Results

[marcus.stephan.schulze]

Hi Lukasz, 

Just want to check that the results are making sense before I start trying to process them - its making sense to me. 

Hope you enjoy your christmas, 

Marcus 

[Lukasz Kaczmraczyk]

Hi,

Looks ok, I don't know about values, but qualitatively it should look like that. Well done.

Lukasz

[marcus.stephan.schulze]

Hi, 

Thank you

Marcus 

[marcus.stephan.schulze]

Hi Lukasz, 

Im trying to graph the compliance as a function of crack length, but the values im getting for crack length are very small. I have attached the results I got from the analysis for D = 500m, could you please have a look when you have time. 

Kind Regards, 

Marcus 

[Lukasz Kaczmraczyk]

Hello,

Check the dimensions, check as well area under force. Check units, note that in the model you are using millimetres, what is the unit of force. You know from geometry what is a maximal crack area, it is a depth minus initial cut times thickness. Moreover, you stop calculations at some point, but it is not a point when the beam is fully broken, i.e. crack is not propagated through full depth,

Regards,

Lukasz

[marcus.stephan.schulze]

Hi Lukasz, 

So just to check, the Griffith number is then in J/mm^2? Because then the external work (W = lambda* disp_vert) is in Nmm (im assuming force is in Newtons because in the mesh it is currently defined as functions of the geometry). So for the equation for delta_a = -0.5(delta_W/Gf) gives unites of mm^2 for delta_a. 

Is that right? 

Also, because the delta_W = (Wn+1 - W), it will give negative values for the decreasing part of the curve? 

Kind Regards, 

Marcus 

[Lukasz Kaczmraczyk]

Marcus,

Looks correct. Only one thing to verify is to check what is a force applied to the structure, i.e. check the area under force and magnitude of force itself. You can do that in Cubit, which can plot the area of the surface.

You have to verify unit of Young modulus, and strength, to be sure that all is all right.

Lukasz

[marcus.stephan.schulze]

Hi lukasz, 

The units of Young's modulus are in MPa, from Peters paper. 

Im a little confused what you mean about the area under the force. In the mesh you defined the force as:

area = h*(s*L)   

force = ((h*D*D)/(1.5*L))/area 

cubit.cmd('create force  on surface 36 43  force value %e direction   y 1' % force)

[marcus.stephan.schulze]

Also, since the units of the crack is mm^2, is this giving the crack area - and to calculate crack length I divide this by the beam thickness? 

And does it make sense that the delta_W switches from positive to negative? Or is the change in external work the absolute value of (Wn+1 - W)? 

Kind Regards, 

Marcus 

[Lukasz Kaczmraczyk]

Hello,

You have to have consistent units, between what you have for fracture energy and young modulus. It is probably right, but please check this, this is fundamental engineering.

Abut force

area = h*(s*L)  

force = ((h*D*D)/(1.5*L))/area

So as you can see overall force magnitude is 

F = lambda * force * area = (((h*D*D)/(1.5*L))/area) * area = lambda * (h*D*D)/(1.5*L)

So it was a small mistake what I write to you before because the force is

F = lambda * (h*D*D)/(1.5*L)

not just lambda.

Regards,

Lukasz

[marcus.stephan.schulze]

Cheers, Ill have a look.

Im still unable to get the compliance function working, I may wait until Im in glasgow before I continue with that part. 

Kind Regards, 

Marcus