Shell Elements Assembly - VALUABLE QUESTION 1

[umcp74]

Dear readers,

I will make the long story short, so consider the sample shell geometry shown: ( 3 x 4 elements - 4 nodes/element)

4-------8--------12-------16--------20
| | | | |
3-------7--------11-------15--------19
| | | | |
2-------6--------10-------14--------18
| | | | |
1-------5---------9-------13--------17

if the horizontal is the tangential direction (r*theta)and the vertical is the axial direction (so you see a top view here). How do you assemble a circular shell?

My answer is that the right end nodes have to take the same numbers as the left end nodes (to form a closed circle). Thus,

4-------8--------12-------16--------4
| | | | |
3-------7--------11-------15--------3
| | | | |
2-------6--------10-------14--------2
| | | | |
1-------5---------9-------13--------1

and the integration limits for an element in the circumferential direction is 2*pi/4, such that each elements has an angle of pi/2, and the 4 elements will form 2*pi to close the circle.

Any thing wrong with what I mentioned here? Please make sure that you know the right answer before replying.

Thanks for your time!

 

[rb1957]

4 elements joined in a circular manner would make a square tube, no?

 

[umcp74]

the strain-displacement relations are for the shell theory ( defined with the radius of curvature and angle - Cylindrical coordinates). The elements will form square tube if they are defined with flat (plate) elements in cartesian coordinates.

 

[rb1957]

but a square is a bad approximation for a circle, if that's what you want ...

 

[johnhors]

Cylindrical or cartesian coordinates, what difference that does make to the element? You have specified 4 node quadrilateral elements, which can only have straight edges, the shape functions of which are defined by the element geometry and not by any coordinate system used in node location. As rb1957 is pointing out, four straight lines don't define a very good circle! You will require a much higher mesh density.

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[umcp74]

The picture is just a sample mesh. The real mesh is different of course and has 100's elements. My question is about the assembly. Do you see any thing wrong with that? is my analogy right?

 

[rb1957]

are you creating a mesh by hand ? man, i haven't done that is about 10 years !!

of course, if the elements form a closed loop then the last set of nodes numbers is the same as the first. if you don't use the same nodes then the loop isn't closed (even if the nodes are coincident).