I’m looking to determine the shear and normal forces on a cylindrical contact surface. My goal is to calculate the stress by dividing these forces by the contact area. Can anyone help me with this?
Thank you in advance for your assistance!
You can request the following output variables, among which there are contact forces but also contact stresses:
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Shouldn’t the contact force values I read on the cylindrical contact surface be in cylindrical coordinates? By default, we see the calculations according to Cartesian coordinates. Since the element shear force outputs will change with the rotation of the element, I think I should see these outputs in cylindrical coordinates? If I select directly from the list you sent, I will see the Cartesian coordinate results. What direction is taking into account in CFS, CF, CFN outputs.
They use surface (contact) normal directions. So CFN is always normal to a contact surface and thus accounts for the curvature of the model’s geometry. CFS follows that and is perpendicular to CFN.
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It seems like there might be a missing point somewhere. When I run the same model in OptiStruct and generate the ‘contf’ output, I obtain the result shown in the image.
And what are the results in CalculiX ? What’s the difference ?
I got Normal Surface Force =0 and Shear Surface Force=274. I can upload the model if you want
Please do. I can try checking it in Abaqus.
I got this from Abaqus:
CONTACT OUTPUT FOR SECONDARY SURFACE SURFACE-2 AND MAIN SURFACE SURFACE-1
FOOT- CFNM CFN1 CFN2 CFN3 CFSM CFS1 CFS2 CFS3
NOTE
5408. 0.2589 1.743 -5408. 5134. -0.2589 -1.743 5134.
while CalculiX shows this:
statistics for slave set SURFACE-2, master set SURFACE-1 and time 0.1000000E+01
total surface force (fx,fy,fz) and moment about the origin (mx,my,mz)
-2.542189E-09 -3.511779E-09 2.740000E+02 5.473039E+04 -2.736623E+04 -9.180131E+00
center of gravity and mean normal
-3.263160E-08 7.777777E-01 -1.453235E-04 -1.000106E-18 7.040626E-01 3.186359E-16
moment about the center of gravity(mx,my,mz)
5.451728E+04 -2.736623E+04 -9.180131E+00
area, normal force (+ = tension) and shear force (size)
8.932875E+01 -2.472425E-09 2.740000E+02
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There are significant differences between the results obtained in Calculix compared to OptiStruct and Abaqus. It appears that CFN and CF outputs do not match. For instance, while Calculix provides a normal force of 274, Abaqus reports 5408.
Additionally, there are noticeable discrepancies in the shear (tangential force) values, indicating that there might be errors in the post-process of the forces.
Optistruct and Abaqus gives same results by the way.
Hi, Could I take a look at the file?. It’s erased.
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Hi,
I have imported this model to MECWAY but there is something different.
Could you please confirm?
1-The model makes 19 mm diameter, and it is located at 0,0 . In your Prepomax picture,
REF node is shown at the origin but in the inp it has coordinates (100mm,200mm,300mm). Is that correct?
2-Is your upper body completely free?
It’s not the REF node, it’s just a symbol of fixed BC at the bottom. The REF node is far away from the model and the screenshots only show the lines going to it.
There’s a tied contact between the two parts.
Not only the contact force results are different. Displacement and stress outputs differ too. Even when the same input file (only with the modifications required by Abaqus syntax) is submitted in Abaqus. So it might be some issue with CalculiX. It would be good to try with different meshes.
Are the Ref and Rot located at the same coordinates in Abaqus?
Abaqus doesn’t use ROT nodes, only REF nodes with 6 DOFs. Apart from that, a rigid body should work in the same way. I suspect contact to be troublesome here.
The fixed support surface is only the bottom surface of the triangle. There won’t be any support on the surface of the cylinder. A 100mm, 200mm, and 300mm reference point is where the force is being applied.
Really???. The rot node is unconstrain .
Try to move the Ref node to coordinates 0,1.5,0. There must be a big difference.
I do not think this is a mesh related think since we cannot get same reaction force values