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Deformation of a Pipe

Deformation of a Pipe

Deformation of a Pipe

(OP)
I am trying to figure out a first principle model for the following:
A steel cannula/pipe (hollow tube) is gripped by a force covering 1/3 of the cannula/pipe length. What is the deformation of the cannula/pipe due to the force.

Cannula Length L = 0.75 inches, gripped length = 0.25inches, Force = (arbitrary; use 1 pound for arguements sake), outside diameter=0.02inches, inside diameter= 0.012inches


RE: Deformation of a Pipe

physics337,
I think your use of the word "cannula" as a synonym for industrial pipe is incorrect.

can·nu·la
[kan-yuh-luh]
noun, plural can·nu·las, can·nu·lae [kan-yuh-lee].
Surgery - a metal tube for insertion into the body to draw off fluid or to introduce medication
- a narrow tube for insertion into a bodily cavity, as for draining off fluid, introducing medication, etc

Origin - from Latin: a small reed, from canna a reed]

prognosis: Lead or Lag

RE: Deformation of a Pipe

(OP)
I am looking at a cannula, but I think when looking at the basic physical properties, it will act similiar to a pipe. Hence the references in the questions. Thanks for the comment

RE: Deformation of a Pipe

At your very, very tiny dimensions (length, OD, ID) and the very, very tiny forces involved in such a small area, industrial "pipe" and industrial "forces" (thousands of pounds over many dozen or hundred feet) are the WRONG analysis to compare results or to compare resistance forces and tube reactions.

Don't do it.

Rather, at 0.75 long, the whole tube will be forced sideways into the tissue or injection site. The tube itself will react as if it were solid pushing the tissue aside like dry spaghetti can be picked up at one end.

. Or, the tube will deform completely (like a wet piece of spaghetti) and completely close up.

What material is the tube inserted into?
For how long?
What material is the tube made up of?
What forces are actually involved at what locations along the tube?

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