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## Q. 3.07

A shaft consisting of a steel tube of 50-mm outer diameter is to transmit 100 kW of power while rotating at a frequency of 20 Hz. Determine the tube thickness that should be used if the shearing stress is not to exceed 60 MPa.

## Verified Solution

The torque exerted on the shaft is given by Eq. (3.21):

$T=\frac{P}{2 \pi f}=\frac{100 \times 10^{3} W }{2 \pi(20 Hz )}=795.8 N \cdot m$

From Eq. (3.22) we conclude that the parameter $J / c_{2}$ must be at least equal to

$\frac{J}{c_{2}}=\frac{T}{\tau_{\max }}=\frac{795.8 N \cdot m }{60 \times 10^{6} N / m ^{2}}=13.26 \times 10^{-6} m ^{3}$                    (3.23)

But, from Eq. (3.10) we have

$\frac{J}{c_{2}}=\frac{\pi}{2 c_{2}}\left(c_{2}^{4}-c_{1}^{4}\right)=\frac{\pi}{0.050}\left[(0.025)^{4}-c_{1}^{4}\right]$                      (3.24)

$\tau=\frac{T \rho}{J}$                  (3.10)

Equating the right-hand members of Eqs. (3.23) and (3.24), we obtain:

$(0.025)^{4}-c_{1}^{4}=\frac{0.050}{\pi}\left(13.26 \times 10^{-6}\right)$

\begin{aligned}&c_{1}^{4}=390.6 \times 10^{-9}-211.0 \times 10^{-9}=179.6 \times 10^{-9} m ^{4} \\&c_{1}=20.6 \times 10^{-3} m =20.6 mm\end{aligned}

The corresponding tube thickness is

$c_{2}-c_{1}=25 mm -20.6 mm =4.4 mm$

A tube thickness of 5 mm should be used.