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While introducing the theory of hyperelastic solids, our professor gave the following example.

"Suppose to have a rubber which is undeformed. Then stretch it, and then release it. The final configuration will be the same as the initial one. Then, the he overall work cannot be negative"

I really cannot understand the overall work can't be negative. While he was writing this, he was mentioning something like "if it were negative, then we have a contradiction". However, I really can't understand what was the idea.

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The argument is probably this: it is assumed that the rubber body returns to the initial state; then its final energy has to be the same as the initial energy because internal energy of a body is function of state; same state means same energy. It is also assumed that no work was lost on heating the environment during this exercise. So if energy of both rubber and its environment did not change, then no total work was done.

In reality, when rubber is stretched, it heats up and some energy is transferred to the cooler environment. During contraction, rubber gets colder and part of this energy is absorbed back by the rubber, but not all of it, so the rubber won't really get back to the initial state and work got back won't be as large as work put in, so there will be some small amount of work lost.

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I think that we could take the definition for a hyperelastic material. From Wikipedia

A hyperelastic or Green elastic material is a type of constitutive model for ideally elastic material for which the stress–strain relationship derives from a strain energy density function.

Then, you will have something as the following

$$S_{ij} = \frac{\partial U}{\partial E_{ij}}\, ,$$

where $S$ is a stress measure and $E$ is a strain measure, and these two are energy conjugate. This means that the product of the two of them represents a density of energy.

Now, you need to assume a sign convention for work.

Hyperelastic materials are a special case of Cauchy elastic materials and, as mentioned by @JánLalinský, you have a state variable and the final and initial configuration are the same. Then the work should be the difference in internal energy from these two states, but they are the same

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  • $\begingroup$ Thanks @nicoguaro. Why did someone downvote your answer? $\endgroup$ Commented Jun 7, 2021 at 7:04
  • $\begingroup$ @bobinthebox, no idea. $\endgroup$
    – nicoguaro
    Commented Jun 7, 2021 at 15:22

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