# slope of deflection of beam at center

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

• The website actually states (correctly) that $\dfrac{d^2y}{dx^2}=0$, not $\dfrac{dy}{dx}$. – 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.
• 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. – Wasabi Nov 30 '16 at 11:39
• @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). – Wasabi Nov 30 '16 at 11:43