Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

188
TORSION
a
b
T
b
T
a
τ max
3T
=
a 2 b
FIGURE 6.20 Equivalent narrow rectangular sections with shear stress distribution.
τ max
3T
=
a 2 b
and
3TL
φ = (6.26)
3
abG
The absolute value of the maximum shear stress in a narrow rectangular bar occurs
on the edge of the bar in the middle of the long side. For a thin-walled member of uniform
thickness and arbitrary shape, the maximum shear stress and the shear stress distribution
are equivalent to those quantities in a rectangular bar with a large b/a ratio.
Thus, Equations (6.25) and (6.26) can be used to compute the maximum shear stress
and the angle of twist for thin-walled shapes such as those shown in Figure 6.20. For use
in these equations, the length a is taken as the thickness of the thin-walled shape. The
length b is equal to the length of the thin-walled shape as measured along the centerline of
the wall.
ExAmpLE 6.12
(a)
1.00 in.
T
(b)
1.875 in.
2.50 in.
T
1.25 in.
The two rectangular polymer bars shown are each subjected to a
torque T = 2,000 lb · in. For each bar, determine
(a) the maximum shear stress.
(b) the rotation angle at the free end if the bar has a length of
12 in. Assume that G = 500 ksi for the polymer material.
Plan the Solution
The aspect ratio b/a for each bar will be computed. On the basis of
this ratio, the constants α and β will be determined from Table 6.1.
The maximum shear stress and rotation angles will be computed
from Equations (6.22) and (6.23), respectively.
SOLUTION
For bar (a), the long side of the bar is b = 2.50 in. and the short side is a = 1.00 in.; therefore,
b/a = 2.5. From Table 6.1, α = 0.258 and β = 0.249.
The maximum shear stress produced in bar (a) by a torque T = 2,000 lb · in. is
T
2,000 lb⋅in.
= =
= 3,100 psi
Ans.
αab
(0.258)(1.00 in.) (2.50 in.)
τ max 2 2