{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MIT2_094S11_lec18

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

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

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

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

View Full Document
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 1 0 L + t u 2 1 t u 2 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 0 t L L t τ 11 = E ˜ · 2 1 1 + 0 t L u 1 (18.11) E ˜ A t u 2 t u t P = 1 + 1 1 + (18.12) t τ 11 A = 2 0 L 0 L This is because of the material-law assumption (18.5) (okay for small strains . . . ) Hyperelasticity t W = f
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### 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
Ask a homework question - tutors are online