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The stiffness tensor for anisotropic linear elasticity stiffness tensor consists of 36 elements, but only 21 are unique.

Are there equations for these elements as a function of the material properties elastic modulus and poisson's ratio? If so, I have not been able to find anything in my literature review (for anisotropic, not isotropic).

user27504
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  • In general, no - if there are 21 independent coefficients of the tensor (lowest symmetry of the material), you cannot derive them from a handful of parameters generally derived for / used for for isotopic or cubic materials - too many unknowns, too few knowns... – Jon Custer May 04 '18 at 21:44

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These ARE the material properties. In some cases they are related to more familiar moduli: Young's modulus, Poisson's ratio, etc. You can measure these tensor components directly or infer them from resonance spectra or wave propagation. Or are you asking how to calculate them about initio? Basically you can always work with the components of the elastic tensor. But most papers will use simplified combinations that are more natural for specific problems.

Here is a link to a technique for measuring the elastic moduli (as many as are required to fit the data) using resonance spectroscopy for macroscopically homogeneous samples. In on example we estimated the 8 moduli for an orthorhombic material. Link to paper.

JohnS
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  • OHH. I did not know that. I was influenced by the isotropic homogeneous case, in which there are analytical relationships between each component of the stiffness matrix and elastic modulus and poisson's ratio. e.g., the first element is $K+4\mu/3$. – user27504 May 04 '18 at 20:41
  • When you say in "some cases", do you mean in less directionally dependent cases such as isotropic, orthotropic cases? – user27504 May 04 '18 at 20:42
  • Exactly. For isotropic or azimuthally isotropic material there are clever combinations that make the calculations easier. No physics though. – JohnS May 04 '18 at 20:46
  • Ah I see. What do you mean by "no physics" in this context? – user27504 May 04 '18 at 20:59
  • It's just notation. The Voigt notation is no better or worse than $\lambda, \mu$, etc. Obviously, for isotropic media, say, the Voigt notation is more cumbersome. But the $C_{ij}$ comprise the linear response function. – JohnS May 04 '18 at 21:18
  • Ahh okay. What about in the orthotropic case? We have 9 independent elements of the stiffness matrix. Is it still the case that those 9 elements cannot analytically be obtained in terms of modulus and poisson's ratio? – user27504 May 04 '18 at 21:24
  • yes, that is correct, you need more parameters for lower symmetry. Of course you may find for some problem that certain combinations of $C_{ij}$ that appear together, or others can be readily interpreted in terms of something simpler, say departure from azimuthal isotropy. Then it makes sense to define new "moduli" to ease calculations or conceptualization. – JohnS May 04 '18 at 21:38
  • Thanks for answering all of my questions! I often see lame's coefficients talked about in terms of linear elasticity, in general, with no mentioning of isotropy. It seems Lame's coefficients would only be relevant in the context of isotropic materials, so I wonder why it seems to be spoken of loosely in a lot of elasticity texts. – user27504 May 04 '18 at 21:54
  • Hmm, I was given a some data and was told that I could compute the 9 stiffness coefficients for an orthotropic material from it. The data that I was given includes elasticic modulus in the 3 orthogonal directions, and 6 poisson's ratios ($\nu_{12},\nu_{21}, \nu_{13}, \nu_{31}, \nu_{23}, \nu_{32}$). I am also given the shear modulus in 3 directions ($G_{xy}, G_{yz}, G_{zx}$. Could the 9 stiffness elements be determined from this? – user27504 May 07 '18 at 15:42