DISPLACEMENT METHOD OF ANALYSIS: SLOPE e! ents! on! nts Span! s Beams! ay! ay DISPLACEMENT METHOD OF ANALYSIS: SLOPE DEFLECTION EQUATIONS

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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

    1