Question

lella ------ loint Coordinates Label = 3/9-- - MOO-> olololo Yin BOX

calculate all support reactions and internal forces of each member by hand and show working.

10lb

Answer 1

2016 lb GI E 14 D I F T prolb 2-625. 5-25 L 10.5 21 sign Convention towards the point > compression (C) - ve away from the junt - tension (T) tve * Solving by method of joints Jemnt S o Joint A os tent (1-142) 3 (Evo) FAGsin 48.81 = 10 FAG - 13.2885(0) FAGGOs 48.81 - FAGI (FAG, = 8.75(T) / (EH-O

Joint E 67 61 A 41.190 (Sv=0) FGE S?n41-19 = FEHSEN 41-19 FGE SER FGE= Fen fen= 0.00666) FEG + FEGC0341-19 = FEH, & FEHC0541-19 FEG, +0.005x2 - Feni 13.295 (FEM = 13.305 (T Joint R, P 13-295 - 49-919 (Eve) FEIGH Los 41-19 = F G H 0 348.81 13.295x6541-19 - Fall Los 48.81 B FEH = 15-19 (6) FGG sin 41 19 = FEco- - FE, H SP148.81 FE, C = 20-1861 €4=0)

Joent G FG, G=0 FAGI = FGIE = 8.75(T) re Joint G 41-19° 41-199 FAG > 13.2885 (Evo) FAG Sim 48.81 = FEG Los 48.818 + FEGC0341-19° evt 10 = FEGr 06586 + 071525FGE, O EUR) FAG 10148.81 - FEG Cos4-01 + EE G Sin 41-19° - 8.751 = 0.7525 FEG + 0.6586 FGG On solving & ③ FEG= 0.0066 (T) FGG = -13.295 an Since obtained -ve, assumed direction is opposte so fag 13- 29517)

Joint Hi Iso FEM 213-305 - FODH, (T) / | Ful, o ont H Joint H 15-19 H 49.41 4-190 / 48.86 4 80 0-00 (Ev=0) FELSO 48.81 t For Sin 41-19 = FcH sin 48-81 + FEH sin 41.19 0.005 + 0.6586 Fon = FcH XO-1595 + 10 (54-0) 0.658650 - 0.7525 FH -9.995=0 FEH C0548.81 +FGHCO 34:19 = For cos 41-19 thn 105 48 81 0.00435 + 11.43 = For X0-1525 + fer x 0-65867 NO 0.7525 fon + 0.6586 FM - 11.43435=0 On solving, Four - 15.186 Fcn = -0.0094

So assinned direction is wrong. (for a 15.186 (T) FH = 0.0094(T) 20 Toint c Joint a | Foga 2010 Now since the trusses symmetric with respect to CD, all those member forces to the left of CD holds good for right of CD members also.