Timber member in compression 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 member under axial compression and bending about both axes, to SIA 265:2021: the cross-section where the member is held against buckling, and buckling with bending where it is not. 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, the strengths fc,0,d, fc,0,k, fm,d, fm,k and E0,05 from your copy of SIA 265, the system factor, the lateral support spacings, the actions at the stabilised section and at the unstabilised section (Nd, My,d, Mz,d for each, since they usually differ), any loss of section, the role of the member and the buckling lengths.

You get the three checks with their utilisations and every intermediate value with its clause, and figures. With the Basic plan, it prints as an A4 calculation note.

Assumptions and limits

Worked example

A glulam member GL24h, b × h = 680 × 280 mm, in service class 2, with N_d = 199 kN, M_y,d = 93 kN·m and M_z,d = 1 kN·m at both sections, and buckling lengths and lateral support spacings of 6.82 m. 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 b680 mm
Section depth h280 mm
Design compressive strength ∥ grain fc,0,d16 N/mm²§ 3.4.2.2 t.9
Characteristic compressive strength ∥ grain fc,0,k24 N/mm²§ Annexe B
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
Lateral support spacing, bending about y-y ay6820 mm§ 4.2.9.3
Lateral support spacing, bending about z-z az6820 mm§ 4.2.9.3
Design compressive force Nd199 kN
Design moment about y-y My,d93 kN·m
Design moment about z-z Mz,d1 kN·m
Section loss determined byNone — gross section
Design compressive force Nd199 kN
Design moment about y-y My,d93 kN·m
Design moment about z-z Mz,d1 kN·m
Section loss determined byNone — gross section
Role of the member λlimMain load-bearing element§ 4.2.8.4
Buckling length given byEntered directly
Buckling length about y-y lk,y6820 mm
Buckling length about z-z lk,z6820 mm
Calculation: Stabilised zone — no buckling
StepValueRef.
Gross section areaA = b · hA = 190,400 mm²
Net section areagross sectionAnet = 190,400 mm²
Section modulus about y-yWy = b · h² / 6Wy = 8,885,333 mm³
Section modulus about z-zWz = h · b² / 6Wz = 21,578,667 mm³
Height factor, y-yuses the dimension perpendicular to the axis, here hkh,y = 1.0792§ 3.4.2.3 (5)
Height factor, z-zfor bending about z-z the relevant dimension is bkh,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-ykm,y = 1.0000§ 4.2.9.2 (39)
Lateral torsional buckling factor, z-zkm,z = 1.0000§ 4.2.9.2 (39)
Compressive resistancefRd,C = fc,0,d · ηt · ηwfRd,C = 12.800 N/mm²§ 4.2.2.1 (10)
Bending resistance, y-yfRd,m,y = km,y · fm,d · kh,y · ksys · ηt · ηwfRd,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
Compressive stressσc,0,d = Nd / Anetσc,0,d = 1.0452 N/mm²
Bending stress, y-yσm,y,d = My,d / Wyσm,y,d = 10.4667 N/mm²
Bending stress, z-zσm,z,d = Mz,d / Wzσm,z,d = 0.0463 N/mm²
Compression with biaxial bending, cross-section(σc/fc)² + σm,y/fm,y + σm,z/fm,z ≤ 1η = 0.768 · Pass77 % of the capacity used§ 4.2.4.2 (22)
Calculation: Common values
StepValueRef.
Net section areagross sectionAnet = 190,400 mm²
Straightness factorβc = 0.10§ 4.2.8.3
Compressive stressσc,0,d = Nd / Anet at this sectionσc,0,d = 1.0452 N/mm²
Independent axesThe verification may be carried out independently for the two principal axes, which is why the two checks below are not combined into one.§4.2.8.5§ 4.2.8.5
Calculation: Check 1 — buckling about y-y
StepValueRef.
Radius of gyration about y-yiy = 0.289 · hiy = 80.92 mm
Buckling length about y-ylk,y = 6,820 mm
Slenderness about y-yλy = lk,y / iyλy = 84.281§ 4.2.8.4
Relative slenderness, y-yλrel = (λ/π)·√(fc,0,k / E0,05), not less than 0.30λrel,y = 1.3414§ 4.2.8.4 (34)
Auxiliary factor, y-yk = 0.5 · [1 + βc·(λrel − 0.3) + λrel²]ky = 1.4517§ 4.2.8.2 (31)
Buckling coefficient, y-ykc = 1 / (k + √(k² − λrel²)), with k from equation (31) abovekc,y = 0.4983§ 4.2.8.2 (30)
Compressive resistance, y-yfRd,C,y = kc,y · fc,0,d · ηt · ηwfRd,C,y = 6.378 N/mm²
Bending stress, y-yσm,y,d = 10.4667 N/mm²
Buckling about y-y with bending about y-yσc/(kc,y·fc) + σm,y/(km,y·fm) ≤ 1η = 0.922 · Pass92 % of the capacity used§ 4.2.8.5 (37)
Calculation: Check 2 — buckling about z-z
StepValueRef.
Radius of gyration about z-ziz = 0.289 · biz = 196.52 mm
Buckling length about z-zlk,z = 6,820 mm
Slenderness about z-zλz = lk,z / izλz = 34.704§ 4.2.8.4
Relative slenderness, z-zλrel = (λ/π)·√(fc,0,k / E0,05), not less than 0.30λrel,z = 0.5523§ 4.2.8.4 (34)
Auxiliary factor, z-zk = 0.5 · [1 + βc·(λrel − 0.3) + λrel²]kz = 0.6652§ 4.2.8.2 (31)
Buckling coefficient, z-zkc = 1 / (k + √(k² − λrel²)), with k from equation (31) abovekc,z = 0.9655§ 4.2.8.2 (30)
Compressive resistance, z-zfRd,C,z = kc,z · fc,0,d · ηt · ηwfRd,C,z = 12.358 N/mm²
Bending stress, z-zσm,z,d = 0.0463 N/mm²
Buckling about z-z with bending about z-zσc/(kc,z·fc) + σm,z/(km,z·fm) ≤ 1η = 0.088 · Pass9 % of the capacity used§ 4.2.8.5 (37)

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 are there two zones?

The section held against buckling and the section where buckling can occur usually carry different actions, and they are verified with different equations: (22) for the cross-section, (37) for buckling with bending.

Why is the compression term squared in the cross-section check?

That is the form of SIA 265 equation (22) for compression with bending on the cross-section; the buckling check, equation (37), is linear in each term.

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: 20 October 2026