The bearings at A and D exert only y and z components of force on the shaft. If \sigma_{\text {allow }}=130 \mathrm{MPa}, determine to the nearest millimeter the smallest-diameter shaft that will support the loading. Use the maximum distortion-energy theory of failure.
The critical moment is at B.
M=\sqrt{(473.7)^{2}+(147.4)^{2}}=496.1 \mathrm{~N} \cdot \mathrm{m}T=150 \mathrm{~N} \cdot \mathrm{m}
Since,
\sigma_{a, b}=\frac{\sigma}{2} \pm \sqrt{\left(\frac{\sigma}{2}\right)^{2}+\tau^{2}}
Let \frac{\sigma}{2}=A \quad and \quad \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}
\quad=\left(\frac{16(496.1)^{2}+12(150)^{2}}{\pi^{2}\left((130)\left(10^{4}\right)\right)^{2}}\right)^{1 / 6}=0.01712 \mathrm{~m}
d = 2 c = 34.3 mm
Use d = 35 mm
