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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): 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?

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Correct me if I am wrong, but just as the work done stretching the band depends on the speed with which it done, but the work available while relaxing the band also depends on the speed, doesn't it. In which case what do you mean by "potential" in this case? The maximum energy available? – dmckee Jun 20 '13 at 20:25
yes i mean energy at the bottom of the stretch – david_adler Jun 22 '13 at 16:18
up vote 0 down vote accepted

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.

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