2-D Linear Elements: Linear Triangles

   

Complete linear polonomial. That is, a constant term and linear terms in x and y

Trianglar elements can take any orientation and satisfy continuity requirements involving adjacent elements

Use CCW convection (Node "i" to "j" to "k" form right-hand rule for outward normal).

   

for local node numbering

 

Solve for coefficients  

   

or

 

Then soln of coefficients

                   only significant entries are in  column 1

            

                  

Recall : Area of triangle is

          ;           ;      

 

   

 

         

 

      

 

Where the

 

CCW ordering gives a positive (+) area

    note: 

 

 

Pictorially, N_i

 

Look at a point (x, y) in

        

 

                    

 

    the  are the "area" coordinates

 

There are multiple ways to calculate the area of a triangle. The following description gives a general 3D solution to the situation. (Triangle Area)

 

The following link provides a table for integration formulaes for linear triangles (Integration Formulae)

 

 

Examine a 2-D Helmholz Eqn

              

Use same procedure as in 1-D case.

1.) Substitute for U a trial fxn

              constant coeff.

                     

           

2.) Force   to zero in "weak form"

    a.) multiply by weight fxn

        

    b.) integrate over entire domain and force to zero

        

3.) Discretize domain into subregions (elements)

4.) Select Basis function:  here we use Linear triangles with N_i just calculated.

5.) Integrate by parts which is the vehicle or mechanism for applying B.C.

      (Use implied summation)

 
Remember that the U_j term is also part of the integrated by parts term.

    

6.) Select Weighting functions and functional coefficients (if used)

      i.e.       

 

7.) Assemble Matrices

   

   

      for  L = 1 : NE
c
c calculate Dx, D y, and area
c
        area = 0;
        for i = 1 : 3
           j = in( mod(i,3) +1, L);
           k = in( mod(i+1,3)+1, L);
           dx(i) = x(j) - x(k);
           dy(i) = y(j) - y(k);
           area = area + x(in(i,L))*dy(i);
        end
        area = area / 2.;
      for    i = 1 : 3
         iGlobal = in(I,L);
         for J = 1 : 3
            jGlobal = in(J,L);
            Bcol = diag + (jGlobal - iGlobal);
           AIJ = -(dy(j)*dy(i) + dx(j)*dx(i))/(4*area);
         if i == j
                 ;
         else
                 ;
         end
      BAND(iGlobal, Bcol) = BAND(iGlobal, Bcol) + AIJ
      End        % (J loop)

      Rhs(iGlobal) = Rhs(iGlobal) + 0

      End          % (I loop)

    End        % (L loop)          

 

8.) Apply the Boundary Conditions