Topic: Engineering Mechanics
A two member truss ABC is configured as shown in figure. The force in the member AB is
- A. 15 kN
- B. 30 kN
- C. 20 kN
- D. 5 kN
Answer: C. 20 kN
Explanation: From the truss geometry, member AB is horizontal (length 1m) and member AC acts as a diagonal. The height at support A and C is 0.5m. Let $\theta$ be the angle of member BC with the horizontal; $\tan(\theta) = 0.5 / 1 = 0.5$, which means $\theta = 26.56^{\circ}$. At joint B, balancing vertical forces gives $F_{BC} \sin(26.56^{\circ}) + 10 = 0$, so $F_{BC} = -10 / 0.447 = 22.37$ kN (Compression). Balancing horizontal forces gives $F_{AB} + F_{BC} \cos(26.56^{\circ}) = 0$. Substituting $F_{BC}$, $F_{AB} - 22.37 \times 0.894 = 0$, leading to $F_{AB} \approx 20$ kN (Tension).