MicromechanicsMacromechanics

FibersLamina

Matrix

Laminate

Structure

Micromechanics

The analysis of relationships between effective composite properties (i.e., stiffness, strength) and the material properties, relative volume contents, and geometric arrangement of the constituent materials.

1. Mechanics of materials models – Simplifying assumptions make it unnecessary to specify details of stress and strain distribution – fiber packing geometry is arbitrary. Use average stresses and strains.

Micromechanics - Stiffness

2. Theory of elasticity models - “Actual” stress and strain distributions are used – fiber packing geometry taken into account.

a) Closed form solutions b) Numerical solutions such as finite elementc) Variational methods (bounds)

Micromechanics - Stiffness

Volume Fractions

c

ff V

Vv

c

mm V

Vv

c

vv V

Vv

fiber volume fraction

void volume fraction

matrix volume fraction

Where 1 vmf vvv vmfc VVVV composite volume

(3.2)

Weight Fractions

c

ff W

Ww

c

mm W

Ww

fiber weight fraction

matrix weight fraction

Where

mfc WWW composite weight

Note: weight of voids neglected

Densities

VW density

mfc WWW

mmffcc VVV

mmffc vv

“Rule of Mixtures” for density

(3.6)

Alternatively,

m

m

f

fc ww

1

(3.8)

Eq. (3.2) can be rearranged as

c

c

m

fc

f

f

v W

)WW(W

v

1 (3.9)

Above formula is useful for void fraction estimation from measured weights and densities.

Typical void fractions:

Autoclaved cured composite: 0.1% - 1%

Press cured w/o vacuum: 2 - 5%

f f f ff f

c c c c

W Vw v

W V

Measurements typically involve weight fractions, which are related to volume fractions by

(3.10)

m m m mm m

c c c c

W Vw v

W V

(3.12)

and

s

s

d

Fiber

ds

s

Representative area elements for idealized square and triangular fiber packing geometries.

Square array Triangular array

Fiber volume fraction – packing geometry relationships

Square array:2

4

sdv f

(3.14)

When s=d, 785.04max

ff vv (3.15)

Fiber volume fraction – packing geometry relationships

Triangular Array:2

32

sdv f

(3.16)

When s=d, 907.032max

ff vv (3.17)

• Real composites:Random fiber packing array

Unidirectional:Chopped:Filament wound: close to

theoretical

Fiber volume fraction – packing geometry relationships

8.05.0 fv4.005.0 fv

Photomicrograph of carbon/epoxy composite showing actual fiber packing geometry at 400X magnification

Voronoi cell and its approximation. (From Yang, H. and Colton, J.S. 1994. Polymer Composites, 51, 34–41. With permission.)

Random nature of fiber packing geometry in real composites can be quantified by the use of the Voronoi cell. Each point within the space of a Voronoi cell for a particular fiber is closer to the center of that fiber than it is to the center of any other fiber

Voronoi cells

Equivalent squarecells, with Voronoicell size, s

s

Typical histogram of Voronoi distances and corresponding Wiebull distribution for a thermoplastic matrix composite. (From Yang, H. and Colton, J.S. 1994. Polymer Composites, 51, 34–41. With permission.)

Elementary Mechanics of Materials Models for Effective Moduli

• Fiber packing array not specified – RVE consists of fiber and matrix blocks.

• Improved mechanics of materials models and elasticity models do take into account fiber packing arrays.

Assumptions:1. Area fractions = volume fractions2. Perfect bonding at fiber/matrix interface –

no slip3. Matrix is isotropic, fiber can be

orthotropic4. Fiber and matrix linear elastic5. Lamina is macroscopically homogeneous,

linear elastic and orthotropic

2 2

d

2 2

2

2

2

2

2 2

Stress Strain

L

Heterogeneous composite under varying stresses and strains

Equivalent homogeneous material under average stresses and strains

Concept of an Effective Modulus of an Equivalent Homogeneous Material.

Stress, Strain,

3x 3x

3x 3x

Representative volume element and simple stress states used in elementary mechanics of materials models

Longitudinal normal stress

In-plane shear stress

Transverse normal stress

Representative volume element and simple stress states used in elementary mechanics of materials models

AV

dAA

dVV

11Average stress over RVE:

(3.19)

AV

dAA

dVV

11Average strain over RVE:

(3.20)

AV

dAA

dVV

11Average displacement over RVE:

(3.21)

Longitudinal ModulusRVE under average stress governed by longitudinal modulus E1.

Equilibrium:

Note: fibers are often orthotropic.Rearranging, we get “Rule of Mixtures” for longitudinal stress

mmffc AAA 1111 (3.22)

mmffc vv 111 (3.23)Static Equilibrium

1c

Hooke’s law for composite, fiber and matrix

111 cc E

111 fff E

11 mmm E

Stress – strain Relations

(3.24)

So that:

mmmfffc vEvEE 11111 (3.25)

Which means that,

Assumption about average strains:

111 mfc (3.26)Geometric Compatibility

mmff vEvEE 11 (3.27)

“Rule of Mixtures” – generally quite accurate – useful for design calculations

Variation of composite moduli with fiber volume fraction

Predicted E1 and E2 from elementary mechanics of materials models

Eq. 3.27

Eq. 3.40

Variation of composite moduli with fiber volume fraction

Comparison of predicted and measured E1 for E-glass/polyester. (From Adams, R.D., 1987. Engineered Materials Handbook, Vol. 1, Composites, 206–217.)

Strain Energy Approachmfc UUU (3.28)

Where strain energy in composite, fiber and matrix are given by,

cc

Vccc VEdVU

c

21111 2

121 (3.29a)

fffV

fff VEdVUf

21111 2

121 (3.29b)

mmmV

mmm VEdVUm

21111 2

121 (3.29c)

Strain energy due to Poisson strain mismatch at fiber/matrix interface is neglected.

Let the stresses in fiber and matrix be defined in terms of the composite stress as:

111 cf a

111 cm b (3.30)

Subst. in “Rule of Mixtures” for longitudinal stress:

mmffc vv 111 (3.23)

1111 cmfc vbva Or 111 mf vbva (3.31)

Combining (3.30), (3.24) & (3.29) in (3.28),

1

21

1

21

1

1

m

m

f

f

Evb

Ev

aE

(3.32)

Solving (3.31) and (3.32) simultaneously for E-glass/epoxy with known properties:

Find a1 and b1, then 00.11

1 m

f

Longitudinal normal stress

In-plane shear stress

Transverse normal stress

Representative volume element and simple stress states used in elementary mechanics of materials models

Transverse ModulusRVE under average stress Response governed by transverse modulus E2

Geometric compatibility:

From definition of normal strain, 222 mfc (3.34)

222 Lcc (3.35)fff L22

mmm L22

2c

Thus, Eq.(3.34) becomes

mmffc LLL 2222 (3.36)Or

mmffc vv 222 (3.37)

Where ,2

ff v

LL ,

2m

m vLL

222 cc E (3.38)222 fff E

22 mmm E

1-D Hooke’s laws for transverse loading:

Where Poisson strains have been neglected. Combining (3.37) and (3.38),

mm

mf

f

fc vE

vEE

2

2

2

2

2 (3.39)

Assuming that 222 mfc We get

m

m

f

f

Ev

Ev

E

22

1(3.40)

- “Inverse Rule of Mixtures” – Not very accurate

- Strain energy approach for transverse loading, Assume,

222 cf a

222 cm b (3.41)

Substituting in the compatibility equation (Rule of mixture for transverse strain), we get

122 mf vbva (3.42)

Then substituting these expressions for and in

2f2m

mfc UUU (3.28)

We get

mmff vEbvEaE 222

222 (3.43)

Solving (3.42) and (3.43) simultaneously for a2 and b2, we get for E-glass/epoxy,

63.52

2 m

f

Longitudinal normal stress

In-plane shear stress

Transverse normal stress

Representative volume element and simple stress states used in elementary mechanics of materials models

In-Plane Shear Modulus, G12

• Using compatibility of shear displacement and assuming equal stresses in fiber and matrix:

(Not very accurate)m

m

f

f

Gv

Gv

G

1212

1(3.47)

Major Poisson’s Ratio, υ12

• Using compatibility in 1 and 2 directions:

(Good enough for design use)(3.45)mmff vv 1212

Design Equations

• Elementary mechanics of materials Equations derived for G12 and E2 are not very useful – need to develop improved models for G12 and E2.

mmff vEvEE 11

mmff vv 1212

Improved Mechanics of Materials

Models for E2 and G12Mechanics of materials models refined by assuming a specific fiber packing array. Example: Hopkins – Chamis method of

sub-regions

RVE

Convert RVE with circular fiber to equivalent RVE having square fiber whose area is the same as the circular fiber.

Division of representative volume element into sub regions based on square fiber having equivalent fiber volume fraction.

d

RVE

s

A

B

A

sf

sf

A

B

A

Sub Region A

Sub Region B

Sub Region A

Equivalent Square Fiber:

ds f 4

(from )22

4ds f

(3.48)

Size of RVE:

dv

sf4

(3.49)

For Sub Region B: s

sf

sf

Following the procedure for the elementary mechanics of materials analysis of transverse modulus:

ss

Ess

EEm

m

f

fB

111

22

(3.50)

but ;f

f vss ;1 f

m vs

s (3.51)

So that

22 11 fmf

mB EEv

EE

(3.52)

For sub regions A and B in parallel,

ssE

ss

EE mm

fB 22

(3.53)

Or finally

22 11

1fmf

ffm EEv

vvEE (3.54)

Similarly,

212 11

1fmf

ffm GGv

vvGG

Simplified Micromechanics Equations (Chamis)

Only used part of the analysis for sub region B in Eq. (3.52):

22 11 fmf

mB EEv

EE

(3.52)

1212 11 fmf

m

GGvGG

Fiber properties Ef2 and Gf12 in tables inferred from these equations.

Semi empirical ModelsUse empirical equations which have a theoretical basis in mechanics

Halpin-Tsai Equations

f

f

m vv

EE

1

12 (3.63)

Where

mf

mf

EEEE 1

(3.64)

And curve-fitting parameter

2 for E2 of square array of circular fibers

1 for G12

As Rule of Mixtures

As Inverse Rule of Mixtures

0

Tsai-Hahn Stress Partitioning Parameters

222 fm let

Get

m

m

f

f

mf Ev

Ev

vvE2

22

11

(3.66)

Where stress partitioning parameter

(when get inverse Rule of Mixtures)

2,0.12

(3.65)

Transverse modulus for glass/epoxy according to Tsai-Hahn equation (Eq. 3.66). (From Tsai, S.W. and Hahn, H.T. 1980. Introduction to Composite Materials. Technomic Publishing Co., Lancaster, PA. With permission from Technomic Publishing Co.)

Eq. 3.66

Micromechanical Analysis of Composite Materials Using Elasticity Theory

Micromechanical analysis of composite materials involve the development of analytical models for predicting macroscopic composite properties in terms of constituent material properties and information on geometry and loading. Analysis begins with the selection of a representative volume element, or RVE, which depends on the assumed fiber packing array in the composite.

RVE

Example: Square packing array

Matrix

Fiber

Due to double symmetry, we only need to consider one quadrant of RVE

MatrixFiber

• The RVE is then subjected to uniform stress or displacement along the boundary. The resulting boundary value problem is solved by either stress functions, finite differences or finite elements.

• Later in this course we will discuss specific examples of finite difference solutions and finite element solutions for micromechanics problems.