For a column of length L hinged at both ends with flexural rigidity EI, what is the Euler critical buckling load?

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)