I need to calculate the maximum moment and reaction forces in an axle with both sides fixed in bearings, the axle is 1010mm long and has downwards forces of:

  • 236kN at 403mm
  • 86kN at 499mm
  • 32kN at 595mm
  • 12kN at 691mm
  • 4kN at 787mm

I need the results to determine what bearings to use and to calculate the minimum axle diameter, I have had lessons in how to calculate these problems but those only explained a single point load or a distributed load. How would I go about calculating this?

EDIT: The A drum will be mounted on the axle to make a winch, the forces are based on the cable windings pulling on this drum.

  • 2
    $\begingroup$ I’m voting to close this question because it looks like a homework exercise. $\endgroup$ – NMech May 10 at 12:34
  • $\begingroup$ It is not a homework question, I'm designing a drum for a winch and these are the forces the individual cable windings exert on the axle of the winch. Sorry I didn't add that in the question but I really do not know how to solve this. $\endgroup$ – David May 10 at 13:07
  • 1
    $\begingroup$ @David, you might want to add details in the comment to the body of the question. $\endgroup$ – user8055 May 10 at 14:26
  • $\begingroup$ @David, I assume you are primarily looking for the results. If so, you might want to give this tool a try: mechanicalc.com/calculators/beam-analysis $\endgroup$ – Andrew May 10 at 17:24
  • $\begingroup$ @Andrew Thank you! I wasn't aware such a calculator existed and this will definitely help me. $\endgroup$ – David May 10 at 22:07

If you know how to calculate the reactions (R, V & M) of a simply supported beam with a single concentrate load, you can repeat that calculation for every single load in a multi-load situation, then super-impose the results at the points of interest.

The graph below shows the process of a beam with two concentrate loads $P_A$ & $P_B$, and how to obtain the reactions $R_L, R_R$, the internal shear $V$ and internal moment $M$, along the span, by the method of superposition, which is valid for a linear elastic beam with any types/numbers of externally applied loads. Note the maximum moment always occurs at where the shear force changes sign and crosses the horizontal beam axis. (Note the graph is not made to scale)

enter image description here

Solution for a fixed end beam with a single concentrate load

enter image description here

  • $\begingroup$ Thanks for the explanation, however in my situation the beam is not supported but fixed in both ends, as far as I know this takes away all degrees of freedom so I'm not sure if this applies? $\endgroup$ – David May 10 at 22:02
  • $\begingroup$ Yes, it applies to beams with any kind of end support. It is tedious, but as long as you know how to calculate the reactions of the beam with a single load, the superposition is the most simple method to use without resorting to the structural analysis program. Note it is applicable to deflection calculation as well. Also, a spreadsheet can be utilized to eliminate the pain of repetition. If programmed correctly, it eliminates the mistake too. $\endgroup$ – r13 May 10 at 22:38
  • $\begingroup$ For a fixed end beam, the only addition to the graph above is the fixed-end moment at each end, again, the total is the sum of the parts. $\endgroup$ – r13 May 10 at 22:44
  • $\begingroup$ The solution for a fixed end beam with a single load is added FYI. $\endgroup$ – r13 May 11 at 2:04

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