COMPUTED DEFLECTION OF CONTINUOUS REINFORCED CONCRETE ...
Transcript of 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
ยฉGarth Roger Christie, 2014
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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 Bischoffโs 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. Bischoffโs work with simply
supported members is reviewed. Bransonโs equation and the S806 method are also
reviewed and used for comparison.
The results indicate that Bischoffโs 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, ๐พ .................................................................. 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, Razaqpurโs Work, and CSA S806 ................. 19
2.5.1 Fibre Reinforced Polymers as Concrete Reinforcing................................ 19
2.5.2 Razaqpurโs Work ....................................................................................... 20
2.5.3 Concrete Deflection in CSA S806 ............................................................ 21
2.6 Bending Deflection in CSA A23.3-04 ....................................................... 22
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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 Bransonโs Work ......................................................................................... 25
2.7.1 Limited Accurate Range for Bransonโs Equation...................................... 26
2.7.2 Modification Factors Examples for Bransonโs Equation .......................... 26
2.8 Bischoffโs Work ........................................................................................ 27
2.8.1 Purpose of Bischoffโs Work ...................................................................... 27
2.8.2 Bischoffโs Equation and Loading Type Factor.......................................... 28
2.8.3 Discussion of Arguments Against Use of Bischoffโs 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 Bransonโs 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 ๐พ for Continuous Linear-Elastic Members ..................... 89 Appendix A
Calculate ๐พ for Point Load at Midspan and Generic End-Moments .................. 90
Calculate ๐พ for Two Equal Third-Point Loads and Generic End-Moments ....... 91
Calculate ๐พ 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 / Razaqpurโs 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 โ Data for Section 3.4 ....................... 143 Appendix M
Third-Point Loaded Examples โ Data for Section 3.5 ........................ 156 Appendix N
Uniformly Distributed Load Examples โ 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 Bransonโs 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 โ Example 3.4.2a โ Page 1 .................. 144
Table M-2 - Data for CPL, ML=MR, Ig/Icr=2.3 โ Example 3.4.2a โ Page 2 .................. 145
Table M-3 - Data for CPL, ML=MR, Ig/Icr=3.9 โ Example 3.4.2b โ Page 1 .................. 147
Table M-4 - Data for CPL, ML=MR, Ig/Icr=3.9 โ Example 3.4.2b โ Page 2 .................. 148
Table M-5 - Data for CPL, ML=MR, Ig/Icr=3.8 โ Example 3.4.2c โ Page 1 .................. 150
Table M-6 - Data for CPL, ML=MR, Ig/Icr=3.8 โ Example 3.4.2c โ Page 2 .................. 151
Table M-7 - Data for CPL, ML=MR, Ig/Icr=12 โ Example 3.4.2d โ Page 1 ................... 153
Table M-8 - Data for CPL, ML=MR, Ig/Icr=12 โ Example 3.4.2d โ Page 2 ................... 154
Table N-1 - Data for 2PL, ML=MR, Ig/Icr=3.0 โ Example 3.5.2a โ Page 1 ................... 157
Table N-2 - Data for 2PL, ML=MR, Ig/Icr=3.0 โ Example 3.5.2a โ Page 2 ................... 158
Table N-3 - Data for 2PL, MR=0, Ig/Icr=3.0 โ Example 3.5.2b โ Page 1 ...................... 160
Table N-4 - Data for 2PL, MR=0, Ig/Icr=3.0 โ Example 3.5.2b โ Page 2 ...................... 161
Table N-5 - Data for 2PL, ML=MR, Ig/Icr=12 โ Example 3.5.2c โ Page 1 .................... 163
Table N-6 - Data for 2PL, ML=MR, Ig/Icr=12 โ Example 3.5.2c โ Page 2 .................... 164
Table N-7 - Data for 2PL, MR=0, Ig/Icr=12 โ Example 3.5.2d โ Page 1 ....................... 166
Table N-8 - Data for 2PL, MR=0, Ig/Icr=12 โ Example 3.5.2d โ Page 2 ....................... 167
Table O-1 - Data for UDL Beam, ML=MR, Ig/Icr=3.0 โ Example 3.6.2a โ Page 1 ........ 171
Table O-2 - Data for UDL Beam, ML=MR, Ig/Icr=3.0 โ Example 3.6.2a โ Page 2 ........ 172
Table O-3 - Data for UDL Beam, MR=0, Ig/Icr=3.0 โ Example 3.6.2b โ Page 1 ........... 174
Table O-4 - Data for UDL Beam, MR=0, Ig/Icr=3.0 โ Example 3.6.2b โ Page 2 ........... 175
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Table O-5 - Data for UDL Beam, ML=MR, Ig/Icr=4.9 โ Example 3.6.2c โ Page 1 ........ 177
Table O-6 - Data for UDL Beam, ML=MR, Ig/Icr=4.9 โ Example 3.6.2c โ Page 2 ........ 178
Table O-7 - Data for UDL Slab, ML=MR, Ig/Icr=4.9 โ Example 3.6.2d โ Page 1 .......... 180
Table O-8 - Data for UDL Slab, ML=MR, Ig/Icr=4.9 โ Example 3.6.2d โ Page 2 .......... 181
Table O-9 - Data for UDL Slab, MR=0, Ig/Icr=4.9 โ Example 3.6.2e โ Page 1 ............. 183
Table O-10 - Data for UDL Slab, MR=0, Ig/Icr=4.9 โ Example 3.6.2e โ Page 2 ........... 184
Table O-11 - Data for UDL Slab, ML=MR, Ig/Icr=18 โ Example 3.6.2f โ Page 1 .......... 186
Table O-12 - Data for UDL Slab, ML=MR, Ig/Icr=18 โ Example 3.6.2f โ Page 2 ......... 187
Table O-13 - Data for UDL Slab, MR=0, Ig/Icr=6 โ Example 3.6.2g โ Page 1 .............. 189
Table O-14 - Data for UDL Slab, MR=0, Ig/Icr=6 โ Example 3.6.2g โ Page 2 .............. 190
Table O-15 - Data for UDL Beam, ML=MR, Ig/Icr=17 โ Example 3.6.2h โ Page 1....... 192
Table O-16 - Data for UDL Beam, ML=MR, Ig/Icr=17 โ Example 3.6.2h โ Page 2....... 193
Table P-1 - Data for UDL Beam, Ig/Icr=3.0, New A23.3 Mcr Example P1 โ Page 1 ..... 197
Table P-2 - Data for UDL Beam, Ig/Icr=3.0, New A23.3 Mcr Example P1 โ Page 2 ..... 198
Table P-3 - Data for UDL Beam, Ig/Icr=4.9, New A23.3 Mcr Example P2 โ Page 1 ..... 200
Table P-4 - Data for UDL Beam, Ig/Icr=4.9, New A23.3 Mcr Example P2 โ Page 2 ..... 201
Table P-5 - Data for UDL Beam, Ig/Icr=17, Reduced Mcr Example P3 โ Page 1 .......... 203
Table P-6 - Data for UDL Beam, Ig/Icr=17, Reduced Mcr Example P3 โ Page 2 .......... 204
Table R-1 - Example Midspan vs Maximum Deflection for UDL โ Page 1 ................. 211
Table R-2 - Example Midspan vs Maximum Deflection for UDL โ Page 2 ................. 212
Table R-3 - Example Midspan vs Maximum Deflection for UDL โ Page 3 ................. 213
Table R-4 - Example Midspan vs Maximum Deflection for UDL โ Page 4 ................. 214
Table R-5 - Example Midspan vs Maximum Deflection for UDL โ Page 5 ................. 215
Table R-6 - Example Midspan vs Maximum Deflection for UDL โ Page 6 ................. 216
Table R-7 - Example Midspan vs Maximum Deflection for UDL โ Summary ............ 217
Table R-8 - Example Midspan vs Maximum Deflection for CPL โ Summary ............. 218
Table R-9 - Example Midspan vs Maximum Deflection for 2PL โ 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 โ 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
Load with Ig/Icr=3.9, Mm/Mcr=1.6, and ML=MR .................................................. 51
Figure 3-3 - Midspan Deflection of FRP Reinforced Beams under Centered Point
Load with Ig/Icr=3.3, Mm/Mcr=2.5, and ML=MR .................................................. 52
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
Figure 3-10 - Midspan and Maximum Deflection of Steel Reinforced Beams under
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
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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 โ Midspan Point Load, Ig/Icr=2.3 and Mm/Mcr=3.0 .. 146
Figure M-2 - Copy of Figure 3-2 โ Midspan Point Load, Ig/Icr=3.9 and Mm/Mcr=1.6 .. 149
Figure M-3 - Copy of Figure 3-3 โ Midspan Point Load, Ig/Icr=3.3 and Mm/Mcr=2.5 .. 152
Figure M-4 - Copy of Figure 3-4 โ Midspan Point Load, Ig/Icr=12 and Mm/Mcr=1.6 ... 155
Figure N-1 - Copy of Figure 3-5 โ Third-Point Loaded, Ig/Icr=3 and Mm/Mcr=2.2....... 159
Figure N-2 - Copy of Figure 3-6 โ Third-Point Loaded, Ig/Icr=3, Mmax/Mcr=2.2,
MR=0 ................................................................................................................. 162
Figure N-3 - Copy of Figure 3-7 โ Third-Point Loaded, Ig/Icr=12 and Mm/Mcr=1.4..... 165
Figure N-4 - Copy of Figure 3-8 โ Third-Point Loaded, Ig/Icr=12, Mmax/Mcr=1.4,
MR=0 ................................................................................................................. 168
Figure O-1 - Copy of Figure 3-9 โ UDL on Beam, Ig/Icr=3, Mm /Mcr=2.2, ML=MR ..... 173
Figure O-2 - Copy of Figure 3-10 โ 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 โ UDL on Slab, Ig/Icr=5, Mm /Mcr=1.3, ML=MR ..... 182
Figure O-5 - Copy of Figure 3-12 โ 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 โ UDL on Beam, Ig/Icr=17, Mm /Mcr=1.3, ML=MR . 194
Figure P-1 - Midspan Deflection Computed using Shrinkage Restraint Mcr โ 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 โ Compare to Figure P-1 ...... 199
Figure P-3 โ Midspan and Maximum Deflection Computed using Shrinkage
Restraint Mcr โ 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 โ Compare to Figure P-3 .... 202
Figure P-5 - Midspan Deflection Computed using Shrinkage Restraint Mcr โ 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 โ 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
๐ = depth of equivalent rectangular stress block
๐ด๐ = area of FRP reinforcement in tension for this member segment
๐ด๐ = area of tension reinforcement, ๐ด๐ or ๐ด๐, at/near midspan for this member
๐ด๐ = area of steel reinforcement in tension for this member segment
๐ด๐ฟ = area of tension reinforcement, ๐ด๐ or ๐ด๐, in the left end of this member
๐ด๐ = area of tension reinforcement, ๐ด๐ or ๐ด๐, in the right end of this member
๐ = width of rectangular beam across the compression face
๐ = distance from the compression face to the neutral axis
๐๐ฟ = ๐ for the design of the left end moment of the member
๐๐ = ๐ for the design of the maximum moment near midspan of the member
๐๐ = ๐ for the design of the right end moment of the member
CPL = one point loads located at midpoint on the relevant member span
๐ = effective depth of tension reinforcement from compression face
๐ธ = elastic modulus of the material being analyzed
๐ธ๐ = elastic modulus of reinforcing bar, ๐ธ๐ or ๐ธ๐
๐ธ๐ = elastic modulus of concrete
๐ธ๐ = design or guaranteed elastic modulus of FRP reinforcement
๐ธ๐ = elastic modulus of steel reinforcing
๐๐ = service load stress in the reinforcing bar
๐๐โฒ = specified compressive strength of concrete
๐๐๐ข = tensile strength at failure, (unfactored ultimate tensile stress)
๐๐๐ฟ๐ = serviceability limit for FRP reinforced concrete design
๐๐,`๐ = creep-rupture stress limit for FRP reinforced concrete design
๐๐ = modulus of rupture of concrete (taken as 0.6โ๐๐โฒ in MPa)
๐๐ฆ = yield strength of steel reinforcement (or similar for FRP)
โ = height of rectangular beam, from compression face to tension face
๐ = symbol/counter for sections, 0 to ๐, or segments, 1 to ๐, for the span
๐ = denotes total number of equal segments used for numerical integration
๐ผ = moment of inertia of the member about the axis the load is applied
๐ผ๐๐ = moment of inertia of the cracked transformed section, ๐ผ๐๐= ๐ผ๐๐๐ unless noted
๐ผ๐๐โ = moment of inertia of appropriate section ( ๐ผ๐๐๐ฟ , ๐ผ๐๐๐ , or ๐ผ๐๐๐ )
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๐ผ๐๐๐ฟ = moment of inertia, ๐ผ๐๐, for negative bending at the left end
๐ผ๐๐๐ = moment of inertia, ๐ผ๐๐, for positive bending for the midspan segment
๐ผ๐๐๐ = moment of inertia, ๐ผ๐๐, for negative bending at the right end
๐ผ๐ = effective moment of inertia for the member using the method indicated
๐ผ๐(๐ฅ) = (local) section-based effective moment of inertia at position ๐ฅ along member
๐ผ๐,๐๐ฃ๐ = average effective moment of inertia using CSA A23.3 clause 9.8.2.4
๐ผ๐๐ = effective moment of inertia with for the midspan segment in method indicated
๐ผ๐๐ฟ = effective moment of inertia with left end-moment in method indicated
๐ผ๐๐ = effective moment of inertia with right end-moment in method indicated
๐ผ๐โฒ = integration-based (constant) equivalent moment of inertia for a member
๐ผ๐โโฒ = approximate equivalent moment of inertia, ๐ผ๐
โฒ , but using โ in lieu of
๐ผ๐ = gross moment of inertia, (๐โ3 12โ for prismatic rectangular sections)
๐๐ = the depth of the compression face for purposes of defining ๐ผ๐๐
๐พ = factor which ratios a simply supported ๐ฅ to a continuous member ๐ฅ
๐พ๐ = variable which gives factored resistance such that ๐๐ = ๐พ๐๐๐2
๐พ๐ ๐ = ๐พ๐ for midspan based on the maximum positive moment (near midspan)
๐พ๐ ๐ฟ = ๐พ๐ for the left end of the member
๐พ๐ ๐ = ๐พ๐ for the right end of the member
๐ฟ = length of member span (measured center-to-center of relevant supports)
๐ฟโ๐๐๐ฅ = length from left end of span to location of maximum deflection
๐ฟ1 = length from left end of span to end of left-negative-cracked segment
๐ฟ2 = length from left end of span to start of positive-cracked segment
๐ฟ3 = length from left end of span to midspan, therefore ๐ฟ3 = ๐ฟ/2
๐ฟ3๐ด = length from left end of span to the 1/3 point, therefore ๐ฟ3๐ด = ๐ฟ/3
๐ฟ3๐ต = length from left end of span to the 2/3 point, therefore ๐ฟ3๐ต = 2๐ฟ/3
๐ฟ4 = length from left end of span to end of positive-cracked segment
๐ฟ5 = length from left end of span to start of right-negative-cracked segment
๐ฟ6 = length of span, therefore ๐ฟ6 = ๐ฟ
๐ฟ๐๐ = distance ๐ฅ to leftmost section where ๐(๐ฅ) > 0 and ๐(๐ฅ) > ๐๐๐
๐ฟ๐ = length of uncracked section(s) in FRP-reinforced concrete (Appendix C)
๐ฟ๐ = length from left end of span to right end of integration segment ๐
๐ฟ๐๐ฟ = distance to the closest point load from closest support
๐ฟ๐ 4 = length from right end of span to right end of positive-cracked segment
๐ฟ๐ 5 = 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
๐(๐ฅ) = internal virtual moment from virtual load, using method of virtual work
๐๐(๐ฅ) = virtual moment, ๐(๐ฅ), at location ๐ on the span
๐ = midspan bending moment for the applied load
๐๐ = the applied bending moment used to compute the effective moment of inertia
๐(๐ฅ) = bending moment at position ๐ฅ for the applied load
๐0 = total static moment (midspan service load moment if end-moments released)
๐0,0 = ๐0 for a simply supported member when producing example members
๐1๐๐ฟ = maximum moment from one point load applied at midspan
๐2๐๐ฟ = maximum moment from two equal loads applied at third points
๐๐๐ = cracking moment of the member, (fr๐โ2/6 for rectangular sections)
๐๐ = factored moment of the member (required to be less than ๐๐)
๐๐ = net midspan moment (with worst case service load applied)
๐๐๐๐ฅ = maximum positive service moment (with worst case service load applied)
๐๐ = the nominal moment capacity (๐๐ without the reductions from ๐๐ & ๐๐ )
๐๐ = factored bending moment resistance in ultimate limit states design
๐๐ = service load bending moment at location indicated
๐๐ ๐ข๐ = sustained service load bending moment at indicated segment
๐๐ฟ = service moment at left support a loading when ๐๐๐๐ฅ occurs
๐๐ = service moment at right support a loading when ๐๐๐๐ฅ occurs
๐๐๐ท๐ฟ = maximum moment from uniformly distributed load, ๐ค๐๐ท๐ฟ๐ฟ2 8โ
๐ = modular ratio (๐ธ๐/๐ธ๐)
๐ = denotes specific point/location along the span to determine deflection at
๐ = load applied at a particular point(s) (short segment(s)) on the span
๐๐๐ = load ๐ which causes tension face to exceed ๐๐
๐0 = ๐1๐๐ฟ or ๐2๐๐ฟ used to produce continuous member based on a simple member
๐s = maximum service load ๐
๐1๐๐ฟ = one point load (๐) applied at midspan
๐2๐๐ฟ = total of two equal loads (๐), each equal ๐2๐๐ฟ 2โ , applied at third points
๐ = time-dependant factor used for calculating sustained-load effects
UDL = a uniformly distributed load located along the relevant member span
๐๐ฟ = shear force resulting from loads ๐ or ๐ค at the left end of the span
๐๐ = shear force resulting from loads ๐ or ๐ค at the right end of the span
๐ค = any distributed load applied on a member; ๐ค๐๐ท๐ฟ in this report
๐ค0 = ๐ค๐๐ท๐ฟ on a simply supported member
๐ค๐๐ท๐ฟ = service load distributed uniformly across full member span
๐ฅ = distance, from left, along a span (to a particular position/cross-section)
xiv
๐ฆ๐ก = distance from section gross centroid to tension face (rectangular = โ/2)
z = denotes load case number for specific example load and end-moments
๐ผ1 = ratio of average stress in rectangular compression block to ๐๐โฒ
๐ผ๐๐ = ratio of cracking moment divided by total moment (๐๐๐/๐0)
๐ผ+ ๐โ = ratio of ๐๐/๐๐ , midspan, where additional bars added to reduce ๐ฅ๐๐๐ฅ
๐ผ๐ ๐โ = service load ratio (factored load divided by service load, ๐๐ /๐๐)
๐ผ๐ฟ = ratio of left end-moment divided by total moment (๐๐ฟ/๐0)
๐ผ๐ฟ/๐๐๐ฅ = ratio of left end-moment divided by maximum moment (๐๐ฟ/๐๐๐๐ฅ)
๐ผ๐ = ratio of right end-moment divided by total moment (๐๐ /๐0)
๐ผ๐ /๐๐๐ฅ= ratio of right end-moment divided by maximum moment (๐๐ /๐๐๐๐ฅ)
๐ฝ = tension stiffening factor for tensile contribution of concrete after cracking
๐ฝ1 = ratio of depth of rectangular compression block to depth of neutral axis
๐ฝ๐ = coefficient to modify Bransonโs equation for FRP-reinforcing
= integration factor to account for changes in stiffness along member span
โ = correction to which improves continuous member effective stiffness
๐ฟ = symbol indicating the change in the subsequent variable(s)
๐ฟ๐ฅ๐๐ = the change in the deflection from the bending-moment related cracking
๐ฅ = midspan deflection resulting from the maximum service load case
๐ฅ(๐ฅ) = deflection at ๐ฅ from the given service load case
๐ฅ1 = analytical integration results, ๐ฅ๐ , for ๐ฅ = 0 ๐ก๐ ๐ฟ1
๐ฅ1+2 = ๐ฅ๐ for ๐ฅ = 0 ๐ก๐ ๐ฟ2 for analytical integration with no left-end cracking
๐ฅ2 = analytical integration results, ๐ฅ๐ , for ๐ฅ = ๐ฟ1 ๐ก๐ ๐ฟ2
๐ฅ3 = analytical integration results, ๐ฅ๐ , for ๐ฅ = ๐ฟ2 ๐ก๐ ๐ฟ/2
๐ฅ3๐ด = third-point loading integration results, ๐ฅ๐ , for ๐ฅ = ๐ฟ2 ๐ก๐ ๐ฟ/3
๐ฅ3๐ต = third-point loading integration results, ๐ฅ๐ , for ๐ฅ = ๐ฟ/3 ๐ก๐ 2๐ฟ/3
๐ฅ3๐ถ = third-point loading integration results, ๐ฅ๐ , for ๐ฅ = ๐ฟ/2 ๐ก๐ 2๐ฟ/3
๐ฅ3๐ท = third-point loading integration results, ๐ฅ๐ , for ๐ฅ = 2๐ฟ/3 ๐ก๐ ๐ฟ4
๐ฅ4 = analytical integration results, ๐ฅ๐ , for ๐ฅ = ๐ฟ/2 ๐ก๐ ๐ฟ4
๐ฅ5 = analytical integration results, ๐ฅ๐ , for ๐ฅ = ๐ฟ4 ๐ก๐ ๐ฟ5
๐ฅ5+6 = ๐ฅ๐ for ๐ฅ = ๐ฟ4 ๐ก๐ ๐ฟ for analytical integration with no right-end cracking
๐ฅ6 = analytical integration results, ๐ฅ๐ , for ๐ฅ = ๐ฟ5 ๐ก๐ ๐ฟ
๐ฅ๐๐ = midspan deflection assuming the member is fully cracked (using ๐ผ๐๐)
๐ฅ๐ = midspan deflection assuming the member is uncracked (using ๐ผ๐)
๐ฅ๐,๐๐๐ฅ = maximum deflection assuming the member is uncracked (using ๐ผ๐)
๐ฅ๐,๐๐๐ = midspan deflection assuming the member is uncracked (using ๐ผ๐)
๐ฅ๐ = deflection at point ๐, found by method of virtual work by using ๐๐(๐ฅ)
xv
๐ฅ๐ = portion of ๐ฅ๐ contributed by respective integration segment: ๐ = 1,2,3,โฆ
๐ฅ๐๐๐ฅ = maximum deflection of a member at the maximum service load case
๐ฅ๐๐๐ฅ,๐ผ๐(๐ฅ) = maximum deflection of a member calculated using ๐ผ๐(๐ฅ)
๐ฅ๐๐๐ฅ,๐ผ๐โฒ = maximum deflection calculated using Bischoffโs ๐ผ๐
โฒ
๐ฅ๐๐๐ = midspan deflection, ๐ฅ, exactly at midspan (๐ฟ/2) for 2-support member
๐ฅ๐ข๐๐๐ = midspan deflection when ๐๐๐๐ฅ โ ๐๐๐ (and assuming no cracking)
๐ฅ๐ผ๐ = midspan deflection determined using Bransonโs ๐ผ๐
๐ฅ๐ผ๐โฒ = midspan deflection calculated using Bischoffโs ๐ผ๐โฒ
๐ฅ๐ผ๐โโฒ = midspan deflection calculated using proposed ๐ผ๐โโฒ (equal third-point loading)
๐ฅ๐ผ๐(๐ฅ) = midspan deflection calculated using section-based moment of inertia, ๐ผ๐(๐ฅ)
๐ฅ๐ผ๐,๐๐ฃ๐ = midspan deflection calculated using the average moment of inertia (see ๐ผ๐,๐๐ฃ๐)
๐ฅ๐ผ๐,๐ฝ๐ = midspan deflection calculated using ๐ผ๐ per ACI440.1R (with a ๐ฝ๐ modifier)
๐ฅ๐๐ฟ = deflection caused by left end-moment
๐ฅ๐๐ = deflection caused by right end-moment
๐ฅ๐๐ท๐ฟ = deflection caused by uniformly distributed load
๐ฅ๐๐ง(๐ฅ) = total deflection at ๐ฅ caused by uniformly distributed load for load case ๐ง
๐ฅ๐ฝ=0 = midspan deflection calculated with tension stiffening neglected (per S806)
๐ฅ๐พ=1 = midspan deflection calculated using the Bischoffโs ๐ผ๐ (equals ๐ผ๐โฒ with = 1)
ํ๐๐ข = maximum strain at extreme compression face at ultimate, 0.0035
๐ = stiffness reduction coefficient (1 โ ๐ผ๐๐/๐ผ๐), use ๐=๐๐ unless noted
๐๐ = stiffness reduction coefficient (1 โ ๐ผ๐๐๐/๐ผ๐) for at/near midspan
๐๐ฟ = stiffness reduction coefficient (1 โ ๐ผ๐๐๐ฟ/๐ผ๐) for the left end
๐๐ = stiffness reduction coefficient (1 โ ๐ผ๐๐๐ /๐ผ๐) for the right end
๐ = 1 โ โ1 โ๐๐๐/๐๐ , ( ๐๐ฟ/2 is the uncracked end length with ๐ค๐๐ท๐ฟ load)
๐ = reinforcement ratio of the tension bars, ๐ด๐ /๐๐ or ๐ด๐/๐๐
๐โฒ = reinforcement ratio of the compression bars
๐๐ = balanced reinforcement ratio for a reinforced member
๐๐ = tension reinforcement ratio, ๐, at/near member midspan
๐๐ฟ = tension reinforcement ratio, ๐, at left end of member
๐๐ = tension reinforcement ratio, ๐, at right end of member
๐ = curvature of member being considered at the point being considered
๐๐ = resistance factor for concrete under CSA A23.3-04 (or as required)
๐๐ = resistance factor for reinforcement bar: ๐๐ or ๐๐
๐๐ = resistance factor for FRP reinforcement per relevant standard
๐๐ = 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, Bransonโs (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 โeffective moment of inertiaโ 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 Bransonโs (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 Bransonโs 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 Bischoffโs 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 Bischoffโs 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 Bischoffโs simply-supported equations (Bischoff and Gross, 2011)
are generally accurate for continuous members, results are compared between
Bischoffโs equations, analytical/numerical integrations, Bransonโs (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 Bischoffโs 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.
Bransonโs (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. Bischoffโs 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 โmoment of inertiaโ 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 ๐ธ๐ or ๐ธ๐) 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 ๐ผ , ๐ผ๐ , ๐ผ๐๐ , ๐ผ๐ , ๐ผ๐โฒ , 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, ๐, and maximum deflection, โ,
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. ๐(๐ฅ), 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. โContinuous membersโ and
โmembers with end-momentsโ 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, โ represents the bending deflection of the member, ๐๐ฟ and ๐๐ represent
the left end and right end bending moment, and ๐๐ is the net midspan moment.
10
Figure 2-1 - Deflected Shape Comparison of Four Different Loads & End-Moments
2.2.3 Continuous Member Factor, ๐ฒ
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, ๐พ, to compute these deflections. The equations for ๐พ
apply to any constant stiffness members with known end-moments. The midspan
deflection of a prismatic continuous member is the ๐พ 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 ๐พ 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 Bransonโs (1965) equation for the constant effective moment of
inertia, ๐ผ๐. 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 ๐ธ๐๐ผ๐ 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, ๐ธ๐๐ผ๐ is computed at the service load which causes the applied moment,
๐๐. The gross moment of inertia, ๐ผ๐, is taken as ๐โ3/12 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, ๐ผ๐๐, 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. ๐๐๐ represents
the bending moment which causes the concrete to crack in bending and ฯ 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. ๐๐๐/๐(๐ฅ) is a rational and accurate tension stiffening factor
(Bischoff 2007); it is used to compute the section-based effective moment of inertia,
๐ผ๐(๐ฅ), in this report. ๐(๐ฅ) is the bending moment at the location along the beam of ๐ฅ.
Tension stiffening has a large effect at the service load depicted in Figure 2-3, for
example. Here, ๐๐ represents the service load and ๐๐๐ 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 ๐ผ๐. An accurate weighted
average effective moment of inertia, based on integration of curvature, has been defined
as the equivalent moment of inertia, ๐ผ๐โฒ (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 ๐๐๐ = 113 kNm, which means
the member is cracked at both ends and at midspan under service loads. The service
load bending moments are: ๐๐ฟ = โ247 kNm (left end-moment), ๐๐๐๐ฅ = 125 kNm
Figure 2-4 - Gross, Local-Effective, Equivalent, and Cracked Moments of Inertia
16
(maximum moment in positive bending), and ๐๐ = โ165 kNm (right end-moment).
The reinforcing steel for the member represented by Figure 2-4 was calculated using a
factored moment resistance of ๐๐=1.6๐๐๐๐ฅ, ๐๐=1.6๐๐ฟ, and ๐๐=1.6๐๐ 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 ๐ผ๐๐โ.
๐ผ๐(๐ฅ) represents the (local) section-based moment of inertia at the position ๐ฅ, and ๐ผ๐โฒ
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, ๐๐ (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 0.5๐๐โฒ (where ๐๐
โฒ 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, Razaqpurโs 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 Bransonโs effective moment of
inertia. ACI 440.1R-06 (ACI Committee 440 2006) provides an empirical modification
factor to the (๐๐๐/๐๐)3 ๐ผ๐ 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 Bischoffโs equation can be used with FRP
reinforcing without any correction factor or other modification.
2.5.2 Razaqpurโs 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 Bransonโs 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 ๐ฆ๐ก is the dimension from the centroid to the tension face):
22
๐๐๐ =๐๐๐ผ๐
๐ฆ๐ก where: ๐๐ = 0.6โ๐๐โฒ (2 โ 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 โS806โ
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 ๐พ factor for
computing deflections of continuous members, as explained in Section 2.2.3 (above).
That chapter also provides an equation for ๐ผ๐๐ 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, โ๐ (defined as ๐ฟ 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 Bransonโs (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 ๐๐๐ for bending deflection
be calculated using one half of ๐๐. The resulting equation, ๐๐๐ = 0.5๐๐๐ผ๐/๐ฆ๐ก, accounts
for shrinkage restraint stresses when using Bransonโs ๐ผ๐ (Scanlon and Bischoff 2008).
For Bischoffโs ๐ผ๐โฒ or ๐ผ๐(๐ฅ) as defined in this report, the use of ๐๐๐ = 0.67๐๐๐ผ๐/๐ฆ๐ก
provides an equivalent adjustment (Scanlon and Bischoff 2008). Until 2009, the
bending deflection was to be calculated using ๐๐๐ = 0.5๐๐๐ผ๐/๐ฆ๐ก, 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 ๐ผ๐ ๐๐ฃ๐). The effective moment of inertia at
25
midspan, ๐ผ๐๐, is rationally defined as the maximum moment in the positive bending
segment of the member. The moment at the supports are ๐ผ๐๐ฟ and ๐ผ๐๐ . This clause states
that one of the following equations be used to account for the contribution of midspan
and end-moment cracked sections:
๐ผ๐,๐๐ฃ๐ = 0.7๐ผ๐๐ + 0.15(๐ผ๐๐ฟ + ๐ผ๐๐ ) if both ends are continuous (2 โ 2)
๐ผ๐,๐๐ฃ๐ = 0.85๐ผ๐๐ + 0.15(๐ผ๐๐ฟ) if only one end is continuous (2 โ 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 +๐
1 + 50๐โฒ) (2 โ 4)
The multiplier accounts for the duration of the sustained load using the factor ๐. There
is a slight reduction in the multiplier from the compression reinforcement ratio, ๐โฒ. The
commentary for this clause, N9.8.2.5 of A23.3 (CSA 2004), explains this calculation in
more detail.
2.7 Bransonโs 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:
๐ผ๐ = (๐๐๐
๐๐)3
๐ผ๐ + [1 โ (๐๐๐
๐๐)3
] ๐ผ๐๐ โค ๐ผ๐ (2 โ 5)
The applied moment, ๐๐, in Bransonโs 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 Bransonโs Equation
Bransonโs (1965) equation was calibrated for steel reinforcing ratio, ๐, of 1% to 2%,
which is also roughly the range of 2 โค ๐ผ๐/๐ผ๐๐ โค 3. Bransonโs 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 ๐ < 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 Bransonโs Equation
Modification factors exist for most member types for which Bransonโs equation
performs poorly. These corrections work well for only their intended range of members.
In the latest update to A23.3 (CSA 2004), ๐๐๐ is reduced by 50% for bending deflection
calculations; this over-accounts for typical shrinkage-restraint so that Bransonโs ๐ผ๐
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:
๐ผ๐ = (๐๐๐
๐๐)3
๐ฝ๐๐ผ๐ + [1 โ (๐๐๐
๐๐)3
] ๐ผ๐๐ โค ๐ผ๐ where: ๐ฝ๐ =1
5
๐
๐๐< 1.0 (2 โ 6)
Here, ๐ is the actual reinforcement ratio of an FRP reinforced member and ๐๐ is the
balanced reinforcement ratio for the same FRP reinforced member. Again, use the
maximum moment in a continuous member as the applied moment, ๐๐.
2.8 Bischoffโs 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 Bischoffโs Work
Bischoffโs 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. Bischoffโs 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 โค ๐ผ๐/๐ผ๐๐ โค 30 including
typical steel-reinforced, lightly steel-reinforced, and FRP-reinforced concrete members.
The section-based form of this equation, ๐ผ๐(๐ฅ), 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 Bransonโs (1965) equation, Bischoffโs 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 ๐ผ๐ as it interpolates between ๐ผ๐๐ and ๐ผ๐, it produces significant
unconservative error when ๐ผ๐/๐ผ๐๐ โซ 3. Bischoffโs 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 Bischoffโs Equation and Loading Type Factor
Bischoffโs equation has a section-based form (๐ผ๐(๐ฅ)), a member-based form (the
equivalent moment of inertia, ๐ผ๐โฒ ), and a simplified form (an effective moment of inertia,
๐ผ๐). The section-based ๐ผ๐(๐ฅ) provides a local moment of inertia corresponding to the
assumed moment-curvature response of the member. The equivalent moment of inertia,
๐ผ๐โฒ , is an exact solution resulting from integration of curvature. The ๐ผ๐
โฒ 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 ๐ฝ = ๐๐๐/๐(๐ฅ) 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, ๐๐. For
loadings in this report, the equations to determine deflection using Bischoffโs 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)
Bischoffโs section-based equation defines the local moment of inertia, ๐ผ๐(๐ฅ), as:
๐ผ๐(๐ฅ) =๐ผ๐๐
1 โ ๐ (๐๐๐
๐(๐ฅ))2 where: ๐ = 1 โ
๐ผ๐๐๐ผ๐ (2 โ 7)
30
Bischoffโs equivalent moment of inertia, ๐ผ๐โฒ , is defined as:
๐ผ๐โฒ =
๐ผ๐๐
1 โ ๐ (๐๐๐
๐๐)2 where: ๐ = 1 โ
๐ผ๐๐๐ผ๐ (2 โ 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 Bransonโs (1965) equation,
setting = 1.0 produces Bischoffโs equation for the effective moment of inertia:
๐ผ๐ =๐ผ๐๐
1 โ ๐ (๐๐๐
๐๐)2 where: ๐ = 1 โ
๐ผ๐๐๐ผ๐ (2 โ 9)
2.8.3 Discussion of Arguments Against Use of Bischoffโs Equation
There are different arguments against prescribing the use of Bischoffโs equations for the
effective moment of inertia in codes and standards. Some engineers have used
Bransonโs 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 Bischoffโs 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 Bischoffโs equation than with other methods
If these issues are correctly accounted for, then using Bischoffโs 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 Bischoffโs equivalent moment of
inertia (Bischoff and Gross 2011), ๐ผ๐โฒ , for simply supported members. This ๐ผ๐
โฒ 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 Bischoffโs ๐ผ๐โฒ works for computing deflection of continuous members. This ๐ผ๐
โฒ , 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 ๐ผ๐โฒ ), 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, ๐ผ๐, 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 Bischoffโs equation for ๐ผ๐โฒ 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 ๐ผ๐โฒ . Deflection is also computed by integration of
curvature using both the S806 (CSA 2012) method and Bischoffโs section-based
moment of inertia (Bischoff and Gross 2011), ๐ผ๐(๐ฅ). 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 Bischoffโs
๐ผ๐(๐ฅ), 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).
โ= โซ๐(๐ฅ)๐(๐ฅ)
๐ธ๐๐ผ๐(๐ฅ)๐๐ฅ
๐ฟ
0
(3 โ 1)
This equation determines the deflection, โ, at a defined location along a member of span
length, ๐ฟ; the virtual bending moment, ๐(๐ฅ), is obtained by applying a unit load at the
location where deflection is being computed. The variable ๐ฅ is defined as the distance
from the left end of the span, so 0 โค ๐ฅ โค ๐ฟ. The curvature function, ๐(๐ฅ)/ ๐ธ๐ ๐ผ๐(๐ฅ),
includes the service load bending moment function, ๐(๐ฅ), the elastic modulus of
concrete, ๐ธ๐, and the section-based effective moment of inertia function, ๐ผ๐(๐ฅ).
Additional explanation of the method of virtual work and Equation (3-1) is given in
Appendix B. For a description of how to calculate ๐ผ๐(๐ฅ) per Bischoffโs 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 Bischoffโs ๐ผ๐(๐ฅ). When a constant moment of inertia is used, the ๐ผ๐(๐ฅ) term is
34
replaced with ๐ผ๐, ๐ผ๐๐, ๐ผ๐, or ๐ผ๐โฒ , or similar. The ๐ผ๐ term is defined as the gross moment of
inertia and the ๐ผ๐๐ 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 ๐พ 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: ๐ผ๐ > ๐ผ๐ > ๐ผ๐๐ and ๐ผ๐ > ๐ผ๐โฒ > ๐ผ๐๐. The latter is indicated in Figure
2-4, which shows ๐ผ๐, ๐ผ๐๐ , ๐ผ๐(๐ฅ), and the equivalent moment of inertia, ๐ผ๐โฒ . In the
regions where a member is uncracked, ๐ผ๐(๐ฅ) = ๐ผ๐.
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:
๐๐ is the midspan moment, determined at the midpoint of the memberโs span
๐๐๐๐ฅ is the maximum positive bending moment of an example member
- ๐๐๐๐ฅ is equal to ๐๐ in this report when end-moments are equal or zero
- ๐๐๐๐ฅ is used to calculate the effective/equivalent moment of inertia
whenever ๐๐ is used to calculate ๐ผ๐ or ๐ผ๐โฒ for a simply supported member
๐๐ฟ is the bending moment at the left end of the member for the loading provided
- ๐๐ฟ 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
๐๐ is the bending moment at the right end of the member for the loading provided
- ๐๐ is assumed to be zero or negative throughout this report
- As per the definition of ๐๐ฟ above, โ๐๐ฟ โฅ โ๐๐ โฅ 0
๐0 is the total static moment caused by the loads applied between supports
- For the loading types provided, ๐0 is equal to ๐๐ 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 ๐ at midspan, two
point loads of ๐/2 at third points, or a uniformly distributed load of ๐ค.
๐๐ is a general reference to the service moment of ๐๐ฟ, ๐๐ , ๐๐, or ๐๐๐๐ฅ.
- ๐๐ :๐๐ 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 ๐๐ฟ or ๐๐ , respectively, and the aforementioned ratio
- The amount of bottom reinforcing in the positive bending region is selected
based on ๐๐๐๐ฅ and the aforementioned ratio
37
The factored bending moments, as used in this chapter, are:
๐๐ is the factored applied bending moment at the indicated location
- In practice, ๐๐ 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, ๐๐ = 1.375๐๐ is assumed
๐๐ 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 ๐๐
- Additional capacity of ๐๐ = 1.145๐๐ is typically provided
- For typical example members, the assumed ๐๐ (above) and the additional
Mr capacity (above) were combined as ๐๐ = 1.575๐๐
- Where deflection governs the amount of positive moment reinforcing, a
larger ratio of ๐๐ :๐๐ is provided, such as ๐๐ = 2.5๐๐ (for Figure 3-4)
- If ๐๐ = 1.575๐๐ is changed to ๐๐ = 1.30๐๐ 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 ๐๐๐:
๐๐๐ is the bending moment at cracking for a member
- When the applied bending moment at a local section of the member reaches
๐๐๐, the tension face of the member will crack at this section
- ๐๐๐ 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 ๐๐๐ 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 (๐ฟ), width-to-height
ratio (๐: โ), reinforcing bar depth to member height ratio (๐: โ), material properties,
resistance factors, factored moment resistance to service moment ratio (๐๐:๐๐ ), and
maximum positive moment to cracking moment ratio (๐๐๐๐ฅ:๐๐๐). After the common
properties are defined, unique left and right end-moment to maximum positive moment
ratios (๐๐ฟ:๐๐๐๐ฅ and ๐๐ :๐๐๐๐ฅ) 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
๐๐๐๐ฅ = 195 kNm and end-moments equal to: 0, โ0.25๐๐๐๐ฅ, โ0.43๐๐๐๐ฅ,
โ0.67๐๐๐๐ฅ, โ๐๐๐๐ฅ, โ1.22๐๐๐๐ฅ, โ1.5๐๐๐๐ฅ, โ1.7๐๐๐๐ฅ, and โ1.94๐๐๐๐ฅ.
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:
๐๐๐๐ฅ is determined for the simply supported member
A unique load is contrived for each continuous member such that the ๐๐๐๐ฅ
computed for each member will be equal for all members of the set
39
๐0, ๐๐ฟ, ๐๐ , ๐๐, and ๐๐๐๐ฅ are computed for each member
๐๐๐, โ, ๐, ๐, and ๐ผ๐ are calculated and confirmed as common
The reinforcing ratio, ฯ, is computed (based on ๐๐:๐๐ ) for ๐๐๐๐ฅ, ๐๐ฟ, and ๐๐
๐ผ๐๐ 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 โค โ๐๐ฟ โค 3๐๐๐๐ฅ, where ๐๐ = ๐๐ฟ (both ends continuous) or
๐๐ = 0 (one end continuous). In general, results are not provided when
โ๐๐ฟ > 3๐๐๐๐ฅ 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 ๐๐๐ 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 ๐ฅ as follows:
All constant stiffness equations provided denote the midspan deflection as ๐ฅ
- ๐ฅ 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 ๐ฅ๐๐๐ฅ
For clarity, midspan deflection is shown as ๐ฅ๐๐๐ in equations that include ๐ฅ๐๐๐ฅ
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:
๐ฅ๐(Gross) is the midspan deflection computed using the gross moment of inertia
- Uses constant moment of inertia set equal to ๐ผ๐ (even if loads exceed ๐๐๐)
๐ฅ๐๐(Cracked) is the midspan deflection computed with the ๐ผ๐๐ moment of inertia
- Computed using entire beam with the fully cracked moment of inertia
- ๐ผ๐๐ is computed using the reinforcing ratio at the location of ๐๐๐๐ฅ
๐ฅ๐ผ๐(Branson) is the midspan deflection computed with the common ๐ผ๐ equation
- Uses Bransonโs (1965) equation for the effective moment of inertia
- ๐ผ๐ is computed using Equation (2-5) with an applied moment of ๐๐๐๐ฅ
๐ฅ๐พ=1(Approx) is the midspan deflection computed per Bischoffโs ๐ผ๐ where = 1
- Uses Bischoffโs effective moment of inertia (Bischoff and Gross 2011)
- ๐ผ๐ is computed using Equation (2-9) with an applied moment of ๐๐๐๐ฅ
๐ฅ๐ผ๐โฒ(Proposed) or ๐ฅ๐ผ๐โฒ (Bischoff) is the midspan deflection computed using ๐ผ๐โฒ
- ๐ผ๐โฒ is Bischoffโs equivalent moment of inertia (Bischoff and Gross 2011)
- ๐ผ๐โฒ is computed using Equation (2-8) with an applied moment of ๐๐๐๐ฅ
Midspan deflection, computed using one (or more) of these additional solutions for
constant moment of inertia, was also plotted as follows:
42
๐ฅ๐ผ๐,๐ฝ๐(ACI440) is the midspan deflection computed per ACI 440.1Rโs 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
๐ฅ๐ผ๐,๐๐ฃ๐(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)
- ๐ผ๐,๐๐ฃ๐ is an approximate average of the Bransonโs ๐ผ๐ computed at the applied
moment locations of ๐๐๐๐ฅ, ๐๐ฟ, and ๐๐
- ๐ผ๐,๐๐ฃ๐ is computed with Equation (2-6) for members designed with GFRP
๐ฅ๐ผ๐โโฒ (Proposed) is the midspan deflection computed using ๐ผ๐โโฒ per Equation (3-10)
- ๐ผ๐ โโฒ is an empirically derived improvement to ๐ผ๐
โฒ
- Only used for third-point loading on a continuous member
- ๐ผ๐ โโฒ is computed with Equation (2-8), but with โ 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:
๐ฅ๐ฝ=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
๐ฅ๐ผ๐(๐ฅ)(Exact) is the midspan deflection using ๐ผ๐(๐ฅ) per Equation (2-7)
- Computed using Bischoffโs section-based moment of inertia (Bischoff and
Gross 2011), ๐ผ๐(๐ฅ), 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:
๐ฅ๐๐๐ฅ,๐ผ๐โฒ(Proposed) or ๐ฅ๐๐๐ฅ,๐ผ๐
โฒ(Bischoff) is ๐ฅ๐๐๐ฅ computed with ๐ผ๐โฒ
- Uses the same ๐ผ๐โฒ , per Equation (2-8), as the ๐ฅ๐ผ๐โฒ 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
๐ฅ๐๐๐ฅ,๐ผ๐(๐ฅ)(Exact) is the maximum deflection computed using Bischoffโs ๐ผ๐(๐ฅ)
- Uses the same ๐ผ๐(๐ฅ), per Equation (2-7), as the ๐ฅ๐ผ๐(๐ฅ) 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 ๐๐ is used in lieu of ๐๐๐๐ฅ, 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 (๐๐ โ ๐๐๐๐ฅ) or if the member is highly cracked at midspan (๐๐ โซ 2๐๐๐).
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 ๐๐ฟ, ๐๐,
and ๐๐ , as provided for each member. Stiffnesses affected by other load patterns could
be included, using integration of curvature, using an ๐ผ๐(๐ฅ) 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 โnegativeโ cracked
segment at each end, with an adjacent uncracked segment, and a โpositiveโ cracked
segment in the middle. The service bending moment function, ๐(๐ฅ) changes at each
point load and the virtual moment function, ๐(๐ฅ), 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
Maplesoftยฎ) 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 Microsoftยฎ) spreadsheets and linked to the correct variables. The
pertinent integration segments are selected with โif-thenโ 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.
Razaqpurโs 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 Razaqpurโs
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 ๐ โฅ 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.
โ=โ[๐(๐ฅ๐)๐(๐ฅ๐)
๐ธ๐๐ผ๐(๐ฅ๐)+๐(๐ฅ๐โ1)๐(๐ฅ๐โ1)
๐ธ๐๐ผ๐(๐ฅ๐โ1)]๐ฟ
2๐
๐=๐
๐=1
where: ๐ฅ๐ =๐๐ฟ
๐ (3 โ 2)
and where the variables not explained previously in this chapter are defined as follows:
๐ - counter for each segment of the span, from 1 ๐ก๐ ๐
๐ - total number of equal length segments used for numerical integration
๐ฅ๐ - distance from the left end of the span to the right end of segment ๐; ๐ฅ0 = 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 ๐(๐ฅ) and ๐(๐ฅ)
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 ๐๐ = ๐๐๐๐ฅ 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, ๐ผ๐โฒ , proposed by Bischoff and Gross
(2011). The equation for ๐ผ๐โฒ , including the factor for a simply supported member, is
used with the midspan moment and midspan properties of the continuous member.
While ๐ผ๐โฒ 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:
๐ผ๐โฒ =
๐ผ๐๐
1 โ ๐ (๐๐๐
๐๐)2 where ๐ = 1 โ
๐ผ๐๐๐ผ๐ (as per Equation 2 โ 8)
For simply supported and continuous members with a point load, ๐, applied at midspan:
= 3 โ 2๐๐๐
๐๐ where ๐0 =
๐๐ฟ
4 and ๐๐ = ๐๐๐๐ฅ = ๐0 +
๐๐ฟ
2+๐๐
2 (3 โ 3)
For a centered point load, the approximate midspan deflection is:
๐ฅ = ๐ฅ๐๐๐ = ๐ฅ๐ผ๐โฒ = ๐พ๐๐๐ฟ
2
12๐ธ๐๐ผ๐โฒ where ๐พ = 1.5 โ 0.5
๐0
๐๐ (3 โ 4)
For unequal end-moments, where โ๐๐ฟ < 2๐๐, maximum deflection is computed as:
๐ฅ๐๐๐ฅ โ ๐ฅ๐๐๐
๐ฅ๐,๐๐๐ฅ
๐ฅ๐,๐๐๐ (3 โ 5)
In Equation 3-5, ๐ฅ๐,๐๐๐ฅ and ๐ฅ๐,๐๐๐ 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 ๐ผ๐ ๐ผ๐๐โ ratio. For the centered point load examples provided,
end-moments are kept equal to each other (๐๐ฟ = ๐๐ ) 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 ๐ผ๐(๐ฅ), 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 ๐ผ๐โฒ 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 ๐๐โฒ = 36 MPa
Reinforcing bar depth, from the compression face, is ๐ = 540 mm
Steel reinforcing bars used have a yield strength of ๐๐ฆ = 400 MPa
GFRP (glass fibre reinforced polymer) reinforcing bars used have an ultimate
strength of ๐๐๐ข = 690 MPa and an elastic modulus of ๐ธ๐ = 44 GPa.
The reinforcing ratio, ๐, is provided for the bottom reinforcing bars; when ๐/๐๐ is
provided for the GFRP reinforced members, ๐๐ is the balanced reinforcing ratio.
Figure 3-1 shows results of a set of example members with equal end-moments where
๐ผ๐/๐ผ๐๐ = 2.28 and ๐๐/๐๐๐ = 3.00. These member are designed with steel reinforcing
such that ๐๐ = 1.575๐๐ and ๐ = 1.2%. In this example, there is significant deflection,
๐ฅ๐ผ๐(๐ฅ), when constant stiffness results are zero (at โ๐๐ฟ = 2๐๐). All effective moment
of inertia approaches are conservative by approximately 10% (which is slightly better
than using the ๐ผ๐๐ results) except where results diverge as end-moments become
significantly larger than 1.2๐๐. Bransonโs (1965) method offers slight improvement on
using ๐ผ๐๐ but the proposed method offers far more accuracy for โ๐๐ฟ < 1.2๐๐. The
50
S806 (2012) curve, which represents integration of curvature while neglecting tension
stiffening, is conservative by at least 10%, even for ๐๐ฟ โซ ๐๐, as expected. Use of
๐ผ๐,๐๐ฃ๐ per Equation (2-2), fails to improve the accuracy of ๐ผ๐, and produces an aberration
at 0.4 โ โ๐๐ฟ/๐๐ โณ ๐๐๐/๐๐. This anomalous point on the ๐ฅ๐ผ๐,๐๐ฃ๐ curve occurs where
the lightly cracked ends contribute disproportionately to the member stiffness.
Figure 3-1 โ 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 ๐ผ๐/๐ผ๐๐ = 3.89, ๐ = 0.6%, ๐๐/๐๐๐ = 1.60, and ๐๐ = 1.575๐๐ . These results
again show that the effective moment of inertia approaches are conservative; results
using Bransonโs (1965) ๐ผ๐ are conservative by about 20% for โ๐๐ฟ < 1.4๐๐. The
proposed equation provides very accurate results when โ๐๐ฟ < 1.6๐๐. If the proposed
method is approximated by using = 1, or if tension stiffening is neglected (๐ฝ = 0),
51
results are more than 40% conservative. ๐ฅ๐ผ๐,๐๐ฃ๐ provides no improvement (see the
discussion for Figure 3-1 regarding the data point near 0.75 โ โ๐๐ฟ/๐๐ โณ ๐๐๐/๐๐).
Figure 3-2 - Midspan Deflection of Steel Reinforced Beams under Centered Point Load
with Ig/Icr=3.9, Mm/Mcr=1.6, and ML=MR
Figure 3-3 is based on members which are reinforced with GFRP reinforcing bars as
follows: ๐ผ๐/๐ผ๐๐ = 3.32, ๐ = 3.2%, ๐/๐๐ = 5.6, ๐๐/๐๐๐ = 2.50, ๐๐ = 2.78๐๐ for
bottom bars, and ๐๐ = 1.575๐๐ for top bars. To control deflection for this set of
members, the amount of additional reinforcing added causes the ๐ผ๐/๐ผ๐๐ 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 โ๐๐ฟ < 0.5๐๐.
The proposed equation offers improvement, but it becomes overly conservative between
โ๐๐ฟ โฅ 0.7๐๐ and โ๐๐ฟ โค 1.7๐๐. 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 ๐ผ๐ or ๐ผ๐โฒ 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 ๐ฅ๐ฝ=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 โ๐๐ฟ โฅ 0.7๐๐ and โ๐๐ฟ โค 1.3๐๐.
Figure 3-3 - Midspan Deflection of FRP Reinforced Beams under Centered Point Load
with Ig/Icr=3.3, Mm/Mcr=2.5, and ML=MR
Figure 3-4 plots computed deflection for GFRP reinforced concrete members where
๐ผ๐/๐ผ๐๐ = 12.3, ๐ = 0.7%, ๐/๐๐ = 1.2, ๐๐/๐๐๐ = 1.60, ๐๐ = 2.50๐๐ for bottom bars,
and ๐๐ = 1.575๐๐ for top bars. The results from the different approaches vary greatly
because the effective moment of inertia at midspan is much larger than ๐ผ๐๐. For this set
of members, the proposed approach offers accurate results when โ๐๐ฟ < ๐๐; it
53
becomes conservative for โ๐๐ฟ > ๐๐ because the cracked length at midspan is very
short for those members in this set. Bransonโs (1965) ๐ผ๐ causes erroneous
unconservative results for typical cases (โ๐๐ฟ < ๐๐), while use of ๐ฝ๐ = 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, ๐ผ๐ โฒ ,
with centered point loading. To provide the ranges of validity shown, results were
reviewed as each of the relevant variables was varied. Values for ๐๐๐ ๐๐โ , ๐ผ๐ ๐ผ๐๐โ ,
๐๐ ๐๐ฟโ , ๐๐ ๐๐โ , 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 (๐๐ฟ = ๐๐ )
# One End-Moment
#
Cracked Ratio
๐ผ๐โฒ Valid If: Iฮณ=1 Valid? ๐ผ๐๐ Valid? Ie
โฒ Valid If:
3 โค๐๐
๐๐๐ โ๐๐ฟ โค 1.3๐๐ Yes Ok โ๐๐ฟ โค 1.5๐๐
1.7 โค๐๐
๐๐๐< 3 โ๐๐ฟ โค 1.4๐๐ Ok No โ๐๐ฟ โค 1.5๐๐
1.3 โค๐๐
๐๐๐< 1.7 โ๐๐ฟ โค 1.5๐๐ No No โ๐๐ฟ โค 2๐๐
1 โค๐๐
๐๐๐< 1.3 โ๐๐ฟ โค 1.4๐๐ No No โ๐๐ฟ โค 1.5๐๐
# the results in this table assume ๐๐ฟ โค ๐๐ โค 0.
There is often a minor conservative error when โ๐๐ฟ โ โ๐๐ < ๐๐, but it appears to be
less than 5% for all cases except with FRP members with โ๐๐ฟ โ โ๐๐ < ๐๐/2 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 ๐ผ๐โฒ are, therefore,
(slightly) larger than the integrated ๐ผ๐(๐ฅ) result when โ๐๐ฟ โ โ๐๐ < ๐๐.
The proposed equations are often unconservative for โ๐๐ฟ โ โ๐๐ โซ ๐๐. If
deflections for โ๐๐ฟ > ๐๐ 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 ๐๐ฟ/๐๐๐ > 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 ๐ผ๐(๐ฅ) at the member
ends is smaller than the ๐ผ๐(๐ฅ) at midspan under the same service moments.
When โ๐๐ฟ โ โ๐๐ > 2๐๐, 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
ฮณ 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 ๐ผ๐ โฒ as per Bischoff and Gross (2011) for simply
supported members gives a good approximation for highly cracked continuous members
(๐๐ ๐๐๐โ > 2). For continuous members that experience less cracking, however, the
original simply supported equations offer poor results. Consequently, an improved
equation for , denoted as โ , is provided and used to compute an improved equivalent
moment of inertia, ๐ผ๐โโฒ . The more accurate deflection values, computed using ๐ผ๐โ
โฒ , are
indicated as ๐ฅ๐ผ๐โโฒ in Figures 3-5 to 3-8 and in Appendix N.
The proposed โ equation attempts to account for two prominent errors in results based
on the ๐ผ๐โฒ calculated using . For third-point loading with equal end-moments, the actual
equivalent moment of inertia is smaller than ๐ผ๐โฒ because the ๐พ 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 ๐๐๐, the exact member deflection is much less than the deflection computed using
๐ผ๐โฒ . 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 โ 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, โ = ,
which is the exact integrated result for a simply supported member. Further
improvement in โ 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 Bischoffโs equivalent moment of inertia is:
๐ผ๐โฒ =
๐ผ๐๐
1 โ ๐ (๐๐๐
๐๐๐๐ฅ)2 where ๐ = 1 โ
๐ผ๐๐๐ผ๐ (as per Equation 2 โ 8)
For members with two equal loads of ๐/2 at the third points and โ๐๐ฟ โฅ โ๐๐ :
= 1.7 โ 0.7 (๐๐๐
๐๐๐๐ฅ) where ๐0 =
๐๐ฟ
6 and ๐๐ = ๐0 +
๐๐ฟ
2+๐๐
2 (3 โ 6)
๐๐๐๐ฅ = ๐0 +1
3๐๐ฟ +
2
3๐๐ where โ ๐๐ฟ > โ๐๐ (3 โ 7)
If ๐๐ฟ = ๐๐ , the bending moment remains constant between the third-points
If โ๐๐ฟ > โ๐๐ , then ๐๐๐๐ฅ will occur at ๐ฅ = 2๐ฟ/3
For two equal loads at the third points, the midspan deflection using Bischoffโs ๐ผ๐โฒ for
simply supported members is computed as:
๐ฅ = ๐ฅ๐ผ๐โโฒ = ๐พ23๐๐๐ฟ
2
216๐ธ๐๐ผ๐โฒ where ๐พ =
27
23โ
4๐0
23๐๐ (3 โ 8)
The proposed solution for midspan deflection is:
โ = โ 0.1(๐๐ฟ โ 1.5๐๐ )/๐๐๐ where โ ๐๐ฟ โฅ โ๐๐ (3 โ 9)
58
๐ผ๐โโฒ =
๐ผ๐๐
1 โ โ ๐ (๐๐๐
๐๐๐๐ฅ)2 (3 โ 10)
๐ฅ = ๐ฅ๐๐๐ = ๐ฅ๐ผ๐โโฒ = ๐พ23๐๐๐ฟ
2
216๐ธ๐๐ผ๐โโฒ where ๐พ =
27
23โ
4๐0
23๐๐ (3 โ 11)
For unequal end-moments where โ๐๐ฟ < 2๐๐๐๐ฅ and โ๐๐ฟ > โ๐๐ , 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 (๐๐ฟ = ๐๐ )
Steel reinforced beams that are continuous at only one end (๐๐ = 0)
GFRP reinforced beams with equal end-moments (๐๐ฟ = ๐๐ )
GFRP reinforced beams that are continuous at only one end (๐๐ = 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 ๐ผ๐(๐ฅ), 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 ๐ผ๐โโฒ 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 ๐๐โฒ = 36 MPa. These beams were 600 mm deep, 300 mm wide, with 10 m
spans and with top and bottom reinforcement at ๐ = 540 mm. Figure 3-5 and Figure
3-6 use steel reinforcing with ๐๐ฆ = 400 MPa. Figure 3-7 and Figure 3-8 use GFRP
reinforcing bars with ๐๐๐ข = 690 MPa and ๐ธ๐ = 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 ๐ผ๐/๐ผ๐๐ = 2.97, ๐๐๐๐ฅ/๐๐๐ = 2.20, ๐ = 0.8%, and ๐๐ = 1.575๐๐ . 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
with Ig/Icr=3.0, Mm/Mcr=2.2, and ML=MR
60
members; significant error occurs only when the end-moments are relatively large:
โ๐๐ฟ = โ๐๐ > 1.5๐๐๐๐ฅ. For this set of examples, no effective moment of inertia
methods are suitable for large end moments (even ๐ฅ๐๐ results are unconservative for
โ๐๐ฟ = โ๐๐ = 3๐๐๐๐ฅ). The utility of the proposed ๐ผ๐โโฒ is negligible for these
members. However, these members are an example of the ๐ฅ๐ผ๐,๐๐ฃ๐ curve failing to match
the integrated results well. The ๐ฅ๐ผ๐,๐๐ฃ๐ 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 (๐ผ๐/๐ผ๐๐, ๐, ๐๐/๐๐ , 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
๐๐๐๐ฅ/๐๐๐) as the members in Figure 3-5. Figure 3-6 indicates that Bransonโs (1965) ๐ผ๐
approach is up to 15% conservative. The constant moment of inertia solutions are also
conservative when โ๐๐ฟ < 2.5๐๐๐๐ฅ, with the exception of the proposed ๐ผ๐โโฒ . The
deflection values computed using ๐ผ๐โโฒ provide the most accurate approximations in this
graph when โ๐๐ฟ < 2๐๐๐๐ฅ, which encompasses most practical members, but this
approach becomes unconservative for larger end-moments. Two curves of computed
maximum deflection, ๐ฅ๐๐๐ฅ,๐ผ๐(๐ฅ) and ๐ฅ๐๐๐ฅ,๐ผ๐โฒ , are also provided in Figure 3-6; these
show that the maximum deflection is significantly larger than the midspan deflection for
members which have โ๐๐ฟ > 1.5๐๐๐๐ฅ.
Figure 3-7 is based on members which are reinforced with GFRP reinforcing bars as
follows: ๐ผ๐/๐ผ๐๐ = 12.2, ๐ = 0.7%, ๐/๐๐ = 1.21, ๐๐/๐๐๐ = 1.40, ๐๐ = 2.857๐๐ for
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
62
bottom bars, and ๐๐ = 1.575๐๐ for top bars. The results using ๐ผ๐โฒ are unconservative,
and become so by more than 10% when โ๐๐ฟ โฅ 2.0๐๐๐๐ฅ . The S806 integration results
are too conservative to be plotted on this graph because tension stiffening is important
when ๐๐ ๐๐๐โ = 1.4. Results using Bransonโs (1965) ๐ผ๐ underpredict deflection by a
factor of about 2.5 when โ๐๐ฟ โค 2.0๐๐๐๐ฅ, and by more for โ๐๐ฟ > 2.0๐๐๐๐ฅ. Use of
๐ฝ๐ = 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 ๐ฅ๐ผ๐โโฒ curve is clearly the most accurate approach when โ๐๐ฟ โค 2.0๐๐๐๐ฅ .
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
Bransonโs (1965) ๐ผ๐ predict only 50% of the actual deflection for typical members
(โ๐๐ฟ < 1.5๐๐๐๐ฅ). The ๐ฅ๐ผ๐,๐ฝ๐ curve, with ๐ฝ๐ = 0.24, is conservative with an error of
30% to 200% for โ๐๐ฟ โค 2.0๐๐๐๐ฅ. The results using ๐ผ๐โฒ are very conservative for
unequal end-moments; this contrasts with the unconservative results in Figure 3-7. The
๐ฅ๐ผ๐โโฒ curve is conservative by up to 25% for โ๐๐ฟ โค ๐๐๐๐ฅ 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, ๐ฅ๐๐๐ฅ,๐ผ๐(๐ฅ) and ๐ฅ๐๐๐ฅ,๐ผ๐โฒ , show a significant difference between ๐ฅ๐๐๐ฅ and ๐ฅ๐๐๐
when โ๐๐ฟ > 1.5๐๐๐๐ฅ.
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, ๐ผ๐ โ โฒ , with two equal loads at third points. Results assume ๐๐ฟ โค ๐๐ โค 0. To
provide these ranges of validity, relevant variables were tested within reasonable ranges.
Values for ๐๐๐ ๐๐โ , ๐ผ๐ ๐ผ๐๐โ , ๐๐ ๐๐ฟโ , ๐๐ ๐๐โ , ๐/โ, 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, โ๐๐ฟ โค 1.5๐๐๐๐ฅ will be true for the load pattern which
governs deflection; therefore the valid ranges provided for ๐ผ๐โโฒ are rarely problematic.
64
To compute deflections in cases outside the valid ranges, and for more accurate
predictions where ๐๐๐๐ฅ ๐๐๐โ < 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 (๐๐ฟ = ๐๐ )
#f One End-Moment
Cracked Ratio ๐ผ๐โโฒ Valid If: ๐ผ๐
โฒ Valid If Iฮณ=1
Valid? ๐ผ๐โโฒ Valid If:
#a
3 โค๐๐๐๐ฅ
๐๐๐ โ๐๐ฟ โค 2.0๐๐๐๐ฅ โ๐๐ฟ โค 2.0๐๐๐๐ฅ Yes โ๐๐ฟ โค 2.0๐๐๐๐ฅ
1.7 โค๐๐๐๐ฅ
๐๐๐< 3 โ๐๐ฟ โค 2.0๐๐๐๐ฅ โ๐๐ฟ โค 1.5๐๐๐๐ฅ Ok
#b
โ๐๐ฟ โค 2.0๐๐๐๐ฅ
1.3 โค๐๐๐๐ฅ
๐๐๐< 1.7 โ๐๐ฟ โค 2.0๐๐๐๐ฅ โ๐๐ฟ โค 1.3๐๐๐๐ฅ Ok
#c
โ๐๐ฟ โค 2.0๐๐๐๐ฅ
1.2 โค๐๐๐๐ฅ
๐๐๐< 1.3 โ๐๐ฟ โค 2.0๐๐๐๐ฅ โ๐๐ฟ โค 1.0๐๐๐๐ฅ Ok
#d
โ๐๐ฟ โค 2.0๐๐๐๐ฅ
1 โค๐๐๐๐ฅ
๐๐๐< 1.2 โ๐๐ฟ โค 1.0๐๐๐๐ฅ โ๐๐ฟ โค 1.5๐๐๐๐ฅ Ok
#e
โ๐๐ฟ โค 0.5๐๐๐๐ฅ #a
If the ๐ผ๐โโฒ equation yields results outside limits 0 < ๐ผ๐โ
โฒ < ๐ผ๐, use of ๐ผ๐ is reasonable.
Prediction with ๐ผ๐โฒ using =1.7-.7(๐๐๐ ๐๐๐๐ฅโ ) reasonable if ๐๐๐๐ฅ ๐๐๐โ < 2.5, ๐๐ โ 0.
Overprediction exceeds 20% with ๐ผ๐โฒ for ๐๐๐๐ฅ ๐๐๐โ > 2.5, โ๐๐ฟ > 0.1๐๐๐๐ฅ, ๐๐ โ 0.
#b Overprediction of deflection exceeding 10% likely if โ๐๐ฟ < 1.0๐๐๐๐ฅ and ๐ผ๐ ๐ผ๐๐โ > 4.
#c Overprediction of deflection exceeding 10% likely when โ๐๐ฟ < 0.5๐๐๐๐ฅ
#d Overprediction of deflection, exceeding 10%, likely if ๐๐๐๐ฅ ๐๐๐โ < 2.5
#e Overprediction, exceeding 10%, likely if โ๐๐ฟ > 0.3๐๐๐๐ฅ; ๐ผ๐ is ok if โ๐๐ฟ > ๐๐๐๐ฅ
#f ๐ผ๐๐ provides reasonable results only when ๐๐๐๐ฅ ๐๐๐โ < 3
The improved result, ๐ผ๐โโฒ , using โ = โ 0.1(๐๐ฟ โ 1.5๐๐ )/๐๐๐, was only intended to
give good results for ๐๐๐๐ฅ ๐๐๐โ > 1.2. For single end-moment cases where
๐๐๐๐ฅ ๐๐๐โ < 1.2, the limits of 0 < ๐ผ๐โโฒ โค ๐ผ๐ gives good results. For equal end-moment
cases where ๐๐๐๐ฅ ๐๐๐โ < 1.2, use of the original ๐ผ๐โฒ with = 1 yields an improvement
65
in the ๐ผ๐โโฒ results. Use of ๐ผ๐โ
โฒ , or assuming ๐๐๐๐ฅ = 1.2๐๐๐ when ๐๐๐๐ฅ < 1.2๐๐๐, is
conservative and may be wise for ๐๐๐๐ฅ โ ๐๐๐. Because ๐ผ๐โโฒ was derived empirically,
the divergence of results where ๐๐๐๐ฅ ๐๐๐โ < 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, ๐ผ๐โฒ , proposed by Bischoff and Gross (2011).
Use of ๐ผ๐โฒ 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 ๐ผ๐โฒ and are computed using the member properties at the location
of maximum positive service bending moment, ๐๐๐๐ฅ, and an applied moment of ๐๐๐๐ฅ.
Although deflection calculations are still reasonably accurate if |๐๐ฟ โ๐๐ | โค 0.5๐๐,
๐๐๐๐ฅ 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:
๐ผ๐โฒ =
๐ผ๐๐
1 โ ๐ (๐๐๐
๐๐๐๐ฅ)2 where ๐ = 1 โ
๐ผ๐๐๐ผ๐ (as per Equation 2 โ 8)
The following equations were used to compute bending moment for a continuous
member with a uniformly distributed load:
๐0 =(๐๐ฟ + ๐๐ )๐ฟ
8=๐ค๐ฟ2
8 (3 โ 13)
๐0 =๐๐๐๐ฅ โ
๐๐ฟ
2 โ๐๐
2 + โ๐๐ฟ๐๐ โ๐๐ฟ๐๐๐๐ฅ โ๐๐ ๐๐๐๐ฅ +๐๐๐๐ฅ2
2 (3 โ 14)
๐๐ = ๐0 +๐๐ฟ
2+๐๐
2 and ๐๐๐๐ฅ = ๐0 +
๐๐ฟ
2+๐๐
2+(๐๐ฟ โ๐๐ )
2
16๐0 (3 โ 15)
To use the equation for ๐ผ๐โฒ with a uniformly distributed load, the integration factor is:
=1.6๐3 โ 0.6๐4
(๐๐๐
๐๐๐๐ฅ)2 + 2.4 ln(2 โ ๐) where ๐ = 1 โ โ1 โ
๐๐๐
๐๐๐๐ฅ (3 โ 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
๐ฅ = ๐ฅ๐๐๐ = ๐ฅ๐ผ๐โฒ = ๐พ5๐๐๐ฟ
2
48๐ธ๐๐ผ๐โฒ where ๐พ = 1.2 โ 0.2
๐0
๐๐ (3 โ 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%:
๐ฅ๐๐๐ฅ
๐ฅ๐๐๐< 1.05 if
๐๐๐๐ฅ
๐๐< 1.05 (3 โ 18)
The maximum deflection can be approximated as follows:
๐ฅ๐๐๐ฅ โ ๐ฅ๐๐๐โ๐๐๐๐ฅ
๐๐=5โ๐๐๐๐๐๐ฅ
48๐ธ๐๐ผ๐โฒ (3 โ 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 (๐๐ฟ = ๐๐ )
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 (๐๐ฟ = ๐๐ )
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 ๐๐ฟ โค ๐๐ โค 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 ๐ผ๐(๐ฅ) 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
๐ผ๐โฒ , 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 ๐๐โฒ = 36 MPa
Figures 3-9 to 3-12 are designed with steel reinforcing bars having ๐๐ฆ = 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 ๐ = 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 ๐ = 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 ๐ = 510 mm with GFRP reinforcing bars
Figure 3-13 GFRP reinforcing bars have an ๐๐๐ข = 690 MPa and an ๐ธ๐ = 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 ๐ฅ๐ผ๐,๐๐ฃ๐ curve has only been provided in Figures 3-9 and 3-11 because other
results showed very similar results; all are unhelpful modifications to ๐ผ๐
The curves provided for ๐ฅ๐ and ๐ฅ๐๐ 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 ๐ผ๐ and ๐ผ๐๐ for concrete members
Continuous concrete members are not bounded by the curves ๐ฅ๐ and ๐ฅ๐๐; 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 ๐ผ๐/๐ผ๐๐ = 2.99, ๐ = 0.8%, ๐๐ = 1.575๐๐ , and ๐๐/๐๐๐ = 2.17. For this
example, all of the effective moments of inertia are used to compute reasonably accurate
values for deflection when โ๐๐ฟ = โ๐๐ < 2.5๐๐๐๐ฅ. These curves are an example of
when effective moment of inertia methods are not suitable for larger end-moments; even
using ๐ผ๐๐ yields unconservative results for โ๐๐ฟ = โ๐๐ > 3๐๐๐๐ฅ. The exact result,
๐ฅ๐ผ๐(๐ฅ), begins to diverge from the constant stiffness results as the end moments increase
beyond โ๐๐ฟ = โ๐๐ = 2๐๐๐๐ฅ; extrapolating these curves is evidence that the, ๐ฅ๐ผ๐(๐ฅ),
will be significantly larger than the constant stiffness member result of 0 mm of
deflection when โ๐๐ฟ = โ๐๐ = 5๐๐. 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 โ๐๐ฟ = โ๐๐ > 3๐๐๐๐ฅ. When using Bransonโs (1965) ๐ผ๐, results
for this set of members are 7% conservative for โ๐๐ฟ = โ๐๐ > 1.5๐๐๐๐ฅ and are
70
unconservative when โ๐๐ฟ = โ๐๐ > 2.2๐๐๐๐ฅ. Using ๐ผ๐,๐๐ฃ๐ per CSA A23.3 (2004), as
shown by the ๐ฅ๐ผ๐,๐๐ฃ๐ curve, modifies the Bransonโs ๐ผ๐ in ways that do not relate to the
exact results, such as becoming very unconservative when โ๐๐ฟ = โ๐๐ > 1.8๐๐๐๐ฅ.
The ๐ฅ๐ผ๐,๐๐ฃ๐ curve includes an aberration near ๐๐ฟ/๐๐ = ๐๐๐/๐๐ that is explained in
the discussion of Figure 3-1. Use of Bischoffโs ๐ผ๐, shown by the curve for ๐ฅ๐พ=1,
provides results that are slightly more conservative than results using Bransonโs ๐ผ๐
approach. The proposed solution, using ๐ผ๐โฒ , is the most accurate approximation for the
deflection in Figure 3-9 when โ๐๐ฟ = โ๐๐ < 2๐๐๐๐ฅ. For โ๐๐ฟ = โ๐๐ < 1.8๐๐๐๐ฅ,
the curve for ๐ฅ๐ผ๐โฒ 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 (โ๐๐ฟ > 2๐๐๐๐ฅ). 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.
Figure 3-10 - Midspan and Maximum Deflection of Steel Reinforced Beams under
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 ๐ผ๐/๐ผ๐๐ = 4.90,
๐ = 0.54%, ๐๐ = 1.575๐๐ , and ๐๐/๐๐๐ = 1.33. Bransonโs 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 (๐๐/๐๐๐ โช 2), as
proven by the ๐ฅ๐ฝ=0 curve values being more than double those of the ๐ฅ๐ผ๐(๐ฅ) curve. For
this set of slabs, the results computed using the proposed ๐ผ๐โฒ have only a 3% error when
โ๐๐ฟ = โ๐๐ < 2.8๐๐๐๐ฅ. The use of ๐ผ๐,๐๐ฃ๐ 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. Bransonโs method
again underestimates deflection, while the ๐ฅ๐ฝ=0 curve again greatly exceeds the
73
deflection results which consider tension stiffening. The ๐ฅ๐ผ๐โฒ curve shows that the
proposed approach is again very accurate if โ๐๐ฟ = โ๐๐ โค 3๐๐๐๐ฅ. 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:
๐ผ๐/๐ผ๐๐ = 16.9, ๐ ๐๐โ = 1.02, ๐๐/๐๐๐ = 1.25, ๐ = 0.6, ๐๐ = 2.674๐๐ for the bottom
bars, and ๐๐ = 1.575๐๐ for the top bars. Bransonโs 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 ๐ผ๐โฒ are within 5% of the exact results
for โ๐๐ฟ = โ๐๐ < ๐๐๐๐ฅ, and within 11% for โ๐๐ฟ = โ๐๐ < 3๐๐๐๐ฅ, in this example.
Appendix O provides three examples for uniformly loaded GFRP members: a set of
example slabs with ๐ฝ๐ = 1.0, a set of example beams with ๐๐/๐๐๐ = 2.0, and the set
of data for the beams in Figure 3-13. Using ฮฒd = 0.204 in accordance with ACI
440.1R yields conservative results with about 80% error for this set of examples.
Where ๐ฝ๐ = 1.0 and deflection is underpredicted using Bransonโs ๐ผ๐ 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, ๐ผ๐โฒ ,
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 ๐๐๐ ๐๐โ ,
๐ผ๐ ๐ผ๐๐โ , ๐๐ ๐๐ฟโ , ๐๐ ๐๐โ , 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 ๐๐ โ 0,
โ๐๐ฟ โค 1.5๐๐๐๐ฅ and โ๐๐ฟ โค 2.5๐๐๐๐ฅ, and ๐๐๐๐ฅ ๐๐๐โ < 1.2. The maximum
deflection found using ๐ฅ๐๐๐ฅ โ ๐ฅ๐๐๐ โ๐๐๐๐ฅ ๐๐โ 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 (๐๐ฟ = ๐๐ ) One End-Moment
#
Cracked Ratio
๐ผ๐โฒ Valid If:
๐ผ =1 Valid? ๐ผ๐๐ Valid?
๐ผ๐โฒ Valid If:
3 โค๐๐๐๐ฅ
๐๐๐ โ๐๐ฟ โค 2.0๐๐๐๐ฅ Yes Ok โ๐๐ฟ โค 2.3๐๐๐๐ฅ
1.7 โค๐๐๐๐ฅ
๐๐๐< 3 โ๐๐ฟ โค 2.2๐๐๐๐ฅ Ok No โ๐๐ฟ โค 2.5๐๐๐๐ฅ
1.1 โค๐๐๐๐ฅ
๐๐๐< 1.7 โ๐๐ฟ โค 2.5๐๐๐๐ฅ No No โ๐๐ฟ โค 2.7๐๐๐๐ฅ
1 โค๐๐๐๐ฅ
๐๐๐< 1.1 โ๐๐ฟ โค 1.5๐๐๐๐ฅ No No โ๐๐ฟ โค 1.5๐๐๐๐ฅ
# Results were tested assuming ๐๐ฟ โค ๐๐ โค 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 โ๐๐ฟ > 2๐๐๐๐ฅ are rare, and are very unlikely to fail service deflection
requirements.
The idealized member testing shows a minor error for deflections computed with ๐ผ๐โฒ
when โ๐๐ฟ โ โ๐๐ < 2๐๐๐๐ฅ. 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 ๐ผ๐โฒ are, therefore, (slightly) larger than the integrated ๐ผ๐(๐ฅ) result when
โ๐๐ฟ โ โ๐๐ < ๐๐. When โ๐๐ฟ < 1.5๐๐๐๐ฅ and ๐๐ = 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
(โ๐๐ฟ โ โ๐๐ > 2๐๐๐๐ฅ). This is especially true for members where ๐๐๐๐ฅ/๐๐๐ > 1.7.
This is evident in Figures 3-9 and 3-10, where exact deflections exceed ๐ผ๐๐ results when
โ๐๐ฟ > 2๐๐๐๐ฅ. 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
โ๐๐ฟ โ โ๐๐ > 3๐๐๐๐ฅ, 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 โ๐๐ฟ and โ๐๐ are larger than 1.5๐๐๐๐ฅ.
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 ๐๐๐ to be calculated using one half of ๐๐, 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 ๐๐ = 0, the
difference in deflection reaches 10% at โ๐๐ฟ = 2.5๐๐๐๐ฅ. For constant stiffness
78
members subject to a centered point load or two equal point loads at the third-points
where ๐๐ = 0, the difference in deflection reaches 10% at โ๐๐ฟ = 2๐๐๐๐ฅ. Appendix
R provides data and more discussion on the comparison of midspan and maximum
deflection of prismatic linear-elastic members where ๐๐ = 0, ๐๐ = ๐๐๐๐ฅ/2, and
where ๐๐ = ๐๐ฟ/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 ๐พ = 0 in Table 2-2. Thus
โ๐๐๐= 0 if: โ๐๐ฟ = โ๐๐ = 2๐๐ for a centered point load, โ๐๐ฟ = โ๐๐ = 5.75๐๐
for equal point loads at third points, and โ๐๐ฟ = โ๐๐ = 5๐๐ 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 โ๐๐ฟ โ โ๐๐ โ 3๐๐๐๐ฅ.
Using an integration method becomes critical where โ๐๐ฟ โ โ๐๐ > 2.5๐๐๐๐ฅ and
๐๐๐๐ฅ/๐๐๐ > 1.3 (outside the valid range) because results from proposed equations
become increasingly unconservative as โ๐๐ฟ and โ๐๐ increase relative to ๐๐๐๐ฅ. The
ends of these members become stiffer than the proposed effective moment of inertia, ๐ผ๐โฒ ,
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 ๐ผ๐ and ๐ผ๐๐, one would expect the solution to
always be such that ๐ผ๐ > ๐ผ๐โฒ > ๐ผ๐๐. In this situation, however, an accurate solution for ๐ผ๐
โฒ
would have to reduce it to less than the midspan cracked moment of inertia, ๐ผ๐๐. A
visual example of why a more robust solution would require ๐ผ๐โฒ to be become less than
๐ผ๐๐ is seen in Figure 3-9; here, when โ๐๐ฟ = โ๐๐ = 3๐๐, the exact deflection exceeds
the deflection of a member with the constant stiffness of ๐ผ๐๐. 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 (๐๐/๐ธ๐ผ) curve in the negative moment region (see example ๐๐/๐ธ๐ผ 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 ๐๐๐ = 0.5๐๐๐ผ๐/๐ฆ๐ก (for use with Bransonโs
๐ผ๐). This change is intended to account for shrinkage restraint stresses. The use of
๐๐๐ = 0.67๐๐๐ผ๐/๐ฆ๐ก provides an equivalent adjustment for calculations based on ๐ผ๐โฒ or
๐ผ๐(๐ฅ) (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 ๐๐๐ = ๐๐๐ผ๐/๐ฆ๐ก to deflections determined using the reduced ๐๐๐ 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
Bransonโs ๐ผ๐ generally shift towards ๐ผ๐๐ results by more than those which use ๐ผ๐โฒ or
๐ผ๐(๐ฅ). This causes deflection calculations to produce conservative results when
๐ผ๐/๐ผ๐๐ < 10. While results using Bransonโs ๐ผ๐ and ๐๐๐ = ๐๐๐ผ๐/๐ฆ๐ก underestimate
deflection when ๐ผ๐/๐ผ๐๐ > 4, results that use this ๐ผ๐ with ๐๐๐ = 0.5๐๐๐ผ๐/๐ฆ๐ก factor are
improved because they only underestimate deflection when ๐ผ๐/๐ผ๐๐ > 12.
3.8 Summary of Results using Bransonโs Method
There are many results provided in Sections 3.4, 3.5, and 3.6 that compare exact
integrations results to results using Bransonโs method. Figures 3-2, 3-3, 3-5, 3-6, 3-9,
and 3-10 demonstrate that Bransonโs method provides reasonably accurate results when
3 โค ๐ผ๐/๐ผ๐๐ โค 4. Figures 3-1, 3-2, and 3-3 demonstrate that Bransonโs method is
sometimes more conservative than proposed methods. Figures 3-4, 3-7, 3-8, 3-11, 3-12,
and 3-13 demonstrate that Bransonโs 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
Bransonโs method, with ๐๐๐ = 0.5๐๐๐ผ๐/๐ฆ๐ก 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 Bransonโs 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, ๐พ (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 ๐ผ๐, ๐ผ๐โฒ ,
or ๐ผ๐โโฒ . 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 ๐ผ๐(๐ฅ)
exceeds ๐ผ๐โฒ 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 โ๐๐ฟ โฅ 2๐๐๐๐ฅ), 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
Bischoffโs 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 ๐ผ๐โฒ ;
consequently, a reasonably accurate and robust equation is proposed as ๐ผ๐โโฒ . 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 ๐ผ๐โฒ or ๐ผ๐โ
โฒ when members and loading meet the valid ranges noted.
The solutions for ๐ผ๐โฒ and ๐ผ๐โ
โฒ 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 Bransonโs
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 Bransonโs 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 ๐ผ๐ 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, ๐๐๐, 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 ๐ผ๐โโฒ , using โ
from equation (3 โ 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
๐๐๐๐ฅ ๐๐๐โ < 1.2 for equal point loads at third points, it should be determined whether
the recommended solution is to use ๐๐๐๐ฅ = 1.2๐๐๐, to use the proposed ๐ผ๐โโฒ results, or
whether a better solution for ๐ผ๐โโฒ 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), 992โ1002.
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), 68โ75.
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 โ 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), 498โ506.
Vesey, S., and Bischoff, P.H. 2011. Designing FRP reinforced concrete for deflection
control. ACI SP-275-03: 1-24.
89
Derivation of ๐ฒ for Continuous Linear-Elastic Members Appendix A
๐พ factors are used to compute deflections with generic elastic deflection equations using
a constant moment of inertia. The following is a derivation of the ๐พ 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 ๐พ factor multiplied by
the midspan deflection of a simply supported member with the same span, properties,
and midspan moment. The equations for ๐พ apply to constant stiffness members with
known end-moments. See List of Symbols for equation variable definitions.
Let ฮ 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, ๐๐ฟ and ๐๐ , 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 ๐0 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 ๐๐ as the midspan moment. This means:
๐0 = ๐๐ โ๐๐ฟ +๐๐
2
The subsequent three sections of this appendix will show the derivation of the ๐พ factor
for the three load cases analyzed in this report.
90
Calculate ๐ฒ 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 (๐๐ฟ = ๐๐ = 0):
๐๐ = ๐0 =๐๐ฟ
4 ; โ=
๐๐ฟ3
48๐ธ๐ผ= ๐พ
๐๐๐ฟ2
12๐ธ๐ผ ; ๐พ = 1
For a point load at midspan on a continuous member:
โ=๐๐ฟ3
48๐ธ๐ผ+๐๐ฟ๐ฟ
2
16๐ธ๐ผ+๐๐ ๐ฟ
2
16๐ธ๐ผ= ๐พ
๐๐๐ฟ2
12๐ธ๐ผ
โ=๐ฟ2
12๐ธ๐ผ(๐๐ฟ
4+12๐๐ฟ
16+12๐๐
16) =
๐ฟ2
12๐ธ๐ผ(๐0 โ
3
2(โ
๐๐ฟ +๐๐
2))
โ=๐ฟ2
12๐ธ๐ผ(๐0 โ
3
2(๐0 โ๐๐)) =
๐ฟ2
12๐ธ๐ผ(1.5๐๐ โ 0.5๐๐)
โ= (1.5 โ 0.5๐0
๐๐)๐๐๐ฟ
2
12๐ธ๐ผ ; ๐พ = 1.5 โ 0.5
๐0
๐๐
For a fixed-fixed member with a point load at midspan, ๐๐ = โ๐๐ฟ = โ๐๐ = ๐๐ฟ/8,
therefore:
๐0 = 2๐๐ ; ๐พ = 1.5 โ 0.5(2๐๐)
๐๐=1
2
โ= ๐พ๐๐๐ฟ
2
12๐ธ๐ผ= (
1
2)๐๐๐ฟ
2
12๐ธ๐ผ= (
1
4)๐0๐ฟ
2
12๐ธ๐ผ ; โ=
1
4 of simply supported deflection
91
Calculate ๐ฒ 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):
๐๐ = ๐0 =๐๐ฟ
6 ; โ=
23๐๐ฟ3
1296๐ธ๐ผ= ๐พ
23๐๐๐ฟ2
216๐ธ๐ผ ; ๐พ = 1
For a continuous member loaded equally at third points:
โ=23๐๐ฟ3
1296๐ธ๐ผ+๐๐ฟ๐ฟ
2
16๐ธ๐ผ+๐๐ ๐ฟ
2
16๐ธ๐ผ= ๐พ
23๐๐๐ฟ2
216๐ธ๐ผ
โ=23๐ฟ2
216๐ธ๐ผ(๐๐ฟ
6+27๐๐ฟ
46+27๐๐
46) =
23๐ฟ2
216๐ธ๐ผ(๐0 โ
27
23(โ
๐๐ฟ +๐๐
2))
โ=23๐ฟ2
216๐ธ๐ผ(๐0 โ
27
23(๐0 โ๐๐)) =
23๐ฟ2
216๐ธ๐ผ(27
23๐๐ โ
4
23๐0)
โ= (27
23โ
4
23
๐0
๐๐)23๐๐๐ฟ
2
216๐ธ๐ผ ; ๐พ =
27
23โ
4
23
๐0
๐๐
For a fixed-fixed member loaded equally at third points, 2๐๐ = โ๐๐ฟ = โ๐๐ = ๐๐ฟ/9,
therefore:
๐0 = 3๐๐ ; ๐พ =27
23โ
4
23
(3๐๐)
๐๐=15
23
โ= ๐พ23๐๐๐ฟ
2
216๐ธ๐ผ= (
15
23)23๐๐๐ฟ
2
216๐ธ๐ผ= (
5
23)23๐0๐ฟ
2
216๐ธ๐ผ ; โ=
5
23 of simply supported deflection
92
Calculate ๐ฒ 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):
๐๐ = ๐0 =๐ค๐ฟ2
8 ; โ=
5๐ค๐ฟ4
384๐ธ๐ผ= ๐พ
5๐๐๐ฟ2
48๐ธ๐ผ ; ๐พ = 1
For a uniformly distributed load on a continuous member:
โ=5๐ค๐ฟ4
384๐ธ๐ผ+๐๐ฟ๐ฟ
2
16๐ธ๐ผ+๐๐ ๐ฟ
2
16๐ธ๐ผ= ๐พ
5๐๐๐ฟ2
48๐ธ๐ผ
โ=5๐ฟ2
48๐ธ๐ผ(๐ค๐ฟ2
8+3๐๐ฟ
5+3๐๐
5) =
5๐ฟ2
48๐ธ๐ผ(๐0 โ
6
5(โ
๐๐ฟ +๐๐
2))
โ=5๐ฟ2
48๐ธ๐ผ(๐0 โ
6
5(๐0 โ๐๐)) =
5๐ฟ2
48๐ธ๐ผ(6
5๐๐ โ
1
5๐0)
โ= (1.2 โ 0.2๐0
๐๐)5๐๐๐ฟ
2
48๐ธ๐ผ ; ๐พ = 1.2 โ 0.2
๐0
๐๐
For a uniform load on a fixed-fixed member, 2๐๐ = โ๐๐ฟ = โ๐๐ = ๐ค๐ฟ2/12,
therefore:
๐0 = 3๐๐ ; ๐พ = 1.2 โ 0.2(3๐๐)
๐๐=3
5
โ= ๐พ5๐๐๐ฟ
2
48๐ธ๐ผ= (
3
5)5๐๐๐ฟ
2
48๐ธ๐ผ= (
1
5)5๐0๐ฟ
2
48๐ธ๐ผ ; โ =
1
5 of simply supported deflection
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:
โ= ๐๐ = โซ๐๐
๐ธ๐ผ
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:
โ๐= โซ ๐๐(๐ฅ) (๐(๐ฅ)
๐ธ๐ผ(๐ฅ))๐๐ฅ
๐ฟ
0
The variable ๐ฅ is used to denote the position along the beam, and the function is
integrated with respect to ๐ฅ, as ๐ฅ increases from 0 to the beam length, ๐ฟ.
โ๐ denotes that integration will determine the deflection at location ๐. The maximum
deflection is typically found at midspan, so that is where point ๐ is typically taken. The
subscript, ๐, 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 โ๐๐๐ฅ.
The function ๐๐(๐ฅ) is the virtual moment function for bending deflection when the
virtual unit load is placed at location ๐. As ๐ฅ progresses from one end of the beam to
the other, ๐๐(๐ฅ) will take into account the effect that the curvature, ๐(๐ฅ)/๐ธ๐ผ(๐ฅ), has
on the deflection at point ๐. 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:
๐๐ฟ 2โ (๐ฅ) = ๐ฅ 2โ for ๐ฅ โค ๐ฟ 2โ and ๐๐ฟ 2โ (๐ฅ) = (๐ฟ โ ๐ฅ) 2โ for ๐ฟ 2โ โค ๐ฅ โค ๐ฟ .
๐(๐ฅ) represents the bending moment as a function of ๐ฅ (for the relevant load case).
The ๐(๐ฅ) function is the typical structural design moment function determined from
statics and moment distribution.
The elastic modulus is denoted as ๐ธ. 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,
๐ผ(๐ฅ). This variation in stiffness can come in the form of varying cross-section shape,
size, cracking, reinforcing, or similar. The ๐ผ(๐ฅ) 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 ๐(๐ฅ)/ ๐ธ ๐ผ(๐ฅ) indicates the curvature of the beam at each point, ๐ฅ, 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
Razaqpurโs 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,
Razaqpurโs equations are converted to the variable notation of this report. Razaqpurโs
format for these equations requires loads, ๐ or ๐ค, and length of the uncracked segments,
๐ฟ๐. Rearranged equations, which are instead based on midspan moment, ๐๐, and the
cracking moment, ๐๐๐, 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 ๐ธ 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, ๐(๐ฅ), a unit load is applied at the point of
maximum deflection, which is midspan. The member midspan is located at ๐ฟ/2.
๐(๐ฅ) =๐ฅ
2 ๐๐๐ ๐ฅ โค
๐ฟ
2
A simply supported member with uniformly distributed loading, ๐ค, has a bending
moment function, ๐(๐ฅ), equal to:
๐(๐ฅ) =๐ค๐ฅ๐ฟ โ ๐ค๐ฅ2
2
Deflection, โ๐, if the member were to remain uncracked, equals:
โ๐= 2โซ ๐(๐ฅ) (๐(๐ฅ)
๐ธ๐๐ผ๐)๐๐ฅ
๐ฟ2
0
= 2โซ ๐ฅ (๐ค๐ฅ๐ฟ โ ๐ค๐ฅ2
2๐ธ๐๐ผ๐)๐๐ฅ
๐ฟ2
0
=5๐ค๐ฟ4
384๐ธ๐๐ผ๐๐(๐ผ๐๐๐ผ๐)
Additional deflection from the cracked segments can be found from the change in
curvature, ๐ฟ๐, relative to the uncracked curvature, as defined by this generic equation:
๐ฟ๐ = ๐ฟ๐
๐ธ๐ผ=
๐(๐ฅ)
๐ธ๐๐ผ๐(๐ฅ)โ๐(๐ฅ)
๐ธ๐๐ผ๐=๐๐(๐ฅ)
๐ธ๐๐ผ๐๐[1 โ (
๐๐๐
๐(๐ฅ))2
] ; ๐คโ๐๐๐ ๐ = 1 โ๐ผ๐๐๐ผ๐
Integrating the change in curvature over the length of the cracked region results in the
additional deflection from cracking, ๐ฟฮ๐๐:
98
๐ฟฮ๐๐ = 2โซ ๐(๐ฅ)(๐ฟ๐)๐๐ฅ
๐ฟ2
๐ฟ๐๐
= 2โซ๐ฅ
2(๐ (
๐ค๐ฅ๐ฟ โ ๐ค๐ฅ2
2 )
๐ธ๐๐ผ๐๐[1 โ (
๐๐๐
(๐ค๐ฅ๐ฟ โ ๐ค๐ฅ2)/2)2
])๐๐ฅ
๐ฟ2
๐ฟ๐๐
๐ฟฮ๐๐ =๐๐ค
2๐ธ๐๐ผ๐๐โซ ๐ฅ2
(
(๐ฟ โ ๐ฅ) [1 โ (๐๐๐
2
๐ค2๐ฅ2
4(๐ฟ โ ๐ฅ)2
)]
)
๐๐ฅ
๐ฟ2
๐ฟ๐๐
๐ฟฮ๐๐ =๐๐ค
2๐ธ๐๐ผ๐๐โซ ๐ฟ๐ฅ2 โ ๐ฅ3 โ
4๐ค2๐๐๐
2
๐ฟ โ ๐ฅ๐๐ฅ
๐ฟ2
๐ฟ๐๐
๐ฟฮ๐๐ =๐๐ค
2๐ธ๐๐ผ๐๐(๐ฟ๐ฅ3
3|๐ฟ๐๐
๐ฟ2
โ๐ฅ4
4|๐ฟ๐๐
๐ฟ2
+4
๐ค2๐๐๐
2 ln(๐ฟ โ ๐ฅ)|๐ฟ๐๐
๐ฟ2) ; let ๐ฟ๐๐ = ๐
๐ฟ
2
๐ฟฮ๐๐ =๐๐ค
2๐ธ๐๐ผ๐๐(8๐ฟ4
192โ8๐3๐ฟ4
192โ3๐ฟ4
192+3๐4๐ฟ4
192+192
1924๐๐๐
2 ๐ฟ4
64๐๐2 ln (
๐ฟ โ๐ฟ2
๐ฟ โ ๐๐ฟ2
))
๐ฟฮ๐๐ =5๐๐ค๐ฟ4
384๐ธ๐๐ผ๐๐(1 +
3๐4
5โ8๐3
5โ12
5
๐๐๐2
๐๐2 ln(2 โ ๐))
These two values are summed to give the midspan deflection, ฮ:
ฮ = ฮ๐ + ๐ฟฮ๐๐
ฮ =5๐ค๐ฟ4
384๐ธ๐๐ผ๐๐(๐ผ๐๐๐ผ๐) +
5๐๐ค
384๐ธ๐๐ผ๐๐(1 + 0.6๐4 โ 1.6๐3 โ 2.4
๐๐๐2
๐๐2 ln(2 โ ๐))
ฮ =5๐ค๐ฟ4
384๐ธ๐๐ผ๐๐[1 โ ๐
๐๐๐2
๐๐2 (
1.6๐3 โ 0.6๐4
๐๐๐2/๐๐
2 + 2.4 ln(2 โ ๐))]
Set ฮ = 5๐ค๐ฟ4/384๐ธ๐๐ผ๐โฒ and solve for the effective moment of inertia, ๐ผ๐
โฒ :
๐ผ๐โฒ =
๐ผ๐๐
1 โ ๐ (๐๐๐
๐๐)2 where =
1.6๐3 โ 0.6๐4
(๐๐๐
๐๐)2 + 2.4 ln(2 โ ๐)
Note: ๐๐ =๐ค๐ฟ2
8 ; ๐ = 1 โ
๐ผ๐๐๐ผ๐ ; ๐ = 1 โ โ1 โ๐๐๐/๐๐
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:
๐ฅ = ๐ฅ1 + ๐ฅ2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5 + ๐ฅ6 where:
๐ฅ1 = โซ๐(๐ฅ)๐(๐ฅ)
๐ธ๐๐ผ๐(๐ฅ)๐๐ฅ
๐ฟ
0
๐ฅ2 = โซ๐(๐ฅ)๐(๐ฅ)
๐ธ๐๐ผ๐๐๐ฅ
๐ฟ
๐ฟ
๐ฅ3 = โซ๐(๐ฅ)๐(๐ฅ)
๐ธ๐๐ผ๐(๐ฅ)๐๐ฅ
๐ฟ
๐ฟ
๐ฅ4 = โซ๐(๐ฅ)๐(๐ฅ)
๐ธ๐๐ผ๐(๐ฅ)๐๐ฅ
๐ฟ
๐ฟ
๐ฅ5 = โซ๐(๐ฅ)๐(๐ฅ)
๐ธ๐๐ผ๐๐๐ฅ
๐ฟ
๐ฟ
๐ฅ6 = โซ๐(๐ฅ)๐(๐ฅ)
๐ธ๐๐ผ๐(๐ฅ)๐๐ฅ
๐ฟ
๐ฟ
The virtual work method used here, and its integration variables, are described in
Appendix B. The lengths ๐ฟ1 through ๐ฟ5 can be calculated as a ratio (based on total
moment and end-moments) of the member length, ๐ฟ.
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 ๐ฟ/3 and 2๐ฟ/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:
๐๐ =๐๐ฟ
2+๐๐
2+๐0
For all loads, virtual moment for midspan deflection is:
for 0 โค ๐ฅ โค๐ฟ
2 โถ ๐(๐ฅ) =
๐ฅ
2
102
for ๐ฟ
2โค ๐ฅ โค ๐ฟ โถ ๐(๐ฅ) =
๐ฟ โ ๐ฅ
2
For single midspan point load:
for 0 โค ๐ฅ โค๐ฟ
2 โถ ๐(๐ฅ) = ๐๐ฟ + (2๐0 โ๐๐ฟ +๐๐ )
๐ฅ
๐ฟ
for ๐ฟ
2โค ๐ฅ โค ๐ฟ โถ ๐(๐ฅ) = ๐๐ + (2๐0 โ๐๐ +๐๐ฟ)
๐ฟ โ ๐ฅ
๐ฟ
For equal third-point loading:
for 0 โค ๐ฅ โค๐ฟ
3 โถ ๐(๐ฅ) = ๐๐ฟ + (3๐0 โ๐๐ฟ +๐๐ )
๐ฅ
๐ฟ
for ๐ฟ
3โค ๐ฅ โค
2๐ฟ
3 โถ ๐(๐ฅ) = ๐0 +๐๐ฟ + (๐๐ โ๐๐ฟ)
๐ฅ
๐ฟ
for 2๐ฟ
3โค ๐ฅ โค ๐ฟ โถ ๐(๐ฅ) = 3๐0 +๐๐ฟ + (๐๐ โ๐๐ฟ โ 3๐0)
๐ฅ
๐ฟ
For a uniformly distributed load:
๐(๐ฅ) = ๐๐ฟ + (4๐0 โ๐๐ฟ +๐๐ )๐ฅ
๐ฟ โ 4๐0
๐ฅ2
๐ฟ2
For all loads, this work uses the following local effective moment of inertia:
for โ๐๐๐ โค ๐(๐ฅ) โค ๐๐๐ โถ ๐ผ๐(๐ฅ) = ๐ผ๐
for โ๐๐๐ < ๐(๐ฅ) or ๐(๐ฅ) > ๐๐๐ โถ ๐ผ๐(๐ฅ) =๐ผ๐๐
1 โ ๐ (๐๐๐
๐(๐ฅ))2 where ๐ = 1 โ
๐ผ๐๐๐ผ๐
Lengths to where the Function being Integrated Changes
Centered Point Load:
๐ฟ1 through ๐ฟ5 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 ๐ผ๐ฟ = ๐๐ฟ/๐0 and ๐ผ๐ = ๐๐ /๐0.
103
๐ฟ1 =โ๐ผ๐ฟ๐0 โ๐๐๐
(2 โ ๐ผ๐ฟ + ๐ผ๐ )๐0๐ฟ ; ๐ฟ2 =
โ๐ผ๐ฟ๐0 +๐๐๐
(2 โ ๐ผ๐ฟ + ๐ผ๐ )๐0๐ฟ ; ๐ฟ3 =
๐ฟ
2
๐ฟ4 = ๐ฟ +๐ผ๐ ๐0 โ๐๐๐
(2 + ๐ผ๐ฟ โ ๐ผ๐ )๐0๐ฟ ; ๐ฟ5 = ๐ฟ +
๐ผ๐ ๐0 +๐๐๐
(2 + ๐ผ๐ฟ โ ๐ผ๐ )๐0๐ฟ
For equal third-point loading:
For the deflection of a member with equal point load at third-points, ๐ฟ1 through ๐ฟ5, per
Figure E-2, are simple ratios (based on total and end-moments) of the member length, ๐ฟ.
These simple ratios can also be substituted in neatly, so they can be removed from the
final equations. The equation for ๐ฟ2 is only valid when ๐(๐ฅ = ๐ฟ 3โ ) > ๐๐๐ and the
equation for ๐ฟ4 is only valid when ๐(๐ฅ = 2๐ฟ 3โ ) > ๐๐๐. The equations for the
alternate situations for ๐ฟ2 and ๐ฟ4 are similar but are not provided.
๐ฟ1 =โ๐๐ฟ โ๐๐๐
๐๐ โ๐๐ฟ + 3๐0๐ฟ ; ๐ฟ2 =
โ๐๐ฟ +๐๐๐
๐๐ โ๐๐ฟ + 3๐0๐ฟ ; ๐ฟ3 =
๐ฟ
2
๐ฟ4 =๐๐ฟ โ 3๐0 +๐๐๐
๐๐ โ๐๐ฟ โ 3๐0๐ฟ ; ๐ฟ5 =
๐๐ฟ โ 3๐0 โ๐๐๐
๐๐ โ๐๐ฟ โ 3๐0๐ฟ
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.
For a uniformly distributed load:
The following are the equations for the lengths, ๐ฟ1 through ๐ฟ5 per Figure E-3, to where
the function being integrated changes for a member with a uniformly distributed load.
In this case, ๐ฟ1, ๐ฟ2, ๐ฟ4, and ๐ฟ5 are complicated expressions, so they remain in the final
equations. Again note that ๐ฟ3 = ๐ฟ/2, ๐ผ๐ฟ = ๐๐ฟ/๐0, and ๐ผ๐ = ๐๐ /๐0.
104
๐ฟ1 = (4 โ ๐ผ๐ฟ + ๐ผ๐ โโ(4 โ ๐ผ๐ฟ + ๐ผ๐ )2 + 16๐๐๐
๐0+ 16๐ผ๐ฟ)
๐ฟ
8
๐ฟ2 = (4 โ ๐ผ๐ฟ + ๐ผ๐ โโ(4 โ ๐ผ๐ฟ + ๐ผ๐ )2 โ 16๐๐๐
๐0+ 16๐ผ๐ฟ)
๐ฟ
8
๐ฟ4 = (4 โ ๐ผ๐ฟ + ๐ผ๐ +โ(4 โ ๐ผ๐ฟ + ๐ผ๐ )2 โ 16๐๐๐
๐0+ 16๐ผ๐ฟ)
๐ฟ
8
๐ฟ5 = (4 โ ๐ผ๐ฟ + ๐ผ๐ +โ(4 โ ๐ผ๐ฟ + ๐ผ๐ )2 + 16๐๐๐
๐0+ 16๐ผ๐ฟ)
๐ฟ
8
Midspan Deflection of Midspan-Point Loaded Member with End-Moments
๐ฅ1 =๐ฟ2 (๐ผ๐ฟ
3๐03 โ 3๐ผ๐ฟ๐๐๐
2๐0 + (6๐๐ฟ โ 2)๐๐๐3 + 6๐๐ฟ๐ผ๐ฟ๐๐๐
2๐0 [1 + ln (โ๐๐๐
๐ผ๐ฟ๐0)])
12๐ธ๐๐ผ๐๐ ๐ฟ(2 โ ๐ผ๐ฟ + ๐ผ๐ )2๐02
๐ฅ2 =๐ฟ2๐๐๐
3
3๐ธ๐๐ผ๐(2 โ ๐ผ๐ฟ + ๐ผ๐ )2๐0
2
๐ฅ3 =๐ฟ2
48๐ธ๐๐ผ๐๐ ๐(๐ผ๐ฟ โ ๐ผ๐ โ 2)2๐02 (12๐ผ๐ฟ๐๐๐
2๐0 + 24๐๐๐๐๐3 โ 8๐๐๐
3
+๐03(8 โ 2๐ผ๐ฟ
3 โ 6๐ผ๐ฟ2 โ 3๐ผ๐ฟ
2๐ผ๐ + 12๐ผ๐ + 6๐ผ๐ 2 + ๐ผ๐
3)
โ 12๐๐๐๐๐2๐0 (๐ผ๐ + 2 + ๐ผ๐ฟ [1 + 2 ln (
2๐๐๐
๐0(๐ผ๐ฟ + ๐ผ๐ + 2))]))
๐ฅ4 =๐ฟ2
48๐ธ๐๐ผ๐๐ ๐(๐ผ๐ โ ๐ผ๐ฟ โ 2)2๐02 (12๐ผ๐ ๐๐๐
2๐0 + 24๐๐๐๐๐3 โ 8๐๐๐
3
+๐03(8 โ 2๐ผ๐
3 โ 6๐ผ๐ 2 โ 3๐ผ๐
2๐ผ๐ฟ + 12๐ผ๐ฟ + 6๐ผ๐ฟ2 + ๐ผ๐ฟ
3)
โ 12๐๐๐๐๐2๐0 (๐ผ๐ฟ + 2 + ๐ผ๐ [1 + 2 ln (
2๐๐๐
๐0(๐ผ๐ฟ + ๐ผ๐ + 2))]))
๐ฅ5 =๐ฟ2๐๐๐
3
3๐ธ๐๐ผ๐(2 + ๐ผ๐ฟ โ ๐ผ๐ )2๐02
๐ฅ6 =๐ฟ2 (๐ผ๐
3๐03 โ 3๐ผ๐ ๐๐๐
2๐0 + (6๐๐ โ 2)๐๐๐3 + 6๐๐ ๐ผ๐ ๐๐๐
2๐0 [1 + ln (โ๐๐๐
๐ผ๐ ๐0)])
12๐ธ๐๐ผ๐๐ ๐ (2 + ๐ผ๐ฟ โ ๐ผ๐ )2๐02
105
When the left end moment < ๐๐๐:
๐ฅ1&2 =๐ฟ2(2๐๐๐
3 โ 3๐ผ๐ฟ๐๐๐2๐0 + ๐ผ๐ฟ
3๐03)
12๐ธ๐๐ผ๐(2 โ ๐ผ๐ฟ + ๐ผ๐ )2๐02
When the right end moment < ๐๐๐:
๐ฅ5&6 =๐ฟ2(2๐๐๐
3 โ 3๐ผ๐ ๐๐๐2๐0 + ๐ผ๐
3๐03)
12๐ธ๐๐ผ๐(2 + ๐ผ๐ฟ โ ๐ผ๐ )2๐02
Total midspan deflection (note these results are only valid if ๐๐ > ๐๐๐):
๐ฅ = ๐ฅ1 + ๐ฅ2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5 + ๐ฅ6 (If both ends and midspan are cracked)
๐ฅ = ๐ฅ1&2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5&6 (If only midspan is cracked)
๐ฅ = ๐ฅ1&2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5 + ๐ฅ6 (If only right end and midspan are cracked)
๐ฅ = ๐ฅ1 + ๐ฅ2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5&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 ๐ฟ4 and ๐ฟ5, as provided previously in this appendix. Therefore use:
๐ฟ๐ 4 = ๐ฟ โ ๐ฟ4 ; ๐ฟ๐ 5 = ๐ฟ โ ๐ฟ5
Then, similar to the centered point load case, define:
106
๐ฅ1 = (๐0๐ฟ
2
48๐ธ๐๐ผ๐๐๐ฟ)
(
3๐๐ฟ (
๐๐๐
๐0)2
ln (๐ผ๐ฟ๐ฟ
2 โ 4๐ฟ12 + 4๐ฟ1๐ฟ โ ๐ผ๐ฟ๐ฟ1๐ฟ + ๐ผ๐ ๐ฟ1๐ฟ
๐ผ๐ฟ๐ฟ2)
+1
๐ฟ4(12๐ผ๐ฟ๐ฟ1
2๐ฟ2 + 32๐ฟ13๐ฟ โ 24๐ฟ1
4 โ 8๐ผ๐ฟ๐ฟ13๐ฟ + 8๐ผ๐ ๐ฟ1
3๐ฟ)
+3๐๐ฟ(๐ผ๐ฟ โ 4 โ ๐ผ๐ )
โ(4 โ ๐ผ๐ + ๐ผ๐ฟ)2 + 16๐ผ๐ (๐๐๐
๐0)2
โ ln
(
โ(4 โ ๐ผ๐ + ๐ผ๐ฟ)2 + 16๐ผ๐ โ (โ4 + ๐ผ๐ฟ โ ๐ผ๐ )
โ(4 โ ๐ผ๐ + ๐ผ๐ฟ)2 + 16๐ผ๐ + (โ4 + ๐ผ๐ฟ โ ๐ผ๐ )
โ(4 โ ๐ผ๐ + ๐ผ๐ฟ)2 + 16๐ผ๐ โ (๐ผ๐ฟ โ 4 + 8๐ฟ1 ๐ฟโ โ ๐ผ๐ )
โ(4 โ ๐ผ๐ + ๐ผ๐ฟ)2 + 16๐ผ๐ + (๐ผ๐ฟ โ 4 + 8๐ฟ1 ๐ฟโ โ ๐ผ๐ ))
)
๐ฅ2 = ๐0๐ฟ
2
2๐ธ๐๐ผ๐((4 โ ๐ผ๐ฟ + ๐ผ๐ ) (
๐ฟ23 โ ๐ฟ1
3
3๐ฟ3) + ๐ผ๐ฟ (
๐ฟ22 โ ๐ฟ1
2
2๐ฟ2) โ (
๐ฟ24 โ ๐ฟ1
4
๐ฟ4))
๐ฅ3 = (๐0๐ฟ
2
96๐ธ๐๐ผ๐๐๐)
(
6๐๐ (
๐๐๐
๐0)2
ln ((โ๐ผ๐ฟ โ 2 โ ๐ผ๐ )๐ฟ
2
2 (โ๐ผ๐ฟ๐ฟ2 + 4๐ฟ22 + ๐ฟ2๐ฟ(๐ผ๐ฟ โ 4 โ ๐ผ๐ ))
)
+1
๐ฟ4(โ24๐ผ๐ฟ๐ฟ2
2๐ฟ2 + 16๐ผ๐ฟ๐ฟ23๐ฟ + 48๐ฟ2
4 โ 64๐ฟ23๐ฟ โ 16๐ผ๐ ๐ฟ2
3๐ฟ
+ 5๐ฟ4 + 2๐ผ๐ ๐ฟ4 + 4๐ผ๐ฟ๐ฟ
4) +6๐๐(๐ผ๐ฟ โ 4 โ ๐ผ๐ )
โ(4 โ ๐ผ๐ฟ + ๐ผ๐ )2 + 16๐ผ๐ฟ(๐๐๐
๐0)2
โ ln
(
โ(4 โ ๐ผ๐ฟ + ๐ผ๐ )2 + 16๐ผ๐ฟ โ (๐ผ๐ฟ โ 4 + 8๐ฟ2 ๐ฟโ โ ๐ผ๐ )
โ(4 โ ๐ผ๐ฟ + ๐ผ๐ )2 + 16๐ผ๐ฟ + (๐ผ๐ฟ โ 4 + 8๐ฟ2 ๐ฟโ โ ๐ผ๐ )
โ(4 โ ๐ผ๐ฟ + ๐ผ๐ )2 + 16๐ผ๐ฟ โ (๐ผ๐ฟ โ ๐ผ๐ )
โ(4 โ ๐ผ๐ฟ + ๐ผ๐ )2 + 16๐ผ๐ฟ + (๐ผ๐ฟ โ ๐ผ๐ ) )
)
107
๐ฅ4 = (๐0๐ฟ
2
96๐ธ๐๐ผ๐๐๐)
(
6๐๐ (
๐๐๐
๐0)2
ln ((โ๐ผ๐ โ 2 โ ๐ผ๐ฟ)๐ฟ
2
2(โ๐ผ๐ ๐ฟ2 + 4๐ฟ๐ 42 + ๐ฟ๐ 4๐ฟ(๐ผ๐ โ 4 โ ๐ผ๐ฟ))
)
+1
๐ฟ4(โ24๐ผ๐ ๐ฟ๐ 4
2 ๐ฟ2 + 16๐ผ๐ ๐ฟ๐ 43 ๐ฟ + 48๐ฟ๐ 4
4 โ 64๐ฟ๐ 43 ๐ฟ โ 16๐ผ๐ฟ๐ฟ๐ 4
3 ๐ฟ
+ 5๐ฟ4 + 2๐ผ๐ฟ๐ฟ4 + 4๐ผ๐ ๐ฟ
4) +6๐๐(๐ผ๐ โ 4 โ ๐ผ๐ฟ)
โ(4 โ ๐ผ๐ + ๐ผ๐ฟ)2 + 16๐ผ๐ (๐๐๐
๐0)2
โ ln
(
โ(4 โ ๐ผ๐ + ๐ผ๐ฟ)2 + 16๐ผ๐ โ (๐ผ๐ โ 4 + 8๐ฟ๐ 4 ๐ฟโ โ ๐ผ๐ฟ)
โ(4 โ ๐ผ๐ + ๐ผ๐ฟ)2 + 16๐ผ๐ + (๐ผ๐ โ 4 + 8๐ฟ๐ 4 ๐ฟโ โ ๐ผ๐ฟ)
โ(4 โ ๐ผ๐ + ๐ผ๐ฟ)2 + 16๐ผ๐ โ (๐ผ๐ โ ๐ผ๐ฟ)
โ(4 โ ๐ผ๐ + ๐ผ๐ฟ)2 + 16๐ผ๐ + (๐ผ๐ โ ๐ผ๐ฟ) )
)
๐ฅ5 = ๐0๐ฟ
2
2๐ธ๐๐ผ๐((4 โ ๐ผ๐ + ๐ผ๐ฟ) (
๐ฟ๐ 43 โ ๐ฟ๐ 5
3
3๐ฟ3) + ๐ผ๐ (
๐ฟ๐ 42 โ ๐ฟ๐ 5
2
2๐ฟ2) โ (
๐ฟ๐ 44 โ ๐ฟ๐ 5
4
๐ฟ4))
๐ฅ6 = (๐0๐ฟ
2
48๐ธ๐๐ผ๐๐๐ )
(
1
๐ฟ4(12๐ผ๐ ๐ฟ๐ 5
2 ๐ฟ2 + 32๐ฟ๐ 53 ๐ฟ โ 24๐ฟ๐ 5
4 โ 8๐ผ๐ ๐ฟ๐ 53 ๐ฟ + 8๐ผ๐ฟ๐ฟ๐ 5
3 ๐ฟ)
+ 3๐๐ (๐๐๐
๐0)2
ln (๐ผ๐ ๐ฟ
2 โ 4๐ฟ๐ 52 + 4๐ฟ๐ 5๐ฟ โ ๐ผ๐ ๐ฟ๐ 5๐ฟ + ๐ผ๐ฟ๐ฟ๐ 5๐ฟ
๐ผ๐ ๐ฟ2)
+3๐๐ (๐ผ๐ โ 4 โ ๐ผ๐ฟ)
โ(4 โ ๐ผ๐ฟ + ๐ผ๐ )2 + 16๐ผ๐ฟ(๐๐๐
๐0)2
โ ln
(
โ(4 โ ๐ผ๐ฟ + ๐ผ๐ )2 + 16๐ผ๐ฟ โ (โ4 + ๐ผ๐ โ ๐ผ๐ฟ)
โ(4 โ ๐ผ๐ฟ + ๐ผ๐ )2 + 16๐ผ๐ฟ + (โ4 + ๐ผ๐ โ ๐ผ๐ฟ)
โ(4 โ ๐ผ๐ฟ + ๐ผ๐ )2 + 16๐ผ๐ฟ โ (๐ผ๐ โ 4 + 8๐ฟ๐ 5 ๐ฟโ โ ๐ผ๐ฟ)
โ(4 โ ๐ผ๐ฟ + ๐ผ๐ )2 + 16๐ผ๐ฟ + (๐ผ๐ โ 4 + 8๐ฟ๐ 5 ๐ฟโ โ ๐ผ๐ฟ))
)
108
When the left end moment < ๐๐๐ โถ
๐ฅ1&2 =๐0๐ฟ
2
2๐ธ๐๐ผ๐((4 โ ๐ผ๐ฟ + ๐ผ๐ )๐ฟ2
3
3๐ฟ3 โ
๐ฟ24
๐ฟ4+๐ผ๐ฟ๐ฟ2
2
2๐ฟ2)
When the right end moment < ๐๐๐ โถ
๐ฅ5&6 =๐0๐ฟ
2
2๐ธ๐๐ผ๐((4 โ ๐ผ๐ + ๐ผ๐ฟ)๐ฟ๐ 4
3
3๐ฟ3 โ
๐ฟ๐ 44
๐ฟ4+๐ผ๐ ๐ฟ๐ 4
2
2๐ฟ2)
Total midspan deflection (note these results are only valid if ๐๐ > ๐๐๐):
๐ฅ = ๐ฅ1 + ๐ฅ2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5 + ๐ฅ6 (If both ends and midspan are cracked)
๐ฅ = ๐ฅ1&2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5&6 (If only midspan is cracked)
๐ฅ = ๐ฅ1&2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5 + ๐ฅ6 (If only right end and midspan are cracked)
๐ฅ = ๐ฅ1 + ๐ฅ2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5&6 (If only left end and midspan are cracked)
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 ๐๐ = โ๐๐ฟ = โ๐๐ , cracked
(๐๐ > ๐๐๐), not previously loaded, reinforced with equal top and bottom reinforcing,
and far more simplifiable than typical.
Equations: ๐0 = 2๐๐ ; ๐ = ๐๐ฟ = ๐๐ = 1 โ ๐ผ๐๐/๐ผ๐ ; ๐ผ๐๐ = ๐ผ๐๐ ๐ฟ = ๐ผ๐๐ ๐ = ๐ผ๐๐ ๐
๐1 = 1 + ln(2๐๐๐
๐0) ; ๐2 = 1 โ ln (
2๐๐๐
๐0) ; ๐1 + ๐2 = 2
๐ฅ1 = ๐ฅ6 =โ๐0
8
3
+ 3๐0
2 ๐๐๐
2
+ 6๐๐๐๐3 โ 2๐๐๐
3 โ 6๐๐0
2 ๐๐๐2๐
1
48๐ธ๐๐ผ๐๐๐02/๐ฟ2
๐ฅ2 = ๐ฅ5 =๐ฟ2๐๐๐
3
12๐ธ๐๐ผ๐๐02
๐ฅ3 = ๐ฅ4 =โ6๐๐๐
2๐0 โ 18๐๐๐๐2๐0 + 24๐๐๐๐
3 โ 8๐๐๐3 +
52๐0
3 โ 12๐๐๐๐2๐0๐2
192๐ธ๐๐ผ๐๐๐02/๐ฟ2
๐ฅ =16๐ฟ2๐๐๐
3
96๐ธ๐๐ผ๐๐02
+๐ฟ2
96๐ธ๐๐ผ๐๐๐02 (โ
๐0
2
3
+ 6๐๐๐2๐0 + 24๐๐๐๐
3 โ 8๐๐๐3
โ 12๐๐๐๐2๐0๐1)
+๐ฟ2
96๐ธ๐๐ผ๐๐๐02 (5
2๐0
3 โ 6๐๐๐2๐0 โ 12๐๐๐๐
2๐0๐2 + 24๐๐๐๐3
โ 8๐๐๐3)
๐ฅ =๐ฟ2
96๐ธ๐๐ผ๐๐๐02 (2๐0
3 โ 24๐๐๐๐2๐0 + 48๐๐๐๐
3 โ 16๐๐๐3) +
16๐ฟ2๐๐๐3
96๐ธ๐๐ผ๐๐02
110
๐ฅ =๐ฟ2(2๐0
3 โ 24๐๐๐๐2๐0)
96๐ธ๐๐ผ๐๐๐02 +
16๐ฟ2๐๐๐3
96๐ธ๐๐02 (
3๐ โ 1
๐ผ๐๐+1
๐ผ๐)
๐ฅ =(๐0
3 โ 12๐๐๐๐2๐0 + 16๐๐๐๐
3)๐ฟ2
48๐ธ๐๐ผ๐๐๐02
Interestingly, this case simplifies to using a identical to the simply supported solution:
๐ฅ = ๐พ๐๐๐ฟ
2
12๐ธ๐๐ผ๐โฒ where ๐พ =
1
2 ; ๐ผ๐
โฒ =๐ผ๐๐
(1 โ ๐(๐๐๐ ๐๐โ )2) ; = 3 โ 2
๐๐๐
๐๐
111
Integration using CSA S806 / Razaqpurโs 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, ๐ผ๐๐ is used when the moment exceeds ๐๐๐. This is equivalent to
providing a tension stiffening factor of ๐ฝ = 0; this is drastically different than
๐ฝ = ๐๐๐/๐(๐ฅ) as provided by Bischoff and Gross (2011). This assumption is
reasonable when ๐๐๐๐ฅ ๐๐๐โ > 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. Razaqpurโs 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 Bischoffโs 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, ๐ผ๐๐, is used for the full cracked segments of
112
the member. In this appendix, ๐(๐ฅ) is as defined in Appendix E and the virtual load
function, ๐(๐ฅ), 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 ๐ฟ1, ๐ฟ2, ๐ฟ4, and ๐ฟ5 with this loading.
๐ฅ1 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐ฟ๐๐ฅ
๐ฟ
0
=๐ฟ2(๐ผ๐ฟ
3๐03 โ 3๐ผ๐ฟ๐๐๐
2๐0 โ 2๐๐๐3)
12๐ธ๐๐ผ๐๐๐ฟ(2 โ ๐ผ๐ฟ + ๐ผ๐ )2๐02
๐ฅ2 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐ฅ
๐ฟ
๐ฟ
=๐ฟ2๐๐๐
3
3๐ธ๐๐ผ๐(2 โ ๐ผ๐ฟ + ๐ผ๐ )2๐02
๐ฅ3 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐๐๐ฅ
๐ฟ2
๐ฟ
=๐ฟ2 (12๐ผ๐ฟ๐๐๐
2๐0 โ 8๐๐๐3 +๐0
3(8 โ 2๐ผ๐ฟ3 โ 6๐ผ๐ฟ
2 โ 3๐ผ๐ฟ2๐ผ๐ + 12๐ผ๐ + 6๐ผ๐
2 + ๐ผ๐ 3))
48๐ธ๐๐ผ๐๐๐(2 โ ๐ผ๐ฟ + ๐ผ๐ )2๐02
๐ฅ4 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐๐๐ฅ
๐ฟ
๐ฟ2
=๐ฟ2 (12๐ผ๐ ๐๐๐
2๐0 โ 8๐๐๐3 +๐0
3(8 โ 2๐ผ๐ 3 โ 6๐ผ๐
2 โ 3๐ผ๐ 2๐ผ๐ฟ + 12๐ผ๐ฟ + 6๐ผ๐ฟ
2 + ๐ผ๐ฟ3))
48๐ธ๐๐ผ๐๐๐(2 + ๐ผ๐ฟ โ ๐ผ๐ )2๐02
๐ฅ5 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐ฅ
๐ฟ
๐ฟ
=๐ฟ2๐๐๐
3
3๐ธ๐๐ผ๐(2 + ๐ผ๐ฟ โ ๐ผ๐ )2๐02
๐ฅ6 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐ ๐๐ฅ
๐ฟ
๐ฟ
=๐ฟ2(๐ผ๐
3๐03 โ 3๐ผ๐ ๐๐๐
2๐0 โ 2๐๐๐3)
12๐ธ๐๐ผ๐๐๐ (2 + ๐ผ๐ฟ โ ๐ผ๐ )2๐02
๐ฅ1&2 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐ฅ
๐ฟ
0
=๐ฟ2(๐ผ๐ฟ
3๐03 โ 3๐ผ๐ฟ๐๐๐
2๐0 + 2๐๐๐3)
12๐ธ๐๐ผ๐(2 โ ๐ผ๐ฟ + ๐ผ๐ )2๐02
๐ฅ5&6 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐ฅ
๐ฟ
๐ฟ
=๐ฟ2(๐ผ๐
3๐03 โ 3๐ผ๐ ๐๐๐
2๐0 + 2๐๐๐3)
12๐ธ๐๐ผ๐(2 + ๐ผ๐ฟ โ ๐ผ๐ )2๐02
Total midspan deflection:
๐ฅ = ๐ฅ1 + ๐ฅ2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5 + ๐ฅ6 (If both ends and midspan are cracked)
๐ฅ = ๐ฅ1&2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5&6 (If only midspan is cracked)
๐ฅ = ๐ฅ1&2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5 + ๐ฅ6 (If only right end and midspan are cracked)
๐ฅ = ๐ฅ1 + ๐ฅ2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5&6 (If only left end and midspan are cracked)
114
S806 method for equal third-point loading:
Figure L.2 โ 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 ๐ฟ1, ๐ฟ2, ๐ฟ4, and ๐ฟ5 with this loading.
๐ฅ1 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐ฟ๐๐ฅ
๐ฟ
0
=๐ฟ2(๐ผ๐ฟ
3๐03 โ 3๐ผ๐ฟ๐๐๐
2๐0 โ 2๐๐๐3)
12๐ธ๐๐ผ๐๐๐ฟ(32 โ ๐ผ๐ฟ + ๐ผ๐ )2๐02
๐ฅ2 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐ฅ
๐ฟ
๐ฟ
=๐ฟ2๐๐๐
3
3๐ธ๐๐ผ๐(3 โ ๐ผ๐ฟ + ๐ผ๐ )2๐02
115
๐ฅ3๐ด = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐๐๐ฅ
๐ฟ3
๐ฟ
=๐ฟ2(81๐ผ๐ฟ๐๐๐
2๐0 โ 54๐๐๐3)
324๐ธ๐๐ผ๐๐๐(3 โ ๐ผ๐ฟ + ๐ผ๐ )2๐02
+๐ฟ2๐0
3(54 โ 36๐ผ๐ฟ2 โ 20๐ผ๐ฟ
3 โ 12๐ผ๐ฟ2๐ผ๐ + 18๐ผ๐ฟ๐ผ๐ + 3๐ผ๐ฟ๐ผ๐
2)
324๐ธ๐๐ผ๐๐๐(3 โ ๐ผ๐ฟ + ๐ผ๐ )2๐02
+๐ฟ2๐0
3(27๐ผ๐ฟ โ 54๐ผ๐ + 18๐ผ๐ 2 + 2๐ผ๐
3)
324๐ธ๐๐ผ๐๐๐(3 โ ๐ผ๐ฟ + ๐ผ๐ )2๐0
2
๐ฅ3๐ต = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐๐๐ฅ
๐ฟ2
๐ฟ3
=๐ฟ2๐0(26๐ผ๐ฟ + 19๐ผ๐ + 45)
1296๐ธ๐๐ผ๐๐๐
๐ฅ3๐ถ = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐๐๐ฅ
2๐ฟ3
๐ฟ2
=๐ฟ2๐0(19๐ผ๐ฟ + 26๐ผ๐ + 45)
1296๐ธ๐๐ผ๐๐๐
๐ฅ3๐ท = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐๐๐ฅ
๐ฟ
2๐ฟ3
=๐ฟ2(324๐ผ๐ฟ๐๐๐(๐ผ๐ โ 2๐ผ๐ฟ)๐0
2 + 81(๐ผ๐ โ 4๐ผ๐ฟ)๐๐๐2๐0 โ 54๐๐๐
3)
324๐ธ๐๐ผ๐๐๐(3 + ๐ผ๐ฟ โ ๐ผ๐ )2๐02
+๐ฟ2๐0
3(54 + 18๐ผ๐ฟ2 โ 430๐ผ๐ฟ
3 + 327๐ผ๐ฟ2๐ผ๐ + 18๐ผ๐ฟ๐ผ๐ โ 12๐ผ๐ฟ๐ผ๐
2)
324๐ธ๐๐ผ๐๐๐(3 + ๐ผ๐ฟ โ ๐ผ๐ )2๐02
+๐ฟ2๐0
3(54๐ผ๐ฟ + 27๐ผ๐ โ 36๐ผ๐ 2 โ 20๐ผ๐
3)
324๐ธ๐๐ผ๐๐๐(3 + ๐ผ๐ฟ โ ๐ผ๐ )2๐02
๐ฅ5 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐ฅ
๐ฟ
๐ฟ
=๐ฟ2(12๐ผ๐ฟ๐0
2 +๐๐๐2 โ 6๐ผ๐ฟ๐ผ๐ ๐0
2)๐๐๐
3๐ธ๐๐ผ๐(3 + ๐ผ๐ฟ โ ๐ผ๐ )2๐02
116
๐ฅ6 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐ ๐๐ฅ
๐ฟ
๐ฟ
=๐ฟ2 ((16๐ผ๐ฟ
3 โ 12๐ผ๐ฟ2๐ผ๐ + ๐ผ๐
3)๐03 โ 3(๐ผ๐ โ 4๐ผ๐ฟ)๐๐๐
2๐0)
12๐ธ๐๐ผ๐๐๐ (3 + ๐ผ๐ฟ โ ๐ผ๐ )2๐02
+๐ฟ2(12(๐ผ๐ฟ๐ผ๐ โ 2๐ผ๐ฟ
2)๐02๐๐๐ โ 2๐๐๐
3)
12๐ธ๐๐ผ๐๐๐ (3 + ๐ผ๐ฟ โ ๐ผ๐ )2๐02
๐ฅ1&2 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐ฅ
๐ฟ
0
=๐ฟ2(๐ผ๐ฟ
3๐03 โ 3๐ผ๐ฟ๐๐๐
2๐0 + 2๐๐๐3)
12๐ธ๐๐ผ๐(3 โ ๐ผ๐ฟ + ๐ผ๐ )2๐0
2
๐ฅ5&6 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐ฅ
๐ฟ
๐ฟ
=๐ฟ2 ((16๐ผ๐ฟ
3 โ 12๐ผ๐ฟ2๐ผ๐ + ๐ผ๐
3)๐03 โ 3(๐ผ๐ โ 4๐ผ๐ฟ)๐๐๐
2๐0)
12๐ธ๐๐ผ๐(3 + ๐ผ๐ฟ โ ๐ผ๐ )2๐02
+๐ฟ2(12(2๐ผ๐ฟ
2 โ ๐ผ๐ฟ๐ผ๐ )๐02๐๐๐ + 2๐๐๐
3)
12๐ธ๐๐ผ๐(3 + ๐ผ๐ฟ โ ๐ผ๐ )2๐02
Total midspan deflection:
Note these results are only valid if ๐(๐ฟ 3โ ) > ๐๐๐ and ๐(2๐ฟ 3โ ) > ๐๐๐.
If both ends and midspan are cracked:
๐ฅ = ๐ฅ1 + ๐ฅ2 + ๐ฅ3๐ด + ๐ฅ3๐ต + ๐ฅ3๐ถ + ๐ฅ3๐ท + ๐ฅ5 + ๐ฅ6
If only midspan is cracked:
๐ฅ = ๐ฅ1&2 + ๐ฅ3๐ด + ๐ฅ3๐ต + ๐ฅ3๐ถ + ๐ฅ3๐ท + ๐ฅ5&6
If only right end and midspan are cracked:
๐ฅ = ๐ฅ1&2 + ๐ฅ3๐ด + ๐ฅ3๐ต + ๐ฅ3๐ถ + ๐ฅ3๐ท + ๐ฅ5 + ๐ฅ6
If only left end and midspan are cracked:
๐ฅ = ๐ฅ1 + ๐ฅ2 + ๐ฅ3๐ด + ๐ฅ3๐ต + ๐ฅ3๐ถ + ๐ฅ3๐ท + ๐ฅ5&6
117
S806 method for a uniformly distributed load:
Figure L.3 โ 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 ๐ฟ1, ๐ฟ2, ๐ฟ4, and ๐ฟ5 with this loading.
๐ฟ3 = ๐ฟ 2โ ; ๐ฟ๐ 4 = ๐ฟ โ ๐ฟ4 ; ๐ฟ๐ 5 = ๐ฟ โ ๐ฟ5.
๐ฅ1 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐ฟ๐๐ฅ
๐ฟ
0
=๐0๐ฟ
2
2๐ธ๐๐ผ๐๐๐ฟ((4 โ ๐ผ๐ฟ + ๐ผ๐ )๐ฟ1
3
3๐ฟ3 โ
๐ฟ14
๐ฟ4+๐ผ๐ฟ๐ฟ1
2
2๐ฟ2)
118
๐ฅ2 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐ฅ
๐ฟ
๐ฟ
=๐0๐ฟ
2
2๐ธ๐๐ผ๐((4 โ ๐ผ๐ฟ + ๐ผ๐ ) (
๐ฟ23 โ ๐ฟ1
3
3๐ฟ3) + ๐ผ๐ฟ (
๐ฟ22 โ ๐ฟ1
2
2๐ฟ2)
โ (๐ฟ2
4 โ ๐ฟ14
๐ฟ4))
๐ฅ3 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐๐๐ฅ
๐ฟ2
๐ฟ
=๐0๐ฟ
2
2๐ธ๐๐ผ๐๐๐((4 โ ๐ผ๐ฟ + ๐ผ๐ ) (
๐ฟ33 โ ๐ฟ2
3
3๐ฟ3) + ๐ผ๐ฟ (
๐ฟ32 โ ๐ฟ2
2
2๐ฟ2)
โ (๐ฟ3
4 โ ๐ฟ24
๐ฟ4))
๐ฅ4 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐๐๐ฅ
๐ฟ
๐ฟ2
=๐0๐ฟ
2
2๐ธ๐๐ผ๐๐๐((4 โ ๐ผ๐ + ๐ผ๐ฟ) (
๐ฟ33 โ ๐ฟ๐ 4
3
3๐ฟ3) + ๐ผ๐ (
๐ฟ32 โ ๐ฟ๐ 4
2
2๐ฟ2)
โ (๐ฟ3
4 โ ๐ฟ๐ 44
๐ฟ4))
๐ฅ5 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐ฅ
๐ฟ
๐ฟ
=๐0๐ฟ
2
2๐ธ๐๐ผ๐((4 โ ๐ผ๐ + ๐ผ๐ฟ) (
๐ฟ๐ 43 โ ๐ฟ๐ 5
3
3๐ฟ3) + ๐ผ๐ (
๐ฟ๐ 42 โ ๐ฟ๐ 5
2
2๐ฟ2)
โ (๐ฟ๐ 44 โ ๐ฟ๐ 5
4
๐ฟ4))
119
๐ฅ6 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐ ๐๐ฅ
๐ฟ
๐ฟ
=๐0๐ฟ
2
2๐ธ๐๐ผ๐๐๐ ((4 โ ๐ผ๐ฟ + ๐ผ๐ )๐ฟ๐ 5
3
3๐ฟ3 โ
๐ฟ๐ 54
๐ฟ4+๐ผ๐ฟ๐ฟ๐ 5
2
2๐ฟ2)
๐ฅ1&2 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐ฅ
๐ฟ
0
=๐0๐ฟ
2
2๐ธ๐๐ผ๐((4 โ ๐ผ๐ฟ + ๐ผ๐ )๐ฟ2
3
3๐ฟ3 โ
๐ฟ24
๐ฟ4+๐ผ๐ฟ๐ฟ2
2
2๐ฟ2)
๐ฅ5&6 = โซm(๐ฅ)๐(๐ฅ)
E ๐ผ๐๐๐ฅ
๐ฟ
๐ฟ
=๐0๐ฟ
2
2๐ธ๐๐ผ๐((4 โ ๐ผ๐ + ๐ผ๐ฟ)๐ฟ๐ 4
3
3๐ฟ3 โ
๐ฟ๐ 44
๐ฟ4+๐ผ๐ ๐ฟ๐ 4
2
2๐ฟ2)
Total midspan deflection (note these results are only valid if ๐๐ > ๐๐๐ :
๐ฅ = ๐ฅ1 + ๐ฅ2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5 + ๐ฅ6 (If both ends and midspan are cracked)
๐ฅ = ๐ฅ1&2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5&6 (If only midspan is cracked)
๐ฅ = ๐ฅ1&2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5 + ๐ฅ6 (If only right end and midspan are cracked)
๐ฅ = ๐ฅ1 + ๐ฅ2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5&6 (If only left end and midspan are cracked)
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:
๐ผ =๐โ3
12 ๐ธ = 4500โ๐๐โฒ ๐1๐๐ฟ =
๐1๐๐ฟ๐ฟ
4 ๐2๐๐ฟ =
๐2๐๐ฟ๐ฟ
6 ๐๐๐ท๐ฟ =
๐ค๐๐ท๐ฟ๐ฟ2
8
For a midspan point load:
โ(๐ฅ) =
{
๐๐ (๐ฅ โค๐ฟ
2) โถ
๐1๐๐ฟ ๐ฅ(3๐ฟ2 โ 4๐ฅ2)
48๐ธ๐ผ
๐๐๐ ๐ โถ ๐1๐๐ฟ(๐ฟ โ ๐ฅ)(3๐ฟ2 โ 4(๐ฟ โ ๐ฅ)2)
48๐ธ๐ผ }
For two point load at third points:
โ(๐ฅ) =
{
๐๐ (๐ฅ โค๐ฟ
3) โถ
2๐2๐๐ฟ ๐ฅ (59๐ฟ2 โ ๐ฅ2) + ๐2๐๐ฟ๐ฅ (
89๐ฟ2 โ ๐ฅ2)
36๐ธ๐ผ
๐๐ (๐ฟ
3< ๐ฅ โค
2๐ฟ
3) โถ
๐2๐๐ฟ(๐ฟ โ ๐ฅ) (8๐ฟ2
9 โ (๐ฟ โ ๐ฅ)2) + ๐2๐๐ฟ๐ฅ (89 ๐ฟ
2 โ ๐ฅ2)
36๐ธ๐ผ
๐๐๐ ๐ โถ ๐2๐๐ฟ(๐ฟ โ ๐ฅ) (
8๐ฟ2
9 โ (๐ฟ โ ๐ฅ)2) + 2๐2๐๐ฟ(๐ฟ โ ๐ฅ) (5๐ฟ2
9 โ (๐ฟ โ ๐ฅ)2)
36๐ธ๐ผ }
For a uniformly distributed load:
โ(๐ฅ)๐๐๐ ๐๐ท๐ฟ =๐ค๐๐ท๐ฟ ๐ฅ(๐ฅ
3 โ 2๐ฟ๐ฅ2 + ๐ฟ3)
24๐ธ๐ผ
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 MPaฮ lef t ฮright
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 ฮ(x) for 1 PL ฮ(x) for 2 PL ฮ(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
ฮ(D
efl
ecti
on
)
x (Position)
ฮ(x) for 1 PL
ฮ(x) for 2 PL
ฮ(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, ๐๐ is set to โ๐0. For case 3 and case 4, ๐๐ฟ
equals ๐๐ and they are set to โ๐0/2 and โ2๐0/3 for those cases, respectively. See
List of Symbols for symbol definitions. The spreadsheet equations include:
๐ผ๐ =๐โ3
12 ; ๐ธ๐ = 4500โ๐๐โฒ ; ๐๐๐ =
๐๐๐ผโโ2
=๐โ2โ๐๐โฒ
10 ; ๐0 =
๐ค๐๐ท๐ฟ๐ฟ2
8
โ๐๐ฟ(๐ฅ) =
๐๐ฟ๐ฅ(๐ฅ2 โ 3๐ฟ๐ฅ + 2๐ฟ2)
6๐ฟ๐ธ๐๐ผ๐ for
โ๐๐ (๐ฅ) =
๐๐ (๐ฟ โ ๐ฅ)((๐ฟ โ ๐ฅ)2 โ 3๐ฟ(๐ฟ โ ๐ฅ) + 2๐ฟ2)
6๐ฟ๐ธ๐๐ผ๐ for
โ๐๐ท๐ฟ(๐ฅ) =๐ค๐๐ท๐ฟ ๐ฅ(๐ฅ
3 โ 2๐ฟ๐ฅ2 + ๐ฟ3)
24๐ธ๐๐ผ๐ for
โ1(๐ฅ) = โ๐๐ท๐ฟ(๐ฅ)
โ2(๐ฅ) = โ๐๐ (๐ฅ) + โ๐๐ท๐ฟ(๐ฅ)
โ3(๐ฅ) = โ4(๐ฅ) = โ๐๐ฟ(๐ฅ) + โ๐๐
(๐ฅ) + โ๐๐ท๐ฟ(๐ฅ)
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 ฮM_R ฮUDL ฮ2(x) ฮM_L ฮM_R ฮUDL ฮ3(x) ฮM_L ฮM_R ฮUDL ฮ4(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
ฮ1(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
ฮ(D
efl
ecti
on
)
x (Position)
ฮ with ML = MR = 0
ฮ with ML = 0 ; MR = -Mm
ฮ with ML = MR = -Mm
ฮ with ML = MR = -2Mmฮ with ๐๐ฟ = ๐๐ =โ2๐๐
ฮ with ๐๐ฟ = ๐๐ =โ๐๐
ฮ with ๐๐ฟ = 0 ; ๐๐ = โ๐๐
ฮ with ๐๐ฟ = ๐๐ =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
Bransonโs (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 ๐๐โฒ 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 ๐๐ = โ๐๐ฟ and ๐๐ = 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:
๐๐ = 0.65 ; ๐๐โฒ = 36 MPa ; ๐ธ๐ = 4500โ๐๐โฒ = 27 GPa
ฮฑ1 = 0.85 โ 0.0015๐๐โฒ = 0.796 ; ฮฒ1 = 0.97 โ 0.0025๐๐
โฒ = 0.880
๐๐ = 0.85 ; ๐๐ฆ = 400 MPa ; ๐ธ๐ = 200 GPa ; ๐ = ๐ธ๐ /๐ธ๐ = 7.407
The example span length, ๐ฟ, and simply supported example load, ๐ค0, for this member
are arbitrarily defined as:
๐ฟ = 10.0 m ; ๐ค0 = 10 kN mโ ; therefore ๐0,0 =๐ค0๐ฟ
2
8= 125 kNm
The uniformly distributed load used to produce member properties is ๐ค0. The simply
supported midspan bending moment, ๐0,0, applies to each set of compared members.
Set the following ratios for this example beam:
๐๐๐
๐0,0= 0.7 ; ๐ = 0.5 โ ; ๐ = 0.85 โ
Produce a member concrete size using above information and A23.3 (CSA 2004)
definitions:
๐๐๐ = 0.7๐0,0 = 87.5 kNm ; ๐๐ = 0.6โ๐๐โฒ = 3.6 MPa
๐๐๐ =๐๐๐โ
2
6= 0.3 (โ3) ; โ = โ
87,500,000
0.3
= 663 mm
๐ = 0.5 โ = 332 mm ; ๐ = 0.85 โ = 564 mm ; ๐ผ๐ =๐โ3
12= 8.06 x 109 mm4
126
Next, a service load to moment resistance ratio is defined as follows:
๐๐ /๐๐ = 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
๐๐๐๐ฅ = ๐๐ = ๐0,0. The reinforcing steel design at midspan is:
๐พ๐ =๐๐
๐๐2=๐0,0
๐๐2๐๐
๐๐ = 1.87 MPa ; ๐พ๐ = (1 โ
๐๐๐ ๐๐ฆ
2๐ผ1๐๐๐๐โฒ)๐๐๐ ๐๐ฆ
๐ =๐ผ1๐๐๐๐
โฒ โ โ(๐ผ1๐๐๐๐โฒ)2 โ 2๐ผ1๐๐๐๐โฒ๐พ๐๐๐ ๐๐ฆ
= 0.00580 ; ๐ด๐ = ๐๐๐ = 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 ๐๐ > ๐๐. The GFRP reinforcement design for midspan is:
๐๐๐ข = 690 MPa ; ๐๐ = 0.75 ; ํ๐๐ข = 0.0035 ; ๐ธ๐ = 44 GPa ; ๐ = 1.63 ; ๐๐ ๐๐โ = 2.0
๐ =๐
๐ฝ1โโ
๐2
๐ฝ12 โ
2๐๐
๐ผ1๐ฝ12๐๐๐๐โฒ๐
= 144.9 mm ; ๐ด๐ =๐ผ1๐ฝ1๐๐๐๐
โฒ๐๐2
๐๐ํ๐๐ข(๐ โ ๐)๐ธ๐= 2555 mm2
๐ =๐ด๐
๐๐= 0.014 < ๐๐๐ =
๐ผ1๐ฝ1๐๐๐๐โฒํ๐๐ข
๐๐๐๐๐ข (ํ๐๐ข +๐๐๐ข๐ธ๐)
= 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 ๐๐ ๐ข๐ = 80 kNm and calculate:
๐ = โ๐2๐2 + 2๐๐ โ ๐๐ = 0.202
127
๐๐(๐๐ฟ๐) =๐๐
๐ด๐ (1 โ๐3) ๐
= 81 MPa โค ๐๐๐ฟ๐ = 0.25๐๐๐ข = 172 MPa (therefore ok)
๐๐(๐ ๐ข๐ ) =๐๐ ๐ข๐
๐ด๐ (1 โ๐3)๐
= 52 MPa โค ๐๐,๐ = 0.002๐ธ๐ = 88 MPa (therefore ok)
For FRP, ๐ = ๐ธ๐/๐ธ๐ and ๐ด๐ (not ๐ด๐ ) 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):
๐๐ = (โ2๐ (๐
๐๐ด๐ ) + 1 โ 1) (
๐
๐๐ด๐ )โ = 142.8 mm
๐ผ๐๐ =๐(๐๐)3
3+ ๐๐ด๐ (๐ โ ๐๐)2 = 1.75 x 109 mm4
Deflection results using different methods can now be compared. Some constant
stiffness methods assume the ๐ธ๐ผ term is equal to ๐ธ๐๐ผ๐, ๐ธ๐๐ผ๐๐, ๐ธ๐๐ผ๐ (๐ต๐๐๐๐ ๐๐), or
๐ธ๐๐ผ๐ (๐ต๐๐ ๐โ๐๐๐)โฒ . 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).
โ๐ข๐๐๐๐๐๐๐๐=5๐ค๐ฟ4
384๐ธ๐๐ผ๐= 5.98 mm ; โ๐๐ข๐๐๐ฆ ๐๐๐๐๐๐๐=
5๐ค๐ฟ4
384๐ธ๐๐ผ๐๐= 27.64 mm
Referring to results from Bransonโs (1965) work:
๐ผ๐ = (๐๐๐
๐๐)3
๐ผ๐ + [1 โ (๐๐๐
๐๐)3
] ๐ผ๐๐ = 3.91 x 109 mm4 ; โ=5๐ค๐ฟ4
384๐ธ๐๐ผ๐= 12.3 mm
Referring to results from Bischoffโs (Bischoff and Gross 2011) work:
128
๐ = 1 โ๐ผ๐๐๐ผ๐
= 0.783 ; ๐ = 1 โ โ1 โ๐๐๐
๐๐= 0.452
=1.6๐3 โ 0.6๐4
(๐๐๐
๐๐)2 + 2.4 ln(2 โ ๐) = 1.30
๐ผ๐โฒ =
๐ผ๐๐
1 โ ๐ (๐๐๐
๐๐)2 = 3.48 x 109 mm4 ; โ=
5๐ค๐ฟ4
384๐ธ๐๐ผ๐โฒ=5๐๐๐ฟ
2
48๐ธ๐๐ผ๐โฒ= 13.85 mm
Next, use the analytical equations per Appendix E (see List of Symbols where required):
๐ผ๐ฟ = ๐๐ฟ ๐0โ = 0 and ๐ผ๐ = ๐๐ ๐0โ = 0 (Simply Supported Member)
๐ฟ2 = (4 โ ๐ผ๐ฟ + ๐ผ๐ โโ(4 โ ๐ผ๐ฟ + ๐ผ๐ )2 โ 16๐ผ๐๐ + 16๐ผ๐ฟ)๐ฟ
8= 2261 mm
๐ฟ4 = (4 โ ๐ผ๐ฟ + ๐ผ๐ +โ(4 โ ๐ผ๐ฟ + ๐ผ๐ )2 โ 16๐ผ๐๐ + 16๐ผ๐ฟ)๐ฟ
8= 7739 mm
๐ฅ1+2 =๐0๐ฟ
2
2๐ธ๐๐ผ๐((4 โ ๐ผ๐ฟ + ๐ผ๐ )๐ฟ2
3
3๐ฟ3 โ
๐ฟ24
๐ฟ4+๐ผ๐ฟ๐ฟ2
2
2๐ฟ2) = 0.37 mm
๐ฅ3 = 6.56 mm (see Appendix E for full equation)
๐ฅ4 = 6.56 mm (see Appendix E for full equation)
๐ฅ5+6 =๐0๐ฟ
2
2๐ธ๐๐ผ๐((4 โ ๐ผ๐ + ๐ผ๐ฟ)๐ฟ๐ โ4
3
3๐ฟ3 โ
๐ฟ๐ โ44
๐ฟ4+๐ผ๐ ๐ฟ๐ โ4
2
2๐ฟ2) = 0.37 mm
๐ฅ = ๐ฅ1+2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5+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, ๐0, will offset the
129
negative bending moment(s) and the maximum positive bending moment will be
maintained. Variables ๐ผ๐ฟ = ๐๐ฟ ๐0โ and ๐ผ๐ = ๐๐ ๐0โ 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 โฅ ๐ผ๐ฟ โฅ โ3 and ๐ผ๐ = 0.
Calculations are provided for ๐ผ๐ฟ = โ1 and ๐ผ๐ = 0, so ๐๐ฟ = โ๐๐๐๐ฅ and ๐๐ = 0. The
calculation of the uniform load required, ๐ค, is determined based on common structural
analysis equations and the end-moment to total moment ratio as follows:
๐ผ๐ฟ =๐๐ฟ
๐0= โ0.686 ; ๐ผ๐ =
๐๐
๐0= 0 ; ๐ค =
๐ค0
1 +๐ผ๐ฟ2 +
๐ผ๐ 2 +
(๐ผ๐ฟ โ ๐ผ๐ )2
16
= 14.58 N
mm
Similarly, to maintain the same midspan moment with a centered point load:
๐ =๐0
1 +๐ผ๐ฟ2 +
๐ผ๐ 2
Likewise, to maintain the same maximum moment with equal third-point loading:
๐ =๐0
1 +๐ผ๐ฟ3 +
2๐ผ๐ 3
(requires โ๐ผ๐ฟ > โ๐ผ๐ )
For the example beam with a uniformly distributed beam and end continuous, the load
๐ค is then used to determine the total service, end, midspan, and maximum moments:
๐0 =๐ค๐ฟ2
8= 182 kNm ; ๐๐ฟ = ๐ผ๐ฟ๐0 = โ125 kNm ; ๐๐ = ๐ผ๐ ๐0 = 0 kNm
๐๐ = ๐0 +(๐๐ฟ +๐๐ )
2= 120 kNm ; ๐๐๐๐ฅ = ๐๐ +
(๐๐ฟ +๐๐ )2
16๐0= 125 kNm
130
Proceed to calculate the properties for this continuous member. As intended, the
maximum positive moment, concrete dimensions, ๐๐๐, ๐ผ๐, positive ๐ด๐ , ๐ผ๐๐, and such
remain the same as the simply supported example. For both positive and negative
bending, again use ๐๐ = ๐๐. The calculations preformed above using ๐0,0 must now
use ๐๐๐๐ฅ for the continuous member. At the right end of the member, where ๐๐ = 0,
simply use ๐ผ = ๐ผ๐ = 8.06 x 109 mm4. Again determine the cross-section as follows:
๐๐๐ = 87.5 kNm ; โ = 663 mm ; ๐ = 332 mm ; ๐ผ๐ = 8.06 x 109 mm4
๐ = 564 mm ; ๐พ๐ = 1.87 MPa ; ๐ = 0.00580 ; ๐ด๐ +๐ฃ๐ = 1085 mm2
๐๐ = 142.8 mm ; ๐ผ๐๐๐ = 1.75 x 109 mm4
Because the left end negative moment is the same magnitude as the maximum positive
moment, as ๐๐๐๐ฅ = โ๐๐ฟ and ๐๐ = ๐๐, the same properties occur at the left end:
๐พ๐ โ๐ฃ๐ =โ๐๐ฟ
๐๐2๐๐
๐๐ = 1.87 MPa ; ๐ = 0.00580 ; ๐ด๐ โ๐ฃ๐ = 1085 mm2
๐๐ = 142.8 mm ; ๐ผ๐๐๐ฟ = 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 ๐๐๐/๐0,0 = 0.7, increase this value as required, and revise subsequent calculations.
As indicated in the Concrete Design Handbook (CAC 2005) and in Appendix A:
โ= ๐พ5๐๐๐ฟ
2
48๐ธ๐ผ where ๐พ = 1.2 โ 0.2
๐0
๐๐ ; hence ๐พ = 0.896
131
Calculate midspan deflection using different assumptions for ๐ธ๐ผ:
โ๐ข๐๐๐๐๐๐๐๐= ๐พ5๐๐๐ฟ
4
48๐ธ๐๐ผ๐= 5.36 mm ; โ๐๐ข๐๐๐ฆ ๐๐๐๐๐๐๐= ๐พ
5๐๐๐ฟ4
48๐ธ๐๐ผ๐๐= 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 0.85๐ผ๐๐ + 0.15๐ผ๐๐ฟ = ๐ผ๐.
Therefore the result using Bransonโs equation is:
๐ผ๐ = (๐๐๐
๐๐๐๐ฅ)3
๐ผ๐ + [1 โ (๐๐๐
๐๐๐๐ฅ)3
] ๐ผ๐๐ = 3.91x109 mm4 ; โ= ๐พ5๐๐๐ฟ
4
48๐ธ๐๐ผ๐= 11.04 mm
Referring to results using Bischoffโs equations, base the ๐ผ๐โฒ calculation on the maximum
positive moment, so the results are also the same as with the simply supported example:
๐ = 0.783 ; ๐ = 0.452 ; = 1.299 ; ๐ผ๐โฒ = 3.48 x 109 mm4
Unlike Bischoffโs 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 ๐๐ = โ๐๐ฟ = โ๐๐ 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.
โ= ๐พ5๐๐๐ฟ
4
48๐ธ๐๐ผ๐โฒ= 11.87 mm
Again obtaining results using exact analytical equations from Appendix E:
๐ฟ1 = (4 โ ๐ผ๐ฟ + ๐ผ๐ โโ(4 โ ๐ผ๐ฟ + ๐ผ๐ )2 + 16๐๐๐/๐๐ + 16๐ผ๐ฟ)๐ฟ
8= 457 mm
132
๐ฟ2 = (4 โ ๐ผ๐ฟ + ๐ผ๐ โโ(4 โ ๐ผ๐ฟ + ๐ผ๐ )2 โ 16๐๐๐/๐๐ + 16๐ผ๐ฟ)๐ฟ
8= 3588 mm
๐ฟ4 = (4 โ ๐ผ๐ฟ + ๐ผ๐ +โ(4 โ ๐ผ๐ฟ + ๐ผ๐ )2 โ 16๐๐๐/๐๐ + 16๐ผ๐ฟ)๐ฟ
8= 8125 mm
๐ฅ1 = โ0.04 mm (see Appendix E for full equation)
๐ฅ2 = 0.50 mm (see Appendix E for full equation)
๐ฅ3 = 3.24 mm (see Appendix E for full equation)
๐ฅ4 = 7.59 mm (see Appendix E for full equation)
๐ฅ5+6 = 0.25 mm (see Appendix E for full equation)
๐ฅ = ๐ฅ1 + ๐ฅ2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5+6 = 11.54 mm
For comparison, calculate numerical integration results for this example member (using
the method shown in Appendix K):
Using 10 segments: ๐ฅ = โซ๐(๐ฅ)๐(๐ฅ)
๐ธ๐๐ผ๐(๐ฅ)๐๐ฅ
๐ฟ
0
โโ ๐ฅ๐๐=10
๐=1= 11.75 mm
Using 100 segments: ๐ฅ = โซ๐(๐ฅ)๐(๐ฅ)
๐ธ๐๐ผ๐(๐ฅ)๐๐ฅ
๐ฟ
0
โโ ๐ฅ๐๐=100
๐=1= 11.54 mm
Using the S806 method, introduced in Appendix G:
๐ฟ1 = 457 mm ; ๐ฟ2 = 3588 mm ; ๐ฟ3 = 5000 mm ; ๐ฟ4 = 8125 mm
๐ฅ1 = โ0.11 mm (see Appendix G for full equation)
๐ฅ2 = 0.50 mm (see Appendix G for full equation)
๐ฅ3 = 6.88 mm (see Appendix G for full equation)
๐ฅ4 = 13.44 mm (see Appendix G for full equation)
๐ฅ5+6 = 0.25 mm (see Appendix G for full equation)
๐ฅ = ๐ฅ1 + ๐ฅ2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5+6 = 20.96 mm
133
As indicated throughout the report, failing to account for tension stiffening is often
overly conservative. The error using Bransonโs equation for this particular situation is
unconservative but is reasonably small because 2 < ๐ผ๐ ๐ผ๐๐โ < 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 ๐ฅ % difference
๐ผ๐ 5.36 โ46.4%
๐ผ๐๐ 24.75 +114% ๐ผ๐ ๐ต๐๐๐๐ ๐๐ 11.04 โ4.3%
๐ผ๐ ๐ต๐๐ ๐โ๐๐๐โฒ 11.87 +2.9%
๐ผ๐(๐ฅ)๐๐๐ง๐๐๐๐ข๐ 20.96 +82%
๐ผ๐(๐ฅ)๐๐ข๐๐๐๐๐๐๐:10 11.75 +1.8%
๐ผ๐(๐ฅ)๐๐ข๐๐๐๐๐๐๐:100 11.543 +0.001%
๐ผ๐(๐ฅ)๐๐๐๐๐ฆ๐ก๐๐๐๐ 11.542 โ
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, ๐, 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 ๐ โฅ 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.
๐ฅ = โซ๐(๐ฅ)๐(๐ฅ)
๐ธ๐๐ผ๐(๐ฅ)๐๐ฅ
๐ฟ
0
โโ [๐(๐ฅ๐)๐(๐ฅ๐)
๐ธ๐๐ผ๐(๐ฅ๐)+๐(๐ฅ๐โ1)๐(๐ฅ๐โ1)
๐ธ๐๐ผ๐(๐ฅ๐โ1)]๐ฟ
2๐
๐=๐
๐=1
where ๐ฅ๐ =๐๐ฟ
๐ and ๐ฅ๐โ1 =
๐ โ 1
๐๐ฟ
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:
๐๐ = 0.65 ; ๐๐โฒ = 36 MPa ; ๐ธ๐ = 27 GPa ; ฮฑ1 = 00.796 ; ฮฒ1 = 0.880
๐๐ = 0.85 ; ๐๐ฆ = 400 MPa ; ๐ธ๐ = 200 GPa ; ๐ =๐ธ๐ ๐ธ๐
= 7.407 ; ๐๐ = 3.6 MPa
๐ฟ = 10 m ; ๐ค = 14.58 ; ๐0 = 182 kNm ; ๐๐ = 0 ; ๐๐ฟ
๐0= โ0.686 ; ๐๐ฟ = โ125 kNm
๐๐ = ๐0 +(๐๐ฟ +๐๐ )
2= 120 kNm ; ๐๐๐๐ฅ = ๐๐ +
(๐๐ฟ +๐๐ )2
16๐0= 125 kNm
๐๐
๐๐= 0.635 ;
๐0
๐๐๐= 1.43 ; ๐๐๐ = 87.5 kNm ; โ = 663 mm
๐ = โ 2โ = 332 mm ; ๐ = 0.85(โ) = 564 mm ; ๐ผ๐ = 8.06 x 109 mm4
135
At the location of maximum positive moment (near midspan):
๐พ๐ =๐๐๐๐ฅ
๐๐2๐๐
๐๐ = 1.87 MPa ; ๐ = 0.00580 ; ๐ด๐ +๐ฃ๐ = 1085 mm2
๐๐ = 142.8 mm ; ๐ผ๐๐๐ = 1.75 x 109 mm4 ; ๐๐ = 1 โ๐ผ๐๐๐๐ผ๐
= 0.783
At member span ends, the right end is uncracked and the design at the left end is:
๐พ๐ =๐๐ฟ
๐๐2๐๐
๐๐ = 1.87 MPa ; ๐ = 0.00580 ; ๐ด๐ โ๐ฃ๐ = 1085 mm2
๐๐ = 142.8 mm ; ๐ผ๐๐๐ฟ = 1.75 x 109 mm4 ; ๐๐ฟ = 1 โ๐ผ๐๐๐ฟ๐ผ๐
= 0.783
The same functions as for analytical integration are required for numerical integration:
๐(๐ฅ) = ๐๐ฟ + (4๐0 โ๐๐ฟ +๐๐ )๐ฅ
๐ฟโ 4๐0
๐ฅ2
๐ฟ2
๐๐๐
{
๐ฅ <๐ฟ
2 and ๐(๐ฅ) < โ๐๐๐ ๐ผ๐(๐ฅ) =
๐ผ๐๐๐ฟ
1 โ ๐๐ฟ (๐๐๐
๐(๐ฅ))2
โ๐๐๐ โค ๐(๐ฅ) โค ๐๐๐ ๐ผ๐(๐ฅ) = ๐ผ๐
๐(๐ฅ) > ๐๐๐ ๐ผ๐(๐ฅ) =๐ผ๐๐๐
1 โ ๐๐ (๐๐๐
๐(๐ฅ))2
For midspan deflection, set the virtual moment function as follows:
๐๐๐
{
๐ฅ โค๐ฟ
2๐(๐ฅ) =
๐ฅ
2
๐ฅ >๐ฟ
2 ๐(๐ฅ) =
๐ฟ โ ๐ฅ
2
136
Generate 10 equal segments for this example as follows:
๐ฅ๐๐๐ = โซ๐(๐ฅ)๐(๐ฅ)
๐ธ๐๐ผ๐(๐ฅ)๐๐ฅ
๐ฟ
0
โโ ๐ฅ๐๐=10
๐=1
๐ฅ๐ =๐ฅ๐ โ ๐ฅ๐โ12๐ธ๐
[๐(๐ฅ๐)๐(๐ฅ๐)
๐ผ๐(๐ฅ๐)+๐(๐ฅ๐โ1)๐(๐ฅ๐โ1)
๐ผ๐(๐ฅ๐โ1)]
Table K-1 - Midspan Deflection Example using 10 Segment Numerical Integration
๐ ๐ฅ๐ ๐(๐ฅ๐) ๐(๐ฅ๐) ๐ผ๐(๐ฅ๐) ๐(๐ฅ๐)๐(๐ฅ๐) ๐ฅ๐
mm kN m mm4 ๐ผ๐(๐ฅ๐) mm
0 0 0 โ125 2.83x109 0 โ1 1000 500 โ47 8.06x109 โ2.91 โ0.052 2000 1000 17 8.06x109 2.06 โ0.023 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
๐ฅ๐๐๐ โโ ๐ฅ๐๐=10
๐=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
๐ ๐ฅ๐ ๐(๐ฅ๐) ๐(๐ฅ๐) ๐ผ๐(๐ฅ๐) ๐(๐ฅ๐)๐(๐ฅ๐) ๐ฅ๐
mm kN m mm4 ๐ผ๐(๐ฅ๐) mm
0 0 0 โ125 2.83x109 0 โ1 100 50 โ117 3.13x109 โ1.86 โ0.003 2 200 100 โ108 3.58x109 โ3.02 โ0.009 โฆ โฆ โฆ โฆ โฆ โฆ โฆ50 5000 2500 120 3.00x109 99.6 0.360โฆ โฆ โฆ โฆ โฆ โฆ โฆ59 5900 2050 125 2.83x109 90.4 0.339โฆ โฆ โฆ โฆ โฆ โฆ โฆ99 9900 50 6 8.06x109 0.04 0.0003 100 10000 0 0 8.06x109 0.00 0.0000
137
๐ฅ๐๐๐ = โซ๐(๐ฅ)๐(๐ฅ)
๐ธ๐๐ผ๐(๐ฅ)๐๐ฅ
๐ฟ
0
โโ ๐ฅ๐๐=100
๐=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 ๐ฟโ๐๐๐ฅ= 5.5 m. Most
other functions remain the same, but the virtual moment function must be set as follows:
๐๐๐
{
๐ฅ โค ๐ฟโ๐๐๐ฅ๐(๐ฅ) = (1 โ
๐ฟโ๐๐๐ฅ
๐ฟ)๐ฅ
2
๐ฅ > ๐ฟโ๐๐๐ฅ ๐(๐ฅ) = (
๐ฟโ๐๐๐ฅ
๐ฟ)๐ฟ โ ๐ฅ
2
Use the same 100 equal segments for the maximum deflection as follows:
Table K-3 - Maximum Deflection Example using 100 Segment Numerical Integration
๐ ๐ฅ๐ ๐(๐ฅ๐) ๐(๐ฅ๐) ๐ผ๐(๐ฅ๐) ๐(๐ฅ๐)๐(๐ฅ๐) ๐ฅ๐
mm kN m mm4 ๐ผ๐(๐ฅ๐) mm
0 0 0 โ125 2.83x109 0 โ1 100 45 โ117 3.13x109 โ1.68 โ0.003 2 200 90 โ108 3.58x109 โ2.72 โ0.008 โฆ โฆ โฆ โฆ โฆ โฆ โฆ50 5000 2250 120 3.00x109 89.6 0.324โฆ โฆ โฆ โฆ โฆ โฆ โฆ55 5500 2475 124 2.86x109 107.4 0.392โฆ โฆ โฆ โฆ โฆ โฆ โฆ99 9900 55 6 8.06x109 0.04 0.0004 100 10000 0 0 8.06x109 0.00 0.0000
๐ฅ๐๐๐ฅ = โซ๐(๐ฅ)๐(๐ฅ)
๐ธ๐๐ผ๐(๐ฅ)๐๐ฅ
๐ฟ
0
โโ ๐ฅ๐๐=100
๐=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:
๐๐ = 0.65 ; ๐๐โฒ = 36 MPa ; ๐๐ = 3.6 MPa ; ๐ธ๐ = 27 GPa ; ฮฑ1 = 0.796
๐๐ = 0.85 ; ๐๐ฆ = 400 MPa ; ๐ธ๐ = 200 GPa ; ๐ = ๐ธ๐ /๐ธ๐ = 7.407
๐ฟ = 10 m ; ๐ = 1000 mm ; โ = 178 mm ; ๐ = 146 mm
๐ผ๐ = 470 x 106 mm4 ; ๐๐๐ = 19.0 kNm
The first example presents the fully fixed-fixed case, ๐๐ = โ๐๐ฟ = โ๐๐ , as follows:
๐๐
๐๐๐= 1.515 ; ๐ = 23 kN ;
๐๐ฟ
๐0= โ0.5 ;
๐๐
๐0= โ0.5 ; ๐พ = 0.50
๐0 = 57.5 kNm ; ๐๐ฟ = โ28.8 kNm ; ๐๐ = 28.8 kNm ; ๐๐ = โ28.8 kNm
Select 15Mโs at 200 mm (top & bottom). Then: ๐ด๐ = 1000 mm2 ; ๐ = 0.685%
๐๐ = 39.7 mm ; ๐ผ๐๐ = 104 x 106 mm4 ; ๐พ๐ = 2.18
๐๐ = ๐พ๐๐๐2 = 46.5 kNm ; Assume ๐๐ โ 1.4๐๐ = 40.3 kNm (โด ok)
139
The proposed equations, based on results from work by Bischoff and Gross (2011):
๐ = 1 โ๐ผ๐๐๐ผ๐
= 0.778 ; = 3 โ 2๐๐๐
๐๐= 1.68
๐ผ๐โฒ =
๐ผ๐๐
1 โ ๐ (๐๐๐
๐๐)2 = 243 x 106 mm4 ; โ = ๐พ
๐๐๐ฟ2
12๐ธ๐๐ผ๐โฒ= 18.24 mm
Using 100 section numerical integration, โ๐ผ๐(๐ฅ)= 18.28 mm and โ๐ผ๐โฒ= 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, โ,
shows the proposed solution (indicated using ๐ผ๐โฒ ). 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
๐(๐ฅ)๐(๐ฅ) ๐ธ๐ผ๐(๐ฅ)โ and ๐(๐ฅ)๐(๐ฅ) ๐ธ๐ผ๐โฒโ , are integrated to obtain the midspan
deflection results, โ๐ผ๐(๐ฅ) and โ๐ผ๐โฒ . 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 (๐๐/๐๐๐ = 2.22) and
end-moments (1.86๐๐ = โ๐๐ฟ = โ๐๐ ). Other properties that are different include:
๐ = 48 kN ; ๐๐ฟ ๐0โ = โ0.65 ; ๐๐ ๐0โ = โ0.65 ; ๐พ = 0.071
๐0 = 120 kNm ; ๐๐ฟ = โ78 kNm ; ๐๐ = 42 kNm ; ๐๐ = โ78 kNm
For bottom steel (midspan), select 20Mโs at 200 mm. Then: ๐ด๐ = 1500 mm2
๐ = 1.03% ; ๐๐ = 46.9 mm ; ๐ผ๐๐ = 143 x 106 mm4 ; ๐พ๐ = 3.17
Assume ๐๐ = 1.4๐๐ = 58.8 kNm ; ๐๐ = 67.5 kNm (ok)
For top steel (ends), select 20Mโs at 100 mm. Then: ๐ด๐ = 3000 mm2
๐ = 2.06% ; ๐๐ = 61.3 mm ; ๐ผ๐๐ = 236 x 106 mm4 ; ๐พ๐ = 5.68
Assume ๐๐ = 1.4๐๐ = 109 kNm ; ๐๐ = 121 kNm (ok)
141
The proposed equations, again based on relevant midspan values, are:
๐ = 1 โ๐ผ๐๐๐ผ๐
= 0.695 ; = 3 โ 2๐๐๐
๐๐= 2.095
๐ผ๐โฒ =
๐ผ๐๐
1 โ ๐ (๐๐๐
๐๐)2 = 204 x 106 mm4 ; โ = ๐พ
๐๐๐ฟ2
12๐ธ๐๐ผ๐โฒ= 4.53 mm
Using 100 sections for numerical integration, midspan results are:
โ๐ผ๐(๐ฅ)= 9.88 mm ; โ๐ผ๐โฒ= 4.57 mm ; โ๐ผ๐๐= 6.5 mm
.
Figure L-2 - Integrated Function and Inaccurate Deflection Example
142
The graphs for the second example show the deflected shapes using ๐ผ๐(๐ฅ), ๐ผ๐โฒ , and ๐ผ๐๐.
The proposed approximation is far less accurate for this example. The midspan
deflection found using ๐ผ๐โฒ is only half of the real member deflection found using ๐ผ๐(๐ฅ).
This is not surprising considering that actual deflections are larger than deflections
found using ๐ผ๐๐. 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 ๐(๐ฅ)๐(๐ฅ) ๐ธ๐ผโ graph.
Another example (not shown in this appendix) of when all constant stiffness approaches
will fail is a centered point load where 2๐๐ = โ๐๐ฟ = โ๐๐ . For this case, constant
stiffness equations result in โ๐๐๐ฅ= 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 โ 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 (๐๐ฟ = ๐๐ ) 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 < โ๐๐ฟ ๐๐โ โค 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 ๐๐/๐๐๐ = 3.0. The second set carries less
load and contains less reinforcement: ๐๐/๐๐๐ = 1.6.
The third and fourth sets of examples are GFRP reinforced members. The third set has
๐๐/๐๐๐ = 2.5 and the fourth set has ๐๐/๐๐๐ = 1.6. There is an increase in
deflection because of the reduced ๐ผ๐๐. For end-moments with ๐๐ฟ/๐๐ > 0.4, short-term
deflection of these members is generally less than ๐ฟ/ฮ = 560.
144
Table M-1 - Data for CPL, ML=MR, Ig/Icr=2.3 โ Example 3.4.2a โ Page 1
Example 3.4.2a, pg 1 of 2 ฮฆc = 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 ฮฑ1 = 0.796
d = 0.9 * h mm ฮฒ1 = 0.880
M0,0 = P0 L/4 = 1.95E+8 N mm ฮฆb = 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
ฮฑL = ML/M0 = -0.66 -0.63 -0.6 -0.55 -0.5 -0.4 -0.3 -0.2 0
ฮฑR = 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
ฮฑcr = 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
ฮฑL/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
ฯ L = 0.0263 0.0219 0.0186 0.0145 0.0115 0.0074 0.0046 0 0
AL=ฯ 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
ฮทL=1 โ 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
ฯ m = 0.0115 0.0115 0.0115 0.0115 0.0115 0.0115 0.01151 0.0115 0.0115
Am=ฯ 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
ฮทm=1 โ 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
ฯ R = 0.0263 0.0219 0.0186 0.0145 0.0115 0.0074 0.0046 0 0
AR=ฯ 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
ฮทR=1 โ 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
๐๐โฒ
โ3 ๐๐โฒ
145
Table M-2 - Data for CPL, ML=MR, Ig/Icr=2.3 โ Example 3.4.2a โ 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 ฯ m = 0.0115 fr = 0.6 *Mmax/Mcr = 3.00 Ig/Icr m = 2.28 Eb = 200000 MPa
ฮฑL/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
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
K=1.5โ.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
ฮg(Gross) 0.33 1.65 2.78 4.33 5.56 7.41 8.74 9.73 11.12 mm
ฮcr(Cracked) 0.74 3.76 6.33 9.84 12.65 16.87 19.88 22.14 25.30 mm
Max Uncrack โuncr = 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
โฮฒ=0(S806) 4.92 6.43 7.89 10.26 12.40 16.23 19.29 21.51 24.77 mm
Exact Integration โIe(x)
Analytical ฮ1= -2.67 -2.28 -1.92 -1.37 -0.92 -0.29 -0.02 0.00 0.00 mm
Analytical ฮ2 or ฮ1+2= 0.05 0.06 0.07 0.08 0.10 0.15 0.20 0.25 0.21 mm
Analytical ฮ3= 4.43 4.78 5.12 5.68 6.22 7.25 8.20 9.08 10.61 mm
Analytical ฮ4= 4.43 4.78 5.12 5.68 6.22 7.25 8.20 9.08 10.61 mm
Analytical ฮ5 or ฮ5+6= 0.05 0.06 0.07 0.08 0.10 0.15 0.20 0.25 0.21 mm
Analytical ฮ6= -2.67 -2.28 -1.92 -1.37 -0.92 -0.29 -0.02 0.00 0.00 mm
ฮIe(x)(Exact) 3.63 5.11 6.55 8.79 10.82 14.22 16.75 18.65 21.63 mm
numerical ฮmax = 3.63 5.12 6.55 8.79 10.82 14.22 16.76 18.66 21.63 mm
numerical ฮmid = 3.63 5.12 6.55 8.79 10.82 14.22 16.76 18.66 21.63 mm
Proposed Method ฮI'e I'e=Icr/[1-ฮณฮทm (Mcr/Mmax)2] ฮmid=K(MmL2)/(12EcI'e)
Ie Bischoff (ฮณ=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
ฮฮณ=1(Approx) 0.70 3.53 5.93 9.23 11.87 15.82 18.65 20.76 23.73 mm
ฮณ=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
ฮI'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/ฮ 2758 1956 1527 1138 925 703 597 536 462
Branson's Method ฮIe 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
ฮIe(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
ฮIe,avg(A23.3) 0.587 3.114 5.455 8.985 12.08 16.801 17.591 21.145 24.17 mm
๐๐โฒ
146
Figure M-1 - Copy of Figure 3-1 โ 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 โ Example 3.4.2b โ Page 1
Example 3.4.2b, pg 1 of 2 ฮฆc = 0.65
Midspan Point Load P0 = 41480 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 ฮฑ1 = 0.796
d = 0.9 * h mm ฮฒ1 = 0.880
M0,0 = P0 L/4 = 1.04E+8 N mm ฮฆb = 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 = 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
ฮฑL = ML/M0 = -0.66 -0.63 -0.58 -0.53 -0.48 -0.42 -0.35 -0.25 0
ฮฑR = 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
Mm = Mmax = 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 N mm
ฮฑcr = Mcr/Mmax= 0.625 0.625 0.625 0.625 0.625 0.625 0.625 0.625 0.625
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
ฮฑL/max=ML/Mmax= -1.94 -1.70 -1.38 -1.13 -0.92 -0.72 -0.54 -0.33 0.00
Member Properties Determined with Factored Loads
Left End Kr L = 3.62 3.18 2.58 2.10 1.72 1.35 0 0 0 MPa
ฯ L = 0.0120 0.0103 0.0082 0.0066 0.0053 0.0041 0 0 0
AL=ฯ Lbd= 1938 1672 1328 1067 863 669 0 0 0 mm2
Icr L = 2.44E+9 2.19E+9 1.83E+9 1.54E+9 1.29E+9 1.05E+9 5.40E+9 5.40E+9 5.40E+9 mm4
ฮทL=1 โ 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
ฯ m = 0.0058 0.0058 0.0058 0.0058 0.0058 0.0058 0.0058 0.0058 0.0058
Am=ฯ mbd= 939 939 939 939 939 939 939 939 939 mm2
Icr m = 1.39E+9 1.39E+9 1.39E+9 1.39E+9 1.39E+9 1.39E+9 1.39E+9 1.39E+9 1.39E+9 mm4
ฮทm=1 โ 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
ฯ R = 0.0120 0.0103 0.0082 0.0066 0.0053 0.0041 0 0 0
AR=ฯ Rbd= 1938 1672 1328 1067 863 669 0 0 0 mm2
Icr R = 2.44E+9 2.19E+9 1.83E+9 1.54E+9 1.29E+9 1.05E+9 5.40E+9 5.40E+9 5.40E+9 mm4
ฮทR=1 โ 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
๐๐โฒ
โ3 ๐๐โฒ
148
Table M-4 - Data for CPL, ML=MR, Ig/Icr=3.9 โ Example 3.4.2b โ 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 ฯ m = 0.0058 fr = 0.6 *Mmax/Mcr = 1.60 Ig/Icr m = 3.89 Eb = 200000 MPa
ฮฑL/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
Mm = Mmax = 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 N mm
K=1.5โ.5(M0/Mm)= 0.029 0.149 0.310 0.436 0.538 0.638 0.731 0.833 1.000
Constant Stiffness Results Using Constant Stiffness Equations
ฮg(Gross) 0.17 0.88 1.83 2.58 3.19 3.78 4.33 4.94 5.93 mm
ฮcr(Cracked) 0.68 3.43 7.14 10.06 12.42 14.72 16.86 19.23 23.07 mm
Max Uncrack โuncr = 0.11 0.55 1.15 1.62 1.99 2.36 2.71 3.09 3.70 mm
Deflection using the S806 Integration Method with Numerical Integration
โฮฒ=0(S806) 2.84 4.19 6.36 8.34 10.12 11.82 13.40 15.26 18.89 mm
Exact Integration โIe(x)
Analytical ฮ1= -1.66 -1.28 -0.75 -0.38 -0.14 -0.01 0.00 0.00 0.00 mm
Analytical ฮ2 or ฮ1+2= 0.17 0.20 0.26 0.32 0.39 0.49 0.60 0.70 0.72 mm
Analytical ฮ3= 1.88 2.04 2.29 2.54 2.79 3.08 3.42 3.88 4.95 mm
Analytical ฮ4= 1.88 2.04 2.29 2.54 2.79 3.08 3.42 3.88 4.95 mm
Analytical ฮ5 or ฮ5+6= 0.17 0.20 0.26 0.32 0.39 0.49 0.60 0.70 0.72 mm
Analytical ฮ6= -1.66 -1.28 -0.75 -0.38 -0.14 -0.01 0.00 0.00 0.00 mm
ฮIe(x)(Exact) 0.77 1.91 3.59 4.97 6.09 7.12 8.04 9.16 11.35 mm
numerical ฮmax = 0.77 1.91 3.59 4.98 6.09 7.12 8.04 9.16 11.35 mm
numerical ฮmid = 0.77 1.91 3.59 4.98 6.09 7.12 8.04 9.16 11.35 mm
Proposed Method ฮI'e I'e=Icr/[1-ฮณฮทm (Mcr/Mmax)2] ฮmid=K(MmL2)/(12EcI'e)
Ie Bischoff (ฮณ=1) = 1.95E+9 1.95E+9 1.95E+9 1.95E+9 1.95E+9 1.95E+9 1.95E+9 1.95E+9 1.95E+9 mm4
ฮฮณ=1(Approx) 0.48 2.43 5.07 7.14 8.82 10.45 11.97 13.65 16.38 mm
ฮณ=3-2(Mcr/Mmax)= 1.75 1.75 1.75 1.75 1.75 1.75 1.75 1.75 1.75
Bischoff's I'e = 2.82E+9 2.82E+9 2.82E+9 2.82E+9 2.82E+9 2.82E+9 2.82E+9 2.82E+9 2.82E+9 mm4
ฮI'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/ฮ 13063 5238 2789 2010 1642 1405 1244 1092 881
Branson's Method ฮIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3
Branson's Ie= 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
ฮIe(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
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)= 2.54E+9 2.35E+9 2.16E+9 2.20E+9 2.57E+9 3.85E+9 2.37E+9 2.37E+9 2.37E+9 mm4
Ie R (Bransons)= 2.54E+9 2.35E+9 2.16E+9 2.20E+9 2.57E+9 3.85E+9 2.37E+9 2.37E+9 2.37E+9 mm4
Ie 9.8.2 (Bransons)= 2.42E+9 2.36E+9 2.31E+9 2.32E+9 2.43E+9 2.81E+9 2.37E+9 2.37E+9 2.37E+9 mm4
ฮIe,avg(A23.3) 0.389 2.016 4.297 6.028 7.10 7.263 9.881 11.267 13.52 mm
๐๐โฒ
149
Figure M-2 - Copy of Figure 3-2 โ Midspan Point Load, Ig/Icr=3.9 and Mm/Mcr=1.6
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 โ Example 3.4.2c โ Page 1
Example 3.4.2c, pg 1 of 2 ฮฆc = 0.65 ฮตcu = 0.0035 mm/mm
Midspan Point Load P0 = 64800 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 ฮฑ1 = 0.796
d = 0.9 * h mm ฮฒ1 = 0.880 ฯ b = 0.00578
M0,0 = P0 L/4 = 1.62E+8 N mm ฮฆb = 0.75 ฯ b=ฮฑ1ฮฒ1ฯcf'cฮตcu/(ฯbffu(ฮตcu+ffu/Ef))
End Moment Ms/Mr = 0.635 ffu = 690 MPa
+ve Moment Ms/Mr = 0.38 Eb = 44000 MPa n=Eb/Ec= 1.62963
Member Properties Determined from Provided Info (Primairly Servicability) Units
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
ฮฑL = ML/M0 = -0.66 -0.63 -0.6 -0.56 -0.5 -0.45 -0.34 -0.2 0
ฮฑR = 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
Mm = Mmax = 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 N mm
ฮฑcr = Mcr/Mmax= 0.4 0.4 0.4 0.4 0.4 0.4 0.4 0.4 0.4
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
ฮฑL/max=ML/Mmax= -1.94 -1.70 -1.50 -1.27 -1.00 -0.82 -0.52 -0.25 0.00
Member Properties Determined with Factored Loads
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
ฯ L =AL/bd= 0.0445 0.0287 0.0196 0.0124 0.0067 0.0041 0 0 0
Icr L = 2.10E+9 1.49E+9 1.08E+9 7.33E+8 4.23E+8 2.72E+8 1.01E+8 5.40E+9 5.40E+9 mm4
ฮทL=1 โ 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
ฯ m =Am/bd= 0.0271 0.0271 0.0271 0.0271 0.0271 0.0271 0.0271 0.0271 0.0271
Icr m = 1.42E+9 1.42E+9 1.42E+9 1.42E+9 1.42E+9 1.42E+9 1.42E+9 1.42E+9 1.42E+9 mm4
ฮทm=1 โ 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
ฯ R =AR/bd= 0.0445 0.0287 0.0196 0.0124 0.0067 0.0041 0.0014 0.0000 0.0000
Icr R = 2.10E+9 1.49E+9 1.08E+9 7.33E+8 4.23E+8 2.72E+8 1.01E+8 5.40E+9 5.40E+9 mm4
ฮทR=1 โ 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
๐๐โฒ
โ3 ๐๐โฒ
151
Table M-6 - Data for CPL, ML=MR, Ig/Icr=3.8 โ Example 3.4.2c โ Page 2
Ex. 3.4.2c, pg 2 of 2 P0 = 64800 N fc' = 36 MPa ฯ b = 0.00578
L = 10000 mm ffu = 690 MPa ฯ m = 0.0271
+ve Moment Ms/Mr = 0.380 fr = 0.6 *Mmax/Mcr = 2.50 Ig/Icr m = 3.81 Eb = 44000 MPa
ฮฑL/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
Mm = Mmax = 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 N mm
K=1.5โ.5(M0/Mm)= 0.029 0.149 0.250 0.364 0.500 0.591 0.742 0.875 1.000
Constant Stiffness Results Using Constant Stiffness Equations
ฮg(Gross) 0.27 1.38 2.31 3.37 4.63 5.47 6.87 8.10 9.26 mm
ฮcr(Cracked) 1.04 5.24 8.81 12.82 17.63 20.83 26.17 30.84 35.25 mm
Max Uncrack โuncr = 0.11 0.55 0.93 1.35 1.85 2.19 2.75 3.24 3.70 mm
Deflection using the S806 Integration Method with Numerical Integration
โฮฒ=0(S806) 4.98 5.30 6.09 7.14 10.38 13.80 23.01 28.99 33.65 mm
Exact Integration โIe(x)
Analytical ฮ1= -3.86 -3.95 -3.88 -3.57 -2.72 -1.82 -0.17 0.00 0.00 mm
Analytical ฮ2 or ฮ1+2= 0.07 0.08 0.09 0.11 0.15 0.18 0.26 0.34 0.30 mm
Analytical ฮ3= 5.18 5.59 6.00 6.53 7.31 7.94 9.26 10.80 12.75 mm
Analytical ฮ4= 5.18 5.59 6.00 6.53 7.31 7.94 9.26 10.80 12.75 mm
Analytical ฮ5 or ฮ5+6= 0.07 0.08 0.09 0.11 0.15 0.18 0.26 0.34 0.30 mm
Analytical ฮ6= -3.86 -3.95 -3.88 -3.57 -2.72 -1.82 -0.17 0.00 0.00 mm
ฮIe(x)(Exact) 2.76 3.43 4.42 6.16 9.48 12.61 18.69 22.29 26.10 mm
numerical ฮmax = 2.76 3.44 4.43 6.17 9.49 12.61 18.70 22.29 26.10 mm
numerical ฮmid = 2.76 3.44 4.43 6.17 9.49 12.61 18.70 22.29 26.10 mm
Proposed Method ฮI'e I'e=Icr/[1-ฮณฮทm (Mcr/Mmax)2] ฮmid=K(MmL2)/(12EcI'e)
Ie Bischoff (ฮณ=1) = 1.61E+9 1.61E+9 1.61E+9 1.61E+9 1.61E+9 1.61E+9 1.61E+9 1.61E+9 1.61E+9 mm4
ฮฮณ=1(Approx) 0.91 4.62 7.77 11.31 15.55 18.37 23.08 27.21 31.09 mm
ฮณ=3-2(Mcr/Mmax)= 2.20 2.20 2.20 2.20 2.20 2.20 2.20 2.20 2.20
Bischoff's I'e = 1.92E+9 1.92E+9 1.92E+9 1.92E+9 1.92E+9 1.92E+9 1.92E+9 1.92E+9 1.92E+9 mm4
ฮI'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/ฮ 3625 2912 2261 1622 1055 793 535 449 383
Branson's Method ฮIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3
Branson's Ie= 1.67E+9 1.67E+9 1.67E+9 1.67E+9 1.67E+9 1.67E+9 1.67E+9 1.67E+9 1.67E+9 mm4
ฮIe(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+(ฮฒdIg-Icr)(Mcr/Mmax)3 ฮฒd=0.2(ฯm/ฯ b)<1 ฯ b=.85ฮฒ1(f'c/ffu)Efฮตcu/(Efฮตcu+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
ฮIe,ฮฒd(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
ฮIe,avg(A23.3) 0.820 4.598 8.310 12.836 18.22 21.021 19.713 26.489 30.27 mm
๐๐โฒ
152
Figure M-3 - Copy of Figure 3-3 โ Midspan Point Load, Ig/Icr=3.3 and Mm/Mcr=2.5
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 โ Example 3.4.2d โ Page 1
Example 3.4.2d, pg 1 of 2 ฮฆc = 0.65 ฮตcu = 0.0035 mm/mm
Midspan Point Load P0 = 41480 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 ฮฑ1 = 0.796
d = 0.9 * h mm ฮฒ1 = 0.880 ฯ b = 0.00578
M0,0 = P0 L/4 = 1.04E+8 N mm ฮฆb = 0.75 ฯ b=ฮฑ1ฮฒ1ฯcf'cฮตcu/(ฯbffu(ฮตcu+ffu/Ef))
End Moment Ms/Mr = 0.635 ffu = 690 MPa
+ve Moment Ms/Mr = 0.4 Eb = 44000 MPa n=Eb/Ec= 1.62963
Member Properties Determined from Provided Info (Primairly Servicability) Units
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
ฮฑL = ML/M0 = -0.66 -0.63 -0.6 -0.56 -0.5 -0.45 -0.34 -0.2 0
ฮฑR = 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
Mm = Mmax = 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 N mm
ฮฑcr = Mcr/Mmax= 0.625 0.625 0.625 0.625 0.625 0.625 0.625 0.625 0.625
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
ฮฑL/max=ML/Mmax= -1.94 -1.70 -1.50 -1.27 -1.00 -0.82 -0.52 -0.25 0.00
Member Properties Determined with Factored Loads
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
ฯ L =AL/bd= 0.0116 0.0083 0.0060 0.0041 0.0023 0.0015 0 0 0
Icr L = 6.93E+8 5.13E+8 3.86E+8 2.69E+8 1.60E+8 1.05E+8 5.40E+9 5.40E+9 5.40E+9 mm4
ฮทL=1 โ 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
ฯ m =Am/bd= 0.0069 0.0069 0.0069 0.0069 0.0069 0.0069 0.0069 0.0069 0.0069
Icr m = 4.38E+8 4.38E+8 4.38E+8 4.38E+8 4.38E+8 4.38E+8 4.38E+8 4.38E+8 4.38E+8 mm4
ฮทm=1 โ 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
ฯ R =AR/bd= 0.0116 0.0083 0.0060 0.0041 0.0023 0.0015 0.0000 0.0000 0.0000
Icr R = 6.93E+8 5.13E+8 3.86E+8 2.69E+8 1.60E+8 1.05E+8 5.40E+9 5.40E+9 5.40E+9 mm4
ฮทR=1 โ 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
๐๐โฒ
โ3 ๐๐โฒ
154
Table M-8 - Data for CPL, ML=MR, Ig/Icr=12 โ Example 3.4.2d โ Page 2
Ex. 3.4.2d, pg 2 of 2 P0 = 41480 N fc' = 36 MPa ฯ b = 0.00578
L = 10000 mm ffu = 690 MPa ฯ m = 0.0069
+ve Moment Ms/Mr = 0.400 fr = 0.6 *Mmax/Mcr = 1.60 Ig/Icr m = 12.32 Eb = 44000 MPa
ฮฑL/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
Mm = Mmax = 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 1.04E+8 N mm
K=1.5โ.5(M0/Mm)= 0.029 0.149 0.250 0.364 0.500 0.591 0.742 0.875 1.000
Constant Stiffness Results Using Constant Stiffness Equations
ฮg(Gross) 0.17 0.88 1.48 2.15 2.96 3.50 4.40 5.18 5.93 mm
ฮcr(Cracked) 2.15 10.85 18.25 26.55 36.51 43.14 54.20 63.88 73.01 mm
Max Uncrack โuncr = 0.11 0.55 0.93 1.35 1.85 2.19 2.75 3.24 3.70 mm
Deflection using the S806 Integration Method with Numerical Integration
โฮฒ=0(S806) 6.76 8.63 12.14 15.51 23.22 30.48 40.02 47.89 56.64 mm
Exact Integration โIe(x)
Analytical ฮ1= -4.97 -4.47 -3.82 -2.79 -1.25 -0.34 0.00 0.00 0.00 mm
Analytical ฮ2 or ฮ1+2= 0.17 0.20 0.23 0.28 0.36 0.44 0.62 0.73 0.72 mm
Analytical ฮ3= 4.83 5.23 5.63 6.16 6.94 7.57 8.94 10.61 12.85 mm
Analytical ฮ4= 4.83 5.23 5.63 6.16 6.94 7.57 8.94 10.61 12.85 mm
Analytical ฮ5 or ฮ5+6= 0.17 0.20 0.23 0.28 0.36 0.44 0.62 0.73 0.72 mm
Analytical ฮ6= -4.97 -4.47 -3.82 -2.79 -1.25 -0.34 0.00 0.00 0.00 mm
ฮIe(x)(Exact) 0.06 1.91 4.09 7.29 12.09 15.34 19.11 22.67 27.15 mm
numerical ฮmax = 0.07 1.92 4.10 7.30 12.10 15.35 19.12 22.67 27.16 mm
numerical ฮmid = 0.07 1.92 4.10 7.30 12.10 15.35 19.12 22.67 27.16 mm
Proposed Method ฮI'e I'e=Icr/[1-ฮณฮทm (Mcr/Mmax)2] ฮmid=K(MmL2)/(12EcI'e)
Ie Bischoff (ฮณ=1) = 6.84E+8 6.84E+8 6.84E+8 6.84E+8 6.84E+8 6.84E+8 6.84E+8 6.84E+8 6.84E+8 mm4
ฮฮณ=1(Approx) 1.38 6.96 11.70 17.02 23.40 27.66 34.75 40.95 46.81 mm
ฮณ=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
ฮI'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/ฮ 171386 5223 2446 1372 827 652 523 441 368
Branson's Method ฮIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3
Branson's Ie= 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
ฮIe(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+(ฮฒdIg-Icr)(Mcr/Mmax)3 ฮฒd=0.2(ฯm/ฯ b)<1 ฯ b=.85ฮฒ1(f'c/ffu)Efฮตcu/(Efฮตcu+ffu)
Ie m (ACI440.1R) = 6.48E+8 6.48E+8 6.48E+8 6.48E+8 6.48E+8 6.48E+8 6.48E+8 6.48E+8 6.48E+8 mm4
ฮIe,ฮฒd(ACI440) 1.453 7.342 12.348 17.961 24.70 29.187 36.670 43.218 49.39 mm
Ie L (ACI440.1R) = 7.14E+8 5.52E+8 4.52E+8 3.91E+8 4.38E+8 6.36E+8 6.48E+8 6.48E+8 6.48E+8 mm4
Ie R (ACI440.1R) = 7.14E+8 5.52E+8 4.52E+8 3.91E+8 4.38E+8 6.36E+8 6.48E+8 6.48E+8 6.48E+8 mm5
Ie 9.8.2 (& ACI440.1R)= 6.68E+8 6.19E+8 5.89E+8 5.71E+8 5.85E+8 6.44E+8 6.48E+8 6.48E+8 6.48E+8 mm4
ฮIe,avg(A23.3) 1.410 7.683 13.579 20.389 27.36 29.348 36.670 43.218 49.39 mm
๐๐โฒ
155
Figure M-4 - Copy of Figure 3-4 โ 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 โ 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 (๐๐ฟ = ๐๐ ) 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 < โ๐๐ฟ ๐๐โ โค 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 ๐๐๐๐ฅ ๐๐๐โ = 2.2. The other two
examples are similar to the first two example, but are designed using GFRP and
๐๐๐๐ฅ ๐๐๐โ = 1.4. In order for these members to meet deflection requirements, the
bottom GFRP bars are increase to ๐๐ = 2.44๐๐ .
Results using the for simply supported members are shown separately from results
using the proposed equation, โ = โ 0.1(๐๐ฟ โ 1.5๐๐ )/๐๐๐, for equal third-point
loading of continuous members.
157
Table N-1 - Data for 2PL, ML=MR, Ig/Icr=3.0 โ Example 3.5.2a โ Page 1
Example 3.5.2a, pg 1 of 2 ฮฆc = 0.65
โ L & โ L Point Loads P0/2 = 42720 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 ฮฑ1 = 0.796
d = 0.9 * h mm ฮฒ1 = 0.880
M0,0 = P0 L/6 = 1.42E+8 N mm ฮฆb = 0.85
End Moment Ms/Mr = 0.635 fy = 400 MPa ฮฑ R = 1 *ฮฑ L
+ve Moment Ms/Mr = 0.635 Eb = 200000 MPa n=Eb/Ec= 7.40741
Member Properties Determined from Provided Info (Primairly Servicability) Units
ฮฑL/max=ML/Mmax= -3.00 -2.50 -2.00 -1.50 -1.00 -0.80 -0.60 -0.30 0
ฮฑR/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
ฮฑL = ML/M0 = -0.75 -0.71 -0.67 -0.60 -0.50 -0.44 -0.38 -0.23 0
ฮฑR = 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
ฮฑcr = Mcr/Mmax= 0.455 0.455 0.455 0.455 0.455 0.455 0.455 0.455 0.455
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
Member Properties Determined with Factored Loads
Left End Kr L = 7.69 6.41 5.13 3.85 2.56 2.05 1.54 0 0 MPa
ฯ L = 0.0319 0.0242 0.0181 0.0128 0.0081 0.0064 0.0047 0 0
AL=ฯLbd= 5171 3919 2925 2075 1320 1038 766 0 0 mm2
Icr L = 4.76E+9 3.99E+9 3.28E+9 2.57E+9 1.82E+9 1.50E+9 1.17E+9 5.40E+9 5.40E+9 mm4
ฮทL=1 โ 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
ฯ m = 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081
Am=ฯmbd= 1320 1320 1320 1320 1320 1320 1320 1320 1320 mm2
Icr m = 1.82E+9 1.82E+9 1.82E+9 1.82E+9 1.82E+9 1.82E+9 1.82E+9 1.82E+9 1.82E+9 mm4
ฮทm=1 โ 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
ฯ R = 0.0319 0.0242 0.0181 0.0128 0.0081 0.0064 0.0047 0 0
AR=ฯRbd= 5171 3919 2925 2075 1320 1038 766 0 0 mm2
Icr R = 4.76E+9 3.99E+9 3.28E+9 2.57E+9 1.82E+9 1.50E+9 1.17E+9 5.40E+9 5.40E+9 mm4
ฮทR=1 โ 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
๐๐โฒ
โ3 ๐๐โฒ
158
Table N-2 - Data for 2PL, ML=MR, Ig/Icr=3.0 โ Example 3.5.2a โ 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 ฯ m = 0.0081 fr = 0.6 *
Mmax/Mcr = 2.20 Ig/Icr m = 2.97 Eb = 200000 MPa
ฮฑL/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(โ L) = 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(โ L) = 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
Constant Stiffness Results Using Constant Stiffness Equations
ฮg(Gross) 4.97 5.88 6.78 7.69 8.59 8.95 9.32 9.86 10.40 mm
ฮcr(Cracked) 14.75 17.43 20.12 22.80 25.48 26.55 27.63 29.24 30.85 mm
Max Uncrack โuncr = 2.26 2.68 3.09 3.50 3.91 4.07 4.24 4.49 4.73 mm
Deflection using the S806 Method with Numerical Integration
โฮฒ=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)
ฮmax,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/ฮmax= 609 581 548 511 472 457 442 422 400
ฮIe(x)(Exact) 16.41 17.20 18.24 19.56 21.18 21.90 22.62 23.71 25.01 mm
ฮmax,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-ฮณฮทm (Mcr/Mmax)2] ฮmid=K(23MmL2)/(216EcI'e)
Ie Bischoff (ฮณ=1) = 2.11E+9 2.11E+9 2.11E+9 2.11E+9 2.11E+9 2.11E+9 2.11E+9 2.11E+9 2.11E+9 mm4
ฮฮณ=1(Approx) 12.73 15.04 17.36 19.67 21.99 22.91 23.84 25.23 26.61 mm
ฮณ=1.7-.7(Mcr/Mmax)= 1.38 1.38 1.38 1.38 1.38 1.38 1.38 1.38 1.38
Bischoff's I'e = 2.25E+9 2.25E+9 2.25E+9 2.25E+9 2.25E+9 2.25E+9 2.25E+9 2.25E+9 2.25E+9 mm4
ฮI'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
ฮณ*=ฮณ-.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 ฮณ* )= 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
ฮI'e*(Proposed) 12.62 14.79 16.91 18.99 21.04 21.84 22.64 23.83 25.00 mm
Branson's Method ฮIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3
Branson's Ie= 2.16E+9 2.16E+9 2.16E+9 2.16E+9 2.16E+9 2.16E+9 2.16E+9 2.16E+9 2.16E+9 mm4
ฮIe(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
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)= 4.76E+9 4.00E+9 3.31E+9 2.65E+9 2.16E+9 2.22E+9 3.02E+9 2.16E+9 2.16E+9 mm4
Ie R (Bransons)= 4.76E+9 4.00E+9 3.31E+9 2.65E+9 2.16E+9 2.22E+9 3.02E+9 2.16E+9 2.16E+9 mm4
Ie 9.8.2 (Bransons)= 2.94E+9 2.71E+9 2.50E+9 2.30E+9 2.16E+9 2.18E+9 2.42E+9 2.16E+9 2.16E+9 mm4
ฮIe,avg(A23.3) 9.14 11.71 14.64 18.01 21.50 22.21 20.83 24.67 26.03 mm
๐๐โฒ
159
Figure N-1 - Copy of Figure 3-5 โ 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 โ Example 3.5.2b โ Page 1
Example 3.5.2b, pg 1 of 2 ฮฆc = 0.65
โ L & โ L Point Loads P0/2 = 42720 N fc' = 36 MPa fr = 0.6 * MPa
Vary ML, Mm constant L = 10000 mm Ec = 27000 MPa Mcr= 0.05 * N mm
Let MR=0 b = 0.5* h mm ฮฑ1 = 0.796
d = 0.9 * h mm ฮฒ1 = 0.880
M0,0 = P0 L/6 = 1.42E+8 N mm ฮฆb = 0.85
End Moment Ms/Mr = 0.635 fy = 400 MPa ฮฑ R = 0 *ฮฑ L
+ve Moment Ms/Mr = 0.635 Eb = 200000 MPa n=Eb/Ec= 7.40741
Member Properties Determined from Provided Info (Primairly Servicability) Units
ฮฑL/max=ML/Mmax= -3.00 -2.50 -2.00 -1.50 -1.00 -0.80 -0.60 -0.30 0
ฮฑR/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
ฮฑL = ML/M0 = -1.50 -1.36 -1.20 -1.00 -0.75 -0.63 -0.50 -0.27 0
ฮฑR = 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
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
ฮฑcr = Mcr/Mmax= 0.455 0.455 0.455 0.455 0.455 0.455 0.455 0.455 0.455
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
Member Properties Determined with Factored Loads
Left End Kr L = 7.69 6.41 5.13 3.85 2.56 2.05 1.54 0 0 MPa
ฯ L = 0.0319 0.0242 0.0181 0.0128 0.0081 0.0064 0.0047 0 0
AL=ฯLbd= 5171 3919 2925 2075 1320 1038 766 0 0 mm2
Icr L = 4.76E+9 3.99E+9 3.28E+9 2.57E+9 1.82E+9 1.50E+9 1.17E+9 5.40E+9 5.40E+9 mm4
ฮทL=1 โ 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
ฯ m = 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081
Am=ฯmbd= 1320 1320 1320 1320 1320 1320 1320 1320 1320 mm2
Icr m = 1.82E+9 1.82E+9 1.82E+9 1.82E+9 1.82E+9 1.82E+9 1.82E+9 1.82E+9 1.82E+9 mm4
ฮทm=1 โ 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
ฯ R = 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0
AR=ฯRbd= 0 0 0 0 0 0 0 0 0 mm2
Icr R = 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
ฮทR=1 โ 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
๐๐โฒ
โ3 ๐๐โฒ
161
Table N-4 - Data for 2PL, MR=0, Ig/Icr=3.0 โ Example 3.5.2b โ 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
+ve Moment Ms/Mr = 0.635 ฯ m = 0.0081 fr = 0.6 *
Mmax/Mcr = 2.20 Ig/Icr m = 2.97 Eb = 200000 MPa
ฮฑL/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(โ L) = 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(โ L) = 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
Constant Stiffness Results Using Constant Stiffness Equations
ฮg(Gross) 2.49 3.81 5.13 6.44 7.76 8.29 8.82 9.61 10.40 mm
ฮcr(Cracked) 7.38 11.29 15.20 19.11 23.02 24.59 26.15 28.50 30.85 mm
Max Uncrack โuncr = 2.26 2.97 3.50 3.91 4.24 4.35 4.46 4.60 4.73 mm
Deflection using the S806 Method with Numerical Integration
โฮฒ=0(S806) 10.53 12.81 15.60 19.11 22.44 23.88 25.46 27.72 30.15 mm
Exact Integration Ie(x) (uses numerical integration; analytical integration not performed)
ฮmax,Ie(x)(Exact) 9.16 10.50 12.37 14.86 17.90 19.19 20.50 22.63 25.01 mm
Length:Defl, L/ฮmax= 1092 952 809 673 559 521 488 442 400
ฮIe(x)(Exact) 7.18 8.96 11.25 14.24 17.61 19.02 20.44 22.63 25.01 mm
numerical ฮI'e = 5.98 9.15 12.32 15.49 18.66 19.93 21.20 23.10 25.00 mm
ฮmax,I'e(Bischoff) 8.66 10.95 13.42 16.02 18.90 20.06 21.22 23.10 25.00 mm
Proposed Method I'e=Icr/[1-ฮณฮทm (Mcr/Mmax)2] ฮmid=K(23MmL2)/(216EcI'e)
Ie Bischoff (ฮณ=1) = 2.11E+9 2.11E+9 2.11E+9 2.11E+9 2.11E+9 2.11E+9 2.11E+9 2.11E+9 2.11E+9 mm4
ฮฮณ=1(Approx) 6.36 9.74 13.11 16.49 19.86 21.21 22.56 24.59 26.61 mm
ฮณ=1.7-.7(Mcr/Mmax)= 1.38 1.38 1.38 1.38 1.38 1.38 1.38 1.38 1.38
Bischoff's I'e = 2.25E+9 2.25E+9 2.25E+9 2.25E+9 2.25E+9 2.25E+9 2.25E+9 2.25E+9 2.25E+9 mm4
ฮI'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
ฮณ*=ฮณ-.1(ML-1.5MR)/Mcr 2.04 1.93 1.82 1.71 1.60 1.56 1.51 1.45 1.38
I'e* (using ฮณ* )= 2.53E+9 2.48E+9 2.43E+9 2.38E+9 2.33E+9 2.32E+9 2.30E+9 2.27E+9 2.25E+9 mm4
ฮI'e*(Proposed) 5.31 8.30 11.40 14.62 17.96 19.33 20.72 22.84 25.00 mm
Branson's Method ฮIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3
Branson's Ie= 2.16E+9 2.16E+9 2.16E+9 2.16E+9 2.16E+9 2.16E+9 2.16E+9 2.16E+9 2.16E+9 mm4
ฮIe(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
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)= 4.76E+9 4.00E+9 3.31E+9 2.65E+9 2.16E+9 2.22E+9 3.02E+9 2.16E+9 2.16E+9 mm4
Ie R (Bransons)= 2.16E+9 2.16E+9 2.16E+9 2.16E+9 2.16E+9 2.16E+9 2.16E+9 2.16E+9 2.16E+9 mm4
Ie 9.8.2 (Bransons)= 2.55E+9 2.43E+9 2.33E+9 2.23E+9 2.16E+9 2.17E+9 2.29E+9 2.16E+9 2.16E+9 mm4
ฮIe,avg(A23.3) 5.27 8.44 11.88 15.59 19.43 20.66 20.82 24.05 26.03 mm
๐๐โฒ
162
Figure N-2 - Copy of Figure 3-6 โ 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 โ Example 3.5.2c โ Page 1
Example 3.5.2c, pg 1 of 2 ฮฆc = 0.65 ฮตcu = 0.0035 mm/mm
โ L & โ L Point Loads P0/2 = 27300 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 ฮฑ1 = 0.796
d = 0.9 * h mm ฮฒ1 = 0.880 ฯ b=ฮฑ1ฮฒ1ฯcf'cฮตcu/(ฯbffu(ฮตcu+ffu/Ef))
M0,0 = P0 L/6 = 9.10E+7 N mm ฮฆb = 0.75 ฯ b = 0.00578
End Moment Ms/Mr = 0.635 ffu = 690 MPa ฮฑ R = 1 *ฮฑ L
+ve Moment Ms/Mr = 0.35 Eb = 44000 MPa n=Eb/Ec= 1.62963
Member Properties Determined from Provided Info (Primairly Servicability) Units
ฮฑL/max=ML/Mmax= -3.00 -2.50 -2.00 -1.50 -1.00 -0.80 -0.60 -0.30 0
ฮฑR/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
ฮฑL = ML/M0 = -0.75 -0.71 -0.67 -0.60 -0.50 -0.44 -0.38 -0.23 0
ฮฑR = 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
Mmax = 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
ฮฑcr = Mcr/Mmax= 0.712 0.712 0.712 0.712 0.712 0.712 0.712 0.712 0.712
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
Member Properties Determined with Factored Loads
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
ฯ L =AL/bd= 0.0278 0.0162 0.0089 0.0044 0.0017 0.0011 0 0 0
Icr L = 1.45E+9 9.25E+8 5.50E+8 2.90E+8 1.21E+8 7.59E+7 5.40E+9 5.40E+9 5.40E+9 mm4
ฮทL=1 โ 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
ฯ m =Am/bd= 0.0070 0.0070 0.0070 0.0070 0.0070 0.0070 0.0070 0.0070 0.0070
Icr m = 4.41E+8 4.41E+8 4.41E+8 4.41E+8 4.41E+8 4.41E+8 4.41E+8 4.41E+8 4.41E+8 mm4
ฮทm=1 โ 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
ฯ R =AR/bd= 0.0278 0.0162 0.0089 0.0044 0.0017 0.0011 0.0000 0.0000 0.0000
Icr R = 1.45E+9 9.25E+8 5.50E+8 2.90E+8 1.21E+8 7.59E+7 5.40E+9 5.40E+9 5.40E+9 mm4
ฮทR=1 โ 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
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164
Table N-6 - Data for 2PL, ML=MR, Ig/Icr=12 โ Example 3.5.2c โ Page 2
Ex. 3.5.2c, pg 2 of 2 P0 = 27300 N fc' = 36 MPa ฯ b = 0.00578
L = 10000 mm ffu = 690 MPa ฯ m = 0.0070
+ve Moment Ms/Mr = 0.350 fr = 0.6
Mmax/Mcr = 1.40 Ig/Icr m = 12.23 Eb = 44000 *
ฮฑL/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(โ L) = 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(โ L) = 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
Constant Stiffness Results Using Constant Stiffness Equations
ฮg(Gross) 3.18 3.76 4.34 4.91 5.49 5.72 5.95 6.30 6.65 mm
ฮcr(Cracked) 38.89 45.96 53.03 60.10 67.17 70.00 72.83 77.07 81.31 mm
Max Uncrack โuncr = 2.26 2.68 3.09 3.50 3.91 4.07 4.24 4.49 4.73 mm
Deflection using the S806 Method with Numerical Integration
โฮฒ=0(S806) 51.80 52.11 52.80 55.01 60.62 63.75 65.38 68.15 72.24 mm
Exact Integration Ie(x) (uses numerical integration; analytical integration not performed)
ฮmax,Ie(x)(Exact) 25.27 25.54 26.47 28.51 31.71 32.75 33.36 34.43 35.88 mm
Length:Defl, L/ฮmax= 396 391 378 351 315 305 300 290 279
ฮIe(x)(Exact) 25.27 25.54 26.47 28.51 31.71 32.75 33.36 34.43 35.88 mm
ฮmax,I'e(Bischoff) 17.14 20.26 23.37 26.49 29.60 30.85 32.09 33.96 35.83 mm
Proposed Method I'e=Icr/[1-ฮณฮทm (Mcr/Mmax)2] ฮmid=K(23MmL2)/(216EcI'e)
Ie Bischoff (ฮณ=1) = 8.26E+8 8.26E+8 8.26E+8 8.26E+8 8.26E+8 8.26E+8 8.26E+8 8.26E+8 8.26E+8 mm4
ฮฮณ=1(Approx) 20.79 24.57 28.34 32.12 35.90 37.42 38.93 41.19 43.46 mm
ฮณ=1.7-.7(Mcr/Mmax)= 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20
Bischoff's I'e = 1.00E+9 1.00E+9 1.00E+9 1.00E+9 1.00E+9 1.00E+9 1.00E+9 1.00E+9 1.00E+9 mm4
ฮI'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
ฮณ*=ฮณ-.1(ML-1.5MR)/Mcr 0.99 1.03 1.06 1.10 1.13 1.15 1.16 1.18 1.20
I'e* (using ฮณ* )= 8.19E+8 8.45E+8 8.72E+8 9.01E+8 9.32E+8 9.45E+8 9.59E+8 9.80E+8 1.00E+9 mm4
ฮI'e*(Proposed) 20.95 24.01 26.84 29.43 31.80 32.68 33.52 34.72 35.83 mm
Branson's Method ฮIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3
Branson's Ie= 2.23E+9 2.23E+9 2.23E+9 2.23E+9 2.23E+9 2.23E+9 2.23E+9 2.23E+9 2.23E+9 mm4
ฮIe(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+(ฮฒdIg-Icr)(Mcr/Mmax)3 ฮฒd=0.2(ฯm/ฯ b)<1 ฮฒd= 0.242
Ie m (ACI440.1R) = 7.54E+8 7.54E+8 7.54E+8 7.54E+8 7.54E+8 7.54E+8 7.54E+8 7.54E+8 7.54E+8 mm4
ฮIe,ฮฒd(ACI440) 22.77 26.91 31.05 35.19 39.33 40.99 42.65 45.13 47.61 mm
Ie L (ACI440.1R) = 1.45E+9 9.34E+8 5.84E+8 3.98E+8 5.49E+8 9.44E+8 7.54E+8 7.54E+8 7.54E+8 mm4
Ie R (ACI440.1R) = 1.45E+9 9.34E+8 5.84E+8 3.98E+8 5.49E+8 9.44E+8 7.54E+8 7.54E+8 7.54E+8 mm5
Ie 9.8.2 (& ACI440.1R)= 9.62E+8 8.08E+8 7.03E+8 6.47E+8 6.92E+8 8.11E+8 7.54E+8 7.54E+8 7.54E+8 mm4
ฮIe,avg(A23.3) 17.85 25.11 33.30 40.99 42.82 38.11 42.65 45.13 47.61 mm
๐๐โฒ
165
Figure N-3 - Copy of Figure 3-7 โ 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 โ Example 3.5.2d โ Page 1
Example 3.5.2d, pg 1 of 2 ฮฆc = 0.65 ฮตcu = 0.0035 mm/mm
โ L & โ L Point Loads P0/2 = 27300 N fc' = 36 MPa fr = 0.6 * MPa
Vary ML, Mm constant L = 10000 mm Ec = 27000 MPa Mcr= 0.05 * N mm
Let MR=0 b = 0.5* h mm ฮฑ1 = 0.796
d = 0.9 * h mm ฮฒ1 = 0.880 ฯ b=ฮฑ1ฮฒ1ฯcf'cฮตcu/(ฯbffu(ฮตcu+ffu/Ef))
M0,0 = P0 L/6 = 9.10E+7 N mm ฮฆb = 0.75 ฯ b = 0.00578
End Moment Ms/Mr = 0.635 ffu = 690 MPa ฮฑ R = 0 *ฮฑ L
+ve Moment Ms/Mr = 0.35 Eb = 44000 MPa n=Eb/Ec= 1.62963
Member Properties Determined from Provided Info (Primairly Servicability) Units
ฮฑL/max=ML/Mmax= -3.00 -2.50 -2.00 -1.50 -1.00 -0.80 -0.60 -0.30 0
ฮฑR/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
ฮฑL = ML/M0 = -1.50 -1.36 -1.20 -1.00 -0.75 -0.63 -0.50 -0.27 0
ฮฑR = 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
Mmax = 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
ฮฑcr = Mcr/Mmax= 0.712 0.712 0.712 0.712 0.712 0.712 0.712 0.712 0.712
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
Member Properties Determined with Factored Loads
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
ฯ L =AL/bd= 0.0278 0.0162 0.0089 0.0044 0.0017 0.0011 0 0 0
Icr L = 1.45E+9 9.25E+8 5.50E+8 2.90E+8 1.21E+8 7.59E+7 5.40E+9 5.40E+9 5.40E+9 mm4
ฮทL=1 โ 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
ฯ m =Am/bd= 0.0070 0.0070 0.0070 0.0070 0.0070 0.0070 0.0070 0.0070 0.0070
Icr m = 4.41E+8 4.41E+8 4.41E+8 4.41E+8 4.41E+8 4.41E+8 4.41E+8 4.41E+8 4.41E+8 mm4
ฮทm=1 โ 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
ฯ R =AR/bd= 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
Icr R = 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
ฮทR=1 โ 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
๐๐โฒ
โ3 ๐๐โฒ
167
Table N-8 - Data for 2PL, MR=0, Ig/Icr=12 โ Example 3.5.2d โ Page 2
Ex. 3.5.2d, pg 2 of 2 P0 = 27300 N fc' = 36 MPa ฯ b = 0.00578
L = 10000 mm ffu = 690 MPa ฯ m = 0.0070
+ve Moment Ms/Mr = 0.350 fr = 0.6 *
Mmax/Mcr = 1.40 Ig/Icr m = 12.23 Eb = 44000 MPa
ฮฑL/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(โ L) = 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(โ L) = 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
Constant Stiffness Results Using Constant Stiffness Equations
ฮg(Gross) 1.59 2.43 3.28 4.12 4.96 5.30 5.64 6.14 6.65 mm
ฮcr(Cracked) 19.44 29.76 40.07 50.38 60.69 64.82 68.94 75.13 81.31 mm
Max Uncrack โuncr = 2.26 2.97 3.50 3.91 4.24 4.35 4.46 4.60 4.73 mm
Deflection using the S806 Method with Numerical Integration
โฮฒ=0(S806) 17.15 19.15 23.73 32.37 48.47 55.91 59.84 65.46 72.24 mm
Exact Integration Ie(x) (uses numerical integration; analytical integration not performed)
ฮmax,Ie(x)(Exact) 7.42 8.52 10.33 13.40 18.90 21.96 25.05 30.02 35.88 mm
Length:Defl, L/ฮmax= 1347 1174 968 746 529 455 399 333 279
ฮIe(x)(Exact) 4.41 5.62 7.77 11.59 18.03 21.41 24.75 30.02 35.88 mm
numerical ฮI'e = 8.57 13.11 17.66 22.20 26.75 28.56 30.38 33.11 35.83 mm
ฮmax,I'e(Bischoff) 12.42 15.69 19.23 22.96 27.09 28.75 30.42 33.11 35.83 mm
Proposed Method I'e=Icr/[1-ฮณฮทm (Mcr/Mmax)2] ฮmid=K(23MmL2)/(216EcI'e)
Ie Bischoff (ฮณ=1) = 8.26E+8 8.26E+8 8.26E+8 8.26E+8 8.26E+8 8.26E+8 8.26E+8 8.26E+8 8.26E+8 mm4
ฮฮณ=1(Approx) 10.39 15.90 21.42 26.93 32.44 34.64 36.85 40.16 43.46 mm
ฮณ=1.7-.7(Mcr/Mmax)= 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20
Bischoff's I'e = 1.00E+9 1.00E+9 1.00E+9 1.00E+9 1.00E+9 1.00E+9 1.00E+9 1.00E+9 1.00E+9 mm4
ฮI'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
ฮณ*=ฮณ-.1(ML-1.5MR)/Mcr 1.62 1.55 1.48 1.41 1.34 1.31 1.29 1.24 1.20
I'e* (using ฮณ* )= 1.81E+9 1.59E+9 1.42E+9 1.29E+9 1.18E+9 1.14E+9 1.10E+9 1.05E+9 1.00E+9 mm4
ฮI'e*(Proposed) 4.75 8.25 12.42 17.26 22.78 25.17 27.67 31.63 35.83 mm
Branson's Method ฮIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3
Branson's Ie= 2.23E+9 2.23E+9 2.23E+9 2.23E+9 2.23E+9 2.23E+9 2.23E+9 2.23E+9 2.23E+9 mm4
ฮIe(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+(ฮฒdIg-Icr)(Mcr/Mmax)3 ฮฒd=0.2(ฯm/ฯ b)<1 ฮฒd= 0.242
Ie m (ACI440.1R) = 7.54E+8 7.54E+8 7.54E+8 7.54E+8 7.54E+8 7.54E+8 7.54E+8 7.54E+8 7.54E+8 mm4
ฮIe,ฮฒd(ACI440) 11.39 17.42 23.46 29.50 35.54 37.95 40.37 43.99 47.61 mm
Ie L (ACI440.1R) = 1.45E+9 9.34E+8 5.84E+8 3.98E+8 5.49E+8 9.44E+8 7.54E+8 7.54E+8 7.54E+8 mm4
Ie R (ACI440.1R) = 7.54E+8 7.54E+8 7.54E+8 7.54E+8 7.54E+8 7.54E+8 7.54E+8 7.54E+8 7.54E+8 mm5
Ie 9.8.2 (& ACI440.1R)= 8.58E+8 7.81E+8 7.28E+8 7.00E+8 7.23E+8 7.82E+8 7.54E+8 7.54E+8 7.54E+8 mm4
ฮIe,avg(A23.3) 10.00 16.82 24.28 31.74 37.05 36.57 40.37 43.99 47.61 mm
๐๐โฒ
168
Figure N-4 - Copy of Figure 3-8 โ 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 โ 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 ๐๐๐ = 0.6โ๐๐โฒ(๐ผ๐/๐ฆ๐ก); the use of ๐๐๐ = 0.3โ๐๐โฒ(๐ผ๐/๐ฆ๐ก), 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 ๐๐๐๐ฅ/๐๐๐ = 2.17 and ๐ผ๐/๐ผ๐๐ = 3, show that the provided methods
which account for tension stiffening work well if 2๐๐๐๐ฅ < โ๐๐ฟ. Examples 3.6.2c and
3.6.2d are a beam and a slab which demonstrate that, if ratios such as ๐๐๐๐ฅ/๐๐๐ and
๐ผ๐/๐ผ๐๐ are held constant, then span, width, height, and load are inputs that donโt 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,
๐๐๐๐ฅ/๐๐๐ = 1.333 and ๐ผ๐/๐ผ๐๐ = 4.9, therefore Bransonโs (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 ๐๐๐๐ฅ ๐๐๐โ > 0.5 and bottom bars
exceeding those required for ๐๐ = ๐๐. Example 3.6.2f demonstrates a slab with a
single layer of reinforcing, ๐๐๐๐ฅ/๐๐๐ = 1.22, and ๐๐/๐๐ = 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 ๐ฝ๐ 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, ๐ผ๐/๐ผ๐๐ = 6,
๐๐๐๐ฅ/๐๐๐ = 2.0, and ๐๐/๐๐ = 1.6; this contrasts with the beam in Example 3.6.2h
which has two ends continuous, ๐ผ๐/๐ผ๐๐ = 14, ๐๐๐๐ฅ/๐๐๐ = 1.25, and ๐๐/๐๐ = 1.29.
The second and third GFRP examples show that the ๐ฝ๐ factor often yields reasonable
results. As expected, the GFRP results show Bransonโs (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 < โ๐๐ฟ ๐๐โ โค 0. For
results with ๐๐๐๐ฅ/๐๐๐ โ 1.25, 2.5๐๐ > โ๐๐ฟ โฅ 0 is the proposed valid range. For
๐๐๐๐ฅ/๐๐๐ โ 2.0, the proposed valid range is reduced to 2.0๐๐ > โ๐๐ฟ โฅ 0.
After determining the midspan deflection using proposed equations, the equation for the
approximate maximum deflection is โ๐๐๐ฅโ โ๐๐๐โ๐๐๐๐ฅ โ ๐๐ . 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 โ Example 3.6.2a โ Page 1
Example 3.6.2a, pg 1 of 2 ฮฆc = 0.65
UDL Continuous w0 = 11.27 N/mm 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 ฮฑ1 = 0.796
d = 0.9 * h mm ฮฒ1 = 0.880
M0,0 = w0 L2/8 = 1.41E+8 N mm ฮฆb = 0.85
End Moment Ms/Mr = 0.635 fy = 400 MPa ฮฑ R = 1 *ฮฑ L
+ve Moment Ms/Mr = 0.635 Eb = 200000 MPa n=Eb/Ec= 7.40741
Member Properties Determined from Provided Info (Primairly Servicability) Units
ฮฑL/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
ฮฑL = ML/M0 = -0.75 -0.70 -0.64 -0.63 -0.50 -0.40 -0.30 -0.20 0
ฮฑR = 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
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 N mm
ฮฑcr = Mcr/Mmax= 0.46 0.46 0.46 0.46 0.46 0.46 0.46 0.46 0.46
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
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
Member Properties Determined with Factored Loads
Left End Kr L = 7.61 5.92 4.51 4.32 2.54 1.69 0.00 0 0 MPa
ฯ L = 0.0313 0.0217 0.0154 0.0147 0.0081 0.0052 0.0000 0 0
AL=ฯLbd= 5077 3516 2500 2375 1304 846 0 0 0 mm2
Icr L = 4.71E+9 3.72E+9 2.94E+9 2.83E+9 1.80E+9 1.27E+9 5.40E+9 5.40E+9 5.40E+9 mm4
ฮทL=1 โ 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
ฯ m = 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081
Icr m = 1.80E+9 1.80E+9 1.80E+9 1.80E+9 1.80E+9 1.80E+9 1.80E+9 1.80E+9 1.80E+9 mm4
ฮทm=1 โ Icr m/Ig = 0.666 Mmax/Mcr = 2.17 Ig/Icr m = 2.99 Am=ฯmbd= 1304 mm2
Right End Kr R = 7.61 5.92 4.51 4.32 2.54 1.69 0.00 0 0 MPa
ฯ R = 0.0313 0.0217 0.0154 0.0147 0.0081 0.0052 0.0000 0 0
AR=ฯRbd= 5077 3516 2500 2375 1304 846 0 0 0 mm2
Icr R = 4.71E+9 3.72E+9 2.94E+9 2.83E+9 1.80E+9 1.27E+9 5.40E+9 5.40E+9 5.40E+9 mm4
ฮทR=1 โ 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
๐๐โฒ
โ3 ๐๐โฒ
172
Table O-2 - Data for UDL Beam, ML=MR, Ig/Icr=3.0 โ Example 3.6.2a โ 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
+ve Moment Ms/Mr = 0.635 ฯ m = 0.0081 fr = 0.6 *
Mmax/Mcr = 2.17 Ig/Icr m = 2.99 Eb = 200000 MPa
ฮฑL/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 ฮmid=K(5MmL2)/(48EcI)
ฮg(Gross) 4.03 5.37 6.49 6.64 8.05 8.72 9.20 9.56 10.06 mm
ฮcr(Cracked) 12.05 16.07 19.41 19.87 24.10 26.11 27.54 28.62 30.13 mm
Branson's Method ฮIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3
Branson's Ie= 2.15E+9 2.15E+9 2.15E+9 2.15E+9 2.15E+9 2.15E+9 2.15E+9 2.15E+9 2.15E+9 mm4
ฮIe(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
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)= 4.71E+9 3.73E+9 2.98E+9 2.88E+9 2.15E+9 2.63E+9 2.15E+9 2.15E+9 2.15E+9 mm4
Ie R (Bransons)= 4.71E+9 3.73E+9 2.98E+9 2.88E+9 2.15E+9 2.63E+9 2.15E+9 2.15E+9 2.15E+9 mm4
Ie 9.8.2 (Bransons)= 2.92E+9 2.63E+9 2.40E+9 2.37E+9 2.15E+9 2.30E+9 2.15E+9 2.15E+9 2.15E+9 mm4
ฮIe,avg(A23.3) 7.44 11.03 14.58 15.10 20.18 20.51 23.07 23.97 25.23 mm
Deflection using the S806 Method with Numerical Integration
โฮฒ=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 ฮ1= -1.27 -0.95 -0.62 -0.57 -0.15 -0.02 0.00 0.00 0.00 mm
Analytical ฮ2 or ฮ1+2= 0.11 0.12 0.13 0.13 0.15 0.17 0.18 0.18 0.14 mm
Analytical ฮ3= 7.21 7.77 8.35 8.44 9.45 10.08 10.62 11.09 11.85 mm
Analytical ฮ4= 7.21 7.77 8.35 8.44 9.45 10.08 10.62 11.09 11.85 mm
Analytical ฮ5 or ฮ5+6= 0.11 0.12 0.13 0.13 0.15 0.17 0.18 0.18 0.14 mm
Analytical ฮ6= -1.27 -0.95 -0.62 -0.57 -0.15 -0.02 0.00 0.00 0.00 mm
ฮIe(x)(Exact) 12.09 13.87 15.71 15.98 18.89 20.44 21.60 22.52 23.97 mm
Proposed Method ฮI'e I'e=Icr/[1-ฮณฮทm(Mcr/Mmax)2] ฮณ=(1.6ฮพ 3-0.6ฮพ 4)/(Mcr/Mmax)
2+2.4ln(2-ฮพ )
I'e (ฮณ=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
ฮฮณ=1(Approx) 10.35 13.80 16.68 17.07 20.70 22.43 23.66 24.59 25.88 mm
ฮพ =1-โ(1-Mcr/Mmax)= 0.265 0.265 0.265 0.265 0.265 0.265 0.265 0.265 0.265
ฮณ= 1.45 1.45 1.45 1.45 1.45 1.45 1.45 1.45 1.45
Bischoff's I'e = 2.27E+9 2.27E+9 2.27E+9 2.27E+9 2.27E+9 2.27E+9 2.27E+9 2.27E+9 2.27E+9 mm4
ฮI'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 ฮmax โ ฮI'e โ(Mmax/Mm)
ฮmax,Ie(x)(Exact) 12.09 13.87 15.71 15.99 18.89 20.44 21.60 22.53 23.97 mm
Length:Defl, L/ฮmax= 827 721 637 626 529 489 463 444 417
ฮmax,I'e(Proposed) 9.59 12.79 15.45 15.81 19.18 20.78 21.92 22.78 23.97 mm
๐๐โฒ
173
Figure O-1 - Copy of Figure 3-9 โ 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 โ Example 3.6.2b โ Page 1
Example 3.6.2b, pg 1 of 2 ฮฆc = 0.65
UDL Continuous w0 = 11.27 N/mm fc' = 36 MPa fr = 0.6 * MPa
Vary ML, Mm constant L = 10000 mm Ec = 27000 MPa Mcr= 0.05 * N mm
Let MR=0 b = 0.5* h mm ฮฑ1 = 0.796
d = 0.9 * h mm ฮฒ1 = 0.880
M0,0 = w0 L2/8 = 1.41E+8 N mm ฮฆb = 0.85
End Moment Ms/Mr = 0.635 fy = 400 MPa ฮฑ R = 0 *ฮฑ L
+ve Moment Ms/Mr = 0.635 Eb = 200000 MPa n=Eb/Ec= 7.40741
Member Properties Determined from Provided Info (Primairly Servicability) Units
ฮฑL/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
ฮฑL = ML/M0 = -1.33 -1.10 -0.9 -0.7 -0.50 -0.40 -0.30 -0.20 0
ฮฑR = 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
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 N mm
ฮฑcr = Mcr/Mmax= 0.46 0.46 0.46 0.46 0.46 0.46 0.46 0.46 0.46
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
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
Member Properties Determined with Factored Loads
Left End Kr L = 7.57 5.31 3.80 2.61 1.66 1.25 0.00 0 0 MPa
ฯ L = 0.0311 0.0189 0.0126 0.0083 0.0051 0.0038 0.0000 0 0
AL=ฯLbd= 5035 3054 2047 1345 828 618 0 0 0 mm2
Icr L = 4.68E+9 3.38E+9 2.54E+9 1.85E+9 1.25E+9 9.79E+8 5.40E+9 5.40E+9 5.40E+9 mm4
ฮทL=1 โ 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
ฯ m = 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081
Icr m = 1.80E+9 1.80E+9 1.80E+9 1.80E+9 1.80E+9 1.80E+9 1.80E+9 1.80E+9 1.80E+9 mm4
ฮทm=1 โ Icr m/Ig = 0.666 Mmax/Mcr = 2.17 Ig/Icr m = 2.99 Am=ฯmbd= 1304 mm2
Right End Kr R = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0 0 MPa
ฯ R = 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0
AR=ฯRbd= 0 0 0 0 0 0 0 0 0 mm2
Icr R = 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
ฮทR=1 โ 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
๐๐โฒ
โ3 ๐๐โฒ
175
Table O-4 - Data for UDL Beam, MR=0, Ig/Icr=3.0 โ Example 3.6.2b โ 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
+ve Moment Ms/Mr = 0.635 ฯ m = 0.0081 fr = 0.6 *
Mmax/Mcr = 2.17 Ig/Icr m = 2.99 Eb = 200000 MPa
ฮฑL/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
Constant Stiffness Results Using Constant Stiffness Equations ฮmid=K(5MmL2)/(48EcI)
ฮg(Gross) 4.56 6.51 7.71 8.58 9.20 9.44 9.65 9.81 10.06 mm
ฮcr(Cracked) 13.66 19.49 23.07 25.67 27.54 28.27 28.87 29.37 30.13 mm
Branson's Method ฮIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3
Branson's Ie= 2.15E+9 2.15E+9 2.15E+9 2.15E+9 2.15E+9 2.15E+9 2.15E+9 2.15E+9 2.15E+9 mm4
ฮIe(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
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)= 4.68E+9 3.40E+9 2.63E+9 2.17E+9 2.70E+9 4.55E+9 2.15E+9 2.15E+9 2.15E+9 mm4
Ie R (Bransons)= 2.15E+9 2.15E+9 2.15E+9 2.15E+9 2.15E+9 2.15E+9 2.15E+9 2.15E+9 2.15E+9 mm4
Ie 9.8.2 (Bransons)= 2.53E+9 2.34E+9 2.22E+9 2.16E+9 2.24E+9 2.51E+9 2.15E+9 2.15E+9 2.15E+9 mm4
ฮIe,avg(A23.3) 9.73 15.01 18.71 21.48 22.22 20.29 24.18 24.60 25.23 mm
Deflection using the S806 Method with Numerical Integration
โฮฒ=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 ฮ1= -2.25 -1.32 -0.67 -0.23 -0.03 0.00 0.00 0.00 0.00 mm
Analytical ฮ2 or ฮ1+2= 0.19 0.20 0.20 0.21 0.21 0.21 0.21 0.19 0.14 mm
Analytical ฮ3= 2.04 3.88 5.55 7.18 8.70 9.42 10.09 10.72 11.85 mm
Analytical ฮ4= 12.83 13.08 13.07 12.92 12.68 12.53 12.38 12.21 11.85 mm
Analytical ฮ5 or ฮ5+6= 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.12 0.14 mm
Analytical ฮ6= 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 mm
ฮIe(x)(Exact) 12.86 15.92 18.24 20.17 21.67 22.27 22.79 23.24 23.97 mm
Proposed Method ฮI'e I'e=Icr/[1-ฮณฮทm(Mcr/Mmax)2] ฮณ=(1.6ฮพ 3-0.6ฮพ 4)/(Mcr/Mmax)
2+2.4ln(2-ฮพ )
I'e (ฮณ=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
ฮฮณ=1(Approx) 11.73 16.74 19.82 22.05 23.66 24.28 24.80 25.24 25.88 mm
ฮพ =1-โ(1-Mcr/Mmax)= 0.265 0.265 0.265 0.265 0.265 0.265 0.265 0.265 0.265
ฮณ= 1.45 1.45 1.45 1.45 1.45 1.45 1.45 1.45 1.45
Bischoff's I'e = 2.27E+9 2.27E+9 2.27E+9 2.27E+9 2.27E+9 2.27E+9 2.27E+9 2.27E+9 2.27E+9 mm4
ฮI'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
Maximum Deflection Results using numerical and approximation methods ฮmax โ ฮI'e โ(Mmax/Mm)
ฮmax,Ie(x)(Exact) 14.48 16.90 18.76 20.46 21.76 22.27 22.79 23.24 23.97 mm
Length:Defl, L/ฮmax= 691 592 533 489 460 449 439 430 417
ฮmax,I'e(Proposed) 12.53 16.76 19.19 20.91 22.15 22.63 23.05 23.41 23.97 mm
๐๐โฒ
176
Figure O-2 - Copy of Figure 3-10 โ 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 โ Example 3.6.2c โ Page 1
Example 3.6.2c, pg 1 of 2 ฮฆc = 0.65
UDL Continuous w0 = 12.285 N/mm fc' = 36 MPa fr = 0.6 * MPa
Vary ML, Mm constant L = 7500 mm Ec = 27000 MPa Mcr= 0.05 * N mm
Let ML=MR b = 0.5* h mm ฮฑ1 = 0.796
d = 0.85 * h mm ฮฒ1 = 0.880
M0,0 = w0 L2/8 = 8.64E+7 N mm ฮฆb = 0.85
End Moment Ms/Mr = 0.635 fy = 400 MPa ฮฑ R = 1 *ฮฑ L
+ve Moment Ms/Mr = 0.635 Eb = 200000 MPa n=Eb/Ec= 7.40741
Member Properties Determined from Provided Info (Primairly Servicability) Units
ฮฑL/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
ฮฑL = ML/M0 = -0.75 -0.70 -0.65 -0.55 -0.50 -0.45 -0.38 -0.20 0
ฮฑR = 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
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 N mm
ฮฑcr = Mcr/Mmax= 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75
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 = 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
Member Properties Determined with Factored Loads
Left End Kr L = 5.23 4.07 3.24 2.13 1.74 1.43 0.00 0 0 MPa
ฯ L = 0.0185 0.0137 0.0105 0.0067 0.0054 0.0044 0.0000 0 0
AL=ฯLbd= 2832 2092 1612 1021 825 669 0 0 0 mm2
Icr L = 2.81E+9 2.27E+9 1.87E+9 1.31E+9 1.10E+9 9.25E+8 5.40E+9 5.40E+9 5.40E+9 mm4
ฮทL=1 โ 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
ฯ m = 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054
Icr m = 1.10E+9 1.10E+9 1.10E+9 1.10E+9 1.10E+9 1.10E+9 1.10E+9 1.10E+9 1.10E+9 mm4
ฮทm=1 โ Icr m/Ig = 0.796 Mmax/Mcr = 1.33 Ig/Icr m = 4.90 Am=ฯmbd= 825 mm2
Right End Kr R = 5.23 4.07 3.24 2.13 1.74 1.43 0.00 0 0 MPa
ฯ R = 0.0185 0.0137 0.0105 0.0067 0.0054 0.0044 0.0000 0 0
AR=ฯRbd= 2832 2092 1612 1021 825 669 0 0 0 mm2
Icr R = 2.81E+9 2.27E+9 1.87E+9 1.31E+9 1.10E+9 9.25E+8 5.40E+9 5.40E+9 5.40E+9 mm4
ฮทR=1 โ 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
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178
Table O-6 - Data for UDL Beam, ML=MR, Ig/Icr=4.9 โ Example 3.6.2c โ 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
+ve Moment Ms/Mr = 0.635 ฯ m = 0.0054 fr = 0.6 *
Mmax/Mcr = 1.33 Ig/Icr m = 4.90 Eb = 200000 MPa
ฮฑL/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
Constant Stiffness Results Using Constant Stiffness Equations ฮmid=K(5MmL2)/(48EcI)
ฮg(Gross) 1.39 1.85 2.18 2.62 2.78 2.90 3.05 3.30 3.47 mm
ฮcr(Cracked) 6.81 9.08 10.70 12.86 13.62 14.24 14.94 16.17 17.02 mm
Branson's Method ฮIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3
Branson's Ie= 2.91E+9 2.91E+9 2.91E+9 2.91E+9 2.91E+9 2.91E+9 2.91E+9 2.91E+9 2.91E+9 mm4
ฮIe(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
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)= 2.85E+9 2.37E+9 2.10E+9 2.25E+9 2.91E+9 4.37E+9 2.91E+9 2.91E+9 2.91E+9 mm4
Ie R (Bransons)= 2.85E+9 2.37E+9 2.10E+9 2.25E+9 2.91E+9 4.37E+9 2.91E+9 2.91E+9 2.91E+9 mm4
Ie 9.8.2 (Bransons)= 2.90E+9 2.75E+9 2.67E+9 2.72E+9 2.91E+9 3.35E+9 2.91E+9 2.91E+9 2.91E+9 mm4
ฮIe,avg(A23.3) 2.59 3.63 4.41 5.21 5.15 4.68 5.64 6.11 6.43 mm
Deflection using the S806 Method with Numerical Integration
โฮฒ=0(S806) 7.43 8.43 9.34 10.86 11.51 12.06 12.60 13.76 14.88 mm
Exact Integration Ie(x)Analytical ฮ1= -0.55 -0.36 -0.21 -0.05 -0.01 0.00 0.00 0.00 0.00 mm
Analytical ฮ2 or ฮ1+2= 0.19 0.21 0.23 0.26 0.27 0.29 0.30 0.31 0.28 mm
Analytical ฮ3= 1.94 2.10 2.25 2.50 2.61 2.71 2.85 3.15 3.42 mm
Analytical ฮ4= 1.94 2.10 2.25 2.50 2.61 2.71 2.85 3.15 3.42 mm
Analytical ฮ5 or ฮ5+6= 0.19 0.21 0.23 0.26 0.27 0.29 0.30 0.31 0.28 mm
Analytical ฮ6= -0.55 -0.36 -0.21 -0.05 -0.01 0.00 0.00 0.00 0.00 mm
ฮIe(x)(Exact) 3.16 3.91 4.53 5.42 5.74 6.00 6.30 6.90 7.40 mm
Proposed Method ฮI'e I'e=Icr/[1-ฮณฮทm(Mcr/Mmax)2] ฮณ=(1.6ฮพ 3-0.6ฮพ 4)/(Mcr/Mmax)
2+2.4ln(2-ฮพ )
I'e (ฮณ=1) (M(x)=Mmax) = 1.99E+9 1.99E+9 1.99E+9 1.99E+9 1.99E+9 1.99E+9 1.99E+9 1.99E+9 1.99E+9 mm4
ฮฮณ=1(Approx) 3.76 5.01 5.91 7.10 7.52 7.86 8.25 8.93 9.40 mm
ฮพ =1-โ(1-Mcr/Mmax)= 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500
ฮณ= 1.26 1.26 1.26 1.26 1.26 1.26 1.26 1.26 1.26
Bischoff's I'e = 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
ฮI'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
Maximum Deflection Results using numerical and approximation methods ฮmax โ ฮI'e โ(Mmax/Mm)
ฮmax,Ie(x)(Exact) 3.16 3.91 4.53 5.42 5.74 6.00 6.30 6.90 7.40 mm
Length:Defl, L/ฮmax= 2376 1917 1657 1383 1307 1251 1191 1086 1013
ฮmax,I'e(Proposed) 2.96 3.95 4.65 5.59 5.92 6.19 6.50 7.03 7.40 mm
๐๐โฒ
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 โ Example 3.6.2d โ Page 1
Example 3.6.2d, pg 1 of 2 ฮฆc = 0.65
UDL Continuous w0 = 8.605 N/mm fc' = 36 MPa fr = 0.6 * MPa
Vary ML, Mm constant L = 7500 mm Ec = 27000 MPa Mcr= 0.3636 * N mm
Let ML=MR b = 3.636* h mm ฮฑ1 = 0.796
d = 0.85 * h mm ฮฒ1 = 0.880
M0,0 = w0 L2/8 = 6.05E+7 N mm ฮฆb = 0.85
End Moment Ms/Mr = 0.635 fy = 400 MPa ฮฑ R = 1 *ฮฑ L
+ve Moment Ms/Mr = 0.635 Eb = 200000 MPa n=Eb/Ec= 7.40741
Member Properties Determined from Provided Info (Primairly Servicability) Units
ฮฑL/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
ฮฑL = ML/M0 = -0.75 -0.70 -0.65 -0.55 -0.50 -0.45 -0.38 -0.20 0
ฮฑR = 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
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 N mm
ฮฑcr = Mcr/Mmax= 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75
Mcr = 4.54E+7 4.54E+7 4.54E+7 4.54E+7 4.54E+7 4.54E+7 4.54E+7 4.54E+7 4.54E+7 N mm
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
Member Properties Determined with Factored Loads
Left End Kr L = 5.23 4.07 3.24 2.13 1.74 1.43 0.00 0 0 MPa
ฯ L = 0.0185 0.0137 0.0105 0.0067 0.0054 0.0044 0.0000 0 0
AL=ฯLbd= 4328 3196 2463 1560 1261 1022 0 0 0 mm2
Icr L = 9.03E+8 7.29E+8 6.00E+8 4.20E+8 3.54E+8 2.97E+8 1.73E+9 1.73E+9 1.73E+9 mm4
ฮทL=1 โ 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
ฯ m = 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054
Icr m = 3.54E+8 3.54E+8 3.54E+8 3.54E+8 3.54E+8 3.54E+8 3.54E+8 3.54E+8 3.54E+8 mm4
ฮทm=1 โ Icr m/Ig = 0.796 Mmax/Mcr = 1.33 Ig/Icr m = 4.90 Am=ฯmbd= 1261 mm2
Right End Kr R = 5.23 4.07 3.24 2.13 1.74 1.43 0.00 0 0 MPa
ฯ R = 0.0185 0.0137 0.0105 0.0067 0.0054 0.0044 0.0000 0 0
AR=ฯRbd= 4328 3196 2463 1560 1261 1022 0 0 0 mm2
Icr R = 9.03E+8 7.29E+8 6.00E+8 4.20E+8 3.54E+8 2.97E+8 1.73E+9 1.73E+9 1.73E+9 mm4
ฮทR=1 โ 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
๐๐โฒ
โ3 ๐๐โฒ
181
Table O-8 - Data for UDL Slab, ML=MR, Ig/Icr=4.9 โ Example 3.6.2d โ 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
+ve Moment Ms/Mr = 0.635 ฯ m = 0.0054 fr = 0.6 *
Mmax/Mcr = 1.33 Ig/Icr m = 4.90 Eb = 200000 MPa
ฮฑL/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
Constant Stiffness Results Using Constant Stiffness Equations ฮmid=K(5MmL2)/(48EcI)
ฮg(Gross) 3.03 4.04 4.76 5.72 6.06 6.34 6.65 7.20 7.58 mm
ฮcr(Cracked) 14.86 19.81 23.34 28.06 29.71 31.06 32.59 35.28 37.14 mm
Branson's Method ฮIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3
Branson's Ie= 9.36E+8 9.36E+8 9.36E+8 9.36E+8 9.36E+8 9.36E+8 9.36E+8 9.36E+8 9.36E+8 mm4
ฮIe(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
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)= 9.16E+8 7.62E+8 6.75E+8 7.24E+8 9.36E+8 1.40E+9 9.36E+8 9.36E+8 9.36E+8 mm4
Ie R (Bransons)= 9.16E+8 7.62E+8 6.75E+8 7.24E+8 9.36E+8 1.40E+9 9.36E+8 9.36E+8 9.36E+8 mm4
Ie 9.8.2 (Bransons)= 9.30E+8 8.84E+8 8.57E+8 8.72E+8 9.36E+8 1.08E+9 9.36E+8 9.36E+8 9.36E+8 mm4
ฮIe,avg(A23.3) 5.65 7.93 9.63 11.38 11.23 10.21 12.31 13.33 14.03 mm
Deflection using the S806 Method with Numerical Integration
โฮฒ=0(S806) 16.20 18.40 20.38 23.70 25.10 26.32 27.48 30.01 32.47 mm
Exact Integration Ie(x)Analytical ฮ1= -1.21 -0.78 -0.46 -0.10 -0.03 0.00 0.00 0.00 0.00 mm
Analytical ฮ2 or ฮ1+2= 0.42 0.46 0.50 0.57 0.60 0.63 0.66 0.67 0.62 mm
Analytical ฮ3= 4.23 4.58 4.90 5.45 5.69 5.92 6.21 6.86 7.46 mm
Analytical ฮ4= 4.23 4.58 4.90 5.45 5.69 5.92 6.21 6.86 7.46 mm
Analytical ฮ5 or ฮ5+6= 0.42 0.46 0.50 0.57 0.60 0.63 0.66 0.67 0.62 mm
Analytical ฮ6= -1.21 -0.78 -0.46 -0.10 -0.03 0.00 0.00 0.00 0.00 mm
ฮIe(x)(Exact) 6.89 8.53 9.87 11.83 12.52 13.08 13.74 15.06 16.15 mm
Proposed Method ฮI'e I'e=Icr/[1-ฮณฮทm(Mcr/Mmax)2] ฮณ=(1.6ฮพ 3-0.6ฮพ 4)/(Mcr/Mmax)
2+2.4ln(2-ฮพ )
I'e (ฮณ=1) (M(x)=Mmax) = 6.40E+8 6.40E+8 6.40E+8 6.40E+8 6.40E+8 6.40E+8 6.40E+8 6.40E+8 6.40E+8 mm4
ฮฮณ=1(Approx) 8.20 10.94 12.89 15.50 16.41 17.15 18.00 19.48 20.51 mm
ฮพ =1-โ(1-Mcr/Mmax)= 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500
ฮณ= 1.26 1.26 1.26 1.26 1.26 1.26 1.26 1.26 1.26
Bischoff's I'e = 8.13E+8 8.13E+8 8.13E+8 8.13E+8 8.13E+8 8.13E+8 8.13E+8 8.13E+8 8.13E+8 mm4
ฮI'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
Maximum Deflection Results using numerical and approximation methods ฮmax โ ฮI'e โ(Mmax/Mm)
ฮmax,Ie(x)(Exact) 6.89 8.53 9.87 11.83 12.52 13.08 13.74 15.06 16.15 mm
Length:Defl, L/ฮmax= 1089 879 760 634 599 573 546 498 464
ฮmax,I'e(Proposed) 6.46 8.61 10.15 12.20 12.92 13.51 14.17 15.34 16.15 mm
๐๐โฒ
182
Figure O-4 - Copy of Figure 3-11 โ 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 โ Example 3.6.2e โ Page 1
Example 3.6.2e, pg 1 of 2 ฮฆc = 0.65
UDL Continuous w0 = 8.605 N/mm fc' = 36 MPa fr = 0.6 * MPa
Vary ML, Mm constant L = 7500 mm Ec = 27000 MPa Mcr= 0.3636 * N mm
Let MR=0 b = 3.636* h mm ฮฑ1 = 0.796
d = 0.85 * h mm ฮฒ1 = 0.880
M0,0 = w0 L2/8 = 6.05E+7 N mm ฮฆb = 0.85
End Moment Ms/Mr = 0.635 fy = 400 MPa ฮฑ R = 0 *ฮฑ L
+ve Moment Ms/Mr = 0.635 Eb = 200000 MPa n=Eb/Ec= 7.40741
Member Properties Determined from Provided Info (Primairly Servicability) Units
ฮฑL/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
ฮฑL = ML/M0 = -1.33 -1.10 -1 -0.85 -0.69 -0.55 -0.40 -0.20 0
ฮฑR = 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
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 N mm
ฮฑcr = Mcr/Mmax= 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75
Mcr = 4.54E+7 4.54E+7 4.54E+7 4.54E+7 4.54E+7 4.54E+7 4.54E+7 4.54E+7 4.54E+7 N mm
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
Member Properties Determined with Factored Loads
Left End Kr L = 5.21 3.65 3.10 2.39 1.76 0.00 0.00 0 0 MPa
ฯ L = 0.0184 0.0121 0.0100 0.0075 0.0054 0.0000 0.0000 0 0
AL=ฯLbd= 4301 2819 2346 1765 1271 0 0 0 0 mm2
Icr L = 8.99E+8 6.64E+8 5.79E+8 4.64E+8 3.56E+8 1.73E+9 1.73E+9 1.73E+9 1.73E+9 mm4
ฮทL=1 โ 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
ฯ m = 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054
Icr m = 3.54E+8 3.54E+8 3.54E+8 3.54E+8 3.54E+8 3.54E+8 3.54E+8 3.54E+8 3.54E+8 mm4
ฮทm=1 โ Icr m/Ig = 0.796 Mmax/Mcr = 1.33 Ig/Icr m = 4.90 Am=ฯmbd= 1261 mm2
Right End Kr R = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0 0 MPa
ฯ R = 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0
AR=ฯRbd= 0 0 0 0 0 0 0 0 0 mm2
Icr R = 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
ฮทR=1 โ 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
๐๐โฒ
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184
Table O-10 - Data for UDL Slab, MR=0, Ig/Icr=4.9 โ Example 3.6.2e โ 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
+ve Moment Ms/Mr = 0.635 ฯ m = 0.0054 fr = 0.6 *
Mmax/Mcr = 1.33 Ig/Icr m = 4.90 Eb = 200000 MPa
ฮฑL/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
Constant Stiffness Results Using Constant Stiffness Equations ฮmid=K(5MmL2)/(48EcI)
ฮg(Gross) 3.43 4.90 5.39 5.99 6.48 6.82 7.11 7.39 7.58 mm
ฮcr(Cracked) 16.84 24.02 26.41 29.34 31.78 33.45 34.85 36.21 37.14 mm
Branson's Method ฮIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3
Branson's Ie= 9.36E+8 9.36E+8 9.36E+8 9.36E+8 9.36E+8 9.36E+8 9.36E+8 9.36E+8 9.36E+8 mm4
ฮIe(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
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)= 9.12E+8 7.14E+8 6.65E+8 6.72E+8 9.24E+8 9.36E+8 9.36E+8 9.36E+8 9.36E+8 mm4
Ie R (Bransons)= 9.36E+8 9.36E+8 9.36E+8 9.36E+8 9.36E+8 9.36E+8 9.36E+8 9.36E+8 9.36E+8 mm4
Ie 9.8.2 (Bransons)= 9.32E+8 9.02E+8 8.95E+8 8.96E+8 9.34E+8 9.36E+8 9.36E+8 9.36E+8 9.36E+8 mm4
ฮIe,avg(A23.3) 6.39 9.41 10.43 11.58 12.03 12.64 13.17 13.68 14.03 mm
Deflection using the S806 Method with Numerical Integration
โฮฒ=0(S806) 16.82 20.85 22.60 24.91 27.01 28.41 29.88 31.19 32.47 mm
Exact Integration Ie(x)Analytical ฮ1= -2.13 -1.00 -0.64 -0.25 -0.04 0.00 0.00 0.00 0.00 mm
Analytical ฮ2 or ฮ1+2= 0.75 0.78 0.79 0.81 0.82 0.83 0.81 0.73 0.62 mm
Analytical ฮ3= 0.01 0.93 1.47 2.36 3.37 4.26 5.19 6.37 7.46 mm
Analytical ฮ4= 8.43 8.94 9.01 8.98 8.83 8.63 8.36 7.93 7.46 mm
Analytical ฮ5 or ฮ5+6= 0.27 0.32 0.35 0.38 0.42 0.46 0.50 0.56 0.62 mm
Analytical ฮ6= 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 mm
ฮIe(x)(Exact) 7.33 9.97 10.98 12.28 13.40 14.17 14.85 15.59 16.15 mm
Proposed Method ฮI'e I'e=Icr/[1-ฮณฮทm(Mcr/Mmax)2] ฮณ=(1.6ฮพ 3-0.6ฮพ 4)/(Mcr/Mmax)
2+2.4ln(2-ฮพ )
I'e (ฮณ=1) (M(x)=Mmax) = 6.40E+8 6.40E+8 6.40E+8 6.40E+8 6.40E+8 6.40E+8 6.40E+8 6.40E+8 6.40E+8 mm4
ฮฮณ=1(Approx) 9.30 13.27 14.58 16.20 17.55 18.47 19.24 20.00 20.51 mm
ฮพ =1-โ(1-Mcr/Mmax)= 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500
ฮณ= 1.26 1.26 1.26 1.26 1.26 1.26 1.26 1.26 1.26
Bischoff's I'e = 8.13E+8 8.13E+8 8.13E+8 8.13E+8 8.13E+8 8.13E+8 8.13E+8 8.13E+8 8.13E+8 mm4
ฮI'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
Maximum Deflection Results using numerical and approximation methods ฮmax โ ฮI'e โ(Mmax/Mm)
ฮmax,Ie(x)(Exact) 8.95 10.93 11.71 12.72 13.69 14.34 14.90 15.59 16.15 mm
Length:Defl, L/ฮmax= 838 686 641 590 548 523 503 481 464
ฮmax,I'e(Proposed) 8.45 11.29 12.18 13.25 14.13 14.74 15.25 15.77 16.15 mm
๐๐โฒ
185
Figure O-5 - Copy of Figure 3-12 โ 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 โ Example 3.6.2f โ Page 1
Example 3.6.2f, pg 1 of 2 ฮฆc = 0.65 ฮตcu = 0.0035 mm/mm
UDL Continuous w0 = 5.268 N/mm fc' = 36 MPa fr = 0.6 * MPa
Vary ML, Mm constant L = 5000 mm Ec = 27000 MPa Mcr= 0.6667 * N mm
Let ML=MR b = 6.667* h mm ฮฑ1 = 0.796
d = 0.5 * h mm ฮฒ1 = 0.880 ฯ b=ฮฑ1ฮฒ1ฯcf'cฮตcu/(ฯbffu(ฮตcu+ffu/Ef))
M0,0 = w0 L2/8 = 1.65E+7 N mm ฯb = 0.75 ฯ b = 0.00578
End Moment Ms/Mr = 0.635 ffu = 690 MPa ฮฑ R = 1 *ฮฑ L
+ve Moment Ms/Mr = 0.552 Eb = 44000 MPa n=Eb/Ec= 1.63
Member Properties Determined from Provided Info (Primairly Servicability) Units
ฮฑL/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
ฮฑL = ML/M0 = -0.75 -0.66 -0.61 -0.55 -0.50 -0.46 -0.30 -0.20 0
ฮฑR = 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
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 N mm
ฮฑcr = Mcr/Mmax= 0.82 0.82 0.82 0.82 0.82 0.82 0.82 0.82 0.82
Mcr = 1.35E+7 1.35E+7 1.35E+7 1.35E+7 1.35E+7 1.35E+7 1.35E+7 1.35E+7 1.35E+7 N mm
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
Member Properties Determined with Factored Loads
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
ฯ 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
ฮทL=1 โ 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
ฯ m =Am/bd= 0.0355 0.0355 0.0355 0.0355 0.0355 0.0355 0.0355 0.0355 0.0355
Icr m = 1.57E+7 1.57E+7 1.57E+7 1.57E+7 1.57E+7 1.57E+7 1.57E+7 1.57E+7 1.57E+7 mm4
ฮทm=1 โ Icr m/Ig = 0.944 Mmax/Mcr = 1.22 Ig/Icr m = 17.86 Am=ฯmbd= 2665 mm2
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
ฯ 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
ฮทR=1 โ 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
๐๐โฒ
โ3 ๐๐โฒ
187
Table O-12 - Data for UDL Slab, ML=MR, Ig/Icr=18 โ Example 3.6.2f โ Page 2
Ex. 3.6.2f, pg 2 of 2 w0 = 5.27 N/mm fc' = 36 MPa ฯ b = 0.00578
L = 5000 mm ffu = 690 MPa ฯ m = 0.0355
Ms /Mr (+ve) = 1.150 Mmax/Mcr = 1.22 Ig/Icr m = 17.86 Eb = 44000 MPa
ฮฑL/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
Constant Stiffness Results Using Constant Stiffness Equations ฮmid=K(5MmL2)/(48EcI)
ฮg(Gross) 2.26 3.45 3.88 4.27 4.52 4.68 5.16 5.36 5.65 mm
ฮcr(Cracked) 40.35 61.71 69.31 76.21 80.69 83.68 92.22 95.82 100.87 mm
Branson's Method ฮIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3
Branson's Ie= 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 mm4
ฮIe(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+(ฮฒdIg-Icr)(Mcr/Mmax)3 ฮฒd= 1.000 ฮฒd=0.2(ฯm/ฯ b)<1 Ie R = Ie L
Ie m (ACI440.1R) = 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 1.62E+8 mm4
ฮIe,ฮฒd(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
ฮIe,avg(A23.3) #NUM! 6.50 8.05 8.40 8.35 8.65 8.95 9.30 9.79 mm
Deflection using the S806 Method with Numerical Integration
โฮฒ=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 ฮ1= #NUM! -0.32 -0.30 -0.13 -0.03 0.00 0.00 0.00 0.00 mm
Analytical ฮ2 or ฮ1+2= 0.45 0.52 0.56 0.60 0.63 0.66 0.71 0.71 0.68 mm
Analytical ฮ3= 6.17 7.08 7.52 8.01 8.38 8.67 9.67 10.22 11.19 mm
Analytical ฮ4= 6.17 7.08 7.52 8.01 8.38 8.67 9.67 10.22 11.19 mm
Analytical ฮ5 or ฮ5+6= 0.45 0.52 0.56 0.60 0.63 0.66 0.71 0.71 0.68 mm
Analytical ฮ6= #NUM! -0.32 -0.30 -0.13 -0.03 0.00 0.00 0.00 0.00 mm
ฮIe(x)(Exact) #NUM! 14.55 15.56 16.95 17.98 18.64 20.76 21.86 23.72 mm
Length:Defl, L/ฮmid= #NUM! 344 321 295 278 268 241 229 211
Proposed Method ฮI'e I'e=Icr/[1-ฮณฮทm(Mcr/Mmax)2] ฮณ=(1.6ฮพ
3-0.6ฮพ
4)/(Mcr/Mmax)
2+2.4ln(2-ฮพ )
I'e (ฮณ=1) (M(x)=Mmax) = 4.31E+7 4.31E+7 4.31E+7 4.31E+7 4.31E+7 4.31E+7 4.31E+7 4.31E+7 4.31E+7 mm4
ฮฮณ=1(Approx) 14.74 22.54 25.32 27.83 29.47 30.56 33.68 35.00 36.84 mm
ฮพ =1-โ(1-Mcr/Mmax)= 0.576 0.576 0.576 0.576 0.576 0.576 0.576 0.576 0.576
ฮณ= 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20
Bischoff's I'e = 6.69E+7 6.69E+7 6.69E+7 6.69E+7 6.69E+7 6.69E+7 6.69E+7 6.69E+7 6.69E+7 mm4
ฮI'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/ฮ exact #NUM! 343.62 321.40 294.92 278.10 268.17 240.82 228.71 210.75
Maximum Deflection Results using numerical and approximation methods ฮmax โ ฮI'e โ(Mmax/Mm)
ฮmax,Ie(x)(Exact) #NUM! 14.55 15.56 16.96 17.98 18.65 20.76 21.86 23.73 mm
ฮmax,I'e(Proposed) 9.49 14.51 16.30 17.93 18.98 19.68 21.69 22.54 23.72 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 โ Example 3.6.2g โ Page 1
Example 3.6.2g, pg 1 of 2 ฮฆc = 0.65 ฮตcu = 0.0035 mm/mm
UDL Continuous w0 = 10.370 N/mm fc' = 36 MPa fr = 0.6 * MPa
Vary ML, Mm constant L = 10000 mm Ec = 27000 MPa Mcr= 0.05 * N mm
Let MR=0 b = 0.5* h mm ฮฑ1 = 0.796
d = 0.9 * h mm ฮฒ1 = 0.880 ฯ b=ฮฑ1ฮฒ1ฯcf'cฮตcu/(ฯbffu(ฮตcu+ffu/Ef))
M0,0 = w0 L2/8 = 1.30E+8 N mm ฯb = 0.75 ฯ b = 0.00578
End Moment Ms/Mr = 0.635 ffu = 690 MPa ฮฑ R = 0 *ฮฑ L
+ve Moment Ms/Mr = 0.365 Eb = 44000 MPa n=Eb/Ec= 1.63
Member Properties Determined from Provided Info (Primairly Servicability) Units
ฮฑL/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
ฮฑL = ML/M0 = -1.33 -1.10 -0.9 -0.8 -0.69 -0.50 -0.30 -0.20 0
ฮฑR = 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
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 N mm
ฮฑcr = Mcr/Mmax= 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5
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
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
Member Properties Determined with Factored Loads
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
ฯ L =As/bd= 0.1045 0.0272 0.0106 0.0067 0.0040 0.0015 0 0 0
Icr L = 3.87E+9 1.43E+9 6.38E+8 4.23E+8 2.63E+8 1.04E+8 5.40E+9 5.40E+9 5.40E+9 mm4
ฮทL=1 โ 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
ฯ m =Am/bd= 0.0158 0.0158 0.0158 0.0158 0.0158 0.0158 0.0158 0.0158 0.0158
Icr m = 9.06E+8 9.06E+8 9.06E+8 9.06E+8 9.06E+8 9.06E+8 9.06E+8 9.06E+8 9.06E+8 mm4
ฮทm=1 โ Icr m/Ig = 0.832 Mmax/Mcr = 2.00 Ig/Icr m = 5.96 Am=ฯmbd= 2564 mm2
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
ฯ R =AR/bd= 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
Icr R = 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
ฮทR=1 โ 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
๐๐โฒ
โ3 ๐๐โฒ
190
Table O-14 - Data for UDL Slab, MR=0, Ig/Icr=6 โ Example 3.6.2g โ Page 2
Ex. 3.6.2g, pg 2 of 2 w0 = 10.37 N/mm fc' = 36 MPa ฯ b = 0.00578
L = 10000 mm ffu = 690 MPa ฯ m = 0.0158
Ms /Mr (+ve) = 1.740 Mmax/Mcr = 2.00 Ig/Icr m = 5.96 Eb = 44000 MPa
ฮฑL/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
Constant Stiffness Results Using Constant Stiffness Equations ฮmid=K(5MmL2)/(48EcI)
ฮg(Gross) 4.20 5.99 7.09 7.52 7.92 8.47 8.87 9.03 9.26 mm
ฮcr(Cracked) 25.03 35.71 42.28 44.85 47.24 50.47 52.90 53.83 55.20 mm
Branson's Method ฮIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3
Branson's Ie= 1.47E+9 1.47E+9 1.47E+9 1.47E+9 1.47E+9 1.47E+9 1.47E+9 1.47E+9 1.47E+9 mm4
ฮIe(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
ACI 440.1R clause 8.3.2.2 Ie=Icr+(ฮฒdIg-Icr)(Mcr/Mmax)3 ฮฒd= 0.548 ฮฒd=0.2(ฯm/ฯ b)<1 Ie R = Ie L
Ie m (ACI440.1R) = 1.16E+9 1.16E+9 1.16E+9 1.16E+9 1.16E+9 1.16E+9 1.16E+9 1.16E+9 1.16E+9 mm4
ฮIe,ฮฒd(ACI440) 19.50 27.83 32.95 34.95 36.82 39.33 41.23 41.95 43.02 mm
Ie L (ACI440.1R) = 3.87E+9 1.48E+9 7.03E+8 4.97E+8 3.49E+8 2.26E+8 1.16E+9 1.16E+9 1.16E+9 mm4
Ie 9.8.2 (& ACI440.1R)= 1.57E+9 1.21E+9 1.09E+9 1.06E+9 1.04E+9 1.02E+9 1.16E+9 1.16E+9 1.16E+9 mm4
ฮIe,avg(A23.3) 14.45 26.74 35.02 38.24 41.13 44.74 41.23 41.95 43.02 mm
Deflection using the S806 Method with Numerical Integration
โฮฒ=0(S806) 31.69 35.82 39.78 41.67 44.18 48.19 50.90 52.02 53.53 mm
Exact Integration Ie(x)Analytical ฮ1= -2.38 -2.43 -1.86 -1.40 -0.85 -0.08 0.00 0.00 0.00 mm
Analytical ฮ2 or ฮ1+2= 0.23 0.24 0.24 0.25 0.25 0.25 0.25 0.23 0.17 mm
Analytical ฮ3= 2.62 5.57 8.34 9.73 11.23 13.72 16.13 17.24 19.24 mm
Analytical ฮ4= 21.03 21.56 21.55 21.44 21.25 20.81 20.24 19.92 19.24 mm
Analytical ฮ5 or ฮ5+6= 0.07 0.09 0.10 0.11 0.11 0.13 0.14 0.15 0.17 mm
Analytical ฮ6= 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 mm
ฮIe(x)(Exact) 21.57 25.02 28.38 30.12 32.00 34.83 36.75 37.53 38.82 mm
Length:Defl, L/ฮmid= 464 400 352 332 312 287 272 266 258
Proposed Method ฮI'e I'e=Icr/[1-ฮณฮทm(Mcr/Mmax)2] ฮณ=(1.6ฮพ 3-0.6ฮพ 4)/(Mcr/Mmax)
2+2.4ln(2-ฮพ )
I'e (ฮณ=1) (M(x)=Mmax) = 1.14E+9 1.14E+9 1.14E+9 1.14E+9 1.14E+9 1.14E+9 1.14E+9 1.14E+9 1.14E+9 mm4
ฮฮณ=1(Approx) 19.82 28.28 33.48 35.52 37.41 39.97 41.90 42.63 43.72 mm
ฮพ =1-โ(1-Mcr/Mmax)= 0.293 0.293 0.293 0.293 0.293 0.293 0.293 0.293 0.293
ฮณ= 1.43 1.43 1.43 1.43 1.43 1.43 1.43 1.43 1.43
Bischoff's I'e = 1.29E+9 1.29E+9 1.29E+9 1.29E+9 1.29E+9 1.29E+9 1.29E+9 1.29E+9 1.29E+9 mm4
ฮI'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/ฮ exact 463.55 399.63 352.35 332.03 312.49 287.14 272.08 266.44 257.62
Maximum Deflection Results using numerical and approximation methods ฮmax โ ฮI'e โ(Mmax/Mm)
ฮmax,Ie(x)(Exact) 24.15 26.86 29.41 30.93 32.58 35.02 36.75 37.53 38.82 mm
ฮmax,I'e(Proposed) 20.30 27.14 31.07 32.57 33.96 35.86 37.32 37.90 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
O-13 and Table O-14.
192
Table O-15 - Data for UDL Beam, ML=MR, Ig/Icr=17 โ Example 3.6.2h โ Page 1
Example 3.6.2h, pg 1 of 2 ฮฆc = 0.65 ฮตcu = 0.0035 mm/mm
UDL Continuous w0 = 6.480 N/mm 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 ฮฑ1 = 0.796
d = 0.85 * h mm ฮฒ1 = 0.880 ฯ b=ฮฑ1ฮฒ1ฯcf'cฮตcu/(ฯbffu(ฮตcu+ffu/Ef))
M0,0 = w0 L2/8 = 8.10E+7 N mm ฯb = 0.75 ฯ b = 0.00578
End Moment Ms/Mr = 0.635 ffu = 690 MPa ฮฑ R = 1 *ฮฑ L
+ve Moment Ms/Mr = 0.374 Eb = 44000 MPa n=Eb/Ec= 1.63
Member Properties Determined from Provided Info (Primairly Servicability) Units
ฮฑL/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
ฮฑL = ML/M0 = -0.75 -0.70 -0.61 -0.55 -0.50 -0.46 -0.30 -0.20 0
ฮฑR = 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
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 N mm
ฮฑcr = Mcr/Mmax= 0.8 0.8 0.8 0.8 0.8 0.8 0.8 0.8 0.8
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 = 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
Member Properties Determined with Factored Loads
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
ฯ L =As/bd= 0.0276 0.0133 0.0048 0.0027 0.0017 0.0012 0 0 0
Icr L = 1.21E+9 6.59E+8 2.66E+8 1.56E+8 1.02E+8 7.26E+7 5.40E+9 5.40E+9 5.40E+9 mm4
ฮทL=1 โ 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
ฯ m =Am/bd= 0.0059 0.0059 0.0059 0.0059 0.0059 0.0059 0.0059 0.0059 0.0059
Icr m = 3.19E+8 3.19E+8 3.19E+8 3.19E+8 3.19E+8 3.19E+8 3.19E+8 3.19E+8 3.19E+8 mm4
ฮทm=1 โ Icr m/Ig = 0.941 Mmax/Mcr = 1.25 Ig/Icr m = 16.92 Am=ฯmbd= 904 mm2
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
ฯ R =AR/bd= 0.0276 0.0133 0.0048 0.0027 0.0017 0.0012 0.0000 0.0000 0.0000
Icr R = 1.21E+9 6.59E+8 2.66E+8 1.56E+8 1.02E+8 7.26E+7 5.40E+9 5.40E+9 5.40E+9 mm4
ฮทR=1 โ 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
๐๐โฒ
โ3 ๐๐โฒ
193
Table O-16 - Data for UDL Beam, ML=MR, Ig/Icr=17 โ Example 3.6.2h โ Page 2
Ex. 3.6.2h, pg 2 of 2 w0 = 6.48 N/mm fc' = 36 MPa ฯ b = 0.00578
L = 10000 mm ffu = 690 MPa ฯ m = 0.0059
Ms /Mr (+ve) = 1.698 Mmax/Mcr = 1.25 Ig/Icr m = 16.92 Eb = 44000 MPa
ฮฑL/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
Constant Stiffness Results Using Constant Stiffness Equations ฮmid=K(5MmL2)/(48EcI)
ฮg(Gross) 2.31 3.09 3.98 4.37 4.63 4.80 5.29 5.50 5.79 mm
ฮcr(Cracked) 39.17 52.23 67.29 73.99 78.34 81.24 89.53 93.03 97.92 mm
Branson's Method ฮIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3
Branson's Ie= 2.92E+9 2.92E+9 2.92E+9 2.92E+9 2.92E+9 2.92E+9 2.92E+9 2.92E+9 2.92E+9 mm4
ฮIe(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
ACI 440.1R clause 8.3.2.2 Ie=Icr+(ฮฒdIg-Icr)(Mcr/Mmax)3 ฮฒd= 0.204 ฮฒd=0.2(ฯm/ฯ b)<1 Ie R = Ie L
Ie m (ACI440.1R) = 7.21E+8 7.21E+8 7.21E+8 7.21E+8 7.21E+8 7.21E+8 7.21E+8 7.21E+8 7.21E+8 mm4
ฮIe,ฮฒd(ACI440) 17.34 23.12 29.79 32.75 34.68 35.96 39.63 41.18 43.34 mm
Ie L (ACI440.1R) = 1.29E+9 7.47E+8 3.81E+8 2.94E+8 2.64E+8 2.58E+8 7.21E+8 7.21E+8 7.21E+8 mm4
Ie 9.8.2 (& ACI440.1R)= 8.92E+8 7.29E+8 6.19E+8 5.93E+8 5.84E+8 5.82E+8 7.21E+8 7.21E+8 7.21E+8 mm4
ฮIe,avg(A23.3) 14.02 22.87 34.69 39.82 42.81 44.53 39.63 41.18 43.34 mm
Deflection using the S806 Method with Numerical Integration
โฮฒ=0(S806) 39.95 43.22 49.54 55.95 58.87 61.37 68.68 72.66 77.97 mm
Exact Integration Ie(x)Analytical ฮ1= -1.81 -1.60 -0.82 -0.32 -0.07 0.00 0.00 0.00 0.00 mm
Analytical ฮ2 or ฮ1+2= 0.41 0.45 0.52 0.56 0.59 0.61 0.66 0.66 0.62 mm
Analytical ฮ3= 6.83 7.41 8.32 8.85 9.26 9.57 10.67 11.27 12.32 mm
Analytical ฮ4= 6.83 7.41 8.32 8.85 9.26 9.57 10.67 11.27 12.32 mm
Analytical ฮ5 or ฮ5+6= 0.41 0.45 0.52 0.56 0.59 0.61 0.66 0.66 0.62 mm
Analytical ฮ6= -1.81 -1.60 -0.82 -0.32 -0.07 0.00 0.00 0.00 0.00 mm
ฮIe(x)(Exact) 10.86 12.53 16.03 18.18 19.56 20.36 22.66 23.85 25.87 mm
Length:Defl, L/ฮmid= 921 798 624 550 511 491 441 419 387
Proposed Method ฮI'e I'e=Icr/[1-ฮณฮทm(Mcr/Mmax)2] ฮณ=(1.6ฮพ 3-0.6ฮพ 4)/(Mcr/Mmax)
2+2.4ln(2-ฮพ )
I'e (ฮณ=1) (M(x)=Mmax) = 8.02E+8 8.02E+8 8.02E+8 8.02E+8 8.02E+8 8.02E+8 8.02E+8 8.02E+8 8.02E+8 mm4
ฮฮณ=1(Approx) 15.58 20.78 26.77 29.43 31.17 32.32 35.62 37.01 38.96 mm
ฮพ =1-โ(1-Mcr/Mmax)= 0.553 0.553 0.553 0.553 0.553 0.553 0.553 0.553 0.553
ฮณ= 1.22 1.22 1.22 1.22 1.22 1.22 1.22 1.22 1.22
Bischoff's I'e = 1.21E+9 1.21E+9 1.21E+9 1.21E+9 1.21E+9 1.21E+9 1.21E+9 1.21E+9 1.21E+9 mm4
ฮI'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/ฮ exact 920.65 797.90 623.87 550.13 511.12 491.20 441.36 419.27 386.51
Maximum Deflection Results using numerical and approximation methods ฮmax โ ฮI'e โ(Mmax/Mm)
ฮmax,Ie(x)(Exact) 10.87 12.54 16.03 18.18 19.57 20.36 22.66 23.85 25.87 mm
ฮmax,I'e(Proposed) 10.35 13.80 17.78 19.55 20.70 21.46 23.65 24.58 25.87 mm
194
Figure O-8 - Copy of Figure 3-13 โ 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
O-15 and Table O-16.
195
Results Using New Mcr per CSA A23.3-04 (R2010) Appendix P
The use of ๐๐๐ = 0.5๐๐๐ผ๐/๐ฆ๐ก, 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 Bransonโs ๐ผ๐ equation, as provided in CSA A23.3.
Alternatively, use of ๐๐๐ = 0.67๐๐๐ผ๐/๐ฆ๐ก provides an equivalent adjustment for ๐ผ๐โฒ or
๐ผ๐(๐ฅ) 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 ๐ผ๐(๐ฅ) with ๐๐๐ = 0.67๐๐๐ผ๐/๐ฆ๐ก 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 ๐ผ๐/๐ผ๐๐ ratio, but
the ๐๐๐๐ฅ/๐๐๐ 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 ๐๐๐๐ฅ/๐๐๐ increase and
results are conservative when using Bransonโs ๐ผ๐ with the 0.5 factor for ๐๐๐. Bransonโs
๐ผ๐ 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 ๐๐๐๐ฅ/๐๐๐ increase and results
using Bransonโs ๐ผ๐ with the 0.5 factor for ๐๐๐ 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 ๐๐๐๐ฅ/๐๐๐ increase and
results using Bransonโs ๐ผ๐ with the 0.5 factor for ๐๐๐ 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 ๐ผ๐/๐ผ๐๐ < 12, then using Bransonโs ๐ผ๐ with the 0.5 factor
for ๐๐๐ will result in higher deflection predictions than using ๐ผ๐โฒ or ๐ผ๐(๐ฅ) with 0.67
factor for ๐๐๐. Results using Bransonโs ๐ผ๐ are unconservative when ๐ผ๐/๐ผ๐๐ > 12. In
spreadsheet testing that is not provided, the ๐ผ๐ reduced for shrinkage restraint was found
to highly underestimate deflection when ๐ผ๐/๐ผ๐๐ is much larger than 12. In other omitted
spreadsheet testing, ๐ผ๐ results sometimes highly overestimate deflection when shrinkage
restraint is included, ๐๐๐๐ฅ/๐๐๐ < 1.5, and ๐ผ๐/๐ผ๐๐ < 10.
197
Table P-1 - Data for UDL Beam, Ig/Icr=3.0, New A23.3 Mcr Example P1 โ Page 1
Example P1 (Example 3.6.2a modified to New Mcr) ฮฆc = 0.65 pg 1/2
UDL Continuous w0 = 11.27 N/mm fc' = 36 MPa fr = 0.6 * MPa
Vary ML, Mm constant L = 10000 mm Ec = 27000 MPa use: Mcr = 0.67 frIg/yt
Let ML=MR b = 0.5* h mm ฮฑ1 = 0.796 Mcr= 0.0335 * N mm
d = 0.9 * h mm ฮฒ1 = 0.880
M0,0 = w0 L2/8 = 1.41E+8 N mm ฮฆb = 0.85
End Moment Ms/Mr = 0.635 fy = 400 MPa ฮฑ R = 1 *ฮฑ L
+ve Moment Ms/Mr = 0.635 Eb = 200000 MPa n=Eb/Ec= 7.40741
Member Properties Determined from Provided Info (Primairly Servicability) Units
ฮฑL/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
ฮฑL = ML/M0 = -0.75 -0.70 -0.65 -0.55 -0.50 -0.40 -0.30 -0.20 0
ฮฑR = 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
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 N mm
ฮฑcr = Mcr/Mmax= 0.3082 0.3082 0.3082 0.3082 0.3082 0.3082 0.3082 0.3082 0.3082
Mcr = 4.34E+7 4.34E+7 4.34E+7 4.34E+7 4.34E+7 4.34E+7 4.34E+7 4.34E+7 4.34E+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
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
Member Properties Determined with Factored Loads
Left End Kr L = 7.61 5.92 4.71 3.10 2.54 1.69 1.09 0 0 MPa
ฯ L = 0.0313 0.0217 0.0163 0.0100 0.0081 0.0052 0.0033 0 0
AL=ฯLbd= 5077 3516 2635 1626 1304 846 534 0 0 mm2
Icr L = 4.71E+9 3.72E+9 3.05E+9 2.14E+9 1.80E+9 1.27E+9 8.64E+8 5.40E+9 5.40E+9 mm4
ฮทL=1 โ 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
ฯ m = 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081
Icr m = 1.80E+9 1.80E+9 1.80E+9 1.80E+9 1.80E+9 1.80E+9 1.80E+9 1.80E+9 1.80E+9 mm4
ฮทm=1 โ Icr m/Ig = 0.666 Mmax/Mcr = 3.24 Ig/Icr m = 2.99 Am=ฯmbd= 1304 mm2
Right End Kr R = 7.61 5.92 4.71 3.10 2.54 1.69 1.09 0 0 MPa
ฯ R = 0.0313 0.0217 0.0163 0.0100 0.0081 0.0052 0.0033 0 0
AR=ฯRbd= 5077 3516 2635 1626 1304 846 534 0 0 mm2
Icr R = 4.71E+9 3.72E+9 3.05E+9 2.14E+9 1.80E+9 1.27E+9 8.64E+8 5.40E+9 5.40E+9 mm4
ฮทR=1 โ 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
๐๐โฒ
โ3 ๐๐โฒ
198
Table P-2 - Data for UDL Beam, Ig/Icr=3.0, New A23.3 Mcr Example P1 โ Page 2
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
+ve Moment Ms/Mr = 0.635 ฯ m = 0.0081 fr = 0.6 *
Mmax/Mcr = 3.24 Ig/Icr m = 2.99 Eb = 200000 MPa
ฮฑL/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
Constant Stiffness Results Using Constant Stiffness Equations ฮmid=K(5MmL2)/(48EcI)
ฮg(Gross) 4.03 5.37 6.33 7.60 8.05 8.72 9.20 9.56 10.06 mm
ฮcr(Cracked) 12.05 16.07 18.94 22.76 24.10 26.11 27.54 28.62 30.13 mm
Branson's Method ฮIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3 Mcr =(0.3โf'c)Ig/y= N mm
Branson's Ie= 1.85E+9 1.85E+9 1.85E+9 1.85E+9 1.85E+9 1.85E+9 1.85E+9 1.85E+9 1.85E+9 mm4
ฮIe(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
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)= 4.71E+9 3.72E+9 3.06E+9 2.16E+9 1.85E+9 1.44E+9 1.57E+9 5.40E+9 1.85E+9 mm4
Ie R (Bransons)= 4.71E+9 3.72E+9 3.06E+9 2.16E+9 1.85E+9 1.44E+9 1.57E+9 5.40E+9 1.85E+9 mm4
Ie 9.8.2 (Bransons)= 2.71E+9 2.41E+9 2.21E+9 1.94E+9 1.85E+9 1.73E+9 1.76E+9 2.91E+9 1.85E+9 mm4
ฮIe,avg(A23.3) 8.04 12.03 15.46 21.15 23.53 27.29 28.18 17.72 29.41 mm
Deflection using the S806 Method with Numerical Integration
โฮฒ=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 ฮ1= -1.36 -1.06 -0.79 -0.40 -0.26 -0.08 -0.01 0.00 0.00 mm
Analytical ฮ2 or ฮ1+2= 0.03 0.03 0.04 0.04 0.04 0.05 0.05 0.05 0.04 mm
Analytical ฮ3= 8.49 9.13 9.68 10.61 11.01 11.71 12.29 12.78 13.57 mm
Analytical ฮ4= 8.49 9.13 9.68 10.61 11.01 11.71 12.29 12.78 13.57 mm
Analytical ฮ5 or ฮ5+6= 0.03 0.03 0.04 0.04 0.04 0.05 0.05 0.05 0.04 mm
Analytical ฮ6= -1.36 -1.06 -0.79 -0.40 -0.26 -0.08 -0.01 0.00 0.00 mm
ฮIe(x)(Exact) 14.34 16.21 17.85 20.51 21.59 23.34 24.66 25.68 27.21 mm
Proposed Method ฮI'e I'e=Icr/[1-ฮณฮทm(Mcr/Mmax)2] ฮณ=(1.6ฮพ 3-0.6ฮพ 4)/(Mcr/Mmax)
2+2.4ln(2-ฮพ )
I'e (ฮณ=1) (M(x)=Mmax) = 1.93E+9 1.93E+9 1.93E+9 1.93E+9 1.93E+9 1.93E+9 1.93E+9 1.93E+9 1.93E+9 mm4
ฮฮณ=1(Approx) 11.29 15.05 17.74 21.32 22.58 24.46 25.80 26.81 28.22 mm
ฮพ =1-โ(1-Mcr/Mmax)= 0.168 0.168 0.168 0.168 0.168 0.168 0.168 0.168 0.168
ฮณ= 1.53 1.53 1.53 1.53 1.53 1.53 1.53 1.53 1.53
Bischoff's I'e = 2.00E+9 2.00E+9 2.00E+9 2.00E+9 2.00E+9 2.00E+9 2.00E+9 2.00E+9 2.00E+9 mm4
ฮI'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
Maximum Deflection Results using numerical and approximation methods ฮmax โ ฮI'e โ(Mmax/Mm)
ฮmax,Ie(x)(Exact) 14.34 16.21 17.85 20.51 21.59 23.34 24.66 25.68 27.21 mm
Length:Defl, L/ฮmax= 697 617 560 488 463 428 405 389 367
ฮmax,I'e(Proposed) 10.89 14.51 17.11 20.56 21.77 23.59 24.88 25.85 27.21 mm
3.24E+7
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199
Figure P-1 - Midspan Deflection Computed using Shrinkage Restraint Mcr โ 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 โ Compare to Figure P-1
๐๐๐ = ๐๐๐ผ๐/๐ฆ๐ก
(all)
200
Table P-3 - Data for UDL Beam, Ig/Icr=4.9, New A23.3 Mcr Example P2 โ Page 1
Example P2 (Example 3.6.2e modified to New Mcr) ฮฆc = 0.65 pg 1/2
UDL Continuous w0 = 8.605 N/mm fc' = 36 MPa fr = 0.6 * MPa
Vary ML, Mm constant L = 7500 mm Ec = 27000 MPa use: Mcr = 0.67 frIg/yt
Let MR=0 b = 3.636* h mm ฮฑ1 = 0.796 Mcr= 0.24361 * N mm
d = 0.85 * h mm ฮฒ1 = 0.880
M0,0 = w0 L2/8 = 6.05E+7 N mm ฮฆb = 0.85
End Moment Ms/Mr = 0.635 fy = 400 MPa ฮฑ R = 0 *ฮฑ L
+ve Moment Ms/Mr = 0.635 Eb = 200000 MPa n=Eb/Ec= 7.40741
Member Properties Determined from Provided Info (Primairly Servicability) Units
ฮฑL/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
ฮฑL = ML/M0 = -1.33 -1.10 -1 -0.85 -0.69 -0.55 -0.40 -0.20 0
ฮฑR = 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
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 N mm
ฮฑcr = Mcr/Mmax= 0.5025 0.5025 0.5025 0.5025 0.5025 0.5025 0.5025 0.5025 0.5025
Mcr = 3.04E+7 3.04E+7 3.04E+7 3.04E+7 3.04E+7 3.04E+7 3.04E+7 3.04E+7 3.04E+7 N mm
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
Member Properties Determined with Factored Loads
Left End Kr L = 5.21 3.65 3.10 2.39 1.76 1.29 0.00 0 0 MPa
ฯ L = 0.0184 0.0121 0.0100 0.0075 0.0054 0.0039 0.0000 0 0
AL=ฯLbd= 4301 2819 2346 1765 1271 919 0 0 0 mm2
Icr L = 8.99E+8 6.64E+8 5.79E+8 4.64E+8 3.56E+8 2.72E+8 1.73E+9 1.73E+9 1.73E+9 mm4
ฮทL=1 โ 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
ฯ m = 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054 0.0054
Icr m = 3.54E+8 3.54E+8 3.54E+8 3.54E+8 3.54E+8 3.54E+8 3.54E+8 3.54E+8 3.54E+8 mm4
ฮทm=1 โ Icr m/Ig = 0.796 Mmax/Mcr = 1.99 Ig/Icr m = 4.90 Am=ฯmbd= 1261 mm2
Right End Kr R = 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0 0 MPa
ฯ R = 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0
AR=ฯRbd= 0 0 0 0 0 0 0 0 0 mm2
Icr R = 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
ฮทR=1 โ 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
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201
Table P-4 - Data for UDL Beam, Ig/Icr=4.9, New A23.3 Mcr Example P2 โ Page 2
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
+ve Moment Ms/Mr = 0.635 ฯ m = 0.0054 fr = 0.6 *
Mmax/Mcr = 1.99 Ig/Icr m = 4.90 Eb = 200000 MPa
ฮฑL/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
Constant Stiffness Results Using Constant Stiffness Equations ฮmid=K(5MmL2)/(48EcI)
ฮg(Gross) 3.43 4.90 5.39 5.99 6.48 6.82 7.11 7.39 7.58 mm
ฮcr(Cracked) 16.84 24.02 26.41 29.34 31.78 33.45 34.85 36.21 37.14 mm
Branson's Method ฮIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3 Mcr =(0.3โf'c)Ig/y= N mm
Branson's Ie= 4.26E+8 4.26E+8 4.26E+8 4.26E+8 4.26E+8 4.26E+8 4.26E+8 4.26E+8 4.26E+8 mm4
ฮIe(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
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)= 9.01E+8 6.71E+8 5.89E+8 4.90E+8 4.27E+8 4.62E+8 1.73E+9 4.26E+8 4.26E+8 mm4
Ie R (Bransons)= 4.26E+8 4.26E+8 4.26E+8 4.26E+8 4.26E+8 4.26E+8 4.26E+8 4.26E+8 4.26E+8 mm4
Ie 9.8.2 (Bransons)= 4.97E+8 4.63E+8 4.51E+8 4.36E+8 4.26E+8 4.32E+8 6.22E+8 4.26E+8 4.26E+8 mm4
ฮIe,avg(A23.3) 11.97 18.35 20.71 23.80 26.35 27.39 19.80 30.03 30.80 mm
Deflection using the S806 Method with Numerical Integration
โฮฒ=0(S806) 20.21 24.48 26.09 28.49 30.65 32.23 33.59 35.05 36.06 mm
Exact Integration Ie(x)Analytical ฮ1= -2.57 -1.43 -1.02 -0.54 -0.20 -0.04 0.00 0.00 0.00 mm
Analytical ฮ2 or ฮ1+2= 0.19 0.20 0.20 0.20 0.21 0.21 0.21 0.19 0.14 mm
Analytical ฮ3= 1.82 3.84 4.78 6.20 7.69 8.94 10.21 11.75 13.11 mm
Analytical ฮ4= 14.32 14.67 14.69 14.63 14.46 14.25 13.98 13.57 13.11 mm
Analytical ฮ5 or ฮ5+6= 0.06 0.07 0.08 0.09 0.09 0.10 0.11 0.12 0.14 mm
Analytical ฮ6= 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 mm
ฮIe(x)(Exact) 13.82 17.35 18.73 20.58 22.26 23.47 24.51 25.63 26.50 mm
Proposed Method ฮI'e I'e=Icr/[1-ฮณฮทm(Mcr/Mmax)2] ฮณ=(1.6ฮพ
3-0.6ฮพ
4)/(Mcr/Mmax)
2+2.4ln(2-ฮพ )
I'e (ฮณ=1) (M(x)=Mmax) = 4.42E+8 4.42E+8 4.42E+8 4.42E+8 4.42E+8 4.42E+8 4.42E+8 4.42E+8 4.42E+8 mm4
ฮฮณ=1(Approx) 13.45 19.19 21.10 23.45 25.39 26.73 27.84 28.93 29.67 mm
ฮพ =1-โ(1-Mcr/Mmax)= 0.295 0.295 0.295 0.295 0.295 0.295 0.295 0.295 0.295
ฮณ= 1.43 1.43 1.43 1.43 1.43 1.43 1.43 1.43 1.43
Bischoff's I'e = 4.95E+8 4.95E+8 4.95E+8 4.95E+8 4.95E+8 4.95E+8 4.95E+8 4.95E+8 4.95E+8 mm4
ฮI'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
Maximum Deflection Results using numerical and approximation methods ฮmax โ ฮI'e โ(Mmax/Mm)
ฮmax,Ie(x)(Exact) 15.76 18.55 19.60 21.14 22.60 23.64 24.53 25.64 26.50 mm
Length:Defl, L/ฮmax= 476 404 383 355 332 317 306 293 283
ฮmax,I'e(Proposed) 13.86 18.53 19.99 21.74 23.19 24.18 25.02 25.87 26.50 mm
2.27E+7
๐๐โฒ
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Figure P-3 โ Midspan and Maximum Deflection Computed using Shrinkage Restraint
Mcr โ Slab with Ig/Icr=5, Mmax/Mcr=2, and MR=0
The lines plotted in Figure P-3 use data from Example P2 per Table P-3 and Table P-4.
Figure P-4 - Copy of Figure O-5, Ig/Icr=5, Mmax/Mcr=1.3 โ Compare to Figure P-3
๐๐๐ = ๐๐๐ผ๐/๐ฆ๐ก
(all)
203
Table P-5 - Data for UDL Beam, Ig/Icr=17, Reduced Mcr Example P3 โ Page 1
Example P3 (Example 3.6.2h modified to New Mcr) ฮฆc = 0.65 ฮตcu = 0.0035 mm/mm pg 1/2
UDL Continuous w0 = 6.480 N/mm fc' = 36 MPa fr = 0.6 * MPa
Vary ML, Mm constant L = 10000 mm Ec = 27000 MPa use: Mcr = 0.67 frIg/yt
Let ML=MR b = 0.5* h mm ฮฑ1 = 0.796 Mcr= 0.0335 * N mm
d = 0.85 * h mm ฮฒ1 = 0.880 ฯ b=ฮฑ1ฮฒ1ฯcf'cฮตcu/(ฯbffu(ฮตcu+ffu/Ef))
M0,0 = w0 L2/8 = 8.10E+7 N mm ฯb = 0.75 ฯ b = 0.00578
End Moment Ms/Mr = 0.635 ffu = 690 MPa ฮฑ R = 1 *ฮฑ L
+ve Moment Ms/Mr = 0.374 Eb = 44000 MPa n=Eb/Ec= 1.63
Member Properties Determined from Provided Info (Primairly Servicability) Units
ฮฑL/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
ฮฑL = ML/M0 = -0.75 -0.70 -0.61 -0.55 -0.50 -0.46 -0.30 -0.20 0
ฮฑR = 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
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 N mm
ฮฑcr = Mcr/Mmax= 0.536 0.536 0.536 0.536 0.536 0.536 0.536 0.536 0.536
Mcr = 4.34E+7 4.34E+7 4.34E+7 4.34E+7 4.34E+7 4.34E+7 4.34E+7 4.34E+7 4.34E+7 N mm
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
Member Properties Determined with Factored Loads
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
ฯ L =As/bd= 0.0276 0.0133 0.0048 0.0027 0.0017 0.0012 0 0 0
Icr L = 1.21E+9 6.59E+8 2.66E+8 1.56E+8 1.02E+8 7.26E+7 5.40E+9 5.40E+9 5.40E+9 mm4
ฮทL=1 โ 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
ฯ m =Am/bd= 0.0059 0.0059 0.0059 0.0059 0.0059 0.0059 0.0059 0.0059 0.0059
Icr m = 3.19E+8 3.19E+8 3.19E+8 3.19E+8 3.19E+8 3.19E+8 3.19E+8 3.19E+8 3.19E+8 mm4
ฮทm=1 โ Icr m/Ig = 0.941 Mmax/Mcr = 1.87 Ig/Icr m = 16.92 Am=ฯmbd= 904 mm2
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
ฯ R =AR/bd= 0.0276 0.0133 0.0048 0.0027 0.0017 0.0012 0.0000 0.0000 0.0000
Icr R = 1.21E+9 6.59E+8 2.66E+8 1.56E+8 1.02E+8 7.26E+7 5.40E+9 5.40E+9 5.40E+9 mm4
ฮทR=1 โ 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
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204
Table P-6 - Data for UDL Beam, Ig/Icr=17, Reduced Mcr Example P3 โ Page 2
New Mcr Example P3 w0 = 6.48 N/mm fc' = 36 MPa ฯ b = 0.00578 pg 2/2
L = 10000 mm ffu = 690 MPa ฯ m = 0.0059
Ms /Mr (+ve) = 1.698 Mmax/Mcr = 1.87 Ig/Icr m = 16.92 Eb = 44000 MPa
ฮฑL/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
Constant Stiffness Results Using Constant Stiffness Equations ฮmid=K(5MmL2)/(48EcI)
ฮg(Gross) 2.31 3.09 3.98 4.37 4.63 4.80 5.29 5.50 5.79 mm
ฮcr(Cracked) 39.17 52.23 67.29 73.99 78.34 81.24 89.53 93.03 97.92 mm
Branson's Method ฮIe Ie=Icr+(Ig-Icr)(Mcr/Mmax)3 Mcr =(0.3โf'c)Ig/y= N mm
Branson's Ie= 6.44E+8 6.44E+8 6.44E+8 6.44E+8 6.44E+8 6.44E+8 6.44E+8 6.44E+8 6.44E+8 mm4
ฮIe(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
ACI 440.1R clause 8.3.2.2 Ie=Icr+(ฮฒdIg-Icr)(Mcr/Mmax)3 ฮฒd= 0.204 ฮฒd=0.2(ฯm/ฯ b)<1 Ie R = Ie L
Ie m (ACI440.1R) = 3.69E+8 3.69E+8 3.69E+8 3.69E+8 3.69E+8 3.69E+8 3.69E+8 3.69E+8 3.69E+8 mm4
ฮIe,ฮฒd(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
Ie 9.8.2 (& ACI440.1R)= 6.26E+8 4.59E+8 3.43E+8 3.11E+8 2.95E+8 2.87E+8 1.88E+9 3.69E+8 3.69E+8 mm4
ฮIe,avg(A23.3) 19.98 36.28 62.66 76.04 84.70 90.25 15.21 80.38 84.61 mm
Deflection using the S806 Method with Numerical Integration
โฮฒ=0(S806) 52.06 55.53 62.28 67.08 71.32 74.98 83.99 87.67 93.86 mm
Exact Integration Ie(x)Analytical ฮ1= -2.34 -2.38 -1.85 -1.26 -0.75 -0.39 0.00 0.00 0.00 mm
Analytical ฮ2 or ฮ1+2= 0.10 0.11 0.13 0.14 0.14 0.15 0.17 0.16 0.13 mm
Analytical ฮ3= 17.87 19.29 21.46 22.71 23.66 24.36 26.80 28.08 30.23 mm
Analytical ฮ4= 17.87 19.29 21.46 22.71 23.66 24.36 26.80 28.08 30.23 mm
Analytical ฮ5 or ฮ5+6= 0.10 0.11 0.13 0.14 0.14 0.15 0.17 0.16 0.13 mm
Analytical ฮ6= -2.34 -2.38 -1.85 -1.26 -0.75 -0.39 0.00 0.00 0.00 mm
ฮIe(x)(Exact) 31.26 34.04 39.47 43.17 46.11 48.24 53.92 56.48 60.71 mm
Length:Defl, L/ฮmid= 320 294 253 232 217 207 185 177 165
Proposed Method ฮI'e I'e=Icr/[1-ฮณฮทm(Mcr/Mmax)2] ฮณ=(1.6ฮพ 3-0.6ฮพ 4)/(Mcr/Mmax)
2+2.4ln(2-ฮพ )
I'e (ฮณ=1) (M(x)=Mmax) = 4.37E+8 4.37E+8 4.37E+8 4.37E+8 4.37E+8 4.37E+8 4.37E+8 4.37E+8 4.37E+8 mm4
ฮฮณ=1(Approx) 28.58 38.11 49.10 53.99 57.16 59.28 65.33 67.88 71.45 mm
ฮพ =1-โ(1-Mcr/Mmax)= 0.319 0.319 0.319 0.319 0.319 0.319 0.319 0.319 0.319
ฮณ= 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41
Bischoff's I'e = 5.15E+8 5.15E+8 5.15E+8 5.15E+8 5.15E+8 5.15E+8 5.15E+8 5.15E+8 5.15E+8 mm4
ฮI'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/ฮ exact 319.89 293.77 253.33 231.63 216.88 207.31 185.45 177.06 164.70
Maximum Deflection Results using numerical and approximation methods ฮmax โ ฮI'e โ(Mmax/Mm)
ฮmax,Ie(x)(Exact) 31.26 34.05 39.48 43.18 46.11 48.24 53.92 56.48 60.72 mm
ฮmax,I'e(Proposed) 24.29 32.38 41.72 45.87 48.57 50.37 55.51 57.68 60.71 mm
3.24E+7
205
Figure P-5 - Midspan Deflection Computed using Shrinkage Restraint Mcr โ Slab with
Ig/Icr=17, Mm/Mcr=1.9, and ML=MR
The lines plotted in Figure P-5 use data from Example P3 per Table P-5 and Table P-6.
Figure P-6 - Copy of Figure O-8, Ig/Icr=17, Mm/Mcr=1.3 โ Compare to Figure P-5
๐๐๐ = ๐๐๐ผ๐/๐ฆ๐ก
(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 ๐๐/๐ธ๐ผ, 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, ๐ผ, 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, ๐, 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 ๐๐ฟ, ๐๐, and ๐๐ 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 ๐๐ฟ โซ ๐๐ , 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 ๐พ 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, โ๐ ๐ง (๐ฅ), denotes the load case number for the end-moments with
the subscript ๐ง. Deflection is determined using superposition.
โ๐ ๐ง(๐ฅ) = โ๐๐ฟ(๐ฅ) + โ๐๐
(๐ฅ) + โ๐๐ท๐ฟ(๐ฅ)
For cases with ๐๐ = 0 and ๐๐ =๐๐ฟ
2 , no new calculations required
For UDL, if ๐๐ =๐๐๐๐ฅ
2 then
๐๐
๐0=๐๐ฟ
๐0โ 20 + โ384 โ 48
๐๐ฟ
๐0
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 ๐๐ฟ ๐0โ and
๐๐ ๐0โ are strategically selected for each case and the following equations are used:
where ๐๐ฟ = ๐๐ = 0: ๐ค0 = ๐ค๐๐ท๐ฟ
other cases: ๐ค๐๐ท๐ฟ =๐ค0
1 +๐๐ฟ
2๐0+
๐๐
2๐0+ (
๐๐ฟ
4๐0โ
๐๐
4๐0)2
๐0 =๐ค๐๐ท๐ฟ๐ฟ
2
8 ; ๐๐ = ๐0 +
๐๐ฟ+๐๐
2 ; ๐๐๐๐ฅ = ๐0 +
๐๐ฟ+๐๐
2+(๐๐ฟ โ๐๐ )
2
16๐0
211
Table R-1 - Example Midspan vs Maximum Deflection for UDL โ 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 ฮM_L ฮUDL ฮU1(x) ฮM_L ฮUDL ฮU2(x) ฮM_L ฮUDL ฮU3(x) ฮM_L ฮUDL ฮU4(x) ฮM_L ฮUDL ฮU5(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
ฮ(D
efl
ecti
on
)
x (Position)
ฮ๐4(๐ฅ)
ฮ๐3(๐ฅ)
ฮ๐2(๐ฅ)
ฮ๐1(๐ฅ)
ฮ๐5(๐ฅ)
212
Table R-2 - Example Midspan vs Maximum Deflection for UDL โ Page 2
Spreadsheet Function: Compare maximum and midspan deflection for a uniformly dustributed udl pg
h = 400 mm 2 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 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 ฮM_L ฮUDL ฮU6(x) ฮM_L ฮUDL ฮU7(x) ฮM_L ฮUDL ฮU8(x) ฮM_L ฮUDL ฮU9(x) ฮM_L ฮUDL ฮU10(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
load on a uniform elastic beam (I g) with different end-moment conditions
-0.5
0
0.5
1
1.5
2
0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000
ฮ(D
efl
ecti
on
)
x (Position)
ฮ๐6(๐ฅ)
ฮ๐ (๐ฅ)ฮ๐ (๐ฅ)ฮ๐9(๐ฅ)
ฮ๐10 (๐ฅ)
213
Table R-3 - Example Midspan vs Maximum Deflection for UDL โ Page 3
Spreadsheet Function: Compare maximum and midspan deflection for a uniformly dustributed udl pg
h = 400 mm 3 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
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 ฮM_R ฮUDL ฮU11(x) ฮM_L ฮM_R ฮUDL ฮU12(x) ฮM_L ฮM_R ฮUDL ฮU13(x) ฮM_L ฮM_R ฮUDL ฮU14(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
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
ฮ(D
efl
ecti
on
)
x (Position)
ฮ๐11 (๐ฅ)
ฮ๐12 (๐ฅ)
ฮ๐13 (๐ฅ)
ฮ๐14 (๐ฅ)
214
Table R-4 - Example Midspan vs Maximum Deflection for UDL โ Page 4
Spreadsheet Function: Compare maximum and midspan deflection for a uniformly dustributed udl pg
h = 400 mm 4 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
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
MR= -5.00 kN m MR= -5.50 kN m MR= -5.85 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 ฮM_L ฮM_R ฮUDL ฮU15(x) ฮM_L ฮM_R ฮUDL ฮU16(x) ฮM_L ฮM_R ฮUDL ฮU17(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
load on a uniform elastic beam (I g) with different end-moment conditions
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
ฮ(D
efl
ecti
on
)
x (Position)
ฮ๐15 (๐ฅ)
ฮ๐16 (๐ฅ)
ฮ๐1 (๐ฅ)
215
Table R-5 - Example Midspan vs Maximum Deflection for UDL โ Page 5
Spreadsheet Function: Compare maximum and midspan deflection for a uniformly dustributed udl pg
h = 400 mm 5 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
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 ฮM_R ฮUDL ฮU21(x) ฮM_L ฮM_R ฮUDL ฮU22(x) ฮM_L ฮM_R ฮUDL ฮU23(x) ฮM_L ฮM_R ฮUDL ฮU24(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
load on a uniform elastic beam (I g) with different end-moment conditions
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
ฮ(D
efl
ecti
on
)
x (Position)
ฮ๐21 (๐ฅ)
ฮ๐22 (๐ฅ)
ฮ๐23 (๐ฅ)
ฮ๐24 (๐ฅ)
216
Table R-6 - Example Midspan vs Maximum Deflection for UDL โ Page 6
Spreadsheet Function: Compare maximum and midspan deflection for a uniformly dustributed udl pg
h = 400 mm 6 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
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
MR= -2.85 kN m MR= -2.62 kN m MR= -2.41 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 ฮM_L ฮM_R ฮUDL ฮU25(x) ฮM_L ฮM_R ฮUDL ฮU26(x) ฮM_L ฮM_R ฮUDL ฮU27(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
load on a uniform elastic beam (I g) with different end-moment conditions
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
ฮ(D
efl
ecti
on
)
x (Position)
ฮ๐25 (๐ฅ)
ฮ๐26 (๐ฅ)
ฮ๐2 (๐ฅ)
217
Table R-7 - Example Midspan vs Maximum Deflection for UDL โ Summary
Spreadsheet Function: Compare maximum and midspan deflection for a uniformly dustributed udl pg
7 of 7
Cases with MR=0 ML / ฮmax / Cases with MR=ML/2 ML / ฮmax /
ML Mmid Mmax ฮmid ฮmax Mmax ฮmid ML Mmid Mmax ฮmid ฮmax Mmax ฮmid
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 / ฮmax /
ML Mmid Mmax ฮmid ฮmax Mmax ฮmid
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
load on a uniform elastic beam (I g) with different end-moment conditions
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
๐๐ = 0
๐๐ = ๐๐๐๐ฅ 2โ
๐๐ = ๐๐ฟ /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 ๐๐ฟ/๐0 and
๐๐ /๐0 ratios. To achieve identical midspan moment, point loads are varied as follows:
when ๐๐ฟ = ๐๐ = 0 โถ ๐0 = point load ; when ๐๐ = 0 โถ ๐s =๐0
1 + 0.5๐๐ฟ ๐0โ
when ๐๐ =๐๐ฟ
2โถ ๐s =
๐0
1 +0.75๐๐ฟ
๐0
; when ๐๐ =๐๐
2โถ ๐s =
2.5๐0
2 +๐๐ฟ
๐0
and ๐๐
๐0=
๐02๐s
For all cases: ๐1๐๐ฟ =๐s๐ฟ
4 ; ๐๐ = ๐๐๐๐ฅ = ๐1๐๐ฟ +
๐๐ฟ
2+๐๐
2
Table R-8 - Example Midspan vs Maximum Deflection for CPL โ Summary
cpl pg
1 of 1
Midspan PL Cases with MR=0 ML / ฮmax / Midspan PL Cases with MR=ML/2 ML / ฮmax /
ML Mmid Mmax ฮmid ฮmax Mmax ฮmid ML Mmid Mmax ฮmid ฮmax Mmax ฮmid
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 ฮmid ฮmax ML/max ฮmax/mid ML Mmid Mmax ฮmid ฮmax ML/max ฮmax/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
Ratio of End-Moment Relative to Maximum Positive Moment
๐๐ = ๐๐๐๐ฅ 2โ
๐๐ = 0
๐๐ = ๐๐ฟ 2โ
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 ๐๐ฟ/๐๐๐๐ฅ and
๐๐ /๐๐๐๐ฅ ratios. To achieve identical midspan moment, loads are varied as follows:
when ๐๐ฟ = ๐๐ = 0 โถ ๐0 = total point load ; else ๐s = ๐0 (1 โ1
3
๐๐ฟ
๐๐๐๐ฅโ2
3
๐๐
๐๐๐๐ฅ)
๐๐ฟ
๐2๐๐ฟ=
๐๐ฟ
๐๐๐๐ฅ
1 โ13
๐๐ฟ
๐๐๐๐ฅโ23
๐๐
๐๐๐๐ฅ
; ๐๐
๐2๐๐ฟ=
๐๐
๐๐๐๐ฅ
1 โ13
๐๐ฟ
๐๐๐๐ฅโ23
๐๐
๐๐๐๐ฅ
As usual: ๐2๐๐ฟ =๐s๐ฟ
3 ; ๐๐ = ๐2๐๐ฟ +
๐๐ฟ
2+๐๐
2 ; ๐๐๐๐ฅ = ๐2๐๐ฟ +
๐๐ฟ
3+2๐๐
3
Table R-9 - Example Midspan vs Maximum Deflection for 2PL โ Summary
2pl pg
1 of 1
2 PL @ 3rd pts Cases with MR=0 ML / ฮmax / 2 PL @ 3rd pts Cases with MR=ML/2 ML / ฮmax /
ML Mmid Mmax ฮmid ฮmax Mmax ฮmid ML Mmid Mmax ฮmid ฮmax Mmax ฮmid
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 ฮmid ฮmax ML/max ฮmax/mid ML Mmid Mmax ฮmid ฮmax ML/max ฮmax/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
Ratio of End-Moment Relative to Maximum Positive Moment
๐๐ = ๐๐๐๐ฅ 2โ
๐๐ = 0
๐๐ = ๐๐ฟ 2โ
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 Bransonโs Equation in CSA A23.3-04
CSA A23.3-04 (CSA 2004) employs Bransonโs (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 Bransonโs 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 ๐๐๐ for slabs is a correction for Bransonโs (1965) equation. Results
indicate that use of Bransonโs equation underpredicts deflection in some slabs. This
clause over-accounts for shrinkage-restraint because some members are outside the
valid range for Bransonโs 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, ๐ผ๐๐, as the standard suggests, the maximum deflection will often be under
predicted. This importance of using the maximum moment to compute ๐ผ๐ or ๐ผ๐โฒ is
discussed further in Section 3.3. This report finds it rational to define ๐ผ๐๐ 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 ๐ผ๐ ๐ 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 ๐๐ฟ โซ ๐๐ , 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
Candidateโs 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)