buggy sway bar

superpile

superpile

Probably running in circles
Joined
Jan 17, 2006
Location
Asheville nc
Anyone know what the weight resistence is on the currie bar. I have access to some circle track bars/arms for free just need to know what weight to get. I figure it must be pretty light but don't know where to begin. My buggy weighs 2240 lbs, based on a sami motor/trans/tcase with yota axles and 42's with recentered H1 wheels. The wheels/tires weigh in around 600 lbs and the axles are 400 lbs together. So the rest is unsprung if that helps with numbers.
 
Brandon at RIDES knows a place in Cali that sells a similar setup for about half the price. I can't remember it right now. I know that's not what you asked but it could possibly help you out. As far as the Currie goes I think you would be fine with that but you could always call them and ask.
Alex
 
Yeah not really the info I'm looking for. I'm looking for the rating of the currie bar so I can get a similar bar instead of spending 400 on the kit. It's got to be light just don't know how light.
 
Sorry about that, forgot I posted on here. Actually I'm still waiting to hear back, but I'll post up once I know something.
 
From my experience most race guys label there bars by diameter. Assuming the race car bars are of the same material as the currie bar. Just go by diameter. The last currie bar I saw appeared to be around 9/16" to 5/8". I am running a 5/8" dirt track bar with a 14" arm from center of the bar to the end link. It works great for all around wheeling but is a little too stiff for full on rock crawling. Just pick a bar and play around with the end link mounting points on the arms till you get it where you want it. I would say with your rigs weight you probably should start with a smaller bar than mine.
 
MarsFab said:
From my experience most race guys label there bars by diameter. Assuming the race car bars are of the same material as the currie bar. Just go by diameter. The last currie bar I saw appeared to be around 9/16" to 5/8". I am running a 5/8" dirt track bar with a 14" arm from center of the bar to the end link. It works great for all around wheeling but is a little too stiff for full on rock crawling. Just pick a bar and play around with the end link mounting points on the arms till you get it where you want it. I would say with your rigs weight you probably should start with a smaller bar than mine.
Click to expand...
Thanks, I'll measure the arms he is giving me and see what diameter bars he has. How much does your rig weigh? I may start with a 1/2 and go from there or just swap a sami ft sway bar on, there 1/2 and I have one laying here with the heims and all.
 
I actually just read that the currie bar is 1". That seems a little larger than the one I remember but that's what it says.
 
The currie bar is right around 7/8. I tried a 3/4, 7/8, and 1" bar in the back of my bronco and ended up going with the 7/8. If you've got any weight at all to your rig then I can't imagine the 5/8 bar making much of a difference, especially if it's a 37" bar.
 
You may be right he should just go with a little larger bar and play with arm lengths till he gets it where he wants it. I mean the longer the arm the more left from right travel you will have.
 
I can get a bar 26, 28, or 36 inch wide. Will have to measure the arm length. I'll try to measure them this weekend if I can get out to the race shop and someones around.
Braxton what are you up to, hows the twins running?
 
Just remember if you get a bar thats too big you can only make the arms but so long. If you get a bar too small you can always drill a new hole closer to the pivot point to make it stiffer.
 
We've been building sway bars for a while now. We have a new system due out of the machine shop in a few weeks. It seems that there is every dynamic possible for the rest of the suspension system, but a "one size fits all" mentality for anti sway systems.

When our system is finished I'll get a yellow star. Until then if you want to know "how stiff" you can use the math that we've been wading through.


Symbols: Some of these are standard and some are made up for this particular problem.
T= Torsion on the bar
L= Length of bar
R= radius of the outside of the bar
r= radius of the inside of the bar if hallow, =0 if solid
J= Polar moment of Inertia
pi= 3.14159
G= Shear modulus, = 11,000,000 psi for nearly all steels
phi= Rotation of the bar in radians over its length.
t= Shear stress in the outside surface of for the bar
d= The length of arms on either end of the bar
F= The force applied to the ends of the arms to cause rotation, assumed to act perpendicular to the arm
z= The displacement of one are with respect to the other.
k= The stiffness or the bar as seen from the force it takes to move one arm one unit distance (say one inch) relative to the other.
f= Flexibility, this is the reciprocal of k, stiffness and is defined as the movement of one arm relative to the other due to a unit force (one pound)
Basic Formulas: These come from physics and mechanics of materials and will be combined to find the answer we need.
t= T*R/J
J= pi/2*(R^4 - r^4) assumes a round bar, if not round then the problem is way more complex as we must consider warping, and we do not want to do that trust me.
phi= T*L/(J*G)
T= F*d
Assuming small angles, in radians:
phi = sin(phi) = tan(phi)
z=phi*d
In the end we do not care about how stiff the bar is, what we care about is how stiff one arm is relative to the other. Or as I defined it k. To figure this we will place a force of one unit of force (say one pound) on the end of the arms and find out the distance they move this will give use the flexibility, which can then be inverted to get stiffness, or spring rate.
therefore,
T= 1*d
combined with,
phi= T*L/(J*G)
gives us,
phi= d*L/(J*G)
put into,
z=phi*d
gives us,
z=d^2*L/(J*G)
Remember this is the distance moved due to a unit load so, it is actually flexibility,
f=d^2*L/(J*G)
inverting gives us stiffness,
k=J*G/(d^2*L)
Okay, now we know the stiffness from side to side the question now is will we put to much force into the bar and bend it beyond recovery. This will depend on the shear strength of the bar relative t the shear stress. So lets figure t shear stress.
t= T*R/J
combined with,
T=F*d
gives,
t= F*d*R/J
Okay, that is all well and good but what should we use for F (Force)? This should be the largest force to expect from one side to the other of the jeep. What this should be is a whole other discussion.
Now, that we know t the shear stress it the steel, what is to high? The answer here depends on the type of steel. I have no clue what type of steel is used for sway bars but I would assume it is some kind of spring steel. Given that assumption I would say a max t of 60,000 psi would be okay.
One needs to be careful with units when doing these kinds of calculations. So take your time and write every number down including units.
A quick example:
Suppose we have a solid bar 1.25" in diameter. 30" long between arms. Arms are 8" long. Max expected force is 1000 lbs (I have given this number absolutely no thought, it may well be completely unreasonable).
First lets Figure J, remembering that the radius is 0.625"
J= pi/2*(R^4 - r^4)
J= pi/2*((0.625 in)^4 - (0 in)^4)
J= 0.2397 in^4 (notice I am keeping the units with the number)
Now lets figure the stiffness.
k= J*G/(d^2*L)
k= 0.2397 in^4 * 11000000 lb/in^2 /((8 in)^2 *30 in)
k= 1373 lb/in (note, this is not saying the force will ever be this high, it is saying that provided the bar does not permanently bend it would take this much force to move one inch.)
Now let make sure that we will not cause the bar to permanently bend with 1000 lbs of force on the arms. ie, shear stress is not to high.
t= F*d*R/J
t= 1000 lb * 8 in * 0.625 in/ 0.2397 in^4
t= 20860 lbs/in^2 or psi
less then 60000 psi so OK
Some general points: Notice that when figuring stiffness k, that the length of the arms is squared and the radius of the bar is to the 4th power. The length is just its self.
So,
If we double the radius of the bar the stiffness increases by 16 times.
If we double the length of the arms the stiffness decreases by 4 times.
If we halve the length of the bar stiffness increases by 2 times.
Clearly the mesurements of bar diameter need to be rather precise.
If we redid the example using a 1 inch diameter we would get.
J=0.0982 in ^4
k= 562 lb/in
t= 40700 psi
 
Aaron871 said:
We've been building sway bars for a while now. We have a new system due out of the machine shop in a few weeks. It seems that there is every dynamic possible for the rest of the suspension system, but a "one size fits all" mentality for anti sway systems.

When our system is finished I'll get a yellow star. Until then if you want to know "how stiff" you can use the math that we've been wading through.


Symbols: Some of these are standard and some are made up for this particular problem.
T= Torsion on the bar
L= Length of bar
R= radius of the outside of the bar
r= radius of the inside of the bar if hallow, =0 if solid
J= Polar moment of Inertia
pi= 3.14159
G= Shear modulus, = 11,000,000 psi for nearly all steels
phi= Rotation of the bar in radians over its length.
t= Shear stress in the outside surface of for the bar
d= The length of arms on either end of the bar
F= The force applied to the ends of the arms to cause rotation, assumed to act perpendicular to the arm
z= The displacement of one are with respect to the other.
k= The stiffness or the bar as seen from the force it takes to move one arm one unit distance (say one inch) relative to the other.
f= Flexibility, this is the reciprocal of k, stiffness and is defined as the movement of one arm relative to the other due to a unit force (one pound)
Basic Formulas: These come from physics and mechanics of materials and will be combined to find the answer we need.
t= T*R/J
J= pi/2*(R^4 - r^4) assumes a round bar, if not round then the problem is way more complex as we must consider warping, and we do not want to do that trust me.
phi= T*L/(J*G)
T= F*d
Assuming small angles, in radians:
phi = sin(phi) = tan(phi)
z=phi*d
In the end we do not care about how stiff the bar is, what we care about is how stiff one arm is relative to the other. Or as I defined it k. To figure this we will place a force of one unit of force (say one pound) on the end of the arms and find out the distance they move this will give use the flexibility, which can then be inverted to get stiffness, or spring rate.
therefore,
T= 1*d
combined with,
phi= T*L/(J*G)
gives us,
phi= d*L/(J*G)
put into,
z=phi*d
gives us,
z=d^2*L/(J*G)
Remember this is the distance moved due to a unit load so, it is actually flexibility,
f=d^2*L/(J*G)
inverting gives us stiffness,
k=J*G/(d^2*L)
Okay, now we know the stiffness from side to side the question now is will we put to much force into the bar and bend it beyond recovery. This will depend on the shear strength of the bar relative t the shear stress. So lets figure t shear stress.
t= T*R/J
combined with,
T=F*d
gives,
t= F*d*R/J
Okay, that is all well and good but what should we use for F (Force)? This should be the largest force to expect from one side to the other of the jeep. What this should be is a whole other discussion.
Now, that we know t the shear stress it the steel, what is to high? The answer here depends on the type of steel. I have no clue what type of steel is used for sway bars but I would assume it is some kind of spring steel. Given that assumption I would say a max t of 60,000 psi would be okay.
One needs to be careful with units when doing these kinds of calculations. So take your time and write every number down including units.
A quick example:
Suppose we have a solid bar 1.25" in diameter. 30" long between arms. Arms are 8" long. Max expected force is 1000 lbs (I have given this number absolutely no thought, it may well be completely unreasonable).
First lets Figure J, remembering that the radius is 0.625"
J= pi/2*(R^4 - r^4)
J= pi/2*((0.625 in)^4 - (0 in)^4)
J= 0.2397 in^4 (notice I am keeping the units with the number)
Now lets figure the stiffness.
k= J*G/(d^2*L)
k= 0.2397 in^4 * 11000000 lb/in^2 /((8 in)^2 *30 in)
k= 1373 lb/in (note, this is not saying the force will ever be this high, it is saying that provided the bar does not permanently bend it would take this much force to move one inch.)
Now let make sure that we will not cause the bar to permanently bend with 1000 lbs of force on the arms. ie, shear stress is not to high.
t= F*d*R/J
t= 1000 lb * 8 in * 0.625 in/ 0.2397 in^4
t= 20860 lbs/in^2 or psi
less then 60000 psi so OK
Some general points: Notice that when figuring stiffness k, that the length of the arms is squared and the radius of the bar is to the 4th power. The length is just its self.
So,
If we double the radius of the bar the stiffness increases by 16 times.
If we double the length of the arms the stiffness decreases by 4 times.
If we halve the length of the bar stiffness increases by 2 times.
Clearly the mesurements of bar diameter need to be rather precise.
If we redid the example using a 1 inch diameter we would get.
J=0.0982 in ^4
k= 562 lb/in
t= 40700 psi
Click to expand...

I concur
 
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