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

Downloaded From : www.EasyEngineering.net
Design of Wood Members for Formwork 97
From Eq. (5-11) the required section modulus S = M/F b
= (18,225 in.-lb)/(1,080 lb per sq in.)
= 16.88 in. 3
Consider 8-in. nominal depths from Table 4-1:
For a 2 × 8, S = 10.88 < 16.88, therefore not acceptable
For a 3 × 8, S = 18.13 > 16.88, therefore acceptable
Therefore, a 3 × 8 member is satisfactory for bending. Because
the nominal dimension d/b ratio of 8/3 is 2.7, 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 1,080 lb per sq in. Before the final selection
is made, this member should be checked for the allowable unit
stress in shear and the permissible deflection by methods that are
presented in the following sections.
ww.EasyEngineering.n
Example 5-3
A 2 × 4 beam of No. 2 grade Hem-Fir is supported over multiple
supports that are spaced at 30 in. on center. The beam must carry a
500 lb per lin ft uniformly distributed load over the entire length.
The beam will be used in a dry condition with normal load-duration.
Check the bending stress in the beam and compare it to the allowable
stress.
The maximum bending moment for a beam with multiple supports
can be calculated as:
From Eq. (5-6), M = wl 2 /120
= 500 lb/ft(30 in.) 2 /120
= 3,750 in.-lb
The section modulus for a 2 × 4 beam from Table 4-1 is 3.06 in. 3 , or
the section modulus for bending can be calculated as follows:
From Eq. (5-9), S = bd 2 /6
= 1.5 in.(3.5 in.) 2 /6
= 3.06 in. 3
Downloaded From : www.EasyEngineering.net