## Lecture 6: Laminate theory for papermakers

Lecture 6: Laminate theory for papermakers Sören Östlund After Lecture 6 you should be able to ... • describe the basic assumptions in laminate theor...
Author: Alexis Gordon
Lecture 6: Laminate theory for papermakers Sören Östlund

After Lecture 6 you should be able to ... • describe the basic assumptions in laminate theory • relate the features of laminate theory to basic concepts in solid mechanics • use laminate theory to predict stresses and strains in paper sheets subjected to mechanical loading and moisture

1

Literature • Pulp and Paper Chemistry and Technology - Volume 4, Paper Products Physics and Technology, Chapter 11

With the help of laminate theory it is possible to … • Make sensitivity investigations of the causes of curl and twist • Calculate bending stiffness. • Calculate strains and stresses in paper sheets

2

Examples of packaging problems where moisture has an influence: • Multi-colour offset printing • The Th twist t i t and d curll off paperboard b d and d corrugated t d fibreboard • Moisture influences tensile and bending stiffness

Definition of a laminate • A laminate consists of a number of uniformly thick layers which do not move relative to each other.

1 ϕ

x

• The laminate is ordered in a coordinate system 1-2, with the 3-axis, the Zdirection, directed downwards.

y

2

• Each layer is assumed to be orthotropic. • Each layer is ordered in a coordinate system x-y (on-axis), which lies parallel with the 1-2 system (off-axis). • The positive angle is measured clockwise from the 1-system to the xsystem.

1 z0

2

z1

z2 zk

zN

t/2 /2

3 k

t t/2

n Z, 3

3

The thickness coordinates n

t = ∑ tk

1

k =1

z0

t z0 = − 2

2

z1

z2 zk

zN

t/2

3

t t/2

k n

z k = z k −1 + t k

Z 3 Z,

tk = the thickness of layer k

Definition of positive stress and strain 1 N1

2

Stress N1 and strain ε1 in the 1-direction 0

N2 1

Stress N2 and strain ε 2 in the 2-direction 0

2 1

N6

Shear stress N6 and shear strain

ε 60

2

4

Curl of paper can be split into three curvature components 1 2

M1

1

o

w

∂2w ∂x12

κ1 =

1 R1

⎡1⎤ ⎢m⎥ ⎣ ⎦

κ2 = −

∂2w ∂x 22

κ2 =

1 R2

⎡1⎤ ⎢m⎥ ⎣ ⎦

3 M2

o

3

κ1 = −

2

M6

κ 6 = −2

∂2w ∂x1 x 2

(h )

κ 6 = −2 b l

⎡1⎤ ⎢m⎥ ⎣ ⎦

Twist

The twist, κ6, can be illustrated as the change in the inclination of the surface from point A to point B 1 l B h

h

κ 6 = −2

A

∂2w ∂x1 x 2

(h )

κ 6 = −2 b l

⎡1⎤ ⎢m⎥ ⎣ ⎦

2 b

5

Hooke’s law for plane stress ⎡ 1 ⎢ E ⎧εx ⎫ ⎢ x ⎢ ν yx ⎪ ⎪ ⎨ ε y ⎬ = ⎢− Ey ⎢ ⎪γ ⎪ ⎩ xy ⎭ ⎢ ⎢ 0 ⎢⎣

ν xy Ex

1 Ey 0

⎤ 0 ⎥ ⎥ ⎧σ ⎫ ⎥⎪ x ⎪ 0 ⎥ ⎨σ y ⎬ ⎥⎪ ⎪ ⎥ ⎩σ xy ⎭ 1 ⎥ Gxy ⎥⎦

because

or

⎡ 1 ⎢ Ex ⎧εx ⎫ ⎢ ⎢ ν xy ⎪ ⎪ ⎨ ε y ⎬ = ⎢− ⎪ ⎪ ⎢ Ex ⎩γ xy ⎭ ⎢ ⎢ 0 ⎢⎣

ν yx Ey

1 Ey 0

⎤ 0 ⎥ ⎥ ⎧σ ⎫ ⎥⎪ x ⎪ 0 ⎥ ⎨σ y ⎬ ⎥⎪ ⎪ ⎥ ⎩σ xy ⎭ 1 ⎥ Gxy ⎥⎦

E x ν xy = E y ν yx

Stiffness relations ⎡σ x ⎤ ⎡Qxx ⎢ ⎥ ⎢ ⎢σ y ⎥ = ⎢Qyx ⎢τ xy ⎥ ⎢⎣ 0 ⎣ ⎦

Qxx = where

Qyx =

Ex

1 −ν xyν yx

ν yx E x 1 −ν xyν yx

Qxy Qyy 0

0 ⎤ ⎡εx ⎤ ⎢ ⎥ 0 ⎥⎥ ⎢ ε y ⎥ Qss ⎥⎦ ⎢⎣γ xy ⎥⎦

Qxy =

ν xy E y 1 −ν xyν yx

Qyy =

Ey 1 −ν xyν yx Qss = Gxy

6

Comparison between isotropic and orthotropic elastic materials G=

Isotropic

E 2(1+ ν )

Orthotropic paper materials (Baum)

Gxy =

Ex E y 2 ⎡⎣1 + ν xyν yx ⎤⎦

ν xyν yx = 0, 0 293 Gxy = 0.387 E x E y

E x ν xy = E y ν yx

ν xy = 0.293

Ex Ey

Strain in the plane due to moisture absorption ε H = β H ΔH

βH = hygroexpansion coefficient ΔH = the change in the moisture content of the paper

ε H = β RH ΔRH

RH = relative humidity

ε H = β mc Δmc

mc = moisture content

Total strain in a layer due to both moisture sorption and mechanical forces ε =εM +εH

7

Strain due to bending

M

R

ε B = zκ =

M

z R

Compression

Tension

Total strain

ε = ε 0 + εB

Neutral

where the strain of the mean surface is

ε0

Total strain in a given layer due to moisture changes, mechanical influences and bending strain due to change in moisture

ε M = ε 0 + zκ − β H ΔH membrane strain

bending strain

8

Forces and bending moments Not σ because 2D-structure

(

)

N = ∫ Q ε 0 + zκ − β H ΔH dz

(

)

M = ∫ Qz ε + zk − β ΔH dz H

0

n 1 ⎛ ⎞ N = ∑ ⎜ Qε 0 ( zi − zi −1 ) + Qk zi2 − zi2−1 − Q β H ΔH ( zi − zi −1 ) ⎟ 2 ⎠ i =1 ⎝ n 1 1 ⎛1 0 2 ⎞ 2 3 3 H 2 M = ∑ ⎜ Qε zi − zi −1 + Qκ zi − zi −1 − Q β ΔH zi − zi2−1 ⎟ 3 2 ⎠ i =1 ⎝ 2

(

(

)

)

(

)

(

)

Compact notation ⎡ N ⎤ ⎡ A B ⎤ ⎡ε 0 ⎤ ⎡ N H ⎤ ⎢ ⎥=⎢ ⎥⎢ ⎥−⎢ H ⎥ ⎣ M ⎦ ⎣ B D ⎦ ⎢⎣ κ ⎥⎦ ⎢⎣ M ⎥⎦

or

⎡ N ⎤ ⎡ N H ⎤ ⎡ A B ⎤ ⎡ε 0 ⎤ ⎢ ⎥+⎢ H⎥ =⎢ ⎥⎢ ⎥ ⎣ M ⎦ ⎢⎣ M ⎥⎦ ⎣ B D ⎦ ⎢⎣ κ ⎥⎦

where n

A = ∑ Qi [ zi − zi −1 ] i =1

[

B=

1 n ∑ Qi z i2 − z i2−1 2 i =1

D=

1 n ∑ Qi z i3 − i3−1 3 i =1

[

]

]

Note! Coupling between tension and bending!

9

Internal forces and moments caused by hygroexpansion n

N H = ∑ ΔH i β iH Qi [ zi − zi −1 ] i=1

MH =

1 n ΔH i β iH Qi ⎡⎣ zi2 − zi2−1 ⎤⎦ ∑ 2 i =1

External forces, bending moments strains and curvatures have three components p each: ⎡ N1 ⎤ ⎢ ⎥ ⎢ N2 ⎥ ⎢ N6 ⎥ ⎣ ⎦

⎡ N1H ⎤ ⎢ H⎥ ⎢ N2 ⎥ ⎢ N 6H ⎥ ⎣ ⎦

⎡M1 ⎤ ⎢ ⎥ ⎢M 2 ⎥ ⎢M 6 ⎥ ⎣ ⎦

⎡ M 1H ⎤ ⎢ H⎥ ⎢M 2 ⎥ ⎢ M 6H ⎥ ⎣ ⎦

⎡ε1 ⎤ ⎢ ⎥ ⎢ε 2 ⎥ ⎢ε 6 ⎥ ⎣ ⎦

⎡κ1 ⎤ ⎢ ⎥ ⎢κ 2 ⎥ ⎢κ 6 ⎥ ⎣ ⎦

Transformation of stresses and strains ⎡ ⎤ ⎢ ε1 ⎥ ⎢ ⎥ −1 ⎢ ε 2 ⎥ = [T ] ⎢1 ⎥ ⎢ γ 12 ⎥ ⎣2 ⎦

⎡ ⎤ ⎢ εx ⎥ ⎢ ⎥ ⎢ εy ⎥ ⎢1 ⎥ ⎢ γ xy ⎥ ⎣2 ⎦

where

[T ]

−1

⎡c 2 ⎢ = ⎢s 2 ⎢ cs ⎣

− 2cs ⎤ s2 ⎥ c2 2cs ⎥ 2 2⎥ − cs c − s ⎦

⎧c = cos ϕ ⎨ i ϕ ⎩ s = sin

⎡σ x ⎤ ⎡σ 1 ⎤ ⎢σ ⎥ = [T ]−1 ⎢σ ⎥ ⎢ y⎥ ⎢ 2⎥ ⎢τ xy ⎥ ⎢⎣ τ 6 ⎥⎦ ⎣ ⎦

⎡ ⎤ ⎢ β1H ⎥ ⎢ H ⎥ −1 ⎢ β 2 ⎥ = [T ] ⎢1 H ⎥ ⎢ β12 ⎥ ⎣2 ⎦

⎡ ⎤ ⎢ β xH ⎥ ⎢ H ⎥ ⎢ βy ⎥ ⎢1 H ⎥ ⎢ β12 ⎥ ⎣2 ⎦

Vanishes in general!

10

Transformation of the stiffness matrix, Q ⎡σ 1 ⎤ ⎡Q11 ⎢σ ⎥ = ⎢Q ⎢ 2 ⎥ ⎢ 12 ⎢⎣τ 12 ⎦⎥ ⎢⎣Q16 ⎡ Q11 ⎤ ⎡ c 4 ⎢Q ⎥ ⎢ 4 ⎢ 22 ⎥ ⎢ s ⎢Q12 ⎥ ⎢c 2 s 2 ⎢ ⎥=⎢ 2 2 ⎢Q66 ⎥ ⎢c s ⎢Q16 ⎥ ⎢ c 3 s ⎢ ⎥ ⎢ 3 ⎣⎢Q26 ⎥⎦ ⎢⎣ cs

Q12 Q22 Q26

Q16 ⎤ ⎡ ε 1 ⎤ Q26 ⎥⎥ ⎢⎢ ε 2 ⎥⎥ Q66 ⎦⎥ ⎣⎢γ 12 ⎥⎦

⎤ ⎥ 2c 2 s 2 4c 2 s 2 ⎥ ⎡Q xx ⎤ c4 ⎢ ⎥ − 4c 2 s 2 ⎥ ⎢Q yy ⎥ c2s2 c4 + s4 ⎥ c 2 s 2 − 2c 2 s 2 c 2 − s 2 ⎥ ⎢Q xy ⎥ ⎢ ⎥ − cs 3 cs 3 − c 3 s 2 cs 3 − c 3 s ⎥ ⎣ Qss ⎦ ⎥ − c 3 s c 3 s − cs 3 2 c 3 s − cs 3 ⎥⎦ s4

2c 2 s 2

4c 2 s 2

( ( (

)

) )

Forces and moments Expanded equations H ⎡ N 1 ⎤ ⎡ N 1 ⎤ ⎡ A11 ⎢N ⎥ + ⎢N H ⎥ = ⎢ A ⎢ 2 ⎥ ⎢ 2 ⎥ ⎢ 12 H ⎣⎢ N 6 ⎦⎥ ⎢⎣ N 6 ⎥⎦ ⎣⎢ A16 H ⎡ M 1 ⎤ ⎡ M 1 ⎤ ⎡ B11 ⎢M ⎥ + ⎢M H ⎥ = ⎢ B ⎢ 2 ⎥ ⎢ 2 ⎥ ⎢ 12 ⎢⎣ M 6 ⎥⎦ ⎢⎣ M 6H ⎥⎦ ⎢⎣ B16

A12 A22 A26 B12 B22 B26

A16 ⎤ ⎡ε 10 ⎤ ⎡ B11 ⎢ ⎥ A26 ⎥⎥ ⎢ε 20 ⎥ + ⎢⎢ B12 A66 ⎦⎥ ⎢⎣ε 60 ⎦⎥ ⎣⎢ B16 B16 ⎤ ⎡ε 10 ⎤ ⎡ D11 ⎢ ⎥ B26 ⎥⎥ ⎢ε 20 ⎥ + ⎢⎢ D12 B66 ⎥⎦ ⎢⎣ε 60 ⎥⎦ ⎢⎣ D16

B12 B22 B26 D12 D22 D26

B16 ⎤ ⎡κ 1 ⎤ B26 ⎥⎥ ⎢⎢κ 2 ⎥⎥ B66 ⎦⎥ ⎢⎣κ 6 ⎦⎥ D16 ⎤ ⎡κ 1 ⎤ D26 ⎥⎥ ⎢⎢κ 2 ⎥⎥ D66 ⎥⎦ ⎢⎣κ 6 ⎥⎦

⎡ N ⎤ ⎡ N H ⎤ ⎡ A B ⎤ ⎡ε 0 ⎤ ⎢M ⎥ + ⎢ H ⎥ = ⎢B D⎥ ⎢ ⎥ ⎣ ⎦ ⎢⎣ M ⎥⎦ ⎣ ⎦ ⎢⎣ κ ⎥⎦

11

Bending stiffness and tensile stiffness NH = M H = 0 Strains and curvatures give external forces and external moments

⎡ N ⎤ ⎡ A B ⎤ ⎡ε °⎤ ⎢M ⎥ = ⎢ B D⎥ ⎢ κ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦ Forces and curvatures give external strains and external moments

⎡ε 0 ⎤ ⎡ A∗ ⎢ ⎥=⎢ ∗ ⎢⎣ M ⎥⎦ ⎢⎣C

B∗ ⎤ ⎡ N ⎤ ⎥⎢ ⎥ D∗ ⎥⎦ ⎣ κ ⎦

Forces and moments give external strains and external curvatures

⎡ε 0 ⎤ ⎡ A′ B′ ⎤ ⎡ N ⎤ ⎢ ⎥=⎢ ′ ⎥⎢ ⎥ ⎢⎣ κ ⎥⎦ ⎣C D′⎦ ⎣ M ⎦

Expanded equations ⎡ε 10 ⎤ ⎡ A11′ ⎢ 0⎥ ⎢ ⎢ε 2 ⎥ = ⎢ A12′ ⎢ 0⎥ ⎣ε 6 ⎦ ⎢⎣ A16′

⎡κ 1 ⎤ ⎡C11′ ⎢κ ⎥ = ⎢C ′ ⎢ 2 ⎥ ⎢ 12 ⎢⎣κ 6 ⎦⎥ ⎣⎢C16′

A12′ ′ A22 ′ A26

C12′ ′ C 22 ′ C 26

A16′ ⎤ ⎡ N 1 ⎤ ⎡ B11′ ′ ⎥ ⎢ N 2 ⎥ + ⎢ B12′ A26 ⎥⎢ ⎥ ⎢ ′ ⎦⎥ ⎣⎢ N 6 ⎥⎦ ⎢⎣ B16′ A66

B12′ ′ B22 ′ B26

C16′ ⎤ ⎡ N 1 ⎤ ⎡ D11′ ′ ⎥ ⎢ N 2 ⎥ + ⎢ D12′ C 26 ⎥⎢ ⎥ ⎢ ′ ⎦⎥ ⎣⎢ N 6 ⎦⎥ ⎣⎢ D16′ C 66

D12′ ′ D22 ′ D26

B16′ ⎤ ⎡ M 1 ⎤ ′ ⎥ ⎢M 2 ⎥ B26 ⎥⎢ ⎥ ′ ⎦⎥ ⎣⎢ M 6 ⎥⎦ B66

D16′ ⎤ ⎡ M 1 ⎤ ′ ⎥ ⎢M 2 ⎥ D26 ⎥⎢ ⎥ ′ ⎦⎥ ⎢⎣ M 6 ⎥⎦ D66

12

Tensile stiffness

Eb =

x3

N

x2

N ε°

x1

N

δ = ε 0L

N2 = N6 = 0

⎫ 1 N ⎬ ⇒ E1b = o1 = M1 = M 2 = M 6 = 0⎭ ε1 A11′ N 1 E 2b = o2 = ′ ε2 A22 1 N E 6b = o61 = ′ ε16 A66

shear stiffness

Bending stiffness is important for the stacking strength of boxes ⎛1⎞ M = EI ⎜ ⎟ = EI κ ⎝R⎠

S ≡ EI =

E 3 bt 12

The bending stiffness of the beam is the slope of the moment vs. curvature relation. Bending stiffness per unit width

Sb ≡

S E 3 = t b 12

13

Bending stiffness of a single-ply sheet Sb ≡

S E 3 Eb 2 = t = t b 12 12

where

Eb E= t

introduce

ρ=

w t

E b w2 S = 12 ρ 2

yields

b

Bending stiffness of a single-ply sheet Tensile stiffness index is related to density according to (experimental)

E = kρ w

a

Ew =

Eb w

The bending stiffness per unit width then becomes

S b = kw3 ρ a − 2 /12 D fi b Define bending di stiffness tiff iindex d (i (independent d d t off grammage)) as

Sw =

S b 1 ⎛ Ew ⎞ = ⎜ ⎟ w3 12 ⎝ ρ 2 ⎠

NOTE! Bending stiffness index depends on density

14

Bending stiffness of three-ply sheet

Introduce

E surface = n surface Eref E core = n core Eref

Bending stiffness of three-ply sheet The bending stiffness is then given by

S = Eref I ref = Eref ∑ ni I i i

Calculation of the surface moment of inertia yields

I ref = ∑ n I i = n i

i

surface

2 3 ⎛ btS3 ⎛ tC tS ⎞ ⎞ core btC + btC ⎜ + ⎟ ⎟ 2 + n ⎜⎜ 12 12 ⎝ 2 2 ⎠ ⎟⎠ ⎝

c.f. Steiner’s theorem from the basic course

15

Bending stiffness of three-ply sheet ⎧n surface = 2 ⎪ core ⎪n = 1 ⎨ ⎪tS = t / 3 ⎪t = t / 3 ⎩C

Assume

I ref bt

3

=

13 1 53 + = 81 324 324

surface plies

middle ply

Bending stiffness N1 = N 2 = N 6 = 0 ⎫ ⎬ ⇒ κ1 = D11′ ⋅ M 1 M2 = M6 = 0 ⎭ κ2 ≠ 0 Note!

κ6 ≠ 0

Bending or torsional stiffness in all directions:

M k=1/R

M k

Sb =

M

κ

⎧ b M1 1 = ⎪ S1 = κ1 D11′ ⎪ ⎪ b M2 1 = ⎨ S2 = ′ κ 2 D22 ⎪ ⎪ b M6 1 = ⎪ S6 = κ 6 D66′ ⎩

16

⎡ N ⎤ ⎡ N H ⎤ ⎡ A B ⎤ ⎡ε 0 ⎤ ⎢M ⎥ + ⎢ H ⎥ = ⎢B D⎥ ⎢ ⎥ ⎣ ⎦ ⎢⎣ M ⎥⎦ ⎣ ⎦ ⎢⎣ κ ⎥⎦

⎧⎪ N = Aε 0 + Bκ NH = M H = 0 ⇒ ⎨ 0 ⎪⎩M = Bε + Dκ ⎡ B2 ⎤ M = ⎢D − ⎥ κ A⎦ ⎣

N =0 ⇒

B2 S = D− A b

Example Calculation of bending stiffness for 4-ply sheet Ply k

w g/m2

Ew =

E

ρ MNm/kg

ρ

E = Ew ⋅ ρ

w

t =

ρ

kg/m3

N/mm2 (MPa)

mm

1

80

7.5

800

6000

0.1

2

80

2.5

400

1000

0.2

3

50

4.0

500

2000

0.1

4

70

7.14

700

5000

0.1

t = ∑ tk = 0.1 + 0.2 + 0.1 + 0.1 = 0.5⎫ ⎪ 0.5 ⎪ z0 = − = − 0.25 0 25 ⎪ 2 ⎪ z1 = z0 + t1 = −0.25 + 0.1 = −0.15 ⎬ ⎪ z2 = 0.05 ⎪ z3 = 0.15 ⎪ ⎪ z4 = 0.25 ⎭

[mm]

17

zk mm

Layer k

z k2 mm2

n

A = ∑ Qi [ zi − zi −1 ]

z k3 mm3

i =1

0

-0,25

0,0625

-15,625 · 10-3

1

-0,15

0,0225

-3,375 · 10-3

2

0,05

0,0025

0,125 · 10-3

3

0,15

0,0225

3,374 · 10-3

4

0,25

0,0625

15,625 · 10-3

Layer k

Qk (zk − zk−1 ) ≈ ≈ E k (zk − zk−1 )

1

6000 [-0,15-(-0,25)]= = 600

2 Qk (z k2 − z k−1 )≈

3 Qk (z k3 − z k−1 )≈

2 ≈ E k (z k2 − z k−1 )

3 ≈ E k (z k3 − z k−1 )

6000 (0,0225-0,0625)= = -240

6000 [-3.375-(15,625)]·10-3= = 73,5

2

1000 [0,05-(-0,15)]= = 200

1000 (0,0025-0,0225)= = -20

1000 [0.125-(-3,375)]·10-3= = 3,5

3

2000 [0,15-0,05]= = 200

2000 (0,0225-0,0025)= = 40

2000 [3,375-0.125]·10-3= = 6,5

4

5000 [0,25-0,15]= = 500

5000 (0,0625-0,0225)= = 200

5000 [15,625-3.375]·10-3= = 61,25

A = ∑ = 1500

B=

1 ∑ = − 10 2

D=

[

B=

1 n ∑ Qi z i2 − z i2−1 2 i =1

D=

1 n ∑ Qi z i3 − 3i −1 3 i =1

Sb = D −

[

]

B2 A

]

(−10) 2 S = 48.25 − = 1500 48.18 mNm b

1 ∑ = 48, 25 3

Curl, twist, strain and shear, as a result of a change in moisture N1 = N 2 = N 6 = 0 ⎫ ⎡ε 10 ⎤ ⎡ A11′ ⎬ ⇒ ⎢ 0⎥ ⎢ M1 = M 2 = M 6 = 0⎭ ⎢ε 2 ⎥ = ⎢ A12′ ⎢ 0⎥ ⎣ε 6 ⎦ ⎣⎢ A16′

A12′ ′ A22 ′ A26

A16′ ⎤ ⎡ N 1H ⎤ ⎡ B11′ ⎢ ⎥ ′ ⎥ ⎢ N 2H ⎥ + ⎢ B12′ A26 ⎥ ⎢ ′ ⎦⎥ ⎢⎣ N 6H ⎥⎦ ⎣⎢ B16′ A66

B12′ ′ B22 ′ B26

⎡κ 1 ⎤ ⎡C11′ ⎢κ ⎥ = ⎢C ′ ⎢ 2 ⎥ ⎢ 12 ⎢⎣κ 6 ⎥⎦ ⎢⎣C16′

C12′ ′ C 22 ′ C 26

C16′ ⎤ ⎡ N 1H ⎤ ⎡ D11′ ⎢ ⎥ ′ ⎥ ⎢ N 2H ⎥ + ⎢ D12′ C 26 ⎥ ⎢ H ′ ⎥⎦ ⎢⎣ N 6 ⎥⎦ ⎢⎣ D16′ C 66

D12′ ′ D22 ′ D26

⎫ ⎪ ⎪ i =1 ⎬⇒ n 1 H 2 2 ⎪ ⎡ ⎤ = ∑ ΔH i β i Qi ⎣ zi − zi −1 ⎦ ⎪⎭ 2 i =1 n

N H = ∑ ΔH i β iH Qi [ zi − zi −1 ] MH

B16′ ⎤ ⎡ M 1H ⎤ ⎢ ⎥ ′ ⎥ ⎢ M 2H ⎥ B26 ⎥ ′ ⎦⎥ ⎢⎣ M 6H ⎥⎦ B66 D16′ ⎤ ⎡ M 1H ⎤ ⎢ ⎥ ′ ⎥ ⎢ M 2H ⎥ D26 ⎥ H ′ ⎥⎦ ⎢⎣ M 6 ⎥⎦ D66

κ1 = [C11′ ] ⎡⎣ N1H ⎤⎦ + [C12′ ] ⎡⎣ N 2H ⎤⎦ + [C16′ ] ⎡⎣ N6H ⎤⎦ +

[ D11′ ] ⎡⎣ M1H ⎤⎦ + [ D12′ ] ⎡⎣ M 2H ⎤⎦ + [ D16′ ] ⎡⎣ M 6H ⎤⎦ κ 2 = ... κ 6 = ...

18

Stress and strains in the thickness direction of the laminate due to hygroexpansion

σ

z

Laminate theory program Excel Christer Fellers Innventia/KTH

19

1 z0

2

z1

z2 zk

zN

t/2

3 k

1 t

ϕ

t/2

x

N

y

Z, 3

G xy = 0 , 3 8 7

2

ExEy

20

εy

x

εx

x

εx

y

εy

ν x yν E E

ν

x y

xy

=

yx

ν ν

= 0, 293

xy

y

yx

= 0, 293

Ex Ey

Change in moisture

2004-09-02 / 42

21

Results κ1 2

M2

κ2

1

ε κ

κ6

M1

M6

Twist

1

2 Shear ε 6

Tensile stiffness index

Calculate strains and curvatures

1 2

N2

⎡ε 0 ⎤ ⎡ A′ B′ ⎤ ⎡ N ⎤ ⎢ ⎥=⎢ ′ C D′⎥⎦ ⎢⎣ M ⎥⎦ ⎣⎢ κ ⎦⎥ ⎣

κ1

M1

N1 M2

1

κ2 N6

M6

κ6

2

Shear

22

After Lecture 6 you should be able to ... • describe the basic assumptions in laminate theory • relate the features of laminate theory to basic concepts in solid mechanics • use laminate theory to predict stresses and strains in paper sheets subjected to mechanical loading and moisture

23