Cyndi Nyberg
Former EASA Technical Support Specialist
Have you ever wondered why the shaft of an electric motor is often larger than that of the driven equipment? One reason for this is that the standard shaft sizes specified for the standard NEMA frame machines are larger than the minimum required, as we will see in the examples below. Manufacturers tend to design using an ample safety factor. Given the dire consequences if a shaft breaks, that is understandable.
Even so, the difference between a T and TS shaft can raise questions for those unfamiliar with mechanical design. It is important that the shaft is large enough to (a) transmit the required torque without exceeding the maximum allowable torsional shearing stress for the shaft material, and (b) prevent torsional deflection, or twisting, during service. All this, with a substantial safety factor.
The allowable twisting moment, or torque, is a function of the allowable torsional shearing stress (in psi or kPa) and the polar section modulus, which is a function of the cross-sectional area of the shaft. Using standard values for steel, the Machinery’s Handbook derives the following equations to determine shaft size. The allowable stresses used are 4000 psi (2.86 kg/mm2) for power transmitting shafts, and 6000 psi (4.29 kg/mm2) for lineshafts with a pulley.
Before going through some examples, we need to note that motor shafts are typically keyed, so that needs to be accounted for in the calculations. A typical recommendation is that the shaft should be designed using no more than 75% of the maximum recommended stress for a non-keyed shaft. Note that comparing the equations in this article to the ones in the Machinery’s Handbook, the presence of a keyway has already been factored into the calculations used in this article.
Example 1
Consider a 200 hp (150 kW), 4pole open motor. For a direct couple application, the standard frame size is 445TS. For a belted application, a 445T frame is used, since the shaft diameter is greater. A belted application puts a significantly higher bending stress on the shaft than a direct-coupled application, so the shaft outside diameter is larger, giving it more strength.
The shaft diameter of the 445TS frame is 2.375” (60 mm). Using equation (1), the minimum shaft size is:
From this equation, it can be seen that there is a significant factor of safety inherent in the shaft design. In fact, there is already a safety factor built into the equations. However, just for comparison, let’s look at the maximum torque that the same 445TS frame shaft can transmit. If it was possible to redesign the motor winding to 400 hp (300 kW), then the new minimum shaft size requirement would be:
If it were possible to redesign the stator for double the torque, there is a slight possibility that the shaft is at the absolute minimum diameter, but as I mentioned previously, there is already a safety factor built into the equations. (Note: A typical engineering safety factor is a factor of 5.) This engineer has never heard of a winding change that resulted in too high a shaft stress – although doubling the torque of a larger motor as in the example is rare. The stator core iron would be the limiting factor much more than the shaft size!
Rather than to just consider the size of shaft necessary to transmit a certain torque, another way to calculate the required shaft diameter is to limit the shaft torsional deflection or twisting that may occur as a result of the torsional stress acting on it. A length of steel shaft will have a certain amount of resistance to twisting up to a certain level of twisting force acting on the shaft. A designer can take the approach of setting a limit on the torsional deflection – the larger the shaft diameter, the greater its resistance to twisting will be.
A generally accepted rule of thumb is that the shaft should have a diameter large enough to not deflect more than 1 degree in a length that is 20 times the shaft diameter. The equation to use to determine the shaft diameter is:
Example 2
Using the same 200 hp (150 kW), 4 pole motor, calculate the required shaft size to limit the torsional deflection.
The minimum shaft diameter that we calculate in both examples is essentially the same; however, a good approach would be to use both calculations, and use the larger value as the absolute minimum.
Hollowshaft Designs
Because most of the stress on a torsionally-loaded shaft is near the surface, hollowshaft designs are common for vertical motors.
This simplifies the process of coupling the motor to a heavy pump with a long down-hole pump shaft. For a hollowshaft motor, the calculations for shaft diameter are not so straightforward.
There are two variables – the outside and inside diameters of the hollowshaft. These diameters are not standardized, so it is impossible to use a simple ratio to simplify the calculation. For this reason, it is easier for illustrative purposes to determine if a specific hollowshaft is sufficient for a given power rating.
Example 3
A 200 hp (150 kW), 4-pole vertical hollowshaft is in a 445TP frame. Assume the outside diameter of the shaft to be 3” (76 mm), and the inside diameter to be 2” (51 mm). To determine if this shaft size is sufficient to transmit the required torque, we use the following equation (which incidentally is derived from the same equations as the preceeding ones):
So, for our example, P has to be greater than 200 hp to ensure the shaft is large enough.
Theorectically, this shaft would be capable of transmitting 485 hp, so our 200 hp requirement is sufficient. The thinner the wall of the hollowshaft, the less torque it will be able to transmit.
Example 4
What if the 3” OD shaft has a wall thickness of only 1/4” – what then would be the maximum horsepower possible for the shaft to transmit?
With a thinner wall, the ability to transmit torque is reduced.
For the metric equivalent, use the following equation:
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