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
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
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
RE: Deformation of a Pipe
cheers
RE: Deformation of a Pipe
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?