Page 79 - Fundamentals of The Finite Element Method for Heat and Fluid Flow
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71
                        THE FINITE ELEMENT METHOD
                        where
                                            1
                                       N i =
                                            6V  (a i + b i x + c i y + d i z) with i = 1, 2, 3, 4  (3.148)
                           The volume of the tetrahedron is expressed as
                                                                    
                                                            1 x 1 y 1 z 1
                                                           1 x 2 y 2 z 2 
                                                  6V = det                               (3.149)
                                                            1 x 3 y 3 z 3
                                                                    
                                                            1 x 4 y 4 z 4
                           Also note that
                                                        ∂N 1   b 1
                                                            =
                                                         ∂x   6V
                                                        ∂N 1   c 1
                                                            =
                                                         ∂y   6V
                                                        ∂N 1   d 1
                                                            =                              (3.150)
                                                         ∂z   6V
                           Therefore, the gradient matrix of the shape functions can be written as
                                                                     
                                                       1   b 1 b 2 b 3 b 4
                                                 [B] =      c 1 c 2 c 3 c 4              (3.151)
                                                       6V
                                                           d 1 d 2 d 3 d 4
                        where
                                                                   
                                                              1 y 2 z 2
                                                   b 1 =−det   1 y 3 z 3                 (3.152)
                                                              1 y 4 z 4
                                                                   
                                                              x 2 1 z 2
                                                   c 1 =−det   x 3 1 z 3                 (3.153)
                                                              x 4 1 z 4
                                                                    
                                                              x 2 y 2 1
                                                   d 1 =−det   x 3 y 3 1                 (3.154)
                                                              x 4 y 4 1


                           Similarly, the other terms in Equation 3.151 can also be determined. We therefore
                        summarize all the terms as follows:
                        b-terms
                                          b 1 = (y 2 − y 4 )(z 3 − z 4 ) − (y 3 − y 4 )(z 2 − z 4 )
                                          b 2 = (y 3 − y 4 )(z 1 − z 4 ) − (y 1 − y 4 )(z 3 − z 4 )
                                          b 3 = (y 1 − y 4 )(z 2 − z 4 ) − (y 2 − y 4 )(z 1 − z 4 )

                                          b 4 = b 1 + b 2 + b 3                            (3.155)
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