Question

4 m 4 m 4 m EY EX E F G H 2.5 m D A B С 13 KN 3 kN 3 kN

find all the forces

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Answer #1

Joint E TEY=624 tan 2.5 E FEF Ex=0 0 = 32 FBE FAE = 3kN Efy FBE sino + 3 = 6 EBE 5.6612 KN I Ź Fx =o FEE = FBE Cosa F EF 4.80Toint FEF=4.810/0 kn FFG FAF FBF * FFC =3kN = Eff 3 = FFC sino FFC = 5.6612 KN Efx , 7 (5.6612 Cos 0) + 4.8010 FFCe FFG = 9.6a Joint (9.602 kN) FFG FGH Q Fad Trace FBG = 3 kN = 5.6612 KN ŹFY FAD 5.6612 sino + 3 FCD zero force member E Fx Fut +5.6612

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