Flat spring design FEA 1

[mirelavus]

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.

 

[desertfox]

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

 

[mirelavus]

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.

 

[desertfox]

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

 

[desertfox]

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

 

[desertfox]

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

 

[mirelavus]

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

 

[desertfox]

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

 

[mirelavus]

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.

 

[mirelavus]

Also the coordonate systems are off a little as when I am printed graphics they do not start from 0 point. I also forgot to attach the new dims.

 

[desertfox]

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

 

[megabill]

Not sure if this will help, but here is a link to a website that offers power spring design software.

http://www.spiral-spring.com

There is also some technical information on the about power springs tab.