Hi All
I am trying to obtain SIF value for a stationary crack using XFEM in Abaqus.
I am using the 5th contour value and plotting against distance from crack front using excel.
The values are from History Output.
I am getting a very funny graph that is going up and down.
Forums
Free Tags
i think the first contour
i think the first contour is not correct due to singularity!
why don't u use conventional way through assign crack tip plus plane of it?
In reply to i think the first contour by shzangeneh
Do you mean the first 5th
Do you mean the first 5th contour?
I reason why I am using XFEM is because I have recently started a project on XFEM and am totally new to modelling.
Thanks for your suggestion shzangeneh
sif oscillations
12.00
The problem is associated with the domain definition. The xfem method
is almost mesh independent, however the domain integral is not totally
independent. For avoiding it your mesh --or the domain if you can change its definition--
should be adapted to the crack front. This is not an unknown and new effect, you can find information about
this effect , for example, in the next paper (although the abaqus
implementation is not the same as in my implementation, and the main part of
the paper is the curvilinear correction for the derivatives, the problem is
explained at the domain definition section):
V. F. González–Albuixech, E. Giner, J. E. Tarancón, F. J. Fuenmayor and A.
Gravouil. Domain integral formulation
for 3-D curved and non-planar cracks with the extended finite element method.
Computer Methods in Applied Mechanics and Engineering, 264:129–144, 2013.
How did you get SIF values?
For starters, how did you manage to get the SIF values? What values about the crack does the XFEM method give? I know the contour integral method gives the J integral values