In the symmetrical truss shown (bottom chord A–C–B, apex D, vertical CD, base angle 30°, pin at A, roller at B, 3 t load at D), the force in member BC is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 24/01/2018 (Evening)
Detailed Explanation
The truss and load are symmetrical, so each support takes half of the 3 t load. CD is a zero-force member (two collinear members AC and CB meet at the unloaded joint C).
\(R_A = R_B = 1.5\) t.
Joint B: \(F_{BD}\sin30^\circ = 1.5 \Rightarrow F_{BD} = 3\) t (compression).
\(F_{BC} = F_{BD}\cos30^\circ = 3 \times \tfrac{\sqrt3}{2} = \tfrac{3\sqrt3}{2} \approx 2.6\) t (tension).
Why “(3√3)/2 t tension” Is Correct
The rafter BD pushes outward on joint B, and the tie BC pulls it back, so BC is in tension with force 3√3/2 t — option (c).
Why the Other Options Are Wrong
- A. 3.0 t compression: 3 t is the rafter force (BD), and BC is not in compression.
- B. 3.0 t tension: 3 t is the rafter force, not the tie force.
- D. (3√3)/2 t compression: Right magnitude but wrong sense: the tie is in tension.
Key Point
King-post type truss: rafters in compression, tie in tension; tie force = rafter force × cos θ.
Key Concepts for Students
- With a 30° pitch, rafter force = R/sin30° = 2R.
- The central vertical carries no force when there is no load at C.
- Lami's theorem at D also gives BD = AD = 3 t.
Practice more: this is question 15 in Beams and Trusses MCQs – Part 1. Try the whole set and check every answer instantly.