Micromechanics of Unidirectional Fibre Composites … on mechanics/ppt... · Micromechanics of...

Micromechanics of Unidirectional Fibre Composites

A School on Mechanics of Fibre Reinforced Polymer Composites

Knowledge Incubation for TEQIPIndian Institute of Technology Kanpur

PM Mohite

Department of Aerospace Engineering

Indian Institute of Technology Kanpur

22-25 January 2017

• The 3D constitutive equations of an orthotropic material has 9 independent

constants.

• For a transversely isotropic material there are 5 independent constants.

Constitutive Relations:

• Effective properties depend upon:

• Properties of constituent

• Proportions of constituents

• Distribution and arrangement of constituents

• Shape and size of reinforcing constituent

Assumptions made in micromechanical analyses:

• Constituent obeys Hooke’s law

• Constituents behave linear elastic

• Perfect bonding between constituents

• Fibre are circular in cross section and parallel

• Fibres are of uniform diameter

• The elastic, thermal, hygral properties of fibre and matrix are known

• Fibre and matrix are the only two phases

• There are no voids

Introduction:

• Idealization of cross-section

• Most commonly preferred arrangements are square and hexagonal packed

arrays of fibres in matrix.

Fibre Arrangement in Matrix Material:

• The fibre volume fraction is defined as the ratio of fibre volume to composite

volume and matrix volume fraction is defined as the ratio of matrix volume

fraction to composite volume.

• If the interphase is also present as a third phase then,

Volume Fractions of Constituents:

• In case, there are voids present in composite,

• The mass fractions are defined as the ratio of mass of the respective phase to

the mass of composite.

Volume Fractions of Constituents:

• Now, let us write the mass of each phase in terms of density and volume of

respective phase as

• Now, mass fractions can be written in terms of density and volume fractions

as

Weight Fractions of Constituents:

• Relation between mass and volume fractions is given in terms of individual

constituent properties

• The density of composite is derived in terms of densities and volume fractions

of the individual phases

Weight Fractions, Densities of Constituents:

• This is written using the definition of volume fraction for fibre and matrix as

• We will write the density of composite in terms of mass fraction

Densities of Constituents:

Strength of Materials Approach

• The most popular and easy to apply method

• 4 out of 5 engineering constants can be obtained by this method

• The assumptions made not necessarily satisfy the requirements of exact

elasticity solution

• Properties and volume fractions of the constituents are needed. The

arrangement is not required.

• Volumes represented through cross sectional areas of fibre and matrix

Strength of Material Approximations:

• Unidirectional lamina and RVE of unidirectional composite for prediction of

elastic properties

• Effective Axial Young’s Modulus:

• The strains in the fibre and matrix are same (condition of continuity in axial

direction) can be given as


• Axial stress in the fibre and matrix as

• The forces can be written as

• The total axial force in the composite substituting the expressions for axial

strains in fibre and matrix

• Effective Axial Young’s Modulus:


• Undeformed unit cell under average axial stress and deformed individual

constituents of the unit cell


• The effective axial Poisson’s ratio as

• The effective strain in direction 2

• The axial Poisson’s ratios for fibre and matrix

• The total change in W

Effective Axial (Major) Poisson’s Ratio:

• The strain in direction 2 for the composite

• the effective axial Poisson’s ratio

Effective Axial (Major) Poisson’s Ratio:

• Assumption: Fibre and matrix deform independently. Deformation in

direction 1 is not considered.

• The state of stress

• The total deformation in direction 2 as

• The composite strain in direction 2

Effective Transverse Modulus:

• Undeformed unit cell under uniform stress and deformed individual



• Introducing the volume fractions

• Rule of mixtures equation for effective modulus


• Alternate approach considering the deformations in direction 1:

• The axial and transverse stresses in fibre and matrix

where,


• Undeformed unit cell under uniform stress and deformed individual



• Generally, the stresses are uniform in the fibre and matrix

• For equilibrium in transverse direction

• The axial and transverse strains in fibre and matrix

• Putting these in above

equations


• The transverse composite strain

• Expression for Effective Transverse Modulus

where


• The effective axial shear modulus

• (a) Undeformed unit cell under uniform (b) overall deformation of unit cell

(c) and (d) deformed individual constituents of the unit cell

Effective Axial Shear Modulus :

• The shear strains in the fiber and matrix

• The individual deformations, when the shear strains are small, in fibre and

matrix as

• The total axial deformation is given as


• The overall shear strain of the RVE

• The effective axial shear modulus of the composite


• Deformation same as case with

• The deformation in fibre and matrix in transverse direction

• Define: The coefficient of thermal expansion for composite in transverse

direction

• From the volume fraction definitions, we get

Coefficients of Thermal Expansion:

*

2E 2 0σ =

• The stresses in fibre can be given for the temperature change of ȂT

• The stresses in matrix can be given for the temperature change of ȂT

• The stresses in transverse direction in fibre and matrix should be identically

zero. The first condition leads to


• The stresses in fibre and matrix are uniform

• The second condition that the stresses in transverse direction are zero leads to

• On solving the above equation, we get


• The coefficients of thermal expansions for composites as

and


• The force in axial direction in composite should be zero and this condition

leads to (constraints exactly similar to thermal expansion coefficients)

• The axial stresses in fibre and matrix due to moisture absorption

• The axial strain in fibre and matrix are equal, that is

Coefficients of Hygral Expansion:

• Define: the effective coefficient of hygral expansion in axial direction

where, ȂM is the per weight % moisture absorption for composite. Thus, the

above equation becomes

• The percentage weight moisture absorbed by composite to the moisture

absorbed by fibre and matrix


• The above equation can be written as

• The mass fractions are written in terms of volume fractions with the use of

densities of composite, fibre and matrix.

• The coefficient of thermal expansion in transverse direction


Statistical Homogeneity

• Scale of inhomogeneity and averaging of properties along with an RVE

(a) continuous variation of heterogeneity

(b) abrupt variation of heterogeneity

λ – characteristic dimension of

inhomogeneity

δ – length scale of averaging

δ>> λ

δ – must be large enough to represent the microstructural details and

interaction between fibres and matrix.

Other terms: effective, equivalent, macroscopic or statistical homogeneity

The Concept of Equivalent Homogeneity:

• The average stress

• The average strain

• The effective linear stiffness tensor

• The inverse of the above relation

The Concept of Volumetric Averaging:

Concept of Energy Equivalence:

• Contracting the average stresses with average strains

• Left side can be obtained in terms of surface tractions and displacements as

• Using the divergence theorem and equilibrium equations for the body

without any body forces

• (a) A beam representing alternate fibre and matrix (b) force distribution and

(c) strain distribution along the length of the beam

• The axial strain in each of the segment

of the bar

• The force in equivalent material

Concept of Averaging: An Example

• The average or equivalent axial strain in the homogeneous material

• The average axial strain in the element

• The equivalent axial Young’s modulus as


• The strain energies of the non-homogeneous and homogeneous materials

which upon simplification gives


Standard Mechanics Approach

• The average strain in terms of displacements using divergence theorem

ui - are the displacements

• The average stress

Ti - are the surface tractions

Standard Mechanics Approach:

• Boundary conditions that produce average strain or stress

• Uniform boundary conditions on an RVE

• In case of the applied displacements the weak form of the RVE equilibrium

equations


• In case of applied tractions the weak form the equilibrium equation to be

solved for each applied stress component is

• The relation between the average strain and local or microstructural strains

• The local structure tensor is known then local strains at any point

• The effective stiffness tensor


ijklA

• Using the definition of average stress the above equation can be written as

the above equation becomes,

• The effective stiffness tensor


Hills Concentration Factors Approach

• Separation of phasewise quantities

• The average stress in terms of the stresses in individual phases are

• The volume averaged stress in fibre

• The volume averaged stress in matrix

Hill’s Concentration Factors Approach:

• Average stress in composite

• The volume averaged strains in fibre and matrix

• Average strain in composite


• The average stress in fibre and matrix using the pointwise constitutive

equation for fibre and matrix

• The above equation can be written in terms of compliance of fibre and matrix

material


• The average stress in composite in terms of volume fractions, stiffness tensor

and phase averaged strains

• The average composite strain in terms of phase averaged stresses in the fiber

and matrix, respective compliances and volume fractions

• The pointwise strain and stress in fibre and matrix


• The phase averaged concentration factors

• The phase averaged strains in fibre and matrix


• The phase averaged concentration factors

• Composite average stress is given as

• composite average strain is given as


• The bracketed term in above equation is an identity tensor of fourth order

we can write,


• Voigt assumed that the strains are constants throughout the composite

this leads to,

• We can write,

Voigt Approximation:

• Reuss assumed that the stresses are constant throughout the composite

which upon substitution leads to the relation,

Reuss Approximation:

Mathematical Theory of

Homogenization

• (a) A bar representing alternate fibre and matrix (b) actual and average

strain (c) actual and average displacement and (d) periodic nature of

displacement difference over RVE length

Concept of Homogenization:

• The global level or macro coordinates and the micro level coordinates are

related through

• The field variables involved in this study are approximated by an asymptotic

expansion as

• The small deformations the strain tensor

Theory of Mathematical Homogenization:

• We can define strain tensor as

• The virtual strain is given as

• This strain can be written as average and microscopic strain due to virtual

displacement


• The weak form of the equilibrium equations is given by

• The resulting expanded form of the above equation is then given as

• The macroscopic equilibrium equation as


• The above equation may be simplified assuming η approaching zero in the

limit as

and

• In the above equation the integration term over the RVE should be zero, this

leads to the following condition:


• The unit strains in the following equation are given for 3D case.

• Using these strains, a stress tensor is formed as


• The resulting auxilliary problem

• The periodicity of the strain field is obtained by constraining equal

displacements on opposite sides of RVE and is determined by

• When the relationship between local RVE strain and the average strain is

substituted into the macroscopic equilibrium equations, we get


• The effective stiffness tensor is defined

• Using this effective stiffness tensor, the final form of macroscopic equilibrium

equation may be written as


Concentric Cylinder Assemblage

Model

• RVE:

• All the space is filled by concentric cylinders of different radii

• Ratio of fibre and matrix annulus are such that they represent volume

fractions and same for all cylinders.

Concentric Cylinder Assemblage Model:

Background to Concentric Cylinder Assemblage Model:

• Constitutive relations:

• Assume that and all other

components are zero.

Gives us

11 0σ ≠


• From the relation

• Define

to give

• Similarly, define

which gives

Further, we get


• Assume

• That is, plane strain condition in plane 23

this leads to

• Define plane strain bulk modulus as

• Finally, we get all constants

in terms of four engineering

constants!

• Other engineering constants can be obtained indirectly.

Let and all other

stresses be zero.


22 0σ ≠

• From the second equation define

• Similarly, define

which gives

• Further,



• Thus, in terms of constants

• Strain energy for a cylinder:

• Subjected to axial load or uniform radial stress then no shear stresses are

developed

• For infinite cylinder, away from ends the stresses are independent of x

• Equilibrium equation

• Normal stresses in cylindrical coordinates

• For fibre and matrix materials:

Analysis of Concentric Cylinder s:

• Stain displacement relations in cylindrical coordinate system:

• u, v, w – radial, circumferential/tangential and radial displacement

1. With no shear stress and axisymmetry tangential displacement is zer

2. Strains are also independent of x

3. W can be function of r

4. U can be function of r and at most linear function of x


• Thus,

• And the equilibrium equation becomes

• With the solution of the form

• For fibre

• For matrix


• Axial displacement from axial strain displacement relation:

• Boundary conditions to be ensured:

• Displacement must be bounded at

the centre of the fibre


Condition 1:

• Continuity of the displacements

at the interface of fibre and matrix

This leads to the conditions

Further, on equating the terms in x and r, this leads to


Condition 2:

• Continuity of radial stress at interface

That is,

• The other constants will be obtained from

additional conditions that will result from specific loading conditions


Condition 3:


• Strains in fibre and matrix:

• Stresses in fibre and matrix:

• Define

• The outer surface must be radial stress free:

That is,

Effective Axial Modulus:

New Condition:


• The axial force is obtained as

which gives

Unknowns are solved as

where,



with,

Finally giving

• One may solve for in terms of

giving

Then find the effective axial modulus


• For isotropic phases:

• Stresses in a phase

• Stresses in fibres and matrix:

• Radial stresses

• Hoop’s stresses


• Continuity of radial stresses at interface:

• Radial stress free condition at outer surface of matrix

• Effective axial force:

• Solution for unknowns


• where,

• and


• When are expressed in terms of

• Constants are

• And modulus is


• Can be obtained from equivalence of strain energy

• Strain energy of concentric cylinder


• For transversely isotropic phases

• For isotropic phases


• Loading conditions same as for earlier case

• Define:

• Using the expressions

leads to (for transversely isotropic phases)

Effective Axial Poisson’s Ratio:

• For isotropic phases:

Effective Axial Poisson’s Ratio:

• Effective Plane Strain Bulk Modulus:

• For this state of strain, the displacement field in cylindrical coordinates

system on the outer boundary of the cylinders becomes

• The tractions in the Cartesian coordinate system

• Alternatively, for the displacement field on the outer surface of the cylinder,

the tractions in cylindrical coordinates are

Effective Plane Strain Bulk Modulus:

And the strains can be given in terms of stresses as

• The second condition is continuity of radial stresses at the fibre and matrix

interface. With transversely isotropic fibre, this condition leads to

and with isotropic fibre, this condition becomes


• The third condition is that the outer boundary of the cylinder has the radial

stress equal to the σ. In case of fibre with transversely isotropic material, this

leads to

• And in case of fibre with isotropic material, it becomes

• For the case of transversely isotropic fibre the constants are determined by

solving these constants


• For the case of isotropic phases the constants are given as

• The effective plane strain bulk modulus of the equivalent homogeneous

material is then defined as


we can write,

we get,

• The effective plane strain bulk modulus can be also defined as

• the ratio of change in volume per unit volume of the concentric cylinders is

given as


• Thus, for the transversely isotropic fibre, using the constants the effective

plane strain bulk modulus becomes

• Dividing the numerator and denominator, we get


• For isotropic fibre, using the constants the effective plane strain bulk

modulus is given as

• The strain energy of the equivalent homogeneous single cylinder is given as

• When the stresses and strains are substituted in above equation, for a unit

length, it gives


Effective Axial Shear Modulus:

• The outer surface of the cylinder in the plane is subjected to a displacement

field such that the overall strain produced in this plane is equal, that is

• The displacement components on the boundary of the cylinder then becomes

• The displacement components in each constituent are then given with

corresponding simplification in the general solution as

• The continuity of the displacement components at the interface may be

written as

• Undeformed and deformed concentric cylinders under shear


• The first continuity condition (radial displacement) of the above equation

gives the relation

• The remaining two displacement continuity conditions give

• The non-zero stresses resulting from the displacement field


• The continuity of the stresses in radial direction leads the continuity of the

stress at the interface and this condition becomes

• Now, at the outer boundary of the concentric cylinders the displacements

must match the following boundary conditions.

• Note that from the second and the third of the above condition, we get


• The constants can be determined as

• At last, at outer boundary the shear stress must match the shear stress in

coordinate system. Thus the shear stress then becomes


• The right hand side of above equation can be written as

• Thus, the equivalent axial shear modulus can be given as


Three Phase Concentric Cylinder

Assemblage Model

• All cylinders except one replaced by equivalent homogeneous material

(a) Three phase concentric cylinder assemblage model (b) homogeneous single

cylinder

Three-Phase CCA Model for Transverse Shear Modulus:

• The planar stresses for the cylinder can be given in terms of stress function

as

(Stress Function Approach)

• where, the stress function

• a3 and c3 are the constants evaluated using boundary conditions


• Using the stress function, the stresses become

• considering the plane strain condition (with εxx=0) and using the stress-strain

relations, the strains in composite material are written as


• Now using the strain displacement relations, the displacement components in

composite material are obtained as

• In similar way, we can derive the displacement field in matrix phase given as


• The displacement in fibre phase is given as

where,


• The displacement continuity at r = a gives the relations

• The displacement continuity at r = b gives


• The stress continuity at r = a gives the following relations

• And the stress continuity at r = b leads to

• Now the effective properties are determined from the equivalence of strain

energy in heterogeneous media and in the equivalent homogeneous media.

Thus, we write this condition as


• The strain energies in the above equation can be given in terms of Eshelby

formula (with an inclusion)

• the strain energy in the same medium when the medium does not contain any

inclusion (U°).

• we can write,

• Using the definitions of stresses and displacements this equation is written as


• The various quantities in above equation are given as

• This expression is derived from the condition that the displacements must be

bounded at r = 0 . a3 = c3=0

• Similarly, the stresses become


• The stresses at r = b can be given as

• Finally, v and w are evaluated at r = b are given as

• Now using all the required terms and carrying out the integration, we get


• The solution obtained by solving eight simultaneous equations is thus set

equal to zero. This leads to the following equation.


and,

• Further, in the case of dilute suspension, that is, a single inclusion or fibre in

an infinite equivalent homogeneous medium (that is, a condition leading to a

low fibre volume fraction) the following relation is given

• The bulk modulus relates the hydrostatic stress and the change in volume as


Self Consistent Method

• The solution to an auxiliary problem where a single ellipsoidal inclusion is

embedded in an finite medium.

• The bond between the inclusion and infinite medium is assumed to be perfect

• Thus, there is continuity of stresses and tractions across the two phases

• Eshelby showed that the stress and strain fields in the inclusion are uniform

• Effective properties can be found by finding the relation between far field

stresses and strains in homogeneous medium to the stresses and strains in the

inclusion (that is, concentration factors)

• In the following the fibre and matrix phases are transversely isotropic in

nature.

Self Consistent Method:

• The self consistent estimates the overall moduli (Hill’s Moduli), given in the

following relations (* represents the effective).

• The moduli are related through so called universal relations in terms of

overall moduli and phase moduli and their volume fractions

Self Consistent Method:

Mori Tanaka Method

• Assumption: The average strain in the inclusion (fibre) is related to averagestrain in the matrix by a fourth order tensor.

• The material is subjected to a uniform strain at infinity

• The strain concentration factors in fibre are given as

where

Tijkl – forth order tensor

Sijkl – Eshelby tensor

Mori-Tanaka Method:

• Dvorak et al have given the explicit relations in terms of Hill’s moduli as

Mori-Tanaka Method:

• In case of transversely isotropic material in 23 plane, the constitutive

relations are given as

and

Hill’s Moduli:

• The relationship between the Hill’s moduli k, l, n, m and p and the

engineering moduli are given as

• Some additional useful relations are

Hill’s Moduli:

• Further, if the phase is isotropic then with bulk modulus k and shear modulus

G , we have

Hill’s Moduli:

Halpin Tsai Models

• The longitudinal Young’s modulus is same as given by rule of mixtures using

strength of materials approach.

• The axial Poisson’s ratio is same as given by rule of mixtures using strength

of materials approach.

• The Hill’s Moduli are given as

where

Halpin-Tsai Semi-Empirical Relations:

• The measure of reinforcement geometry which depends upon loading

conditions and geometries of the inclusion

this gives,

• It is easy to see that this is a series connected model which gives the lower

bound of a composite modulus.

Halpin-Tsai Semi-Empirical Relations:

Hashin Shtrikman Bounds

• Variational principles involving the polarization tensor is used

• Bounds for quasi isotropic and quasi-homogeneous multiphase materials

• Principles of minimum potential energy and minimum complementary

energy

• Bulk Modulus:

Hashin-Strikman Bounds:

1

1

f

mm

f m f m

mf

f

m f f m

VK K

V

K K K G

VK K

V

K K K G

−

+

= +

+− +

= +

+− +

• Transverse Shear Modulus:

• where


( )( )

( )

23 23

2323

1

23 232 2

1

3

21

2

11

31

1

fm

m m m

f mm m m

fm

mf

f

VG G

K G V

G K GG G

VG G

VV

V

β

βρ

α

−

+

= ++

++−

+

= + + − + +

2 1 21 2

1 1 1 2 2

, ,2 2

G K K

G K G K Gγ β β= = =

+ +

1 2 1

2

,1 1

β γβ γ βα ρ

γβ γ

− += =

+ −

• Axial Modulus:


( )

( )

2

1212

1 1 1

23

2

1212

1 1 1

23

4

1

4

1

f mf mfm

m ffm

mf m

f mf mfm

m ffm

ff m

V VE E V E V

VV

K K G

V VE E V E V

VV

K K G

ν ν

ν ν

−

+

−= + +

+ +

−= + +

+ +

• Major Poisson’s Ratio:


( )

( )

12 12

12 12 12

23

1212

12 12 12

23

14

1

14

1

fmf m

f mfmm f

fmf

f m

f mf m

m ffmm f

fmm

f m

V VK K

V VVV

K K G

V VK K

V VVV

K K G

ν ν

ν ν ν

ν ν

ν ν ν

−

+

−

− = + +

+ +

−

− = + +

+ +

• Axial Shear Modulus:


( )( )

( )

( )

12 12

12 12

1212

1212

12 12

12 12

1

1

1

1

fmm fm

f mm f

f mf mf

fmf m

G V G VG G

G V G V

G V G VG G

G V G V

−

+

+ + = + +

+ + =

+ +

Thank You

Micromechanics of Unidirectional Fibre Composites … on mechanics/ppt... · Micromechanics of...

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