146 Change of Volume due to Deformation Recall that for any vectors a, b, and c, a-bxc = determinant whose rows are components of a, b, andc. Therefore Eq. (iii) becomes Using the definition of transpose of a tensor, Eqs. (ii) become and Eq. (v) becomes T Thus, F n is in the direction of 63, so that Therefore, 1 T* Equation (3.28.4) states that the deformed area has a normal in the direction of (F ) 63 and with a magnitude given by In deriving Eq. (3.28.4), we have chosen the initial area to be the rectangular area formed by the Cartesian base vectors ej and e 2 , it can be shown that the formula remains valid for any material area except that e 3 be replaced by the normal vector of the undeformed area n 0 . That is in general, 3.29 Change of Volume due to Deformation Consider three material elements Kinematics of a Continuum 147 emanating from X The rectangular volume formed by cDP \ (DP ' and cflP ' at the reference time t 0 is given by At time t, dX (1) deforms into da (l} = ¥dX (1 \ rfX (2) deforms into <& (2) = F^X (2) and dX^deforms into dx (3) = FrfX (3) and the volume is That is, Since C = F T F and B = FF r , therefore Thus, Eq. (3.29.3) can also be written as We note that for an incompressible material, dV = dV 0 , so that We note also that due to Eq. (3.29.3), the conservation of mass equation can be written as: Example 3.29.1 Consider the deformation given by (a)Find the deformed volume of the unit cube shown in Fig. 3.14. (b)Find the deformed area of OABC. (c) Find the rotation tensor and the axial vector of the antisymmetric part of the rotation tensor. Solution, (a) From Eq. (i), 148 Change of Volume due to Deformation Thus, Since det F is a constant, from the equation we have, with AF 0 = 1, Fig. 3.14 (b) Using Eq. (3.28.6), with A/4 0 = 1, and n 0 = -e 3 , we have i.e., Kinematics of a Continuum 149 Thus, the area OABC, which was of unit area and having a normal in the direction of -63 becomes an area whose normal is in the direction of 62 and with a magnitude of Aj %• It is easily verified that R corresponds to a 90° rotation about the ej, which is the axial vector of the antisymmetric part of R 3.30 Components of Deformation Tensors in other Coordinates The deformation gradient F transforms a differential material element dX. in the reference configuration into a material element d\ in the current configuration in accordance with the equation where describes the motion. If the same rectangular coordinate system is used for both the reference and the current configurations, then since the set of base vectors (61,62,63) is the same at every point, we have That is 150 Components of Deformation Tensors in other Coordinates Thus i.e., We have already used this matrix for computing the components of F in a few examples above. The situation is more complicated if the base vectors at the reference configuration are different from those at the current configuration. Such situations arise not only in the case where different coordinate systems are used for the two configuration ( for example, a rectangular coordinate system for the reference and a cylindrical coordinates for the current configuration, see (B) below), but also in cases where the same curvilinear coordinates are used for the two configurations. The following are examples. (A) Cylindrical coordinate system for both the reference and the current configuration Let be the pathline equations. We shall show in the following that and Kinematics of a Continuum 151 where e 0 / denotes base vectors at the reference position and e,- those at the current position. Substituting in the equation d\ = FrfX , we obtain etc. Thus, These equations are equivalent to Eqs. (3.30.4). The matrix is based on two sets of bases, one at the reference configuration (e or , e^, e oz ) and the other the current configuration (e r ,60 ,e 2 ). The components in this matrix is called the two point components of the tensor F with respect to ( e r ,e# ,e 2 ) and ( e or , e^, e 02 ). By using the definition of transpose of a tensor, one can easily establish Eqs. (3.30.5 ) from Eq. (3.30.4). [see Prob. 3.73] The components of the left Cauchy-Green tensor, with respect to the basis at the spatial position x. can be obtained as follows. From the definition B = FF , and by using Eqs. (3.30.4) and (3.30.5 ) we obtain 152 Components of Deformation Tensors in other Coordinates Similarly, Other components can be obtained in the same way [see Prob. 3.74], We list all the com- ponents below: The components of B can be obtained either by inverting the tensor B or by inverting the pathline equations. Let be the inverse of Eq. (iii). Then from the equation dX = F l d\ , one can obtain Kinematics of a Continuum 153 The components of the right Cauchy-Green tensor C, with respect to the basis at the T reference position X can be obtained as follows. From the definition C = F F , and by using Eqs. (3.30.4) and (3.30.5 ) we obtain Similarly, Other components can be obtained in the same way [see Prob. 3.75], We list all the com- ponents below: 154 Components of Deformation Tensors in other Coordinates Again, the components of C l can be obtained by using the equation dX = F l d* and Eq. (ix). We list here two of the six components. The other four components can be easily written down following the patterns of these two equations. (B) Cylindrical coordinates (r,0,z ) for the current configuration and rectangular Cartesian coordinates (X,Y,Z) for the reference configuration. Let describe the motions. Then using the same procedure as described for the case where one single cylindrical coordinate is used, it can be derived that [see Prob.3.76]. The matrix Kinematics of a Continuum 155 gives the two point components of F with respect to the two sets of bases, one at the reference configuration, the other at the current configuration. The components of the left Cauchy-Green deformation tensor B with respect to the basis at the current configuration are given by [see Prob.3.77] Again, the components of B l can be obtained by using the equation dX = F l dx and the inverse of Eq. (xii). We list here two of the six components. The other four components can be easily written down following the patterns of these two equations. The components of the right Cauchy-Green deformation tensor C with respect to the basis at the reference configuration are given by: [see Prob. 3.78] [...]... components of the right Cauchy-Green tensor C with respect to the basis at the reference configuration 3. 73 From Eqs. (3. 30.4a), obtain Eqs. (3. 30 .5) 3. 74 Verify Eq. (3. 30.8b) and (3. 30.8d) 3. 75 Verify Eq.( 3. 30.9b) and (3. 30.9d) 3. 76 Derive Eqs. (3. 30.10) 3. 77 Using Eqs. (3. 30.10) derive Eqs. (3. 30.12a) and (3. 30.12d) 3. 78 Verify Eqs (3. 30. 13 a) and (3. 30.13d) 4 Stress In the previous chapter, we considered... what are the strain components £"ii^22»^i 23. 32 (a) Do Problem 3. 31 if the measured strains are 200xlO~ 6 , 50 xlO~ 6 , lOOxlO" 6 , respectively (b) If 33 = 32 = 31 = 0, find the principal strains and directions of part (a) (c) How will the result of part (b) be altered if 33 * 0? 166 Problems 33 3 Repeat Problem 3. 32 except that a = b = c= 1000 x 10~6 3. 34 The unit elongations at a certain point... of 2ej 4- 2e2 + 63 and 3ej - 6 63? 3. 23 Do the previous problem for (a) the unit elongation in the direction 3ej — 4e2, (b) the change in angle between two elements in the direction 3ej - 4 63 and 4e^ + 3e3 3. 24 (a)For Prob .3. 22, determine the principal scalar invariants of the strain tensor, (b) Show that the following matrix cannot represent the same state of strain of Prob .3. 22 3. 25 For the displacement... the 60° strain rosette ) in 1 1 the directions ej, -(ej + V3~e2), and -~(-*i + vTe2) If these elongations are designated by jLt £ a,b,c respectively, what are the strain components E-^^lf1^f f 3. 35 Do Problem 3. 34 if the strain rosette measurements give a = 2x 10 , £ = 1x10 , c= 1.5X10"6 3. 36 Do Problem 3. 35 except that a = b=c = 2000 x 10~6 2 3. 37 For the velocity field, v = (kxfai (a) Find the rate... volume to the initial 170 Problems volume, (h) the deformed area (magnitude and its normal) for the area whose normal was in the direction of 62 and whose magnitude was unity in the undeformed state 3. 62 Do Prob 3. 61 for the following deformation 3. 63 Do Prob 3. 61 for the following deformation 3. 64 Do Prob 3. 61 for the following deformation 3. 65 Given Obtain (a)F,C (b) the eigenvalues and eigenvectors... incompressible fluid undergoes a two-dimensional motion with 3. 51 Are the fluid motions described in (a) Prob .3. 15 and (b) Prob .3. 16 incompressible? 3. 52 In a spatial description, the density of an incompressible fluid is given by p = k%i Find the permissible form for the velocity field with v3 = 0, so that the conservation of mass equation is satisfied 3. 53 Consider the velocity field (a) Find the density if... 3. 54 Given the velocity field determine how the fluid density varies with time, if in a spatial description it is a function of time only 3. 55 Check whether or not the following distribution of the state of strain satisfies the compatibility conditions: where k = 10 3. 56 Check whether or not the following distribution of the state of strain satisfies the compatibility conditions: Kinematics of a Continuum. .. jt/2 Show that the above expression reduces to the results of Section 3. 13 3.46 Let ci, 62, 63 and Dt, D^, #3 be the principal directions and values of the rate of deformation tensor D Further, let be three (D/Df)[cbf^' material line elements (dx . components £"ii^22»^i2- 3. 32. (a) Do Problem 3. 31 if the measured strains are 200xlO~ 6 , 50 xlO~ 6 , lOOxlO" 6 , respectively. (b) If 33 = 32 = 31 = 0, find the principal. by using Eqs. (3. 30.4) and (3. 30 .5 ) we obtain Similarly, Other components can be obtained in the same way [see Prob. 3. 75] , We list all the com- ponents below: 154 Components . to ( e r ,e# ,e 2 ) and ( e or , e^, e 02 ). By using the definition of transpose of a tensor, one can easily establish Eqs. (3. 30 .5 ) from Eq. (3. 30.4). [see Prob. 3. 73] The