Eurocode 5 Timber Column Design Calculator
This expert-grade calculator determines the design axial compression resistance of rectangular timber columns according to Eurocode 5 (EN 1995-1-1), including global buckling about both principal axes. It is intended for structural engineers and advanced students seeking a fast, transparent, and standards-aligned check.
Input Parameters
Material properties
Section geometry
Member length and factors
Effective length factors
Imperfection factors
Safety factor and applied load
Results
Enter values to see results. All outputs update instantly.
Data Source and Methodology
Authoritative Source: EN 1995-1-1:2004 Eurocode 5 — Design of timber structures — Part 1-1: General — Common rules and rules for buildings, Clause 6.3.2, and material properties per EN 338/EN 14080. Direct link to the code text (public copy): EN 1995-1-1:2004 PDF.
Tutti i calcoli si basano rigorosamente sulle formule e sui dati forniti da questa fonte.
The Formula Explained
Design compressive strength: \( f_{c,0,d} = \dfrac{k_{mod}\, f_{c,0,k}}{\gamma_M} \)
Critical buckling load about each axis: \( N_{cr} = \dfrac{\pi^2\, E_{0,mean}\, I}{L_{eff}^2} \), with \(L_{eff} = k \cdot L\)
Relative slenderness: \( \lambda_{rel} = \sqrt{\dfrac{A\, f_{c,0,k}}{N_{cr}}} \)
Reduction factor (Perry–Robertson): \( \phi = \tfrac12 \left[1 + \alpha\left(\lambda_{rel} - 0.3\right) + \lambda_{rel}^2\right] \)
Then: \( \chi = \dfrac{1}{\phi + \sqrt{\phi^2 - \lambda_{rel}^2}} \le 1.0 \)
Design axial resistance per axis: \( N_{c,Rd} = \chi \, A \, f_{c,0,d} \)
Governing resistance: \( N_{c,Rd,\,min} = \min\left(N_{c,Rd,y},\, N_{c,Rd,z}\right) \)
Glossary of Variables
| Symbol | Description | Unit |
|---|---|---|
| fc,0,k | Characteristic compressive strength parallel to grain | N/mm² |
| E0,mean | Mean modulus of elasticity parallel to grain | N/mm² |
| kmod | Modification factor for load duration and service class | — |
| γM | Material partial factor | — |
| b, h | Section dimensions (width, depth) | mm |
| A | Cross-section area = b·h | mm² |
| Iy, Iz | Second moments: Iy=b·h³/12, Iz=h·b³/12 | mm⁴ |
| L | Clear member length | m |
| ky, kz | Effective length factors about y and z axes | — |
| αy, αz | Imperfection factors for each axis | — |
| Ncr | Euler critical load for the axis | N |
| λrel | Relative slenderness | — |
| χ | Buckling reduction factor | — |
| Nc,Rd | Design axial compression resistance | kN (reported) |
| NEd | Applied design axial load (input) | kN |
| η | Utilization ratio = NEd / Nc,Rd | — |
Worked Example
How It Works: A Step-by-Step Example
Given: C24, fc,0,k=21 N/mm², E0,mean=11000 N/mm²; b=h=140 mm; L=3.0 m; ky=kz=1.0; kmod=0.80; γM=1.30; αy=αz=0.20.
- Area and inertia: A=140·140=19600 mm²; Iy=Iz=140·140³/12=32.0×10⁶ mm⁴.
- Effective lengths: L_eff,y=L_eff,z=1.0·3000 mm=3000 mm.
- Critical loads: Ncr=π²·11000·32.0e6 / 3000² ≈ 387,000 N.
- Relative slenderness: λrel=√(A·fc,0,k / Ncr)=√(19600·21 / 387000)≈1.03.
- ϕ=0.5·[1+0.2·(1.03−0.3)+1.03²]=0.5·(1+0.146+1.0609)=1.103.
- χ=1/(ϕ+√(ϕ²−λrel²))≈1/(1.103+√(1.216−1.061))≈0.61.
- fc,0,d=kmod·fc,0,k/γM=0.8·21/1.3=12.92 N/mm².
- Nc,Rd=χ·A·fc,0,d=0.61·19600·12.92 ≈ 154,000 N = 154 kN (both axes; governing).
Therefore, the design axial resistance is approximately 154 kN. If NEd = 120 kN, utilization η ≈ 0.78 (OK).
Frequently Asked Questions (FAQ)
Which Eurocode version and clauses are implemented?
EN 1995-1-1:2004, primarily Clause 6.3.2 for compression members with global buckling. Material properties are aligned with EN 338 (solid timber) and EN 14080 (glulam) where applicable.
Do I need to include ksys or kc system factors?
System or stability factors beyond the Perry–Robertson reduction are not included. If required by your National Annex or design situation, account for them separately.
What about combined bending and compression?
This tool is focused on axial compression with global buckling. Interaction with bending should follow EC5 interaction equations and second-order analysis, which are beyond this calculator’s scope.
How should I choose α (imperfection) and effective length factors?
Use recognized guidance or National Annex tables. Defaults α=0.20 and k=1.0 are indicative for pinned-pinned; different end restraints and member types require different values.
Are service class and load duration covered?
Yes, via kmod. Determine kmod per EC5 §2.3.2 based on service class and load duration class, then input that value.
Why are forces reported in kN but the math uses N/mm²?
Internally, EC units are used for consistency with EC5: stresses in N/mm² and geometry in mm. Final forces are presented in kN for readability.