Direct Method-1.pptx

8/14/2019 Direct Method-1.pptx

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CHATER - 1Direct MethodSpring systems

SUMMARY

Developing finite element equations for asystem of springs using directstiffnessapproach.

Finite element numbers and arrays Global and local axes

Application of boundary conditions

Physical significance of the stiffness matrix Direct assembly of the global stiffness matrix

Examples


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FEM Analysis Scheme:

Step 1: Divide the problem domain into non-overlappingregions (elements) connected to each other

through special points (nodes).

Step 2: Describe the behavior of each element.

Step 3: Describe the behavior of the entire body by

putting together the behavior of each elements

(process known as assembly).

Step 4: Solve the assembled system of equations for

unknowns (Displacements) and deduce other

results (element internal forces).


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PROBLEM:Analyze the behavior of given system composed of the twosprings loaded by external forces as shown.

Given

F1x , F2x,F3xare the external loads. Positive directions of theforces are along the positive x-axis

k1and k2are the stiffnesses of the two springs

NELT = Total number of elements = 2

NNOD = Total number of nodes = 3

k1 k2

F1x F2x F3xx

3

12


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SOLUTION:

Step 1: In order to analyze the system we break it upinto smaller parts, i.e., elements connected

to each other through nodes (discretization).

Unknowns: Nodal displacements d1x, d2x,d3x

A spring behavior is described by an axial forceand displacement (one degree of freedom pernode).

k1k2F1x F2x F3x

x

1 2 3

Element 1 Element 2

Node 1d1x d2x d3x


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Step 2: Analyze the behavior of a single element (spring)

using local coordinate system (local axis )

Element has two nodes: 1, 2 Spring constant: k

Nodal displacements:

Nodal forces:

Local axis = Spring axis oriented from node 1 to node 2

NDEL= Number of nodes per element = 2

NDOF= Number of degrees of freedom per node = 1

2x1x dandd

x

2x1x fandf

x


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Local Node Numbering & Global Node Numbering:

Each element is identified by two nodes 1 and 2. These are

called local numbers and are different from the global(structure) node numbers as shown below.

Connectivity information saved in an integer array such as:

Dimension KONEC(NELT, NDEL)

KONEC array will have rows equal to NELT(Total No. of

elements) and columns equal to NDEL(No of nodes perelement)For example, in the above Fig. shown by Gray part

ELEMENT Node 1 Node 2

1 1 2

2 2 3

Local node number

Global node number


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In general 3D case, local ( ) and global (x ,y ,z)coordinate systems are:

For a spring, only one axis is required.

z,y,x


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Local axes are attached to the element

Global coordinate system is

Unique for entire model. Element behavior is more easily

described in local coordinates

Transform of all elements behavior expressions to global

coordinates prior to assembly process.

Coordinate transformation is required in FEM for all vector

quantities (forces, displacements ..)

Coordinate transformation is not required in FEM for scalar

quantities (temperature, pressure, mass.)

LCS= Local Coordinate System

GCS= Global Coordinate System

For a spring system local axis is same as global axis xx


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BEHAVIOR OF A LINEAR SPRING IN LCS:

F = Force in the spring

d = displacement of the springk= stiffness of the spring

By Hookes law for a spring element with one end displacement:

F = kd


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Hookes law for spring element with two end displacements:

Force equilibrium gives:

(1)Eq)dd(kf 1x2x2x

0ff 2x1x

(2)Eq)dd(kff 1x2x2x1x


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Writing equations (1) and (2) in the matrix form gives:

In compact form:

= Element force vector

= Element displacement vector

= Element stiffness matrix

d

2x

1x

kf

2x

1x

d

d

kk-

k-k

f

f

f

d

k

dkf

(2a)Eqdkdkf(1a)Eqdkdkf

2x1x2x

2x1x1x


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Note:

1. The element stiffness matrix is symmetric, i.e.

2. The element stiffness matrix is singular, i.e.

The consequence is that the matrix is NOT invertible. It is not

possible to invert it to obtain the displacements. Why?

The spring is not constrained in space and hence it can attainmultiple positions in space for the same nodal forces

Example:

kk T

0kk)k(det 22

2

2-

4

3

22-

2-2

f

f2

2-

2

1

22-

2-2

f

f

2x

1x

2x

1x


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Absence of restraints = Singular stiffness matrix

Mechanical instability = Numerical instability

Step 3: After describing the behavior of each element, let us obtain the behavior of the structure by assembly.

Split the original structure into component elements

)3(Eq

dd

kk-k-k

ff

)1()1()1(d

(1)

2x

(1)

1x

k

11

11

f

(1)

2x

(1)

1x

)4(Eq

dd

kk-k-k

ff

)2()2()2(d

(2)

2x

(2)

1x

k

22

22

f

(2)

2x

(2)

1x


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To assemble these two results into a single description of the

response of the entire structure we need to link between the

localand globalvariables.

Local(element) displacementsrelated to global (structure)

displacements as shown:

k1k2F1x F2x F3x

x

1 2 3

Element 1 Element 2

Node 1d1x d2x d3x

3x

(2)

2x

2x

(2)

1x

(1)

2x

1x(1)1x

dd

)5(Eqddddd


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Hence, equations (3) and (4) may be rewritten as:

Or, we may expandthe matrices and vectors to obtain:

Expanded element stiffness matrix of element 1 (local)

Expanded nodal force vector for element 1 (local)

Nodal load vector for the entire structure (global)

2x

1x

11

11

(1)2x

(1)

1x

d

d

kk-

k-k

f

f

3x

2x

22

22

(2)2x

(2)

1x

d

d

kk-

k-k

f

f

)6(Eq

d

d

d

000

0kk-

0kk

0

ff

d

3x

2x

1x

k

11

11

f

(1)

2x

(1)

1x

)1()1(

e xe x

)7(Eq

d

d

d

kk-0

kk0

000

f

f0

d

x3

2x

1x

k

22

22

f

(2)

2x

(2)1x

)2()2(

e xe x

ex)1(

k

ex)1(

f

d


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Local(element) nodal forcesrelated to global (structure)

forces as shown (using free body diagrams FBD) :

k1k

2

F1x

F2x

F3x x

1 2 3

d1x d2xd3x

A B C D

0f-F:3nodeAt

0ff-F:2nodeAt

0f-F:1nodeAt

(2)

2x3x

(2)

1x

(1)

2x2x

(1)1x1x


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In vector form,

the nodal force vector (global) is:

Recall the previous expanded element force vectors:

Hence, the global force vector is simply the sum of the

expanded element nodal force vectors

(2)

2x

(2)

1x

(1)

2x

(1)

1x

3x

2x

1x

f

ff

f

F

F

F

F

(2)

2x

(2)1x)2((1)2x

(1)

1x

)1(

ff

0

fand

0

ff

f exex

exex )2()1(

3x

2x

1x

ff

F

F

F

F


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Expressions of the expanded local force vectors from Eqs (6)

and (7):

Hence:

In compact form:

dkfanddkf(2)ex)2((1)ex)1(

exex

dkkdkdkffF(2)ex(1)ex(2)ex(1)ex)2()1(

exex

dKF

matricesstiffnesselementexpandedofsummatrixstiffnessGlobalK

vectorntdisplacemenodalGlobald

vectorforcenodalGlobalF


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For the original structure with two springs, the global stiffness

matrixis:

22

2211

11

k

22

22

k

11

11

kk-0

kkkk-

0kk

kk-0

kk0

000

000

0kk-

0kk

K

)2()1(

exex

NOTES

1. The global stiffness matrix is symmetric

2. The global stiffness matrix is singular


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Structure force vector = sum of expanded element force vectors

Structure stiffness matrix = sum of expanded element stiffness

matrices

In practical computer implementation, element assembly is

performed without using vector or matrix expansion.

Various force and matrix components for each element are added

at their appropriate addresses in the structural force vector or

stiffness matrix.

Component address is easily recognized using the equation

number array as will be seen shortly.

Coordinate transformation for the spring system is simple.

In general, cosine vectors between local and global axes are used.


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The system equations implies:dKF

3x22x23x

3x22x211x12x

2x11x11x

3x

2x

1x

22

2211

11

3x

2x

1x

dkd-kF

dkd)kk(d-kF

dkdkF

d

d

d

kk-0

kkkk-

0kk

F

F

F

These are the 3 equilibrium equations at the 3 nodes.


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k1 k2F1x F2x F3xx

1 2 3

d1x d2x d3xA B C D

0f

-F:3Node

0ff-F:2Node

0f-F:1Node

(2)

2x3x

(2)

1x

(1)

2x2x

(1)

1x1x

(1)

1x2x1x11x fddkF

(2)

1x

(1)

2x

3x2x22x1x1

3x22x211x12x

ff

ddkddkdkd)kk(d-kF

(2)

2x3x2x23x f

dd-kF


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Notice that the sum of the forces equal zero, i.e., the

structure is in static equilibrium: F1x+ F2x+ F3x = 0

Given the nodal forces, can we solve for thedisplacements?

Without restraints, the singular matrix cannot be

inverted.

To obtain unique values of the displacements, at least

one of the nodal displacements must be specified

(restrained).


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Global

Local

Direct assembly of the global stiffness matrix

k1 k2F1x F2x F3xx

1 2 3

Element 1 Element 2d1x d2x d3x


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The structure stiffness matrix can be assembled directly using

the connectivity array described before (without using

expanded element matrices)

This array gives the appropriate address for each element

matrix.

ELEMENT Node 1 Node 2

1 1 2

2 2 3

Local node number

Global node number


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Initialize structure stiffness matrix to zero

Assemble element 1

Assemble element 2

000

0kk-0k-k

K 11

11

22

2211

11

kk-0

k-kkk-

0k-k

K

11

11

1kk-

k-kk

22

22

2kk-

k-kk

[1]

[2]

[2]

[3]


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11

11)1(

kk-

k-kk

Stiffness matrix of element 1

d1x

d2x

d2xd1x

Stiffness matrix of element 2

22

22)2(

kk-

k-kk

d2x

d3x

d3xd2x

Global stiffness matrix

22

2211

11

kk-0

k-kkk-

0k-k

K d2x

d3x

d3xd2x

d1x

d1x


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The structure force vector and stiffness matrix, in this

example, were assembled before considering boundary

conditions.

Boundary conditions (B.C.) usually reduce the number of

equations and a better assembly strategy is the one

performed on the reduced system (after considering B.C.)

For this simple spring system with one degree of freedom pernode, direct assembly can be performed using connectivity

array.

For other finite elements with more than one degree of

freedom per node, direct assembly will require another array

(equation number array)

E l 1


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Example 1

Compute the global stiffness matrix of the spring system

shown (3 elements and 4 nodes, NELT = 3, NNOD = 4)

The structure stiffness matrix is easily found as:

d3xd2xd1x d4x

d2x

d3x

d1x

d4x

33

3322

2211

11

kk-00

k-kkk-0

0k-kkk-

00k-k

K


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1000 1000 0 0

1000 1000 2000 2000 0K

0 2000 2000 3000 3000

0 0 3000 3000

d3xd2xd1x d4x

d2x

d3x

d1x

d4x


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Example 2

Compute the global stiffness matrix of the

assemblage of springs shown

Parallel springs 2 and 3 act as a

single spring with a stiffness equal

to the sum of the two stiffnesses.

Hence:

1 1

1 1 2 3 2 3

2 3 2 3

k -k 0

K -k k k k - k k 0 - k k k k


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Imposition of Boundary Conditions (B.C.)

We consider a two-spring system with 2 cases:

Case 1: HomogeneousB.C. (zero displacement, e.g., d1x

= 0)

Case 2: NonhomogeneousB.C. (non-zero displacement, e.g.,

d3x = 0.06 m)

Homogeneous boundary condition at node 1 The spring system is restrained at node 1 and subjected to a

known force at node 3.

k1=500N/mk2=100N/m F3x=5N

x1

2 3

Element 1 Element 2

d1x=0 d2x d3x


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System equations:

After substitution:

Note that F1xis the reaction which is to be computed as part of

the solution and hence is an unknown in the above equation

In general: When displacements are known, forces (reactions)

are unknown and vice versa.

3x

2x

1x

22

2211

11

3x

2x

1x

d

d

d

kk-0

kkkk-

0kk

F

F

F

3x

2x

1x

d

d

0

100001-0

100600005-

0500500

5

0

F


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Writing out the equations explicitly

Eq(2) and (3) are used to find d2xand d3x by solving a

reduced system

Eq(1) is then used to deduce the reaction by substitution

We find

2x 1

2 3

2 3

-500d

600 100 0

100 100 5

x

x x

x x

F

d d

d d

Eq(1)

Eq(2)

Eq(3)

2

3

2

3

600 100 0

100 100 5

0.01

0.06

x

x

x

x

d

d

d m

d m

1 2x=-500d 5xF N


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The matrix system has thus two parts:

One part for solving and finding unknown displacements

The other part for substitution and finding unknown reactions

The two parts may contain many equations each and not

appear in the right order. In that case, rearrangement of

equations is necessary to group known and unknown

displacements separately

NOTICE: The reduced matrix giving the unknown

displacements, may be obtained from the global stiffness

matrix by deleting the first row and column(suppression of

equations corresponding to zero displacements) as:

500 -500 0

-500 600 -100

0 -100 100

600 100

100 100


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NOTICE:

1. Take care of homogeneousboundary conditions by

deleting the appropriate rows and columns from the globalstiffness matrix and solving the reduced set of equations for

the unknown nodal displacements.

2. Both displacements and forces CANNOT be known at the

same node. If the displacement at a node is known, the

reaction force at that node is unknown (and vice versa)

Deletion of rows and columns corresponding to homogeneous

boundary conditions (zero displacement) will lose theequations retrieving the reactions.

The reactions can in fact be retrieved by a different scheme to

be described later.


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Nonhomogeneous boundary condition: spring 2 is pulled at

node 3 by 0.06 m)

The system equation is:

Note that now F1xand F3xare both unknown.

k1=500N/mk2=100N/m

x1

2 3

Element 1

Element 2

d1x=0 d2x d3x=0.06m

0.06

d

0

100001-0

100600005-

0500500

F

0

F

2x

3x

1x


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Writing out the equations explicitly

Use only eq (2) to compute d2x

Use Eq(1) and (3) to compute F1x= -5N and F3x = 5N

In this case (Nonhomogeneous B.C.), we cannot delete

corresponding rows and columns.

2x 1

2

2 3

-500d

600 100(0.06) 0

100 100(0.06)

x

x

x x

F

d

d F

Eq(1)

Eq(2)

Eq(3)

2

2

600 100(0.06)0.01

x

x

d

d m


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Assembly process with suppression

of zero-displacement equations

In practice, only the reduced matrix system is assembled.

There is no need to assemble the full arrays and then suppress

some rows and columns.

To achieve this a special integer array locating all nodeequations is used.

The NUM array has NDOF rows and NNOD columns:

Dimension NUM(NDOF , NNOD)

For the two-spring system with homogeneous B.C. the array is: NUM= [0 1 2] There are two equations corresponding to

displacements at nodes 2 and 3.

NEQ= Total number of equations = 2


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NUM= [0 1 2]

The array has one row only because there is only one DOF per

node.

NUMarray is used to locate the appropriate addresses for

assembly.

The full size of the reduced stiffness matrix is NEQ x NEQ

For element 1 with nodes 1 and 2 (global numbers) the

equation numbers to use for assembly are [0 1].

A zero number means that the component is not assembled.

For element 2 with nodes 2 and 3 (global numbers) the

equation numbers to use for assembly are [1 2].


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Initialize reduced stiffness matrix to zero

Assembly of element 1

with equations [0 1]

Assembly of element 2

with equations [1 2]

00

0kK

1

22

221

kk-

k-kk

K

11

11

1kk-

k-kk

[0]

[1]

22

22

2kk-

k-kk [1]

[2]


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Element internal forces

Having determined the unknown displacements, element

internal forces can now be deduced. Element internal forces are expressed in LCS. Considering

tension as positive, element force is given by eq. (1) at node 2

(local number)

Case of homogeneous B.C.

We found

k1=500N/mk2=100N/m F3x=5N

x1

2 3

Element 1 Element 2

d1x=0 d2x d3x

d2x = 0.01 m and d3x= 0.06 m

(1)Eq)dd(kf 1x2x2x


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For our spring system local and global displacements are the

same.

Applying eq(1) to both elements gives:

Element 1:

Element 2:

At node 1 (local number), we would obtain an opposite value(reflecting a tension force in the opposite direction).

For the case of non homogeneous B.C., element internal

forces are determined the same way.

N50)-500(0.01)d(dkfF 1x2x12x1

N50.01)-100(0.06)d(dkfF 2x3x22x2

l


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Reaction retrieval

Suppression of all rows and columns corresponding to zero-

displacements, loses all information about unknown reactions.

Reactions are not in fact necessary for structural analysis. They can however be deduced using another alternative of

node equilibrium.

The total force in any node is equal to the sum of element

forces, for all elements connected to the node (in GCS).

Node 1(connected to element 1 only):

This is the same reaction result found previously.

e

e

nFnP

NFF 5)5(P 121

11


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In unrestrained DOF (non-support nodes), node equilibrium

will retrieve the known external force (possibly zero)

This will serve as a check of all the finite element process.

Node 2(connected to elements 1 and 2) :

This confirms the absence of forces at node 2.

Node 3(connected to element 2 only) :

This retrieves the external force applied at node 3.

This method delivers unknown reactions and serves as a

check at other DOF.

0)5(5)(P 221

22 FF

NF 5P 233


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Recap of what we did

Step 1: Divide the problem domain into non overlapping

regions (elements) connected to each other through special

points (nodes)

Step 2:Describe the behavior of each element ( ) in LCS

Step 3:Describe the behavior of the entire body (by

assembly).

This consists of the following steps

1. Write the force-displacement relations of each spring in

expandedform in GCS

dkf

dkf ee


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2. Relate the local forces of each element to the global forces at

the nodes (use FBDs and force equilibrium).

Finally obtain

Where the global stiffness matrix

e

f

F

dKF

ekK


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Apply boundary conditions by partitioning the matrix and

vectors

and are the known and unknown displacements

are unknown reactions and are known external forces

Solve for unknown nodal displacements

Compute unknown reaction forces

Deletion of zero displacement (homogeneous) will lose this

last part. Node equilibrium method is used for reactions.

2

1

2

1

2221

1211

F

F

d

d

KK

KK

1212222 dKFdK

2121111 dKdKF

1d

2d

1F 2F


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Step 4:Deduce element internal forces (in LCS).

Step 5:Retrieve reactions and external nodal forces usingnode equilibrium method.

Ph i l i ifi f th tiff t i


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Physical significance of the stiffness matrix

Typical 3 degree of freedom finite element equations are:

The first equation (force equilibrium at node 1) is:

What if d1=1, d2=0, d3=0 ? (nodes 2 and 3 are held fixed)

Force at node 1 due to unit displacement at node 1



3

2

1

3

2

1

333231

232221

131211

F

F

F

d

d

d

kkk

kkk

kkk

1313212111 Fdkdkdk

313

212

111

kF

kF

kF


51/57

Similarly we obtain the physical significance of the other entries

of the global stiffness matrix

In general:

This offers an alternate route to generating the stiffness matrix

of an element or that of the structure.

ijk = Force at node i due to unit displacement

at node j keeping all the other nodes fixed


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Example: Determine the first column of the stiffness matrix

for the two spring system

Set d1=1, d2=0, d3=0

In this case, Element #2 does not have any contribution.

FBD of Element #1

k1k2F1 F2 F3

x

1 2 3

Element 1 Element 2d1 d2 d3

x

k1

(1)

1xd

(1)

1xf (1)

2xf

(1)

2xd


53/57

The forces at the spring ends are:

(1) (1) (1)

2x 1 2x 1x 1 1 f (d d ) (0 1)k k k (1) (1)1x 2x 1

f f k

F1

F1 = k1d1 = k1=k11

F2 = -F1 = -k1=k21F3 = 0= k31

(1)

1xf

Force equilibrium at node 1

(1)

1 1x 1F =f k

Force equilibrium at node 2

(1)

2x

f

F2(1)

2 2x 1F =f k

Of course, F3=0


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To obtain the second columnof the stiffness matrix, calculate

the nodal reactions at nodes 1, 2 and 3 when d1=0, d2=1, d3=0We find:

The third columnis obtained

when d1=0, d2=0, d3=1

Hence the first columnof the stiffness matrix is

1 1

2 1

3 0

F k

F k

F

1 1

2 1 2

3 2

F k

F k k

F k

1

2 2

3 2

0F

F k

F k

Steps in solving a finite element problem


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Steps in solving a finite element problem

Step 1: Write down the node-element connectivity table

linking local and global displacements

Step 2:Write down the stiffness matrix of each element

Step 3:Assemblethe element stiffness matrices to form the

global stiffness matrix for the entire structure using the node

element connectivity table

Step 4:Incorporate appropriate boundary conditions

Step 5:Solve resulting set of reduced equationsfor the

unknown displacements

Step 6:Compute the unknown element nodal forces

Step 7:retrieve unknown reactions (optional)

Th d d f ll


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The same steps and same parameters are used for all

FEM problems, with minor differences.

NELT= Total number of elements NNOD= Total number of nodes

NDEL= Number of nodes per element

NDOF= Number of degrees of freedom per node NEQ= Total number of equations

KONEC= Connectivity array (integer)

NUM= Equation number array (integer)

These numbers and arrays are necessary for computer

implementation of FEM in 1D, 2D, or 3D.

Memory management and cost saving


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Memory management and cost saving

Cost of FEM problems related to memory requirements and

number of computer calculations.

In large FEM problems with thousands and even millions of

equations, the largest array is the structure stiffness matrix.

Method of suppression of zero-displacement equations is

recommended.

Symmetry of the matrix must be exploited to reduce memory

requirements and number of operations.

Other properties of the stiffness matrix will be used to

achieve further drastic savings.

Direct Method-1.pptx

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