Question

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Determine the force in each member of the Howe roof truss shown. Take P1 = P2 = P3 = 600 lb. (Round the final answers to their nearest whole number.) P2. P1 P3 300 lb 300 lb 6 ft 6 ft E 8f-8f8f8ft l b. (Compression) b. (Tension) The force in member AB (FAB) is The force in member AC(Fad is The force in member BC (Fed is O. The force in member CE(Fce is The force in member BD (FBD) is The force in member BE (FBE is The force in member DF(FDA is The force in member DEFDE is The force in member EF (FEA is The force in member EG (FEG) is The force in member FG (FFG) is 0. The force in member FH (FFH) is The force in member GH (FGH) is C b. (Tension) b. (Compression) Ib. (Compression) Ib. (Compression) Ib. (Tension) b. (Compression) Ib. (Tension) b b. (Compression) . (Tension)

Answer 1

Colution - FBD of Given truss Pe 1 2003h Hy → The above diagram is FED Trues. Take Ety=0 thx = 0 As per RA = Hy - Total load) symetry. B Rn - t ( 300 + P + Pg + B + 300) 1 ( 300+ $00 + 600 + 600+ soo) B (2.000) I RA - 120016) y = 1001 from fig we can write Take Joint A - 3003 - FAC The 900 lb Coro EAS Fnc . | FAC - 900x5- 150016/

90oxy - 1200db Joint C - Fec .: BC is a 200o-force member. Foc=0 . FCE = EAC IFCE = 120006 Take Toint B Gool FBD FARV EBE EF = 0; u FBot u FBC + 4 2 1150016)=0 FBO IFBE = - 15001b & Fy=0 5 FBD - 3 Fee +3 (1500lb) - 600 lb=0

FRD - FBE ==50016 - Adding on ca ① and ② weger FBD + FRE - 150016 FBD -/EBE -- 500lb QFBD = -2000 lb FR = - 1000 lb on substituting Foo in eq ③ we get - 1000 lb-Fee = -500lb BE = -50 olb Take Joint D : 600 DX EFx=0 i For DE 4 (rood (b) + 4 / (FOF) = 0 For= -1000db } (10000) - ( -1000 lb) - 600 db - Foto FDE = 600 lb

& gire FBD Thun member of the True And we know that the is symetric so from we can write that FEF = FBE = 500 lb. FEG = Fee = 12000b * = FARE 0 FGH = FAC = 12000 FFH = FAB = 1500db FFG = FBC = 0 The direction of the given truss is Tension and compression Reprasenting shown below forces in as per - 600 300 300 #X-> Ha A