A simply supported beam of span 3 m carrying a UDL of 10 kN/m has a shear force at mid-span of:
Try it first: pick an option, then read the explanation.
Correct answer: D. Zero
Asked in SSC JE 2007
Detailed Explanation
Under a symmetric load, the shear force at the centre of a simply supported beam is zero: the load on the left half exactly uses up the left reaction.
\( R_A = R_B = \dfrac{10 \times 3}{2} = 15\ \text{kN} \)
\( V(x) = 15 - 10x \Rightarrow V(1.5) = 15 - 15 = 0 \)
Why “Zero” Is Correct
The shear at mid-span is zero — option (d). That is exactly where the bending moment is maximum (wL²/8 = 11.25 kN·m).
Why the Other Options Are Wrong
- A. 15 kN: 15 kN is the shear at the supports.
- B. 30 kN: 30 kN is the total load.
- C. 7.5 kN: 7.5 kN is the shear at the quarter points, not mid-span.
Key Point
Symmetric loading on a SS beam: V = 0 at mid-span, M maximum there.
Key Concepts for Students
- V(x) = w(L/2 − x) for a full UDL.
- Max BM where SF = 0.
- Shear at quarter span = wL/4.
Practice more: this is question 37 in Bending Moment and Shear Force MCQs – Part 3. Try the whole set and check every answer instantly.