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Cyclic softening

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.5.

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
$\sigma_Y$:	380	680	1180	[MPa]
QY:	0	330	830	[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

$$E_p=\frac{680-380}{0.015}=\frac{1150-350}{0.04}=2\times10^4\,\mbox{[MPa]}$$

and the kinematic modulus K_p

$$K_p=\left\{\begin{array}{rcl} 330/0.015=2.2\times10^4\,\mbox{... ...{[MPa]} &~\mbox{for}~&\epsilon_p \ge 0.015 \end{array}\right.$$

In the course of the first two and a half cycle the elastic range decreases by $2(K_p-E_p)\epsilon_p$. During third unloading 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	380	-380	380
+400	20	$1.0\times10^{-3}$	378	-356	400
-400	44	$3.2\times10^{-3}$	374	-400	347
+400	53	$5.8\times10^{-3}$	369	-337	400
-400	63	$9.0\times10^{-3}$	362	-400	324
+400	76	$12.8\times10^{-3}$	355	-309	400
-400	91	$17.4\times10^{-3}$	350	-400	300