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Determine the force in members BE, BJ, DJ, and IJ. (ADJ and HDB are not continuous, cach consist of two elements.) 1 m 2N 3 k

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FB R OG AR im I IM IMI IM im im Ilv Fy T w 2KN 3KN Skin SKN m 2kN. 6= tant (+) = 45° MA= o(at) - (Ly x6) + (2x4) + 5x3 + 3x2consider point k:- Efx=0 -RJK + 5 16668 =0 KE RJ K = 5.16668 KNCG) Etyo REKTO RJK 5.16668KN RBE Consider point E:- Efy=0 RBEXconsides section mom indig: - М1 го - FAB x2 - Lyx2=0 FAB = -ly FAB = -68333KN & consider point Bi- & fx=0 6.8333 - RBD x Cos

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