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

Downloaded From : www.EasyEngineering.net
Forms for Walls 225
Calculate the shear deflection from Eq. (5-60a) as follows:
∆ s
= w s
Ct 2 l s2
/1,270E e
I
= 713(120)(3/4) 2 (10.5) 2 /(1,270)(1,500,000)(0.199)
= 0.014 in.
Total deflection = bending deflection + shear deflection:
= 0.017 in. + 0.014 in.
= 0.031 in.
ww.EasyEngineering.n
Compare the calculated total deflection with the permissible
deflection criteria of l/360, not to exceed ¹⁄16 in.:
Permissible deflection for l/360 = 12 in./360 = 0.033 > 0.031, therefore
okay
Permissible deflection for ¹⁄16 in. = 0.0625 in. > 0.031, therefore okay
Therefore, the 12-in. spacing of studs is adequate for bending
stress, shear stress, and deflection in the ¾-in. Class I Plyform
sheathing.
Studs for Support of Plyform
The studs support the Plyform sheathing and the wales must support
the studs. Therefore, the maximum allowable spacing of wales will be
governed by the bending, shear, and deflection in the studs.
Consider using 2 × 4 S4S studs, whose actual size is 1½ by 3½ in.
From Table 9-1. 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.
From Table 9-2, using the values with appropriate adjustments for
short-duration loading, the allowable stresses are as follows:
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 = 1.0(1,600,000) = 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 713 lb per sq ft × 1.0 ft = 713 lb per lin
ft on the lower portion of the stud. Because the studs will extend to
Downloaded From : www.EasyEngineering.net