Determine the force in each member of the Howe roof truss shown. Take P1 = P2 =

Question

Determine the force in each member of the Howe roof truss shown. Take P1 = P2 = P3 = 720 lb. (Round the final answers to their nearest whole number.) P2 P1 P3 300 lb 300 lb 6 ft IF 6 ft s ft s ft 8 ft s ft lb. (Compression) lb. (Tension) The force in member AB (AB) IS The force in member AC (FAC) is The force in member BC (Fac) is 0 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 (FoF) is The force in member DE (FDE) is The force in member EF (FEF) 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 lb. (Tension) lb. (Compression) lb. (Compression) lb. (Compression) lb. (Tension) lb. (Compression) lb. (Tension) lb. (Compression) lb. (Tension)

Explanation / Answer

Since the truss is symmetrically loaded and also symmetric in geometry, the reaction forces will be symmetric and members in symmetry will habe equal forces

Vertical reaction at each support=(300+300+720+720+720)/2=1380 lb

Determination of member forces :

Joint A

Angle of AB with horizontal=arctan(6/8)=36.87 degrees

Fab*sin36.87+1380-300=0

Fab=-1800 lb=1800 lb(compression)

Fab*cos36.87+Fac=0

Fac=1440 lb(tension)

Joint C

Fbc=0

Fce=Fac=1440 lb(tension)

Joint B

Fbe*cos36.87 + Fbd*cos36.87+1800*cos36.87=0

Or, Fbe+Fbd=-1800 ...(i)

-Fbe*sin36.87 + Fbd*sin36.87 + 1800sin36.87-720=0

Or, -Fbe+Fbd=-600 ...(ii)

Solving above two equations,we get

Fbe=-600 lb=600 lb(compression)

Fbd=-1200 lb=1200 lb(compression)

Joint D

-720+(1200*sin36.87)+(1200*sin36.87)-Fde=0

Fde=720 lb(tension)

Due to symmetry, Fab=Ffh,Fbd=Fdf,Fbc=Fgf,Fbe=Fbf,Fac=Fgh,and Fce=Feg

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