Cylinder support pressure in the inner area

[Lirock]

Hi,everyone:
I built a hollow cylinder which was meshed with SOLID95, and I applied pressure on the inner circumferential area. To avoid axial displacement, I constrained the axial degree freedom.
After solving, I set result coordinate system to a cylindrical coordinate system, actually the global cylindrical coordinate system.
Then the result of UX must be equal along the circumference. But it wasn't.
Why?

 

[meshparts]

Hi,

The program will do, just what you say to him...

You wrote, that you have constrained the axial degree freedom. What about the radial DOF? At least one node of the cylinder must should also be constrained in radial direction, so that the cylinder cannot move perpendiculary to the axial direction. It's just a guess...

Regards,
Alex

 

[Lirock]

Hi,Alex

I got a reality things to express the problem. When water goes into a pipe with pressure, what the radial deformation of the pipe? In this case, we can't constrain the pipe's radial freedom, right?

Regards,

Rock Li

 

[meshparts]

Of course not!

What I mean is, that the pipe could move as a RIGID body perpendiculary to the axial direction, if you do not constrain, say, one node at fist end of the pipe and o nother one at the othe end of the pipe. Also the pipe should hav one node rotational axial DOF constrained.

 

[cbrn]

Hi,
in order to completely avoid what Mihaiupb is saying, you'd better set up an axisymmetric model (supposed that the straight pipe is the only object you want to analyze...)

Regards

 

[Lirock]

Hi,Alex and cbrn
Maybe there is something misunderstood. If there is a very long pipe with a constant pressure in it, we usually see the case as a plane strain problem. So we just make a section as a model. Is it necessary to apply a radial constrain? This case is similar with the cylinder.
The following is a example I made. Could you pls check it if there are some mistakes?
/PREP7
ET,1,PLANE42
KEYOPT,1,1,0
KEYOPT,1,2,0
KEYOPT,1,3,2
KEYOPT,1,5,0
KEYOPT,1,6,0
MPTEMP,,,,,,,,
MPTEMP,1,0
MPDATA,EX,1,,2.1e11
MPDATA,PRXY,1,,0.3
PCIRC,0.05,0.04,0,360,
FLST,5,8,4,ORDE,2
FITEM,5,1
FITEM,5,-8
CM,_Y,LINE
LSEL, , , ,P51X
CM,_Y1,LINE
CMSEL,,_Y
LESIZE,_Y1, , ,20, , , , ,1
MSHAPE,0,2D
MSHKEY,0
CM,_Y,AREA
ASEL, , , , 1
CM,_Y1,AREA
CHKMSH,'AREA'
CMSEL,S,_Y
AMESH,_Y1
CMDELE,_Y
CMDELE,_Y1
CMDELE,_Y2
FINISH
/SOL
FLST,2,4,4,ORDE,2
FITEM,2,5
FITEM,2,-8
/GO
SFL,P51X,PRES,20,
allsel
solve
FINISH
/POST1
rsys,1
plnsol,u,x

 

[cbrn]

Hi,
in the plane strain case you are describing, you take a section of the virtually-infinite cylinder which is perpendicular to the axis.
Whenever the assumption of plane state of strain can not be made, then you still can operate in 2D by taking advantage of the foundamental property of the system: axisymmetry. This is a matter of fact, not an assumption.
In this case, X-axis will be the radial direction and Y-axis mandatorily the axial direction. You can use PLANE42, but with Keyopt,,3,1.
Btw, the shape of your model will simply be a rectangle at a certain distance from the y-axis. In this case, no need to constrain anything in the x-direction (you still have to constrain the y-direction, however).

Regards