Example: A column (compression member) in the upper story of a building is subjected to the following loads: Dead load (D=109 kips compression), floor live load (L=46 kips compression), roof live load (L,=19 kips compression), and snow (S=20 kips compression). a. Determine the controlling factored load for LRFD. b. If the resistance factor (Ø=0.9), what is the required nominal strength? Solution: (a) 1. Pu= 1.4D = 1.4 *109 = 152.6 kips. 2. Pu= 1.2D+ 1.6L + 0.5(L, or S or R) = 1.2*109+1.6*46+0.5*20 = 214.4 kips. 3. P₁ = 1.2D + 1.6(L, or S or R) + (0.5L or 0.8W) = 1.2*109+1.6*20+0.5*46 = 185.8 kips. 4. Pu= 1.2D + 1.6W +0.5 L + 0.5 (L, or S or R) = 1.2*109+0.5*46+0.5*20 = 163.8 kips. 5. Pu= 1.2D + 1.0E+0.5 L + 0.2S = 1.2*109+0.5*46+0.2*20 = 163.8 kips. 6. Pu= 0.9D (1.6W or 1.0E) = 0.9*109 = 98.1 kips. From above results, (Pu= 214.4 kips) is control to be used for design. (b) Ru≤ØRn 214.4 0.9*Rn [L=19 kips
Example: A column (compression member) in the upper story of a building is subjected to the following loads: Dead load (D=109 kips compression), floor live load (L=46 kips compression), roof live load (L,=19 kips compression), and snow (S=20 kips compression). a. Determine the controlling factored load for LRFD. b. If the resistance factor (Ø=0.9), what is the required nominal strength? Solution: (a) 1. Pu= 1.4D = 1.4 *109 = 152.6 kips. 2. Pu= 1.2D+ 1.6L + 0.5(L, or S or R) = 1.2*109+1.6*46+0.5*20 = 214.4 kips. 3. P₁ = 1.2D + 1.6(L, or S or R) + (0.5L or 0.8W) = 1.2*109+1.6*20+0.5*46 = 185.8 kips. 4. Pu= 1.2D + 1.6W +0.5 L + 0.5 (L, or S or R) = 1.2*109+0.5*46+0.5*20 = 163.8 kips. 5. Pu= 1.2D + 1.0E+0.5 L + 0.2S = 1.2*109+0.5*46+0.2*20 = 163.8 kips. 6. Pu= 0.9D (1.6W or 1.0E) = 0.9*109 = 98.1 kips. From above results, (Pu= 214.4 kips) is control to be used for design. (b) Ru≤ØRn 214.4 0.9*Rn [L=19 kips
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Steel Construction
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homework at the bottom of the page
![Example: A column (compression member) in the upper story of a building is
subjected to the following loads:
Dead load (D=109 kips compression), floor live load (L-46 kips compression),
roof live load (L₁=19 kips compression), and snow (S=20 kips compression).
a. Determine the controlling factored load for LRFD.
b. If the resistance factor (Ø=0.9), what is the required nominal strength?
Solution:
(a)
1. Pu= 1.4D = 1.4 *109 = 152.6 kips.
2. Pu= 1.2D + 1.6L + 0.5(L, or S or R)
= 1.2*109+1.6*46+0.5*20 = 214.4 kips.
3. P₁ = 1.2D + 1.6(L, or S or R) + (0.5L or 0.8W)
= 1.2*109+1.6*20+0.5*46 = 185.8 kips.
4. Pu= 1.2D + 1.6W + 0.5 L + 0.5 (L, or S or R)
= 1.2*109+0.5*46+0.5*20 = 163.8 kips.
5. Pu= 1.2D + 1.0E+0.5 L + 0.2S = 1.2*109+0.5*46+0.2*20 = 163.8 kips.
6. Pu= 0.9D +(1.6W or 1.0E) = 0.9*109 = 98.1 kips.
From above results, (Pu= 214.4 kips) is control to be used for design.
(b)
Ru≤ ØRn
214.4 ≤ 0.9*Rn
[L, 19 kips <S=20 kips and R=0 →use S]
R₁ = 238 kips
Homework: Determine the controlling service load for ASD, and if
the safety factor (N=1.67), what is the required nominal strength?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9b1d03d5-ad01-4b6d-b55a-55b876b324cc%2F7f95073f-b158-475b-bccf-0f9b51235a8b%2F9ibw7wk_processed.png&w=3840&q=75)
Transcribed Image Text:Example: A column (compression member) in the upper story of a building is
subjected to the following loads:
Dead load (D=109 kips compression), floor live load (L-46 kips compression),
roof live load (L₁=19 kips compression), and snow (S=20 kips compression).
a. Determine the controlling factored load for LRFD.
b. If the resistance factor (Ø=0.9), what is the required nominal strength?
Solution:
(a)
1. Pu= 1.4D = 1.4 *109 = 152.6 kips.
2. Pu= 1.2D + 1.6L + 0.5(L, or S or R)
= 1.2*109+1.6*46+0.5*20 = 214.4 kips.
3. P₁ = 1.2D + 1.6(L, or S or R) + (0.5L or 0.8W)
= 1.2*109+1.6*20+0.5*46 = 185.8 kips.
4. Pu= 1.2D + 1.6W + 0.5 L + 0.5 (L, or S or R)
= 1.2*109+0.5*46+0.5*20 = 163.8 kips.
5. Pu= 1.2D + 1.0E+0.5 L + 0.2S = 1.2*109+0.5*46+0.2*20 = 163.8 kips.
6. Pu= 0.9D +(1.6W or 1.0E) = 0.9*109 = 98.1 kips.
From above results, (Pu= 214.4 kips) is control to be used for design.
(b)
Ru≤ ØRn
214.4 ≤ 0.9*Rn
[L, 19 kips <S=20 kips and R=0 →use S]
R₁ = 238 kips
Homework: Determine the controlling service load for ASD, and if
the safety factor (N=1.67), what is the required nominal strength?
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