MIT2_092F09_lec07

MIT2_092F09_lec07 - • • • • 2.092/2.093 — Finite Element Analysis of Solids& Fluids I Fall ‘09 Lecture 7 Finite Element Solution

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Unformatted text preview: • • • • 2.092/2.093 — Finite Element Analysis of Solids & Fluids I Fall ‘09 Lecture 7- Finite Element Solution Process, cont’d Prof. K. J. Bathe MIT OpenCourseWare Recall that we have established for the general system some expressions: KU = R (1) K = Σ K ( m ) ; R = R B + R S m K ( m ) = B ( m ) T C ( m ) B ( m ) dV ( m ) (2) V ( m ) Then, since C ( m ) T = C ( m ) , T K ( m ) T = B ( m ) T C ( m ) T B ( m ) T dV ( m ) = K ( m ) V ( m ) Therefore, K is symmetric! R ( m ) B = H ( m ) T f B ( m ) dV ( m ) (3) V ( m ) For statics, we use f B ( m ) , but for linear dynamics we must modify it: u ¨ ( m ) f B ( m ) = f ˜ B ( m ) − ρ ( m ) H ( m ) U ¨ (4) where f ˜ B ( m ) is the body force vector excluding inertial loads, and accelerations are interpolated in the same way as displacements. Now, Eq. (1) becomes MU ¨ + KU = R M = Σ M ( m ) ; M ( m ) = H ( m ) T ρ ( m ) H ( m ) dV ( m ) (5) m V ( m ) We need the initial conditions for U ( t = 0) and U ˙ ( t = 0). We will calculate only...
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This note was uploaded on 02/24/2012 for the course MECHANICAL 2.092 taught by Professor Klaus-jürgenbathe during the Fall '09 term at MIT.

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MIT2_092F09_lec07 - • • • • 2.092/2.093 — Finite Element Analysis of Solids& Fluids I Fall ‘09 Lecture 7 Finite Element Solution

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