Checks a doubly symmetric rolled I or H section under a moment about the strong axis and a shear force acting together, to SIA 263:2013: the class from the geometry, the shear resistance of the web and the moment resistance reduced by the shear. 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, tables 5a and 5b: the flange outstand c/t_f, with c = (b − t_w)/2 − r, in uniform compression, and the web c/t_w, with c = h − 2t_f − 2r, in bending (α = 0.5, ψ = −1), with ε = √(235/f_y,k).
Resistance model, table 4: the class decides whether the plastic resistance may be taken. For the class found, the module takes the moment resistance reduced by the shear, by equation (43) or (53), or makes the elastic check on the effective section. The model it 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.
Bending with the shear force, check 2, equation (43): M_y,Ed ≤ M_V,Rd = M_f,Rd + M_w,V,Rd. The flange contribution M_f,Rd = b · t_f · f_y,k · (h − t_f)/γ_M1 is untouched by the shear. The web contribution M_w,V,Rd = h_w² · t_w · f_y,k/(4γ_M1) · [1 − (V_Ed/V_Rd)²], with h_w = h − 2t_f, is reduced only when V_Ed/V_Rd exceeds 0.5.
Elastic comparison stress σ_g,Ed, reported for reference beside the plastic check.

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 dimensions of the beam profile 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 section properties, the class of the flange and of the web, the resistance model, the shear resistance, the two contributions to the reduced moment resistance, 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, decides it under table 4.
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.
f_y,k drops by 20 to 30 N/mm² in every grade once the plate thickness passes 40 mm.
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 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 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
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
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
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, in bending (α = 0,5, ψ = −1): limits 72.0 · 83.0 · 124.0, class 1.	c/t_w = 23.60	§ t.5a
Section class	– = 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 reduces the plastic moment by SIA 263 (43).	§ 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.	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 shearFlange plus web contribution. With no shear the same expression gives 194.00 kN·m, so the shear costs 7.2 %.	M_V,Rd = 180.01 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 = 0.833 · Pass83 % of the capacity used	§ (43)
Elastic comparison stress, for referenceAgainst 223.8 N/mm². Not the check here.not used	σ_g,Ed = 302.45 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

How much moment resistance does the shear take away?

Only the web term is reduced. In the worked example V_Ed/V_Rd = 0.811, so the web contribution is multiplied by 0.3416. M_V,Rd = 180.0 kN·m, against 194.0 kN·m from the same expression with no shear: the shear costs 7.2 %.

Why is the elastic comparison stress not the check?

Table 4 allows a plastic resistance to a class 1 section, so the check rests on equation (43). In the worked example the comparison stress reads 302.5 N/mm² against 223.8 N/mm², while the plastic check passes with η = 0.833.

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

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

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