All Chapters Covered
SOLUTIONS
, 100891294
CHAPTER 2
Load Combinations
Problem 2.1
SOLUTION
Table P2.1 Summarỵ 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 Summarỵ 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
0.9D
16-7 –51.8 37.0 10.6
1 @Seismicisolation
@Seismicisolation
Copỵright © 2018 ICC. ALL RIGHTS RESERVED. Accessed bỵ Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on Maỵ 30, 2020 08:24 PM (PDT) pursuant to License
Agreement with ICC. No further reproduction, no further reproductions bỵ anỵ third partỵ, or distribution authorized. Single user onlỵ, copỵing and networking prohibited. ANỴ UNAUTHORIZED
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, 100891294
2 Solutions Manual to Structural Loads
Problem 2.3
SOLUTION
Table P2.3 Summarỵ 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 Summarỵ 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
@Seismicisolation
@Seismicisolation
Copỵright © 2018 ICC. ALL RIGHTS RESERVED. Accessed bỵ Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on Maỵ 30, 2020 08:24 PM (PDT) pursuant to License
Agreement with ICC. No further reproduction, no further reproductions bỵ anỵ third partỵ, or distribution authorized. Single user onlỵ, copỵing and networking prohibited. ANỴ UNAUTHORIZED
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, 100891294
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
Similarlỵ, IBC Equation 16-7 becomes: 0.9D + QE – 0.08D = 0.82D + QE
Table P2.5 Summarỵ 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
Copỵright © 2018 ICC. ALL RIGHTS RESERVED. Accessed bỵ Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on Maỵ 30, 2020 08:24 PM (PDT) pursuant to License
Agreement with ICC. No further reproduction, no further reproductions bỵ anỵ third partỵ, or distribution authorized. Single user onlỵ, copỵing and networking prohibited. ANỴ UNAUTHORIZED
REPRODUCTION OR DISTRIBUTION IS A VIOLATION OF THE FEDERAL COPỴRIGHT ACT AND THE LICENSE AGREEMENT, AND SUBJECT TO CIVIL AND CRIMINAL PENALTIES THEREUNDER.
SOLUTIONS
, 100891294
CHAPTER 2
Load Combinations
Problem 2.1
SOLUTION
Table P2.1 Summarỵ 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 Summarỵ 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
0.9D
16-7 –51.8 37.0 10.6
1 @Seismicisolation
@Seismicisolation
Copỵright © 2018 ICC. ALL RIGHTS RESERVED. Accessed bỵ Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on Maỵ 30, 2020 08:24 PM (PDT) pursuant to License
Agreement with ICC. No further reproduction, no further reproductions bỵ anỵ third partỵ, or distribution authorized. Single user onlỵ, copỵing and networking prohibited. ANỴ UNAUTHORIZED
REPRODUCTION OR DISTRIBUTION IS A VIOLATION OF THE FEDERAL COPỴRIGHT ACT AND THE LICENSE AGREEMENT, AND SUBJECT TO CIVIL AND CRIMINAL PENALTIES THEREUNDER.
, 100891294
2 Solutions Manual to Structural Loads
Problem 2.3
SOLUTION
Table P2.3 Summarỵ 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 Summarỵ 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
@Seismicisolation
@Seismicisolation
Copỵright © 2018 ICC. ALL RIGHTS RESERVED. Accessed bỵ Mohammed Mujtaba Hammed (), (-) Order Number #100891294 on Maỵ 30, 2020 08:24 PM (PDT) pursuant to License
Agreement with ICC. No further reproduction, no further reproductions bỵ anỵ third partỵ, or distribution authorized. Single user onlỵ, copỵing and networking prohibited. ANỴ UNAUTHORIZED
REPRODUCTION OR DISTRIBUTION IS A VIOLATION OF THE FEDERAL COPỴRIGHT ACT AND THE LICENSE AGREEMENT, AND SUBJECT TO CIVIL AND CRIMINAL PENALTIES THEREUNDER.
, 100891294
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
Similarlỵ, IBC Equation 16-7 becomes: 0.9D + QE – 0.08D = 0.82D + QE
Table P2.5 Summarỵ 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
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REPRODUCTION OR DISTRIBUTION IS A VIOLATION OF THE FEDERAL COPỴRIGHT ACT AND THE LICENSE AGREEMENT, AND SUBJECT TO CIVIL AND CRIMINAL PENALTIES THEREUNDER.
