Radial displacement in hollow cylinders

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SUMMARY

The discussion centers on the radial displacement in axisymmetric long tubes, specifically addressing why radial displacement is independent of the axial coordinate. The simplification is justified by the principles of symmetry and uniform loading along the length of the tube. The participants reference "Mechanics of Fibrous Composites" by Carl T. Herakovich and provide links to additional resources on elasticity in polar coordinates. The consensus emphasizes the importance of understanding the implications of axisymmetry in mechanical analysis.

PREREQUISITES
  • Understanding of axisymmetric structures
  • Familiarity with cylindrical-polar coordinates
  • Knowledge of displacement fields in solid mechanics
  • Basic principles of elasticity and symmetry in mechanics
NEXT STEPS
  • Study the full expression for compatibility equations in polar coordinates
  • Explore the implications of axisymmetric loading in mechanical systems
  • Review section 10.2.1 and equation 10.3 of "Mechanics of Fibrous Composites" by Carl T. Herakovich
  • Practice problems involving radial displacement in cylindrical-polar coordinates
USEFUL FOR

Mechanical engineers, students studying solid mechanics, and researchers focusing on elasticity and displacement analysis in axisymmetric structures.

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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