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

140 Chapter Five
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From Eq. (5-37b), l ∆
= [1,743EI/16w] 1/4 for ∆ =¹⁄16 in.
= [1,743(1,600,000)(5.36)/16(825)] 1/4
= 32.6 in.
Summary for the 2 ë 4 S4S Studs
For bending, the maximum span length of studs = 28.9 in.
For shear, the maximum span length of studs = 24.8 in.
For deflection, the maximum span length of studs = 32.6 in.
For this design, shear governs the maximum span length of the
studs. Because the wales must support the studs, the maximum spacing
of the wales must be no greater than 24.8 in. Therefore, the wales
will be spaced at 24 in. on center for this design.
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Wales for Support of
For this design, double wales will be used. Wales for wall forms
of this size are frequently made of two-member lumber whose
nominal thickness is 2 in., separated by short pieces of 1-in.-thick
blocks.
Determine the unit stress in bearing between the studs and the
wales. If a wale consists of two 2-in. nominal thickness members, the
contact area between a stud and a wale will be:
A = 2 × [1.5 in. × 1.5 in.]
= 4.5 sq. in.
The total load in bearing between a stud and a wale will be the
unit pressure of the concrete acting on an area based on the stud and
wale spacing. For this design, the area will be 12 in. long by 24 in. high.
Therefore, the pressure acting between the stud and wale will be:
P = 825 lb per sq ft[1.0 ft × 2.0 ft]
= 1,650 lb
The calculated unit stress in bearing, perpendicular to grain,
between a stud and a wale will be:
f c⊥
= P/A
= 1,650 lb/4.5 sq in.
= 367 lb per sq in.
From Table 4-2, the indicated allowable unit compression stress
perpendicular to the grain for No. 2 grade Southern Pine is 565 lb per
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