69427

I'm trying to do some A-B comparison between steel and aluminum halfshafts. I recently put a C4 suspension under my '69, and I'm trying to mathematically compare the original 2.5" diameter steel halfshafts and the 3.5" aluminum C4 shafts I have now. I have a C3 shaft that I had cut up for setting up the new suspension, so I can measure the wall thickness of that, and I'm currently trying to find out the wall thickness of C4 shafts. Boiling it down, the main info I need is what sort of "spring" constant value do I use for aluminum? I have a formula in one of my chassis books for determining the torsion rate for tubular steel bars, but I need a value to plug in for aluminum "bars". Can anybody help me out here?

Thanks,
Mike

BrianCunningham

You're going to need the wall thicknesses for this one.

You best bet is to go to a driveshaft shop.

gkmccready

http://www.matweb.com/ likely has what you're looking for.

Kubs

There are different constants for different types of aluminum. I dont know if the half shafts are pure Al.

davidfarmer

you will need to completely re-figure for the change in diameter, the thickness, AND the metal properties (modulus of elasticity). If the diameter and thickness were the same, you could just compare G (modulus), but in your case you'll need to be more specific since the geometric componenet (moment of inertia) goes up with the square of the radius, but more or less linear with the thickness.

I guess REALLY simplifying, if you assume the thickness is the same, the 40% increase in diameter will cause a 100% increase in stiffness.......you can then approximate the overall stiffness proportianally to the Modulus and thickness.

btw, are you more concerned about Stiffness, or strength....they aren't necessarily at all the same!

sothpaw2

The twist angle of the cross section due to a torque is

theta= TL / GJ

If you want a stiffness such that

T= K * theta

Then your stiffness is GJ / L

L=shaft length
G=shear modulus of material (alum. or steel--look up in text book or online)

J= polar moment of inertia of circular cross section. I think it's pi*radius&^4 but you should look it up.

If hollow tube--take J for the outer diameter then calc for the inner diameter and subtract the smaller value.

davidfarmer

I think he's right, is relative to radius raised to the 4th power, NOT squared. Therefore the increase in radias will quadruple your stiffness, not double it. Thickness and modulus should still be more or less linear.

Hard to keep my polar moments, area moments, and mass moments straight after 20 years.

And again, stiffness and strength are NOT the same....

69427

I'm looking to compare the torsional stiffness of these two parts.

StealthLT4

You'd need the exact material, diameter, and wall thickness of each. Then it's pretty easy to do some FEA in a CAD program (I have SolidWorks).

69427

I looked at a couple websites, and got some numbers for steel and aluminum.

Steel 29 x 10e6 psi
Aluminum 10.2 x 10e6 psi

So, once I do the math to determine the stiffness difference/ratio due to the sizes (diameter, wall thickness, and length), I'm assuming I can then just multiply that by the 10.2/29 ratio to see the final comparison ratio. Is that correct?

Thanks for all the help so far. (I'm a EE design engineer, and my mechanics courses were a few decades ago. )

StealthLT4

Those numbers look like the ultimate tensile strength of each material, which doesn't really matter here, unless you want to compare the failure torque on each one. For torsional rigidity, you need the modulus of elasticity (E) of each (IIRC). I'm going out of town for the week tomorrow, but if I had the wall thicknesses of each and the type of Aluminum, it would honestly take about 10 minutes to create a model and test it, and get a torsional rigidity for each shaft, in ft-lb per degree of rotation. Though I'm sure you could just do hand calculations and get the same thing.

Solofast

Those are numbers for modulus, which is what you want.

Doing a model????? This is a quick hand crank, why would you do a model for this, and really these parts are all so stiff relative to the amount of torsional windup that torsional stiffness is of no consequence. That is, there is really no practical amount of torsional deflection that will effect the drive train here. If you look at small diameter solid shaft and crank in the amount of torque in first gear, then you could see some deflection here, but with a large diameter tube, you aren't going to see any measureable twist in the shaft anyway.

Just use it and enjoy. It isn't worth the effort to calculate and compare...

ghoffman

But as always, what usually is the most important is not being considered here, i.e. the boundary conditions. Doing the tube section comparison is an easy hand calc, but the U joint mounts are also a significant contribution to overall stiffness and strength. What needs to be considered is the total "strain energy density", with the key word here being "density". I see this error in so many products designed by non professionals (structurally speaking). For example, to have a huge truss section on a harness bar with sheet metal end attachments and flat stock struts is like having the anchor chain for the Nimitz having one link from a dog chain. All of the strain energy will be dumped into the one small link.

69427

Well, old habits die slowly. Throughout my career I had to prove to my management that any new designs of mine were as robust or better than what I was trying to replace. The original steel halfshafts were troublefree for a lot of years of high speed fun, and, if the aluminum replacements are stronger, I can relax. If they're not as strong, then I have to worry about what was my previous safety margin, and do these new parts drop below that margin.
I got a call from the driveshaft shop manager this afternoon. He said that they had the first shaft cut apart from the yoke, and the shortened tubing is a loose fit on the yoke. He speculated that the tubing is around .095 wall thickness, while the yokes are made for .125 wall tubing. We agreed to change direction and cut out both yokes and then install .125 wall tubing in there. It's a little more hassle and delay, but I've got plenty to keep me busy on the car in the meantime. Right now I'm working on getting the steering rack hooked up, plumbing the brake system, and a hundred other minor assemblies.

OKsweetrides

Sothpaw's post will help you.


(T * r) / J <---- shear stress

theta = (T*L) / (J *G) <---- rotation bar endures


also, T= spring constant * theta <---you can find the spring constant


oh, since it's aluminum, be conservative with respect to the endurance limit.

I second what ghoffman said.

however, we can assume that for the system you have, the part your machining is the "weak link" i.e. the rest of the system exceeds the capability of the part. Else, you won't know the actual performance since something else will break and leave you with little information (will it break? or is it strong enough?)

sothpaw2

Those are the WRONG numbers! You need the shear modulus, G. Those are the Young's (or elastic) modulii, E. Those are used for tension not torsion.

Post the radii and the Lengths and I'll do the comparison when I'm at work. I really will be as simple as (GJ/L)alum vs (GJ/L) steel.
If L is the same, the aluminum will be stiffer if it is just a little bigger in diameter.

69427

I appreciate the help. I got the narrowed/new halfshafts back from the driveline shop the other day, and put them in this weekend.



Instead of .125 wall tubing, they put in .138 (IIRC). I measured a few shafts I have, and here are the numbers:
3.5" Aluminum, .138" wall, and 12 1/2 inch tubing length between yokes. (My custom shafts)
3.0" steel, .098" wall, 10 1/4 inch tubing length between yokes. (C3)
2.5" steel, .098"? wall, 10 1/4 inch tubing length between yokes. (C3)

I am guessing about the early C3 shaft wall thickness, as I don't have a cut apart unit. I made my best estimate of the actual tubing length by measuring from the centerlines of the yoke welds.

Again, I appreciate whatever results and data you can supply.

sothpaw2

I'll do it tomorrow but I almost guarantee the aluminum will win because it's almost 20% larger in dia. and 40% thicker. The Young's modulus (don't recall for shear) is only 3 times higher for steel vs aluminum.