micromechanics
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Micromechanics. Macromechanics. Fibers. Lamina. Laminate. Structure. Matrix. Micromechanics. - PowerPoint PPT PresentationTRANSCRIPT
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MicromechanicsMacromechanics
FibersLamina
Matrix
Laminate
Structure
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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.
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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
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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
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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)
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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
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Densities
VW density
mfc WWW
mmffcc VVV
mmffc vv
“Rule of Mixtures” for density
(3.6)
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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)
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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%
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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
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s
s
d
Fiber
ds
s
Representative area elements for idealized square and triangular fiber packing geometries.
Square array Triangular array
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Fiber volume fraction – packing geometry relationships
Square array:2
4
sdv f
(3.14)
When s=d, 785.04max
ff vv (3.15)
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Fiber volume fraction – packing geometry relationships
Triangular Array:2
32
sdv f
(3.16)
When s=d, 907.032max
ff vv (3.17)
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• 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
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Photomicrograph of carbon/epoxy composite showing actual fiber packing geometry at 400X magnification
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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
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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.)
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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.
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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
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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
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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
Representative volume element and simple stress states used in elementary mechanics of materials models
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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)
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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
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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)
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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
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Variation of composite moduli with fiber volume fraction
Predicted E1 and E2 from elementary mechanics of materials models
Eq. 3.27
Eq. 3.40
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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.)
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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)
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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)
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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
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Longitudinal normal stress
In-plane shear stress
Transverse normal stress
Representative volume element and simple stress states used in elementary mechanics of materials models
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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
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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:
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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)
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- “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)
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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
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Longitudinal normal stress
In-plane shear stress
Transverse normal stress
Representative volume element and simple stress states used in elementary mechanics of materials models
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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
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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 1212
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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
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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
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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
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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)
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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
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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.
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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)
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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
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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)
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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
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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.
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RVE
Example: Square packing array
Matrix
Fiber
Due to double symmetry, we only need to consider one quadrant of RVE
MatrixFiber
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• 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.