Euler’s column theory – Critical load for a pin-ended (hinged-hinged) column According to Euler, what is the crippling (critical) load for a column of length l with both ends hinged?

Introduction / Context:Euler’s buckling formula gives the elastic critical load for long, slender columns. The effective length depends on end conditions.

Given Data / Assumptions:

• Slender prismatic column, linear elastic material.
• Both ends hinged (pin–pin), no end moment restraint.
• Initial imperfections small; loading concentric.

Concept / Approach:The Euler load is P_cr = π^2 * E * I / (L_e^2), where L_e is the effective length. For hinged–hinged ends, L_e = l.

Step-by-Step Solution:Identify end condition factor K = 1.0 for hinged–hinged.Compute L_e = K * l = l.Therefore P_cr = π^2 * E * I / l^2.

Verification / Alternative check:Comparing with other end conditions: fixed–fixed has L_e = l/2 (four times higher P_cr), fixed–free has L_e = 2l (one-quarter P_cr).

Why Other Options Are Wrong:They use incorrect multipliers or dimensions; the only correct expression for pin–pin is π^2 E I / l^2.

Common Pitfalls:Confusing effective length factors; mixing units of E, I, and length.

Final Answer:P_cr = π^2 * E * I / l^2