Crack Tip Stresses - Comparison between FE and Analytical Solutions 1

[MitchCud]

Hi,

I've been puzzling over this particular problem for quite some time now so it was time to get some opinions!

Problem
[ul]
[li]Plate, 100 x 100mm, with a 5mm edge crack on one side.[/li]
[li]Remote load of 100MPa along the top, fixed BC on the right hand side[/li]
[li]Linear elastic, isotropic material[/li]
[/ul]

Should be very simple? I thought so too until I started getting confusing results, in comparison to results obtained using the Westergaard solution for crack tip stresses.

Results

XX stresses:

and YY stresses:

SIFs
The mode I stress intensity factor returned by ABAQUS is 454.8 and the computed SIF used in the analytical solution was 448.8 (about a 1.3% difference..), calculated using a formula given by Rooke and Cartwright, "Compendium of Stress Intensity Factors".

Question:
Why should the stresses initially converge yet ultimately diverge as the distance from the crack tip increases?


My own thoughts are that neither solution provides an inherently correct picture of the crack tip stress state, however a senior colleague (who I'm always inclined to believe) tells me that they aught to agree at pretty much all points. Therefore, by his logic it is a mesh refinement issue. I disagree as the crack tip elements cannot report the singularity adequately.



So any thoughts or suggestions would be much appreciated!


Thanks,
Mitch

 

[rb1957]

could this be a poisson effect (in your FE) ?

"boundary condition fixed on RH side" ... so this an edge crack in a panel with bending restrained ?

you're loading in the YY direction, but you're showing sizable stresses in the XX direction ... how did you account for this in youe Westergaard calc ? don't see why your Westergaard stress would go below 100 MPa ... it might do locally, (so long as the total load is right).

how fine a mesh ?

personally, i'd start with something simple ... a center crack in a finite with panel,
model the whole panel or a 1/4 (using symmetric boundary constraints .. be careful !).

i'd first model the uncracked panel so i could be sure of the stress state the crack would see ... i might think i'm applying 100 MPa, but the model might be doing something different (due to BCs, poisson, ...)


Quando Omni Flunkus Moritati

 

[MitchCud]

rb1957,

Thanks for the reply.
Displacement BC on the RHS, i.e. U1 = 0.
The mesh is very fine, ~13000 quadratic elements, with 100 elements along a 5mm line from the crack tip.

Surely the Westegaard XX stresses would tend to zero as you move towards the right hand side of the plate?
Also, I didn't make any changes to the Westegaard solution -

This formula would indicate that the two stresses will be the exact same at theta = 0.

Is there an area of influence over which this formula makes sense? Because towards the RHS of the plate the YY stresses must equal the applied stress (which the FE model indicates).

Also, the Westegaard solution disagrees with the complex formulation of Sih et al. (1965) by a factor of 1 / sqrt.(pi).
It states:
sxx = K1 / sqrt. (2r) * f(theta, E, mew, G) - and for theta = 0, sxx = syy

 

[rb1957]

i thnk your FEM is showing sizeable stress in the xx direction ... poisson, constraints, ...
i'm not sure how you accounted for these in your westergaard calc (which i think assumes a simple YY stress).

arh, 1/sqrt(pi) factor ... i recall there was a disagreement with stress intensity formulation in the early days of damage tolerance. i think the industry standardised on the "NASA" formulation (with the sqrt(pi) factor). i doun't think there is a difference between the results of Westergaard or Sih ... i think Sih has the factor buried in the "f()".

Quando Omni Flunkus Moritati

 

[MitchCud]

Working from a statement of the complex Airy stress functions, Sih's solution considers an infinite plate with a central crack and to model the single edge notch panel, K has to be specified accordingly.

Therefore theta = 0, it should agree with Westegaard, but won't.

Also, I wouldn't have thought that these are large XX stresses in the FEM solution? Within r = crack length, the XX stresses are tending towards zero - which is the situation for an uncracked plate.

I think my question should be restated as: "what is the effective radius for the theoretical solutions?"

 

[rb1957]

i think my point is that the FEm might have stresses in it that you haven't considered in your calc.

not quite sure what you meant by "to model the single edge notch panel, K has to be specified accordingly". i'd've thought that theoretical slutions would have considered "simple" easily analyzed models (like a center crack in an infinite panel) practical solutions need to consider finite width (which does have a theoretical solution, but ...).

i'd look at your FEM without a crack ... is it giving the uniaxial tension you expect ?

once you start the crack, does the panel behave the way you expect ?? (far field, not local to the crack)

for short crack length you can model the panel as doubly cracked (ie on each side), which is effectively what you're doing with a symmetrical BC.

if you model different panel widths, do they behave as you'd expect ?

again ... model something really simple (center crack in a large panel) to start with.


if your plots are along theta = 0 (ie along the crack direction), i'd've thought stressXX would have been close to zero, assuming you're loading in YY, and that the offset (between the center of the panel (the CG of the load) and the center of the cracked edge (the CG of the reaction) would be taken out as YY forces on your constrained edge. in bending unrestrained, the CGs align and there's a bending stress in YY; in your case maybe there are a bunch of XX forces on your BC balanced by XX forces on the crack plane ?

Quando Omni Flunkus Moritati

 

[MitchCud]

I think I've it figured out.
The FE model is fine, more importantly the Westegaard solution here isn't the full picture of what's going on. In fact it's realistically only applicable for r / a < 1x10^-3.
For values of r greater than that, more of the higher order terms of the Williams expansion need to be included, i.e. T-stress + O*r^n

Thanks for your help.