WORLD FOR CIVIL

Topic 6.3: Torsion - Power Transmission

One important process which involves applying torque to a shaft is power transmission. To transmit power a drive torque is applied to a rotating shaft. A short derivation will gives us a relationship between Torque, rotation rate, and Power transmitted.

In Diagram 1a we have shown a shaft with a drive torque of 1000 ft-lb. at end A and an equal and opposite load torque at end B.

In Diagram 1b we have shown the shaft from end on. Notice that to apply a torque to a shaft we must exert a force, F, usually at the outer edge of the shaft. This force may be applied through use of a belt or gear. The product of the force (F) and the radius (r) is the applied (or load) torque.
The work done as we rotate the shaft will be the product of the force and the distance the force acts through - which is the circumference. That is, for each revolution of the shaft, the force act through a distance of one circumference, or we may write:
Work = F x d = F * (2 p r) * (# revolutions)
The Power sent down the shaft is then the Work per unit time, or if we divide the equation for Work above by the time we can write:
Power = Work/Time = F * (2 p r) * (# rev/time)
If we now rewrite the above equation slightly, as below:
Power = 2 p ( F * r) * (# rev/sec)
Then we recognize the (F * r) term is the torque in the shaft, and we can rewrite as:
Power = 2 p T n (ft-lb./sec) where T = Torque in ft-lb.; n = # rev/sec
This is the formula for power transmitted in foot-pounds/second. It is often more convenient to express it in horsepower (1 hp = 550 ft-lb./sec) as shown below.
Power _hp = [2 p T n / 550 ft-lb/sec/hp]

Next we look at a simple example of determining transmitted horsepower.

Example
In Diagram 2, we have shown a solid shaft with an applied driving torque of 1000 ft-lb. and an equal and opposite load torque, rotating at a speed of 1800 rpm (30 rev/sec). We would like to determine the horsepower being transmitted down the shaft.

Solution: First we mentally note that since this is a simple shaft with equal external driving and load torque, then the internal torque will be equal in value to the external torque of 1000 ft-lb.
Next we apply the Horsepower equation:
Power _hp = [2 p T n / 550 ft-lb/sec/hp] ; where
T = internal torque in shaft (in foot-pounds) = 1000 ft-lb.
n = the number of revolution per second = 30 rev/sec.
So we have: Power _hp = [2* 3.1416* 1000 ft-lb.* 30 rev/sec / 550 ft-lb/sec/hp] = 343 hp.