Inelastic strain and stress

Question

In a viscoelastic medium, the total strain is the sum of elastic strain and inelastic strain:

\begin{align} \mathcal{E}^t_{ij}= \mathcal{E}^e_{ij}+\mathcal{E}^i_{ij} \end{align}

The elastic stress (Cauchy stress) can then be derived linearly as: \begin{align} \sigma^e= C_{ijkl}\mathcal{E}^e_{kl} \end{align}

The inelastic strain is a kind of eigenstrain which is stress-free, so it doesn't appear in the stress-strain relationship. However, can we assume that the elastic stress is indeed the total stress in this medium?

Answer 1

No.

If a metal for example is forced to assume a given geometry, supposing small deformation, we can know the strain tensor as a function of the material coordinates.

Applying the elastic relation between stress and strain tensor, a trial stress tensor can be known. But it is possible that for some points, when the components of the stress tensor are fed in the Von Mises formula, it results above the yield stress. We know then that our trial stress tensor is not correct. And part of the strain corresponds to plastic deformation.

Answer 2

Viscoelastic materials usually does not undergo irreversible strain. So these materials do not admit a decomposition of strain similar to that of plastic or viscoplastic materials. For viscoelastic materials we have the following stress decomposition:

$$\sigma_{ij}^{(t)} = \sigma_{ij}^{(e)} + \sigma_{ij}^{(v)}$$

For a linear viscoelastic material, the above formula reduces to:

$$\sigma_{ij}^{(t)} = \sum_{k,j} \left[ C_{ijkl}\varepsilon_{kl} + \int_{-\infty}^t b_{ijkl}(t-\tau)\dot{\varepsilon}_{kl}(\tau)\ \text{d}\tau \right]$$

so, for constant strain rate $\dot{\varepsilon}_{kl}$:

$$\sigma_{ij}^{(t)} - \sum_{k,l}C_{ijkl}\varepsilon_{kl} = \Phi(\dot{\varepsilon}_{ij})$$