### Table of Contents

# Isotropic hardening

##### by Dr. Jiří Plešek

## Problem description

Consider the rod shown subjected to cyclic loading. Compute elastic-plastic response using linear isotropic hardening.

See also Kinematic hardening, Cyclic hardening and Cyclic softening.

## Material properties

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

$\varepsilon_p$ | $0.000$ | $0.015$ | $0.040$ |
---|---|---|---|

$\sigma_Y\text{ [MPa]}$ | $350$ | $650$ | $1150$ |

$Q_Y\text{ [MPa]}$ | $0$ | $0$ | $0$ |

## Support

Clamped at $x=0.$ Statically determinate.

## Loading

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

## Solution

Under uniaxial stress loadings the kinematic/isotropic yield condition implemented in PMD reduces to $$\vert\sigma-h\vert\le H(\varepsilon_p)=\sigma_Y(\varepsilon_p)-Q_Y(\varepsilon_p)$$ where $\sigma,$ $h,$ $\sigma_Y,$ $Q_Y,$ and $\varepsilon_p$ are the applied stress, backstress, subsequent yield stress, kinematic function, and the cumulated plastic strain, respectively.

Denoting by $\sigma_{Yt}=h+H,$ $\sigma_{Yc}=h-H$ the yield stresses in *tension* and *compression* and lifting the absolute from the above equation, the yield condition can be recast as
$$\sigma_{Yc}\le\sigma\le\sigma_{Yt}.$$
It should be noted, however, that for pronounced Bauschinger's effect situations with $\sigma_{Yc}>0,$ $\sigma_{Yt}<0$ might be encountered. See the picture below.

If for some fixed values of internal variables $\varepsilon_p$ and $h,$ further denoted as *old*, we observe that $\vert\sigma-h_{old}\vert>H_{old},$ these internal variables must change so that $\vert\sigma-h_{new}\vert=H_{new}.$ This is done as follows. Define $\Delta\sigma=\vert\sigma-h_{old}\vert-H_{old}>0$ and compute $\Delta\varepsilon_p$ such that
$$\sigma_Y(\varepsilon_p+\Delta\varepsilon_p)-\sigma_Y(\varepsilon_p)=\Delta\sigma.$$
For linear hardening this equation becomes
$$E_p\Delta\varepsilon_p=\Delta\sigma$$
where $E_p=\mathrm{d}\sigma_Y/\mathrm{d}\varepsilon_p$ is the plastic modulus. New value of backstress $h_{new}$ is computed from
$$\vert\sigma-h_{new}\vert=\sigma_Y(\varepsilon_p+\Delta\varepsilon_p)-Q_Y(\varepsilon_p+\Delta\varepsilon_p).$$

Graphically, the right edge of the elastic domain $\sigma_{Yt}=h+H$ is stretched by $\Delta\sigma$ to the right and its centroid $h$ is then updated according to $Q_Y.$ If $Q_Y\equiv0$ then $h_{new}=h_{old}=0;$ the elastic range $2H$ extends symmetrically by $2\Delta\sigma$ each time and material hardens isotropically. If $Q_Y$ has the same slope as $\sigma_Y,$ the elastic range remains constant and the yield surface moves about in stress space as a rigid body—hardening is said to be kinematic. In general, a proper adjustment of $Q_Y$ allows the elastic domain to move and stretch simultaneously—see Cyclic hardening and Cyclic softening.

In this example we have $$E_p=\frac{650-350}{0.015}=\frac{1150-350}{0.04}=2\times10^4\text{ MPa}.$$

In the course of first loading, the yield stresses in tension and compression increase to $\pm400\text{ MPa}$ so that all the subsequent cycles proceed in the elastic regime.

$\sigma_{xx}$ | $\Delta\sigma$ | $\varepsilon_p$ | $H$ | $\sigma_{Yc}$ | $\sigma_{Yt}$ |
---|---|---|---|---|---|

$0$ | $0$ | $0$ | $350$ | $-350$ | $350$ |

$+400$ | $50$ | $2.5\times10^{-3}$ | $400$ | $-400$ | $400$ |

$-400$ | $0$ | $2.5\times10^{-3}$ | $400$ | $-400$ | $400$ |

$+400$ | $0$ | $2.5\times10^{-3}$ | $400$ | $-400$ | $400$ |

$-400$ | $0$ | $2.5\times10^{-3}$ | $400$ | $-400$ | $400$ |

$+400$ | $0$ | $2.5\times10^{-3}$ | $400$ | $-400$ | $400$ |

$-400$ | $0$ | $2.5\times10^{-3}$ | $400$ | $-400$ | $400$ |