COMPUTED DEFLECTION OF CONTINUOUS REINFORCED CONCRETE ...

COMPUTED DEFLECTION OF CONTINUOUS

REINFORCED CONCRETE FLEXURAL MEMBERS

by

Garth Roger Christie

Bachelor of Science in Engineering (Civil), University of New Brunswick, 2006

Bachelor of Computer Science, University of New Brunswick, 2003

A Report Submitted in Partial Fulfillment

of the Requirements for the Degree of

Masters of Engineering

in the Graduate Academic Unit of Civil Engineering

Supervisor: P.H. Bischoff, PhD, PEng, Department of Civil Engineering

Examining Board: J.H. Rankin, PhD, PEng, Department of Civil Engineering

H.H. El Naggar, PhD, PEng, Department of Civil Engineering

This report is accepted by the Dean of Graduate Studies

THE UNIVERSITY OF NEW BRUNSWICK

May, 2014

©Garth Roger Christie, 2014

ii

ABSTRACT

Bending deflection is important to the design of some concrete members. While

deflection is rarely a safety issue when it governs, deflection limits are always a code

requirement. For beams and slabs, deflection checks are not required if a member meets

the recommended depth-to-span ratio. In design of thinner steel reinforced concrete

slabs and most FRP reinforced members, though, deflection requirements often govern.

Because commonly used deflection calculations, as per ACI 318-05 and CSA A23.3-04,

are often inaccurate in important ways, this work studies improved calculations.

This report extends Bischoff’s method for computing an effective moment of inertia for

simply supported members to an effective moment of inertia for continuous members.

This comparison is done for immediate deflections with a uniformly-distributed load, a

center-point load, and equal loads at third-points. Bischoff’s work with simply

supported members is reviewed. Branson’s equation and the S806 method are also

reviewed and used for comparison.

The results indicate that Bischoff’s equations for simply-supported members also work

well for continuous members. These proposed equations work very well for centered

point loads and uniformly distributed loads (within proposed limits). For a member

with equal point loads at third-points, a minor calculation modification is suggested

which improves its usable range and accuracy. For members with unequal end-

moments, accuracy requires use of the maximum positive bending moment (not the

moment at midspan). For situations where the end-moment magnitude greatly exceeds

the positive moment, a numerical integration approach is recommended.

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TABLE OF CONTENTS

ABSTRACT ...................................................................................................................... ii

TABLE OF CONTENTS ................................................................................................. iii

LIST OF TABLES........................................................................................................... vii

LIST OF FIGURES .......................................................................................................... ix

LIST OF SYMBOLS ....................................................................................................... xi

1.0 INTRODUCTION ................................................................................................. 1

1.1 Project Need ................................................................................................ 2

1.2 Project Objectives ....................................................................................... 2

1.3 Project Scope ............................................................................................... 3

1.4 Report Organization .................................................................................... 4

2.0 BACKGROUND TO DEFLECTION OF REINFORCED CONCRETE ............. 6

2.1 Introduction to Deflection ........................................................................... 7

2.2 Elastic Deflection of Prismatic Members ................................................... 7

2.2.1 Simply Supported Members ........................................................................ 8

2.2.2 Members with Bending Moments at Supports ............................................ 9

2.2.3 Continuous Member Factor, 𝐾 .................................................................. 10

2.3 Bending Deflection of Reinforced Concrete ............................................. 11

2.3.1 Concrete Bending Response ..................................................................... 12

2.3.2 Tension Stiffening of Concrete Bending Members ................................... 13

2.3.3 Constant Stiffness Approach ..................................................................... 14

2.3.4 Integration Approach to Deflection ........................................................... 16

2.4 Effect of Materials and Load History on Deflection ................................. 17

2.4.1 Variation in Mix Materials and Field Conditions ...................................... 17

2.4.2 Effect of Load-History on Deflection ....................................................... 18

2.5 FRP Reinforced Members, Razaqpur’s Work, and CSA S806 ................. 19

2.5.1 Fibre Reinforced Polymers as Concrete Reinforcing................................ 19

2.5.2 Razaqpur’s Work ....................................................................................... 20

2.5.3 Concrete Deflection in CSA S806 ............................................................ 21

2.6 Bending Deflection in CSA A23.3-04 ....................................................... 22

iv

2.6.1 CSA A23.3-04, Clause 9.8.2.1, Minimum Thickness ............................... 23

2.6.2 CSA A23.3-04, Clause 9.8.2.2 and 9.8.2.3, Immediate Deflection........... 24

2.6.3 CSA A23.3-04, Clause 9.8.2.4, Moment of Inertia for Continuous Spans 24

2.6.4 CSA A23.3-04, Clause 9.8.2.5, Sustained Load Deflections .................... 25

2.7 Branson’s Work ......................................................................................... 25

2.7.1 Limited Accurate Range for Branson’s Equation...................................... 26

2.7.2 Modification Factors Examples for Branson’s Equation .......................... 26

2.8 Bischoff’s Work ........................................................................................ 27

2.8.1 Purpose of Bischoff’s Work ...................................................................... 27

2.8.2 Bischoff’s Equation and Loading Type Factor.......................................... 28

2.8.3 Discussion of Arguments Against Use of Bischoff’s Equation ................. 30

3.0 METHODOLOGY AND RESULTS ................................................................... 31

3.1 Virtual Work and Moment of Inertia Methodology .................................. 32

3.1.1 Deflection of Concrete by Integration Using Virtual Work ...................... 33

3.1.2 Deflection of Concrete Using a Constant Moment of Inertia ................... 34

3.2 Generating Idealized Members ................................................................. 35

3.2.1 Definitions of Bending Moment Variables ............................................... 35

3.2.2 Automated Member Generation ................................................................ 38

3.3 Computing Deflection of Idealized Members ........................................... 39

3.3.1 Development and Use of Analytical Integration ....................................... 44

3.3.2 Discussion of Analytical Integration Simplifications ............................... 45

3.3.3 Use of Numerical Integration .................................................................... 46

3.3.4 Comparing Results of Analytical and Numerical Integration ................... 46

3.4 Continuous Beam with a Centered Point Load ......................................... 47

3.4.1 Proposed Solution for a Centered Point Load ........................................... 47

3.4.2 Comparison of Results: Centered Point Load and Equal End-Moments .. 48

3.4.3 Summary of Results for a Centered Point Load........................................ 53

3.5 Continuous Beam with Two Equal Point Loads at Third Points ............... 55

3.5.1 Proposed Solution for Two Equal Loads at Third Points .......................... 56

3.5.2 Comparison of Results for Two Equal Loads at Third Points ................... 58

3.5.3 Summary of Results for Two Equal Loads at Third Points ....................... 63

3.6 Continuous Beam with a Uniformly Distributed Load ............................. 65

3.6.1 Proposed Solution for a Uniformly Distributed Load ............................... 65

3.6.2 Comparison of Results for a Uniformly Distributed Load........................ 67

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3.6.3 Summary of Results for a Uniformly Distributed Load............................ 74

3.7 Additional Findings ................................................................................... 77

3.7.1 When Midspan and Maximum Deflections are Different ......................... 77

3.7.2 Accurate Constant Stiffnesses can be Impossible ..................................... 78

3.7.3 Importance of the Correct Bending Moment Function ............................. 79

3.7.4 Effect on Results of the CSA A23.3 Update to Clause 9.8.2.3 ................. 80

3.8 Summary of Results using Branson’s Method .......................................... 81

4.0 CONCLUSIONS AND RECOMMENDATIONS ............................................... 83

4.1 Conclusions ............................................................................................... 83

4.2 Recommendation for Future Work ............................................................ 85

4.2.1 Improve Deflection Equation Information Provided to Engineers ........... 85

4.2.2 Improve Assumptions for Stiffness ........................................................... 85

4.2.3 Investigation of Other Possible Moment of Inertia Equations .................. 86

REFERENCES ................................................................................................................ 88

Derivation of 𝐾 for Continuous Linear-Elastic Members ..................... 89 Appendix A

Calculate 𝐾 for Point Load at Midspan and Generic End-Moments .................. 90

Calculate 𝐾 for Two Equal Third-Point Loads and Generic End-Moments ....... 91

Calculate 𝐾 for Uniformly Distributed Load and Generic End-Moments .......... 92

Bending Deflection by Integration Using Virtual Work ....................... 93 Appendix B

Deflection for Simply Supported Member without Tension Appendix C

Stiffening ................................................................................................. 95

Derivation of Bischoff's Factor for a Uniformly Distributed Load .... 97 Appendix D

Analytical Integration for Midspan Deflection ..................................... 99 Appendix E

Bending Moment and Virtual Moment Equations ............................................ 101

Lengths to where the Function being Integrated Changes ................................ 102

Midspan Deflection of Midspan-Point Loaded Member with End-Moments .. 104

Midspan Deflection of Third-Point Loaded Member with End-Moments ....... 105

Midspan Deflection of Member with Uniform Load and End-Moments ......... 105

Analytical Results Simplified for Fixed-Fixed Midspan Point Load .. 109 Appendix F

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Integration using CSA S806 / Razaqpur’s Method ............................. 111 Appendix G

Example Simply Supported Constant Stiffness Beam ........................ 120 Appendix H

Example Constant Stiffness Beam with End-Moments .......................... 122 Appendix I

Example Generation and Deflection Computation for an Idealized Appendix J

Concrete Bending Member ................................................................... 124

Methodology and Example using Numerical Integration ................... 134 Appendix K

Examples Graphs of the Integrated Function ...................................... 138 Appendix L

Centered Point Load Examples – Data for Section 3.4 ....................... 143 Appendix M

Third-Point Loaded Examples – Data for Section 3.5 ........................ 156 Appendix N

Uniformly Distributed Load Examples – Data for Section 3.6 ........... 169 Appendix O

Results Using New Mcr per CSA A23.3-04 (R2010) .......................... 195 Appendix P

The Effects of Cracking near Supports ............................................... 206 Appendix Q

Midspan and Maximum Deflection of Linear-Elastic Members ......... 209 Appendix R

Criticisms of CSA A23.3 and the Concrete Handbook ....................... 220 Appendix S

Criticism of Use of Branson’s Equation in CSA A23.3-04 .............................. 220

Criticism of 0.5 Mcr Modifier in CSA A23.3-04 ............................................... 220

Criticism of Use of Midspan Moment in CSA A23.3-04 ................................. 221

Criticism of Concrete Design Handbook Using Midspan Deflection ............... 221

Curriculum Vitae

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LIST OF TABLES

Table 2-1 - Example simply supported members with equations ..................................... 8

Table 2-2 - Deflection of continuous prismatic linear-elastic members ......................... 10

Table 2-3 - Deflection using Bischoff's Equation ........................................................... 29

Table 3-1 - Valid Ranges for I'e for a Centered Point Load ............................................ 54

Table 3-2 - Valid Ranges for I'e* for Equal Point Loads at Third Points ........................ 64

Table 3-3 - Valid Ranges for I'e for Uniformly Distributed Load .................................. 75

Table C-1 - Deflection Equations for Idealized FRP-Reinforced Members ................... 96

Table H-1 - Equal Midspan Deflection Example for CPL, 2PL, and UDL .................. 121

Table I-1 - Equal Midspan Deflection Example for Continuous UDL ......................... 123

Table J-1 - Summary of Appendix J and Appendix K Results for Continuous

Member ............................................................................................................. 133

Table K-1 - Midspan Deflection Example using 10 Segment Numerical Integration .. 136

Table K-2 - Midspan Deflection Example using 100 Segment Numerical Integration 136

Table K-3 - Maximum Deflection Example using 100 Segment Numerical

Integration ......................................................................................................... 137

Table M-1 - Data for CPL, ML=MR, Ig/Icr=2.3 – Example 3.4.2a – Page 1 .................. 144

Table M-2 - Data for CPL, ML=MR, Ig/Icr=2.3 – Example 3.4.2a – Page 2 .................. 145

Table M-3 - Data for CPL, ML=MR, Ig/Icr=3.9 – Example 3.4.2b – Page 1 .................. 147

Table M-4 - Data for CPL, ML=MR, Ig/Icr=3.9 – Example 3.4.2b – Page 2 .................. 148

Table M-5 - Data for CPL, ML=MR, Ig/Icr=3.8 – Example 3.4.2c – Page 1 .................. 150

Table M-6 - Data for CPL, ML=MR, Ig/Icr=3.8 – Example 3.4.2c – Page 2 .................. 151

Table M-7 - Data for CPL, ML=MR, Ig/Icr=12 – Example 3.4.2d – Page 1 ................... 153

Table M-8 - Data for CPL, ML=MR, Ig/Icr=12 – Example 3.4.2d – Page 2 ................... 154

Table N-1 - Data for 2PL, ML=MR, Ig/Icr=3.0 – Example 3.5.2a – Page 1 ................... 157

Table N-2 - Data for 2PL, ML=MR, Ig/Icr=3.0 – Example 3.5.2a – Page 2 ................... 158

Table N-3 - Data for 2PL, MR=0, Ig/Icr=3.0 – Example 3.5.2b – Page 1 ...................... 160

Table N-4 - Data for 2PL, MR=0, Ig/Icr=3.0 – Example 3.5.2b – Page 2 ...................... 161

Table N-5 - Data for 2PL, ML=MR, Ig/Icr=12 – Example 3.5.2c – Page 1 .................... 163

Table N-6 - Data for 2PL, ML=MR, Ig/Icr=12 – Example 3.5.2c – Page 2 .................... 164

Table N-7 - Data for 2PL, MR=0, Ig/Icr=12 – Example 3.5.2d – Page 1 ....................... 166

Table N-8 - Data for 2PL, MR=0, Ig/Icr=12 – Example 3.5.2d – Page 2 ....................... 167

Table O-1 - Data for UDL Beam, ML=MR, Ig/Icr=3.0 – Example 3.6.2a – Page 1 ........ 171

Table O-2 - Data for UDL Beam, ML=MR, Ig/Icr=3.0 – Example 3.6.2a – Page 2 ........ 172

Table O-3 - Data for UDL Beam, MR=0, Ig/Icr=3.0 – Example 3.6.2b – Page 1 ........... 174

Table O-4 - Data for UDL Beam, MR=0, Ig/Icr=3.0 – Example 3.6.2b – Page 2 ........... 175

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Table O-5 - Data for UDL Beam, ML=MR, Ig/Icr=4.9 – Example 3.6.2c – Page 1 ........ 177

Table O-6 - Data for UDL Beam, ML=MR, Ig/Icr=4.9 – Example 3.6.2c – Page 2 ........ 178

Table O-7 - Data for UDL Slab, ML=MR, Ig/Icr=4.9 – Example 3.6.2d – Page 1 .......... 180

Table O-8 - Data for UDL Slab, ML=MR, Ig/Icr=4.9 – Example 3.6.2d – Page 2 .......... 181

Table O-9 - Data for UDL Slab, MR=0, Ig/Icr=4.9 – Example 3.6.2e – Page 1 ............. 183

Table O-10 - Data for UDL Slab, MR=0, Ig/Icr=4.9 – Example 3.6.2e – Page 2 ........... 184

Table O-11 - Data for UDL Slab, ML=MR, Ig/Icr=18 – Example 3.6.2f – Page 1 .......... 186

Table O-12 - Data for UDL Slab, ML=MR, Ig/Icr=18 – Example 3.6.2f – Page 2 ......... 187

Table O-13 - Data for UDL Slab, MR=0, Ig/Icr=6 – Example 3.6.2g – Page 1 .............. 189

Table O-14 - Data for UDL Slab, MR=0, Ig/Icr=6 – Example 3.6.2g – Page 2 .............. 190

Table O-15 - Data for UDL Beam, ML=MR, Ig/Icr=17 – Example 3.6.2h – Page 1....... 192

Table O-16 - Data for UDL Beam, ML=MR, Ig/Icr=17 – Example 3.6.2h – Page 2....... 193

Table P-1 - Data for UDL Beam, Ig/Icr=3.0, New A23.3 Mcr Example P1 – Page 1 ..... 197




Table P-5 - Data for UDL Beam, Ig/Icr=17, Reduced Mcr Example P3 – Page 1 .......... 203

Table P-6 - Data for UDL Beam, Ig/Icr=17, Reduced Mcr Example P3 – Page 2 .......... 204

Table R-1 - Example Midspan vs Maximum Deflection for UDL – Page 1 ................. 211






Table R-7 - Example Midspan vs Maximum Deflection for UDL – Summary ............ 217

Table R-8 - Example Midspan vs Maximum Deflection for CPL – Summary ............. 218

Table R-9 - Example Midspan vs Maximum Deflection for 2PL – Summary ............. 219

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LIST OF FIGURES

Figure 2-1 - Deflected Shape Comparison of Four Different Loads & End-Moments .. 10

Figure 2-2 - Moment-Curvature Response of Reinforced Concrete ............................... 12

Figure 2-3 - Effect of Tension Stiffening on a Reinforced Concrete Member ................ 14

Figure 2-4 - Gross, Local-Effective, Equivalent, and Cracked Moments of Inertia ....... 15

Figure 3-1 – Midspan Deflection of Steel Reinforced Beams under Centered Point

Load with Ig/Icr=2.3, Mm/Mcr=3.0, and ML=MR .................................................. 50

Figure 3-2 - Midspan Deflection of Steel Reinforced Beams under Centered Point


Figure 3-3 - Midspan Deflection of FRP Reinforced Beams under Centered Point


Figure 3-4 - Midspan Deflection of FRP Reinforced Beams under Centered Point

Load with Ig/Icr=12, Mm/Mcr=1.6, and ML=MR ................................................... 53

Figure 3-5 - Midspan Deflection of Steel Reinforced Beams under Third Point

Loading with Ig/Icr=3.0, Mm/Mcr=2.2, and ML=MR ............................................. 59

Figure 3-6 - Midspan and Maximum Deflection of Steel Reinforced Beams under

Third Point Loading with Ig/Icr=3.0, Mmax /Mcr=2.2, and MR=0 ......................... 60

Figure 3-7 - Midspan Deflection of GFRP Reinforced Beams under Third Point

Loading with Ig/Icr=12.2, Mm/Mcr=1.4, and ML=MR ........................................... 61

Figure 3-8 - Midspan and Maximum Deflection of GFRP Reinforced Beams under

Third Point Loading with Ig/Icr=12.2, Mmax /Mcr=1.4, and MR=0 ....................... 62

Figure 3-9 - Midspan Deflection of Steel Reinforced Beams under Uniformly

Distributed Load with Ig/Icr=3.0, Mm /Mcr=2.17, and ML=MR ............................ 70


Uniformly Distributed Load with Ig/Icr=3.0, Mmax /Mcr=2.17, and MR=0 ........... 71

Figure 3-11 - Midspan Deflection of Steel Reinforced Slabs under Uniformly

Distributed Load with Ig/Icr=4.9, Mm /Mcr=1.33, and ML=MR ............................ 72

Figure 3-12 - Midspan and Maximum Deflection of Steel Reinforced Slabs under

Uniformly Distributed Load with Ig/Icr=4.9, Mmax /Mcr=1.33, and MR=0 ........... 73

Figure 3-13 - Midspan and Deflection of GFRP Reinforced Beams under Uniformly

Distributed Load with Ig/Icr=17, Mm /Mcr=1.25, and ML=MR ............................. 74

Figure A-1 - Midspan Point Load on a Continuous Member ......................................... 90

Figure A-2 - Equal Point Load at Third Points on a Continuous Member ..................... 91

Figure A-3 - Uniformly Distributed Load on a Continuous Member ............................. 92

Figure C-1 - Idealized Moment-Curvature for FRP-Reinforced Member ...................... 95

Figure E-1 - Lengths to Integration Segments for Example Midspan Point Load.......... 99

x

Figure E-2 - Lengths to Integration Segments for Example Equal Third-Point Loads 100

Figure E-3 - Lengths to Integration Segments for Example Uniform Load ................. 100

Figure G-1 - Lengths to Integration Segments for Example Centered Point Load ....... 112

Figure L-1 - Integrated Function and Accurate Deflection Example ............................ 139

Figure L-2 - Integrated Function and Inaccurate Deflection Example ......................... 141

Figure M-1 - Copy of Figure 3-1 – Midspan Point Load, Ig/Icr=2.3 and Mm/Mcr=3.0 .. 146



Figure M-4 - Copy of Figure 3-4 – Midspan Point Load, Ig/Icr=12 and Mm/Mcr=1.6 ... 155

Figure N-1 - Copy of Figure 3-5 – Third-Point Loaded, Ig/Icr=3 and Mm/Mcr=2.2....... 159

Figure N-2 - Copy of Figure 3-6 – Third-Point Loaded, Ig/Icr=3, Mmax/Mcr=2.2,

MR=0 ................................................................................................................. 162

Figure N-3 - Copy of Figure 3-7 – Third-Point Loaded, Ig/Icr=12 and Mm/Mcr=1.4..... 165

Figure N-4 - Copy of Figure 3-8 – Third-Point Loaded, Ig/Icr=12, Mmax/Mcr=1.4,

MR=0 ................................................................................................................. 168

Figure O-1 - Copy of Figure 3-9 – UDL on Beam, Ig/Icr=3, Mm /Mcr=2.2, ML=MR ..... 173

Figure O-2 - Copy of Figure 3-10 – UDL on Beam, Ig/Icr=3, Mmax/Mcr=2.2, MR=0 .... 176

Figure O-3 - Midspan Deflection of Steel Reinforced Beams under Uniformly

Distributed Load with Ig/Icr=5, Mm /Mcr=1.3, and ML=MR ............................... 179

Figure O-4 - Copy of Figure 3-11 – UDL on Slab, Ig/Icr=5, Mm /Mcr=1.3, ML=MR ..... 182

Figure O-5 - Copy of Figure 3-12 – UDL on Slab, Ig/Icr=5, Mmax /Mcr=1.3, MR=0 ...... 185

Figure O-6 - Midspan Deflection of FRP Reinforced Slabs under Uniformly

Distributed Load with Ig/Icr=18, Mm/Mcr=1.2, ML=0 ........................................ 188

Figure O-7 - Midspan and Maximum Deflection of FRP Reinforced Slabs under

Uniformly Distributed Load with Ig/Icr=6, Mmax/Mcr=2, ML=0 ........................ 191

Figure O-8 - Copy of Figure 3-13 – UDL on Beam, Ig/Icr=17, Mm /Mcr=1.3, ML=MR . 194

Figure P-1 - Midspan Deflection Computed using Shrinkage Restraint Mcr – Beam

with Ig/Icr=3 Mm/Mcr=3.2, ML=MR .................................................................... 199

Figure P-2 - Copy of Figure O-1, Ig/Icr=3, Mm /Mcr=2.2 – Compare to Figure P-1 ...... 199

Figure P-3 – Midspan and Maximum Deflection Computed using Shrinkage

Restraint Mcr – Slab with Ig/Icr=5, Mmax/Mcr=2, and MR=0 .............................. 202

Figure P-4 - Copy of Figure O-5, Ig/Icr=5, Mmax/Mcr=1.3 – Compare to Figure P-3 .... 202

Figure P-5 - Midspan Deflection Computed using Shrinkage Restraint Mcr – Slab

with Ig/Icr=17, Mm/Mcr=1.9, and ML=MR .......................................................... 205

Figure P-6 - Copy of Figure O-8, Ig/Icr=17, Mm/Mcr=1.3 – Compare to Figure P-5 ..... 205

Figure R-1 - Examples of Differences between Midspan and Maximum Deflection ... 210

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LIST OF SYMBOLS

This report defines the following symbols as:

2PL = two point loads located at third points on the relevant member span

𝑎 = depth of equivalent rectangular stress block

𝐴𝑓 = area of FRP reinforcement in tension for this member segment

𝐴𝑚 = area of tension reinforcement, 𝐴𝑠 or 𝐴𝑓, at/near midspan for this member

𝐴𝑠 = area of steel reinforcement in tension for this member segment

𝐴𝐿 = area of tension reinforcement, 𝐴𝑠 or 𝐴𝑓, in the left end of this member

𝐴𝑅 = area of tension reinforcement, 𝐴𝑠 or 𝐴𝑓, in the right end of this member

𝑏 = width of rectangular beam across the compression face

𝑐 = distance from the compression face to the neutral axis

𝑐𝐿 = 𝑐 for the design of the left end moment of the member

𝑐𝑚 = 𝑐 for the design of the maximum moment near midspan of the member

𝑐𝑅 = 𝑐 for the design of the right end moment of the member

CPL = one point loads located at midpoint on the relevant member span

𝑑 = effective depth of tension reinforcement from compression face

𝐸 = elastic modulus of the material being analyzed

𝐸𝑏 = elastic modulus of reinforcing bar, 𝐸𝑓 or 𝐸𝑠

𝐸𝑐 = elastic modulus of concrete

𝐸𝑓 = design or guaranteed elastic modulus of FRP reinforcement

𝐸𝑠 = elastic modulus of steel reinforcing

𝑓𝑏 = service load stress in the reinforcing bar

𝑓𝑐′ = specified compressive strength of concrete

𝑓𝑓𝑢 = tensile strength at failure, (unfactored ultimate tensile stress)

𝑓𝑆𝐿𝑆 = serviceability limit for FRP reinforced concrete design

𝑓𝑓,`𝑠 = creep-rupture stress limit for FRP reinforced concrete design

𝑓𝑟 = modulus of rupture of concrete (taken as 0.6√𝑓𝑐′ in MPa)

𝑓𝑦 = yield strength of steel reinforcement (or similar for FRP)

ℎ = height of rectangular beam, from compression face to tension face

𝑖 = symbol/counter for sections, 0 to 𝑗, or segments, 1 to 𝑗, for the span

𝑗 = denotes total number of equal segments used for numerical integration

𝐼 = moment of inertia of the member about the axis the load is applied

𝐼𝑐𝑟 = moment of inertia of the cracked transformed section, 𝐼𝑐𝑟= 𝐼𝑐𝑟𝑚 unless noted

𝐼𝑐𝑟∗ = moment of inertia of appropriate section ( 𝐼𝑐𝑟𝐿 , 𝐼𝑐𝑟𝑚 , or 𝐼𝑐𝑟𝑅)

xii

𝐼𝑐𝑟𝐿 = moment of inertia, 𝐼𝑐𝑟, for negative bending at the left end

𝐼𝑐𝑟𝑚 = moment of inertia, 𝐼𝑐𝑟, for positive bending for the midspan segment

𝐼𝑐𝑟𝑅 = moment of inertia, 𝐼𝑐𝑟, for negative bending at the right end

𝐼𝑒 = effective moment of inertia for the member using the method indicated

𝐼𝑒(𝑥) = (local) section-based effective moment of inertia at position 𝑥 along member

𝐼𝑒,𝑎𝑣𝑔 = average effective moment of inertia using CSA A23.3 clause 9.8.2.4

𝐼𝑒𝑚 = effective moment of inertia with for the midspan segment in method indicated

𝐼𝑒𝐿 = effective moment of inertia with left end-moment in method indicated

𝐼𝑒𝑅 = effective moment of inertia with right end-moment in method indicated

𝐼𝑒′ = integration-based (constant) equivalent moment of inertia for a member

𝐼𝑒∗′ = approximate equivalent moment of inertia, 𝐼𝑒

′ , but using ∗ in lieu of

𝐼𝑔 = gross moment of inertia, (𝑏ℎ3 12⁄ for prismatic rectangular sections)

𝑘𝑑 = the depth of the compression face for purposes of defining 𝐼𝑐𝑟

𝐾 = factor which ratios a simply supported 𝛥 to a continuous member 𝛥

𝐾𝑟 = variable which gives factored resistance such that 𝑀𝑟 = 𝐾𝑟𝑏𝑑2

𝐾𝑟 𝑚 = 𝐾𝑟 for midspan based on the maximum positive moment (near midspan)

𝐾𝑟 𝐿 = 𝐾𝑟 for the left end of the member

𝐾𝑟 𝑅 = 𝐾𝑟 for the right end of the member

𝐿 = length of member span (measured center-to-center of relevant supports)

𝐿∆𝑚𝑎𝑥 = length from left end of span to location of maximum deflection

𝐿1 = length from left end of span to end of left-negative-cracked segment

𝐿2 = length from left end of span to start of positive-cracked segment

𝐿3 = length from left end of span to midspan, therefore 𝐿3 = 𝐿/2

𝐿3𝐴 = length from left end of span to the 1/3 point, therefore 𝐿3𝐴 = 𝐿/3

𝐿3𝐵 = length from left end of span to the 2/3 point, therefore 𝐿3𝐵 = 2𝐿/3

𝐿4 = length from left end of span to end of positive-cracked segment

𝐿5 = length from left end of span to start of right-negative-cracked segment

𝐿6 = length of span, therefore 𝐿6 = 𝐿

𝐿𝑐𝑟 = distance 𝑥 to leftmost section where 𝑀(𝑥) > 0 and 𝑀(𝑥) > 𝑀𝑐𝑟

𝐿𝑔 = length of uncracked section(s) in FRP-reinforced concrete (Appendix C)

𝐿𝑗 = length from left end of span to right end of integration segment 𝑗

𝐿𝑃𝐿 = distance to the closest point load from closest support

𝐿𝑅4 = length from right end of span to right end of positive-cracked segment

𝐿𝑅5 = length, from right end of span, of the right-negative-cracked segment

l n = the clear span between supports as defined in CSA A23.3-04

xiii

𝑚(𝑥) = internal virtual moment from virtual load, using method of virtual work

𝑚𝑖(𝑥) = virtual moment, 𝑚(𝑥), at location 𝑖 on the span

𝑀 = midspan bending moment for the applied load

𝑀𝑎 = the applied bending moment used to compute the effective moment of inertia

𝑀(𝑥) = bending moment at position 𝑥 for the applied load

𝑀0 = total static moment (midspan service load moment if end-moments released)

𝑀0,0 = 𝑀0 for a simply supported member when producing example members

𝑀1𝑃𝐿 = maximum moment from one point load applied at midspan

𝑀2𝑃𝐿 = maximum moment from two equal loads applied at third points

𝑀𝑐𝑟 = cracking moment of the member, (fr𝑏ℎ2/6 for rectangular sections)

𝑀𝑓 = factored moment of the member (required to be less than 𝑀𝑟)

𝑀𝑚 = net midspan moment (with worst case service load applied)

𝑀𝑚𝑎𝑥 = maximum positive service moment (with worst case service load applied)

𝑀𝑛 = the nominal moment capacity (𝑀𝑟 without the reductions from 𝜙𝑐 & 𝜙𝑠)

𝑀𝑟 = factored bending moment resistance in ultimate limit states design

𝑀𝑠 = service load bending moment at location indicated

𝑀𝑠𝑢𝑠 = sustained service load bending moment at indicated segment

𝑀𝐿 = service moment at left support a loading when 𝑀𝑚𝑎𝑥 occurs

𝑀𝑅 = service moment at right support a loading when 𝑀𝑚𝑎𝑥 occurs

𝑀𝑈𝐷𝐿 = maximum moment from uniformly distributed load, 𝑤𝑈𝐷𝐿𝐿2 8⁄

𝑛 = modular ratio (𝐸𝑏/𝐸𝑐)

𝑝 = denotes specific point/location along the span to determine deflection at

𝑃 = load applied at a particular point(s) (short segment(s)) on the span

𝑃𝑐𝑟 = load 𝑃 which causes tension face to exceed 𝑓𝑟

𝑃0 = 𝑃1𝑃𝐿 or 𝑃2𝑃𝐿 used to produce continuous member based on a simple member

𝑃s = maximum service load 𝑃

𝑃1𝑃𝐿 = one point load (𝑃) applied at midspan

𝑃2𝑃𝐿 = total of two equal loads (𝑃), each equal 𝑃2𝑃𝐿 2⁄ , applied at third points

𝑆 = time-dependant factor used for calculating sustained-load effects

UDL = a uniformly distributed load located along the relevant member span

𝑉𝐿 = shear force resulting from loads 𝑃 or 𝑤 at the left end of the span

𝑉𝑅 = shear force resulting from loads 𝑃 or 𝑤 at the right end of the span

𝑤 = any distributed load applied on a member; 𝑤𝑈𝐷𝐿 in this report

𝑤0 = 𝑤𝑈𝐷𝐿 on a simply supported member

𝑤𝑈𝐷𝐿 = service load distributed uniformly across full member span

𝑥 = distance, from left, along a span (to a particular position/cross-section)

xiv

𝑦𝑡 = distance from section gross centroid to tension face (rectangular = ℎ/2)

z = denotes load case number for specific example load and end-moments

𝛼1 = ratio of average stress in rectangular compression block to 𝑓𝑐′

𝛼𝑐𝑟 = ratio of cracking moment divided by total moment (𝑀𝑐𝑟/𝑀0)

𝛼+ 𝑓⁄ = ratio of 𝑀𝑟/𝑀𝑓 , midspan, where additional bars added to reduce 𝛥𝑚𝑎𝑥

𝛼𝑠 𝑓⁄ = service load ratio (factored load divided by service load, 𝑀𝑠/𝑀𝑓)

𝛼𝐿 = ratio of left end-moment divided by total moment (𝑀𝐿/𝑀0)

𝛼𝐿/𝑚𝑎𝑥 = ratio of left end-moment divided by maximum moment (𝑀𝐿/𝑀𝑚𝑎𝑥)

𝛼𝑅 = ratio of right end-moment divided by total moment (𝑀𝑅/𝑀0)

𝛼𝑅/𝑚𝑎𝑥= ratio of right end-moment divided by maximum moment (𝑀𝑅/𝑀𝑚𝑎𝑥)

𝛽 = tension stiffening factor for tensile contribution of concrete after cracking

𝛽1 = ratio of depth of rectangular compression block to depth of neutral axis

𝛽𝑑 = coefficient to modify Branson’s equation for FRP-reinforcing

= integration factor to account for changes in stiffness along member span

∗ = correction to which improves continuous member effective stiffness

𝛿 = symbol indicating the change in the subsequent variable(s)

𝛿𝛥𝑐𝑟 = the change in the deflection from the bending-moment related cracking

𝛥 = midspan deflection resulting from the maximum service load case

𝛥(𝑥) = deflection at 𝑥 from the given service load case

𝛥1 = analytical integration results, 𝛥𝑗 , for 𝑥 = 0 𝑡𝑜 𝐿1

𝛥1+2 = 𝛥𝑗 for 𝑥 = 0 𝑡𝑜 𝐿2 for analytical integration with no left-end cracking

𝛥2 = analytical integration results, 𝛥𝑗 , for 𝑥 = 𝐿1 𝑡𝑜 𝐿2

𝛥3 = analytical integration results, 𝛥𝑗 , for 𝑥 = 𝐿2 𝑡𝑜 𝐿/2

𝛥3𝐴 = third-point loading integration results, 𝛥𝑗 , for 𝑥 = 𝐿2 𝑡𝑜 𝐿/3

𝛥3𝐵 = third-point loading integration results, 𝛥𝑗 , for 𝑥 = 𝐿/3 𝑡𝑜 2𝐿/3

𝛥3𝐶 = third-point loading integration results, 𝛥𝑗 , for 𝑥 = 𝐿/2 𝑡𝑜 2𝐿/3

𝛥3𝐷 = third-point loading integration results, 𝛥𝑗 , for 𝑥 = 2𝐿/3 𝑡𝑜 𝐿4

𝛥4 = analytical integration results, 𝛥𝑗 , for 𝑥 = 𝐿/2 𝑡𝑜 𝐿4

𝛥5 = analytical integration results, 𝛥𝑗 , for 𝑥 = 𝐿4 𝑡𝑜 𝐿5

𝛥5+6 = 𝛥𝑗 for 𝑥 = 𝐿4 𝑡𝑜 𝐿 for analytical integration with no right-end cracking

𝛥6 = analytical integration results, 𝛥𝑗 , for 𝑥 = 𝐿5 𝑡𝑜 𝐿

𝛥𝑐𝑟 = midspan deflection assuming the member is fully cracked (using 𝐼𝑐𝑟)

𝛥𝑔 = midspan deflection assuming the member is uncracked (using 𝐼𝑔)

𝛥𝑔,𝑚𝑎𝑥 = maximum deflection assuming the member is uncracked (using 𝐼𝑔)

𝛥𝑔,𝑚𝑖𝑑 = midspan deflection assuming the member is uncracked (using 𝐼𝑔)

𝛥𝑖 = deflection at point 𝑖, found by method of virtual work by using 𝑚𝑖(𝑥)

xv

𝛥𝑗 = portion of 𝛥𝑖 contributed by respective integration segment: 𝑗 = 1,2,3,…

𝛥𝑚𝑎𝑥 = maximum deflection of a member at the maximum service load case

𝛥𝑚𝑎𝑥,𝐼𝑒(𝑥) = maximum deflection of a member calculated using 𝐼𝑒(𝑥)

𝛥𝑚𝑎𝑥,𝐼𝑒′ = maximum deflection calculated using Bischoff’s 𝐼𝑒

′

𝛥𝑚𝑖𝑑 = midspan deflection, 𝛥, exactly at midspan (𝐿/2) for 2-support member

𝛥𝑢𝑛𝑐𝑟 = midspan deflection when 𝑀𝑚𝑎𝑥 ≈ 𝑀𝑐𝑟 (and assuming no cracking)

𝛥𝐼𝑒 = midspan deflection determined using Branson’s 𝐼𝑒

𝛥𝐼𝑒′ = midspan deflection calculated using Bischoff’s 𝐼𝑒′

𝛥𝐼𝑒∗′ = midspan deflection calculated using proposed 𝐼𝑒∗′ (equal third-point loading)

𝛥𝐼𝑒(𝑥) = midspan deflection calculated using section-based moment of inertia, 𝐼𝑒(𝑥)

𝛥𝐼𝑒,𝑎𝑣𝑔 = midspan deflection calculated using the average moment of inertia (see 𝐼𝑒,𝑎𝑣𝑔)

𝛥𝐼𝑒,𝛽𝑑 = midspan deflection calculated using 𝐼𝑒 per ACI440.1R (with a 𝛽𝑑 modifier)

𝛥𝑀𝐿 = deflection caused by left end-moment

𝛥𝑀𝑅 = deflection caused by right end-moment

𝛥𝑈𝐷𝐿 = deflection caused by uniformly distributed load

𝛥𝑈𝑧(𝑥) = total deflection at 𝑥 caused by uniformly distributed load for load case 𝑧

𝛥𝛽=0 = midspan deflection calculated with tension stiffening neglected (per S806)

𝛥𝛾=1 = midspan deflection calculated using the Bischoff’s 𝐼𝑒 (equals 𝐼𝑒′ with = 1)

휀𝑐𝑢 = maximum strain at extreme compression face at ultimate, 0.0035

𝜂 = stiffness reduction coefficient (1 − 𝐼𝑐𝑟/𝐼𝑔), use 𝜂=𝜂𝑚 unless noted

𝜂𝑚 = stiffness reduction coefficient (1 − 𝐼𝑐𝑟𝑚/𝐼𝑔) for at/near midspan

𝜂𝐿 = stiffness reduction coefficient (1 − 𝐼𝑐𝑟𝐿/𝐼𝑔) for the left end

𝜂𝑅 = stiffness reduction coefficient (1 − 𝐼𝑐𝑟𝑅/𝐼𝑔) for the right end

𝜉 = 1 − √1 −𝑀𝑐𝑟/𝑀𝑚 , ( 𝜉𝐿/2 is the uncracked end length with 𝑤𝑈𝐷𝐿 load)

𝜌 = reinforcement ratio of the tension bars, 𝐴𝑠/𝑏𝑑 or 𝐴𝑓/𝑏𝑑

𝜌′ = reinforcement ratio of the compression bars

𝜌𝑏 = balanced reinforcement ratio for a reinforced member

𝜌𝑚 = tension reinforcement ratio, 𝜌, at/near member midspan

𝜌𝐿 = tension reinforcement ratio, 𝜌, at left end of member

𝜌𝑅 = tension reinforcement ratio, 𝜌, at right end of member

𝜙 = curvature of member being considered at the point being considered

𝜙𝑐 = resistance factor for concrete under CSA A23.3-04 (or as required)

𝜙𝑏 = resistance factor for reinforcement bar: 𝜙𝑓 or 𝜙𝑠

𝜙𝑓 = resistance factor for FRP reinforcement per relevant standard

𝜙𝑠 = resistance factor for steel reinforcement under CSA A23.3-04

1

1.0 INTRODUCTION

Building and bridge codes usually prescribe deflection limits. Deflection limits exist so

that all designed structures will meet reasonable serviceability requirements. Poor

serviceability causes unnecessary inconvenience to users, sometimes becoming a safety

concern. Engineers need accurate equations for predicting deflection so that they can

efficiently meet bridge and building code requirements.

Bending deflection of concrete members can govern their design. Steel-reinforced

concrete members can be assumed to meet deflection requirements if they comply with

Table 9.1 of A23.3 (CSA 2004). For other cases, however, deflection must be

calculated. To accurately predict bending deflection, a reasonably accurate effective

moment of inertia is needed. For steel reinforced slabs and FRP reinforced concrete

members, Branson’s (1965) equation often underestimates deflections (Bischoff 2005).

Deflection often governs design in these cases, so this issue needs to be rectified.

This report introduces bending deflection of continuous concrete members and

evaluates different member-based moment of inertia solutions, including a proposed

solution, for use with continuous members. All evaluations provided are based on

immediate deflection caused by dead load plus live load; the deflection computed is the

maximum vertical sag between two supports of a one-way slab or beam. Accurate

solutions for simply-supported prismatic members are provided by Bischoff and Gross

(2011). This report will demonstrate that those solutions also work well for almost all

deflection-critical continuous members.

2

1.1 Project Need

Moment of inertia calculations should progress toward increasing accuracy and this

work intends to aid in that evolution. ACI 318 (ACI Committee 318 2011), CSA A23.3

(CSA 2004), and CSA S806 (CSA 2012) use the “effective moment of inertia” approach

to calculate bending deflection of reinforced concrete members. This approach

calculates deflection using linear-elastic deflection equations and an effective moment

of inertia is used to account for non-linearity after cracking. An improved approach

should also account for variations in member stiffness along the length of the member.

While Branson’s (1965) solution for the effective moment of inertia has been used for

nearly 50 years, it is an empirical equation that was only calibrated using simply

supported beams with steel reinforcing ratios between 1% and 2%. Rationally derived

equations that also work well for other member types without losing accuracy, such as

those proposed by Bischoff and Gross (2011), should replace Branson’s solution.

1.2 Project Objectives

The report will demonstrate the complexity of continuous concrete member bending

deflection by showing the results of analytical integration for midspan deflection and

discussing issues related to the service moment history.

The report will also demonstrate that Bischoff’s equations for the equivalent moment of

inertia for simply-supported members (Bischoff and Gross 2011) are useful for

continuous members. These equations are used in calculations of one-way bending

deflection for cracked reinforced concrete members. The examples provided show that

the equations offer similar or improved results relative to the other available methods.

3

1.3 Project Scope

The unique aspects of the bending deflection of continuous concrete members are

introduced and methods for computing predicted deflections are introduced; this

culminates in a description of Bischoff’s approach (Bischoff and Gross, 2011) for

determining the local (section based) moment of inertia and equivalent moment of

inertia. This report only provides three possible loadings because of the complexity of

the continuous member problem. Only short-term deflection is investigated; long-term

deflection is assumed to be the predicted short-term deflection augmented by a

calculated multiplier.

To demonstrate that Bischoff’s simply-supported equations (Bischoff and Gross, 2011)

are generally accurate for continuous members, results are compared between

Bischoff’s equations, analytical/numerical integrations, Branson’s (1965) equation, and

the S806 (CSA 2012) method. Continuous member clauses and methods from A23.3

(CSA 2004) for steel reinforced concrete and S806 (CSA 2012) for FRP reinforced

concrete are included where relevant. The example members selected are intended to

cover the practical range of steel reinforced members and provide FRP-reinforced

member examples.

Loading cases provided include midspan point load, two equal third-point loads, and a

uniformly distributed load. Examples are prismatic members spanning between

supports with bending moments that are positive near midspan and negative (or zero) at

supports.

4

The results are not compared to beam deflection test data in this report. The results rely

on the accuracy of Bischoff’s method (Bischoff and Gross, 2011), which integrates

curvature while accounting for tension stiffening to calculate an effective stiffness.

Deflections in this report are calculated for a statically determinate span with end

moments. The relative stiffness of the adjacent structural members, end rotations, and

pattern loading may affect the actual end moments. Designers must resolve the

structure and worst case service load to accurately determine deflections using the

provided equations.

The appendices provide detailed information about the equations used in this report,

along with example calculations. The means of producing the example idealized

members within spreadsheets and the means of computing deflection using each method

are presented. The data for the example results shown in Chapter 3 are provided, along

with other data and graphs which compare computed results using the proposed

equations and other methods. The appendices also discuss several interesting and

relevant aspects of concrete bending deflection, such as derivations of equations used

and discussions of some of the complications of computing deflection of concrete

members.

1.4 Report Organization

Chapter 2 of this report introduces bending deflection for linear-elastic members and

concrete members that are cracked in bending. It then introduces relevant published

research, requirements from Canadian standards, and the deflection calculation guide

5

from the Concrete Design Handbook (CAC 2005). Research is presented from Branson

(1965), Razaqpur et al. (2000), and Bischoff and Gross (2011).

Chapter 3 presents the methodology and results of this project. It shows and/or

discusses the following: relevant integration methods, setup for example members,

example deflection results, and proposed deflection equations and limits. Equations,

results, and discussions are provided for members with a midspan point load, equal

third-point loads, and a uniformly distributed load.

Chapter 4 summarizes useful results in this report and recommends relevant future

work.

6

2.0 BACKGROUND TO DEFLECTION OF

REINFORCED CONCRETE

Accurately predicting bending deflection in reinforced concrete is complicated. A

constant stiffness is required to calculate deflection using the standard North American

method. When portions of concrete members are in tension and crack, however, the

stiffness is not constant. Therefore, the use of standard methods for computing

deflection requires the use of an effective constant moment of inertia for a reinforced

concrete member. Research has led to different approaches and solutions for predicting

deflection; these include solutions using a constant moment of inertia. Most solutions,

however, are valid only for a limited range of reinforcement ratios or types of

reinforcement. This work presents four methods for calculating deflection for these

members, with two determining a constant stiffness. It also presents and discusses

relevant Canadian building code requirements.

Branson’s (1965) work introduced the concept of an effective moment of inertia. His

equation, sometimes provided with modification factors, is currently recommended in

most North American concrete member design standards. CSA S806 (2012), however,

employs integration of curvature (while ignoring tension stiffening) and provides

solutions for common cases. Similar solutions that include tension stiffening have been

proposed by Bischoff (2011) for FRP reinforced concrete, and these solutions also work

well with steel reinforcement. Bischoff’s method would complete the evolution to one

common and accurate deflection equation that does not depend on the type or amount of

reinforcement used.

7

2.1 Introduction to Deflection

Deflection is a simple but important concept for people who design structures, even

though it is rarely noticed by the public. Deflection is the movement of a portion of a

structure from its initial or previous position. Even if a structure is strong enough to

resist every required design load, engineers must accurately predict (and sometimes

reduce) deflection because bridge and building codes and standards impose deflection

limits. These limits exist because too much movement can result in structures not being

sufficiently comfortable, aligned, or usable. Examples of deflection include beam sag,

column tilt, and floor vibration.

2.2 Elastic Deflection of Prismatic Members

In North America, the standard equations for deflection assume linear-elastic behaviour

of prismatic members. Elastic deflection occurs if a member returns to its pre-loaded

position after the load is removed. Prismatic members have a uniform cross-section and

moment of inertia for the full member length. For a linear-elastic material, prismatic

members enable accurate deflection prediction using available equations; these

equations enable engineers to meet deflection requirements for most structures.

Concrete undergoes non-linear deflection behavior when it cracks, and also tends to

have additional long-term deflection. Equations for linear-elastic members are,

however, still used to determine reinforced concrete bending deflection.

The second moment of inertia (referred to as the “moment of inertia” in this report) is

important in determining bending deflection. The bending stiffness is a function of the

materials used (specifically their modulus of the elasticity, such as 𝐸𝑠 or 𝐸𝑐) and the

8

moment of inertia for the axis about which the member is being bent. The moment of

inertia will be denoted with the symbols 𝐼 , 𝐼𝑔 , 𝐼𝑐𝑟 , 𝐼𝑒 , 𝐼𝑒′ , and so on, depending on

what is assumed, neglected, or taken into account.

2.2.1 Simply Supported Members

Simply supported members are simple to design and simple to test in the laboratory.

These members are free to rotate at both ends and free to move longitudinally at one

end. The engineering analysis is simple because members are statically determinate and

pattern loadings can be ignored. Because simply supported designs are easier to study

and test, their deflection has been studied more thoroughly.

Table 2-1 provides the equations for bending moment, 𝑀, and maximum deflection, ∆,

for simply supported members that are modelled as linear-elastic and prismatic. The

equations can be found in the Handbook of Steel Construction (CISC 2009) and many

Table 2-1 - Example simply supported members with equations

9

other sources. 𝑀(𝑥), as used throughout this report, is the bending moment at distance

x from the left end of the beam. Bending moments are taken as positive if they act to

cause tension in the bottom face of the member (hence adding downwards deflection

within the span). See Appendix H for example calculations and see the List of Symbols

for other variable definitions.

2.2.2 Members with Bending Moments at Supports

Continuous members are used throughout cast-in-place concrete construction. The use

of continuous members reduces the amount of construction work and concrete required.

Continuous slabs, beams, and columns are typical examples for concrete construction,

but all cantilevers and other moment connections are also, in effect, continuous

members. The negative bending moments (at supports) for continuous prismatic beams

or slabs are typically between one-half and three times the positive (midspan) bending

moments. These large negative moments reduce the member depth required because

they reduce the positive bending moment and deflection. “Continuous members” and

“members with end-moments” are referred to synonymously in this report.

Figure 2-1 shows exaggerated deflected shapes for four different end-moment

conditions. Four different uniformly distributed loads are contrived to give equal

midspan deflection. Note that the midspan deflection is not the maximum deflection for

unequal end-moments. The derivations and data for Figure 2-1 are found in Appendix I.

In Figure 2-1, ∆ represents the bending deflection of the member, 𝑀𝐿 and 𝑀𝑅 represent

the left end and right end bending moment, and 𝑀𝑚 is the net midspan moment.

10

Figure 2-1 - Deflected Shape Comparison of Four Different Loads & End-Moments

2.2.3 Continuous Member Factor, 𝑲

Deflections for prismatic linear-elastic continuous members can by determined using

the equations shown in Table 2-2. The Concrete Design Handbook (CAC 2005),

Chapter 6, introduces the factor, 𝐾, to compute these deflections. The equations for 𝐾

apply to any constant stiffness members with known end-moments. The midspan

deflection of a prismatic continuous member is the 𝐾 factor multiplied by the midspan

Table 2-2 - Deflection of continuous prismatic linear-elastic members

11

deflection of a prismatic simply supported member with the same span, midspan

moment, and midspan properties. A derivation of the 𝐾 factors shown in Table 2-2 can

be found in Appendix A. See the List of Symbols for other equation variable

definitions.

2.3 Bending Deflection of Reinforced Concrete

There are many issues to overcome in order to predict bending deflection for reinforced

concrete. The modulus of rupture, shrinkage stress, crack spacing, and the effect of

concrete between (tension face) cracks must be approximated; they cannot be predicted

with high precision. Input materials, site condition history, load history, and the type of

reinforcement used also affect the member response. Calculations predicting deflection

are therefore only intended to give the engineer an indication of whether deflection is

likely to be of concern.

When a portion of a reinforced concrete member cracks in bending, it is no longer

linear-elastic. Simply supported concrete members develop bending cracks in the

bottom face at a midspan segment and have uncracked end segments. Continuous

members develop positive bending cracks at midspan and negative bending cracks at

supports with uncracked segments separating these three cracked segments. The

moment of inertia also varies within a cracked segment. To accurately determine the

deflection, the variation in the moment of inertia along the member should be accounted

for. This report accepts the values and formulas recommended in Bischoff and Gross

(2011) and explores use of those equivalent moments of inertia for continuous members.

12

As will be discussed in Section 2.5, A23.3 (CSA 2004) provides some guidance for

steel-reinforced concrete members. These members can be assumed to meet deflection

requirements if they meet a recommended minimum depth per A23.3 Table 9.1 (CSA

2004). For other cases, A23.3 (CSA 2004) recommends the linear-elastic equation

approach and provides Branson’s (1965) equation for the constant effective moment of

inertia, 𝐼𝑒. Many standards and codes indicate that integration of curvature can be used

as a robust and practiced method to determine deflection without a limited range of

validity.

2.3.1 Concrete Bending Response

Figure 2-2 shows the typical moment-curvature response of a flexural member, and the

load deflection response looks essentially the same. The steep line on this graph, on the

left, shows the deflection of an uncracked member. The low-slope line, on the right,

shows the deflection of a fully cracked member. The thicker solid line shows the

Figure 2-2 - Moment-Curvature Response of Reinforced Concrete

(Bischoff 2007)

13

response of an initially uncracked concrete member which is loaded until it is heavily

cracked. In the transition between the cracked and uncracked lines, the member is

partially cracked. The line with the 𝐸𝑐𝐼𝑒 label is a linear-elastic representation of a

partially cracked reinforced concrete member; the origin and the point of curvature at

the service load are the two points which define this line.

In this figure, 𝐸𝑐𝐼𝑒 is computed at the service load which causes the applied moment,

𝑀𝑎. The gross moment of inertia, 𝐼𝑔, is taken as 𝑏ℎ3/12 throughout this report (for

rectangular sections); this ignores the contribution of the reinforcing bars when

analyzing bending of the uncracked section (which is a reasonable simplification for

most reinforced concrete members). The cracked moment of inertia, 𝐼𝑐𝑟, is the moment

of inertia assuming that the reinforcing bars resist all of the bending tension (without

yielding) and that the concrete in compression is elastic at service loads. 𝑀𝑐𝑟 represents

the bending moment which causes the concrete to crack in bending and φ represents the

curvature of the member at the point being considered.

While the section is assumed to remain elastic, the response is not linear-elastic (as

Figure 2-2 indicates). Any linear-elastic representation of a cracked member will only

be accurate for a particular service load. Section 2.4.2 discusses other relevant issues.

2.3.2 Tension Stiffening of Concrete Bending Members

Tension stiffening can be important in determining the bending deflection of a

reinforced concrete member. Within the cracked region(s), a concrete bending member

has short segments of uncracked cross-section between the cross-sections at cracks.

14

The moment of inertia at a crack is different from the moment of inertia at uncracked

cross-sections. Tension stiffening accounts for the influence of the uncracked segments

a region with cracks. 𝑀𝑐𝑟/𝑀(𝑥) is a rational and accurate tension stiffening factor

(Bischoff 2007); it is used to compute the section-based effective moment of inertia,

𝐼𝑒(𝑥), in this report. 𝑀(𝑥) is the bending moment at the location along the beam of 𝑥.

Tension stiffening has a large effect at the service load depicted in Figure 2-3, for

example. Here, 𝑃𝑠 represents the service load and 𝑃𝑐𝑟 represents the load at cracking.

Figure 2-3 - Effect of Tension Stiffening on a Reinforced Concrete Member

(Gilbert 2007)

2.3.3 Constant Stiffness Approach

The North American approach to determine deflection for prismatic reinforced concrete

members is to use linear-elastic equations. Linear-elastic equations assume members

have a constant stiffness across the entire span. Thus, they require an effective constant

moment of inertia for concrete members that are cracked in bending. The effective

constant moment of inertia is commonly represented as 𝐼𝑒. An accurate weighted

average effective moment of inertia, based on integration of curvature, has been defined

as the equivalent moment of inertia, 𝐼𝑒′ (Bischoff and Gross, 2011).

15

Linear-elastic equations are simple, can be computed with a calculator, and offer the

benefit of understandable intermediate values for all variables involved. An

experienced engineer can determine if the intermediate values are reasonable, greatly

reducing possible errors. A method using simple equations, even if slightly inaccurate,

is also a good way of checking computer results to see if they are reasonable.

In Figure 2-4, the different moments of inertia are shown for an example continuous

member with known end-moments and properties, and supporting a uniformly

distributed load of 10 kN/m. The cracking moment is 𝑀𝑐𝑟 = 113 kNm, which means

the member is cracked at both ends and at midspan under service loads. The service

load bending moments are: 𝑀𝐿 = −247 kNm (left end-moment), 𝑀𝑚𝑎𝑥 = 125 kNm

Figure 2-4 - Gross, Local-Effective, Equivalent, and Cracked Moments of Inertia

16

(maximum moment in positive bending), and 𝑀𝑅 = −165 kNm (right end-moment).

The reinforcing steel for the member represented by Figure 2-4 was calculated using a

factored moment resistance of 𝑀𝑟=1.6𝑀𝑚𝑎𝑥, 𝑀𝑟=1.6𝑀𝐿, and 𝑀𝑟=1.6𝑀𝑅 for the

midspan, left-end, and right-end segments, respectively. The cracked moment of inertia

for the face of the beam that is in tension in these three segments is denoted as 𝐼𝑐𝑟∗.

𝐼𝑒(𝑥) represents the (local) section-based moment of inertia at the position 𝑥, and 𝐼𝑒′

represents the equivalent moment of inertia.

2.3.4 Integration Approach to Deflection

An integration approach, such as using the method of virtual work, is a logical approach

to calculating deflection because integrating curvature can accurately predict the

deflection if bending moment and stiffness along the member are known. Eurocode 2

(CEN, 2004) and some other codes and standards indicate that integration should be

used to determine deflection in concrete members. While linear-elastic deflection

equation methods offer more understandable intermediate calculations, numerical

integration is a tenable practice and should provide increased accuracy for all possible

cases. Unlike other approaches, integration-based approaches can easily account for

any variation in the quantity of tension reinforcement along of the member. The use of

integration also enables proper modelling of the negative moment regions; most other

approaches can only offer rough approximations of their effects. It is likely that much

of the opposition to this approach occurs simply because integration is rarely used in

practice by North American structural engineers.

17

Analytical integration, such as the formulas obtained for this report, can be performed

by hand or computer. Numerical integration is easily performed with spreadsheets; in

work for this report, these spreadsheets proved to be robust, practical, and relatively

simple.

2.4 Effect of Materials and Load History on Deflection

Unfortunately, the cracking in the tension face of concrete members is not the only

concern related to bending deflection of concrete members. The concrete mix and field

conditions affect concrete cracking. The load history will also affect how much

deflection will occur under future loads. Reinforcing bars made from different

materials also affects deflection calculations; modifications are thus often required to

the empirical equations developed for steel reinforced concrete members. FRP

reinforcing is discussed in Section 2.5. Other potential reinforcing materials also differ

from steel, but are not commonly in use and are not discussed in this report.

2.4.1 Variation in Mix Materials and Field Conditions

Cast-in-place concrete has many different variables that affect its bending deflection.

These variables include the exact materials, mix ratios, batch timeline, and conditions

for mixing/placement/curing. All of these variables will affect the stress at which

concrete cracks in tension, 𝑓𝑟 (the modulus of rupture). Some of these variables will

also affect the modulus of elasticity of the concrete in the compression zone. Most

concrete mixes will shrink when curing and in the long-term. Both curing-shrinkage

and the future concrete material chemical reactions will affect the amount of bending-

18

moment induced tensile stress a member can sustain before cracking. Shrinkage and

creep equations are intended to account for these effects. All these effects, however, are

not perfectly understood and will not be fully controlled in the field. As such, concrete

deflection predictions are imprecise approximations (but are necessary, as noted in the

introduction to Section 2.3).

2.4.2 Effect of Load-History on Deflection

Concrete creep and pre-loading will affect the deflection of concrete members.

Members experience creep as the tension-stiffening effect and the compression face

relax slightly. Creep effects are usually considered to be a function of the initial

deflection of the sustained loads multiplied by a factor for duration (CSA 2004). Creep

is inelastic deflection and results from sustained concrete compression stresses below

the elastic limit of 0.5𝑓𝑐′ (where 𝑓𝑐

′ is the specified compressive strength of concrete).

Shrinkage, creep, and axial effects can be accounted for in other calculations not

discussed in this report. If a pre-loading exceeds the service load, this increases service

load deflection.

Continuous members are also affected by pattern-load history. Different pattern-load

cases will typically be required to give the maximum negative service moment and the

maximum positive service moment. A continuous concrete beam tested with only the

largest positive moment case will produce a different deflection than a beam that has the

largest negative moment case applied to it first. This is discussed further in Section

3.7.3.

19

2.5 FRP Reinforced Members, Razaqpur’s Work, and CSA S806

The behaviour of members reinforced with fibre reinforced polymer (FRP) bars and

steel bars is significantly different. Work by Razaqpur (Razaqpur et al. 2000)

demonstrates that the integration of curvature method, with solutions provided for

common cases, is a practical and effective way to calculate deflection for an FRP-

reinforced concrete member. The Canadian standard for FRP-reinforced concrete, S806

(CSA 2012), accepts this and specifies to use this method.

2.5.1 Fibre Reinforced Polymers as Concrete Reinforcing

Fibre reinforced polymer (FRP) reinforcing bars for concrete are significantly different

from steel bars. Deflection design usually governs the amount of reinforcing in these

members. FRP reinforced concrete members also typically require over-reinforced

strength design. FRP reinforcing consists of fibre polymers and resins; the tensile

strength and stiffness is primarily from the fibre ingredient. FRP materials are

characterized by high tensile strength only in the direction of the reinforcing fibres.

There are three main fibre types for FRP-reinforcement, resulting in three different

types of FRP: glass (GFRP), carbon (CFRP), and aramid (AFRP).

FRP reinforcing bars differ from steel in ways that are critical to strength design. FRP

materials do not yield; rather, they are elastic until failure (typically when the concrete

crushes). Design procedures must account for the brittle failure method. For more

information on strength design criteria, see Chapter 8 of ACI 440.1R (ACI Committee

440 2006). Some typical properties for FRP reinforcement are found in ACI 440.1R

(ACI Committee 440 2006).

20

In FRP reinforced concrete member design, it is common to design for deflection and

then check strength, since the deflection requirements will usually govern. This means

that the amount of FRP reinforcement required for deflection design will usually exceed

the amount required for strength design. An FRP-reinforced concrete slab will typically

achieve the maximum permissible long-term deflection for the member at a service load

of 20% to 30% of its nominal moment resistance; similarly, an FRP-reinforced concrete

beam will have a service load of 35% to 45% of its nominal moment resistance (Vesey

and Bischoff, 2011).

The methods for determining deflection with FRP-reinforced members are very similar

to steel-reinforced members. While the usual structural analysis techniques are

permitted, a correction factor is required in order to use Branson’s effective moment of

inertia. ACI 440.1R-06 (ACI Committee 440 2006) provides an empirical modification

factor to the (𝑀𝑐𝑟/𝑀𝑎)3 𝐼𝑔 term in Branson's equation so that it does not underpredict

actual deflections with FRP-reinforced members. This correction factor is empirical

and is not a logical way to account for the lower elastic modulus of FRP, so will not

compute accurate deflections for all types of members. ACI does mention that other

approaches exist. Integration approaches and Bischoff’s equation can be used with FRP

reinforcing without any correction factor or other modification.

2.5.2 Razaqpur’s Work

Razaqpur et al. (2000) recommend that deflection of FRP-reinforced concrete be

calculated using integration of curvature. They assert that tension stiffening can be

neglected for FRP-reinforced concrete and use test results to support this assertion.

21

They also note that Branson’s method does not work well for FRP-reinforced members.

This leads to the integration of curvature method, which is known to be a robust

method. They justify use of an idealized tri-linear moment-curvature relationship which

makes the integration less complicated and enables them to provide simple solutions for

common simply supported and fixed-cantilever cases. See Appendix C for a brief

introduction to their results and Appendix G for information about integration without

tension stiffening.

Razaqpur and Isgor (2003) published a similar work for continuous members. The

same methodology is recommended with a couple of additional simplifying

assumptions. It also provides three example solutions. Work for this report determined

that two of these example solutions contained minor algebraic errors. It was also found

that the simplifying approximations are accurate whenever tension stiffening can be

ignored. The work by Razaqpur and Isgor (2003) does, however, provide conservative

results while other common deflection calculations often underpredict bending

deflection of FRP-reinforced concrete.

2.5.3 Concrete Deflection in CSA S806

CSA S806 (CSA 2012) specifies how to calculate deflection in FRP-reinforced one-way

bending members. S806 is the Canadian standard for the Design and Construction of

Building Structures with Fibre-Reinforced Polymers. Clause 8.3.2 of this standard

states that one-way bending members must meet the typical requirements for deflection

under service load. This clause defines cracking moment using the following common

equation (where 𝑦𝑡 is the dimension from the centroid to the tension face):

22

𝑀𝑐𝑟 =𝑓𝑟𝐼𝑔

𝑦𝑡 where: 𝑓𝑟 = 0.6√𝑓𝑐′ (2 − 1)

S806 (CSA 2012) provides a different way to determine deflection than A23.3 (CSA

2004) and similar codes. Clause 8.3.2.3 declares that deflection is to be calculated using

methods based on integration of curvature. Clause 8.3.2.4 states that maximum

deflections may be calculated using formulas that are provided. These formulas are the

results determined by Razaqpur et al. (2000) as discussed in Section 2.5.2. The standard

does not provide any formulas for members with end-moments, so integration of

curvature is required for continuous members. Clause 8.3.2.5 states that all integration

of curvature is to be performed using the tri-linear relationship as discussed in Appendix

C. Continuous member equations which use the S806 method are denoted with “S806”

in this report; relevant integration equations are provided in Appendix G.

The method indicated in CSA S806 for predicting deflection warrants criticism. The

assumption that tension stiffening effects are negligible will yield excessively

conservative results if an FRP reinforced member is lightly-cracked because this

approach was originally validated for unrealistically high levels of service load

(Bischoff and Gross 2011). The standard should provide a solution for all levels of

service load and also provide solutions for common continuous members.

2.6 Bending Deflection in CSA A23.3-04

The Canadian standard for the Design of Concrete Structures, A23.3-04 (CSA 2004),

states a few requirements for deflection. Most of the clauses relevant to calculating

deflection are found in Clause 9.8. Clause 9.8 first indicates that a minimum thickness

23

can be sufficient to control deflection. This standard then states the methods which may

be used to compute immediate deflections. After this, the standard provides equations

which combine the stiffness from the positive and negative bending segments of a

continuous member. Finally, clause 9.8 requires the designer to account for the effects

of creep.

The Concrete Design Handbook (CAC 2005) provides a design aid for the A23.3 (CSA

2004) standard. The handbook provides useful examples and explanations of design

using this standard. The Concrete Design Handbook provides the 𝐾 factor for

computing deflections of continuous members, as explained in Section 2.2.3 (above).

That chapter also provides an equation for 𝐼𝑐𝑟 and some other equations required to

calculate bending deflection for reinforced concrete members; these equations are used

for work in this report.

There are a few equations and statements in the A23.3 (CSA 2004) standard and the

Concrete Design Handbook (CAC 2005) that are likely to cause some designers to

commit errors when attempting to compute deflection; these equations and statements

are reviewed in Appendix S.

2.6.1 CSA A23.3-04, Clause 9.8.2.1, Minimum Thickness

The standard provides a table of minimum thicknesses, as a ratio relative to the clear

span, ℓ𝑛 (defined as 𝐿 in this report), and indicates that members meeting this table will

normally meet typical deflection requirements without further calculations. If a member

is sufficiently thick relative to the span length, then the deflection will typically be

24

acceptably small. The standard qualifies this statement by saying this minimum

thickness may not be sufficient if there is especially sensitive construction above the

relevant span or if superimposed loads exceed the self-weight.

2.6.2 CSA A23.3-04, Clause 9.8.2.2 and 9.8.2.3, Immediate Deflection

If the member length-to-thickness ratio or other criteria indicate that immediate

deflections should be computed, the A23.3 (CSA 2004) standard recommends the use of

an effective moment of inertia method and provides Branson’s (1965) equation. The

standard does mention that designers may alternatively use a more comprehensive

method for computing the effective moment of inertia or use integration of curvature to

determine deflection (but no further information is provided for these methods).

Update 3 to A23.3 (CSA 2004), Clause 9.8.2.3, requires that 𝑀𝑐𝑟 for bending deflection

be calculated using one half of 𝑓𝑟. The resulting equation, 𝑀𝑐𝑟 = 0.5𝑓𝑟𝐼𝑔/𝑦𝑡, accounts

for shrinkage restraint stresses when using Branson’s 𝐼𝑒 (Scanlon and Bischoff 2008).

For Bischoff’s 𝐼𝑒′ or 𝐼𝑒(𝑥) as defined in this report, the use of 𝑀𝑐𝑟 = 0.67𝑓𝑟𝐼𝑔/𝑦𝑡

provides an equivalent adjustment (Scanlon and Bischoff 2008). Until 2009, the

bending deflection was to be calculated using 𝑀𝑐𝑟 = 0.5𝑓𝑟𝐼𝑔/𝑦𝑡, except for two-way

slabs (see Clause 13.2.7). Because this update to Clause 9.8.2.3 occurred after work for

this report began, it is only incorporated in Section 3.7.4 and Appendix P.

2.6.3 CSA A23.3-04, Clause 9.8.2.4, Moment of Inertia for Continuous Spans

For continuous prismatic members, the Clause 9.8.2.4 provides an average effective

moment of inertia (indicated in this report as 𝐼𝑒 𝑎𝑣𝑔). The effective moment of inertia at

25

midspan, 𝐼𝑒𝑚, is rationally defined as the maximum moment in the positive bending

segment of the member. The moment at the supports are 𝐼𝑒𝐿 and 𝐼𝑒𝑅. This clause states

that one of the following equations be used to account for the contribution of midspan

and end-moment cracked sections:

𝐼𝑒,𝑎𝑣𝑔 = 0.7𝐼𝑒𝑚 + 0.15(𝐼𝑒𝐿 + 𝐼𝑒𝑅) if both ends are continuous (2 − 2)

𝐼𝑒,𝑎𝑣𝑔 = 0.85𝐼𝑒𝑚 + 0.15(𝐼𝑒𝐿) if only one end is continuous (2 − 3)

2.6.4 CSA A23.3-04, Clause 9.8.2.5, Sustained Load Deflections

Clause 9.8.2.5 of A23.3 (CSA 2004) provides the designer with sustained load

deflection calculations which account for creep and shrinkage. The total deflection is to

be calculated as a multiple of the immediate deflection. The multiplier provided is:

(1 +𝑆

1 + 50𝜌′) (2 − 4)

The multiplier accounts for the duration of the sustained load using the factor 𝑆. There

is a slight reduction in the multiplier from the compression reinforcement ratio, 𝜌′. The

commentary for this clause, N9.8.2.5 of A23.3 (CSA 2004), explains this calculation in

more detail.

2.7 Branson’s Work

Branson's equation (Branson 1965) is the standard equation in North America to model

the effective stiffness of a cracked reinforced concrete bending member. The critical

aspect of this task is that concrete members do not immediately change from uncracked

stiffness to fully cracked stiffness when the cracking moment is exceeded. Branson

26

developed an empirical equation for the transition by testing typical rectangular beams.

His model determines an effective moment of inertia intended to be used with linear-

elastic deflection equations. Branson's equation for the effective moment of inertia is:

𝐼𝑒 = (𝑀𝑐𝑟

𝑀𝑎)3

𝐼𝑔 + [1 − (𝑀𝑐𝑟

𝑀𝑎)3

] 𝐼𝑐𝑟 ≤ 𝐼𝑔 (2 − 5)

The applied moment, 𝑀𝑎, in Branson’s equation, should be defined as the maximum

moment of a continuous member, as this moment has the largest effect on deflection.

2.7.1 Limited Accurate Range for Branson’s Equation

Branson’s (1965) equation was calibrated for steel reinforcing ratio, 𝜌, of 1% to 2%,

which is also roughly the range of 2 ≤ 𝐼𝑔/𝐼𝑐𝑟 ≤ 3. Branson’s equation works well

within this calibrated range (as should be expected because it is an empirical equation).

Tests show it does not, however, work well for steel-reinforced members with 𝜌 < 1%

or for any typical FRP reinforcing ratios (Bischoff and Scanlon 2007). Most designers

use Branson's equation without noting its limitations; this demonstrates the need for a

more robust equation.

2.7.2 Modification Factors Examples for Branson’s Equation

Modification factors exist for most member types for which Branson’s equation

performs poorly. These corrections work well for only their intended range of members.

In the latest update to A23.3 (CSA 2004), 𝑀𝑐𝑟 is reduced by 50% for bending deflection

calculations; this over-accounts for typical shrinkage-restraint so that Branson’s 𝐼𝑒

equation will underpredict deflection less often. Another example is ACI 440.1R (ACI

27

Committee 440 2006), which endorses the following modification factor for FRP-

reinforced concrete:


𝑀𝑎)3

𝛽𝑑𝐼𝑔 + [1 − (𝑀𝑐𝑟

𝑀𝑎)3

] 𝐼𝑐𝑟 ≤ 𝐼𝑔 where: 𝛽𝑑 =1

5

𝜌

𝜌𝑏< 1.0 (2 − 6)

Here, 𝜌 is the actual reinforcement ratio of an FRP reinforced member and 𝜌𝑏 is the

balanced reinforcement ratio for the same FRP reinforced member. Again, use the

maximum moment in a continuous member as the applied moment, 𝑀𝑎.

2.8 Bischoff’s Work

Bischoff's work (Bischoff 2005, Bischoff 2007, Bischoff and Gross 2011, Bischoff and

Scanlon 2007) uses integration of curvature to develop an equation for an effective

moment of inertia of reinforced concrete members. As previously mentioned, Eurocode

2 (CEN 2004) and S806 (CSA 2012) recommend that the deflection of reinforced

concrete members be calculated using integration of curvature. This is a logical

approach to calculating deflection because, to the extent that moment and stiffness are

known, integration of curvature can accurately predict the deflection for any member.

2.8.1 Purpose of Bischoff’s Work

Bischoff’s equation has been derived to offer a rational approach for determining an

effective moment of inertia that accounts for the change in stiffness along the length of a

member. This approach also accounts for tension stiffening and the stiffness of the

reinforcement. Bischoff’s equation has been derived theoretically (to the extent

possible with concrete) and tested against empirical data through a full range of

28

reinforcing ratios. It works well for concrete members with 1 ≤ 𝐼𝑔/𝐼𝑐𝑟 ≤ 30 including

typical steel-reinforced, lightly steel-reinforced, and FRP-reinforced concrete members.

The section-based form of this equation, 𝐼𝑒(𝑥), allows the designer to use integration of

curvature to account for any loading and support conditions. Note that this method

computes immediate deflection. Long-term deflection calculations are still required.

(Bischoff and Gross 2011)

Unlike Branson’s (1965) equation, Bischoff’s equation never requires modification in

order to accurately predict deflection of simply supported members. Without

modification, Branson's equation works for a limited range. Because it is pulls too

strongly toward 𝐼𝑔 as it interpolates between 𝐼𝑐𝑟 and 𝐼𝑔, it produces significant

unconservative error when 𝐼𝑔/𝐼𝑐𝑟 ≫ 3. Bischoff’s equation provides relative

improvement for reasons that are explained in the parallel/series springs discussions in

both Bischoff (2007) and Bischoff and Scanlon (2007).

2.8.2 Bischoff’s Equation and Loading Type Factor

Bischoff’s equation has a section-based form (𝐼𝑒(𝑥)), a member-based form (the

equivalent moment of inertia, 𝐼𝑒′ ), and a simplified form (an effective moment of inertia,

𝐼𝑒). The section-based 𝐼𝑒(𝑥) provides a local moment of inertia corresponding to the

assumed moment-curvature response of the member. The equivalent moment of inertia,

𝐼𝑒′ , is an exact solution resulting from integration of curvature. The 𝐼𝑒

′ equation employs

a factor, , which mathematically accounts for the variation in stiffness along the

member length under a specific loading condition; this removes an approximation from

calculations for predicting the deflection of simply supported concrete bending

29

members. The simplified form provides an approximation based solely on midspan

moment and works well for most situations (Bischoff 2005). All three forms of the

equation use a tension stiffening factor of 𝛽 = 𝑀𝑐𝑟/𝑀(𝑥) and are found in Bischoff and

Gross (2011). Because the maximum positive moment for a continuous member is the

moment with the largest effect on deflection, it is used as the applied moment, 𝑀𝑎. For

loadings in this report, the equations to determine deflection using Bischoff’s method

are summarized in Table 2-3.

Table 2-3 - Deflection using Bischoff's Equation

(Sources for Table 2-3: Bischoff and Gross 2011, CAC 2005, and CISC 2009)

Bischoff’s section-based equation defines the local moment of inertia, 𝐼𝑒(𝑥), as:

𝐼𝑒(𝑥) =𝐼𝑐𝑟

1 − 𝜂 (𝑀𝑐𝑟

𝑀(𝑥))2 where: 𝜂 = 1 −

𝐼𝑐𝑟𝐼𝑔 (2 − 7)

30

Bischoff’s equivalent moment of inertia, 𝐼𝑒′ , is defined as:

𝐼𝑒′ =

𝐼𝑐𝑟


𝑀𝑎)2 where: 𝜂 = 1 −

𝐼𝑐𝑟𝐼𝑔 (2 − 8)

For the centered point load, third-point loading, and uniform loading, is defined as

given in Table 2-3. An independent derivation of the factor for a uniformly

distributed load is provided in Appendix D. Other factors can be derived similarly.

To provide a simplified approximation that improves on Branson’s (1965) equation,

setting = 1.0 produces Bischoff’s equation for the effective moment of inertia:

𝐼𝑒 =𝐼𝑐𝑟


𝑀𝑎)2 where: 𝜂 = 1 −

𝐼𝑐𝑟𝐼𝑔 (2 − 9)

2.8.3 Discussion of Arguments Against Use of Bischoff’s Equation

There are different arguments against prescribing the use of Bischoff’s equations for the

effective moment of inertia in codes and standards. Some engineers have used

Branson’s equation for a long time and found that it works well for their typical

situations (with modification factors as required), so they argue that change is

unnecessary. Other arguments against using Bischoff’s equation, mostly to do with the

difficulties in predicting bending stiffness, are explained and resolved in Bischoff and

Darabi (2012):

Incorporating shrinkage and the correct cracking moment, and resolving other

issues, is not more complicated with Bischoff’s equation than with other methods

If these issues are correctly accounted for, then using Bischoff’s equation will

provide a standard equation for all reinforcing materials and ratios

31

3.0 METHODOLOGY AND RESULTS

This work aims to demonstrate a simple and accurate method to determine the

deflection of a continuous concrete member. Ideally, solutions that are both simple and

exact could be provided. Simple solutions could be worked out in full on a letter-sized

sheet of paper. These solutions would be similar to Bischoff’s equivalent moment of

inertia (Bischoff and Gross 2011), 𝐼𝑒′ , for simply supported members. This 𝐼𝑒

′ yields

exact results that incorporate the variation in stiffness along the length of the member.

Unfortunately, simple solutions for continuous members must be approximate and will

have limits. The original goal for this work, which included finding either exact or

approximate equations suitable for all possible ranges of end-moments, was therefore

discarded.

The scope of work for this report is to provide and demonstrate the limitations within

which Bischoff’s 𝐼𝑒′ works for computing deflection of continuous members. This 𝐼𝑒

′ , as

explained in Section 2.8.2 and provided in Equation (2-8), provides a good

approximation for continuous members under a centered point load or a uniformly

distributed load (within proposed limits). For continuous members with equal point

loads at third points, a revised equation is proposed for the integration factor, (which

is used to compute 𝐼𝑒′ ), in order to improve results and limits. With relatively large end-

moments (outside proposed limits), results often become significantly unconservative,

so integration or another method must be used.

Many new approximate solutions for an effective moment of inertia, 𝐼𝑒, based on logic

or curve-fitting, were attempted in work for this report. These solutions were discarded

32

and the relevant work is not provided because Bischoff’s equation for 𝐼𝑒′ was found to

be more robust and accurate for continuous members.

For this work, computer spreadsheets are used to compute the midspan deflection of

example idealized members assuming a few different values for the constant moment of

inertia, including the proposed 𝐼𝑒′ . Deflection is also computed by integration of

curvature using both the S806 (CSA 2012) method and Bischoff’s section-based

moment of inertia (Bischoff and Gross 2011), 𝐼𝑒(𝑥). The deflection is computed using

the S806 method in order to provide examples which neglect tension stiffening. For

comparison purposes, this work assumes that integration of curvature using Bischoff’s

𝐼𝑒(𝑥), including its tension stiffening factor, computes the exact deflection for an

idealized member. All deflections computed are immediate (short-term) deflections

based on the full dead-plus-live service load. The immediate deflections using the

different approaches are compared and discussed; this leads to the conclusions provided.

3.1 Virtual Work and Moment of Inertia Methodology

In this work, deflection is computed using one of two very different methods. The

method of virtual work is used to compute deflection by integrating curvature along the

full length of the member for all numerical and analytical integration provided.

Deflections are also computed with common linear-elastic equations using different

approaches for computing the effective and equivalent constant moments of inertia.

33

3.1.1 Deflection of Concrete by Integration Using Virtual Work

As indicated in Section 2.3.4, calculating deflection by using the method of virtual work

to integrate curvature is a logical approach because it will accurately predict the

deflection, at any point along the member, if the bending moment and stiffness along the

member are known. While it is often computationally intensive to use the method of

virtual work for reinforced concrete members, it is well suited for use with a computer.

A descriptive form of the equation for calculating deflection of concrete bending

members using the method of virtual work is provided in Equation (3-1).

∆= ∫𝑚(𝑥)𝑀(𝑥)

𝐸𝑐𝐼𝑒(𝑥)𝑑𝑥

𝐿

0

(3 − 1)

This equation determines the deflection, ∆, at a defined location along a member of span

length, 𝐿; the virtual bending moment, 𝑚(𝑥), is obtained by applying a unit load at the

location where deflection is being computed. The variable 𝑥 is defined as the distance

from the left end of the span, so 0 ≤ 𝑥 ≤ 𝐿. The curvature function, 𝑀(𝑥)/ 𝐸𝑐 𝐼𝑒(𝑥),

includes the service load bending moment function, 𝑀(𝑥), the elastic modulus of

concrete, 𝐸𝑐, and the section-based effective moment of inertia function, 𝐼𝑒(𝑥).

Additional explanation of the method of virtual work and Equation (3-1) is given in

Appendix B. For a description of how to calculate 𝐼𝑒(𝑥) per Bischoff’s work, see

Equation (2-7).

Equation (3-1) is used to compute deflection using both numerical and analytical

integration; it is used for the S806 (CSA 2012) integration method and for exact results

with Bischoff’s 𝐼𝑒(𝑥). When a constant moment of inertia is used, the 𝐼𝑒(𝑥) term is

34

replaced with 𝐼𝑔, 𝐼𝑐𝑟, 𝐼𝑒, or 𝐼𝑒′ , or similar. The 𝐼𝑔 term is defined as the gross moment of

inertia and the 𝐼𝑐𝑟 term is defined as the moment of inertia of the cracked transformed

section.

3.1.2 Deflection of Concrete Using a Constant Moment of Inertia

As described in Section 2.3.3, a constant effective moment of inertia can be useful for

engineers. Deflection equations for linear-elastic prismatic members are commonly

available. For members with end-moments, the equations for simply supported

members can be used when modified by the 𝐾 factor as shown in Section 2.2.3. With

an accurate constant effective moment of inertia, these equations are easy to use and

will result in accurate deflection predictions for reinforced concrete members.

For the majority of reinforced concrete members, both the effective moment of inertia

and the equivalent moment of inertia will lie between the gross and fully cracked

moments of inertia: 𝐼𝑔 > 𝐼𝑒 > 𝐼𝑐𝑟 and 𝐼𝑔 > 𝐼𝑒′ > 𝐼𝑐𝑟. The latter is indicated in Figure

2-4, which shows 𝐼𝑔, 𝐼𝑐𝑟 , 𝐼𝑒(𝑥), and the equivalent moment of inertia, 𝐼𝑒′ . In the

regions where a member is uncracked, 𝐼𝑒(𝑥) = 𝐼𝑔.

Methods that use a constant moment of inertia for continuous concrete members carry

intrinsic limitations because the stiffness of concrete members varies in ways that are

sometimes difficult (or impossible) to model with constant stiffness equations; this is

explained in Section 3.7.2. To eliminate approximation and the possibility of large

errors, integration is the simplest solution.

35

3.2 Generating Idealized Members

There were approximately two thousand different idealized members used for this

report. All members were produced with full and realistic properties where possible,

and were based on Canadian design standards. Initially, a few members were modelled

by copying all properties from example members found from other sources. In order to

compare computed deflection results for a huge variety of members, these different

idealized concrete bending members were generated in computer spreadsheets using an

automated member production method developed for this purpose.

For all example graphs provided, specific properties such as loads and lengths are

selected so that deflections could be calculated. However, if two different members

(with different aspect ratios, for example) have a few key properties and ratios in

common, they will have exactly the same normalized moment-deflection diagrams.

This is demonstrated graphically in Appendix O (see Figure O-3 and Figure O-4).

3.2.1 Definitions of Bending Moment Variables

In order to generate the example concrete members, a few key bending moments were

computed and used. The service load moments used for this report are the full service

loads, meaning dead load plus live load without load factors, applied to a member that

has not experienced higher bending moments at their respective locations. The factored

applied moment, factored moment resistance, and cracking moment, are intended to be

as defined in the applicable codes and standards. Most of the bending moments are

defined in automated member generating spreadsheets as a ratio to another bending

moment.

36

The important service load bending moments, as used in this chapter, are:

𝑀𝑚 is the midspan moment, determined at the midpoint of the member’s span

𝑀𝑚𝑎𝑥 is the maximum positive bending moment of an example member

- 𝑀𝑚𝑎𝑥 is equal to 𝑀𝑚 in this report when end-moments are equal or zero

- 𝑀𝑚𝑎𝑥 is used to calculate the effective/equivalent moment of inertia

whenever 𝑀𝑚 is used to calculate 𝐼𝑒 or 𝐼𝑒′ for a simply supported member

𝑀𝐿 is the bending moment at the left end of the member for the loading provided

- 𝑀𝐿 is assumed to be zero or negative throughout this report

- The left end of the member is always taken to be the end of the member

having the end-moment of larger magnitude

𝑀𝑅 is the bending moment at the right end of the member for the loading provided

- 𝑀𝑅 is assumed to be zero or negative throughout this report

- As per the definition of 𝑀𝐿 above, −𝑀𝐿 ≥ −𝑀𝑅 ≥ 0

𝑀0 is the total static moment caused by the loads applied between supports

- For the loading types provided, 𝑀0 is equal to 𝑀𝑚 with the end-moments

released (but the load maintained); see the moment defined in Table 2-1

- Provided load between supports include: a point load of 𝑃 at midspan, two

point loads of 𝑃/2 at third points, or a uniformly distributed load of 𝑤.

𝑀𝑠 is a general reference to the service moment of 𝑀𝐿, 𝑀𝑅, 𝑀𝑚, or 𝑀𝑚𝑎𝑥.

- 𝑀𝑠:𝑀𝑟 is the ratio of service moment to factored moment resistance

- The amount of top reinforcing in the left and right end-moment regions is

selected based an 𝑀𝐿 or 𝑀𝑅, respectively, and the aforementioned ratio

- The amount of bottom reinforcing in the positive bending region is selected

based on 𝑀𝑚𝑎𝑥 and the aforementioned ratio

37

The factored bending moments, as used in this chapter, are:

𝑀𝑓 is the factored applied bending moment at the indicated location

- In practice, 𝑀𝑓 would be calculated based on National Building Code load

factors applied to the actual dead and live loads applied to a member

- For this work, 𝑀𝑓 = 1.375𝑀𝑠 is assumed

𝑀𝑟 is the factored bending moment resistance for a given segment of a member

- CSA A23.3-04 (2004) and CSA S806 (2012) are used to compute 𝑀𝑟

- Additional capacity of 𝑀𝑟 = 1.145𝑀𝑓 is typically provided

- For typical example members, the assumed 𝑀𝑓 (above) and the additional

Mr capacity (above) were combined as 𝑀𝑟 = 1.575𝑀𝑠

- Where deflection governs the amount of positive moment reinforcing, a

larger ratio of 𝑀𝑠:𝑀𝑟 is provided, such as 𝑀𝑟 = 2.5𝑀𝑠 (for Figure 3-4)

- If 𝑀𝑟 = 1.575𝑀𝑠 is changed to 𝑀𝑟 = 1.30𝑀𝑠 while the properties and loads

remain otherwise unchanged, then reinforcing is reduced and deflection

increases, but the normalized results will be essentially identical

The other key bending moment used in this chapter is 𝑀𝑐𝑟:

𝑀𝑐𝑟 is the bending moment at cracking for a member

- When the applied bending moment at a local section of the member reaches

𝑀𝑐𝑟, the tension face of the member will crack at this section

- 𝑀𝑐𝑟 is defined as per Equation (2-1) for all provided results in this report

except for work provided in Appendix P (which explores the effect of

reducing 𝑀𝑐𝑟 to one half of the value computed by Equation (2-1) in order

to account for shrinkage restraint per CSA A23.3-04 (R2010))

38

3.2.2 Automated Member Generation

The automated method used to generate example idealized concrete bending members is

developed in order to enable comparison of midspan deflections between similar

members. Each member of the set is defined using the same length (𝐿), width-to-height

ratio (𝑏: ℎ), reinforcing bar depth to member height ratio (𝑑: ℎ), material properties,

resistance factors, factored moment resistance to service moment ratio (𝑀𝑟:𝑀𝑠), and

maximum positive moment to cracking moment ratio (𝑀𝑚𝑎𝑥:𝑀𝑐𝑟). After the common

properties are defined, unique left and right end-moment to maximum positive moment

ratios (𝑀𝐿:𝑀𝑚𝑎𝑥 and 𝑀𝑅:𝑀𝑚𝑎𝑥) are provided for each member. Each set contains a

simply supported member; the load on this member is adjusted to obtain a desired

height for the members of the set.

Sets of 9 example members are generated with equal maximum positive moments and

various end-moments (where each member has an area of top reinforcing bars suitable

for the end-moments). The end-moments are selected to provide useful data points for a

plot of these similar members. In Figure 3-1, for example, the set of 9 members have

𝑀𝑚𝑎𝑥 = 195 kNm and end-moments equal to: 0, −0.25𝑀𝑚𝑎𝑥, −0.43𝑀𝑚𝑎𝑥,

−0.67𝑀𝑚𝑎𝑥, −𝑀𝑚𝑎𝑥, −1.22𝑀𝑚𝑎𝑥, −1.5𝑀𝑚𝑎𝑥, −1.7𝑀𝑚𝑎𝑥, and −1.94𝑀𝑚𝑎𝑥.

After the ratios and design properties are defined for the set of members, the automated

method then produces each member of the set as follows:

𝑀𝑚𝑎𝑥 is determined for the simply supported member

A unique load is contrived for each continuous member such that the 𝑀𝑚𝑎𝑥

computed for each member will be equal for all members of the set

39

𝑀0, 𝑀𝐿, 𝑀𝑅, 𝑀𝑚, and 𝑀𝑚𝑎𝑥 are computed for each member

𝑀𝑐𝑟, ℎ, 𝑏, 𝑑, and 𝐼𝑔 are calculated and confirmed as common

The reinforcing ratio, ρ, is computed (based on 𝑀𝑟:𝑀𝑠) for 𝑀𝑚𝑎𝑥, 𝑀𝐿, and 𝑀𝑅

𝐼𝑐𝑟 and other properties are determined at midspan and at each end

An example of the computations performed to generate a member, along with example

midspan deflections calculations for that member, is provided in Appendix J. For

members produced using the automated method, the amount of reinforcing required is

based on A23.3 (CSA 2004) or S806 (CSA 2012) and the moment ratios provided as

inputs. The reinforcing ratio is defined by solving typical design equations using other

inputs and these ratios (as demonstrated in Appendix J).

For the provided plots, which contain sets of automatically generated members, the end-

moments vary from 0 ≤ −𝑀𝐿 ≤ 3𝑀𝑚𝑎𝑥, where 𝑀𝑅 = 𝑀𝐿 (both ends continuous) or

𝑀𝑅 = 0 (one end continuous). In general, results are not provided when

−𝑀𝐿 > 3𝑀𝑚𝑎𝑥 because they tend to diverge towards being increasingly erroneous in

this range. At these relative values of end-moment, the proportions of negative

reinforcing needed are larger than permitted and/or deflections near midspan are near

zero (where such small deflections will be within codes/standards limits).

3.3 Computing Deflection of Idealized Members

All plots provided in this chapter show values of deflection computed using different

methods and different solutions for the moment of inertia. Some of the results

computed in this work were compared to laboratory test data and found to be realistic,

40

but this data is not provided. Except for work in Appendix P (where 𝑀𝑐𝑟 is reduced by

one half or one third), shrinkage restraint was neglected for the data presented here.

The spreadsheets used to generate idealized members, as described in Section 3.2, also

compute the deflection for each member using each relevant approach. The input

properties for these spreadsheets were varied in order to determine the limits of the

proposed equations. Only a small sample of the computed results is provided.

In concrete structures, the midspan deflection is usually a good approximation for the

maximum deflection. The midspan deflection, determined using each of the different

approaches, has been plotted for each set of example members. For some load patterns

and end conditions, however, the maximum deflection of continuous members is

significantly larger than the midspan deflection. When applicable, the values of

maximum deflection are computed and plotted with the midspan deflection.

Throughout this report, deflection is denoted by the symbol 𝛥 as follows:

All constant stiffness equations provided denote the midspan deflection as 𝛥

- 𝛥 is computed at exactly the midpoint of the span

If the maximum deflection does not occur at midspan, the computed value of the

maximum deflection is denoted by 𝛥𝑚𝑎𝑥

For clarity, midspan deflection is shown as 𝛥𝑚𝑖𝑑 in equations that include 𝛥𝑚𝑎𝑥

As discussed in Section 3.1.1, deflection can be computed at any point along the

member when the method of virtual work is used to integrate curvature

All example deflection calculations and plots provided are for midspan or

maximum deflections

41

For each set of members generated for this report, deflection for each member was

computed using at least six constant moment of inertia solutions and two integration of

curvature solutions. A graphical comparison of the deflection values determined using

the different approaches is provided to facilitate analysis and discussion.

Midspan deflection, computed using five different constant moment of inertia solutions,

was plotted for all sets of members as follows:

𝛥𝑔(Gross) is the midspan deflection computed using the gross moment of inertia

- Uses constant moment of inertia set equal to 𝐼𝑔 (even if loads exceed 𝑀𝑐𝑟)

𝛥𝑐𝑟(Cracked) is the midspan deflection computed with the 𝐼𝑐𝑟 moment of inertia

- Computed using entire beam with the fully cracked moment of inertia

- 𝐼𝑐𝑟 is computed using the reinforcing ratio at the location of 𝑀𝑚𝑎𝑥

𝛥𝐼𝑒(Branson) is the midspan deflection computed with the common 𝐼𝑒 equation

- Uses Branson’s (1965) equation for the effective moment of inertia

- 𝐼𝑒 is computed using Equation (2-5) with an applied moment of 𝑀𝑚𝑎𝑥

𝛥𝛾=1(Approx) is the midspan deflection computed per Bischoff’s 𝐼𝑒 where = 1

- Uses Bischoff’s effective moment of inertia (Bischoff and Gross 2011)

- 𝐼𝑒 is computed using Equation (2-9) with an applied moment of 𝑀𝑚𝑎𝑥

𝛥𝐼𝑒′(Proposed) or 𝛥𝐼𝑒′ (Bischoff) is the midspan deflection computed using 𝐼𝑒′

- 𝐼𝑒′ is Bischoff’s equivalent moment of inertia (Bischoff and Gross 2011)

- 𝐼𝑒′ is computed using Equation (2-8) with an applied moment of 𝑀𝑚𝑎𝑥

Midspan deflection, computed using one (or more) of these additional solutions for

constant moment of inertia, was also plotted as follows:

42

𝛥𝐼𝑒,𝛽𝑑(ACI440) is the midspan deflection computed per ACI 440.1R’s approach

- Computed as recommended in ACI 440.1R (ACI Committee 440 2006)

- Uses the modified effective moment of inertia provided in Equation (2-6)

- Provided for FRP reinforced concrete members

𝛥𝐼𝑒,𝑎𝑣𝑔(A23.3) is the midspan deflection per A23.3 (CSA 2004), Clause 9.8.2.4

- Computed using effective moment of inertia per Equation (2-2) or (2-3)

- 𝐼𝑒,𝑎𝑣𝑔 is an approximate average of the Branson’s 𝐼𝑒 computed at the applied

moment locations of 𝑀𝑚𝑎𝑥, 𝑀𝐿, and 𝑀𝑅

- 𝐼𝑒,𝑎𝑣𝑔 is computed with Equation (2-6) for members designed with GFRP

𝛥𝐼𝑒∗′ (Proposed) is the midspan deflection computed using 𝐼𝑒∗′ per Equation (3-10)

- 𝐼𝑒 ∗′ is an empirically derived improvement to 𝐼𝑒

′

- Only used for third-point loading on a continuous member

- 𝐼𝑒 ∗′ is computed with Equation (2-8), but with ∗ per Equation (3-9) used in

lieu of

Midspan deflection, computed using integration of curvature (with analytical or

numerical integration) and moment of inertia solutions which vary along the member

length, was also plotted for all sets of members as follows:

𝛥𝛽=0(S806) is the midspan deflection computed neglecting tension stiffening

- Computed using moment of inertia as defined in the S806 (CSA 2012) (as

shown in Appendix G) for the integration method

- Intended only for FRP reinforced members by S806; provided herein for all

members to show the effect of neglecting tension stiffening

43

𝛥𝐼𝑒(𝑥)(Exact) is the midspan deflection using 𝐼𝑒(𝑥) per Equation (2-7)

- Computed using Bischoff’s section-based moment of inertia (Bischoff and

Gross 2011), 𝐼𝑒(𝑥), which accounts for tension stiffening

- Assumed to be the exact midspan deflection (for comparison purposes)

Maximum deflection was plotted for members with unequal end-moments as follows:

𝛥𝑚𝑎𝑥,𝐼𝑒′(Proposed) or 𝛥𝑚𝑎𝑥,𝐼𝑒

′(Bischoff) is 𝛥𝑚𝑎𝑥 computed with 𝐼𝑒′

- Uses the same 𝐼𝑒′ , per Equation (2-8), as the 𝛥𝐼𝑒′ midspan deflection

- Computed by a constant stiffness model for mid-point and third-point loads

- Approximated by Equation (3-19) for a uniformly distributed load

𝛥𝑚𝑎𝑥,𝐼𝑒(𝑥)(Exact) is the maximum deflection computed using Bischoff’s 𝐼𝑒(𝑥)

- Uses the same 𝐼𝑒(𝑥), per Equation (2-7), as the 𝛥𝐼𝑒(𝑥) midspan deflection

- Computed by applying the virtual work method to integrate curvature at

many locations, then selecting the largest deflection as the maximum

- Assumed to be the exact maximum deflection (for comparison purposes)

The maximum positive moment (rather than the midspan moment) should be used to

compute a constant moment of inertia for the member. This is because the member

experiences the worst cracking in the positive bending segment at the location of the

maximum moment. If 𝑀𝑚 is used in lieu of 𝑀𝑚𝑎𝑥, then a smaller constant moment of

inertia will be calculated, which will usually result in unconservative predictions for

deflection. The midspan moment should only be used if it is similar to the maximum

moment (𝑀𝑚 ≈ 𝑀𝑚𝑎𝑥) or if the member is highly cracked at midspan (𝑀𝑚 ≫ 2𝑀𝑐𝑟).

44

Larger end-moments or cracked lengths may result when different load patterns occur

beforehand. This is not directly accounted for in this report. The cracking that is

accounted for is based on the bending moment function which passes through 𝑀𝐿, 𝑀𝑚,

and 𝑀𝑅, as provided for each member. Stiffnesses affected by other load patterns could

be included, using integration of curvature, using an 𝐼𝑒(𝑥) function based on the service

moment envelope. The amount of reinforcing required at the supports should then also

be determined based on the factored moment envelope. These two corrections, for other

load patterns, partially offset each other, as discussed in Appendix P.

3.3.1 Development and Use of Analytical Integration

Unless the result is approximated, complicated formulas are required in order to

calculate the midspan deflection of an idealized continuous member. Analytical

integration is feasible, however, because the exact integral along each segment of the

beam can be found by basic calculus. Members sometimes have a “negative” cracked

segment at each end, with an adjacent uncracked segment, and a “positive” cracked

segment in the middle. The service bending moment function, 𝑀(𝑥) changes at each

point load and the virtual moment function, 𝑚(𝑥), changes at the location at which

deflection is being computed. As a result, a uniformly loaded beam has 6 segments to

integrate, a single point-load beam has 6 segments to integrate, and a third-point-loaded

beam has 7 segments to integrate. The resulting expressions are lengthy, containing 12

input variables in many permutations. When negative moment segments remain

uncracked, different expressions are required. Discussion about analytical integration

and the expressions resulting from its use are shown in Appendix E.

45

Math software was used to maintain error-free integration for this report. Maple (by

Maplesoft®) was used to integrate along each beam segment for each load-type, support,

and cracking situation considered. The resulting equations were subsequently copied

into Excel (by Microsoft®) spreadsheets and linked to the correct variables. The

pertinent integration segments are selected with “if-then” logic in the spreadsheets and

those integration results are added together to obtain non-approximated analytical

results.

3.3.2 Discussion of Analytical Integration Simplifications

Different simplifications of integrated results are possible and can sometimes be useful.

Razaqpur’s work (Razaqpur and Isgor 2003, Razaqpur et al. 2000) indicates some

useful assumptions which provide simplified approximate equations for specific load

types and support conditions. As discussed in Section 2.5, one of Razaqpur’s

simplifications is to neglect the effects of tension stiffening. When tension stiffening is

taken into account, similar approximate equations can be derived and simplified for

specific cases.

Simplified equations which provide exact results (without mathematical

approximations) can be derived for some specific continuous member cases. These

exact equations, however, would be for a very specific combination of loading, cracked

segments, reinforcment ratios, and support conditions. As shown in Appendix F, the

simplest exact equation for a continuous member occurs with a midspan-point load,

where the magnitude of postive moment and end-moments are all equal (along with all

46

respective reinforcing ratios). Very few simplifications are possible for most continuous

members.

3.3.3 Use of Numerical Integration

Numerical integration results for this report are calculated in spreadsheets. For

numerical integration calculations, members were divided into 𝑗 ≥ 100 equal segments.

Integration using the virtual work method was used to determine the deflection at the

desired point (such as midspan) by summing the effect that each segment has on

deflection at that point. Appendix K outlines the methodology and an example of

numerical integration as employed for this report. Equation (3-2) is Equation (3-1)

written in a form which enables accurate numerical integration.

∆=∑[𝑚(𝑥𝑖)𝑀(𝑥𝑖)

𝐸𝑐𝐼𝑒(𝑥𝑖)+𝑚(𝑥𝑖−1)𝑀(𝑥𝑖−1)

𝐸𝑐𝐼𝑒(𝑥𝑖−1)]𝐿

2𝑗

𝑖=𝑗

𝑖=1

where: 𝑥𝑖 =𝑖𝐿

𝑗 (3 − 2)

and where the variables not explained previously in this chapter are defined as follows:

𝑖 - counter for each segment of the span, from 1 𝑡𝑜 𝑗

𝑗 - total number of equal length segments used for numerical integration

𝑥𝑖 - distance from the left end of the span to the right end of segment 𝑖; 𝑥0 = 0

3.3.4 Comparing Results of Analytical and Numerical Integration

The difference between exact analytical integration results and computed numerical

integration results in predicted deflection should be very small. Numerical integration

causes slight approximations because every segment of the integrand is assumed to be

linear (using the effect on deflection computed from the exact properties and bending

47

moments at both ends of the segment). Despite this approximation, work for this report

found that with 100 or more equal-length segments, the error from numerical integration

is negligible for concrete bending deflection calculations. If the number of segments is

selected so that a segment end occurs at each point load for the 𝑚(𝑥) and 𝑀(𝑥)

functions, the error caused by using numerical integration in this work was less than

0.2%. To reduce the possibility of mathematical errors, both analytical and numerical

methods were used for the majority of the work for this report.

3.4 Continuous Beam with a Centered Point Load

In this section, results using the proposed equations are compared to other solutions for

determining deflection with a centered point load. These results are based solely on one

point load at midspan; self-weight is not otherwise accounted for. Maximum moment is

not mentioned because 𝑀𝑚 = 𝑀𝑚𝑎𝑥 for a centered point load with either equal or

unequal end-moments. Plotted members are generated as explained in Section 3.2 and

deflection values are obtained as explained in Section 3.3.

3.4.1 Proposed Solution for a Centered Point Load

The proposed approach for a continuous member with a centered point load is to use the

simply supported equivalent moment of inertia, 𝐼𝑒′ , proposed by Bischoff and Gross

(2011). The equation for 𝐼𝑒′ , including the factor for a simply supported member, is

used with the midspan moment and midspan properties of the continuous member.

While 𝐼𝑒′ is an exact solution for simply supported members, it is an approximation for

continuous members.

48

The proposed equation for the equivalent moment of inertia is:

𝐼𝑒′ =

𝐼𝑐𝑟


𝑀𝑚)2 where 𝜂 = 1 −

𝐼𝑐𝑟𝐼𝑔 (as per Equation 2 − 8)

For simply supported and continuous members with a point load, 𝑃, applied at midspan:

= 3 − 2𝑀𝑐𝑟

𝑀𝑚 where 𝑀0 =

𝑃𝐿

4 and 𝑀𝑚 = 𝑀𝑚𝑎𝑥 = 𝑀0 +

𝑀𝐿

2+𝑀𝑅

2 (3 − 3)

For a centered point load, the approximate midspan deflection is:

𝛥 = 𝛥𝑚𝑖𝑑 = 𝛥𝐼𝑒′ = 𝐾𝑀𝑚𝐿

2

12𝐸𝑐𝐼𝑒′ where 𝐾 = 1.5 − 0.5

𝑀0

𝑀𝑚 (3 − 4)

For unequal end-moments, where −𝑀𝐿 < 2𝑀𝑚, maximum deflection is computed as:

𝛥𝑚𝑎𝑥 ≈ 𝛥𝑚𝑖𝑑

𝛥𝑔,𝑚𝑎𝑥

𝛥𝑔,𝑚𝑖𝑑 (3 − 5)

In Equation 3-5, 𝛥𝑔,𝑚𝑎𝑥 and 𝛥𝑔,𝑚𝑖𝑑 are the maximum and midspan deflections,

respectively, of a linear-elastic member with the same loads, span, bending moments,

and using the gross moment of inertia (although any shared constant moment of inertia

would suffice).

3.4.2 Comparison of Results: Centered Point Load and Equal End-Moments

Each graph in this section provides results computed from many deflection approaches

for continuous members under a centered point load; each graph contains a set of

members with a different 𝐼𝑔 𝐼𝑐𝑟⁄ ratio. For the centered point load examples provided,

end-moments are kept equal to each other (𝑀𝐿 = 𝑀𝑅) in order to simplify the

comparison of results using the different approaches. The magnitude of the end-

moments and the amount of reinforcement at the ends of the members increase as the

49

plots progress from left to right. The deflections computed using integration of

curvature based on 𝐼𝑒(𝑥), per Equation (2-7), are assumed to be exact for comparison

purposes. The data for the properties and deflections is provided in Appendix M. These

examples will show that the proposed 𝐼𝑒′ equation provides an improved moment of

inertia for computing deflection of continuous members with linear-elastic equations.

For Figures 3-1 to 3-4, members were designed and analyzed as follows:

Members are 600 mm deep and 300 mm wide rectangular beams with a 10 m span

Specified concrete strength used was 𝑓𝑐′ = 36 MPa

Reinforcing bar depth, from the compression face, is 𝑑 = 540 mm

Steel reinforcing bars used have a yield strength of 𝑓𝑦 = 400 MPa

GFRP (glass fibre reinforced polymer) reinforcing bars used have an ultimate

strength of 𝑓𝑓𝑢 = 690 MPa and an elastic modulus of 𝐸𝑏 = 44 GPa.

The reinforcing ratio, 𝜌, is provided for the bottom reinforcing bars; when 𝜌/𝜌𝑏 is

provided for the GFRP reinforced members, 𝜌𝑏 is the balanced reinforcing ratio.

Figure 3-1 shows results of a set of example members with equal end-moments where

𝐼𝑔/𝐼𝑐𝑟 = 2.28 and 𝑀𝑚/𝑀𝑐𝑟 = 3.00. These member are designed with steel reinforcing

such that 𝑀𝑟 = 1.575𝑀𝑠 and 𝜌 = 1.2%. In this example, there is significant deflection,

𝛥𝐼𝑒(𝑥), when constant stiffness results are zero (at −𝑀𝐿 = 2𝑀𝑚). All effective moment

of inertia approaches are conservative by approximately 10% (which is slightly better

than using the 𝐼𝑐𝑟 results) except where results diverge as end-moments become

significantly larger than 1.2𝑀𝑚. Branson’s (1965) method offers slight improvement on

using 𝐼𝑐𝑟 but the proposed method offers far more accuracy for −𝑀𝐿 < 1.2𝑀𝑚. The

50

S806 (2012) curve, which represents integration of curvature while neglecting tension

stiffening, is conservative by at least 10%, even for 𝑀𝐿 ≫ 𝑀𝑚, as expected. Use of

𝐼𝑒,𝑎𝑣𝑔 per Equation (2-2), fails to improve the accuracy of 𝐼𝑒, and produces an aberration

at 0.4 ≈ −𝑀𝐿/𝑀𝑚 ≳ 𝑀𝑐𝑟/𝑀𝑚. This anomalous point on the 𝛥𝐼𝑒,𝑎𝑣𝑔 curve occurs where

the lightly cracked ends contribute disproportionately to the member stiffness.

Figure 3-1 – Midspan Deflection of Steel Reinforced Beams under Centered Point Load

with Ig/Icr=2.3, Mm/Mcr=3.0, and ML=MR

The set of example members generated for Figure 3-2 are steel-reinforced members

having 𝐼𝑔/𝐼𝑐𝑟 = 3.89, 𝜌 = 0.6%, 𝑀𝑚/𝑀𝑐𝑟 = 1.60, and 𝑀𝑟 = 1.575𝑀𝑠. These results

again show that the effective moment of inertia approaches are conservative; results

using Branson’s (1965) 𝐼𝑒 are conservative by about 20% for −𝑀𝐿 < 1.4𝑀𝑚. The

proposed equation provides very accurate results when −𝑀𝐿 < 1.6𝑀𝑚. If the proposed

method is approximated by using = 1, or if tension stiffening is neglected (𝛽 = 0),

51

results are more than 40% conservative. 𝛥𝐼𝑒,𝑎𝑣𝑔 provides no improvement (see the

discussion for Figure 3-1 regarding the data point near 0.75 ≈ −𝑀𝐿/𝑀𝑚 ≳ 𝑀𝑐𝑟/𝑀𝑚).

Figure 3-2 - Midspan Deflection of Steel Reinforced Beams under Centered Point Load


Figure 3-3 is based on members which are reinforced with GFRP reinforcing bars as

follows: 𝐼𝑔/𝐼𝑐𝑟 = 3.32, 𝜌 = 3.2%, 𝜌/𝜌𝑏 = 5.6, 𝑀𝑚/𝑀𝑐𝑟 = 2.50, 𝑀𝑟 = 2.78𝑀𝑠 for

bottom bars, and 𝑀𝑟 = 1.575𝑀𝑠 for top bars. To control deflection for this set of

members, the amount of additional reinforcing added causes the 𝐼𝑔/𝐼𝑐𝑟 ratio to be

relatively high compared to a typical case for GFRP reinforcing. Thus, the common

constant stiffness approaches are only conservative by about 20% for –𝑀𝐿 < 0.5𝑀𝑚.

The proposed equation offers improvement, but it becomes overly conservative between

–𝑀𝐿 ≥ 0.7𝑀𝑚 and –𝑀𝐿 ≤ 1.7𝑀𝑚. In this same range, the effective moment of inertia

approaches are even more conservative. In practice, if the inputs bending moments

52

were derived from a constant stiffness model, then the results based on 𝐼𝑒 or 𝐼𝑒′ in that

range are probably much less conservative than this graph suggests because the ends are

much less stiff than the midspan segment (which would normally increase the midspan

moment, as discussed in Section 3.7.3). The plot of 𝛥𝛽=0 for this example shows that

the effects of neglecting tension stiffening, at the ends and at midspan, happen to nearly

offset each other between −𝑀𝐿 ≥ 0.7𝑀𝑚 and −𝑀𝐿 ≤ 1.3𝑀𝑚.

Figure 3-3 - Midspan Deflection of FRP Reinforced Beams under Centered Point Load


Figure 3-4 plots computed deflection for GFRP reinforced concrete members where

𝐼𝑔/𝐼𝑐𝑟 = 12.3, 𝜌 = 0.7%, 𝜌/𝜌𝑏 = 1.2, 𝑀𝑚/𝑀𝑐𝑟 = 1.60, 𝑀𝑟 = 2.50𝑀𝑠 for bottom bars,

and 𝑀𝑟 = 1.575𝑀𝑠 for top bars. The results from the different approaches vary greatly

because the effective moment of inertia at midspan is much larger than 𝐼𝑐𝑟. For this set

of members, the proposed approach offers accurate results when −𝑀𝐿 < 𝑀𝑚; it

53

becomes conservative for −𝑀𝐿 > 𝑀𝑚 because the cracked length at midspan is very

short for those members in this set. Branson’s (1965) 𝐼𝑒 causes erroneous

unconservative results for typical cases (−𝑀𝐿 < 𝑀𝑚), while use of 𝛽𝑑 = 0.24 per ACI

440.1R (ACI Committee 440 2006) provides very conservative results.

Figure 3-4 - Midspan Deflection of FRP Reinforced Beams under Centered Point Load

with Ig/Icr=12, Mm/Mcr=1.6, and ML=MR

3.4.3 Summary of Results for a Centered Point Load

Table 3-1 summarizes the valid ranges for the proposed effective moment of inertia, 𝐼𝑒 ′ ,

with centered point loading. To provide the ranges of validity shown, results were

reviewed as each of the relevant variables was varied. Values for 𝑀𝑐𝑟 𝑀𝑚⁄ , 𝐼𝑔 𝐼𝑐𝑟⁄ ,

𝑀𝑅 𝑀𝐿⁄ , 𝑀𝑚 𝑀𝑟⁄ , depth divided by height, and other relevant ratios have been varied

within reasonable ranges in an attempt to provide valid ranges that are applicable to all

54

realistic members. Concrete members reinforced with steel, AFRP, and GFRP bars have

also been reviewed. The valid ranges were terminated when the proposed

approximation reached 10% error. Results within the provided ranges typically result in

less than 5% error.

Table 3-1 - Valid Ranges for I'e for a Centered Point Load

Equal End-Moments (𝑀𝐿 = 𝑀𝑅)

# One End-Moment

#

Cracked Ratio

𝐼𝑒′ Valid If: Iγ=1 Valid? 𝐼𝑐𝑟 Valid? Ie

′ Valid If:

3 ≤𝑀𝑚

𝑀𝑐𝑟 −𝑀𝐿 ≤ 1.3𝑀𝑚 Yes Ok −𝑀𝐿 ≤ 1.5𝑀𝑚

1.7 ≤𝑀𝑚

𝑀𝑐𝑟< 3 −𝑀𝐿 ≤ 1.4𝑀𝑚 Ok No −𝑀𝐿 ≤ 1.5𝑀𝑚

1.3 ≤𝑀𝑚

𝑀𝑐𝑟< 1.7 −𝑀𝐿 ≤ 1.5𝑀𝑚 No No −𝑀𝐿 ≤ 2𝑀𝑚

1 ≤𝑀𝑚

𝑀𝑐𝑟< 1.3 −𝑀𝐿 ≤ 1.4𝑀𝑚 No No −𝑀𝐿 ≤ 1.5𝑀𝑚

# the results in this table assume 𝑀𝐿 ≤ 𝑀𝑅 ≤ 0.

There is often a minor conservative error when −𝑀𝐿 ≈ −𝑀𝑅 < 𝑀𝑚, but it appears to be

less than 5% for all cases except with FRP members with −𝑀𝐿 ≈ −𝑀𝑅 < 𝑀𝑚/2 and

which have the amount of reinforcing bars increased in order to control deflection. This

error occurs because more of the member is uncracked for the members with end-

moment than for similar simply supported members. The results using 𝐼𝑒′ are, therefore,

(slightly) larger than the integrated 𝐼𝑒(𝑥) result when −𝑀𝐿 ≈ −𝑀𝑅 < 𝑀𝑚.

The proposed equations are often unconservative for −𝑀𝐿 ≈ −𝑀𝑅 ≫ 𝑀𝑚. If

deflections for −𝑀𝐿 > 𝑀𝑚 are worth calculating, it may be necessary to use an

55

integration method to account for the additional stiffness provided by the top reinforcing

bars at the ends of the member. The second example in Appendix L explains the error

in more detail.

The proposed equations are often overly conservative when 𝑀𝐿/𝑀𝑐𝑟 > 1 and there is

additional reinforcing in the midspan region to control deflection. It may be necessary

to use an integration method to compute deflection in these cases because it will account

for both the shortened cracked length at midspan and because the 𝐼𝑒(𝑥) at the member

ends is smaller than the 𝐼𝑒(𝑥) at midspan under the same service moments.

When −𝑀𝐿 ≈ −𝑀𝑅 > 2𝑀𝑚, concrete members undergoing centered point loading

cannot be accurately modelled as a constant stiffness member. For such large end-

moments to occur, the concrete member will typically require more depth at the ends,

therefore it will be a non-prismatic member. Alternatively, the member will be

uncracked at midspan and the midspan deflection will be near zero or upwards, so

preliminary deflection checks would indicate that an in-depth deflection analysis is not

required.

3.5 Continuous Beam with Two Equal Point Loads at Third Points

In this section, results using the proposed equations are compared to other solutions for

determining deflection with two equal point loads at the third points. As with the

Section 3.4, self-weight is not otherwise accounted for. Plotted members are generated

as explained in Section 3.2 and deflection values are obtained as explained in Section

3.3. Unequal end-moments often cause significant errors for deflection computations

γ Continuous

56

for third-point loaded members; this error is very significant for all known effective and

equivalent moment of inertia approaches, so an improved equation is proposed.

3.5.1 Proposed Solution for Two Equal Loads at Third Points

The proposed solution for a continuous member with equal loading at third points

incorporates the end-moment values to determine the equivalent moment of inertia. Use

of the integration factor, , to calculate 𝐼𝑒 ′ as per Bischoff and Gross (2011) for simply

supported members gives a good approximation for highly cracked continuous members

(𝑀𝑚 𝑀𝑐𝑟⁄ > 2). For continuous members that experience less cracking, however, the

original simply supported equations offer poor results. Consequently, an improved

equation for , denoted as ∗ , is provided and used to compute an improved equivalent

moment of inertia, 𝐼𝑒∗′ . The more accurate deflection values, computed using 𝐼𝑒∗

′ , are

indicated as 𝛥𝐼𝑒∗′ in Figures 3-5 to 3-8 and in Appendix N.

The proposed ∗ equation attempts to account for two prominent errors in results based

on the 𝐼𝑒′ calculated using . For third-point loading with equal end-moments, the actual

equivalent moment of inertia is smaller than 𝐼𝑒′ because the 𝐾 factor for constant

stiffness members will reduce deflection by too much. This occurs because the middle

third of a member is the main contributor to the deflection and the integrated area of this

segment sees no effect from the end-moments. For a similar reason, when unequal end-

moments cause the smaller of the bending moments at the third points to be near or less

than 𝑀𝑐𝑟, the exact member deflection is much less than the deflection computed using

𝐼𝑒′ . The proposed equations provided account for some of this error within the proposed

limits, but an integration approach must be used outside these limits.

57

The equation for ∗ was empirically derived for equal third-point loading using a ratio

of the cracking moment to each end-moment. When there are no end-moments, ∗ = ,

which is the exact integrated result for a simply supported member. Further

improvement in ∗ is likely possible for continuous members. However, such solutions

are unlikely to be straightforward because the member response for equal third-point

loads and various end-moments is quite unlike that of a constant stiffness member.

The equation for Bischoff’s equivalent moment of inertia is:

𝐼𝑒′ =

𝐼𝑐𝑟


𝑀𝑚𝑎𝑥)2 where 𝜂 = 1 −


For members with two equal loads of 𝑃/2 at the third points and −𝑀𝐿 ≥ −𝑀𝑅:

= 1.7 − 0.7 (𝑀𝑐𝑟

𝑀𝑚𝑎𝑥) where 𝑀0 =

𝑃𝐿

6 and 𝑀𝑚 = 𝑀0 +

𝑀𝐿

2+𝑀𝑅

2 (3 − 6)

𝑀𝑚𝑎𝑥 = 𝑀0 +1

3𝑀𝐿 +

2

3𝑀𝑅 where − 𝑀𝐿 > −𝑀𝑅 (3 − 7)

If 𝑀𝐿 = 𝑀𝑅, the bending moment remains constant between the third-points

If −𝑀𝐿 > −𝑀𝑅, then 𝑀𝑚𝑎𝑥 will occur at 𝑥 = 2𝐿/3

For two equal loads at the third points, the midspan deflection using Bischoff’s 𝐼𝑒′ for

simply supported members is computed as:

𝛥 = 𝛥𝐼𝑒∗′ = 𝐾23𝑀𝑚𝐿

2

216𝐸𝑐𝐼𝑒′ where 𝐾 =

27

23−

4𝑀0

23𝑀𝑚 (3 − 8)

The proposed solution for midspan deflection is:

∗ = − 0.1(𝑀𝐿 − 1.5𝑀𝑅)/𝑀𝑐𝑟 where − 𝑀𝐿 ≥ −𝑀𝑅 (3 − 9)

58

𝐼𝑒∗′ =

𝐼𝑐𝑟

1 − ∗ 𝜂 (𝑀𝑐𝑟

𝑀𝑚𝑎𝑥)2 (3 − 10)

𝛥 = 𝛥𝑚𝑖𝑑 = 𝛥𝐼𝑒∗′ = 𝐾23𝑀𝑚𝐿

2

216𝐸𝑐𝐼𝑒∗′ where 𝐾 =

27

23−

4𝑀0

23𝑀𝑚 (3 − 11)

For unequal end-moments where −𝑀𝐿 < 2𝑀𝑚𝑎𝑥 and −𝑀𝐿 > −𝑀𝑅, the maximum

deflection is computed as per Equation (3-5).

3.5.2 Comparison of Results for Two Equal Loads at Third Points

The graphs in this section compare deflection calculation approaches for third-point

loading with four example sets of reinforced concrete members comprised of:

Steel reinforced beams with equal end-moments (𝑀𝐿 = 𝑀𝑅)

Steel reinforced beams that are continuous at only one end (𝑀𝑅 = 0)

GFRP reinforced beams with equal end-moments (𝑀𝐿 = 𝑀𝑅)

GFRP reinforced beams that are continuous at only one end (𝑀𝑅 = 0)

Data and additional discussion for these four members are provided in Appendix N.

Each graph contains a set of members with the same maximum moment (and relevant

properties); the magnitude of the end-moment(s) and the amount of reinforcement at the

end(s) of the members increase as the plots progress from left to right. The deflections

computed using integration of curvature based on 𝐼𝑒(𝑥), per Equation (2-7), are

assumed to be exact for comparison purposes. The S806 integration method is again

used for all members in order to provide results that neglect tension stiffening. The

59

examples will show that the proposed 𝐼𝑒∗′ equation provides an improved moment of

inertia for computing deflection of continuous members with linear-elastic equations.

The examples provided in Figures 3-5 to 3-8 were designed and analyzed as rectangular

beams with 𝑓𝑐′ = 36 MPa. These beams were 600 mm deep, 300 mm wide, with 10 m

spans and with top and bottom reinforcement at 𝑑 = 540 mm. Figure 3-5 and Figure

3-6 use steel reinforcing with 𝑓𝑦 = 400 MPa. Figure 3-7 and Figure 3-8 use GFRP

reinforcing bars with 𝑓𝑓𝑢 = 690 MPa and 𝐸𝑏 = 44 GPa. The S806 (2012) integration

method, which neglects tension stiffening, is conservative in each set of members.

Figure 3-5 shows results for a set of steel-reinforced members with equal end-moments

and 𝐼𝑔/𝐼𝑐𝑟 = 2.97, 𝑀𝑚𝑎𝑥/𝑀𝑐𝑟 = 2.20, 𝜌 = 0.8%, and 𝑀𝑟 = 1.575𝑀𝑠. For this

example, all of the effective moment of inertia methods work reasonably well for most

Figure 3-5 - Midspan Deflection of Steel Reinforced Beams under Third Point Loading


60

members; significant error occurs only when the end-moments are relatively large:

−𝑀𝐿 = −𝑀𝑅 > 1.5𝑀𝑚𝑎𝑥. For this set of examples, no effective moment of inertia

methods are suitable for large end moments (even 𝛥𝑐𝑟 results are unconservative for

−𝑀𝐿 = −𝑀𝑅 = 3𝑀𝑚𝑎𝑥). The utility of the proposed 𝐼𝑒∗′ is negligible for these

members. However, these members are an example of the 𝛥𝐼𝑒,𝑎𝑣𝑔 curve failing to match

the integrated results well. The 𝛥𝐼𝑒,𝑎𝑣𝑔 curve also shows that the aberration discussed for

Figure 3-1 also occurs for equal third-point loaded members. The S806 (2012)

integration method shows that is it very conservative to neglect tension stiffening for

these members (even for large end-moments).

The set of members generated for Figure 3-6 are steel reinforced beams with one end

continuous with the same maximum positive moment properties (𝐼𝑔/𝐼𝑐𝑟, 𝜌, 𝑀𝑟/𝑀𝑠, and

Figure 3-6 - Midspan and Maximum Deflection of Steel Reinforced Beams under Third

Point Loading with Ig/Icr=3.0, Mmax /Mcr=2.2, and MR=0

61

𝑀𝑚𝑎𝑥/𝑀𝑐𝑟) as the members in Figure 3-5. Figure 3-6 indicates that Branson’s (1965) 𝐼𝑒

approach is up to 15% conservative. The constant moment of inertia solutions are also

conservative when −𝑀𝐿 < 2.5𝑀𝑚𝑎𝑥, with the exception of the proposed 𝐼𝑒∗′ . The

deflection values computed using 𝐼𝑒∗′ provide the most accurate approximations in this

graph when −𝑀𝐿 < 2𝑀𝑚𝑎𝑥, which encompasses most practical members, but this

approach becomes unconservative for larger end-moments. Two curves of computed

maximum deflection, 𝛥𝑚𝑎𝑥,𝐼𝑒(𝑥) and 𝛥𝑚𝑎𝑥,𝐼𝑒′ , are also provided in Figure 3-6; these

show that the maximum deflection is significantly larger than the midspan deflection for

members which have −𝑀𝐿 > 1.5𝑀𝑚𝑎𝑥.

Figure 3-7 is based on members which are reinforced with GFRP reinforcing bars as

follows: 𝐼𝑔/𝐼𝑐𝑟 = 12.2, 𝜌 = 0.7%, 𝜌/𝜌𝑏 = 1.21, 𝑀𝑚/𝑀𝑐𝑟 = 1.40, 𝑀𝑟 = 2.857𝑀𝑠 for

Figure 3-7 - Midspan Deflection of GFRP Reinforced Beams under Third Point Loading


62

bottom bars, and 𝑀𝑟 = 1.575𝑀𝑠 for top bars. The results using 𝐼𝑒′ are unconservative,

and become so by more than 10% when −𝑀𝐿 ≥ 2.0𝑀𝑚𝑎𝑥 . The S806 integration results

are too conservative to be plotted on this graph because tension stiffening is important

when 𝑀𝑚 𝑀𝑐𝑟⁄ = 1.4. Results using Branson’s (1965) 𝐼𝑒 underpredict deflection by a

factor of about 2.5 when −𝑀𝐿 ≤ 2.0𝑀𝑚𝑎𝑥, and by more for −𝑀𝐿 > 2.0𝑀𝑚𝑎𝑥. Use of

𝛽𝑑 = 0.24, in accordance with ACI 440.1R, yields conservative results with this

example, but a similar AFRP (aramid FRP) reinforced member gives unconservative

results for this approach, as expected based on its limitations (Bischoff and Gross 2011).

The 𝛥𝐼𝑒∗′ curve is clearly the most accurate approach when −𝑀𝐿 ≤ 2.0𝑀𝑚𝑎𝑥 .

The set of example members used to produce Figure 3-8 have only one end continuous

but have otherwise identical properties to those used for Figure 3-7. Results using

Figure 3-8 - Midspan and Maximum Deflection of GFRP Reinforced Beams under Third

Point Loading with Ig/Icr=12.2, Mmax /Mcr=1.4, and MR=0

63

Branson’s (1965) 𝐼𝑒 predict only 50% of the actual deflection for typical members

(−𝑀𝐿 < 1.5𝑀𝑚𝑎𝑥). The 𝛥𝐼𝑒,𝛽𝑑 curve, with 𝛽𝑑 = 0.24, is conservative with an error of

30% to 200% for −𝑀𝐿 ≤ 2.0𝑀𝑚𝑎𝑥. The results using 𝐼𝑒′ are very conservative for

unequal end-moments; this contrasts with the unconservative results in Figure 3-7. The

𝛥𝐼𝑒∗′ curve is conservative by up to 25% for −𝑀𝐿 ≤ 𝑀𝑚𝑎𝑥 and up to 60% conservative

for larger end-moments, but it is clearly the most accurate constant moment of inertia

approach in Figure 3-8. As they did with Figure 3-6, the two maximum deflection

curves, 𝛥𝑚𝑎𝑥,𝐼𝑒(𝑥) and 𝛥𝑚𝑎𝑥,𝐼𝑒′ , show a significant difference between 𝛥𝑚𝑎𝑥 and 𝛥𝑚𝑖𝑑

when −𝑀𝐿 > 1.5𝑀𝑚𝑎𝑥.

3.5.3 Summary of Results for Two Equal Loads at Third Points

Table 3-2 summarizes the valid ranges for the corrected proposed effective moment of

inertia, 𝐼𝑒 ∗ ′ , with two equal loads at third points. Results assume 𝑀𝐿 ≤ 𝑀𝑅 ≤ 0. To

provide these ranges of validity, relevant variables were tested within reasonable ranges.

Values for 𝑀𝑐𝑟 𝑀𝑚⁄ , 𝐼𝑔 𝐼𝑐𝑟⁄ , 𝑀𝑅 𝑀𝐿⁄ , 𝑀𝑚 𝑀𝑟⁄ , 𝑑/ℎ, and other relevant ratios were

varied to catch major divergence. Generally, the valid ranges were terminated when the

proposed approximation reached 10% error. More error is prominent for the one end-

moment cases, so overprediction errors of up to 50% are presented as valid (as noted).

All errors described are between approximate results using the noted constant moment

of inertia and the exact idealized deflection as explained in Section 3.3. For the

majority of practical members, −𝑀𝐿 ≤ 1.5𝑀𝑚𝑎𝑥 will be true for the load pattern which

governs deflection; therefore the valid ranges provided for 𝐼𝑒∗′ are rarely problematic.

64

To compute deflections in cases outside the valid ranges, and for more accurate

predictions where 𝑀𝑚𝑎𝑥 𝑀𝑐𝑟⁄ < 1.2, it may be necessary to use an integration method.

Table 3-2 - Valid Ranges for I'e* for Equal Point Loads at Third Points

Equal End-Moments (𝑀𝐿 = 𝑀𝑅)

#f One End-Moment

Cracked Ratio 𝐼𝑒∗′ Valid If: 𝐼𝑒

′ Valid If Iγ=1

Valid? 𝐼𝑒∗′ Valid If:

#a

3 ≤𝑀𝑚𝑎𝑥

𝑀𝑐𝑟 −𝑀𝐿 ≤ 2.0𝑀𝑚𝑎𝑥 −𝑀𝐿 ≤ 2.0𝑀𝑚𝑎𝑥 Yes −𝑀𝐿 ≤ 2.0𝑀𝑚𝑎𝑥

1.7 ≤𝑀𝑚𝑎𝑥

𝑀𝑐𝑟< 3 −𝑀𝐿 ≤ 2.0𝑀𝑚𝑎𝑥 −𝑀𝐿 ≤ 1.5𝑀𝑚𝑎𝑥 Ok

#b

−𝑀𝐿 ≤ 2.0𝑀𝑚𝑎𝑥


𝑀𝑐𝑟< 1.7 −𝑀𝐿 ≤ 2.0𝑀𝑚𝑎𝑥 −𝑀𝐿 ≤ 1.3𝑀𝑚𝑎𝑥 Ok

#c




#d




#e

−𝑀𝐿 ≤ 0.5𝑀𝑚𝑎𝑥 #a

If the 𝐼𝑒∗′ equation yields results outside limits 0 < 𝐼𝑒∗

′ < 𝐼𝑔, use of 𝐼𝑔 is reasonable.

Prediction with 𝐼𝑒′ using =1.7-.7(𝑀𝑐𝑟 𝑀𝑚𝑎𝑥⁄ ) reasonable if 𝑀𝑚𝑎𝑥 𝑀𝑐𝑟⁄ < 2.5, 𝑀𝑅 ≈ 0.

Overprediction exceeds 20% with 𝐼𝑒′ for 𝑀𝑚𝑎𝑥 𝑀𝑐𝑟⁄ > 2.5, −𝑀𝐿 > 0.1𝑀𝑚𝑎𝑥, 𝑀𝑅 ≈ 0.

#b Overprediction of deflection exceeding 10% likely if −𝑀𝐿 < 1.0𝑀𝑚𝑎𝑥 and 𝐼𝑔 𝐼𝑐𝑟⁄ > 4.

#c Overprediction of deflection exceeding 10% likely when −𝑀𝐿 < 0.5𝑀𝑚𝑎𝑥

#d Overprediction of deflection, exceeding 10%, likely if 𝑀𝑚𝑎𝑥 𝑀𝑐𝑟⁄ < 2.5

#e Overprediction, exceeding 10%, likely if −𝑀𝐿 > 0.3𝑀𝑚𝑎𝑥; 𝐼𝑔 is ok if −𝑀𝐿 > 𝑀𝑚𝑎𝑥

#f 𝐼𝑐𝑟 provides reasonable results only when 𝑀𝑚𝑎𝑥 𝑀𝑐𝑟⁄ < 3

The improved result, 𝐼𝑒∗′ , using ∗ = − 0.1(𝑀𝐿 − 1.5𝑀𝑅)/𝑀𝑐𝑟, was only intended to

give good results for 𝑀𝑚𝑎𝑥 𝑀𝑐𝑟⁄ > 1.2. For single end-moment cases where

𝑀𝑚𝑎𝑥 𝑀𝑐𝑟⁄ < 1.2, the limits of 0 < 𝐼𝑒∗′ ≤ 𝐼𝑔 gives good results. For equal end-moment

cases where 𝑀𝑚𝑎𝑥 𝑀𝑐𝑟⁄ < 1.2, use of the original 𝐼𝑒′ with = 1 yields an improvement

65

in the 𝐼𝑒∗′ results. Use of 𝐼𝑒∗

′ , or assuming 𝑀𝑚𝑎𝑥 = 1.2𝑀𝑐𝑟 when 𝑀𝑚𝑎𝑥 < 1.2𝑀𝑐𝑟, is

conservative and may be wise for 𝑀𝑚𝑎𝑥 ≈ 𝑀𝑐𝑟. Because 𝐼𝑒∗′ was derived empirically,

the divergence of results where 𝑀𝑚𝑎𝑥 𝑀𝑐𝑟⁄ < 1.2 is not easily explained.

3.6 Continuous Beam with a Uniformly Distributed Load

Results using equations for deflection of a continuous beam under a uniformly

distributed load are compared in the following section. The exact predicted deflection is

determined using numerical integration (as shown in Appendix K) and analytical

equations (as provided in Appendix E). Plotted members are generated as explained in

Section 3.2 and deflection values are obtained as explained in Section 3.3.

3.6.1 Proposed Solution for a Uniformly Distributed Load

The proposed solution for a continuous member with a uniformly distributed load is to

employ the equivalent moment of inertia, 𝐼𝑒′ , proposed by Bischoff and Gross (2011).

Use of 𝐼𝑒′ is an approximation for continuous members whereas it is the exact integrated

result for simply supported members. The most accurate and robust results were found

using the integration factor, , as provided in Equation (3-16). This accounts for the

variation in stiffness along the length of a simply supported member without

approximation. Both 𝐼𝑒′ and are computed using the member properties at the location

of maximum positive service bending moment, 𝑀𝑚𝑎𝑥, and an applied moment of 𝑀𝑚𝑎𝑥.

Although deflection calculations are still reasonably accurate if |𝑀𝐿 −𝑀𝑅| ≤ 0.5𝑀𝑚,

𝑀𝑚𝑎𝑥 is more accurate and is used for all computations in this report. The stiffness at

66

the ends of the span was found to have little effect on the calculation of an accurate

constant moment of inertia.

For a uniformly distributed load, using as provided in Equation (3-16), the proposed

equation for the equivalent moment of inertia is:

𝐼𝑒′ =

𝐼𝑐𝑟


𝑀𝑚𝑎𝑥)2 where 𝜂 = 1 −


The following equations were used to compute bending moment for a continuous

member with a uniformly distributed load:

𝑀0 =(𝑉𝐿 + 𝑉𝑅)𝐿

8=𝑤𝐿2

8 (3 − 13)

𝑀0 =𝑀𝑚𝑎𝑥 −

𝑀𝐿

2 −𝑀𝑅

2 + √𝑀𝐿𝑀𝑅 −𝑀𝐿𝑀𝑚𝑎𝑥 −𝑀𝑅𝑀𝑚𝑎𝑥 +𝑀𝑚𝑎𝑥2

2 (3 − 14)

𝑀𝑚 = 𝑀0 +𝑀𝐿

2+𝑀𝑅

2 and 𝑀𝑚𝑎𝑥 = 𝑀0 +

𝑀𝐿

2+𝑀𝑅

2+(𝑀𝐿 −𝑀𝑅)

2

16𝑀0 (3 − 15)

To use the equation for 𝐼𝑒′ with a uniformly distributed load, the integration factor is:

=1.6𝜉3 − 0.6𝜉4

(𝑀𝑐𝑟

𝑀𝑚𝑎𝑥)2 + 2.4 ln(2 − 𝜉) where 𝜉 = 1 − √1 −

𝑀𝑐𝑟

𝑀𝑚𝑎𝑥 (3 − 16)

This factor is also provided in Table 2-3. The approximate provided in Table 2-3 was

not used in the presented results as it was found to cause a significant decrease in

accuracy for some examples.

For a uniformly distributed load, the approximate midspan deflection is:

67

𝛥 = 𝛥𝑚𝑖𝑑 = 𝛥𝐼𝑒′ = 𝐾5𝑀𝑚𝐿

2

48𝐸𝑐𝐼𝑒′ where 𝐾 = 1.2 − 0.2

𝑀0

𝑀𝑚 (3 − 17)

Calculations for this report indicate that maximum deflection exceeds midspan

deflection by less than 5% if the maximum moment does not exceed the midspan

moment by more than 5%:

𝛥𝑚𝑎𝑥

𝛥𝑚𝑖𝑑< 1.05 if

𝑀𝑚𝑎𝑥

𝑀𝑚< 1.05 (3 − 18)

The maximum deflection can be approximated as follows:

𝛥𝑚𝑎𝑥 ≈ 𝛥𝑚𝑖𝑑√𝑀𝑚𝑎𝑥

𝑀𝑚=5√𝑀𝑚𝑀𝑚𝑎𝑥

48𝐸𝑐𝐼𝑒′ (3 − 19)

3.6.2 Comparison of Results for a Uniformly Distributed Load

The graphs in this section compare deflection calculation approaches for uniformly

distributed loading for five example sets of reinforced concrete members comprising:

Steel reinforced beams with equal end-moments (𝑀𝐿 = 𝑀𝑅)

Steel reinforced beams that are continuous at only one end (MR = 0)

Steel reinforced one-way slabs with equal end-moments (ML = MR)

Steel reinforced one-way slabs that are continuous at only one end (MR = 0)

GFRP reinforced beams with equal end-moments (𝑀𝐿 = 𝑀𝑅)

Data and additional discussion for these five members are provided in Appendix O.

Three additional examples provided in Appendix O may offer other useful information.

All data for this section have been created with 𝑀𝐿 ≤ 𝑀𝑅 ≤ 0, but this does not appear

to be required by the proposed solution.

68

Each graph contains a set of members with the same maximum moment (and relevant

properties); the magnitude of the end-moment(s) and the amount of reinforcement at the

end(s) of the members increases as the plots progress from left to right. The deflections

computed using integration of curvature, based on 𝐼𝑒(𝑥) per Equation (2-7), are

assumed to be exact for comparison purposes. The S806 (2012) integration method is

again used for all members in order to provide results that neglect tension stiffening.

The example graphs will show that deflection computed using the proposed equation for

𝐼𝑒′ , as found in Equation (2-8) and computed using as per Equation (3-16), provides an

improved moment of inertia for computing deflection of continuous members with

linear-elastic equations.

Examples in this section are designed as follows:

Rectangular beams and slabs with 𝑓𝑐′ = 36 MPa

Figures 3-9 to 3-12 are designed with steel reinforcing bars having 𝑓𝑦 = 400 MPa

Figures 3-9 and 3-10 are 600 mm deep, 300 mm wide beams with 10 m spans and

with top and bottom reinforcement at 𝑑 = 540 mm

Figures 3-11 and 3-12 are 1 m strips of slabs which are 275 mm deep, span 7.5 m,

and have top and bottom reinforcing at 𝑑 = 233.8 mm

Figure 3-13 is designed as a 600 mm deep and 300 mm wide beam which spans

10 m and has reinforcing at 𝑑 = 510 mm with GFRP reinforcing bars

Figure 3-13 GFRP reinforcing bars have an 𝑓𝑓𝑢 = 690 MPa and an 𝐸𝑏 = 44 GPa

The curves depicted in Figures 3-10 and 3-12 provide results for both midspan

and maximum deflections for members with unequal end moments

69

The 𝛥𝐼𝑒,𝑎𝑣𝑔 curve has only been provided in Figures 3-9 and 3-11 because other

results showed very similar results; all are unhelpful modifications to 𝐼𝑒

The curves provided for 𝛥𝑔 and 𝛥𝑐𝑟 are normally thought to be lower-bound and

upper-bound solutions for the deflection of concrete members because moment on

inertia should always vary between 𝐼𝑔 and 𝐼𝑐𝑟 for concrete members

Continuous concrete members are not bounded by the curves 𝛥𝑔 and 𝛥𝑐𝑟; these

curves are provided for reference purposes only

Explanation provided for a common occurrence for these five plots is not repeated

on subsequent graphs. The most thorough explanation is provided for Figure 3-9

Figure 3-9 shows results for a set of steel-reinforced beams with equal end-moments

where 𝐼𝑔/𝐼𝑐𝑟 = 2.99, 𝜌 = 0.8%, 𝑀𝑟 = 1.575𝑀𝑠, and 𝑀𝑚/𝑀𝑐𝑟 = 2.17. For this

example, all of the effective moments of inertia are used to compute reasonably accurate

values for deflection when −𝑀𝐿 = −𝑀𝑅 < 2.5𝑀𝑚𝑎𝑥. These curves are an example of

when effective moment of inertia methods are not suitable for larger end-moments; even

using 𝐼𝑐𝑟 yields unconservative results for −𝑀𝐿 = −𝑀𝑅 > 3𝑀𝑚𝑎𝑥. The exact result,

𝛥𝐼𝑒(𝑥), begins to diverge from the constant stiffness results as the end moments increase

beyond −𝑀𝐿 = −𝑀𝑅 = 2𝑀𝑚𝑎𝑥; extrapolating these curves is evidence that the, 𝛥𝐼𝑒(𝑥),

will be significantly larger than the constant stiffness member result of 0 mm of

deflection when −𝑀𝐿 = −𝑀𝑅 = 5𝑀𝑚. The S806 (2012) integration method, which is

conservative by more than 25% for all members plotted in this set, is the only method to

be conservative when −𝑀𝐿 = −𝑀𝑅 > 3𝑀𝑚𝑎𝑥. When using Branson’s (1965) 𝐼𝑒, results

for this set of members are 7% conservative for −𝑀𝐿 = −𝑀𝑅 > 1.5𝑀𝑚𝑎𝑥 and are

70

unconservative when −𝑀𝐿 = −𝑀𝑅 > 2.2𝑀𝑚𝑎𝑥. Using 𝐼𝑒,𝑎𝑣𝑔 per CSA A23.3 (2004), as

shown by the 𝛥𝐼𝑒,𝑎𝑣𝑔 curve, modifies the Branson’s 𝐼𝑒 in ways that do not relate to the

exact results, such as becoming very unconservative when −𝑀𝐿 = −𝑀𝑅 > 1.8𝑀𝑚𝑎𝑥.

The 𝛥𝐼𝑒,𝑎𝑣𝑔 curve includes an aberration near 𝑀𝐿/𝑀𝑚 = 𝑀𝑐𝑟/𝑀𝑚 that is explained in

the discussion of Figure 3-1. Use of Bischoff’s 𝐼𝑒, shown by the curve for 𝛥𝛾=1,

provides results that are slightly more conservative than results using Branson’s 𝐼𝑒

approach. The proposed solution, using 𝐼𝑒′ , is the most accurate approximation for the

deflection in Figure 3-9 when −𝑀𝐿 = −𝑀𝑅 < 2𝑀𝑚𝑎𝑥. For −𝑀𝐿 = −𝑀𝑅 < 1.8𝑀𝑚𝑎𝑥,

the curve for 𝛥𝐼𝑒′ maintains less than 2% error.

Figure 3-9 - Midspan Deflection of Steel Reinforced Beams under Uniformly

Distributed Load with Ig/Icr=3.0, Mm /Mcr=2.17, and ML=MR

The set of beams for Figure 3-10 has the same properties as the set of members in

Figure 3-9, except that Figure 3-10 has only one end continuous. These figures show

71

that, for this set of members, the accuracy for members with one end continuous is

similar to both ends continuous. Again, large errors occur when end-moments become

relatively large (−𝑀𝐿 > 2𝑀𝑚𝑎𝑥). Figure 3-10 demonstrates the difference between

midspan deflection and maximum deflection for uniformly distributed loading. The

approximate factor between midspan and maximum deflection, provided in Equation (3-

19), accurately computes the maximum deflection when supplied with an accurate

midspan deflection.


Uniformly Distributed Load with Ig/Icr=3.0, Mmax /Mcr=2.17, and MR=0

Figure 3-11 depicts the deflection values for a set of slabs with uniformly distributed

load and equal end-moments. These slabs are generated with 𝐼𝑔/𝐼𝑐𝑟 = 4.90,

𝜌 = 0.54%, 𝑀𝑟 = 1.575𝑀𝑠, and 𝑀𝑚/𝑀𝑐𝑟 = 1.33. Branson’s method underestimates

deflection for these slabs by at least 10% for all values of end-moment. Tension

72

stiffening should not be neglected where a member is lightly cracked (𝑀𝑚/𝑀𝑐𝑟 ≪ 2), as

proven by the 𝛥𝛽=0 curve values being more than double those of the 𝛥𝐼𝑒(𝑥) curve. For

this set of slabs, the results computed using the proposed 𝐼𝑒′ have only a 3% error when

−𝑀𝐿 = −𝑀𝑅 < 2.8𝑀𝑚𝑎𝑥. The use of 𝐼𝑒,𝑎𝑣𝑔 as the effective moment of inertia shows

the same problems noted for Figure 3-9. Appendix O also provides a full example for a

set of beams, with the same property ratios mentioned for the slabs used in Figure 3-11,

where the beams have smaller deflections that these slabs, but the plot of deflections has

the exact same shape for both sets.

Figure 3-11 - Midspan Deflection of Steel Reinforced Slabs under Uniformly Distributed

Load with Ig/Icr=4.9, Mm /Mcr=1.33, and ML=MR

The slabs used in Figure 3-12 are identical to those from Figure 3-11, except that these

slabs have one end continuous rather than both ends continuous. Branson’s method

again underestimates deflection, while the 𝛥𝛽=0 curve again greatly exceeds the

73

deflection results which consider tension stiffening. The 𝛥𝐼𝑒′ curve shows that the

proposed approach is again very accurate if −𝑀𝐿 = −𝑀𝑅 ≤ 3𝑀𝑚𝑎𝑥. This curve

maintains less than a 5% error when compared to the exact result. Figure 3-12 shows

the midspan deflection for most of the outlined approaches, and shows the maximum

deflection for the proposed approach and the exact approach. In this graph, Equation

(3-19) accurately computes maximum deflection because midspan results are accurate.

Figure 3-12 - Midspan and Maximum Deflection of Steel Reinforced Slabs under

Uniformly Distributed Load with Ig/Icr=4.9, Mmax /Mcr=1.33, and MR=0

The set of example members generated for Figure 3-13 are GFRP reinforced beams

under a uniformly distributed load with equal end-moments. For these beams:

𝐼𝑔/𝐼𝑐𝑟 = 16.9, 𝜌 𝜌𝑏⁄ = 1.02, 𝑀𝑚/𝑀𝑐𝑟 = 1.25, 𝜌 = 0.6, 𝑀𝑟 = 2.674𝑀𝑠 for the bottom

bars, and 𝑀𝑟 = 1.575𝑀𝑠 for the top bars. Branson’s approach severely underestimates

deflection. This clearly demonstrates that it was not intended for use with GFRP

74

reinforcing bars. The results based on the proposed 𝐼𝑒′ are within 5% of the exact results

for −𝑀𝐿 = −𝑀𝑅 < 𝑀𝑚𝑎𝑥, and within 11% for −𝑀𝐿 = −𝑀𝑅 < 3𝑀𝑚𝑎𝑥, in this example.

Appendix O provides three examples for uniformly loaded GFRP members: a set of

example slabs with 𝛽𝑑 = 1.0, a set of example beams with 𝑀𝑚/𝑀𝑐𝑟 = 2.0, and the set

of data for the beams in Figure 3-13. Using βd = 0.204 in accordance with ACI

440.1R yields conservative results with about 80% error for this set of examples.

Where 𝛽𝑑 = 1.0 and deflection is underpredicted using Branson’s 𝐼𝑒 equation,

deflection calculations in accordance with ACI 440.1R do not improve results.

Figure 3-13 - Midspan and Deflection of GFRP Reinforced Beams under Uniformly

Distributed Load with Ig/Icr=17, Mm /Mcr=1.25, and ML=MR

3.6.3 Summary of Results for a Uniformly Distributed Load

Table 3-3 summarizes the valid ranges for the proposed effective moment of inertia, 𝐼𝑒′ ,

with a uniformly distributed load. To provide the ranges of validity shown, results were

75

reviewed as the value of each relevant variables was changed. Values for 𝑀𝑐𝑟 𝑀𝑚⁄ ,

𝐼𝑔 𝐼𝑐𝑟⁄ , 𝑀𝑅 𝑀𝐿⁄ , 𝑀𝑚 𝑀𝑟⁄ , depth divided by height, and other relevant ratios have been

varied within reasonable ranges in an attempt to provide valid ranges that are applicable

to all realistic members. Concrete members reinforced with steel, AFRP, and GFRP

bars have also been reviewed. Results within the provided ranges typically result in less

than 5% error. The valid ranges were terminated when the proposed approximation

reached 10% error, with one exception. The error was permitted to exceed 10%, as

conservative, for midspan deflection for a one-end continuous case where 𝑀𝑅 ≈ 0,

−𝑀𝐿 ≤ 1.5𝑀𝑚𝑎𝑥 and −𝑀𝐿 ≤ 2.5𝑀𝑚𝑎𝑥, and 𝑀𝑚𝑎𝑥 𝑀𝑐𝑟⁄ < 1.2. The maximum

deflection found using 𝛥𝑚𝑎𝑥 ≈ 𝛥𝑚𝑖𝑑 √𝑀𝑚𝑎𝑥 𝑀𝑚⁄ will result in less than 10% error in

this case.

Table 3-3 - Valid Ranges for I'e for Uniformly Distributed Load

Equal End-Moments (𝑀𝐿 = 𝑀𝑅) One End-Moment

#

Cracked Ratio

𝐼𝑒′ Valid If:

𝐼 =1 Valid? 𝐼𝑐𝑟 Valid?

𝐼𝑒′ Valid If:


𝑀𝑐𝑟 −𝑀𝐿 ≤ 2.0𝑀𝑚𝑎𝑥 Yes Ok −𝑀𝐿 ≤ 2.3𝑀𝑚𝑎𝑥


𝑀𝑐𝑟< 3 −𝑀𝐿 ≤ 2.2𝑀𝑚𝑎𝑥 Ok No −𝑀𝐿 ≤ 2.5𝑀𝑚𝑎𝑥


𝑀𝑐𝑟< 1.7 −𝑀𝐿 ≤ 2.5𝑀𝑚𝑎𝑥 No No −𝑀𝐿 ≤ 2.7𝑀𝑚𝑎𝑥


𝑀𝑐𝑟< 1.1 −𝑀𝐿 ≤ 1.5𝑀𝑚𝑎𝑥 No No −𝑀𝐿 ≤ 1.5𝑀𝑚𝑎𝑥

# Results were tested assuming 𝑀𝐿 ≤ 𝑀𝑅 ≤ 0.

If end-moments exceed proposed limits, even the proposed equations will often

underpredict deflection, so integration or another reliable method must be used. These

76

approximate service load deflection equations are very useful, however, because

situations where −𝑀𝐿 > 2𝑀𝑚𝑎𝑥 are rare, and are very unlikely to fail service deflection

requirements.

The idealized member testing shows a minor error for deflections computed with 𝐼𝑒′

when −𝑀𝐿 ≈ −𝑀𝑅 < 2𝑀𝑚𝑎𝑥. This error appears to be less than 5% for all cases except

with FRP members having equal end-moments and at least 50% more reinforcement

than is required for ultimate limit states. This error occurs because more of the member

is uncracked for the continuous members than for similar simply supported members.

The results using 𝐼𝑒′ are, therefore, (slightly) larger than the integrated 𝐼𝑒(𝑥) result when

−𝑀𝐿 ≈ −𝑀𝑅 < 𝑀𝑚. When −𝑀𝐿 < 1.5𝑀𝑚𝑎𝑥 and 𝑀𝑅 = 0, this minor conservative

error occurs for the same reason.

Constant stiffness solutions for concrete members often begin to diverge from exact

solutions as the end-moments increase beyond twice the maximum positive moment

(−𝑀𝐿 ≈ −𝑀𝑅 > 2𝑀𝑚𝑎𝑥). This is especially true for members where 𝑀𝑚𝑎𝑥/𝑀𝑐𝑟 > 1.7.

This is evident in Figures 3-9 and 3-10, where exact deflections exceed 𝐼𝑐𝑟 results when

−𝑀𝐿 > 2𝑀𝑚𝑎𝑥. In cases like this, if the deflection requirements cannot be shown to be

met without an in-depth analysis, it may be necessary to use an integration method to

account for the heavily reinforced member ends.

A constant stiffness analysis would normally not be attempted when

−𝑀𝐿 ≈ −𝑀𝑅 > 3𝑀𝑚𝑎𝑥, because the concrete member depth will be uncracked at

midspan or will require more member depth at the ends. Typically, deflection for these

members will also be near zero. Accurate deflection predictions in this range require

77

the use of an integration method. Section 3.7.2 provides discussion as to why it

becomes increasingly difficult, and sometimes impossible, to model a concrete beams as

a constant stiffness member when −𝑀𝐿 and −𝑀𝑅 are larger than 1.5𝑀𝑚𝑎𝑥.

3.7 Additional Findings

When preparing this report, four things were found that are critical to accurately

compute deflection and which lacked explanation in most other relevant literature. One

thing is that the difference between midspan and maximum deflection can be

significant, even exceeding 20%. Secondly, it is impossible for any constant stiffness

method to be correct for a wide range of continuous concrete members with large

negative end-moments. The third finding is that concrete members must be modelled

carefully because use of incorrect bending moments values to compute member

stiffness will often result in underpredicting deflection. Finally, the recent update to

A23.3 (2004), which changes 𝑀𝑐𝑟 to be calculated using one half of 𝑓𝑟, causes

significant changes to predicted deflections for reinforced concrete members.

3.7.1 When Midspan and Maximum Deflections are Different

It is important to use maximum deflection when it is not similar to midspan deflection.

Building codes (and other similar requirements) limit the permitted maximum

deflection, not the midspan deflection, so it is necessary to determine a good

approximation for the maximum deflection when the difference is significant. For

constant stiffness members undergoing uniformly distributed loads where 𝑀𝑅 = 0, the

difference in deflection reaches 10% at −𝑀𝐿 = 2.5𝑀𝑚𝑎𝑥. For constant stiffness

78

members subject to a centered point load or two equal point loads at the third-points

where 𝑀𝑅 = 0, the difference in deflection reaches 10% at −𝑀𝐿 = 2𝑀𝑚𝑎𝑥. Appendix

R provides data and more discussion on the comparison of midspan and maximum

deflection of prismatic linear-elastic members where 𝑀𝑅 = 0, 𝑀𝑅 = 𝑀𝑚𝑎𝑥/2, and

where 𝑀𝑅 = 𝑀𝐿/2.

3.7.2 Accurate Constant Stiffnesses can be Impossible

For certain end-moments, midspan deflection for all constant stiffness (prismatic and

linear-elastic) members is zero. This is also true for maximum deflection, which is

different from midspan deflection if end-moments are unequal. To achieve zero

midspan deflection for a constant stiffness member, solve for 𝐾 = 0 in Table 2-2. Thus

∆𝑚𝑖𝑑= 0 if: −𝑀𝐿 = −𝑀𝑅 = 2𝑀𝑚 for a centered point load, −𝑀𝐿 = −𝑀𝑅 = 5.75𝑀𝑚

for equal point loads at third points, and −𝑀𝐿 = −𝑀𝑅 = 5𝑀𝑚 for a uniformly

distributed load. In most circumstances, the midspan deflection of a concrete beam will

not be zero in such situations because the concrete beams will not have a constant

stiffness. This makes it impossible to create an accurate effective moment of inertia for

near-zero deflection. The limits to the proposed equations reflect this fact.

Neither typical nor proposed constant stiffness solutions are accurate for continuous

concrete members with a uniformly distributed load having −𝑀𝐿 ≈ −𝑀𝑅 ≈ 3𝑀𝑚𝑎𝑥.

Using an integration method becomes critical where −𝑀𝐿 ≈ −𝑀𝑅 > 2.5𝑀𝑚𝑎𝑥 and

𝑀𝑚𝑎𝑥/𝑀𝑐𝑟 > 1.3 (outside the valid range) because results from proposed equations

become increasingly unconservative as −𝑀𝐿 and −𝑀𝑅 increase relative to 𝑀𝑚𝑎𝑥. The

ends of these members become stiffer than the proposed effective moment of inertia, 𝐼𝑒′ ,

79

because of the additional reinforcement required for the larger negative bending

moments. This additional stiffness in the member ends will, in turn, reduce the

curvature and rotation at the member ends and therefore increase the midspan deflection

(relative to solutions based on simply supported members). Because stiffness

throughout these members is between 𝐼𝑔 and 𝐼𝑐𝑟, one would expect the solution to

always be such that 𝐼𝑔 > 𝐼𝑒′ > 𝐼𝑐𝑟. In this situation, however, an accurate solution for 𝐼𝑒

′

would have to reduce it to less than the midspan cracked moment of inertia, 𝐼𝑐𝑟. A

visual example of why a more robust solution would require 𝐼𝑒′ to be become less than

𝐼𝑐𝑟 is seen in Figure 3-9; here, when −𝑀𝐿 = −𝑀𝑅 = 3𝑀𝑚, the exact deflection exceeds

the deflection of a member with the constant stiffness of 𝐼𝑐𝑟. More robust effective

moment of inertia solutions, despite increased complexity, would still be limited to

certain ranges of validity for the reasons noted in the first paragraph of this section.

3.7.3 Importance of the Correct Bending Moment Function

The deflection equations provided for continuous members assume the designer has

determined the correct value for the bending moment at the supports. The equations

provided also do not take any pre-loading or pattern loading into account. If a pattern

load results in a small reduction in negative bending moment, this will in turn result in

increased positive bending moments and increased midspan deflection. Modelling the

actual stiffnesses of the member will often reduce the negative bending moments under

the worst-case loading for positive bending.

The following situation describes an example where the midspan deflection is larger

than computed. First, imagine a new uncracked beam (Beam A) loaded only to the

80

positive (midspan) bending service moment case. Now, imagine an identical new

uncracked beam (Beam B) that is first loaded to the worst negative bending service

moment case, and then loaded to the same maximum positive bending service moment

case. In most cases that occur in concrete buildings, Beam B will have more midspan

deflection because the bending moment will shift to the positive bending. Notably,

there is actually less deflection in Beam B if the same bending moment function is used

to model both beams. This decrease in the model occurs because the model will

increase the rotation at the ends of the member in order to achieve the negative bending

moments that are provided to it. Mathematically, the decrease in deflection occurs

because the increased cracking at the ends of Beam B will increase the area under the

integrated (𝑚𝑀/𝐸𝐼) curve in the negative moment region (see example 𝑚𝑀/𝐸𝐼 graph

that is provided in Appendix L).

More discussion about this phenomenon is given in Appendix P. One possible method

of determining the worst case bending moment function is also provided in Appendix P.

This appendix also explains why it appears to be reasonable, typically, to use only the

worst-case positive moment function.

3.7.4 Effect on Results of the CSA A23.3 Update to Clause 9.8.2.3

As mentioned previously, the prescribed cracking moment in Clause 9.8.2.3 of A23.3

(CSA 2004), R2010 version, was reduced to 𝑀𝑐𝑟 = 0.5𝑓𝑟𝐼𝑔/𝑦𝑡 (for use with Branson’s

𝐼𝑒). This change is intended to account for shrinkage restraint stresses. The use of

𝑀𝑐𝑟 = 0.67𝑓𝑟𝐼𝑔/𝑦𝑡 provides an equivalent adjustment for calculations based on 𝐼𝑒′ or

𝐼𝑒(𝑥) (Scanlon and Bischoff 2008).

81

Results based on a cracking moment that is reduced for shrinkage restraint stresses are

substantially different from the results presented throughout this report. Appendix P has

been added to provide discussion and examples that compare deflections determined

using 𝑀𝑐𝑟 = 𝑓𝑟𝐼𝑔/𝑦𝑡 to deflections determined using the reduced 𝑀𝑐𝑟 values. The

appendix shows that the section-based, effective, and equivalent moment of inertia

values become closer to the cracked moment of inertia value; thus, all deflection results

increase and become closer to the fully cracked results. It appears that results which use

Branson’s 𝐼𝑒 generally shift towards 𝐼𝑐𝑟 results by more than those which use 𝐼𝑒′ or

𝐼𝑒(𝑥). This causes deflection calculations to produce conservative results when

𝐼𝑔/𝐼𝑐𝑟 < 10. While results using Branson’s 𝐼𝑒 and 𝑀𝑐𝑟 = 𝑓𝑟𝐼𝑔/𝑦𝑡 underestimate

deflection when 𝐼𝑔/𝐼𝑐𝑟 > 4, results that use this 𝐼𝑒 with 𝑀𝑐𝑟 = 0.5𝑓𝑟𝐼𝑔/𝑦𝑡 factor are

improved because they only underestimate deflection when 𝐼𝑔/𝐼𝑐𝑟 > 12.

3.8 Summary of Results using Branson’s Method

There are many results provided in Sections 3.4, 3.5, and 3.6 that compare exact

integrations results to results using Branson’s method. Figures 3-2, 3-3, 3-5, 3-6, 3-9,

and 3-10 demonstrate that Branson’s method provides reasonably accurate results when

3 ≤ 𝐼𝑔/𝐼𝑐𝑟 ≤ 4. Figures 3-1, 3-2, and 3-3 demonstrate that Branson’s method is

sometimes more conservative than proposed methods. Figures 3-4, 3-7, 3-8, 3-11, 3-12,

and 3-13 demonstrate that Branson’s method often underpredicts deflection.

Unconservative errors in predictions also occur for large negative end-moments in

Figures 3-5, 3-6, 3-9, and 3-10. When accounting for shrinkage restraint while using

82

Branson’s method, with 𝑀𝑐𝑟 = 0.5𝑓𝑟𝐼𝑔/𝑦𝑡 as described in Section 3.7.4, accuracy is

improved and underpredicting deflection is much less common. A complete adoption of

the proposed equations is recommended, nonetheless, because Branson’s equation is

empirical and less robust than the proposed rational equations.

83

4.0 CONCLUSIONS AND RECOMMENDATIONS

4.1 Conclusions

To determine midspan deflection of continuous members using the effective moment of

inertia method, deflection at midspan can be calculated using a factor, 𝐾 (as provided in

Section 2.2.3), in conjunction with a generalized linear-elastic deflection equation.

In this report, exact deflections obtained by integration are compared to deflections

obtained by the approximate solutions for the effective moment of inertia, such as 𝐼𝑒, 𝐼𝑒′ ,

or 𝐼𝑒∗′ . Because the proposed solutions are approximations for continuous members,

minor errors are unavoidable. As negative end-moments become slightly larger than the

midspan moment, proposed solutions usually underpredict deflections because 𝐼𝑒(𝑥)

exceeds 𝐼𝑒′ at member ends. When negative moments increase beyond the proposed

limits (which generally occurs when the larger end-moment is at least double the

magnitude of the maximum positive moment, i.e. when −𝑀𝐿 ≥ 2𝑀𝑚𝑎𝑥), the member is

no longer suitable for constant stiffness member deflection calculations. Because of the

complexity of the problem, only a reliable integration method can provide solutions for

all possible limitations and situations. Integration using virtual work is explained in

Appendices B, E, and K; it is demonstrated in Appendix J.

The proposed approximate solutions require adherence to the noted valid ranges.

Accurate results are only obtainable for members if the negative end-moment(s) do not

greatly exceed the maximum positive moments. When end-moments become relatively

large, such that they are outside the noted valid range, deflections will rarely be

significant.

84

Bischoff’s equivalent moment of inertia for simply-supported members with centered

point loading and uniformly distributed loading also works well for continuous

members. Third-point loaded members require a modified equation for 𝐼𝑒′ ;

consequently, a reasonably accurate and robust equation is proposed as 𝐼𝑒∗′ . While other

effective moment of inertia solutions often provide adequate accuracy, results found

using the proposed equations are generally more accurate and more robust. It is

recommended that deflections be predicted using the proposed effective moment of

inertia equations for 𝐼𝑒′ or 𝐼𝑒∗

′ when members and loading meet the valid ranges noted.

The solutions for 𝐼𝑒′ and 𝐼𝑒∗

′ are provided in Sections 3.4.1, 3.5.1, and 3.6.1. The valid

ranges for these solutions are provided for these sections in Tables 3-1, 3-2, and 3-3,

respectively.

Branson's equation can continue to be used effectively within its limitations. If the

effects of shrinkage restraint are fully mitigated, the limitations for use of Branson’s

equation are much more severe than the limitations for the proposed equations. The

proposed equations are of similar complexity, are rationally derived, and apply for all

concrete reinforcing ratios and to both steel and FRP reinforcing bars. The limitations

for use of Branson’s equation are significantly reduced when shrinkage restraint is taken

into account, but the proposed solutions are more robust.

The requirement to include 15% of the 𝐼𝑒 for negative bending, at each end of each

continuous member per A23.3 clause 9.8.2.4 (CSA 2004), should be ignored for all

methods because work for this report shows that it typically reduces the accuracy of

deflection calculations.

85

4.2 Recommendation for Future Work

4.2.1 Improve Deflection Equation Information Provided to Engineers

Concrete standards and handbooks should provide more information about deflection

calculations. CSA A23.3, for example, provides poor recommendations for what to do

when adjacent span lengths are not similar, and only provides a means of calculating

deflection for uniformly distributed loads. For effective moment of inertia methods, it

is crucial to state and explain the relevant limitations. Assistance should also be

provided about what to do when limitations are not met. Providing equations,

methodology, and examples of how to determine deflection using integration would

enable engineers to have a robust method at their disposal.

To encourage more accurate calculation of deflections, it would be useful to determine

what moment of inertia values, functions, and assumptions are used in common

structural engineering practice. This knowledge could subsequently be used to provide

improved instruction to engineers on accurate stiffness assumptions. Accurate

assumptions, based on load cases, would improve the ability for engineers to correctly

predict the required moment resistance and maximum deflection.

4.2.2 Improve Assumptions for Stiffness

A sequence of pattern loads on a concrete member will often cause more deflection than

selecting the worst load case for a member of constant stiffness. Load-history does not

affect constant stiffness members. Reinforced concrete members are unique and

complex because the amount of cracking and the amount of reinforcement both affect

curvature and deflection. As such, it is unlikely that a constant stiffness member

86

analysis will provide the worst case deflection of a concrete member. The real worst

case for deflection will often require worst case negative bending pattern loading to

occur before the maximum positive bending service pattern.

A thorough investigation of assumptions for stiffness, including load history, should be

performed in order to determine more accurate prediction of deflections. Appendix P

provides a discussion of the effects of cracking at supports. It introduces some ways in

which concrete cracking can affect deflection. For this report, there was little work

done to study the effects of pre-loading on maximum deflections. The study performed

on the idealized members used in this work suggests that when the negative bending

moment envelope is taken into account for both the amount of cracking at supports and

the amount of negative bending reinforcement provided, then effects on deflections are

small.

The reduced cracking moment, 𝑀𝑐𝑟, to account for shrinkage-restraint may dramatically

reduce the errors caused by poor stiffness assumptions. This should also be

investigated.

4.2.3 Investigation of Other Possible Moment of Inertia Equations

Other equations for a constant moment of inertia were developed, investigated, and

dismissed in preliminary work for this report. Effort was concentrated on uniformly

distributed load cases, because that is the most typical loading for concrete members.

The omitted equations, based on cracked length ratios, various bending moment ratios,

and etc., were investigated in preliminary work for this report. These equations had a

87

smaller valid range than proposed solutions or were far more complicated and offered

little improvement. It may be worthwhile to use the techniques provided in this report

to investigate whether any minor changes would improve the proposed equations for

uniformly distributed load or centered point load.

For equal point loads at third points, the improved/corrected equation for 𝐼𝑒∗′ , using ∗

from equation (3 − 11), was derived empirically except that it intentionally includes

the simply supported . It is likely that a simple but more robust and accurate equation

for could be derived. For the case of a single large end-moment, considering the

shape of the positive-cracked portion of the moment diagram, improved results might be

derived by adapting the simply-supported equations for a single third-point load, a

centered point load, or a uniformly distributed load with the same total load. Where

𝑀𝑚𝑎𝑥 𝑀𝑐𝑟⁄ < 1.2 for equal point loads at third points, it should be determined whether

the recommended solution is to use 𝑀𝑚𝑎𝑥 = 1.2𝑀𝑐𝑟, to use the proposed 𝐼𝑒∗′ results, or

whether a better solution for 𝐼𝑒∗′ is required.

88

REFERENCES

ACI Committee 318. 2011. Building code requirements for structural concrete and

commentary. ACI 318-05. American Concrete Institute, Farmington Hills, Mich.

ACI Committee 440. 2006. Guide for the design and construction for concrete

reinforced with FRP bars. ACI 440.1R-06. American Concrete Institute,

Farmington Hills, Mich.

Bischoff, P.H. 2007. Rational model for calculating deflection of reinforced concrete

beams and slabs. Canadian Journal of Civil Engineering, 34(8), 992–1002.

Bischoff, P.H. 2005. Re-evaluation of deflection prediction for concrete beams

reinforced with steel and fiber reinforced polymer bars. Journal of Structural

Engineering, ASCE, 131(5): 752-767.

Bischoff, P.H., and Darabi, M. 2012. Unified approach for computing deflection of steel

and FRP reinforced concrete. ACI SP-284-16: 1-20.

Bischoff, P.H., and Gross, S.P. 2011. Equivalent moment of inertia based on integration

of curvature. Journal of Composites for Construction, ASCE, 15(3): 263-273.

Bischoff, P.H., and Scanlon, A. 2007. Effective moment of inertia for calculating

deflections of concrete members containing steel reinforcement and fiber-

reinforced polymer reinforcement. ACI Structural Journal, 104(1), 68–75.

Branson, D.E. 1965. Instantaneous and time-dependant deflections of simple and

continuous reinforced concrete beams. Alabama Highway Department, Bureau of

Public Roads, Ala. HPR Report No. 7, Part 1.

CAC. 2005. Concrete Design Handbook. 3rd

Edition. Cement Association of Canada,

Ottawa, Ont.

CEN. 2004. Eurocode 2: design of concrete structures – Part 1-1: general rules for

buildings. European prestandard, DD ENV 1992-1-1: 2004, European Committee

for Standardization (CEN), Brussels, Belgium.

CISC. 2009. Handbook of steel construction. 9th

Edition. Canadian Institute of Steel

Construction, Markham, Ont.

CSA. 2012. Design and construction of building structures with fibre-reinforced

polymers. Standard S806-12, Canadian Standards Association (CSA), Toronto,

Ont.

CSA. 2004. Design of concrete structures. Standard A23.3-04, Canadian Standards

Association (CSA), Toronto, Ont.

Gilbert, R.I. 2007. Tension stiffening in lightly reinforced concrete slabs. Journal of

Structural Engineering, ASCE, 133(6): 899-903.

Razaqpur, A.G., Isgor, O.B. 2003. Rational method for calculating deflection of

continuous FRP R/C beams.SP-210: Deflection Control for the Future. ACI

International: 191-208.

Razaqpur, A.G., Svecova, D., and Cheung, M.S. 2000. Rational method for calculating

deflection of fibre-reinforced polymer reinforced beams. ACI Structural Journal,

97(1): 175-195.

Scanlon, A., and Bischoff, P.H. 2008. Shrinkage restraint and loading history effects on

deflections of flexural members. ACI Structural Journal, 105(4), 498–506.

Vesey, S., and Bischoff, P.H. 2011. Designing FRP reinforced concrete for deflection

control. ACI SP-275-03: 1-24.

89

Derivation of 𝑲 for Continuous Linear-Elastic Members Appendix A

𝐾 factors are used to compute deflections with generic elastic deflection equations using

a constant moment of inertia. The following is a derivation of the 𝐾 factors as shown in

Table 2-2 and as introduced in Chapter 6 of the Concrete Design Handbook (CAC

2005). The midspan deflection of a continuous member is the 𝐾 factor multiplied by

the midspan deflection of a simply supported member with the same span, properties,

and midspan moment. The equations for 𝐾 apply to constant stiffness members with

known end-moments. See List of Symbols for equation variable definitions.

Let Δ be the midspan deflection caused by the load applied (as indicated in Figure A-1,

Figure A-2, and Figure A-3). Let downward deflection and downward load on the span

be positive. For the purposes of this report, the end-moments, 𝑀𝐿 and 𝑀𝑅, are zero or

negative, meaning they reduce downwards deflection at midspan. This follows the sign

convention for reinforced concrete design, where bottom reinforcing is considered

positive reinforcing, and top reinforcing is considered negative reinforcing.

Let 𝑀0 be the total static moment, which is the difference between the average end-

moments and the midspan moment; this equals the midspan moment for a simply

support member. Also define 𝑀𝑚 as the midspan moment. This means:

𝑀0 = 𝑀𝑚 −𝑀𝐿 +𝑀𝑅

2

The subsequent three sections of this appendix will show the derivation of the 𝐾 factor

for the three load cases analyzed in this report.

90

Calculate 𝑲 for Point Load at Midspan and Generic End-Moments

Figure A-1 - Midspan Point Load on a Continuous Member

For a point load at midspan on a simply supported span (𝑀𝐿 = 𝑀𝑅 = 0):

𝑀𝑚 = 𝑀0 =𝑃𝐿

4 ; ∆=

𝑃𝐿3

48𝐸𝐼= 𝐾

𝑀𝑚𝐿2

12𝐸𝐼 ; 𝐾 = 1

For a point load at midspan on a continuous member:

∆=𝑃𝐿3

48𝐸𝐼+𝑀𝐿𝐿

2

16𝐸𝐼+𝑀𝑅𝐿

2

16𝐸𝐼= 𝐾

𝑀𝑚𝐿2

12𝐸𝐼

∆=𝐿2

12𝐸𝐼(𝑃𝐿

4+12𝑀𝐿

16+12𝑀𝑅

16) =

𝐿2

12𝐸𝐼(𝑀0 −

3

2(−

𝑀𝐿 +𝑀𝑅

2))

∆=𝐿2

12𝐸𝐼(𝑀0 −

3

2(𝑀0 −𝑀𝑚)) =

𝐿2

12𝐸𝐼(1.5𝑀𝑚 − 0.5𝑀𝑜)

∆= (1.5 − 0.5𝑀0

𝑀𝑚)𝑀𝑚𝐿

2

12𝐸𝐼 ; 𝐾 = 1.5 − 0.5

𝑀0

𝑀𝑚

For a fixed-fixed member with a point load at midspan, 𝑀𝑚 = −𝑀𝐿 = −𝑀𝑅 = 𝑃𝐿/8,

therefore:

𝑀0 = 2𝑀𝑚 ; 𝐾 = 1.5 − 0.5(2𝑀𝑚)

𝑀𝑚=1

2

∆= 𝐾𝑀𝑚𝐿

2

12𝐸𝐼= (

1

2)𝑀𝑚𝐿

2

12𝐸𝐼= (

1

4)𝑀0𝐿

2

12𝐸𝐼 ; ∆=

1

4 of simply supported deflection

91

Calculate 𝑲 for Two Equal Third-Point Loads and Generic End-Moments

Figure A-2 - Equal Point Load at Third Points on a Continuous Member

For an equal point load at third points on a simply supported span (no end-moments):

𝑀𝑚 = 𝑀0 =𝑃𝐿

6 ; ∆=

23𝑃𝐿3

1296𝐸𝐼= 𝐾

23𝑀𝑚𝐿2

216𝐸𝐼 ; 𝐾 = 1

For a continuous member loaded equally at third points:

∆=23𝑃𝐿3


2


2

16𝐸𝐼= 𝐾

23𝑀𝑚𝐿2

216𝐸𝐼

∆=23𝐿2

216𝐸𝐼(𝑃𝐿

6+27𝑀𝐿

46+27𝑀𝑅

46) =

23𝐿2

216𝐸𝐼(𝑀0 −

27

23(−

𝑀𝐿 +𝑀𝑅

2))

∆=23𝐿2

216𝐸𝐼(𝑀0 −

27

23(𝑀0 −𝑀𝑚)) =

23𝐿2

216𝐸𝐼(27

23𝑀𝑚 −

4

23𝑀0)

∆= (27

23−

4

23

𝑀0

𝑀𝑚)23𝑀𝑚𝐿

2

216𝐸𝐼 ; 𝐾 =

27

23−

4

23

𝑀0

𝑀𝑚

For a fixed-fixed member loaded equally at third points, 2𝑀𝑚 = −𝑀𝐿 = −𝑀𝑅 = 𝑃𝐿/9,

therefore:

𝑀0 = 3𝑀𝑚 ; 𝐾 =27

23−

4

23

(3𝑀𝑚)

𝑀𝑚=15

23

∆= 𝐾23𝑀𝑚𝐿

2

216𝐸𝐼= (

15

23)23𝑀𝑚𝐿

2

216𝐸𝐼= (

5

23)23𝑀0𝐿

2

216𝐸𝐼 ; ∆=

5


92

Calculate 𝑲 for Uniformly Distributed Load and Generic End-Moments

Figure A-3 - Uniformly Distributed Load on a Continuous Member

For a uniformly distributed load on a simply supported span (no end-moments):

𝑀𝑚 = 𝑀0 =𝑤𝐿2

8 ; ∆=

5𝑤𝐿4

384𝐸𝐼= 𝐾

5𝑀𝑚𝐿2

48𝐸𝐼 ; 𝐾 = 1

For a uniformly distributed load on a continuous member:

∆=5𝑤𝐿4


2


2

16𝐸𝐼= 𝐾

5𝑀𝑚𝐿2

48𝐸𝐼

∆=5𝐿2

48𝐸𝐼(𝑤𝐿2

8+3𝑀𝐿

5+3𝑀𝑅

5) =

5𝐿2

48𝐸𝐼(𝑀0 −

6

5(−

𝑀𝐿 +𝑀𝑅

2))

∆=5𝐿2

48𝐸𝐼(𝑀0 −

6

5(𝑀0 −𝑀𝑚)) =

5𝐿2

48𝐸𝐼(6

5𝑀𝑚 −

1

5𝑀0)

∆= (1.2 − 0.2𝑀0

𝑀𝑚)5𝑀𝑚𝐿

2

48𝐸𝐼 ; 𝐾 = 1.2 − 0.2

𝑀0

𝑀𝑚

For a uniform load on a fixed-fixed member, 2𝑀𝑚 = −𝑀𝐿 = −𝑀𝑅 = 𝑤𝐿2/12,

therefore:

𝑀0 = 3𝑀𝑚 ; 𝐾 = 1.2 − 0.2(3𝑀𝑚)

𝑀𝑚=3

5


2

48𝐸𝐼= (

3

5)5𝑀𝑚𝐿

2

48𝐸𝐼= (

1

5)5𝑀0𝐿

2

48𝐸𝐼 ; ∆ =

1


93

Bending Deflection by Integration Using Virtual Work Appendix B

Deflection from flexure can be determined by integrating curvature along the beam span

using the principle of virtual work. This method can be understood and replicated

without difficulty, as explained in the following paragraphs. The generic form of this

equation used for integrating curvature using virtual work is:

∆= 𝑀𝜙 = ∫𝑚𝑀

𝐸𝐼

This equation can be more descriptive if it is expanded to denote where each variable is

applicable and which variables are functions of the position along the beam, as follows:

∆𝑝= ∫ 𝑚𝑝(𝑥) (𝑀(𝑥)

𝐸𝐼(𝑥))𝑑𝑥

𝐿

0

The variable 𝑥 is used to denote the position along the beam, and the function is

integrated with respect to 𝑥, as 𝑥 increases from 0 to the beam length, 𝐿.

∆𝑝 denotes that integration will determine the deflection at location 𝑝. The maximum

deflection is typically found at midspan, so that is where point 𝑝 is typically taken. The

subscript, 𝑝, indicating location is usually omitted for midspan deflection because the

midspan deflection is typical, so denoting it is considered redundant. When maximum

deflection is not at midspan, the report denotes maximum deflection as ∆𝑚𝑎𝑥.

The function 𝑚𝑝(𝑥) is the virtual moment function for bending deflection when the

virtual unit load is placed at location 𝑝. As 𝑥 progresses from one end of the beam to

the other, 𝑚𝑝(𝑥) will take into account the effect that the curvature, 𝑀(𝑥)/𝐸𝐼(𝑥), has

on the deflection at point 𝑝. The virtual moment function is explained in the virtual

94

work section of most introductory structural engineering textbooks. For the midspan

deflection of a beam, the virtual moment function is:

𝑚𝐿 2⁄ (𝑥) = 𝑥 2⁄ for 𝑥 ≤ 𝐿 2⁄ and 𝑚𝐿 2⁄ (𝑥) = (𝐿 − 𝑥) 2⁄ for 𝐿 2⁄ ≤ 𝑥 ≤ 𝐿 .

𝑀(𝑥) represents the bending moment as a function of 𝑥 (for the relevant load case).

The 𝑀(𝑥) function is the typical structural design moment function determined from

statics and moment distribution.

The elastic modulus is denoted as 𝐸. It is taken as a constant for almost all materials,

including concrete, so is considered a constant in this integration.

If there is any variation in stiffness, it is accounted for in the moment of inertia term,

𝐼(𝑥). This variation in stiffness can come in the form of varying cross-section shape,

size, cracking, reinforcing, or similar. The 𝐼(𝑥) term is simple for most materials; for

cracked concrete, however, it must be approximated by an effective local (section-

based) moment of inertia. A constant moment of inertia can often be used accurately for

reinforced concrete members and would not require use of the virtual work method.

The function 𝑀(𝑥)/ 𝐸 𝐼(𝑥) indicates the curvature of the beam at each point, 𝑥, along

the beam. For a linear-elastic material there is a known moment curvature response; for

concrete there are assumed moment-curvature responses.

95

Deflection for Simply Supported Member without Appendix C

Tension Stiffening

Razaqpur (Razaqpur et al. 2000, Razaqpur and Isgor 2003) developed equations and

simplifications for FRP-reinforced members. These are used in this report. His work

assumes no tension stiffening effect and no permanent deflection. As such, members

are assumed to follow the idealized moment-curvature relationship seen in Figure C-1:

Figure C-1 - Idealized Moment-Curvature for FRP-Reinforced Member

Razaqpur’s method requires integration of curvature using virtual work or a similar

technique. The solutions by Razaqpur et al. (2000) are provided for simply supported

members in S806 (CSA 2012) and in the following table, Table C-1. In this table,

Razaqpur’s equations are converted to the variable notation of this report. Razaqpur’s

format for these equations requires loads, 𝑃 or 𝑤, and length of the uncracked segments,

𝐿𝑔. Rearranged equations, which are instead based on midspan moment, 𝑀𝑚, and the

cracking moment, 𝑀𝑐𝑟, are also provided in Table C-1. Cantilever cases are not

provided in this table. For continuous members, the relevant equations for the three

load cases used in this report are provided in Appendix G.

96

Table C-1 - Deflection Equations for Idealized FRP-Reinforced Members

See List of Symbols for variables not defined in this appendix.

97

Derivation of Bischoff's 𝜸 Factor for a Uniformly Appendix D

Distributed Load

The following is an independent check of the derivation of Bischoff's factor (Bischoff

and Gross 2011) for a uniformly distributed load. The method of virtual work is used,

as described in Appendix B, to determine the deflection of the member. Because the

member is symmetric about midspan, the derivation integrates the virtual work to

midspan and doubles this value.

To determine the virtual moment function, 𝑚(𝑥), a unit load is applied at the point of

maximum deflection, which is midspan. The member midspan is located at 𝐿/2.

𝑚(𝑥) =𝑥

2 𝑓𝑜𝑟 𝑥 ≤

𝐿

2

A simply supported member with uniformly distributed loading, 𝑤, has a bending

moment function, 𝑀(𝑥), equal to:

𝑀(𝑥) =𝑤𝑥𝐿 − 𝑤𝑥2

2

Deflection, ∆𝑔, if the member were to remain uncracked, equals:

∆𝑔= 2∫ 𝑚(𝑥) (𝑀(𝑥)

𝐸𝑐𝐼𝑔)𝑑𝑥

𝐿2

0

= 2∫ 𝑥 (𝑤𝑥𝐿 − 𝑤𝑥2

2𝐸𝑐𝐼𝑔)𝑑𝑥

𝐿2

0

=5𝑤𝐿4

384𝐸𝑐𝐼𝑐𝑟(𝐼𝑐𝑟𝐼𝑔)

Additional deflection from the cracked segments can be found from the change in

curvature, 𝛿𝜙, relative to the uncracked curvature, as defined by this generic equation:

𝛿𝜙 = 𝛿𝑀

𝐸𝐼=

𝑀(𝑥)

𝐸𝑐𝐼𝑒(𝑥)−𝑀(𝑥)

𝐸𝑐𝐼𝑔=𝜂𝑀(𝑥)

𝐸𝑐𝐼𝑐𝑟[1 − (

𝑀𝑐𝑟

𝑀(𝑥))2

] ; 𝑤ℎ𝑒𝑟𝑒 𝜂 = 1 −𝐼𝑐𝑟𝐼𝑔

Integrating the change in curvature over the length of the cracked region results in the

additional deflection from cracking, 𝛿Δ𝑐𝑟:

98

𝛿Δ𝑐𝑟 = 2∫ 𝑚(𝑥)(𝛿𝜙)𝑑𝑥

𝐿2

𝐿𝑐𝑟

= 2∫𝑥

2(𝜂 (

𝑤𝑥𝐿 − 𝑤𝑥2

2 )

𝐸𝑐𝐼𝑐𝑟[1 − (

𝑀𝑐𝑟

(𝑤𝑥𝐿 − 𝑤𝑥2)/2)2

])𝑑𝑥

𝐿2

𝐿𝑐𝑟

𝛿Δ𝑐𝑟 =𝜂𝑤

2𝐸𝑐𝐼𝑐𝑟∫ 𝑥2

(

(𝐿 − 𝑥) [1 − (𝑀𝑐𝑟

2

𝑤2𝑥2

4(𝐿 − 𝑥)2

)]

)

𝑑𝑥

𝐿2

𝐿𝑐𝑟


2𝐸𝑐𝐼𝑐𝑟∫ 𝐿𝑥2 − 𝑥3 −

4𝑤2𝑀𝑐𝑟

2

𝐿 − 𝑥𝑑𝑥

𝐿2

𝐿𝑐𝑟


2𝐸𝑐𝐼𝑐𝑟(𝐿𝑥3

3|𝐿𝑐𝑟

𝐿2

−𝑥4

4|𝐿𝑐𝑟

𝐿2

+4

𝑤2𝑀𝑐𝑟

2 ln(𝐿 − 𝑥)|𝐿𝑐𝑟

𝐿2) ; let 𝐿𝑐𝑟 = 𝜉

𝐿

2


2𝐸𝑐𝐼𝑐𝑟(8𝐿4

192−8𝜉3𝐿4

192−3𝐿4

192+3𝜉4𝐿4

192+192

1924𝑀𝑐𝑟

2 𝐿4

64𝑀𝑚2 ln (

𝐿 −𝐿2

𝐿 − 𝜉𝐿2

))

𝛿Δ𝑐𝑟 =5𝜂𝑤𝐿4

384𝐸𝑐𝐼𝑐𝑟(1 +

3𝜉4

5−8𝜉3

5−12

5

𝑀𝑐𝑟2

𝑀𝑚2 ln(2 − 𝜉))

These two values are summed to give the midspan deflection, Δ:

Δ = Δ𝑔 + 𝛿Δ𝑐𝑟

Δ =5𝑤𝐿4

384𝐸𝑐𝐼𝑐𝑟(𝐼𝑐𝑟𝐼𝑔) +

5𝜂𝑤

384𝐸𝑐𝐼𝑐𝑟(1 + 0.6𝜉4 − 1.6𝜉3 − 2.4

𝑀𝑐𝑟2

𝑀𝑚2 ln(2 − 𝜉))

Δ =5𝑤𝐿4

384𝐸𝑐𝐼𝑐𝑟[1 − 𝜂

𝑀𝑐𝑟2

𝑀𝑚2 (

1.6𝜉3 − 0.6𝜉4

𝑀𝑐𝑟2/𝑀𝑚

2 + 2.4 ln(2 − 𝜉))]

Set Δ = 5𝑤𝐿4/384𝐸𝑐𝐼𝑒′ and solve for the effective moment of inertia, 𝐼𝑒

′ :

𝐼𝑒′ =

𝐼𝑐𝑟


𝑀𝑚)2 where =

1.6𝜉3 − 0.6𝜉4

(𝑀𝑐𝑟

𝑀𝑚)2 + 2.4 ln(2 − 𝜉)

Note: 𝑀𝑚 =𝑤𝐿2

8 ; 𝜂 = 1 −

𝐼𝑐𝑟𝐼𝑔 ; 𝜉 = 1 − √1 −𝑀𝑐𝑟/𝑀𝑚

99

Analytical Integration for Midspan Deflection Appendix E

When mathematically possible, analytical integration can be used to generate exact-

result equations for deflection; this is referred to as analytical integration in this report.

Analytical integration can be used to find the deflection at any point on a cracked

concrete member if the local stiffness and bending moment along the member are

known. This appendix provides detail of the analytical integration setup and results for

midspan deflection; these results are used throughout this report. The results provide

the midspan deflection for the given load case, using only the bending moment function

provided, with no pre-loading effects. Figure E-1, Figure E-3, and Figure E-2 each

show examples for the bending moment function, the various moments of inertia, and

special lengths required for integration.

Figure E-1 - Lengths to Integration Segments for Example Midspan Point Load

100

Figure E-2 - Lengths to Integration Segments for Example Equal Third-Point Loads

Figure E-3 - Lengths to Integration Segments for Example Uniform Load

101

For cases as indicated Figure E-1 and Figure E-3, the total midspan deflection is:

𝛥 = 𝛥1 + 𝛥2 + 𝛥3 + 𝛥4 + 𝛥5 + 𝛥6 where:

𝛥1 = ∫𝑚(𝑥)𝑀(𝑥)


𝐿

0

𝛥2 = ∫𝑚(𝑥)𝑀(𝑥)

𝐸𝑐𝐼𝑔𝑑𝑥

𝐿

𝐿

𝛥3 = ∫𝑚(𝑥)𝑀(𝑥)


𝐿

𝐿

𝛥4 = ∫𝑚(𝑥)𝑀(𝑥)


𝐿

𝐿

𝛥5 = ∫𝑚(𝑥)𝑀(𝑥)

𝐸𝑐𝐼𝑔𝑑𝑥

𝐿

𝐿

𝛥6 = ∫𝑚(𝑥)𝑀(𝑥)


𝐿

𝐿

The virtual work method used here, and its integration variables, are described in

Appendix B. The lengths 𝐿1 through 𝐿5 can be calculated as a ratio (based on total

moment and end-moments) of the member length, 𝐿.

As shown in Figure E-2, third-point loading is more complicated than the other two

situations. The cases with third-point loading requires more than six integration terms

because the integrating function also changes at 𝐿/3 and 2𝐿/3. These additional terms

are shown with the S806 (CSA 2012) method in Appendix G. A second complication is

that more variables would actually need to be added to account for situations where

midspan cracking begins or ends between the third-points. Because of this second

complication, the exact results for third-point loading were obtained using only

numerical integration for data provided in this report.

Bending Moment and Virtual Moment Equations

For all loads, the net midspan moment is:

𝑀𝑚 =𝑀𝐿

2+𝑀𝑅

2+𝑀0

For all loads, virtual moment for midspan deflection is:

for 0 ≤ 𝑥 ≤𝐿

2 ∶ 𝑚(𝑥) =

𝑥

2

102

for 𝐿

2≤ 𝑥 ≤ 𝐿 ∶ 𝑚(𝑥) =

𝐿 − 𝑥

2

For single midspan point load:


2 ∶ 𝑀(𝑥) = 𝑀𝐿 + (2𝑀0 −𝑀𝐿 +𝑀𝑅)

𝑥

𝐿

for 𝐿

2≤ 𝑥 ≤ 𝐿 ∶ 𝑀(𝑥) = 𝑀𝑅 + (2𝑀0 −𝑀𝑅 +𝑀𝐿)

𝐿 − 𝑥

𝐿

For equal third-point loading:


3 ∶ 𝑀(𝑥) = 𝑀𝐿 + (3𝑀0 −𝑀𝐿 +𝑀𝑅)

𝑥

𝐿

for 𝐿

3≤ 𝑥 ≤

2𝐿

3 ∶ 𝑀(𝑥) = 𝑀0 +𝑀𝐿 + (𝑀𝑅 −𝑀𝐿)

𝑥

𝐿

for 2𝐿

3≤ 𝑥 ≤ 𝐿 ∶ 𝑀(𝑥) = 3𝑀0 +𝑀𝐿 + (𝑀𝑅 −𝑀𝐿 − 3𝑀0)

𝑥

𝐿

For a uniformly distributed load:

𝑀(𝑥) = 𝑀𝐿 + (4𝑀0 −𝑀𝐿 +𝑀𝑅)𝑥

𝐿 − 4𝑀0

𝑥2

𝐿2

For all loads, this work uses the following local effective moment of inertia:

for −𝑀𝑐𝑟 ≤ 𝑀(𝑥) ≤ 𝑀𝑐𝑟 ∶ 𝐼𝑒(𝑥) = 𝐼𝑔

for −𝑀𝑐𝑟 < 𝑀(𝑥) or 𝑀(𝑥) > 𝑀𝑐𝑟 ∶ 𝐼𝑒(𝑥) =𝐼𝑐𝑟


𝑀(𝑥))2 where 𝜂 = 1 −

𝐼𝑐𝑟𝐼𝑔

Lengths to where the Function being Integrated Changes

Centered Point Load:

𝐿1 through 𝐿5 are the lengths to where the function being integrated changes for a

member with a centered point load, as per Figure E-1. For this case, these lengths can

be substituted neatly and are removed from the final deflection equations. The

equations for the end/total moment ratios are 𝛼𝐿 = 𝑀𝐿/𝑀0 and 𝛼𝑅 = 𝑀𝑅/𝑀0.

103

𝐿1 =−𝛼𝐿𝑀0 −𝑀𝑐𝑟

(2 − 𝛼𝐿 + 𝛼𝑅)𝑀0𝐿 ; 𝐿2 =

−𝛼𝐿𝑀0 +𝑀𝑐𝑟

(2 − 𝛼𝐿 + 𝛼𝑅)𝑀0𝐿 ; 𝐿3 =

𝐿

2

𝐿4 = 𝐿 +𝛼𝑅𝑀0 −𝑀𝑐𝑟

(2 + 𝛼𝐿 − 𝛼𝑅)𝑀0𝐿 ; 𝐿5 = 𝐿 +

𝛼𝑅𝑀0 +𝑀𝑐𝑟

(2 + 𝛼𝐿 − 𝛼𝑅)𝑀0𝐿

For equal third-point loading:

For the deflection of a member with equal point load at third-points, 𝐿1 through 𝐿5, per

Figure E-2, are simple ratios (based on total and end-moments) of the member length, 𝐿.

These simple ratios can also be substituted in neatly, so they can be removed from the

final equations. The equation for 𝐿2 is only valid when 𝑀(𝑥 = 𝐿 3⁄ ) > 𝑀𝑐𝑟 and the

equation for 𝐿4 is only valid when 𝑀(𝑥 = 2𝐿 3⁄ ) > 𝑀𝑐𝑟. The equations for the

alternate situations for 𝐿2 and 𝐿4 are similar but are not provided.

𝐿1 =−𝑀𝐿 −𝑀𝑐𝑟

𝑀𝑅 −𝑀𝐿 + 3𝑀0𝐿 ; 𝐿2 =

−𝑀𝐿 +𝑀𝑐𝑟

𝑀𝑅 −𝑀𝐿 + 3𝑀0𝐿 ; 𝐿3 =

𝐿

2

𝐿4 =𝑀𝐿 − 3𝑀0 +𝑀𝑐𝑟

𝑀𝑅 −𝑀𝐿 − 3𝑀0𝐿 ; 𝐿5 =

𝑀𝐿 − 3𝑀0 −𝑀𝑐𝑟

𝑀𝑅 −𝑀𝐿 − 3𝑀0𝐿

In this report, these lengths are used in Appendix G. The length to cracking if positive-

bending cracking begins (or ends) between the third-points is not difficult, but those

lengths are not used in this report.


The following are the equations for the lengths, 𝐿1 through 𝐿5 per Figure E-3, to where

the function being integrated changes for a member with a uniformly distributed load.

In this case, 𝐿1, 𝐿2, 𝐿4, and 𝐿5 are complicated expressions, so they remain in the final

equations. Again note that 𝐿3 = 𝐿/2, 𝛼𝐿 = 𝑀𝐿/𝑀0, and 𝛼𝑅 = 𝑀𝑅/𝑀0.

104

𝐿1 = (4 − 𝛼𝐿 + 𝛼𝑅 −√(4 − 𝛼𝐿 + 𝛼𝑅)2 + 16𝑀𝑐𝑟

𝑀0+ 16𝛼𝐿)

𝐿

8

𝐿2 = (4 − 𝛼𝐿 + 𝛼𝑅 −√(4 − 𝛼𝐿 + 𝛼𝑅)2 − 16𝑀𝑐𝑟

𝑀0+ 16𝛼𝐿)

𝐿

8

𝐿4 = (4 − 𝛼𝐿 + 𝛼𝑅 +√(4 − 𝛼𝐿 + 𝛼𝑅)2 − 16𝑀𝑐𝑟

𝑀0+ 16𝛼𝐿)

𝐿

8

𝐿5 = (4 − 𝛼𝐿 + 𝛼𝑅 +√(4 − 𝛼𝐿 + 𝛼𝑅)2 + 16𝑀𝑐𝑟

𝑀0+ 16𝛼𝐿)

𝐿

8

Midspan Deflection of Midspan-Point Loaded Member with End-Moments

𝛥1 =𝐿2 (𝛼𝐿

3𝑀03 − 3𝛼𝐿𝑀𝑐𝑟

2𝑀0 + (6𝜂𝐿 − 2)𝑀𝑐𝑟3 + 6𝜂𝐿𝛼𝐿𝑀𝑐𝑟

2𝑀0 [1 + ln (−𝑀𝑐𝑟

𝛼𝐿𝑀0)])

12𝐸𝑐𝐼𝑐𝑟 𝐿(2 − 𝛼𝐿 + 𝛼𝑅)2𝑀02

𝛥2 =𝐿2𝑀𝑐𝑟

3

3𝐸𝑐𝐼𝑔(2 − 𝛼𝐿 + 𝛼𝑅)2𝑀0

2

𝛥3 =𝐿2

48𝐸𝑐𝐼𝑐𝑟 𝑚(𝛼𝐿 − 𝛼𝑅 − 2)2𝑀02 (12𝛼𝐿𝑀𝑐𝑟

2𝑀0 + 24𝜂𝑚𝑀𝑐𝑟3 − 8𝑀𝑐𝑟

3

+𝑀03(8 − 2𝛼𝐿

3 − 6𝛼𝐿2 − 3𝛼𝐿

2𝛼𝑅 + 12𝛼𝑅 + 6𝛼𝑅2 + 𝛼𝑅

3)

− 12𝜂𝑚𝑀𝑐𝑟2𝑀0 (𝛼𝑅 + 2 + 𝛼𝐿 [1 + 2 ln (

2𝑀𝑐𝑟

𝑀0(𝛼𝐿 + 𝛼𝑅 + 2))]))

𝛥4 =𝐿2

48𝐸𝑐𝐼𝑐𝑟 𝑚(𝛼𝑅 − 𝛼𝐿 − 2)2𝑀02 (12𝛼𝑅𝑀𝑐𝑟

2𝑀0 + 24𝜂𝑚𝑀𝑐𝑟3 − 8𝑀𝑐𝑟

3

+𝑀03(8 − 2𝛼𝑅

3 − 6𝛼𝑅2 − 3𝛼𝑅

2𝛼𝐿 + 12𝛼𝐿 + 6𝛼𝐿2 + 𝛼𝐿

3)

− 12𝜂𝑚𝑀𝑐𝑟2𝑀0 (𝛼𝐿 + 2 + 𝛼𝑅 [1 + 2 ln (

2𝑀𝑐𝑟

𝑀0(𝛼𝐿 + 𝛼𝑅 + 2))]))

𝛥5 =𝐿2𝑀𝑐𝑟

3

3𝐸𝑐𝐼𝑔(2 + 𝛼𝐿 − 𝛼𝑅)2𝑀02

𝛥6 =𝐿2 (𝛼𝑅

3𝑀03 − 3𝛼𝑅𝑀𝑐𝑟

2𝑀0 + (6𝜂𝑅 − 2)𝑀𝑐𝑟3 + 6𝜂𝑅𝛼𝑅𝑀𝑐𝑟

2𝑀0 [1 + ln (−𝑀𝑐𝑟

𝛼𝑅𝑀0)])

12𝐸𝑐𝐼𝑐𝑟 𝑅(2 + 𝛼𝐿 − 𝛼𝑅)2𝑀02

105

When the left end moment < 𝑀𝑐𝑟:

𝛥1&2 =𝐿2(2𝑀𝑐𝑟

3 − 3𝛼𝐿𝑀𝑐𝑟2𝑀0 + 𝛼𝐿

3𝑀03)

12𝐸𝑐𝐼𝑔(2 − 𝛼𝐿 + 𝛼𝑅)2𝑀02

When the right end moment < 𝑀𝑐𝑟:

𝛥5&6 =𝐿2(2𝑀𝑐𝑟

3 − 3𝛼𝑅𝑀𝑐𝑟2𝑀0 + 𝛼𝑅

3𝑀03)

12𝐸𝑐𝐼𝑔(2 + 𝛼𝐿 − 𝛼𝑅)2𝑀02

Total midspan deflection (note these results are only valid if 𝑀𝑚 > 𝑀𝑐𝑟):

𝛥 = 𝛥1 + 𝛥2 + 𝛥3 + 𝛥4 + 𝛥5 + 𝛥6 (If both ends and midspan are cracked)

𝛥 = 𝛥1&2 + 𝛥3 + 𝛥4 + 𝛥5&6 (If only midspan is cracked)

𝛥 = 𝛥1&2 + 𝛥3 + 𝛥4 + 𝛥5 + 𝛥6 (If only right end and midspan are cracked)

𝛥 = 𝛥1 + 𝛥2 + 𝛥3 + 𝛥4 + 𝛥5&6 (If only left end and midspan are cracked)

Midspan Deflection of Third-Point Loaded Member with End-Moments

The equations for midspan deflection of equal third-point loading are not provided as

part of this report. While solutions with the added complication are practicable, this

complication does not affect numerical integration, so the numerical solution was used

exclusively for third-point loading throughout this report.

Midspan Deflection of Member with Uniform Load and End-Moments

For a uniformly distributed load, the lengths from the right end of the beam to the end of

negative cracking and to the beginning of positive cracking are more convenient than

using 𝐿4 and 𝐿5, as provided previously in this appendix. Therefore use:

𝐿𝑅4 = 𝐿 − 𝐿4 ; 𝐿𝑅5 = 𝐿 − 𝐿5

Then, similar to the centered point load case, define:

106

𝛥1 = (𝑀0𝐿

2

48𝐸𝑐𝐼𝑐𝑟𝐿)

(

3𝑛𝐿 (

𝑀𝑐𝑟

𝑀0)2

ln (𝛼𝐿𝐿

2 − 4𝐿12 + 4𝐿1𝐿 − 𝛼𝐿𝐿1𝐿 + 𝛼𝑅𝐿1𝐿

𝛼𝐿𝐿2)

+1

𝐿4(12𝛼𝐿𝐿1

2𝐿2 + 32𝐿13𝐿 − 24𝐿1

4 − 8𝛼𝐿𝐿13𝐿 + 8𝛼𝑅𝐿1

3𝐿)

+3𝑛𝐿(𝛼𝐿 − 4 − 𝛼𝑅)

√(4 − 𝛼𝑅 + 𝛼𝐿)2 + 16𝛼𝑅(𝑀𝑐𝑟

𝑀0)2

∗ ln

(

√(4 − 𝛼𝑅 + 𝛼𝐿)2 + 16𝛼𝑅 − (−4 + 𝛼𝐿 − 𝛼𝑅)

√(4 − 𝛼𝑅 + 𝛼𝐿)2 + 16𝛼𝑅 + (−4 + 𝛼𝐿 − 𝛼𝑅)

√(4 − 𝛼𝑅 + 𝛼𝐿)2 + 16𝛼𝑅 − (𝛼𝐿 − 4 + 8𝐿1 𝐿⁄ − 𝛼𝑅)

√(4 − 𝛼𝑅 + 𝛼𝐿)2 + 16𝛼𝑅 + (𝛼𝐿 − 4 + 8𝐿1 𝐿⁄ − 𝛼𝑅))

)

𝛥2 = 𝑀0𝐿

2

2𝐸𝑐𝐼𝑔((4 − 𝛼𝐿 + 𝛼𝑅) (

𝐿23 − 𝐿1

3

3𝐿3) + 𝛼𝐿 (

𝐿22 − 𝐿1

2

2𝐿2) − (

𝐿24 − 𝐿1

4

𝐿4))

𝛥3 = (𝑀0𝐿

2

96𝐸𝑐𝐼𝑐𝑟𝑚)

(

6𝜂𝑚 (

𝑀𝑐𝑟

𝑀0)2

ln ((−𝛼𝐿 − 2 − 𝛼𝑅)𝐿

2

2 (−𝛼𝐿𝐿2 + 4𝐿22 + 𝐿2𝐿(𝛼𝐿 − 4 − 𝛼𝑅))

)

+1

𝐿4(−24𝛼𝐿𝐿2

2𝐿2 + 16𝛼𝐿𝐿23𝐿 + 48𝐿2

4 − 64𝐿23𝐿 − 16𝛼𝑅𝐿2

3𝐿

+ 5𝐿4 + 2𝛼𝑅𝐿4 + 4𝛼𝐿𝐿

4) +6𝑛𝑚(𝛼𝐿 − 4 − 𝛼𝑅)

√(4 − 𝛼𝐿 + 𝛼𝑅)2 + 16𝛼𝐿(𝑀𝑐𝑟

𝑀0)2

∗ ln

(

√(4 − 𝛼𝐿 + 𝛼𝑅)2 + 16𝛼𝐿 − (𝛼𝐿 − 4 + 8𝐿2 𝐿⁄ − 𝛼𝑅)

√(4 − 𝛼𝐿 + 𝛼𝑅)2 + 16𝛼𝐿 + (𝛼𝐿 − 4 + 8𝐿2 𝐿⁄ − 𝛼𝑅)

√(4 − 𝛼𝐿 + 𝛼𝑅)2 + 16𝛼𝐿 − (𝛼𝐿 − 𝛼𝑅)

√(4 − 𝛼𝐿 + 𝛼𝑅)2 + 16𝛼𝐿 + (𝛼𝐿 − 𝛼𝑅) )

)

107

𝛥4 = (𝑀0𝐿

2

96𝐸𝑐𝐼𝑐𝑟𝑚)

(

6𝜂𝑚 (

𝑀𝑐𝑟

𝑀0)2

ln ((−𝛼𝑅 − 2 − 𝛼𝐿)𝐿

2

2(−𝛼𝑅𝐿2 + 4𝐿𝑅42 + 𝐿𝑅4𝐿(𝛼𝑅 − 4 − 𝛼𝐿))

)

+1

𝐿4(−24𝛼𝑅𝐿𝑅4

2 𝐿2 + 16𝛼𝑅𝐿𝑅43 𝐿 + 48𝐿𝑅4

4 − 64𝐿𝑅43 𝐿 − 16𝛼𝐿𝐿𝑅4

3 𝐿

+ 5𝐿4 + 2𝛼𝐿𝐿4 + 4𝛼𝑅𝐿

4) +6𝑛𝑚(𝛼𝑅 − 4 − 𝛼𝐿)

√(4 − 𝛼𝑅 + 𝛼𝐿)2 + 16𝛼𝑅(𝑀𝑐𝑟

𝑀0)2

∗ ln

(

√(4 − 𝛼𝑅 + 𝛼𝐿)2 + 16𝛼𝑅 − (𝛼𝑅 − 4 + 8𝐿𝑅4 𝐿⁄ − 𝛼𝐿)

√(4 − 𝛼𝑅 + 𝛼𝐿)2 + 16𝛼𝑅 + (𝛼𝑅 − 4 + 8𝐿𝑅4 𝐿⁄ − 𝛼𝐿)

√(4 − 𝛼𝑅 + 𝛼𝐿)2 + 16𝛼𝑅 − (𝛼𝑅 − 𝛼𝐿)

√(4 − 𝛼𝑅 + 𝛼𝐿)2 + 16𝛼𝑅 + (𝛼𝑅 − 𝛼𝐿) )

)

𝛥5 = 𝑀0𝐿

2

2𝐸𝑐𝐼𝑔((4 − 𝛼𝑅 + 𝛼𝐿) (

𝐿𝑅43 − 𝐿𝑅5

3

3𝐿3) + 𝛼𝑅 (


2

2𝐿2) − (


4

𝐿4))

𝛥6 = (𝑀0𝐿

2

48𝐸𝑐𝐼𝑐𝑟𝑅)

(

1

𝐿4(12𝛼𝑅𝐿𝑅5

2 𝐿2 + 32𝐿𝑅53 𝐿 − 24𝐿𝑅5

4 − 8𝛼𝑅𝐿𝑅53 𝐿 + 8𝛼𝐿𝐿𝑅5

3 𝐿)

+ 3𝑛𝑅 (𝑀𝑐𝑟

𝑀0)2

ln (𝛼𝑅𝐿

2 − 4𝐿𝑅52 + 4𝐿𝑅5𝐿 − 𝛼𝑅𝐿𝑅5𝐿 + 𝛼𝐿𝐿𝑅5𝐿

𝛼𝑅𝐿2)

+3𝑛𝑅(𝛼𝑅 − 4 − 𝛼𝐿)

√(4 − 𝛼𝐿 + 𝛼𝑅)2 + 16𝛼𝐿(𝑀𝑐𝑟

𝑀0)2

∗ ln

(

√(4 − 𝛼𝐿 + 𝛼𝑅)2 + 16𝛼𝐿 − (−4 + 𝛼𝑅 − 𝛼𝐿)

√(4 − 𝛼𝐿 + 𝛼𝑅)2 + 16𝛼𝐿 + (−4 + 𝛼𝑅 − 𝛼𝐿)

√(4 − 𝛼𝐿 + 𝛼𝑅)2 + 16𝛼𝐿 − (𝛼𝑅 − 4 + 8𝐿𝑅5 𝐿⁄ − 𝛼𝐿)

√(4 − 𝛼𝐿 + 𝛼𝑅)2 + 16𝛼𝐿 + (𝛼𝑅 − 4 + 8𝐿𝑅5 𝐿⁄ − 𝛼𝐿))

)

108

When the left end moment < 𝑀𝑐𝑟 ∶

𝛥1&2 =𝑀0𝐿

2

2𝐸𝑐𝐼𝑔((4 − 𝛼𝐿 + 𝛼𝑅)𝐿2

3

3𝐿3 −

𝐿24

𝐿4+𝛼𝐿𝐿2

2

2𝐿2)

When the right end moment < 𝑀𝑐𝑟 ∶

𝛥5&6 =𝑀0𝐿

2

2𝐸𝑐𝐼𝑔((4 − 𝛼𝑅 + 𝛼𝐿)𝐿𝑅4

3

3𝐿3 −

𝐿𝑅44

𝐿4+𝛼𝑅𝐿𝑅4

2

2𝐿2)

Total midspan deflection (note these results are only valid if 𝑀𝑚 > 𝑀𝑐𝑟):





109

Analytical Results Simplified for Fixed-Fixed Midspan Appendix F

Point Load

This appendix provides an example of simplifying the equations provided in Appendix

E for the midspan deflection of a midspan point-loaded member. This prismatic

member example is assumed to be: fixed-fixed such that 𝑀𝑚 = −𝑀𝐿 = −𝑀𝑅, cracked

(𝑀𝑚 > 𝑀𝑐𝑟), not previously loaded, reinforced with equal top and bottom reinforcing,

and far more simplifiable than typical.

Equations: 𝑀0 = 2𝑀𝑚 ; 𝜂 = 𝜂𝐿 = 𝜂𝑅 = 1 − 𝐼𝑐𝑟/𝐼𝑔 ; 𝐼𝑐𝑟 = 𝐼𝑐𝑟 𝐿 = 𝐼𝑐𝑟 𝑚 = 𝐼𝑐𝑟 𝑅

𝜉1 = 1 + ln(2𝑀𝑐𝑟

𝑀0) ; 𝜉2 = 1 − ln (

2𝑀𝑐𝑟

𝑀0) ; 𝜉1 + 𝜉2 = 2

𝛥1 = 𝛥6 =−𝑀0

8

3

+ 3𝑀0

2 𝑀𝑐𝑟

2

+ 6𝜂𝑀𝑐𝑟3 − 2𝑀𝑐𝑟

3 − 6𝜂𝑀0

2 𝑀𝑐𝑟2𝜉

1

48𝐸𝑐𝐼𝑐𝑟𝑀02/𝐿2

𝛥2 = 𝛥5 =𝐿2𝑀𝑐𝑟

3

12𝐸𝑐𝐼𝑔𝑀02

𝛥3 = 𝛥4 =−6𝑀𝑐𝑟

2𝑀0 − 18𝜂𝑀𝑐𝑟2𝑀0 + 24𝜂𝑀𝑐𝑟

3 − 8𝑀𝑐𝑟3 +

52𝑀0

3 − 12𝜂𝑀𝑐𝑟2𝑀0𝜉2

192𝐸𝑐𝐼𝑐𝑟𝑀02/𝐿2

𝛥 =16𝐿2𝑀𝑐𝑟

3


+𝐿2

96𝐸𝑐𝐼𝑐𝑟𝑀02 (−

𝑀0

2

3

+ 6𝑀𝑐𝑟2𝑀0 + 24𝜂𝑀𝑐𝑟

3 − 8𝑀𝑐𝑟3

− 12𝜂𝑀𝑐𝑟2𝑀0𝜉1)

+𝐿2

96𝐸𝑐𝐼𝑐𝑟𝑀02 (5

2𝑀0

3 − 6𝑀𝑐𝑟2𝑀0 − 12𝜂𝑀𝑐𝑟

2𝑀0𝜉2 + 24𝜂𝑀𝑐𝑟3

− 8𝑀𝑐𝑟3)

𝛥 =𝐿2

96𝐸𝑐𝐼𝑐𝑟𝑀02 (2𝑀0

3 − 24𝜂𝑀𝑐𝑟2𝑀0 + 48𝜂𝑀𝑐𝑟

3 − 16𝑀𝑐𝑟3) +

16𝐿2𝑀𝑐𝑟3


110

𝛥 =𝐿2(2𝑀0

3 − 24𝜂𝑀𝑐𝑟2𝑀0)

96𝐸𝑐𝐼𝑐𝑟𝑀02 +

16𝐿2𝑀𝑐𝑟3

96𝐸𝑐𝑀02 (

3𝜂 − 1

𝐼𝑐𝑟+1

𝐼𝑔)

𝛥 =(𝑀0

3 − 12𝜂𝑀𝑐𝑟2𝑀0 + 16𝜂𝑀𝑐𝑟

3)𝐿2

48𝐸𝑐𝐼𝑐𝑟𝑀02

Interestingly, this case simplifies to using a identical to the simply supported solution:

𝛥 = 𝐾𝑀𝑚𝐿

2

12𝐸𝑐𝐼𝑒′ where 𝐾 =

1

2 ; 𝐼𝑒

′ =𝐼𝑐𝑟

(1 − 𝜂(𝑀𝑐𝑟 𝑀𝑚⁄ )2) ; = 3 − 2

𝑀𝑐𝑟

𝑀𝑚

111

Integration using CSA S806 / Razaqpur’s Method Appendix G

The S806 (CSA 2012) method was intended only for FRP reinforced concrete members,

but it is used in this report to demonstrate the effect of tension stiffening on both steel

and FRP reinforced members. The method is explained and some equations for

deflection of continuous members, neglecting tension stiffening, are provided in this

appendix.

The S806 method does not account for tension stiffening in the cracked region of the

member; thus, 𝐼𝑐𝑟 is used when the moment exceeds 𝑀𝑐𝑟. This is equivalent to

providing a tension stiffening factor of 𝛽 = 0; this is drastically different than

𝛽 = 𝑀𝑐𝑟/𝑀(𝑥) as provided by Bischoff and Gross (2011). This assumption is

reasonable when 𝑀𝑚𝑎𝑥 𝑀𝑐𝑟⁄ > 3 and is overly conservative otherwise. Razaqpur et al.

(2000) provide simplified results for simply supported members using the S806 method.

Razaqpur and Isgor (2003) also provide simplified equations for three typical

continuous members. Razaqpur’s work includes other simplifying assumptions that are

not used in this report.

The equations which follow in this appendix (referred to in this report as the S806

integration method) are derived by integrating curvature, while ignoring tension

stiffening, to provide deflection equations for the cases indicated. Derivation and use of

the S806 method are similar to Bischoff’s method (Bischoff and Gross 2011). Using

virtual work, the deflection is found by using the same steps as described in Appendix

E, except the moment of inertia at a crack, 𝐼𝑐𝑟, is used for the full cracked segments of

112

the member. In this appendix, 𝑀(𝑥) is as defined in Appendix E and the virtual load

function, 𝑚(𝑥), is again defined to provide midspan deflection.

For certain continuous member cases, such as most cases with equal end-moments,

equations resulting from using the S806 method can be simplified. Razaqpur also

recommends simplifying equations by removing some portions of the calculation which

approximately cancel each other out. Razaqpur also assumes that the reinforcing bar

area for the end-moment and the midspan moment are equal. The differences caused by

these simplifying assumptions are typically much less significant than ignoring tension

stiffening.

S806 method for centered point load:

Figure G-1 - Lengths to Integration Segments for Example Centered Point Load

113

Results using the S806 method for a continuous member with a centered point load

follow. Appendix E provides equations for 𝐿1, 𝐿2, 𝐿4, and 𝐿5 with this loading.

𝛥1 = ∫m(𝑥)𝑀(𝑥)

E 𝐼𝑐𝑟𝐿𝑑𝑥

𝐿

0

=𝐿2(𝛼𝐿


2𝑀0 − 2𝑀𝑐𝑟3)

12𝐸𝑐𝐼𝑐𝑟𝐿(2 − 𝛼𝐿 + 𝛼𝑅)2𝑀02

𝛥2 = ∫m(𝑥)𝑀(𝑥)

E 𝐼𝑔𝑑𝑥

𝐿

𝐿

=𝐿2𝑀𝑐𝑟

3


𝛥3 = ∫m(𝑥)𝑀(𝑥)

E 𝐼𝑐𝑟𝑚𝑑𝑥

𝐿2

𝐿

=𝐿2 (12𝛼𝐿𝑀𝑐𝑟

2𝑀0 − 8𝑀𝑐𝑟3 +𝑀0

3(8 − 2𝛼𝐿3 − 6𝛼𝐿

2 − 3𝛼𝐿2𝛼𝑅 + 12𝛼𝑅 + 6𝛼𝑅

2 + 𝛼𝑅3))

48𝐸𝑐𝐼𝑐𝑟𝑚(2 − 𝛼𝐿 + 𝛼𝑅)2𝑀02

𝛥4 = ∫m(𝑥)𝑀(𝑥)


𝐿

𝐿2

=𝐿2 (12𝛼𝑅𝑀𝑐𝑟

2𝑀0 − 8𝑀𝑐𝑟3 +𝑀0

3(8 − 2𝛼𝑅3 − 6𝛼𝑅

2 − 3𝛼𝑅2𝛼𝐿 + 12𝛼𝐿 + 6𝛼𝐿

2 + 𝛼𝐿3))

48𝐸𝑐𝐼𝑐𝑟𝑚(2 + 𝛼𝐿 − 𝛼𝑅)2𝑀02

𝛥5 = ∫m(𝑥)𝑀(𝑥)

E 𝐼𝑔𝑑𝑥

𝐿

𝐿

=𝐿2𝑀𝑐𝑟

3


𝛥6 = ∫m(𝑥)𝑀(𝑥)

E 𝐼𝑐𝑟𝑅𝑑𝑥

𝐿

𝐿

=𝐿2(𝛼𝑅


2𝑀0 − 2𝑀𝑐𝑟3)

12𝐸𝑐𝐼𝑐𝑟𝑅(2 + 𝛼𝐿 − 𝛼𝑅)2𝑀02

𝛥1&2 = ∫m(𝑥)𝑀(𝑥)

E 𝐼𝑔𝑑𝑥

𝐿

0

=𝐿2(𝛼𝐿


2𝑀0 + 2𝑀𝑐𝑟3)

12𝐸𝑐𝐼𝑔(2 − 𝛼𝐿 + 𝛼𝑅)2𝑀02

𝛥5&6 = ∫m(𝑥)𝑀(𝑥)

E 𝐼𝑔𝑑𝑥

𝐿

𝐿

=𝐿2(𝛼𝑅


2𝑀0 + 2𝑀𝑐𝑟3)

12𝐸𝑐𝐼𝑔(2 + 𝛼𝐿 − 𝛼𝑅)2𝑀02

Total midspan deflection:





114

S806 method for equal third-point loading:

Figure L.2 – Lengths to Integration Segments for Example Third-Point Loads

Results using the S806 method for an equal third-point loaded continuous member

follow. Appendix E provides equations for 𝐿1, 𝐿2, 𝐿4, and 𝐿5 with this loading.

𝛥1 = ∫m(𝑥)𝑀(𝑥)


𝐿

0

=𝐿2(𝛼𝐿


2𝑀0 − 2𝑀𝑐𝑟3)

12𝐸𝑐𝐼𝑐𝑟𝐿(32 − 𝛼𝐿 + 𝛼𝑅)2𝑀02

𝛥2 = ∫m(𝑥)𝑀(𝑥)

E 𝐼𝑔𝑑𝑥

𝐿

𝐿

=𝐿2𝑀𝑐𝑟

3


115

𝛥3𝐴 = ∫m(𝑥)𝑀(𝑥)


𝐿3

𝐿

=𝐿2(81𝛼𝐿𝑀𝑐𝑟

2𝑀0 − 54𝑀𝑐𝑟3)


+𝐿2𝑀0

3(54 − 36𝛼𝐿2 − 20𝛼𝐿

3 − 12𝛼𝐿2𝛼𝑅 + 18𝛼𝐿𝛼𝑅 + 3𝛼𝐿𝛼𝑅

2)


+𝐿2𝑀0

3(27𝛼𝐿 − 54𝛼𝑅 + 18𝛼𝑅2 + 2𝛼𝑅

3)


2

𝛥3𝐵 = ∫m(𝑥)𝑀(𝑥)


𝐿2

𝐿3

=𝐿2𝑀0(26𝛼𝐿 + 19𝛼𝑅 + 45)

1296𝐸𝑐𝐼𝑐𝑟𝑚

𝛥3𝐶 = ∫m(𝑥)𝑀(𝑥)


2𝐿3

𝐿2

=𝐿2𝑀0(19𝛼𝐿 + 26𝛼𝑅 + 45)

1296𝐸𝑐𝐼𝑐𝑟𝑚

𝛥3𝐷 = ∫m(𝑥)𝑀(𝑥)


𝐿

2𝐿3

=𝐿2(324𝛼𝐿𝑀𝑐𝑟(𝛼𝑅 − 2𝛼𝐿)𝑀0

2 + 81(𝛼𝑅 − 4𝛼𝐿)𝑀𝑐𝑟2𝑀0 − 54𝑀𝑐𝑟

3)


+𝐿2𝑀0

3(54 + 18𝛼𝐿2 − 430𝛼𝐿

3 + 327𝛼𝐿2𝛼𝑅 + 18𝛼𝐿𝛼𝑅 − 12𝛼𝐿𝛼𝑅

2)


+𝐿2𝑀0

3(54𝛼𝐿 + 27𝛼𝑅 − 36𝛼𝑅2 − 20𝛼𝑅

3)


𝛥5 = ∫m(𝑥)𝑀(𝑥)

E 𝐼𝑔𝑑𝑥

𝐿

𝐿

=𝐿2(12𝛼𝐿𝑀0

2 +𝑀𝑐𝑟2 − 6𝛼𝐿𝛼𝑅𝑀0

2)𝑀𝑐𝑟


116

𝛥6 = ∫m(𝑥)𝑀(𝑥)


𝐿

𝐿

=𝐿2 ((16𝛼𝐿

3 − 12𝛼𝐿2𝛼𝑅 + 𝛼𝑅

3)𝑀03 − 3(𝛼𝑅 − 4𝛼𝐿)𝑀𝑐𝑟

2𝑀0)


+𝐿2(12(𝛼𝐿𝛼𝑅 − 2𝛼𝐿

2)𝑀02𝑀𝑐𝑟 − 2𝑀𝑐𝑟

3)


𝛥1&2 = ∫m(𝑥)𝑀(𝑥)

E 𝐼𝑔𝑑𝑥

𝐿

0

=𝐿2(𝛼𝐿


2𝑀0 + 2𝑀𝑐𝑟3)


2

𝛥5&6 = ∫m(𝑥)𝑀(𝑥)

E 𝐼𝑔𝑑𝑥

𝐿

𝐿

=𝐿2 ((16𝛼𝐿

3 − 12𝛼𝐿2𝛼𝑅 + 𝛼𝑅

3)𝑀03 − 3(𝛼𝑅 − 4𝛼𝐿)𝑀𝑐𝑟

2𝑀0)

12𝐸𝑐𝐼𝑔(3 + 𝛼𝐿 − 𝛼𝑅)2𝑀02

+𝐿2(12(2𝛼𝐿

2 − 𝛼𝐿𝛼𝑅)𝑀02𝑀𝑐𝑟 + 2𝑀𝑐𝑟

3)

12𝐸𝑐𝐼𝑔(3 + 𝛼𝐿 − 𝛼𝑅)2𝑀02

Total midspan deflection:

Note these results are only valid if 𝑀(𝐿 3⁄ ) > 𝑀𝑐𝑟 and 𝑀(2𝐿 3⁄ ) > 𝑀𝑐𝑟.

If both ends and midspan are cracked:

𝛥 = 𝛥1 + 𝛥2 + 𝛥3𝐴 + 𝛥3𝐵 + 𝛥3𝐶 + 𝛥3𝐷 + 𝛥5 + 𝛥6

If only midspan is cracked:

𝛥 = 𝛥1&2 + 𝛥3𝐴 + 𝛥3𝐵 + 𝛥3𝐶 + 𝛥3𝐷 + 𝛥5&6

If only right end and midspan are cracked:

𝛥 = 𝛥1&2 + 𝛥3𝐴 + 𝛥3𝐵 + 𝛥3𝐶 + 𝛥3𝐷 + 𝛥5 + 𝛥6

If only left end and midspan are cracked:

𝛥 = 𝛥1 + 𝛥2 + 𝛥3𝐴 + 𝛥3𝐵 + 𝛥3𝐶 + 𝛥3𝐷 + 𝛥5&6

117

S806 method for a uniformly distributed load:

Figure L.3 – Lengths to Integration Segments for Example Uniform Load

Results using the S806 method for a continuous member with a uniformly distributed

load follow. Appendix E provides equations for 𝐿1, 𝐿2, 𝐿4, and 𝐿5 with this loading.

𝐿3 = 𝐿 2⁄ ; 𝐿𝑅4 = 𝐿 − 𝐿4 ; 𝐿𝑅5 = 𝐿 − 𝐿5.

𝛥1 = ∫m(𝑥)𝑀(𝑥)


𝐿

0

=𝑀0𝐿

2

2𝐸𝑐𝐼𝑐𝑟𝐿((4 − 𝛼𝐿 + 𝛼𝑅)𝐿1

3

3𝐿3 −

𝐿14

𝐿4+𝛼𝐿𝐿1

2

2𝐿2)

118

𝛥2 = ∫m(𝑥)𝑀(𝑥)

E 𝐼𝑔𝑑𝑥

𝐿

𝐿

=𝑀0𝐿

2

2𝐸𝑐𝐼𝑔((4 − 𝛼𝐿 + 𝛼𝑅) (

𝐿23 − 𝐿1

3

3𝐿3) + 𝛼𝐿 (

𝐿22 − 𝐿1

2

2𝐿2)

− (𝐿2

4 − 𝐿14

𝐿4))

𝛥3 = ∫m(𝑥)𝑀(𝑥)


𝐿2

𝐿

=𝑀0𝐿

2

2𝐸𝑐𝐼𝑐𝑟𝑚((4 − 𝛼𝐿 + 𝛼𝑅) (

𝐿33 − 𝐿2

3

3𝐿3) + 𝛼𝐿 (

𝐿32 − 𝐿2

2

2𝐿2)

− (𝐿3

4 − 𝐿24

𝐿4))

𝛥4 = ∫m(𝑥)𝑀(𝑥)


𝐿

𝐿2

=𝑀0𝐿

2

2𝐸𝑐𝐼𝑐𝑟𝑚((4 − 𝛼𝑅 + 𝛼𝐿) (

𝐿33 − 𝐿𝑅4

3

3𝐿3) + 𝛼𝑅 (

𝐿32 − 𝐿𝑅4

2

2𝐿2)

− (𝐿3

4 − 𝐿𝑅44

𝐿4))

𝛥5 = ∫m(𝑥)𝑀(𝑥)

E 𝐼𝑔𝑑𝑥

𝐿

𝐿

=𝑀0𝐿

2

2𝐸𝑐𝐼𝑔((4 − 𝛼𝑅 + 𝛼𝐿) (


3

3𝐿3) + 𝛼𝑅 (


2

2𝐿2)

− (𝐿𝑅44 − 𝐿𝑅5

4

𝐿4))

119

𝛥6 = ∫m(𝑥)𝑀(𝑥)


𝐿

𝐿

=𝑀0𝐿

2

2𝐸𝑐𝐼𝑐𝑟𝑅((4 − 𝛼𝐿 + 𝛼𝑅)𝐿𝑅5

3

3𝐿3 −

𝐿𝑅54

𝐿4+𝛼𝐿𝐿𝑅5

2

2𝐿2)

𝛥1&2 = ∫m(𝑥)𝑀(𝑥)

E 𝐼𝑔𝑑𝑥

𝐿

0

=𝑀0𝐿

2


3

3𝐿3 −

𝐿24

𝐿4+𝛼𝐿𝐿2

2

2𝐿2)

𝛥5&6 = ∫m(𝑥)𝑀(𝑥)

E 𝐼𝑔𝑑𝑥

𝐿

𝐿

=𝑀0𝐿

2

2𝐸𝑐𝐼𝑔((4 − 𝛼𝑅 + 𝛼𝐿)𝐿𝑅4

3

3𝐿3 −

𝐿𝑅44

𝐿4+𝛼𝑅𝐿𝑅4

2

2𝐿2)

Total midspan deflection (note these results are only valid if 𝑀𝑚 > 𝑀𝑐𝑟 :





120

Example Simply Supported Constant Stiffness Beam Appendix H

The three loadings used in this report each produce slightly different deflected shapes,

so they have been provided here, with identical midspan deflection, for comparison.

This example models a simply supported uncracked concrete beam of uniform cross-

section as a perfect prismatic linear-elastic member. The load for each case is selected

to result in the equal midspan deflection for: a single midspan point load (1PL), two

equal point loads at third points (2PL), and a uniformly distributed load (UDL). The

small variation in deflection along the beams, as seen in the following deflection-

exaggerated graph, results from the different curvature of the three load configurations.

See List of Symbols for symbol definitions. Equations used are as follows:

𝐼 =𝑏ℎ3

12 𝐸 = 4500√𝑓𝑐′ 𝑀1𝑃𝐿 =

𝑃1𝑃𝐿𝐿

4 𝑀2𝑃𝐿 =

𝑃2𝑃𝐿𝐿

6 𝑀𝑈𝐷𝐿 =

𝑤𝑈𝐷𝐿𝐿2

8

For a midspan point load:

∆(𝑥) =

{

𝑖𝑓 (𝑥 ≤𝐿

2) ∶

𝑃1𝑃𝐿 𝑥(3𝐿2 − 4𝑥2)

48𝐸𝐼

𝑒𝑙𝑠𝑒 ∶ 𝑃1𝑃𝐿(𝐿 − 𝑥)(3𝐿2 − 4(𝐿 − 𝑥)2)

48𝐸𝐼 }

For two point load at third points:

∆(𝑥) =

{

𝑖𝑓 (𝑥 ≤𝐿

3) ∶

2𝑃2𝑃𝐿 𝑥 (59𝐿2 − 𝑥2) + 𝑃2𝑃𝐿𝑥 (

89𝐿2 − 𝑥2)

36𝐸𝐼

𝑖𝑓 (𝐿

3< 𝑥 ≤

2𝐿

3) ∶

𝑃2𝑃𝐿(𝐿 − 𝑥) (8𝐿2

9 − (𝐿 − 𝑥)2) + 𝑃2𝑃𝐿𝑥 (89 𝐿

2 − 𝑥2)

36𝐸𝐼

𝑒𝑙𝑠𝑒 ∶ 𝑃2𝑃𝐿(𝐿 − 𝑥) (

8𝐿2

9 − (𝐿 − 𝑥)2) + 2𝑃2𝑃𝐿(𝐿 − 𝑥) (5𝐿2

9 − (𝐿 − 𝑥)2)

36𝐸𝐼 }


∆(𝑥)𝑓𝑜𝑟 𝑈𝐷𝐿 =𝑤𝑈𝐷𝐿 𝑥(𝑥

3 − 2𝐿𝑥2 + 𝐿3)

24𝐸𝐼

121

Table H-1 - Equal Midspan Deflection Example for CPL, 2PL, and UDL

Spreadsheet Function: Compare deflected shapes for a constant stiffness beam with three

h = 800 mm example load types; load selected for equal midspan deflection

b = 400 mm L = 10000 mm fc' = 36 MPa

Ig = mm4 # of beam sections : 20 Ec = 27000 MPaΔ lef t Δright

For One PL at Midspan For Two Equal PL at 1/3 pts Uniformly Distributed

P1PL= 59725 N P2PL / 2 = 35050 N wUDL= 9.556 N / mm

M1PL= 149 kN m M2PL= 117 kN m MUDL= 119 kN m

x Δ(x) for 1 PL Δ(x) for 2 PL Δ(x) for UDL

0 0.00 0.00 0.00 0.00 0.00

500 0.40 0.23 0.19 0.42 0.43

1000 0.80 0.46 0.37 0.83 0.85

1500 1.18 0.68 0.55 1.22 1.24

2000 1.53 0.87 0.72 1.59 1.60

2500 1.86 1.04 0.87 1.91 1.92

3000 2.14 1.18 1.01 2.19 2.20

3500 2.37 1.28 1.13 2.41 2.41

4000 2.55 1.34 1.23 2.57 2.57

4500 2.66 1.36 1.31 2.67 2.67

5000 2.70 1.35 1.35 2.70 2.70

5500 2.66 1.31 1.36 2.67 2.67

6000 2.55 1.23 1.34 2.57 2.57

6500 2.37 1.13 1.28 2.41 2.41

7000 2.14 1.01 1.18 2.19 2.20

7500 1.86 0.87 1.04 1.91 1.92

8000 1.53 0.72 0.87 1.59 1.60

8500 1.18 0.55 0.68 1.22 1.24

9000 0.80 0.37 0.46 0.83 0.85

9500 0.40 0.19 0.23 0.42 0.43

10000 0.00 0.00 0.00 0.00 0.00

1.71E+010

Deflection of simply supported constant stiffness member

0

0.5

1

1.5

2

2.5

3

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

Δ(D

efl

ecti

on

)

x (Position)

Δ(x) for 1 PL

Δ(x) for 2 PL

Δ(x) for UDL

122

Example Constant Stiffness Beam with End-Moments Appendix I

The following linear-elastic prismatic beam is loaded with different uniformly

distributed loads to provide a comparison of deflected shapes with different end-

moment conditions. The deflected shapes can be compared easily because the load was

strategically set to different values for each end-moment condition such that the

midspan deflections are equal. Solutions with end-moments are determined using the

principle of superposition. For case 2, 𝑀𝑅 is set to −𝑀0. For case 3 and case 4, 𝑀𝐿

equals 𝑀𝑅 and they are set to −𝑀0/2 and −2𝑀0/3 for those cases, respectively. See

List of Symbols for symbol definitions. The spreadsheet equations include:

𝐼𝑔 =𝑏ℎ3

12 ; 𝐸𝑐 = 4500√𝑓𝑐′ ; 𝑀𝑐𝑟 =

𝑓𝑟𝐼ℎℎ2

=𝑏ℎ2√𝑓𝑐′

10 ; 𝑀0 =

𝑤𝑈𝐷𝐿𝐿2

8

∆𝑀𝐿(𝑥) =

𝑀𝐿𝑥(𝑥2 − 3𝐿𝑥 + 2𝐿2)

6𝐿𝐸𝑐𝐼𝑔 for

∆𝑀𝑅(𝑥) =

𝑀𝑅(𝐿 − 𝑥)((𝐿 − 𝑥)2 − 3𝐿(𝐿 − 𝑥) + 2𝐿2)

6𝐿𝐸𝑐𝐼𝑔 for

∆𝑈𝐷𝐿(𝑥) =𝑤𝑈𝐷𝐿 𝑥(𝑥

3 − 2𝐿𝑥2 + 𝐿3)

24𝐸𝑐𝐼𝑔 for

∆1(𝑥) = ∆𝑈𝐷𝐿(𝑥)

∆2(𝑥) = ∆𝑀𝑅(𝑥) + ∆𝑈𝐷𝐿(𝑥)

∆3(𝑥) = ∆4(𝑥) = ∆𝑀𝐿(𝑥) + ∆𝑀𝑅

(𝑥) + ∆𝑈𝐷𝐿(𝑥)

123

Table I-1 - Equal Midspan Deflection Example for Continuous UDL

Spreadsheet Function: Show deflected shapes for a uniform beam under 4 example loads

h = 400 mm (with end-moments) set to equal midspan deflection

b = 200 mm # of beam sections : 20 fc' = 36 MPa Mcr = 19.2 kNm

Ig = mm4 L = 10000 mm Ec = 27000 MPa

wUDL= 0.442 wUDL= 1.105 N / mm wUDL= 1.105 N / mm wUDL= 2.21 N / mm

MUDL= 6 M0= 13.8 kN m M0= 13.8 kN m M0= 27.6 kN m

ML= 0.0 kN m ML= -6.9 kN m ML= -18.4 kN m

MR= -13.8 kN m MR= -6.9 kN m MR= -18.4 kN m

x ΔM_R ΔUDL Δ2(x) ΔM_L ΔM_R ΔUDL Δ3(x) ΔM_L ΔM_R ΔUDL Δ4(x)

0 0 0 0.00 0 0 0 0.00 0 0 0 0.00

500 -0.40 0.80 0.40 -0.37 -0.20 0.80 0.23 -0.99 -0.53 1.59 0.07

1000 -0.79 1.57 0.78 -0.68 -0.40 1.57 0.49 -1.82 -1.06 3.14 0.26

1500 -1.17 2.30 1.13 -0.94 -0.59 2.30 0.77 -2.51 -1.56 4.60 0.52

2000 -1.53 2.97 1.43 -1.15 -0.77 2.97 1.05 -3.07 -2.05 5.93 0.82

2500 -1.87 3.56 1.69 -1.31 -0.94 3.56 1.31 -3.50 -2.50 7.12 1.12

3000 -2.18 4.06 1.88 -1.43 -1.09 4.06 1.54 -3.80 -2.91 8.12 1.41

3500 -2.45 4.46 2.01 -1.50 -1.23 4.46 1.74 -4.00 -3.27 8.93 1.65

4000 -2.69 4.76 2.07 -1.53 -1.34 4.76 1.88 -4.09 -3.58 9.52 1.84

4500 -2.87 4.94 2.07 -1.53 -1.43 4.94 1.97 -4.09 -3.82 9.87 1.96

5000 -3.00 5.00 2.00 -1.50 -1.50 5.00 2.00 -4.00 -4.00 9.99 2.00

5500 -3.07 4.94 1.87 -1.43 -1.53 4.94 1.97 -3.82 -4.09 9.87 1.96

6000 -3.07 4.76 1.69 -1.34 -1.53 4.76 1.88 -3.58 -4.09 9.52 1.84

6500 -3.00 4.46 1.46 -1.23 -1.50 4.46 1.74 -3.27 -4.00 8.93 1.65

7000 -2.85 4.06 1.21 -1.09 -1.43 4.06 1.54 -2.91 -3.80 8.12 1.41

7500 -2.62 3.56 0.94 -0.94 -1.31 3.56 1.31 -2.50 -3.50 7.12 1.12

8000 -2.30 2.97 0.67 -0.77 -1.15 2.97 1.05 -2.05 -3.07 5.93 0.82

8500 -1.89 2.30 0.41 -0.59 -0.94 2.30 0.77 -1.56 -2.51 4.60 0.52

9000 -1.37 1.57 0.20 -0.40 -0.68 1.57 0.49 -1.06 -1.82 3.14 0.26

9500 -0.74 0.80 0.06 -0.20 -0.37 0.80 0.23 -0.53 -0.99 1.59 0.07

10000 0 0 0.00 0 0 0 0.00 0 0 0 0.00

0.32

0.00

3) Uniformly Distributed

Small Equal End-Moments

2) UDL with Large

Right End-Moment

1.62

1.42

1.19

0.92

0.63

1.97

2.00

1.97

1.90

1.79

1.07E+009

Deflection of member with increasing load and end-moments

4) Uniformly Distributed

Large Equal End-Moments

1) Graph

Simple UDL

Δ1(x)

0.00

0.32

0.63

0.92

1.19

1.42

1.62

1.79

1.90

0

0.5

1

1.5

2

2.5

0 2000 4000 6000 8000 10000

Δ(D

efl

ecti

on

)

x (Position)

Δ with ML = MR = 0

Δ with ML = 0 ; MR = -Mm

Δ with ML = MR = -Mm

Δ with ML = MR = -2MmΔ with 𝑀𝐿 = 𝑀𝑅 =−2𝑀𝑚

Δ with 𝑀𝐿 = 𝑀𝑅 =−𝑀𝑚

Δ with 𝑀𝐿 = 0 ; 𝑀𝑅 = −𝑀𝑚

Δ with 𝑀𝐿 = 𝑀𝑅 =0

124

Example Generation and Deflection Computation for an Appendix J

Idealized Concrete Bending Member

In order to compare the deflection for a wide range of concrete bending members,

idealized members were produced throughout this project. The predicted deflection was

calculated using the following methods: the full analytical approach per Appendix E,

the full numerical approach per Appendix K, and simplified approaches based on

Branson’s (1965) work, the S806 (CSA 2012) method, and the proposed equations.

Most of the example concrete bending members were produced using the steps shown

in this appendix.

In order to indicate practical results throughout the research, all members are produced

with full properties. The intention is that all members have realistic properties. This

work is based primarily on Canadian design standards, so ultimate limits states

methodology and calculations from A23.3 (CSA 2004) and S806 (CSA 2012) are used.

Some input values were assumed throughout, but generally these inputs can be changed

with negligible effect on the relative results presented in the report. A specified

compressive strength of concrete of 36 MPa is used typically, but 25 MPa to 64 MPa

was used where convenient. For the example below, changing 𝑓𝑐′ from 36 MPa to

25 MPa results in more depth and the midspan deflection changing from 14.35 mm to

13.05 mm. However, this change does not change non-dimensionalized results and

only causes a slight change of scale in the vertical axis for the graphs which compare

approximate and exact deflections.

125

The following is an example of a steel reinforced concrete bending member with a

uniformly distributed load where the 𝑀𝑚 = −𝑀𝐿 and 𝑀𝑅 = 0, but it begins by

designing an example simply supported member with the same maximum moment.

For this example, the concrete properties and ultimate limit states design constants are:

𝜙𝑐 = 0.65 ; 𝑓𝑐′ = 36 MPa ; 𝐸𝑐 = 4500√𝑓𝑐′ = 27 GPa

α1 = 0.85 − 0.0015𝑓𝑐′ = 0.796 ; β1 = 0.97 − 0.0025𝑓𝑐

′ = 0.880

𝜙𝑠 = 0.85 ; 𝑓𝑦 = 400 MPa ; 𝐸𝑠 = 200 GPa ; 𝑛 = 𝐸𝑠/𝐸𝑐 = 7.407

The example span length, 𝐿, and simply supported example load, 𝑤0, for this member

are arbitrarily defined as:

𝐿 = 10.0 m ; 𝑤0 = 10 kN m⁄ ; therefore 𝑀0,0 =𝑤0𝐿

2

8= 125 kNm

The uniformly distributed load used to produce member properties is 𝑤0. The simply

supported midspan bending moment, 𝑀0,0, applies to each set of compared members.

Set the following ratios for this example beam:

𝑀𝑐𝑟

𝑀0,0= 0.7 ; 𝑏 = 0.5 ℎ ; 𝑑 = 0.85 ℎ

Produce a member concrete size using above information and A23.3 (CSA 2004)

definitions:

𝑀𝑐𝑟 = 0.7𝑀0,0 = 87.5 kNm ; 𝑓𝑟 = 0.6√𝑓𝑐′ = 3.6 MPa

𝑀𝑐𝑟 =𝑓𝑟𝑏ℎ

2

6= 0.3 (ℎ3) ; ℎ = √

87,500,000

0.3

= 663 mm

𝑏 = 0.5 ℎ = 332 mm ; 𝑑 = 0.85 ℎ = 564 mm ; 𝐼𝑔 =𝑏ℎ3

12= 8.06 x 109 mm4

126

Next, a service load to moment resistance ratio is defined as follows:

𝑀𝑠/𝑀𝑟 = 0.635

The required reinforcement bars can now be calculated. The maximum positive

moment is at midspan for this first member, which is simply supported, therefore

𝑀𝑚𝑎𝑥 = 𝑀𝑚 = 𝑀0,0. The reinforcing steel design at midspan is:

𝐾𝑟 =𝑀𝑟

𝑏𝑑2=𝑀0,0

𝑏𝑑2𝑀𝑓

𝑀𝑠= 1.87 MPa ; 𝐾𝑟 = (1 −

𝜌𝜙𝑠𝑓𝑦

2𝛼1𝜙𝑐𝑓𝑐′)𝜌𝜙𝑠𝑓𝑦

𝜌 =𝛼1𝜙𝑐𝑓𝑐

′ − √(𝛼1𝜙𝑐𝑓𝑐′)2 − 2𝛼1𝜙𝑐𝑓𝑐′𝐾𝑟𝜙𝑠𝑓𝑦

= 0.00580 ; 𝐴𝑠 = 𝜌𝑏𝑑 = 1085 mm2

For FRP design in this work, GFRP reinforcement was used and designed for concrete

crushing failure. To achieve reasonably low deflections, extra reinforcing was added to

the bottom steel by using 𝑀𝑟 > 𝑀𝑓. The GFRP reinforcement design for midspan is:

𝑓𝑓𝑢 = 690 MPa ; 𝜙𝑏 = 0.75 ; 휀𝑐𝑢 = 0.0035 ; 𝐸𝑏 = 44 GPa ; 𝑛 = 1.63 ; 𝑀𝑟 𝑀𝑓⁄ = 2.0

𝑐 =𝑑

𝛽1−√

𝑑2

𝛽12 −

2𝑀𝑟

𝛼1𝛽12𝜙𝑐𝑓𝑐′𝑏

= 144.9 mm ; 𝐴𝑓 =𝛼1𝛽1𝜙𝑐𝑓𝑐

′𝑏𝑐2

𝜙𝑏휀𝑐𝑢(𝑑 − 𝑐)𝐸𝑏= 2555 mm2

𝜌 =𝐴𝑓

𝑏𝑑= 0.014 < 𝜌𝑓𝑏 =

𝛼1𝛽1𝜙𝑐𝑓𝑐′휀𝑐𝑢

𝜙𝑏𝑓𝑓𝑢 (휀𝑐𝑢 +𝑓𝑓𝑢𝐸𝑓)

= 0.0062 (concrete crushes)

The serviceability stress limit and sustained load (creep-rupture) stress limit should be

checked. For the report, it is assumed that serviceability and creep-rupture stress limits

are met, but this should be confirmed for real cases (example calculations follow). For

this example, assume 𝑀𝑠𝑢𝑠 = 80 kNm and calculate:

𝑘 = √𝜌2𝑛2 + 2𝜌𝑛 − 𝜌𝑛 = 0.202

127

𝑓𝑏(𝑆𝐿𝑆) =𝑀𝑚

𝐴𝑓 (1 −𝑘3) 𝑑

= 81 MPa ≤ 𝑓𝑆𝐿𝑆 = 0.25𝑓𝑓𝑢 = 172 MPa (therefore ok)

𝑓𝑏(𝑠𝑢𝑠) =𝑀𝑠𝑢𝑠

𝐴𝑓 (1 −𝑘3)𝑑

= 52 MPa ≤ 𝑓𝑓,𝑠 = 0.002𝐸𝑓 = 88 MPa (therefore ok)

For FRP, 𝑛 = 𝐸𝑓/𝐸𝑐 and 𝐴𝑓 (not 𝐴𝑠) would be used in subsequent calculations. Instead,

the steel reinforced member example is continued in this appendix.

Based on equations from page 6-31 of the Concrete Design Handbook (CAC 2005):

𝑘𝑑 = (√2𝑑 (𝑏

𝑛𝐴𝑠) + 1 − 1) (

𝑏

𝑛𝐴𝑠)⁄ = 142.8 mm

𝐼𝑐𝑟 =𝑏(𝑘𝑑)3

3+ 𝑛𝐴𝑠(𝑑 − 𝑘𝑑)2 = 1.75 x 109 mm4

Deflection results using different methods can now be compared. Some constant

stiffness methods assume the 𝐸𝐼 term is equal to 𝐸𝑐𝐼𝑔, 𝐸𝑐𝐼𝑐𝑟, 𝐸𝑐𝐼𝑒 (𝐵𝑟𝑎𝑛𝑠𝑜𝑛), or

𝐸𝑐𝐼𝑒 (𝐵𝑖𝑠𝑐ℎ𝑜𝑓𝑓)′ . The exact midspan deflection can be computed using the analytical

results indicated in Appendix E. Alternatively, or for a check, numerical integration can

also be used (example calculations provided in Appendix K).

∆𝑢𝑛𝑐𝑟𝑎𝑐𝑘𝑒𝑑=5𝑤𝐿4

384𝐸𝑐𝐼𝑔= 5.98 mm ; ∆𝑓𝑢𝑙𝑙𝑦 𝑐𝑟𝑎𝑐𝑘𝑒𝑑=

5𝑤𝐿4

384𝐸𝑐𝐼𝑐𝑟= 27.64 mm

Referring to results from Branson’s (1965) work:


𝑀𝑚)3

𝐼𝑔 + [1 − (𝑀𝑐𝑟

𝑀𝑚)3

] 𝐼𝑐𝑟 = 3.91 x 109 mm4 ; ∆=5𝑤𝐿4

384𝐸𝑐𝐼𝑒= 12.3 mm

Referring to results from Bischoff’s (Bischoff and Gross 2011) work:

128

𝜂 = 1 −𝐼𝑐𝑟𝐼𝑔

= 0.783 ; 𝜉 = 1 − √1 −𝑀𝑐𝑟

𝑀𝑚= 0.452

=1.6𝜉3 − 0.6𝜉4

(𝑀𝑐𝑟

𝑀𝑚)2 + 2.4 ln(2 − 𝜉) = 1.30

𝐼𝑒′ =

𝐼𝑐𝑟


𝑀𝑚)2 = 3.48 x 109 mm4 ; ∆=

5𝑤𝐿4

384𝐸𝑐𝐼𝑒′=5𝑀𝑚𝐿

2

48𝐸𝑐𝐼𝑒′= 13.85 mm

Next, use the analytical equations per Appendix E (see List of Symbols where required):

𝛼𝐿 = 𝑀𝐿 𝑀0⁄ = 0 and 𝛼𝑅 = 𝑀𝑅 𝑀0⁄ = 0 (Simply Supported Member)

𝐿2 = (4 − 𝛼𝐿 + 𝛼𝑅 −√(4 − 𝛼𝐿 + 𝛼𝑅)2 − 16𝛼𝑐𝑟 + 16𝛼𝐿)𝐿

8= 2261 mm

𝐿4 = (4 − 𝛼𝐿 + 𝛼𝑅 +√(4 − 𝛼𝐿 + 𝛼𝑅)2 − 16𝛼𝑐𝑟 + 16𝛼𝐿)𝐿

8= 7739 mm

𝛥1+2 =𝑀0𝐿

2


3

3𝐿3 −

𝐿24

𝐿4+𝛼𝐿𝐿2

2

2𝐿2) = 0.37 mm

𝛥3 = 6.56 mm (see Appendix E for full equation)


𝛥5+6 =𝑀0𝐿

2

2𝐸𝑐𝐼𝑔((4 − 𝛼𝑅 + 𝛼𝐿)𝐿𝑅−4

3

3𝐿3 −

𝐿𝑅−44

𝐿4+𝛼𝑅𝐿𝑅−4

2

2𝐿2) = 0.37 mm

𝛥 = 𝛥1+2 + 𝛥3 + 𝛥4 + 𝛥5+6 = 13.85 mm

After a simply supported member has been designed and analyzed, the next step

performed is to design continuous members with the same midspan properties as the

simply supported member. To generate this set of continuous members, the load is

increased relative to the simply supported situation. The load is increased in a precise

manner so that the increase in the total static bending moment, 𝑀0, will offset the

129

negative bending moment(s) and the maximum positive bending moment will be

maintained. Variables 𝛼𝐿 = 𝑀𝐿 𝑀0⁄ and 𝛼𝑅 = 𝑀𝑅 𝑀0⁄ are introduced and used for each

member because they are useful mathematically. Typically, a set of 9 to 15 example

members are generated using the subsequent equations, with 0 ≥ 𝛼𝐿 ≥ −3 and 𝛼𝑅 = 0.

Calculations are provided for 𝛼𝐿 = −1 and 𝛼𝑅 = 0, so 𝑀𝐿 = −𝑀𝑚𝑎𝑥 and 𝑀𝑅 = 0. The

calculation of the uniform load required, 𝑤, is determined based on common structural

analysis equations and the end-moment to total moment ratio as follows:

𝛼𝐿 =𝑀𝐿

𝑀0= −0.686 ; 𝛼𝑅 =

𝑀𝑅

𝑀0= 0 ; 𝑤 =

𝑤0

1 +𝛼𝐿2 +

𝛼𝑅2 +

(𝛼𝐿 − 𝛼𝑅)2

16

= 14.58 N

mm

Similarly, to maintain the same midspan moment with a centered point load:

𝑃 =𝑃0

1 +𝛼𝐿2 +

𝛼𝑅2

Likewise, to maintain the same maximum moment with equal third-point loading:

𝑃 =𝑃0

1 +𝛼𝐿3 +

2𝛼𝑅3

(requires −𝛼𝐿 > −𝛼𝑅)

For the example beam with a uniformly distributed beam and end continuous, the load

𝑤 is then used to determine the total service, end, midspan, and maximum moments:

𝑀0 =𝑤𝐿2

8= 182 kNm ; 𝑀𝐿 = 𝛼𝐿𝑀0 = −125 kNm ; 𝑀𝑅 = 𝛼𝑅𝑀0 = 0 kNm

𝑀𝑚 = 𝑀0 +(𝑀𝐿 +𝑀𝑅)

2= 120 kNm ; 𝑀𝑚𝑎𝑥 = 𝑀𝑚 +

(𝑀𝐿 +𝑀𝑅)2

16𝑀0= 125 kNm

130

Proceed to calculate the properties for this continuous member. As intended, the

maximum positive moment, concrete dimensions, 𝑀𝑐𝑟, 𝐼𝑔, positive 𝐴𝑠, 𝐼𝑐𝑟, and such

remain the same as the simply supported example. For both positive and negative

bending, again use 𝑀𝑟 = 𝑀𝑓. The calculations preformed above using 𝑀0,0 must now

use 𝑀𝑚𝑎𝑥 for the continuous member. At the right end of the member, where 𝑀𝑅 = 0,

simply use 𝐼 = 𝐼𝑔 = 8.06 x 109 mm4. Again determine the cross-section as follows:

𝑀𝑐𝑟 = 87.5 kNm ; ℎ = 663 mm ; 𝑏 = 332 mm ; 𝐼𝑔 = 8.06 x 109 mm4

𝑑 = 564 mm ; 𝐾𝑟 = 1.87 MPa ; 𝜌 = 0.00580 ; 𝐴𝑠 +𝑣𝑒 = 1085 mm2

𝑘𝑑 = 142.8 mm ; 𝐼𝑐𝑟𝑚 = 1.75 x 109 mm4

Because the left end negative moment is the same magnitude as the maximum positive

moment, as 𝑀𝑚𝑎𝑥 = −𝑀𝐿 and 𝑀𝑟 = 𝑀𝑓, the same properties occur at the left end:

𝐾𝑟 −𝑣𝑒 =−𝑀𝐿

𝑏𝑑2𝑀𝑓

𝑀𝑠= 1.87 MPa ; 𝜌 = 0.00580 ; 𝐴𝑠 −𝑣𝑒 = 1085 mm2

𝑘𝑑 = 142.8 mm ; 𝐼𝑐𝑟𝐿 = 1.75 x 109 mm4

Note that if the end-moment design requires more than the maximum allowable steel,

this means the original midspan design section makes a uniform concrete cross-section

impossible. If this occurs, one solution may be to return to the step where the example

sets 𝑀𝑐𝑟/𝑀0,0 = 0.7, increase this value as required, and revise subsequent calculations.

As indicated in the Concrete Design Handbook (CAC 2005) and in Appendix A:


2

48𝐸𝐼 where 𝐾 = 1.2 − 0.2

𝑀0

𝑀𝑚 ; hence 𝐾 = 0.896

131

Calculate midspan deflection using different assumptions for 𝐸𝐼:

∆𝑢𝑛𝑐𝑟𝑎𝑐𝑘𝑒𝑑= 𝐾5𝑀𝑚𝐿

4

48𝐸𝑐𝐼𝑔= 5.36 mm ; ∆𝑓𝑢𝑙𝑙𝑦 𝑐𝑟𝑎𝑐𝑘𝑒𝑑= 𝐾

5𝑀𝑚𝐿4

48𝐸𝑐𝐼𝑐𝑟= 24.75 mm

Since the midspan (maximum positive) moment and end-moment (and reinforcing) are

the same, CSA A23.3-04 clause 9.8.2.4 will simplify to 0.85𝐼𝑒𝑚 + 0.15𝐼𝑒𝐿 = 𝐼𝑒.

Therefore the result using Branson’s equation is:


𝑀𝑚𝑎𝑥)3

𝐼𝑔 + [1 − (𝑀𝑐𝑟

𝑀𝑚𝑎𝑥)3

] 𝐼𝑐𝑟 = 3.91x109 mm4 ; ∆= 𝐾5𝑀𝑚𝐿

4

48𝐸𝑐𝐼𝑒= 11.04 mm

Referring to results using Bischoff’s equations, base the 𝐼𝑒′ calculation on the maximum

positive moment, so the results are also the same as with the simply supported example:

𝜂 = 0.783 ; 𝜉 = 0.452 ; = 1.299 ; 𝐼𝑒′ = 3.48 x 109 mm4

Unlike Bischoff’s results for simply supported members, this report introduces an

approximation by using the for simply supported members. Work for this report

determined that the exact for a continuous member was excessively complicated

(except for very specific cases, such as 𝑀𝑚 = −𝑀𝐿 = −𝑀𝑅 with relevant reinforcement

being equal). The simply supported works almost as well as the exact result for any

realistic member for which it is possible to obtain an effective constant moment of

inertia.


4

48𝐸𝑐𝐼𝑒′= 11.87 mm

Again obtaining results using exact analytical equations from Appendix E:

𝐿1 = (4 − 𝛼𝐿 + 𝛼𝑅 −√(4 − 𝛼𝐿 + 𝛼𝑅)2 + 16𝑀𝑐𝑟/𝑀𝑚 + 16𝛼𝐿)𝐿

8= 457 mm

132

𝐿2 = (4 − 𝛼𝐿 + 𝛼𝑅 −√(4 − 𝛼𝐿 + 𝛼𝑅)2 − 16𝑀𝑐𝑟/𝑀𝑚 + 16𝛼𝐿)𝐿

8= 3588 mm

𝐿4 = (4 − 𝛼𝐿 + 𝛼𝑅 +√(4 − 𝛼𝐿 + 𝛼𝑅)2 − 16𝑀𝑐𝑟/𝑀𝑚 + 16𝛼𝐿)𝐿

8= 8125 mm

𝛥1 = −0.04 mm (see Appendix E for full equation)




𝛥5+6 = 0.25 mm (see Appendix E for full equation)

𝛥 = 𝛥1 + 𝛥2 + 𝛥3 + 𝛥4 + 𝛥5+6 = 11.54 mm

For comparison, calculate numerical integration results for this example member (using

the method shown in Appendix K):

Using 10 segments: 𝛥 = ∫𝑚(𝑥)𝑀(𝑥)


𝐿

0

≈∑ 𝛥𝑗𝑗=10

𝑗=1= 11.75 mm

Using 100 segments: 𝛥 = ∫𝑚(𝑥)𝑀(𝑥)


𝐿

0

≈∑ 𝛥𝑗𝑗=100

𝑗=1= 11.54 mm

Using the S806 method, introduced in Appendix G:

𝐿1 = 457 mm ; 𝐿2 = 3588 mm ; 𝐿3 = 5000 mm ; 𝐿4 = 8125 mm

𝛥1 = −0.11 mm (see Appendix G for full equation)

𝛥2 = 0.50 mm (see Appendix G for full equation)



𝛥5+6 = 0.25 mm (see Appendix G for full equation)

𝛥 = 𝛥1 + 𝛥2 + 𝛥3 + 𝛥4 + 𝛥5+6 = 20.96 mm

133

As indicated throughout the report, failing to account for tension stiffening is often

overly conservative. The error using Branson’s equation for this particular situation is

unconservative but is reasonably small because 2 < 𝐼𝑔 𝐼𝑐𝑟⁄ < 5. The following table

summarizes the results from this appendix and Appendix K.

Table J-1 - Summary of Appendix J and Appendix K Results for Continuous Member

Moment of Inertia 𝛥 % difference

𝐼𝑔 5.36 −46.4%

𝐼𝑐𝑟 24.75 +114% 𝐼𝑒 𝐵𝑟𝑎𝑛𝑠𝑜𝑛 11.04 −4.3%

𝐼𝑒 𝐵𝑖𝑠𝑐ℎ𝑜𝑓𝑓′ 11.87 +2.9%

𝐼𝑒(𝑥)𝑟𝑎𝑧𝑎𝑞𝑝𝑢𝑟 20.96 +82%

𝐼𝑒(𝑥)𝑛𝑢𝑚𝑒𝑟𝑖𝑐𝑎𝑙:10 11.75 +1.8%

𝐼𝑒(𝑥)𝑛𝑢𝑚𝑒𝑟𝑖𝑐𝑎𝑙:100 11.543 +0.001%

𝐼𝑒(𝑥)𝑎𝑛𝑎𝑙𝑦𝑡𝑖𝑐𝑎𝑙 11.542 −

134

Methodology and Example using Numerical Integration Appendix K

Numerical integration was often used to determine deflection in work for this report.

Like analytical integration, numerical integration is performed using the method of

virtual work to calculate the exact result. For this method, the member is cut into a

number, 𝑗, of segments and the deflection effects from all segments are added together.

Rounding errors inherent to numerical integration can be safely assumed to be

negligible for 𝑗 ≥ 100 (accuracy within 0.2% as long as a section is cut at the location

of each large point load). Other symbol definitions remain the same as with the

analytical approach and as defined in the List of Symbols.

𝛥 = ∫𝑚(𝑥)𝑀(𝑥)


𝐿

0

≈∑ [𝑚(𝑥𝑖)𝑀(𝑥𝑖)

𝐸𝑐𝐼𝑒(𝑥𝑖)+𝑚(𝑥𝑖−1)𝑀(𝑥𝑖−1)

𝐸𝑐𝐼𝑒(𝑥𝑖−1)]𝐿

2𝑗

𝑖=𝑗

𝑖=1

where 𝑥𝑖 =𝑖𝐿

𝑗 and 𝑥𝑖−1 =

𝑖 − 1

𝑗𝐿

To further explain this method, the following example uses it for the continuous

member provided in Appendix J. The member has the following properties and loads:

𝜙𝑐 = 0.65 ; 𝑓𝑐′ = 36 MPa ; 𝐸𝑐 = 27 GPa ; α1 = 00.796 ; β1 = 0.880

𝜙𝑠 = 0.85 ; 𝑓𝑦 = 400 MPa ; 𝐸𝑠 = 200 GPa ; 𝑛 =𝐸𝑠𝐸𝑐

= 7.407 ; 𝑓𝑟 = 3.6 MPa

𝐿 = 10 m ; 𝑤 = 14.58 ; 𝑀0 = 182 kNm ; 𝑀𝑅 = 0 ; 𝑀𝐿

𝑀0= −0.686 ; 𝑀𝐿 = −125 kNm

𝑀𝑚 = 𝑀0 +(𝑀𝐿 +𝑀𝑅)

2= 120 kNm ; 𝑀𝑚𝑎𝑥 = 𝑀𝑚 +

(𝑀𝐿 +𝑀𝑅)2

16𝑀0= 125 kNm

𝑀𝑠

𝑀𝑓= 0.635 ;

𝑀0

𝑀𝑐𝑟= 1.43 ; 𝑀𝑐𝑟 = 87.5 kNm ; ℎ = 663 mm

𝑏 = ℎ 2⁄ = 332 mm ; 𝑑 = 0.85(ℎ) = 564 mm ; 𝐼𝑔 = 8.06 x 109 mm4

135

At the location of maximum positive moment (near midspan):

𝐾𝑟 =𝑀𝑚𝑎𝑥

𝑏𝑑2𝑀𝑓

𝑀𝑠= 1.87 MPa ; 𝜌 = 0.00580 ; 𝐴𝑠 +𝑣𝑒 = 1085 mm2

𝑘𝑑 = 142.8 mm ; 𝐼𝑐𝑟𝑚 = 1.75 x 109 mm4 ; 𝜂𝑚 = 1 −𝐼𝑐𝑟𝑚𝐼𝑔

= 0.783

At member span ends, the right end is uncracked and the design at the left end is:

𝐾𝑟 =𝑀𝐿

𝑏𝑑2𝑀𝑓

𝑀𝑠= 1.87 MPa ; 𝜌 = 0.00580 ; 𝐴𝑠 −𝑣𝑒 = 1085 mm2

𝑘𝑑 = 142.8 mm ; 𝐼𝑐𝑟𝐿 = 1.75 x 109 mm4 ; 𝜂𝐿 = 1 −𝐼𝑐𝑟𝐿𝐼𝑔

= 0.783

The same functions as for analytical integration are required for numerical integration:

𝑀(𝑥) = 𝑀𝐿 + (4𝑀0 −𝑀𝐿 +𝑀𝑅)𝑥

𝐿− 4𝑀0

𝑥2

𝐿2

𝑓𝑜𝑟

{

𝑥 <𝐿

2 and 𝑀(𝑥) < −𝑀𝑐𝑟 𝐼𝑒(𝑥) =

𝐼𝑐𝑟𝐿

1 − 𝜂𝐿 (𝑀𝑐𝑟

𝑀(𝑥))2

−𝑀𝑐𝑟 ≤ 𝑀(𝑥) ≤ 𝑀𝑐𝑟 𝐼𝑒(𝑥) = 𝐼𝑔

𝑀(𝑥) > 𝑀𝑐𝑟 𝐼𝑒(𝑥) =𝐼𝑐𝑟𝑚

1 − 𝜂𝑚 (𝑀𝑐𝑟

𝑀(𝑥))2

For midspan deflection, set the virtual moment function as follows:

𝑓𝑜𝑟

{

𝑥 ≤𝐿

2𝑚(𝑥) =

𝑥

2

𝑥 >𝐿

2 𝑚(𝑥) =

𝐿 − 𝑥

2

136

Generate 10 equal segments for this example as follows:

𝛥𝑚𝑖𝑑 = ∫𝑚(𝑥)𝑀(𝑥)


𝐿

0

≈∑ 𝛥𝑖𝑖=10

𝑖=1

𝛥𝑗 =𝑥𝑖 − 𝑥𝑖−12𝐸𝑐

[𝑚(𝑥𝑖)𝑀(𝑥𝑖)

𝐼𝑒(𝑥𝑖)+𝑚(𝑥𝑖−1)𝑀(𝑥𝑖−1)

𝐼𝑒(𝑥𝑖−1)]

Table K-1 - Midspan Deflection Example using 10 Segment Numerical Integration

𝑖 𝑥𝑖 𝑚(𝑥𝑖) 𝑀(𝑥𝑖) 𝐼𝑒(𝑥𝑖) 𝑚(𝑥𝑖)𝑀(𝑥𝑖) 𝛥𝑖

mm kN m mm4 𝐼𝑒(𝑥𝑖) mm

0 0 0 −125 2.83x109 0 ∗1 1000 500 −47 8.06x109 −2.91 −0.052 2000 1000 17 8.06x109 2.06 −0.023 3000 1500 66 8.06x109 12.2 0.264 4000 2000 100 4.38x109 45.6 1.075 5000 2500 120 3.00x109 99.6 2.696 6000 2000 125 2.84x109 88 3.477 7000 1500 115 3.17x109 54.6 2.648 8000 1000 92 6.15x109 14.9 1.299 9000 500 53 8.06x109 3.29 0.3410 10000 0 0 8.06x109 0 0.06

𝛥𝑚𝑖𝑑 ≈∑ 𝛥𝑖𝑖=10

𝑖=1 = 11.75 mm

Generating 100 equal segments provides the results partially shown in Table K-2.

Table K-2 - Midspan Deflection Example using 100 Segment Numerical Integration



0 0 0 −125 2.83x109 0 ∗1 100 50 −117 3.13x109 −1.86 −0.003 2 200 100 −108 3.58x109 −3.02 −0.009 … … … … … … …50 5000 2500 120 3.00x109 99.6 0.360… … … … … … …59 5900 2050 125 2.83x109 90.4 0.339… … … … … … …99 9900 50 6 8.06x109 0.04 0.0003 100 10000 0 0 8.06x109 0.00 0.0000

137

𝛥𝑚𝑖𝑑 = ∫𝑚(𝑥)𝑀(𝑥)


𝐿

0

≈∑ 𝛥𝑖𝑖=100

𝑖=1 = 11.543 mm

Maximum deflection, if required, can be found by trial-and-error or other methods. For

this example, the maximum deflection can be found at position 𝐿∆𝑚𝑎𝑥= 5.5 m. Most

other functions remain the same, but the virtual moment function must be set as follows:

𝑓𝑜𝑟

{

𝑥 ≤ 𝐿∆𝑚𝑎𝑥𝑚(𝑥) = (1 −

𝐿∆𝑚𝑎𝑥

𝐿)𝑥

2

𝑥 > 𝐿∆𝑚𝑎𝑥 𝑚(𝑥) = (

𝐿∆𝑚𝑎𝑥

𝐿)𝐿 − 𝑥

2

Use the same 100 equal segments for the maximum deflection as follows:

Table K-3 - Maximum Deflection Example using 100 Segment Numerical Integration



0 0 0 −125 2.83x109 0 −1 100 45 −117 3.13x109 −1.68 −0.003 2 200 90 −108 3.58x109 −2.72 −0.008 … … … … … … …50 5000 2250 120 3.00x109 89.6 0.324… … … … … … …55 5500 2475 124 2.86x109 107.4 0.392… … … … … … …99 9900 55 6 8.06x109 0.04 0.0004 100 10000 0 0 8.06x109 0.00 0.0000

𝛥𝑚𝑎𝑥 = ∫𝑚(𝑥)𝑀(𝑥)


𝐿

0

≈∑ 𝛥𝑖𝑖=100

𝑖=1 = 11.768 mm

In this case, there is only 2% error if the midspan deflection is assumed to equal the

maximum deflection.

138

Examples Graphs of the Integrated Function Appendix L

Two examples are given in this appendix; one example features a good approximation

and lies within the proposed range of validity while the other example features a poor

approximation and lies outside of the proposed limits. These members are each a 1 m

wide strip of steel reinforced concrete slab with a centered point load and equal end-

moments. The midspan deflection is the maximum deflection for both members.

Graphs for both demonstrate the complexity of attempting to approximate the variable

stiffness of a cracked concrete bending member as a constant stiffness member.

Both examples include the following properties:

𝜙𝑐 = 0.65 ; 𝑓𝑐′ = 36 MPa ; 𝑓𝑟 = 3.6 MPa ; 𝐸𝑐 = 27 GPa ; α1 = 0.796

𝜙𝑠 = 0.85 ; 𝑓𝑦 = 400 MPa ; 𝐸𝑠 = 200 GPa ; 𝑛 = 𝐸𝑠/𝐸𝑐 = 7.407

𝐿 = 10 m ; 𝑏 = 1000 mm ; ℎ = 178 mm ; 𝑑 = 146 mm

𝐼𝑔 = 470 x 106 mm4 ; 𝑀𝑐𝑟 = 19.0 kNm

The first example presents the fully fixed-fixed case, 𝑀𝑚 = −𝑀𝐿 = −𝑀𝑅, as follows:

𝑀𝑚

𝑀𝑐𝑟= 1.515 ; 𝑃 = 23 kN ;

𝑀𝐿

𝑀0= −0.5 ;

𝑀𝑅

𝑀0= −0.5 ; 𝐾 = 0.50

𝑀0 = 57.5 kNm ; 𝑀𝐿 = −28.8 kNm ; 𝑀𝑚 = 28.8 kNm ; 𝑀𝑅 = −28.8 kNm

Select 15M’s at 200 mm (top & bottom). Then: 𝐴𝑠 = 1000 mm2 ; 𝜌 = 0.685%

𝑘𝑑 = 39.7 mm ; 𝐼𝑐𝑟 = 104 x 106 mm4 ; 𝐾𝑟 = 2.18

𝑀𝑟 = 𝐾𝑟𝑏𝑑2 = 46.5 kNm ; Assume 𝑀𝑓 ≈ 1.4𝑀𝑠 = 40.3 kNm (∴ ok)

139

The proposed equations, based on results from work by Bischoff and Gross (2011):


= 0.778 ; = 3 − 2𝑀𝑐𝑟

𝑀𝑚= 1.68

𝐼𝑒′ =

𝐼𝑐𝑟


𝑀𝑚)2 = 243 x 106 mm4 ; ∆ = 𝐾

𝑀𝑚𝐿2

12𝐸𝑐𝐼𝑒′= 18.24 mm

Using 100 section numerical integration, ∆𝐼𝑒(𝑥)= 18.28 mm and ∆𝐼𝑒′= 18.26 mm.

Figure L-1 - Integrated Function and Accurate Deflection Example

140

The graphs in the first example show that the midspan deflection is accurate despite the

differences in the functions that are integrated to obtain it. The graph for deflection, ∆,

shows the proposed solution (indicated using 𝐼𝑒′ ). There are small errors locally, using

the approximation, near the ends of the member and again each side of midspan. The

approximate solution is only intended to determine the maximum deflection, so these

errors are typically irrelevant. The lines shown in the lower graph, for both

𝑚(𝑥)𝑀(𝑥) 𝐸𝐼𝑒(𝑥)⁄ and 𝑚(𝑥)𝑀(𝑥) 𝐸𝐼𝑒′⁄ , are integrated to obtain the midspan

deflection results, ∆𝐼𝑒(𝑥) and ∆𝐼𝑒′ . The approximation works well despite these pre-

integration functions being significantly different because the average stiffness of the

simply supported member with the same midspan moment is very close to the average

stiffness of the actual continuous member.

The second example is a case with increased midspan cracking (𝑀𝑚/𝑀𝑐𝑟 = 2.22) and

end-moments (1.86𝑀𝑚 = −𝑀𝐿 = −𝑀𝑅). Other properties that are different include:

𝑃 = 48 kN ; 𝑀𝐿 𝑀0⁄ = −0.65 ; 𝑀𝑅 𝑀0⁄ = −0.65 ; 𝐾 = 0.071

𝑀0 = 120 kNm ; 𝑀𝐿 = −78 kNm ; 𝑀𝑚 = 42 kNm ; 𝑀𝑅 = −78 kNm

For bottom steel (midspan), select 20M’s at 200 mm. Then: 𝐴𝑠 = 1500 mm2

𝜌 = 1.03% ; 𝑘𝑑 = 46.9 mm ; 𝐼𝑐𝑟 = 143 x 106 mm4 ; 𝐾𝑟 = 3.17

Assume 𝑀𝑓 = 1.4𝑀𝑠 = 58.8 kNm ; 𝑀𝑟 = 67.5 kNm (ok)

For top steel (ends), select 20M’s at 100 mm. Then: 𝐴𝑠 = 3000 mm2

𝜌 = 2.06% ; 𝑘𝑑 = 61.3 mm ; 𝐼𝑐𝑟 = 236 x 106 mm4 ; 𝐾𝑟 = 5.68

Assume 𝑀𝑓 = 1.4𝑀𝑠 = 109 kNm ; 𝑀𝑟 = 121 kNm (ok)

141

The proposed equations, again based on relevant midspan values, are:


= 0.695 ; = 3 − 2𝑀𝑐𝑟

𝑀𝑚= 2.095

𝐼𝑒′ =

𝐼𝑐𝑟


𝑀𝑚)2 = 204 x 106 mm4 ; ∆ = 𝐾

𝑀𝑚𝐿2

12𝐸𝑐𝐼𝑒′= 4.53 mm

Using 100 sections for numerical integration, midspan results are:

∆𝐼𝑒(𝑥)= 9.88 mm ; ∆𝐼𝑒′= 4.57 mm ; ∆𝐼𝑐𝑟= 6.5 mm

.

Figure L-2 - Integrated Function and Inaccurate Deflection Example

142

The graphs for the second example show the deflected shapes using 𝐼𝑒(𝑥), 𝐼𝑒′ , and 𝐼𝑐𝑟.

The proposed approximation is far less accurate for this example. The midspan

deflection found using 𝐼𝑒′ is only half of the real member deflection found using 𝐼𝑒(𝑥).

This is not surprising considering that actual deflections are larger than deflections

found using 𝐼𝑐𝑟. This member cannot be accurately modelled as a constant stiffness

member. In this example, the source of the large error is within the negative bending

segments. This error can be visually identified by the additional area near the ends of

the 𝑚(𝑥)𝑀(𝑥) 𝐸𝐼⁄ graph.

Another example (not shown in this appendix) of when all constant stiffness approaches

will fail is a centered point load where 2𝑀𝑚 = −𝑀𝐿 = −𝑀𝑅. For this case, constant

stiffness equations result in ∆𝑚𝑎𝑥= 0 mm. Because of the varying stiffness, the real

midspan deflection is typically non-zero and hence unattainable by linear-elastic

deflection equations. Alternatively, the actual deflection can be near zero when the

constant stiffness equations inherently give deflections that are much larger in

magnitude.

143

Centered Point Load Examples – Data for Section 3.4 Appendix M

This appendix provides the data and graphs for the example prismatic concrete members

with centered point loads that were provided in Section 3.4 of this report. The provided

members are beams because they are more likely to be controlled by a centered point

load. The calculations in this appendix use the general methodology indicated in

Appendix J.

All members in this appendix are shown for the same concrete cross-section and the

same range of end-moments (relative to midspan moment). The concrete cross-section

used is 300 mm wide by 600 mm deep with tension reinforcing at a depth of 540 mm

and compression reinforcing neglected. The end-moments are kept equal (𝑀𝐿 = 𝑀𝑅) in

order to eliminate errors caused by the actual maximum deflection not being at midspan

and to simplify discussion of differences in results from different methods. The range

of end-moments provided is 2 < −𝑀𝐿 𝑀𝑚⁄ ≤ 0; this range exceeds both the proposed

valid range and the range required for the majority of center point loaded members.

The first two sets of examples are steel reinforced members. The first set is developed

with a cracking to service moment ratio of 𝑀𝑚/𝑀𝑐𝑟 = 3.0. The second set carries less

load and contains less reinforcement: 𝑀𝑚/𝑀𝑐𝑟 = 1.6.

The third and fourth sets of examples are GFRP reinforced members. The third set has

𝑀𝑚/𝑀𝑐𝑟 = 2.5 and the fourth set has 𝑀𝑚/𝑀𝑐𝑟 = 1.6. There is an increase in

deflection because of the reduced 𝐼𝑐𝑟. For end-moments with 𝑀𝐿/𝑀𝑚 > 0.4, short-term

deflection of these members is generally less than 𝐿/Δ = 560.

144

Table M-1 - Data for CPL, ML=MR, Ig/Icr=2.3 – Example 3.4.2a – Page 1

Example 3.4.2a, pg 1 of 2 Φc = 0.65

Midspan Point Load P0 = 77850 N fc' = 36 MPa fr = 0.6 * MPa

Vary ML, Mm constant L = 10000 mm Ec = 27000 MPa Mcr= 0.05 * N mm

Let ML=MR b = 0.5 * h mm α1 = 0.796

d = 0.9 * h mm β1 = 0.880

M0,0 = P0 L/4 = 1.95E+8 N mm Φb = 0.85

End Moment Ms/Mr = 0.635 fy = 400 MPa

+ve Moment Ms/Mr = 0.635 Eb = 200000 MPa n=Eb/Ec= 7.40741

Member Properties Determined from Provided Info (Primairly Servicability) Units

P1PL = 228971 210405 194625 173000 155700 129750 111214 97312.5 77850 N

M0 = 5.72E+8 5.26E+8 4.87E+8 4.33E+8 3.89E+8 3.24E+8 2.78E+8 2.43E+8 1.95E+8 N mm

αL = ML/M0 = -0.66 -0.63 -0.6 -0.55 -0.5 -0.4 -0.3 -0.2 0

αR = MR/M0 = -0.66 -0.63 -0.6 -0.55 -0.5 -0.4 -0.3 -0.2 0

ML = -3.78E+8 -3.31E+8 -2.92E+8 -2.38E+8 -1.95E+8 -1.30E+8 -8.34E+7 -4.87E+7 0.00E+0 N mm

MR = -3.78E+8 -3.31E+8 -2.92E+8 -2.38E+8 -1.95E+8 -1.30E+8 -8.34E+7 -4.87E+7 0.00E+0 N mm

Mm = Mmax = 1.95E+8 1.95E+8 1.95E+8 1.95E+8 1.95E+8 1.95E+8 1.95E+8 1.95E+8 1.95E+8 N mm

αcr = Mcr/Mmax= 0.333 0.333 0.333 0.333 0.333 0.333 0.333 0.333 0.333

Mcr = 6.48E+7 6.48E+7 6.48E+7 6.48E+7 6.48E+7 6.48E+7 6.48E+7 6.48E+7 6.48E+7 N mm

h = 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 mm

d = 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 mm

b = 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 mm

Ig = 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 mm4

αL/max=ML/Mmax= -1.94 -1.70 -1.50 -1.22 -1.00 -0.67 -0.43 -0.25 0.00

Member Properties Determined with Factored Loads

Left End Kr L = 6.80 5.96 5.25 4.28 3.50 2.34 2 0 0 MPa

ρ L = 0.0263 0.0219 0.0186 0.0145 0.0115 0.0074 0.0046 0 0

AL=ρ Lbd= 4265 3554 3017 2352 1865 1193 747 0 0 mm2

Icr L = 4.22E+9 3.75E+9 3.35E+9 2.81E+9 2.37E+9 1.68E+9 1.15E+9 5.40E+9 5.40E+9 mm4

ηL=1 – Icr L/Ig = 0.219 0.306 0.379 0.479 0.560 0.689 0.787 0 0

ML/Mcr = -5.83 -5.11 -4.50 -3.67 -3.00 -2.00 -1.29 -0.75 0.00

Ig/Icr L = 1.28 1.44 1.61 1.92 2.28 3.21 4.70 1.00 1.00

Midspan Kr m = 3.50 3.50 3.50 3.50 3.50 3.50 3.50 3.50 3.50 MPa

ρ m = 0.0115 0.0115 0.0115 0.0115 0.0115 0.0115 0.01151 0.0115 0.0115

Am=ρ mbd= 1865 1865 1865 1865 1865 1865 1865 1865 1865 mm2

Icr m = 2.37E+9 2.37E+9 2.37E+9 2.37E+9 2.37E+9 2.37E+9 2.37E+9 2.37E+9 2.37E+9 mm4

ηm=1 – Icr m/Ig = 0.560 0.560 0.560 0.560 0.560 0.560 0.560 0.560 0.560

Mmax/Mcr = 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00

Ig/Icr m = 2.28 2.28 2.28 2.28 2.28 2.28 2.28 2.28 2.28

Right End Kr R = 6.80 5.96 5.25 4.28 3.50 2.34 2 0 0 MPa

ρ R = 0.0263 0.0219 0.0186 0.0145 0.0115 0.0074 0.0046 0 0

AR=ρ Rbd= 4265 3554 3017 2352 1865 1193 747 0 0 mm2

Icr R = 4.22E+9 3.75E+9 3.35E+9 2.81E+9 2.37E+9 1.68E+9 1.15E+9 5.40E+9 5.40E+9 mm4

ηR=1 – Icr R/Ig = 0.219 0.306 0.379 0.479 0.560 0.689 1 0 0

MR/Mcr = -5.83 -5.11 -4.50 -3.67 -3.00 -2.00 -1.29 -0.75 0.00

Ig/Icr R = 1.28 1.44 1.61 1.92 2.28 3.21 4.70 1.00 1.00

Simply

Supprt'd

𝑓𝑐′

ℎ3 𝑓𝑐′

145

Table M-2 - Data for CPL, ML=MR, Ig/Icr=2.3 – Example 3.4.2a – Page 2

Ex. 3.4.2a, pg 2 of 2 P0 = 77850 N fc' = 36 MPa b = 0.5 * h

L = 10000 mm fy = 400 MPa d = 0.9 * h

+ve Moment Ms/Mr = 0.635 ρ m = 0.0115 fr = 0.6 *Mmax/Mcr = 3.00 Ig/Icr m = 2.28 Eb = 200000 MPa

αL/max=ML/Mmax = -1.94 -1.70 -1.50 -1.22 -1.00 -0.67 -0.43 -0.25 0.00

ML = -3.78E+8 -3.31E+8 -2.92E+8 -2.38E+8 -1.95E+8 -1.30E+8 -8.34E+7 -4.87E+7 0.00E+0 N mm

MR = -3.78E+8 -3.31E+8 -2.92E+8 -2.38E+8 -1.95E+8 -1.30E+8 -8.34E+7 -4.87E+7 0.00E+0 N mm


K=1.5–.5(M0/Mm)= 0.029 0.149 0.250 0.389 0.500 0.667 0.786 0.875 1.000

Constant Stiffness Results Using Constant Stiffness Equations

Δg(Gross) 0.33 1.65 2.78 4.33 5.56 7.41 8.74 9.73 11.12 mm

Δcr(Cracked) 0.74 3.76 6.33 9.84 12.65 16.87 19.88 22.14 25.30 mm

Max Uncrack ∆uncr = 0.11 0.55 0.93 1.44 1.85 2.47 2.91 3.24 3.70 mm

Deflection using the S806 Integration Method with Numerical Integration

∆β=0(S806) 4.92 6.43 7.89 10.26 12.40 16.23 19.29 21.51 24.77 mm

Exact Integration ∆Ie(x)

Analytical Δ1= -2.67 -2.28 -1.92 -1.37 -0.92 -0.29 -0.02 0.00 0.00 mm

Analytical Δ2 or Δ1+2= 0.05 0.06 0.07 0.08 0.10 0.15 0.20 0.25 0.21 mm

Analytical Δ3= 4.43 4.78 5.12 5.68 6.22 7.25 8.20 9.08 10.61 mm

Analytical Δ4= 4.43 4.78 5.12 5.68 6.22 7.25 8.20 9.08 10.61 mm


Analytical Δ6= -2.67 -2.28 -1.92 -1.37 -0.92 -0.29 -0.02 0.00 0.00 mm

ΔIe(x)(Exact) 3.63 5.11 6.55 8.79 10.82 14.22 16.75 18.65 21.63 mm

numerical Δmax = 3.63 5.12 6.55 8.79 10.82 14.22 16.76 18.66 21.63 mm

numerical Δmid = 3.63 5.12 6.55 8.79 10.82 14.22 16.76 18.66 21.63 mm

Proposed Method ΔI'e I'e=Icr/[1-γηm (Mcr/Mmax)2] Δmid=K(MmL2)/(12EcI'e)

Ie Bischoff (γ=1) = 2.53E+9 2.53E+9 2.53E+9 2.53E+9 2.53E+9 2.53E+9 2.53E+9 2.53E+9 2.53E+9 mm4

Δγ=1(Approx) 0.70 3.53 5.93 9.23 11.87 15.82 18.65 20.76 23.73 mm

γ=3-2(Mcr/Mmax)= 2.33 2.33 2.33 2.33 2.33 2.33 2.33 2.33 2.33

Bischoff's I'e = 2.78E+9 2.78E+9 2.78E+9 2.78E+9 2.78E+9 2.78E+9 2.78E+9 2.78E+9 2.78E+9 mm4

ΔI'e(Proposed) 0.64 3.22 5.41 8.41 10.82 14.42 17.00 18.93 21.63 mm

% error, proposed 82.45 37.10 17.40 4.26 0.00 1.45 1.45 1.47 0.00

Length:Defl, L/Δ 2758 1956 1527 1138 925 703 597 536 462

Branson's Method ΔIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3

Branson's Ie= 2.49E+9 2.49E+9 2.49E+9 2.49E+9 2.49E+9 2.49E+9 2.49E+9 2.49E+9 2.49E+9 mm4

ΔIe(Branson) 0.71 3.59 6.04 9.40 12.08 16.11 18.99 21.14 24.17 mm

% error, Branson 80.40 29.74 7.73 6.95 11.71 13.32 13.32 13.35 11.71

CSA A23.3 Clause 9.8.2.4 Ie avg=.7Ie max+.15(IeL+IeR) or Ie =0.85Ie max + 0.15 IeL

Ie L (Bransons)= 4.23E+9 3.76E+9 3.38E+9 2.87E+9 2.49E+9 2.15E+9 3.14E+9 2.49E+9 2.49E+9 mm4

Ie R (Bransons)= 4.23E+9 3.76E+9 3.38E+9 2.87E+9 2.49E+9 2.15E+9 3.14E+9 2.49E+9 2.49E+9 mm4

Ie 9.8.2 (Bransons)= 3.01E+9 2.87E+9 2.75E+9 2.60E+9 2.49E+9 2.38E+9 2.68E+9 2.49E+9 2.49E+9 mm4

ΔIe,avg(A23.3) 0.587 3.114 5.455 8.985 12.08 16.801 17.591 21.145 24.17 mm

𝑓𝑐′

146

Figure M-1 - Copy of Figure 3-1 – Midspan Point Load, Ig/Icr=2.3 and Mm/Mcr=3.0

The lines plotted in Figure M-1 use data in bold from Example 3.4.2a as found in Table

M-1 and Table M-2.

147

Table M-3 - Data for CPL, ML=MR, Ig/Icr=3.9 – Example 3.4.2b – Page 1

Example 3.4.2b, pg 1 of 2 Φc = 0.65



Let ML=MR b = 0.5 * h mm α1 = 0.796

d = 0.9 * h mm β1 = 0.880

M0,0 = P0 L/4 = 1.04E+8 N mm Φb = 0.85

End Moment Ms/Mr = 0.635 fy = 400 MPa



P1PL = 122000 112108 98761.9 88255.3 79769.2 71517.2 63815.4 55306.7 41480 N

M0 = 3.05E+8 2.80E+8 2.47E+8 2.21E+8 1.99E+8 1.79E+8 1.60E+8 1.38E+8 1.04E+8 N mm

αL = ML/M0 = -0.66 -0.63 -0.58 -0.53 -0.48 -0.42 -0.35 -0.25 0

αR = MR/M0 = -0.66 -0.63 -0.58 -0.53 -0.48 -0.42 -0.35 -0.25 0

ML = -2.01E+8 -1.77E+8 -1.43E+8 -1.17E+8 -9.57E+7 -7.51E+7 -5.58E+7 -3.46E+7 0.00E+0 N mm

MR = -2.01E+8 -1.77E+8 -1.43E+8 -1.17E+8 -9.57E+7 -7.51E+7 -5.58E+7 -3.46E+7 0.00E+0 N mm


αcr = Mcr/Mmax= 0.625 0.625 0.625 0.625 0.625 0.625 0.625 0.625 0.625


h = 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 mm

d = 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 mm

b = 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 mm

Ig = 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 mm4

αL/max=ML/Mmax= -1.94 -1.70 -1.38 -1.13 -0.92 -0.72 -0.54 -0.33 0.00


Left End Kr L = 3.62 3.18 2.58 2.10 1.72 1.35 0 0 0 MPa

ρ L = 0.0120 0.0103 0.0082 0.0066 0.0053 0.0041 0 0 0

AL=ρ Lbd= 1938 1672 1328 1067 863 669 0 0 0 mm2


ηL=1 – Icr L/Ig = 0.548 0.595 0.661 0.715 0.760 0.806 0 0 0

ML/Mcr = -3.11 -2.72 -2.21 -1.80 -1.48 -1.16 -0.86 -0.53 0.00

Ig/Icr L = 2.21 2.47 2.95 3.51 4.17 5.16 1.00 1.00 1.00

Midspan Kr m = 1.87 1.87 1.87 1.87 1.87 1.87 1.87 1.87 1.87 MPa

ρ m = 0.0058 0.0058 0.0058 0.0058 0.0058 0.0058 0.0058 0.0058 0.0058

Am=ρ mbd= 939 939 939 939 939 939 939 939 939 mm2


ηm=1 – Icr m/Ig = 0.743 0.743 0.743 0.743 0.743 0.743 0.743 0.743 0.743

Mmax/Mcr = 1.60 1.60 1.60 1.60 1.60 1.60 1.60 1.60 1.60

Ig/Icr m = 3.89 3.89 3.89 3.89 3.89 3.89 3.89 3.89 3.89

Right End Kr R = 3.62 3.18 2.58 2.10 1.72 1.35 0 0 0 MPa

ρ R = 0.0120 0.0103 0.0082 0.0066 0.0053 0.0041 0 0 0

AR=ρ Rbd= 1938 1672 1328 1067 863 669 0 0 0 mm2


ηR=1 – Icr R/Ig = 0.548 0.595 0.661 0.715 0.760 0.806 0 0 0

MR/Mcr = -3.11 -2.72 -2.21 -1.80 -1.48 -1.16 -0.86 -0.53 0.00

Ig/Icr R = 2.21 2.47 2.95 3.51 4.17 5.16 1.00 1.00 1.00

Simply

Supprt'd

𝑓𝑐′

ℎ3 𝑓𝑐′

148

Table M-4 - Data for CPL, ML=MR, Ig/Icr=3.9 – Example 3.4.2b – Page 2

Ex. 3.4.2b, pg 2 of 2 P0 = 41480 N fc' = 36 MPa b = 0.5 * h

L = 10000 mm fy = 400 MPa d = 0.9 * h

+ve Moment Ms/Mr = 0.635 ρ m = 0.0058 fr = 0.6 *Mmax/Mcr = 1.60 Ig/Icr m = 3.89 Eb = 200000 MPa

αL/max=ML/Mmax = -1.94 -1.70 -1.38 -1.13 -0.92 -0.72 -0.54 -0.33 0.00

ML = -2.01E+8 -1.77E+8 -1.43E+8 -1.17E+8 -9.57E+7 -7.51E+7 -5.58E+7 -3.46E+7 0.00E+0 N mm

MR = -2.01E+8 -1.77E+8 -1.43E+8 -1.17E+8 -9.57E+7 -7.51E+7 -5.58E+7 -3.46E+7 0.00E+0 N mm


K=1.5–.5(M0/Mm)= 0.029 0.149 0.310 0.436 0.538 0.638 0.731 0.833 1.000


Δg(Gross) 0.17 0.88 1.83 2.58 3.19 3.78 4.33 4.94 5.93 mm

Δcr(Cracked) 0.68 3.43 7.14 10.06 12.42 14.72 16.86 19.23 23.07 mm



∆β=0(S806) 2.84 4.19 6.36 8.34 10.12 11.82 13.40 15.26 18.89 mm


Analytical Δ1= -1.66 -1.28 -0.75 -0.38 -0.14 -0.01 0.00 0.00 0.00 mm


Analytical Δ3= 1.88 2.04 2.29 2.54 2.79 3.08 3.42 3.88 4.95 mm

Analytical Δ4= 1.88 2.04 2.29 2.54 2.79 3.08 3.42 3.88 4.95 mm


Analytical Δ6= -1.66 -1.28 -0.75 -0.38 -0.14 -0.01 0.00 0.00 0.00 mm

ΔIe(x)(Exact) 0.77 1.91 3.59 4.97 6.09 7.12 8.04 9.16 11.35 mm





Δγ=1(Approx) 0.48 2.43 5.07 7.14 8.82 10.45 11.97 13.65 16.38 mm

γ=3-2(Mcr/Mmax)= 1.75 1.75 1.75 1.75 1.75 1.75 1.75 1.75 1.75


ΔI'e(Proposed) 0.33 1.69 3.51 4.95 6.11 7.24 8.30 9.46 11.35 mm

% error, proposed 56.39 11.61 1.99 0.48 0.36 1.78 3.18 3.29 0.00

Length:Defl, L/Δ 13063 5238 2789 2010 1642 1405 1244 1092 881



ΔIe(Branson) 0.40 2.01 4.19 5.90 7.28 8.63 9.88 11.27 13.52 mm

% error, Branson 48.05 5.28 16.74 18.55 19.54 21.23 22.90 23.03 19.11





ΔIe,avg(A23.3) 0.389 2.016 4.297 6.028 7.10 7.263 9.881 11.267 13.52 mm

𝑓𝑐′

149


The lines plotted in Figure M-2 use data in bold from Example 3.4.2b as found in Table

M-3 and Table M-4.

150

Table M-5 - Data for CPL, ML=MR, Ig/Icr=3.8 – Example 3.4.2c – Page 1

Example 3.4.2c, pg 1 of 2 Φc = 0.65 εcu = 0.0035 mm/mm



Let ML=MR b = 0.5 * h mm α1 = 0.796

d = 0.9 * h mm β1 = 0.880 ρ b = 0.00578

M0,0 = P0 L/4 = 1.62E+8 N mm Φb = 0.75 ρ b=α1β1φcf'cεcu/(φbffu(εcu+ffu/Ef))

End Moment Ms/Mr = 0.635 ffu = 690 MPa



P1PL = 190588 175135 162000 147273 129600 117818 98181.8 81000 64800 N

M0 = 4.76E+8 4.38E+8 4.05E+8 3.68E+8 3.24E+8 2.95E+8 2.45E+8 2.03E+8 1.62E+8 N mm

αL = ML/M0 = -0.66 -0.63 -0.6 -0.56 -0.5 -0.45 -0.34 -0.2 0

αR = MR/M0 = -0.66 -0.63 -0.6 -0.56 -0.5 -0.45 -0.34 -0.2 0

ML = -3.14E+8 -2.76E+8 -2.43E+8 -2.06E+8 -1.62E+8 -1.33E+8 -8.35E+7 -4.05E+7 0.00E+0 N mm

MR = -3.14E+8 -2.76E+8 -2.43E+8 -2.06E+8 -1.62E+8 -1.33E+8 -8.35E+7 -4.05E+7 0.00E+0 N mm


αcr = Mcr/Mmax= 0.4 0.4 0.4 0.4 0.4 0.4 0.4 0.4 0.4


h = 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 mm

d = 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 mm

b = 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 mm

Ig = 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 mm4

αL/max=ML/Mmax= -1.94 -1.70 -1.50 -1.27 -1.00 -0.82 -0.52 -0.25 0.00


Left End c L = 229.37 194.37 166.78 137.74 105.07 84.41 52 0 0 mm

AL= 7211 4654 3173 2008 1081 666 233 0 0 mm2

ρ L =AL/bd= 0.0445 0.0287 0.0196 0.0124 0.0067 0.0041 0 0 0


ηL=1 – Icr L/Ig = 0.611 0.725 0.799 0.864 0.922 0.950 1 0 0

ML/Mcr = -4.85 -4.26 -3.75 -3.18 -2.50 -2.05 -1.29 -0.63 0.00

Ig/Icr L = 2.57 3.63 4.98 7.36 12.77 19.88 53.46 1.00 1.00

Midspan c m = 189.95 189.95 189.95 189.95 189.95 189.95 189.95 189.95 189.95 mm

Am= 4388 4388 4388 4388 4388 4388 4388 4388 4388 mm2

ρ m =Am/bd= 0.0271 0.0271 0.0271 0.0271 0.0271 0.0271 0.0271 0.0271 0.0271


ηm=1 – Icr m/Ig = 0.737 0.737 0.737 0.737 0.737 0.737 0.737 0.737 0.737

Mmax/Mcr = 2.50 2.50 2.50 2.50 2.50 2.50 2.50 2.50 2.50

Ig/Icr m = 3.81 3.81 3.81 3.81 3.81 3.81 3.81 3.81 3.81

Right End cR = 229.37 194.37 166.78 137.74 105.07 84.41 51.67 0 0 mm

AR= 7211 4654 3173 2008 1081 666 233 0 0 mm2

ρ R =AR/bd= 0.0445 0.0287 0.0196 0.0124 0.0067 0.0041 0.0014 0.0000 0.0000


ηR=1 – Icr R/Ig = 0.611 0.725 0.799 0.864 0.922 0.950 1 0 0

MR/Mcr = -4.85 -4.26 -3.75 -3.18 -2.50 -2.05 -1.29 -0.63 0.00

Ig/Icr R = 2.57 3.63 4.98 7.36 12.77 19.88 53.46 1.00 1.00

Simply

Supprt'd

𝑓𝑐′

ℎ3 𝑓𝑐′

151

Table M-6 - Data for CPL, ML=MR, Ig/Icr=3.8 – Example 3.4.2c – Page 2

Ex. 3.4.2c, pg 2 of 2 P0 = 64800 N fc' = 36 MPa ρ b = 0.00578

L = 10000 mm ffu = 690 MPa ρ m = 0.0271

+ve Moment Ms/Mr = 0.380 fr = 0.6 *Mmax/Mcr = 2.50 Ig/Icr m = 3.81 Eb = 44000 MPa

αL/max=ML/Mmax = -1.94 -1.70 -1.50 -1.27 -1.00 -0.82 -0.52 -0.25 0.00

ML = -3.14E+8 -2.76E+8 -2.43E+8 -2.06E+8 -1.62E+8 -1.33E+8 -8.35E+7 -4.05E+7 0.00E+0 N mm

MR = -3.14E+8 -2.76E+8 -2.43E+8 -2.06E+8 -1.62E+8 -1.33E+8 -8.35E+7 -4.05E+7 0.00E+0 N mm


K=1.5–.5(M0/Mm)= 0.029 0.149 0.250 0.364 0.500 0.591 0.742 0.875 1.000


Δg(Gross) 0.27 1.38 2.31 3.37 4.63 5.47 6.87 8.10 9.26 mm

Δcr(Cracked) 1.04 5.24 8.81 12.82 17.63 20.83 26.17 30.84 35.25 mm



∆β=0(S806) 4.98 5.30 6.09 7.14 10.38 13.80 23.01 28.99 33.65 mm


Analytical Δ1= -3.86 -3.95 -3.88 -3.57 -2.72 -1.82 -0.17 0.00 0.00 mm


Analytical Δ3= 5.18 5.59 6.00 6.53 7.31 7.94 9.26 10.80 12.75 mm

Analytical Δ4= 5.18 5.59 6.00 6.53 7.31 7.94 9.26 10.80 12.75 mm


Analytical Δ6= -3.86 -3.95 -3.88 -3.57 -2.72 -1.82 -0.17 0.00 0.00 mm

ΔIe(x)(Exact) 2.76 3.43 4.42 6.16 9.48 12.61 18.69 22.29 26.10 mm





Δγ=1(Approx) 0.91 4.62 7.77 11.31 15.55 18.37 23.08 27.21 31.09 mm

γ=3-2(Mcr/Mmax)= 2.20 2.20 2.20 2.20 2.20 2.20 2.20 2.20 2.20


ΔI'e(Proposed) 0.77 3.88 6.53 9.49 13.05 15.42 19.38 22.84 26.10 mm

% error, proposed 72.17 12.98 47.54 53.99 37.67 22.34 3.68 2.45 0.00

Length:Defl, L/Δ 3625 2912 2261 1622 1055 793 535 449 383



ΔIe(Branson) 0.88 4.44 7.47 10.87 14.94 17.66 22.19 26.15 29.88 mm

ACI 440.1R clause 8.3.2.2 Ie=Icr+(βdIg-Icr)(Mcr/Mmax)3 βd=0.2(ρm/ρ b)<1 ρ b=.85β1(f'c/ffu)Efεcu/(Efεcu+ffu)

Ie m (ACI440.1R) = 1.65E+9 1.65E+9 1.65E+9 1.65E+9 1.65E+9 1.65E+9 1.65E+9 1.65E+9 1.65E+9 mm4

ΔIe,βd(ACI440) 0.890 4.500 7.568 11.009 15.14 17.889 22.476 26.489 30.27 mm

Ie L (ACI440.1R) = 2.13E+9 1.53E+9 1.16E+9 8.68E+8 7.20E+8 8.31E+8 2.42E+9 1.65E+9 1.65E+9 mm4

Ie R (ACI440.1R) = 2.13E+9 1.53E+9 1.16E+9 8.68E+8 7.20E+8 8.31E+8 2.42E+9 1.65E+9 1.65E+9 mm5

Ie 9.8.2 (& ACI440.1R)= 1.79E+9 1.62E+9 1.50E+9 1.42E+9 1.37E+9 1.41E+9 1.88E+9 1.65E+9 1.65E+9 mm4

ΔIe,avg(A23.3) 0.820 4.598 8.310 12.836 18.22 21.021 19.713 26.489 30.27 mm

𝑓𝑐′

152


The lines plotted in Figure M-3 use data in bold from Example 3.4.2c as found in Table

M-5 and Table M-6.

153

Table M-7 - Data for CPL, ML=MR, Ig/Icr=12 – Example 3.4.2d – Page 1

Example 3.4.2d, pg 1 of 2 Φc = 0.65 εcu = 0.0035 mm/mm



Let ML=MR b = 0.5 * h mm α1 = 0.796

d = 0.9 * h mm β1 = 0.880 ρ b = 0.00578

M0,0 = P0 L/4 = 1.04E+8 N mm Φb = 0.75 ρ b=α1β1φcf'cεcu/(φbffu(εcu+ffu/Ef))

End Moment Ms/Mr = 0.635 ffu = 690 MPa



P1PL = 122000 112108 103700 94272.7 82960 75418.2 62848.5 51850 41480 N

M0 = 3.05E+8 2.80E+8 2.59E+8 2.36E+8 2.07E+8 1.89E+8 1.57E+8 1.30E+8 1.04E+8 N mm

αL = ML/M0 = -0.66 -0.63 -0.6 -0.56 -0.5 -0.45 -0.34 -0.2 0

αR = MR/M0 = -0.66 -0.63 -0.6 -0.56 -0.5 -0.45 -0.34 -0.2 0

ML = -2.01E+8 -1.77E+8 -1.56E+8 -1.32E+8 -1.04E+8 -8.48E+7 -5.34E+7 -2.59E+7 0.00E+0 N mm

MR = -2.01E+8 -1.77E+8 -1.56E+8 -1.32E+8 -1.04E+8 -8.48E+7 -5.34E+7 -2.59E+7 0.00E+0 N mm


αcr = Mcr/Mmax= 0.625 0.625 0.625 0.625 0.625 0.625 0.625 0.625 0.625


h = 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 mm

d = 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 mm

b = 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 mm

Ig = 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 mm4

αL/max=ML/Mmax= -1.94 -1.70 -1.50 -1.27 -1.00 -0.82 -0.52 -0.25 0.00


Left End c L = 134.00 115.59 100.46 84.01 64.93 52.56 0 0 0 mm

AL= 1883 1340 978 659 378 241 0 0 0 mm2

ρ L =AL/bd= 0.0116 0.0083 0.0060 0.0041 0.0023 0.0015 0 0 0


ηL=1 – Icr L/Ig = 0.872 0.905 0.928 0.950 0.970 0.981 0 0 0

ML/Mcr = -3.11 -2.72 -2.40 -2.04 -1.60 -1.31 -0.82 -0.40 0.00

Ig/Icr L = 7.79 10.52 13.98 20.08 33.76 51.65 1.00 1.00 1.00

Midspan c m = 106.94 106.94 106.94 106.94 106.94 106.94 106.94 106.94 106.94 mm

Am= 1124 1124 1124 1124 1124 1124 1124 1124 1124 mm2

ρ m =Am/bd= 0.0069 0.0069 0.0069 0.0069 0.0069 0.0069 0.0069 0.0069 0.0069


ηm=1 – Icr m/Ig = 0.919 0.919 0.919 0.919 0.919 0.919 0.919 0.919 0.919

Mmax/Mcr = 1.60 1.60 1.60 1.60 1.60 1.60 1.60 1.60 1.60

Ig/Icr m = 12.32 12.32 12.32 12.32 12.32 12.32 12.32 12.32 12.32

Right End cR = 134.00 115.59 100.46 84.01 64.93 52.56 0.00 0 0 mm

AR= 1883 1340 978 659 378 241 0 0 0 mm2

ρ R =AR/bd= 0.0116 0.0083 0.0060 0.0041 0.0023 0.0015 0.0000 0.0000 0.0000


ηR=1 – Icr R/Ig = 0.872 0.905 0.928 0.950 0.970 0.981 0 0 0

MR/Mcr = -3.11 -2.72 -2.40 -2.04 -1.60 -1.31 -0.82 -0.40 0.00

Ig/Icr R = 7.79 10.52 13.98 20.08 33.76 51.65 1.00 1.00 1.00

Simply

Supprt'd

𝑓𝑐′

ℎ3 𝑓𝑐′

154

Table M-8 - Data for CPL, ML=MR, Ig/Icr=12 – Example 3.4.2d – Page 2

Ex. 3.4.2d, pg 2 of 2 P0 = 41480 N fc' = 36 MPa ρ b = 0.00578

L = 10000 mm ffu = 690 MPa ρ m = 0.0069

+ve Moment Ms/Mr = 0.400 fr = 0.6 *Mmax/Mcr = 1.60 Ig/Icr m = 12.32 Eb = 44000 MPa

αL/max=ML/Mmax = -1.94 -1.70 -1.50 -1.27 -1.00 -0.82 -0.52 -0.25 0.00

ML = -2.01E+8 -1.77E+8 -1.56E+8 -1.32E+8 -1.04E+8 -8.48E+7 -5.34E+7 -2.59E+7 0.00E+0 N mm

MR = -2.01E+8 -1.77E+8 -1.56E+8 -1.32E+8 -1.04E+8 -8.48E+7 -5.34E+7 -2.59E+7 0.00E+0 N mm


K=1.5–.5(M0/Mm)= 0.029 0.149 0.250 0.364 0.500 0.591 0.742 0.875 1.000


Δg(Gross) 0.17 0.88 1.48 2.15 2.96 3.50 4.40 5.18 5.93 mm

Δcr(Cracked) 2.15 10.85 18.25 26.55 36.51 43.14 54.20 63.88 73.01 mm



∆β=0(S806) 6.76 8.63 12.14 15.51 23.22 30.48 40.02 47.89 56.64 mm


Analytical Δ1= -4.97 -4.47 -3.82 -2.79 -1.25 -0.34 0.00 0.00 0.00 mm


Analytical Δ3= 4.83 5.23 5.63 6.16 6.94 7.57 8.94 10.61 12.85 mm

Analytical Δ4= 4.83 5.23 5.63 6.16 6.94 7.57 8.94 10.61 12.85 mm


Analytical Δ6= -4.97 -4.47 -3.82 -2.79 -1.25 -0.34 0.00 0.00 0.00 mm

ΔIe(x)(Exact) 0.06 1.91 4.09 7.29 12.09 15.34 19.11 22.67 27.15 mm





Δγ=1(Approx) 1.38 6.96 11.70 17.02 23.40 27.66 34.75 40.95 46.81 mm

γ=3-2(Mcr/Mmax)= 1.75 1.75 1.75 1.75 1.75 1.75 1.75 1.75 1.75

Proposed I'e = 1.18E+9 1.18E+9 1.18E+9 1.18E+9 1.18E+9 1.18E+9 1.18E+9 1.18E+9 1.18E+9 mm4

ΔI'e(Proposed) 0.80 4.04 6.79 9.87 13.58 16.04 20.16 23.76 27.15 mm

% error, proposed 1268.65 110.81 66.01 35.46 12.32 4.57 5.46 4.80 0.00

Length:Defl, L/Δ 171386 5223 2446 1372 827 652 523 441 368



ΔIe(Branson) 0.57 2.88 4.85 7.05 9.70 11.46 14.40 16.97 19.40 mm

ACI 440.1R clause 8.3.2.2 Ie=Icr+(βdIg-Icr)(Mcr/Mmax)3 βd=0.2(ρm/ρ b)<1 ρ b=.85β1(f'c/ffu)Efεcu/(Efεcu+ffu)


ΔIe,βd(ACI440) 1.453 7.342 12.348 17.961 24.70 29.187 36.670 43.218 49.39 mm




ΔIe,avg(A23.3) 1.410 7.683 13.579 20.389 27.36 29.348 36.670 43.218 49.39 mm

𝑓𝑐′

155

Figure M-4 - Copy of Figure 3-4 – Midspan Point Load, Ig/Icr=12 and Mm/Mcr=1.6

The lines plotted in Figure M-4 use data in bold from Example 3.4.2d as found in Table

M-7 and Table M-8.

156

Third-Point Loaded Examples – Data for Section 3.5 Appendix N

This appendix provides the data and graphs for the example prismatic concrete members

with equal point loads at third points from Section 3.5 of this report. The concrete

members provided are beams as they are more likely to incur third point loading. The

calculations in this appendix use the general methodology indicated in Appendix J.

All members shown in this appendix have the same concrete cross-section. The point

loads vary with the end-moments such that all members for one graph are generated

with equal maximum moment. The concrete cross-section used is 300 mm wide by 600

mm deep with tension reinforcing at a depth of 540 mm and compression reinforcing

neglected. The end-moments are equal (𝑀𝐿 = 𝑀𝑅) for the first and third sets of

examples and the right end-moment is set equal to zero for two other sets. The end-

moments range provided is 3 < −𝑀𝐿 𝑀𝑚⁄ ≤ 0; this exceeds the proposed valid range.

The first two examples are selected as steel reinforced members to demonstrate typical

steel reinforced members with two ends continuous and with one end continuous.

These members reach the cracking moment at 𝑀𝑚𝑎𝑥 𝑀𝑐𝑟⁄ = 2.2. The other two

examples are similar to the first two example, but are designed using GFRP and

𝑀𝑚𝑎𝑥 𝑀𝑐𝑟⁄ = 1.4. In order for these members to meet deflection requirements, the

bottom GFRP bars are increase to 𝑀𝑟 = 2.44𝑀𝑠.

Results using the for simply supported members are shown separately from results

using the proposed equation, ∗ = − 0.1(𝑀𝐿 − 1.5𝑀𝑅)/𝑀𝑐𝑟, for equal third-point

loading of continuous members.

157

Table N-1 - Data for 2PL, ML=MR, Ig/Icr=3.0 – Example 3.5.2a – Page 1


⅓L & ⅔L Point Loads P0/2 = 42720 N fc' = 36 MPa fr = 0.6 * MPa


Let ML=MR b = 0.5* h mm α1 = 0.796

d = 0.9 * h mm β1 = 0.880

M0,0 = P0 L/6 = 1.42E+8 N mm Φb = 0.85

End Moment Ms/Mr = 0.635 fy = 400 MPa α R = 1 *α L



αL/max=ML/Mmax= -3.00 -2.50 -2.00 -1.50 -1.00 -0.80 -0.60 -0.30 0

αR/max=MR/Mmax= -3.00 -2.50 -2.00 -1.50 -1.00 -0.80 -0.60 -0.30 0

P2PL/2 = 170880 149520 128160 106800 85440 76896 68352 55536 42720 N

M0 = 5.70E+8 4.98E+8 4.27E+8 3.56E+8 2.85E+8 2.56E+8 2.28E+8 1.85E+8 1.42E+8 N mm

αL = ML/M0 = -0.75 -0.71 -0.67 -0.60 -0.50 -0.44 -0.38 -0.23 0

αR = MR/M0 = -0.75 -0.71 -0.67 -0.60 -0.50 -0.44 -0.38 -0.23 0

ML = -4.27E+8 -3.56E+8 -2.85E+8 -2.14E+8 -1.42E+8 -1.14E+8 -8.54E+7 -4.27E+7 0.00E+0 N mm

MR = -4.27E+8 -3.56E+8 -2.85E+8 -2.14E+8 -1.42E+8 -1.14E+8 -8.54E+7 -4.27E+7 0.00E+0 N mm

Mm = 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 N mm

Mmax = 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 N mm

αcr = Mcr/Mmax= 0.455 0.455 0.455 0.455 0.455 0.455 0.455 0.455 0.455


h = 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 mm

d = 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 mm

b = 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 mm

Ig = 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 mm4


Left End Kr L = 7.69 6.41 5.13 3.85 2.56 2.05 1.54 0 0 MPa

ρ L = 0.0319 0.0242 0.0181 0.0128 0.0081 0.0064 0.0047 0 0

AL=ρLbd= 5171 3919 2925 2075 1320 1038 766 0 0 mm2


ηL=1 – Icr L/Ig = 0.119 0.260 0.392 0.524 0.663 0.721 0.783 0 0

ML/Mcr = -6.59 -5.49 -4.40 -3.30 -2.20 -1.76 -1.32 -0.66 0.00

Ig/Icr L = 1.13 1.35 1.65 2.10 2.97 3.59 4.61 1.00 1.00

Midspan Kr m = 2.56 2.56 2.56 2.56 2.56 2.56 2.56 2.56 2.56 MPa

ρ m = 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081

Am=ρmbd= 1320 1320 1320 1320 1320 1320 1320 1320 1320 mm2


ηm=1 – Icr m/Ig = 0.663 0.663 0.663 0.663 0.663 0.663 0.663 0.663 0.663

Mmax/Mcr = 2.20 2.20 2.20 2.20 2.20 2.20 2.20 2.20 2.20

Ig/Icr m = 2.97 2.97 2.97 2.97 2.97 2.97 2.97 2.97 2.97

Right End Kr R = 7.69 6.41 5.13 3.85 2.56 2.05 1.54 0 0 MPa

ρ R = 0.0319 0.0242 0.0181 0.0128 0.0081 0.0064 0.0047 0 0

AR=ρRbd= 5171 3919 2925 2075 1320 1038 766 0 0 mm2


ηR=1 – Icr R/Ig = 0.119 0.260 0.392 0.524 0.663 0.721 0.783 0 0

MR/Mcr = -6.59 -5.49 -4.40 -3.30 -2.20 -1.76 -1.32 -0.66 0.00

Ig/Icr R = 1.13 1.35 1.65 2.10 2.97 3.59 4.61 1.00 1.00

Simply

Supprt'd

𝑓𝑐′

ℎ3 𝑓𝑐′

158

Table N-2 - Data for 2PL, ML=MR, Ig/Icr=3.0 – Example 3.5.2a – Page 2

Ex. 3.5.2a, pg 2 of 2 P0 = 42720 N fc' = 36 MPa b = 0.5 * h

L = 10000 mm fy = 400 MPa d = 0.9 * h

+ve Moment Ms/Mr = 0.635 ρ m = 0.0081 fr = 0.6 *

Mmax/Mcr = 2.20 Ig/Icr m = 2.97 Eb = 200000 MPa

αL/max=ML/Mmax = -3.00 -2.50 -2.00 -1.50 -1.00 -0.80 -0.60 -0.30 0.00

ML = -4.27E+8 -3.56E+8 -2.85E+8 -2.14E+8 -1.42E+8 -1.14E+8 -8.54E+7 -4.27E+7 0.00E+0 N mm

M(⅓L) = 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8

Mm = 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 N mm

M(⅔L) = 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8

MR = -4.27E+8 -3.56E+8 -2.85E+8 -2.14E+8 -1.42E+8 -1.14E+8 -8.54E+7 -4.27E+7 0.00E+0 N mm

K=27/23-4M0/23Mm= 0.478 0.565 0.652 0.739 0.826 0.861 0.896 0.948 1.000


Δg(Gross) 4.97 5.88 6.78 7.69 8.59 8.95 9.32 9.86 10.40 mm

Δcr(Cracked) 14.75 17.43 20.12 22.80 25.48 26.55 27.63 29.24 30.85 mm


Deflection using the S806 Method with Numerical Integration

∆β=0(S806) 19.85 20.76 21.77 23.26 25.12 26.03 27.08 28.49 30.15 mm

Exact Integration Ie(x) (uses numerical integration; analytical integration not performed)

Δmax,Ie(x)(Exact) 16.41 17.20 18.24 19.56 21.18 21.90 22.62 23.71 25.01 mm

Length:Defl, L/Δmax= 609 581 548 511 472 457 442 422 400

ΔIe(x)(Exact) 16.41 17.20 18.24 19.56 21.18 21.90 22.62 23.71 25.01 mm

Δmax,I'e(Bischoff) 11.96 14.13 16.31 18.48 20.65 21.52 22.39 23.70 25.00 mm

Proposed Method I'e=Icr/[1-γηm (Mcr/Mmax)2] Δmid=K(23MmL2)/(216EcI'e)


Δγ=1(Approx) 12.73 15.04 17.36 19.67 21.99 22.91 23.84 25.23 26.61 mm

γ=1.7-.7(Mcr/Mmax)= 1.38 1.38 1.38 1.38 1.38 1.38 1.38 1.38 1.38


ΔI'e(Bischoff) 11.96 14.13 16.30 18.48 20.65 21.52 22.39 23.69 25.00 mm

% error, Bischoff's 27.13 17.87 10.62 5.54 2.52 1.73 1.02 0.04 0.04

γ*=γ-.1(ML-1.5MR)/Mcr 1.05 1.11 1.16 1.22 1.27 1.29 1.32 1.35 1.38

I'e* (using γ* )= 2.13E+9 2.15E+9 2.17E+9 2.19E+9 2.21E+9 2.21E+9 2.22E+9 2.23E+9 2.25E+9 mm4

ΔI'e*(Proposed) 12.62 14.79 16.91 18.99 21.04 21.84 22.64 23.83 25.00 mm



ΔIe(Branson) 12.45 14.71 16.97 19.24 21.50 22.41 23.31 24.67 26.03 mm

% error, Branson 24.13 14.49 6.94 1.65 1.49 2.31 3.05 4.07 4.07





ΔIe,avg(A23.3) 9.14 11.71 14.64 18.01 21.50 22.21 20.83 24.67 26.03 mm

𝑓𝑐′

159

Figure N-1 - Copy of Figure 3-5 – Third-Point Loaded, Ig/Icr=3 and Mm/Mcr=2.2

The lines plotted in Figure N-1 use data in bold from Example 3.5.2a as found in Table

N-1 and Table N-2.

160

Table N-3 - Data for 2PL, MR=0, Ig/Icr=3.0 – Example 3.5.2b – Page 1




Let MR=0 b = 0.5* h mm α1 = 0.796

d = 0.9 * h mm β1 = 0.880

M0,0 = P0 L/6 = 1.42E+8 N mm Φb = 0.85




αL/max=ML/Mmax= -3.00 -2.50 -2.00 -1.50 -1.00 -0.80 -0.60 -0.30 0

αR/max=MR/Mmax= 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0

P2PL/2 = 85440 78320 71200 64080 56960 54112 51264 46992 42720 N

M0 = 2.85E+8 2.61E+8 2.37E+8 2.14E+8 1.90E+8 1.80E+8 1.71E+8 1.57E+8 1.42E+8 N mm

αL = ML/M0 = -1.50 -1.36 -1.20 -1.00 -0.75 -0.63 -0.50 -0.27 0

αR = MR/M0 = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0

ML = -4.27E+8 -3.56E+8 -2.85E+8 -2.14E+8 -1.42E+8 -1.14E+8 -8.54E+7 -4.27E+7 0.00E+0 N mm

MR = 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 N mm

Mm = 7.12E+7 8.31E+7 9.49E+7 1.07E+8 1.19E+8 1.23E+8 1.28E+8 1.35E+8 1.42E+8 N mm


αcr = Mcr/Mmax= 0.455 0.455 0.455 0.455 0.455 0.455 0.455 0.455 0.455


h = 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 mm

d = 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 mm

b = 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 mm

Ig = 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 mm4


Left End Kr L = 7.69 6.41 5.13 3.85 2.56 2.05 1.54 0 0 MPa

ρ L = 0.0319 0.0242 0.0181 0.0128 0.0081 0.0064 0.0047 0 0

AL=ρLbd= 5171 3919 2925 2075 1320 1038 766 0 0 mm2


ηL=1 – Icr L/Ig = 0.119 0.260 0.392 0.524 0.663 0.721 0.783 0 0

ML/Mcr = -6.59 -5.49 -4.40 -3.30 -2.20 -1.76 -1.32 -0.66 0.00

Ig/Icr L = 1.13 1.35 1.65 2.10 2.97 3.59 4.61 1.00 1.00

Midspan Kr m = 2.56 2.56 2.56 2.56 2.56 2.56 2.56 2.56 2.56 MPa

ρ m = 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081

Am=ρmbd= 1320 1320 1320 1320 1320 1320 1320 1320 1320 mm2


ηm=1 – Icr m/Ig = 0.663 0.663 0.663 0.663 0.663 0.663 0.663 0.663 0.663

Mmax/Mcr = 2.20 2.20 2.20 2.20 2.20 2.20 2.20 2.20 2.20

Ig/Icr m = 2.97 2.97 2.97 2.97 2.97 2.97 2.97 2.97 2.97

Right End Kr R = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0 0 MPa

ρ R = 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0

AR=ρRbd= 0 0 0 0 0 0 0 0 0 mm2


ηR=1 – Icr R/Ig = 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0 0

MR/Mcr = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

Ig/Icr R = 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00

Simply

Supprt'd

𝑓𝑐′

ℎ3 𝑓𝑐′

161

Table N-4 - Data for 2PL, MR=0, Ig/Icr=3.0 – Example 3.5.2b – Page 2

Ex. 3.5.2b, pg 2 of 2 P0 = 42720 N fc' = 36 MPa b = 0.5 * h

L = 10000 mm fy = 400 MPa d = 0.9 * h



αL/max=ML/Mmax = -3.00 -2.50 -2.00 -1.50 -1.00 -0.80 -0.60 -0.30 0.00

ML = -4.27E+8 -3.56E+8 -2.85E+8 -2.14E+8 -1.42E+8 -1.14E+8 -8.54E+7 -4.27E+7 0.00E+0 N mm

M(⅓L) = 0.00E+0 2.37E+7 4.75E+7 7.12E+7 9.49E+7 1.04E+8 1.14E+8 1.28E+8 1.42E+8

Mm = 7.12E+7 8.31E+7 9.49E+7 1.07E+8 1.19E+8 1.23E+8 1.28E+8 1.35E+8 1.42E+8 N mm

M(⅔L) = 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8 1.42E+8

MR = 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 N mm

K=27/23-4M0/23Mm= 0.478 0.627 0.739 0.826 0.896 0.920 0.942 0.973 1.000


Δg(Gross) 2.49 3.81 5.13 6.44 7.76 8.29 8.82 9.61 10.40 mm

Δcr(Cracked) 7.38 11.29 15.20 19.11 23.02 24.59 26.15 28.50 30.85 mm



∆β=0(S806) 10.53 12.81 15.60 19.11 22.44 23.88 25.46 27.72 30.15 mm


Δmax,Ie(x)(Exact) 9.16 10.50 12.37 14.86 17.90 19.19 20.50 22.63 25.01 mm


ΔIe(x)(Exact) 7.18 8.96 11.25 14.24 17.61 19.02 20.44 22.63 25.01 mm

numerical ΔI'e = 5.98 9.15 12.32 15.49 18.66 19.93 21.20 23.10 25.00 mm




Δγ=1(Approx) 6.36 9.74 13.11 16.49 19.86 21.21 22.56 24.59 26.61 mm

γ=1.7-.7(Mcr/Mmax)= 1.38 1.38 1.38 1.38 1.38 1.38 1.38 1.38 1.38


ΔI'e(Bischoff) 5.98 9.15 12.32 15.49 18.66 19.93 21.19 23.10 25.00 mm

% error, Bischoff's 16.71 2.11 9.50 8.79 5.94 4.77 3.67 2.07 0.04

γ*=γ-.1(ML-1.5MR)/Mcr 2.04 1.93 1.82 1.71 1.60 1.56 1.51 1.45 1.38


ΔI'e*(Proposed) 5.31 8.30 11.40 14.62 17.96 19.33 20.72 22.84 25.00 mm



ΔIe(Branson) 6.22 9.52 12.83 16.13 19.43 20.75 22.07 24.05 26.03 mm

% error, Branson 13.28 6.31 14.00 13.27 10.30 9.08 7.94 6.27 4.07





ΔIe,avg(A23.3) 5.27 8.44 11.88 15.59 19.43 20.66 20.82 24.05 26.03 mm

𝑓𝑐′

162

Figure N-2 - Copy of Figure 3-6 – Third-Point Loaded, Ig/Icr=3, Mmax/Mcr=2.2, MR=0

The lines plotted in Figure N-2 use data in bold from Example 3.5.2b as found in Table

N-3 and Table N-4.

163

Table N-5 - Data for 2PL, ML=MR, Ig/Icr=12 – Example 3.5.2c – Page 1

Example 3.5.2c, pg 1 of 2 Φc = 0.65 εcu = 0.0035 mm/mm



Let ML=MR b = 0.5* h mm α1 = 0.796

d = 0.9 * h mm β1 = 0.880 ρ b=α1β1φcf'cεcu/(φbffu(εcu+ffu/Ef))

M0,0 = P0 L/6 = 9.10E+7 N mm Φb = 0.75 ρ b = 0.00578

End Moment Ms/Mr = 0.635 ffu = 690 MPa α R = 1 *α L



αL/max=ML/Mmax= -3.00 -2.50 -2.00 -1.50 -1.00 -0.80 -0.60 -0.30 0

αR/max=MR/Mmax= -3.00 -2.50 -2.00 -1.50 -1.00 -0.80 -0.60 -0.30 0

P2PL/2 = 109200 95550 81900 68250 54600 49140 43680 35490 27300 N

M0 = 3.64E+8 3.19E+8 2.73E+8 2.28E+8 1.82E+8 1.64E+8 1.46E+8 1.18E+8 9.10E+7 N mm

αL = ML/M0 = -0.75 -0.71 -0.67 -0.60 -0.50 -0.44 -0.38 -0.23 0

αR = MR/M0 = -0.75 -0.71 -0.67 -0.60 -0.50 -0.44 -0.38 -0.23 0

ML = -2.73E+8 -2.28E+8 -1.82E+8 -1.37E+8 -9.10E+7 -7.28E+7 -5.46E+7 -2.73E+7 0.00E+0 N mm

MR = -2.73E+8 -2.28E+8 -1.82E+8 -1.37E+8 -9.10E+7 -7.28E+7 -5.46E+7 -2.73E+7 0.00E+0 N mm

Mm = 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 N mm


αcr = Mcr/Mmax= 0.712 0.712 0.712 0.712 0.712 0.712 0.712 0.712 0.712


h = 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 mm

d = 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 mm

b = 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 mm

Ig = 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 mm4


Left End c L = 191.94 154.34 119.60 87.15 56.58 44.81 0 0 0 mm

AL= 4506 2630 1449 714 282 173 0 0 0 mm2

ρ L =AL/bd= 0.0278 0.0162 0.0089 0.0044 0.0017 0.0011 0 0 0


ηL=1 – Icr L/Ig = 0.732 0.829 0.898 0.946 0.978 0.986 0 0 0

ML/Mcr = -4.21 -3.51 -2.81 -2.11 -1.40 -1.12 -0.84 -0.42 0.00

Ig/Icr L = 3.73 5.83 9.81 18.64 44.53 71.16 1.00 1.00 1.00

Midspan c m = 107.30 107.30 107.30 107.30 107.30 107.30 107.30 107.30 107.30 mm

Am= 1133 1133 1133 1133 1133 1133 1133 1133 1133 mm2

ρ m =Am/bd= 0.0070 0.0070 0.0070 0.0070 0.0070 0.0070 0.0070 0.0070 0.0070


ηm=1 – Icr m/Ig = 0.918 0.918 0.918 0.918 0.918 0.918 0.918 0.918 0.918

Mmax/Mcr = 1.40 1.40 1.40 1.40 1.40 1.40 1.40 1.40 1.40

Ig/Icr m = 12.23 12.23 12.23 12.23 12.23 12.23 12.23 12.23 12.23

Right End cR = 191.94 154.34 119.60 87.15 56.58 44.81 0.00 0 0 mm

AR= 4506 2630 1449 714 282 173 0 0 0 mm2

ρ R =AR/bd= 0.0278 0.0162 0.0089 0.0044 0.0017 0.0011 0.0000 0.0000 0.0000


ηR=1 – Icr R/Ig = 0.732 0.829 0.898 0.946 0.978 0.986 0 0 0

MR/Mcr = -4.21 -3.51 -2.81 -2.11 -1.40 -1.12 -0.84 -0.42 0.00

Ig/Icr R = 3.73 5.83 9.81 18.64 44.53 71.16 1.00 1.00 1.00

Simply

Supprt'd

𝑓𝑐′

ℎ3 𝑓𝑐′

164

Table N-6 - Data for 2PL, ML=MR, Ig/Icr=12 – Example 3.5.2c – Page 2

Ex. 3.5.2c, pg 2 of 2 P0 = 27300 N fc' = 36 MPa ρ b = 0.00578

L = 10000 mm ffu = 690 MPa ρ m = 0.0070

+ve Moment Ms/Mr = 0.350 fr = 0.6

Mmax/Mcr = 1.40 Ig/Icr m = 12.23 Eb = 44000 *

αL/max=ML/Mmax = -3.00 -2.50 -2.00 -1.50 -1.00 -0.80 -0.60 -0.30 0.00

ML = -2.73E+8 -2.28E+8 -1.82E+8 -1.37E+8 -9.10E+7 -7.28E+7 -5.46E+7 -2.73E+7 0.00E+0 N mm

M(⅓L) = 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7

Mm = 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 N mm

M(⅔L) = 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7

MR = -2.73E+8 -2.28E+8 -1.82E+8 -1.37E+8 -9.10E+7 -7.28E+7 -5.46E+7 -2.73E+7 0.00E+0 N mm

K=27/23-4M0/23Mm= 0.478 0.565 0.652 0.739 0.826 0.861 0.896 0.948 1.000


Δg(Gross) 3.18 3.76 4.34 4.91 5.49 5.72 5.95 6.30 6.65 mm

Δcr(Cracked) 38.89 45.96 53.03 60.10 67.17 70.00 72.83 77.07 81.31 mm



∆β=0(S806) 51.80 52.11 52.80 55.01 60.62 63.75 65.38 68.15 72.24 mm


Δmax,Ie(x)(Exact) 25.27 25.54 26.47 28.51 31.71 32.75 33.36 34.43 35.88 mm


ΔIe(x)(Exact) 25.27 25.54 26.47 28.51 31.71 32.75 33.36 34.43 35.88 mm




Δγ=1(Approx) 20.79 24.57 28.34 32.12 35.90 37.42 38.93 41.19 43.46 mm

γ=1.7-.7(Mcr/Mmax)= 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20


ΔI'e(Bischoff) 17.14 20.25 23.37 26.48 29.60 30.85 32.09 33.96 35.83 mm

% error, Bischoff's 32.18 20.71 11.72 7.10 6.66 5.81 3.81 1.37 0.14

γ*=γ-.1(ML-1.5MR)/Mcr 0.99 1.03 1.06 1.10 1.13 1.15 1.16 1.18 1.20


ΔI'e*(Proposed) 20.95 24.01 26.84 29.43 31.80 32.68 33.52 34.72 35.83 mm



ΔIe(Branson) 7.69 9.09 10.49 11.89 13.29 13.85 14.41 15.25 16.09 mm

ACI 440.1R clause 8.3.2.2 Ie=Icr+(βdIg-Icr)(Mcr/Mmax)3 βd=0.2(ρm/ρ b)<1 βd= 0.242


ΔIe,βd(ACI440) 22.77 26.91 31.05 35.19 39.33 40.99 42.65 45.13 47.61 mm




ΔIe,avg(A23.3) 17.85 25.11 33.30 40.99 42.82 38.11 42.65 45.13 47.61 mm

𝑓𝑐′

165

Figure N-3 - Copy of Figure 3-7 – Third-Point Loaded, Ig/Icr=12 and Mm/Mcr=1.4

The lines plotted in Figure N-3 use data in bold from Example 3.5.2c as found in Table

N-5 and Table N-6.

166

Table N-7 - Data for 2PL, MR=0, Ig/Icr=12 – Example 3.5.2d – Page 1

Example 3.5.2d, pg 1 of 2 Φc = 0.65 εcu = 0.0035 mm/mm



Let MR=0 b = 0.5* h mm α1 = 0.796


M0,0 = P0 L/6 = 9.10E+7 N mm Φb = 0.75 ρ b = 0.00578




αL/max=ML/Mmax= -3.00 -2.50 -2.00 -1.50 -1.00 -0.80 -0.60 -0.30 0

αR/max=MR/Mmax= 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0

P2PL/2 = 54600 50050 45500 40950 36400 34580 32760 30030 27300 N

M0 = 1.82E+8 1.67E+8 1.52E+8 1.37E+8 1.21E+8 1.15E+8 1.09E+8 1.00E+8 9.10E+7 N mm

αL = ML/M0 = -1.50 -1.36 -1.20 -1.00 -0.75 -0.63 -0.50 -0.27 0

αR = MR/M0 = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0

ML = -2.73E+8 -2.28E+8 -1.82E+8 -1.37E+8 -9.10E+7 -7.28E+7 -5.46E+7 -2.73E+7 0.00E+0 N mm

MR = 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 N mm

Mm = 4.55E+7 5.31E+7 6.07E+7 6.83E+7 7.58E+7 7.89E+7 8.19E+7 8.65E+7 9.10E+7 N mm


αcr = Mcr/Mmax= 0.712 0.712 0.712 0.712 0.712 0.712 0.712 0.712 0.712


h = 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 mm

d = 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 mm

b = 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 mm

Ig = 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 mm4


Left End c L = 191.94 154.34 119.60 87.15 56.58 44.81 0 0 0 mm

AL= 4506 2630 1449 714 282 173 0 0 0 mm2

ρ L =AL/bd= 0.0278 0.0162 0.0089 0.0044 0.0017 0.0011 0 0 0


ηL=1 – Icr L/Ig = 0.732 0.829 0.898 0.946 0.978 0.986 0 0 0

ML/Mcr = -4.21 -3.51 -2.81 -2.11 -1.40 -1.12 -0.84 -0.42 0.00

Ig/Icr L = 3.73 5.83 9.81 18.64 44.53 71.16 1.00 1.00 1.00

Midspan c m = 107.30 107.30 107.30 107.30 107.30 107.30 107.30 107.30 107.30 mm

Am= 1133 1133 1133 1133 1133 1133 1133 1133 1133 mm2

ρ m =Am/bd= 0.0070 0.0070 0.0070 0.0070 0.0070 0.0070 0.0070 0.0070 0.0070


ηm=1 – Icr m/Ig = 0.918 0.918 0.918 0.918 0.918 0.918 0.918 0.918 0.918

Mmax/Mcr = 1.40 1.40 1.40 1.40 1.40 1.40 1.40 1.40 1.40

Ig/Icr m = 12.23 12.23 12.23 12.23 12.23 12.23 12.23 12.23 12.23

Right End cR = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0 0 mm

AR= 0 0 0 0 0 0 0 0 0 mm2

ρ R =AR/bd= 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000


ηR=1 – Icr R/Ig = 0.000 0.000 0.000 0.000 0.000 0.000 0 0 0

MR/Mcr = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

Ig/Icr R = 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00

Simply

Supprt'd

𝑓𝑐′

ℎ3 𝑓𝑐′

167

Table N-8 - Data for 2PL, MR=0, Ig/Icr=12 – Example 3.5.2d – Page 2

Ex. 3.5.2d, pg 2 of 2 P0 = 27300 N fc' = 36 MPa ρ b = 0.00578

L = 10000 mm ffu = 690 MPa ρ m = 0.0070

+ve Moment Ms/Mr = 0.350 fr = 0.6 *


αL/max=ML/Mmax = -3.00 -2.50 -2.00 -1.50 -1.00 -0.80 -0.60 -0.30 0.00

ML = -2.73E+8 -2.28E+8 -1.82E+8 -1.37E+8 -9.10E+7 -7.28E+7 -5.46E+7 -2.73E+7 0.00E+0 N mm

M(⅓L) = 0.00E+0 1.52E+7 3.03E+7 4.55E+7 6.07E+7 6.67E+7 7.28E+7 8.19E+7 9.10E+7

Mm = 4.55E+7 5.31E+7 6.07E+7 6.83E+7 7.58E+7 7.89E+7 8.19E+7 8.65E+7 9.10E+7 N mm

M(⅔L) = 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7 9.10E+7

MR = 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 N mm

K=27/23-4M0/23Mm= 0.478 0.627 0.739 0.826 0.896 0.920 0.942 0.973 1.000


Δg(Gross) 1.59 2.43 3.28 4.12 4.96 5.30 5.64 6.14 6.65 mm

Δcr(Cracked) 19.44 29.76 40.07 50.38 60.69 64.82 68.94 75.13 81.31 mm



∆β=0(S806) 17.15 19.15 23.73 32.37 48.47 55.91 59.84 65.46 72.24 mm


Δmax,Ie(x)(Exact) 7.42 8.52 10.33 13.40 18.90 21.96 25.05 30.02 35.88 mm


ΔIe(x)(Exact) 4.41 5.62 7.77 11.59 18.03 21.41 24.75 30.02 35.88 mm

numerical ΔI'e = 8.57 13.11 17.66 22.20 26.75 28.56 30.38 33.11 35.83 mm




Δγ=1(Approx) 10.39 15.90 21.42 26.93 32.44 34.64 36.85 40.16 43.46 mm

γ=1.7-.7(Mcr/Mmax)= 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20


ΔI'e(Bischoff) 8.57 13.11 17.66 22.20 26.74 28.56 30.38 33.10 35.83 mm

% error, Bischoff's 94.45 133.45 127.21 91.61 48.37 33.37 22.76 10.29 0.14

γ*=γ-.1(ML-1.5MR)/Mcr 1.62 1.55 1.48 1.41 1.34 1.31 1.29 1.24 1.20


ΔI'e*(Proposed) 4.75 8.25 12.42 17.26 22.78 25.17 27.67 31.63 35.83 mm



ΔIe(Branson) 3.85 5.89 7.93 9.97 12.01 12.82 13.64 14.86 16.09 mm

ACI 440.1R clause 8.3.2.2 Ie=Icr+(βdIg-Icr)(Mcr/Mmax)3 βd=0.2(ρm/ρ b)<1 βd= 0.242


ΔIe,βd(ACI440) 11.39 17.42 23.46 29.50 35.54 37.95 40.37 43.99 47.61 mm




ΔIe,avg(A23.3) 10.00 16.82 24.28 31.74 37.05 36.57 40.37 43.99 47.61 mm

𝑓𝑐′

168

Figure N-4 - Copy of Figure 3-8 – Third-Point Loaded, Ig/Icr=12, Mmax/Mcr=1.4, MR=0

The lines plotted in Figure N-4 use data in bold from Example 3.5.2d as found in Table

N-7 and Table N-8.

169

Uniformly Distributed Load Examples – Data for Appendix O

Section 3.6

The calculations in this appendix provide diverse sets of example continuous prismatic

members undergoing a uniformly distributed load. These examples are produced using

the methodology from Appendix I. In each set of examples, the uniform load varies

with the end-moments such that all members for one graph are generated with equal

maximum positive bending moments and reinforcing. These examples use a cracking

moment of 𝑀𝑐𝑟 = 0.6√𝑓𝑐′(𝐼𝑔/𝑦𝑡); the use of 𝑀𝑐𝑟 = 0.3√𝑓𝑐′(𝐼𝑔/𝑦𝑡), per the R2010

version of A23.3 (CSA 2004), is discussed and compared in Appendix P.

The first five examples demonstrate results for steel reinforced members. Examples

3.6.2a and 3.6.2b are similar steel reinforced beams that demonstrate typical steel

members with two end-moments and one end-moment, respectively. These two

examples, where 𝑀𝑚𝑎𝑥/𝑀𝑐𝑟 = 2.17 and 𝐼𝑔/𝐼𝑐𝑟 = 3, show that the provided methods

which account for tension stiffening work well if 2𝑀𝑚𝑎𝑥 < −𝑀𝐿. Examples 3.6.2c and

3.6.2d are a beam and a slab which demonstrate that, if ratios such as 𝑀𝑚𝑎𝑥/𝑀𝑐𝑟 and

𝐼𝑔/𝐼𝑐𝑟 are held constant, then span, width, height, and load are inputs that don’t affect

the essential results because they can be changed without affecting the normalized

moment-deflection graphs. Examples 3.6.2c and 3.6.2d have equal end-moments with

both ends continuous while Example 3.6.2e demonstrates a slab comparable to Example

3.6.2d but with one end continuous. For Examples 3.6.2c, 3.6.2d, and 3.6.2e,

𝑀𝑚𝑎𝑥/𝑀𝑐𝑟 = 1.333 and 𝐼𝑔/𝐼𝑐𝑟 = 4.9, therefore Branson’s (1965) method underpredicts

deflection by more than 10%, as expected.

170

Examples 3.6.2f, 3.6.2g, and 3.6.2h are designed with GFRP reinforcing. In order to

limit deflection, these GFRP examples required 𝑀𝑚𝑎𝑥 𝑀𝑐𝑟⁄ > 0.5 and bottom bars

exceeding those required for 𝑀𝑟 = 𝑀𝑓. Example 3.6.2f demonstrates a slab with a

single layer of reinforcing, 𝑀𝑚𝑎𝑥/𝑀𝑐𝑟 = 1.22, and 𝑀𝑟/𝑀𝑓 = 1.15; portions of the

results, where the required negative moment capacity cannot be achieved, were omitted.

Example 3.6.2f also shows the futility of the 𝛽𝑑 factor from ACI 440.1R (ACI

Committee 440 2006) when reinforcing ratios are more than three times the balanced

reinforcing ratio. Example 3.6.2g is a beam with one end continuous, 𝐼𝑔/𝐼𝑐𝑟 = 6,

𝑀𝑚𝑎𝑥/𝑀𝑐𝑟 = 2.0, and 𝑀𝑟/𝑀𝑠 = 1.6; this contrasts with the beam in Example 3.6.2h

which has two ends continuous, 𝐼𝑔/𝐼𝑐𝑟 = 14, 𝑀𝑚𝑎𝑥/𝑀𝑐𝑟 = 1.25, and 𝑀𝑟/𝑀𝑠 = 1.29.

The second and third GFRP examples show that the 𝛽𝑑 factor often yields reasonable

results. As expected, the GFRP results show Branson’s (1965) method underpredicts

deflection and show that the S806 (CSA 2012) method is overly conservative when

tension stiffening is significant.

For the example graphs, the end-moments range provided is 3 < −𝑀𝐿 𝑀𝑚⁄ ≤ 0. For

results with 𝑀𝑚𝑎𝑥/𝑀𝑐𝑟 ≈ 1.25, 2.5𝑀𝑚 > −𝑀𝐿 ≥ 0 is the proposed valid range. For

𝑀𝑚𝑎𝑥/𝑀𝑐𝑟 ≈ 2.0, the proposed valid range is reduced to 2.0𝑀𝑚 > −𝑀𝐿 ≥ 0.

After determining the midspan deflection using proposed equations, the equation for the

approximate maximum deflection is ∆𝑚𝑎𝑥≈ ∆𝑚𝑖𝑑√𝑀𝑚𝑎𝑥 ⁄ 𝑀𝑚 . This equation is only

intended to account for the difference between the midspan and maximum deflection, so

it only gives an accurate maximum deflection result when the input midspan deflection

is accurate. Examples 3.6.2b, 3.6.2e, and 3.6.2g demonstrate use of this equation.

171

Table O-1 - Data for UDL Beam, ML=MR, Ig/Icr=3.0 – Example 3.6.2a – Page 1


UDL Continuous w0 = 11.27 N/mm fc' = 36 MPa fr = 0.6 * MPa


Let ML=MR b = 0.5* h mm α1 = 0.796

d = 0.9 * h mm β1 = 0.880

M0,0 = w0 L2/8 = 1.41E+8 N mm Φb = 0.85




αL/max=ML/Mmax= -3.00 -2.33 -1.78 -1.70 -1.00 -0.67 -0.43 -0.25 0

wUDL = 45.08 37.57 31.31 30.46 22.54 18.78 16.10 14.09 11.27 N/mm

M0 = 5.64E+8 4.70E+8 3.91E+8 3.81E+8 2.82E+8 2.35E+8 2.01E+8 1.76E+8 1.41E+8 N mm

αL = ML/M0 = -0.75 -0.70 -0.64 -0.63 -0.50 -0.40 -0.30 -0.20 0

αR = MR/M0 = -0.75 -0.70 -0.64 -0.63 -0.50 -0.40 -0.30 -0.20 0

ML = -4.23E+8 -3.29E+8 -2.50E+8 -2.40E+8 -1.41E+8 -9.39E+7 -6.04E+7 -3.52E+7 0.00E+0 N mm

MR = -4.23E+8 -3.29E+8 -2.50E+8 -2.40E+8 -1.41E+8 -9.39E+7 -6.04E+7 -3.52E+7 0.00E+0 N mm

Mm = 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 N mm


αcr = Mcr/Mmax= 0.46 0.46 0.46 0.46 0.46 0.46 0.46 0.46 0.46


h = 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 mm

d = 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 mm

b = 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 mm

Ig = 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 mm4

L1 = 1979 1691 1375 1325 728 320 -55 -404 -1042 mm

L2 = 3163 2988 2795 2765 2402 2154 1926 1714 1326 mm

LR4 = 3163 2988 2795 2765 2402 2154 1926 1714 1326 mm

LR5 = 1979 1691 1375 1325 728 320 -55 -404 -1042 mm


Left End Kr L = 7.61 5.92 4.51 4.32 2.54 1.69 0.00 0 0 MPa

ρ L = 0.0313 0.0217 0.0154 0.0147 0.0081 0.0052 0.0000 0 0

AL=ρLbd= 5077 3516 2500 2375 1304 846 0 0 0 mm2


ηL=1 – Icr L/Ig = 0.129 0.311 0.456 0.475 0.666 0.764 0.000 0 0

ML/Mcr = -6.52 -5.07 -3.86 -3.70 -2.17 -1.45 -0.93 -0.54 0.00

Ig/Icr L = 1.15 1.45 1.84 1.91 2.99 4.24 1.00 1.00 1.00

Midspan Kr m = 2.54 2.54 2.54 2.54 2.54 2.54 2.54 2.54 2.54 MPa

ρ m = 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081


ηm=1 – Icr m/Ig = 0.666 Mmax/Mcr = 2.17 Ig/Icr m = 2.99 Am=ρmbd= 1304 mm2

Right End Kr R = 7.61 5.92 4.51 4.32 2.54 1.69 0.00 0 0 MPa

ρ R = 0.0313 0.0217 0.0154 0.0147 0.0081 0.0052 0.0000 0 0

AR=ρRbd= 5077 3516 2500 2375 1304 846 0 0 0 mm2


ηR=1 – Icr R/Ig = 0.129 0.311 0.456 0.475 0.666 0.764 0.000 0 0

MR/Mcr = -6.52 -5.07 -3.86 -3.70 -2.17 -1.45 -0.93 -0.54 0.00

Ig/Icr R = 1.15 1.45 1.84 1.91 2.99 4.24 1.00 1.00 1.00

Simply

Supprt'd

𝑓𝑐′

ℎ3 𝑓𝑐′

172

Table O-2 - Data for UDL Beam, ML=MR, Ig/Icr=3.0 – Example 3.6.2a – Page 2

Ex. 3.6.2a, pg 2 of 2 w0 = 11.27 N/mm fc' = 36 MPa b = 0.5* h mm

L = 10000 mm fy = 400 MPa d = 0.9 * h mm



αL/max=ML/Mmax = -3.00 -2.33 -1.78 -1.70 -1.00 -0.67 -0.43 -0.25 0.00

ML = -4.23E+8 -3.29E+8 -2.50E+8 -2.40E+8 -1.41E+8 -9.39E+7 -6.04E+7 -3.52E+7 0.00E+0 N mm

Mm = 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 N mm

Mmax = 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8

MR = -4.23E+8 -3.29E+8 -2.50E+8 -2.40E+8 -1.41E+8 -9.39E+7 -6.04E+7 -3.52E+7 0.00E+0 N mm

K=1.2-0.2M0/Mm= 0.400 0.533 0.644 0.659 0.800 0.867 0.914 0.950 1.000

Constant Stiffness Results Using Constant Stiffness Equations Δmid=K(5MmL2)/(48EcI)

Δg(Gross) 4.03 5.37 6.49 6.64 8.05 8.72 9.20 9.56 10.06 mm

Δcr(Cracked) 12.05 16.07 19.41 19.87 24.10 26.11 27.54 28.62 30.13 mm



ΔIe(Branson) 10.09 13.46 16.26 16.64 20.18 21.87 23.07 23.97 25.23 mm

% error, Branson 16.52 3.00 3.50 4.09 6.87 6.96 6.82 6.41 5.24





ΔIe,avg(A23.3) 7.44 11.03 14.58 15.10 20.18 20.51 23.07 23.97 25.23 mm


∆β=0(S806) 15.79 17.81 19.87 20.10 23.45 25.38 26.83 27.90 29.59 mm

Exact Integration Ie(x)Analytical Δ1= -1.27 -0.95 -0.62 -0.57 -0.15 -0.02 0.00 0.00 0.00 mm


Analytical Δ3= 7.21 7.77 8.35 8.44 9.45 10.08 10.62 11.09 11.85 mm

Analytical Δ4= 7.21 7.77 8.35 8.44 9.45 10.08 10.62 11.09 11.85 mm


Analytical Δ6= -1.27 -0.95 -0.62 -0.57 -0.15 -0.02 0.00 0.00 0.00 mm

ΔIe(x)(Exact) 12.09 13.87 15.71 15.98 18.89 20.44 21.60 22.52 23.97 mm

Proposed Method ΔI'e I'e=Icr/[1-γηm(Mcr/Mmax)2] γ=(1.6ξ 3-0.6ξ 4)/(Mcr/Mmax)

2+2.4ln(2-ξ )

I'e (γ=1) (M(x)=Mmax) = 2.10E+9 2.10E+9 2.10E+9 2.10E+9 2.10E+9 2.10E+9 2.10E+9 2.10E+9 2.10E+9 mm4

Δγ=1(Approx) 10.35 13.80 16.68 17.07 20.70 22.43 23.66 24.59 25.88 mm

ξ =1-√(1-Mcr/Mmax)= 0.265 0.265 0.265 0.265 0.265 0.265 0.265 0.265 0.265

γ= 1.45 1.45 1.45 1.45 1.45 1.45 1.45 1.45 1.45


ΔI'e(Proposed) 9.59 12.79 15.45 15.81 19.18 20.78 21.92 22.78 23.97 mm

% error, proposed 20.68 7.83 1.65 1.09 1.55 1.64 1.50 1.11 0.00

Maximum Deflection Results using numerical and approximation methods Δmax ≈ ΔI'e √(Mmax/Mm)

Δmax,Ie(x)(Exact) 12.09 13.87 15.71 15.99 18.89 20.44 21.60 22.53 23.97 mm


Δmax,I'e(Proposed) 9.59 12.79 15.45 15.81 19.18 20.78 21.92 22.78 23.97 mm

𝑓𝑐′

173

Figure O-1 - Copy of Figure 3-9 – UDL on Beam, Ig/Icr=3, Mm /Mcr=2.2, ML=MR

The lines plotted in Figure O-1 use data in bold from Example 3.6.2a as found in Table

O-1 and Table O-2.

174

Table O-3 - Data for UDL Beam, MR=0, Ig/Icr=3.0 – Example 3.6.2b – Page 1




Let MR=0 b = 0.5* h mm α1 = 0.796

d = 0.9 * h mm β1 = 0.880

M0,0 = w0 L2/8 = 1.41E+8 N mm Φb = 0.85




αL/max=ML/Mmax= -2.99 -2.09 -1.50 -1.03 -0.65 -0.49 -0.35 -0.22 0

wUDL = 25.29 21.44 18.76 16.56 14.72 13.91 13.17 12.49 11.27 N/mm

M0 = 3.16E+8 2.68E+8 2.35E+8 2.07E+8 1.84E+8 1.74E+8 1.65E+8 1.56E+8 1.41E+8 N mm

αL = ML/M0 = -1.33 -1.10 -0.9 -0.7 -0.50 -0.40 -0.30 -0.20 0

αR = MR/M0 = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0

ML = -4.21E+8 -2.95E+8 -2.11E+8 -1.45E+8 -9.20E+7 -6.96E+7 -4.94E+7 -3.12E+7 0.00E+0 N mm

MR = 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 N mm

Mm = 1.06E+8 1.21E+8 1.29E+8 1.35E+8 1.38E+8 1.39E+8 1.40E+8 1.40E+8 1.41E+8 N mm


αcr = Mcr/Mmax= 0.46 0.46 0.46 0.46 0.46 0.46 0.46 0.46 0.46


h = 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 mm

d = 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 mm

b = 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 mm

Ig = 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 mm4

L1 = 2630 1995 1443 891 339 63 -213 -489 -1042 mm

L2 = 4210 3711 3277 2844 2410 2193 1976 1759 1326 mm

LR4 = 885 961 1027 1094 1160 1193 1226 1259 1326 mm

LR5 = -695 -755 -807 -859 -911 -937 -963 -989 -1042 mm


Left End Kr L = 7.57 5.31 3.80 2.61 1.66 1.25 0.00 0 0 MPa

ρ L = 0.0311 0.0189 0.0126 0.0083 0.0051 0.0038 0.0000 0 0

AL=ρLbd= 5035 3054 2047 1345 828 618 0 0 0 mm2


ηL=1 – Icr L/Ig = 0.133 0.374 0.529 0.658 0.768 0.819 0.000 0 0

ML/Mcr = -6.49 -4.55 -3.26 -2.24 -1.42 -1.07 -0.76 -0.48 0.00

Ig/Icr L = 1.15 1.60 2.12 2.92 4.32 5.51 1.00 1.00 1.00

Midspan Kr m = 2.54 2.54 2.54 2.54 2.54 2.54 2.54 2.54 2.54 MPa

ρ m = 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081



Right End Kr R = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0 0 MPa

ρ R = 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0

AR=ρRbd= 0 0 0 0 0 0 0 0 0 mm2


ηR=1 – Icr R/Ig = 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0 0

MR/Mcr = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

Ig/Icr R = 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00

Simply

Supprt'd

𝑓𝑐′

ℎ3 𝑓𝑐′

175

Table O-4 - Data for UDL Beam, MR=0, Ig/Icr=3.0 – Example 3.6.2b – Page 2

Ex. 3.6.2b, pg 2 of 2 w0 = 11.27 N/mm fc' = 36 MPa b = 0.5* h mm

L = 10000 mm fy = 400 MPa d = 0.9 * h mm



αL/max=ML/Mmax = -2.99 -2.09 -1.50 -1.03 -0.65 -0.49 -0.35 -0.22 0.00

ML = -4.21E+8 -2.95E+8 -2.11E+8 -1.45E+8 -9.20E+7 -6.96E+7 -4.94E+7 -3.12E+7 0.00E+0 N mm

Mm = 1.06E+8 1.21E+8 1.29E+8 1.35E+8 1.38E+8 1.39E+8 1.40E+8 1.40E+8 1.41E+8 N mm

Mmax = 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8

MR = 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 N mm

K=1.2-0.2M0/Mm= 0.603 0.756 0.836 0.892 0.933 0.950 0.965 0.978 1.000


Δg(Gross) 4.56 6.51 7.71 8.58 9.20 9.44 9.65 9.81 10.06 mm

Δcr(Cracked) 13.66 19.49 23.07 25.67 27.54 28.27 28.87 29.37 30.13 mm



ΔIe(Branson) 11.44 16.32 19.32 21.50 23.07 23.67 24.18 24.60 25.23 mm

% error, Branson 11.07 2.54 5.95 6.61 6.44 6.30 6.10 5.84 5.24





ΔIe,avg(A23.3) 9.73 15.01 18.71 21.48 22.22 20.29 24.18 24.60 25.23 mm


∆β=0(S806) 16.82 20.21 22.86 25.12 26.88 27.65 28.23 28.73 29.59 mm

Exact Integration Ie(x)Analytical Δ1= -2.25 -1.32 -0.67 -0.23 -0.03 0.00 0.00 0.00 0.00 mm


Analytical Δ3= 2.04 3.88 5.55 7.18 8.70 9.42 10.09 10.72 11.85 mm

Analytical Δ4= 12.83 13.08 13.07 12.92 12.68 12.53 12.38 12.21 11.85 mm


Analytical Δ6= 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 mm

ΔIe(x)(Exact) 12.86 15.92 18.24 20.17 21.67 22.27 22.79 23.24 23.97 mm


2+2.4ln(2-ξ )


Δγ=1(Approx) 11.73 16.74 19.82 22.05 23.66 24.28 24.80 25.24 25.88 mm

ξ =1-√(1-Mcr/Mmax)= 0.265 0.265 0.265 0.265 0.265 0.265 0.265 0.265 0.265

γ= 1.45 1.45 1.45 1.45 1.45 1.45 1.45 1.45 1.45


ΔI'e(Proposed) 10.87 15.51 18.36 20.43 21.92 22.49 22.98 23.38 23.97 mm

% error, proposed 15.50 2.57 0.68 1.31 1.14 1.00 0.82 0.57 0.00


Δmax,Ie(x)(Exact) 14.48 16.90 18.76 20.46 21.76 22.27 22.79 23.24 23.97 mm



𝑓𝑐′

176

Figure O-2 - Copy of Figure 3-10 – UDL on Beam, Ig/Icr=3, Mmax/Mcr=2.2, MR=0

The lines plotted in Figure O-2 use data in bold from Example 3.6.2b as found in Table

O-3 and Table O-4.

177

Table O-5 - Data for UDL Beam, ML=MR, Ig/Icr=4.9 – Example 3.6.2c – Page 1

Example 3.6.2c, pg 1 of 2 Φc = 0.65



Let ML=MR b = 0.5* h mm α1 = 0.796

d = 0.85 * h mm β1 = 0.880

M0,0 = w0 L2/8 = 8.64E+7 N mm Φb = 0.85




αL/max=ML/Mmax= -3.00 -2.33 -1.86 -1.22 -1.00 -0.82 -0.61 -0.25 0

wUDL = 49.14 40.95 35.10 27.30 24.57 22.34 19.81 15.36 12.29 N/mm

M0 = 3.46E+8 2.88E+8 2.47E+8 1.92E+8 1.73E+8 1.57E+8 1.39E+8 1.08E+8 8.64E+7 N mm

αL = ML/M0 = -0.75 -0.70 -0.65 -0.55 -0.50 -0.45 -0.38 -0.20 0

αR = MR/M0 = -0.75 -0.70 -0.65 -0.55 -0.50 -0.45 -0.38 -0.20 0

ML = -2.59E+8 -2.02E+8 -1.60E+8 -1.06E+8 -8.64E+7 -7.07E+7 -5.29E+7 -2.16E+7 0.00E+0 N mm

MR = -2.59E+8 -2.02E+8 -1.60E+8 -1.06E+8 -8.64E+7 -7.07E+7 -5.29E+7 -2.16E+7 0.00E+0 N mm

Mm = 8.64E+7 8.64E+7 8.64E+7 8.64E+7 8.64E+7 8.64E+7 8.64E+7 8.64E+7 8.64E+7 N mm


αcr = Mcr/Mmax= 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75


h = 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 mm

d = 510.0 510.0 510.0 510.0 510.0 510.0 510.0 510.0 510.0 mm

b = 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 mm

Ig = 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 mm4

L1 = 1270 1033 815 422 242 71 -156 -687 -1211 mm

L2 = 2813 2723 2641 2492 2424 2359 2274 2073 1875 mm

LR4 = 2813 2723 2641 2492 2424 2359 2274 2073 1875 mm

LR5 = 1270 1033 815 422 242 71 -156 -687 -1211 mm


Left End Kr L = 5.23 4.07 3.24 2.13 1.74 1.43 0.00 0 0 MPa

ρ L = 0.0185 0.0137 0.0105 0.0067 0.0054 0.0044 0.0000 0 0

AL=ρLbd= 2832 2092 1612 1021 825 669 0 0 0 mm2


ηL=1 – Icr L/Ig = 0.479 0.580 0.654 0.757 0.796 0.829 0.000 0 0

ML/Mcr = -4.00 -3.11 -2.48 -1.63 -1.33 -1.09 -0.82 -0.33 0.00

Ig/Icr L = 1.92 2.38 2.89 4.12 4.90 5.84 1.00 1.00 1.00

Midspan Kr m = 1.74 1.74 1.74 1.74 1.74 1.74 1.74 1.74 1.74 MPa

ρ m = 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054



Right End Kr R = 5.23 4.07 3.24 2.13 1.74 1.43 0.00 0 0 MPa

ρ R = 0.0185 0.0137 0.0105 0.0067 0.0054 0.0044 0.0000 0 0

AR=ρRbd= 2832 2092 1612 1021 825 669 0 0 0 mm2


ηR=1 – Icr R/Ig = 0.479 0.580 0.654 0.757 0.796 0.829 0.000 0 0

MR/Mcr = -4.00 -3.11 -2.48 -1.63 -1.33 -1.09 -0.82 -0.33 0.00

Ig/Icr R = 1.92 2.38 2.89 4.12 4.90 5.84 1.00 1.00 1.00

Simply

Supprt'd

𝑓𝑐′

ℎ3 𝑓𝑐′

178

Table O-6 - Data for UDL Beam, ML=MR, Ig/Icr=4.9 – Example 3.6.2c – Page 2

Ex. 3.6.2c, pg 2 of 2 w0 = 12.29 N/mm fc' = 36 MPa b = 0.5* h mm

L = 7500 mm fy = 400 MPa d = 0.85 * h mm



αL/max=ML/Mmax = -3.00 -2.33 -1.86 -1.22 -1.00 -0.82 -0.61 -0.25 0.00

ML = -2.59E+8 -2.02E+8 -1.60E+8 -1.06E+8 -8.64E+7 -7.07E+7 -5.29E+7 -2.16E+7 0.00E+0 N mm

Mm = 8.64E+7 8.64E+7 8.64E+7 8.64E+7 8.64E+7 8.64E+7 8.64E+7 8.64E+7 8.64E+7 N mm

Mmax = 8.64E+7 8.64E+7 8.64E+7 8.64E+7 8.64E+7 8.64E+7 8.64E+7 8.64E+7 8.64E+7

MR = -2.59E+8 -2.02E+8 -1.60E+8 -1.06E+8 -8.64E+7 -7.07E+7 -5.29E+7 -2.16E+7 0.00E+0 N mm

K=1.2-0.2M0/Mm= 0.400 0.533 0.629 0.756 0.800 0.836 0.877 0.950 1.000


Δg(Gross) 1.39 1.85 2.18 2.62 2.78 2.90 3.05 3.30 3.47 mm

Δcr(Cracked) 6.81 9.08 10.70 12.86 13.62 14.24 14.94 16.17 17.02 mm



ΔIe(Branson) 2.57 3.43 4.04 4.86 5.15 5.38 5.64 6.11 6.43 mm

% error, Branson 18.48 12.28 10.65 10.34 10.33 10.29 10.37 11.47 13.12





ΔIe,avg(A23.3) 2.59 3.63 4.41 5.21 5.15 4.68 5.64 6.11 6.43 mm


∆β=0(S806) 7.43 8.43 9.34 10.86 11.51 12.06 12.60 13.76 14.88 mm



Analytical Δ3= 1.94 2.10 2.25 2.50 2.61 2.71 2.85 3.15 3.42 mm

Analytical Δ4= 1.94 2.10 2.25 2.50 2.61 2.71 2.85 3.15 3.42 mm


Analytical Δ6= -0.55 -0.36 -0.21 -0.05 -0.01 0.00 0.00 0.00 0.00 mm

ΔIe(x)(Exact) 3.16 3.91 4.53 5.42 5.74 6.00 6.30 6.90 7.40 mm


2+2.4ln(2-ξ )


Δγ=1(Approx) 3.76 5.01 5.91 7.10 7.52 7.86 8.25 8.93 9.40 mm

ξ =1-√(1-Mcr/Mmax)= 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500

γ= 1.26 1.26 1.26 1.26 1.26 1.26 1.26 1.26 1.26


ΔI'e(Proposed) 2.96 3.95 4.65 5.59 5.92 6.19 6.50 7.03 7.40 mm

% error, proposed 6.18 0.97 2.84 3.20 3.20 3.26 3.16 1.90 0.00


Δmax,Ie(x)(Exact) 3.16 3.91 4.53 5.42 5.74 6.00 6.30 6.90 7.40 mm



𝑓𝑐′

179

Figure O-3 - Midspan Deflection of Steel Reinforced Beams under Uniformly

Distributed Load with Ig/Icr=5, Mm /Mcr=1.3, and ML=MR

The lines plotted in Figure O-3 use data in bold from Example 3.6.2c as found in Table

O-5 and Table O-6.

180

Table O-7 - Data for UDL Slab, ML=MR, Ig/Icr=4.9 – Example 3.6.2d – Page 1

Example 3.6.2d, pg 1 of 2 Φc = 0.65



Let ML=MR b = 3.636* h mm α1 = 0.796

d = 0.85 * h mm β1 = 0.880

M0,0 = w0 L2/8 = 6.05E+7 N mm Φb = 0.85




αL/max=ML/Mmax= -3.00 -2.33 -1.86 -1.22 -1.00 -0.82 -0.61 -0.25 0

wUDL = 34.42 28.68 24.59 19.12 17.21 15.65 13.88 10.76 8.61 N/mm

M0 = 2.42E+8 2.02E+8 1.73E+8 1.34E+8 1.21E+8 1.10E+8 9.76E+7 7.56E+7 6.05E+7 N mm

αL = ML/M0 = -0.75 -0.70 -0.65 -0.55 -0.50 -0.45 -0.38 -0.20 0

αR = MR/M0 = -0.75 -0.70 -0.65 -0.55 -0.50 -0.45 -0.38 -0.20 0

ML = -1.82E+8 -1.41E+8 -1.12E+8 -7.39E+7 -6.05E+7 -4.95E+7 -3.71E+7 -1.51E+7 0.00E+0 N mm

MR = -1.82E+8 -1.41E+8 -1.12E+8 -7.39E+7 -6.05E+7 -4.95E+7 -3.71E+7 -1.51E+7 0.00E+0 N mm

Mm = 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 N mm


αcr = Mcr/Mmax= 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75


h = 275.0 275.0 275.0 275.0 275.0 275.0 275.0 275.0 275.0 mm

d = 233.8 233.8 233.8 233.8 233.8 233.8 233.8 233.8 233.8 mm

b = 1000.0 1000.0 1000.0 1000.0 1000.0 1000.0 1000.0 1000.0 1000.0 mm

Ig = 1.73E+9 1.73E+9 1.73E+9 1.73E+9 1.73E+9 1.73E+9 1.73E+9 1.73E+9 1.73E+9 mm4

L1 = 1270 1033 815 422 242 71 -156 -687 -1211 mm

L2 = 2813 2723 2641 2492 2424 2359 2274 2073 1875 mm

LR4 = 2813 2723 2641 2492 2424 2359 2274 2073 1875 mm

LR5 = 1270 1033 815 422 242 71 -156 -687 -1211 mm


Left End Kr L = 5.23 4.07 3.24 2.13 1.74 1.43 0.00 0 0 MPa

ρ L = 0.0185 0.0137 0.0105 0.0067 0.0054 0.0044 0.0000 0 0

AL=ρLbd= 4328 3196 2463 1560 1261 1022 0 0 0 mm2


ηL=1 – Icr L/Ig = 0.479 0.580 0.654 0.757 0.796 0.829 0.000 0 0

ML/Mcr = -4.00 -3.11 -2.48 -1.63 -1.33 -1.09 -0.82 -0.33 0.00

Ig/Icr L = 1.92 2.38 2.89 4.12 4.90 5.84 1.00 1.00 1.00

Midspan Kr m = 1.74 1.74 1.74 1.74 1.74 1.74 1.74 1.74 1.74 MPa

ρ m = 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054



Right End Kr R = 5.23 4.07 3.24 2.13 1.74 1.43 0.00 0 0 MPa

ρ R = 0.0185 0.0137 0.0105 0.0067 0.0054 0.0044 0.0000 0 0

AR=ρRbd= 4328 3196 2463 1560 1261 1022 0 0 0 mm2


ηR=1 – Icr R/Ig = 0.479 0.580 0.654 0.757 0.796 0.829 0.000 0 0

MR/Mcr = -4.00 -3.11 -2.48 -1.63 -1.33 -1.09 -0.82 -0.33 0.00

Ig/Icr R = 1.92 2.38 2.89 4.12 4.90 5.84 1.00 1.00 1.00

Simply

Supprt'd

𝑓𝑐′

ℎ3 𝑓𝑐′

181

Table O-8 - Data for UDL Slab, ML=MR, Ig/Icr=4.9 – Example 3.6.2d – Page 2

Ex. 3.6.2d, pg 2 of 2 w0 = 8.61 N/mm fc' = 36 MPa b = 3.636* h mm

L = 7500 mm fy = 400 MPa d = 0.85 * h mm



αL/max=ML/Mmax = -3.00 -2.33 -1.86 -1.22 -1.00 -0.82 -0.61 -0.25 0.00

ML = -1.82E+8 -1.41E+8 -1.12E+8 -7.39E+7 -6.05E+7 -4.95E+7 -3.71E+7 -1.51E+7 0.00E+0 N mm

Mm = 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 N mm

Mmax = 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7

MR = -1.82E+8 -1.41E+8 -1.12E+8 -7.39E+7 -6.05E+7 -4.95E+7 -3.71E+7 -1.51E+7 0.00E+0 N mm

K=1.2-0.2M0/Mm= 0.400 0.533 0.629 0.756 0.800 0.836 0.877 0.950 1.000


Δg(Gross) 3.03 4.04 4.76 5.72 6.06 6.34 6.65 7.20 7.58 mm

Δcr(Cracked) 14.86 19.81 23.34 28.06 29.71 31.06 32.59 35.28 37.14 mm



ΔIe(Branson) 5.61 7.48 8.82 10.60 11.23 11.74 12.31 13.33 14.03 mm

% error, Branson 18.48 12.28 10.65 10.34 10.33 10.29 10.37 11.47 13.12





ΔIe,avg(A23.3) 5.65 7.93 9.63 11.38 11.23 10.21 12.31 13.33 14.03 mm


∆β=0(S806) 16.20 18.40 20.38 23.70 25.10 26.32 27.48 30.01 32.47 mm



Analytical Δ3= 4.23 4.58 4.90 5.45 5.69 5.92 6.21 6.86 7.46 mm

Analytical Δ4= 4.23 4.58 4.90 5.45 5.69 5.92 6.21 6.86 7.46 mm


Analytical Δ6= -1.21 -0.78 -0.46 -0.10 -0.03 0.00 0.00 0.00 0.00 mm

ΔIe(x)(Exact) 6.89 8.53 9.87 11.83 12.52 13.08 13.74 15.06 16.15 mm


2+2.4ln(2-ξ )


Δγ=1(Approx) 8.20 10.94 12.89 15.50 16.41 17.15 18.00 19.48 20.51 mm

ξ =1-√(1-Mcr/Mmax)= 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500

γ= 1.26 1.26 1.26 1.26 1.26 1.26 1.26 1.26 1.26


ΔI'e(Proposed) 6.46 8.61 10.15 12.20 12.92 13.51 14.17 15.34 16.15 mm

% error, proposed 6.18 0.97 2.84 3.20 3.20 3.26 3.16 1.90 0.00


Δmax,Ie(x)(Exact) 6.89 8.53 9.87 11.83 12.52 13.08 13.74 15.06 16.15 mm



𝑓𝑐′

182

Figure O-4 - Copy of Figure 3-11 – UDL on Slab, Ig/Icr=5, Mm /Mcr=1.3, ML=MR

The lines plotted in Figure O-4 use data in bold from Example 3.6.2d as found in Table

O-7 and Table O-8.

183

Table O-9 - Data for UDL Slab, MR=0, Ig/Icr=4.9 – Example 3.6.2e – Page 1

Example 3.6.2e, pg 1 of 2 Φc = 0.65



Let MR=0 b = 3.636* h mm α1 = 0.796

d = 0.85 * h mm β1 = 0.880

M0,0 = w0 L2/8 = 6.05E+7 N mm Φb = 0.85




αL/max=ML/Mmax= -2.99 -2.09 -1.78 -1.37 -1.01 -0.74 -0.49 -0.22 0

wUDL = 19.31 16.37 15.30 13.88 12.57 11.57 10.62 9.53 8.61 N/mm

M0 = 1.36E+8 1.15E+8 1.08E+8 9.76E+7 8.84E+7 8.13E+7 7.47E+7 6.70E+7 6.05E+7 N mm

αL = ML/M0 = -1.33 -1.10 -1 -0.85 -0.69 -0.55 -0.40 -0.20 0

αR = MR/M0 = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0

ML = -1.81E+8 -1.27E+8 -1.08E+8 -8.29E+7 -6.10E+7 -4.47E+7 -2.99E+7 -1.34E+7 0.00E+0 N mm

MR = 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 N mm

Mm = 4.55E+7 5.18E+7 5.38E+7 5.61E+7 5.79E+7 5.90E+7 5.98E+7 6.03E+7 6.05E+7 N mm


αcr = Mcr/Mmax= 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75


h = 275.0 275.0 275.0 275.0 275.0 275.0 275.0 275.0 275.0 mm

d = 233.8 233.8 233.8 233.8 233.8 233.8 233.8 233.8 233.8 mm

b = 1000.0 1000.0 1000.0 1000.0 1000.0 1000.0 1000.0 1000.0 1000.0 mm

Ig = 1.73E+9 1.73E+9 1.73E+9 1.73E+9 1.73E+9 1.73E+9 1.73E+9 1.73E+9 1.73E+9 mm4

L1 = 1686 1185 967 640 292 -13 -340 -775 -1211 mm

L2 = 3745 3422 3281 3070 2845 2648 2438 2156 1875 mm

LR4 = 1252 1359 1406 1477 1552 1617 1688 1781 1875 mm

LR5 = -808 -878 -908 -953 -1002 -1044 -1090 -1150 -1211 mm


Left End Kr L = 5.21 3.65 3.10 2.39 1.76 0.00 0.00 0 0 MPa

ρ L = 0.0184 0.0121 0.0100 0.0075 0.0054 0.0000 0.0000 0 0

AL=ρLbd= 4301 2819 2346 1765 1271 0 0 0 0 mm2


ηL=1 – Icr L/Ig = 0.481 0.617 0.666 0.732 0.795 0.000 0.000 0 0

ML/Mcr = -3.98 -2.79 -2.37 -1.83 -1.34 -0.99 -0.66 -0.30 0.00

Ig/Icr L = 1.93 2.61 3.00 3.74 4.87 1.00 1.00 1.00 1.00

Midspan Kr m = 1.74 1.74 1.74 1.74 1.74 1.74 1.74 1.74 1.74 MPa

ρ m = 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054



Right End Kr R = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0 0 MPa

ρ R = 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0

AR=ρRbd= 0 0 0 0 0 0 0 0 0 mm2


ηR=1 – Icr R/Ig = 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0 0

MR/Mcr = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

Ig/Icr R = 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00

Simply

Supprt'd

𝑓𝑐′

ℎ3 𝑓𝑐′

184

Table O-10 - Data for UDL Slab, MR=0, Ig/Icr=4.9 – Example 3.6.2e – Page 2

Ex. 3.6.2e, pg 2 of 2 w0 = 8.61 N/mm fc' = 36 MPa b = 3.636* h mm

L = 7500 mm fy = 400 MPa d = 0.85 * h mm



αL/max=ML/Mmax = -2.99 -2.09 -1.78 -1.37 -1.01 -0.74 -0.49 -0.22 0.00

ML = -1.81E+8 -1.27E+8 -1.08E+8 -8.29E+7 -6.10E+7 -4.47E+7 -2.99E+7 -1.34E+7 0.00E+0 N mm

Mm = 4.55E+7 5.18E+7 5.38E+7 5.61E+7 5.79E+7 5.90E+7 5.98E+7 6.03E+7 6.05E+7 N mm

Mmax = 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7

MR = 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 N mm

K=1.2-0.2M0/Mm= 0.603 0.756 0.800 0.852 0.895 0.924 0.950 0.978 1.000


Δg(Gross) 3.43 4.90 5.39 5.99 6.48 6.82 7.11 7.39 7.58 mm

Δcr(Cracked) 16.84 24.02 26.41 29.34 31.78 33.45 34.85 36.21 37.14 mm



ΔIe(Branson) 6.36 9.08 9.98 11.09 12.01 12.64 13.17 13.68 14.03 mm

% error, Branson 13.16 8.99 9.09 9.73 10.39 10.81 11.36 12.24 13.12





ΔIe,avg(A23.3) 6.39 9.41 10.43 11.58 12.03 12.64 13.17 13.68 14.03 mm


∆β=0(S806) 16.82 20.85 22.60 24.91 27.01 28.41 29.88 31.19 32.47 mm



Analytical Δ3= 0.01 0.93 1.47 2.36 3.37 4.26 5.19 6.37 7.46 mm

Analytical Δ4= 8.43 8.94 9.01 8.98 8.83 8.63 8.36 7.93 7.46 mm


Analytical Δ6= 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 mm

ΔIe(x)(Exact) 7.33 9.97 10.98 12.28 13.40 14.17 14.85 15.59 16.15 mm


2+2.4ln(2-ξ )


Δγ=1(Approx) 9.30 13.27 14.58 16.20 17.55 18.47 19.24 20.00 20.51 mm

ξ =1-√(1-Mcr/Mmax)= 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500

γ= 1.26 1.26 1.26 1.26 1.26 1.26 1.26 1.26 1.26


ΔI'e(Proposed) 7.32 10.45 11.49 12.76 13.82 14.55 15.16 15.75 16.15 mm

% error, proposed 0.05 4.74 4.63 3.90 3.14 2.66 2.02 1.01 0.00


Δmax,Ie(x)(Exact) 8.95 10.93 11.71 12.72 13.69 14.34 14.90 15.59 16.15 mm



𝑓𝑐′

185

Figure O-5 - Copy of Figure 3-12 – UDL on Slab, Ig/Icr=5, Mmax /Mcr=1.3, MR=0

The lines plotted in Figure O-5 use data in bold from Example 3.6.2e as found in Table

O-9 and Table O-10.

186

Table O-11 - Data for UDL Slab, ML=MR, Ig/Icr=18 – Example 3.6.2f – Page 1

Example 3.6.2f, pg 1 of 2 Φc = 0.65 εcu = 0.0035 mm/mm



Let ML=MR b = 6.667* h mm α1 = 0.796


M0,0 = w0 L2/8 = 1.65E+7 N mm φb = 0.75 ρ b = 0.00578




αL/max=ML/Mmax= -3.00 -1.94 -1.56 -1.22 -1.00 -0.85 -0.43 -0.25 0

wUDL = 21.07 15.49 13.51 11.71 10.54 9.76 7.53 6.59 5.27 N/mm

M0 = 6.59E+7 4.84E+7 4.22E+7 3.66E+7 3.29E+7 3.05E+7 2.35E+7 2.06E+7 1.65E+7 N mm

αL = ML/M0 = -0.75 -0.66 -0.61 -0.55 -0.50 -0.46 -0.30 -0.20 0

αR = MR/M0 = -0.75 -0.66 -0.61 -0.55 -0.50 -0.46 -0.30 -0.20 0

ML = -4.94E+7 -3.20E+7 -2.57E+7 -2.01E+7 -1.65E+7 -1.40E+7 -7.06E+6 -4.12E+6 0.00E+0 N mm

MR = -4.94E+7 -3.20E+7 -2.57E+7 -2.01E+7 -1.65E+7 -1.40E+7 -7.06E+6 -4.12E+6 0.00E+0 N mm

Mm = 1.65E+7 1.65E+7 1.65E+7 1.65E+7 1.65E+7 1.65E+7 1.65E+7 1.65E+7 1.65E+7 N mm


αcr = Mcr/Mmax= 0.82 0.82 0.82 0.82 0.82 0.82 0.82 0.82 0.82


h = 150.0 150.0 150.0 150.0 150.0 150.0 150.0 150.0 150.0 mm

d = 75.0 75.0 75.0 75.0 75.0 75.0 75.0 75.0 75.0 mm

b = 1000.0 1000.0 1000.0 1000.0 1000.0 1000.0 1000.0 1000.0 1000.0 mm

Ig = 2.81E+8 2.81E+8 2.81E+8 2.81E+8 2.81E+8 2.81E+8 2.81E+8 2.81E+8 2.81E+8 mm4

L1 = 814 533 394 238 115 22 -322 -517 -873 mm

L2 = 1970 1882 1838 1788 1750 1721 1613 1551 1439 mm

LR4 = 1970 1882 1838 1788 1750 1721 1613 1551 1439 mm

LR5 = 814 533 394 238 115 22 -322 -517 -873 mm


Left End c L = #NUM! 68.33 44.72 31.65 24.66 20.41 0 0 0 mm

AL= #NUM! 99384 9372 3281 1714 1083 0 0 0 mm2

ρ L =As/bd= #NUM! 1.3251 0.1250 0.0437 0.0228 0.0144 0 0 0

Icr L = #NUM! 1.07E+8 3.87E+7 1.85E+7 1.10E+7 7.47E+6 2.81E+8 2.81E+8 2.81E+8 mm4

ηL=1 – Icr L/Ig = #NUM! 0.621 0.862 0.934 0.961 0.973 0.000 0 0

ML/Mcr = -3.66 -2.37 -1.91 -1.49 -1.22 -1.04 -0.52 -0.30 0.00

Ig/Icr L = #NUM! 2.64 7.26 15.20 25.53 37.63 1.00 1.00 1.00

Midspan c m = 29.30 29.30 29.30 29.30 29.30 29.30 29.30 29.30 29.30 mm

ρ m =Am/bd= 0.0355 0.0355 0.0355 0.0355 0.0355 0.0355 0.0355 0.0355 0.0355



Right End cR = #NUM! 68.33 44.72 31.65 24.66 20.41 0.00 0 0 mm

AR= #NUM! 99384 9372 3281 1714 1083 0 0 0 mm2

ρ R =AR/bd= #NUM! 1.3251 0.1250 0.0437 0.0228 0.0144 0.0000 0.0000 0.0000

Icr R = #NUM! 1.07E+8 3.87E+7 1.85E+7 1.10E+7 7.47E+6 2.81E+8 2.81E+8 2.81E+8 mm4

ηR=1 – Icr R/Ig = #NUM! 0.621 0.862 0.934 0.961 0.973 0.000 0 0

MR/Mcr = -3.66 -2.37 -1.91 -1.49 -1.22 -1.04 -0.52 -0.30 0.00

Ig/Icr R = #NUM! 2.64 7.26 15.20 25.53 37.63 1.00 1.00 1.00

Simply

Supprt'd

𝑓𝑐′

ℎ3 𝑓𝑐′

187

Table O-12 - Data for UDL Slab, ML=MR, Ig/Icr=18 – Example 3.6.2f – Page 2

Ex. 3.6.2f, pg 2 of 2 w0 = 5.27 N/mm fc' = 36 MPa ρ b = 0.00578

L = 5000 mm ffu = 690 MPa ρ m = 0.0355

Ms /Mr (+ve) = 1.150 Mmax/Mcr = 1.22 Ig/Icr m = 17.86 Eb = 44000 MPa

αL/max=ML/Mmax = -3.00 -1.94 -1.56 -1.22 -1.00 -0.85 -0.43 -0.25 0.00

ML = -4.94E+7 -3.20E+7 -2.57E+7 -2.01E+7 -1.65E+7 -1.40E+7 -7.06E+6 -4.12E+6 0.00E+0 N mm

Mm = 1.65E+7 1.65E+7 1.65E+7 1.65E+7 1.65E+7 1.65E+7 1.65E+7 1.65E+7 1.65E+7 N mm

Mmax = 1.65E+7 1.65E+7 1.65E+7 1.65E+7 1.65E+7 1.65E+7 1.65E+7 1.65E+7 1.65E+7

MR = -4.94E+7 -3.20E+7 -2.57E+7 -2.01E+7 -1.65E+7 -1.40E+7 -7.06E+6 -4.12E+6 0.00E+0 N mm

K=1.2-0.2M0/Mm= 0.400 0.612 0.687 0.756 0.800 0.830 0.914 0.950 1.000


Δg(Gross) 2.26 3.45 3.88 4.27 4.52 4.68 5.16 5.36 5.65 mm

Δcr(Cracked) 40.35 61.71 69.31 76.21 80.69 83.68 92.22 95.82 100.87 mm



ΔIe(Branson) 3.92 5.99 6.73 7.40 7.84 8.13 8.95 9.30 9.79 mm

% error, Branson #NUM! 58.82 56.74 56.35 56.42 56.42 56.87 57.44 58.72

ACI 440.1R clause 8.3.2.2 Ie=Icr+(βdIg-Icr)(Mcr/Mmax)3 βd= 1.000 βd=0.2(ρm/ρ b)<1 Ie R = Ie L


ΔIe,βd(ACI440) 3.92 5.99 6.73 7.40 7.84 8.13 8.95 9.30 9.79 mm

Ie L (ACI440.1R) = #NUM! 1.20E+8 7.37E+7 9.78E+7 1.29E+8 1.29E+8 1.62E+8 1.62E+8 1.62E+8 mm4

Ie 9.8.2 (& ACI440.1R)= #NUM! 1.49E+8 1.36E+8 1.43E+8 1.52E+8 1.52E+8 1.62E+8 1.62E+8 1.62E+8 mm4

ΔIe,avg(A23.3) #NUM! 6.50 8.05 8.40 8.35 8.65 8.95 9.30 9.79 mm


∆β=0(S806) #NUM! 49.30 52.42 56.21 59.97 61.97 68.32 71.39 78.15 mm

Exact Integration Ie(x)Analytical Δ1= #NUM! -0.32 -0.30 -0.13 -0.03 0.00 0.00 0.00 0.00 mm


Analytical Δ3= 6.17 7.08 7.52 8.01 8.38 8.67 9.67 10.22 11.19 mm

Analytical Δ4= 6.17 7.08 7.52 8.01 8.38 8.67 9.67 10.22 11.19 mm


Analytical Δ6= #NUM! -0.32 -0.30 -0.13 -0.03 0.00 0.00 0.00 0.00 mm

ΔIe(x)(Exact) #NUM! 14.55 15.56 16.95 17.98 18.64 20.76 21.86 23.72 mm

Length:Defl, L/Δmid= #NUM! 344 321 295 278 268 241 229 211

Proposed Method ΔI'e I'e=Icr/[1-γηm(Mcr/Mmax)2] γ=(1.6ξ

3-0.6ξ

4)/(Mcr/Mmax)

2+2.4ln(2-ξ )


Δγ=1(Approx) 14.74 22.54 25.32 27.83 29.47 30.56 33.68 35.00 36.84 mm

ξ =1-√(1-Mcr/Mmax)= 0.576 0.576 0.576 0.576 0.576 0.576 0.576 0.576 0.576

γ= 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20


ΔI'e(Proposed) 9.49 14.51 16.30 17.93 18.98 19.68 21.69 22.54 23.72 mm

% error, proposed #NUM! 0.25 4.80 5.73 5.57 5.57 4.48 3.10 0.00

L/Δ exact #NUM! 343.62 321.40 294.92 278.10 268.17 240.82 228.71 210.75


Δmax,Ie(x)(Exact) #NUM! 14.55 15.56 16.96 17.98 18.65 20.76 21.86 23.73 mm


188

Figure O-6 - Midspan Deflection of FRP Reinforced Slabs under Uniformly Distributed

Load with Ig/Icr=18, Mm/Mcr=1.2, ML=MR

The lines plotted in Figure O-6 use data in bold from Example 3.6.2f as found in Table

O-11 and Table O-12.

189

Table O-13 - Data for UDL Slab, MR=0, Ig/Icr=6 – Example 3.6.2g – Page 1

Example 3.6.2g, pg 1 of 2 Φc = 0.65 εcu = 0.0035 mm/mm



Let MR=0 b = 0.5* h mm α1 = 0.796


M0,0 = w0 L2/8 = 1.30E+8 N mm φb = 0.75 ρ b = 0.00578




αL/max=ML/Mmax= -2.99 -2.09 -1.50 -1.25 -1.01 -0.65 -0.35 -0.22 0

wUDL = 23.27 19.73 17.27 16.20 15.14 13.54 12.12 11.49 10.37 N/mm

M0 = 2.91E+8 2.47E+8 2.16E+8 2.03E+8 1.89E+8 1.69E+8 1.51E+8 1.44E+8 1.30E+8 N mm

αL = ML/M0 = -1.33 -1.10 -0.9 -0.8 -0.69 -0.50 -0.30 -0.20 0

αR = MR/M0 = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0

ML = -3.87E+8 -2.71E+8 -1.94E+8 -1.62E+8 -1.31E+8 -8.47E+7 -4.54E+7 -2.87E+7 0.00E+0 N mm

MR = 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 N mm

Mm = 9.75E+7 1.11E+8 1.19E+8 1.22E+8 1.24E+8 1.27E+8 1.29E+8 1.29E+8 1.30E+8 N mm


αcr = Mcr/Mmax= 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5


h = 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 mm

d = 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 mm

b = 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 mm

Ig = 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 mm4

L1 = 2575 1935 1379 1101 795 267 -289 -568 -1124 mm

L2 = 4303 3812 3385 3172 2937 2531 2105 1891 1464 mm

LR4 = 978 1062 1135 1172 1212 1281 1355 1391 1464 mm

LR5 = -750 -815 -871 -899 -930 -983 -1039 -1068 -1124 mm


Left End c L = 305.48 190.40 128.67 105.08 83.08 52.44 0 0 0 mm

AL= 16939 4414 1714 1081 643 240 0 0 0 mm2

ρ L =As/bd= 0.1045 0.0272 0.0106 0.0067 0.0040 0.0015 0 0 0


ηL=1 – Icr L/Ig = 0.284 0.736 0.882 0.922 0.951 0.981 0.000 0 0

ML/Mcr = -5.97 -4.19 -3.00 -2.50 -2.02 -1.31 -0.70 -0.44 0.00

Ig/Icr L = 1.40 3.79 8.46 12.77 20.54 51.90 1.00 1.00 1.00

Midspan c m = 152.73 152.73 152.73 152.73 152.73 152.73 152.73 152.73 152.73 mm

ρ m =Am/bd= 0.0158 0.0158 0.0158 0.0158 0.0158 0.0158 0.0158 0.0158 0.0158



Right End cR = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0 0 mm

AR= 0 0 0 0 0 0 0 0 0 mm2

ρ R =AR/bd= 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000


ηR=1 – Icr R/Ig = 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0 0

MR/Mcr = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

Ig/Icr R = 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00

Simply

Supprt'd

𝑓𝑐′

ℎ3 𝑓𝑐′

190

Table O-14 - Data for UDL Slab, MR=0, Ig/Icr=6 – Example 3.6.2g – Page 2

Ex. 3.6.2g, pg 2 of 2 w0 = 10.37 N/mm fc' = 36 MPa ρ b = 0.00578

L = 10000 mm ffu = 690 MPa ρ m = 0.0158


αL/max=ML/Mmax = -2.99 -2.09 -1.50 -1.25 -1.01 -0.65 -0.35 -0.22 0.00

ML = -3.87E+8 -2.71E+8 -1.94E+8 -1.62E+8 -1.31E+8 -8.47E+7 -4.54E+7 -2.87E+7 0.00E+0 N mm

Mm = 9.75E+7 1.11E+8 1.19E+8 1.22E+8 1.24E+8 1.27E+8 1.29E+8 1.29E+8 1.30E+8 N mm

Mmax = 1.30E+8 1.30E+8 1.30E+8 1.30E+8 1.30E+8 1.30E+8 1.30E+8 1.30E+8 1.30E+8

MR = 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 N mm

K=1.2-0.2M0/Mm= 0.603 0.756 0.836 0.867 0.895 0.933 0.965 0.978 1.000


Δg(Gross) 4.20 5.99 7.09 7.52 7.92 8.47 8.87 9.03 9.26 mm

Δcr(Cracked) 25.03 35.71 42.28 44.85 47.24 50.47 52.90 53.83 55.20 mm



ΔIe(Branson) 15.45 22.04 26.09 27.68 29.16 31.15 32.65 33.22 34.07 mm

% error, Branson 28.40 11.93 8.06 8.09 8.89 10.56 11.16 11.49 12.23



ΔIe,βd(ACI440) 19.50 27.83 32.95 34.95 36.82 39.33 41.23 41.95 43.02 mm



ΔIe,avg(A23.3) 14.45 26.74 35.02 38.24 41.13 44.74 41.23 41.95 43.02 mm


∆β=0(S806) 31.69 35.82 39.78 41.67 44.18 48.19 50.90 52.02 53.53 mm



Analytical Δ3= 2.62 5.57 8.34 9.73 11.23 13.72 16.13 17.24 19.24 mm

Analytical Δ4= 21.03 21.56 21.55 21.44 21.25 20.81 20.24 19.92 19.24 mm


Analytical Δ6= 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 mm

ΔIe(x)(Exact) 21.57 25.02 28.38 30.12 32.00 34.83 36.75 37.53 38.82 mm

Length:Defl, L/Δmid= 464 400 352 332 312 287 272 266 258


2+2.4ln(2-ξ )


Δγ=1(Approx) 19.82 28.28 33.48 35.52 37.41 39.97 41.90 42.63 43.72 mm

ξ =1-√(1-Mcr/Mmax)= 0.293 0.293 0.293 0.293 0.293 0.293 0.293 0.293 0.293

γ= 1.43 1.43 1.43 1.43 1.43 1.43 1.43 1.43 1.43


ΔI'e(Proposed) 17.60 25.11 29.73 31.54 33.22 35.49 37.20 37.85 38.82 mm

% error, proposed 18.42 0.34 4.75 4.72 3.81 1.90 1.22 0.84 0.00

L/Δ exact 463.55 399.63 352.35 332.03 312.49 287.14 272.08 266.44 257.62


Δmax,Ie(x)(Exact) 24.15 26.86 29.41 30.93 32.58 35.02 36.75 37.53 38.82 mm


191

Figure O-7 - Midspan and Maximum Deflection of FRP Reinforced Slabs under

Uniformly Distributed Load with Ig/Icr=6, Mmax/Mcr=2, MR=0

The lines plotted in Figure O-7 use data in bold from Example 3.6.2g as found in Table


192

Table O-15 - Data for UDL Beam, ML=MR, Ig/Icr=17 – Example 3.6.2h – Page 1

Example 3.6.2h, pg 1 of 2 Φc = 0.65 εcu = 0.0035 mm/mm



Let ML=MR b = 0.5* h mm α1 = 0.796


M0,0 = w0 L2/8 = 8.10E+7 N mm φb = 0.75 ρ b = 0.00578




αL/max=ML/Mmax= -3.00 -2.33 -1.56 -1.22 -1.00 -0.85 -0.43 -0.25 0

wUDL = 25.92 21.60 16.62 14.40 12.96 12.00 9.26 8.10 6.48 N/mm

M0 = 3.24E+8 2.70E+8 2.08E+8 1.80E+8 1.62E+8 1.50E+8 1.16E+8 1.01E+8 8.10E+7 N mm

αL = ML/M0 = -0.75 -0.70 -0.61 -0.55 -0.50 -0.46 -0.30 -0.20 0

αR = MR/M0 = -0.75 -0.70 -0.61 -0.55 -0.50 -0.46 -0.30 -0.20 0

ML = -2.43E+8 -1.89E+8 -1.27E+8 -9.90E+7 -8.10E+7 -6.90E+7 -3.47E+7 -2.03E+7 0.00E+0 N mm

MR = -2.43E+8 -1.89E+8 -1.27E+8 -9.90E+7 -8.10E+7 -6.90E+7 -3.47E+7 -2.03E+7 0.00E+0 N mm

Mm = 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 N mm


αcr = Mcr/Mmax= 0.8 0.8 0.8 0.8 0.8 0.8 0.8 0.8 0.8


h = 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 mm

d = 510.0 510.0 510.0 510.0 510.0 510.0 510.0 510.0 510.0 mm

b = 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 mm

Ig = 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 mm4

L1 = 1646 1326 811 500 257 70 -612 -1000 -1708 mm

L2 = 3882 3775 3604 3500 3419 3357 3129 3000 2764 mm

LR4 = 3882 3775 3604 3500 3419 3357 3129 3000 2764 mm

LR5 = 1646 1326 811 500 257 70 -612 -1000 -1708 mm


Left End c L = 180.79 134.23 85.93 65.92 53.32 45.08 0 0 0 mm

AL= 4227 2041 741 417 265 186 0 0 0 mm2

ρ L =As/bd= 0.0276 0.0133 0.0048 0.0027 0.0017 0.0012 0 0 0


ηL=1 – Icr L/Ig = 0.775 0.878 0.951 0.971 0.981 0.987 0.000 0 0

ML/Mcr = -3.75 -2.92 -1.96 -1.53 -1.25 -1.06 -0.54 -0.31 0.00

Ig/Icr L = 4.45 8.20 20.28 34.64 53.11 74.41 1.00 1.00 1.00

Midspan c m = 93.98 93.98 93.98 93.98 93.98 93.98 93.98 93.98 93.98 mm

ρ m =Am/bd= 0.0059 0.0059 0.0059 0.0059 0.0059 0.0059 0.0059 0.0059 0.0059



Right End cR = 180.79 134.23 85.93 65.92 53.32 45.08 0.00 0 0 mm

AR= 4227 2041 741 417 265 186 0 0 0 mm2

ρ R =AR/bd= 0.0276 0.0133 0.0048 0.0027 0.0017 0.0012 0.0000 0.0000 0.0000


ηR=1 – Icr R/Ig = 0.775 0.878 0.951 0.971 0.981 0.987 0.000 0 0

MR/Mcr = -3.75 -2.92 -1.96 -1.53 -1.25 -1.06 -0.54 -0.31 0.00

Ig/Icr R = 4.45 8.20 20.28 34.64 53.11 74.41 1.00 1.00 1.00

Simply

Supprt'd

𝑓𝑐′

ℎ3 𝑓𝑐′

193

Table O-16 - Data for UDL Beam, ML=MR, Ig/Icr=17 – Example 3.6.2h – Page 2

Ex. 3.6.2h, pg 2 of 2 w0 = 6.48 N/mm fc' = 36 MPa ρ b = 0.00578

L = 10000 mm ffu = 690 MPa ρ m = 0.0059


αL/max=ML/Mmax = -3.00 -2.33 -1.56 -1.22 -1.00 -0.85 -0.43 -0.25 0.00

ML = -2.43E+8 -1.89E+8 -1.27E+8 -9.90E+7 -8.10E+7 -6.90E+7 -3.47E+7 -2.03E+7 0.00E+0 N mm

Mm = 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 N mm

Mmax = 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7

MR = -2.43E+8 -1.89E+8 -1.27E+8 -9.90E+7 -8.10E+7 -6.90E+7 -3.47E+7 -2.03E+7 0.00E+0 N mm

K=1.2-0.2M0/Mm= 0.400 0.533 0.687 0.756 0.800 0.830 0.914 0.950 1.000


Δg(Gross) 2.31 3.09 3.98 4.37 4.63 4.80 5.29 5.50 5.79 mm

Δcr(Cracked) 39.17 52.23 67.29 73.99 78.34 81.24 89.53 93.03 97.92 mm



ΔIe(Branson) 4.28 5.71 7.35 8.08 8.56 8.88 9.78 10.17 10.70 mm

% error, Branson 60.60 54.47 54.13 55.52 56.25 56.40 56.82 57.38 58.64



ΔIe,βd(ACI440) 17.34 23.12 29.79 32.75 34.68 35.96 39.63 41.18 43.34 mm



ΔIe,avg(A23.3) 14.02 22.87 34.69 39.82 42.81 44.53 39.63 41.18 43.34 mm


∆β=0(S806) 39.95 43.22 49.54 55.95 58.87 61.37 68.68 72.66 77.97 mm



Analytical Δ3= 6.83 7.41 8.32 8.85 9.26 9.57 10.67 11.27 12.32 mm

Analytical Δ4= 6.83 7.41 8.32 8.85 9.26 9.57 10.67 11.27 12.32 mm


Analytical Δ6= -1.81 -1.60 -0.82 -0.32 -0.07 0.00 0.00 0.00 0.00 mm

ΔIe(x)(Exact) 10.86 12.53 16.03 18.18 19.56 20.36 22.66 23.85 25.87 mm



2+2.4ln(2-ξ )


Δγ=1(Approx) 15.58 20.78 26.77 29.43 31.17 32.32 35.62 37.01 38.96 mm

ξ =1-√(1-Mcr/Mmax)= 0.553 0.553 0.553 0.553 0.553 0.553 0.553 0.553 0.553

γ= 1.22 1.22 1.22 1.22 1.22 1.22 1.22 1.22 1.22


ΔI'e(Proposed) 10.35 13.80 17.78 19.55 20.70 21.46 23.65 24.58 25.87 mm

% error, proposed 4.72 10.10 10.92 7.54 5.79 5.43 4.40 3.05 0.00

L/Δ exact 920.65 797.90 623.87 550.13 511.12 491.20 441.36 419.27 386.51


Δmax,Ie(x)(Exact) 10.87 12.54 16.03 18.18 19.57 20.36 22.66 23.85 25.87 mm


194

Figure O-8 - Copy of Figure 3-13 – UDL on Beam, Ig/Icr=17, Mm /Mcr=1.3, ML=MR

The lines plotted in Figure O-8 use data in bold from Example 3.6.2h as found in Table


195

Results Using New Mcr per CSA A23.3-04 (R2010) Appendix P

The use of 𝑀𝑐𝑟 = 0.5𝑓𝑟𝐼𝑔/𝑦𝑡, per Update no. 3 to A23.3 (CSA 2004) and the R2010

version of A23.3 (CSA 2004), has a significant effect on results for this report. This

change generally provides a reasonably accurate account for shrinkage restraint stresses

and construction pre-loading for Branson’s 𝐼𝑒 equation, as provided in CSA A23.3.

Alternatively, use of 𝑀𝑐𝑟 = 0.67𝑓𝑟𝐼𝑔/𝑦𝑡 provides an equivalent adjustment for 𝐼𝑒′ or

𝐼𝑒(𝑥) as defined in this report (Scanlon and Bischoff 2008). Discussion and graphs

showing the effects on deflection for simply supported members are also provided by

Scanlon and Bischoff (2008). This discussion explains that rational and integration-

based solutions provide a simple and robust way to account for shrinkage restraint when

calculating deflection. Use of 𝐼𝑒(𝑥) with 𝑀𝑐𝑟 = 0.67𝑓𝑟𝐼𝑔/𝑦𝑡 should provide the most

accurate results for all simply supported and continuous concrete members.

For the examples provided in this appendix, three sets of members from Appendix O are

repeated but modified to account for shrinkage restraint. Each set maintains the same

member dimensions and positive moment reinforcing, with the same 𝐼𝑔/𝐼𝑐𝑟 ratio, but

the 𝑀𝑚𝑎𝑥/𝑀𝑐𝑟 ratio increases with the reduced cracking moment.

The first example in this appendix, Example P1, is based on Example 3.6.2a from

Appendix O. The graph for Example 3.6.2a is also provided, with discussion, as Figure

3-9. For Example P1, deflection results increase with the 𝑀𝑚𝑎𝑥/𝑀𝑐𝑟 increase and

results are conservative when using Branson’s 𝐼𝑒 with the 0.5 factor for 𝑀𝑐𝑟. Branson’s

𝐼𝑒 also produced conservative results for Example 3.6.2a.

196

The second example in this appendix, Example P2, is based on Example 3.6.2e from

Appendix O. The graph for Example 3.6.2e is also provided, with discussion, Figure

3-12. For Example P2, deflection increase with the 𝑀𝑚𝑎𝑥/𝑀𝑐𝑟 increase and results

using Branson’s 𝐼𝑒 with the 0.5 factor for 𝑀𝑐𝑟 becomes conservative by about 15%

instead of being unconservative by about 15% for Example 3.6.2e.

The final example in this appendix, Example P3, is based on Example 3.6.2h from

Appendix O. The graph for Example 3.6.2h is also provided, with discussion, in Figure

3-13. For Example P3, deflection results increase with the 𝑀𝑚𝑎𝑥/𝑀𝑐𝑟 increase and

results using Branson’s 𝐼𝑒 with the 0.5 factor for 𝑀𝑐𝑟 remain unconservative, but

improve significantly, relative to Example 3.6.2h.

The main result of reducing the cracking moment, as prescribed for CSA A23.3, is that

the section-based, effective, and equivalent moments of inertia all shift towards the

value of the cracked moment of inertia. This shift causes all deflection results to

increase and become much closer to the fully cracked results; this can be seen by

comparing Figure P-1 to Figure P-2, Figure P-3 to Figure P-4, and Figure P-5 to Figure

P-6. These results show that if 𝐼𝑔/𝐼𝑐𝑟 < 12, then using Branson’s 𝐼𝑒 with the 0.5 factor

for 𝑀𝑐𝑟 will result in higher deflection predictions than using 𝐼𝑒′ or 𝐼𝑒(𝑥) with 0.67

factor for 𝑀𝑐𝑟. Results using Branson’s 𝐼𝑒 are unconservative when 𝐼𝑔/𝐼𝑐𝑟 > 12. In

spreadsheet testing that is not provided, the 𝐼𝑒 reduced for shrinkage restraint was found

to highly underestimate deflection when 𝐼𝑔/𝐼𝑐𝑟 is much larger than 12. In other omitted

spreadsheet testing, 𝐼𝑒 results sometimes highly overestimate deflection when shrinkage

restraint is included, 𝑀𝑚𝑎𝑥/𝑀𝑐𝑟 < 1.5, and 𝐼𝑔/𝐼𝑐𝑟 < 10.

197

Table P-1 - Data for UDL Beam, Ig/Icr=3.0, New A23.3 Mcr Example P1 – Page 1

Example P1 (Example 3.6.2a modified to New Mcr) Φc = 0.65 pg 1/2


Vary ML, Mm constant L = 10000 mm Ec = 27000 MPa use: Mcr = 0.67 frIg/yt

Let ML=MR b = 0.5* h mm α1 = 0.796 Mcr= 0.0335 * N mm

d = 0.9 * h mm β1 = 0.880

M0,0 = w0 L2/8 = 1.41E+8 N mm Φb = 0.85




αL/max=ML/Mmax= -3.00 -2.33 -1.86 -1.22 -1.00 -0.67 -0.43 -0.25 0

wUDL = 45.08 37.57 32.20 25.04 22.54 18.78 16.10 14.09 11.27 N/mm

M0 = 5.64E+8 4.70E+8 4.03E+8 3.13E+8 2.82E+8 2.35E+8 2.01E+8 1.76E+8 1.41E+8 N mm

αL = ML/M0 = -0.75 -0.70 -0.65 -0.55 -0.50 -0.40 -0.30 -0.20 0

αR = MR/M0 = -0.75 -0.70 -0.65 -0.55 -0.50 -0.40 -0.30 -0.20 0

ML = -4.23E+8 -3.29E+8 -2.62E+8 -1.72E+8 -1.41E+8 -9.39E+7 -6.04E+7 -3.52E+7 0.00E+0 N mm

MR = -4.23E+8 -3.29E+8 -2.62E+8 -1.72E+8 -1.41E+8 -9.39E+7 -6.04E+7 -3.52E+7 0.00E+0 N mm

Mm = 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 N mm


αcr = Mcr/Mmax= 0.3082 0.3082 0.3082 0.3082 0.3082 0.3082 0.3082 0.3082 0.3082


h = 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 mm

d = 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 540.0 mm

b = 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 mm

Ig = 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 mm4

L1 = 2141 1868 1617 1164 956 570 215 -115 -719 mm

L2 = 2921 2722 2540 2210 2059 1779 1521 1280 841 mm

LR4 = 2921 2722 2540 2210 2059 1779 1521 1280 841 mm

LR5 = 2141 1868 1617 1164 956 570 215 -115 -719 mm


Left End Kr L = 7.61 5.92 4.71 3.10 2.54 1.69 1.09 0 0 MPa

ρ L = 0.0313 0.0217 0.0163 0.0100 0.0081 0.0052 0.0033 0 0

AL=ρLbd= 5077 3516 2635 1626 1304 846 534 0 0 mm2


ηL=1 – Icr L/Ig = 0.129 0.311 0.435 0.604 0.666 0.764 0.840 0 0

ML/Mcr = -9.73 -7.57 -6.03 -3.97 -3.24 -2.16 -1.39 -0.81 0.00

Ig/Icr L = 1.15 1.45 1.77 2.52 2.99 4.24 6.25 1.00 1.00

Midspan Kr m = 2.54 2.54 2.54 2.54 2.54 2.54 2.54 2.54 2.54 MPa

ρ m = 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081



Right End Kr R = 7.61 5.92 4.71 3.10 2.54 1.69 1.09 0 0 MPa

ρ R = 0.0313 0.0217 0.0163 0.0100 0.0081 0.0052 0.0033 0 0

AR=ρRbd= 5077 3516 2635 1626 1304 846 534 0 0 mm2


ηR=1 – Icr R/Ig = 0.129 0.311 0.435 0.604 0.666 0.764 0.840 0 0

MR/Mcr = -9.73 -7.57 -6.03 -3.97 -3.24 -2.16 -1.39 -0.81 0.00

Ig/Icr R = 1.15 1.45 1.77 2.52 2.99 4.24 6.25 1.00 1.00

Simply

Supprt'd

𝑓𝑐′

ℎ3 𝑓𝑐′

198


New Mcr Example P1 w0 = 11.27 N/mm fc' = 36 MPa b = 0.5* h mm pg 2/2

L = 10000 mm fy = 400 MPa d = 0.9 * h mm



αL/max=ML/Mmax = -3.00 -2.33 -1.86 -1.22 -1.00 -0.67 -0.43 -0.25 0.00

ML = -4.23E+8 -3.29E+8 -2.62E+8 -1.72E+8 -1.41E+8 -9.39E+7 -6.04E+7 -3.52E+7 0.00E+0 N mm

Mm = 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 N mm

Mmax = 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8 1.41E+8

MR = -4.23E+8 -3.29E+8 -2.62E+8 -1.72E+8 -1.41E+8 -9.39E+7 -6.04E+7 -3.52E+7 0.00E+0 N mm

K=1.2-0.2M0/Mm= 0.400 0.533 0.629 0.756 0.800 0.867 0.914 0.950 1.000


Δg(Gross) 4.03 5.37 6.33 7.60 8.05 8.72 9.20 9.56 10.06 mm

Δcr(Cracked) 12.05 16.07 18.94 22.76 24.10 26.11 27.54 28.62 30.13 mm

Branson's Method ΔIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3 Mcr =(0.3√f'c)Ig/y= N mm


ΔIe(Branson) 11.76 15.69 18.49 22.22 23.53 25.49 26.89 27.94 29.41 mm

% error, Branson 17.94 3.21 3.59 8.36 9.00 9.21 9.04 8.83 8.08





ΔIe,avg(A23.3) 8.04 12.03 15.46 21.15 23.53 27.29 28.18 17.72 29.41 mm


∆β=0(S806) 16.35 18.28 20.01 22.76 23.90 25.84 27.31 28.41 29.99 mm

Exact Integration Ie(x)Analytical Δ1= -1.36 -1.06 -0.79 -0.40 -0.26 -0.08 -0.01 0.00 0.00 mm


Analytical Δ3= 8.49 9.13 9.68 10.61 11.01 11.71 12.29 12.78 13.57 mm

Analytical Δ4= 8.49 9.13 9.68 10.61 11.01 11.71 12.29 12.78 13.57 mm


Analytical Δ6= -1.36 -1.06 -0.79 -0.40 -0.26 -0.08 -0.01 0.00 0.00 mm

ΔIe(x)(Exact) 14.34 16.21 17.85 20.51 21.59 23.34 24.66 25.68 27.21 mm


2+2.4ln(2-ξ )


Δγ=1(Approx) 11.29 15.05 17.74 21.32 22.58 24.46 25.80 26.81 28.22 mm

ξ =1-√(1-Mcr/Mmax)= 0.168 0.168 0.168 0.168 0.168 0.168 0.168 0.168 0.168

γ= 1.53 1.53 1.53 1.53 1.53 1.53 1.53 1.53 1.53


ΔI'e(Proposed) 10.89 14.51 17.11 20.56 21.77 23.59 24.88 25.85 27.21 mm

% error, proposed 24.07 10.44 4.15 0.26 0.86 1.05 0.89 0.70 0.00


Δmax,Ie(x)(Exact) 14.34 16.21 17.85 20.51 21.59 23.34 24.66 25.68 27.21 mm



3.24E+7

𝑓𝑐′

199

Figure P-1 - Midspan Deflection Computed using Shrinkage Restraint Mcr – Beam with

Ig/Icr=3 Mm/Mcr=3.2, ML=MR

The lines plotted in Figure P-1 use data from Example P1 per Table P-1 and Table P-2.

Figure P-2 - Copy of Figure O-1, Ig/Icr=3, Mm /Mcr=2.2 – Compare to Figure P-1

𝑀𝑐𝑟 = 𝑓𝑟𝐼𝑔/𝑦𝑡

(all)

200


Example P2 (Example 3.6.2e modified to New Mcr) Φc = 0.65 pg 1/2



Let MR=0 b = 3.636* h mm α1 = 0.796 Mcr= 0.24361 * N mm

d = 0.85 * h mm β1 = 0.880

M0,0 = w0 L2/8 = 6.05E+7 N mm Φb = 0.85




αL/max=ML/Mmax= -2.99 -2.09 -1.78 -1.37 -1.01 -0.74 -0.49 -0.22 0

wUDL = 19.31 16.37 15.30 13.88 12.57 11.57 10.62 9.53 8.61 N/mm

M0 = 1.36E+8 1.15E+8 1.08E+8 9.76E+7 8.84E+7 8.13E+7 7.47E+7 6.70E+7 6.05E+7 N mm

αL = ML/M0 = -1.33 -1.10 -1 -0.85 -0.69 -0.55 -0.40 -0.20 0

αR = MR/M0 = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0

ML = -1.81E+8 -1.27E+8 -1.08E+8 -8.29E+7 -6.10E+7 -4.47E+7 -2.99E+7 -1.34E+7 0.00E+0 N mm

MR = 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 N mm

Mm = 4.55E+7 5.18E+7 5.38E+7 5.61E+7 5.79E+7 5.90E+7 5.98E+7 6.03E+7 6.05E+7 N mm


αcr = Mcr/Mmax= 0.5025 0.5025 0.5025 0.5025 0.5025 0.5025 0.5025 0.5025 0.5025


h = 275.0 275.0 275.0 275.0 275.0 275.0 275.0 275.0 275.0 mm

d = 233.8 233.8 233.8 233.8 233.8 233.8 233.8 233.8 233.8 mm

b = 1000.0 1000.0 1000.0 1000.0 1000.0 1000.0 1000.0 1000.0 1000.0 mm

Ig = 1.73E+9 1.73E+9 1.73E+9 1.73E+9 1.73E+9 1.73E+9 1.73E+9 1.73E+9 1.73E+9 mm4

L1 = 1929 1449 1240 927 593 301 -12 -429 -847 mm

L2 = 3231 2864 2704 2464 2208 1984 1744 1425 1105 mm

LR4 = 738 801 829 870 914 953 994 1050 1105 mm

LR5 = -565 -614 -635 -667 -701 -730 -762 -804 -847 mm


Left End Kr L = 5.21 3.65 3.10 2.39 1.76 1.29 0.00 0 0 MPa

ρ L = 0.0184 0.0121 0.0100 0.0075 0.0054 0.0039 0.0000 0 0

AL=ρLbd= 4301 2819 2346 1765 1271 919 0 0 0 mm2


ηL=1 – Icr L/Ig = 0.481 0.617 0.666 0.732 0.795 0.843 0.000 0 0

ML/Mcr = -5.94 -4.16 -3.54 -2.73 -2.01 -1.47 -0.98 -0.44 0.00

Ig/Icr L = 1.93 2.61 3.00 3.74 4.87 6.38 1.00 1.00 1.00

Midspan Kr m = 1.74 1.74 1.74 1.74 1.74 1.74 1.74 1.74 1.74 MPa

ρ m = 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054



Right End Kr R = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0 0 MPa

ρ R = 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0

AR=ρRbd= 0 0 0 0 0 0 0 0 0 mm2


ηR=1 – Icr R/Ig = 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0 0

MR/Mcr = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

Ig/Icr R = 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00

Simply

Supprt'd

𝑓𝑐′

ℎ3 𝑓𝑐′

201


New Mcr Example P2 w0 = 8.61 N/mm fc' = 36 MPa b = 3.636* h mm pg 2/2

L = 7500 mm fy = 400 MPa d = 0.85 * h mm



αL/max=ML/Mmax = -2.99 -2.09 -1.78 -1.37 -1.01 -0.74 -0.49 -0.22 0.00

ML = -1.81E+8 -1.27E+8 -1.08E+8 -8.29E+7 -6.10E+7 -4.47E+7 -2.99E+7 -1.34E+7 0.00E+0 N mm

Mm = 4.55E+7 5.18E+7 5.38E+7 5.61E+7 5.79E+7 5.90E+7 5.98E+7 6.03E+7 6.05E+7 N mm

Mmax = 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7 6.05E+7

MR = 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 0.00E+0 N mm

K=1.2-0.2M0/Mm= 0.603 0.756 0.800 0.852 0.895 0.924 0.950 0.978 1.000


Δg(Gross) 3.43 4.90 5.39 5.99 6.48 6.82 7.11 7.39 7.58 mm

Δcr(Cracked) 16.84 24.02 26.41 29.34 31.78 33.45 34.85 36.21 37.14 mm



ΔIe(Branson) 13.96 19.92 21.90 24.34 26.36 27.74 28.90 30.03 30.80 mm

% error, Branson 1.02 14.81 16.96 18.27 18.43 18.21 17.89 17.15 16.23





ΔIe,avg(A23.3) 11.97 18.35 20.71 23.80 26.35 27.39 19.80 30.03 30.80 mm


∆β=0(S806) 20.21 24.48 26.09 28.49 30.65 32.23 33.59 35.05 36.06 mm



Analytical Δ3= 1.82 3.84 4.78 6.20 7.69 8.94 10.21 11.75 13.11 mm

Analytical Δ4= 14.32 14.67 14.69 14.63 14.46 14.25 13.98 13.57 13.11 mm


Analytical Δ6= 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 mm

ΔIe(x)(Exact) 13.82 17.35 18.73 20.58 22.26 23.47 24.51 25.63 26.50 mm

Proposed Method ΔI'e I'e=Icr/[1-γηm(Mcr/Mmax)2] γ=(1.6ξ

3-0.6ξ

4)/(Mcr/Mmax)

2+2.4ln(2-ξ )


Δγ=1(Approx) 13.45 19.19 21.10 23.45 25.39 26.73 27.84 28.93 29.67 mm

ξ =1-√(1-Mcr/Mmax)= 0.295 0.295 0.295 0.295 0.295 0.295 0.295 0.295 0.295

γ= 1.43 1.43 1.43 1.43 1.43 1.43 1.43 1.43 1.43


ΔI'e(Proposed) 12.01 17.14 18.84 20.94 22.68 23.87 24.86 25.84 26.50 mm

% error, proposed 13.09 1.22 0.63 1.76 1.90 1.71 1.43 0.79 0.00


Δmax,Ie(x)(Exact) 15.76 18.55 19.60 21.14 22.60 23.64 24.53 25.64 26.50 mm



2.27E+7

𝑓𝑐′

202

Figure P-3 – Midspan and Maximum Deflection Computed using Shrinkage Restraint

Mcr – Slab with Ig/Icr=5, Mmax/Mcr=2, and MR=0


Figure P-4 - Copy of Figure O-5, Ig/Icr=5, Mmax/Mcr=1.3 – Compare to Figure P-3


(all)

203

Table P-5 - Data for UDL Beam, Ig/Icr=17, Reduced Mcr Example P3 – Page 1

Example P3 (Example 3.6.2h modified to New Mcr) Φc = 0.65 εcu = 0.0035 mm/mm pg 1/2



Let ML=MR b = 0.5* h mm α1 = 0.796 Mcr= 0.0335 * N mm


M0,0 = w0 L2/8 = 8.10E+7 N mm φb = 0.75 ρ b = 0.00578




αL/max=ML/Mmax= -3.00 -2.33 -1.56 -1.22 -1.00 -0.85 -0.43 -0.25 0

wUDL = 25.92 21.60 16.62 14.40 12.96 12.00 9.26 8.10 6.48 N/mm

M0 = 3.24E+8 2.70E+8 2.08E+8 1.80E+8 1.62E+8 1.50E+8 1.16E+8 1.01E+8 8.10E+7 N mm

αL = ML/M0 = -0.75 -0.70 -0.61 -0.55 -0.50 -0.46 -0.30 -0.20 0

αR = MR/M0 = -0.75 -0.70 -0.61 -0.55 -0.50 -0.46 -0.30 -0.20 0

ML = -2.43E+8 -1.89E+8 -1.27E+8 -9.90E+7 -8.10E+7 -6.90E+7 -3.47E+7 -2.03E+7 0.00E+0 N mm

MR = -2.43E+8 -1.89E+8 -1.27E+8 -9.90E+7 -8.10E+7 -6.90E+7 -3.47E+7 -2.03E+7 0.00E+0 N mm

Mm = 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 N mm


αcr = Mcr/Mmax= 0.536 0.536 0.536 0.536 0.536 0.536 0.536 0.536 0.536


h = 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 600.0 mm

d = 510.0 510.0 510.0 510.0 510.0 510.0 510.0 510.0 510.0 mm

b = 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 300.0 mm

Ig = 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 5.40E+9 mm4

L1 = 1902 1606 1130 843 618 446 -185 -543 -1197 mm

L2 = 3297 3135 2873 2715 2592 2497 2150 1954 1594 mm

LR4 = 3297 3135 2873 2715 2592 2497 2150 1954 1594 mm

LR5 = 1902 1606 1130 843 618 446 -185 -543 -1197 mm


Left End c L = 180.79 134.23 85.93 65.92 53.32 45.08 0 0 0 mm

AL= 4227 2041 741 417 265 186 0 0 0 mm2

ρ L =As/bd= 0.0276 0.0133 0.0048 0.0027 0.0017 0.0012 0 0 0


ηL=1 – Icr L/Ig = 0.775 0.878 0.951 0.971 0.981 0.987 0.000 0 0

ML/Mcr = -5.60 -4.35 -2.92 -2.28 -1.87 -1.59 -0.80 -0.47 0.00

Ig/Icr L = 4.45 8.20 20.28 34.64 53.11 74.41 1.00 1.00 1.00

Midspan c m = 93.98 93.98 93.98 93.98 93.98 93.98 93.98 93.98 93.98 mm

ρ m =Am/bd= 0.0059 0.0059 0.0059 0.0059 0.0059 0.0059 0.0059 0.0059 0.0059



Right End cR = 180.79 134.23 85.93 65.92 53.32 45.08 0.00 0 0 mm

AR= 4227 2041 741 417 265 186 0 0 0 mm2

ρ R =AR/bd= 0.0276 0.0133 0.0048 0.0027 0.0017 0.0012 0.0000 0.0000 0.0000


ηR=1 – Icr R/Ig = 0.775 0.878 0.951 0.971 0.981 0.987 0.000 0 0

MR/Mcr = -5.60 -4.35 -2.92 -2.28 -1.87 -1.59 -0.80 -0.47 0.00

Ig/Icr R = 4.45 8.20 20.28 34.64 53.11 74.41 1.00 1.00 1.00

Simply

Supprt'd

𝑓𝑐′

ℎ3 𝑓𝑐′

204

Table P-6 - Data for UDL Beam, Ig/Icr=17, Reduced Mcr Example P3 – Page 2

New Mcr Example P3 w0 = 6.48 N/mm fc' = 36 MPa ρ b = 0.00578 pg 2/2

L = 10000 mm ffu = 690 MPa ρ m = 0.0059


αL/max=ML/Mmax = -3.00 -2.33 -1.56 -1.22 -1.00 -0.85 -0.43 -0.25 0.00

ML = -2.43E+8 -1.89E+8 -1.27E+8 -9.90E+7 -8.10E+7 -6.90E+7 -3.47E+7 -2.03E+7 0.00E+0 N mm

Mm = 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 N mm

Mmax = 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7 8.10E+7

MR = -2.43E+8 -1.89E+8 -1.27E+8 -9.90E+7 -8.10E+7 -6.90E+7 -3.47E+7 -2.03E+7 0.00E+0 N mm

K=1.2-0.2M0/Mm= 0.400 0.533 0.687 0.756 0.800 0.830 0.914 0.950 1.000


Δg(Gross) 2.31 3.09 3.98 4.37 4.63 4.80 5.29 5.50 5.79 mm

Δcr(Cracked) 39.17 52.23 67.29 73.99 78.34 81.24 89.53 93.03 97.92 mm



ΔIe(Branson) 19.40 25.87 33.33 36.65 38.80 40.24 44.34 46.08 48.50 mm

% error, Branson 37.94 24.01 15.56 15.12 15.84 16.58 17.76 18.41 20.11



ΔIe,βd(ACI440) 33.84 45.12 58.14 63.93 67.69 70.19 77.36 80.38 84.61 mm

Ie L (ACI440.1R) = Ie R = 1.22E+9 6.70E+8 2.81E+8 1.73E+8 1.22E+8 9.58E+7 5.40E+9 3.69E+8 3.69E+8 mm4


ΔIe,avg(A23.3) 19.98 36.28 62.66 76.04 84.70 90.25 15.21 80.38 84.61 mm


∆β=0(S806) 52.06 55.53 62.28 67.08 71.32 74.98 83.99 87.67 93.86 mm



Analytical Δ3= 17.87 19.29 21.46 22.71 23.66 24.36 26.80 28.08 30.23 mm

Analytical Δ4= 17.87 19.29 21.46 22.71 23.66 24.36 26.80 28.08 30.23 mm


Analytical Δ6= -2.34 -2.38 -1.85 -1.26 -0.75 -0.39 0.00 0.00 0.00 mm

ΔIe(x)(Exact) 31.26 34.04 39.47 43.17 46.11 48.24 53.92 56.48 60.71 mm



2+2.4ln(2-ξ )


Δγ=1(Approx) 28.58 38.11 49.10 53.99 57.16 59.28 65.33 67.88 71.45 mm

ξ =1-√(1-Mcr/Mmax)= 0.319 0.319 0.319 0.319 0.319 0.319 0.319 0.319 0.319

γ= 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41


ΔI'e(Proposed) 24.29 32.38 41.72 45.87 48.57 50.37 55.51 57.68 60.71 mm

% error, proposed 22.31 4.87 5.70 6.25 5.34 4.42 2.94 2.13 0.00

L/Δ exact 319.89 293.77 253.33 231.63 216.88 207.31 185.45 177.06 164.70


Δmax,Ie(x)(Exact) 31.26 34.05 39.48 43.18 46.11 48.24 53.92 56.48 60.72 mm


3.24E+7

205

Figure P-5 - Midspan Deflection Computed using Shrinkage Restraint Mcr – Slab with

Ig/Icr=17, Mm/Mcr=1.9, and ML=MR


Figure P-6 - Copy of Figure O-8, Ig/Icr=17, Mm/Mcr=1.3 – Compare to Figure P-5


(all)

206

The Effects of Cracking near Supports Appendix Q

The amount of cracking near supports does affect deflection. When the stiffness at

supports is changed, the result will normally be a different bending moment distribution

and a different deflection. In idealized testing, or for design, these different results can

be surprising.

Deflections can be reduced, in theory, if end segments experience more cracking while

the negative end-moments are kept constant. This result makes sense when one looks at

the effect that each segment of the beam has on deflection. The figures in Appendix L

show graphs of 𝑚𝑀/𝐸𝐼, the virtual moment function which is integrated to determine

deflection. The graphical area above the negative moment segments reduces deflection

at midspan. Increased cracking at the ends will decrease the moment of inertia, 𝐼, in

these segments; thus, the area in the negative moment region is increased and deflection

reduced. This full scenario could occur if negative-moment pre-loading is performed

when determining the midspan deflection of a beam which has cantilever segments past

the supports on both ends. This is because the ends of the member must be able to

rotate in the manner that reduces midspan deflection. In most other cases, however,

deflection will increase if negative-moment pre-loading occurs because the reduced

end-stiffness will cause bending moments to redistribute such that the negative

moments will reduce and the positive moments will increase.

Conversely, if the stiffness at the supports is increased without shifting the

corresponding bending moments of a continuous member, then deflection will increase.

For example, if a beam has been analyzed with a particular set of cracked stiffnesses

207

and the only change made to calculations is that more top reinforcing is added, then the

stiffer end-segments will result in more midspan deflection. To accurately determine

the deflection, however, the moment diagram would have to be determined again using

the new stiffnesses.

The effect of cracking in the negative moment region and the amount of rotation at

supports are important. Using the bending moment function, 𝑀, from a constant

stiffness model is simple, but can often cause deflection to be underpredicted. It may be

difficult to accurately determine the bending moments required in order to calculate

continuous member deflections. In general, when cracking occurs near the beam

supports, the bending moment will shift towards midspan. Also, pattern loading or

changing adjacent span lengths will change the rotation at the supports, which will shift

the positive and negative bending moments. The deflection equations provided for

continuous members assume the designer has determined the bending moment correctly.

The correct worst case moment for deflection can be found by putting the different pre-

loading possibilities and load cases on a member; it also requires use of the correct

moment distribution and cracked stiffnesses all along the member. A reasonably simple

method can be used to determine a very good approximation for the load sequence

which causes maximum deflection. First, use a constant stiffness model to determine

the reduced end stiffness using the worst negative bending load-pattern(s). Again using

a constant stiffness model, determine the reduced midspan stiffness using the worst

positive bending load-pattern. Next, model the member with the appropriate segments

having these reduced end and midspan stiffnesses. Apply the worst case positive

208

bending load-pattern to this reduced stiffness model and to determine the bending

moments and deflection. While that answer may underpredict or overpredict deflection,

it is likely to be quite accurate. To obtain an alternate answer that is likely to be slightly

conservative, calculate deflection using a new stiffness model where cracking is based

on only the bending moments obtained from the reduced stiffness model mentioned

above.

For work in this report, it is assumed that the simple bending moment function which

passes through 𝑀𝐿, 𝑀𝑚, and 𝑀𝑅 is the only moment function that is relevant to

deflection. Some spreadsheet model tests were performed to test this assumption by

modelling increased the end-moments, to account for pre-loading, before adding the

normal load case. If the end-segments are more heavily cracked, without reinforcing

being added or bending moments being shifted, there will be less midspan deflection.

However, if increased end-moments are possible under service loads, then increased

end-moments will also occur under ultimate loads, so the designer must increase

reinforcing at supports. This added stiffness appears to produce an effect which

approximately offsets the additional cracking and produces sufficiently accurate

deflection results. If any particular case if believed to be an exception, the engineer

should determine deflection using integration with an appropriate stiffness function.

209

Midspan and Maximum Deflection of Linear- Appendix R

Elastic Members

There can be a significant different between midspan and maximum deflection. Section

6.3 of the Canadian Concrete Design Handbook (CAC 2005) provides deflection

calculations for prismatic linear-elastic members that are loaded primarily with a

uniformly distributed load. For end-moments where 𝑀𝐿 ≫ 𝑀𝑅, there is a relatively

large difference between deflection results as indicated in that section and the maximum

deflection determined using integration. This difference was subsequently investigated

and determined to be the difference between midspan deflection and maximum

deflection, predominantly. As such, the same issue applies to all equations which use

the midspan deflection 𝐾 factor provided in Table 2-2 and derived in Appendix A.

To summarize the relationship between midspan and maximum deflection, non-

dimensionalized results are produced and indicated in Figure R-1, Table R-7, Table R-8,

and Table R-9. These centered point load and equal third-point loaded results were

obtained using numerical integration (although the same plots could be obtained using

analytical integration). An example derivation of the full uniformly distributed loading

graph was produced for this appendix using common linear-elastic beam formulas. For

these results, the deflection was determined at 0.5 m intervals on 10 m beam and the

maximum deflection was selected. The difference between midspan and maximum

deflection, comparing the different loading types on the same graphs, is provided in

Figure R-1. See List of Symbols for symbol definitions.

210

Figure R-1 - Examples of Differences between Midspan and Maximum Deflection

Equations used in Table R-1 to Table R-7 are as provided in Appendix I, except for new

equations as provided below. The total deflection from a uniformly distributed load

with end-moments, ∆𝑈 𝑧 (𝑥), denotes the load case number for the end-moments with

the subscript 𝑧. Deflection is determined using superposition.

∆𝑈 𝑧(𝑥) = ∆𝑀𝐿(𝑥) + ∆𝑀𝑅

(𝑥) + ∆𝑈𝐷𝐿(𝑥)

For cases with 𝑀𝑅 = 0 and 𝑀𝑅 =𝑀𝐿

2 , no new calculations required

For UDL, if 𝑀𝑅 =𝑀𝑚𝑎𝑥

2 then

𝑀𝑅

𝑀0=𝑀𝐿

𝑀0− 20 + √384 − 48

𝑀𝐿

𝑀0

An identical load and beam is used for Table R-1 to Table R-7. Similar plots would be

produced if midspan moment was plotted instead of maximum moment. To determine

identical midspan moments for those graphs, values for the ratios of 𝑀𝐿 𝑀0⁄ and

𝑀𝑅 𝑀0⁄ are strategically selected for each case and the following equations are used:

where 𝑀𝐿 = 𝑀𝑅 = 0: 𝑤0 = 𝑤𝑈𝐷𝐿

other cases: 𝑤𝑈𝐷𝐿 =𝑤0

1 +𝑀𝐿

2𝑀0+

𝑀𝑅

2𝑀0+ (

𝑀𝐿

4𝑀0−

𝑀𝑅

4𝑀0)2

𝑀0 =𝑤𝑈𝐷𝐿𝐿

2

8 ; 𝑀𝑚 = 𝑀0 +

𝑀𝐿+𝑀𝑅

2 ; 𝑀𝑚𝑎𝑥 = 𝑀0 +

𝑀𝐿+𝑀𝑅

2+(𝑀𝐿 −𝑀𝑅)

2

16𝑀0

211

Table R-1 - Example Midspan vs Maximum Deflection for UDL – Page 1

Spreadsheet Function: Compare maximum and midspan deflection for a uniformly dustributed udl pg

h = 400 mm 1 of 7

b = 200 mm fc' = 36 MPa Mcr = 19.2 kN m wUDL= 1.0 N/mm

Ig = mm4 Ec = 27000 MPa L = 10000 mm M0= 12.5 kN m

1) Uniformly Distributed Load with varying Left End-Moment and Right End-Moment = 0

UDL Case 1: UDL Case 2: UDL Case 3: UDL Case 4: UDL Case 5:

ML= 0.0 kN m ML= -3.0 kN m ML= -6.25 kN m ML= -8.5 kN m ML= -11.0 kN m

MR= 0.0 kN m MR= 0.0 kN m MR= 0.0 kN m MR= 0.0 kN m MR= 0.0 kN m

Mm= 12.5 kN m Mm= 11.0 kN m Mm= 9.4 kN m Mm= 8.3 kN m Mm= 7.0 kN m

Mmax= 12.5 kN m Mmax= 11.0 kN m Mmax= 9.6 kN m Mmax= 8.6 kN m Mmax= 7.6 kN m

x ΔM_L ΔUDL ΔU1(x) ΔM_L ΔUDL ΔU2(x) ΔM_L ΔUDL ΔU3(x) ΔM_L ΔUDL ΔU4(x) ΔM_L ΔUDL ΔU5(x)

0 0 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

500 0 0.72 0.72 -0.16 0.72 0.56 -0.34 0.72 0.38 -0.46 0.72 0.26 -0.59 0.72 0.13

1000 0 1.42 1.42 -0.30 1.42 1.12 -0.62 1.42 0.80 -0.84 1.42 0.58 -1.09 1.42 0.33

1500 0 2.08 2.08 -0.41 2.08 1.67 -0.85 2.08 1.23 -1.16 2.08 0.92 -1.50 2.08 0.58

2000 0 2.69 2.69 -0.50 2.69 2.19 -1.04 2.69 1.64 -1.42 2.69 1.27 -1.83 2.69 0.85

2500 0 3.22 3.22 -0.57 3.22 2.65 -1.19 3.22 2.03 -1.61 3.22 1.61 -2.09 3.22 1.13

3000 0 3.68 3.68 -0.62 3.68 3.06 -1.29 3.68 2.38 -1.76 3.68 1.92 -2.27 3.68 1.40

3500 0 4.04 4.04 -0.65 4.04 3.39 -1.36 4.04 2.68 -1.85 4.04 2.19 -2.39 4.04 1.65

4000 0 4.31 4.31 -0.67 4.31 3.64 -1.39 4.31 2.92 -1.89 4.31 2.42 -2.44 4.31 1.86

4200 0 4.38 4.38 -0.67 4.38 3.71 -1.39 4.38 2.99 -1.89 4.38 2.49 -2.45 4.38 1.93

4400 0 4.44 4.44 -0.67 4.44 3.78 -1.39 4.44 3.05 -1.89 4.44 2.55 -2.45 4.44 2.00

4600 0 4.49 4.49 -0.66 4.49 3.82 -1.38 4.49 3.10 -1.88 4.49 2.60 -2.44 4.49 2.05

4800 0 4.51 4.51 -0.66 4.51 3.85 -1.37 4.51 3.14 -1.87 4.51 2.65 -2.42 4.51 2.10

5000 0 4.52 4.52 -0.65 4.52 3.87 -1.36 4.52 3.16 -1.84 4.52 2.68 -2.39 4.52 2.13

5200 0 4.51 4.51 -0.64 4.51 3.87 -1.34 4.51 3.18 -1.82 4.51 2.70 -2.35 4.51 2.16

5400 0 4.49 4.49 -0.63 4.49 3.86 -1.31 4.49 3.17 -1.78 4.49 2.70 -2.31 4.49 2.18

5600 0 4.44 4.44 -0.62 4.44 3.83 -1.28 4.44 3.16 -1.75 4.44 2.70 -2.26 4.44 2.18

5800 0 4.38 4.38 -0.60 4.38 3.78 -1.25 4.38 3.13 -1.70 4.38 2.68 -2.20 4.38 2.18

6000 0 4.31 4.31 -0.58 4.31 3.72 -1.22 4.31 3.09 -1.65 4.31 2.65 -2.14 4.31 2.17

6200 0 4.21 4.21 -0.56 4.21 3.65 -1.18 4.21 3.04 -1.60 4.21 2.61 -2.07 4.21 2.14

6400 0 4.10 4.10 -0.54 4.10 3.56 -1.13 4.10 2.97 -1.54 4.10 2.56 -1.99 4.10 2.11

6600 0 3.98 3.98 -0.52 3.98 3.45 -1.09 3.98 2.89 -1.48 3.98 2.50 -1.91 3.98 2.06

6800 0 3.83 3.83 -0.50 3.83 3.33 -1.04 3.83 2.79 -1.41 3.83 2.42 -1.83 3.83 2.00

7000 0 3.68 3.68 -0.47 3.68 3.20 -0.99 3.68 2.69 -1.34 3.68 2.33 -1.74 3.68 1.94

7200 0 3.50 3.50 -0.45 3.50 3.06 -0.93 3.50 2.57 -1.27 3.50 2.24 -1.64 3.50 1.86

7500 0 3.22 3.22 -0.41 3.22 2.81 -0.85 3.22 2.37 -1.15 3.22 2.07 -1.49 3.22 1.73

8000 0 2.69 2.69 -0.33 2.69 2.35 -0.69 2.69 1.99 -0.94 2.69 1.74 -1.22 2.69 1.46

8500 0 2.08 2.08 -0.25 2.08 1.83 -0.53 2.08 1.55 -0.72 2.08 1.36 -0.93 2.08 1.15

9000 0 1.42 1.42 -0.17 1.42 1.25 -0.36 1.42 1.06 -0.49 1.42 0.93 -0.63 1.42 0.79

9500 0 0.72 0.72 -0.09 0.72 0.63 -0.18 0.72 0.54 -0.25 0.72 0.47 -0.32 0.72 0.40

10000 0 0.00 0.00 0 0.00 0.00 0 0.00 0.00 0 0.00 0.00 0 0.00 0.00

load on a uniform elastic beam (I g) with different end-moment conditions

1.07E+009

0

1

2

3

4

5

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

Δ(D

efl

ecti

on

)

x (Position)

Δ𝑈4(𝑥)

Δ𝑈3(𝑥)

Δ𝑈2(𝑥)

Δ𝑈1(𝑥)

Δ𝑈5(𝑥)

212



h = 400 mm 2 of 7



1) Uniformly Distributed Load with varying Left End-Moment and Right End-Moment = 0

UDL Case 6: UDL Case 7: UDL Case 8: UDL Case 9: UDL Case 10:

ML= -12.5 kN m ML= -13.7 kN m ML= -14.5 kN m ML= -15.5 kN m ML= -16.7 kN m

MR= 0.0 kN m MR= 0.0 kN m MR= 0.0 kN m MR= 0.0 kN m MR= 0.0 kN m

Mm= 6.3 kN m Mm= 5.7 kN m Mm= 5.3 kN m Mm= 4.8 kN m Mm= 4.2 kN m

Mmax= 7.0 kN m Mmax= 6.6 kN m Mmax= 6.3 kN m Mmax= 6.0 kN m Mmax= 5.5 kN m

x ΔM_L ΔUDL ΔU6(x) ΔM_L ΔUDL ΔU7(x) ΔM_L ΔUDL ΔU8(x) ΔM_L ΔUDL ΔU9(x) ΔM_L ΔUDL ΔU10(x)

0 0 0.00 0.00 0 0.00 0.00 0 0.00 0.00 0 0.00 0.00 0 0.00 0.00

500 -0.67 0.72 0.05 -0.73 0.72 -0.01 -0.78 0.72 -0.06 -0.83 0.72 -0.11 -0.90 0.72 -0.18

1000 -1.24 1.42 0.18 -1.36 1.42 0.06 -1.43 1.42 -0.02 -1.53 1.42 -0.11 -1.65 1.42 -0.23

1500 -1.71 2.08 0.37 -1.87 2.08 0.21 -1.98 2.08 0.10 -2.12 2.08 -0.04 -2.28 2.08 -0.20

2000 -2.08 2.69 0.60 -2.28 2.69 0.40 -2.42 2.69 0.27 -2.58 2.69 0.10 -2.78 2.69 -0.10

2500 -2.37 3.22 0.85 -2.60 3.22 0.62 -2.75 3.22 0.47 -2.94 3.22 0.28 -3.17 3.22 0.05

3000 -2.58 3.68 1.09 -2.83 3.68 0.85 -3.00 3.68 0.68 -3.20 3.68 0.47 -3.45 3.68 0.23

3500 -2.72 4.04 1.32 -2.98 4.04 1.06 -3.15 4.04 0.89 -3.37 4.04 0.67 -3.63 4.04 0.41

4000 -2.78 4.31 1.53 -3.04 4.31 1.26 -3.22 4.31 1.08 -3.44 4.31 0.86 -3.71 4.31 0.59

4200 -2.78 4.38 1.60 -3.05 4.38 1.33 -3.23 4.38 1.15 -3.45 4.38 0.93 -3.72 4.38 0.66

4400 -2.78 4.44 1.66 -3.05 4.44 1.40 -3.23 4.44 1.22 -3.45 4.44 1.00 -3.71 4.44 0.73

4600 -2.77 4.49 1.72 -3.03 4.49 1.45 -3.21 4.49 1.28 -3.43 4.49 1.06 -3.70 4.49 0.79

4800 -2.74 4.51 1.77 -3.01 4.51 1.50 -3.18 4.51 1.33 -3.40 4.51 1.11 -3.67 4.51 0.85

5000 -2.71 4.52 1.81 -2.97 4.52 1.55 -3.15 4.52 1.37 -3.36 4.52 1.16 -3.62 4.52 0.90

5200 -2.67 4.51 1.84 -2.93 4.51 1.58 -3.10 4.51 1.41 -3.31 4.51 1.20 -3.57 4.51 0.94

5400 -2.62 4.49 1.86 -2.88 4.49 1.61 -3.04 4.49 1.44 -3.25 4.49 1.23 -3.50 4.49 0.98

5600 -2.57 4.44 1.88 -2.81 4.44 1.63 -2.98 4.44 1.47 -3.18 4.44 1.26 -3.43 4.44 1.01

5800 -2.50 4.38 1.88 -2.74 4.38 1.64 -2.90 4.38 1.48 -3.10 4.38 1.28 -3.34 4.38 1.04

6000 -2.43 4.31 1.88 -2.66 4.31 1.64 -2.82 4.31 1.49 -3.01 4.31 1.29 -3.25 4.31 1.06

6200 -2.35 4.21 1.86 -2.58 4.21 1.63 -2.73 4.21 1.48 -2.92 4.21 1.30 -3.14 4.21 1.07

6400 -2.27 4.10 1.83 -2.48 4.10 1.62 -2.63 4.10 1.47 -2.81 4.10 1.29 -3.03 4.10 1.07

6600 -2.18 3.98 1.80 -2.38 3.98 1.59 -2.52 3.98 1.45 -2.70 3.98 1.28 -2.91 3.98 1.07

6800 -2.08 3.83 1.76 -2.28 3.83 1.56 -2.41 3.83 1.42 -2.58 3.83 1.26 -2.78 3.83 1.06

7000 -1.97 3.68 1.70 -2.16 3.68 1.51 -2.29 3.68 1.39 -2.45 3.68 1.23 -2.64 3.68 1.04

7200 -1.87 3.50 1.64 -2.05 3.50 1.46 -2.17 3.50 1.34 -2.31 3.50 1.19 -2.49 3.50 1.01

7500 -1.70 3.22 1.53 -1.86 3.22 1.36 -1.97 3.22 1.25 -2.10 3.22 1.12 -2.27 3.22 0.96

8000 -1.39 2.69 1.30 -1.52 2.69 1.16 -1.61 2.69 1.07 -1.72 2.69 0.96 -1.86 2.69 0.83

8500 -1.06 2.08 1.02 -1.16 2.08 0.92 -1.23 2.08 0.85 -1.32 2.08 0.76 -1.42 2.08 0.66

9000 -0.72 1.42 0.70 -0.78 1.42 0.63 -0.83 1.42 0.59 -0.89 1.42 0.53 -0.96 1.42 0.46

9500 -0.36 0.72 0.36 -0.40 0.72 0.32 -0.42 0.72 0.30 -0.45 0.72 0.27 -0.48 0.72 0.24

10000 0 0.00 0.00 0 0.00 0.00 0 0.00 0.00 0 0.00 0.00 0 0.00 0.00

1.07E+009


-0.5

0

0.5

1

1.5

2

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

Δ(D

efl

ecti

on

)

x (Position)

Δ𝑈6(𝑥)

Δ𝑈 (𝑥)Δ𝑈 (𝑥)Δ𝑈9(𝑥)

Δ𝑈10 (𝑥)

213



h = 400 mm 3 of 7



2) Uniformly Distributed Load with varying Left End-Moment and Right End-Moment = Left End-Moment /2

UDL Case 11: UDL Case 12: UDL Case 13: UDL Case 14:

ML= 0.0 kN m ML= -3.0 kN m ML= -7.18 kN m ML= -9.0 kN m

MR= 0.00 kN m MR= -1.50 kN m MR= -3.59 kN m MR= -4.50 kN m

Mm= 12.5 kN m Mm= 10.3 kN m Mm= 7.1 kN m Mm= 5.8 kN m

Mmax= 12.50 kN m Mmax= 10.26 kN m Mmax= 7.18 kN m Mmax= 5.85 kN m

x ΔM_R ΔUDL ΔU11(x) ΔM_L ΔM_R ΔUDL ΔU12(x) ΔM_L ΔM_R ΔUDL ΔU13(x) ΔM_L ΔM_R ΔUDL ΔU14(x)

0 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

500 0.00 0.72 0.72 -0.16 -0.04 0.72 0.52 -0.38 -0.10 0.72 0.23 -0.48 -0.13 0.72 0.11

1000 0.00 1.42 1.42 -0.30 -0.09 1.42 1.04 -0.71 -0.21 1.42 0.50 -0.89 -0.26 1.42 0.27

1500 0.00 2.08 2.08 -0.41 -0.13 2.08 1.54 -0.98 -0.30 2.08 0.80 -1.23 -0.38 2.08 0.47

2000 0.00 2.69 2.69 -0.50 -0.17 2.69 2.02 -1.20 -0.40 2.69 1.09 -1.50 -0.50 2.69 0.69

2500 0.00 3.22 3.22 -0.57 -0.20 3.22 2.45 -1.36 -0.49 3.22 1.37 -1.71 -0.61 3.22 0.90

3000 0.00 3.68 3.68 -0.62 -0.24 3.68 2.82 -1.48 -0.57 3.68 1.63 -1.86 -0.71 3.68 1.11

3500 0.00 4.04 4.04 -0.65 -0.27 4.04 3.12 -1.56 -0.64 4.04 1.84 -1.96 -0.80 4.04 1.29

4000 0.00 4.31 4.31 -0.67 -0.29 4.31 3.35 -1.60 -0.70 4.31 2.01 -2.00 -0.88 4.31 1.43

4200 0.00 4.38 4.38 -0.67 -0.30 4.38 3.41 -1.60 -0.72 4.38 2.06 -2.00 -0.90 4.38 1.48

4400 0.00 4.44 4.44 -0.67 -0.31 4.44 3.47 -1.60 -0.74 4.44 2.11 -2.00 -0.92 4.44 1.52

4600 0.00 4.49 4.49 -0.66 -0.31 4.49 3.51 -1.59 -0.75 4.49 2.14 -1.99 -0.94 4.49 1.55

4800 0.00 4.51 4.51 -0.66 -0.32 4.51 3.53 -1.58 -0.77 4.51 2.17 -1.98 -0.96 4.51 1.57

5000 0.00 4.52 4.52 -0.65 -0.33 4.52 3.54 -1.56 -0.78 4.52 2.18 -1.95 -0.98 4.52 1.59

5200 0.00 4.51 4.51 -0.64 -0.33 4.51 3.54 -1.53 -0.79 4.51 2.19 -1.92 -0.99 4.51 1.60

5400 0.00 4.49 4.49 -0.63 -0.33 4.49 3.52 -1.51 -0.79 4.49 2.18 -1.89 -1.00 4.49 1.60

5600 0.00 4.44 4.44 -0.62 -0.33 4.44 3.49 -1.47 -0.80 4.44 2.17 -1.85 -1.00 4.44 1.59

5800 0.00 4.38 4.38 -0.60 -0.33 4.38 3.45 -1.44 -0.80 4.38 2.15 -1.80 -1.00 4.38 1.58

6000 0.00 4.31 4.31 -0.58 -0.33 4.31 3.39 -1.40 -0.80 4.31 2.11 -1.75 -1.00 4.31 1.56

6200 0.00 4.21 4.21 -0.56 -0.33 4.21 3.32 -1.35 -0.79 4.21 2.07 -1.69 -0.99 4.21 1.52

6400 0.00 4.10 4.10 -0.54 -0.33 4.10 3.23 -1.30 -0.79 4.10 2.01 -1.63 -0.98 4.10 1.49

6600 0.00 3.98 3.98 -0.52 -0.32 3.98 3.13 -1.25 -0.77 3.98 1.95 -1.57 -0.97 3.98 1.44

6800 0.00 3.83 3.83 -0.50 -0.32 3.83 3.02 -1.19 -0.76 3.83 1.88 -1.50 -0.95 3.83 1.39

7000 0.00 3.68 3.68 -0.47 -0.31 3.68 2.89 -1.13 -0.74 3.68 1.80 -1.42 -0.93 3.68 1.32

7200 0.00 3.50 3.50 -0.45 -0.30 3.50 2.76 -1.07 -0.72 3.50 1.71 -1.34 -0.90 3.50 1.26

7500 0.00 3.22 3.22 -0.41 -0.28 3.22 2.53 -0.97 -0.68 3.22 1.57 -1.22 -0.85 3.22 1.15

8000 0.00 2.69 2.69 -0.33 -0.25 2.69 2.10 -0.80 -0.60 2.69 1.29 -1.00 -0.75 2.69 0.94

8500 0.00 2.08 2.08 -0.25 -0.20 2.08 1.62 -0.61 -0.49 2.08 0.98 -0.76 -0.61 2.08 0.70

9000 0.00 1.42 1.42 -0.17 -0.15 1.42 1.10 -0.41 -0.36 1.42 0.65 -0.52 -0.45 1.42 0.46

9500 0.00 0.72 0.72 -0.09 -0.08 0.72 0.55 -0.21 -0.19 0.72 0.32 -0.26 -0.24 0.72 0.22

10000 0.00 0.00 0.00 0 0.00 0.00 0.00 0 0.00 0.00 0.00 0 0.00 0.00 0.00


1.07E+009

0

1

2

3

4

5

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

Δ(D

efl

ecti

on

)

x (Position)

Δ𝑈11 (𝑥)

Δ𝑈12 (𝑥)

Δ𝑈13 (𝑥)

Δ𝑈14 (𝑥)

214



h = 400 mm 4 of 7



2) Uniformly Distributed Load with varying Left End-Moment and Right End-Moment = Left End-Moment /2

UDL Case 15: UDL Case 16: UDL Case 17:

ML= -10.0 kN m ML= -11.0 kN m ML= -11.7 kN m


Mm= 5.0 kN m Mm= 4.3 kN m Mm= 3.7 kN m

Mmax= 5.1 kN m Mmax= 4.4 kN m Mmax= 3.9 kN m

x ΔM_L ΔM_R ΔUDL ΔU15(x) ΔM_L ΔM_R ΔUDL ΔU16(x) ΔM_L ΔM_R ΔUDL ΔU17(x)

0 0 0.00 0.00 0.00 0 0.00 0.00 0.00 0.00 0.00 0.00 0.00

500 -0.54 -0.14 0.72 0.04 -0.59 -0.16 0.72 -0.03 -0.63 -0.17 0.72 -0.08

1000 -0.99 -0.29 1.42 0.14 -1.09 -0.32 1.42 0.02 -1.16 -0.34 1.42 -0.07

1500 -1.37 -0.42 2.08 0.29 -1.50 -0.47 2.08 0.11 -1.60 -0.50 2.08 -0.01

2000 -1.67 -0.56 2.69 0.46 -1.83 -0.61 2.69 0.24 -1.95 -0.65 2.69 0.09

2500 -1.90 -0.68 3.22 0.64 -2.09 -0.75 3.22 0.39 -2.22 -0.79 3.22 0.21

3000 -2.07 -0.79 3.68 0.82 -2.27 -0.87 3.68 0.53 -2.42 -0.92 3.68 0.33

3500 -2.17 -0.89 4.04 0.98 -2.39 -0.98 4.04 0.67 -2.54 -1.04 4.04 0.46

4000 -2.22 -0.97 4.31 1.11 -2.44 -1.07 4.31 0.79 -2.60 -1.14 4.31 0.57

4200 -2.23 -1.00 4.38 1.15 -2.45 -1.10 4.38 0.83 -2.61 -1.17 4.38 0.61

4400 -2.22 -1.03 4.44 1.19 -2.45 -1.13 4.44 0.87 -2.60 -1.20 4.44 0.64

4600 -2.21 -1.05 4.49 1.22 -2.44 -1.15 4.49 0.90 -2.59 -1.23 4.49 0.67

4800 -2.20 -1.07 4.51 1.25 -2.42 -1.18 4.51 0.92 -2.57 -1.25 4.51 0.69

5000 -2.17 -1.09 4.52 1.27 -2.39 -1.19 4.52 0.94 -2.54 -1.27 4.52 0.71

5200 -2.14 -1.10 4.51 1.28 -2.35 -1.21 4.51 0.95 -2.50 -1.28 4.51 0.73

5400 -2.10 -1.11 4.49 1.28 -2.31 -1.22 4.49 0.96 -2.46 -1.30 4.49 0.74

5600 -2.05 -1.11 4.44 1.28 -2.26 -1.22 4.44 0.96 -2.40 -1.30 4.44 0.74

5800 -2.00 -1.11 4.38 1.27 -2.20 -1.23 4.38 0.96 -2.34 -1.30 4.38 0.74

6000 -1.94 -1.11 4.31 1.25 -2.14 -1.22 4.31 0.94 -2.28 -1.30 4.31 0.73

6200 -1.88 -1.10 4.21 1.23 -2.07 -1.21 4.21 0.93 -2.20 -1.29 4.21 0.72

6400 -1.81 -1.09 4.10 1.19 -1.99 -1.20 4.10 0.90 -2.12 -1.28 4.10 0.70

6600 -1.74 -1.08 3.98 1.16 -1.91 -1.19 3.98 0.88 -2.04 -1.26 3.98 0.68

6800 -1.66 -1.06 3.83 1.11 -1.83 -1.16 3.83 0.84 -1.94 -1.24 3.83 0.65

7000 -1.58 -1.03 3.68 1.06 -1.74 -1.14 3.68 0.80 -1.85 -1.21 3.68 0.62

7200 -1.49 -1.00 3.50 1.01 -1.64 -1.10 3.50 0.76 -1.75 -1.17 3.50 0.58

7500 -1.36 -0.95 3.22 0.92 -1.49 -1.04 3.22 0.68 -1.59 -1.11 3.22 0.52

8000 -1.11 -0.83 2.69 0.74 -1.22 -0.92 2.69 0.55 -1.30 -0.98 2.69 0.41

8500 -0.85 -0.68 2.08 0.55 -0.93 -0.75 2.08 0.40 -0.99 -0.80 2.08 0.29

9000 -0.57 -0.49 1.42 0.35 -0.63 -0.54 1.42 0.24 -0.67 -0.58 1.42 0.17

9500 -0.29 -0.27 0.72 0.16 -0.32 -0.29 0.72 0.11 -0.34 -0.31 0.72 0.07

10000 0 0.00 0.00 0.00 0 0.00 0.00 0.00 0 0.00 0.00 0.00


1.07E+009

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

Δ(D

efl

ecti

on

)

x (Position)

Δ𝑈15 (𝑥)

Δ𝑈16 (𝑥)

Δ𝑈1 (𝑥)

215



h = 400 mm 5 of 7



3) Uniformly Distributed Load with varying Left End-Moment and Right End-Moment = Maximum Moment /2

UDL Case 21: UDL Case 22: UDL Case 23: UDL Case 24:

ML= 0.0 kN m ML= -3.0 kN m ML= -7.18 kN m ML= -9.5 kN m

MR= -5.05 kN m MR= -4.40 kN m MR= -3.59 kN m MR= -3.18 kN m

Mm= 10.0 kN m Mm= 8.8 kN m Mm= 7.1 kN m Mm= 6.2 kN m

Mmax= 10.10 kN m Mmax= 8.81 kN m Mmax= 7.18 kN m Mmax= 6.36 kN m

x ΔM_R ΔUDL ΔU21(x) ΔM_L ΔM_R ΔUDL ΔU22(x) ΔM_L ΔM_R ΔUDL ΔU23(x) ΔM_L ΔM_R ΔUDL ΔU24(x)

0 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

500 -0.15 0.72 0.57 -0.16 -0.13 0.72 0.43 -0.38 -0.10 0.72 0.23 -0.51 -0.09 0.72 0.12

1000 -0.29 1.42 1.13 -0.30 -0.25 1.42 0.87 -0.71 -0.21 1.42 0.50 -0.94 -0.18 1.42 0.30

1500 -0.43 2.08 1.65 -0.41 -0.37 2.08 1.30 -0.98 -0.30 2.08 0.80 -1.30 -0.27 2.08 0.51

2000 -0.56 2.69 2.12 -0.50 -0.49 2.69 1.70 -1.20 -0.40 2.69 1.09 -1.58 -0.35 2.69 0.75

2500 -0.69 3.22 2.54 -0.57 -0.60 3.22 2.05 -1.36 -0.49 3.22 1.37 -1.80 -0.43 3.22 0.99

3000 -0.80 3.68 2.88 -0.62 -0.70 3.68 2.36 -1.48 -0.57 3.68 1.63 -1.96 -0.50 3.68 1.21

3500 -0.90 4.04 3.14 -0.65 -0.78 4.04 2.61 -1.56 -0.64 4.04 1.84 -2.06 -0.57 4.04 1.41

4000 -0.98 4.31 3.32 -0.67 -0.86 4.31 2.78 -1.60 -0.70 4.31 2.01 -2.11 -0.62 4.31 1.58

4200 -1.01 4.38 3.37 -0.67 -0.88 4.38 2.83 -1.60 -0.72 4.38 2.06 -2.12 -0.64 4.38 1.63

4400 -1.04 4.44 3.41 -0.67 -0.90 4.44 2.87 -1.60 -0.74 4.44 2.11 -2.11 -0.65 4.44 1.68

4600 -1.06 4.49 3.43 -0.66 -0.92 4.49 2.90 -1.59 -0.75 4.49 2.14 -2.10 -0.67 4.49 1.72

4800 -1.08 4.51 3.43 -0.66 -0.94 4.51 2.91 -1.58 -0.77 4.51 2.17 -2.09 -0.68 4.51 1.75

5000 -1.10 4.52 3.42 -0.65 -0.96 4.52 2.91 -1.56 -0.78 4.52 2.18 -2.06 -0.69 4.52 1.77

5200 -1.11 4.51 3.40 -0.64 -0.97 4.51 2.90 -1.53 -0.79 4.51 2.19 -2.03 -0.70 4.51 1.78

5400 -1.12 4.49 3.37 -0.63 -0.97 4.49 2.88 -1.51 -0.79 4.49 2.18 -1.99 -0.70 4.49 1.79

5600 -1.12 4.44 3.32 -0.62 -0.98 4.44 2.85 -1.47 -0.80 4.44 2.17 -1.95 -0.71 4.44 1.79

5800 -1.13 4.38 3.26 -0.60 -0.98 4.38 2.80 -1.44 -0.80 4.38 2.15 -1.90 -0.71 4.38 1.77

6000 -1.12 4.31 3.18 -0.58 -0.98 4.31 2.74 -1.40 -0.80 4.31 2.11 -1.85 -0.71 4.31 1.75

6200 -1.12 4.21 3.10 -0.56 -0.97 4.21 2.67 -1.35 -0.79 4.21 2.07 -1.79 -0.70 4.21 1.72

6400 -1.10 4.10 3.00 -0.54 -0.96 4.10 2.59 -1.30 -0.78 4.10 2.01 -1.72 -0.70 4.10 1.68

6600 -1.09 3.98 2.89 -0.52 -0.95 3.98 2.50 -1.25 -0.77 3.98 1.95 -1.65 -0.69 3.98 1.64

6800 -1.07 3.83 2.76 -0.50 -0.93 3.83 2.40 -1.19 -0.76 3.83 1.88 -1.58 -0.67 3.83 1.58

7000 -1.04 3.68 2.63 -0.47 -0.91 3.68 2.29 -1.13 -0.74 3.68 1.80 -1.50 -0.66 3.68 1.52

7200 -1.01 3.50 2.49 -0.45 -0.88 3.50 2.17 -1.07 -0.72 3.50 1.71 -1.42 -0.64 3.50 1.45

7500 -0.96 3.22 2.26 -0.41 -0.84 3.22 1.98 -0.97 -0.68 3.22 1.57 -1.29 -0.60 3.22 1.33

8000 -0.84 2.69 1.84 -0.33 -0.73 2.69 1.62 -0.80 -0.60 2.69 1.29 -1.06 -0.53 2.69 1.10

8500 -0.69 2.08 1.39 -0.25 -0.60 2.08 1.22 -0.61 -0.49 2.08 0.98 -0.81 -0.43 2.08 0.84

9000 -0.50 1.42 0.92 -0.17 -0.44 1.42 0.81 -0.41 -0.36 1.42 0.65 -0.54 -0.31 1.42 0.56

9500 -0.27 0.72 0.45 -0.09 -0.24 0.72 0.40 -0.21 -0.19 0.72 0.32 -0.27 -0.17 0.72 0.28

10000 0.00 0.00 0.00 0 0.00 0.00 0.00 0 0.00 0.00 0.00 0 0.00 0.00 0.00


1.07E+009

0

0.5

1

1.5

2

2.5

3

3.5

4

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

Δ(D

efl

ecti

on

)

x (Position)

Δ𝑈21 (𝑥)

Δ𝑈22 (𝑥)

Δ𝑈23 (𝑥)

Δ𝑈24 (𝑥)

216



h = 400 mm 6 of 7



3) Uniformly Distributed Load with varying Left End-Moment and Right End-Moment = Maximum Moment /2

UDL Case 25: UDL Case 26: UDL Case 27:

ML= -11.5 kN m ML= -13.0 kN m ML= -14.4 kN m


Mm= 5.3 kN m Mm= 4.7 kN m Mm= 4.1 kN m

Mmax= 5.7 kN m Mmax= 5.2 kN m Mmax= 4.8 kN m

x ΔM_L ΔM_R ΔUDL ΔU25(x) ΔM_L ΔM_R ΔUDL ΔU26(x) ΔM_L ΔM_R ΔUDL ΔU27(x)

0 0 0.00 0.00 0.00 0 0.00 0.00 0.00 0.00 0.00 0.00 0.00

500 -0.62 -0.08 0.72 0.02 -0.70 -0.08 0.72 -0.05 -0.77 -0.07 0.72 -0.12

1000 -1.14 -0.16 1.42 0.12 -1.29 -0.15 1.42 -0.02 -1.43 -0.14 1.42 -0.14

1500 -1.57 -0.24 2.08 0.27 -1.77 -0.22 2.08 0.08 -1.97 -0.20 2.08 -0.09

2000 -1.92 -0.32 2.69 0.45 -2.17 -0.29 2.69 0.23 -2.40 -0.27 2.69 0.02

2500 -2.18 -0.39 3.22 0.65 -2.47 -0.35 3.22 0.40 -2.73 -0.33 3.22 0.16

3000 -2.38 -0.45 3.68 0.85 -2.69 -0.41 3.68 0.58 -2.98 -0.38 3.68 0.32

3500 -2.50 -0.51 4.04 1.04 -2.82 -0.46 4.04 0.75 -3.13 -0.43 4.04 0.48

4000 -2.56 -0.55 4.31 1.20 -2.89 -0.51 4.31 0.91 -3.20 -0.47 4.31 0.64

4200 -2.56 -0.57 4.38 1.25 -2.90 -0.52 4.38 0.96 -3.21 -0.48 4.38 0.69

4400 -2.56 -0.59 4.44 1.30 -2.89 -0.54 4.44 1.01 -3.20 -0.49 4.44 0.75

4600 -2.55 -0.60 4.49 1.34 -2.88 -0.55 4.49 1.06 -3.19 -0.51 4.49 0.79

4800 -2.52 -0.61 4.51 1.38 -2.85 -0.56 4.51 1.10 -3.16 -0.51 4.51 0.84

5000 -2.50 -0.62 4.52 1.41 -2.82 -0.57 4.52 1.13 -3.13 -0.52 4.52 0.87

5200 -2.46 -0.63 4.51 1.43 -2.78 -0.57 4.51 1.16 -3.08 -0.53 4.51 0.91

5400 -2.41 -0.63 4.49 1.44 -2.73 -0.58 4.49 1.18 -3.02 -0.53 4.49 0.93

5600 -2.36 -0.63 4.44 1.45 -2.67 -0.58 4.44 1.19 -2.96 -0.54 4.44 0.95

5800 -2.30 -0.63 4.38 1.45 -2.60 -0.58 4.38 1.20 -2.88 -0.54 4.38 0.96

6000 -2.24 -0.63 4.31 1.44 -2.53 -0.58 4.31 1.20 -2.80 -0.54 4.31 0.97

6200 -2.16 -0.63 4.21 1.42 -2.45 -0.58 4.21 1.19 -2.71 -0.53 4.21 0.97

6400 -2.09 -0.62 4.10 1.39 -2.36 -0.57 4.10 1.17 -2.61 -0.53 4.10 0.96

6600 -2.00 -0.61 3.98 1.36 -2.26 -0.56 3.98 1.15 -2.51 -0.52 3.98 0.95

6800 -1.91 -0.60 3.83 1.32 -2.16 -0.55 3.83 1.12 -2.39 -0.51 3.83 0.93

7000 -1.82 -0.59 3.68 1.27 -2.05 -0.54 3.68 1.08 -2.28 -0.50 3.68 0.90

7200 -1.72 -0.57 3.50 1.22 -1.94 -0.52 3.50 1.04 -2.15 -0.48 3.50 0.87

7500 -1.56 -0.54 3.22 1.12 -1.76 -0.50 3.22 0.96 -1.95 -0.46 3.22 0.81

8000 -1.28 -0.47 2.69 0.93 -1.44 -0.44 2.69 0.80 -1.60 -0.40 2.69 0.68

8500 -0.98 -0.39 2.08 0.72 -1.10 -0.36 2.08 0.62 -1.22 -0.33 2.08 0.53

9000 -0.66 -0.28 1.42 0.48 -0.74 -0.26 1.42 0.42 -0.83 -0.24 1.42 0.36

9500 -0.33 -0.15 0.72 0.24 -0.38 -0.14 0.72 0.20 -0.42 -0.13 0.72 0.18

10000 0 0.00 0.00 0.00 0 0.00 0.00 0.00 0 0.00 0.00 0.00


1.07E+009

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

Δ(D

efl

ecti

on

)

x (Position)

Δ𝑈25 (𝑥)

Δ𝑈26 (𝑥)

Δ𝑈2 (𝑥)

217

Table R-7 - Example Midspan vs Maximum Deflection for UDL – Summary


7 of 7

Cases with MR=0 ML / Δmax / Cases with MR=ML/2 ML / Δmax /

ML Mmid Mmax Δmid Δmax Mmax Δmid ML Mmid Mmax Δmid Δmax Mmax Δmid

Case 1 0.0 12.5 12.5 4.5 4.5 0.00 1.00 Case11 0.0 12.5 12.5 4.5 4.5 0.00 1.00

Case 2 -3.0 11.0 11.0 3.9 3.9 -0.27 1.00 Case12 -3.0 10.3 10.3 3.5 3.5 -0.29 1.00

Case 3 -6.3 9.4 9.6 3.2 3.2 -0.65 1.00 Case13 -7.2 7.1 7.2 2.2 2.2 -1.00 1.00

Case 4 -8.5 8.3 8.6 2.7 2.7 -0.99 1.01 Case14 -9.0 5.8 5.9 1.6 1.6 -1.54 1.01

Case 5 -11.0 7.0 7.6 2.1 2.2 -1.45 1.02 Case15 -10.0 5.0 5.1 1.3 1.3 -1.95 1.01

Case 6 -12.5 6.3 7.0 1.8 1.9 -1.78 1.04 Case16 -11.0 4.3 4.4 0.9 1.0 -2.50 1.02

Case 7 -13.7 5.7 6.6 1.5 1.6 -2.08 1.06 Case17 -11.7 3.7 3.9 0.7 0.7 -3.00 1.04

Case 8 -14.5 5.3 6.3 1.4 1.5 -2.30 1.08

Case 9 -15.5 4.8 6.0 1.2 1.3 -2.60 1.12

Case10 -16.7 4.2 5.5 0.9 1.1 -3.01 1.20

Cases with MR=Mmax/2 ML / Δmax /

ML Mmid Mmax Δmid Δmax Mmax Δmid

Case21 0.0 10.0 10.1 3.4 3.4 0.00 1.00

Case22 -3.0 8.8 8.8 2.9 2.9 -0.34 1.00

Case23 -7.2 7.1 7.2 2.2 2.2 -1.00 1.00

Case24 -9.5 6.2 6.4 1.8 1.8 -1.49 1.01

Case25 -11.5 5.3 5.7 1.4 1.4 -2.02 1.03

Case26 -13.0 4.7 5.2 1.1 1.2 -2.49 1.06

Case27 -14.4 4.1 4.8 0.9 1.0 -2.99 1.11


1.00

1.05

1.10

1.15

1.20

-3.00-2.50-2.00-1.50-1.00-0.500.00Rat

io o

f Max

imu

m t

o

Mid

span

Def

lect

ion

Ratio of End-Moment Relative to Maximum Positive Moment

𝑀𝑅 = 0

𝑀𝑅 = 𝑀𝑚𝑎𝑥 2⁄

𝑀𝑅 = 𝑀𝐿 /2

218

To produce graphs comparing midspan and maximum deflection for centered point load

beams, identical maximum moments and concrete sections are used for all cases. Each

(numbered) load case provides different end-moments by selecting different 𝑀𝐿/𝑀0 and

𝑀𝑅/𝑀0 ratios. To achieve identical midspan moment, point loads are varied as follows:

when 𝑀𝐿 = 𝑀𝑅 = 0 ∶ 𝑃0 = point load ; when 𝑀𝑅 = 0 ∶ 𝑃s =𝑃0

1 + 0.5𝑀𝐿 𝑀0⁄

when 𝑀𝑅 =𝑀𝐿

2∶ 𝑃s =

𝑃0

1 +0.75𝑀𝐿

𝑀0

; when 𝑀𝑅 =𝑀𝑚

2∶ 𝑃s =

2.5𝑃0

2 +𝑀𝐿

𝑀0

and 𝑀𝑅

𝑀0=

𝑃02𝑃s

For all cases: 𝑀1𝑃𝐿 =𝑃s𝐿

4 ; 𝑀𝑚 = 𝑀𝑚𝑎𝑥 = 𝑀1𝑃𝐿 +

𝑀𝐿

2+𝑀𝑅

2

Table R-8 - Example Midspan vs Maximum Deflection for CPL – Summary

cpl pg

1 of 1

Midspan PL Cases with MR=0 ML / Δmax / Midspan PL Cases with MR=ML/2 ML / Δmax /


Case 1 -375.0 125.0 125.0 0.8 1.1 -3.00 1.41 Case11 -312.9 125.0 125.0 0.2 0.3 -2.50 1.29

Case 2 -311.8 125.0 125.0 1.2 1.4 -2.49 1.19 Case12 -283.8 125.0 125.0 0.5 0.5 -2.27 1.10

Case 3 -250.0 125.0 125.0 1.6 1.7 -2.00 1.09 Case13 -250.0 125.0 125.0 0.8 0.8 -2.00 1.04

Case 4 -188.6 125.0 125.0 2.0 2.1 -1.51 1.04 Case14 -184.2 125.0 125.0 1.4 1.4 -1.47 1.01

Case 5 -125.9 125.0 125.0 2.4 2.4 -1.01 1.01 Case15 -124.5 125.0 125.0 2.0 2.0 -1.00 1.00

Case 6 -62.5 125.0 125.0 2.8 2.8 -0.50 1.00 Case16 -71.4 125.0 125.0 2.5 2.5 -0.57 1.00

Case 7 -44.1 125.0 125.0 2.9 2.9 -0.35 1.00 Case17 -48.4 125.0 125.0 2.7 2.7 -0.39 1.00

Case 8 0.0 125.0 125.0 3.2 3.2 0.00 1.00 Case18 0.0 125.0 125.0 3.2 3.2 0.00 1.00

Midspan PL Cases with MR=Mmax/2

ML Mmid Mmax Δmid Δmax ML/max Δmax/mid ML Mmid Mmax Δmid Δmax ML/max Δmax/mid

Case21 -381.9 125.0 125.0 0.4 0.6 -3.06 1.70 Case25 -168.3 125.0 125.0 1.7 1.7 -1.35 1.01

Case22 -312.5 125.0 125.0 0.8 0.9 -2.50 1.18 Case26 -124.6 125.0 125.0 2.0 2.0 -1.00 1.00

Case23 -255.7 125.0 125.0 1.2 1.2 -2.05 1.07 Case27 -78.1 125.0 125.0 2.3 2.3 -0.63 1.00

Case24 -208.3 125.0 125.0 1.5 1.5 -1.67 1.03 Case28 0.0 125.0 125.0 2.8 2.8 0.00 1.00

-1.35 1.01

on a uniform elastic beam with different end-moment conditions

Spreadsheet Function: Compare maximum and midspan deflection for midspan point loads

1.00

1.05

1.10

1.15

1.20

-3.00-2.50-2.00-1.50-1.00-0.500.00

Rat

io o

f M

axim

um

to

M

idsp

an D

efl

ect

ion



𝑀𝑅 = 0

𝑀𝑅 = 𝑀𝐿 2⁄

219

To produce graphs for equal third-point loaded beams comparing midspan and

maximum deflection, identical maximum moments and beams are used for all cases.

Each load case provides different end-moments by selecting different 𝑀𝐿/𝑀𝑚𝑎𝑥 and

𝑀𝑅/𝑀𝑚𝑎𝑥 ratios. To achieve identical midspan moment, loads are varied as follows:

when 𝑀𝐿 = 𝑀𝑅 = 0 ∶ 𝑃0 = total point load ; else 𝑃s = 𝑃0 (1 −1

3

𝑀𝐿

𝑀𝑚𝑎𝑥−2

3

𝑀𝑅

𝑀𝑚𝑎𝑥)

𝑀𝐿

𝑀2𝑃𝐿=

𝑀𝐿

𝑀𝑚𝑎𝑥

1 −13

𝑀𝐿


𝑀𝑅

𝑀𝑚𝑎𝑥

; 𝑀𝑅

𝑀2𝑃𝐿=

𝑀𝑅

𝑀𝑚𝑎𝑥

1 −13

𝑀𝐿


𝑀𝑅

𝑀𝑚𝑎𝑥

As usual: 𝑀2𝑃𝐿 =𝑃s𝐿

3 ; 𝑀𝑚 = 𝑀2𝑃𝐿 +

𝑀𝐿

2+𝑀𝑅

2 ; 𝑀𝑚𝑎𝑥 = 𝑀2𝑃𝐿 +

𝑀𝐿

3+2𝑀𝑅

3

Table R-9 - Example Midspan vs Maximum Deflection for 2PL – Summary

2pl pg

1 of 1

2 PL @ 3rd pts Cases with MR=0 ML / Δmax / 2 PL @ 3rd pts Cases with MR=ML/2 ML / Δmax /


Case 1 -343.8 67.7 125.0 2.3 2.9 -2.75 1.30 Case11 -375.0 93.8 125.0 2.7 2.9 -3.00 1.06

Case 2 -312.5 72.9 125.0 2.7 3.3 -2.50 1.20 Case12 -343.8 96.4 125.0 3.1 3.2 -2.75 1.05

Case 3 -281.3 78.1 125.0 3.2 3.7 -2.25 1.13 Case13 -312.5 99.0 125.0 3.5 3.6 -2.50 1.03

Case 4 -250.0 83.3 125.0 3.7 4.0 -2.00 1.09 Case14 -281.3 101.6 125.0 3.9 4.0 -2.25 1.02

Case 5 -187.5 93.8 125.0 4.7 4.8 -1.50 1.04 Case15 -250.0 104.2 125.0 4.3 4.4 -2.00 1.02

Case 6 -125.0 104.2 125.0 5.6 5.7 -1.00 1.01 Case16 -187.5 109.4 125.0 5.1 5.1 -1.50 1.01

Case 7 -62.5 114.6 125.0 6.6 6.6 -0.50 1.00 Case17 -125.0 114.6 125.0 5.9 5.9 -1.00 1.00

Case 8 0.0 125.0 125.0 7.5 7.5 0.00 1.00 Case18 0.0 125.0 125.0 7.5 7.5 0.00 1.00

2 PL @ 3rd pts Cases with MR=Mmax/2

ML Mmid Mmax Δmid Δmax ML/max Δmax/mid ML Mmid Mmax Δmid Δmax ML/max Δmax/mid

Case21 -375.0 72.9 125.0 2.1 2.6 -3.00 1.26 Case25 -250.0 93.8 125.0 4.0 4.2 -2.00 1.04

Case22 -343.8 78.1 125.0 2.6 3.0 -2.75 1.17 Case26 -187.5 104.2 125.0 5.0 5.0 -1.50 1.01

Case23 -312.5 83.3 125.0 3.0 3.4 -2.50 1.11 Case27 -125.0 114.6 125.0 5.9 5.9 -1.00 1.00

Case24 -281.3 88.5 125.0 3.5 3.8 -2.25 1.07 Case28 0.0 135.4 125.0 7.8 7.8 0.00 1.00

-2.00 1.04

on a uniform elastic beam with different end-moment conditions

Spreadsheet Function: Compare maximum and midspan deflection for 2 point load at 3rd points

1.00

1.05

1.10

1.15

1.20

-3.00-2.50-2.00-1.50-1.00-0.500.00

Rat

io o

f M

axim

um

to

M

idsp

an D

efl

ect

ion



𝑀𝑅 = 0

𝑀𝑅 = 𝑀𝐿 2⁄

220

Criticisms of CSA A23.3 and the Concrete Handbook Appendix S

The following paragraphs provide some criticisms, as relevant to work in this report, of

A23.3 (CSA 2004) and the Concrete Design Handbook (CAC 2005).

Criticism of Use of Branson’s Equation in CSA A23.3-04

CSA A23.3-04 (CSA 2004) employs Branson’s (1965) effective moment of inertia, an

empirically derived equation which has many limitations. Section 2.7 of this report

describes the limitations, which do not include lightly reinforced members and FRP

reinforced concrete members. The recent changes which mandate inclusion of

shrinkage restraint and pre-loading effect into cracking moment calculations do result in

fewer limitations for Branson’s equation. Nonetheless, use of the effective moment of

inertia provided in CSA A23.3-04 is prone to unnecessary error when compared to the

proposed rationally derived equations.

Criticism of 0.5 Mcr Modifier in CSA A23.3-04

The modified 𝑀𝑐𝑟 for slabs is a correction for Branson’s (1965) equation. Results

indicate that use of Branson’s equation underpredicts deflection in some slabs. This

clause over-accounts for shrinkage-restraint because some members are outside the

valid range for Branson’s equation. An improved solution would:

Accurately account for additional deflection in all lightly reinforced members or

FRP reinforced members and only those members,

Would accurately and rationally account shrinkage-restraint where required.

221

Criticism of Use of Midspan Moment in CSA A23.3-04

In equations provided by A23.3 (CSA 2004), there is an inherent error in using the

equations provided if the midspan bending moment of a member is not its maximum

moment under worst case service loads. If designers use the effective moment of inertia

at midspan, 𝐼𝑒𝑚, as the standard suggests, the maximum deflection will often be under

predicted. This importance of using the maximum moment to compute 𝐼𝑒 or 𝐼𝑒′ is

discussed further in Section 3.3. This report finds it rational to define 𝐼𝑒𝑚 as the

effective moment of inertia based on the maximum moment in the positive bending

segment of the member. Engineers who have less experience with bending deflection

calculations for concrete members may, however, read the definition in the standard and

decide 𝐼𝑒 𝑚 should be based on the moment at precisely the mid-point between the

member supports.

Criticism of Concrete Design Handbook Using Midspan Deflection

The Concrete Design Handbook (CAC 2005) provides equations for the deflection in

Chapter 6. For end-moments where 𝑀𝐿 ≫ 𝑀𝑅, it is important to calculate the maximum

deflection. The handbook should, but does not, clearly indicate that these equations

compute only midspan deflection. This is discussed further in Section 3.7.1.

0

Curriculum Vitae

Candidate’s full name: Garth Roger Christie

Universities attended: University of New Brunswick

Bachelor of Computer Science (1998-2003)

Bachelor of Science in Engineering (2003-2006)

Professional Experience: Eastern Designers & Company Ltd

Structural Engineer (2006-current)

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