A column of length L has flexural rigidity EI; one end is fixed and the loaded end is free. The critical load will be:
Try it first: pick an option, then read the explanation.
Correct answer: B. Pc = π²EI/(4L²)
Asked in SSC JE 27/01/2018 (Morning)
Detailed Explanation
The column is fixed at its base and free at the loaded top — the flagpole case with le = 2L.
\[ P_c = \frac{\pi^2 EI}{l_e^2} = \frac{\pi^2 EI}{(2L)^2} = \frac{\pi^2EI}{4L^2} \]
Why “Pc = π²EI/(4L²)” Is Correct
The critical load is π²EI/(4L²).
Why the Other Options Are Wrong
- A. Pc = πEI/(4L²): π appears squared in Euler's formula; πEI/4L² is dimensionally the same but numerically wrong.
- C. Pc = πEI²/(4L²): EI² is dimensionally wrong — the formula uses EI once.
- D. None of these: Option (b) is the correct Euler load.
Key Point
Fixed–free: Pc = π²EI/4L² — one quarter of the pin-ended load.
Key Concepts for Students
- Check units: π²EI/L² has units of force.
- Always use π², not π, in Euler's formula.
- Use the least EI of the section.
Practice more: this is question 9 in Columns and Struts MCQs – Part 1. Try the whole set and check every answer instantly.