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 [21]).

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.

See also: ParameterLib::CoordinateSystem.

Child parameters, attributes and cases

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Used in the following test data files

Used in no end-to-end test cases.