Checks the torsional resistance of a circular tube at the ultimate limit state, to SIA 263:2013, closed by Bredt's formula or slit along its length as an open section. Every step shows its formula and the clause it comes from.

What it checks

Torsion constant of the closed tube, K = 2π · r³ · t: Bredt's formula applied to a circle, with Ω = π · r² and the perimeter 2π · r.
Ultimate torsional resistance T_R = 2 · Ω · t · τ_y,k = 2π · r² · t · τ_y,k, from τ = T_v/(2Ω · t), constant round the wall.
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 tube: M_x,Ed ≤ T_Rd, the same shape as the general condition E_d ≤ R_d. The reserve T_Rd/M_x,Ed is shown beside it.
Slit tube, as an option: a single longitudinal slit turns the closed shear flow into an open one, and the tube is then treated as an open section.

What you enter and what you get

You enter τ_y,k from table 1 of your copy of SIA 263, whether the tube is closed or slit along its length, the radius to the mid-wall r, the wall thickness t, γ_M1 and the design torsional moment M_x,Ed as a magnitude.

The radius is measured to the middle of the wall, not the outside: for a tube 355.6 × 12.5 that is (355.6 − 12.5)/2 = 171.6 mm.

You get K, the governing wall t_min, T_R, T_Rd, τ_max and the utilisation with a pass or fail. 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.
One wall thickness all round, and uniform torsion only.
A closed tube is the best section there is in torsion for its weight: the flow runs the whole way round at a constant stress and every part of the wall is fully used.
Keep it closed. Any cut, hole or unwelded seam drops it to the slit case, a factor of several hundred.
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 closed tube 355.6 × 12.5 mm, with r = 171.6 mm to the mid-wall, in τ_y,k = 205 N/mm², under M_x,Ed = 20 kNm. 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
The tube is	Closed
Radius to the mid-wall r	171.6 mm
Wall thickness t	12.5 mm
Resistance factor γ_M1	1.05	§ 4.1.3
Design torsional moment M_x,Ed	20 kNm

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
Torsion constantK = 2π·r³·t, Bredt applied to a circle: Ω = π·r² and the perimeter is 2π·r.	K = 396,864,024 mm⁴

Calculation: Check 1 — uniform torsion
Step	Value	Ref.
Governing wall — thinnest	t_min = 12.5 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 = 8.65 N/mm²
Ultimate torsional resistanceT_R = 2·Ω·t·τ_y,k = 2π·r²·t·τ_y,k, from τ = T_v/(2Ω·t), constant round the wall.	T_R = 474.11 kNm
Design torsional resistanceT_Rd = T_R/γ_M1	T_Rd = 451.53 kNm
Torsional resistance of the tubeM_x,Ed ≤ T_Rd	η = 0.044 · Pass4 % 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

How much does a slit cost?

Almost everything. On a tube 355.6 × 12.5, K = 396.9 · 10⁶ mm⁴ closed becomes 0.702 · 10⁶ mm⁴ slit, a factor of 565. Choose the slit case for any tube whose wall is cut along its length or has an unwelded seam.

Why is the stress the same everywhere in the wall?

The shear flow of a closed cell is constant round it. With one thickness all round, τ = T_v/(2Ω · t) is the same at every point, so the whole wall works at once.

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

Circular steel tube in torsion to SIA 263

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