Plywood diaphragm analysis and chord splice design This example illustrates an analysis of a flexible diaphragm and splice connector of a diaphragm chord. Wind pressure = 23 psf. The building shown in Figure 8.1 is a three-story wood-framed building. The building is 45 ft × 75 ft. The main level

Question

Plywood diaphragm analysis and chord splice design
This example illustrates an analysis of a flexible diaphragm and splice connector of a diaphragm chord.

Wind pressure = 23 psf.

The building shown in Figure 8.1 is a three-story wood-framed building. The building is 45 ft × 75 ft. The main level is 12 ft tall and the second and third levels are 10 ft tall. The wind load applied in the N–S direction is 23 pounds per square foot.
1. Determine the design shear in pounds per liner foot for the second and third levels.
Consider the floors and the roof to be flexible diaphragms. There are no internal shear walls (Figure 8.2).

Accepted Answer

The design shear for the roof, the third- and the second-level diaphragms are calculated by multiplying the wind force by the associated tributary width as shown as follows (Figures 8.3 through 8.5):
Roof:
[latex]F_r=23(5)=115^{\#}[/latex]/linear foot
Third level:
[latex]F_3=23(10)=230^{\#}[/latex]/linear foot
Second level:
[latex]F_2=23(11)=253^{\#}[/latex]/linear foot
To calculate the chord force,first calculate the maximum bending moment in the diaphragm. The chord force (tension or compression) is calculated by dividing the moment by the width of the diaphragm (d).
[latex]T_{chord}=C_{chord}=\frac{M}{d}=\frac{wl^2/8}{d}=\frac{wl^2}{8d}=\frac{253(75)^2}{8(45)}=3953^{\#}[/latex]
This building is a platform-framed structure, see Figure 8.6, and as such is built with a perimeter rim board, which if sized properly, can function as the chord and strut members of the diaphragm.
The 2 × 10 chord is 75 ft long on each side of the building and must be spliced together at several locations along its length. The splice connectors are typically constructed by end-butting two members together and overlapping another 2 × 12 member over the joint, see Figure 8.7. The connection is usually performed by bolting or nailing to fasten the splice on each side of the butt joint.
Select a 2 × 12 Douglas Fin-Larch (No. 1 and better) for the chord member. We will splice with the same material.
From the Natural Design Specifications (NDS) supplement, 2012 edition
Table 4A
Douglas Fin-Larch
No. 1 and better
[latex]F_t = 800^{\#}[/latex] (tension parallel to grain)
From 2012 NDS Table 2.3.2, the load duration factor [latex]C_D[/latex] = 1.6 because the chord is subject to the wind load.
In addition, Table 4A of the NDS supplement 2012 notes additional adjustment factors that must be considered when using the values from this table.
They are as follows:
Size factor [latex]C_F[/latex] applies to bending tension and compression parallel to grain design values for dimensional lumber 2″ to 4″ thick:
Grade: No.1 and better
Width: 12″
Thick: 2″
For the tension parallel to the grain [latex]F_t: C_f = 1.0[/latex]
Repetitive member factor, [latex]C_r[/latex]: does not apply
Wet surface factor, [latex]C_m[/latex]: does not apply
Flat use factor, [latex]C_{fu}[/latex]: does not apply
Incising factor, [latex]C_i[/latex]: does not apply
NDS 2012 Table 4.3.1
The tension equation is as follows:
[latex]F_t^\prime=F_t C_DC_mC_tC_fC_i[/latex]
All adjustment factors are then equal to 1.0, with the exception of the duration factor, so the equation reduces to
[latex]F_t^\prime=F_t C_D=800(1.6)=1280^{\#/in^2}[/latex]
The actual tension parallel to the grains:
[latex]f_t=\frac{3594^{\#}}{1.5 imes 11.25}=213^{\#/in^2}[/latex]

[latex]f_t\lt F_t^\prime=1280^{\#/in^2}[/latex]
Therefore, the 2 × 12 Doug Fir-Larch is more than adequate to sustain the tension force in the chord.
While the splice connector in Figure 8.7 illustrates bolts as the fasteners of the splice, because the tension force is not that large, nails may be a possibility. Therefore, try 10d common nails:
From 2012 NDS, Table 11N
Side member thickness = 1 ½ inches
Doug Fir-Larch
10d Common nails, [latex]Z = 118^{\#}[/latex]
From NDS 2012 Table 10.3.1.
The connection equation is as follows:
[latex]Z^\prime=ZC_DC_{d1}[/latex]

[latex]C_{d1}=[/latex] diaphragm factor
Section 11.5.3 NDS 2012
[latex]C_{d1} = 1.1[/latex]
Therefore,
Z′ = 118 (1.6) 1.1 = [latex]207.7^{\#}[/latex]
Then, calculate the number of nails per side of splice,
No. of nails [latex]=\frac{3953^{\#}}{208^{\#}/nail}=19[/latex]
Use 19 10d common nails per side of splice.

Question 8.1: Plywood diaphragm analysis and chord splice design This exam...

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