Page 66 - Fundamentals of The Finite Element Method for Heat and Fluid Flow
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                                                           (x 3 ,y 3 )
                                                y
                                                            3      THE FINITE ELEMENT METHOD
                                                  (x 4 ,y 4 )
                                                      4
                                                        1          2
                                                                     (x 2 ,y 2 )
                                                       (x 1 ,y 1 )
                                                                        x
                                          Figure 3.13  A typical quadrilateral element


                                                         y
                                                      b      b
                                                  4  (−b,a)      3  (b,a)
                                                                   a
                                                                       x
                                                        (0,0)
                                                                   a
                                                  1(−b,−a)       2(b,−a)
                                           Figure 3.14 A simple rectangular element

                        and thus the temperature gradients may be written as
                                                      ∂T
                                                         = α 2 + α 4 y
                                                      ∂x
                                                      ∂T
                                                         = α 3 + α 4 x                      (3.87)
                                                      ∂y
                           Therefore, the gradient varies within the element in a linear way. On substituting the
                        values of T 1 ,T 2 ,T 3 and T 4 into Equation 3.86 for the nodes (x 1 ,y 1 )... (x 4 ,y 4 ) and solving,
                        we obtain the values of α 1 ,α 2 ,α 3 and α 4 . Substituting these relationships into Equation 3.86
                        and collating the coefficients of T 1 ,T 2 ,... ,T 4 ,weget

                                              T = N 1 T 1 + N 2 T 2 + N 3 T 3 + N 4 T 4     (3.88)
                        where for a rectangular element (Figure 3.14),
                                                        1
                                                  N 1 =    (b − x)(a − y)
                                                       4ab
                                                        1
                                                  N 2 =    (b + x)(a − y)
                                                       4ab
                                                        1
                                                  N 3 =    (b + x)(a + y)
                                                       4ab
                                                        1
                                                  N 4 =    (b − x)(a + y)                   (3.89)
                                                       4ab
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