Question
Using an allowable shearing stress of $5.4 \mathrm{ksi}$, design a solid steel shaft to transmit 16 hp at a speed of (a) 1200 $\mathrm{rpm}$ (b) 2400 rpm.
Step 1
Step 1: Calculate the torque (T) for each speed using the formula: T = (hp * 63,025) / rpm (a) For 1200 rpm: T = (16 * 63,025) / 1200 T = 1,008,400 / 1200 T = 840 in-lb (b) For 2400 rpm: T = (16 * 63,025) / 2400 T = 1,008,400 / 2400 T = 420 in-lb Show more…
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