displacement method of analysis: slope deflection...

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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 MjiwP

    θj

    θi

    ψ

    i j

  • 3

    Degrees of Freedom

    L

    θΑ

    A B

    M

    1 DOF: θΑ

    PθΑ

    θΒΒΒΒA B C 2 DOF: θΑ , θΒΒΒΒ

  • 4

    L

    A B

    1

    Stiffness

    kBAkAA

    LEIkAA

    4=

    LEIkBA

    2=

  • 5

    L

    A B1

    kBBkAB

    LEIkBB

    4=

    LEIkAB

    2=

  • 6

    Fixed-End ForcesFixed-End Moments: Loads

    P

    L/2 L/2

    L

    w

    L

    8PL

    8PL

    2P

    2P

    12

    2wL12

    2wL

    2wL

    2wL

  • 7

    General Case

    settlement = ∆j

    Pi j kw Cj

    Mij MjiwP

    θj

    θi

    ψ

    i j

  • 8

    wP

    settlement = ∆j

    (MFij)∆ (MFji)∆

    (MFij)Load (MF

    ji)Load

    +

    Mij Mji

    θi

    θj

    +

    i jwPMijMji

    settlement = ∆jθj

    θi

    ψ

    =+ ji LEI

    LEI θθ 24

    ji LEI

    LEI θθ 42 +=

    ,)()()2()4( LoadijF

    ijF

    jiij MMLEI

    LEIM +++= ∆θθ Loadji

    Fji

    Fjiji MML

    EILEIM )()()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

    LEIkii

    4=

    LEIk ji

    2=

    LEIkij

    2=

    LEIk jj

    4=

    iθ×

    jθ×

    Stiffness Coefficients

    L

  • 11

    [ ]

    =

    jjji

    ijii

    kkkk

    k

    Stiffness Matrix

    )()2()4( ijFjiij MLEI

    LEIM ++= θθ

    )()4()2( jiFjiji MLEI

    LEIM ++= θθ

    +

    =

    F

    ji

    Fij

    j

    iI

    ji

    ij

    MM

    LEILEILEILEI

    MM

    θθ

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

    Matrix Formulation

  • 12

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

    Displacementmatrix

    Stiffness matrix

    Force matrixwP(MFij)Load (MFji)Load

    +

    +

    i jwPMijMji

    θj

    θi

    ψ ∆j

    Mij Mji

    θi

    θj

    (MFij)∆ (MFji)∆

    Fixed-end momentmatrix

    ][]][[][ FEMKM += θ

    ]][[])[]([ θKFEMM =−

    ][][][][ 1 FEMMK −= −θ

    L

  • 13

    L

    Real beam

    Conjugate beam

    Stiffness Coefficients Derivation: Fixed-End Support

    MjMi

    L/3

    θi

    LMM ji +

    EIM j

    EIMi

    θι

    EILM j

    2

    EILMi

    2

    )1(2

    0)3

    2)(2

    ()3

    )(2

    (:0'

    −−−=

    =+−=Σ+

    ji

    jii

    MM

    LEI

    LMLEI

    LMM

    )2(0)2

    ()2

    (:0 −−−=+−=Σ↑+EI

    LMEI

    LMF jiiy θ ij

    ii

    LEIM

    LEIM

    andFrom

    θ

    θ

    )2(

    )4(

    );2()1(

    =

    =

    LMM ji +

    i j

  • 14

    Conjugate beam

    L

    Real beam

    Stiffness Coefficients Derivation: Pinned-End Support

    Mi θi

    θj

    LM i

    LMi

    EIM i

    EILMi

    2

    32L

    θi θj

    0)3

    2)(2

    (:0' =−=Σ+ LLEI

    LMM iij θ

    i j

    0)2

    ()3

    (:0 =+−=Σ↑+ jiiy EILM

    EILMF θ

    LEIM

    EILM

    ii

    i3)

    3(1 =→==θ

    )6

    (EI

    LM ij

    −=θ)3

    (EI

    LM ii =θ

  • 15

    Fixed end moment : Point Load

    Real beam

    8,0

    162

    22:0

    2 PLMEI

    PLEI

    MLEI

    MLFy ==+−−=Σ↑+

    P

    M

    M EIM

    Conjugate beamA

    EIM

    B

    L

    P

    A B

    EIM

    EIML2

    EIM

    EIML2

    EIPL

    16

    2

    EIPL4 EI

    PL16

    2

  • 16

    L

    P

    841616PLPLPLPL

    =+−

    +−

    8PL

    8PL

    2P

    2PP/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

    EIM

    EIM

    B

    12,0

    242

    22:0

    23 wLMEI

    wLEI

    MLEI

    MLFy ==+−−=Σ↑+

    EIwL8

    2

    EIwL

    24

    3

    EIwL

    24

    3

    EIM

    EIML2

    EIM

    EIML2

  • 18

    Settlements

    M

    M

    L

    MjMi = Mj

    LMM ji +L

    MM ji +

    Real beam

    2

    6LEIM ∆=

    Conjugate beam

    EIM

    A B

    EIM

    ,0)3

    2)(2

    ()3

    )(2

    (:0 =+−∆−=Σ+ LEI

    MLLEI

    MLM B

    EIM

    EIML2

    EIMEI

    ML2

  • 19

    wP

    A B

    A B+θA θB

    (FEM)AB (FEM)BA

    Pin-Supported End Span: Simple Case

    BA LEI

    LEI θθ 24 + BA L

    EILEI 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

    AB

    L

  • 20

    AB

    wP

    A B

    A BθA θB

    (MF AB)load (MF

    BA)load

    Pin-Supported End Span: With End Couple and Settlement

    BA LEI

    LEI θθ 24 + BA L

    EILEI θθ 42 +

    L

    (MF AB)∆ (MF

    BA) ∆

    )1()()(24 −−−+++== ∆FABload

    FABBAAAB MML

    EILEIMM θθ

    )2()()(42 −−−+++= ∆FBAload

    FBABABA MML

    EILEIM θθ

    2)(

    21])(

    21)[(3:

    2)1()2(2lim AFBAload

    FABload

    FBABBAA

    MMMMLEIMbyinateE ++−+=− ∆θθ

    MAwP

    AB

  • 21

    Fixed-End MomentsFixed-End Moments: Loads

    P

    L/2 L/2

    P

    L/2 L/2

    8PL

    8PL

    163)]

    8()[

    21(

    8PLPLPL

    =−−+

    12

    2wL12

    2wL

    8)]

    12()[

    21(

    12

    222 wLwLwL=−−+

  • 22

    Typical Problem

    0

    0

    0

    0

    A C

    B

    P1 P2

    L1 L2

    wCB

    P

    8PL

    8PL w12

    2wL

    12

    2wL

    8024 11

    11

    LPLEI

    LEIM BAAB +++= θθ

    8042 11

    11

    LPLEI

    LEIM BABA −++= θθ

    128024

    2222

    22

    wLLPLEI

    LEIM CBBC ++++= θθ

    128042

    2222

    22

    wLLPLEI

    LEIM CBCB −

    −+++= θθ

    L L

  • 23

    MBA MBC

    A C

    B

    P1 P2

    L1 L2

    wCB

    B

    CB

    8042 11

    11

    LPLEI

    LEIM BABA −++= θθ

    128024

    2222

    22

    wLLPLEI

    LEIM CBBC ++++= θθ

    BBCBABB forSolveMMCM θ→=−−=Σ+ 0:0

  • 24

    A C

    B

    P1 P2

    L1 L2

    wCB

    Substitute θB in MAB, MBA, MBC, MCB

    MABMBA

    MBCMCB

    0

    0

    0

    0

    8024 11

    11

    LPLEI

    LEIM BAAB +++= θθ

    8042 11

    11

    LPLEI

    LEIM BABA −++= θθ

    128024

    2222

    22

    wLLPLEI

    LEIM CBBC ++++= θθ

    128042

    2222

    22

    wLLPLEI

    LEIM CBCB −

    −+++= θθ

  • 25

    A BP1

    MABMBA

    L1

    A CB

    P1 P2

    L1 L2

    wCB

    ByR CyAy ByL

    Ay Cy

    MABMBA

    MBCMCB

    By = ByL + ByR

    B CP2

    MBCMCB

    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 qualitativedeflected curve for the beam shown. EI is constant.

  • 28

    PPL8 wwL

    2

    30FEM

    PL8

    wL220

    MBA MBC

    B

    Substitute θB in the moment equations:

    MBC = 8.8 kNïm

    MCB = -10 kNïm

    MAB = 10.6 kNïm,

    MBA = - 8.8 kNïm,

    [M] = [K][Q] + [FEM]

    10 kN 6 kN/m

    A C

    B 6 m4 m4 m

    0

    0

    0

    0

    8)8)(10(

    82

    84

    ++= BAABEIEIM θθ

    8)8)(10(

    84

    82

    −+= BABAEIEIM θθ

    30)6)(6(

    62

    64 2

    ++= CBBCEIEIM θθ

    20)6)(6(

    64

    62 2

    −+= CBCBEIEIM θθ

    0:0 =−−=Σ+ BCBAB MMM

    EI

    EIEI

    B

    B

    4.2

    030

    )6)(6(10)6

    48

    4(2

    =

    =+−+

    θ

    θ

  • 29

    10 kN 6 kN/m

    A C

    B 6 m4 m4 m

    MBC = 8.8 kNïm

    MCB= -10 kNïm

    MAB = 10.6 kNïm,

    MBA = - 8.8 kNïm,

    = 5.23 kN = 4.78 kN = 5.8 kN = 12.2 kN

    10 kNïm8.8 kNïm

    10.6 kNïm8.8 kNïm

    10 kN

    10.6 kNïm

    8.8 kNïm

    A B

    ByLAy

    2 m

    6 kN/m8.8 kNïm

    10 kNïmB

    CyByR

    18 kN

  • 30

    10 kN 6 kN/m

    A C

    B 6 m4 m4 m

    10 kNïm10.6 kNïm

    5.23 kN 12.2 kN

    4.78 + 5.8 = 10.58 kN

    V (kN)x (m)

    5.23

    - 4.78

    5.8

    -12.2

    +-

    +

    -

    M (kNïm) x (m)

    -10.6

    10.3

    -8.8 -10- -

    +-

    EIB4.2

    Deflected shape x (m)

  • 31

    10 kN 6 kN/m

    A C

    B 6 m4 m4 m

    Example 2

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

  • 32

    PPL8 wwL

    2

    30FEM

    PL8

    wL220

    [M] = [K][Q] + [FEM]

    10 kN 6 kN/m

    A C

    B 6 m4 m4 m

    )1(8

    )8)(10(8

    28

    4−−−++= BAAB

    EIEIM θθ

    )2(8

    )8)(10(8

    48

    2−−−−+= BABA

    EIEIM θθ

    )3(30

    )6)(6(6

    26

    4 2−−−++= CBBC

    EIEIM θθ

    )4(20

    )6)(6(6

    46

    2 2−−−−+= CBCB

    EIEIM θθ

    10

    10

    0

    0

    0

    308

    62:)1()2(2 −=− BBAEIM θ

    )5(158

    3−−−−= BBA

    EIM θ

  • 33

    MBA MBC

    B

    )5(158

    3−−−−= BBA

    EIM θ

    )3(30

    )6)(6(6

    4 2−−−+= BBC

    EIM θ

    0:0 =−−=Σ+ BCBAB MMM

    EI

    EIEI

    B

    B

    488.7

    )6(030

    )6)(6(15)6

    48

    3(2

    =

    −−−=+−+

    θ

    θ

    EI

    EIEIinSubstitute

    A

    BAB

    74.23

    108

    28

    40:)1(

    −=

    −+=

    θ

    θθθ

    Substitute θA and θB in (5), (3) and (4):

    MBC = 12.19 kNïm

    MCB = - 8.30 kNïm

    MBA = - 12.19 kNïm)4(

    20)6)(6(

    62 2

    −−−−= BCBEIM θ

  • 34

    10 kN 6 kN/m

    A C

    B 6 m4 m4 m

    MBA = - 12.19 kNïm, MBC = 12.19 kNïm, MCB = - 8.30 kNïm

    = 3.48 kN = 6.52 kN = 6.65 kN = 11.35 kN

    12.19 kNïm

    12.19 kNïm8.30 kNïm

    10 kN

    12.19 kNïm

    A B

    ByLAy

    2 m

    6 kN/m12.19 kNïm

    8.30 kNïmC

    CyByR

    18 kN

    B

  • 35

    Deflected shape x (m)EIB49.7

    10 kN 6 kN/m

    A C

    B 6 m4 m4 m

    V (kN)x (m)

    3.48

    - 6.52

    6.65

    -11.35

    M (kNïm) x (m)

    14

    -12.2-8.3

    11.35 kN3.48 kN

    6.52 + 6.65 = 13.17 kN

    EIA74.23−

  • 36

    10 kN 4 kN/m

    A C

    B6 m4 m4 m

    2EI 3EI

    Example 3

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

  • 37

    10 kN 4 kN/m

    A C

    B6 m4 m4 m

    2EI 3EI(4)(62)/12 (4)(62)/12 (10)(8)/8(10)(8)/8

    15

    12

    )1(8

    )8)(10(8

    )2(28

    )2(4−−−++= BAAB

    EIEIM θθ

    )2(8

    )8)(10(8

    )2(48

    )2(2−−−−+= BABA

    EIEIM θθ

    0 10

    10

    )2(8

    )8)(10)(2/3(8

    )2(3:2

    )1()2(2 aEIM BBA −−−−=− θ

    )3(12

    )6)(4(6

    )3(4 2−−−+= BBC

    EIM θ

  • 38

    10 kN 4 kN/m

    A C

    B6 m4 m4 m

    2EI 3EI(4)(62)/12 (4)(62)/12(3/2)(10)(8)/8

    EIEIMM

    B

    BBCBA

    /091.13151275.2:0

    ==+−==−−

    θθ

    15

    12

    )2(8

    )8)(10)(2/3(8

    )2(3 aEIM BBA −−−−= θ

    )3(12

    )6)(4(6

    )3(4 2−−−+= BBC

    EIM θ

    mkNEIM

    mkNEI

    EIM

    mkNEI

    EIM

    BCB

    BC

    BA

    •−=−=

    •=−=

    •−=−=

    91.10126

    )3(2

    18.1412)091.1(6

    )3(4

    18.1415)091.1(8

    )2(3

    θ

  • 39

    10 kN 4 kN/m

    A C

    B6 m4 m4 m

    2EI 3EI

    MBA = - 14.18 kNïm, MBC = 14.18 kNïm, MCB = -10.91 kNïm

    14.18

    140.18 kNïm

    10 kN

    A B

    ByLAy = 6.73 kN= 3.23 kN

    14.18 kNïm

    10.91 kNïm

    4 kN/m

    C

    CyByR

    24 kN

    14.18 10.91

    = 11.46 kN= 12.55 kN

  • 40

    10 kN 4 kN/m

    A C

    B6 m4 m4 m

    2EI 3EI

    11.46 kN3.23 kN

    10.91 kNïm

    V (kN)x (m)

    3.23

    -6.73

    12.55

    -11.46

    +-

    +-

    2.86

    M (kNïm) x (m)

    12.91

    -14.18

    5.53

    -10.91

    +

    -+

    -

    6.77 + 12.55 = 19.32 kN

    θB = 1.091/EIDeflected shape x (m)

  • 41

    10 kN 4 kN/m

    A C

    B6 m4 m4 m

    2EI 3EI

    Example 4

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

    12 kNïm

  • 42

    10 kN 4 kN/m

    A C

    B6 m4 m4 m

    2EI 3EI

    12 kNïm

    wL2/12 = 12 wL2/12 = 121.5PL/8 = 15

    MBA

    MBCB

    12 kNïmMBA

    MBC

    EIB273.3

    −=θ

    EIA21.7

    −=θ

    )1(158

    )2(3−−−−= BBA

    EIM θ

    )2(126

    )3(4−−−+= BBC

    EIM θ

    )3(126

    )3(2−−−−= BCB

    EIM θ

    8)8)(10(

    8)3(2

    8)2(4

    ++= BAABEIEIM θθ

    0 -3.273/EI

    012:int =−−− BCBA MMBJo

    012)122()1575.0( =−+−−− BEIEI θ

    mkNEI

    EIM BA •−=−−= 45.1715)273.3(75.0

    mkNEI

    EIM BC •=+−= 45.512)273.3(2

    mkNEI

    EIM CB •−=−−= 27.1512)273.3(

  • 43

    10 kN

    A B

    4 kN/m

    C

    24 kN

    17.45 kNïm5.45 kNïm

    15.27 kNïm

    2.82 kN 13.64 kN10.36 kN7.18 kN

    12 kNïm10 kN 4 kN/m

    A C

    B13.64 kN2.82 kN

    15.27 kNïm

    17.54 kN

    mkNEI

    EIM BA •−=−−= 45.1715)273.3(75.0

    mkNEI

    EIM BC •=+−= 45.512)273.3(2

    mkNEI

    EIM CB •−=−−= 27.1512)273.3(

  • 44

    10 kN 4 kN/m

    A C

    B6 m4 m4 m

    2EI 3EI

    12 kNïm

    13.64 kN2.82 kN

    15.27 kNïm

    17.54 kN

    V (kN)x (m)3.41 m+

    -+

    -

    2.82

    -7.18

    10.36

    -13.64

    M (kNïm) x (m)+

    - -+

    11.28

    -17.45

    -5.45-15.27

    7.98

    Deflected shape x (m)

    EIB273.3

    =θEIA

    21.7−=θ

    EIB273.3

    −=θ

    EIA21.7

    −=θ

  • 45

    Example 5

    Draw the quantitative shear, bending moment diagrams, and qualitativedeflected curve for the beam shown. Support B settles 10 mm, and EI isconstant. Take E = 200 GPa, I = 200x106 mm4.

    12 kNïm 10 kN 6 kN/m

    A CB

    6 m4 m4 m

    2EI 3EI10 mm

  • 46

    12 kNïm 10 kN 6 kN/m

    A CB

    6 m4 m4 m

    2EI 3EI10 mm

    [FEM]∆

    A

    B

    2

    6LEI∆

    2

    6LEI∆

    BC

    2

    6LEI∆

    2

    6LEI∆

    P w

    [FEM]load

    8PL

    8PL

    30

    2wL30

    2wL

    )1(8

    )8)(10(8

    )01.0)(2(68

    )2(28

    )2(42 −−−+++=

    EIEIEIM BAAB θθ

    )2(8

    )8)(10(8

    )01.0)(2(68

    )2(48

    )2(22 −−−−++=

    EIEIEIM BABA θθ

    )3(30

    )6)(6(6

    )01.0)(3(66

    )3(26

    )3(4 22 −−−+−+=

    EIEIEIM CBBC θθ

    )4(30

    )6)(6(6

    )01.0)(3(66

    )3(46

    )3(2 22 −−−−−+=

    EIEIEIM CBCB θθ

    -12

    0

    0

  • 47

    12 kNïm 10 kN 6 kN/m

    A CB

    6 m4 m4 m

    2EI 3EI10 mm

    Substitute EI = (200x106 kPa)(200x10-6 m4) = 200x200 kNï m2 :

    )1(8

    )8)(10(8

    )01.0)(2(68

    )2(28

    )2(42 −−−+++=

    EIEIEIM BAAB θθ

    )2(8

    )8)(10(8

    )01.0)(2(68

    )2(48

    )2(22 −−−−++=

    EIEIEIM BABA θθ

    )1(10758

    )2(28

    )2(4−−−+++= BAAB

    EIEIM θθ

    )2(10758

    )2(48

    )2(2−−−−++= BABA

    EIEIM θθ

    )2(2/12)2/10(10)2/75(758

    )2(3:2

    )1()2(2 aEIM BBA −−−−−−−+=− θ

    16.5

  • 48

    + ΣMB = 0: - MBA - MBC = 0 (3/4 + 2)EIθB + 16.5 - 192.8 = 0

    θB = 64.109/ EI

    Substitute θB in (1): θA = -129.06/EI

    Substitute θA and θB in (5), (3), (4):

    MBC = -64.58 kNïm

    MCB = -146.69 kNïm

    MBA = 64.58 kNïm,

    MBA MBC

    B

    12 kNïm 10 kN 6 kN/m

    A CB

    6 m4 m4 m

    2EI 3EI10 mm

    MBC = (4/6)(3EI)θB - 192.8

    MBA = (3/4)(2EI)θB + 16.5

  • 49

    12 kNïm 10 kN 6 kN/m

    A CB

    6 m4 m4 m

    64.58 kNïm 64.58 kNïm

    = -1.57 kN= 11.57 kN

    146.69 kNïm

    = 47.21 kN= -29.21 kN

    10 kN

    A B

    ByLAy

    12 kNïm64.58 kNïm

    2 m

    6 kN/m

    C

    CyByR

    18 kN

    B

    64.58 kNïm

    146.69 kNïm

    MBC = -64.58 kNïm

    MCB = -146.69 kNïm

    MBA = 64.58 kNïm,

  • 50

    12 kNïm 10 kN 6 kN/m

    A C

    B6 m4 m4 m

    2EI 3EI

    V (kN)x (m)

    11.57 1.57

    -29.21-47.21

    -+

    M (kNïm) x (m)

    1258.29 64.58

    -146.69

    +

    -

    47.21 kN

    146.69 kNïm

    11.57 kN

    1.57 + 29.21 = 30.78 kN

    10 mmθA = -129.06/EI

    θB = 64.109/ EIθA = -129.06/EI

    Deflected shape x (m)

    θB = 64.109/ EI

  • 51

    Example 6

    For the beam shown, support A settles 10 mm downward, use the slope-deflectionmethod to(a)Determine all the slopes at supports(b)Determine all the reactions at supports(c)Draw its quantitative shear, bending moment diagrams, and qualitativedeflected shape. (3 points)Take E= 200 GPa, I = 50(106) mm4.

    6 kN/m

    B A C

    3 m 3 m2EI 1.5EI

    12 kNïm

    10 mm

  • 52

    6 kN/m

    B A C

    3 m 3 m2EI 1.5EI

    12 kNïm

    )1(3

    )2(4−−−= CCB

    EIM θ

    )2(1005.43

    )5.1(23

    )5.1(4−−−+++= ACCA

    EIEIM θθ

    )2(2

    122

    1002

    )5.4(33

    )5.1(3:2

    )2()2(2 aEIM CCA −−−+++=− θ

    )3(1005.43

    )5.1(43

    )5.1(2−−−+−+= ACAC

    EIEIM θθ12

    0.01 m

    C

    AmkN •=

    ××

    1003

    )01.0)(502005.1(62

    mkN •100MF∆

    10 mm

    6 kN/m

    A C

    5.412

    )3(6 2= 4.5

    MFw

  • 53

    6 kN/m

    B A C

    3 m 3 m2EI 1.5EI

    12 kNïm

    )1(3

    )2(4−−−= CCB

    EIM θ

    )2(2

    122

    1002

    )5.4(33

    )5.1(3 aEIM CCA −−−+++= θ

    10 mm

    MCB MCA

    C

    0=+ CACB MM

    02

    122

    1002

    )5.4(33

    )5.48(=+++

    +C

    EI θ

    radEIC

    0015.006.15 −=−=θ

    )3(1005.43

    )5.1(4)06.15(3

    )5.1(212 −−−+−+−= AEI

    EIEI θSubstitute θC in eq.(3)

    radEIA

    0034.022.34 −=−=θ

    ï Equilibrium equation:

  • 54

    6 kN/m

    B A C

    3 m 3 m2EI 1.5EI

    12 kNïm

    10 mm

    radEIC

    0015.006.15 −=−=θ radEIA

    0034.022.34 −=−=θ

    mkNEI

    EIEIM CBC •−=−

    == 08.20)06.15(3

    )2(23

    )2(2 θ

    mkNEI

    EIEIM CCB •−=−

    == 16.40)06.15(3

    )2(43

    )2(4 θ

    kN08.203

    08.2016.40=

    +kN08.20

    B C40.16 kNïm20.08 kNïm

    6 kN/m

    AC

    12 kNïm

    40.16 kNïm

    18 kN

    8.39 kN26.39 kN

  • 55

    6 kN/m

    B A C

    3 m 3 m2EI 1.5EI

    12 kNïm

    10 mm

    6 kN/m

    AC

    12 kNïm

    40.16 kNïm8.39 kN26.39 kN

    B C40.16 kNïm20.08 kNïm

    20.08 kN20.08 kN

    V (kN)

    x (m)

    26.398.39

    -20.08

    +-

    M (kNïm)

    x (m)20.08

    -40.16

    12

    radC 0015.0−=θradA 0034.0−=θ

    Deflected shapex (m)

    radC 0015.0=θ

    radA 0034.0=θ

  • 56

    Example 7

    For the beam shown, support A settles 10 mm downward, use theslope-deflection method to(a)Determine all the slopes at supports(b)Determine all the reactions at supports(c)Draw its quantitative shear, bending moment diagrams, and qualitativedeflected shape.Take E= 200 GPa, I = 50(106) mm4.

    6 kN/m

    B A C

    3 m 3 m2EI 1.5EI

    12 kNïm

    10 mm

  • 57

    6 kN/m

    B A C

    3 m 3 m2EI 1.5EI

    12 kNïm

    10 mm

    5.412

    )3(6 2=

    6 kN/m

    AC4.5

    mkN •=

    ××

    1003

    )01.0)(502005.1(620.01 m

    C

    A

    100

    34 CEI∆

    ∆C

    CB

    34

    3)2(6

    2CC EIEI ∆=∆

    )2(3

    43

    )2(4−−−∆−= CCCB

    EIEIM θ

    )3(1005.43

    )5.1(23

    )5.1(4−−++∆++= CACCA EI

    EIEIM θθ

    )4(1005.43

    )5.1(43

    )5.1(2−−+−∆++= CACAC EI

    EIEIM θθ12

    )1(3

    43

    )2(2−−−∆−= CCBC

    EIEIM θ

    )3(2

    122

    1002

    )5.4(323

    )5.1(3:2

    )4()3(2 aEIEIM CCCA −−−+++∆+=− θ

    CC EIEI ∆=∆23

    )5.1(6∆C

    C

    A

    CEI∆

  • 58

    6 kN/m

    B A C

    3 m 3 m2EI 1.5EI

    12 kNïm

    10 mm

    ï Equilibrium equation:

    )3

    ()( CBBCCByMMC +−=

    6 kN/m

    AC

    12 kNïm

    MCA

    18 kN

    AyB C

    MCBMBC

    By3

    393

    )5.1(1812)( +=++= CACACAyMMC

    C

    MCB MCA

    (Cy)CA(Cy)CB

    *)1(0:0 −−−=+=Σ CACBC MMM

    *)2(0)()(:0 −−−=+=Σ CAyCByy CCC

    )5(15.628333.0167.4*)1( −−−−=∆− CC EIEIinSubstitute θ

    )6(75.101167.35.2*)2( −−−−=∆+− CC EIEIinSubstitute θ

    mmEIradEIandFrom CC 227.5/27.5200255.0/51.25)6()5( −=−=∆−=−=θ

  • 59

    6 kN/m

    B A C

    3 m 3 m2EI 1.5EI

    12 kNïm

    10 mm

    ï Solve equation

    radEIC

    00255.051.25 −=−=θ

    mmEIC

    227.527.52 −=−=∆

    Substitute θC and ∆C in (4)

    radEIA

    000286.086.2 −=−=θ

    Substitute θC and ∆C in (1), (2) and (3a)

    mkNM BC •= 68.35

    mkNMCB •= 67.1

    mkNMCA •−= 67.1

    6 kN/m

    B A C

    3 m 3 m2EI 1.5EI

    12 kNïm

    35.68 kNïm

    kN

    Ay

    55.56

    68.3512)5.4(18

    =

    −−=

    kNBy

    45.12

    55.518

    =

    −=

  • 60

    10 mm

    6 kN/m

    B A C

    3 m 3 m2EI 1.5EI

    12 kNïm

    35.68 kNïm

    12.45 kN 5.55 kN

    Deflected shape

    x (m)

    radC 00255.0−=θ

    mmC 227.5−=∆

    radA 000286.0−=θ

    radC 00255.0=θ radA 000286.0=θ

    mmC 227.5=∆

    M (kNïm)

    x (m)1214.57

    1.67

    -35.68

    -+

    V (kN)

    x (m)0.925 m12.45

    -5.55

    +

  • 61

    Example of Frame: No Sidesway

  • 62

    Example 6

    For the frame shown, use the slope-deflection method to (a) Determine the end moments of each member and reactions at supports(b) Draw the quantitative bending moment diagram, and also draw the qualitative deflected shape of the entire frame.

    10 kN

    C

    12 kN/m

    A

    B

    6 m

    40 kN

    3 m

    3 m

    1 m

    2EI

    3EI

  • 63

    10 kN

    C

    12 kN/m

    A

    B

    6 m

    40 kN

    3 m

    3 m

    1 m

    2EI

    3EIPL/8 = 30

    PL/8 = 30

    ï Equilibrium equations

    10

    MBC

    MBA

    )3(18366

    )2(3−−−++= BBC

    EIM θ

    *)1(010 −−−=−− BCBA MM

    05430310 =−+− BEIθ

    EIEIB667.4

    )3(14 −

    =−

    =θ)1(30

    6)3(2

    −−−+= BABEIM θ

    mkNMmkNM

    mkNM

    BC

    BA

    AB

    •=•−=

    •=

    33.4933.39

    33.25

    36/2 = 18

    (wL2/12 ) =3636

    ï Slope-Deflection Equations

    )2(306

    )3(4−−−−= BBA

    EIM θ

    Substitute (2) and (3) in (1*)

    )3()1(667.4 toinEI

    Substitute B−

  • 64

    10 kN

    C

    12 kN/m

    A

    B

    6 m

    40 kN

    3 m

    3 m

    1 m

    2EI

    3EI39.33

    25.33

    49.33

    A

    B

    40 kN

    39.33

    25.33

    C

    12 kN/m

    B49.33

    17.67 kN

    27.78 kN

    Bending moment diagram

    -25.33

    27.7

    -39.3

    Deflected curve

    -49.33

    20.58

    10

    θB = -4.667/EIθB

    θB

    MAB = 25.33 kNïm

    MBA = -39.33 kNïm

    MBC = 49.33 kNïm

  • 65

    A

    B

    C

    D

    E

    25 kN

    5 m

    5 kN/m

    60(106) mm4

    240(106) mm4 180(106)

    120(106) mm4

    3 m 4 m3 m

    Example 7

    Draw the quantitative shear, bending moment diagrams and qualitativedeflected curve for the frame shown. E = 200 GPa.

  • 66

    A

    B

    C

    D

    E

    25 kN

    5 m

    5 kN/m

    60(106) mm4

    240(106) mm4 180(106)

    120(106) mm4

    3 m 4 m3 m0

    BAABEIEIM θθ

    5)2(2

    5)2(4

    +=0

    BABAEIEIM θθ

    5)2(4

    5)2(2

    +=

    75.186

    )4(26

    )4(4++= CBBC

    EIEIM θθ

    75.186

    )4(46

    )4(2−+= CBCB

    EIEIM θθ

    CCDEIM θ5

    )(3=

    104

    )3(3+= CCE

    EIM θ

    0=+ BCBA MM

    )1(75.18)68()

    616

    58( −−−−=++ CB EIEI θθ

    0=++ CECDCB MMM

    )2(75.8)49

    53

    616()

    68( −−−=+++ CB EIEI θθ

    EIEIandFrom CB

    86.229.5:)2()1( =−= θθ

    PL/8 = 18.75 18.75

    6.667 (wL2/12 ) = 6.667+ 3.333

  • 67

    MAB = −4.23 kNïm

    MBA = −8.46 kNïm

    MBC = 8.46 kNïm

    MCB = −18.18 kNïm

    MCD = 1.72 kNïm

    MCE = 16.44 kNïm

    Substitute θB = -1.11/EI, θc = -20.59/EI below

    0

    0BAAB

    EIEIM θθ5

    )2(25

    )2(4+=

    BABAEIEIM θθ

    5)2(4

    5)2(2

    +=

    75.186

    )4(26

    )4(4++= CBBC

    EIEIM θθ

    75.186

    )4(46

    )4(2−+= CBCB

    EIEIM θθ

    CCDEIM θ5

    )(3=

    104

    )3(3+= CCE

    EIM θ

  • 68

    MAB = -4.23 kNïm, MBA = -8.46 kNïm, MBC = 8.46 kNïm, MCB = -18.18 kNïm,MCD = 1.72 kNïm, MCE = 16.44 kNïm

    A

    B

    5 m

    C E

    20 kN

    A

    B

    5 m

    4.23 kNïm

    2.54 kN

    (8.46 + 4.23)/5 = 2.54 kN

    8.46 kNïm

    C

    25 kN

    B 3 m 3 m2.54 kN 2.54 kN

    8.46 kNïm(25(3)+8.46-18.18)/6 = 10.88 kN

    14.12 kN18.18 kNïm

    16.44 kNïm

    (20(2)+16.44)/4= 14.11 kN

    5.89 kN

    0.34 kN

    28.23 kN

    14.12+14.11=28.23 kN1.72 kNïm

    (1.72)/5 = 0.34 kN

    10.88 kN

    10.88 kN

    2.54-0.34=2.2 kN

    2.2 kN

  • 69

    Shear diagram

    -2.54

    -2.54

    10.88

    -14.12

    14.11

    -5.89

    0.34

    +

    -

    -

    +-

    +

    2.82 m

    1.18 m

    Deflected curve

    1.18 m

    0.78 m2.33 m1.29 m

    1.67m

    1.29 m

    0.78 m

    Moment diagram

    1.18 m

    3.46

    24.18

    1.72

    -18.18 -16.44

    4.23

    -8.46-8.46

    2.33 m

    +

    -

    +

    - -

    +

    θB = −5.29/EI

    θC = 2.86/EI1.67m

  • 70

    Example of Frames: Sidesway

  • 71

    A

    B C

    D

    3m

    10 kN

    3 m

    1 m

    Example 8

    Determine the moments at each joint of the frame and draw the quantitativebending moment diagrams and qualitative deflected curve . The joints at A andD are fixed and joint C is assumed pin-connected. EI is constant for each member

  • 72

    A

    B C

    D

    3m

    10 kN

    3 m

    1 m

    MABMDC

    Ax Dx

    Ay Dy

    ï Boundary Conditions

    θA = θD = 0

    ï Equilibrium Conditions

    ï Unknowns

    θB and ∆

    - Entire Frame

    *)2(010:0 −−−=−−=Σ→+

    xxx DAF

    *)1(0:0 −−−=+=Σ BCBAB MMM

    ï Overview

    B- Joint B

    MBCMBA

  • 73

    )3(3

    )(3−−−= BBC

    EIM θ

    )4(1875.0375.021375.0 −−−∆=∆−∆= EIEIEIM DC

    )1(4

    64

    )1)(3(104

    )(222

    2

    −−−∆

    ++=EIEIM BAB θ

    5.625 0.375EI∆

    A

    B C

    D3m

    10 kN

    3 m

    1 m

    (5.625)load

    (1.875)load

    (0.375EI∆)∆

    (0.375EI∆)∆

    (0.375EI∆)∆

    (0.375EI∆)∆

    :0=Σ+ CM

    MDC

    D

    C

    4 m

    Dx

    MAB

    MBA

    A

    B

    4 m

    Ax

    10 kN

    ∆ ∆

    (1/2)(0.375EI∆)∆

    :0=Σ+ BM

    )5(563.11875.0375.0 −−−+∆+= EIEIA Bx θ4

    )( BAABx

    MMA +=

    )6(0468.04

    −−−∆== EIMD DCx

    )2(4

    64

    )1)(3(104

    )(422

    2

    −−−∆

    +−=EIEIM BBA θ

    5.625 0.375EI∆

    ï Slope-Deflection Equations

  • 74

    ï Solve equation

    *)2(010 −−−=−− xx DA*)1(0 −−−=+ BCBA MM

    )3(3

    )(3−−−= BBC

    EIM θ

    )4(1875.0 −−−∆= EIM DC

    )1(375.0625.54

    )(2−−−∆++= EIEIM BAB θ

    )5(563.11875.0375.0 −−−+∆+= EIEIA Bx θ

    )6(0468.0 −−−∆= EIDx

    )2(375.0625.54

    )(4−−−∆+−= EIEIM BBA θ

    Equilibrium Conditions:

    Slope-Deflection Equations:

    Horizontal reaction at supports:

    Substitute (2) and (3) in (1*)

    2EI θB + 0.375EI ∆ = 5.625 ----(7)

    Substitute (5) and (6) in (2*)

    )8(437.8235.0375.0 −−−−=∆−− EIEI Bθ

    From (7) and (8) can solve;

    EIEIB8.446.5

    =∆−

    )6()1(8.446.5 toinEI

    andEI

    Substitute B =∆−

    MAB = 15.88 kNïmMBA = 5.6 kNïmMBC = -5.6 kNïmMDC = 8.42 kNïmAx = 7.9 kNDx = 2.1 kN

  • 75

    A

    BC

    D

    Deflected curve

    A

    BC

    D

    Bending moment diagram

    MAB = 15.88 kNïm, MBA = 5.6 kNïm, MBC = -5.6 kNïm, MDC = 8.42 kNïm, Ax = 7.9 kN, Dx = 2.1 kN,

    A

    B C

    D

    3m

    10 kN

    3 m

    1 m

    15.88

    5.6

    8.42

    7.9 kN 2.1 kN

    15.88

    5.6

    8.42

    ∆ = 44.8/EI ∆ = 44.8/EI

    θB = -5.6/EI5.6

    7.8

  • 76

    B C

    DA

    3m4 m

    10 kN4 m

    2 m

    pin

    2 EI

    2.5 EI EI

    Example 9

    From the frame shown use the slope-deflection method to:(a) Determine the end moments of each member and reactions at supports(b) Draw the quantitative bending moment diagram, and also draw thequalitative deflected shape of the entire frame.

  • 77

    B

    C

    DA

    3m4 m

    10 kN4 m

    2 m

    2EI

    2.5EI EI

    Ax DxAy Dy

    MDCMAB

    ∆ CD C¥

    ∆BC

    ï Overview

    ï Boundary Conditions

    θA = θD = 0

    ï Equilibrium Conditions

    ï Unknowns

    θB and ∆

    - Entire Frame

    *)2(010:0 −−−=−−=Σ→+

    xxx DAF

    *)1(0:0 −−−=+=Σ BCBAB MMM

    B- Joint B

    MBCMBA

  • 78

    B

    C

    DA 3m4 m

    10 kN4 m

    2 m

    2EI

    2.5EI EI36.87c = ∆ tan 36.87c = 0.75 ∆

    = ∆ / cos 36.87c = 1.25 ∆

    ∆∆BC

    ∆ CD C¥

    ∆BC∆CD

    C

    36.87c

    )1(5375.04

    )(2−−−+∆+= EIEIM BAB θ

    )2(5375.04

    )(4−−−−∆+= EIEIM BBA θ

    )3(2813.04

    )2(3−−−∆−= EIEIM BBC θ

    )4(375.0 −−−∆= EIM DC

    C

    BB¥

    ∆BC= 0.75 ∆

    0.375EI∆

    6EI∆/(4) 2 = 0.375EI∆

    B

    A

    ∆¥ B¥

    (6)(2EI)(0.75∆)/(4) 2 = 0.5625EI∆0.5625EI∆

    D

    C

    C¥∆CD= 1.25 ∆

    (1/2) 0.75EI∆

    (1/2) 0.5625EI∆

    PL/8 = 5

    5

    (6)(2.5EI)(1.25∆)/(5)2 = 0.75EI∆

    0.75EI∆

    ï Slope-Deflection Equation

  • 79

    B

    C

    DA

    3m4 m

    10 kN4 m

    2 m

    pin2 EI

    2.5 EI EI

    Dx= (MDC-(3/4)MBC)/4 ---(6)

    MBC4

    Ax = (MBA+ MAB-20)/4 -----(5)

    B CMBC

    C

    D

    MBC/4MDC

    MBC4

    A

    B

    MBA

    10 kN

    MAB

    ï Horizontal reactions

  • 80

    ï Solve equations

    *)2(010 −−−=−− xx DA*)1(0 −−−=+ BCBA MM

    Equilibrium Conditions:

    Slope-Deflection Equation:

    Horizontal reactions at supports:

    Substitute (2) and (3) in (1*)

    Substitute (5) and (6) in (2*)

    From (7) and (8) can solve;

    EIEIB56.1445.1 −

    =∆=θ

    )6()1(56.1445.1 toinEI

    andEI

    Substitute B−

    =∆=θ

    MAB = 15.88 kNïmMBA = 5.6 kNïmMBC = -5.6 kNïmMDC = 8.42 kNïmAx = 7.9 kNDx = 2.1 kN

    )1(4

    654

    )(22 −−−∆

    ++=EIEIM BAB θ

    )2(4

    654

    )(42 −−−∆

    +−=EIEIM BBA θ

    )3(4

    )75.0)(2(34

    )2(32 −−−

    ∆−=

    EIEIM BBC θ

    )4(5

    )25.1)(5.2(32 −−−

    ∆=

    EIM DC

    )5(4

    )20(−−−

    −+= ABBAx

    MMA

    )6(443

    −−−−

    =BCDC

    x

    MMD

    )7(050938.05.2 −−−=−∆+ EIEI Bθ

    )8(05334.00938.0 −−−=−∆+ EIEI Bθ

  • 81

    BC

    DA

    Deflected shape

    θB=1.45/EI

    θB=1.45/EI

    Bending-moment diagram

    BC

    DA11.19

    1.91

    5.35

    5.46

    1.91

    ∆ ∆

    MAB = 11.19 kNïm MBA = 1.91 kNïm MBC = -1.91 kNïm MDC = 5.46 kNïm

    Ax = 8.28 kNïm Dx = 1.72 kNïm

    B

    C

    DA

    3m4 m

    10 kN4 m

    2 m

    pin2 EI

    2.5 EI EI11.19 kNïm

    8.27 kN

    5.46

    1.91

    1.91

    0.478 kN1.73

    0.478 kN

  • 82

    Example 10

    From the frame shown use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected curve.

    A

    BC

    D

    3 m

    3m

    4m

    20 kN/mpin

    2EI

    3EI

    4EI

  • 83

    A

    BC

    D

    3 m

    3m

    4m

    20 k

    N/m

    2EI

    3EI

    4EI

    [FEM]load

    ï Overview

    ∆ ∆

    ï Boundary Conditions

    θA = θD = 0

    ï Equilibrium Conditions

    ï Unknowns

    θB and ∆

    - Entire Frame

    *)2(060:0 −−−=−−=Σ→+

    xxx DAF

    *)1(0:0 −−−=+=Σ BCBAB MMM

    B- Joint B

    MBCMBA

  • 84

    A

    BC

    D

    3 m

    3m

    4m

    20 k

    N/m

    2EI

    3EI

    4EI

    [FEM]load

    wL2/12 = 15

    wL2/12 = 15

    A

    B C

    D

    ∆ ∆

    [FEM]∆∆∆∆

    6(2EI∆)/(3) 2= 1.333EI∆

    6(2EI∆)/(3) 2= 1.333EI∆ 6(4EI∆)/(4) 2

    = 1.5EI∆

    1.5EI∆

    BBCEIM θ3

    )3(3=

    ∆++= EIEI B 333.115333.1 θ ----------(1)

    ∆+−= EIEI B 333.115667.2 θ ----------(2)

    BEIθ3= ----------(3)

    ∆= EI75.0 ----------(4)

    ∆+++= EIEIEIM BAAB 333.1153)2(2

    3)2(4 θθ

    0

    ∆+−+= EIEIEIM BABA 333.1153)2(4

    3)2(2 θθ

    0

    ∆+= EIEIM DDC 75.04)4(3 θ

    0

    (1/2)(1.5EI∆)

    ï Slope-Deflection Equation

  • 85

    C

    DxMDC

    4 m

    D

    MAB

    A

    B

    60 kN

    MBA

    1.5 m

    1.5 m

    Ax + ΣMC = 0:

    :0=Σ+ BM

    )6(188.04

    −−−∆== EIMD DCx

    )5(30889.0333.13

    )5.1(60

    −−−+∆+=

    ++=

    EIEIA

    MMA

    Bx

    ABBAx

    θ

    ï Horizontal reactions

  • 86

    ï Solve equation

    *)1(0 −−−=+ BCBA MM

    Equilibrium Conditions

    Equation of moment

    Horizontal reaction at support

    Substitute (2) and (3) in (1*)

    Substitute (5) and (6) in (2*)

    From (7) and (8), solve equations;

    EIEIB67.3451.5

    =∆−

    )6()1(67.3451.5 toinEI

    andEI

    Substitute B =∆−

    )7(15333.1667.5 −−−=∆+ EIEI Bθ

    )8(30077.1333.1 −−−−=∆−− EIEI Bθ

    *)2(060 −−−=−− xx DA

    )1(333.115333.1 −−−∆++= EIEIM BAB θ

    )2(333.115667.2 −−−∆+−= EIEIM BBA θ

    )3(3 −−−= BBC EIM θ

    )4(75.0 −−−∆= EIM DC

    )6(188.0 −−−∆= EIDx

    )5(30889.0333.1 −−−+∆+= EIEIA Bx θ

    mkNM AB •= 87.53mkNM BA •= 52.16

    mkNM BC •−= 52.16mkNM DC •= 0.26

    Ax = 53.48 kNDx = 6.52 kN

  • 87

    A

    C

    DMoment diagram

    D

    A

    B C

    Deflected shape

    ∆ ∆

    A

    BC

    D

    3 m

    3m

    4m20

    kN/m 16.52 kNïm

    53.87 kNïm

    53.48 kN

    6.52 kN26 kNïm

    5.55 kN

    5.55 kN

    26.

    16.52

    mkNM AB •= 87.53mkNM BA •= 52.16

    mkNM BC •−= 52.16mkNM DC •= 0.26

    Ax = 53.48 kNDx = 6.52 kN

    53.87

    B16.52