Design Timber Structures Using Eurocode 5

(1)

Seminar on

Sustainable Future through

Timber Design

UITM, Dec. 16.12.2014

Simon Aicher

Design Timber Structures

using

(2)

Aicher Eurocode 5 Timber Structures UITM 2014, Malaysia

Contents of lecture

2

B

asics 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)

(3)

(4)

100 years old

glulam beams,

train repair hall,

Bellinzona, Italy

(5)

Olympic Ice rink

Hammar, Norway,

1994

glulam truss beams,

span:97m

(6)

(7)

(8)

Aicher Eurocode 5 Timber Structures UITM 2014, Malaysia 8

(9)

(10)

7-storey timber

(11)

10-storey timber

building, Melbourne,

Australia 2013

(12)

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,…..

(13)

Permissible stress concept

13

σ

_act

= σ

₉₅

acting loads, hence resulting section forces E and stresses σ

represent in general 95% quantiles of the distributions

Design verification

σ

_act

≤ σ

_permissible

where in case of structural timber (roughly)

σ

_permissible

= f

₅₀

/3

(14)

Semiprobabilistic design concept with partial factors

14

σ

_act

= σ

₉₅

as in permissible stress concept the loads / section forces/ stress

_{distributions represent 95% quantiles of the distributions}

Design verification

σ

_d

≤ f

_d

σ

_d

design stress

f

_d

design strength

σ

_d

= σ

_act

· γ

_L

γ

_L

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

f

_d

= f

_k

· k

_mod

/ γ

_M

f

_k

characteristic strength property (5% quantile)

k

_mod

modification factor (time, climate)

(15)

Semiprobabilistic vs. permissible stress design concept

15

σ

_act

≤

σ

_d

= σ

_act

· γ

_L

= σ

_act

· 1,5 ≤ f

_d

=

γ

_L

= 1,5 partial factor for load

f

₀₅

= f

₅₀

(1 - 1,64 · COV)

assuming COV = 0,12

f

₀₅

= f

₅₀

(1 – 0,2) = f

₅₀

/ 1,25

f

₀₅

· k

_mod

f

₅₀

· k

_mod

γ

_M

_{1,25 γ}

_M

with γ

_M

= 1,3 and k

_mod

= 0,8

f

₀₅

· k

_mod

f

₅₀

· 0,8

γ

_M

1,25 · 1,3

=

≈ f

50

2 f

₀₅

· k

_mod

f

₅₀

γ

_M

=

f

₅₀

2 · 1,5 =

_{3 =}

f

50

σ

permissible

2

(16)

Graphical illustration of semiprobabilistic design concept

Probability

density

m

_s

f

_s

m

_{s 95}

m

_{s 95}

· γ

_s

k

_mod

· m

_{R 05}

/ γ

_R

k

_mod

· m

_{R 05}

k

_mod

· m

_R

f

_R

R, s

=

β · σ

_z

= m

_z

= k

_mod

· m

_R

- m

_s

f

_z

p

_f

= 10

-6

(17)

(18)

(19)

(20)

(21)

(22)

(23)

(24)

(25)

(26)

(27)

(28)

(29)

(30)

(31)

(32)

(33)

(34)

(35)

(36)

(37)

2.4 Verification by the partial factor method

(38)

Recommended partial factors

γ

_M

for material properties

EC 5 – Table 2.3

(39)

(40)

2.4.2 Design value of a resistance

(41)

(42)

(43)

(44)

EC 5 – Table 3.1 Strength modification values k

_mod

(45)

Accumulated duration of load [hours]

Str

eng

th

mo

dif

ica

tio

n

fa

ct

or

k

_mod 0 0.2 0.4 0.6 0.8 1 1.2

0.

001

0.

01

0.

1

1 ₁₀

100 ₁₀

00

10

000 ₁₀

0000

10 0000

0

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 k

_mod

= f( time; moisture)

Service class 1 and2

Service class 3

short

very

short

medium

long

permanent

short

_{1 week}

6 months

10 years

10

(46)

(47)

EC 5 – Table 3.2 Deformation modification values k

_def

(48)

EC 5 – 3.2: Solid timber

(49)

(50)

EC 5 – 3.3: Glued laminated timber

EN 14080

Now large finger joints are directly regulated in the

harmonized product standard for glulam,

(51)

(52)

(53)

(54)

(55)

(56)

(57)

(58)

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.

(59)

(60)

(61)

(62)

(63)

(64)

(65)

(66)

(67)

(68)

(69)

(70)

Examples of assumed initial geometry deviations

geometry of

frames

initial geometry

deviation corresponding

to symmetrical load

initial geometry

deviation corresponding

to non-symmetrical load

(71)

(72)

Tension

6.1.2 Tension parallel to the grain

(73)

Compression

(74)

Compression

6.1.4 Compression perpendicular to the grain

σ

_c,90,d

is the design compressive stress in the effective contact area

perpendicular to the grain;

F

_c,90,d

is the design compressive load perpendicular to the grain;

A

_ef

is the effective contact area in compression perpendicular to

the grain;

F

_c,90,d

is the design compressive strength perpendicular to the

grain;

k

_c,90

is a factor taking into account the load configuration, the

possibility of splitting and the degree of compressive

deformation.

(75)

Compression

6.1.4 Compression perpendicular to the grain

The effective contact area perpendicular to the grain, A

_ef

, 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 ℓ

₁

/2, see Figure 6.2.

2. The value of k

_c,90

should be taken as 1,0 unless the conditions in the following paragraphs

apply. In these cases the higher value of k

_c,90

specified may be taken, with a limiting value

of k

_c,90

= 1,75.

3. For members on continuous supports, provided that ℓ

₁

≥ 2h, see Figure 6.2a, the value

of k

_c,90

should be taken as:

– k

_c,90

= 1,25 for solid softwood timber

– k

_c,90

= 1,5 for glued laminated softwood timber

(76)

Compression

6.1.4 Compression perpendicular to the grain

4. For members on discrete supports, provided that ℓ

₁

≥ 2h, see Figure 6.2b, the value of k

_c,90

should be taken as:

– k

_c,90

= 1,5 for solid softwood timber

– k

_c,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.

(77)

(78)

(79)

(80)

6.1.7 Shear (crack factor issue)

k

_cr

= 0,67

for solid timber

k

_cr

= 0,67

for glued laminated timber

k

_cr

= 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:

b

_ef

= k

_cr

b

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

NOTE: The recommended value for k

_cr

is given as

(81)

(82)

(83)

(84)

(85)

(86)

(87)

6.4 Members with varying cross-section or curved shape

(88)

(89)

(90)

(91)

6.4 Members with varying cross-section or curved shape

(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

(92)

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

(93)

(94)

(95)

(96)

(97)

6.4 Members with varying cross-section or curved shape

(98)

(99)

Design of straight glulam member

- comparison of

Eurocode 5 vs. DIN 1052

(100)

GL 24 / BS 11

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

10 m

16x80 c

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

_mm

4

(101)

Property

permissible concept

semi-probabilistic concept

Bending strength

σ

_m,perm

= 11 N/mm²

f

_m,k

= 24 N/mm²

Shear strength

τ

_v,perm

= 1.2 N/mm²

f

_v,k

= 3.5 N/mm²

MOE

E

_m

= 11000 N/mm²

E

_m,mean

= 11000 N/mm²

crack factor

-

k

_cr

= 0.67

modification factor for duration

of load and moisture content

k

medium-term)

mod

= 0.6 (Service Class I/II,

Partial factor for material

properties

γ

(glulam, EC 5)

M

= 1.25

Deformation factor

k

_def

= 0.8 (Service Class I)

Partial factor for permanent

actions

γ

G

= 1.35

Partial factor for variable

actions

γ

G

= 1.5

Factor for quasi-permanent

value of a variable action

ψ

2,1

= 0.3

(102)

Result

permissible concept

semi-probabilistic concept

distributed load

F = g + q = 15 kN/m

F

_d

=

γ

_G

g +

γ

_Q

q = 21.6 kN/m

bending moment M

M = F l² / 8 = 188 kNm

M

_d

= F

_d

⋅ l² / 8 = 270 kNm

bending stress

σ

_m

= M/S = 11 N/mm²

σ

_m

= M

_d

/S = 15.8 N/mm²

utilization (bending)

11 / 11 = 1.00

f

15.8 / f

m,d

= f

m,k _m,d

⋅

= 1.03

k

mod

/

γ

M

= 15.4 N/mm²

shear force V

V = F l/2 = 75 kN

V

_d

= F

_d

l/2 = 108 kN

shear stress

τ

_v

1.5 V / (b h) = 0.88 N/mm²

1.5 V

_d

/ (b h) = 1.89 N/mm²

utilization (shear)

1.2 / 0.88 = 0.73

f

1.89 / f

v,d

= f

v,k _v,d

⋅

k

= 0.84

mod

/

γ

M

= 2.24 N/mm²

(103)

deflection

𝑢 =

_{384𝐸𝐸 = 26𝑚𝑚}

5 𝐹 𝑙

4

_5𝑔𝑙

𝑢

𝑖𝑖𝑖𝑖₄

= 𝑢

𝑖𝑖𝑖𝑖,𝑔

+ 𝑢

𝑖𝑖𝑖𝑖,𝑞=

384𝐸𝐸 +

5𝑞𝑙

4

384𝐸𝐸 = 10.4 + 15.6

= 26𝑚𝑚

𝑢

_𝑓𝑖𝑖

= 𝑢

_{𝑖𝑖𝑖𝑖,𝑔}

(1 + 𝑘

_𝑑𝑑𝑓

) +

𝑢

_{𝑖𝑖𝑖𝑖,𝑞}

(1 +

ψ

2,1

𝑘

𝑑𝑑𝑓

)=

16.7 + 18.4 = 35.1mm

utilitization (deflection)

𝑢

𝑙/300 = 0.78

𝑢

_{𝑖𝑖𝑖𝑖}

𝑙/300 = 0.78

𝑢

_𝑓𝑖𝑖

𝑙/150 = 0.53

(104)

Design of curved glulam beam

- comparison of

Eurocode 5 vs. DIN 1052

(105)

(106)

(107)

W

M

R

H

,

max ,

=

0

25 σ

_⊥ 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

























+

=

σ

2

6

0

35

0

1

+ - 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

(108)

Aicher Eurocode 5 Timber Structures UITM 2014, Malaysia 108 h stress perp. to grain

-

+ h

-

+

Stress

σ

_t,90

of curved and tapered beams with line loads

stress perp. to grain

(109)

(110)

(111)

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

Geometry, dimensions and quality /strength class

of example beam

EN 14080 GL 28: f

_m,k

= 28 N/mm

2

(112)

Design for bending:

k_r = 0,96

h

_ap

/ r = 0,118

EC5

F = 23,31 kN

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

r

_in

/t = 200,

k

_l

= 1,05

(113)

Design for bending:

k_r = 0,96 k_l = 1,05

EC5

fm,d = fm,k × kmod /γm GL28: f_m,k = 28 N/mm2 load duration:

„medium“, k_mod = 0,8 glulam: γm = 1,25

σ_m,d = 6,85 N/mm2 M_ap,d = γ_f× M_ap combined loading: γ_f= 1,4 f_m,d= 17,92 N/mm2

ratio = 0,38

F = 23,31 kN

(114)

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

k_p = 0,0294

k

₅

= 0; k

₆

= 0,25, k

₇

= 0

h

_ap

/ r = 0,118

(115)

Example: Curved Beam with pure moment loading

k_p = 0,0294

EC5

f_t,90,d = f_t,90,k × k_mod /γ_m glulam: f_t,90,k = 0,5 N/mm2 load duration:

„medium“, k_mod = 0,8 glulam: γm = 1,25

σ_t,90,d_{= 0,19 N/mm}2 M_ap,d = γ_f× M_ap combined loading: γ_f= 1,4 f_t,90,d= 0,32 N/mm2