Question 6.14: The rectangular beam of width b and height h (Fig. 1) is sub......

Plan the Solution This is a straightforward application of the shear-stress formula, Eq. 6.67. Since V, I, and b are constant, τ_{max} will occur at the neutral axis, where Q has its maximum value.

τ= \frac{VQ} {It}               (6.67)

(a) Determine an expression for the transverse shear stress. The shear-stress formula, Eq. 6.67, is

As indicated in Fig. 2, the neutral axis of the rectangular cross section is at its mid-depth, and the area to be used in calculating Q is the area above level y.

Since Q = A′\bar{y}^′, t = b, and I = bh³/12,

or

τ = \frac{6V} {bh^3}\left(\frac{h^2} {4} – y^2\right) = \frac{3} {2} \frac{V} {A}\left(\frac{c^2 – y^2} {c^2}\right)          (a) (1)

(b) Sketch the shear-stress distribution. From Eq. 1 we see that the distribution of τ is parabolic. As we noted earlier when discussing Fig. 6.38d.
τ vanishes at the top edge (y = +c) and at the bottom edge (y = -c).
The parabolic distribution of τ is illustrated in Fig. 3.
(c) Determine the maximum shear-stress. From Eq. 1 and Fig. 3 it is obvious that τ_{max} occurs at the neutral axis, as expected. Therefore.

τ_{max} = \frac{3} {2} \frac{V} {A}    (c)

Review the Solution Unlike the case of direct shear (Section 2.7), where the shear stress is just the shear force divided by the area on which the shear stress acts, we have found a parabolic distribution of shear stress on the cross section of a rectangular beam. Since we know (Fig. 6.40d) that the shear stress must vanish at the top and bottom surfaces (y = ±c), it is reasonable for the maximum shear stress to be 50% greater than the overall average shear stress, V/A.