Question

The twisting of a beam with rectangular cross-section is described by the inhomogeneous partial differential equation (PDE) below: 024 049 = -2 əx2 + ayż Eqn 2.1 where x and y are the coordinates of the cross-section and p(x,y) is the warp or distortion of the cross-section. The cross-section is bounded by –p sx sp and —q sy sq. The boundary conditions are given by: 0(p,y) = 0, 4(-p,y) = 0, 4(x,q) = 0 and 4(x,-q) = 0. Using the inhomogeneous PDE (Equation 2.1) and the general solution: Q(x, y) = u(x,y) +p2 – x2 Eqn 2.2 i) Show that the inhomogeneous PDE (Equation 2.1) reduces to the homogeneous PDE (Equation 2.3) given below. om + on = 0 + = 0 Egn 2.3 Eqn 2.3 axlay2 ii) Show that u(x, y) satisfies the boundary conditions: u(p,y) = 0; u(-p,y) = 0; u(x,q) = x2 – p?; u(x,-q) = x2 – p2

Answer 1

o o 2-2 0 Put a(0, y) = u(x,y) +37-xx? 22 1) to be a lip 2 u 10,7) * #*992) = 1 35 3 be from equm o , = -2 cs comes with letter Scanned with CamScanner 07 (Proved) Poved) 2 x 2

© ¢Cb, y) = 0 / 0 (-6, y) = 0, zu (B,y) + B2 $2 = 0 ; u(-B,y) +bº-EB)?=D => ) u (boy)-o] ] = luk-t, y) = 0] $ (x,) = 0 / 0 (2,-4) = 0 > u (8,9)+6²-20 u (m-a) +6²-2² 0 »Juca, q) = x² p? Tu (26-2) = 2262 CS Scanned with CamScanner