Question

90 kN/m 165 KN 120 KN.m ܠܠܠܠ 2m 3m 1m 1m А в 2m С DE F

which among the choices almost give the reaction at the fix support? use any method

A. 8.312
B. 52.660
C. 255.653
D. 85.427

Answer 1

1. EMC F 2 =G 2 Ey 120 8 By + 2 + ME +16589 +[(4 83x90) *4] 8 By + 2 Ey = 2145 Eq.(1) + ME = LEY By + Ey + Fy 165 +( *3*90) • 6 By 300 + Ey & Fy E9.12) singularity Function - Moment eq EI d²Y Fy (re-oy ME <x-oy ol 2 da & Ey <x-2)' + By <x-8) Il +165 (se - 97' 120 <3 1) + 120<x-67 ol 3 -67 3 31 To Tx + go <x- 3 3 2 + 90<x-3) 21 3)

Integrate 2 EI dy - y <x) 2) ME <x>' + Ey <x-27 !!! dx 2) + By <*-»>" +120 64-6 2) G + 165 <x-gy 21 go <x-67 3 4) +90 <x-37 3 41 + go <x-37 3) CH At x =0 slope dy da from above equation c = 0 once Integrate again EIY Fy <xy3 - METRY + Ey ME<xy + Ey <2-27 = 31 31 3 + 120 x-672 +165(x-g7 3 + By <x-87 3:) 21 3) - go <.2-675 3 57 + go <x-3) tgo <x-3) 3 5) t cx + C2 zero Eq. (1)

x=0 At No downward deflection from Fq? ci) C2 Also, At x = 2 m ; y=0 Eq? co Becomes EI (o) Fy (2) - MF (2) 2 2) 3) 1 - 3333 Fy 2 ME .. Eq.(3) x = 8 m t Also, At Eq cid becomes EICO) Fy (8)" - MF (8) ²+ Ey (6) 3) 2) 3) 3 3) 31 + By Co + 120 (2)2 +165(8-9) 2) Negative.-cancel 2)5 +90 (5) 51 4) - 90 3 tgo (6) 4. 3 5! OC

je; o = 85.3333 Fy 32 ME +36EY + 240 8 + 781.25 +2343.75 j.e. 85.3333 Fy -32 MF +36 Ey = - 3357 Eq. (4) We have 4 Eq. (I to 4) Voda Unknowns 4 Solve We get Ey - 244 3451 KN -ve Fy 244.3451 KN close to option ME - 162.8927 KNor We have assumed 2 ME 162.8927 G