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Modeling a Sliding Constraint

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I am trying to model a sliding constraint, basically a pin in a hole.

To try to understand how best to apply the contraints I created a simple model of a tube and rod. The tube and rode are both 1 m long, the rod is inserted .1m into the tube the Tube ID and Rod OD are the same. I fix the free ends of the tube and rod. Then I run an Eigenfrequency study on the assembly.

If the geometry is a union (or assembly with continuity at the sliding interface), the first mode of the two parts is as expected a single wave bow of the system.
If the geometry is an assembly with a roller pair at the sliding interface, the model behaves as if the roller is fixed in space. i.e. the area where the rod is in the tube does not deflect. Rather the first couple modes are of the rod bowing then the tube bowing.

So how does one model a sliding contraint that can move with the surrounding geometry?

I have attached a version 4.3a model that I have been querying. Switch between the continuity 3 and Roller 1 constraint to see the differences as descriped above.


18 Replies Last Post Feb 6, 2013, 3:09 p.m. EST
Josh Thomas Certified Consultant

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Posted: 1 decade ago Jan 30, 2013, 3:40 p.m. EST
Roy -

The roller condition is constraining displacement in an absolute sense as zero. What I believe you want is to constrain displacement of the parts relative to each other.

You are on the right track with selecting Form an assembly. But, I believe with what you are looking to do you want to set the pair type as Contact pair instead of Identity pair. Contact means that the parts can slide past each other but not interpenetrate domains.

Once you define the borders where the pin touches the hole as contact pairs, you can right-click on the Solid Mechanics node and select Pairs>Contact and set up the settings window settings properly. You can also specify friction between your parts if you'd like by right-clicking on the Contact node.

Best regards,
Josh Thomas
AltaSim Technologies
Roy - The roller condition is constraining displacement in an absolute sense as zero. What I believe you want is to constrain displacement of the parts relative to each other. You are on the right track with selecting Form an assembly. But, I believe with what you are looking to do you want to set the pair type as Contact pair instead of Identity pair. Contact means that the parts can slide past each other but not interpenetrate domains. Once you define the borders where the pin touches the hole as contact pairs, you can right-click on the Solid Mechanics node and select Pairs>Contact and set up the settings window settings properly. You can also specify friction between your parts if you'd like by right-clicking on the Contact node. Best regards, Josh Thomas AltaSim Technologies

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Posted: 1 decade ago Jan 30, 2013, 4:04 p.m. EST
I did try with a contact pair, but the solution showed no connection betweenthe rod and the tube. i.e. the domains passed through each other.

I attached the model for anyone to check if the contact pair is setup correctly.

I did try with a contact pair, but the solution showed no connection betweenthe rod and the tube. i.e. the domains passed through each other. I attached the model for anyone to check if the contact pair is setup correctly.


Josh Thomas Certified Consultant

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Posted: 1 decade ago Jan 30, 2013, 4:57 p.m. EST
Roy -

Your contact set-up looks good and works for a stationary problem with a boundary load.

Is an Eigenfrequency analysis your final goal? If not, try using the contact set-up but then applying a load condition and solving the stationary form of the equations. In this case, COMSOL will engage the contact.

My understanding of the Eigenfrequency analysis is that by definition all sources go to zero. Since, contact is formulated with a penalized contact pressure defined on the destination boundary then contact is not engaged during an eigenfrequency analysis.

Maybe someone with a better handle on the formulation could comment, but I don't believe it's possible to solve a natural frequency problem with contact in any FEA code.

Regards,
Josh Thomas
AltaSim Technologies
Roy - Your contact set-up looks good and works for a stationary problem with a boundary load. Is an Eigenfrequency analysis your final goal? If not, try using the contact set-up but then applying a load condition and solving the stationary form of the equations. In this case, COMSOL will engage the contact. My understanding of the Eigenfrequency analysis is that by definition all sources go to zero. Since, contact is formulated with a penalized contact pressure defined on the destination boundary then contact is not engaged during an eigenfrequency analysis. Maybe someone with a better handle on the formulation could comment, but I don't believe it's possible to solve a natural frequency problem with contact in any FEA code. Regards, Josh Thomas AltaSim Technologies

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Posted: 1 decade ago Jan 31, 2013, 1:02 p.m. EST

Ok, good to know about natural frequency problem and contact. This problem is just a small subset of a larger case study. I am trying to determine the natrual frequency of a structure which has 1 sliding constraint. I'll just have to continue without the sliding constraint and assume some inaccuracy i nthe Eigenfrequency Analysis. But when I do the static analysis I can use the contact pair and deal with the convergense issues. I've used contact pairs before and usually try to avoid them, as solving the problem becomes more about getting the contact problem to solve than solve the actual problem.
Ok, good to know about natural frequency problem and contact. This problem is just a small subset of a larger case study. I am trying to determine the natrual frequency of a structure which has 1 sliding constraint. I'll just have to continue without the sliding constraint and assume some inaccuracy i nthe Eigenfrequency Analysis. But when I do the static analysis I can use the contact pair and deal with the convergense issues. I've used contact pairs before and usually try to avoid them, as solving the problem becomes more about getting the contact problem to solve than solve the actual problem.

Nagi Elabbasi Facebook Reality Labs

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Posted: 1 decade ago Feb 1, 2013, 12:23 a.m. EST
Hi Roy, Josh,

Interesting problem!

There is no theoretical reason for contact constraints to not be active in an eigenfrequency analysis. You have to setup a stationary analysis step before the eigenfrequency one. Only the nodes that are in contact in the stationary analysis should remain so in the frequency analysis. You can think of contact in this case as providing a displacement constraint only between the nodes in contact (Augmented Lagrange method), or adding stiff springs between the same nodes (penalty method).

In my opinion the roller option between identity pairs should work as well. It should constrain the motion of the parts relative to each other as Josh mentioned. Instead it seems like it is setting all pair displacements to zero.The mesh through the cross-section is very coarse but I don’t think this should affect results.

Nagi Elabbasi
Veryst Engineering
Hi Roy, Josh, Interesting problem! There is no theoretical reason for contact constraints to not be active in an eigenfrequency analysis. You have to setup a stationary analysis step before the eigenfrequency one. Only the nodes that are in contact in the stationary analysis should remain so in the frequency analysis. You can think of contact in this case as providing a displacement constraint only between the nodes in contact (Augmented Lagrange method), or adding stiff springs between the same nodes (penalty method). In my opinion the roller option between identity pairs should work as well. It should constrain the motion of the parts relative to each other as Josh mentioned. Instead it seems like it is setting all pair displacements to zero.The mesh through the cross-section is very coarse but I don’t think this should affect results. Nagi Elabbasi Veryst Engineering

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago Feb 1, 2013, 4:32 a.m. EST
Hi

probably the "most correct" is to work in union mode, and replace the roller condition by a thin elastic Layer with a high stiffness in the transverse direction only, but one might get scaling and numerical issues due to the high spring stiffness required

I was a bit astonished why the "roller" condition should block in space, indeed it looks like that on the graphs, but its probably linked to the absolute reference frame so it remains "stuck in space

my plots are also slightly bizarre, it looks like the parts dissociate, I hope its only a display issue, else I'll have some doubts about the solution

Then you Inertias are so different that its not that easy to see the effect of the sliding

More I try less I'm convinced that the Thin Elastic Layer is working as a sliding element, probably due to limitations in the "linear" theory

--
Good luck
Ivar
Hi probably the "most correct" is to work in union mode, and replace the roller condition by a thin elastic Layer with a high stiffness in the transverse direction only, but one might get scaling and numerical issues due to the high spring stiffness required I was a bit astonished why the "roller" condition should block in space, indeed it looks like that on the graphs, but its probably linked to the absolute reference frame so it remains "stuck in space my plots are also slightly bizarre, it looks like the parts dissociate, I hope its only a display issue, else I'll have some doubts about the solution Then you Inertias are so different that its not that easy to see the effect of the sliding More I try less I'm convinced that the Thin Elastic Layer is working as a sliding element, probably due to limitations in the "linear" theory -- Good luck Ivar

Henrik Sönnerlind COMSOL Employee

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Posted: 1 decade ago Feb 1, 2013, 5:47 a.m. EST
Hi,

Running an eigenfrequency analysis where the boundaries are constrained based on the contact state is possible, but not at all easy. Since the contact formulation is not implemented as actual constraints, the contact condition will not be linearized as you would want in the subsequent eigenfrequency analysis. It is possible to pick up the gap variable from the initial contact analysis, and base extra constraints in the eigenfrequency analysis on it.

So unless a contact analysis it really needed to determine the size of the sliding boundaries, i would suggest a type of roller condition.

The roller condition (as well as other constraints) on pairs is however 'absolute'. i.e. a constraint with respect to 'ground'. So it cannot be used here. There are however different methods through which you can implement it:

a) If the slip condition is on a flat surface, aligned with the global axes, the quickest way is to create a an Identity Pair with Continuity, and remove the unwanted in-plane constraints in Equation View mode.

b) More general: Create a general extrusion mapping (say 'genext1') operator on one of the boundaries. Then add a Prescribed Displacement in a local system (e.g. a Boundary system) on the other boundary. Prescribe the displacement in the normal direction with something like -genext1(nx*u+ny*v+nz*w).

Beware however that the rotating motion between two cylindrical surfaces may easily become more constrained than you intend, just because of small numerical errors in the computation of the normals.


Regards,
Henrik
Hi, Running an eigenfrequency analysis where the boundaries are constrained based on the contact state is possible, but not at all easy. Since the contact formulation is not implemented as actual constraints, the contact condition will not be linearized as you would want in the subsequent eigenfrequency analysis. It is possible to pick up the gap variable from the initial contact analysis, and base extra constraints in the eigenfrequency analysis on it. So unless a contact analysis it really needed to determine the size of the sliding boundaries, i would suggest a type of roller condition. The roller condition (as well as other constraints) on pairs is however 'absolute'. i.e. a constraint with respect to 'ground'. So it cannot be used here. There are however different methods through which you can implement it: a) If the slip condition is on a flat surface, aligned with the global axes, the quickest way is to create a an Identity Pair with Continuity, and remove the unwanted in-plane constraints in Equation View mode. b) More general: Create a general extrusion mapping (say 'genext1') operator on one of the boundaries. Then add a Prescribed Displacement in a local system (e.g. a Boundary system) on the other boundary. Prescribe the displacement in the normal direction with something like -genext1(nx*u+ny*v+nz*w). Beware however that the rotating motion between two cylindrical surfaces may easily become more constrained than you intend, just because of small numerical errors in the computation of the normals. Regards, Henrik

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago Feb 1, 2013, 6:31 a.m. EST
Hi Henrik

When I use the "thin elastic layer" (in union mode) for the sliding only on the boundary normal direction, and increase the spring stiffness, I tend towards the fully "continuity" values (perhaps due to numerical issues too spring stiffness 1E12), if I add a slight damping along the tangents I get complex eigenvalues, more or less indicating the some "sliding" is seen by the eigenfrequency solver.

Then comes the question, turn on or not the Solver "Include geometric non-linearity" I do not see any significant effects, why ?

This normally separate the Frames, but then if I define my Thin elastic layer on the "boundary coordinate" or on a "cylindrical coordinate", attached either to the "material" or to the "spatial" frame, should I see any difference ?

A nice example to get a better view of the coupled effects and the frames + geometrical non linearities

--
Good luck
Ivar
Hi Henrik When I use the "thin elastic layer" (in union mode) for the sliding only on the boundary normal direction, and increase the spring stiffness, I tend towards the fully "continuity" values (perhaps due to numerical issues too spring stiffness 1E12), if I add a slight damping along the tangents I get complex eigenvalues, more or less indicating the some "sliding" is seen by the eigenfrequency solver. Then comes the question, turn on or not the Solver "Include geometric non-linearity" I do not see any significant effects, why ? This normally separate the Frames, but then if I define my Thin elastic layer on the "boundary coordinate" or on a "cylindrical coordinate", attached either to the "material" or to the "spatial" frame, should I see any difference ? A nice example to get a better view of the coupled effects and the frames + geometrical non linearities -- Good luck Ivar

Nagi Elabbasi Facebook Reality Labs

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Posted: 1 decade ago Feb 1, 2013, 10:28 a.m. EST
Henrik, so the Identify Pair Continuity condition is relative but the Identity Pair Roller condition is absolute?
Thanks,
Nagi
Henrik, so the Identify Pair Continuity condition is relative but the Identity Pair Roller condition is absolute? Thanks, Nagi

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago Feb 4, 2013, 2:40 a.m. EST
Hello

I tried a 2D simpler case (solves quicker etc). Here the Thin elastic Layer works OK, and is not tending to the same frequency values as the fixed "continuity" case. Furthermore you see the difference in the shape (eigenvalues) of the first mode, between sliding and fixed central part. I modified slightly the dimensions to make them easier to see on the screen.

Furthermore, I see no frequency difference with our without "Include geometrical nonlinearities", this I need to understand better, so I'll have to dig into the doc again ;)

The fact I'm not getting results as expected in 3D could be a meshing density issue too

Nice example to clean up our understanding of linear / non-linear solid analysis

--
Good luck
Ivar
Hello I tried a 2D simpler case (solves quicker etc). Here the Thin elastic Layer works OK, and is not tending to the same frequency values as the fixed "continuity" case. Furthermore you see the difference in the shape (eigenvalues) of the first mode, between sliding and fixed central part. I modified slightly the dimensions to make them easier to see on the screen. Furthermore, I see no frequency difference with our without "Include geometrical nonlinearities", this I need to understand better, so I'll have to dig into the doc again ;) The fact I'm not getting results as expected in 3D could be a meshing density issue too Nice example to clean up our understanding of linear / non-linear solid analysis -- Good luck Ivar


Henrik Sönnerlind COMSOL Employee

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Posted: 1 decade ago Feb 5, 2013, 2:10 a.m. EST

Henrik, so the Identify Pair Continuity condition is relative but the Identity Pair Roller condition is absolute?
Thanks,
Nagi


Hi Nagi,

Yes, the continuity condition makes the displacements equal (determined by the solution) whereas all the "copies of boundary conditions" just applies that boundary condition to both sides of the pair.

Regards,
Henrik
[QUOTE] Henrik, so the Identify Pair Continuity condition is relative but the Identity Pair Roller condition is absolute? Thanks, Nagi [/QUOTE] Hi Nagi, Yes, the continuity condition makes the displacements equal (determined by the solution) whereas all the "copies of boundary conditions" just applies that boundary condition to both sides of the pair. Regards, Henrik

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago Feb 5, 2013, 5:46 a.m. EST
Hi again

I have added a 3D example with "thin elastic layer" contrary to the 2D example, I'm clearly converging towards the fixed case, perhaps it's my mesh ?

Why is the 3D case so different from the 2D case above ?

Is this really only numerics, and how to identify / prove that ?

And no differences however I define the "Thin Elastic Layer", attached to boundary or a cylindrical frame, to the spatio or the Material frame, in geometrical linear or non-linear developments.
--
Good luck
Ivar

File attached,
I have some issues with my Opera explorer, gets an Java error when trying to upload files, but IE works fine ;)
Hi again I have added a 3D example with "thin elastic layer" contrary to the 2D example, I'm clearly converging towards the fixed case, perhaps it's my mesh ? Why is the 3D case so different from the 2D case above ? Is this really only numerics, and how to identify / prove that ? And no differences however I define the "Thin Elastic Layer", attached to boundary or a cylindrical frame, to the spatio or the Material frame, in geometrical linear or non-linear developments. -- Good luck Ivar File attached, I have some issues with my Opera explorer, gets an Java error when trying to upload files, but IE works fine ;)


Nagi Elabbasi Facebook Reality Labs

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Posted: 1 decade ago Feb 6, 2013, 12:17 a.m. EST
Hi Ivar,

I believe your thin elastic layer modeling approach is working!

It just happens that the lowest 6 frequencies that you are requesting are all bending modes and for those a radial constraint is pretty much the same as a full constraint. If you request more frequencies you will see the torsional and axial modes converge to values different from the case without the thin elastic layer.

Nagi Elabbasi
Veryst Engineering
Hi Ivar, I believe your thin elastic layer modeling approach is working! It just happens that the lowest 6 frequencies that you are requesting are all bending modes and for those a radial constraint is pretty much the same as a full constraint. If you request more frequencies you will see the torsional and axial modes converge to values different from the case without the thin elastic layer. Nagi Elabbasi Veryst Engineering

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago Feb 6, 2013, 1:08 a.m. EST
Hi Nagi

Well I was rather convinced that it should works, apart for the huge numerical values for the spring that can easily lead to numerical issues.

But I'm still astonished of the differences between 2D and in 3D. Even for bending modes the slippage should (from my guts feelings) be more different, but you might be right.
And as in 3D all modes are de-doubled indeed I should rather go for 18 rather than 6 modes.

And I'm also astonished there are not larger differences from the "non-linear Geometry" settings, this slipping makes the model non linear and allows for larger bending. But I do not have any bench mark to compare it with, and as COMSOL is mostly giving "correct" results ...

--
Good luck
Ivar
Hi Nagi Well I was rather convinced that it should works, apart for the huge numerical values for the spring that can easily lead to numerical issues. But I'm still astonished of the differences between 2D and in 3D. Even for bending modes the slippage should (from my guts feelings) be more different, but you might be right. And as in 3D all modes are de-doubled indeed I should rather go for 18 rather than 6 modes. And I'm also astonished there are not larger differences from the "non-linear Geometry" settings, this slipping makes the model non linear and allows for larger bending. But I do not have any bench mark to compare it with, and as COMSOL is mostly giving "correct" results ... -- Good luck Ivar

Nagi Elabbasi Facebook Reality Labs

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Posted: 1 decade ago Feb 6, 2013, 9:48 a.m. EST
Hi Ivar,

True there should be a small difference in the bending modes, but since the radius is much smaller than the length that difference should be very small. In the 2D case both sides of the “pipe” can move independently that’s why you get a bigger difference in natural frequencies.

Now when you active non-linear geometry you should see a difference in response had this been a time dependent analysis. Not so in a frequency analysis though since it a linear perturbation analysis and the base state in this case is the stress-free initial configuration. Do you agree?

Nagi Elabbasi
Veryst Engineering
Hi Ivar, True there should be a small difference in the bending modes, but since the radius is much smaller than the length that difference should be very small. In the 2D case both sides of the “pipe” can move independently that’s why you get a bigger difference in natural frequencies. Now when you active non-linear geometry you should see a difference in response had this been a time dependent analysis. Not so in a frequency analysis though since it a linear perturbation analysis and the base state in this case is the stress-free initial configuration. Do you agree? Nagi Elabbasi Veryst Engineering

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago Feb 6, 2013, 10:25 a.m. EST
Hi

not necessarily, as you can linearise the stationary stressed solution as starting point for the eigenfrequency (what is also done in the buckling analysis).
And if you check the eigenfrequency analysis formulas/equations, these change when you tick the geometrical non-linearity on

--
Good luck
Ivar
Hi not necessarily, as you can linearise the stationary stressed solution as starting point for the eigenfrequency (what is also done in the buckling analysis). And if you check the eigenfrequency analysis formulas/equations, these change when you tick the geometrical non-linearity on -- Good luck Ivar

Nagi Elabbasi Facebook Reality Labs

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Posted: 1 decade ago Feb 6, 2013, 10:39 a.m. EST
True Ivar, but in this model there is no stationary stressed solution, that’s why there should be no difference in results when you activate non-linear geometry.
True Ivar, but in this model there is no stationary stressed solution, that’s why there should be no difference in results when you activate non-linear geometry.

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago Feb 6, 2013, 3:09 p.m. EST
Hi Nagi

You are probably right there.

Anyhow in 2D it works nicely (perhaps also because my boundaries for the thin elastic layer are nicely ligned up along the Cartesian coordiante system, this should reduce numerical locking from my experience)

--
Good luck
Ivar
Hi Nagi You are probably right there. Anyhow in 2D it works nicely (perhaps also because my boundaries for the thin elastic layer are nicely ligned up along the Cartesian coordiante system, this should reduce numerical locking from my experience) -- Good luck Ivar

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