Material and cross-section, shared by every verification. Change the verification above and these stay as they are.
Checks a rectangular solid-timber or glulam beam in torsion at the ultimate limit state, to SIA 265:2021: torsion alone and torsion combined with shear. Every step shows its formula and the clause it comes from.
You enter the member (wood type, strength class, load duration, service class, b, h), the design shear strength fv,d from your copy of SIA 265, whether the piece contains the pith, the system factor and the design torsional moment Td. To combine with shear, you also enter the design shear forces Vz,d and Vy,d.
You get the side ratio, the shape factor, the torsional shear stress, every intermediate value with its clause, and the utilisation of each check with a pass or fail. With the Basic plan, it prints as an A4 calculation note.
A glulam beam GL24h, b × h = 280 × 520 mm, in service class 1 under short to permanent actions, with ksys = 0.8, a design torsional moment of 0.1 kN·m and a design shear force of 172 kN. Torsion alone passes; in this example torsion combined with shear does not pass. These are the module's own results for its worked example, computed when this page was built.
| Input | Value | Ref. |
|---|---|---|
| Wood type | Glulam | |
| Strength class | GL24h | |
| Load duration ηt | Short to permanent | § 2.2.6 |
| Service class ηw | 1 | § 3.2.1.3 t.4 |
| Section width b | 280 mm | |
| Section depth h | 520 mm | |
| Design shear strength fv,d | 1.8 N/mm² | § 3.4.2.2 t.9 |
| Piece contains the pith? | No | § 3.3.2.1 t.8 n.6 |
| System factor ksys | 0.8 | § 5.7.2.2 |
| Design torsional moment Td | 0.1 kN·m | |
| Combine with shear? | Yes | § 4.2.5.3 |
| Design shear force, z direction Vz,d | 172 kN | |
| Design shear force, y direction Vy,d | 0 kN |
| Step | Value | Ref. |
|---|---|---|
| Side ratio of the sectionratio of the shorter to the longer half-side, min(b,h) / max(b,h). The norm gives no expression for the torsional shear stress of a rectangle, so the elasticity solution is used. | b/a = 0.5385 | |
| Shape factorκ = 1 + 0.6095·(b/a) + 0.8865·(b/a)² − 1.8023·(b/a)³ + 0.9100·(b/a)⁴ | κ = 1.3803 | |
| Load duration factor | ηt = 1.00 | § 2.2.6 |
| Moisture factor | ηw = 1.00 | § 3.2.1.3 t.4 |
| Design shear resistancefRd = fv,d · ksys · ηt · ηw | fRd = 1.440 N/mm² | § 4.2.5.2 (24) |
| Design torsional shear stressτtor,d = 3·Td / (8·a·b²) · κ, with a and b the half-sides, a ≥ b. Elasticity solution for a rectangular section; acts at the midpoint of each longer side. The norm gives no expression for it. | τtor,d = 0.0102 N/mm² |
| Step | Value | Ref. |
|---|---|---|
| Torsionτtor,d ≤ fv,d | η = 0.007 · Pass1 % of the capacity used | § 4.2.5.2 (24) |
| Step | Value | Ref. |
|---|---|---|
| Shear stress, z directionτxz,d = 1.5 · Vz,d / A | τxz,d = 1.7720 N/mm² | § 4.2.3 |
| Torsion combined with shearτtor,d/fv,d + (τd/fv,d)² ≤ 1 | η = 1.521 · Fail152 % of the capacity used | § 4.2.5.3 (25) |
Verified clause by clause against SIA 265:2021 and against independent calculations used on real projects, before release.
SIA 265 gives no expression for it. The module uses the elasticity solution for a rectangular section, with a shape factor κ that depends on the ratio of the shorter to the longer side.
By equation (25): the torsional term enters linearly and the shear term squared.
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 265:2021 · Page updated: 21 October 2026