The critical load for a column of length L hinged at both ends and having flexural rigidity EI is given by:
Try it first: pick an option, then read the explanation.
Correct answer: B. Pc = π²EI/L²
Asked in SSC JE 29/01/2018 (Evening)
Detailed Explanation
Solving EI y″ + Py = 0 with y = 0 at both hinges gives the lowest non-trivial load.
\[ P_c = \frac{\pi^2EI}{L^2} \]
Why “Pc = π²EI/L²” Is Correct
The critical load is π²EI/L².
Why the Other Options Are Wrong
- A. Pc = πEI/L²: π must be squared.
- C. Pc = πEI²/L²: EI² is dimensionally wrong.
- D. None of these: Option (b) is correct.
Key Point
Pin-ended Euler load: Pc = π²EI/L².
Key Concepts for Students
- The next buckling mode would need 4π²EI/L² (full sine wave).
- Use the minimum I.
- Euler stress = π²E/λ².
Practice more: this is question 13 in Columns and Struts MCQs – Part 1. Try the whole set and check every answer instantly.