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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
Force at node 2 due to unit displacement at node 1
Force at node 3 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
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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
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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.
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