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Mathematical representation of fixedDisplacementZeroShear boundary condition 

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May 22, 2013, 00:11 
Mathematical representation of fixedDisplacementZeroShear boundary condition

#1 
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Sangeeta
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Hello all,
I am using fixedDisplacementZeroShear (in OpenFOAMext) for uniaxial compression loading problem. In the problem, there is a 3D solid cube geometry which is fixed at the bottom surface, where I use fixedDisplacementZeroShear boundary condition and traction force on the top surface. As I found fixedDisplacementZeroShear boundary condition is type of Directionmixed boundary condition which has fixed value and fixed gradient i.e. combination of Neumann and Dirichlet boundary conditions. I am using this boundary condition in the bottom surface of solid cube. I am wondering about mathematical representation of fixedDisplacementZeroShear boundary condition. Please find the mathematical representation of fixedDisplacementZeroShear boundary condition as attachment. if l = width= length of the cube . Am I representing it correctly? I'll be very grateful if one can give me advice. Thank you 

May 22, 2013, 05:06 

#2 
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Philip Cardiff
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Hi Sangeeta,
If your boundary condition is this: Code:
type fixedDisplacementZeroShear; value uniform (4 5 6); then the normal displacement of the patch Un is set to n & U, where U is (4 5 6) and '&' is a dot product. The shear components on the patch are then set such that the shear traction is zero (so essentially zero gradient). For example, if your patch is flat and has the unit normal (1 0 0), then the xdisplacement Ux (normal displacement) is set to (1 0 0) & ( 4 5 6) which is equal to 4. The shear gradients are then zero, dUy/dx == dUz/dx == 0. This means on the patch we know that: Ux = 4, sigma_xy = sigma_xz = 0. And we don't know anything about the other sigma components. Hope it helps, Philip 

May 22, 2013, 08:33 

#3 
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Sangeeta
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Hi Philip,
Thank you so much for the explanation. Actually I am using following code at bottom surface: type fixedDisplacementZeroShear; I am not mentioning the value uniform (0 0 0), but still getting the answer. Does it is okay to not to mention value uniform in case of fixed surface (i.e fixed displacement without shear)? Best regards, Sangeeta 

May 22, 2013, 09:13 

#4 
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Philip Cardiff
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Hi Sangeeta,
Hmnn actually I think fixedDisplacementZeroShear was meant to give an error if the value is not specified  I suppose it assumes a value of ( 0 0 0 ) if the value is not specified so it is probably OK. To be safe, I would specify "value uniform ( 0 0 0 );". Best regards, Philip 

May 23, 2013, 10:19 

#5 
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Sangeeta
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Hi Philip,
Thank you so very much for the information! Best regards, Sangeeta 

May 23, 2013, 10:35 

#6 
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Sangeeta
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Location: Kingston, Canada
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Hi Philip,
I have one more concern about boundary condition. As you mentioned in the http://www.cfdonline.com/Forums/ope...structure.html post: "fixedDisplacementZeroShear which applies a displacement in the normal direction and enforces zero shear stress in the tangential direction (equivalent to directionMixed). " It seems that fixedDisplacementZeroShear boundary condition works as symmetry boundary condition works? When I am using Symmetry and fixedDisplacementZeroShear boundary conditions to run microstructure, fixedDisplacementZeroShear is working well and giving converge solution but symmetry condition is giving diverge solution. I do not know why this is happening. I also checked my geometry and it is okay. As I know symmetry boundary condition assumes mirror symmetry of the geometry and shear stresses are zero in the plane of symmetry. What is the different between fixedDisplacementZeroShear and symmetry conditions? Best regards, Sangeeta 

May 23, 2013, 10:45 

#7 
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Philip Cardiff
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Sangeeta,
Yes fixedDisplacementZeroShear should work as a symmetry plane. One difference between them is that the shear gradients are always zero for symmetry plane and never changed, but they shear gradient are constantly being changed in fixedDisplacementZeroShear (in updateCoeffs functions) to enforce the traction to be zero > this must help convergence in your case. Actually on a related note, I just realised nonorthogonal correction has not be added to fixedDisplacementZeroShear, this will improve results and possibly convergence. I will added the correction and send it to you when I have some free time. Philip 

May 23, 2013, 11:15 

#8 
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Sangeeta
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Hi Philip,
Thank you for the quick reply and help! I appreciate it! Best regards, Sangeeta 

May 23, 2013, 23:49 

#9 
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Sangeeta
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Hi Philip,
I have run some more problems with symmetry and fixedDisplacementZeroShear boundary conditions to check the difference between these two. For simple dense geometries both conditions give same answers but not for microstructure (as I mentioned previously). As you mentioned that symmetry condition always consider shear gradient to be zero that mean it also enforces traction to be zero. As you mentioned: "shear gradient are constantly being changed in fixedDisplacementZeroShear (in updateCoeffs functions) to enforce the traction to be zero" Does it mean this boundary condition calculates new shear stress values (with some value not zero) in each step and then use the new calculated value to enforce the traction to be zero? As I understand if shear gradient have some value then it will change traction force with some value not zero (based on formula Traction = sigma . n). Best regards, Sangeeta 

May 24, 2013, 04:58 

#10 
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Philip Cardiff
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Hi Sangeeta,
probably the easiest way to check is to look at the tractions on the surface. I have attached a post processing utility called surfaceTractions which reads in the stress tensor sigma and calculate three tractions fields for visualisation in ParaView: totalTraction, shearTraction and normalTraction. Check the shear tractions for both of your cases and see which is zero (or closest to zero). Philip 

May 25, 2013, 17:50 

#11 
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Sangeeta
Join Date: Jul 2012
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Hi Philip,
Thank you for the valuable suggestions. I have run cases for both the boundary conditions and checked it for shear traction values. I am emailing these cases to you because files are larger in size and I cannot attach these here. Best regards, Sangeeta 

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