Diaphragm design and collector analysis The second floor diaphragm of the building in Example 8.1 is subject to the seismic load, FP, based on anchorage to flexible diaphragm force shown in Figure 8.8. Open web wood trusses span, N–S, 45 ft. They are spaced at 2 ft O.C., and have a 2″ nominal truss

Question

Diaphragm design and collector analysis
The second floor diaphragm of the building in Example 8.1 is subject to the seismic load, [latex]F_P[/latex], based on anchorage to flexible diaphragm force shown in Figure 8.8. Open web wood trusses span, N–S, 45 ft. They are spaced at 2 ft O.C., and have a 2″ nominal truss chord, which the sheathing is attached. Blocking is not required.
This example will explain the steps required to determine the appropriate sheathing thickness and nailing pattern for the diaphragm.

Accepted Answer

Use basic load combinations from IBC ASD:
[latex]U=0.7E=0.7(420)= 294^{\#/ft}[/latex]
The shear at the east and west sides of the building is
[latex]V =\frac{294^{\#/ft}(75 ft)}{2}=11,025^{\#}[/latex]

[latex]V=11,025^{\#}/45[/latex] ft-long wall
[latex]=245^{\#/ft}[/latex]
From AWC SDPWS 2008
Table 4.2C:
Unblocked wood structural panel diaphragms
Case I sheathing pattern
Section: 4.2.3:
ASD allowable unit shear capacities have a reduction factor of 2.0
Enter Table 4.2C:
Seismic loading
Case 1
2″ Min. nominal width of nailed face (2″ truss chord)
[latex]V_s=\frac{510}{2.0 reduction factor}=255^{\#/ft}\gt 245^{\#/ft}[/latex], therefore OK
Use sheathing 15/32
10d common nail size, 6 in nail spacing at diaphragm boundary, and supporting members
The main-level east and west elevations of the building are shown here with second the floor diaphragm unit shear forces shown (Figures 8.9 and 8.10)
The collector “gathers” the shear from the diaphragm and “pulls” the shear force into the wall or each side of the collector depending on the direction of the force.
Collectors are also called drag struts
The collector’s force is shared between the two adjacent walls
Hence, the force in the collector is
[latex]F_{collector}=\frac{245^{\#/ft}(15 ft)}{2 walls}=1837.5^{\#}[/latex]
Therefore, each connection between the collector and the wall has to be designed for this force
The shear at the base of the two 15 ft walls is
[latex]V_{base}=\frac{245^{\#/ft}(40 ft)}{2 walls}=4900^{\#}[/latex]

Question 8.2: Diaphragm design and collector analysis The second floor dia...

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