By James G Knudsen
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20) After simplifications, we obtain F= 1 ∂r ∂r ∂r gr ⊗ GR + gr ⊗ GΦ + gr ⊗ GZ ∂R R ∂Φ ∂Z ∂ϕ r ∂ϕ ∂ϕ gϕ ⊗ GR + gϕ ⊗ GΦ + r gϕ ⊗ GZ +r ∂R R ∂Φ ∂Z ∂z 1 ∂z ∂z gz ⊗ GR + gz ⊗ GΦ + gz ⊗ GZ . 23) where is the right Cauchy–Green tensor. 24) where ζi is the corresponding eigenvalue of C. The latter equation means that eigenvalues of the right Cauchy–Green tensor are equal to the squared stretches in principal directions. Thus, we can write the spectral decomposition of C in the form C = λ21 m(1) ⊗ m(1) + λ22 m(2) ⊗ m(2) + λ23 m(3) ⊗ m(3) .
24) where ζi is the corresponding eigenvalue of C. The latter equation means that eigenvalues of the right Cauchy–Green tensor are equal to the squared stretches in principal directions. Thus, we can write the spectral decomposition of C in the form C = λ21 m(1) ⊗ m(1) + λ22 m(2) ⊗ m(2) + λ23 m(3) ⊗ m(3) . 26) where all principal stretches are positive. We assume then that any deformation gradient can be multiplicatively decomposed as F = RU. 28) R = FU−1 . Let us analyze properties of R. First, we observe that it is orthogonal RT R = (FU−1 )T FU−1 = U−T FT FU−1 = U−T U2 U−1 = 1.
49) is the referential mass density. Mass balance takes form Linear momentum balance is where a = ρ0 ∂v/∂t is the acceleration vector. Angular momentum balance is PFT = FPT . 50) Since the first Piola-Kirchhoff stress tensor is not symmetric it is convenient to introduce the second Piola-Kirchhoff stress tensor (2PK), which is symmetric, S = F−1 P = J F−1 σF−T = ST . 52) where n0 is the unit vector outward normal to the body surface ∂Ω0 in the referential configuration and t¯0 is the prescribed Lagrangean traction on this surface.