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

98 Chapter Five
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The applied bending stress can be calculated as follows:
From Eq. (5-7), f b
= M/S
= 3750 in.-lb/3.06 in. 3
= 1225 lb per sq in.
Determine the allowable bending stress for a No. 2 grade Hem-
Fir as follows:
Table 4-3, reference design value for bending stress = 850 lb per
sq in.
Table 4-3a, size adjustment factor for a 2 × 4 beam, C F
= 1.5
Allowable bending stress F b
= (850 lb per sq in.)(C F
)
= (850 lb per sq in.)(1.5)
= 1,275 lb per sq in.
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Because the allowable bending stress F b
of 1,275 lb per sq in. is
greater than the applied bending stress f b
of 1225 lb per sq in., the No. 2
grade Hem-Fir beam is satisfactory for bending. The d/b ratio of a
2 × 4 is 4/2 = 2.0; therefore, no lateral support is required. Although
the 2 × 4 is adequate for bending, it must be checked for shear and
deflection as shown in the following sections.
Horizontal Shearing Stress in Beams
As presented in the preceding sections, loads acting transverse to the
long axis of the beam cause bending moments. These loads also produce
shear forces in the beam that tend to separate adjacent parts of the
beam in the vertical direction. The shear stresses at any point in a beam
are equal in magnitude and at right angles, perpendicular to the axis of
the beam, and parallel to the axis of the beam. Thus, there are both vertical
and horizontal shear stresses in a beam subjected to bending. Because
the strength of wood is stronger across grain than parallel to grain, the
shear failure of a wood member is higher in the horizontal direction
(parallel to grain) along the axis of the beam. Thus, in the design of
wood beams for shear, the horizontal shear stress is considered.
The maximum applied horizontal shear stress in a rectangular
wood beam is calculated by the following equation:
f v
= 3V/2bd (5-13)
For a single-span beam with one support at each end and a uniformly
load distributed over its full length, the maximum total shear
will occur at one end, and it will be:
V = wL/2
= wl/24 lb (5-14)
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