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

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

Uniaxial strain.

$\varepsilon_0=-2.666\times10^{-3}.$

• Quadrilaterals, BEAM6P1
• Hexahedra, BEAM56P1

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.