Mechanics of aircraft materials 7 laminate plates theory cdio

Transverse shear stresses xz and yz vanish on plate surfaces defined by z = t/2 and z = -t/2 Basic Assumptions... Symmetric Cross-Ply Laminate... Problem 1 Determine the stiffness matr

Classical Lamination Theory

Dr Ly Hung Anh

Department of Aerospace Engineering – Faculty of Transportation Engineering

Basic Assumptions

1 Each ply (or layer or lamina) is in

state of plane stress

2 Interlaminar stresses are ignored

3 Lamina are perfectly bonded to each

other

a. No interfacial slip

b. No interfacial shear deformation

x

Coordinate system and stress resultants

for a laminated plate

Basic Assumptions

1. Plate consists of orthotropic laminae

bonded together Principal material axes of laminae oriented along arbitrary directions

w.r.t x-y axes

2. Thickness of the plate, t, is much smaller

than lengths along edges, a and b

3. Displacement u,v, and w are small

compared to thickness, t

Basic Assumptions

4 In-plane strains, x , y , and xy are

small compared to unity

5 Transverse shear strains, xz and yz

are negligible (=0)

6 Tangential displacement, u and v are

linear functions of the z coordinate

7 Transverse normal strain, z are negligible

(=0)

Basic Assumptions

8 Each ply obeys Hooke’s law

9 The plate thickness t is constant

10 Transverse shear stresses xz and yz

vanish on plate surfaces defined by z = t/2 and z = -t/2

Basic Assumptions

Middle Surface

1 2 3

k N

z 

k

z

1 k

z 

Laminate Geometry and ply numbering system

y , x F

z y

, x v

v

y , x F

z y

, x u

u

0

2 0

1 0

in surface

middle of

nt displaceme

direction

x in

surface

middle of

nt displaceme

z w w

y , x v

y , x u

0 0







xy

0 xy xy

y

0 y y

x

0 x x

z x

v y

u

z y

v

z x

Mid-plane Strains

x

v y

u y v x u

0

0 0

xy

0 0

y

0 0

y x

w 2

y w x w

2 xy

2

2 y

2

2 x

(in global axes)

Q

 

 

(6)

The curvatures (the  ’s) are

hard to measure We seek to

express stresses in terms of

loads and moments.

t

k 1 2

z N

k 1 z

z N

from distance

layer k

of surface inner

to surface middle

from distance

laminate k

in stress

thickness laminate

t

: Where

th th

k x

z z

k 1

z N

x 11 k x x 12 k y y 16 k xy xy

k 1 z

z N

x 11 k x 12 k y 16 k xy

z N

t 2

t 2

Stress Resultants

Similar Results for

N y N xy M y and M xy

Complete Set of Equations

0 xy

0 y

0 x

66 26

16

26 22

12

16 12

11

66 26

16

26 22

12

16 12

11

66 26

16

26 22

12

16 12

11

66 26

16

26 22

12

16 12

11

κ κ κ γ ε ε

D D

D

D D

D

D D

D

B B

B

B B

B

B B

B

B B

B

B B

B

B B

B

A A

A

A A

A

A A

A

M M M N N N

xy y x xy y x

B A

M N

(10)

Extensional Stiffnesses Matrix [A]

k ij

k ij

ij

z z

Q

dz Q

Bending Stiffnesses Matrix [D]

k ij

k ij ij

z z

Q

dz z

Q D

t

t

1

3 1 3

2

3 1

2

2

“Coupling” Stiffnesses Matrix [B]

k ij

k ij ij

z z

Q

zdz Q

2

2 1

2

2

Classifying Laminates

1 Symmetric

2 Antisymmetric

3 Quasi-Isotropic

Both geometric and material properties are symmetric about the mid-plane

Types of symmetric laminates include

angle-ply and cross-ply laminates No

bending-extensional coupling, Bij=0

Symmetric Laminate

-Same thickness -Same Materials -Same Orientation

Help to simplify the computing!!!

Symmetric Cross-Ply Laminate

Problem 1 Determine the stiffness matrix of a Laminate

Determine the stiffness matrix for a 45/-45/+45] symmetric angle-ply laminate consisting of 0.25 mm thick unidirectional

[+45/-AS/3501 graphite epoxy laminate.

Determine the stiffness matrix of a Laminate

solution

-1 Find the value of the reduced stiffness matrix [Q] for

each ply using its four elastic moduli E11, E22, G12

2 Find the value of the “transformed reduced stiffness

matrix” for each ply by using the [Q] and the angle

of each ply

3 Find the coordinate of the top and bottom surface, zi,

i=1…n of each ply

4 Find the three stiffness matrices [A], [B] and [D] using

the results obtained from step 2 and 3

Q

 

 

0.0196 ν

E ν

0.3 ν

GPa 6.9

G

GPa 9.0

E

GPa 138.0

E

: Properties Material

12

2 21

   

12 66

66

21 12

2 12

2 12 22

11

12 21

12

21 12

2 2

12 22

11

11 22

21 12

1 2

12 22

11

22 11

G S

1 Q

1

E S

S S

S Q

Q

1

E S

S S

S Q

1

E S

S S

S Q

  GPa

6.9 0

0

0 9.05

2.72

0 2.72

138.8 Q

θ sin Q

Q

θ cos θ

sin Q

θ sin θ

cos Q

4 Q

Q Q

θ cos θ

sin Q

2 Q

2

θ sin Q

θ cos Q

Q

2 2

66 12

4 22

4 11

22

4 4

12

2 2

66 22

11 12

2 2

66 12

4 22

4 11

Q 2 Q

2 Q

Q Q

sinθ θ

cos Q

2 Q

Q

θ sin cosθ

Q 2 Q

Q Q

θ sin cosθ

Q 2 Q

Q

sinθ θ

cos Q

2 Q

Q Q

4 4

66

2 2

66 12

22 11

66

3 66

12 22

3 66

12 11

26

3 66

12 22

3 66

12 11

  GPa

35.6 32.44

32.44

32.44 45.22

31.42

32.44 31.42

45.22 Q

mm 0.50

z

mm 0.250

z

mm 0.0

z

mm 0.250

z

mm 0.50

z

43210

3 k k

ij

2 k

ij ij

N

1 k

2 1 k

2 k k

ij k

ij ij

N

1 k

1 k k

k ij k

ij ij

z z

Q 3

1 dz

z Q

D

z z

Q 2

1 zdz

Q B

z z

Q dz

Q A

2 t

2 t

2 t

2 t

2 t

Step 4

97 2 03 2 03 2

03 2 77 3 62 2

03 2 62 2 77 3 D

mm GPa

0 0 0

0 0 0

0 0 0 B

mm GPa

6 35 0

0

0 22

45 42

31

0 42

31 22

45 A

Antisymmetric Laminates

Plies of identical material and thickness at equal positive and negative distances

from middle surface Orientation is

antisymmetric, that is if at distance +z

the orientation is  then at distance -z the

orientation is -.

o 45



o 45



o 45



o 45



o 45



o 45



o 45



o 45



o 45



Antisymmetric Angle-Ply Laminate

o 45



Antisymmetric Angle-Ply Laminate

Determine the stiffness matrix for a

[-45/+45/-45/+45] anti-symmetric ply laminate consisting of 0.25 mm thick unidirectional AS/3501 graphite epoxy

angle-laminae.

y x

Exploded view of a [-45/+45/-45/+45] antisymmetric laminate

0196

0 E

3 0

GPa 9

6 G

GPa 0

9 E

GPa 0

138 E

12

2 21

0 77

3 62 2

0 62

2 77 3

0 055

4 055 4

055 4 0

0

055 4 0

0

6 35 0

0

0 22

45 42

31

0 42

31 22

45

mm GPa

D

mm GPa

B mm

GPa A

One example:

Angle between adjacent laminae = /N where N is the number of lamina

10 /

70

60 /

0 / 60







0 E

.

6

G

GPa 0

.

9

E

GPa 0

138

E

12 1

A

12 11

Laminate Compliances

Inverse of stiffness relationship!

B A

M

N

 

1 0

[B*]: coupling compliance matrix

[D*] : bending compliance matrix [C*] = [B*]T

(11)

Analyze a laminated composite

subjected to the applied forces and

moments

Problem 2

1 Find the three stiffness matrices [A], [B] and [D] (problem 1)

2 Knowing the applied forces {N} and moments {M}, finding the mid-plane

strains and curvatures (by using Eqs (11))

3 Knowing the location of each ply, find the global strains in each (by using

Eqs (3))

4 Find the global stresses in each ply by using the stress-strain relationship

in global axes

5 Find the local strains, the local stresses in each ply ; Use the failure criteria

to study the strength of the laminate (“first ply failure” theory)

xy

0 xy xy

y

0 y y

x

0 x x

z x

v y u

z y

v

z x

The [-45/+45/-45/+45] antisymmetric angle-ply laminate consisting of 0.25 mm thick unidirectional AS/3501 graphite epoxy laminae.

The laminate is subjected to a single axial force per unit

associated with the x and y axes in each lamina.

Step 1 – Step 4

0 77

3 62 2

0 62

2 77 3 D

mm GPa

0 055

4 055 4

055 4 0

0

055 4 0

0 B

mm GPa

6 35 0

0

0 22

45 42

31

0 42

31 22

45 A

001430

0

002193

0

y x

0 xy

0 y

0 x

Step 2

Global strains in each ply

xy

0 xy xy

y

0 y y

x

0 x x

z z z

mm 50

0 z

mm 250

0 z

mm 0

0 z

mm 250

0 z

mm 50

0 z

4 3 2 1 0

Location (mm) x y xy

-0.50 0.002193 -0.00143 0.000521-0.25 0.002193 -0.00143 0.000261

0.25 0.002193 -0.00143 -0.0002610.50 0.002193 -0.00143 -0.000521

Strains

Step 3

     

MPa 2

6

7 12

3 37

000521

0

001430

0

002193

0

6 35 44

32 44

32

44 32 22

45 42

31

44 32 42

31 22

45

Q

y x xy

y x

1 2 3 4

Step 4

In-plane and flexural modulus of

a symmetric laminate

In-plane Engineering Constants

Engineering Constants

11

1 E

modulus

12 xy

11

A A

Flexural Engineering Constants

Symmetric laminate, bending loads only

fx 3

11

12 E

t D



Flexural Engineering Constants

Effective Flexural longitudinal modulus

22

12 E



