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

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Forms for Concrete Bridge Decks 435
M= (2,916 lb)(33 in.) – (1,458 lb)(30 in.) – (1,458 lb)(10 in.)
= (96,228 in.-lb) – (43,740 in. lb) – (14,580 in.-lb)
= 37,908 in.-lb
An alternate method of calculating the maximum bending moment
is to use the equation for maximum moment in Beam 4 in Table 5-2,
combined with the technique of superposition as follows:
M = Pa 1
+ Pa 2
= 1,458 lb (3-in.) + 1,458 lb (23-in.)
= 4,374 in.-lb + 33,534 in.-lb
= 37,908 in.-lb
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Consider using double 2 × 8 S4S lumber for stringers. From
Table 4-1, the section modulus for a double 2 × 8 is double the value
of a single 2 × 8 as follows:
Section modulus for double 2 × 8 stringers, S = 2(13.14 in. 3 )
= 26.28 in. 3
The applied bending stress, from Eq. (5-7),
f b
= M/S
= (37,908 in.-lb)/26.28 in. 3
= 1,442 lb per sq in.
The allowable bending stresses for 2 × 8 No. 2 grade Southern
Pine for can be determined from Tables 4-2 and 4-4 as follows:
Allowable bending stress, F b
= C D
× (reference bending stress)
= 1.25(1,200)
= 1,500 lb per sq in.
The allowable bending stress of F b
= 1,500 lb per sq in. is greater
than the applied bending stress of f b
= 1,442 lb per sq in. Therefore, the
double 2 × 8 stringer is adequate for bending.
The maximum shear force at the end support of the stringer is
2,916 lb, which will occur at the support hanger. Since the stringer is
a double member, the shear area of bd will be two times the area of a
single 2 × 8, or 2 × 10.88 in. 2 = 21.76 in. 2 , reference Table (4-1). The
applied shear stress can be calculated as follows:
The applied shear stress, from Eq. (5-13) can be calculated as follows:
f v
= 3V/2bd
= 3(2,916)/2(21.76)
= 201.0 lb per sq in.
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