Questions: 25. Compute the torsional shear stress in a solid circular shaft having a diameter of 1.25 in that is transmitting 110 hp at a speed of 560 rpm .

Solution Steps

First, we need to convert the power from horsepower (hp) to torque (T) in inch-pounds (in-lb). The formula to convert power to torque is:

$$T = \frac{5252 \times P}{N}$$

where $P$ is the power in horsepower and $N$ is the speed in revolutions per minute (rpm).

Given:

• $P = 110$ hp
• $N = 560$ rpm

$$T = \frac{5252 \times 110}{560} = 1031.8571 \, \text{in-lb}$$

Next, we calculate the polar moment of inertia $J$ for a solid circular shaft. The formula for $J$ is:

$$J = \frac{\pi d^4}{32}$$

Given the diameter $d = 1.25$ in:

$$J = \frac{\pi (1.25)^4}{32} = 0.1915 \, \text{in}^4$$

The torsional shear stress $\tau$ in a solid circular shaft is given by:

$$\tau = \frac{T \cdot c}{J}$$

where $c$ is the radius of the shaft, which is half of the diameter.

$$c = \frac{1.25}{2} = 0.625 \, \text{in}$$

Substituting the values:

$$\tau = \frac{1031.8571 \times 0.625}{0.1915} = 3366.1974 \, \text{psi}$$

Final Answer