Poisson's Ratio - Transversely Isotropic Materials

Transversely Isotropic Materials

Transversely isotropic materials have a plane of symmetry in which the elastic properties are isotropic. If we assume that this plane of symmetry is, then Hooke's law takes the form

$\begin{bmatrix} \epsilon_{{\rm xx}} \\ \epsilon_{\rm yy} \\ \epsilon_{\rm zz} \\ 2\epsilon_{\rm yz} \\ 2\epsilon_{\rm zx} \\ 2\epsilon_{\rm xy} \end{bmatrix} = \begin{bmatrix} \tfrac{1}{E_{\rm x}} & - \tfrac{\nu_{\rm yx}}{E_{\rm y}} & - \tfrac{\nu_{\rm yx}}{E_{\rm y}} & 0 & 0 & 0 \\ -\tfrac{\nu_{\rm xy}}{E_{\rm x}} & \tfrac{1}{E_{\rm y}} & - \tfrac{\nu_{\rm zy}}{E_{\rm y}} & 0 & 0 & 0 \\ -\tfrac{\nu_{\rm xy}}{E_{\rm x}} & - \tfrac{\nu_{\rm yz}}{E_{\rm y}} & \tfrac{1}{E_{\rm y}} & 0 & 0 & 0 \\ 0 & 0 & 0 & \tfrac{1}{G_{\rm yz}} & 0 & 0 \\ 0 & 0 & 0 & 0 & \tfrac{1}{G_{\rm xy}} & 0 \\ 0 & 0 & 0 & 0 & 0 & \tfrac{1}{G_{\rm xy}} \\ \end{bmatrix} \begin{bmatrix} \sigma_{\rm xx} \\ \sigma_{\rm yy} \\ \sigma_{\rm zz} \\ \sigma_{\rm yz} \\ \sigma_{\rm zx} \\ \sigma_{\rm xy} \end{bmatrix}$

where we have used the plane of symmetry to reduce the number of constants, i.e., .

The symmetry of the stress and strain tensors implies that

$\cfrac{\nu_{\rm xy}}{E_{\rm x}} = \cfrac{\nu_{\rm yx}}{E_{\rm y}} ~,~~ \nu_{\rm yz} = \nu_{\rm zy} ~.$

This leaves us with six independent constants . However, transverse isotropy gives rise to a further constraint between and which is

$G_{\rm yz} = \cfrac{E_{\rm y}}{2(1+\nu_{\rm yz})} ~.$

Therefore, there are five independent elastic material properties two of which are Poisson's ratios. For the assumed plane of symmetry, the larger of and is the major Poisson's ratio. The other major and minor Poisson's ratios are equal.

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