Cyclic softening

by Dr. Jiří Plešek

Problem description

Consider the rod shown subjected to cyclic loading. Compute elastic-plastic response using linear mixed-mode hardening.

Material properties

$E=2\times10^5\text{ MPa},$ $\nu=0.3.$ Prandtl–Reuss–von Mises model with mixed hardening.

$\varepsilon_p$	$0.000$	$0.015$	$0.040$
$\sigma_Y\text{ [MPa]}$	$380$	$680$	$1180$
$Q_Y\text{ [MPa]}$	$0$	$330$	$830$

Support

Clamped at $x=0.$ Statically determinate.

Loading

$\sigma_{xx}=\pm400\text{ MPa},$ 3 cycles.

Solution

For detailed explanation see Isotropic hardening. The plastic modulus $E_p$ is computed as $E_p=\frac{680-380}{0.015}=\frac{1180-380}{0.04}=2\times10^4\text{ MPa}$

and the kinematic modulus $K_p$ is $K_p=\left\{\begin{array}{r} 330/0.015 = 2.2\times10^4\text{ MPa}\quad\text{for}\quad\varepsilon_p\le0.015 \\ (830-330)/(0.04-0.015) = 2.0\times10^4\text{ MPa}\quad\text{for}\quad\varepsilon_p\ge0.015 \end{array}\right..$

In the course of the first two and a half cycle the elastic range decreases by $2(K_p-E_p)\varepsilon_p.$

During the third unloading the threshold value $\varepsilon_p=0.015$ is exceeded and $K_p=E_p.$ Further hardening is of the kinematic type, which causes the hysteresis loop to close as in Kinematic hardening. The response is said to be saturated.

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