1 / 28

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.

grovem
Télécharger la présentation

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

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related