Finite Element Method-3

8/10/2019 Finite Element Method-3

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Finite

Element

MethodBy the Direct Stiffness Method (DSM)


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General ProcedurePre-processing

3. Globalisation: therefore, for the element, the global stiffness equation is

Recall from previous

lecture


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General ProcedurePre-processing3. Globalisation: where the global stiffness

matrix for the element is

()

Recall from

previous lecture


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General ProcedurePre-processing3. Globalisation: Now we write the

global stiffness equation for eachelement

To achieve this, we first set a tableof values of relevant data as shownbelow


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3. Globalisation: Table of values

() () () 1 1 22 2 33 2 4

22

4590135

2 2 0 2 2

2 2 12 2

0.500.5

0.510.5

0.500.5


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3. Globalisation: Element stiffness matrices.

For member (1)

,

,

,,

2

0.5 0.50.5 0.5

0.5 0.50.5 0.50.5 0.5

0.50.5 0.50.5 0.5

,

,

,,


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General ProcedurePre-processing3. Globalisation: Element stiffness matrices.

For member (1)

0.3536 0.3536

0.3536 0.3536

0.3536 0.3536

0.3536 0.35360.3536 0.35360.3536 0.3536 0.3536 0.35360.3536 0.3536

,

,

,,


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For member (2)

,

,

,

,

0 00 1

0 00 10 0

0 10 00 1

,

,

,

,


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For member (3)

,,

,,

2

0.5 0.50.5 0.5

0.5 0.50.5 0.5

0.5 0.50.5 0.5 0.5 0.50.5 0.5

,,

,,


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4.Assembly: Once the globalised

stiffness equation for each element of

the given structure is determined, the

next task is to merge these elements

stiffness equations into a singlemaster st i ffness equat ion.


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4.Assembly: The merging

operation must satisfy two

requirements

Equilibrium of forcesCompatibility of

displacements


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4.AssemblyEquilibrium of forces:This implies

that the algebraic sum of all externally

applied forces at a joint are balanced bythe sum of the reactive forces exerted

by all the members meeting at that

joint


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4.Assembly

Equilibrium of forces: For example, the

static equilibrium equation for joint 2 of

the example truss is given by:

() + () + ()


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General ProcedurePre-processing4.Assembly

Compatibility of displacement: Thismeans that the joint displacement of allmembers meeting at a joint is the same.

Again as an example, the compatibilityequations for joint 2 of the exampletruss is

,() ,() , and ,() ,() ,()


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4.Assembly: thus theequilibrium and compatibility

equations for nodes 1, 2, 3and 4 are:


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General ProcedurePre-processing4.Assembly:

1 ()

() 0

2 () + () + (),() ,() , and ,() ,() ,()


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4.Assembly:

3 ()

()

0

4 ()

()

0


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General ProcedurePre-processing4.Assembly: Thereare basically

two approaches to theassembly operation.

A. Manual assembly, and

B. Computer oriented assemblyapproach.


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4.AssemblyA. Manual assembly is suitable

for hand computation andconsist of two key steps


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4.AssemblyA. Manual assembly

I. Augmentation, andII. merging


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A. Manual assemblyI. Augmentation: This is the process of

expanding the force and displacement

vectors to an order that covers all degreeof freedom in the structure. This isrequired for compatibility during themerging operation to create the globalsystem stiffness equation.


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4.Assembly

A. Manual assemblyI. Augmentation: since the truss hasa total of eight (8) DoF, theequilibrium equations for each

element is expanded such that theforce and displacement fieldsbecome vectors containing eightelements each.


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A. Manual assembly

I. Augmentation:the stiffness

matrix for each element in theexample truss is thus expanded

to an square matrix


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4.Assembly

A. Manual assembly

I. Augmentation:consequently, theequilibrium equations for elements (1),(2), (3) and (4) becomes

()

+ ()

+ ()

() + () + () () + () + ()

()

+ ()

+ ()


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General Procedure Pre-processing

4. Assembly

A. Manual assembly

I. Augmentation:Thus, the stiffness equation for member (1)become

,()

,(),(),()

,

()

,(),(),()

0.3536 0.35360.3536 0.3536

0.3536 0.35360.3536 0.3536

0.3536 0.35360.3536 0.3536

0.3536 0.35360.3536 0.3536

0 00 0

0 00 0

0 00 0

0 00 0

0 00 0

0 00 00 00 0 0 00 0

0 00 0

0 00 00 00 0 0 00 0

,

,,,

,

,,,


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4. Assembly

A. Manual assembly

I. Augmentation:The stiffness equation for member (2)becomes

,(),(),()

,

()

,(),(),()

,()

0 00 0

0 00 0

0 0

0 0

0 0

0 1

0 00 0

0 00 0

0 0

0 1

0 0

0 00 00 0 0 00 10 00 0

0 00 0

0 00 1 0 00 00 00 0

0 00 0

,,,

,

,,,

,


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4. Assembly

A. Manual assembly

I. Augmentation:And finally the stiffness equation for

member (3) becomes

,(),(),(),(),()

,(),(),()

0 00 0

0 00 0

0 00 0

0.3536 0.35360.3536 0.3536

0 00 0

0 00 0

0 00 0

0.3536 0.35360.3536 0.3536

0 0

0 0

0 0

0 00 00 0 0.3536 0.35360.3536 0.3536

0 0

0 0

0 0

0 00 00 0 0.3536 0.35360.3536 0.3536

,,,,,

,,,


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4. Assembly

A. Manual assembly

I. Augmentation:Considering the requirement

for compatibility of displacement, the

member identification index can be dropped

from the equations above.


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4.Assembly

A. Manual assemblyII. Merging:in qualitative sense, this is

actually the process of reconnecting

the elements back together. This isachieved by simply superimposing

the stiffness equations of the

individual elements. This yields


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4. Assembly

A. Manual assembly

II. Merging

,,,,,,

,,

0.3536 0.3536

0.3536 0.35360.3536 0.3536

0.3536 0.35360.3536 0.3536

0.3536 0.3536

0.7072 0

0 1.7072

0 00 0

0 00 0

0 00 1

0.3536 0.3536

0.3536 0.35360 00 0

0 00 1

0 0

0 0

0.3536 0.3536

0.3536

0.3536

0 00 1

0 00 0

0 0

0 0

0.3536 0.3536

0.3536 0.3536

,,,,,,

,,


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4. AssemblyA. Manual assembly

II. Merging: where the global system stiffness matrix is given by

0.3536 0.3536

0.3536 0.3536

0.3536 0.3536

0.3536 0.3536

0.3536

0.35360.3536 0.3536 0.7072 00 1.7072

0 00 0

0 00 0

0 00 1

0.3536 0.35360.3536 0.35360 0

0 0

0 0

0 10 0

0 0

0.3536 0.3536

0.3536 0.3536

0 0

0 1

0 0

0 00 0

0 0

0.3536 0.3536

0.3536 0.3536


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4.Assembly

B. Computer oriented assembly:

There are different approaches to thecomputer oriented assembly operation,depending on the complexity of the structureinvolved.

The simplified assembler as the namesuggest is the easiest assembly procedureand can be adopted where the followingproperties hold true for the DSM

implementation.


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B. Computer oriented assembly:All elements are of the same type. For example, all

elements are made up of 2-node plane barNumber and configuration of the Degree of Freedom

for all nodes is the same.

The sequence of joint numbering proceeds withoutbreaks

The support conditions are single freedomconstraints: that is, they can be expressed asconstraints on individual degree of freedom.

The master stiffness matrix is stored a full symmetricmatrix.


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4.Assembly

B. Computer oriented assembly: Other

assemblers such as the Multi Element Type(MET) assembler and the Multi ElementType and Variable Freedom Configuration(MET-VFC) assembler are higher

assembling procedures for more complexstructural forms and are beyond the scopeof this module and hence would not becovered here.


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4.Assembly

B. Computer oriented assembly:in the simplified assembler procedure,

instead of the stiffness equation being

augmented, the entries from the elementsstiffness matrix are mapped directly unto

the global system stiffness matrix by the

use of Freedom Pointers


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4.Assembly

B. Computer oriented assembly

An array of the set of these freedom

pointers for an element is called the

Element Freedom Table (EFT).


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4.Assembly


This technique expresses the entries to themaster stiffness matrix as the sum ofentries in the elements stiffness matrix ()using the following relationship

=

and ()=

for 1, 4, 1, 4, EFT , E F T


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4.Assembly


The example truss will be used to illustrate the

application of the simplified assembler

procedure.

We will begin by setting the entries of the 8 8master stiffness matrix to zero. Thus


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4. Assembly


0 00 0

0 00 00 0

0 00 00 0

0 00 0

0 00 00 0

0 00 00 0

0 00 0 0 00 00 00 0

0 00 0

0 00 0 0 00 00 00 0

0 00 0

1234

5678

The numbers written against each row are the global

DoF numbers.


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General Procedure

Pre-processing4. Assembly


Next we assemble the FET for all elements by

computing the freedom pointers (FP)corresponding to all DoF associated with theelement. The FP for elements in the exampletruss are obtained from the relation

() 2 12 The value 2 in the formula represents the number of

DoF in the node under consideration and is thenode number.


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4.Assembly

B. Computer oriented assembly:The FP

the element are computed thus;

() 2 1 1 1

2 1 2 for node 1, and

() 2 2 1 32 2 4 for node 2


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B. Computer oriented assembly This procedureis repeated for elements 2 and 3. The

resulting for each element are collectedas shown below () 1,2,3,4 , () 3,4,5,6, , ()

3,4,7,8

The are then used to map entries in theelement stiffness matrix ()to the master stiffnessmatrix as follows


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For element 1

()

0.3536 0.35360.3536 0.3536

0.3536 0.35360.3536 0.3536

0.3536 0.35360.3536 0.3536 0.3536 0.35360.3536 0.3536

12

34

Upon merging with, we have


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4.Assembly


For element 1

0.3536 0.35360.3536 0.3536

0.3536 0.35360.3536 0.3536

0.3536 0.35360.3536 0.3536

0.3536 0.35360.3536 0.3536

0 00 0

0 00 0

0 00 0

0 00 0

0 00 0 0 00 00 00 0

0 00 0

0 00 0 0 00 00 00 0

0 00 0

1234


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4.Assembly


For element 2

()

0 0

0 1

0 0

0 10 00 1 0 00 1

3

456,

and upon merging with, we have


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For element 2

0.3536 0.35360.3536 0.3536

0.3536 0.35360.3536 0.3536

0.3536 0.35360.3536 0.3536 0.3536 0.35360.3536 1.3536

0 00 0

0 00 0

0 00 1 0 00 00 00 0

0 00 1

0 00 0

0 00 0

0 00 1

0 00 0

0 00 0

0 00 0

3456


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4.Assembly


For element 3

()

0.3536 0.3536

0.3536 0.3536

0.3536 0.3536

0.3536 0.35360.3536 0.35360.3536 0.3536 0.3536 0.35360.3536 0.3536

3

478

When merged with , will result to


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4. Assembly


For element 3

0.3536 0.35360.3536 0.3536

0.3536 0.35360.3536 0.3536

0.3536 0.35360.3536 0.3536

0.7072 00 1.7072

0 00 0

0 00 0

0 00 1

0.3536 0.35360.3536 0.3536

0 00 0

0 00 10 0

0 00.3536 0.3536

0.3536 0.3536

0 00 1

0 00 00 0

0 00.35360 0.35360.3536 0.3536

34

78


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General ProcedurePre-processing4. Assembly

B. Computer oriented assembly:The global stiffnessmatrixfor the truss system is therefore

0.3536 0.35360.3536 0.3536

0.3536 0.35360.3536 0.3536

0.3536 0.35360.3536 0.3536

0.7072 00 1.7072

0 00 0

0 00 0

0 00 1

0.3536 0.35360.3536 0.3536

0 00 0 0 00 10 00 0

0.3536 0.35360.3536 0.3536

0 00 1 0 00 00 00 0

0.35360 0.35360.3536 0.3536


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5. Boundary Conditions (BCs)

a) Reduction Methodb) Modification Method


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a) Reduction Method,, 0

0.7072 00 1.7072 ,,


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b) Modification Method

0000000

1 00 1 0 00 00 00 0

0.7072 00 1.7072

0 00 0 0 00 00 00 0

0 00 0

0 00 0

0 00 00 0

0 00 00 0

1 00 1

0 00 00 0

0 01 00 1

,,,,,,,,


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General Procedure

6.Joint Displacement Solution

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Finite Element Method-3

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