Analytical techniques and solutions for
linear elastic solids
5.5 Solutions to generalized plane
problems for anisotropic linear elastic solids
such as wood, laminated composites, and single crystal metals are stiffer when
loaded along some material directions than others. Such materials are said to be anisotropic, and cannot be modeled using
the procedures described in the preceding sections. In this chapter, we describe briefly the
most widely used method for calculating elastic deformation and stress in two
dimensional anisotropic solids. As you
might expect, these calculations are difficult, and while the solutions can be
expressed in a surprisingly compact form, the resulting expressions can usually
only be evaluated using a computer. In
many practical situations it is simplest to calculate solutions for anisotropic
materials using direct numerical computations (e.g. using the finite element
method, discussed in Chapters 7 and 8).
Nevertheless, analytical solutions are useful: for example, the finite
element method cannot easily be applied to problems involving cracks,
dislocations, or point forces, because they contain singularities; in addition
exact calculations can show how the solutions vary parametrically with elastic
constants and material orientation.
5.5.1 Governing Equations of elasticity
for anisotropic solids
A typical plane elasticity problem is illustrated in the
picture. The solid is two dimensional:
in this case we are concerned with plane strain solutions, which means that the
solid is very long in the direction, and every cross section is loaded
identically and only in the plane. The material is an anisotropic, linear elastic
solid, whose properties can be characterized by the elasticity tensor (or an equivalent matrix) as discussed in
simplify calculations, we shall assume that (i) The solid is free of body
forces; (ii) thermal strains can be neglected.
Under these conditions the general equations of elasticity listed in
Section 5.1.2 reduce to
to the usual boundary conditions. In
subsequent discussions, it will be convenient to write the equilibrium equations in matrix form as
Conditions necessary for strict plane
strain deformation of anisotropic solids.
For Plane strain deformations the displacement field has the form .
Under these conditions the equilibrium equations reduce to
In this case, can be chosen to satisfy two, but not all
three, of the three equations. The
elastic constants must satisfy . Consequently, the third equation can only be
satisfied by setting
Strict plane deformations
therefore only exist in a material with elastic constants and orientation
most common class of crystals cubic materials satisfies these conditions for appropriate orientations.
Generalized plane strain deformations. A generalized plane strain displacement
field can exist in any general anisotropic crystal. In this case the displacement field has the
the displacement is independent of position along the length of the cylindrical
solid, but points may move out of their original plane when the solid is
5.5.2 Stroh representation for fields in
The Stroh solution is a
compact, complex variable, representation for generalized plane strain
solutions to elastically anisotropic solids.
To write the solution, we need to define several new quantities:
We define three new
3x3 matrices of elastic constants, as follows
three complex valued eigenvalues (i=1…3)
and eigenvectors which satisfy
eigenvalues can be computed by solving the equation
Since Q, R and T are 3x3 matrices, this
is a sextic equation for p, with 6 roots. It is possible to show that for a material
with physically admissible elastic constants p is always complex, so the 6 roots are pairs of complex conjugates
. Each pair of complex roots has a
corresponding pair of complex valued eigenvectors , We define to be the roots with positive imaginary part,
and to be the corresponding eigenvector.
3. To calculate the stresses, it is helpful to introduce
three further vectors defined as
It is often convenient
to collect the eigenvectors and and the eigenvalues into matrices as follows
Note also that, as always, while the eigenvalues are
uniquely defined for a particular set of elastic constants, the eigenvectors (and consequently the vectors ) are not unique, since they may be multiplied
by any arbitrary complex number and will remain eigenvectors. It is helpful to normalize the eigenvectors
so that the matrices A and B satisfy
where I is
the identity matrix.
General representation of displacements: The displacement at a point in the solid is
are the three pairs of complex roots of the
characteristic equation; are the corresponding eigenvalues, and are analytic functions, which are analogous to
the complex potentials for isotropic solids.
General representation of stresses: The stresses can be expressed in terms of a vector
valued stress function (you can think of this as a generalized Airy
function) defined as
The stresses can be
calculated from the three components of as
Combined matrix representation for displacement and
stresses: The solution for the displacement field and
stress function can be expressed in the form
Simpler representation for stresses and
displacements: The solutions given
above are the most general form of the generalized plane strain solution to the
governing equations of linear elasticity.
However, not all the solutions of this form are of practical interest,
since the displacements and stresses must be real valued. In practice most solutions can be expressed
in a much simpler the form as
where Re(z) denotes the
real part of z,
5.5.3 Demonstration that the Stroh representation
satisfies the governing equations
Our first objective is to
show that a displacement field of the form ,
) are any one of the eigenvalues and eigenvectors defined in the preceding section, satisfy the
To see this,
Note that ,
where is the Kronecker delta. Therefore, it follows that
result into the governing equation shows that
This can be
in matrix form as
are the matrices defined in Section
5.5.2. The eigenvalue/eigenvector pairs
(p, ) satisfy this equation by definition, which
shows that the governing equation is indeed satisfied.
Our next objective is to
show that stresses can be computed from the formulas given in Section
5.5.2. To see this,
Note that the stresses
can be obtained from the constitutive equation
Recall that for
each of the six characteristic solutions we may obtain displacements as ,
Q, R and T are the matrices defined in the preceding section.
To simplify this
note that the governing equations require that
results of (2) and (3) shows that stresses can be computed from
that the stress function has components ,
and . Consequently, the stresses are related to the
stress function by as required.
5.5.4 Stroh eigenvalues and anisotropy
matrices for cubic materials
the eigenvalues p for a general
anisotropic material involve the solution to a sextic equation, an explicit
general solution cannot be found. Even
monoclinic materials (which have a single symmetry plane) give solutions that
are so cumbersome that many symbolic manipulation programs cannot handle them. The solution for cubic materials is
manageable, as long as one of the coordinate axes is parallel to the direction.
If the cube axes coincide with the coordinate directions, the elasticity
matrix reduces to
The characteristic equation
therefore has the form
For the eigenvalues are purely imaginary. The special case corresponds to an isotropic material.
The matrices A and B can be expressed as
5.5.5 Degenerate Materials
are some materials for which the general procedure outlined in the preceding
sections breaks down. We can illustrate
this by attempting to apply it to an isotropic material. In this case we find that ,
and there only two independent eigenvectors a associated with the repeated eigenvalue .
In addition, if you attempt to substitute material constants representing an
isotropic material into the formulas for A
and B given in the preceding
section you will find that the terms in the matrices are infinite.
physical significance of this degeneracy is not known. Although isotropic materials are degenerate,
isotropy does not appear to be a necessary condition for degeneracy, as fully
anisotropic materials may exhibit the same degeneracy for appropriate values of
T. Choi, H. Shin and Y. Y. Earmme, Int J. Solids Structures 40, (6) 1411-1431 (2003) have found a
way to re-write the complex variable formulation for isotropic materials into a
form that is identical in structure to the Stroh formulation. This approach is very useful, because it
enables us to solve problems involving interfaces between isotropic and
anisotropic materials, but it does not provide any fundamental insight into the
cause of degeneracy, nor does it provide a general fix for the problem.
many practical situations the problems associated with degeneracy can be
avoided by re-writing the solution in terms of special tensors (to be defined
below) which can be computed directly from the elastic constants, without
needing to determine A and B.
5.5.6 Fundamental Elasticity Matrix
vector and corresponding eigenvector can be shown to be the right eigenvectors and
eigenvalues of a real, unsymmetric matrix known as the fundamental elasticity matrix, defined as
where the matrices are the elasticity matrices defined in Section
5.5.2. Similarly, can be shown to be the left eigenvector of N.
To see this, note that the
expressions relating vectors a and b
can be expressed as
Since T is positive definite and symmetric its inverse can always be
computed. Therefore we may write
This is an eigenvalue
equation, and multiplying out the matrices gives the required result.
The second identity may be
proved in exactly the same way. Note
again, giving the required
For non-degenerate materials N
has six distinct eigenvectors. A matrix
of this kind is called simple. For some materials N has repeated eigenvalues, but still has six distinct
eigenvectors. A matrix of this kind is
called semi-simple. For degenerate materials N does not have six distinct eigenvectors. A matrix of this kind is called non semi-simple.
5.5.7 Orthogonal properties of Stroh
matrices A and B
observation that and are right and left eigenvectors of N has an important consequence. If the eigenvalues are distinct (i.e. the
material is not degenerate), the left and right eigenvectors of a matrix are
orthogonal. This implies that
In addition. the vectors
can always be normalized so that
If this is done, we see
that the matrices A and B must satisfy
Clearly the two matrices
are inverses of each other, and therefore we also have that
These results give the
following relations between A and B
5.5.8 Barnett-Lothe tensors and the
In this section we define
four important tensors that can be calculated from the Stroh matrices A and B. Specifically, we
Tensor with properties ( )
following relations between the Barnett-Lothe tensors and the impedance tensor
are also useful
solutions can be expressed in terms of S,
H and L directly, rather than in terms of A and B. In addition, Barnett and Lothe devised a
procedure for computing S, H and L without needing to calculate A
and B (See Sect. 5.5.11). Consequently, these tensors can be calculated
even for degenerate materials.
As an example, for cubic
materials, with coordinate axes aligned with coordinate directions,
5.5.9 Useful properties of matrices in
collect below various useful algebraic relations between the various matrices
that were introduced in the preceding sections.
definition, a matrix satisfying is Hermitian. A matrix satisfying is skew-Hermitian.
is skew Hermitian. To see this, note that the orthogonality
relations for A and B require that
This follows trivially from the preceding expression.
and are both Hermitian. To see this, note and use the preceding result.
The matrices are Hermitian.
To show the first expression, note that and recall that L is real. A similar
technique shows the second.
are both orthogonal matrices. To see this for the first matrix, note that ,
where we have used the orthogonality properties of B. A similar procedure shows
the second result.
tensors are real (i.e. they have zero imaginary part). To see this, note that the orthogonality of A and B (see sect. 5.5.7) implies that
and are pure imaginary, while the real part of .
tensor can be expressed in terms of the Barnett Lothe tensors as
To see the first result, note that and use the definitions of H and S. The second result follows
in the same way. Note that H, L
and S are all real, so this gives a
decomposition of M and its inverse
into real and imaginary parts. In
addition, since we can compute the Barnett-Lothe tensors for degenerate
materials, M can also be determined
without needing to compute A and B explicitly.
. To see these, note that M and its inverse are Hermitian, note that the imaginary part of a
Hermitian matrix is skew symmetric, and use the preceding result.
where . To see this, recall that the fundamental
elasticity tensor satisfies
second row of this equation is .
5.5.10 Basis Change formulas for
matrices used in anisotropic elasticity
The various tensors and matrices defined in the preceding
sections are all functions of the elastic constants for the material. Since the elastic constants depend on the
orientation of the material with respect to the coordinate axes, the matrices
are functions of the direction of the coordinate system.
and be two Cartesian bases, as indicated in the
denote the components of in ,
be the components of the elasticity tensor in ,
and let matrices Q, R and T be matrices
of elastic constants defined in Section 5.5.2.
Let denote any one of the three Stroh eigenvalues
and the matrices of Stroh eigenvectors, computed for the coordinate system ;
denote the Barnett-Lothe tensors and impedance
tensor in the basis;
let , ,
etc denote the various matrices and tensors in the basis.
addition, define rotation matrices as follows
The following alternative
expressions for are also useful
The basis change formulas
can then be expressed as
results can be derived as follow:
Note that the
displacements transform as vectors, so that . Consequently,
shows that anddirectly gives the basis change formula for
To find the
expression for p, we note that
we may write with and
The basis change
formulas for Q, R and T follow directly from the definitions
of these matrices.
The basis change
formula for B is a bit more
cumbersome. By definition
finally recalling that we obtain the required result.
The basis change
formulas for the Barnett-Lothe tensors and impedance tensor follow trivially
from their definitions. The basis change
formulas justify our earlier assertion that these quantities are tensors.
5.5.11 Barnett-Lothe integrals
basis change formulas in the preceding section lead to a remarkable direct
procedure for computing the Barnett-Lothe tensors, without needing to calculate
A and B. The significance of this
result is that, while A and B break down for degenerate materials, S, H,
and L are well-behaved. Consequently, if a solution can be expressed
in terms of these tensors, it can be computed for any combination of material
we shall show that S, H, and L can be computed by integrating the sub-matrices of the
fundamental elasticity matrix over orientation space, as follows. Let
see this, we show first that can be diagonalized as
and was defined earlier. From the preceding
section, we note that
which can be expressed as
as before, we can arrange
this into an Eigenvalue problem by writing
This shows that [a,b] are eigenvectors of the rotated
elasticity matrix. Following standard
procedure, we obtain the diagonalization stated.
Now, we examine more closely.
(the sign of the integral
is determined by Im(p) because the branch cut for is taken to lie along the negative real axis).
5.5.12 Stroh representation for a state
of uniform stress
uniform state of stress (with generalized plane strain deformation) provides a
very simple example of the Stroh representation. The solution can be expressed in several
different forms. Note that for a uniform
state of stress and corresponding strain we may write
In terms of these vectors
the Stroh representation is given by
or, in matrix form
see this, recall that a and b form eigenvectors of the fundamental
elasticity matrix N as
therefore we can write (for
each pair of eigenvectors/values)
and finally, defining
gives the required result.
5.5.13 Line load and Dislocation in an Infinite
figure illustrates the problem to be solved.
We consider an infinite, anisotropic, linear elastic solid, whose
elastic properties will be characterized using the Stroh matrices A and B.
solid contains a straight dislocation, with line direction ,
perpendicular to the plane of the figure.
The dislocation has Burger’s vector .
the same time, the solid is subjected to a line of force (with line direction
extending out of the plane of the figure). The force per unit length acting on
the solid will be denoted by .
displacement and stress function can be expressed in terms of the Stroh
where diag denotes a diagonal matrix,
The solution can also be expressed as
Derivation: We must show
that the solution satisfies the following conditions:
1. The displacement field for a dislocation with burgers
vector b must satisfy (this corresponds to taking a counterclockwise
Burgers circuit around the dislocation, as described in Section 5.3.4).
2. The resultant force exerted by the stresses acting on
any contour surrounding the point force must balance the external force F.
For example, taking a circular contour with radius r centered at the origin, we see that
We can create the
required solution using properties of log(z).
We try a solution of the form
is a vector to be determined. Recall
that we may write ,
whence . This, in turn, implies that .
orthogonality properties of A and B
can solve for q
5.5.14 Line load and dislocation below
the surface of an anisotropic half-space
figure shows an anisotropic, linear elastic half-space. The elastic properties of the solid are
characterized by the Stroh matrices A,
B and P defined in Section 5.5.2.
The solid contains a dislocation with Burgers vector b and is also subjected to a linear load with force per unit length
F at a point ,
while the surface of the solid is traction free.
The solution can be
computed from the simplified Stroh representation
The first term in the
expression for f will be recognized
as the solution for a dislocation and point force in an infinite solid; the
second term corrects this solution for the presence of the free surface.