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HAUNCH

CalculationsTimber Post

Timber Post

EN 1995-1-1

Checks a timber post under axial compression and combined bending to EN 1995-1-1.

Diagram

NEd=30kNh=2700mm

Example inputs

Height2700 mm
Axial load (NEd)30 kN
Width b97 mm
Depth h97 mm
Timber gradeC24
Service class1
Eff. length factor (y)1
Eff. length factor (z)1
National annexUK

Results

Timber Post — 97×97mm

C24 · SC1 · 2700mm height · EN 1995-1-1

Results

Axial Compression PASS
util 0.253.19EN 1995-1-1 §6.1.4
kmod (modification factor)0.8EN 1995-1-1 Table 3.1
γM (partial factor)1.3EN 1995-1-1 Table 2.3
fc,0,k (char. compressive strength)21N/mm²EN 338
fc,0,d (design compressive strength) = kmod × fc,0,k / γM12.92N/mm²EN 1995-1-1 §2.4.1
Cross-section area A = b × h9409mm²
σc,0,d (design compressive stress) = NEd × 1000 / (b × h)3.19N/mm²EN 1995-1-1 §6.1.4
Buckling Y PASS
util 0.763.19EN 1995-1-1 §6.3.2
Effective length Lcr,y = ky × height2700mmEN 1995-1-1 §6.3.2
Radius of gyration iy = h / √1228mm
Slenderness λy = Lcr,y / iy96.42
λrel,y (relative slenderness) = (λ/π) × √(fc,0,k / E0,05)1.635EN 1995-1-1 §6.3.2
ky instability factor = 0.5 × (1 + βc × (λrel − 0.3) + λrel²)1.97EN 1995-1-1 §6.3.2
kc,y (buckling factor) = 1 / (ky + √(ky² − λrel,y²))0.326EN 1995-1-1 §6.3.2
kc,y × fc,0,d (buckling capacity) = kc,y × fc,0,d4.21N/mm²
σc,0,d (design compressive stress) = NEd × 1000 / (b × h)3.19N/mm²
Buckling Z PASS
util 0.763.19EN 1995-1-1 §6.3.2
Effective length Lcr,z = kz × height2700mmEN 1995-1-1 §6.3.2
Radius of gyration iz = b / √1228mm
Slenderness λz = Lcr,z / iz96.42
λrel,z (relative slenderness) = (λ/π) × √(fc,0,k / E0,05)1.635EN 1995-1-1 §6.3.2
kz instability factor = 0.5 × (1 + βc × (λrel − 0.3) + λrel²)1.97EN 1995-1-1 §6.3.2
kc,z (buckling factor) = 1 / (kz + √(kz² − λrel,z²))0.326EN 1995-1-1 §6.3.2
kc,z × fc,0,d (buckling capacity) = kc,z × fc,0,d4.21N/mm²
σc,0,d (design compressive stress) = NEd × 1000 / (b × h)3.19N/mm²

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