 31 74

definition 5

Index 581

effects on internal forces analysis byconventional computer programs177ndash206

effect on internal forces in staticallyindeterminate structures 102 120146 174

intrinsic 6variation with time 536

reduction of 7ndash9reduction coefficient 8 534step-by-step analysis of effect of 136

147temperature effect on 7

Secant modulus of elasticity of concrete 3Segmental construction 146ndash147 172

174Serviceability of members reinforced

with fibre-reinforced polymer seeFibre-reinforced polymers

Settlement of supports 101 121ndash128136

example of calculation of reactionsdue to gradual 125ndash127

Shear deflections 293Shrinkage of concrete

curvature due to 243 303 309 327346 564 565

deflection due to 309 327in a composite section 44 49 64in continuous members 311ndash313in simple beams 309ndash310

description of the phenomenon and itseffects 4


equations for the values of 479 480483 486

in a fully cracked section effects of 237in a partially prestressed section

effects of 243in a prestressed section effects of 35

44 49 60 64 75 95in a reinforced concrete section

without prestressing effects of79 81

relative humidity effect on 490restraining effect of the reinforcement

on the deformations due to 242step-by-step analysis of the effect of

136 147 172

stress and strain due to 30 74 309 310thickness of member effect on 479time function for 479 483 486

Sign convention 22 209see also Notation

Slabs see Two-way slab systemsSolar radiation see TemperatureSpecific heat see TemperatureStates 1 and 2 definitions 208Statically indeterminate forces

analysis by the displacement method146

analysis by the force method 100due to gradual settlement of supports

101 121ndash128 136due to shrinkage 312due to temperature 352 361ndash366step-by-step procedure of time-

dependent 136 146 172Step-by-step analysis

by the displacement method 147 172of the effects of creep 14ndash18of the effects of relaxation 18of the effects of shrinkage 18by the force method 136of thermal stresses 370ndash371

example of calculation 371Stiffness matrix definition 148Stiffness method of analysis see

Displacement methodStiffness reduction due to cracking 261

295 547Strain

axial due to temperature 28 361 362in composite section 22 30in cracked sections 210 237effect of presence of non-prestressed

steel on 78 94 95in homogeneous sections 22instantaneous due to post-tensioning

35 44 64 75 94instantaneous due to pretensioning 33

42 72mean value

due to axial tension 266example of calculation of 271

due to temperature 27ndash30in uncracked sections 20ndash22 30ndash67

74 128Strength of concrete

development with time of 476tensile 477

582 Index

Stressin composite sections 22in cracked sections 207

time-dependent change 237in homogeneous sections 22instantaneous at prestress transfer 32

43 72non-prestressed steel effect on concrete

78ndash95temperature due to

continuity stresses 27 28 352ndash353361 363

eigen-stresses 27 28 352ndash353 360uncracked sections in 20ndash26

time-dependent changes in 20 3057 60ndash97 128 144

Temperatureabsorptivity of surface 351 358coefficient of thermal expansion 358conductivity 352 355 358continuity stresses 27 28 352ndash353 360

example of calculation 363convection 251 355ndash356distribution over bridge cross-sections

354 367 369effect on creep 374 475effect of creep on stress due to 28 350effect on maturity of concrete 475effect on relaxation of prestressed

steel 7eigen-stresses 27 28 252ndash253 360

see also Self-equilibrating stressesemissivity of surface 298 302 304

351 356358heat of hydration of cement 351 370

371example of calculation of stress due

to 371internal forces in indeterminate

structures due to 27ndash28 363 non-linear variation 27ndash30radiation solar 351 355 356re-radiation 351 355self-equilibrating forces restraining

expansion of a member 358self-equilibrating stresses 27 28

352ndash353 359example of calculation 29 363 371

specific heat 351 355 358statically indeterminate forces in

continuous beams 362

StefanndashBoltzmann constant 356step-by-step analysis of stress due to

370mdash371stress and strain due to 27ndash30

in a fully-cracked section 376stress relief by cracking 350 374stresses in transverse direction in a box

girder 359turbidity of atmosphere 351

Tension stiffening definition 266Thermal effects see TemperatureTime-dependent changes

in creep coefficient 374 475in deformations of cracked members

284examples of calculation 285 299

in fixed-end forces 149in internal forces accounting for the

reinforcement 128in internal forces due to alteration of

support conditions 121 149 152154

in internal forces in compositestructures 154 160 164

examples of calculation 141 160164 175 301

in internal forces in cracked structures136 175 301

in internal forces in indeterminatestructures

by the displacement method 144145

by conventional linear computerprograms 176ndash206

examples of calculation 108 109113 116 125 141 152

by the force method 21ndash22 100 128in internal forces in structures built in

stages 105 109 113 116 128141 152 155

in internal forces due to supportsettlement 101 121ndash128136

in modulus of elasticity of concrete476

in shrinkage values 479 480 483 486in stress and strain in composite

sections 22 27 30ndash35in stress and strain in cracked sections

237in stress and strain in uncracked

sections 20 30 57 63 95 128144

Index 583

Transformed sectionage-adjusted 17ndash18 40definition 17ndash18fully cracked definition 209properties of calculation examples 38

40 46properties of a rectangle graphs 86

Truss idealization of cracked members294 296

Turbidity of atmosphere seeTemperature

Twistingof fully-cracked members 295of uncracked members 294

Two-way floors see Two-way slabsystems

Two-way slab systemscurvature-bending relations 335deflection due to loads 332ndash343deflection due to shrinkage 345examples of deflection calculations

338 341 345geometric relationship curvature-

deflection 333

Unit load theory 88ndash89United States units see British

units

Virtual work principle 70 88ndash89119ndash120

584 Index

Book Cover

Title

Copyright

Contents


Acknowledgements

Note

The SI system of units and British equivalents

Notation

Chapter 1 Creep and shrinkage of concrete and relaxation of steel

Chapter 2 Stress and strain of uncracked sections

Chapter 3 Special cases of uncracked sections and calculation of displacements

Chapter 4 Time-dependent internal forces in uncracked structures analysis by the force method

Chapter 5 Time-dependent internal forces in uncracked structures analysis by the displacement method

Chapter 6 Analysis of time-dependent internal forces with conventional computer programs

Chapter 7 Stress and strain of cracked sections

Chapter 8 Displacements of cracked members

Chapter 9 Simplified prediction of deflections

Chapter 10 Effects of temperature

Chapter 11 Control of cracking

Chapter 12 Design for serviceability of prestressed concrete

Chapter 13 Non-linear analysis of plane frames

Chapter 14 Serviceability of members reinforced with fibrereinforced polymers

Appendix A Time functions for modulus of elasticity creep shrinkage and aging coefficient of concrete

Appendix B Relaxation reduction coefficient Xr

Appendix C Elongation end rotation and central deflection of a beam in terms of the values of axial strain and curvature at a number of sections

Appendix D Depth of compression zone in a fully cracked T section

Appendix E Crack width and crack spacing

Appendix F Values of curvature coefficients κs κ and κcs

Appendix G Description of computer programs provided at

Further reading

Index


Third Edition




















Third Edition




London and New York















(Print Edition)

Contents





112 General 18





















vi Contents













48 General 144





Contents vii








69 General 205






viii Contents







79 General 262












Contents ix




811 General 301












x Contents


910 General 348













Contents xi












xii Contents
















Contents xiii













xiv Contents




Contents xv













A GhaliR Favre




Acknowledgements





Note




1 m = 3281 ft


1m2 = 1076 ft2








1MPa = 01450ksi




1W = 3416 Btuh



Notation

























Notation xxiii





xxiv Notation


11 Introduction



Chapter 1








εc(t0) =σc(t0)

Ec(t0)(11)






εc(t) =σc(t0)

Ec(t0) [1 + φ(t t0)] (12)
















∆σpr

σp0


10 σp0

fpy

minus 055 (14)






∆σprinfin

σp0


where


λ =σp0

fptk

(16)
















σp0 (18)



055 060 065 070 075 080

00 1000 1000 1000 1000 1000 100001 06492 06978 07282 07490 07642 0775702 04168 04820 05259 05573 05806 0598703 02824 03393 03832 04166 04425 0463004 02118 02546 02897 03188 03429 0362705 01694 02037 02318 02551 02748 02917









εc(t) = σc(t0) 1 + φ(t t0)

Ec(t0)+

∆σc(t)

0

1 + φ(t τ)


where











εc(t) = σc(t0) 1 + φ(t t0)

Ec(t0)+ ∆σc(t)

1 + χφ(t t0)

Ec(t0)+ εcs(t t0) (110)










∆σc(t)(111)



dσc(τ)

dτ= ∆σc(t)

dξ1

dτ(112)


εc(t) = σc(t0)1 + φ(t t0)

Ec(t0)+ ∆σc(t)

t

t0

1 + φ(t τ)

Ec(τ) dξ1

dτ dτ

+ εcs (t t0) (113)


χ(t t0) =Ec(t0)

φ(t t0)

t

t0

1 + φ(t τ)

Ec(τ) dξ1

dτ dτ minus1

φ(t t0)(114)











σc(t0) = εcEc(t0) (115)


σc(t) = εcr(t t0) (116)



∆σc(τ) = ξ[∆σc(t)] (117)







ξ =∆σc(τ)

∆σc(t)(120)



εc = σc(t0) 1 + φ (t t0)

Ec(t0)+ ∆σc(t)

1 + χφ(t t0)

Ec(t0)(121)





Ec(t0)(122)


χ(t t0) =1


1

φ(t t0)(123)







2 tj and tj + 1

2 are used to refer




εc(ti + 12) =

i

j = 1(∆σc)j

1 + φ(ti + 12 tj)


2 t0) (124)


σc(ti + 12) =

i

j = 1

(∆σc)j (125)





σc(ti + 12) = εcr(ti + 1

2 t0) (126)


r(ti + 12 t0) =

1

εc

i

j = 1

(∆σc)j (127)


εc(ti + 12) = (∆σc)i

1 + φ(ti + 12 ti)

Ec(ti)+

i minus 1

j = 1

(∆σc)j 1 + φ(ti + 1

2 tj)

Ec(tj)

+ εcs(ti + 12 t0) (128)



(∆σc)i =Ec(ti)

1 + φ(ti + 12 ti)


2 t0)

minus i minus 1

j = 1

(∆σc)j 1 + φ(ti + 1

2 tj)

Ec(tj) (129)


2 t0) = 0 and







εc(t) = σc(t0) 1 + φ(t t0)

Ec(t0)+

∆σc(t)

Ec(t t0)+ εcs(t t0) (130)

where

Ec(t t0) =Ec(t0)

1 + χφ(t t0)(131)



Ec(t t0) (132)





Ec(t0)(133)




Ec(t t0) (134)



112 General












21 Introduction



Chapter 2










22 Sign convention









ε = εO + ψy (21)





σ = Ei(εO + ψy) (22)


N = σdA (23)

M = σydA (24)


N = εO m

i = 1

Ei dA + ψ m

i = 1

Ei ydA (25)

M = εO m

i = 1

Ei y dA + ψ m

i = 1

Ei y2dA (26)






A = m

i = 1

Ei

Eref

Ai (29)

B = m

i = 1

Ei

Eref

Bi (210)

I = m

i = 1

Ei

Eref

Ii (211)




NM = Eref ABB

I εO

ψ (212)


εO

ψ =1

ErefAB

B

I minus1

NM (213)


ABB

I minus1

=1

(AI minus B2) I

minusB

minusB

A (214)


εO

ψ =1

Eref(AI minus B2) I

minusB

minusB

A NM (215)


εO

ψ =1

Eref

NA

MI (216)

231 Basic equations


ε = εO + ψy σ = E(εO + ψy) (217)

N = E(AεO + Bψ) M = E(BεO + Iψ) (218)


εO =IN minus BM

E(AI minus B2)ψ =

minusBN + AM

E(AI minus B2)(219)

σO =IN minus BM

AI minus B2γ =

minusBN + AM

AI minus B2(220)

where













εf = αtT (221)













∆εO

∆ψ =1

E(AI minus B2) I

minusB

minusB

A minus∆N

minus∆M (227)

∆σ = E[∆εO + (∆ψ)y] (228)


∆εO

∆ψ =1

E minus∆NA

minus∆MI (229)


σ = E [minus εf + ∆εO + (∆ψ)y] (230)










∆N

∆M = αtTtopE

minusbh

m + 1

bh2m

2(m + 1)(m + 2)


εO =αtTtop

m + 1

ψ = minusαtTtop

h

6m

(m + 1)(m + 2)













Nequivalent





εO(t0)

ψ(t0) =

1

Eref(AI minus B2) I

minusB

minusB

A NMequivalent

(232)



εO(t0)

ψ(t0) =

1

Eref

Nequivalent

A

Mequivalent

I

(233)



εc(t0) = εO(t0) + ψ(t0)y (234)

σc(t0) = [Ec(t0)]i[εO(t0) + ψ(t0)y] (235)
















∆εO

∆ψ =1

Ec(AI minus B2) I

minusB

minusB

A minus∆N

minus∆M (240)


∆N

∆M = ∆N

∆Mcreep

+ ∆N

∆Mshrinkage

+ ∆N

∆Mrelaxation

(241)


∆N

∆Mcreep

= minusm

i = 1

EcφAc

Bc

Bc

Ic εO(t0)

ψ(t0)

i

(242)






∆N

∆Mshrinkage

= minus m

i = 1

EcεcsAc

Bc

i

(243)





∆N

∆Mrelaxation

= Aps∆σ-pr

Apsyps∆σ-pr

i

(244)


















NMequivalent





εO(t0)

ψ(t0) =

1






minus240 times 103

= 10minus6 minus126

minus170mminus1










Tabl

e 2

1C

alcu

latio

n of

A B

and

I of

tra

nsfo

rmed

sec

tion

at t

ime

t 0

Prop

ertie

s of

are

aPr

oper

ties

of t

rans

form

ed a

rea

AB

IAE

Ere

fBE

Ere

fIE

Ere

f

(m2)

(m3)

(m4)

(m2)

(m3)

(m4)

Conc

rete

035

45minus1

625

times 1

0minus341

84

times 1

0minus30

3545

minus16

25 times

10minus3

418

4 times

10minus3

Non

-pre

stre

ssed

ste

el25

00 times

10minus6

027

5 times

10minus3

075

6 times

10minus3

001

671

833

times 1

0minus35

04 times

10minus3

Pres

tres

sed

stee

lmdash

mdashmdash

mdashmdash

mdash

Prop

ertie

s of

037

120

208

times 1

0minus346

88

times 1

0minus3

trans

form

ed s

ectio

nA

BI




1 + 08 times 3= 882GPa



= 2741MPa (0398ksi)


= 8145MPa (1181ksi)


∆N

∆Mcreep




4184 times 10minus3

times minus126

minus170 10minus6 = 106 1175N

01828N-m

∆N

∆Mshrinkage



= 106 0750N

minus00034N-m

∆N

∆Mrelaxation



= 106minus0090N

minus00403N-m

∆N

∆M = 106 1175 + 0750 minus 0090

01828 minus 00034 minus 00403 =1061835 N

0139 N-m



Tabl

e 2

2C

alcu

latio

n of

A B

and

I of

the

age

-adj

uste

d tr

ansf

orm

ed s

ectio

n

Prop

ertie

s of

are

aPr

oper

ties

of tr

ansf

orm

ed a

rea

AB

IAE

Ere

fBE

Ere

fIE

Ere

f

(m2)

(m3)

(m4)

(m2)

(m3)

(m4)

Conc

rete

035

45minus1

625

times 1

0minus341

84

times 1

0minus30

3545

minus16

25 times

10minus3

418

4 times

10minus3

Non

-pre

stre

ssed

ste

el25

00 times

10minus6

027

5 times

10minus3

075

6 times

10minus3

005

676

236

times 1

0minus317

24

times 1

0minus3

Pres

tres

sed

stee

l11

20 times

10minus6

050

4 times

10minus3

022

7 times

10minus3

002

5411

429

times 1

0minus35

15 times

10minus3

Prop

ertie

s of

age

-adj

uste

d0

4366

160

40 times

10minus3

641

2 times

10minus3

trans

form

ed s

ectio

nA

BI


∆εO

∆ψ =

1





04366 minus1835

minus0139 106

= 10minus6minus470

minus128mminus1





= 3313MPa (0481ksi)



∆σns1 = 200[minus471 + 055(minus128)]103







































(m2) (m3) (m4) (m2) (m3) (m4)






NM = 02800 times 103 N-m


εO(t0)

ψ(t0) =

1


times 21367

16788

16788

16474 0

2800 times 103

= 10minus6 223

219mminus1





1 + 075 times 25= 10435GPa



= 824MPa


= 699MPa




∆N

∆Mcreep


times 13081

minus15828

minus15828

19205 113

219 10minus6

= 106 5187 N



∆N

∆Mshrinkage


minus15828

= 106 4777 N

minus5781 N-m

∆N

∆Mrelaxation



= 106 minus0351 N


∆N

∆M = 106 5187 + 4777 minus 0351

minus6306 minus 5781 + 0425 = 106 9613 N



A = 2284m2 B = minus1859m3 I = 2542m4





∆εO

∆ψ =

106


1859

1859

2284minus 9613

11662

= 10minus6 minus112

357mminus1




















NMequivalent





εO (t1)

ψ(t1) =

1


minus00117

minus00117

05583



minus218mminus1




Ec(60 3) =25 times 109

1 + 086 times 120= 1230GPa (1780ksi)





(m2) (m3) (m4) (m2) (m3) (m4)










∆N

∆Mcreep


0

0

01090 minus289

minus218 10minus6

= 106 2171 N

0351 N-m

∆N

∆Mshrinkage


0 = 106 0357 N

0

∆N

∆Mrelaxation




∆N

∆M = 106 2171 + 0357 minus 0047

00351 + 0 minus 0025 = 106 2481 N




∆εO(t2 t1)






∆εO(t2)


415mminus1






1 + 08 times 227= 1314GPa (1900ksi)


1 + 078 times 240= 801GPa (1160ksi)



1

Ec(t)[1 + φ(60 t)] =

1

Ec(3) [1 + χ(60 3) φ(60 3)]





= 14304MPa




= 0106MPa





∆N

∆Mcreep


0

0

0109

times

110minus289 +095 times 1012

25 times 109 + 227(minus6)

110minus218 +161 times 1012

25 times 109 + 227(415)

10minus6

= 106 1937 N

minus1108 N-m


∆N

∆Mshrinkage


0


minus04307

= 106 2437 N

minus0928 N-m


∆N

∆Mrelaxation



minus0117N-m





(m2) (m3) (m4) (m2) (m3) (m4)



Concrete inbeam 05090 00 01090 05090 00 01090






∆N

∆M = 106 1937 + 2437 minus 0221


minus2153 N-m



∆εO(t3 t2)


195mminus1




Notation






Subscripts


Four analysis steps














Commentary




















αns = 806 αps = 750



A = 1158 in2 B = 19819 in3 I = 547200 in4










Ec(t t0) = 1059ksi


Ac = 1023 in2 Bc = 16000 in3 Ic = 410800 in4










αns = 2739 αps = 2550



A = 1483 in2 B = 28950 in3 I = 874600 in4




∆σ(t t0)top bot = 0047 0485 ksi










































28 General




Note





Chapter 3

31 Introduction






















1 +Ast

Ac

Est

Ec(t t0)1 +

y2st

r2c

(34)























∆σns

Est


Est

=σcst(t0)φ(t t0)


1

Ec(t t0)∆Pc

Ac

+∆Pc y2

st

Ic (38)














Ec(t t0)Ac

(310)


Ec(t t0)Ic

(311)


∆σc =∆Pc

Ac

+∆Pc yst

Ic

y


2c in the last

equation gives

∆σc =∆Pc

Ac

1 +yyst

r2c (312)


















Figu

re3

2A

naly

sis

of s

tres

s an

d st

rain

in a

sec

tion

with

one

laye

r of

pre

stre

ssed

ste

el (E

xam

ple

31)

(a)

cro

ss-s

ectio

n (b

) con

ditio

nsim

med

iate

ly a

fter

pre

stre

ss (

c) c

hang

es d

ue t

o cr

eep

shr

inka

ge a

nd r

elax

atio

n




0357

200

8824 1 +

02059

01193minus1

= 2427kN (5456kip)


∆σps =∆Pps

Aps

= minus∆Pc

Aps




σp0 =1400 times 103



λ =σp0

fptk

=1250

1770= 0706 Ω =

2167 minus 115

1250= 0081






0357 1 +

(minus0596)0454

01193 Pa




0357 1 +

0604 times 0454

01193 Pa = 2241MPa (0325ksi)




















r2c

+ εcs(t t0)yc

r2c (316)



Symbolused












c(t)]dAc



η = AcA (317)

κ = IcI (318)












∆N

∆Mcreep


Ac yc

Ac yc

Ic εO(t0)

ψ(t0) (320)

∆N

∆Mshrinkage


Ac yc (321)


∆N




r2c + εcs(t t0)

yc

r2cAcr

2c (322)


∆εO

∆ψ =1

Ec(t t0) minus∆NA

minus∆MI (323)







1 + 08 times 3= 8824GPa (1280ksi)

α(t t0) =200

8824= 22665






A = 03811m2 I = 3750 times 10minus3 m4


η =02963

03811= 0777 κ =

2526

3750= 0674




8445 times 10minus3






= 1102MPa (0159ksi)








= 4163kN-m (3685kip-in)







∆ψ = 0674 3428 + (minus120)minus 0054




+ (minus334) + 1021(minus0551)

= 1508MPa (0219ksi)






∆εO = ηφ(t t0)[εO(t0) + ψ(t0)yc]) (324)

∆ψ = κφ(t t0) ψ(t0) + εO(t0) yc

r2c (325)





∆εO ηεO(t0)φ(t t0) (326)


∆ψ κψ(t0)φ(t t0) (327)


2c]



and κ







α = α(t0) = EsEc(t0) (328)

α = α(t t0) = Es[1 + χφ(t t0)]Ec(t0) (329)





κ =Ic

I (or I )(330)


Ic bh3

12 1 + 12y2

c

h2 (331)










































Fig 38(b) )

Q =

7l48

l24

minusl48

l8

5l12

l8

minusl48

l24

7l48

ψ (a)




D3 = [1 05 0]Q (c)

D4 = l [0 025 0] Q (d)






D1 = εO dl (e)


D1 = l

6

2l

3

l

6 εO






(186)2

96 [1 10 1]

64minus298





186

6 [1 4 1]





186

6 [1 2 0]

64minus298














4 times10minus6



60 times10minus6




























between steel





ns







310 General






Notes








Chapter 4

41 Introduction





















[ f ] F = minusD (41)





A = As + [Au] F (42)














Ec =Ec

1 + χφ(44)





[ f ] ∆F = minus∆D (45)


∆A = ∆As + [∆Au] ∆F (46)













Ec(t t0) =Ec(t0)

1 + χφ(t t0)





φ(infin 7) = 27 χ(infin 7) = 074 φ(60 7) = 11

φ(infin 60) = 23 χ(infin 60) = 078


Ec(60)Ec(7) = 126




D1 = (D1)AB + (D1)BC

= 0 +ql3

24Ec(7)Ic




f11 = ( f11)AB + ( f11)BC

=l

3Ec(60)Ic

+l

3Ec(7)Ic


F1 = minus D1 f11




∆D1 = (∆D1)AB + (∆D1)BC


(∆D1)AB =ql 3

24Ec(7)Ic


3Ec(60)Ic

φ(infin 60)


(∆D1)BC =ql 3

24Ec(7)Ic


3Ec(7)Ic

φ(infin 7)


∆D1 = 00720 ql 3

Ec(7)Ic



f11 = ( f11)AB + ( f11)BC

( f11)AB =l

3Ec (infin 60)Ic

( f11)BC =l

3Ec(infin 7)Ic



1 + χφ(infin 60)= 1

1 + 078 times 23 Ec(60) = 045Ec(7)

Ec(infin 7) =Ec(7)

1 + χφ(infin 7)= 1

1 + 074 times 27 Ec(7) = 034Ec(7)

Thus

f11 =l

3 times 045Ec(7)Ic

+l

3 times 034Ec(7)Ic

= 1724 l

Ec(7)Ic









2

3q =

8Pa

l 2













Ec(t0)Ic

2 timesl


ql 3

Ec(t0)Ic


Ec(t0)Ic


Ec(t0)Ic


Ec(t t0) = 1

1 + 08 times 25 Ec(t0) =1

3Ec(t0)



[Ec(t0)3]Ic

2 timesl



Ec(t0)Ic

(∆D1) = (1390 + 125)10minus3 ql 3

Ec(t0)Ic


Ec(t0)Ic


f11 =l

[Ec(t0)3]Ic

13 +1

2 = 25 l

Ec(t0)Ic


∆F1 = minus1515





Example 44



φ(t1 t0) = 11 χ(t1 t0) = 079 φ(t2 t0) = 27 χ(t2 t0) = 074

φ(t2 t1) = 23 χ(t2 t1) = 078 Ec(t1)Ec(t0) = 126









1 + 078 times 23= 036Ec(t1) = 045 Ec(t0)


1 + 074 times 27= 034Ec(t0)



(∆D1)loads =1

Ec(t0)Ic

D


+1

Ec(t1)Ic

D

AMeMu1 dl φ(t2 t1)

+1

Ec(t0)Ic

C

DMeMu1 dl φ(t2 t0)




Ec(t0)Ic


timesql 3

Ec(t1)Ic


Ec(t0)Ic


Ec(t0)Ic







(Ec)ADIc

D

AMiMu1 dl +

1

(Ec)DCIc

C

DMiMu1 dl





ql 3

Ic +

1

034Ec(t0)



ql 3

Ec(t0)Ic

∆D1 = (277 + 70)10minus3ql 3

Ec(t0)Ic


Ec(t0)Ic


f 11 =1

(Ec)ADIc

D

AM2

u1 dl +1

(Ec)DCIc

C

DM 2

u1 dl =251l

Ec(t0)Ic


∆F1 = minus347











F(t1) = Fsudden

1

1 + χ φ(t1 t0)(47)


Ec(te)


1 + χφ(t2 t1) (48)




1

Ec(te)[1 + φ(t1 te)] =

1

Ec(t0)[1 + χφ (t1t0)] (49)






1 + χφ(t t0) (410)



κ = Ic I (411)



Fsudden = 6

l 3Ec(t0)Icδ (412)




δ = l 3

6Ec(t1 t0)Ic F(t1) (413)


Ec(t1 t0) =Ec(t0)

1 + χφ(t1 t0)(414)




∆δ = l 3




∆δ + l 3

6Ec(t2 t1)Ic ∆F = 0 (416)

where

Ec(t2 t1) =Ec(t1)

1 + χφ(t2 t1)(417)


F(t2) = F(t1) + ∆F (418)















1414141423232323

104104104

104500

200010000

104500

200010000

5002000

10000

114179226257101172220255117170196

079076076076

081080079






δ(t)


t095 minus t0 (419)































Tabl

e 4

1C

ross

-sec

tion

prop

ertie

s1 and

cal

cula

tion

of in

stan

tane

ous

axia

l str

ain

and

curv

atur

e fo

r a

cont

inuo

us b

ridg

e (E

xam

ple

46)

Forc

es a

pplie

d at

time

t 0 e

quiva

lent

sof

pre

stre

ss fo

rce

Inst

anta

neou

sSe

ctio

n nu

mbe

rTr

ansf

orm

ed s

ectio

nAg

e-ad

just

ed tr

ansf

orm

edan

d de

ad-lo

adax

ial s

train

and

(see

Co

ncre

te s

ectio

npr

oper

ties

at ti

me

t 0se

ctio

n pr

oper

ties

bend

ing

mom

ent

curv

atur

eFi

g 4

10(b

))pr

oper

ties

E ref

=E c

(t0)

=28

GPa

E ref

=E c

(t t

0)=

909

GPa

(Equ

atio

n (2

31)

)(E

quat

ion(

232

))

A cB c

I cA

BI

AB

IN

M O

(t0)

(t

0)

10

780

015

90

8068

00

1645

091

60minus0

003

60

1835

minus28

20

189

minus125

410

20

780

015

90

8068

00

1645

091

600

0393

020

47minus2

82

024

5minus1

2553

23

078

00

159

080

680

016

450

9160

minus00

036

018

35minus2

82

008

8minus1

2519

14

078

00

159

080

680

016

450

9160

003

930

2047

minus28

20

195

minus125

423

Mul

tiplie

rsm

2m

3m

4m

2m

3m

4m

2m

3m

410

6 N10

6 N-m

10minus6

10minus6

mminus1

1R

efer

ence

poi

nt O

is c

hose

n at

the

com

mon

cen

troi

d of

Ac o

r A n

s

Tabl

e 4

2C

hang

es in

axi

al s

trai

n an

d in

cur

vatu

re o

f the

rel

ease

d st

ruct

ure

duri

ng t

he p

erio

d t 0

to

t in

Exam

ple

46

Calcu

latio

n of

rest

rain

ing

forc

es

Tota

lCh

ange

s in

axi

al s

train

Sect

ion

num

ber

Cree

pSh

rinka

geRe

laxa

tion

rest

rain

ing

forc

esan

d in

cur

vatu

re(s

ee F

ig 4

10(

b))

(Equ

atio

n (2

42)

)(E

quat

ion

(24

3))

(Equ

atio

n (2

44)

)(E

quat

ion

(24

1))

(Equ

atio

n (2

40)

)

N

M

N

M

N

M

N

M

O

12

304

minus01

541

170

20

minus02

190

0148

378

7minus0

139

3minus4

5574

62

230

4minus0

199

91

702

0minus0

219

minus01

607

378

7minus0

360

6minus4

6728

34

32

304

minus00

718

170

20

minus02

190

0148

378

7minus0

057

0minus4

5525

34

230

4minus0

159

01

702

0minus0

219

minus01

607

378

7minus0

319

7minus4

6626

13

Mul

tiplie

rs10

6 N10

6 N-m

106 N

106 N

-m10

6 N10

6 N-m

106 N

106 N

-m10

minus610

minus6 m

minus1


to t

∆D1 =25

6 [0 2 1]

7462834253

10minus6 +25

6 [1 2 0]

2532613253



(25)2

96[1 10 1]

410532191




(25)2

96 [1 10 1]

7462834253


= 191mm (0752 in)




A





f 11 =25



f 12 =25



(25)2




( f 11 + f 12)∆F1 = minus ∆D1

Thus








∆A = ∆As + [∆Au] ∆F



= 1039mm (0409 in)









[ f ] F = minusD (420)






2 tj)]







Fi + 12=

i

j = 1

∆Fj (421)

Di + 12=

i

j = 1

∆Dj (422)


i

j = 1


i

j = 1

∆Dj (423)


[ f (ti + 12 ti)] ∆Fi = minus Di + 1

2minus

i minus 1

j = 1

[f(ti + 12 tj)]∆Fj (424)










6EcIc

(425)




2 a displacement

f(ti + 12 tj) = C

1

Ec(tj) [1 + φ(ti + 1

2 tj)] (426)


C =l 3

6Ic

(427)



2) (428)



f (ti + 12 ti)(∆F)i = δ(ti + 1

2) minus C

i minus 1

j = 11 + φ(ti + 1

2 tj)

Ec(tj) (∆F)j (429)



2) + (∆F)i (430)



2)

+ [ f (ti + 12 ti)]


i minus 1

j = 11 + φ(ti + 1

2 tj)

Ec(tj) (∆F)j (431)




















∆As =

01770296

minus77851324

ksi








f 11 = ∆ψB 2l




[∆Au] = 10minus6

minus3059minus1539

minus2765988428

ksi

kip-in





∆A =

01770296

minus77851324

+ 10minus6

minus3059minus1539

minus2765988428

(minus117700) =

05370427

minus4530minus9084

ksi


48 General



Notes









51 Introduction



Chapter 5

































∆D = D(t0)[φ(t2 t0) minus φ(t1 t0)] (53)



[ f ] = [ f ][1 + χφ(t2 t1)] (54)



[ f ]∆F = minus∆D (55)



1 + χφ(t2 t1) F (56)

where











F = minusql2



∆F = 26 minus 09




























[ f ] ∆F = minus ∆D (58)







[ f ]minus1 = Ec

l

A

0

0

0

4I

2I

0

2I

4I

(510)




∆F = Ec

l

A

0

0

0

4I

2I

0

2I

4I

minus∆D (511)


∆D =int(∆εO)

int(∆ψ)

int(∆ψ)

Nul

Mu2

Mu3

dl

dl

dl

(512)




∆εO

∆ψ = 1

Ec(AI minus B2) I

minusB

minusB

A minus∆N

minus∆M (513)



∆εO

∆ψ = minus1

Ec

∆NA

∆MI (514)








p = minusd

dx(∆N) (515)

q = minusd2

dx2(∆M) (516)
















Ec(t1 t0) = 30 times 109

1 + 08 times 25 = 10GPa α(t1 t0) =

200

10 = 20



A = 210m2 B = 0 I = 10232m4


Ec(t0) = 30GPa α(t0) = 200

30 = 6667 Eref = Ec(t0)


A = 158m2 B = minus03947m3 I = 05221m4



Ac = 132m2 Bc = minus05927m3 Ic = 02714m4



∆N


minus05927 = 3564 times 106 N























∆F = 10 times 109

33

210

0

0

0

4(10232)

2(10232)

0

2(10232)

4(10232)

1691

807

minus807

10minus6

=

10761MN

05004MN-m

minus05004MN-m



Tabl

e 5

1In

stan

tane

ous

axia

l str

ain

and

curv

atur

e at

t 0 i

mm

edia

tely

aft

er a

pplic

atio

n of

the

load

q (E

xam

ple

53

Fig

51

1)

Prop

ertie

s of

Defl

ectio

ntra

nsfo

rmed

1 sec

tion

atAx

ial s

train

and

at G

age

t 0In

tern

al fo

rces

curv

atur

e at

t 0Pr

oper

ties

of(E

quat

ion

(Ere

f=

E c(t 0

)=30

GPa

)in

trodu

ced

at t 0

(Equ

atio

n (2

32)

)co

ncre

te a

rea

(C8

))

Mem

ber

Sect

ion

AB

IN

M O

(t 0)

(t 0

)A c

B cI c

D(t

0)

B1

58minus0

394

70

5221

minus02

431

minus18

21minus4

21

minus148

11

32minus0

592

70

2714

BCG

158

minus03

947

052

21minus0

243

13

624

649

280

51

32minus0

592

70

2714

284

6C

158

minus03

947

052

21minus0

243

1minus1

821

minus42

1minus1

481

132

minus05

927

027

14

Mul

tiplie

rm

2m

3m

410

6 N10

6 N-m

10minus6

10minus6

mminus1

m2

m3

m4

10minus3

m

1T

he r

efer

ence

poi

nt O

is a

t th

e ce

ntro

id o

f age

-adj

uste

d tr

ansf

orm

ed s

ectio

n (F

ig 5

8(c

))

Tabl

e 5

2C

hang

es in

axi

al s

trai

n an

d in

cur

vatu

re a

nd c

orre

spon

ding

elo

ngat

ion

and

end

rota

tions

of t

he r

elea

sed

stru

ctur

e in

Fig

51

1(c)

Chan

ges

inCh

ange

inCh

ange

s in

axi

aldi

spla

cem

ents

at t

hede

flect

ion

Inte

rnal

forc

es to

Prop

ertie

s of

age

-adj

uste

dst

rain

and

inco

ordi

nate

s in

Fig

at

Gre

stra

in c

reep

trans

form

ed s

ectio

ncu

rvat

ure

511

(c) (

Equa

tions

(Equ

atio

n(E

quat

ion

(24

2))

(Ere

f=

E c(t 1

t0)

=10

GPa

)(E

quat

ion

(24

0))

(C5

ndash7))

(C8

))

Mem

ber

Sect

ion

N

M

AB

I

O

D1

D

2

D3

D

Bminus0

805

20

3810

210

01

0232

383

minus37

2BC

G2

015

minus09

415

210

01

0232

minus96

092

0minus1

691

minus807

807

959

Cminus0

805

20

3810

210

01

0232

383

minus37

2

Mul

tiplie

rs10

6 N10

6 N-m

m2

m3

m4

10minus6

10minus6

mminus1

10minus6

m10

minus6 1

0minus610

minus3 m

radi

anra

dian

Figu

re5

12A

naly

sis

of s

tatic

ally

inde

term

inat

e fo

rces

cau

sed

by c

reep

in a

com

posi

te fr

ame

(Exa

mpl

e 5

3) (

a) fi

xed-

end

forc

es d

ue t

ocr

eep

Ar

(b) a

ssem

blag

e of

fixe

d-en

d fo

rces

and

rev

ersa

l of d

irec

tion

minusF

(c)

mem

ber-

end

forc

es d

ue to

join

t dis

plac

emen

t[A

u]

D

(d) t

otal

mem

ber-

end

forc

es d

ue t

o cr

eep

= su

m (a

) and

(c)

(e) b

endi

ng m

omen

t di

agra

m a

t t 1














Tabl

e 5

3St

ress

dis

trib

utio

n at

sec

tion

G (E

xam

ple

53)

Stre

ss in

sta

ges

(MPa

)Cr

eep

effe

ct =

Stre

ss a

t tim

e t 1

=At

tim

e t 0

Cree

p ef

fect

(b)+

(c)+

(d)

(a)+

(b)+

(c)+

(d)

(a)

(b)

(c)

(d)

MPa

MPa

ksi

Top

of c

oncr

ete

minus27

562

297

minus14

740

140

096

31

793

minus02

60Bo

ttom

of c

oncr

ete

minus09

060

755

minus12

720

078

minus04

39minus1

345

minus01

95To

p of

ste

elminus6

04

0minus2

544

155

minus23

89minus2

993

minus43

4Bo

ttom

of s

teel

893

30

584

minus80

4minus2

20

871

312

64


= 3421mm (1347 in)





2 the end of


εc(ti + 12) =

i

j = 11 + φ(ti + 1

2 tj)


2 t0) (517)

εps(ti + 12) =

1

Eps

i

j = 1


εns(ti + 12) =

1

Ens

i

j = 1

(∆σns)j (519)



2 t0)





(∆εc)i = 1 + φ(ti + 1

2 ti)

Ec(ti) (∆σc)i

+ i minus 1

j = 1

(∆σc)j

Ec(tj)[φ(ti + 1


2 tj)] + (∆εcs)i (520)


Eps

minus (∆σpr)i

Eps

(521)


Ens

(522)



(Ece)i

+ (∆εc)i (523)


Eps

+ (∆εps)i (524)


(Ece)i = Ec(ti)

1 + φ(ti + 12 ti)

(525)







αi = Es

(Ece)i

(526)






59 General




Notes






Chapter 6

61 Introduction






































A = Ar + [Au] D (61)


AD = [Au] D (62)









Ar(tj)prestress =




Ec(tk tj) = Ec(tj)

1 + χφ(tk tj)(65)

















TO

T prime1T prime2

= 1

Ecα

1A00

0l(6I)5l(6I)

0minus1I

minus1I

Ar1

Ar2

Ar3

(69)



TO = Ar (tk tj)

αEc (tk tj)A(610)








= minusEc (tk tj)




68 Examples



















Ec (t2 t1) = Ec(t1)

1 + χφ(t2 t1) =

Ec(t1)

1 + 08(245) = 03378 Ec(t0)





12

u00000E+0000000E+00

v41668Eminus0712500E+00

10417Eminus0616667E+00


1Fx

00000E+00Fy

minus10000E+01Mz

minus50000E+00


1F1

00000E+00F2

minus10000E+01F3

minus50000E+00F4

00000E+00F5


55511Eminus15





A(t2 t1) = 0 02155 ql 02154 ql 2 0 minus02155 ql 0





12



00000E+00 00000E+00 00000E+00

00000E+00 00000E+00 00000E+00



1 1Ar1

0000E+00Ar2

2872E+00Ar3

2393E+00Ar4

0000E+00Ar5

minus2872E+00Ar6

4786Eminus01



12

u00000E+0000000E+00

v17688Eminus07

minus17688Eminus07

17688Eminus07

minus35377Eminus01


12

Fx

00000E+00

00000E+00

Fy

21550E+00minus21550E+00

Mz

21540E+00

00000E+00


1F1

00000E+00F2

21550E+00F3

21540E+00F4

00000E+00F5

minus21550E+00F6

00000E+00









u v θ u v θ2 0 1 0 00 00 00










12

x y00 00

100 00


11st node

12nd node

2a

10000E+01I

10000E+00


2



00000E+00 00000E+00 00000E+00


1Node

1Fx

17890E+06Fy

minus89440E+06Mz

00000E+00


1Member

1Ar1

0000E+00Ar2

minus1250E+06Ar3

minus2083E+06Ar4

0000E+00Ar5

minus1250E+06Ar6

2083E+06


Node12



12200Eminus03

minus14730Eminus09


2Fx

minus17890E+06Fy

minus16056E+06Mz

35560E+06


1F1

17890E+06F2

minus89440E+05F3


minus17890E+06F5

minus16056E+06F6

35560E+06








Ec(t2 t1) = 25GPa

1 + 08(20) = 9615GPa



250 (20)AD(t1)AB










123

x y00 00

100 00100 minus50


12

1st node11

2nd node23

a10000E+0152000Eminus02

I10000E+0010000Eminus06


23



00000E+00 00000E+00 00000E+00

00000E+00 00000E+00 00000E+00


1Node

1Fx

00000E+00Fy

00000E+00Mz

00000E+00


111

Member111

Ar1

minus1376E+06minus2885E+071250E+05

Ar2

minus2740E+050000E+000000E+00

Ar3

minus1602E+060000E+000000E+00

Ar4

1376E+06

2885E+07minus1250E+05

Ar5

2740E+05

0000E+00

0000E+00

Ar6

minus1133E+050000E+000000E+00



123



minus24022Eminus08



23

Fx


Fy

minus77388E+0477388E+04

Mz

77890E+05

19580E+01


12

F1minus15478E+0517305E+05

F277388E+0432547E+00

F3minus16808E+0116808E+01

F415478E+05

minus17305E+05


F677890E+0519580E+01
















Ec (t2 t1) = 25GPa

1 + 08(225) = 8929GPa





Ec (t1) φ(t2 t1) AD(t1)





123456789

101112

83225Eminus03




minus23736Eminus08





44022Eminus08

44022Eminus08

14765Eminus01

29254Eminus01

29254Eminus01

39067Eminus01

48840Eminus01

48840Eminus01

52367Eminus01

55758Eminus01

55758Eminus01

00000E+00


129

101112

00000E+00

00000E+00

00000E+00minus72000E+06minus72000E+0614400E+07

00000E+00




1234567

31

19209E+06

51887E+06

67612E+06

19209E+06

51887E+06

67612E+06

96046E+05

43425E+04


minus43425E+04





111111111111

123456123456


1544E+06

4170E+06

5433E+06

1544E+06

4170E+06

5433E+06

1429E+07

1429E+07

1429E+07

1429E+07

1429E+07

1429E+07


Node u v w123456789

101112

87131Eminus02




minus28656Eminus23







123456789

10 262728293031


71626E+05

71626E+05

80999E+05

80999E+05

86603E+05

86603E+05

64064E+05

72448E+05

77460E+05

64064E+05

72448E+05

77460E+05

32032E+05

68256E+05

74954E+05

38730E+05




250 (225) 19209 minus19209




= minus14286 14286kN











Ec(t2 t1) = 25GPa

1 + 08(20) = 9615GPa



123456789

10111213






17

13

Fx

25047E+05minus26649E+0726399E+07

Fy

minus31200E+0600000E+0000000E+00

Mz

35933E+05minus38885E+06minus51617E+00

Member end forces

Member F1 F2 F3 F4 F5 F6123456789

1011

31200E+06

26544E+07

26586E+07

26618E+07

26639E+07



35933E+05

10001E+06

20538E+06

28612E+06

34321E+06

37732E+06

11827Eminus02

33526Eminus02

50243Eminus02

61547Eminus02

67716Eminus02








11111111111111111

123456123456789

1011




2400E+06

2042E+07

2045E+07

2048E+07

2049E+07

2050E+07

4615E+06

2700E+07

2700E+07

2700E+07

2700E+07





Nodal displacements

Node123456789

10111213





Node17

13

Fx

86860E+03

19104E+06minus19190E+06

Fy

26096Eminus04

00000E+00

00000E+00

Mz


Member end forces

Member F1 F2 F3 F4 F5 F6123456789

1011





26096Eminus04

33753E+05

29902E+06

24999E+06

21136E+06








= minus9615


442kN minus2060kNm




= 2700kN




69 General




Notes






Chapter 7

71 Introduction














73 Sign convention









e = MN (71)


and ys




ε = εO + yψ (72)




σc = Ec1 minus y

ynεO y lt yn (74)

0 y yn (75)


σs = Es1 minus ys

ynεO (76)


εOEcyn

yt1 minus

y


ys

yn = N (77)

εOEcyn

yt

y1 minus y


ys

yn = M (78)

where




yn

yt





yn

yt


yn

yt


minus e = 0 (710)











σt1 0 while σb1 0 (711)




I1 minus ytB1

B1 minus ytA1

M

N

I1 minus ybB1

B1 minus ybA1

(712)





yt M

N yb (713)


ΣIs minus yt ΣBs

ΣBs minus yt ΣAs

M

N

ΣIs minus yb ΣBs

ΣBs minus yb ΣAs

(714)

















2 minus 4a1a3)

2a1

(716)

where

a1 = bw2 (717)




a3 = minus12h




















d = ΣαiAsidi

ΣαiAsi

(722)






Figu

re7

4D

epth

of t

he c

ompr

essi

on z

one

in a

fully

cra

cked

rec

tang

ular

sec

tion

subj

ecte

d to

ecc

entr

ic n

orm

al fo

rce


es = MN (723)



ρ = Asbd (724)

and









Figu

re7

5D

ista

nce

from

the

top

fibr

e to

the

cen

troi

d of

the

tra

nsfo

rmed

are

a of

a fu

lly c

rack

ed r

ecta

ngul

ar s

ectio

n

Figu

re7

6M

omen

t of

iner

tia a

bout

an

axis

thr

ough

the

cen

troi

d of

the

tra

nsfo

rmed

are

a of

a fu

lly c

rack

ed r

ecta

ngul

ar s

ectio

n

Tabl

e 7

1D

epth

of c

ompr

essi

on z

one

in a

fully

cra

cked

T s

ectio

n su

bjec

ted

to e

ccen

tric

com

pres

sive

forc

e c

= (c

oeffi

cien

t fr

om t

able

times 1

0minus3)d

b wb

= 0

05

e sd

= minus

09

e sd

= minus

10

e sd

= minus

15

e sd

= minus

20

e sd

= minus

50

e sd

= minus

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

000

582

090

982

958

632

024

615

013

713

713

752

5252

5252

4343

4343

4334

3434

3434

3131

3131

310

010

828

912

836

613

339

333

214

173

170

170

7772

7272

7261

6060

6060

4848

4848

4843

4343

4343

003

085

392

385

968

340

349

037

228

324

523

916

212

312

012

012

012

410

210

210

210

291

8282

8282

7974

7474

740

050

870

931

875

725

453

567

461

361

304

278

227

164

151

151

151

177

133

129

129

129

129

105

105

105

105

111

9595

9595

007

588

693

989

176

250

462

953

443

336

331

429

220

918

418

118

123

216

815

515

515

517

113

012

712

712

714

711

611

511

511

50

100

898

946

902

788

544

671

585

488

412

345

345

249

214

204

204

278

200

178

176

176

208

153

145

145

145

180

136

131

131

131

012

590

895

191

180

957

870

362

553

145

237

338

928

424

122

622

431

722

920

019

419

424

117

516

116

016

020

915

414

614

614

60

150

916

955

919

825

606

729

656

566

487

398

426

316

266

246

242

352

256

220

210

210

271

196

176

174

174

236

172

159

158

158

020

092

896

193

185

065

276

770

362

154

344

348

637

131

228

327

141

030

425

824

023

732

323

420

519

819

828

420

518

318

018

0

b wb

= 0

1

e sd

= minus

09

e sd

= minus

10

e sd

= minus

15

e sd

= minus

20

e sd

= minus

50

e sd

= minus

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

001

064

770

065

352

433

829

220

817

317

017

077

7272

7272

6160

6060

6048

4848

4848

4343

4343

430

020

670

717

676

562

371

373

290

231

212

211

120

100

100

100

100

9384

8484

8470

6868

6868

6261

6161

610

060

731

766

734

651

467

519

452

378

325

294

242

181

165

164

164

192

147

140

140

140

143

115

114

114

114

124

103

103

103

103

010

076

879

777

070

153

159

153

446

440

334

432

224

521

320

420

426

319

817

817

617

620

015

314

514

514

517

413

513

113

113

10

150

799

824

801

742

588

649

600

535

473

397

394

308

264

246

242

329

251

220

210

210

257

194

176

174

174

226

171

159

158

158

020

082

284

382

477

263

068

964

558

652

544

044

835

930

828

227

138

229

725

624

023

730

423

120

419

819

826

920

318

318

018

00

250

840

859

841

794

663

720

679

624

566

477

491

401

346

315

296

424

335

289

268

260

343

263

230

219

218

306

232

206

199

199

030

085

487

185

581

269

074

470

765

560

050

952

743

737

934

431

846

036

931

929

328

037

729

225

423

923

633

825

922

721

721

60

400

875

890

876

840

731

780

748

702

651

560

583

496

435

395

358

518

426

370

338

315

434

343

297

275

267

392

306

266

249

245

Tabl

e 7

1co

ntin

ued

b wb

= 0

2

e sd

= minus

09

e sd

= minus

10

e sd

= minus

15

e sd

= minus

20

e sd

= minus

50

e sd

= minus

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

002

054

256

954

648

536

831

826

922

721

121

111

610

010

010

010

092

8484

8484

7068

6868

6862

6161

6161

004

058

260

458

453

341

739

535

130

427

226

017

714

213

713

713

714

111

711

611

611

610

794

9494

9493

8585

8585

012

066

968

467

063

453

453

450

245

941

736

531

826

123

222

122

126

621

519

519

119

120

916

815

815

715

718

414

914

314

314

30

200

717

730

718

688

599

604

577

539

499

436

400

340

302

281

271

343

285

253

240

237

277

225

203

198

198

247

199

183

180

180

030

075

776

875

873

265

465

963

660

356

650

047

041

036

834

031

841

235

031

229

128

034

128

225

123

923

630

725

122

521

721

60

400

785

794

786

764

693

698

678

648

614

548

521

463

419

388

357

464

402

360

334

315

390

328

292

274

267

354

294

263

248

245

050

080

781

580

878

772

372

870

968

265

058

656

150

546

142

839

150

544

340

037

134

543

036

732

830

629

339

433

129

527

727

00

600

824

831

825

806

746

751

734

709

679

618

593

540

496

462

421

538

478

434

404

372

464

400

359

334

316

427

363

324

303

291

080

085

085

685

083

478

278

777

275

072

366

664

459

455

251

847

259

253

449

045

841

951

945

541

238

335

748

141

637

434

932

8

b wb

= 0

5

e sd

= minus

09

e sd

= minus

10

e sd

= minus

15

e sd

= minus

20

e sd

= minus

50

e sd

= minus

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

005

047

248

047

345

841

833

532

130

328

827

817

415

715

115

115

114

413

112

912

912

911

410

510

510

510

510

195

9595

950

100

525

531

526

513

477

408

396

381

364

342

242

221

209

204

204

205

185

177

176

176

165

149

145

145

145

147

133

131

131

131

030

063

263

663

262

359

554

353

552

350

848

038

336

134

333

031

833

731

429

728

628

028

225

924

523

823

625

623

422

121

721

60

500

687

690

688

680

656

611

604

594

581

555

460

440

422

407

388

413

390

372

358

344

353

329

311

300

293

324

299

283

274

270

075

073

273

573

372

670

566

666

065

164

061

552

650

648

947

445

247

945

643

842

340

441

639

237

336

034

638

536

034

233

031

91

000

764

766

764

758

740

704

699

691

681

658

573

555

538

523

500

527

506

488

472

451

464

440

421

406

389

431

407

388

374

360

125

078

879

078

878

376

673

372

872

171

269

161

059

357

756

253

956

554

552

751

249

050

247

946

044

542

646

944

542

641

139

41

500

807

808

807

802

786

756

752

745

737

717

639

623

608

594

571

596

577

560

545

522

534

511

492

477

457

501

477

458

443

424

200

083

583

683

583

181

779

178

778

177

475

668

567

165

764

462

264

462

761

159

757

458

456

354

553

050

855

252

951

049

547

4

Tabl

e 7

1co

ntin

ued b w

b =

10

e sd

minus09

minus10

minus15

minus20

minus50

minusinfin

010

044

834

020

417

614

513

10

200

507

412

271

237

198

180

060

062

054

640

836

731

629

11

000

678

614

483

441

385

358

150

072

466

854

650

444

741

72

000

757

706

592

550

493

463

250

078

173

562

758

753

050

03

000

801

758

656

617

560

530

400

083

079

270

066

360

957

9

Tabl

e 7

2D

epth

of t

he c

ompr

essi

on z

one

in a

fully

cra

cked

T s

ectio

n su

bjec

ted

to e

ccen

tric

ten

sile

forc

e c

= (c

oeffi

cien

t fr

om t

able

times 1

0minus3)d

b wb

= 0

05

e sd

= 1

e sd

= 5

e sd

= 1

0e s

d =

20

e sd

= 5

0e s

d =

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

000

59

99

99

1818

1818

1822

2222

2222

2525

2525

2528

2828

2828

3131

3131

310

010

1313

1313

1325

2525

2525

3131

3131

3135

3535

3535

4040

4040

4043

4343

4343

003

023

2323

2323

4343

4343

4353

5353

5353

6261

6161

6171

6868

6868

7974

7474

740

050

2929

2929

2956

5656

5656

7168

6868

6885

7878

7878

9987

8787

8711

195

9595

950

075

3636

3636

3671

6868

6868

9283

8383

8311

295

9595

9513

010

610

610

610

614

711

611

511

511

50

100

4242

4242

4286

7979

7979

112

9595

9595

137

110

109

109

109

159

123

121

121

121

180

136

131

131

131

012

546

4646

4646

9988

8888

8813

110

610

610

610

616

012

312

112

112

118

613

913

413

413

420

915

414

614

614

60

150

5151

5151

5111

396

9696

9614

911

711

611

611

618

113

713

213

213

221

015

514

614

614

623

617

215

915

815

80

200

5959

5959

5913

811

011

011

011

018

213

713

313

313

322

016

215

115

115

125

418

516

816

716

728

420

518

318

018

0

b wb

= 0

1

e sd

= 1

e sd

= 5

e sd

= 1

0e s

d =

20

e sd

= 5

0e s

d =

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

001

013

1313

1313

2525

2525

2531

3131

3131

3535

3535

3540

4040

4040

4343

4343

430

020

1818

1818

1835

3535

3535

4343

4343

4350

5050

5050

5656

5656

5662

6161

6161

006

032

3232

3232

6261

6161

6179

7474

7474

9585

8585

8511

095

9595

9512

410

310

310

310

30

100

4242

4242

4285

7979

7979

111

9595

9595

134

110

109

109

109

155

123

121

121

121

174

135

131

131

131

015

051

5151

5151

111

9696

9696

145

117

116

116

116

175

136

132

132

132

202

154

146

146

146

226

171

159

158

158

020

059

5959

5959

135

110

110

110

110

176

137

133

133

133

211

161

151

151

151

242

183

168

167

167

269

203

183

180

180

025

067

6565

6565

156

124

122

122

122

203

156

147

147

147

243

184

168

168

168

276

209

188

185

185

306

232

206

199

199

030

075

7171

7171

176

137

133

133

133

228

174

161

160

160

271

205

185

182

182

307

234

207

201

201

338

259

227

217

216

040

090

8282

8282

213

162

152

152

152

272

206

186

183

183

320

244

215

208

207

359

277

242

229

228

392

306

266

249

245

Tabl

e 7

2co

ntin

ued

b wb

= 0

2

e sd

= 1

e sd

= 5

e sd

= 2

0e s

d =

10

e sd

= 5

0e s

d =

infin

h fd

h fd

h fd

h fd

h f d

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

002

018

1818

1818

3535

3535

3543

4343

4343

5050

5050

5056

5656

5656

6261

6161

610

040

2626

2626

2650

5050

5050

6261

6161

6173

7070

7070

8378

7878

7893

8585

8585

012

045

4545

4545

9486

8686

8612

110

410

410

410

414

512

011

911

911

916

513

513

213

213

218

414

914

314

314

30

200

5959

5959

5913

011

011

011

011

016

713

613

313

313

319

716

015

115

115

122

418

116

816

716

724

719

918

318

018

00

300

7571

7171

7116

713

713

313

313

321

217

116

116

016

024

920

118

518

218

228

122

820

620

120

130

725

122

521

721

60

400

8982

8282

8219

916

115

215

215

225

120

218

518

318

329

223

721

420

820

732

626

824

022

922

835

429

426

324

824

50

500

102

9292

9292

227

183

170

169

169

283

230

208

203

203

328

269

241

230

229

364

302

270

255

251

394

331

295

277

270

060

011

510

010

010

010

025

220

318

618

418

431

225

522

922

122

035

929

726

625

224

839

633

329

727

927

142

736

332

430

329

10

800

138

116

115

115

115

295

240

217

210

210

361

299

268

253

250

410

346

309

290

281

449

384

344

322

306

481

416

374

349

328

b wb

= 0

5

e sd

= 1

e sd

= 5

e sd

= 1

0e s

d =

20

e sd

= 5

0e s

d =

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

005

029

2929

2929

5656

5656

5670

6868

6868

8178

7878

7892

8787

8787

101

9595

9595

010

042

4242

4242

8279

7979

7910

295

9595

9511

910

910

910

910

913

412

212

112

112

114

713

313

113

113

10

300

7371

7171

7114

913

513

313

313

318

416

616

116

016

021

319

218

418

218

223

621

520

420

120

125

623

422

121

721

60

500

9792

9292

9219

617

717

016

916

923

921

720

620

320

327

324

923

623

022

930

027

626

125

325

132

429

928

327

427

00

750

122

112

111

111

111

240

218

207

204

204

290

266

251

244

243

328

303

287

278

273

359

334

317

306

298

385

360

342

330

319

100

014

312

912

812

812

827

625

223

923

323

233

130

528

928

027

537

234

632

931

730

840

537

936

134

833

643

140

738

837

436

01

250

161

145

142

142

142

306

282

267

259

256

365

339

321

310

302

408

382

364

351

339

442

417

398

384

369

469

445

426

411

394

150

017

716

015

515

515

533

330

829

128

227

739

436

834

933

732

743

941

339

438

036

647

344

842

941

539

850

147

745

844

342

42

000

206

186

178

177

177

378

352

334

322

313

442

416

397

383

369

489

464

444

429

412

524

500

481

466

446

552

529

510

495

474

Tabl

e 7

2co

ntin

ued b w

b =

10

e sd

15

10

20

50

infin

010

042

7995

109

121

131

020

059

110

133

151

167

180

060

010

018

422

024

827

129

11

000

128

232

275

308

335

358

150

015

527

732

636

439

341

72

000

177

313

367

406

438

463

250

019

734

440

044

247

450

03

000

214

370

429

472

504

530

400

024

341

447

652

055

357

9

Tabl

e 7

3D

ista

nce

from

top

fibr

e to

cen

troi

d of

tra

nsfo

rmed

1 fully

cra

cked

T s

ectio

n y

= (c

oeffi

cien

t fr

om t

able

times 1

0minus3)d

b wb

= 0

05

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

000

536

5454

5454

4659

7910

210

262

6785

105

151

108

9510

311

915

918

414

513

914

717

727

220

718

718

520

30

010

4559

5959

5954

6382

104

104

6972

8710

815

211

499

106

121

160

188

148

142

149

178

276

210

189

186

204

003

079

7777

7777

8581

9411

311

398

8899

116

158

138

113

116

129

166

207

160

151

156

183

290

221

197

193

209

005

011

195

9595

9511

497

105

121

121

125

104

110

125

163

160

127

127

137

171

224

173

160

163

188

304

231

206

199

213

007

514

811

611

611

611

614

911

711

913

213

215

612

312

313

517

018

614

513

914

717

724

518

717

117

219

432

024

421

520

821

90

100

182

136

136

136

136

180

136

133

142

142

185

141

136

145

177

211

161

151

156

184

265

202

182

181

200

336

256

225

215

225

012

521

415

515

515

515

520

915

514

615

215

221

215

914

915

518

423

417

716

316

519

028

421

519

318

920

635

126

823

522

323

00

150

243

173

173

173

173

237

172

159

162

162

237

176

161

164

190

256

192

174

174

196

302

229

203

198

212

366

279

244

231

236

020

029

520

820

820

820

828

620

518

318

118

128

420

718

518

320

329

622

119

719

220

933

525

422

321

422

439

330

126

224

624

7

b wb

= 0

1

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

001

046

5959

5959

6267

8410

410

487

8294

111

152

152

127

127

136

168

250

202

186

184

201

357

290

259

245

246

002

063

6868

6868

7675

8910

810

899

9099

115

155

161

133

131

140

170

256

207

189

187

203

361

294

262

248

248

006

012

510

310

310

310

312

810

711

212

612

614

311

912

013

116

619

415

714

915

418

027

922

520

419

921

237

830

827

425

825

60

100

178

136

136

136

136

174

137

134

142

142

183

146

140

147

177

224

179

167

168

190

301

243

219

211

221

394

322

286

268

264

015

023

717

317

317

317

322

617

115

916

216

222

917

716

416

619

026

020

618

818

520

232

726

423

622

623

241

333

930

028

127

40

200

288

208

208

208

208

272

203

183

181

181

269

207

187

184

203

292

231

207

201

214

351

284

253

240

242

431

354

314

293

283

025

033

224

024

024

024

031

223

320

619

919

930

623

420

820

221

532

125

422

621

722

637

430

326

925

425

344

736

932

730

429

20

300

372

269

269

269

269

348

260

227

217

217

339

259

228

218

227

348

276

244

232

237

394

321

284

267

262

463

384

340

316

301

040

043

832

132

132

132

141

030

926

725

025

039

630

626

625

025

039

731

627

826

125

843

235

431

229

128

249

241

036

433

731

8

1T

he s

ame

tabl

e m

ay b

e us

ed fo

r ag

e-ad

just

ed t

rans

form

ed s

ectio

n re

plac

ing

b

y ndash

= E

s[1 +

(t

t0)

]E c

(t0)

Tabl

e 7

3co

ntin

ued

b wb

= 0

2

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

002

064

6868

6868

8581

9210

810

811

710

510

912

115

519

717

016

216

418

430

826

824

723

724

042

537

534

432

531

20

040

9386

8686

8610

796

103

117

117

134

118

119

129

161

208

179

169

170

188

315

274

252

242

244

430

380

348

329

315

012

019

415

115

115

115

118

415

114

515

015

019

616

415

615

918

225

021

319

819

420

734

229

827

326

125

944

839

736

434

432

70

200

274

208

208

208

208

250

200

183

181

181

250

206

189

187

203

287

244

224

217

224

367

320

293

279

273

465

413

379

357

339

030

035

526

926

926

926

931

825

322

621

721

730

725

222

821

922

732

928

025

524

424

539

634

731

730

029

148

543

239

737

435

30

400

419

321

321

321

321

375

299

264

250

250

357

294

263

250

250

366

313

284

269

265

422

371

339

320

307

503

449

413

389

366

050

047

236

636

636

636

642

334

129

927

927

939

933

129

527

727

139

934

331

129

328

444

639

336

033

932

352

046

642

940

437

90

600

516

406

406

406

406

464

377

331

307

307

437

365

324

303

291

429

370

335

315

302

468

414

379

357

338

536

482

444

419

391

080

058

547

247

247

247

253

143

938

735

735

749

942

237

634

932

848

141

938

035

533

550

845

241

439

036

656

551

147

244

541

5

b wb

= 0

5

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h f d

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

005

010

195

9595

9512

011

211

412

112

115

614

614

414

716

324

322

922

121

822

236

134

433

232

431

748

246

344

943

842

30

100

154

136

136

136

136

152

140

138

142

142

179

166

162

163

177

256

241

233

229

231

368

352

339

331

323

487

468

454

442

428

030

031

526

926

926

926

926

223

622

221

721

725

923

922

722

322

730

528

727

626

926

739

838

036

735

734

750

548

747

246

044

40

500

424

366

366

366

366

346

312

291

279

279

324

299

284

274

271

348

328

314

306

299

424

406

392

382

370

522

504

489

476

460

075

052

045

745

745

745

742

838

836

234

534

539

236

334

333

032

039

437

335

734

733

645

543

642

141

039

654

252

450

849

647

91

000

589

524

524

524

524

491

449

420

400

400

447

416

394

378

362

434

412

395

383

369

482

463

447

435

420

560

542

526

514

496

125

064

057

755

755

757

754

250

046

844

644

649

346

143

741

939

947

044

742

941

539

950

748

747

145

944

357

755

954

353

151

21

500

680

620

620

620

620

584

541

509

485

485

532

500

474

456

433

501

477

459

444

427

529

509

493

480

463

593

574

559

546

528

200

073

868

368

368

368

364

860

757

454

954

959

456

253

651

649

055

352

951

049

547

456

854

853

251

950

062

160

358

757

455

5

Tabl

e 7

3co

ntin

ued b w

b =

10

cd

12

35

75

10

010

013

614

217

726

438

350

40

200

208

181

203

278

391

509

060

040

630

729

133

042

152

81

000

524

400

362

375

448

545

150

061

948

543

342

347

956

52

000

683

549

489

464

506

583

250

072

859

953

650

053

159

93

000

762

640

574

531

553

615

400

080

969

963

558

359

264

2

Tabl

e 7

4M

omen

t of

iner

tia o

f a t

rans

form

ed fu

lly c

rack

ed T

sec

tion

abou

t ce

ntro

idal

axi

s I =

(coe

ffici

ent

from

tab

le times

10minus4

)bd3

b wb

= 0

05

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

000

54

55

55

55

710

107

78

1126

1718

1820

3246

5050

5159

9911

111

411

411

80

010

99

99

99

1011

1414

1111

1215

2921

2222

2435

5054

5454

6210

211

511

711

712

10

030

2727

2727

2727

2728

3030

2828

2931

4336

3838

3949

6268

6869

7511

212

713

013

113

40

050

4343

4343

4343

4344

4646

4445

4547

5851

5353

5463

7582

8383

8912

213

914

314

314

60

075

6263

6363

6362

6364

6565

6264

6465

7568

7272

7280

8999

100

100

105

133

153

159

159

161

010

079

8282

8282

7982

8383

8379

8383

8492

8490

9090

9710

311

517

711

712

114

516

717

417

517

60

125

9510

110

110

110

195

101

101

101

101

9610

110

210

210

999

107

108

108

114

116

130

133

134

137

155

181

189

190

191

015

011

011

811

811

811

811

011

811

911

911

911

111

811

912

012

511

312

312

512

513

012

914

514

915

015

316

619

420

320

520

60

200

137

151

151

151

151

138

151

153

153

153

138

151

154

154

157

139

155

158

159

162

152

174

180

181

183

185

219

231

234

235

b wb

= 0

1

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

001

09

99

99

1010

1114

1413

1314

1629

2931

3132

4175

8587

8791

157

185

195

197

198

002

018

1818

1818

1919

2022

2221

2222

2436

3639

3940

4880

9194

9497

161

190

200

203

204

006

051

5151

5151

5152

5253

5352

5454

5564

6368

6869

7510

211

612

012

012

217

720

922

222

522

60

100

7982

8282

8280

8283

8383

8084

8484

9288

9697

9710

212

213

914

414

514

719

222

824

224

724

80

150

111

118

118

118

118

111

118

119

119

119

112

119

120

120

125

117

128

131

131

134

145

167

174

176

177

210

250

267

273

275

020

013

815

115

115

115

114

015

115

315

315

314

015

115

415

415

714

315

916

316

316

516

719

320

320

520

622

727

229

129

930

10

250

162

181

181

181

181

165

182

186

186

186

165

182

186

186

189

167

187

193

194

196

188

218

230

233

234

242

292

314

323

326

030

018

320

920

920

920

918

721

021

621

721

718

821

021

721

821

918

921

422

222

422

520

624

225

626

126

225

731

133

634

735

10

400

218

258

258

258

258

225

261

273

276

276

228

261

273

276

277

229

264

277

281

282

241

286

305

313

315

284

348

378

392

398

Tabl

e 7

4co

ntin

ued

b wb

= 0

2

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h f d

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

002

018

1818

1818

1919

2022

2224

2425

2636

4853

5354

5911

813

614

314

514

624

829

031

332

332

70

040

3535

3535

3536

3636

3838

3940

4041

5161

6667

6772

128

147

155

157

158

255

298

321

332

337

012

093

9797

9797

9497

9798

9895

9910

010

010

510

811

812

112

112

316

418

819

820

220

228

032

835

436

737

40

200

140

151

151

151

151

143

151

153

153

153

143

152

154

154

157

151

165

170

171

173

197

226

239

244

245

303

356

386

401

409

030

018

720

920

920

920

919

421

121

721

721

719

521

121

721

821

919

922

022

823

023

123

527

028

729

529

733

139

042

344

145

20

400

224

258

258

258

258

236

263

273

276

276

239

264

274

276

277

241

269

281

285

286

270

311

332

342

346

356

421

458

479

493

050

025

530

130

130

130

127

231

032

533

033

027

831

132

533

033

227

931

433

033

733

930

234

937

538

739

338

045

049

251

553

20

600

280

339

339

339

339

303

351

372

380

380

312

354

373

381

383

313

355

376

385

389

331

385

414

429

438

402

478

524

550

570

080

032

040

140

140

140

135

442

045

346

946

936

842

745

747

147

837

242

945

947

448

138

344

948

750

852

144

352

858

261

464

1

b wb

= 0

5

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

005

043

4343

4343

4445

4546

4651

5252

5358

9299

101

102

103

225

244

255

261

264

483

523

551

569

585

010

081

8282

8282

8283

8383

8385

8888

8892

120

128

131

132

133

245

265

278

284

287

497

537

566

584

602

030

019

720

920

920

920

920

721

421

721

721

720

721

521

821

821

922

423

624

224

524

532

134

636

137

037

654

759

262

364

566

50

500

276

301

301

301

301

303

319

327

330

330

307

321

328

331

332

314

332

341

346

348

390

419

438

449

458

595

643

677

701

725

075

034

438

738

738

738

739

642

444

044

844

841

043

344

645

245

541

343

745

145

946

446

850

352

654

055

364

970

274

076

779

61

000

394

452

452

452

452

469

508

532

546

546

493

525

545

556

564

498

529

548

560

568

539

578

606

624

640

699

756

798

828

861

125

043

150

250

250

250

252

757

760

962

962

956

360

463

064

665

957

361

063

565

066

360

364

767

970

072

074

680

785

288

592

21

500

459

542

542

542

542

575

634

674

700

700

622

671

704

725

744

639

682

712

731

749

661

710

746

770

795

789

853

902

938

980

200

050

160

260

260

260

264

872

477

981

681

671

778

182

685

788

975

080

584

487

189

976

282

186

489

592

986

693

899

310

3410

85

Tabl

e 7

4co

ntin

ued

b wb

= 1

0

cd

12

35

75

10

010

082

8392

159

390

858

020

015

115

315

721

242

788

20

600

339

380

383

405

568

974

100

045

254

656

457

269

610

601

500

542

700

744

753

839

1159

200

060

281

688

990

796

812

492

500

645

906

1007

1041

1083

1333

300

067

797

811

0611

5811

8814

104

000

722

1086

1261

1354

1370

1547





α = Es

Ec

= 200

30 = 6667






2 times 015

= 0200m (79 in)


I = 0302003

3 + (15 minus 03)0120122

12 + 0142

+ 6667(0004)10002 + 5667(00006)0152

= 3054 times 10minus3 m4 (353 ft4)


ψ = 1000 times 103





= 2182 MPa (3165ksi)






c = 0444m (175 in)


A = 03073m2 I = 3173 times 10minus3 m4




N = minus800kN


= 3832kN m (3400kip in)


εO = 1




ψ = 1

30 times 109 3832 times 103

3173 times 10minus3












yc

r2c + εcs(t t0)

yc

r2c (727)


∆σs = Es(∆εO + ∆ψys) (729)

where

r2c = IcAc (730)








Figu

re7

9A

naly

sis

of c

hang

es in

str

ain

and

stre

ss d

ue t

o cr

eep

and

shri

nkag

e of

a fu

lly c

rack

ed r

einf

orce

d co

ncre

te s

ectio

n (E

xam

ple

73)

(a

) effe

ctiv

e ar

ea (

b) s

trai

n an

d st

ress

dis

trib

utio

ns a

t t0

(c) c

hang

es in

str

ain

and

stre

ss d

ue to

cre

ep a

nd s

hrin

kage

(d)

str

ain

and

stre

ss a

t t

Tabl

e 7

5C

urva

ture

red

uctio

n fa

ctor

κ fo

r a

fully

cra

cked

T s

ectio

n (fo

r us

e in

Equ

atio

n (6

23)

) κ =

coe

ffici

ent

from

tab

le times

10minus3

b wb

= 0

05

cd

= 1

0c

d=

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

ndash

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

000

543

160

160

160

160

170

236

420

623

623

408

426

522

661

862

774

773

781

809

891

928

927

927

929

942

973

971

971

971

973

001

036

9494

9494

105

140

270

455

455

266

276

356

496

758

636

632

643

681

804

868

865

864

868

892

948

945

944

943

947

003

059

5858

5858

8075

126

229

229

146

135

171

258

516

394

381

387

425

582

700

691

686

691

736

865

857

852

851

860

005

090

6565

6565

9874

9816

516

513

611

112

818

539

731

228

928

931

846

160

058

457

658

063

080

178

878

077

778

90

075

127

8181

8181

128

8692

134

134

149

110

112

149

315

274

240

233

252

372

524

500

488

489

539

741

721

710

705

717

010

016

110

010

010

010

015

710

196

123

123

170

118

110

134

267

262

219

206

217

318

479

447

431

429

474

697

670

655

649

660

012

519

311

811

811

811

818

611

810

312

012

019

213

011

412

923

726

321

119

319

828

245

141

239

238

842

766

363

161

360

461

40

150

223

136

136

136

136

213

134

112

121

121

215

143

121

128

218

270

210

187

187

257

435

389

366

358

392

637

600

579

568

576

020

027

617

117

117

117

126

216

713

313

013

025

817

113

813

519

629

221

918

718

022

842

036

333

332

134

460

255

652

951

551

8

b wb

= 0

1

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

ndash

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

001

041

9494

9494

160

179

286

455

455

387

387

430

530

758

758

759

760

771

834

925

925

924

924

930

973

972

971

971

971

002

047

6363

6363

107

111

175

300

300

256

250

281

366

613

618

617

616

631

717

863

863

860

860

870

949

947

945

944

944

006

010

271

7171

7111

184

9814

914

916

413

914

418

335

839

037

737

038

046

969

669

168

468

169

786

986

385

785

485

30

100

156

100

100

100

100

149

103

9712

312

317

413

412

514

526

732

430

028

629

036

060

259

057

957

358

980

979

879

078

478

20

150

215

136

136

136

136

198

133

113

121

121

204

150

130

135

218

301

265

245

241

292

535

516

499

490

501

754

738

725

717

712

020

026

717

117

117

117

124

316

413

313

013

023

917

214

313

919

630

325

723

022

125

649

947

145

143

844

471

469

367

766

665

90

250

313

203

203

203

203

284

193

154

142

142

273

196

160

148

186

314

260

228

214

237

479

445

420

405

405

686

660

640

626

616

030

035

323

323

323

323

332

122

117

415

715

730

521

917

716

018

433

026

823

221

422

646

942

940

138

337

866

463

461

259

658

30

400

421

288

288

288

288

385

271

214

186

186

362

265

213

186

189

365

292

249

224

220

466

417

382

359

345

638

601

573

553

534

Tabl

e 7

5co

ntin

ued

b wb

= 0

2

cd

= 1

0c

d =

20

cd

= 3

0c

d =

75

cd

= 5

0c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

ndash

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

002

049

6363

6363

150

148

194

300

300

356

355

366

417

613

733

741

739

741

775

919

921

921

920

921

973

973

972

971

971

004

071

6161

6161

114

103

124

189

189

241

233

238

274

448

590

596

593

594

637

853

857

856

854

855

949

948

947

945

944

012

016

911

411

411

411

415

311

510

512

012

018

515

814

615

424

237

837

336

335

939

168

468

668

167

667

586

986

786

385

985

40

200

251

171

171

171

171

214

157

132

130

130

215

174

152

147

196

329

313

296

287

305

595

592

583

575

571

811

807

800

794

786

030

033

423

323

323

323

328

221

017

215

715

726

320

917

816

418

432

229

527

325

826

353

652

751

450

249

376

075

274

273

372

20

400

400

288

288

288

288

340

257

211

186

186

310

247

208

187

189

336

301

273

254

248

507

492

475

461

446

723

712

700

689

674

050

045

533

533

533

533

539

030

024

621

621

635

328

323

821

220

035

631

528

326

024

649

347

445

443

741

769

868

466

965

663

80

600

500

376

376

376

376

433

338

280

244

244

391

316

268

236

215

378

332

297

271

250

489

466

443

423

400

680

663

647

632

611

080

057

144

544

544

544

550

340

333

829

529

545

737

632

128

324

742

437

133

029

926

749

646

643

841

538

465

963

861

759

957

3

b wb

= 0

5

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

ndash

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

005

074

6565

6565

135

130

135

165

165

305

309

307

316

397

691

701

702

701

707

909

912

912

911

972

972

972

972

971

010

012

310

010

010

010

012

411

411

012

312

321

521

320

921

226

754

255

255

355

155

683

784

284

384

284

194

794

794

794

694

50

300

287

233

233

233

233

211

184

166

157

157

206

192

180

172

184

354

356

352

347

344

661

667

667

665

660

866

866

865

864

861

050

040

133

533

533

533

529

626

023

421

621

625

923

822

020

720

032

432

131

330

529

657

658

057

957

556

780

880

880

780

479

90

750

500

429

429

429

429

382

340

307

283

283

325

298

276

258

239

335

326

315

304

289

525

526

523

517

506

758

758

755

752

744

100

057

250

050

050

050

044

940

536

934

134

138

235

232

730

627

936

034

733

432

130

250

350

249

649

047

672

472

371

971

570

51

250

625

556

556

556

556

504

458

421

391

391

431

400

372

349

318

388

374

358

344

321

496

493

486

478

462

701

699

694

689

678

150

066

760

060

060

060

054

950

346

543

443

447

444

141

338

835

341

740

038

436

834

349

749

248

447

545

868

568

267

767

165

92

000

727

667

667

667

667

618

574

536

504

504

542

509

480

454

415

470

451

433

415

387

512

504

494

484

463

669

664

658

650

636

Tabl

e 7

5co

ntin

ued

b wb

= 1

0

cd

ndash

12

35

75

10

010

010

012

326

766

090

297

10

200

171

130

196

510

826

945

060

037

624

421

533

664

686

31

000

500

341

279

318

563

805

150

060

043

435

333

751

575

52

000

667

504

415

367

497

722

250

071

455

846

639

949

369

93

000

750

602

510

431

496

685

400

080

066

857

948

751

567

0


η = AcA (731)

κ = IcI (732)
















Ec(t t0) = 30 times 109

1 + 075 times 25 = 1043GPa (1500ksi)

α(t t0) = 200

1043 = 1917


A = 03648m2 (560 in2)







η = 02766

03648 = 07582 κ =

1756

7301 = 02404










∆ψ = 0240425403 minus 35 (minus022)






= 0876MPa (0127ksi)

















M = M1 + M2 (733)

N = N1 + N2 (734)








(∆ε)1 = (∆εO)1 + (∆ψ)1y (735)

(∆ε)2 = (∆εO)2 + (∆ψ)2 y (736)


∆ε = (∆ε)1 + (∆ε)2 (737)












γ = γ(t) = d

dyσ(t) (742)



M1 = minusIγ (744)


(∆εO)1 = minus1

Ec

σO (745)

(∆ψ)1 = minus1

Ec

γ (746)




77 Flow chart












α = 20024 = 833

A = 00881 + 833(930 + 1000)10minus6 = 01042m2










= 10948MPa (1588ksi)




Ec = 24 times 109

1 + 24 times 08 = 8215GPa (1192ksi)

α = 200

8215 = 2433


A = 00881 + 2433(930 + 1000)10minus6 = 01351m2





= 08759 times 106 N (1969kip)







= 4407MPa (06392ksi)








α = 20035 = 571


A = 00881 + 571(930 + 1000)10minus6 = 00991m2





(∆ε)1 = 6152 times 106





Transformed area is

A = 571(930 + 1000)10minus6 = 00110m2


N2 = 1200 minus 6098 = 5902kN (113kip)


(∆ε)2 = 5902 times 103




= 3060MPa (444ksi)



(∆ε)2 = 5902 times 103





10948 minus 2479 + 352 + 3060 = 11881MPa (1723ksi)














α = 20030 = 6667



A = 02768 + 6667(1600 + 1200 + 400)10minus6 = 02981m2













= 0955 times 106N




(∆εO)1 = 3490 times 106












es = 6299 times 103




c = 0263m


















and curvature














(∆εO)1 = minus1


(∆ψ)1 = minus1






A = 635 in2 B = 5824 in3 I = 1418 times 103 in4












79 General





Note





Chapter 8

81 Introduction





















σsr = NrAs (82)








σs2 = NAs (83)

εs2 = NEsAs (84)


εsm = ∆ll (85)




εs1 = εc1 = N

Ec(Ac + αAs) =

N

EcA1

(87)






σs2

(88)



σs2

(89)




ζ = 1 minus σsr

σs2

2




or

ζ = 1 minus Nr

N2





σs2

2













σsr = Nr

As

= 111MPa σs2 = 249MPa









Mr = W1 fct (815)






where


σs2

2

= 1 minus β1β2Mr

M2

(817)





ψ = M

EI(819)



or


d(820)


ψm = (1 minus ζ)ψ1 + ζψ2 (821)



(EI)m = M

ψm

(822)




ψ1 = M

EcI1

(823)

ψ2 = M

EcI2

(824)
















I1 = 00191m4 W1 = 00488m3



c = 0191m (752 in)


I2 = 000543m4














e = MN (825)



Nr = fct 1

A +

e

Wbot

minus1

1(826)

Mr = eNr (827)







where


σs2

2

(829)

or


M2

(830)

or




N2

(831)



σsr

σs2

= Mr

M =

Nr

N(832)




ψm = (1 minus ζ)ψ1 + ζψ2 (834)







A1 = 0336m2



e = 250



Mr = 138kN-m


ζ = 085




c = 0241m (949 in)


y = 0195m (768 in)


A2 = 0115m2 I2 = 000544m4



ψ2 = 209 times 103
















Mr = fct minus N

A1W1 (835)












where


N2

(837)


α = EsEc (839)








ψm = (1 minus ζ)ψ1 + ζψ2 (840)

where


M2

(841)

Mr = fctW1 (842)




ψm = (1 minus ζ)ψ1 + ζψ2 (844)

where


M2

= 1 minus β1β2Nr

N2

(845)






e = M

N =

Mr

Nr

(846)


Mr = efct 1

A1

+ e

W1

minus1

(847)






σ1 max

2

(849)




























136 2

= 082





136 2

= 091


State 1

ψ1(t0) = 136 times 103


State 2

ψ2(t0) = 136 times 103







Ec(t t0) = 30 times 109

1 + 08 times 25 = 10GPa

α = Es

Ec(t t0) =

200

10 = 20







8724 times 10minus3 = 0795




κ2 = 1190

4277 = 0278





3534 times 10minus3









2762 times 10minus3











96

= 00239m

= 239mm (0948 in)

















σ1 max = 2323 + 8580 = 10903MPa




σ1 max

2


109032

= 095










= 1012mm (399 in)















θ1 = T

GcJ1

(850)



c 1 minus b4

12c4 (851)


τmax = T

microbc2(852)






cb 10 15 175 20 25 30 40 60 80 100 infin

0208 0231 0239 0246 0258 0267 0282 0299 0307 0313 0333


τt = T2A0 (854)







τt = T2hb (855)






F1 = F3 = 1

2hF2 = F4 =

1

2b (856)


F5 = F6 = F7 = F8 = x

2hb (857)


F9 = F11 = minus 1

2b sin α1

F10 = F12 = minus 1

2h sin α2

(858)



θ = T

x 12

i = 1F

2l

AEi

(859)


Ai = Av

x

s(860)



A9 = A11 = th cos α1 (861)

and

A10 = A12 = tb cos α2 (862)


















ζ = 1 minus 10(05) 0500

10962

= 0896





96 [0 + 10(3547 times 10minus6) + 0] = 341 in























811 General






Notes











91 Introduction



Chapter 9








ψ(t0) = κsψc (91)

(∆ψ)φ = ψ(t0)φκφ (92)



d κcs (93)


ψc = M

Ec(t0)Ig

(94)






















10s1 09 0 to 02 ndash F1

08

10s2 09 0 to 02 ndash F2

08

1010 0 to 02 20 F3

3040

101 09 0 to 02 20 F4

3040

1020

08 0 to 02 30 F540

100 20 F6

3040

1020

2 08 to 10 01 30 F740

1002 20 F8

3040


0cs2 08 to 10 20 F10

02













D1(t0) = κs1Dc (95)

D2(t0) = κs2Dc (96)



(∆D1)φ = D1(t0)φκφ1 (97)

(∆D2)φ = D2(t0)φκφ2 (98)





I (99)

or


d(910)







l2

8(911)





A (912)



A (913)






l2

2(914)





8d(915)


8d(916)


κcs = minusAcd

Iyc (917)





A (918)









Ac

A (920)






D1 = D1(t0) + (∆D1)φ + (∆D1)cs (921)

D2 = D2(t0) + (∆D2)φ + (∆D2)cs (922)





D = (1 minus ζ)D1 + ζD2 (923)



M(924)


Mr = fctW1 (925)









Ie = Mr

Mm

Ig + 1 minus Mr

Mm

I2 with M Mr (926)

where








ρ = As


Aprimesbd

= 015 per cent



Ig = bh3

12 =

03 times (065)3


Basic deflection

Dc = 5

384

ql4

Ec(t0)Ig

= 5

384




κs1 = 092 κφ1 = 079 κcs1 = 027

κs2 = 380 κφ2 = 014 κcs2 = 097


D1(t0) = 092 times 440 = 405mm

D2(t0) = 380 times 440 = 1672mm


(∆D1)φ = 405 times 25 times 079 = 800mm

(∆D2)φ = 1672 times 25 times 014 = 585mm





8 times 06 = 090mm


8 times 06 = 323mm


D1 = 405 + 800 + 090 = 1295mm

D2 = 1672 + 585 + 323 = 258mm


Mr bh2

6fct =

03 times (065)2

6 25 times 106 = 528kN-m


M = 17 times 82

8 = 1360kN-m



1360 = 081


(1 minus 081)1295 + 081 times 258 = 234mm (0920 in)















κs1(1 + κφ1)(927)



κs2(1 + κφ2)(928)


MO minusN |y12| 1




ψ2N = minusMO 1

(slope)OD

minus 1

(slope)AB (930)












D = (1 minus ζ)D1 + ζD2 (931)



ζ = 1 minus β1β2

Mr

M(932)

0 for M lt Mr (933)




ζ = 1 minusMO

M(934)

0 for M lt MO (935)


Mr = fct minus N

A1W1 (936)














κs1 = 092 κφ1 = 079 κs2 = 38 κφ2 = 014


D1(t0) = 405mm D2(t0) = 1672mm


(∆D1)φ = 800mm (∆D2)φ = 585mm



D1 = 405 + 800 = 1205mm


D2 = 1672 + 585 = 2257mm


M = 1360kN-m



03 times 065 03(065)2

6 = 961kN-m


|y12| = 331 minus 145 = 186mm



1 minus (09238) = 981kN-m




ζ = 1 minus 981

1360 = 028











approximation

D Dcκth

d 3






Figu

re9

8G

loba

l coe

ffici

ent

κt f

or c

alcu

latio

n of

inst

anta

neou

s pl

us c

reep

def

lect

ions

of u

ncra

cked

or

crac

ked

mem

bers

by

Equa

tion

(93

8)















M(940)






l 2

8dκtcs (942)




l 2

2dκtcs (942a)










Mr = 528kN-m M = 1360kN-m ζ = 081

αρ = 200

30 times

06

100 = 004

Mr

M =

528

1360 = 039


44 times 38 065

06 3


100 = 206mm



8(06) = 28mm


D = 206 + 28 = 234mm (094 in)







8 times 0275 = 2327kN


M = 20 times 60

100 times

82

8 = 960kN-m



03 times 065 03(065)2

6 = 780kN-m

Mr

M =

78

96 = 081



Total steel ratio

ρ = (500 + 200)10minus6


α = 200

30 = 667 αρ = 0026


960 times 103

30 times 109(03 times 065312)

82

96 = 311mm


03 times 060 = 00011


D 311 times 44065

0603

(1 minus 20 times 00011) = 171mm (0671 in)









δ = l 2

96 (ψ1 + 10ψ2 + ψ3) (943)

where






member


D = δEF + 1

2(δAB + δDC) (944)

or

D = δHI + 1

2(δAD + δBC) (945)









ψx = 12


12



Ig slab = h3

12(1 minus ν2)(948)


ψbeam = M

EcIg beam

(949)

















(EI) slab

(EI) beam cl = 00 cl = 01 cl = 02



08 00 420 378 230 301 262 131 189 155 05702 316 271 149 237 192 088 159 116 04005 146 195 099 191 138 059 136 085 02810 191 134 063 154 095 038 117 058 01820 147 083 036 124 058 022 100 036 01140 116 048 019 103 033 012 089 020 006

06 00 327 321 099 234 228 040 143 137 00802 256 246 063 189 178 027 119 108 00605 201 187 040 150 134 017 098 098 00410 153 135 025 116 096 011 079 059 00320 110 087 013 085 061 006 061 037 00240 077 051 007 063 035 003 048 022 001

04 00 284 284 031 204 204 00402 231 230 020 166 165 00305 183 181 012 131 128 00210 137 134 007 098 094 00120 093 088 004 066 061 00140 059 053 002 042 036 000

(El)slab = Ec

h3









δ = (1 minus ζ)δ1 + ζδ2 (952)









Ig slab = (02)3





000482ql 4

EcIg slab

= 560mm (0221 in)


000342ql 4

EcIg slab

= 397mm (0157 in)








κs1 = 098 κφ1 = 093


κs2 = 70 κφ2 = 014


δ1 = 397(098)[1 + 25(093)] = 1294mm

δ2 = 397(70)[1 + 25(014)] = 3752mm



6 = 133kN-m



1 minus β1β2

Mr

M = 1 minus 10(05)

133

186 = 064


δAB = (1 minus 064)1294 + 064(3752) = 2867mm




κsl = 098 κφ1 = 095


δEF = 163(098)[1 + 25(095)] = 538mm


D = 2867 + 538 = 3405mm (1341 in)














ψ2 = 12[281 minus 02 (minus144)] 103



δABbasic = (655)2



ψ1 = ψ3 = 12[minus144 minus 02(281)]103


ψ2 = 12[101 minus 02(187)] 103



δEFbasic = (700)2



Dbasic = 649 + 072 = 721mm (0284 in)



Tabl

e 9

3D

eflec

tion

at c

entr

e of

pan

el c

alcu

late

d fr

om t

wo

stri

p pa

tter

ns E

xam

ple

96

Strip

Bend

ing

Curv

atur

eCu

rvat

ure

Inte

r-D

eflec

tion

mom

ent a

tG

eom

etric

alco

effic

ient

s an

dco

effic

ient

s an

dpo

latio

nof

Basic

mid

-spa

nCr

acki

ngpr

oper

ties

of s

ectio

nde

flect

ion

inde

flect

ion

inco

effic

ient

strip

defle

ctio

nsp

an

mom

ent

at m

id-s

pan

uncr

acke

d st

ate

fully

-cra

cked

sta

te

basic

MM

r

A sd

κ

s1κ

1

1

κs2

κ

2

2

(m

m)

(kN

-m)

(kN

-m)

(mm

2 )(m

)(1

0minus3)

(mm

)(m

m)

(mm

)

AB

649

281

133

900

017

05

290

960

8819

91

468

016

424

60

7637

05

EF0

7210

113

335

00

155

226

099

097

244

1ndash

ndashndash

244

Firs

t es

timat

e of

defl

ectio

n at

cen

tre

of p

anel

D =

A

B +

EF

= 3

705

+ 2

44

= 39

49

mm

BC3

5518

613

36

500

155

419

098

093

115

77

480

1435

85

064

271

12

003

004

089

AD

21

4641

220

44

top

092

079

400

821

009

146

80

7512

03

kN-m

kN-m

500

045

222

bot

tom

HI

451

187

133

600

017

03

530

970

9114

33

655

013

391

40

6430

21

Seco

nd e

stim

ate

of d

eflec

tion

at c

entr

e of

pan

el D

= ndashsup1 sup2

(BC

+

AD)+

H

I=

ndashsup1 sup2(2

711

+12

03)

+30

21

= 49

78

Defl

ectio

n at

cen

tre

of p

anel

incl

udin

g ef

fect

s of

cre

ep a

nd c

rack

ing

=39

4+

497

8

2=

446

mm

(17

6in

)

1W

hen

M

Mr c

rack

ing

does

not

occ

ur

=

1 and

the

col

umns

for

κs2 κ

2

2 a

nd

are

left

bla

nk

2T

he e

dge

beam

is t

reat

ed a

s a

T-s

ectio

n w

ith fl

ange

wid

th =

05

0m

3T

his

line

give

s Aprime

s dprime

and

prime f

or t

he t

op r

einf

orce

men

t of

str

ip A

D










(∆D1)cs = 300 times 10minus6(030)(655)2

8(017) = 288mm


(∆D2)cs = 300 times 10minus6(112)(655)2

8(017) = 1061mm


(∆D)cs = (1 minus 076)288 + 076(1061) = 875mm


δAB = δDC = 07(875) = 613mm



(∆D1)cs = 300 times 10minus6(010)(700)2

8(0155) = 119mm


(∆D)cs = 119mm


δEF = 05(119) = 060mm


(∆D)cs = 613 + 060 = 673mm



= 1










02207= 0660MPa





910 General






Notes












Chapter 10

101 Introduction
































kpart2T

partx2 + k

part2T

party2 + Q = pc

partT

partt(101)

where




kpartT

partxnx + k

partT

partyny + q = 0 (102)




where


qs = αaIs (104)













cooling










h = hc + hr (108)




where

qcr = qc + qr (1010)












8588603575

1516110613





(1 minus ν2)T(y) (1011)










09

coefficient a


coefficient c

09 08 09



(or per degF)


∆N = thickness


∆M = thickness



p = minusd(∆N)

ds(1014)

q = minus d2(∆M)

ds2(1015)


















∆σ = minus ∆N

A +

∆Mx

Ix

y + ∆My

Iy

x (1021)





∆εO = minus∆N

EA(1023)


EIx

(1024)


EIy

(1025)







2(1026)

D3 = ∆εOl (1027)


∆εO = αt

A T b dy (1028)

∆ψ = αt

I T b ydy (1029)















125




125





∆N = minus(2652 times 106) 035102

0y5 dy + 25

120

102y5 dy


∆M = minus(2652 times 106) 035 102


+2512



∆εo

∆ψ = 1

30 times 109 times

2229 times 106

0877


01615






or


for y = 0769 to 0969m





2 +

24



f11 = l

3EIAB

+ l

3EIBC

= 1


f21 = l

6EIBC



(f11 + f12)F1 = minusD1

F1 = minusD1

f11 + f12

= 8675 times 103 N-m
















Ec(t t0) = Ec(t0)

1 + χφ(t t0)(1031)

where













2 ti and ti+ 1



αt i

j = 1

(∆T)j (1032)



αt i

j = 1

(∆T)j + i

j = 1

(∆σrestraint)j

Ec(tj) [1 + φ(ti + 1

2 tj)] = 0 (1033)







1 + φ(ti + 12 ti)

αti

j = 1

(∆T )j

+ i minus 1

j = 1

(∆σrestraint)j

Ec(tj) [1 + φ (ti + 1

2 tj)] (1034)






φ(2 1) = 056 φ(6 1) = 060 φ(50 1) = 082

φ(6 4) = 055 φ(50 4) = 076 φ(50 28) = 048







1 + 056 (053) = minus0150[αtTc maxEc(28)]


1 + 055 (053 + 047)

minus 0150

044 (1 + 060)



1 + 048


044 (1 + 082) minus

0214

073(1 + 076)

= 0727 [αtTc maxEc(28)]





























f11 = M 2

u1 dl

EcIf12 =

Mu1Mu2dl

EcIf22 =

M 2u2 dl

EcI





1012 General






Notes














Control of cracking

111 Introduction




Chapter 11



















Figu

re11

1A

rei

nfor

ced

conc

rete

mem

ber

subj

ecte

d to

(a)

axi

al fo

rce

N (

b) im

pose

d en

d di

spla

cem

ent

D












Mri = fctiW1 (112)



M = αtEcI1 ∆T

h (0 le ∆T le ∆T1) (113)



∆T1 = Mr1h

αtEcI1

(114)


1

EcIrm

= (1 minus ζ) 1

EcI1

+ ζ 1

EcI2 (115)




Irm = I1I2


(116)



M = αt∆T

h EcI1 1 +

i srm

l I1

Irm

minus 1minus1

(with ∆Ti ∆T ∆Ti + 1) (117)



Mri = αt∆Ti

h EcI11 +

(i minus 1)srm

l I1

Irm

minus1minus1

(118)





εs2 = M

yCTAsEs

(1110)















19758

210172

310582

411090

511698

6122106

7129114

8136122



Nri = fctiA1 (1112)

N = EcA1ε (0 ε ε1) (1113)

ε1 = Nr1

EcA1

(1114)

1

EcArm

= (1 minus ζ) 1

EcA1

+ ζ 1

EcA2 (1115)

Arm = A1A2


(1116)



N = εEcA11 + i srm

l A1

Arm

minus 1 minus1



l A1

Arm

minus 1 minus1

(1118)

where











is formed

Crack number ii


184 times 10minus6

394100

2362 times 10minus6

430246

3700 times 10minus6

476334


534411











Nr = fctA1 (1119)




ρmin y = fct

fy 1









fct = N

A1

+ M

W1

(1121)



fybd(1122)







ρmin y = As min y

bd 024

fct

fy

(1123)












es = M


εs2 = (NesyCT) + N

EsAs

(1126)

σs2 = Esεs2 (1127)

As = (NesyCT) + N

σs2

(1128)

where



















Figu

re11

6C

hang

e in

ste

el s

tres

s

s2 in

a r

ecta

ngul

ar c

rack

ed s

ectio

n du

e to

a n

orm

al fo

rce

N a

nd a

ben

ding

mom

ent

M T

he v

alue

of

s2 is

equ

al t

o th

e va

lue

read

from

the

gra

ph m

ultip

lied

by

c m

ax w

here

c

max

is t

he e

xtre

me

fibre

str

ess

igno

ring

cra

ckin

g Po

sitiv

e si

gn c

onve

ntio

n fo

r N

and

M a

rein

dica

ted





Example 111











σc max = 249

03216 +

40(015)



σs2 = 51(312) = 160MPa (232ksi)




wm = 400(080) 160

200 times 103 = 026mm (0010 in)






























A = 5983 in2 B = 2704 in3 I = 118 500 in4


N1 = 2132kip M1 = 8354kip-in





σs2 = 213(0868) = 185ksi


wm = 16(09) 185

29 000 = 00092 in (023mm)













ρmin y = 360

60 times 103

1

1 minus29000

2900 360

60 times 103= 00063



ρ = Asbar

sts2 =

020

(8 times 8)2 = 00063




(120mm2)





1111 General


Notes













121 Introduction



Chapter 12

122 Permanent state





βD = minusD(Pm)

D(q)(121)





















q(125)





σperm = Mq y


Pm

A(127)





1 + I(8αqA f0 y)(128)


σq = Mq y

I(129)


αq = Mq (ql 2) (1210)


Pm = βD ql 2

8f0

(1211)















951 = 697MPa




1 + 951[8(minus00763)(725)(29)(minus1168)] = 079


Pm = 079 (2065 times 103)(60)2

8(29) = 25 200kN












σq = 773MPa

βD 079

Pm = 21 600kN




Mq = 0125 ql 2 αq = 0125


σq = 125(2065 times 103) (60)2 1832

951 = 1790MPa



1 + 951[8(0125)(725)(29)(1832)] = 089

Pm = 089 (2065 times 103)(60)2

8(29) = 28 500kN




σq = 3405MPa

βD = 085

Pm = 23 300kN





lh l = 30 l = 60m l = 90m


20 094 12 800 079 25 400 075 41 60025 087 14 500 073 27 500 071 45 50030 083 16 400 072 31 100 069 49 700




















128 Water-tightness




























fct = N

A1

+ M

W1

(1213)










2Mmax

ψm Mmax minus

Mmax

EI1 (1214)




1212 General







Notes






above



131 Introduction



Chapter 13



132 Reference axis











[S] D = F (131)






[S] = [S11]

[S21]

[S12]

[S22] (132)




[S21] = [R] [S11] (133)

where

[R] =minus1

0

0

0

minus1

l

0

0

minus1

(134)



[S] = [S11]

[R][S11]

[S11] [R]T

[R] [S11] [R]T (135)


[S11] = [f ]minus1 (136)


f ij = l

0Nui ε Oujdx +

l

0Muiψujdx (137)



f 1j = minusl


l

0ψujx dx f 3j =

l

0ψujdx with j = 1 2 3 (138)



[Smember] = [T]T [S] [T] (139)


[T] = [t][0]

[0]

[t] [t] =c

minuss

0

s

c

0

0

0

1

(1310)










εOu1 = minusI


B



εOu2 = Bx


Ax


For F3 = 1 Nu3 = 0 and Mu3 = 1

εOu3 = minusB


A



I minusBl

2B

[f] = l

E(AI minus B2)minus

Bl

2

Al 2

3minus

Al

2(1315)

B minusAl

2A





2 I =

b h3


1 0 minush

2

[S]cantilever = Ebh

l0

h

l 2

h 2

2l

minush

2

h2

2l

7h2

12(1317)


D = [S]minus1

0

P

0

= l

Ebh3

4h2

minus3hl

6h

minus3hl

4l 2

minus6l

6h

minus6l

12

0

P

0

= Pl 2

Ebh3

minus3h

4l

minus6

(1318)


A

l

012(AI minus B2)

Al 3

symmetrical

minusB

l

6(AI minus B2)

Al 2

4AI minus 3B2

Al



minusA

l

0 B

l


A

l



Al 2


012(AI minus B2)

Al 3

B

l

6(AI minus B 2)

Al 2

2AI minus 3B 2


B


AI 2

4AI minus 3B 2

Al










εO = minus02344


2)minus1

ψ = 03420


3)minus1


εO = minus01582


2)minus1

ψ = 01849


3)minus1


εO

ψ

mean

= (1 minus ζ)εO

ψ

uncracked

+ ζ εO

ψ

fully-cracked

= minus21700 h

25680 (Ec h3)minus1





u1 7886 30130 25680 10minus3 (h3Ec)minus1




u3 15770 41530 36380 10minus3(h4Ec)minus1


21700 minus25680

h l

225680

h

[f ] = 10minus3l

Ech2

minus25680

h l

236380

h2 l2

3 minus36380

h2 l

225680

hminus

36380

h2 l

236380

h2


02799 0 minus01976h


l0 03295

h2

l201647

h2

l

minus01976h 01647h2

l02493 h2


D = [S]minus1 0

P

0 =

Pl2

Ec h4

minus1284h

1213l

minus1819



Ar = ArO1

ArO2 (1319)






D1s = minusl

0εO dx D2s = minus

l

0ψ x dx D3s =

l

0ψ dx (1321)




Q1

Q2


(1322)



QA

QB =

s

6 2

1

1

2 qA

qB (1323)










εcentroid = αt

Al

T dA1 ψ = αt


yO = minusB1

A1


1

A1

(1326)





sψ x dx =

s






l minus xC

l

s ψ x dx =

l minus xC

l s


when xB xC (1328)

xC

l

s ψ(l minus x)dx =

xC

l l s(ψA + ψB)

2 minus

s

6(2ψA xA + ψA xB








[S] D = minusF (1330)


A = Ar + [S] D (1331)






where

[H] =1

0

0

0

1

0

0

0

1

minus1

0

0

0

minus1

0

0

l

minus1

(1333)





[R] [S11] Derror (1334)





cycle 1 (1335)




cycle 1 (1336)








([S]∆D)i = δi F (1337)
























0

0

9380

(b)



minus1119

6922

1530

(c)



1119

minus6922

minus5922

(d)


fcycle 1 = 10minus9

5569

minus16790

minus1685

minus16790

4839

minus5692

minus1685

minus5692

8207

(e)



2162

6908

5235

6908

1343

9454

5235

9454

7883

(f )




1631

minus7165

minus5357


minus1631

7165

minus3242

(g)



1631

minus7165

minus5357

(h)


[S]cycle 1 = 106

2162

6908

5235

6908

1343times106

9454

5235

9454

7883

(i)




1631

minus7165

minus5357

Dcycle 2 = 10minus6

minus1095

0

1406


minus1095

0

2344




AO1cycle 2 = 103

1631

minus7165

minus5357

+ [S]

minus1095

0

1406

10minus6 = 103

0

minus1430

0

AO2cycle 2 = 103

0

1430

minus1717




minus1402

minus9574

3904



minus1095

0

2344

minus

minus1402

minus9574

3904

= 10minus6

3074

9574

minus1559



3117

0340

minus1145





















1 000 minus2939 01509 1000 minus2939 0843







1313 General



Note




141 Introduction



Chapter 14








Reinforcementtype


GPa 103 ksi MPa ksi




















c = 1


a1 = 2hflange

bw


bw


a2 = h2

flange

bw


bw







yCT = d minus c

3with c hf (145)


yCT = d minus c


2 (c + 2hflange)c




d minus c(147)



σf service

(148)









ψm = (1 minus ζ) ψ1 + ζ ψ2 (1410)

ψ1 = M

EcI1

ψ2 = M

EcI2

(1411)

ζ = 1 minus β Mr

M2

(1412)

or

ζ = 1 minus β fct

σ1 max

2

(1413)




ψm = M

EIem

(1414)



Iem = I1I2

I2 + 1 minus 05 Mr

M2

(I1 minus I2)

(1415)





Dcentre = 1

2 l2

0ψx dx +

1

2

l



Dcentre = l 2







Dcentre = 5

48ψm centre l

2 (1418)













ψ2 = εs or εf

βh(1419)

where

β = d minus c

h(1420)


ψm = ηψ2 (1421)

where


η = 1 minus 05 fct

σ1 max

2

1 minus I2

I1 (1422)


Dcentre = 5

48 l 2

h ηβs or f

εs or f (1423)


l

hf = l

hs (ηβ)s

(ηβ)f εs

εf (1424)




l

hf

= l

hs

εs

εf

αd

(1425)


αd = 05 + 3b

100bw

minus b

80hs

(1426)













hs = 8m

16 = 05m


αd = 05 + 3(2)

100(04) minus

2

80(05) = 060

l

hf


2000times10minus606

= 118


hf = 8

118 = 068m take hf = 07 m d = 06m


M = (16kNm) (8m)2

8 = 1280kNm


Af 128kN minus m








= 2740mm2 and M = 128kN-m)



σ1 max = 278MPa ζ = 0742


Dcentre = 184mm

(l Dcentre)f = 80m

00184m = 435


M = (16kN-m)(112)2

8 = 2509kN-m

As = 1830mm2 εs = 1200 times 106


ψ1 = 428 times 10minus6 ψ2 = 2338

σ1 max = 501MPa ζ = 0920


Dcentre = 286mm

l Dcentres = 112m

00286m = 392










1411 General


Notes











εc(t) = σc(t0)

Ec(t0) [1 + φ(t t0)] (A1)



εc(t) = σc(t0)

Ec(t0) 1 +

Ec(t0)



Appendix A

φ = φCEB Ec(t0)

Ec(28)(A3)




h0 = 2Ac

u(A4)









Time functions 475

tT = n

i = 1


273 + T(∆ti) (A6)




Ec(28) = 21 500 ( fcmfcm0)13 (A7)


Ec(28) = 21 500[( fck + ∆f )fcm0]13 (A8)








476 Appendix A

where




Ec(t) = βE(t)Ec(28) (A11)

with










βE(t0) = Ec(t0)

Ec(28)(A15)

Time functions 477







046(h0href)13

(A18)

where href = 100mm

β(fcm) = 53

radicfcmfcmo

(A19)

where fcm0 = 10MPa

β(t0) = 1

01 + t002

(A20)



βH + t minus t0

03

(A21)


βH = 150h0

href

[1 + (0012 RH)18] + 250 le 1500mm (A22)

where href = 100mm

478 Appendix A



t0 = t0 T 9

2 + (t0 T)12+ 1

a

ge 05 (A23)




φ0k = φ0 exp[15(k minus 04)] (A24)


A19 Shrinkage






05

(A26)



Time functions 479

where




1003







Ec(t0) = 095 21 500[(fck(t0) + 8)10]13 (A31)






480 Appendix A





Age atloading


t0 (days) 50 150 600 50 150 600


1 32 26 21 20 18 157 34 28 22 22 19 17

28 32 25 20 19 17 1590 28 23 17 17 15 13

365 22 18 13 13 11 11



le150 600



Time functions 481

A31 Creep



06

10 + (t minus t0)06

φu (A32)

where



φu = 235 γc (A34)


γc = 125 tminus01180 (A35)

or

γc = 1113 tminus00940 (A36)



Ec(t0)

Ec(28) = t0

α + βt0

12 (A37)




482 Appendix A










φ(tinfin t0)


0 (A42)


A32 Shrinkage






55 + (t minus 1 to 3) (εcs)u (A44)

Time functions 483







Value of


(tinfin t0)

10 days 05 0525 0804 0811 080915 0720 0826 0825 0820 027325 0774 0842 0837 083035 0806 0856 0848 0839

102 05 0505 0888 0916 0915days 15 0739 0919 0932 0928 0608

25 0804 0935 0943 093835 0839 0946 0951 0946

103 05 0511 0912 0973 0981days 15 0732 0943 0981 0985 0857

25 0795 0956 0985 098835 0830 0964 0987 0990

104 05 0461 0887 0956 0965days 15 0702 0924 0966 0972 0954

25 0770 0940 0972 097635 0808 0950 0977 0980

(tinfin t0)

(tinfin 7)

0960 0731 0558 0425

Ec(t0)

Ec(28)

0895 1060 1083 1089

484 Appendix A






Ec(28) = K0 + 02 fcu(28) (A47)



Ec(t) = Ec(28) 04 + 06 fcu (t)





Time functions 485



A43 Creep


A44 Shrinkage







20 200 135 220 230 240 25025 250 165 275 290 300 31030 300 200 330 350 360 37040 400 280 440 455 475 50050 500 360 540 555 575 600

486 Appendix A






Time functions 487





488 Appendix A




1 + radict0

(A49)



Time functions 489



490 Appendix A



Time functions 491



MPa psi mm in

100 4 A650 200 8 A7

400 16 A820 3000 1000 40 A9

100 4 A1080 200 8 A11

400 16 A121000 40 A13

100 4 A1450 200 8 A15

400 16 A1630 4500 1000 40 A17

100 4 A1880 200 8 A19

400 16 A201000 40 A21

100 4 A2250 200 8 A23

400 16 A2440 6000 1000 40 A25

100 4 A2680 200 8 A27

400 16 A281000 40 A29

100 4 A3050 200 8 A31

400 16 A3260 9000 1000 40 A33

100 4 A3480 200 8 A35

400 16 A361000 40 A37

100 4 A3850 200 8 A39

400 16 A4080 12000 1000 40 A41

100 4 A4280 200 8 A43

400 16 A441000 40 A45

492 Appendix A


Time functions 493


494 Appendix A


Time functions 495


496 Appendix A


Time functions 497


498 Appendix A


Time functions 499


500 Appendix A


Time functions 501


502 Appendix A


Time functions 503


504 Appendix A


Time functions 505


506 Appendix A


Time functions 507


508 Appendix A


Time functions 509


510 Appendix A


Time functions 511


512 Appendix A


Time functions 513


514 Appendix A


Time functions 515


516 Appendix A


Time functions 517


518 Appendix A


Time functions 519


520 Appendix A


Time functions 521


522 Appendix A


Time functions 523


524 Appendix A


Time functions 525


526 Appendix A


Time functions 527


528 Appendix A


Time functions 529


530 Appendix A


Time functions 531


532 Appendix A

Notes








Time functions 533





0

λ ge 04

λ lt 04(B1)

where

λ = σp0

fptk

(B2)




Appendix B


χr = 1


λ minus 04 2

dξ (B4)



σp0

(B5)




















536 Appendix B

χr = exp[(minus67 + 53 λ)Ω] (B11)



∆σprinfin 1

16 ln τ minus t0

10 + 10 le (τ minus t0) le 1000 (B12)


05 times 10602




Notes







Appendix C









D1 = l

4 [1 2 1] εO (C1)

D2 = minusl

24 [1 6 5] ψ (C2)

D3 = l

24 [5 6 1] ψ (C3)

D4 = l 2

48 [1 4 1] ψ (C4)


D1 = l

6 [1 4 1]εO (C5)

D2 = minusl

6 [0 2 1]ψ (C6)

D3 = l

6 [1 2 0]ψ (C7)

D4 = l 2

96 [1 10 1]ψ (C8)


D1 = l

8 [1 2 2 2 1]εO (C9)

D2 = minusl

96 [1 6 12 18 11]ψ (C10)

D3 = l

96 [11 18 12 6 1]ψ (C11)

540 Appendix C

D4 = l 2

192 [1 6 10 6 1]ψ (C12)


D1 = l

12 [1 4 2 4 1]εO (C13)

D2 = minusl

12 [0 1 1 3 1]ψ (C14)

D3 = l

12 [1 3 1 1 0]ψ (C15)

D4 = l 2

24 [0 1 1 1 0]ψ (C16)


D1 = minusl 2

6 [1 2]ψ (C17)

D2 = l

2 [1 1]ψ (C18)


D1 = minusl 2

24 [1 6 5]ψ (C19)

D2 = l

4 [1 2 1]ψ (C20)


D1 = minusl 2

12 [0 1 1 3 1]ψ (C21)

D2 = l

12 [1 4 2 4 1]ψ (C22)




c3 + a1c2 + a2c + a3 = 0 (D1)

where


a2 = minus6

bw



a3 = 6

bw

[12h







Appendix D

a4 = a2 minus a2

1

3(D5)

a5 = 2 a1

3 3

minus a1a2

3 + a3 (D6)

a6 = a4

3 3

+ a5

2 2

(D7)


c = minus a5

2 + radica6

12

+ minus a5

2 minus radica6

12

minus a1

3(D8)


13 ]


cos θ = minusa5


c1 = 2 minus a4

3 12

cosθ3 minus a1

3(D10)



a1

3 (D11)

Note




E1 Introduction





Appendix E

ρmin = fct

fsy

(E1)






wm = srmζεs2 (E2)





E2 Crack spacing




db (E3)



κ1 = fct

fbm

(E4)




546 Appendix E

sr0 = κ1 db

4ρ(E5)









wk = βwm (E6)




srm = 50 + κ1κ2 db

4ρr

(E7)

where

κ2 = ε1 + ε2

2ε1

(E8)


ρr = As

Acef

(E9)


E4 CEB-FIP 1990 (MC-90)



where




548 Appendix E

εsr2 = fctm(t)

ρrEs

(1 + αρr) (E11)

α = EsEc(t) (E12)




within Acef





ls max = db

36 ρr

(E13)

ls max = σs2db

2τbk(1 + αρr)(E14)


E5 ACI318-89 and ACI318-99



where








τbk τbk



550 Appendix E






Srm = 4te (E18)


te = dc 1 + s

4dc

2

(E19)





s(in) = 540


s(mm) = 95


and

s(in) = 12(36)

σs (ksi)(E22)

s(mm) = 76

σs (MPa)(E23)






552 Appendix E



1 + 2a minus cover

h minus c (E24)



Notes












554 Appendix E






κs = Ig

I(F1)


I(F2)

κcs = minusA

cycd

I(F3)





Appendix F


556 Appendix F




558 Appendix F




560 Appendix F




562 Appendix F




564 Appendix F








566 Appendix F










G1 Introduction








Appendix G













sustained


G22 FORTRAN code










respectively




570 Appendix G


























572 Appendix G












2 3 34 27000



Notes


574 Appendix G

Further reading








Index














474 548















205














255 256 262 391 395control of 380









to bending 384


examples 234 236 243 250 254260



















578 Index



























41 437 43 44 49 61 64 75 8183
















Index 579








146



569






























580 Index








































33 43 72




definition 5

Index 581












curvature due to 243 303 309 327346 564 565











136 147 172











295 547Strain




42 72mean value





582 Index







































stages 105 109 113 116 128141 152 155






sections 20 30 57 63 95 128144

Index 583









deflection 333


units


584 Index

Book Cover

Title

Copyright

Contents


Acknowledgements

Note


Notation






















Further reading

Index




















Third Edition




London and New York















(Print Edition)

Contents





112 General 18





















vi Contents













48 General 144





Contents vii








69 General 205






viii Contents







79 General 262












Contents ix




811 General 301












x Contents


910 General 348













Contents xi












xii Contents
















Contents xiii













xiv Contents




Contents xv













A GhaliR Favre




Acknowledgements





Note




1 m = 3281 ft


1m2 = 1076 ft2








1MPa = 01450ksi




1W = 3416 Btuh



Notation

























Notation xxiii





xxiv Notation


11 Introduction



Chapter 1








εc(t0) =σc(t0)

Ec(t0)(11)






εc(t) =σc(t0)

Ec(t0) [1 + φ(t t0)] (12)
















∆σpr

σp0


10 σp0

fpy

minus 055 (14)






∆σprinfin

σp0


where


λ =σp0

fptk

(16)
















σp0 (18)



055 060 065 070 075 080

00 1000 1000 1000 1000 1000 100001 06492 06978 07282 07490 07642 0775702 04168 04820 05259 05573 05806 0598703 02824 03393 03832 04166 04425 0463004 02118 02546 02897 03188 03429 0362705 01694 02037 02318 02551 02748 02917









εc(t) = σc(t0) 1 + φ(t t0)

Ec(t0)+

∆σc(t)

0

1 + φ(t τ)


where











εc(t) = σc(t0) 1 + φ(t t0)

Ec(t0)+ ∆σc(t)

1 + χφ(t t0)

Ec(t0)+ εcs(t t0) (110)










∆σc(t)(111)



dσc(τ)

dτ= ∆σc(t)

dξ1

dτ(112)


εc(t) = σc(t0)1 + φ(t t0)

Ec(t0)+ ∆σc(t)

t

t0

1 + φ(t τ)

Ec(τ) dξ1

dτ dτ

+ εcs (t t0) (113)


χ(t t0) =Ec(t0)

φ(t t0)

t

t0

1 + φ(t τ)

Ec(τ) dξ1

dτ dτ minus1

φ(t t0)(114)











σc(t0) = εcEc(t0) (115)


σc(t) = εcr(t t0) (116)



∆σc(τ) = ξ[∆σc(t)] (117)







ξ =∆σc(τ)

∆σc(t)(120)



εc = σc(t0) 1 + φ (t t0)

Ec(t0)+ ∆σc(t)

1 + χφ(t t0)

Ec(t0)(121)





Ec(t0)(122)


χ(t t0) =1


1

φ(t t0)(123)







2 tj and tj + 1

2 are used to refer




εc(ti + 12) =

i

j = 1(∆σc)j

1 + φ(ti + 12 tj)


2 t0) (124)


σc(ti + 12) =

i

j = 1

(∆σc)j (125)





σc(ti + 12) = εcr(ti + 1

2 t0) (126)


r(ti + 12 t0) =

1

εc

i

j = 1

(∆σc)j (127)


εc(ti + 12) = (∆σc)i

1 + φ(ti + 12 ti)

Ec(ti)+

i minus 1

j = 1

(∆σc)j 1 + φ(ti + 1

2 tj)

Ec(tj)

+ εcs(ti + 12 t0) (128)



(∆σc)i =Ec(ti)

1 + φ(ti + 12 ti)


2 t0)

minus i minus 1

j = 1

(∆σc)j 1 + φ(ti + 1

2 tj)

Ec(tj) (129)


2 t0) = 0 and







εc(t) = σc(t0) 1 + φ(t t0)

Ec(t0)+

∆σc(t)

Ec(t t0)+ εcs(t t0) (130)

where

Ec(t t0) =Ec(t0)

1 + χφ(t t0)(131)



Ec(t t0) (132)





Ec(t0)(133)




Ec(t t0) (134)



112 General












21 Introduction



Chapter 2










22 Sign convention









ε = εO + ψy (21)





σ = Ei(εO + ψy) (22)


N = σdA (23)

M = σydA (24)


N = εO m

i = 1

Ei dA + ψ m

i = 1

Ei ydA (25)

M = εO m

i = 1

Ei y dA + ψ m

i = 1

Ei y2dA (26)






A = m

i = 1

Ei

Eref

Ai (29)

B = m

i = 1

Ei

Eref

Bi (210)

I = m

i = 1

Ei

Eref

Ii (211)




NM = Eref ABB

I εO

ψ (212)


εO

ψ =1

ErefAB

B

I minus1

NM (213)


ABB

I minus1

=1

(AI minus B2) I

minusB

minusB

A (214)


εO

ψ =1

Eref(AI minus B2) I

minusB

minusB

A NM (215)


εO

ψ =1

Eref

NA

MI (216)

231 Basic equations


ε = εO + ψy σ = E(εO + ψy) (217)

N = E(AεO + Bψ) M = E(BεO + Iψ) (218)


εO =IN minus BM

E(AI minus B2)ψ =

minusBN + AM

E(AI minus B2)(219)

σO =IN minus BM

AI minus B2γ =

minusBN + AM

AI minus B2(220)

where













εf = αtT (221)













∆εO

∆ψ =1

E(AI minus B2) I

minusB

minusB

A minus∆N

minus∆M (227)

∆σ = E[∆εO + (∆ψ)y] (228)


∆εO

∆ψ =1

E minus∆NA

minus∆MI (229)


σ = E [minus εf + ∆εO + (∆ψ)y] (230)










∆N

∆M = αtTtopE

minusbh

m + 1

bh2m

2(m + 1)(m + 2)


εO =αtTtop

m + 1

ψ = minusαtTtop

h

6m

(m + 1)(m + 2)













Nequivalent





εO(t0)

ψ(t0) =

1

Eref(AI minus B2) I

minusB

minusB

A NMequivalent

(232)



εO(t0)

ψ(t0) =

1

Eref

Nequivalent

A

Mequivalent

I

(233)



εc(t0) = εO(t0) + ψ(t0)y (234)

σc(t0) = [Ec(t0)]i[εO(t0) + ψ(t0)y] (235)
















∆εO

∆ψ =1

Ec(AI minus B2) I

minusB

minusB

A minus∆N

minus∆M (240)


∆N

∆M = ∆N

∆Mcreep

+ ∆N

∆Mshrinkage

+ ∆N

∆Mrelaxation

(241)


∆N

∆Mcreep

= minusm

i = 1

EcφAc

Bc

Bc

Ic εO(t0)

ψ(t0)

i

(242)






∆N

∆Mshrinkage

= minus m

i = 1

EcεcsAc

Bc

i

(243)





∆N

∆Mrelaxation

= Aps∆σ-pr

Apsyps∆σ-pr

i

(244)


















NMequivalent





εO(t0)

ψ(t0) =

1






minus240 times 103

= 10minus6 minus126

minus170mminus1










Tabl

e 2

1C

alcu

latio

n of

A B

and

I of

tra

nsfo

rmed

sec

tion

at t

ime

t 0

Prop

ertie

s of

are

aPr

oper

ties

of t

rans

form

ed a

rea

AB

IAE

Ere

fBE

Ere

fIE

Ere

f

(m2)

(m3)

(m4)

(m2)

(m3)

(m4)

Conc

rete

035

45minus1

625

times 1

0minus341

84

times 1

0minus30

3545

minus16

25 times

10minus3

418

4 times

10minus3

Non

-pre

stre

ssed

ste

el25

00 times

10minus6

027

5 times

10minus3

075

6 times

10minus3

001

671

833

times 1

0minus35

04 times

10minus3

Pres

tres

sed

stee

lmdash

mdashmdash

mdashmdash

mdash

Prop

ertie

s of

037

120

208

times 1

0minus346

88

times 1

0minus3

trans

form

ed s

ectio

nA

BI




1 + 08 times 3= 882GPa



= 2741MPa (0398ksi)


= 8145MPa (1181ksi)


∆N

∆Mcreep




4184 times 10minus3

times minus126

minus170 10minus6 = 106 1175N

01828N-m

∆N

∆Mshrinkage



= 106 0750N

minus00034N-m

∆N

∆Mrelaxation



= 106minus0090N

minus00403N-m

∆N

∆M = 106 1175 + 0750 minus 0090

01828 minus 00034 minus 00403 =1061835 N

0139 N-m



Tabl

e 2

2C

alcu

latio

n of

A B

and

I of

the

age

-adj

uste

d tr

ansf

orm

ed s

ectio

n

Prop

ertie

s of

are

aPr

oper

ties

of tr

ansf

orm

ed a

rea

AB

IAE

Ere

fBE

Ere

fIE

Ere

f

(m2)

(m3)

(m4)

(m2)

(m3)

(m4)

Conc

rete

035

45minus1

625

times 1

0minus341

84

times 1

0minus30

3545

minus16

25 times

10minus3

418

4 times

10minus3

Non

-pre

stre

ssed

ste

el25

00 times

10minus6

027

5 times

10minus3

075

6 times

10minus3

005

676

236

times 1

0minus317

24

times 1

0minus3

Pres

tres

sed

stee

l11

20 times

10minus6

050

4 times

10minus3

022

7 times

10minus3

002

5411

429

times 1

0minus35

15 times

10minus3

Prop

ertie

s of

age

-adj

uste

d0

4366

160

40 times

10minus3

641

2 times

10minus3

trans

form

ed s

ectio

nA

BI


∆εO

∆ψ =

1





04366 minus1835

minus0139 106

= 10minus6minus470

minus128mminus1





= 3313MPa (0481ksi)



∆σns1 = 200[minus471 + 055(minus128)]103







































(m2) (m3) (m4) (m2) (m3) (m4)






NM = 02800 times 103 N-m


εO(t0)

ψ(t0) =

1


times 21367

16788

16788

16474 0

2800 times 103

= 10minus6 223

219mminus1





1 + 075 times 25= 10435GPa



= 824MPa


= 699MPa




∆N

∆Mcreep


times 13081

minus15828

minus15828

19205 113

219 10minus6

= 106 5187 N



∆N

∆Mshrinkage


minus15828

= 106 4777 N

minus5781 N-m

∆N

∆Mrelaxation



= 106 minus0351 N


∆N

∆M = 106 5187 + 4777 minus 0351

minus6306 minus 5781 + 0425 = 106 9613 N



A = 2284m2 B = minus1859m3 I = 2542m4





∆εO

∆ψ =

106


1859

1859

2284minus 9613

11662

= 10minus6 minus112

357mminus1




















NMequivalent





εO (t1)

ψ(t1) =

1


minus00117

minus00117

05583



minus218mminus1




Ec(60 3) =25 times 109

1 + 086 times 120= 1230GPa (1780ksi)





(m2) (m3) (m4) (m2) (m3) (m4)










∆N

∆Mcreep


0

0

01090 minus289

minus218 10minus6

= 106 2171 N

0351 N-m

∆N

∆Mshrinkage


0 = 106 0357 N

0

∆N

∆Mrelaxation




∆N

∆M = 106 2171 + 0357 minus 0047

00351 + 0 minus 0025 = 106 2481 N




∆εO(t2 t1)






∆εO(t2)


415mminus1






1 + 08 times 227= 1314GPa (1900ksi)


1 + 078 times 240= 801GPa (1160ksi)



1

Ec(t)[1 + φ(60 t)] =

1

Ec(3) [1 + χ(60 3) φ(60 3)]





= 14304MPa




= 0106MPa





∆N

∆Mcreep


0

0

0109

times

110minus289 +095 times 1012

25 times 109 + 227(minus6)

110minus218 +161 times 1012

25 times 109 + 227(415)

10minus6

= 106 1937 N

minus1108 N-m


∆N

∆Mshrinkage


0


minus04307

= 106 2437 N

minus0928 N-m


∆N

∆Mrelaxation



minus0117N-m





(m2) (m3) (m4) (m2) (m3) (m4)



Concrete inbeam 05090 00 01090 05090 00 01090






∆N

∆M = 106 1937 + 2437 minus 0221


minus2153 N-m



∆εO(t3 t2)


195mminus1




Notation






Subscripts


Four analysis steps














Commentary




















αns = 806 αps = 750



A = 1158 in2 B = 19819 in3 I = 547200 in4










Ec(t t0) = 1059ksi


Ac = 1023 in2 Bc = 16000 in3 Ic = 410800 in4










αns = 2739 αps = 2550



A = 1483 in2 B = 28950 in3 I = 874600 in4




∆σ(t t0)top bot = 0047 0485 ksi










































28 General




Note





Chapter 3

31 Introduction






















1 +Ast

Ac

Est

Ec(t t0)1 +

y2st

r2c

(34)























∆σns

Est


Est

=σcst(t0)φ(t t0)


1

Ec(t t0)∆Pc

Ac

+∆Pc y2

st

Ic (38)














Ec(t t0)Ac

(310)


Ec(t t0)Ic

(311)


∆σc =∆Pc

Ac

+∆Pc yst

Ic

y


2c in the last

equation gives

∆σc =∆Pc

Ac

1 +yyst

r2c (312)


















Figu

re3

2A

naly

sis

of s

tres

s an

d st

rain

in a

sec

tion

with

one

laye

r of

pre

stre

ssed

ste

el (E

xam

ple

31)

(a)

cro

ss-s

ectio

n (b

) con

ditio

nsim

med

iate

ly a

fter

pre

stre

ss (

c) c

hang

es d

ue t

o cr

eep

shr

inka

ge a

nd r

elax

atio

n




0357

200

8824 1 +

02059

01193minus1

= 2427kN (5456kip)


∆σps =∆Pps

Aps

= minus∆Pc

Aps




σp0 =1400 times 103



λ =σp0

fptk

=1250

1770= 0706 Ω =

2167 minus 115

1250= 0081






0357 1 +

(minus0596)0454

01193 Pa




0357 1 +

0604 times 0454

01193 Pa = 2241MPa (0325ksi)




















r2c

+ εcs(t t0)yc

r2c (316)



Symbolused












c(t)]dAc



η = AcA (317)

κ = IcI (318)












∆N

∆Mcreep


Ac yc

Ac yc

Ic εO(t0)

ψ(t0) (320)

∆N

∆Mshrinkage


Ac yc (321)


∆N




r2c + εcs(t t0)

yc

r2cAcr

2c (322)


∆εO

∆ψ =1

Ec(t t0) minus∆NA

minus∆MI (323)







1 + 08 times 3= 8824GPa (1280ksi)

α(t t0) =200

8824= 22665






A = 03811m2 I = 3750 times 10minus3 m4


η =02963

03811= 0777 κ =

2526

3750= 0674




8445 times 10minus3






= 1102MPa (0159ksi)








= 4163kN-m (3685kip-in)







∆ψ = 0674 3428 + (minus120)minus 0054




+ (minus334) + 1021(minus0551)

= 1508MPa (0219ksi)






∆εO = ηφ(t t0)[εO(t0) + ψ(t0)yc]) (324)

∆ψ = κφ(t t0) ψ(t0) + εO(t0) yc

r2c (325)





∆εO ηεO(t0)φ(t t0) (326)


∆ψ κψ(t0)φ(t t0) (327)


2c]



and κ







α = α(t0) = EsEc(t0) (328)

α = α(t t0) = Es[1 + χφ(t t0)]Ec(t0) (329)





κ =Ic

I (or I )(330)


Ic bh3

12 1 + 12y2

c

h2 (331)










































Fig 38(b) )

Q =

7l48

l24

minusl48

l8

5l12

l8

minusl48

l24

7l48

ψ (a)




D3 = [1 05 0]Q (c)

D4 = l [0 025 0] Q (d)






D1 = εO dl (e)


D1 = l

6

2l

3

l

6 εO






(186)2

96 [1 10 1]

64minus298





186

6 [1 4 1]





186

6 [1 2 0]

64minus298














4 times10minus6



60 times10minus6




























between steel





ns







310 General






Notes








Chapter 4

41 Introduction





















[ f ] F = minusD (41)





A = As + [Au] F (42)














Ec =Ec

1 + χφ(44)





[ f ] ∆F = minus∆D (45)


∆A = ∆As + [∆Au] ∆F (46)













Ec(t t0) =Ec(t0)

1 + χφ(t t0)





φ(infin 7) = 27 χ(infin 7) = 074 φ(60 7) = 11

φ(infin 60) = 23 χ(infin 60) = 078


Ec(60)Ec(7) = 126




D1 = (D1)AB + (D1)BC

= 0 +ql3

24Ec(7)Ic




f11 = ( f11)AB + ( f11)BC

=l

3Ec(60)Ic

+l

3Ec(7)Ic


F1 = minus D1 f11




∆D1 = (∆D1)AB + (∆D1)BC


(∆D1)AB =ql 3

24Ec(7)Ic


3Ec(60)Ic

φ(infin 60)


(∆D1)BC =ql 3

24Ec(7)Ic


3Ec(7)Ic

φ(infin 7)


∆D1 = 00720 ql 3

Ec(7)Ic



f11 = ( f11)AB + ( f11)BC

( f11)AB =l

3Ec (infin 60)Ic

( f11)BC =l

3Ec(infin 7)Ic



1 + χφ(infin 60)= 1

1 + 078 times 23 Ec(60) = 045Ec(7)

Ec(infin 7) =Ec(7)

1 + χφ(infin 7)= 1

1 + 074 times 27 Ec(7) = 034Ec(7)

Thus

f11 =l

3 times 045Ec(7)Ic

+l

3 times 034Ec(7)Ic

= 1724 l

Ec(7)Ic









2

3q =

8Pa

l 2













Ec(t0)Ic

2 timesl


ql 3

Ec(t0)Ic


Ec(t0)Ic


Ec(t0)Ic


Ec(t t0) = 1

1 + 08 times 25 Ec(t0) =1

3Ec(t0)



[Ec(t0)3]Ic

2 timesl



Ec(t0)Ic

(∆D1) = (1390 + 125)10minus3 ql 3

Ec(t0)Ic


Ec(t0)Ic


f11 =l

[Ec(t0)3]Ic

13 +1

2 = 25 l

Ec(t0)Ic


∆F1 = minus1515





Example 44



φ(t1 t0) = 11 χ(t1 t0) = 079 φ(t2 t0) = 27 χ(t2 t0) = 074

φ(t2 t1) = 23 χ(t2 t1) = 078 Ec(t1)Ec(t0) = 126









1 + 078 times 23= 036Ec(t1) = 045 Ec(t0)


1 + 074 times 27= 034Ec(t0)



(∆D1)loads =1

Ec(t0)Ic

D


+1

Ec(t1)Ic

D

AMeMu1 dl φ(t2 t1)

+1

Ec(t0)Ic

C

DMeMu1 dl φ(t2 t0)




Ec(t0)Ic


timesql 3

Ec(t1)Ic


Ec(t0)Ic


Ec(t0)Ic







(Ec)ADIc

D

AMiMu1 dl +

1

(Ec)DCIc

C

DMiMu1 dl





ql 3

Ic +

1

034Ec(t0)



ql 3

Ec(t0)Ic

∆D1 = (277 + 70)10minus3ql 3

Ec(t0)Ic


Ec(t0)Ic


f 11 =1

(Ec)ADIc

D

AM2

u1 dl +1

(Ec)DCIc

C

DM 2

u1 dl =251l

Ec(t0)Ic


∆F1 = minus347











F(t1) = Fsudden

1

1 + χ φ(t1 t0)(47)


Ec(te)


1 + χφ(t2 t1) (48)




1

Ec(te)[1 + φ(t1 te)] =

1

Ec(t0)[1 + χφ (t1t0)] (49)






1 + χφ(t t0) (410)



κ = Ic I (411)



Fsudden = 6

l 3Ec(t0)Icδ (412)




δ = l 3

6Ec(t1 t0)Ic F(t1) (413)


Ec(t1 t0) =Ec(t0)

1 + χφ(t1 t0)(414)




∆δ = l 3




∆δ + l 3

6Ec(t2 t1)Ic ∆F = 0 (416)

where

Ec(t2 t1) =Ec(t1)

1 + χφ(t2 t1)(417)


F(t2) = F(t1) + ∆F (418)















1414141423232323

104104104

104500

200010000

104500

200010000

5002000

10000

114179226257101172220255117170196

079076076076

081080079






δ(t)


t095 minus t0 (419)































Tabl

e 4

1C

ross

-sec

tion

prop

ertie

s1 and

cal

cula

tion

of in

stan

tane

ous

axia

l str

ain

and

curv

atur

e fo

r a

cont

inuo

us b

ridg

e (E

xam

ple

46)

Forc

es a

pplie

d at

time

t 0 e

quiva

lent

sof

pre

stre

ss fo

rce

Inst

anta

neou

sSe

ctio

n nu

mbe

rTr

ansf

orm

ed s

ectio

nAg

e-ad

just

ed tr

ansf

orm

edan

d de

ad-lo

adax

ial s

train

and

(see

Co

ncre

te s

ectio

npr

oper

ties

at ti

me

t 0se

ctio

n pr

oper

ties

bend

ing

mom

ent

curv

atur

eFi

g 4

10(b

))pr

oper

ties

E ref

=E c

(t0)

=28

GPa

E ref

=E c

(t t

0)=

909

GPa

(Equ

atio

n (2

31)

)(E

quat

ion(

232

))

A cB c

I cA

BI

AB

IN

M O

(t0)

(t

0)

10

780

015

90

8068

00

1645

091

60minus0

003

60

1835

minus28

20

189

minus125

410

20

780

015

90

8068

00

1645

091

600

0393

020

47minus2

82

024

5minus1

2553

23

078

00

159

080

680

016

450

9160

minus00

036

018

35minus2

82

008

8minus1

2519

14

078

00

159

080

680

016

450

9160

003

930

2047

minus28

20

195

minus125

423

Mul

tiplie

rsm

2m

3m

4m

2m

3m

4m

2m

3m

410

6 N10

6 N-m

10minus6

10minus6

mminus1

1R

efer

ence

poi

nt O

is c

hose

n at

the

com

mon

cen

troi

d of

Ac o

r A n

s

Tabl

e 4

2C

hang

es in

axi

al s

trai

n an

d in

cur

vatu

re o

f the

rel

ease

d st

ruct

ure

duri

ng t

he p

erio

d t 0

to

t in

Exam

ple

46

Calcu

latio

n of

rest

rain

ing

forc

es

Tota

lCh

ange

s in

axi

al s

train

Sect

ion

num

ber

Cree

pSh

rinka

geRe

laxa

tion

rest

rain

ing

forc

esan

d in

cur

vatu

re(s

ee F

ig 4

10(

b))

(Equ

atio

n (2

42)

)(E

quat

ion

(24

3))

(Equ

atio

n (2

44)

)(E

quat

ion

(24

1))

(Equ

atio

n (2

40)

)

N

M

N

M

N

M

N

M

O

12

304

minus01

541

170

20

minus02

190

0148

378

7minus0

139

3minus4

5574

62

230

4minus0

199

91

702

0minus0

219

minus01

607

378

7minus0

360

6minus4

6728

34

32

304

minus00

718

170

20

minus02

190

0148

378

7minus0

057

0minus4

5525

34

230

4minus0

159

01

702

0minus0

219

minus01

607

378

7minus0

319

7minus4

6626

13

Mul

tiplie

rs10

6 N10

6 N-m

106 N

106 N

-m10

6 N10

6 N-m

106 N

106 N

-m10

minus610

minus6 m

minus1


to t

∆D1 =25

6 [0 2 1]

7462834253

10minus6 +25

6 [1 2 0]

2532613253



(25)2

96[1 10 1]

410532191




(25)2

96 [1 10 1]

7462834253


= 191mm (0752 in)




A





f 11 =25



f 12 =25



(25)2




( f 11 + f 12)∆F1 = minus ∆D1

Thus








∆A = ∆As + [∆Au] ∆F



= 1039mm (0409 in)









[ f ] F = minusD (420)






2 tj)]







Fi + 12=

i

j = 1

∆Fj (421)

Di + 12=

i

j = 1

∆Dj (422)


i

j = 1


i

j = 1

∆Dj (423)


[ f (ti + 12 ti)] ∆Fi = minus Di + 1

2minus

i minus 1

j = 1

[f(ti + 12 tj)]∆Fj (424)










6EcIc

(425)




2 a displacement

f(ti + 12 tj) = C

1

Ec(tj) [1 + φ(ti + 1

2 tj)] (426)


C =l 3

6Ic

(427)



2) (428)



f (ti + 12 ti)(∆F)i = δ(ti + 1

2) minus C

i minus 1

j = 11 + φ(ti + 1

2 tj)

Ec(tj) (∆F)j (429)



2) + (∆F)i (430)



2)

+ [ f (ti + 12 ti)]


i minus 1

j = 11 + φ(ti + 1

2 tj)

Ec(tj) (∆F)j (431)




















∆As =

01770296

minus77851324

ksi








f 11 = ∆ψB 2l




[∆Au] = 10minus6

minus3059minus1539

minus2765988428

ksi

kip-in





∆A =

01770296

minus77851324

+ 10minus6

minus3059minus1539

minus2765988428

(minus117700) =

05370427

minus4530minus9084

ksi


48 General



Notes









51 Introduction



Chapter 5

































∆D = D(t0)[φ(t2 t0) minus φ(t1 t0)] (53)



[ f ] = [ f ][1 + χφ(t2 t1)] (54)



[ f ]∆F = minus∆D (55)



1 + χφ(t2 t1) F (56)

where











F = minusql2



∆F = 26 minus 09




























[ f ] ∆F = minus ∆D (58)







[ f ]minus1 = Ec

l

A

0

0

0

4I

2I

0

2I

4I

(510)




∆F = Ec

l

A

0

0

0

4I

2I

0

2I

4I

minus∆D (511)


∆D =int(∆εO)

int(∆ψ)

int(∆ψ)

Nul

Mu2

Mu3

dl

dl

dl

(512)




∆εO

∆ψ = 1

Ec(AI minus B2) I

minusB

minusB

A minus∆N

minus∆M (513)



∆εO

∆ψ = minus1

Ec

∆NA

∆MI (514)








p = minusd

dx(∆N) (515)

q = minusd2

dx2(∆M) (516)
















Ec(t1 t0) = 30 times 109

1 + 08 times 25 = 10GPa α(t1 t0) =

200

10 = 20



A = 210m2 B = 0 I = 10232m4


Ec(t0) = 30GPa α(t0) = 200

30 = 6667 Eref = Ec(t0)


A = 158m2 B = minus03947m3 I = 05221m4



Ac = 132m2 Bc = minus05927m3 Ic = 02714m4



∆N


minus05927 = 3564 times 106 N























∆F = 10 times 109

33

210

0

0

0

4(10232)

2(10232)

0

2(10232)

4(10232)

1691

807

minus807

10minus6

=

10761MN

05004MN-m

minus05004MN-m



Tabl

e 5

1In

stan

tane

ous

axia

l str

ain

and

curv

atur

e at

t 0 i

mm

edia

tely

aft

er a

pplic

atio

n of

the

load

q (E

xam

ple

53

Fig

51

1)

Prop

ertie

s of

Defl

ectio

ntra

nsfo

rmed

1 sec

tion

atAx

ial s

train

and

at G

age

t 0In

tern

al fo

rces

curv

atur

e at

t 0Pr

oper

ties

of(E

quat

ion

(Ere

f=

E c(t 0

)=30

GPa

)in

trodu

ced

at t 0

(Equ

atio

n (2

32)

)co

ncre

te a

rea

(C8

))

Mem

ber

Sect

ion

AB

IN

M O

(t 0)

(t 0

)A c

B cI c

D(t

0)

B1

58minus0

394

70

5221

minus02

431

minus18

21minus4

21

minus148

11

32minus0

592

70

2714

BCG

158

minus03

947

052

21minus0

243

13

624

649

280

51

32minus0

592

70

2714

284

6C

158

minus03

947

052

21minus0

243

1minus1

821

minus42

1minus1

481

132

minus05

927

027

14

Mul

tiplie

rm

2m

3m

410

6 N10

6 N-m

10minus6

10minus6

mminus1

m2

m3

m4

10minus3

m

1T

he r

efer

ence

poi

nt O

is a

t th

e ce

ntro

id o

f age

-adj

uste

d tr

ansf

orm

ed s

ectio

n (F

ig 5

8(c

))

Tabl

e 5

2C

hang

es in

axi

al s

trai

n an

d in

cur

vatu

re a

nd c

orre

spon

ding

elo

ngat

ion

and

end

rota

tions

of t

he r

elea

sed

stru

ctur

e in

Fig

51

1(c)

Chan

ges

inCh

ange

inCh

ange

s in

axi

aldi

spla

cem

ents

at t

hede

flect

ion

Inte

rnal

forc

es to

Prop

ertie

s of

age

-adj

uste

dst

rain

and

inco

ordi

nate

s in

Fig

at

Gre

stra

in c

reep

trans

form

ed s

ectio

ncu

rvat

ure

511

(c) (

Equa

tions

(Equ

atio

n(E

quat

ion

(24

2))

(Ere

f=

E c(t 1

t0)

=10

GPa

)(E

quat

ion

(24

0))

(C5

ndash7))

(C8

))

Mem

ber

Sect

ion

N

M

AB

I

O

D1

D

2

D3

D

Bminus0

805

20

3810

210

01

0232

383

minus37

2BC

G2

015

minus09

415

210

01

0232

minus96

092

0minus1

691

minus807

807

959

Cminus0

805

20

3810

210

01

0232

383

minus37

2

Mul

tiplie

rs10

6 N10

6 N-m

m2

m3

m4

10minus6

10minus6

mminus1

10minus6

m10

minus6 1

0minus610

minus3 m

radi

anra

dian

Figu

re5

12A

naly

sis

of s

tatic

ally

inde

term

inat

e fo

rces

cau

sed

by c

reep

in a

com

posi

te fr

ame

(Exa

mpl

e 5

3) (

a) fi

xed-

end

forc

es d

ue t

ocr

eep

Ar

(b) a

ssem

blag

e of

fixe

d-en

d fo

rces

and

rev

ersa

l of d

irec

tion

minusF

(c)

mem

ber-

end

forc

es d

ue to

join

t dis

plac

emen

t[A

u]

D

(d) t

otal

mem

ber-

end

forc

es d

ue t

o cr

eep

= su

m (a

) and

(c)

(e) b

endi

ng m

omen

t di

agra

m a

t t 1














Tabl

e 5

3St

ress

dis

trib

utio

n at

sec

tion

G (E

xam

ple

53)

Stre

ss in

sta

ges

(MPa

)Cr

eep

effe

ct =

Stre

ss a

t tim

e t 1

=At

tim

e t 0

Cree

p ef

fect

(b)+

(c)+

(d)

(a)+

(b)+

(c)+

(d)

(a)

(b)

(c)

(d)

MPa

MPa

ksi

Top

of c

oncr

ete

minus27

562

297

minus14

740

140

096

31

793

minus02

60Bo

ttom

of c

oncr

ete

minus09

060

755

minus12

720

078

minus04

39minus1

345

minus01

95To

p of

ste

elminus6

04

0minus2

544

155

minus23

89minus2

993

minus43

4Bo

ttom

of s

teel

893

30

584

minus80

4minus2

20

871

312

64


= 3421mm (1347 in)





2 the end of


εc(ti + 12) =

i

j = 11 + φ(ti + 1

2 tj)


2 t0) (517)

εps(ti + 12) =

1

Eps

i

j = 1


εns(ti + 12) =

1

Ens

i

j = 1

(∆σns)j (519)



2 t0)





(∆εc)i = 1 + φ(ti + 1

2 ti)

Ec(ti) (∆σc)i

+ i minus 1

j = 1

(∆σc)j

Ec(tj)[φ(ti + 1


2 tj)] + (∆εcs)i (520)


Eps

minus (∆σpr)i

Eps

(521)


Ens

(522)



(Ece)i

+ (∆εc)i (523)


Eps

+ (∆εps)i (524)


(Ece)i = Ec(ti)

1 + φ(ti + 12 ti)

(525)







αi = Es

(Ece)i

(526)






59 General




Notes






Chapter 6

61 Introduction






































A = Ar + [Au] D (61)


AD = [Au] D (62)









Ar(tj)prestress =




Ec(tk tj) = Ec(tj)

1 + χφ(tk tj)(65)

















TO

T prime1T prime2

= 1

Ecα

1A00

0l(6I)5l(6I)

0minus1I

minus1I

Ar1

Ar2

Ar3

(69)



TO = Ar (tk tj)

αEc (tk tj)A(610)








= minusEc (tk tj)




68 Examples



















Ec (t2 t1) = Ec(t1)

1 + χφ(t2 t1) =

Ec(t1)

1 + 08(245) = 03378 Ec(t0)





12

u00000E+0000000E+00

v41668Eminus0712500E+00

10417Eminus0616667E+00


1Fx

00000E+00Fy

minus10000E+01Mz

minus50000E+00


1F1

00000E+00F2

minus10000E+01F3

minus50000E+00F4

00000E+00F5


55511Eminus15





A(t2 t1) = 0 02155 ql 02154 ql 2 0 minus02155 ql 0





12



00000E+00 00000E+00 00000E+00

00000E+00 00000E+00 00000E+00



1 1Ar1

0000E+00Ar2

2872E+00Ar3

2393E+00Ar4

0000E+00Ar5

minus2872E+00Ar6

4786Eminus01



12

u00000E+0000000E+00

v17688Eminus07

minus17688Eminus07

17688Eminus07

minus35377Eminus01


12

Fx

00000E+00

00000E+00

Fy

21550E+00minus21550E+00

Mz

21540E+00

00000E+00


1F1

00000E+00F2

21550E+00F3

21540E+00F4

00000E+00F5

minus21550E+00F6

00000E+00









u v θ u v θ2 0 1 0 00 00 00










12

x y00 00

100 00


11st node

12nd node

2a

10000E+01I

10000E+00


2



00000E+00 00000E+00 00000E+00


1Node

1Fx

17890E+06Fy

minus89440E+06Mz

00000E+00


1Member

1Ar1

0000E+00Ar2

minus1250E+06Ar3

minus2083E+06Ar4

0000E+00Ar5

minus1250E+06Ar6

2083E+06


Node12



12200Eminus03

minus14730Eminus09


2Fx

minus17890E+06Fy

minus16056E+06Mz

35560E+06


1F1

17890E+06F2

minus89440E+05F3


minus17890E+06F5

minus16056E+06F6

35560E+06








Ec(t2 t1) = 25GPa

1 + 08(20) = 9615GPa



250 (20)AD(t1)AB










123

x y00 00

100 00100 minus50


12

1st node11

2nd node23

a10000E+0152000Eminus02

I10000E+0010000Eminus06


23



00000E+00 00000E+00 00000E+00

00000E+00 00000E+00 00000E+00


1Node

1Fx

00000E+00Fy

00000E+00Mz

00000E+00


111

Member111

Ar1

minus1376E+06minus2885E+071250E+05

Ar2

minus2740E+050000E+000000E+00

Ar3

minus1602E+060000E+000000E+00

Ar4

1376E+06

2885E+07minus1250E+05

Ar5

2740E+05

0000E+00

0000E+00

Ar6

minus1133E+050000E+000000E+00



123



minus24022Eminus08



23

Fx


Fy

minus77388E+0477388E+04

Mz

77890E+05

19580E+01


12

F1minus15478E+0517305E+05

F277388E+0432547E+00

F3minus16808E+0116808E+01

F415478E+05

minus17305E+05


F677890E+0519580E+01
















Ec (t2 t1) = 25GPa

1 + 08(225) = 8929GPa





Ec (t1) φ(t2 t1) AD(t1)





123456789

101112

83225Eminus03




minus23736Eminus08





44022Eminus08

44022Eminus08

14765Eminus01

29254Eminus01

29254Eminus01

39067Eminus01

48840Eminus01

48840Eminus01

52367Eminus01

55758Eminus01

55758Eminus01

00000E+00


129

101112

00000E+00

00000E+00

00000E+00minus72000E+06minus72000E+0614400E+07

00000E+00




1234567

31

19209E+06

51887E+06

67612E+06

19209E+06

51887E+06

67612E+06

96046E+05

43425E+04


minus43425E+04





111111111111

123456123456


1544E+06

4170E+06

5433E+06

1544E+06

4170E+06

5433E+06

1429E+07

1429E+07

1429E+07

1429E+07

1429E+07

1429E+07


Node u v w123456789

101112

87131Eminus02




minus28656Eminus23







123456789

10 262728293031


71626E+05

71626E+05

80999E+05

80999E+05

86603E+05

86603E+05

64064E+05

72448E+05

77460E+05

64064E+05

72448E+05

77460E+05

32032E+05

68256E+05

74954E+05

38730E+05




250 (225) 19209 minus19209




= minus14286 14286kN











Ec(t2 t1) = 25GPa

1 + 08(20) = 9615GPa



123456789

10111213






17

13

Fx

25047E+05minus26649E+0726399E+07

Fy

minus31200E+0600000E+0000000E+00

Mz

35933E+05minus38885E+06minus51617E+00

Member end forces

Member F1 F2 F3 F4 F5 F6123456789

1011

31200E+06

26544E+07

26586E+07

26618E+07

26639E+07



35933E+05

10001E+06

20538E+06

28612E+06

34321E+06

37732E+06

11827Eminus02

33526Eminus02

50243Eminus02

61547Eminus02

67716Eminus02








11111111111111111

123456123456789

1011




2400E+06

2042E+07

2045E+07

2048E+07

2049E+07

2050E+07

4615E+06

2700E+07

2700E+07

2700E+07

2700E+07





Nodal displacements

Node123456789

10111213





Node17

13

Fx

86860E+03

19104E+06minus19190E+06

Fy

26096Eminus04

00000E+00

00000E+00

Mz


Member end forces

Member F1 F2 F3 F4 F5 F6123456789

1011





26096Eminus04

33753E+05

29902E+06

24999E+06

21136E+06








= minus9615


442kN minus2060kNm




= 2700kN




69 General




Notes






Chapter 7

71 Introduction














73 Sign convention









e = MN (71)


and ys




ε = εO + yψ (72)




σc = Ec1 minus y

ynεO y lt yn (74)

0 y yn (75)


σs = Es1 minus ys

ynεO (76)


εOEcyn

yt1 minus

y


ys

yn = N (77)

εOEcyn

yt

y1 minus y


ys

yn = M (78)

where




yn

yt





yn

yt


yn

yt


minus e = 0 (710)











σt1 0 while σb1 0 (711)




I1 minus ytB1

B1 minus ytA1

M

N

I1 minus ybB1

B1 minus ybA1

(712)





yt M

N yb (713)


ΣIs minus yt ΣBs

ΣBs minus yt ΣAs

M

N

ΣIs minus yb ΣBs

ΣBs minus yb ΣAs

(714)

















2 minus 4a1a3)

2a1

(716)

where

a1 = bw2 (717)




a3 = minus12h




















d = ΣαiAsidi

ΣαiAsi

(722)






Figu

re7

4D

epth

of t

he c

ompr

essi

on z

one

in a

fully

cra

cked

rec

tang

ular

sec

tion

subj

ecte

d to

ecc

entr

ic n

orm

al fo

rce


es = MN (723)



ρ = Asbd (724)

and









Figu

re7

5D

ista

nce

from

the

top

fibr

e to

the

cen

troi

d of

the

tra

nsfo

rmed

are

a of

a fu

lly c

rack

ed r

ecta

ngul

ar s

ectio

n

Figu

re7

6M

omen

t of

iner

tia a

bout

an

axis

thr

ough

the

cen

troi

d of

the

tra

nsfo

rmed

are

a of

a fu

lly c

rack

ed r

ecta

ngul

ar s

ectio

n

Tabl

e 7

1D

epth

of c

ompr

essi

on z

one

in a

fully

cra

cked

T s

ectio

n su

bjec

ted

to e

ccen

tric

com

pres

sive

forc

e c

= (c

oeffi

cien

t fr

om t

able

times 1

0minus3)d

b wb

= 0

05

e sd

= minus

09

e sd

= minus

10

e sd

= minus

15

e sd

= minus

20

e sd

= minus

50

e sd

= minus

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

000

582

090

982

958

632

024

615

013

713

713

752

5252

5252

4343

4343

4334

3434

3434

3131

3131

310

010

828

912

836

613

339

333

214

173

170

170

7772

7272

7261

6060

6060

4848

4848

4843

4343

4343

003

085

392

385

968

340

349

037

228

324

523

916

212

312

012

012

012

410

210

210

210

291

8282

8282

7974

7474

740

050

870

931

875

725

453

567

461

361

304

278

227

164

151

151

151

177

133

129

129

129

129

105

105

105

105

111

9595

9595

007

588

693

989

176

250

462

953

443

336

331

429

220

918

418

118

123

216

815

515

515

517

113

012

712

712

714

711

611

511

511

50

100

898

946

902

788

544

671

585

488

412

345

345

249

214

204

204

278

200

178

176

176

208

153

145

145

145

180

136

131

131

131

012

590

895

191

180

957

870

362

553

145

237

338

928

424

122

622

431

722

920

019

419

424

117

516

116

016

020

915

414

614

614

60

150

916

955

919

825

606

729

656

566

487

398

426

316

266

246

242

352

256

220

210

210

271

196

176

174

174

236

172

159

158

158

020

092

896

193

185

065

276

770

362

154

344

348

637

131

228

327

141

030

425

824

023

732

323

420

519

819

828

420

518

318

018

0

b wb

= 0

1

e sd

= minus

09

e sd

= minus

10

e sd

= minus

15

e sd

= minus

20

e sd

= minus

50

e sd

= minus

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

001

064

770

065

352

433

829

220

817

317

017

077

7272

7272

6160

6060

6048

4848

4848

4343

4343

430

020

670

717

676

562

371

373

290

231

212

211

120

100

100

100

100

9384

8484

8470

6868

6868

6261

6161

610

060

731

766

734

651

467

519

452

378

325

294

242

181

165

164

164

192

147

140

140

140

143

115

114

114

114

124

103

103

103

103

010

076

879

777

070

153

159

153

446

440

334

432

224

521

320

420

426

319

817

817

617

620

015

314

514

514

517

413

513

113

113

10

150

799

824

801

742

588

649

600

535

473

397

394

308

264

246

242

329

251

220

210

210

257

194

176

174

174

226

171

159

158

158

020

082

284

382

477

263

068

964

558

652

544

044

835

930

828

227

138

229

725

624

023

730

423

120

419

819

826

920

318

318

018

00

250

840

859

841

794

663

720

679

624

566

477

491

401

346

315

296

424

335

289

268

260

343

263

230

219

218

306

232

206

199

199

030

085

487

185

581

269

074

470

765

560

050

952

743

737

934

431

846

036

931

929

328

037

729

225

423

923

633

825

922

721

721

60

400

875

890

876

840

731

780

748

702

651

560

583

496

435

395

358

518

426

370

338

315

434

343

297

275

267

392

306

266

249

245

Tabl

e 7

1co

ntin

ued

b wb

= 0

2

e sd

= minus

09

e sd

= minus

10

e sd

= minus

15

e sd

= minus

20

e sd

= minus

50

e sd

= minus

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

002

054

256

954

648

536

831

826

922

721

121

111

610

010

010

010

092

8484

8484

7068

6868

6862

6161

6161

004

058

260

458

453

341

739

535

130

427

226

017

714

213

713

713

714

111

711

611

611

610

794

9494

9493

8585

8585

012

066

968

467

063

453

453

450

245

941

736

531

826

123

222

122

126

621

519

519

119

120

916

815

815

715

718

414

914

314

314

30

200

717

730

718

688

599

604

577

539

499

436

400

340

302

281

271

343

285

253

240

237

277

225

203

198

198

247

199

183

180

180

030

075

776

875

873

265

465

963

660

356

650

047

041

036

834

031

841

235

031

229

128

034

128

225

123

923

630

725

122

521

721

60

400

785

794

786

764

693

698

678

648

614

548

521

463

419

388

357

464

402

360

334

315

390

328

292

274

267

354

294

263

248

245

050

080

781

580

878

772

372

870

968

265

058

656

150

546

142

839

150

544

340

037

134

543

036

732

830

629

339

433

129

527

727

00

600

824

831

825

806

746

751

734

709

679

618

593

540

496

462

421

538

478

434

404

372

464

400

359

334

316

427

363

324

303

291

080

085

085

685

083

478

278

777

275

072

366

664

459

455

251

847

259

253

449

045

841

951

945

541

238

335

748

141

637

434

932

8

b wb

= 0

5

e sd

= minus

09

e sd

= minus

10

e sd

= minus

15

e sd

= minus

20

e sd

= minus

50

e sd

= minus

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

005

047

248

047

345

841

833

532

130

328

827

817

415

715

115

115

114

413

112

912

912

911

410

510

510

510

510

195

9595

950

100

525

531

526

513

477

408

396

381

364

342

242

221

209

204

204

205

185

177

176

176

165

149

145

145

145

147

133

131

131

131

030

063

263

663

262

359

554

353

552

350

848

038

336

134

333

031

833

731

429

728

628

028

225

924

523

823

625

623

422

121

721

60

500

687

690

688

680

656

611

604

594

581

555

460

440

422

407

388

413

390

372

358

344

353

329

311

300

293

324

299

283

274

270

075

073

273

573

372

670

566

666

065

164

061

552

650

648

947

445

247

945

643

842

340

441

639

237

336

034

638

536

034

233

031

91

000

764

766

764

758

740

704

699

691

681

658

573

555

538

523

500

527

506

488

472

451

464

440

421

406

389

431

407

388

374

360

125

078

879

078

878

376

673

372

872

171

269

161

059

357

756

253

956

554

552

751

249

050

247

946

044

542

646

944

542

641

139

41

500

807

808

807

802

786

756

752

745

737

717

639

623

608

594

571

596

577

560

545

522

534

511

492

477

457

501

477

458

443

424

200

083

583

683

583

181

779

178

778

177

475

668

567

165

764

462

264

462

761

159

757

458

456

354

553

050

855

252

951

049

547

4

Tabl

e 7

1co

ntin

ued b w

b =

10

e sd

minus09

minus10

minus15

minus20

minus50

minusinfin

010

044

834

020

417

614

513

10

200

507

412

271

237

198

180

060

062

054

640

836

731

629

11

000

678

614

483

441

385

358

150

072

466

854

650

444

741

72

000

757

706

592

550

493

463

250

078

173

562

758

753

050

03

000

801

758

656

617

560

530

400

083

079

270

066

360

957

9

Tabl

e 7

2D

epth

of t

he c

ompr

essi

on z

one

in a

fully

cra

cked

T s

ectio

n su

bjec

ted

to e

ccen

tric

ten

sile

forc

e c

= (c

oeffi

cien

t fr

om t

able

times 1

0minus3)d

b wb

= 0

05

e sd

= 1

e sd

= 5

e sd

= 1

0e s

d =

20

e sd

= 5

0e s

d =

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

000

59

99

99

1818

1818

1822

2222

2222

2525

2525

2528

2828

2828

3131

3131

310

010

1313

1313

1325

2525

2525

3131

3131

3135

3535

3535

4040

4040

4043

4343

4343

003

023

2323

2323

4343

4343

4353

5353

5353

6261

6161

6171

6868

6868

7974

7474

740

050

2929

2929

2956

5656

5656

7168

6868

6885

7878

7878

9987

8787

8711

195

9595

950

075

3636

3636

3671

6868

6868

9283

8383

8311

295

9595

9513

010

610

610

610

614

711

611

511

511

50

100

4242

4242

4286

7979

7979

112

9595

9595

137

110

109

109

109

159

123

121

121

121

180

136

131

131

131

012

546

4646

4646

9988

8888

8813

110

610

610

610

616

012

312

112

112

118

613

913

413

413

420

915

414

614

614

60

150

5151

5151

5111

396

9696

9614

911

711

611

611

618

113

713

213

213

221

015

514

614

614

623

617

215

915

815

80

200

5959

5959

5913

811

011

011

011

018

213

713

313

313

322

016

215

115

115

125

418

516

816

716

728

420

518

318

018

0

b wb

= 0

1

e sd

= 1

e sd

= 5

e sd

= 1

0e s

d =

20

e sd

= 5

0e s

d =

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

001

013

1313

1313

2525

2525

2531

3131

3131

3535

3535

3540

4040

4040

4343

4343

430

020

1818

1818

1835

3535

3535

4343

4343

4350

5050

5050

5656

5656

5662

6161

6161

006

032

3232

3232

6261

6161

6179

7474

7474

9585

8585

8511

095

9595

9512

410

310

310

310

30

100

4242

4242

4285

7979

7979

111

9595

9595

134

110

109

109

109

155

123

121

121

121

174

135

131

131

131

015

051

5151

5151

111

9696

9696

145

117

116

116

116

175

136

132

132

132

202

154

146

146

146

226

171

159

158

158

020

059

5959

5959

135

110

110

110

110

176

137

133

133

133

211

161

151

151

151

242

183

168

167

167

269

203

183

180

180

025

067

6565

6565

156

124

122

122

122

203

156

147

147

147

243

184

168

168

168

276

209

188

185

185

306

232

206

199

199

030

075

7171

7171

176

137

133

133

133

228

174

161

160

160

271

205

185

182

182

307

234

207

201

201

338

259

227

217

216

040

090

8282

8282

213

162

152

152

152

272

206

186

183

183

320

244

215

208

207

359

277

242

229

228

392

306

266

249

245

Tabl

e 7

2co

ntin

ued

b wb

= 0

2

e sd

= 1

e sd

= 5

e sd

= 2

0e s

d =

10

e sd

= 5

0e s

d =

infin

h fd

h fd

h fd

h fd

h f d

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

002

018

1818

1818

3535

3535

3543

4343

4343

5050

5050

5056

5656

5656

6261

6161

610

040

2626

2626

2650

5050

5050

6261

6161

6173

7070

7070

8378

7878

7893

8585

8585

012

045

4545

4545

9486

8686

8612

110

410

410

410

414

512

011

911

911

916

513

513

213

213

218

414

914

314

314

30

200

5959

5959

5913

011

011

011

011

016

713

613

313

313

319

716

015

115

115

122

418

116

816

716

724

719

918

318

018

00

300

7571

7171

7116

713

713

313

313

321

217

116

116

016

024

920

118

518

218

228

122

820

620

120

130

725

122

521

721

60

400

8982

8282

8219

916

115

215

215

225

120

218

518

318

329

223

721

420

820

732

626

824

022

922

835

429

426

324

824

50

500

102

9292

9292

227

183

170

169

169

283

230

208

203

203

328

269

241

230

229

364

302

270

255

251

394

331

295

277

270

060

011

510

010

010

010

025

220

318

618

418

431

225

522

922

122

035

929

726

625

224

839

633

329

727

927

142

736

332

430

329

10

800

138

116

115

115

115

295

240

217

210

210

361

299

268

253

250

410

346

309

290

281

449

384

344

322

306

481

416

374

349

328

b wb

= 0

5

e sd

= 1

e sd

= 5

e sd

= 1

0e s

d =

20

e sd

= 5

0e s

d =

infin

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

005

029

2929

2929

5656

5656

5670

6868

6868

8178

7878

7892

8787

8787

101

9595

9595

010

042

4242

4242

8279

7979

7910

295

9595

9511

910

910

910

910

913

412

212

112

112

114

713

313

113

113

10

300

7371

7171

7114

913

513

313

313

318

416

616

116

016

021

319

218

418

218

223

621

520

420

120

125

623

422

121

721

60

500

9792

9292

9219

617

717

016

916

923

921

720

620

320

327

324

923

623

022

930

027

626

125

325

132

429

928

327

427

00

750

122

112

111

111

111

240

218

207

204

204

290

266

251

244

243

328

303

287

278

273

359

334

317

306

298

385

360

342

330

319

100

014

312

912

812

812

827

625

223

923

323

233

130

528

928

027

537

234

632

931

730

840

537

936

134

833

643

140

738

837

436

01

250

161

145

142

142

142

306

282

267

259

256

365

339

321

310

302

408

382

364

351

339

442

417

398

384

369

469

445

426

411

394

150

017

716

015

515

515

533

330

829

128

227

739

436

834

933

732

743

941

339

438

036

647

344

842

941

539

850

147

745

844

342

42

000

206

186

178

177

177

378

352

334

322

313

442

416

397

383

369

489

464

444

429

412

524

500

481

466

446

552

529

510

495

474

Tabl

e 7

2co

ntin

ued b w

b =

10

e sd

15

10

20

50

infin

010

042

7995

109

121

131

020

059

110

133

151

167

180

060

010

018

422

024

827

129

11

000

128

232

275

308

335

358

150

015

527

732

636

439

341

72

000

177

313

367

406

438

463

250

019

734

440

044

247

450

03

000

214

370

429

472

504

530

400

024

341

447

652

055

357

9

Tabl

e 7

3D

ista

nce

from

top

fibr

e to

cen

troi

d of

tra

nsfo

rmed

1 fully

cra

cked

T s

ectio

n y

= (c

oeffi

cien

t fr

om t

able

times 1

0minus3)d

b wb

= 0

05

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

000

536

5454

5454

4659

7910

210

262

6785

105

151

108

9510

311

915

918

414

513

914

717

727

220

718

718

520

30

010

4559

5959

5954

6382

104

104

6972

8710

815

211

499

106

121

160

188

148

142

149

178

276

210

189

186

204

003

079

7777

7777

8581

9411

311

398

8899

116

158

138

113

116

129

166

207

160

151

156

183

290

221

197

193

209

005

011

195

9595

9511

497

105

121

121

125

104

110

125

163

160

127

127

137

171

224

173

160

163

188

304

231

206

199

213

007

514

811

611

611

611

614

911

711

913

213

215

612

312

313

517

018

614

513

914

717

724

518

717

117

219

432

024

421

520

821

90

100

182

136

136

136

136

180

136

133

142

142

185

141

136

145

177

211

161

151

156

184

265

202

182

181

200

336

256

225

215

225

012

521

415

515

515

515

520

915

514

615

215

221

215

914

915

518

423

417

716

316

519

028

421

519

318

920

635

126

823

522

323

00

150

243

173

173

173

173

237

172

159

162

162

237

176

161

164

190

256

192

174

174

196

302

229

203

198

212

366

279

244

231

236

020

029

520

820

820

820

828

620

518

318

118

128

420

718

518

320

329

622

119

719

220

933

525

422

321

422

439

330

126

224

624

7

b wb

= 0

1

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

001

046

5959

5959

6267

8410

410

487

8294

111

152

152

127

127

136

168

250

202

186

184

201

357

290

259

245

246

002

063

6868

6868

7675

8910

810

899

9099

115

155

161

133

131

140

170

256

207

189

187

203

361

294

262

248

248

006

012

510

310

310

310

312

810

711

212

612

614

311

912

013

116

619

415

714

915

418

027

922

520

419

921

237

830

827

425

825

60

100

178

136

136

136

136

174

137

134

142

142

183

146

140

147

177

224

179

167

168

190

301

243

219

211

221

394

322

286

268

264

015

023

717

317

317

317

322

617

115

916

216

222

917

716

416

619

026

020

618

818

520

232

726

423

622

623

241

333

930

028

127

40

200

288

208

208

208

208

272

203

183

181

181

269

207

187

184

203

292

231

207

201

214

351

284

253

240

242

431

354

314

293

283

025

033

224

024

024

024

031

223

320

619

919

930

623

420

820

221

532

125

422

621

722

637

430

326

925

425

344

736

932

730

429

20

300

372

269

269

269

269

348

260

227

217

217

339

259

228

218

227

348

276

244

232

237

394

321

284

267

262

463

384

340

316

301

040

043

832

132

132

132

141

030

926

725

025

039

630

626

625

025

039

731

627

826

125

843

235

431

229

128

249

241

036

433

731

8

1T

he s

ame

tabl

e m

ay b

e us

ed fo

r ag

e-ad

just

ed t

rans

form

ed s

ectio

n re

plac

ing

b

y ndash

= E

s[1 +

(t

t0)

]E c

(t0)

Tabl

e 7

3co

ntin

ued

b wb

= 0

2

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

002

064

6868

6868

8581

9210

810

811

710

510

912

115

519

717

016

216

418

430

826

824

723

724

042

537

534

432

531

20

040

9386

8686

8610

796

103

117

117

134

118

119

129

161

208

179

169

170

188

315

274

252

242

244

430

380

348

329

315

012

019

415

115

115

115

118

415

114

515

015

019

616

415

615

918

225

021

319

819

420

734

229

827

326

125

944

839

736

434

432

70

200

274

208

208

208

208

250

200

183

181

181

250

206

189

187

203

287

244

224

217

224

367

320

293

279

273

465

413

379

357

339

030

035

526

926

926

926

931

825

322

621

721

730

725

222

821

922

732

928

025

524

424

539

634

731

730

029

148

543

239

737

435

30

400

419

321

321

321

321

375

299

264

250

250

357

294

263

250

250

366

313

284

269

265

422

371

339

320

307

503

449

413

389

366

050

047

236

636

636

636

642

334

129

927

927

939

933

129

527

727

139

934

331

129

328

444

639

336

033

932

352

046

642

940

437

90

600

516

406

406

406

406

464

377

331

307

307

437

365

324

303

291

429

370

335

315

302

468

414

379

357

338

536

482

444

419

391

080

058

547

247

247

247

253

143

938

735

735

749

942

237

634

932

848

141

938

035

533

550

845

241

439

036

656

551

147

244

541

5

b wb

= 0

5

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h f d

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

005

010

195

9595

9512

011

211

412

112

115

614

614

414

716

324

322

922

121

822

236

134

433

232

431

748

246

344

943

842

30

100

154

136

136

136

136

152

140

138

142

142

179

166

162

163

177

256

241

233

229

231

368

352

339

331

323

487

468

454

442

428

030

031

526

926

926

926

926

223

622

221

721

725

923

922

722

322

730

528

727

626

926

739

838

036

735

734

750

548

747

246

044

40

500

424

366

366

366

366

346

312

291

279

279

324

299

284

274

271

348

328

314

306

299

424

406

392

382

370

522

504

489

476

460

075

052

045

745

745

745

742

838

836

234

534

539

236

334

333

032

039

437

335

734

733

645

543

642

141

039

654

252

450

849

647

91

000

589

524

524

524

524

491

449

420

400

400

447

416

394

378

362

434

412

395

383

369

482

463

447

435

420

560

542

526

514

496

125

064

057

755

755

757

754

250

046

844

644

649

346

143

741

939

947

044

742

941

539

950

748

747

145

944

357

755

954

353

151

21

500

680

620

620

620

620

584

541

509

485

485

532

500

474

456

433

501

477

459

444

427

529

509

493

480

463

593

574

559

546

528

200

073

868

368

368

368

364

860

757

454

954

959

456

253

651

649

055

352

951

049

547

456

854

853

251

950

062

160

358

757

455

5

Tabl

e 7

3co

ntin

ued b w

b =

10

cd

12

35

75

10

010

013

614

217

726

438

350

40

200

208

181

203

278

391

509

060

040

630

729

133

042

152

81

000

524

400

362

375

448

545

150

061

948

543

342

347

956

52

000

683

549

489

464

506

583

250

072

859

953

650

053

159

93

000

762

640

574

531

553

615

400

080

969

963

558

359

264

2

Tabl

e 7

4M

omen

t of

iner

tia o

f a t

rans

form

ed fu

lly c

rack

ed T

sec

tion

abou

t ce

ntro

idal

axi

s I =

(coe

ffici

ent

from

tab

le times

10minus4

)bd3

b wb

= 0

05

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

000

54

55

55

55

710

107

78

1126

1718

1820

3246

5050

5159

9911

111

411

411

80

010

99

99

99

1011

1414

1111

1215

2921

2222

2435

5054

5454

6210

211

511

711

712

10

030

2727

2727

2727

2728

3030

2828

2931

4336

3838

3949

6268

6869

7511

212

713

013

113

40

050

4343

4343

4343

4344

4646

4445

4547

5851

5353

5463

7582

8383

8912

213

914

314

314

60

075

6263

6363

6362

6364

6565

6264

6465

7568

7272

7280

8999

100

100

105

133

153

159

159

161

010

079

8282

8282

7982

8383

8379

8383

8492

8490

9090

9710

311

517

711

712

114

516

717

417

517

60

125

9510

110

110

110

195

101

101

101

101

9610

110

210

210

999

107

108

108

114

116

130

133

134

137

155

181

189

190

191

015

011

011

811

811

811

811

011

811

911

911

911

111

811

912

012

511

312

312

512

513

012

914

514

915

015

316

619

420

320

520

60

200

137

151

151

151

151

138

151

153

153

153

138

151

154

154

157

139

155

158

159

162

152

174

180

181

183

185

219

231

234

235

b wb

= 0

1

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

001

09

99

99

1010

1114

1413

1314

1629

2931

3132

4175

8587

8791

157

185

195

197

198

002

018

1818

1818

1919

2022

2221

2222

2436

3639

3940

4880

9194

9497

161

190

200

203

204

006

051

5151

5151

5152

5253

5352

5454

5564

6368

6869

7510

211

612

012

012

217

720

922

222

522

60

100

7982

8282

8280

8283

8383

8084

8484

9288

9697

9710

212

213

914

414

514

719

222

824

224

724

80

150

111

118

118

118

118

111

118

119

119

119

112

119

120

120

125

117

128

131

131

134

145

167

174

176

177

210

250

267

273

275

020

013

815

115

115

115

114

015

115

315

315

314

015

115

415

415

714

315

916

316

316

516

719

320

320

520

622

727

229

129

930

10

250

162

181

181

181

181

165

182

186

186

186

165

182

186

186

189

167

187

193

194

196

188

218

230

233

234

242

292

314

323

326

030

018

320

920

920

920

918

721

021

621

721

718

821

021

721

821

918

921

422

222

422

520

624

225

626

126

225

731

133

634

735

10

400

218

258

258

258

258

225

261

273

276

276

228

261

273

276

277

229

264

277

281

282

241

286

305

313

315

284

348

378

392

398

Tabl

e 7

4co

ntin

ued

b wb

= 0

2

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h f d

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

002

018

1818

1818

1919

2022

2224

2425

2636

4853

5354

5911

813

614

314

514

624

829

031

332

332

70

040

3535

3535

3536

3636

3838

3940

4041

5161

6667

6772

128

147

155

157

158

255

298

321

332

337

012

093

9797

9797

9497

9798

9895

9910

010

010

510

811

812

112

112

316

418

819

820

220

228

032

835

436

737

40

200

140

151

151

151

151

143

151

153

153

153

143

152

154

154

157

151

165

170

171

173

197

226

239

244

245

303

356

386

401

409

030

018

720

920

920

920

919

421

121

721

721

719

521

121

721

821

919

922

022

823

023

123

527

028

729

529

733

139

042

344

145

20

400

224

258

258

258

258

236

263

273

276

276

239

264

274

276

277

241

269

281

285

286

270

311

332

342

346

356

421

458

479

493

050

025

530

130

130

130

127

231

032

533

033

027

831

132

533

033

227

931

433

033

733

930

234

937

538

739

338

045

049

251

553

20

600

280

339

339

339

339

303

351

372

380

380

312

354

373

381

383

313

355

376

385

389

331

385

414

429

438

402

478

524

550

570

080

032

040

140

140

140

135

442

045

346

946

936

842

745

747

147

837

242

945

947

448

138

344

948

750

852

144

352

858

261

464

1

b wb

= 0

5

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

005

043

4343

4343

4445

4546

4651

5252

5358

9299

101

102

103

225

244

255

261

264

483

523

551

569

585

010

081

8282

8282

8283

8383

8385

8888

8892

120

128

131

132

133

245

265

278

284

287

497

537

566

584

602

030

019

720

920

920

920

920

721

421

721

721

720

721

521

821

821

922

423

624

224

524

532

134

636

137

037

654

759

262

364

566

50

500

276

301

301

301

301

303

319

327

330

330

307

321

328

331

332

314

332

341

346

348

390

419

438

449

458

595

643

677

701

725

075

034

438

738

738

738

739

642

444

044

844

841

043

344

645

245

541

343

745

145

946

446

850

352

654

055

364

970

274

076

779

61

000

394

452

452

452

452

469

508

532

546

546

493

525

545

556

564

498

529

548

560

568

539

578

606

624

640

699

756

798

828

861

125

043

150

250

250

250

252

757

760

962

962

956

360

463

064

665

957

361

063

565

066

360

364

767

970

072

074

680

785

288

592

21

500

459

542

542

542

542

575

634

674

700

700

622

671

704

725

744

639

682

712

731

749

661

710

746

770

795

789

853

902

938

980

200

050

160

260

260

260

264

872

477

981

681

671

778

182

685

788

975

080

584

487

189

976

282

186

489

592

986

693

899

310

3410

85

Tabl

e 7

4co

ntin

ued

b wb

= 1

0

cd

12

35

75

10

010

082

8392

159

390

858

020

015

115

315

721

242

788

20

600

339

380

383

405

568

974

100

045

254

656

457

269

610

601

500

542

700

744

753

839

1159

200

060

281

688

990

796

812

492

500

645

906

1007

1041

1083

1333

300

067

797

811

0611

5811

8814

104

000

722

1086

1261

1354

1370

1547





α = Es

Ec

= 200

30 = 6667






2 times 015

= 0200m (79 in)


I = 0302003

3 + (15 minus 03)0120122

12 + 0142

+ 6667(0004)10002 + 5667(00006)0152

= 3054 times 10minus3 m4 (353 ft4)


ψ = 1000 times 103





= 2182 MPa (3165ksi)






c = 0444m (175 in)


A = 03073m2 I = 3173 times 10minus3 m4




N = minus800kN


= 3832kN m (3400kip in)


εO = 1




ψ = 1

30 times 109 3832 times 103

3173 times 10minus3












yc

r2c + εcs(t t0)

yc

r2c (727)


∆σs = Es(∆εO + ∆ψys) (729)

where

r2c = IcAc (730)








Figu

re7

9A

naly

sis

of c

hang

es in

str

ain

and

stre

ss d

ue t

o cr

eep

and

shri

nkag

e of

a fu

lly c

rack

ed r

einf

orce

d co

ncre

te s

ectio

n (E

xam

ple

73)

(a

) effe

ctiv

e ar

ea (

b) s

trai

n an

d st

ress

dis

trib

utio

ns a

t t0

(c) c

hang

es in

str

ain

and

stre

ss d

ue to

cre

ep a

nd s

hrin

kage

(d)

str

ain

and

stre

ss a

t t

Tabl

e 7

5C

urva

ture

red

uctio

n fa

ctor

κ fo

r a

fully

cra

cked

T s

ectio

n (fo

r us

e in

Equ

atio

n (6

23)

) κ =

coe

ffici

ent

from

tab

le times

10minus3

b wb

= 0

05

cd

= 1

0c

d=

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

ndash

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

000

543

160

160

160

160

170

236

420

623

623

408

426

522

661

862

774

773

781

809

891

928

927

927

929

942

973

971

971

971

973

001

036

9494

9494

105

140

270

455

455

266

276

356

496

758

636

632

643

681

804

868

865

864

868

892

948

945

944

943

947

003

059

5858

5858

8075

126

229

229

146

135

171

258

516

394

381

387

425

582

700

691

686

691

736

865

857

852

851

860

005

090

6565

6565

9874

9816

516

513

611

112

818

539

731

228

928

931

846

160

058

457

658

063

080

178

878

077

778

90

075

127

8181

8181

128

8692

134

134

149

110

112

149

315

274

240

233

252

372

524

500

488

489

539

741

721

710

705

717

010

016

110

010

010

010

015

710

196

123

123

170

118

110

134

267

262

219

206

217

318

479

447

431

429

474

697

670

655

649

660

012

519

311

811

811

811

818

611

810

312

012

019

213

011

412

923

726

321

119

319

828

245

141

239

238

842

766

363

161

360

461

40

150

223

136

136

136

136

213

134

112

121

121

215

143

121

128

218

270

210

187

187

257

435

389

366

358

392

637

600

579

568

576

020

027

617

117

117

117

126

216

713

313

013

025

817

113

813

519

629

221

918

718

022

842

036

333

332

134

460

255

652

951

551

8

b wb

= 0

1

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

ndash

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

001

041

9494

9494

160

179

286

455

455

387

387

430

530

758

758

759

760

771

834

925

925

924

924

930

973

972

971

971

971

002

047

6363

6363

107

111

175

300

300

256

250

281

366

613

618

617

616

631

717

863

863

860

860

870

949

947

945

944

944

006

010

271

7171

7111

184

9814

914

916

413

914

418

335

839

037

737

038

046

969

669

168

468

169

786

986

385

785

485

30

100

156

100

100

100

100

149

103

9712

312

317

413

412

514

526

732

430

028

629

036

060

259

057

957

358

980

979

879

078

478

20

150

215

136

136

136

136

198

133

113

121

121

204

150

130

135

218

301

265

245

241

292

535

516

499

490

501

754

738

725

717

712

020

026

717

117

117

117

124

316

413

313

013

023

917

214

313

919

630

325

723

022

125

649

947

145

143

844

471

469

367

766

665

90

250

313

203

203

203

203

284

193

154

142

142

273

196

160

148

186

314

260

228

214

237

479

445

420

405

405

686

660

640

626

616

030

035

323

323

323

323

332

122

117

415

715

730

521

917

716

018

433

026

823

221

422

646

942

940

138

337

866

463

461

259

658

30

400

421

288

288

288

288

385

271

214

186

186

362

265

213

186

189

365

292

249

224

220

466

417

382

359

345

638

601

573

553

534

Tabl

e 7

5co

ntin

ued

b wb

= 0

2

cd

= 1

0c

d =

20

cd

= 3

0c

d =

75

cd

= 5

0c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

ndash

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

002

049

6363

6363

150

148

194

300

300

356

355

366

417

613

733

741

739

741

775

919

921

921

920

921

973

973

972

971

971

004

071

6161

6161

114

103

124

189

189

241

233

238

274

448

590

596

593

594

637

853

857

856

854

855

949

948

947

945

944

012

016

911

411

411

411

415

311

510

512

012

018

515

814

615

424

237

837

336

335

939

168

468

668

167

667

586

986

786

385

985

40

200

251

171

171

171

171

214

157

132

130

130

215

174

152

147

196

329

313

296

287

305

595

592

583

575

571

811

807

800

794

786

030

033

423

323

323

323

328

221

017

215

715

726

320

917

816

418

432

229

527

325

826

353

652

751

450

249

376

075

274

273

372

20

400

400

288

288

288

288

340

257

211

186

186

310

247

208

187

189

336

301

273

254

248

507

492

475

461

446

723

712

700

689

674

050

045

533

533

533

533

539

030

024

621

621

635

328

323

821

220

035

631

528

326

024

649

347

445

443

741

769

868

466

965

663

80

600

500

376

376

376

376

433

338

280

244

244

391

316

268

236

215

378

332

297

271

250

489

466

443

423

400

680

663

647

632

611

080

057

144

544

544

544

550

340

333

829

529

545

737

632

128

324

742

437

133

029

926

749

646

643

841

538

465

963

861

759

957

3

b wb

= 0

5

cd

= 1

0c

d =

20

cd

= 3

0c

d =

50

cd

= 7

5c

d =

100

h fd

h fd

h fd

h fd

h fd

h fd

ndash

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

05

10

15

20

30

005

074

6565

6565

135

130

135

165

165

305

309

307

316

397

691

701

702

701

707

909

912

912

911

972

972

972

972

971

010

012

310

010

010

010

012

411

411

012

312

321

521

320

921

226

754

255

255

355

155

683

784

284

384

284

194

794

794

794

694

50

300

287

233

233

233

233

211

184

166

157

157

206

192

180

172

184

354

356

352

347

344

661

667

667

665

660

866

866

865

864

861

050

040

133

533

533

533

529

626

023

421

621

625

923

822

020

720

032

432

131

330

529

657

658

057

957

556

780

880

880

780

479

90

750

500

429

429

429

429

382

340

307

283

283

325

298

276

258

239

335

326

315

304

289

525

526

523

517

506

758

758

755

752

744

100

057

250

050

050

050

044

940

536

934

134

138

235

232

730

627

936

034

733

432

130

250

350

249

649

047

672

472

371

971

570

51

250

625

556

556

556

556

504

458

421

391

391

431

400

372

349

318

388

374

358

344

321

496

493

486

478

462

701

699

694

689

678

150

066

760

060

060

060

054

950

346

543

443

447

444

141

338

835

341

740

038

436

834

349

749

248

447

545

868

568

267

767

165

92

000

727

667

667

667

667

618

574

536

504

504

542

509

480

454

415

470

451

433

415

387

512

504

494

484

463

669

664

658

650

636

Tabl

e 7

5co

ntin

ued

b wb

= 1

0

cd

ndash

12

35

75

10

010

010

012

326

766

090

297

10

200

171

130

196

510

826

945

060

037

624

421

533

664

686

31

000

500

341

279

318

563

805

150

060

043

435

333

751

575

52

000

667

504

415

367

497

722

250

071

455

846

639

949

369

93

000

750

602

510

431

496

685

400

080

066

857

948

751

567

0


η = AcA (731)

κ = IcI (732)
















Ec(t t0) = 30 times 109

1 + 075 times 25 = 1043GPa (1500ksi)

α(t t0) = 200

1043 = 1917


A = 03648m2 (560 in2)







η = 02766

03648 = 07582 κ =

1756

7301 = 02404










∆ψ = 0240425403 minus 35 (minus022)






= 0876MPa (0127ksi)

















M = M1 + M2 (733)

N = N1 + N2 (734)








(∆ε)1 = (∆εO)1 + (∆ψ)1y (735)

(∆ε)2 = (∆εO)2 + (∆ψ)2 y (736)


∆ε = (∆ε)1 + (∆ε)2 (737)












γ = γ(t) = d

dyσ(t) (742)



M1 = minusIγ (744)


(∆εO)1 = minus1

Ec

σO (745)

(∆ψ)1 = minus1

Ec

γ (746)




77 Flow chart












α = 20024 = 833

A = 00881 + 833(930 + 1000)10minus6 = 01042m2










= 10948MPa (1588ksi)




Ec = 24 times 109

1 + 24 times 08 = 8215GPa (1192ksi)

α = 200

8215 = 2433


A = 00881 + 2433(930 + 1000)10minus6 = 01351m2





= 08759 times 106 N (1969kip)







= 4407MPa (06392ksi)








α = 20035 = 571


A = 00881 + 571(930 + 1000)10minus6 = 00991m2





(∆ε)1 = 6152 times 106





Transformed area is

A = 571(930 + 1000)10minus6 = 00110m2


N2 = 1200 minus 6098 = 5902kN (113kip)


(∆ε)2 = 5902 times 103




= 3060MPa (444ksi)



(∆ε)2 = 5902 times 103





10948 minus 2479 + 352 + 3060 = 11881MPa (1723ksi)














α = 20030 = 6667



A = 02768 + 6667(1600 + 1200 + 400)10minus6 = 02981m2













= 0955 times 106N




(∆εO)1 = 3490 times 106












es = 6299 times 103




c = 0263m


















and curvature














(∆εO)1 = minus1


(∆ψ)1 = minus1






A = 635 in2 B = 5824 in3 I = 1418 times 103 in4












79 General





Note





Chapter 8

81 Introduction





















σsr = NrAs (82)








σs2 = NAs (83)

εs2 = NEsAs (84)


εsm = ∆ll (85)




εs1 = εc1 = N

Ec(Ac + αAs) =

N

EcA1

(87)






σs2

(88)



σs2

(89)




ζ = 1 minus σsr

σs2

2




or

ζ = 1 minus Nr

N2





σs2

2













σsr = Nr

As

= 111MPa σs2 = 249MPa









Mr = W1 fct (815)






where


σs2

2

= 1 minus β1β2Mr

M2

(817)





ψ = M

EI(819)



or


d(820)


ψm = (1 minus ζ)ψ1 + ζψ2 (821)



(EI)m = M

ψm

(822)




ψ1 = M

EcI1

(823)

ψ2 = M

EcI2

(824)
















I1 = 00191m4 W1 = 00488m3



c = 0191m (752 in)


I2 = 000543m4














e = MN (825)



Nr = fct 1

A +

e

Wbot

minus1

1(826)

Mr = eNr (827)







where


σs2

2

(829)

or


M2

(830)

or




N2

(831)



σsr

σs2

= Mr

M =

Nr

N(832)




ψm = (1 minus ζ)ψ1 + ζψ2 (834)







A1 = 0336m2



e = 250



Mr = 138kN-m


ζ = 085




c = 0241m (949 in)


y = 0195m (768 in)


A2 = 0115m2 I2 = 000544m4



ψ2 = 209 times 103
















Mr = fct minus N

A1W1 (835)












where


N2

(837)


α = EsEc (839)








ψm = (1 minus ζ)ψ1 + ζψ2 (840)

where


M2

(841)

Mr = fctW1 (842)




ψm = (1 minus ζ)ψ1 + ζψ2 (844)

where


M2

= 1 minus β1β2Nr

N2

(845)






e = M

N =

Mr

Nr

(846)


Mr = efct 1

A1

+ e

W1

minus1

(847)






σ1 max

2

(849)




























136 2

= 082





136 2

= 091


State 1

ψ1(t0) = 136 times 103


State 2

ψ2(t0) = 136 times 103







Ec(t t0) = 30 times 109

1 + 08 times 25 = 10GPa

α = Es

Ec(t t0) =

200

10 = 20







8724 times 10minus3 = 0795




κ2 = 1190

4277 = 0278





3534 times 10minus3









2762 times 10minus3











96

= 00239m

= 239mm (0948 in)

















σ1 max = 2323 + 8580 = 10903MPa




σ1 max

2


109032

= 095










= 1012mm (399 in)















θ1 = T

GcJ1

(850)



c 1 minus b4

12c4 (851)


τmax = T

microbc2(852)






cb 10 15 175 20 25 30 40 60 80 100 infin

0208 0231 0239 0246 0258 0267 0282 0299 0307 0313 0333


τt = T2A0 (854)







τt = T2hb (855)






F1 = F3 = 1

2hF2 = F4 =

1

2b (856)


F5 = F6 = F7 = F8 = x

2hb (857)


F9 = F11 = minus 1

2b sin α1

F10 = F12 = minus 1

2h sin α2

(858)



θ = T

x 12

i = 1F

2l

AEi

(859)


Ai = Av

x

s(860)



A9 = A11 = th cos α1 (861)

and

A10 = A12 = tb cos α2 (862)


















ζ = 1 minus 10(05) 0500

10962

= 0896





96 [0 + 10(3547 times 10minus6) + 0] = 341 in























811 General






Notes











91 Introduction



Chapter 9








ψ(t0) = κsψc (91)

(∆ψ)φ = ψ(t0)φκφ (92)



d κcs (93)


ψc = M

Ec(t0)Ig

(94)






















10s1 09 0 to 02 ndash F1

08

10s2 09 0 to 02 ndash F2

08

1010 0 to 02 20 F3

3040

101 09 0 to 02 20 F4

3040

1020

08 0 to 02 30 F540

100 20 F6

3040

1020

2 08 to 10 01 30 F740

1002 20 F8

3040


0cs2 08 to 10 20 F10

02













D1(t0) = κs1Dc (95)

D2(t0) = κs2Dc (96)



(∆D1)φ = D1(t0)φκφ1 (97)

(∆D2)φ = D2(t0)φκφ2 (98)





I (99)

or


d(910)







l2

8(911)





A (912)



A (913)






l2

2(914)





8d(915)


8d(916)


κcs = minusAcd

Iyc (917)





A (918)









Ac

A (920)






D1 = D1(t0) + (∆D1)φ + (∆D1)cs (921)

D2 = D2(t0) + (∆D2)φ + (∆D2)cs (922)





D = (1 minus ζ)D1 + ζD2 (923)



M(924)


Mr = fctW1 (925)









Ie = Mr

Mm

Ig + 1 minus Mr

Mm

I2 with M Mr (926)

where








ρ = As


Aprimesbd

= 015 per cent



Ig = bh3

12 =

03 times (065)3


Basic deflection

Dc = 5

384

ql4

Ec(t0)Ig

= 5

384




κs1 = 092 κφ1 = 079 κcs1 = 027

κs2 = 380 κφ2 = 014 κcs2 = 097


D1(t0) = 092 times 440 = 405mm

D2(t0) = 380 times 440 = 1672mm


(∆D1)φ = 405 times 25 times 079 = 800mm

(∆D2)φ = 1672 times 25 times 014 = 585mm





8 times 06 = 090mm


8 times 06 = 323mm


D1 = 405 + 800 + 090 = 1295mm

D2 = 1672 + 585 + 323 = 258mm


Mr bh2

6fct =

03 times (065)2

6 25 times 106 = 528kN-m


M = 17 times 82

8 = 1360kN-m



1360 = 081


(1 minus 081)1295 + 081 times 258 = 234mm (0920 in)















κs1(1 + κφ1)(927)



κs2(1 + κφ2)(928)


MO minusN |y12| 1




ψ2N = minusMO 1

(slope)OD

minus 1

(slope)AB (930)












D = (1 minus ζ)D1 + ζD2 (931)



ζ = 1 minus β1β2

Mr

M(932)

0 for M lt Mr (933)




ζ = 1 minusMO

M(934)

0 for M lt MO (935)


Mr = fct minus N

A1W1 (936)














κs1 = 092 κφ1 = 079 κs2 = 38 κφ2 = 014


D1(t0) = 405mm D2(t0) = 1672mm


(∆D1)φ = 800mm (∆D2)φ = 585mm



D1 = 405 + 800 = 1205mm


D2 = 1672 + 585 = 2257mm


M = 1360kN-m



03 times 065 03(065)2

6 = 961kN-m


|y12| = 331 minus 145 = 186mm



1 minus (09238) = 981kN-m




ζ = 1 minus 981

1360 = 028











approximation

D Dcκth

d 3






Figu

re9

8G

loba

l coe

ffici

ent

κt f

or c

alcu

latio

n of

inst

anta

neou

s pl

us c

reep

def

lect

ions

of u

ncra

cked

or

crac

ked

mem

bers

by

Equa

tion

(93

8)















M(940)






l 2

8dκtcs (942)




l 2

2dκtcs (942a)










Mr = 528kN-m M = 1360kN-m ζ = 081

αρ = 200

30 times

06

100 = 004

Mr

M =

528

1360 = 039


44 times 38 065

06 3


100 = 206mm



8(06) = 28mm


D = 206 + 28 = 234mm (094 in)







8 times 0275 = 2327kN


M = 20 times 60

100 times

82

8 = 960kN-m



03 times 065 03(065)2

6 = 780kN-m

Mr

M =

78

96 = 081



Total steel ratio

ρ = (500 + 200)10minus6


α = 200

30 = 667 αρ = 0026


960 times 103

30 times 109(03 times 065312)

82

96 = 311mm


03 times 060 = 00011


D 311 times 44065

0603

(1 minus 20 times 00011) = 171mm (0671 in)









δ = l 2

96 (ψ1 + 10ψ2 + ψ3) (943)

where






member


D = δEF + 1

2(δAB + δDC) (944)

or

D = δHI + 1

2(δAD + δBC) (945)









ψx = 12


12



Ig slab = h3

12(1 minus ν2)(948)


ψbeam = M

EcIg beam

(949)

















(EI) slab

(EI) beam cl = 00 cl = 01 cl = 02



08 00 420 378 230 301 262 131 189 155 05702 316 271 149 237 192 088 159 116 04005 146 195 099 191 138 059 136 085 02810 191 134 063 154 095 038 117 058 01820 147 083 036 124 058 022 100 036 01140 116 048 019 103 033 012 089 020 006

06 00 327 321 099 234 228 040 143 137 00802 256 246 063 189 178 027 119 108 00605 201 187 040 150 134 017 098 098 00410 153 135 025 116 096 011 079 059 00320 110 087 013 085 061 006 061 037 00240 077 051 007 063 035 003 048 022 001

04 00 284 284 031 204 204 00402 231 230 020 166 165 00305 183 181 012 131 128 00210 137 134 007 098 094 00120 093 088 004 066 061 00140 059 053 002 042 036 000

(El)slab = Ec

h3









δ = (1 minus ζ)δ1 + ζδ2 (952)









Ig slab = (02)3





000482ql 4

EcIg slab

= 560mm (0221 in)


000342ql 4

EcIg slab

= 397mm (0157 in)








κs1 = 098 κφ1 = 093


κs2 = 70 κφ2 = 014


δ1 = 397(098)[1 + 25(093)] = 1294mm

δ2 = 397(70)[1 + 25(014)] = 3752mm



6 = 133kN-m



1 minus β1β2

Mr

M = 1 minus 10(05)

133

186 = 064


δAB = (1 minus 064)1294 + 064(3752) = 2867mm




κsl = 098 κφ1 = 095


δEF = 163(098)[1 + 25(095)] = 538mm


D = 2867 + 538 = 3405mm (1341 in)














ψ2 = 12[281 minus 02 (minus144)] 103



δABbasic = (655)2



ψ1 = ψ3 = 12[minus144 minus 02(281)]103


ψ2 = 12[101 minus 02(187)] 103



δEFbasic = (700)2



Dbasic = 649 + 072 = 721mm (0284 in)



Tabl

e 9

3D

eflec

tion

at c

entr

e of

pan

el c

alcu

late

d fr

om t

wo

stri

p pa

tter

ns E

xam

ple

96

Strip

Bend

ing

Curv

atur

eCu

rvat

ure

Inte

r-D

eflec

tion

mom

ent a

tG

eom

etric

alco

effic

ient

s an

dco

effic

ient

s an

dpo

latio

nof

Basic

mid

-spa

nCr

acki

ngpr

oper

ties

of s

ectio

nde

flect

ion

inde

flect

ion

inco

effic

ient

strip

defle

ctio

nsp

an

mom

ent

at m

id-s

pan

uncr

acke

d st

ate

fully

-cra

cked

sta

te

basic

MM

r

A sd

κ

s1κ

1

1

κs2

κ

2

2

(m

m)

(kN

-m)

(kN

-m)

(mm

2 )(m

)(1

0minus3)

(mm

)(m

m)

(mm

)

AB

649

281

133

900

017

05

290

960

8819

91

468

016

424

60

7637

05

EF0

7210

113

335

00

155

226

099

097

244

1ndash

ndashndash

244

Firs

t es

timat

e of

defl

ectio

n at

cen

tre

of p

anel

D =

A

B +

EF

= 3

705

+ 2

44

= 39

49

mm

BC3

5518

613

36

500

155

419

098

093

115

77

480

1435

85

064

271

12

003

004

089

AD

21

4641

220

44

top

092

079

400

821

009

146

80

7512

03

kN-m

kN-m

500

045

222

bot

tom

HI

451

187

133

600

017

03

530

970

9114

33

655

013

391

40

6430

21

Seco

nd e

stim

ate

of d

eflec

tion

at c

entr

e of

pan

el D

= ndashsup1 sup2

(BC

+

AD)+

H

I=

ndashsup1 sup2(2

711

+12

03)

+30

21

= 49

78

Defl

ectio

n at

cen

tre

of p

anel

incl

udin

g ef

fect

s of

cre

ep a

nd c

rack

ing

=39

4+

497

8

2=

446

mm

(17

6in

)

1W

hen

M

Mr c

rack

ing

does

not

occ

ur

=

1 and

the

col

umns

for

κs2 κ

2

2 a

nd

are

left

bla

nk

2T

he e

dge

beam

is t

reat

ed a

s a

T-s

ectio

n w

ith fl

ange

wid

th =

05

0m

3T

his

line

give

s Aprime

s dprime

and

prime f

or t

he t

op r

einf

orce

men

t of

str

ip A

D










(∆D1)cs = 300 times 10minus6(030)(655)2

8(017) = 288mm


(∆D2)cs = 300 times 10minus6(112)(655)2

8(017) = 1061mm


(∆D)cs = (1 minus 076)288 + 076(1061) = 875mm


δAB = δDC = 07(875) = 613mm



(∆D1)cs = 300 times 10minus6(010)(700)2

8(0155) = 119mm


(∆D)cs = 119mm


δEF = 05(119) = 060mm


(∆D)cs = 613 + 060 = 673mm



= 1










02207= 0660MPa





910 General






Notes












Chapter 10

101 Introduction
































kpart2T

partx2 + k

part2T

party2 + Q = pc

partT

partt(101)

where




kpartT

partxnx + k

partT

partyny + q = 0 (102)




where


qs = αaIs (104)













cooling










h = hc + hr (108)




where

qcr = qc + qr (1010)












8588603575

1516110613





(1 minus ν2)T(y) (1011)










09

coefficient a


coefficient c

09 08 09



(or per degF)


∆N = thickness


∆M = thickness



p = minusd(∆N)

ds(1014)

q = minus d2(∆M)

ds2(1015)


















∆σ = minus ∆N

A +

∆Mx

Ix

y + ∆My

Iy

x (1021)





∆εO = minus∆N

EA(1023)


EIx

(1024)


EIy

(1025)







2(1026)

D3 = ∆εOl (1027)


∆εO = αt

A T b dy (1028)

∆ψ = αt

I T b ydy (1029)















125




125





∆N = minus(2652 times 106) 035102

0y5 dy + 25

120

102y5 dy


∆M = minus(2652 times 106) 035 102


+2512



∆εo

∆ψ = 1

30 times 109 times

2229 times 106

0877


01615






or


for y = 0769 to 0969m





2 +

24



f11 = l

3EIAB

+ l

3EIBC

= 1


f21 = l

6EIBC



(f11 + f12)F1 = minusD1

F1 = minusD1

f11 + f12

= 8675 times 103 N-m
















Ec(t t0) = Ec(t0)

1 + χφ(t t0)(1031)

where













2 ti and ti+ 1



αt i

j = 1

(∆T)j (1032)



αt i

j = 1

(∆T)j + i

j = 1

(∆σrestraint)j

Ec(tj) [1 + φ(ti + 1

2 tj)] = 0 (1033)







1 + φ(ti + 12 ti)

αti

j = 1

(∆T )j

+ i minus 1

j = 1

(∆σrestraint)j

Ec(tj) [1 + φ (ti + 1

2 tj)] (1034)






φ(2 1) = 056 φ(6 1) = 060 φ(50 1) = 082

φ(6 4) = 055 φ(50 4) = 076 φ(50 28) = 048







1 + 056 (053) = minus0150[αtTc maxEc(28)]


1 + 055 (053 + 047)

minus 0150

044 (1 + 060)



1 + 048


044 (1 + 082) minus

0214

073(1 + 076)

= 0727 [αtTc maxEc(28)]





























f11 = M 2

u1 dl

EcIf12 =

Mu1Mu2dl

EcIf22 =

M 2u2 dl

EcI





1012 General






Notes














Control of cracking

111 Introduction




Chapter 11



















Figu

re11

1A

rei

nfor

ced

conc

rete

mem

ber

subj

ecte

d to

(a)

axi

al fo

rce

N (

b) im

pose

d en

d di

spla

cem

ent

D












Mri = fctiW1 (112)



M = αtEcI1 ∆T

h (0 le ∆T le ∆T1) (113)



∆T1 = Mr1h

αtEcI1

(114)


1

EcIrm

= (1 minus ζ) 1

EcI1

+ ζ 1

EcI2 (115)




Irm = I1I2


(116)



M = αt∆T

h EcI1 1 +

i srm

l I1

Irm

minus 1minus1

(with ∆Ti ∆T ∆Ti + 1) (117)



Mri = αt∆Ti

h EcI11 +

(i minus 1)srm

l I1

Irm

minus1minus1

(118)





εs2 = M

yCTAsEs

(1110)















19758

210172

310582

411090

511698

6122106

7129114

8136122



Nri = fctiA1 (1112)

N = EcA1ε (0 ε ε1) (1113)

ε1 = Nr1

EcA1

(1114)

1

EcArm

= (1 minus ζ) 1

EcA1

+ ζ 1

EcA2 (1115)

Arm = A1A2


(1116)



N = εEcA11 + i srm

l A1

Arm

minus 1 minus1



l A1

Arm

minus 1 minus1

(1118)

where











is formed

Crack number ii


184 times 10minus6

394100

2362 times 10minus6

430246

3700 times 10minus6

476334


534411











Nr = fctA1 (1119)




ρmin y = fct

fy 1









fct = N

A1

+ M

W1

(1121)



fybd(1122)







ρmin y = As min y

bd 024

fct

fy

(1123)












es = M


εs2 = (NesyCT) + N

EsAs

(1126)

σs2 = Esεs2 (1127)

As = (NesyCT) + N

σs2

(1128)

where



















Figu

re11

6C

hang

e in

ste

el s

tres

s

s2 in

a r

ecta

ngul

ar c

rack

ed s

ectio

n du

e to

a n

orm

al fo

rce

N a

nd a

ben

ding

mom

ent

M T

he v

alue

of

s2 is

equ

al t

o th

e va

lue

read

from

the

gra

ph m

ultip

lied

by

c m

ax w

here

c

max

is t

he e

xtre

me

fibre

str

ess

igno

ring

cra

ckin

g Po

sitiv

e si

gn c

onve

ntio

n fo

r N

and

M a

rein

dica

ted





Example 111











σc max = 249

03216 +

40(015)



σs2 = 51(312) = 160MPa (232ksi)




wm = 400(080) 160

200 times 103 = 026mm (0010 in)






























A = 5983 in2 B = 2704 in3 I = 118 500 in4


N1 = 2132kip M1 = 8354kip-in





σs2 = 213(0868) = 185ksi


wm = 16(09) 185

29 000 = 00092 in (023mm)













ρmin y = 360

60 times 103

1

1 minus29000

2900 360

60 times 103= 00063



ρ = Asbar

sts2 =

020

(8 times 8)2 = 00063




(120mm2)





1111 General


Notes













121 Introduction



Chapter 12

122 Permanent state





βD = minusD(Pm)

D(q)(121)





















q(125)





σperm = Mq y


Pm

A(127)





1 + I(8αqA f0 y)(128)


σq = Mq y

I(129)


αq = Mq (ql 2) (1210)


Pm = βD ql 2

8f0

(1211)















951 = 697MPa




1 + 951[8(minus00763)(725)(29)(minus1168)] = 079


Pm = 079 (2065 times 103)(60)2

8(29) = 25 200kN












σq = 773MPa

βD 079

Pm = 21 600kN




Mq = 0125 ql 2 αq = 0125


σq = 125(2065 times 103) (60)2 1832

951 = 1790MPa



1 + 951[8(0125)(725)(29)(1832)] = 089

Pm = 089 (2065 times 103)(60)2

8(29) = 28 500kN




σq = 3405MPa

βD = 085

Pm = 23 300kN





lh l = 30 l = 60m l = 90m


20 094 12 800 079 25 400 075 41 60025 087 14 500 073 27 500 071 45 50030 083 16 400 072 31 100 069 49 700




















128 Water-tightness




























fct = N

A1

+ M

W1

(1213)










2Mmax

ψm Mmax minus

Mmax

EI1 (1214)




1212 General







Notes






above



131 Introduction



Chapter 13



132 Reference axis











[S] D = F (131)






[S] = [S11]

[S21]

[S12]

[S22] (132)




[S21] = [R] [S11] (133)

where

[R] =minus1

0

0

0

minus1

l

0

0

minus1

(134)



[S] = [S11]

[R][S11]

[S11] [R]T

[R] [S11] [R]T (135)


[S11] = [f ]minus1 (136)


f ij = l

0Nui ε Oujdx +

l

0Muiψujdx (137)



f 1j = minusl


l

0ψujx dx f 3j =

l

0ψujdx with j = 1 2 3 (138)



[Smember] = [T]T [S] [T] (139)


[T] = [t][0]

[0]

[t] [t] =c

minuss

0

s

c

0

0

0

1

(1310)










εOu1 = minusI


B



εOu2 = Bx


Ax


For F3 = 1 Nu3 = 0 and Mu3 = 1

εOu3 = minusB


A



I minusBl

2B

[f] = l

E(AI minus B2)minus

Bl

2

Al 2

3minus

Al

2(1315)

B minusAl

2A





2 I =

b h3


1 0 minush

2

[S]cantilever = Ebh

l0

h

l 2

h 2

2l

minush

2

h2

2l

7h2

12(1317)


D = [S]minus1

0

P

0

= l

Ebh3

4h2

minus3hl

6h

minus3hl

4l 2

minus6l

6h

minus6l

12

0

P

0

= Pl 2

Ebh3

minus3h

4l

minus6

(1318)


A

l

012(AI minus B2)

Al 3

symmetrical

minusB

l

6(AI minus B2)

Al 2

4AI minus 3B2

Al



minusA

l

0 B

l


A

l



Al 2


012(AI minus B2)

Al 3

B

l

6(AI minus B 2)

Al 2

2AI minus 3B 2


B


AI 2

4AI minus 3B 2

Al










εO = minus02344


2)minus1

ψ = 03420


3)minus1


εO = minus01582


2)minus1

ψ = 01849


3)minus1


εO

ψ

mean

= (1 minus ζ)εO

ψ

uncracked

+ ζ εO

ψ

fully-cracked

= minus21700 h

25680 (Ec h3)minus1





u1 7886 30130 25680 10minus3 (h3Ec)minus1




u3 15770 41530 36380 10minus3(h4Ec)minus1


21700 minus25680

h l

225680

h

[f ] = 10minus3l

Ech2

minus25680

h l

236380

h2 l2

3 minus36380

h2 l

225680

hminus

36380

h2 l

236380

h2


02799 0 minus01976h


l0 03295

h2

l201647

h2

l

minus01976h 01647h2

l02493 h2


D = [S]minus1 0

P

0 =

Pl2

Ec h4

minus1284h

1213l

minus1819



Ar = ArO1

ArO2 (1319)






D1s = minusl

0εO dx D2s = minus

l

0ψ x dx D3s =

l

0ψ dx (1321)




Q1

Q2


(1322)



QA

QB =

s

6 2

1

1

2 qA

qB (1323)










εcentroid = αt

Al

T dA1 ψ = αt


yO = minusB1

A1


1

A1

(1326)





sψ x dx =

s






l minus xC

l

s ψ x dx =

l minus xC

l s


when xB xC (1328)

xC

l

s ψ(l minus x)dx =

xC

l l s(ψA + ψB)

2 minus

s

6(2ψA xA + ψA xB








[S] D = minusF (1330)


A = Ar + [S] D (1331)






where

[H] =1

0

0

0

1

0

0

0

1

minus1

0

0

0

minus1

0

0

l

minus1

(1333)





[R] [S11] Derror (1334)





cycle 1 (1335)




cycle 1 (1336)








([S]∆D)i = δi F (1337)
























0

0

9380

(b)



minus1119

6922

1530

(c)



1119

minus6922

minus5922

(d)


fcycle 1 = 10minus9

5569

minus16790

minus1685

minus16790

4839

minus5692

minus1685

minus5692

8207

(e)



2162

6908

5235

6908

1343

9454

5235

9454

7883

(f )




1631

minus7165

minus5357


minus1631

7165

minus3242

(g)



1631

minus7165

minus5357

(h)


[S]cycle 1 = 106

2162

6908

5235

6908

1343times106

9454

5235

9454

7883

(i)




1631

minus7165

minus5357

Dcycle 2 = 10minus6

minus1095

0

1406


minus1095

0

2344




AO1cycle 2 = 103

1631

minus7165

minus5357

+ [S]

minus1095

0

1406

10minus6 = 103

0

minus1430

0

AO2cycle 2 = 103

0

1430

minus1717




minus1402

minus9574

3904



minus1095

0

2344

minus

minus1402

minus9574

3904

= 10minus6

3074

9574

minus1559



3117

0340

minus1145





















1 000 minus2939 01509 1000 minus2939 0843







1313 General



Note




141 Introduction



Chapter 14








Reinforcementtype


GPa 103 ksi MPa ksi




















c = 1


a1 = 2hflange

bw


bw


a2 = h2

flange

bw


bw







yCT = d minus c

3with c hf (145)


yCT = d minus c


2 (c + 2hflange)c




d minus c(147)



σf service

(148)









ψm = (1 minus ζ) ψ1 + ζ ψ2 (1410)

ψ1 = M

EcI1

ψ2 = M

EcI2

(1411)

ζ = 1 minus β Mr

M2

(1412)

or

ζ = 1 minus β fct

σ1 max

2

(1413)




ψm = M

EIem

(1414)



Iem = I1I2

I2 + 1 minus 05 Mr

M2

(I1 minus I2)

(1415)





Dcentre = 1

2 l2

0ψx dx +

1

2

l



Dcentre = l 2







Dcentre = 5

48ψm centre l

2 (1418)













ψ2 = εs or εf

βh(1419)

where

β = d minus c

h(1420)


ψm = ηψ2 (1421)

where


η = 1 minus 05 fct

σ1 max

2

1 minus I2

I1 (1422)


Dcentre = 5

48 l 2

h ηβs or f

εs or f (1423)


l

hf = l

hs (ηβ)s

(ηβ)f εs

εf (1424)




l

hf

= l

hs

εs

εf

αd

(1425)


αd = 05 + 3b

100bw

minus b

80hs

(1426)













hs = 8m

16 = 05m


αd = 05 + 3(2)

100(04) minus

2

80(05) = 060

l

hf


2000times10minus606

= 118


hf = 8

118 = 068m take hf = 07 m d = 06m


M = (16kNm) (8m)2

8 = 1280kNm


Af 128kN minus m








= 2740mm2 and M = 128kN-m)



σ1 max = 278MPa ζ = 0742


Dcentre = 184mm

(l Dcentre)f = 80m

00184m = 435


M = (16kN-m)(112)2

8 = 2509kN-m

As = 1830mm2 εs = 1200 times 106


ψ1 = 428 times 10minus6 ψ2 = 2338

σ1 max = 501MPa ζ = 0920


Dcentre = 286mm

l Dcentres = 112m

00286m = 392










1411 General


Notes











εc(t) = σc(t0)

Ec(t0) [1 + φ(t t0)] (A1)



εc(t) = σc(t0)

Ec(t0) 1 +

Ec(t0)



Appendix A

φ = φCEB Ec(t0)

Ec(28)(A3)




h0 = 2Ac

u(A4)









Time functions 475

tT = n

i = 1


273 + T(∆ti) (A6)




Ec(28) = 21 500 ( fcmfcm0)13 (A7)


Ec(28) = 21 500[( fck + ∆f )fcm0]13 (A8)








476 Appendix A

where




Ec(t) = βE(t)Ec(28) (A11)

with










βE(t0) = Ec(t0)

Ec(28)(A15)

Time functions 477







046(h0href)13

(A18)

where href = 100mm

β(fcm) = 53

radicfcmfcmo

(A19)

where fcm0 = 10MPa

β(t0) = 1

01 + t002

(A20)



βH + t minus t0

03

(A21)


βH = 150h0

href

[1 + (0012 RH)18] + 250 le 1500mm (A22)

where href = 100mm

478 Appendix A



t0 = t0 T 9

2 + (t0 T)12+ 1

a

ge 05 (A23)




φ0k = φ0 exp[15(k minus 04)] (A24)


A19 Shrinkage






05

(A26)



Time functions 479

where




1003







Ec(t0) = 095 21 500[(fck(t0) + 8)10]13 (A31)






480 Appendix A





Age atloading


t0 (days) 50 150 600 50 150 600


1 32 26 21 20 18 157 34 28 22 22 19 17

28 32 25 20 19 17 1590 28 23 17 17 15 13

365 22 18 13 13 11 11



le150 600



Time functions 481

A31 Creep



06

10 + (t minus t0)06

φu (A32)

where



φu = 235 γc (A34)


γc = 125 tminus01180 (A35)

or

γc = 1113 tminus00940 (A36)



Ec(t0)

Ec(28) = t0

α + βt0

12 (A37)




482 Appendix A










φ(tinfin t0)


0 (A42)


A32 Shrinkage






55 + (t minus 1 to 3) (εcs)u (A44)

Time functions 483







Value of


(tinfin t0)

10 days 05 0525 0804 0811 080915 0720 0826 0825 0820 027325 0774 0842 0837 083035 0806 0856 0848 0839

102 05 0505 0888 0916 0915days 15 0739 0919 0932 0928 0608

25 0804 0935 0943 093835 0839 0946 0951 0946

103 05 0511 0912 0973 0981days 15 0732 0943 0981 0985 0857

25 0795 0956 0985 098835 0830 0964 0987 0990

104 05 0461 0887 0956 0965days 15 0702 0924 0966 0972 0954

25 0770 0940 0972 097635 0808 0950 0977 0980

(tinfin t0)

(tinfin 7)

0960 0731 0558 0425

Ec(t0)

Ec(28)

0895 1060 1083 1089

484 Appendix A






Ec(28) = K0 + 02 fcu(28) (A47)



Ec(t) = Ec(28) 04 + 06 fcu (t)





Time functions 485



A43 Creep


A44 Shrinkage







20 200 135 220 230 240 25025 250 165 275 290 300 31030 300 200 330 350 360 37040 400 280 440 455 475 50050 500 360 540 555 575 600

486 Appendix A






Time functions 487





488 Appendix A




1 + radict0

(A49)



Time functions 489



490 Appendix A



Time functions 491



MPa psi mm in

100 4 A650 200 8 A7

400 16 A820 3000 1000 40 A9

100 4 A1080 200 8 A11

400 16 A121000 40 A13

100 4 A1450 200 8 A15

400 16 A1630 4500 1000 40 A17

100 4 A1880 200 8 A19

400 16 A201000 40 A21

100 4 A2250 200 8 A23

400 16 A2440 6000 1000 40 A25

100 4 A2680 200 8 A27

400 16 A281000 40 A29

100 4 A3050 200 8 A31

400 16 A3260 9000 1000 40 A33

100 4 A3480 200 8 A35

400 16 A361000 40 A37

100 4 A3850 200 8 A39

400 16 A4080 12000 1000 40 A41

100 4 A4280 200 8 A43

400 16 A441000 40 A45

492 Appendix A


Time functions 493


494 Appendix A


Time functions 495


496 Appendix A


Time functions 497


498 Appendix A


Time functions 499


500 Appendix A


Time functions 501


502 Appendix A


Time functions 503


504 Appendix A


Time functions 505


506 Appendix A


Time functions 507


508 Appendix A


Time functions 509


510 Appendix A


Time functions 511


512 Appendix A


Time functions 513


514 Appendix A


Time functions 515


516 Appendix A


Time functions 517


518 Appendix A


Time functions 519


520 Appendix A


Time functions 521


522 Appendix A


Time functions 523


524 Appendix A


Time functions 525


526 Appendix A


Time functions 527


528 Appendix A


Time functions 529


530 Appendix A


Time functions 531


532 Appendix A

Notes








Time functions 533





0

λ ge 04

λ lt 04(B1)

where

λ = σp0

fptk

(B2)




Appendix B


χr = 1


λ minus 04 2

dξ (B4)



σp0

(B5)




















536 Appendix B

χr = exp[(minus67 + 53 λ)Ω] (B11)



∆σprinfin 1

16 ln τ minus t0

10 + 10 le (τ minus t0) le 1000 (B12)


05 times 10602




Notes







Appendix C









D1 = l

4 [1 2 1] εO (C1)

D2 = minusl

24 [1 6 5] ψ (C2)

D3 = l

24 [5 6 1] ψ (C3)

D4 = l 2

48 [1 4 1] ψ (C4)


D1 = l

6 [1 4 1]εO (C5)

D2 = minusl

6 [0 2 1]ψ (C6)

D3 = l

6 [1 2 0]ψ (C7)

D4 = l 2

96 [1 10 1]ψ (C8)


D1 = l

8 [1 2 2 2 1]εO (C9)

D2 = minusl

96 [1 6 12 18 11]ψ (C10)

D3 = l

96 [11 18 12 6 1]ψ (C11)

540 Appendix C

D4 = l 2

192 [1 6 10 6 1]ψ (C12)


D1 = l

12 [1 4 2 4 1]εO (C13)

D2 = minusl

12 [0 1 1 3 1]ψ (C14)

D3 = l

12 [1 3 1 1 0]ψ (C15)

D4 = l 2

24 [0 1 1 1 0]ψ (C16)


D1 = minusl 2

6 [1 2]ψ (C17)

D2 = l

2 [1 1]ψ (C18)


D1 = minusl 2

24 [1 6 5]ψ (C19)

D2 = l

4 [1 2 1]ψ (C20)


D1 = minusl 2

12 [0 1 1 3 1]ψ (C21)

D2 = l

12 [1 4 2 4 1]ψ (C22)




c3 + a1c2 + a2c + a3 = 0 (D1)

where


a2 = minus6

bw



a3 = 6

bw

[12h







Appendix D

a4 = a2 minus a2

1

3(D5)

a5 = 2 a1

3 3

minus a1a2

3 + a3 (D6)

a6 = a4

3 3

+ a5

2 2

(D7)


c = minus a5

2 + radica6

12

+ minus a5

2 minus radica6

12

minus a1

3(D8)


13 ]


cos θ = minusa5


c1 = 2 minus a4

3 12

cosθ3 minus a1

3(D10)



a1

3 (D11)

Note




E1 Introduction





Appendix E

ρmin = fct

fsy

(E1)






wm = srmζεs2 (E2)





E2 Crack spacing




db (E3)



κ1 = fct

fbm

(E4)




546 Appendix E

sr0 = κ1 db

4ρ(E5)









wk = βwm (E6)




srm = 50 + κ1κ2 db

4ρr

(E7)

where

κ2 = ε1 + ε2

2ε1

(E8)


ρr = As

Acef

(E9)


E4 CEB-FIP 1990 (MC-90)



where




548 Appendix E

εsr2 = fctm(t)

ρrEs

(1 + αρr) (E11)

α = EsEc(t) (E12)




within Acef





ls max = db

36 ρr

(E13)

ls max = σs2db

2τbk(1 + αρr)(E14)


E5 ACI318-89 and ACI318-99



where








τbk τbk



550 Appendix E






Srm = 4te (E18)


te = dc 1 + s

4dc

2

(E19)





s(in) = 540


s(mm) = 95


and

s(in) = 12(36)

σs (ksi)(E22)

s(mm) = 76

σs (MPa)(E23)






552 Appendix E



1 + 2a minus cover

h minus c (E24)



Notes












554 Appendix E






κs = Ig

I(F1)


I(F2)

κcs = minusA

cycd

I(F3)





Appendix F


556 Appendix F




558 Appendix F




560 Appendix F




562 Appendix F




564 Appendix F








566 Appendix F










G1 Introduction








Appendix G













sustained


G22 FORTRAN code










respectively




570 Appendix G


























572 Appendix G












2 3 34 27000



Notes


574 Appendix G

Further reading








Index














474 548















205














255 256 262 391 395control of 380









to bending 384


examples 234 236 243 250 254260



















578 Index



























41 437 43 44 49 61 64 75 8183
















Index 579








146



569






























580 Index








































33 43 72




definition 5

Index 581












curvature due to 243 303 309 327346 564 565











136 147 172











295 547Strain




42 72mean value





582 Index







































stages 105 109 113 116 128141 152 155






sections 20 30 57 63 95 128144

Index 583









deflection 333


units


584 Index

Book Cover

Title

Copyright

Contents


Acknowledgements

Note


Notation






















Further reading

Index

concrete structures: stresses and deformations, third edition

Documents