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## Introduction to Finite Element Methods

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**Numerical Methods – Definition and Advantages**• Definition: Methods that seek quantitative approximations to the solutions of mathematical problems • Advantages:**What is a Numerical Method – An Example**Example 1:**What is a Numerical Method – An Example**Example 1:**What is a Numerical Method – An Example**Example 2:**What is a Numerical Method – An Example**Example 2:**What is a Numerical Method – An Example**Example 3:**Discretization**1-D 2-D ?-D 3-D Hybrid**Approximation**Numerical Interpolation Non-exact Boundary Conditions**Applications of Finite Element Methods**• Structural & Stress Analysis • Thermal Analysis • Dynamic Analysis • Acoustic Analysis • Electro-Magnetic Analysis • Manufacturing Processes • Fluid Dynamics**Lecture 2**Review**Matrix Algebra**• Row and column vectors • Addition and Subtraction – must have the same dimensions • Multiplication – with scalar, with vector, with matrix • Transposition – • Differentiation and Integration**Matrix Algebra**• Determinant of a Matrix: • Matrix inversion - • Important Matrices • diagonal matrix • identity matrix • zero matrix • eye matrix**Numerical Integration**Calculate: • Newton – Cotes integration • Trapezoidal rule – 1st order Newton-Cotes integration • Trapezoidal rule – multiple application**Numerical Integration**Calculate: • Newton – Cotes integration • Simpson 1/3 rule – 2nd order Newton-Cotes integration**Numerical Integration**Calculate: • Gaussian Quadrature Trapezoidal Rule: Gaussian Quadrature: Choose according to certain criteria**Numerical Integration**Calculate: • Gaussian Quadrature • 2pt Gaussian Quadrature • 3pt Gaussian Quadrature Let:**Numerical Integration - Example**Calculate: • Trapezoidal rule • Simpson 1/3 rule • 2pt Gaussian quadrature • Exact solution**Linear System Solver**Solve: • Gaussian Elimination: forward elimination + back substitution Example:**Linear System Solver**Solve: • Gaussian Elimination: forward elimination + back substitution Pseudo code: Forward elimination: Back substitution: Do k = 1, n-1 Do i = k+1,n Do j = k+1, n Do ii = 1, n-1 i = n – ii sum = 0 Do j = i+1, n sum = sum +**UNIT II**Finite Element Analysis (F.E.A.) of 1-D Problems**Historical Background**• Hrenikoff, 1941 – “frame work method” • Courant, 1943 – “piecewise polynomial interpolation” • Turner, 1956 – derived stiffness matrice for truss, beam, etc • Clough, 1960 – coined the term “finite element” Key Ideas: - frame work method piecewise polynomial approximation**Axially Loaded Bar**Review: Stress: Stress: Strain: Strain: Deformation: Deformation:**Axially Loaded Bar**Review: Stress: Strain: Deformation:**Axially Loaded Bar – Governing Equations and Boundary**Conditions • Differential Equation • Boundary Condition Types • prescribed displacement (essential BC) • prescribed force/derivative of displacement (natural BC)**Axially Loaded Bar –Boundary Conditions**• Examples • fixed end • simple support • free end**Potential Energy**• Elastic Potential Energy (PE) - Spring case Unstretched spring Stretched bar x - Axially loaded bar undeformed: deformed: - Elastic body**Potential Energy**• Work Potential (WE) f P f: distributed force over a line P: point force u: displacement B A • Total Potential Energy • Principle of Minimum Potential Energy For conservative systems, of all the kinematically admissible displacement fields, those corresponding to equilibrium extremize the total potential energy. If the extremum condition is a minimum, the equilibrium state is stable.**Potential Energy + Rayleigh-Ritz Approach**Example: f P B A Step 1: assume a displacement field f is shape function / basis function n is the order of approximation Step 2: calculate total potential energy**Potential Energy + Rayleigh-Ritz Approach**Example: f P B A Step 3:select ai so that the total potential energy is minimum**Galerkin’s Method**Example: f P B A Seek an approximation so In the Galerkin’s method, the weight function is chosen to be the same as the shape function.**Galerkin’s Method**Example: f P B A 3 2 1 1 2 3**FEM Formulation of Axially Loaded Bar – Governing**Equations • Differential Equation • Weighted-Integral Formulation • Weak Form**Approximation Methods – Finite Element Method**Example: Step 1: Discretization Step 2: Weak form of one element P2 P1 x1 x2**Approximation Methods – Finite Element Method**Example (cont): Step 3: Choosing shape functions - linear shape functions x x x=0 x=-1 x=1 x2 x1 l**Approximation Methods – Finite Element Method**Example (cont): Step 4: Forming element equation E,A are constant Let , weak form becomes Let , weak form becomes**Approximation Methods – Finite Element Method**Example (cont): Step 5: Assembling to form system equation Approach 1: Element 1: Element 2: Element 3:**Approximation Methods – Finite Element Method**Example (cont): Step 5: Assembling to form system equation Assembled System:**Approximation Methods – Finite Element Method**Example (cont): Step 5: Assembling to form system equation Approach 2: Element connectivity table global node index (I,J) local node (i,j)**Approximation Methods – Finite Element Method**Example (cont): Step 6: Imposing boundary conditions and forming condense system Condensed system:**Approximation Methods – Finite Element Method**Example (cont): Step 7: solution Step 8: post calculation**Summary - Major Steps in FEM**• Discretization • Derivation of element equation • weak form • construct form of approximation solution over one element • derive finite element model • Assembling – putting elements together • Imposing boundary conditions • Solving equations • Postcomputation**Exercises – Linear Element**Example 1: E = 100 GPa, A = 1 cm2**Linear Formulation for Bar Element**f(x) u2 u u1 x L = x2-x1 x= x2 x=x1 1 1 f1 f2 x=x2 x=x1**Higher Order Formulation for Bar Element**u2 u3 u u u u1 x x x 3 2 1 u4 u2 u3 u1 3 2 1 4 un u1 u4 u3 …………… u2 1 n …………… 2 3 4**Natural Coordinates and Interpolation Functions**x x x x x=-1 x=1 x=-1 x=1 2 1 3 1 2 x=-1 x=1 4 2 1 3 x=1 x=-1 x x=x1 x= x2 Natural (or Normal) Coordinate:**Quadratic Formulation for Bar Element**f3 f1 f2 x=1 x=0 x=-1**Quadratic Formulation for Bar Element**u2 f(x) u3 u1 P2 P1 P3 x=1 x=0 x=-1