STUDY NOTES & REFERENCE GUIDE
All Chapters Covered
SOLUTIONS
Acadexas Stuvia | Page 1 of 173
, 100891294
STUDY NOTES & REFERENCE GUIDE
CHAPTER 2
Load Combinations
Problem 2.1
SOLUTION
Table P2.1 Summary of Load Combinations Using Strength Design for Beam in Problem 2.1
Load Combination
IBC Equation No. Equation Exterior Interior
Positive
Negative Negative
16-1 1.4D –18.6 61.5 –74.5
16-2 1.2D + 1.6L –36.6 120.7 –146.4
16-3, 16-4, 16-5 1.2D + 0.5L –22.4 73.9 –89.6
16-6, 16-7 0.9D –12.0 39.5 –47.9
Problem 2.2
SOLUTION
Table P2.2 Summary of Load Combinations Using Strength Design for Beam in
Problem 2.2
Load Combination
IBC Equation
Equation Bending Moment Shear Force
No.
Support Midspan Support
16-1 1.4D –80.6 57.5 16.5
16-2 1.2D + 1.6L –105.1 75.2 21.5
1.2D + 0.5L –80.4 57.4 16.5
16-3 1.2D + 0.5W –42.1 49.3 11.8
1.2D – 0.5W –96.1 49.3 16.6
1.2D + 1.0W + 0.5L –26.4 57.4 11.7
16-4
1.2D – 1.0W + 0.5L –134.4 57.4 21.3
16-5 1.2D + 0.5L –80.4 57.4 16.5
0.9D + 1.0W 2.2 37.0 5.8
16-6
0.9D – 1.0W –105.8 37.0 15.4
16-7 0.9D –51.8 37.0 10.6
1 @Seismicisolation
@Seismicisolation
Acadexas
Copyright © 2018 ICC. ALL RIGHTS Stuvia
RESERVED. Accessed | Page
by Mohammed 2 of
Mujtaba 173(), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
Hammed
Agreement with ICC. No further reproduction, no further reproductions by any third party, or distribution authorized. Single user only, copying and networking prohibited. ANY UNAUTHORIZED
REPRODUCTION OR DISTRIBUTION IS A VIOLATION OF THE FEDERAL COPYRIGHT ACT AND THE LICENSE AGREEMENT, AND SUBJECT TO CIVIL AND CRIMINAL PENALTIES THEREUNDER.
, 100891294
STUDY NOTES & REFERENCE GUIDE
2 Solutions Manual to Structural Loads
Problem 2.3
SOLUTION
Table P2.3 Summary of Load Combinations Using Basic Allowable Stress Design for
Beam in Problem 2.3
Load Combination
IBC Equation
Equation Bending Moment Shear Force
No.
Support Midspan Support
16-8, 16-10 D –57.6 41.1 11.8
16-9 D+L –80.1 57.3 16.4
16-11, 16-14 D + 0.75L –74.5 53.3 15.3
D + 0.6W –25.2 41.1 8.9
16-12
D – 0.6W –90.0 41.1 14.7
D + 0.75(0.6W) + 0.75L –50.2 53.3 13.1
16-13
D – 0.75(0.6W) + 0.75L –98.8 53.3 17.4
0.6D + 0.6W –2.2 24.7 4.2
16-15
0.6D – 0.6W –67.0 24.7 10.0
16-16 0.6D –34.6 24.7 7.1
Problem 2.4
SOLUTION
Table P2.4 Summary of Load Combinations Using Alternative Basic Allowable Stress
Design for Beam in Problem 2.4
Load Combination
IBC Equation
Equation Bending Moment Shear Force
No.
Support Midspan Support
16-17, 16-21 D+L –80.1 57.3 16.4
D + L + 0.6 W –38.0 57.3 12.7
16-18, 16-19
D + L – 0.6 W –122.2 57.3 20.1
D + L + 0.6 W/2 –59.0 57.3 14.5
16-20
D + L – 0.6 W/2 –101.2 57.3 18.3
16-22 0.9D –51.8 37.0 10.6
@Seismicisolation
@Seismicisolation
Acadexas
Copyright © 2018 ICC. ALL RIGHTS Stuvia
RESERVED. Accessed | Page
by Mohammed 3 of
Mujtaba 173(), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
Hammed
Agreement with ICC. No further reproduction, no further reproductions by any third party, or distribution authorized. Single user only, copying and networking prohibited. ANY UNAUTHORIZED
REPRODUCTION OR DISTRIBUTION IS A VIOLATION OF THE FEDERAL COPYRIGHT ACT AND THE LICENSE AGREEMENT, AND SUBJECT TO CIVIL AND CRIMINAL PENALTIES THEREUNDER.
, 100891294
STUDY NOTES & REFERENCE GUIDE
Chapter 2 3
Problem 2.5
SOLUTION
Because the live loads on the floors are equal to 100 psf, f1 =
0.5. The seismic load effect, E, is determined as follows:
For use in IBC Equation 16-5: E = Eh + Ev = QE + 0.2SDSD
= (1.0 × QE) + (0.2 × 0.41 × D) = QE + 0.08D
For use in IBC Equation 16-7: E = Eh – Ev = QE – 0.2SDSD
= (1.0 × QE) – (0.2 × 0.41 × D) = QE – 0.08D
Substituting for E, IBC Equation 16-5 becomes: 1.2D + QE + 0.08D + 0.5L = 1.28D + QE
+ 0.5L Similarly, IBC Equation 16-7 becomes: 0.9D + QE – 0.08D = 0.82D + QE
Table P2.5 Summary of Load Combinations Using Strength Design for Column in
Problem 2.5
Load Combination
IBC Equation
Equation
No. Axial Force Bending Moment Shear Force
16-1 1.4D 235.1 29.8 3.2
16-2 1.2D + 1.6L + 0.5Lr 275.3 59.2 6.3
1.2D + 1.6 Lr + 0.5L 246.1 36.1 3.9
16-3 1.2D + 1.6 Lr + 0.5W 232.1 86.1 8.3
1.2D + 1.6 Lr – 0.5W 218.5 –34.9 –2.8
1.2D + 1.0W + 0.5L + 0.5Lr 243.3 157.1 15.0
16-4
1.2D – 1.0W + 0.5L + 0.5Lr 216.1 –84.9 –7.2
1.28D + QE + 0.5L 272.1 469.9 46.2
16-5
1.28D – QE + 0.5L 199.3 –394.3 –38.2
0.9D + 1.0W 164.7 140.2 13.2
16-6
0.9D – 1.0W 137.5 –101.8 –9.0
0.82D + QE 174.1 449.6 44.1
16-7
0.82D – QE 101.3 –414.6 –40.3
@Seismicisolation
@Seismicisolation
Acadexas
Copyright © 2018 ICC. ALL RIGHTS Stuvia
RESERVED. Accessed | Page
by Mohammed 4 of
Mujtaba 173(), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
Hammed
Agreement with ICC. No further reproduction, no further reproductions by any third party, or distribution authorized. Single user only, copying and networking prohibited. ANY UNAUTHORIZED
REPRODUCTION OR DISTRIBUTION IS A VIOLATION OF THE FEDERAL COPYRIGHT ACT AND THE LICENSE AGREEMENT, AND SUBJECT TO CIVIL AND CRIMINAL PENALTIES THEREUNDER.
All Chapters Covered
SOLUTIONS
Acadexas Stuvia | Page 1 of 173
, 100891294
STUDY NOTES & REFERENCE GUIDE
CHAPTER 2
Load Combinations
Problem 2.1
SOLUTION
Table P2.1 Summary of Load Combinations Using Strength Design for Beam in Problem 2.1
Load Combination
IBC Equation No. Equation Exterior Interior
Positive
Negative Negative
16-1 1.4D –18.6 61.5 –74.5
16-2 1.2D + 1.6L –36.6 120.7 –146.4
16-3, 16-4, 16-5 1.2D + 0.5L –22.4 73.9 –89.6
16-6, 16-7 0.9D –12.0 39.5 –47.9
Problem 2.2
SOLUTION
Table P2.2 Summary of Load Combinations Using Strength Design for Beam in
Problem 2.2
Load Combination
IBC Equation
Equation Bending Moment Shear Force
No.
Support Midspan Support
16-1 1.4D –80.6 57.5 16.5
16-2 1.2D + 1.6L –105.1 75.2 21.5
1.2D + 0.5L –80.4 57.4 16.5
16-3 1.2D + 0.5W –42.1 49.3 11.8
1.2D – 0.5W –96.1 49.3 16.6
1.2D + 1.0W + 0.5L –26.4 57.4 11.7
16-4
1.2D – 1.0W + 0.5L –134.4 57.4 21.3
16-5 1.2D + 0.5L –80.4 57.4 16.5
0.9D + 1.0W 2.2 37.0 5.8
16-6
0.9D – 1.0W –105.8 37.0 15.4
16-7 0.9D –51.8 37.0 10.6
1 @Seismicisolation
@Seismicisolation
Acadexas
Copyright © 2018 ICC. ALL RIGHTS Stuvia
RESERVED. Accessed | Page
by Mohammed 2 of
Mujtaba 173(), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
Hammed
Agreement with ICC. No further reproduction, no further reproductions by any third party, or distribution authorized. Single user only, copying and networking prohibited. ANY UNAUTHORIZED
REPRODUCTION OR DISTRIBUTION IS A VIOLATION OF THE FEDERAL COPYRIGHT ACT AND THE LICENSE AGREEMENT, AND SUBJECT TO CIVIL AND CRIMINAL PENALTIES THEREUNDER.
, 100891294
STUDY NOTES & REFERENCE GUIDE
2 Solutions Manual to Structural Loads
Problem 2.3
SOLUTION
Table P2.3 Summary of Load Combinations Using Basic Allowable Stress Design for
Beam in Problem 2.3
Load Combination
IBC Equation
Equation Bending Moment Shear Force
No.
Support Midspan Support
16-8, 16-10 D –57.6 41.1 11.8
16-9 D+L –80.1 57.3 16.4
16-11, 16-14 D + 0.75L –74.5 53.3 15.3
D + 0.6W –25.2 41.1 8.9
16-12
D – 0.6W –90.0 41.1 14.7
D + 0.75(0.6W) + 0.75L –50.2 53.3 13.1
16-13
D – 0.75(0.6W) + 0.75L –98.8 53.3 17.4
0.6D + 0.6W –2.2 24.7 4.2
16-15
0.6D – 0.6W –67.0 24.7 10.0
16-16 0.6D –34.6 24.7 7.1
Problem 2.4
SOLUTION
Table P2.4 Summary of Load Combinations Using Alternative Basic Allowable Stress
Design for Beam in Problem 2.4
Load Combination
IBC Equation
Equation Bending Moment Shear Force
No.
Support Midspan Support
16-17, 16-21 D+L –80.1 57.3 16.4
D + L + 0.6 W –38.0 57.3 12.7
16-18, 16-19
D + L – 0.6 W –122.2 57.3 20.1
D + L + 0.6 W/2 –59.0 57.3 14.5
16-20
D + L – 0.6 W/2 –101.2 57.3 18.3
16-22 0.9D –51.8 37.0 10.6
@Seismicisolation
@Seismicisolation
Acadexas
Copyright © 2018 ICC. ALL RIGHTS Stuvia
RESERVED. Accessed | Page
by Mohammed 3 of
Mujtaba 173(), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
Hammed
Agreement with ICC. No further reproduction, no further reproductions by any third party, or distribution authorized. Single user only, copying and networking prohibited. ANY UNAUTHORIZED
REPRODUCTION OR DISTRIBUTION IS A VIOLATION OF THE FEDERAL COPYRIGHT ACT AND THE LICENSE AGREEMENT, AND SUBJECT TO CIVIL AND CRIMINAL PENALTIES THEREUNDER.
, 100891294
STUDY NOTES & REFERENCE GUIDE
Chapter 2 3
Problem 2.5
SOLUTION
Because the live loads on the floors are equal to 100 psf, f1 =
0.5. The seismic load effect, E, is determined as follows:
For use in IBC Equation 16-5: E = Eh + Ev = QE + 0.2SDSD
= (1.0 × QE) + (0.2 × 0.41 × D) = QE + 0.08D
For use in IBC Equation 16-7: E = Eh – Ev = QE – 0.2SDSD
= (1.0 × QE) – (0.2 × 0.41 × D) = QE – 0.08D
Substituting for E, IBC Equation 16-5 becomes: 1.2D + QE + 0.08D + 0.5L = 1.28D + QE
+ 0.5L Similarly, IBC Equation 16-7 becomes: 0.9D + QE – 0.08D = 0.82D + QE
Table P2.5 Summary of Load Combinations Using Strength Design for Column in
Problem 2.5
Load Combination
IBC Equation
Equation
No. Axial Force Bending Moment Shear Force
16-1 1.4D 235.1 29.8 3.2
16-2 1.2D + 1.6L + 0.5Lr 275.3 59.2 6.3
1.2D + 1.6 Lr + 0.5L 246.1 36.1 3.9
16-3 1.2D + 1.6 Lr + 0.5W 232.1 86.1 8.3
1.2D + 1.6 Lr – 0.5W 218.5 –34.9 –2.8
1.2D + 1.0W + 0.5L + 0.5Lr 243.3 157.1 15.0
16-4
1.2D – 1.0W + 0.5L + 0.5Lr 216.1 –84.9 –7.2
1.28D + QE + 0.5L 272.1 469.9 46.2
16-5
1.28D – QE + 0.5L 199.3 –394.3 –38.2
0.9D + 1.0W 164.7 140.2 13.2
16-6
0.9D – 1.0W 137.5 –101.8 –9.0
0.82D + QE 174.1 449.6 44.1
16-7
0.82D – QE 101.3 –414.6 –40.3
@Seismicisolation
@Seismicisolation
Acadexas
Copyright © 2018 ICC. ALL RIGHTS Stuvia
RESERVED. Accessed | Page
by Mohammed 4 of
Mujtaba 173(), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
Hammed
Agreement with ICC. No further reproduction, no further reproductions by any third party, or distribution authorized. Single user only, copying and networking prohibited. ANY UNAUTHORIZED
REPRODUCTION OR DISTRIBUTION IS A VIOLATION OF THE FEDERAL COPYRIGHT ACT AND THE LICENSE AGREEMENT, AND SUBJECT TO CIVIL AND CRIMINAL PENALTIES THEREUNDER.
