A steel round bar Ø20 mm, 800 mm long, pinned at both ends, carries 5 kN of compressive load. Is that safe against buckling?
Slenderness, not thickness, sets the behaviour
Slenderness λ = L_eff / i, with radius of gyration i = √(I/A) and effective length L_eff = k · L. The end-condition factor k depends on how the bar is clamped: pinned-pinned k = 1, fixed-free k ≈ 2.1, fixed-fixed k ≈ 0.65. These design values (AISC/Shigley) sit slightly less favourably than the theoretically ideal ones, because perfect fixity doesn't exist in practice.
Worked example Ø20 mm, 800 mm, pinned-pinned
For a solid round bar, i = D/4, so i = 5 mm and λ = 800/5 = 160. With I = π·20⁴/64 = 7,854 mm⁴, E = 210,000 N/mm² and L_eff = 800 mm, F_cr = π²EI/L_eff² ≈ 25.4 kN. At a 5 kN load, the ideal theoretical load ratio F_cr/F ≈ 5.1. This is not demonstrated design safety: imperfections, eccentricity and applicable design rules require separate assessment.
Know where Euler stops applying
The limiting slenderness λ_limit = π·√(E/Rp0.2) is the intersection of ideal Euler stress and yield strength. For steel (E = 210,000, Rp0.2 = 235), that limit sits around 94. With λ = 160, this example sits well above it: the Euler result remains an ideal theoretical load, not a verified capacity. This threshold is only a screening check; it does not establish proportional elastic behaviour. Below the limit, Mechify reports an upper bound (A·Rp0.2) instead of a pass/fail answer — that calls for a separate column check such as a Johnson parabola, Tetmajer, or a national design standard.
Assumptions and applicability limits
The model assumes an ideally straight bar under centric loading, with no initial curvature or eccentricity. Local buckling of thin-walled sections, dynamic loading and fatigue are not included. Use the design values of k, not the theoretically ideal ones, since they account for imperfect fixity.