Timber bar in tension and biaxial bending to SIA 265

The code governs which inputs appear and how every value is derived.

Member

Material and cross-section, shared by every verification. Change the verification above and these stay as they are.

About this check

clauses, worked example and questions

Checks a rectangular solid-timber or glulam bar under axial tension and bending about both principal axes at the ultimate limit state, to SIA 265:2021, including lateral torsional buckling. Every step shows its formula and the clause it comes from.

What it checks

What you enter and what you get

You enter the member (wood type, strength class, load duration, service class, b, h), the strengths ft,0,d and fm,d from your copy of SIA 265 (clause 3.4.2.2, table 9), fm,k and E0,05 (annex B), the system factor ksys, the design forces Nd, My,d and Mz,d, the method for the net section and the fastener type, and the lateral support spacings ay and az.

You get the areas, section moduli, factors, resistances and stresses, and the utilisation of equation (21) with a pass or fail, each with its clause. With the Basic plan, it prints as an A4 calculation note.

Assumptions and limits

Worked example

A glulam bar GL24h, b × h = 280 × 440 mm, in service class 2 under short to permanent actions, with Nd = 33 kN, My,d = 103 kN·m and Mz,d = 2 kN·m, the net section from the indicative coefficient for wood–wood dowels, the compression edge held for bending about y-y and lateral supports every 5.15 m for bending about z-z. These are the module's own results for its worked example, computed when this page was built.

Inputs
InputValueRef.
Wood typeGlulam
Strength classGL24h
Load duration ηtShort to permanent§ 2.2.6
Service class ηw2§ 3.2.1.3 t.4
Section width b280 mm
Section depth h440 mm
Design tensile strength ∥ grain ft,0,d12.8 N/mm²§ 3.4.2.2 t.9
Design bending strength fm,d16 N/mm²§ 3.4.2.2 t.9
Characteristic bending strength fm,k24 N/mm²§ Annexe B
Modulus of elasticity, 5% fractile E0,059600 N/mm²§ Annexe B
System factor ksys1.0§ 5.7.2.2
Design tensile force Nd33 kN
Design moment about y-y My,d103 kN·m
Design moment about z-z Mz,d2 kN·m
Section loss determined byIndicative coefficient (pre-design)
Fastener type Anet/ADowel, wood–wood
Lateral support spacing, bending about y-y ay0 mm§ 4.2.9.3
Lateral support spacing, bending about z-z az5150 mm§ 4.2.9.3
Calculation: Ultimate limit state
StepValueRef.
Gross section areaA = b · hA = 123,200 mm²
Net section areaAnet = 0.65 · AAnet = 80,080 mm²§ 6.1.1.8
Section modulus about y-yWy = b · h² / 6Wy = 9,034,667 mm³
Section modulus about z-zWz = h · b² / 6Wz = 5,749,333 mm³
Height factor, y-ykh uses the dimension perpendicular to the axis considered, here hkh,y = 1.0315§ 3.4.2.3 (5)
Height factor, z-zfor bending about z-z the relevant dimension is bkh,z = 1.0792§ 3.4.2.3 (5)
Relative slenderness, y-yλrel,m = 1.15 · (√(ay·h) / b) · √(fm,k / E0,05); 0 when the edge is heldλrel,m,y = 0.0000§ 4.2.9.3 (42)
Relative slenderness, z-zλrel,m = 1.15 · (√(az·b) / h) · √(fm,k / E0,05)λrel,m,z = 0.1569§ 4.2.9.3 (42)
Lateral torsional buckling factor, y-ykm,y = 1.0000§ 4.2.9.2 (39)
Lateral torsional buckling factor, z-zkm,z = 1.0000§ 4.2.9.2 (39)
Tensile resistancefRd,T = ft,0,d · ηt · ηwfRd,T = 10.240 N/mm²§ 4.2.1.1 (6)
Bending resistance, y-yfRd,m,y = km,y · fm,d · kh,y · ksys · ηt · ηwfRd,m,y = 13.203 N/mm²§ 4.2.9.1 (38)
Bending resistance, z-zfRd,m,z = km,z · fm,d · max(kh,z ; ksys) · ηt · ηw. For bending about z-z, §3.4.2.2 forbids applying kh and ksys together, so the larger of the two is taken.fRd,m,z = 13.814 N/mm²§ 3.4.2.2
Tensile stressσt,0,d = Nd / Anetσt,0,d = 0.4121 N/mm²
Bending stress, y-yσm,y,d = My,d / Wyσm,y,d = 11.4005 N/mm²
Bending stress, z-zσm,z,d = Mz,d / Wzσm,z,d = 0.3479 N/mm²
Tension with biaxial bendingσt/ft + σm,y/fm,y + σm,z/fm,z ≤ 1η = 0.929 · Pass93 % of the capacity used§ 4.2.4.1 (21)

How this module is verified

Verified clause by clause against SIA 265:2021 and against independent calculations used on real projects, before release.

Questions

Why does the net section reduce only the tensile stress?

The module takes the section moduli from the full section and applies the connection loss to the axial term only. Where the connection needs a detailed net-section calculation, the tension module provides it.

Why is the larger of k_h and k_sys taken about z-z?

Clause 3.4.2.2 forbids applying both together for bending about z-z, so the module uses the larger of the two.

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.

Related

Norm: SIA 265:2021 · Page updated: 21 October 2026