Leaf spring design 4

[claver]

Hi,

Can anyone help with pointing me in the right direction for design calculation of leaf spring capable of taking loads of 1000kg to 3000kg - For use on slow moving vehicules

Jan

 

[israelkk]

MANUAL ON DESIGN AND APPLICATION OF LEAF SPRINGD - SAE HS J788 FOURTH EDITION 1982. There maybe a newer one.

 

[GregLocock]

That would be a great place to start. It is also fairly easy to design them from scratch, using a spreadsheet and classic beam theory.

However, to get you going, here is the formula for a multileaf spring, from the Bosch Blue Book

Bending stress= 3*length*Force/(width*N*thickness^2)

N is the number of leaves in the middle, n is the number at the end

rate=(2+n/N)*E*N*width*thickness^3/(6*length^3)

This leaf spring is triangular in side view, ie each leaf is a fixed amount shorter than the one above.

The Devil is definitely in the details, which is where the SAE book will help.

Cheers

Greg Locock

 

[claver]

Hi Greg,

I have had a go at the formula you have shown. Just a few questions ….

If length, width and thickness is in m, Force in Newton’s. N is the number of leaves at the end. i.e. next to the eye … With 8 springs and 2 at the end (1 is the top spring with the eye) The other 6 are then evenly shorter …right ??

The calculated rate is then the movement in m (from the centre of the spring) for each N applied to the spring ??

E = Youngs Modulus for the material used.

Or have I got myself barking up the wrong tree here ….

Appreciate you comments on this …

Jan

 

[GregLocock]

N=8 n=2, everything else you say is correct.





Cheers

Greg Locock

 

[claver]

Hi Creg

Thanks for your reply. I think I must have a decimal error. With ….

L = 0.89m
F = 24516N
W = 0.05m
N = 8
n = 2
t = 0.04m
E = 210000 MN/m^2

The rate is 2.86m … clearly not right ….?? I must have missed something ….

Regards

Jan

 

[NormPeterson]

Maybe t = 0.004m?

Norm

 

[NormPeterson]

And I think that the formula assumes a symmetrical spring package with the axle located at the midpoint as seen in side view and defines the length "L" as being from the axle to either eye rather than being the overall length of the spring between the eyes.

Norm

 

[GregLocock]

Yes, L is perversely the half length of the spring.

I don't see how you got 2.86, I think you've got an arithmetic error.

I'd use a t of more like .015, and more leaves, for this design.

You probably want 400-2000 N/mm (roughly) as a rate

Cheers

Greg Locock

 

[claver]

It looks like I am using to many “0” in my value for E …. The error with the thickness did not help matters ! Thanks for point out that L is ½ the physical length of the spring

Regards

Jan

 

[claver]

Hi,

Can you please explain why I can only get a sensible answer when I use 210x10^6 for E. I thought E should have been 210x10^11 (in Newtons)

Regards

Jan

 

[desertfox]

Hi claver

E for steel should be 210*10^9 N/m^2 or 210*10^9 N/m^2.

regards desertfox