Me Chap Handouts 1

1

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 solutionsc) 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

2

Densities

==VWρ density

mfc WWW +=

mmffcc VVV ρρρ +=∴

mmffc vv ρρρ +=∴

“Rule of Mixtures” for density

Alternatively,

m

m

f

fc ww

ρρ

ρ+

=1

(3.8)

Eq. (3.2) can be rearranged as

c

c

m

fc

f

f

v w

www

vρ

ρρ)(

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% s

s

d

Fiber

ds

s

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

Square array Triangular array

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

Fiber volume fraction – packing geometry relationships

Square array:2

4

=

sdv f

π(3.10)

When s=d, 785.04max ===π

ff vv (3.11)

3


Triangular Array:2

32

=

sdv f

π(3.12)

When s=d, 907.032max ===

πff vv (3.13)

• Real composites:Random fiber packing arrayUnidirectional:Chopped:Filament wound: close to theoretical


8.05.0 ≤≤ fv4.005.0 ≤≤ fv

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

2σ2σ

d

2σ 2σ

2σ

2σ

2ε

2ε

2σ 2ε

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

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Longitudinal normal stress

In-plane shear stress

Transverse normal stress


∫∫ ==AV

dAA

dVV

σσσ 11Average stress over RVE:

(3.14)

∫∫ ==AV

dAA

dVV

εεε 11Average strain over RVE:

(3.15)

∫∫ ==AV

dAA

dVV

δδδ 11Average displacement over RVE:

(3.16)

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 σσσ += (3.17)

mmffc vv 111 σσσ += (3.18)Static Equilibrium

1cσHooke’s law for composite, fiber and matrix

111 cc E εσ =

111 fff E εσ =

11 mmm E εσ =

Stress – strain Relations

(3.19)

So that:

mmmfffc vEvEE 11111 εεε += (3.20)

Which means that,

Assumption about average strains:

111 mfc εεε == (3.21)Geometric Compatibility

mmff vEvEE += 11 (3.22)

“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

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Variation of composite moduli with fiber volume fraction

Comparison of predicted and measured E1 for E-glass/polyester. From Adams(3.29)

Strain Energy Approachmfc UUU += (3.23)

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

cc

Vccc VEdVU

c

21111 2

121 εεσ == ∫ (3.24a)

fffV

fff VEdVUf

21111 2

121 εεσ == ∫ (3.24b)

mmmV

mmm VEdVUm

21111 2

121 εεσ == ∫ (3.24c)

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 σσ =

111 cm b σσ =(3.25)

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

mmffc vv 111 σσσ += (3.18)

( ) 1111 cmfc vbva σσ +=Or 111 =+ mf vbva (3.26)

Combining (3.25), (3.19) + (3.23),

1

21

1

21

1

1

m

m

f

f

Evb

Ev

aE

+= (3.27)

Solving (3.26) and (3.27) simultaneously for E-glass/epoxy with known properties:

Find a1 and b1, then 00.11

1 =m

f

εε




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

RVE under average stress Response governed by transverse modulus E2

Geometric compatibility:

From definition of normal strain, 222 mfc δδδ += (3.29)

222 Lcc εδ =(3.30)fff L22 εδ =

mmm L22 εδ =

2cσ

6

Thus, Eq.(3.29) becomes

mmffc LLL 2222 εεε += (3.31)Or

mmffc vv 222 εεε += (3.32)

Where ,2

ff v

LL

= ,2

mm v

LL

=

222 cc E εσ =(3.33)222 fff E εσ =

22 mmm E εσ =

1-D Hooke’s laws for transverse loading:

Where Poisson strains have been neglected. Combining (3.32) and (3.33),

mm

mf

f

fc vE

vEE

2

2

2

2

2 σσσ+= (3.34)

Assuming that 222 mfc σσσ ==We get

m

m

f

f

Ev

Ev

E+=

22

1(3.35)

- “Inverse Rule of Mixtures” – Not very accurate

- Strain energy approach for transverse loading, Assume,

222 cf a εε =

222 cm b εε =(3.36)

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

111 =+ mf vbva (3.37)

Then substituting these expressions for and in

2fε2mε

mfc UUU += (3.23)

We get

mmff vEbvEaE 222

222 += (3.38)

Solving (3.37) and (3.38) simultaneously for a2and b2, we get for E-glass/epoxy,

63.52

2 =m

f

σσ





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(3.42)

Major Poisson’s Ratio, υ12

• Using compatibility in 1and 2 directions:

(Good enough for design use)(3.40)mmff vv υυυ += 1212

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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 υυυ += 1212

Improved Mechanics of Materials Models for E2 and G12

Mechanics 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 4π

= (from )22

4ds f

π= (3.43)

Size of RVE:

dv

sf4

π= (3.44)

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

+= (3.45)

but ;f

f vs

s= ;1 f

m vs

s−= (3.46)

So that

( )22 11 fmf

mB EEv

EE−−

= (3.47)

For sub regions A and B in parallel,

ssE

ss

EE mm

fB += 22

(3.48)

Or finally

( ) ( )

−−+−=

22 11

1fmf

ffm EEv

vvEE (3.49)

Similarly,

( ) ( )

−−+−=

212 11

1fmf

ffm GGv

vvGG

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Simplified Micromechanics Equations (Chamis)

Only used part of the analysis for sub region B in Eq. (3.47):

( )22 11 fmf

mB EEv

EE−−

≅∴ (3.47)

( )1212 11 fmf

m

GGvGG−−

≅

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

Semi empirical Models

Use empirical equations which have a theoretical basis in mechanics

Halpin-Tsai Equations

f

f

m vv

EE

ηξη−

+=

112 (3.57)

Where ( )( ) ξ

η+

−=

mf

mf

EEEE 1

(3.58)

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

=ξ=ξ

=ξ

⇒∞→ξ⇒→ 0ξ

Tsai-Hahn Stress Partitioning Parameters

222 fm σησ =let

Get

+

+=

m

m

f

f

mf Ev

Ev

vvE2

22

11 ηη

(3.60)

Where stress partitioning parameter

(when get inverse Rule of Mixtures)

=2η,0.12 =η

9

Transverse modulus for glass/epoxy according to Tsai-Hahn equation (Eq. 3.60). From Tsai and Hahn (3.8) Micromechanical analysis of Composites

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

FiberDue 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.

• We will now discuss specific examples of finite difference solutions, and stress function solutions for micromechanics problems.

One quadrant of representative volume element from Adams and Doner elasticity solution for shear modulus G12. From Adams and Doner [3.19].

x

y

z

xzτa

b

*w

• Problem: For a square array of circular fibers, find the longitudinal shear modulus Gxz or Gyz.

z

x

y

Reference: “Longitudinal shear loading of a Unidirectional Composite”, D. F. Adams and D. R. Doner, J. Composite Materials, Vol.1, 1967, pp. 4-17

10

Solution: Solve displacement boundary value problems on one quadrant of RVE as shown:

x

y

z

zxτa

b

*w

=zyzx ττ , Average shear stresses along boundaries

u, v, w = displacement along x, y, z axes

Displacement: u=v=0, w= w(x,y)

where w = wf for fiber and w = wm for matrix

Strain-Displacement Relations:

(3.52)

,0=∂∂

=xuex ,0=

∂∂

=yvey 0=xyγ

,0=∂∂

=zwez x

wzu

xw

xy ∂∂

=∂∂

+∂∂

=γ

yw

zv

yxw

yz ∂∂

=∂∂

+∂∂

=γ

Hooke’s Law:

,0== xyxy Gγτ

xwGG zxzx ∂∂

== γτ

ywGG zyzy ∂∂

== γτ

0=== zyx σσσ

(3.53)

where G = Gf or Gm for fiber or matrix

0=∂∂

+∂

∂+

∂∂

yyxxzxyx ττσ

Equilibrium: No body force

0=∂

∂+

∂

∂+

∂

∂

zxyyzxyy ττσ

0=∂

∂+

∂∂

+∂∂

yxzyzxzz ττσ

only non-trivial equation is,

0=∂

∂+

∂∂

yxyzxz ττ

∴

Thus, the governing partial differential equation is,

02

2

2

2

=

∂∂

+∂∂

yw

xwG (3.54)

Or , Laplacian Equation subject to boundary conditions.

Boundary conditions: specify uniform displacement along

1. @

2. @

3. @

4. @

02 =∇ w

∗= ww ax =,0=x 0=w,ax = ∗= ww

,0=y 0=∂∂

=ywGyzτ 0=

∂∂

∴yw

,by = 0=∂∂

=ywGyzτ 0=

∂∂

∴yw

(3.55)

Finite difference solution of ,02 =∇ w

h

h – mesh size2

104

3

“Central” differences:

( ) 0412

043212

2 =−+++≈∇hwwwww

hw

at node point 0.

11

• Get similar difference equations for each node point and solve the set of simultaneous equations using matrix methods.

• Also need finite difference approximations for other partial derivatives like .

• In addition to the shear moduli, the stress concentration factors were found – see reference for details.

yw

xw

∂∂

∂∂ ,

Continuity conditions at fiber/matrix interface displacement:

Shear stress:

mf ww =

mmff nwG

nwG

∂∂

=∂∂

f

m n = normal at point on interface

The effective shear modulus for loading is then

xzτ

awG xz

xz *

τ= (3.56)

where is found by solving using finite difference, then using resulting displacements w to find along x = a,

(where ) then calculating

average value .

Similarly, for loading only,

where is found by solving analogous problem for w = w** along y = b.The problem of combined loading can be solved using superposition.

xzτ

xzτ

xwGxz ∂∂

=τxzτ

xzτbw

G yzyz **

τ=

,02 =∇ w

xzτ

• For similar analysis of transverse modulus, Ex or Ey see “Transverse Normal Loading of a Unidirectional Composite”, D. F. Adams and D.R. Doner, J. Composite Materials, Vol. 1, 1967, pp. 152-164.

Normalized composite shear stiffness (G12/Gm) versus shear modulus ratio Gf/Gm for circular fibers in a square array. From Adams and Doner [3.19]

Normalized composite transverse stiffness (E2/Em) versus modulus ratio Ef/Em for circular fibers in a

square array. From Adams and Doner [3.19]

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Stress concentration factor (SCF) for circular filaments in a square array subjected to longitudinal

shear loading (τXZ).

Composite shear stiffness (G/Gm) for circular filaments in a square array subjected to longitudinal

shear loading (τXZ).

Stress concentration factors (SCF) for boron filaments of various shapes in an epoxy matrix (Gf/Gm=120)

subjected to longitudinal shear loading (τXZ).

Composite shear stiffness (G/Gm) for boron filaments of various shapes in an epoxy matrix (Gf/Gm=120)

subjected to longitudinal shear loading (τXZ).

Two dimensional finite element models of representative volume elements. From Schroeder

[3.22].

Model 1

Two dimensional finite element models of representative volume elements. From Schroeder

[3.22].

Model 2

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Comparison of predicted transverse modulus for E-glass/ epoxy from two-dimensional finite element models with

other predictions. From Schroeder [3.22].

Three dimensional finite element models of representative volume elements. From Caruso and

Chamis [3.17].

The single cell model for the SC calculation

Three dimensional finite element models of representative volume elements. From Caruso and

Chamis [3.17].

The multi cell model of which only the center cell is used for the CCMC calculation

The multi cell model which is used for the MC calculation

Comparison of three-dimensional finite element results for lamina elastic constants with predictions

from simplified Micromechanics Equations for graphite / epoxy. From Caruso and Chamis [3.17].





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RVEs for other Finite Element Micromechanical Models

RVE for composite withwoven fabric reinforcement

Quarter domain of RVEfor composite with fiber coating or interphase

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Me Chap Handouts 1