Question

Problem 1 Determine the force in each member of the truss and state if the members are in tension or compression. (Please use Method of Joint) Given: P, = 500 lb P, = 1500 lb a=10 ft b=10 ft Problem 2

Answer 1

Pi=500l6 P2 = 15oolb toch pe loft loft & loft of - O= Tan ( 10 ) = 45 RA+ Rp = pi + P2 = 500+ 1500 - RA+ Rp=2000db Now take moment about D, Mo=o RAx30 - Pix 20-P2x10=0 30 RA - 500×20 - 15008 lo=0 - RA= 833.33 lb .: Ro=2000-833.33 - RB = 1166.67 lb take summation of horizontal - HA= 0 forces

Take Joint A, (by method of joints) FAF FAB Summation of vertical RA + FAF sin 45=0 forces FAF = - RA - 833:33 - Sinus 1178.507 lb sin us :: FAF = 1178.507 lb (compressive) summation of horizontal forces: HA+ FAB & FAF 20345=0 O+ FAB - 1178.507 cosys=0 FAB= 833.33 lb (Tensile Now take Joint B. AFBF O FAR AFBF FBC FBC Summation of vertical- forces Pi-FBF = 0 FBF = 500 lb (Tensile)

summation of horizontal forces FAB + FBc =0 FBc = - FAB = - 833.3366 FBc = 833.33 lb (compressive) Take Joint F - FFE . Summation of vertical forces: - FAF Cos 45 - Ffc cos 45 + FBf = 0 - 1178.507 C0545-ffc cosus + 500 = 0 ffc =-471.4 lb ffc = 471.4 lb (compressive) Summation of horizontal forces - FAF Sin 45 + Ffc sin 45 + FFE = 0 - 1178.507 Sin 45-471.4 sin 45 + FFE=0 FFE = 1166.66 lb (Tensile)

Take Joint C: FA AFLE et Eco 45 (Ф. Fco . summation of vertical forces: fle - Pz -ffc Sin 45 = 0 FLE - 1500 + 471.4 si 45=0 FLE= 1166.67 lb (Tensile) Summation of horizontal forces: FBc + fot ffc cos45=0 - 833.33 + Fcp -471-4 20545=0 - Fo = 1166.6% lb (Tensile ) Take Joint D: FDE Rot FD E Sin 45=0 c e a FDE = 1166.67 =-165016 FDE = 16 50 lb ( compressive) O sun 45