8 CHS Joints 2011

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    Hollow sections in structural applications Course

    Wardenier part 8 CHS Joints page

    Wardenier Tubular Structure Course - 2011

    Tubular Structures

    Circular Hollow Section Joints

    em. Prof.dr.ir. J. Wardenier

    Delft University of Technology

    National University of Singapore

    Wardenier Tubular Structure Course - 2011

    Now many Applications

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    Hollow sections in structural applications Course

    Wardenier part 8 CHS Joints page

    Wardenier Tubular Structure Course - 2011

    Basic Types of welded Joints

    Wardenier Tubular Structure Course - 2011

    Symbols used (e.g. for K Joints)d

    b

    h

    t

    e

    g

    01

    2

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    Hollow sections in structural applications Course

    Wardenier part 8 CHS Joints page

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    Joint parameters

    0

    1

    d

    d=

    0

    21

    2d

    dd +=

    for T and X

    for K-joint

    0

    02t

    d=

    0

    'tgg =

    00

    'y

    op

    fANn

    =

    0

    0

    yfn =

    Wardenier Tubular Structure Course - 2011

    Failure modes (basic)

    Chord punching shear

    Chord plastification

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    Wardenier part 8 CHS Joints page

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    Further possible failure modes (additional)

    Chord shear failure

    Brace effective width

    Local Buckling in chord or brace

    Wardenier Tubular Structure Course - 2011

    Analytical models

    Ring model

    Punching shear model

    Shear failure of the chord

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    Hollow sections in structural applications Course

    Wardenier part 8 CHS Joints page

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    Ring model (elastic stress distribution)

    1,Ed

    joint joint

    N1

    N1

    1,Ed

    joint joint

    N1

    N1

    Wardenier Tubular Structure Course - 2011

    Ring model (assume line loads)

    Be

    2,5 to 3d0

    11 sin

    2

    N

    11 sin

    2N

    N1

    N1

    c1d11

    1 sin2

    Ne

    11

    B2

    sinN

    Be

    2,5 to 3d0

    11 sin

    2

    N

    11 sin

    2N

    N1

    N1

    c1d11

    1 sin2

    Ne

    11

    B2

    sinN

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    Ring model (plastic hinges)

    0y20p ft

    4

    1m =

    Wardenier Tubular Structure Course - 2011

    Ring model

    =

    2

    dc

    2

    d

    2

    sinNBm2 11011ep

    1

    0y20

    1

    0e1

    sin

    ft

    )c1(

    d/B2N

    =

    C0

    Qf= f(n)f

    1

    0y20

    1

    01 Q

    sin

    ft

    )c1(

    cN

    =

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    Wardenier part 8 CHS Joints page

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    Ring model

    some effects have been neglected, e.g.:

    - influence axial and shear force on mp

    and

    - chord stress effect

    Wardenier Tubular Structure Course - 2011

    Effect chord loading (be careful different def. in codes)

    Function f(n) = Qfbased on n (IIW 2009)

    (In Eurocode based on n)

    0,0

    0,2

    0,4

    0,6

    0,8

    1,0

    1,2

    -1,0 -0,8 -0,6 -0,4 -0,2 0,0 0,2 0,4 0,6 0,8 1,0

    2 = 63,5

    2 = 25,4

    2 = 63,5

    2 = 50,8

    2 = 25,4

    N1usin1

    /

    (fy0

    t02Q

    u)

    N0/ Npl,0

    N0

    N2

    N0p

    A

    A

    1

    N1

    2N0

    N2

    N0p

    A

    A

    1

    N1

    2

    p011gap,0

    2

    1i

    p0ii0

    NcosNN

    NcosNN

    +=

    +==

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    Wardenier part 8 CHS Joints page

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    Punching shear model

    N1sin 1 = d1t0. f(ellips )vp with vp = 0,58fy0

    Assumption:

    at failure about uniform

    punching shear yield

    stress distribution

    (check by experiments)

    validity !

    Wardenier Tubular Structure Course - 2011

    f(ellips )

    Punching shear model

    12

    10y011

    sin2

    sin1ftd58,0N

    +=

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    Chord shear model

    Based on yield criterion of Huber-Hencky-Von Mises

    Ngap,0Ngap,0

    Wardenier Tubular Structure Course - 2011

    Chord shear model

    )f58,0(A2

    3

    fAV 0y0

    0y

    v0,pl

    ==

    Npl,0 = A0 fy0 = (d0 - t0) t0 fy0

    The shear load capacity is given by:

    The axial load capacity is given by:

    )f58,0(A2

    3

    fAV 0y0

    0y

    v0,pl

    ==

    The shear load capacity is given by:

    The axial load capacity is given by:

    )f58,0(A2

    3

    fAV 0y0

    0y

    v0,pl

    ==

    The shear load capacity is given by:

    Npl,0 = A0 fy0 = (d0 - t0) t0 fy0

    The axial load capacity is given by:

    )f58,0(A2

    3

    fAV 0y0

    0y

    v0,pl

    ==

    The shear load capacity is given by:

    )f58,0(A2

    3

    fAV 0y0

    0y

    v0,pl

    ==

    The shear load capacity is given by:

    The axial load capacity is given by:

    )f58,0(A2

    3

    fAV 0y0

    0y

    v0,pl

    ==

    The shear load capacity is given by:

    Npl,0 = A0 fy0 = (d0 - t0) t0 fy0

    The axial load capacity is given by:

    )f58,0(A2

    3

    fAV 0y0

    0y

    v0,pl

    ==

    The shear load capacity is given by:

    Ngap,0

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    Hollow sections in structural applications Course

    Wardenier part 8 CHS Joints page

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    Chord shear model

    0,1N

    N

    V

    sinN2

    0,pl

    0,gap

    2

    0,pl

    ii

    +

    2

    v0y

    ii

    0y00,gap Af58,0

    sinN

    1fAN

    )f58,0(A2

    3

    fAV 0y0

    0y

    v0,pl

    ==

    Npl,0 = A0 fy0 = (d0 - t0) t0 fy0

    The shear load capacity is given by:

    The axial load capacity is given by:

    or

    2

    v0y

    ii

    0y00,gap Af58,0

    sinN

    1fAN

    or

    0,1N

    N

    V

    sinN2

    0,pl

    0,gap

    2

    0,pl

    ii

    +

    2

    v0y

    ii

    0y00,gap Af58,0

    sinN

    1fAN

    or

    Ngap,0

    Wardenier Tubular Structure Course - 2011

    Experiments

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    Finite element simulation

    0

    100

    200

    300

    400

    500

    0 20 40 60 80 100

    Ovalisation [mm]

    Load[kN]

    EX-03 - Collar, Compression

    Experiment = 0.54

    Numerical 2= 50.6

    Calibration with experiments

    (Choo/Van de Vegte)

    Wardenier Tubular Structure Course - 2011

    Verification with experiments

    For example:

    X-joints

    analysis for

    IIW (2009)

    0

    2

    4

    6

    8

    0,0 0,2 0,4 0,6 0,8 1,0

    f(N1u

    )Xj

    oints

    +

    7,01

    1

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    Evaluation to design rules

    Analytical models

    ExperimentsStrength formulae

    Scatter in test results

    scatter in mechanical properties

    tolerances in dimensions

    tolerances in fabrication

    failure mode

    Design strength

    N1,Rd =Nk/m

    Wardenier Tubular Structure Course - 2011

    Strength formulae for CHS joints(IIW 2009 = Draft ISO = CIDECT)

    or

    f

    1

    200y

    Rd,1 Qsin

    tf)'g(f)(f)(fN

    =

    or

    i

    2

    0y0fuRdi,

    sintfQQN =

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    Wardenier part 8 CHS Joints page

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    Criterion

    to be checked

    Axially loaded joints with CHS

    Braces and Chord

    Design strength:

    chord plastification

    Design strength:

    chord punching shear

    (only for di d0 2t0)

    Limit states criteria IIW (2009)

    i

    20y0

    fuRdi,sin

    tfQQN =

    12

    101y0Rd1,

    sin2

    sin1tdf0,58=N

    +

    i

    20y0

    fuRdi,sin

    tfQQN =

    12

    101y0Rd1,

    sin2

    sin1tdf0,58=N

    +

    Criterion

    to be checked

    Axially loaded joints with CHS

    Braces and Chord

    Design strength:

    chord plastification

    Design strength:

    chord punching shear

    (only for di d0 2t0)

    i

    20y0

    fuRdi,sin

    tfQQN =

    12

    101y0Rd1,

    sin2

    sin1tdf0,58=N

    +

    Wardenier Tubular Structure Course - 2011

    Function Qu

    Notes: 1) When cos1 >, check shear

    ( )8.612.6Q 0.22u +=

    ]

    )tg(1.2

    1[1)8(165.1Q

    0.8

    0

    0.31.6u

    +

    ++=

    1

    N1

    t1

    d1

    N1

    d0 t0

    0.15u

    0.7 -1

    12.6Q

    +=

    1

    N1

    t1

    d1d

    0 t0

    2

    N2

    t2

    d2N1

    1

    t1

    d1g

    +e

    d0

    N0

    X joints 1)

    T and Y joints

    K gap joints

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    Wardenier part 8 CHS Joints page

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    Function Qf

    Compression tension

    T, Y, X joints C1 = 0,45 0,25 C1 = 0,20

    K gap joints C1 = 0,25

    Qfequations X, T and K gap joints (IIW 2009)

    0,pl

    0

    0,pl

    0

    M

    M

    N

    Nn +=( ) 1Cf |n|1Q =

    Wardenier Tubular Structure Course - 2011

    Range of validity based on:

    - test evidence

    - validity of formulae

    - limitation of failure criteria (e.g. avoid local buckling)

    - deformation- deformation capacity

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    Comparison Qu for T, X and K gap joints for

    2=25 (IIW, 2008)

    0

    10

    20

    30

    40

    0 0.2 0.4 0.6 0.8 1

    Qu

    K gap jointg'=2

    K gap jointg'=infinite

    T joint

    X joint

    g'=g/t0

    Relation Qu between joint capacities

    9

    Comparison Qu for T, X and K gap joints for

    2=25 (IIW, 2008)

    0

    10

    20

    30

    40

    0 0.2 0.4 0.6 0.8 1

    Qu

    K gap jointg'=2

    K gap jointg'=infinite

    T joint

    X joint

    g'=g/t0

    9

    Wardenier Tubular Structure Course - 2011

    Overlap joints IIW (2009)

    Brace effective width (1)

    Chord M-N yield (2)

    Brace shear (3)

    brace i = overlapping member; brace j = overlapped member

    di

    d0

    dj

    ti

    t0

    Ni

    Nop

    Nj

    No

    ij

    4.8

    4.11

    4.10

    4.11

    4.10

    tj

    brace i =overlapping member; brace j = overlapped member

    brace i = overlapping member; brace j = overlapped member

    di

    d0

    dj

    ti

    t0

    Ni

    Nop

    Nj

    No

    ij

    4.8

    4.11

    4.10

    4.11

    4.10

    tj

    brace i = overlapping member; brace j = overlapped member

    (2) (2)

    (1)

    (3)

    N0

    brace i = overlapping member; brace j = overlapped member

    di

    d0

    dj

    ti

    t0

    Ni

    Nop

    Nj

    No

    ij

    4.8

    4.11

    4.10

    4.11

    4.10

    tj

    brace i =overlapping member; brace j = overlapped member

    brace i = overlapping member; brace j = overlapped member

    di

    d0

    dj

    ti

    t0

    Ni

    Nop

    Nj

    No

    ij

    4.8

    4.11

    4.10

    4.11

    4.10

    tj

    brace i = overlapping member; brace j = overlapped member

    (2) (2)

    (1)

    (3)

    brace i = overlapping member; brace j = overlapped member

    di

    d0

    dj

    ti

    t0

    Ni

    Nop

    Nj

    No

    ij

    4.8

    4.11

    4.10

    4.11

    4.10

    tj

    brace i = overlapping member; brace j = overlapped member

    di

    d0

    dj

    ti

    t0

    Ni

    Nop

    Nj

    No

    ij

    4.8

    4.11

    4.10

    4.11

    4.10

    tj

    brace i =overlapping member; brace j = overlapped member

    brace i = overlapping member; brace j = overlapped member

    di

    d0

    dj

    ti

    t0

    Ni

    Nop

    Nj

    No

    ij

    4.8

    4.11

    4.10

    4.11

    4.10

    tj

    brace i = overlapping member; brace j = overlapped member

    (2) (2)

    (1)

    (3)

    N0

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    Relation CHS vs RHS overlap Joints

    IIW (2009)

    - criteria for RHS overlap joints multiplied by /4

    (ratio between the cross sectional areas for d=b)

    - all b and h dimensions in formulae replaced by d.

    Wardenier Tubular Structure Course - 2011

    Simplification

    Ce in graph

    i

    f

    iyi

    0y0

    e

    yii

    Rdi,

    sin

    Q

    tf

    tfC

    fA

    Neff. ==

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    Design graphs for X joints

    X joint efficiency

    0,0

    0,1

    0,2

    0,3

    0,4

    0,5

    0,6

    0,7

    0,8

    0,9

    1,0

    0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1

    effciencyCX

    2=10

    2=15

    2=20

    2=30

    2=40

    1

    f

    11y

    00yX

    1y1

    Rd,1

    s in

    Q

    tf

    tfC

    fA

    N

    =

    Wardenier Tubular Structure Course - 2011

    Design graph for T joints

    T joint efficiency

    0,0

    0,10,2

    0,3

    0,4

    0,5

    0,6

    0,7

    0,8

    0,9

    1,0

    0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1

    effciencyCT

    2=10

    2=15

    2=20

    2=30

    2=40

    2=50

    1

    f

    11y

    00yT

    1y1

    Rd,1

    s in

    Q

    tf

    tfC

    fA

    N

    =

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    Design graphs for K gap joints

    Compare the strength for various d0/t0 ratios and various gaps

    1

    f

    11y

    00yK

    1y1

    *1

    sin

    Q

    tf

    tfC

    fA

    N

    =

    K gap joint efficiency g'=2

    0,00,1

    0,2

    0,3

    0,4

    0,5

    0,6

    0,7

    0,8

    0,9

    1,0

    0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1

    effciencyCK

    2=10

    2=15

    2=20

    2=30

    2=40

    2=50

    i

    21

    i

    f

    iyi

    00yK

    yii

    Rd,i

    d2

    dd

    sin

    Q

    tf

    tfC

    fA

    N +

    =

    1

    f

    11y

    00yK

    1y1

    *1

    sin

    Q

    tf

    tfC

    fA

    N

    =

    K gap joint efficiency g'=10

    0,0

    0,1

    0,2

    0,3

    0,4

    0,5

    0,6

    0,7

    0,8

    0,9

    1,0

    0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1

    effciencyCK

    2=10

    2=15

    2=20

    2=30

    2=40

    2=50

    i

    21

    i

    f

    iyi

    00yK

    yii

    Rd,i

    d2

    dd

    sin

    Q

    tf

    tfC

    fA

    N +

    =

    Wardenier Tubular Structure Course - 2011

    Joints related to basic types (IIW 2009)

    N2 N1N2 N11N2N

    X

    X

    KK

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    Joints related to basic types IIW (2009)

    h1

    d0

    t0

    N1

    N1

    1t1

    d0

    b1

    t0

    N1

    N1

    h

    1

    d0

    1

    t0

    t

    N1

    N1

    h1

    d0

    t 0

    N1

    b1

    N1

    1

    f20y01 Qtf)(f)(f)(fN =

    Wardenier Tubular Structure Course - 2011

    Multi-planar Joints

    Storm Surge Barrier near Hook of Holland, The Netherlands

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    Multiplanar joints

    Geometrical effectGeometrical effect

    Loading effectLoading effect

    Effect depends on: - type of joint (T, X, K, CHS, RHS)

    - type of loading (axial, b.i.p., b.o.p.)

    Effect for strength, stiffness and deformation capacity

    Wardenier Tubular Structure Course - 2011

    Example:XX joint of CHS

    multiplanar effects

    N2

    = N1

    N2 = 0.6 N1

    N2 = 0.0

    N2 = -0.6 N1

    Uniplanar joint = 16.0 = 0.60

    2 = 50.8

    1111 / d0

    N1

    /(fy

    o*to)2

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    Moment joints

    Bending in plane

    Bending out of plane

    In frames and Vierendeel girders

    (see beam to column joints)

    In principle the

    same philosophy

    Wardenier Tubular Structure Course - 2011

    Interaction between N, Mip and Mop

    0,1M

    M

    M

    M

    N

    N

    Rd,i,op

    Ed,i,op

    2

    Rd,i,ip

    Ed,i,ip

    Rd,i

    Ed,i +

    +

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    Special items

    Concrete filling of the chord

    complete

    or

    only between tubulars e.g. between pile and leg ina jacket

    Punching shear check

    Wardenier Tubular Structure Course - 2011

    Joints with a Can

    Can length for X joints L > 2d0

    llcancan/d/d00

    f(Nf(N))

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    Semi-flattened end joints

    T and X-joint:

    replace d1 by d1,min.

    K-joint with gap:

    replace di by

    0,5(di +di,min.)

    Wardenier Tubular Structure Course - 2011

    Slotted Gusset Plate Joints

    Exposed roof structure

    of

    Toronto Convention Centre

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    Slotted plate joints

    Lw 1,3d An

    Lw

    An=A

    g

    a)

    crack

    slot

    GussetPlate b)

    CHS

    crack

    CHSLw

    Wardenier Tubular Structure Course - 2011

    Tee end to CHS Joints

    N1,Rd = 2fy1 t1 (tw + 5tp) A1 fy1

    N1,Rd

    = 2fyw

    tw

    (t1

    + 2.5tp

    + s)

    2fyw tw (t1 + 5tp)

    2.51

    tp

    N1

    d1

    tw

    5tp+ t

    w

    s s

    t1

    2.51

    tp

    N1

    d1

    tw

    5tp+ t

    w

    s s

    2.51

    tp

    N1

    d1

    tw

    5tp+ t

    w

    s s

    t1

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    Hollow sections in structural applications Course

    Wardenier Tubular Structure Course - 2011

    The end ofthe lectures

    on

    CHS joints

    Wardenier Tubular Structure Course - 2011