Design of flexural members such as timber beams principally involves consideration of the effects of actions such as bending, deflection, vibration, lateral buckling, shear, and bearing. The process of design of such structures is described in Eurocode 5 (EN 1995-1-1:2004), and a design example is shown in this article.

**Design Example**

A 75 mm by 200 mm deep sawn timber beam in a domestic residence supports the characteristic loading shown above. The beam has a clear span of 2.75 m, the bearing length has been restricted to 100 mm at each end, is of strength class C24 to BS EN 338:2003, and functions in service class 2 conditions. The beam is laterally restrained against lateral buckling along its length.

Given that;

*G*_{k}_{.}_{udl} = 1*.*3 kN/m (characteristic uniformly distributed permanent action)*Q*_{k}_{.}_{udl }= 1*.*5 kN/m (characteristic uniformly distributed medium-term action)*G*_{k}_{.}_{p }= 1*.*00 kN characteristic point load at mid-span

**1. Beam geometric properties**

Breadth of the beam *b *= 75 mm

Depth of the beam* h *= 200 mm

Clear span of the beam, *l _{c}* = 2750 mm

Bearing length of the beam at each end,

*b*= 100 mm

_{l}Design span of the beam l = (

*l*+

_{c}*l*) = 2750 + 100 = 2850 = 2.85 m

_{b}Section modulus of the beam about the

*y*–

*y*axis,

*W*=

_{y}*bh*

^{2}/6 = (75 × 200

^{2})/6 = 5

*× 10*

^{5}mm

^{3}

**2. Timber properties**

Strength class C24 (BS EN 338:2003, *Table 1*):

Characteristic bending strength, *f*_{m}_{.}_{k} = 24 N/mm^{2}

Characteristic shear strength, *f*_{v}_{.}_{k }= 2*.*5 N/mm^{2}

Characteristic bearing strength, *f*_{c}_{,}_{90}_{,}_{k }= 2*.*5 N/mm^{2}

Fifth-percentile modulus of elasticity parallel to the grain, *E*_{0}_{.}_{05 }= 7*.*4 kN/mm^{2}

Mean modulus of elasticity parallel to the grain, *E*_{0}_{,}_{mean }= 11 kN/mm^{2}

Mean shear modulus, *G*_{0}_{,}_{mean }= 0*.*69 kN/mm^{2}

Mean density of the beam timber, *ρ*_{m} = 420 kg/m^{3}

**3. Partial safety factors**

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

Permanent actions, *γ _{G.}*

_{ULS }= 1

*.*35

Variable actions,

*γ*

_{Q.}_{ULS }= 1

*.*5

(UKNA to BS EN 1990:2002, *Table NA.A1.1 *– Category A)

Factor for the quasi-permanent value of the variable action, *ψ*_{2} = 0*.*3

(UKNA to EC5, *Table NA.3*)

Material factor for solid timber at the ULS, *γ*_{M} = 1*.*3

**4. Actions**

**(i) ULS***(a) Characteristic self-weight of the beam, G _{k,swt}*

*G*

_{k}

_{,}_{swt }=

*b*·

*h*·

*g*·

*ρ*

_{m}= (0.075 × 0.2 × 9.81 × 420)/1000 = 0

*.*062 kN/m

Design action from the selfweight of the beam,

*F*

_{d}

_{,}_{swt}

*F*

_{d}

_{,}_{swt }=

*γ*

_{G.}_{ULS }·

*G*

_{k}

_{.}_{swt }

*=*1.35 × 0.062 = 0

*.*0837 kN/m

*(b) Characteristic permanent action due to the point load, G _{k,p}*

*G*

_{k}

_{,}_{p }= 1

*.*00 kN

Design permanent action due to the point load,

*F*

_{d}

_{.}_{p}

*F*

_{d}

_{,}_{p }=

*γ*

_{G.}_{ULS }·

*G*

_{k}

_{.}_{p }= 1.35 × 1.0 = 1

*.*35 kN

*(c) Characteristic permanent action due to the UDL, G _{k,udl}*

*G*

_{k}

_{,}_{udl }= 1

*.*3 kN/m

Design action due to the permanent action UDL,

*F*

_{d}

_{,}_{p}

_{,}_{udl}

*F*

_{d}

_{,}_{p}

_{,}_{udl }=

*γ*

_{G.}_{ULS }·

*G*

_{k}

_{.}_{udl }=1.35 x 1.3 = 1.755 kN/m

*(d) Characteristic medium-term action due to the UDL, Q _{k,udl}*

*Q*

_{k}

_{,}_{udl }= 1

*.*5 kN/m

Design action due to the variable action UDL,

*F*

_{d}

_{,}_{q}

_{,}_{udl}

*F*

_{d}

_{,}_{q}

_{,}_{udl }=

*γ*

_{Q.}_{ULS }·

*Q*

_{k}

_{.}_{udl }= 1.5 × 1.5 = 2.25 kN/m

Total UDL @ ULS = 0*.*0837 + 1.755 + 2.25 = 4.1 kN/m

Total concentrated action @ ULS = 1.35 kN

**5. Modification factors**

Factor for medium-duration loading and service class 2, *k*_{mod}_{.}_{med} _{ }= 0*.*8 (EC5, *Table 3.1)*

Size factor for depth greater than 150 mm, *k*_{h} = 1*.*0 (EC5,* equation (3.1)*)

Lateral stability of the beam: *k*_{crit} = 1 (EC5, *6.3.3*))

Bearing factor k_{c,90 }= (taken as 1.0) (EC5, clause 6.1.5(2) )

Deformation factor for service class 2, *k*_{def} = 0*.*8 (EC5, *Table 3.2*)

Load sharing factor, *k*_{sys} is not relevant *k*_{sys} = 1.0

**(6) Bending strength**

The design bending moment;

M_{d} =*ql ^{2}/8 + PL/4*= (4.1 × 2.85

^{2})/8 + (1.35 × 2.85)/4 = 4.162 + 0.96 = 5.122 kNm

Design bending stress, *σ*_{m}_{,}_{y}_{,}_{d }= *M*_{d}/*W _{y }= *(5.122 × 10

^{6})/( 5

*× 10*

^{5}) = 10.244 N/mm

^{2}

Design bending strength, *f*_{m}_{,y,}_{d }= (*k*_{mod}_{.}_{med}·*k*_{sys}·*k*_{h}· *f*_{m}_{.}_{k})/*γ*_{M} = (0.8 × 1.0 × 1.0 × 24)/1.3 = 14*.*77 N/mm^{2}

*σ*_{m}_{,y,}_{d }< *f*_{m}_{,y,}_{d } Section is okay in bending

**(7) Shear Strength**

Design shear force *V _{d }= ql/2 + P/2* = (4.1 × 2.85)/2 + (1.35/2) = 6.52 kN

Design shear stress,

*τ*

_{v}

_{.}_{d }(EC5,

*equation (6.60)*)

*τ*

_{v}

_{.}_{d }=

*1.5V*= (1.5 × 6.52 × 1000)/(75 × 200) = 0.652 N/mm

_{d}/bh_{ef}^{2}

Design shear strength, *f*_{v}_{,}_{d }= (*k*_{mod}_{.}_{med}·*k*_{sys}· *f*_{v}_{.}_{k})/*γ*_{M} = (0.8 × 1.0 × 2.5)/1.3 = 1.54 N/mm^{2}

*τ*_{v}_{.}_{d }<* f*_{v}_{,}_{d }Section is okay in shear

**(8) Bearing Strength**

The design bearing force will equal the design shear force in the beam, *V*_{d}

Design bearing stress, *σ*_{c}_{,}_{90}_{,}_{d }= *V*_{d}*/b*·*l _{b} =* (6.52 × 1000)/(75 x 100) = 0.833 N/mm

^{2}

Design bearing strength, (EC5,*equation (6.3)*))

*f*_{c}_{.}_{90}_{.}_{d }= (*k*_{mod}_{.}_{med }· *k*_{sys} · *k*_{c}_{.}_{90 }· *f*_{c}_{.}_{90}_{.}_{k})/* γ*_{M} = (0.8 × 1.0 × 2.5)/1.3 = 1.54 N/mm^{2}

*σ*_{c}_{,}_{90}_{,}_{d }<* f*_{c}_{.}_{90}_{.}_{d }Section is okay in bearing

**(9) Deflection**

**Instantaneous deflection due to permanent actions**

u_{inst,point,G }= (1/4) × [1/(11 × 75 × 200^{3})] × 2850^{3} × [1 + 1.2 × (11/0*.*69) × (200/2850)^{2}] = 0.959 mm

U_{inst,udl,G} = (5/32) × [(1.3 × 10^{-3} + 0.062 × 10^{-3})/(11 × 75 × 200^{3})] × 2850^{4} × [1 + 0.96 × (11/0*.*69) × (200/2850)^{2}] = 2.287 mm

U_{inst.G }= u_{inst,point,G }+ U_{inst,udl,G }= 0.959 + 2.287 = 3.246 mm

*Instantaneous deflection due to variable action*

U_{inst,Q} = (5/32) × [(1.5 × 10^{-3})/(11 × 75 × 200^{3})] × 2850^{4} × [1 + 0.96 × (11/0*.*69) × (200/2850)^{2}] = 2.519 mm

Combined permanent and variable instantaneous deflection = u_{inst} = u_{inst,G }+ u_{inst,Q }= 3.246 + 2.519 = 5.765 mm

Eurocode 5 limit on deflection (Table 7.2, EC5) *w _{inst}* =

*l*/300 = 2850/300 = 9.5 mm (u

_{inst}<

*w*Instantaneous deflection is okay)

_{inst}**Final deflection**

Final deflection due to permanent actions u_{fin,G }= u_{inst,G }(1 + k_{def}) (Equation 2.3, EC5)

u_{fin,G }= 3.246 (1 + 0.8) = 5.843 mm

Final deflection due to the variable and quasi-permanent actions, u_{fin,Q }= u_{inst,Q }(1 + *ψ*_{2}k_{def}) (Equation 2.4, EC5)

u_{fin,Q }= u_{inst,Q }(1 + *ψ*_{2}k_{def}) = 2.519 (1 + 0.3 × 0.8* *) = 3.12 mm

Final deflection due to the permanent and quasi-permanent actions actions

u_{net,fin }= u_{fin,G }+ u_{fin,Q }= 5.843 + 3.12 = 8.963 mm

Deflection limit w_{net,fin }= *l*/150 = 2850/150 = 19 mm

The deflection of the beam is satisfactory

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good work, keep the work flying.