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MIT2_094S11_lec12 - 2.094 Finite Element Analysis of Solids...

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2.094 Finite Element Analysis of Solids and Fluids Fall ‘08 Lecture 12 - Total Lagrangian formulation Prof. K.J. Bathe MIT OpenCourseWare We discussed: t t X = x x i j d t x = t X d 0 x , d 0 x = t X 1 d t x (12.1) 0 0 0 0 t C = t X T t X (12.2) 0 0 0 d 0 x = 0 t X d t x where 0 t X = 0 t X 1 = t 0 x i (12.3) ∂ x j The Green-Lagrange strain: 1 1 0 t = 0 t X T 0 t X I = 0 t C I (12.4) 2 2 Polar decomposition: 1 2 0 t X = 0 t R 0 t U 0 t = 2 0 t U I (12.5) We see, physically that: where d t t x and d t x are the same lengths the components of the G-L strain do not change. Note in FEA 0 0 k x i = h k x i k for an element (12.6) t t k u i = h k u i k t 0 t x i = x i + u i for any particle (12.7) Hence for the element t 0 k t k x i = h k x i + h k u i (12.8) k k = h k 0 x i k + t u i k (12.9) k = h k t x k k (12.10) k 49
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MIT 2.094 12. Total Lagrangian formulation E.g., k = 4 2nd Piola-Kirchhoff stress 0 ρ 0 t S = t ρ 0 t X t τ 0 t X T components also
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