Modeling a spring

3D Design Volume Model
1

Hello

I would like to create a fairly simple spring.

Øfil 0.38, Outer Ø2.71, Nb Sprire Usable 2.75, Nb Total Spire 4.75, Free Length 4.6.

Modeling it is not too much of a problem for me. Except that I want the assembly to be flexible.

It doesn't keep me the right number of turns when it's in compression.

Have any of you ever had this problem?

Thank you

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2

What software is it with? 

The subject interests me, I have never managed to get it compressed properly

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I work with Creo elements/pro 5.0.

My spring keeps the dead sprires of the beginning and the end, but it diminishes the useful sprires.

So to check if there is a joint sprire it's not practical.

Thank you

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4

There is a tutorial to model the movement of a spring.

 

I'm giving you this;)

5

If it's on Solidworks.

 

Here is the link

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And for the settings:

 

See this link

 

 

EDIT: Sorry, I hadn't seen that it was under Creo....

 

So much for me!

7

Hello

Here's a tutorial for Pro e, but it doesn't have to be much different:

http://www.proengineertips.com/index.php/proe-wildfire/assembly/38-tips-restricted-parameters-in-proengineer25

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By varying the compressed length, does the number of turns always change?

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@Bart, the man who responds faster than his shadow!! Speed should not be confused with haste. That's what my teachers kept telling me! lol :-) 

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Look at this link

Variable pitch spring tutorial

http://proewildfire.free.fr/tutoriaux-fichiers/tutorial04%20composants%20flexibles.pdf

http://proewildfire.free.fr/tutoriaux-fichiers/tutorial02%20ressort%20pas%20variable.pdf

https://lh4.googleusercontent.com/UDVo1u6LTkkA4hJKeWWiZdOlTwk3VA_eaaLgJoL8lxw=w159-h204-p-no

everything is there, just read and watch

@+ ;-))

PS: @ jm savoyat it's not to defend @ Bart our colleague but it is visual since the new format to detect for which log is the question

@+ ;-((

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Yes Jm! totally agree

 

My teachers told me:

 

"It's by going slowly that we go fast" =)

 

And I wasn't the first to answer!! :p

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Thanks Gt22, it is not noted anywhere in the question that the software used is Creo.

 

Hence my misunderstanding....

 

Inadmissible!!! =)

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13

Hello

 

I made a spring model under PRO/E (it still works under CREO).

On this spring you have parameters.

You enter the length, the outer diameter, the diameter of the wire and the number of turns directly.

It then builds it on its own (with the right relationships).

And of course it is flexible.

It just has one flaw, see below.

I'll explain the design principle and if you want, I'll give it to you on Thursday:

  • Start with a helical scan for the body. (we'll see later for the quoted values). Constructs the function to have the outside diameter dimension and to be offset from the plane (d5 below)
  • Do a 2nd helical scan for a joint coil. (start on the map). pitch = wire diameter
  • ditto for the second conjoined whorls.

The defect is between the body and the joined coils, the change of pitch is brutal and not in accordance with reality.

  • Then you make a cut for the 2 grinding wheels of the spring. (put the length rating well)

That's it for the functions.

4 parameters mentioned above real numbers to create. (Dia-Wire, Length, Dia-Exter, Nb-Turns)

Let's move on to the relationships:

For the length of the 1st function:

  • D1 = Dia-Wire (Wire Diameter Dimension)
  • d2 = length-(3*dia-wire) (side length function)
  • d3 = d1/nb-coils (steps)
  • d4 = dia-exter
  • d5 = 1.5*dia-fil

2nd function:

  • d6 = dia-fil
  • d7 = dia-fil (pitch)

3rd function:

  • d8 = dia-wire
  • d9 = dia-fil (pitch)
  • d10 = d5 (length helical function)

4th function:

  • d11 = length

And that's it. For flexibility, you change the length parameter in the assembly constraints.

good construction.

 

S.B

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14

Thanks to the person who answered.

Here is what I managed to do but there is still a value that I don't understand.

I created a helical scan protusion.
Attributes: Variable pitch, Right hand, By axis.
Scan profile: "see scan profile image"
Not: "See graph-pas image"
Section: 'see image section'

Then here are the relationships:
X=(d49-1.9)*2000/20
/*X= FACTOR DETERMINING THE POSITION OF THE SPRING HEIGHT ACCORDING TO ITS COMPRESSION RATIO
/*(IF d38=4.6 THEN X=100 AND IF d38=1.9 THEN X=0. IF d38=1.35 THEN X=50. \
/*4.6 = FREE LENGTH AND 1.9 = LENGTH OF JOINED TURNS)

d44=d16+(X/100)*(d16-d16)
d41=d16+(X/100)*(1.9/2-d16)
d40=d16+(X/100)*(1.9/2-d16)

/*Dead Whorl=SM
SM=1
d34=SM*d16
d36=SM*d16

Now I don't understand the 2000/20 value of the first line.

Thank you


profil-balayage.png
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Image Graph-pas


graph-pas.png
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Section image


section.png
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Here is an image of my spring with values to better understand relationships


ressort.png
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Hello

Here is the spring with the parameters and relationships.

It is the size you want.

I made a modification from what I had marked to have 2 non-useful turns instead of 3.

 

S.B


spring.prt
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an image for the many non-Creo users.

S.B


ressort.png
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What I can conclude from this is that the helical function is useless for a spring?