Cyclic softening

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.

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

Clamped at $x=0.$ Statically determinate.

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

• Quadrilaterals, BEAM6P6
• Hexahedra, BEAM56P6

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.