Question

4. Determine the shape of the deflection curve of a uniform horizontal beam of length 2 and weight per unit length w that fixed at both ends r 0, and x 2.

Answer 1

Given fixed end moment for udl M_AB = -w l^2/12 M_BA = w l^2/12 Due to symmetry R_A = R_B = wl/2 from deflection curve equation M/EI = d^2y/dx^2 M_x = wl/2 x - wl^2/12 - wx^2/2 d^2y/dx^2 = 1/EI [wlx/2 - wl^2/12 - wx^2/2] dy/dx = 1/I [wlx^2/4 - wl^2 x/12 - wx^3/6 + c_1] (1) y = 1/EI [wlx^3/12 - wl^2x^2/24 - wx^4/24 + 4x + c_2] (2) at fixed end (because x = 0) implies dy/dx = 0, y = 0 (1) implies at x = 0 (2) implies x = 0 0 = c_1 0 = c_2 therefore y = 1/EI [wlx^3/12 - wl^2x^2/24 - wx^4/24] Therefore @ x = 0, y = 0 @ x = l/z \r\n y = 1/EI [wl/12 (l/3)^3 - wl^2/24 (l/3)^2 - w/24 (l/3)^4] = 1/EI [wl^4/324 - wl^4/216 - wl^4/1944] = -wl^4/486 EI (because sign indicate downwards)