Continuum Mechanics – Understanding Inelastic Strain and Stress

continuum-mechanicsstress-strainviscosity

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?

Best Answer

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.

Related Solutions

[Physics] Strain energy density in index notation

Your original math was correct. The strain energy density should have those factors of two in your original answer, when defined in terms of the tensorial definitions of the shear strains. $$ U=\frac{1}{2}\sigma_{ij}\epsilon_{ij}=\frac{1}{2} ( \sigma_{11}\epsilon_{11} + \sigma_{22}\epsilon_{22} + \sigma_{33}\epsilon_{33} + 2\sigma_{12}\epsilon_{12} + 2\sigma_{13} \epsilon_{13}+ 2\sigma_{23}\epsilon_{23} ) $$ The key is to realize that in switching from tensorial notation:

$$ U=\frac{1}{2}\sigma_{ij}\epsilon_{ij}=\frac{1}{2} \left[ \begin{array}{ccc} \sigma_{11} & \sigma_{12} & \sigma_{13} \\ \sigma_{21} & \sigma_{22} & \sigma_{23}\\ \sigma_{31} & \sigma_{32} & \sigma_{33} \end{array} \right]\ \left[ \begin{array}{ccc} \epsilon_{11} & \epsilon_{12} & \epsilon_{13} \\ \epsilon_{21} & \epsilon_{22} & \epsilon_{23}\\ \epsilon_{31} & \epsilon_{32} & \epsilon_{33} \end{array} \right]\ $$

to engineering (i.e. Voigt) notation, one must account for a change in definition of the shear strains. Explicitly, in engineering notation, we write: $$ \left[ \begin{array} {ccc} \sigma_{1} \\ \sigma_{2} \\ \sigma_{3} \\ \sigma_{4} \\ \sigma_{5} \\ \sigma_{6} \end{array} \right]\ = C_{ij} \left[ \begin{array} {ccc} \epsilon_{1} \\ \epsilon_{2} \\ \epsilon_{3} \\ \epsilon_{4} \\ \epsilon_{5} \\ \epsilon_{6} \end{array} \right]\, where \left[ \begin{array} {ccc} \epsilon_{1} \\ \epsilon_{2} \\ \epsilon_{3} \\ \epsilon_{4} \\ \epsilon_{5} \\ \epsilon_{6} \end{array} \right]\ = \left[ \begin{array} {ccc} \epsilon_{11} \\ \epsilon_{22} \\ \epsilon_{33} \\ 2\epsilon_{23} \\ 2\epsilon_{13} \\ 2\epsilon_{12} \end{array} \right]\ = \left[ \begin{array} {ccc} \epsilon_{11} \\ \epsilon_{22} \\ \epsilon_{33} \\ \gamma_{23} \\ \gamma_{13} \\ \gamma_{12} \end{array} \right]\ $$

In engineering notation, we can also write: $U=\frac{1}{2}\sigma_i \epsilon_i$, and this is equivalent to the tensorial definition; you get the same factors of two when you switch from $\epsilon_i$ to $\epsilon_{ij}$ notation. The various defintions of shear strains can get confusing!

[Physics] Origin of the major symmetry property of the elasticity tensor

You should start with the strain energy density $\psi$, then define: $$ C_{ijkl} = \frac{\partial^2 \psi}{\partial \epsilon_{ij}\partial \epsilon_{kl}}, $$

and then define $$ \sigma_{ij} = C_{ijkl} \epsilon_{kl} $$

The remainder of my answer will be about explaining why you have to do it that way. Firstly it is physical, there really is energy associated with strain, and if there weren't there would not be any stress. Secondly, it is linear exactly because we are considering the changes in energy due to small strains.

But let's go back to the minor symmetries. We need $C_{ijkl}=C_{jikl}$ because otherwise $\sigma_{ij}\neq\sigma_{ji}$ (and then we get arbitrarily large, hence unphysical, angular velocities for smaller and smaller regions). But the other minor symmetry is not required. If someone handed you a random rank four tensor, let's call it $B_{ijkl}$, and called it an elastic tensor and it didn't have the second minor symmetry, you can define $C_{ijkl}=(B_{ijkl}+B_{ijlk})/2$ $D_{ijkl}=(B_{ijkl}-B_{ijlk})/2$ and then $B=C+D$ but when D is contracted with a symmetric rank two tensor (like the strain tensor) it gives zero. So the part of the rank four tensor without that second minor symmetry simply doesn't contribute, as an actor it does nothing (when you think all the elasticity does is give you stress from strain). So you might as well assume your tensor has both minor symmetries because it acts like it has the second ($B$ and $C$ act the same on symmetric tensors) and it has to have the first.

Did I bring that up to be pedantic? No, I brought it up because the same thing happens if you contract the elasticity tensor with a rank four symmetric combination of the strain tensor. The part of the elasticity tensor without the major symmetry doesn't contribute to the strain energy density. So a random tensor needs the first minor symmetry to be physical. But you might as well assume it has the second minor symmetry since it doesn't affect the stress-strain relationship. And you might as well assume it has the major symmetry because the part that doesn't will not contribute to the strain energy density.

But it is the strain energy density that is physical, and how it changes is what elasticity is. So you aren't really deriving these symmetries as much as saying that only the symmetric ones generate the physical things you want, energy when given strain. And a real derivation should start with strain energy density and strain, and then just define elasticity from that.

Best Answer

Related Solutions

[Physics] Strain energy density in index notation

[Physics] Origin of the major symmetry property of the elasticity tensor

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