Mechanics of Composite Materials

Constitutive Relationships for Composite Materials

Ⅰ. Material Behavior in Principal Material Axes• Isotropic materials

– uniaxial loading

12

1

E G

12

EG

– 2-D loading

S

G

EE

EE

xy

y

x

xy

y

x

1 , 0 , 0

0 ,1

,

0 ,, 1

Where [ S ]: compliance matrix

Q

G

EE

EE

xy

y

x

xy

y

x

, 0 , 0

0 ,1 ,

1

0 ,1

, 1

22

22

Where [Q]: stiffness matrix

Isotropic Materials

Note:1. Only two independent material constants in the

constitutive equation.2. No normal stress and shear strain coupling, or no

shear stress and normal strain coupling.

Examples: polycrystalline metals,PolymersRandomly oriented fiber-reinforced

compositesParticulate-reinforced composites

Transversely isotropic materials

Principal material axesL: longitudinal directionT: transverse direction

In L–T plane

LT

T

L

LT

TT

TL

L

LT

L

LT

T

L

G

EE

EE

1 , 0 , 0

0 ,1

,

0 ,, 1

LT

T

L

LT

TLLT

T

TLLT

TLT

TLLT

LTL

TLLT

L

LT

T

L

G

EE

EE

, 0 , 0

0 ,1 ,

1

0 ,1

, 1

Transversely isotropic materials

Principal material axesL: longitudinal directionT: transverse direction

In T1, T2 plane

21

2

1

21

2

1

1 , 0 , 0

0 ,1

,

0 ,, 1

TT

T

T

TT

TT

TT

T

TT

T

TT

T

T

G

EE

EE

Same as those for isotropic materials:

TT

TTT

EG

12

Transversely isotropic materials

Where EL: elastic modulus in longitudinal direction

ET: elastic modulus in transverse direction

GLT: shear modulus in L – T plane

GTT: shear modulus in transverse plane

LT: major Poisson’s ratio

(strain in T – direction caused by stress in L – direction)

TL : minor Poisson’s ratio

And

Note: 1. 4 independent material constants (EL, ET, GLT, LT ) in L – T plane while 5 (EL, ET, GLT, LT, GTT) for 3-D state.

2. No normal stress and shear strain coupling in L – T axes or no shear stress and normal strain coupling in L – T axes

T

TL

L

LT

EE

Orthotropic materials

1.2.3: principal material axes

For example in 1-2 plane

12

2

1

12

22

21

1

12

1

12

2

1

1 , 0 , 0

0 ,1

,

0 ,, 1

G

EE

EE

12

2

1

12

2112

2

2112

212

2112

121

2112

1

12

2

1

, 0 , 0

0 ,1 ,

1

0 ,1

, 1

G

EE

EE

2

21

1

12

EE

Orthotropic Materials

Note:

1. 4 independent constants in 2-D state (e.g. 1-2 plane, E1,

E2, G12, 12 )while 9 in 3-D state (E1, E2, E3, G12, G13, G23, 12 ,

13 , 23 )

2. No coupling between normal stress and shear strain or no coupling between shear stress and normal strain

Question

Ex. Find the deformed shape of the following composite:

Possible answers?

Off-axis loading of unidirectional composite

For orthotropic material in principal material axes (1-2 axes)

12

2

1

66

2221

1211

12

2

1

0 0

0

0

Q

QQ

QQ

By coordinate transformation

12

2

11

12

2

1

22

22

22

sin-cos ,sincos- ,sinos

sin2cos cos sin

sin2cos- sin cos

T

cxy

y

x

12

2

11

T

xy

y

x

, xyxy are tensorial shear strains

Let

12

2

1

12

2

1

12

2

1

2 0 0

0 1 0

0 0 1

R

Then

xy

y

x

xy

y

x

xy

y

x

xy

y

x

RTRQTTRQT

RQTQTT

Q

111

12

2

11

12

2

11

12

2

11

Transformed stiffness matrix

Where 11 RTRQTQ = transformed stiffness matrix

sin ,cos

22

22

22

4

22

22

3662212

366121126

3662212

366121116

4466

226612221166

4412

2266221112

422

226612

41122

422

226612

41111

nm

nmQQQmnQQQQ

mnQQQnmQQQQ

nmQnmQQQQQ

nmQnmQQQQ

mQnmQQnQQ

nQnmQQmQQ

Transformed compliance matrix

xy

y

x

xy

y

x

xy

y

x

SQ

1

S : transformed compliance matrix

Off-axis loading - deformation

xy

y

x

xy

y

x

QQQ

QQQ

QQQ

662616

262212

161211

1. 4 material constants in 1-2 plane.2. There is normal stress and shear strain coupling (forθ≠0, 90˚ ), or

shear stress and normal strain coupling.

Transformation of engineering constants

For uni-axial tensile testing in x-direction 0 ,0 xyyx

∴ stresses in L – T axes

cossin

sin

cos

0

0 2

2

x

x

xx

LT

T

L

T

Strains in L – T axes

LT

LLT

T

TTL

L

x

x

x

x

LT

TT

TL

L

LT

L

LT

T

L

LT

T

L

G

EE

EE

G

EE

EE

S

SS

SS

cossin

cossin

sincos

cossin

sin

cos

1, 0 , 0

0 ,1

,

0 ,, 1

0 0

0

0

22

22

2

2

66

2212

1211

And strains in x – y axes

LT

LLT

T

TTL

L

x

LT

T

L

xy

y

x

G

EE

EE

TT

2

cossin

cossin

sincos

2

1

2

1

22

22

11

LTTL

LT

LLTTL

LT

LTTL

LT

LL

LT

L

LT

LTTL

x

xy

y

x

GEEEGEE

GEEEE

EGEE

1121cos

2

112sin

2

1

2sin1121

4

1

2sin21

4

1sincos

2

1

2

2

244

Recall for uni-axial tensile testing

2sin1121

4

1

and

2sin21

4

1sincos1

2

244

LTTL

LT

LL

LT

x

y

x

xy

xx

y

x

yxy

xxyy

L

LT

LTTLx

x

xx

GEEEEE

E

EGEEE

E

Define cross-coefficient, mx

LT

L

T

LLT

LT

L

T

LLT

x

Lxyx

L

xxxy

G

E

E

E

G

E

E

E

Em

Em

21cos2

2sin

2

Similarly, for uni-axial tensile testing in y-direction

LT

L

T

LLT

LT

L

T

LLTy

x

xy

LTTT

TL

LT

TL

y

yx

L

LT

LTTLy

G

E

E

E

G

E

E

Em

EGEEEEE

EGEEE

21sin2

2sin

2sin1121

4

1

2sin21

4

1cossin1

2

2

244

For simple shear testing in x – y plane

0 ,0 xyyx

stresses in L – T axes

22 sincos

cossin2

cossin2

0

0

xy

xy

xy

xyLT

T

L

T

Strains in L – T axes

22

22

sincos1

1cossin2

1cossin2

sincos

cossin2

cossin2

1, 0 , 0

0 , 1

,

0 ,,1

1, 0 , 0

0 , 1

,

0 , , 1

LT

L

LT

T

T

TL

L

xy

xy

xy

xy

LT

TT

TL

L

LT

L

LT

T

L

LT

TT

TL

L

LT

L

LT

T

L

G

EE

EE

G

EE

EE

G

EE

EE

Strains in x – y axes

2cos1121121

where

22

2

1

LTTL

LT

LTL

LT

Lxyxy

L

xyyy

L

xyxx

LT

T

L

xy

y

x

GEEEEEE

Em

Em

T

2cos11211211

2

LTLL

LT

LTL

LT

Lxy

xy

xyxy

GEEEEEEG

G

In summary, for a general planar loading, by principle of superposition

xy

y

x

xyL

y

L

x

L

y

yy

yx

L

x

x

xy

x

xy

y

x

GE

m

E

m

E

m

EE

E

m

EE

1 ,,

, 1

,

,, 1

Micromechanics of Unidirectional Composites

• Properties of unidirectional lamina is determined by

– volume fraction of constituent materials (fiber, matrix, void, etc.)

– form of the reinforcement (fiber, particle, …)– orientation of fibers

Volume fraction & Weight fraction

• Vi=volume, vi=volume fraction=

• Wi=weight, wi=weight fraction=

Where subscripts i = c: composite

f: fiber

m: matrix

c

i

i

i

V

V

V

V

i i

i c

W W

W W

1

1

c f m

f m

c c

f m

W W W

W W

W W

w w

Conservation of mass:

Assume composite is void-free:

1

1

c f m

f m

c c

f m

V V V

V V

V V

v v

Density of composite

m

m

f

f

c

m

m

f

f

c

c

mfc

mmffc

c

mmff

c

mf

c

cc

ww

g

W

g

W

g

W

VVV

vv

gV

gVgV

gV

WW

gV

W

1

or

Generalized equations for n – constituent composite

1

1

1n

c i i ni i

i i

vw

Void content determination

ce f f m m v v

f f m m

v v v

v v

1

t

c f m

ct v f f m m

ct f f m m c v

W W W

v v v

v v v

Experimental result (with voids):

Theoretical calculation (excluding voids):

: ct cev

ct

void content v

In general, void content < 1% Good composite

> 5% Poor composite

Burnout test of glass/epoxy composite Weight of empty crucible = 47.6504 gWeight of crucible +composite = 50.1817 gWeight of crucible +glass fibers = 49.4476 g

33 2.1 ,5.2cmg

cmg

mf

386.1 if cmgv cev Find

3

49.4476 47.65040.71

50.1817 47.6504

1 1 0.71 0.29

1 11.902

0.71 0.292.5 1.2

1.9020.71 0.54

2.5

1.9020.29 0.46

1.2

ff

c

m f

c ctf m

f m

cf f

f

cm m

m

Ww

W

w w

gw cmw

v w

v w

Sol:

1.902 1.86

1.902

0.0221 2.21%

ct cev

ct

v

Longitudinal Stiffness

For linear fiber and matrix:

Lmmffc EvEvEE

Generalized equation for composites with n constituents:

n

iiic vEE

1

Rule-of-mixture

Longitudinal Strength

fmff

mmffc

vv

vv

1

Modes of Failurematrix-controlled failure: 11 fmucu v

fiber-controlled failure:

1

[ ] 2

fu

fu fu

cu fu f m f

fu m f m

v v

v

21 ,max cucucu

Critical fiber volume fractionFor fiber-controlled failure to be valid:

For matrix is to be reinforced:

min

12

1

vv

vv

fu

fu

fufu

mmufu

mmu

f

fmumfmfu

cucu

crit

vv

v

fu

fu

fufu

mfu

mmu

f

mumfmfu

mucu

Factors influencing EL and cu

• mis-orientation of fibers

• fibers of non-uniform strength due to variations in diameter, handling and surface treatment, fiber length

• stress concentration at fiber ends (discontinuous fibers)

• interfacial conditions

• residual stresses

Transverse Stiffness, ET

Assume all constituents are in linear elastic range:

1 f m

c f m

v v

E E E

Generalized equation for n – constituent composite:

modulus) e(transvers 1

or

1

1

1

Tn

i i

i

c

n

i i

i

c

E

E

vE

E

v

E

Transverse Strength

Due to stress (strain) concentration

mucu

Factors influence cu:• properties of fiber and matrix• the interface bond strength• the presence and distribution of voids (flaws)• internal stress and strain distribution (shape of fib

er, arrangement of fibers)

In-plane Shear Modulus

For linearly elastic fiber and matrix: 1

or

f m

c f m

f mc LT

m f f m

v v

G G G

G GG G

G v G v

Major Poisson’s Ratio

LT f f m mv v

Analysis of Laminated Composites

• Classical Laminate Theory (CLT)

yxwzyxw

y

wzyxvzyxv

x

wzyxuzyxu

,,,

,,,

,,,

0

00

00

Displacement field:

Resultant Forces and Moments

2

2

d 1

xx x xh

hy y y y

xy xy xyxy

N k

N z A B k

N k

2

2

d 1

x

y

xy

x x xh

hy y y

xy xy xy

M k

M z z B D k

M k

11 11

d

kk

n

k

h

h

n

kkijkijij hhQzQA k

k

2

d22

1 1

1

1

kkk

k

hhQzzQB

n

k

h

h

n

kkijkijij

3

d

33

1 1

2 1

1

kkk

k

hhQzzQD

n

k

h

h

n

kkijkijij

Resultant forces:

Resultant moments:

[A]: extensional stiffness matrix

[B]: coupling stiffness matrix

[D]: bending stiffness matrix

Laminates of Special Configurations

• Symmetric laminates

• Unidirectional (UD) laminates– specially orthotropic – off-axis

• Cross-ply laminates

• Angle-ply laminates

• Quasi-isotropic laminates

Strength of Laminates

Maximum Stress Criterion

• Lamina fails if one of the following inequalities is satisfied:

LTLT

TcT

TtT

LcL

LtL

ˆ

ˆ

ˆ

ˆ

ˆ

Maximum Strain Criterion

• Lamina fails if one of the following inequalities is satisfied:

LTLT

TcT

TtT

LcL

LtL

ˆ

ˆ

ˆ

ˆ

ˆ

Tsai – Hill Criterion

• Lamina fails if the following inequality is satisfied:

1ˆˆˆˆ

222

LT

LT

T

T

L

TL

L

L

0 if ˆ

0 if ˆˆ

0 if ˆ

0 if ˆˆ

TTc

TTtT

LLc

LLtL

Where :

Comparison among Criteria

• Maximum stress and strain criteria can tell the mode of failure

• Tsai-Hill criterion includes the interaction among stress components

Strength of Off-Axis Lamina in Uni-axial Loading

Maximum stress criterionTsai-Hill criterion

Strength of a Laminate

• First-ply failure

• Last-ply failure