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Chapter 9 Torsion of Thin-Walled Tubes Summary of Saint-Venant Torsion Theory Warping function, Ψ shear stress ( ) xy G z y Ψ τ = β , ( ) xz G y z ∂Ψ τ = β + compatibility relationship automatic equilibrium equations 2 0 Ψ= boundary conditions 0 xy xz dz dy ds ds τ −τ = 2 2 1 ( ) 2 dz dy d y z y ds z ds ds Ψ ∂Ψ = + torque ( ) xz xy A T y z dA = τ −τ ∫∫ ( ) A T G y z dA G J z y Ψ ∂Ψ = β + β ∫∫ Prandtl stress function, φ shear stress xy z φ τ = , xz y φ τ =− compatibility relationship 2 2 G ∇φ =− β equilibrium equations automatic boundary conditions 0 d φ = torque 2 A T dA = φ ∫∫
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Transcript of Chapter 9, Torsion of Thin-Walled Tubes - ase.uc.edupnagy/ClassNotes/AEEM6001 Advanced Strength...

  • Chapter 9 Torsion of Thin-Walled Tubes

    Summary of Saint-Venant Torsion Theory

    Warping function,

    shear stress ( )xy G zy

    =

    , ( )xz G yz

    = +

    compatibility relationship automatic

    equilibrium equations 2 0 =

    boundary conditions 0xy xzdz dyds ds

    =

    2 21 ( )2

    dz dy d y zy ds z ds ds

    = +

    torque ( )xz xy

    AT y z dA=

    ( )A

    T G y z dA G Jz y

    = +

    Prandtl stress function,

    shear stress xy z

    =

    , xz y

    =

    compatibility relationship 2 2G =

    equilibrium equations automatic

    boundary conditions 0d =

    torque 2

    AT dA=

  • Membrane Analogy

    y

    z

    S dy

    S dz

    yx

    S dzS dz

    p dy dzy

    dyy y y

    +

    (y,z)

    y

    z

    S dy

    S dz

    yx

    S dzS dz

    p dy dzy

    dyy y y

    +

    (y,z)

    sin sin

    sin sin 0

    xdF S dz S dz dyy y y y

    S dy S dy dz p dy dzz z z z

    = + +

    + + + =

    2 2

    2 2 0S dz dy S dy dz p dy dzy z

    + + =

    2 2

    22 2

    pSy z

    = + =

    compatibility relationship:

    2 2G =

    boundary condition: 0d =

  • Examples of Solid Cross Sections

    elliptical cross section rectangular cross section

    triangular cross section I-beam cross section

  • Discontinuous Boundary

    y

    z

    yx (y,z)

    = 0

    = 1p

    y

    z

    yx (y,z)

    = 0

    = 1p

    elliptical cross section with hole rectangular cross section with hole

  • Narrow Rectangular Cross Section

    b >> t

    b

    t y

    z

    b

    t y

    z

    equilibrium in x-direction

    2

    2 0xdF S dy dz p dy dzz

    = + =

    2

    2pSz

    =

    0p zz S

    = +

    2

    2p z CS

    = +

    2( ) 0

    2 8t pz t C

    S = = + =

    2

    8pC tS

    =

    2

    22 4p t zS

    =

    22

    4tG z

    =

  • torque

    2/ 2 2

    / 22 2

    4

    t

    A z t

    tT dA G b z dz=

    = =

    / 22 3

    / 2

    24 3

    t

    t

    t z zT G b

    =

    313

    T G bt=

    shear stress

    2xy G zz

    = =

    , 0xz y

    =

    max 23TG tbt

    = =

    torsional stiffness

    ( )tTK JG

    = =

    31

    3K bt=

    torque contributions

    bulk end( )xy xzA

    T z y dA T T= + = +

    / 2 2

    bulk/ 2

    2t

    xyA z t

    T z dA G b z dz=

    = =

    3bulk

    1 16 2

    T G bt T= =

  • End Correction

    h

    y = b/2 y = +b/2

    hh

    y = b/2 y = +b/2

    h

    1D approximation:

    22 if and 0 else

    4 2 2t b bG z y

    = =

    0 if and if2 2 2xzb b by y

    y y

    = = < < = =

    2D tapered approximation:

    2 22 2 2if and else

    4 2 2 4

    b yt b b tG z h y h G zh

    = + =

    end xzdT y dA=

    2/ 2 / 2

    2end

    / 2 / 24 2

    t bxz

    A z t b h

    t b bT y dA G z dz y dyh=

    =

    3end

    1 16 2

    T G bt T= =

  • General Thin-Walled Shapes

    TKG

    =

    R

    t

    R

    t

    3 31 23 3

    K bt R t=

    b2

    b 1

    t1

    t 2

    bt = t(s)

    s

    b2

    b 1

    t1

    t 2

    b2

    b 1

    t1

    t 2

    bt = t(s)

    s

    bt = t(s)

    s

    31

    13

    ni i

    iK b t

    == 3

    0

    1 ( )3

    bK t s ds=

  • Single-Cell Thin-Walled Tube

    y

    L

    zx

    T

    A

    D

    B

    C

    t

    1 2A

    D

    B

    C

    dx

    t1

    t2t

    P

    d

    ds

    q ds

    r

    y

    L

    zx

    T

    A

    D

    B

    C

    ty

    L

    zx

    T

    A

    D

    B

    C

    t

    1 2A

    D

    B

    C

    dx

    t1

    t2

    1 2A

    D

    B

    C

    dx

    t1t1

    t2t2t

    P

    d

    ds

    q ds

    rt

    P

    d

    ds

    q ds

    r

    1 1 2 2 0xdF t dx t dx= + =

    constant shear flow q

    1 1 2 2q t t= =

    twisting moment T

    dT dF r t ds r q ds r= = = 12

    d ds r =

    2dT q d=

    2T q=

    2Tq =

    surf( ) 2T

    t = =

  • Rate of Twist, Torsional Stiffness

    total strain energy over length L

    20

    12 2V

    W T L U U dV Lt dsG

    = = = =

    2

    212 8

    T L dsT LtG

    =

    Bredts formula

    24T ds

    tG =

    2q dsG t

    =

    Example: closed tube of circular cross section:

    R

    t

    R

    t

    closed24

    TG t

    =

    , 2 R= , 2R =

    closed32

    TG R t

    =

    closed 32K R t=

  • Warping Displacement

    z

    y

    x, uT

    T

    Q

    axis of twist

    z

    y

    x, uT

    T

    Qaxis of twist

    closed cell open cell

    z

    y

    x, uT

    T

    Q

    axis of twist

    z

    y

    x, uT

    T

    Qaxis of twist

    closed cell open cell

    closed 32K R t=

    open 3 closed23

    K Rt K=

  • Multicell Tubes

    1 2 0axial wF q L q L q L= + + = 1 2wq q q=

    1 2 w

    CBA ADC ACT q r ds q r ds q r ds= + +

    1 1 2 22( ) 2( ) 2w w w wT q q q= + +

    1 1 2 22 2T q q= +

    For general n-cell cross section:

    12

    ni i

    iT q

    ==

    shear flow qi in the ith loop:

    12 i ii i

    qG dst

    =

  • Restraint of Warping

    Example: I Sections

    symmetric loading u(x = 0, y, z) = 0

    top view of the upper flange with differential element

    axial view

    ( )x =

  • bending of the upper flange: 2

    2f fvM E I

    x

    =

    3

    12f

    ft b

    I =

    2hv =

    2

    21 12 2f f f

    M E I h E I hxx

    = =

    2

    212

    ff f

    MV E I h

    x x

    = =

    torque due to shear force in the flanges:

    f fT V h=

    total torque:

    2

    22

    12f f

    T V h K G E I h K Gx

    = + = +

    3 22

    2 24f

    ft b hhJ I = =

    (called the sectorial area of inertia)

    2

    2T E J K Gx

    = +

    2

    2K G TE J E Jx

    =

    2 2

    22

    k TkG Kx

    =

    where 2

    2(1 )K G KkE J J

    = =+

    1 2( ) sinh coshTx C kx C kx G K = + +

  • boundary conditions:

    ( 0) 0x = = (no warping at the center)

    0x Lx =

    =

    (no bending moment at the end)

    2TC G K= and 1 tanh

    TC kLG K=

    ( )( ) tanh sinh cosh 1Tx kL kx kxG K = +

    surf( ) G t = =

    0

    tanh( ) ( ) 1 ( )L T L kL T LL x dx kLG K kL G K

    = = =

    0

    0.2

    0.4

    0.6

    0.8

    1

    0 10 20 30 40 50Normalized Length, kL

    Nor

    mal

    ized

    Tw

    ist, (

    kL)

    0

    0.2

    0.4

    0.6

    0.8

    1

    0

    0.2

    0.4

    0.6

    0.8

    1

    0 10 20 30 40 500 10 20 30 40 50Normalized Length, kL

    Nor

    mal

    ized

    Tw

    ist, (

    kL)

    ( )( ) tanh cosh sinh2f

    fE I hT k

    M x kL kx kxG K=

    ( )2

    ( ) tanh sinh cosh2f

    fE I hT k

    V x kL kx kxG K=