All Chapters Covered
SOLUTIONS
, 100891
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
16-7 0.9D –51.8 37.0 10.6
1 @Seismicisolation
@Seismicisolation
Copyright © 2018 ICC. ALL RIGHTS RESERVED. Accessed by Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
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
, 100891
0.9D – 1.0W –105.8 37.0 15.4
16-7 0.9D –51.8 37.0 10.6
1 @Seismicisolation
@Seismicisolation
Copyright © 2018 ICC. ALL RIGHTS RESERVED. Accessed by Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
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
, 100891
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
Copyright © 2018 ICC. ALL RIGHTS RESERVED. Accessed by Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
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
, 100891
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
Copyright © 2018 ICC. ALL RIGHTS RESERVED. Accessed by Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
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
, 100891
4 Solutions Manual to Structural Loads
Problem 2.6
SOLUTION
Because the shear wall is in a parking garage, f1 =
1.0. 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 × 1.0 × D) = QE + 0.2D
For use in IBC Equation 16-7: E = Eh – Ev = QE – 0.2SDSD
= (1.0 × QE) – (0.2 × 1.0 × D) = QE – 0.2D
Substituting for E, IBC Equation 16-5 becomes: 1.2D + QE + 0.2D + 1.0L = 1.4D + QE +
1.0L. Similarly, IBC Equation 16-7 becomes: 0.9D + QE – 0.2D = 0.7D + QE
Table P2.6 Summary of Load Combinations Using Strength Design for Shear Wall in
Problem 2.6
Load Combination
IBC Equation
Equation
No. Axial Force Bending Moment Shear Force
16-1 1.4D 903.0 0 0
16-2 1.2D + 1.6L 1,012.4 0 0
16-3, 16-4 1.2D + 1.0L 923.0 0 0
1.4D + QE + 1.0L 1,052.0 4,280.0 143.0
16-5
1.4D – QE + 1.0L 1,052.0 –4,280.0 –143.0
16-6 0.9D 580.5 0 0
0.7D + QE 451.5 4,280.0 143.0
16-7
0.7D – QE 451.5 –4,280.0 –143.0
Problem 2.7
SOLUTION
The governing load combination in IBC 1605.2 is Equation 16-2:
Negative bending moment:
1.2D + 1.6L = (1.2 × 80.6) + (1.6 × 42.1) = 164.1 ft-kips
Positive bending moment:
1.2D + 1.6L = (1.2 × 53.7) + (1.6 × 30.4) = 113.1 ft-kips
@Seismicisolation
@Seismicisolation
Copyright © 2018 ICC. ALL RIGHTS RESERVED. Accessed by Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
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
, 100891
Shear force:
1.2D + 1.6L = (1.2 × 29.7) + (1.6 × 19.0) = 66.0 kips
@Seismicisolation
@Seismicisolation
Copyright © 2018 ICC. ALL RIGHTS RESERVED. Accessed by Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
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
, 100891
Chapter 2 5
The following basic combinations for strength design with overstrength are also applicable:
• (1.2 + 0.2SDS) D + O QE + 1.0L
Axial force: QE = 2.0
O 241 = 482 kips tension or compression
Negative bending moment:
(1.2 + 0.2SDS)D + 1.0L = (1.38 80.6) + (1.0 42.1) = 153.3 ft-kips
Positive bending moment:
(1.2 + 0.2SDS)D + 1.0L = (1.38 53.7) + (1.0 30.4) = 104.5 ft-kips
Shear force:
(1.2 + 0.2SDS)D + 1.0L = (1.38 29.7) + (1.0 19.0) = 60.0 kips
Note that the load factor on L must be equal to 1.0 because of the assembly occupancy.
• (0.9 – 0.2SDS)D + O QE
Axial force: QE = 2.0
O 241 = 482 kips tension or compression
Negative bending moment:
(0.9 – 0.2SDS)D = 0.72 80.6 = 58.0 ft-kips
Positive bending moment:
(0.9 – 0.2SDS)D = 0.72 53.7 = 38.7 ft-kips
Shear force:
(0.9 – 0.2SDS)D = 0.72 29.7 = 21.4 kips
Problem 2.8
SOLUTION
The governing load combination in IBC 1605.3.1 is Equation
16-9: Negative bending moment:
D + L = 80.6 + 42.1 = 122.7 ft-kips
Positive bending moment:
D + L = 53.7 + 30.4 = 84.1 ft-kips
Shear force:
D + L = 29.7 + 19.0 = 48.7 kips
@Seismicisolation
@Seismicisolation
Copyright © 2018 ICC. ALL RIGHTS RESERVED. Accessed by Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
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
, 100891
6 Solutions Manual to Structural Loads
The following basic allowable stress design load combinations with overstrength are also applicable:
• (1.0 + 0.14SDS)D + 0.7 O QE
Axial force:
0.7 O QE = 0.7 2.0 241 = 337.4 kips tension or compression
Negative bending moment:
(1.0 + 0.14SDS)D = 1.13 80.6 = 91.1 ft-kips
Positive bending moment:
(1.0 + 0.14SDS)D = 1.13 53.7 = 60.7 ft-kips
Shear force:
(1.0 + 0.14SDS)D = 1.13 29.7 = 33.6 kips
• (1.0 + 0.105SDS)D + 0.525 OQE + 0.75L
Axial force:
0.525 QE = 0.525
O 2.0 241 = 253.1 kips tension or
compression Negative bending moment:
1.1D + 0.75L = (1.1 80.6) + (0.75 42.1) = 120.2 ft-kips
Positive bending moment:
1.1D + 0.75L = (1.1 53.7) + (0.75 30.4) = 81.9 ft-kips
Shear force:
1.1D + 0.75L = (1.1 29.7) + (0.75 19.0) = 46.9 kips
• (0.6 – 0.14SDS)D + 0.7 O QE
Axial force:
0.7 O QE = 0.7 2.0 241 = 337.4 kips tension or compression
Negative bending moment:
(0.6 – 0.14SDS)D = 0.47 80.6 = 37.9 ft-kips
Positive bending moment:
(0.6 – 0.14SDS)D = 0.47 53.7 = 25.2 ft-kips
Shear force:
(0.6 – 0.14SDS)D = 0.47 29.7 = 14.0 kips
@Seismicisolation
@Seismicisolation
Copyright © 2018 ICC. ALL RIGHTS RESERVED. Accessed by Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
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
, 100891
Chapter 2 7
Problem 2.9
SOLUTION
Table P2.9 Summary of Load Combinations Using Strength Design for Beam in
Problem 2.9
IBC Equation No. Equation Load Combination
16-1 1.4D 105.0
1.2D + 0.5Lr 140.0
16-2, 16-4 1.2D + 0.5S 152.5
1.2D + 0.5R 190.0
1.2D + 1.6Lr 250.0
16-3 1.2D + 1.6S 290.0
1.2D + 1.6R 410.0
16-5 1.2D + 0.2S 115.0
16-6, 16-7 0.9D 67.5
Problem 2.10
SOLUTION
Table P2.10 Summary of Load Combinations Using Basic Allowable Stress Design for Beam in
Problem 2.10
IBC Equation No. Equation Load Combination
16-8, 16-9, 16-12 D 75.0
D + Lr 175.0
16-10 D+S 200.0
D+R 275.0
D + 0.75Lr 150.0
16-11, 16-13 D + 0.75S 168.8
D + 0.75R 225.0
16-14 D + 0.75S 168.8
16-15, 16-16 0.6D 45.0
@Seismicisolation
@Seismicisolation
Copyright © 2018 ICC. ALL RIGHTS RESERVED. Accessed by Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
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
SOLUTIONS
, 100891
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
16-7 0.9D –51.8 37.0 10.6
1 @Seismicisolation
@Seismicisolation
Copyright © 2018 ICC. ALL RIGHTS RESERVED. Accessed by Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
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
, 100891
0.9D – 1.0W –105.8 37.0 15.4
16-7 0.9D –51.8 37.0 10.6
1 @Seismicisolation
@Seismicisolation
Copyright © 2018 ICC. ALL RIGHTS RESERVED. Accessed by Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
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
, 100891
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
Copyright © 2018 ICC. ALL RIGHTS RESERVED. Accessed by Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
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
, 100891
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
Copyright © 2018 ICC. ALL RIGHTS RESERVED. Accessed by Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
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
, 100891
4 Solutions Manual to Structural Loads
Problem 2.6
SOLUTION
Because the shear wall is in a parking garage, f1 =
1.0. 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 × 1.0 × D) = QE + 0.2D
For use in IBC Equation 16-7: E = Eh – Ev = QE – 0.2SDSD
= (1.0 × QE) – (0.2 × 1.0 × D) = QE – 0.2D
Substituting for E, IBC Equation 16-5 becomes: 1.2D + QE + 0.2D + 1.0L = 1.4D + QE +
1.0L. Similarly, IBC Equation 16-7 becomes: 0.9D + QE – 0.2D = 0.7D + QE
Table P2.6 Summary of Load Combinations Using Strength Design for Shear Wall in
Problem 2.6
Load Combination
IBC Equation
Equation
No. Axial Force Bending Moment Shear Force
16-1 1.4D 903.0 0 0
16-2 1.2D + 1.6L 1,012.4 0 0
16-3, 16-4 1.2D + 1.0L 923.0 0 0
1.4D + QE + 1.0L 1,052.0 4,280.0 143.0
16-5
1.4D – QE + 1.0L 1,052.0 –4,280.0 –143.0
16-6 0.9D 580.5 0 0
0.7D + QE 451.5 4,280.0 143.0
16-7
0.7D – QE 451.5 –4,280.0 –143.0
Problem 2.7
SOLUTION
The governing load combination in IBC 1605.2 is Equation 16-2:
Negative bending moment:
1.2D + 1.6L = (1.2 × 80.6) + (1.6 × 42.1) = 164.1 ft-kips
Positive bending moment:
1.2D + 1.6L = (1.2 × 53.7) + (1.6 × 30.4) = 113.1 ft-kips
@Seismicisolation
@Seismicisolation
Copyright © 2018 ICC. ALL RIGHTS RESERVED. Accessed by Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
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
, 100891
Shear force:
1.2D + 1.6L = (1.2 × 29.7) + (1.6 × 19.0) = 66.0 kips
@Seismicisolation
@Seismicisolation
Copyright © 2018 ICC. ALL RIGHTS RESERVED. Accessed by Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
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
, 100891
Chapter 2 5
The following basic combinations for strength design with overstrength are also applicable:
• (1.2 + 0.2SDS) D + O QE + 1.0L
Axial force: QE = 2.0
O 241 = 482 kips tension or compression
Negative bending moment:
(1.2 + 0.2SDS)D + 1.0L = (1.38 80.6) + (1.0 42.1) = 153.3 ft-kips
Positive bending moment:
(1.2 + 0.2SDS)D + 1.0L = (1.38 53.7) + (1.0 30.4) = 104.5 ft-kips
Shear force:
(1.2 + 0.2SDS)D + 1.0L = (1.38 29.7) + (1.0 19.0) = 60.0 kips
Note that the load factor on L must be equal to 1.0 because of the assembly occupancy.
• (0.9 – 0.2SDS)D + O QE
Axial force: QE = 2.0
O 241 = 482 kips tension or compression
Negative bending moment:
(0.9 – 0.2SDS)D = 0.72 80.6 = 58.0 ft-kips
Positive bending moment:
(0.9 – 0.2SDS)D = 0.72 53.7 = 38.7 ft-kips
Shear force:
(0.9 – 0.2SDS)D = 0.72 29.7 = 21.4 kips
Problem 2.8
SOLUTION
The governing load combination in IBC 1605.3.1 is Equation
16-9: Negative bending moment:
D + L = 80.6 + 42.1 = 122.7 ft-kips
Positive bending moment:
D + L = 53.7 + 30.4 = 84.1 ft-kips
Shear force:
D + L = 29.7 + 19.0 = 48.7 kips
@Seismicisolation
@Seismicisolation
Copyright © 2018 ICC. ALL RIGHTS RESERVED. Accessed by Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
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
, 100891
6 Solutions Manual to Structural Loads
The following basic allowable stress design load combinations with overstrength are also applicable:
• (1.0 + 0.14SDS)D + 0.7 O QE
Axial force:
0.7 O QE = 0.7 2.0 241 = 337.4 kips tension or compression
Negative bending moment:
(1.0 + 0.14SDS)D = 1.13 80.6 = 91.1 ft-kips
Positive bending moment:
(1.0 + 0.14SDS)D = 1.13 53.7 = 60.7 ft-kips
Shear force:
(1.0 + 0.14SDS)D = 1.13 29.7 = 33.6 kips
• (1.0 + 0.105SDS)D + 0.525 OQE + 0.75L
Axial force:
0.525 QE = 0.525
O 2.0 241 = 253.1 kips tension or
compression Negative bending moment:
1.1D + 0.75L = (1.1 80.6) + (0.75 42.1) = 120.2 ft-kips
Positive bending moment:
1.1D + 0.75L = (1.1 53.7) + (0.75 30.4) = 81.9 ft-kips
Shear force:
1.1D + 0.75L = (1.1 29.7) + (0.75 19.0) = 46.9 kips
• (0.6 – 0.14SDS)D + 0.7 O QE
Axial force:
0.7 O QE = 0.7 2.0 241 = 337.4 kips tension or compression
Negative bending moment:
(0.6 – 0.14SDS)D = 0.47 80.6 = 37.9 ft-kips
Positive bending moment:
(0.6 – 0.14SDS)D = 0.47 53.7 = 25.2 ft-kips
Shear force:
(0.6 – 0.14SDS)D = 0.47 29.7 = 14.0 kips
@Seismicisolation
@Seismicisolation
Copyright © 2018 ICC. ALL RIGHTS RESERVED. Accessed by Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
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
, 100891
Chapter 2 7
Problem 2.9
SOLUTION
Table P2.9 Summary of Load Combinations Using Strength Design for Beam in
Problem 2.9
IBC Equation No. Equation Load Combination
16-1 1.4D 105.0
1.2D + 0.5Lr 140.0
16-2, 16-4 1.2D + 0.5S 152.5
1.2D + 0.5R 190.0
1.2D + 1.6Lr 250.0
16-3 1.2D + 1.6S 290.0
1.2D + 1.6R 410.0
16-5 1.2D + 0.2S 115.0
16-6, 16-7 0.9D 67.5
Problem 2.10
SOLUTION
Table P2.10 Summary of Load Combinations Using Basic Allowable Stress Design for Beam in
Problem 2.10
IBC Equation No. Equation Load Combination
16-8, 16-9, 16-12 D 75.0
D + Lr 175.0
16-10 D+S 200.0
D+R 275.0
D + 0.75Lr 150.0
16-11, 16-13 D + 0.75S 168.8
D + 0.75R 225.0
16-14 D + 0.75S 168.8
16-15, 16-16 0.6D 45.0
@Seismicisolation
@Seismicisolation
Copyright © 2018 ICC. ALL RIGHTS RESERVED. Accessed by Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on May 30, 2020 08:24 PM (PDT) pursuant to License
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
