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Chapter 3

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.