· PDF file Point Load wad 1100 N m:= ⋅ Distributed Load Start wld 1200 N m:= ⋅...

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Transcript of · PDF file Point Load wad 1100 N m:= ⋅ Distributed Load Start wld 1200 N m:= ⋅...

  • Td 30 C⋅:= Design Temperature

    NL 100 N⋅ 10 3

    ⋅:= Installation Preload

    SY 450 MPa⋅:= Yield Stress

    EE 207 10 9

    ⋅ N⋅ m 2−

    ⋅:= Beam Elastic Modulus

    α 11.6 10 6−

    ⋅ 1

    C ⋅:= Thermal Expansion Coefficient

    mvp 7850 kg⋅ m 3−

    ⋅:= Beam Density

    mvi 730 kg⋅ m 3−

    ⋅:= Internal Fluid Density

    mve 1027 kg⋅ m 3−

    ⋅:= External Fluid Density

    The effective length is used for soft ends, eg beams on soil. For hard ends fe = 1

    The bending moment and deflection tend to infinity as the compressive axial load tends to ther buckling load

    Buckling will occur in the orientation with the lowest buckling load

    The selected orientation for bending might not be the orientation for buckling, depending on constraints etc.

    Depending on beam geometry and axial load:

    The maximum stress can occur either at the top of the beam or at the base of the beam

    PIPENG.COM : Circular Beam Bending Combined Load

    This example should be used with the Pipeng Toolbox :

    http://pipeng.com/index.php/ts/itdmotbeam003a : Pipeng beam bending calculators

    Copyright Pipeng Ltd : www.pipeng.com : Creative Commons attribution license

    Data Values

    OD 0.3 m⋅:= Beam Diameter

    tn 0.015 m⋅:= Beam Wall Thickness

    Lo 10.5 m⋅:= Nominal Length

    fe 1.05:= Effective Length Factor

    Le fe Lo⋅:= Le 1.1025 10 1

    × m= Effective Length

    x 6.5 m⋅:= Distance From Left End

    Ts 19.5 C⋅:= Installation Temperature

    itdhebeam003201 9/09/2019 9:28 AM Page 1 Of 58

  • Point Load

    wad 1100 N

    m ⋅:= Distributed Load Start

    wld 1200 N

    m ⋅:= Distributed Load End

    dm 2800− N⋅ m⋅:= Concentrated Moment

    do 0.0015− rad⋅:= Angular Displacement

    dy 0.015− m⋅:= Lateral Displacement

    dt 35 C⋅:= Temperature Difference

    waw 732 N

    m ⋅:= Weight Load Start

    For beams with axial load calculated from temperature, and also with a temperature difference load:

    The design temperature is asumed to be the average temperature across the beam (along the whole beam).

    Note: Large displacements with slope greater than 20-25 degrees or 0.4-0.5 radians are not valid.

    Combined loads:

    p 3 -4000 point load -4000 N (upwards) at a = 3 m

    d 2 1100 1200 partial distributed load 1100 N/m downwards at 2 m 1200 N/m downwards at end

    m 4 -2800 concentrated moment at 4 m 2800 N.m anti clockwise

    a 4 -0.0015 angular displacement at 4 m left end rotated anti clockwise 0.0015 rad

    y 3 -0.015 lateral displacement at 3 m right side displaced downwards 0.015 m

    t 2 35 uniform delta temperature of 35 C from 2 m

    w 732 weight load 732 N/m downwards

    ap 3 m⋅:= Point Load Location

    ad 2 m⋅:= Distributed load Start

    am 4 m⋅:= Concentrated Moment Location

    ao 1.5 m⋅:= Angular Displacement Location

    ay 2.5 m⋅:= Lateral Displacement Location

    at 3.5 m⋅:= Temperature Load Start

    aw 0 m⋅:= Weight Load Start

    wp 4000− N⋅:=

    itdhebeam003201 9/09/2019 9:28 AM Page 2 Of 58

  • lor 9.910589 10 3−

    × m 2

    = Slenderness ratio

    ya OD

    2 := ya 1.5 10

    1− × m= Outer Fiber Distance

    yb OD

    2 := yb 1.5 10

    1− × m= Outer Fiber Distance

    ZM I

    ya := ZM 9.115822 10

    4− × m

    3 = Section Modulus

    h ya yb+:= h 3 10 1−

    × m= Section Height In Plane Of Bending

    Mass And Weight

    MP AX mvp⋅:= MP 1.054279 10 2

    × kg

    m = Beam Unit Mass

    MC AC mvi⋅:= MC 4.179653 10 1

    × kg

    m = Contents Fluid Unit Mass

    MD AD mve⋅:= Displaced Fluid Unit Mass

    ML MP MC+:= ML 1.472245 10 2

    × kg

    m = Unit Mass Full

    waw g MP MC+ MD−( )⋅:= waw 7.318713 10 2

    × N

    m = Unit Weight Full Submerged

    Cross Section Area : Hollow Section

    ID OD 2 tn⋅−:= ID 2.7 10 1−

    × m= Internal Diameter

    AX π

    4 OD

    2 ID

    2 −( )⋅:= AX 1.343031 10 2−× m2= Beam Cross Section Area

    AD π

    4 OD

    2( )⋅:= AD 7.068583 10 2−× m2= Displaced Fluid Cross Section Area

    AC π

    4 ID

    2( )⋅:= AC 5.725553 10 2−× m2= Contents Cross Section Area

    I π

    64 OD

    4 ID

    4 −( )⋅:= I 1.367373 10 4−× m4= Beam Second Area Moment

    rr I

    AX := rr 1.009022 10

    1− × m= Radius Of Gyration

    EA EE AX⋅:= EA 2.780074 10 9

    × N= Axial Stiffness

    EAA EE α⋅ AX⋅:= EAA 3.224886 10 4

    × N

    C = Thermal Expansion Modulus

    EI EE I⋅:= EI 2.830463 10 7

    × N m 2

    ⋅= Bending Stiffness

    lor L

    rr :=

    itdhebeam003201 9/09/2019 9:28 AM Page 3 Of 58

  • The Johnson buckling load is assumed to be >= 0

    Buckling Load From LengthFB 1( ) 2.298264 10 6

    × N=FB cc( ) lt LBT cc( )←

    FBE cc( ) Le lt≥if

    FBJ cc( ) otherwise

    :=

    Johnson Buckling Load From LengthFBJ 1( ) 2.07047 10 6

    × N=FBJ cc( ) max 0 SY AX⋅ SY AX⋅ Le⋅

    2 π⋅  

     

    2 1

    cc EI⋅ ⋅−,

      

      

    :=

    Euler Buckling Load From LengthFBE 1( ) 2.298264 10 6

    × N=FBE cc( ) cc π

    2 ⋅ EI⋅

    Le 2

    :=

    Transition Buckling LengthLBT 1( ) 9.61489 10 0

    × m=LBT cc( ) 2 π

    2 ⋅ cc⋅ EI⋅

    SY AX⋅ :=

    UX 3 1, 2,( ) 4=

    Unit Step FunctionUX 1 2, 0,( ) 0=UX xx aa, n,( )

    0 n 0=if

    aa aa−( ) n

    otherwise

    xx aa

  • YA 0

    3.799698 10 2−

    × m= Deflection At A

    RB 0

    wp:= RB 0

    4− 10 3

    × N= Reaction At B

    MB 0

    wp− Le ap−( )⋅:= MB 0

    3.21 10 4

    × N m⋅= Moment At B

    OB 0

    0 rad⋅:= OB 0

    0 10 0

    ×= Angle At B

    YB 0

    0 m⋅:= YB 0

    0 10 0

    × m= Deflection At B

    RX 0

    RA 0

    wp UX x ap, 0,( )⋅−:= RX 0

    4 10 3

    × N= Shear At X

    MX 0

    MA 0

    RA 0

    x⋅+ wp UX x ap, 1,( )⋅−:= MX 0

    1.4 10 4

    × N m⋅= Moment At X

    OX 0

    OA 0

    MA 0

    x⋅

    EI +

    RA 0

    x 2

    2 EI⋅ +

    wp

    2 EI⋅ UX x ap, 2,( )⋅−:=

    OX 0

    3.684954− 10 3−

    ×= Slope At X

    YX 0

    YA 0

    OA 0

    x⋅+ MA

    0 x 2

    2 EI⋅ +

    RA 0

    x 3

    6 EI⋅ +

    wp

    6 EI⋅ UX x ap, 3,( )⋅−:=

    YX 0

    9.42834 10 3−

    × m= Deflection At X

    Beam Bending - Free-Fix

    kk 0.25:= Buckling Load Factor

    lt LBT kk( ):= lt 4.807445 10 0

    × m= Buckle Transition Length

    fbe FBE kk( ):= fbe 5.745661 10 5

    × N= Euler Buckling Load

    fbj FBJ kk( ):= fbj 0 10 0

    × N= Johnson Buckling Load

    fb FB kk( ):= fb 5.745661 10 5

    × N= Buckling Load

    Free-Fix : Point Load

    RA 0

    0 N⋅:= RA 0

    0 10 0

    × N= Reaction At A

    MA 0

    0 N⋅ m⋅:= MA 0

    0 10 0

    × N m⋅= Moment At A

    OA 0

    wp

    2 EI⋅ Le ap−( )

    2 ⋅:= OA

    0 4.550537− 10

    3− ×= Angle At A

    YA 0

    wp−

    6 EI⋅ 2 Le

    3 ⋅ 3 Le

    2 ⋅ ap⋅− ap

    3 +( )⋅:=

    itdhebeam003201 9/09/2019 9:28 AM Page 5 Of 58

  • OB 1

    0 rad⋅:= OB 1

    0 10 0

    ×= Angle At B

    YB 1

    0 m⋅:= YB 1

    0 10 0

    × m= Deflection At B

    RX 1

    RA 1

    wad UX x ad, 1,( )⋅− wld wad−

    2 Le ad−( )⋅ UX x ad, 2,( )⋅−:=

    RX 1

    5.062188− 10 3

    × N= Shear At X

    MX 1

    MA 1

    RA 1

    x⋅+ wad

    2 UX x ad, 2,( )⋅−

    wld wad−

    6 Le ad−( )⋅ UX x ad, 3,( )⋅−:=

    MX 1

    1.130578− 10 4

    × N m⋅= Moment At X

    OX 1

    OA 1

    MA 1

    x⋅

    EI +

    RA 1

    x 2

    2 EI⋅ +

    wad

    6 EI⋅ UX x ad, 3,( )⋅−

    wld wad−

    24 EI⋅ Le ad−( )⋅ UX x ad, 4,( )⋅−:=

    OX 1

    4.272593 10 3−

    ×= Slope At X

    YX 1

    YA 1

    OA 1

    x⋅+ MA

    1 x 2

    2 EI⋅ +

    RA 1

    x 3

    6 EI⋅ +

    wad

    24 EI⋅ UX x ad, 4,( )⋅−

    wld wad−

    120 EI⋅ Le ad−( )⋅ UX x ad, 5,( )⋅−:=

    YX 1

    1.176657− 10 2−

    × m= Deflection At X

    Free-Fix : Distributed Load

    RA 1

    0 N⋅:= RA 1

    0 10 0

    × N= Reaction At A

    MA 1

    0 N⋅ m⋅:= MA 1

    0 10 0

    × N m⋅= Moment At A

    OA 1

    wad

    6 EI⋅ Le ad−( )

    3 ⋅

    wld wad−

    24 EI⋅ Le ad−( )

    3 ⋅+:= OA

    1 4.869512 10

    3− ×= Angle At A

    YA 1

    wad−

    24 EI⋅ Le ad−( )

    3 ⋅ 3 Le⋅ ad+( )⋅

    wld wad−

    120 EI⋅ Le ad−( )

    3 ⋅ 4 Le⋅ ad+( )⋅−:=

    YA 1

    4.274837− 10 2−

    × m= Deflection At A

    RB 1

    wad wld+