Checks the interaction of an axial force with a shear force on a section described by its properties, to SIA 263:2013: the class you enter decides between the plastic and the elastic check. Every step shows its formula and the clause it comes from.

What it checks

Resistance model, table 4, from the class you enter: class 1 or 2 gives the plastic model, class 3 or 4 the elastic comparison stress.
Largest shear stress τ_max = V_Ed · S/(I_y · t), formed from the properties or entered directly. It is the elastic peak, reported for comparison.
Shear alone: τ_Ed = V_Ed/A ≤ τ_y,k/γ_M1. The plastic model works on the average stress, the shear being idealised as uniform over the area.
Reduced yield strength (model 1): f_yr = √(f_y,k² − 3τ²), with τ = V_Ed/A, and N_pl,V = f_yr · A. The loss to the shear is shown.
Axial force with the shear acting: N_Ed ≤ N_pl,V/γ_M1.

What you enter and what you get

You enter f_y,k and τ_y,k from table 1 of your copy of SIA 263, for the grade and the plate thickness, the gross area A, the section class (1 to 4, from the classification module), the largest shear stress or the values S, I_y and the thickness t at the neutral axis, γ_M1 (clause 4.1.3), and N_Ed and V_Ed as magnitudes.

You get the resistance model, the shear stresses, the reduced resistance, the loss to the shear and the utilisation of each check with a pass or fail. With the Basic plan, it prints as an A4 calculation note.

Assumptions and limits

Nothing here knows what section you have, so nothing checks what you enter: the area, the class and the shear stress are taken on trust. Where the rectangle or rolled I module fits your section, use it.
No section is drawn: a set of properties says nothing about the shape that produced them.
Only model 1 is available. Model 2, the central shear strip, is a construction on the geometry of a rectangle and is not defined for a general section.
τ_y,k is entered from table 1, not derived as f_y,k/√3.
The interaction is the same whether the axial force pulls or pushes. A section check only: buckling under compression is the subject of the compression modules, and bending with these two forces is not covered here.

Worked example

A class 1 section with A = 6,000 mm², S = 225,000 mm³, I_y = 45 · 10⁶ mm⁴ and t = 20 mm in f_y,k = 235 N/mm² and τ_y,k = 135 N/mm², with N_Ed = 400 kN and V_Ed = 200 kN. These are the module's own results for its worked example, computed when this page was built.

Inputs
Input	Value	Ref.
Characteristic yield strength f_y,k	235 N/mm²	§ t.1
Characteristic shear yield strength τ_y,k	135 N/mm²	§ t.1
Gross area A	6000 mm²
Section class	1	§ t.4
Largest shear stress τ_max	Computed from S, I and t
First moment above the neutral axis S	225000 mm³
Second moment of area I_y	45000000 mm⁴
Thickness at the neutral axis t	20 mm
Resistance factor γ_M1	1.05	§ 4.1.3
Design axial force N_Ed	400 kN
Design shear force V_Ed	200 kN

Calculation: Common values
Step	Value	Ref.
Characteristic yield strength	f_y,k = 235 N/mm²	§ t.1
Characteristic shear yield strengthEntered from table 1, not derived.	τ_y,k = 135 N/mm²	§ t.1
Resistance factor	γ_M1 = 1.05	§ 4.1.3
Section classEntered. Section properties carry no information about the slenderness of the plates that produced them.	– = 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
Gross area	A = 6,000 mm²
Largest shear stress at V_Edτ = V_Ed·S/(I·t)	τ_max = 50.00 N/mm²

Calculation: Check 1 — the shear force alone
Step	Value	Ref.
Average shear stressModel 1 works on the AVERAGE stress, not the peak: it idealises the shear as uniform over the area. The peak in the common values is the elastic one and is reported for comparison.	τ_Ed = V_Ed/A = 33.33 N/mm²
Shear alone, before any interactionτ_Ed ≤ τ_y,k/γ_M1	η = 0.259 · Pass26 % of the capacity used

Calculation: Check 2 — the axial force with the shear force acting
Step	Value	Ref.
Model 1 — shear stress, taken as uniform	τ = 33.33 N/mm²
Model 1 — reduced yield strengthf_yr = √(f_y,k² − 3τ²)	f_yr = 227.80 N/mm²
Model 1 — reduced plastic axial resistanceN_pl,V = f_yr·A	N_pl,V = 1,366.8 kN
Loss to the shear	– = 3.1 %
Axial force with the shear force actingN_Ed ≤ N_pl,V/γ_M1	η = 0.307 · Pass31 % of the capacity used

How this module is verified

Verified clause by clause against SIA 263:2013 and against independent calculations used on real projects, before release.

Questions

Why does the rectangle module give a lower resistance for the same numbers?

The worked example has the properties of a plate 20 × 300 mm. Here only model 1 applies, giving N_pl,V = 1,366.8 kN. The rectangle module also computes the central shear strip, 1,061.9 kN, and verifies the lower of the two. For a rectangle, use that module.

Why is the class entered and not computed?

Section properties carry no information about the slenderness of the plates that produced them. The class decides the method, so take it from the classification module.

Can I print the calculation?

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

Steel section under axial force and shear from its properties to SIA 263

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