concrete structures: stresses and deformations, third edition
TRANSCRIPT
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
Concrete StructuresStresses and Deformations
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
4 Concrete Structures
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
Creep and shrinkage of concrete and relaxation of steel 5
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
6 Concrete Structures
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
Creep and shrinkage of concrete and relaxation of steel 7
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
8 Concrete Structures
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
Creep and shrinkage of concrete and relaxation of steel 9
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
10 Concrete Structures
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
Creep and shrinkage of concrete and relaxation of steel 11
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
12 Concrete Structures
∆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)
Creep and shrinkage of concrete and relaxation of steel 13
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
14 Concrete Structures
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
Creep and shrinkage of concrete and relaxation of steel 15
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)
16 Concrete Structures
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
Creep and shrinkage of concrete and relaxation of steel 17
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
18 Concrete Structures
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
Creep and shrinkage of concrete and relaxation of steel 19
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) )
22 Concrete Structures
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
Stress and strain of uncracked sections 23
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)
24 Concrete Structures
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)
Stress and strain of uncracked sections 25
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
26 Concrete Structures
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
Stress and strain of uncracked sections 27
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
28 Concrete Structures
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)
Stress and strain of uncracked sections 29
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
30 Concrete Structures
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
Stress and strain of uncracked sections 31
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
32 Concrete Structures
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) )
Stress and strain of uncracked sections 33
∆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
34 Concrete Structures
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
Stress and strain of uncracked sections 35
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
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
= minus6830MPa (minus0991ksi)
The stress distribution is shown in Fig 26(b)
36 Concrete Structures
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
Stress and strain of uncracked sections 37
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
minus1625 times 10minus3
minus1625 times 10minus3
4184 times 10minus3
times minus126
minus170 10minus6 = 106 1175N
01828N-m
∆N
∆Mshrinkage
= minus882 times 109(minus 240 times 10minus6) 03545
minus1625 times 10minus3
= 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) )
Stress and strain of uncracked sections 39
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
882 times 109[04366 times 6412 times 10minus3 minus (1604 times 10minus3)2]
times 6412 times 10minus3
minus16040 times 10minus3
minus16040 times 10minus3
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
= minus0736MPa (minus0107ksi)
(∆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
= minus1083MPa (minus157ksi)
∆σps = minus80 times 106 + 200[minus471 + 045(minus128)]103
= minus1857MPa (minus269ksi)
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
Stress and strain of uncracked sections 41
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)
42 Concrete Structures
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)
(a) Stress and strain at age t0
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)
Stress and strain of uncracked sections 43
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
44 Concrete Structures
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
Stress and strain of uncracked sections 45
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
46 Concrete Structures
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
Ec(t t0) =30 times 109
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
Stress and strain of uncracked sections 47
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(minus90 times 106)
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
48 Concrete Structures
∆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)
Stress and strain of uncracked sections 49
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)
50 Concrete Structures
NMequivalent
= minus4100 times 103
1400 times 103 minus 4100 times 103 times 053 = minus4100 times 103 N
minus773 times 103 N-m
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
The age-adjusted elasticity modulus of concrete (Equation (131) ) is
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
A B I AEEref BEEref IEEref
(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
Stress and strain of uncracked sections 51
(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
52 Concrete Structures
∆N
∆Mcreep
= minus1230 times 109 times 12 05090
0
0
01090 minus289
minus218 10minus6
= 106 2171 N
0351 N-m
∆N
∆Mshrinkage
= minus123 times 109(minus57 times 10minus6) 05090
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
Stress and strain of uncracked sections 53
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)
Ec2 (infin 3) =23 times 109
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
54 Concrete Structures
(∆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
= minus1314 times 1090509
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
= minus1314 times 109(minus205 times 10minus6) 0509
0
minus801 times 109(minus269 times 10minus6) 0495
minus04307
= 106 2437 N
minus0928 N-m
Stress and strain of uncracked sections 55
∆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
A B I AEEref BEEref IEEref
(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
56 Concrete Structures
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)
∆ψ(t3 t2) = 10minus6 minus261
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
Stress and strain of uncracked sections 57
∆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
58 Concrete Structures
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
Stress and strain of uncracked sections 59
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
60 Concrete Structures
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
Stress and strain of uncracked sections 61
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
62 Concrete Structures
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
Stress and strain of uncracked sections 63
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
64 Concrete Structures
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
Stress and strain of uncracked sections 65
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
66 Concrete Structures
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
∆εO(t t0) = minus8666 times 10minus6 ∆ψ(t t0) = 4784 times 10minus6 inminus1
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
Stress and strain of uncracked sections 67
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
68 Concrete Structures
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
70 Concrete Structures
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
72 Concrete Structures
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
∆σpr = χr∆σpr (39)
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
Special cases of uncracked sections and calculation of displacements 73
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
74 Concrete Structures
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
Special cases of uncracked sections and calculation of displacements 75
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
= minus2427 times 103
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
= minus0863MPa (minus0125ksi)
Special cases of uncracked sections and calculation of displacements 77
(∆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) )
78 Concrete Structures
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
Special cases of uncracked sections and calculation of displacements 79
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
80 Concrete Structures
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) )
Special cases of uncracked sections and calculation of displacements 81
Ec(t t0) =30 times 109
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
82 Concrete Structures
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
= minus0036MPa (minus0005ksi)
(∆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
Special cases of uncracked sections and calculation of displacements 83
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)
84 Concrete Structures
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 )
Special cases of uncracked sections and calculation of displacements 85
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)
86 Concrete Structures
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
Special cases of uncracked sections and calculation of displacements 87
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
88 Concrete Structures
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) )
Special cases of uncracked sections and calculation of displacements 89
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
90 Concrete Structures
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
Special cases of uncracked sections and calculation of displacements 91
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)
92 Concrete Structures
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
Special cases of uncracked sections and calculation of displacements 93
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
minus470 times10minus6minus126 times10minus6
minus479 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
94 Concrete Structures
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
Special cases of uncracked sections and calculation of displacements 95
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
96 Concrete Structures
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
Special cases of uncracked sections and calculation of displacements 97
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
98 Concrete Structures
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
Special cases of uncracked sections and calculation of displacements 99
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
102 Concrete Structures
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
Time-dependent internal forces in uncracked structures 103
[ 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 ]
104 Concrete Structures
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
Time-dependent internal forces in uncracked structures 105
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
106 Concrete Structures
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
Time-dependent internal forces in uncracked structures 107
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
108 Concrete Structures
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
Time-dependent internal forces in uncracked structures 109
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
110 Concrete Structures
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
Time-dependent internal forces in uncracked structures 111
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
112 Concrete Structures
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
Time-dependent internal forces in uncracked structures 113
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
114 Concrete Structures
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
= 1515 times 10minus3 ql 3
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
Time-dependent internal forces in uncracked structures 115
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
116 Concrete Structures
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
(∆D1)loads = 277 times 10minus3 ql 3
Ec(t0)Ic
Time-dependent internal forces in uncracked structures 117
118 Concrete Structures
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
Time-dependent internal forces in uncracked structures 119
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
(∆D1)prestress loss =1
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
= 347 times 10minus3 ql 3
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
120 Concrete Structures
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
Time-dependent internal forces in uncracked structures 121
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
122 Concrete Structures
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)
Time-dependent internal forces in uncracked structures 123
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
124 Concrete Structures
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
Time-dependent internal forces in uncracked structures 125
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
126 Concrete Structures
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)
Time-dependent internal forces in uncracked structures 127
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-
128 Concrete Structures
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
Time-dependent internal forces in uncracked structures 129
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
130 Concrete Structures
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
Time-dependent internal forces in uncracked structures 131
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
134 Concrete Structures
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
Time-dependent internal forces in uncracked structures 135
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-
136 Concrete Structures
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
Time-dependent internal forces in uncracked structures 137
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
138 Concrete Structures
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
Time-dependent internal forces in uncracked structures 139
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
F(ti + 12) = F(ti minus 1
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)
140 Concrete Structures
47 Example worked out in British units
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
Time-dependent internal forces in uncracked structures 141
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
142 Concrete Structures
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
Step 3 The age-adjusted elasticity modulus of concrete (Equation(44) ) Ec = 1558ksi
Select the reference point O as shown in Fig 216(a) properties ofthe age-adjusted section are
A = 10170 in2 B = minus2421 times 103 in3 I = 1629 times 106 in4
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) )
Time-dependent internal forces in uncracked structures 143
∆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
144 Concrete Structures
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 internal forces in uncracked structures 145
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
Time-dependent internal forces in uncracked structures 147
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
148 Concrete Structures
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
Time-dependent internal forces in uncracked structures 149
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
150 Concrete Structures
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
Time-dependent internal forces in uncracked structures 151
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
152 Concrete Structures
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
Time-dependent internal forces in uncracked structures 153
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
154 Concrete Structures
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)
Time-dependent internal forces in uncracked structures 155
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)
156 Concrete Structures
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
Time-dependent internal forces in uncracked structures 157
∆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
158 Concrete Structures
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
Time-dependent internal forces in uncracked structures 159
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
160 Concrete Structures
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
Time-dependent internal forces in uncracked structures 161
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
minus1600 times 106 N-m
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
162 Concrete Structures
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)
Time-dependent internal forces in uncracked structures 163
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
164 Concrete Structures
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
Time-dependent internal forces in uncracked structures 165
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
166 Concrete Structures
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
170 Concrete Structures
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)
172 Concrete Structures
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)
(∆εps)i = (∆σps)i
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
Time-dependent internal forces in uncracked structures 173
(∆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
174 Concrete Structures
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
Time-dependent internal forces in uncracked structures 175
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
178 Concrete Structures
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
Analysis of time-dependent internal forces 179
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
180 Concrete Structures
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
Analysis of time-dependent internal forces 181
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
182 Concrete Structures
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
Analysis of time-dependent internal forces 183
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
184 Concrete Structures
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
Analysis of time-dependent internal forces 185
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
186 Concrete Structures
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
Analysis of time-dependent internal forces 187
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
188 Concrete Structures
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
Analysis of time-dependent internal forces 189
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
190 Concrete Structures
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
Analysis results load case No 1
Nodal displacementsNode
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
Member end forcesMember
1F1
00000E+00F2
21550E+00F3
21540E+00F4
00000E+00F5
minus21550E+00F6
00000E+00
Analysis of time-dependent internal forces 191
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
192 Concrete Structures
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
Analysis of time-dependent internal forces 193
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
Prescribed displacementsu v
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
Forces at the supported nodesNode
2Fx
minus17890E+06Fy
minus16056E+06Mz
35560E+06
Member end forces Member
1F1
17890E+06F2
minus89440E+05F3
minus29104Eminus10F4
minus17890E+06F5
minus16056E+06F6
35560E+06
194 Concrete Structures
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
Analysis of time-dependent internal forces 195
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
Input dataNumber of joints = 3 Number of members = 2 Number of load cases = 1Number of joints with prescribed displacements = 2 Elasticity modulus = 9615E+09
Nodal coordinatesNode
123
x y00 00
100 00100 minus50
Element informationElement
12
1st node11
2nd node23
a10000E+0152000Eminus02
I10000E+0010000Eminus06
Support conditions Node
23
Restraint indicatorsu v 0 0 00 0 0
Prescribed displacementsu v
00000E+00 00000E+00 00000E+00
00000E+00 00000E+00 00000E+00
Forces applied at the nodesLoad case
1Node
1Fx
00000E+00Fy
00000E+00Mz
00000E+00
Member end forces with nodal displacement restrainedLdcase
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
Analysis results load case No 1
Nodal displacementsNode
123
u31270Eminus0231270Eminus0812014Eminus08
v38513Eminus0230455Eminus08
minus24022Eminus08
minus16116Eminus03minus49711Eminus09minus56920Eminus09
Forces at the supported nodesNode
23
Fx
minus15478E+05minus15478E+05
Fy
minus77388E+0477388E+04
Mz
77890E+05
19580E+01
Member end forcesMember
12
F1minus15478E+0517305E+05
F277388E+0432547E+00
F3minus16808E+0116808E+01
F415478E+05
minus17305E+05
F5minus77388E+04minus32547E+00
F677890E+0519580E+01
196 Concrete Structures
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
Analysis of time-dependent internal forces 197
The age-adjusted elasticity modulus is (Equation (65) )
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
198 Concrete Structures
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
minus23736Eminus0241362Eminus0941362Eminus09
minus23736Eminus08
60029Eminus03minus60029Eminus0310672Eminus24
minus27823Eminus0327823Eminus0332702Eminus23
minus26566Eminus0326566Eminus0399262Eminus23
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
Analysis of time-dependent internal forces 199
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
87131Eminus02minus11291Eminus1661614Eminus0261614Eminus02
minus82952Eminus1731826Eminus0231826Eminus02
minus28189Eminus1729462Eminus0829462Eminus08
minus28656Eminus23
minus20020Eminus0320020Eminus03
minus14889Eminus22minus42660Eminus0342660Eminus0349631Eminus23
minus46846Eminus0346846Eminus03
minus12550Eminus45minus48412Eminus0348412Eminus0300000E+00
28363Eminus22minus13222Eminus2286130Eminus0214988Eminus0114988Eminus0120936Eminus0124353Eminus0124353Eminus0127301Eminus0127543Eminus0127543Eminus0100000E+00
Member end forcesMember F1 F2
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
200 Concrete Structures
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
Analysis of time-dependent internal forces 201
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
202 Concrete Structures
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
Nodal displacementsNode
123456789
10111213
uminus48997Eminus0813645Eminus0210922Eminus0281954Eminus0354654Eminus0327332Eminus0313086Eminus1315449Eminus0276927Eminus0325185Eminus03
minus55401Eminus05minus58490Eminus0442656Eminus20
v39000Eminus0939000Eminus0361245Eminus0211027Eminus0114764Eminus0117100Eminus0117894Eminus0139001Eminus0361245Eminus0211027Eminus0114764Eminus0117100Eminus0117894Eminus01
wminus84349Eminus0925057Eminus0222427Eminus0218195Eminus0212780Eminus0265836Eminus0320542Eminus1325057Eminus0222427Eminus0218195Eminus0212780Eminus0265836Eminus0327272Eminus19
Forces at the supported nodesNode
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
89303E+05minus20599E+06minus28711E+06minus34412E+06minus37785E+06minus38885E+06minus42733Eminus02minus54424Eminus02minus62442Eminus02minus67479Eminus02minus69436Eminus02
Analysis of time-dependent internal forces 203
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
2700E+07minus1320E+06minus1320E+06minus1320E+06minus1320E+06minus1320E+06
1927E+05minus3397E+05minus2619E+05minus1859E+05minus1111E+05minus3693E+040000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+00
minus6870E+051585E+062209E+062647E+062907E+062991E+060000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+000000E+00
Analysis results load case No 1
Nodal displacements
Node123456789
10111213
uminus10241Eminus0759852Eminus0248107Eminus0236251Eminus0224258Eminus0212157Eminus0258205Eminus1364525Eminus0239544Eminus0221176Eminus0297055Eminus0334761Eminus0376495Eminus15
v22799Eminus0822799Eminus0217343Eminus0130379Eminus0140286Eminus0146433Eminus0148512Eminus0122799Eminus0217343Eminus0130379Eminus0140286Eminus0146433Eminus0148512Eminus01
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
86860E+03minus30456E+05minus20993E+05minus12543E+05minus63670E+04minus19190E+04minus19886Eminus02minus13754Eminus02minus82658Eminus03minus42217Eminus03minus12775Eminus03
minus38882E+04minus32595E+0534956E+0570033E+0583440E+0588486E+05
minus22704Eminus0328021Eminus0248621Eminus0260855Eminus0267470Eminus02
26096Eminus04
33753E+05
29902E+06
24999E+06
21136E+06
19103E+06minus33977E+06minus30063E+06minus25117E+06minus21233E+06minus19191E+06
minus86860E+0330456E+0520993E+0512543E+0563670E+0419190E+0419886Eminus0213754Eminus0282658Eminus0342217Eminus0312775Eminus03
82302E+04minus40501E+05minus85331E+05minus10013E+06minus98733E+05minus93086E+05minus45648Eminus02minus61112Eminus02minus68483Eminus02minus70992Eminus02minus70536Eminus02
204 Concrete Structures
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
Analysis of time-dependent internal forces 205
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
206 Concrete Structures
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
208 Concrete Structures
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
210 Concrete Structures
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
Stress and strain of cracked sections 211
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
212 Concrete Structures
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)
Stress and strain of cracked sections 213
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
214 Concrete Structures
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
Stress and strain of cracked sections 215
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
216 Concrete Structures
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
Stress and strain of cracked sections 217
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
Stress and strain of cracked sections 219
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