Linear Imperfect Interface
Description
This imperfect interface contact law assumes the normal and tangential tractions are linear and depend only on the normal and tangentatial displacement discontinuities, respectively:
[math]\displaystyle{ T_n = D_n[u_n] }[/math]
[math]\displaystyle{ T_t = D_t[u_t] }[/math]
where Dn and Dt are two interface parameters, which are infinite for a perfect interface (zero displacement discontinuity) and 0 for a debonded interface (zero traction). But this linear law would allow the two materials to interpenetrate, especially if Dn was low. To correct this issue, the traction law in the normal direction is allowed to be bilinear:
[math]\displaystyle{ T_n = \left\{ \begin{array}{ll} D_{nt}[u_n] & {\rm \ if\ }[u_n]\gt 0 \\ D_{nc}[u_n] & {\rm \ otherwise} \end{array} \right. }[/math]
where Dnt and Dnc are separate interface parameters for the interface being in tension or compression.
Properties
The properties for this law are:
| Property | Description | Units | Default |
|---|---|---|---|
| Dnt (or Dn) | Imperfect interface parameter for normal direction when in tension | pressure/length units | -1 |
| Dnc | Imperfect interface parameter for normal direction when in compression. If Dnc is not specified, it set equal to Dnt. | pressure/length units | none |
| Dt | Imperfect interface parameter for tangential direction | pressure/length units | -1 |
In theory, very large interface parameters will result in a perfect interface, but they also make the numerical equations very "stiff" and therefore potentially unstable. To solve this issue, you can set any interface parameter to -1 to indicate an infinite interface parameter or a perfect interface. When a parameter is -1, the equations are handled differently to avoid stability issues.
Examples
Material "interfaceID","My Imperfect Interface","LinearInterface" Dn 500 Dt 0 Done