Euler buckling (both ends hinged): For a column of length l with least moment of inertia I and modulus of elasticity E, the Euler critical buckling load is:

Introduction / Context:
Long, slender columns fail by elastic buckling at loads much lower than material crushing strength. Euler's formula predicts the critical load for ideal columns with various end conditions.

Given Data / Assumptions:

• Both ends hinged (pinned).
• Prismatic column; elastic behavior up to buckling.
• Initial imperfections and eccentricities neglected (ideal case).

Concept / Approach:
For a column with effective length l_e, Euler's load is P_cr = π^2 * E * I / l_e^2. For both ends hinged, l_e = l. The least I governs buckling in the weakest axis.

Step-by-Step Solution:
Identify end condition: hinged–hinged ⇒ l_e = l.Apply Euler formula: P_cr = π^2 * E * I / l^2.Select the option matching this expression.

Verification / Alternative check:
Other end conditions: fixed–free (l_e = 2l), fixed–fixed (l_e = l/2), fixed–hinged (l_e = l/√2). Substituting these recovers standard results.

Why Other Options Are Wrong:
Dimensional inconsistency or wrong placement of I and l (e.g., π E l^2 / I is incorrect).Correct dependence is inversely proportional to l^2 and directly proportional to I and E.

Common Pitfalls:
Using the wrong effective length for the boundary condition.

Final Answer:
(π^2 * E * I) / l^2.