Physics-Based Simulation

General Slip DBC

Fortunately, for constraints like slip DBCs that are decoupled per node, SVD simply results in block-diagonal $U$ and $V$ which could be constructed procedurally in an efficient way. 3D planar slip DBC at node $i$ can be expressed as $${\mathbf{n}}_i^T({\mathbf{x}}_i-{\mathbf{x}}_i^{\prime })=0,$$ where ${\mathbf{n}}_i$ is the normal of the plane that node $i$ is slipping, and ${\mathbf{x}}_i^{\prime }$ is an arbitrary point on that plane. As discussed near Equation (5.1.2), if at the beginning of the time step node $i$ is already on the plane, the constraint simplifies to $${\mathbf{n}}_i^T\Delta {\mathbf{x}}_i=0.$$ Then performing SVD on the row vector ${\mathbf{n}}_i^T$, we obtain $${\mathbf{n}}_i^T={\mathbf{U}}_i{\mathbf{S}}_i{\mathbf{V}}_i^T=1\left[\begin{array}{ccc}1& 0& 0\end{array}\right]\left[\begin{array}{c}{\mathbf{n}}_i^T\\ {\text{m}}_i^T\\ {\mathbf{l}}_i^T\end{array}\right],\text{\hspace{1em}(6.3.1)}$$ where unit vectors ${\mathbf{n}}_i$, ${\text{m}}_i$, and ${\mathbf{l}}_i$ together form an orthonormal basis in 3D.

Then it becomes clear that globally, $U$ is simply a $m\times m$ identity matrix, $S$ is a $m\times dn$ matrix where every row contains exactly one unit-valued entry in the column corresponding to the first DOF of the slip BC node, and $V$ is a $dn\times dn$ block-diagonal matrix with the $d\times d$ orthonormal blocks only on those corresponding to BC nodes, and $d\times d$ identity matrix elsewhere.

To compute ${\text{m}}_i$ and ${\mathbf{l}}_i$ from ${\mathbf{n}}_i$, we first note that there are an infinite number of possible solutions. Therefore, we can simply first construct ${\text{m}}_i={\mathbf{n}}_i\times {\left[\begin{array}{ccc}1& 0& 0\end{array}\right]}^T$, or ${\text{m}}_i={\mathbf{n}}_i\times {\left[\begin{array}{ccc}0& 1& 0\end{array}\right]}^T$ if ${\mathbf{n}}_i$ is almost colinear with ${\left[\begin{array}{ccc}1& 0& 0\end{array}\right]}^T$, and then construct ${\mathbf{l}}_i={\mathbf{n}}_i\times {\text{m}}_i$. To obtain $V^T(-g)$, one only needs to left-multiply each ${\mathbf{V}}_i^T=\left[\begin{array}{c}{\mathbf{n}}_i^T\\ {\text{m}}_i^T\\ {\mathbf{l}}_i^T\end{array}\right]$ to $-{\mathbf{g}}_i$. As for $V^{T}HV$, first left-multiply each ${\mathbf{V}}_i^T$ to every block on the $i$-th block row of $H$ to obtain $V^{T}H$. Then for the $i$-th block column of $V^{T}H$, left-multiply ${\mathbf{V}}_i=\left[\begin{array}{ccc}{\mathbf{n}}_i& {\text{m}}_i& {\mathbf{l}}_i\end{array}\right]$ to every block. Finally, after solving for $y$ by applying the DOF elimination method on the modified system (Equation (6.2.3)), $\Delta x$ can be obtained by $\Delta x=Vy$ with similar block(node)-wise operations.

Example 6.3.1 (General Slip DBC). For the same two-node system in 2D as mentioned in the slip DBC example (Example 5.1.2), to constrain the first node $(x_{11},x_{12})$ inside the $3x+4y=2$ line, the slip DBC can be expressed as $$\left[\begin{array}{cccc}3& 4& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_{11}\\ x_{12}\\ x_{21}\\ x_{22}\end{array}\right]=2$$ and we can build $$U=1,S=\left[\begin{array}{cccc}1& 0& 0& 0\end{array}\right],V^T=\left[\begin{array}{cccc}0.6& 0.8& & \\ -0.8& 0.6& & \\ & & 1& \\ & & & 1\end{array}\right]$$ for changing the basis. Then in a time step where this slip DBC is already satisfied, assume we have $$H=\left[\begin{array}{cccc}4& -1& -1& -1\\ -1& 4& -1& -1\\ -1& -1& 4& -1\\ -1& -1& -1& 4\end{array}\right],\text{and}g=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right],$$ we can compute $$V^{T}HV=\left[\begin{array}{cccc}4.96& 0.28& 0.2& 0.2\\ 0.28& 3.04& -1.4& -1.4\\ 0.2& -1.4& 4& -1\\ 0.2& -1.4& -1& 4\end{array}\right],\text{and}{V}^{T}g=\left[\begin{array}{c}-1\\ 2\\ 3\\ 4\end{array}\right],$$ and solve the system $$\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 3.04& -1.4& -1.4\\ 0& -1.4& 4& -1\\ 0& -1.4& -1& 4\end{array}\right]\left[\begin{array}{c}{y}_{11}\\ y_{12}\\ y_{21}\\ y_{22}\end{array}\right]=\left[\begin{array}{c}0\\ -2\\ -3\\ -4\end{array}\right]$$ for $y$. Then the search direction can be obtained by $\Delta x=Vy$ so that $3\Delta x_{11}+4\Delta x_{12}=0$ and so the first node will stay on the $3x+4y=2$ line for arbitrary step size.