Column buckling – asymmetric end restraint: For a compression member that is effectively held in position and restrained in direction at one end but neither held in position nor restrained in direction at the other end (i.e., fixed–free), the effective length equals which of the following?

Introduction / Context:
The effective length factor K captures end restraint conditions for columns. The most flexible case is the cantilever (fixed–free), which has the highest effective length and therefore the lowest buckling strength for a given member length and section.

Given Data / Assumptions:

• One end fixed (no translation or rotation), the other end free (no restraint).
• Prismatic member, elastic buckling considered.

Concept / Approach:
Standard K values: pinned–pinned K = 1.0; fixed–pinned K ≈ 0.7–0.8; fixed–fixed K ≈ 0.65–0.70; fixed–free K = 2.0. Therefore, the effective length Le = K * L = 2L for the fixed–free (cantilever) case.

Step-by-Step Solution:
Identify boundary condition as fixed–free.Use K = 2.0.Compute Le = 2.0 * L = 2 L.

Verification / Alternative check:
Euler buckling solution for a cantilever gives the first buckling mode with a node at the fixed end and a maximum deflection at the free end, corresponding to Le = 2 L when compared to the pinned–pinned baseline.

Why Other Options Are Wrong:

• L, 0.67 L, 0.85 L, 1.5 L: all correspond to more restrained conditions, which would produce higher buckling loads than the cantilever case.

Common Pitfalls:

• Mistaking a partially braced top (e.g., lateral support only) as “free”; true free end has no translational or rotational restraint.

Final Answer:
2 L