A simply supported beam of span L carries a clockwise couple M at mid-span. The shear force at one-fourth of the span from the left support is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 10/10/2023 (5 pm–7 pm)
Detailed Explanation
A couple has no net force, so the two reactions must be equal and opposite. Together they form a couple that balances the applied one. Because no transverse load acts on the span, the shear is the same everywhere.
Take moments about A (clockwise +):
\[ M - R_B \cdot L = 0 \Rightarrow R_B = \frac{M}{L}\ (\uparrow),\qquad R_A = -\frac{M}{L}\ (\downarrow) \]
Shear at \(x = L/4\) (forces to the left, upward +): \( V = R_A = -\dfrac{M}{L} \)
Why “M/L (−ve)” Is Correct
The only force to the left of the section is the downward reaction at A, so the shear is M/L, negative — option (d). The same value holds at every section; the position of the couple does not matter for the shear.
Why the Other Options Are Wrong
- A. M/(2L) (+ve): M/2L would arise only if one reaction carried half the couple — it does not.
- B. M/L (+ve): Correct magnitude but wrong sign: the left reaction acts downward.
- C. M/(2L) (−ve): Magnitude is wrong: the reactions are M/L, not M/2L.
Key Point
Couple on a SS beam: reactions ±M/L, shear constant = M/L throughout.
Key Concepts for Students
- An applied couple causes a jump in the BMD but no change in the SFD.
- Reversing the couple reverses the signs of both reactions.
- The point of application of the couple decides the BMD shape, not the shear value.
Practice more: this is question 9 in Bending Moment and Shear Force MCQs – Part 1. Try the whole set and check every answer instantly.