unit5_notes

unit5_notes - 16.21 Techniques of Structural Analysis and...

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Unformatted text preview: 16.21 Techniques of Structural Analysis and Design Spring 2005 Unit #5 - Constitutive Equations Ra´ul Radovitzky February 23, 2005 Constitutive Equations For elastic materials: σ ij = σ ij ( ) = ρ ∂U 0 ∂ ij (1) If the relation is linear: σ ij = C ijkl kl , Generalized Hooke’s Law (2) In this expression: C ijkl fourth-order tensor of material properties or Elastic moduli (How many material constants?). Making use of the symmetry of the stress tensor: σ ij = σ ji ⇒ C jikl = C ijkl (3) Proof by (generalizable) example: σ 21 = C 21 kl kl , σ 12 = C 12 kl kl σ 21 = σ 12 C 21 kl kl = C 12 kl kl ⇒ C 21 kl − C 12 kl kl = 0 ⇒ C 21 kl = C 12 kl 1 which generalizes to the statement. This reduces the number of material constants from 81 to 54. In a similar fashion we can make use of the symmetry of the strain tensor ij = ji ⇒ C ijlk = C ijkl (4) This further reduces the number of material constants to 36. To further reduce the number of material constants consider the conclusion from the first law for elastic materials, equation ( 1 ): ∂U σ ij = , U 0 : strain energy density per unit volume (5) ∂ ij ∂U C ijkl kl = (6) ∂ ij ∂ 2 U C ijkl kl = (7) ∂ mn ∂ mn ∂ ij ∂ 2 U C ijkl δ km δ ln = (8) ∂ mn ∂ ij ∂ 2 U C ijmn = (9) ∂ mn ∂ ij Assuming equivalence of the mixed partials: C ijkl = ∂ 2 U 0 ∂ kl ∂ ij = ∂ 2 U 0 ∂ ij ∂ kl = C klij (10) This further reduces the number of material constants to 21. The most general anisotropic linear elastic material therefore has 21 material constants. general anisotropic linear elastic material therefore has 21 material constants....
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unit5_notes - 16.21 Techniques of Structural Analysis and...

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