A beam ABC is supported at A and B (span L) and has an overhang BC with a couple M applied at the free end C. The shear force at the mid-span point D of AB is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 2013
Detailed Explanation
A couple anywhere on the beam — even on the overhang — must be balanced by the two support reactions acting as a counter-couple. Between the supports, the shear equals the reaction at A.
(The book printed “2M/L” twice among the options; the fourth option is shown here as M/2L.)
\[ \Sigma M_A = 0:\ R_B L - M = 0 \Rightarrow R_B = \frac{M}{L}\ (\uparrow),\quad R_A = \frac{M}{L}\ (\downarrow) \]
Shear on AB: \( |V| = \dfrac{M}{L} \); shear on BC: 0.
Why “M/L” Is Correct
The shear at D is M/L — option (c). It is the same at every point between A and B; on the overhang it is zero.
Why the Other Options Are Wrong
- A. zero: The reactions are not zero, so the shear between supports is not zero.
- B. 2M/L: 2M/L doubles the true reaction.
- D. M/(2L): M/2L halves it.
Key Point
A couple anywhere ⇒ reactions ±M/L ⇒ |V| = M/L between the supports.
Key Concepts for Students
- A couple produces no shear on the overhang where it acts.
- The BMD over AB is linear, reaching M at B.
- Direction of M sets the signs; magnitudes stay M/L.
Practice more: this is question 53 in Bending Moment and Shear Force MCQs – Part 4. Try the whole set and check every answer instantly.