Four columns are identical in all parameters except end conditions: (i) both ends hinged, (ii) both ends fixed, (iii) one end fixed and the other free, (iv) one end fixed and the other hinged. Arrange them in ascending order of load-carrying capacity.
Try it first: pick an option, then read the explanation.
Asked in SSC JE 29/10/2020 (Morning)
Detailed Explanation
Euler's load P = π²EI/le² depends only on the effective length when E, I and L are equal. Write each case as a multiple of π²EI/L².
\[ \text{iii: } \frac{\pi^2EI}{(2L)^2} = 0.25 \quad \text{i: } \frac{\pi^2EI}{L^2} = 1 \quad \text{iv: } \frac{\pi^2EI}{(L/\sqrt2)^2} = 2 \quad \text{ii: } \frac{\pi^2EI}{(L/2)^2} = 4 \]
Why “iii, i, iv, ii” Is Correct
Ascending order of capacity: iii, i, iv, ii (fixed–free, hinged–hinged, fixed–hinged, fixed–fixed).
| 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. i, ii, iii, iv: Lists the cases in question order, not by capacity — fixed–fixed (ii) is strongest, not second.
- B. i, iv, iii, ii: Puts fixed–free (iii) third, but it is the weakest.
- C. ii, iv, i, iii: This is the descending order with i and iv swapped — not ascending.
Key Point
Capacity ratio fixed–free : hinged : fixed–hinged : fixed–fixed = 1/4 : 1 : 2 : 4.
Key Concepts for Students
- More end restraint → shorter le → higher buckling load.
- Ratio of extremes (fixed–fixed / fixed–free) is 16.
- Order of effective length is the reverse of order of capacity.
Practice more: this is question 6 in Columns and Struts MCQs – Part 1. Try the whole set and check every answer instantly.