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Computer (matrix) version of the stiffness · PDF fileComputer (matrix) version of the stiffness method 1. The computer version of the stiffness matrix is a generalization of the classical

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  • Computer (matrix) version of the stiffness method

    1. The computer version of the stiffness matrix is a generalization of the classical version leading towards the computer applications and the finite element method. The basic ideas remain the same, though some assumptions are released. The computer version is based on the following assumptions displacements and strains remain small compared to the dimensions of the structure the

    equilibrium is related to the undeformed configuration (1st order theory) material is linearly elastic with classical Hookes law plane cross-sections remain plane during bending from these three conditions it follows that the superposition rule is valid Compared to the classical approach the assumption of no deformability in the axial direction of the bars is released. Hence, the influence of the axial forces on the displacements is taken into account. The main concept of the method remains, i.e. the bar structure is subdivided (discretised) into a finite number of bars (elements). The ends of the elements are called nodes. e element number i, k node numbers The deformed state of the structure is described by the generalized nodal displacements, i.e. linear displacements and angles of cross-section rotations. Type and number of generalized displacements in the node depends on the type of structure. The nodal displacements are also called the nodal degrees of freedom. They can be assembled into a vector of nodal displacements qn, those for the element into the vector of element displacements qe and, finally, those for the entire structure into the global vector of displacements q.

    =

    i

    in

    v

    uq

    =

    i

    i

    i

    n

    w

    v

    u

    q

    i Truss Frame

    k e

    Plane truss

    i

    y

    x

    ui

    vi

    i

    y

    x

    ui vi

    z

    wi 3D truss

  • =

    i

    i

    i

    n v

    u

    q

    =

    z

    y

    x

    i

    i

    i

    n

    w

    v

    u

    q

    The displacements presented in the above figures are referred to the global sets of co-ordinates xy or xyz. In practice it is much easier to form the basic equations on the level of elements, hence, the local systems of co-ordinates are introduced. The values corresponding to the local co-ordinates are additionally denoted with tildes. The slope-defection formulae from the classical stiffness method are replaced by the element stiffness matrix, which relates the element displacements in local co-ordinates to the reactions at the element supports in local co-ordinates:

    eee qKR~~~ = [ ]

    nneijek

    =

    ~~K

    where n is the number of element degrees of freedom. The elements of the stiffness matrix represent reactions at the element supports created by unit displacements. For instance, the

    element eijk~

    represents the reaction number i created by the action of the displacement 1~ =jq

    First, let us consider the plane truss element. In this case the vectors of element displacements and reactions have four components

    =

    =

    4

    3

    2

    1

    ~

    ~

    ~

    ~

    ~

    ~

    ~

    ~

    ~

    q

    q

    q

    q

    v

    u

    v

    u

    k

    k

    i

    i

    eq

    =

    =

    k

    k

    i

    i

    e

    T

    N

    T

    N

    R

    R

    R

    R

    4

    3

    2

    1

    ~

    ~

    ~

    ~

    ~R

    Plane frame

    i

    y

    x ui

    vi

    i

    i y

    x

    ui vi

    z

    wi 3D frame y

    z

    x

    Interpretation of the rotation angle

    x

    Plane yz

    y~

    x~

    i

    k

    11~ ,

    ~qR

    22~ ,

    ~qR

    44~ ,

    ~qR

    33~ ,

    ~qR

    e

    l

  • From the Hookes law

    l

    EAR

    l

    EAR

    l

    EAN === 31

    ~ ,

    ~ 1

    From the Hookes law

    l

    EAR

    l

    EAR

    l

    EAN === 31

    ~ ,

    ~ 1

    No internal forces are created. The same

    situation occurs in the state 1~4 =q .

    Summarising of these results yields the following element stiffness matrix

    =

    0000

    0101

    0000

    0101

    ~

    l

    EAeK

    From the physical interpretation of the element stiffness matrix it follows, that for instance the first column of this matrix represents the vector of reactions in the element created by the action of the

    displacement 1~1 =q .

    Now, let us consider the stiffness matrix for the plane bar element under bending (i.e. the plane beam element).

    =

    =

    6

    5

    4

    3

    2

    1

    ~

    ~

    ~

    ~

    ~

    ~

    ~

    ~

    ~

    ~

    ~

    ~

    ~

    q

    q

    q

    q

    q

    q

    v

    u

    v

    u

    k

    k

    k

    i

    i

    i

    e

    q

    =

    =

    ki

    ki

    ki

    ik

    ik

    ik

    e

    M

    T

    N

    M

    T

    N

    R

    R

    R

    R

    R

    R

    6

    5

    4

    3

    2

    1

    ~

    ~

    ~

    ~

    ~

    ~

    ~R

    x~

    y~

    i

    k

    1~ 1 =q

    e

    State 1~1 =q

    N

    N

    x~

    y~

    i

    k

    1~ 3 =q e

    State 1~3 =q

    N

    N

    x~

    y~

    i

    k

    1~ 2 =q

    e

    State 1~2 =q

    y~

    x~

    i

    k

    11~ ,

    ~qR

    22~ ,

    ~qR

    44~ ,

    ~qR

    33~ ,

    ~qR

    e

    55~ ,

    ~qR

    66~ ,

    ~qR

    l

  • Reactions number 1 and 4 created by the action of displacements number 1 and 4 are obtained in the same way as in the case of the plane truss element. Note, that these displacements do not yield any other reactions. Also, these reactions are zero under the action of all the other displacements. Hence, in the classical plane beam element the action of axial forces and displacements is fully decoupled from the combined action of shear and bending. The reactions due to the transverse displacements and cross-section rotations can be found using the slope-deflection formulae from the classical stiffness method. This yields:

    ( ) 65232225

    633~2~6~4~6

    ~~3~~2

    232

    2~q

    l

    EIq

    l

    EIq

    l

    EIq

    l

    EI

    l

    qqqq

    l

    EI

    l

    EIMR ikkiik ++=

    +=+==

    ( ) 65232225

    366~4~6~2~6

    ~~3~~2

    232

    2~q

    l

    EIq

    l

    EIq

    l

    EIq

    l

    EI

    l

    qqqq

    l

    EI

    l

    EIMR ikikki ++=

    +=+==

    ( ) 6253322325

    63222~6~12~6~12

    ~~2~~

    62

    6~q

    l

    EIq

    l

    EIq

    l

    EIq

    l

    EI

    l

    qqqq

    l

    EI

    l

    EITR ikkiik ++=

    +=+==

    ( )

    62533223

    2563225

    ~6~12~6~12

    ~~2~~

    62

    6~

    ql

    EIq

    l

    EIq

    l

    EIq

    l

    EI

    l

    qqqq

    l

    EI

    l

    EITR ikkiki

    +=

    =

    +=+==

    Summarising these calculations yield the following stiffness matrix for the plane beam element

    =

    l

    EI

    l

    EI

    l

    EI

    l

    EIl

    EI

    l

    EI

    l

    EI

    l

    EIl

    EA

    l

    EAl

    EI

    l

    EI

    l

    EI

    l

    EIl

    EI

    l

    EI

    l

    EI

    l

    EIl

    EA

    l

    EA

    e

    460

    260

    6120

    6120

    0000

    260

    460

    6120

    6120

    0000

    ~

    22

    2323

    22

    2323

    K

    It is worth to note, that the element stiffness matrices are symmetric. This is the consequence of the reactions reciprocity theorem (Rayleighs theorem) stating that

    jiij kk =

    The physical interpretation of the entries in this matrix is unchanged, i.e. for instance the second

    column collects the values of reactions created by the action of the displacement 1~2 =q .

    After the derivation of these two examples of the element stiffness matrices let us now deal with the loading. Generally, the loading is split into actions directly at the nodes and the actions along the elements. The former one will be considered at the later stages. The latter one must be transferred to the nodes in the form of appropriate reactions. They are assembled into a vector of reactions due to the span loading

    [ ] 6,...,2,1 ~~1600

    ==

    iR ieR

    This vector enters the equation of element equilibrium as an additional term

  • eeee 0

    ~~~~ RqKR +=

    For instance, in the case of the uniformly distributed load on the beam we have

    =

    =

    12

    2

    012

    2

    0

    ~

    ~

    ~

    ~

    ~

    ~

    ~

    2

    2

    06

    05

    04

    03

    02

    01

    0

    ql

    ql

    ql

    ql

    R

    R

    R

    R

    R

    R

    eR

    and in the case of the non-uniform heating t with warmer upper side and the uniform heating t0

    =

    =

    h

    tEI

    tEAh

    tEI

    tEA

    R

    R

    R

    R

    R

    R

    t

    t

    t

    t

    e

    0

    0

    ~

    ~

    ~

    ~

    ~

    ~

    ~

    0

    0

    06

    05

    04

    03

    02

    01

    0R

    In the latter case note that the distribution of the bending mome

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