Some things to discuss...not sure how to interpret this actually.

Heres what I've learned.

Torsional yield strength of circular shafts is based on shear yield strength.

http://www.engineeringtoolbox.com/to...fts-d_947.html (equation 4)

Shear yield strength for all steels is about 0.6 of the yield tensile strength.

http://en.wikipedia.org/wiki/Shear_strength

If we choose 4340, there is an enormous variation in tensile strength depending on how its hardened. But lets say 180,000psi tensile yield strength.

https://books.google.com/books?id=S0...%20psi&f=false

So 0.6 of that means a shear yield strength of 108,000 psi

I measured my 90 Daytona's axle shafts and they appear to be 1.037" diameter.

So using the equation #4 linked above, we get a yield torque of 1972 ft/lbs. Thats the theoretical point a perfect shaft in perfect torsion will start to permanently deform, not break.

The breaking point would be use the ultimate shear strength, which is about 0.75 ultimate tensile strength. So if we say UTS is 230000psi (should be close), that gives an USS of 172500, and a breaking torque of 3150 ft/lbs.

In 1st gear an A568 has an overall ratio of about 11.5.

http://www.thedodgegarage.com/trans_guide.html

So if our theoretical launch is taking place with say at least 200ft lbs of torque at the flywheel, thats about 2300 ft lbs at the wheels. Split in two assuming equal traction. So 1150 ft/lbs per wheel.

If you crazy racer guys are somehow getting near full engine power at launch and its more like 400 ft lbs, thats more like 4600 ft lbs. Split in two around 2300 ft lbs per wheel.

No we throw in the reality of stress risers in our non-ideal splined axles/hub interfaces, and its starting to look like even very good 4340 axles + very high power launches are close enough to start breaking.

If we downgrade to 1040 axles, which I'm guessing the stock ones are made of, the breaking strength goes down to about 2250 ft lbs, or about 40% less than the 4340.

http://www.alloyusa.com/front-axle-shaft-kits.html

2250ft/lbs times 2, and then divided by 11.5 is 391 ft lbs...and that before stress risers. I think there are plenty of TD's putting out that much torque. And thats the breaking torque, not the yield torque.

Even if traction is definitely not 100%, during a controlled wheel-spin sub 2 second 60ft launch, I would imagine that the engine output is high enough to put out that much torque. Some 60ft to horsepower calculators could prove it as far as minimum required torque to launch so much weight 60 ft in so much time. Its got to be way up there for some of the nastier builds.

Has anyone figured out how to make the axles bigger? I'm guessing there is some kind of knuckle upgrade to something with a bigger hub, but what about the splined ends in the diff?