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Procedures of Finite Element Analysis Two-Dimensional Elasticity Problems Professor M. H. Sadd

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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Procedures of Finite Element Analysis Two-Dimensional Elasticity Problems Professor M. H. Sadd

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  1. Procedures of Finite Element Analysis Two-Dimensional Elasticity Problems Professor M. H. Sadd

  2. Two Dimensional Elasticity Element EquationOrthotropic Plane Strain/Stress Derivation Using Weak Form – Ritz/Galerin Scheme Displacement Formulation Orthotropic Case

  3. 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

  4. Two Dimensional Elasticity Ritz-Galerkin Method

  5. (x3,y3)  3 e 1  2 (x1,y1)  (x2,y2) Two Dimensional Elasticity Element EquationTriangular Element N = 3

  6. 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

  7. Two Dimensional Elasticity Element EquationInterpolation Scheme

  8. Two Dimensional Elasticity Element Equation

  9. 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

  10. Triangular Element With Linear Approximation

  11. 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

  12. Rectangular Element Interpolation y 4 3 b x 1 2 a

  13. Two Dimensional Elasticity Element EquationRectangular Element N = 4 (x3,y3) (x4,y4)   4 3 e 2 1   (x2,y2) (x1,y1)

  14. Rectangular Element With BiLinear Approximation

  15. Two Dimensional Elasticity Rectangular Element Equation - Orthotropic Case (x4,y4) (x3,y3)   4 3 e 2 1   (x2,y2) (x1,y1)

  16. 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

  17. 3 4   2 3 3 2 T 1 1 2 1   2 1 FEA of Elastic Plate Boundary Conditions U1 = V1 = U4 = V4 = 0

  18. 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

  19. z • = constant plane 4 3 1 2 r Axisymmetric Formulation

  20. Axisymmetric Formulation

  21. 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

  22. 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

  23. 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

  24. 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

  25. Two-Dimensional FEA ExampleUsing MATLAB PDE ToolboxPlate With Circular Hole Contours of Horizontal Stress x Stress Concentration Factor: K  2.7Theoretical Value: K = 3

  26. Two-Dimensional FEA ExampleUsing MATLAB PDE ToolboxPlate With Circular Hole Contours of Horizontal Stress x Stress Concentration Factor: K  3.5Theoretical Value: K = 4

  27. 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

  28. 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

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