Simply supported beam—fourth-order ODE

[Equations in this post may not look right (or appear at all) in your RSS reader. Go to the original article to see them rendered properly.] In the introductory post to this series, I mentioned that the shear is the derivative of the moment. It’s easy to see that for our specific problem, the simply supported beam with a uniform distributed load, but it’s also true in general. A further relationship is that the distributed load function—let’s call it q—is the derivative of the shear (with a minus sign according to the usual sign convention). Therefore V=dMdxandq=−dVdx which can be combined to give q=−d2Mdx2 Again, for our specific problem with a uniform load, q is just the constant value w, and it’s easy to see that the slope of the shear diagram is −w. In yesterday’s post, we used the differential relationship between the moment and displacement, M=−EId2ydx2 and the fact that we already knew that the moment was a parabola to get a solution quickly. In general, though, people usually…