Member Force in a Two-Member Truss
Easytrussmethod of joints
A symmetric two-member truss carries a 20 kN vertical load at its apex. Each member is inclined 30° above the horizontal and is pinned at its base. The force in each member is most nearly:
- A10.0 kN
- B11.5 kN
- C40.0 kN
- D20.0 kN
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Answer: D. 20.0 kN
Equation: vertical equilibrium at the apex, 2F sin θ = P. Values: P = 20 kN, θ = 30° (each member above the horizontal). Step 1: sin 30° = 0.5. Step 2: F = 20/(2 × 0.5) = 20 kN, in compression. 10 kN is P/2 without the angle; 11.5 kN uses cos 30°; 40 kN divides by sin 30° twice.
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