I am looking to use FEA (specifically, Creo Simulate) to re-create the results of a static strength test on a part my company will be testing in our lab. Because the lab results are the ones we will be using to determine the utility and capability of the part, I'm doing this mostly as an exercise, to strengthen my FEA skills and to see how close to reality I can get the results of the analysis to be.

For a couple different reasons (write access to model files, investigation of the bolt functionality of Creo, more realistic application of load) I am attempting to model part of the test stand in addition to the part of interest. The part itself has a flange with a bolt pattern as part of its design, and we are using that bolt pattern to attach it to an adapter. The adapter will attach to the output of the motor, with the torque being applied through a key that sits between the adapter and the motor output.

Below I have a picture of the keyway in the adapter for clarity. The part of interest is attached to the back side of this adapter, the key fits along the entire length of the keyway, and the adapter connects to the test stand output using the four larger bolt holes (these don't transmit any significant torque, they simply keep the assembly together.)

Flange with keyway

My issue is determining how best to model the applied load on the walls of the keyway. My intuition tells me that the load will vary along the length of the keyway, as what we're really applying is a torque that gets distributed throughout the key.

  • Given that $T=r \times F$, I'd expect there to be an inverse relationship between the force I should apply and the position along the keyway at which I apply it. Does the fact that this line of action is not radial complicate things?

  • In addition, I have two coordinate systems defined, the standard Cartesian, and a cylindrical one with the z-axis going through the center hole of the adapter as you might expect. Will applying the load in one coordinate system or the other significantly alter the result, and is one preferable? Will the distributed force equation be easier to define in one coordinate system?

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    $\begingroup$ So to be clear, you are looking to apply a distributed force along the keyway that will mimic the force applied from the bar? $\endgroup$
    – hazzey
    Apr 30, 2015 at 15:19
  • $\begingroup$ Yes. I'm expecting two loads, one on the top right and one on the bottom left, that mirror each other. And because of singularity issues, I'm fairly certain the loads won't be defined at the point y=0, but I guess that depends on the equation. $\endgroup$ Apr 30, 2015 at 15:30
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    $\begingroup$ Like a flat-head screwdriver on a screw. $\endgroup$ Apr 30, 2015 at 15:55
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    $\begingroup$ Quick note for anyone searching for relevant literature, "chord" may be an useful alternative search term for "non-radial." $\endgroup$
    – Air
    Apr 30, 2015 at 19:44
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    $\begingroup$ Trevor, if I were in your position, I would choose to not model this because I would be worried that possible errors in this step would actually decrease, rather than enhance, model fidelity and would be difficult to validate. $\endgroup$
    – regdoug
    Apr 30, 2015 at 20:20

1 Answer 1


I would ignore the fact that the keyway makes the actual forces slightly off-center at least when calculating the forces themselves. The forces can be placed at their true locations once they are calculated.

I would also assume that the load is continuously distributed from the center to the outer edge. This assumes that the key is machined with a tight enough tolerance (or deflects/deforms enough) to equalize the distribution. This also assumes that a given amount of rotation produces more force at the extremes of the key than at the center (where it is zero).

From there, it is a matter of finding the maximum force per unit length at the end.

Simplified Forces

$P = \frac{1}{2}\frac{T}{\frac{2}{3}R} \\ and \\ P = \frac{1}{2}R *w_\text{max}\\ gives:\\ w_\text{max} = \frac{T}{\frac{2}{3}R^2}$

As I have gone through this, I have the nagging feeling that I am missing the point of your question, so let me know if I missed it.

Note also that I am used to working with models where the finer details of the exact load placement does not affect the final results, so I may be making assumptions that are not correct for your model.

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    $\begingroup$ I'm certainly okay with ignoring the fact that the loads are off-center, if it simplifies the calculations. The thing that still irks me about a continuous load distribution that varies directly with r is that the torque would then vary directly with r^2, which seems counter-intuitive to me, but I don't have a good reason why. $\endgroup$ May 1, 2015 at 14:11

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