Question

Part A - Transmitted power A solid circular rod is used to transmit power from a motor to a machine. The diameter of the rod is D = 2.5 in and the machine operates at ω = 170 rpm . If the allowable shear stress in the shaft is 10.4 ksi , what is the maximum power transmissible to the machine? Express your answer with appropriate units to three significant figures.

Answer 1

Answer:

Explanation:

Given that solid circular rod rotates at constant speed and neglecting losses throughout the system, power is calculated as the product of torque and angular speed. That is to say:

$\dot W = T \cdot \omega$

There is a formula that relates torque with shear stress:

$\tau = \frac{T \cdot D}{2 \cdot J}$

Where $J$ is the torsion module, whose formula for a solid circular cross section is:

$J = \frac{\pi \cdot D^{4}}{32}$

The tension module is calculated herein:

$J = 3.835 in^{4}$

Maximum allowed torsion is found by isolating it from shear stress equation:

$T_{max} = \frac{2 \cdot J \cdot \tau_{max}}{D}$

$T_{max} = 31.907 kip \cdot in\\T_{max} = 2.659 kip \cdot ft$

Then, maximum transmissible power is determined directly:

$\dot W = (2.659 kip \cdot ft) \cdot (2)\cdot (\pi) \cdot (170 rpm)\\\dot W \approx 2840.188 \frac{kip \cdot ft}{min} \\\dot W \approx 86.066 h.p.$