Checks any steel beam section described by its properties under a moment and a shear force acting together, 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, plastic resistance; class 3, elastic; class 4, elastic on the effective section. The model is named beside the result.
Plastic moment resistance: M_pl,Rd = f_y,k · W_y,pl/γ_M1.
Shear force, check 1, equation (41): V_Ed ≤ V_Rd = τ_y,k · A_v/γ_M1.
Moment resistance reduced by the shear, equation (43) in a generalised form: M_V,Rd = f_y,k · (W_y,pl − W_v,pl · (V_Ed/V_Rd)²)/γ_M1. The part of W_y,pl in the shear area is reduced only when V_Ed/V_Rd exceeds 0.5.
Bending with the shear force, check 2, equation (43): M_y,Ed ≤ M_V,Rd.

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 section class of the beam (table 4), the gross area A, the plastic section modulus W_y,pl, the shear area A_v (equation (42a)), the plastic modulus of the shear area W_v,pl (equation (43)) and γ_M1 (clause 4.1.3). You enter the design moment M_y,Ed about the strong axis and the design shear force V_Ed in the same plane, both as magnitudes.

You get the resistance model, the plastic moment resistance, the shear utilisation, the moment resistance reduced by 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 can check what you enter. Where the rectangle or the rolled I module fits your section, use it: those derive the properties from the dimensions and draw the section.
The class is entered, because section properties carry no information about the slenderness of the plates. Classify the section under these forces with the classification module.
The shear reduction of equation (43) is written for a doubly symmetric I: the flange term is untouched and the web term is reduced by (V_Ed/V_Rd)². Here it is applied in the same way to W_v,pl. For an I with W_v,pl = h_w² · t_w/4 this is equation (43) itself.
τ_y,k is entered from table 1, not derived as f_y,k/√3. The tabulated values are not exactly f_y,k/√3: over the eight cells of table 1 the division gives the higher, unconservative value in six.
A section check only. Lateral-torsional buckling, the introduction of concentrated forces and torsion are not covered. An axial force acting with the moment and the shear is the subject of the bending, axial force and shear modules.

Worked example

A section of class 1 with A = 8,680 mm², W_y,pl = 920,000 mm³, A_v = 2,876 mm² and W_v,pl = 94,922 mm³, in f_y,k = 235 N/mm² and τ_y,k = 135 N/mm², under M_y,Ed = 150 kN·m and V_Ed = 300 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
Section class	1	§ t.4
Gross area A	8680 mm²
Plastic section modulus W_y,pl	920000 mm³
Shear area A_v	2876 mm²	§ (42a)
Plastic modulus of the shear area W_v,pl	94922 mm³	§ (43)
Resistance factor γ_M1	1.05	§ 4.1.3
Design bending moment M_y,Ed	150 kN·m
Design shear force V_Ed	300 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.	– = 1	§ t.4
Resistance model	PLASTIC. Table 4 allows a plastic resistance to a class 1 section.	§ t.4
Plastic moment resistancef_y,k·W_y,pl/γ_M1.	M_pl,Rd = 205.90 kN·m	§ 4.1.3

Calculation: Common values — the moment resistance reduced by the shear
Step	Value	Ref.
Shear utilisationV_Rd = τ_y,k·A_v/γ_M1 = 369.8 kN. The shear-area part of W_pl is reduced only above 0,5.	V_Ed/V_Rd = 0.8113
Moment resistance reduced by the shearf_y,k·[W_y,pl − W_v,pl·(V_Ed/V_Rd)²]/γ_M1, (V_Ed/V_Rd)² = 0.6582.	M_V,Rd = 191.92 kN·m	§ (43)

Calculation: Check 1 — shear force
Step	Value	Ref.
Shear forceV_Ed ≤ V_Rd	η = 0.811 · Pass81 % of the capacity used	§ (41)

Calculation: Check 2 — bending with the shear force
Step	Value	Ref.
Bending with a shear forceM_y,Ed ≤ M_V,Rd	η = 0.782 · Pass78 % of the capacity used	§ (43)

How this module is verified

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

Questions

What is the plastic modulus of the shear area?

It is the part of W_y,pl that lies in the shear area: for a doubly symmetric I, the web, h_w² · t_w/4. Equation (43) reduces this part by (V_Ed/V_Rd)² once V_Ed passes half of V_Rd. In the worked example (V_Ed/V_Rd)² = 0.6582, and M_V,Rd = 191.9 kN·m against M_pl,Rd = 205.9 kN·m.

Why must I enter the section class?

A list of section properties says nothing about the slenderness of the plates, so the module cannot classify the beam. The class you enter decides between the plastic and the elastic check under table 4.

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: 5 November 2026

Steel section in bending and shear from its properties to SIA 263

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