Handout 1.3 FEA BigIdeas WithNotes
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Transcript of Handout 1.3 FEA BigIdeas WithNotes
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7/26/2019 Handout 1.3 FEA BigIdeas WithNotes
1/12
Pre-Analysis
1. Mathematical model
2. Numerical solution procedure
3. Hand-calculations of expected results/trends
Example: Steady One-Dimensional
Heat Conduction in a Bar
L
x
y
z
We are interested in finding the temperature
distribution in the bar due to heat conduction
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x
Energy Conservation for an
Infinitesimal Control VolumeInfinitesimal
Control Volume
Mathematical Model: Governing
Equation and Boundary Conditions
Governing equation
+ = 0 , 0 Boundary conditions
0 = q L = q = Exact solution is straightforward
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Numerical Solution:
Discretization Reduce the problem to determining temperature
values at selected locations (nodes)
31 42
We have assumed a shape for () consisting of piecewisepolynomials
x
T
x
Post
processing
How to Find Nodal
Temperatures ?Mathematical Model
(Boundary Value
Problem)
System of
algebraic
equations in nodal
temperatures
Nodal
temperaturesInvert
= {}Piecewisepolynomialapproximation for T
()Each algebraicequation willrelate a nodaltemperature toits neighbors
1 3 42
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How to Derive System of Algebraic
Equations?
Weighted Integral
Form
+ = 0
()is an
arbitrary function
+ = 0(x) is an
arbitrary piecewise
polynomial function
Piecewise polynomial
approximation for T
System of algebraic eqs.
in nodal temperatures
+ = 0
Piecewise
polynomial
approximation for T
System of algebraic
eqs. in nodal
temperatures
How to Derive System of Algebraic
Equations?
+ dx = 0(x) is an
arbitrary piecewise
polynomial function
Piecewise polynomial
approximation for T
System of algebraic eqs.
in nodal temperatures
x
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Integration by Parts
+ = 0 wk
k + = 0
wk dT
k dT +
= 01 3 42
w + + 0.5 +
+ + + +
+ + + +w + + 0.5 + = 0
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wk dT
k dT +
= 0
1 3 42
+ w + + 0.5 +
wk dT
k dT +
= 01 3 42
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wk dT
k dT +
= 0
1 3 42
w + + 0.5 +
+ + + + + + + +
w + + 0.5 + = 0 = {}
wk dT
k dT +
= 01 3 42
= {}
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wk dT
k dT +
= 0
1 3 42
w + + 0.5 +
+ + + + + + + +
w + + 0.5 + = 0
wk dT
k dT +
= 01 3 42
+ + = + = 0.5
+ + = + = 0.5 +
= {}
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Essential Boundary Conditions
+ =0.5Q = T + + =
+ + = + = 0.5 +
Comparison of Finite-Element and Exact
Solutions
Nodal temperature values are exact Unusual property of 1D FE solution
Temperature boundary condition issatisfied exactly
Flux boundary condition is satisfiedapproximately
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Comparison of / between Finite-Element and Exact Solutions
Error in /> Error in Energy is not conserved for
each element
Reaction at Left Boundary
=
= 5.5 W/m Energy isconserved forthe bar
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How to Improve the Polynomial
Approximation?
Increase no. of elements
Increase order of polynomialwithin each element
Use more nodes perelement
Original Mesh
1 3 42
Refined Mesh
1 2 3 4 5 6 7
Second-Order Element
Error Reduction: Results
3 elements 6 elements1 element, second-
order polynomial
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Finite-Element Analysis: Summary of
the Big Ideas
Mathematical model to be solved is usually a boundary valueproblem
Reduce the problem to solving selected variable(s) at selectedlocations (nodes)
Assume a shape for selected variable(s) within each element
Derive system of algebraic equations relating neighboring nodalvalues
Invert this system to determine selected variable(s) at nodes
Derive everything else from selected variable(s) at nodes
Finite-Element Analysis: Summary of
the Big Ideas
Reduce error by using more elements and/orincreasing the order of interpolation
Finite-element solution doesnt satisfy thedifferential equation(s) Satisfies a special weighted integral form
Essential boundary conditions are satisfiedexactly
Natural or gradient boundary conditions aresatisfied approximately
