Member Force in a Symmetric Pin Joint
Mediumtrussesmethod of joints
A symmetric pin joint is supported by two ideal two-force members, each at 45° to the horizontal, and carries a 20 kN downward load. What is the axial force magnitude in each member?
- A7.07 kN
- B10 kN
- C14.14 kN
- D20 kN
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Answer: C. 14.14 kN
The horizontal force components cancel. Vertical equilibrium gives 2F sin 45° = 20, so F = 20/(2 × 0.7071) = 14.14 kN. Whether the members are in tension or compression depends on their arrangement.
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