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    ! 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 Deflection Equations

    settlement = j

    Pi j kw Cj

    Mij MjiwP

    j

    i

    i j

  • 3

    Degrees of Freedom

    L

    A B

    M

    1 DOF:

    P

    A B C 2 DOF: ,

  • 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

    j

    i

    i j

  • 8

    wP

    settlement = j

    (MFij) (MFji)

    (MFij)Load (MF

    ji)Load

    +

    Mij Mji

    i

    j

    +

    i jwPMijMji

    settlement = jj

    i

    =+ ji LEI

    LEI 24

    ji LEI

    LEI 42 +=

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

    ijF

    jiij MMLEI

    LEIM +++= Loadji

    Fji

    Fjiji MML

    EILEIM )()()4()2( +++=

    L

  • 9

    Mji Mjk

    Pi j kw Cj

    Mji Mjk

    Cj

    j

    Equilibrium Equations

    0:0 =+=+ jjkjij CMMM

  • 10

    +

    1

    1

    i jMij Mji

    i

    j

    LEIkii

    4=

    LEIk ji

    2=

    LEIkij

    2=

    LEIk jj

    4=

    i

    j

    Stiffness Coefficients

    L

  • 11

    [ ]

    =

    jjji

    ijii

    kkkk

    k

    Stiffness Matrix

    )()2()4( ijFjiij MLEI

    LEIM ++=

    )()4()2( jiFjiji MLEI

    LEIM ++=

    +

    =

    F

    ji

    Fij

    j

    iI

    ji

    ij

    MM

    LEILEILEILEI

    MM

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

    Matrix Formulation

  • 12

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

    Displacementmatrix

    Stiffness matrix

    Force matrixwP(MFij)Load (MFji)Load

    +

    +

    i jwPMijMji

    j

    i

    j

    Mij Mji

    i

    j

    (MFij) (MFji)

    Fixed-end momentmatrix

    ][]][[][ FEMKM +=

    ]][[])[]([ KFEMM =

    ][][][][ 1 FEMMK =

    L

  • 13

    L

    Real beam

    Conjugate beam

    Stiffness Coefficients Derivation: Fixed-End Support

    MjMi

    L/3

    i

    LMM ji +

    EIM j

    EIMi

    EILM j

    2

    EILMi

    2

    )1(2

    0)3

    2)(2

    ()3

    )(2

    (:0'

    =

    =+=+

    ji

    jii

    MM

    LEI

    LMLEI

    LMM

    )2(0)2

    ()2

    (:0 =+=+EI

    LMEI

    LMF jiiy ij

    ii

    LEIM

    LEIM

    andFrom

    )2(

    )4(

    );2()1(

    =

    =

    LMM ji +

    i j

  • 14

    Conjugate beam

    L

    Real beam

    Stiffness Coefficients Derivation: Pinned-End Support

    Mi i

    j

    LM i

    LMi

    EIM i

    EILMi

    2

    32L

    i j

    0)3

    2)(2

    (:0' ==+ LLEI

    LMM iij

    i j

    0)2

    ()3

    (:0 =+=+ jiiy EILM

    EILMF

    LEIM

    EILM

    ii

    i3)

    3(1 ===

    )6

    (EI

    LM ij

    =)3

    (EI

    LM ii =

  • 15

    Fixed end moment : Point Load

    Real beam

    8,0

    162

    22:0

    2 PLMEI

    PLEI

    MLEI

    MLFy ==+=+

    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 ==+=+

    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 =+=+ LEI

    MLLEI

    MLM B

    EIM

    EIML2

    EIMEI

    ML2

  • 19

    wP

    A B

    A B+A B

    (FEM)AB (FEM)BA

    Pin-Supported End Span: Simple Case

    BA LEI

    LEI 24 + BA L

    EILEI 42 +

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

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

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

    2)()()/3( BABABBA

    FEMFEMLEIM +=

    wP

    AB

    L

  • 20

    AB

    wP

    A B

    A BA B

    (MF AB)load (MF

    BA)load

    Pin-Supported End Span: With End Couple and Settlement

    BA LEI

    LEI 24 + BA L

    EILEI 42 +

    L

    (MF AB) (MF

    BA)

    )1()()(24 +++== FABload

    FABBAAAB MML

    EILEIMM

    )2()()(42 +++= FBAload

    FBABABA MML

    EILEIM

    2)(

    21])(

    21)[(3:

    2)1()2(2lim AFBAload

    FABload

    FBABBAA

    MMMMLEIMbyinateE +++=

    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 +++=

    8042 11

    11

    LPLEI

    LEIM BABA ++=

    128024

    2222

    22

    wLLPLEI

    LEIM CBBC ++++=

    128042

    2222

    22

    wLLPLEI

    LEIM CBCB

    +++=

    L L

  • 23

    MBA MBC

    A C

    B

    P1 P2

    L1 L2

    wCB

    B

    CB

    8042 11

    11

    LPLEI

    LEIM BABA ++=

    128024

    2222

    22

    wLLPLEI

    LEIM CBBC ++++=

    BBCBABB forSolveMMCM ==+ 0:0

  • 24

    A C

    B

    P1 P2

    L1 L2

    wCB

    Substitute B in MAB, MBA, MBC, MCB

    MABMBA

    MBCMCB

    0

    0

    0

    0

    8024 11

    11

    LPLEI

    LEIM BAAB +++=

    8042 11

    11

    LPLEI

    LEIM BABA ++=

    128024

    2222

    22

    wLLPLEI

    LEIM CBBC ++++=

    128042

    2222

    22

    wLLPLEI

    LEIM CBCB

    +++=

  • 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 B in the moment equations:

    MBC = 8.8 kNm

    MCB = -10 kNm

    MAB = 10.6 kNm,

    MBA = - 8.8 kNm,

    [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

    8)8)(10(

    84

    82

    += BABAEIEIM

    30)6)(6(

    62

    64 2

    ++= CBBCEIEIM

    20)6)(6(

    64

    62 2

    += CBCBEIEIM

    0:0 ==+ BCBAB MMM

    EI

    EIEI

    B

    B

    4.2

    030

    )6)(6(10)6

    48

    4(2

    =

    =++

  • 29

    10 kN 6 kN/m

    A C

    B 6 m4 m4 m

    MBC = 8.8 kNm

    MCB= -10 kNm

    MAB = 10.6 kNm,

    MBA = - 8.8 kNm,

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

    10 kNm8.8 kNm

    10.6 kNm8.8 kNm

    10 kN

    10.6 kNm

    8.8 kNm

    A B

    ByLAy

    2 m

    6 kN/m8.8 kNm

    10 kNmB

    CyByR

    18 kN

  • 30

    10 kN 6 kN/m

    A C

    B 6 m4 m4 m

    10 kNm10.6 kNm

    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 (kNm) 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

    )2(8

    )8)(10(8

    48

    2+= BABA

    EIEIM

    )3(30

    )6)(6(6

    26

    4 2++= CBBC

    EIEIM

    )4(20

    )6)(6(6

    46

    2 2+= CBCB

    EIEIM

    10

    10

    0

    0

    0

    308

    62:)1()2(2 = BBAEIM

    )5(158

    3= BBA

    EIM

  • 33

    MBA MBC

    B

    )5(158

    3= BBA

    EIM

    )3(30

    )6)(6(6

    4 2+= BBC

    EIM

    0:0 ==+ BCBAB MMM

    EI

    EIEI

    B

    B

    488.7

    )6(030

    )6)(6(15)6

    48

    3(2

    =

    =++

    EI

    EIEIinSubstitute

    A

    BAB

    74.23

    108

    28

    40:)1(

    =

    +=

    Substitute A and B in (5), (3) and (4):

    MBC = 12.19 kNm

    MCB = - 8.30 kNm

    MBA = - 12.19 kNm)4(

    20)6)(6(

    62 2

    = BCBEIM

  • 34

    10 kN 6 kN/m

    A C

    B 6 m4 m4 m

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

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

    12.19 kNm

    12.19 kNm8.30 kNm

    10 kN

    12.19 kNm

    A B

    ByLAy

    2 m

    6 kN/m12.19 kNm

    8.30 kNmC

    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 (kNm) 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

    )2(8

    )8)(10(8

    )2(48

    )2(2+= BABA

    EIEIM

    0 10

    10

    )2(8

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

    )2(3:2

    )1()2(2 aEIM BBA =

    )3(12

    )6)(4(6

    )3(4 2+= BBC

    EIM

  • 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

    ==+==

    15

    12

    )2(8

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

    )2(3 aEIM BBA =

    )3(12

    )6)(4(6

    )3(4 2+= BBC

    EIM

    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

  • 39

    10 kN 4 kN/m

    A C

    B6 m4 m4 m

    2EI 3EI

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

    14.18

    140.18 kNm

    10 kN

    A B

    ByLAy = 6.73 kN= 3.23 kN

    14.18 kNm

    10.91 kNm

    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 kNm

    V (kN)x (m)

    3.23

    -6.73

    12.55

    -11.46

    +-

    +-

    2.86

    M (kNm) x (m)

    12.91

    -14.18

    5.53

    -10.91

    +

    -+

    -

    6.77 + 12.55 = 19.32 kN

    B = 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 kNm

  • 42

    10 kN 4 kN/m

    A C

    B6 m4 m4 m

    2EI 3EI

    12 kNm

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

    MBA

    MBCB

    12 kNmMBA

    MBC

    EIB273.3

    =

    EIA21.7

    =

    )1(158

    )2(3= BBA

    EIM

    )2(126

    )3(4+= BBC

    EIM

    )3(126

    )3(2= BCB

    EIM

    8)8)(10(

    8)3(2

    8)2(4

    ++= BAABEIEIM

    0 -3.273/EI

    012:int = BCBA MMBJo

    012)122()1575.0( =+ BEIEI

    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 kNm5.45 kNm

    15.27 kNm

    2.82 kN 13.64 kN10.36 kN7.18 kN

    12 kNm10 kN 4 kN/m

    A C

    B13.64 kN2.82 kN

    15.27 kNm

    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 kNm

    13.64 kN2.82 kN

    15.27 kNm

    17.54 kN

    V (kN)x (m)3.41 m+

    -+

    -

    2.82

    -7.18

    10.36

    -13.64

    M (kNm) x (m)+

    - -+

    11.28

    -17.45

    -5.45-15.27

    7.98

    Deflected shape x (m)

    EIB273.3

    =EIA

    21.7=

    EIB273.3

    =

    EIA21.7

    =

  • 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 kNm 10 kN 6 kN/m

    A CB

    6 m4 m4 m

    2EI 3EI10 mm

  • 46

    12 kNm 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

    )2(8

    )8)(10(8

    )01.0)(2(68

    )2(48

    )2(22 ++=

    EIEIEIM BABA

    )3(30

    )6)(6(6

    )01.0)(3(66

    )3(26

    )3(4 22 ++=

    EIEIEIM CBBC

    )4(30

    )6)(6(6

    )01.0)(3(66

    )3(46

    )3(2 22 +=

    EIEIEIM CBCB

    -12

    0

    0

  • 47

    12 kNm 10 kN 6 kN/m

    A CB

    6 m4 m4 m

    2EI 3EI10 mm

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

    )1(8

    )8)(10(8

    )01.0)(2(68

    )2(28

    )2(42 +++=

    EIEIEIM BAAB

    )2(8

    )8)(10(8

    )01.0)(2(68

    )2(48

    )2(22 ++=

    EIEIEIM BABA

    )1(10758

    )2(28

    )2(4+++= BAAB

    EIEIM

    )2(10758

    )2(48

    )2(2++= BABA

    EIEIM

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

    )2(3:2

    )1()2(2 aEIM BBA +=

    16.5

  • 48

    + MB = 0: - MBA - MBC = 0 (3/4 + 2)EIB + 16.5 - 192.8 = 0

    B = 64.109/ EI

    Substitute B in (1): A = -129.06/EI

    Substitute A and B in (5), (3), (4):

    MBC = -64.58 kNm

    MCB = -146.69 kNm

    MBA = 64.58 kNm,

    MBA MBC

    B

    12 kNm 10 kN 6 kN/m

    A CB

    6 m4 m4 m

    2EI 3EI10 mm

    MBC = (4/6)(3EI)B - 192.8

    MBA = (3/4)(2EI)B + 16.5

  • 49

    12 kNm 10 kN 6 kN/m

    A CB

    6 m4 m4 m

    64.58 kNm 64.58 kNm

    = -1.57 kN= 11.57 kN

    146.69 kNm

    = 47.21 kN= -29.21 kN

    10 kN

    A B

    ByLAy

    12 kNm64.58 kNm

    2 m

    6 kN/m

    C

    CyByR

    18 kN

    B

    64.58 kNm

    146.69 kNm

    MBC = -64.58 kNm

    MCB = -146.69 kNm

    MBA = 64.58 kNm,

  • 50

    12 kNm 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 (kNm) x (m)

    1258.29 64.58

    -146.69

    +

    -

    47.21 kN

    146.69 kNm

    11.57 kN

    1.57 + 29.21 = 30.78 kN

    10 mmA = -129.06/EI

    B = 64.109/ EIA = -129.06/EI

    Deflected shape x (m)

    B = 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 kNm

    10 mm

  • 52

    6 kN/m

    B A C

    3 m 3 m2EI 1.5EI

    12 kNm

    )1(3

    )2(4= CCB

    EIM

    )2(1005.43

    )5.1(23

    )5.1(4+++= ACCA

    EIEIM

    )2(2

    122

    1002

    )5.4(33

    )5.1(3:2

    )2()2(2 aEIM CCA +++=

    )3(1005.43

    )5.1(43

    )5.1(2++= ACAC

    EIEIM 12

    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 kNm

    )1(3

    )2(4= CCB

    EIM

    )2(2

    122

    1002

    )5.4(33

    )5.1(3 aEIM CCA +++=

    10 mm

    MCB MCA

    C

    0=+ CACB MM

    02

    122

    1002

    )5.4(33

    )5.48(=+++

    +C

    EI

    radEIC

    0015.006.15 ==

    )3(1005.43

    )5.1(4)06.15(3

    )5.1(212 ++= AEI

    EIEI Substitute C in eq.(3)

    radEIA

    0034.022.34 ==

    Equilibrium equation:

  • 54

    6 kN/m

    B A C

    3 m 3 m2EI 1.5EI

    12 kNm

    10 mm

    radEIC

    0015.006.15 == radEIA

    0034.022.34 ==

    mkNEI

    EIEIM CBC =

    == 08.20)06.15(3

    )2(23

    )2(2

    mkNEI

    EIEIM CCB =

    == 16.40)06.15(3

    )2(43

    )2(4

    kN08.203

    08.2016.40=

    +kN08.20

    B C40.16 kNm20.08 kNm

    6 kN/m

    AC

    12 kNm

    40.16 kNm

    18 kN

    8.39 kN26.39 kN

  • 55

    6 kN/m

    B A C

    3 m 3 m2EI 1.5EI

    12 kNm

    10 mm

    6 kN/m

    AC

    12 kNm

    40.16 kNm8.39 kN26.39 kN

    B C40.16 kNm20.08 kNm

    20.08 kN20.08 kN

    V (kN)

    x (m)

    26.398.39

    -20.08

    +-

    M (kNm)

    x (m)20.08

    -40.16

    12

    radC 0015.0=radA 0034.0=

    Deflected shapex (m)

    radC 0015.0=

    radA 0034.0=

  • 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 kNm

    10 mm

  • 57

    6 kN/m

    B A C

    3 m 3 m2EI 1.5EI

    12 kNm

    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

    )3(1005.43

    )5.1(23

    )5.1(4++++= CACCA EI

    EIEIM

    )4(1005.43

    )5.1(43

    )5.1(2+++= CACAC EI

    EIEIM 12

    )1(3

    43

    )2(2= CCBC

    EIEIM

    )3(2

    122

    1002

    )5.4(323

    )5.1(3:2

    )4()3(2 aEIEIM CCCA ++++=

    CC EIEI =23

    )5.1(6C

    C

    A

    CEI

  • 58

    6 kN/m

    B A C

    3 m 3 m2EI 1.5EI

    12 kNm

    10 mm

    Equilibrium equation:

    )3

    ()( CBBCCByMMC +=

    6 kN/m

    AC

    12 kNm

    MCA

    18 kN

    AyB C

    MCBMBC

    By3

    393

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

    C

    MCB MCA

    (Cy)CA(Cy)CB

    *)1(0:0 =+= CACBC MMM

    *)2(0)()(:0 =+= CAyCByy CCC

    )5(15.628333.0167.4*)1( = CC EIEIinSubstitute

    )6(75.101167.35.2*)2( =+ CC EIEIinSubstitute

    mmEIradEIandFrom CC 227.5/27.5200255.0/51.25)6()5( ====

  • 59

    6 kN/m

    B A C

    3 m 3 m2EI 1.5EI

    12 kNm

    10 mm

    Solve equation

    radEIC

    00255.051.25 ==

    mmEIC

    227.527.52 ==

    Substitute C and C in (4)

    radEIA

    000286.086.2 ==

    Substitute C 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 kNm

    35.68 kNm

    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 kNm

    35.68 kNm

    12.45 kN 5.55 kN

    Deflected shape

    x (m)

    radC 00255.0=

    mmC 227.5=

    radA 000286.0=

    radC 00255.0= radA 000286.0=

    mmC 227.5=

    M (kNm)

    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

    Equilibrium equations

    10

    MBC

    MBA

    )3(18366

    )2(3++= BBC

    EIM

    *)1(010 = BCBA MM

    05430310 =+ BEI

    EIEIB667.4

    )3(14

    =

    =)1(30

    6)3(2

    += BABEIM

    mkNMmkNM

    mkNM

    BC

    BA

    AB

    ==

    =

    33.4933.39

    33.25

    36/2 = 18

    (wL2/12 ) =3636

    Slope-Deflection Equations

    )2(306

    )3(4= BBA

    EIM

    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

    B = -4.667/EIB

    B

    MAB = 25.33 kNm

    MBA = -39.33 kNm

    MBC = 49.33 kNm

  • 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

    5)2(2

    5)2(4

    +=0

    BABAEIEIM

    5)2(4

    5)2(2

    +=

    75.186

    )4(26

    )4(4++= CBBC

    EIEIM

    75.186

    )4(46

    )4(2+= CBCB

    EIEIM

    CCDEIM 5

    )(3=

    104

    )3(3+= CCE

    EIM

    0=+ BCBA MM

    )1(75.18)68()

    616

    58( =++ CB EIEI

    0=++ CECDCB MMM

    )2(75.8)49

    53

    616()

    68( =+++ CB EIEI

    EIEIandFrom CB

    86.229.5:)2()1( ==

    PL/8 = 18.75 18.75

    6.667 (wL2/12 ) = 6.667+ 3.333

  • 67

    MAB = 4.23 kNm

    MBA = 8.46 kNm

    MBC = 8.46 kNm

    MCB = 18.18 kNm

    MCD = 1.72 kNm

    MCE = 16.44 kNm

    Substitute B = -1.11/EI, c = -20.59/EI below

    0

    0BAAB

    EIEIM 5

    )2(25

    )2(4+=

    BABAEIEIM

    5)2(4

    5)2(2

    +=

    75.186

    )4(26

    )4(4++= CBBC

    EIEIM

    75.186

    )4(46

    )4(2+= CBCB

    EIEIM

    CCDEIM 5

    )(3=

    104

    )3(3+= CCE

    EIM

  • 68

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

    A

    B

    5 m

    C E

    20 kN

    A

    B

    5 m

    4.23 kNm

    2.54 kN

    (8.46 + 4.23)/5 = 2.54 kN

    8.46 kNm

    C

    25 kN

    B 3 m 3 m2.54 kN 2.54 kN

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

    14.12 kN18.18 kNm

    16.44 kNm

    (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 kNm

    (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

    +

    -

    +

    - -

    +

    B = 5.29/EI

    C = 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

    Boundary Conditions

    A = D = 0

    Equilibrium Conditions

    Unknowns

    B and

    - Entire Frame

    *)2(010:0 ==+

    xxx DAF

    *)1(0:0 =+= BCBAB MMM

    Overview

    B- Joint B

    MBCMBA

  • 73

    )3(3

    )(3= BBC

    EIM

    )4(1875.0375.021375.0 == EIEIEIM DC

    )1(4

    64

    )1)(3(104

    )(222

    2

    ++=EIEIM BAB

    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=+ CM

    MDC

    D

    C

    4 m

    Dx

    MAB

    MBA

    A

    B

    4 m

    Ax

    10 kN

    (1/2)(0.375EI)

    :0=+ BM

    )5(563.11875.0375.0 ++= EIEIA Bx 4

    )( BAABx

    MMA +=

    )6(0468.04

    == EIMD DCx

    )2(4

    64

    )1)(3(104

    )(422

    2

    +=EIEIM BBA

    5.625 0.375EI

    Slope-Deflection Equations

  • 74

    Solve equation

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

    )3(3

    )(3= BBC

    EIM

    )4(1875.0 = EIM DC

    )1(375.0625.54

    )(2++= EIEIM BAB

    )5(563.11875.0375.0 ++= EIEIA Bx

    )6(0468.0 = EIDx

    )2(375.0625.54

    )(4+= EIEIM BBA

    Equilibrium Conditions:

    Slope-Deflection Equations:

    Horizontal reaction at supports:

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

    2EI B + 0.375EI = 5.625 ----(7)

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

    )8(437.8235.0375.0 = EIEI B

    From (7) and (8) can solve;

    EIEIB8.446.5

    =

    =

    )6()1(8.446.5 toinEI

    andEI

    Substitute B =

    =

    MAB = 15.88 kNmMBA = 5.6 kNmMBC = -5.6 kNmMDC = 8.42 kNmAx = 7.9 kNDx = 2.1 kN

  • 75

    A

    BC

    D

    Deflected curve

    A

    BC

    D

    Bending moment diagram

    MAB = 15.88 kNm, MBA = 5.6 kNm, MBC = -5.6 kNm, MDC = 8.42 kNm, 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

    B = -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

    B

    CD C

    BC

    Overview

    Boundary Conditions

    A = D = 0

    Equilibrium Conditions

    Unknowns

    B and

    - Entire Frame

    *)2(010:0 ==+

    xxx DAF

    *)1(0:0 =+= 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

    B

    BC

    CD C

    BCCD

    C

    C

    36.87c

    )1(5375.04

    )(2++= EIEIM BAB

    )2(5375.04

    )(4+= EIEIM BBA

    )3(2813.04

    )2(3= EIEIM BBC

    )4(375.0 = EIM DC

    C

    C

    BB

    BC= 0.75

    0.375EI

    6EI/(4) 2 = 0.375EI

    B

    A

    B

    (6)(2EI)(0.75)/(4) 2 = 0.5625EI0.5625EI

    D

    C

    CCD= 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

    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

    Horizontal reactions

  • 80

    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

    ==

    )6()1(56.1445.1 toinEI

    andEI

    Substitute B

    ==

    MAB = 15.88 kNmMBA = 5.6 kNmMBC = -5.6 kNmMDC = 8.42 kNmAx = 7.9 kNDx = 2.1 kN

    )1(4

    654

    )(22

    ++=EIEIM BAB

    )2(4

    654

    )(42

    +=EIEIM BBA

    )3(4

    )75.0)(2(34

    )2(32

    =

    EIEIM BBC

    )4(5

    )25.1)(5.2(32

    =

    EIM DC

    )5(4

    )20(

    += ABBAx

    MMA

    )6(443

    =BCDC

    x

    MMD

    )7(050938.05.2 =+ EIEI B

    )8(05334.00938.0 =+ EIEI B

  • 81

    BC

    DA

    Deflected shape

    B=1.45/EI

    B=1.45/EI

    Bending-moment diagram

    BC

    DA11.19

    1.91

    5.35

    5.46

    1.91

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

    Ax = 8.28 kNm Dx = 1.72 kNm

    B

    C

    DA

    3m4 m

    10 kN4 m

    2 m

    pin2 EI

    2.5 EI EI11.19 kNm

    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

    Overview

    Boundary Conditions

    A = D = 0

    Equilibrium Conditions

    Unknowns

    B and

    - Entire Frame

    *)2(060:0 ==+

    xxx DAF

    *)1(0:0 =+= 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 3

    )3(3=

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

    += EIEI B 333.115667.2 ----------(2)

    BEI3= ----------(3)

    = EI75.0 ----------(4)

    +++= EIEIEIM BAAB 333.1153)2(2

    3)2(4

    0

    ++= EIEIEIM BABA 333.1153)2(4

    3)2(2

    0

    += EIEIM DDC 75.04)4(3

    0

    (1/2)(1.5EI)

    Slope-Deflection Equation

  • 85

    C

    DxMDC

    4 m

    D

    MAB

    A

    B

    60 kN

    MBA

    1.5 m

    1.5 m

    Ax + MC = 0:

    :0=+ BM

    )6(188.04

    == EIMD DCx

    )5(30889.0333.13

    )5.1(60

    ++=

    ++=

    EIEIA

    MMA

    Bx

    ABBAx

    Horizontal reactions

  • 86

    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 B

    )8(30077.1333.1 = EIEI B

    *)2(060 = xx DA

    )1(333.115333.1 ++= EIEIM BAB

    )2(333.115667.2 += EIEIM BBA

    )3(3 = BBC EIM

    )4(75.0 = EIM DC

    )6(188.0 = EIDx

    )5(30889.0333.1 ++= EIEIA Bx

    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 kNm

    53.87 kNm

    53.48 kN

    6.52 kN26 kNm

    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