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

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Forms for Walls 227
Determine the unit stress in bearing between the studs and the
wales. If a wale consists of two 2-in. nominal thick 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:
ww.EasyEngineering.n
P = 713 lb per sq ft[1.0 ft × 2.0 ft]
= 1,426 lb
The calculated unit stress in bearing, perpendicular to grain, between
a stud and a wale will be:
f c
⊥= 1,426 lb/4.5 sq in.
= 317 lb per sq. in.
From Table 9-2, the indicated allowable unit compression stress
perpendicular to the grain for mixed southern pine is 565 lb per sq in.
The allowable bearing stress of 565 lb per sq in. is greater than the
applied bearing stress of 317 lb per sq in. Therefore, the unit bearing
stress is within the allowable value for this species and grade of lumber.
If another species of lumber, having a lower allowable unit stress,
is used, it may be necessary to reduce the spacing of the wales, or to
use wales whose members are thicker than 2 in. The wales will be
spaced 24 in. apart, center to center.
Size of Wale Based on Selected 24 in. Spacing of Studs
Although the loads transmitted from the studs to the wales are concentrated,
it is generally sufficiently accurate to treat them as uniformly
distributed loads, having the same total values as the concentrated
loads, when designing formwork for concrete walls. However, in
some critical situations, it may be desirable to design the forms using
concentrated loads, as they actually exist.
With a 24 in. spacing and a 713 lb per sq ft pressure, the load on a
wale will be 713 lb per sq ft × 2.0 ft = 1,426 lb per ft. Consider double
2 × 4 S4S wales with the following physical properties:
From Table 9-1, for a double 2 × 4: A = 2(5.25 in. 2 ) = 10.5 in. 2
S = 2(3.06 in. 3 ) = 6.12 in. 3
I = 2(5.36 in. 4 ) = 10.72 in. 4
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