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Relaxation, BEAM6C2

: Problem description: The rod shown is subjected to the initial temperature change $\Delta T$ after which the temperature has been kept constant. Compute the relaxation history of the thermal stress.

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

Mesh: Four quadrilaterals--see appendix A.1.

Material properties: $\alpha=10^{-5}$ 1/K, $E=2\times 10^5$ MPa, $\nu=0.3$ .


Norton's law:

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


u=v=0 nodes:	 1 9
u=0 nodes: 	 2 10 11 23

Loading: $T_0=650^\circ\mbox{C},~T=600^\circ\mbox{C},~ \Delta T=-50^\circ\mbox{C}$

Solution: Norton's law with the whole exponent number n=3 can be written in the polynomial form

$\begin{displaymath}\dot\epsilon_c=a_1 +a_2\sigma_e+a_3\sigma_e^2+a_4\sigma_e^3 \end{displaymath}$

where

$\begin{displaymath}a_1=a_2=a_3=0~,~~~a_4=2\times 10^{-28}\,\mbox{1/h} \end{displaymath}$

which is directly inserted into the PMD input file.

Numerical results may be compared with the theoretical solution. The equation of relaxation for $\sigma>0$ reads

$\begin{displaymath}\dot\epsilon=\dot\sigma/E+\gamma\sigma^n=0 \end{displaymath}$

with the initial condition

$\begin{displaymath}\sigma(0)=-E\alpha\Delta T \end{displaymath}$

Hence

$\begin{displaymath}\sigma(t)=\left[(n-1)E\gamma t+(-E\alpha\Delta T)^{1-n}\right] ^{\frac{1}{1-n}} \end{displaymath}$

The results of computation (black points) and the theoretical curve (full line) are shown in the next figure.

$\begin{figure} \centering\hspace{0pt}\rotate{ \epsfclipon\epsfxsize=9cm\epsffile{bea6c2.ps}} \end{figure}$

Execute from prompt:
>rmd2 beam6c2.I1
>rpd2 beam6c2.I2
>srh2 beam6c2.I3
>fefs beam6c2.I4
>hpp2 beam6c2.IP
>hpls beam6c2.IL
>str2 beam6c2.I5

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