Question 4.14: Impact Loading on a Shaft A shaft of diameter d and length L...

The area properties of the shaft are

A=\frac{\pi(3)^2}{4}=7.069  in ^2, \quad J=\frac{\pi(3)^4}{32}=7.952  in ^4

The angular velocity equals

\omega=n\left(\frac{2 \pi}{60}\right)=150\left(\frac{2 \pi}{60}\right)=5 \pi  rad / s

a. The kinetic energy of the flywheel must be absorbed by the shaft. So, substituting Equation 4.43 into Equation 4.41, we have

E_k=\frac{1}{2} I \omega^2                (4.41)

I = mr^{2}                     (4.43)

\begin{aligned} E_k &=\frac{W \omega^2 r^2}{2 g} \\ &=\frac{120(5 \pi)^2(10)^2}{2(386)}=3835  in \cdot lb \end{aligned}                     (a)

From Equation 4.39a,

\begin{aligned} \phi_{\max } &=\sqrt{\frac{2 E_k L}{G J}}=\left[\frac{2(3835)(2.5 \times 12)}{\left(11.5 \times 10^6\right)(7.952)}\right]^{1 / 2} \\ &=0.05 rad =2.87^{\circ} \end{aligned}

b. Through the use of Equation 4.40,

\begin{aligned} \tau_{\max } &=2 \sqrt{\frac{E_k G}{A L}}=2\left[\frac{(3835 \left(11.5 \times 10^6\right)}{(7.069)(2.5 \times 12)}\right]^{1 / 2} \\ &=28.84  ksi \end{aligned}

Comment: The stress is within the elastic range, 28.84 < 30, and hence assumption 2 of Section 4.7 is satisfied.