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On Wikipedia derivation on Timoshenko beam equation we arrive at this:

$$ \delta U = \int_L \left[-M_{xx}\frac{\partial (\delta\varphi)}{\partial x} + Q_{x}\left(-\delta\varphi + \frac{\partial (\delta w)}{\partial x}\right)\right]~\mathrm{d}L $$

then:

Integration by parts, and noting that because of the boundary conditions the variations are zero at the ends of the beam, leads to

$$ \delta U = \int_L \left[\left(\frac{\partial M_{xx}}{\partial x} - Q_x\right)~\delta\varphi - \frac{\partial Q_{x}}{\partial x}~\delta w\right]~\mathrm{d}L $$

But why are the variations on the angle $\delta\varphi$ and displacement $\delta w$ zero at the ends? If we have a beam resting on two supports at the ends, then I can understand why the displacement variation would be zero since the beam is bound there, but I don't see how the angle variation could be zero, since more the beam is displaced along its length, bigger the change in angle at the ends. And if we are dealing with a beam, say, attached to a wall with one end free, the one of the ends experiences a difference in both the angle and the vertical displacement.

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    $\begingroup$ Have you studied the calculus of variations? Your question suggest that you have read the Wikipedia page but you don't really understand what a "variation" is. $\endgroup$ – alephzero Jun 24 at 9:09
  • $\begingroup$ @alephzero I think I do have an understanding. The principle of virtual work says that if the beam is in equilibrium, then when the beam is subject to virtual displacements the internal and external work are equal. What I don't understand is why the virtual angle change and virtual displacement are necessarily zero at the endpoints. $\endgroup$ – S. Rotos Jun 24 at 9:36
  • $\begingroup$ The virtual angle is not 0 but the contribution of the infitesimally small sliver at the end is zero. The one does not follow the other. $\endgroup$ – joojaa Jun 24 at 13:05
  • $\begingroup$ @joojaa I think I'm close to understanding it, but you could you explain more about what you mean by 'contribution'? Contribution to what? $\endgroup$ – S. Rotos Jun 25 at 8:22
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but I don't see how the angle variation could be zero

The angle variation would be zero if the beam is simply supported at each end, because angle variations come from bending, bending comes from moment, and a simple support does not exert a moment on the beam.

I didn't see that called out in the problem statement on the Wikipedia page, but that pretty much has to be it.

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  • $\begingroup$ Try to bend a beam yourself (a ruler for example) and you will see that the angle of the ends will change. $\endgroup$ – S. Rotos Jun 24 at 20:07
  • $\begingroup$ I'm pretty sure that they mean the variation in angle over distance, which is a function of the moment on the beam divided by its stiffness. $\endgroup$ – TimWescott Jun 24 at 22:53
  • $\begingroup$ @TimWescott assuming small rotations, the fundamental beam equation would state that the derivative of the angle is the moment of the beam divided by its stiffness. $\endgroup$ – Wasabi Jun 27 at 0:46
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    $\begingroup$ @Wasabi thanks. I'm a little rusty at this, so I stuck to what I know that I know. $\endgroup$ – TimWescott Jun 27 at 14:56

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