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 rolled 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 from five 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 five profile 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 the section carries (simple bending about the strong axis, uniform compression, or α and ψ for the web), or entered directly. Av can be computed or entered from a table.
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 profile. With the Basic plan, it prints as an A4 calculation note.
A rolled section with h = 250 mm, b = 260 mm, tw = 7.5 mm, tf = 12.5 mm and r = 24 mm (an HEA 260) in fy,k = 235 N/mm² and τy,k = 135 N/mm², under VEd = 350 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 |
| Characteristic shear yield strength τy,k | 135 N/mm² | § t.1 |
| Modulus of elasticity E | 210000 N/mm² | § 2.2 |
| Overall depth h | 250 mm | § (42a) |
| Flange width b | 260 mm | § (42a) |
| Web thickness tw | 7.5 mm | § (42a) |
| Flange thickness tf | 12.5 mm | § (42a) |
| Root radius r | 24 mm | § (42a) |
| 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 | 350 kN | § 5.1.4 |
| Step | Value | Ref. |
|---|---|---|
| Characteristic yield strength | fy,k = 235 N/mm² | § t.1 |
| Characteristic shear yield strengthEntered from table 1, not derived. | τy,k = 135 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 = 135.7 N/mm² | § 1.2.3 |
| Resistance factor | γM1 = 1.05 | § 4.1.3 |
| AreaPlates plus four fillets of (1 − π/4)·r² each. Against the published section tables this is within 0,06 % on an HEA 340, an HEA 260 and an IPE 300 — the fillets are what makes that possible, and leaving them out costs 4 to 5 %. | A = 8,682 mm² | § 5.1.4.2 |
| Clear web depthh − 2·tf, between the flange faces. | hw = 225.0 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 = 1,781 mm² | § 5.2.4 |
| Second moment of area | Iy = 104,539,541 mm⁴ | § 5.3.4 |
| First moment above the neutral axisWhat (57) needs. Wpl = 2·S for a doubly symmetric section, which is 920·10³ mm³ here. | S = 459,886 mm³ | |
| Elastic section modulusReported so the section can be recognised against a table row. Not used by this check.not used | Wel = 836,316 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 = 1.0000 | § t.5a |
| Step | Value | Ref. |
|---|---|---|
| Web — flat widthc = h − 2·tf − 2·r, between the fillet toes. | cw = 177.0 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 = 23.60 | § t.5a |
| Web — is it class 3 or bettercw/tw ≤ 124.0 | η = 0.190 · Pass19 % of the capacity used | § t.5a |
| Step | Value | Ref. |
|---|---|---|
| Flange — flat widthc = (b − tw)/2 − r, from the fillet toe to the free tip. | cf = 102.3 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 = 8.18 | § t.5b |
| Flange outstand — is it class 3 or bettercf/tf ≤ 14.0 | η = 0.584 · Pass58 % 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 = 31.67 | § (42) |
| Limit√(4·210,000/235). 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) = 59.79 | § (42) |
| Web stability — the condition (41) is granted under(h − tf)/tw ≤ √(4E/fy,k) | η = 0.530 · Pass53 % of the capacity used | § (42) |
| Step | Value | Ref. |
|---|---|---|
| Junction and fillet termWhat (42a) adds back after taking the two flanges out: the strip of web running through the flange, plus both root fillets. It is 24.1 % of Av here — a quarter of the shear area on an HEA 260 — which is why the web alone will not do for a class 1 or 2 rolled section. | (tw + 2r)·tf = 693.75 mm² | § (42a) |
| Shear areaAv = A − 2·b·tf + (tw + 2r)·tf, equation (42a). The term it adds back is the web strip through the flange plus the two root fillets. | Av = 2,876 mm² | § 5.1.4.2 |
| Aw, were the section class 3The two differ by a factor of 1.61 here. That factor is what the class is worth in shear, and it is larger than what the class is worth in bending.not used | Aw = 1,781 mm² | § 5.2.4 |
| Shear resistanceVR = τy,k·Av, the plastic model: the stress is taken as uniform over the area at τy,k. | VR = 388.2 kN | § (41) |
| Design shear resistanceVRd = VR/γM1 | VRd = 369.7 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 = 2,722 mm² | § (41) |
| Elastic shear resistance, for comparisonThe force that would bring the ELASTIC peak τ = V·S/(I·t) to τy,k. It is 1.69 times smaller than VR, and that ratio is the redistribution the plastic model of (41) relies on.not used | Vel = 230.2 kN | § (57) |
| Shear resistance of the sectionVEd ≤ VRd | η = 0.947 · Pass95 % of the capacity used | § (41) |
Verified clause by clause against SIA 263:2013 and against independent calculations used on real projects, before release.
Equation (42a) draws part of the flange and both root fillets into Av. On an HEA 260 this gives 2,874 mm² against 1,781 mm² for the web area Aw, a factor of 1.61. For a class 3 section, clause 5.2.4 puts Aw in its place, so the module reports it alongside.
Vel is the force that brings the elastic peak τ = V · S/(I · t) of equation (57) to τy,k. It is for comparison only: the ratio VR/Vel is the redistribution the plastic model of equation (41) relies on.
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