Is steel relaxation plastic?

Question

Hooke's Law defines a linear-elastic relationship between stress and strain.

$$ \sigma = E\epsilon $$

Steel behaves very much like a linear-elastic material, following Hooke's Law closely. It does, however, display non-elastic behaviors such as relaxation. Relaxation is the behavior wherein a member under constant strain displays variable (and reducing) stress over time.

My question is: is relaxation plastic? If the relaxed member were released, how would it behave? Would it follow a path defined by its elastic modulus? If this is the case, then it will end with a plastic deformation, no? After all, when stressed, the member will have reached $\left(\sigma_1, \epsilon_1\right)$. After relaxation, it will reach $\left(\sigma_2, \epsilon_1\right)$. Once released, it would have to reach $\sigma =0$, which implies in $\epsilon = \epsilon_1 - \dfrac{\sigma_2}{E}$ and since $\sigma_2 <\sigma_1$, that implies in a nonzero $\epsilon$.

Or is there some other behavior? Does the elastic modulus change to allow for a return without plastic deformations?

Answer 1

In short, yes, relaxation should probably be considered plastic deformation, as plastic strain is defined as non-recoverable deformation when applied stresses are removed.

Definitional Explanation

Assume you have a sample of some material, in this case steel, and want to apply a load for an extended period of time, sufficiently long for noticeable relaxation to occur. The load is not sufficient to leave the elastic regime. Just after the load is applied but before relaxation begins, the strain in the material due to the load is $\varepsilon_{0}$. If the load is immediately removed, again before relaxation proceeds, all $\varepsilon_{0}$ is recovered and the steel material returns to its original shape.

If instead the material experiences the load sufficiently long so that relaxation proceeds, and the load is removed, then $|\varepsilon_{1}|<|\varepsilon_{0}|$ is recovered. As a result, not all of the strain is recovered. Therefore, there must have been non-recoverable strain of $|\varepsilon_{0}-\varepsilon_{1}|>0$ due to relaxation. Therefore, by definition, relaxation is plastic deformation.

Thermodynamic & Kinetic Explanation

If the definitional explanation is insufficient, we can also look at this from a thermodynamic and kinetic point of view. Suppose for the moment the steel is instead a single crystal of pure iron. Elastic strain stores energy in the crystal lattice. Because the energy is higher than its rest state, there is free energy available to do work, and thus a driving force for reorganization of atoms in the crystal lattice. There are also point defects in the lattice in the form of vacancies, or missing atoms. Random fluctuations cause neighboring atoms to fill the vacancies, which results in the vacancies moving around the lattice. The vacancies provide a means for reorganization of the atoms.

Note that if the strain is not isotropic (i.e. is not purely hydrostatic), then the lattice strain field makes vacancies slightly larger in tensile-strain directions than in compressive-strain directions. As a result, the energy barrier to moving in the tensile directions will be lower than in the compressive directions. Think of the atoms being squeezed out from between their compressive-direction neighbors along the tensile directions. There will thus be a net flow of atoms in the crystal, with atoms tending to move from directions of high compression to directions of high tension. The overall, long term effect is to extend the crystal in the directions of tension and shorten the crystal in the directions of compression, causing a non-recoverable deformation. The same effects occur with multiple grains, except the mechanics are complicated by the presence of grain boundaries and varying crystal orientations. The same effects also occur with the presence of interstitial atoms like carbon, and they probably have negligible effect on vacancy motion as they don't get in the way (though I am not 100% sure of this part, see note below).

The above is a most-likely theory based on theories of vacancy flow and grain boundary migration due to thermal stresses (e.g. creep and grain-growth) and from dislocation motion, which have been observed directly. The described behavior for relaxation, however, has not been observed directly to the best of my knowledge (i.e. with a tunneling electron microscope).

Note

*Interstitial atoms will have lower energy in interstitial sites aligned with the tensile directions, as those sites are slightly increased in volume. This is related to anelastic strain and martensite formation, but may or may not have an impact on relaxation. However it is worth noting that purely axial strain may induce anisotropic properties in the steel.