Simply supported beam—energy minimization with a polynomial

[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.] Today we’ll use the Rayleigh-Ritz method again, but this time we’ll avoid dealing with an infinite sum. In case you’ve forgotten, this is our problem: We’ll express the shape as a polynomial. The form of the governing differential equation, EIyiv=w tells us that our solution won’t have any terms of higher power than x4. That gets rid of the infinity problem, but a generic fourth-order polynomial still has five parameters, y=a0+a1x+a2x2+a3x3+a4x4 which is more than we want to deal with. Let’s cut down further on the number of parameters by taking advantage of some other things we know. First, the solution will have to be symmetric, so we can express it with symmetric terms right from the start. Second, we know the solution will have to meet these boundary conditions: y(0)=y(L)=y″(0)=y″(L)=0 Here’s the form we’ll start with: y=ax(L−x)+bx2(L−x)2…