The equivalent length of a column of length L having one end fixed and the other end hinged is:
Try it first: pick an option, then read the explanation.
Correct answer: D. L/√2
Asked in SSC JE 27/01/2018 (Morning)
Detailed Explanation
The fixed end adds restraint at one end only, so the effective length lies between the hinged–hinged value (L) and the fixed–fixed value (L/2).
\[ \tan kL = kL \Rightarrow kL = 4.493 \Rightarrow P = \frac{20.19\,EI}{L^2} \approx \frac{2\pi^2EI}{L^2} \Rightarrow l_e = \frac{L}{\sqrt2} \]
Why “L/√2” Is Correct
The equivalent length is L/√2 (about 0.7L).
Why the Other Options Are Wrong
- A. 2L: 2L is fixed–free.
- B. L: L is hinged–hinged.
- C. L/2: L/2 is fixed–fixed.
Key Point
Fixed–hinged: le = L/√2, P ≈ 2π²EI/L².
Key Concepts for Students
- The exact solution gives 20.19EI/L², close to 2π²EI/L² = 19.74EI/L².
- IS 800 recommends 0.8L for this case.
- Effective length is between L/2 and L.
Practice more: this is question 14 in Columns and Struts MCQs – Part 1. Try the whole set and check every answer instantly.