Micromechanics: Homogenization

3/2/15

Objectives

• Review Voigt, Reuss and Hybrid

• Introduce Semi-Emperical Halpin-Tsai relationships

• Introduce Self-Consistent Field Relationships

• Introduce codes

Voigt and Reuss

Voigt Reuss

Hybrid

Halpin-Tsai micromechanics: transverse modulus

Perhaps the most widely used micromechanics model was developed entirely empirically. The insightfulness of Halpin and Tsai has been admired by all for many years. The coefficient x ranges from 1-2 to represent the various microstructures from hexagonal (x =1) to square (x =2) arrays. Note that the model allows for non-isotropic fibers, E2f.

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arrayhexagonal1=x

arraysquare2=x

Comparison of Voigt and Reuss with Halpin-Tsai predictions and data

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Halpin-Tsai micromechanics: shear modulus

The Halpin-Tsai equations use the same form for prediction of the in-plane shearing modulus as it used for the transverse Young’s modulus. While the Halpin-Tsai equations provide good estimates of the “matrix dominated” properties of the composite. The “fiber dominated” properties such as E1 and n12are adequately predicted by the simple rule of mixtures.

P = VfPf + VmPm

Note that it is the volume fraction of each constituent that is important. Yet one encounters typical measures of mixtures in weight fraction. Use of density is required for the conversion.

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arrayhexagonal1=x

Halpin-Tsai limiting conditions

The limiting conditions for the Halpin-Tsai equations are the Voigt and Reuss models. By expanding the expression for the property:

Then the limiting values of the term x are x = 0 and x = ∞:

Thus, the Halpin-Tsai equations provide a way to interpolate between the Voigt and Reuss estimates for elastic properties of the equivalent homogeneous and to account for microstructural geometry ranging from square to hexagonal arrays. Note that the Voigt model (x = ∞ )is volume averaging of the stiffnesseswhile Reuss model (x = 0 )is volume averaging of the compliances, thereby giving the inverse relationships for moduli.

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)]([

)]([

mPfPfVmPfP

mPfPfVmPfPmPP

=

x

xx

Self consistent field analysis geometry

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See: Micromechanics of Composite Materials by George Dvorak, Springer Solid Mechanicsand its Applications, Vol. 186, (2013), XVII, 442 pages.

Self-Consistent field representations

The self-consistent field equation for the primary Young’s modulus was developed using a more complex mathematical formulation as shown below:

Yet, by examining the relative magnitude of the last term for most materials, it is clear that the first two terms dominate the prediction.

The same is true for the primary Poisson’s ratio. The first two terms typically dominate this prediction as well.

These results clearly show that the linear “rule of mixtures” is appropriate for these properties:

P = VfPf + VmPm

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=

fmmfmmf

ffmfmfmmmff

VGkkkGk

VVGkkVEVEE

)1(411

nn

E1f=fiber primary Young’s modulus

Em=matrix (resin) Young’s modulus

km=matrix (resin) plane strain bulk modulus

kf=fiber plane strain bulk modulus

Gm=matrix shearing modulus

Vf=volume fraction fiber

Vm=volume fraction matrix (resin)

==

fmmfmmf

ffmfmfmfmff

VGkkkGk

VVGkkVV

)1(1

2

1312

nnnnnn

Self-consistent field equations for transverse modulus

The self-consistent field model produces a complex equation for the inter-relationships of the composite properties:

Where kT is the plane strain bulk modulus of the composite:

And the transverse shearing modulus, G23 is:

Thus, the determination of E2 require that the values of kT, G23 and E1 be first determined in this approach. Given the potential for cumulative error in each calculation, this approach can be problematic. Finally n23 is:

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fmfmf

fmmfmmfT

VkkGk

VGkkkGkk

=

1

2

12

23

2

4

1

4

1

1

EGk

E

T

n

=

fmfmmmffmm

fmfmmffmmm

VGGGkGGGGk

VGGkGGGGkGG

=

22

)(223

T

TT

kE

EkEEkE

1

2212211

232

42 nn

=

Weight fraction and volume fraction

It is common for mixtures of materials to be reported as weight fractions since one need only measure weight on a scale. Volume fraction, however, is the important

measure. Conversion between weight and volume fraction is an important calculation as shown:

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Hexagonal and square arrays

To determine fiber volume fraction, one only need determine the cross-sectional area occupied by fiber in the representative volume element (RVE).

Hexagonal array:

Where 2R is the center-to-center fiber spacing, 2r is the fiber diameter an d is the smallest distance between fibers

Square array

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Fiber volume fraction

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Example Problems and Solutions

Example problem 1

a) For the same fiber volume fraction, how different is the fiber center-to-center fiber spacing for hexagonal and square array geometries?

Sol: Set the volume fraction equations equal:

b) For a volume fraction of 0.5, what is the difference in minimum spacing for the two geometries?

Sol:

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91.1

63.33

2

432

2

22

=

==

=

sq

hex

sq

hex

sqhex

R

R

R

R

R

r

R

r

37.1253.0

347.0

12

13

2/1

2/1

==

=

sq

hex

h

h

Example problem 2

a) What are the maximum packing fractions for hexagonal and square arrays?Solution:

b) If the fiber diameter is 5 microns, what is the area of the RVE for Vf = 0.6?

Solution:

The area containing the fibers is made up of 6 equilateral triangles with sides 2R for a total area of 6R2(3)1/2.

If the number of fibers in the hexagonal RVE is 7, then the total fibers in the RVE is [1 + 6(120/360)] = 3, with a fiber area of 3r2

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785.044

907.03232

2

2

==

=

==

=

=

sq

sq

f

hex

hex

f

R

rV

R

rV

rRLet

)10(39336

15.632.15

5

326.0

1222

2

2

mRA

RR

RV

hexRVE

hex

hex

hex

hex

f

==

==

==

22 235)5(3 ==fA

Example problem 3

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Problem 3 - Solution

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Problem 3 - Solution - 1

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Problem 3 - Solution-2

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Problem 3 - Solution-3

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Problem 3 - Solution-4

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Problem 3 - Solution-5

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Problem 3 - Solution-6

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Problem 4 – Solution: Halpin-Tsai model

Consider the properties used in the previous example and use the Halpin-Tsai equations to determine the continuous fiber composite material property G12 for volume fractions of 0.3 and 0.6 for square and hexagonal arrays.

Solution:

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3.02668

1.0172380

===

===

mmm

fff

GPaGGPaE

GPaGGPaE

n

n

652.0

262172

26172

22

737.026172

261721

1

1

12

12

12

12

12

12

12

=

=

==

=

=

==

=

=

mf

mf

mf

mf

mf

mf

f

fm

GG

GGLet

GG

GGLet

GG

GG

V

VGG

x

x

x

x

mf

mf

f

fm

GG

GG

V

VGG

x

x

=

=

12

12

121

1

arrayhexagonal1=x

arraysquare2=x

GPaG sq 0.45

804.0

2.36

)3.0(652.01

3.0304.1126)30(12 ==

=

f

f

f

fmsq

V

V

V

VGG

652.01

652.02126

1

112

=

=

x

f

f

f

fmhex

V

V

V

VGG

737.01

737.0126

1

112

=

=

x

GPaG sq 2.67

558.0

5.34

)6.0(737.01

6.0737.0126)60(12 ==

=

GPaG sq 8.40

779.0

8.31

)3.0(737.01

3.0737.0126)30(12 ==

=

GPaG sq 2.76

608.0

3.46

)6.0(652.01

6.0304.1126)60(12 ==

=

Problem 4 – Solution: Halpin-Tsai model - 2

Now compare these values to the estimates given previously for the other models:

The Halpin-Tsai model compares favorably with the square fiber model

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*H-T (60)

*H-T (30)

*H-T (60)