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

100 Chapter Five
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Douglas Fir-Larch would be required for bending. Will this beam satisfy
the requirements for the allowable unit stress in horizontal shear?
The uniform load and dimensions of the beam are:
ww.EasyEngineering.n
w= 150 lb per lin ft
L = 9 ft, or l = 108 in.
b= 2.5 in.
d= 7.25 in.
The shear force can be calculated as follows:
From Eq. (5-14), V = wl/24
= [(150 lb per ft) x (108 in.)]/24
= 675 lb
The applied horizontal shear stress can be calculated as follows:
From Eq. (5-13), f v
= 3V/2bd
= [3 × 675 lb)]/[2 × 2.5 in. × 7.25 in.]
= 55.9 lb per sq in.
Table 4-3 indicates an allowable shear stress for a 3 × 8 beam of
No. 2 grade Douglas Fir-Larch as 180 lb per sq in., which is greater
than the required 55.9 lb per sq in. Therefore, the 3 × 8 beam will be
satisfactory for shear.
Modified Method of Determining the Unit Stress in
Horizontal Shear in a Beam
When calculating the shear force V in bending members with uniformly
distributed loads, the NDS allows neglecting all loads within
a distance d from the supports provided the beam is supported by full
bearing on one surface and loads are applied to the opposite surface.
For this condition, the shear force at a reaction or support point of
single-span and multiple-span beams that sustain a uniformly distributed
load will be as follows.
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