Checks a doubly symmetric rolled I or H section under a moment about the strong axis, an axial force and a shear force acting together, to SIA 263:2013: the class from the geometry and the forces, the moment resistance reduced by the shear, then its interaction with the axial force. Every step shows its formula and the clause it comes from.

What it checks

Section properties from h, b, t_w, t_f and r: A with four fillets of (1 − π/4) · r², I_y, W_y,el, W_y,pl and the shear area A_v = A − 2b · t_f + (t_w + 2r) · t_f, equation (42a).
Class under these forces, tables 5a and 5b, with ε = √(235/f_y,k): the flange outstand c/t_f, with c = (b − t_w)/2 − r, in uniform compression; the web c/t_w, with c = h − 2t_f − 2r, classified with the fraction α of its depth in compression under N and M (plastic) and the stress ratio ψ (elastic), not with the pure-bending values α = 0.5 and ψ = −1.
Resistance model, table 4: the class decides between the shear-reduced plastic moment with the axial force and the elastic check. The model the module settles on is named beside the result.
Web in shear, equation (42): (h − t_f)/t_w against √(4E/f_y,k). Below that limit the web yields in shear before it buckles, and equation (41) applies.
Shear force, check 1, equation (41): V_Ed ≤ V_Rd = τ_y,k · A_v/γ_M1. Equation (43) holds only up to V_Rd.
Moment resistance reduced by the shear, equation (43): M_V,Rd = M_f,Rd + M_w,V,Rd. The flange term is untouched by the shear; the web term, h_w² · t_w · f_y,k/(4γ_M1) · [1 − (V_Ed/V_Rd)²], is reduced only above V_Ed/V_Rd = 0.5.
Bending with an axial force and a shear force, check 2: N_Ed/N_pl,Rd + M_y,Ed/M_V,Rd ≤ 1, with N_pl,Rd = f_y,k · A/γ_M1 unreduced: SIA 263 takes no reduction of N_pl for the shear on a rolled I. The linear interaction is a chord across the true interaction surface, so it is safe but can be conservative.
Elastic comparison stress σ_g,Ed, reported for reference and not used for a class 1 section.

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, E (clause 3.2.2.1), the five profile dimensions and γ_M1 (clause 4.1.3). You choose whether the axial force is compression or tension, and enter N_Ed, the moment M_y,Ed about the strong axis and V_Ed in the same plane, all as magnitudes.

You get the section properties, the class of the flange and of the web with the α and ψ used, the resistance model, the shear resistance, the two contributions to M_V,Rd, N_pl,Rd, the two terms of the interaction 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

A doubly symmetric rolled I or H, bent about the strong axis in the plane of the shear force.
No calculation method is asked for: the class, computed from the geometry and the forces, decides it under table 4.
The resistances are the same for either sense of the axial force. The sense matters for the class: a compressive force puts more of the web in compression.
τ_y,k is entered from table 1, not derived as f_y,k/√3. Over the eight cells of table 1 the division gives the higher, unconservative value in six.
A section check only. Buckling under a compressive force, lateral-torsional buckling and torsion are not covered.

Worked example

A rolled section with h = 250 mm, b = 260 mm, t_w = 7.5 mm, t_f = 12.5 mm and r = 24 mm, in f_y,k = 235 N/mm² and τ_y,k = 135 N/mm², under N_Ed = 400 kN in compression, M_y,Ed = 80 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
Modulus of elasticity E	210000 N/mm²	§ 3.2.2.1
Overall depth h	250 mm
Flange width b	260 mm
Web thickness t_w	7.5 mm
Flange thickness t_f	12.5 mm
Root radius r	24 mm
Resistance factor γ_M1	1.05	§ 4.1.3
The axial force is	Compression
Design axial force N_Ed	400 kN
Design bending moment M_y,Ed	80 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
AreaPlates plus four fillets of (1 − π/4)·r².	A = 8,682 mm²
Second moment of area	I_y = 104,539,541 mm⁴
Elastic section modulus	W_y,el = 836,316 mm³
Plastic section modulus	W_y,pl = 919,771 mm³
Shear areaA_v = A − 2·b·t_f + (t_w + 2r)·t_f	A_v = 2,876 mm²	§ (42a)
Material factorε = √(235/f_y,k)	ε = 1.0000	§ t.5a
Compressed flangec = (b − t_w)/2 − r. Limits 9.0 · 10.0 · 14.0 in uniform compression: class 1.	c/t_f = 8.18	§ t.5b
Webc = h − 2t_f − 2r. Under these forces α = 1.000 (plastic, from N and M) and ψ = -0.190 (elastic): limits 33.0 · 38.0 · 69.2, class 1.	c/t_w = 23.60	§ t.5a
Section class under these forces	– = Class 1	§ t.5a

Calculation: Common values — shear resistance of the web
Step	Value	Ref.
Web slendernessAgainst √(4E/f_y,k) = 59.8: the web yields in shear before it buckles, so (41) applies.	(h − t_f)/t_w = 31.67	§ (42)
Design shear resistanceV_Rd = τ_y,k·A_v/γ_M1 = 369.7 kN, (41).	V_Rd = 369.7 kN	§ (41)
Resistance model	PLASTIC. Table 4 allows a plastic resistance to a class 1 section. The shear first reduces the plastic moment by SIA 263 (43); the axial force then interacts with what is left.	§ t.4

Calculation: Common values — the moment resistance reduced by the shear
Step	Value	Ref.
Shear utilisationThe web term is reduced only above 0,5.	V_Ed/V_Rd = 0.8114
Flange contributionb·t_f·f_y,k·(h − t_f)/γ_M1. Untouched by the shear: the flanges carry the moment and take no part in the shear.	M_f,Rd = 172.75 kN·m	§ (43)
Web contributionh_w²·t_w·f_y,k/(4γ_M1)·[1 − (V_Ed/V_Rd)²], h_w = h − 2t_f. Reduced by the factor 0.3416.	M_w,V,Rd = 7.26 kN·m	§ (43)
Moment resistance reduced by the shearThe plate model of (43), without the fillets, so at V = 0 it sits 5.8 % below f_y·W_pl/γ_M1.	M_V,Rd = 180.01 kN·m	§ (43)
Plastic axial resistancef_y,k·A/γ_M1, unreduced: SIA 263 takes no reduction of N_pl for the shear on a rolled I.	N_pl,Rd = 1,943.1 kN	§ 4.1.3

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, axial force and shear
Step	Value	Ref.
Axial term	N_Ed/N_pl,Rd = 0.2059
Moment termThe moment resistance already reduced by the shear goes into the denominator.	M_y,Ed/M_V,Rd = 0.4444
Bending with an axial force and a shear forceN_Ed/N_pl,Rd + M_y,Ed/M_V,Rd ≤ 1	η = 0.650 · Pass65 % of the capacity used
Elastic comparison stress, for referenceAgainst 223.8 N/mm². Not the check for a class 1 section.not used	σ_g,Ed = 293.46 N/mm²

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 axial force change the class?

The web is classified with the part of its depth that the axial force and the moment put in compression. In the worked example the 400 kN of compression gives α = 1.000 and ψ = −0.190, so the web limits are 33.0, 38.0 and 69.2. With c/t_w = 23.6 the web is still class 1.

How do the three forces combine?

The shear first reduces the moment resistance: V_Ed/V_Rd = 0.811 gives M_V,Rd = 180.0 kN·m. The axial force then interacts linearly with what is left: 400/1,943.1 = 0.206 plus 80/180.0 = 0.444, so η = 0.650.

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: 8 November 2026

Rolled I or H section in bending, axial force and shear to SIA 263

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