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  • 1

    STRUCTURAL ANALYSIS

    By

    Assoc. Prof. Dr. Sittichai SeangatithSCHOOL OF CIVIL ENGINEERING

    INSTITUTE OF ENGINEERINGSURANAREE UNIVERSITY OF TECHNOLOGY

    2

    3Moment Distribution

    1. concept moment distribution2. frame statically

    indeterminate moment distribution shear diagram, moment diagram elastic curve

    3.1 moment distribution simultaneous slope-deflection ( step 3)

    3

    : moment distribution

    3.1, 3.7 3.9 +

    4

    Sign Convention

    slope-deflection

    Fixed-End Moments (FEM)

    slope-deflection

  • 5

    1. stiffness factors

    3K - Far End PinnedEIL

    =

    4 K - Far End FixedEIL

    =

    2. distribution factors KDF=K

    DF = 1 - For pinned end

    DF = 0 - For fixed end

    3. fixed-end moment FEM

    4. moment distribution5. FBD shear diagram moment diagram

    6. elastic curve 6

    Member Stiffness Factor Carry-Over Factormember stiffness factor moment 1 radian

    = 1 radian

    2 (2 3 ) FEMAB A B ABIM EL L

    = + +

    2 (2 3 ) FEMBA B A BAIM EL L

    = + +

    4A

    EIL

    =

    2A

    EIL

    =

    4 K

    Far End Fixed

    EIL

    =

    7

    Carry-over factor moment moment

    20.54

    ABA

    AB A

    EIM L

    EIML

    = = +

    Joint Stiffness Factor member stiffness factor joint

    K KT =

    8

    Joint Stiffness Factor

    K K K KT AD AB AC= + + 10000 kN-m=1000 4000 5000= + +

    JSF = moment joint A 1 radian

    ADAD L

    EI

    =

    4K

    ABAB L

    EI

    =

    4K

    ADAD L

    EI

    =

    4K

  • 9

    Distribution Factor (DF) moment M joint A AD, AB, AC moment

    AD AB ACM M M M= + +

    slope-deflection4 KAD AD A

    AD

    EIML

    = = KAB AB AM = KAC AC AM =

    moment joint A

    2 (2 3 ) FEMAD A D ADAD

    EIML L

    = + +

    10

    (K K K )AD AB AC AM = + +

    (K K K )A AD AB AC

    M =+ +

    A moment joint A K DF

    KAD

    AD ADM M M= =K DF

    KAB

    AB ABM M M= =K DF

    KAC

    AC ACM M M= =

    distribution factor

    KDF=K

    KM

    = KAD AD AM =

    KAB AB AM =

    KAC AC AM =

    11

    DF 1000/10000 0.1AD = =

    DF 4000 /10000 0.4AB = =

    DF 5000 /10000 0.5AC = =

    = 1.0

    0.1(2000) 200 N-mADM = =

    0.4(2000) 800 N-mABM = =

    0.5(2000) 1000 N-mACM = =

    = 2,000 N-m

    12

    3.2 Moment Distribution1. member stiffness factor K

    4( )K ABABAB

    EIL

    =

    4( )K BCBCBC

    EIL

    =

    A C fixed support K KA C = =

    6 440 (10 ) m /mE =

    6 460 (10 ) m /mE =

    64 [30(10 )]3

    E =

    64 [60(10 )]4

    E =

    4 K

    Far End Fixed

    EIL

    =

  • 13

    2. distribution factor DF joint

    DF ABBAAB BC

    KK K

    =+

    A:

    B:

    C:

    DF BCBCAB BC

    KK K

    =+

    6

    6

    40 (10 )DF 040 (10 )ABE

    E

    = =+

    6

    6

    60 (10 )DF 060 (10 )CBE

    E

    = =+

    KDF=K

    6

    6

    40 10 0.4040 60 10

    EE E

    = =+6

    6

    60 10 0.6040 60 10

    EE E

    = =+

    14

    3. fixed-end moment FEM

    AB (FEM) (FEM) 0AB BA= =

    BC w = 3 kN/m 2

    (FEM)12BCwL

    =

    2

    (FEM)12CBwL

    = +

    23(4) 4 kN-m12

    = =

    23(4) 4 kN-m12

    = =

    FEMCB

    FEMBC

    15

    4. moment4.a joint B AB BC

    fixed-end moment BC

    0 0

    16

    moment joint B BC =

    moment joint B BA = 0.4( 4) 1.6 kN-m+ = +

    0.6( 4) 2.4 kN-m+ = +

    4.b joint B MB = -4 kN-m joint B MB = + 4 kN-m AB BC (distribution)

    4 kN-m

    1.6 kN-m 2.4 kN-m

  • 17

    4.c joint B moment joint B (carry-over) carry-over factor +0.5

    moment BC CB =

    moment BA AB =0.5( 2.4) 1.2 kN-m+ = +

    0.5( 1.6) 0.8 kN-m+ = +

    18

    4.d A C fixed support moment joint B

    4.e moment joint moment

    19

    5. shear diagram moment diagram

    20

  • 21

    6. elastic curve

    22

    3.3 Stiffness-Factor Modifications1.) Member Pin-Supported at Far End

    2 (2 )AB A BIM EL

    = +

    2 ( 2 ) 0BA A BIM EL = + =

    2A

    B =

    3AB A

    EIML

    =

    3K EIL

    =

    MBA

    MAB

    A B

    23

    2.) Symmetric Structure and Loading

    B C = BC CBM M =

    2 (2 )BC B CIM EL

    = +

    2 (2 )CB C BIM EL

    = +

    2EIM

    L=

    2K EIL

    =

    2 ( 2 )IEL

    = + 2EI ML

    = =

    2 (2 )IEL

    = 2EI ML

    = =

    MBC

    MCB

    24

    3.) Symmetric Structure with Antisymmetric Loading

    6EIM

    L=

    6K EIL

    =

    B C = = BC CBM M M= =

    2 (2 ) 2 (2 )BC B CI IM E EL L

    = + = +

    2 (2 ) 2 (2 )CB C BI IM E EL L

    = + = +

    MBC

    MCB

    6EI ML

    = =

    6EI ML

    = =

  • 25

    3-1 moment distribution shear diagram, moment diagram elastic curve rotation B EIAB = EI EIBC = 2EI

    DOF = 1 B degree of indeterminacy = 7-3 = 4

    26

    1. stiffness factors4( )K ABAB

    AB

    EIL

    =

    4( )K BCBCBC

    EIL

    =

    43EI

    =

    85EI

    =

    2. distribution factors A C fixed end

    4 / 3(DF) 04 / 3ABEI

    EI= =

    +8 / 5(DF) 0

    8 / 5CBEI

    EI= =

    +

    4 / 3(DF)4 / 3 8 / 5BA

    EIEI EI

    =+

    (DF) 1 0.4545 0.5455BC = =

    0.4545=

    27

    2

    2(FEM)ABPab

    L=

    2

    2(FEM)BAPba

    L= +

    2

    (FEM)12BCwL

    =

    (FEM) 4.167 kN-mCB = +

    3. fixed-end moment

    FEMBA FEMBCFEMAB FEMCB

    2

    210(1)2 4.444 kN-m

    3= =

    2

    210(2)1 2.222 kN-m

    3= = +

    22(5) 4.167 kN-m12

    = =

    28

    4. moment distribution

    +4.697-3.106+3.106-4.003M+0.5304+0.4419COM

    +1.0607+0.8838DM+4.1667-4.1667+2.2222-4.4444FEM

    00.54550.45450DFCBBCBAABMemberCBAJoint

    (DF) (DF) 0AB BC= = (DF) 0.4545CB =(DF) 0.5455BC =

    (FEM) 4.4444 kN-mAB = (FEM) 2.2222 kN-mBA = +

    (FEM) 4.1667 kN-mBC = (FEM) 4.1667 kN-mCB = +

  • 29

    5. FBD shear diagram moment diagram

    +4.697-3.106+3.106-4.003 MCBBCBAABMember

    30

    2.341 m

    31

    6. elastic curve

    32

    rotation B

    slope-deflection

    2 (2 3 ) (FEM)ABBA B A BAABAB

    EIML L

    = + +

    23.106 (2 ) 2.22223 B

    EI = +

    0.663B EI

    =

  • 33

    moment distribution shear diagram, moment diagram elastic curve rotation B EIAB = 0.75EIBC

    DOF = 3 A, B C degree of indeterminacy = 4-3 = 1

    34

    1. stiffness factors3K ABAB

    AB

    EIL

    =

    3K BCBCBC

    EIL

    =

    3(0.75 )5

    0.45

    BC

    BC

    EI

    EI

    =

    =

    0.5 BCEI=2. distribution factors

    A C pinned end 0.45(DF) 1.0

    0 0.45BC

    ABBC

    EIEI

    = =+

    0.5(DF) 1.00 0.5

    BCCB

    BC

    EIEI

    = =+

    0.45(DF)0.45 0.5

    BCBA

    BC BC

    EIEI EI

    =+

    (DF) 1 0.4737BC =

    0.4737=

    0.5263=

    0.75AB BCEI EI=

    35

    2

    (FEM)8BC

    wL=

    3. fixed-end moment

    FEMBA FEMBC

    212(6) 54.0 kN-m8

    = =

    2

    (FEM)8BA

    wL= +

    28(5) 25.0 kN-m8

    = + = +

    36

    4. moment distribution

    0-38.737+38.7370M+15.263+13.737DM

    0-54.0+25.00FEM1.00.52630.47371.0DFCBBCBAABMemberCBAJoint

    (DF) (DF) 1.0AB CB= = (DF) 0.4737BA = (DF) 0.5263BC =

    (FEM) 40.0 kN-mBA = +

    (FEM) 54.0 kN-mBC =

  • 37

    5. FBD shear diagram moment diagram

    0-38.737+38.7370 MCBBCBAABMember

    38.737 38.737

    12.25327.747 42.456

    29.5445 m 6 m

    38

    39

    6. elastic curve

    40

    rotation B

    slope-deflection3 ( ) (FEM)ABBA B BA

    ABAB

    EIML L

    = +

    338.737 ( ) 25.05 B

    EI + = +

    22.895B EI

    = +

  • 41

    3-2 moment distribution shear diagram, moment diagram elastic curve

    AB moment AB

    1.2EI EI

    42

    4K BCBCBC

    EIL

    =

    4K CDCDCD

    EIL

    =

    1. stiffness factors4(1.2 ) 0.96

    5EI EI= =

    44EI EI= =

    1.2EI EI

    43

    2. distribution factors

    D fixed end

    AB (DF) 0BA = (DF) 1.00

    BCBC

    BC

    KK

    = =+

    (DF) 0CDDCCD

    K EIK EI

    = = =+ +

    0.96(DF)0.96CB

    EIEI EI

    =+

    (DF) 1 (DF) 0.5102CD CB= =0.4898=

    K 0.96BC EI=

    KCD EI=

    44

    3. fixed-end moment

    FEMCB

    FEMBA

    FEMBC

    (FEM) 20(2) 40 kN-mBA = = +2

    (FEM)12BCwL

    = 240(5) 83.333 kN-m

    12= =

    AB

    2

    (FEM)12CBwL

    = + 83.333 kN-m= +

    C FEM FEM

  • 45-2.336+2.653-2.154COM-4.672-4.308+5.306DM

    -5.528+8.980-5.306COM-11.055-10.612+17.960DM

    -18.707+21.667-17.960COM-37.414-35.919+43.333DM

    -10+83.333-83.333+40FEM00.51020.48981.00DF

    DCCDCBBCBAMemberDCBJoint

    4. moment distribution(DF) 0BA = (DF) 1BC = (DF) 0DC = (DF) 0.4898CB = (DF) 0.5102CD =

    (FEM) 40 kN-mBA = + (FEM) 83.333 kN-mBC = (FEM) 83.333 kN-mCB = +

    46-27.639-10-55.297+65.297-4040M-0.020-0.020+0.033DM

    -0.033+0.040-0.033COM-0.067-0.065+0.080DM

    -0.083+0.132-0.080COM-0.166-0.159+0.264DM

    -0.275+0.325-0.264COM-0.549-0.528+0.650DM

    -0.667+1.077-0.650COM-1.354-1.299+2.154DM

    -2.336+2.653-2.154COM00.51020.489810DF

    DCCDCBBCBAMember

    47

    5. FBD shear diagram moment diagram

    -27.639-10-55.297+65.297-4040MDC

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