0
$\begingroup$

In isotropic elastic materials, the shear modulus, G, and the Young's modulus, E, are related via

$$ G = \frac{E}{2 (1 + \nu)} $$

where $\nu$ is the Poisson ratio. As a consequence, as long as $\nu > -0.5$, $G < E$. I am interested in this relationship for transversely isotropic materials. Such materials have 5 instead of two independent constants: two Poisson ratios, two Young's moduli, and one "mixed" shear modulus. I am interested in the latter, as I need to reduce the number of independent constants in a fitting procedure. Are there any reasonable bounds or approximations for the mixed shear modulus? For example, one could use

$$ G_{m} \approx \frac{E_1 + E_2}{2} $$

in rough analogy to the expression for the isotropic case (of course, any other form of mean/including the Poisson ratios may be equally (un)suited). Could this at all be justified? If not/alternatively, can the values of $E_1$ and $E_2$ be used to set bounds on $G_m$? For example, would $G_m$ need to be smaller than the larger of the two Young's moduli? I am grateful for any hint/opinion.

$\endgroup$
  • $\begingroup$ Instead of arithmetic mean, it is quite common in homogenization techniques to use the harmonic mean $2E_1E_2/(E_1+E_2)$, maybe that could have some interest here? $\endgroup$ – user130529 Jan 2 '17 at 17:59
1
$\begingroup$

your suspicion is justified: The stiffness tensor (or matrix if you will) has to be positive definite. In the transversal isotropic case you have 4 Eigenvalues and one "angle" (5 parameters in total), thus there are 4 inequalitis restricting your components, requiring positive eigenvalues

The full spectral decomposition of the transversal isotropic stiffness tensor can be found in Appendix A1 in this article

Kalisch and Glüge 2015

In terms of the usual 6x6 Matrix representation, these inequalities hold:

$$C_{11} - C_{12}>0 $$ $$1/2 C_{11} + C_{12} - \sqrt{ 8 C_{13}^2 + (C_{11} + C_{12} - C_{33})^2} + C_{33}>0$$ $$1/2 C_{11} + C_{12} + \sqrt{ 8 C_{13}^2 + (C_{11} + C_{12} - C_{33})^2} + C_{33}>0$$ $$C_{44}>0$$

where $C_{ij}$ are the components of the stiffness tensor w.r.t. the non-normalized 6D basis for symmetric second order tensors. We see that the third equation is satisfied when the second one holds, so in effect there are 3 inequalities that constrain the 5 independent components.

This of course only limits the ranges of your components, probably just telling you that the moduli should be positive if expressed in components $C_{ij}$. AFAIK there are no approximations for the shear moduli except that they should be of the same order of magnitude as the in-plane and out-of-plane Young moduli. In terms of these components, the moduli are:

The in-plane Young modulus: $E_{inplane}=((C_{11}- C_{12}) (-2 C_{13}^2 + (C_{11} + C_{12}) C_{33}))/(-C_{13}^2 + C_{11} C_{33})$

The out-of-plane Young modulus: $E_{outofplane}=-((2 C_{13}^2)/(C_{11} + C_{12})) + C_{33}$

The shear modulus for in-plane shear: $G_{inplane}=C_{44}$

The shear modulus for out-of-plane shear $G_{outofplane}=(C_{11}-C_{22})/2$.

Maybe you can try to use these equations to relate your shear and Young moduli.

PS. I can provide more material if you need, just send a message.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.