Steel Design · 8 min read
How to size a steel column (compression and buckling)
A clear step-by-step for sizing a steel column to Eurocode 3 — squash load, slenderness, buckling curves, and effective length.
Steel columns fail in two ways: by squashing (yielding the cross-section) or by buckling (the column bows out under axial load). Eurocode 3 handles both with a single equation that scales the squash load down by a buckling reduction factor χ. Here's the practical sequence.
Step 1 — Get the design axial load NEd
Sum tributary loads from floors above, factor with 1.35·Gₖ + 1.5·Qₖ, and add the column self-weight. For multi-storey columns, do the take-down storey by storey — the column gets heavier the lower you go.
Step 2 — Effective length Lcr
The effective length depends on end restraints. For pinned-pinned, Lcr = L. For fixed-fixed in a non-sway frame, Lcr ≈ 0.7·L. For a sway column fixed at the base only, Lcr ≈ 2·L. Get this wrong and your buckling resistance is wildly off — it's the single biggest source of error in column design.
Step 3 — Trial section and squash load Npl,Rd
Pick a trial UC. Most building columns are UCs because they have similar Iy and Iz, which is good for axial-dominated members. Calculate the plastic squash load:
Npl,Rd = A · fy ÷ γₘ₀Step 4 — Non-dimensional slenderness λ̄
λ̄ = √(A · fy ÷ Ncr) Ncr = π² · E · I ÷ Lcr²Ncr is the Euler critical load. λ̄ is the ratio of yield strength to elastic buckling strength expressed as a slenderness. λ̄ ≤ 0.2 means buckling is irrelevant. λ̄ > 1.5 means the column is slender and buckling dominates.
Step 5 — Pick a buckling curve and find χ
EN 1993 Table 6.2 maps section type, axis, and steel grade to one of five buckling curves (a₀, a, b, c, d). Each curve gives a different reduction factor χ for a given λ̄. For S275 UCs about the minor axis, you'll usually be on curve c.
Nb,Rd = χ · A · fy ÷ γₘ₁The check: NEd ÷ Nb,Rd ≤ 1.0. If not, increase the section, reduce the effective length (add bracing), or both.
Quick worked example
203×203×60 UC, S275, height 4.0 m, pinned-pinned, NEd = 1200 kN. A = 76.4 cm², iz = 5.20 cm. λ̄z = (Lcr ÷ iz) ÷ λ₁ = (4000 ÷ 52.0) ÷ 86.8 ≈ 0.886. From curve c, χ ≈ 0.61. Nb,Rd = 0.61 × 7640 × 275 ÷ 1.0 ≈ 1281 kN. Utilisation = 1200 ÷ 1281 = 0.94. OK — but tight. A 203×203×71 UC gives more headroom.
Frequently asked questions
When is buckling not a concern?
When λ̄ ≤ 0.2, EC3 lets you ignore buckling and design for the squash load alone. This is rare for real columns — short, stocky members like pedestals or stub columns.
Should I use UB or UC for columns?
UCs almost always. They have similar properties about both axes, so weak-axis buckling doesn't dominate. UBs are very weak about the minor axis and only suit columns with continuous restraint to the minor axis (e.g. cladding rails).
How do I handle combined axial and bending?
Use the interaction equations in EN 1993-1-1 clause 6.3.3. The simplified Method B (Annex B) is usually enough for typical building columns. The full Method A is more accurate but more work.