design timber structures using eurocode 5

Seminar on Sustainable Future through Timber Design UITM, Dec. 16.12.2014

Simon Aicher

Design Timber Structures using

Eurocode 5

Aicher Eurocode 5 Timber Structures UITM 2014, Malaysia

Contents of lecture

2

Basics of permissible stress and semi-probabilistic partial factor concept Interrelationship of - Eurocodes, - harmonized (timber) product standards, - classification standards, calculation standards and - test test standards Basics of Eurocode 5 structure and contents Design example: straight glulam beam (EC 5 vs. permissible concept) Design example: curved glulam beam (EC 5 vs. permissible concept)

Aicher Eurocode 5 Timber Structures UITM 2014, Malaysia 3

100 years old glulam beams, train repair hall, Bellinzona, Italy

Olympic Ice rink Hammar, Norway, 1994 glulam truss beams, span:97m

Manufacture of timber parasols for Expo 2000, Hannover


HESS – Limitless –Verbindung (22)

7-storey timber

building, Berlin, 2011

10-storey timber

building, Melbourne,

Australia 2013

Eurocodes and supporting product and test standards

Eurocodes regulate design of timber, steel, concrete structures in conjunction with national application documents but give no provisions on material properties

Harmonized product standards regulate material properties of harmonized building products (e.g. not adhesives) such as EN 14080 glulam EN 14081-1 solid timber in conjunction with national grading rules and classification standard EN 1912 and strength class standard EN 338 EN 15497 finger jointed lumber EN 16351 cross laminated timber EN 14374 LVL EN 13986 panel products in conjunction with product / production standards, e.g. EN 300 for OSB Test standards, e.g. EN 408, EN 789,….. Calculation standards, e.g. EN 14358 on characteristic values


Permissible stress concept

13

σact = σ95 acting loads, hence resulting section forces E and stresses σ represent in general 95% quantiles of the distributions

Design verification

σact ≤ σpermissible where in case of structural timber (roughly) σpermissible = f50 /3 f50 mean value of strength


Semiprobabilistic design concept with partial factors

14

σact = σ95 as in permissible stress concept the loads / section forces/ stress distributions represent 95% quantiles of the distributions

Design verification σd ≤ fd

σd design stress fd design strength

σd = σact · γL

γL partial factor for load (1,5 for live load; 1,35 for perm. load)

f d = fk · kmod / γM fk characteristic strength property (5% quantile)

kmod modification factor (time, climate) γM partial factor for strength (material dependant; 1,1 to 1,3)


Semiprobabilistic vs. permissible stress design concept

15

σact ≤

σd = σact · γL = σact · 1,5 ≤ fd =

γL = 1,5 partial factor for load f05 = f50 (1 - 1,64 · COV) assuming COV = 0,12 f05 = f50 (1 – 0,2) = f50 / 1,25

f05 · kmod f50 · kmod

γM 1,25 γM

with γM = 1,3 and kmod = 0,8

f05 · kmod f50 · 0,8

γM 1,25 · 1,3

=

= ≈ f50

2 f05 · kmod f50

γM =

f50 2 · 1,5 = f50

3 = σpermissible

2


Graphical illustration of semiprobabilistic design concept

Probability density

ms

fs

ms 95

ms 95· γs kmod · mR 05 / γR

kmod · mR 05

kmod · mR

fR

R, s

=

β · σz = mz = kmod · mR - ms

fz

pf = 10 -6


Eurocode 5: Design of Timber Structures – Part 1-1


Structural Eurocode Program comprises


Scope of EN 1995


Structure of Eurocode 5 ( = EN 1995)


Subjects / Topics of EN 1995-1-1


Normative References


Normative References (continued)


Section 2 of EC 5: Basis of design


Section 2.2 of EC 5: Principles of limit state design


2.2.2 Ultimate limit states


2.2.3 Serviceability limit states


2.3 Basic variables


2.3.1.2 Load-duration classes


2.3.1.3 Service classes


2.3.2 Materials and product properties


2.4 Verification by the partial factor method

5%- quantile value (lognormal dist.)


Recommended partial factors γM for material properties

EC 5 – Table 2.3


2.4.2 Design values of geometrical data


2.4.2 Design value of a resistance

Example: Rk = Xk · relevant cross-sectional quantity


EC 5 –Section 3 – Materials properties


3.1.3/4 Strength and deformation modification factors


EC 5 – Table 3.1 Strength modification values kmod


EC 5 – Table 3.1 Strength modification values kmod

(continued)


Accumulated duration of load [hours]

Stre

ngth

mod

ifica

tion

fact

or

kmod

0

0.2

0.4

0.6

0.8

1

1.2

0.0010.01 0.1 1 10 100

100010000

100000

1000000

1 min 1 Woche 6 Monate 10 Jahre 50 Jahre

sehr kurz kurz mittel lang ständig

Nutzungsklasse 1/2

Nutzungsklasse 3

Madison-Kurve

Strength modification values kmod = f( time; moisture)

Service class 1 and2

Service class 3

short very short

medium long permanent

short 1 week 6 months 10 years 10 years


EC 5 – Table 3.2 Deformation modification values kdef


EC 5 – Table 3.2 Deformation modification values kdef

(continued)


EC 5 – 3.2: Solid timber

EN 15497


EC 5 – 3.3: Glued laminated timber



EN 14080 Now large finger joints are directly regulated in the harmonized product standard for glulam,


Example of large finger joint (single joint line)


Example of large finger joint (two joint lines)


EC 5 – 3.4: Laminated veneer lumber (LVL)


EC 5 – 3.5: Wood-based panels


EC 5 – 3.6: Adhesives

Note: As permissible structural adhesive families and respective classifications have been profoundly changed in conjunction with introduction of one-component Polyurethane (1K-PU) and polymer isocyanate (EPI) adhesives according to EN 15425 and EN 16351 principle P (2) is no more throughout valid because of EPI definitions.


EC 5 – 3.7: Metal fasteners


EC 5 – Section 4: Durability


EC 5 – Table 4.1 Corrosion protection of fasteners


EC 5 - Section 5: Basis of structural analysis


5.2 Members


5.4 Assemblies


5.4.2 Frame structures


5.4.4 Plane frames and arches


Examples of assumed initial geometry deviations

geometry of frames

initial geometry deviation corresponding to symmetrical load

initial geometry deviation corresponding to non-symmetrical load


EC 5 - Section 6: Ultimate limit states


Tension 6.1.2 Tension parallel to the grain

6.1.2 Tension perpendicular to the grain


Compression 6.1.4 Compression parallel to the grain


Compression 6.1.4 Compression perpendicular to the grain

σc,90,d is the design compressive stress in the effective contact area perpendicular to the grain;

Fc,90,d is the design compressive load perpendicular to the grain;

Aef is the effective contact area in compression perpendicular to the grain;

Fc,90,d is the design compressive strength perpendicular to the grain;

kc,90 is a factor taking into account the load configuration, the possibility of splitting and the degree of compressive deformation.

where



The effective contact area perpendicular to the grain, Aef, should be determined taking into account an effective contact length parallel to the grain, where the actual contact length, ℓ, at each side is increased by 30 mm, but not more than a, ℓ or ℓ1/2, see Figure 6.2. 2. The value of kc,90 should be taken as 1,0 unless the conditions in the following paragraphs apply. In these cases the higher value of kc,90 specified may be taken, with a limiting value of kc,90 = 1,75. 3. For members on continuous supports, provided that ℓ1 ≥ 2h, see Figure 6.2a, the value of kc,90 should be taken as: – kc,90 = 1,25 for solid softwood timber – kc,90 = 1,5 for glued laminated softwood timber where h is the depth of the member and ℓ is the contact length.



4. For members on discrete supports, provided that ℓ1 ≥ 2h, see Figure 6.2b, the value of kc,90 should be taken as: – kc,90 = 1,5 for solid softwood timber – kc,90 = 1,75 for glued laminated softwood timber provided that I ℓ ≤ 400 mm where h is the depth of the member and ℓ is the contact length.


6.1.6 Bending


6.1.7 Shear


6.1.7 Shear (crack factor issue)

kcr = 0,67 for solid timber

kcr = 0,67 for glued laminated timber

kcr = 1,0 for other wood-based products in accordance with EN 13986 and EN 14374.

2. For the verification of shear resistance of members in bending, the influence of cracks should be taken into account using an effective width of the member given as: bef = kcr b where b is the width of the relevant section of the member. NOTE: The recommended value for kcr is given as


6.1.7 Shear


6.1.8 Torsion


6.2.2 Compression stresses at an angle to grain


6.2.3 Combined bending and axial tension


6.2.3 Combined bending and axial compression


6.4 Members with varying cross-section or curved shape



Figure 6.8 Single tapered beam



(a)

Figure 6.9 – Double tapered (a) and curved (b) beams with the fibre direction parallel to the lower edge of the beam

Note: In curved beams the apex zone extends over the curved parts of the beam


Figure 6.9 – Pitched cambered beam (c) beam with the fibre direction parallel to the lower edge of the beam

Note: In pitched cambered beams the apex zone extends over the curved parts of the beam




or


Design of straight glulam member

- comparison of Eurocode 5 vs. DIN 1052

Design examples


GL 24 / BS 11

q = 9 kN/m, g = 6 kN/m

10 m

16x8

0 cm

Geometry: l = 10 m b = 160 mm h = 800 mm S = b h²/6 = 17 ⋅ 10-6 mm³ I = b h³/12 = 6.8 ⋅ 10-9 mm4

Straight beam design comparison – EC 5 vs. perm. stress concept


Property permissible concept semi-probabilistic concept

Bending strength σm,perm = 11 N/mm² fm,k = 24 N/mm²

Shear strength τv,perm = 1.2 N/mm² fv,k = 3.5 N/mm²

MOE Em = 11000 N/mm² Em,mean = 11000 N/mm²

crack factor - kcr = 0.67

modification factor for duration of load and moisture content

kmod = 0.6 (Service Class I/II, medium-term)

Partial factor for material properties

γM = 1.25 (glulam, EC 5)

Deformation factor kdef = 0.8 (Service Class I)

Partial factor for permanent actions

γG = 1.35

Partial factor for variable actions

γG = 1.5

Factor for quasi-permanent value of a variable action

ψ2,1 = 0.3

Design comparison – EC 5 vs. perm. stress concept


Result permissible concept semi-probabilistic concept

distributed load F = g + q = 15 kN/m Fd = γG g + γQ q = 21.6 kN/m

bending moment M M = F l² / 8 = 188 kNm Md = Fd ⋅ l² / 8 = 270 kNm

bending stress σm = M/S = 11 N/mm² σm = Md/S = 15.8 N/mm²

utilization (bending)

11 / 11 = 1.00

fm,d = fm,k ⋅kmod /γM = 15.4 N/mm² 15.8 / fm,d = 1.03

shear force V V = F l/2 = 75 kN Vd = Fd l/2 = 108 kN

shear stress τv 1.5 V / (b h) = 0.88 N/mm² 1.5 Vd / (b h) = 1.89 N/mm²

utilization (shear)

1.2 / 0.88 = 0.73

fv,d = fv,k ⋅kmod /γM = 2.24 N/mm² 1.89 / fv,d = 0.84



deflection 𝑢 =

5 𝐹 𝑙4

384𝐸𝐸 = 26𝑚𝑚 𝑢𝑖𝑖𝑖𝑖 = 𝑢𝑖𝑖𝑖𝑖,𝑔 + 𝑢𝑖𝑖𝑖𝑖,𝑞=

5𝑔𝑙4

384𝐸𝐸 +5𝑞𝑙4

384𝐸𝐸 = 10.4 + 15.6= 26𝑚𝑚

𝑢𝑓𝑖𝑖 = 𝑢𝑖𝑖𝑖𝑖,𝑔(1 + 𝑘𝑑𝑑𝑓) +𝑢𝑖𝑖𝑖𝑖,𝑞(1 +ψ2,1 𝑘𝑑𝑑𝑓)= 16.7 + 18.4 = 35.1mm

utilitization (deflection) 𝑢𝑙/300 = 0.78

𝑢𝑖𝑖𝑖𝑖𝑙/300 = 0.78 𝑢𝑓𝑖𝑖𝑙/150 = 0.53



Design of curved glulam beam

- comparison of Eurocode 5 vs. DIN 1052

Design examples


WM

RH,max, 250=σ⊥

R H

R2 / H = 11

R1 / H = 2,5

R H

H/R2 = 0,09

H/R1 = 0,4

WM

RH,

RH,II

++=σ

2

603501

+

-

H/R2 = 0,09

H/R1 = 0,4

tension stresses perpendicular to grain

Stress distributions in curved beams with const. moment

bending stresses parallel to grain

R1 < R2

R1 < R2


h

stress perp. to grain

- +

h

- +

Stress σt,90 of curved and tapered beams with line loads

stress perp. to grain


Curved beam design comparison – EC 5 vs. perm. stress concept

Geometry, dimensions and quality /strength class of example beam

EN 14080 GL 28: fm,k = 28 N/mm2

DIN 1052 BS 14: σm,permissible = 14 N/mm2


Design for bending:

kr = 0,96

hap / r = 0,118

EC5 F = 23,31 kN


rin/t = 200,

kl = 1,05

k1 = 1; k2 = 0,35, k3 = 0,6


Design for bending:

kr = 0,96

kl = 1,05

EC5

fm,d = fm,k × kmod /γm

GL28: fm,k = 28 N/mm2

load duration: „medium“, kmod = 0,8

glulam: γm = 1,25

σm,d = 6,85 N/mm2 Map,d = γf × Map

combined loading: γf = 1,4

fm,d = 17,92 N/mm2

ratio = 0,38

F = 23,31 kN



Design for tension perp.:

kdis = 1,4 kVol = (V0/V)0,2 = 0,43

EC5

glulam: V0 = 0,01 m3 V = 0,691 m3

curved beam design comparison – EC 5 vs. perm. stress concept

kp = 0,0294

k5 = 0; k6 = 0,25, k7 = 0

hap / r = 0,118


Example: Curved Beam with pure moment loading

kp = 0,0294

EC5

ft,90,d = ft,90,k × kmod /γm

glulam: ft,90,k = 0,5 N/mm2

load duration: „medium“, kmod = 0,8

glulam: γm = 1,25

σt,90,d = 0,19 N/mm2 Map,d = γf × Map

combined loading: γf = 1,4

ft,90,d = 0,32 N/mm2

ratio = 1,0

F = 23,31 kN

kdis = 1,4 , kVol = 0,43,

1,4 x 0,43 x 0,32 = 0,19 N/mm2

Design for tension perp.:


Design for bending:

hap / r = 0,118, kl = 1,05

DIN 1052

F = 23,31 kN

σm ≤ σm,permissible

σm = kl × 6 Map/b h2

σm,permissible = 14 N/mm2

σm = 4,66 N/mm2

ratio = 0,33



Design für tension perp.:

DIN 1052

F = 23,31 kN

σt,90 ≤ σt,90,permissible

σt,90 = kp× 6 Map/b h2

σt,90,permissible = 0,2 N/mm2

σt,90 = 0,14 N/mm2

ratio = 0,69

hap / r = 0,118, kp = 0,0294



DIN 1052

F = 23,31 kN

EC5

bending tension perp.

1,00

0,69

0,40

0,33

no pre-stress effect no size effect



Now ist time to finish!

Thank you very much for your patient listening!

design timber structures using eurocode 5

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