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

110 Chapter Five
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Substituting W = wl/12 in Eq. (5-25), the maximum deflection of a
single span beam with a uniformly distributed load can be calculated
by the following equation.
For a single-span beam; ∆ = 5wl 4 /4,608EI (5-26)
The uniform load that will produce a deflection equal to ∆ will be
as follows:
For a single-span beam; w = 4,608EI∆/5l 4
= 921.6EI∆/l 4 (5-27)
For ∆ = l/270, w = 4,608EI/1350l 3
For ∆ =l/360, w = 4,608EI/1800l 3
For ∆ = ¹⁄8 in., w = 4,608EI/40l 4
For ∆ = ¹⁄16 in., w = 4,608EI/80l 4
(5-27a)
(5-27b)
(5-27c)
(5-27d)
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Example 5-9
In Example 5-2, it was determined that the bending moment
required a No. 2 grade Douglas Fir-Larch 3 × 8 beam to support a
uniformly distributed load of 150 lb per lin ft over the 9 ft length
of the beam. Determine the deflection of this beam. Because the
beam is a single-span beam with a uniform load, the deflection is
calculated using Eq. (5-26). From Table 4-1, the moment of inertia
I = 79.39 in. 4
∆= 5wl 4 /4,608EI
= [5(150 lb per lin ft)(108 in.) 4 ]/[4,608(1,600,000 lb per sq in.)
(79.39 in. 4 )]
= 0.17 in.
Assume the allowable deflection is l/360 = 108/360 = 0.30 in.
Because the calculated deflection of 0.17 in. is less than the allowable
deflection of 0.30 in., then this beam is acceptable for deflection.
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