Question

Solve by Flexibility-Method Thank You,
Given:

Por - ש (6L2 24EI 4Lx + x) Ро Por 6EI (3Ľ2 – 3Lx + x2) Pol4 Pol 8B OB SEI 6EI v(x = 1) 172(P.)L4 1875EI Pr? P - ע (3a - x) 6EI Pa? (3x – a) 6EI Pa? (3L - a) 6EI Px (2a - x) 0 Sisa 2ΕΙ Pa? a Srl 2E1 V dB өв Pa? 2EI Pa3 v(x = a) ЗЕІ

Answer 1

at that Principal of virtual Work : applying unit load point & de flection A A- smmudu manents due to Houd and uwit load respectively, Where m&mare ЕТ esternal

FBD let Rg as a redundant reaction VIKN u Knim с ( peaming A gay d MA B↑ RB ht For CB 0<x<2m for BA m=urnir =222 me 0 m= 480.32 - RBX +4x2 (+1) M= 242 +8818 -RBU m= line o<uc8m AB= 0 = 5x2+2x+8–R8u) udbe d EI 8 sroom + 5% extent (2x V +82 +84? –R$ 25) -2 yg +8 #ę} +8 xq? – Reveg? =0 2 3 O = 2 X84

RB 215 KN [Fy=0 Ay UX10 + 2115 Ay = 18.5 kn -MA tuX1085 2115X8 = 0 EMA=o MA= 28 KN-m Shear 18.5 kn SKN + A B 2 135KN for a 18.5-42=0 ->241625 KN Moment M = -28+18.5X41625 – 4X4162,5 m= 14.78 KN-m + 28 kn-m 4*2*1 = 8 KN-