Introduction to Continuum Mechanics 3 Episode 2 doc

26 Tensors therefore, From the previous examples we can see that the value of 4-1 corresponds to rotation and -1 corresponds to reflection. 2B11 Transformation Matrix Between Two Rectangular Cartesian Coordinate Systems. Suppose {e,-} and {ej} are unit vectors corresponding to two rectangular Cartesian coor- dinate systems (see Fig. 2B.3). It is clear that {e,-} can be made to coincide with {e^ } through either a rigid body rotation (if both bases are same handed) or a rotation followed by a reflection (if different handed). That is {e/} and {e, : } can be related by an orthogonal tensor Q through the equations i.e., where or We note that Qn - e^-Qej = ej-ei = cosine of the angle between *i and ei, 012 = e i" Q e 2 = e i' e 2 = cosine of the angle between ej and 63, etc. In general, Qij = cosine of the angle between e,- and e. : which may be written: The matrix of these directional cosines, i.e., the matrix "011 012 013 [Q]= 021 022 023 031 032 033 Part B Transformation Matrix Between Two Rectangular Cartesian Coordinate Systems. 27 is called the transformation matrix between {e/} and {e/}. Using this matrix, we shall obtain, in the following sections, the relationship between the two sets of components, with respect to these two sets of base vectors, of either a vector or a tensor. Fig.2B3 Example 2B 11.1 Let {e/ } be obtained by rotating the basis {e/} about the 63 axis through 30° as shown in Fig. 2B.4. We note that in this figure, e 3 and e 3 coincide. Solution, We can obtain the transformation matrix in two ways. (i) Using Eq. (2B11.2), we have //jr j2 11 =cos(e 1 ,ei)=cos30°=—, (2i2 =cos ( e i» e 2) =cosl2 0 0 = , (2i3 = cos(e 1 ,e 3 )=cos90 0 =G ^2i=cos(e2,ei)=cos60 0 =-,j322 == cos(e 2 ,ei)=cos30 0 =—,j223 ==c os(e2,e 3 )=cos90 0 =0 (Q 3 i=cos(e 3 ,ei)=cos90°=0, £> 3 2=cos(e 3 ,e 2 )=cos90°=0, j233 = cos(e 3 ,e 3 )=cosO°= 1 (ii) It is easier to simply look at Fig. 2B.4 and decompose each of the e/ 's into its components in the {e^e^} directions, i.e., 28 Tensors C 3 = e 3 Thus, by either method, the transformation matrix is '£- I n 2 ~2 ° rol _- 1^3 [Q] ~ 2 2 ° 0 0 1 Fig.2B.4 2B12 Transformation Laws for Cartesian Components of Vectors Consider any vector a, then the components of a with respect to {e/} are a/ = a-e/ and its components with respect to {e/ }are a'j = a • e'i Now, e/ = Q m i* m , [Eq. (2Bll.la)], therefore, i.e., In matrix notation, Eq. (2B12.1a) is Part B Transformation Laws for Cartesian Components of Vectors 29 or Equation (2B12.1) is the transformation law relating components of the same vector with respect to different rectangular Cartesian unit bases. It is very important to note that in Eq. (2B12.1c), [a]' denote the matrix of the vector a with respect to the primed basis e,' and [a] denote that with respect to the unprimed basis e/. Eq. (2B12.1) is not the same as 7* a'=Q a. The distinction is that [a] and [a]' are matrices of the same vector, where a and a' are T two different vectors; a' being the transformed vector of a (through the transformation Q ). If we premultiply Eq. (2B12.1c) with [Q], we get The indicial notation equation for Eq.(2B12.2a) is Example 2B 12.1 Given that the components of a vector a with respect to {e/} are given by (2,0,0), (i.e., a = 2ei), find its components with respect to {e/}, where the e/ axes are obtained by a 90° counter-clockwise rotation of the e/ axes about the C3-axis. Solution. The answer to the question is obvious from Fig. 2B.5, that is a = 2ex = -2ei We can also obtain the answer by using Eq. (2B12.2a). First we find the transformation matrix. With e'i - e 2 , ej> = -*i and €3 = e 3 , by Eq. (2Bll.lb), we have 0 -1 0" [Q]= 1 0 0 0 0 1 Thus, I" 0 1 O] |~2] To" [a]' = [Q] [a]= -100 0 = -2 0010 0 i.e., a = -2e' 2 30 Tensors Fig. 2B.5 2B13 Transformation Law for Cartesian Components of a Tensor Consider any tensor T, then the components of T with respect to the basis {e,- }are: T = c -Te L i] e » ie ; Its components with respect to {e/ }are: T.'. = e : -Te ; 1 ij e i * v j With e, : = Q mi e m , ' ij ~ Qtnfim' *vj/i/ e /i ~ \2m&£nj\*m' * ®n) i.e., In matrix notation, Eq. (2B13.1a) reads Tn T'n T^\ \Q n Q 2l 0 31 1 |"r n T n T 13 \ \Q n Q n Qi3\ < 2B13 - lb ) T 2l T 22 T 23 ~ Ql2 Qn 032 T 21 T 22 T 23 Qzi Ql2 Q23 TII T^ 2 7*33 ^13 023 033 T 31 T 32 T 33 031 032 033 or Part B Transformation Law for Cartesian Components of a Tensor 31 We can also express the unprimed components in terms of the primed components. Indeed, premultiply Eq. (2B13.1c) with [Q] and postmultiply it with [Q] , we obtain, since [QMQf=[Qf[Q] = m> Using indicia! notation, Eq. (2B13.2a) reads Equations (2B13.1& 2B13.2) are the transformation laws relating the components of the same tensor with respect to different Cartesian unit bases. It is important to note that in these equations, [T] and [TJ'are different matrices of the same tensor T. We note that the equation [T]' = [Q] T [T][Q] differs from the equation T' = Q r TQ in that the former relates the com- ponents of the same tensor T whereas the latter relates the two different tensors T and T '. Example 2B 13.1 Given the matrix of a tensor T in respect to the basis {e/}: "0 1 0~ [T] = 120 L° 0 1 Find [T] e :, i.e., find the matrix of T with respect to the {e/} basis, where {e/} is obtained by rotating {e/} about €3 through 90°. (see Fig. 2B.5). Solution. Since ei — 62,62 = -ej and 63 = 63, by Eq. (2Bll.lb), we have 0 -1 0" [Q]= 1 00 0 0 1 Thus, Eq.(2B13.1c) gives " 0 1 O] [0 1 0] |~0 -1 Ol [2 -1 0" [T]' =-100 120 1 00= -1 00 [ 0 0 IJ [0 0 Ij [0 0 IJ [0 01 i.e., TU = 2, T{ 2 = -1, T{ 3 = 0,r 2 '! = -1, etc. Example 2B13.2 Given a tensor T and its components Tjy and Tjj with respect to two sets of bases {e/} and {e/ }. Show that 7}/ is invariant with respect to this change of bases, i.e., 7}/ = 7//. 32 Tensors Solution. The primed components are related to the unprimed components by Eq. (2B13.1a) ' ij ~ \2mv>2ni*mn Thus, 'ii = QmiQni* mn But, Q mi Q ni = d mn (Eq. (2B10.2c)), therefore, 'ii ~ ®mn*mn ~ 'mm i.e., T\\ + 7*22+ TB = 7ll+ T 22 + ^33 We see from Example 2B13.1, that we can calculate all nine components of a tensor T with respect to e,' from the matrix [T] e ., by using Eq. (2B13.1c). However, there are often times when we need only a few components. Then it is more convenient to use the Eq. (2B2.2) (TIJ = e/ -Tej) which defines each of the specific components. In matrix form this equation is written as: T where [e'] denotes a row matrix whose elements are the components of e/ with respect to the basis {e/}. Example 2B13.3 Obtain T[i for the tensor T and the bases e/ and e/ given in Example 2B13.1 Solution. Since ej = 62, and 62 = -e l5 thus TU = ei-Tei = e 2 -T(- ei ) =-e 2 -T ei = -T 2l = -1 Alternatively, using Eq. (2B13.4) TO i oi [-ii r o" 7i2 = [«il mtel = [0,1,0] 1 2 0 0 = [0,1,0] -1 = -1 0 0 IJ [ OJ I 0 2B14 Defining Tensors by Transformation Laws Equations (2B12.1) or (2B13.1) state that when the components of a vector or a tensor with respect to {e,-} are known, then its components with respect to any {e, : } are uniquely deter- mined from them. In other words, the components a,- or 7^- with respect to one set of {e/} Part B Defining Tensors by Transformation Laws 33 completely characterizes a vector or a tensor. Thus, it is perfectly meaningful to use a statement such as "consider a tensor T/,-" meaning consider the tensor T whose components with respect to some set of {e,-} are 7)y. In fact, an alternative way of defining a tensor is through the use of transformation laws relating the components of a tensor with respect to different bases. Confining ourselves to only rectangular Cartesian coordinate systems and using unit vectors along positive coordinate directions as base vectors, we now define Cartesian components of tensors of different orders in terms of their transformation laws in the following where the primed quantities are referred to basis {e/ } and unprimed quantities to basis {e,-}, the e/ and e, are related by e,'-Qe/, Q being an orthogonal transformation a' = a zeroth-order tensor(or scalar) a - ~ Q m ia m first-order tensor (or vector) TJJ = QmiQnjTmn second-order tensor(or tensor) T/jk = QmiQnjQrkTmnr third-order tensor etc. Using the above transformation laws, one can easily establish the following three rules (a)the addition rule (b) the multiplication rule and (c) the quotient rule. (a)The addition rule: If TJ; and Sy are components of any two tensors, then TJJ+SJJ are components of a tensor. Similarly if TpandS,-^ are components of any two third order tensors, then Tp 1-Sp. are components of a third order tensor. To prove this rule, we note that since Tl jk =Q mi Q nj Q rk T mnr and S; jk =Q mi Q nj Q rk S mnr we have, *ijk + Sijk = QmiQnjQrk*mnr+QmiQnjQrkTmnr ~ QmiQn}Qrk(^mnr + ^nmr) Letting W- jk = T^+S^ and W mnr =T mnr +S mnr , we have, ™ijk — QmiQnjQrkTmnr i.e, Wfjff are components of a third order tensor. (b)The multiplication rule: Let a/ be components of any vector and Tjy be components of any tensor. We can form many kinds of products from these components. Examples are (a)a/a,« (b)a/(3ya^ (c) TijT kl , etc. It can be proved that each of these products are components of a tensor, whose order is equal to the number of the free indices. For example, a/a/ is a scalar (zeroth order tensor), a^ija k are components of a third order tensor, 7]y7]y are components of a fourth order tensor. To prove that T^Tjy are components of a fourth-order tensor, let M / y W =^r w , then 34 Tensors MIJU ~ TijTu=Q m iQ n jr mn Q r kQslT r s = QmiQnjQr1<QslTmnTrs i.e., Mijkl = QmiQnjQrkQslMmnrs which is the transformation law for a fourth order tensor. It is quite clear from the proof given above that the order of the tensor whose components are obtained from the multiplication of components of tensors is determined by the number of free indices; no free index corresponds to a scalar, one free index corresponds to a vector, two free indices correspond a second-order tensor, etc. (c) The quotient rule: If a,- are components of an arbitrary vector and 7^- are components of an arbitrary tensor and a,- = 7^6y for all coordinates, then £>/ are components of a vector. To prove this, we note that since a,- are components of a vector, and T)y are components of a second-order tensor, therefore, and Now, substituting Eqs. (i) and (ii) into the equation a,- = Tybj, we have But, the equation a,- = Tqbj is true for all coordinates, thus, we also have Thus, Eq. (iii) becomes Multiplying the above equation with Q ik and noting that Q^Qi m ~ <5fcm» we 8 et i.e., Since the above equation is to be true for any tensor T, therefore, the parenthesis must be identically zero. Thus, PartB Symmetric and Antisymmetric Tensors 35 This is the transformation law for the components of a vector. Thus, fy are components of a vector. Another example which will be important later when we discuss the relationship between stress and strain for an elastic body is the following: If 7^- and EJJ are components of arbitrary second order tensors T and E then T ij = CijklEkl for all coordinates, then C^ are components of a fourth order tensor. The proof for this example follows that of the previous example. 2B15 Symmetric and Antisymmetric Tensors 7* A tensor is said to be symmetric if T = T . Thus, the components of a symmetric tensor have the property, i.e., A tensor is said to be antisymmetic if T = -T r . Thus, the components of an antisymmetric tensor have the property i.e., and Any tensor T can always be decomposed into the sum of a symmetric tensor and an antisymmetric tensor. In fact, where and It is not difficult to prove that the decomposition is unique (see Prob. 2B27) [...]... dual vector (or axial vector ) of the antisymmetric tensor The form of the dual vector is given below: From Eq.(2B16.1), we have, since a-bxc = b-cxa, 7i2 = e1-Te2 = e 1 -f 4 Xe 2 = t 4 -e 2 Xe 1 = -f 4 -e 3 = -f$ T31 = e3-TCl = e 3 -f 4 xe 1 = t^xes = -f 4 ^ = -$ ^ 23 = *2- Te3 = e^f^ = . (2B 13. 1a) reads Tn T'n T^ Q n Q 2l 0 31 1 |"r n T n T 13 Q n Q n Qi3 < 2B 13 - lb ) T 2l T 22 T 23 ~ Ql2 Qn 0 32 T 21 T 22 T 23 Qzi Ql2 Q 23 TII T^ 2 . have //jr j2 11 =cos(e 1 ,ei)=cos30°=—, (2i2 =cos ( e i» e 2) =cosl2 0 0 = , (2i3 = cos(e 1 ,e 3 )=cos90 0 =G ^2i=cos(e2,ei)=cos60 0 =-,j 322 == cos(e 2 ,ei)=cos30 0 =—,j2 23 ==c os(e2,e 3 )=cos90 0 =0 (Q 3 i=cos(e 3 ,ei)=cos90°=0, . T 22 T 23 Qzi Ql2 Q 23 TII T^ 2 7 *33 ^ 13 0 23 033 T 31 T 32 T 33 031 0 32 033 or Part B Transformation Law for Cartesian Components of a Tensor 31 We can also express the unprimed