MIT2_094S11_lec18

MIT2_094S11_lec18 - 2.094 Finite Element Analysis of Solids...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
± 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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 6

MIT2_094S11_lec18 - 2.094 Finite Element Analysis of Solids...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online