Thermo-plasticity

Consider the rod shown subjected to uniaxial stress and thermal loadings. Compute elastic-plastic response using isotropic nonlinear hardening with temperature-dependent yield stress.

$E=2\times10^5\text{ MPa},$ $\nu=0.3,$ $\alpha=10^{-5}\text{ 1/K}.$ Prandtl–Reuss–von Mises model with isotropic hardening.

Initial yield stress $\sigma_Y^0$ varies linearly with temperature.

Clamped at $x=0.$ Statically determinate.

$\sigma_{xx}=300\text{ MPa},$ $T_0=0\text{ $^\circ$C},$ $T=200\text{ $^\circ$C}.$

• Quadrilaterals, BEAM6P8
• Hexahedra, BEAM56P8

The yielding function satisfying the data given in figures can be expressed in a polynomial form as $$\sigma_Y(\varepsilon_p,T)=a_1+a_2\varepsilon_p+a_3T+a_4\varepsilon_p^2+a_6\varepsilon_pT+a_9\varepsilon_p^2T$$ where $$\begin{align} a_1 &= 3.5\times10^8\text{ Pa} & a_4 &= -1.0\times10^{14}\text{ Pa} \\ a_2 &= 2.0\times10^{11}\text{ Pa} & a_5 &= 1.0\times10^9\text{ Pa/K} \\ a_3 &= -5.0\times10^5\text{ Pa/K} & a_6 &= -5.0\times10^{11}\text{ Pa/K} \\ \end{align}.$$

Loading is prescribed as a series:

1. Initially $\sigma_{xx}=0,$ $T=0.$
2. Increase stress to $\sigma_{xx}=300$ at constant temperature $T=0.$ The material remains in an elastic state. Total strain $\varepsilon=\varepsilon^e=1.5\times10^{-3}.$
3. Heating from $T=0$ to $T=200$ under constant stress $\sigma_{xx}=300.$ The initial yield stress reduces to $\sigma_Y^0=250.$ Then $\sigma_{xx}>\sigma_Y^0$ and material starts to yield until $\varepsilon_p=0.134\times10^{-3}$ when the yield stress will harden to $\sigma_Y(\varepsilon_p,T)=300.$ Total strain $\varepsilon=\varepsilon^e+\varepsilon^p+\varepsilon^0=3.634\times10^{-3}.$
4. Return to state 2. $\sigma_{xx}=300,$ $T=0.$ Total strain $\varepsilon=\varepsilon^e+\varepsilon^p=1.634\times10^{-3}.$

Numerical computation gives $\varepsilon_p=0.135\times10^{-3}.$ A more precise result can be achieved by changing the default value $\mathtt{NINT}=10$ to $\mathtt{NINT}=100$ on the IP line in the .iL input file.