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Cyclic hardening

Problem description: Consider the rod shown subjected to cyclic loading. Compute elastic-plastic response using linear mixed-mode hardening. See also examples V.3, V.4 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 mixed hardening.

$\epsilon_p$:	0.000	0.015	0.040
$\sigma_Y$:	350	650	1150	[MPa]
QY:	0	270	770	[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

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

and the kinematic modulus K_p

$$K_p=\left\{\begin{array}{rcl} 270/0.015=1.8\times10^4\,\mbox{... ...{[MPa]} &~\mbox{for}~&\epsilon_p \ge 0.015 \end{array}\right.$$

In the course of the first two cycles the elastic range increases by $2(E_p-K_p)\epsilon_p$. During third loading up to 400 [MPa] the threshold value $\epsilon_p=0.015$ is exceeded and K_p=E_p. Further hardening is of the kinematic type, which causes the hysteresis loop to close as in example V.4. The response is said to be saturated.

Loading	$\Delta\sigma$	$\epsilon_p$	H	$\sigma_{Yc}$	$\sigma_{Yt}$
0	0	0	350	-350	350
+400	50	$2.5\times10^{-3}$	355	-310	400
-400	90	$7.0\times10^{-3}$	364	-400	328
+400	72	$10.6\times10^{-3}$	371	-342	400
-400	58	$13.5\times10^{-3}$	377	-400	354
+400	46	$15.8\times10^{-3}$	380	-360	400
-400	40	$17.8\times10^{-3}$	380	-400	360