Valid Basis Functions for Triangular 2D Elements: Solving for Coefficients

[ThurmanMurman]

Howdy,

I am trying to formulate a proof to show that the shape function

[N(x,y)] = [Ni(x,y) Nj(x,y) Nk(x,y)]

and the the basis functions

Ni(x,y) = (1/2A)(ai + bix + ciz)
Nj(x,y) = (1/2A)(aj + bjx + cjy)
Nk(x,y) = (1/2A)(ak + bkx + cky)

are valid for triangular, 2 dimensional elements. "A" is the element area, and I would like to solve for "A", "a", "b", and "c" in terms of x and y.

Can anyone help? Is there a better sub-forum in which to place this thread?

 

[AlephZero]

Sorry, I don't understand the question.

You seem to have written down some linear functions (though it's not obvious what the factor of (1/2A) is for).

What do you mean by "proving the shape functions are valid"?

What do you mean by "solving for A, a, b, c in terms of x and y"?

 

[ThurmanMurman]

That's how the question was posed to me. I wish I knew more about FEM so that I could provide more insightful detail.

The way I understand things, [N(x,y)] = [Ni(x,y) Nj(x,y) Nk(x,y)] is the shape function, and the other three functions are the basis functions. These shape/basis functions should prove to be true when dealing with a 2 dimensional linear triangle plot. "A" is the area of each finite element, but I'm not sure why (1/2A) it is included in front of each of the basis functions. I think that the shape function describes three, non-linear points in x-y space (defining a triangle, right?), and the basis functions link these points in some way?

I don't know if I said anything new there, or just reiterated what I wrote before...

 

[SteamKing]

The equations you have shown are for a linear triangular finite element. The derivation of the shape functions can be found from many different sources; one is included here:
http://www.rose-hulman.edu/~fine/FE2004/Class2/Notes2.pdf

In calculating with these quantities, one must know the (x,y) coordinates of the three vertices of the triangle. This allows for the calculation of the area A of the triangle and the determination of the shape functions N(x,y). The solution of a finite element problem determines the unknown nodal values of a function at each vertex of the triangle; values of the function within the interior of the triangle can be calculated using the shape functions and the coordinates of the point inside the triangle.

 

[ThurmanMurman]

Thanks a lot for the response SteamKing! The link you posted was very helpful.

I just want to make sure that I understand the notation of the functions correctly:

Does "[N(x,y)] = [Ni(x,y) Nj(x,y) Nk(x,y)]" translate to "Triangle N" with vertices at "Ni(x,y), Nj(x,y) and Nk(x,y)? Is the "z" in "Ni(x,y) = (1/2A)(ai + bix + ciz)" most likely actually a "y"? From the link you posted, I assume that the "a", "b", and "c" variable are linear equations of "x" and "y".

Sound about right?

Thanks again...

 

[SteamKing]

No, you do not have the correct interpretation of N(x,y).

The vertices of the triangle are at (x1,y1), (x2,y2), and (x3,y3). The triangle itself lies in the x-y plane and is not otherwise positioned in 3-D space.

In Finite Element Analysis, you are trying to determine some function u(x,y) which is specified on a general 2-D region which has been broken up into several triangular pieces. The functions N(x,y) can be thought of as a means to find by interpolation the value of the function u(x,y) at a particular location within a certain triangle. The solution by the FEM gives values of u(x,y) at the vertices of the triangle, and the N(x,y) functions allow one to approximate the value of u(x,y) inside the triangle. See pages 1 and 2 of the attachment above. The functions N(x,y) are given by the equations listed at the top of page 2 and depend on the area of the triangle and the coordinates of the vertices of the triangle only. The a, b, and c values represent the assumed form of the function u(x,y) specified on the triangle. u(x,y) = a + b*x + c*y

 

[ThurmanMurman]

Thanks again SteamKing... Do you have any idea how the "tedious" algebra would flow to solve for "a","b", and "c"? I have taken quite a few paths, and I quickly end up in "trivial solution land"...

 

[AlephZero]

You know the shape function at the nodes of the element (zero at two nodes and 1 at the other node). You know the (x,y) positions of the nodes.

That gives you 3 equations to solve for a,b,c.

This is tedious and longwinded, because a,b,c are different for every shape function in every element.

That is one reason why "real world" finite elements have not been formulated this way since about the 1960s. It is much neater to use a local coordinate system defined by the element geometry instead of the global (x,y) coordinates. (That's part of the reason why I didn't understand what your original question was about!).

If your course includes isoparametric elements later on, you will learn about this.