If stress is applied to a material but does not exceed the yield point, when the stress is relieved, the material such as steel returns to its original size and does not experience any permanent deformation. However, if the stress exceeds the yield stress, the material deforms plastically. The method I see for determining this deformation is to shift the $\sigma=E\varepsilon$ line horizontally until it aligns with the maximum stress, then find where this line intercepts the $\sigma=0$ axis, as shown below. This is considered the plastic deformation and the new 0 stress size of the material.

Plastic Deformation Found Using Stress Strain Curve

From here, I would reason that this same stress-strain curve could be used again. A simple shift of by $\varepsilon_{plastic}$ and scaling by $\frac{1}{1+\varepsilon_{plastic}}$ of strain should suffice to obtain a new stress strain curve from the original, like so:

New Stress-Strain Curve

Is this a valid method for determining a new stress strain curve, or does the releasing of the material allow the material to restructure itself in some way to create a different, non-trivial change in the stress-strain curve?

Trivial Versus Non-Trivial Change

More generally, materials with a given chemical composition and temperature history will, all else equal, have identical stress strain curves. If multiple samples of this material with identical starting conditions are deformed plastically through multiple deformations and releases, will their final stress strain curves be identical as long as the maximum stress each endured was the same, or is the process path dependent in that releasing a material alters its stress strain curve in some non-trivial manner, and therefor materials which experienced the same maximum stress but different intermediate stresses will have different stress strain curves?

  • $\begingroup$ You are considering a plasticity model, but take into account that there are a lot of non-reversible deformation models in addition to plasticity. So the affirmation that beyond a yield/limit stress "the material deforms plastically" is not in general de case. For example a material can undergo continuous damage models: en.wikipedia.org/wiki/Damage_mechanics $\endgroup$
    – Davius
    Commented Jul 4, 2022 at 10:00

1 Answer 1


It depends on the material and temperature. For steel at room temperature, the process can be said most path independent.

What can affect the outcome are thermally activated processes as diffusion of interstitial impurity atoms, that can migrate to stressed regions around dislocations. The dislocation density increases during plastic deformation, and that migration tends to increase the yield point.

Hydrogen for instance can diffuse even at room temperatures.

On the other hand, increasing temperature makes the self difusion of the lattice atoms relevant, changing the morphology of the dislocation arrays and releasing elastic energy, what tends to soften the material.


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