Need help from mechanical engineer on Custom Caltrac bars

[BrazilianZ28Camaro]

Hello Gents

A few months ago I built my own set of Caltracs to my street strip 71' Z28. It was much cheaper than import the originals , everything imported is way expensive here in Brazil.

I'm a dye and tool maker, so milling isn't a problem to me. I made the brackets of mid steel and the bar I "wiselly" made of duraluminum ( aeronautic applications , I believe its a 5051 alloy) to save weight.

The Caltracs worked really well with my old 370 HP 283" engine, but now I'm building a stronger engine and I'm wondering if those alloy bars will take the beating. I run with true slicks at the strip.

I haven't mechanical engineering degree and I tried to calculate the bar's critical moment of deflection with no luck.

So, I'd like to ask if somebody can calculate this to me and say: those bars will take XXX lbs before they bend. (hope I'm not asking too much )

The bar is solid, 21.5" long and 1" diameter. The rotular terminals are bolted exactly in the center of the bar.

This is the bar, please let me know if more details are needed.

http://imageshack.com/a/img23/8241/bkbx.jpg

http://imagizer.imageshack.us/v2/800x60 ... 2/d25w.jpg

http://imageshack.com/a/img198/3420/4omk.jpg


Thanks in advance.

[PackardV8]

Absolutely beautiful workmanship. There's no import duty coming this way, but if shipping weren't so prohibitive, I'd be sending you a check for some work today.

Yes, I know there's more working room below the spring than above, but the problem with all the CalTrac, TractionMaster type bars is they place the bar in compression rather than tension.

However, even in compression, I'd say your bars will be unlikely to bend before the rod ends.

[BrazilianZ28Camaro]

Absolutely beautiful workmanship. There's no import duty coming this way, but if shipping weren't so prohibitive, I'd be sending you a check for some work today.

Yes, I know there's more working room below the spring than above, but the problem with all the CalTrac, TractionMaster type bars is they place the bar in compression rather than tension.

However, even in compression, I'd say your bars will be unlikely to bend before the rod ends.

Thank you very much for the nice words PackardV8, I appreciate. I love to modify muscle cars to perform better, If I have time and machinery, I do the best possible.

I read the tension and compression loads have about the same values before metal deformation, but the force to bend a bar is considerably smaller.

Yes, those rod ends were the only ones I found with CW and CCW threads, and I think they are a bit small for the job also, but the bar is also long.... lets see how much they will withstand.

Thanks for reply.

[ijames]

Ignoring the rod ends and the taper, a 1" dia long round rod has:

area .7853981625 in^2
moment of inertia (I) .0490873852 in^4
section modulus (Z) .0981747703 in^3
radius of gyration (k) .25 in

The length of 21.5" gives a slenderness ratio of 86 and I used the Amer. Institute of Steel Construction formula,
and assuming ends free to rotate:

Assuming a yield tensile strength for mild steel of 36,000 psi and modulus of elasticity of 29,000,000, the critical load will be 19,155 lbs
with a max stress of 24,400 psi, well below the yield strength, and a factor of 5 safety margin gives a max rated load of 3831 lbs.

US alloy 6061T6 aluminum has a yield tensile strength of 40,000 psi and a modulus of elasticity of 10,000,000 so the critical load will be 9,191 lbs
with a max stress of 11,700 psi, also well below the yield strength, and a factor of 5 safety margin gives a max rated load of 1,838 lbs.

Wikipedia says duraluminum is similar to 2024 in the US which has (in the T3 temper) a yield tensile strength of 50,000 psi and a modulus of elasticity of
10,500,000 psi so the critical load will be 9,650 lbs, max stress of 12,300 psi, and max rated load of 1,930 lbs.

So even though 2024 is slightly stiffer and stronger, because of the shape it doesn't buy you much more. Mild steel gives double the load rating. What rating
do you need? I have no clue and I'm not a professional engineer. Oh, if you go to www.pwr-tools.com you can download the program Engineering
Power Tools and run these numbers for yourself. The basic version is fully functional so you don't have to register it if you don't want to, but I think the plus
version is worth every penny of the $50 price.

Oh, look up the radial load rating on the rod ends you used and compare that to the column rating of 9,000 lbs to see if your ends are big enough.

[Back to lurking now ]

[BrazilianZ28Camaro]

Ignoring the rod ends and the taper, a 1" dia long round rod has:

area .7853981625 in^2
moment of inertia (I) .0490873852 in^4
section modulus (Z) .0981747703 in^3
radius of gyration (k) .25 in

The length of 21.5" gives a slenderness ratio of 86 and I used the Amer. Institute of Steel Construction formula,
and assuming ends free to rotate:

Assuming a yield tensile strength for mild steel of 36,000 psi and modulus of elasticity of 29,000,000, the critical load will be 19,155 lbs
with a max stress of 24,400 psi, well below the yield strength, and a factor of 5 safety margin gives a max rated load of 3831 lbs.

US alloy 6061T6 aluminum has a yield tensile strength of 40,000 psi and a modulus of elasticity of 10,000,000 so the critical load will be 9,191 lbs
with a max stress of 11,700 psi, also well below the yield strength, and a factor of 5 safety margin gives a max rated load of 1,838 lbs.

Wikipedia says duraluminum is similar to 2024 in the US which has (in the T3 temper) a yield tensile strength of 50,000 psi and a modulus of elasticity of
10,500,000 psi so the critical load will be 9,650 lbs, max stress of 12,300 psi, and max rated load of 1,930 lbs.

So even though 2024 is slightly stiffer and stronger, because of the shape it doesn't buy you much more. Mild steel gives double the load rating. What rating
do you need? I have no clue and I'm not a professional engineer. Oh, if you go to http://www.pwr-tools.com you can download the program Engineering
Power Tools and run these numbers for yourself. The basic version is fully functional so you don't have to register it if you don't want to, but I think the plus
version is worth every penny of the $50 price.

Oh, look up the radial load rating on the rod ends you used and compare that to the column rating of 9,000 lbs to see if your ends are big enough.

[Back to lurking now ]

Thank you very much for take the time to calculate this to me! Helped alot!

I'll download the software for sure!

To be honest, I'll try to calculate the compression load those bars will be subjected to when the car launch, but probably they will hold well.

Thanks

[ijames]

You're welcome. By big enough on the rod ends, I just meant to make sure that they are rated at about the same as the rod so that neither was a glaring weak spot.

[BrazilianZ28Camaro]

You're welcome. By big enough on the rod ends, I just meant to make sure that they are rated at about the same as the rod so that neither was a glaring weak spot.

I just searched on the google and the rod ends are rated at 10.000 lbs. Should be ok I guess.

Thanks again.

[Brian P]

Judging by the picture, it looks like the axle-end attachment point of the new rod is about halfway between the leaf spring and the ground. I will continue with that ASSumption in the interest of making the following discussion simple and it looks like it won't be too far from the real situation. I will also make the ASSumption that the leaf spring is now acting like a tension member (sort of like an upper link on a link-type suspension).

If we draw a free-body diagram then the tension on the leaf spring will be the same as the forward thrust at the ground (split between the two sides) and the compression on the rod (split between the two sides) will be double whatever the forward thrust is.

If the car weighs 3000 lbs and can accelerate at 1 g then the two rods have to take up 6000 lbs between them i.e. 3000 lbs each.

Spike loads caused by wheel-hopping etc will be outside this (so don't do that).

Obviously there are some approximations here but given that the above discussion included a built-in safety factor of 5, I think it will be okay.

P.S. Looks really good, too.

[BrazilianZ28Camaro]

Judging by the picture, it looks like the axle-end attachment point of the new rod is about halfway between the leaf spring and the ground. I will continue with that ASSumption in the interest of making the following discussion simple and it looks like it won't be too far from the real situation. I will also make the ASSumption that the leaf spring is now acting like a tension member (sort of like an upper link on a link-type suspension).

If we draw a free-body diagram then the tension on the leaf spring will be the same as the forward thrust at the ground (split between the two sides) and the compression on the rod (split between the two sides) will be double whatever the forward thrust is.

If the car weighs 3000 lbs and can accelerate at 1 g then the two rods have to take up 6000 lbs between them i.e. 3000 lbs each.

Spike loads caused by wheel-hopping etc will be outside this (so don't do that).

Obviously there are some approximations here but given that the above discussion included a built-in safety factor of 5, I think it will be okay.

P.S. Looks really good, too.

Brian P,

I'm far from a suspension expert but I'm having a hard time to see the leaf spring foward section being in tension while it accelerate the car and supporting the car weight.

I imagine the leaf spring being compressed at launch, both by the car weight, acceleration forces and longitudinal twist the rear axle impart on them.

The Caltracs are suppose to cancel this longitudinal axle twist and the load they will take will be the difference between the torque applied to the axle at launch and the force necessary to twist the springs .

Thanks for reply!

[Brian P]

The springs will be in tension because the rods will be in (more or less) twice as much compression ...

Another way to think about it is this. Imagine the axle trying to twist backwards from torque reaction. The link bar (on the bottom) stops this happening because between the links and the leaf spring, they apply equal and opposite counter-torque to the axle. Link in compression ... spring in tension.

Doing it this way stops the torque reaction from having to be absorbed by the leaf spring. The leaf spring no longer has to "wrap up" in order to take up that torque reaction.

Overall, removing the responsibility of "wrapping up" from the leaf springs and taking it up instead by direct compression and tension loads is A Very Good Thing - as long as the geometry is right, so that the links are not fighting with the way the springs want to deflect. (For example, you would not want to pivot those links from the shackle end of the spring, nor would you want the radius to be too different from the path the spring wants to follow.)

The torque reaction is not the only thing acting on the axle; there is also the net forward force applied at the tire contact patches, that is why the compression loads being applied to those links are higher than the tension loads on the spring (the difference between the two being exactly the net forward force being applied at the tires).

[2seater]

Lovely work. Not really related, but isn't that directional tire running backwards?

[BrazilianZ28Camaro]

The springs will be in tension because the rods will be in (more or less) twice as much compression ...

Another way to think about it is this. Imagine the axle trying to twist backwards from torque reaction. The link bar (on the bottom) stops this happening because between the links and the leaf spring, they apply equal and opposite counter-torque to the axle. Link in compression ... spring in tension.

Doing it this way stops the torque reaction from having to be absorbed by the leaf spring. The leaf spring no longer has to "wrap up" in order to take up that torque reaction.

In other words, you're telling me the axle twist force is equal to the foward force the tires are aplying to the spring/bars at the exact moment the car is accelerated ?

I tought the leaf spring alone would absorb some of the twist force (like a pinnion snubber) and make the car go foward. The diference between those forces would be the compression the bars are subject to.

Thanks for the insight Brian, much appreciated.

[BrazilianZ28Camaro]

Lovely work. Not really related, but isn't that directional tire running backwards?

I was pretty sure somebody would ask about the tire direction.

Long story short, at that time I was racing on a crappy prepped -asphalt- 1/8 mile drag strip.

Initially I had these tires on the correct direction, and after several attempts to improve out the hole traction, I took a close look at the tires and noticed the edges of the treads wore forming a radius.I tought that won't do any good and inverted the tires direction to get new edges exposed to the rotational direction that matters. I didn't drive or ran on the rain, so It should be ok.

Too bad the car ran the same sixty foot time as before! At least I learned something.(Trunk ballast is king with street radials)

Thanks for reply

[BrazilianZ28Camaro]

Looks like you friends liked my work, maybe you'dlike more pics ?

http://imageshack.com/a/img691/7995/q5um.jpg
http://imageshack.com/a/img811/9429/s3e2.jpg
http://imageshack.com/a/img834/2483/0h62.jpg
http://imageshack.com/a/img607/3982/igun.jpg
http://imageshack.com/a/img534/8559/up67.jpg


Thanks for replies again!

[John Wallace]

Try this web site:

Billy Shope's Calculators


Here's a web page made by HO Racing back in the day, on working with leaf spring suspension.

Leaf Spring Calculation