Trapezoidal steel sheeting in bending to SIA 263
Checks the bending resistance of a trapezoidal steel sheet on its effective section, per metre of width, in the span or over a support, to SIA 263:2013. 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 the bending resistance of a trapezoidal steel sheet on its effective section, per metre of width, in the span or over a support, to SIA 263:2013. Every step shows its formula and the clause it comes from.
What it checks
- Reduction factor ε = √(235/fy), table 5a, with the edge stress of every flat taken as σmax = fy.
- Web geometry of one rib: horizontal projection sw = (p − bt − bb)/2, length between the bends sw,L = √(sw² + (h − t)²), and the inclination φ.
- Slenderness and class of each flat, b/t, table 5a: both webs as internal flats in bending, the compressed flange as an internal flat in compression.
- Effective widths of both webs and the compressed flange, from the stress ratio ψ, the buckling coefficient k, the critical plate stress σcr,P and the plate slenderness λp, each in two equal parts at the ends of the compressed zone.
- Effective section per metre: zeff by iteration, Aeff, Ieff, Weff,c, Weff,t, and Weff as the smaller of the two moduli.
- Bending resistance, clause 4.1.3: MEd ≤ MRd = fy · Weff / γM1, with the gross-section value for comparison.
What you enter and what you get
You enter fy,k (table 1), E (clause 2.2), and one rib: the overall height h (out-to-out), the core thickness t without coating, the top and bottom flange widths bt and bb (at mid-thickness between the bends) and the rib pitch p. You choose sagging (top flanges compressed) or hogging (bottom flanges compressed), and enter γM1 (clause 4.1.3) and the design bending moment per metre width MEd in kNm/m.
You get the web geometry, the class of each flat, the effective widths, the section properties, the resistance and the utilisation with a pass or fail. With the Pro plan, it prints as an A4 calculation note.
Assumptions and limits
- Rib values are converted to one metre of width by the factor 1000/p.
- Ultimate check of the section: the most compressed fibre reaches the yield strength, and the plastic reserve of the tension zone is not counted.
- A flange with a stiffening groove is not covered; the groove would divide it in two.
- The usual range is 0.6 to 1.5 mm for the thickness and 30 to 180 mm for the height.
- Not covered: shear and web crippling at the supports, the fixings, deflection and the construction stage. Sheeting is also the formwork of a composite slab, the support of the roof insulation and, during construction, the working platform; those states can govern.
Worked example
A sheet 106 mm high and 0.75 mm thick, with bt = 40 mm, bb = 112 mm and a pitch of 250 mm, fy,k = 320 N/mm² and γM1 = 1.05, in sagging under 5.0 kNm/m; the webs and the top flange are class 4. 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 | 320 N/mm² | § t.1 |
| Modulus of elasticity E | 210000 N/mm² | § 2.2 |
| Overall height h | 106 mm | |
| Sheet thickness t | 0.75 mm | |
| Top flange width bt | 40 mm | |
| Bottom flange width bb | 112 mm | |
| Rib pitch p | 250 mm | |
| The section is | Sagging — in the span, top flanges compressed | |
| Resistance factor γM1 | 1.05 | § 4.1.3 |
| Design bending moment per metre width MEd | 5.0 kNm/m |
| Step | Value | Ref. |
|---|---|---|
| Characteristic yield strength | fy,k = 320 N/mm² | § t.1 |
| Reduction factorε = √(235/fy) | ε = 0.8570 | § t.5a |
| Edge stress for the effective widthsfy: the ultimate check, the most compressed fibre at yield. Not asked for. | σmax = 320 N/mm² | |
| ThicknessUsual range 0.6 to 1.5 mm for profiled sheeting, within a base material range of 0.5 to 5.0 mm. | t = 0.75 mm | |
| HeightUsual range 30 to 180 mm for profiled sheeting. | h = 106.0 mm | |
| Web — horizontal projection(p − bt − bb)/2 | sw = 49.0 mm | |
| Web — length between the bends√(sw² + (h − t)²) | sw,L = 116.1 mm | |
| Web inclination | φ = 65.0 ° | |
| Web, left — slendernessMid-line flat width 116.1 mm, internal (held along both edges), in bending. Class limits 61.7 · 71.1 · 106.3 — class 4, past the class 3 limit by a factor of 1.46. | b/t = 154.8 | § t.5a |
| Top flange — slendernessMid-line flat width 40.0 mm, internal (held along both edges), compressed. Class limits 28.3 · 32.6 · 36.0 — class 4, past the class 3 limit by a factor of 1.48. | b/t = 53.3 | § t.5a |
| Web, right — slendernessMid-line flat width 116.1 mm, internal (held along both edges), in bending. Class limits 61.7 · 71.1 · 106.3 — class 4, past the class 3 limit by a factor of 1.46. | b/t = 154.8 | § t.5a |
| 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 |
| Step | Value | Ref. |
|---|---|---|
| Web, left — effective widthψ = -0.460, k = 12.775, σcr,P = 101.2 N/mm², λp = 1.778. Winter, on the compressed width 79.5 mm: 49.3 % of it works, in two equal parts at the ends of the compressed zone. | beff = 39.2 mm | |
| Top flange — effective widthψ = 1.000, k = 4.000, σcr,P = 266.9 N/mm², λp = 1.095. Winter, on the compressed width 40.0 mm: 73.0 % of it works, in two equal parts at the ends of the compressed zone. | beff = 29.2 mm | |
| Web, right — effective widthψ = -0.460, k = 12.775, σcr,P = 101.2 N/mm², λp = 1.778. Winter, on the compressed width 79.5 mm: 49.3 % of it works, in two equal parts at the ends of the compressed zone. | beff = 39.2 mm | |
| Neutral axis of the gross section, from the bottom mid-line | zG = 42.76 mm | |
| Neutral axis of the effective sectionMoved by 9.59 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 = 33.17 mm | |
| Gross area | A = 1,153 mm²/m | |
| Effective area76.2 % of the gross area works. | Aeff = 878 mm²/m | |
| Second moment of area, gross | I = 1,793,803 mm⁴/m | |
| Second moment of area, effective | Ieff = 1,390,143 mm⁴/m | |
| Elastic section modulus, grossThe smaller of the two extreme fibres. | Wel = 28,535 mm³/m | |
| Effective modulus, compressed sideIeff over the distance from the effective axis to the outer face of the compressed side. | Weff,c = 19,186 mm³/m | |
| Effective modulus, tensioned side | Weff,t = 41,443 mm³/m | |
| Effective section modulusThe smaller of the two: 67.2 % of the gross elastic modulus. The plastic reserve of the tension zone is not counted. | Weff = 19,186 mm³/m |
| Step | Value | Ref. |
|---|---|---|
| Design bending resistanceMRd = fy·Weff/γM1 = 320 × 19,186 mm³/m / 1.05. | MRd = 5.847 kNm/m | § 4.1.3 |
| On the gross sectionWhat the section would carry if no flat buckled — local buckling costs 33 % of it here.for comparison | fy·Wel/γM1 = 8.697 kNm/m | |
| Bending on the effective sectionMEd ≤ MRd | η = 0.855 · Pass86 % of the capacity used | § 4.1.3 |
How this module is verified
Verified clause by clause against SIA 263:2013 and against independent calculations used on real projects, before release.
Questions
Sagging or hogging: which should I check?
Both. In the span of a continuous sheet the top flanges are compressed; over an intermediate support the bottom ones are. The narrower of the two flanges usually decides which case is worse.
Why is the neutral axis found by iteration?
The effective widths depend on the stress distribution, which depends on the position of the axis. In the worked example the axis moves 9.59 mm away from the compressed side.
Which plan includes this module?
The Pro plan, with its calculation note.
Related
- Cold-formed channel in bending
- Cold-formed lipped channel in bending
- Steel bending: all sections
- SIA 263: modules and clauses
Norm: SIA 263:2013 · Page updated: 26 October 2026