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Constraining a Tube

L

Loftarasa

New member
Hello,

I am trying to do a simulation on an aluminum tube. It is meant to be placed in the ocean, so I've just got a pressure load on it.

I am not sure how to constrain it, however. Picture the tube just dropped into the water and slowing heading towards the bottom. I want to know at what depth it will fail, so I am adjusting the pressure to represent different depths. In my mind, the tube is not really constrained anywhere - but I need to select something in order to do a simulation. Any ideas?
 
Use the minimum constraints that eliminate rigid-body motion. Typically three points with different constraints on them. One is a 123, then a 12, and finally a 3 usually works. Draw the FBD to make sure your reaction loads are zero.
 
Just dropping in. Could you please explain what do you mean by 123, 12 and 3? Also, what is supposed to be FBD?


Thanks.
Edited by: skraba
 
Why not divide the tube and use symmetric constraints?


Most of mine analysis is with pressure loads and I try to use symmetry as much as possible to avoid over constraining and creating false stress levels.
Same problem, just need to stop rigid body motion.


/Anders
 
Hi skraba,


For locating/aligning any object in space, amethod known as 123 method is usually used. For example, to locate the position of centre of hole on a flat rectangular plate, the steps are used.

  1. <LI>The plate is made parallel to the machine table by using 3 points</LI>
    <LI>then the plate is aligned with the transverse axis of the machine table using 2 points on the edge of the plate. One of the 2 palnessharing this edge would be parallel to the machine table plane and the other would be perpendicular to the machine table</LI>
    <LI>Finally, the translation of the plate in the third direction can be arrested using 1 point on the edge of the plate which is perpendicular to the previous edge.</LI>


FBD means Free Body Diagram. It is one of the most common method of analysing any physical system. A representative figure of the component under analysis is drawing which is isolated from its physical surroundings. However, all interactions with the surroundings are drawn in the form of force, moment, translation and rotation vectors.
 
FBD = Free Body Diagram - a common technique for finding the equations for equilibrium.


123456 etc. are the designations for the degrees of freedom to be removed from a node. A 123 constraint removes the 3 translational degrees of freedom (leaving essentially a ball joint). This is common terminology in FEM and structural analysis.
 
Thanks for the help guys. I was unfamiliar with the 123 constraint because I am a FEM noob and have not covered it in school yet, but I am on a work term and trying to work something out in Mechanica.

So as I understand it, I should insert three constrained points into my model. 1 point will restrict translational motion in xyz, one will restrict translational motion in xy, and one will restrict translational motion in z. Is this what you mean by 123, 12, 3?
 
Hi Loftarasa,


In case you are dealing with tube with open ends, then there is no need to do any FE analysis. This is because, the stress developed inside the tube is directly proportional to the pressure difference across its wall and not dependent on the the absolute pressure of the surroundings. The pressure difference across this tube would be corresponding pressure eaquivalent to a water column of height equal to the wall thickness. Please remember that 10.3 m of water column corresponds to 1 bar of pressure and the the wall thickness would much much lesser compared to 10.3m, the pressure difference across its wall would be very small. Hence, for all practicle purposes, the tube would never fail.


In case, the tube ends are closed then mode of failure would be implosion. And before this could happen, buckling takes place. Depending on the ration of diameter to the wall thickness, you can choose suitable formula to calculate the buckling load factor which gives the failure limit for the tube.


Regards,


Anand
 
Thanks Anand. In realty, the tube has closed ends. However, I am currently modeling it with open ends just because I want to learn how to properly constrain it. I tried the 123, 12, 3 thing but it doesn't want to work properly. Is there any specific place I should be putting these three points?
 
Your problem can be assumed to be a plane stress problem. What you need is a quarter section of the tube and you need to constrain the horizontal and vertical faces for symmetricity in the following manner


Horizontal face&n bsp;&n bsp; Vertical Face


Tx=free&nbsp ;&nbsp ;&nbsp ; Tx=0


Ty=0&n bsp;&n bsp;&n bsp;&n bsp;Ty=Free


Tz=0&n bsp;&n bsp;&n bsp;&n bsp;Tz=0


Rx=0&n bsp;&n bsp;&n bsp;&n bsp; Rx=0


Ry=0&n bsp;&n bsp;&n bsp;&n bsp;Ry=0


Rz=Free&nbsp ;&nbsp ;&nbsp ;&nbsp ;Rz=Free


This is assuming that, XY plane of your co-ordinate system corresponds to the plane of the tube section and X axis coinides with East-West line and Y axis coincides with North-South line.


Regards,


Anand
 
I think this is a prime opportunity to use the inertia relief option. I do not have a lot of experience with this option but this is what it is designed to do. If there are others who do not agree please let me know.


When you are setting up your model, do not apply any constraints but keep your load. When you setup your analysis there is a check box for inertia relief. Thats all it should take.


An easy way to find the pressure to cause yielding: Create a parameter called "PRESSURE" and give it a start value. When you create your pressure load, rather than punching in a value, type in PRESSURE. After you run your initial analysis, run a sensitivity study and vary your pressure through a set up values. You will then be able to identify a stress value for yielding much easier than re-running several analyses.


Here is some background information for inertia relief via PTC:
Using this option, Mechanica analyzes your model as if it were floating freely in space, without any constraints, but with the loads applied.
Note: You can also use the Inertia Relief option to analyze an underconstrained model. However, if the model already has some constraint sets, then Mechanica ignores these constraints during the analysis.
During an analysis with the Inertia Relief option selected, Mechanica internally creates a new Cartesian coordinate system (UCS) and defines a constraint set containing three-point constraints with respect to the newly created UCS. Mechanica also automatically applies body loads that balance your applied loads.
The three-point constraints affect the displacement solution but not the stress solution.
An analysis with inertia relief should always run fine, as long as there is non-zero stiffness for all the six degrees of freedom between two bodies.
Before you run an analysis with inertia relief, Mechanica asks you whether you want error detection to be performed. If you do, Mechanica checks for various modeling conditions including the presence of multiple bodies in the model. If Mechanica encounters more than one body, then it displays a message indicating the number of separate or disjoint bodies that it finds. You can use this information to determine whether parts you thought were connected are truly connected. If you see an unexpected number of bodies, you may want to cancel the analysis or study and correct the assembly.
For an analysis with Inertia relief ensure that the model does not have more than one disjoint body. If multiple disjoint bodies exist in your model, then the analysis fails with an underconstrained error. To run an inertia relief analysis with multiple disjoint bodies, ensure that all the disjoint bodies are connected in such a way that there is no relative motion between the bodies. If the connections are such that relative motion exists between any two bodies in the model, then the inertia relief analysis fails with an underconstrained error. For example, if two bodies are connected by bolts, then the inertia relief analysis fails if the bolts have no stiffness for some degrees of freedom.
 
Regardless of the constraint & load sets, one still needs to know their success criteria! What constitutes failure for this analysis?
 
Sorry to hi-jack this thread but where can the "inertia relief" option be found (or in what version of mechanica was it introduced)?


/Anders
 
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