Effective length of compression members: A column of length L is held in position (no translation) at both ends but is not restrained against rotation at either end. What is its effective length for buckling calculations?

Introduction / Context:
The effective length of a compression member models end restraints for Euler buckling. Different end conditions translate into different effective length factors K, which multiply the physical length L to give the column's buckling length K*L.

Given Data / Assumptions:

• Both ends held in position (no lateral translation).
• No rotational restraint: ends are free to rotate (pinned).
• Prismatic member, elastic buckling idealization.

Concept / Approach:

Classical end condition — pinned–pinned: both ends prevent translation but allow rotation. The effective length factor K for this case is 1.0, so the effective length equals the actual length.

Step-by-Step Solution:

Verification / Alternative check:

Euler buckling tables list K = 1.0 for pinned–pinned, K ≈ 0.7 for fixed–fixed, K ≈ 2.0 for fixed–free (cantilever), etc. Our case matches K = 1.0.

Why Other Options Are Wrong:

• 0.67 L, 0.85 L: represent stiffer end restraints (partial or full fixity).
• 1.5 L, 2 L: represent weaker restraint or cantilever conditions.

Common Pitfalls:

• Confusing “held in position” (no sway) with “fixed” (no rotation).
• Using incorrect K values, leading to unconservative capacity estimates.

Final Answer:

L.