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Finite Element Analysis in Structural Mechanics

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en:example:plas:5:start

Cyclic hardening

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 softening.

Material properties

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

$\varepsilon_p$ $0.000$ $0.015$ $0.040$
$\sigma_Y\text{ [MPa]}$ $350$ $650$ $1150$
$Q_Y\text{ [MPa]}$ $0$ $270$ $770$

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 $$E_p=\frac{650-350}{0.015}=\frac{1150-350}{0.04}=2\times10^4\text{ MPa}$$

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.

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

en/example/plas/5/start.txt · Last modified: 2022-02-11 14:25 by Petr Pařík