Euler buckling – effect of end conditions on critical load For columns of the same length, material, and cross-section, which end condition yields the <em>highest</em> Euler buckling load (i.e., the strongest column against buckling)?

Introduction / Context:
Euler's critical load shows how end restraints influence a column's stability. Understanding end conditions helps designers choose supports that maximize buckling resistance without unnecessary material.

Given Data / Assumptions:

• Same material (modulus E) and cross-section (second moment I).
• Same clear length L.
• Elastic buckling, slender column (Euler regime).

Concept / Approach:
Euler's formula: P_cr = (pi^2 * E * I) / (L_e)^2, where L_e is the effective length depending on end conditions. Smaller L_e gives larger P_cr. Effective length factors K are: fixed–fixed K = 0.5; fixed–hinged K ≈ 0.699; hinged–hinged K = 1.0; fixed–free (cantilever) K = 2.0.

Step-by-Step Solution:
Compute L_e = K * L for each case.Compare K: the smallest K is 0.5 (fixed–fixed).Therefore, fixed–fixed columns have the largest P_cr for identical L, E, I.

Verification / Alternative check:
Ratios: P_cr(fixed–fixed) : P_cr(hinged–hinged) = (1/0.5^2) : (1/1^2) = 4 : 1 → fixed–fixed supports quadruple Euler capacity relative to pin–pin for the same column.

Why Other Options Are Wrong:
Fixed–hinged and hinged–hinged have larger K than fixed–fixed, giving lower P_cr. Fixed–free has the largest K and the smallest P_cr.

Common Pitfalls:
Confusing effective length K values; forgetting that stiffer end fixity reduces effective length and increases buckling load.

Final Answer:
Both ends fixed (fixed–fixed)