Application of load combinations For the steel special moment frame building shown in Figure 5.1, determine the maximum and minimum column loads on the main level (Figure 5.13).

Question

Application of load combinations
For the steel special moment frame building shown in Figure 5.1, determine the maximum and minimum column loads on the main level (Figure 5.13).

Service loads:	Seismic lateral loads shown in figure
[latex]W_{DR}=1.2^{k/ft}[/latex]
[latex]W_{LR}=0.6^{k/ft}[/latex]	ρ = 1.0
[latex]W_{D}=1.2^{k/ft}[/latex]	[latex]S_{DS}[/latex] = 1.0
[latex]W_{L}=1.5^{k/ft}[/latex]	[latex]F_1[/latex]= 0.5

Factored load combinations as per ASCE 7, Section 2.3.2:
1. 1.4 D
2. 1.2 D + 1.6L + 0.5([latex]L_r[/latex] or S or R)
3. 1.2 D + 1.6 ([latex]L_r[/latex] or S or R) + [latex]F_1[/latex]L or 0.5W
4. 1.2 D + 1.0W + [latex]F_1[/latex]L + 0.5([latex]L_r[/latex] or S or R)
5. 1.2 D + 1.0E +[latex]F_1[/latex]L +0.2S
6. 0.9 D + 1.0W
7. 0.9 D + 1.0E
where:
D is the dead load
L is the live load
[latex]L_r[/latex] is the live load roof
S is the snow load
R is the rain load
E is the earthquake load
W is the wind load
[latex]F_1[/latex] is the load factor on L
[latex]F_1[/latex] = 0.5 when L ≤ 100 psf, else
[latex]F_1[/latex] = 1.0
The governing load cases for D, L, [latex]L_r[/latex], and E are 2, 5, and 7.
Wind loads and seismic loads need not be considered to act simultaneously.

Accepted Answer

The gravity loads are collected as follows (Figures 5.14 and 5.15):
[latex]D=\frac{\left[w_{DR}+w_{D}(3 floors) ight]\left(width of building ight)}{2 columns}=\frac{(1.2+1.2(3))26}{2}=62.4^k[/latex]

[latex]L_r=\frac{w_{LR} imes 26^\prime}{2}=\frac{0.6(26)}{2}=7.8^k[/latex]

[latex]L=\frac{\left(w_L imes 26^\prime ight)3 floors}{2}=\frac{1.5(26)(3)}{2}=58.5^k[/latex].
The seismic loads at the main level columns are found using the portal method.
Shear at base:
x + x = 20 + 30 + 40 + 50
2x = 140
x = [latex]70^k[/latex]
Therefore, [latex]A_x=B_x=70^k[/latex].
The moment at base:
[latex]M_A=M_B=70^k\left(\frac{10^\prime}{2} ight)=350^{ft\cdot k}[/latex]
Then, the seismic compression force in the column is found by summing moments about A:
[latex]+⤸ ΣM_A =0 =20 (10) +30(20)+40(30)+50(40)-350-350-B_y(26)\ B_y = 126.9^k[/latex]
The seismic load combination from ASCE 7, Section 12.4.2.3 are used in lieu of those of Section 2.3.2
[latex]5. (1.2+ 0.2S_{DS})D+ ho Q_E+F_1L+0.2S\7. (0.9 −0.2S_{DS})D+ ho Q_E[/latex]
then,
5. [latex][1.2+0.2(1.0)]62.4+1.0(126.9)+0.5(58.5)=243.5^k[/latex]
6. [latex][0.9−0.2(1.0)]62.4+1.0(−126.9)=−83.22^k[/latex]
7. [latex](1.2)(62.4)+(1.6)(85.5)+0.5(7.8)=172.38^k[/latex]
Therefore, maximum [latex]P_u = 243.5^k[/latex] (compression)
and minimum [latex]P_u = 83.22^k[/latex] (tension).

Service loads:	Seismic lateral loads shown in figure
$W_{DR}=1.2^{k/ft}$
$W_{LR}=0.6^{k/ft}$	ρ = 1.0
$W_{D}=1.2^{k/ft}$	$S_{DS}$ = 1.0
$W_{L}=1.5^{k/ft}$	$F_1$ = 0.5

Question 5.1: Application of load combinations For the steel special momen...

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