3 micron of dispacements? Maybe there is something wrong. Try to calculate reaction forces on this way: tools-sum-select Fz and the element set relative to the fixed face on z direction. Take into account that considering the spring as a hollow section (same material and same dimension) the displacement on z direction is about 16 micron (0,016 mm). So I think that there are two options: load is wrong or dimensions are wrong. Imagine that you are loading a very small spring (2 mm diameter-7.5 mm long) with 2 1/2 bricks of milk (1 litre/brick).
FYI I went back to simple to prove out my Boundary conditions. I took a part that was .0787 inches diameter (2mm) x .500 inches Modeled in SW and save it as a stl. Saving as Stl is a verification of how it seems I can best important this tiny spring made of PEEK into Mecway. I had to bring it into mecway in MM units and using the tape measure I confirmed that it was indeed .0787inch dia x .500inch length. I 3d meshed it using .05 inches. hand calculation of stress += load/area, 5lbs/.00486 in squared = 1026 psi. I applied material condition of 572000 psi for YM, and .391 for PR. I fixed one end of this solid rod and applied 5 lbs load on the other end. I ran the part. There were stress riser associated with the BCs that held the part fixed but showed a stress of 1043.00 psi which is under 2% error relative to hand calculations. This prove to me many different things. 1st that that stress is close to the simple calculations. That the 5lbs load is applied correctly relative to Mecway and my part. That the fix end is doing what it is supposed to.
Andrea Thank you for your help! The stress riser associated with the spring ends is not important to me. As I know what to do to deal with this. If you would be so kind as to report to me from your model the stress at the inside corner of the middle of the spring coil the stress in PSI. this would be helpful as a doubled check to my work. I am getting somewhere between 75-90 ksi.
About 140 ksi vM stress. But is linear elastic material model. I don't know the mechanical properties od PEEK but I think that is too high. Above all is to high the elongation of the spring under 5 lbf load. The spring become two times long
Load was wrong in my model. Thanks Andrea. Mecway divide the applied force load by the number of selected nodes. I made the same operation before imposing BC so my TOTAL load was less than required.
First Principal Stress 202 Ksi Von Misses 187Ksi Displacements are 0,07mm
NL analysis ccx 2.13 MT tet10 elements What I can’t understand is why twilliams55 you are still working on the model knowing you have most of the spring over the Tensile Strength of PEEK (14.9 Ksi).
Andrea Thank you so much! I know all the answers a outrageous relative to the design. All I am trying to do here is establish a bench mark for future designs! By knowing that the BC and Load are good and that I have consistency in out put. I can now apply these constraints to a new model. One that will maybe work with in the current design. You guys have been a great help!
...but don't forget classical theory. For my opinion is better to start by hand calculation (or spreadsheet) because the basic theory of spring is not difficult and the calculus of nominal shear stress is fast also for rectangular section (See Timoshenko-Strength of materials or Shigley-Mechanical engineering design). Other aspect is that for my opinion is better to avoid building a mesh from STL file.
Andrea Thank you so much! I know all the answers a outrageous relative to the design. All I am trying to do here is establish a bench mark for future designs! By knowing that the BC and Load are good and that I have consistency in out put. I can now apply these constraints to a new model. One that will maybe work with in the current design. You guys have been a great help! Ps much more use to using the old machine design handbook. by Shigley xref with esbach and the machinist handbook
Relative to the use of the normal text. The aspect ratios relative to the K values of this square spring design are off of the charts. the problem is the aspect ratio of thickness over height and thickness over mean diameter all start around 3 and go up from there where as the aspect ratios of the current square spring design we have been working with is 1.3 to 1.8 reapectively which put me off of charts relative to Shigley and MHbook. I have found some military info that give me a little larger range to work with
disla: When you specify a force, it's the total force, which Mecway then distributes equally to the nodes. So use 5 lbf for 5 lb total. Usually it's better to use faces instead of nodes because that'll give a uniform distribution of force. Especially for quadratic elements where node forces are a bit counterintuitive and should not be uniform for accurate results near the force.
Those red points at where the forces are indicate a small reaction force. Are there some constraints there too? It shouldn't produce reaction forces where there are no constraints. EDIT It looks like red includes zero so they may zero forces - click a node to see its value.
Comments
Try to calculate reaction forces on this way: tools-sum-select Fz and the element set relative to the fixed face on z direction.
Take into account that considering the spring as a hollow section (same material and same dimension) the displacement on z direction is about 16 micron (0,016 mm). So I think that there are two options: load is wrong or dimensions are wrong. Imagine that you are loading a very small spring (2 mm diameter-7.5 mm long) with 2 1/2 bricks of milk (1 litre/brick).
Regards
I went back to simple to prove out my Boundary conditions. I took a part that was .0787 inches diameter (2mm) x .500 inches Modeled in SW and save it as a stl. Saving as Stl is a verification of how it seems I can best important this tiny spring made of PEEK into Mecway. I had to bring it into mecway in MM units and using the tape measure I confirmed that it was indeed .0787inch dia x .500inch length. I 3d meshed it using .05 inches. hand calculation of stress += load/area, 5lbs/.00486 in squared = 1026 psi. I applied material condition of 572000 psi for YM, and .391 for PR. I fixed one end of this solid rod and applied 5 lbs load on the other end. I ran the part. There were stress riser associated with the BCs that held the part fixed but showed a stress of 1043.00 psi which is under 2% error relative to hand calculations. This prove to me many different things. 1st that that stress is close to the simple calculations. That the 5lbs load is applied correctly relative to Mecway and my part. That the fix end is doing what it is supposed to.
Tom
Thank you for your help!
The stress riser associated with the spring ends is not important to me. As I know what to do to deal with this. If you would be so kind as to report to me from your model the stress at the inside corner of the middle of the spring coil the stress in PSI. this would be helpful as a doubled check to my work. I am getting somewhere between 75-90 ksi.
Thank you
Tom
Regards.
Mecway divide the applied force load by the number of selected nodes. I made the same operation before imposing BC so my TOTAL load was less than required.
First Principal Stress 202 Ksi
Von Misses 187Ksi
Displacements are 0,07mm
NL analysis ccx 2.13 MT tet10 elements
What I can’t understand is why twilliams55 you are still working on the model knowing you have most of the spring over the Tensile Strength of PEEK (14.9 Ksi).
¿Which is your acceptance criteria?
Thank you so much! I know all the answers a outrageous relative to the design. All I am trying to do here is establish a bench mark for future designs! By knowing that the BC and Load are good and that I have consistency in out put. I can now apply these constraints to a new model. One that will maybe work with in the current design.
You guys have been a great help!
Tom
Other aspect is that for my opinion is better to avoid building a mesh from STL file.
Regards
Thank you so much! I know all the answers a outrageous relative to the design. All I am trying to do here is establish a bench mark for future designs! By knowing that the BC and Load are good and that I have consistency in out put. I can now apply these constraints to a new model. One that will maybe work with in the current design.
You guys have been a great help!
Ps much more use to using the old machine design handbook. by Shigley xref with esbach and the machinist handbook
When you specify a force, it's the total force, which Mecway then distributes equally to the nodes. So use 5 lbf for 5 lb total. Usually it's better to use faces instead of nodes because that'll give a uniform distribution of force. Especially for quadratic elements where node forces are a bit counterintuitive and should not be uniform for accurate results near the force.
Those red points at where the forces are indicate a small reaction force. Are there some constraints there too? It shouldn't produce reaction forces where there are no constraints. EDIT It looks like red includes zero so they may zero forces - click a node to see its value.
for a more quick calculus there is an approximate formula on "Roark's formulas for stress and strain"