Effective length of compression members: For a compression member of physical length L that is held in position and restrained in direction at both ends (i.e., effectively fixed at both ends), the effective length is approximately equal to which value?

Introduction / Context:
Column buckling strength depends on effective length, which captures end restraints. Better end restraint reduces the effective length factor and increases buckling capacity. Recognizing standard effective length factors is essential in steel design.

Given Data / Assumptions:

• Prismatic compression member.
• End conditions: held in position and restrained in rotation at both ends (fixed–fixed idealization).
• Euler-type buckling concept for effective length.

Concept / Approach:
Effective length Le = K * L, where K is the effective length factor. Typical values: pinned–pinned K ≈ 1.0, fixed–free K ≈ 2.0, fixed–pinned K ≈ 0.7–0.8, fixed–fixed K ≈ 0.5–0.7 depending on code conservatism and real restraint. Many steel design guides adopt about 0.65–0.7 for ideal fixed–fixed; 0.67 L is a common rounded value used in problems.

Step-by-Step Solution:
Identify end condition: both ends restrained in position and direction.Adopt K ≈ 0.65–0.7 for fixed–fixed.Select the closest listed value: 0.67 L.

Verification / Alternative check:
Euler buckling load is proportional to 1 / Le^2; reducing Le to about two-thirds increases capacity by roughly (1 / 0.67^2) ≈ 2.2 times compared to a pinned–pinned column, which aligns with expectations.

Why Other Options Are Wrong:

• L: corresponds to pinned–pinned, less restraint.
• 0.85 L: closer to fixed–pinned, not fixed–fixed.
• 1.5 L or 2 L: represent poor restraint or cantilever case; not applicable.

Common Pitfalls:

• Assuming perfect fixity in practice; real joints may justify K values slightly above the ideal.

Final Answer:
0.67 L