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

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Design of Wood Members for Formwork 139
Consider using 2 × 4 S4S studs whose actual size is 1½ by 3½ in.
From Table 4-1, the physical properties can be obtained for 2 × 4 S4S
lumber:
Area, A = 5.25 in. 2
Section modulus, S = 3.06 in. 3
Moment of inertia, I = 5.36 in. 4
The lumber selected for the forms is No. 2 grade Southern Pine with
no splits. From Table 4-2, using the values that are increased by 25% for
short-duration load conditions, the allowable stresses are as follows:
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Allowable bending stress, F b
= 1.25(1,500) = 1,875 lb per sq in.
Allowable shear stress, F v
= 1.25(175) = 218 lb per sq in.
Modulus of elasticity, E = 1,600,000 lb per sq in.
Bending in 2 ë 4 S4S Studs
Each stud will support a vertical strip of sheathing 12-in. wide, which
will produce a uniform load of 825 lb per sq ft × 1.0 ft = 825 lb per lin ft
on the lower portion of the stud. Because the studs will extend tothe full
height of the wall, the design should be based on continuous beam
action for the studs. The allowable span length of studs, the distance
between wales, based on bending can be determined as follows:
From Eq. (5-34), l b
= [120F b
S/w] 1/2
= [120(1,875)(3.06)/825] 1/2
= 28.9 in.
Shear in 2 ë 4 S4S Studs
The allowable span length based on shear in the studs can be calculated
as follows.
From Eq. (5-36), l v
= 192F v
bd/15w + 2d
= 192(218)(1.5)(3.5)/15(825) + 2(3.5)
= 24.8 in.
Deflection in 2 ë 4 S4S Studs
The permissible deflection is l/360, but not greater than ¹⁄16 in. Using
Eqs. (5-37a) and (5-37b), the allowable span lengths due to deflection
can be calculated as follows:
From Eq. (5-37a), l ∆
= [1,743EI/360w] 1/3 for ∆ =l/360
= 1,743(1,600,000)(5.36)/360(825)] 1/3
= 36.9 in.
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