Question 5.4: Design a hollow steel shaft to transmit 300 hp at 75 rpm wit......
Design a hollow steel shaft to transmit 300 hp at 75 rpm without exceeding a shear stress of 8000 psi. Use 1.2:1 as the ratio of the outside to the inner diameter. What solid shaft could be used instead?
Given: Desired performance of hollow shaft.
Find: Inner and outer diameters of shaft; dimensions of equivalent solid shaft.
Assume: Hooke’s law applies.
We can obtain the torque required with
T=\frac{P}{\omega},and the conversions, necessary to find a torque in U.S. units of in-lb are built into the following version of this relationship, where N is the number of rotations per minute: T(\mathrm{in}-\mathrm{lb})=\frac{63,000 \times(\mathrm{hp})}{N(\mathrm{rpm})} . We have
T=\frac{63,000 \times 300 \mathrm{hp}}{75 \mathrm{rpm}}=252,000 \mathrm{in}-\mathrm{lb} .Since the maximum shear stress induced by this torque is given by τ_{max} = Tc/J , we obtain the value of J /c needed to transmit 600 hp without exceeding the stated stress limit:
\begin{aligned} & \frac{J}{c}=\frac{T}{\tau_{\max }}=\frac{252,000 \mathrm{in}-\mathrm{lb}}{8000 \mathrm{psi}}=31.5 \mathrm{in}^3, \\ & \frac{J}{c}=\frac{\pi}{2} \frac{\left(c^4-(c / 1.2)^4\right)}{c}=0.813 c^3=31.5 \mathrm{in}^3 . \end{aligned}This has the solution c = 3.4 in, so the outer diameter necessary is D_o = 2c = 6.8 in, and the inner diameter is D_i = D_o/1.2 = 5.6 in.
For a solid shaft, J /c has a simpler form, and we require only
\frac{J}{c}=\frac{\pi}{2} c^3=31.5 \mathrm{in}^3,Solving for c, the radius of a solid shaft capable of transmitting 600 hp without exceeding a shear stress of 8000 psi, we have D = 2c = 5.4 in.