## Benchmark UP5

### Cylinder Torsion

Let us test the forces required to sustain a torsion of a cylinder. The cylinder is supposed to have length $l$, diameter $D$ and to be made of the Neo-Hookean material. The cylinder radius is denoted by $a=D}{2}$.

##### Parameters

$\begin{array}{ccc}\hfill D& \hfill =\hfill & 1\mathrm{m}\hfill \\ \hfill l& \hfill =\hfill & 1\mathrm{m}\hfill \\ \hfill \kappa & \hfill =\hfill & {10}^{7}\mathrm{Pa}\hfill \\ \hfill \mu & \hfill =\hfill & 280709\mathrm{Pa}\hfill \end{array}$

##### Theory

The cylinder lower base is fully fixed and the cylinder upper base is subjected to the surface load according to the theoretical formula given in [Taber, 2004]

$\begin{array}{ccc}\hfill {\sigma }_{\mathrm{zz}}& \hfill =\hfill & -2{\theta }^{2}\underset{r}{\overset{a}{\int }}\stackrel{˜}{r}\frac{\partial {\Psi }_{\text{iso}}}{\partial {I}_{1}}{|}_{J=1}d\stackrel{˜}{r}\hfill \\ \hfill {\sigma }_{\theta z}& \hfill =\hfill & 2\theta r\frac{\partial {\Psi }_{\text{iso}}}{\partial {I}_{1}}{|}_{J=1}\hfill \end{array}$

After exerting coresponding surface loading on the upper base of the cylinder, a value of $\theta$ is numerically measured and compared with its assumed value.

##### Results

Theory femCalc (BRICK24) femCalc (BRICK24)
K=4, N=960 K=6, N=3240
$\theta \text{\hspace{0.17em}}\left[\mathrm{rad}\right]$ $\theta \text{\hspace{0.17em}}\left[\mathrm{rad}\right]$ $\text{Ratio}\text{\hspace{0.17em}}\left[\mathrm{-}\right]$ $\theta \text{\hspace{0.17em}}\left[\mathrm{rad}\right]$ $\text{Ratio}\text{\hspace{0.17em}}\left[\mathrm{-}\right]$
$0.1$ $0.1007$ $1.007$ $0.0997$ $0.997$
$0.2$ $0.2019$ $1.010$ $0.1998$ $1.001$
$0.3$ $0.3035$ $1.012$ $0.3003$ $1.001$
$0.4$ $0.4053$ $1.013$ $0.4014$ $1.003$
$0.5$ $0.5072$ $1.014$ $0.5026$ $1.005$
$0.6$ $0.6082$ $1.014$ $0.6034$ $1.006$
$0.7$ $0.7071$ $1.010$ $0.7024$ $1.003$
$0.8$ $0.8021$ $1.003$ $0.7975$ $0.997$
$0.9$ $0.8912$ $0.990$ $0.8864$ $0.985$
$1.0$ $0.9723$ $0.972$ $0.9671$ $0.967$