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Belleville Springs Mechanica 1
- Thread starter tazengr
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seymours2571
Mechanical
Tazengr,
Can you post the geometry? I just got back from the Technical Committee meetings at PTC and we discussed this exact simulation. Unfortunately, because Mechanica can only model frictionless contact (or infinite friction in newer versions) one will never get exact results. However, the frictionless approach will be more accurate.
You also may need to enable large displacements to account for large strains during the deformation of the spring. Unfortunately, I have never found a good definition for what constitutes large strains to the point non-linear deformations are required, but the practice I have seen others take is to run both simulations and compare results.
Post the geometry and I will give it a try on my end.
Steve
Can you post the geometry? I just got back from the Technical Committee meetings at PTC and we discussed this exact simulation. Unfortunately, because Mechanica can only model frictionless contact (or infinite friction in newer versions) one will never get exact results. However, the frictionless approach will be more accurate.
You also may need to enable large displacements to account for large strains during the deformation of the spring. Unfortunately, I have never found a good definition for what constitutes large strains to the point non-linear deformations are required, but the practice I have seen others take is to run both simulations and compare results.
Post the geometry and I will give it a try on my end.
Steve
Tazengr,
1) Your question regarding the arc length: This appears to be a chart of all the deflections around the circumference of the washer. Ideally, this graph would be a straight horizontal line (same deflection at all points), but it's a numerical estimate.
2) The first thing to check are your units and the material stiffness. We've had trouble with imported material props in WF2 at work. It's worth checking, at least.
3) Seymours2571 is correct about this being a contact/friction problem. However, a simple substitute you can try is to delete the displacement constraint and replace it with a 60 lbf load. Sometimes, when you constrain a surface, you can inadvertently add rotational stiffness on a surface that needs to be free to rotate.
You might get some fake stresses in the area, but if all you want is displacement, this is a good first step (using contact will give more realistic stresses). You probably want to add a very weak (e.g., 0.1 lbf/in in x,y,& z) spring to ground to prevent numberical "flyaway".
4) This problem lends itself nicely to using radial symmetry. Effectively, you are doing a 2D problem on the cross-section. It takes care of some of the global constraint issues.
5) If possible, can you post the dimensions of your washer (ID, OD, thickness, & height)? I'd like to try modelling this myself, and it sounds like seymours would also.
Good luck,
1) Your question regarding the arc length: This appears to be a chart of all the deflections around the circumference of the washer. Ideally, this graph would be a straight horizontal line (same deflection at all points), but it's a numerical estimate.
2) The first thing to check are your units and the material stiffness. We've had trouble with imported material props in WF2 at work. It's worth checking, at least.
3) Seymours2571 is correct about this being a contact/friction problem. However, a simple substitute you can try is to delete the displacement constraint and replace it with a 60 lbf load. Sometimes, when you constrain a surface, you can inadvertently add rotational stiffness on a surface that needs to be free to rotate.
You might get some fake stresses in the area, but if all you want is displacement, this is a good first step (using contact will give more realistic stresses). You probably want to add a very weak (e.g., 0.1 lbf/in in x,y,& z) spring to ground to prevent numberical "flyaway".
4) This problem lends itself nicely to using radial symmetry. Effectively, you are doing a 2D problem on the cross-section. It takes care of some of the global constraint issues.
5) If possible, can you post the dimensions of your washer (ID, OD, thickness, & height)? I'd like to try modelling this myself, and it sounds like seymours would also.
Good luck,
- Thread starter
- #6
Sure.
I show a cross section of the washer in the attached file. Note that 0.695 and 0.598 are the OD and ID dimensions respectively. The thickness is .008 and the overall length is .042. Dimensions are in inches. See attachment.
Regards,
TazEngr
I show a cross section of the washer in the attached file. Note that 0.695 and 0.598 are the OD and ID dimensions respectively. The thickness is .008 and the overall length is .042. Dimensions are in inches. See attachment.
Regards,
TazEngr
TazEngr,
Ran a few cases in Pro/Mech WF4 and had relatively good success - see the file.
I would have liked to have closer agreement with your hand calcs, but can't spend forever on it. So, it's a start. If you need some tips on setup, let me know.
Best of luck
PetieSmo
Ran a few cases in Pro/Mech WF4 and had relatively good success - see the file.
I would have liked to have closer agreement with your hand calcs, but can't spend forever on it. So, it's a start. If you need some tips on setup, let me know.
Best of luck
PetieSmo
- Thread starter
- #8
Your results look like they are pretty accurate. I am trying to duplicate them, however I get an error saying:
This point-to-ground spring is
not associated with any element.
Do you know what the cause of this is?
Was the origin of the cylindrical or the cartesian coordinate system used when defining the constraints and the loads?
I used the cylindrical.
Were the surfaces or edges used as constraints? I have tried both and did not have much luck.
Also for the spring constraint did you use the simple, point to point, or ground.
PS I am using WF 3.0 I Don't know if that makes a difference or not,
TazEngr
This point-to-ground spring is
not associated with any element.
Do you know what the cause of this is?
Was the origin of the cylindrical or the cartesian coordinate system used when defining the constraints and the loads?
I used the cylindrical.
Were the surfaces or edges used as constraints? I have tried both and did not have much luck.
Also for the spring constraint did you use the simple, point to point, or ground.
PS I am using WF 3.0 I Don't know if that makes a difference or not,
TazEngr
Based on your comments, it appears you are setting up the 3D simulation.
With regards to the coordinate system, I used the cylindrical coordinate system for the 3D simulation.
With regards to constraints, I ran one model with the inner surface constrained in Z (cylindrical) and a second model with the edge constrained in Z. Both yielded similar displacement results for 60lbf (Surface: 0.0055" max, Edge: 0.0061" max)
The only reason I use what I call "flyaway" springs is a poor man's way of faking Mechanica into limiting rigid body motion of the part due to unbalanced loads (without introducing hard point constraints). If someone has a better process for this, I'd be interested in finding out about it. Here is how I set up the flyaway springs:
1) Defined 8 evenly spaced points along the top edge (when I tried using only the two vertexes of the arc ends, I got weird results)
2) Defined a spring property with 0.1 lbf/in in each direction (Kxx, Kyy, Kzz)
3) Created a spring to ground referenced at 8 points, relative to the cylindrical coordinate system.
I am curious - did you obtain your force vs displacement graph from the manufacturer, from test data, or from hand calcs? If hand calcs, I'm interested to know what equations you used (or what book/reference you got it from). In your graph, it appears that after 0.015, the washer yields or something - just wanted to understand that better from reading your reference source. (As a side note, I tried applying a fixed displacement to the washer, in hopes of generating a graph identical to yours - the linear material in mechanica of course didn't yield)
With regards to the coordinate system, I used the cylindrical coordinate system for the 3D simulation.
With regards to constraints, I ran one model with the inner surface constrained in Z (cylindrical) and a second model with the edge constrained in Z. Both yielded similar displacement results for 60lbf (Surface: 0.0055" max, Edge: 0.0061" max)
The only reason I use what I call "flyaway" springs is a poor man's way of faking Mechanica into limiting rigid body motion of the part due to unbalanced loads (without introducing hard point constraints). If someone has a better process for this, I'd be interested in finding out about it. Here is how I set up the flyaway springs:
1) Defined 8 evenly spaced points along the top edge (when I tried using only the two vertexes of the arc ends, I got weird results)
2) Defined a spring property with 0.1 lbf/in in each direction (Kxx, Kyy, Kzz)
3) Created a spring to ground referenced at 8 points, relative to the cylindrical coordinate system.
I am curious - did you obtain your force vs displacement graph from the manufacturer, from test data, or from hand calcs? If hand calcs, I'm interested to know what equations you used (or what book/reference you got it from). In your graph, it appears that after 0.015, the washer yields or something - just wanted to understand that better from reading your reference source. (As a side note, I tried applying a fixed displacement to the washer, in hopes of generating a graph identical to yours - the linear material in mechanica of course didn't yield)
- Thread starter
- #10
Thanks for your help. I got the equations from the following website.
I used a Modulus of Elasticity of 29600000 psi and a Poisson's ratio of 0.28.
I also compared my hand calculation results to Universal Technical Systems' Advanced Spring Design Software (trial version) and the results were consistent up to a deflection of .015.
Attached is the results from the Advanced Spring Design Software. Note that I drew in the results I got from the equation I found online.
TazEngr
I used a Modulus of Elasticity of 29600000 psi and a Poisson's ratio of 0.28.
I also compared my hand calculation results to Universal Technical Systems' Advanced Spring Design Software (trial version) and the results were consistent up to a deflection of .015.
Attached is the results from the Advanced Spring Design Software. Note that I drew in the results I got from the equation I found online.
TazEngr
- Thread starter
- #11
Thanks PetieSmo for the explanation.
I have a question however on how to set up the springs in the correct direction. The springs in my model are 90° off from the direction of the force.
In the attachment I show the spring definition dialog box for one of the eight springs that I set up. Here I set up a spring at a point and the type is ground. Note the direction of the spring.
I also tried setting up using different coordinate systems but no luck.
TazEngr
I have a question however on how to set up the springs in the correct direction. The springs in my model are 90° off from the direction of the force.
In the attachment I show the spring definition dialog box for one of the eight springs that I set up. Here I set up a spring at a point and the type is ground. Note the direction of the spring.
I also tried setting up using different coordinate systems but no luck.
TazEngr
Taz,
I've found that the graphical image of the spring does not rotate its direction relative to the spring properties. So, if your spring property is set up correctly, I think you'll be ok.
Here's 2 things to check:
1) On your displacement constraint, I think you want to release all the rotational DOF, so that dZ is the only restrained DOF.
2) Are you using the IPS (inch-lbf-sec) unit system? The Pro/E default is inch-lbm-sec, and this can mess up your entry of the loads. One way to check this is to open/edit the definition of your 60 pound load - at the bottom of the dialog box, it will show the units of your entry. Hopefully it will read 'lbf'; however if it says 'lbm*s^2/in', you can do one of two things:
a) Multiply your input by 386 in/s^2 (i.e., gravity) (60*286 = 23160 lbf) OR
b) Change the system of units in the Pro/E Setup menu
I've found that the graphical image of the spring does not rotate its direction relative to the spring properties. So, if your spring property is set up correctly, I think you'll be ok.
Here's 2 things to check:
1) On your displacement constraint, I think you want to release all the rotational DOF, so that dZ is the only restrained DOF.
2) Are you using the IPS (inch-lbf-sec) unit system? The Pro/E default is inch-lbm-sec, and this can mess up your entry of the loads. One way to check this is to open/edit the definition of your 60 pound load - at the bottom of the dialog box, it will show the units of your entry. Hopefully it will read 'lbf'; however if it says 'lbm*s^2/in', you can do one of two things:
a) Multiply your input by 386 in/s^2 (i.e., gravity) (60*286 = 23160 lbf) OR
b) Change the system of units in the Pro/E Setup menu
seymours2571
Mechanical
Taz,
Here is my attempt:
I modeled the problem as a 3D contact problem with two seperate cases: A) frictionless contact and B) infinite friction contact. I created a measure on the edge of the spring where the force load is applied. For the analysis definition, I enabled contact with 10 load steps equally spaced. I set the convergence as MPA @ 5% on the measure only. Here are the results:
Case A: 0.0057 inches
Case B: 0.0021 inches
According to the spring design program output you provided the axial displacement at 60 lbf loading would be approx .008 inches. While the results are different from one another by greater that 10%, we must keep in mind these are very small numbers. Percentage comparison of small numbers is difficult.
Notice how the 3D results from my model agree very well with the 3D results from Pete. This to me demonstrates that the problem can be simulated either as a linear static or non-linear with contacts. However, it would be more efficient using Pete's method because linear simulations run faster.
And to answer Pete's question on better ways to model the problem while not using springs...there are two methods to acomplish the same effect more efficiently:
1) You can turn on the intertial relief in the analysis definition dialog box. This only works for linear analysis (i.e. no contacts)
2) You can run the model as a quarter symmetry model (This is what I did for my attempt at this problem). By using the quarter symmetry model the radial translations are constrained along the symmetry planes. Therefore, the only DOF is along the axis of the spring.
Also, Pete in your soluton the reason there may be so much disagreement between your 2D and 3D models might be based on the results output. In your 2D results you are displaying the displacement magnitude (vector sum of disp in X and Y), while in your 3D model you are displaying the displacement along the Z axis corresponding to the centerline of the spring.
Hope all this helps,
Steve
Here is my attempt:
I modeled the problem as a 3D contact problem with two seperate cases: A) frictionless contact and B) infinite friction contact. I created a measure on the edge of the spring where the force load is applied. For the analysis definition, I enabled contact with 10 load steps equally spaced. I set the convergence as MPA @ 5% on the measure only. Here are the results:
Case A: 0.0057 inches
Case B: 0.0021 inches
According to the spring design program output you provided the axial displacement at 60 lbf loading would be approx .008 inches. While the results are different from one another by greater that 10%, we must keep in mind these are very small numbers. Percentage comparison of small numbers is difficult.
Notice how the 3D results from my model agree very well with the 3D results from Pete. This to me demonstrates that the problem can be simulated either as a linear static or non-linear with contacts. However, it would be more efficient using Pete's method because linear simulations run faster.
And to answer Pete's question on better ways to model the problem while not using springs...there are two methods to acomplish the same effect more efficiently:
1) You can turn on the intertial relief in the analysis definition dialog box. This only works for linear analysis (i.e. no contacts)
2) You can run the model as a quarter symmetry model (This is what I did for my attempt at this problem). By using the quarter symmetry model the radial translations are constrained along the symmetry planes. Therefore, the only DOF is along the axis of the spring.
Also, Pete in your soluton the reason there may be so much disagreement between your 2D and 3D models might be based on the results output. In your 2D results you are displaying the displacement magnitude (vector sum of disp in X and Y), while in your 3D model you are displaying the displacement along the Z axis corresponding to the centerline of the spring.
Hope all this helps,
Steve
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