Yes on to the next phase of the problem, the first goal was to eliminate the rigid body modes. I think that has been done two different ways.
The next problem is that it does not appear that
Normal-Tangential Displacement = True
Force 2 = -100E6
is working for a curved surface, the results look to me like the load is in global cartesian direction 2
not tangential to the boundary as implied.
I think this may be due to the curved boundary
Residual stress on the inner pipe wall and boundary conditions
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Re: Residual stress on the inner pipe wall and boundary conditions
I took your posted case as is. Plotting the un-deformed with the deformed shows that the load is in the Cartesian Y direction and not in the tangential direction, which is what was expected from
Boundary Condition 3
Target Boundaries(1) = 3
Name = "BoundaryCondition 3"
Normal-Tangential Displacement = True
Force 2 = -100E6
End
However I recall this having issues with curved surfaces.
Boundary Condition 3
Target Boundaries(1) = 3
Name = "BoundaryCondition 3"
Normal-Tangential Displacement = True
Force 2 = -100E6
End
However I recall this having issues with curved surfaces.
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Re: Residual stress on the inner pipe wall and boundary conditions
I did it with writing my own tangential load function
Boundary Condition 3
Target Boundaries(1) = 3
Name = "BoundaryCondition 3"
Force 2 = variable coordinate 1;real MATC "sqrt(1.E8^2-(sin(acos(tx/0.08))*1.E8)^2)*(-1.0)"
Force 1 = variable coordinate 1;real MATC "sqrt(1.E8^2-(cos(acos(tx/0.08))*1.E8)^2)"
End and with using
Boundary Condition 3
Target Boundaries(1) = 3
Name = "BoundaryCondition 3"
Normal-Tangential Displacement = True
Force 2 = -1.E8
End The results are reasonably the same
Boundary Condition 3
Target Boundaries(1) = 3
Name = "BoundaryCondition 3"
Force 2 = variable coordinate 1;real MATC "sqrt(1.E8^2-(sin(acos(tx/0.08))*1.E8)^2)*(-1.0)"
Force 1 = variable coordinate 1;real MATC "sqrt(1.E8^2-(cos(acos(tx/0.08))*1.E8)^2)"
End and with using
Boundary Condition 3
Target Boundaries(1) = 3
Name = "BoundaryCondition 3"
Normal-Tangential Displacement = True
Force 2 = -1.E8
End The results are reasonably the same