Page 17 - Mechanics of Asphalt Microstructure and Micromechanics
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10   Ch a p t e r  O n e


                 Therefore, it does not vary with the frame changes. In analogy, it is equivalent to
              one of the scalars.

              Fourth Order Isotropic Tensor
              Any of the fourth order isotropic tensors can be represented as the combinations of the
              following three fourth order istropic tensors:





                 or
                                                                                 (1-26)


              1.5.5 Matrix Operations
              The second order tensors are often represented as 3×3 matrices and all the matrix op-
              erations are valid for the tensors. Among many of the operations, the operation to find
              the principal directions and principal values of matrices has significances in determin-
              ing the principal directions and values of stresses. Considering a tensor A, its projection
              onto the u direction is a vector v.
                                         Au =  v  or  Au•=  v                    (1-27)
                                          ij  j  i
                 A special case is that the projection on to the n direction ends up with a vector that
              is in line with n direction.

                                        An = λ n or  A n =• ˆ  λ n ˆ             (1-28)
                                         ij  j  i
                 or

                                    (A − λδ  )n = 0 or  (A − λ I ) ˆ n =•  0     (1-29)
                                      ij   ij  j
                 Therefore, one has the following equations:
                                      (A − λ )n +  A n + A n =  0
                                        11    1   12  2  13 3
                                                 λ
                                      An + ( A − ) n +  A n =  0
                                       21 1   22    2   23 3
                                      An +  A n + ( A − )λ  n =  0               (1-30)
                                       31 1  32  2  33    3
                 The solution to l is equivalent to the following matrix equation:


                                             A − λδ  = 0                         (1-31)
                                              ij   ij
                 or

                                               2
                                          3
                                         λ − I  λ + I  λ − I  =  0               (1-32)
                                             1     2   3
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