Difficulty: Easy
Correct Answer: The axis with the least radius of gyration (weak axis)
Explanation:
Introduction / Context:
Elastic buckling of columns is a stability phenomenon. The critical buckling load depends on flexural stiffness EI about a given axis and the effective length. Designers must anticipate the likely buckling mode to size members and arrange bracing correctly.
Given Data / Assumptions:
Concept / Approach:
Euler load Pcr = π^2 * E * I / (K * L)^2. For a given cross-section, I = A * r^2, where r is the radius of gyration about the axis. The axis with smaller r (weaker bending stiffness) gives a lower I and thus a smaller Pcr, so the member will preferentially buckle about that weak axis.
Step-by-Step Solution:
Relate I to r: I = A * r^2.Substitute into Pcr: smaller r → smaller I → smaller Pcr.Therefore, buckling initiates about the axis of least radius of gyration.
Verification / Alternative check:
Practical observation of angle, channel, and I-sections confirms weak-axis buckling unless braced. Design manuals specify different slenderness ratios (KL/r) for the two axes accordingly.
Why Other Options Are Wrong:
(b) and (c) ignore section properties; (d) is incorrect—buckling prefers the weakest stiffness path; (e) is opposite of reality.
Common Pitfalls:
Neglecting different end conditions (K), residual stresses, or initial crookedness; assuming symmetry when sections are monosymmetric.
Final Answer:
The axis with the least radius of gyration (weak axis)
Discussion & Comments