In the truss shown (bottom chord A–C–E with AC = CE = 4 m, top chord B–D, end members inclined at 60°, pin at A, roller at E; loads 10 t at B and 5 t at D), the force in member AC is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 24/01/2018 (Evening)
Detailed Explanation
First find the reactions, then go to joint A, where only two members (AB and AC) meet.
B is 2 m from A and D is 6 m from A (60° end members, height \(2\sqrt3\) m).
\(\Sigma M_A = 0:\ R_E \times 8 = 10 \times 2 + 5 \times 6 = 50 \Rightarrow R_E = 6.25\) t, so \(R_A = 15 - 6.25 = 8.75\) t.
Joint A: \(F_{AB}\sin 60^\circ = 8.75 \Rightarrow F_{AB} = \dfrac{17.5}{\sqrt3}\) t (compression).
\(\Sigma F_x = 0:\ F_{AC} = F_{AB}\cos60^\circ = \dfrac{17.5}{\sqrt3} \times \tfrac12 = \dfrac{8.75}{\sqrt3} \approx 5.05\) t (tension).
Why “(8.75/√3) t tensile” Is Correct
The inclined member AB pushes the joint outward, and the bottom chord AC must pull back, so AC is in tension with force 8.75/√3 t — option (c).
Why the Other Options Are Wrong
- A. 6.25 t compressive: 6.25 t is the reaction at E, not a member force.
- B. 8.75 t tensile: 8.75 t is the reaction at A; the member force is 8.75/√3.
- D. (8.75/√3) t compressive: Right magnitude, but the bottom chord is in tension.
Key Point
Find reactions → go to the support joint → resolve vertically for the inclined member, horizontally for the chord.
Key Concepts for Students
- Bottom chords of simply supported trusses are usually in tension.
- End rafters (inclined members) at supports are in compression.
- tan 60° = √3 gives the height 2√3 m for a 2 m half-panel.
Practice more: this is question 14 in Beams and Trusses MCQs – Part 1. Try the whole set and check every answer instantly.