A small question for the community. I work in SOLIDWORKS 2017 and I have to calculate the force of an axis. My system allows you to compact and pressurize filter cartridges. It is subjected on one side to a torque tightening torque of 15 N/m, and on the other side to a pressure of 6 bar or 0.6 N/mm^2. (Photo attached). The results of my static study seem weak to me, I apply my forces at the end of the axis on each side (on one side the tightening torque and on the other the pressure suffered) and in imposed displacements I fix the body right?
I need to be a little more clear about how to do this and have distortion results that are close to reality.
I have looked at your image and I do not share - unfortunately - the method you want to use.
I'll give you in order after we'll sort it out together
You put pressure instead of force. The use of pressure is used for compressed air for a tank for example. In your case, it is a force, or rather a counter-force. There seems to be, if I understood the resistance force of [[ compacting and pressurizing filter cartridges ]] which seems to be two distinct actions, one that would be counter-force and the other of pressure generating counter-force.
Putting a couple at the end of a tree is not correct for three reasons. - 2a _ The forces are made on the threads of the screw and not at the end of the shaft as you have represented. - 2b _ given the use I also assume that these are screws with square or trapezoidal threads and not isometric threads like ordinary screws. - 2c _ In addition, you probably have a ball bearing, made of bronze or plastic at the end of the shaft so the anti-slip coeff at the end should not be very important.
Don't forget the multiplier of a screw since it acts like a wedge. On the other hand, you have the binding force which has a significant coefficient of friction depending on the profile of the screw. As you have programmed your simulation, you obviously do not have the corner effect.
In two words, if I were you ;-) I would use the known and standardized abacus of screws (depending on the type of screw, Ø of screws, etc...). The supplier of the special screw must provide you with the charts that fit well because you are closer to a pressure screw than to a fixing screw whose deformation is done by elongation.
Thank you for your answer, on the functioning of my system it's exactly that. Indeed in a 1st time I changed my parameters and I put a force instead of a pressure (I specify Metric trapezoidal screw Tr 36*6 ). On my small Ø20mm section and with a pressure of 6 bar I get 188.5 Newton, it already seems more logical to me (lack of reflection).
Yes, there is no doubt that the effort will be made on the net, but the torque wrench tightening is imposed on us (previously we designed a torque limiter but too expensive).
I don't know how to proceed on solidworks, yesterday I wanted to run the calculation on my net but I have the impression that it may take time and I don't even know if I'm doing things correctly....
I would just like to have a view of the effort I would have and the risks that can be there (compression, flexion, shear...)
Coefficient of friction to be taken into account?.
And the charts won't help me more than that I think.
In general, I buy this kind of thing from a catalog and it's up to the supplier to provide me with the elements of guarantee in relation to what he offers. These are both simple and relatively complex calculations, so I save time.
For my part , I am absolutely sorry that I cannot help you with SW simulation for your problem because the number of parameters to master can lead to important errors.
If you want to continue anyway; It is certain that you need to draw the screw in volume since the shears will be on the threads.