MIT2_092F09_lec03

MIT2_092F09_lec03 - 2.092/2.093 — Finite Element Analysis...

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Unformatted text preview: 2.092/2.093 — Finite Element Analysis of Solids & Fluids I Fall ‘09 Lecture 3- Analysis of Solids/Structures and Fluids Prof. K. J. Bathe MIT OpenCourseWare The fundamental conditions to be satisfied are: I. Equilibrium: in solids, F = m a ; in ﬂuids, conservation of momentum II. Compatibility: continuity and boundary conditions III. Constitutive relations: Stress/strain law Each joint and each element must be in equilibrium. From last lecture: We want to solve KU = R for this system. We know that ⎤ ⎡ R = ⎢ ⎢ ⎢ ⎢ ⎣ − P ⎥ ⎥ ⎥ ⎥ ⎦ To solve for K , assume u 5 = 1, u 1 = u 2 = u 3 = u 4 = 0. Then, the left hand side becomes ⎤ ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ k 15 k 25 k 35 k 45 k 55 ⎥ ⎥ ⎥ ⎥ ⎦ = R where R are the external applied forces corresponding to the imposed displacement. Example Consider a simple bar: 1 Lecture 3 Analysis of Solids/Structures and Fluids 2.092/2.093, Fall ‘09 EA EA F = L × [ Displacement u ] → K ˜ = L Truss bars can only resist...
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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_lec03 - 2.092/2.093 — Finite Element Analysis...

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