Find the force in all the members of the truss shown by the method of joint

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 Find the force in all the members of the truss shown by the method of joint










MA = 0,

60*2-RD*4=0

=120 - RD*4 =0

RD = 30 KN

Fy = 0, RAV +30 -60 = 0

RAV = 30 KN 

fy =0

FAB sin60 = -30

FAB = -34.641 KN(C)

FX = 0

FAF +FAB cos60 =0

FAF - 34.641 cos60 =0

FAF -17.32 kn (T)

FBD of Joint F

Fx = 0 

FEF - FAF =0

FEF = 17.32 KN (T)

FBD OF JOINT B

FY = 0

-FBESIN60-FABSIN60-FFB =0

-FBESIN60-(-34.641SIN60)-0=0

-FBE SIN60=-34.641SIN60

FBE = 34.641KN(T)

FX = 0

FBC + FBE COS60 - FAB COS60=0

FBC + 34.641 COS 60+34.641 COS60=0

FBC+34.641=0

FBC = -34.641 KN(C)

DUE TO SYMMETRY OF STRUCTURE FCD = FAB

FGD=FAF

FGE=FEF

FCG=FFB

FCE=FBE

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