To have exactly zero tensile stress at one extreme fibre of a solid circular section (diameter D) under combined direct compressive and bending stress, a normal point load must be placed ______.
Try it first: pick an option, then read the explanation.
Asked in SSC JE 07/06/2024 (9 am–11 am)
Detailed Explanation
The book places this core (kern) question in the torsion chapter. A compressive load P at eccentricity e gives P/A on the whole section and ±Pe/Z from bending. The far fibre just reaches zero when P/A = Pe/Z.
\[ e = \frac{Z}{A} = \frac{\pi D^3/32}{\pi D^2/4} = \frac{D}{8} \qquad \text{distance from the periphery} = \frac{D}{2} - \frac{D}{8} = \frac{3D}{8} \]
Why “beyond a distance of 3D/8 measured towards the centre from the periphery” Is Correct
The load must sit within D/8 of the centre, i.e. at least 3D/8 in from the periphery. At exactly 3D/8 the extreme fibre has zero stress; further in, it stays in compression.
Why the Other Options Are Wrong
- A. at a radial distance D/4 from the centre: D/4 from the centre is outside the core (D/8) — tension develops.
- B. at a radial distance D/6 from the centre: D/6 is the rectangle's limit idea; for a circle D/6 > D/8, so tension develops.
- D. at a radial distance D/3 from the centre: D/3 from the centre is well outside the core.
Key Point
Solid circle: e ≤ D/8 from the centre (core diameter D/4) = 3D/8 or more from the edge.
Key Concepts for Students
- Core boundary e = Z/A.
- Rectangle core: rhombus b/3 × d/3.
- Hollow circle core radius: (D² + d²)/(8D).
Practice more: this is question 1 in Torsion MCQs – Part 1. Try the whole set and check every answer instantly.