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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Hello snouzy13
What you have to understand is that it's a bit like with your fingers pressing on a spring. If you go from 90 N to 200 N you will have a greater deformation since the pins will flex more at 200 N.
It is in your interest to keep this reference for the various tests you will have to carry out since this is what your tutor asks of you.
By the way, in your tests, put your two pins in an X, obviously by crossing them it will be there that you will have the least deformation. In other words, it will be where your shin will be best supported. If you have the courage to do the X-shaped system, look at what it looks like with 200 N in the axis of the tibia. :-) :-)
Kind regards
Hello Zozo_mp
I don't see what has changed in the assembly you modified, you added connectors, mesh control but I still get 444 N for an imposed displacement of 1 mm.
I have experimental values that are around 50-100 N for 1mm and there I find 444 N, I don't understand anything anymore.