Chapter 2.4, Problem 7SA

If we also fix the free end of the diving board, we have a “clamped -clamped” beam, obeying

identical boundary conditions at each end: $y(0)=y\text{'}(0)=y(L)=y\text{'}(L)=0$ . This version is used to model the sag in a structure, like a bridge. Begin with the slightly different evenly spaced grid $0=x_0<x_1<\mathrm{...}<x_n<x_{n+1}=0$

. where $h=x_i-x_{i-1}$ for $i=1,\mathrm{...},n,$, and find the system of n equations in n unknowns that determine $y_1,\mathrm{...},y_n$ . (It should be similar to the clamped-free version. except that the last two rows of the coefficient matrix A should be the first two rows reversed.) Solve for a sinusoidal load and answer the questions of Step 5 for the center $x=L/2$ of the beam. The exact solution for the clamped-clamped beam under a sinusoidal load is $y(x)=\frac{f}{24EI}{x}^2{\left(L-x\right)}^2-\frac{pg{l}^2}{{\pi }^{4}EI}\left(L^2\sin \frac{\pi }{L}x+\pi x\left(x-L\right)\right)$.