Residual stress on the inner pipe wall and boundary conditions

[Frogga]

Good evening,

I want to calculate the stress state in a cross-section of a non-circular pipe in a 2D model incorporating residual stress. Unluckily, I have an issue setting the boundary conditions correctly and would like to ask you for your help.

The model is working fine if I apply inner pressure only with ElasticSolver since we have high pressure and large deformations. We know that we may have residual compressive stresses of the magnitude of 100MPa in circumferential direction which are experimentally determined on the inner surface and I fail applying these ones.

As a first step, I want to apply these residual stress only along the whole circumference and see what that means to the pipe's deformation. So I removed all other forces from the boundaries and set the Normal-Tangential Displacement flag and applied my stress. I use "Linear System Solver = Direct" because it is a small system and it worked well for normal forces.

My boundary conditions are defined like this. Target boundary 2 (boundary conditions 1) is the outer surface of the pipe, target boundary 1 (boundary condition 2) is the inner surface.

Code: Select all

Boundary Condition 1
  Target Boundaries(1) = 2
  Name = "pressure_outer"
  Normal Force = 0
End

Boundary Condition 2
  Target Boundaries(1) = 1 
  Name = "pressure_inner"
!  Normal Force = -643.9e5
!  Normal Force = 0e5
Normal-Tangential Displacement = True
Force 2 = -100e6
End

Sadly, after a quite long calculation with big values for NRM and RELC (compared to the case when I use normal forces only), I receive a stress state which is obviously wrong, seeming that I did not apply the Normal-Tangential command correctly.

So, how do I apply a residual stress on a curved surface which is not in normal direction with the ElasticSolve module?

Thank you!

[kevinarden]

Would be better if you could post the whole sif file. Generally Elasticsolver does not converge well if there are rigid body modes possible.

[Frogga]

Hi Kevin,

thanks for taking care for my problem! I also found that this solver is a bit harder to handle but in comparison to StressSolver it yields 10% less stresses (which is confirmed by other software), so it would be nice to get it running on this solver. General advice how to make the sif shorter and more stable and accelerate the calculation are very much appreciated.

The sif is as follows (large portions are taken over from default values from the GUI):

Code: Select all

Header
  CHECK KEYWORDS Warn
  Mesh DB "." "mesh"
  Include Path ""
  Results Directory "results"
End

Simulation
  Partition Mesh = Logical True
  Partitioning Method = String "zoltan"
  Max Output Level = 5
  Coordinate System = Cartesian 2D
  Coordinate Mapping(3) = 1 2 3
  Simulation Type = Steady state
  Steady State Max Iterations = 10
  Output Intervals = 1
  Solver Input File = case.sif
!  Post File = case.vtu
End

Constants
  Gravity(4) = 0 -1 0 9.82
  Stefan Boltzmann = 5.670374419e-08
  Permittivity of Vacuum = 8.85418781e-12
  Permeability of Vacuum = 1.25663706e-6
  Boltzmann Constant = 1.380649e-23
  Unit Charge = 1.6021766e-19
End

Body 1
  Target Bodies(1) = 3
  Name = "Wall"
  Equation = 1
  Material = 1
End

Solver 1
  Equation = Linear elasticity
  Procedure = "ElasticSolve" "ElasticSolver"
Calculate Principal = True
!Calculate Stresses = True
!Calculate Strains = True
  Exec Solver = Always
  Stabilize = True
  Optimize Bandwidth = True
  Steady State Convergence Tolerance = 1.0e-5
  Nonlinear System Convergence Tolerance = 1.0e-7
  Nonlinear System Max Iterations = 15
  Nonlinear System Newton After Iterations = 20
  Nonlinear System Newton After Tolerance = 1.0e-3
  Nonlinear System Relaxation Factor = 1
Nonlinear System Convergence Measure = Residual
  Linear System Solver = Direct
!For 2D systems only, probably, see ElmerSolver Manual 4.2
!  Linear System Solver = Iterative
  Linear System Iterative Method = BiCGStab
  Linear System Max Iterations = 1000
  Linear System Convergence Tolerance = 1.0e-10
  BiCGstabl polynomial degree = 2
  Linear System Preconditioning = ILU9
  Linear System ILUT Tolerance = 1.0e-3
  Linear System Abort Not Converged = False
  Linear System Residual Output = 10
  Linear System Precondition Recompute = 1
End

Solver 2
Exec Solver = After Simulation
Equation = SaveLine
Procedure = "SaveData" "SaveLine"
Filename = "line.dat"
End


Equation 1
  Name = "linelastic"
  Plane Stress = True
  Calculate Stresses = True
  Active Solvers(1) = 1
End

Material 1
  Name = "Q125"
  Youngs modulus = 207.e9 
  Mesh Poisson ratio = 0.3
  Poisson ratio = 0.3
End

Boundary Condition 1
  Target Boundaries(1) = 2
  Name = "pressure_outer"
  Normal Force = 0
End

Boundary Condition 2
  Target Boundaries(1) = 1 
  Name = "pressure_inner"
!  Normal Force = -643.9e5
!  Normal Force = 0e5
Normal-Tangential Displacement = True
Force 2 = -100e6
End

If there is need I am happy to post the mesh tonight, too!

Thanks!

[Frogga]

...and here comes the mesh. I tried it with
Procedure = "StressSolve" "StressSolver"
,too, and with the iterative approach. Both to no avail. Obviously, I need to improve something in the sif file.

Thanks for the help!

mesh.zip

(407.13 KiB) Downloaded 15 times

[kevinarden]

That is what I assumed the mesh would be. So I had already created one and did some testing. There is a problem with rigid body modes. The solid mechanics solvers are expecting there to be some resistance to the load. You have no boundary conditions that restrain the cross section from moving in space, so applying a load to the inside surface just means it will move as a rigid body in space to infinity. This condition generally cause the solvers to diverge unless you are doing a transient case.

I used a 1/4 symmetry mesh, because that way you can remove the rigid body modes, at least the solver comes to a solution.

pipe.zip

(22.87 KiB) Downloaded 17 times

[Frogga]

Good afternoon,

thank you, Kevin, great help indeed. Now I understand the problem. I am going to have a look at it once I am at home.

Unfortunately, my pipe cross section is not circular and exhibits a small cut-out, so the symmetry is broken and reduced symmetry won't do it. I try to add two points to the original structure and fix the cross-section in one space direction at each of those, hoping that that will also do. I will keep you posted.

Thanks, and have a nice day!

[kevinarden]

I was able to stop rotation of a whole cross section by using the target node BCs

targetnode.PNG

(310.06 KiB) Not downloaded yet

case.sif

(1.89 KiB) Downloaded 27 times

[Frogga]

Thank you, I will give it a shot. I tried a bit this evening (apart from joining the start of the carneval season) and I admit I failed miserably. I tried to take two points from the mesher and fix one in x, y, the other just in x. That did not work too well, high stresses around (possibly the distance was far enough apart), and allowing for a re-adjustment of the mesh with deformation led to huge deformation.

I try to apply your idea and come back with results. But, apart from it all, thanks so much for the program and the impressive support!

[Frogga]

Hm, that was not very successful.

In the example below I took your very first example of a (centro-symmetric) quarter pipe, applied a stress of -100MPa along the inner surface in circumferential direction. The stress distribution looks like this:

pic.png

(120.08 KiB) Not downloaded yet

Th eproblem is this: In addition the the high magnitude of stress, the stresses are all tensile stresses (even if we apply a compressive stress on the inner side). In addition, the principal stress distribution is inhomogenous, which makes no physical sense, I'd say.

If I set the Displace Mesh parameter to True, the calculations converge, but that does not change the overall picture.

Any idea what is going on here? What do I do wrong?

For clarity, I attach the mesh and case once more.

Thanks for any help!

[raback]

Hi,

To fix rigid body motion one could perhaps also add a small "Spring" to the BCs. So small that it will not affect the results but yet so big that the linear solvers converge to a unique solution.

-Peter