# Cohesive debonding using traction separation law

Hi Everyone,

I am trying to model peeling of polymeric film over metal substrate. I am uisng a zero thickness cohesive element layer between the film and substrate. In abaqus the quads mode is what I use and mode independent. I have a maximum displacement of 0.0001 m for damage evolution. I also have nominal stress normal only mode as 100 and Normal stress first direction as 50.

I am not able to obtain crack for any of the elements after analysis. Although I do find that for some cohesive elements the damage SDEG reaches a value of nearly 1 and  normal stresses observed in them reach almost upto normal only mode limit specified. After reaching the limit there is a sharp fall in the normal stress as observed in the traction separation response.

I have the following questions

1. Is it necessary to specify normal stress first direction even if I use mode independent damage evolution

2. If required should the shear stess in my element exceed the normal stress first direction for damage for damage initiation and subsequently dip till zero similar to normal stress for damage evolution. I need to evaluate the reason for no damage in my model.

3. In a 2D model I assume the normal stress second direction can be ignored since there is only one shear stress component or is it also required.

Regards,

Shouvik

### Cohesive Law

Hi Shouvik,

I assume you are using bilinear cohesive law.What is your initial stiffness K? You need K and interfacial strength(maximum traction) Τ12 and T22 (100,50 in your case)to determine the critical displacement for damage initiation δc=T/K.

δc should be much smaller than the maximum displacement δm of 0.0001 m. Otherwise, it will give you problem.

J.X.

### Hi,

Hi,

Regarding the stiffness I am using E/Enn = 5000 and G/Ess = 1500. are these same as K. I thought Enn = Kn*element thickness. Should I input stiffness for E/Enn or take a ratio of elastic modulus and stiffness. For example

if E = 2.5e9 and k = 2.5e7 then should I use E/Enn = 100 or 2.5e7 if my element thickness is unit.

As you mention I need interfacial strength T12 and T22. Does the normal only mode signify T12 i.e shear and first direction signify T22 i.e normal stress. I would like to have my maximum normal traction as 100 and max shear traction as 50. So do I need T12(shear) = 100 instead of 50.

My option for damage evolution are as

Type - Displacement

Softening- Linear

Mixed mode behavior - Mode independent

Mode mix ratio- energy

displacement at failure = 0.0001.

Is this specifying bilinear cohesive law?

Finally if I am using mode independent mixed mode behavior are both maximum traction T12 and T22 and E/Enn and G/Ens required. I thought mode independent meant that damage only depended on traction in one direction.

Regards,

Shouvik

### Cohesive parameters

Normal
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false
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Hi Shouvik,

Be careful with the units you are using. We are talking about N, mm and MPa, right

Initial stiffness K is usually in the order of 1e7 to 1e5, when assuming the thickness is one. K is not necessarily the Young's Modulus of the film or the substrate.

Let’s use  E=G1=G2=1e6 for both mode I (Normal Only) and mode II (Nominal stress First/Second direction).

So Maximum traction  should be 50 MPa (N/mm2) for Mode I and 100 Mpa for mode II.

The maximum displacement is 0.0001m, which should be 0.01mm in the input, for both Mode I and Mode II. The critical energy release rate (fracture toughness) GIc=1/2 * 50 * 0.01=0.25N/mm for mode I and GIIc= 0.5N/mm for mode II, which is lower than normally expected, but I guess it is OK. For such small Gc, the cohesive zone length would be short. You would need to use really fine mesh or artificially reduce the interfacial strength.

Also, in your case of peeling “soft” film over metal substrate, you may have convergence problems for standard/implicit analysis. Viscous regulation for cohesive element is recommended, or you can just use explicit solver.

Regards,

Jia

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### Hi Jia,

Hi Jia,

thanks for resolving many doubts. I wasnt sure that traction was N/mm2 and E also N/mm2 and that displacements are in mm. Just a few last questions using E = G1 = G2 = 1e6 as you suggested and say k = 1e5.

1. What should be the value E/Enn and G1/Ess  input to abaqus for the adhesive. Should this be 1e6/1e5 = 10 or  1e6 or 1e5 for this set of data .

2. I see that the displacement at failure should be in mm (0.01 in this case). I am using an initial cohesive layer thickness of 10-2 mm. In abaqus the cohesive section uses the "'analysis default"' thickness option. Will abaqus assume thickness 1 mm and hence strain at failure = 0.01/1 = 0.01. Or will the strain be 0.01/10-2 = 1 at failure. Also does this change if option of cohesive section is set as ''üse nodal coordinates".

Thanks again for the most valuable inputs.

Regards,

Shouvik

### Cohesive parameters input in Abaqus

Hi Shouvik,
You are welcome. I am glad to help.
1.    E/Enn in Abaqus means E or Enn, not E divided by Enn. You should input 1e6.
2.    In section control, “analysis default” assume the thickness of the cohesive elements to be 1mm. Thus, strain equals displacement (separation) in value.  It’s the traction-separation law that we defined for the cohesive elements after all. You can also specify the initial thickness to be 1 in the section control, just to be sure.
The Abaqus user’s guide is very helpful, you should check it out.

Regards,
Jia

### Hi Jia and all,

Hi Jia and all,

what should be units for density. i find in abaqus documentation that tom/mm3 is used in SI(mm). Hence should density for steel (8700 kg/m3) be input as 8.7 e-9 ton/mm3 . Also I am creating a unit thickness adhesive layer. Later I edit the nodes to make it zero thickness layer. However since 1 mm thickness is magnified many times (real thickness of adhesive would be in microns probably). Hence how to input  the density of the adhesive. If in real adhesive density is 2000 kg/m3 would it also be input as  2 e-9 ton/mm3.

Regards,

Shouvik

### Cohesive Density

Hi,

For denstiy of the cohesive layer, input of 2 e-9 ton/mm3 is OK. Actually, it won't effect your calculation too much.You can also artifically reduce the  "real" denstity, so that you can cut down the step time and speed up the calculation in Explicit analysis.

For further questions, please feel free to contact me at xuejia@comac.cc

Regards,

Jia.

### units

Dear Jia,
I want to model my part in mm. So what will be the units for these two materials?

First for cohesive: Nominal stress normal-only mode, E or Enn

Second for Engineering constants: E1, G12

Thanks