Cyclic hardening

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 softening.

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

Clamped at $x=0.$ Statically determinate.

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

• Quadrilaterals, BEAM6P5
• Hexahedra, BEAM56P5

For detailed explanation see Isotropic hardening. The plastic modulus $E_p$ is computed as $$E_p=\frac{650-350}{0.015}=\frac{1150-350}{0.04}=2\times10^4\text{ MPa}$$

and the kinematic modulus $K_p$ is $$K_p=\left\{\begin{array}{r} 270/0.015 = 1.8\times10^4\text{ MPa}\quad\text{for}\quad\varepsilon_p\le0.015 \\ (770-270)/(0.04-0.015) = 2.0\times10^4\text{ MPa}\quad\text{for}\quad\varepsilon_p\ge0.015 \end{array}\right..$$

In the course of the first two cycles the elastic range increases by $2(E_p-K_p)\varepsilon_p.$

During the third loading up to 400 MPa 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.