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Unformatted text preview: 2.092/2.093 — Finite Element Analysis of Solids & Fluids I Fall ‘09 Lecture 6- Finite Element Solution Process Prof. K. J. Bathe MIT OpenCourseWare In the last lecture, we used the principle of virtual displacements to obtain the following equations: KU = R (1) K = Σ K ( m ) ; K ( m ) = B ( m ) T C ( m ) B ( m ) dV ( m ) m V ( m ) R = R B + R S R B = Σ R B ( m ) ; R B ( m ) = H ( m ) T f B ( m ) dV ( m ) m V ( m ) f f m S S i i ( m ) f S R S = Σ R ( m ) ; R ( m ) = Σ H S i ( m ) T f S i ( m ) dS i ( m ) f u ( m ) = H ( m ) U (2) ↓ ε ( m ) = B ( m ) U (3) Note that the dimension of u ( m ) is in general not the same as the dimension of ε ( m ) . Example: Static Analysis Reading assignment: Example 4.5 1 Lecture 6 Finite Element Solution Process 2.092/2.093, Fall ‘09 Assume: i. Plane sections remain plane ii. Static analysis no vibrations/no transient response → iii. One-dimensional problem; hence, only one degree of freedom per node Elements 1 and 2 are compatible because...
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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.
- Fall '09
- Finite Element Analysis