Column effective length: For a compression member with one end fixed and the other end free (a classical cantilever column), what is the equivalent length used in Euler's buckling formula, expressed as a multiple of the actual length l?

Introduction / Context:
In structural engineering, the slenderness and end restraints of a compression member control its elastic buckling load. Euler's theory uses an effective (equivalent) length factor to model different boundary conditions. A column that is fixed at one end and free at the other is the most flexible common case.

Given Data / Assumptions:

• Prismatic column, homogeneous, linear elastic.
• One end perfectly fixed (zero rotation and zero translation), the other end perfectly free (no restraint).
• Euler buckling (small deflection, elastic instability) is applicable.

Concept / Approach:
Euler buckling load P_cr = (π^2 * E * I) / (L_e)^2, where L_e is the effective length. Different end conditions are captured by L_e = K * l with a coefficient K. For fixed–free, K is largest (most flexible), so the critical load is the smallest among common cases.

Step-by-Step Solution:
End condition: fixed–free ⇒ effective length factor K = 2.0.Therefore, L_e = K * l = 2 * l.Thus the equivalent length is 2 l.

Verification / Alternative check:
Classical tables (or solving the eigenvalue problem y'' + (P/(E I)) y = 0 with the appropriate boundary conditions) yield the first buckling mode giving K = 2.0 for fixed–free.

Why Other Options Are Wrong:
0.5 l and 0.7 l correspond to much stiffer end conditions (e.g., fixed–fixed or fixed–pinned), not fixed–free.l corresponds to pinned–pinned.1.5 l is not a standard Euler end condition.

Common Pitfalls:
Confusing effective length (a stability concept) with actual length, and mixing up end-condition factors K.

Final Answer:
2 l.