Consider a rectangular column cross-section of size 4 m × 2 m. Identify the correct statement.
Try it first: pick an option, then read the explanation.
Asked in SSC JE 05/06/2024 (9 am–11 am)
Detailed Explanation
A compressive load P at eccentricity e produces direct stress P/A plus bending stress Pe/Z. For no tension anywhere, P/A ≥ Pe/Z, which limits e to b/6 along one axis and d/6 along the other. Joining these four limits gives the core (kern) — a rhombus.
\[ e_x \le \frac{b}{6} = \frac{4}{6} = 0.667\ \text{m} \quad\Rightarrow\quad \text{diagonal} = \frac{b}{3} = 1.33\ \text{m} \]\[ e_y \le \frac{d}{6} = \frac{2}{6} = 0.333\ \text{m} \quad\Rightarrow\quad \text{diagonal} = \frac{d}{3} = 0.67\ \text{m} \]
Why “The point load shall be applied within the rhombus located at the centre with diagonals 1.33 m × 0.66 m to ensure no tensile stress develops in the column cross-section.” Is Correct
The no-tension zone is the central rhombus with diagonals 1.33 m and 0.66 m (middle-third rule in each direction). Statement (a) describes it exactly.
Why the Other Options Are Wrong
- B. The point load shall be applied within 1.5 m from the centre to avoid tension in the cross-section.: 1.5 m from the centre is far outside the core — tension would develop.
- C. The limit of eccentricity for an axial compressive load is 1 m from the C.G. of the section for no tensile stress.: The limit along the 4 m side is only b/6 = 0.667 m, and along the 2 m side just 0.333 m.
- D. The core of the section is a square of size 0.6 m.: The core of a rectangle is a rhombus, not a square, and its size depends on b and d.
Key Point
Rectangle b × d: core = rhombus, diagonals b/3 and d/3 (e ≤ b/6, e ≤ d/6).
Key Concepts for Students
- This is the middle-third rule applied in both directions.
- A solid circle's core is a circle of diameter D/4.
- Masonry and brittle columns should have the load inside the core.
Practice more: this is question 2 in Columns and Struts MCQs – Part 1. Try the whole set and check every answer instantly.