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 for bending about both principal axes to SIA 265:2021, at the ultimate limit state, including lateral torsional buckling. Every step shows its formula and the clause it comes from.
You enter the wood type (solid timber or glulam), the strength class, the load duration, the service class, the section width b and depth h, the design bending strength fm,d from your copy of SIA 265, the system factor, the design moments My,d and Mz,d, and the lateral support spacings ay and az. With a spacing greater than zero, the characteristic bending strength fm,k and the modulus of elasticity E0,05 are also needed.
You get the utilisation of equation (14) with a pass or fail, every intermediate value with its clause (section moduli, height factors, slenderness, lateral torsional buckling factors, resistances and stresses), and figures: where the utilisation comes from, the cross-section, and the lateral torsional buckling curve with both axes marked. With the Basic plan, the same calculation prints as an A4 calculation note.
A glulam beam GL24h in service class 2 under short to permanent actions, section b × h = 680 × 280 mm, with lateral supports every 6.82 m for both axes. 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 | 2 | § 3.2.1.3 t.4 |
| Section width b | 680 mm | |
| Section depth h | 280 mm | |
| Design bending strength fm,d | 16 N/mm² | § 3.4.2.2 t.9 |
| Characteristic bending strength fm,k | 24 N/mm² | § Annexe B |
| Modulus of elasticity, 5% fractile E0,05 | 9600 N/mm² | § Annexe B |
| System factor ksys | 1.0 | § 5.7.2.2 |
| Design moment about y-y My,d | 93 kN·m | |
| Design moment about z-z Mz,d | 1 kN·m | |
| Lateral support spacing, bending about y-y ay | 6820 mm | § 4.2.9.3 |
| Lateral support spacing, bending about z-z az | 6820 mm | § 4.2.9.3 |
| Step | Value | Ref. |
|---|---|---|
| Section modulus about y-yWy = b · h² / 6 | Wy = 8,885,333 mm³ | |
| Section modulus about z-zWz = h · b² / 6 | Wz = 21,578,667 mm³ | |
| Height factor, y-yuses the dimension perpendicular to the axis, here h | kh,y = 1.0792 | § 3.4.2.3 (5) |
| Height factor, z-zfor bending about z-z the relevant dimension is b | kh,z = 1.0000 | § 3.4.2.3 (5) |
| Relative slenderness, y-yλrel,m = 1.15 · (√(ay·h) / b) · √(fm,k / E0,05) | λrel,m,y = 0.1169 | § 4.2.9.3 (42) |
| Relative slenderness, z-zλrel,m = 1.15 · (√(az·b) / h) · √(fm,k / E0,05) | λrel,m,z = 0.4422 | § 4.2.9.3 (42) |
| Lateral torsional buckling factor, y-y | km,y = 1.0000 | § 4.2.9.2 (39) |
| Lateral torsional buckling factor, z-z | km,z = 1.0000 | § 4.2.9.2 (39) |
| Bending resistance, y-yfRd,m,y = km,y · fm,d · kh,y · ksys · ηt · ηw | fRd,m,y = 13.814 N/mm² | § 4.2.9.1 (38) |
| Bending resistance, z-zfRd,m,z = km,z · fm,d · max(kh,z ; ksys) · ηt · ηw. §3.4.2.2 forbids applying kh and ksys together for bending about z-z. | fRd,m,z = 12.800 N/mm² | § 3.4.2.2 |
| Bending stress, y-yσm,y,d = My,d / Wy | σm,y,d = 10.4667 N/mm² | |
| Bending stress, z-zσm,z,d = Mz,d / Wz | σm,z,d = 0.0463 N/mm² | |
| Biaxial bendingσm,y/fm,y + σm,z/fm,z ≤ 1 | η = 0.761 · Pass76 % of the capacity used | § 4.2.3.1 (14) |
Verified clause by clause against SIA 265:2021 and against independent calculations used on real projects, before release.
SIA 265:2021. The clause and equation numbers in the calculation refer to that edition.
The site reproduces no table from any norm. You enter fm,d (and, for lateral torsional buckling, fm,k and E0,05) from your own copy of SIA 265; the strength class you choose is recorded with the calculation.
Linearly: the utilisations about the two axes add, as in SIA 265 equation (14). Eurocode 5 combines them differently, with a reduction factor on one of the two terms for rectangular sections. The two forms should not be mixed; in SIA 265, km is the lateral torsional buckling factor.
Yes. For each axis the module derives the relative slenderness from the lateral support spacing and reduces the bending resistance with km. Enter 0 where the compression edge is held continuously, which gives km = 1.
No. This module checks biaxial bending at the ultimate limit state. The fire situation and compression with biaxial bending have their own modules; shear and serviceability are separate checks.
The full calculation is shown on screen, free and without an account. A print-ready A4 calculation note, with your title block, is part of the Basic plan.
Norm: SIA 265:2021 · Page updated: 20 October 2026