Euler’s column theory – Comparing end conditions\nAccording to Euler’s theory, how does the critical (crippling) load of a column of length l fixed at both ends compare with that of an identical column hinged at both ends?

Introduction / Context:
Euler’s elastic buckling load depends on the effective length L_e determined by end restraints. Stronger end fixity reduces L_e and increases the critical load dramatically.

Given Data / Assumptions:

• Same material (E), moment of inertia (I), and clear length l.
• One column is fixed–fixed; the other is hinged–hinged.
• Elastic Euler buckling applies (long, slender columns).

Concept / Approach:
Euler formula: P_cr = π^2 * E * I / (L_e^2). For hinged–hinged, L_e = l. For fixed–fixed, L_e = l / 2. The ratio of critical loads follows from the square of the effective length ratio.

Step-by-Step Solution:
Hinged–hinged: P_cr,HH = π^2 * E * I / l^2.Fixed–fixed: P_cr,FF = π^2 * E * I / (l/2)^2 = π^2 * E * I / (l^2/4) = 4 * π^2 * E * I / l^2.Ratio: P_cr,FF / P_cr,HH = 4.

Verification / Alternative check:
Effective length factors K: K_HH = 1.0, K_FF = 0.5. Since P_cr ∝ 1/K^2, the factor is 1/(0.5^2) = 4.

Why Other Options Are Wrong:
Two times or eight times contradict the 1/L_e^2 relationship. “Equal” ignores end restraint effects.

Common Pitfalls:
Confusing clear length with effective length; misapplying K factors for mixed end conditions (e.g., fixed–free or fixed–hinged).

Final Answer:
Four times