Apparent elastic modulus formula and meaning of squared Poisson's ratio

Question

I come from a maths background but I am currently engaged in a project which involves viscoelasticity modelling, so stress and strain come up a lot. In some papers I've read (here and here) people mention the "apparent elastic modulus" $K$, and give its formula as: $$\begin{equation} K=\frac{E}{1-\nu^2} \text{,} \end{equation}$$ where $E$ is Young's modulus and $\nu$ is Poisson's ratio.

The trouble here is that when I try to look up this formula or the apparent elastic modulus in general, I can't seem to find anything related to this topic. Even more confusingly, different names and notations are used for different things. For example, Wikipedia lists $K$ as the Bulk modulus, which has a different physical meaning and wouldn't suit my context, but regardless, the formulae listed at the bottom of the page show nothing similar to what I am looking for.

This paper mentions a formula for "dimensionless torsional rigidity": $\alpha=2\rho(1-\nu^2/\beta)$, where $\beta=E_y/E_x$ and $\rho=G/E_x$, $E_x$ and $E_y$ being Young's moduli in the x and y direction and $G$ being the shear modulus. Even though this is again a different physical context (torsion as opposed to mine being unilateral traction), this is the formula most "resembling" my own and the only one I have found so far that mentions $\nu^2$.

On that note, is there any physical meaning to $\nu^2$, or $1-\nu^2$ for that matter? If its purpose is only to make the Poisson's ratio argument positive, $1-\nu$ is already positive, as the theoretical range of $\nu$ is from $-1$ to $0.5$, correct?

In summary, is anyone familiar with calculating the apparent elastic modulus from Young's modulus and Poisson's ratio? Could someone recommend reading material on the matter?

Answer 1

Rather than viscoelasticity, the application to retina, which can be modelled as a thin shell structure may be important. In such structures, normal stress ($\sigma_z$) can often be neglected in comparison to in-plane stresses ($\sigma_x$ and $\sigma_y$), so Hooke's law simplifies (biaxial stress):

$$\epsilon_x = \frac{1}{E}\cdot\left(\sigma_x-\nu\cdot \sigma_y\right)$$

$$\epsilon_y = \frac{1}{E}\cdot\left(\sigma_y-\nu\cdot\sigma_x\right)$$

Deriving $\sigma_x$ from the first one: $$\sigma_x = \nu\cdot \sigma_y+E\cdot \epsilon_x$$ and substituting this to the second one: $$\epsilon_y = \frac{1}{E}\cdot\left(\sigma_y-\nu\cdot\left(\nu\cdot \sigma_y+E\cdot \epsilon_x\right)\right) = \sigma_y\cdot\frac{1-\nu^2}{E}-\nu\cdot\epsilon_x$$ Finally, you can express equation for $\sigma_y$: $$\sigma_y = \frac{E}{1-\nu^2}\cdot\left(\epsilon_y+\nu\cdot\epsilon_x\right)$$ Analogically, you can get: $$\sigma_x = \frac{E}{1-\nu^2}\cdot\left(\epsilon_x+\nu\cdot\epsilon_y\right)$$ Now in case where lateral deformation is restrained: $$\sigma_x\left(\epsilon_y=0\right) = \frac{E}{1-\nu^2}\cdot\epsilon_x = K\cdot \epsilon_x$$ and similarly: $$\sigma_y\left(\epsilon_x=0\right) = \frac{E}{1-\nu^2}\cdot\epsilon_y = K\cdot \epsilon_y$$