Answer to Question #143650 in Differential Equations for Nikhil Singh

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\n\\textsf{Consider the differential equation}\\\\\n\nay" = py- bx^2\\hspace{1cm}(1)\\\\\n\nay" - py = -bx^2\\\\\n\n\\textsf{The auxiliary equation is}\\\\\nak^2 - p = 0\\\\\n\nk^2 = \\frac{p}{a}\\\\\n\nk = \\pm\\sqrt{\\frac{p}{a}}\\\\\n\n\\textsf{Since}\\,\\, p, a > 0\\\\\n\ny = CF = A\\cosh\\left(\\sqrt{\\frac{p}{a}}x\\right) + B\\sinh\\left(\\sqrt{\\frac{p}{a}}x\\right)\\\\\n\nay" - py = -bx^2\\\\\n\n\ny = PI = lx^2 + mx + n\\\\\n\ny" = 2l\\\\\n\n\n\\implies 2la - p(lx^2 + mx + n) = -bx^2\\\\\n\n\npl = b, l = \\frac{b}{p}\\\\\n\npm = 0, m = 0\\\\\n\n2la - pn = 0, \\\\\n\nn = \\frac{2la}{p} = \\frac{2ab}{p^2}\\\\\n\n\n\\therefore y = PI = \\frac{b}{p}x^2 + \\frac{2ab}{p^2}\\\\\n\ny = CF + PI\\\\\n\n\n\\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}\\\\\n\n\n0 = A + \\frac{2ab}{p^2}\\\\\n\nA = -\\frac{2ab}{p^2}\\\\\n\ny' = 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}\\\\\n\ny'(0) = 0 = B\\sqrt{\\frac{p}{a}}\\\\\n\n\\therefore B = 0\\\\\n\n\ny = -\\frac{2ab}{p^2}\\cosh\\left(\\sqrt{\\frac{p}{a}}x\\right)+ \\frac{b}{p}x^2 + \\frac{2ab}{p^2} \\\\\n\ny = \\frac{2ab}{p^2}\\left(1 - \\cosh\\left(\\sqrt{\\frac{p}{a}}x\\right)\\right) + \\frac{b}{p}x^2 \\\\\n\n\\textsf{Comparing the DE in}\\, (1)\\, \\, \\textsf{with the given DE.}\\\\\n\nay" = py- bx^2\\\\\ny"= P.y- 1\/2(W.x^2)\\\\\n\n\nb = \\frac{W}{2}, p = P, a = El, v = \\sqrt{\\frac{P}{a}}\\\\\n\ny = \\frac{2El.W}{2P^2}\\left(1 - \\cosh(vx)\\right) + \\frac{W}{2P}x^2 \\\\\n\n\\therefore y = \\frac{El.W}{P^2}\\left(1 - \\cosh(vx)\\right) + \\frac{W}{2P}x^2\\\\"