fem l1(c)

Dr.S.Rasool MohideenDepartment of l Engineering Mechanics

Faculty of Mechanical and Manufacturing EngineeringUniversity Tun Hussein Onn Malaysia

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FINITE ELEMENT METHODFINITE ELEMENT METHOD

( BDA 4033 )( BDA 4033 )

Lecture #03Lecture #03



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SYLLABUSCHAPTER 1- INTRODUCTION

Introduction and history of FEM Basic steps in the Finite Element

Methods Direct Formulation Minimum Total Potential Energy

Formulation

• Weighted Residual formulations



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Weighted Residual Formulation• Direct formulation is useful for simple geometries• Minimum Potential Energy Formulation is meant for

complex structural applications• For Non- Structural applications , Weighted residual

method is preferred• The method of weighted residuals is

– an approximate technique – for solving boundary value problems – that utilizes trial functions – satisfying the prescribed boundary conditions and – an integral formulation to minimize error, in an average

sense, over the problem domain.



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Weighted Residual:• Given a differential equation of the general form

D [y (x ), x ] = 0 a < x < b• subject to homogeneous boundary conditions

y(a) = y(b) = 0 • the method of weighted residuals seeks an approximate solution in the form

where y* is the approximate solution expressed as the product of ci unknown,constant parameters to be determined and Ni (x ) trial functions.

• On substitution of the assumed solution into the given differential Equation, a residual error ( or residual) results such that

where R(x) is the residual which is also a function of the unknown parameters ci.

Weighted Residual Formulation (Contd.)



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• The method of weighted residuals requires that the unknown parameters ci be evaluated such that

where wi (x ) represents n arbitrary weighting functions. • On integration, the above equation results in n algebraic

equations, which can be solved for the n values of ci. • The above equation expresses that the sum (integral) of

the weighted residual error over the domain of the problem is zero.

• The solution is exact at the end points (the boundary conditions must be satisfied) but, in general, at any interior point the residual error is nonzero.



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• Several techniques of MWR exist

• they vary primarily in how the weighting factors are determined or selected.

• The most common techniques are – point collocation, – sub domain collocation, – least squares, and– Galerkin’s



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Galerkin’s Weighted Residual Formulation

• In Galerkin’s weighted residual method, the weighting functions are chosen to be identical to the trial functions

• That is,• Unknown parameters are thus determined via

• Results in n algebraic equations for evaluation of the unknown parameters.



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Galerkin’s Weighted Residual Formulation

• Consider the problem

• where xj and xj+1 are contained in (a, b) and define the nodes of a finite element.

• The appropriate boundary conditions applicable to the above equation are

these are the unknown values of the solution at the end points of the sub-domain.



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Galerkin’s Weighted Residual Formulation (Contd.)

• The proposed approximate solution be of the form

• Where the trial functions are

• satisfy the conditions



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• Substitution of the assumed solution into the equation gives the residual as

where the superscript ‘e’ is used to indicate that the residual is for the element.

• Applying the Galerkin’s weighted residual criterion results in



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• The element residual equation

• Applying integration by parts to the first integral results in



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• After evaluation of the non-integral term and rearranging gives the two equations

• Setting j = 1 for notational simplicity and substituting, the equation yields



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• The above equation is of the form

• The terms of the coefficient (element stiffness) matrix are defined by

• the element nodal forces are given by the right-hand side of the Equation.



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Galerkin’s Weighted Residual Formulation for a Bar Element

• For a bar element , the governing differential equation is given ( from constant strain & stress) as

• Denoting element length by L, the displacement field is discretized by the Equation

• Galerkin residual equations is then given as



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Galerkin’s Weighted Residual Formulation for a Bar Element (Contd..)

• Integrating by parts and rearranging

• Substituting the trial functions,

right side of the equation represents the applied nodal force since σA = F.



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Galerkin’s Weighted Residual Formulation for a Bar Element (Contd..)

• The above equation can be readily combined into matrix form as

• Carrying out the indicated differentiations and integrations,



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One Dimensional Heat Conduction Problem

• Considering a surface insulated body as shown below

• The principle of conservation of energy is applied to obtain the governing equation for steady-state, one-dimensional conduction as



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One Dimensional Heat Conduction Problem (Contd..)

• Applying the Galerkin finite element method , the trial function equation is given as

where T1 and T2 are the temperatures at nodes 1 and 2, which define the element and N1 and N2 are the interpolation functions

• The residual integral is

• Integrating the first term by parts



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• Evaluating the first term at the limits and substituting the trial function Equation and rearranging,

• Equations 5.63 and 5.64 are of the form

where [k] is the element conductance (“stiffness”) matrix having terms defined by



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• The first term on the right-hand side of the Equation is the nodal “force” vector arising from internal heat generation with values defined by

and vector {fg} represents the gradient boundary conditions at the element nodes.

• Performing the integrations, conductance matrix is obtained as shown below



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• constant internal heat generation (Q) matrix and the element gradient matrix are given as

and



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Problem 1

• The circular rod shown in Figure has – an outside diameter of 60 mm, – length of 1 m and– is perfectly insulated on its circumference.

• The left half of the cylinder is – aluminum, for which kx = 200 W/m-°C and – the right half is copper having kx = 389 W/m-°C. – The extreme right end of the cylinder is maintained at a temperature

of 80°C, – while the left end is subjected to a heat input rate 4000 W/m2.

• Using four equal-length elements, determine the steady-state temperature distribution in the cylinder.



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Solution

• The elements and nodes are chosen as shown in the bottom of Figure.

• For aluminum elements 1 and 2, the conductance matrices are

• For copper elements 3 and 4, the conductance matrices are



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Solution (Contd..)

• Applying the end conditions T5 = 80°C and q1 = 4000 W/m2, the assembled system equations are



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Solution (Contd..)



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Solution (Contd..)• Applying the end conditions T5 = 80°C and q1 = 4000 W/m2, the assembled system

equations are

• Accounting for the known temperature at node 5, the first four equations can be written as

• Solving for unknowns, the temperatures areobtained as

• Substituting the unknown (T4) in 5th equation,

fem l1(c)

Documents

weighted residual method

weighted residual error

residual results

residual formulationconsider

residual aswhere

element residual equationappl

integral formulation

unknown parameters ci