ad2000 s3-2 vessel on 3 saddles or rings

CALCULATION FOR REACTION FORCES ON VESSEL - AD 2000 S3/0

Da = 3.95 L = 17.50 Ht = 3.90 H = 2.00

E = 3.70

Lt = 19.50 l1 = 6.95

STRESS CASES: ACTUAL VALUES

DEAD LOADS (z)

STRUCTURE 1.00 1.00 1.00 1.00 1.00 1.00 508900.00 NINTERNALS 1.00 1.00 1.00 1.00 1.00 1.00 30550.00 NPLATFORMS 1.00 1.00 1.00 1.00 1.00 1.00 103687.50 NINSULATION 1.00 0.00 1.00 1.00 1.00 1.00 37500.00 N

A = Ambient, D = Design, PT = Test Pressure, T = Temperature, P = PressureLIVE LOADS (z) ACTUAL VALUES

PROCESS FLUIDS 0.00 0.00 1.00 1.00 0.00 0.00 3377193.75 NTEST FLUIDS 0.00 1.00 0.00 0.00 0.00 0.00 2144250.00 NPLATFORMS 0.00 0.50 1.00 0.00 1.00 0.00 98750.00 N

GLOBAL LOADS (z) ACTUAL VALUES

SEISMIC OPE. 0.00 0.00 0.00 1.00 0.00 0.00 3505966.20 NSEISMIC ERE. 0.00 0.00 0.00 0.00 0.00 1.00 588070.80 NSNOW 0.00 0.00 1.00 0.00 0.00 1.00 61620.00 N

GLOBAL LOADS (x,y) ACTUAL VALUES

LONG. WIND (x) 1.00 0.70 1.00 0.00 1.00 0.00 18486.00 NTRANSV. WIND (y) 1.00 0.70 1.00 0.00 1.00 0.00 92430.00 NSEISMIC OPE. (x,y) 0.00 0.00 0.00 1.00 0.00 0.00 2597012.00 NSEISMIC ERE. (x,y) 0.00 0.00 0.00 0.00 0.00 1.00 435608.00 NFRICTION OPE. (x) 0.00 0.00 1.00 0.00 0.00 0.00 632730.19 NFRICTION SHUT.(x) 0.00 0.00 0.00 0.00 1.00 0.00 116908.13 N

NOTE: THE VALUES IN THE MATRIXES ABOVE, ARE MULTIPLICATION FACTORS ON THE ACTUAL LOADS.------------------------------------------------------------

ANALYSIS FOR SNOW LOAD: EUROCODE 1VESSEL EXTERNAL DIAMETER (INCLUDING INSULATION): Da = 3.95 mVESSEL TOTAL LENGTH: Lt = 19.50 mVESSEL AREA EXPOSED TO SNOW: As = Lt * Da 77.03 m²SNOW PRESSURE: So = 800.00 PaSNOW LOAD SHALL BE: Fs = So * As = 61620.00 N

------------------------------------------------------------ESTIMATION FOR PLATFORM DEAD LOADS

VESSEL EXTERNAL DIAMETER (INCLUDING INSULATION): Da = 3.95 mVESSEL TAN-TAN LENGTH: L = 17.50 m

ERECTION (P=0, T=A)

SITE TEST (P=PT, T=A)

OPERATION (P=D, T=D)

SEISMIC (P=D, T=D)

SHUT DOWN (P=0, T=D)

ERECTION SEISMIC (P=0, T=A)

PLATFORM AREA SHALL BE COVERING ALL OF THE VESSEL: Ap = L * Da = 69.13 m²PLATFORM DEAD LOAD IS Pp: 1500.00 Pa => Fp = Pp * Ap = 103687.50 N

------------------------------------------------------------

ESTIMATION FOR PLATFORM LIVE LOADSVESSEL EXTERNAL DIAMETER (INCLUDING INSULATION): Da = 3.95 mVESSEL LENGTH WHERE LIVE LOAD WILL BE APPLICABLE: Ll = 5.00 mPLATFORM LIVE LOAD AREA SHALL BE: Apl = Ll * Da = 19.75 m²PLATFORM LIVE LOAD IS Ppl: 5000.00 Pa => Fpl = Ppl*Apl = 98750.00 N

------------------------------------------------------------ANALYSIS FOR LONGITUDINAL WIND LOAD: EUROCODE 1

VESSEL EXTERNAL DIAMETER (INCLUDING INSULATION); Da = 3.95 mVESSEL TOTAL HEIGHT: Ht = 3.90 mVESSEL AREA EXPOSED TO WIND, LONGITUDINAL Aw = Da Ht = 15.41 m²WIND PRESSURE EXCERTING ON VESSEL: q = 1000.00 PaSHAPE COEFFICIENT: c = 0.7 simple vessel, c = 1.2 vessel with platforms) c = 1.20 PaLONGITUDINAL WIND LOAD SHALL BE: Fwx = c q Aw = 18486.00 N

------------------------------------------------------------ANALYSIS FOR TRANSVERSE WIND LOAD: EUROCODE 1

VESSEL EXTERNAL DIAMETER (INCLUDING INSULATION); Da = 3.95 mVESSEL TOTAL LENGTH: Lt = 19.50 mVESSEL AREA EXPOSED TO WIND, TRANSVERSE: Aw = Da Lt = 77.03 m²WIND PRESSURE EXCERTING ON VESSEL: q = 1000.00 PaSHAPE COEFFICIENT: c = 0.7 simple vessel, c = 1.2 vessel with platforms) c = 1.20 PaTRANSVERSE WIND LOAD SHALL BE: Fwy = c q Aw = 92430.00 N

------------------------------------------------------------ANALYSIS FOR SEISMIC LOAD - HORIZONTAL x,y FORCE: EUROCODE 8

SEISMIC DESIGN GROUND ACCELARATION: ag = 2.40 m/s²LOWER LIMIT OF PERIOD OF SPECTRAL ACCELERATION: TB = 0.15 secUPPER LIMIT OF PERIOD OF SPECTRAL ACCELERATION: TC = 0.40 secVALUE OF CONSTANT DISPLACEMENT RANGE: TD = 2.00 secSOIL / GROUND FACTOR: S = 1.00

DAMPING CORRECTION FACTOR: βο = 2.50

BEHAVIOUR FACTOR: q = 1.50

CRITICALITY FACTOR: γ = 1.60

TOTAL MASS IN EXAMINED CASE - OPERATION: Mo = 405783.13 KGTOTAL MASS IN EXAMINED CASE - ERECTION: Mo = 68063.75 KGMAXIMUM (SIMPLIFIED) HORIZONTAL SEISMIC FORCE: Feh = Mo * ag * γ * S * 2,5 / q = 2597012.00 NMAXIMUM (SIMPLIFIED) HORIZONTAL SEISMIC FORCE: Feh = Mo * ag * γ * S * 2,5 / q = 435608.00 N

------------------------------------------------------------ANALYSIS FOR SEISMIC LOAD - VERTICAL FORCE: EUROCODE 8

SEISMIC DESIGN GROUND ACCELARATION: ag,v = 0.9 ag = 2.16 m/s²LOWER LIMIT OF PERIOD OF SPECTRAL ACCELERATION: TB = 0.05 secUPPER LIMIT OF PERIOD OF SPECTRAL ACCELERATION: TC = 0.15 secVALUE OF CONSTANT DISPLACEMENT RANGE: TD = 1.00 secSOIL / GROUND FACTOR: S = 1.00

DAMPING CORRECTION FACTOR: βο = 0.20

BEHAVIOUR FACTOR: q = 1.00

CRITICALITY FACTOR: γ = 1.60

TOTAL MASS IN EXAMINED CASE - OPERATION: Mo = 405783.13 KGTOTAL MASS IN EXAMINED CASE - ERECTION: Mo = 68063.75 KGMAXIMUM (SIMPLIFIED) VERTICAL SEISMIC FORCE: Fev = Mo * ag * γ * S * 2,5 / q = 3505966.20 N

MAXIMUM (SIMPLIFIED) VERTICAL SEISMIC FORCE: Fev = Mo * ag * γ * S * 2,5 / q = 588070.80 N------------------------------------------------------------

NOTE: THE COLOR REPRESENTS INPUT VALUES BY USER

ANALYSIS FOR FRICTION LOADSFRICTION COEFFICIENT (WHEN THE VESSEL IS IN DESIGN TEMPER.): μ = 0.15

VERTICAL WEIGHT (OPERATION) - WITH LIVE + GLOBAL LOADS Wo = 4218201.25 NVERTICAL WEIGHT (SHUT DOWN) - WITH LIVE + GLOBAL LOADS Wo = 779387.50 NLONGITUDINAL FRICTION FORCE (OPERATION): Ff = μ Wo 632730.1875 NLONGITUDINAL FRICTION FORCE (SHUT DOWN): Ff = μ Wo 116908.125 N

------------------------------------------------------------

SETTING OF ALLOWABLE STRESSES - AD 2000 S3/0 ( & SUPPLEMENTED BY ASME VIII DIV. 1)

MATERIAL YIELD STRENGTH AT TEMPERATURE / AMBIENT K K20SHELL MATERIAL: ASME II: SA516-60 189.00 220.00 MpaSADDLE PAD MATERIAL: ASME II: SA516-60 189.00 220.00 MpaSADDLE WEB/RIB MATERIAL: ASME II: SA516-70 225.00 260.00 MpaSADDLE BASE MATERIAL: ASME II: SA516-70 225.00 260.00 MpaSTIFFENING RING MATERIAL: ASME II: SA516-70 225.00 260.00 MpaELASTICITY MODULOUS (GENERAL): Ey = 200000.00 Mpa

SETTING OF ALLOWABLE STRESSESALLOWABLE STRESSES SHALL BE PER ASME VIII DIV. 1 IN NORMAL OPERATING MODES

90% YIELD AT DESIGN / ROOM TEMPERATURE IN TEST CASES OR IN EXCEPTIONAL CASES (SEISMIC E.T.C)ALLOWABLE STRESSES ERECTION SITE TEST OPERATION SEISMIC SHUT DOWN SEISMIC ER.

SHELL: f 118.00 198.00 118.00 170.10 118.00 198.00 MpaSADDLE PAD: fv 118.00 198.00 118.00 170.10 118.00 198.00 MpaSADDLE WEB/RIB: fs 138.00 234.00 138.00 202.50 138.00 234.00 MpaSADDLE BASE: fp 138.00 234.00 138.00 202.50 138.00 234.00 MpaSTIFFENING RING: fr 138.00 234.00 138.00 202.50 138.00 234.00 Mpa

------------------------------------------------------------SADDLE GEOMETRY DATA PER AD 2000 S3/2

SHELL INTERNAL CORRODED DIAMETER: D 3806.00 mmSHELL CORRODED THICKNESS e 13.00 mmSADDLE TO SHELL TANGENT LINE DISTANCE a1 1800.00 mmSHELL TAN-TAN DISTANCE L 17500.00 mmHEAD DEPTH h2 950.00 mmSADDLE TO HEAD 2/3 DEPTH DISTANCE a3 = a1 + 2/3h2 2433.33 mmSADDLE TO SADDLE DISTANCE: l1 6950.00 mmSADDLE EXTERNAL RIBS WIDTH: b1 625.00 mmSADDLE REINFORCEMENT PLATE WIDTH b2 725.00 mmSADDLE REINFORCEMENT PLATE THICKNESS e2 (ev) 20.00 mmSADDLE ANGLE: δ1 165.00 deg 2.88 radSADDLE REINFORCEMENT PLATE ANGLE: δ2 180.00 deg 3.14 radSADDLE LENGTH PORTION OF EXTENT: b3 250.68 mmSADDLE BASE PLATE THICKNESS: ts 32.00 mmSADDLE WEB & RIB THICKNESS: es 32.00 mm

FIGURE 9 RESULTS

a3 / l1 = 0.35n = 3.00factor w1 0.90

factor w2 = 1.20

factor w3 = w1 = 0.90

Examination for F1 (F3) & F2

Vessel section area A =155891.58 mm²

CHECK OF CALCULATION APPLICABILITYSIMPLIFIED CONSIDERATION OF SADDLE PAD, PER PAR. 5.2.2.1:

WHERE K11 = 5 / { 6 (D/e) ^ 1/3 δ1 } = 0.04 THUS: K11 D +1.5 b1 = 1103.45 mmSADDLE PLATE SHALL NOT BE CONSIDERED AS REINFORCEMENT TO THE SHELL SINCE b2 < K11 D + 1.5 b1

NOT APPLICABLE

NOT APPLICABLE

CHECK OF SADDLE ANGLE: OK CHECK δ2 > δ1 OK

OK

FOR VESSELS NOT STIFFENED BY RINGS:FOR SAKE OF SIMPLICITY, IN CASE THE SADDLE PAD IS NOT CONSIDERED AS REINFORCMENT, THEN THE SADDLESWILL BE CALCULATED ACCORDING TO PAR. 5.2.1.1 & 5.2.1.2. PARAGRAPHS 5.2.2.2 SHALL NOT BE CONSIDERED.

------------------------------------------------------------CALCULATION OF VARIOUS GEOMETRY FACTORS:β = 2.56 K5 = 0.74 K9 = 0.68γ = 0.08 K6 = 0.00 K10 = 0.35K3 = 0.25 K7 = 0.21 K12 = 79.50 (TABLE 2)K4 = 0.42 K8 = 0.24 K13 = 1.30 (FIG. 7)

be = 1750.23 mm be/l2 = 0.46 x = L/D = 4.60l2 = 3773.08 mm y = D/e = 292.77 K14 = 1.42 (n/a for 3 saddles)

hA = 781.37 mmψ = 46.65 deg

Ixx Beam moment of inertia calculation:I1 = b2 * e2^3 / 12 = 483333.33I2 = es * ha^3 / 12 = 1272142269.45I3 = b1 * ts^3 / 12 = 1706666.67

Assuming with relative accuracy that the center mass is at midway of ha =>center mass distance a1 = 400.68 mm

SADDLE REINFORCEMENT PLATE SHALL BE CONSIDERED AS REINFORCEMENT TO THE SHELL IF b2 ≥ K11 D + 1.5 b1

IN CASE SADDLE PLATE IS CONSIDERED REINFORCEMENT, GEOMETRY CHECK, e ≤ ev ≤ 1.5e =>IN CASE SADDLE PLATE IS CONSIDERED REINFORCEMENT, GEOMETRY CHECK, b3 ≥ 0.1 D =>

60°≤ δ1 ≤ 180°

CHECK OF THICKNESS TO SHELL DIAMETER RATIO e/D ≤ 0,05

mm4

mm4

mm4

center mass distance a2 = 0.00 mmcenter mass distance a3 = 406.68 mmIxx,total = I1 + a1^2 * e2 * b2 + I2 + I3 + a3^2 * ts * b1 = 6910111699.42Wxx = 2 * Ixx,total / (hA + ts + e2) = Wa = 16583577.16 mm³

mm4

EQUIVALENT VERTICAL FORCES ON VESSEL - AD 2000 S3/2 (ASSUMING SYMMETRICAL VESSEL)CALCULATION OF REACTION FORCES: SEISMIC CASE

CASE DATA:EXAMINATION COEFFICIENT ( SERVICE / SPECIAL / TEST CONDITION) K2 = 1.00 S = 1.10INTERNAL OVERPRESSURE: p = 0.50 bargIMPACT FACTOR γ (SERVICE = 1.5 , TEST = 1.3 , SPECIAL CONDITIONS = 1.15) γ = 1.15VESSEL ALLOWABLE STRESS: f = 170.10 MpaSADDLES ALLOWABLE STRESS: fs = 202.50 Mpa

------------------------------------------------------------FORCES DATA:TOTAL VERTICAL FORCES ON C.M: Fz (Gz) = 7563797.45 NTOTAL TRANSVERSE FORCES ON C.M: Fy = 2597012.00 NTOTAL LONGITUDINAL FORCES ON C.M: Fx = 2597012.00 N

SADDLE BASE TO SHELL CENTERLINE: H = 2000.00 mmSADDLE TO SADDLE DISTANCE: l1 = 6950.00 mmSADDLE BASE PLATE LENGTH: E = 3700.00 mm

Gy = Fy * 3 H / E = 4211370.81 NGx = Fx * H / l1 = 747341.58 N TOTAL G = Gz + Gy + Gx = 12522510 N

q = G / ( L + 4/3 h2) = 667 N/mm ACCORDING TO PAR. 4.2.2, MID-SPAN MOMENTS DO NOT GOVERNMo = q D² / 16 = 604118071 Nmm WHEN THERE ARE 3 SADDLES. => PAR. 4.3 NEED NOT BE APPLIED.

FORCES ON SADDLE POSITIONS: MOMENTS ON SADDLE POSITIONS:F1 = w1 * G / n = 3756752.95 N M1 = max { q l1² / 8 ; q a3² / 2 - Mo } = 4028875655 NmmF2 = w2 * G / n 5009003.94 N M2 = q l1² / 8 = 4028875655 NmmF3 = F1 = 3756752.95 N M3 = M1 = 4028875655 Nmm

------------------------------------------------------------VERIFICATION OF SHELL LOAD CARRYING CAPACITY IN SADDLE REGION - PROOF OF STRENGTH

SADDLE1 & SADDLE3 POSITIONS - WITH STIFFENING RINGSsmx = ABS(4M1/πD²e) = 27.25 sp = p D / 40 e + smx = 30.91 Mpa OK

SADDLE2 POSITION - WITH STIFFENING RINGSsmx = ABS(4M2/πD²e) = 27.25 sp = p D / 40 e + smx = 30.91 Mpa OK

------------------------------------------------------------VERIFICATION OF SHELL LOAD CARRYING CAPACITY IN SADDLE REGION - PROOF OF STABILITY

SADDLE 1 & SADDLE 3 POSITIONSEQUIVALENT AXIAL FORCE FROM LOCAL BENDING: Fe = 0 (per PAR 5.3.1.1) 0.00 N

27.07 MpaSHEAR STRESS: t = 2 * Q1 / A , where Q1 = F1/2 => t = F1 / A = 24.10 MpaEQUIVALENT STRESS: sv = γ √ s² + 3t² = 57.21 MpaCHECK sv < f => OK

SADDLE 2 POSITIONEQUIVALENT AXIAL FORCE FROM LOCAL BENDING: Fe = 0 (per PAR 5.3.1.1) 0.00 N

27.07 MpaSHEAR STRESS: t = 2 * Q2 / A , where Q1 = F2 / 2 => t = F2 / A = 32.13 MpaEQUIVALENT STRESS: sv = γ √ s² + 3t² = 38.38 MpaCHECK sv < f => OK

------------------------------------------------------------VERIFICATION OF SADDLE LOAD CARRYING CAPACITY

VARIOUS FACTORS:εs = 1000 fs / Ey = 1.0125 0.28FF4,adm = l2 es fs φ = 6725874 N CHECK MAX (F1;F2) < FF4,adm => OK

CHECK sp ≤ f

CHECK sp ≤ f

PRIMARY EQUIVALENT MEMBRANE STRESS: s = 4 M1 / π ( D + e)² e + Fe / ( π (D + e) e ) =

PRIMARY EQUIVALENT MEMBRANE STRESS: s = 4 M2 / π ( D + e)² e + Fe / ( π (D + e) e ) =

φ = 1 / √ 1 + { (0.15 εs / K13) (be / 10es)^2 } ^2 =

FF5,adm = 4 fs Wa sin(0,5δ1) / D ( 1 - cosψ) = 11169423 N CHECK MAX (F1;F2) < FF5,adm => OKFF6,1 = 690560 NFF6,2 = 2 fs b1 e2 sin (0.5 δ2) = 5062498 NFF6,adm = max (FF6,1 ; FF6,2) = 5062498 N CHECK MAX (F1;F2) < FF6,adm => OK

1,4 fs D e2² sin (0.5 δ2) / b1 =

GEOMETRY OF STIFFENING RING SEISMIC CASE APPLICABLE p = 0.50 bar f = 170.10 MpaSTIFFENING RING WELD JOINT COEFFICIENCY v = 0.70 fr = 202.50 Mpa

TYPE1 (IGNORE VALUES IF NOT SELECTED)t = 15.00 mm eef = 10.45 mmh = 500.00 mm le = 717.69 mmAr = 7500.00 a2 = 0.00 mm

Mp = 387623062.50 Nmm

TYPE2 (IGNORE VALUES IF NOT SELECTED)t7 = 22.00 mm eef = 10.45 mmb4 = 200.00 mm le = 783.15 mmt6 = 22.00 mm a2 = 0.00 mmh = 172.00 mm Mp = 237610645.80 NmmAr = 8184.00



USER SELECTED RING TYPE: 2STIFFENING RING INTERNAL / EXTERNAL (1 OR 2): 2

APPLICABLE TO VERIFICATION EQUATION: Mp = 237610645.80 Nmmv = 0.70a2 = 0.00 mm

VERIFICATION OF STIFFENING RING

FROM THE APPLICABLE CASE WE HAVEF1 = F3 = 3756752.95 N Fr,adm = K12 Mp v / 0.5 d + a2 = 6948519.41 NF2 = 3756752.95 N CHECK MAX (F1;F2;F3) < Fr,adm => OK

mm²

mm²

mm²

mm²

EQUIVALENT VERTICAL FORCES ON VESSEL - AD 2000 S3/2 (ASSUMING SYMMETRICAL VESSEL)CALCULATION OF REACTION FORCES: SEISMIC CASE

CASE DATA:EXAMINATION COEFFICIENT ( SERVICE / SPECIAL / TEST CONDITION) K2 = 1.00 S = 1.10INTERNAL OVERPRESSURE: p = 0.50 bargIMPACT FACTOR γ (SERVICE = 1.5 , TEST = 1.3 , SPECIAL CONDITIONS = 1.15) γ = 1.15VESSEL ALLOWABLE STRESS: f = 170.10 MpaSADDLE WEB/RIB ALLOWABLE STRESS: fs = 202.50 Mpa

------------------------------------------------------------FORCES DATA:TOTAL VERTICAL FORCES ON C.M: Fz (Gz) = 7563797.45 NTOTAL TRANSVERSE FORCES ON C.M: Fy = 2597012.00 NTOTAL LONGITUDINAL FORCES ON C.M: Fx = 2597012.00 N

SADDLE BASE TO SHELL CENTERLINE: H = 2000.00 mmSADDLE TO SADDLE DISTANCE: l1 = 6950.00 mmSADDLE BASE PLATE LENGTH: E = 3700.00 mm

Gy = Fy * 3 H / E = 4211370.81 NGx = Fx * H / l1 = 747341.58 N TOTAL G = Gz + Gy + Gx = 12522510 N

q = G / ( L + 4/3 h2) = 667 N/mm ACCORDING TO PAR. 4.2.2, MID-SPAN MOMENTS DO NOT GOVERNMo = q D² / 16 = 604118071 Nmm WHEN THERE ARE 3 SADDLES. => PAR. 4.3 NEED NOT BE APPLIED.

FORCES ON SADDLE POSITIONS: MOMENTS ON SADDLE POSITIONS:F1 = w1 * G / n = 3756752.95 N M1 = max { q l1² / 8 ; q a3² / 2 - Mo } = 4028875655 NmmF2 = w2 * G / n 5009003.94 N M2 = q l1² / 8 = 4028875655 NmmF3 = F1 = 3756752.95 N M3 = M1 = 4028875655 Nmm

------------------------------------------------------------VERIFICATION OF SHELL LOAD CARRYING CAPACITY IN SADDLE REGION - PROOF OF STRENGTH

SADDLE1 & SADDLE3 POSITIONS - NO STIFFENING RINGS MODIFICATION PER PAR 5.1smx = ABS(4M1/πD²e) = 27.25 Θ1 Θ2,1 Θ2,2 Θ1(2,1) Θ1(2,2) Θ2,1 MOD Θ2,2 MODLOCATION 2 (SADDLE BOTTOM) 0.00000 -0.14566 -0.12610 0.0000 0.0000 0.14566 0.12610LOCATION 3 (HORN OF SADDLE) -4.41486 0.00000 0.03912 -4.4149 -4.4149 0.00000 0.03912

LOCATION 2 (SADDLE BOTTOM) A - Θ1(2,1) A - Θ1(2,2)FACTOR A = 1 + 3Θ1 Θ2 / 3 Θ1^2 , Α = 1,5 FOR Θ1 = 0 1.5000 1.5000FACTOR B1 = SQRT ( 9 Θ1^2 ( 1 - Θ2^2) / (1+3Θ1 Θ2)^2 + 1 ) - 1 , Β1 = 1 - Θ2^2 FOR Θ1 = 0 0.9788 0.9841FACTOR B2 = - SQRT ( 9 Θ1^2 ( 1 - Θ2^2) / (1+3Θ1 Θ2)^2 + 1 ) - 1 , Β1 = 1 - Θ2^2 FOR Θ1 = 0 0.9788 0.9841

LOCATION 3 (HORN OF SADDLE) A - Θ1(2,1) A - Θ1(2,2)FACTOR A = 1 + 3Θ1 Θ2 / 3 Θ1^2 , Α = 1,5 FOR Θ1 = 0 0.0171 0.0082FACTOR B1 = SQRT ( 9 Θ1^2 ( 1 - Θ2^2) / (1+3Θ1 Θ2)^2 + 1 ) - 1 , Β1 = 1 - Θ2^2 FOR Θ1 = 0 12.2823 26.4808FACTOR B2 = - SQRT ( 9 Θ1^2 ( 1 - Θ2^2) / (1+3Θ1 Θ2)^2 + 1 ) - 1 , Β1 = 1 - Θ2^2 FOR Θ1 = 0 -14.2823 -28.4808

for Θ1(2,1) for Θ1(2,2) MIN K1

LOCATION 2 (SADDLE BOTTOM)Κ1 = A*B1 1.47 1.48 MIN K1 > 0 => 1.47

1.47Κ1 = A*B2 1.47 1.48 MIN K1 > 0 => 1.47


LOCATION 3 (HORN OF SADDLE)Κ1 = A*B1 0.21 0.22 MIN K1 > 0 => 0.21

0.21Κ1 = A*B2 -0.24 -0.23 MIN K1 > 0 => 10000000000000000000.00

sgr2 = K1, LOC.2 * f * S / K2 = 274.71 Mpa 2988526.38 Nsgr3 = K1, LOC.3 * f * S / K2 = 39.30 Mpa 2028540.61 N

SADDLE2 POSITION - NO STIFFENING RING MODIFICATION PER PAR 5.1smx = ABS(4M2/πD²e) = 27.25 Θ1 Θ2,1 Θ2,2 Θ1(2,1) Θ1(2,2) Θ2,1 MOD Θ2,2 MODLOCATION 2 (SADDLE BOTTOM) 0.00000 -0.14566 -0.12610 0.0000 0.0000 0.14566 0.12610LOCATION 3 (HORN OF SADDLE) -4.41486 0.00000 0.03912 -4.4149 -4.4149 0.00000 0.03912

LOCATION 2 (SADDLE BOTTOM) A - Θ1(2,1) A - Θ1(2,2)FACTOR A = 1 + 3Θ1 Θ2 / 3 Θ1^2 , Α = 1,5 FOR Θ1 = 0 1.5000 1.5000FACTOR B1 = SQRT ( 9 Θ1^2 ( 1 - Θ2^2) / (1+3Θ1 Θ2)^2 + 1 ) - 1 , Β1 = 1 - Θ2^2 FOR Θ1 = 0 0.9788 0.9841FACTOR B2 = - SQRT ( 9 Θ1^2 ( 1 - Θ2^2) / (1+3Θ1 Θ2)^2 + 1 ) - 1 , Β1 = 1 - Θ2^2 FOR Θ1 = 0 0.9788 0.9841

LOCATION 3 (HORN OF SADDLE) A - Θ1(2,1) A - Θ1(2,2)FACTOR A = 1 + 3Θ1 Θ2 / 3 Θ1^2 , Α = 1,5 FOR Θ1 = 0 0.0171 0.0082FACTOR B1 = SQRT ( 9 Θ1^2 ( 1 - Θ2^2) / (1+3Θ1 Θ2)^2 + 1 ) - 1 , Β1 = 1 - Θ2^2 FOR Θ1 = 0 12.2823 26.4808FACTOR B2 = - SQRT ( 9 Θ1^2 ( 1 - Θ2^2) / (1+3Θ1 Θ2)^2 + 1 ) - 1 , Β1 = 1 - Θ2^2 FOR Θ1 = 0 -14.2823 -28.4808


LOCATION 2 (SADDLE BOTTOM)Κ1 = A*B1 1.47 1.48 MIN K1 > 0 => 1.47

1.47Κ1 = A*B2 1.47 1.48 MIN K1 > 0 => 1.47


LOCATION 3 (HORN OF SADDLE)Κ1 = A*B1 0.21 0.22 MIN K1 > 0 => 0.21

0.21Κ1 = A*B2 -0.24 -0.23 MIN K1 > 0 => 10000000000000000000.00

sgr2 = K1, LOC.2 * f * S / K2 = 274.71 Mpa 2988526.38 Nsgr3 = K1, LOC.3 * f * S / K2 = 39.30 Mpa 2028540.61 N

CHECK OF APPLICABLE FORCES:SADDLE 1 & 3 - AT LOCATION 2: F1 & F3 < (1) OR (1.5) * FF2,adm => NOT OKSADDLE 1 & 3 - AT LOCATION 3: F1 & F3 < (1) OR (1.5) * FF3,adm => NOT OKSADDLE 2 - AT LOCATION 2: F2 < (1) OR (1.5) * FF2,adm => NOT OKSADDLE 2- AT LOCATION 3: F2 < (1) OR (1.5) * FF3,adm => NOT OK

MULTIPLICATION FACTOR 1.5 AT ADMISSIBLE FORCES APPLIES ONLY IF THE SADDLE PAD IS CONSIDERED AS REINFORCEMENT ------------------------------------------------------------

VERIFICATION OF SHELL LOAD CARRYING CAPACITY IN SADDLE REGION - PROOF OF STABILITY

SADDLE 1 & SADDLE 3 POSITIONS0.00 N

27.07 MpaSHEAR STRESS: t = 2 * Q1 / A , where Q1 = F1/2 => t = F1 / A = 24.10 MpaEQUIVALENT STRESS: sv = γ √ s² + 3t² = 57.21 MpaCHECK sv < f => OK

SADDLE 2 POSITION0.00 N

27.07 MpaSHEAR STRESS: t = 2 * Q2 / A , where Q1 = F2 / 2 => t = F2 / A = 32.13 MpaEQUIVALENT STRESS: sv = γ √ s² + 3t² = 38.38 MpaCHECK sv < f => OK

------------------------------------------------------------VERIFICATION OF SADDLE LOAD CARRYING CAPACITY

FF2,adm = 0.7 sgr2√ D e * e / K3 K5 =

FF3,adm = 0.9 sgr3√ D e * e / K7 K9 K10 =

FF2,adm = 0.7 sgr2√ D e * e / K3 K5 =

FF3,adm = 0.9 sgr3√ D e * e / K7 K9 K10 =

EQUIVALENT AXIAL FORCE FROM LOCAL BENDING: Fe = F1 π/4 √ D / e K6 K8 =PRIMARY EQUIVALENT MEMBRANE STRESS: s = 4 M1 / π ( D + e)² e + Fe / ( π (D + e) e ) =

EQUIVALENT AXIAL FORCE FROM LOCAL BENDING: Fe = F2 π/4 √ D / e K6 K8 =PRIMARY EQUIVALENT MEMBRANE STRESS: s = 4 M2 / π ( D + e)² e + Fe / ( π (D + e) e ) =

VARIOUS FACTORS:εs = 1000 fs / Ey = 1.0125 0.28

FF4,adm = l2 es fs φ = 6725874 N CHECK MAX (F1;F2) < FF4,adm => OKFF5,adm = 4 fs Wa sin(0,5δ1) / D ( 1 - cosψ) = 11169423 N CHECK MAX (F1;F2) < FF5,adm => OKFF6,1 = 690560 NFF6,2 = 2 fs b1 e2 sin (0.5 δ2) = 5062498 NFF6,adm = max (FF6,1 ; FF6,2) = 5062498 N CHECK MAX (F1;F2) < FF6,adm => OK

φ = 1 / √ 1 + { (0.15 εs / K13) (be / 10es)^2 } ^2 =

1,4 fs D e2² sin (0.5 δ2) / b1 =

ad2000 s3-2 vessel on 3 saddles or rings

Documents