Columns and effective length: Which end condition gives the maximum equivalent (effective) length and therefore the least buckling load for the same physical length?

Introduction / Context:
Euler buckling load depends inversely on the square of the effective length. Recognizing which boundary condition produces the largest effective length is crucial for stability design of slender columns.

Given Data / Assumptions:

• Slender, straight column, initially perfect, elastic behavior.
• Same physical length L and cross-section for all cases.
• Classical Euler effective length factors K apply.

Concept / Approach:
Euler buckling load P_cr = (pi^2 * E * I) / (Le^2), where Le = K * L. Larger Le gives a smaller buckling load. Standard K values: both ends fixed → K = 0.5; both hinged → K = 1.0; one end fixed, other hinged → K ≈ 0.7; one end fixed, other free (cantilever) → K = 2.0. Thus, the fixed–free case has the largest effective length (Le = 2L).

Step-by-Step Solution:

Verification / Alternative check:
Compare Euler loads qualitatively: P_cr ∝ 1 / K^2, so fixed–free has the smallest P_cr and is the easiest to buckle, aligning with intuition for a cantilevered column.

Why Other Options Are Wrong:
Hinged–hinged (K = 1) and fixed–hinged (K ≈ 0.7) have smaller Le; fixed–fixed (K = 0.5) is the stiffest (smallest Le). The guided case is not a standard basic boundary condition in Euler's classic set.

Common Pitfalls:
Mixing up K for fixed–hinged vs hinged–hinged; assuming fixed–fixed is worst, when it is actually best for buckling resistance.

Final Answer:
One end fixed and the other end free