3
$\begingroup$

Equivalent plastic strain rate is defined as

$$ \dot{\bar{\epsilon}}=\sqrt{\frac{2}{3}\dot{\epsilon_{ij}}^{p}\dot{\epsilon_{ij}}^{p} } $$

Where, $ \dot{\bar{\epsilon}}$ is equivalent plastic strain rate

$\dot{\epsilon_{ij}}^{p}$ is plastic strain rate. My question is how to find $\dot{\epsilon_{ij}}^{p}$ ?

A similar relation holds true for equivalent stress:

$$ \bar{\sigma}=\sqrt{\frac{2}{3}s_{ij}s_{ij} } $$ But there is relation between $s_{ij}$ and $\sigma_{ij}$ $$ s_{ij}=\sigma_{ij}-\frac{1}{3}\sigma_{kk}\delta_{ij} $$ $\sigma$ is Cauchy stress tensor, $\bar\sigma$ is equivalent plastic stress, $s$ is deviatoric stress defined by above relation, $\delta$ is Kronecker delta. Is there any such relation for strain?

$\endgroup$
  • $\begingroup$ In general, it depends. Can I assume that you are using rate-independent plasticity with isotropic hardening? If so, you can find $\dot{\bar{\epsilon}}^p$ by using the Kuhn-Tucker conditions. It's fairly straightforward, albeit with a fair bit of algebra. If you'd like, I can write out a long-form derivation below. $\endgroup$ – Tyler Olsen Jun 28 '15 at 22:35
  • $\begingroup$ Yes, I want to use rate independent plasticity. If you refer me to a source, it would be great. $\endgroup$ – 343_458 Jun 29 '15 at 13:34
4
$\begingroup$

First, I'm going to define some notation. I prefer the direct notation of modern continuum mechanics as presented in this book by Gurtin et al., so I won't be carrying around the indices that you use. Lastly, the following holds for small deformation, rate-independent plasticity with isotropic hardening.

  • $\mathbf{T}$ = Cauchy stress tensor
  • $\mathbf{T}_0 = \mathbf{T} - \frac{1}{3}\mathrm{tr}(\mathbf{T}) \mathbf{1}$ = Deviatoric part of Cauchy stress
  • $\bar{\sigma} = \sqrt{\frac{2}{3} \mathbf{T}_0 : \mathbf{T}_0}$ = Equivalent tensile stress
  • $\mathbf{N}^p = \sqrt{\frac{3}{2}}\frac{\mathbf{T}_0}{\bar{\sigma}}$ = Direction of plastic flow (assume co-directionality)
  • $\mathbf{E} = \mathbf{E}^e + \mathbf{E}^p$ = Additive strain decomposition
  • $\mathbf{T} = \mathbb{C}:\mathbf{E}^e = (2G)\mathbf{E}^e + \lambda\mathrm{tr}(\mathbf{E}^e)\mathbf{1}$ = Linearized Elasticity constitutive equation
  • $\dot{\mathbf{E}}^p = \sqrt{\frac{3}{2}}\dot{\bar{\epsilon}}^p \mathbf{N}^p$
  • $\dot{\bar{\epsilon}}^p = \sqrt{\frac{2}{3}} |\dot{\mathbf{E}}^p|$ = equivalent plastic strain rate
  • $\bar{\epsilon}^p = \int_0^t \dot{\bar{\epsilon}}^p(t')\,dt'$ = accumulated plastic strain
  • $Y(\bar{\epsilon}^p)$ = "flow resistance" = 1D tensile yield stress $\sigma_y$ (function of $\bar{\epsilon}^p$)
  • $H(\bar{\epsilon}^p) = \frac{dY}{d\bar{\epsilon}^p}$ = strain-hardening rate
  • $f(\mathbf{T}_0, Y(\bar{\epsilon}^p)) = \bar{\sigma} - Y(\bar{\epsilon}^p) \le 0 \; \forall \,\mathbf{T}, \,Y$ <-- Yield Function

Next, we write out the Kuhn-Tucker consistency condition at yield (which occurs at $f=0$):

  • If, $f = 0$ then $\dot{\bar{\epsilon}}^p \dot{f} = 0$.

This may seem like a strong statement, and it is, but if you think about all of the possible types of loading from a point on the "yield surface" (defined as all stress states that give $f=0$), you will find this to be true.


Now, we apply this condition to find $\dot{\bar{\epsilon}}^p$. First, compute $\dot{f}$.

\begin{align} \dot{f} &= \sqrt{\frac{3}{2}} \dot{\bar{\sigma}} - \dot{Y}\\ &= \sqrt{\frac{3}{2}} \frac{\dot{\mathbf{T}}_0 : \mathbf{T}_0}{|\mathbf{T}_0|} - H(\bar{\epsilon}^p)\dot{\bar{\epsilon}}^p\\ &= \sqrt{\frac{3}{2}} \dot{\mathbf{T}}_0 : \mathbf{N}^p - H(\bar{\epsilon}^p)\dot{\bar{\epsilon}}^p \\ &= \left(\sqrt{\frac{3}{2}}\right)(2G)[\dot{\mathbf{E}} - \dot{\mathbf{E}}^p] : \mathbf{N}^p - H(\bar{\epsilon}^p)\dot{\bar{\epsilon}}^p \\ &= \left(\sqrt{\frac{3}{2}}\right)(2G)\dot{\mathbf{E}} : \mathbf{N}^p - (3G + H)\dot{\bar{\epsilon}}^p \end{align}

Now, we have to consider three cases: 1) Elastic unloading, 2) Neutral Loading, 3) Plastic Loading

  1. Elastic Unloading: This is like releasing the stress on a body that is at its yield point. Thus, $\dot{\bar{\epsilon}}^p = 0$, and $\dot{\mathbf{E}}:\mathbf{N}^p < 0$. Therefore, $\dot{f} < 0$.
  2. Neutral Loading: This one is a bit trickier. Here, the loading is tangent to the yield surface. For this case, the strain rate is orthogonal to the direction of plastic flow, thus $\dot{\mathbf{E}}:\mathbf{N}^p = 0$. Here, still, we have not loaded the body plastically, so $\dot{\bar{\epsilon}}^p = 0$. In codes, this is usually handled without special consideration by either the above or below cases, with the same result.
  3. Plastic Loading: Here, we are loading the body plastically. This means that $\dot{\mathbf{E}}:\mathbf{N}^p > 0$. Since $f=0$ at the yield surface, and $f$ cannot have a positive value, this means that $\dot{f} = 0$. This is how we can find the value of $\dot{\bar{\epsilon}}^p$.

$$ \dot{\bar{\epsilon}}^p = \frac{\sqrt{3/2} (2G) (\dot{\mathbf{E}}:\mathbf{N}^p)}{3G + H(\bar{\epsilon}^p)} $$

I glossed over some of the more tedius algebra and buried the assumption that $(3G + H)>0$, which is true for every case that I have encountered. You can go on from here to compute the so-called "elasto-plastic" modulus, which is necessary to do implicit integration in a finite element method.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ (+1) Good answer. Do you have any preferred references on this topic? $\endgroup$ – OnStrike Nov 17 '15 at 5:53
  • $\begingroup$ This answer came from a continuum mechanics text called "The Mechanics and Thermodynamics of Continua" by Gurtin, Fried, and Anand. This is a general continuum mechanics text, and it has a large section on several plasticity theories (most of which are large deformation theories, unlike this one). It is very much a THEORY book though. Might be worth looking through it on Amazon before you buy it to make sure it's what you're looking for. $\endgroup$ – Tyler Olsen Nov 17 '15 at 15:07
  • $\begingroup$ Thanks, Ill check it out. This topic is few and far between on Physics.SE. I recently migrated my question on Impact Force to Engineering.SE for lack of interest- seems like it might be in your wheel house? My apologies if this breaks convention. $\endgroup$ – OnStrike Nov 17 '15 at 20:01
  • $\begingroup$ Contact mechanics isn't really my specialty, but I can say that analytical solutions don't really exist in the general case. For idealized geometries (eg, sphere contacting half-space) and idealized constitutive response (linearized elasticity, no plasticity) you may be able to integrate the linear momentum ODE to find the sphere position vs time, from which you could extract the peak contact force and contact time. For general geometries, this would be nearly impossible without approaching the problem numerically (not necessarily FEM, just Herzian approximation). $\endgroup$ – Tyler Olsen Nov 17 '15 at 21:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.