v 2154 101 a 216 c_mechanical calculation (v 510, v 511)

7/16/2019 V 2154 101 a 216 C_Mechanical Calculation (v 510, V 511)

http://slidepdf.com/reader/full/v-2154-101-a-216-cmechanical-calculation-v-510-v-511 1/25

Project: NOEV LUBE OIL BLENDING PLANT

Job No.: AL-2499

Document No.: V-2154-101-A-216

Reference Drawing: V-2154-101-A-203_Rev.E & V-2154-101-A-207_Rev.B

Vessel Name: Multi Production Tank

Vessel Tag No.: V-510 & V-511

C 5/27/2013

B 5/10/2013

A 4/9/2013

Rev Date

MECHANICAL CALCULATION SHEET

Description Prepared Approval

Issue for review / approval

Issue for review / approval

Issue for review / approval L.D.T

L.D.T

L.D.T

L.A.V

L.A.V

L.A.V

Checked

L.N.B

L.N.B

L.N.B



Project: NOEV LUBE OIL BLENDING PLANT Job No.: AL-2499 Rev. No.: C

INDEX Page

1. Design Data 2

2. Shell Thickness Calculation 2

3. Bottom Head Thickness Calculation 3

4. Top Head Thickness Calculation 4

5. Auxiliary Stiffener Calculation 5

6. Main Stiffener Calculation 6

7. Weight Calculation Sheet 7

8. Lug Support 89. Nozzle Calculation 11

10. Welding 14

11. Wind Load Analysis 17

12. Seismic Analysis 18

13. Support Analysis for Wind/Seismic 19

14. Stresses in Shell by Wind/Seismic 21

15. Lifting Lug Calculation 22

16. Conclusion 25

MECHANICAL CALCULATION SHEET

Page 1 of 25



Project: NOEV LUBE OIL BLENDING PLANT Job No.: AL-2499 Rev.

1. Design Data

Design Code : None

Service: Multi Production Tank (Atmospheric Tank)

Design pressure

Max. Internal pressure - (Full 3.79 meters of Water) P = 0.372 barg = 0.0372 MPa G

External pressure 0.0 barg = 0.0 MPa GWorking pressure 0.0 barg = 0.0 MPa G

Design temperature 105 degrees C

Working temperature 60 ~ 80 degrees C

Corrosion allowance 0.0 mm

Vessel inside diameter 2400 mm (O/D = 2412 mm)

Vessel length (Flat Head to T.L) 2800 mm

Material S : Maximum allowable stress value

Shell SA-240 TP304 / 304L 112 MPa

Flat Top Head SA-240 TP304 / 304L 112 MPa

Bottom Cone Head SA-240 TP304 / 304L 112 MPa

Nozzle Neck SA-312 TP304 / 304L 115 MPa

Support SA-240 TP304 / 304L 112 MPa

Nozzle Flange: ASME B16.5 Standard

SCH THK.

N1A/N1B 150 80S 5.54

N2 150 80S 5.08

N3 150 80S 5.54

N4 150 40S 6.02

N5 150 40S 6.02

N6 150 40S 5.49

N7 150 40S 6.02

2. Shell Thickness Calculation (Refer to ASME Section VIII, Division 1, UG-27)

2.1 Minimum required thickness of shell exclusive corrosion allowance (t):

Circumferential Stress (Longitudinal Joints)

0.0372 x 1200.0

112.0 x 0.85 - 0.6 x 0.0372

where:

P : internal design pressure = 0.0372 MPa < = 36.652 MPa

R : Inside radius of the shell = 1200.0 mmS : Maximum allowable stress value = 112.0 MPa

E : Joint efficiency = 0.85

Longitudinal Stress (Circumferential Joints)

0.0372 x 1200.0

2.0 x 112.0 x 0.85 + 0.4 x 0.0372

where:


Degree of Radiographic Examination: 10%

2.2 Minimum Thickness of Pressure retaining Components (UG-16 (b)) = 2.5 mm

2.3 Minimum required thickness of shell included corrosion allowance

tr = 0.47 + 0.0 = 0.47 mm

2.4 Choose Nominal thickness of shell, ts = 6.0 mm

=

= 0.23

95.2

=

=44.6

mm

mm

0.47

0.385SE

=44.6

190.4

=

Nozzle Size Flange Type ClassNozzle Neck Nozzle outside

diameter, mm

Nozzle inside

diameter, mm

DN 50 (2") SO RF 60.3 49.22

DN 40 (1-1/2") SO RF 48.3 38.14

DN 50 (2") SO RF 60.3 49.22

DN 100 (4") SO RF 114.3 102.26

DN 100 (4") SO RF 114.3 102.26

DN 80 (3") SO RF 88.9 77.92

DN 100 (4") SO RF 114.3 102.26

Page 2




3. Bottom Head Thickness Calculation (Refer to ASME Section VIII, Division 1, UG-32)

Type of head: Toriconical Head (with Knuckle)

3.1 Conical Section

3.1.1 Minimum required thickness of Conical Section exclusive corrosion allowance (t):

0.0372 x 2400.0

where:

P : internal design pressure = 0.0372 MPa < = 36.65 MPa

D : Inside diameter of the head skirt = 2400.0 mm

S : Maximum allowable stress value = 112.0 MPa


α : one-half of the included angle of = 60 degrees

the cone at the centerline of head

3.1.2 Minimum required thickness of Head included corrosion allowance

= 0.94 + 0.0 = 0.94 mm

3.1.3 Choose nominal thickness of Head, tb = 8.0 mm

3.2 Knuckle Section

r : inside knuckle radius = 250 mm

Di : inside cone diameter at point of tangency = = 2150.0

to knuckle

3.2.1 Minimum required thickness of Knuckle Section exclusive corrosion allowance (t):

where:

3.2.2 Minimum required thickness of Knuckle included corrosion allowance

= 0.62 + 0.0 = 0.62 mm

3.2.3 Choose nominal thickness of Knuckle, tK = 8.0 mm

3.3 Stress Relief (Refer to UCS-79)

Type of head: Toriconical Head (Double Curvature)

=> Percent extreme fiber elongation is not exceeded by 5% so that a stress relief is not required

where:

t : Nominal Straight Flange thickness t = 8.0 mm

Rf : Final centerline radius (mean knuckle radius) Rf = 250.0 mm

Ro : Original centerline radius (Crown radius) Ro = 2150.0 mm

4. Top Head Thickness Calculation (Refer to Roark's Formulas for Stress and Strain)

Type of head: Flat Head

= mm0.6227

= 2.12 %

0.385SE

89.295.22 x cos(60) x (112 x 0.85 - 0.6 x 0.0372)

= mm0.94= =

118.6

190.4=

: Crown radius = 2150

: Factor = 1.48

0.0372 x 2150 x 1.48

2 x 112 x 0.85 - 0.2 x 0.0372=

mm

Page 3




Assume square plate (axb) 500x500 mm with all edges simply supported

and uniform loads over entire plate.

Top Head Self-Weight = 219.324 kg

F = m x g = 2151.57 N

Area (A) = πD2/4 = 4.57 m

2

P1 = F/A = 0.471 kPa

Structural Weight = 100 kg (Including weight of nozzles, manhole on Top Head)

F = m x g = 981 N

Area (A) = πD2/4 = 4.57 m

2

P2 = F/A = 0.215 kPa

Concentrated Load = 100 kg (Assumed)

F = m x g = 981 N

Area (A) = πD2/4 = 4.57 m

2

P3 = F/A = 0.215 kPa

Total Dead Load (P) = P1+P2+P3 = 0.90 kPa

Total Live Load (L) = 1.0 kPa (As per API 650 11th Ed. Errata, Oct. 2011, Clause 5.2.1)

Total Uniform Load (q) = P + L = 1.90 kPa

Edges of Plate (a x b) = 500 x 500 mm

a/b = 1

β = 0.2874

α = 0.0444

Elastic Modulus (E) = 1.890E+08 kPa (Interpolate from API 650 11th Ed. Errata, October 2011, Appendix S)

Allowable Stress ([ϭ]) = 112000 kPa (Refer to ASME Section II, Part D, Table 1A)

Required Plate thickness (Refer to Roark's Formulas, Table 11.4, Case 1a)

Choose thickness of Top Head Plate = 6 mm

Max. Deflection (Refer to Roark's Formulas, Table 11.4, Case 1a)

Max. Stress in plate

5. Auxiliary Stiffener Calculation

Length of stiffener L = 500 mm

Width of Plate that using stiffener Wp = 500 mm (Assumed)

Uniform load wa = q x Wp = 0.950 kN/m

mmt/2 = 3

< [ϭ] 112000

= 1.10 mm

= 0.13 mm

= 3793 kPa

(ACCEPTED)<

kPa (ACCEPTED)=

=

=

=

L

Page 4




Select stiffener properties as below

Area moment of inertia (As per Roark's Formulas, Table A.1, Case 4: Tee section):

Effective area = 1.1 x (D x t)0.5 = 22 x t (Base on ASME Section VIII, Division 1, Appendix 1-8)

where:

t : The thickness of Top Head = 6 mm

b : Effective area (22 x t) = 132 mm

tw : The thickness of Stiffener = 8 mm

d : Stiffener height = 50 mm

= 1192 mm2

Elastic modulus (E) = kPa

Plastic section modulus (Zx)

Max. deflection at the center of stiffener (Refer to Roark's Formulas, Table 8.1, Case 2e )

-5 x 0.95 x 10004

Unity Check (UC) ratio calculation

M : Maximum bending moment caused by uniform load wa = 0.030 KN.m

S : Bending Stress caused by M = 2456.9163 KPa

UC ratio : = 0.022 < 1

=> (ACCEPTED)

6. Main Stiffener Calculation

Length of stiffener L = 2400 mm

Width of Plate that using stiffener Wp = 340 mm (Assumed)

= -0.014 mm

I = = 294074 mm4

= 12.396 mm

= 12085 mm3

189000000

=384 x 189000 x 294074

M =

8

S =

=

L

t = 8

Page 5



Project: NOEV LUBE OIL BLENDING PLANT Job No.: AL-2499

Uniform load wa = q x Wp = 0.646 kN/m

Select stiffener properties as below

b3 : Effective area (22 x d3) = 132 mm

d3 : The thickness of Top Head = 6 mm

d2 : Web height = 150 mm

b2 : Web thickness = 6 mm

b1 : Bottom flange length = 70 mm

d1 : Bottom flange thickness = 8 mm

PART Area (a) y a x y h h2

bd3/12

mm mm mm mm mm mm

1 560 4 2240 86.79 7531.96 2986.6667

2 900 83 74700 7.79 60.64 1687500

3 792 161 127512 -70.21 4929.89 2376

TOTAL 2252 204452 1692862.7

Therefore,

Distance from bottom to Neutral axis

C = 204452 / 2252 = 90.787 mm

Area moment of inertia

I = 8176937.691 + 1692863 = 9869800.4 mm

Section Modulus

Z = I /C = 108713.98 mm3

Max. deflection at the center of stiffener (Refer to Roark's Formulas, Table 8.1, Case 2e )

-5 x 0.646 x 24004

Unity Check (UC) ratio calculation

M : Maximum bending moment caused by uniform load W = 0.465 KN.m

S : Bending Stress caused by M = 4278.9964 KPa

UC ratio : = 0.038 < 1

=> (ACCEPTED)

7. Weight Calculation Sheet

No.Thickness

(mm)Q'ty

Unit Weight

(kg)

Total

Weight

k

1 Shell 6.0 1 1006.3 1006.26

2 Bottom Cone 8.0 1 418.64 418.64

3 Roof 6 1 248.22 248.22

4 Stiffenner T-150x6+70x8 - 4 27.68 110.73

5 Stiffenner PL.50x8 - 2 19.47 38.94

6 Top Angle Bar L-70x70x8 - 1 68.01 68.01

= -0.150 mm

8176937.691

mm4

a x h2

3904469.373

54571.61552

4217896.702

Description Unit

=384 x 189000 x 9869800

Remark

Ea

Ea

Ea

Ea

Ea

Ea

M =

8

S =

=

Pa



Project: NOEV LUBE OIL BLENDING PLANT Job No.: AL-2499

7 Support Lug - Reinforcement PL. 8 4 18.20 72.79

8 Support Lug - Base Plate 25 4 30.31 121.26

9 Support Lug - Compress Plate 18 4 13.33 53.32

10 Support Lug - Gusset Plate 1 16 8 3.03 24.26

11 Support Lug - Gusset Plate 2 16 8 13.07 104.55

12 Lifting Lug 25 2 6.33 12.66

13 Nozzle N1A - Neck - 0.15 7.63 1.14 2" SCH 80S14 Nozzle N1A - Flange - 1 2.30 2.30 2" 150#

15 Nozzle N1B - Neck - 0.15 7.63 1.14 2" SCH 80S

16 Nozzle N1B - Flange - 1 2.30 2.30 2" 150#

17 Nozzle N2 - Neck - 0.172 5.51 0.95 1-1/2" SCH 80S

18 Nozzle N2 - Flange - 1 1.40 1.40 1-1/2" 150#

19 Nozzle N3 - Neck - 0.15 7.63 1.14 2" SCH 80S

20 Nozzle N3 - Flange - 1 2.30 2.30 2" 150#


22 Nozzle N4 - Flange - 1 5.90 5.90 4" 150#


24 Nozzle N5 - Flange - 1 5.90 5.90 4" 150#


26 Nozzle N6 - Flange - 1 3.70 3.70 3" 150#


28 Nozzle N7 - Flange - 1 5.90 5.90 4" 150#

29 Nozzle N7 - Spray Nozzle - 1 13.08 13.08

30 MANWAY M1 - Reinforcement 6 1 30.33 30.33

31 MANWAY M1 - Shell 6 1 24.00 24.00

32 MANWAY M1 - Cap 6 1 24.61 24.61

Empty Weight 2420.0

Weight of Liquid at Operating Level 10080.0

Weight of Full Water 14340.0

Total Weight of Liquid at Operating Level 12500.0

Total Weight of Full Water 16760.0

8. Lug Support (Refer to Pressure Vessel Handbook 10th Edition - By Eugene F. Megyesy)

8.1 Wear plate

PCD : Point of Force Application 2920 mm

Circle Diameter

BCD : Bolt Circle Diameter 3020 mm

OD : Vessel Outside Diameter 2412 mm

W : Weight of Vessel (Full of Water) = 16760.0 kg

n : Number of lugs = 4

Q = W/n : Load on one lug = 4190 kg

R : Radius of shell = 1206 mm

H : Lever arm of load = 254 mm

2A : 1st Dimension of wear plate = 569 mm

2B : 2nd Dimension of wear plate = 550 mm

t : Wall thickness of shell = 6.0 mm

t : Wear plate thickness 8.0

Ea

Ea

Ea

Ea

Ea

mEa

m

Ea

m

Ea

m

Ea

m

Ea

Ea

m

Ea

Ea

Ea

m

Ea

m

Ea

EaPipe & Flange 2",

B. Flange 4", Bolt/Nut

Ea



Project: NOEV LUBE OIL BLENDING PLANT Job No.: AL-2499 Rev. N

P : Internal Pressure at Wear plate location = 0.00 MPa

Shell material : SA-240 TP304 / 304L

Allowable stress value : 112.0 MPa

Joint Efficiency : 0.85

Shape factors C :

1206 275

6.0 284.5

C1 = 1

C2 = 1

C3 = 1

C4 = 1

The factors K

K1 = 3.9

K2 = 0.0075

K3 = 6.5

K4 = 0.005

Longitudinal Stress :

= 72.03 MPa

Stress due to internal pressure:

PR

2t

The sum of tensional stresses:

72.03 + 0.00 = 72.03 MPa

It does not exceed the stress value of the girth seam:

112.0 x 0.85 = 95.2 MPa (ACCEPTED)

Circumferential Stress:

= 64.27 MPa

Stress due to internal pressure:

PR

t

The sum of tensional stresses:

64.27 + 0.00 = 64.27 MPa

It does not exceed the stress value of shell material multiplied by 1.5 :

R/t ;= = 201

=

= =0 x 1206

2 x 6

= 0.00 MPa6

= 0.9666

= 0.23

B/A =

0 x 1206

0.00 MPa

=

=284.5

1208

275

284.5

= ±

+ 6

+

2 1.17+ /×

= ±

+ 6

Page 8 o




112.0 x 1.50 = 168 MPa (ACCEPTED)

Choose wear plate = 569x550x8 mm

8.2 Gussets (Refer to Pressure Vessel Design Manual 3rd Ed. 2004 - Dennis R. Moss, Procedure 3-13)

Q : Vertical load per lug = 4190 kg = 41104 N

n : Number of gussets per lug = 2 (Double Gusset)

b = 344 mm

h = 396.5 mm

tg : Gusset thickness = 16 mm

θ = 63 degrees

Fy : Minimum Yield Strength of gusset material = 170 MPa

Fa : Allowable axial Stress = 68 MPa (0.4 Fy)

Fb : Allowable bending Stress = 102 MPa (0.6 Fy)

Qa = Q.sinθ : Axial load on gusset = 36623.8 N

Qb = Q.cosθ : Bending load on gusset = 18660.8 N

A = tg.C : Cross-sectional area of assumed column = 2452.05 mm2

Axial stress

=> PASS

Bending Stress

=> PASS

Choose dimensions of Double Gusset as shown above

8.3 Base Plate for Double Gusset (Refer to Pressure Vessel Design Manual 3rd Ed. 2004 - Dennis R. Moss, Procedure 3-13)

7.468 MPa < Fa = 68

: Bending moment = 4152046 Nmm

= 62630.7 mm3

MPa

Mpa= 66.2941 MPa < Fb = 102

=

= 153.253 mm

: Section modulus

= 445.002 mm

Page 9




l : Base plate width = 410 mm

a : Bearing width = 200 mm

l1 = 340 mm

φ : Bolt hole diameter = 38 mm

8.3.1 Bending (Assume to be between simply supported with two equal spans)

8.3.2 Bearing

8.3.3 Thickness required base plate

where Mb is bending moment

Choose the actual thickness base plate tb = 25 mm

8.4 Compression Plate for Double Gusset (Refer to Pressure Vessel Design Manual 3rd Ed. 2004 - Dennis R. Moss, Procedure 3-13)

e = 254 (Refer to figure of 8.2 Gussets)

Ev : Modulus of elasticity of vessel = 189000 MPa

Es : Modulus of elasticity of compression plate = 189000 MPa

t : Shell Vessel thickness = 6.0 mm

R : Inside Radius of vessel = 1200.0 mm

tc : Assumed Compression Plate thickness = 18.0 mm

y : Compression Plate width = 200.0 mmx : Distance between loads = 340.0 mm

Fy : Minimum Yield Strength of gusset material = 170 MPa

Fb : Allowable bending Stress = 102 MPa (0.6 Fy)

Concentrated load on Double Gusset

Foundation modulus

Moment of inertial of Compression plate

Section modulus of Compression plate

Damping factor

= 13165.7 N (Assumed)

Bending

(Uniform load on base plate)

= 1.2E+07 mm

4

= 120000 mm3

= 0.00054

= 20.12 mm

= 0.7875 N/mm3

= 2106575 Nmm

= 0.50127 Nmm2

= 5794.65 Nmm

=

8=

8

Page 10




Internal bending moment in Compression plate

Bending stress

=> PASS

Choose Compression Plate thickness, tc = 18.0 mm

9. Nozzle calculation

9.1 Shell Nozzle N2 DN40 (1-1/2") (Refer to ASME Section VIII, Division 1, UG-45)

Minimum Nozzle neck thickness

where:

S : Maximum allowable stress value, S = 115.0 MPa

E : Joint efficiency E = 1.00

Nozzle inside radius Rn = 19.1 mm

tn (min) : minimum required thickness of Nozzle

9.1.1. ta : minimum neck thickness required for internal and external pressure using UG-27 and UG-28

(plus corrosion allowance), as applicable. The effects of external forces and moments from supplemental

loads (see UG-22) shall be considered. Shear stresses caused by UG-22 loadings shall not exceed 70%

of the allowable tensile stress for the nozzle material.

0.0372 x 19.1 0.71

115.0 x 1.00 - 0.6 x 0.0372 115.0

ta = 0.01 + 0.0 = 0.01 mm

9.1.2. = min ( 3.22 , 2.5 ) = 2.5 mm

where:

tb1 : for vessels under internal pressure, the thickness (plus corrosion allowance) required for pressure

(assuming E = 1.0) for the shell or head at the location where the nozzle neck or other connection

attaches to the vessel but in no case less than the minimum thickness specified for the material in UG-16(b).

tb1 = 2.5 mm

tb2 : for vessels under external pressure

tb2 = 0.0 mm

Max (tb1, tb2) = Max ( 2.5 , 0.0 ) = 2.5 mm

tb3 : the thickness given in Table UG-45 plus the thickness added for corrosion allowance.

(for standard wall pipe)

tb3 = 3.22 + 0.0 = 3.22 mm

=> = max ( 0.01 , 2.50 ) = 2.50 mm

9.1.3. Choose Nozzle actual thickness = 5.08 mm (SCH 80S)

9.1.4. Nozzle actual thickness is compared with the minimum thickness provided which for pipe

material would include a 12.5% undertolerance

= 0.875 x 5.08 = 4.45 > tn (min) = 2.50 mm

Result: the actual thickness provided meets the rules of UG-45 <PASS>

9.2 Flat Top Head Nozzle N1A, N1B, N3 DN50 (2") (Refer to API 650 11th Ed. Errata, October 2011, Appendix S)

= = = 0.01 mm

Mpa

= 8239177 Nmm

= 68.6598 MPa < Fb = 102

bant t t ,max(min)

bant t t ,max(min)

Page 11




9.2.1 Minimum Nozzle Neck thickness

Base on table S.3.3.1

Minimum thickness of Nozzle Neck = 5.54 mm (SCH 80S)

Corrosion Allowance (C.A) = 0.00 mm

Minimum thickness of Nozzle Neck including C.A = 5.54 mm

9.2.2 Choose Nozzle actual thickness = 5.54 mm (SCH 80S)

9.3 Flat Top Head Nozzle N6 DN80 (3") (Refer to API 650 11th Ed. Errata, October 2011, Appendix S)







9.4 Flat Top Head Nozzle N5, N7 DN100 (4") (Refer to API 650 11th Ed. Errata, October 2011, Appendix S)







9.5. Cone Bottom Head Nozzle N4 DN100 (4") (Refer to ASME Section VIII, Division 1 & API 650 11th Ed. Errata, October 2011, Appendix S)

9.5.1. Minimum Nozzle neck thickness (Refer to ASME Section VIII, Division 1)

where:

S : Maximum allowable stress value, S = 115 MPa

E : Joint efficiency E = 1.00

Nozzle inside radius Rn = 51.13 mm

tn (min) : minimum required thickness of Nozzle

9.5.1.1 ta : minimum neck thickness required for internal and external pressure using UG-27 and UG-28

(plus corrosion allowance), as applicable. The effects of external forces and moments from supplemental

loads (see UG-22) shall be considered. Shear stresses caused by UG-22 loadings shall not exceed 70%

of the allowable tensile stress for the nozzle material.

0.0372 x 51.13 1.901

115.0 x 1.00 - 0.6 x 0.0372 115.0= = = 0.02 mm

bant t t ,max(min)

Page 12




ta = 0.02 + 0.0 = 0.02 mm

9.5.1.2 = min ( 5.27 , 2.5 ) = 2.50 mm

where:

tb1 : for vessels under internal pressure, the thickness (plus corrosion allowance) required for pressure

(assuming E = 1.0) for the shell or head at the location where the nozzle neck or other connection

attaches to the vessel but in no case less than the minimum thickness specified for the material in UG-16(b).

tb1 = 2.50 mm

tb2 : for vessels under external pressure

tb2 = 0.00 mm

Max (tb1, tb2) = Max ( 2.50 , 0.00 ) = 2.5 mm

tb3 : the thickness given in Table UG-45 plus the thickness added for corrosion allowance.

(for standard wall pipe)

tb3 = 5.27 + 0.0 = 5.27 mm

=> = max ( 0.02 , 2.5 ) = 2.50 mm

9.5.2. Minimum Nozzle neck thickness (Refer to API 650 11th Ed. Errata, October 2011, Appendix S)


Minimum thickness of Nozzle Neck = 6.02 mm



9.5.3. Choose Nozzle actual thickness = 6.02 mm (SCH 40S)

10. Welding

10.1 Shell Nozzle N2 DN40 (1-1/2") - (Refer to ASME Section VIII, Division 1, UW-16)

10.1.1 Size of weld / Shell thickness

tn (actual) = 5.08 mm

Fillet Leg Length = 6.00 mm

=> tc (actual) = 4.20 mm

t = 6.00 mm

10.1.2 Check for full penetration groove weld and fillet cover weld shown in Fig above

tc (min) = Min ( 6 , 0.7 tmin )

where:

tmin = the smaller of 19 mm or the thickness of the thinner of the parts joined by a fillet, single-bevel, or single-J weld

tmin = Min ( 19.0 , 5.08 , 6.00 ) = 5.08 mm

0.7tmin = 0.7 x 5.08 = 3.56 mm

=> tc (min) = Min ( 6 , 0.7 tmin ) = Min ( 6.00 , 3.56 ) = 3.56 < tc (actual) = 4.2 mm

=> PASS

10.2 Flat Top Head Nozzle N1A, N1B, N3 DN50 (2") - (Refer to API 650 11th Ed. Errata, October 2011, Clause 5.9)

As per Fig. 5-19 - Flanged Roof Nozzles

bant t t ,max(min)

Page 1



Project: NOEV LUBE OIL BLENDING PLANT Job No.: AL-2499 Rev

where :

DP : Diameter of Hole in Roof Plate (Table 5-14a)

DP = 65 mm

10.3 Flat Top Head Nozzle N6 DN80 (3") - (Refer to API 650 11th Ed. Errata, October 2011, Clause 5.9)


where :


DP = 92 mm

10.4 Flat Top Head Nozzle N5, N7 DN100 (4") - (Refer to API 650 11th Ed. Errata, October 2011, Clause 5.9)


where :


DP = 120 mm

10.5 Cone Bottom Head Nozzle N4 DN100 (4")

10.5.1 Size of weld, Nozzle thickness & Shell thickness

tn = 6.02 mm (Nozzle thickness)

Fillet Leg Length = 7.00 mm

tc (actual) = 4.90 mm (Refer to fabrication detail drawing)

t = 8.00 mm (Min. Bottom Head thickness after forming)

10.5.2 Check for full penetration groove weld and fillet cover weld shown in Fig above

tc (min) = Min ( 6 , 0.7 tmin )

where:

tmin = the smaller of 19 mm or the thickness of the thinner of the parts joined by a fillet, single-bevel, or single-J weld

tmin = Min ( 19.0 , 6.02 ) = 6.02 mm

0.7tmin = 0.7 x 6.02 = 4.21 mm

=> tc (min) = Min ( 6 , 0.7 tmin ) = Min ( 6.00 , 4.21 ) = 4.21 < tc (actual) = 4.90 mm

=> PASS

10.5.3 Reinforcement Calculation (Refer to ASME Section VIII, Division 1, UG-37)

Page 1



Project: NOEV LUBE OIL BLENDING PLANT Job No.: AL-2499 R

A : Area of Reinforcement required

A = dtr F + 2tn tr F (1-f r1) = 95.569 mm2

A1 : area in excess thickness in the vessel wall available for reinforcement

(includes consideration of nozzle area through shell if Sn /Sv<1.0)

A1 = larger of below value = 724 mm2

d(E1t - Ftr ) - 2tn(E1t - Ftr )(1 - f r1) = 724.49 mm2

2(t + tn)(E1t - Ftr ) - 2tn(E1t - Ftr )(1 - f r1) = 200.31 mm2

A2 : area in excess thickness in the nozzle wall available for reinforcement (see Fig. UG-37.1)

A2 = smaller of below value = 108.79 mm2

5(tn - tr n) f r2t = 144.57 mm2

5(tn - tr n) f r2tn = 108.79 mm2

A3 : area available for reinforcement when the nozzle extends inside the vessel wall

A3 = min (5t ti f r2, 5ti ti f r2, 2h ti f r2) = 0 mm2

A41 : cross-sectional area available in outward weld

A41 = (leg)2f r2 = 50.31 mm

2

A42 : cross-sectional area available in inward weld = 0 mm2

where:

leg : Fillet weld size = 7.00 mm

d : finished diameter of circular opening or finished dimension = 102.26 mm

h : distance nozzle projects = 0 mm

t : specified vessel wall thickness = 8.0 mm

tn : nozzle wall thickness = 6.02 mm

tr : required thickness of a seamless shell = 0.94 mm

tr n : required thickness of a seamless Nozzle wall = 2.50 mm

ti : nominal thickness of internal projection of nozzle wall = 0 mm

Sn : allowable stress in nozzle = 115 MPa

Sv : allowable stress in vessel = 112 MPa

F : correction factor = 1

E1 = 1

f r1 = 1.027

f r2 = Sn /Sv = 1.027

A1 + A2 + A3 + A41 + A42 = 884 mm2 > A = 95.569 mm

2

RESULT : Reinforcement for Nozzle N4 is not needed

10.6 Cone Bottom Head Welded Joint

Weld type : Butt weld, Full Penetration

Critical weld length K = 6.0 mm (Assumed equal to the thickness of shell)

Inside Diameter of Tank D = 2400 mm

Area of weld Aw = 45238.9 mm2 Aw = πDK

Allowable Stress of Shell material = 112.0 MPa

Applied by force Total Weight Load (TWL)

Weight Load of Water inside the tank Ww = 140675 N (Estimate)

Page




Weight Load of Cone Bottom Head Wb = 4106.9 N (Estimate)

TWL = 144782 N TWL = Ww + Wb

Tensile Stress on Welded Joint

Tensile = 3.20 MPa Tensile = TWL / Aw

Allowable = 112.0 MPa

Safety Factor = 35.0 => PASS

11. Wind Load Analysis (Refer to Pressure Vessel Design Manual 3rd Ed. 2004 - Dennis R. Moss, Procedure 3-1)

11.1 Design Specification

Vessel Support Type Support Lug

N : Number of Support Lugs = 4

Wo : Operating Weight of Vessel = 12500.0 kg = 122625 N

h : Overall Height of Vessel = 3796 mm = 12.454 ft.

D : Outside Vessel Diameter = 2412 mm

De : Vessel Effective Diameter from Table 3-4 = 1.4D = 3376.8 mm

Le : Vessel Effective Length = 3796 mm

l : Distance to the center of the projected area = 1049 mm

P : Point of Force Application Circle Diameter = 2920 mm

Structure Category = IV

Exposure Category = B

Cf : Force Coefficient (shape factor) = 0.8 (for cylindrical Vessel)

G : Gust effect factor = 0.8

KZ : Velocity pressure exposure coefficient from Table 3-3a

= 0.57

KZT : Topographic factor = 1V : Basic Wind Speed = 39 m/sec = 87.24 mph

I : Importance factor = 1.15

Ct : Period Coefficient (0.02 - 0.035) = 0.035 (Assume Max. Value)

11.2 Design Calculation

h/D ratio

h/D = = 1.57 < 4

Approximate fundamental period

T = = 0.232 sec < 1 sec

=> The Vessel is considered rigid

Projected area of vessel

Af = Le x De = 3796 x 3376.8 = 12818333 mm2 = 12.818 m

2

3796 / 2412

Ct x h3/4

P

Page 1




Velocity pressure at height z above the ground

qz = = 12.772439 psf = 611.54 N/m2

Design Wind force (Shear force)

= 5016.9469 N

Moment at the base plate

M = F x l = 5262777.3 Nmm

Max. Vertical force per lug

Q = (Wo/N) + (Fl/B) = 32458.571 N

12. Seismic Analysis (Refer to Pressure Vessel Design Manual 3rd Ed. 2004 - Dennis R. Moss, Procedure 3-8)

Neutral Axis Calculation

A = 200.0 mm

B = 344.0 mm

C = 100 mm

D = 25 mm

H = 275 mm

Distance from base to neutral axis

= 95.1 mm

where:

S1 = B x D = 8600 mm2

S2 = B x C = 34400 mm2

S3 = (A+B) x H / 2 = 74800 mm2

12.1 Design Specification

Wo : Operating Weight of Vessel = 12500 kg = 122625 N

L : Level Arm to Neutral axis = 1144 mm

P : Point of Force Application Circle Diameter = 2920 mm

a : Force arm of Vertical Load = 254 mm

b ~ Y : Force arm of Horizontal Load = 95 mm

=

12

+12

+(2+ )3( + )

+ +

P

Page 17




Ch : Horizontal seismic factor = 0.1291 (Assumed)

Cv : Vertical seismic factor = 0.1291 (Assumed)

N : Number of lugs = 4

12.2 Design Calculation

Horizontal force

Fh = Ch x Wo = = 15830.888 N

Horizontal shear per lug

Vertical force

Fv = (1 + Cv)Wo = = 138455.89 N

Vertical shear per lug

Max. Vertical load Q per lug

Max. moment M per lug

M = ML3 = = 10743935 Nmm

13. Support Analysis for Wind/Seismic (Refer to Pressure Vessel Design Handbook 2nd Ed. 1986 - Henry H. Bednar, Chapter 5)

Maximum Vertical force per lug in Wind Case

Fw = = 32458.571 N

Maximum Moment per lug in Wind Case

Mw = = 5262777.3 Nmm

Maximum Vertical force per lug in Seismic Case

Fs = = 40816.854 N

Maximum Moment per lug in Seismic Case

Ms = = 10743935 Nmm

Maximum Vertical force per lug in Support Lug Analysis

F = = 40816.854 N

Maximum Moment per lug in Support Lug Analysis

M = = 10743935 Nmm

M

ML3

Max(Mw , Ms)

3957.7219 N

(1 + 0.1291) x 122625

0.1291 x 122625

=15830.8875

4=

=138455.8875

4= N34613.972

=Q = Q3 34613.97 +15830.89 x 1144

292040816.854= N

40816.85 x 254 + 3957.722 x 95

Q

Q3

Max(Fw , Fs)

Page 18




w : = 200 mm

c : Compression Plate width = 200.0 mm

a : Base plate width = 410 mm

ta : Actual Compression Plate thickness = 18.0 mm

tb : Actual Base plate thickness = 25 mm

tg : Actual Gusset thickness = 16 mm

α : = 63 degrees

[σ] : Allowable stress of support material = 112 MPa

d : Level arm of load F = 254 mm

Sb : Allowable stress in bending for top bar material = 102 MPa (0.6 Fy)

Sa : Allowable stress in compression

h = 396.5 mm

b = 344 mm

13.1 Base plate

Bearing Pressure

q = F/(w x a) = = 0.49777 MPa

Maximum stress in base plate (Refer to Roark's Formulas, Table 11.4, Case 2a)

=> PASS

where

β : Factor from (a/w) = 0.79

13.2 Gusset Plate

Allowable Compressive Stress

where

= 445.0 mm

r : Least radius of gyration of Gusset = 0.289 x tg = 4.624 mm

Maximum compressive stress

=> PASS

where= 22905 N

= 73.1 mm

MPa <25.1671 [σ] = MPa= 112

40816.854 / (200 x 410)

= 11.4 MPa < Sa = 81.95 MPa

= 11885 psi = 81.946 MPa

=

=

+6

= /(2sin ())

= −

2sin ()

=18000

1 +

18000

= ℎ/sin ()

Page 19




= 306.5 mm

13.3 Anchor Bolts (Refer to Pressure Vessel Design Handbook 2nd Ed. 1986 - Henry H. Bednar, Chapter 5, Size of Anchor Bolts)

=> No Uplift exists and the minimum bolt size is about 3/4 to 1 in.

`

where:

Mb : Overturning moment at base ~ Maximum moment M = 10743935 Nmm

Db : Bolt Diameter Circle = 3020 mm

W : Operating Weight of Vessel = 12500.0 kg

N : Number of Support Legs = 4

Choose Anchor Bolt Diameter M36 is satisfactory in this case.

14. Stresses in Shell by Wind/Seismic

(Refer to Pressure Vessel Design Handbook, Chapter 7-Local Stresses in Shell Due to Loads on Attachments)

: Half-length of the loaded square area = 201.60 mm

β = c/r : Attachment parameter = 0.17

ϒ =r/t : Shell parameter = 200.5

r : Mean Shell Radius = 1203.0 mm

t : Shell thickness = 6.0 mm

Maximum bending stress

where

CLt : Bending Stress factor = 0.11

ML = Fd = 10367481 Nmm

Internal pressure Stress

σt = pr/t = 7.45457 MPa

where

p : Internal pressure = 0.0372 MPa

Combined Stress

σ=σb + σt = 164.592 MPa < 2 x [σ] = 224 MPa

=> PASS

15. Lifting Lug Calculation

Equipment weight We = 2420.0 kg

Lifting Lug material SA-240 TP304 / 304L

15.1 Check for β = 90 degrees

Angle β = 90.0 degree

Considered a load factor of 2.0 applied to the structure gravity loads

Design Load P = 2x9.81xWe = 47480.4 N

Force

Fz = 0.5 P = 23740 N

Fx = Fz / tg β = 0 N

Max tensile force in Wire Rope

Ps = Fz / sin β = 23740 N

MPa

0= -27099 <

= 157.13752

= ()

= (ℎ)//2

Page 20




Lifting lug configuration

where :

SWL = Safe working load

Rh = Hole radius

R = Main plate radius

T = Main plate thickness

h = Base widthb = Distance from edge of taper to center of hole

c = Distance from base of plate to center of hole

a = Taper angle

D = Shackle pin diameter

Fy = Yield Strength of lifting lug material

The dimension T should equal 60 - 85% of shackle jaw width.

The pin hole diameter should be 3 mm greater than the selected shackle pin size

The main plate radius is approximately R = 3 R h

Choose Shackle

Shackle load Ps = 23740 N = 2.420 tonne

Choose Shackle with SWL = 4.7 tonne

As per Table shown above :

Shackle jaw width W = 32 mm

Shackle pin size D = 22 mm

Choose Lug Configuration

Rh = 14 mm

R = 50 mm

T = 25 mm

h = 168 mm

b = 200 mm

c = 122 mm

a = 76 degrees

Fy = 170 MPa

Stress in Lifting Lug

Bearing Stress

Bearing = 43.16 MPa Bearing = Ps/(T x D)

Allowable = 153 MPa Allowable = 0.9 x Fy

Page 21





Shear Stress

Shear = 13.19 MPa Shear = Ps/(2(R-Rh)*T)



Tensile Stress

From Section D3.2 of AISC, the distance used in calculations, across the hole, is the minimum of 4 times the plate

thickness at the pinhole or 0.8 times the hole diameter.

Effective width = 22.4 mm

Plate thickness = 25 mm

Tensile = 42.39 MPa Tensile = Ps/(Effective width*plate thickness)

Allowable = 76.5 MPa Allowable = 0.45 Fy (AISC Section D3.2)


Bending Stress

Section modulus Z = 117600 mm3

Z = h2

x T / 6

Area of lug base A = 4200 mm

2

A = h x T

Bending = 24.63 MPa Bending = (Fz*c / Z) + (Fx / A)

Allowable = 102 MPa Allowable = 0.6 Fy


Stress in Weld Joint

Weld type : T-Butt weld, Full Penetration

Critial weld length K = 25 mm (Assumed equal to the thickness of lug)

Section modulus of weld Zw = 235200 mm Zw = h x K / 3

Area of weld Aw = 8400 mm Aw = 2 x K x h

Applied by force Fz

Bending S1 = 12.3 MPa Bending S1 = Fz*c/Zw

Shear S2 = 2.8 MPa Shear S2 = Fz/Aw

Combined = 12.63 MPa Combined = (S1 + S2 ).



Applied by force Fx

Tensile S3 = 0.00 MPa Tensile S3 = Fx/Aw


=> PASS

15.2 Check for β = 85 degrees (Considering Tolerance 5 degrees)

Angle β = 85.0 degree

Considered a load factor of 2.0 applied to the structure gravity loads

Design Load P = 2x9.81xWe = 47480.4 N

Force

Fz = 0.5 P = 23740 N

Fx = Fz / tg β = 2077 N

Max tensile force in Wire Rope

Ps = Fz / sin β = 23831 N

Choose Shackle

Shackle load Ps = 23831 N = 2.43 tonne

Choose Shackle with SWL = 4.7 tonne

Page 2




Stress in Lifting Lug

Bearing Stress

Bearing = 43.33 MPa Bearing = Ps/(T x D)



Shear StressShear = 13.24 MPa Shear = Ps/(2(R-Rh)*T)



Tensile Stress

From Section D3.2 of AISC, the distance used in calculations, across the hole, is the minimum of 4 times the plate

thickness at the pinhole or 0.8 times the hole diameter.

Effective width = 22.4 mm

Plate thickness = 25 mm

Tensile = 42.56 MPa Tensile = Ps/(Effective width*plate thickness)

Allowable = 76.5 MPa Allowable = 0.45 Fy (AISC Section D3.2)Safety Factor = 1.80 => PASS

Bending Stress

Section modulus Z = 117600 mm3

Z = h2

x T / 6

Area of lug base A = 4200 mm2

A = h x T

Bending = 25.12 MPa Bending = (Fz*c / Z) + (Fx / A)



Stress in Weld Joint

Weld type : T-Butt weld, Full Penetration

Critial weld length K = 25 mm (Assumed equal to the thickness of lug)

Section modulus of weld Zw = 235200 mm Zw = h2

x K / 3

Area of weld Aw = 8400 mm Aw = 2 x K x h

Applied by force Fz

Bending S1 = 12.3 MPa Bending S1 = Fz*c/Zw

Shear S2 = 2.8 MPa Shear S2 = Fz/Aw

Combined = 12.63 MPa Combined = (S12

+ S22)0.5



Applied by force Fx

Tensile S3 = 0.25 MPa Tensile S3 = Fx/Aw



Choose Lug Configuration as shown above is satisfactory

16. Conclusion

Shell thickness:

Thickness required: 2.50 mm

Thickness actual: 6.0 mm

Bottom Head thickness:

Nominal Thickness required: 8.00 mm

Top Head thickness:

Thickness actual: 6.00 mm

Page 2




Auxiliary Stiffener size: 50 x 8 mm

Main Stiffener size: T-150x6 + 70x8 mm (Refer to GA Drawing for more details)

Nozzle thickness:

Shell Nozzle N2 DN40 (1-1/2")

Thickness required: 2.50 mmThickness actual: 5.08 mm (SCH 80S)

Flat Top Head Nozzle N1A, N1B, N3 DN50 (2")


Thickness actual: 5.54 mm (SCH 80S)

Flat Top Head Nozzle N6 DN80 (3")



Flat Top Head Nozzle N5, N7 DN100 (4")



Cone Bottom Head Nozzle N4 DN100 (4")



v 2154 101 a 216 c_mechanical calculation (v 510, v 511)

Documents