When both ends of a column are fixed, the crippling load is F. If one end of the column is made free, the crippling load will change to:
Try it first: pick an option, then read the explanation.
Correct answer: C. F/16
Asked in SSC JE 02/03/2017 (Morning)
Detailed Explanation
Euler's load varies as 1/le². Compare the effective lengths of the two cases.
\[ F = \frac{\pi^2EI}{(L/2)^2} = \frac{4\pi^2EI}{L^2} \qquad F' = \frac{\pi^2EI}{(2L)^2} = \frac{\pi^2EI}{4L^2} \qquad \frac{F'}{F} = \frac{1/4}{4} = \frac{1}{16} \]
Why “F/16” Is Correct
The new crippling load is F/16.
| 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. F/4: F/4 is the change from hinged–hinged to fixed–free (or fixed–fixed to hinged–hinged).
- B. F/2: F/2 is the change from fixed–fixed to fixed–hinged.
- D. 4F: Releasing an end can only reduce the buckling load, never increase it.
Key Point
Fixed–fixed → fixed–free: le × 4, load ÷ 16.
Key Concepts for Students
- P ∝ 1/le².
- Fixed–fixed is the strongest, fixed–free the weakest standard case.
- Load ratios 4 : 2 : 1 : 1/4.
Practice more: this is question 21 in Columns and Struts MCQs – Part 2. Try the whole set and check every answer instantly.