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Introduction and Fundamentals for Mathematics and Continuum Mechanics   17


              1.6.2.6  Changes of Segment, Area, and Volume
              1.6.2.6.1  Line Segment Change

                                                   •
                                             dx =  F dX                          (1-81)
                                                                                 (1-82)

                                                       •
                                            L F dX =
                                         dx =•  •     L dx                       (1-83)
                 The material derivative of the dot product of dx:
                                                                                 (1-84)


              1.6.2.6.2  Area Change
              The area change can be evaluated through an area element composed of two segments
                        (2)
              dX  and dX . In the reference configuration the area is:
                (1)
                                          dS = ξ  dX dX  ( )                     (1-85)
                                                    ()
                                            o
                                                         2
                                                     1
                                            A   ABC  B  C
                 In the current configuration:
                                   dS = ξ  dx dx ( )  = ξ  x dX x dX  (22)
                                                         ()
                                           1
                                               2
                                                         1
                                           ()
                                     i  ijk  j  k  ijk  j B ,  B  k C ,  C
                            x dS = ξ  x x x dX dX   ( ) 2  = ξ  F F F dX dX  ( )
                                                                      2
                                                                  1
                                                                  ()
                                                () 1
                              ,
                             i A  i  ijk  i A  j B k C  B  C  ijk iAAjB kC  B  C
                                            ,
                                       ,
                                          ,
                                      ξ FFF =   ξ   det F = ξ  J
                                       ijk iA jB kC  ABC   ABC
                                         x dS = ξ  JdX dX  ( ) 2
                                                      () 1
                                         iA  i  ABC   B   C
                                          ,
                 Multiply both sides of the above equation by X A,q , one obtains:
                                                      () 1
                                                          ( ) 2
                                     xX    dS = ξ  JdX dX X
                                      iA  A q ,  i  ABC  B  C  A q ,
                                       ,
                                             xX    = δ
                                              iA  A q ,  iq
                                              ,
                                            dS =  X  JdS o
                                              q   A q ,  A                       (1-86)
                                       ·      ·
                                                 –1
                 Considering the identity detB = tr(B  B ) detB and replace B with F, one can obtain:
                                               ●
                                   .      .                .
                                 det F =  tr (F F•  −1 )det F =  JtrL or  J =  Jdivv  (1-87)

                 This relationship is important for defining stress in the reference configuration:
                                           dS = (  −1  T •  dS o
                                               J F )

                                             dS F =  JdS  o
                                               •
                                                                                 (1-88)
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