Mathematical definition of elastic materials

Question

Physically, elastic materials are materials which return to their original state upon complete removal of applied mechanical loads under isothermal conditions.

In the book "Mechanics of Laminated Composite Plates and Shells 2nd Edition" by JN Reddy", the author defines

Materials for which the constitutive behavior is only a function of the current state of deformation are known as elastic. In the special case in which the work done by stresses during a deformation is dependent only the initial state and the current configuration, the material is called hyperelastic.

I am unable to relate the mathematical definition of elastic materials with the physical one. How is the dependence of constitutive relations on current state of deformation related to the material returning to its original state upon complete removal of loads?

Answer 1

It is probably beneficial to list different other types of material.

This figure summarizes, the most common behaviour of the materials and their models, with wrt time (not deformation).

As other have mentioned, the idea of the purely elastic model is that at any instant its behaviour is dependent on the deformation. Most materials in the majority of structural engineering applications are considered elastic (or at least they were before the progress in FEM codes).

In reality, no material behaves as purely elastic, however, it is a very, very good approximation for a great number of calculations.

Answer 2

It means the material has no "memory" of what previously happened to it. The deformation for any given loading is always the same (including zero deformation for zero load, as a particular case).

An inelastic material model needs some additional variables to "remember" the past history of the stress and/or strain. For example, to model plasticity you need to remember the accumulated plastic strain. To model creep or viscoelasticity you also need include the time history of the applied loads.

Answer 3

Here's one way in which the two definitions are not related: falling within OP's "physical definition" is not a sufficient condition for falling within OP's "mathematical definition". For example, the Kelvin/Voigt model, as illustrated in the answer by @NMech , falls within OP's "physical definition" but not within OP's "mathematical definition". I think the real answer is that different people mean subtly different things by the word "elastic", and the reader must take care.

Answer 4

In the case of tensile test of metals, below the yield stress for example, for a given strain there is a given stress. We can say that the stress is a function a strain $\sigma = \sigma(\epsilon)$.

After the yield point, plastic deformation begins, and if the sample is unloaded, it doesn't return to its original length $L_0$, but to a bigger length $L_1$.

If we compute the total strain (after unloading) as $\frac{L_1 - L_0}{L_0}$, it was previously (in the loading step) correspondent to a non zero stress. So, for this type of material we can not say that $\sigma = \sigma(\epsilon)$ anymore, it not an elastic behaviour.