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Finding the B matrix for constant strain triangle

Nasser M. Abbasi

JulyΒ 2, 2015 page compiled on JulyΒ 2, 2015 at 1:37am

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1 Solution

PIC

1 Solution

The problem is first solve for scalar field πœƒ  with the interpolating polynomial a1 + a2x + a3y  . Writing

              ⌊  βŒ‹
    [       ] |a1|
πœƒ =  1  x  y  |a2|
              ⌈  βŒ‰
               a3
(1)

Evaluating the field πœƒ  at each node gives

⌊   βŒ‹   ⌊          βŒ‹ ⌊   βŒ‹
  πœƒ1      1  x1  y1   a1
||   ||   ||          || ||   ||
⌈ πœƒ2βŒ‰ = ⌈ 1  x2  y2βŒ‰ ⌈a2 βŒ‰
  πœƒ3      1  x3  y3   a3

Hence

pict

Where Ξ”  is the determinant x1y2 βˆ’ x2y1 βˆ’ x1y3 + x3y1 + x2y3 βˆ’ x3y2

Substituting (2) into (1) gives

pict

Where

pict

For constant stress triangle, the field is a vector field. Hence replacing πœƒ  with [ ]
 u

 v equation (3) becomes

                                   ⌊   βŒ‹
                                     u1
                                   ||   ||
[ ]    [                          ]| v1|
 u      N    0   N     0  N     0  || u ||
    =     1        2        3      ||  2||
 v       0   N1   0   N2   0   N3  | v2|
                                   ||   ||
                                   ⌈ u3βŒ‰
                                     v3

From the above

pict

Hence

pict

From (4) all of the βˆ‚Ni, βˆ‚Nj
 βˆ‚x  βˆ‚y  terms are evaluated. Substituting the result into (5) gives the BΒ matrix

pict

And

pict

Hence B  becomes

       ⌊                                                     βŒ‹
         y2 βˆ’ y3     0     y3 βˆ’ y1     0     y1 βˆ’ y2     0
     1 |                                                     |
B =  --|⌈    0     x3 βˆ’ x2     0     x1 βˆ’ x3     0     x2 βˆ’ x1|βŒ‰
     Ξ”   x βˆ’  x   y  βˆ’ y   x  βˆ’ x   y  βˆ’ y   x  βˆ’ x   y  βˆ’ y
          3    2   2    3   1    3   3    1   2    1   1    2
(6)

Letting yi βˆ’ yj = yij  and xi βˆ’ xj = xij  , the above becomes

|--------------------------------------|
|       ⌊                            βŒ‹ |
|        y23   0   y31   0   y12   0   |
B  = -1 || 0   x     0   x     0   x  || |
|    Ξ”  ⌈      32        13        21βŒ‰ |
|        x32  y23  x13  y31  x21  y12  |
----------------------------------------

But the area of triangle is give by

pict

And the determinant Ξ”  was found above to be x1y2 βˆ’ x2y1 βˆ’ x1y3 + x3y1 + x2y3 βˆ’ x3y2   , hence

2A  = Ξ”

Substituting the above into B  found above in equation (6) gives

         ⌊                                                     βŒ‹
          y2 βˆ’ y3     0      y3 βˆ’ y1    0      y1 βˆ’ y2     0
     -1- ||                                                     ||
B  = 2A  ⌈   0     x3 βˆ’ x2     0     x1 βˆ’ x3     0      x2 βˆ’ x1βŒ‰
          x3 βˆ’ x2   y2 βˆ’ y3 x1 βˆ’ x3   y3 βˆ’ y1 x2 βˆ’ x1   y1 βˆ’ y2