Question

The differential equation satisfied by a beam uniformly loaded at one end that is fixed and the other end 8s subjected to a tensile force P, is given by EI.d^2y/dx^2= P.y- 1/2(W.x^2),where E,I,P,W are constants. Show that the elastic curve for the beam under the conditions y=0, dy/dx=0 at x=0, is given by y(x)=(W/P^2)[1-cosh(vx)]+ (W/2P)(x^2+2/n^2),where EI= P/n^2

Answer 1

$\displaystyle \textsf{Consider the differential equation}\\ ay" = py- bx^2\hspace{1cm}(1)\\ ay" - py = -bx^2\\ \textsf{The auxiliary equation is}\\ ak^2 - p = 0\\ k^2 = \frac{p}{a}\\ k = \pm\sqrt{\frac{p}{a}}\\ \textsf{Since}\,\, p, a > 0\\ y = CF = A\cosh\left(\sqrt{\frac{p}{a}}x\right) + B\sinh\left(\sqrt{\frac{p}{a}}x\right)\\ ay" - py = -bx^2\\ y = PI = lx^2 + mx + n\\ y" = 2l\\ \implies 2la - p(lx^2 + mx + n) = -bx^2\\ pl = b, l = \frac{b}{p}\\ pm = 0, m = 0\\ 2la - pn = 0, \\ n = \frac{2la}{p} = \frac{2ab}{p^2}\\ \therefore y = PI = \frac{b}{p}x^2 + \frac{2ab}{p^2}\\ y = CF + PI\\ \therefore y = A\cosh\left(\sqrt{\frac{p}{a}}x\right) + B\sinh\left(\sqrt{\frac{p}{a}}x\right) + \frac{b}{p}x^2 + \frac{2ab}{p^2}\\ 0 = A + \frac{2ab}{p^2}\\ A = -\frac{2ab}{p^2}\\ y' = A\sqrt{\frac{p}{a}}\sinh\left(\sqrt{\frac{p}{a}}x\right) + B\sqrt{\frac{p}{a}}\cosh\left(\sqrt{\frac{p}{a}}x\right) + \frac{2bx}{p}\\ y'(0) = 0 = B\sqrt{\frac{p}{a}}\\ \therefore B = 0\\ y = -\frac{2ab}{p^2}\cosh\left(\sqrt{\frac{p}{a}}x\right)+ \frac{b}{p}x^2 + \frac{2ab}{p^2} \\ y = \frac{2ab}{p^2}\left(1 - \cosh\left(\sqrt{\frac{p}{a}}x\right)\right) + \frac{b}{p}x^2 \\ \textsf{Comparing the DE in}\, (1)\, \, \textsf{with the given DE.}\\ ay" = py- bx^2\\ y"= P.y- 1/2(W.x^2)\\ b = \frac{W}{2}, p = P, a = El, v = \sqrt{\frac{P}{a}}\\ y = \frac{2El.W}{2P^2}\left(1 - \cosh(vx)\right) + \frac{W}{2P}x^2 \\ \therefore y = \frac{El.W}{P^2}\left(1 - \cosh(vx)\right) + \frac{W}{2P}x^2\\$