Question

Question 4 10 poin Calculate the force in member AB by the method of joints in kN. [ Compression: negative. Tension: positive] E 4 meters F 4 meters G 2 meters H →30 kN 3 meters U 3 meters A 2 meters 4 meters 4 meters 30 kN 30 KN 60 kN

Calculate the force in member FG,EF and FC by the method of joints in kN

Answer 1

let all members are in Tension. H. Now: free body diagram of point For Equilibrium of Point H. E 30 kn {fx=0 and {fyro. FA FHD .. {fx=0; fra 30 KN {fy =o. FHD = 0 free body diagram of Point D. on fon 8 = fan '[] an '[] = 33.69 fbc грн = 0 6 مہ GORN

D. for Equilibrium of Point Point D. Efn=0; Efyo. at Eryso; For Cos 33.69° = 60 KN fon = 72.11 KN and foc + fou sin 33.69"=0 Fc = - 40 KN Now free body diagram of point for > FGH = 30kn 56.31 52.319 for Equilibrium of Point 4. {fy = 0 For Sin 56.31 + Fuck fac Ford = 72.11 kN Sim 56.310 =0. Fuc - - 72.11 KN Efn=0; Fon & Foo Cosst+31° - fuc Cos 56031 - Fup to 30+ 40+40 - fuf fuf - 110 Tension.

FBD of point Doint. C. For fon = - 72.11 KN Sty=0); stih 56.31 C For - -40KN FCB E Sid 56.31 it for sin 56.31-30=0 30 KN OF [for 108. 165 KN Tension FCB + Fof COS 56.31 fon Cos 56.31 - Foot =0 -) Fcp + 108.16 x Co566-3) - (-72.11 Cos 36-31) - (-40)=0 fee 140 kn Now FBD of point F- F FREE Fee Sin Sb-31 + foc sims 2.31'= 0. {ty=0; 56.310 56.31 ffy = 110kn Fre=108.105 ffo = -108. 165 KN FFB {fx=0, FFE tff W$ 56.31 - for afre Cos 56.31*20 [free gension. = 220 IN

Force in member FG = 110 KN tensile

Force in member EF = 220 KN tensile

Force in member FC = 108.165 KN tensile