Question 4.2: 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 diameter to the inside 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 easily obtain the torque required for 300 hp of power at a rotational frequency of 75 rpm:
T=\frac{P}{\omega }and the conversions necessary to find a torque in U.S. units of in. · lb are built in to the following version of this relationship:
T[\textrm{in.-lb}]=\frac{63,000\times [\textrm{hp}]}{\textrm{N [rpm]}} .
We have
T=\frac{63,000\times 300 \textrm{hp}}{75 \textrm{rpm}} =252,000 \textrm{in.-lb.}Since the maximum shear stress induced by this torque is given by
\tau _{max}=\frac{Tc}{J},we obtain the value of J/c needed to transmit 600 hp without exceeding the stated stress limit:
\frac{J}{c}=\frac{T}{\tau _{max}}=\frac{252,000 \textrm{in. -lb}}{8000 \textrm{psi}}=31.5 \textrm{in.}^3\frac{J}{c}=\frac{\pi }{2} \frac{(c^4-(c/1.2)^4)}{c}=0.813 c^3=31.5 \textrm{in.}^3
This has the solution c = 3.38 in., so the outer diameter necessary is D_o = 2c = 6.77 in., and the inner diameter is D_i = D_i /1.2 = 5.64 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 \textrm{in.}^3Solving 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 = 5.44 in.