Structural Engineering Tools

Cold-formed channel under bending and axial force to SIA 263

Checks a cold-formed channel without lips under a moment about its major axis and an axial force in compression or tension, on its effective section, to SIA 263:2013: the section, flexural buckling and their interaction. Every step shows its formula and the clause it comes from.

Pro SIA 263:2013

Part of the Pro plan

This module is included in the Pro plan and above, with its calculation note.

About this check

clauses, worked example and questions

Checks a cold-formed channel without lips under a moment about its major axis and an axial force in compression or tension, on its effective section, to SIA 263:2013: the section, flexural buckling and their interaction. Every step shows its formula and the clause it comes from.

What it checks

  • Class of each flat, with ε = √(235/fy): the web in bending (table 5a) and the compressed flange as an outstand (table 5b), on the mid-line widths hp = h − t and bp = b − t/2.
  • Effective section in bending, at σmax = fy: effective widths from ψ, k, σcr,P and λp, zeff by iteration, and Weff, the smaller of Weff,c and Weff,t.
  • Effective section in compression, EN 1993-1-3 6.1.3: every flat compressed at fy,k, giving Aeff and βA = Aeff/A.
  • Shift of the effective centroid eN, EN 1993-1-3 6.1.9: the channel is symmetric about y only, so NEd adds ΔMz,Ed = NEd · eN, resisted by Mz,Rd = fy,k · Wz,eff/γM1.
  • Resistances, clause 4.1.3: Nc,Rd = Aeff · fy,k/γM1 and MRd = fy,k · Weff/γM1.
  • Flexural buckling about both axes, clause 4.5.1, curve b: NK,Rd = χK · Aeff · fy,k/γM1, with λ̄K = (lK/i)/λE · √βA and i on the gross section.
  • Check 1, section, EN 1993-1-3 6.1.9: NEd/Nc,Rd + My,Ed/MRd + ΔMz,Ed/Mz,Rd ≤ 1.
  • Check 2, stability: NEd/NK,Rd, with the smaller buckling resistance, plus the moment terms amplified by 1/(1 − NEd/Ncr,y), Ncr,y on Ieff, and 1/(1 − NEd/Ncr,z).

What you enter and what you get

You enter fy,k (table 1), E (clause 2.2), the out-to-out depth h and flange width b, the core thickness t, γM1 (clause 4.1.3), the sense of the axial force, NEd, the largest moment My,Ed and the moment diagram. For buckling, the length l with the end conditions and the intermediate supports ny and nz in each plane, or the buckling lengths directly.

You get both effective sections, the resistances and the two utilisations with a pass or fail. With the Pro plan, it prints as an A4 calculation note.

Assumptions and limits

  • SIA 263 refers to SN EN 1993-1-3 for thin-walled members; the rows that follow it name the EN clause.
  • ω = 0.6 + 0.4 · Mmin/Mmax, at least 0.4, for end moments only; otherwise ω = 1.0. No shape factor ξ: local buckling has taken the plastic reserve.
  • Under tension the member checks fall away, since the tension would stabilise the member.
  • The shear centre of a C lies outside the web. The member is assumed held against twist, by the sheeting or by pairing the channels; lateral-torsional, torsional and flexural-torsional buckling are not checked.
  • Not covered: shear and web crippling at the supports, the fixings, deflection and the construction stage.

Worked example

A channel 200 × 70 × 2 mm with fy,k = 350 N/mm² and γM1 = 1.05, 4,000 mm long and pinned at both ends, with three intermediate supports about z, under a compression of 8 kN and a moment of 3.0 kNm from a transverse load. These are the module's own results for its worked example, computed when this page was built.

Inputs
InputValueRef.
Characteristic yield strength fy,k350 N/mm²§ t.1
Modulus of elasticity E210000 N/mm²§ 2.2
Overall depth h200 mm
Flange width b70 mm
Thickness t2 mm
Resistance factor γM11.05§ 4.1.3
The axial force isCompression
Design axial force NEd8 kN
Largest moment about y along the member My,Ed3.0 kN·m
Moment diagramTransverse load between the ends — ω = 1,0
Buckling lengthsFrom the end conditions and intermediate supports
Length of the member l4000 mm
Ends, buckling about the strong axis yHeld against sway — both ends pinned
Intermediate supports against buckling about y ny0
Ends, buckling about the weak axis zHeld against sway — both ends pinned
Intermediate supports against buckling about z nz3
Calculation: Common values
StepValueRef.
Characteristic yield strengthfy,k = 350 N/mm²§ t.1
Reduction factorε = √(235/fy)ε = 0.8194§ t.5a
Edge stress for the effective widthsfy: the ultimate check, the most compressed fibre at yield. Not asked for.σmax = 350 N/mm²
ThicknessUsual range 1.0 to 5.0 mm for a cold-formed profile, within a base material range of 0.5 to 5.0 mm.t = 2.00 mm
HeightUsual range 80 to 250 mm for a cold-formed profile.h = 200.0 mm
Web — mid-line heighth − thp = 198.0 mm
Flange — mid-line widthb − t/2bp = 69.0 mm
Web — slendernessMid-line flat width 198.0 mm, internal (held along both edges), in bending. Class limits 59.0 · 68.0 · 101.6 — class 3.b/t = 99.0§ t.5a
Top flange — slendernessMid-line flat width 69.0 mm, outstand (one edge free), compressed. Class limits 7.4 · 8.2 · 11.5 — class 4, past the class 3 limit by a factor of 3.01.b/t = 34.5§ t.5b
Section classClass 4, which is the normal condition of a thin-walled section: the flats buckle locally before the yield strength is reached, and only an EFFECTIVE section works — computed below.– = Class 4§ t.5a
Calculation: Common values — the effective section
StepValueRef.
Web — effective widthψ = -0.665, k = 16.312, σcr,P = 315.9 N/mm², λp = 1.053. Winter, on the compressed width 118.9 mm: 75.1 % of it works, in two equal parts at the ends of the compressed zone.beff = 89.4 mm
Top flange — effective widthψ = 1.000, k = 0.426, σcr,P = 67.9 N/mm², λp = 2.270. Winter, on the compressed width 69.0 mm: 39.8 % of it works, next to the supported edge.beff = 27.5 mm
Neutral axis of the gross section, from the bottom mid-linezG = 99.00 mm
Neutral axis of the effective sectionMoved by 19.94 mm away from the compressed side. Found by iteration — the effective widths depend on the stress distribution, which depends on the axis: 8 rounds here.zeff = 79.06 mm
Gross areaA = 672 mm²
Effective area78.8 % of the gross area works.Aeff = 530 mm²
Second moment of area, grossI = 3,998,900 mm⁴
Second moment of area, effectiveIeff = 2,877,127 mm⁴
Elastic section modulus, grossThe smaller of the two extreme fibres.Wel = 39,989 mm³
Effective modulus, compressed sideIeff over the distance from the effective axis to the outer face of the compressed side.Weff,c = 23,988 mm³
Effective modulus, tensioned sideWeff,t = 35,937 mm³
Effective section modulusThe smaller of the two: 60.0 % of the gross elastic modulus. The plastic reserve of the tension zone is not counted.Weff = 23,988 mm³
Calculation: Common values — the effective section in compression
StepValueRef.
Bottom flange — effective width in compressionOutstand, ψ = 1, k = 0.426, λp = 2.270. 39.8 % of it works.beff,N = 27.5 mm
Web — effective width in compressionInternal, ψ = 1, k = 4.000, λp = 2.126. 42.2 % of it works.beff,N = 83.5 mm
Top flange — effective width in compressionOutstand, ψ = 1, k = 0.426, λp = 2.270. 39.8 % of it works.beff,N = 27.5 mm
Effective area in compressionEvery flat compressed at σ = fy,k. 41.2 % of the gross area A = 672 mm² works.Aeff = 277 mm²EN 1993-1-3 6.1.3
Thinness factorβA = Aeff/A.βA = 0.4119
Shift of the effective centroidTowards the web: the channel is symmetric about y only, so losing parts of its flanges moves the centroid across the web. NEd stays at the gross centroid and acts on the effective section with this eccentricity about z.eN = 8.72 mmEN 1993-1-3 6.1.9
Additional moment about zNEd·eN.ΔMz,Ed = 0.0698 kN·mEN 1993-1-3 6.1.9
Effective modulus about z, section in compressionIz of the effective section, 19,433 mm⁴, over 22.0 mm from its centroid to the fibre ΔMz compresses: the end of the effective flanges, towards the tips.Wz,eff = 883 mm³
Bending resistance about zfy,k·Wz,eff/γM1.Mz,Rd = 0.294 kN·m
Compression resistance of the effective sectionAeff·fy,k/γM1.Nc,Rd = 92.3 kN§ 4.1.3
Bending resistance of the effective sectionfy,k·Weff/γM1, Weff = 23,988 mm³ — the section in bending alone.MRd = 7.996 kN·m§ 4.1.3
Calculation: Common values — flexural buckling
StepValueRef.
Elastic limit slendernessπ·√(E/fy,k).λE = 76.95
Buckling length, strong axisk·l with k = 1.00.lK,y = 4,000 mm
Relative slenderness, strong axis(lK/i)/λE·√βA, i = 77.1 mm on the gross section: local buckling enters through √βA.λ̄K,y = 0.4325
Buckling resistance, strong axisχK·Aeff·fy,k/γM1, curve b (plain or lipped channel), χK = 0.9130.NK,Rd,y = 84.2 kN§ 4.5.1
Buckling length, weak axisl/(nz + 1).lK,z = 1,000 mm
Relative slenderness, weak axis(lK/i)/λE·√βA, i = 21.2 mm on the gross section: local buckling enters through √βA.λ̄K,z = 0.3926
Buckling resistance, weak axisχK·Aeff·fy,k/γM1, curve b (plain or lipped channel), χK = 0.9290.NK,Rd,z = 85.7 kN§ 4.5.1
Critical load of the section reduced for bendingπ²·E·Ieff/lK,y², Ieff = 2,877,127 mm⁴.Ncr,y = 372.7 kN
Critical load about zπ²·E·Iz/lK,z² on the gross section, Iz = 303,221 mm⁴; it amplifies ΔMz.Ncr,z = 628.5 kN
Calculation: Check 1 — section, compression with bending
StepValueRef.
Section, compression with bendingNEd/Nc,Rd + My,Ed/MRd + ΔMz,Ed/Mz,Rd ≤ 1η = 0.699 · Pass70 % of the capacity usedEN 1993-1-3 6.1.9
Calculation: Check 2 — stability, compression with bending
StepValueRef.
Moment factor1,0 for this moment diagram.ω = 1.0000
Amplification factor1/(1 − NEd/Ncr,y) = 1.0219
Axial termThe smaller of the two buckling resistances.NEd/NK,Rd = 0.0950
Moment term, amplifiedNo shape factor ξ: local buckling has taken the plastic reserve of a thin-walled section.(ω·MEd + ΔM)/((1 − NEd/Ncr)·MRd) = 0.3834
Term of ΔMz, amplifiedΔMz,Ed/((1 − NEd/Ncr,z)·Mz,Rd) = 0.2402
Stability, compression with bendingNEd/NK,Rd + amplified moment terms ≤ 1η = 0.719 · Pass72 % of the capacity used

How this module is verified

Verified clause by clause against SIA 263:2013 and against independent calculations used on real projects, before release.

Questions

Why does a small eccentricity weigh so much?

ΔMz,Ed acts about the weak axis, where the effective channel has little resistance. In the worked example eN = 8.72 mm gives only 0.070 kNm, yet it takes 0.237 of the section check, against Mz,Rd = 0.294 kNm.

How does local buckling enter the buckling check?

Through βA: in the area of NK,Rd, and under the root in λ̄K. In the worked example only 41.2 % of the gross area works in compression.

Which plan includes this module?

The Pro plan, with its calculation note.

Related

Norm: SIA 263:2013 · Page updated: 28 October 2026