Sheet metal Spring, Force to Compress To a smaller Diameter 2

• Friday at 9:25 PM

• #21

mfgenggear

Aerospace

Jan 23, 2008

3,666

US

Op
How many, quantity?
Low volume 10 pcs, or 10000 parts.
This is a great brain teaser.
My thoughts and opinion.
Thus a very small part. I would first of make a new design.
OK but if you are set on this.
It will require developing.
Trial and error to allow for spring back.
A two or three stage die. Reason the leg
Pushed into the center.
Create the two small bends on b each end.
Then the large radii
Then bend the inner leg.
Depending on hardness, or tensile strength.
For small lot a small fixture with removable pins. Similar use for bending tubing.
HTH

 

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• Saturday at 12:32 AM

• #22

GregLocock

Automotive

Apr 10, 2001

23,931

Orbiting a small yellow star

@rothers was that a non linear contact analysis? Please could you plot the contact force?

 

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• Saturday at 10:43 PM

• #23

Stress_Eng

Aerospace

Jun 15, 2019

259

GB

Out of curiosity, I revisited the original method and updated it to be in-line with the polar method (same assumed forced radial displacement, point loading at start of contact not in middle of angular increment, etc). The result gives a bending stress of 196.7 MPa. Although it hasn't been checked, and as rothers has highlighted, the stress is close to FEA. Attached is the revisited file. Looks like this particular analytical method has been finalized!

 

Attachments

• Clip 45deg Revisited.pdf

  88 KB · Views: 2

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• Sunday at 12:46 PM

• #24

rothers

Mechanical

May 1, 2008

189

GB

My FEA is not great at post processing contact forces but for what it's worth here goes:

Reaction force contour plot:
Interestingly reactions concentrated in three places, sum of reactions are:
Left hand contact point = 15.37 N
Bottom contact point = 15.37 N
Right hand contact point = 14.33 N



Reaction Force Vector plot:
Ignore the exact position of the vector, it plots the vector start points at the final contact point of part 1 (the ring) but at the initial contact point of part 2 (the spring).
Anyway I'm sure you get the idea and is shows the directions nicely.


Make of this what you like !!

 

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• Monday at 11:34 PM

• #25

Stress_Eng

Aerospace

Jun 15, 2019

259

GB

Just thought I'd post this last adjusted calculation. By altering the forced displacement, the polar method now calculates a maximum bending stress of 212.3 MPa and an initial contact load of 14.47 N. This compares well to the posted FEA results. Attached is the adjusted file.

 

Attachments

• Clip 45deg Polar Adjusted.pdf

  85.5 KB · Views: 5

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• Tuesday at 12:17 AM

• #26

GregLocock

Automotive

Apr 10, 2001

23,931

Orbiting a small yellow star

Thanks for the plots. Just 3 contacts all in the obvious places. I like it when FE actually agrees with common sense.

 

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• Tuesday at 1:59 PM

• #27

rb1957

Aerospace

Apr 15, 2005

16,176

CA

maybe something like a jubilee clip ... to compress the OD ?

or a tool to pick up the two obvious hooks, and shorten the distance between them ? Option 1 sounds easier ....

 

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• Tuesday at 7:01 PM

• #28

Stress_Eng

Aerospace

Jun 15, 2019

259

GB

Just to simplify things, here's a two pager showing the results for a single point load at the free end, based on an assumed forced displacement and at varying contact angles.

 

Attachments

• Clip - Point Load.pdf

  76.4 KB · Views: 6

Last edited: Tuesday at 7:14 PM

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• Yesterday at 11:17 AM

• #29

prex

Structural

Jul 4, 2000

1,932

IT

This problem should be solved by a simple analysis.
Assume the force needed to compress the spring is appllied by a circlip plier between the two hooks. This force F has a vertical component V=~0.93F and an horizontal component H=~0.36F (estimated from geometry).
Now assume also, for simplicity, that F acts onto the outer hook, while the inner hook is connected to a pin.
The bending moment in the spring due to V is -Vr sin(theta) (theta from 0 to 1.5pi), and the moment due to H is Hr(1-cos(theta))
So the average bending moment in the spring is M_ave=Fr*integral(0.36*(1-cos(theta))-0.93*sin(theta))/1.5pi=~-0.24Fr
From the relationship (delta r)/r²=-M_ave/EI, with delta r=0.2 mm, one obtains F=46 N
This seems a relatively high value, so please correct me if I'm mistaken

 

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