Question

please solve it as soon as possible

Q3) For the frame shown. Determine the member end moments and draw the bending moment diagram (with value). EI - 5000 KN.m ta

Answer 1

DF of CD Goldsf 30 IN E Solue 34 20kn/h Com mo 44 conuenting 60 kn into moment of 180 knn and thus FB is neglected Ayten By 4 N. 7 fixed and morcents - Fee => -180 kan.(6003) FAB 0 2 - 26.667 lena > we? => -2004x4 fod 12 => 26.667 KNM. foc we? - 20x4x4 12 12 FCA o ICNN. FAC o INM, Distribution Pastor Stiffness of Toint az t BEL 3 BE + BEL + YE? to 5 4 3) ET + 0.75 Es + 0.8 ET => 2.55 EL Dr. of CB => E = 0.392 2.55et 0.75 EL Q.SSEL 08EC 2SS ET >>0.294 => 0.313 DF of CA =

Joint B Joint DC Joint A BC Jointo CB CA 0.392 0.313 10.294 Member DF AC FEM -180 0 0 |-26.667 26.667 o + 180 go 119.933 -26.667 Balancing 0 o FEN o go -40 - 19.6 - 15.651-14.7 o o Balancing 0 -7.826 C.OOF -7.820 tihgt 99.91-Thot LEM 0

final fixed and moments - FBC = O INN FCB = 70.4 ICNM FAC = -7.825 KNM FD 2-54.7 Kenny FDC - Olent, fCE FEC O INM. FCA - 15.65 INN N O ENN B.N.D 180 INH 40 KNM 170.4 KNM KNN 15.65 KNNY 7.825 kwn