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Find the axial forces of the members 2-3, 9-3 of the truss for the given loads.
Solution
From the equilibrium of the forces on the truss we have:
ΣFx = 0 -> H1 = 0
ΣM1 = 0 -> 200 * 4 - B5 * 8 = 0 -> B5 = 100KN
ΣFy = 0 -> B1 + B5 = 200KN -> B1 = 100KN
We consider the cross section T so as to calculate the axial forces on member 2-3 and 3-9. We consider the equilibrium of moments to the node 9.
ΣM9 = 0 -> 2 * 100 - S2-3 * 2 = 0 -> S2-3 = 100KN
From the quilibrium of forces at Y direction
ΣFy = 0 -> -S9-3 * sin45 + 100 = 0 -> S9-3 = 100 / sin45 KN
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