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Imagine a 3D problem where the goal is to calculate the max. cantilever beam deflection. There are 4 equal point loads A, B,C,D separated equally from the beam's major axis.

Is there a way to represent these 3D point load pairs as equivalent single point loads at the centreline of the beam so it becomes solvable by 2D deflection calculation methods? I.e :

Forces A and B , C and D represented as E and F respectively along the centreline

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  • $\begingroup$ FYI - it appears you have the same picture up twice $\endgroup$
    – Forward Ed
    May 8, 2022 at 15:58
  • $\begingroup$ and yes you can sum the transverse point loads. $\endgroup$
    – Forward Ed
    May 8, 2022 at 16:00
  • $\begingroup$ @ForwardEd, Thank you for your advice. I'm glad that there is a way, yet are there any academic sources that can be referenced for it? $\endgroup$
    – Noideas
    May 8, 2022 at 18:25

4 Answers 4

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Yes, it is called the "equivalent load" method, which is to replace multiple loads with a single load to simplify the analysis. The essence is that both systems will yield the same reactions at the support. Both sketches below are valid (the equivalent load is shown in red).

enter image description here enter image description here

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    $\begingroup$ this is wrong. it is not going to deflect the same. it well deflect less. $\endgroup$
    – kamran
    May 8, 2022 at 23:17
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    $\begingroup$ I agree., for the second case, the internal forces are different from the first which would give the wrong result of deflection, but the reactions are equal for both cases. Also, please read the sentence in bold face. $\endgroup$
    – r13
    May 9, 2022 at 0:32
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To get a more exact answer for deflection you can use linear superposition.

This method involves breaking apart the real beam and loading into a series of conjugate beams.

So you could have a beam model for each point load along. Solve the deflection and bending moment for each one, then add them all up together.

For linear elastic systems that are determinate (like your cantilever) you can obtain exact solutions using this approach.

Perhaps you could neglect the twisting effect of each one observing that if the loads are equal spacing and magnitude the torsion would cancel out.

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Yes, it is okay to replace two equal transverse loads with one combined load placed at the neutral axis of the beam. Unless the beam's transverse strength is not enough to support the individual loads without transverse deflection. Or the ratio of width to Height is bigger than say 5.

Then the cantilever beam acts as a cantilever plate and has to be dealt with using plate formulas. They usually result in a bit less stress and deflection!

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  • $\begingroup$ Thank you for your reply. I'm glad there is a way, yet are there any academic sources (books, articels etc) that can be referenced regarding this method of replacing 2 loads with 1? $\endgroup$
    – Noideas
    May 8, 2022 at 18:30
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Seems like you're after statically equivalent nodal loads. There's two things at play here:

  • Load equivalency: the total of your equivalent loads will be 4P (sum of all loads), and
  • Displacement compatibility. The equivalent loads need to be located in spots such that the overall displacement is equal to your original loading situation.

There's already several ways to do this in FEA. Here's an excerpt from Appendix D of Daryl Logan's excellent book "First Course in Finite Element Method" :

enter image description here

The table splits loads into their statically equivalent counterparts at either ends. Don't worry about the end conditions - the equivalent loads will still come out correct. The main thing here is that you'll add extra moments to your node locations.

On to your query - what you want to do is use point 3 in the table and get the equivalent loads at the start and end of the cantilever. Not quite what you set out to do but this will still give you the right results.

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