Euler buckling – Equivalent length comparison A column of clear length l is fixed at both ends. It is equivalent, for buckling strength, to which effective length if the end conditions were hinged at both ends?

Introduction / Context:Critical load for elastic buckling depends on the effective length, which is a function of end restraints. Comparing different end conditions using effective length is central to column design.

Given Data / Assumptions:

• Same cross-section and material for both cases.
• Case 1: fixed–fixed ends; Case 2: hinged–hinged ends.
• Euler buckling regime (long, slender column), linear elastic behavior.

Concept / Approach:Euler load: P_cr = π^2 * E * I / (L_e^2), where L_e = K * l. For fixed–fixed, K = 0.5; for hinged–hinged, K = 1.0. Equivalence of buckling strength requires equal L_e values.

Step-by-Step Solution:Fixed–fixed: L_e = 0.5 * l.Hinged–hinged: L_e = 1.0 * l.Thus a fixed–fixed column of length l has the same buckling strength as a hinged–hinged column of effective length l/2.

Verification / Alternative check:For the same P_cr, set π^2 E I /(0.5 l)^2 = π^2 E I /(L_eq)^2, giving L_eq = l/2.

Why Other Options Are Wrong:Values like l/8 or l/4 overstate restraint; l or 2l ignore the increased stiffness due to fixity.

Common Pitfalls:Confusing clear length with effective length factor K; mixing end conditions (fixed–free, fixed–hinged, etc.).

Final Answer:l/2