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Real formula for gamma(deformation) in torsion of a rod? 
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#1
Apr612, 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 pseudoright 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 


#2
Apr612, 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.



#3
Apr912, 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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