Checks the interaction of an axial force with a shear force on a doubly symmetric rolled I or H section, to SIA 263:2013: the class from the geometry, then the axial force and the shear 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: the area A with four fillets of (1 − π/4) · r², within 0.06 % of the published section tables on an HEA 340, an HEA 260 and an IPE 300, and the shear area A_v = A − 2b · t_f + (t_w + 2r) · t_f, equation (42a).
Web slenderness, table 5a: c_w/t_w, with c_w = h − 2t_f − 2r between the fillet toes. For the class 3 limit the module uses 42ε/(0.67 + 0.33ψ), ε = √(235/f_y,k).
Flange outstand slenderness, table 5b: c_f/t_f, with c_f = (b − t_w)/2 − r, against the 14ε row for class 3.
Resistance model, table 4: the class decides whether the plastic resistance may be taken. The method the module settles on is named beside the result.
Axial force: N_Ed ≤ N_pl/γ_M1, with N_pl = f_y,k · A, unreduced for the shear.
Shear force, equation (41): V_Ed ≤ V_R/γ_M1, with V_R = τ_y,k · A_v, unreduced for the axial force. With no interaction term, these two checks together are the interaction verification.

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 five profile dimensions, the direct stress on the section for the class (uniform compression, simple bending about the strong axis, or α and ψ for the web), γ_M1 (clause 4.1.3), and N_Ed and V_Ed as magnitudes.

You get the section properties, the class, the resistance model, both resistances and the utilisation of each check with a pass or fail, and a drawing of the profile. With the Basic plan, it prints as an A4 calculation note.

Assumptions and limits

A doubly symmetric rolled I or H.
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 interaction is the same whether the axial force pulls or pushes. A section check only: a compressive force is also a stability problem, and buckling is the subject of the compression modules.
Bending acting with these two forces is not covered here.

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 (an HEA 260) in f_y,k = 235 N/mm², classified under uniform compression, with N_Ed = 1,200 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
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
Direct stress on the section	Uniform compression	§ t.5a
Resistance factor γ_M1	1.05	§ 4.1.3
Design axial force N_Ed	1200 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
AreaPlates plus four fillets of (1 − π/4)·r². Within 0,06 % of the published section tables on an HEA 340, an HEA 260 and an IPE 300.	A = 8,682 mm²
Shear areaA_v = A − 2·b·t_f + (t_w + 2r)·t_f	A_v = 2,876 mm²	§ (42a)
Second moment of area	I_y = 104,539,541 mm⁴
Material factorε = √(235/f_y,k)	ε = 1.0000	§ t.5a
Section classComputed from the flat widths and uniform compression in checks 1 and 2. The web and flange equally governs.	– = 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. For a rolled I that plastic resistance is the UNREDUCED one: no interaction is applied, so N and V are verified separately in checks 3 and 4.	§ t.4

Calculation: Check 1 — web slenderness
Step	Value	Ref.
Web — slendernessc = h − 2·t_f − 2·r, between the fillet toes. class 3: 42ε/(0,67+0,33ψ) [ψ ≥ −1]	c_w/t_w = 23.60	§ t.5a
Web — is it class 3 or betterc_w/t_w ≤ 42.0	η = 0.562 · Pass56 % of the capacity used	§ t.5a

Calculation: Check 2 — flange outstand slenderness
Step	Value	Ref.
Flange — slendernessc = (b − t_w)/2 − r, from the fillet toe to the free tip.	c_f/t_f = 8.18	§ t.5b
Flange outstand — is it class 3 or betterc_f/t_f ≤ 14.0	η = 0.584 · Pass58 % of the capacity used	§ t.5b

Calculation: Check 3 — axial force, no reduction for the shear
Step	Value	Ref.
Plastic axial resistanceN_pl = f_y,k·A, unreduced — nothing is taken off it for the shear.	N_pl = 2,040.3 kN
Axial force, with no reduction for the shearN_Ed ≤ N_pl/γ_M1	η = 0.618 · Pass62 % of the capacity used

Calculation: Check 4 — shear force, no reduction for the axial force
Step	Value	Ref.
Shear resistanceV_R = τ_y,k·A_v, likewise unreduced.	V_R = 388.2 kN	§ (41)
Shear force, with no reduction for the axial forceV_Ed ≤ V_R/γ_M1	η = 0.541 · Pass54 % of the capacity used	§ (41)
Elastic comparison stress, for referenceWhat the elastic check would give at the governing point, against 223.8 N/mm². Not the check for a class 1 section, but a useful sanity reading: if this is comfortable, the "no reduction" rule is not being leaned on at all.not used	σ_g,Ed = 236.16 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 is no reduction taken for the interaction?

The plastic calculation, worked through, gives a reduction so small that SIA 263 does not apply one. The shear lives in the web and the axial force is carried largely by the flanges, so the two barely compete for the same material. On a solid rectangle they compete everywhere, and the rectangle module reduces the resistance.

What is the elastic comparison stress for?

It is reported for reference and not used for a class 1 section. In the worked example it reads 236.2 N/mm² against 223.8 N/mm², so the result rests on the plastic resistance the class permits.

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

Rolled I or H section under axial force and shear to SIA 263

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