This is really a general question, but my interest is primarily for structural mechanics.

Say you are modeling an elastic solid material that is orthotropic. You have the orthotropic properties (elastic modulus in 3 directions, the 6 poissons ratios, and the shear modulus in the 3 directions), but you only have a code that can take in isotropic properties. So if you want to model an orthotropic material with an isotropic code, you'd have to assume the orthotropic material exhibits isotropic properties. In doing so, would you just be averaging the properties to obtain a hypothetical isotropic property? For example, you would average the 3 elastic moduli in the 3 principal directions to obtain one single modulus for use in the isotropic code?

This is probably not a great example because it is pretty easy to change a linear elastic solver code to take in orthotropic properties as opposed to only isotropic properties.


1 Answer 1


First, I would like to mention that orthotropic materials have 9 parameters, and not 12, as can be implied from your question.

The simplest approach would be to compute the arithmetic/geometric mean between your parameters. Another one would identify which of the moduli is more important in your problem of interest, although that means that you should know your solution already, and that it can vary from one problem to another.

An approach that is more theoretically sound is to find the closest isotropic tensor to your orthotropic one. That implies the following optimization problem

$$\min_{\lambda, \mu}\quad d(C_\text{iso}, C_\text{ortho})\, .$$

For that you can use three different metrics, that I know of. Let's consider two tensors $C_1$ and $C_2$, the different metrics are defined by:

  1. The Frobenius metric

$$d_F(C_1, C_2) = \Vert C_1 - C_2 \Vert\, .$$

  1. The log-Euclidean metric

$$d_L(C_1, C_2) = \left\Vert \log(C_1) - \log(C_2)\right\Vert\, .$$

  1. The Riemannian metric

$$d_R(C_1, C_2) = \left\Vert(C_1^{-1/2} C_2 C_1^{-1/2}\right\Vert\, .$$

The logarithm and square root used above are the matrix logarithm and the square root of a matrix. Also, it should be noted that the log-Euclidean and Riemmanian norms that induce these metrics are better suited in the sense that they are invariant under inversion and return the same result for stiffness and compliance tensors distances.

As a last comment, in the case of Frobenius metric you can obtain an analytic solution easily, and the solution is (I would double check these)

\begin{align} &\lambda + 2\mu = \frac{1}{33}[9(C_{11} + C_{22} + C_{33}) + 4(C_{44} + C_{55} + C_{66}) + 2(C_{12} + C_{13} + C_{23})]\, ,\\ &\mu = \frac{1}{33}[4(C_{11} + C_{22} + C_{33}) + 3(C_{44} + C_{55} + C_{66}) - 4(C_{12} + C_{13} + C_{23})]\, . \end{align}


  1. Moakher, Maher, and Andrew N. Norris. "The closest elastic tensor of arbitrary symmetry to an elasticity tensor of lower symmetry." Journal of Elasticity 85.3 (2006): 215-263. Preprint: https://arxiv.org/pdf/cond-mat/0608311
  2. Norris, Andrew. "The isotropic material closest to a given anisotropic material." Journal of Mechanics of Materials and Structures 1.2 (2006): 223-238. Preprint: https://arxiv.org/pdf/cond-mat/0509705
  • $\begingroup$ Ah yes. Do you know how computational people obtain the full stiffness coefficient matrix for a fully anisotropic material? I'm assuming they usually get something from experiments? $\endgroup$
    – user27504
    Commented May 27, 2018 at 19:24
  • $\begingroup$ @user27504, I don't have a definite answer to this. But I think that question is better asked separately. $\endgroup$
    – nicoguaro
    Commented May 27, 2018 at 19:33
  • 1
    $\begingroup$ @user27504: "how computational people obtain the full stiffness coefficient matrix ...". They guess. It's extremely difficult to experimentally measure these quantities (particularly Poisson's ratios) even for transversely isotropic materials. $\endgroup$ Commented May 27, 2018 at 23:36
  • $\begingroup$ Ah interesting, maybe I should start a new thread on this, but not sure if it would fit in the scicomp question. What would their guess be based on though? Transversely isotropic is still a lot better for full anisotropic. I wonder how people do it for full anisotropic mateirals $\endgroup$
    – user27504
    Commented May 28, 2018 at 2:36
  • $\begingroup$ @user27504, yes, please start a new question. $\endgroup$
    – nicoguaro
    Commented May 28, 2018 at 4:44

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