Table of Contents
Bina's model
by Dr. Jiří Plešek
Problem description
Consider the rod shown for a transient creep analysis. The initial stress $\sigma_{xx}=\sigma_{xx}^{(1)}$ abruptly changes to $\sigma_{xx}=\sigma_{xx}^{(2)}$ at time $t=t_1,$ after which the straining continues at the new load level until $t=t_2.$ The temperature $T$ is assumed to be constant throughout.
Material properties
$E=2\times10^5\text{ MPa},$ $\nu=0.3,$ $\alpha=10^{-5}\text{ 1/K}.$
Parameters for Bina's model:
$E_1$ | $0.2152947\times10^6$ | $M_1$ | $0.1455834\times10^1$ |
---|---|---|---|
$E_2$ | $-0.2502203\times10^6$ | $M_2$ | $-0.1260596\times10^1$ |
$E_3$ | $0.1434375\times10^4$ | $M_3$ | $0.4886680\times10^2$ |
$A_1$ | $-0.2079980\times10^3$ | $M_4$ | $-0.5002250\times10^{-1}$ |
$A_2$ | $-0.8215359\times10^2$ | $M_5$ | $0.1825170\times10^1$ |
$A_3$ | $-0.3412273\times10^0$ | $N$ | $0.3260447\times10^0$ |
$A_4$ | $-0.1391403\times10^1$ | $M$ | $-0.6814129\times10^0$ |
$A_5$ | $0.2730000\times10^3$ | $K$ | $1$ |
$A_6$ | $0.1000000\times10^{-3}$ |
Support
Clamped at $x=0.$ Statically determinate.
Loading
$T=450\text{ }^\circ\text{C},$ $t_1=1.001\times10^5\text{ h},$ $t_2=1.998\times10^5\text{ h},$ $\sigma_{xx}=\left\{\begin{align} \sigma_{xx}^{(1)} &= 230\text{ MPa} && \text{for} && 0<t\le t_1 \\ \sigma_{xx}^{(2)} &= 250\text{ MPa} && \text{for} && t_1<t\le t_2 \end{align}\right..$
Solution
It is useful, at this point, to overview some important features of the Bina's material model, omitting, however, unnecessary details. The most important issue concerns the compatibility between the model's data collected in a separate .dat
input file and the elastic parameters that have been entered independently in the .i2
input file.
According to Bina's model, the functional dependence of the Young modulus on temperature is supposed to have the form
$$E\left(T\right)=E_1+E_2\exp\left(\frac{-E_3}{T}\right)$$
where the numerical values of the constants $E_1,$ $E_2,$ $E_3$ shown in the table above correspond to physical units MPa and K for the Young modulus and the thermodynamic temperature, respectively. It should be pointed out that the above expression applies to the evaluation of the uniaxial creep curves only, leaving the elastic constants used for the formation of the stiffness matrix unaltered. In order to achieve compatibility, it is necessary to describe the Young modulus in the .i2
input file as a function of temperature that approximates the exponential relation of the creep model as closely as possible.
To this end, we may, for instance, estimate the upper and lower bounds for temperatures occurring in the body and define a piecewise linear function that will cover up the entire range. Examples of this technique can be found in the .i2
input files shown in the subsections. Of course, in these problems with a homogeneous temperature field we could have simply put
$$E = 0.2152947\times10^6-0.2502203\times10^6\times\exp\left(\displaystyle\frac{-0.1434375\times10^4}{273+450}\right) = 1.81\times10^5\text{ MPa}$$
just as well.
Next, we turn our attention toward the analytical solution. In general, it is difficult to obtain the closed form solution for a complex material model but in this particular case at least one variable—the one describing material damage—can be calculated rather easily. In the Bina's model the damage parameter is defined as $$\pi(t)=\frac{t}{t_r}~\in~(0,1)$$ where $t_r$ is the time elapsed at rupture, estimated by the empiric formula $$\begin{align} \log t_r &= A_1+A_2\log\left|\displaystyle\frac{1}{T}-\frac{1}{A_5}\right|+\\ &+ A_3\log\left|\displaystyle\frac{1}{T}-\frac{1}{A_5}\right|\log\Bigl(\sinh(A_6\sigma_e T)\Bigr)+A_4\log\Bigl(\sinh(A_6\sigma_e T)\Bigr) \end{align}$$ in which $\sigma_e$ is the effective stress in MPa, $T$ is the thermodynamic temperature in K, $t_r$ the time of rupture in hours, and the coefficients $A_1\dots A_6$ are given in the table above. Thus we can calculate the rupture times $t_r^{(1)},$ $t_r^{(2)}$ corresponding to the two stress levels $\sigma_{xx}^{(1)},$ $\sigma_{xx}^{(2)}$ as $$t_r^{(1)}=466518\text{ h},\quad t_r^{(2)}=229746\text{ h}.$$
These rupture times pertain to situations with constant stress levels. In order to obtain the total damage for varying loading the increments of $\pi$ must be summed up according to a simple rule $$\pi(t_2)=\frac{t_1}{t_r^{(1)}}+\frac{t_2-t_1}{t_r^{(2)}}=\frac{100100}{466518}+\frac{99700}{229746}=0.6485.$$
This result ($64.85\text{ %}$) should be compared with the $\mathtt{DMG}$ damage output variables shown in the subsections.
Finally, we must account for a sudden stress change at time $t_1$ when the stress point traverses from one creep curve to another one in the $\varepsilon_c$–$t$ plot. In principle, we have three approaches at hand to accomplish this transition: the strain hardening method, characterized by the constitutive equation for the equivalent creep strain, having the form $$\dot\varepsilon_c=\dot\varepsilon_c^{sh}(\varepsilon_c,\sigma_e,T),$$ the time hardening method $$\dot\varepsilon_c=\dot\varepsilon_c^{th}(t,\sigma_e,T),$$ and the damage hardening (softening) method $$\dot\varepsilon_c=\dot\varepsilon_c^{dh}(\pi,\sigma_e,T).$$
One may expect that $\varepsilon_c^{sh}<\varepsilon_c^{dh}<\varepsilon_c^{th}$ since the stress change had occurred later in the softening stage when the creep curves were convex. This was indeed confirmed by the computational results as well as the fact the damage remained the same, independent of the choice of a transition method.