Effective (equivalent) length for a column with both ends fixed: choose the correct relation between equivalent length L and actual length l.

Introduction / Context:
End restraint increases the critical buckling load by reducing the effective length of a column. Knowing the correct effective length for each end condition is vital for safe, economical column design.

Given Data / Assumptions:

• Straight, prismatic column under axial compression.
• Both ends fixed (built-in), elastic buckling per Euler theory.
• No initial imperfections considered in the basic formula.

Concept / Approach:
The Euler critical load uses effective length L_e = K * l, where K depends on end conditions. For both ends fixed, K = 0.5. Thus L_e = 0.5 * l. The reduced length raises the critical load to four times that of a pinned–pinned column of the same actual length.

Step-by-Step Solution:

Verification / Alternative check:
From buckling mode shapes, a fixed–fixed column has half-wavelength equal to l, implying effective pinned length of l/2 for the same mode.

Why Other Options Are Wrong:
L = l corresponds to pinned–pinned; L = 2l to fixed–free (cantilever); fractional roots do not match standard K-factors.

Common Pitfalls:
Confusing K-factors among end conditions; misapplying effective length in slenderness ratio and Euler load calculations.

Final Answer:

L = l/2