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- 1. Finite Element Methods (FEM) Suzanne Vogel COMP 259 Spring, 2002
- 2. <ul><li>The finite element method is the formulation of a global model to simulate static or dynamic response to applied forces. </li></ul><ul><li>Models: energy, force, volume,… </li></ul><ul><li>This differs from a mass spring system , which is a local model . </li></ul>Definition of FEM
- 3. 1. Set up a global model in terms of the world coordinates of mass points of the object. These equations will be continuous . 2. Discretize the object into a nodal mesh. 3. Discretize the equations using finite differences and summations (rather than derivatives and integrals). 4. Use (2) and (3) to write the global equations as a stiffness matrix times a vector of (unknown) nodal values. Top-Down: Steps in FEM
- 4. Top-Down: Steps in FEM <ul><li>6. Solve for the nodal values. </li></ul><ul><li>Static – nodal values at equilibrium </li></ul><ul><li>Dynamic – nodal values at next time step </li></ul><ul><li>7. Interpolate values between nodal coordinates. </li></ul>discretize interpolate + global model object nodal mesh interpolate values between nodes + local model 5 2 3 1 4 6 7 8 u
- 5. Bottom-Up: Steps in FEM Nodes are point masses connected with springs. A continuum equation is solved for the nodes, and intermediate points are interpolated. A collection of nodes forms an element . A collection of elements forms the object. 5 2 3 1 4 6 7 8 u
- 6. Elements and Interpolations Interpolating equations for an element are determined by the number and distribution of nodes within the element. More nodes mean higher degree, for smoother simulation.
- 7. Example: Hermite as 1D Cubic Interpolation Equation 1. Assume cubic equation equation using shape (blending) functions and u r
- 8. Example: Hermite as 1D Cubic Interpolation Equation 2. Normalize the element to [0,1] and rewrite as a matrix equation or
- 9. Example: Hermite as 1D Cubic Interpolation Equation 3. Solve for the coefficients Q 4. Plug the coefficients into the cubic equation 5. Rewrite the cubic equation in the form
- 10. Example: Hermite as 1D Cubic Interpolation Equation 4 + 5. are equivalent to the steps values at the 4 nodes of the element shape (blending) functions
- 11. Example: Hermite as 1D Cubic Interpolation Equation shape (blending) functions within one element Let u r
- 12. 1D Elements (x) (x) (x) Example: bungee
- 13. 2D Elements (x,y) (x,y) (x,y) Example: cloth
- 14. 3D Elements (x,y,z) (x,y,z) Example: skin
- 15. Static analysis is good for engineering, to find just the end result. Dynamic analysis is good for simulation, to find all intermediate steps. Static vs. Dynamic FEM
- 16. Types of Global Models [6] Variational - Find the position function, w ( t ) that minimizes the some variational integral. This method is valid only if the position computed satisfies the governing differential equations. Rayleigh-Ritz - Use the variational method assuming some specific form of w ( t ) and boundary conditions. Find the coefficients and exponents of this assumed form of w ( t ).
- 17. Example of Variational Method [6] Minimizing the variation w.r.t. w of the variational function under the conditions satisfies the governing equation, Lagrange’s Equation
- 18. Galerkin ( weighted residual ) - Minimize the residual of the governing differential equation, F ( w,w’,w’’,…,t ) = 0. The residual is the form of F that results by plugging a specific form of the position function w ( t ) into F . Find the coefficients and exponents of this assumed form of w ( t ). Types of Global Models [6]
- 19. We can approximate w ( t ) using Hooke’s Law Example of Galerkin Method [6] If we use that equation to compute the 1st and 2nd time derivatives of w , then we can compute the residual as
- 20. Example of Static, Elastic FEM Problem : If you apply the pressure shown, what is the resulting change in length? Object
- 21. <ul><li>First step . Set up a continuum model: </li></ul><ul><li>F = force </li></ul><ul><li>P = pressure </li></ul><ul><li>A = area </li></ul><ul><li>L = initial length </li></ul><ul><li>E = Young’s modulus </li></ul>Entire length: Infinitessimal length: Example of Static, Elastic FEM
- 22. Since the shape is regular, we can integrate to find the solution analytically. But suppose we want to find the solution numerically. Next step . Discretize the object. Example of Static, Elastic FEM
- 23. Example of Static, Elastic FEM Discretization of object into linear elements bounded by nodes 1 2 3 4 n 1 n 2 n 3 n 4 n 5
- 24. Example of Static, Elastic FEM Next step . Set up a local model. Stress-Strain Relationship (like Hooke’s Law) Young’s modulus distance between adjacent nodes stress (elastic force)
- 25. Example of Static, Elastic FEM Next step . Set up a local (element) stiffness matrix. Rewrite the above as a matrix equation. Same for the adjacent element. element stiffness matrix nodal stresses nodal coordinates
- 26. Example of Static, Elastic FEM Now, all of the element stiffness matrices are as follows. r i is the x -coordinate of node u i 1 2 3 4 n 1 n 2 n 3 n 4 n 5
- 27. Example of Static, Elastic FEM Next step . Set up a global stiffness matrix. Pad the element stiffness matrices with zeros and sum them up. Example:
- 28. Example of Static, Elastic FEM Final step . Solve the matrix equation for the nodal coordinates. Global stiffness matrix. Captures material properties. Nodal coordinates. Solve for these! Applied forces
- 29. Elastic FEM <ul><li>A material is elastic if its behavior depends only on its state during the previous time step. </li></ul><ul><ul><li>Think: Finite state machine </li></ul></ul><ul><li>The conditions under which an “elastic” material behaves elastically are: </li></ul><ul><li>Force is small. </li></ul><ul><li>Force is applied slowly and steadily. </li></ul>
- 30. Inelastic FEM <ul><li>A material is inelastic if its behavior depends on all of its previous states. </li></ul><ul><li>A material may behave inelastically if: </li></ul><ul><li>Force is large - fracture, plasticity. </li></ul><ul><li>Force is applied suddenly and released, i.e., is transient - viscoelasticity. </li></ul><ul><li>Conditions for elastic vs. inelastic depend on the material. </li></ul>
- 31. Examples of Elasticity <ul><li>Elasticity </li></ul><ul><ul><li>Springs, rubber, elastic, with small, slowly-applied forces </li></ul></ul>
- 32. Examples of Inelasticity <ul><li>Inelasticity </li></ul><ul><li>Viscoelasticity </li></ul><ul><ul><li>Silly putty bounces under transient force (but flows like fluid under steady force) </li></ul></ul><ul><li>Plasticity </li></ul><ul><ul><li>Taffy pulls apart much more easily under more force (material prop.) </li></ul></ul><ul><li>Fracture </li></ul><ul><ul><li>Lever fractures under heavy load </li></ul></ul>
- 33. Linear and Nonlinear FEM Similarly to elasticity vs. inelasticity, there are conditions for linear vs. nonlinear deformation. Often these coincide, as in elastoplastic . = e
- 34. <ul><li>Hooke’s Law </li></ul><ul><li>Describes spring without damping </li></ul><ul><li>Linear range of preceding stress vs. strain graph </li></ul>Elastic Deformation Elastic vs. Inelastic FEM e e t loading unloading or stress strain Young’s modulus
- 35. Elastic vs. Inelastic FEM Damped Elastic Deformation e e t loading unloading viscous linear stress Rate of deformation is constant. a 1 e . a 1 e .
- 36. Viscoelastic Deformation Elastic vs. Inelastic FEM e e t loading unloading . viscous new term! This graph is actually viscous, but viscoelastic is probably similar Rate of deformation is greatest immediately after starting loading or unloading. depends on time t linear stress
- 37. Elastoplastic Deformation Elastic vs. Inelastic FEM e This graph is actually plastic, but viscoelastic is probably similar f e x x compare loading unloading loading x elastic plastic depends on force f e
- 38. Elastic vs. Inelastic FEM Fracture <ul><li>Force response is locally discontinuous </li></ul><ul><li>Fracture will propogate if energy release rate is greater than a threshold </li></ul>e x x loading unloading depends on force f
- 39. 1. World coordinates w in inertial frame (a frame with constant velocity) 2. Object (material) coordinates r in non-inertial frame r(w,t) = r ref (w,t) + e(w,t) Elastic vs. Inelastic FEM 4,5 world, or inertial frame ref r object, or non-inertial frame origin of = center of mass in
- 40. <ul><li>Transform </li></ul><ul><li>reference component r ref </li></ul><ul><li>elastic component e </li></ul><ul><li>object frame w.r.t. world frame </li></ul>r(w,t) = r ref (w,t) + e(w,t) Elastic vs. Inelastic FEM 4,5 ref r
- 41. Elastic vs. Inelastic FEM <ul><li>All these equations are specific for: </li></ul><ul><li>Elasticity </li></ul><ul><li>Viscosity </li></ul><ul><li>Viscoelasticity </li></ul><ul><li>Plasticity </li></ul><ul><li>Elastoplasticity </li></ul><ul><li>Fracture </li></ul><ul><li>( not mentioned ) “Elastoviscoplasticity” </li></ul><ul><li>Ideally : We want a general equation that will fit all these cases. </li></ul>
- 42. Elastic vs. Inelastic FEM 4,5 A More General Approach To simulate dynamics we can use Lagrange ’s equation of strain force . At each timestep, the force is calculated and used to update the object’s state (including deformation). stress component of force mass density damping density elastic potential energy Lagrange’s Equation
- 43. Elastic vs. Inelastic FEM 4,5 Given : Mass density and damping density are known. Elastic potential energy derivative w.r.t. r can be approximated using one of various equations. The current position w t of all nodes of the object are known. Unknown : The new position w t + dt of nodes is solved for at each timestep. vector vector matrices Lagrange’s Equation next slide
- 44. Elastic vs. Inelastic FEM 4,5 For both elastic and inelastic deformation, express elastic potential energy as an integral in terms of elastic potential energy density . elastic potential energy density elastic potential energy
- 45. Elastic vs. Inelastic FEM 4,5 <ul><li>Elastic potential energy density can be approximated using one of various equations which incorporate material properties. </li></ul><ul><li>Elastic deformation : Use tensors called metric (1D, 2D, 3D stretch), curvature (1D, 2D bend), and “ twist ” (1D twist). </li></ul><ul><li>Inelastic deformation : Use controlled-continuity splines . </li></ul>
- 46. Elastic FEM 4 <ul><li>For elastic potential energy density in 2D, use </li></ul><ul><li>metric tensors G (for stretch) </li></ul><ul><li>curvature tensors B (for bend) </li></ul>|| M || = weighted norm of matrix M
- 47. Elastic FEM 4 Overview of derivation of metric tensor Since the metric tensor G represents stretch, it incorporates distances between adjacent points. world coordinates object coordinates
- 48. Elastic FEM 4 Overview of metric and curvature tensors . From the previous slides, we found: Similarly: represents stretch represents bend Theorem . G and B together determine shape.
- 49. Elastic FEM 4 For elastic FEM, e lastic potential energy density in 2D incorporates changes in the metric tensor G and the curvature tensor B . || M || = weighted norm of matrix M weights = material properties
- 50. Inelastic FEM 5 For inelastic FEM, elastic potential energy density is represented as a controlled-continuity spline . For some degree p , dimensionality d , compute the sum of sums of all combinations of weighted 1 st , 2 nd ,…, m th derivatives of strain e w.r.t. node location r , where m <= p . weighting function = material property
- 51. Inelastic FEM 5 Then the elastic potential energy density derivative w.r.t. strain e is: weighting function = material property Example : p = 2, d = 3
- 52. Elastic vs. Inelastic FEM 4,5 Inelastic Elastic Recap Lagrange’s Eq’n total force (includes stress) elastic potential energy elastic potential energy density 4 5 5 material properties How it has been expanded and is continuing to be expanded...
- 53. Elastic FEM 4 Continuing elastic potential energy >0: surface wants to shrink <0: surface wants to expand >0: surface wants to flatten <0: surface wants to bend
- 54. Inelastic FEM 5 Continuing Deformation has been modeled by approximating elastic potential energy. elastic potential energy elastic potential energy density strain
- 55. Inelastic FEM 5 Continuing Now rigid-body motion and other aspects of deformation must be computed using physics equations of motion. In this way, both (in)elastic deformation and rigid-body motion can be modeled, providing a very general framework. r(w,t) = r ref (w,t) + e(w,t)
- 56. Inelastic FEM 5 Motion of object (non-inertial) frame w.r.t. world (inertial) frame Combines dynamics of deformable and rigid bodies elastic rot trans
- 57. Inelastic FEM 5 Velocity of node of object (non-inertial) frame w.r.t. world (inertial) frame (radians / sec) x (radius) Identically, in another coordinate system, r(w,t) = r ref (w,t) + e(w,t) w.r.t. object velocity of reference component velocity of elastic component w.r.t. world
- 58. Inelastic FEM 5 rot angular momentum inertia tensor Angular momentum is conserved in the absense of force. So a time-varying angular momentum indicates the presence of foce.
- 59. Inelastic FEM 5 rot indicates changing angle between position and direction of stretch
- 60. Inelastic FEM 5 If the reference component has no translation or rotation, then Furthermore, if the elastic component has no acceleration, then elastic inertial centripetal Coriolis transverse damping elastic potential energy strain restoring
- 61. Inelastic FEM 5 Recall that non-elastic behavior is characterized by acceleration of the elastic component (strain)... And elastic behavior is characterized by constant velocity of strain. loading x e
- 62. <ul><li>Now Lagrange’s equation has been expanded. </li></ul><ul><li>Final Steps </li></ul><ul><li>Discretize using finite differences (rather than derivatives). </li></ul><ul><li>Write as a matrix times a vector of nodal coordinates (rather than a single mass point). </li></ul><ul><li>Solve for the object’s new set of positions of all nodes. </li></ul>Elastic vs. Inelastic FEM 4,5
- 63. Discretization of FEM 4,5 Discretize Lagrange’s equation over all nodes Procedure described in [4] but not [5]
- 64. Discretization of Elastic FEM 4
- 65. Results of Elastic FEM 4
- 66. Results of Elastic FEM 4
- 67. Results of Elastic FEM 4
- 68. 3D plasticine bust of Victor Hugo. 180 x 127 mesh; 68,580 equations. Results of Inelastic FEM 5
- 69. Results of Inelastic FEM 5 Sphere pushing through 2D mesh. 23 x 23 mesh; 1,587 equations. Yield limit is uniform, causing linear tears.
- 70. Results of Inelastic FEM 5 2D paper tearing by opposing forces. 30 x 30 mesh; 2,700 equations. Yield limit is perturbed stochastically, causing randomly-propogating tears.
- 71. References 0. David Baraff. Rigid Body Simulation. Physically Based Modeling, SIGGRAPH Course Notes, August 2001. 1. George Buchanan. Schaum’s Outlines: Finite Element Analysis. McGraw-Hill, 1995. 2. Peter Hunter and Andrew Pullan. FEM/BEM Notes . The University of Auckland, New Zealand, February 21 2001.
- 72. References 3. Tom Lassanske. [Slides from class lecture] 4. Demetri Terzopoulost, John Platt, Alan Barr, and Kurt Fleischert. Elastically Deformable Models . Computer Graphics, Volume 21, Number 4, July 1987. 5. Demetri Terzopoulos and Kurrt Fleiseher. Modeling Inelastic Deformation: Viscoelasticity, Plasticity, Fracture. Computer Graphics, Volume 22, Number 4, August 1988
- 73. Notation

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