A cantilever of length L carries a UDL of intensity w over its middle one-third. The shear force at mid-span and at the fixed end is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 14/11/2022
Detailed Explanation
For a cantilever, the shear at a section is the load lying between that section and the free end. Here the load occupies the middle third, from L/3 to 2L/3 measured from the fixed end, and mid-span is at its centre.
Mid-span (L/2): loaded length beyond it \( = \dfrac{2L}{3} - \dfrac{L}{2} = \dfrac{L}{6} \Rightarrow V = \dfrac{wL}{6} \)
Fixed end: whole load \( = w \cdot \dfrac{L}{3} \Rightarrow V = \dfrac{wL}{3} \)
Why “Mid-span = wL/6, fixed end = wL/3” Is Correct
The values are wL/6 at mid-span and wL/3 at the fixed end — option (c).
Why the Other Options Are Wrong
- A. Mid-span = wL/3, fixed end = 2wL/3: 2wL/3 would need a load over two-thirds of the length.
- B. Mid-span = wL/3, fixed end = wL: wL is the total load only if the whole length were loaded.
- D. Mid-span = wL, fixed end = zero: The fixed end always carries the full load; it cannot be zero.
Key Point
Cantilever shear at a section = load between that section and the free end.
Key Concepts for Students
- Under a UDL the SFD is a sloping line; outside it, horizontal.
- Fixed-end moment here = (wL/3) × (L/2) = wL²/6.
- The resultant of the UDL acts at the middle of the loaded length.
Practice more: this is question 20 in Bending Moment and Shear Force MCQs – Part 2. Try the whole set and check every answer instantly.