# Radial displacement in hollow cylinders

1. May 7, 2013

### farrukh.hafeez

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. May 7, 2013

### Simon Bridge

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).

3. May 12, 2013

### farrukh.hafeez

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

4. May 12, 2013

### Simon Bridge

Last edited by a moderator: May 6, 2017