The two pulleys attached to the shaft are loaded as shown. If the bearings at A and B exert only vertical forces on the shaft, determine the required diameter of the shaft to the nearest \frac{1}{8} in. using the maximum-distortion energy theory. \sigma_{\text {allow }}=67 \mathrm{ksi}.
Section just to the left of point C is the most critical. Both states of stress will yield the same result.
\sigma_{a, b}=\frac{\sigma}{2} \pm \sqrt{\left(\frac{\sigma}{2}\right)^{2}+\tau^{2}}Let \frac{\sigma}{2}=A and \sqrt{\left(\frac{\sigma}{2}\right)^{2}+\tau^{2}}=B
\sigma_{a}^{2}=(A+B)^{2} \quad \sigma_{b}^{2}=(A-B)^{2}\sigma_{a} \sigma_{b}=(A+B)(A-B)
\sigma_{a}^{2}-\sigma_{a} \sigma_{b}+\sigma_{b}^{2} =A^{2}+B^{2}+2 A B-A^{2}+B^{2}+A^{2}+B^{2}-2 A B
=A^{2}+3 B^{2}
=\frac{\sigma^{2}}{4}+3\left(\frac{\sigma^{2}}{4}+\tau^{2}\right)
=\sigma^{2}+3 \tau^{2}
\sigma_{a}^{2}-\sigma_{a} \sigma_{b}+\sigma_{b}^{2}=\sigma_{\text {allow }}^{2}
\sigma^{2}+3 \tau^{2}=\sigma_{\text {allow }}^{2} \quad(1)
\sigma=\frac{M c}{I}=\frac{M c}{\frac{\pi}{4} c^{4}}=\frac{4 M}{\pi c^{3}}
\tau=\frac{T c}{J}=\frac{T c}{\frac{\pi}{2} c^{4}}=\frac{2 T}{\pi c^{3}}
From Eq (1)
\frac{16 M^{2}}{\pi^{2} c^{6}}+\frac{12 T^{2}}{\pi^{2} c^{6}}=\sigma_{\text {allow }}^{2}c=\left(\frac{16 M^{2}+12 T^{2}}{\pi^{2} \sigma_{\text {allow }}^{2}}\right)^{1 / 6}=\left(\frac{16((700)(12))^{2}+12((90)(12))^{2}}{\pi^{2}\left((67)\left(10^{3}\right)\right)^{2}}\right)^{1 / 6}
c = 0.544 in .
d = 2 c = 1.087 in
Use d = 1 \frac{1}{8} \mathrm{in} .
