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calculating torsion - EDIT Got a bar and basic calcs now

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calculating torsion - EDIT Got a bar and basic calcs now

Post by rockcrawler31 »

I'm designing a currie sway bar for the red devil, and trying to calculate what length and thickness to make the arms.

Can any one tell me how to calculate the rate of a torsion bar for a given length and wire thickness? I need to reduce it to a force per degrees of twist rate so i can calculate what length arms to make and what kind of force it will transfer back to the chassis. I.E. Nm per degree

It seem nobody makes torsion bars in this country or know jack shit about them as they all seem to be imported. I've tried Carrols EFS, dobinsons, king, and a bunch of boutique spring manufacturers but they all either don't know, don't care or import only.

I'm thinking of making the arms from 12 -15 mm laser cut plate for a guesstimated length of about 400mm. does anyone have any input on how to calculate what is sufficient without going over board.

Also, any suggestions what size tube and material type to make the links to the chassis out of? I'm thinking something around 20mm OD with a 5mm wall thickness so i can tap and fit a M14 Heim joint at each end.
Last edited by rockcrawler31 on Sun May 02, 2010 4:17 pm, edited 1 time in total.
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Post by KiwiBacon »

Formulas for torsion of tubes and bars are what you need. Getting a good connection without any stress risers (i.e. good places to break) to your side arms will be difficult.

2/3rds down this page is the formule you need:
http://www.eformulae.com/engineering/st ... erials.php

If you know the forces you need at the end of the links (i.e. roll stiffness) and maximum deflection it'll be reasonably straightforward to calculate. Plenty of people here can check your calcs once you've made a start.
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Post by Slunnie »

KiwiBacon wrote:Formulas for torsion of tubes and bars are what you need. Getting a good connection without any stress risers (i.e. good places to break) to your side arms will be difficult.

2/3rds down this page is the formule you need:
http://www.eformulae.com/engineering/st ... erials.php

If you know the forces you need at the end of the links (i.e. roll stiffness) and maximum deflection it'll be reasonably straightforward to calculate. Plenty of people here can check your calcs once you've made a start.
Surely you could just use standard axle design to prevent stress raisers.
Cheers
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Post by rockcrawler31 »

KiwiBacon wrote:Formulas for torsion of tubes and bars are what you need. Getting a good connection without any stress risers (i.e. good places to break) to your side arms will be difficult.

2/3rds down this page is the formule you need:
http://www.eformulae.com/engineering/st ... erials.php

If you know the forces you need at the end of the links (i.e. roll stiffness) and maximum deflection it'll be reasonably straightforward to calculate. Plenty of people here can check your calcs once you've made a start.
Thanks heaps for that

I'm more or less totally plagiarising the currie so i'm hoping to have the spline broached into the arms assuming the engineering company can work off an existing spline to copy.

I'm thinking that since none of the spring companies i have approached are capable or even interested in making a torsion bar i will appropriate an existing bar from an IFS vehicle. That takes care of the stress riser issue since i would assume that they would already be designed appropriately

Kiwi - the one thing i can't figure out how to work out is how much force i want induced as roll stiffness. I mean once i know the rate i can calculate a linear force on the links and arms, but how would i translate that into an appropriate roll stiffness for my particular set up?
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Post by rockcrawler31 »

I just checked out that formula

Seemed a bit gibberish to me.

How does it differentiate between material types such as mild, tensile and spring steel.?
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Post by rockcrawler31 »

I found this little gem but i'm wondering if a spring is a spring is a spring as it doesn't take into acount material type.

http://www.alternativecarparts.com/util ... nbars.html

I worked out that a bar just wider than my chassis at 44 inches, 30mm thick with 14in (355mm) arms would give a linear force of 270Lb/Ft.

Given that the axle moves down 12 inches and up by about the same is it safe to say that full articulation would give me the equivalent of 540Lbs sitting on the side of the car?
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Post by KiwiBacon »

Slunnie. Yes you can just use a standard axle type design. Spline each and and reduce the diameter along the length to less than the root of the splines.

But that's a lot of work requiring machinery not many people have. That's what I meant by difficult. I should have said difficult or expensive unless as already noted you can find one that works.

Rockcrawler, all steels are pretty much the same elasticity (~207 GPa) regardless of strength. The strength determines how far they can bend and still return to their original shape.

I'll run your figures tomorrow.
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Post by DamTriton »

rockcrawler31 wrote:I found this little gem but i'm wondering if a spring is a spring is a spring as it doesn't take into acount material type.

http://www.alternativecarparts.com/util ... nbars.html

I worked out that a bar just wider than my chassis at 44 inches, 30mm thick with 14in (355mm) arms would give a linear force of 270Lb/Ft.

Given that the axle moves down 12 inches and up by about the same is it safe to say that full articulation would give me the equivalent of 540Lbs sitting on the side of the car?
270lb.ft seems a bit low at a guestimate from the figures you supplied. (Hint - keep all measurements the same ie, metres and kg and Newtons)

You also need to look at the plastic and elastic limits of the metal you are dealing with, ie the ability to return to it's original form without being "bent" by the forces you applied. Just because it is spring steel, doesn't mean it can't be bent. 12" of movement off a 14" arm connected to a 1.25" solid bar is going to break something as it is about 50 degrees of twist along the length of the torsion bar.

Contact a spring maker (Kings/Dobinsons/Ironman/etc) who should be able to do the maths for you and specify the correct grade of steel, or conversely may have the rates of some IFS torsion bars that you could use as a guide.
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Post by bender4865 »

Undergrad engineer here, whats vehicle weight and I'll do up some spreadsheet for ya mate. :)

EDIT: started on spreadsheet, just let me know exactly what you want to know and i'll make sure it's got everything you want.
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Post by rockcrawler31 »

bender4865 wrote:Undergrad engineer here, whats vehicle weight and I'll do up some spreadsheet for ya mate. :)

EDIT: started on spreadsheet, just let me know exactly what you want to know and i'll make sure it's got everything you want.

cheers mate, what kind of data do you need? I can go out tomorrow and get measurements off the truck.
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Post by pyrohamish »

T=27.416 x10^-9 (a x d^4 x G)/L

T is torsion in kNm
a is angle of twist in degrees
d is diameter of shaft in mm
G is shear modulus of the material in GPa (steel is typically around 80GPa)
L is length of bar in metres
note the units
a quick example

a= 5 degrees
d= 32 mm
G= 80 GPa
L=1.5m

T= 27.416*10^-9*(5*32^4*80)/1.5
T= 7.67 kNm
This equation is in theory are correct. This won't tell you if you have gone past the elastic limit of the material.

Use at your own risk,I am not liable for anything to do with this. Please consult a professional.
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Post by pyrohamish »

oops forgot to mention, set a=1 degree if you want T as kNm/degree
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Post by rockcrawler31 »

bloody awesome mate. A couple of questions

in that equation what does the symbol ^ mean? Is that 'to the power of'?
If so how do you work ^-9? Is that the 9th sqare root?

Also is there a way to determine the elastic limit for spring steel in this application for a given set of dimensions??
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Post by KiwiBacon »

Right, crunching some numbers. Working in metric. I'm including my working so if I get anything wrong others can check it.

With 350mm arms and 300mm of travel you need 51 deg of twist on your torsion bar.
Bar is 1110mm long, 30mm diameter,
Picked a spring steel of Matweb for material properties:
http://www.matweb.com/search/DataSheet. ... 849&ckck=1
Elasticity (E) = 210 GPa
Poissons ratio (v) = 0.313

Torsion angle (A) = Torque (T)*Length (L)/Polar Moment of Inertia (J)/Shear modulus (G)

Rearranged so we can find the torque.

T = A*J*G/L

J = pi*d^4/64
J = pi*30^4/64
J = 39760 mm^4

G = E/(2*(1-v))
G = 210/(2*(1-0.313)
G = 153 GPa
G = 153,000 N/mm^2

A is measured in radians, we have it in degrees.
51/360*2*pi = 0.89 radians

Now we can run the torsion formula

T = 0.89*39760*153,000/1110
T = 4,880,000 Nmm
T = 4,880 Nm

At the end of your 350mm link that would create a force of:
F = T/L
F = 4880/0.35
F = 13,940N
This is about 1.4 ton of force. Way out of what you need so I won't bother checking the stress in the torsion bar.

To reduce this you need much longer links to gain more leverage on the torsion rod and reduce the angle of twist you need from it.
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Post by rockcrawler31 »

holy shit kiwi you've gone to a lot of trouble there, i appreciate it. I still need to figure out what the ^ symbol in the calcs means so i can substitute values to play around with configurations
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Post by pyrohamish »

^ means to the power of.
If you set the calculator in windows to scientific mode you can have x^y
the reason for the constant at the start accounts for all the non SI units.

Tell me the steel you are going to use and I should be able to find out the elastic limit and get another equation for it
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Post by KiwiBacon »

rockcrawler31 wrote:holy shit kiwi you've gone to a lot of trouble there, i appreciate it. I still need to figure out what the ^ symbol in the calcs means so i can substitute values to play around with configurations
That's simply power of. It's easier to put in than finding the alt-codes to draw superscripts properly.
x^2 means x², x^3 means x³ etc.
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Post by rockcrawler31 »

kiwi just a question on how that force would tanslate to real life effects. That 1.4 tonne would that translate to the same amount weight sitting on a point where th arm joins the torsion bar? i'm just trying to work backwards to dtermine how much torque (roll stiffness) i want transferred back into the chassis.
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Post by rockcrawler31 »

pyrohamish wrote:
Tell me the steel you are going to use and I should be able to find out the elastic limit and get another equation for it
erm, roundy roundy longy springy steel i reckons :)

seriously, unless i can find a spring manufacturer in this country who makes torsion bars i'm going to be appropriating one out of an existing IFS system so whatever standard spring steel is if there is such a thing
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Post by KiwiBacon »

rockcrawler31 wrote:kiwi just a question on how that force would tanslate to real life effects. That 1.4 tonne would that translate to the same amount weight sitting on a point where th arm joins the torsion bar? i'm just trying to work backwards to dtermine how much torque (roll stiffness) i want transferred back into the chassis.
If the support bearings/bushings were near the outside of that bar, then there would be 1.4 ton difference in upward force on them.
I don't have any reference for what size the forces need to be in a 4wd swaybar. I do have some rangerover sway bars kicking around somewhere, but they'd need FEA to get a good idea of the rates as the geometry makes hand calculations difficult. I do have FEA but I really should do some paying work today. :lol:
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Post by DamTriton »

rockcrawler31 wrote:kiwi just a question on how that force would tanslate to real life effects. That 1.4 tonne would that translate to the same amount weight sitting on a point where th arm joins the torsion bar? i'm just trying to work backwards to dtermine how much torque (roll stiffness) i want transferred back into the chassis.
Put simply, you would have to put 1.4 ton of pressure on that 350mm (14") arm in order to get it to twist your desired 51 degrees (12") across its overall length.
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Post by rockcrawler31 »

KiwiBacon wrote:
rockcrawler31 wrote:kiwi just a question on how that force would tanslate to real life effects. That 1.4 tonne would that translate to the same amount weight sitting on a point where th arm joins the torsion bar? i'm just trying to work backwards to dtermine how much torque (roll stiffness) i want transferred back into the chassis.
If the support bearings/bushings were near the outside of that bar, then there would be 1.4 ton difference in upward force on them.
I don't have any reference for what size the forces need to be in a 4wd swaybar. I do have some rangerover sway bars kicking around somewhere, but they'd need FEA to get a good idea of the rates as the geometry makes hand calculations difficult. I do have FEA but I really should do some paying work today. :lol:
awesome mate. thanks heaps for everything everyone. It's given me a bit of a starting point
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Post by rockcrawler31 »

Ok

So i now have a torsion bar. I believe it's from a 100 series and looks like a stocker.

The main bar in the middle is 1020mm long and 29mm diameter.

I ramped the truck and found that with my taller coils the travel measured from level with the bottom of the chassis rails to the top of the axle tubes is 180mm on compression and 660mm on droop. There is still about 100mm of up travel on compression if i had lower rate coils and if i move my shocks outboard a little i may get a little more droop, but it's a pretty good indication. So lets say for ease of calcs i have 500mm of linear movement that the arms have to move through.

I can fit arms 580mm in length. so that means the bar will move through 51 degrees of twist.

So knowing this is it a matter of substitution in the above calculations to work out the effective linear rate? I'm going to try and find out how much difference in rate having multiple holes 25mm apart would make for adjustability.

can anyone tell me if 51-55 degrees will push it past the plastic point of the steel?

I'm thinking that even if the arms are too long at 580mm to give an effective torsion amount i can sort that by going to a larger bar from standard

Can anyone tell me if it's possible for and engineering mob to effectively produce a female spline from a male example or do they really need original specifications? If they can't is there any reason i can't have the spline milled off and a new one put in its place?
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Post by rockcrawler31 »

I've just run kiwi's calcs based on what i have using substitution

with 580mm arms, 51 degrees of twist i get 814kg/force

555mm = 850kg/force
530mm = 891kg/force

does that still seem like too much?
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Post by KiwiBacon »

rockcrawler31 wrote:I've just run kiwi's calcs based on what i have using substitution

with 580mm arms, 51 degrees of twist i get 814kg/force

555mm = 850kg/force
530mm = 891kg/force

does that still seem like too much?
I think a good starting point would be 100-200kg of force at maximum articulation. Even that amount is going to reduce the articulation you get significantly.

Anyway you can put a big torque wrench on your chassis somewhere? Doing that and pulling down on the torque wrench will give you an idea of how much the extra roll-stiffness will affect your suspension travel.
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Post by rockcrawler31 »

KiwiBacon wrote:
rockcrawler31 wrote:I've just run kiwi's calcs based on what i have using substitution

with 580mm arms, 51 degrees of twist i get 814kg/force

555mm = 850kg/force
530mm = 891kg/force

does that still seem like too much?
I think a good starting point would be 100-200kg of force at maximum articulation. Even that amount is going to reduce the articulation you get significantly.

Anyway you can put a big torque wrench on your chassis somewhere? Doing that and pulling down on the torque wrench will give you an idea of how much the extra roll-stiffness will affect your suspension travel.
100kg doesn't seem like a lot. I'm 110kg and when it's flexed up if i hang like buggery on the corner of the tray it's no real difference.

I think i'll mock up some plate to sit under the chassis and mount the bar through that temporarily so i can test it all out before going and holesawing mounts through the chassis.

I have found a mob in sydney who can supply spring barstock at 1mm increments but i'll have to have the splines cut into it and there'll be no thicker sections at the end for more splines like an automotive torsion bar. But if worst comes to worst i can use that to get a thinner bar.
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Post by DamTriton »

rockcrawler31 wrote:I've just run kiwi's calcs based on what i have using substitution

with 580mm arms, 51 degrees of twist i get 814kg/force

555mm = 850kg/force
530mm = 891kg/force

does that still seem like too much?
Something's not right in the initial setup going by you OP request of 12" of articulation up and down ie, 24" total (assumes attached to very end of axle, ie no extra leverage arms in the equations)

I make a total deflection with the 580mm arm and 51 degrees deflection of about 500mm (19.2")

2 x sin 25.5 x 580

Bisecting your 51 degree deflection and joining the two ends of the arm (one 25.5 degree rightangle triangle reflected)

With the 530mm arm it would be proportionately less at 456mm (17.9")

-------------

You can increase the effective leverage force applied to the torsion bar by moving the attatchment points on the axle inboard.

Have you included the proportionate difference in axle (wheel track) length vs torsion bar pickup points (additional leverage and reduction of necesary travel) into the calculations?
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Post by rockcrawler31 »

DAMKIA wrote:
rockcrawler31 wrote:I've just run kiwi's calcs based on what i have using substitution

with 580mm arms, 51 degrees of twist i get 814kg/force

555mm = 850kg/force
530mm = 891kg/force

does that still seem like too much?
Something's not right in the initial setup going by you OP request of 12" of articulation up and down ie, 24" total (assumes attached to very end of axle, ie no extra leverage arms in the equations)

I make a total deflection with the 580mm arm and 51 degrees deflection of about 500mm (19.2")

2 x sin 25.5 x 580

Bisecting your 51 degree deflection and joining the two ends of the arm (one 25.5 degree rightangle triangle reflected)

With the 530mm arm it would be proportionately less at 456mm (17.9")

-------------

You can increase the effective leverage force applied to the torsion bar by moving the attatchment points on the axle inboard.

Have you included the proportionate difference in axle (wheel track) length vs torsion bar pickup points (additional leverage and reduction of necesary travel) into the calculations?
The original post of 12" was a guesstimate of travel. today i went out and did some proper measurements of travel on the ramp. The figures just indicated were measured on the axle tube vertically below where the splines of the torsion bar and the arms would be. the arms have to go outside the chassis rails as there is just not the space to go inside the rails as well as i'd like the bar to go through the chassis as i see it being the strongest and simplest way to mount and transmit force back to the chassis rails. Therefore the links and arms really have to be square and vertical instead of kicking inwards more to thbe centre of the axle tube.
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Post by KiwiBacon »

rockcrawler31 wrote:
KiwiBacon wrote:
rockcrawler31 wrote:I've just run kiwi's calcs based on what i have using substitution

with 580mm arms, 51 degrees of twist i get 814kg/force

555mm = 850kg/force
530mm = 891kg/force

does that still seem like too much?
I think a good starting point would be 100-200kg of force at maximum articulation. Even that amount is going to reduce the articulation you get significantly.

Anyway you can put a big torque wrench on your chassis somewhere? Doing that and pulling down on the torque wrench will give you an idea of how much the extra roll-stiffness will affect your suspension travel.
100kg doesn't seem like a lot. I'm 110kg and when it's flexed up if i hang like buggery on the corner of the tray it's no real difference.

I think i'll mock up some plate to sit under the chassis and mount the bar through that temporarily so i can test it all out before going and holesawing mounts through the chassis.

I have found a mob in sydney who can supply spring barstock at 1mm increments but i'll have to have the splines cut into it and there'll be no thicker sections at the end for more splines like an automotive torsion bar. But if worst comes to worst i can use that to get a thinner bar.
Do you know your current spring rates?
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Post by rockcrawler31 »

Christ i don't know. I'd have to contact Shane from suspension stuff who sold them to me.

The rears are 5" single rate heavy duty coils for an 80 series. I could go and measure the wire size to get a bit of an idea.

The front are 5" flexi coils

Dobinsons all round, and probably too heavy a rate really, i originally had 5" flexi coils on the rear but they didn't provide nearly enough body roll control on cross slopes so i went heavier single rate coils. The problem was that on the flat the ride was compromised and hard as buggery as a result - Way more than enough to hold up the weight of the car on the flat but nowhere near enough (due to inboarding) to stiffen the rear on any kind of side slopes or articulation.

I'm hoping that with the sway bar i can run longer - lower rate coils all round so the rear stops dislocating and banging every time they do that, and so that the front is free to compress and flex instead of being the main roll stiffness for the truck

why do you ask by the way?
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