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

  1. Apr 6, 2012 #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
     
  2. jcsd
  3. Apr 6, 2012 #2

    OldEngr63

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    Gold Member

    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.
     
  4. Apr 9, 2012 #3
    Yes, nice observation, but as OldEngr63 pointed out, this is for small rotations. So sin(gamma) would approximately be gamma.
     
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