For a column of length L hinged at both ends with flexural rigidity EI, what is the Euler critical buckling load?
Civil Engineering
Steel Structure Design
Difficulty: Medium
Choose an option
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A(pi^2 * EI) / (L^2)
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B(pi^2 * EI) / (2L^2)
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C(pi^2 * EI) / (4L^2)
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D(pi^2 * EI) / L
Answer
Correct Answer: (pi^2 * EI) / (L^2)
Explanation
Introduction / Context:Euler's theory predicts the elastic buckling load for long, slender columns. End conditions control the effective length and therefore the capacity.
Given Data / Assumptions:
- Length L; end condition: hinged–hinged.
- Flexural rigidity EI; material remains elastic; load is concentric.
Concept / Approach:The general form is P_cr = (pi^2 * EI) / (L_eff^2). For hinged–hinged, L_eff = L.
Step-by-Step Solution:Use L_eff = L.P_cr = (pi^2 * EI) / (L^2).
Verification / Alternative check:Other common cases: fixed–free → L_eff = 2L; fixed–fixed → L_eff = L/2; fixed–hinged → L_eff ≈ 0.699L. Pinned–pinned is the baseline.
Why Other Options Are Wrong:
- 2L^2 or 4L^2 in denominator correspond to different end restraints.
- /L is dimensionally inconsistent with load.
Common Pitfalls:Confusing P_cr (load) with σ_cr (stress) and mixing up effective length factors.
Final Answer:(pi^2 * EI) / (L^2)