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FE Course Lecture II – Outline UCSD - 10/09/03 Review of Last Lecture (I) Formal Definition of FE: Basic FE Concepts Ba

FE Course Lecture II – Outline UCSD - 10/09/03 Review of Last Lecture (I) Formal Definition of FE: Basic FE Concepts Basic FE Illustration Some Examples of the Second Order Equations in 1- Dimension

JasminFlorian
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FE Course Lecture II – Outline UCSD - 10/09/03 Review of Last Lecture (I) Formal Definition of FE: Basic FE Concepts Ba

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  1. FE Course Lecture II – Outline • UCSD - 10/09/03 • Review of Last Lecture (I) • Formal Definition of FE: • Basic FE Concepts • Basic FE Illustration • Some Examples of the Second Order Equations in 1- Dimension • Some Examples of the Poisson Equation – . (ku) = f and Some Examples of Coupled Systems • Intro to 1-Dimensional FEs [Beams and Bars]. • Fluid Mechanics Problem • Heat Transfer (Thermal) Problem • Beam/Bar problem Finite Elements Principles and Practices - Fall 03

  2. 1-Dimensional Finite Elements • Stiffness and Load Vector Formulations for mechanical, heat transfer and fluid flow problems. • The system equation to be solved can be written in matrix form as: • [K] {D} = {q} • Where • [K] is traditional known as the ‘stiffness’ or ‘coefficient’ matrix (conductance matrix for heat transfer, flow-resistance matrix for fluid flow), • {D}is the displacement (or temperature, or velocity) vector and • {q} is the force (or thermal load, or pressure gradient) vector. Finite Elements Principles and Practices - Fall 03

  3. Tbase=100oC Tamb=20oC 5 • A) For heat transfer problem in 1-dimensional, we have: • fx = -Kdt/dx [Fourier Heat Conduction Equation] • Q = -KAdt/dx (where Q=A fx) • [KT}{T} = {Q} [applicable for steady-state heat transfer problems] 1 5 Finite Elements Principles and Practices - Fall 03

  4. B) For fluid flow problem in 1-dimensional, we have: • md2u/dy2 – dp/dx = 0 • [KF}{u} = {P} [applicable for steady-state flow problems]. P – pressure gradient Finite Elements Principles and Practices - Fall 03

  5. C) For stress problem in 1-dimensional, we have: • -kd2u/dx2 – q = 0 • [KF}{u} = {F}. F – joint force. u=uo = 0 How about for a tube under pure torsion? How will the coefficients look like? Finite Elements Principles and Practices - Fall 03

  6. Review of Analysis Results. E.g., stress distribution. Exact Vs FE solution. Error Estimation. • SOFTWARE-Specific Session: • Intro to software-specific issues. h-elements, p-Elements, adoptive meshing. • Build 1D problem on ANSYS. Go through all steps. • Thermal problem on ANSYS • Bar problem on ANSYS • Flow problem on ANSYS/FEMLAB. • Homework 1 and Reading Assignments. Finite Elements Principles and Practices - Fall 03

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