Material properties, shared by every verification. The cross-section is not shared: SIA 263 classifies it from the internal forces, so it belongs to each verification and appears with it.
Checks a steel member of any section under an axial force at the ultimate limit state, to SIA 263:2013, from the area, the radii of gyration and the buckling curves you enter: flexural buckling about both principal axes, section resistance and the slenderness limit. Every step shows its formula and the clause it comes from.
You enter fy,k and fu,k (table 1 of your copy of SIA 263), E (clause 2.2), the area A, the radii of gyration iy and iz, and the buckling curve for each axis (a₀, a, b, c or d; a₀ is for Eurocode 3 only). Then the member length, the end conditions and intermediate supports about each axis, or the buckling lengths directly, the inclination α, NEd as a magnitude, γM1 and γM2 (clause 4.1.3), any holes deducted, and the member category.
You get every intermediate value with its clause, the governing axis and the utilisation of each check with a pass or fail. With the Basic plan, it prints as an A4 calculation note.
A member with A = 11,250 mm², iy = 127.4 mm and iz = 74.9 mm on curves b and c, in fy,k = 235 N/mm², 4.00 m long with pinned ends held against sway about both axes, under NEd = 1,500 kN. These are the module's own results for its worked example, computed when this page was built.
| Input | Value | Ref. |
|---|---|---|
| Characteristic yield strength fy,k | 235 N/mm² | § t.1 |
| Modulus of elasticity E | 210000 N/mm² | § 2.2 |
| Area of the section A | 11250 mm² | § 4.5.1.3 |
| Radius of gyration, strong axis iy | 127.4 mm | |
| Radius of gyration, weak axis iz | 74.9 mm | |
| Curve for buckling about the strong axis | b | |
| Curve for buckling about the weak axis | c | |
| Length of the member l | 4000 mm | |
| Buckling lengths | From the end conditions and intermediate supports | |
| Ends, buckling about the strong axis y | Held against sway — both ends pinned | |
| Intermediate supports against buckling about y ny | 0 | |
| Ends, buckling about the weak axis z | Held against sway — both ends pinned | |
| Intermediate supports against buckling about z nz | 0 | |
| Inclination of the member to the horizontal α | 90 ° | |
| Design compressive force NEd | 1500 kN | |
| Resistance factor γM1 | 1.05 | § 4.1.3 |
| Section resistance | Considered | |
| Characteristic tensile strength fu,k | 360 N/mm² | § t.1 |
| Resistance factor of the net section γM2 | 1.25 | § 4.1.3 |
| Reduction of the section | None — the area entered | |
| Member category | Main structural members — λK ≤ 200 |
| Step | Value | Ref. |
|---|---|---|
| Characteristic yield strength | fy,k = 235 N/mm² | § t.1 |
| Elastic limit slendernessλE = π·√(E/fy,k) — the slenderness at which the Euler load and the plastic axial force coincide: 94, 87, 76 and 67 for S 235, S 275, S 355 and S 460. | λE = 93.91 | |
| Plastic axial forceNpl = fy,k·A — the resistance with no buckling at all. Everything below is a fraction of it. | Npl = 2,643.8 kN | |
| Resistance factor | γM1 = 1.05 | § 4.1.3 |
| Step | Value | Ref. |
|---|---|---|
| Buckling lengthk·l with k = 1.00 — held against sway — both ends pinned. | lK,y = 4,000 mm | |
| SlendernessλK = lK/i. | λK,y = 31.4 | |
| Relative slendernessλ̄K = λK/λE. Referring the slenderness to λE is what lets one set of curves serve every grade of steel. | λ̄K,y = 0.3343 | |
| Buckling curveCurve b. Sections whose extreme fibres carry residual tension AND compression — an I section buckling about its STRONG axis, a welded box — and every case that falls in neither a nor c. | α = 0.34 | |
| Auxiliary termφ = 0,5[1 + α(λ̄K − 0,2) + λ̄K²]. The 0,2 is the plateau: below it a member is stocky enough that buckling takes nothing away. | φ = 0.5787 | § 4.5.1 |
| Buckling coefficientχK = 1/(φ + √(φ² − λ̄K²)), not more than 1. | χK,y = 0.9514 | § 4.5.1 |
| Ultimate buckling resistanceNK = χK·A·fy,k. | NK,y = 2,515.2 kN | § 4.5.1 |
| Design buckling resistanceNK/γM1. | NK,Rd,y = 2,395.4 kN | § 4.5.1 |
| Elastic critical loadNcr = π²·E·I/lK², the Euler load — bifurcation buckling of a perfect bar. A real bar diverges instead, because of its initial bow and its residual stresses, and reaches only 11 % of it here.the perfect bar | Ncr,y = 23,653.2 kN |
| Step | Value | Ref. |
|---|---|---|
| Buckling lengthk·l with k = 1.00 — held against sway — both ends pinned. | lK,z = 4,000 mm | |
| SlendernessλK = lK/i. | λK,z = 53.4 | |
| Relative slendernessλ̄K = λK/λE. Referring the slenderness to λE is what lets one set of curves serve every grade of steel. | λ̄K,z = 0.5687 | |
| Buckling curveCurve c. Sections whose extreme fibres carry residual COMPRESSION — an I section buckling about its WEAK axis, angles, round and square bars, cold-formed hollow sections. | α = 0.49 | |
| Auxiliary termφ = 0,5[1 + α(λ̄K − 0,2) + λ̄K²]. The 0,2 is the plateau: below it a member is stocky enough that buckling takes nothing away. | φ = 0.7520 | § 4.5.1 |
| Buckling coefficientχK = 1/(φ + √(φ² − λ̄K²)), not more than 1. | χK,z = 0.8038 | § 4.5.1 |
| Ultimate buckling resistanceNK = χK·A·fy,k. | NK,z = 2,125.0 kN | § 4.5.1 |
| Design buckling resistanceNK/γM1. | NK,Rd,z = 2,023.9 kN | § 4.5.1 |
| Elastic critical loadNcr = π²·E·I/lK², the Euler load — bifurcation buckling of a perfect bar. A real bar diverges instead, because of its initial bow and its residual stresses, and reaches only 26 % of it here.the perfect bar | Ncr,z = 8,175.5 kN |
| Step | Value | Ref. |
|---|---|---|
| Force a restraint must carryThe cheapest way to raise the buckling resistance of a compressed bar is almost always to SHORTEN ITS BUCKLING LENGTH with intermediate supports. The force such a support has to carry to count as a fixed point is small: 1 % of the compression in the bar. | ≈ 0,01·NEd = 15.00 kN | |
| Halving the governing buckling length would giveOne support at mid-height on the governing axis, everything else unchanged — a gain of 19 % for a support that has to hold about 15.0 kN.what a support buys | NK,Rd = 2,409.9 kN |
| Step | Value | Ref. |
|---|---|---|
| Governing axisBoth principal axes are checked, each with its own buckling length. The smaller resistance governs. | – = Weak axis (z–z) | § 4.5.1.2 |
| BucklingNEd ≤ NK/γM1 | η = 0.741 · Pass74 % of the capacity used | § 4.5.1 |
| Step | Value | Ref. |
|---|---|---|
| Section, yielding | A·fy,k/γM1 = 2,517.9 kN | § 4.1.3 |
| Net areaNo reduction: the area entered. | Anet = 11,250 mm² | |
| Net section, fracture | 0,9·Anet·fu,k/γM2 = 2,916.0 kN | § 4.1.3 |
| Section resistanceNEd ≤ NRd | η = 0.596 · Pass60 % of the capacity used |
| Step | Value | Ref. |
|---|---|---|
| Largest slenderness | λK = 53.4 | |
| Slenderness limitλK ≤ 200 | η = 0.267 · Pass27 % of the capacity used |
Verified clause by clause against SIA 263:2013 and against independent calculations used on real projects, before release.
It depends on where the residual stresses sit relative to the direction of buckling. An I section takes curve b about its strong axis and curve c about its weak axis; angles, round and square bars and cold-formed hollow sections take curve c.
A real bar has an initial bow and residual stresses, so it diverges instead of bifurcating. In the worked example it reaches 26 % of Ncr about the weak axis.
The full calculation is shown on screen, free and without an account. A print-ready A4 calculation note is part of the Basic plan.
Norm: SIA 263:2013 · Page updated: 27 October 2026