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

436 Chapter Sixteen
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The allowable shear stress from Tables 4-2 and 4-4 can be calculated
as follows:
Allowable shear stress, F v
= C D
× (reference shear stress)
= 1.25(175)
= 218 lb per sq in.
The allowable shear stress F v
= 218 lb per sq in. is greater than the
calculated applied shear stress f v
= 201.0. Therefore, the double 2 × 8
stringer is satisfactory for shear.
The deflection due to the four 1,458 lb concentrated loads can be
calculated by using the deflection equation for Beam 4 in Table 5-2,
combined with the technique of superposition. From Table 4-1, the
moment of inertia of the double 2 × 8 stringer can be calculated as
2(47.63 in. 4 ), or 95.26 in. 4 From Table 4-2, the modulus of elasticity
for a 2 × 8 No. 2 grade Southern Pine is 1,600,000 lb per sq in. The
deflection can be calculated using the technique of superposition as
follows:
ww.EasyEngineering.n
2
2
2
2
Pa [( 3 l )/ 4
1
− ( a )]
1
Pa [( 3
2
l )/
4−
( a )]
2
∆=
+
6EI
6EI
2 2
2 2
1, 458( 3)[ 3( 66) / 4 − ( 3) ] 1, 458( 23)[ 3( 66) / 4 − ( 23) ]
=
+
6( 1, 600, 000)( 95.
26)
6( 1, 600, 000)(
95. 26)
= 0.016 in. + 0.10 in.
= 0.12 in.
The permissible deflection is l/360, or 66/360 = 0.18 in. Thus, the
calculated deflection of 0.12 in. is less than the permissible deflection
of 0.18 in. Therefore, the double 2 × 8 stringer is satisfactory for
deflection.
An approximate deflection can be calculated by transforming the
four concentrated loads into an equivalent uniformly distributed
load having the same total value as the sum of the concentrated loads.
Using this method, the deflection can be approximated as follows:
The equivalent uniformly distributed load on stringers is
w = 4(1,458 lb)/(66/12 ft)
= 1,060 lb per lin ft
Because the stringer is a single-span member, the deflection for a
single-span beam can be calculated using Eq. (5-26) as follows:
The deflection, from Eq. (5-26):
∆= 5wl 4 /4,608EI
= 5(1,060)(66) 4 /(4,608)(1,600,000)(95.26)
= 0.14 in.
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