# Constraint counting

The basic idea is the ratio, R, of the equilibrium equations divided by the number of locations imposing the incompressibility condition should equal the ratio of the number of continuum equilibrium equations divided by the number of continuum constraints (which is simply 1 for the incompressibility condition).

## Example: The Triangular Element

Assume that the number of nodes in Figure 3 is extended to n + 1 in each direction. The number of unconstrained nodes is therefore n², for 2n² equilibrium equations, and the number of elements is 2n², with each element contributing one volume constraint. Therefore, R = 1, implying that the number of degrees of freedom per volume constraint is too low. A value of 1 doesnêt imply the rigid locking exhibited by this example; many formulations having R = 1 are simply overly sti . For the crossed triangle mesh, the number of nodes in the mesh is doubled to 4n², as is the number of elements. Based on this superìcial level of analysis, R = 1, and therefore this mesh should also lock up. However, it is well known that crossed triangle meshes donêt lock. A closer examination shows that that quadrilateral composed of four triangles has only three independent volume constraints, and therefore R = 4/3. To understand this, consider node C in Figure 4, which belongs to triangles DEC and CEB, and assume that nodes D, E and B are ìxed. The areas of the triangles are conserved as long as C moves parallel to the edges opposite it in each triangle, line segments DE and EB. These two line segments are parallel, and therefore node C can move along their common direction. In e ect, the two triangles only impose one constraint on the motion of C. This situation is in contrast to the one in Figure 2, which also has two volume constraints, but which locks up completely. The underlying reason that the crossed triangular mesh doesnêt lock is each pair of adjacent elements has a common edge direction along one of the quadrilateralês diagonals. If node E is moved so that DE is no longer parallel to EB, then node C can no longer move (and similarly for node A).

With the crossed triangles, each of the four vertex nodes (A, B, C, and D) of the quadrilateral has one degree of freedom. There are an additional three degrees of freedom associated with the rigid body modes, which exactly satisfy the volume constraints. The total number of degrees of freedom is therefore seven, and since the ìve nodes have a potential of ten degrees of freedom, the four volume constraints impose 10-7=3 actual constraints.

## Example: Quadrilateral Elements

Again assuming that the number of nodes in each direction is n + 1, the number of elements and unconstrained nodes is n². Imposing the volume constraint at each integration point, the number of constraints is 4n² for full integration. The ratio R = 1/2, and it would appear that the quadrilateral element should lock up like the triangular element in Figure 3. Since this isnêt the case, constraint counting is at best a rough guide. If the incompressibility constraint is imposed only at the center of the element, R = 2, the ideal ratio. In fact, this formulation works very well for incompressible problems whether the constraint is imposed via selective reduced integration, uniformly reduced integration, or the **B** formulation.

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