Column effective length: For a column fixed at both ends (both ends fully restrained against rotation and translation), the equivalent length used in Euler’s formula is:

Introduction / Context:
End conditions drastically affect a column’s critical buckling load. A column with both ends fixed is significantly stiffer against buckling than one that is pinned or free.

Given Data / Assumptions:

• Prismatic, elastic column.
• Both ends fixed (zero slope and zero translation at ends).
• Euler elastic buckling governs.

Concept / Approach:
Euler's critical load P_cr = (π^2 E I)/(L_e)^2. The effective length L_e = K * l captures end conditions via K. For fixed–fixed, K = 0.5, giving the highest P_cr among the common ideal end restraints.

Step-by-Step Solution:
End condition: fixed–fixed ⇒ K = 0.5.L_e = K * l = 0.5 l.

Verification / Alternative check:
Mode-shape solutions to the column differential equation yield K = 0.5 for the lowest buckling mode with both ends fixed.

Why Other Options Are Wrong:
0.7 l is not a standard ideal end factor.l is for pinned–pinned; 2 l is for fixed–free; 1.5 l is nonstandard in Euler's ideal set.

Common Pitfalls:
Using pinned–pinned formulas or forgetting the strong influence of rotational fixity.

Final Answer:
0.5 l.