3. Modeling A#

3.1. Characteristics of modeling#

In this modeling, we use the operator DYNA_VIBRA (see [U4.53.03]) with the relationship DIS_CHOC. The integration diagram is DIFF_CENTRE.

A DIS_T element on a POI1 mesh is used to model the system.

Relationships between degrees of freedom are used to force the movement to be unidirectional in the \(\theta\) direction:

LIAISON_DDL = _F (NOEUD = ('NO1', 'NO1'),

DDL = ('DX' 'DY'),

COEF_MULT = (0.707, -0.707),

COEF_IMPO = 0.)

An obstacle of type PLAN_Z (two parallel planes separated by a game) is used to simulate the sliding plane. We choose to use the \(\mathrm{Oy}\) axis as the generator of this plane, i.e. NORM_OBST = (0., 1., 0.). The origin of the obstacle is ORIG_OBST = (0.,0.,1.). It remains to define his game, which gives the halfway gap between the planes.

For there to be a reaction force from the plane on the system, the system must be slightly pressed into the plane obstacle by a distance \(\delta n\) such as: \({F}_{n}={K}_{n}\cdot \mathrm{\delta }n\).

Like \({F}_{n}=\mathrm{mg}\), we then have \(\delta n=\mathrm{mg}/{K}_{n}\).

We considered a normal shock stiffness of \(20N/m\) (fictional stiffness that only makes sense to generate a reaction force from the plane on the system), so we have \(\delta n=\mathrm{0,5}\). The obstacle PLAN_Z having its origin in \(Z=1\) and the solid being in \(Z=0\); a game of \(\mathrm{0,5}m\) will create a hole \(\delta n=\mathrm{0,5}m\) from where JEU: 0.5

Tangential shock stiffness: \({K}_{T}=400000N/m\): it is greater than the stiffness of the oscillator for the stopping phase to be modelled correctly.

No time used for time integration: \({5.10}^{-4}\mathit{sec}\).

3.2. Characteristics of the mesh#

Number of knots: 1

Number of meshes and types: 1 POI1

3.3. Tested sizes and results#

Displacement values (in meters) in the \(\theta\) direction for the times when the speed sign changed over the \((0;0.3s)\) time period.

Identification

moment (s)

Reference

\(\mathrm{DY}=\mathrm{r2}\mathrm{cos45}\)

\(\pi \times {10}^{-2}\)

—4.596E—4

\(\mathrm{DY}=\mathrm{r3}\mathrm{cos45}\)

\(2\pi \times {10}^{-2}\)

3.182E—4

\(\mathrm{DY}=\mathrm{r4}\mathrm{cos45}\)

\(3\pi \times {10}^{-2}\)

—1.768E—4

\(\mathrm{DY}=\mathrm{r5}\mathrm{cos45}\)

\(4\pi \times {10}^{-2}\)

3.536E—5