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Finding increased length after deformation

  1. Nov 29, 2009 #1
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    the strain, as a function of the angle is K*sin2(x)

    now i know that the change in length is the integral of the strain

    =[tex]\int[/tex]K*sin2(x)dx from 0->2pi

    =K/2*[tex]\int[/tex]1-cos(2x)dx

    =K/2*(2pi - 0.5*sin(4pi) )

    =K*pi


    but the answer says K*pi*R

    where does the R come from? i realise that the change in length should be dependant on the radius, but mathematically how do i come to that?
     
  2. jcsd
  3. Nov 29, 2009 #2

    Mapes

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    You're working with [itex]\epsilon_\theta[/itex], an angular strain. When you integrate it, you get an increase in angle, not length. Know what I mean?
     
  4. Nov 29, 2009 #3
    thats what i thought, but what does that mean, are there not still 360 degrees?? does it mean that each radian is now longer than 1/2pi of the circumference of the original circle, sort of like the length of an arc??
     
  5. Nov 29, 2009 #4

    Mapes

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    Right. The equation doesn't know that the material is connected in a hoop. To fix this, use the differential element for segment length, [itex]R\,d\theta[/itex].
     
  6. Nov 29, 2009 #5
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