Checks the torsional resistance of a rolled profile at the ultimate limit state, to SIA 263:2013, from the torsion constant of the section tables, with the warping of a double tee as an option. Every step shows its formula and the clause it comes from.

What it checks

Section from the shape (double tee I or H, tee, channel U or angle L) and its dimensions: the area A, the second moment I_y and the thickest plate t_max, derived from the section rather than entered.
Torsion constant K, taken from the section tables. K_plate, the plate sum of the same outline, is shown beside it as a cross-check.
Ultimate torsional resistance T_R = τ_y,k · K/t_max. The shear flow turns back within the thickness of each plate, so the stress is τ_v = T_v · t/K and peaks in the thickest one.
Design torsional resistance T_Rd = T_R/γ_M1, with γ_M1 from clause 4.1.3.
Peak shear stress τ_max = τ_y,k · M_x,Ed/T_R, since the stress is linear in the torque.
Torsional resistance of the rolled profile: M_x,Ed ≤ T_Rd, with the reserve T_Rd/M_x,Ed beside it.
Warping, optional: a prismatic double tee built in at one end with a torque at the other, the one case with a closed-form solution.

What you enter and what you get

You enter τ_y,k from table 1 of your copy of SIA 263, the tabulated K of your profile, the shape and its dimensions h, b, t_w and t_f, γ_M1 and the design torsional moment M_x,Ed as a magnitude. For the warping case, only the bar length is added: the section is already known.

You get the section values, T_R, T_Rd, τ_max and the utilisation with a pass or fail, and a drawing of the section to check against the profile you mean. With the Basic plan, it prints as an A4 calculation note.

Assumptions and limits

τ_y,k is entered from table 1, not derived as f_y/√3.
K is the one quantity that cannot be computed from the dimensions. The root fillets carry torsion and the plate model leaves them out: on an HEA 340 the tables give 1.29 · 10⁶ mm⁴ against 0.988 · 10⁶ mm⁴ for the plates, 23 % less. A gap far from 20 to 30 % is worth a second look at the table row and the dimensions.
Flanges are taken as constant in thickness. A tapered-flange channel is idealised with parallel flanges.
Uniform torsion unless the warping case is chosen. At a built-in end, a moment connection or a stiffened support, warping can carry the whole torque.
Not covered: torsion combined with axial force, bending or shear (the interaction modules do that), distortion of the section, local buckling of a wall under the torsional shear, the connections and fatigue.

Worked example

A double tee with h = 330 mm, b = 300 mm, t_w = 9.5 mm, t_f = 16.5 mm and a tabulated K = 1.29 · 10⁶ mm⁴, in τ_y,k = 205 N/mm², under M_x,Ed = 20 kNm in uniform torsion; it does not pass the torsional resistance check, with η = 1.31. These are the module's own results for its worked example, computed when this page was built.

Inputs
Input	Value	Ref.
Characteristic shear yield strength τ_y,k	205 N/mm²	§ t.1
Torsion constant K	1290000 mm⁴
Shape	Double tee — I or H
Overall depth h	330 mm
Flange width b	300 mm
Web thickness t_w	9.5 mm
Flange thickness t_f	16.5 mm
Resistance factor γ_M1	1.05	§ 4.1.3
Design torsional moment M_x,Ed	20 kNm
Warping	Not considered — uniform torsion only

Calculation: Common values
Step	Value	Ref.
Characteristic shear yield strengthEntered from table 1, not derived.	τ_y,k = 205 N/mm²	§ t.1
Resistance factor	γ_M1 = 1.05	§ 4.1.3
Overall depth	h = 330.0 mm
Overall width	b = 300.0 mm
AreaThe sum of the rectangles of the outline. Exact for a section built of plates; for a ROLLED profile it is short by the root fillets, 4,7 % on an HEA 340.	A = 12,722 mm²
Thickest plateDerived from the section rather than entered. The peak shear stress of an open section is T_v·t/K, so it occurs here.	t_max = 16.5 mm
Second moment of area about yReported so the section can be recognised. Not used by the torsion check: the plate model is 4,6 % below the tabulated value for I_y, because the fillets sit near the web where they count for I_y and not for I_z.not used	I_y = 264,213,316 mm⁴
K from the plate model of these dimensionsWhat the plate sum would give for this outline, 23.4 % below the tabulated K entered above. A figure far from 20 to 30 % is worth a second look at the table row and at the dimensions.not used	K_plate = 988,021 mm⁴
Torsion constantTaken from the section tables.	K = 1,290,000 mm⁴

Calculation: Check 1 — uniform torsion
Step	Value	Ref.
Governing element — thickest	t_max = 16.5 mm
Peak shear stress at M_x,EdThe stress is linear in the torque, so τ_max = τ_y,k·M_x,Ed/T_R.	τ_max = 255.81 N/mm²
Ultimate torsional resistanceT_R = τ_y,k·K/t_max. The flow turns back within the thickness of each plate, so the stress is τ_v = T_v·t/K and peaks in the THICKEST one, which these dimensions give as 16.5 mm.	T_R = 16.03 kNm
Design torsional resistanceT_Rd = T_R/γ_M1	T_Rd = 15.26 kNm
Torsional resistance of the rolled profileM_x,Ed ≤ T_Rd	η = 1.310 · Fail131 % 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 is K entered and not computed?

The root fillets carry torsion and a plate sum leaves them out. The module still shows K_plate beside it, so a wrong table row stands out.

Why does the thickest plate govern?

In an open section the flow turns back within each plate, and the stress T_v · t/K grows with the plate thickness. In the worked example the flange, 16.5 mm thick, sets T_R.

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 steel profile in torsion to SIA 263

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