AllChaptersCovered
i i
SOLUTIONS
, 1008912
CHAPTER i 2
LoadCombinations i
Problem 2.1 i
SOLUTION
Table P2.1
i Summary of Load Combinations Using Strength Design for Beam in Problem 2.1
i i i i i i i i i i i
Load Combination i
IBC Equation No.
i i Equation Exterior Interior
Positive
i Negative i Negative
16-1 1.4D –18.6 61.5 –74.5
16-2 1.2D + 1.6L i i –36.6 120.7 –146.4
16-3,16-4,16-5
i i 1.2D + 0.5L i i –22.4 73.9 –89.6
16-6,16-7 i 0.9D –12.0 39.5 –47.9
Problem 2.2 i
SOLUTION
Table P2.2
i Summary of Load Combinations Using Strength Design for Beam in
i i i i i i i i i
Problem 2.2
i i
Load Combination i
IBCEquation
Equation Bending Moment Shear Force
i
No.
i i
i
Support Midspan Support
16-1 1.4D –80.6 57.5 16.5
16-2 1.2D + 1.6L i i –105.1 75.2 21.5
1.2D + 0.5L i i –80.4 57.4 16.5
16-3 1.2D + 0.5W i i –42.1 49.3 11.8
1.2D – 0.5W i i –96.1 49.3 16.6
1.2D + 1.0W + 0.5L
i i i i –26.4 57.4 11.7
16-4
1.2D – 1.0W + 0.5L
i i i i –134.4 57.4 21.3
16-5 1.2D + 0.5L i i –80.4 57.4 16.5
0.9D + 1.0W i i 2.2 37.0 5.8
16-6
0.9D – 1.0W i i –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
i i i i i i i i i i i i i i i i i i i i i i i i i i
iAgreement with ICC. No further reproduction, no further reproductions by any third party, or distribution authorized. Single user only, copying and networking prohibited. ANY UNAUTHORIZED
i i i i i i i i i i i i i i i i i i i i i i i i
iREPRODUCTION OR DISTRIBUTION IS A VIOLATION OF THE FEDERAL COPYRIGHT ACT AND THE LICENSE AGREEMENT, AND SUBJECT TO CIVIL AND CRIMINAL PENALTIES THEREUNDER.
i i i i i i i i i i i i i i i i i i i i i i
, 1008912
2 Solutions Manual to Structural Loads i i i i
Problem 2.3 i
SOLUTION
Table P2.3
i Summary of Load Combinations Using Basic Allowable Stress Design for
i i i i i i i i i
Beam in Problem 2.3
i i i i
Load Combination i
IBCEquation
Equation Bending Moment Shear Force
i
No.
i i
i
Support Midspan Support
16-8,16-10 i D –57.6 41.1 11.8
16-9 D+L i i –80.1 57.3 16.4
16-11, 16-14 i D + 0.75Li i –74.5 53.3 15.3
D + 0.6W i i –25.2 41.1 8.9
16-12
D – 0.6W i i –90.0 41.1 14.7
D + 0.75(0.6W) + 0.75L
i i i i –50.2 53.3 13.1
16-13
D – 0.75(0.6W) +0.75L
i i i i –98.8 53.3 17.4
0.6D + 0.6W i i –2.2 24.7 4.2
16-15
0.6D – 0.6W i i –67.0 24.7 10.0
16-16 0.6D –34.6 24.7 7.1
Problem 2.4 i
SOLUTION
Table P2.4
i Summary of Load Combinations Using Alternative Basic Allowable Stress
i i i i i i i i
Design for Beam in Problem 2.4
i i i i i i
Load Combination i
IBCEquation
Equation Bending Moment Shear Force
i
No.
i i
i
Support Midspan Support
16-17,16-21 i D+L i i –80.1 57.3 16.4
D + L + 0.6+W
i i i i –38.0 57.3 12.7
16-18,16-19 i
D + L – 0.6+W
i i i i –122.2 57.3 20.1
D + L + 0.6+W/2
i i i i –59.0 57.3 14.5
16-20
D + L – 0.6+W/2
i i i i –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
i i i i i i i i i i i i i i i i i i i i i i i i i i
iAgreement with ICC. No further reproduction, no further reproductions by any third party, or distribution authorized. Single user only, copying and networking prohibited. ANY UNAUTHORIZED
i i i i i i i i i i i i i i i i i i i i i i i i
iREPRODUCTION OR DISTRIBUTION IS A VIOLATION OF THE FEDERAL COPYRIGHT ACT AND THE LICENSE AGREEMENT, AND SUBJECT TO CIVIL AND CRIMINAL PENALTIES THEREUNDER.
i i i i i i i i i i i i i i i i i i i i i i
, 1008912
Chapter 2 i 3
Problem 2.5 i
SOLUTION
Because the live loads on the floors are equal to 100 psf, f1 = 0.5. The
i i i i i i i i i i i i i i i
i seismic load effect, E, is determined as follows:
i i i i i i i
For use in IBC Equation 16-5: E = Eh + Ev = QE + 0.2SDSD
i i i i i i i i
i
i
i
i
i
i
= (1.0 × QE) + (0.2 × 0.41 × D) = QE + 0.08D i i i i i i i i i i i
i
i
For use in IBC Equation 16-7: E = Eh – Ev = QE – 0.2SDSD
i i i i i i i i
i
i
i
i
i
i
= (1.0 × QE) – (0.2 × 0.41 × D) = QE – 0.08D
i i i i i i i i i i i
i
i
Substituting for E, IBC Equation 16-5 becomes: 1.2D + QE + 0.08D + 0.5L = 1.28D + QE + 0.5L
i i i i i i i i i
i
i i i i i i i
i
i
i Similarly, IBC Equation 16-7 becomes: 0.9D + QE – 0.08D = 0.82D + QE
i i i i i i i
i
i i i i i
Table P2.5i Summary of Load Combinations Using Strength Design for Column in i i i i i i i i i
Problem 2.5
i i
Load Combination i
IBCEquation
Equation
i
No. i
Axial Force i Bending Moment i Shear Force i
16-1 1.4D 235.1 29.8 3.2
16-2 1.2D + 1.6L + 0.5Lr
i i i i 275.3 59.2 6.3
1.2D + 1.6 Lr + 0.5L
i i i
i
i 246.1 36.1 3.9
16-3 1.2D + 1.6 Lr + 0.5W
i i i
i
i 232.1 86.1 8.3
1.2D + 1.6 Lr – 0.5W
i i i
i
i 218.5 –34.9 –2.8
1.2D + 1.0W + 0.5L + 0.5Lr
i i i i i i 243.3 157.1 15.0
16-4
1.2D – 1.0W + 0.5L + 0.5Lr
i i i i i i 216.1 –84.9 –7.2
1.28D + QE + 0.5L i i i 272.1 469.9 46.2
16-5
i
1.28D – QE + 0.5L i i
i
i 199.3 –394.3 –38.2
0.9D + 1.0W i i 164.7 140.2 13.2
16-6
0.9D – 1.0W i i 137.5 –101.8 –9.0
0.82D + QE i i 174.1 449.6 44.1
16-7
0.82D– QE i i 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
i i i i i i i i i i i i i i i i i i i i i i i i i i
iAgreement with ICC. No further reproduction, no further reproductions by any third party, or distribution authorized. Single user only, copying and networking prohibited. ANY UNAUTHORIZED
i i i i i i i i i i i i i i i i i i i i i i i i
iREPRODUCTION OR DISTRIBUTION IS A VIOLATION OF THE FEDERAL COPYRIGHT ACT AND THE LICENSE AGREEMENT, AND SUBJECT TO CIVIL AND CRIMINAL PENALTIES THEREUNDER.
i i i i i i i i i i i i i i i i i i i i i i
i i
SOLUTIONS
, 1008912
CHAPTER i 2
LoadCombinations i
Problem 2.1 i
SOLUTION
Table P2.1
i Summary of Load Combinations Using Strength Design for Beam in Problem 2.1
i i i i i i i i i i i
Load Combination i
IBC Equation No.
i i Equation Exterior Interior
Positive
i Negative i Negative
16-1 1.4D –18.6 61.5 –74.5
16-2 1.2D + 1.6L i i –36.6 120.7 –146.4
16-3,16-4,16-5
i i 1.2D + 0.5L i i –22.4 73.9 –89.6
16-6,16-7 i 0.9D –12.0 39.5 –47.9
Problem 2.2 i
SOLUTION
Table P2.2
i Summary of Load Combinations Using Strength Design for Beam in
i i i i i i i i i
Problem 2.2
i i
Load Combination i
IBCEquation
Equation Bending Moment Shear Force
i
No.
i i
i
Support Midspan Support
16-1 1.4D –80.6 57.5 16.5
16-2 1.2D + 1.6L i i –105.1 75.2 21.5
1.2D + 0.5L i i –80.4 57.4 16.5
16-3 1.2D + 0.5W i i –42.1 49.3 11.8
1.2D – 0.5W i i –96.1 49.3 16.6
1.2D + 1.0W + 0.5L
i i i i –26.4 57.4 11.7
16-4
1.2D – 1.0W + 0.5L
i i i i –134.4 57.4 21.3
16-5 1.2D + 0.5L i i –80.4 57.4 16.5
0.9D + 1.0W i i 2.2 37.0 5.8
16-6
0.9D – 1.0W i i –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
i i i i i i i i i i i i i i i i i i i i i i i i i i
iAgreement with ICC. No further reproduction, no further reproductions by any third party, or distribution authorized. Single user only, copying and networking prohibited. ANY UNAUTHORIZED
i i i i i i i i i i i i i i i i i i i i i i i i
iREPRODUCTION OR DISTRIBUTION IS A VIOLATION OF THE FEDERAL COPYRIGHT ACT AND THE LICENSE AGREEMENT, AND SUBJECT TO CIVIL AND CRIMINAL PENALTIES THEREUNDER.
i i i i i i i i i i i i i i i i i i i i i i
, 1008912
2 Solutions Manual to Structural Loads i i i i
Problem 2.3 i
SOLUTION
Table P2.3
i Summary of Load Combinations Using Basic Allowable Stress Design for
i i i i i i i i i
Beam in Problem 2.3
i i i i
Load Combination i
IBCEquation
Equation Bending Moment Shear Force
i
No.
i i
i
Support Midspan Support
16-8,16-10 i D –57.6 41.1 11.8
16-9 D+L i i –80.1 57.3 16.4
16-11, 16-14 i D + 0.75Li i –74.5 53.3 15.3
D + 0.6W i i –25.2 41.1 8.9
16-12
D – 0.6W i i –90.0 41.1 14.7
D + 0.75(0.6W) + 0.75L
i i i i –50.2 53.3 13.1
16-13
D – 0.75(0.6W) +0.75L
i i i i –98.8 53.3 17.4
0.6D + 0.6W i i –2.2 24.7 4.2
16-15
0.6D – 0.6W i i –67.0 24.7 10.0
16-16 0.6D –34.6 24.7 7.1
Problem 2.4 i
SOLUTION
Table P2.4
i Summary of Load Combinations Using Alternative Basic Allowable Stress
i i i i i i i i
Design for Beam in Problem 2.4
i i i i i i
Load Combination i
IBCEquation
Equation Bending Moment Shear Force
i
No.
i i
i
Support Midspan Support
16-17,16-21 i D+L i i –80.1 57.3 16.4
D + L + 0.6+W
i i i i –38.0 57.3 12.7
16-18,16-19 i
D + L – 0.6+W
i i i i –122.2 57.3 20.1
D + L + 0.6+W/2
i i i i –59.0 57.3 14.5
16-20
D + L – 0.6+W/2
i i i i –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
i i i i i i i i i i i i i i i i i i i i i i i i i i
iAgreement with ICC. No further reproduction, no further reproductions by any third party, or distribution authorized. Single user only, copying and networking prohibited. ANY UNAUTHORIZED
i i i i i i i i i i i i i i i i i i i i i i i i
iREPRODUCTION OR DISTRIBUTION IS A VIOLATION OF THE FEDERAL COPYRIGHT ACT AND THE LICENSE AGREEMENT, AND SUBJECT TO CIVIL AND CRIMINAL PENALTIES THEREUNDER.
i i i i i i i i i i i i i i i i i i i i i i
, 1008912
Chapter 2 i 3
Problem 2.5 i
SOLUTION
Because the live loads on the floors are equal to 100 psf, f1 = 0.5. The
i i i i i i i i i i i i i i i
i seismic load effect, E, is determined as follows:
i i i i i i i
For use in IBC Equation 16-5: E = Eh + Ev = QE + 0.2SDSD
i i i i i i i i
i
i
i
i
i
i
= (1.0 × QE) + (0.2 × 0.41 × D) = QE + 0.08D i i i i i i i i i i i
i
i
For use in IBC Equation 16-7: E = Eh – Ev = QE – 0.2SDSD
i i i i i i i i
i
i
i
i
i
i
= (1.0 × QE) – (0.2 × 0.41 × D) = QE – 0.08D
i i i i i i i i i i i
i
i
Substituting for E, IBC Equation 16-5 becomes: 1.2D + QE + 0.08D + 0.5L = 1.28D + QE + 0.5L
i i i i i i i i i
i
i i i i i i i
i
i
i Similarly, IBC Equation 16-7 becomes: 0.9D + QE – 0.08D = 0.82D + QE
i i i i i i i
i
i i i i i
Table P2.5i Summary of Load Combinations Using Strength Design for Column in i i i i i i i i i
Problem 2.5
i i
Load Combination i
IBCEquation
Equation
i
No. i
Axial Force i Bending Moment i Shear Force i
16-1 1.4D 235.1 29.8 3.2
16-2 1.2D + 1.6L + 0.5Lr
i i i i 275.3 59.2 6.3
1.2D + 1.6 Lr + 0.5L
i i i
i
i 246.1 36.1 3.9
16-3 1.2D + 1.6 Lr + 0.5W
i i i
i
i 232.1 86.1 8.3
1.2D + 1.6 Lr – 0.5W
i i i
i
i 218.5 –34.9 –2.8
1.2D + 1.0W + 0.5L + 0.5Lr
i i i i i i 243.3 157.1 15.0
16-4
1.2D – 1.0W + 0.5L + 0.5Lr
i i i i i i 216.1 –84.9 –7.2
1.28D + QE + 0.5L i i i 272.1 469.9 46.2
16-5
i
1.28D – QE + 0.5L i i
i
i 199.3 –394.3 –38.2
0.9D + 1.0W i i 164.7 140.2 13.2
16-6
0.9D – 1.0W i i 137.5 –101.8 –9.0
0.82D + QE i i 174.1 449.6 44.1
16-7
0.82D– QE i i 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
i i i i i i i i i i i i i i i i i i i i i i i i i i
iAgreement with ICC. No further reproduction, no further reproductions by any third party, or distribution authorized. Single user only, copying and networking prohibited. ANY UNAUTHORIZED
i i i i i i i i i i i i i i i i i i i i i i i i
iREPRODUCTION OR DISTRIBUTION IS A VIOLATION OF THE FEDERAL COPYRIGHT ACT AND THE LICENSE AGREEMENT, AND SUBJECT TO CIVIL AND CRIMINAL PENALTIES THEREUNDER.
i i i i i i i i i i i i i i i i i i i i i i
