MPC type SLIDER keeps a node on a straight line defined by two other nodes but allows the possibility of moving along the line and allows the line to change length.

The sliding constraint has a variety of uses. For example, this MPC is used in conjunction with other MPC types to constrain a shell element mesh to a solid element mesh. The MPC maintains consistency with standard shell theory by forcing initially straight lines through the thickness to remain straight despite rotation and displacement. When applied to solid element nodes on the shell-solid interface, this MPC enforces a kinematic approximation of compatibility with the shell model.

The theory of this constraint is as follows:

Let , be the points defining the line; and let be the node that must lie on this line. The direction of the line is given by

where Let be base vectors in the x-, y-, z-directions in the global coordinate system. Then, define a unit vector normal to the line as unless , in which case we use Now we can define an orthogonal normal as , and now form a set of orthonormal base vectors with and normal to the line joining and . The constraint can be imposed by the condition that the line joining the node m to node 1 be perpendicular to and ; that is, and We now choose a local coordinate numbering system such that i is the global direction on which has its largest projection: Likewise, we choose global direction j such that and Using this definition of the constraint conditions can be written explicitly in terms of coordinate components of node m as and These equations can be used to eliminate (note that the numbering of avoids dividing through by zero in this elimination): and

The above equations will enforce the desired constraint. We also need the derivatives of these constraints. These are

and where and

These equations reduce to

and can be obtained from the definition of to give and, therefore, and The incremental constraint equations become and Let . Then, the above equations, when written out in full with the same ordering of used above, are and Solving for we obtain and

In the two-dimensional case lies in the x–y or r–z plane and . This implies that the second constraint equation is satisfied automatically. The remaining constraint equation is

and its derivative is