Question

6. Using both the method of sections and the method of joints, calculate all the bar forces in the truss below and indicate whether the bar is in tension or compression. 100 KN

Answer 1

sol Kumum 100Kn Tano = 2 >5000s to Coso = 1/2 1 Using Method of Joints : Toint - D' 100 m ✓ FDE EFR=0 a Fept fog lodo = 0 - 0 E fu=0 FED Sino +100=0 => FED= -10045 = -23366N · From 0, F10 :-223.6 C09O . : 2236x 2 [c]. = 200 KN (I) Joint - E- FEC E E A FED FED & Fy=0 AFECE O FED = FFE = -223.6 kN [c ] FFE

Fus Toint - cir ' Fuqe Fers = 200 kN (T EFX=O ... FIF FCE FCB + FCF cosusº + Fo=D * FLF Cosu5° = 0 aff=0 > EFy=0 FLE=0 KN Joint -Bi- Efq=0 . . FBAL FBC FBF = 0 > EFx=0 FBA = Foc e) FBA = 200kN [+] Joint -F:- FFB Fc FFE For ao FFA A A FBF=0 & FBar =0 * Far = FFE = -223.6kN [c] Joint-Gi- . Fatafuf "EFU=0 → FaA =-Fur sino --1-223.6) es) = IDOKN (T}

Efrio Ra == far loso = + (423-6) (5) = 200 K . Hoint -A- Efr= → FAB FAG (Ralu = FAB= 200KN (RA)v = FAG = 100kn Using Method of sections - FED D , b = 160 Sino 100KN IDO KR =) FED = -100 Vs IN (c) FED=_FED coso = 4(1075) tons) 7200 KN (T ) Bro Foots * YokN , LE W . EFx=0 . . FER CODE = -200 offp = -200 x 15 =-100 Vs [c]

EFq=0 Fer Sino + 100 = Fee * (_10015)(ts) + 100 = Fee FCE = 0 KN Br Fubat Here, Fort Fee = (100 v5)coso = !00 16 ) = 200KN 07) Noco, 1005 Sing = 100+ FFs : :=) Fcp = 100v5) (+)-100 = OKN FAB в с. E&x=0 FBc = FBA = 200 KN EFyzo . ) Гру - oks Efx=0 * 200+FA+ Sinus NE = For leo : As Faroo q ra od Fap = -20065 = -223.6kN [C] FOF

(RA)v EMA=0 (100 65 Cabo ) 6 = RG (6) | » Ra = (po 150s). - 200KN » AF TO *10085 Ra - 200KN I (RA) = 100kn 100KN Ray = 200 kn Efy=0 > Fag = - Far Sing +(+2006)(*) EN = 100 KN (T)