Design of Axially Loaded Timber Columns to EC5

Eurocode 5 deals with the limit state design of timber structures. In this post, we are going to present an example on how we can design axially loaded timber columns according to the requirements of Eurocode 5.



Example
Verify the capacity of 200 mm x 100 mm timber section of strength class C18 to satisfy ultimate limit state requirements under the following conditions:

Service Condition: Class 2
Characteristic permanent action Gk = 10 kN
Variable medium term action = 25 kN
Height of column = 3.0m
Support condition: Effectively held in position but nor in direction in both axes.

Solution
Geometric properties
Depth of member h = 200 mm
Width of member b = 100 mm

Area of member  A =  bh = 200 x 100 = 20000 mm²
Second moment of area about the y-y axis I_y = bh³/12 = (100 x 200³)/12 = 6.667 x 10⁷ mm⁴
Radius of gyration about the y-y axis i_y = sqrt(I_y/A) = sqrt(6.667 x 10⁷/20000 ) = 57.74 mm

Second moment of area about the z-z axis I_z = hb³/12 = (200 x 100³)/12 = 1.667 x 10⁷ mm⁴
Radius of gyration about the z-z axis i_z = sqrt(I_z/A) = sqrt(1.667 x 10⁷/20000 ) = 28.86 mm

Effective length about the y-y axis L_E,y = Effective length about the z-z axis L_E,z = 1.0 L = 3000 mm

Therefore slenderness ratio about the y-y axis;
λ_y = L_E,y/i_y =3000/57.74 = 51.957

The slenderness ratio about the z-z axis;
λ_z = L_E,z/i_z =3000/28.86 = 103.95

Timber Properties (Table 1 BS EN 338:2003(E))
For strength class C18;
Characteristic compression strength parallel to the grain f_c.0.k = 18 N/mm²

Fifth percentile modulus of the elasticity parallel to the grain E_0.05 = 6.0 kN/mm²

Partial safety factors
UKNA to BS EN 1990:2002, Table NA.A1.2(B)) for the ULS
Permanent actions, γ_G = 1.35
Variable actions, γ_Q = 1.5

Material factor for solid timber, γ_M = 1.3 (UKNA to EC5, Table NA.3)

Design Action
Design Action at ultimate limit state  N_Ed = 1.35Gk + 1.5Qk
N_Ed = 1.35(10) + 1.5(25) = 51 kN = 51000 N

Modification factors
Factor for medium duration loading and service class 2, k_mod.med = 0.8 (EC5, Table 3.1)
System strength factor, (not relevant) k_sys = 1.0

Therefore, the design compression stress α_c,0,d = N_Ed/A = 51000/20000 = 2.55 N/mm²




The design compression strength f_c,0,d = (k_mod.med k_sys f_c.0.k)/ γ_M 
f_c,0,d = (0.8 x 1.0 x 18)/1.3 = 11.08 N/mm²

Buckling Resistance (clause 6.3.2 EC5)
Relative slenderness about the y-y axis
λ_rel,y = (λ_y /π) x sqrt(f_c.0.k/E_0.05) = (51.957/π) x sqrt(18/6000) = 0.9058 (equation 6.21 EC5)

Relative slenderness about the z-z axis
λ_rel,z = (λ_z /π) x sqrt(f_c.0.k/E_0.05) = (103.95/π) x sqrt(18/6000) = 1.81 (equation 6.21 EC5)

As both relative slenderness ratios are greater than 0.3, the conditions in clause 6.3.2(3):

Maximum relative slenderness ratio of the column = λ_rel,z = 1.81

Factor β_c for solid timber β_c = 0.2 (EC5, equation (6.29))

The factor k_z (equation 6.28 EC5)

k_z = 0.5[1 + 0.2(1.81 - 0.3) + 1.81²] = 2.289

Instability factor (equation 6.26 EC5)

k_zc = 1/[2.289 + sqrt(2.289² - 1.81²)] = 0.27

Design buckling strength k_zc .f_c,0,d = 0.27 x 11.08 = 2.9916 N/mm²

Therefore, design stress/design buckling strength ratio = 2.55/2.9916 = 0.852 < 1.0 Ok

Therefore, the section is satisfactory.