fea-mechanical 2010 student version

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FINITE ELEMENT ANALYSIS

(ME 413)

Course TeacherCourse Teacher

Dr.Dr.--IngIng. Amir. Amir SiddiqSiddiqAssociate Professor

Automotive & Marine Engineering Department

NED University of Engg. & Technoogy

Consultation Hours

Every Tuesday: 1 pm – 2 pm

Office Location

Room # B2, First Floor



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1. Introduction to Computational Mechanics

????

What is Computational Mechanics?What is Computational Mechanics?



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What is Computational Mechanics?What is Computational Mechanics?

• Sub-discipline of Theoretical &Applied Mechanics

Some of the examples will be discussed briefly today

• Computational methods anddevices to study events governedby the principles of mechanics

• Strong impact on manufacturing,medicine, defence, energy and

transportation industries, etc.



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Automotive IndustryAutomotive Industry

• Crashworthiness of a vehicle

• Design and optimization of the automobilecomponents

• Including Dynamic and fracture behaviour

of a single component or an assembly.

••Why such simulation??Why such simulation??

••What advantages?What advantages?



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Automotive IndustryAutomotive Industry

Computer-generated simulation of a full-size

car in oblique impact with a cylindrical pole

•Simulations are cheaper than experiments

•Impact event (thousandth of a second) can bestudied in detail

•All the information regarding damage to the car,passenger, deformation, forces, etc. can beobtained to design a safer and reliable structure.



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MedicineMedicine

• A very new and one of the most challenging field• Development and implementation of different

mathematical models to describe livingarganism‘s behaviour

• Used in studying from a small unit cell to acomplete body

• Designing of implants, such as tooth, hip, knee,

etc



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Defence ApplicationsDefence Applications

Simulation of a metallic rod impacting andpiercing a collection of armor plates

• Fluid-Structure Interactions

Computatoinal Fluid Dynamics

Computational Fracture Mechanics

Thermodynamics



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Reliability of ComputationalReliability of Computational

Mechanics?Mechanics?

• How to justify the reliablility of acertain computational method,model and tool??

????????



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FEM & FVM: APPLICATION IN AUTOMOTIVEENGINEERING (AU 413)

1. Introduction to Computational Mechanics

2. Computational Modelling – Basic Concepts

3. Review of Mechanics of Solids and Structures

4. Fundamentals of Finite Element Methods

5. Numerical Integration (Analysis) Techniques

6. FEM for Trusses, Beams, 2D Solids, 3D Solids

7. Modelling Techniques

8. Specialized Topics in FEA

Sessional Marks

1. Test

2. Individual Project

3. Assignments

Proposal of Individual Project Due by28th February, 2010



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Individual Project Proposal

1. Title of the Project

2. Student Name, Roll Number, Department

3. Write one page proposal with three

paragraphs

4. First Paragraph should give an introduction

to the problem

5. Second Paragraph should discuss what is

planned to be done

6. Third paragraph should explain the expected

outcome of the project

Consultation Hours

Every Tuesday: 1 pm – 2 pm

Individual Project Report Due by 13th

August, 2010



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2. Computational Modelling – Basic Concepts

Processes leading to fabrication of advanced

engineering systems

• We shall address the computational aspects,

which are also underlined in the abovefigure.

• Focus will be on techniques of physical,

mathematical and computational modelling.

• A good understanding of these techniques

ensure a rapid and cost effective fabrication

of and advanced engineering system.

2.1 Introduction



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2.2 What is FEM?

• Analyst seeks to determine the distribution of some

field variable, such as displacement in stress analysis , temperature or heat flux in thermal

analysis , the electrical charge in electrical analysis ….

• A numerical method seeking an approximatedsolution of the distribution of field variables in theproblem domain that is difficult to obtain analytically.

• It is done by dividing the problem domain intoseveral elements.



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• Known physical laws are then applied to eachsmall element.

• Figure shows the finite element

approximations for a one-dimensional case.• A continuous function of an unknown fieldvariable is approximated using piecewise

linear functions in each sub-domain, called

an element formed by nodes.

• Proper principles are followed to establishequations for the elements.

• This process leads to a set of linearalgebraic simultaneous equations for theentire system that can be solved to yield

the required field variable.

Aim of this course is to study various concepts,methods and principles used in the formulation

of FE equations.



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2.4 Computational Modelling using the FEM

The behaviour of a phenomenon in a system

depends upon:

• geometry or domain of the system

• property of the material or medium

• boundary, initial and loading conditions

For an engineering system, the geometry or

domain can be very complex. Also, the boundaryand initial conditions can be complicated.

Therefore, very difficult to solve the governing

differential equation using analytical means.

In practice, such problems are solved usingnumerical methods, one of them is FEM.

The procedure of computational modelling usingthe FEM broadly consists of four steps:

Modelling of the geometry

Meshing (discretization)

Specification of material property

Specification of boundary, initial andloading conditions



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The procedure of computational modelling using theFEM broadly consists of four steps:

• Modelling of the geometry• Meshing (discretization)

• Specification of material property• Specification of boundary, initial and loading

conditions



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2.5 Simulation

• Based on the mesh, a set of discrete simultaneous

system of equations can be formulated using differentapproaches.

• There are a few types of approaches for

establishing the simultaneous equations.

− Based on energy principles, such as

Hamilton’s principle, minimum potential energy

principle, and so on. [FEM]

− Based on the weighted residual method. [FEM]

− Based on Taylor series. [FDM]

− Based on control of conservation laws on eachfinite volume in the domain. [FVM]

• After formulation of discrete simultaneous system ofequations, it is fed to a solver to solve for the fieldvariables at the node of the mesh.

• This is the most computer hardware demanding

process.

• Different software package use different algorithms.

• Two important considerations when choosing analgorithm are storage required and CPU timeneeded.



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2.6 Visualization

• The result generated after solving the system of

equations is usually a vast volume of digital data.• The results have to be visualized in such a waythat it is easy to interpolate, analyze and present.

• The visualization is performed through a so-called post-processor.

• Most of these post processors allow the user to

display 3D objects in many convenient andcolourful ways on-screen.

• Field variables can be plotted on the object in

the form of contours, fringes, wire-frames anddeformations.

Simulate a Real Life Phenomenon

Summary

• Briefly discussed steps involved in computermodelling and simulation

• Subsequent sections will discuss what goes intothe computer when performing FEM



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3. Review of Mechanics of Solids

3.1 Equations for Truss Members

• A typical truss structure is shown in the figure.

• Each truss member in a truss structure is a

solid whose dimension in one direction is much

larger than in the other two directions.

• Force is applied in only one direction (x-direction).

• Therefore, a truss member is actually a 1D solid.



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u

• Lets assume that force is only acting in x-

direction.

• Due to this force the displacement of theend of the bar is u.

• The force fx will cause stress and strain in

only x-direction, i.e. σx, and εx, respectively.

• For an infinitesimal element of the truss thestrain-displacement relationship is simplygiven by:

Kinematic Equation

• Hooke’s law for 1D solids has the following form:

Constitutive Equation

• Dynamic equilibrium equation for 1 solid is givenby:

x x E ε σ =



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• Substituting Constitutive and Kinematicequations into dynamic equilibrium equation:

• The static equilibrium equation for trusses is

obtained by eliminating the inertial term:

3.2 Equations for Beam Members

• A beam possesses geometrically similardimensional characteristics as a truss member.

• The difference is that the forces applied on

beams are transversal.

• Therefore, a beam experiences bending, i.e. the

deflecion in the y direction as a function of x .



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• There are several theories for analysing beamdeflections.

• These theories can be divided into two broadcategories:

Thin Beam Theory (Euler-Bernoulli beam)

Thick Beam Theory (Timoschenko beam)

• Euler-Bernoulli beam theory assumes that the

plane cross-sections remain plane after bending

and remain normal to the deformed centroidalaxis.

• With this assumption, we have

• which means that shear stress is negligible.

• Axial displacement, u , at a distance z from thecentroidal axis can be given by



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• θ is the rotation in the x-z plane.

• Rotation can be obtained from the deflection ofthe centroidal axis of the beam, w, in z direction:

• The relationship between the normal strain and

the deflection can be given by

• where L is the differential operator given by

Laplace Operator

KinematicEquation

Constitutive

Equation

xx xxE ε σ =

• Original Hooke’s law is applicable for beams

• Because the beam is loaded in the transversedirection, there will be moments andcorresponding shear forces.



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• A small representative cell of length dx ofthe beam is shown above.

• The beam cell is subjected to external force,f z , moment, M , shear force, Q , and inertialforce, .

• The moment on the cross-section at x

results from the distributed normal stress σxx.

w A && ρ

• Normal stress σxx varies linearly in thevertical direction on the cross-section.



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• Moments resulting from the normal stress onthe cross-section can be calculated by thefollowing integration over the area of the cross-section:

where I is the second moment of area of thecross-section with respect to y -axis.

• Now consider the force equilibrium of the smallbeam cell in the z direction.



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• We would also need to consider the momentequilibrium of the small beam cell with respect to

any point at the right surface of the cell,

Neglecting the second order small term

containing (dx )2 leads to

Substituting following equation of M into theabove equation

yields,

Above equation gives the relationship betweenthe moments and shear forces in a beam withdeflection of the Euler-Bernoulli beam.



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• Dynamic equilibrium equation for beams can beobtained by simply substituting the moment and

shear forces relationship with the deflection.

into the force equilibrium relation

to give,

For static equilibrium equation for beam:



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3.3 Equations for Three-Dimensional Solids

• Consider a continuous three-dimensional elastic

solid with volume V and a surface S ,

• Surface of the solid is further divided into twotypes of surfaces: on which external forces are

prescribed (S F ); and on which the displacementsare prescribed (S d ).

• The solid can be loaded by body force fb andsurface force fs in any distributed fashion.

• At any point in the solid, the components ofstress are indicated on the surface of an infinitely

small cubic volume.



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• On each surface, there will be a normal stresscomponent, and two components of shearing

stress.

• The sign convention for the subscript is that the

first letter represents the surface on which thestress is acting, and the second letter representsthe direction of the stress.

• For the state of equilibrium, summation of all the

moments of forces about the central axis must bezero.



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• There are six stress components in total at apoint in solids. These stresses are often termed as

stress tensor, denoted by,

• Corresponding to the six stress tensors, thereare six strain components at any point in a sold,given by,

• Strain is the change of displacement per unitlength, and therefore the components of strain canbe obtained from the derivatives of thedisplacement:

X

Y

Z

εzz

εyy

εxx

εxy

εxzεyx

εyzεzx

εzy

du

dvdw

ElementaryVolume dV

dx

dy

dzY

Z

εzz

εyy

εxx

εxy

εxzεyx

εyzεzx

εzy

du

dvdw

ElementaryVolume dV

dx

dy

dz



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• The six strain-displacement relationships can berewritten in the following matrix form:

where U is the displacement vector, and has the

form of

and L is a matrix of partial differential operators,

Kinematic Equation

• The constitutive equation gives the relationship

between stress and strain in solid materials. It isoften termed Hooke’s law.

• The generalized Hooke’s law for general

anisotropic materials can be given by:

where c is a matrix of material constants, which

are normally obtained through experiments.



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• The constitutive equation can be written explicitlyas

• There are 21 independent material constants c ij ,

which is the case for a fully anisotropic material.

• For isotropic materials, c can be reduced to

where

• There are only two independent constants

among E , υ and G . The relationship among thesethree constants is



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• To formulate the dynamic equilibrium equations, letus consider an infinitely small block of solid,

• The equilibrium forces in the x -direction gives

where the term on the right-hand side of theequation is the intertial force term, and f x is theexternal body force applied at the centre of theblock.



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• Note that

• Hence, previous equation becomes one of theequilibrium equation, given as

• Similarly, the equilibrium of forces in y and zdirections result into two other equilibrium

equations:

• The equilibrium equations in x, y and zdirections can be written in concise matrix form:

where fb is the vector of external body forces inx, y and z directions:



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• Using kinematic equation

and constitutive equation

The equilibrium equation can be further written indisplacements:

Above relation is the general form of the dynamicequilibrium equation in matrix from.

For static equilibrium,

• Equations obtained in this section are applicable

to 3D solids.

• The objective of most analysts is to solve theequilibrium equations and obtain the solution of thefield variable, e.g. displacement.

• Theoretically, these equations can be applied toall other types of structures such as trusses,

beams, plates, and shells.• However, treating all the structural componentsas 3D solids makes computation very expensive,and sometimes practically impossible.

• That is why, theories for specific geometries havebeen developed to reduce computational effort.



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3.3 Equations for Two-Dimensional Solids

• Three-dimensional problems can be drastically

simplified if they can be treated as a two-dimensional solid.

• In general, thickness direction (e.g. z direction) istried to be removed.

• Assume that all the dependent variables areindependent of the thickness direction and all

external loads are independent of the z coordinate,and applied only in the x-y plane.

• Therefore, we are left with a system with only twocoordinates, the x and the y coordinates.

• There are primarily two types of 2D solids.

1. Plane stress solid

2. Plane strain solid

• Plane stress solids are those with thickness in z direction very small compared with the x and

Plane stress solids are those with thickness in zdirection very small compared with the x and y directions.

• External forces are applied only in x-y plane.

• Stresses in z direction (σzz,σxz, σyz) are all zero.

• Plane strain solids are those solids whose

thickness in z direction is very large compared withdimensions in x and y directions.



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• For plane strain, external forces are applied evenlyalong the z axis, and the movement in z direction at

any point is constrained.

• The strain components in the z direction ( εzz,εxz,εyz)are all zero.

• Note that for the plane stress problem, the strains

εxz and εyz are zero, but εzz will not be zero.



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• Constitutive equations for 2D in matrix form wouldbe

where c is a matrix of material constants.

• For plane stress, isotropic materials, we have

•To obtain the plane stress c matrix above, the

conditions of σzz=σxz=σyz=0 are imposed on thegeneralized Hooke’s law for isotropic materials.

• For plane strain case, εzz=εxz=εyz=0 areimposed to obtain

• The dynamic equilibrium equations for 2D solidscan be easily obtained by removing the termsrelated to the z coordinate from the 3D

counterparts.



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Above 3D equations will become

• These equilibrium equations can be written ina concise matrix form of

where fb is the external force vector given by

• For static problems,

• Above equations will be much easier to solveand computationally less expensive ascompared with the equations for the 3D solids.



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4. Fundamentals for Finite Element Method

4.1 Introduction

• First Step when using FEM is to formulate a setof partial differential equations.

• The problem domain is first discretized intosmall elements.

• In each of these elements, the profile of thedisplacements is assumed in simple forms to

obtain element equations.

• The equations obtained for each element arethen assembled together with adjoining elements

to form the global FE equation for the wholeproblem domain.

• Equations thus created for the global problem

domain can be solved easily for the entiredisplacement field.

Above mentioned FEM process does not seemto be a difficult task. However, if one goes intothe detail,

?? How can one simply assume the profile of theHow can one simply assume the profile of the

solution of the displacement in any simple form?solution of the displacement in any simple form?

?? How can one ensure that the governing partialHow can one ensure that the governing partial

differential equations will be satisfied by thedifferential equations will be satisfied by the

assumed displacement?assumed displacement?



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4.2 Strong and Weak Forms

• The partial differential equations developed in

previous chapter, are strong forms of the governing

system of equations for solids.

• Obtaining the exact solution for a strong form of the

system equation is usually very difficult for practicalengineering problems.

• The finite difference method can be used to solve

system equations of the strong form to obtain an

approximated solution.

• However, the method usually works well for

problems with simple and regular geometry andboundary conditions.

• A weak form of the system equations is usually

created using one of the following widely used

methods:

Energy principles

Weighted residual mehods

• The energy principle is a kind of variationalprinciple which is particularly suited for problems of

the mechanics of solids and structures.

• The weighted residual method is a more general

mathematical tool applicable, in principle, for solvingall kinds of partial differential equations.

• Weak form is often an integral form and requires aweaker continuity on the field variables.



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4.3 Hamilton’s Principle

• Hamilton’s principle is a simple yet powerful tool that

can be used to derive discretized dynamic systemequations.

• It states, “Of all the admissible time histories of

displacement the most accurate solution makes the Lagrangian functional a minimum.”

• An admissible displacement must satisfy the following

conditions:

(a) the compatibility equations

(b) the essential or the kinematic boundary conditions

(c) the conditions at initial (t1) and final time (t2).

• Mathematically, Hamilton’s principle states:

The Langrangian functional, L, is obtained using a setof admissible time histories of displacement and itsconsists of

where T is the kinetic energy, Π is the potential energy,and W f is the work done by the external forces.



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• The kinetic energy of the entire problem domain isdefined in integral form

where V represents the whole volume of the solid, and

U is the set of admissible time histories of

displacements.

• Potential energy in our case is elastic strain energy inthe entire domain of elastic solids and structures can

be expressed as

where ε εε ε are the strains obtained using the set ofadmissible time histories of displacements.

• The work done by the external forces over the set of

admissible time histories of displacements can beobtained by

where S f

represents the surface of the solid on whichthe surface forces are prescribed.

• Hamilton’s principle allows one to simply assume anyset of displacements, as long as it satisfies the threeadmissible conditions.

• The assumed set of displacements will not usuallysatisfy the strong form of governing system equations

unless we are extremely lucky, or the problem is

extremely simple and we know the exact solution.



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4.4 Standard FEM Procedure Using Hamilton’sPrinciple

4.4.1 Domain Discretization

• The solid body is divided into N e elements.

• The procedure is often called meshing, which is

usually performed using so-called pre-processors.

• An element is formed by connecting its nodes in a

pre-defined consistent fashion to create theconnectivity of the element.

• The density of the mesh depends upon the accuracyrequirement of the analysis and the computational

resources available.



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4.4.2 Displacement Interpolation

• The FEM formulation has to be based on acoordinate system.

• In formulating FEM equations for elements, it is oftenconvenient to used a local coordinate system.

• Based on the local coordinate system defined on an

element, the displacement within the element is nowassumed simply by polynomial interpolation using thedisplacements at its node as

where h stands for approximation, n d is the number ofnodes forming the element, and di is the nodal

displacement at i th node.



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where n f is the number of Degrees of Freedom

(DOF) at a node. For 3D solids, n f = 3.

Therefore, the total DOF for the entire element is n d xn f

N is a matrix of shape functions for the nodes in theelement,



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4.4.3 Requisites for the Shape Functions

A. Delta Function Properties

• The delta function property implies that the shape

function N i should be unit at its home node i ,andvanishes at remote nodes j ≠ i of the element.B. Partitions of unity property

• Shape functions are partitions of unity:

4.4.4 Formation of FE Equations in Local Coordinate

System

Once the shape functions are constructed, the FEequation for an element can be formulated.

What we have until now:

- Interpolation of the nodes:

- Strain-displacemet equation (Kinematic Equation):

- Strain energy term:

Substituting the interpolation of nodes and kinematic

equations into strain energy term.



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Substituting the interpolation of nodes and kinematicequations into strain energy term.

where B is often called the strain matrix, defined by

• By denoting

which is called the element stiffness matrix, abovestrain energy equation can be rewritten as

Note that the stiffness matrix ke is symmetric!

Making use of the symmetry of the stiffnessmatrix, only half of the terms in the matrix needto be evaluated and stored.

• Similarly, by substituting the relationship for the

interpolation of nodes into the kinetic energyequation



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In the above equation, mass matrix of an element is

generally denoted by

Therefore, kinetic energy can be rewritten as:

• Finally, to obtain the work done by the externalforces, interpolation of nodes is substituted intothe following relation:

yields

By denoting

and

Therefore, total work done on the system isgiven by:



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If we now summarize,

• Strain energy of an element is given by

• Kinetic energy of an element is given by

• Work done by external forces

• Recall the Langrangian function L in theHamilton’s principle

• Substituting the relations for kinetic energy, strain

energy and work done by external forces into the

above equation:

• Applying Hamilton’s principle, using

we have

Note that the variation and integration operatorsare interchangeable, hence we obtain



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• Integrating the first term by parts, we obtain

A

Substituting the above above relationship in

equation (A)

• To have the integration in above equation asZERO for an arbitrary integrand, the integranditself has to vanish, i.e.

• The only insurance for above equation to besatisfied is

• Above equation is the FEM equation for an

element, while ke and me are the stiffness and

mass matrices for the element, and fe is theelement force vector of the total external forcesacting on the nodes of the element.

• All these element matrices and vectors can beobtained simply by integration for the given shapefunctions of displacements.



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4.4.5 Coordinate Transformation

• The element equation derived above is

formulated based on the local coordinate system

defined on an element.• In genera, structure is divided into many elements

with different orientations, as shown below.

• To assemble all the element equations to form theglobal system equations, a coordinatetransformation has to be performed for each

element.• The coordinate gives the relationship between the

displacement vector de based on the localcoordinate system and the displacement vector De

for the same element, based on the global

coordinate system:



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T is the transformation matrix, which has differentforms depending upon the type of element and will

be discussed in later lectures.

• The transformation matrix can also be applied to

the force vectors between the local and globalcoordinate systems:

in which Fe stands for the force vector at node i inthe global coordinate system.

• Substitution of above transformed equations inglobal coordinate system into element equation willyield:

where,

4.4.6 Assembly of Global FE Equation

• The FE equations for all the individual elementscan be assembled together to form the global FE

system equation:

where K and M are the global stiffness and massmatrix, D is a vector of all the displacements at all

the nodes in the entire problem domain, and F isa vector of all the equivalent nodal force vectors.



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4.4.7 Solving the Global FE Equation

• By solving the global FE equation, displacements

at the nodes can be obtained.

• The strain and stress in any element can then beretrieved using kinematic and constitutive

equations.

4.5 Static Analysis

• Static analysis involves the solving of FE equation

without the inertia terms. Hence, for the staticsystem the equations take the form

• There are numerous methods and algorithms tosolve the above matrix equation.

• The methods often used are Gauss elimination and LU decompositions for small systems, and

iterative methods for large systems.

4.6 Analysis of Free Vibration

• Free vibration analysis is the solution of thehomogeneous equation, i.e. F = 0.

• For a solid or structure that undergoes a free

vibration, the discretized system equation becomes

• The commonly used methods to obtain the eigen

frequencies and eigen vectors are:

Jacobi’s method, bisection method, QR method,

supspace iteration, Lanczos’ method.



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4.7 Transient Response

• Structural systems are very often subjected totransient excitation.

• A transient excitation is a highly dynamic, time-dependent force exerted on the solid or structure,

such as earthquake, impact and shocks.

• The widely used methods to solve such problems

use direct integration method.

• The direct integration method basically uses the

finite difference method for time stepping to solve FEequations.

• There are two main types of direct integrationmethod: implicit and explicit .

• Implicity methods are generally more efficient for arelatively slow phenomenon.

• Explicit methods are more efficient for a very fastphenomenon, such as impact and explosion.

Recap of FEM Procedure

• In FEM, the displacement field is expressed bydisplacements at nodes using shape functions defined

over elements.

• Once the shape functions are found, the mass matrixand force vector can be obtained.

• The stiffness matrix can also be obtained, once theshape functions and the strain matrix have been found.

• Therefore, to develop FE equations for any type ofstructure components, all one need to do is to formulatthe shape function.



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5. FEM for Trusses

5.1 Introduction

• A truss is one of the simplest and most widelyused structural members.

• It is a straight bar that is designed to take onlyaxial forces, i.e. it deforms only in its axial direction.

• Finite element equations for truss members willbe developed in this section.

• The element developed is commonly known astruss element or bar element.

• Types of Trusses: Planar trusses

Space trusses



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5.2 Shape Function Construction

• Consider a structure consisting of a number of

trusses or bar members.

• Each of the members can be considered as atruss/bar element of uniform cross-section bounded

by two nodes (n d =2).

• Consider a bar element with nodes 1 and 2 at eachend of the element, as shown in the figure.

• The length of the element is l e

.

• The local x -axis is taken in the axial direction ofthe element with the origin at node 1.

• In the local coordinate system, there is only oneDOF at each node of the element, and that is theaxial displacement.

• Therefore, there are 2 DOF for the element.



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• In the FEM discussed previously, the displacementin an element should be written in the form

Where u h is the approximation of the axialdisplacement within the element, N is a matrix of

shape functions, and de should be the vector of thedisplacements at the two nodes of the element:

How can we determine the shape functions for How can we determine the shape functions for

the truss elements? the truss elements?

• Standard procedure for constructing shapefunctions is discussed in the following.

• Assume that the axial displacement in the truss

element can be given in a general form

where u h is the approximation of the displacement,

α is the vector of two unknown constants, α0 and α1,and p is the vector of polynomial basis functions.

• For this particular problem, we use up to the first

order of polynomial basis.

• Depending upon the problem, we could use higherorder polynomial basis functions.



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• The number of terms of basis functions depend uponthe number of nodes and degrees of freedom in theelement.

• Polynomial basis functions usually be complete oforders, i.e. any lower terms is not skipped. This is to

ensure the consistency.

• In deriving the shape function, we use the fact that

• Using,

• Substituting the boundary conditions, we then have

• Solving the above equation for α, we have

• Substituting the above equation into

will yield



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• The matrix of shape functions is then obtained in theform

where the shape functions for a truss element can bewritten as

Note that two shape functions have been obtained

because we have two nodes in the truss element.

• It is easy to confirm that these two shape functionssatisfy the delta function property and the partitions ofunity.

• It is clearly shown in the figure that N i gives theshape of the contribution from nodal displacement at

node I , i.e. why it is called a shape function.• In this case, the shape functions vary linearly acrossthe element, and they are termed linear shapefunctions.

• Substitution of these shape function will show thatdisplacement in the element varies linearly, therefore

called a linear element.



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5.3 Strain Matrix

• As discussed previously, the strain can be

computed using kinematic equation, given by

With U=Nde, above equation becomes

where the strain matrix B has the following form

for the truss element:

5.4 Element Matrices in the Local Coordinate System

• Once the strain matrix B has been obtained, thestiffness matrix for truss elements can be obtainedusing,

where A is the area of cross-section of the trusselement.

• The mass matrix for truss elements can be

obtained using

Symmetric Symmetric



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Symmetric Symmetric

• The nodal force vector for truss elements can be

obtained using

• Suppose the element is loaded by an evenly

distributed force f x along the x -axis, and twoconcentrated forces f s 1 and f s2 , respectively, at twonodes 1 and 2.

• The total nodal force vector becomes



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5.5 Element Matrices in the Global Coord. System

• Element matrices (stiffness, mass, nodal force)

were formulated based on local coordinate system,

where x -axis coincides with the mid axis of the bar.• In practical trusses, there are many bars of

different orientations and at different locations.

• To assemble all the element matrices to form theglobal system matrices, a coordinate transformation

has to be performed for each element.

• For a planar truss, the global coordinates X-Y canbe employed to represent the plane of the truss.

• The displacement at the global node i should have

two components in the X and Y directions only: D 2i-1

and D 2i .

• The coordinate transformation, which gives the

relationship between the displacement vector de

based on the local coordinate system and the

displacement vector De, has the same form as

discussed previously,

with



66

and the transformation matrix T given by

in which

are the direction cosines of the axial axis of the

element.

• The force vector in the global coordinate system is

• Equations for global stiffness Ke and mass Me

matrix are given by

5.6 Boundary Conditions

• The stiffness matrix Ke is usually singular,because the whole structure can perform rigid

body movements.

• These rigid body movements are constrainedby supports or displacement constraints.

• The imposition of these constraints lead to asymmetric positive definite, if sufficient

displacements are constrained.



67

5.7 Recovering Stress and Strain

• Following equilibrium equation can be solved

using standard routines and the displacements at

all the nodes can be obtained after imposing theboundary conditions.

• The stress in a truss element can also berecovered using the following equation:

5.8 Worked Examples

1. A uniform bar subjected to an axial force

Consider a bar of uniform cross-sectional area,shown in figure. The bar is fixed at one end and

is subjected to a horizontal load of P at the freeend. The dimensions of the bar are shown in thefigure, and the beam is made of an isotropic

material with Young’s modulus E.



68

Exact Solution

We first derive the exact solution, as this problem isvery simple. From Chapter 2, the strong form of a

truss member is given by

Since the bar is free of body forces and hencef x =0. Hence,

The general form of solution for above equationcan be obtained easily as

where c 0 and c 1 are unknown constants to be

determined by boundary conditions.The displacement boundary condition for this

problem can be given as

Therefore, we have c 0 =0.



69

Therefore, equation for u now becomes

Using,

we have,

The force boundary condition for this bar can be

given as

Equating the right-hand side of above two σx

equations, we obtain

The stress in the bar is obtained by substituting

the above equation back into

we have

Similarly, u(x) can be obtained as,



70

FEM solution

Using one element, the bar is modelled as shown inthe figure above.

Using

*There is no need to perform coordinatetransformation, as the local and global coordinatesystems are the same.

*There is also no need to perform assembly,because there is only one element.

The finite element equation becomes,

where F 1 is the reaction force applied at node 1,

which is unknown at this stage.

Instead, what we know is the displacementboundary condition at node 1.

We can then simply remove the first equation in

above, i.e.



71

which leads to

This is the FE solution of the bar, which is exactlythe same as the exact solution obtained.

The distribution of the displacement in the bar canbe obtained by substituting,

into

yielding

which is exactly the same as the exact solution.

Using,

we obtain the stress in the bar



72

2. A triangular truss structure subjected to a vertical force

Consider the plane truss structure shown in the

figure. The structure is made of three planar truss

members as shown, and a vertical downward force of1000 N is applied at node 2. The figure also showsthe numbering of the elements used, as well as thenumbering of the nodes.

StepStep--11: Define Local and Global Coordinate

System



73

•The local coordinates of the three truss elements areshown in the figure.

•The figure also shows the numbering of the global

degrees of freedom, D 1, D 2 ,…,D 6 , corresponding to thethree nodes.

• There are six global degrees of freedom altogether,

with each node having two DOFs.

• However, there is actually only one degree of

freedom in each node in the local coordinate system

for each element.• DOF at each node have contributions from more thanone element. For e.g. at node 1, the global degrees offreedom D 1 and D 2 have a contribution from elements

1 and 2.



74

Following table shows the dimensions and materialproperties of the truss members in the structure.

StepStep--22: Obtaining the direction cosines of theelements

• Knowing the coordinates of the nodes in the

global coordinate system, next step would be totake into account the orientation of the elementsw.r.t. the global coord. system.

• This can be done using,



75

StepStep--33: Calculation of element matrices in globalcoordinate system

• After obtaining the direction cosines, the element

matrices in the global coordinate system can beobtained.

• Note that, this problem is a static problem, hence

element mass matrices need not be computed.

• In the local coordinate system, the element

stiffness matrix is a 2 x 2 matrix.

• However, in the transformation to the globalcoordinate system, the DOF for each element

become four, i.e. stiffness matrix is a 4x4 matrix.

• Global stiffness matrix can be computed using,

with

and



77

Similarly, for element 2:

And for element 3:



79

•At the beginning of the assembly, the entire globalstiffness matrix is zeroed.

=

000000

000000

000000

000000

000000

000000

K

D 1 D 2 D 3 D 4 D 5 D 6

D 1

D 2

D 3

D 4

D 5

D 6

•As shown in the figure element 1 contributes to D 1,D 2 ,D 3 and D 4 .

•Therefore, the contribution in the global form forelement 1 is given by

−

−

×=

000000

000000

000000

000707

000000

000707

1091e

K

D 1 D 2 D 3 D 4

D 1

D 2

D 3

D 4



80

•Similarly, the contribution in the global form forelement 2 is given by

−

−

×=

700070

000000

000000

000000

700070

000000

1092e

K

D 1 D 2 D 5 D 6

D 1

D 2

D 5

D 6



81

•Similarly, the contribution in the global form forelement 3 is given by

−−

−−

−−

−−=

22 / 722 / 722 / 722 / 700

22 / 722 / 722 / 722 / 700

22 / 722 / 722 / 722 / 700

22 / 722 / 722 / 722 / 700

000000

000000

3eK

D 3 D 4 D 5 D 6

D 3

D 4

D 5

D 6



82

• In summary, we can get the final global stiffnessmatrix by adding the above discussed elemental

matrices.321 eee

K K K K ++=

which yields

Step 5Step 5: Applying boundary conditions

• The global matrix can normally be reduced in sizeafter applying boundary conditions.

• In this case, D 1, D 2 and D 5 are constrained, andthus

•This implies that the first, second and fifth rows andcolumns will actually have no effect on the solving of

the matrix equation.



83

Hence,

• The condensed global matrix becomes a 3 x 3matrix, given as follows:

• The constrained global FE equation is

where

and the force vector F is given as



84

This implies that we have three simultaneousequations with three unknowns,

Step 6Step 6: Solving the FE matrix equation

• The final step would be to solve the FE equation,to obtain the solution for D 3 , D 4 , and D 6 .

• In the present problem, it is possible to solve these

equations manually to obtain,

• To obtain the stresses in the elements, we use theconstitutive equation given by,



85

• In engineering practice, the problem can be of a

much larger scale, and thus the unknowns ornumber of DOFs will also be very much more.

• Therefore, numerical methods, or so-called solvers

for solving the FEM, have to be used.



86

5.9 Higher Order One-Dimensional Elements

• For truss members that are free of body forces,

there is no need to use higher order elements, as

the linear element can already give the exactsolution.

• However, for truss members subjected to bodyforces arbitrarily distributed in the truss elements

along its axial direction, higher order elementscan be used for more accurate analysis.

• The procedure for developing such high orderone-dimensional elements is the same as for thelinear elements. The only difference is the shapefunctions.

• In deriving high order shape functions, we

usually use the natural coordinate ξ, instead of

the physical coordinate x . The natural coordinateξ is defined as,

where x c is the physical coordinate of the mid pointof the one-dimensional element.

• In the natural coordinate system, the element is

defined in the range of -1 ≤ ξ ≤ 1.



87

• Figure shows a one-dimensional element of n thorder with (n +1) nodes.

• The shape function of the element can be writtenin the following form using so-called Lagrangeinterpolants:

where is the well known Lagrange interpolants,defined as

)( xln

k

From above equation it is clear that



88

One-dimensional QUADRATIC Shape Function:

Using the lagrange interpolant equation, the

quadratic one-dimensional element with threenodes shown in figure above can be obtainedexplicitly as,

One-dimensional CUBIC Shape Function:



Assignment # 1Assignment # 1

m

m2

1.

2. Do the same with the following figure. All the truss members are of the samematerial (E = 70 GPa ) and with the same cross-sectional area of 0.1 m2.

fea-mechanical 2010 student version

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