Member Force in a Triangular Truss
Mediumtrussesmethod of joints
A triangular truss has joints A (pin support, left), C (roller support, right) and B (apex). The span from A to C is 8 m, B is directly above midspan, and the apex height is 3 m. A 10 kN vertical load acts downward at B. Using the method of joints, the force in member AB is most nearly:
- A8.3 kN compression
- B8.3 kN tension
- C6.7 kN tension
- D5.0 kN compression
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Answer: A. 8.3 kN compression
By symmetry each reaction is 10/2 = 5 kN upward. Member AB runs 4 m horizontally and 3 m vertically, so its length is 5 m, sin θ = 3/5 = 0.6 and cos θ = 4/5 = 0.8. At joint A, ΣFy = 0 gives 5 + FAB (0.6) = 0, so FAB = −8.33 kN: the negative sign means compression, which is what you expect for the sloping top chord of a loaded triangle. Then ΣFx = 0 gives FAC = 8.33 × 0.8 = 6.67 kN tension in the bottom chord. Answering 5.0 kN uses the reaction as if it were the member force and ignores the member's slope.
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