Buckling versus crushing: For long (slender) columns, how does the Euler buckling load compare with the material's crushing (compressive) load?

Introduction / Context:
Long columns fail by elastic instability at stresses far below the compressive strength of the material. Recognizing this distinction prevents unsafe designs that rely solely on compressive strength without checking Euler stability limits.

Given Data / Assumptions:

• Slender column, length and end conditions such that Euler theory applies.
• Elastic behavior up to buckling.
• Crushing load refers to A * σ_c (material strength), far higher than elastic buckling load for slender members.

Concept / Approach:
Euler buckling load is P_cr = (π^2 * E * I) / (Le^2). As slenderness increases (Le large), P_cr drops rapidly. For sufficiently slender members, P_cr is much less than A * σ_c; thus buckling precedes material crushing.

Step-by-Step Solution:

Verification / Alternative check:
Numerical example: For a steel strut with large slenderness, Euler critical stress may be a fraction (say 10–20%) of σ_y, confirming that buckling governs well before crushing.

Why Other Options Are Wrong:

• “Equal to” applies only to a special transition slenderness, not generally for long columns.
• “More than” contradicts the inverse square dependence on Le.
• “Cannot be compared” ignores established formulas.

Common Pitfalls:
Using material compressive strength alone; neglecting effective length factor; overlooking imperfections that further reduce actual buckling load.

Final Answer:
less than