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Implementation of traction BC in tractionDisplacementFvPatchVectorField.C 

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July 12, 2011, 12:58 
Implementation of traction BC in tractionDisplacementFvPatchVectorField.C

#1 
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Johan Roenby
Join Date: May 2011
Location: Denmark
Posts: 10
Rep Power: 7 
Hello,
I am trying to understand how the implementation of the traction boundary condition in tractionDisplacementFvPatchVectorField.C works. On line 176180, it says:  gradient() = ( (traction_ + pressure_*n)/rho.value() + twoMuLambda*fvPatchField<vector>::snGrad()  (n & sigmaD) )/twoMuLambda;  When we apply this to a displacement field, D, the code sets the value of the normal gradient, n.grad(D), on the relevant boundary (right?). What I don't understand is why n.grad(D) is set to the value appearing on the right hand side in the code. As I understand it fvPatchField<vector>::snGrad() means n.grad(), and so if we mutliply by twoMuLambda on both sides of the equation above, we get something like (2*mu + lambda) n.grad(D) = (t + p*n)/rho + (2*mu + lambda) n.grad(D)  n.sigma Then the left hand side and the second term on the right hand side cancel out, and we are left with (t + p*n)/rho = n.sigma Which is physically sound... but I am still puzzled about how this is used to set n.grad(D) on the boundary, since the n.grad(D)terms apparenly cancel out. Can anyone give a hint as to how this works? Cheers, Johan 

July 31, 2011, 19:21 

#2 
Senior Member
Hisham El Safti
Join Date: Apr 2011
Location: Braunschweig, Germany
Posts: 248
Blog Entries: 10
Rep Power: 8 
Hi Johan
I'm trying to understand this BC as well. I'm not really sure but I think this calculation is called for each time (or iteration) the governing equation is to be solved, and therefore, the current normal gradient fvPatchField<vector>::snGrad() is cancelled with the current stress sigmaD and the relation is more like: gradient() = (traction + p * n) / (rho (2mu+lamda)) which is also sound if you can consider the strain=grad(u) I don't know if that is 100% correct. But thanks for the post (saved me some time while going through the code) Best regards, Hisham 

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