According to Euler's column theory, the crippling load for a column of length l fixed at both ends is ______ the crippling load for a similar column hinged at both ends.
Try it first: pick an option, then read the explanation.
Correct answer: B. 4 times
Asked in SSC JE 25/01/2018 (Evening); SSC JE 03/03/2017 (Morning)
Detailed Explanation
Halving the effective length (from l to l/2) multiplies the Euler load by 2² = 4.
\[ \frac{P_{ff}}{P_{hh}} = \frac{\pi^2EI/(l/2)^2}{\pi^2EI/l^2} = 4 \]
Why “4 times” Is Correct
The fixed–fixed column carries 4 times the load.
Why the Other Options Are Wrong
- A. equal to: Fixing the ends shortens le, so the loads cannot be equal.
- C. 2 times: 2 times is fixed–hinged vs hinged–hinged.
- D. 8 times: 8 times does not match any standard pair.
Key Point
P(fixed–fixed) = 4 × P(hinged–hinged).
Key Concepts for Students
- P ∝ 1/le².
- le = l/2 for fixed–fixed.
- In practice codes use 0.65l, so the real gain is smaller (≈ 2.4×).
Practice more: this is question 23 in Columns and Struts MCQs – Part 2. Try the whole set and check every answer instantly.