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

$E=2\times10^5\text{ MPa},$ $\nu=0.3,$ $\alpha=10^{-5}\text{ 1/K},$ $\sigma_Y=350\text{ MPa}.$ Prandtl–Reuss–von Mises elastic–perfectly plastic model.

Clamped at $x=0,$ $x=l.$ Statically indeterminate.

$\Delta T=200\ ^\circ\text{C}.$

• Quadrilaterals, BEAM6P7
• Hexahedra, BEAM56P7

Thermal strain $\varepsilon^0$ is computed as $$\varepsilon^0=\alpha\Delta T=2\times10^{-3}.$$

Elastic strain $\varepsilon_Y^e$ on the onset of yielding $$\varepsilon_Y^e=\frac{\sigma_Y}{E}=1.75\times10^{-3}.$$

Since $\varepsilon=\varepsilon^e+\varepsilon^p+\varepsilon^0=0$ and $\varepsilon^e=-\varepsilon_Y^e$ at $T=T_0+\Delta T,$ we have $$\varepsilon^p=\varepsilon_Y^e-\varepsilon^0=-0.25\times10^{-3}.$$

After cooling the thermal strain disappears, therefore $\varepsilon_0^e=-\varepsilon^p$ and the residual stress $$\sigma_0=E\varepsilon_0^e=50\text{ MPa}.$$