2. Benchmark solution#
2.1. Calculation method used for reference solutions#
The whole of this demonstration can be read in more detail in document \([1]\)
The stress tensor is written as:
\(\sigma =(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& {\sigma }_{z}\end{array})\)
As a result of loading, we have:
\(\{\begin{array}{c}{\varepsilon }_{z}-{\varepsilon }_{\nu z}=\frac{{\sigma }_{z}}{E}\\ {\varepsilon }_{\theta }-{\varepsilon }_{\nu \theta }=-\frac{\nu }{E}{\sigma }_{z}\\ {\varepsilon }_{r}-{\varepsilon }_{\nu r}=-\frac{\nu }{E}{\sigma }_{z}\end{array}\) and \(\dot{{\varepsilon }_{\nu }}=\frac{3}{2}g(\sigma \text{*})\frac{{\sigma }^{D}}{\sigma \text{*}}\)
therefore
\(\{\begin{array}{c}\dot{{\varepsilon }_{\nu z}}=g({\sigma }_{z})\\ \dot{{\varepsilon }_{\nu \theta }}=-\frac{1}{2}g({\sigma }_{z})\\ \dot{{\varepsilon }_{\nu r}}=-\frac{1}{2}g({\sigma }_{z})\end{array}\)
If \(t\le {t}_{0}\), we have \({\varepsilon }_{z}=\frac{{\varepsilon }_{0}}{{t}_{0}}t\),
Be \(a=\sqrt{\frac{{\varepsilon }_{0}}{{t}_{0}}}\)
We get \({\varepsilon }_{z}={a}^{2}t\)
Substituting, we find:
\(\dot{{\varepsilon }_{\nu z}}=g(({a}^{2}t-{\varepsilon }_{\nu z})E)\)
We set \(E=1\) and \(z={a}^{2}t-{\varepsilon }_{\nu z}\), we get: \(\dot{z}={a}^{2}–{z}^{2}\)
Integrating with \(z(0)=0\) we get:
\(z=a\mathrm{tanh}(\mathrm{at})\)
For \(t\le {t}_{0}\)
\(\{\begin{array}{c}{\sigma }_{r}={\sigma }_{\theta }=0\\ {\sigma }_{z}=a\mathrm{tanh}(\mathrm{at})\\ {\varepsilon }_{r}={\varepsilon }_{\theta }=a[(\frac{1}{2}-\nu )\mathrm{tanh}(\mathrm{at})-\frac{1}{2}\mathrm{at}]\\ {\varepsilon }_{z}={a}^{2}t\\ w=\mathrm{ar}[(\frac{1}{2}-\nu )\mathrm{tanh}(\mathrm{at})-\frac{1}{2}\mathrm{at}]\end{array}\)
If \(t\ge {t}_{0}\)
\({\varepsilon }_{z}={a}^{2}{t}_{0}\)
\(\dot{{\varepsilon }_{\nu z}}=g({a}^{2}{t}_{0}-{\varepsilon }_{\nu z})={({a}^{2}{t}_{0}-{\varepsilon }_{\nu z})}^{2}\)
What results when integrating:
\({\varepsilon }_{\nu z}={a}^{2}{t}_{0}–\frac{1}{\frac{1}{a\mathrm{tanh}({\mathrm{at}}_{0})}+t–{t}_{0}}\)
So we finally have
\(\{\begin{array}{c}{\sigma }_{r}={\sigma }_{\theta }=0\\ {\sigma }_{z}=\frac{1}{\frac{1}{a\mathrm{tanh}({\mathrm{at}}_{0})}+t–{t}_{0}}\\ {\varepsilon }_{r}={\varepsilon }_{\theta }=(\frac{1}{2}-\nu )\frac{1}{\frac{1}{a\mathrm{tanh}({\mathrm{at}}_{0})}+t–{t}_{0}}–\frac{1}{2}{a}^{2}{t}_{0}\\ {\varepsilon }_{z}={a}^{2}{t}_{0}\\ w=r\left[(\frac{1}{2}-\nu )\frac{1}{\frac{1}{a\mathrm{tanh}({\mathrm{at}}_{0})}+t–{t}_{0}}–\frac{1}{2}{a}^{2}{t}_{0}\right]\end{array}\)
2.2. Reference quantities#
Move \(\mathrm{DX}\) to node \(B\)
Constraints \(\mathrm{SIXX}\), \(\mathrm{SIYY}\), and \(\mathrm{SIZZ}\) at node \(B\)
2.3. Benchmark results#
Size |
Point |
Moments |
Reference |
\(\mathrm{DX}\) |
|
4 |
|
\(\mathrm{SIXX}\) |
|
4 |
|
\(\mathrm{SIYY}\) |
|
4 |
|
\(\mathrm{SIZZ}\) |
|
4 |
|
2.4. Uncertainty about the solution#
Analytical solution
2.5. Bibliographical references#
[1] Ph. De BONNIERES, M. ZIDI: Introduction of viscoplasticity into the Cyrano3 thermomechanics module: Principle, description and validation, Note HI-71/8334.