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1
Derivation of Boundary Conditions for Micromechanics Analyses of Plain and Satin Weave
Composites
John D. Whitcomb
Clinton D. Chapman
Xiaodong Tang
Abstract
Introduction
Composite materials consist of a combination of materials, often fibers and matrix. The basic
challenge of micromechanics is to determine how the properties and spatial distribution of the
constituents and the spatial arrangement of the fibers affect the overall material response. The
overall response is often characterized in terms of the effective engineering properties, such as
extensional moduli and Poisson's ratio. These are also referred to as the homogenized
engineering properties. To obtain these properties a representative volume element (RVE) or unit
cell is identified that includes all characteristics of the composite. The literature includes a wide
variety of analyses for various unit cells. [ Xiaodong: get references] Recently, there has been
increased interest in developing robust micromechanics analyses of textile composites
[references…]. Although a variety of analyses exist, including some very detailed threedimensional finite element analyses [references], there has not been a systematic description of
how one obtains the boundary conditions to perform the analysis. Accordingly, the goal of this
paper is to present a systematic procedure for deriving the boundary conditions for both full unit
cell and partial unit cell analyses of plain and satin weave composites.
Background
Implicit in micomechanics analyses is that a small region of the microstructure fully represents
the behavior of a much larger region (usually an infinite domain). There is no standard definition
of this region in the literature. Herein, the term "unit cell" will be referred to as the smallest
region that represents the behavior of the larger region without any mirroring transformations.
For such a region to exist there must be a basic pattern (of geometry and response, such as
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strains, stresses, etc.) that is repeated periodically throughout a domain. In particular, the larger
region can be synthesized by replication and translation of the unit cell in the three coordinate
directions. Figure 1 show examples of periodic microstructures. In each case, the region is built
from a single building block or unit cell (shaded regions in the figures). Note that there can be
more than one definition of the unit cell for a single microstructure.
<figure 1>
To determine effective engineering properties, the periodic array is subjected a series of loads
consisting of either macroscopically constant strain or stress. The term macroscopically constant
indicates that the volume averaged strain for every unit cell is identical. (The same can be said
for the stresses.) Although one cannot analyze an infinite number of unit cells, it is possible to
derive boundary conditions for a single unit cell that will make it behave as though it is buried
within an infinite array. How this is done will be described herein.
It is convenient to think of the challenge as consisting of two parts. The first part consists of
identifying the appropriate boundary conditions for a single full unit cell. The second challenge
is to exploit the symmetries in the microstructure of the unit cell and the loading to reduce the
analysis region to just a fraction of the full cell in order to reduce the computational costs.
Derivation of basic equations
The derivation of the boundary conditions is based on two conditions. The first is periodicity and
the second is equivalence of coordinate system. It should be noted that that these concepts can be
combined, but the authors choose not to do so. The periodic conditions state that the
displacements in the various unit cells differ only by constant offsets that depend on the volume
averaged displacement gradients. Further, the strains and stresses are identical in all of the unit
cells.
This can be expressed as
b g b g xu
 b
x d g
 b
x g
 b
x d g
 b
x g
ui x  d  ui x 
i
d

ij
ij





ij
ij

(1)
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where d  is a vector of periodicity. This vector is a vector from a point in one unit cell to an
equivalent point in another unit cell.
These conditions are sufficient for deriving the boundary conditions for the full unit cell.
However, it is usually desirable to exploit symmetries in order to be able to analyze a smaller
region. The concept of "Equivalent Coordinate Systems" (ECS) is useful in identifying the
symmetries and constraint conditions. Coordinate systems are equivalent if the geometry,
loading, and the various fields which describe the response (eg. displacement, strains, etc.) are
identical in the two systems.
For example, xi and xi are equivalent coordinate systems if
bg bg
 b
x g
 b
x g
 b
x g
 b
x g
ui x  ui x
ij
ij


ij

ij

(2)

etc.
Exploiting equivalent coordinate systems requires identification of coordinate transformations
that provide useful constraint conditions. Consider the equivalent coordinate system xi as a
potentially useful ECS. Define: aij  direction cosines for transformation from the original ( x j )
to the equivalent ( xi ) coordinate system….i.e. xi  aij x j
bg bg
 b
x g
a a  b
x g
 b
x g
a a  b
x g
ui x  aij u j x
ij
ij


im
im
jn mn
jn
mn

(3)

Combine equations (2) and (3) to give
bg
 b
x g
a a  b
x g
 b
x g
a a  b
x g
ui ( x )  aij u j x
ij
ij


im
im
jn mn
jn
mn


Finally, replace x with a j x j so that everything is now expressed in terms of a single
coordinate system.
(4)
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d i
 b
x g
a a  d
a x i
 b
x g
a a  d
a x i
ui ( x )  aij u j aj x j

ij
im

ij
im
jn
j
mn
jn
j
mn
(5)
j
j
Sometimes it is necessary to switch the sign of the loading to make two coordinate systems
equivalent. If tension and compression properties are different, one must not switch the sign of
the load to obtain an equivalent coordinate system. To generalize equations (5), a factor  is
introduced.
  1 if load reversal is not required
  1 if load reversal is required
Equation (5) then becomes
d i
 b
x g
 a a  d
a x i
 b
x g
 a a  d
a x i
ui ( x )   aij u j aj x j

ij
im

ij
im
j
jn mn
jn
mn
j
(6)
j
j
The constraints that are implied by equations (6) are very useful in deriving the required
boundary conditions. This will be demonstrated later through examples.
The value of the factor  is easily determined from the loading, since only simple loading is
considered. In particular, the model is subjected to either macroscopically constant strain or
stress. For example, one might specify a volume-averaged value of  12 (with the other volumeaveraged stresses=0). The sign of  can be obtained by examining the effect of the coordinate
transformation on the sign of  12 . In fact, we can consider all six load cases at once. For
example, if the coordinate transformation is mirroring about the x1 =0 plane,
aij
1
L
M
M
0
M
N0
O
0P
which gives
P
1P
Q
0 0
1
0

L
M
M

M

N
11
aima jn mn
12
13
 12
 13
 22
 23
 23
 33
O
P
P
P
Q
Accordingly, the values of  are as follows for the six different load conditions for mirroring
about the x1 =0 plane.
Stress  11  22  33  12  23  13

1
1
1
1
1
1
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The values of  for six different coordinate transformations and the six load cases are
summarized below. The values for strain loading are the same as for stress loading.
Table 1 Values of  for various transformations and load cases.
Transformation
Load Case
 11  22  33  12  23  13
Mirror about x1 = 0
1
1
1
-1
1
-1
Mirror about x2 = 0
Mirror about x3 = 0
 rotation about x1
1
1
1
1
1
1
1
1
1
-1
1
-1
-1
-1
1
1
-1
-1
 rotation about x2
 rotation about x3
1
1
1
1
1
1
-1
1
-1
-1
1
-1
Boundary conditions for symmetrically stacked plain weave
If one is willing to analyze the entire unit cell, the periodic conditions for the displacements can
be used to specify the boundary conditions. However, to reduce the region to be analyzed
requires a bit more effort. The following describes a systematic way to go about accomplishing
this task. Although the focus is on the plain weave, the techniques are quite general. Below is a
schematic of a symmetrically stacked plain weave composite. As can be seen from the figure, the
planes xi  0 are planes of geometric symmetry. (uc_1.cdr and subcells.cdr)
The first step will be to develop relationships for the planes x1 a .
( vector of periodicity
d  2 a, 0,0 ) Substituting x1 
a into the periodic relations gives
b g b
g xu
 b
a, x , x g
 b
a , x , x g
 b
a, x , x g
 b
a , x , x g
ui a , x 2 , x 3  ui  a , x 2 , x 3 
i
2a
1
ij
ij
2
2
3
3
ij
2
ij
3
2
3
(7)
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Consider the equivalent coordinate system (ECS) x that is obtained by mirroring about x1 0 .
The direction cosines for this transformation are
aij
1
L
M
M
0
M
N0
O
0P
P
1P
Q
0 0
1
0
(8)
Substituting the a ij above into equations (6) yields the following relationships for displacements
and stresses,
u O
L
M
u P
M
P
M
u
NP
Q
(a , x
1
1
2
3

L
M

M
M

N
11
12
13
O
P
P
P
Q
u O
L
M
 M
u P
P
M
u
NP
Q
,x )
( a , x
2
2
3
3
L
M
M
M
N
 12  13
 11
 22  23
   12
 23  33 (a , x , x )
 13
2
3
2
,x )
3
 12
 13
 22
 23
 23
 33
O
P
P
P
Q
( a , x
(9)
2
,x )
3
Tractions will transform like the displacements and strains like the stresses, so the details are not
shown. These periodic and ECS conditions will now be combined … first for displacements and
then for the stresses.
Displacements
Combining the constraints in equations (7) and (9) yields:
L
u
M
x
M
M
u
M
x
M
M
u
M
Nx
1
u O
L
M
u P
M
P
M
u
NP
Q
( a , x
1
1

2
3
2
1
2
,x )
3
3
1
O
P
P
u O
L
P
M
2a   M
u P
P
P
P
M
u
NP
Q
( a , x , x )
P
2 3
P
Q
1
2
3
(10)
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L
u
M
x
M
Mu
M
x
M
M
u
M
Nx
1
u (1   ) O
L
M
u (1   ) P
M
P
M
u
(
1


)
N P
Q
( a , x
1
1
Rearranging, yields
2
,x )
2 3
3

2
1
3
1
O
P
P
P
2a
P
P
P
P
Q
(11)
Recall that the value of  depends on whether the loads must be reversed for the new coordinate
system to be equivalent. The constraints indicated by equation (11) are listed below. The ones
that provide useful information are boxed. (switch rows and columns?)
  1
 1

b
g
u1 a , x2 , x3  

u1
a
x1
u1
0
x1
(12)
b
g
u2
a
x1
b
g
u3
a
x1
u2
0
x1
u2  a , x 2 , x 3  
u 3
0
x1
u3  a , x 2 , x 3  
This can be expressed quite succinctly as follows
If   1 normal displacements on x1   a are equal. Because of periodicity, the normal
displacements on x1  a are also equal and opposite those on x1   a .
If   1 tangential displacements on x1   a are equal. Because of periodicity, the tangential
displacements in each direction on x1  a are also equal and opposite the corresponding
displacements on x1   a .
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Stresses
Now consider the constraints on the stresses. Combine equations (7) and (9) to obtain the
following useful relations (the remainder are trivially satisfied or do not affect the boundary
conditions.)
  1:
b
g
 b
a , x , x g
0
b
 12 a , x 2 , x 3  0
  1:
11
2
g
 13  a , x 2 , x 3  0
(13)
3
These can be summarized as follows
If 1
shear traction on x1 a is zero
If 
1
normal traction on x1 a is zero
Because of the symmetries in the microstructure, we will obtain analogous results for
x2   b and x 3   c .
<add summary of results thus far…in words>
Next we proceed to exploit symmetries within the cell to reduce the region which must be
modeled. Again, we will consider an ECS obtained by mirroring about the x1  0 plane.
However, this time we examine the displacements, etc., along the plane x1  0 . Equations 6
become
u O
L
M
u P
M
P
M
u
NP
Q
b
1
1
2
3
u O
L
M
M
u P
P
M
u
Q
g NP
b
11
and
2
0 , x2 , x3
3
0 , x2 , x3

L
M

M
M

N
g
12
13
O
P
P
P
Q
 12  13
 22  23
 23  33 b0, x
2 , x3

L
M
M


N
g M
11
12
13
 12
 13
 22
 23
 23
 33
O
P
P
P
Q
b
(14)
0 , x2 , x3
g
Therefore, on x1  0 we have the following constraints
 1
normal displacement = 0
tangential tractions = 0
 
1
tangential displacements = 0
normal tranctions = 0
(15)
If we repeat this exercise for the planes x 2  0 and x3  0 (which are also planes of
microstructural symmetry), we will obtain analogous results. We are now down to 18 of the
original unit cell. (See Figure subcells. cdr. show 1/8 unit cell with boundary conditions) The
plain weave in Figure _ is a balanced weave, ie. the warp and fill tows are identical. For such
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weaves the analysis region can be reduced to 1/32 of the unit cell. To simplify the derivation of
the boundary conditions, the coordinate system origin is shifted to the center of the 1/8 unit cell.
Consider a rotation about the x1 axis of 180 to obtain an ECS.
aij
1
L
M
M
0
M
N0
O
0P
P
1P
Q
0
0
1
0
(16)
Equation 6 from periodicAnalysis.doc yields (specialized for x 2  0 ).
u O
L
M
u P
M
P
M
u
NP
Q
b
1

L
M

M
M

N
1
2
3
u O
L
M
u P
M
P
M

u
Q
g N P
b
and
2
x1 , 0 , x3
3
O
P
P
P
Q
x1 , 0 ,  x3
12
g
L
M
M
M
N
 12  13
 11
 22  23
   12
 23  33 bx , 0, x g
 13
11
13
1
3
 12
 13
 22
 23
 23
 33
O
P
P
P
Q
b
(17)
x1 , 0 ,  x3
g
The results in the following displacement constraints
 1
u1 x1 , 0, x 3  u1 x1 , 0,  x 3
b g b
g
u b
x , 0, x g
 u b
x , 0,  x g
u b
x , 0, x g
 u b
x , 0,  x g
2
1
3
3
1
3
2
3
1
3
1
3
  1
u1 x1 , 0, x 3  u1 x1 , 0,  x 3
b g b
g
u b
x , 0, x g
u b
x , 0,  x g
ub
x , 0, x g
u b
x , 0,  x g
2
1
3
3
1
3
2
3
1
1
3
(18)
3
Note that constraints are derived for all three-displacement components. Now one can replace
the region x 2  0 with these constraints. The analysis region is now 1/16 of the unit cell.
Now consider a rotation about the x 2 axis of 180 to obtain an ECS and specialize equation 6 for
the plane x1  0 .
a ij
1
L
M
M
0
M
N0
O
P
P
P
Q
0 0
1 0
0 1
(19)
The result is
u O
L
M
u P
M
P
M
u
NP
Q
b
1
2
3
0 , x2 , x3
u O
L
M
M
u P
P
M

u
Q
g N P
b
1
2
3
0 , x2 ,  x3

L
M

and M

N
g M
11
12
13
O
L
P
M
P
M
P
QM
N
 12  13
 11
 22  23    12
 23  33
 13
 12
 22
 23
O
P
P
P
Q
 13
 23
 33 b0, x
2 ,  x3
(20)
g
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This results in the following displacement constraints
 1
u1 0, x 2 , x 3   u 1 0, x 2 ,  x 3
  1
u1 0, x 2 , x 3  u 1 0, x 2 ,  x 3
b g b
g
u b
0, x , x g
u b
0, x ,  x g
ub
0, x , x g
 u b
0, x ,  x g
2
2
3
2
3
3
2
2
3
3
2
3
b g b
g
u b
0, x , x g
 u b
0, x ,  x g
u b
0, x , x g
u b
0, x ,  x g
2
2
3
3
2
3
2
2
3
3
2
3
(21)
Again, one obtains constraints for all three-displacement components. Hence, the region x1  0
can be reduced with these constraints, leaving an analysis region of only 1/32 of the unit cell.
Boundary conditions for satin weave
Conclusion
Miscellaneous
 Need to show region which must be analyzed if tension and compression properties differ!
 mention need to have mesh generator put nodes in “right” place for MPC’s
 Find Dasgupta papers
1
b g
ba, x , x g 
ba, x , x g 
ba, x , x g
ba, x , x g 
ba, x , x g

1
b g
ba, x , x gNA
ba, x , x gNA
ba, x , x gU
ba, x , x gNA
ba, x , x gU
b g
ba, x , x g
ba, x , x g
ba, x , x g 
ba, x , x g
ba, x , x g 
b g
ba, x , x gNA
ba, x , x gNA
ba, x , x gE
ba, x , x gNA
ba, x , x gE
 11 
a, x2 , x3   11 a, x2 , x3 E
 11 
a, x2 , x3 
 11 a, x2 , x3 U
 22
 33
 12
 23
 13
2
3
22
2
3
 22
2
3
33
2
3
 33
2
3
12
2
3
 12
2
3
23
2
3
 23
2
3
13
2
3
 13
2
3
22
2
3
2
3
33
2
3
2
3
12
2
3
2
3
23
2
3
2
3
13
2
3
Warning: It is important to remember that the displacement gradient specified in the periodic
conditions is a volume-averaged value. Theses volume averaged values are the same throughout
the entire region!
It should be pointed out that the constraints need not be limited to planes which are perpendicular
to the axes. For example, consider the following configuration and assume the loading is
extension in the x1 direction. Rotation about the x1 axis of 180 degrees results in the following
relations for equivalent coordinate systems:
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u O
L
M
u P
M
P
M
Pb
u Q
N
1
2
3
x1 , x2 , x3
u O
L
M
M
u P
P
M
P

u
N
Q
g
b
1
2
3
x1 ,  x2 ,  x3
g
One could use this relation to impose constraints on the plane x 2  0 or x 3  0 . But it is equally
valid to use these constraints along the plane whose edge is indicated by the line AB. Figure!