Question

AB length is 4000mm

AD length is 3000mm

A pin jointed frame ABCD is supported by a pinned support at A, a roller at B and is subjected to the loading indicated in Figure Q1. All members have circular cross-section and all are made of steel materials with same cross-section. Determine the support reactions at A and B Determine all the member forces Find out the horizontal displacement at point C using a table template as shown in Table Q1. Total (15) marks. D 40 KN I 3000 mm B 4000 mm F 1 Figure Q1: Simply supported pin-jointed frame E-210 kN/mm² : A = 300 mm for all members. Table Q1: Unit load template Member Length L Cross-section Modulus of Elasticity Member force for Member force for FFLE (mm) A (mm') (N/mm') real system P (kN) virtual systemu (N) EA Imm) AB BC AD BO CD EA

Answer 1

Hokn V 3 sm BD=134+42 3m B A 4 -5m B HA um Te f Re . RA take Tension as positive compression as negative لا P-Force system & Fu=0 HA YOKN EMA = 0 yox 3 - Roxy=0 RB = 30kn SFyzo RAT RB=0 30KW { negative means downward din so i joint a FLD FID = 40kn) [rension] FBC =0 Tree į FBC Fcd = 40 ni joint D haber Food FBD FAP 8=tant - 368199 Coso=0.8 g sino=o.b Fcd _FBD Soso =0 40- FB DX og ABD = 50KN [ compressive] so put negative T FBD sino = AAD FADZ SOXO 6 FAD= 30kn ] [ Tension] (iin joint A Efr=0 FAB = YOKN [tension ] FAB HA=40 RA=30 a Force system D apply unit load atc sm Bm ΣΥΝτο HA B HA ym 2fү - RA= RB RA IMARO RB X Y - 1+3=0 RB-075 PA

vointa Erzo FLA= | KN Feo {tension] T IFBC Eryzo FBC → FD = 1 Jointo sinosol coso -0.8 FBD EF=0 I - FBD Oso FBP = 1.25 kn] [compressive so negative] Efy FAD = FBD sino FADE 125xoof FAD=075 kn) [tension] iii) Joint A FAD = 0.75 EF=0 FAB= IKH FAB HA=1 (tension) RA=0.75 member Length (L) canarea Elka) u(kr) P(KN) mm A (mm2) mm PYL AE (mm) Youo AB 300 210 yo 2.53968 BL 210 o 3 VO 300 300 AD 20 30 0.75 3 ovo 1:0714 300 210 --50 -125 BA 4,96 03 Sovo 300 qoro 210 yo 2:53968 žpul AE sol = 1.llimm De= llell mm (in direction of unit load] .