Chapter 5

 

Analytical techniques and solutions for linear elastic solids

 

 

 

5.5 Solutions to generalized plane problems for anisotropic linear elastic solids

 

Materials 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 $e_3$ direction, and every cross section is loaded identically and only in the $\{e_1,e_2\}$ plane.   The material is an anisotropic, linear elastic solid, whose properties can be characterized by the elasticity tensor $C_{ijkl}$ (or an equivalent matrix) as discussed in Chapter 3.

 

To 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

$C_{ijkl}\frac{{\partial }^2{u}_k}{\partial x_i\partial x_l}=0$

subject to the usual boundary conditions.  In subsequent discussions, it will be convenient to write the  equilibrium equations in matrix form as

$\left[\begin{array}{cccccc}\frac{\partial }{\partial x_1}& 0& 0& \frac{\partial }{\partial x_2}& \frac{\partial }{\partial x_3}& 0\\ 0& \frac{\partial }{\partial x_2}& 0& \frac{\partial }{\partial x_1}& 0& \frac{\partial }{\partial x_3}\\ 0& 0& \frac{\partial }{\partial x_3}& 0& \frac{\partial }{\partial x_1}& \frac{\partial }{\partial x_2}\end{array}\right]\left[\begin{array}{cccccc}{c}_{11}& c_{12}& c_{13}& c_{14}& c_{15}& c_{16}\\ c_{12}& c_{22}& c_{23}& c_{24}& c_{25}& c_{26}\\ c_{13}& c_{23}& c_{33}& c_{34}& c_{35}& c_{36}\\ c_{14}& c_{24}& c_{34}& c_{44}& c_{45}& c_{46}\\ c_{15}& c_{25}& c_{35}& c_{45}& c_{55}& c_{56}\\ c_{16}& c_{26}& c_{36}& c_{46}& c_{56}& c_{66}\end{array}\right]\left[\begin{array}{c}\partial u_1/\partial x_1\\ \partial u_2/\partial x_2\\ \partial u_3/\partial x_3\\ \partial u_2/\partial x_3+\partial u_3/\partial x_2\\ \partial u_1/\partial x_3+\partial u_3/\partial x_1\\ \partial u_1/\partial x_2+\partial u_2/\partial x_1\end{array}\right]=0$

 

 

Conditions necessary for strict plane strain deformation of anisotropic solids.  For Plane strain deformations the displacement field has the form $u=u_1(x_1,x_2)e_1+u_2(x_1,x_2)e_2$. Under these conditions the equilibrium equations reduce to

$\begin{array}{l}{c}_{11}\frac{{\partial }^2{u}_1}{\partial x_1^2}+c_{66}\frac{{\partial }^2{u}_1}{\partial x_2^2}+2c_{16}\frac{{\partial }^2{u}_1}{\partial x_1\partial x_2}+c_{16}\frac{{\partial }^2{u}_2}{\partial x_1^2}+c_{26}\frac{{\partial }^2{u}_2}{\partial x_2^2}+(c_{12}+c_{66})\frac{{\partial }^2{u}_2}{\partial x_1\partial x_2}=0\\ c_{16}\frac{{\partial }^2{u}_1}{\partial x_1^2}+c_{26}\frac{{\partial }^2{u}_1}{\partial x_2^2}+(c_{66}+c_{12})\frac{{\partial }^2{u}_1}{\partial x_1\partial x_2}+c_{66}\frac{{\partial }^2{u}_2}{\partial x_1^2}+c_{22}\frac{{\partial }^2{u}_2}{\partial x_2^2}+2c_{26}\frac{{\partial }^2{u}_2}{\partial x_1\partial x_2}=0\\ c_{15}\frac{{\partial }^2{u}_1}{\partial x_1^2}+c_{46}\frac{{\partial }^2{u}_1}{\partial x_2^2}+(c_{56}+c_{14})\frac{{\partial }^2{u}_1}{\partial x_1\partial x_2}+c_{56}\frac{{\partial }^2{u}_2}{\partial x_1^2}+c_{24}\frac{{\partial }^2{u}_2}{\partial x_2^2}+(c_{25}+c_{46})\frac{{\partial }^2{u}_2}{\partial x_1\partial x_2}=0\end{array}$

In this case, $u_{\alpha \beta }$ can be chosen to satisfy two, but not all three, of the three equations.  The elastic constants must satisfy $c_{11}>0,c_{22}>0,c_{66}>0$.  Consequently, the third equation can only be satisfied by setting

$c_{15}=c_{46}=c_{14}=c_{56}=c_{24}=c_{25}=0$

Strict plane deformations therefore only exist in a material with elastic constants and orientation satisfying

$\left[\begin{array}{cccccc}{c}_{11}& c_{12}& c_{13}& 0& 0& c_{16}\\ c_{12}& c_{22}& c_{23}& 0& 0& c_{26}\\ c_{13}& c_{23}& c_{33}& c_{34}& c_{35}& c_{36}\\ 0& 0& c_{34}& c_{44}& c_{45}& 0\\ 0& 0& c_{35}& c_{45}& c_{55}& 0\\ c_{16}& c_{26}& c_{36}& 0& 0& c_{66}\end{array}\right]$

The 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 form

$u=u_1(x_1,x_2)e_1+u_2(x_1,x_2)e_2+u_3(x_1,x_2)e_3$

i.e. 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 loaded.

 

 

 

5.5.2 Stroh representation for fields in anisotropic solids

 

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:

1.      We define three new 3x3 matrices of elastic constants, as follows

$Q=\left[\begin{array}{ccc}{c}_{11}& c_{16}& c_{15}\\ c_{16}& c_{66}& c_{56}\\ c_{15}& c_{56}& c_{55}\end{array}\right]R=\left[\begin{array}{ccc}{c}_{16}& c_{12}& c_{14}\\ c_{66}& c_{26}& c_{46}\\ c_{56}& c_{25}& c_{45}\end{array}\right]T=\left[\begin{array}{ccc}{c}_{66}& c_{26}& c_{46}\\ c_{26}& c_{22}& c_{24}\\ c_{46}& c_{24}& c_{44}\end{array}\right]$

2.      We introduce three complex valued eigenvalues $p_i$ (i=1…3) and eigenvectors $a^{(i)}$ which satisfy

$\left[Q+p(R+R^T)+p^{2}T\right]a^{(i)}=0$

The eigenvalues can be computed by solving the equation

$\det \left[Q+p(R+R^T)+p^{2}T\right]=0$

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 $(p,\overline{p})$.  Each pair of complex roots has a corresponding pair of complex valued eigenvectors $(a,\overline{a})$,   We define $p_i$ to be the roots with positive imaginary part, and $a^{(i)}$ to be the corresponding eigenvector.

3.      To calculate the stresses, it is helpful to introduce three further vectors $b^{(i)}$ defined as

$\left[R^T+p_{i}T\right]a^{(i)}=b^{(i)}$

4.      It is often convenient to collect the eigenvectors $a^{(i)}$ and $b^{(i)}$ and the eigenvalues $p_i$ into matrices $A,B,P$ as follows

$A\equiv \left[\begin{array}{ccc}{a}_1^{(1)}& a_1^{(2)}& a_1^{(3)}\\ a_2^{(1)}& a_2^{(2)}& a_2^{(3)}\\ a_3^{(1)}& a_3^{(2)}& a_3^{(3)}\end{array}\right]B\equiv \left[\begin{array}{ccc}{b}_1^{(1)}& b_1^{(2)}& b_1^{(3)}\\ b_2^{(1)}& b_2^{(2)}& b_2^{(3)}\\ b_3^{(1)}& b_3^{(2)}& b_3^{(3)}\end{array}\right]P=\left[\begin{array}{ccc}{p}_1& 0& 0\\ 0& p_2& 0\\ 0& 0& p_3\end{array}\right]$

Note also that, as always, while the eigenvalues $p_i$ are uniquely defined for a particular set of elastic constants, the eigenvectors $a^{(i)}$ (and consequently the vectors $b^{(i)}$ ) 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

$B^{T}A+A^{T}B=I{\overline{B}}^T\overline{A}+{\overline{A}}^T\overline{B}=I$

where I is the identity matrix.

 

 

General representation of displacements: The displacement $u=u_1(x_1,x_2)e_1+u_2(x_1,x_2)e_2+u_3(x_1,x_2)e_3$ at a point $(x_1,x_2)$ in the solid is

$u(x_1,x_2)={\displaystyle \sum _{i=1}^3{a}^{(i)}{f}_i(z_i)+}{\overline{a}}^{(i)}\overline{g_i({\overline{z}}_i)}$

where $p_i,{\overline{p}}_i$ are the three pairs of complex roots of the characteristic equation; $a^{(i)}$ are the corresponding eigenvalues, $z_i=x_1+p_i{x}_2$ and $f(z),g(z)$ are analytic functions, which are analogous to the complex potentials $\Omega (z),\omega \left(z\right)$ for isotropic solids.

General representation of stresses: The stresses can be expressed in terms of a vector valued stress function $\phi$ (you can think of this as a generalized Airy function) defined as

$\phi ={\displaystyle \sum _{i=1}^3{b}^{(i)}{f}_i(z_i)+}{\overline{b}}^{(i)}\overline{g_i({\overline{z}}_i)}$

The stresses can be calculated from the three components of $\phi$ as

${\sigma }_{i1}=-\frac{\partial {\varphi }_i}{\partial x_2}{\sigma }_{i2}=\frac{\partial {\varphi }_i}{\partial x_1}$

 

Combined matrix representation for displacement and stresses:  The solution for the displacement field and stress function can be expressed in the form

$\left[\begin{array}{c}u\\ \phi \end{array}\right]=\left[\begin{array}{cc}A& \overline{A}\\ B& \overline{B}\end{array}\right]\left[\begin{array}{c}f\\ g\end{array}\right]$

where $g\equiv {[\overline{g_1({\overline{z}}_1)},\overline{g_2({\overline{z}}_2)},\overline{g_3({\overline{z}}_3)}]}^T$

 

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

$u=2\mathrm{Re}(Af)\phi =2\mathrm{Re}(Bf)t_2=2\mathrm{Re}(Bf\text{'})t_1=-2\mathrm{Re}(BPf\text{'})$

where Re(z) denotes the real part of z,

$t_1=\left[\begin{array}{c}{\sigma }_{11}\\ {\sigma }_{21}\\ {\sigma }_{31}\end{array}\right]t_2=\left[\begin{array}{c}{\sigma }_{12}\\ {\sigma }_{22}\\ {\sigma }_{32}\end{array}\right]f=\left[\begin{array}{c}{f}_1(z_1)\\ f_2(z_2)\\ f_3(z_3)\end{array}\right]f\text{'}=\left[\begin{array}{c}{f^{\prime }}_1(z_1)\\ {f^{\prime }}_2(z_2)\\ {f^{\prime }}_3(z_3)\end{array}\right]$

and $f\text{'}(z)\equiv \partial f/\partial z$.

 

 

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 $u_i=a_{i}f(z)$, with $z=x_1+p{x}_2$, and  (p, $a_i$ ) are any one of the eigenvalues $p_i$ and eigenvectors $a^{(i)}$ defined in the preceding section, satisfy the governing equations

$C_{ijkl}\frac{{\partial }^2{u}_k}{\partial x_i\partial x_l}=0$

To see this,

1.      Note that  $\partial z/\partial x_i={\delta }_{i1}+p{\delta }_{i2}$, where ${\delta }_{ij}$ is the Kronecker delta.  Therefore, it follows that

$\frac{\partial u_k}{\partial x_l}=a_k({\delta }_{l1}+p{\delta }_{l2})f\text{'}(z)\frac{\partial u_k}{\partial x_{lj}}=a_k({\delta }_{l1}+p{\delta }_{l2})({\delta }_{j1}+p{\delta }_{j2})f\text{'}\text{'}(z)$

2.      Substituting this result into the governing equation shows that

$C_{ijkl}({\delta }_{l1}+p{\delta }_{l2})({\delta }_{j1}+p{\delta }_{j2})a_{k}f\text{'}\text{'}(z)=0$

3.      This can be re-written as

$\left(C_{i1k1}+p(C_{i1k2}+C_{i2k1})+p^2{C}_{i2k2}\right)a_k=0$

or in matrix form as

$\left[Q+p(R+R^T)+p^{2}T\right]a=0$

where $Q,R,T$ are the matrices defined in Section 5.5.2.  The eigenvalue/eigenvector pairs (p, $a$ ) 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,

1.      Note that the stresses can be obtained from the constitutive equation ${\sigma }_{ij}=C_{ijkl}\frac{\partial u_k}{\partial x_l}$

2.      Recall that for each of the six characteristic solutions we may obtain displacements as $\partial u_k/\partial x_l=({\delta }_{l1}+p{\delta }_{l2})a_{k}f\text{'}(z)$, so that

$\begin{array}{l}{\sigma }_{i1}=\left(C_{i1k1}+p{C}_{i1k2}\right)a_{k}f\text{'}(z)=[Q_{ik}+p{R}_{ik}]a_{k}f\text{'}(z)\\ {\sigma }_{i2}=\left(C_{i2k1}+p{C}_{i2k2}\right)a_{k}f\text{'}(z)=[R_{ki}+p{T}_{ik}]a_{k}f\text{'}(z)\end{array}$

where Q, R and T are the matrices defined in the preceding section.

3.      To simplify this result, define

$\left[R^T+pT\right]a=b\left[Q+pR\right]a=c$

and note that the governing equations require that

$\left[Q+p(R+R^T)+p^{2}T\right]a=c+pb=0$

4.      Combining the results of (2) and (3) shows that stresses can be computed from

${\sigma }_{i1}=-p{b}_{i}f\text{'}(z){\sigma }_{i2}=b_{i}f\text{'}(z)$

5.      Finally, recall that the stress function $\phi$ has components ${\varphi }_i=b_{i}f(z)$, and $\partial z/\partial x_i={\delta }_{i1}+p{\delta }_{i2}$.  Consequently, the stresses are related to the stress function by ${\sigma }_{i1}=-\partial {\varphi }_i/\partial x_2{\sigma }_{i2}=\partial {\varphi }_i/\partial x_1$ as required.

 

 

 

5.5.4 Stroh 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 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 $x_3$ direction.  If the cube axes coincide with the coordinate directions, the elasticity matrix reduces to

$\left[\begin{array}{cccccc}{c}_{11}& c_{12}& c_{12}& 0& 0& 0\\ c_{12}& c_{11}& c_{12}& 0& 0& 0\\ c_{12}& c_{12}& c_{11}& 0& 0& 0\\ 0& 0& 0& c_{44}& 0& 0\\ 0& 0& 0& 0& c_{44}& 0\\ 0& 0& 0& 0& 0& c_{44}\end{array}\right]$

whence

$Q=\left[\begin{array}{ccc}{c}_{11}& 0& 0\\ 0& c_{44}& 0\\ 0& 0& c_{44}\end{array}\right]R=\left[\begin{array}{ccc}0& c_{12}& 0\\ c_{44}& 0& 0\\ 0& 0& 0\end{array}\right]T=\left[\begin{array}{ccc}{c}_{44}& 0& 0\\ 0& c_{11}& 0\\ 0& 0& c_{44}\end{array}\right]$

The characteristic equation therefore has the form

$\det \left[\begin{array}{ccc}{c}_{11}+p^2{c}_{44}& p\left(c_{12}+c_{44}\right)& 0\\ p\left(c_{12}+c_{44}\right)& c_{44}+c_{11}{p}^2& 0\\ 0& 0& c_{44}(1+p^2)\end{array}\right]=0$

giving

$(1+p^2)(p^4+\eta p^2+1)=0\eta =\frac{c_{11}^2-c_{12}^2-2c_{12}{c}_{44}}{c_{11}{c}_{44}}$

whence

$p_1=\frac{1}{2}\left(\sqrt{2-\eta }+i\sqrt{2+\eta }\right)p_2=\frac{1}{2}\left(-\sqrt{2-\eta }+i\sqrt{2+\eta }\right)p_3=i$

For $\eta >2$ the eigenvalues are purely imaginary.  The special case $\eta =2$ corresponds to an isotropic material.

 

The matrices A and B can be expressed as

$\begin{array}{l}A=\left[\begin{array}{ccc}-p_1(c_{12}+c_{44})/{\beta }_1& -p_2(c_{12}+c_{44})/{\beta }_2& 0\\ (c_{11}+p_1^2{c}_{44})/{\beta }_1& (c_{11}+p_2^2{c}_{44})/{\beta }_2& 0\\ 0& 0& (1-i)/2\end{array}\right]\begin{array}{c}{\beta }_i=\sqrt{2c_{11}{p}_i\left({(c_{11}+p_i^2{c}_{44})}^2-{(c_{12}+c_{44})}^2\right)}\\ \end{array}\\ B=\left[\begin{array}{ccc}{c}_{44}(c_{11}-c_{12}{p}_1^2)/{\beta }_1& c_{44}(c_{11}-c_{12}{p}_2^2)/{\beta }_2& 0\\ p_1(c_{11}{c}_{44}{p}_1^2+c_{11}^2-c_{12}^2-c_{12}{c}_{44})/{\beta }_1& p_2(c_{11}{c}_{44}{p}_2^2+c_{11}^2-c_{12}^2-c_{12}{c}_{44})/{\beta }_2& 0\\ 0& 0& (1+i)/2\end{array}\right]\end{array}$

 

 

 

5.5.5 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 $p_1=p_2=p_3=i$, and there only two independent eigenvectors a associated with the repeated eigenvalue $p_i$. 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.

 

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.

 

S. 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.

 

In 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

 

The vector $[a^{(i)},b^{(i)}]$ and corresponding eigenvector $p_i$ can be shown to be the right eigenvectors and eigenvalues of a real, unsymmetric matrix known as the fundamental elasticity matrix, defined as

$N=\left[\begin{array}{cc}{N}_1& N_2\\ N_3& N_1^T\end{array}\right]N_1=-T^{-1}{R}^T{N}_2=T^{-1}{N}_3=R{T}^{-1}{R}^T-Q$

where the matrices $Q,R,T$ are the elasticity matrices defined in Section 5.5.2.  Similarly, $[b^{(i)},a^{(i)}]$ can be shown to be the left eigenvector of N.

 

To see this, note that the expressions relating vectors a and b

$\left[R^T+pT\right]a=b\left[Q+pR\right]a=-pb$

can be expressed as

$\left[\begin{array}{cc}-Q& 0\\ -R^T& I\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=p\left[\begin{array}{cc}R& I\\ T& 0\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]$

Since T is positive definite and symmetric its inverse can always be computed.  Therefore we may write

$\left[\begin{array}{cc}0& T^{-1}\\ I& -R{T}^{-1}\end{array}\right]\left[\begin{array}{cc}R& I\\ T& 0\end{array}\right]=\left[\begin{array}{cc}I& 0\\ 0& I\end{array}\right]$

and therefore

$\left[\begin{array}{cc}0& T^{-1}\\ I& -R{T}^{-1}\end{array}\right]\left[\begin{array}{cc}-Q& 0\\ -R^T& I\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=p\left[\begin{array}{c}a\\ b\end{array}\right]$

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

$\begin{array}{l}\left[\begin{array}{cc}b& a\end{array}\right]\left[\begin{array}{cc}0& I\\ -Q& -R\end{array}\right]=p\left[\begin{array}{cc}b& a\end{array}\right]\left[\begin{array}{cc}I& 0\\ R^T& T\end{array}\right]\\ \left[\begin{array}{cc}I& 0\\ R^T& T\end{array}\right]\left[\begin{array}{cc}I& 0\\ -T^{-1}{R}^T& T^{-1}\end{array}\right]=\left[\begin{array}{cc}I& 0\\ 0& I\end{array}\right]\end{array}$

so

$\left[\begin{array}{cc}b& a\end{array}\right]\left[\begin{array}{cc}0& I\\ -Q& -R\end{array}\right]\left[\begin{array}{cc}I& 0\\ -T^{-1}{R}^T& T^{-1}\end{array}\right]=p\left[\begin{array}{cc}b& a\end{array}\right]$

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.

 

 

 

5.5.7 Orthogonal properties of Stroh matrices A and B

 

The observation that $[a^{(i)},b^{(i)}]$ and $[b^{(i)},a^{(i)}]$ 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

$\begin{array}{l}[b^{(i)},a^{(i)}]\left[\begin{array}{c}{a}^{(j)}\\ b^{(j)}\end{array}\right]=0i\ne j\\ [b^{(i)},a^{(i)}]\left[\begin{array}{c}{\overline{a}}^{(j)}\\ {\overline{b}}^{(j)}\end{array}\right]=[{\overline{b}}^{(i)},{\overline{a}}^{(i)}]\left[\begin{array}{c}{a}^{(j)}\\ b^{(j)}\end{array}\right]=0\end{array}$

In addition. the vectors can always be normalized so that

$[b^{(i)},a^{(i)}]\left[\begin{array}{c}{a}^{(i)}\\ b^{(i)}\end{array}\right]=1$

 

If this is done, we see that the matrices A and B must satisfy

$\left[\begin{array}{cc}{B}^T& A^T\\ {\overline{B}}^T& {\overline{A}}^T\end{array}\right]\left[\begin{array}{cc}A& \overline{A}\\ B& \overline{B}\end{array}\right]=\left[\begin{array}{cc}I& 0\\ 0& I\end{array}\right]$

Clearly the two matrices are inverses of each other, and therefore we also have that

$\left[\begin{array}{cc}A& \overline{A}\\ B& \overline{B}\end{array}\right]\left[\begin{array}{cc}{B}^T& A^T\\ {\overline{B}}^T& {\overline{A}}^T\end{array}\right]=\left[\begin{array}{cc}I& 0\\ 0& I\end{array}\right]$

These results give the following relations between A and B

$\begin{array}{l}{B}^{T}A+A^{T}B={\overline{B}}^T\overline{A}+{\overline{A}}^T\overline{B}=A{B}^T+\overline{A}{\overline{B}}^T=B{A}^T+\overline{B}{\overline{A}}^T=I\\ B^T\overline{A}+A^T\overline{B}={\overline{B}}^{T}A+{\overline{A}}^{T}B=A{A}^T+\overline{A}{\overline{A}}^T=B{B}^T+\overline{B}{\overline{B}}^T=0\end{array}$

 

 

 

5.5.8 Barnett-Lothe tensors and the Impedance Tensor.

 

In this section we define four important tensors that can be calculated from the Stroh matrices A and B.  Specifically, we introduce:

The Barnett-Lothe tensors $S=i(2A{B}^T-I)H=2iA{A}^{T}L=-2iB{B}^T$

The Impedance Tensor with properties $M=-iB{A}^{-1}{M}^{-1}=iA{B}^{-1}B=iMA$ ( $i=\sqrt{-1}$ )

The following relations between the Barnett-Lothe tensors and the impedance tensor are also useful

$M=-iB{A}^{-1}=H^{-1}+i{H}^{-1}S{M}^{-1}=iA{B}^{-1}=L^{-1}+iS{L}^{-1}$

 

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 (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,

 $M=\left[\begin{array}{ccc}\gamma & \frac{-i{c}_{44}(c_{11}-c_{12})}{c_{11}+c_{44}}& 0\\ \frac{i{c}_{44}(c_{11}-c_{12})}{c_{11}+c_{44}}& \gamma & 0\\ 0& 0& 1\end{array}\right]\gamma =\frac{\sqrt{c_{11}{c}_{44}(c_{11}-c_{12})(c_{11}+2c_{44}+c_{12})}}{c_{11}+c_{44}}$

 

 

 

5.5.9 Useful properties of matrices in anisotropic elasticity

 

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

 

By definition, a matrix $M$ satisfying $M^T=\overline{M}$ is Hermitian.  A matrix satisfying $M^T=-\overline{M}$ is skew-Hermitian.

 

·         ${\overline{B}}^{T}A$ is skew Hermitian.  To see this, note that the orthogonality relations for A and B require that ${\overline{B}}^{T}A+{\overline{A}}^{T}B=0$

·         $i{\overline{B}}^{T}A$ is Hermitian.  This follows trivially from the preceding expression.

·         $M$ and $M^{-1}$ are both Hermitian. To see this, note $M=B{(-i{\overline{B}}^{T}A)}^{-1}{\overline{B}}^T{M}^{-1}=A{(i{\overline{A}}^{T}B)}^{-1}{\overline{A}}^T$ and use the preceding result.

·         The matrices $i{B}^{-1}\overline{B}-i{A}^{-1}\overline{A}$ are Hermitian.  To show the first expression, note that $i{B}^{-1}\overline{B}=i{B}^T{(B{B}^T)}^{-1}\overline{B}=2B^T{L}^{-1}\overline{B}$ and recall that L is real.  A similar technique shows the second.

·         $i{B}^{-1}\overline{B}-i{A}^{-1}\overline{A}$ are both orthogonal matrices.  To see this for the first matrix, note that $i{B}^{-1}\overline{B}{\left(i{B}^{-1}\overline{B}\right)}^T=-B^{-1}\overline{B}{\overline{B}}^T{B}^{-T}=B^{-1}B{B}^T{B}^{-T}=I$, where we have used the orthogonality properties of B.  A similar procedure shows the second result.

·         The Barnett-Lothe 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

$A{B}^T+\overline{A}{\overline{B}}^T=IA{A}^T+\overline{A}{\overline{A}}^T=B{B}^T+\overline{B}{\overline{B}}^T=0$

Therefore $A{A}^T$ and $B{B}^T$ are pure imaginary, while the real part of $A{B}^T=1/2I$.   

·         The impedance tensor can be expressed in terms of the Barnett Lothe tensors as

$M=-iB{A}^{-1}=H^{-1}+i{H}^{-1}S{M}^{-1}=iA{B}^{-1}=L^{-1}+iS{L}^{-1}$

To see the first result, note that $B{A}^{-1}={(A{B}^T)}^T{(A{A}^T)}^{-1}$ 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.

·         $H^{-1}S+S^T{H}^{-1}=0S{L}^{-1}+L^{-1}{S}^T=0$.  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.

·         $BP=N_{3}A+N_1^{T}B$, where $P=\mathrm{diag}(p_1,p_2,p_3)$.  To see this, recall that the fundamental elasticity tensor satisfies

$\left[\begin{array}{cc}{N}_1& N_2\\ N_3& N_1^T\end{array}\right]\left[\begin{array}{c}{a}^{(i)}\\ b^{(i)}\end{array}\right]=p_i\left[\begin{array}{c}{a}^{(i)}\\ b^{(i)}\end{array}\right]\Rightarrow \left[\begin{array}{cc}{N}_1& N_2\\ N_3& N_1^T\end{array}\right]\left[\begin{array}{c}A\\ B\end{array}\right]=\left[\begin{array}{cc}A& 0\\ 0& B\end{array}\right]\left[\begin{array}{c}P\\ P\end{array}\right]$

The second row of this equation is $N_{3}A+N_1^{T}B=BP$.

 

 

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.  

 

To this end:

1.      Let $\{e_1,e_2,e_3\}$ and $\{{\widehat{e}}_1,{\widehat{e}}_2,{\widehat{e}}_3\}$ be two Cartesian bases, as indicated in the figure.

2.      Let $n_i,m_i$ denote the components of ${\widehat{e}}_1,{\widehat{e}}_2$ in $\{e_1,e_2,e_3\}$, i.e. ${\widehat{e}}_1=n_i{e}_i$   ${\widehat{e}}_2=m_i{e}_i$

3.      Let $C_{ijkl}$ be the components of the elasticity tensor in $\{e_1,e_2,e_3\}$, and let matrices Q, R and T be matrices of elastic constants defined in Section 5.5.2.

4.      Let  $p,A,B$ denote any one of the three Stroh eigenvalues and the matrices of Stroh eigenvectors, computed for the coordinate system $\{e_1,e_2,e_3\}$;

5.      Let $S,H,L,M$ denote the Barnett-Lothe tensors and impedance tensor in the $\{e_1,e_2,e_3\}$ basis;

6.      Similarly, let $\widehat{Q},\widehat{R},\widehat{T}$,  $\widehat{p},\widehat{A},\widehat{B}$, etc denote the various matrices and tensors in the $\{{\widehat{e}}_1,{\widehat{e}}_2,{\widehat{e}}_3\}$ basis.

 

In addition, define rotation matrices $\Omega ,$ $Q(\theta ),R(\theta ),T(\theta )$ as follows

$\Omega \equiv \left[\begin{array}{ccc}\cos \theta & \sin \theta & 0\\ -\sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right]$     $\left[\begin{array}{cc}Q(\theta )& R(\theta )\\ R^T(\theta )& T(\theta )\end{array}\right]=\left[\begin{array}{cc}\cos \theta I& \sin \theta I\\ -\sin \theta I& \cos \theta I\end{array}\right]\left[\begin{array}{cc}Q& R\\ R^T& T\end{array}\right]\left[\begin{array}{cc}\cos \theta I& -\sin \theta I\\ \sin \theta I& \cos \theta I\end{array}\right]$

The following alternative expressions for $Q(\theta ),R(\theta ),T(\theta )$ are also useful

$Q_{ij}(\theta )=C_{ikjl}{n}_k{n}_l{R}_{ij}(\theta )=C_{ikjl}{n}_k{m}_l{T}_{ij}(\theta )=C_{ikjl}{m}_k{m}_l$

$\begin{array}{l}Q(\theta )=Q{\cos }^2\theta +(R+R^T)\sin \theta \cos \theta +T{\sin }^2\theta \\ R(\theta )=R{\cos }^2\theta +(T-Q)\sin \theta \cos \theta -R^T{\sin }^2\theta \\ T(\theta )=T{\cos }^2\theta -(R+R^T)\sin \theta \cos \theta +Q{\sin }^2\theta \end{array}$

 

The basis change formulas can then be expressed as

$\begin{array}{l}\widehat{p}\equiv p(\theta )=\frac{p\cos \theta -\sin \theta }{p\sin \theta +\cos \theta }\\ \widehat{A}=\Omega A\widehat{B}=\Omega B\\ \widehat{Q}=\Omega Q(\theta ){\Omega }^T\widehat{R}=\Omega R(\theta ){\Omega }^T\widehat{T}=\Omega T(\theta ){\Omega }^T\\ \widehat{S}=\Omega S{\Omega }^T\widehat{H}=\Omega H{\Omega }^T\widehat{L}=\Omega L{\Omega }^T\widehat{M}=\Omega M{\Omega }^T\end{array}$

 

 

Derivation: These results can be derived as follow:

1.      Note that the displacements transform as vectors, so that $\widehat{u}=\Omega u$.  Consequently,

$\widehat{u}={\displaystyle \sum _{i=1}^3{\widehat{a}}^{(i)}{f}_i(z_i)+}{\overline{\widehat{a}}}^{(i)}\overline{g_i({\overline{z}}_i)}=\Omega \left({\displaystyle \sum _{i=1}^3{a}^{(i)}{f}_i(z_i)+}{\overline{a}}^{(i)}\overline{g_i({\overline{z}}_i)}\right)$

which shows that ${\widehat{a}}^{(i)}=\Omega a^{(i)}$ anddirectly gives the basis change formula for A.

2.      To find the expression for p, we note that

$\begin{array}{l}\widehat{z}={\widehat{x}}_1+\widehat{p}{\widehat{x}}_2=x_1+p{x}_2={\widehat{x}}_1\cos \theta -{\widehat{x}}_2\sin \theta +p({\widehat{x}}_1\sin \theta +{\widehat{x}}_2\cos \theta )\\ =(\cos \theta +p\sin \theta )\left\{{\widehat{x}}_1+\frac{(p\cos \theta -\sin \theta )}{\cos \theta +p\sin \theta }{\widehat{x}}_2\right\}\end{array}$

Therefore, we may write ${\widehat{f}}_i({\widehat{z}}_i)=f_i\left(\left[\cos \theta +p_i\sin \theta \right]{\widehat{z}}_i\right)$ with ${\widehat{z}}_i={\widehat{x}}_1+\widehat{p}{}_i{\widehat{x}}_2$ and

${\widehat{p}}_i\equiv p_i(\theta )=\frac{p_i\cos \theta -\sin \theta }{p_i\sin \theta +\cos \theta }$

as required.

3.      The basis change formulas for Q, R and T follow directly from the definitions of these matrices. 

4.      The basis change formula for B is a bit more cumbersome.  By definition

$\widehat{b}=({\widehat{R}}^T+\widehat{p}\widehat{T})\widehat{a}=\left(\Omega R{(\theta )}^T{\Omega }^T+\widehat{p}\Omega T(\theta ){\Omega }^T\right)\Omega a=\Omega \left(R{(\theta )}^T+\widehat{p}T(\theta )\right)a$

Substituting for $R(\theta ),T(\theta ),\widehat{p}$ gives

$\begin{array}{l}\widehat{b}=\Omega \left(R^T{\cos }^2\theta +(T-Q)\sin \theta \cos \theta -R{\sin }^2\theta +\frac{p\cos \theta -\sin \theta }{p\sin \theta +\cos \theta }\left[T{\cos }^2\theta -(R+R^T)\sin \theta \cos \theta +Q{\sin }^2\theta \right]\right)a\\ =\frac{\Omega }{p\sin \theta +\cos \theta }\left(R^T\cos \theta +pT\sin \theta -pR\sin \theta -Q\cos \theta \right)a\end{array}$

and finally recalling that $\left[R^T+pT\right]a=b\left[Q+pR\right]a=-pb$ we obtain the required result.

5.      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

 

The 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 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, as follows. Let

$\begin{array}{l}Q(\theta )=Q{\cos }^2\theta +(R+R^T)\sin \theta \cos \theta +T{\sin }^2\theta \\ R(\theta )=R{\cos }^2\theta +(T-Q)\sin \theta \cos \theta -R^T{\sin }^2\theta \\ T(\theta )=T{\cos }^2\theta -(R+R^T)\sin \theta \cos \theta +Q{\sin }^2\theta \end{array}$

and define

$\begin{array}{l}N(\theta )=\left[\begin{array}{cc}{N}_1(\theta )& N_2(\theta )\\ N_3(\theta )& N_1^T(\theta )\end{array}\right]\\ N_1(\theta )=-T^{-1}(\theta )R^T(\theta )N_2(\theta )=T^{-1}(\theta )N_3(\theta )=R(\theta )T^{-1}(\theta )R^T(\theta )-Q(\theta )\end{array}$

Then

$\begin{array}{l}\left[\begin{array}{cc}S& H\\ -L& S^T\end{array}\right]=\frac{1}{\pi }{\displaystyle \underset{0}{\overset{\pi }{\int }}N(\theta )d\theta }\\ S=-\frac{1}{\pi }{\displaystyle \underset{0}{\overset{\pi }{\int }}{T}^{-1}(\theta )R^T(\theta )d\theta }H=\frac{1}{\pi }{\displaystyle \underset{0}{\overset{\pi }{\int }}{T}^{-1}(\theta )d\theta }L=-\frac{1}{\pi }{\displaystyle \underset{0}{\overset{\pi }{\int }}\left\{R(\theta )T^{-1}(\theta )R^T(\theta )-Q(\theta )\right\}d\theta }\end{array}$

 

Derivation: To see this, we show first that $N(\theta )$ can be diagonalized as

$N(\theta )=\left[\begin{array}{cc}A& \overline{A}\\ B& \overline{B}\end{array}\right]\left[\begin{array}{cc}P(\theta )& 0\\ 0& \overline{P}(\theta )\end{array}\right]\left[\begin{array}{cc}{B}^T& A^T\\ {\overline{B}}^T& {\overline{A}}^T\end{array}\right]$

where

$P(\theta )=\left[\begin{array}{ccc}{p}_1(\theta )& 0& 0\\ 0& p_2(\theta )& 0\\ 0& 0& p_2(\theta )\end{array}\right]$

and $p(\theta )$ was defined earlier. From the preceding section, we note that

$\begin{array}{l}b=\left(R{(\theta )}^T+p(\theta )T(\theta )\right)a\\ pb=-\left(Q(\theta )+p(\theta )R(\theta )\right)a\end{array}$

which can be expressed as

$\left[\begin{array}{cc}-Q(\theta )& 0\\ -R^T(\theta )& I\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=p(\theta )\left[\begin{array}{cc}R(\theta )& I\\ T(\theta )& 0\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]$

as before, we can arrange this into an Eigenvalue problem by writing

$\left[\begin{array}{cc}0& T^{-1}(\theta )\\ I& -R(\theta )T^{-1}(\theta )\end{array}\right]\left[\begin{array}{cc}R(\theta )& I\\ T(\theta )& 0\end{array}\right]=\left[\begin{array}{cc}I& 0\\ 0& I\end{array}\right]$

whence

$\begin{array}{l}\left[\begin{array}{cc}{N}_1(\theta )& N_2(\theta )\\ N_3(\theta )& N_1^T(\theta )\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=p(\theta )\left[\begin{array}{c}a\\ b\end{array}\right]\\ N_1(\theta )=-T^{-1}(\theta )R^T(\theta )N_2(\theta )=T^{-1}(\theta )N_3(\theta )=R(\theta )T^{-1}(\theta )R^T(\theta )-Q(\theta )\end{array}$

This shows that [a,b] are eigenvectors of the rotated elasticity matrix.  Following standard procedure, we obtain the diagonalization stated.

 

Now, we examine $p(\theta )$ more closely.  Recall that

$p(\theta )=\frac{p\cos \theta -\sin \theta }{p\sin \theta +\cos \theta }=\frac{1}{p\sin \theta +\cos \theta }\frac{d}{d\theta }(p\sin \theta +\cos \theta )$

Integrating gives

$\begin{array}{l}{\displaystyle \underset{0}{\overset{\theta }{\int }}p(\theta )d\theta }=\ln (\cos \theta +p\sin \theta )\\ {\displaystyle \underset{0}{\overset{\pi }{\int }}p(\theta )d\theta }=\{\begin{array}{c}i\pi \mathrm{Im}(p)>0\\ -i\pi \mathrm{Im}(p)<0\end{array}\end{array}$

(the sign of the integral is determined by Im(p) because the branch cut for $\ln (\cos \theta +p\sin \theta )$ is taken to lie along the negative real axis). Thus,

$\frac{1}{\pi }{\displaystyle \underset{0}{\overset{\pi }{\int }}N(\theta )d\theta }=\left[\begin{array}{cc}A& \overline{A}\\ B& \overline{B}\end{array}\right]\left[\begin{array}{cc}iI& 0\\ 0& -iI\end{array}\right]\left[\begin{array}{cc}{B}^T& A^T\\ {\overline{B}}^T& {\overline{A}}^T\end{array}\right]=\left[\begin{array}{cc}iA{B}^T-i\overline{A}{\overline{B}}^T& iA{A}^T-i\overline{A}{\overline{A}}^T\\ iB{B}^T-i\overline{B}{\overline{B}}^T& iB{A}^T-i\overline{B}{\overline{A}}^T\end{array}\right]=\left[\begin{array}{cc}S& H\\ -L& S^T\end{array}\right]$

 

 

 

5.5.12 Stroh representation for a state of uniform stress

 

A 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 ${\sigma }_{ij}$ and corresponding strain ${\epsilon }_{ij}$ we may write

$\begin{array}{l}u={\epsilon }_1{x}_1+{\epsilon }_2{x}_2\phi =t_2{x}_1-t_1{x}_2\\ u=\left[\begin{array}{c}{u}_1\\ u_2\\ u_3\end{array}\right]{\epsilon }_1=\left[\begin{array}{c}{\epsilon }_{11}\\ {\epsilon }_{12}\\ 2{\epsilon }_{31}\end{array}\right]=\frac{\partial u}{\partial x_1}{\epsilon }_2=\left[\begin{array}{c}{\epsilon }_{12}\\ {\epsilon }_{22}\\ 2{\epsilon }_{32}\end{array}\right]=\frac{\partial u}{\partial x_2}\\ t_1=\left[\begin{array}{c}{\sigma }_{11}\\ {\sigma }_{12}\\ {\sigma }_{31}\end{array}\right]=-\frac{\partial \phi }{\partial x_2}{t}_2=\left[\begin{array}{c}{\sigma }_{12}\\ {\sigma }_{22}\\ {\sigma }_{32}\end{array}\right]=\frac{\partial \phi }{\partial x_1}\end{array}$

In terms of these vectors the Stroh representation is given by

$\begin{array}{l}u=2\mathrm{Re}(AZq)\phi =2\mathrm{Re}(BZq)\\ Z=\mathrm{diag}(z_1,z_2,z_3)q=A^T{t}_2+B^T{\epsilon }_1\\ z_i=x_1+p_i{x}_2\end{array}$

or, in matrix form

$\left[\begin{array}{c}u\\ \phi \end{array}\right]=\left[\begin{array}{cc}A& \overline{A}\\ B& \overline{B}\end{array}\right]\left[\begin{array}{cc}Z& 0\\ 0& \overline{Z}\end{array}\right]\left[\begin{array}{cc}{B}^T& A^T\\ {\overline{B}}^T& {\overline{A}}^T\end{array}\right]\left[\begin{array}{c}{\epsilon }_1\\ t_2\end{array}\right]$

 

Derivation: To see this, recall that a and b form eigenvectors of the fundamental elasticity matrix N as

$N\left[\begin{array}{c}a\\ b\end{array}\right]=p\left[\begin{array}{c}a\\ b\end{array}\right]$

therefore we can write (for each pair of eigenvectors/values)

$\left[\begin{array}{c}a\\ b\end{array}\right]\left(x_1+p{x}_2\right)=\left[\begin{array}{c}a\\ b\end{array}\right]x_1+N\left[\begin{array}{c}a\\ b\end{array}\right]x_2$

Hence

$\left[\begin{array}{cc}AZ& \overline{A}\overline{Z}\\ BZ& \overline{B}\overline{Z}\end{array}\right]=x_1\left[\begin{array}{cc}A& \overline{A}\\ B& \overline{B}\end{array}\right]+x_{2}N\left[\begin{array}{cc}A& \overline{A}\\ B& \overline{B}\end{array}\right]$

 

Recall that

$\left[\begin{array}{cc}A& \overline{A}\\ B& \overline{B}\end{array}\right]\left[\begin{array}{cc}{B}^T& A^T\\ {\overline{B}}^T& {\overline{A}}^T\end{array}\right]=\left[\begin{array}{cc}I& 0\\ 0& I\end{array}\right]$

so

$\left[\begin{array}{cc}AZ& \overline{A}\overline{Z}\\ BZ& \overline{B}\overline{Z}\end{array}\right]\left[\begin{array}{cc}{B}^T& A^T\\ {\overline{B}}^T& {\overline{A}}^T\end{array}\right]=x_1\left[\begin{array}{cc}I& 0\\ 0& I\end{array}\right]+x_{2}N$

$\left[\begin{array}{cc}AZ& \overline{A}\overline{Z}\\ BZ& \overline{B}\overline{Z}\end{array}\right]\left[\begin{array}{cc}{B}^T& A^T\\ {\overline{B}}^T& {\overline{A}}^T\end{array}\right]\left[\begin{array}{c}{\epsilon }_1\\ t_2\end{array}\right]=x_1\left[\begin{array}{c}{\epsilon }_1\\ t_2\end{array}\right]+x_{2}N\left[\begin{array}{c}{\epsilon }_1\\ t_2\end{array}\right]$

and finally, defining

$N\left[\begin{array}{c}{\epsilon }_1\\ t_2\end{array}\right]=\left[\begin{array}{c}{\epsilon }_2\\ -t_1\end{array}\right]$

gives the required result.

 

 

 

 

5.5.13 Line load and Dislocation in an Infinite Anisotropic Solid

 

The 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.

 

The solid contains a straight dislocation, with line direction $e_3$, perpendicular to the plane of the figure.  The dislocation has Burger’s vector $b=b_i{e}_i$.

 

At 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  $F=F_i{e}_i$.

 

The displacement and stress function can be expressed in terms of the Stroh matrices as

$\left[\begin{array}{c}u\\ \phi \end{array}\right]=\frac{1}{2\pi i}\left[\begin{array}{cc}A& \overline{A}\\ B& \overline{B}\end{array}\right]\left[\begin{array}{cc}\lambda & 0\\ 0& \overline{\lambda }\end{array}\right]\left[\begin{array}{cc}{B}^T& A^T\\ {\overline{B}}^T& {\overline{A}}^T\end{array}\right]\left[\begin{array}{c}b\\ F\end{array}\right]$

where $\lambda =diag(\ln (z_1),\ln (z_2),\ln \left(z_3\right))$, where diag denotes a diagonal matrix, and

$b={[b_1,b_2,b_3]}^{T}F={[F_1,F_2,F_3]}^T$

 The solution can also be expressed as

$u=2\mathrm{Re}(Af)\phi =2\mathrm{Re}(Bf)t_2=2\mathrm{Re}(Bf\text{'})t_1=-2\mathrm{Re}(BPf\text{'})f=\lambda \left(B^{T}b+A^{T}F\right)/(2\pi i)$

 

 

Derivation:  We must show that the solution satisfies the following conditions:

1.      The displacement field for a dislocation with burgers vector b must satisfy $u(r,\theta =\pi )-u(r,\theta =-\pi )=b$ (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

$\begin{array}{l}{F}_i+{\displaystyle \underset{-\pi }{\overset{\pi }{\int }}{\sigma }_{ij}{n}_j}rd\theta =0\Rightarrow F_i+{\displaystyle \underset{-\pi }{\overset{\pi }{\int }}\left(-\frac{\partial {\varphi }_i}{\partial x_2}r\cos \theta +\frac{\partial {\varphi }_i}{\partial x_1}r\sin \theta \right)d\theta }=0\\ \Rightarrow F_i+{\displaystyle \underset{-\pi }{\overset{\pi }{\int }}\left(-\frac{\partial {\varphi }_i}{\partial x_2}\frac{\partial x_2}{\partial \theta }-\frac{\partial {\varphi }_i}{\partial x_1}\frac{\partial x_1}{\partial \theta }\right)d\theta }=0\Rightarrow F_i-\left({\varphi }_i(\theta =\pi )-{\varphi }_i(\theta =-\pi )\right)=0\end{array}$

3.      We can create the required solution using properties of log(z).  We try a solution of the form

$\left[\begin{array}{c}u\\ \phi \end{array}\right]=\left[\begin{array}{cc}A& \overline{A}\\ B& \overline{B}\end{array}\right]\left[\begin{array}{cc}\lambda & 0\\ 0& \overline{\lambda }\end{array}\right]\left[\begin{array}{c}q\\ \overline{q}\end{array}\right]$

where $\lambda =diag(\ln (z_1),\ln (z_2),\ln \left(z_3\right))$ and q is a vector to be determined.  Recall that we may write $z=r{e}^{i\theta }$, whence $\log (z)=\log (r)+i\theta$.  This, in turn, implies that $\log (z(r,\pi ))-\log (z(r,-\pi ))=2\pi i$. Therefore

$\left[\begin{array}{c}b\\ F\end{array}\right]=\left[\begin{array}{cc}A& \overline{A}\\ B& \overline{B}\end{array}\right]\left[\begin{array}{cc}\lambda (r,\pi )-\lambda (r,-\pi )& 0\\ 0& \overline{\lambda }(r,\pi )-\overline{\lambda }(r,-\pi )\end{array}\right]\left[\begin{array}{c}q\\ \overline{q}\end{array}\right]\Rightarrow \left[\begin{array}{c}b\\ F\end{array}\right]=\left[\begin{array}{cc}A& \overline{A}\\ B& \overline{B}\end{array}\right]\left[\begin{array}{cc}2\pi iI& 0\\ 0& -2\pi iI\end{array}\right]\left[\begin{array}{c}q\\ \overline{q}\end{array}\right]$

4.      Recalling the orthogonality properties of A and B

$\left[\begin{array}{cc}{B}^T& A^T\\ {\overline{B}}^T& {\overline{A}}^T\end{array}\right]\left[\begin{array}{cc}A& \overline{A}\\ B& \overline{B}\end{array}\right]=\left[\begin{array}{cc}I& 0\\ 0& I\end{array}\right]$

we can solve for q

$\frac{1}{2\pi i}\left[\begin{array}{cc}{B}^T& A^T\\ {\overline{B}}^T& {\overline{A}}^T\end{array}\right]\left[\begin{array}{c}b\\ F\end{array}\right]=\left[\begin{array}{c}q\\ \overline{q}\end{array}\right]$

giving

$\left[\begin{array}{c}u\\ \phi \end{array}\right]=\frac{1}{2\pi i}\left[\begin{array}{cc}A& \overline{A}\\ B& \overline{B}\end{array}\right]\left[\begin{array}{cc}\lambda & 0\\ 0& \overline{\lambda }\end{array}\right]\left[\begin{array}{cc}{B}^T& A^T\\ {\overline{B}}^T& {\overline{A}}^T\end{array}\right]\left[\begin{array}{c}b\\ F\end{array}\right]$

 

 

 

5.5.14 Line load and dislocation below the surface of an anisotropic half-space

 

The 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 $(d_1,d_2)$, while the surface of the solid is traction free.

 

The solution can be computed from the simplified Stroh representation

$u=2\mathrm{Re}(Af)\phi =2\mathrm{Re}(Bf)t_2=2\mathrm{Re}(Bf\text{'})t_1=-2\mathrm{Re}(BPf\text{'})$

where

$f=\lambda (z_i)(B^{T}b+A^{T}F)-B^{-1}\overline{B}\overline{\lambda ({\overline{z}}_i)}({\overline{B}}^{T}b+{\overline{A}}^{T}F)$

and

$\lambda (z_i)=\frac{1}{2\pi i}\left[\begin{array}{ccc}\log (z_1-(d_1+p_1{d}_2))& 0& 0\\ 0& \log (z_2-(d_1+p_2{d}_2))& 0\\ 0& 0& \log (z_3-(d_1+p_3{d}_2))\end{array}\right]$

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.