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 solid rectangle under a moment, an axial force and a shear force acting together, to SIA 263:2013: the shear alone first, then the moment with 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 b across the plate, the depth h in the plane of bending, which is also the direction of the shear force, and γM1 (clause 4.1.3). You choose whether the axial force is compression or tension, and enter NEd, My,Ed and VEd as magnitudes.
You get the resistances with no shear, the shear check, every step of both models, the model verified and the utilisation of each check with a pass or fail. With the Basic plan, it prints as an A4 calculation note.
A plate b = 20 mm thick and h = 300 mm deep in fy,k = 235 N/mm² and τy,k = 135 N/mm², under NEd = 300 kN in compression, My,Ed = 30 kN·m and VEd = 150 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 b | 20 mm | |
| Depth in the plane of bending h | 300 mm | |
| Resistance factor γM1 | 1.05 | § 4.1.3 |
| The axial force is | Compression | |
| Design axial force NEd | 300 kN | |
| Design bending moment My,Ed | 30 kN·m | |
| Design shear force VEd | 150 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 classClass 1 by construction: a solid section has no plate to classify. | – = 1 | § t.4 |
| Resistance model | PLASTIC. Table 4 allows a plastic resistance to a class 1 section. | § t.4 |
| AreaA = b·h | A = 6,000 mm² | |
| Plastic section modulusWpl = b·h²/4 | Wpl = 450,000 mm³ | |
| Plastic axial resistance, no shearfy,k·A/γM1. | Npl,Rd = 1,342.9 kN | § 4.1.3 |
| Plastic moment resistance, no shearfy,k·Wpl/γM1. | Mpl,Rd = 100.71 kN·m | § 4.1.3 |
| Step | Value | Ref. |
|---|---|---|
| Average shear stress | τEd = VEd/A = 25.00 N/mm² | |
| Shear alone, before any interactionτEd ≤ τy,k/γM1 | η = 0.194 · Pass19 % of the capacity used |
| Step | Value | Ref. |
|---|---|---|
| Model 1 — reduced yield strengthfyr = √(fy,k² − 3τ²), τ = VEd/A = 25.00 N/mm²: what is left for a direct stress at every point once the shear has taken its share. | fyr = 230.98 N/mm² | |
| Model 1 — axial and moment resistance with the shearfyr·A/γM1 and fyr·b·h²/4/γM1: the whole rectangle at the reduced yield strength. | Npl,V · Mpl,V = 1,319.9 kN · 98.99 kN·m | |
| Model 1 — moment resistance with NEdMpl,V·[1 − (NEd/Npl,V)²], n = 0.2273: the exact interaction of a rectangle, at fyr. | MR,1 = 93.88 kN·m | |
| Model 2 — depth of the shear strip at τy,kFrom τ = VEd/(2a·b) = τy,k. The strip carries all the shear and nothing else. | 2a = 55.6 mm | |
| Model 2 — band next to the strip that carries NEdd = NEd/(2·b·fy,k/γM1): the axial force takes the fibres closest to the axis, where they cost the moment least. | d = 33.5 mm | |
| Model 2 — moment resistance with NEdfy,k·b·[h²/4 − (a + d)²]/γM1. At N = 0 it is fy·b·(h²/4 − a²), at M = 0 it gives N = fy·b·(h − 2a). | MR,2 = 83.90 kN·m | |
| Which model is verified | The LOWER of the two, here model 2, the central shear strip at 83.90 kN·m against 93.88 kN·m for the other. | |
| Moment with the axial force and the shear force actingMy,Ed ≤ MR | η = 0.358 · Pass36 % of the capacity used | |
| Elastic comparison stress, for referenceWhat the elastic check would give at its governing point, against 223.8 N/mm². Not the check for a class 1 section; the gap is the redistribution the class permits.not used | σg,Ed = 150.00 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: the shear spread over the whole rectangle, or packed into a central strip. In the worked example the strip gives MR,2 = 83.90 kN·m against MR,1 = 93.88 kN·m, and Mpl,Rd with neither force is 100.71 kN·m. The module verifies model 2, η = 0.358.
If the shear alone exceeds τy,k/γM1, no model leaves anything for the moment and the axial force. In the worked example τEd = 25 N/mm², η = 0.194.
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: 8 November 2026