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Deformation of axially loaded bar

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine the extension, due to its own weight, of the conical bar show in Fig 5.13. The bar is made of aluminum alloy [E=10,600ksi] and [tex]\gamma[/tex]=.100 lb/in3
    The bar has a 2in radius at its upper end and a length of L=20ft assume the taper of the bar is slight enough for the assumption of a uniform axial stress distribution over a cross section to be valid.

    So its basically an cone hanging off the ceiling and I need to the extension of it.

    2. Relevant equations
    [tex]\delta[/tex]=[tex]\int[/tex] [tex]\frac{Fdy}{AE}[/tex]
    3. The attempt at a solution
    Since the only force acting on the cone is its own weight:
    F=weight=[tex]\gamma[/tex]*A
    because the density times the area

    Since the it's a cone, the radius changes and therefor so does the area
    Area=[tex]\pi[/tex]*r2
    To find r, I used the similar triangles which gave me:
    2/10=r/y => r=y/10
    so: Area=[tex]\pi[/tex]*(y/10)2

    So when I plug it all in I get:
    [tex]\delta[/tex]=[tex]\int[/tex] [tex]\frac{ \gamma Aydy}{AE}[/tex]

    And the area's cancel out. I continued the problem even though I'm quite sure that the area's should not cancel out. I ended up with:

    [tex]\delta[/tex]=[tex]\frac{ \gamma y^2}{2E}[/tex]

    Upon plugging in the numbers i got an answer of: 2.7*10-4inches
    When the correct answer (according to the book) is: 9.06*10-5

    EDIT: I figured out what I did wrong, silly mistake
     
    Last edited: Sep 23, 2009
  2. jcsd
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