Table of Contents
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.
See also Isotropic hardening, Kinematic hardening and Cyclic 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.
| $\sigma_{xx}$ | $\Delta\sigma$ | $\varepsilon_p$ | $H$ | $\sigma_{Yc}$ | $\sigma_{Yt}$ |
|---|---|---|---|---|---|
| $0$ | $0$ | $0$ | $350$ | $-350$ | $350$ |
| $+400$ | $50$ | $2.5\times10^{-3}$ | $355$ | $-310$ | $400$ |
| $-400$ | $90$ | $7.0\times10^{-3}$ | $364$ | $-400$ | $328$ |
| $+400$ | $72$ | $10.6\times10^{-3}$ | $371$ | $-342$ | $400$ |
| $-400$ | $58$ | $13.5\times10^{-3}$ | $377$ | $-400$ | $354$ |
| $+400$ | $46$ | $15.8\times10^{-3}$ | $380$ | $-360$ | $400$ |
| $-400$ | $40$ | $17.8\times10^{-3}$ | $380$ | $-400$ | $360$ |
