Design Timber Structures Using Eurocode 5

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

Design Timber Structures using Eurocode 5

Simon Aicher

Contents of lecture 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

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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)

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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 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 Aicher

mean value of strength


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Semiprobabilistic design concept with partial factors

σact = σ95

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

Design verification σd design stress

σd ≤ fd

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) Aicher


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Semiprobabilistic vs. permissible stress design concept f05 = f50 (1 - 1,64 · COV) assuming COV = 0,12 f05 = f50 (1 – 0,2) = f50 / 1,25 f05 · kmod γM

=

γL = 1,5 partial factor for load

f50 · kmod 1,25 γM

with γM = 1,3 and kmod = 0,8 f05 · kmod γM

=

f50 · 0,8 1,25 · 1,3

≈

f50 2

f05 · kmod f50 = σd = σact · γL = σact · 1,5 ≤ fd = γM 2 f50 σact ≤ f50 2 · 1,5 = 3 = σpermissible Aicher


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Graphical illustration of semiprobabilistic design concept Probability density

ms fs

β · σz = mz = kmod · mR - ms fz

ms 95 pf = 10 -6 Aicher

ms 95· γs

kmod · mR 05

=

kmod · mR fR

R, s

kmod · mR 05 / γR


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

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Structural Eurocode Program comprises

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Scope of EN 1995

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Structure of Eurocode 5 ( = EN 1995)

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Subjects / Topics of EN 1995-1-1

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Normative References

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Normative References (continued)

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Normative References (continued)

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Section 2 of EC 5: Basis of design

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Section 2.2 of EC 5: Principles of limit state design

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2.2.2 Ultimate limit states

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2.2.3 Serviceability limit states

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2.2.3 Serviceability limit states

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2.3 Basic variables

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2.3.1.2 Load-duration classes

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2.3.1.2 Load-duration classes

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2.3.1.3 Service classes

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2.3.2 Materials and product properties

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2.3.2 Materials and product properties

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2.4 Verification by the partial factor method

5%- quantile value (lognormal dist.)

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Recommended partial factors γM for material properties EC 5 – Table 2.3

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2.4.2 Design values of geometrical data

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2.4.2 Design value of a resistance

Example: Rk = Xk · relevant cross-sectional quantity

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EC 5 –Section 3 – Materials properties

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3.1.3/4 Strength and deformation modification factors

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EC 5 – Table 3.1 Strength modification values kmod

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EC 5 – Table 3.1 Strength modification values kmod (continued)

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Strength modification values kmod = f( time; moisture) short 1 min

Strength modification factor

1.2 kmod

10Jahre years50 Jahre 10 1 1 week Woche 66months Monate 10 years

Madison-Kurve

Service class 11/2 and2 Nutzungsklasse

1

0.8

Service class 3

Nutzungsklasse 3

0.6

0.4

very short

short kurz

sehr kurz

medium mittel

permanent ständig

long lang

0.2

0

0 0 .0

1

1 0 . 0

0 .1

1

10

10

0

0 10

0 1

0 00

0 10

0 00

Accumulated duration of load [hours] Aicher


0 10

0

0 00

0

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EC 5 – Table 3.2 Deformation modification values kdef

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EC 5 – Table 3.2 Deformation modification values kdef (continued)

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EC 5 – 3.2: Solid timber

EN 15497 Aicher


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EC 5 – 3.3: Glued laminated timber

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

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EN 14080

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Example of large finger joint (single joint line)

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Example of large finger joint (two joint lines)

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Example of large finger joint (two joint lines)

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EC 5 – 3.4: Laminated veneer lumber (LVL)

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EC 5 – 3.4: Laminated veneer lumber (LVL)

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EC 5 – 3.5: Wood-based panels

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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.

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EC 5 – 3.7: Metal fasteners

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EC 5 – Section 4: Durability

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EC 5 – Table 4.1 Corrosion protection of fasteners

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EC 5 - Section 5: Basis of structural analysis

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5.2 Members

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5.4 Assemblies

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5.4 Assemblies

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5.4.2 Frame structures

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5.4.4 Plane frames and arches

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Examples of assumed initial geometry deviations geometry of frames

initial geometry deviation corresponding to symmetrical load

initial geometry deviation corresponding to non-symmetrical load

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EC 5 - Section 6: Ultimate limit states

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Tension 6.1.2 Tension parallel to the grain

6.1.2 Tension perpendicular to the grain

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Compression 6.1.4 Compression parallel to the grain

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Compression

6.1.4 Compression perpendicular to the grain

where

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.

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Compression


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.

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Compression


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.

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6.1.6 Bending

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6.1.6 Bending

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6.1.7 Shear

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6.1.7 Shear (crack factor issue) 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 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.

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6.1.7 Shear

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6.1.8 Torsion

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6.2.2 Compression stresses at an angle to grain

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6.2.3 Combined bending and axial tension

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6.2.3 Combined bending and axial compression

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6.4 Members with varying cross-section or curved shape

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Figure 6.8 Single tapered beam

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(a)

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

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


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6.4 Members with varying cross-section or curved shape Note: In pitched cambered 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 Aicher


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or Aicher


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Design examples

Design of straight glulam member - comparison of Eurocode 5 vs. DIN 1052

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

16x80 cm

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

GL 24 / BS 11 10 m

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

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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

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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

1.2 / 0.88 = 0.73

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

utilization (shear)

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Design comparison – EC 5 vs. perm. stress concept

deflection

utilitization (deflection)

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5 𝐹 𝑙4 𝑢= = 26𝑚𝑚 384𝐸𝐸

𝑢 = 0.78 𝑙/300


𝑢𝑖𝑖𝑖𝑖 = 𝑢𝑖𝑖𝑖𝑖,𝑔 + 𝑢𝑖𝑖𝑖𝑖,𝑞= 5𝑔𝑙 4 5𝑞𝑙 4 + = 10.4 + 15.6 384𝐸𝐸 384𝐸𝐸 = 26𝑚𝑚 𝑢𝑓𝑓𝑓 = 𝑢𝑖𝑖𝑖𝑖,𝑔 (1 + 𝑘𝑑𝑑𝑑 ) + 𝑢𝑖𝑖𝑖𝑖,𝑞 (1 +ψ2,1 𝑘𝑑𝑑𝑑 )= 16.7 + 18.4 = 35.1mm 𝑢𝑖𝑖𝑖𝑖 = 0.78 𝑙/300 𝑢𝑓𝑓𝑓 = 0.53 𝑙/150

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Design examples

Design of curved glulam beam - comparison of Eurocode 5 vs. DIN 1052

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Stress distributions in curved beams with const. moment

H

R

R1 < R2

H

R

R1 < R2 H/R2 = 0,09

= 0,09 RH/R 2 / H2 = 11 RH/R =2,5 0,4 1 / H1 =

tension stresses perpendicular to grain

σ ⊥ ,max Aicher

H M = 0,25 R W Eurocode 5 Timber Structures

+

H/R1 = 0,4

bending stresses parallel to grain 2  H H  M σ II = 1+ 0,35 + 0,6    R  R   W 

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

stress perp. to grain h

-

+

stress perp. to grain

h

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

DIN 1052 BS 14: σm,permissible = 14 N/mm2 EN 14080 GL 28: fm,k = 28 N/mm2 Geometry, dimensions and quality /strength class of example beam Aicher


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EC5

F = 23,31 kN

Design for bending:

rin/t = 200, kr = 0,96

hap / r = 0,118 k1 = 1; k2 = 0,35, k3 = 0,6 kl = 1,05 Aicher


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F = 23,31 kN

EC5

Design for bending:

GL28: fm,k = 28 N/mm2

load duration: „medium“, kmod = 0,8

fm,d = fm,k × kmod /γm

kr = 0,96

fm,d =

glulam: γm = 1,25

17,92 N/mm2

ratio = 0,38 kl = 1,05

Map,d = γf × Map

σm,d = 6,85 N/mm2

combined loading: γf = 1,4

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

EC5

Design for tension perp.:

glulam: V0 = 0,01 m3 kdis = 1,4

V = 0,691 m3

kVol = (V0/V)0,2 = 0,43

hap / r = 0,118 k5 = 0; k6 = 0,25, k7 = 0 kp = 0,0294 Aicher


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Example: Curved Beam with pure moment loading F = 23,31 kN

EC5

Design for tension perp.:

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

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

load duration: „medium“, kmod = 0,8

ft,90,d = ft,90,k × kmod /γm

glulam: γm = 1,25

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

1,4 x 0,43 x 0,32 = 0,19 N/mm2 kp = 0,0294

Map,d = γf × Map

σt,90,d = 0,19 N/mm2

ratio = 1,0

combined loading: γf = 1,4

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DIN 1052

F = 23,31 kN

Design for bending: σm ≤ σm,permissible

σm,permissible = 14 N/mm2

ratio = 0,33 σm = kl × 6 Map/b h2 hap / r = 0,118, kl = 1,05 Aicher

σm = 4,66 N/mm2


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DIN 1052

F = 23,31 kN

Design für tension perp.: σt,90 ≤ σt,90,permissible

σt,90,permissible = 0,2 N/mm2

ratio = 0,69

σt,90 = kp× 6 Map/b h2 hap / r = 0,118, kp = 0,0294

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σt,90 = 0,14 N/mm2

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F = 23,31 kN

bending

tension perp.

EC5

0,40

1,00

DIN 1052

0,33

0,69

no pre-stress effect Aicher


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Now ist time to finish!

Thank you very much for your patient listening!

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Design Timber Structures Using Eurocode 5

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