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Gundrilled De Dion drive shafts


edmandsd

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Dave E, I keep seeing messages about bulk buy of shafts and then messages about drilling shafts. Which are we doing? If Elite are looking for a pair of shafts to work off of to create a new ligher drilled version, great you can use mine.

 

If they want a pair of the newest to cost out how much to drill them then.... I would have to side with Chelmspeed on the cost of the new shafts then adding the cost of drilling being a tad OTT.

 

Either way they are welcolme to my shafts to use as prototypes. Won't be the newest but 30% savings in unsprung weight is huge!.

 

 

 

Whoever said ... If it aint broke, Don't fix it...

 

Just doesn't get the point!

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Anthony - This is the ideal solution - Graham's right though 1 will suffice if both are equal length etc..........and Elite will only need it long enough to take measurements etc. This has got to be the best solution as the later shafts are apparantly a lighter/improved design to start with.

 

I was never assuming someone would go and buy a new pair of latest spec shafts just for these purposes - Obviously a new build such as Anthony's car is the ideal solution or someone who's prepared to remove an existing (new) shaft for a couple of weeks.

 

Thanks Anthony I'll be in touch offline.

 

 

 

Home of BDR700

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They are different part numbers but as far as I know the only difference is one has LH thread on the hub stub axle and one has RH thread. Perhaps someone can lay a tape across a pair to confirm the lengths are the same. Mine are in Bromsgrove at present so I can't.
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Stick them on the scales too then we have a baseline for the newer shafts. These are the newer ones I presume, the ones with three notches in the metal sleeves at each end rather than plain cylindrical metal sleeves.

 

My diff's in the chassis as is the de dion so I guess measuring diff to ear is as good as measuring the shafts themselves. First job tonight then.

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Have been away for a while but the simple equation needed to calculate shear stress in a tubular components is:

 

Shear stress = 16 x T x D / Pi * (D^4 - d^4)

 

T is torque is outer diameter and d is inner diameter.

 

Good Luck with shafts.

 

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Another way of getting to the same answer is:

 

Ssmax = 2T/(Pi * r1^3) for a solid shaft and...

Ssmax = 2Tr1/(Pi * (r1^4 - r2^4)) for a tubular shaft where...

 

Ssmax = maximum sheer stress (Psi)

T = Applied torque (in-lb)

r1 = Outside radius of bar

r2 = Inside radius of bar

 

Therefore for a solid shaft with 1.125 inch diameter and 400 ft lb's (4,800 in-lb) applied to it 17,169 psi sheer stress will result, and for the same shaft with a 5/8th inch hole puched through it 18,977 psi sheer stress will result....or only 10.5% more.

 

Cheers

David.

 

Home of BDR700

 

Edited by - edmandsd on 28 Feb 2003 16:20:00

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