A lever ABDC is supported on hinges at A and C (AB = BD = DC = 1 m). A vertical arm 1 m high is fixed at B, and a horizontal force of 3 kN acts at its top. The bending moment at B is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 03/03/2017 (Morning)
Detailed Explanation
A horizontal force acting at the top of an arm attached to a beam does two things at the joint: it pushes the beam axially and it applies a couple equal to force × arm height. That couple is what bends the beam.
Moment at joint B from the arm: \( M_B = 3 \times 1 = 3\ \text{kN·m} \)
Reactions (vertical): \( R_C \times 3 = 3 \Rightarrow R_C = 1\ \text{kN} \), \( R_A = 1\ \text{kN} \) (opposite direction).
Beam moment just left of B: \( 1 \times 1 = 1\ \text{kN·m} \); just right of B: \( 1 \times 2 = 2\ \text{kN·m} \). The jump \( 1 + 2 = 3\ \text{kN·m} \) equals the applied couple.
Why “3 kN·m” Is Correct
The bending moment at B — the moment delivered by the arm at the joint — is 3 kN·m, the book's answer (a). Note that it splits between the two sides of the beam: 1 kN·m on the AB side and 2 kN·m on the BC side. If a question asks for the moment just to one side of B, the answer would be 1 or 2 kN·m instead.
Why the Other Options Are Wrong
- B. 2 kN·m: 2 kN·m is the beam moment just to the right of B only.
- C. 1 kN·m: 1 kN·m is the beam moment just to the left of B only.
- D. None of the above: The joint moment, 3 kN·m, is listed.
Key Point
Horizontal force F on an arm of height h ⇒ couple F·h applied to the beam at the joint.
Key Concepts for Students
- An applied couple makes the BMD jump by the value of the couple.
- Reactions from a pure couple are equal and opposite: M/span.
- Hinges also take the 3 kN horizontally, as axial force in the lever.
Practice more: this is question 49 in Bending Moment and Shear Force MCQs – Part 4. Try the whole set and check every answer instantly.