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96 Mechanical Engineering Design
Figure 3–18 y y y
A
Transverse shear stresses in a dy b = 3V
dA max 2A
rectangular beam.
M V
c
y
y 1
x z h x
O O
A
(a) (b) (c)
y
dx
c
y
1 x
(d)
If we now use this value of I for Eq. (3–32) and rearrange, we get
3V y 1 2
τ = 1 − (3–33)
2A c 2
We note that the maximum shear stress exists when y 1 = 0, which is at the bending neu-
tral axis. Thus
3V
τ max = (3–34)
2A
for a rectangular section. As we move away from the neutral axis, the shear stress
decreases parabolically until it is zero at the outer surfaces where y 1 =±c, as shown
in Fig. 3–18c. Horizontal shear stress is always accompanied by vertical shear stress
of the same magnitude, and so the distribution can be diagrammed as shown in
Fig. 3–18d. Figure 3–18c shows that the shear τ on the vertical surfaces varies with
y. We are almost always interested in the horizontal shear, τ in Fig. 3–18d, which is
nearly uniform over dx with constant y y 1 . The maximum horizontal shear occurs
where the vertical shear is largest. This is usually at the neutral axis but may not be
if the width b is smaller somewhere else. Furthermore, if the section is such that b
can be minimized on a plane not horizontal, then the horizontal shear stress occurs
on an inclined plane. For example, with tubing, the horizontal shear stress occurs on
a radial plane and the corresponding “vertical shear” is not vertical, but tangential.
The distributions of transverse shear stresses for several commonly used cross sec-
tions are shown in Table 3–2. The profiles represent the VQ/Ib relationship, which is a
function of the distance y from the neutral axis. For each profile, the formula for the
maximum value at the neutral axis is given. Note that the expression given for the
I beam is a commonly used approximation that is reasonable for a standard I beam with
a thin web. Also, the profile for the I beam is idealized. In reality the transition from the
web to the flange is quite complex locally, and not simply a step change.