2
$\begingroup$

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.

enter image description here

Improve this question
$\endgroup$
4
  • $\begingroup$ 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
2
$\begingroup$

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?

Improve this answer
$\endgroup$
2
  • $\begingroup$ Yes, exactly that. But why do we need - before Q1x ? $\endgroup$
    – MechNovice
    Aug 7 '18 at 16:33
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.