Element Assembly11
Transcript of Element Assembly11
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The Finite Element Method
Element assembly
Element assemblyComputational Mechanics, AAU, EsbjergThe Finite Element Method
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Structure Stiffness matrix
Number of elements NELE=3
Number of FE nodes NNODE=4
Total Number of Degrees of Freedom NDOF=4x2=8
Number of Degrees of Freedom per node NDOF=2
K1 L1 K2
N4
Element assemblyComputational Mechanics, AAU, EsbjergThe Finite Element Method
N1 N2N3 E2E1 E3
i j
i j
i j
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Element assemblyComputational Mechanics, AAU, EsbjergThe Finite Element Method
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Element assemblyComputational Mechanics, AAU, EsbjergThe Finite Element Method
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Element assemblyComputational Mechanics, AAU, EsbjergThe Finite Element Method
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Element assemblyComputational Mechanics, AAU, EsbjergThe Finite Element Method
1
3
4
1
2
2
x
y
5000 N
5000 N
Two Element Model with Equivalent Nodal Loads
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Element assemblyComputational Mechanics, AAU, EsbjergThe Finite Element Method
i = 1
1
j = 3m = 2
Element 1
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=
435130356040070
1302407010060140
3570350070601000100600
400600604000
701407000140
91.075000k
)1(
Element assemblyComputational Mechanics, AAU, EsbjergThe Finite Element Method
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Element assemblyComputational Mechanics, AAU, EsbjergThe Finite Element Method
i = 1
2
j = 4
m = 3Element 2
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=
400040060060
014070140700
4007043513035606014013024070100
0703570350
600601000100
91.075000k
)2(
Element assemblyComputational Mechanics, AAU, EsbjergThe Finite Element Method
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Assembly
[ ]
++
++
=
)2(
22
)2(
23
)2(
21
)2(
32
)2(
33
)1(
22
)1(
23
)2(
31
)1(
21
)1(
32
)1(
33
)1(
31
)2(12
)2(13
)1(12
)1(13
)2(11
)1(11
0
0
kkk
kkkkkk
kkk
kkkkkk
K
Element assemblyComputational Mechanics, AAU, EsbjergThe Finite Element Method
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Systematic Assembly of the System
Stiffness Matrix Assembly of the stiffness matrix, [K] follows a pattern
based on element-node number connectivity shown
in the table.
1 2 3
1 k1 2 k2 3 k3 4P
Element (e) Nodes 1e, 2e
1 1, 2
2 2, 3
3 3, 4
k1
-k1
-k1
k1+ k2
-k2
-k2
-k3
k3-k3
k2+ k3
[ ]=K
1 2 3 4
2
1
3
4
Element assemblyComputational Mechanics, AAU, EsbjergThe Finite Element Method
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Systematic Assembly of the
System Stiffness MatrixRepeat previous problem with a different
node numbering system:
1 2 3
3 k1
1 k2
4 k3
2
P
Element assemblyComputational Mechanics, AAU, EsbjergThe Finite Element Method
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Systematic Assembly of the
System Stiffness MatrixRepeat previous problem with different node
numbering system:
Element(e) Nodes 1e, 2e
1 3, 1
2 1, 4
3 4, 2
1 2 3
3 k1 1 k2 4 k3 2P
Element assemblyComputational Mechanics, AAU, EsbjergThe Finite Element Method
k1-k1
-k1k1+ k2 -k2
-k2
-k3k3
-k3 k2+ k3
1 2 3 4
2
1
3
4
[ ]=K
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Example
1 2 31 2
11 u,X 33 u,X
1k 2k22 u,X
Force equilibrium w/ load-displacement relationship:
=
3
2
1
22
2211
11
3
2
1
3222112
3223
2111
u
u
u
kk0
kkkk
0kk
X
X
X
or
uukuukX
uukX
uukX
The stiffness matrix of a system without constraints is singular.
Symmetric
+ diagonal elements
Simple way to assemble
stiffness matrix
Element assemblyComputational Mechanics, AAU, EsbjergThe Finite Element Method
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Example: Assembly of the Stiffness Matrix
1 2 31 2
11 u,X 33 u,X
1k 2k22 u,X
Step 1: Define the element stiffness matrices.
=
=
11
11kK,
11
11kK 2
21
1
[ ]
=
=
22
2211
11
21
kk0
kkkk0kk
110
110000
k
000
011011
kK
(Step 2: Align the element coordinates with the global coord.)
Step 3: Expand into the global DOF.
Banded:
Bandwidth
of 3
Force equilibrium and displacement compatibility is maintained.Element assembly
Computational Mechanics, AAU, EsbjergThe Finite Element Method
E l A bl f h S iff M i
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Example: Assembly of the Stiffness MatrixOrdering of the nodes and its impact
1 311 u,X 22 u,X
2
33 u,X
1 2
1k 2k
Step 1: Define the element stiffness matrices.
=
=
11
11kK,
11
11kK 2
21
1
[ ]
=
=
2121
22
11
21
kkkk
kk0
k0k
110
110
000
k
101
000
101
kK
(Step 2: Align the element coordinates with the global coord.)
Step 3: Expand into the global DOF.
Full:Bandwidth
of 5
The node ordering influences the stiffness matrix characteristics.
Element assemblyComputational Mechanics, AAU, EsbjergThe Finite Element Method
E l S i f l l ti
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Example: Spring force calculation
1 2 31 2
1k 2kR P
The stiffness equation:
=
=
P
0
R
X
X
X
u
u
u
kk0
kkkk
0kk
3
21
3
21
22
221111
0u1=
=
P
0
u
u
kk
kkk
3
2
22
221
21
3
1
2k
P
k
Pu,
k
Pu
Apply the prescribed displacement or the boundary condition
The stiffness equation becomes
Solve the system of equations to get (the fundamental solution)
Either displacement or load
(not both) is known at each DOF
Element assemblyComputational Mechanics, AAU, EsbjergThe Finite Element Method
E l S i f l l ti ( t )
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Example: Spring force calculation (cont.)
1 2 31 2
1k 2k
R P
The stiffness equation:
=
=
P
0
R
X
X
X
u
u
u
kk0
kkkk
0kk
3
2
1
3
2
1
22
2211
11
P
k
PkukukR
1
12111
=
=
=
=
P
P
kP
0
kk
kk
u
u
kk
kk
u
u
kk
kk
X
X
111
11
2
1
11
111
2
111
2
1
Spring 1:
The reaction:
Internal spring forces:
=
=
=
=
P
P
k
P
k
P
kP
kk
kk
u
u
kk
kk
u
u
kk
kk
X
X
21
1
22
22
3
2
22
222
2
122
2
1
Spring 2:
X1 X2= P
P
Element assemblyComputational Mechanics, AAU, EsbjergThe Finite Element Method