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

228 Chapter Nine
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From Table 9-2, 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 = (1.0)(1,600,000) = 1,600,000 lb per sq in.
Bending in Double 2 ë 4 Wales
Based on bending, the span length of the wales must not exceed the
value calculated from Eq. (5-34) in Table 9-5:
l b
= [120F b
S/w] 1/2
= [120(1,875)(6.12)/1,426] 1/2
= 31.1 in.
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Shear in Double 2 ë 4 Wales
Based on shear, the span length of the wales must not exceed the
value calculated from Eq. (5-36) in Table 9-5:
l v
= 192F v
bd/15w + 2d
= 192(218)(10.5)/15(1,426) + 2(3.5)
= 27.5 in.
Deflection in Double 2 ë 4 Wales
Based on the deflection criteria of less than l/360, the span length of the
wales must not exceed the value calculated from Eq. (5-37a) in Table 9-5:
l ∆
= [1,743EI/360w] 1/3
= [1,743(1,600,000)(10.72)/360(1,426)] 1/3
= 38.8 in.
Based on the deflection criteria of less than ¹⁄16 in., the span length
of the wales must not exceed the value calculated form Eq. (5-37b) in
Table 9-5:
l ∆
= [1,743EI/16w] 1/4
= [1,743(1,600,000)(10.72)/16(1,426)] 1/4
= 33.8 in.
Summary for Double 2 ë 4 Wales
For bending, the maximum span length of wales = 31.1 in.
For shear, the maximum span length of wales = 27.5 in.
For deflection, the maximum span length of wales = 33.8 in.
For this design, shear governs the maximum span length of the
wales. Because the wales are supported by the ties, the spacing of the ties
must not exceed 27.5 in. However, the ties must have adequate strength
to support the load from the wales. For uniformity and constructability,
choose a maximum span length of 24 in. for wales and determine the
required strength of ties to support the wales.
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