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  • 1

    STRUCTURAL ANALYSIS

    By

    Assoc. Prof. Dr. Sittichai SeangatithSCHOOL OF CIVIL ENGINEERING

    INSTITUTE OF ENGINEERINGSURANAREE UNIVERSITY OF TECHNOLOGY

    2

    2Slope-Deflection Method

    1. concept slope-deflection2. frame statically

    indeterminate slope-deflection shear diagram, moment diagram elastic curve

    3

    equilibrium equations force-displacement relationships

    displacementsDisplacement method

    compatibility equations force-displacement relationships

    forcesForce methodUnknown

    2.1 Displacement Method: General Procedures

    :1. linear elastic2.

    4

    1. DOF (joint) (node) B C

    displacement method

    DOF(force-displacement relations)

    2 2 3 (FEM)BA B A BAIM EL L

    = + +

  • 5

    2. (DOF)

    4. force-displacement relations /

    0532 =++ MMM

    074 =+MM

    3.

    6

    Degrees of FreedomNodes - Degree of freedom [] node

    a.) Degree of indeterminacy = 3 degree of freedom = 1

    b.) Degree of indeterminacy = 1 degree of freedom = 3

    /

    7

    2 node DOF 3 2 1 etc. DOF

    DOF = 1

    DOF = 4

    8

    DOF = 3 DOF = 9

  • 9 10

    2.2 Slope-Deflection Equations

    moment AB(MAB MBA) DOF (A, B )

    AB (continuous beam)

    bending moment (joint) rigid joint

    11

    Sign conventionMAB MBA + A B + tangent

    deflection curve + ( ) 12

    principle of superposition MAB MBA

    1. A(A) B

    2. B(B) A

    3. B A ()

    4. tangent cord

  • 13

    =

    +

    +

    +

    principle of superposition

    14

    1: A

    0;AM + =12 3

    ABM LLEI

    0;BM + =12 3

    BAM LLEI

    4AB A

    EIML

    =

    2BA A

    EIML

    =

    1 2 02 3

    BAM LLEI

    =

    1 2 02 3

    ABA

    M LL LEI

    + =

    15

    2: B4

    BA BEIML

    =

    2AB B

    EIML

    =

    16

    3:

    0;BM + =

    1 22 3M LLEI

    moment M

    2

    6AB BA

    EIM M ML

    = = =

    12 3M LLEI

    0 =

  • 17

    4: Fixed-End Moment (FEM)

    0;yF+ =1 12 02 4 2PL ML LEI EI

    =

    8PLM =

    FEM node A - ()FEM node B + ()

    18

    Slope-Deflection Equation

    2 2 3 (FEM)AB A B ABIM EL L

    = + +

    2 2 3 (FEM)BA B A BAIM EL L

    = + +

    4AB A

    EIML

    = 2BA AEIML

    =

    4BA B

    EIML

    = 2AB BEIML

    =

    2

    6AB BA

    EIM ML

    = =

    19

    Pin-Supported End Span

    2 2 3 (FEM)AB A B ABIM EL L

    = + +

    0 2 2 3B AIEL L

    = +

    3 (FEM)AB A ABIM EL L

    = +

    32 2A

    B L = +

    2 2 3 (FEM)AB A B ABIM EL L

    = + +

    2 2 3 (FEM)BA B A BAIM EL L

    = + +

    20

    2 2 3 (FEM)AB A B ABIM EL L

    = + +

    2 2 3 (FEM)BA B A BAIM EL L

    = + +

    3 (FEM)AB A ABIM EL L

    = +

    Slope-Deflection Equation:

  • 21

    Fixed-End Moment (FEM)

    22

    Fixed-End Moment ()

    23

    2.3 1. FEM 2. / (E, I, L, , FEM) slope-

    deflection 3. joint / support slope-deflection ( 2) (simultaneous equation) slope deflection joint / support

    4. slope deflection slope-deflection ( 2) end-moment M

    5. shear diagram, moment diagram elastic curve24

    : slope-deflection2.2, 2.7, 2.9 +

  • 25

    2-1 slope-deflection shear moment diagram elastic curve E = 200 GPa, IAB = 0.006 m4 IBC = 0.008 m4

    DOF = 3 A, B, C A C pin roller DOF B

    degree of Indeterminacy = 4-3 = 1

    26

    1. fixed-end moment A C pin roller

    2

    (FEM)8BAwL

    = +

    2

    (FEM)8BCwL

    =

    28(5) 25 kN-m8

    = =

    212(6) 54 kN-m8

    = =

    FEMBA FEMBC

    3 (FEM)AB A ABIM EL L

    = +

    27

    2. slope-deflection A C pin roller

    3 (FEM)AB A ABIM EL L

    = +

    3 ( ) 25ABBA BAB AB

    IM EL L

    = +

    3 ( ) 255 AB BEI = +MBA MBC

    3 ( ) 546BC BC B BC

    M EIL

    =

    1 ( ) 542 BC BEI =

    28

    3. 0BA BCM M+ =

    MBA MBC

    3 (0.006) 255 BE + +

    1 (0.008) 54 02 BE =

    3815.79B E

    =51.9079(10 ) radian=

    4. 3 (0.006) 255BA B

    M E = + 38.74 kN-m=

    1 (0.008) 542BC B

    M E = 38.74 kN-m=

    3 ( ) 255BA AB B

    M EI = + 1 ( ) 542BC BC B

    M EI =

  • 29

    5. FBD shear diagram moment diagram

    5 m 6 m

    30

    31

    6. elastic curve

    32

    2-2 slope-deflection shear moment diagram D elastic curve B = 10 mm EIAB = 100 kN-m2 EIBC = 200 kN-m2

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

  • 33

    2

    2(FEM)ABPabL

    =

    2

    2(FEM)ABPbaL

    = +

    2

    (FEM)12BCwL

    =

    2

    (FEM) 4.167 kN-m12CBwL

    = + = +

    1. 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

    = =

    slope-deflection 1 why???

    34

    2. slope-deflection

    MBA MBCMAB MCB

    2 2 3 (FEM)AB A B ABIM EL L

    = + + 2 2 3 (FEM)BA B A BA

    IM EL L

    = + +

    ABEI EI= 2BCEI EI=

    2 0.0102 3 4.4443 3AB A BEIM = +

    2 0.020 4.4443 3BEI =

    2 0.0102 3 2.2223 3BA B AEIM = + +

    2(2 ) 0.0102 3 4.1675 5BC B CEIM = +

    8 0.120 4.1675 25BEI = +

    2(2 ) 0.0102 3 4.1675 5CB C BEIM = + +

    4 0.120 4.1675 25BEI = + +

    4 0.020 2.2223 3BEI = +

    35

    3.

    MBA MBCMAB MCB

    0BA BCM M+ =

    [ ]2.933 0.001867 1.945BEI =

    37.27(10 ) radianB=

    4 0.020 2.2223 3B

    BAM EI = + 8 0.120 4.1675 25B

    BCM EI = +

    2100 kN-mEI =

    36

    4. 37.27(10 ) radianB=

    2 0.020 4.4443 3B

    ABM EI =

    4 0.020 2.2223 3B

    BAM EI = +

    4 0.120 4.1675 25B

    CBM EI = + +

    8 0.120 4.1675 25B

    BCM EI = +

    4.626 kN-mABM =

    2.525 kN-mBAM = +

    2.525 kN-mBCM =

    5.229 kN-mCBM = +

    2.525 2.5254.626 5.229

    2100 kN-mEI =

  • 37

    5. FBD shear diagram moment diagram 4.626 kN-mABM = 2.525 kN-mBAM = + 2.525 kN-mBCM = 5.229 kN-mCBM = +

    1 m 2 m 5 m

    38

    39

    6. elastic curve

    40

    7. D

    0;DM =1 2.741 0.372(0.372)2 3D

    MEI

    = =

    1 4.626 2(0.628) 0.372 (0.628)2 3EI

    + 1.085EI

    = 0.011 m=

    D conjugate beam D'

  • 41

    2-3 slope-deflection shear moment diagram elastic curve EI = 10,000 kN-m2

    DOF = 7 A, B, C, D, E, F F EF A DOF 4 B, C, D E

    degree of Indeterminacy = 6-3 = 3

    42

    1. fixed-end moment A

    2

    (FEM)8BAwL

    = +23(3) 3.375 kN-m

    8= =

    FEMBA FEMBC

    2

    (FEM)12BCwL

    = 23(3) 2.25 kN-m

    12= =

    FEMCB

    (FEM) 2.25 kN-mCB =

    FEMCD

    (FEM)8CDPL

    = 8(3) 3.0 kN-m8

    = =

    FEMDC

    (FEM) 3.0 kN-mDC =

    FEMDE

    2

    (FEM)12DEwL

    = 23(3) 2.25 kN-m

    12= =

    FEMED

    (FEM) 2.25 kN-mED =

    43

    2. slope-deflection2 2 3 (FEM)AB A B AB

    IM EL L

    = + + 2 2 3 (FEM)BA B A BA

    IM EL L

    = + +

    MBA MBC MCB

    3 ( 0) 3.375ABBA BAB

    IM EL

    = + 3.375BEI= +

    2 (2 0) 2.25BCBC B CBC

    IM EL

    = + 4 2 2.253 3B CEI EI = +

    2 (2 0) 2.25BCCB C BBC

    IM EL

    = + +

    MCD

    2 (2 0) 3.0CDCD C DCD

    IM EL

    = +

    7.5 kN-m

    2 4 2.253 3B CEI EI = + +

    4 2 3.03 3C DEI EI = +

    44

    MBA MBC MCB MCD

    4 2 3.03 3CD C D

    M EI EI = +

    MDC

    2 (2 0) 3.0CDDC D CCD

    IM EL

    = + +

    MDE

    4 2 2.253 3DE D E

    M EI EI = +

    MED

    2 4 2.253 3ED D E

    M EI EI = + +

    7.5 kN-m

    2 4 3.03 3C DEI EI = + +

  • 45

    3.

    MBA MBC MCB MCD

    B

    C

    0BA BCM M+ =

    7 2 1.1253 3B CEI EI + = 3.375BA BM EI= +

    4 2 2.253 3BC B C

    M EI EI = +

    0CB CDM M+ =

    2 8 2 0.753 3 3B C DEI EI EI + + =

    2 4 2.253 3CB B C

    M EI EI = + +

    4 2 33 3CD C D

    M EI EI = +

    7.5 kN-m

    46

    MDC MDE MED

    D

    E

    7.5 kN-m

    0DC DEM M+ =2 4 33 3DC C D

    M EI EI = + +

    4 2 2.253 3DE D E

    M EI EI = +

    2 8 2 0.753 3 3C D EEI EI EI + + =

    7.5 0EDM =

    2 4 5.253 3D EEI EI + =2 4 2.25

    3 3ED D EM EI EI = + +

    47

    simultaneous equation

    0.7366B EI

    = 0.8906C EI =

    1.7009D EI

    = 4.7879E EI =

    7 2 0 0 1.1253 3B CEI EI + + + =

    2 8 2 0 0.753 3 3B C DEI EI EI + + + =

    2 8 20 0.753 3 3C D EEI EI EI + + + =

    2 40 0 5.253 3D EEI EI + + + =

    57.366(10 ) radianB=

    58.906(10 ) radianC=

    517.009(10 ) radianD=

    547.879(10 ) radianE=

    210000 kN-mEI =

    48

    4. 57.366(10 ) radianB

    = 58.906(10 ) radianC=

    517.009(10 ) radianD= 547.879(10 ) radianE

    =

    3.375BA BM EI= +

    4 2 2.253 3BC B C

    M EI EI = + 2 4 2.253 3CB B C

    M EI EI = + +

    4 2 33 3CD C D

    M EI EI = +

    2.638 kN-mBAM = 2.638 kN-mBCM = 2.946 kN-mCBM =

    2.946 kN-mCDM = 1.326 kN-mDCM =

    2 4 33 3DC C D

    M EI EI = + +

    4 2 2.253 3DE D E

    M EI EI = + 2 4 2.253 3ED D E

    M EI EI = + +

    1.326 kN-mDEM = 7.50 kN-mEDM =

    MBA MBC MCB MCD MDC MDE MED7.5 kN-m

  • 49

    5. shear diagram moment diagram

    50

    6. elastic curve

    F ?

    51

    2.4 Frames (No Sidesway)

    52

    /

  • 53

    2-5 slope-deflection shear moment diagram elastic curve D 10 mm IAB = 0.0008 m4 IBC = IBD = 0.0006 m4

    DOF = 2 B C degree of Indeterminacy = 8-3 = 5

    54

    1. fixed-end moment

    FEMAB BC BD FEM why?

    FEMBA

    2

    (FEM)12 8ABwL PL

    = 215(4) 15(4)

    12 8= 27.5 kN-m=

    (FEM) 27.5 kN-mBA = +

    55

    2. slope-deflection2 2 3 (FEM)AB A B AB

    IM EL L

    = + + 2 2 3 (FEM)BA B A BA

    IM EL L

    = + +

    radian4AB BAL L

    = = +

    radian3BC CBL L

    = =

    0.0008 0.0102 2 3 27.54 4AB A B

    M E = +

    MAB MBA

    [ ]0.0004 0.0075 27.5BE = 0.0008 0.0102 2 3 27.5

    4 4BA B AM E = + +

    MBC

    0.0006 0.0103 ( ) 03 3B

    E = + 3 ( ) (FEM)BCBC B BC

    BC BC

    IM EL L

    = +

    0.0006 ( 0.003333)BE = +

    [ ]0.0004 2 0.0075 27.5BE = +

    56

    MAB

    MBA MBC

    MBD

    MDB

    0.0006 02 2 03 3BD B D

    M E = + 0.0008 BE=

    0.0006 02 2 03 3DB D B

    M E = +

    0.0004 BE=

  • 57

    3.

    MAB

    MBA MBC

    MBD

    MDB

    B

    0BA BC BDM M M+ + =

    [ ] 5.270075.020004.0 += BBA EM )003333.0(0006.0 += BBC EM

    BBD EM 0008.0=

    3172.5

    440(10 )B = 30.39205(10 ) radian=

    58

    4. 30.39205(10 ) radianB

    =

    [ ]0.0004 2 0.0075 27.5BA BM E = +0.0006 ( 0.003333)BC BM E = +0.0008BD BM E=0.0004DB BM E=

    [ ]0.0004 0.0075 27.5AB BM E =

    596.136 kN-mABM =

    509.772 kN-mBAM =

    447.046 kN-mBCM =

    62.726 kN-mBDM =31.363 kN-mDBM =

    MAB

    MBA MBC

    MBD

    MDB

    59

    5. shear diagram moment diagram

    596.136

    509.772 447.046

    62.726

    31.363

    60

    2.5 Frames (Sidesway) /

  • 61

    DOF

    31 2

    sin sin(90 ) sin90o o

    = =

    1 2 3tan sin = =

    62

    31 2

    2 1 2 1sin sin( ) sin

    = =+

    63

    DOF

    31 2 4

    1 2 2 1sin( ) sin(90 ) sin(90 )o o +

    = =+

    31 2 4

    1 2 2 1sin( ) cos cos +

    = =+

    64

    31 2 4

    1 2 2 1sin( ) sin(90 ) sin(90 )o o

    = =+

    31 2 4

    1 2 2 1sin( ) cos cos

    = =+

  • 65 66

    2-6 slope-deflection shear moment diagram elastic curve EI

    DOF () = 3 C, D degree of indeterminacy = 6-3 = 3

    67

    1. fixed-end moment

    FEMCD

    2

    2(FEM)CDPabL

    =

    2

    2

    40(3)4 39.184 kN-m7

    = =

    FEMDC

    2

    2(FEM)DCPbaL

    = +

    2

    2

    40(4)3 29.388 kN-m7

    = + = +

    68

    2. slope-deflection2 2 3 (FEM)AB A B AB

    IM EL L

    = + +

    2 2 3 (FEM)BA B A BAIM EL L

    = + +

    MAC

    radian5AC CA BD DBL L L L

    = = = =

    2 2 35 5AC A CEIM = +

    2 65 25C

    EI = +

    2 2 35 5CA C AEIM = +

    4 65 25C

    EI = +

    MCA

    2 2 35 5BD B DEIM = +

    2 65 25D

    EI = +

    MBD

    MDB

    2 2 35 5DB D BEIM = +

    4 65 25D

    EI = +

  • 69

    MCD MDC

    MAC

    MCA

    MBD

    MDB

    [ ]2 2 39.1847CD C DEIM = +

    4 2[ ] 39.1847 7C D

    EI = +

    [ ]2 2 29.3887DC D CEIM = + +

    2 4[ ] 29.3887 7C D

    EI = + +

    70

    3.

    MCD MDC

    MAC

    MCA

    MBD

    MDB

    C

    D

    4 65 25CA C

    M EI = + 2 65 25AC C

    M EI = +

    2 65 25BD D

    M EI = + 4 65 25DB D

    M EI = + 4 2[ ] 39.1847 7CD C D

    M EI = +

    2 4[ ] 29.3887 7DC C D

    M EI = + +

    0CA CDM M+ =48 2 6 39.18435 7 25C D

    EI + + =

    0DB DCM M+ =

    2 48 6 29.3887 35 25C D

    EI + + =

    71

    0A BV V+ =

    VA VB FBD C

    5AC CA

    AM MV +=

    D

    5BD DB

    BM MV +=

    0AC CA BD DBM M M M+ + + =6 6 24 05 5 25C D

    EI + + = 72

    simultaneous equation48 2 6 39.18435 7 25C D

    EI + + = 2 48 6 29.3887 35 25C D

    EI + + = 6 6 24 05 5 25C D

    EI + + =

    36.212CEI =

    26.946DEI =

    11.583EI =

  • 73

    4. MCD MDC

    MAC

    MCA

    MBD

    MDB

    36.212CEI =26.946DEI =

    11.583EI = 4 65 25CA C

    M EI = + 2 65 25AC C

    M EI = +

    2 65 25BD D

    M EI = + 4 65 25DB D

    M EI = + 4 2[ ] 39.1847 7CD C D

    M EI = +

    2 4[ ] 29.3887 7DC C D

    M EI = + +

    11.705 kN-mACM =

    26.190 kN-mCAM =13.558 kN-mBDM =

    24.337 kN-mDBM = 26.190 kN-mCDM =

    24.337 kN-mDCM =

    26.190 24.337

    11.705

    26.190

    13.558

    24.337

    74

    5. FBD shear diagram moment diagram

    75

    axial-force diagram frame ???

    (axial-force diagram) ??76

    6. elastic curve 11.583EI =

  • 77

    2-7 slope-deflection shear moment diagram elastic curve EI

    degree of indeterminacy = (6-1)-3 = 2 DOF = 4

    B

    B

    ( )C CB ( )C CD

    joint C internal hinge DOF 2

    78

    1. fixed-end moment 6 kN internal hinge C FEM

    2. slope-deflection

    radian4AB BAL L

    = = +

    radian2CD DCL L

    = = +

    79

    2 2 3 (FEM)AB A B ABIM EL L

    = + + 2 2 3 (FEM)BA B A BA

    IM EL L

    = + +

    MAB

    MBA

    MDC

    MBC radian4AB BAL L

    = = +

    radian2CD DCL L

    = = +

    22 2 34 4AB A BIM E = +

    [ ]2 0.75BEI = 23 ( 0)3BC BIM E =

    3 02 2DCIM E =

    [ 0.75 ]BEI =

    2 BEI=

    0.75EI=

    22 2 34 4BA B AIM E = +

    80

    3.

    B

    MAB

    MBA

    MDC

    MBC

    [ 0.75 ]AB BM EI =

    [ ]2 0.75BA BM EI = 2BC BM EI=

    0.75DCM EI=

    0BA BCM M+ =

    316B

    =

  • 81

    VA VB FBD B

    C

    6A DV V+ =

    4AB BA

    AM MV += (3 1.5 )4 B

    EI =

    2DC

    DMV = 3

    8EI=

    3(3 1.5 ) 64 8BEI EI + =

    82

    3(3 1.5 ) 64 8BEI EI + =

    316B

    =

    12813

    EI =

    3 12816 13B

    EI = 2413

    =

    4. 5.54 kN-mABM =

    3.69 kN-mBAM =

    3.69 kN-mBCM =

    7.39 kN-mDCM =

    2.31 kNAV =

    3.69 kNDV =

    [ 0.75 ]AB BM EI =

    [ ]2 0.75BA BM EI = 2BC BM EI=

    0.75DCM EI=

    (3 1.5 )4A BEIV =

    2DC

    DMV =

    MAB

    MBA

    MDC

    MBC

    83

    5.54 kN-m

    3.69 kN-m

    7.39 kN-m

    3.69 kN-m

    5.54 kN-mABM =

    3.69 kN-mBAM =

    3.69 kN-mBCM =

    7.39 kN-mDCM =

    2.31 kN

    2.31 kN

    1.23 kN

    1.23 kN

    3.69 kN

    3.69 kN

    1.23 kN

    1.23 kN

    1.23 kN

    2.31 kN 2.31 kN

    2.31 kN

    3.69 kN

    84

  • 85

    5. FBD shear diagram moment diagram

    (axial-force diagram) ??86

    6. elastic curve

    87

    End of Chapter 2