# How to determine plastic strain rate

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?

• 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. Jun 28 '15 at 22:35
• Yes, I want to use rate independent plasticity. If you refer me to a source, it would be great. Jun 29 '15 at 13:34

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}

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.