Rect. Tanks Sample

CONTENT PAGE

1 DESIGN DATA 1

3 WALL DESIGN 2

4 STIFFENER PROPERTIES 20

5 NOZZLES & OPENING 21

7 WEIGHT SUMMARY 38

8 WIND LOADING 39

9 TRANSPORTATION LOAD 40

10 LOAD AT BASE 41

11 LEG DESIGN 42

12 LEG BASEPLATE DESIGN 43

13 LIFTING LUG DESIGN CALCULATION 44

APPENDIX

ROARK'S FORMULA 47

PRESSURE VESSEL HANDBOOK 48

SHACKLE 49

( Cd' x qz x Az ) x 103

DESIGN DATA

DESCRIPTION :TAG NO. : T-6500MANUFAC'S. SERIAL NO : PV-725DIMENSION ( mm ) : 2520 mm (W) x 2520 mm (L) x 2020 mm (H)DESIGN CODE : ASME SECT. VIII. DIV. I, (2004 EDITION) + ROARK'S FORMULA CODE STAMPED : NOTHIRD PARTY : NO

PROPERTIES UNIT DATA

CAPACITY

CONTENT - SEAWATER / OIL

FLUID SPECIFIC GRAVITY - 1.00

DESIGN PRESSURE bar g FULL WATER + 0.05 / - 0.02 BARG

DESIGN TEMPERATURE, 60

HYDROSTATIC TEST PRESSURE bar g FULL OF WATER + 300mm STAND PIPE

IMPACT TEST - NO

RADIOGRAPHY - 10%

CORROSION ALLOWANCE mm in 3.0 0.12

mm3 1.25 x 1010

oC

ROARK'S FORMULA

SIDE WALL DESIGN

ITEM NAME : T-1060,T-1070,T-1080,T-1090 RECT. TANKS

Tank Height, H = 39.37 in 1000 mmTank Width, W = 39.37 in 1000 mmTank Length, L = 59.06 in 1500 mm

Design Pressure = FULL WATER + 0.0075 bar g

Design Temp. = 65Material = SA 516M GR 485

As per Table 26 Case No.1a Chapter 10 of Roark'sRectangular plate, all edges simply supported, with uniform loads over entire plate.

For Section , A (Worst Case)

g = 9.81

1000 a = 19.69 in 500.0 mm

b = 19.69 in 500.0 mma/b = 1.0000

b = 0.2994 Loading q= + Pa

a = 0.0462 = 9810 + 750g = 0.4320 = 1.4225 + 0.1088 psiE = 2.84E+07 psi = 1.5312 psit = 0.1969 in 5.0 mm

c.a = 0.1181 in 3 mmt (corr) = 0.3150 in 8.0 mm

At Center,

Maximum Deflection, == -1.25= 1.25 mm < t/2 then O.K

= 4584 psi < 25081 psi. then OK

Material SA 516M GR 48538002 psi0.121

At center of long side,Maximum reaction force per unit length normal to the plate surface,

R == 13.02 lb/in= 1471.24 N/mm

o C

m/s2

ρwater = kg/m3

ρwater gH ρwater gh

N/m2

-(aqb4)/Et3

Maximum Bending stress, s = (bqb2)/ t2

σallowable

Yield Stress, sy =Stress Ratio, s/sy =

g qb

S

a

S

S

Sb

B

1000

1500

500

500 x 3

500

A

ROARK'S FORMULA

SIDE WALL DESIGN


Tank Height, H = 39.37 in 1000 mmTank Width, W = 39.37 in 1000 mmTank Length, L = 59.06 in 1500 mm




For Section , B (Worst Case)

g = 9.81

1000 a = 24.80 in 630.0 mm

b = 26.50 in 673.0 mma/b = 0.9361




At Center,


= 3890 psi < 25081 psi. then OK

Material SA 36M38001.5 psi

0.102


R == 17.23 lb/in= 1946.61 N/mm

o C

m/s2

ρwater = kg/m3


N/m2

-(aqb4)/Et3


σallowable


g qb

S

a

S

S

Sb

B

6742020

2520

673

630 X 4

A

673

STIFFENER CALCULATIONFor Horizontal Maximum bending moment occurs at the point where dM/dx = 0 and shear force is zero,that is, at the middle of the beam.

Stiffener No. 1 (typical)L = 500 mm = 19.69 in

30.14 lb/in ґ = 500 mm = 19.7 inLoad q = 1.5312 psiunit load W = q x ґ psi

= 30.14 lb/in

FB 50 x 9X

I = 0.733319.69 in

Bending MomentAs per Table 8.1 Case 2e of Roark's (Uniform load on entire span) At x = L/2 = 9.84 in

Maximum moment, == 1460 lb-in

M/I =

=

= 0.088Use FB 50 x 9

I/y = 1.296 > then O.K

Therefore, = 1127 psi < 16500 psi. then O.K

DeflectionAs per Table 8.1 Case 2e of Roark's (Uniform load on entire span)

At x =L/2= 9.84 in

384EI

= 0.003 < L/360 = 0.0547 in. then O.K

Therefore the size used is adequate.

in4

Mmax WL2/8

s/y

(I/y)required M/s

in3

in3 (I/y)required

s σallowable

dmax = (5WL4)

WbWa

STIFFENER CALCULATIONFor Vertical



= 30.14 lb/in

FB 50 x 9X

I = 0.733319.69 in

Bending MomentAs per Table 8.1 Case 2d of Roark's (Uniformly increasing load) At x = 0.548L = 10.79 in


M/I =

=

= 0.015Use FB 50 x 9

I/y = 1.296 > then O.K


DeflectionAs per Table 8.1 Case 2d of Roark's (Uniformly increasing load)

At x = 0.525L = 10.33 in

=EI

= 0.0003 < L/360 = 0.0547 in. then O.K


in4

Mmax 0.0215WL2

s/y

(I/y)required M/s

in3

in3 (I/y)required

s σallowable

dmax 0.001309WL4

WbWa




= 40.57 lb/in

FB 50 x 9X

I = 0.733324.80 in



M/I =

=

= 0.189Use FB 50 x 9

I/y = 1.296 > then O.K



At x =L/2= 12.40 in

384EI

= 0.010 < L/360 = 0.0689 in. then O.K


in4

Mmax WL2/8

s/y

(I/y)required M/s

in3

in3 (I/y)required

s σallowable

dmax = (5WL4)

WbWa




= 37.98 lb/in

FB 50 x 9X

I = 0.733326.50 in



M/I =

=

= 0.035Use FB 50 x 9

I/y = 1.296 > then O.K



At x = 0.525L = 13.91 in

=EI

= 0.0012 < L/360 = 0.0736 in. then O.K


in4

Mmax 0.0215WL2

s/y

(I/y)required M/s

in3

in3 (I/y)required

s σallowable

dmax 0.001309WL4

WbWa

STIFFENER PROPERTIES

Size FB 50 x 9

Material, SS 316L

Yield Stress, 25000 psi

Allowable Stress, 16500 psi

Where :

d1 = 6 mmd2 = 50 mmb1 = 110 mm *b2 = 9 mmC

PART area (a) y a x y h

mm mm1 657.8879 3 1973.664 11.37 129.3449 85094.47 1973.6642 450 31 13950 16.63 276.4574 124405.8 93750

TOTAL 1107.888 15923.66 209500.3 95723.66

Therefore,

14.37 mm

I = 305224 = 0.7333

Z (I/C) = 21235.94 = 1.2959

"*"b1 =

b1 = 110 mm take min. L = 1000 mm, so R:R = 500 mm

σy

σallowable

h2 a x h2 bd3/12

mm2 mm3 mm2 mm4 mm4

C =

mm4 in4

mm3 in3

h1

b2

b1

h2

y2 C

d1

d2

y1 1

2

Plate

Stiffener

tr+1 .56√R∗ts

ROARK'S FORMULA

BOTTOM PLATE DESIGN


Tank Height, H 39.4 in 1000 mmTank Width, W 39.37 in 1000 mmTank Length, L 59.06 in 1500 mm

Design Pressure = FULL WATER + 0.0075 BARG


As per Table 26 Case No.1a Chapter 10 of Roark'sAssume rectangular plate, all edges simply supported, with uniform loads over entire plate

For Section , Each Section (Largest area)

g = 9.81

1000 a = 19.69 in 500 mm

b = 19.69 in 500 mma/b = 1.0000

b = 0.2874 Loading q= + 0.0075 BARG


c.a = 0.1181 in 3 `t (corr) = 0.3150 in 8.0 mm

At Center,


= 4401 psi < 25081 psi. then OK

Material SA 516M GR 48538001.5 psi

0.116


R == 12.66 lb/in

o C

m/s2

ρwater = kg/m3

ρwater gh

N/m2

-(aqb4)/Et3


σallowable


g qb

S

a

S

S

Sb

A1000

1500

500 x 2

500 x 3

STIFFENER CALCULATION (at Bottom Plate)For Short BeamMaximum bending moment occurs at the point where dM/dx = 0 and shear force is zero,that is, at the middle of the beam.

L = 500 mm = 19.69 in30.14 lb/in ґ = 500 mm = 19.7 in

Load q = 1.5312 psiunit load W = q x ґ psi

= 30.14 lb/in

FB 50 x 9X

I = 0.73319.69 in

Maximum bending moment, At x = L/2 = 9.84 in

== 1460 lb-in

M/I =

=

= 0.088Use FB 50 x 9

I/y = 1.296 > then O.K

Therefore, = 1127 psi < 16500 psi. then OK

Maximum Deflection at Center of Beam At x = L/2 = 9.84 in

=384EI

= 0.0028 < L/360 = 0.0547 in. then O.K


in4

Mmax WL2/8

s/y

(I/y)required M/s

in3

in3 (I/y)required

s σallowable

dmax (5WL4)

WbWa

ROARK'S FORMULA

ROOF CALCULATION


Assume rectangular plate, all edges simply supported, with uniform loads over entire plate

Live load, LL = 0.2846 psiRoof weight = 206.353 lbStructure weight = 470 lbConcentrated weight = 661 lbTotal dead load,TDL = 0.2910 psiTotal conc. load, CL = 0.2845 psi

Tank Width, W 39.37 in 1000 mm

Tank Length, L 59.06 in 1500 mm

g = 9.81

1000 a = 19.69 in 500.0 mm

b = 19.69 in 500.0 mma/b = 1.0000

b = 0.2874

a = 0.0444 Loading q = LL + CL + TDLg = 0.4200 = 0.860 psiE = 2.80E+07 psit = 0.11811 in 3.0 mm

c.a = 0.11811 in 3 mm

t (corr) = 0.23622 in 6.0 mm

At Center,

Maximum Deflection, = All. Deflection =1500/300= 5.00 (max) mm= -3.16 mm= 3.16 mm < All. Deflection. O.K = 5.00

= 6866 psi < 25081 psi then OK

Material SA 516M GR 485

38001.5 psi0.181


R == 7.11 lb/in

m/s2

ρwater = kg/m3

-(aqb4)/Et3


σallowable


g qb

S

a

S

S

Sb

A

1000

1280

500 x 2

500 x 3

1500

STIFFENER CALCULATION(at Roof Plate)

Short Beam

Maximum bending moment occurs at the point where dM/dx = 0 and shear force is zero,that is, at the middle of the beam.

L = 500 mm = 19.69 in16.93 lb/in ґ = 500 mm = 19.7 in

Load q = 0.860 psiunit load W = q x ґ psi

= 16.93 lb/in

FB 50 x 9X

I = 0.733319.69 in

Maximum moment, At x = L/2 = 9.84 in

== 2 lb-in

M/I =

=

= 9.487E-05Use FB 50 x 9

I/y = 1.296 > then O.K

Therefore, = 1.208 psi < 16500 psi. then O.K

Maximum deflection at center of beam At x = L/2 = 9.84 in

=384EI

= 0.002 < L/360 = 0.0547 in. then O.K


in4

Mmax WL2/8

s/y

(I/y)required M/s

in3

in3 (I/y)required

s σallowable

dmax (5WL4)

WbWa

ROARK'S FORMULA

BASE WEIR PLATE DESIGN


Weir Plate Height, H = 35.43 in 900 mmWeir Plate Width, W = 98.43 in 2500 mm





g = 9.81

1000 a = 24.61 in 625.0 mm

b = 17.72 in 450.0 mma/b = 1.3889




At Center,


= 5048 psi < 25081 psi. then OK


0.133


R == 11.73 lb/in= 1325.45 N/mm

o C

m/s2

ρwater = kg/m3


N/m2

-(aqb4)/Et3


σallowable


g qb

S

a

S

S

Sb

900

2500

450

625 X 4

A

450




= 27.13 lb/in

FB 50 x 9X

I = 0.733324.61 in



M/I =

=

= 0.124Use FB 50 x 9

I/y = 1.296 > then O.K



At x =L/2= 12.30 in

384EI

= 0.006 < L/360 = 0.0684 in. then O.K


in4

Mmax WL2/8

s/y

(I/y)required M/s

in3

in3 (I/y)required

s σallowable

dmax = (5WL4)

WbWa




= 37.68 lb/in

FB 50 x 9X

I = 0.733317.72 in



M/I =

=

= 0.015Use FB 50 x 9

I/y = 1.296 > then O.K



At x = 0.525L = 9.30 in

=EI

= 0.0002 < L/360 = 0.0492 in. then O.K


in4

Mmax 0.0215WL2

s/y

(I/y)required M/s

in3

in3 (I/y)required

s σallowable

dmax 0.001309WL4

WbWa

ROARK'S FORMULA

ADJUSTABLE WEIR PLATE DESIGN


Weir Plate Height, H = 23.62 in 600 mmWeir Plate Width, W = 98.43 in 2500 mm





g = 9.81

1000 a = 16.46 in 418.0 mm

b = 12.60 in 320.0 mma/b = 1.3063




At Center,


= 4535 psi < 25081 psi. then OK


0.119


R == 5.66 lb/in= 639.95 N/mm

o C

m/s2

ρwater = kg/m3


N/m2

-(aqb4)/Et3


σallowable


g qb

S

a

S

S

Sb

600

2500

315

416

A

285

416

416416

418418




= 19.29 lb/in

FB 50 x 9X

I = 0.733316.46 in



M/I =

=

= 0.040Use FB 50 x 9

I/y = 1.296 > then O.K



At x =L/2= 8.23 in

384EI

= 0.001 < L/360 = 0.0457 in. then O.K


in4

Mmax WL2/8

s/y

(I/y)required M/s

in3

in3 (I/y)required

s σallowable

dmax = (5WL4)

WbWa




= 25.20 lb/in

FB 50 x 9X

I = 0.733312.60 in



M/I =

=

= 0.005Use FB 50 x 9

I/y = 1.296 > then O.K



At x = 0.525L = 6.61 in

=EI

= 0.0000 < L/360 = 0.0350 in. then O.K


in4

Mmax 0.0215WL2

s/y

(I/y)required M/s

in3

in3 (I/y)required

s σallowable

dmax 0.001309WL4

WbWa

WIND LOADING - BS 6399 - PART 2 -1997

Terrain Category = 1

Region = D

Basic Wind Speed Vb = 50.00 m/s

Shielding Factor Ms = 1

Topographic Factor Sa = 1

Direction Factor Sd = 1

Probability Factor Sp = 1

Seasonal Factor Ss = 1

Terrain and Building Factor Sb = 1

Design Wind Speed Vz = 50.00 m/s ( Vb x Sa x Sd x Sp x Ss )

Effective (Design) Wind speed Ve = 50.00 m/s ( Vz x Sb )

Dynamic Pressure qz = 1.5325

Drag Coefficient Cd = 1

H = 1000.000 mm

D = 1000.000 mm

Az = 1000000.000

10

00

H / D = 1.00

1000.000 Kar = 1

Cd' = 1 ( Cd x Kar )

Wind Force Fw = 1532.5 N

Height to COA h = 500.000 mm ( H / 2 )

Overturning Moment Mw = 766250 Nmm ( Fw x h )

Moment at the joint of the leg to the tank

550 mm = 998040 Nmm

kPa ( 0.613 x Ve2 x 10-3 )

mm2

( Cd' x qz x Az ) / 103

hT = Mw1 Mw - hT ( Fw - 0.5*qz*D*hT )

WEIGHT SUMMARY

ITEM : T-1060,T-1070,T-1080,T-1090 RECT. TANKS JOB NO. JN05-320

QTY or ITEM DESCRIPTION UNIT WT. WEIGHT

SIDE PLATE 2.520 m x 2.000 m x ### thk 4 1562.5 kgBASE PLATE 2.520 m x 2.520 m x ### thk 1 590.6 kgROOF PLATE 2.520 m x 2.520 m x 8 thk 1 393.7 kgPARTITION / WEIR PLATE - - - kg

STIFFENER SIDE WALL FB 50 x 9 x 44 m 1 154.9 kg

ROOF PLATE FB 50 x 9 x 15 m 1 52.7 kg

BOTTOM PLATE FB 50 x 9 x 15 m 1 52.7 kg

WEIR PLATE 2.500 m x 1.400 m x 8 thk 1 217.0 kg

ANGLE 75 x 75 x 9t x 3.0 m 1 29.9 kg

NOZZLE / OPENINGS 500.0 kgAND OTHERS

TOTAL WEIGHT 3554 kg

Liquid Weight 12701 kgWater Weight 12701 kg

EMPTY WEIGHT 3554 kgOPERATING WEIGHT 16255 kgFULL WATER WEIGHT 16255 kg

TRANSPORTATION LOADS

TRANSPORTATION ACCELERATIONS

WEIGHTS

ERECTED ……………..… We = 3554 kg -> 34865 N

OPERATING …………….. Wo = 16255 kg -> 159460 N

FLOODED ………….…… Wf = 16255 kg -> 159460 N

VERTICAL = 13.73 m/s ( 1.4 x g )

LONGITUDINAL = 4.91 m/s ( 0.5 x g )

TRANSVERSE = 4.91 m/s ( 0.5 x g )

TRANSPORTATION FORCES

VERTICAL = 48811.4 N

HORIZONTAL = 17432.6 N

TRANSVERSE = 17432.6 N

AtV

AtH

AtT

FtV ( We x AtV )

FtH ( We x AtH )

FtT ( We x AtT )

LOADS AT BASE

WEIGHTS

Erected We = 3554 kg ------> 34865 NOperating Wo = 16255 kg ------> 159460 NFlooded Wf = 16255 kg ------> 159460 N

EXTERNAL LOADS

Wind Force Fw = 1533 NEarthquake Force Feq = 0 N

Blasting Force Fb = 0 N

Client Specified Dynamic Force = 51831 N( during tow-out and installation )

Wind Moment Mw = 766250 NmmEarthquake Moment Meq = 0 Nmm

Blasting Moment Mb = 0 Nmm

Client Specified Moment Mc = 68676 Nmm 1325 mm( during tow-out and installation ) (from base)

Maximun Shear Force F = 51831 N >>> P = F/n = 12957.7 N Maximun O/T Moment M = 766250 Nmm n = 4

HOLD DOWN BOLTS

Bolt Material…………….………………. = SA 193M GR B7

Bolt Yield Stress……… Sy = 207 MPa

Bolt UTS…….…..………… Su = 507 MPa

Allowable Tensile…………… Ft = 124.2 MPa

Allowable Shear…………… Fs = 69 MPa

Bolt Size…………………………………… = M20

Bolt Number………………… N = 4

Tensile Area………….…… = 245

Shear Area…………………… = 225

Bolt PCD……………………… PCD = 2258 mm

AXIAL STRESS IN BOLT SHEAR STRESS IN BOLT

Load / Bolt, P = 4Mw - We Shear / Bolt, S = FPCD.N N N x As

Load / Bolt = -8377 N fs = 57.59 MPa OK

** Since the value is -ve, therefore no axial stress Fs = 69 MPa

since fs < Fs the shear stress is OK

FD [( 0.5 x We )2 + ( 1.4 x We )2 ]0.5

( FD x COGerected ) COGerected =

where, n = no of leg.

AT mm2

AS mm2

LEG DESIGN

LEG DATA

Material……………...………………..= SA 36M

Yield Stress, Sy………….…………..= 248.2

Allowable Axial Stress, fall.…...……= 148.9

Allowable Bending Stress, fball.......= 165.5

LEG GEOMETRY :- ANGLE 90 x 90 x 8t

A = 1390

Ixx = 1040000d = 50 mme = 25 mmL = 450 mmr = 11 mm

AXIAL STRESS

Axial Stress, fa = F / A = 28.68

BENDING STRESS

Bending Stress, fb = P x L x e = 16.58Ixx

COMBINED STRESS

Combined Stress, f = (fa/fall + fb/fball) = 0.29

Since Combined Stress is < 1.00 The Leg Design is OK!

N/mm2

N/mm2 ( 0.6 x Sy )

N/mm2 ( 2/3 x Sy )

mm2

mm4

N/mm2

N/mm2

e

d

X X

LEG BASEPLATE DESIGN

Refer Dennis R Moss Procedure 3-10

tb = 3 x Q x F4 x A x Fb

Q = Maximum Load / Support = 16255 NF = Baseplate Width = 150 mmA = Baseplate Length = 150 mmFb = Allowable Bending Stress = 163.68 MPa ( 0.66 Fy )

tb = 8.6 mm

Use Tb = 16 mm OK

BASE PLATE WELD CHECKING

Maximum stress due to Q & F = max(Q, F)/Aw = 10.80

< 86.9 OK

Weld leg size, g = 8.0 mmLength of weld, l = 2*( 2*F + 2*A ) = 1200 mm

Area of weld, Aw = 0.5*g*l = 4800Joint efficiency for fillet weld, E = 0.6 -

Welding stress for steel, fw = 144.8

Allowable stress for weld, fw = E*fw = 86.9Maximum vertical force, Q = 16254.9 N Maximum horizontal force, F = 51831.0 N

N/mm2

N/mm2

mm2

N/mm2

N/mm2

LIFTING LUG DESIGN CALCULATION

tL rL d

Weight of tank, We = 3,554 kg= 34,865 N

Number of lifting lug, N = 4

1.1 LIFTING LUGDistance, hc = 85 mmDistance k = 106 mmDistance J = 48 mmDistance M = 50 mmLug radius, rL = 50 mmDiameter of hole, d = 40 mmLug thickness, tL = 15 mmPlate thickness, tM = 9 mmLength a = 100 mmLength b = 60 mmPad length, Lp = 150 mmPad width, Wp = 100 mmPad thickness, tp = 6 mmAngle, U (max) = 15 °

Shackle S.W.L : 4.75 tonsType of shackle : Chain ShacklesPin size, Dp = 22.225 mm

2.0 LIFTING LUG MATERIAL & MECHANICAL PROPERTIESMaterial used =SA 240M GR 316 LSpecified yield stress, Sy = 248.22 N/mm²Impact load factor, p = 3.00

3.0 ALLOWABLE STRESSESAllowable tensile stress, St.all ( = 0.6 Sy ) = 148.93 N/mm²Allowable bearing stress, Sbr.all ( = 0.9 Sy ) = 223.40 N/mm²Allowable bending stress, Sbn.all ( = 0.66 Sy ) = 163.83 N/mm²Allowable shear stres = 99.29 N/mm²( Cd' x qz x Az ) x 103

tp

Wpb

a

Lp

5

5

10

tM

hc

k

Fy

Pa

U

J

5

M

A A

4.0 LIFTING LUG DESIGN - VERTICAL LIFTING4.1 DESIGN LOAD

Design load , Wt ( = p.We ) = 104596 NDesign load per lug, W ( = Wt / N ) = 26149 NVertical component force, Fy = 26149 N

4.2 STRESS CHECK AT PIN HOLE(a) Tensile Stress

Vertical component force, Fy = 26149 NCross sectional area of lug eye, Ae ( = 2*[ rL - d/2 ] x tL ) = 900 mm²Tensile stress, St ( = Fy / Ae ) = 29.05 N/mm²Since St < St.all, therefore the lifting lug size is satisfactory.

(b) Bearing StressVertical component force, Fy = 26149 NCross sectional area of lug eye, Ae ( = Dp x tL ) = 333 mm²Bearing stress, Sbr ( = Fy / Ae ) = 78.44 N/mm²Since Sbr < Sbr.all,therefore the lifting lug size is satisfactory.

(c) Shear StressVertical component force, Fy = 26149 NCross sectional area of lug eye, Ae ( = 2.(rL-d/2).tL ) = 900 mm²Shear stress, Ss ( = Fy / Ae ) = 29.05 N/mm²Since Ss < Ss.all,therefore the lifting lug size is satisfactory.

5.0 STRESS CHECK AT SECTION A-A(a) Bending Stress

Bending stress due to Pa ( = Fy x tan U ) = 7007 NBending moment, Mb ( = Pa x J ) = 336316 Nmm

= 3750Bending stress, Sb ( = Mb/Z ) = 89.68 N/mm²Since Sb < Sb.all, therefore the lifting lug size is satisfactory.

(b) Tensile Stress due to FyCross section area, Ae (=2rL x tL) = 1500 mm²Tensile Stress, St (=Fy/Ae) = 17.43 N/mm²Since St < St.all, therefore the lifting lug size is satisfactory.

Combine Stress Ratio, CS (= St/St.all + Sb/Sbn.all) = 0.66Since CS <= 1.0, therefore lifting lug size is satisfactory.

6.0 DESIGN OF WELD SIZE AT LUG TO PAD JOINT6.1 GENERAL

Weld leg , w1 = 10 mmWeld throat thickness, tr1 = 7.1 mmWeld leg , w1 = 10 mmWeld throat thickness, tr2 = 7.1 mmFillet weld joint efficiency, E = 0.6Allowable welding stress for steel grade 43 ( E-43 ) = 125 N/mm²

6.2 CRITICAL WELD CROSS-SECTIONAL PROPERTIESArea of weld, Aw1 ( = tr1 ( a + 2b ) = 1555 mm²Area of weld, Aw2 ( = tr2 ( (2a-tM) + (2M+tM ))) = 2121 mm²Total area of weld, Aw ( = Aw1 + Aw2 ) = 3676 mm²

6.3 STRESS DUE TO FORCE FyComponent force, Fy = 26149 NShear stress, Ssx ( = Fy / Aw ) = 7.11 N/mm²

Section modulus, Z ( = 2rL*tL2/6 mm3

Allowable welding stress, Sa ( = E.Sa ) = 75.00 N/mm²Since Ssx < Sa, therefore the selected weld size i satisfactory .

7.0 DESIGN OF WELD SIZE AT PAD TO TANK JOINT7.1 GENERAL

Weld leg , w = 6 mmWeld throat thickness, tr = 4.2 mmFillet weld joint efficiency, E = 0.6Allowable welding stress for steel grade 43 ( E-43 ) = 125 N/mm²

7.2 CRITICAL WELD CROSS-SECTIONAL PROPERTIESArea of weld, Aw ( = 2 tr ( Wp + Lp ) ) = 2121 mm²

7.3 STRESS DUE TO FORCE FyComponent force, Fy = 26149 NShear stress, Ssx ( = Fy / Aw ) = 12.33 N/mm²Allowable welding stress, Sa ( = E.Sa ) = 75.00 N/mm²Since Ssx < Sa, therefore the selected weld size i satisfactory .

INTERPOLATION TABLE

RANGEa/b 1.20 1.3333 1.40β 0.3762 unknown 0.4530α 0.0616 unknown 0.0770γ 0.4550 unknown 0.4780

β = 0.4274α = 0.0719γ = 0.4703

Rect. Tanks Sample

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