Consider the rod shown for an elastic-plastic analysis. The yield stress $\sigma_Y$ is assumed to be a function of the cumulated plastic strain $\varepsilon_p.$

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

Clamped at $x=0.$ Statically determinate.

$\sigma_{xx}=375\text{ MPa}.$

• Quadrilaterals, BEAM6P2
• Hexahedra, BEAM56P2

The total strain is easily computed as a sum of elastic and plastic parts $$\varepsilon_{xx}=\frac{\sigma_{xx}}{E}+\varepsilon_p$$ for the uniaxial monotonic tensile loading. The magnitude of plastic component $\varepsilon_p,$ responsible for hardening, follows from the material table. For the loading given, the initial yield stress $\sigma_Y=350\text{ MPa}$ must be increased to $\sigma_Y=375\text{ MPa},$ which neccesitates plastic straining $\varepsilon_p=0.03.$ Therefore $$\varepsilon_{xx}=\frac{375}{2\times10^5}+0.03=0.031875.$$

It is interesting to note that the ratio $$\bar\nu=\frac{-\varepsilon_{yy}}{\varepsilon_{xx}}=\frac{\displaystyle\frac{\nu\sigma_{xx}}{E}+\frac{1}{2}\varepsilon_p}{\displaystyle\frac{\sigma_{xx}}{E}+\varepsilon_p}$$ approaches $\frac{1}{2}$ as $\varepsilon_p\to\infty,$ whereas for $\varepsilon_p=0$ it becomes the Poisson's ratio $\nu.$ In this example $\bar\nu=0.488.$