Appendix F.1-JOINT 1561 1563

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  • RING DESIGN - SACSAnalysis: InplaceJoint No: 1561-1565

    86.4cm BRACE

    6.5 cm

    45.0 cm

    182.9 cmCHORD

    0.0 cm

    3.0 cm 0.0 cm

    A) PROPERTIES DATA

    Chord Outer Diameter (OD) = 182.9 cm @ 72.01 inchChord Wall Thickness (WT) = 6.5 cm @ 2.56 inchBrace Outer Diameter (OD) = 86.4 cm @ 34.02 inchBrace Wall Thickness (WT) = 2.5 cm @ 0.98 inch

    Chord Effective Diameter = 1.1*(D*t)^0.5= 37.93 cm @ 14.932

    Ring Stiffener Depth = 45.0 cm @ 17.72 inchRing Stiffener Thickness = 3.0 cm @ 1.18 inch

    Flange Thickness= 0.0 @ 0.00 inchFlange Height= 0.0 @ 0.00 inch

    No of ring required = 2Yield Strength, Fy = 345 Mpa @ 50.04 ksi

    Young's Modulus, E = 200 Gpa @ 29.01 kksiPoisson Ratio = 0.3

    B) REACTION FORCES

    Axial Force, Fx = -5987.4 kNMoment, Mz = -11.80 kNm Moment, My = -126.90 kNm

    Lever arm, L1 = 64.4 cm Ring distance, L2 = 54.6 cmForce from Moment = 9.2 kN Force from Moment = -232.4176 kN

    MyBrace

    MzBrace

    Lever armLever arm Ring

    Flange

    Ring

    Section A-A Section I-I

    x

    z

    Reaction Force

    x

    y

    I

    I

    A

    A

    x

    z

  • Since 2 nos of ring required, Fx need to be distributed equally

    Thus, Fx = -2993.7 kN

    3 cases to be considered for ring design Case 8 , Case 18 and Case 16

    Case 18 Point Load, F = -2993.68 kN

    Moment = -126.90 kNmForce, Ft = Moment

    Ring distance, L2

    = -232.42 kNTotal Force, W1 = 3226.10 kN @ 725.23 kips

    Case 8 Uniform Load,w =

    = 24.21 kN/cm @ 13.82 kips/inch

    Alpha, = 46.77 degree

    Theta, = 133.23 degree

    Case 16

    Moment = -11.80 kNmTotal Force,W2 = Moment

    Lever arm, L1= 9.16 kN @ 2.06 kips

    For case 16,Theta, = 133.23 degreePhi, = -133.23 degree

    W1 = 3226.10 kN w = 24.21 kN/cm W2 = 9.16 kN

    Initially, the jacket leg has been checked without ring stiffener. The result is as follow;

    Allowable Ring Stress = 0.66 x Fy= 228 Mpa

    Based on the ring stiffener analysis of the jacket leg by considering the ring stiffener, the result is as follow;

    Maximum Outer Fibre Stress = -7.88 ksi @ 54.33 MpaMaximum Inner Fibre Stress = 26 ksi @ 179.26 Mpa

    Stress Interaction Ratio = Actual Stress / Allowable Stress= 0.787 < 1, ==>> OK!

    W1/(2*R*sin )

    +

    =

    CASE-18

    w

    2 w R sin

    CASE-8

    w

    W1

    +

    W2

    CASE-16

    +

    W2 W2

    W2

  • RING DESIGN - SACSAnalysis: InplaceJoint No: 1563-1565

    86.4cm BRACE

    6.5 cm

    45.0 cm

    182.9 cmCHORD

    0.0 cm

    3.0 cm 0.0 cm

    A) PROPERTIES DATA

    Chord Outer Diameter (OD) = 182.9 cm @ 72.01 inchChord Wall Thickness (WT) = 6.5 cm @ 2.56 inchBrace Outer Diameter (OD) = 86.4 cm @ 34.02 inchBrace Wall Thickness (WT) = 2.5 cm @ 0.98 inch

    Chord Effective Diameter = 1.1*(D*t)^0.5= 37.93 cm @ 14.932

    Ring Stiffener Depth = 45.0 cm @ 17.72 inchRing Stiffener Thickness = 3.0 cm @ 1.18 inch

    Flange Thickness= 0.0 @ 0.00 inchFlange Height= 0.0 @ 0.00 inch

    No of ring required = 2Yield Strength, Fy = 345 Mpa @ 50.04 ksi

    Young's Modulus, E = 200 Gpa @ 29.01 kksiPoisson Ratio = 0.3

    B) REACTION FORCES

    Axial Force, Fx = -5683.7 kNMoment, Mz = 2.28 kNm Moment, My = -91.85 kNm

    Lever arm, L1 = 64.4 cm Ring distance, L2 = 54.6 cmForce from Moment = 1.8 kN Force from Moment = -168.2234 kN

    MyBrace

    MzBrace

    Lever armLever arm Ring

    Flange

    Ring

    Section A-A Section I-I

    x

    z

    Reaction Force

    x

    y

    I

    I

    A

    A

    x

    z

  • Since 2 nos of ring required, Fx need to be distributed equally

    Thus, Fx = -2841.9 kN

    3 cases to be considered for ring design Case 8 , Case 18 and Case 16

    Case 18 Point Load, F = -2841.87 kN

    Moment = -91.85 kNmForce, Ft = Moment

    Ring distance, L2

    = -168.22 kNTotal Force, W1 = 3010.09 kN @ 676.67 kips

    Case 8 Uniform Load,w =

    = 22.59 kN/cm @ 12.90 kips/inch

    Alpha, = 46.77 degree

    Theta, = 133.23 degree

    Case 16

    Moment = 2.28 kNmTotal Force,W2 = Moment

    Lever arm, L1= 1.77 kN @ 0.40 kips

    For case 16,Theta, = 133.23 degreePhi, = -133.23 degree

    W1 = 3010.09 kN w = 22.59 kN/cm W2 = 1.77 kN

    Initially, the jacket leg has been checked without ring stiffener. The result is as follow;

    Allowable Ring Stress = 0.66 x Fy= 228 Mpa

    Based on the ring stiffener analysis of the jacket leg by considering the ring stiffener, the result is as follow;

    Maximum Outer Fibre Stress = -7.35 ksi @ 50.68 MpaMaximum Inner Fibre Stress = 24.25 ksi @ 167.20 Mpa

    Stress Interaction Ratio = Actual Stress / Allowable Stress= 0.734 < 1, ==>> OK!

    W1/(2*R*sin )

    +

    =

    CASE-18

    w

    2 w R sin

    CASE-8

    w

    W1

    +

    W2

    CASE-16

    +

    W2 W2

    W2

    RING DESIGN (1561-1565-F)RING DESIGN (1563-1565-F)