A 1.25-in. -diameter shaft contains a 0.25-in.-deep U-shaped groove that has a 1/8-in. radius at the bottom of the groove. The maximum shear stress in the shaft must be limited to 12,000 psi. If the shaft rotates at a constant angular speed of 6 Hz, determine the maximum power that may be delivered by the shaft.
Question 6.99: A 1.25-in. -diameter shaft contains a 0.25-in.-deep U-shaped...
Groove: From Fig. 6.17a:
\begin{aligned}&d=1.25 \text { in. }-2(0.25 in .)=0.75 in . \\&\frac{D}{d}=\frac{1.25 in .}{0.75 in .}=1.67 \quad \quad \frac{r}{d}=\frac{0.125 in .}{0.75 in .}=0.167 \quad \therefore K \cong 1.39\end{aligned}Section Properties at Minimum Diameter Section:
J=\frac{\pi}{32}(0.75 in .)^{4}=0.031063 in .^{4}Maximum Torque:
\begin{aligned}&\tau=K \frac{T c}{J} \\&\therefore T_{\max }=\frac{\tau_{\text {allow }} J}{K c}=\frac{(12,000 psi )\left(0.031063 in .^{4}\right)}{(1.39)(0.75 in . / 2)}=715.1219 lb – in .=59.5935 lb – ft\end{aligned}Power transmission: The maximum power that can be transmitted at 6 Hz is:
\begin{aligned}P_{\max } &=T_{\max } \omega=(59.5935 lb – ft )\left(\frac{6 rev }{ s }\right)\left(\frac{2 \pi rad }{1 rev }\right) \\&=2,246.6219 lb – ft / s\end{aligned}or in units of horsepower,
P_{\max }=\frac{2,246.6219 lb – ft / s }{\frac{550 lb – ft / s }{1 hp }}=4.08 hp
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