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. MANUFAC'S. SERIAL NO DIMENSION ( mm ) DESIGN CODE CODE STAMPED THIRD PARTY

: : : : : : :

T-6500 PV-725 2520 mm (W) x 2520 mm (L) x 2020 mm (H) ASME SECT. VIII. DIV. I, (2004 EDITION) + ROARK'S FORMULA NO NO

PROPERTIES

UNIT

DATA

CAPACITY

mm3

1.25 x 1010

CONTENT

-

SEAWATER / OIL

FLUID SPECIFIC GRAVITY

-

1.00

DESIGN PRESSURE

bar g

DESIGN TEMPERATURE,

HYDROSTATIC TEST PRESSURE

FULL WATER + 0.05 / - 0.02 BARG

C

60

o

bar g

FULL OF WATER + 300mm STAND PIPE

IMPACT TEST

-

NO

RADIOGRAPHY

-

10%

CORROSION ALLOWANCE

mm

in

3.0

0.12

ROARK'S FORMULA 1500

SIDE WALL DESIGN ITEM NAME :

T-1060,T-1070,T-1080,T-1090 RECT. TANKS 500

Tank Height, H Tank Width, W Tank Length, L

= = =

39.37 in 39.37 in 59.06 in

1000 1000 1500

mm mm mm

Design Pressure Design Temp. Material

= FULL WATER + o = 65 C = SA 516M GR 485

0.01

bar g

1000

A

B 500 x 3

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 , g = 9.81 ρwater = 1000 a= b= a/b = β= α= γ= Ε= t= c.a = t (corr) =

S

A (Worst Case) m/s2

19.69 19.69 1.0000 0.2994 0.0462 0.4320 2.84E+07 0.1969 0.1181 0.3150

kg/m in in

a

3

500.0 500.0

psi in in in

5.0 3 8.0

mm mm

b

S

Loading q= = = = mm mm mm

ρwater gH 9810 1.4225 1.5312

S + + + psi

S

ρwater gh

Pa 750 N/m2 0.1088 psi

At Center, = -(αqb4)/Et3 = -1.25 = 1.25

Maximum Deflection,

Maximum Bending stress, σ = = Material Yield Stress, σy = Stress Ratio, σ/σy =

4584

mm

< t/2 then O.K

(βqb2)/ t2 psi

<

σallowable

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

= = =

γ qb 13.02 1471.24

lb/in N/mm

25081

psi. then OK

500

ROARK'S FORMULA 2520

SIDE WALL DESIGN ITEM NAME :

T-1060,T-1070,T-1080,T-1090 RECT. TANKS

Tank Height, H Tank Width, W Tank Length, L

= = =

39.37 in 39.37 in 59.06 in

1000 1000 1500

mm mm mm

Design Pressure Design Temp. Material

= FULL WATER + o = 65 C = SA 516M GR 485

0.01

bar g

673 2020 674

A

B 630 X 4

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 , g = 9.81 ρwater = 1000 a= b= a/b = β= α= γ= Ε= t= c.a = t (corr) =

S

B (Worst Case) m/s2

24.80 26.50 0.9361 0.2749 0.0413 0.4247 2.84E+07 0.2756 0.1181 0.3937

kg/m in in

a

3

psi in in in

b

S

630.0 673.0

mm mm

7.0 3 10.0

Loading q= = = = mm mm mm

ρwater gH 9810 1.4225 1.5312

S + + + psi

S

ρwater gh

Pa 750 N/m2 0.1088 psi

At Center, = -(αqb4)/Et3 = -1.33 = 1.33

Maximum Deflection,

Maximum Bending stress, σ = = Material Yield Stress, σy = Stress Ratio, σ/σy =

3890

mm

< t/2 then O.K

(βqb2)/ t2 psi

<

σallowable

SA 36M 38001.5 psi 0.102

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

= = =

γ qb 17.23 1946.61

lb/in N/mm

25081

psi. then OK

673

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 ґ= 500 mm = 19.7 in Load q = 1.5312 psi unit load W = q x ґ psi = 30.14 lb/in

30.14 lb/in

FB 50 x 9 X I 19.69 in

=

0.7333 in4

Wb

Wa

Bending Moment As per Table 8.1 Case 2e of Roark's (Uniform load on entire span) At x = L/2 = 9.84 in Mmax Maximum moment, = WL2/8 = 1460 lb-in M/I

=

(I/y)required = =

σ/y M/σ 0.088 in3

Use FB 50 x 9

Therefore,

I/y

=

1.296

in3

>

(I/y)required

then O.K

σ

=

1127

psi

<

σallowable

16500

Deflection As per Table 8.1 Case 2e of Roark's (Uniform load on entire span) At x =L/2= δmax

9.84 in = (5WL4) 384EI

=

0.003

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

Therefore the size used is adequate.

psi. then O.K

STIFFENER CALCULATION For Vertical Stiffener No.

1 (typical) L= 500 ґ= 500 Load q = unit load W = =

30.14 lb/in

mm = mm = 1.5312 qxґ 30.14

19.69 19.7 psi psi lb/in

FB 50 x 9 X I 19.69 in

=

0.7333

in4

Wb

Wa

Bending Moment As per Table 8.1 Case 2d of Roark's (Uniformly increasing load) At x = 0.548L = 10.79 in Mmax Maximum moment, = 0.0215WL2 = 251 lb-in M/I

=

(I/y)required

= =

σ/y M/σ 0.015 in3

Use FB 50 x 9

Therefore,

I/y

=

1.296

in3

>

σ

=

194

psi

<

(I/y)required then O.K σallowable

Deflection As per Table 8.1 Case 2d of Roark's (Uniformly increasing load) At x = 0.525L = δmax

=

=

10.33 in 0.001309WL4 EI

0.0003 < L/360 =

0.0547 in. then O.K

Therefore the size used is adequate.

16500 psi. then O.K

in in

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= 630 mm = 24.80 in ґ= 673 mm = 26.5 in Load q = 1.5312 psi unit load W = q x ґ psi = 40.57 lb/in

40.57 lb/in

FB 50 x 9 X I 24.80 in

=

0.7333 in4

Wb

Wa

Bending Moment As per Table 8.1 Case 2e of Roark's (Uniform load on entire span) At x = L/2 = 12.40 in Mmax Maximum moment, = WL2/8 = 3120 lb-in M/I

=

(I/y)required = =

σ/y M/σ 0.189 in3

Use FB 50 x 9

Therefore,

I/y

=

1.296

in3

>

(I/y)required

then O.K

σ

=

2408

psi

<

σallowable

16500

Deflection As per Table 8.1 Case 2e of Roark's (Uniform load on entire span) At x =L/2= δmax

12.40 in = (5WL4) 384EI

=

0.010

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

Therefore the size used is adequate.

psi. then O.K

STIFFENER CALCULATION For Vertical Stiffener No.

1 (typical) L= 673 ґ= 630 Load q = unit load W = =

37.98 lb/in

mm = mm = 1.5312 qxґ 37.98

26.50 24.8 psi psi lb/in

FB 50 x 9 X I 26.50 in

=

0.7333

in4

Wb

Wa

Bending Moment As per Table 8.1 Case 2d of Roark's (Uniformly increasing load) At x = 0.548L = 14.52 in Mmax Maximum moment, = 0.0215WL2 = 573 lb-in M/I

=

(I/y)required

= =

σ/y M/σ 0.035 in3

Use FB 50 x 9

Therefore,

I/y

=

1.296

in3

>

σ

=

442

psi

<

(I/y)required then O.K σallowable

Deflection As per Table 8.1 Case 2d of Roark's (Uniformly increasing load) At x = 0.525L = δmax

=

=

13.91 in 0.001309WL4 EI

0.0012 < L/360 =

0.0736 in. then O.K

Therefore the size used is adequate.

16500 psi. then O.K

in in

STIFFENER PROPERTIES Size

FB 50 x 9

Material, Yield Stress,

SS 316L σy

25000 psi

Allowable Stress,

σallowable

16500 psi

Stiffener

b2

Where :

h2 d2

2 y2

h1

C

1

y1

d1 d2 b1 b2 C

= = = =

6 mm 50 mm 110 mm * 9 mm

d1

b1 Plate

PART 1 2 TOTAL

area (a) mm2 657.89 450 1107.89

y mm 3 31

axy mm3 1973.66 13950 15923.66

h mm 11.37 16.63

h2 mm2 129.34 276.46

a x h2 bd3/12 mm4 mm4 85094.47 1973.66 124405.83 93750 209500.3 95723.66

305223.97 mm4

=

0.7333

in4

Z (I/C) = 21235.94 mm3

=

1.2959

in3

Therefore, C =

14.37 I =

"*" b1 = b1 =

mm

tr1.5 6 R∗ts 110 mm

take min. L = R=

1000 mm, so R: 500 mm

ROARK'S FORMULA BOTTOM PLATE DESIGN ITEM NAME :

1500

T-1060,T-1070,T-1080,T-1090 RECT. TANKS

Tank Height, H Tank Width, W Tank Length, L Design Pressure Design Temp. Material =

39.4 in 39.37 in 59.06 in = FULL WATER o = 65 C SA 516M GR 485

+

500 x 2

1000 1000 1500

mm mm mm

0.01

BARG

1000

A

500 x 3

As per Table 26 Case No.1a Chapter 10 of Roark's Assume rectangular plate, all edges simply supported, with uniform loads over entire plate For Section , g= 9.81 ρwater = 1000 a= 19.69 b = 19.69 a/b = 1.0000 β = 0.2874 α = 0.0444 γ = 0.4200 Ε = 2.80E+07 t = 0.1969 c.a = 0.1181 t (corr) = 0.3150

S

Each Section (Largest area) m/s2 kg/m in in

a

3

psi in in in

b

S

500 500

mm mm

5.0 3 8.0

Loading q= = = = mm ` mm

ρwater gh 9810 1.4225 1.5312

S

S + + + psi

0.01 750 0.1088

BARG N/m2 psi

At Center, = -(αqb4)/Et3 = -1.21 = 1.21

Maximum Deflection,

mm

< t/2 then O.K

Maximum Bending stress, σ = (βqb2)/ t2 =

4401

psi

<

σallowable

Material SA 516M GR 485 Yield Stress, σy = 38001.5 psi Stress Ratio, σ/σy = 0.116 At center of long side, Maximum reaction force per unit length normal to the plate surface, R

= =

γ qb 12.66

lb/in

25081

psi. then OK

STIFFENER CALCULATION (at Bottom Plate) For 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 ґ= 500 Load q = unit load W = =

30.14 lb/in

mm = 19.69 in mm = 19.7 in 1.5312 psi q x ґ psi 30.14 lb/in

FB 50 x 9 X I 19.69 in

0.733 in4

Wb

Wa

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

=

in Mmax

= =

M/I

=

(I/y)required

= =

WL2/8 1460 lb-in σ/y M/σ 0.088 in3

Use FB 50 x 9

Therefore,

I/y

=

1.296

in3

>

σ

=

1127

psi

<

(I/y)required then O.K σallowable

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

=

=

0.0028

(5WL4) 384EI < L/360 = 0.0547 in. then O.K

Therefore the size used is adequate.

16500 psi. then OK

ROARK'S FORMULA ROOF CALCULATION ITEM NAME :

T-1060,T-1070,T-1080,T-1090 RECT. TANKS

Assume rectangular plate, all edges simply supported, with uniform loads over entire plate S 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 1280 1500

a b

S

S

S Tank Width, W Tank Length, L g= 9.81 ρwater = 1000

39.37 in 59.06 in

1000 1500

mm mm

m/s2

500 x 2

kg/m3 in in

a= 19.69 b = 19.69 a/b = 1.0000 β= 0.29 α = 0.0444 γ = 0.4200 Ε = 2.80E+07 psi t= 0.12 in c.a = 0.12 in t (corr) = 0.24 in

1000 500.0 500.0

mm mm

Loading q

3.0 3 6.0

A = =

LL + CL + TDL 0.860 psi

500 x 3

mm mm mm

At Center, = -(αqb4)/Et3 All. Deflection =1500/300= = -3.16 mm = 3.16 mm < All. Deflection. O.K =

Maximum Deflection,

5.00 (max) mm 5.00

Maximum Bending stress, σ = (βqb2)/ t2 =

6866

psi

<

σallowable

Material SA 516M GR 485 Yield Stress, σy = 38001.5 psi Stress Ratio, σ/σy = 0.181 At center of long side, Maximum reaction force per unit length normal to the plate surface, R

= =

γ qb 7.11

lb/in

25081

psi then OK

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 ґ= 500 Load q = unit load W = =

16.93 lb/in

mm = mm = 0.860 qxґ 16.93

19.69 in 19.7 in psi psi lb/in

FB 50 x 9 X =

I 19.69 in

Wb

Wa

Maximum moment, At x = L/2 = 9.84

0.7333 in4

in Mmax

= =

WL2/8 2

M/I

=

σ/y

(I/y)required

= =

lb-in

M/σ 9.487E-05 in3

Use FB 50 x 9 I/y Therefore, σ

=

1.296

in3

>

(I/y)required then O.K

=

1.208

psi

<

σallowable

16500 psi. then O.K

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

=

=

0.002

(5WL4) 384EI < L/360 = 0.0547

Therefore the size used is adequate.

in. then O.K

ROARK'S FORMULA 2500

BASE WEIR PLATE DESIGN ITEM NAME : Weir Plate Height, H Weir Plate Width, W

T-1060,T-1070,T-1080,T-1090 RECT. TANKS = =

35.43 in 98.43 in

900 2500

mm mm

450

A

900

450 Design Pressure Design Temp. Material

= FULL WATER + o = 65 C = SA 516M GR 485

0.01

bar g 625 X 4

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 m/s2 ρwater = 1000 kg/m3 a= 24.61 in 625.0 b= 17.72 in 450.0 a/b = 1.3889 β = 0.4487 α= 0.0761 γ = 0.4767 Ε = 2.84E+07 psi t= 0.1969 in 5.0 c.a = 0.1181 in 3 t (corr) = 0.3150 in 8.0

S a mm mm

b

S

Loading q= = = = mm mm mm

ρwater gH 8829 1.2802 1.3890

S + + + psi

S

ρwater gh

Pa 750 N/m2 0.1088 psi

At Center, = -(αqb4)/Et3 = -1.22 = 1.22

Maximum Deflection,

Maximum Bending stress, σ = = Material Yield Stress, σy = Stress Ratio, σ/σy =

mm

< t/2 then O.K

(βqb2)/ t2

5048

psi

<

σallowable

SA 36M 38001.5 psi 0.133

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

= = =

γ qb 11.73 1325.45

lb/in N/mm

25081

psi. then OK

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= 625 mm = 24.61 in ґ= 450 mm = 17.7 in Load q = 1.5312 psi unit load W = q x ґ psi = 27.13 lb/in

27.13 lb/in

FB 50 x 9 X I 24.61 in

=

0.7333 in4

Wb

Wa

Bending Moment As per Table 8.1 Case 2e of Roark's (Uniform load on entire span) At x = L/2 = 12.30 in Mmax Maximum moment, = WL2/8 = 2053 lb-in M/I

=

(I/y)required = =

σ/y M/σ 0.124 in3

Use FB 50 x 9

Therefore,

I/y

=

1.296

in3

>

(I/y)required

then O.K

σ

=

1584

psi

<

σallowable

16500

Deflection As per Table 8.1 Case 2e of Roark's (Uniform load on entire span) At x =L/2= δmax

12.30 in = (5WL4) 384EI

=

0.006

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

Therefore the size used is adequate.

psi. then O.K

STIFFENER CALCULATION For Vertical Stiffener No.

1 (typical) L= 450 ґ= 625 Load q = unit load W = =

37.68 lb/in

mm = mm = 1.5312 qxґ 37.68

17.72 24.6 psi psi lb/in

FB 50 x 9 X I 17.72 in

=

0.7333

in4

Wb

Wa

Bending Moment As per Table 8.1 Case 2d of Roark's (Uniformly increasing load) At x = 0.548L = 9.71 in Mmax Maximum moment, = 0.0215WL2 = 254 lb-in M/I

=

(I/y)required

= =

σ/y M/σ 0.015 in3

Use FB 50 x 9

Therefore,

I/y

=

1.296

in3

>

σ

=

196

psi

<

(I/y)required then O.K σallowable

Deflection As per Table 8.1 Case 2d of Roark's (Uniformly increasing load) At x = 0.525L = δmax

=

=

9.30 in 0.001309WL4 EI

0.0002 < L/360 =

0.0492 in. then O.K

Therefore the size used is adequate.

16500 psi. then O.K

in in

ROARK'S FORMULA 2500

ADJUSTABLE WEIR PLATE DESIGN ITEM NAME :

T-1060,T-1070,T-1080,T-1090 RECT. TANKS

Weir Plate Height, H Weir Plate Width, W

= =

23.62 in 98.43 in

600 2500

mm mm

Design Pressure Design Temp. Material

= FULL WATER + o = 65 C = SA 516M GR 485

0.01

bar g

315

A 600

285 416

416

418 418 416

416

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 m/s2 ρwater = 1000 kg/m3 a= 16.46 in 418.0 b= 12.60 in 320.0 a/b = 1.3063 β = 0.4170 α= 0.0698 γ = 0.4672 Ε = 2.84E+07 psi t= 0.1185 in 3.0 c.a = 0.1181 in 3 t (corr) = 0.2366 in 6.0

S a mm mm

b

S

Loading q= = = = mm mm mm

ρwater gH 5886 0.8535 0.9622

S + + + psi

S

ρwater gh

Pa 750 N/m2 0.1088 psi

At Center, = -(αqb4)/Et3 = -0.91 = 0.91

Maximum Deflection,

Maximum Bending stress, σ = = Material Yield Stress, σy = Stress Ratio, σ/σy =

mm

< t/2 then O.K

(βqb2)/ t2

4535

psi

<

σallowable

SA 36M 38001.5 psi 0.119

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

= = =

γ qb 5.66 639.95

lb/in N/mm

25081

psi. then OK

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= 418 mm = 16.46 in ґ= 320 mm = 12.6 in Load q = 1.5312 psi unit load W = q x ґ psi = 19.29 lb/in

19.29 lb/in

FB 50 x 9 X I 16.46 in

=

0.7333 in4

Wb

Wa

Bending Moment As per Table 8.1 Case 2e of Roark's (Uniform load on entire span) At x = L/2 = 8.23 in Mmax Maximum moment, = WL2/8 = 653 lb-in M/I

=

(I/y)required = =

σ/y M/σ 0.040 in3

Use FB 50 x 9

Therefore,

I/y

=

1.296

in3

>

(I/y)required

then O.K

σ

=

504

psi

<

σallowable

16500

Deflection As per Table 8.1 Case 2e of Roark's (Uniform load on entire span) At x =L/2= δmax

8.23 in = (5WL4) 384EI

=

0.001

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

Therefore the size used is adequate.

psi. then O.K

STIFFENER CALCULATION For Vertical Stiffener No.

1 (typical) L= 320 ґ= 418 Load q = unit load W = =

25.20 lb/in

mm = mm = 1.5312 qxґ 25.20

12.60 16.5 psi psi lb/in

FB 50 x 9 X I 12.60 in

=

0.7333

in4

Wb

Wa

Bending Moment As per Table 8.1 Case 2d of Roark's (Uniformly increasing load) At x = 0.548L = 6.90 in Mmax Maximum moment, = 0.0215WL2 = 86 lb-in M/I

=

(I/y)required

= =

σ/y M/σ 0.005 in3

Use FB 50 x 9

Therefore,

I/y

=

1.296

in3

>

σ

=

66

psi

<

(I/y)required then O.K σallowable

Deflection As per Table 8.1 Case 2d of Roark's (Uniformly increasing load) At x = 0.525L = δmax

=

=

6.61 in 0.001309WL4 EI

0.0000 < L/360 =

0.0350 in. then O.K

Therefore the size used is adequate.

16500 psi. then O.K

in in

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

kPa ( 0.613 x Ve2 x 10-3 )

Drag Coefficient

Cd

=

1

H

=

1000.000

mm

D

=

1000.000

mm

Az

=

H/D

=

1.00

Kar

=

1

Cd'

=

1

1000

1000000.000 mm2

1000.000

Wind Force Height to COA Overturning Moment

Fw

=

1532.5

N

( Cd' x qz x Az ) / 103

h

=

500.000

mm

(H/2)

Mw

=

766250

Nmm

( Fw x h )

998040

Nmm

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

Moment at the joint of the leg to the tank hT = 550 Mw1 mm =

( Cd x Kar )

WEIGHT SUMMARY ITEM : JOB NO.

T-1060,T-1070,T-1080,T-1090 RECT. TANKS JN05-320

ITEM

QTY or UNIT WT. WEIGHT

DESCRIPTION

SIDE PLATE 2.520 m x 2.000 m BASE PLATE 2.520 m x 2.520 m ROOF PLATE 2.520 m x 2.520 m PARTITION / WEIR PLATE STIFFENER SIDE WALL

x x x

10 thk 12 thk 8 thk

4 1 1 -

1562.5 590.6 393.7 -

kg kg kg kg

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

1

217.0

kg

1

29.9

kg

NOZZLE / OPENINGS AND OTHERS

500.0

kg

TOTAL WEIGHT

3554

kg

Liquid Weight Water Weight

12701 12701

kg kg

3554 16255 16255

kg kg kg

WEIR PLATE

2.500 m x 1.400 m

ANGLE

75 x 75 x 9t x 3.0 m

x

8 thk

EMPTY WEIGHT OPERATING WEIGHT FULL WATER WEIGHT

TRANSPORTATION LOADS TRANSPORTATION ACCELERATIONS

WEIGHTS ERECTED ……………..……. We

=

3554

kg ->

34865

N

OPERATING ……………..

Wo

=

16255

kg ->

159460

N

FLOODED ………….……

Wf

=

16255

kg ->

159460

N

VERTICAL

AtV

=

13.73

m/s

( 1.4 x g )

LONGITUDINAL

AtH

=

4.91

m/s

( 0.5 x g )

TRANSVERSE

AtT

=

4.91

m/s

( 0.5 x g )

TRANSPORTATION FORCES VERTICAL

FtV

=

48811.4

N

( We x AtV )

HORIZONTAL

FtH

=

17432.6

N

( We x AtH )

TRANSVERSE

FtT

=

17432.6

N

( We x AtT )

LOADS AT BASE WEIGHTS Erected Operating Flooded

We Wo Wf

= = =

3554 16255 16255

kg ------> kg ------> kg ------>

Fw Feq Fb FD

= = =

1533 0 0

N N N

=

51831

N

Mw Meq Mb

= = =

766250 0 0

Nmm Nmm Nmm

Client Specified Moment ( during tow-out and installation )

Mc

=

68676

Nmm

Maximun Shear Force Maximun O/T Moment

F M

34865 159460 159460

N N N

EXTERNAL LOADS Wind Force Earthquake Force Blasting Force Client Specified Dynamic Force ( during tow-out and installation ) Wind Moment Earthquake Moment Blasting Moment

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

( FD x COGerected )

COGerected =

1325 mm

(from base) = =

51831 766250

N >>> Nmm

P = F/n = 12957.7 N n= 4 where, n = no of leg.

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

AT = Tensile Area………….……………..….……

245

mm2

AS Shear Area……………………………..…

=

225

mm2

Bolt PCD………………………………….. PCD

=

2258

mm

AXIAL STRESS IN BOLT Load / Bolt,

Load / Bolt

SHEAR STRESS IN BOLT

P=

4Mw PCD.N

=

-8377

-

We N

N

** Since the value is -ve, therefore no axial stress

Shear / Bolt, S =

F N x As

fs

=

57.59

MPa

Fs

=

69

MPa

since fs < Fs the shear stress is

OK

OK

LEG DESIGN LEG DATA Material……………...………………..= Yield Stress, Sy………….…………..= Allowable Axial Stress, fall.…...……= Allowable Bending Stress, fball.......= LEG GEOMETRY :-

SA 36M 248.2 N/mm2 148.9 N/mm2 ( 0.6 x Sy ) 165.5 N/mm2 ( 2/3 x Sy )

ANGLE 90 x 90 x 8t A= Ixx = d= e= L= r=

d X

X e

AXIAL STRESS Axial Stress, fa =

F/A =

28.68

N/mm2

PxLxe= Ixx

16.58

N/mm2

BENDING STRESS Bending Stress, fb =

COMBINED STRESS Combined Stress, f = (fa/fall + fb/fball) = Since Combined Stress is

0.29

< 1.00 The Leg Design is OK!

1390 mm2 1040000 mm4 50 mm 25 mm 450 mm 11 mm

LEG BASEPLATE DESIGN Refer Dennis R Moss Procedure 3-10

tb

=

3xQxF 4 x A x Fb

Q = Maximum Load / Support F = Baseplate Width A = Baseplate Length Fb = Allowable Bending Stress

= = = =

16255 150 150 163.68

N mm mm MPa

tb

=

8.6

mm

Use Tb

=

16

mm

BASE PLATE WELD CHECKING Maximum stress due to Q & F = max(Q, F)/Aw

= <

Weld leg size, g Length of weld, l = 2*( 2*F + 2*A ) Area of weld, Aw = 0.5*g*l Joint efficiency for fillet weld, E Welding stress for steel, fw Allowable stress for weld, fw = E*fw Maximum vertical force, Q Maximum horizontal force, F

= = = = = = = =

( 0.66 Fy )

OK

10.80 86.9

N/mm2 N/mm2

8.0 1200 4800 0.6 144.8 86.9 16254.9 51831.0

mm mm mm2 N/mm2 N/mm2 N N

OK

LIFTING LUG DESIGN CALCULATION Fy U tL rL

d 5

A

A

J

5

tM

k M

b

hc

Wp a Lp

tp 10 5 Pa

Weight of tank, We Number of lifting lug, N 1.1 LIFTING LUG Distance, hc Distance k Distance J Distance M Lug radius, rL Diameter of hole, d Lug thickness, tL Plate thickness, tM Length a Length b Pad length, Lp Pad width, Wp Pad thickness, tp Angle, U (max) Shackle S.W.L Type of shackle Pin size, Dp

= = =

= = = = = = = = = = = = = =

3,554 kg 34,865 N 4

85 mm 106 mm 48 mm 50 mm 50 mm 40 mm 15 mm 9 mm 100 mm 60 mm 150 mm 100 mm 6 mm 15 °

: 4.75 tons : Chain Shackles = 22.225 mm

2.0 LIFTING LUG MATERIAL & MECHANICAL PROPERTIES Material used = SA 240M GR 316 L Specified yield stress, Sy = 248.22 N/mm² Impact load factor, p = 3.00

3.0 ALLOWABLE STRESSES Allowable tensile stress, St.all ( = 0.6 Sy ) Allowable bearing stress, Sbr.all ( = 0.9 Sy ) Allowable bending stress, Sbn.all ( = 0.66 Sy ) 3 Allowable shear stress, = 0.4 Sy ) ( Cd'Ss.all x qz x( Az ) x 10

= = = =

148.93 N/mm² 223.40 N/mm² 163.83 N/mm² 99.29 N/mm²

4.0 LIFTING LUG DESIGN - VERTICAL LIFTING 4.1 DESIGN LOAD Design load , Wt ( = p.We ) Design load per lug, W ( = Wt / N ) Vertical component force, Fy

= = =

4.2 STRESS CHECK AT PIN HOLE (a) Tensile Stress Vertical component force, Fy Cross sectional area of lug eye, Ae ( = 2*[ rL - d/2 ] x tL ) Tensile stress, St ( = Fy / Ae ) Since St < St.all, therefore the lifting lug size is

= 26149 N = 900 mm² = 29.05 N/mm² satisfactory.

104596 N 26149 N 26149 N

(b) Bearing Stress Vertical component force, Fy = 26149 N Cross sectional area of lug eye, Ae ( = Dp x tL ) = 333 mm² Bearing stress, Sbr ( = Fy / Ae ) = 78.44 N/mm² Since Sbr < Sbr.all,therefore the lifting lug size is satisfactory. (c) Shear Stress Vertical component force, Fy Cross sectional area of lug eye, Ae ( = 2.(rL-d/2).tL ) Shear stress, Ss ( = Fy / Ae ) Since Ss < Ss.all,therefore the lifting lug size is

= 26149 N = 900 mm² = 29.05 N/mm² satisfactory.

5.0 STRESS CHECK AT SECTION A-A (a) Bending Stress Bending stress due to Pa ( = Fy x tan U ) Bending moment, Mb ( = Pa x J ) Section modulus, Z ( = 2rL*tL2/6 Bending stress, Sb ( = Mb/Z ) Since Sb < Sb.all, therefore the lifting lug size is

= 7007 N = 336316 Nmm = 3750 mm3 = 89.68 N/mm² satisfactory.

(b) Tensile Stress due to Fy Cross section area, Ae (=2rL x tL) = 1500 mm² Tensile Stress, St (=Fy/Ae) = 17.43 N/mm² Since St < St.all, therefore the lifting lug size is satisfactory. Combine Stress Ratio, CS (= St/St.all + Sb/Sbn.all) Since CS