Question

10tes CC Question 3 10 pts First solve member BC and GC using Method of Sections, then solve the remaining through Method of Joints. 6-6. Determine the force in each member of the platform truss and state if the members are in tension or compression. Approximate each joint as a pin. 16 KN Prob. 6-6

Answer 1

free body diagram a truss Kim 3m ) le 3m (6 см 8KN 8KN 3m IRE k am sowing for * the 4m reactions fa=tant (6) L = 18.435° o=tan' (3²) L = 36.870 Taking moment about a +5 EMA =0 RE (8)-8(8-1)-16 (4)-8(1) =0 RE = 16 KN for symmetric truss RAZRE = 16 KN using method or joints sections, cutting through Be & GC consider a section of 3FBC 7 Fund BG is ter to AC so that link isG is zero force 1. member - lfben ao For equilibrium or section {fyzo - RAT for sino =0 An RA I forc 5-16 = sino - sin (3687)

fire =-26.67 KN (compression) IfGC = 26.67 kN E Fa=0 Foreloso = FBC FBC = 26.67 Cos (36.87) (FBc = 29.333 kN (Tensions , By using joints method consider joint at A FBA FAG Efy=0 lay > Fan Coisin B6-87) - FAI3 W08 (18.435) Ko ARA =0 FAB S (18.435) - FAG sin(36-87) = 160 Zfx=0 - FAG ROS (36.87) + FAB Sin (18.435) = 0 soling equations ① & ② FAR = 22.487 KN (compressive) IFAG = 8.89 KN Tensile for symmetric loading of truss FED = FAB = 22.487 KN FER = FAG = 8.89 KN FLD = For = 29.33 KN forc = FCF = 26.67 kN (compressive) (Tensile) ( Tensile) (compressive) FBen - FFD = 0