rect tanks sample calculation

8/6/2019 Rect Tanks Sample Calculation

1/33

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


2/33

( Cd' x qz x Az ) x 103


3/33

DESIGN DATA

DESCRIPTION :

TAG NO. : T-6500MANUFAC'S. SERIAL NO : PV-725

DIMENSION ( 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 : NO

THIRD 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

mm

3

1.25 x 10

10

oC


4/33

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 mm

Tank Width, W = 39.37 in 1000 mm

Tank Length, L = 59.06 in 1500 mm

Design Pressure = FULL WATER + 0.01 bar g

Design Temp. = 65

Material = SA 516M GR 485

As per Table 26 Case No.1a Chapter 10 of Roark's

Rectangular 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 mm

a/b = 1.0000

= 0.2994 Loading q= + Pa

= 0.0462 = 9810 + 750 = 0.4320 = 1.4225 + 0.1088 psi = 2.84E+07 psi = 1.5312 psi

t = 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 485

38002 psi

0.121

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

water

= kg/m3

water

gH water

gh

N/m2

-(qb4)/Et3

Maximum Bending stress, = (qb2)/ t2

allowable

Yield Stress, y =

Stress Ratio, /y =

qb

S

a

S

S

Sb

B

1000

1500

500

500 x 3

500

A


5/33

ROARK'S FORMULA

SIDE WALL DESIGN


Tank Height, H = 39.37 in 1000 mm

Tank Width, W = 39.37 in 1000 mm

Tank Length, L = 59.06 in 1500 mm


Design Temp. = 65




For Section , B (Worst Case)g = 9.81

1000

a = 24.80 in 630.0 mm

b = 26.50 in 673.0 mm

a/b = 0.9361


= 0.0413 = 9810 + 750 = 0.4247 = 1.4225 + 0.1088 psi = 2.84E+07 psi = 1.5312 psi

t = 0.2756 in 7.0 mm


At Center,


= -1.33

= 1.33 mm < t/2 then O.K

= 3890 psi < 25081 psi. then OK

Material SA 36M

38001.5 psi

0.102



R =

= 17.23 lb/in

= 1946.61 N/mm

o C

m/s2

water

= kg/m3

water

gH water

gh

N/m2

-(qb4)/Et3


allowable

Yield Stress, y =

Stress Ratio, /y =

qb

S

a

S

S

Sb

B

674

2020

2520

673

630 X 4

A

673


6/33

STIFFENER CALCULATION

For 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 = 500 mm = 19.7 in

Load q = 1.5312 psi

unit load W = q x psi

= 30.14 lb/in

FB 50 x 9

X

I = 0.7333

19.69 in

Bending Moment

As 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 =

=

= 0.088

Use FB 50 x 9I/y = 1.296 > then O.K

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

Deflection


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

/y

(I/y)required M/

in3

in3 (I/y)required

allowable

max

(5WL4)

WbWa


7/33


For Vertical


L = 500 mm = 19.69 in

30.14 lb/in = 500 mm = 19.7 in

Load q = 1.5312 psi


= 30.14 lb/in

FB 50 x 9

X

I = 0.7333

19.69 in

Bending Moment

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

Maximum moment, =

= 251 lb-in

M/I =

=

= 0.015

Use FB 50 x 9

I/y = 1.296 > then O.K


Deflection

As 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

/y

(I/y)required M/

in3

in3 (I/y)required

allowable

max

0.001309WL4

WbWa


8/33


For Horizontal




L = 630 mm = 24.80 in

40.57 lb/in = 673 mm = 26.5 in

Load q = 1.5312 psi


= 40.57 lb/in

FB 50 x 9

X

I = 0.7333

24.80 in

Bending Moment


At x = L/2 = 12.40 in

Maximum moment, =

= 3120 lb-in

M/I =

=

= 0.189



Deflection


At x =L/2= 12.40 in

384EI

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


in4

Mmax WL2/8

/y

(I/y)required M/

in3

in3 (I/y)required

allowable

max

(5WL4)

WbWa


9/33


For Vertical


L = 673 mm = 26.50 in

37.98 lb/in = 630 mm = 24.8 in

Load q = 1.5312 psi


= 37.98 lb/in

FB 50 x 9

X

I = 0.7333

26.50 in

Bending Moment


Maximum moment, =

= 573 lb-in

M/I =

=

= 0.035

Use FB 50 x 9

I/y = 1.296 > then O.K


Deflection


At x = 0.525L = 13.91 in

=

EI

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


in4

Mmax 0.0215WL2

/y

(I/y)required M/

in3

in3 (I/y)required

allowable

max

0.001309WL4

WbWa


10/33

STIFFENER PROPERTIES

Size FB 50 x 9

Material, SS 316L

Yield Stress, 25000 psi

Allowable Stress, 16500 psi

Where :

d1 = 6 mm

d2 = 50 mm

b1 = 110 mm *

b2 = 9 mm

C

PART area (a) y a x y h

mm mm

1 657.89 3 1973.66 11.37 129.34 85094.47 1973.66

2 450 31 13950 16.63 276.46 124405.83 93750

TOTAL 1107.89 15923.66 209500.3 95723.66

Therefore,

14.37 mm

I = 305223.97 = 0.7333

Z (I/C) = 21235.94 = 1.2959

"*"b1 =

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

R = 500 mm

y

allowable

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

tr1 .56Rts


11/33

ROARK'S FORMULA

BOTTOM PLATE DESIGN


Tank Height, H 39.4 in 1000 mm

Tank Width, W 39.37 in 1000 mm

Tank Length, L 59.06 in 1500 mm

Design Pressure = FULL WATER + 0.01 BARG

Design Temp. = 65



Assume 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 mm

a/b = 1.0000

= 0.2874 Loading q= + 0.01 BARG

= 0.0444 = 9810 + 750 = 0.4200 = 1.4225 + 0.1088 psi = 2.80E+07 psi = 1.5312 psi

t = 0.1969 in 5.0 mm

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

At Center,


= -1.21

= 1.21 mm < t/2 then O.K

= 4401 psi < 25081 psi. then OK


38001.5 psi

0.116



R =

= 12.66 lb/in

o C

m/s2

water

= kg/m3

water

gh

N/m2

-(qb4)/Et3


allowable

Yield Stress, y =

Stress Ratio, /y =

qb

S

a

S

S

Sb

A1000

1500

500 x 2

500 x 3


12/33


13/33

STIFFENER CALCULATION (at Bottom Plate)

For Short Beam



L = 500 mm = 19.69 in

30.14 lb/in = 500 mm = 19.7 in

Load q = 1.5312 psi


= 30.14 lb/in

FB 50 x 9

X

I = 0.733

19.69 in

Maximum bending moment,

At x = L/2 = 9.84 in

=

= 1460 lb-in

M/I =

=

= 0.088

Use 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

/y

(I/y)required M/

in3

in3 (I/y)required

allowable

max

(5WL4)

WbWa


14/33

ROARK'S FORMULA

ROOF CALCULATION


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

Live load, LL = 0.2846 psi

Roof weight = 206.353 lb

Structure weight = 470 lb

Concentrated weight = 661 lb

Total dead load,TDL = 0.2910 psi

Total 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 mm

a/b = 1.0000

= 0.29

= 0.0444 Loading q = LL + CL + TDL = 0.4200 = 0.860 psi

= 2.80E+07 psit = 0.12 in 3.0 mm

c.a = 0.12 in 3 mm

t (corr) = 0.24 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


38001.5 psi

0.181



R =

= 7.11 lb/in

m/s2

water

= kg/m3

-(qb4)/Et3


allowable

Yield Stress, y =

Stress Ratio, /y =

qb

S

a

S

S

Sb

A

1000

1280

500 x 2

500 x 3

1500


15/33


16/33

STIFFENER CALCULATION(at Roof Plate)

Short Beam



L = 500 mm = 19.69 in

16.93 lb/in = 500 mm = 19.7 in

Load q = 0.860 psi


= 16.93 lb/in

FB 50 x 9

X

I = 0.7333

19.69 in

Maximum moment,

At x = L/2 = 9.84 in

=

= 2 lb-in

M/I =

== 9.487E-05

Use 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

/y

(I/y)required

M/

in3

in3 (I/y)required

allowable

max

(5WL4)

WbWa


17/33

ROARK'S FORMULA

BASE WEIR PLATE DESIGN


Weir Plate Height, H = 35.43 in 900 mm

Weir Plate Width, W = 98.43 in 2500 mm


Design Temp. = 65





1000

a = 24.61 in 625.0 mm

b = 17.72 in 450.0 mm

a/b = 1.3889


= 0.0761 = 8829 + 750 = 0.4767 = 1.2802 + 0.1088 psi = 2.84E+07 psi = 1.3890 psi

t = 0.1969 in 5.0 mm


At Center,


= -1.22

= 1.22 mm < t/2 then O.K

= 5048 psi < 25081 psi. then OK

Material SA 36M

38001.5 psi

0.133



R =

= 11.73 lb/in

= 1325.45 N/mm

o C

m/s2

water

= kg/m3

water

gH water

gh

N/m2

-(qb4)/Et3


allowable

Yield Stress, y =

Stress Ratio, /y =

qb

S

a

S

S

Sb

900

2500

450

625 X 4

A

450


18/33


For Horizontal




L = 625 mm = 24.61 in

27.13 lb/in = 450 mm = 17.7 in

Load q = 1.5312 psi


= 27.13 lb/in

FB 50 x 9

X

I = 0.7333

24.61 in

Bending Moment


At x = L/2 = 12.30 in

Maximum moment, =

= 2053 lb-in

M/I =

=

= 0.124



Deflection


At x =L/2= 12.30 in

384EI

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


in4

Mmax WL2/8

/y

(I/y)required M/

in3

in3 (I/y)required

allowable

max

(5WL4)

WbWa


19/33


20/33

ROARK'S FORMULA

ADJUSTABLE WEIR PLATE DESIGN


Weir Plate Height, H = 23.62 in 600 mm

Weir Plate Width, W = 98.43 in 2500 mm


Design Temp. = 65





1000

a = 16.46 in 418.0 mm

b = 12.60 in 320.0 mm

a/b = 1.3063


= 0.0698 = 5886 + 750 = 0.4672 = 0.8535 + 0.1088 psi = 2.84E+07 psi = 0.9622 psi

t = 0.1185 in 3.0 mm


At Center,


= -0.91

= 0.91 mm < t/2 then O.K

= 4535 psi < 25081 psi. then OK

Material SA 36M

38001.5 psi

0.119



R =

= 5.66 lb/in

= 639.95 N/mm

o C

m/s2

water

= kg/m3

water

gH water

gh

N/m2

-(qb4)/Et3


allowable

Yield Stress, y =

Stress Ratio, /y =

qb

S

a

S

S

Sb

600

2500

315

416

A

285

416

416416

418418


21/33


For Horizontal




L = 418 mm = 16.46 in

19.29 lb/in = 320 mm = 12.6 in

Load q = 1.5312 psi


= 19.29 lb/in

FB 50 x 9

X

I = 0.7333

16.46 in

Bending Moment


At x = L/2 = 8.23 in

Maximum moment, =

= 653 lb-in

M/I =

=

= 0.040



Deflection


At x =L/2= 8.23 in

384EI

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


in4

Mmax WL2/8

/y

(I/y)required M/

in3

in3 (I/y)required

allowable

max

(5WL4)

WbWa


22/33


For Vertical


L = 320 mm = 12.60 in

25.20 lb/in = 418 mm = 16.5 in

Load q = 1.5312 psi


= 25.20 lb/in

FB 50 x 9

X

I = 0.7333

12.60 in

Bending Moment


Maximum moment, =

= 86 lb-in

M/I =

=

= 0.005

Use FB 50 x 9

I/y = 1.296 > then O.K


Deflection


At x = 0.525L = 6.61 in

=

EI

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


in4

Mmax 0.0215WL2

/y

(I/y)required M/

in3

in3 (I/y)required

allowable

max

0.001309WL4

WbWa


23/33

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

1000

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*h

T)


24/33

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 10 thk 4 1562.5 kg

BASE PLATE 2.520 m x 2.520 m x 12 thk 1 590.6 kg

ROOF PLATE 2.520 m x 2.520 m x 8 thk 1 393.7 kg

PARTITION / WEIR PLATE - - - kg

STIFFENER

SIDE WALL FB 50 x 9 x 44.4 m 1 154.9 kg

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

BOTTOM PLATE FB 50 x 9 x ### 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 kg

AND OTHERS

TOTAL WEIGHT 3554 kg

Liquid Weight 12701 kgWater Weight 12701 kg

EMPTY WEIGHT 3554 kg

OPERATING WEIGHT 16255 kg

FULL WATER WEIGHT 16255 kg


25/33

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

)


26/33

LOADS AT BASE

WEIGHTS

Erected We = 3554 kg ------> 34865 N

Operating Wo = 16255 kg ------> 159460 N

Flooded Wf = 16255 kg ------> 159460 N

EXTERNAL LOADS

Wind Force Fw = 1533 N

Earthquake Force Feq = 0 N

Blasting Force Fb = 0 N

Client Specified Dynamic Force = 51831 N

( during tow-out and installation )

Wind Moment Mw = 766250 Nmm

Earthquake 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 = F

PCD.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


27/33

LEG DESIGN

LEG DATA

Material.....= SA 36M

Yield Stress, Sy...= 248.2Allowable Axial Stress, fall....= 148.9

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

LEG GEOMETRY :- ANGLE 90 x 90 x 8t

A = 1390

Ixx = 1040000

d = 50 mm

e = 25 mm

L = 450 mm

r = 11 mm

AXIAL STRESS

Axial Stress, fa = F / A = 28.68

BENDING STRESS

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

Ixx

COMBINED STRESS

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

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

N/mm

2

N/mm2 ( 0.6 x Sy )

N/mm2 ( 2/3 x Sy )

mm2

mm4

N/mm2

N/mm2

e

d

X X


28/33

LEG BASEPLATE DESIGN

Refer Dennis R Moss Procedure 3-10

tb = 3 x Q x F

4 x A x Fb

Q = Maximum Load / Support = 16255 N

F = Baseplate Width = 150 mm

A = Baseplate Length = 150 mm

Fb = 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 mm

Length of weld, l = 2*( 2*F + 2*A ) = 1200 mm

Area of weld, Aw = 0.5*g*l = 4800

Joint 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/mm

2


29/33


30/33

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 LUG

Distance, hc = 85 mm

Distance k = 106 mm

Distance J = 48 mmDistance M = 50 mm

Lug radius, rL = 50 mm

Diameter of hole, d = 40 mm

Lug thickness, tL = 15 mm

Plate thickness, tM = 9 mm

Length a = 100 mm

Length b = 60 mm

Pad length, Lp = 150 mm

Pad width, Wp = 100 mm

Pad thickness, tp = 6 mm

Angle, U (max) = 15

Shackle S.W.L : 4.75 tons

Type of shackle : Chain Shackles

Pin size, Dp = 22.225 mm

2.0 LIFTING LUG MATERIAL & MECHANICAL PROPERTIES

Material used =SA 240M GR 316 L

Specified yield stress, Sy = 248.22 N/mm

Impact load factor, p = 3.00

3.0 ALLOWABLE STRESSESAllowable tensile stress, St.all ( = 0.6 Sy ) = 148.93 N/mm

Allowable bearing stress, Sbr.all ( = 0.9 Sy ) = 223.40 N/mm

Allowable bending stress, Sbn.all ( = 0.66 Sy ) = 163.83 N/mm

Allowable shear stre = 99.29 N/mm( 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


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4.0 LIFTING LUG DESIGN - VERTICAL LIFTING

4.1 DESIGN LOAD

Design load , Wt ( = p.We ) = 104596 N

Design load per lug, W ( = Wt / N ) = 26149 N

Vertical component force, Fy = 26149 N

4.2 STRESS CHECK AT PIN HOLE

(a) Tensile Stress


Cross sectional area of lug eye, Ae ( = 2*[ rL - d/2 ] x tL ) = 900 mm

Tensile stress, St ( = Fy / Ae ) = 29.05 N/mm

Since St < St.all, therefore the lifting lug size is satisfactory.

(b) Bearing Stress


Cross sectional area of lug eye, Ae ( = Dp x tL ) = 333 mm

Bearing stress, Sbr ( = Fy / Ae ) = 78.44 N/mm

Since Sbr < Sbr.all,therefore the lifting lug size is satisfactory.

(c)Shear Stress


Cross sectional area of lug eye, Ae ( = 2.(rL-d/2).tL ) = 900mm

Shear stress, Ss ( = Fy / Ae ) = 29.05 N/mm

Since 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 N

Bending moment, Mb ( = Pa x J ) = 336316 Nmm

= 3750

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

(b) Tensile Stress due to Fy

Cross section area, Ae (=2rL x tL) = 1500 mm

Tensile Stress, St (=Fy/Ae) = 17.43 N/mm

Since St < St.all, therefore the lifting lug size is satisfactory.

Combine Stress Ratio, CS (= St/St.all + Sb/Sbn.all) = 0.66

Since CS


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Allowable welding stress, Sa ( = E.Sa ) = 75.00 N/mm

Since Ssx < Sa, therefore the selected weld size i satisfactory .

7.0 DESIGN OF WELD SIZE AT PAD TO TANK JOINT

7.1 GENERAL

Weld leg , w = 6 mm

Weld throat thickness, tr = 4.2 mm

Fillet weld joint efficiency, E = 0.6

Allowable welding stress for steel grade 43 ( E-43 ) = 125 N/mm

7.2 CRITICAL WELD CROSS-SECTIONAL PROPERTIES

Area of weld, Aw ( = 2 tr ( Wp + Lp ) ) = 2121 mm

7.3 STRESS DUE TO FORCE Fy

Component force, Fy = 26149 N

Shear stress, Ssx ( = Fy / Aw ) = 12.33 N/mm

Allowable welding stress, Sa ( = E.Sa ) = 75.00 N/mm

Since Ssx < Sa, therefore the selected weld size i satisfactory .


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INTERPOLATION TABLE

RANGE

a/b 1.20 1.3333 1.40

0.3762 unknown 0.4530

0.0616 unknown 0.0770

0.4550 unknown 0.4780

= 0.4274

= 0.0719

= 0.4703

rect tanks sample calculation

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