Mar APR May 5 2013 2014 2015
1 capture 5 Apr 14 - 5 Apr 14

Close Help

  
Thermo-plasticity

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

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, $\nu=0.3$.
Prandtl-Reuss-von Mises model with isotropic hardening.

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

Support: Clamped at x=0. Statically determinate.

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

Solution: The yielding function satisfying the data given in figures can be expressed in a polynomial form as

$$\sigma_Y(\epsilon_p,T)=a_1+a_2\epsilon_p+a_3T+a_4\epsilon_p^2 +a_6\epsilon_pT+a_9\epsilon_p^2T$$

where

$$\begin{array}{rclrcl} a_1 &=& 3.5\times10^8 \,\mbox{Pa}~~~~~~... ...a/K}~~~~~~& a_9 &=& -5.0\times10^{11} \,\mbox{Pa/K} \end{array}$$

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. Material remains in an elastic state. Total strain $\epsilon=\epsilon^e=1.5\times10^{-3}$.

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 $\epsilon_p=0.134\times10^{-3}$ when the yield stress will harden to $\sigma_Y(\epsilon_p,T)=300$. Total strain $\epsilon=\epsilon^e+\epsilon^p+\epsilon^0=3.634\times10^{-3}$.

4.

Return to state 2. $\sigma_{xx}=300$, T=0. Total strain $\epsilon=\epsilon^e+\epsilon^p=1.634\times10^{-3}$.

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