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Long-term creep, BEAM6C1

Problem description: Consider the rod shown for a steady-state creep analysis with the temperature T and the uniaxial stress $\sigma_{xx}$ being constant. The effective creep strain rate $\dot\epsilon_c$ is assumed to be a function of the cumulated creep strain  $\epsilon_c$.

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

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

Creep curve: 

$\epsilon_c=a+b\tan(ct-d)$


$a=0.949,~b=0.6322435755,~c=0.01312784\,\mbox{1/h},~d=0.9831024372$

Support: Clamped at x=0. Statically determinate.

u=v=0 node:	 1 
u=0 nodes: 	 2 11      

Loading: $\sigma_{xx}=30$ MPa, $T=800^\circ$C

Solution: The strain rate, which description is required in the input file, is obtained by differentiating the creep curve

$$\dot\epsilon_c=\frac{bc}{\cos^2(ct-d)}$$

In order to derive the strain hardening form in terms of $\epsilon_c$ we must eliminate parameter t from the simultaneous expressions $\epsilon_c(t)$ and $\dot\epsilon_c(t)$. To this end we use the indentity

$$\tan x =\frac{\sin x}{\cos x} = \frac{\sqrt{1-\cos^2 x}}{\cos x} =\sqrt{\frac{1}{\cos^2 x}-1}$$

and

$$\frac{1}{\cos^2(ct-d)}=\frac{\dot\epsilon_c}{bc}$$

Substituting for $\tan(ct-d)=\sqrt{\dot\epsilon_c/bc-1}$ into $\epsilon_c$ we arrive at

$$\epsilon_c=a+b\sqrt{\frac{\dot\epsilon_c}{bc}-1}$$

and finally

$$\dot\epsilon_c=bc+a^2\frac{c}{b} -2a\frac{c}{b}\epsilon_c+ \frac{c}{b}\epsilon_c^2$$

This polynomial can be written in the PMD format as

$$\dot\epsilon_c=a_1 +a_2\epsilon_c+a_3\epsilon_c^2$$

where

$$a_1=2.7\times 10^{-2}\,\mbox{1/h},~~ a_2=-3.94099\times 10^{-2}\,\mbox{1/h},~~ a_3=2.07639\times 10^{-2}\,\mbox{1/h}$$

Execute from prompt:
>rmd2 beam6c1.I1
>rpd2 beam6c1.I2
>srh2 beam6c1.I3
>fefs beam6c1.I4
>hpp2 beam6c1.IP
>hpls beam6c1.IL
>str2 beam6c1.I5