Euler–column idealization: What is the equivalent (effective) length for a column of actual length L when both ends are fixed against rotation and translation?

Introduction / Context:
Effective (equivalent) length is central to Euler buckling and slenderness calculations. End restraints reduce the buckling half-wave, effectively shortening the buckling length compared with the actual length. This question focuses on the standard end condition of a column with both ends fixed.

Given Data / Assumptions:

• Prismatic column of length L.
• Both ends fully fixed (no rotation, no translation).
• Elastic buckling per Euler theory.

Concept / Approach:
For Euler buckling, the critical load is:
P_cr = (π^2 * E * I) / (L_e)^2where L_e is the effective length determined by end conditions. Standard effective-length factors K give L_e = K * L. For both ends fixed, K = 0.5, hence L_e = L/2.

Step-by-Step Solution:
1) Identify end condition: both ends fixed.2) Use standard K-value table: K = 0.5.3) Compute effective length L_e = K * L = 0.5 * L = L/2.

Verification / Alternative check:
Compare with other end conditions: both hinged (K = 1 ⇒ L_e = L), one fixed–one free (K = 2 ⇒ L_e = 2L). The fixed–fixed case is the stiffest, so it must have the smallest effective length, L/2, and largest P_cr.

Why Other Options Are Wrong:
L, 2L, √2 L, 3L/2 correspond to other end conditions, not the fixed–fixed case.

Common Pitfalls:

• Confusing fixed–fixed with hinged–hinged.
• Using unsupported K-values without checking end fixity details.
• Forgetting that greater fixity reduces effective length.

Final Answer:
L / 2