Radial displacement in hollow cylinders

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Discussion Overview

The discussion revolves around the justification for the assumption that radial displacement in an axisymmetric long tube is independent of the axial coordinate. Participants explore the implications of this simplification within the context of solid mechanics, specifically focusing on the behavior of hollow cylinders under various loading conditions.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the justification for the assumption that radial displacement is independent of the axial coordinate, suggesting that symmetry plays a role.
  • Another participant expresses skepticism about the independence of displacements, indicating that there may be additional factors to consider.
  • A later reply acknowledges that the assumption may hold under conditions of axisymmetry and uniform loading along the length of the tube.
  • Participants suggest exploring problems without assuming independence to better understand the implications of this simplification.
  • There is a request for references that provide full expressions for compatibility equations in polar coordinates.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the justification for the simplification. There are competing views regarding the independence of radial displacement from the axial coordinate, and the discussion remains unresolved.

Contextual Notes

Participants note the importance of axisymmetry and uniform loading conditions but do not fully resolve the implications of these assumptions. There is also a mention of the need for further exploration of problems to clarify the discussion.

farrukh.hafeez
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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
 
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Welcome to PF;

For an axisymmetric long tube, it is said that radial displacement is independent of axial coordinate. What is the justification for this simplification?.
Symmetry.
http://homepages.engineering.auckla...lasticityPolar/ElasticityPolars_01_Polars.pdf

Due to axisymmetry all displacements are independent of θ.
Hmmm... I don't think so... did you leave something out?

For a long tube ,also only radial displacement is independent of axial coordinate i.e. x axis. Why?
Try a few problems without assuming the independence and see what happens (do still use cylindrical-polar coordinates though).
 
Simon Bridge said:
Welcome to PF;

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

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

Yes, you are right. I think I didnot mention axisymmetric and uniform loading along the length

Try a few problems without assuming the independence and see what happens (do still use cylindrical-polar coordinates though).

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
 

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