# Inelastic strain and stress

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?

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.

• thanks. Could you please see my updates on the question? In a viscoelastic body, if we know the distribution of an inelastic strain, we can calculate the elastic stress generated by that strain. Can we assume the elastic stress is the total stress in the medium? Sep 21, 2021 at 3:45
• Yes, but we don't know it. All we can measure is the total strain tensor. And the stress tensor is not a function of the strain tensor out of the elastic range. Sep 21, 2021 at 11:23
• my understanding is that the stress tensor can be a function of the strain rate tensor out of the elastic range in a viscoelastic body. if we want to calculate the effective viscosity of the viscoelastc body, should we use the elastic stress or total stress or the stress only related to the inelastic strain? Or, the elastic stress is the total stress and there aren't any 'inelastic stress'? Sep 21, 2021 at 18:53
• I think that it is better to ask another questions. It is too much for a comment. Sep 21, 2021 at 22:10
• please see this question: physics.stackexchange.com/questions/667185/… Sep 21, 2021 at 22:41

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 $$\dot{\varepsilon}_{kl}$$:

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