Package for Machine Design

Finite Element Analysis in Structural Mechanics

User Tools

Site Tools


Uniaxial strain

by Dr. Jiří Plešek

Problem description

An elastic–perfectly plastic material is compressed under uniaxial strain conditions. Compute the total stress-strain response.

Material properties

$E=2\times10^5\text{ MPa},$ $\nu=0.3,$ $\sigma_Y=200\text{ MPa}.$ Prandtl–Reuss–von Mises model.


Uniaxial strain.




Owing to symmetry there are only two independent components of the stress and strain tensors; the axial component denoted by subscript $0$ and the radial one designated by $r.$ The deviatoric stress tensor takes the form $$\mathbf{S}=\left[\begin{array}{ccc} S_0 & 0 & 0 \\ 0 & S_r & 0 \\ 0 & 0 & S_r \\ \end{array}\right].$$

Because $\mathrm{tr}(\mathbf{S})=S_0+S_r+S_r=0,$ we have $S_r=-\frac{1}{2}S_0.$ Furthermore, at the plastic state von Mises' yield condition must be satisfied, i.e., $$\sigma_e=\sqrt{\textstyle\frac{3}{2}(S_0^2+S_r^2+S_r^2)}=\textstyle\frac{3}{2}\vert S_0\vert=\sigma_Y.$$

Thus, in the course of plastic compression $$\mathbf{S}=\frac{\sigma_Y}{3}\left[\begin{array}{rcc} -2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right].$$

It should be noted that the deviatoric stress remains constant during plastic flow, which simplifies the integration of constitutive relations. According to Prandtl–Reuss' equations the increment of plastic strain $$\Delta\mathbf{\varepsilon}^p=\lambda\mathbf{S}=\lambda\frac{\sigma_Y}{3}\left[\begin{array}{rcc} -2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right],\quad\lambda\ge 0$$ or $\Delta\varepsilon_0^p=-2\lambda\sigma_Y/3$ and $\Delta\varepsilon_r^p=-\frac{1}{2}\Delta\varepsilon_0^p,$ where $\lambda$ is the plastic multiplier to be determined. Employing boundary conditions in the radial direction $$\Delta\varepsilon_r=\Delta\varepsilon_r^e+\Delta\varepsilon_r^p=0$$ therefore, $\Delta\varepsilon_r^e=-\Delta\varepsilon_r^p=\frac{1}{2}\Delta\varepsilon_0^p.$ Since $\mathbf{S}$ is constant, any possible increment of the stress tensor involves its hydrostatic part only, hence $$\Delta\varepsilon_0^e=\Delta\varepsilon_r^e=\textstyle\frac{1}{2}\Delta\varepsilon_0^p.$$

Obviously, the total strain increment can now be decomposed into two parts as $$\Delta\varepsilon_0=\Delta\varepsilon_0^e+\Delta\varepsilon_0^p,\quad \text{where}\quad \Delta\varepsilon_0^e=\textstyle\frac{1}{3}\Delta\varepsilon_0,\quad \Delta\varepsilon_0^p=\textstyle\frac{2}{3}\Delta\varepsilon_0$$ and the corresponding change of hydrostatic stress $$\Delta\sigma_0=\frac{E}{1-2\nu}\Delta\varepsilon_0^e=\frac{E}{3(1-2\nu)}\Delta\varepsilon_0.$$ It is interesting to note that the tangent modulus $$E_t(\text{elastic-plastic})=\frac{E}{3(1-2\nu)} > 0$$ although there is no actual material hardening ($\sigma_Y=\text{const.}$).

The initial slope of the stress-strain curve is obtained from Hooke's law applied to the uniaxial state of strain $$E_t(\text{elastic})=\frac{(1-\nu)E}{(1+\nu)(1-2\nu)}$$ while the critical state on the onset of plastic yielding is computed from von Mises' criterion as $$\begin{align} \vert\sigma_0(\text{yield})\vert &= \frac{1-\nu}{1-2\nu}\sigma_Y=350\text{ MPa}, \\ \vert\varepsilon_0(\text{yield})\vert &= \frac{1+\nu}{E}\sigma_Y=1.333\times10^{-3}. \end{align}$$ The total strain was given as $\varepsilon_0=-2.666\times10^{-3},$ thus we may define an increment $\Delta\varepsilon_0=\varepsilon_0-\varepsilon_0(\text{yield})=-1.333\times10^{-3}$ and using $$\Delta\sigma_0=E_t(\text{elastic-plastic})\Delta\varepsilon_0=-216.67\text{ MPa}$$ we find that $\sigma_0=\sigma_0(\text{yield})+\Delta\sigma_0=-567\text{ MPa}.$ The complete solution is plotted below.

en/example/plas/1/start.txt · Last modified: 2022-02-11 14:25 by Petr Pařík