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

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

 

[mirelavus]

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

 

[btrueblood]

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!

 

[mirelavus]

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.

 

[mirelavus]

sorry, I forgot to attach the picture like I said

 

[desertfox]

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

 

[mirelavus]

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.

 

[mirelavus]

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

 

[mirelavus]

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

 

[desertfox]

hi mirelavus

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

desertfox

 

[desertfox]

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

 

[mirelavus]

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?

 

[mirelavus]

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

 

[mirelavus]

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

 

[desertfox]

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

 

[mirelavus]

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.

 

[mirelavus]

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.

 

[desertfox]

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

 

[mirelavus]

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.