A column has effective length L when both its ends are fixed. What will be the new effective length if one end becomes hinged?
Try it first: pick an option, then read the explanation.
Correct answer: C. 1.41L
Asked in SSC JE 02/03/2017 (Morning)
Detailed Explanation
Careful: L here is the effective length of the fixed–fixed column, not its actual length. Find the actual length first.
\[ \frac{l}{2} = L \Rightarrow l = 2L \qquad l_{e,new} = \frac{l}{\sqrt2} = \frac{2L}{\sqrt2} = \sqrt2\,L \approx 1.41L \]
Why “1.41L” Is Correct
The new effective length is 1.41L.
| 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. L: L would mean no change — but releasing one end always lengthens le.
- B. 0.5L: 0.5L confuses L with the actual length and reduces le, which is impossible.
- D. 2L: 2L would be the hinged–hinged value (le = l = 2L), not fixed–hinged.
Key Point
Convert given effective length to actual length first: l = 2L, then le(fixed–hinged) = l/√2 = 1.41L.
Key Concepts for Students
- Releasing restraint increases effective length.
- Effective-length ratios: fixed–hinged/fixed–fixed = √2.
- Buckling load drops to half (∝ 1/le²).
Practice more: this is question 18 in Columns and Struts MCQs – Part 2. Try the whole set and check every answer instantly.