
12022006, 06:33 PM

#1

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Driveshaft stresses
For a project I am analyzing the forces on a drive shaft (just the shaft no joints or splines), wikipedia states the shaft is under shear and torsion. I see torsion but i dont see how the shaft is under shear.
Comments?
(see the reponses already posted)
http://forum.miata.net/vb/showthread.php?t=210368



12022006, 07:06 PM

#2

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Other than shearing the splines, which you say aren't included, I, too, am only seeing torsion.
Maybe the wiki was written by one of those lowrider guys, dump the hydraulics too fast on one of those parking curb things, and there could be some shearing.
(I'm only in my 1st semester, my engineering materials class didn't focus on forces all that much)



12022006, 07:23 PM

#3

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Im graduating in the fall, get accustomed for forces and stresses you'll play with them alot



12022006, 09:01 PM

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Torsion is realized by shearing stresses, but not the same shearing stresses as beam shear. These are the other offdiagonal terms in the shear tensor. If you want more details, let me know. If you need derivations or just end results for failure mechanisms of shear in a solid cylinder, it's pretty easy to do...just use the shear relationship to the Laplace of the stres function of the cross section, solve the eigenproblem for the fleshed stress tensor, apply an easy failure mechanism theory, probably Mises, then boom, you're done. I can go on and on about solid and continuum mechanics, but I won't bore everybody (one of the guys on my phd committee is a continuum mechanics freak).



12022006, 09:08 PM

#5

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**** this ****...out!
I'm going back to the "hey look at this car" threads, no more stephen hawking for me this evening.



12022006, 09:44 PM

#6

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heh, i can't tell you much about that stuff either, but i can tell you from experiance that the 1.6L driveshaft doesn't like turbos much. my driveshaft went to poo poo after i turboed... heres the link to what was wrong
http://www.clubroadster.net/forum/viewtopic.php?t=4004



12032006, 12:51 AM

#7

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Forutnatly i dont have to deal with matrices just simple Tr/j junk.
Atlanta, thanks for the information, I understand what your saying (i understand the words, i dont have experience with that)
If theres only a torque applied how can you have a shear stress?



12032006, 12:53 AM

#8

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Or how can you have shear stress with U joints in place?
Remember wiki can be written by any jackass, and the editors are most likely not engineers.



12032006, 09:10 AM

#9

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Loki, there are only two types of stresses, axial and shear. For any stress state in 3D, there are 3 axial shear stresses, and 6 shear stresses. As I said above, the shear stresses due to an applied torsion is not along the same axis as the traditional shear stresses resulting from forces on a beam or similar.
The easiest (albeit slightly wrong and oversimplified) way to see this is by drawing out the infinitesimal element in 3D. Then, apply a "torque" to the axis you want to take as the axial axis (the axis not chosen as the strong/major and weak/minor axis in the crosssection). As you can see, this "torque" is nothing but a moment in a different axis. Draw in the stresses that will counteract the torque (keep in mind you have two shear stresses and one axial stress on each of the cube's plane (but 3 planes are enough to define the shear tensor), and that this is an infinitesimal element). Now if you apply a horizontal force on the element and draw in those resulting shears, you'll see that they are along different axes. Now, you can't really do anything with this, but it helps see how an external torsion is resisted by an internal shear.
I just looked at the wikipedia article for Torsion (mechanics) and there is nothing wrong with it, aside from being a generalized case for objects with dual planes of symmetry about the crosssection (which your cylinder is, so it's ok).
Don't make the mistake confusing matrices and tensors. Very different animals. While you're not dealing with matrices, you ARE dealing with tensors, whether it looks like it or not. Stresses, in reality, cannot be defined without using tensors. Yours is a 3x3 stress tensor though, with only 4 nonzero entries (stress symmetry implies each stress has a pair, so only 2 unique stresses), and because of dualaxis symmetry of your crosssection, both of those entries are equal. The important thing to realize here is that any nondiagonal stress in the stress tensor is a shear stress, but they cannot be idealized in the infinitesimal biaxial stress element you're used to seeing; you need a triaxial element.
Is it confusing you that there is a shear stress without a shear force? If you want, you can use Mohr's circle to transform the equated shear stresses to principle stresses and work from there. But don't make the mistake of using the 2D Mohr's circle here, as that has traditional shear as the yaxis. You'd have to use a 3D Mohr's circle, which I'm not sure is touched on in undergraduate.
If this still isn't clear, let me know.



12032006, 09:30 AM

#10

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I'm with Hustler. Too much
The only thing I will worry about in shear is stockings(pantyhose for you younger guys)
I'll let Lola worry about torsion in his/her kinky world.:gay:
Anybody need a 1.6 driveshaft? When this one sells:
http://www.miataturbo.net/forums/showthread.php?t=5917
I've got one also.



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