Difficulty: Easy
Correct Answer: In the direction of least radius of gyration
Explanation:
Introduction / Context:
Column stability depends on flexural stiffness EI about potential buckling axes. The effective slenderness ratio determines the critical Euler load; weaker bending axes buckle first.
Given Data / Assumptions:
Concept / Approach:
The radius of gyration r = sqrt(I/A). For a given axial load and member, the slenderness ratio λ = L_e / r governs buckling. Smaller r means larger λ and lower critical load, so buckling occurs about the axis with least r (and least I).
Step-by-Step Explanation:
Compute r_x and r_y for principal axes.Compare slenderness ratios λ_x and λ_y; the larger one indicates the weaker direction.Expect buckling about the axis with least r (i.e., least flexural stiffness).
Verification / Alternative check:
Euler load P_cr = π^2 * E * I / (L_e^2). For the axis with smaller I, P_cr is smaller; initial imperfections trigger buckling there first.
Why Other Options Are Wrong:
Buckling is not along the load axis (axial shortening). “Any direction” ignores anisotropy of section stiffness. Minimum area alone is not decisive unless it correlates with least I and least r. Greatest I would be the strongest direction, not the weakest.
Common Pitfalls:
Ignoring end-restraint effects on L_e; overlooking weak-axis bracing requirements in design.
Final Answer:
In the direction of least radius of gyration
Discussion & Comments