Question F15.2: Determine the minimum diameter d to the nearest 1/8 in. of t......
Determine the minimum diameter d to the nearest \frac{1}{8} \mathrm{in}. of the rod to safely support the load. The rod is made of a material having an allowable normal stress of \sigma_{\text {allow }}=20 \mathrm{ksi} and an allowable shear stress of \tau_{\text {allow }}=10 \mathrm{ksi}.
At support,
V_{\max }=3 \mathrm{kip} \quad M_{\max }=12\ \mathrm{kip} \cdot \mathrm{ft}
I=\frac{\pi}{4}\left(\frac{d}{2}\right)^{4}=\frac{\pi d^{4}}{64}
\sigma_{\text {allow }}=\frac{M_{\max } c}{I} ; \quad 20=\frac{12(12)\left(\frac{d}{2}\right)}{\frac{\pi d^{4}}{64}}
Use d=4 \frac{1}{4} in.
d=4.19 in
I=\frac{\pi}{64}\left(4.25^{4}\right)=16.015\ \mathrm{in}^{4}
Semicircle,
\begin{aligned} Q_{\max } & =\frac{4(4.25 / 2)}{3 \pi}\left[\frac{1}{2}\left(\frac{\pi}{4}\right)\left(4.25^{2}\right)\right]=6.397 \mathrm{in}^{3} \\ \tau_{\max } & =\frac{V_{\max } Q_{\max }}{I t}=\frac{3(6.397)}{16.015(4.25)} \\ & =0.282 \mathrm{ksi}<\tau_{\text {allow }}=10 \mathrm{ksi}(\mathrm{OK}) \end{aligned}