Formwork for Concrete Structures by R.L.Peurifoy and G.D- By EasyEngineering.net

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Forms for Columns 299
Table 4-1 indicates that a 2 × 4 S4S member with the 4-in. face
perpendicular to the sheathing will have a section modulus of 3.06 in. 3 ,
which is greater than the required 2.19 in. 3 Therefore, a 2 × 4 S4S
member will be adequate. Also, a 3 × 4 S4S member, with the 3-in.
face perpendicular to the sheathing, will be adequate.
Example 10-4
Determine the minimum size wood yoke member A required for the
S4S lumber with an allowable bending stress F b
of 1,093 lb per sq in.
and an allowable shear stress F v
of 237 lb per sq in. for the following
conditions:
P = 900 lb per sq ft
l = 12 in.
x = 16 in.
y = 28 in.
The required section modulus can be calculated as follows:
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From Eq. (10-12), S = Plx[2y – x]/1,152F b
= 900(12)(16)[2(28) – 16)]/1,152(1,093)
= 5.48 in. 3
Table 4-1 indicates that a 3 × 4 S4S member, with a section modulus
of 7.15 in. 3 , will be required. Check the member for the unit stress
in shear. The shear force is:
V = [Plx]/[(2)(12)(12)]
= 900(12)(16)/288
= 600 lb
The applied shear stress is:
f v
= 3V/2bd
= 3(600)/2(12.25)
= 73 lb per sq in.
Because the allowable shear stress F v
of 237 lb per sq in. is
greater than the applied shear stress f v
, the S4S member is adequate
in shear.
Each bolt is subjected to a tensile stress equal to 600 lb. A ³⁄8-in.-
diameter bolt is strong enough, but a ½-in.-diameter bolt should be
used to provide sufficient area in bearing between the bolt and the
wood wedges.
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