Stiffness coefficients for nozzles in API 650 tanks.pdf

Proceedings of PVP2002 2002 ASME Pressure Vessels and Piping Conference August 5-9, 2002, Vancouver, BC, Canada

STIFFNESS COEFFICIENTS FOR NOZZLES IN API 650 TANKS

PVP2002-1279

Manfred Lengsfeld Fluor Daniel Inc. manfred.lengsfeld @fluor.com

Kanhaiya L. Bardia Fluor Daniel Inc. [email protected]

Jaan Taagepera Valero Refining Co. [email protected]

Kanajett Hathaitham Fluor Signature Services Inc. ken.hathaitham @fluor.com

Donald G. LaBounty Fluor Daniel Inc. [email protected]

Mark C. Lengsfeld Crane Valves lengsfeld @yahoo.com

ABSTRACT

this paper stiffness coefficients are provided for nozzles away from a structural discontinuity. The distance at which a discontinuity has no influence on the spring rate of a nozzle is defined by Welding Research Council (WRC) Bulletin 297 [11]. Height factors are used to calculate stiffness coefficients for nozzles located close to a gross structural discontinuity. With the height factors provided, the engineer is able to arrive at stiffness coefficients for nozzles at any location on the tank shell, which in turn helps to predict more accurately the piping loads at the nozzle.

The analysis of tank nozzles for API Standard 650 [1] tanks is a complex problem. Appendix P of API 650 provides a method for determining the allowable external loads on tank shell openings. The method in Appendix P is based on two papers, one by Billimoria and Hagstrom [2] and the other by Billimoria and Tam [3]. Although Appendix P is optional, industry has used it for a number of years for large diameter tanks. For tanks less than 120 feet (33.6 m) in diameter, Appendix P is not applicable. In previously published papers [4-10], the authors used finite element analysis (FEA) to verify the experimental results reported by Billimoria and Tam for shell nozzles. The analysis showed the variance between stiffness coefficients and stresses obtained by FEA and API 650 methods for tanks. In this follow-up paper, the authors present stiffness coefficients for tank nozzles located away from a structural discontinuity. Factors to establish spring rates for nozzles varying from 6 to 48 inches and tank diameters from 30 feet to 300 feet and for nozzles at different elevations on the shell are provided. Mathematical equations are provided together with graphs for the stiffness coefficient factors.

NOMENCLATURE B

=

D Do FR KBc

= = = =

Kc

=

KBL =

INTRODUCTION KL

In Appendix P of API 650 a procedure has been established to determine the allowable loads on tank shell openings. This procedure is a practical solution to a complex problem, especially since low-type nozzles, as defined in API 650, are close to the bottom and thus are affected by the bottom-to-shell junction (See Fig. 1). As mentioned by Billimoria and Hagstrom, this procedure is conservative. Users in industry have questioned the need for such conservatism. Even though Appendix P is not mandatory, many designers use this method for lack of any other guidance. In previously published papers, the authors used FEA to verify the experimental results reported by Billimoria and Tam. In papers by Lengsfeld, et. al [4-7] various degrees of conservatism were reported for different nozzle sizes attached to tanks. Stress factors and stiffness coefficients for low-type nozzles were published by the authors [8,9]. Stress factors for varying nozzles heights were published by the authors [10]. In

=

KBR = KR

=

L

=

LB

=

Mc ME a d h mc

= = = = = = =

2(12*Dt) in, height from tank bottom per WRC, Bulletin 297 where tank bottom has no influence on stiffness on nozzles (see Figure 1), in nominal diameter of tank, ft outside diameter of reinforcing pad, in radial load, lbs stiffness coefficient due to circumferential moment at distance B, in-lbs/radian stiffness coefficient for circumferential moment, in-lbs/radian stiffness coefficient due to longitudinal moment at distance B, in-lbs/radian stiffness coefficient for longitudinal moment, inlbs/radian stiffness coefficient due to radial force at distance B, lbs/in stiffness coefficient for shell thrust (radial) load, lbs/in vertical distance from nozzle centerline to tank bottom (see Figure 1), in vertical distance of nozzle centerline where tank bottom has no influence on nozzle stiffness B + ½Do circumferential moment, in-lbs longitudinal moment, in-lbs outside radius of opening connection, in outside diameter of nozzle (2a), in height factorL/LB stiffness ratio for circumferential moment

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Y

..... ~ l

:-----~LONGrlIlJDINAL 1 MOMENTM~, .

X

-

+M~ Figure 1: Dimensions for nozzles per API 650 mL = mR t = t. = tp =

three-dimensional models of the tanks and the nozzles. Larger tanks (D>30 feet) were modeled with 4 node shell elements. Smaller tanks were modeled using 8 node solid elements. Each variation of tank and nozzle diameter had different numbers of elements. The analyses were performed on a Silicon Graphics Workstation and PC's. Figure 3a represents stiffness coefficients due to a radial force for tanks from 30 feet to 300 feet in diameter with a wall thickness of 3/4". Figure 3b gives the stiffness ratio due to a radial force to be used for nozzles located closer to a structural discontinuity. Figures 4a and 4b are for circumferential moments where as Figures 5a and 5b for longitudinal moments. Actual stiffness ratios mi conform to a relative narrow scatter band. For simplicity these bands were combined into single lines in Figures 3b, 4b and 5b. Stiffness coefficients for above figures 3a, 4a and 5a are for nozzles 6" to 48" in diameter located away from a gross structural discontinuity. Nozzles were chosen to have reinforcing pads with equivalent

stiffness ratio for longitudinal moment stiffness ratio for radial force shell thickness of tank, in thickness of nozzle wall, in thickness of reinforcing pad, in

DESCRIPTION

Figure 2 shows a detail of the nozzle area. Each tank was assumed to be at ambient temperature of 70 ° Fahrenheit. The bottom of the shell course for each model had the nodes fixed in all displacements while rotations were not fixed. This assumes that the annular ring provides little resistance to shell rotation due to imposed piping loads. Only an 180 ° section of each tank was modeled, utilizing symmetry to reduce model size. Stiffness coefficients were calculated from the deflection of the nozzle after the loads were applied. For the FEA, COSMOS software developed by Structural Research and Analysis Corporation was used to construct

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Figure 2: Detail of a typical nozzle and shell area The loadings were applied independently. In excess of 100 combinations of loading, tank, thickness, and nozzle sizes were evaluated.

dimensions of Table 3-6, column 5 of API 650. The tanks modeled were 30 feet to 300 feet in diameter and 64 feet high. Several shell-thickness were investigated for each tank. The mathematical method of least-square fits for polynomial curves was used to smooth these curves and derive mathematical equations. Table 1 lists the equations for tank diameters and thickness presently available. These equations were then used to produce the graphs of Figures 3a through 5b. Using the mathematical equations will simplify the creation of computer programs for the calculation of nozzle stiffness coefficient at the nozzle-to-shell junction.

RESULTS Stiffness coefficients vary with the location of the nozzle in height on the tank wall. Stiffness coefficients increase as the nozzle moves closer to a gross structural discontinuity. Factors have been established to calculate spring rates for nozzles at any location on the tank wall using as a basis the rates for nozzles away from a discontinuity. Stiffness coefficients are inverse proportional to the height, the lower the location on the tank, the higher is the spring rate. Depending on the location of the nozzle, the value for the stiffness coefficients from Figures 3a, 4a or 5a will be divided by the height factor from Figures 3b, 4b or 5b respectively.

LOADING The same loadings were applied to all finite element models, namely P = 1,000 lbs Radial, ML = 10,000 in-lbs Longitudinal Moment, Mc = 10,000 in-lbs Circumferential Moment,

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Time and size constrains prevent the authors from investigation of several more shell thickness.

ANALYSIS PROCEDURE The procedure for the evaluation of stiffness coefficients is as follows: 1. Calculate the distance B 2. Establish the height LB 3. Establish the height factor h 4. For a given nozzle on a tank with established wall thickness read the stiffness coefficient from Figures 3a, 4a or 5a for the corresponding loading 5. From Figures 3b, 4b or 5b establish the stiffness ratio factors mi 6. Divide the value of the spring rate from (4) by the stiffness ratio factor from (5) to arrive at the stiffness coefficient for the nozzle under consideration

ACKNOWLEDGMENTS The authors gratefully acknowledge the support of the managements of Fluor Daniel, Valero Refining Co and CCI to prepare and publish this paper. Special thanks to Gilbert Chen, Avtar S. Mann and Dennis Mitchell for their review of the manuscript and their encouragement.

REFERENCES [1] American Petroleum Institute, API Standard 650 tenth Edition, November 1998 "Welded Steel Tanks for Oil Storage" [2] Billimoria, H.D., and Hagstrom, K.K, "Stiffness Coefficients and Allowable Loads for Nozzles in Flat Bottom Storage Tanks" Paper 77-PVP-19. ASME 1977 [3] Billimoria, H.D., Tam, K.K., 1980, "Experimental Investigation of Stiffness Coefficients and Allowable Loads for a Nozzle in a Flat Bottom Tank" ASME Publication 80C2/PVP-5 [4] Lengsfeld, M., Bardia, K.L, Taagepera, J., 1995 "Nozzle Stresses Resulting from Piping Loads at Low-Type Nozzles in API 650 Storage Tanks" ASME PVP-Vol. 315 [5] Lengsfeld, M., Bardia, K.L, Taagepera, J., 1996 "FEA vs. API 650 for Low Tank Nozzles" ASME PVP-Vol. 336 [6] Lengsfeld, M., Bardia, K.L, Taagepera, J., 1997 "FEA vs. API 650 for Low Tank Nozzles (2)" ASME PVP-Vol. 359 [7] Lengsfeld, M., Bardia, K.L, Taagepera, J., 1998 "Spring Rates for Low Tank Nozzles" ASME PVP-Vol. 368 [8] Lengsfeld, M., Bardia, K.L, Taagepera, J., Hathaitham, K., 1999 "Stress Factors for Low-Type Nozzles in API 650 Tanks" ASME PVP-Vol. 388 [9] Lengsfeld, M., Bardia, K.L, Taagepera, J., Hathaitham, K., 1999 "Spring Rates for Low Tank Nozzles in API 650 Tanks" ASME PVP-Vol. 388 [10] Lengsfeld, M., Bardia, K.L., Taagepera, J., Hathaitham, K., LaBounty, D.G., Lengsfeld, M.C., 2001 " Analysis of Loads for Nozzles in API 650 Tanks" ASME PVP-Vol. 430 Ill]Bulletin 297, September1987, "Local Stresses in Cylindrical Shells due to External Loadings on Nozzles" Welding Research Council (WRC), New York

DISCUSSION The graphs presented in this paper were constructed to be on the conservative side. The presented results may be interpolated to establish stiffness coefficients for other nozzles, tank diameters and shell thickness. The purpose of this paper is to give the designer a simple means to arrive at a spring rate at a nozzle to tank shell connection. For more complex or critical applications, it is recommended to perform an FEA analysis including the complete piping system.

CONCLUSION The method presented in this paper provides the design engineer a means to calculate stiffness coefficients at the shell to nozzle junction. With these rates applied piping loads can be established. Once accurate piping loads have been established, stresses at the nozzle-shell junction can be calculated using the methods published previously [10]. The use of the finite element analysis models in determining the stiffness coefficients for tank nozzles is recommended when piping loads indicated by the method provided in this paper are excessive and would result in expensive piping systems. Additional data for other tank sizes are being developed

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= 32.86 in 2) Establish the centerline distance of the nozzle of no discontinuity influence LB = B+0.5*D0 = 32.86 + 0.5 * 49.5 = 57.61 in 3) Establish the height factor h = L/LB 28/57.61 0.486 4) From Figure 3a, 4a, 5a read the stiffness coefficients KBR = 280,000 lbs/in (Fig. 3a) KBC = 80,000 in-lbs/rad (Fig. 4a) KBL = 200,000 in-lbs/rad (Fig. 5a) 5) From Figure 4, 5, 6 read the coefficient factor mr = 0.6 for Kr ( Fig. 3b) mc = 0.68 for Kc (Fig. 4b) mE = 0.88 for KL (Fig. 5b) 6) Divide the stiffness coefficients established in step 4 by the factors from step 5 to arrive at the stiffness coefficients for the nozzle under investigation.

Table 1 Size

Formulae

30' Tank, 1" Wall

KBC = 0.0529x 2 + 4.0405X - 26.146

30' Tank, 1" Wall 30' Tank, 1" Wall

KBL = 0.2662X 2 + 4.3491X - 35.678 KBR = 0.0079X + 0.1919

30' Tank, 1/2" Wall

KBC = 0.0421X 2 - 0.2723X + 9.9571

30' Tank, 1/2" Wall 30' Tank, 1/2" Wall

KBL = 0.15X 2 " 1.0719X + 6.4632 KBR = 4.1196X + 50.97

30' Tank, 3/4" Wall

Kac = 0.0421X 2 - 0.2723x + 9.9571

30' Tank, 3/4" Wall 30' Tank, 3/4" Wall

KBL = 0 . 1 5 X 2 " 1.0719x + 6.4632 KBR = 4.1196X + 50.97

120' Tank 1" Wall

KBC = - 0 . 0 5 8 9 X 2 +

120' Tank. 1" Wall 120' Tank 1" Wall

KBL = 0.1462X 2 - 3.7726X + 54.295 KBR = 0.5828x + 88.103

4.9447x + 2.4016

120' Tank. 1/2" Wall KBc = -010199x 2 + 1.6614x - 4.0694 120' Tank. 1/2" Wall KBL = 0.0215X 2 + 0.2208X + 5.3213 1 2 0 ' T a n k 1/2" Wall KBR = 0.3677X + 38.396 120' Tank 3/4" Wall KBC = 0.0364X 2 + 0.106X " 3.3327 120' Tank. 3/4" Wall KBL = 0.1373X 2 - 0.798X - 9.0685 120' Tank, 3/4" Wall KBR = 3.4804X + 70.876 300' Tank, 1" Wall

KBC = 0-0089X 2 + 0.771X + 13.236

300' Tank, 1" Wall 300' Tank, 1" Wall

KBL = 0.0393X 2 + 0.5196X + 13.023 KBR = 0.1211X + 34.977

KR

=

Kc

280,000 / 0.6 466,000 lbs/in = KBc/mc 80,000 / 0.68 117,600 in-lbs/rad

KL

=

300' Tank, 3/4" Wall KBC = 0.0421X 2 - 0.2723X + 9.9571 300' Tank, 3/4" Wall KBt_ = 0 . 1 5 X 2 - 1.0719x + 6.4632 300' Tank, 3/4" Wall KBR = 4.1196X + 50.97

KBL/me

200,000 / 0.88 227,000 in-lbs/rad

SAMPLE PROBLEM Calculate Material: D = t = d = t, = L =

KBR/mR

the spring rate for the following tank: A36 30 feet (360 in) %in 30 in 3/4 in 28 in (regular p e r A P I 6 5 0 )

Thus for this sample problem the above calculated stiffness coefficients should be used when establishing the loads for the piping system attached to the nozzle, namely: KR= 466,000 lbs/in for a radial load Kc = 117,600 in-lbs/rad for a circumferential moment KL = 227,000 in-lbs/rad for a longitudinal moment

The nozzle has a reinforcing plate in accordance with API 650. Do = 49.5 in t = ½ in

SOLUTION 1) Calculate the distance where the bottom discontinuity has no influence on the spring rate B = 2(12*Dt) °5 = 2* (12"30"0.75) 0.5

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STIFFNESS C O E F F I C I E N T DUE T O C I R C U M F E R E N T I A L M O M E N T 200 -

180

T .

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3O'Oia.x3/4"

K ~ = 0.0622x 2 + 0.451 l x + 9.1617

.

160 180'Dia.x,3/4"

140

KBC = 0.0243x = + 0.0707x + 2.2218 o

8~

12o

7,

1oo.

ff,-

~.~

3O'Dia,xl/2" P KBc = 0.0421 x2 - 0 . 2 7 2 3 x + 9.9571

80120'Dia.x3/4"

60 40 .

KBC = 0.0364x ~ + 0.106x + 3 3 3 2 7

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~

I

=.

. t

0

300'Dia.x3/4"

Kec= 0.0075x 2 + 0.2921x + 4.1626 i ........

. . . . . . . 10

20

30

x Nozzle Size (in)

40

50

Note 3/4" thicknesses are solid line

Figure 4a

S T I F F N E S S RATIO DUE TO C I R C U M F E R E N T I A L M O M E N T

0.9

0.8

07

0.6

~

05

E

0.4

0.3

0.2

0.1

mc = -0.2947h2 + 1.0205h + 0.2751 0 0.1

0.2

03

0.4

0.5

06

0.7

0.8

0.9

h=L/LB

Figure 4b

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SPRING RATE DUE TO LONGITUDINAL MOMENT

:eli

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-

400

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"~ 3001 2'°1

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K,L = o.2~oo,' - o.o883,+ 2o.,~

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....

300'Dia.x3/4"

o|

~,L= o o249x~ + o 2o8,x + , , , ~ 9

0

t0

20

30

40

Nozzle Size (In)

50

Note 3/4" thicknesses are solid line

Figure 5a

STIFFNESS RATIO DUE TO LONGITUDINAL MOMENT

0.9 0.8 0.7 .j 0.6 0.5

E

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................................................................

0.4 0.3 0.2 0.1 m L = -0.2512h 2 + 0.641h + 0.6034 0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

h=L/LB

Figure 5b

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= 32.86 in 2) Establish the centerline distance of the nozzle of no discontinuity influence LB = B+0.5*D0 = 32.86 + 0.5 * 49.5 = 57.61 in 3) Establish the height factor h = L/LB 28/57.61 0.486 4) From Figure 3a, 4a, 5a read the stiffness coefficients KBR = 280,000 lbs/in (Fig. 3a) KBC = 80,000 in-lbs/rad (Fig. 4a) KBL = 200,000 in-lbs/rad (Fig. 5a) 5) From Figure 4, 5, 6 read the coefficient factor mr = 0.6 for Kr ( Fig. 3b) mc = 0.68 for Kc (Fig. 4b) mE = 0.88 for KL (Fig. 5b) 6) Divide the stiffness coefficients established in step 4 by the factors from step 5 to arrive at the stiffness coefficients for the nozzle under investigation.

Table 1 Size

Formulae

30' Tank, 1" Wall

KBC = 0.0529x 2 + 4.0405X - 26.146

30' Tank, 1" Wall 30' Tank, 1" Wall

KBL = 0.2662X 2 + 4.3491X - 35.678 KBR = 0.0079X + 0.1919

30' Tank, 1/2" Wall

KBC = 0.0421X 2 - 0.2723X + 9.9571

30' Tank, 1/2" Wall 30' Tank, 1/2" Wall

KBL = 0.15X 2 " 1.0719X + 6.4632 KBR = 4.1196X + 50.97

30' Tank, 3/4" Wall

Kac = 0.0421X 2 - 0.2723x + 9.9571

30' Tank, 3/4" Wall 30' Tank, 3/4" Wall

KBL = 0 . 1 5 X 2 " 1.0719x + 6.4632 KBR = 4.1196X + 50.97

120' Tank 1" Wall

KBC = - 0 . 0 5 8 9 X 2 +

120' Tank. 1" Wall 120' Tank 1" Wall

KBL = 0.1462X 2 - 3.7726X + 54.295 KBR = 0.5828x + 88.103

4.9447x + 2.4016

120' Tank. 1/2" Wall KBc = -010199x 2 + 1.6614x - 4.0694 120' Tank. 1/2" Wall KBL = 0.0215X 2 + 0.2208X + 5.3213 1 2 0 ' T a n k 1/2" Wall KBR = 0.3677X + 38.396 120' Tank 3/4" Wall KBC = 0.0364X 2 + 0.106X " 3.3327 120' Tank. 3/4" Wall KBL = 0.1373X 2 - 0.798X - 9.0685 120' Tank, 3/4" Wall KBR = 3.4804X + 70.876 300' Tank, 1" Wall

KBC = 0-0089X 2 + 0.771X + 13.236

300' Tank, 1" Wall 300' Tank, 1" Wall

KBL = 0.0393X 2 + 0.5196X + 13.023 KBR = 0.1211X + 34.977

KR

=

Kc

280,000 / 0.6 466,000 lbs/in = KBc/mc 80,000 / 0.68 117,600 in-lbs/rad

KL

=

300' Tank, 3/4" Wall KBC = 0.0421X 2 - 0.2723X + 9.9571 300' Tank, 3/4" Wall KBt_ = 0 . 1 5 X 2 - 1.0719x + 6.4632 300' Tank, 3/4" Wall KBR = 4.1196X + 50.97

KBL/me

200,000 / 0.88 227,000 in-lbs/rad

SAMPLE PROBLEM Calculate Material: D = t = d = t, = L =

KBR/mR

the spring rate for the following tank: A36 30 feet (360 in) %in 30 in 3/4 in 28 in (regular p e r A P I 6 5 0 )

Thus for this sample problem the above calculated stiffness coefficients should be used when establishing the loads for the piping system attached to the nozzle, namely: KR= 466,000 lbs/in for a radial load Kc = 117,600 in-lbs/rad for a circumferential moment KL = 227,000 in-lbs/rad for a longitudinal moment

The nozzle has a reinforcing plate in accordance with API 650. Do = 49.5 in t = ½ in

SOLUTION 1) Calculate the distance where the bottom discontinuity has no influence on the spring rate B = 2(12*Dt) °5 = 2* (12"30"0.75) 0.5

8

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