DISPLACEMENT METHOD OF ANALYSIS: SLOPE ... ...

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  • 1

    ! General Case ! Stiffness Coefficients ! Stiffness Coefficients Derivation ! Fixed-End Moments ! Pin-Supported End Span ! Typical Problems ! Analysis of Beams ! Analysis of Frames: No Sidesway ! Analysis of Frames: Sidesway

    DISPLACEMENT METHOD OF ANALYSIS: SLOPE DEFLECTION EQUATIONS

  • 2

    Slope ñ Deflection Equations

    settlement = ∆j

    Pi j kw Cj

    Mij Mji wP

    θj

    θi

    ψ

    i j

  • 3

    Degrees of Freedom

    L

    θΑ

    A B

    M

    1 DOF: θΑ

    P θΑ

    θΒΒΒΒ A B C 2 DOF: θΑ , θΒΒΒΒ

  • 4

    L

    A B

    1

    Stiffness

    kBAkAA

    L EIkAA

    4 =

    L EIkBA

    2 =

  • 5

    L

    A B1

    kBBkAB

    L EIkBB

    4 =

    L EIkAB

    2 =

  • 6

    Fixed-End Forces Fixed-End Moments: Loads

    P

    L/2 L/2

    L

    w

    L

    8 PL

    8 PL

    2 P

    2 P

    12

    2wL 12

    2wL

    2 wL

    2 wL

  • 7

    General Case

    settlement = ∆j

    Pi j kw Cj

    Mij Mji wP

    θj

    θi

    ψ

    i j

  • 8

    wP

    settlement = ∆j

    (MFij)∆ (MFji)∆

    (MFij)Load (M F

    ji)Load

    +

    Mij Mji

    θi

    θj

    +

    i jwPMij Mji

    settlement = ∆jθj

    θi

    ψ

    =+ ji L EI

    L EI θθ 24

    ji L EI

    L EI θθ 42 +=

    ,)()()2()4( Loadij F

    ij F

    jiij MML EI

    L EIM +++= ∆θθ Loadji

    F ji

    F jiji MML

    EI L EIM )()()4()2( +++= ∆θθ

    L

  • 9

    Mji Mjk

    Pi j kw Cj

    Mji Mjk

    Cj

    j

    Equilibrium Equations

    0:0 =+−−=Σ+ jjkjij CMMM

  • 10

    +

    1

    1

    i jMij Mji

    θi

    θj

    L EIkii

    4 =

    L EIk ji

    2 =

    L EIkij

    2 =

    L EIk jj

    4 =

    iθ×

    jθ×

    Stiffness Coefficients

    L

  • 11

    [ ]  

      

     =

    jjji

    ijii

    kk kk

    k

    Stiffness Matrix

    )()2()4( ijFjiij ML EI

    L EIM ++= θθ

    )()4()2( jiFjiji ML EI

    L EIM ++= θθ

      

      

     +

      

      

      

     =

      

     F

    ji

    F ij

    j

    iI

    ji

    ij

    M M

    LEILEI LEILEI

    M M

    θ θ

    )/4()/2( )/2()/4(

    Matrix Formulation

  • 12

    [D] = [K]-1([Q] - [FEM])

    Displacement matrix

    Stiffness matrix

    Force matrix wP(MFij)Load (MFji)Load

    +

    +

    i jwPMij Mji

    θj

    θi

    ψ ∆j

    Mij Mji

    θi

    θj

    (MFij)∆ (MFji)∆

    Fixed-end moment matrix

    ][]][[][ FEMKM += θ

    ]][[])[]([ θKFEMM =−

    ][][][][ 1 FEMMK −= −θ

    L

  • 13

    L

    Real beam

    Conjugate beam

    Stiffness Coefficients Derivation: Fixed-End Support

    MjMi

    L/3

    θi

    L MM ji +

    EI M j

    EI Mi

    θι

    EI LM j

    2

    EI LMi

    2

    )1(2

    0) 3

    2)( 2

    () 3

    )( 2

    (:0'

    −−−=

    =+−=Σ+

    ji

    ji i

    MM

    L EI

    LML EI

    LMM

    )2(0) 2

    () 2

    (:0 −−−=+−=Σ↑+ EI

    LM EI

    LMF jiiy θ ij

    ii

    L EIM

    L EIM

    andFrom

    θ

    θ

    )2(

    )4(

    );2()1(

    =

    =

    L MM ji +

    i j

  • 14

    Conjugate beam

    L

    Real beam

    Stiffness Coefficients Derivation: Pinned-End Support

    Mi θi

    θj

    L M i

    L Mi

    EI M i

    EI LMi

    2

    3 2L

    θi θj

    0) 3

    2)( 2

    (:0' =−=Σ+ LL EI

    LMM iij θ

    i j

    0) 2

    () 3

    (:0 =+−=Σ↑+ jiiy EI LM

    EI LMF θ

    L EIM

    EI LM

    i i

    i 3)

    3 (1 =→==θ

    ) 6

    ( EI

    LM i j

    − =θ)3

    ( EI

    LM i i =θ

  • 15

    Fixed end moment : Point Load

    Real beam

    8 ,0

    16 2

    22 :0

    2 PLM EI

    PL EI

    ML EI

    MLFy ==+−−=Σ↑+

    P

    M

    M EI M

    Conjugate beam A

    EI M

    B

    L

    P

    A B

    EI M

    EI ML 2

    EI M

    EI ML 2

    EI PL

    16

    2

    EI PL 4 EI

    PL 16

    2

  • 16

    L

    P

    841616 PLPLPLPL

    =+ −

    + −

    8 PL

    8 PL

    2 P

    2 PP/2

    P/2

    -PL/8 -PL/8

    PL/8

    -PL/16 -PL/8

    -

    PL/4 +

    -PL/16-PL/8 -

  • 17

    Uniform load

    L

    w

    A B

    w

    M

    M

    Real beam Conjugate beam

    A

    EI M

    EI M

    B

    12 ,0

    24 2

    22 :0

    23 wLM EI

    wL EI

    ML EI

    MLFy ==+−−=Σ↑+

    EI wL 8

    2

    EI wL

    24

    3

    EI wL

    24

    3

    EI M

    EI ML 2

    EI M

    EI ML 2

  • 18

    Settlements

    M

    M

    L

    MjMi = Mj

    L MM ji +L

    MM ji +

    Real beam

    2

    6 L EIM ∆=

    Conjugate beam

    EI M

    A B

    EI M

    ,0) 3

    2)( 2

    () 3

    )( 2

    (:0 =+−∆−=Σ+ L EI

    MLL EI

    MLM B

    EI M

    EI ML 2

    EI MEI

    ML 2

  • 19

    wP

    A B

    A B+ θA θB

    (FEM)AB (FEM)BA

    Pin-Supported End Span: Simple Case

    BA L EI

    L EI θθ 24 + BA L

    EI L EI θθ 42 +

    )1()()/2()/4(0 −−−++== ABBAAB FEMLEILEIM θθ

    )2()()/4()/2(0 −−−++== BABABA FEMLEILEIM θθ

    BABABBA FEMFEMLEIM )()(2)/6(2:)1()2(2 −+=− θ

    2 )()()/3( BABABBA

    FEMFEMLEIM −+= θ

    wP

    A B

    L

  • 20

    A B

    wP

    A B

    A B θA θB

    (MF AB)load (M F

    BA)load

    Pin-Supported End Span: With End Couple and Settlement

    BA L EI

    L EI θθ 24 + BA L

    EI L EI θθ 42 +

    L

    (MF AB)∆ (M F

    BA) ∆

    )1()()(24 −−−+++== ∆ F ABload

    F ABBAAAB MML

    EI L EIMM θθ

    )2()()(42 −−−+++= ∆ F BAload

    F BABABA MML

    EI L EIM θθ

    2 )(

    2 1])(

    2 1)[(3:

    2 )1()2(2lim AFBAload

    F ABload

    F BABBAA

    MMMM L EIMbyinateE ++−+=− ∆θθ

    MA wP

    A B

  • 21

    Fixed-End Moments Fixed-End Moments: Loads

    P

    L/2 L/2

    P

    L/2 L/2

    8 PL

    8 PL

    16 3)]

    8 ()[

    2 1(

    8 PLPLPL

    =−−+

    12

    2wL 12

    2wL

    8 )]

    12 ()[

    2 1(

    12

    222 wLwLwL =−−+

  • 22

    Typical Problem

    0

    0

    0

    0

    A C

    B

    P1 P2

    L1 L2

    w CB

    P

    8 PL

    8 PL w12

    2wL

    12

    2wL

    8 024 11

    11

    LP L EI

    L EIM BAAB +++= θθ

    8 042 11

    11

    LP L EI

    L EIM BABA −++= θθ

    128 024

    2 222

    22

    wLLP L EI

    L EIM CBBC ++++= θθ

    128 042

    2 222

    22

    wLLP L EI

    L EIM CBCB −

    − +++= θθ

    L L

  • 23

    MBA MBC

    A C

    B

    P1 P2

    L1 L2

    w CB

    B

    CB

    8 042 11

    11

    LP L EI

    L EIM BABA −++= θθ

    128 024

    2 222

    22

    wLLP L EI

    L EIM CBBC ++++= θθ

    BBCBABB forSolveMMCM θ→=−−=Σ+ 0:0

  • 24

    A C

    B

    P1 P2

    L1 L2

    w CB

    Substitute θB in MAB, MBA, MBC, MCB

    MAB MBA

    MBC MCB

    0

    0

    0

    0

    8 024 11

    11

    LP L EI

    L EIM BAAB +++= θθ

    8 042 11

    11

    LP L EI

    L EIM BABA −++= θθ

    128 024

    2 222

    22

    wLLP L EI

    L EIM CBBC ++++= θθ

    128 042

    2 222

    22

    wLLP L EI

    L EIM CBCB −

    − +++= θθ

  • 25

    A B P1

    MAB MBA

    L1

    A CB

    P1 P2

    L1 L2

    w CB

    ByR CyAy ByL

    Ay Cy

    MAB MBA

    MBC MCB

    By = ByL + ByR

    B C P2

    MBC MCB

    L2

  • 26

    Example of Beams

  • 27

    10 kN 6 kN/m

    A C

    B 6 m4 m4 m

    Example 1

    Draw the quantitative shear , bending moment diagrams and qualitative deflected curve for the beam shown. EI is constant.

  • 28

    PPL 8 wwL

    2