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  • 1

    CHAP 4 FINITE ELEMENT ANALYSIS OF BEAMS AND FRAMES

    2

    INTRODUCTION

    We learned Direct Stiffness Method in Chapter 2 Limited to simple elements such as 1D bars

    we will learn Energy Method to build beam finite element Structure is in equilibrium when the potential energy is minimum

    Potential energy: Sum of strain energy and potential of applied loads

    Interpolation scheme:

    ( ) ( ) { }v x x N q

    Beam deflection

    Interpolationfunction

    NodalDOF

    Potential of applied loads

    Strain energy

    U V

  • 3

    BEAM THEORY Euler-Bernoulli Beam Theory

    can carry the transverse load slope can change along the span (x-axis) Cross-section is symmetric w.r.t. xy-plane The y-axis passes through the centroid Loads are applied in xy-plane (plane of loading)

    LF

    x

    y

    F

    Plane of loading

    y

    z

    Neutral axis

    A

    4

    BEAM THEORY cont. Euler-Bernoulli Beam Theory cont.

    Plane sections normal to the beam axis remain plane and normal to the axis after deformation (no shear stress)

    Transverse deflection (deflection curve) is function of x only: v(x) Displacement in x-dir is function of x and y: u(x, y)

    y

    y(dv/dx)

    = dv/dx

    v(x)L

    F

    x

    y

    Neutral axis

    0( , ) ( )dvu x y u x ydx

    2

    02xx

    duu d vyx dx dx

    dvdx

  • 5

    BEAM THEORY cont. Euler-Bernoulli Beam Theory cont.

    Strain along the beam axis: Strain xx varies linearly w.r.t. y; Strain yy = 0 Curvature: Can assume plane stress in z-dir basically uniaxial status

    Axial force resultant and bending moment

    20

    2xxduu d vy

    x dx dx

    0 0 /du dx

    2 2/d v dx

    2

    0 2xx xxd vE E Eydx

    2

    0 2

    22

    0 2

    xxA A A

    xxA A A

    d vP dA E dA E ydAdx

    d vM y dA E ydA E y dAdx

    Moment of inertia I(x)

    02

    2

    P EAd vM EIdx

    EA: axial rigidityEI: flexural rigidity

    6

    BEAM THEORY cont. Beam constitutive relation

    We assume P = 0 (We will consider non-zero P in the frame element) Moment-curvature relation:

    Sign convention

    Positive directions for applied loads

    2

    2d vM EIdx

    Moment and curvature is linearly dependent

    +P +P+M+M

    +Vy

    +Vy

    yx

    p(x)

    F1 F2 F3

    C1 C2 C3

    y

    x

  • 7

    GOVERNING EQUATIONS Beam equilibrium equations

    Combining three equations together: Fourth-order differential equation

    yy

    dVV dx

    dxyyy

    dMM dx

    dxyVM

    dx

    p

    0 ( ) 0yy y ydV

    f p x dx V dx Vdx

    ( )ydV p x

    dx

    02 y

    dM dxM M dx pdx V dxdx

    4

    4 ( )d vEI p xdx

    ydMVdx

    8

    STRESS AND STRAIN Bending stress

    This is only non-zero stress component for Euler-Bernoulli beam

    Transverse shear strain

    Euler beam predicts zero shear strain (approximation) Traditional beam theory says the transverse shear stress is However, this shear stress is in general small compared to

    the bending stress

    2

    2xxd vEydx

    2

    2d vM EIdx

    ( )( , )xxM x yx yI

    0xyu v v vy x x x

    0

    ( , ) ( ) dvu x y u x ydx

    Bending stress

    xyVQIb

  • 9

    POTENTIAL ENERGY Potential energy Strain energy

    Strain energy density

    Strain energy per unit length

    Strain energy

    U V

    2 22 22 2

    0 2 21 1 1 1( )2 2 2 2xx xx xx

    d v d vU E E y Eydx dx

    2 22 22 2

    0 2 21 1( ) ( , , )2 2L A A A

    d v d vU x U x y z dA Ey dA E y dAdx dx

    Moment of inertia

    22

    21( )2L

    d vU x EIdx

    22

    20 0

    1( )2

    L L

    Ld vU U x dx EI dxdx

    10

    POTENTIAL ENERGY cont. Potential energy of applied loads

    Potential energy

    Potential energy is a function of v(x) and slope The beam is in equilibrium when has its minimum value

    01 1

    ( )( ) ( ) ( )CF NNL

    ii i i

    i i

    dv xV p x v x dx Fv x Cdx

    22

    20 01 1

    ( )1 ( ) ( ) ( )2

    CF NNL Li

    i i ii i

    dv xd vU V EI dx p x v x dx Fv x Cdx dx

    vv*

    0v

  • 11

    RAYLEIGH-RITZ METHOD1. Assume a deflection shape

    Unknown coefficients ci and known function fi(x) Deflection curve v(x) must satisfy displacement boundary conditions

    2. Obtain potential energy as function of coefficients

    3. Apply the principle of minimum potential energy to determine the coefficients

    1 1 2 2( ) ( ) ( )..... ( )n nv x c f x c f x c f x

    1 2( , ,... )nc c c U V

    12

    EXAMPLE SIMPLY SUPPORTED BEAM Assumed deflection curve

    Strain energy

    Potential energy of applied loads (no reaction forces)

    Potential energy

    PMPE:

    p0

    E,I,L( ) sin xv x C

    L

    22 2 4

    2 30

    12 4

    L d v C EIU EI dxdx L

    00

    0 0

    2( ) ( ) sinL L p LxV p x v x dx p C dx C

    L

    42 0

    3

    24

    p LEIU V C CL

    440 0

    3 5

    2 402

    p L p Ld EI C CdC L EI

  • 13

    EXAMPLE SIMPLY SUPPORTED BEAM cont. Exact vs. approximate solutions

    Approximate bending moment and shear force

    Exact solutions

    4 40 0

    approx exact76.5 76.8p L p LC CEI EI

    22 20

    2 2 3

    3 30

    3 3 2

    4( ) sin sin

    4( ) cos cosy

    p Ld v x xM x EI EICdx L L L

    p Ld v x xV x EI EICdx L L L

    33 40 0 0

    20 0

    00

    1( )24 12 24

    ( )2 2

    ( )2y

    p L p L pv x x x xEIp L pM x x x

    p LV x p x

    14

    EXAMPLE SIMPLY SUPPORTED BEAM cont. Deflection

    Bendingmoment

    Shear force

    0.0

    0.2

    0.4

    0.6

    0.8

    1.0

    0 0.2 0.4 0.6 0.8 1x

    v(x)

    /v_m

    ax

    v-exact

    v-approx.

    -0.14

    -0.12

    -0.10

    -0.08

    -0.06

    -0.04

    -0.02

    0.000 0.2 0.4 0.6 0.8 1

    x

    Bend

    ing

    Mom

    ent M

    (x)

    M_exactM_approx

    -0.6

    -0.4

    -0.2

    0.0

    0.2

    0.4

    0.6

    0 0.2 0.4 0.6 0.8 1x

    Shea

    r For

    ce V

    (x)

    V_exactV_approx

    Error increases

  • 15

    EXAMPLE CANTILEVERED BEAM

    Assumed deflection

    Need to satisfy BC

    Strain energy

    Potential of loads

    F

    C

    p0

    E,I,L2 31 2( )v x a bx c x c x

    (0) 0, (0) / 0v dv dx 2 31 2( )v x c x c x

    21 20

    2 62

    LEIU c c x dx

    1 2 00

    3 42 3 20 0

    1 2

    , ( ) ( ) ( )

    2 33 4

    L dvV c c p v x dx Fv L C Ldx

    p L p Lc FL CL c FL CL

    16

    EXAMPLE CANTILEVERED BEAM cont. Derivatives of U:

    PMPE:

    Solve for c1 and c2:

    Deflection curve:

    Exact solution:

    21 2 1 2

    1 0

    2 31 2 1 2

    2 0

    2 2 6 4 6

    6 2 6 6 12

    L

    L

    U EI c c x dx EI Lc L cc

    U EI c c x xdx EI L c L cc

    1

    2

    0

    0

    c

    c

    32 20

    1 2

    42 3 3 20

    1 2

    4 6 23

    6 12 34

    p LEI Lc L c FL CL

    p LEI L c L c FL CL

    3 31 223.75 10 , 8.417 10c c

    3 2 3( ) 10 23.75 8.417v x x x 2 3 41( ) 5400 800 30024v x x x xEI

  • 17

    EXAMPLE CANTILEVERED BEAM cont. Deflection

    Bendingmoment

    Shear force

    Error increases

    0.0

    0.0

    0.0

    0.0

    0.0

    0 0.2 0.4 0.6 0.8 1x

    v(x)

    /v_m

    ax

    v-exact

    v-approx.

    -100.00

    0.00

    100.00

    200.00

    300.00

    400.00

    500.00

    0 0.2 0.4 0.6 0.8 1x

    Bend

    ing

    Mom

    ent M

    (x) M_exact

    M_approx

    200.0

    300.0

    400.0

    500.0

    600.0

    0 0.2 0.4 0.6 0.8 1x

    Shea

    r For

    ce V

    (x)

    V_exactV_approx

    18

    FINITE ELEMENT INTERPOLATION Rayleigh-Ritz method approximate solution in the entire beam

    Difficult to find approx solution that satisfies displacement BC

    Finite element approximates solution in an element Make it easy to satisfy displacement BC using interpolation technique

    Beam element Divide the beam using a set of elements Elements are connected to other elements at nodes Concentrated forces and couples can only be applied at nodes Consider two-node bean element Positive directions for forces and couples Constant or linearly

    distributed loadF1 F2

    C2C1

    p(x)

    x

  • 19

    FINITE ELEMENT INTERPOLATION cont. Nodal DOF of beam element

    Each node has deflection v and slope

    Positive directions of DOFs Vector of nodal DOFs

    Scaling parameter s Length L of the beam is scaled to 1 using scaling parameter s

    Will write deflection curve v(s) in terms of s

    v1 v2

    2

    1

    Lx1s = 0

    x2s = 1

    x

    1 1 2 2{ } { }Tv v q

    1 1, ,

    1,

    x xs ds dxL L

    dsdx Ldsdx L

    20

    FINITE ELEMENT INTERPOLATION cont. Deflection interpolation

    Interpolate the deflection v(s) in terms of four nodal DOFs Use cubic function: Relation to the slope:

    Apply four conditions:

    Express four coefficients in terms of nodal DOFs

    2 30 1 2 3( )v s a a s a s a s

    21 2 3

    1 ( 2 3 )dv dv ds a a s a sdx ds dx L

    1 1 2 2(0) (1)(0) (1)dv dvv v v vdx dx

    1 0

    1 1

    2 0 1 2 3

    2 1 2 3

    (0)1(0)

    (1)1(1) ( 2 3 )

    v v adv adx L

    v v a a a adv a a adx L

    0 1

    1 1

    2 1 1 2 2

    3 1 1 2 2

    3 2 32 2

    a va La v L v La v L v L

  • 21

    FINITE ELEMENT INTERPOLATION cont. Deflection interpolation cont.

    Shape functions

    Hermite polynomials Interpolation property

    2 3 2 3 2 3 2 31 1 2 2( ) (1 3 2 ) ( 2 ) (3 2 ) ( )v s s s v L s s s s s v L s s

    2 31

    2 32

    2 33

    2 34

    ( ) 1 3 2

    ( ) ( 2 )

    ( ) 3 2

    ( ) ( )

    N s s sN s L s s sN s s s

    N s L s s

    1

    11 2 3 4

    2

    2

    ( ) [ ( ) ( ) ( ) ( )]

    v

    v s N s N s N s N sv

    ! !! ! " #! !! !$ %

    ( ) { }v s N q

    -0.2

    0.0

    0.2

    0.4

    0.6

    0.8

    1.0

    0.0 0.2 0.4 0.6 0.8 1.0

    N1 N3

    N2/L

    N4/L

    22

    FINITE ELEMENT INTERPOLATION cont. Properties of interpolation

    Deflection is a cubic polynomial (discuss accuracy and limitation) Interpolation is valid within an element, not outside of the element Adjacent elements have continuous deflection and slope

    Approximation of curvature Curvature is second derivative and related to strain and stress

    B is linear function of s and, thus, the strain and stress Alternative expression:

    If the given problem is linearly varying curvature, the approximation is accurate; if higher-order variation of curvature, then it is approximate

    12 2

    12 2 2 2

    2

    2

    1 1 [ 6 12 , ( 4 6 ), 6 12 , ( 2 6 )]

    vd v d v s L s s L s

    vdx L ds L

    ! !! ! " #! !! !$ %

    2

    2 24 11 4

    1 { }d vdx L

    B q B: strain-displacement vector

    2

    2 24 11 4

    1 { }T Td vdx L

    Bq

  • 23

    FINITE ELEMENT INTERPOLATION cont. Approximation of bending moment and shear force

    Stress is proportional to M(s); M(s) is linear; stress is linear, too Maximum stress always occurs at the node Bending moment and shear force are not continuous between adjacent

    elements

    2

    2 2( ) { }d v EIM s EIdx L

    B q

    3

    3 3 [ 12 6 12 6 ]{ }ydM d v EIV EI L Ldx dx L

    q

    Linear

    Constant

    24

    EXAMPLE INTERPOLATION Cantilevered beam Given nodal DOFs

    Deflection and slope at x = 0.5L Parameter s = 0.5 at x = 0.5L Shape functions: Deflection at s = 0.5:

    Slope at s = 0.5:

    { } {0, 0, 0.1, 0.2}T qL

    v1v2

    21

    1 1 1 11 2 3 42 2 2 2

    1 1( ) , ( ) , ( ) , ( )2 8 2 8

    L LN N N N

    1 1 1 1 11 1 2 1 3 2 4 22 2 2 2 2

    2 22 2

    ( ) ( ) ( ) ( ) ( )1 10 0 0.0252 8 2 8 2 8

    v N v N N v NL L v Lv

    31 2 41 1 2 2

    2 2 2 21 1 2 2

    1 1

    1 1( 6 6 ) 1 4 3 (6 6 ) 2 3 0.1

    dNdv dv dN dN dNv vdx L ds L ds ds ds ds

    v s s s s v s s s sL L

  • 25

    EXAMPLE A beam finite element with length L

    Calculate v(s)

    Bending moment

    3 2

    1 1 2 20, 0, ,3 2L Lv vEI EI

    1 1 2 1 3 2 4 2( ) ( ) ( ) ( ) ( )v s N s v N s N s v N s

    2 3 2 32 2( ) (3 2 ) ( )v s s s v L s s

    & '2 2

    2 22 2 2 2

    3 2

    2

    ( ) (6 12 ) ( 2 6 )

    (6 12 ) ( 2 6 )3 2

    (1 ) ( )

    d v EI d v EIM s EI s v L sdx L ds L

    EI L Ls L sL EI EIL s L x

    ( )

    L F

    Bending moment cause by unit force at the tip

    26

    FINITE ELEMENT EQUATION FOR BEAM Finite element equation using PMPE

    A beam is divided by NEL elements with constant sections

    Strain energy Sum of each elements strain energy

    Strain energy of element (e)

    p(x)

    F1 F2 F3

    C1 C2 C3

    y

    x

    F4 F5

    C4 C51 2 3 4 5

    11x

    1 22 1x x

    221x

    2 32 1x x

    331x

    3 42 1x x

    441x

    42x

    2

    10 1 1( ) ( )

    eT

    e

    NEL NELL x eL Lx

    e eU U x dx U x dx U

    2

    1

    2 22 21

    2 3 20

    1 12 2

    e

    e

    xe

    x

    d v EI d vU EI dx dsdx L ds

  • 27

    FE EQUATION FOR BEAM cont. Strain energy cont.

    Approximate curvature in terms of nodal DOFs

    Approximate element strain energy in terms of nodal DOFs

    Stiffness matrix of a beam element

    22 2 2

    2 2 21 4 4 14 1 1 4

    { } { }T Te ed v d v d v

    ds ds ds

    q B B q

    1( )3 0

    1 1{ } { } { } [ ]{ }2 2

    eTe e e e ee T TEIU ds

    L( ) q B B q q k q

    & '1

    3 0

    6 12( 4 6 )

    [ ] 6 12 ( 4 6 ) 6 12 ( 2 6 )6 12( 2 6 )

    e

    sL sEI s L s s L s ds

    sLL s

    ( )

    k

    28

    FE EQUATION FOR BEAM cont. Stiffness matrix of a beam element

    Strain energy cont.

    Assembly

    2 2

    3

    2 2

    12 6 12 66 4 6 2

    [ ]12 6 12 6

    6 2 6 4

    e

    L LL L L LEI

    L LLL L L L

    ( )

    kSymmetric, positive semi-definiteProportional to EIInversely proportional to L

    ( )

    1 1

    1 { } [ ]{ }2

    NEL NELe e ee T

    e eU U

    q k q

    1{ } [ ]{ }2

    Ts s sU Q K Q

  • 29

    EXAMPLE ASSEMBLY Two elements Global DOFs

    F3F2

    y

    x

    1 2 3

    2EI EI

    2L L 1 1 2 2 3 3{ } { }T

    s v v v Q

    1 1 2 2

    12 2

    1 13

    22 2

    2

    3 3 3 33 4 3 2

    [ ]3 3 3 3

    3 2 3 4

    v v

    L L vL L L LEI

    L LL vL L L L

    ( )

    k

    2 2 3 3

    22 2

    2 23

    32 2

    3

    12 6 12 66 4 6 2

    [ ]12 6 12 6

    6 2 6 4

    v v

    L L vL L L LEI

    L LL vL L L L

    ( )

    k

    2 2

    2 2 23

    2 2

    3 3 3 3 0 03 4 3 2 0 0

    3 3 15 3 12 6[ ]

    3 2 3 8 6 20 0 12 6 12 60 0 6 2 6 4

    s

    L LL L L L

    L L LEIL L L L L LL

    L LL L L L

    ( )

    K

    30

    FE EQUATION FOR BEAM cont. Potential energy of applied loads

    Concentrated forces and couples

    Distributed load (Work-equivalent nodal forces)

    1

    ND

    i i i ii

    V Fv C

    1

    1

    1 1 2 2...... { } { }T

    ND s s

    ND

    FC

    V v v F

    C

    ! !! !! ! " # ! !! !! !$ %

    Q F

    2

    1

    ( )

    1 1( ) ( )

    e

    e

    NEL NELx e

    xe e

    V p x v x dx V

    2

    1

    1( ) ( )

    0

    ( ) ( ) ( ) ( )e

    e

    xe e

    xV p x v x dx L p s v s ds

    1

    ( ) ( )1 1 1 2 2 3 2 4

    0

    1 1 1 1( ) ( ) ( ) ( )

    1 1 1 2 2 3 2 40 0 0 0

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

    ( )

    ( ) ( ) ( ) ( )

    e e

    e e e e

    e e e e

    V L p s v N N v N N ds

    v L p s N ds L p s N ds v L p s N ds L p s N ds

    v F C v F C

  • 31

    EXAMPLE WORK-EQUIVALENT NODAL FORCES Uniformly distributed load

    pL/2 pL/2

    pL2/12 pL2/12

    p

    Equivalent

    1 1 2 31 10 0

    ( ) (1 3 2 )2pLF pL N s ds pL s s ds

    21 12 2 31 20 0

    ( ) ( 2 )12pLC pL N s ds pL s s s ds

    1 1 2 32 30 0

    ( ) (3 2 )2pLF pL N s ds pL s s ds

    21 12 2 32 40 0

    ( ) ( )12pLC pL N s ds pL s s ds

    2 2

    { }2 12 2 12

    T pL pL pL pL " #$ %

    F

    32

    FE EQUATION FOR BEAM cont. Finite element equation for beam

    One beam element has four variables

    When there is no distributed load, p = 0

    Applying boundary conditions is identical to truss element

    At each DOF, either displacement (v or ) or force (F or C) must be known, not both

    Use standard procedure for assembly, BC, and solution

    1 12 2 2

    1 13

    2 22 2 2

    2 2

    12 6 12 6 / 26 4 6 2 /1212 6 12 6 / 2

    6 2 6 4 /12

    v FL L pLCL L L L pLEI

    v FL L pLLCL L L L pL

    ( ) ! ! ! ! ! ! ! ! ! ! ! ! " # " # " # ! ! ! ! ! ! ! ! ! ! ! ! $ %$ % $ %

  • 33

    PRINCIPLE OF MINIMUM POTENTIAL ENERGY Potential energy (quadratic form)

    PMPE Potential energy has its minimum when

    Applying BC The same procedure with truss elements (striking-the-rows and

    striking-he-columns)

    Solve for unknown nodal DOFs {Q}

    1{ } [ ]{ } { } { }2

    T Ts s s s sU V Q K Q Q F

    [ ]{ } { }s s sK Q F

    [ ]{ } { }K Q F

    [Ks] is symmetric & PSD

    [K] is symmetric & PD

    34

    BENDING MOMENT & SHEAR FORCE Bending moment

    Linearly varying along the beam span

    Shear force

    Constant When true moment is not linear and true shear is not constant, many

    elements should be used to approximate it

    Bending stress Shear stress for rectangular section

    2 2

    2 2 2 2( ) { }d v EI d v EIM s EIdx L ds L

    B q

    13 3

    13 3 3 3

    2

    2

    ( ) [ 12 6 12 6 ]y

    vdM d v EI d v EIV s EI L L

    vdx dx L ds L

    ! !! ! " #! !! !$ %

    xMyI

    2

    2

    1.5 4( ) 1yxyV yybh h

  • 35

    EXAMPLE CLAMPED-CLAMPED BEAM Determine deflection &

    slope at x = 0.5, 1.0, 1.5 m Element stiffness matrices

    F2 = 240 N

    y

    x

    1 21 m

    31 m

    1 1 2 2

    1

    1(1)

    2

    2

    12 6 12 66 4 6 2

    [ ] 100012 6 12 66 2 6 4

    v v

    v

    v

    ( )

    k

    2 2 3 3

    2

    2(2)

    3

    3

    12 6 12 66 4 6 2

    [ ] 100012 6 12 66 2 6 4

    v v

    v

    v

    ( )

    k

    11

    11

    2

    2

    33

    33

    12 6 12 6 0 06 4 6 2 0 0

    24012 6 24 0 12 61000

    06 2 0 8 6 20 0 12 6 12 60 0 6 2 6 4

    FvC

    v

    FvC

    ( )! ! ! ! ! ! ! ! ! ! ! !

    " # " # ! ! ! ! ! ! ! ! ! ! ! ! $ %$ %

    36

    EXAMPLE CLAMPED-CLAMPED BEAM cont. Applying BC

    At x = 0.5 s = 0.5 and use element 1

    At x = 1.0 either s = 1 (element 1) or s = 0 (element 2)

    2

    2

    24 0 2401000

    0 8 0v

    ( )

    " # " # $ %$ %

    2

    2

    0.010.0

    v

    12

    1 1 1 1 1 11 1 1 2 2 3 2 4 32 2 2 2 2 2

    3122 (1)

    ( ) ( ) ( ) ( ) ( ) 0.01 ( ) 0.005m1( ) 0.015rad

    s

    v v N N v N N NdNv

    L ds

    2 3 3

    32(1)

    1

    (1) (1) 0.01 (1) 0.01m1(1) 0.0rad

    s

    v v N NdNv

    L ds

    2 1 1

    12(2)

    0

    (0) (0) 0.01 (0) 0.01m1(0) 0.0rad

    s

    v v N NdNv

    L ds

    Will this solution be accurate or approximate?

  • 37

    EXAMPLE CANTILEVERED BEAM One beam element No assembly required Element stiffness

    Work-equivalent nodal forces

    C = 50 N-m

    p0 = 120 N/m

    EI = 1000 N-m2

    1

    1

    2

    2

    12 6 12 66 4 6 2

    [ ] 100012 6 12 66 2 6 4

    s

    v

    v

    ( )

    K

    2 31

    2 3110 02 30

    22 3

    2

    1/ 2 601 3 2/12 10( 2 )

    1/ 2 603 2/12 10( )

    e

    e

    e

    e

    F s sC Ls s s L

    p L ds p LF s sC Ls s L

    ! !! ! ! ! ! ! ! ! ! ! ! ! ! ! " # " # " # " #

    ! ! ! ! ! ! ! !! ! ! ! ! ! ! ! $ % $ %$ % $ %

    L = 1m

    38

    EXAMPLE CANTILEVERED BEAM cont. FE matrix equation

    Applying BC

    Deflection curve: Exact solution:

    1 1

    1 1

    2

    2

    12 6 12 6 606 4 6 2 10

    100012 6 12 6 606 2 6 4 10 50

    v FC

    v

    ( ) ! ! ! ! ! ! ! ! " # " # ! ! ! ! ! ! ! ! $ %$ %

    2

    2

    12 6 601000

    6 4 60v

    ( ) " # " # $ %$ %

    2

    2

    0.010.03

    v

    m

    rad

    33 4( ) 0.01 ( ) 0.03 ( ) 0.01v s N s N s s 4 3 2( ) 0.005( 4 )v x x x x

  • 39

    EXAMPLE CANTILEVERED BEAM cont. Support reaction (From assembled matrix equation)

    Bending moment

    Shear force

    2 2 1

    2 2 1

    1000 12 6 60

    1000 6 2 10

    v F

    v C

    1

    1

    120 N10 N m

    FC

    & '

    2

    1 1 2 22

    ( ) { }

    ( 6 12 ) ( 4 6 ) (6 12 ) ( 2 6 )

    1000[ 0.01(6 12 ) 0.03( 2 6 )]60 N m

    EIM sLEI s v L s s v L sL

    s ss

    B q

    & '1 1 2 23 12 6 12 61000[ 12 ( 0.01) 6( 0.03)]

    60 N

    yEIV v L v LL

    40

    EXAMPLE CANTILEVERED BEAM cont. Comparisons

    -0.010

    -0.008

    -0.006

    -0.004

    -0.002

    0.000

    0 0.2 0.4 0.6 0.8 1x

    v

    FEMExact

    -0.030

    -0.025

    -0.020

    -0.015

    -0.010

    -0.005

    0.000

    0 0.2 0.4 0.6 0.8 1x

    FEMExact

    -60

    -50

    -40

    -30

    -20

    -10

    0

    10

    0 0.2 0.4 0.6 0.8 1x

    M

    FEMExact

    -120

    -100

    -80

    -60

    -40

    -20

    0

    0 0.2 0.4 0.6 0.8 1x

    Vy

    FEMExact

    Deflection Slope

    Bending moment Shear force

  • 41

    PLANE FRAME ELEMENT Beam

    Vertical deflection and slope. No axial deformation

    Frame structure Can carry axial force, transverse shear force, and bending moment

    (Beam + Truss)

    Assumption Axial and bending effects

    are uncoupled Reasonable when deformation

    is small

    3 DOFs per node

    Need coordinate transfor-mation like plane truss

    F

    p

    1

    2 3

    4

    1u2u

    1v 2v

    11

    22

    1u

    2u

    1v

    2v

    11

    22

    1u

    2u

    1v

    2v

    11

    22 1

    2 3

    { , , }i i iu v

    42

    PLANE FRAME ELEMENT cont. Element-fixed local coordinates Local DOFs Local forces Transformation between local and global coord.

    1

    2*

    x

    y

    Local coordinates

    Global coordinates

    xy

    1u

    2u

    1v

    2v

    11

    22

    x y{ , , }u v { , , }x yf f c

    1 1

    1 1

    1 1

    2 2

    2 2

    2 2

    cos sin 0 0 0 0sin cos 0 0 0 00 0 1 0 0 00 0 0 cos sin 00 0 0 sin cos 00 0 0 0 0 1

    x x

    y y

    x x

    y y

    f ff fc cf ff fc c

    * ** *

    * ** *

    ( )! ! ! ! ! ! ! ! ! ! ! !

    " # " # ! ! ! ! ! ! ! ! ! ! ! !

    $ % $ %

    { } [ ]{ }f T f

    { } [ ]{ }q T q

  • 43

    PLANE FRAME ELEMENT cont. Axial deformation (in local coord.)

    Beam bending

    Basically, it is equivalent to overlapping a beam with a bar A frame element has 6 DOFs

    11

    22

    1 11 1

    x

    x

    fuEAfuL

    ( )" # " # $ % $ %

    1 12 2

    1 13

    2 22 2

    2 2

    12 6 12 66 4 6 212 6 12 6

    6 2 6 4

    y

    y

    vL L fL L L L cEI

    vL LL fL L L L c

    ( )! !! ! ! ! ! ! " # " # ! ! ! ! ! ! ! ! $ % $ %

    44

    PLANE FRAME ELEMENT cont. Element matrix equation (local coord.)

    Element matrix equation (global coord.)

    Same procedure for assembly and applying BC

    1 1 11

    2 2 2 2 112 2

    2 2 2 2 11

    1 1 22

    2 2 2 2 222 2

    2 2 2 2 22

    0 0 0 00 12 6 0 12 60 6 4 0 6 2

    0 0 0 00 12 6 0 12 60 6 2 0 6 4

    x

    y

    x

    y

    a a fua La a La fvLa L a La L a c

    a a fua La a La fv

    La L a La L a c

    ( )! ! ! ! ! ! ! ! ! ! ! !

    " # " # ! ! ! ! ! ! ! ! ! ! ! ! $ %$ %

    1

    2 3

    EAaLEIaL

    [ ]{ } { }k fq

    [ ][ ]{ } [ ]{ }k T q T f [ ] [ ][ ]{ } { }T T k T q f [ ]{ } { }k q f

    [ ] [ ] [ ][ ]Tk T k T

  • 45

    PLANE FRAME ELEMENT cont. Calculation of element forces

    Element forces can only be calculated in the local coordinate Extract element DOFs {q} from the global DOFs {Qs} Transform the element DOFs to the local coordinate Then, use 1D bar and beam formulas for element forces

    Axial force

    Bending moment

    Shear force

    Other method:

    { } [ ]{ }q T q

    2 1AEP u uL

    2( ) { }EIM sL

    B q

    + ,3( ) [ 12 6 12 6 ]yEIV s L LL

    q

    1 12 2

    1 13

    2 22 2

    2 2

    12 6 12 66 4 6 212 6 12 6

    6 2 6 4

    y

    y

    V vL LM L L L LEIV vL LLM L L L L

    ( )! ! ! ! ! ! ! ! " # " # ! ! ! ! ! ! ! ! $ % $ %