A simply supported beam of length 4 m carries a UDL of 10 kN/m over the entire span. The maximum shear force anywhere in the beam is:
Try it first: pick an option, then read the explanation.
Correct answer: B. 20 kN
Asked in SSC JE 10/10/2023 (5 pm–7 pm)
Detailed Explanation
For a simply supported beam, the shear force is largest just next to a support, where it equals the reaction. Under a symmetric load the two reactions are equal.
\[ R_A + R_B = 10 \times 4 = 40\ \text{kN},\quad R_A = R_B = 20\ \text{kN} \]
\[ V(x) = 20 - 10x \Rightarrow V_{max} = 20\ \text{kN at } x = 0 \]
Why “20 kN” Is Correct
The shear falls linearly from +20 kN at A to −20 kN at B. Its maximum magnitude is 20 kN — option (b) — that is, wL/2.
Why the Other Options Are Wrong
- A. 40/6 kN: 40/6 kN is not a reaction for this loading.
- C. 40 kN: 40 kN is the total load, shared between two supports.
- D. 10 kN: 10 kN is the load per metre, not a shear force.
Key Point
SS beam with full UDL: V_max = wL/2 at the supports; M_max = wL²/8 at mid-span.
Key Concepts for Students
- Under a UDL the SFD is linear.
- Shear is zero at mid-span, where the bending moment is maximum.
- Design for shear is checked near the supports.
Practice more: this is question 8 in Bending Moment and Shear Force MCQs – Part 1. Try the whole set and check every answer instantly.