70 kN 5 m B 50 kN 50 kN 4 at 6 m = 24 m-

Using the method of joints, determine the force in each member of the roof truss shown. Take the load P = 1.4 kN. (Round the final answers to a single decimal place.) 6 m- 2.4 kN B 9 m 6 m 2.4 kN O 6 m 9 m The force in member DF (FDF) is 0. The force in member EF (FEF) is The force in member AB (FAB) is The force in member AC (FAC) is The force in member BC (FBC) is The force in member BD (FBD) is The force in member CD (FCD) is The force in member CE (FCE) is The force in member DE (FDE) is P 7.5 m kN. (Compression) kN. (Compression) KN. (Tension) kN. (Compression) kN. (Compression) kN. (Tension) kN. (Tension) kN. (Compression)

Part B As shown, the truss is loaded by the forces P₁= 207 lb and P₂ = 398 lb and has the dimension a = 3.10 ft. FAB, FBC, FBD = VAΣ vec Using the method of joints, determine FAB FBC, and FBD, the magnitude of the force in each of the members connected to joint B. Assume for your calculations that each member is in tension, and include in your response the sign of each force that you obtain by applying this assumption. Express your answers numerically in pounds to three significant figures separated by commas. ► View Available Hint(s) ? P²1₁ B A lb P₁ C

All two-force members of the loaded structure are of the same length. Member BCD is a rigid beam. Determine the forces in members BG and CG. The forces are positive if in tension, negative if in compression. A Answers: BG= CG = i i B G L L L E