Columns — Effective (Equivalent) Length For a column with one end fixed and the other end hinged (pin), what is the correct relation between equivalent length L and actual length l?

Introduction:
Euler buckling strength depends on the effective (equivalent) length, which incorporates end restraints. Different end conditions lead to different buckling half-wavelengths, captured by L = K * l, where K is an end-condition factor.

Given Data / Assumptions:

• Straight prismatic column, elastic behavior.
• One end fixed (no rotation, no translation), other end hinged (rotation free, no translation).
• Axial concentric compression.

Concept / Approach:
For fixed–pinned columns, the end-condition factor K is approximately 0.7. In many exam standards, this is exactly taken as 1/√2 ≈ 0.707. Thus, the equivalent length L = K * l = l/√2.

Step-by-Step Solution:
General Euler: P_cr = (pi^2 * E * I) / (L^2)Introduce L = K * l for end restraints.For fixed–pinned: K ≈ 0.7; widely tabulated as K = 1/√2.Therefore L = l/√2 gives the correct effective length.

Verification / Alternative check:
Compare to other cases: pinned–pinned K = 1, fixed–fixed K = 0.5, fixed–free K = 2. The value for fixed–pinned fits this sequence.

Why Other Options Are Wrong:
L = l/2: corresponds to fixed–fixed, not fixed–pinned.L = l: corresponds to pinned–pinned.L = 2l: corresponds to fixed–free (cantilever).

Common Pitfalls:
Memorizing formulas without linking them to end restraints and mode shapes.

Final Answer:
L = l/√2