Question
If a solid shaft having a diameter $d$ is subjected to a torque $\mathbf{T}$ and moment $\mathbf{M}$, show that by the maximumshear-stress theory the maximum allowable shear stress is $\tau_{\text {allow }}=\left(16 / \pi d^{3}\right) \sqrt{M^{2}+T^{2}} .$ Assume the principal stresses to be of opposite algebraic signs.
Step 1
From the normal stress formula, we have \[\sigma = \frac{M}{I} = \frac{M}{\frac{\pi d^{4}}{32}} = \frac{32M}{\pi d^{4}}.\] And for the shear stress, we have \[\tau = \frac{T}{J} = \frac{T}{\frac{\pi d^{4}}{32}} = \frac{32T}{\pi d^{4}}.\] Show more…
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