Determination of a diaphragm design seismic force low-rise building For the building discussed in Example 3.4 the following seismic data apply: • Seismic design category D • Design spectral acceleration SDS = 1.0 • Risk category II (ASCE 7 T 1.5-1) • Importance factor Ie = 1.0 (ASCE 7 T 1.5-2)

Question

Determination of a diaphragm design seismic force low-rise building
For the building discussed in Example 3.4 the following seismic data apply:
• Seismic design category D
• Design spectral acceleration [latex]S_{DS}[/latex] = 1.0
• Risk category II (ASCE 7 T 1.5-1)
[latex]\begin{matrix} & \end{matrix}[/latex]• Importance factor [latex]I_e[/latex] = 1.0 (ASCE 7 T 1.5-2)
• Seismic-force-resisting system (ASCE 7 T 12.2-1)
[latex]\begin{matrix} & \end{matrix}[/latex]• Building frame system B
[latex]\begin{matrix} & \end{matrix}[/latex]• Intermediate precast shear walls response modification coefficient R = 5.0
• Redundancy factor ρ = 1 0.
• The weight of the wood diaphragm construction = [latex]20^{\#/ft^2}[/latex]
• The weight of the precast panel =[latex]100^{\#/ft^2}[/latex]

Accepted Answer

The seismic weight [latex]w_{px}[/latex] includes the weight of the diaphragm as well as the weight of the walls tributary to the diaphragm that are normal to the direction of the seismic force applied to the building (see Figure 3.20).
The walls that are parallel to the applied seismic force are not typically considered in determining the tributary diaphragm weight, as the diaphragm does not support these walls.
The weight of the wall, which is tributary to the diaphragm (Figure 3.21) is calculated as follows based on a 1 ft width of wall.
Summing moments about the base of the wall to determine the seismic weight of the wall tributary to the diaphragm.
[latex]+⤸ ΣM_A = 0[/latex]

[latex]100 (25)\frac{25}{2}−B(21)= 0[/latex]

[latex]B = 1,488.1^{\#/ft}[/latex] of wall
The weight of the two walls normal to a seismic force, applied in the E–W direction, is
[latex]W_{wall} =2 x1,488.1^{\#/ft}=2976.2^{\#/ft}[/latex]
The diaphragm is 90′ wide,
[latex]W_{diaphragm}=90^\prime (20^{\#/ft^2})= 1800^{\#/ft}[/latex]
Then, the total weight of the diaphragm walls, [latex]w_{px}[/latex], is
[latex]W_{px} = (2976.2 + 1800)180 = 859.714^{\#}[/latex]

[latex]W_{px}=\frac{859.714^{\#}}{1000^{\#/k}}=859.7^k[/latex]
The diaphragm force equation for a single story building is developed as follows:
[latex]F_{px}=\frac{\sum\limits_{i=x}^{n}F_i}{\sum\limits_{i=x}^{n}w_i}w_{px}[/latex]
when x = 1, [latex]\sum\limits_{i=x}^{n}w_i=w\F_{px}=\frac{F_i}{w}w_{pi}[/latex] (3.27)
From [latex]F_x = C_{vx}V[/latex]

[latex]C_{vx}=\frac{w_x h_x^k}{\sum\limits_{i=x}^{n}w_i h_x^k}[/latex]
For short periods T < 0.5 sec, k = 1.0
Then [latex]C_v[/latex] = 1.0
And [latex]F_1 = C_vV = F_1 = V[/latex] (3.28)
[latex]V=e_sw=\frac{S_{DS}I_e}{R}w[/latex] (3.29)
Then combining (3.27) and (3.28)
[latex]F_{p1}=\frac{S_{DS}I_e}{R}w_{p1}[/latex]
Then
[latex]F_{p1}=\frac{(1.0)(1.0)}{5.0}859.7=172^k[/latex]
Check limits:
[latex]F_{p1}[/latex] minimum
[latex]0.2S_{DS}I_ew_{px} = (0.2)(1.0)(1.0)859.7 = 172^k[/latex]
Therefore [latex]F_{p1} = 172^k[/latex] OK

Question 3.7: Determination of a diaphragm design seismic force low-rise b...

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