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Question 6.100: A torque of magnitude T = 1.5 kip-in. is applied to each of ...

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 \tau_{\text {allow }}=8 ksi, determine the minimum required dimension b for each bar.

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(a) Circular Section
Rearrange the elastic torsion formula to group terms with d on the left-hand side:

\frac{\pi}{32} \frac{d^{4}}{(d / 2)}=\frac{T}{\tau}  which can be simplified to  \frac{\pi d^{3}}{16}=\frac{T}{\tau}

From this equation, the unknown diameter of the solid shaft can be expressed as

d=\sqrt[3]{\frac{16 T}{\pi \tau}}

To 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

b_{\min } \geq \sqrt[3]{\frac{16 T}{\pi \tau}}=\sqrt[3]{\frac{16(1.5  kip – in .)}{\pi(8  ksi )}}=0.985  in.

(b) Square Section
From Table 6.1,

a=b   \quad \text { aspect ratio }  \frac{b}{a}=1    \quad \Rightarrow \quad \alpha=0.208

The maximum shear stress in a rectangular section is given by Eq. (6.22):

\tau_{\max }=\frac{T}{\alpha a^{2} b}

For a square section where a = b,

b^{3}=\frac{T}{\alpha \tau_{\max }}=\frac{1.5  kip – in .}{(0.208)(8  ksi )}=0.901442  in .{ }^{3}    \quad \therefore b_{\min }=0.966  in .

(c) Rectangular Section
From Table 6.1,

\text { aspect ratio } \frac{2 b}{b}=2 \quad \Rightarrow \quad \alpha=0.246

For a rectangular section where a = 2b,

b^{3}=\frac{T}{2 \alpha \tau_{\max }}=\frac{1.5  kip – in .}{2(0.246)(8  ksi )}=0.381098  in .{ }^{3}    \quad \therefore b_{\min }=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

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