Timber Design Using Eurocode

7/25/2019 Timber Design Using Eurocode

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Seminar on

Sustainable Future through

Timber Design

UITM, Dec. 16.12.2014

Simon Aicher

Design Timber Structures

usingEurocode 5

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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)

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Aicher Eurocode 5 Timber Structures UITM 2014, Malaysia 3

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100 years old

glulam beams,

train repair hall,

Bellinzona, ItalyPage 4 of 119


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Olympic Ice rink

Hammar, Norway,

1994

glulam truss beams,

span:97m

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Manufacture of timber parasols for Expo 2000, Hannover

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8/119Aicher Eurocode 5 Timber Structures UITM 2014, Malaysia 8

HESS Limitless Verbindung (22)

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7-storey timber

building, Berlin, 2011Page 10 of 119


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10-storey timber

building, Melbourne,

Australia 2013Page 11 of 119


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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 standardEN 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

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Permissible stress concept

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act = 95acting loads, hence resulting section forces E and stresses

represent in general 95% quantiles of the distributions

Design verification

act

permissible

where in case of structural timber (roughly)

permissible = f50/3

f50 mean value of strength

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Semiprobabilistic design concept with partial factors

14

act = 95as in permissible stress concept the loads / section forces/ stress

distributions represent 95% quantiles of the distributions

Design verificationd fd

d design stress fd design strength

d = act L

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

fd = fk kmod/ M

fk characteristic strength property (5% quantile)

kmod modification factor (time, climate)

M

partial factor for strength (material dependant; 1,1 to 1,3)Page 14 of 119



Semiprobabilistic vs. permissible stress design concept

15

act

d = act L= act 1,5 fd=

L = 1,5 partial factor for loadf05= f50(1 - 1,64 COV)

assuming COV = 0,12

f05= f50(1 0,2) = f50 / 1,25

f05 kmod f50 kmod

M

1,25 M

with M = 1,3 and kmod= 0,8

f05 kmod f50 0,8

M 1,25 1,3

=

= f50

2f05 kmod f50

M=

f50

2 1,5

=f50

3= permissible

2

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Graphical illustration of semiprobabilistic design concept

Probability

densityms

fs

ms 95

ms 95 s kmod mR 05 / R

kmod mR 05

kmod mR

fR

R, s

=

z = mz= kmod mR - ms

fz

pf = 10-6

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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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S f



Scope of EN 1995

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St t f E d 5 ( EN 1995)



Structure of Eurocode 5 ( = EN 1995)

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S bj t / T i f EN 1995 1 1


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Subjects / Topics of EN 1995-1-1

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N ti R f


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Normative References

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N ti R f ( ti d)


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Normative References (continued)

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N ti R f ( ti d)


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Normative References (continued)

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Aicher Eurocode 5 Timber Structures UITM 2014, Malaysia 25Page 25 of 119

S ti 2 f EC 5 B i f d i


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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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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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2.2.2 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 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 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 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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2.3.1.3 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 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


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2.4 Verification by the partial factor method

5%- quantile value (lognormal dist.)

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Recommended partial factors M for material properties


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Recommended partial factors Mfor 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 values of geometrical data

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2 4 2 Design value of a resistance


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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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EC 5 Section 3 Materials properties

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3.1.3/4 Strength and deformation modification factors


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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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C 5 ab e 3 S e g od ca o a ues mod

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EC 5 Table 3.1 Strength modification values kmod


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g mod

(continued)

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Strength modification values kmod = f( time; moisture)


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Accumulated duration of load [hours]

St

rengthmodif

icationfactor

kmod

0

0.2

0.4

0.6

0.8

1

1.2

.001

0.01 0.

1 1 10 100

1000

1000

0

1000

00

1000

000

1 min 1 Woche 6 Monate 10 Jahre 50 Jahre

sehr kurz kurz mittel lang stndig

Nutzungsklasse 1/2

Nutzungsklasse 3

Madison-Kurve

g mod ( ; )

Service class 1 and2

Service class 3

shortvery

shortmedium long permanent

short 1 week 6 months 10 years 10

years

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EC 5 Table 3.2 Deformation modification values kdef


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def

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EC 5 Table 3.2 Deformation modification values kdef


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def

(continued)

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EC 5 3.2: Solid timber


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EN 15497Page 48 of 119

EC 5 3.3: Glued laminated timber


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EN 14080Now large finger joints are directly regulated in the

harmonized product standard for glulam,

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Example of large finger joint (single joint line)


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p g g j ( g j )

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Example of large finger joint (two joint lines)


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p g g j ( j )

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Example of large finger joint (two joint lines)


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p g g j ( j )

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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


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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


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geometry of

frames

initial geometry

deviation corresponding

to symmetrical load

initial geometry

deviation correspondingto non-symmetrical load

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EC 5 - Section 6: Ultimate limit states


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Tension


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6.1.2 Tension parallel to the grain

6.1.2 Tension perpendicular to the grain

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Compression


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6.1.4 Compression parallel to the grain

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Compression 6.1.4 Compression perpendicular to the grain


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c,90,d is the design compressive stress in the effective contact areaperpendicular 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 thegrain;

kc,90 is a factor taking into account the load configuration, the

possibility of splitting and the degree of compressive

deformation.

where

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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, , ateach side is increased by 30 mm, but not more than a, or 1/2, 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,90specified may be taken, with a limiting value

of kc,90= 1,75.

3. For members on continuous supports, provided that 1 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 is the contact length.

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4. For members on discrete supports, provided that 1 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 400 mm

where h is the depth of the member and 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)


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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= kcrb

where b is the width of the relevant section of the member.

NOTE: The recommended value for kcris given as

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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)

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 partsof the beam

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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 curvedparts of the beam

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or

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Design examples


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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


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GL 24 / BS 11

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

10 m

16x80cm

Geometry:

l = 10 m

b = 160 mm

h = 800 mm

S = b h/6 = 17 10-6mm

I = b h/12 = 6.8 10-9mm4

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Design comparison EC 5 vs. perm. stress concept


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Property permissible concept semi-probabilistic concept

Bending strength m,perm= 11 N/mm fm,k= 24 N/mm

Shear strength v,perm= 1.2 N/mm fv,k= 3.5 N/mm

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

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

M= 1.25

(glulam, EC 5)

Deformation factor kdef= 0.8 (Service Class I)

Partial factor for permanent

actions

G= 1.35

Partial factor for variable

actions

G= 1.5

Factor for quasi-permanent

value of a variable action

2,1= 0.3

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Result permissible concept semi-probabilistic concept

distributed load F= g+ q = 15 kN/m Fd = Gg+ Qq = 21.6 kN/m

bending moment M M= F l / 8 = 188 kNm Md= Fdl / 8 = 270 kNm

bending stress m= M/S= 11 N/mm m= Md/S= 15.8 N/mm

utilization (bending) 11 / 11 = 1.00

fm,d=fm,k kmod/M= 15.4 N/mm

15.8 /fm,d = 1.03shear force V V= F l/2 = 75 kN Vd= Fdl/2 = 108 kN

shear stress v 1.5 V/ (b h) = 0.88 N/mm 1.5 Vd/ (b h) = 1.89 N/mm

utilization (shear) 1.2 / 0.88 = 0.73

fv,d=fv,k kmod/M= 2.24 N/mm1.89 /fv,d = 0.84

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deflection = 5

384=26

=, +,=

5

384+

5

384= 10.4 + 15.6

=26

=,(1 +) +

,(1 +2,1)=

16.7 + 18.4 = 35.1mm

utilitization (deflection)

/300= 0.78

/300= 0.78

/150= 0.53

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Design examples


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Design of curved glulam beam

- comparison of

Eurocode 5 vs. DIN 1052

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HESS Limitless Verbindung (7)


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Stress distributions in curved beams with const. moment


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W

M

R

H,

max,250=

R H

R2 / H = 11

R1 / H = 2,5

R H

H/R2= 0,09

H/R1= 0,4

W

M

R

H,

R

H,

II

++=

2

603501

+

-

H/R2= 0,09

H/R1= 0,4

tension stressesperpendicular to grain

bending stressesparallel to grain

R1< R2

R1< R2

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


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h

stress perp.

to grain

- +

h

- +

stress perp.

to grain

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


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Geometry, dimensions and quality /strength class

of example beam

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

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

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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

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Design for bending:

kr= 0,96

kl= 1,05

EC5

fm,d= fm,k kmod/m

GL28:

fm,k = 28 N/mm2

load duration:

medium, kmod= 0,8

glulam: m = 1,25

m,d= 6,85 N/mm2Map,d= f Map

combined loading: f = 1,4

fm,d = 17,92 N/mm2

ratio = 0,38

F = 23,31 kN

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


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Design for tension perp.:

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

EC5

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

kp= 0,0294

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

hap/ r = 0,118

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Example: Curved Beam with pure moment loading

kp= 0,0294

EC5

ft,90,d= ft,90,k kmod/m

glulam:

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

load duration:

medium, kmod= 0,8glulam:

m = 1,25

t,90,d= 0,19 N/mm2Map,d= f Map

combined loading: f = 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.:

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Design for bending:

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

DIN

1052

F = 23,31 kN

m m,permissible

m= kl 6 Map/b h2

m,permissible= 14 N/mm2

m= 4,66 N/mm2

ratio = 0,33

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Design fr tension perp.:

DIN

1052

F = 23,31 kN

t,90 t,90,permissible

t,90= kp 6 Map/b h2

t,90,permissible= 0,2 N/mm2

t,90= 0,14 N/mm2

ratio = 0,69

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

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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

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Now ist time to finish!


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Thank you very much for your patient listening!

Documents

Timber Design Using Eurocode