Effective (equivalent) length factor for a pin–pin (both ends hinged) column\n\nFor a column with both ends hinged (pinned), are the effective length L and actual length l the same?

Introduction / Context:
Euler buckling load depends on effective length, which accounts for end conditions. Knowing the effective length factor K allows quick evaluation of buckling capacity and slenderness.

Given Data / Assumptions:

• Column with ideal hinged (pinned) ends allowing rotation but no translation.
• Straight prismatic member; elastic buckling analysis.
• No intermediate bracing.

Concept / Approach:
Effective length L equals K * l, where l is the actual unsupported length and K is the effective length factor determined by end restraints. For a pin–pin column, K = 1.0, so L = l.

Step-by-Step Solution:
Euler load: P_cr = π^2 E I / L^2.End condition pin–pin → K = 1 → L = K l = l.Thus, the effective length equals the actual length.

Verification / Alternative check:
Buckling half-sine mode shape for pin–pin has a node at each end and a single antinode at midheight. The half-wave equals the actual length, confirming L = l.

Why Other Options Are Wrong:
“No” contradicts K = 1.0. Conditions on prismatic shape or frictionless pins are not needed for the idealized definition—though real connections may modify K slightly, the standard model uses K = 1.

Common Pitfalls:
Confusing fixed–fixed (K = 0.5) or fixed–free (K = 2.0) with pin–pin; using braced lengths instead of unsupported lengths; mixing effective length with effective area.

Final Answer:
Yes