2
$\begingroup$

I have a material for which I know both the tensile storage modulus $E'$, and loss modulus $E''$ at a frequency of 1 Hz. As I understand it, $E'$ conceptually describes the elastic properties of the material, while $E''$ is related to viscous properties. Of course these can be combined to form the complex modulus $E^* = E' + iE''$. It appears that people will commonly plot $E^*$ as a vector in the complex plane because the phase angle tells you where on the scale of purely elastic ($0^\circ$) to purely viscous ($90^\circ$) the material is.

That is all starting to make a little sense conceptually, but for my material with known $E'$ and $E''$ at 1 Hz, I would like to be able to figure out its dynamic viscosity, $\eta$. To cut a long story short, I have some equations in terms of $\eta$, and I want to know whether or not it is justifiable to assume that the viscosity is negligible when the material is driven at 1 Hz. But the only data I have are $E'$ and $E''$ at 1 Hz.

Is there any mathematical way of connecting these concepts?

Update: I have found one possible clue. In these lecture notes (p14), the author notes in passing that $$ \eta' = \frac{G''}{\omega} $$ where $G''$ is the shear loss modulus, and $\eta'$ is the "effective viscosity." I wonder if this is the answer?

I am fairly new to studying viscoelastic materials, so any guidance would be greatly appreciated!

$\endgroup$
1
  • $\begingroup$ it may very well be that this is your answer, but be aware that shear modulus is not the same thing as tensile or Young's modulus. I've seen the equation you wrote above which has shear modulus in it, but I've not seen it for E, which is Young's modulus. I'm just saying it is probably worth your time to continue to look into this. Good Luck! $\endgroup$
    – CGS
    Commented Jan 1, 2021 at 23:26

2 Answers 2

2
$\begingroup$

I want to know whether or not it is justifiable to assume that the viscosity is negligible when the material is driven at 1 Hz

The typical approach is to compare the magnitudes of the storage modulus $G^\prime(1\,\text{Hz})$ and the loss modulus $G^{\prime\prime}(1\,\text{Hz})$. If the latter (former) is relatively small, then the material resembles an ideal spring (damper), and the viscosity is negligible (predominant). Equivalently, you could check whether the so-called loss tangent $\tan \delta=G^{\prime\prime}/G^\prime$ is small at 1 Hz.

The dynamic viscosity is typically defined in terms of an ideal damper (typical constitutive equations: $\sigma=\eta\dot\varepsilon$ or $\tau=\eta\dot\gamma$). For the more general case of a viscoelastic material with complex modulus ${G^\star}(\omega)=G^\prime(\omega)+iG^{\prime\prime}(\omega)$, we have the complex viscosity $\eta^\star={G^\star}(\omega)/\omega=\eta^\prime(\omega)+i\eta^{\prime\prime}(\omega)$. If you wish, you could call $\eta^{\prime\prime}$ the dynamic or effective viscosity, but that might invite confusion, as $\eta^\star$ is also sometimes called the dynamic viscosity.

$\endgroup$
0
$\begingroup$

I don't think there is a mathematical relation between the two. The modulus, E, is related to stress/strain, while viscosity, $\eta$, is related to stress/strain rate.

Here's a good link that talks about these concepts. You'll find more good information here.

$\endgroup$
2
  • $\begingroup$ "I don't think there is a mathematical relation between the two." Your link specifically mentions a mathematical relation between the two. The rate/frequency connection comes in because the modulus is a function of the rate/frequency. $\endgroup$ Commented Jan 1, 2021 at 0:49
  • $\begingroup$ @Chemomechanics . I am familiar with the equation you wrote in your answer relating the complex viscosity to the complex shear modulus, but not to the complex Young's modulus. The same equation applies? Thanks! $\endgroup$
    – CGS
    Commented Jan 1, 2021 at 13:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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