API 650 Tank Design

950 m3 (TYPE-3) TANK CALCULATIONS A) SYSTEM AND DESIGN DATA Design pressure

Atmospheric

Tank inner diameter (m):

Di  11.5

Tank height (m):

H  11

Freeboard (m):

fb  0.5

Liquid level (m):

Hliq  H  fb

Discharge pipe level (m):

Hd  0

Tank usefull volume:

V 

Stored material:

Su

Density of stored material:

  1000

Hliq  10.5

m

m 2

  Di 4

3

 ( Hliq  Hd)

V  1.091  10

m

kg m

3



Specific gravity:

G 

Wind Velocity:

Vwm  36

G1

1000 m

Vw  Vwm  3.6

s

Vw  129.6

km h

Tank is outside the building. Design temperature:

Td  30

Snow load (kg/m2):

Sn  100

C kg m

Live Load on Roof (kg/m2):

2

Lr  250

kg m

Seismic Zone : (Turkis h Earthquake Code)

1

Corrosion allowance:

CA  6

Material:

ST37-2

Height of courses (m):

h0  1.5

Minimum Yield Strenght (MPa):

Sy  235

Minimum Tensile Strenght (MPa):

Sut  485

2

mm

The Maximum Allowable Product Design Stress (MPa): Sd1 

2 3

 Sy

Sd1  156.667

1

MPa

3

Sd2 

Sd 

 Sd1     Sd2 

2 5

 Sut

Sd2  194

Sd  min ( Sd)

MPa

Sd  156.667

Mpa

The Maximum Allowable Hydrostatic Test Stress (MPa): St1 

St2 

St 

 St1     St2 

3 4 3 7

 Sy

St1  176.25

MPa

 Sut

St2  207.857

MPa

St  min ( St)

Reference Standard:

St  176.25

Mpa

API Standard 650 12th Edition, 2013

B) SHELL DESIGN 1) 1 FOOT METHOD: Di  11.5 m

API 650 Section 5.6.3

60 m



1 Foot method can be used

Design shell thickness ( mm):

td 

Hydrostatic test shell thickness ( mm):

tt 

4.9  Di  ( Hliq  0.3)  G Sd

 CA

4.9  Di  ( Hliq  0.3) St

td  9.669

mm

tt  3.261

mm

2) VARIABLE DESIGN POINT METHOD: L 

( 500  Di  td)

L  235.787

mm API 650 Section 5.6.4

L Hliq

 22.456



1000

Variable Design Point Method can be used.

6

a) The bottom course thickness (t1): Design shell thickness (mm):

 

t1d   1.06 

0.0696  Di Hliq



 Hliq  G     4.9  Hliq  Di  G   CA     Sd  Sd    

t1d  9.929 mm

Hydrostatic test shell thickness (in):

 

t1t   1.06 

0.0696  Di Hliq



 Hliq     4.9  Hliq  Di      St  St    

t1t  3.496 mm

2

t 

 t1d     t1t 

max ( t)  9.929

t1  max ( t)

t1  9.929

mm

b) The second course thickness (t2): Ratio for the bottom course:

h0  1000

ratio 

ratio  6.278

Ri  1000  t1 Calculation of t2a:

H1  H

H2  H1  h0

H2  9.5 m

First trial for second course: t2d 

t2t 

t 

 t2d     t2t 

max ( t )  9.669

Thickness of lower course:

4.9  Di  ( Hliq  0.3)  G Sd

 CA

t2d  9.669 mm

4.9  Di  ( Hliq  0.3)

t2t  3.261 mm

St

tu  max ( t)

tu  9.669

Ratio:

tL  t1

tL

K 

K  1.027

tu

K  (K  1)

C 

mm

1  K1.5

C  0.013

Distance of the variable design point from the bottom of the course: (x) x1  0.61  ( Ri  1000  tu)  320  C  H2

x1  184.421

x2  1000  C  H2

x2  126.849

x3  1.22  ( Ri  1000  tu)

x3  287.66

 x1  x   x2  min ( x )  126.849 xe  min ( x )  x3   

 

4.9  Di   H2  t2d1 

Sd

 

4.9  Di   H2  t2t1 

t 

 t2d1     t2t1 

max ( t)  9.371

xe

 G  1000   CA t2d1  9.371 mm xe

 

1000 

St

t2a  max ( t)

t2t1  2.997

t2a  9.371 mm

t2  9.371

c) The third course thickness (t3): Ratio for the lower course:

ratio 

h0  1000

ratio  6.462

( Ri  1000  t2) Calculation of t3a:

H3  H2  h0

H3  8 m

3

mm

First trial for third course: t3d 

t3t 

t 

 t3d     t3t 

max ( t)  8.77


4.9  Di  ( H3  0.3)  G Sd

 CA

t3d  8.77

4.9  Di  ( H3  0.3)

t3t  2.462 mm

St

tu  max ( t)

tu  8.77

Ratio:

tL  t2

mm

K 

C 

mm

tL

K  1.069

tu K  (K  1)

C  0.034

1  K1.5


x1  223.265

x2  1000  C  H3

x2  269.644

x3  1.22  ( Ri  1000  tu)

x3  273.957

 x1  x   x2  min ( x )  223.265  x3   

xe  min ( x )

 

4.9  Di   H3  t3d1 

Sd

 

4.9  Di   H3  t3t1 

t 

 t3d1     t3t1 

t3  8.797

max ( t)  8.797

xe

 G  1000   CA

t3d1  8.797 mm

xe

  1000 

St

t3a  max ( t)

t3t1  2.486

mm

t3a  8.797

mm

mm

d) The fourth course thickness (t4): Ratio for the lower course:

ratio 

h0  1000

ratio  6.669


H4  H3  h0

First trial for fourth course:

t4d 

t4t 

t 

 t4d     t4t 

max ( t)  8.23

H4  6.5

4.9  Di  ( H4  0.3)  G Sd 4.9  Di  ( H4  0.3) St

tu  max ( t)

 CA

t4d  8.23

mm

t4t  1.982 mm

tu  8.23

4

m

mm


Ratio:

tL  t3

tL

K 

K  1.069

tu K  (K  1)

C 

1  K1.5

C  0.034


x1  203.091

x2  1000  C  H4

x2  219.979

x3  1.22  ( Ri  1000  tu)

x3  265.396

 x1  x   x2  min ( x )  203.091  x3   

xe  min ( x ) xe

 

4.9  Di   H4  t4d1 

 

t4t1 

 t4d1     t4t1 

t4  8.265

max ( t)  8.265

 CA

Sd 4.9  Di   H4 

t 

 G 

1000 

xe

t4d1  8.265 mm

 

1000 

St

t4a  max ( t)

t4t1  2.013

mm

t4a  8.265

mm

mm

e) The fifth course thickness (t5): Ratio for the lower course:

ratio 

h0  1000

ratio  6.881


H5  H4  h0


t5d 

t5t 

t 

 t5d     t5t 

max ( t)  7.691


tL  t4

H5  5

4.9  Di  ( H5  0.3)  G Sd

 CA

4.9  Di  ( H5  0.3)

t5d  7.691 mm

t5t  1.503 mm

St

tu  max ( t)

tu  7.691

Ratio:

K 

C 

Distance of the variable design point from the bottom of the course: (x)

5

m

tL

mm

K  1.075

tu K  (K  1)

1  K1.5

C  0.037

x1  0.61  ( Ri  1000  tu)  320  C  H5

x1  186.872

x2  1000  C  H5

x2  183.117

x3  1.22  ( Ri  1000  tu)

x3  256.549

 x1  x   x2  min ( x )  183.117  x3   

xe  min ( x )

 

4.9  Di   H5  t5d1 

 

t5t1 

 t5d1     t5t1 

t5  7.733

max ( t)  7.733

 G 

Sd 4.9  Di   H5 

t 

xe

1000 

 CA

t5d1  7.733 mm

xe

  1000 

St

t5a  max ( t)

t5t1  1.54

mm

t5a  7.733

mm

mm

f) The sixth course thickness (t6): Ratio for the lower course:

ratio 

h0  1000

ratio  7.114


H6  H5  h0


t6d 

t6t 

t 

 t6d     t6t 

max ( t)  7.151


tL  t5

H6  3.5

4.9  Di  ( H6  0.3)  G Sd

 CA

4.9  Di  ( H6  0.3)

t6d  7.151 mm

t6t  1.023 mm

St

tu  max ( t)

tu  7.151

Ratio:

K 

C 

tL

mm

K  1.081

tu

K  (K  1)

1  K1.5


x1  168.278

x2  1000  C  H6

x2  139.326

x3  1.22  ( Ri  1000  tu)

x3  247.387

6

m

C  0.04

 x1  x   x2  min ( x )  139.326  x3   

xe  min ( x )

 

4.9  Di   H6  t6d1 

Sd

 

4.9  Di   H6  t6t1  t 

 t6d1     t6t1 

t6  7.209

max ( t)  7.209

xe

 G  1000   CA t6d1  7.209 mm xe

 

1000 

St

t6a  max ( t)

t6t1  1.074

mm

t6a  7.209

mm

mm

g) The seventh course thickness (t7): Ratio for the lower course:

h0  1000

ratio 

ratio  7.368


H7  H6  h0


t7d 

t7t 

t 

 t7d     t7t 

max ( t)  6.611


H7  2

4.9  Di  ( H7  0.3)  G Sd

 CA

t7d  6.611 mm

4.9  Di  ( H7  0.3)

t7t  0.544 mm

St

tu  max ( t)

tL  t6

tu  6.611

Ratio:

m

K 

tL

K  1.09

tu

C 

mm

K  (K  1)

1  K  1.5

C  0.044


x1  147.168

x2  1000  C  H7

x2  88.227

x3  1.22  ( Ri  1000  tu)

x3  237.871

 x1  x   x2  min ( x )  88.227  x3   

xe  min ( x )

 

4.9  Di   H7  t7d1 

Sd

7

xe

 G  1000   CA t7d1  6.688 mm

 

4.9  Di   H7  t7t1  t 

 t7d1     t7t1 

t7  6.688

max ( t)  6.688

xe

  1000 

St

t7a  max ( t)

t7t1  0.611

mm

t7a  6.688

mm

mm

h) The eighth course thickness (t8): Ratio for the lower course:

h0  1000

ratio 

ratio  7.649


H8  H7  h0


t8d 

t8t 

t 

 t8d     t8t 

max ( t)  6.072


H8  0.5

4.9  Di  ( H8  0.3)  G Sd

 CA

t8d  6.072 mm

4.9  Di  ( H8  0.3)

t8t  0.064 mm

St

tu  max ( t)

tu  6.072

Ratio:

tL  t7

m

K 

tL

K  1.101

tu

C 

mm

K  (K  1)

1  K1.5

C  0.049


x1  121.877

x2  1000  C  H8

x2  24.68

x3  1.22  ( Ri  1000  tu)

x3  227.959

 x1  x   x2  min ( x )  24.68  x3   

xe  min ( x )

 

4.9  Di   H8  t8d1 

 

t8t1 

 t8d1     t8t1 

max ( t)  6.171

 G 

1000 

Sd 4.9  Di   H8 

t 

xe

St

t8a  max ( t)

8

xe

 CA

t8d1  6.171 mm

 

1000 

t8t1  0.152

mm

t8a  6.171

mm

t8  6.171

mm

3) THICKNESSES OF ALL SHELL COURSES: Minimum shell thickness: According to API 650 Section 5.6.1.1. minimum shell thickness can not be less than this values: Tank Diameter (m): Di

Di  15

15  Di  36

36  Di  60

5

6

8

Plate Thickness (mm): t Di  11.5

60  Di 10

m

tmin  5

Selected Thicness of Shell Courses: Number of Shell Courses:

nsh  8

i  1  nsh

Course No

Thickness (mm)

Selected Thickness (mm)

Course Height (m)

1

t1  9.929

th1  12

h1  h0

h1  1.5

2

t2  9.371

th2  12

h2  h0

h2  1.5

3

t3  8.797

th3  10

h3  h0

h3  1.5

4

t4  8.265

th4  10

h4  h0

h4  1.5

5

t5  7.733

th5  8

h5  h0

h5  1.5

6

t6  7.209

th6  8

h6  h0

h6  1.5

7

t7  6.688

th7  8

h7  h0

h7  1.5

8

t8  6.171

th8  8

h8  H8

h8  0.5

Mid Elevations of Shell Courses: Course No

Mid Elevations of Shell Courses h1

1

hm1 

2

hm2  h1 

3

hm3  hm2 

4

hm4  hm3 

5

hm5  hm4 

6

hm6  hm5 

7

hm7  hm6 

8

hm8  hm7 

Mid Elevations (m) hm1  0.75

2 h2

hm2  2.25

2 h2 2 h3 2 h4 2 h5 2 h6 2 h7 2













h3

hm3  3.75

2 h4

hm4  5.25

2 h5

hm5  6.75

2 h6

hm6  8.25

2 h7

hm7  9.75

2 h8

hm8  10.75

2

9

mm

 hi thi Average Thickness of Tank Shell (mm):

i

tav 

tav  9.636



mm

hi

i 4

 hi thi

Wsh  Di    7.85 

Weight of Shell Courses (kg):

Wsh  3.006  10

i

 hi thi hmi Center of Gravity of Shell Courses (m):

i

Hs 

Hs  4.991



hi  thi

i

C) BOTTOM PLATES td  CA

Product Stress (MPa):

PS 

Hydrostatic Test Stress (MPa):

HTS 

Stress in First Shell Course (MPa):

th1  CA tt th1

 Sd

PS  95.795

 St

MPa

HTS  47.898 MPa

  max ( PS  HTS)

  95.795

MPa

According to API 650 Table 5.1 Annular Bottom Plate Thickness (tb): Plate Thickness of First Shell Course, t (mm)

tb  6

Stress in First Shell Course,  (MPa)   190

  210

  230

  250

t  19

6

6

7

19  t  25

6

7

10

11

25  t  32

6

9

12

14

32  t  40

8

11

14

17

40  t  45

9

13

16

19

9

mm Selected Annular Bottom Plate Thickness (including Corrosion Allowance):

tbs  12 mm

Selected Bottom Plate Thickness:

tbs  8

If annular plates are used, minimum radial width of annular plates:

w 

215  tbs ( Hliq  G)

D) TOP AND INTERMEDIATE WIND GIRDERS 1) TOP WIND GIRDER:

10

mm

w  530.804 mm

2

Required minimum section modulus (cm3):

Z 

Di  H 17

 Vw    190 

2



3

Z  39.815

cm

H1  94.477

m

Profile UNP100 can be selected with section Z = 41.2 cm3.

2) INTERMEDIATE WIND GIRDER: The top shell course plate thickness:

t  8

The maximum height of the unstiffened shell :

H1  9.47  t 

mm

 t    Di 

3

 190    Vw 

2



Vertical distance between the intermediate wind girder and top wind girder H1:

H1  94.477

m

If the height of the transformed shell, Wtr, is greater than the maximum height H1, an intermediate wind girder is required.

H1  94.477

m



 Wtr  8.806

m

The intermediate wind girder is not required.

E) ROOF PLATES Loads Dead Load (the weight of the roof):

DL = t x (7.85) x 0.01

kPa

Design External Pressure:

Pe  0.25

kPa

Roof Live Load:

LR  Lr  0.01

LR  2.5

kPa

Snow Load:

S  Sn  0.01

S1

kPa

Self supporting cone roof Self supporting cone roofs should conform to the following requirements: Angle of the cone roof elements to the horizontal (degree):

9.5    37

Assume an angle for plate thickness calculation:

  18

Dead Load (with plate thickness assumption):

DL  12  ( 7.85)  0.01

DL  0.942

kPa

1) DL + (Lr or S) + 0.4Pe

T1  DL  LR  0.4  Pe

T1  3.542

kPa

2) DL + Pe + 0.4(Lr or S)

T2  DL  Pe  0.4  LR

T2  2.192

kPa

deg

deg

Greater of load combinations:

11

T  max ( T1  T2)

Minimum roof plate thickness:

trmin 

T  3.542

Di

    4.8  sin    180 



T 2.2

kPa

trmin  11.838 mm

2

Calculated minimum roof plate thickness should not be greater than 13 mm according to API 650. Therefore supported cone roof will not be considered. Selected plate thickness of the supported cone roof:

tr  12

mm

F) OVERTURNING STABILITY UNDER WINDLOAD The wind pressure on projected areas of cylindrical surfaces for 100 miles/h wind velocity: fw  0.86

kPa 2

 Vw   Di  H  1000  9.81  190 

The wind load acting on tank:

Fw  fw  

Overturning moment from wind load:

Mw  Fw 

3

Fw  5.16  10

H

kg 4

kg  m

3

kg

Mw  2.838  10

2

Weight of tank: Weight of Bottom Plates:

Wb 

 Di  0.001 th1  0.52   4

 tbs  ( 7.85) Wb  7.117  10

4

Weight of Shell Courses:

Wsh  3.006  10

kg

2

( Di  0.5)  

Weight of Roof (with stiffeners):

Wro 

Resisting weight:

Wres  Wsh  Wro

Overturning moment from wind load:

Mw  2.838  10

4

4

4

 ( tr  1)  ( 7.85)

kg  m

Wro  1.154  10

kg

4

Wres  4.16  10



 Wres  Di   1.595  105  3  2  2



kg

kg  m

There is no overturning due to wind load. Therefore anchor bolts are not required.

G) SEISMIC DESIGN OF TANK (for MCE - Maximum Considered Earthquake) Reference Standard:

API Standard 650, ASCE 7

SEISMIC DESIGN FACTORS Seismic Use Group: Effective Ground Acceleration Coefficient: (for Seismic Zone 1 according to TEC 2007)

SUG  3 Seismic Zone

A0  0.4

Acceleration Coefficient

12

1 2 3 4     0.4 0.3 0.2 0.1 

Importance Factor: (API 650 Table E-5)

I  1.5

Response Modification Factor - impulsive: (API 650 Table E-4)

Ri  4

(mechanically anchored)

Response Modification Factor - convective: (API 650 Table E-4)

Rc  2

(mechanically anchored)

        1.0 1.25 1.5 

Seismic Use Group Importance Factor

SITE GROUND MOTION Acceleration Parameters For sites not addressed by ASCE methods, the peak ground acceleration method shall be used. The peak ground acceleration parameter will be calculated by using the effective ground acceleration coefficient in TE C 2007. With a conservative approach, the effective ground acceleration coefficient in TEC 2007 will be multiplied by two. Peak Ground Acceleration Parameter:

Sp  A0  2

Sp  0.8

Mapped MCE, 5% damped, spectral response acceleration parameter at short periods (0.2 sec), %g

Ss  2.5  Sp

Ss  2

Mapped MCE, 5 percent damped, spectral response acceleration parameter at a period of 1 sec, %g

S1  1.25  Sp

S1  1

Modifications for Site Soil Conditions Site Class based on the Site Soil Properties:

E

Acceleration Based Site Coefficient - at 0.2 sec period: (API 650 Table E-1)

Fa  0.9

Velocity Based Site Coefficient - at 1.0 sec period: (API 650 Table E-1

Fv  2.4

Adjusted Maximum Considered Earthquake (MCE) Spectral Response Acceleration Parameters: (According to ASCE 7-05 Section 11.4.3) For short periods:

Sms  Ss  Fa

Sms  1.8

For 1 second:

Sm1  S1  Fv

Sm1  2.4

Design Spectral Response Acceleration Parameters: (According to ASCE 7-05 Section 11.4.4) For short periods:

Sds 

For 1 second:

Sd1 

2 3 2 3

 Sms

Sds  1.2

 Sm1

Sd1  1.6

Design Response Spectrum (DRS): (According to ASCE 7-05 Section 11.4.5) Characteristic Periods:

T0  0.2 

Sd1 Sds

13

T0  0.267 s

Sd1

Ts 

Ts  1.333 s

Sds

Regional Dependent Transition Period for Longer Period Ground Motion: TL  4

(Regions outside the USA)

s

Natural Vibration Period (s):

T  0.01  0.015  6

Design Responce Spectrum

When

T  T0

Sa ( T)  Sds   0.4  0.6 

When

T0  T  Ts

Sa ( T)  Sds

When

Ts  T  TL

Sa ( T) 

TL  T

Sa ( T)   Sd1 

When

 

 

Sd1 T



TL 



T

2



Spectral Response Acceleration

1.2

1

0.8

Sa ( T) 0.6

0.4

0.2

0

1

2

3

4

T Period (s)

STRUCTURAL PERIOD OF VIBRATION Impulsive Natural Period Density of Fluid:

  1  10

3

kg m

Height to Diameter Ratio:

Hliq Di

 0.913

Coefficient Ci: (API 650 Figure E-1)

Ci  7.2

Elastic Modulus of Tank Material (MPa):

E  2.1  10

5

14

3

5

6

T

  T0 

Equivalent Uniform Thickness of Tank Shell: (mm) (Average thickness)

tu  tav

Impulsive Natural Period (s): (API 650 Eq. E.4.5.1)

Ti 

1



Ci  Hliq

2000



tu



tu  9.636

mm

Ti  0.127

s

E

Di

Convective (Sloshing) Period Sloshing Period Coefficient:

0.578

Ks 

Ks  0.579

 3.68  Hliq 

tanh 



The First Mode Sloshing Wave Period (s ): (API 650 Eq. E.4.5.2)

Di

 

Tc  1.8  Ks  Di

Tc  3.532

s

DESIGN SPECTRAL RESPONSE ACCELERATIONS Impulsive Spectral Acceleration Parameter

 I   Ri 

Ai  Sds  

Ai  0.45

%g

Convective Spectral Acceleration Parameter Coefficient to adjust the spectral acceleration from 5% - 0.5% damping:

K  1.5

 1   I     Tc   Ri 

When

Tc  TL

Ac  K  Sd1  

When

Tc  TL

Ac  K  Sd1  

 TL    I    2  Rc  Tc    Ac  0.255

%g

DESIGN LOADS Effective Weight of Product Diameter to Height Ratio:

Di Hliq

 1.095 2

Total weight of tank contents (N):

Wp 

  Di 4

 Hliq    9.81

15

Wp  1.07  10

7

N

Effective Impulsive Weight (N): (API 650 Eq. E.6.1.1)

Selection of Effective Impulsive Weight Equation:

When

Di Hliq

When

 

tanh  0.866 

Di Hliq

Wi 

 1.333

0.866 

  Hliq   Wp

Di

Hliq

 

 1.333

Di

Wi   1.0  0.218 

Di

  Wp 

Hliq 

6

N

Wi  8.144  10

Effective Convective Weight (N): (API 650 Eq. E.6.1.1)

Wc  0.230 

Di Hliq

 3.67  Hliq   Wp   Di 

 tanh 

6

Wc  2.689  10

N

Center of Action for Ringwall Overturning Moment The ringwall overturning moment is the portion of the total overturning moment that acts at the base of the tank shell perimeter. This moment is used to determine loads on a ringwall foundation, the tank anchorage forces, and to check the longitudinal shell compression.

Height of the Lateral Seismic Force: Applied to Wi (m) (API 650 Eq. E.6.1.2.1)

Selection of Height Equation:

When

Di Hliq

When

Di Hliq

 1.333

 1.333

Xi  0.375  Hliq

 

Xi   0.5  0.094 

Di

  Hliq  Hliq  m

Xi  4.169

  3.67  Hliq   1  cosh    Di    Hliq Xc   1.0    3.67  Hliq   sinh  3.67  Hliq         Di   Di  

Height of the Lateral Seismic Force: Applied to Wc (m) (API 650 Eq. E.6.1.2.1)

Xc  7.579

m

Center of Action for Slab Overturning Moment The slab overturning moment is the total overturning moment acting across the entire tank base cross section. This overturning moment is used to design slab and pile cap foundation (if any). Height of the Lateral Seismic Force: Applied to Wi (m) (API 650 Eq. E.6.1.2.2) When

Selection of Height Equation:

Di Hliq

 1.333

  0.866  Di   Hliq Xis  0.375   1.0  1.333    1.0   Hliq    Di     tanh  0.866  Hliq    16

When

Di Hliq

 

 1.333

Xis   0.5  0.06 

Di

  Hliq  Hliq  m

Xis  5.94

Height of the Lateral Seismic Force: Applied to Wc (m) (API 650 Eq. E.6.1.2.2)

 Xcs   1.0   

 3.67  Hliq   1.937    Di    Hliq 3.67  Hliq 3.67  Hliq    sinh        Di   Di   cosh 

Xcs  7.785

m

Overturning Moment The seismic overturning moment at the base of the tank is evaluated as the SRSS summation of the impulsive and convective components multiplied by the respective moment arms to the center of action of these forces. 5

Total weight of tank shell (N):

Ws  Wsh  9.81

Ws  2.949  10

Height of Shell's Center of Gravity (m)

Xs  Hs

Xs  4.991

Weight of Roof (N):

Wr  Wro  9.81

Wr  1.132  10

Height of Roof's Center of Gravity (m)

Xr  H 

Ringwall Overturning Moment (Nm): (API 650 Eq. E.6.1.5) for global evaluations

Mrw 

Slab Overturning Moment (Nm): (API 650 Eq. E.6.1.5)

Ms 

m 5

 Di  tan         3 2  180   1



Xr  11.623

2

[ Ai  ( Wi  Xi  Ws  Xs  Wr  Xr) ]  [ Ac  ( Wc  Xc) ]

2

7

2

[ Ai  ( Wi  Xis  Ws  Xs  Wr  Xr) ]  [ Ac  ( Wc  Xcs ) ] 7

Vertical Seismic Effects The vertical seismic acceleration parameter Av is defined as 0.14*Sds in API 650 and as 0.2*Sds in ASCE 7 method. Conservatively 0.2*Sds is choosen in calculations. Av  0.24

Dynamic Liquid Hoop Forces Dynamic hoop tensile stress due to seismic motion of the liquid is calculated by the following formulas. Calculation for the 1.st shell course: Distance from liquid surface to analysis point (m):

Y  Hliq

17

Y  10.5 m

Nm

2

Ms  2.363  10

Av  0.2  Sds

N

m

Mrw  1.733  10

Vertical Seismic Acceleration Coeff. (%g):

N

Nm

Impulsive Hoop Membrane Force in Tank Shell (N/mm): (API 650 Eq. E.6.1.4)

When

Di Hliq

When

Di Hliq

When

Di Hliq

Selection of Force Equation:

2  Y  Y    tanh  0.866  Di   0.5      Hliq   Hliq  Hliq   

Ni  8.48  Ai  G  Di  Hliq  

 1.333

 1.333

and

Y  0.75  Di

2  Y Y     Ni  5.22  Ai  G  Di   0.5      0.75  Di  0.75  Di  

 1.333

and

Y  0.75  Di

Ni  2.6  Ai  G  Di

2

2

N

Ni  154.732

mm

Convective Hoop Membrane Force in Tank Shell (N/mm): (API 650 Eq. E.6.1.4)

 

2

1.85  Ac  G  Di  cosh  3.68  Nc 

( Hliq  Y)  Di

 3.68  Hliq    Di 

 

N

Nc  4.325

mm

cosh 

Liquid Hydrostatic Membrane Force in Tank Shell (N/mm): Y  G  Di

Nh 

2

Thickness of the shell ring under consideration (mm):

Total Combined Hoop Stress (MPa):

t 

ts  th1  CA 2

Nh 

N

Nh  592.279

 9.81

ts  6

2

Ni  Nc  ( Av  Nh) ts

mm

mm

2

t  133.74

MPa

The maximum allowable hoop tension membrane stress for the combination of hydrostatic and dynamic membrane hoop effects should be less than allowable design stress of the shell increased by 33%. Allowable Stress for MCE seismic design:

Comparison:

t  133.74

Hoop Stress Ratio:

all  1.33  Sd

MPa



all  208.367

SRhs 

t all

all  208.367

MPa

SRhs  0.642

OK

MPa

Calculation for the 2.nd shell course: Distance from liquid surface to analysis point (m):

Y  Hliq  h0

18

Y9

m


When

Di Hliq

When

Di Hliq

When

Di Hliq


2  Y  Y    tanh  0.866  Di   0.5      Hliq   Hliq  Hliq   

Ni  8.48  Ai  G  Di  Hliq  

 1.333

 1.333

and

 1.333

and

2



Y  0.75  Di

Ni  5.22  Ai  G  Di  

Y  0.75  Di

Ni  2.6  Ai  G  Di

Y

 0.75  Di

2  Y     0.75  Di  

 0.5  

2

N

Ni  154.732

mm


 

2

1.85  Ac  G  Di  cosh  3.68  Nc 


 3.68  Hliq    Di 

 

N

Nc  4.833

mm

cosh 


Nh 

2



t 

ts  th2  CA 2

Nh 

N

Nh  507.668

 9.81

ts  6

2


mm

mm

2

t  117.445

MPa


Comparison:

t  117.445

Hoop Stress Ratio:

all  1.33  Sd

MPa



all  208.367

SRhs 

t all

all  208.367

MPa

SRhs  0.564

OK

MPa

Calculation for the 3.rd shell course: Distance from liquid surface to analysis point (m):

Y  Hliq  2  h0

19

Y  7.5

m


When

Di Hliq

When

Di Hliq

When

Di Hliq


2  Y Y   Di    Ni  8.48  Ai  G  Di  Hliq    0.5      tanh  0.866   Hliq   Hliq  Hliq   

 1.333

 1.333

and

 1.333

and

2



Y  0.75  Di

Ni  5.22  Ai  G  Di  

Y  0.75  Di

Ni  2.6  Ai  G  Di

Y

 0.75  Di

2  Y     0.75  Di  

 0.5  

2

N

Ni  152.685

mm


 

2

1.85  Ac  G  Di  cosh  3.68  Nc 


 3.68  Hliq    Di 

 

N

Nc  6.476

mm

cosh 


Nh 

2



t 

ts  th3  CA 2

Nh 

N

Nh  423.056

 9.81

ts  4

2


mm

mm

2

MPa

t  151.633


Comparison:

t  151.633

Hoop Stress Ratio:

MPa

all  1.33  Sd



all  208.367

SRhs 

t all

all  208.367

MPa

SRhs  0.728

OK

MPa

FOUNDATION LOADS Dead Load per Unit Length (N/m): (Shell and Roof)

DL 

Ws  Wr

20

Di  

DL  1.13  10

4

N m

2

Lr  9.81 

( Di  0.5)   4

Live Load per Unit Length (N/m): (Live Load on Roof)

LL 

Total Dead Weight (N): (Shell, Roof and Liquid)

Wt  Ws  Wr  Wp

Total Load per Unit Area during Operation (N/m2): (Shell, Roof and Liquid)

Wo 

LL  7.677  10

Di  

N

3

m 7

Wt  1.111  10

Wt

N N

5

Wo  1.069  10

 Di2       4 

m

2

Seismic loads: The equivalent lateral seismic forces are calculated by considering the effective mass and dynamic liquid pressures. The seismic base shear is evaluated as the SRSS summation of the impulsive and convective components.

Base Shear due to Seismic Load (N):

Seq 

2

[ Ai  ( Wi  Ws  Wr) ]  ( Ac  Wc )

2 6

N

7

Seq  3.909  10 Ringwall Overturning Moment due to Seismic Load (Nm):

Mrw  1.733  10

Nm

Slab Overturning Moment due to Seismic Load (Nm):

Ms  2.363  10

7

Nm

4

Vertical Seismic Force (N): (Shell and Roof)

Fvs  Av  ( Ws  Wr)

Vertical Seismic Force per Unit Length (N/m): (Shel and Roof)

VSF 

Total Vertical Seismic Force (N): (Shell, Roof and Liquid)

Fvst  Av  Wt

Fvst  2.666  10

N

Total Vertical Load (N): (Total Vertical Seismic and Total Dead W.)

Fvt  Fvst  Wt

Fvt  1.377  10

7

N

Fvs ( Di   )

Fvs  9.795  10

3

VSF  2.711  10

6

N

N m

ANCHORAGE LOADS Resistance to the overturning (ringwall) moment at the base of the shell is provided by mechanical anchorage devices (anchor bolts). The resisting weight of the liquid is neglected in the calculation of the uplift load on the anchors. The anchors are sized to provide at least the minimum anchorage resistance calculated as follows: Distributed Compression Force due to Roof (N/m): wr 

Distributed Compression Force due to Shell (N/m): ws  Total Distributed Compression Force (N/m):

Wr Di   Ws Di  

wt  wr  ws

Vertical Seismic Acceleration (g's):

Minimum Anchorage Resistance (N/m): (API 650 Eq. E.6.2.1.2)

3

wr  3.134  10

N m

3

ws  8.163  10 4

wt  1.13  10

N m N m

Av  0.24

wab 

 1.273  Mrw  wt  ( 1  0.4  Av)  wab  1.566  105   2  Di  21

N m

ANCHOR BOLT VERIFICATION (LRFD CRITERION) Due to the adoption of shear keys, anchor bolts are subjected to traction loads only. Max applied tractions are evaluated from above calculated anchorage loads and anchor bolt capacity is determined according to ACI 318-05 Appendix D Requirements according to API 650 E.7.1.2:

- Minimum 6 anchors should be provided. - The spacing between anchors should be less than 3 m. - Anchors should have a minimum diameter of 25 mm.

Number of Equally Spaced Anchors Around the Tank Circumference:

nb  24

Distance from bolt center to shell (mm):

Dbs  92

Bolt Circle Diameter (m):

Db  Di  2 

Bolt Spacing Angle:

 

Bolt Spacing (m):

Dbs  th1 1000

360

Db  11.708

Db  

m

degrees

  15

nb

Bsp 

mm

Bsp  1.533

nb

Concrete strength (MPa):

flc  25

m

MPa

Anchor Bolt Characteristics Cast in headed stud anchor Nominal Diameter of Anchor (mm):

db  48

Threaded Area of Bolt (mm2):

Ath 

mm 2

0.75  db   4

3

Ath  1.357  10

Anchor bolt material: S275JR (St44-2) or equivalent Ultimate Tensile Strenght (MPa):

Sub  430

Yield Strenght (MPa):

Syb  275

Maximum traction As LRFD design method is used for anchor bolt verification, following load combination will be adopted U = 0.9 x D + E Bsp  1000

Bolt Spacing to Diameter ratio

db Max traction on single bolt (kN)

Tb 

wab  Bsp 1000

22

Tb  240

 31.929

kN

2

mm

Bolts tension capacity (according to clause D.5) Reduction Factor (according to clause D.4.4.a)

t  0.75

Additional seismic strength reduction factor

s  0.75

Design tensile strength (ACI 318 D.5.1.2) (MPa)

futa  min ( Sub  860  1.9  Syb)

Nominal bolt strength in tension (kN)

Nsa  Ath  min ( futa  860)  10

Nsa  583.582

Bolt tension capacity (kN)

Nsa  s  t  Nsa

Nsa  328.265 kN

Bolt tension demand (kN)

Nua  Tb

Nua  240

Comparison:

futa  430

3

Nsa  328.265 kN

Nua  240



Bolt usage ratio:

kN

kN

kN

Nua

FUt 

MPa

OK

FUt  0.731

Nsa

Pullout strength in tension (according to clause D.5.3) Modification Factor:

cp  1.4

Reduction Factor:

p  0.75

Bearing area at head of anchor bolt (mm2):

 db2     Abrg  160    4 

Pull out strength in tension of an headed bolt (kN):

Np  8  Abrg  flc  10

Nominal pull out strength (kN):

Npn  Np  cp

Npn  6.661  10 kN

Design pull out strength (kN):

Np  p  s  Npn

Np  3.747  10 kN

Comparison:

3

Np  3.747  10 kN

2



Nsa  328.265

2

4

Abrg  2.379  10 mm

3

3

Np  4.758  kN 10

3

3

OK

kN

Bolt adequacy for uplift loads According to Table 3.21 of API 650 Dead load of shell minus any corrosion allowance and any dead load including roof plate acting on the shell minus any corrosion allowance (N):

 

 tav  CA   Wro  tr  CA    9.81     tav   tr  

W2   Wsh  

Seismic uplift loads (N):

U  4 

Ms Di

5

N

7

Nm

W2  1.679  10

 W2  ( 1  0.4  Av)

As Ms is used for a verification based on ASD criterion a new evaluation can be made as follows: Slab Overturning Moment (Nm): Ms 

2

[ Ai  0.7  ( Wi  Xis  Ws  Xs  Wr  Xr) ]  [ Ac  ( Wc  Xcs ) ]

23

2

Ms  1.698  10

Levhali

Seismic uplift loads (N):

Uplift load per anchor (N):

Uasd  4 

tb 

Ms Di

Uasd

al  0.8  Syb

Average induced stress (MPa):

ub 

5

6

Nm

N

MPa

al  220

tb

ub  176.627 MPa

Ath ub

SRu 

Uasd  5.753  10

tb  2.397  10

nb

Allowable Ancher Bolt Stress (MPa): according to Table 3.21 of API 650

Uplift stress ratio

 W2  ( 1  0.4  Av)

OK

SRu  0.803

al

SHEAR KEY VERIFICATION (ASD CRITERION) Shear keys characteristics

Depth of shear key (mm):

dp  100 mm

Width of shear key (mm):

wsk  100 mm

Thickness of shear key (mm):

tsk  20

Number:

nsk  24

Material:

S275 JRG2

Plate minimum yield stress (MPa)

ysk  275

mm

Verification procedure The shear keys are verified for the bending moment and shear stresses in the plates produced by the concrete bearing reaction in the contact area, assumed as uniformly distributed. Two verifications are performed: A global verification at the shear key connection to the annular plate A local verification at the connection of the two vertical plates forming the shear key. 6

Total Base Shear due to seismic load (N):

0.7  Seq  2.736  10

Shear for each shear key (N):

Ssk 

Concrete compression (MPa):

fc 

0.7  Seq nsk Ssk

wsk  dp

Concrete allowable compression (MPa):

fcall  0.65  0.85  flc

Concrete compression ratio

SRck 

fc fcall

N 5

Ssk  1.14  10

fc  11.402 MPa

fcall  13.813

MPa OK

SRck  0.825

Global verification Shear area (mm2):

Assk  tsk  wsk

24

Assk  2  10

3

2

mm

Shear stress (MPa):

 

Ssk

Shear key allowable bending stress (MPa):

allsk 

Shear key allowable shear stress (MPa):

allsk 

MPa

  57.009

Assk 2 3

allsk  183.333 MPa

 ysk

allsk

MPa

allsk  129.636

2 Shear stress ratio:



SR 

dp

Arm of the global concrete reaction (mm):

afc 

Global bending moment (Nmm):

Mgk  Ssk  afc

Global inertia moment (mm4):

Igk 

afc  50

2

1

mm 6

Mgk  5.701  10

3

12

OK

SR  0.44

allsk

  ( wsk)  tsk  ( wsk  tsk)  tsk

3

 4

6

Igk  1.72  10 Global section modulus (mm3):

Shear key global bending stress (MPa):

Shear key global bending stress ratio.

Igk

Wgk 

gk 

wsk

mm

3

4

Wgk  3.44  10

2

Mgk

mm

gk  165.723 MPa

Wgk

SRgk 

Nmm

gk

OK

SRgk  0.904

allsk

Local verification Conservatively we consider a simple cantilever beam of unit width

Shear key overhang (mm):

esk 

wsk  tsk

Bending moment due to concrete reaction (Nmm/mm): Mlk  fc  esk  1

Shear key section modulus per unit depth (mm3/mm): Wlk 

Shear key bending stress (MPa):

Shear key local bending stress ratio:

esk  40

2

lk 

6

2

Mlk Wlk

SRlk 

25

 tsk

esk

lk allsk

2

mm 3 N  mm

Mlk  9.121  10

mm 3

Wlk  66.667

lk  136.821

SRlk  0.746

mm

mm MPa

OK

MAXIMUM LONGITUDILAN SHELL MEMBRANE COMPRESSION STRESS Shell Compression in Mechanically Anchored Tanks The maximum longitudinal shell compression stress at the bottom of the shell for mechanically anchored tanks is evaluated according to API 650 E.6.2.2.2 Thickness of Bottom Shell Course less CA (mm): tsb  th1  CA



c  wt  ( 1  0.4  Av) 

1.273  Mrw

1

  1000  tsb 

2

Di



tsb  6

mm

MPa

c  29.865

Allowable Longitudinal Shell Membrane Compression Stress The seismic allowable stress Fc is evaluated according to API 650 E.6.2.2.3 2

The Parameter:

Para 

G  Hliq  Di

Para  38.573

2

tsb The Allowable Compression Stress (MPa): (API 650 Eq. E.6.2.2.3)

Selection of Stress Equation: 2

G  Hliq  Di

When

 44

2

Fc 

tsb

83  tsb Di

2

G  Hliq  Di

When

2

 44

tsb

Fc 

83  tsb 2.5  Di

 7.5  ( G  Hliq)  0.5  Sy Fc  41.625

Comparison:

c  29.865

Compression Stress Ratio:

Rcs 

MPa



c

Fc  41.625

MPa

MPa

OK

Rcs  0.717

Fc

ANCHOR CHAIR VERIFICATION (ASD CRITERION) The tank is anchored to the foundation by mean of anchor bolts and chairs. The verification of various components of the chair (top plate and gussets) is performed according to procedure 3-14 "Design of base details for vertical vessels" of Pressure Vessel Design Manual by D. Moss. Used symbols are shown in next figure. Input data Material S235 JRG2 Plate minimum yield stress (MPa):

y  Sy

y  235

Plate allowable stress (MPa):

ball  Sd

ball  156.667 MPa

Bolt eccentricity (mm):

a  Dbs

a  92

26

MPa

mm

Height from top of annular plate (mm):

h  250 mm

Distance between gussets (mm):

b  100 mm

Thickness of bottom shell (mm):

ts  th1

Bolt diameter (mm):

ts  12

mm

db  48

mm

Bolt hole in the top plate (mm)

dbh  db  24

Top plate thickness (mm):

tc  30

Top plate width (mm):

A  400 mm

Top plate edge distance from bolt axis (mm):

c  85

Top plate width ouside bolt hole (mm):

e  c 

Thickness of gussets (mm):

tg  25 mm

Bolt pitch (mm):

bp  Bsp  1000

bp  1.533  10

3

mm

Base plate span between chairs (mm):

bs  bp  ( b  2  tg)

bs  1.383  10

3

mm

Number of gussets per chair:

ng  2

Shell reinforcement plate thickness (mm:)

rpt  20 mm

Shell reinforcement plate halfwidth (mm):

rpw  200 mm

dbh  72

mm

mm

mm

dbh

e  49 mm

2

Design loads Bolt traction As ASD design method is used for anchor chair verification, a new evaluation of max bolt traction is done as follows:

Maximum traction on single bolt (N):

Tbc 

 1.273  Mrw  wt  ( 1  0.4  Av  0.7)   bp   1000 2  Db  Tbc  2.305  10

5

N

For additional conservatism we consider the max between the computed traction and the ASD bolt capacity Maximum load considered for the chair verification (N): Tbc max Tbc0.7Nsa1000 Tbc  2.305  10

Maximum compression per unit length (N/m):

C  wt  ( 1  0.4  Av  0.7) 

5

N

1.273  Mrw 2

Di

5

C  1.789  10

27

N m

Annular bottom plate characteristics Selected bottom plate thickness (mm):

tb  tbs

tb  8

Annular plate width (mm):

mm

w  530.804

mm

Top plate verification The top plate is assumed as a beam, with dimensions e x A, with partially fixed ends, and a portion (1/3) of the total anchor bolt force Tbc, distributed along part of the span.

Maximum induced bending stress (MPa):

Tbc

tp 

2

 ( 0.375  b  0.22db)

tp  140.808 MPa

e  tc

Top plate bending stress ratio

SRtp 

tp

OK

SRtp  0.899

ball

Gusset verification Gusset maximum axial compression force (N):

Tbc

Cg 

5

Cg  1.152  10

ng

Gusset width at bottom edge (mm):

wo  15 mm

Gusset mean width (mm):

bg 

( a  c )  wo

bg  96

2

mm

Gusset thickness (mm):

tg  25

Shell reinforcement plate thickness (mm):

rpt  20 mm

Shell reinforcement plate halfwidth (mm):

rpw  200

Section total area (mm2):

Neutral axis distance from midsurface of reinforcement plate (mm):

N

mm

mm 2

3

Ag  bg  tg  rpt  rpw

Ag  6.4  10

 bg  rpt    2  2 na  tg  bg 

na  21.75

Ag

mm

mm

Longitudinal inertia moment (mm4): 3

Il 

tg  bg 12

2

3

 bg  rpt  na  rpw  rpt  rpt  rpw  ( na) 2  2 12 2 

6

 tg  bg  

Transv ersal inertia moment (mm4):

It 

Inertia radius (mm):

rl 

 12 1

Ag

28

3

 bg  tg  rpt  rpw

Il

rt 

Il  7.023  10

It Ag



3

7

It  1.346  10

rl  33.125

mm

rt  45.857

mm

4

mm

4

mm

rmin  min ( rl  rt) Instability Factor:

IF  1

Young's modulus (MPa):

E  210000 MPa

Yield s tress (MPa):

rmin  33.125 mm

y  235

Cc factor:

2 E

Cc 

2  

y

MPa

Cc  132.813

Allowable compression stress (MPa):

cgall 

2   IF  h        rmin   1  2  2  rmin   3 5  h    IF  h   1     3  IF     3 3  8  Cc  rmin   rmin  8  Cc  

Max compression stress (MPa)

Compression stress ratio

cg 

Cg Ag

SRcg 

29

 y

cg cgall

cgall  135.608 MPa

cg  18.008

SRcg  0.133

MPa

OK

API 650 Tank Design

Recommend Documents