Question

The simply supported beam has length L, elasticity modulus E, and cross-section with moment of inertia I. A concentrated force is applied at half point, as illustrated below 1/2 1/2 o The deflection curve for the the first half of the beam is given by: 21 (2) = + (- +) Obtain the equation for the deflection curve y(x) for L/2 < x < L, where: y2(x) = (Ao + A1 x + A2 x2 + A3 x3) When solving this problem symbolically, you will notice that the coefficients above depend only on the length of the beam L. Determine the remaining coefficients when L = 8 m. y2(x) = Fil numbe + numbe æ+ numbe x2 +-0.08333 x3) [m]

Answer 1

42- Ey free body diagram Ay = Reaction force 12-- at A Ayxx Eye Reachon force ato take moment C! - Ay XL = PX 1 / [Ay = 9/2] Bending moment equation as a function of ze (o measured from A) Mx = Ay x |- P(x-4) Me = fx ) - P(x-4) include of a742 include ef 27th from Er det, - Me El d2y = 5 x I include ef azt on entegration, ald |_ P(x-7) ² 2 include ef x > 1

on integrating xt6 a ? ? ? ? + x + b, - P(x -50° enclude fazt for first hall :x=0 yzo deflection at A=0 E160)= p code + GX0+62 >[(250) at x=2 yzo deflection at Ca=0 EI (0)= {x (432 +0,XL - P (4) 3 C = -- PL² hence deflection equation és - enclude yur EI L 12 [ 3 - P1 x 1 x 2 ) for fint hofle y cws for 19 - Homex] for second half 42<x<L You) = f ( tamata :2x - 4673307 - [ - Lave -[ 23 - 2x2L+&223 1123 |2x - -L 16 L22-I mit

5,00- [- S - | Js(x) = P. [ As + H x + Agx² + A,3 ] 야롭로 뺨 3일 22 for 2=8m | Aos = 1066667 , A, 0x82 = -12 | A2 - 2 -2 As -J = -008 33733 | 95 (3) [ 1966667+ / I x + [3] x2 +0.8 33277