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Question 3.62:  A 40-mm-diameter solid steel shaft, used as a torque tr mit...

A 40-mm-diameter solid steel shaft, used as a torque tr mitter, is replaced with a hollow

shaft having a 40-mm OD and a 36-mm ID. If both materials have the same strength, what is

the percentage reduction in torque tr mission? What is the percentage reduction in shaft

weight?

 

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\begin{aligned}&\text { (a) } T_{\text {solid }}=\frac{J \tau_{\max }}{r}=\frac{\pi d_{o}^{4} \tau_{\max }}{16 d_{o}} \quad T_{\text {hollow }}=\frac{J \tau_{\max }}{r}=\frac{\pi\left(d_{o}^{4}-d_{i}^{4}\right) \tau_{\max }}{16 d_{o}} \\&\% \Delta T=\frac{T_{\text {solid }}-T_{\text {hollow }}}{T_{\text {solid }}}(100 \%)=\frac{d_{i}^{4}}{d_{o}^{4}}(100 \%)=\frac{\left(36^{4}\right)}{\left(40^{4}\right)}(100 \%)=65.6 \% \text {  . } \\&\text { (b) } W_{\text {solid }}=k d_{o}^{2}, \quad W_{\text {hollow }}=k\left(d_{o}^{2}-d_{i}^{2}\right) \\&\% \Delta W=\frac{W_{\text {solid }}-W_{\text {hollow }}}{W_{\text {solid }}}(100 \%)=\frac{d_{i}^{2}}{d_{o}^{2}}(100 \%)=\frac{\left(36^{2}\right)}{\left(40^{2}\right)}(100 \%)=81.0 \% \quad \text {   }\end{aligned}

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