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Tension in bracket-truss.

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Half of mechanics is about balancing forces. Some trusses can be treated like pin-jointed structures and decomposed into axial forces, i.e., without any bending or shear forces. The solution is therefore obtained through trigonometry.

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Balancing vertical forces at B: $P_{BC}\\sin(\\var{theta})+\\var{force}=0$,

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where units are kN. Rearranging:

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$P_{BC}=-\\var{force}/\\sin(\\var{theta}) = \\var{siground(Pbc,3)}$ [units: kN].

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Angle to vertical of Bar AB.

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Applied vertical (downward) force.

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Tension in Bar BC.

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A pin-jointed truss is shown in the figure below. The pivots at A and C are fixed.

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\"Pin-jointed

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If the applied force, $F$, is $\\var{force}$ kN vertically down, and the angle between Bar AB and Bar BC, $\\theta$, is $\\var{theta}^\\circ$, what is the tension in Bar BC?

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[[0]] [Units: kN]

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