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Finite Element Analysis in Structural Mechanics

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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.

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


Clamped at $x=0.$ Statically determinate.


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


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.

$\sigma_{xx}$ $\Delta\sigma$ $\varepsilon_p$ $H$ $\sigma_{Yc}$ $\sigma_{Yt}$
$0$ $0$ $0$ $350$ $-350$ $350$
$+400$ $50$ $2.5\times10^{-3}$ $400$ $-400$ $400$
$-400$ $0$ $2.5\times10^{-3}$ $400$ $-400$ $400$
$+400$ $0$ $2.5\times10^{-3}$ $400$ $-400$ $400$
$-400$ $0$ $2.5\times10^{-3}$ $400$ $-400$ $400$
$+400$ $0$ $2.5\times10^{-3}$ $400$ $-400$ $400$
$-400$ $0$ $2.5\times10^{-3}$ $400$ $-400$ $400$

en/example/plas/3/start.txt · Last modified: 2022-02-11 14:25 by Petr Pařík