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

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Design of Wood Members for Formwork 141
sq in., which is greater than the applied stress of 367 lb per sq in.
Therefore, a 24-in. spacing of wales will be satisfactory.
Size of Wale Based on Selected 24-in. Spacing
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. Assuming
a uniformly distributed load from a pressure of 825 lb per sq ft on
the wales that are spaced at 24 in. (2.0 ft), the value of w can be calculated
as follows:
ww.EasyEngineering.n
w = 825 lb per sq ft (2.0 ft)
= 1,650 lb per lin ft
With 24-in. spacing and 825 lb per sq ft pressure on the wales,
consider double 2 × 4 S4S wales.
From Table 4-1, 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
From Table 4-2, for a 2 × 4, F b
= 1.25(1,500) = 1,875 lb per sq in.
F v
= 1.25(175) = 218 lb per sq in.
E = 1,600,000 lb per sq in.
Bending in Wales
Based on bending, the span length of the wales must not exceed the
value calculated.
From Eq. (5-34), l b
= [120F b
S/w] 1/2
= [120(1,875)(6.12)/1,650] 1/2
= 28.8 in.
Shear in Wales
Based on shear, the span length of the wales must not exceed the
value calculated.
From Eq. (5-36), l v
= 192F v
bd/15w + 2d
= 192(218)(10.5)/15(1,650) + 2(3.5)
= 24.7 in.
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