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

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 Ritz-Galerkin Method

Two Dimensional Elasticity Element Equation Triangular Element N =    ee (x1,y1)(x1,y1) (x2,y2)(x2,y2) (x3,y3)(x3,y3)

Two Dimensional Elasticity Element Equation Plane Strain/Stress Derivation Using Virtual Work Statement x y    ee (Element Geometry)  e =  12 +  23 +  31 (x1,y1)(x1,y1) (x2,y2)(x2,y2) (x3,y3)(x3,y3) h e = thickness

Two Dimensional Elasticity Element Equation Interpolation Scheme

Two Dimensional Elasticity Element Equation

Triangular Element With Linear Approximation    1 1   1  2    1 33 Lagrange Interpolation Functions x y (x1,y1)(x1,y1) (x2,y2)(x2,y2) (x3,y3)(x3,y3) u1u1 u2u u3u3 v1v1 v2v2 v3v3

Triangular Element With Linear Approximation

Loading Terms for Triangular Element With Uniform Distribution x y    ee (Element Geometry)  e =  12 +  23 +  31 (x1,y1)(x1,y1) (x2,y2)(x2,y2) (x3,y3)(x3,y3) h e = thickness

Rectangular Element Interpolation a b x y

Two Dimensional Elasticity Element Equation Rectangular Element N = 4 (x3,y3)(x3,y3)   ee (x1,y1)(x1,y1) (x2,y2)(x2,y2) 4  (x4,y4)(x4,y4) 

Rectangular Element With BiLinear Approximation

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

FEA of Elastic 1x1 Plate Under Uniform Tension x    T  y Element 1:  1 = -1,  2 = 1,  3 = 0,  1 = 0,  2 = -1,  3 = 1, A 1 = ½. Element 2:  1 = 0,  2 = 1,  3 = -1,  1 = -1,  2 = 0,  3 = 1, A 1 = ½

FEA of Elastic Plate Boundary Conditions U 1 = V 1 = U 4 = V 4 = 0    T 

   T  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  constant plane z r

Axisymmetric Formulation

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 L = 2 g1=0 g2=100 2c = 0.4 Mesh: 4864 Elements, 2537 Nodes L/2c = 5

FEA MATLAB PDE Toolbox Example Cantilever Beam Problem Stress Results L = 2 g1=0 g2=100 E = 10x10 6, v = 0.3 2c = 0.4 Contours of s x FEA Result: s max = 3200

FEA MATLAB PDE Toolbox Example Cantilever Beam Problem Displacement Results L = 2 g1=0 g2=100 E = 10x10 6, v = 0.3 Contours of Vertical Displacement v FEA Result: v max = c = 0.4

Two-Dimensional FEA Example Using MATLAB PDE Toolbox Plate With Circular Hole Contours of Horizontal Stress  x Stress Concentration Factor: K  2.7 Theoretical Value: K = 3

Contours of Horizontal Stress  x Stress Concentration Factor: K  3.5 Theoretical Value: K = 4 Two-Dimensional FEA Example Using MATLAB PDE Toolbox Plate With Circular Hole

FEA MATLAB Example Plate with Elliptical Hole (Finite Element Mesh: 3488 Elements, 1832 Nodes) (Contours of Horizontal Stress  x ) Stress Concentration Factor K  3.3 Theoretical Value: K = 5 Aspect Ratio b/a = 2

FEA Example Diametrical Compression of Circular Disk (FEM Mesh: 1112 Elements, 539 Nodes) (Contours of Max Shear Stress) (FEM Mesh: 4448 Elements, 2297 Nodes) (Contours of Max Shear Stress) Theoretical Contours of Maximum Shear Stress Experimental Photoelasticity Isochromatic Contours