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 welded I or H section under a shear force in the plane of the web, to SIA 263:2013: section class, web stability and the plastic shear resistance on the web area, from four dimensions. Every step shows its formula and the clause it comes from.
You enter fy,k and τy,k from table 1 of your copy of SIA 263, for the grade and the plate thickness, E (clause 2.2), the four plate dimensions, γM1 (clause 4.1.3) and the design shear force VEd as a magnitude. The class is computed from the dimensions and the direct stress on the section (simple bending about the strong axis, uniform compression, or α and ψ for the web), or entered directly. Av can also be entered directly.
You get the section properties, the class, the resistance model, the shear resistance and the utilisation of each check with a pass or fail, and a drawing of the section. With the Basic plan, it prints as an A4 calculation note.
A welded section with h = 700 mm, flanges 300 × 25 mm and a web 16 mm thick in fy,k = 355 N/mm² and τy,k = 205 N/mm², under VEd = 1,200 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 | 355 N/mm² | § t.1 |
| Characteristic shear yield strength τy,k | 205 N/mm² | § t.1 |
| Modulus of elasticity E | 210000 N/mm² | § 2.2 |
| Overall depth h | 700 mm | § 5.1.4 |
| Flange width b | 300 mm | § 5.1.4 |
| Web thickness tw | 16 mm | § 5.1.4 |
| Flange thickness tf | 25 mm | § 5.1.4 |
| Section class | Computed from the dimensions | § t.4 |
| Direct stress on the section | Simple bending about the strong axis | § t.5a |
| Shear area Av | Computed from the dimensions | § 5.1.4.2 |
| Resistance factor γM1 | 1.05 | § 4.1.3 |
| Design shear force VEd | 1200 kN | § 5.1.4 |
| Step | Value | Ref. |
|---|---|---|
| Characteristic yield strength | fy,k = 355 N/mm² | § t.1 |
| Characteristic shear yield strengthEntered from table 1, not derived. | τy,k = 205 N/mm² | § t.1 |
| The definition in §1.2.3§1.2.3 defines τy as fy/√3 and table 1 then tabulates values that disagree with it in all eight cells. The table governs.not used | fy,k/√3 = 205.0 N/mm² | § 1.2.3 |
| Resistance factor | γM1 = 1.05 | § 4.1.3 |
| AreaThe sum of the rectangles, exactly. | A = 25,400 mm² | § 5.1.4.2 |
| Clear web depthh − 2·tf, between the flange faces. | hw = 650.0 mm | § 5.1.4.2 |
| Web areaAw = (h − tf)·tw, the notation of figure 1. Note it is taken between the flange MID-PLANES and is therefore a little larger than hw·tw. | Aw = 10,800 mm² | § 5.2.4 |
| Second moment of area | Iy = 2,075,541,667 mm⁴ | § 5.3.4 |
| First moment above the neutral axisWhat (57) needs. Wpl = 2·S for a doubly symmetric section, which is 6,753·10³ mm³ here. | S = 3,376,250 mm³ | |
| Elastic section modulusReported so the section can be recognised against a table row. Not used by this check.not used | Wel = 5,930,119 mm³ | § 5.1.4 |
| Section classComputed from the flat widths and simple bending. The web and the flange equally governs. | – = 1 | § t.4 |
| Resistance model | PLASTIC. §5.1.4.1 (41) takes the whole shear area at τy,k: the section is stocky enough to yield right through Av before anything buckles, so the stress is taken as uniform over it. | § 5.1.4.1 |
| Material factorε = √(235/fy,k) | ε = 0.8136 | § t.5a |
| Step | Value | Ref. |
|---|---|---|
| Web — flat widthc = h − 2·tf, between the flange faces. | cw = 650.0 mm | § t.5a |
| Web — slendernessclass 1: 36ε/α class 2: 41,5ε/α [α ≤ 0,5] class 3: 62ε(1−ψ)√(−ψ) [ψ ≤ −1] | cw/tw = 40.63 | § t.5a |
| Web — is it class 3 or bettercw/tw ≤ 100.9 | η = 0.403 · Pass40 % of the capacity used | § t.5a |
| Step | Value | Ref. |
|---|---|---|
| Flange — flat widthc = (b − tw)/2, from the face of the web to the free tip. | cf = 142.0 mm | § t.5b |
| Flange — slendernessclass 1: 9ε/α class 2: 10ε/α [free edge in compression] class 3: 14ε, the tabulated uniform-compression row of table 5b | cf/tf = 5.68 | § t.5b |
| Flange outstand — is it class 3 or bettercf/tf ≤ 11.4 | η = 0.499 · Pass50 % of the capacity used | § t.5b |
| Step | Value | Ref. |
|---|---|---|
| Web slendernessThe ratio (42) is written against. It is NOT the flat width of table 5a: this one runs between the flange mid-planes and takes no account of the fillets. | (h − tf)/tw = 42.19 | § (42) |
| Limit√(4·210,000/355). 59,8 in S 235 and 48,6 in S 355 — the stronger the steel, the sooner its web buckles in shear rather than yielding. | √(4E/fy,k) = 48.64 | § (42) |
| Web stability — the condition (41) is granted under(h − tf)/tw ≤ √(4E/fy,k) | η = 0.867 · Pass87 % of the capacity used | § (42) |
| Step | Value | Ref. |
|---|---|---|
| Shear areaAv = Aw = (h − tf)·tw. (42a) adds the fillets of a rolled section, and a welded one has none. | Av = 10,800 mm² | § 5.1.4.2 |
| Shear resistanceVR = τy,k·Av, the plastic model: the stress is taken as uniform over the area at τy,k. | VR = 2,214.0 kN | § (41) |
| Design shear resistanceVRd = VR/γM1 | VRd = 2,108.6 kN | § (41) |
| Shear area required(41) turned round: Av,req = γM1·VEd/τy,k. The sizing form, the one used to CHOOSE a section rather than to check one. | Av,req = 6,146 mm² | § (41) |
| Elastic shear resistance, for comparisonThe force that would bring the ELASTIC peak τ = V·S/(I·t) to τy,k. It is 1.10 times smaller than VR, and that ratio is the redistribution the plastic model of (41) relies on.not used | Vel = 2,016.4 kN | § (57) |
| Shear resistance of the sectionVEd ≤ VRd | η = 0.569 · Pass57 % of the capacity used | § (41) |
Verified clause by clause against SIA 263:2013 and against independent calculations used on real projects, before release.
In the worked example the web reaches 87 % of the limit of equation (42). It passes, but a thinner web, a higher grade or a smaller section would take it past, and past it the shear area model of equation (41) no longer applies.
They answer different clauses. Table 5a uses the flat width between the flange faces; equation (42) is written against (h − tf)/tw, between the flange mid-planes.
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