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 bending about its y-y axis at the ultimate limit state, to SIA 265:2021, including lateral torsional buckling. 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 system factor ksys, the design bending strength fm,d from your copy of SIA 265 (clause 3.3.2.1, table 8, for ηt = ηw = 1.0 and kh = 1), the characteristic bending strength fm,k and the modulus of elasticity E0,05 (annex B), the design bending moment My,d, and how the compression edge is held: continuously, or by lateral supports at a spacing a.
You get the section modulus, the slenderness, every factor, the design resistance and stress, and the utilisation with a pass or fail, each with its clause. With the Basic plan, it prints as an A4 calculation note.
A solid-timber beam C24, b × h = 80 × 120 mm, in service class 1 under short to permanent actions, with ksys = 1.1, fm,d = 14.0 N/mm², fm,k = 24 N/mm², E0,05 = 7,300 N/mm², a design moment of 2.655 kN·m and lateral supports every 2.00 m. These are the module's own results for its worked example, computed when this page was built.
| Input | Value | Ref. |
|---|---|---|
| Wood type | Solid | |
| Strength class | C24 | |
| Load duration ηt | Short to permanent | § 2.2.6 |
| Service class ηw | 1 | § 3.2.1.3 t.4 |
| Section width b | 80 mm | |
| Section depth h | 120 mm | |
| System factor ksys | 1.1 | § 5.7.2.2 |
| Design bending strength fm,d | 14.0 N/mm² | § 3.3.2.1 t.8 |
| Characteristic bending strength fm,k | 24 N/mm² | § Annexe B |
| Modulus of elasticity, 5% fractile E0,05 | 7300 N/mm² | § Annexe B |
| Design bending moment My,d | 2.655 kN·m | |
| Compression edge | Lateral supports at spacing a | § 4.2.9 |
| Spacing of lateral supports a | 2000 mm | § 4.2.9.3 |
| Step | Value | Ref. |
|---|---|---|
| Section modulus about y-yWy = b · h² / 6 | Wy = 192,000 mm³ | |
| Relative slendernessλrel,m = 1.15 · (√(a·h) / b) · √(fm,k / E0,05) | λrel,m = 0.4038 | § 4.2.9.3 (42) |
| Lateral torsional buckling factorkm = 1 for λrel,m ≤ 0.75 · 1.56 − 0.75·λrel,m for 0.75 < λrel,m ≤ 1.4 · 1/λrel,m² above | km = 1.0000 | § 4.2.9.2 (39) |
| Height factorkh = 1.0 for solid timber | kh = 1.0000 | § 3.4.2.3 (5) |
| System factor | ksys = 1.10 | § 5.7.2.2 |
| Load duration factor | ηt = 1.00 | § 2.2.6 |
| Moisture factor | ηw = 1.00 | § 3.2.1.3 t.4 |
| Design resistancefRd = km · kh · ksys · fm,d · ηt · ηw | fRd = 15.400 N/mm² | § 4.2.9.1 (38) |
| Design stressσm,y,d = My,d / Wy | σm,y,d = 13.828 N/mm² | |
| Bending parallel to the grainσm,d ≤ km · fm,d | η = 0.898 · Pass90 % of the capacity used | § 4.2.9.1 (38) |
Verified clause by clause against SIA 265:2021 and against independent calculations used on real projects, before release.
When λrel,m exceeds 0.75. Below that, km = 1; with a continuously held compression edge it is 1 by definition.
Clause 4.2.9.3 forms the critical bending stress from the characteristic stiffness, so E0,05 is entered as the characteristic value.
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