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

96 Chapter Five
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Because the nominal dimension d/b ratio of 6/2 is 3.0, which is in the
range of 2 < d/b < 4 in Table 5-1, the ends of the beam must be held in
position to prevent displacement or rotation. The beam must have
this end condition in order to satisfy beam stability and to justify
using the allowable bending stress of 1,250 lb per sq in.
An alternate method to design a beam is to assume an allowable
bending stress F b
and calculate the required section modulus S using
Eq, (5-11), S = M/F b
. Then, searching Table 4-1, an available section
modulus is selected that is greater than the calculated required section
modulus. However, the allowable stress F b
will vary depending
on the depth of the beam. Therefore, after the section modulus is
selected from Table 4-1, the designer must then verify that the originally
assumed allowable stress F b
is correct for the particular depth of
the member that was selected from Table 4-1.
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Example 5-2
Determine the minimum-size joist, 9 ft long, required to support a uniform
load of 150 lb per lin ft. The joist will be used in a dry condition
with normal load-duration and adequate bracing and end supports.
For this condition of a uniformly distributed load over a singlespan
beam, the maximum bending moment can be calculated from
Eq. (5-5) as follows:
Using Eq. (5-5), M = wl 2 /96
= (150 lb per ft)(108 in.) 2 /96
= 150(108) 2 /96
= 18,225 in.-lb
Consider a joist of No. 2 Douglas Fir-Larch with 8 in. nominal
depth. Determine the allowable bending stress as follows:
Table 4-3, reference a design value for bending stress = 900 lb per sq in.
Table 4-3a, size adjustment for 8-in. lumber C F
= 1.2
Allowable bending stress F b
= 900 lb per sq in. × 1.2
= 1,080 lb per sq in.
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