A simply supported beam carries a couple somewhere in the span. It would produce: 1. a rectangular SF diagram; 2. parabolic BM diagrams; 3. both +ve and −ve BM, which are maximum at the point of application of the couple. Of these statements:

Asked in SSC JE 2008

Detailed Explanation

A couple gives reactions M/L up and down, so the shear is constant. With no transverse load on the span, the BMD is made of straight lines; the couple makes it jump by M at its point of application.

\( R_A = -\dfrac{M}{L},\ R_B = +\dfrac{M}{L} \Rightarrow V = -\dfrac{M}{L} \) everywhere (rectangle).

Just left of the couple: \( -\dfrac{Ma}{L} \); just right: \( -\dfrac{Ma}{L} + M = +\dfrac{Mb}{L} \).

Why “1 and 3 are correct” Is Correct

Statement 1 is true (rectangular SFD). Statement 2 is false — the BM lines are straight, not parabolic. Statement 3 is true — the BM is negative on one side and positive on the other, both largest at the couple. So 1 and 3 — option (d).

Simply supported beam with a clockwise couple M at distance a from A. The reactions form an equal and opposite couple: M/L down at A, M/L up at B. So the SFD is a rectangle of height M/L over the whole span. The BMD is two straight lines: −Ma/L just left of the couple and +Mb/L just right of it, jumping by M at the couple and passing through zero there (point of contraflexure).

Why the Other Options Are Wrong

A. 1, 2 and 3 are correct: Statement 2 is false: there is no distributed load, so the BMD is linear.
B. 1 and 2 are correct: Statement 2 is false.
C. 2 and 3 are correct: Statement 2 is false, and statement 1 is true.

Key Point

Couple on SS beam: rectangular SFD (M/L), linear BMD with a jump M at the couple.

Key Concepts for Students

Parabolic BMD needs a UDL.
The point of application of the couple is also a point of contraflexure.
If the couple acts at a support, the BMD is a single triangle.

Practice more: this is question 51 in Bending Moment and Shear Force MCQs – Part 4. Try the whole set and check every answer instantly.