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Isotropic hardening

Problem description: Consider the rod shown subjected to cyclic loading. Compute elastic-plastic response using linear isotropic hardening. See also examples V.4, V.5 and V.6.

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, $\nu=0.3$.
Prandtl-Reuss-von Mises model with linear isotropic hardening.

$\epsilon_p$:	0.000	0.015	0.040
$\sigma_Y$:	350	650	1150	[MPa]
QY:	0	0	0	[MPa]

Support: Clamped at x=0. Statically determinate.

Loading: $\sigma_{xx}=\pm 400$ MPa, 3 cycles.

Solution: Under uniaxial stress loadings the kinematic/isotropic yield condition implemented in the PMD package reduces to

$$\vert\sigma-h\vert\le H(\epsilon_p)=\sigma_Y(\epsilon_p)-Q_Y(\epsilon_p)$$

where $\sigma$, h, $\sigma_Y$, Q_Y, $\epsilon_p$ are the applied stress, backstress, subsequent yield stress, kinematic function, and the cumulated plastic strain, respectively. Denoting by $\sigma_{Yt}=h+H$, $\sigma_{Yc}=h-H$ the yield stresses in `tension' and `compression' and lifting the absolute from the above equation, the yield condition can be recast as

$$\sigma_{Yc} \le \sigma \le \sigma_{Yt}$$

It should be noted, however, that for pronounced Bauschinger's effect situations with $\sigma_{Yc}>0$, $\sigma_{Yt}<0$ might be encountered. See picture on next page.

If for some fixed values of internal variables $\epsilon_p$, h, further denoted as `old', we observe that $\vert\sigma-h_{old}\vert>H_{old}$, these internal variables must change so that $\vert\sigma-h_{new}\vert=H_{new}$. This is done as follows. Define $\Delta\sigma=\vert\sigma-h_{old}\vert-H_{old}>0$ and compute $\Delta\epsilon_p$ such that

$$\sigma_Y(\epsilon_p+\Delta\epsilon_p)-\sigma_Y(\epsilon_p)=\Delta\sigma$$

For linear hardening this equation becomes

$$E_p\Delta\epsilon_p=\Delta\sigma$$

where $E_p=\,\mbox{\rm d}\sigma_Y/\,\mbox{\rm d}\epsilon_p$ is the plastic modulus. New value of backstress h_new is computed from

$$\vert\sigma-h_{new}\vert=\sigma_Y(\epsilon_p+\Delta\epsilon_p) -Q_Y(\epsilon_p+\Delta\epsilon_p)$$

Graphically, the right edge of the elastic domain $\sigma_{Yt}=h+H$ is stretched by $\Delta\sigma$ to the right and its centroid h is then updated according to Q_Y. If $Q_Y\equiv 0$ then h_new=h_old=0; the elastic range 2H extends symmetrically by $2\Delta\sigma$ each time and material hardens isotropically. If Q_Y has the same slope as $\sigma_Y$, the elastic range remains constant and the yield surface moves about in stress space as a rigid body--hardening is said to be kinematic. In general, a proper adjustment of Q_Y allows the elastic domain to move and stretch simultaneously--see sections V.5 and V.6.

In this example we have

$$E_p=\frac{650-350}{0.015}=\frac{1150-350}{0.04}=2\times10^4\,\mbox{[MPa]}$$

In the course of first loading the yield stresses in tension and compression increase to $\pm 400\,$MPa so that all the subsequent cycles proceed in elastic regime.

Loading	$\Delta\sigma$	$\epsilon_p$	H	$\sigma_{Yc}$	$\sigma_{Yt}$
0	0	0	350	-350	350
+400	50	$2.5\times10^{-3}$	400	-400	400
-400	0	$2.5\times10^{-3}$	400	-400	400
+400	0	$2.5\times10^{-3}$	400	-400	400
-400	0	$2.5\times10^{-3}$	400	-400	400
+400	0	$2.5\times10^{-3}$	400	-400	400
-400	0	$2.5\times10^{-3}$	400	-400	400