EN224: Linear Elasticity

  

 





 

 

 

 





 

Division of Engineering

    Brown University

 

 

2D plane solutions for Anisotropic Elasticity

 

We now consider the more complex case of 2D deformation.  The formulation of anisotropic elasticity is still evolving – a recent contribution is Choi et al International Journal of Solids and Structures 40 (2003)1411 –1431.  There are strong indications that the solution techniques used today are not the best one – for example with today’s representations, coordinate transformations lead to very strange contortions in the solutions; and the invariance of quantities such as hydrostatic stress are not apparent in the algebraic solutions.  Moreover, there are some materials (including isotropic and transversely isotropic materials) for which the formulation blows up.  Issues such as completeness of representations need to be addressed.  Nevertheless, the Stroh formulation – presented below with some modifications - is the most commonly used version of anisotropic elasticity

 

 

Stroh representation of solutions

 

As always, we are looking for solutions to the Navier equation

We have already seen that strict plane strain deformation solutions  can only exist for special materials and orientations.   For a general anisotropic solid we must therefore relax the plane strain constraint, and seek generalized plane strain solutions of the form

 

The Stroh solution proceeds along the lines we followed to obtain anti-plane shear solutions.  We set  with p a complex number to be determined, and seek solutions of the form

where  is a vector to be determined.  We see that

whence the governing equation can be expressed as

This can be re-written as

or in matrix form as

where

 

The matrix equations have nontrivial solutions if

Since Q, R and T are 3x3 matrices, this is a sextic equation for p, with 6 roots.  For positive definite materials p is always complex, so the 6 roots are pairs of complex conjugates.  The corresponding eigenvectors  - which also play a central role in the following development – must also appear as pairs of complex conjugates.  

 

The most general solution to the displacement field may therefore be expressed as

where  are the three pairs of complex roots of the characteristic equation;  are the corresponding eigenvalues,  and  are analytic functions.

 

The original Stroh formulation, as extended and described in Ting’s book on anisotropic elasticity, uses the full expansion for u. However, since u must be real, not all the solutions obtained in this way are of interest.  Most known solutions have the form

where

and   use the eigenvalues  with positive real part, and  are the corresponding eigenvectors.  Although known solutions do seem to have this structure, to my knowledge the completeness of this representation has not been proved.

 

Stresses

 

The stresses can be obtained from the constitutive equation

where we recall that for each of the six characteristic solutions we may obtain displacements as , so that

where Q, R and T were defined earlier.  Define

and note that the governing equations require that

Therefore, for each member of the family of solutions, we can obtain stresses from

The symmetry of the stress tensor requires that

 

Ting simplifies the expression for stresses by defining a vector valued stress function

whence

but this is not particularly helpful.

 

The general expression for stresses then follows by summing the six separate eigensolutions:

 

Resultant force on an arc

 

The stress function is related to the resultant force exerted by tractions acting on an arc in the solid

The resultant force (per unit out of plane distance) acting on an arc AB due to tractions acting on the outward normal of the arc can be computed as

Matrix form of solutions

 

There are various matrix representations for displacements and stresses, as follows.  Following the procedure outlined above, we define

and define the characteristic equation for the system through

 are the three pairs of complex roots of the characteristic equation with positive and negative imaginary part, respectively;   are the corresponding eigenvalues, and let .  We then introduce normalized matrices of combinations of the eigenvectors as follows

In addition, define

 

Then

Most solutions have the form .  In this case, we may write

where

are the tractions acting on planes with normals in the  and  directions, but again I am not aware of a proof that this representation is complete.

 

Obviously, while the eigenvalues p are uniquely defined for a particular set of elastic constants, the eigenvectors  are not unique, since they may be multiplied by any arbitrary complex number and will remain eigenvectors.  It is sometimes helpful to normalize the eigenvectors so that the matrices A and B satisfy

but this can result in very cumbersome expressions for A and B (although it generally simplifies expressions for the solutions) and is not absolutely necessary. 

 

 

Eigenvalues and anisotropy matrices for cubic materials

 

Since 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 Maple 8 simply bombs.  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

whence

The characteristic equation therefore has the form

giving

whence

For  the eigenvalues are purely imaginary.  The special case  corresponds to an isotropic material.

 

The matrices A and B can be expressed as

 

These matrices have not been normalized to ensure that .  Instead

 

 

 

These quantities can also be expressed in terms of the engineering constants.  Recall that for cubic materials the constitutive equation can be expressed as

and in terms of , we have

and we may define an anisotropy factor .  For A=1 the material is isotropic.  In terms of these parameters

 

while

 

 

 

 

Degenerate Materials

 

There 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

The eigenvalues are repeated, and there only two independent eigenvectors a associated with the repeated eigenvalue , and so the representation of displacements and stress is not complete.

 

The 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 their stiffnesses.

 

Choi et al 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 to the problem.

 

In some cases 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.

Fundamental Elasticity Matrix

 

The vector  and corresponding eigenvector  can be shown to be the right eigenvectors and eigenvalues of an unsymmetric matrix known as the fundamental elasticity matrix, defined as

where the matrices

were introduced earlier.  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

and therefore

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 that

so

again, giving the required answer.

 

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.

 

 

 

Orthogonal properties of A and B

 

The 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

 

 

Barnett-Lothe tensors

 

From the preceding section, we observe that

Therefore  and  are pure imaginary, while the real part of .  The three matrices defined by

are therefore purely real.  These matrices will be shown to transform as tensors under a change of basis.  They are known as Barnett-Lothe tensors.

 

Many 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.  Consequently, these tensors can be calculated even for degenerate materials.

 

For cubic materials, with coordinate axes aligned with coordinate directions, one can show that

where

 

In terms of engineering constants  defined earlier, we can set

 

 

The Impedance Tensor

 

The impedance tensor M relates A and B through the definition

M or its inverse appears in the solution to many problems.  It can be expressed in terms of Barnett-Lothe tensors using formulas given in the next section, and so can be calculated for degenerate materials.

 

 

Useful relations between matrices in anisotropic elasticity

 

We collect below various useful algebraic relations between the various matrices that were introduced in the preceding sections.

 

Recall that 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

·         is Hermitian.  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.

·        The impedance 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.

 

 

Basis Changes

 

 

We obtain a number of additional useful results by finding basis change formulas for the various elasticity matrices defined in preceding sections.

 

To this end, consider a basis change that involves rotating the  axes through angle  about the  axis.

 

It is straightforward to show that

 

where

The transformation relations for elasticity matrices Q, R and T can be expressed as

where  are computed as

where ,  and  are the components (in the  basis) of the stiffness tensor and unit vectors parallel to   and , respectively. Alternatively

or

 

To see these, note that coordinates and displacements transform as vectors, so that

consequently

which directly gives the basis change formula for A.

 

To find the expression for p, we note that

so that   can thus both functions of    and .  Since   we conclude that

as required.

 

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

Substituting for  gives

and 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.

 

 

 

Barnett-Lothe integrals

 

The basis change formulas in the preceding section lead to a remarkable direct procedure for computing the Barnett-Lothe tensors directly, 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 parameters.

 

Specifically, we shall show that S, H, and L can be computed by integrating the sub-matrices of the fundamental elasticity matrix over orientation space, namely, let

and define

Then

 

 

To see this, we show first that  can be diagonalized as

where

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

whence

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.  Recall that

Integrating gives

(the sign of the integral is determined by Im(p) because the branch cut for  is taken to lie along the negative real axis).

 

Thus,