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Residual stress

: Problem description: The rod shown is subjected to thermal loading $\Delta T$ . Compute residual stress after cooling to the initial temperature.

$\begin{figure} \centering\hspace{0pt} \epsfclipon\epsfxsize=6cm\epsffile{beam6c2.ps} \end{figure}$

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

Material properties: $\alpha=10^{-5}$ , $E=2\times 10^5$ MPa, $\nu=0.3$ , $\sigma_Y=350$ MPa.
Prandtl-Reuss-von Mises elastic-perfectly plastic model.

Support: Clamped at x=0, x=l. Statically indeterminate.

Loading: $\Delta T=200^\circ\mbox{C}$

Solution: Thermal strain $\epsilon^0$ is computed as

$\begin{displaymath}\epsilon^0=\alpha\Delta T = 2\times10^{-3} \end{displaymath}$

Elastic strain $\epsilon_Y^e$ on the onset of yielding

$\begin{displaymath}\epsilon_Y^e=\frac{\sigma_Y}{E} = 1.75\times10^{-3} \end{displaymath}$

Since $\epsilon=\epsilon^e+\epsilon^p+\epsilon^0=0$ and $\epsilon^e=-\epsilon_Y^e$ at $T=T_0+\Delta T$ , we have

$\begin{displaymath}\epsilon^p=\epsilon_Y^e-\epsilon^0=-0.25\times10^{-3} \end{displaymath}$

After cooling the thermal strain disappears, therefore $\epsilon_0^e=-\epsilon^p$ and the residual stress

$\begin{displaymath}\sigma_0=E\epsilon_0^e=50\,\mbox{[MPa]} \end{displaymath}$