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
16-7 0.9D –51.8 37.0 10.6
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1.2D – 1.0W + 0.5L –134.4 57.4 21.3
16-4
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
, 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 + 0.6W –38.0 57.3 12.7
16-18, 16-19
D + L – 0.6W –122.2 57.3 20.1
D + L + 0.6W/2 –59.0 57.3 14.5
16-20 D + L – 0.6W/2 –101.2 57.3 18.3
16-22 0.9D –51.8 37.0 10.6
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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
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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
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Shear force:
1.2D + 1.6L = (1.2 × 29.7) + (1.6 × 19.0) = 66.0 kips
, 100891
Chapter 2 5
The following basic combinations for strength design with overstrength are also applicable:
• (1.2 + 0.2S
) D + OQE + 1.0L
D
Axial force: OQE = 2.0 241 = 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 + OQE
Axial force: OQE = 2.0 241 = 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
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6 Solutions Manual to Structural Loads
The following basic allowable stress design load combinations with overstrength are also applicable:
• (1.0 + 0.14S
)D + 0.7 Q
DS O E
Axial force:
0.7OQE = 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 + 0.105S
)D + 0.525 Q + 0.75L
DS O E
Axial force:
0.525OQE = 0.525 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 – 0.14S )D + 0.7OQE
D
Axial force:
0.7OQE = 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
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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
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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
16-7 0.9D –51.8 37.0 10.6
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1.2D – 1.0W + 0.5L –134.4 57.4 21.3
16-4
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
, 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 + 0.6W –38.0 57.3 12.7
16-18, 16-19
D + L – 0.6W –122.2 57.3 20.1
D + L + 0.6W/2 –59.0 57.3 14.5
16-20 D + L – 0.6W/2 –101.2 57.3 18.3
16-22 0.9D –51.8 37.0 10.6
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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
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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
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Shear force:
1.2D + 1.6L = (1.2 × 29.7) + (1.6 × 19.0) = 66.0 kips
, 100891
Chapter 2 5
The following basic combinations for strength design with overstrength are also applicable:
• (1.2 + 0.2S
) D + OQE + 1.0L
D
Axial force: OQE = 2.0 241 = 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 + OQE
Axial force: OQE = 2.0 241 = 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
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6 Solutions Manual to Structural Loads
The following basic allowable stress design load combinations with overstrength are also applicable:
• (1.0 + 0.14S
)D + 0.7 Q
DS O E
Axial force:
0.7OQE = 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 + 0.105S
)D + 0.525 Q + 0.75L
DS O E
Axial force:
0.525OQE = 0.525 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 – 0.14S )D + 0.7OQE
D
Axial force:
0.7OQE = 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
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, 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
@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
