Question

Determine the force in each member of the Fink roof truss shown using Method of Joints. State whether each member is in tension or compression. Forces A= 5 kN, B= 10kN, D= 10 kN , F = 10 kN ,and G = 5kN

2.25 m 2.25 m KN A 3 m3 m3 m

Answer 1

Draw the FBD of the truss. TO KN 2.25m TO KN OKN 5KN SKN . Efx=0: Ax=0 AX 3m ć 3m E 3m 1 Since the loading is symmetric w Ay - Gy = /2 (5+10+10+10+5) = 20 KN .. Using method of Joints! SKN FAB Joint ® : VL225 > FAC FAB FAC 20KN 2-462 2.25 FAR 2463 -- FAO = 36-93 kw/(comp.) FAC - 33-75 kN (Tension) 15 KN k2.2 FAC

10KN Joint | wy? Foo 2.25 7+ S FBC FAB=36-93 KN + E Fx = 0 + 2.25 S Fan + 2-25 3.4620 2.462 (36.33) - 0.6 Foc +0.913 8 FA - - 33.75 - Efyzo: FBc + Foot 2.462 36-33 2.462 -lo = 0 -0.8 F8C + 0.4061 FBD =-5 — ☺ solving we get : FBC = -9.374 KN os Foc = 9.374 KN f F80 = -30.77 KN FB0 = 30-77 kN) (comp) Føe =3.374 kW Joint ©: { Fyzo: FACE33.75 KN Fae= 33.75kN ? FCE 415 Feo - 475(9.374) 20 → FCO = 9.374 kN) (Tension)

EFxzo Fce - 33.75 + 3/5 (9.374) + 3% (9.374) 20 Fee = 22.50 kn) (tention) Due to Symmetry! . FoE - Fco =) FDE = 9.374 KN (rention) FOF = FBO FEF = Foc → FEG - FAC FFG - FAB For = 30.77 kn (comb) FeF = 9.374 KN (comp.) FEG = 33.75 KN (Tension ) FFG = 36-83 kN (combo)