带加强管的裙座的尾部吊耳计算

ATTACHMENT 4 (701-V-01)

TAILING LUG CALCULATION

1. CALCULATION OF LIFTING FORCE

(1) DESIGN DATA

Wo=49870kg(ERECTION WEGHT OF VESSEL)

Df= 1.5(FACTOT)

Y=9590mm(FROM CENTER LINE OF TALING LUG TO C.O.G) X=9690mm(FROM C.O.G TO LUG CENTERLING)

R=2358mm(OUTSIDE RADIUS OF C.L TO TAILING LUG)

** FORMULAS **

LL=Df*Wo*(Y*cosθ+Rsinθ)/((X+Y)*cosθ+R*sinθ)

TL=Df*Wo*(X*cosθ)/((X+Y)*cosθ+R*sinθ)

LV=LL*cosθ

LH=LL*sinθ

TV=TL*cosθ

TH=TL*sinθ

PV=0.5*LH

PH=PV*tan15

2. RESULT OF LIFTING FORCE (θ = 0°to 90°)

MAX. FORCE SUMMARY AT LIFTING AND TAILING LUG

3. STRENGTH OF TAILING LUG SA-516 Gr60USED MATERIAL

Sa =12.02

kg/mm2ALLOWABLE STRESS OF TAILING LUG Sy =22.5kg/mm2YIELD STRESS OF TAILING LUG N =2

TAILING LUG QTY.

TV =37596.496kg LIFTING LOAD (MAX. AT HORIZONTAL COND.)

TV′= TV/N 18798.248kg LIFTING LOAD PER TAILING LUG (VERTICAL FORCE)TH =26994.021kg LIFTING LOAD (MAX. AT VERTICAL COND.) AT 65 deg TH′= TH/N

13497.011

kg

LIFTING LOAD PER TAILING LUG (HORIZONTAL FORCE)

DIMENSIONS

t =42mm R =126mm d =120mm L =240mm L1=252mm W =21mm L3=

122mm TOTAL WELD LENGTH:= L4496

mm

** STRENGTH CALCULATION **

3-1) SHEAR STRESS AT HORIZONTAL STATE = 6.7814748kg/mm2So, S1 = 6.7814748kg/mm2<0.4Sy =9kg/mm2OK!

3-2) TENSION STRESS AT HORIZONTAL STATE = 3.3907374kg/mm2So, S2 = 3.3907374kg/mm2<0.6Sy =13.5kg/mm2OK!

3-3) BENDING STRESS AT MAX. SLOPED LOAD STATE, AT 60 deg.=7.2870157kg/mm2So, S3 = 7.2870157kg/mm2<0.66Sy =14.85kg/mm2OK!

3-4) TENSION STRESS AT MAX. SLOPED LOAD STATE, AT 60 deg.= 2.4345257kg/mm2So, S4 = 2.4345257kg/mm2<0.6Sy =13.5kg/mm2OK!3-5) COMBINE STRESS AT MAX. SLOPED STATE, AT 60 deg.=0.6710434<1

OK!

3-6) SHEAR FORCE IN WELDMENT =18.949847kg/mm FILLET WELD SIZE = 2.8664116mm =21OK!

S4 = TH′/[(2*R-d)*t]Scomb = S3/(0.66Sy)+S4/(0.6Sy)Ws = TV′/(2*L4)W = Ws/(0.55*Sa)USED WELD LEG LENGTH S1 = 2* TV′/[(2*R-d)*t]S2 = TV′/[(2*R-d)*t]S3 = 6*TH′*L/[t*(2*R)^2

]

4. BASE BLOCK UNDER ERECTION CONDITION 4-1) SECTIONAL PROPERTIES OF BASE BLOCK T3 = t1=18mm T1 = t2=28mm T2 = t3=28mm J3 = L2a =79.5mm L2b =93mm J1 = L3=122mm K1 = L =252mm Di =4200mm SA-516 Gr6022.5kg/mm24-2) EFFECTIVE LENGTH OF SKIRT =2109mm 288mm

4-3) SECTION AREA, A =10224mm2=4830mm2=3416mm2=18470mm24-4) CENTROID, A h1 = L2a+0.5*t1=88.5mm h2 = (L2a+L2b)/2=86.25mm h3 = 0.5*L3-(L3-L2a)=18.5mm 74.965214mm

4-5) MOMENT OF INERTIAL OF AREA, I

2148987mm412591974mm415128281mm4

I = I1+I2+I3=29869242mm44-6) MIN. SECTION MODULUS, Z Z = I/C =398441.36

4-7) STRESS IN BASE BLOCK UEO TO TV

a) BENDING MOMENT & DEFLECTION OF BASE BLOCK Ko*W*R^2=2182.5mm = 2.7416578kg/mm Where, TV = PT =37596.496kg

42.564397mm 21000kg/mm2

b) STRESS IN BASE BLOCK Mmax =19589034kg-mm S1 = M/Z =49.164158kg/mm2>0.66Sy =14.85kg/mm2NOT OK!

THEREFORE, STIFFENER BEAM IS REQUIRED!!

DEFLECTION Dx = -W*R ^4/(E*I)*0.4292 = Where, E = modulue of elas. =I3 = (t3*L3^3)/12+A3*(h3-C)^2 =M = W*R 2(1+0.5*cos α-π*sin α+αsinα) =W = TV/(2*π*R)R TOTAL AREA A C = [(A1*h1)+(A2*h2)+(A3*h3)]/A =I1 = [(Le+t3+L)*t1^3]/12+A1*(h1-C)^2 =I2 = [t2*(L2a+L2b)^3]/12+A2*(h2-C)^2 =A1 = t1*(L+Le+t3)Le = Max.(0.78*(Rm*t1)^0.5,16*t1) =A2 = t2*(L2a+L2b)A3 = t3*L3YIELD STRESS OF BASE BLOCK =Rm = (Di+t1)/2

BASE BLOCK MATERIAL: