Question

Solve equation (4) in Section 5.2 ΕΙ = w(x) (4) dx4 subject to the appropriate boundary conditions. The beam is of length L, and wo is a constant. (a) The beam is embedded at its left end and free at its right end, and w(x) = Wo, 0<x<L. Wo y(x) = - 2Lx + 3 L’x) 24EI x (b) Use a graphing utility to graph the deflection curve when wo = 24EI and L = 1. X X 0.2 0.4 0.6 0.8 0 0.2 0.4 0.6 0,8 1.0 1 1 .2 2 031 y X 02 0.4 0.6 0,8 1.0 0.2 0.4 0.6 0.8 1.0 2 .2 31 34

Answer 1

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wo EI 2 da3 EL ولے Wo + EI wo xt +C, x3 + C2 x ₂ +(3x + C4 EI 24 EI dty zw(2) (4-) dat > day W(x) dut ΕΙ Now w(a)= Wo UCULL we have day 1 dnt integrating both sides with respect tox, continuously, wo atd x² +cix+C2 da EI x3 cix² + (₂x + da ayin Now y(o)=0 » cy=0 y' lol bow dany conditime: "(L)=y"(L)0 uning yo) = 0 =C3=0 t y" (L)=0 -> C2 + 6LX 2 EI y!''(L)20 3 6 x + To EI C1- WO L EI 2 so C2 Wol? Z EI howe have cq = -Wo L, C = 1/2 (3=0,4=0 EI putting trere values in (5), we get wol? a? you = Amwar (.: beam is free at its right end) wo 13 L - O I wol? EI! + wo 24 2 <4 4 n Wo L x 3 EI 24 6 EL 4 EI 319x)= wo - 4603 +622x²) 24 EI (b) Wo = 24 EI 41=1 = y(n) = 6x2403+ .6.81 con Curing 8) . 2 .4 y = x4 403 +642 3 in correct. 1 so optio 2 CSscanned hvith CamScanner.