@LECTJULIESOLUTIONSSTUVIA
All Chapters Covered
SOLUTIONS
, Solutions Manual
to Structural Loads
2012 IBC and ASCE/SEI 7-10
This Solutions Manual was developed as a companion to the Structural Loads: 2012 IBC and
ASCE/SEI 7-10 textbook. To increase understanding of this material and for its most
effective use, readers should study the Structural Loads textbook and reference the
material as they read through this Solutions Manual.
To order additional copies of the Structural Loads textbook, click here.
, @LECTJULIESOLUTIONSSTUVIA
CHAPTER 2
Load Combinations
2.1. Determine the strength design load combinations for a reinforced concrete beam
on a typical floor of a multistory residential building using the nominal bending
moments in Table 2.11. All bending moments are in foot-kips. Assume the live
load on the floor is less than 100 psf.
Table 2.11 Design Data for Problem 2.1
External Negative Positive Interior Negative
Dead load, D –13.3 43.9 –53.2
Live load, L –12.9 42.5 –51.6
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
2.2. Determine the strength design load combinations for a steel beam that is part of an
ordinary moment frame in an office building using the nominal bending
moments and shear forces in Table 2.12. All bending moments are in foot-kips
and all shear forces are in kips. Assume the live load on the floor is less than 100
psf.
Table 2.12 Design Data for Problem 2.2
Bending Moment Shear Force
Support Midspan Support
Dead load, D –57.6 41.1 11.8
Live load, L –22.5 16.2 4.6
Wind, W 54.0 --- 4.8
,2 Solutions Manual to Structural Loads
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
0.9D – 1.0W –105.8 37.0 15.4
16-7 0.9D –51.8 37.0 10.6
2.3. Given the information in Problem 2.2, determine the basic allowable stress design
load combinations.
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
,Chapter 2 3
2.4. Given the information in Problem 2.2, determine the alternative basic allowable
stress design load combinations. Assume the wind loads have been determined
using ASCE/SEI 7.
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
2.5. Determine the strength design load combinations for a reinforced concrete
column that is part of an intermediate moment frame in an office building using
the nominal axial forces, bending moments and shear forces in Table 2.13. All
axial forces are in kips, all bending moments are in foot-kips and all shear forces
are in kips. Assume the live loads on the floors are equal to 100 psf and SDS =
0.41g.
Table 2.13 Design Data for Problem 2.5
Bending
Axial Force Shear Force
Moment
Dead load, D 167.9 21.3 2.3
Live load, L 41.5 21.0 2.2
Roof live load, Lr 14.9 --- ---
Wind, W 13.6 121.0 11.1
Seismic, QE 36.4 432.1 42.2
SOLUTION
Since the live loads on the floors are equal to 100 psf, f1 = 0.5.
Since an intermediate moment frame is used, the SDC is C or lower. Thus,
= 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 × 0.41 × D) = QE + 0.08D
For use in IBC Equation 16-7: E = Eh – Ev = QE – 0.2SDSD
,4 Solutions Manual to Structural Loads
= (1.0 × QE) – (0.2 × 0.41 × D) = QE – 0.08D
,Chapter 2 5
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
2.6. Determine the strength design load combinations for a reinforced concrete shear
wall in a parking garage using the nominal axial forces, bending moments and
shear forces in Table 2.14. All axial forces are in kips, all bending moments are in
foot-kips and all shear forces are in kips. Assume that ρ = 1.0 and SDS = 1.0g.
Table 2.14 Design Data for Problem 2.6
Axial Force Bending Moment Shear Force
Dead load, D 645 0 0
Live load, L 149 0 0
Seismic, QE 0 4,280 143
SOLUTION
Since 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
,6 Solutions Manual to Structural Loads
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
,Chapter 2 7
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 1012.4 0 0
16-3, 16-4 1.2D + 1.0L 923.0 0 0
1.4D + QE + 1.0L 1052.0 4280.0 143.0
16-5
1.4D – QE + 1.0L 1052.0 –4280.0 –143.0
16-6 0.9D 580.5 0 0
0.7D + QE 451.5 4280.0 143.0
16-7
0.7D – QE 451.5 –4280.0 –143.0
2.7. Determine the strength design load combinations and the basic combinations for
strength design with overstrength factor for a simply supported steel collector
beam in an assembly building using the nominal axial forces, bending moments
and shear forces in Table 2.15. All axial forces are in kips, all bending moments
are in foot-kips and all shear forces are in kips. Assume SDS = 0.9g and Ω0 = 2.0.
Table 2.15 Design Data for Problem 2.7
Bending Moment
Axial Force Shear Force
Negative Positive
Dead load, D 0 80.6 53.7 29.7
Live load, L 0 42.1 30.4 19.0
Seismic, QE 241 0 0 0
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
Shear force:
1.2D + 1.6L = (1.2 × 29.7) + (1.6 × 19.0) = 66.0 kips
The following basic combinations for strength design with overstrength factor
are also applicable (see IBC 1605.1 and 1605.2; ASCE/SEI 12.4.3.2 and 12.10.2.1):
• IBC Equation 16-5: (1.2 + 0.2SDS) D + 0 QE + 1.0L
Axial force: 0QE = 2.0 241 = 482 kips tension or compression
, 8 Solutions Manual to Structural Loads
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.
• IBC Equation 16-7: (0.9 – 0.2SDS)D + 0QE
Axial force: 0QE = 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
2.8. Given the information in Problem 2.7, determine the basic allowable stress design
load combinations.
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
The following basic combinations for strength design with overstrength factor
are also applicable (see IBC 1605.1 and 1605.3.1; ASCE/SEI 12.4.3.2):
• IBC Equation 16-12: (1.0 + 0.14SDS)D + 0.7 0QE
Axial force: 0.7 0 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
• IBC Equation 16-14: (1.0 + 0.105SDS)D + 0.525 0QE + 0.75L
Axial force: 0.525 0QE = 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
• IBC Equation 16-16: (0.6 – 0.14SDS)D + 0.7 0QE
All Chapters Covered
SOLUTIONS
, Solutions Manual
to Structural Loads
2012 IBC and ASCE/SEI 7-10
This Solutions Manual was developed as a companion to the Structural Loads: 2012 IBC and
ASCE/SEI 7-10 textbook. To increase understanding of this material and for its most
effective use, readers should study the Structural Loads textbook and reference the
material as they read through this Solutions Manual.
To order additional copies of the Structural Loads textbook, click here.
, @LECTJULIESOLUTIONSSTUVIA
CHAPTER 2
Load Combinations
2.1. Determine the strength design load combinations for a reinforced concrete beam
on a typical floor of a multistory residential building using the nominal bending
moments in Table 2.11. All bending moments are in foot-kips. Assume the live
load on the floor is less than 100 psf.
Table 2.11 Design Data for Problem 2.1
External Negative Positive Interior Negative
Dead load, D –13.3 43.9 –53.2
Live load, L –12.9 42.5 –51.6
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
2.2. Determine the strength design load combinations for a steel beam that is part of an
ordinary moment frame in an office building using the nominal bending
moments and shear forces in Table 2.12. All bending moments are in foot-kips
and all shear forces are in kips. Assume the live load on the floor is less than 100
psf.
Table 2.12 Design Data for Problem 2.2
Bending Moment Shear Force
Support Midspan Support
Dead load, D –57.6 41.1 11.8
Live load, L –22.5 16.2 4.6
Wind, W 54.0 --- 4.8
,2 Solutions Manual to Structural Loads
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
0.9D – 1.0W –105.8 37.0 15.4
16-7 0.9D –51.8 37.0 10.6
2.3. Given the information in Problem 2.2, determine the basic allowable stress design
load combinations.
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
,Chapter 2 3
2.4. Given the information in Problem 2.2, determine the alternative basic allowable
stress design load combinations. Assume the wind loads have been determined
using ASCE/SEI 7.
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
2.5. Determine the strength design load combinations for a reinforced concrete
column that is part of an intermediate moment frame in an office building using
the nominal axial forces, bending moments and shear forces in Table 2.13. All
axial forces are in kips, all bending moments are in foot-kips and all shear forces
are in kips. Assume the live loads on the floors are equal to 100 psf and SDS =
0.41g.
Table 2.13 Design Data for Problem 2.5
Bending
Axial Force Shear Force
Moment
Dead load, D 167.9 21.3 2.3
Live load, L 41.5 21.0 2.2
Roof live load, Lr 14.9 --- ---
Wind, W 13.6 121.0 11.1
Seismic, QE 36.4 432.1 42.2
SOLUTION
Since the live loads on the floors are equal to 100 psf, f1 = 0.5.
Since an intermediate moment frame is used, the SDC is C or lower. Thus,
= 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 × 0.41 × D) = QE + 0.08D
For use in IBC Equation 16-7: E = Eh – Ev = QE – 0.2SDSD
,4 Solutions Manual to Structural Loads
= (1.0 × QE) – (0.2 × 0.41 × D) = QE – 0.08D
,Chapter 2 5
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
2.6. Determine the strength design load combinations for a reinforced concrete shear
wall in a parking garage using the nominal axial forces, bending moments and
shear forces in Table 2.14. All axial forces are in kips, all bending moments are in
foot-kips and all shear forces are in kips. Assume that ρ = 1.0 and SDS = 1.0g.
Table 2.14 Design Data for Problem 2.6
Axial Force Bending Moment Shear Force
Dead load, D 645 0 0
Live load, L 149 0 0
Seismic, QE 0 4,280 143
SOLUTION
Since 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
,6 Solutions Manual to Structural Loads
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
,Chapter 2 7
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 1012.4 0 0
16-3, 16-4 1.2D + 1.0L 923.0 0 0
1.4D + QE + 1.0L 1052.0 4280.0 143.0
16-5
1.4D – QE + 1.0L 1052.0 –4280.0 –143.0
16-6 0.9D 580.5 0 0
0.7D + QE 451.5 4280.0 143.0
16-7
0.7D – QE 451.5 –4280.0 –143.0
2.7. Determine the strength design load combinations and the basic combinations for
strength design with overstrength factor for a simply supported steel collector
beam in an assembly building using the nominal axial forces, bending moments
and shear forces in Table 2.15. All axial forces are in kips, all bending moments
are in foot-kips and all shear forces are in kips. Assume SDS = 0.9g and Ω0 = 2.0.
Table 2.15 Design Data for Problem 2.7
Bending Moment
Axial Force Shear Force
Negative Positive
Dead load, D 0 80.6 53.7 29.7
Live load, L 0 42.1 30.4 19.0
Seismic, QE 241 0 0 0
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
Shear force:
1.2D + 1.6L = (1.2 × 29.7) + (1.6 × 19.0) = 66.0 kips
The following basic combinations for strength design with overstrength factor
are also applicable (see IBC 1605.1 and 1605.2; ASCE/SEI 12.4.3.2 and 12.10.2.1):
• IBC Equation 16-5: (1.2 + 0.2SDS) D + 0 QE + 1.0L
Axial force: 0QE = 2.0 241 = 482 kips tension or compression
, 8 Solutions Manual to Structural Loads
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.
• IBC Equation 16-7: (0.9 – 0.2SDS)D + 0QE
Axial force: 0QE = 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
2.8. Given the information in Problem 2.7, determine the basic allowable stress design
load combinations.
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
The following basic combinations for strength design with overstrength factor
are also applicable (see IBC 1605.1 and 1605.3.1; ASCE/SEI 12.4.3.2):
• IBC Equation 16-12: (1.0 + 0.14SDS)D + 0.7 0QE
Axial force: 0.7 0 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
• IBC Equation 16-14: (1.0 + 0.105SDS)D + 0.525 0QE + 0.75L
Axial force: 0.525 0QE = 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
• IBC Equation 16-16: (0.6 – 0.14SDS)D + 0.7 0QE
