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