Transverse isotropy

A transversely isotropic material is symmetric about an axis that is normal to a plane of isotropy. This transverse plane has infinite planes of symmetry and thus, within this plane, the material properties are same in all directions. With this type of material symmetry, the number of independent constants in the elasticity tensor are reduced to 5 from a total of 21 independent constants in case of fully anisotropic solid.

Example of transversely isotropic materials

An example of a transversely isotropic material is the so-called on-axis unidirectional fiber composite lamina where the fibers are circular in cross section. In a unidirectional composite, the plane normal to fiber direction can be considered isotropic. In the figure at left, the fibers would be aligned with the $x_3$ axis, which is normal to the plane of isotropy.

In terms of effective properties, geological layers of rocks are often interpreted as being transversally isotropic. Calculating the effective elastic properties of such layers in petrology have been coined Backus upscaling, which is described below.

Elasticity tensor

: $egin{pmatrix}C_{11}&C_{12}&C_{13}&0&0&0\&C_{11}&C_{13}&0&0&0\&&C_{33}&0&0&0\&&&C_{44}&0&0\&symmetric&&&C_{44}&0\&&&&&(C_{11}-C_{12})/2end{pmatrix}$

The elasticity tensor $C_{ij}$ has 5 independent constants, which are related to well known engineering elastic moduli in the following way. These engineering moduli are experimentally determined.

Longitudinal Elastic Modulus, $E_L=C_{11}-C_{12}^2/C_{22}$

Transverse Elastic Modulus, $E_T=C_{22}-C_{12}^2/C_{11}$

Inplane Shear Modulus, $G_{LT}=C_{66}$

Poisson's ratio, $Nu_{LT}=C_{11}/C_{22}$

Here, L represents the longitudinal direction and T represents the transverse direction; 1 is the fiber direction.

Backus upscaling

A layered model of homogeneous and isotropic material, can be upscaled to a transverse isotropic medium, proposed by Backus [Backus, G. E. (1962), Long-Wave Elastic Anisotropy Produced by Horizontal Layering, J. Geophys. Res., 67(11), 4427–4440] .

Backus presented an equivalent medium theory, a heterogeneous medium can be replaced by a homogeneous one which will predict the wave propagation in the actual medium. [ Ikelle, Luc T. and Amundsen, Lasse (2005),Introduction to petroleum seismology, SEG Investigations in Geophysics No. 12] Backus showed that layering on a scale much finer than the wave length has an impact and that a number of isotropic layers can be replaced by a homeogenous transverely isotropic medium that behaves exactly in the same manner as the actual medium under static load in the infinite wave length limit.

If each layer $i$ are described by 5 transversally isotropic parameters $(a_i, b_i, c_i, d_i, e_i)$ , representing the tensor: $egin{pmatrix}a_i & a_i - 2e_i & b_i & 0 & 0 & 0 \a_i-2e_i & a_i & b_i & 0 & 0 & 0 \ b_i & b_i & c_i & 0 & 0 & 0 \ 0 & 0 & 0 & d_i & 0 & 0\ 0 & 0 & 0 & 0 & d_i & 0\ 0 & 0 & 0 & 0 & 0 & e_i\end{pmatrix}$ The elastic moduli for the effective medium will be: $egin{pmatrix} A & A-2E & B & 0 & 0 & 0 \ A-2E & A & B & 0 & 0 & 0 \ B & B & C & 0 & 0 & 0 \ 0 & 0 & 0 & D & 0 & 0 \ 0 & 0 & 0 & 0 & D & 0 \ 0 & 0 & 0 & 0 & 0 & Eend{pmatrix}$ where : $egin{align}A &= +^{-1}^2 \B &=^{-1} \C &=^{-1} \D &=^{-1} \E &= \end{align}$

denotes the volume weighted average over all layers.

This includes isotropic layers, as the layer is isotropic if $b_i = a_i - 2e_i$ , $a_i = c_i$ and $d_i = e_i$ .

Assumptions

* All materials are linearly elastic
* No sources of intrinsic energy dissipation (e.g. friction)
* Valid in the infinite wave length limit, hence good results only if layer thichness is much smaller than wave length

Wave speeds

The direction dependent wave speeds for elastic waves through the material can be found by using the Christoffel equation and are given by [G. Mavko, T. Mukerji, J. Dvorkin. "The Rock Physics Handbook". Cambridge University Press 2003 (paperback). ISBN 0-521-54344-4] : $egin{align} V_P( heta) &= sqrt{frac{C_{11} sin^2( heta) + C_{33} cos^2( heta)+C_{44}+sqrt{M( heta){2
ho \ V_{SV}( heta) &= sqrt{frac{C_{11} sin^2( heta) + C_{33} cos^2( heta)+C_{44}-sqrt{M( heta){2
ho \ V_{SH} &= sqrt{frac{C_{66} sin^2( heta) + C_{44}cos^2( heta)}{
ho \ M( heta) &= left [left(C_{11}-C_{44}
ight) sin^2( heta) - left(C_{33}-C_{44}
ight)cos^2( heta)
ight] + left(C_{13} + C_{44}
ight) sin^2(2 heta) \ end{align}$ where $egin{align} hetaend{align}$ is the angle between the $TI$ symmetry axis and the wave propagation direction.

References

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