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Uniaxial strain

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

Mesh: Use the following element types:
ITE=6--see appendix A.1, four quadrilaterals
ITE=56--see appendix B.1, four hexahedra.

Material properties: $E=2\times 10^5$ MPa, $\nu=0.3$, $\sigma_Y=200$ MPa,
Prandtl-Reuss-von Mises model.

Support: Uniaxial strain.

Loading: $\epsilon_0=-2.666\times10^{-3}$.

Solution: 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

$${\bf S} = \left[\begin{array}{ccc} S_0 & 0 & 0 \\ 0 & S_r & 0 \\ 0 & 0 & S_r \\ \end{array}\right]$$

Because $\mbox{tr}({\bf S})=S_0+S_r+S_r=0$, 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

$${\bf S} = \frac{\sigma_Y}{3}\left[\begin{array}{rcc} -2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right]$$

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

$$\Delta\mbox{\boldmath$\epsilon$ }^p=\lambda{\bf S}=\lambda\fr... ... & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right]~,~~~\lambda\ge 0$$

or $\Delta\epsilon_0^p=-2\lambda\sigma_Y/3$ and $\Delta\epsilon_r^p=-\frac{1}{2}\Delta\epsilon_0^p$, where $\lambda$ is the plastic multiplier to be determined. Employing boundary conditions in the radial direction

$$\Delta\epsilon_r=\Delta\epsilon_r^e + \Delta\epsilon_r^p = 0$$

therefore, $\Delta\epsilon_r^e=-\Delta\epsilon_r^p=\frac{1}{2}\Delta\epsilon_0^p$. Since ${\bf S}$ is constant, any possible increment of the stress tensor involves its hydrostatic part only, hence

$$\Delta\epsilon_0^e= \Delta\epsilon_r^e= \textstyle\frac{1}{2}\Delta\epsilon_0^p$$

Obviously, the total strain increment can now be decomposed into two parts as

$$\Delta\epsilon_0=\Delta\epsilon_0^e + \Delta\epsilon_0^p~,~\m... ...~,~~~ \Delta\epsilon_0^p=\textstyle\frac{2}{3}\Delta\epsilon_0$$

and the corresponding change of hydrostatic stress

$$\Delta\sigma_0=\frac{E}{1-2\nu}\Delta\epsilon_0^e= \frac{E}{3(1-2\nu)}\Delta\epsilon_0$$

It is interesting to note that the tangent modulus

$$E_t(\mbox{elastic-plastic})=\frac{E}{3(1-2\nu)} > 0$$

although there is no actual material hardening ( $\sigma_Y=\mbox{const.}$).

The initial slope of the stress-strain curve is obtained from Hooke's law applied to the uniaxial state of strain

$$E_t(\mbox{elastic}) = \frac{(1-\nu)E}{(1+\nu)(1-2\nu)}$$

while the critical state on the onset of plastic yielding is computed from von Mises' criterion as

$$\begin{array}{rcl} \vert\sigma_0(\mbox{yield})\vert &=& \disp... ...aystyle\frac{1+\nu}{E}\sigma_Y = 1.333\times10^{-3} \end{array}$$

The total strain was given as $\epsilon_0=-2.666\times10^{-3}$, thus we may define an increment $\Delta\epsilon_0=\epsilon_0-\epsilon_0(\mbox{yield}) =-1.333\times10^{-3}$ and using

$$\Delta\sigma_0=E_t(\mbox{elastic-plastic})\Delta\epsilon_0= -216.67\,\mbox{MPa}$$

we find that $\sigma_0=\sigma_0(\mbox{yield})+\Delta\sigma_0= -567\,\mbox{MPa}$. The complete solution is plotted on next page.