Buckling of columns — parameters affecting Euler load: The critical (buckling) load for a given column depends on which set of properties?

Introduction / Context:
Euler’s formula for long columns shows how geometry, material stiffness, and end conditions jointly govern buckling strength. Understanding each parameter's role is essential in member sizing.

Given Data / Assumptions:

• Pcr = π^2 * E * I / (K * L)^2 (long, straight columns).
• I = A * k^2, where k is radius of gyration.
• Least radius of gyration controls the weakest axis buckling.

Concept / Approach:
Critical load scales linearly with E and I, and inversely with the square of effective length (K * L). Since I involves area A and radius k, area influences Pcr through the second moment of area. Thus multiple properties simultaneously affect buckling.

Step-by-Step Solution:

Verification / Alternative check:
Comparing two sections with same area but different k (e.g., solid vs thin-walled tube) shows different Pcr, confirming dependence on k in addition to A.

Why Other Options Are Wrong:

• Single-factor answers ignore key contributors.
• “Only end load magnitude” is irrelevant; buckling capacity is a property of the member.

Common Pitfalls:
Equating larger area with higher buckling load without considering how area is distributed (k).

Final Answer:
All of the above (area through I, length, least radius of gyration, and E)