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Flat spring design FEA

Flat spring design FEA

Flat spring design FEA

(OP)
Hi all,
I am new to the forum and I have a problem to solve in FEA.
I am starting using it since last week and my hand calculation is not giving the same answer with FEA.
Here is the spring desing (I attached it) and this has a preload force of 16 LBS at a preload length of .16". What will be the constant k of the spring and the free length?
The main problem is to find out if my spring design works without failing. Thank you for your answer.

RE: Flat spring design FEA

hi mirelavus

Perhaps you can tell us how we can answer your problem without you giving us any dimensions of the spring or material and what does the preload length of 0.16" mean, does it mean 0.16" off the flattened state or 0.16" off some arc shape.

desertfox

RE: Flat spring design FEA

(OP)
Sorry about that...I will attach a drawing of the spring, that includes the flat dimensions of the spring. The spring has to sit in a tube, so 0,16" will be from the flattened state. The space is really small. Material is 1075 steel or similar, hardened, marquench and tempered to Rc 42-48, thickness is .015-.017". Thank you very much

RE: Flat spring design FEA

I'm not sure any hand calc would give a valid answer, that appears to be a doubly-curved section, and likely undergoes partial buckling when deformed (flattened).  If so, I'd have to be a pretty darn good FEA user to get valid results, and wouldn't believe them until validated by testing.  Good luck!

RE: Flat spring design FEA

(OP)
btueblood, I used F=4Efbt^2/L^3 (for load in LBS) and S=3FL/2bt^2 (for stress in PSI)formulas.
I considered B-B section as flat in my calc, just to get a close answer. In FEA, I considered it as an unsupported beam, I did not treat this as a spring. In fact I do not know how should I treat it in FEA, as I am a new user and I only use this ones in a while since we have just 1 license of it at work. I really think that I should rely on testing it, but I would first have to get close answers from both hand calc and FEA, which I did not get. Here is another picture of where is used. This sits on the curved base with two pins and all assembly sits inside a tube. There are three of these assemblies around inside the tube. The distance between the tube ID and the curved base is .16", so this will be the preload length of the spring. In this position spring will have to have 16 LBS of force, so this is the preload force of spring. I got a k=56LBS/in as a spring rate and a free length of .443" but I am not sure about my answers.

RE: Flat spring design FEA

Hi mirelavus
Thanks for the drawing.
Why is the beam curved across the the .825" width ie 1" rad on section B-B? if it was flat I have a method that would get you a stiffness figure assuming its not over stressed.
Alternatively if I can assume its flat then I'll have a look and see if I can get a ballpark figure.
Let me know if you want to go down this road.

Regards

desertfox

RE: Flat spring design FEA

(OP)
desertfox, it is curved because it sits on a curved surface that has 1" ID (a tube). You can assume is flat, this is what I assumed too. I know it will be a difference on testing, but still, that will give me a closer answer than nothing at all.

RE: Flat spring design FEA

(OP)
Sorry, I ment to say R=1" not diameter.

RE: Flat spring design FEA

(OP)
I got area from the spring model as it is now: A=2.537in^2, so I know this is correct as it is taken automatically from the model. I know the pressure inside the black rubber tube is 80PSI and that pushes the spring from under, and compresses it as it sits inside another tube, so that will give me a force of 202.96 LBSf and 20.71 LBS load on spring at the preload length=.16".
See the picture

RE: Flat spring design FEA

hi mirelavus

I am looking at your spring now I'll post what I find later,
thanks for the information.

desertfox

RE: Flat spring design FEA

Hi mirelavus

I need to check my figures again, but if they are correct your spring is far to weak, with a 16lb load according to my figures it will be flattened.
What is the f symbol in your formula for force is it allowable stress?

desertfox

RE: Flat spring design FEA

(OP)
the f will be the deformation of the spring. E=30x10^6 PSI (young Modulus), b=width, t=thickness, L=length. Here I am not sure if I have to use the L and b from the curled view or from the flat view. Considered as a supported beam, my FEA shows that the spring is too stiff (it will not move at all under load of 21 LBS or force of 202.96 lbm*in/sec^2). I did not use any moment. I considered only translation on X (free) and rotation about z (.8116 rad). What did you used?

RE: Flat spring design FEA

(OP)
I ment to say unsuported beam (what is wrong with me?)

RE: Flat spring design FEA

(OP)
can you tell me how should I approach this problem? What do you think it should be approached?

RE: Flat spring design FEA

Hi mirelavus

I can't get the value of F that you get using your formula.
I started to model the spring with two ends on the ground and applying a load at the centre of the arch.
If you can post your workings out it may help, as soon as I re-check my figures I'll post again.

desertfox

RE: Flat spring design FEA

(OP)
That means I have to change the geometry of the spring.
A=2.537in^2 (area of the curved spring)--from model
P=80PSI (pressure inside the tube)--> F (force)=202.96 lbf--> Load = 21 LB. That is the minium force or load that I need at the preload length (.16"). At this point the spring has to support a weight of 880LB (they are 96 springs, so for each one it will be another 9LB)
Preload length will be the entire space that I have between tubes: .16"
So the total load will be 21+9=30 LB for each spring. That means the spring will be compressed from the preload length another amount (I don't know how much, but I don't want this to fail or to be flattened). With this geometry, what will be the spring rate and the stroke? Will this going to work? FEA shows no movement at all. Is it too stiff? I am also in the green area.

RE: Flat spring design FEA

(OP)
I got my FEA to work, it was a unit problem, I had it wrong. The force was calculated wrong too. It does not have to support any load. The force is P*A=202.96lbf, so the load will be 21 LB. In FEA, I need to apply the foce at the center at the arch, not the load. Sorry about the confusion.

RE: Flat spring design FEA

Hi mirelavus

Well its bad news, I checked my figures and your spring will go flat, its far to weak.
If you consider a simply supported flat beam of 2.25" long, 0.016" thick and 0.825" wide and apply a 1 lbf at centre and calculate the deflection it comes to 1.75".
I don't understand your comments:- It doesn't have to support any load. The force is 202.96lbf, so it will be 21 lb. I need to apply the force at the center of the arch.
Can you post your workings out so I can see exactly what you have done.

Regards

desertfox

RE: Flat spring design FEA

(OP)
I applied mass instead of force, meaning I applied 21 LBM instead of 202.96 LBF in my FEA calculations that's why my wrong calc.
As soon I got a chance I can post pictures with FEA, but now I am at home.
Thank you.

RE: Flat spring design FEA

Hi mirelavus

I found an error in my calculations to, however I have corrected this now and I can tell you that your spring stiffness is 37 lbf/inch which means that to get 0.16" deflection you need only 5.92 lbf which suggests your spring is to soft.
I checked my figures in mathcad using calculus and strain energy.
When I did the mathcad calculation, I didn't allow for the cutouts to go round the tubes so the spring came out marginally stiffer at 40 lbf/inch but in the graphical intergration I allowed for the cutouts, anyway the figures from both methods are within 10% when compared.
More importantly I used 65000lbs/in for the tensile stress of your steel and according to my calculations if the load at the centre of the arch exceeds 2 lbf your stress will be over the tensile strength.
I haven't allowed for the curvature across the spring width as we agreed to analsye it as a flat, neither have I allowed for any stress raisers which may occur at the bends or the cutouts.
Another piece of advice I used 0.016" thickness material for the calculation which was the average of the sizes you have given, however if you use 0,017" thick and compare it with 0.015" thick the stiffness goes up by a cubed law and not surprisingly a spring made from 0.017" is almost 1.5 times stiffer than one made from 0.015".
Your major problem with this spring is stress so its back to the drawing board.

regards

desertfox

RE: Flat spring design FEA

(OP)
desertfox, I've done FEA all day today and I modified the geometry of the spring including the free height, until I came up in FEA with a spring rate of 100 lb/in. I set the material thickness to .016 too, because this is the thickness of material I can get. I scooped it more in so now looks more like a wrench and I changed the free height. The job that this spring has is to collapse the lugs inside the tube when is no air in the black rubber tube. The rubber tube tends to stretch so the spring has to push it back. The area the pressure applies to is around 2.5in^2, like I said earlier in my postings, so that being said, with the known pressure of 80 PSI inside the rubber tube, will give me a force of F=P*A=2.5*80=200 lbf. I will use just 10% of this force because the tubes are supported both ends, so that will give me a boundary condition for force in FEA on Y of -20. For constraits, on X the condition will be on edges, translation on X of +.72 and -.72in. I found this by taken the difference between the length of the flattened spring and the length of the curved spring and divided by two, because it will translate on X and -X. On Y there will be no constraints, just load of -20 lbf.
Then I plotted the graphic of displacements for several different heights. Idon't have the space here to explain what I did, but I think this time will be OK. I still don't know how you concluded that the spring is too soft, because I got it too stiff and I have to scoop it out and modify the geometry many times to get the rate down to 100lb/in which is close to the goal of 70 lb/in which i initially had. Thank you for yoour posting. You were the only one that took the time to think about my spring. It wasn't a simple analysis.

RE: Flat spring design FEA

Hi mirelavus

I have uploaded my calculations assuming there is no curvature across the width of the spring,also I have ignored the cutouts in the spring where the tubes sit and that allowed for a constant "I" value across the spring.
The spring being symmetrical you only need to analyse one half of the spring which is what I have done.
The fact I have not allowed for the cutouts means the spring should be stiffer than with the cutouts but it still only comes to around 40 lbs/in for stiffness.
I have used metric units and converted the stiffness at the end,you will also find I have omitted the results of the intergration, but I have stated what needs to be intergrated and I checked them through Mathcad and you could probably do the same.
My graphical intergration method including the cutouts results in a stiffness of 37 lb/in stiffness which is in the same ball park, so unless I have made the same mistake in each method I am convinced my calculations are correct.

Regards

desertfox

RE: Flat spring design FEA

Hi mirelavus

In addition the bending stress on this spring can be obtained using:-

stress = M*y/I

if you consider the stress to be the tensile strength of the material you will see if you put more than 2 lbf on the centre of the arch you will exceed your tensile strength.
I don't know what changes you have made to your spring since you posted the original drawing, however I hope you look at my posts before you go ahead with your design if only to serve as a reference.

regards

desertfox

RE: Flat spring design FEA

Hi mirelavus

If you can please feed back if you could and let me now if you have solved your problem or if you have any questions about my calculations.

cheers

desertfox

RE: Flat spring design FEA

(OP)
I agree your calculations are fine, but I think we have asumed too much idealization to the real model.
Thank you for your postings, very valuable. I concludes 36 lbs/in without cutouts myself by using some simple supported spring calculations and some beam idealizations. I think I have to start a new thread about what will be the bounderies conditions in Pro Mechanica for this application. Maybe I did not put them right.
Thank you again and I attached my conclusion apart from FEA

RE: Flat spring design FEA

Hi mirelavus

Thanks for the feedback, I have looked at your conclusions which as you say closely reflect my findings.
So you believe your constraints within the original model may have given you the false results.
I don't know what your latest spring looks like or how different it is from the original posted, however if you post your latest one I'll look at it for you.

regards

desertfox

RE: Flat spring design FEA

(OP)
Hi desertfox. I work with Pro Mechanica. Maybe my boundary conditions are wrong and/or edges that I selected for constraints are wrong, something is got to be wrong in FEA as we both have close foundings by different ways of hand calculations. Here is the last model.I used metric units (mm, N, KG, s, C. I had to scoope out the model as it was giving me an enormous stiffness. Now, as it is, it gives me 100 LB/in stiffness. Something is wrong here.

RE: Flat spring design FEA

Hi mirelavus

I can see you scalloped out the spring centre but I can't see any dimensions for those centre scallops?
I agree there is no way that spring is 100lb/in on spring rate.
Now if you need a preload of say 20lbf at 0.16 in deflection, then you need a stiffness of 125lbf/in.
So my advice would be to go back to your original spring and increase the thickness and do the calculations again and increase thickness till we get to the rate you need.

desertfox

RE: Flat spring design FEA

Not sure if this will help, but here is a link to a website that offers power spring design software.  www.spiral-spring.com
There is also some technical information on the about power springs tab.

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