Is there a difference between, static loading and fast loading of a polymer (non linear elastic material) in terms of elastic potential generated?

Question

Imagine a non-linear elastic material such as a rubber band, nylon webbing or polyester webbing tensioned between two points.

Scenario 1: A large mass is statically (no acceleration) loaded onto the centre of the material depressing it amount $H$ (shown by this diagram):

Scenario 2: A smaller mass is loaded onto the centre of the material with enough acceleration to also depress the material by amount $H$.

Question: Given that the material has non-linear stretch ie does not obey Hooke's law, will the elastic potential energy generated in the material, by scenario 1 and scenario 2, be equivalent?

Answer 1

For an isothermal elastic material, the strain energy density is a function of the deformation alone:

$$ \psi = \hat \psi(E) $$ where $E$ is a measure of the strain and $\psi$ is the strain energy density.

Therefore the short answer is yes, they will have the same potential given that they both reach a maximum deflection of $H$. However, the work required to cause the deflection may differ from scenario 1 to scenario 2.

If the material's constitutive law dictates that the stress is a function of the strain alone:

$$T = \hat T(E)$$

then the work required will be the same in either case.

If however, the material's constitutive behavior has a viscous term, then the rate of deformation will affect the work required:

$$ T = \hat T(E, \dot E)$$

The dependence on $\dot E$ will dissipate energy, so the case that was loaded more quickly will require more work. There are detailed thermodynamic arguments that guarantee that the dependence of the stress on $\dot E$ will always dissipate energy, but they are well beyond the scope of this question.