A flexible diaphragm analysis A roof plan and elevation of a single-story precast concrete tilt-up building with panelized wood roof (flexible roof diaphragm) is shown in Figure 3.14. A wind load of 25 (#/ft² ) is applied in the east-west direction (Figure 3.15). 1. Determine the wind force

Question

A flexible diaphragm analysis
A roof plan and elevation of a single-story precast concrete tilt-up building with panelized wood roof (flexible roof diaphragm) is shown in Figure 3.14. A wind load of 25 (#/ft² ) is applied in the east-west direction (Figure 3.15).
1. Determine the wind force transmitted to the roof diaphragm from the tilt-up wall panels.
2. Determine the chord forces and the shear at the struts of the diaphragm.

Accepted Answer

1. Summing moments about A to determine the wind load transmitted to the roof diaphragm
[latex]+⤸\sum M_A=0[/latex]

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

[latex]B=372^{\#/ft}[/latex]
2. The diaphragm moment is calculated as follows (Figure 3.16):
[latex]M=\frac{wl^2}{8}=\frac{372^{\#/ft}(180^\prime)^2}{8}=1,506,600^{ft\cdot \#}[/latex]
Then, the tension and compression chord forces are as follows:
[latex]T=C=\frac{M}{b}=\frac{1,506,600^{ft\cdot \#}}{90^\prime}=16,740^{\#}[/latex]
Shear in the struts at end walls along grid lines ① and ➃ is as follows:
[latex]V=\frac{wl}{2}=\frac{372^{\#/ft}(180^\prime)}{2}=33,480^{\#}[/latex]
The building is modified and a shear wall is added along grid line ② as shown in Figure 3.17. The wall does not extend to the full width of the building. Determine the following:
1. The shear at each of the walls parallel to the applied wind load
2. The force in the drag strut along grid line ②
3. Assume that the wall panels are connected at vertical wall joints. The in-plane design strength of the connections [latex]\phi V_n = 4.8 k.[/latex] Using a load factor of 1.6 for the wind applied to the diaphragm, determine the number of connections required between two adjacent 8′ wide panels along gridline ④
1. For a flexible diaphragm, the wind loading at each of the walls is equal to the tributary width between adjacent walls. Based on this, the shear at each wall is
Wall A: [latex]V_A=372^{\#/ft}(40^\prime)=14,880^{\#}[/latex]
Wall B: [latex]V_B=372^{\#/ft}(40^\prime+50^\prime)=33,480^{\#}[/latex]
Wall C: [latex]V_C=372^{\#/ft}(50^\prime)=18,600^{\#}[/latex]
2. The unit shear across the width of the building along gridline ② is
[latex]V=\frac{V_B}{b}=\frac{33,480^{\#}}{90^\prime}=372^{\#/ft}[/latex]
Then the force that the drag strut must collect and transfer to the shear wall on gridline ② is
[latex]F=372^{\#/ft}(30^\prime)=11,160^{\#}[/latex]
3. The horizontal shear force at the top of the wall at gridline ➃, taking into account the 1.6 load factor is
[latex]V_h=\frac{1.6(18,600^{\#})}{1000^{\#/k}}=29.76^k[/latex]
The unit shear along the length of the wall is
[latex]V_h=\frac{29.46^k}{90^\prime}=0.327^{k/ft}[/latex]
Then, the shear force at each wall panel is
[latex]V_{h_{panel}}=0.327^{k/ft}[/latex] (8′ wide panel)
[latex]=2.62^k[/latex]
The free body diagram of a wall panel is shown in Figure 3.18.
Neglecting the weight of the panel and summary moments about 0 yields:
[latex]+⤸ \sum{M_O}=0[/latex]
[latex]2.62^k(21^\prime) − V_v(8^\prime) = 0[/latex]
[latex]V_v = 6.87^k[/latex]
Then the number of connections required between two panels is
Number of connections [latex]=\frac{V_v}{\phi V_h}=\frac{6.87^k}{4.8^k}=1.43[/latex]
Therefore, use two connections between panels.

Question 3.4: A flexible diaphragm analysis A roof plan and elevation of a...

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