Consider a rectangular prism of steel subject to a compressive pressure. When the pressure exceeds the yield point, the prism will experience plastic deformation, permanently changing the length of the prism. Does this change in length depend on the original length of the prism, and, if so, how?

I know elastic deformation is thickness dependent, in that the change in thickness will be proportional to the original thickness. This comes from the stress-strain equation $\sigma=E\epsilon$ where strain $\epsilon=\frac{\Delta L}{L_0}$. Solving for $\Delta L$ produces $\Delta L=\frac{\sigma L_0}{E}$, indicating that the change in length is proportional to the original length.

Most literature I find provides merely a qualitative analysis of plastic deformation, particularly that the plastic region is nonlinear and has a certain tensile strength. I do not find this description useful for modeling behavior in the plastic deformation region. Does a similar relation for length and change in length exist for plastic deformation?

Edit: As noted by @alephzero, work hardening occurs in the deformation of metals. As such, a constantly applied pressure which exceeds the initial yield point plastically deforms the material, at which point work hardening strengthens the metal, the yield point changes, and static equilibrium is reached.

That being said, I am interested in how this total plastic deformation relates to the applied pressure, length of steel in compression, and intrinsic properties of the steel - particularly, is the total plastic deformation related linearly or not at all to the length of steel.

An acceptable answer should at least describe what factors (length/hardness/structure) affect the amount of plastic deformation and would preferably include equations, rather than only qualitative analysis.


3 Answers 3


Edit(Correction of grammar etc. to clarify statements): In short, plasticity can be rate dependent and therefore more or less time dependent. But generally history dependent. Time dependence is described as viscoplasticity. For a basic understanding, I will take a very simple rheological model called prandtl and will derive some equations. The left picture is the prandtl model and the right one is the expansion with work hardening.

rheological models The basic idea is that the rheological elements show the behavior that is desired to model a specific material behaviour effect. We have a hooke spring for elasticity and a friction block. To move the friction block (St. Venant model) you need to overcome the stiction. The total strain is described as $\varepsilon_t = \varepsilon_{el} +\varepsilon_{pl}$ and tension will be $\sigma\leq\sigma_{Y}$ and $\sigma=E_1 \varepsilon_{el}$. With $\dot\varepsilon_{pl}=0$ for $\sigma<\sigma_{Y}$ and $\varepsilon_{el}=const$ for $\sigma_Y$. In this model it is not defined how big the plastic deformation will be.

If we take the second model, which has some kind of hardening effect, and do the same, we get $\sigma=\sigma_l + \sigma_r=E_1 \varepsilon_{el}+E_2 \varepsilon_{t}$. Now at yield stress we get $\sigma=\sigma_Y+\Delta\sigma_r=\sigma_Y+E_2 \varepsilon_{pl}$ and therefore $\varepsilon_{pl}=\frac{\sigma-\sigma_Y}{E_2}$. In this model we could actually calculate some kind of displacement for a metal prism.

At yield stress: $\varepsilon_t=\varepsilon_{el}=\frac{\sigma_Y}{E1+E2}$

And for the total strain for $\sigma>\sigma_Y$:


For your example, you could simply multiply the strain with the length of the prism to get what you want. For more complicated situations, you would have to determine the exact state of stress in every point of your structure and then get your strains and displacements. That being said, plastic deformation is dependend on the stress and this can be influenced for a given loadcase by either the material choice, or structural design.

Actually used models for plasticity are usually way more complicated and are often written down in a way that is defined in rates. In these models we talk about so called history variables that need to taken into account to describe a specific moment in the load history of a structure and are needed to get to the next point in the load scenario.

  • $\begingroup$ I appreciate the equations as well as reference to a model, which I will look more into. One thing I notice, is that these equations imply that upon reaching the yield point, the material experiences plastic deformation equal to $-\frac{\sigma_Y}{E_1}$, resulting in a discontinuous leap. Perhaps this is modeled by a kinetic friction component (which has been left out), but could you explain that peculiarity? $\endgroup$
    – Groger
    Commented Apr 23, 2020 at 19:48
  • $\begingroup$ Looking at some sources, in particular homepages.engineering.auckland.ac.nz/~pkel015/… I am finding that plastic deformation is rate-dependent only for viscous materials, therefore non-viscous materials like metals would experience rate-independent plastic deformation. $\endgroup$
    – Groger
    Commented Apr 23, 2020 at 20:05
  • $\begingroup$ Im am sorry for the inconvenience, I wrote this yesterday in a haste. Corrected now quite a few mistakes. You are correct, this kinetic friction would be a dashpod in the case of rheological models. But in this case there is no rate dependency. But for metals at higher temperature, there can be viscoplasticity. Also corrected the plastic and elastic strains. As you already saw the models presented dont actually have a rate dependency and therefore you are able to calculate your total deformation for this simple model by multiplication with the length of the prism. $\endgroup$
    – scheepan
    Commented Apr 24, 2020 at 13:12
  • $\begingroup$ Thanks for the edit. If you could, please qualify what a "higher temperature" is. I am particularly interested in this subject relating to hot-rolling applications, around 1600F. Would viscoplasticity play a role here? $\endgroup$
    – Groger
    Commented Apr 24, 2020 at 14:13
  • 1
    $\begingroup$ No problem. It could play a role. In my case I am working with temperatures around 700-900°C and in this range we have viscoplastic behaviour for a nickel based alloy. But this really depends on the metal you are using. A rule of thumb for when viscoplasticity begins is around a third of the melting temperature. $\endgroup$
    – scheepan
    Commented Apr 24, 2020 at 14:33

For an idealized elastic - perfectly plastic material, theoretically when the stress reaches the yield stress the plastic deformation will increase indefinitely for as long as the stress is applied.

In practice ductile metals are not perfectly plastic, and the plastic deformation causes "work hardening" or "strain hardening" which raises the yield stress as the plastic deformation increases.

Also, a real material is not a homogeneous continuum but has an internal grain structure, and as the plastic deformation increases, eventually the cracks between the grains will grow enough to cause the object to break.

  • $\begingroup$ Thank you for your answer. Based on this, am I right to understand that plastic deformation has a time dependent component to it? $\endgroup$
    – Groger
    Commented Apr 21, 2020 at 19:28

Suppose a metal sample of squared cross section $S = b*b$ and length $L$ being plastically compressed by an small $\Delta L$ in the $L$ direction, (that is called $z$) and assuming no friction between the contact surfaces.

Considering constancy of volume and equal spread of the sides due to the symmetry of the piece, and negleting the products of small $\Delta$'s, we get a relation for infinitesimal deformations:

$$(L+\Delta L)(b+\Delta b)^2 = Lb^2$$ $$2Lb\Delta b + b^2\Delta L$ = 0$$ $$\frac{\Delta L}{L} + 2\frac{\Delta b}{b} = 0 => d\epsilon_z = -2d\epsilon_i$$ where $i = x, y$

It is an uniaxial situation, and only $\sigma_z \neq 0$. The average stress: $$\sigma_{av} = \frac{0 + 0 + \sigma _z}{3} = \frac{\sigma_z}{3} $$

If we take the difference between the components of the stresses and the average, we have the so called deviatoric stress tensor: $$\sigma^d_x = -\frac{1}{3}\sigma_z$$ $$\sigma^d_y = -\frac{1}{3}\sigma_z$$ $$\sigma^d_z = \frac{2}{3}\sigma_z$$

As it can be observed, $\sigma^d_i = k d\epsilon_i$

The basic postulate to modelling plastic deformation on metals is extending that uniaxial situation to a generic one:

$$\boldsymbol{ \sigma^d} = k \boldsymbol{ d\epsilon}$$

The infinitesimal plastic deformation tensor is proportional to the deviatoric stress tensor, what means that the directions of the infinitesimal flow of material are proportional to the stress tensor in each point.

  • $\begingroup$ For infinitesimal deformations, perhaps, but this does not extend to larger deformations, where nonlinear behavior takes over. This seems more kin to elastic deformation with perhaps a different constant. $\endgroup$
    – Groger
    Commented Apr 28, 2020 at 11:40
  • $\begingroup$ But that infinitesimal equation allows finite element calculation for hot rolling or forging. If I made in the example one side expands more that the other, (keeping the same volume), the equation would require side tensions. As they don't exist by the boundary conditions, an algorithm could correct the small deformation tensor. $\endgroup$ Commented Apr 28, 2020 at 14:41
  • $\begingroup$ That's not what I find fault with. I'm okay with equal expansion; however, under plastic deformation, stress is not linearly proportional to strain. Perhaps what I'm missing in this example is that $\sigma=kd\epsilon$ should be integrated to get the relation $\sigma^2=2k\epsilon+c$, which is more profound of a statement. Is this the intended application of the formula? $\endgroup$
    – Groger
    Commented Apr 28, 2020 at 14:58
  • $\begingroup$ It is always possible to get an uniaxial equation $\sigma x \epsilon$ for a given material, temperature and strain rate, But it is only useful in a 3-D situation to have the Von Mises stress as a function of the total plastic strain. $\endgroup$ Commented Apr 29, 2020 at 0:46

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