by วิธีของ castigliano assoc. prof. dr. sittichai...

Download By วิธีของ Castigliano Assoc. Prof. Dr. Sittichai ...eng.sut.ac.th/ce/oldce/theory50/…

Post on 10-Nov-2018

213 views

Category:

Documents

0 download

Embed Size (px)

TRANSCRIPT

  • THEORY OF STRUCTURES

    By

    Assoc. Prof. Dr. Sittichai SeangatithSCHOOL OF CIVIL ENGINEERING

    INSTITUTE OF ENGINEERINGSURANAREE UNIVERSITY OF TECHNOLOGY

    7Deflection

    1. truss frame statically determinate double integration, moment-area, conjugate-beam, virtual work ( unit load), Castigliano

    Deflection statically indeterminate

    /deflection v

    /slope

  • frame

    sidesway/

    7.1 Deflection Diagram and Elastic CurveElastic curve /Deflection diagram - centroid / deflection diagram elastic curve

    moment diagram

    +-

    elastic curve1.

    support (slope) (deflection) ?

  • 0y =

    0y =0x =

    0x

    0y =0x =

    0 =

    0

    0

    elastic curve

  • elastic curve frame moment diagram sign convention

    /

    MM

    / MM

    elastic curve moment diagram

    +-

    elastic curve moment diagram

    +-

  • elastic curve moment diagram

    +-

    elastic curve frame moment diagram

    elastic curve frame moment diagram 7.2 Elastic Beam Theory ( 10 )

    homogeneous and isotropic material ( ) linear elastic

  • Transverse shear stress (shear strain)

    Curvature Differential Elementdv/dx = (slope) d2v/dx2 =

    (curvature)v =

    (deflection)

    2

    21 M d v

    EI dx= =

    flexural rigidity

    curvature differential element1 d

    ds

    =

    slope neutral axistan dv

    dx=

    arctan dvdx

    =

    s x

    1 d dxdx ds

    = (arctan )d dv dx

    dx dx ds=

    2 2 + ds dx dv=

    1/ 221 ( )ds dv

    dx dx = +

    1/ 22

    1

    1 ( )

    dxds dv

    dx

    = +

    calculus,

    (arctan )d dvdx dx

    2

    2

    21 ( )

    d vdx

    dvdx

    = +

  • (curvature) differential element

    1 (arctan )d dv dxdx dx ds

    =

    2

    2

    2 3/ 2

    1

    [1 ( ) ]

    d vdxdvdx

    =

    +

    2

    2

    1/ 22 2

    1

    1 ( ) 1 ( )

    d vdx

    dv dvdx dx

    = + +

    Moment-Curvature Relationships:

    : dx ds d = =

    ( - )ds y d =

    ds dsds

    =

    1y

    =

    ( - ) -

    y d d

    d

    =

    y

    =

    isotropic and homogenous material linear elastic

    1y Ey

    = =

    flexural formula,1

    ( )My

    Ey Ey I

    = =

    flexural rigidity1 MEI

    =

    y +

    Differential Equations Deflection Curve2

    2

    2 3/ 2

    1

    [1 ( ) ]

    d vMdx

    dv EIdx

    = =

    +

    nonlinear second order differential equation elastica deflection L/240 L/360 slope 1

    2( ) 0dvdx

    2

    21 M d v

    EI dx= =

    flexural rigidity

  • 7.3 Double Integration

    2

    2 ( )d vEI M xdx

    =

    V = dM/dx 3

    3 ( )d vEI V xdx

    =

    EI v = deflection dv/dx = slope d2v/dx2 = curvature

    -w = dV/dx 4

    4 ( )d vEI w xdx

    =

    slope deflection integration integration constant of integration boundary conditions continuity conditions

    Sign Convention

    Boundary Continuity Conditions (boundary conditions) slope / deflection

    (continuity condition)

    1 2( ) ( )a a =

    1 2( ) ( )v a v a=

    1 2( ) ( )a b =

    1 2( ) ( )v a v b=

  • EXAMPLE slope B C P flexural rigidity EI

    ( ) ( )M x Px PL P L x= =

    FBD

    elastic beam2

    2d vEI Px PLdx

    =

    2

    12dv xEI P PLx Cdx

    = +

    3 2

    1 26 2x xEIv P PL C x C= + +

    2

    12dv xEI P PLx Cdx

    = +

    2

    1(0)(0) (0)

    2EI P PL C= +

    1 0C =

    3 2

    26 2x xEIv P PL C= +

    3 2

    2(0) (0)(0)

    6 2EI P PL C= +

    2 0C =

    2

    02

    dv xEI P PLxdx

    = +

    21 ( )2xP PLx

    EI =

    3 2

    06 2x xEIv P PL= +

    3 21 ( )6 2x xv P PL

    EI=

    B, x = 2a

    C, x = L3

    3CPLvEI

    =

    )(2 aLEIPa

    =

    x = 2ax = L

    slope 2

    ( )2 2

    wLx wxM x =

    elastic beam2 2

    2 2 2d v wLx wxEIdx

    =

    2 3

    14 6dv wLx wxEI Cdx

    = +

    3 4

    1 212 24wLx wxEIv C x C= + +

    EXAMPLE

  • boundary condition v = 0 x = 0

    3 4

    1 2(0) (0)(0) (0)

    12 24wL wEI C C= + +

    2 0C = = 0 x = L/2

    2 3

    1( / 2) ( / 2)(0)

    4 6wL L w LEI C= +

    3

    1 24wLC =

    2 3

    14 6dv wLx wxEI Cdx

    = +

    3 4

    1 212 24wLx wxEIv C x C= + +

    x = L/2

    2 0C =3

    1 24wLC =

    3 2 3(4 6 )24

    w x Lx LEI

    = +

    3 4 3

    12 24 24wLx wx wLEIv x = +

    4 3 3( 2 )

    24wv x Lx L xEI

    = +

    3

    max 24wL

    EI =

    4

    max5

    384wLv

    EI=

    3 4

    1 212 24wLx wxEIv C x C= + +

    2 3

    14 6dv wLx wxEI Cdx

    = +

    2 3 3

    4 6 24dv wLx wx wLEIdx

    =

    slope deflection

  • 1. FBD

    2. FBD moment x

    3. elastic beam integrate slope deflection

    4. boundary condition / continuity condition slope deflection integration

    5. integration slope deflection

    slope EI EXAMPLE

    slope A B a = 2b

    slope

    10 ;x a 1 1PbM xL

    =

    2 ;a x b 2 2 2

    2

    ( )

    (1 )

    PbM x P x aL

    xPaL

    =

    =

    elastic beam2

    112

    1

    d v PbEI xdx L

    =10 ;x a

    211 1

    1 2dv PbEI x Cdx L

    = +

    31 1 1 1 2 6

    PbEI v x C x CL

    = + +

  • 2 ;a x b 22 22 (1 )

    dv xEI Padx L

    =

    22 2

    2 3( )2dv xEI Pa x Cdx L

    = +

    2 32 2

    2 3 2 4 ( )2 6x xEI v Pa C x C

    L= + +

    integration elastic beam 4 boundary condition

    1 10; 0 :x v= =3

    1 1 1 1 2 6PbEI v x C x C

    L= + + 2 0C =

    2 2; 0 :x L v= =2 3

    3 40 ( )2 6L LPa C L C

    L= + +

    2

    3 40 3PaL C L C= + +

    continuity condition 1 1 2 2( ) ( ) :v x a v x a= = =

    2 33

    1 3 4( )6 2 6Pb a aa C a Pa C a C

    L L+ = + +

    2 32 2

    2 3 2 4 ( )2 6x xEI v Pa C x C

    L= + +

    31 1 1 1 6

    PbEI v x C xL

    = +

    1 21 2

    1 2

    ( ) ( ) :dv dvx a x adx dx

    = = = 21 1 11 2

    dv PbEI x Cdx L

    = +

    22 2

    2 3( )2dv xEI Pa x Cdx L

    = +

    22

    1 3( )2 2Pb aa C Pa a C

    L L+ = +

    2

    3 40 3PaL C L C= + +

    2 33

    1 3 4( )6 2 6Pb a aa C a Pa C a C

    L L+ = + +

    22

    1 3( )2 2Pb aa C Pa a C

    L L+ = +

    2 21 ( )6

    PbC L bL

    = 2 23 (2 )6PaC L a

    L= +

    3

    4 6PaC =

    slope 211 1

    1 2dv PbEI x Cdx L

    = +

    31 1 1 1 2 6

    PbEI v x C x CL

    = + +

    22 2

    2 3( )2dv xEI Pa x Cdx L

    = +

    2 32 2

    2 3 2 4 ( )2 6x xEI v Pa C x C

    L= + +

    2 ;a x b

    10 ;x a 2 2 211 11

    ( 3 )6

    dv Pb L b xdx EIL

    = =

    2 2 211 1( )6

    Pbxv L b xEIL

    =

    2 2 222 2 2

    2

    (3 2 6 )6

    dv Pa x L a Lxdx EIL

    = = + +

    3 2 2 2 22 2 2 2( (2 ) 3 )6

    Pav x L a x Lx a LEIL

    = + +

  • slope A B A, x1 = 0:

    B, x2 = L:

    2 21 ( ) ( )( ) ( )6 6 6

    Pb Pb PabL b L b L b L bEIL EIL EIL

    = = + = +

    2 22 ( ) ( )( ) )6 6 6

    Pa Pa PabL a L a L a L aEIL EIL EIL

    = = + = +

    a = 2b C slope

    2 2 21 1( 3 )6

    Pb L b xEIL

    = 3b

    2 2 21(3 ) 3 0b b x =

    1 1.633x b=

    3

    max 0.48385PbvEI

    =

    2 2 21 1( 3 )6

    Pb L b xEIL

    =

    7.4 Moment-Area Theorems 1 moment-area method2

    2

    d v Mdx EI

    =

    2

    2 ( )d v d dv ddx dx dx dx

    = =

    d Mdx EI=

    Md dxEI

    =

    d slope dx M/EI

    /

    B

    B AA

    M dxEI

    =

  • 2 moment-area methoddt xd=

    Md dxEI

    = integrate A B

    /

    B

    A BA

    Mt x dxEI

    =

    xd A xdA= centroid

    /

    B

    A BA

    Mt x dxEI

    =

    A elastic curve B M/EI (A B) (A)

    slope B C W200x36EXAMPLE

    6 434.4(10 ) mmI =1. moment diagram

    2. IAB = 1.5IBC M/EI diagram

    3. (elastic curve)

    4. slope B B A B/A 1

    /

    6.667 1 13.333 (2) [ ](2)2

    B A B

    EI EI

    =

    = +

    226.667 kN.mB EI

    =

    23

    6 2 6 4

    26.667kN.m 3.88(10 ) rad[200(10 ) kN/m ][34.4(10 )m ]B

    = =

  • 4. C C A C tC/A 2

    /6.667 1 13.333 4 1 10 2[ (2)](2) [ ](2)(1 ) [ ](1)( )

    2 3 2 3C A Ct

    EI EI EI= = + + +

    360.944 kN.mC EI

    =

    2

    6 2 6 4

    60.944 kN.m 0.0089 m 8.9 mm[200(10 ) kN/m ][34.4(10 )m ]C

    = = =

    W360x64 E = 200 GPa I = 176(106) mm4 conjugate beam

    EXAMPLE

    1. FBD

    2. moment diagram

    3. moment diagram elastic weight

    4. ( C) slope = 0 tan C

    /

    2A Bt =

    //2

    A BC C B

    t t =

    2

    /160 8 28 53.33 213.33( ) (6)

    3 33520

    A Bt EI EI EI

    EI

    = + + +

    =

    /53.33 2 213.33 160 10( ) (1) ( )2 3 2 3

    657.78

    C Bt EI EI EI

    EI

    = + +

    =

    3520 657.78 1102.222C EI EI EI

    = =

  • 3

    max 6 2 6 12 4

    1102.2 kN.m 0.0308 m 30.8 mm[200(10 )kN/m ][179(10 )(10 )m ]

    = = =

    12000 389.6 36030.8

    L= = >

    span linear elastic

    40.2 mm12000 298.5 36040.2

    L= =

Recommended

View more >