calculation deformations

[Villengel]

Hello, everyone. I just got into this fascinating subject. Following the catia guide I could perform some simulations, but now I would like to go to the bottom and that is I have 2 questions:
1-know the actual deformation of the product I designed and verified in mm or probably in percentage, don't know?
2-How to view a real deformation and maybe measure it? Is that possible?

Obviously if I didn't misinterpret the images in the red zone, my structure should resist up to an 88900 n_m2. Right? but after deformation then a moment before the material starts to crack and break? and therefore the yellow zone, does it highlight the deformation already in place? highlighting that the green area is an area without any deformation?

I'll get you some results from my simulation and images.
ps. If some pious soul would give me a hand I would be really grateful, maybe I could return that he was with some plastic material for the moment. Thank you very much.

 

[meccanicamg]

hi villengel,
First of all, I recommend that you carefully read the cat's guide which explains how to measure deformation states and make animation, change the measuring units in mpa for efforts and mm for deformations.

It's been a while since I haven't taken a cat's hand, but as long as you look at us in forms and options and find everything.

analyzing the stress analysis you posted as an image, you have that the red zones have a maximum internal tension while the blue ones are the minimum/zero. However always looking from the same image you have that you see internal stresses and not deformation.

Surely in the red zones, as I see the design you have imposed or of the constraints or forces, therefore you will have a localization of concentration of efforts, and it is necessary to evaluate very well if in those areas it is really so or if the effort is less and therefore it is to prefer other systems modeling of the load/forces and to verify by hand with the regulations the recroaching of the sheet when I throw to couple note the screws etc.

this voltage must be compared with the admissible sigma of your material, which if it were in s235jr, being a ductile material has a yielding load equal to rs=235 mpa, then it is identified as admissible yield limit rlim=rs/1.5= 156 mpa.

Considering that the maximum voltages smax= 8.9*10^4 (n/m^2)= 8.9*10^4 (pa) are 0.089 (n/mm^2)=0.089 mpa is:

that the oversized safety factor is s=rlim/smax= 156/0.089 = 1753 which is huge. we say that normally it is enough 3 or 10 depending on the application unless you have to keep the deformation under control.

 

[stef_design]

this voltage must be compared with the admissible sigma of your material, which if it were in s235jr, being a ductile material has a yielding load equal to rs=235 mpa, then it is identified as admissible yield limit rlim=rs/1.5= 156 mpa.

hello mechanicalmg, how come you divide for 1.5 and not for 2?
if the rs is 235 mpa, don't you have to stay about under 1/2 of the rs or 1/3 rs for repeated cycles?

Considering that the maximum voltages smax= 8.9*10^4 (n/m^2)= 8.9*10^4 (pa) are 0.089 (n/mm^2)=0.089 mpa is:

that the oversized safety factor is s=rlim/smax= 156/0.089 = 1753 which is huge. we say that normally it is enough 3 or 10 depending on the application unless you have to keep the deformation under control.

Therefore, from the calculation of the safety coeff., the higher and the more protected you are, right?
Thank you:

 

[meccanicamg]

hi stef, then I give you directions to your questions:
static resistance test casefor the ductile materials it has that admissible sigma = yielding load/1.5
for fragile materials it has that admissible sigma = breaking load/3

the ductile materials are those that have yielding/breaking between 0.6 and 0.8 and have a percentage extension greater than 5%.

Fragile materials are those that have a higher yield/rotation of 0.8 and a percentage extension of less than 5%.

use von mises where its equivalent voltage must be lower than the admissible voltage above calculated so that there is positive verification.
case test fatigueIn the case of fatigue testing, different conditions are regulated depending on whether it is ductile or fragile, if the stresses are variable or not. it has to do with all the coefficients of carving, shape, fatigue relationships between sigmafa and rm that vary depending on the stresses (traction, bending, twist) and depending on materials from 0.19 to 0.55, so it would be to say rm/5.2 or rm/1.8.

the oversized factor also called safety coefficient is the oversized index. It's 1 if you're at the limit. greater than 1 you are oversized and you are on the security side. less than 1 you are undersized, so you are at risk of breaking/damaging permanently.

 

[Villengel]

Well, guys, thank you in advance. So if I have not misinterpreted your kind answers, of course you tell me to go to compare the stresses with the values of the admissible sigma. I have not found the guide totally translated into Italian, so to find in English a module type measurement deformation, it becomes difficult to find (even if I look better thanks) the animation of the deformation I have already made it, but there is a variable mooolto parameter that is called amplification value, that is to say it is a parameter that accentuates at will the deformation, the curiosity is that by setting maximum amplitude, automatic the program attributes a value nylon I took into account the two sleighs on the surface (guide columns) and an enclosure in the rear hole (axis z madrevite). so while I will study and convert some data, in my post I put all the files you need to x view the analysis, if you want to browse, because I exported a report with the calculation of nodes, masses, constraints etc. then I wondered where and how to search in the guide a command, which measures the approximate deformation of the structure. Do you have an idea? Thank you. excellent and precious your suggestions on safety parameters, I must also add that the spindle must remain fixed there forever then for calculation of the stresses.... I will break you later.... because I would like to do that too, but first I would start with an easy static calculation, then I start looking and experimenting also dynamics, then I will break you again. Hihih. Thanks again

 

[Fulvio Romano]

Obviously if I didn't misinterpret the images in the red zone, my structure should resist up to an 88900 n_m2. Right? but after deformation then a moment before the material starts to crack and break? and therefore the yellow zone, does it highlight the deformation already in place? highlighting that the green area is an area without any deformation?

No, it's wrong.
the color does not have to do with the state of the material, it is only referred to the scale. the fem does not know when breaking the material, for him it is like rubber, it is hyperelastic.

 

[meccanicamg]

look that there must be a button between those of the annalisy stress that makes the measure and however you have to be able to view the deformation in mm or m. you pass the buttons, I don't remember what it is but there is. it could be pipette shape that you go to probe each point where you click.. .

 

[Villengel]

but where I am a c...... look a bit by clicking on the deformation icon what jumped out?? the translation in mm!!!!!! !
Just a doubt comes to me, but he doesn't consider it the security load according to you? :smile: p.s. show!

 

[Fulvio Romano]

hi stef, then I give you directions to your questions:
static resistance test casefor the ductile materials it has that admissible sigma = yielding load/1.5
for fragile materials it has that admissible sigma = breaking load/3

the ductile materials are those that have yielding/breaking between 0.6 and 0.8 and have a percentage extension greater than 5%.

Fragile materials are those that have a higher yield/rotation of 0.8 and a percentage extension of less than 5%.

use von mises where its equivalent voltage must be lower than the admissible voltage above calculated so that there is positive verification.
case test fatigueIn the case of fatigue testing, different conditions are regulated depending on whether it is ductile or fragile, if the stresses are variable or not. it has to do with all the coefficients of carving, shape, fatigue relationships between sigmafa and rm that vary depending on the stresses (traction, bending, twist) and depending on materials from 0.19 to 0.55, so it would be to say rm/5.2 or rm/1.8.

the oversized factor also called safety coefficient is the oversized index. It's 1 if you're at the limit. greater than 1 you are oversized and you are on the security side. less than 1 you are undersized, so you are at risk of breaking/damaging permanently.

All right, but I'd like to make a couple of considerations.
a. is the piece really nylon or is the first material that is finished in the analysis?
b. Are we sure that the von mises model will marry nylon well? If the nylon test doesn't split to cup and cone, I would trust more than the tresca pattern
c. do I think that Villengel is quite confusing deformation with stress, or is it just my impression?

attention with powerful tools such as fem...the risk of taking abyssal chanters is around the corner, we are not commercialists (citing myself)

 

[Fulvio Romano]

but where I am a c...... look a bit by clicking on the deformation icon what jumped out?? the translation in mm!!!!!! !
Just a doubt comes to me, but he doesn't consider it the security load according to you? :smile: p.s. show!

What is the safety load? You mean the security factor? absolutely no! that must be an evaluation of the designer, the software must not for any reason put it in the calculations, otherwise you no longer understand anything.
Although those of mechanicalmg are in a very correct line, the values to be used as a safety factor depend on the cases. often for fem analysis with countercaxxi and for non-life structures, it is used r=1.1. for materials that respond to the model of rankine, such as ferrocement, is often used r=7, both because the buildings are the case that do not collapse, both because the composition of sand and cement that makes the masonry is not said to be exactly the same as that hypothesized project.

As you can see, there cannot be a number that goes well a priori. you choose and you ponde case by case. the more uncertain, the more the number rises.

 

[Villengel]

In fact once I found the translation key I just came the abyssal doubt of the admissible safety factor/sigma. It is a material (nylon 6) that I created by reading the data on the technical data sheet of a seller, so I decided to repeat the test considering the form of elasticity with attached the safety load suitable to that material (ductile), finally I wanted to verify if with that mass, with that material and with that form at the time of loading the structure of 2 kg, I would have had a deformation equal to x mm, or if my instrument should not immediately great and thank you. further ahead you expect other questions is obvious, hihih.

 

[meccanicamg]

but where I am a c...... look a bit by clicking on the deformation icon what jumped out?? the translation in mm!!!!!! !
Just a doubt comes to me, but he doesn't consider it the security load according to you? :smile: p.s. show!

Since there was!!!! ! !

the safety load does not know him and I do not remember if there is the possibility to insert it and to show you the scheme of safety factors (this I do not remember if catia does it, but since solidworks does it... should do it).

However for nylon it is better to tread than von mises... But always attention to what you do. fem is not a gospel but must be interpreted and set very well before trusting.

to take our hand you should start making simple patterns, draws bind and compare the data by hand.... then you climb up of difficulty.

 

[Villengel]

Well... let's start again!

 

[Fulvio Romano]

then, since my suggestions were not taken (yet) into consideration (except from mechanicalmg), do a good thing.
doubles the mesh and raises the simulation, then halves it and raises the simulation.
compare the results, get scared and then we'll talk about it!

 

[Villengel]

I didn't think I'd take it into consideration. I was doing some research to find the security factor, the command. for the tresca method I probably have to set a formula. Right?

 

[Fulvio Romano]

I didn't think I'd take it into consideration. I was doing some research to find the security factor, the command. for the tresca method I probably have to set a formula. Right?

what does a formula mean?
In monoaxial arrangements, the equivalent sigma of tresca coincides with the only existing sigma.

 

[Villengel]

what does a formula mean?
In monoaxial arrangements, the equivalent sigma of tresca coincides with the only existing sigma.

You're right sorry, I'm niubbo, I meant if in catia I have to set a formula, vary a parameter, I don't think for the shaking model there's just a button. Maybe I'll put an eye on the guide first.

 

[stef_design]

thanks mechanicalmg for the explanation :finger: