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

298 Chapter Ten
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Equating (a) and (b) gives:
F b
S = Plx 2 /1,152
S = Plx 2 /1,152F b
(10-11)
where S is the required section modulus of end yoke B, in. 3 .
Side yoke A, illustrated in Figure 10-5(b), is a simple beam, subjected
to a uniform load equal to w lb per lin ft over a distance equal
to x in the midsection of the beam. The two bolts, spaced y in. apart,
are the end supports, each subjected to a load equal to wx/24 lb. Summing
moments about Point 1,
ww.EasyEngineering.n
M = [wx/24][y/2] – [wx/24][x/4]
Substituting Pl/12 for w in the above equation, the applied
moment is:
M = [Plx/12(24)][y/2] – [Plx/(12)(24)][(x/4)]
= Plxy/576 – Plx 2 /1,152
= Plx(2y – x)/1,152 (c)
The resisting moment will be:
M= F b
S (d)
Equating (c) and (d) gives:
F b
S = Plx(2y – x)/1,152
Solving for S, we get:
S = Plx(2y – x)/1,152F b
(10-12)
where S is the required section modulus for side yoke A, in. 3
Example 10-3
Determine the minimum size wood yoke member B required for the
given conditions:
P = 900 lb per sq ft.
l = 12 in.
x = 16 in.
Using S4S lumber with an allowable bending stress F b
of 1,093 lb
per sq in., the required section modulus can be calculated as follows:
From Eq. (10-11), S = Plx 2 /1,152 F b
= (900)(12)(16) 2 /1,152(1,093)
= 2.19 in. 3
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