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Radial displacement in hollow cylinders

  1. May 7, 2013 #1
    For an axisymmetric long tube, it is said that radial displacement is independent of axial coordinate. What is the justification for this simplification?.
    If u, v and w are axial , tangential and radial displacements then
    u=u(x,θ,r)
    v=v(x,θ,r)
    w=w(x,θ,r)
    Where x,θ,r are polar coordinates.
    Due to axisymmetry all displacements are independent of θ. For a long tube ,also only radial displacement is independent of axial coordinate i.e. x axis. Why?

    u=u(x,r)
    v=v(x,r)
    w=w(r)?
    For example
    See section 10.2.1 and equation 10.3 of "Mechanics of Fibrous composites" by Carl T Herakovich 1999 ISBN: 978-0-471-10636-4
     
  2. jcsd
  3. May 7, 2013 #2

    Simon Bridge

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    Welcome to PF;

    Symmetry.
    http://homepages.engineering.auckla...lasticityPolar/ElasticityPolars_01_Polars.pdf

    Hmmm... I don't think so... did you leave something out?

    Try a few problems without assuming the independence and see what happens (do still use cylindrical-polar coordinates though).
     
  4. May 12, 2013 #3
    Underprogress....will ask for help if required.

    Do you know any reference which provides full expression for compatibility equations in polar coordinates by any chance?[/QUOTE]

    Thanks
     
  5. May 12, 2013 #4

    Simon Bridge

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    Last edited by a moderator: May 6, 2017
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