A simply supported beam of length a carries a load varying from zero at one end to w at the other. The shear force is zero at a distance x from the least loaded end, where x is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 04/03/2017 (Evening)
Detailed Explanation
Write the shear at a section x from the zero-load end: the reaction there minus the triangular load on the length x. Set it to zero.
\( R_A = \dfrac{wa}{6} \); load on length x \( = \tfrac12 \cdot x \cdot \dfrac{wx}{a} = \dfrac{wx^2}{2a} \)
\[ \frac{wa}{6} - \frac{wx^2}{2a} = 0 \Rightarrow x^2 = \frac{a^2}{3} \Rightarrow x = \frac{a}{\sqrt3} \]
Why “a/√3” Is Correct
Zero shear occurs at a/√3 (≈ 0.577a) from the lightly loaded end — option (c). The maximum moment, wa²/(9√3), acts there.
Why the Other Options Are Wrong
- A. a/2: a/2 is the zero-shear point for a symmetric load.
- B. a/3: a/3 is where the load's resultant acts, measured from the heavy end.
- D. a√3/2: a√3/2 ≈ 0.866a lies too close to the heavy end.
Key Point
Triangular load on SS beam: V = 0 at L/√3 from the zero-load end.
Key Concepts for Students
- The zero-shear point lies past mid-span, towards the heavier end.
- Max BM always occurs where shear changes sign.
- Reactions: wL/6 and wL/3.
Practice more: this is question 44 in Bending Moment and Shear Force MCQs – Part 3. Try the whole set and check every answer instantly.