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 mm

^{2}

Second moment of area about the y-y axis

*I*= bh

_{y}^{3}/12 = (100 x 200

^{3})/12 = 6.667 x 10

^{7}mm

^{4}

Radius of gyration about the y-y axis

*i*= sqrt(I

_{y }_{y}/A) = sqrt(6.667 x 10

^{7}/20000 ) = 57.74 mm

Second moment of area about the z-z axis

*I*= hb

_{z}^{3}/12 = (200 x 100

^{3})/12 = 1.667 x 10

^{7}mm

^{4}

Radius of gyration about the z-z axis

*i*= sqrt(I

_{z }_{z}/A) = sqrt(1.667 x 10

^{7}/20000 ) = 28.86 mm

Effective length about the y-y axis

*L*= Effective length about the z-z axis

_{E,y}*L*= 1.0 L = 3000 mm

_{E,z}Therefore slenderness ratio about the y-y axis;

Î»

_{y}=

*L*/

_{E,y}*i*=3000/57.74 = 51.957

_{y}The slenderness ratio about the z-z axis;

Î»

_{z}=

*L*/

_{E,z}*i*=3000/28.86 = 103.95

_{z}**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/mm

^{2}

Fifth percentile modulus of the elasticity parallel to the grain E

_{0.05}= 6.0 kN/mm

^{2}

**Partial safety factors**

UKNA to BS EN 1990:2002, Table NA.A1.2(B)) for the ULS

Permanent actions, Î³

_{G}= 1.35

Variable actions, Î³

_{Q}= 1.5

Material factor for solid timber, Î³

_{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 Î±

_{c,0,d}= N

_{Ed}/A = 51000/20000 = 2.55 N/mm

^{2}

The design compression strength f

_{c,0,d}= (k

_{mod.med}k

_{sys }f

_{c.0.k})/ Î³

_{M }

f

_{c,0,d}= (0.8 x 1.0 x 18)/1.3 = 11.08 N/mm

^{2}

**Buckling Resistance (clause 6.3.2 EC5)**

Relative slenderness about the y-y axis

Î»

_{rel,y}= (Î»

_{y}/Ï€) x sqrt(f

_{c.0.k}/E

_{0.05}) = (51.957/Ï€) x sqrt(18/6000) = 0.9058 (equation 6.21 EC5)

Relative slenderness about the z-z axis

Î»

_{rel,z}= (Î»

_{z}/Ï€) x sqrt(f

_{c.0.k}/E

_{0.05}) = (103.95/Ï€) 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 = Î»

Factor Î²_{rel,z}= 1.81_{c}for solid timber Î²

_{c}= 0.2 (EC5, equation (6.29))

The factor

*k*(equation 6.28 EC5)

_{z}*k*= 0.5[1 + 0.2(1.81 - 0.3) + 1.81

_{z}^{2}] = 2.289

Instability factor (equation 6.26 EC5)

*k*= 1/[2.289 + sqrt(2.289

_{zc}^{2}- 1.81

^{2})] = 0.27

Design buckling strength

*k*f

_{zc}._{c,0,d }= 0.27 x 11.08 = 2.9916 N/mm

^{2}

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

Therefore, the section is satisfactory.

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