Find the value of shear force at one-fourth of the given span from the left support as shown in the diagram (Simply supported beam of length L with clockwise couple M applied at mid-span).
- A(M / 2L) (-ve)
- B(M / L) (+ve)
- C(M / L) (-ve)Correct
- D(M / 2L) (+ve)
Solution
Concept
A pure couple on a simply supported beam makes the two reactions equal and opposite, so shear force is constant along the whole span.
Formula
RB = M/L (up), RA = M/L (down), F = RA = -M/L
Step-by-step solution
1. Take moments about left support A. The clockwise couple M is resisted by the moment of RB: RB × L = M, so RB = M/L (upward). 2. Vertical equilibrium: RA + RB = 0, so RA = -M/L (acts downward). 3. There is no vertical load on the beam, so shear force is the same at every section. 4. At x = L/4 from the left: F = RA = M/L, acting downward on the left part. 5. With the usual sign convention (upward force on left side = +ve), F = -M/L, i.e. M/L (-ve).
Explanation
A couple does not add any vertical force. It only creates two equal and opposite reactions of value M/L. So the shear force is M/L everywhere on the beam, and with a clockwise couple it is negative.
Why the other options are wrong
A and D: M/2L is wrong because a couple does not split equally between the supports; the full value M/L acts at each support. B: The value is correct, but the sign is wrong for a clockwise couple, since the left reaction acts downward.
Shortcut
Couple only on a simply supported beam: SF = M/L, constant over the full span. Only the sign depends on the direction of the couple.
Correct answer: C) (M / L) (-ve)