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
Join Date: Jul 2012
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Hello all,

I am using fixedDisplacementZeroShear (in OpenFOAM-ext) 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
Attached Files
 fixedDisplacementZeroShearBC.docx (25.5 KB, 19 views)

 May 22, 2013, 05:06 #2 Senior Member     Philip Cardiff Join Date: Mar 2009 Location: Dublin,Ireland Posts: 570 Rep Power: 19 Hi Sangeeta, If your boundary condition is this: Code: ```type fixedDisplacementZeroShear; value uniform (4 5 6);``` And the patch has the unit surface normal n, 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 x-displacement 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 Member   Sangeeta Join Date: Jul 2012 Location: Kingston, Canada Posts: 70 Rep Power: 5 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 Senior Member     Philip Cardiff Join Date: Mar 2009 Location: Dublin,Ireland Posts: 570 Rep Power: 19 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 Member   Sangeeta Join Date: Jul 2012 Location: Kingston, Canada Posts: 70 Rep Power: 5 Hi Philip, Thank you so very much for the information! Best regards, Sangeeta

 May 23, 2013, 10:35 #6 Member   Sangeeta Join Date: Jul 2012 Location: Kingston, Canada Posts: 70 Rep Power: 5 Hi Philip, I have one more concern about boundary condition. As you mentioned in the OpenFOAM solution is diverging for stress analysis in two-pahse microstructure. 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 Senior Member     Philip Cardiff Join Date: Mar 2009 Location: Dublin,Ireland Posts: 570 Rep Power: 19 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 non-orthogonal 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 Member   Sangeeta Join Date: Jul 2012 Location: Kingston, Canada Posts: 70 Rep Power: 5 Hi Philip, Thank you for the quick reply and help! I appreciate it! Best regards, Sangeeta

 May 23, 2013, 23:49 #9 Member   Sangeeta Join Date: Jul 2012 Location: Kingston, Canada Posts: 70 Rep Power: 5 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
Join Date: Mar 2009
Location: Dublin,Ireland
Posts: 570
Rep Power: 19
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
Attached Files
 surfaceTractions.zip (2.8 KB, 12 views)

 May 25, 2013, 17:50 #11 Member   Sangeeta Join Date: Jul 2012 Location: Kingston, Canada Posts: 70 Rep Power: 5 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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