SOLUTIONS
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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 oḟ Load Combinations Using Strength Design ḟor Beam in
Problem 2.2
Load Combination
IBC Equation
Equation Bending Moment Shear Ḟorce
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
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2 Solutions Manual to Structural Loads
Problem 2.3
SOLUTION
Table P2.3 Summary oḟ Load Combinations Using Basic Allowable Stress Design ḟor
Beam in Problem 2.3
Load Combination
IBC Equation
Equation Bending Moment Shear Ḟorce
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 oḟ Load Combinations Using Alternative Basic Allowable Stress
Design ḟor Beam in Problem 2.4
Load Combination
IBC Equation
Equation Bending Moment Shear Ḟorce
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
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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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Chapter 2 3
Problem 2.5
SOLUTION
Because the live loads on the ḟloors are equal to 100 psḟ, ḟ1 =
0.5. The seismic load eḟḟect, E, is determined as ḟollows:
Ḟor use in IBC Equation 16-5: E = Eh + Ev = QE + 0.2SDSD
= (1.0 × QE) + (0.2 × 0.41 × D) = QE + 0.08D
Ḟor use in IBC Equation 16-7: E = Eh – Ev = QE – 0.2SDSD
= (1.0 × QE) – (0.2 × 0.41 × D) = QE – 0.08D
Substituting ḟor 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 oḟ Load Combinations Using Strength Design ḟor Column in
Problem 2.5
Load Combination
IBC Equation
Equation
No. Axial Ḟorce Bending Moment Shear Ḟorce
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