Material Models
The aim of this issue is to guide a path through different material models. Thereby, the start is made with a "simpler" model for plastic material behaviour which should be extended incrementally in order to reach the material model of Johnson-Cook.
1. Linear Elastic - Linear Plastic
Until the yield point, linear elastic material behaviour is assumed. After this point, linear isotropic strain-hardening takes place, using the von-Mises-Criterion and assuming "associated" flow.
This constitutive law is the most commonly used model of inelastic deformation. It has the following properties:
- It will correctly predict the conditions necessary to initiate yield under multiaxial loading
- It will correctly predict the plastic strain rate under an arbitrary multiaxial stress state
- It can model accurately any uniaxial stress-strain curve
It has the following limitations:
- It is valid only for modest plastic strains (<10%)
- It will not predict creep behavior or strain rate sensitivity
- It does not predict behavior under cyclic loading correctly
- It will not predict plastic strains accurately if the principal axes of stress rotate significantly (more than about 30 degrees) during inelastic deformation
Ref.: Bower, Allan F. (2010) "Applied Mechanics of Solids", CRC Press, p. 128
For modelling plastic material behaviour, we assume to split the strain into an elastic and a plastic part. In a uniaxial case, this would be represented by
$$\varepsilon=\varepsilon^{(el)}+\varepsilon^{(p l)}. \tag{1}$$
The introduction of the von-Mises-Equivalent-Stress enables the further application of this stress in the yield criterion. Firstly, the equivalent stress:
$$\sigma_{eqv, M}=\sqrt{\sigma_x^2+\sigma_y^2+\sigma_z^2-\sigma_x \sigma_y-\sigma_x \sigma_z-\sigma_y \sigma_z+3\left(\tau_{x y}^2+\tau_{x z}^2+\tau_{y z}^2\right)} \tag{2}$$
The stress deviator is responsible for the plastic deformation of the solid. Therefore, we consider the hydrostatic pressure $\sigma_m$ and subtract it from the stress tensor. The result is the 2. invariant of the stress tensor, the so-called stress deviator $S$:
$$\sigma_m=\frac{\sigma_x+\sigma_y+\sigma_z}{3} \tag{3}$$
$$\underline{\underline{S}}=\left[\begin{array}{ccc}
\sigma_x & \tau_{x y} & \tau_{x z} \\\
\tau_{y x} & \sigma_y & \tau_{y z} \\\
\tau_{z x} & \tau_{z y} & \sigma_z
\end{array}\right]-\left[\begin{array}{ccc}
\sigma_m & 0 & 0 \\\
0 & \sigma_m & 0 \\\
0 & 0 & \sigma_m
\end{array}\right]=\left[\begin{array}{ccc}
\sigma_x-\sigma_m & \tau_{x y} & \tau_{x z} \\\
\tau_{y x} & \sigma_y-\sigma_m & \tau_{y z} \\\
\tau_{z x} & \tau_{z y} & \sigma_z-\sigma_m
\end{array}\right] \tag{4}$$
In another formulation, the von-Mises-Equivalent-Stress can be written as in the following equation:
$$\sigma_{eqv, M}=\sqrt{\frac{3}{2}S_{ij}S_{ij}} \tag{5}$$
The yield criterion describes that the stress within a plastically deformed solid cannot surpass a certain yield stress $Y$. Either, the occurring stress (represented by the von-Mises-Equivalent-Stress) is lower than the yield stress, or the occurring stress is exactly the yield stress. If we consider a stress lower than the yield stress, we have exclusively elastic strain. However, if we assume having a stress laying on the so-called yield surface, we see plastic deformation of the solid. In order to describe this phenomenon mathematically, we transform the von-Mises-Equivalent-Stress into the principal-normal-stress-form with $\sigma_1$, $\sigma_2$ and $\sigma_3$ and insert it in the yield criterion.
$$f\left(\sigma_{i j}, \bar{\varepsilon}^p\right)=\sqrt{\frac{1}{2}\left[\left(\sigma_1-\sigma_2\right)^2+\left(\sigma_1-\sigma_3\right)^2+\left(\sigma_2-\sigma_3\right)^2\right]}-Y\left(\bar{\varepsilon}^p\right)=0 \tag{6}$$
The accumulated plastic strain magnitude is defined in the following way, can, however, be computed in another way too, as we'll see in equation 11:
$$\bar{\varepsilon}^p=\int \sqrt{\frac{2}{3} d \varepsilon_{i j}^p d \varepsilon_{i j}^p} \tag{7}$$
The yield stress is dependent of the accumulated plastic strain magnitude. As we assume a linear strain hardening solid, we have to take the following equation into consideration when talking about the current yield stress of the solid. With $h$ representing the gradient of the yield stress dependent of the accumulated plastic strain, we receive:
$$Y(\bar{\varepsilon}^p)=Y_0+h \cdot \bar{\varepsilon}^p \tag{8}$$
In order to receive the plastic strain, the derivation of $f$ in respect to $\sigma_{ij}$ is neccessary. After applying some maths, the following term results:
$$\frac{\partial f}{\partial \sigma_{i j}}=\frac{3}{2} \frac{S_{i j}}{\sqrt{\frac{3}{2} S_{k l} S_{k l}}}=\frac{3}{2} \frac{S_{i j}}{Y} \tag{9}$$
Since we assume an associated flow (the flow potential equals the yield function $f$) and isotropic hardening, this is the equation we get for plastic strain:
$$d \varepsilon_{i j}^p=d \bar{\varepsilon}^p \frac{\partial f}{\partial \sigma_{i j}}=d \bar{\varepsilon}^p \frac{3}{2} \frac{S_{i j}}{Y} \tag{10}$$
As we remain on the yield surface during deformation, the yield criterion demands the following equation:
$$\begin{gathered}
\dot{f}=0 \\\
\frac{\partial f}{\partial \sigma_{i j}} d \sigma_{i j}-\frac{\partial Y}{\partial \bar{\varepsilon}^p} d \bar{\varepsilon}^p=0 \\\
d \bar{\varepsilon}^p=\frac{1}{h} \frac{\partial f}{\partial \sigma_{i j}} d \sigma_{i j}=\frac{1}{h} \frac{3}{2} \frac{S_{i j} d \sigma_{i j}}{Y} \quad \text{with } h=\frac{d Y}{d \bar{\varepsilon}^p}
\end{gathered} \tag{11}$$
Insert this into equation 10 and we get:
$$d \varepsilon_{i j}= \begin{cases}0\\ \frac{1}{h} \frac{3}{2} \frac{\left\langle S_{k l} d \sigma_{k l}\right\rangle}{Y} \frac{3}{2} \frac{S_{i j}}{Y} & \sqrt{\frac{3}{2} S_{i j} S_{i j}}-Y\left(\bar{\varepsilon}^p\right)=0\end{cases} \tag{12}$$
$$\text { where }\langle x\rangle= \begin{cases}x & x \geq 0 \\ 0 & x \leq 0\end{cases}$$
The total strain is assembled by an elastic and a plastic part of the strain. Therefore, we need to consider the elastic strain as well:
$$d \varepsilon_{i j}^e=\frac{1+\nu}{E}\left(d \sigma_{i j}-\frac{\nu}{1+\nu} d \sigma_{k k} \delta_{i j}\right) \tag{13}$$
Now, let's combine the elastic and the plastic part of the strain:
$$d \varepsilon_{i j}= \begin{cases}\frac{1+\nu}{E}\left(d \sigma_{i j}-\frac{\nu}{1+\nu} d \sigma_{k k} \delta_{i j}\right) & \sqrt{\frac{3}{2} S_{i j} S_{i j}}-Y\left(\bar{\varepsilon}^p\right)<0 \\ \frac{1+\nu}{E}\left(d \sigma_{i j}-\frac{\nu}{1+\nu} d \sigma_{k k} \delta_{i j}\right)+\frac{1}{h} \frac{3}{2} \frac{\left\langle S_{k l} d \sigma_{k l}\right\rangle}{Y} \frac{3}{2} \frac{S_{i j}}{Y} & \sqrt{\frac{3}{2} S_{i j} S_{i j}}-Y\left(\bar{\varepsilon}^p\right)=0\end{cases} \tag{14}$$
In order to receive the stress increment, the previous equation (14) needs to be inverted and as a result, the following equation remains:
$$d \sigma_{i j}= \begin{cases}\frac{E}{1+\nu}\left\{d \varepsilon_{i j}+\frac{\nu}{1-2 \nu} d \varepsilon_{k k} \delta_{i j}\right\} & \sqrt{\frac{3}{2} S_{i j} S_{i j}}-Y\left(\bar{\varepsilon}^p\right)<0 \\ \frac{E}{1+\nu}\left\{d \varepsilon_{i j}+\frac{\nu}{1-2 \nu} d \varepsilon_{k k} \delta_{i j}-\frac{3 E}{3 E+2(1+\nu) h} \frac{3}{2} \frac{\left\langle S_{k l} d \varepsilon_{k l}\right\rangle}{Y} \frac{3}{2} \frac{S_{i j}}{Y}\right\} & \sqrt{\frac{3}{2} S_{i j} S_{i j}}-Y\left(\bar{\varepsilon}^p\right)=0\end{cases} \tag{15}$$
Sources:
- Bower, Allan F. (2010) "Applied Mechanics of Solids", CRC Press
- Krabbenhoft, Kristian (2002) "Basic Computational Plasticity", Department of Civil Engineering
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