MIT2_094S11_lec4

# MIT2_094S11_lec4 - 2.094 — Finite Element Analysis of...

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Unformatted text preview: 2.094 — Finite Element Analysis of Solids and Fluids Fall ‘08 Lecture 4 - Finite element formulation for solids and structures Prof. K.J. Bathe MIT OpenCourseWare We considered a general 3D body, Reading: Ch. 4 The exact solution of the mathematical model must satisfy the conditions: • Equilibrium within t V and on t Sf , • Compatibility • Stress-strain law(s) I. Diﬀerential formulation II. Variational formulation (Principle of virtual displacements) (or weak formulation) We developed the governing F.E. equations for a sheet or bar We obtained t F = tR (4.1) where t F is a function of displacements/stresses/material law; and t R is a function of time. Assume for now linear analysis: Equilibrium within 0 V and on 0 Sf , linear stress-strain law and small displacements yields t F = K · tU (4.2) We want to establish, K U (t) = R(t) (4.3) 14 MIT 2.094 4. Finite element formulation for solids and structures Consider ˆ UT = � U1 V1 W1 U2 · · · WN � (N nodes) (4.4) ˆ where U T is a distinct nodal point displacement vector. Note: for the moment “remove Su ” We also say � ˆ U T = U1 U2 U3 · · · Un � (n = 3N ) (4.5) We now assume ⎤(m) u = ⎣ v ⎦ w ⎡ ˆ u(m) = H (m) U , u(m) (4.6a) ˆ where H (m) is 3 x n and U is n x 1. ˆ �(m) = B (m) U (4.6b) where B (m) is 6 x n, and T �(m) = e.g. γxy � �xx �yy ∂v ∂ u = + ∂x ∂y �zz γxy γyz γzx � We also assume u(m) � (m) = ˆ H (m) U (4.6c) = (m) ˆ U (4.6d) B 15 MIT 2.094 4. Finite element formulation for solids and structures Principle of Virtual Work: � � T T ˆ � τ dV = U f B dV V (4.7) V (4.7) can be rewritten as �� m T �(m) τ (m) dV (m) = �� V (m) m ˆ U (m) T fB (m) dV (m) (4.8) V (m) Substitute (4.6a) to (4.6d). � � � � T (m) T (m) (m) ˆ U B τ dV = V (m) m ˆ U T � �� H (m) T f B (m) (4.9) � dV (m) V (m) m ˆ τ (m) = C (m) �(m) = C (m) B (m) U Finally, � � T �� ˆ U � B (m) T (4.10) � C (m) B (m) dV (m) ˆ U= V (m) m � � T �� ˆ U � m H (m) T (4.11) � fB (m) dV (m) V (m) with T T ˆ �(m) = U B (m) T (4.12) ˆ KU = RB (4.13) where K is n x n, and RB is n x 1. Direct stiﬀness method: � K= K (m) (4.14) m RB = � (m) RB (4.15) m K (m) (m) RB � T B (m) C (m) B (m) dV (m) = �V V (4.16) (m) (m) T H (m) f B = (m) dV (m) (4.17) 16 MIT 2.094 4. Finite element formulation for solids and structures Example 4.5 textbook E = Young’s Modulus Mathematical model Plane sections remain plane: F.E. model ⎡ ⎤ U1 U = ⎣ U2 ⎦ U3 (4.18) Element 1 ⎤ U1 0 ⎣ U2 ⎦ �U 3 ⎡ u(1) (x) = � � 1− x 100 x �� 100 H (1) � (4.19) 17 MIT 2.094 4. Finite element formulation for solids and structures ⎤ U1 0 ⎣ U2 ⎦ �U 3 (4.20) x 80 (4.21) ⎡ �(1) (x) = xx � − 11 00 � 1 ��100 B (1) � Element 2 u(2) (x) = � 0 � �(2) (x) xx = � � 0 x 1 − 80 �� H (2) − 810 810 �� B (2) � U � � U (4.22) � Then, ⎡ ⎤ 1 −1 0 0 13E ⎣ E⎣ −1 1 0 ⎦ + 0 K= 240 100 0 00 0 ⎡ ⎤ 0 0 1 −1 ⎦ −1 1 (4.23) where, � � A� < η =0 1 � � AE E (1) ≡ L 100 �� 13 E E · 13 = 3 80 3 · 80 � �� � A∗ � � A∗ < A� (4.24) (4.25) (4.26) η =80 < 4.333 < 9 18 MIT OpenCourseWare http://ocw.mit.edu 2.094 Finite Element Analysis of Solids and Fluids II Spring 2011 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. ...
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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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