Boundary Layer Model wrt SSY in Abaqus 2

[renbo86]

Hi All,

I am new to Abaqus (now 1 week experience! ) and I have been tasked with creating a boundary layer model in Abaqus to investigate small scale yeilding effects in steel. I have created the model and the mesh now. After speaking to my boss he told me that I must input a nodel displacement as each of the nodes on the outer edge of the model, as oppose to using a load/pressure that is described in the tutorial. I am a bit stuck on how to carry this out. Does anyone know how I can make Abaqus displace a node by a set amount?

Thanks

 

[mrgoldthorpe]

The nodes on the outer boundary of the model need to be displaced according to the 1/root(r) displacement field for the leading term in the crack tip field (Williams' expansion).

You need to use the BOUNDARY keyword of ABAQUS.

So in your input file you'll need lines as follows:

** apply x displacements to outer boundary
*BOUNDARY, OP=MOD, AMPLITUDE=AMP1
136361 , 1 , 1 , 0.000000
136362 , 1 , 1 , 0.839847
136363 , 1 , 1 , 1.637398
** and so on

** apply x displacements to outer boundary
*BOUNDARY, OP=MOD, AMPLITUDE=AMP1
136361 , 2 , 2 , 8.627576
136362 , 2 , 2 , 8.527111
136363 , 2 , 2 , 8.231758
** and so on

The above are from a boundary layer model of mine from a few years ago.

I used Excel to work out those displacements and form the appropriate lines. These lines are simply pasted into the .inp file.

MRG

 

[renbo86]

Hi mrgoldthorpe,

thanks very much for your reply.

I am quite new to the fracture mechanics field so I am just getting my head around the Williams expansion etc. I have only started to learn Abaqus in the viewpoint mode so I will have to get used to using the keywords now.

Can I ask what values you used for Ki and Poisson's Ratio? I Would like to do the calculation to match the values to yours given so that I can make sure I am on the right track.

Thanks again for the help so far.

 

[mrgoldthorpe]

I have done BL analyses for a variety of materials; so the results won't be consistent with yours, I'm afraid.

You need to apply a value of K that gives a plastic zone size which is much smaller than the radius of your BL model. This ensures that the response in the model is predominantly elastic: consistent with the elastic boundary conditions you put on the outside of the model.

Now the (plane strain) plastic zone size "R_p" is

R_p = (1/ 6 pi) * (K/sig_y)^2,

where "sig_y" is the yield stress of your material: defined as departure from linear elastic response.

The radius of the model "R" must be much greater than R_p; say 10 times. So whatever your model R ensure that the maximum K applied does not make R_p bigger than 0.1R,

email me off-board if you'd like an example .xls with the calculation of the prescribed displacements on the outer boundary.

MRG

 

[renbo86]

Hi MRG,

Thanks so much for your help, I was previously advised that the R_p could only be 20% of R, so this will help me avoid a potential slip up.

Wrt the example excel file, I would like to have a look at it if that's ok. It will give me a good idea of the way in which I should be headed. I am not sure how to email you off thread tho, as I cannot find an email for you and am new to these boards, is there any particular place I should be looking?

Thanks for your continued support.

Peter

 

[mrgoldthorpe]

Peter,

I've uploaded the xls (below). It should be self-explanatory.

The xls includes sheets for displacements for combined K and T-stress. If you are doing a pure BL analyses them use T=0.

It's obviously best to ensure that the plastic zone size is as small as possible. As it increases in size with increasing load, the model becomes less stiff and so the effective K applied will drop below the intended value. You'll see this manifest as a reduction of the outer results for J compared with the intended elastic J (K^2/E').

Incidentally, if you examine negative values of T stress (say up to 80% of yield) the actual plastic zone size is larger than the T=0 value which I use in all instances (though you can estimate it by taking account of the T-stress).

Regards

MRG

 

[renbo86]

Thank you!! Really appreciate this and I'm sure it will help me to understand better the BLM.

Thanks again!!

Pete

 

[renbo86]

Hi MRG,

I'm really sorry to bother you again but I just have one more question regarding the excel spreadsheet.

In your spreadsheet you have a value that you use in your displacement equation, 2G (cells D/E9). The displacement equation that I was using is the same but instead of using the 2G value you have used, I was using E/(1-poisson's ratio), from a journal paper that I have read from Mark Kirk "contraint effects in fracture symposium". I was just wondering what this value 2G represented? My first thought was the energy release rate but I'm not sure this is correct.

Thanks again.

Pete

 

[mrgoldthorpe]

Pete,

G is the shear modulus of the material, G=E/{2(1+ν)}

 

[loki3000]

hey renbo are you working on a phd or in the industry?

 

[renbo86]

Hey Loki,

I am doing a PhD currently in collaboration with industry, why do you ask?