Question: The one-dimensional beam bending problem [136] takes the form \[(pu'')''(x) = f(x) ; 0 < x < L\] \[u(0) = 0 = u(L)\] \[ \frac{d^2u}{dx^2}(0) = 0 = \frac{d^2u}{dx^2}(L). \] Here u(x) is the displacement of the beam, p(x) is the flexural rigidity, f(x) is the applied load, and L is the length of the beam. The boundary conditions correspond to zero displacement at the ends of the beam. Determine the weak form of the beam bending problem. Which of the boundary conditions are essential, and which are natural?
Discussion: For this problem the essential boundary conditions are \[ u(0) = 0 = u(L). \] And the natural boundary conditions are \[ \frac{d^2u}{dx^2}(0) = 0 = \frac{d^2u}{dx^2}(L). \] When we multiply the differential equation by a test function $\delta u$ such that $\delta u(0) = \delta u(L) = 0$ and integrate by parts twice, we obtain \begin{align*} \mathscr{B}(\delta u, u) &= \int_0^L \delta u \frac{d^2}{dx^2}(p(x)\frac{d^2u}{dx^2})dx \\ & = \int_0^L \delta u d(\frac{d}{dx}(p(x)\frac{d^2u}{dx^2})) \\ & = \delta u \frac{d}{dx}(p(x)\frac{d^2u}{dx^2})|_0^L - \int_0^L \frac{d}{dx}(p(x)\frac{d^2u}{dx^2})\delta u' dx \\ & = - \int_0^L \delta u' d(p(x)\frac{d^2u}{dx^2}) \\ & = -\delta u' d(p(x)\frac{d^2u}{dx^2})|_0^L + \int_0^L p(x)\frac{d^2 u}{dx^2}\delta u'' dx \\ & = \int_0^L p(x)\frac{d^2 u}{dx^2}\delta u'' dx = \int_0^L f(x) \delta u dx. \end{align*} Therefore we get the weak form of the beam bending problem \[ \int_0^L p(x)\frac{d^2 u}{dx^2}\delta u'' dx = \int_0^L f(x) \delta u dx. \]
Last modified: