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

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=1.5in and the machine operates at \omega =130rpm. If the allowable shear
stress in the shaft is 11.6 ksi , what is the maximum power transmissible to the machine?
Express your answer with appropriate units to three significant figures.
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Part B - Minimum diameter
A solid circular shaft is required to transmit P=43kW from a motor to a machine. The
maximum allowable stress in the shaft is \tau _(allow )=80MPa. If the machine runs at \omega =11
ra(d)/(s), what is the minimum diameter for the shaft?
Express your answer in cm to three significant figures.
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Learning Goal:
To determine the maximum possible power
transmission for a circular shaft and the minimum
diameter required to avoid exceeding shear stress
limits.
A rotating shaft can be used to transmit mechanical
power from one place to another. For example, a
shaft can be connected to a motor at one end and a
pulley at the other. The shaft can then transmit
power from the motor to the pulley. When the
machine is operating, the shaft will rotate at some
angular velocity \omega and will be subject to a torque
T. The power transmitted by the shaft is P=T\omega .
The shaft's rotation can also be expressed as a
frequency, f, which represents the number of
revolutions of the shaft per unit time. The frequency
and angular velocity are related by \omega =2\pi f, so
the power can also be calculated using
P=2\pi fT.
When designing a shaft, one consideration is limits
on shear stress. The torsion formula \tau _(allow )=(Tc)/(J)
relates the allowable shear stress in a circular shaft
with outer radius c to the torque T. The polar
moment of inertia is J=(\pi )/(2)c^(4)N*m 1W=1N*(m)/(s) ft*lb ft*l(b)/(s)1hp=550ft*l(b)/(s).
Part A - Transmitted power A solid circular rod

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