Consider the diagram below
∑MB=0⟹(10∗2cos60)+(10∗(2+2cos60))=2RA⟹RA=20kN↑
∑Fy=0⟹RA+RB=10+10⟹RA+RB=20⟹RB=20−20=0
Consider point C
FBCsin60=10⟹FBC=11.55kN
∑Fx=0⟹FBCcos60=FDC⟹FDC=11.55cos60=5.78kN
∑Fy=0⟹FADsin60=20⟹FAD=23.09kN
∑Fx=0⟹FAB=FADcos60⟹FAB=11.55kN
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