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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 PowerPoint Presentation
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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 Basic FE Illustration Some Examples of the Second Order Equations in 1- Dimension

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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
        • 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

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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

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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

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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

slide5

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

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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