Checks the torsional resistance of a rectangular hollow section b × h × t at the ultimate limit state, to SIA 263:2013, by Bredt's formula. Every step shows its formula and the clause it comes from.

What it checks

Area enclosed by the mid-line Ω = (b − t)(h − t), and the perimeter over thickness Σ s_i/t_i, both measured to the mid-line of the wall.
Torsion constant K = 4Ω²/Σ(s_i/t_i), Bredt's formula.
Ultimate torsional resistance T_R = 2 · Ω · t · τ_y,k, from τ(s) = T_v/(2Ω · t(s)). The shear flow is constant round the cell, so with one thickness throughout the stress is the same in every 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 hollow section: 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.

What you enter and what you get

You enter τ_y,k from table 1 of your copy of SIA 263, the outside width b, the outside depth h, the wall thickness t, γ_M1 and the design torsional moment M_x,Ed as a magnitude. M_x,Ed is the applied moment; T_R is the resistance the section develops against it.

You get Ω, 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. For walls of differing thickness, use the closed-cell module.
Bredt's formula assumes the wall is thin against the cell. At a thickness approaching a quarter of the smaller outside dimension the formula is already optimistic by a few per cent, since the real flow is not uniform across so thick a wall.
The section must stay closed. Any cut, service hole or discontinuity in the wall that breaks the closed shear flow removes almost all of the resistance. Detail the penetrations, and treat a slit member as an open section.
Uniform torsion only.
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 square hollow section 100 × 100 × 6.3 mm 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
Outside width b	100 mm
Outside depth h	100 mm
Wall thickness t	6.3 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
Area enclosed by the mid-lineΩ = (b − t)(h − t)	Ω = 8,780 mm²
Perimeter over thickness	Σ s_i/t_i = 59.49
Torsion constantK = 4Ω²/Σ(s_i/t_i), Bredt's formula.	K = 5,182,739 mm⁴

Calculation: Check 1 — uniform torsion
Step	Value	Ref.
Governing wall — thinnest	t_min = 6.3 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 = 180.79 N/mm²
Ultimate torsional resistanceT_R = 2·Ω·t·τ_y,k, from τ(s) = T_v/(2Ω·t(s)). The flow is constant round the cell, so with one thickness throughout the stress is the same in every wall.	T_R = 22.68 kNm
Design torsional resistanceT_Rd = T_R/γ_M1	T_Rd = 21.60 kNm
Torsional resistance of the hollow sectionM_x,Ed ≤ T_Rd	η = 0.926 · Pass93 % 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 are b and h reduced by t?

Ω and the wall lengths are taken to the mid-line, so a b × h × t section encloses (b − t)(h − t). This is the area the shear flow runs around, not the area of the material.

How much does a closed section gain over an open one?

A closed cell carries torsion far better than an open one, and the difference is not marginal. A circular tube 355.6 × 12.5 has K = 396.9 · 10⁶ mm⁴ closed and 0.702 · 10⁶ mm⁴ slit, a factor of 565.

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

Rectangular hollow steel section in torsion to SIA 263

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