## Benchmark UP6

### Longitudinal Vibrations of a Mass on a Rubber Spring

##### Description

Let us consider a cantilever of length $l$ and width and depth ${z}_{0}$, fully fixed at $x=0$. The cantilever is made of the Neo-Hookean material of density $\rho$ (considered zero). At position $x=l$ an additional mass $m$ is added, equally distributed to all four nodes located at $x=l$. The cantilever, initally at rest, is subjected to acceleration $g$ in $-x$ direction. The numerical solution is tested on motion of the point at position $x=l$.

##### Parameters

$\begin{array}{ccc}\hfill m& \hfill =\hfill & 0.1\mathrm{kg}\hfill \\ \hfill g& \hfill =\hfill & 1\mathrm{m}·{s}^{-2}\hfill \\ \hfill l& \hfill =\hfill & 1\mathrm{m}\hfill \\ \hfill {z}_{0}& \hfill =\hfill & 0.01\mathrm{m}\hfill \\ \hfill \kappa & \hfill =\hfill & {10}^{8}\mathrm{Pa}\hfill \\ \hfill \mu & \hfill =\hfill & 3.\overline{33}·{10}^{5}\mathrm{Pa}\hfill \\ \hfill \rho & \hfill =\hfill & 0\mathrm{kg}·{m}^{-3}\hfill \end{array}$

##### Theory

The analytical solution of the problem, considering small strain limit, yields
$\frac{{\text{d}}^{2}\lambda }{\text{d}{t}^{2}}+\frac{3\mu {{z}_{0}}^{2}}{\mathrm{ml}}\lambda +\frac{g}{l}=0$
which gives us frequency $f=\frac{{z}_{0}}{2\pi }\sqrt{\frac{3\mu }{\mathrm{ml}}}=5.033\mathrm{Hz}$.

##### FE Model

The model consists of twenty-five BRICK24 elements.

##### Results
Theory ANSYS (HYPER58) femCalc (BRICK24)
$f\text{\hspace{0.17em}}\left[\mathrm{Hz}\right]$ $f\text{\hspace{0.17em}}\left[\mathrm{Hz}\right]$ $\text{Ratio}\text{\hspace{0.17em}}\left[\mathrm{-}\right]$ $f\text{\hspace{0.17em}}\left[\mathrm{Hz}\right]$ $\text{Ratio}\text{\hspace{0.17em}}\left[\mathrm{-}\right]$
$5.033$ $5.038$ $1.001$ $5.045$ $1.002$