A torque of magnitude T = 1.5 kip-in. is applied to each of the bars shown in Figure P6.100/101. If the allowable shear stress is specified as τallow =8 ksi, determine the minimum required dimension b for each bar.
A torque of magnitude T = 1.5 kip-in. is applied to each of the bars shown in Figure P6.100/101. If the allowable shear stress is specified as τallow =8 ksi, determine the minimum required dimension b for each bar.
(a) Circular Section
Rearrange the elastic torsion formula to group terms with d on the left-hand side:
32π(d/2)d4=τT which can be simplified to 16πd3=τT
From this equation, the unknown diameter of the solid shaft can be expressed as
d=3πτ16TTo support a torque of T = 1.5 kip-in. without exceeding the maximum shear stress of 8 ksi, the solid shaft must have a diameter (i.e., dimension b shown in the problem statement) of
bmin≥3πτ16T=3π(8 ksi)16(1.5 kip–in.)=0.985 in.(b) Square Section
From Table 6.1,
The maximum shear stress in a rectangular section is given by Eq. (6.22):
τmax=αa2bTFor a square section where a = b,
b3=ατmaxT=(0.208)(8 ksi)1.5 kip–in.=0.901442 in.3 ∴bmin=0.966 in.(c) Rectangular Section
From Table 6.1,
For a rectangular section where a = 2b,
b3=2ατmaxT=2(0.246)(8 ksi)1.5 kip–in.=0.381098 in.3 ∴bmin=0.725 in.
Table 6.1 Table of Constants for Torsion of a Rectangular Bar |
||
Ratio b/a | α | β |
1.0 | 0.208 | 0.1406 |
1.2 | 0.219 | 0.166 |
1.5 | 0.231 | 0.196 |
2.0 | 0.246 | 0.229 |
2.5 | 0.258 | 0.249 |
3.0 | 0.267 | 0.263 |
4.0 | 0.282 | 0.281 |
5.0 | 0.291 | 0.291 |
10.0 | 0.312 | 0.312 |
∞ | 0.333 | 0.333 |