Common Variable Types in Elasticity Elasticity theory is a mathematical model of material deformation. Using principles of continuum mechanics, it is formulated in terms of many different types of field variables specified at spatial points in the body under study. Some examples include: Scalars - Single magnitude mass density , temperature T, modulus of elasticity E, . . . Vectors – Three components in three dimensions displacement vector u ue1 ve 2 we 3 , e1, e2, e3 are unit basis vectors Matrices – Nine components in three dimensions stress matrix x xy xz [] yx y yz zx zy z Other – Variables with more than nine components Index/Tensor Notation With the wide variety of variables, elasticity formulation makes use of a tensor formalism using index notation. This enables efficient representation of all variables and governing equations using a single standardized method. a1 a11 a12 a13 Index notation is a shorthand scheme whereby a whole set of numbers or components can be a i a 2 , a ij a 21 a 22 a 23 represented by a single symbol with subscripts a 3 a 31 a 32 a 33 In general a symbol aij…k with N distinct indices represents 3N distinct numbers Addition, subtraction, multiplication and equality of index symbols are defined in the normal fashion; e.g. a1 b1 a11 b11 a12 b12 ai bi a2 b2 , aij bij a21 b21 a22 b22 a3 b3 a31 b31 a32 b32 a1 a11 a12 ai a2 , aij a21 a22 a3 a31 a32 a13 a23 a33 a13 b13 a23 b23 a33 b33 a1b1 a1b2 ai b j a2 b1 a2 b2 a3b1 a3b2 a1b3 a2 b3 a3b3 Notation Rules and Definitions Summation Convention - if a subscript appears twice in the same term, then summation over that subscript from one to three is implied; for example 3 aii aii a11 a 22 a33 i 1 3 aij b j aij b j ai1b1 ai 2 b2 ai 3b3 i 1 A symbol aij…m…n…k is said to be symmetric with respect to index pair mn if aij... m... n... k aij... n... m... k A symbol aij…m…n…k is said to be antisymmetric with respect to index pair mn if aij... m... n... k aij... n... m... k 1 2 1 2 Useful Identity aij (aij a ji ) (aij a ji ) a( ij) a[ij] a ( ij ) 1 ( aij a ji ) . . . symmetric 2 a[ij ] 1 ( aij a ji ) . . . antisymmetric 2 Example 1-1: Index Notation Examples The matrix aij and vector bi are specified by 1 2 0 2 aij 0 4 3 , bi 4 2 1 2 0 Determine the following quantities: aii , aij aij , aij a jk , aij b j , aij bi b j , bi bi , bi b j , a(ij ) , a[ij ] Indicate whether they are a scalar, vector or matrix. Following the standard definitions given in section 1.2, aii a11 a 22 a33 7 (scalar) aij aij a11 a11 a12 a12 a13 a13 a 21 a 21 a 22 a 22 a 23 a 23 a31 a31 a32 a32 a33 a33 1 4 0 0 16 9 4 1 4 39 (scalar) 1 10 6 aij a jk ai1 a1k ai 2 a 2 k ai 3 a3k 6 19 18 (matrix) 6 10 7 10 1 2 aij b j ai1b1 ai 2 b2 ai 3b3 16 (vector) 1 1 a ( ij ) aij a ji 0 4 8 2 2 2 1 aij bi b j a11b1b1 a12 b1b2 a13b1b3 a 21b2 b1 84 (scalar) 1 2 bi bi b1b1 b2 b2 b3b3 4 16 0 20 (scalar) 1 1 a[ ij ] aij a ji 0 4 2 2 4 8 0 2 1 bi b j 8 16 0 (matrix) 0 0 0 0 1 1 3 2 2 0 2 0 1 1 3 2 2 0 2 0 2 1 1 1 4 1 1 4 2 3 2 1 2 2 0 2 0 1 4 1 1 0 3 2 1 1 (matrix) 1 1 (matrix) 0 Special Indexed Symbols Kronecker Delta 1 0 0 1 , if i j (no sum ) ij 0 1 0 0 , if i j 0 0 1 ij ji ii 3 , i i 1 Properties: ij a j ai , ij ai a j ij a jk aik , jk aik aij ij aij aii , ij ij 3 Alternating or Permutation Symbol ijk 1 , if ijk is an even permutatio n of 1,2,3 1 , if ijk is an odd permutatio n of 1,2,3 0 , otherwise 123 = 231 = 312 = 1, 321 = 132 = 213 = -1, 112 = 131 = 222 = . . . = 0 a11 a12 a13 Useful in evaluating determinants det[ aij ] | aij | a21 a22 a23 ijk a1i a2 j a3k ijk ai1a j 2 ak 3 and vector cross-products a31 a32 a33 Coordinate Transformations x3 x 3 v e3 e3 e2 e2 e1 e1 x1 x 2 x2 We wish to express elasticity variables in different coordinate systems. This requires development of transformation rules for scalar, vector, matrix and higher order variables – a concept connected with basic definitions of tensor variables. The two Cartesian frames (x1,x2,x3) and ( x 1, x 2 , x 3 ) differ only by orientation x 1 Using Rotation Matrix Qij cos( xi, x j ) e1 Q11e1 Q12 e 2 Q13e3 e 2 Q21e1 Q22 e 2 Q23e3 e3 Q31e1 Q32 e 2 Q33e3 v v1e1 v2 e 2 v3e3 vi ei v1e1 v2 e 2 v3 e3 viei ei Qij e j ei Q ji e j transformation laws for Cartesian vector vi Q ji v j components vi Qij v j Cartesian Tensors General Transformation Laws Scalars, vectors, matrices, and higher order quantities can be represented by an index notational scheme, and thus all quantities may then be referred to as tensors of different orders. The transformation properties of a vector can be used to establish the general transformation properties of these tensors. Restricting the transformations to those only between Cartesian coordinate systems, the general set of transformation relations for various orders are: a a , zero order (scalar) ai Qip a p , first order (vector) aij QipQ jq a pq , second order (matrix) QipQ jqQkr a pqr , third order aijk QipQ jqQkr Qls a pqrs , fourth order aijkl ... m QipQ jqQkr Qmt a pqr... t general order aijk Example 1-2 Transformation Examples The components of a first and second order tensor in a particular coordinate frame are given by 1 1 0 3 ai 4 , aij 0 2 2 2 3 2 4 x3 x 3 Determine the components of each tensor in a new coordinate system found through a rotation of 60o (/6 radians) about the x3-axis. Choose a counterclockwise rotation when viewing down the negative x3-axis, see Figure 1-2. x 2 The original and primed coordinate systems are shown in Figure 1-2. The solution starts by determining the rotation matrix for this case cos 60 cos 30 cos 90 1 / 2 Qij cos 150 cos 60 cos 90 3 / 2 cos 90 cos 90 cos 0 0 3 / 2 0 1 / 2 0 0 1 60 x2 o x 1 x1 The transformation for the vector quantity follows from equation (1.5.1)2 1/ 2 3 / 2 0 1 1 / 2 2 3 ai Qij a j 3 / 2 1 / 2 0 4 2 3 / 2 0 2 0 1 2 and the second order tensor (matrix) transforms according to (1.5.1)3 aij Qip Q jq a pq 1/ 2 3 / 2 0 3 / 2 0 1 0 3 1 / 2 1 / 2 0 0 2 2 3 / 2 0 1 3 2 4 0 T 7/4 3 / 2 0 3/4 3/ 2 3 1 / 2 0 3 / 4 5/ 4 1 3 3 / 2 3 / 2 3 1 3 3 / 2 0 1 4 Principal Values and Directions for Symmetric Second Order Tensors The direction determined by unit vector n is said to be a principal direction or eigenvector of the symmetric second order tensor aij if there exists a parameter (principal value or eigenvalue) such that aij n j ni ( aij ij )n j 0 which is a homogeneous system of three linear algebraic equations in the unknowns n1, n2, n3. The system possesses nontrivial solution if and only if determinant of coefficient matrix vanishes det[ aij ij ] 3 I a 2 II a III a 0 scalars Ia, IIa and IIIa are called the fundamental invariants of the tensor aij I a aii a11 a 22 a33 II a a a12 a 22 1 ( aii a jj aij aij ) 11 a 21 a 22 a32 2 III a det[ aij ] a 23 a33 a11 a13 a31 a33 Principal Axes of Second Order Tensors It is always possible to identify a right-handed Cartesian coordinate system such that each axes lie along principal directions of any given symmetric second order tensor. Such axes are called the principal axes of the tensor, and the basis vectors are the principal directions {n(1), n(2) , n(3)} x3 x3 a11 aij a21 a31 a12 a22 a32 a13 a23 a33 1 aij 0 0 n(3) 0 2 0 0 0 3 n(2) x2 n(1) x1 x1 Original Given Axes Principal Axes x2 Example 1-3 Principal Value Problem Determine the invariants, and principal values and directions of 2 0 0 aij 0 3 4 0 4 3 First determine the principal invariants 2 0 3 4 2 0 I a aii 2 3 3 2 , II a 6 25 6 25 0 3 4 3 0 3 2 0 III a 0 3 0 4 2(9 16) 50 0 4 3 The characteristic equation then becomes det[aij ij ] 3 22 25 50 0 ( 2)(2 25) 0 1 5 , 2 2 , 3 5 Thus for this case all principal values are distinct For the 1 = 5 root, equation (1.6.1) gives the system 3n1(1) 0 2n2(1) 4n3(1) 0 4n2(1) 8n3(1) 0 1 (1) ( 2e 2 e 3 ) which gives a normalized solution n 5 1 In similar fashion the other two principal directions are found to be n( 2 ) e1 n( 3) ( e 2 2e 3 ) 5 It is easily verified that these directions are mutually orthogonal. Note for this case, the transformation matrix Qij defined by (1.4.1) becomes 0 2 / 5 1 / 5 Qij 1 0 0 0 1 / 5 2 / 5 5 0 0 aij 0 2 0 0 0 5 Vector, Matrix and Tensor Algebra Scalar or Dot Product a b a1b1 a2 b2 a3b3 ai bi e1 Vector or Cross Product e2 a b a1 a2 b1 b2 e3 a3 ijk a j bk ei b3 Common Matrix Products Aa [ A]{a} Aij a j a j Aij aT A {a}T [ A] ai Aij Aij ai AB [ A][ B] Aij B jk ABT Aij Bkj AT B A ji B jk tr( AB) Aij B ji tr( ABT ) tr( AT B) Aij Bij Second Order Transformation Law aij Qip Q jq a pq a QaQ T Calculus of Cartesian Tensors Field concept for tensor components a a ( x1 , x2 , x3 ) a ( xi ) a ( x ) ai ai ( x1 , x 2 , x3 ) ai ( xi ) ai ( x ) aij aij ( x1 , x 2 , x3 ) aij ( xi ) aij ( x ) Comma notation for partial differentiation a,i a , ai , j ai , aij,k aij , xi x j xk If differentiation index is distinct, order of the tensor will be increased by one; e.g. derivative operation on a vector produces a second order tensor or matrix ai , j a1 x 1 a 2 x1 a 3 x1 a1 x 2 a 2 x 2 a 3 x 2 a1 x3 a 2 x3 a 3 x3 Vector Differential Operations Directional Derivative of Scalar Field df f dx f dy f dz n f ds x ds y ds z ds n unit normal vector in direction of s vector differenti al operator e1 dx dy dz e1 e 2 e 3 ds ds ds e2 e3 x y z f grad f gradient of scalar function f e1 f f f e2 e3 x y z Common Differential Operations Gradient of a Scalar i e i Gradient of a Vector u ui , j e i e j Laplacian of a Scalar 2 ,ii Divergence of a Vector u ui ,i Curl of a Vector u ijk uk , j e i Laplacian of a Vector 2 u ui ,kk e i Example 1-4: Scalar/Vector Field Example Scalar and vector field functions are given by x 2 y 2 u 2 xe1 3 yze 2 xye 3 Calculate the following expressions, , 2, ∙ u, u, u. Using the basic relations 2 xe1 2 ye 2 2 2 2 0 - (satisfies Laplace equation) ∙ u 2 3z 0 2 3z 2 0 0 u ui , j 0 3z 3 y y x 0 e1 e2 e3 u / x / y / z ( x 3 y )e1 ye 2 2x 3 yz xy Contours =constant and vector distributions of Note vector field is orthogonal to -contours, a result trueGradient in general for all scalar fields Vector Distribution 10 8 6 4 y 2 0 x -2 -4 -6 -8 -10 -10 -5 0 5 10 Vector/Tensor Integral Calculus Divergence Theorem Stokes Theorem S u n dS u dV u dr C V S ( u) n dS C S aij... k nk dS aij... k ,k dV V aij... k dxt rstaij... k ,s nr dS S Green’s Theorem in the Plane g f S x y dxdy C ( fdx gdy) Zero-Value Theorem V g S x dxdy C gn x ds , f ij... k dV 0 f ij... k 0 V f S y dxdy C fny ds Orthogonal Curvilinear Coordinate Systems x3 x3 z ê z êr ê e1 r Cylindrical Coordinate System (r,,z) x1 r cos , x 2 sin , x3 z x12 x 22 , tan 1 x2 , z x3 x1 ê e3 x2 e2 x1 r êr e3 ê e1 R x2 e2 x1 Spherical Coordinate System (R,,) x1 R cos sin , x2 R sin sin , x3 R cos R x12 x22 x32 cos 1 x3 x12 x22 x32 tan 1 x2 , x1 General Curvilinear Coordinate Systems Common Differential Forms x3 3 eˆ1 1 1 1 1 eˆ 2 eˆ3 eˆ i 1 2 3 h1 h2 h3 hi i i f eˆ1 1 f 1 f 1 f 1 f eˆ 2 eˆ3 eˆ i 1 2 3 h1 h2 h3 hi i i 2 ê3 ê2 ê1 e3 1 x2 e2 e1 2 1 h1h2 h3 i ( x , x , x ) , x x ( , , ) m i i u x1 m 1 2 3 m m 1 2 3 (ds ) 2 (h1d1 ) 2 (h2 d 2 ) 2 (h3 d3 ) 2 h1h2 h3 u i i hi 1 h1h2 h3 u u i j h1h2 h3 i 2 i ( h ) i ijk (u k hk )eˆi h j hk j j k eˆ i hi eˆ j u j eˆ j u j i i eˆ j eˆ eˆ u k j eˆ j u j 2 u i i k k h h i j k i k Example 1-5: Polar Coordinates From relations (1.9.5) or simply using the geometry shown in Figure eˆ r cos e1 sin e 2 eˆ sin e1 cos e 2 x2 eˆ eˆ eˆ r eˆ eˆ , eˆ r , r 0 r r ê êr The basic vector differential operations then follow to be 1 eˆ r eˆ r r 1 eˆ r eˆ r r 1 1 u u (ru r ) r r r 1 1 2 2 r r r r r 2 2 1 u r 1 u (ru ) eˆ z r r r u u 1 u 1 u u r eˆ r eˆ r eˆ r eˆ r u eˆ eˆ r u r eˆ eˆ r r r r u u 2 u 2 u 2 u 2 u r 2 2r eˆ r 2 u 2 r 2 eˆ r r r r where u ur eˆ r ueˆ θ , eˆ z eˆ r eˆ θ e2 r e1 x1 (ds ) 2 (dr ) 2 (rd) 2 h1 1 , h2 r