elastic resilience

[travisr34]

I am designing a flat spring from sheet metal, a sheet metal clip. I am having trouble choosing material for my spring. My spring will require a lot of force for its size, so much so that I am worried about yielding. In short I am looking for a high yield strength, ductile, and cheap material (like everyone else designing a spring). I believe what I am looking for is either a spring steel or a spring tempered stainless steel. When I was looking over materials I noticed the fully hardened stainless steel with its huge yield stress and was like oh heck yeah give me that, then I noticed its % elongation and said to myself....ahhh what the heck does that mean? Can anyone give me a better read on what that means in terms of the elasticity of the material. I am looking more for how the elastic modulus has changed. If this is the strain at yield I assume I can calculate it using E=stress/strain, which should still be valid at the yield point, just not after it. However I assume the elongation is all plastic deformation which means I cannot use it to determine my new youngs modulus. Long story short, what do you guys make your clips out of and why? To put some numbers out there I am looking to generate about 30 netwons of force with a cantilever length of about 15mm, and a thickness of around .5-1mm. The base would be about 12mm.

 

[GregLocock]

If it is steel then its E will be 210 GPa in the usable range, plus or minus 15%

% elongation is the elongation at UTS, and is an indication of the ductility of the steel. As an example, various components I deal with have a minimum elongation of 12%, but that is application specific. I have seen spring clips that were yielded during the assembly process, due to a dimensional stuff up. They didn't work. So if I were you I'd design for elastic deformation only.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[drawoh]

travisr34,

I believe that the percent elongation is at failure. You can easily work out the elongation at yield, since you know the yield stress and the elastic modulus.

Elongation at failure gives you some idea of how ductile the material is. Spring materials typically are not ductile. If your material is heat treatable, you can do elaborate fabrication of the material in its annealed condition and then heat treat it.

--
JHG

 

[travisr34]

Thank you both for your response. So are you telling me that my modulus of elasticity will not change after the material has been hardened/heat treated? That the reduction of strain is a factor of an increased yield strength and not a reduction ductility? Basically for a material is stress/strain=constant until yield true no matter what hardness process you perform on a material. That is to say is the elastic modulus for 301 SS half hard, the same as elastic modulus for 301 SS Full hard?

The way I always thought of it is that if you temper something you will make it stronger but less ductile. So while the material can handle more stress, it will also be easier to stress the material. So for a hardened material the same beam deflection will cause the material to feel more stress then an annealed material.

 

[Tmoose]

You say you need a particular force, around 30 newtons.

I'm guessing there is a limit on allowable deflection, but did not see it in your posts

 

[MintJulep]

You know the required force.

You know the geometry.

You (presumably) know the allowable deflection.

Solve the relevant equations for the necessary value of E.

Find material with that E.

Check that stress is below yield.

 

[GregLocock]

I'm hoping this is not a safety related part.

Yes that is correct, E in the elastic range is not affected by the composition (much) or temper (not very much at all) of the steel.

This is a very elementary design problem, you appear to be designing a bulldog clip. They are made from a hard spring steel i suspect.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[Tmoose]

only valid for stresses < yield, and for "small" deflections, but ...................
deflection at end =

http://www.engineersedge.com/beam_bending/images/beam_b95.gif

note the only material property is E

 

[travisr34]

Wow I really like all the interest on this forum. Thank you for your response greg that is exactly what I wanted to know. To break down what I am doing a little further.
I do not know my allowable deflection. My allowable deflection is going to be determined by the materials yield stress, which is why I am having a hard time choosing my material. I am looking for something with a good elastic resilience, elastic resilence=yieldstress^2/2E. My goal is to maximize my force from the clip.
This is a rough picture of my clip.
-- - <--point of deflection
/ \ |
/ |
/ L
/ |
/ _ <--- Force on clip. Vector should point in other direction so the clip bends out of the curve.

Despite the fact that there is a curve in my clip I am going to use the equations for a linear beam to determine my allowable deflection. Because I dont want to sum the moment around the radius. For a single fixed end beam my deflection is given by
Dy=(3F^2/6EI)(3L-x) plugging in initial conditions for beam and I you get the classic equation
Dy=4*F*L^3/E*b*t^3 eq 1
Max stress will occur at the point of greatest tension. And is given by solving eq 1 for force and x=0 and y =t/2 and plugging it into stress equation Stress=F*L/y
MaxStress=(3*E*t/2*L^2)Dy eq 2

Substitute eq 1 into eq 2 and you get.
Fmax=w*t^2*MaxStress/6L

This means in order to maximize my force without hitting yield stress I should:

Increase my width w as much as possible, as it increasing force without increasing stress
Increase my Length L as much as possible as it slightly decreases my force while greatly decreasing stress
Increase my thickness as much as possible as it increases greatly increase force while slightly increases stress.

Also looking at my stress equation I see that an increase in youngs modulus will increase my stress thus I want a low elastic modulus with a high yield stress. Therefore I want a material with a large elastic resilience. Elastic resilience being YieldStress^2/2E. With gregs help I can now calculate this, thank you good sir. By the way greg the safety of the world will depend on this clip functioning properly. Now all I have to worry about is the cost of hardened stainless steel vs the cost of spring steel...and plating spring steel.

 

[travisr34]

Sorry my clip picture got ruined dont even try to follow that.
...--.....<point of deflection
./..../...../
...../
..../
.../
../...<---point of force

 

[drawoh]

travisr34,

Titanium and beryllium copper have yield strengths comparable to spring steel, with much lower elastic moduluses (moduli?). These materials will work better with large deflections.

--
JHG

 

[Tmoose]

"In short I am looking for a high yield strength, ductile, and cheap material
"I do not know my allowable deflection."

If i didn't need a large deflection i'm not sure I would bother with a premium material with a low E.
If this is part is to be subjected to repetitive loading then the endurance limit may prove to more important than the yield strength.
AND the geometry and surface conditions where the cantilever is anchored will be subjected to some potentially deadly stress concentrations.
I think you mentioned using a spring tapered in the "top" view and wider at the anchored end. That could vastly improve the high stress condition at the wall.
So could this -

http://image.streetrodderweb.com/f/24775198+w750+st0/0909sr_20_z+1934_plymouth_coupe+axle.jpg