OGS
[case] LinearElasticTransverseIsotropic

Linear transverse isotropic elastic model.

The parameters of the linear transverse isotropic elastic model are

• $$E_{i}$$, the Young’s modulus within the plane of isotropy,
• $$E_{a}$$, the Young’s modulus w.r.t. the direction of anisotropy,
• $$\nu_{ii}$$, the Poisson’s ratio within the plane of isotropy,
• $$\nu_{ia}$$, the Poisson ratio perpendicular to the plane of isotropy, due to strain in the plane of isotropy,
• $$G_{ia}$$, the shear modulus between directions of isotropy and anisotropy.

where the subscript $$i$$ means isotropy, and the subscript $$a$$ means anisotropy.

With the given parameter, the in-plane shear modulus, $$G_{ii}$$ is computed as

$G_{ii} = \frac{E_{i}}{2(1+\nu_{ii})},$

while the in-plane Poisson ratio, $$\nu_{ai}$$, which is due to the strain perpendicular to the plane of isotropy, is calculated by the following equation:

$\nu_{ai} = \nu_{ia} \frac{E_{a}}{E_{i}}.$

For 3D problems, assuming the plane of isotropy to be spanned by the basis vectors $$\mathbf{e}_1$$ and $$\mathbf{e}_2$$, respectively, and the direction of anisotropy is defined by the basis vector $$\mathbf{e}_3$$, the following relations hold:

\begin{eqnarray*} E_{i} &=& E_1 & =& E_2, \\ E_{a} &=& E_3, & &\\ \nu_{ii} &=& \nu_{12} & =& \nu_{21}, \\ \nu_{ia} &=& \nu_{13} & =& \nu_{23}, \\ \nu_{ai} &=& \nu_{31} & =& \nu_{32}, \\ G_{ia} &=& G_{13} & =& G_{23},\\ G_{ai} &=& G_{ia}. & & \end{eqnarray*}

Under such assumption, the matrix form of the elastic tensor for strain and stress in the Kelvin vector in the local system is

$\begin{bmatrix} a_{ii} & b_{ii} &b_{ai} & 0 & 0 & 0\\ b_{ii} & a_{ii} &b_{ai} & 0 & 0 & 0\\ b_{ai} & b_{ai} &a_{ai} & 0 & 0 & 0\\ 0 & 0 & 0 & 2 c_{ii} & 0 & 0\\ 0 & 0 & 0 & 0 & 2 c_{ai} & 0\\ 0 & 0 & 0 & 0 & 0 & 2 c_{ai} \end{bmatrix}.$

The matrix elements are:

\begin{eqnarray*} a_{ii} &=& \frac{1-\nu_{ia}\nu_{ai}}{E_{i} E_{a} D}, \\ a_{ai} &=& \frac{1-\nu_{ii}^2}{E_{i}^2 D}, \\ b_{ii} &=& \frac{\nu_{ii}+\nu_{ia}\nu_{ai}}{E_{i} E_{a} D}, \\ b_{ai} &=& \frac{\nu_{ia}(1+\nu_{ii})}{E_{i}^2 D}, \\ c_{ii} &=& \frac{E_{i}}{2(1+\nu_{ii})}, \\ c_{ai} &=& G_{ia}, \end{eqnarray*}

with

$D = \frac{(1+\nu_{ii})(1-\nu_{ii}-2\nu_{ia}\nu_{ai})}{E_{i}^2E_{a}}.$

(also see Chapter 9.1 in [22]).

For plane strain problems, assuming the direction of anisotropy is defined by the basis vector $$\mathbf{e}_1$$, the plane of isotropy to be spanned by the basis vector $$\mathbf{e}_0$$ and the unit off-plane direction $$\mathbf{e}_2$$, the following relations hold:

\begin{eqnarray*} E_{i} &=& E_1 & =& E_3, \\ E_{a} &=& E_2, & &\\ \nu_{ii} &=& \nu_{13} & =& \nu_{31}, \\ \nu_{ia} &=& \nu_{32} & =& \nu_{12}, \\ \nu_{ai} &=& \nu_{23} & =& \nu_{21}, \\ G_{ia} &=& G_{32} & =& G_{12},\\ G_{ai} &=& G_{ia}. & & \end{eqnarray*}

Based on this assumption, the matrix form of the elastic tensor for strain and stress in the Kelvin vector in the local system is

$\begin{bmatrix} a_{ii} & b_{ai} &b_{ii} & 0\\ b_{ai} & a_{ai} &b_{ai} & 0\\ b_{ii} & b_{ai} &a_{ii} & 0\\ 0 & 0 & 0 & 2 c_{ai} \end{bmatrix}$

for plane strain problems.

Note
This model requires the definition of a local coordinate system. A local coordinate system can be defined with explicit or implicit bases. The unit norm to the transverse isotropy plane must be input as the parameter value for the base "basis_vector_1" for a 2D local coordinate system, or for the base "basis_vector_2" for a 3D local coordinate system.