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Procedures of Finite Element Analysis Two-Dimensional Elasticity Problems Professor M. H. Sadd. Two Dimensional Elasticity Element Equation Orthotropic Plane Strain/Stress Derivation Using Weak Form – Ritz/Galerin Scheme. Displacement Formulation Orthotropic Case.

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slide1

Procedures of Finite Element Analysis

Two-Dimensional Elasticity Problems

Professor M. H. Sadd

slide2
Two Dimensional Elasticity Element EquationOrthotropic Plane Strain/Stress Derivation Using Weak Form – Ritz/Galerin Scheme

Displacement Formulation Orthotropic Case

two dimensional elasticity weak form
Two Dimensional Elasticity Weak Form

Mulitply Each Field Equation by Test Function & Integrate Over Element

Use Divergence Theorem to Trade Differentiation On To Test Function

two dimensional elasticity element equation triangular element n 3

(x3,y3)

3

e

1

2

(x1,y1)

(x2,y2)

Two Dimensional Elasticity Element EquationTriangular Element N = 3
slide6

y

(x3,y3)

e = 12 + 23 + 31

3

e

1

he = thickness

2

(x1,y1)

(x2,y2)

x

(Element Geometry)

Two Dimensional Elasticity Element EquationPlane Strain/Stress Derivation Using Virtual Work Statement
triangular element with linear approximation

v3

u3

(x3,y3)

v2

3

y

u2

2

(x2,y2)

v1

1

u1

(x1,y1)

x

Lagrange Interpolation Functions

1

3

3

3

2

3

1

1

1

1

1

1

2

2

2

Triangular Element With Linear Approximation
loading terms for triangular element with uniform distribution

y

(x3,y3)

e = 12 + 23 + 31

3

e

1

he = thickness

2

(x1,y1)

(x2,y2)

x

(Element Geometry)

Loading Terms for Triangular Element With Uniform Distribution
two dimensional elasticity element equation rectangular element n 4
Two Dimensional Elasticity Element EquationRectangular Element N = 4

(x3,y3)

(x4,y4)

4

3

e

2

1

(x2,y2)

(x1,y1)

two dimensional elasticity rectangular element equation orthotropic case
Two Dimensional Elasticity Rectangular Element Equation - Orthotropic Case

(x4,y4)

(x3,y3)

4

3

e

2

1

(x2,y2)

(x1,y1)

fea of elastic 1 x 1 plate under uniform tension

y

3

4

2

3

3

T

1

x

2

1

2

1

2

1

FEA of Elastic 1x1 Plate Under Uniform Tension

Element 1:1 = -1, 2 = 1, 3 = 0, 1 = 0, 2 = -1, 3 = 1, A1 = ½.

Element 2:1 = 0, 2 = 1, 3 = -1, 1 = -1, 2 = 0, 3 = 1, A1 = ½

fea of elastic plate

3

4

2

3

3

2

T

1

1

2

1

2

1

FEA of Elastic Plate

Boundary Conditions

U1 = V1 = U4 = V4 = 0

solution of elastic plate problem

3

4

2

3

3

2

T

1

1

2

1

2

1

Solution of Elastic Plate Problem

Choose Material Properties:

E = 207GPa and v = 0.25

Note the lack of symmetry in the displacement solution

axisymmetric formulation

z

  • = constant plane

4

3

1

2

r

Axisymmetric Formulation
two dimensional fea code matlab pde toolbox
Two-Dimensional FEA Code MATLAB PDE Toolbox
  • - Simple Application Package For Two-Dimensional Analysis Initiated by Typing “pdetool” in Main MATLAB Window
  • Includes a Graphical User Interface (GUI) to: - Select Problem Type - Select Material Constants - Draw Geometry - Input Boundary Conditions - Mesh Domain Under Study - Solve Problem - Output Selected Results
two dimensional fea example using matlab pde toolbox cantilever beam problem
Two-Dimensional FEA ExampleUsing MATLAB PDE ToolboxCantilever Beam Problem

L/2c = 5

g1=0

g2=100

2c = 0.4

L = 2

Mesh: 4864 Elements, 2537 Nodes

fea matlab pde toolbox example cantilever beam problem stress results
FEA MATLAB PDE Toolbox ExampleCantilever Beam ProblemStress Results

E = 10x106 , v = 0.3

Contours of sx

g1=0

g2=100

2c = 0.4

L = 2

FEA Result: smax = 3200

fea matlab pde toolbox example cantilever beam problem displacement results
FEA MATLAB PDE Toolbox ExampleCantilever Beam ProblemDisplacement Results

Contours of Vertical Displacement v

E = 10x106 , v = 0.3

g1=0

g2=100

2c = 0.4

L = 2

FEA Result: vmax = 0.00204

two dimensional fea example using matlab pde toolbox plate with circular hole
Two-Dimensional FEA ExampleUsing MATLAB PDE ToolboxPlate With Circular Hole

Contours of Horizontal Stress x

Stress Concentration Factor: K  2.7Theoretical Value: K = 3

two dimensional fea example using matlab pde toolbox plate with circular hole26
Two-Dimensional FEA ExampleUsing MATLAB PDE ToolboxPlate With Circular Hole

Contours of Horizontal Stress x

Stress Concentration Factor: K  3.5Theoretical Value: K = 4

fea matlab example plate with elliptical hole
FEA MATLAB ExamplePlate with Elliptical Hole

(Finite Element Mesh: 3488 Elements, 1832 Nodes)

Aspect Ratio b/a = 2

(Contours of Horizontal Stress x)

Stress Concentration Factor K 3.3

Theoretical Value: K = 5

fea example diametrical compression of circular disk
FEA ExampleDiametrical Compression of Circular Disk

Theoretical Contours of Maximum Shear Stress

(Contours of Max Shear Stress)

(FEM Mesh: 1112 Elements, 539 Nodes)

(Contours of Max Shear Stress)

(FEM Mesh: 4448 Elements, 2297 Nodes)

Experimental Photoelasticity

Isochromatic Contours