1. X1X1 X2X2  Basic Kinematics Real Applications http://www.newswise.com/ Simple Shear Trivial geometry Proscribed homogenous deformations Linear constitutive equations No equilibrium considerations = Analytical solutions Aortic Graft for Aneurysms Complex geometry Complex boundary conditions Nonlinear constitutive equations Solve equilibrium equations = No analytical solutions NEED NUMERICAL METHODS

2. Objective: Numerically approximate the solutions to PDEs History: Evolved first from the matrix methods of structural analysis in the early 1960’s Uses the algorithms of linear algebra Later found to have a more fundamental mathematical foundation The Finite Element Method

3. Steps: 1.Define the problem 2.Reformulate the governing equation into the weak form 3.Discretization 4.Evaluate element integrals 5.Assembly global matrices 6.Apply boundary condition 7.Solve linear system 8.Compute strains & stresses The Finite Element Method

4. Steps: 1.Define the problem 2.Reformulate the governing equation into the weak form 3.Discretization 4.Evaluate element integrals 5.Assembly global matrices 6.Apply boundary condition 7.Solve linear system 8.Compute strains & stresses The Finite Element Method Next class…

5. The Finite Element Method The domain is discretized into a mesh consisting of: –a finite number of elements –each element approximates the solution using a sum of polynomial interpolation functions, Ψ –Continuity between elements is ensured by defining the polynomial coefficients at nodes (element vertices) 5 6 8 9 4 1 2 3 4 5 6 7 8 9 1 2 3 4

6. Piecewise Interpolation Polynomials are convenient, and easily differentiated or integrated But if we increase the polynomial order too much to allow more complex variations, they can oscillate unrealistically Instead, divide the domain into elements and use low-order piecewise polynomials over each element, e.g. piecewise linear: Problem: piecewise polynomials are not continuous at the boundaries Temperature Distance along a metal rod

7. Piecewise Interpolation Rather than constrain the intercepts of neighboring element linear functions to ensure u matches at the boundaries, we reformulate the linear equation in each element with parameters u 1 and u 2, which are the values of u at the nodes at each end of that element: x1x1 x2x2 x3x3 x4x4 u1u1 u2u2 u3u3 u4u4 h is a normalized measure or distance within an element

8. 1-D Linear Lagrange Elements u = (1-  )u 1 +  u 2 =  1 u 1 +  2 u 2 x u 0 1  element 1 element 2element 3 nodes u1u1 u2u2 u3u3 u4u4 + + + + + + + + + + + + + 0 1  1     0 1 1    x = (1-  )x 1 +  x 2 =  1 x 1 +  2 x 2   u x u1u1 u2u2 x2x2 x1x1 1 1 u1u1 u2u2 u x2x2 x1x1 x  1 and  2 are the 1-D linear Lagrange basis functions

9. 1 0 1  1  2 u y x x =  n x n u =  n u n y =  n y n 1 1  1  2  0 2-D Bilinear Lagrange Elements 2D interpolation functions can be constructed by defining two local coordinates (ξ 1, ξ 2 ) and multiplying the 1D interpolation functions together: where  1 –  4 are the 2-D bilinear Lagrange basis functions

10. 3-D Trilinear Lagrange Elements 1 2 3 4 5 6 7 8 11 22 33 The trilinear element has 8 nodes with basis functions formed from products of of 1-D linear Lagrange functions  1 =1- , and  2 =  as functions of three local coordinates (  1,  2,  3 

11. 1-D Quadratic Lagrange Elements Use three nodal parameters u 1, u 2 and u 3    0 0.5 1.0 0 0.5 1.0 0 0.5 1.0 φ1φ1 φ2φ2 φ3φ3  1,  2,  3 are the 1-D quadratic Lagrange basis functions

12. Review Domain Mesh composed of elements that meet at nodes (vertices) Interpolation Approximate solution using piecewise functions over elements Define parameters at nodes to enforce continuity –Solution values are continuous – “Lagrangian interpolation” –Solution derivatives are also continuous – “Hermite interpolation” Basis Functions φ Polynomials used for interpolation Can be used to interpolate the geometry and the solution Weighting functions Can be linear or higher ordered - Quadratic - Cubic

13. Computing Strain Distributions found by solving governing equations with FEM

15. If running Mac OSX, first ensure that Xcode is installed and if not, install it from the CD that shipped with your Mac or from the AppStore for Mac OSX Lion. Go to http://www.continuity.ucsd.edu/Continuity/Download Run installer if it doesn’t run automatically. Accept license and install in default location. Windows install will also install Python if not found; on Mac OS, Python 2.5 will be installed Run Continuity, click on Register Now. Complete the Registration Form making sure to use your ucsd.edu email address, and your own key will be emailed to you Click OK, then File>Exit If you are having trouble, check herehere for known bugs *** ALSO AVAILABLE IN PFBH 161 Install Continuity

16. Install Continuity – Known Bugs

18. Work through Biomechanics Tutorial 4 at: http://www.continuity.ucsd.edu/Continuity/Documentation/Tutorials You will make a model of an artery using cylindrical 3-D trilinear Lagrange mesh with ten elements (blue) and then inflate it (red) Inflate a Cylindrical Tube

20. 1234 U 1 =0 2 4 6 8 x u U 4 =9 U 3 =? U 2 =? Galerkin FEM: Simple 1-D Example

21. 2.Integrate by parts 1.Formulate the (Galerkin) weighted residual form 0 (“natural” i.e. derivative boundary conditions)

22. 00.51 0 1 22 11 3.Discretize the problem 4 global nodes with nodal solution values U 1, U 2, U 3, U 4 3 linear elements each with 2 element nodal values u 1, u 2. Adjacent elements share global nodes, e.g., global variable U 2 is element variable u 2 of element 1 & u 1 of element 2. Two linear Lagrange interpolation functions for each element, Ψ 1 and Ψ 2 where in each element: u(x) = u 1 Ψ 1 + u 2 Ψ 2 = Σu i Ψ i (i=1,2) Global Node: 1234 x1x1 x2x2 x3x3 x4x4 U1U1 U2U2 U3U3 U4U4 x Element 1 Nodes: 2 u1u1 u2u2 1 Element 2 Nodes: 2 u1u1 u2u2 1 Element 3 Nodes: 2 u1u1 u2u2 1

23. In each element, let u(x) = u 1 Ψ 1 + u 2 Ψ 2 and ψ = Ψ i (x) 4.Derive Finite Element equations For Element 1 (no derivative boundary conditions): [k] = [(k ij )] is the element stiffness matrix f = (f i ) is the element load vector

24. [k]u = f Element stiffness matrix, [k] and load (RHS) vector, f 5. Compute element stiffness matrices

25. In this problem, each element is the same size and thus: [k] (ele 1) = [k] (ele 2) = [k] (ele 3) and: f (ele 1) = f (ele 2) = f (ele 3) 6. Compute element RHS vectors

26. 7. Assemble global stiffness matrix and RHS vector

27. That leaves global equations 2 and 3 7. Apply boundary conditions Exact! 8. Solve remaining global equations