# How is the slip calculated in Abaqus small-sliding interaction?

I’m doing research on the sliding of 2 fracture surfaces.

To do that, I’m trying to understand the Small-sliding deformable interaction in the Abaqus (2018), and I use the Node-to-surface discretisation method.

Specifically, I’m interested in how exactly the ‘slip’ amount is calculated and imposed to the Abaqus.

I understood the derivation, that is presented in the Abaqus Theory Manual (6.11), where the calculation of an ‘overclosure’ and a ‘slip’ is shown:

To check how ‘slip’ is calculated I created in Abaqus a very simple example with the slip upwards of the LHS body and downwards of the RHS body:

However, the problem I have is the slip in Abaqus is quite different from the one I understand. If we analyse the lower contacting cells with nodes 13 and 168 (see picture above), then using the logic in the derivation we can decompose it to this schematics:

As in the derivation, we have slave node13 penetrating RHS body which becomes node Xnplus1. Then we draw the anchor node ‘xo’ that coincides with the normal direction from the surface of RHS body to node13. Consequently, xi node is drawn to the normal from the Xnplus1 to the face. Also, unplus1 = xnplus1 as in (5.1.1-4) as in the manual. I understand that the slip is then the distance from xi to xo in the vertical ‘y’ direction, i.e. slip_y = 0.352.

However, Abaqus gives me slip 0.28 all various configurations e.g. small or finite sliding, instantaneous load variation, fixed 1 increment size, surface-to-surface, and all other options are similar.

I’d be glad if you please could help me to understand this.

Sincerely, AMAN