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Unformatted text preview: Section 1.11 Solid Mechanics Part III Kelly 96 1.11 The Eigenvalue Problem and Polar Decomposition 1.11.1 Eigenvalues, Eigenvectors and Invariants of a Tensor Consider a secondorder tensor A . Suppose that one can find a scalar λ and a (nonzero) normalised, i.e. unit, vector n ˆ such that n n A ˆ ˆ λ = (1.11.1) In other words, A transforms the vector n ˆ into a vector parallel to itself, Fig. 1.11.1. If this transformation is possible, the scalars are called the eigenvalues (or principal values ) of the tensor, and the vectors are called the eigenvectors (or principal directions or principal axes ) of the tensor. It will be seen that there are three vectors n ˆ (to each of which corresponds some scalar λ ) for which the above holds. Figure 1.11.1: the action of a tensor A on a unit vector Equation 1.11.1 can be solved for the eigenvalues and eigenvectors by rewriting it as ( ) ˆ = − n I A λ (1.11.2) or, in terms of a Cartesian coordinate system, ( ) ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ = − → = − → = ⊗ − ⊗ i i j ij r r i j ij r r q p pq k k j i ij n n A n n A n n A e e e e e e e e e λ λ λδ In full, [ ] [ ] [ ] ˆ ) ( ˆ ˆ ˆ ˆ ) ( ˆ ˆ ˆ ˆ ) ( 3 3 33 2 32 1 31 2 3 23 2 22 1 21 1 3 13 2 12 1 11 = − + + = + − + = + + − e e e n A n A n A n A n A n A n A n A n A λ λ λ (1.11.3) Dividing out the base vectors, this is a set of three homogeneous equations in three unknowns (if one treats λ as known). From basic linear algebra, this system has a solution (apart from ˆ = i n ) if and only if the determinant of the coefficient matrix is zero, i.e. if A n ˆ n ˆ λ Section 1.11 Solid Mechanics Part III Kelly 97 det ) det( 33 32 31 23 22 21 13 12 11 = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − − = − λ λ λ λ A A A A A A A A A I A (1.11.4) Evaluating the determinant, one has the following cubic characteristic equation of A , III II I 2 3 = − + − A A A λ λ λ Tensor Characteristic Equation (1.11.5) where ( ) [ ] A A A A A A A det III ) tr( ) (tr II tr I 3 2 1 2 2 2 1 2 1 = = − = − = = = k j i ijk ij ji jj ii ii A A A A A A A A ε (1.11.6) It can be seen that there are three roots 3 2 1 , , λ λ λ , to the characteristic equation. Solving for λ , one finds that 3 2 1 1 3 3 2 2 1 3 2 1 III II I λ λ λ λ λ λ λ λ λ λ λ λ = + + = + + = A A A (1.1.7) The eigenvalues (principal values) i λ must be independent of any coordinate system and, from Eqn. 1.11.5, it follows that the functions A A A III , II , I are also independent of any coordinate system. They are called the principal scalar invariants (or simply invariants ) of the tensor. Once the eigenvalues are found, the eigenvectors (principal directions) can be found by solving ˆ ) ( ˆ ˆ ˆ ˆ ) ( ˆ ˆ ˆ ˆ ) ( 3 33 2 32 1 31 3 23 2 22 1 21 3 13 2 12 1 11 = − + + = + − + = + + − n A n A n A n A n A n A n A n A n A λ λ λ (1.11.8) for the three components of the principal direction vector 3 2 1 ˆ , ˆ , ˆ n n n , in addition to the...
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This note was uploaded on 01/20/2012 for the course ENGINEERIN 3 taught by Professor Staff during the Fall '11 term at Auckland.
 Fall '11
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