MIT2_094S11_lec18

# MIT2_094S11_lec18 - 2.094 Finite Element Analysis of Solids...

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2.094 Finite Element Analysis of Solids and Fluids Fall ‘08 Lecture 18 - Solution of F.E. equations Prof. K.J. Bathe MIT OpenCourseWare In structures, Reading: Sec. 8.4 F ( u , p ) = R . (18.1) In heat transfer, F ( θ ) = Q (18.2) In ﬂuid ﬂow, F ( v , p , θ ) = R (18.3) In structures/solids F ( m ) ± t ( m ) T t ˆ ( m ) 0 ( m ) F = = 0 B L 0 S d V (18.4) 0 V ( m ) m m Elastic materials Example p. 590 textbook 76

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± MIT 2.094 18. Solution of F.E. equations Material law t S ˜ 0 11 = E 0 t 11 (18.5) In isotropic elasticity: E ˜ = E (1 ν ) , ( ν = 0 . 3) (18.6) (1 + ν ) (1 2 ν ) 1 ² ³ ´ 2 µ 1 · 0 L + t u ¸ 2 ¹ 1 · t u ¸ 2 ¹ 0 t = 2 0 t U I 0 t 11 = 2 0 L 1 = 2 1 + 0 L 1 (18.7) where 0 t U is the stretch tensor. 0 ρ t S = 0 X t τ 0 X T (18.8) 0 11 t 11 11 t 11 t ρ with 0 L 0 t X 11 = , 0 ρ 0 L = t ρ t L (18.9) 0 L + t u · ¸ 2 t L 0 L 0 L 0 t S 11 = 0 L t L t τ 11 = t L t τ 11 (18.10) · ¸ 2 ¹ 0 t L L t τ 11 = E ˜ · 2 1 1 + 0 t L u 1 (18.11) E ˜ A · t u ¸ 2 ¹ · t u ¸ t P = 1 + 1 1 + (18.12) t τ 11 A = 2 0 L 0 L This is because
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## This note was uploaded on 12/29/2011 for the course ENGINEERIN 2.094 taught by Professor Prof.klaus-jürgenbathe during the Spring '11 term at MIT.

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MIT2_094S11_lec18 - 2.094 Finite Element Analysis of Solids...

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