Crack tip modelling 1

[yuwimech]

Hi,

I have an axially loaded square plate which has a crack in the middle. I am using a double symmetry to model this plate that is, I took only quarter of it. I defined my crack using Interaction->Special->Crack but I havent done anything for singularity. Do I have to define a singularity?

The aim of what I am doing is to determine the first stress intensity factor in the elastic region. The theoritical value is 0.43 times the applied stress but I havent been able to find this result yet.

I will be so glad if one tells me what else I have to do for the crack tip. Thank you.

 

[xerf]

Your mesh looks kind of coarse.

If you compute K from J-integral, it is not necessary to use singular elements. You should request a suficient number of contours such that the values of J-integral, from evaluation on subsequent contours, are close.

I tested ABAQUS capability for computing K for different FE models of fracture mechanics specimens (M(T) and C(T)) and it gave results very close to analytical solutions.

 

[yuwimech]

I am only demanding K's from history output using KII=0, I do not know what abaqus is using to extract these values. The problem is that our professor Eric. B. Becker found 0.47 which has to be 0.43 theoratically using CAE and I found 0.43. I must not find a better result than EB.Becker :-D That is I made a mistake somewhere and I couldnt find where. Anyways thank you for your help.

Just one thing: CPS8 or CPS8R for crack anlysis??

 

[xerf]

I think is more important to know how the computational method works rather than obtaining a specific value for K. If you do not know the method, I think is difficult to understand where the mistake is.

Depending on the numerical method used for computing SIF, you can get better or worse accuracies with respect to the analytical value.

I guess you have a model of an middle-crack specimen with tensile stress (i.e. normal outward tractions) applied on the upper edge. For this case, the analytical K1 can be computed using Feddersen's formula:

K1=stress*Sqrt(Pi*a)*Sqrt(Sec(Pi*a/W))

where W=width of the specimen
a=half of the crack length
stress=applied tensile stress.

Therefore is very easy to check the value of the analytical solution.

Formulas for SIF, for different loading conditions and specimen geometries, can be found in most of the Fracture Mechanics textbook.

In order to decide upon using a specific element type for a specific problem, the wise approach is to check the recommendations in the ABAQUS documentation.