Difficulty: Medium
Correct Answer: 1.5 L
Explanation:
Introduction / Context:
Effective length captures the influence of boundary restraints on the buckling strength of columns. Real-world columns seldom have identical end restraints, so design tables present K-factors for various combinations of translation and rotation restraints at ends.
Given Data / Assumptions:
Concept / Approach:
For the mixed case where one end is fully fixed and the other end cannot rotate but is free to translate (sway), standard alignment charts give a relatively large K due to the permitted sway mode. The widely used value for this end pair is K = 1.5, hence L_eff = 1.5 L. This reflects a much lower buckling capacity than fixed–fixed and even lower than pin–pin in many alignment conditions because sway increases the effective length.
Step-by-Step Solution:
Identify end A (fixed) and end B (rotation restrained, translation free).Select corresponding K-factor from end-condition table: K ≈ 1.5.Compute effective length: L_eff = K * L = 1.5 L.Use in design checks: λ = L_eff / r and P_cr = (pi^2 * E * I) / (L_eff^2).
Verification / Alternative check:
Compare with common cases: fixed–fixed (0.5L), fixed–pin (≈0.7L), pin–pin (L), fixed–free (2L). The given arrangement reasonably lies between pin–pin and cantilever, thus 1.5L is consistent.
Why Other Options Are Wrong:
Common Pitfalls:
Final Answer:
1.5 L
Discussion & Comments