I don't understand how bending moment is being calculated in the problem mentioned in the above link. I understand the parts after that i.e. calculating elastic potential and partial derivatives but I am just not able to understand the bending moment part. I would be glad to receive some responses on this.

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  • $\begingroup$ Hello Engineering! I migrated the question here because I thought it is better suited for your site. If you have any questions or comments about the move, please drop me a line in Robotics Chat! $\endgroup$ – Chuck Aug 7 '18 at 14:43
  • $\begingroup$ Well, show some effort in what you have attempted and people may look at it and offer comments. $\endgroup$ – Solar Mike Aug 7 '18 at 15:55
  • $\begingroup$ Thank you for your input. Actually, solution is already there. But I dont get it how in this numerical have they calculated bending moment. $\endgroup$ – MechNovice Aug 7 '18 at 16:00
  • $\begingroup$ @MechNovice - The answer is poorly labeled, which makes it difficult to read/interpret. For example, what is $M(x)$? Which $M$? Which $x$? When it gives $M_1$, what does that mean? $\endgroup$ – Chuck Aug 7 '18 at 16:16

When you constrain a beam end to point in the same direction as the clamped end (which is the case with the lower beams), then you have to apply a moment equal to $Ql$ (where $Q$ is the end load and $l$ is the length) to maintain this angle. You'll note that this results in an end-moment magnitude of $Ql$ at each end, which produces symmetric curvature.

In this case, though, $Q_1$ is split between two beams, so the applied end moment for each is $Q_1l_1/2$, and the moment along each is $-Q_1x+Q_1l_1/2$.

No such end moment is needed in the free beam on top, so the moment along that beam is simply $-Q_2x$.

Is this what you're asking about?

  • $\begingroup$ Yes, exactly that. But why do we need - before Q1x ? $\endgroup$ – MechNovice Aug 7 '18 at 16:33
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    $\begingroup$ Because counterclockwise moments are taken as positive. This is arbitrary; you'll get the same results if clockwise moments are taken as positive (corresponding to $M(x)=Q_1x-Q_1l_1/2$). $\endgroup$ – Chemomechanics Aug 7 '18 at 16:54

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