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Kinematic hardening

Problem description: Consider the rod shown subjected to cyclic loading. Compute elastic-plastic response using linear kinematic hardening. See also examples V.3, 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 kinematic hardening.

$\epsilon_p$:	0.000	0.015	0.040
$\sigma_Y$:	350	650	1150	[MPa]
QY:	0	300	800	[MPa]

Support: Clamped at x=0. Statically determinate.

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

Solution: For detailed explanation see section V.3. The plastic modulus E_p is computed as

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

Reloading the bar to 350 MPa the hysteresis loop closes and the cycle will stabilize.

Loading	$\Delta\sigma$	$\epsilon_p$	H	$\sigma_{Yc}$	$\sigma_{Yt}$
0	0	0	350	-350	350
+400	50	$2.5\times10^{-3}$	350	-300	400
-400	100	$7.5\times10^{-3}$	350	-400	300
+400	100	$12.5\times10^{-3}$	350	-300	400
-400	100	$17.5\times10^{-3}$	350	-400	300
+400	100	$22.5\times10^{-3}$	350	-300	400
-400	100	$27.5\times10^{-3}$	350	-400	300