Checks the torsional resistance of a solid circular bar at the ultimate limit state, to SIA 263:2013, by the exact membrane solution. Every step shows its formula and the clause it comes from.

What it checks

Torsion constant K = π · r⁴/2, the one section the membrane analogy solves without approximation.
Ultimate torsional resistance T_R = τ_y,k · π · r³/2, from τ_max = 2T_v/(π · r³) at the surface.
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 solid round bar: M_x,Ed ≤ T_Rd, the same shape as the general condition E_d ≤ R_d. The reserve T_Rd/M_x,Ed, how many times over the design resistance covers the design torque, 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 radius r, 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 K, 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.
The cross-section does not warp under torsion: every radius stays straight and the section merely rotates. No support condition can prevent a warping that does not occur, so uniform torsion is the whole of the problem and nothing is omitted.
The stress is zero on the axis and peaks at the surface, varying linearly with the radius. Most of the material is lightly stressed.
For the same weight, a tube of the same outer radius is far stronger and far stiffer in torsion. Where torsion governs, a closed section is the answer.
Not covered: torsion combined with axial force, bending or shear (the interaction modules do that), the connections and fatigue.

Worked example

A solid round bar with r = 65 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
Radius r	65 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 = π·r⁴/2, the one section the membrane analogy solves without approximation.	K = 28,039,696 mm⁴

Calculation: Check 1 — uniform torsion
Step	Value	Ref.
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 = 46.36 N/mm²
Ultimate torsional resistanceT_R = τ_y,k·π·r³/2, from τ_max = 2T_v/(π·r³) at the surface.	T_R = 88.43 kNm
Design torsional resistanceT_Rd = T_R/γ_M1	T_Rd = 84.22 kNm
Torsional resistance of the solid round barM_x,Ed ≤ T_Rd	η = 0.237 · Pass24 % 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 there no warping option?

The round bar is axisymmetric and does not warp, so there is nothing for a support to restrain. Unlike the open sections, where warping is a real second case, uniform torsion covers it completely.

Is the radius or the diameter entered?

The radius r. K grows with its fourth power and T_R with its cube.

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

Solid circular steel bar in torsion to SIA 263

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