in the first photo , it's stated that the dy/dx at center = 0 , however , in the second photo (website) , it's stated that the dy/dx at the boundary = 0 ,which is correct ? I'm confused now http://www.geom.uiuc.edu/education/calc-init/static-beam/support.html

  • $\begingroup$ The website actually states (correctly) that $\dfrac{d^2y}{dx^2}=0$, not $\dfrac{dy}{dx}$. $\endgroup$ – Wasabi Nov 30 '16 at 11:37

If $\theta$ is supposed to be the same quantity as $dy/dx$, the picture is both drawn and labelled wrongly.

The equations correctly state that at $x=2$, $[dy/dx]_\text{AC} = [dy/dx]_\text{BC}$, but that does not mean $[dy/dx]_\text{AC} = [dy/dx]_\text{BC} = 0$!

On the other hand, $\theta$ might be some other quantity that isn't defined in your post - in which case we can't guess what it means.

  • $\begingroup$ Then , in what circumstances , will theta = 0?? $\endgroup$ – kelvinmacks Nov 30 '16 at 0:08
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    $\begingroup$ I find it hard to believe that $\theta$ is anything other than $\dfrac{dy}{dx}$, since that's the standard assumption for small deflections and rotations. So yeah, I think the question is just wrong. $\endgroup$ – Wasabi Nov 30 '16 at 11:39
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    $\begingroup$ @kelvinmacks: I apologize for sounding pedantic, but $\theta = \dfrac{dy}{dx} = 0$ happens when the deflection is horizontal (or, more precisely, parallel to the beam's original longitudinal axis). For a simply-supported beam such as the one you've shown us, that only happens at the mid-span when the loading is symmetric. If the loading is asymmetric (such as in this case), then there's no way to know a priori. You need to do math and figure out where $\theta = 0$ (meaning $\theta=0$ is not one of your boundary conditions). $\endgroup$ – Wasabi Nov 30 '16 at 11:43
  • $\begingroup$ To there's no way to determine dy/ dx =0 in this case . Only doing the Maths can find the dy/dx =0... But , from the shape of graph , we can say that at the middle of graph , local minimum occur , so that's the point where dy/dx =0, right ? $\endgroup$ – kelvinmacks Nov 30 '16 at 11:57
  • $\begingroup$ @Wasabi , when will the deflection is horizontal ? I cant imagine it $\endgroup$ – kelvinmacks Nov 30 '16 at 13:50

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