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Real formula for gamma(deformation) in torsion of a rod?

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KalShaen
#1
Apr6-12, 06:23 PM
P: 1
Okay, so I'm studying for a Mechanics of Materials final at the moment, and I am reviewing the chapter on torsion.
I was reading through the given formulae, and I stumbled across one that I could not fully visualize (or simplify): the formula for gamma.

Based on my understanding and from what I know from before, shouldn't the formula:

[itex]\gamma[/itex]=[itex]\rho[/itex][itex]\phi[/itex] / L (for the deformation of a rod under torsion)

actualy be sin([itex]\gamma[/itex]) = [itex]\rho[/itex][itex]\phi[/itex]/L ?

Correct me if I am wrong, but aren't we considering rho*phi to be the arc length of the end deformation, and considering that arc length to be the opposite side of the angle gamma in the pseudo-right triangle formed when the rod deforms?

I have come with two possible reasons for my confusion
1) The book is not using the sine function because the angle is very very small.
2) I am failing to realize what kind of geometric scenario is occuring between the angle gamma and the arc length rho*phi.

Please get back to me with a response as soon as possible so I can move on from being miserably confused! :P
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OldEngr63
#2
Apr6-12, 08:08 PM
P: 343
The answer is in your first possible reason. This is a small angle approximation, and actually, it is a very good approximation, provided the length L is several times rho.
Pooty
#3
Apr9-12, 05:26 PM
P: 34
Yes, nice observation, but as OldEngr63 pointed out, this is for small rotations. So sin(gamma) would approximately be gamma.


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