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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.

Step-by-Step

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.