The 1st material I use is a custom material for the pins (small cylinders) and the 2nd acetal POM (large cylinder)
@ Aliende
it would not rather be ............... 448 / 2 = 224 N/mm²
With a safety factor of 2, your tensile or compressive stress must not exceed 120 N/mm².
@+ ;-)
Hello
Be careful , the tensile limit or more generally resistance does not mean rupture (excuse me) but indicates the beginning of the contraction, which means that we enter the plastic zone that will end with the rupture (breaking moment quite difficult to determine by the way, depends in particular on the tensile speed, etc...).
The Simu PEF remains in the elastic domain but says nothing about the plastic domain except to indicate that you are getting closer or that you are already at the limit of the elastic or that you are getting closer to it.
Be careful not to confuse tensile testing with compression.
Kind regards
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No @gt22,
The normal tensile stress (the Greek letter Sigma lowercase) equal to the force (N) by the cross-section (mm² or m²).
If we want the part to resist for sure, this stress must be smaller than the elastic resistance (Re) that we have in the graph that @A.Combier has put.
Rp = Re/s is generally used, where s is a safety factor varying from 2 to 5 for steels.
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And we check that the normal stress (Sigma) is less than or equal to Rp (Practical Resistance)
@Snouzy13,
I think that the values of the custom material have been changed and there is an error, the tensile strength is very low for stainless steel and must be greater in any way than the elastic limit. Check by looking at the values of other non-customized stainless steels.
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Hello Souzy13
As we are always on your shin and pin problem
You say """ My objective is to know the force exerted to make my 1mm, 2mm piece, etc ... without of course breaking my coin"""
In your case, what matters is the distance between the fixed point of the spindle and the tangent of the big cylinder The longer the spindle, the more flexing it will be. If it's less than three mm, then you're getting into shear instead, which would be another calculation.
In your case, your pin is relatively long, so there is no chance that you will reach the yield limit, especially since for the test, your tibia is made of POM and it is the one that will go to hell well before the stainless steel (if I have followed correctly what you want to do).
POM has nothing to do with the resistance of a tibia but that's another subject.
Kind regards
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@Aliende,
Indeed, my referent gave me bad values about stainless steel, I told him about it. The steel in question: 1.4441/AISI 316L
Elas limit: 190 and Breaking strength = lim tension: 460
Hello, while changing the materials of the pins, a problem appears. I get a force of 444 N for 1mm imposed, it's just impossible, put 44 Kg to move 1mm, it's incoherent. The same goes for the constraints that exceed the elastic limit of the sudden. See attached
capture1.png
To really move forward, you would have to see all the constraints and connectors you use for your simulation.
I think it comes from the way you constrained the two big cylinders.
Remember that your two pins functions as a parallelogram which limits bending a lot. Remind us of the diameter of your pins and especially the length of them. Look at the bending of a spindle on its own and you'll see some of what I'm sensing.
Can you post your ASM with the latest simulation to check what I say above.
Thank you
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Hello Snouzy13
That's what I thought, you misuse the imposed displacement.
I think it comes from a small misunderstanding on your part.
In addition, you have no load in your simu.
By forced displacement, it must be understood that you allow the part to move according to X or Y or Z by a certain value.
This makes it possible to control the slides (the animated image is explicit)
There are other things that are wrong
I'll do the simu and get back to you later.
Kind regards
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In compression, I want a displacement of 1mm downwards, I select the top face and then the normal to the plane, which makes a z-displacement of 1mm.
I allow the part to move by 1mm and I want to know the force to make it move, which is wrong here.
Hello Snouzy13
Here are the results without imposing a limit on the displacement, but with a load of 90 N we obtain a displacement of about 3 mm (see image)
On the other hand, if we impose that the displacement is limited to 1mm, the result is indeed 1mm with an identical load of 90 N or 200 N if we want to do the test.
Be careful because in the case of a unipodal support the tibia will support the entire weight of the body (Suffice to say that your pins (only 5 mm in diameter) are useless to limit the displacement in compression) in this case it only serves to keep the two parts of the rejoined tibia in line (as in the external fixators ilisarov).
An Ilisarov is mainly used to reduce or fix bone fragments in the case of complex fractures of long bones (tibia, umerus, etc.).
At your disposal to tell you more about what you really want to test.
Kind regards
deplacement_limite_a_1mm.jpg
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The suite in ZIP for screenshots for the simu attention 2018 version
Kind regards
PS you have two interlocking zips ( = 1 ASM with PRT + 1 zip of photos
information_et_assemblae_en_v_2018.zip
Snouzy
I understand what you want, but take the trouble to read what you said an hour ago.
I tell you again that your way of doing things is not the right one. In the results of my previous email you have the expected result 'with the wrong method and with the right one in relation to the goal to be achieved. Using the right method you will know that to move the two tiny pins by 1mm it takes less than !!!!!!
Kind regards
Hello, thank you for your help.
"Here are the results without imposing a limit on the displacement but with a load of 90 N we obtain a displacement of about 3 mm" isn't it 0.22 mm?
"200 N if you want to do the test" where does this value come from and what do you mean?
I don't have solidworks 2018, so I would look at the assembly when installing the student version. I saw that you have added connectors, I don't know what it's for but I'll see in the assembly.
In my case, I study basic external fixatives. The objective is to find the force for each movement, then I will change the angle of inclination of the pins, their diameters etc... In the end, I would compare these forces. That's what my internship supervisor has given me to do for the moment.
Kind regards
The problem is first of all vocabulary which can have several meanings, common sense and the meaning used in a particular field, which is here simulation
How to explain
Imposed travel (I'm only talking about the Standards)
To perform a static analysis, the model must be properly fixed to avoid displacement.
A) Fixed geometry all degrees of freedom including rotation are blocked.
B) Plane support Indicates that the object can move freely on this plane but for example that it cannot move relative to an edge. A bit like your computer mouse where you would say that it can't get out of the mouse pad.
C) Fixed Pivot Indicates that a cylindrical face can only move at the very end of its axis (cannot move radially) but can move in the axial direction (like a door hinge). The radius and length do not change under load.
THE STRENGTHS
Force, Pressure, Gravity, Centrifugal Force, Bearing Loading, Remote Mass Loading, Distributed Mass.
In your case, you apply a force preferably.
Kind regards
@Zozo_mp
Hello
I understood what you mean by forces and imposed displacements, I had already understood this part. Could you explain to me what I quoted above, the 90 N 3mm and the 200 N?
Thank you
Beyond the elastic limit there is plastic deformation, then breakage...
May the force be with you ; )
Edit: oops I hadn't seen the number of dsl pages
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