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 thermal stress.

$E=2\times10^5\text{ MPa},$ $\nu=0.3,$ $\alpha=10^{-5}\text{ 1/K}.$

Norton's law: $\dot\varepsilon_c=\gamma(\sigma_e/\sigma_0)^n$ with $\gamma=2\times10^{-28}\text{ 1/h},$ $n=3,$ $\sigma_0=1\text{ Pa}.$

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

$T_0=650\text{ $^\circ$C},$ $T=600\text{ $^\circ$C},$ $\Delta T=-50\text{ $^\circ$C}.$

• Quadrilaterals, BEAM6C2

Norton's law with the whole exponent number $n=3$ can be written in the polynomial form $$\dot\varepsilon_c=a_1 +a_2\sigma_e+a_3\sigma_e^2+a_4\sigma_e^3$$ where $$a_1=a_2=a_3=0,\quad a_4=2\times10^{-28}\text{ 1/h}$$ 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 $$\dot\varepsilon=\frac{\dot\sigma}{E}+\gamma\sigma^n=0$$ with the initial condition $$\sigma(0)=-E\alpha\Delta T.$$

Hence $$\sigma(t)=\left[(n-1)E\gamma t+(-E\alpha\Delta T)^{1-n}\right]^{\frac{1}{1-n}}.$$ The results of computation (black points) and the theoretical curve (full line) are shown below.