en:example:plas:4:start
Table of Contents
Kinematic hardening
by Dr. Jiří Plešek
Problem description
Consider the rod shown subjected to cyclic loading. Compute elastic-plastic response using linear kinematic hardening.
See also Isotropic hardening, Cyclic hardening and Cyclic softening.
Material properties
$E=2\times10^5\text{ MPa},$ $\nu=0.3.$ Prandtl–Reuss–von Mises model with linear kinematic hardening.
| $\varepsilon_p$ | $0.000$ | $0.015$ | $0.040$ |
|---|---|---|---|
| $\sigma_Y\text{ [MPa]}$ | $350$ | $650$ | $1150$ |
| $Q_Y\text{ [MPa]}$ | $0$ | $300$ | $800$ |
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 $$E_p=\frac{650-350}{0.015}=\frac{1150-350}{0.04}=2\times10^4\text{ MPa}.$$
Reloading the bar to 350 MPa the hysteresis loop closes and the cycle will stabilize.
| $\sigma_{xx}$ | $\Delta\sigma$ | $\varepsilon_p$ | $H$ | $\sigma_{Yc}$ | $\sigma_{Yt}$ |
|---|---|---|---|---|---|
| $0$ | $0$ | $0$ | $350$ | $-350$ | $350$ |
| $+400$ | $50$ | $2.5\times10^{-3}$ | $350$ | $-300$ | $400$ |
| $-400$ | $100$ | $7.5\times10^{-3}$ | $350$ | $-400$ | $300$ |
| $+400$ | $100$ | $12.5\times10^{-3}$ | $350$ | $-300$ | $400$ |
| $-400$ | $100$ | $17.5\times10^{-3}$ | $350$ | $-400$ | $300$ |
| $+400$ | $100$ | $22.5\times10^{-3}$ | $350$ | $-300$ | $400$ |
| $-400$ | $100$ | $27.5\times10^{-3}$ | $350$ | $-400$ | $300$ |
en/example/plas/4/start.txt · Last modified: 2022-02-11 14:25 by Petr Pařík