Compliance function

[marcus.stephan.schulze]

Hi,

For the mesh given for the three point bending, the span to depth ratio is 2.177. I am struggling to find a compliance function for such a ratio as most of what i found applies either for s/d = 4 only, or s/d > 2.5. Is there anywhere that I can look to find the compliance function?

This is just so I can also graph the equivalent LEFM load displacement graph and the Gr v crack length.

Also, is there a way I can write a script that converts all the out_values files to vtk for paraview?

Kind Regards,

Marcus

[Karol Lewandowski]

This will do the job:

for f in out_*h5m; do mbconvert $f ${f%h5m}vtk; done

But be careful as it will also overwrite your previous output files.

[Lukasz Kaczmraczyk]

Hello,

1) You have a load-displacement path for LFEM in files which I send you. 

2) What do you understand by compliance function?

Beam has the same ratio what you find in Peter paper, I think that we should compare results to his and LFEM which you get from me,

Kind regards,

Lukasz

[ignat...@gmail.com]

Markus,

I am in office today, you could pass by and look at this together if you like.

Kind regards,

Ignatios

[marcus.stephan.schulze]

Hi Ignatios,

I am only back in Glasgow on the 1st of January, but thank you for the offer.

Kind Regards,

Marcus 

[marcus.stephan.schulze]

Hi Karol,

Thank you

Marcus

[marcus.stephan.schulze]

Hi Lukasz,

Compliance function is the inverse of stiffness (displacement/force) which is a function of the crack length. In the paper i attached, the energy release rate is calculated by using a dimensionless energy release rate which is dependent on the compliance function. I did the same thing for the double cantilever beam to produce the graphs I showed you a couple weeks ago, however the compliance function for three point bending is more complicated.

Im just looking to graph the energy release functions to further discuss the decreasing impact of ft with increase size - and discuss the equivalent LEFM size effect (also in the paper by Morel).

I will also be comparing it to the load displacement path for LEFM, which you sent.

Kind Regards,

Marcus

[Lukasz Kaczmraczyk]

Marcus,

Good plan. 

Thanks for the update. How big sample you were able to calculate?

BTW I had to remove the paper from your message, the discussion group is public, and we have no right to publish it.

Kind regards,

Lukasz

[marcus.stephan.schulze]

Hi Lukasz,

Apologies, didn't realise i couldnt upload the paper,

It looks like ive managed to run the mesh sizes for all of Peter's beams (biggest being D = 500mm)

Also, any ideas how i can calculate the compliance function? Im still struggling to find a function for a three point bending set up.

Kind Regards,

Marcus

[Lukasz Kaczmraczyk]

If you have a load-displacement path, stiffness is

stiffness =  force/displacement, 

so

compliance = 1/stiffness.

All data are from the load-displacement path. 

[marcus.stephan.schulze]

Hi lukasz, 

Im trying to write compliance as a function of crack length, a, as the energy release function is a function of crack length. Is there any way i can use the step reduction and number of steps in the code to get the size of the change in crack length (a - a0) for each iteration? 

Kind Regards, 

Marcus  

[Lukasz Kaczmraczyk]

Marcus,

Directly not, in particular for smaller samples. For cohesive elements is not clear where the crack ends. Indirectly you can try to calculate crack area if you know the change of dissipated energy. In your particular case, the work of external can be calculated, this is displacement times the pressure times the area where force is applied, i.e. Delta W, where W_n = lambda_n * u_n * (A * pressure). A is the area under pressure, p pressure, lambda_n is load factor,  and u_n vertical displacement, at step n. This is assuming that displacements under pressure are uniform, that is a good assumption in your case. So to the calculated amount of dissipated energy, you subtract value at step t_n+1 from step t_n; so crack area increment is Delta a = - (1/2) * Delta W/ Gf.

Kind regards,

Lukasz

[marcus.stephan.schulze]

Lukasz,

Ok thank you, I will give that a go 

Kind Regards, 

Marcus 

[marcus.stephan.schulze]

Lukasz, 

Actually, what is the area under pressure in this case? is it the small square of the mesh on top of the beam where the load is applied or is it the whole thickness * length of the beam?

Kind regards, 

Marcus 

[Lukasz Kaczmraczyk]

Hello,

Look to the cubit model, if I remember the pressure is scaled like that, that F=A*p = lambda. So you can use directly lambda. The answer is in the journal file.

Lukasz

[marcus.stephan.schulze]

Hi Lukasz, 

So in this case the External Work W = lambda_n * u_n ?? 

[Lukasz Kaczmraczyk]

Yes,