Chapter 9 Method of Weighted Residuals

Similar documents
THE ACCURATE ELEMENT METHOD: A NEW PARADIGM FOR NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATIONS

Shape functions in 1D

Table of Common Derivatives By David Abraham

3.7 Implicit Differentiation -- A Brief Introduction -- Student Notes

1D Heat Transfer. Background. Local Equations. The temperature assumption. Balance law. Lecture 7

Lecture 3: Development of the Truss Equations.

d dx But have you ever seen a derivation of these results? We ll prove the first result below. cos h 1

Calculus of Variations

12.11 Laplace s Equation in Cylindrical and

A. Incorrect! The letter t does not appear in the expression of the given integral

Lectures - Week 10 Introduction to Ordinary Differential Equations (ODES) First Order Linear ODEs

3.6. Implicit Differentiation. Implicitly Defined Functions

Math 1271 Solutions for Fall 2005 Final Exam

Chapter 2 Derivatives

Derivative of a Constant Multiple of a Function Theorem: If f is a differentiable function and if c is a constant, then

x = c of N if the limit of f (x) = L and the right-handed limit lim f ( x)

Unit #6 - Families of Functions, Taylor Polynomials, l Hopital s Rule

The derivative of a function f(x) is another function, defined in terms of a limiting expression: f(x + δx) f(x)

II. First variation of functionals

Thermal conductivity of graded composites: Numerical simulations and an effective medium approximation

Chapter 2 Governing Equations

Diagonalization of Matrices Dr. E. Jacobs

MATH2231-Differentiation (2)

Convergence of Random Walks

5.4 Fundamental Theorem of Calculus Calculus. Do you remember the Fundamental Theorem of Algebra? Just thought I'd ask

MATH 205 Practice Final Exam Name:

Nonlinear Dielectric Response of Periodic Composite Materials

Polynomial Inclusion Functions

Calculus BC Section II PART A A GRAPHING CALCULATOR IS REQUIRED FOR SOME PROBLEMS OR PARTS OF PROBLEMS

m (ft-lb/ft). Using the point-slope

Calculus in the AP Physics C Course The Derivative

3.2 Differentiability

IMPLICIT DIFFERENTIATION

Math 180, Exam 2, Fall 2012 Problem 1 Solution. (a) The derivative is computed using the Chain Rule twice. 1 2 x x

Planar sheath and presheath

WJEC Core 2 Integration. Section 1: Introduction to integration

Lectures HD#14 Dynamic response of continuous systems. Date: April

One Dimensional Convection: Interpolation Models for CFD

Formulation of the Equilibrium Equations of Transversely Loaded Elements Taking Beam-Column Effect into Consideration

Conservation laws a simple application to the telegraph equation

FE Modeling and Analysis of Isotropic and Orthotropic Beams Using First Order Shear Deformation Theory

11.7. Implicit Differentiation. Introduction. Prerequisites. Learning Outcomes

Quantum Mechanics in Three Dimensions

water adding dye partial mixing homogenization time

'HVLJQ &RQVLGHUDWLRQ LQ 0DWHULDO 6HOHFWLRQ 'HVLJQ 6HQVLWLYLW\,1752'8&7,21

OPG S. LIST OF FORMULAE [ For Class XII ] OP GUPTA. Electronics & Communications Engineering. Indira Award Winner

Chapter 1 Overview: Review of Derivatives

Application of the homotopy perturbation method to a magneto-elastico-viscous fluid along a semi-infinite plate

Dissipative numerical methods for the Hunter-Saxton equation

Using Quasi-Newton Methods to Find Optimal Solutions to Problematic Kriging Systems

Module FP2. Further Pure 2. Cambridge University Press Further Pure 2 and 3 Hugh Neill and Douglas Quadling Excerpt More information

CHAPTER 1 INTRODUCTION TO THE FINITE ELEMENT METHOD AND BAR ELEMENTS...

Chapter 2 Lagrangian Modeling

Euler equations for multiple integrals

Separation of Variables

Analytic Scaling Formulas for Crossed Laser Acceleration in Vacuum

Linear First-Order Equations

Two Dimensional Numerical Simulator for Modeling NDC Region in SNDC Devices

Lecture XII. where Φ is called the potential function. Let us introduce spherical coordinates defined through the relations

MMJ1153 COMPUTATIONAL METHOD IN SOLID MECHANICS PRELIMINARIES TO FEM

The Kepler Problem. 1 Features of the Ellipse: Geometry and Analysis

Math 342 Partial Differential Equations «Viktor Grigoryan

Summary of the Class before Exam1

A Novel Decoupled Iterative Method for Deep-Submicron MOSFET RF Circuit Simulation

Assignment 1. g i (x 1,..., x n ) dx i = 0. i=1

Applications of the Wronskian to ordinary linear differential equations

. Using a multinomial model gives us the following equation for P d. , with respect to same length term sequences.

Heat Transfer from Arbitrary Nonisothermal Surfaces in a Laminar Flow

Summary: Differentiation

there is no special reason why the value of y should be fixed at y = 0.3. Any y such that

Static analysis of beams on elastic foundation by the method of discrete singular convolution

inflow outflow Part I. Regular tasks for MAE598/494 Task 1

Math Notes on differentials, the Chain Rule, gradients, directional derivative, and normal vectors

The Sokhotski-Plemelj Formula

7.1 Support Vector Machine

SECTION 3.2 THE PRODUCT AND QUOTIENT RULES 1 8 3

Non-Conservative Stability Analysis of Hauger Types of Columns with Different Boundary Conditions

CHAPTER 4. INTEGRATION 68. Previously, we chose an antiderivative which is correct for the given integrand 1/x 2. However, 6= 1 dx x x 2 if x =0.

Introduction to the Vlasov-Poisson system

Section 2.7 Derivatives of powers of functions

C6-1 Differentiation 2

Modeling time-varying storage components in PSpice

AN INTRODUCTION TO AIRCRAFT WING FLUTTER Revision A

Chapter Primer on Differentiation

SYNCHRONOUS SEQUENTIAL CIRCUITS

A Short Note on Self-Similar Solution to Unconfined Flow in an Aquifer with Accretion

Optimal Control of Spatially Distributed Systems

1. Filling an initially porous tube under a constant head imposed at x =0

FUNDAMENTALS OF STOCHASTIC NETWORKS

Damage identification based on incomplete modal data and constrained nonlinear multivariable function

RETROGRADE WAVES IN THE COCHLEA

Mathematical Review Problems

SOLVING QUADRATICS. Copyright - Kramzil Pty Ltd trading as Academic Teacher Resources

ALGEBRAIC AND ANALYTIC PROPERTIES OF ARITHMETIC FUNCTIONS

The Sokhotski-Plemelj Formula

Problems Governed by PDE. Shlomo Ta'asan. Carnegie Mellon University. and. Abstract

REDUCING the cost of launching a space vehicle into space is

Consider for simplicity a 3rd-order IIR filter with a transfer function. where

ON THE OPTIMALITY SYSTEM FOR A 1 D EULER FLOW PROBLEM

Section The Chain Rule and Implicit Differentiation with Application on Derivative of Logarithm Functions

Transcription:

Chapter 9 Metho of Weighte Resiuals

9- Introuction Metho of Weighte Resiuals (MWR) is an approimate technique for solving bounary value problems. It utilizes a trial functions satisfying the prescribe bounary conitions, an integral formulation to minimize the error, over the problem omain. The general concept is escribe for a one-imensional case. But the concept can easily be etene to two-imensional an threeimensional cases.

9- General Concept Given a ifferential equation of the general form, [ ] D y ( ), = a< < b subect to homogeneous bounary conitions y ( a ) = y ( b ) = (i) (ii) The metho of weighte resiuals seeks an approimate solution in the form n ( ) = ( ) * y ci N i i= (iii) where y* is the approimate solution epresse as the prouct of c i unknown (i.e. constant parameters to be etermine), an N i ( ) are trial functions. Note: The solution represente in Eq.(iii) is not an eact one!

When the assume solution of eq.(iii) is substitute into the ifferential equation of eq.(i), a resiual error R( ) (resiual) will result, which is given by R D y * ( ) = ( ), Note: Resiual R( ) is also a function of the unknown parameters, c i. The metho of weighte resiuals (MWR) requires that the unknown parameters c i be evaluate such that, b w ( ) R i ( ) = i=, n a Where w i ( ) represents n arbitrary weighting functions. Note: On integration, Eq.(v) results in n algebraic equations, which can be solve for the n values of c i. (iv) (v)

b w ( ) R i ( ) = i =, n a (v) - R e p e a t Eq.(v) epresses that the sum (integral) of the weighte resiual error over the omain of the problem is zero. The solution is eact at the en points (the bounary conitions must be satisfie) but, in general, at any interior point the resiual error is nonzero. Note: Several variations of the MWR eist an the technique vary primarily on how the weighting factors are selecte. The most common techniques are point collocation, sub-omain collocation, least squares an the Galerkin s metho. We will only iscuss the Galerkin s metho as it is quite simple to use an reaily aaptable to the finite element metho.

9-3 Galerkin s Metho In Galerkin s weighte resiual metho, the weighting functions are chosen to be ientical to the trial functions, i.e. w ( ) = N ( ) i =, n i i Therefore, the unknown parameters are etermine via b a b w ( ) R ( ) = N ( ) R ( ) = i =, n i a Again, the above integration results in n algebraic equations for evaluation of the n unknown parameters. i

Eample 9- Use Galerkin s weighte resiuals to obtain an approimate solution for the ifferential equation, y with bounary conitions y( ) = y( ) = Solution = 5 Since the integral equation have quaratic terms, the suitable trial functions will be polynomial. For homogeneous bounary conitions at = a an = b, the general form is p N ( ) = ( ) ( ) a where p an q are positive integers, that satisfy the bounary conitions. Using a single trial function, the simplest form will be ( ) = ( ) N b q

Using this trial function, the approimate solution is * y c an the first an secon erivatives are * y y ( ) = ( ) * = c ( ) = c Note: The selecte trial function oes not satisfy the physics of the problem because the secon erivative is a constant, whereas accoring to the original ifferential equation, the secon erivative must be a quaratic. Nevertheless, we will continue with the eample to illustrate the proceure.

Substitution of the secon erivative of y*() into the ifferential equation yiels the resiual, ( ; c ) = c 5 which is clearly a nonzero value. R Substituting the resiual into the integral equation we get N ( ) R ( ; c ) = which, after integration yiels c = 4 c ( ) ( 5) = So the approimate solution is obtaine as ( ) = 4 ( ) y *

For this eample, the eact solution can be obtaine by integrating the ifferential equation twice, i.e. ( ) ( ) 4 3 3 5 6 5 5 3 5 3 5 C C C y y C y y + + + = + + = = + + = + = = Applying the bounary conitions i.e. y() = an y() =, we get 3 an = = C C Therefore the eact solution is ( ) y 3 5 6 5 4 + =

A graphical comparison between the two solutions is shown below. The approimate solution agrees reasonably well with the eact solution.

Eample 9- Obtain two-term Galerkin s solution for y = 5 (i) with bounary conitions y( ) = y( ) = an using the following trial functions ( ) ( ) N = ; N ( ) = ( ) Solution Using these trial functions, the approimate solution is y ( ) = c ( ) + c ( ) * The secon erivatives of the above is y * = c + c (3 ) (ii) (iii)

Substitution of the secon erivative of y*() ifferential equation yiels the resiual into the ( ) R ; c ; c = c + c (3 ) 5 Substituting the resiual into the integral equation we get [ 5] ( ) c + c ( 3 ) [ 5] ( ) c + c ( 3 ) which, after integration an solving the equations yiels c 9 5 = ; c = 6 3 = =

So the two-term approimation solution is an the eact solution is 9 5 y * 6 3 5 3 3 9 y *( ) = + 3 6 y ( ) = ( ) + ( ) Recall, the one-term approimation solution is ( ) = 4 ( ) y * 5 5 3 4 ( ) = + 6

A graphical comparison between the three solutions is shown. We see that the two-term approimate solution matches quite closely to the eact solution.

9-4 The Galerkin s Finite Element Metho Element Formulation Consier a problem represente by a ifferential equation y + f ( ) = ; subecte to bounary conitions y ( ) = y ; y ( ) = y + + The problem omain is represente by + e +

The trial or interpolation functions are chosen such that N N ( ) = ( ) = + + + + (i) At the bounaries of the element omain, these trial functions have the values N ( ) = ; N ( + ) = (ii) N ( ) = ; N ( ) = + Using the above trial functions, the approimate solution will be ( e) y ( ) = y N ( ) + y N ( ) + (iii)

Substituting the approimate solution into the ifferential equation yiels the resiual, given by ( e ) ( e) y R = + f ( ) ( e) R = y N ( ) + y + N ( ) + f ( ) (iv) Applying the Galerkin s weighte resiual criterion results in + ( e ) i ( ) (, ) ( e ) y N i ( ) + N ( ) ( ) i f = + + N R = i= (v) Applying integration by part on the first integral term, we get N e t p a g e

( e ) + ( e ) y + N i y N i ( ) + + N i f = i = ( ) ( ) (, ) (vi) Evaluate the non-integral terms at the bounaries an rearranging the equations we get N y y = N ( ) f ( ) + ( e ) ( e ) + + N y y = N ( ) f ( ) + ( e ) ( e ) + + Setting =, an substitute the approimate solution y ( e ) ( ) into the above eq.(vii), we get + (vii)

N N N y y y N ( ) f ( ) + = + N N N y y y N ( ) f ( ) + = + which can be written in the matri form as k k y f k k = y f ( e ) ( e ) (viii) (i) where, N N = (, =, ) i ki i () Note: The RHS of eq. (viii) represents the forces acting at the two ens of the elements omain.

Assembly of Elements Since the finite element solution is not eact, then if two elements are connecte at a noe, we have 3 4 3 4 5 y ( ) = y ( ) ( 3 ) ( 4 ) 4 4 For eact solution, y y ( 3 ) ( 4 ) 4 4 y y = ( 3 ) ( 4 ) 4 4

9-5 Application of Galerkin s FE Metho. One-Dimensional Problem Consier a prismatic bar uner an aial loaing. The ifferential equation governing this problem is given by σ Eε u E ( ) = ( ) = = (i) The approimate solution for isplacement u( ) is u u N u N * ( ) = ( ) + ( ) (ii) e u u L Note: The omain of our solution is the volume of the element.

The trial functions use in Eq.(ii) are chosen such that N ( ) = ; N = L L Hence, the approimate solution can be written as u u u L L * ( ) = + Substituting the approimate solution of Eq.(iv) into the ifferential equation of Eq.(i) results in a resiual given by * ( e ) u ( ) R = E = E u u + L L (iii) (iv) Applying the Galerkin s weighte resiual criterion results in (v) V * u ( ) N i ( ) E V = (vi)

Since V = A, where A is cross-sectional area of the bar, which is uniform, Eq.(iv) can then be epresse as L * u ( ) N i ( ) E A L = Integrating Eq.(vii) by part an rearranging the result, we get L N i u ( ) u ( ) AE = N i ( ) EA * * L (vii) (viii) Substituting the approimate solution of Eq.(iv) into Eq.(viii) an solving the RHS of the equation, we obtain N L A E ( u N+ u N ) = Aσ = N L A E ( u N+ u N ) = A σ = L (i)

Combining Eq.(i) an write in matri form, we get N N N N L u f A E = N N N N u f () Integrating iniviual term within the square bracket inepenently yiels, A E u f L = (i) u f which is the system of linear equations for a -D bar element. Note: The square matri represents the stiffness matri for the element.

. Beam in Bening Consier a portion of beam loae with uniformly istribute loa q( ) as shown. The ifferential equation for this problem is represente by Galerkin s finite element metho is applie by using an approimate solution for v( ) in the form

Substituting the approimate solution into the original ifferential equation yiels the resiuals, i.e. Integrating the erivative terms by parts an assuming a constant E I z, we obtain We observe that the first term of the above equation represents the shear force conitions at the element noes, i.e.

Integrating again by part an rearranging gives, This equation can be written in the matri form, ( ) [ k] e { δ } = { F} where the terms of the stiffness matri are efine by

The terms for the element force vector are efine by Where the integral term represents the equivalent noal forces an moments prouce by the istribute loa. If q( ) = q = constant (positive upwar), then substitution of the interpolation function into the above equation yiels the element noal force vector given by

3. One-Dimensional Heat Conuction Application of the Galerkin s finite element metho to the problem of one-imensional, steay-state heat conuction is evelope base on the conition epicte in figure below. Note: Surfaces of the boy normal to the -ais are assume to be perfectly insulate.

Performing an energy balance across a small cubic element, we obtain the ifferential equation governing the steay-state heat conuction through the boy, given by where Q is an internal heat generation rate in W/m 3. The approimate solution for the temperature istribution T * ( ) in the element is epresse as, where T an T are the temperatures a noes an. The linear interpolation functions N an N are given by N N + ( ) = ; ( ) = + +

Substituting the approimate solution into the original ifferential equation yiels the resiuals, i.e. Integrating the first terms by parts, we obtain Evaluating the first term of the above equation, substituting the approimate solution T * ( ) into the secon term, an rearranging the results, we obtain two equations which can be epresse as ( see net page )

These equations can be written in a conense matri form as ( ) [ k ] e T { T} { f Q} { f g} = + (i) where [ k T ] (e) is the element conuctivitiy matri. The elements of this matri are efine by (ii)

The first RHS term of eq.(i) is noal force vector arising from internal heat generation with values efine by (iii) an vector {f g } represents the graient bounary conition at the element noes. Performing the integrations inicate in eq.(ii) gives the element conuctivity matri as

For constant internal heat generation Q, eq.(iii) results in the noal vector, The element graient bounary conition {f g } is escribe by T q f k A A T q { } g = =

En of Chapter 9