Euler's crippling load for a column of length L with one end fixed and the other hinged is:
Try it first: pick an option, then read the explanation.
Correct answer: D. 2π²EI/L²
Asked in UP Jal Nigam JE 2013; SSC JE 2008
Detailed Explanation
Substitute the fixed–hinged effective length into Euler's formula.
\[ P = \frac{\pi^2EI}{(L/\sqrt2)^2} = \frac{2\pi^2EI}{L^2} \]
Why “2π²EI/L²” Is Correct
The crippling load is 2π²EI/L².
| End conditions | Effective length le (Euler) | Crippling load P = π²EI/le² | Relative to hinged–hinged |
|---|---|---|---|
| Both ends hinged (pinned) | L | π²EI/L² | 1 |
| One end fixed, other free | 2L | π²EI/4L² | 1/4 |
| One end fixed, other hinged | L/√2 ≈ 0.707L | 2π²EI/L² | 2 |
| Both ends fixed | L/2 | 4π²EI/L² | 4 |
Longest effective length → smallest buckling load (fixed–free); shortest effective length → largest load (fixed–fixed).
Why the Other Options Are Wrong
- A. π²EI/L²: π²EI/L² is hinged–hinged.
- B. 4π²EI/L²: 4π²EI/L² is fixed–fixed.
- C. π²EI/(4L²): π²EI/4L² is fixed–free.
Key Point
Fixed–hinged: P = 2π²EI/L² (twice the pin-ended value).
Key Concepts for Students
- Exact value 20.19EI/L² ≈ 2π²EI/L².
- Order of loads: 1/4 : 1 : 2 : 4.
- Effective length L/√2.
Practice more: this is question 19 in Columns and Struts MCQs – Part 2. Try the whole set and check every answer instantly.