# Kinematics¶

A kinematic model defines an allowed type of deformation for an individual body. Solfec-1.0 includes three kinematic models:

• rigid body
• pseudo–rigid body
• finite–element body

A kinematic model is selected by specifying the kind paramter of the BODY command.

## Rigid body¶

The motion of a rigid body reads

(12)$\mathbf{x}\left(\mathbf{X},t\right)=\mathbf{\Lambda}\left(t\right)\left(\mathbf{X}-\bar{\mathbf{X}}\right)+\bar{\mathbf{x}}\left(t\right)$

where $$\mathbf{\Lambda}\left(t\right)$$ is a $$3\times3$$ rotation matrix, $$\bar{\mathbf{X}}$$ is a selected referential point, and $$\bar{\mathbf{x}}\left(t\right)$$ is a spatial point. Hence $$\bar{\mathbf{x}}\left(t\right)=\mathbf{x}\left(\bar{\mathbf{X}},t\right)$$ is the motion of the selected point $$\bar{\mathbf{X}}$$. The term $$\mathbf{\Lambda}\left(t\right)\left(\mathbf{X}-\bar{\mathbf{X}}\right)$$ represents the rotation of $$\mathbf{X}$$ about the point $$\bar{\mathbf{X}}$$. Representing rotations, $$\mathbf{\Lambda}$$ must be orthogonal: $$\mathbf{\Lambda}^{T}\mathbf{\Lambda}=\mathbf{I}$$, where $$\mathbf{I}$$ is the $$3\times3$$ identity matrix. Thus, the rigidity condition follows $$\left\Vert \mathbf{x}-\bar{\mathbf{x}}\right\Vert =\left\Vert \mathbf{X}-\bar{\mathbf{X}}\right\Vert$$, that is the length of a line segment before and after rotation is preserved.

The velocity of the spatial point reads

(13)$\dot{\mathbf{x}}\left(\mathbf{X},t\right)=\mathbf{\Lambda}\left(t\right)\hat{\mathbf{\Omega}}\left(t\right) \left(\mathbf{X}-\bar{\mathbf{X}}\right)+\dot{\bar{\mathbf{x}}}\left(t\right)=\hat{\mathbf{\omega}}\left(t\right) \mathbf{\Lambda}\left(t\right)\left(\mathbf{X}-\bar{\mathbf{X}}\right)+\dot{\bar{\mathbf{x}}}\left(t\right)$

where $$\mathbf{\omega}$$ is the spatial angular velocity vector, $$\mathbf{\Omega}$$ is the referential angular velocity vector, and $$\dot{\bar{\mathbf{x}}}$$ is the velocity of the spatial image of $$\bar{\mathbf{X}}$$. W note that

$\mathbf{\omega}=\mathbf{\Lambda}\mathbf{\Omega}$

The hat operator $$\hat{\mathbf{y}}$$ creates an anti–symmetric $$3\times3$$ matrix out of a 3–vector as follows

$\begin{split}\hat{\mathbf{y}}=\left[\begin{array}{ccc} 0 & -y_{3} & y_{2}\\ y_{3} & 0 & -y_{1}\\ -y_{2} & y_{1} & 0 \end{array}\right]\end{split}$

In vector notation the configuration $$\mathbf{q}$$ and the velocity $$\mathbf{u}$$ can be expressed as

(15)$\begin{split}\mathbf{q}=\left[\begin{array}{c} \Lambda_{11}\\ \Lambda_{21}\\ ...\\ \bar{x}_{1}\\ \bar{x}_{2}\\ \bar{x}_{3} \end{array}\right],\,\,\,\mathbf{u}=\left[\begin{array}{c} \Omega_{1}\\ \Omega_{2}\\ \Omega_{3}\\ \dot{\bar{x}}_{1}\\ \dot{\bar{x}}_{2}\\ \dot{\bar{x}}_{3} \end{array}\right]\end{split}$

The configuration $$\mathbf{q}$$ has 12 components and the velocity $$\mathbf{u}$$ has 6 components.

## Pseudo–rigid body¶

The motion of a pseudo–rigid body reads

(16)$\mathbf{x}\left(t\right)=\mathbf{F}\left(t\right)\left(\mathbf{X}-\bar{\mathbf{X}}\right)+\bar{\mathbf{x}}\left(t\right)$

where $$\mathbf{x}$$ is the current image of a referential point $$\mathbf{X}$$, $$\mathbf{F}$$ is a spatially homogeneous deformation gradient $$\left(\mathbf{F}=\partial\mathbf{x}/\partial\mathbf{X}\right), \bar{\mathbf{X}}$$ is a selected referential point and $$\bar{\mathbf{x}}=\bar{\mathbf{x}}\left(t\right)$$ is the current image of $$\bar{\mathbf{X}}$$. Deformation gradient $$\mathbf{F}$$ is an invertible and orientation preserving $$\left(\det\left(\mathbf{F}\right)>0\right)$$ $$3\times3$$ matrix.

The velocity of the spatial point reads

(17)$\dot{\mathbf{x}}\left(t\right)=\dot{\mathbf{F}}\left(t\right)\left(\mathbf{X}-\bar{\mathbf{X}}\right)+\dot{\bar{\mathbf{x}}}\left(t\right)$

where $$\dot{\mathbf{F}}$$ is the velocity of the deformation gradient and $$\dot{\bar{\mathbf{x}}}$$ is the velocity of the spatial image of $$\bar{\mathbf{X}}$$.

In vector notation the configuration $$\mathbf{q}$$ and the velocity $$\mathbf{u}$$ can be expressed as

(18)$\begin{split}\mathbf{q}=\left[\begin{array}{c} F_{11}\\ F_{12}\\ ...\\ \bar{x}_{1}\\ \bar{x}_{2}\\ \bar{x}_{3} \end{array}\right],\,\,\,\mathbf{u}=\left[\begin{array}{c} \dot{F}_{11}\\ \dot{F}_{12}\\ ...\\ \dot{\bar{x}}_{1}\\ \dot{\bar{x}}_{2}\\ \dot{\bar{x}}_{3} \end{array}\right]\end{split}$

Both, the configuration $$\mathbf{q}$$ and the velocity $$\mathbf{u}$$ are 12–component vectors.

## Finite–element body¶

(19)$\mathbf{x}\left(\mathbf{X},t\right)=\mathbf{X}+\mathbf{N}\left(\mathbf{X}\right)\mathbf{q}\left(t\right)$

where $$\mathbf{x}$$ is the current image of the referential point $$\mathbf{X}$$, $$\mathbf{N}\left(\mathbf{X}\right)$$ is a matrix of shape functions, and $$\mathbf{q}\left(t\right)$$ is a vector of nodal displacements.

The velocity of the spatial point reads

(20)$\dot{\mathbf{x}}\left(\mathbf{X},t\right)=\mathbf{N}\left(\mathbf{X}\right)\dot{\mathbf{q}}\left(t\right)$

where $$\dot{\mathbf{q}}$$ is the vector of nodal $$x,y,z$$ velocities.

In vector notation, the configuration $$\mathbf{q}$$ and the velocity $$\mathbf{u}$$ can be expressed as

(21)$\begin{split}\mathbf{q}=\left[\begin{array}{c} q_{1x}\\ q_{1y}\\ q_{1z}\\ ...\\ q_{nx}\\ q_{ny}\\ q_{nz} \end{array}\right],\,\,\,\mathbf{u}=\dot{\mathbf{q}}\end{split}$

Both, the configuration $$\mathbf{q}$$ and the velocity $$\mathbf{u}$$ have size $$3\times n$$, where $$n$$ is the number of nodes in a finite–element mesh.

The matrix $$\mathbf{N}$$ in (19) and (20) has the following form

$\begin{split}\mathbf{N}=\left[\begin{array}{ccccccc} N_{1} & & & ... & N_{n}\\ & N_{1} & & ... & & N_{n}\\ & & N_{1} & ... & & & N_{n} \end{array}\right]\end{split}$

where $$N_{i}$$ are nodal shape functions, juxtaposed from element shape functions meeting at coincident mesh nodes.

### Shape functions¶ Fig. 4 Element types in Solfec-1.0: tetrahedron, pyramid, wedge, hexahedron.

Finite–element types available in Solfec-1.0 are depicted in Fig. 4. Shape functions for those elements are summarised in Table 25.

Isoparametric mapping is used to go back and forth between the natural element coordinates $$\xi_{1}$$, $$\xi_{2}$$, $$\xi_{3}$$, local to individual element domains, and the referential mesh coordinates $$\mathbf{Y}$$

$\begin{split}\mathbf{Y}\left(\xi_{1},\xi_{2},\xi_{3}\right)=\mathbf{N}\left(\xi_{1},\xi_{2},\xi_{3}\right)\left[\begin{array}{c} \mathbf{Y}_{1}\\ ...\\ \mathbf{Y}_{n} \end{array}\right]\end{split}$

where $$\mathbf{Y}_{i}$$ are referential coordinates of mesh nodes. The function $$\mathbf{N}\left(\mathbf{X}\right)$$ in (19) and (20) has the following form

$\mathbf{N}\left(\mathbf{X}\right)=\mathbf{N}\left(\mathbf{Y}^{-1}\left(\mathbf{X}\right)\right)$

Newton iterations are required to solve $$\mathbf{X}=\mathbf{Y}\left(\xi_{1},\xi_{2},\xi_{3}\right)$$ and find $$\xi_{1}$$, $$\xi_{2}$$, $$\xi_{3}$$ for a given referential point $$\mathbf{X}$$.

 Tetrahedron $$N_{1}=1-\left(\xi_{1}+\xi_{2}+\xi_{3}\right)$$ $$N_{2}=\xi_{1}$$ $$N_{3}=\xi_{2}$$ $$N_{4}=\xi_{3}$$ Pyramid $$r= \xi_{1}\xi_{2}\xi_{3}/\left(1-\xi_{3}\right)\text{ if }\xi_{3}\ne1\text{ or }0\text{ otherwise}$$ $$N_{1}=0.25\left(\left(1+\xi_{1}\right)\left(1+\xi_{2}\right)-\xi_{3}+r\right)$$ $$N_{2}=0.25\left(\left(1-\xi_{1}\right)\left(1+\xi_{2}\right)-\xi_{3}-r\right)$$ $$N_{3}=0.25\left(\left(1-\xi_{1}\right)\left(1-\xi_{2}\right)-\xi_{3}+r\right)$$ $$N_{4}=0.25\left(\left(1+\xi_{1}\right)\left(1-\xi_{2}\right)-\xi_{3}-r\right)$$ $$N_{5}=\xi_{3}$$ Wedge $$N_{1}=0.5\left(1-\xi_{1}-\xi_{2}\right)\left(1-\xi_{3}\right)$$ $$N_{2}=0.5\xi_{1}\left(1-\xi_{3}\right)$$ $$N_{3}=0.5\xi_{2}\left(1-\xi_{3}\right)$$ $$N_{4}=0.5\left(1-\xi_{1}-\xi_{2}\right)\left(1+\xi_{3}\right)$$ $$N_{5}=0.5\xi_{1}\left(1+\xi_{3}\right)$$ $$N_{6}=0.5\xi_{2}\left(1+\xi_{3}\right)$$ Hexahedron $$N_{1}=0.125\left(1-\xi_{1}\right)\left(1-\xi_{2}\right)\left(1-\xi_{3}\right)$$ $$N_{2}=0.125\left(1+\xi_{1}\right)\left(1-\xi_{2}\right)\left(1-\xi_{3}\right)$$ $$N_{3}=0.125\left(1+\xi_{1}\right)\left(1+\xi_{2}\right)\left(1-\xi_{3}\right)$$ $$N_{4}=0.125\left(1-\xi_{1}\right)\left(1+\xi_{2}\right)\left(1-\xi_{3}\right)$$ $$N_{5}=0.125\left(1-\xi_{1}\right)\left(1-\xi_{2}\right)\left(1+\xi_{3}\right)$$ $$N_{6}=0.125\left(1+\xi_{1}\right)\left(1-\xi_{2}\right)\left(1+\xi_{3}\right)$$ $$N_{7}=0.125\left(1+\xi_{1}\right)\left(1+\xi_{2}\right)\left(1+\xi_{3}\right)$$ $$N_{8}=0.125\left(1-\xi_{1}\right)\left(1+\xi_{2}\right)\left(1+\xi_{3}\right)$$

## Implementation¶

Kinematic models are implement in bod.c (rigid, pseudo–rigid) and fem.c (finite–element) files. Configuration and velocity vectors are declared in bod.h as follows:

struct general_body
{
enum {OBS, RIG, PRB, FEM} kind; /* obstacle, rigid, pseudo-rigid, finite element */

/* ... */

double *conf,    /* configuration */
*velo;    /* velocity */

/* ... */
}