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