Sway bar spring rate formula

[AlexEsteves]

I'd like to get verification on whether this is the correct formula to calculate sway bar spring rate.

spring rate for solid bar =
Pi x G x D^4
--------------
16 x B^2 X L

where G= the modulus of elasticity
D= diameter
B= length of lever arm
L= length of bar

spring rate for hollow bar=
D^4 is substituted for (OD^4 - ID^4)

If these formulas are true how would one accomodate for a bend in the center section of the bar? for example: an upward (half cirlce) bend in the sway bar placed for exhaust pipe clearance?

Online research has provided conflicting formulas, I'd like to get it straight from the creditable and knowledgable engineers here. Thank you in advance for your help.

Alex Esteves, Pasadena CA.

 

[NormPeterson]

A truly precise formula would be too cumbersome to use in manual calculations even if you aren't going to be concerned with local elbow flexibilities.

Fred Puhn's formula is probably good enough for most cases - evidently it was developed from civil/structural simple beam deflection formulae, and it seems to be about 95% of the way to what considering several more deflection effects beyond what his center section torsion and arm bending give (Puhn's formula erring on the side of being a little stiffer). I'd only like to see the Young's modulus broken out separately instead of being buried in a constant that IIRC includes pi and some other constants.

Considering only the center section torsion effect could be ignoring as much as 25% of the actual straight element flexibilities that are present in the bar.

Puhn's formula

[tt]Formula for sway bar stiffness of a solid steel bar

500,000 D^4
K (lbs/in) = --------------------------------------------------
(0.4244 x A^2 x B) + (0.2264 x C^3)


B
---------------------
A | / \ C
| / \

A - Length of end perpendicular to B (torque arm - inches)
B - Length of center section (inches)
C - Length of end (inches)
D - Diameter bar (inches)[/tt]


Don't overlook things like endlink bushing compliance and motion ratio once you've got a bar-only stiffness.


Norm

 

[AlexEsteves]

Thanks for the insight Norm. endlinks will be adjustable to relieve pretension, joints will be spherical, bracket bushings will be sourced from "energy suspension" polyurethane with integrated zerk fittings in the housings. Motion ratio will be
minimal since the application will have a low degree of body roll but it will still be considered.

My application calls for tubular 4130 chrome molybdenum. since the modulus of elasticity remains the same for 4130 chrome-moly and conventional spring steel alloys like 5160, 1065 im assuming that the numerator value of 500,000 remains unchanged. However im not sure how the tubular variant would tie in. Would I replace D^4 with (OD^4 - ID^4) as stated before? Again thank you for sharing your expertise.

Alex Esteves, Pasadena CA.

 

[NormPeterson]

im assuming that the numerator value of 500,000 remains unchanged

The moduli among the various steels won't be exactly the same, but ought to be close enough to get you in the ballpark. It's not terribly difficult to to reverse-engineer Puhn's formula and find out what he did assume, probably 29E6 or 30E6 psi. I think some alloys are less (I only use 27.9E6 for certain piping materials, 28.3E6 for others), but if you need much greater accuracy than that, you'll actually need to drive the thing, observe its behavior, and tweak it from there.


Structurally, a solid round bar is a pipe with a zero diameter hole running down the middle of it.


Norm

 

[GregLocock]

Yes you have the correct modifier for a circular tube.

I'm not wild about cromoly in a sta bar, make sure it isn't going to yield.

Correcting for sensible looking bends in the main section by hand is going to be quite tricky, given that you don't know how much bar you really need proably best to go 1mm up in size from a straight tube and take it from there.

However, you could consider it as being a single thro crankshaft, and spend a bit of time working out the bending in each of the 3 cylinders, and adding that to the torsion you already have.

These days we use non linear FEA to work that out.





Cheers

Greg Locock


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[AlexEsteves]

Thank you Norm and Greg, great information as always!


Alex Esteves, Pasadena CA.

 

[sbullet86]

when i was taking vehicle dynamics in my undergrad, we used the first equation for just the torsional stiffness.

Do not forget there is also bending moment as if the sway bar is a beam. You would then add the two stiffness equations as if they were two springs:
(k1+k2)/(k1*k2)

Like the OP mentioned, there are bends in a real life application sway bar. Those are hard to accomedate for. Our motorsports instructor concluded that the equations are not practical and the best thing to do is to test for the sway bar's resulting stiffness.

This is how we tested the stiffness:
-strap the chassis down to the ground. It can not move.
-Use a corner weight scale and place it under one of the wheels of the sway bar you are testing for.
-Place a peice of wood of which you know the thickness of, and place it between the wheel, and record the weight
-Repeat with an additional peice.. repeat as needed.

You now have force/distance data points to compulate an avg spring rate.

Something things to consider: do no forget the peices of wood also have weight, and also the tire flexes so it is best to have a tireless rim, or just the hub against the scale.

Good luck.

 

[AlexEsteves]

Thanks sbullet, thats a great idea. Relatively simple and accurate enough for the application. Now I just have to find a means to strap the chassis down, perhaps commandeer a chassis dyno.

Alex Esteves, Pasadena CA.

 

[GregLocock]

rawlbolts and a few bits of steel might be a cheaper solution

Cheers

Greg Locock


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[desertfox]

Hi AlexEsteves

I found this relative to your formula:-



desertfox