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Uniaxial stress

Problem description: 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  $\epsilon_p$.

Mesh: Use the following element types:
ITE=6--see appendix A.1, four quadrilaterals
ITE=56--see appendix B.1, four hexahedra.

Material properties: $E=2\times 10^5$ MPa, $\nu=0.3$.
Prandtl-Reuss-von Mises model with piece-wise linear isotropic hardening.

$\epsilon_p$:	0	0.02	0.03	0.04
$\sigma_Y$:	350	350	375	390   [MPa]

Support: Clamped at x=0. Statically determinate.

Loading: $\sigma_{xx}=375$ MPa.

Solution: The total strain is easily computed as a sum of elastic and plastic parts

$$\epsilon_{xx}=\frac{\sigma_{xx}}{E}+\epsilon_p$$

for the uniaxial monotonic tensile loading. The magnitude of plastic component $\epsilon_p$, responsible for hardening, follows from the material table. For the loading given, the initial yield stress $\sigma_Y=350$ MPa must be increased to $\sigma_Y=375$ MPa, which neccesitates plastic straining $\epsilon_p=0.03$. Therefore

$$\epsilon_{xx}=\frac{375}{2\times10^5}+0.03=0.031875$$

It is interesting to note that the ratio

$$\bar\nu=\frac{-\epsilon_{yy}}{\epsilon_{xx}}= \frac{\nu\sigma_{xx}/E+\frac{1}{2}\epsilon_p} {\sigma_{xx}/E+\epsilon_p}$$

approaches 1/2 as $\epsilon_p\to\infty$, whereas for $\epsilon_p=0$ it becomes the Poisson's ratio $\nu$. In this example $\bar\nu=0.488$.