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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 :
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).
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.
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!
Seems like you're after statically equivalent nodal loads. There's two things at play here:
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" :
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.
