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 the interaction of an axial force with a shear force on a solid rectangle, to SIA 263:2013: the shear alone first, then the axial force on the lower of two plastic models. 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, the thickness t across the plate and the depth h along the shear, γM1 (clause 4.1.3), and NEd and VEd as magnitudes.
You get both models side by side, the one verified and why, the loss to the shear, the design resistance and the utilisation of each check with a pass or fail. With the Basic plan, it prints as an A4 calculation note.
A plate 20 mm thick and 300 mm deep in fy,k = 235 N/mm² and τy,k = 135 N/mm², with NEd = 400 kN and VEd = 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 | 235 N/mm² | § t.1 |
| Characteristic shear yield strength τy,k | 135 N/mm² | § t.1 |
| Thickness t | 20 mm | |
| Depth h | 300 mm | |
| Resistance factor γM1 | 1.05 | § 4.1.3 |
| Design axial force NEd | 400 kN | |
| Design shear force VEd | 200 kN |
| 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 |
| Resistance factor | γM1 = 1.05 | § 4.1.3 |
| Section classNot asked for and not computed: a solid rectangle has no slender plate to classify, so the class is 1 by construction. | – = 1 | § t.4 |
| Resistance model | PLASTIC. Table 4 allows a plastic resistance to a class 1 section, so the section may redistribute and the plastic models apply. | § t.4 |
| AreaA = t·h | A = 6,000 mm² | |
| Plastic axial resistance, no shearNpl = fy,k·A, what the section would carry with VEd = 0. Everything in check 2 is what the shear costs against it. | Npl = 1,410.0 kN |
| Step | Value | Ref. |
|---|---|---|
| Average shear stress | τEd = VEd/A = 33.33 N/mm² | |
| Shear alone, before any interactionτEd ≤ τy,k/γM1 | η = 0.259 · Pass26 % of the capacity used |
| Step | Value | Ref. |
|---|---|---|
| Model 1 — shear stress, taken as uniformτ = VEd/A | τ = 33.33 N/mm² | |
| Model 1 — reduced yield strengthfyr = √(fy,k² − 3τ²). The von Mises criterion read backwards: what is left for a direct stress once the shear has taken its share at every point of the section. | fyr = 227.80 N/mm² | |
| Model 1 — reduced plastic axial resistanceNpl,V = fyr·A | Npl,V = 1,366.8 kN | |
| Model 2 — depth of the strip at τy,kFrom τ = VEd/(2a·t) = τy,k. The whole shear is put into this central band at full shear yield; at τ = τy a point has nothing left for a direct stress, so the band is simply removed from the area available to N. | 2a = 74.1 mm | |
| Model 2 — fraction of the depth taken by the strip | 2a/h = 24.7 % | |
| Model 2 — reduced plastic axial resistanceNpl,V = fy,k·(A − 2a·t) | Npl,V = 1,061.9 kN | |
| Which model is verified | The LOWER of the two, which here is model 2, the central shear strip at 1,061.9 kN against 1,366.8 kN for the other, a ratio of 1.29. The two disagree because they idealise the same section differently, and both are safe lower bounds. | |
| Loss to the shearHow much of Npl the shear has taken, on the governing model. | – = 24.7 % | |
| Design resistance | Npl,V/γM1 = 1,011.3 kN | |
| Axial force with the shear force actingNEd ≤ Npl,V/γM1 | η = 0.396 · Pass40 % of the capacity used | |
| Elastic comparison stress, for referenceWhat the elastic check would give at the governing point, against the limit 223.8 N/mm². The elastic route is always available and always the more conservative; the gap between it and the plastic answer above is the redistribution the class permits. On a rectangle the elastic shear peak is 1,5·V/A.not used | σg,Ed = 109.29 N/mm² |
Verified clause by clause against SIA 263:2013 and against independent calculations used on real projects, before release.
They idealise the same section differently. In the worked example the central strip gives Npl,V = 1,061.9 kN against 1,366.8 kN for the reduced yield strength, a ratio of 1.29. The module does not pick the more favourable one on your behalf.
It is reported for reference and not used. The elastic route is always available and always the more conservative; the gap to the plastic result is the redistribution the class permits.
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