# The University of North Carolina – Chapel Hill COMP259-2005 Finite Elements A Theory-lite Intro Jeremy Wendt April 2005.

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The University of North Carolina – Chapel Hill COMP259-2005 Finite Elements A Theory-lite Intro Jeremy Wendt April 2005

The University of North Carolina – Chapel Hill COMP259-2005 Overview Numerical Integration Finite Differences Finite Elements –Terminology –1D FEM –2D FEM 1D output –2D FEM 2D output –Dynamic Problem

The University of North Carolina – Chapel Hill COMP259-2005 Numerical Integration You’ve already seen simple integration schemes: particle dynamics –In that case, you are trying to solve for position given initial data, a set of forces and masses, etc. –Simple Euler  rectangle rule –Midpoint Euler  trapezoid rule –Runge-Kutta 4  Simpson’s rule

The University of North Carolina – Chapel Hill COMP259-2005 Numerical Integration II However, those techniques really only work for the simplest of problems Note that particles were only influenced by a fixed set of forces and not by other particles, etc. Rigid body dynamics is a step harder, but still quite an easy problem –Calculus shows that you can consider it a particle at it’s center of mass for most calculations

The University of North Carolina – Chapel Hill COMP259-2005 Numerical Integration III Harder problems (where neighborhood must be considered, etc) require numerical solvers –Harder Problems: Heat Equation, Fluid dynamics, Non-rigid bodies, etc. –Solver types: Finite Difference, Finite Volume, Finite Element, Point based (Lagrangian), Hack (Spring-Mass), Extensive Measurement

The University of North Carolina – Chapel Hill COMP259-2005 Numerical Integration IV What I won’t go over at all: –How to solve Systems of Equations Linear Algebra, MATH 191,192,221,222

The University of North Carolina – Chapel Hill COMP259-2005 Finite Differences This is probably the easiest solution technique Usually computed on a fixed width grid Approximate stencils on the grid with simple differences

The University of North Carolina – Chapel Hill COMP259-2005 Finite Differences (Example) How we can solve Heat Equation on fixed width grid –Derive 2 nd derivative stencil on white board Boundary Conditions See Numerical Simulation in Fluid Dynamics: A Practical Introduction –By Griebel, Dornseifer and Neunhoeffer

The University of North Carolina – Chapel Hill COMP259-2005 Finite Elements Terminology We want to solve the same problem on a non-regular grid FEM also has some different strengths than Finite Difference Node Element

The University of North Carolina – Chapel Hill COMP259-2005 Problem Statement 1D STRONG FORM –Given f: OMEGA  R 1 and constants g and h –Find u: OMEGA  R 1 such that u xx + f = 0 u x (at 0) = h u(at 1) = g –(Write this on the board)

The University of North Carolina – Chapel Hill COMP259-2005 Problem Statement (cont) Weak Form (AKA Equation of Virtual Work) –Derived by multiplying both sides by weighting function w and integrating both sides Remember Integration by parts? Integral(f*g x ) = f*g - Integral(g*f x )

The University of North Carolina – Chapel Hill COMP259-2005 Galerkin’s Approximation Discretize the space Integrals  sums Weighting Function Choices –Constant (used by radiosity) –Linear (used by Mueller, me (easier, faster)) –Non-Linear (I think this is what Fedkiw uses)

The University of North Carolina – Chapel Hill COMP259-2005 Definitions w h = SUM(c A *N A ) u h = SUM(d A *N A ) + g*N A c A, d A, g – defined on the nodes N A, u h, w h – defined in whole domain Shape Functions

The University of North Carolina – Chapel Hill COMP259-2005 Zoom in We’ve been considering the whole domain, but the key to FEM is the element Zoom in to “The Element Point of View”

The University of North Carolina – Chapel Hill COMP259-2005 Element Point of View Don’t construct an NxN matrix, just a matrix for the nodes this element effects (in 1D it’s 2x2) –Integral(N Ax *N Bx ) –Reduces to width*slopeA*slopeB for linear 1D

The University of North Carolina – Chapel Hill COMP259-2005 Now for RHS We are stuck with an integral over varying data (instead of nice constants from before) Fortunately, these integrals can be solved by hand once and then input into the solver for all future problems (at least for linear shape functions)

The University of North Carolina – Chapel Hill COMP259-2005 Change of Variables Integral(f(y)dy) domain = T = Integral(f(PHI(x))*PHI x *dx) domain = S Write this on the board so it makes some sense

The University of North Carolina – Chapel Hill COMP259-2005 Creating Whole Picture We have solved these for each element Individually number each node Add values from element matrix to corresponding locations in global node matrix

The University of North Carolina – Chapel Hill COMP259-2005 Example Draw even spaced nodes on board –dx = h –Each element matrix = (1/h)*[[1 -1] [-1 1]] –RHS = (h/6)*[[2 1] [1 2]]

The University of North Carolina – Chapel Hill COMP259-2005 Show Demo 1D FEM

The University of North Carolina – Chapel Hill COMP259-2005 2D FEM 1D output Heat equation is an example here Linear shape functions on triangles  Barycentric coordinates Kappa joins the party –Integral(N Ax *Kappa*N Bx ) –If we assume isotropic material, Kappa = K*I

The University of North Carolina – Chapel Hill COMP259-2005 2D Per-Element This now becomes a 3x3 matrix on both sides –Anyone terribly interested in knowing what it is/how to get it?

The University of North Carolina – Chapel Hill COMP259-2005 Demo 2D FEM - 1D output

The University of North Carolina – Chapel Hill COMP259-2005 2D FEM – 2D Out Deformation in 2D requires 2D output –Need an x and y offset Doesn’t handle rotation properly Each element now has a 6x6 matrix associated with it Equation becomes –Integral(B A T *D*B B ) for Stiffness Matrix –B A/B – a matrix containing shape function derivatives –D – A matrix specific to deformation Contains Lame` Parameters based on Young’s Modulus and Poisson’s Ratio (Anyone interested?)

The University of North Carolina – Chapel Hill COMP259-2005 Demo 2D Deformation

The University of North Carolina – Chapel Hill COMP259-2005 Dynamic Version The stiffness matrix (K) only gives you the final resting position –Ku xx = f Dynamics is a different equation –Mu xx + Cu x + Ku = f K is still stiffness matrix M = diagonal mass matrix C = aM + bK (Rayliegh damping)

The University of North Carolina – Chapel Hill COMP259-2005 Demo 2D Dynamic Deformation

The University of North Carolina – Chapel Hill COMP259-2005 Questions

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