Question

Solve equation (4) in Section 5.2 E = w(x) (4) 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 simply supported at its right end, and w(x) = wg. 0<x<L. y(x) = (b) Use a graphing utility to graph the deflection curve when wo = 48E1 and L = 1. y + 0.2 0.4 0.6 0.8 1.0 0.2 0.4 0.6 0.8 1 LLLLLLLLLLLLLLX 0.2 0.4 0.6 0.8 1.0 0.2 0.4 0.6 0.8 10

Answer 1

EI dx4 Wo 0L L Beam Embedded ak eft and Ainply yord at end ight Boundary conditio nb úght end At At Jeft end y" ( ET d Wo EJ dx3 C 2 EI dy dx Es y) C3 Cu C2 6 C 24

yo)=0 O Cu O C3 *(4) g 0 + C Ca WD L t 디 2. 0 Wo L CI C- 2u Ca -Wa C + Ca 2 4 + t20 Wel 96 Wol L CI -5 CI= 96 -5 2 WoL+ 2 10-8 3-

y)= a4 ET -- Wo L Wo L + Wox ye) ET Wo L Wox EI ga) = ET 2 24 When E Wo L m uEx) L udET 2 - 5 3 wi7 2 2a- 5x 3 Guragh Graph from y) 0 to {=x 2

3G.11 4G 22 7:23 AM 11.3KB/s m.wolframalpha.com/input/?i WolframAlpha plot 2x^4-5x^3+3x^2 from x=0 to x Input interpretation: 2 x4 -5x33 x plot x = 0 to 1 + Plot: 0.25 0.20 0.15 0.10 0.05 0.4 0.2 0.6 0.8 1.0 Arc length of curve: Step-by-step solution More digits II 121
answer plot b