hi Jamaleddin

after finding the displacements of the corner nodes of an element, you can calculate the center point displacement by using the center Gauss point coordinates in reference coordinates and the corner displacements and the other variables of Heaviside and crack tip functions.

there aren't any difference between 4 and 10 node modeling. both should result in same values.

Either in conventional FEM or XFEM, the displacement of any one point in an element can be calculated with the nodal values through interpolation. The only difference in XFEM is that the additional DoFs in the enriched nodes need to be considered to calculate the displacement. So you can use the element nodes to calculate the displacement of a point in the element. It doesn't matter how many nodes are used to model the crack.

I have 4 elements and one edge crack. this crack is common between two elements. how can I calculate the displacement of one point around crack? I attached picture of my problem.

thanks

It would be better if you label all the nodes and elements, which will make it easy to explain. Basically, you can use only the nodes of the element in which the point is located to calculate the displacement. This is the general element-based interpolation scheme with standard shape functions in the element local coordinate system in the FEM.

for Example I want to calculate displacement of one point in the middle of element 1 and one point in the middle of element 4. first for point in the element one i have to use general finite element formula to calculate displacement of point. However, how crack effects to our solution?

I read this paper "A finite element method for crack growth without remeshing"

in equation 9 It is referred to one formula:

uh = sum(u(i)*phi(i)), i from 1 to 10, phi is a bilinear shape function.

how can I use this formula for my calculating?

I'm little confused. I can't understand, how crack effects to calculating of displacements one point?

Firstly, if you use XFEM model the edge crack, nodes 9 and 10 in your mesh figure should be one same node (e.g., labeled as 9). Then the crack can be implicitly modeled by enrichments at nodes 5 and 9 (of cause you can enrich more nodes around the crack).

The equation in the paper is just an illustrative formulation. It should be noted that the interpolation in this equation is based on the whole domain rather than an element. You can refer to equation 15 in this paper, which indicates the crack effect with the additional DoFs.

For your example, if you want to calculate the displacement at the center point of element 1, you can use only nodes 1,2,5 9 of the element, with considering the additional DoFs which account for the crack effect. The detailed formulation for calculating the displacement is same as equation 15 in the paper, where the summation only covers nodes 1,2,5,9. The calculation of displacement of a point in element 4 is similar, only interpolated with nodes 5,7,8,9.

Equation 15 in the paper is a general displacement interpolation formulation in XFEM, which can be used to calculate the displacement of any point in an element. Just keep in mind that the calculation is easily implemented only when it is based on an element and then the shape functions are the standard formulations. If the interpolation is based on a domain rather an element, you have to find proper shape functions.

The element-based calculation of displacement of any point with XFEM generally can be divided into two parts: one is the interpolation from the standard DoFs which is same as that in the conventional FEM, the other is the interpolation from the additional DoFs which only covers those enriched nodes in the element (the nodes may not all eriched). If no enriched nodes in the element, the calculation of displacement is just same as the conventional FEM. It is easily implemented for both interpolations because same standard shape functions are used.