Timber design to EC5 Timber design to EC5 –– BeamsBeams

Dr. Keerthi Ranasinghe Dr. Keerthi Ranasinghe Senior Structural EngineerSenior Structural EngineerTRADA TechnologyTRADA Technology

The approach The approach –– BS 5268BS 5268

� Loads� BS 6399 etc.

� Effects of loads� Bending stress, Shear stress, Bearing stress,

� Deflection,

� Vibration ???

� Material properties� Grade stresses (strengths),

� k factors,

� Permissible stresses

� Verifications (Permissible stress design)� Bending, Shear, Bearing,

� Deflection,

� Vibration – a notional check.

The approach The approach –– EC5EC5

� Actions (Loads) – action combinations, partial factors � BS EN 1990: 2002 Eurocode basis of structural design. – Action combinations, PSI

factors, Gamma factors

� BS EN 1991: 2002 etc. - Loads

� Effects of actions� Design bending stress,

� Design shear stress,

� Design bearing stress,� Design bearing stress,

� Final deflection (including creep)

� Vibration – a more fundamental design approach.

� Material properties – partial factors� BS EN 338:2003 (Solid timber), BS EN 1194:1999 (Glulam) etc.,

� Design strength(s)

� Verifications (Limit state design)� ULS (Bending, Shear, Bearing)

� SLS (deflection, Vibration)

ExampleExample

Check the adequacy of 47 x 200 mm regularised domestic floor joists of strength class C24 timber on an effective span of 3.6 m spaced at 600 mm centres.

References:

Regularised timber sizes: BS EN 336:2003

Solid timber strength

Breadth of the joist: b: 47 mm

Depth of the joist: h: 195 mm

Strength class:Solid timber strength classes – BS EN 338:2003

Glulam strength classes – BS EN 1194:1999

System strength (Load sharing factor) – Cl. 6.6 of EC5

Strength class: C24

Effective span: Le: 3600 mm

Joist spacing: sp: 600 mm

System strength,(load sharing ?):

ksys: 1.1

Example Example –– Loads (EC1 the NA referred)Loads (EC1 the NA referred)

References:

Imposed loads – Tables NA.2 & NA.3 of NA to EC1

Service classes – Cl. 2.3.1.3 of EC5 & Table NA.2 of NA to EC5

Dead load, excludingthe joist self weight:

F1: 0.25 kN/m2

Floor imposed distributedload (domestic):

F2: 1.5 kN/m2

Floor imposed point load(domestic):

F3: 2.0 kN Noticed ?Noticed ?

Service class: Service Class 1

Bearing length Lb 40mm≥ : Lb: 100 mm

Example Example –– Section properties (mind the Section properties (mind the axis system!)axis system!)

References:

Axis system – Figure 6.1 of EC5

Cross sectional area:

A b h⋅ 9.16 103× mm2⋅=:=Section modulus, symbol changed from Z to W –Cl. 1.6 of EC5 (Symbols)

Further references:

Design value of geometrical data – Cl. 2.4.2 of EC5

A b h⋅ 9.16 103× mm2⋅=:=

Elastic section moduli about the stronger axis (y-y):

Wyb h2⋅

6297.86 103× mm3⋅=:=

Second moment of area about the stronger axis (y-y):

Iyb h3⋅12

29.04 106× mm4⋅=:=

Example Example –– Material properties (£ vs. Material properties (£ vs. flexibility)flexibility)

References:

Characteristic material strength properties for solid timber – Table 1 of BS EN 338:2003

Characteristic material strength properties for Glulam – Table 1 (for

Cha. bending strength: fm.k 24 N mm 2−⋅⋅=

Cha. compression strength perpendicular to grain:

fc.90.k 2.5 N mm 2−⋅⋅=

Cha. shear strength: fv.k 2.5 N mm 2−⋅⋅=

Mean modulus of elasticity parallel:Glulam – Table 1 (for homogenous glulam) and table 2 (for combined glulam) of BS EN 1194:1994

Characteristic material strength properties (generic) for OSB, particleboards and fibreboards – BS EN 12369-1:2001 and for plywood – BS EN 12369-2:2004

E0.mean 11 kN mm 2−⋅⋅=

Mean shear modulus: Gmean 0.69 kN mm 2−⋅⋅=

Mean density: ρ mean 420 kg m 3−⋅⋅=

Minimum density: ρ k 350 kg m 3−⋅⋅=

Example Example –– Material related modification Material related modification factors (remember the old k factors?)factors (remember the old k factors?)

References:

Partial factors for timber related materials – Table NA.3 of NA to EC5

Depth modification factor for solid timber –Cl. 3.2(3) of EC5 (look

Partial factor for timber: γ M 1.3=

kh min150mm

h

0.2

1.3,

h 150mm<( ) ρ k 700kg

m3≤

∧if

1.0 otherwise

:=

Note: ^ refers to a logical "AND" operand. kh 1.0=Cl. 3.2(3) of EC5 (look at other materials as well) – Not using is conservative

kmod factors – Table 3.1 of EC5 (must use without fail)

kmod value for permanent duration, kmod,perm :

kmod.perm 0.6=

kmod value for medium-term duration, kmod,mt :

kmod.mt 0.8=

kmod value for medium-term duration, kmod,st :

kmod.st 0.9=

Note: only the permanent, medium-term & short-term values areobtained here as we only have these loads.

Example Example –– Material related modification Material related modification factors (contd..)factors (contd..)

References:

System strength factor, ksys - Cl. 6.6 of EC5

Lateral torsional stability factor, kcrit - Cl. 6.3.3 of EC5

Spacing between domestic floor joists is generallyless than 600 mm for conventional floor build-upsand are assumed to share loads with other joistsin the system. Since the spacing is 600 mm forthis calculation load sharing can be assumed forthis calculation.

ksys 1=

For conventional floor build-ups, lateraldisplacement of compressive edge and torsional

kcrit 1.0:=

kdef factor – (not immediately required here under the derivation of material strength properties etc, but used further down in the calculations) – Table 3.2 of EC5

displacement of compressive edge and torsionalrotation at supports of the joists are bothprevented due to the timber decking. Therefore,kcrit is assumed to be 1.

kdef 0.6=

Example Example –– Design values of material Design values of material properties (the fun begins)properties (the fun begins)

References:

Design values of material properties – Cl. 6.3.3(1) of EC0 for general expression

OR

Cl. 2.4.1 of EC5

Xd ηXk

γ M⋅:=

Notes:Xk is the characteristic value of the1.

material or product property.γ M is the partial factor for materials. 2.

Partial factors for timber related materials – Table NA.3 of NA to EC5

M is the partial factor for materials.

η represents conversion factors taking3.

into account, where.

volume and scale effects, such as•depth modification factor kh,

effects of moisture and temperature,•such as kmod, and

any other relevant factors, such as•ksys, kcrit, kc,90 etc, are taken care

of.

Example Example –– Design strengths (end result Design strengths (end result on the strength side)on the strength side)

References:

Design values of material properties – Cl. 6.3.3(1) of EC0 for general expression

OR

DesignBendingStrength:

Permanent:fm.d.perm

fm.k kh⋅ kmod.perm⋅ ksys⋅ kcrit⋅

γ M11.08 N mm 2−⋅⋅=:=

Medium-term:fm.d.mt

fm.k kh⋅ kmod.mt⋅ ksys⋅ kcrit⋅

γ M14.77 N mm 2−⋅⋅=:=

Short-term:fm.d.st

fm.k kh⋅ kmod.st⋅ ksys⋅ kcrit⋅

γ M16.62 N mm 2−⋅=:=

Cl. 2.4.1 of EC5

Crack factor for shear resistance, kcr -Cl. 6.1.7(2) of EC5

γ M

DesignShearStrength:

Permanent:fv.d.perm 0.67

fv.k kmod.perm⋅ ksys⋅

γ M0.77 N mm 2−⋅⋅=:=

Medium-term:fv.d.mt 0.67

fv.k kmod.mt⋅ ksys⋅

γ M1.03 N mm 2−⋅⋅=:=

Short-term:fv.d.st 0.67

fv.k kmod.st⋅ ksys⋅

γ M1.16 N mm 2−⋅⋅=:=

Example Example –– Design strengths (contd..)Design strengths (contd..)

References:

Compression perpendicular to the grain factor, kc,90 - Cl. 6.1.5 & Ex. 6.3 of EC5

DesignBearingStrength:

Permanent:

fc.90.d.permfc.90.k kmod.perm⋅ ksys⋅ kc.90⋅

γ M1.15 N mm 2−⋅⋅=:=

Medium-term:

Assuming simply supported joists, conservatively:

kc.90 1.0:=

Medium-term:

fc.90.d.mtfc.90.k kmod.mt⋅ ksys⋅ kc.90⋅

γ M1.54 N mm 2−⋅⋅=:=

Short-term:

fc.90.d.stfc.90.k kmod.st⋅ ksys⋅ kc.90⋅

γ M1.73 N mm 2−⋅⋅=:=

Example Example –– Actions (derivation of design Actions (derivation of design loads)loads)

References:

Partial safety factor for actions – Table NA.A1.2(B) of NA to EC0

Self-weight – Cl.

γ G 1.35=Unfavourable permanent actions:

Unfavourable variable actions: γ Q 1.5=

Load case 1: Permanent loads only:

The dead load per unit length:

Gk.1 sp F1⋅ 0.15 kN m 1−⋅⋅=:=Self-weight – Cl. 4.1.2(5) of EC0 Self-weight of the joist as a load per unit length:

Acceleration due to gravity: g 9.807m

s2=

Gk.2 ρ mean g⋅ A⋅ 37.749 N m 1−⋅⋅=:=

Total permanent load as a load per unit length:

ΣG k Gk.1 Gk.2+ 0.188 kN m 1−⋅⋅=:=

Permanent design load as a load per unit length:

Gd γ G ΣG k⋅ 0.253 kN m 1−⋅⋅=:=

Example Example –– Derivation of design loads Derivation of design loads (contd..)(contd..)

References:

Load duration classes – Table NA.1 of NA to EC5

For loaded area of category A1, domestic self-containeddwelling units, the worst effects from either a uniformly

distributed area load of 1.5 kN/m2 or a point load of 2.0 kNshould be adopted as the imposed load effects. Theseloads were input previously and are discussed in detailbelow under load cases 2 and 3 respectively.

Load case 2 (permanent + imposeddistributed load, medium-term duration):distributed load, medium-term duration):

Hence, the floor load per unit length:

Qk.1 sp F2⋅ 0.9 kN m 1−⋅⋅=:=

Variable design load 1:

Qd.1 γ Q Qk.1⋅ 1.35 kN m 1−⋅⋅=:=

Total design load for Load case 2:

Gd Qd.1+ 1.6 kN m 1−⋅⋅=

Example Example –– Derivation of design loads Derivation of design loads (contd..)(contd..)

References:

Load duration classes – Table NA.1 of NA to EC5

Load case 3 (permanent + imposed pointload, short-term duration):

Floor imposed point load: Qk.2 F3 2 kN⋅=:=

Variable design load 2: Qd.2 γ Q Qk.2⋅ 3 kN⋅=:=

Design loads for Load case 3:

Load per unit length: Gd 0.253 kN m 1−⋅⋅=d

Point load: Qd.2 3 kN⋅=

Example Example –– Effects of loads (stresses, the Effects of loads (stresses, the easier route to combinations)easier route to combinations)

Bending Stresses:

Bending stress due to the permanent design load, Gd:

σ m.y.d.GdGd Le

2⋅

8 Wy⋅1.38 N mm 2−⋅⋅=:=

Bending stress due to the variable design load 1, Qd.1:

σ m.y.d.Qd1Qd.1 Le

2⋅

8 Wy⋅7.34 N mm 2−⋅⋅=:=m.y.d.Qd1 8 Wy⋅

Bending stress due to the variable design load 2, Qd.2:

σ m.y.d.Qd2Qd.2 Le⋅

4 Wy⋅9.06 N mm 2−⋅⋅=:=

Hence, permanent duration design bending stress:

σ m.y.d.perm σ m.y.d.Gd 1.38 N mm 2−⋅⋅=:=

Medium-term design bending stress:

σ m.y.d.mt σ m.y.d.Gd σ m.y.d.Qd1+ 8.721 N mm 2−⋅=:=

Short-term design bending stress:

σ m.y.d.st σ m.y.d.Gd σ m.y.d.Qd2+ 10.443 N mm 2−⋅=:=

Example Example –– Derivation of stresses (contd..)Derivation of stresses (contd..)

Shear Stresses: Shear stress due to the permanent design load, Gd:

τ d.Gd1.5 Gd Le⋅( )

2 A⋅0.07 N mm 2−⋅⋅=:=

Shear stress due to the variable design load 1, Qd.1:

τ d.Qd11.5 Qd.1 Le⋅( )

2A0.4 N mm 2−⋅⋅=:=

Shear stress due to the variable design load 2, Qd.2:Shear stress due to the variable design load 2, Qd.2:

τ d.Qd21.5 Qd.2( )

A0.49 N mm 2−⋅⋅=:=

Hence, permanent duration design shear stress:

τ d.perm τ d.Gd 0.07 N mm 2−⋅⋅=:=

Medium-term design shear stress:

τ d.mt τ d.Gd τ d.Qd1+ 0.472 N mm 2−⋅=:=

Short-term design shear stress:

τ d.st τ d.Gd τ d.Qd2+ 0.566 N mm 2−⋅=:=

Example Example –– Derivation of stresses (contd..)Derivation of stresses (contd..)

Bearing Stresses:

Effect of effective contact area in compression, Aef

– Ex. 6.4 of EC5

Bearing stress due to the permanent design load, Gd:

σ c.90.d.GdGd Le⋅

2 b⋅ Lb⋅0.1 N mm 2−⋅⋅=:=

Bearing stress due to the variable design load 1, Qd.1:

σ c.90.d.Qd1Qd.1 Le⋅

2 b⋅ Lb⋅0.52 N mm 2−⋅⋅=:=

Bearing stress due to the variable design load 2, Q :Bearing stress due to the variable design load 2, Qd.2:

σ c.90.d.Qd2Qd.2

b Lb⋅0.64 N mm 2−⋅⋅=:=

Hence, permanent duration design bearing stress:

σ c.90.d.perm σ c.90.d.Gd 0.1 N mm 2−⋅⋅=:=

Medium-term design shear stress:

σ c.90.d.mt σ c.90.d.Gd σ c.90.d.Qd1+ 0.614 N mm 2−⋅=:=

Short-term design shear stress:

σ c.90.d.st σ c.90.d.Gd σ c.90.d.Qd2+ 0.735 N mm 2−⋅=:=

Example Example –– Deflections (instantaneous vs. Deflections (instantaneous vs. final)final)

References:

The PSI factors – Table NA.A1.1 of NA to EC0

Values of psi factors for category A,domestic self contained areas:

ψ 0 0.7=

ψ 1 0.5=

ψ 2 0.3=

psi factor for variable load 1, Qk.1: ψ 2.1 ψ 2 0.3=:=psi factor for variable load 2, Qk.2: ψ 2.2 ψ 2 0.3=:=

Deflections – Cl. 2.2.3(5) of EC5

Instantaneous deflection due to the permanent load, SGk:

uinst.G ΣG k5

384

Le4

E0.mean Iy⋅⋅

1.2

8

Le2

Gmean A⋅⋅+

⋅ 1.34 mm⋅=:=

Final deflection due to the permanent load, SGk:

ufin.G uinst.G 1 kdef+( )⋅ 2.15 mm⋅=:=

Example Example –– Deflections (contd..)Deflections (contd..)

References:

Deflections – Cl. 2.2.3(5) of EC5

Instantaneous deflection due to the variable load 1, Qk.1:

uinst.Qk1 Qk.15

384

Le4

E0.mean Iy⋅⋅

1.2

8

Le2

Gmean A⋅⋅+

⋅ 6.44 mm⋅=:=

Final deflection due to the variable load 1, Qk.1:

ufin.Qk1 uinst.Qk1 1 ψ 2.1 kdef⋅+( )⋅ 7.6 mm⋅=:=

Instantaneous deflection due to the variable load 2, Qk.2:

uinst.Qk2 Qk.21

48

Le3

E0.mean Iy⋅⋅

1.5

5

Le

Gmean A⋅⋅+

⋅ 6.43 mm⋅=:=

Final deflection due to the variable load 2, Qk.2:

ufin.Qk2 uinst.Qk2 1 ψ 2.2 kdef⋅+( )⋅ 7.58 mm⋅=:=

Example Example –– Deflections (limits, for Deflections (limits, for guidance only?)guidance only?)

References:

Deflections – Cl. 2.2.3(5) of EC5

Allowable deflection –Table NA.4 of NA to EC5

The total final deflection:

ufin max ufin.G ufin.Qk1+ ufin.G ufin.Qk2+, ( ) 9.746 mm=:=

Allowable final deflection for floor members with aplastered or plasterboard ceiling:

uallowableLe

25014.4 mm⋅=:=

NA to EC5 (Guidance only)

Example Example –– Verifications (the easier bit)Verifications (the easier bit)

ULS verifications

Bending

Bendingratio.permσ m.y.d.perm

fm.d.perm0.124=:=

Bendingratio.mtσ m.y.d.mt

fm.d.mt0.59=:=

Bendingratio.stσ m.y.d.st

fm.d.st0.629=:= Bending "PASSED"=

ShearShearratio.perm

τ d.perm

fv.d.perm0.097=:=

Shearratio.mtτ d.mt

fv.d.mt0.458=:=

Shearratio.stτ d.st

fv.d.st0.488=:= Shear "PASSED"=

Example Example –– Verifications (contd.. But, not Verifications (contd.. But, not finished)finished)

ULS verifications

Bearing

Bearingratio.permσ c.90.d.perm

fc.90.d.perm0.084=:=

Bearingratio.mtσ c.90.d.mt

fc.90.d.mt0.399=:=

Bearingratio.stσ c.90.d.st

fc.90.d.st0.425=:= Bearing "PASSED"=

SLS verifications

Deflection

c.90.d.stDeflection Check:

Deflectionratioufin

uallowable0.677=:= Deflection "PASSED"=

Vibration checks are discussed in section 7 of the code where stiffness parameters for the floor as a system are defined. These checks are not included here, but an in-depth discussion on vibration is given in TRADA guidance document, GD6.

Finally Finally -- Tick the boxes….Tick the boxes….

� Basis of timber design – remains unchanged

� Loadings – some values changed

� Design value of loadings – new concept under limit states

� Partial factors – complex, but not too difficult to grasp the concept

� Material properties – not given, buy other codes, manufacturer literature

� (Material) modification factors – similar to previous� (Material) modification factors – similar to previous

� Axis system – changed, beware!

� Notations, symbols – some changes, beware!

� Suffix system – looks more systematic once got used

� Long winded? YES

� Difficult? NO

� Economic? GENERALLY YES