Column effective length: One end fixed and the other end hinged—what is the Euler equivalent length?

Introduction:
Euler buckling analysis uses an effective length Le = K*L that depends on end restraints. Knowing K for common end conditions is essential in column design to estimate critical load and slenderness.

Given Data / Assumptions:

• Prismatic column of length L.
• One end perfectly fixed (no rotation/translation), the other end hinged (pinned, free to rotate, no translation).
• Euler elastic buckling regime, small deflection theory.

Concept / Approach:
For standard end conditions, the effective length factor K is tabulated. Values include: pinned–pinned K = 1.0, fixed–free K = 2.0, fixed–fixed K = 0.5, and fixed–pinned K ≈ 0.699 (often rounded to 0.7). The Euler critical load is Pcr = π^2 * E * I / (Le^2) = π^2 * E * I / (K^2 * L^2).

Step-by-Step Solution:
Identify end condition: fixed–pinned.From standard tables, K ≈ 0.699.Compute Le = K * L ≈ 0.699L, commonly rounded to 0.7L.Select the option stating 0.7L (approximately 0.699L).

Verification / Alternative check:
Energy or exact solution methods yield the same K. Many design codes provide this value directly; rounding to 0.7 is accepted practice for hand calculations.

Why Other Options Are Wrong:
2L: Fixed–free cantilever, not fixed–pinned.L: Pinned–pinned case.0.5L: Fixed–fixed case.1.5L: Not a standard Euler end condition factor.

Common Pitfalls:
Memorizing only the extreme cases (pinned–pinned or fixed–free) and guessing for mixed end conditions; mixing up effective length with unsupported physical length.

Final Answer:
0.7L (approximately 0.699L).