little Help

Victor
This is a Spring I trying to design. Used in extension. Made Of PEEK plastic. .08" dia .03125" hole, .010" slot, .03125 pitch, I am using a standard linear static youngs modulus of 206000psi, P ratio of .39, 6 turns of the helix. the spring has a square wire section. The values I am getting seem low. What would you suggest?
Tom
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Comments

  • Stress usually increases with mesh refinement so make sure it has converged to a reasonably constant value as the mesh density changes.

    If the stress is too low then the load might be too low. Confirm it by checking the sum of reaction forces. Click the Z component under the reaction force branch of the outline tree to display the sum.

    If reaction force is OK then perhaps the geometry is wrong. Check that Tools -> Volume gives a reasonable value in case of a units/decimal place problem.

    If that doesn't help, you can attach the liml file and show how you concluded that it's too low and I'll look for problems.
  • edited December 2017
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  • edited December 2017
    "standard linear static young’s modulus of 206000psi, P ratio of .39"
    Is the Young Modulus Right?.
    I have found some internet references for Peek. P ratio seems in the same line but Young values are more than double (545342-572899) psi. Seems you have introduce Shear Modulus value not Young Modulus.
    See attached file. ¿Is it the same Peek are you using.?
    ¿Do you impose BC as force or displacement ?
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  • For metal springs, normally stresses are very low. Have you turn on geometric no lineal? And what about the stiffness that you have, does match with the predicted value?

    Regards
  • Disla, you are correct I have been using the value show 206000 in the verification calcs I have been doing to try and match outputs on Mecway. This number is the shear modulus of PEEK I will switch to 572900 which is the youngs modulus. Youngs modulus has nothing to do with the stress found in a material only the deflection present.

    Victor I refined the mesh 2x and got a 25% increase tried to refine it again and went over 3 million nodes. I am using a 5lb load all thought this may seem low the dia. is .0787".
    if one where to look at it in a P/A =stress and for a solid piece at 50lbs /,004=10kpsi.
    When I run this model with 50lbs with 2x refinement is mess I get 2800psi max with helix. When I try to check the reaction force a get values much lower then my applied value. Is the Max value show the max reaction force of the Boundary condition apply or is that the sum of all the reaction forces?
    I just checked the volume and and think this might be the problem as it give me the volume of all elements as 13in cubed. that seems to me to be 2 order of magnitude off. Time to go back and check the solidworks model.
    Thank you guys so much for your help!!!
    Tom
  • Also, I forgot I would like to apply a boundary condition that would be equivalent to having a plan bearing wrap around the OD of this spring at one end of it. I tried doinf this use the frictionless setup on the faces on the od at one end of the part but did not seem to work out to well. Any suggestions?
    Thank you
    Tom
  • "Youngs modulus has nothing to do with the stress found in a material only the deflection present".

    Mmmm.. not agree. Equalities should be read in both directions. If you impose displacement as BC Young Modulus affects directly to the resulting stresses.
    Any way, let us know if you solve your problem. ;)
  • Victor
    I am use solid works as my modeler. I have checked all units in solidworks and they seem correct. I convert the files into stls and bring them into Mecway this way. After Imported I use the Tape measure tool to measure and found that it the imported model was in meters and the part which should be .297" long was 7.5 inch long this is the problem where did I go wrong? BTW I love this software! I started with Algor in 1989.
    Yes I know I am an old guy. I dislike the solidworks FEA.

    Thanx
    Tom
  • Disla
    sorry of my wording I come from a time when all we had was equation to calculate such as this. I have only been on Mecway a few days. Still in the middle of the learning curve.
    I put a five pound load as a Force boundary condition, not as a displacement BC. Is this wrong? I always stress is a function of geometry no matter the material used. What is the advantage of putting the loads on as a displacement and how would one do that?
    Tom
  • For the boundary condition you could create a reference node at the center of the face, then link all the nodes of the faces to this reference node by means of the RIGID BODY card (not present in Mecway solver or GUI, so you should change the solver and add as a custom CCX content). The rigid body feature will let the attached nodes rotate but not separate. Then you can add other bc to this reference node to create the traction on the spring. Feel free to attach or send me the model and I could try during the afternoon.

    Probably you will need to add two references nodes to fix the unwated rotations.
  • twilliams55, you can export directly as step in Solidworks, and then Mecway should be able to import it with the right size. Anyway if you don't want to recreate the model, you can scale the mesh to the right size in Mecway directly.
  • Sergio
    Can I correct correct my imported Volume in Mecway? Or should I model everything in Meters in Solid works and import in that way? I will take you up on the offer. I am just now looking at the CCX stuff now. Here is the file attached it is the solidworks STL done in Inches.

    Thank YOu
    Tom
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  • Tom, what you can do in Mecway is after meshing, scale the mesh (so, if you have created bc, loads, named selections... it will be keeped). Can you attach directly the .liml file? I will have a little of time to see it during the night.
  • Sergio
    The spring is used in extension with a 5lb load applied to one end the other end is held captive. The complete structure (which has been simplefied here for mean of model fit inside a cannula.
    Tom
  • There is a trick to properly re-scale STEP file before import to MECWAY editing directly the step file with a common editor.(Notepad for example)

    There is a section in STEP file telling translator what unit and what conversion factor to use. Open Step file and Find-> "UNIT" . Note that the line number may vary depending on the STEP file and output system.

    #672=(LENGTH_UNIT()NAMED_UNIT(*)SI_UNIT(.MILLI.,.METRE.));
    #673=LENGTH_MEASURE_WITH_UNIT(LENGTH_MEASURE(3.048E2),#672);
    #674=(CONVERSION_BASED_UNIT('FOOT',#673)LENGTH_UNIT()NAMED_UNIT(#671));

    In line 672 (Depends on the step file), it says the system unit (millimetre in this case) used by outputting CAD system. Check Yours.
    In line 674, the outputting unit is set to ft. In line 673, there is
    a factor LENGTH_MEASURE(3.048E2), which is the multiplier. In this case to convert millimetre to ft.
    I now regularly do it with my step files from Salome to convert from "m" to "mm" and it works. For example in my case is simple and direct without scaling factors.

    #Whatever number= ( LENGTH_UNIT() NAMED_UNIT(*) SI_UNIT($,.METRE.) );
    #Whatever number = ( LENGTH_UNIT() NAMED_UNIT(*) SI_UNIT(.MILLI.,.METRE.) );

    Regards,


  • Sergio
    I tried the step file to start with but Mecway did not like the hole thru the middle of the piece, or so it seemed. The stl came in beautifully or so I thought.
    Tom
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  • Sergio
    I have been trying to find a scale command . Where might I find that?
    Tom
  • There is an elegant way to impose this boundary condition as you want directly in MECWAY. I discover it not too long ago. I'm also rookie in Mecway. I use it for the tank bases or when I need a stiffening ring in a pressure vessel.
    Check the liml file attached in the recent post ccx2.13.
    both extremes has this kind of bc you are asking. It is based on the tangent and normal vectors equations on top of a the circle.
  • sergio
    it is too big! 93mb 371000 elements
  • Yes, I supposse that. I will add the PEEK material as per the above spec and then fix one side and apply the load on the other by means of a rigid body simulating a plain bearing (or the plain bearing should be in the two ends???). Just to be sure the outside diameter of the spring is 1mm?

    Regards
  • Sergio

    OD of spring is 2 mm I was able to scale mine the scale it needed was 1/25.4
    the peek is 572000psi with possions of .391 after 20 or so iteration it did converge. Nonlinear the Von mises stress showed 280,000 psi at a stress riser that will have to be dealt with. at an inside edge of the spring it showed 91,000 psi. this seems a little high but could be getting closer. This model has 392,000 elements
    Thank again for your help!

    Tom
  • Sergio
    the plan bearing would actually run the length of the part allowing translation and rotation about the z axis. So this is an extension spring that fits in a tube. held at one end with a five lbs load at the other.
    Tom
  • Your model looks right to me.
    I think you are missreading the units.

    After Imported I use the Tape measure tool to measure and found that it the imported model was in meters and the part which should be .297" long was 7.5 inch long this is the problem where did I go wrong

    The part in the model measures 7.5mm in MECWAY wich is right (.297" = 7.5 mm)
  • edited December 2017
    I tried to apply 5lbf to the model and, without convergence study, I found a peak stress of 349.8 Ksi (first principal stress - NL analysis).

    The problem is to know the right conditions of work. Due to the geometry the spring loaded with a distribuited force bends. Is the bending possible during the normal conditions? If not is necessary to apply correct BC to the upper parts (where the force is applied). This aspect is confirmed visually plotting the deformed shape but also checking the peak stress which is different for upper an lower part.
    Generally forces are guided. I think that the actual model doesn't respect the real conditions.
    If I'm right is possible to refine the model applying the force by RBE locking the upper face rotations around the xy axes.
    Other aspect is the peak stress. Is located on a edge and so I think that reducing the mesh size the stress will growth. So, for my opinion, is necessary to create a fillet.

    Regards
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  • edited December 2017
    Hello,
    I'm getting different values. Two orders of magnitude difference but in line with PEEK compressive strength of PEEK. I have applied the 5lb distributed in some of the upper nodes 5lb/24 each one of the 24 nodes. I have left the spring free to expand and rotate but vertically centred.

    Von Misses 7.86 Ksi = 54 MPa
    Compressive Strength of PEEK 18.8 ksi= 130 MPa.

    5Lbf seems an acceptable load.
    ¿Am I wrong.?
    Like twilliams55 I have been unable to recover resultant force on the base to check force equilibrium in the vertical component.
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  • Units
    The STEP file imports with correct units and is about 0.08 inch diameter as shown by Mecway's tape measure tool. It might also show 2 mm which is the same thing.

    The STL file is in mm, not inches. So when you open it, you have to choose mm from the dropdown box. SolidWorks allows you to specify the units for exporting to STL, which are not necessarily the same as the units you defined the geometry with. The STL file format doesn't contain units itself so you have to remember what they are.

    Reaction forces
    To see the sum of reaction forces in the Z direction, click the Z component of reaction force in the outline tree. The numbers on the color scale are not really meaningful because they're individual node forces which depend on the mesh. The sum is shown at the bottom of the graphics area. That's only available directly if you're using the internal solver.

    Boundary conditions
    Frictionless support is meant for sliding surfaces like this and it should be OK. Disla's suggestion of using formulas in displacement is also OK but I don't think it's necessary here where a 2D surface is being constrained.
  • Spring is in extension not in compression. In compression, with 5lbf, there will be a contact and the classic study is not applicable.
    Try to solve with classical theory of spring. I found a nominal shear stress about 90 ksi which is very near to the zz stress on the inner wall (far from the peak stress) obtained from the FEA. vMises will be about 1,732 times the shear stress.


  • You are right Andrea twilliams55 comment is working in extension. Just comment I do not get contact between spirals in compression.
    Run again with new conditions ,ccx2.13MT, NL second order tet elements and obtained:

    -Von Misses 7.86 Ksi
    -Tensile Strength of PEEK is 14.9 ksi

    According to your results 90 Ksi , 5Lb load this spring is completely destroyed .
    Why so different values.

    Have a doubt why my sum of reaction forces is wrong.
    I have impose the load as nodal forces. 0.208 lbf/node on 24 nodes= 5 lbf TOTAL force.

    Victor, I run with internal solver and sum of z reaction force shows 0.208 lbf. Seems it is only considering I have introduce 0.208 lbf.
    I also found some coloured areas around the force application area. ¿ Why?
    Sorry being persistent but I take any opportunity in the forum to use it as exercise and learn.

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