Since the title is a mouthful, here goes an explanation and a sketch I've drawn a so that I can describe it correctly:

I'm trying to design a rotating arm. The rotational part will be a small pipe, installed vertically. It will be spatially restricted by two bearing, each in an end and they will be fixed to a larger structure. The arm itself is an I beam that's relatively long. I'm trying to calculate the minimum cross section of the pipe as well as the correct bearing forces. The rotating arm will be subjected to a vertical force in its end.

FBD of my problem

What I already have: I believe that once the arm is deflected, each bearing will suffer horizontal and vertical forces. I also believe that the horizontal reactions are such that they neutralize the vertical force where the bodies meet, such as that: $2*R_F*H = F*arm_{length}$.

I suppose the vertical reaction will be just $F$ in the lower bearing.

I've tried using a software to determine the support reactions and fooling around with it I've figured that sometimes the support can have a moment reaction in it and I have no clue what would this situation would be and where is this moment coming from.

Am I correct in my assumptions?


2 Answers 2


You are correct that there will be both vertical and horizontal forces on the bearings. These do not cancel each other out, though, and must be accounted for separately.

The vertical will be equal in total to F (I will take your 6kN to be F). This force will be borne by one or both bearings, as long as one or both have thrust bearing capability. Don't forget the weight of the shaft and the I beam cantilever.

There will also be horizontal forces caused by the moment. The moment is equal to L * F located at the H/2 point on the shaft. This will result in a force of (L * F) / (H/2) on each bearing, in opposite directions. Again, we need to also account for the weight of the cantilever beam at L/2.

At this point we have the data needed to construct a load diagram showing forces downward on the bearing(s), a moment on the shaft that will try to impart an S shape into it, and the horizontal loads on the bearings from the moment. Wiki has a fine explanation of the process.

  • $\begingroup$ Thank you, Scott. It was mostly a matter of confirming if the support would be considered fixed or "pinned". Because once the force is applied, I figured the wall of the topmost bearing would suffer a vertical force as well, but I figured: "well, if I remove the bottom bearing, then the top one won't hold the shaft. Thus, it shouldn't have any vertical forces." The rest, I think it's okay. Fortunately (or unfortunately?) I have already studied this part of statics! $\endgroup$ Jan 28, 2020 at 11:45

If the rotation is gentle enough not to cause any considerable amount of centripetal force on the shaft or swinging motion both horizontally and vertically on the end of the beam or pendulum vibration of the beam or there is no dynamic loading, which are all very likely according to the configuration of your I beam and the swivel mechanism of the shaft to happen, then your approach is ok.

I would check even in the most basic case for the natural frequency of the beam and the shaft and make sure they don't resonate. I can see the beam under certain conditions flailing wildly and become unstable.

  • $\begingroup$ Thank you kamran for giving me something to think about. In reality, the beam will be mostly unstressed. A fall arrest will be installed in the end of the beam and the beam itself is quite rigid. The loading will be applied once someone do fall, though it's not expected nor will be happening frequently. I'll also be installing a inclined arm that will follow the beam and a wire rope that will be attached to the top of the structure that will alleviate stress and there shouldn't be a problem with flailing. $\endgroup$ Jan 28, 2020 at 11:50

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