PVP BUTANE STORAGE BULLET CALCULATION AND FEA VERIFICATION

Transcription

1 Proceedings of PVP ASME Pressure Vessels and Piping Division Conference July 17-21, 2005, Denver, Colorado USA PVP BUTANE STORAGE BULLET CALCULATION AND FEA VERIFICATION Zhanghai Wang MaXfield Inc. Crossfield AB T0M 0S0 CANADA Daryl Bast MaXfield Inc. Crossfield AB T0M 0S0 CANADA David Shen MaXfield Inc. Crossfield AB T0M 0S0 CANADA ABSTRACT This paper presents a comparative study of FEA with the Zick s method of analysis for a two saddles-supported horizontal cylindrical tank. Zick s method is an analytical method commonly used in horizontal vessel support design. We used two methods to calculate local stresses in selected areas with the findings that the results of these two methods produced very agreement. These results are used to verify the FEA for this application. We then used FEA to get more detailed information about stress distribution, which cannot be obtained using the Zick s method. FEA was further used to study the buckling behavior of the object and to determine some critical parameters of the object, e.g., the vacuum ring weld size with consideration of both external and internal loads. Keywords: FEA verification, pressure vessel calculation INTRODUCTION Horizontal cylindrical pressure vessels supported with two saddles are widely used in industries. An earlier method to calculate stress distribution of such a vessel support dates back to 1951 when L.P. Zick proposed an approach for this problem [Zick, 1951]. Zick s method is a simplified calculation to approximate stresses, such as assumed shell thickness: uniform and supports: point support. These calculations often provide solutions only for the maximum stress at certain points in the structure. To have good knowledge about the degree of approximation associated with Zick s method, we calculated stress distribution for a special type of horizontal cylindrical pressure vessel (butane storage bullet) using both FEA and Zick s method. Fig. 1 shows the schematic of the butane storage bullet. The vessel object has the following geometrical parameters: the tangent-to-tangent length: 2,068 in; the thickness of the shell at the place supported by the saddle: 0.75 in; the thickness of the shell in other places: in. The material of the vessel object is SA N. In Fig. 1, the vertical red line represents the design datum. Two saddle rings were designed at each saddle, and six vacuum rings were welded internally between the two saddles and between the saddles to the heads (respectively). All these rings were designed with the same thickness of 1.25 in. The saddle parameters are illustrated as follows: NOMENCLATURE Saddle material: SA N Saddle construction is: Centered web Saddle able stress: S s = 20,000 psi Saddle yield stress: S y = 38,000 psi Saddle distance to datum: 402 in Tangent to tangent length: L = 2,068 in Saddle separation: L s = 1,260 in Vessel radius: R = in Tangent distance left: A l = 404 in Tangent distance right: A r = 404 in Saddle height: H s = in Saddle contact angle: θ = 160 Wear plate thickness: t p = 0.75 in Wear plate width: W p = 36 in Wear plate contact angle: θ w = 180 Web plate thickness: t s = 0.75 in 1 Copyright 2005 by ASME

2 Base plate length: E = 144 in Base plate width: F = 24 in Base plate thickness: t b = 1.5 in Number of stiffener ribs: n = 7 Largest stiffener rib spacing: d i = in Stiffener rib thickness: t w = 0.75 in Saddle width: B = 22 in Anchor bolt size & type: 2.25 inch series 8 threaded Anchor bolt material: SA 193 B7M Anchor bolt able shear: 15,000 psi Anchor bolts per saddle: 4 Base coefficient of friction: m = 0.45 Weight on left saddle: operating = 432,531 lbs test = 678,174 lbs Weight on right saddle: operating = 432,691 lbs test = 678,344 lbs Fig ,900 USWG 140 ID Butane Storage Bullet After that, we used FEA to study the buckling problem with the vessel object and to determine the parameters of the ring. STRESS DISTRIBUTION OF THE VESSEL OBJECT The storage bullet is evaluated using the following conditions that are consistent to the load cases in L. P. Zick s method. They are: Load Case 1: Internal pressure of 170 psi with loaded product (Weight, operating) Load Case 2: Internal pressure of 221 psi with loaded water (Weight, test) Load Case 3: Full vacuum with loaded product (Weight, vacuum) In FEA, the vessel and saddle were modeled with plate elements. The plate element can be viewed as a flat shell element and can provide both flexural and membrane stiffness [2]. It is noted that the plate element has been widely used to model thin wall objects such as pressure vessels and automobile body parts. In the FEA model, in order to reduce the number of parts and to assign component properties, components with the same material and thickness were modeled as one part. Product and water hydrostatic heads were applied to the internal surface of a vessel, corresponding to different load cases. In the definition for the plate element, the internal pressure load and hydrostatic head was considered in a normal direction of an internal point on a cylindrical surface. Further, considering potential vessel expansion, one of saddles can slide upon the foundation in the vessel s axial direction assuming a friction coefficient of The other saddle was fully constrained on the foundation. The results calculated with Zick s method are shown in Tables 1 and 2. Table 3 presents a comparison of results calculated with Zick s method and FEA. From table 3 it can be seen that the stress ( ) between two saddles, calculated with FEA is less than that calculated with Zick s method. The small discrepancy is due to some simplification of both the structure and boundary in the Zick s method. It can be seen from Table 3 the results with the same boundary condition for both FEA and Zick s method are pretty close; yet, the small discrepancy is due to geometry simplification with Zick s method. It is noted that the shear stress at the lowest point of the vessel outside is zero. This is in consistent with the fact that the stress listed in Table 3 is the projection of the maximum principal stress and minimum principal stress in the axial direction of the vessel object. Fig. 2 shows the stress contour and the stress at selected lowest point under the load case 2. Fig. 3 to Fig. 6 further show the stress distribution for the whole vessel (Fig. 3), saddle and saddle ring (Fig. 4), the horn of the saddle (Fig. 5), and the vacuum ring (Fig. 6). Load Vessel condition Table 1 Calculated Result by L. P. Zick s Method Bending + pressure between saddles (psi) Bending + pressure at the saddle (psi) 14,243 17,000 4,558 12,661 11,234 20,000 3,164 12,661 Weight Test 19,831 29,070 7,146 12,661 15,531 34,200 4,960 12,661 Weight Vacuum 4,558 17,000 5,398 12,661 3,164 20,000 3,864 12,661 2 Copyright 2005 by ASME

3 Load Vessel condition Tangential shear (psi) Table 2 Calculated Result by L. P. Zick s Method Stress over saddle (psi) Splitting (psi) S 3 S 5 S 6 S 7 (shell) Stiffener ring stress (psi) (+/-) S 7 (ring) 2,818 16,000 11,209 19,000 4,045 13,333-6,629 30,000 6,881 24,900 Weight Test 4,418 27,360 17,572 34,200 6,341 34,200-10,392 34,200 10,788 32,400 Weight Vacuum 2,818 16,000 11,209 19,000 4,045 13,333-10,143 30,000 9,076 24,900 Load Vessel condition Weight Test Weight Vacuum Table 3 Calculated Stresses by L. P. Zick s Method versus FEA Simulation Calc method Bending + pressure between saddles (psi) (+/-) Bending + pressure at the saddle (psi) LPZ 0 14,243 4,558 11,234 3,164 FEA 1 14,004 17,000 4,863 12,661 14,065 20,000 4,140 FEA 2 12,114 4,863 14,075 4,140 LPZ 0 19,831 7,146 15,531 4,960 FEA 1 19,452 29,070 7,191 12,661 19,072 34,200 6,136 FEA 2 16,510 7,191 17,135 6,136 LPZ 0 4,558 5,398 3,164 3,864 FEA 1 3,619 17,000 4,863 12,661 2,021 20,000 3,650 FEA 2 3,619 4,863 2,021 3,650 0: calculation by using L. P. Zick s method 1: simulation under the same constrains as the calculation by L. P. Zick s method 2: simulation under real boundary condition with friction force between one saddle and foundation 12,661 12,661 12,661 Fig. 2 Stress Contour and Selected Compared Point under Load Case: Weight, Test Fig. 3 Stress Contour of Whole Vessel under Load Case: 3 Copyright 2005 by ASME

4 the critical load factor of Therefore, the design of the saddle and vessel have sufficient margin to buckling. Fig. 4 Saddle and Rings Stress Contour under Load Case: Fig. 7 Saddle Buckling Mode Shape and Critical Load Factor Fig. 5 Stress Contour at Horn of Saddle under Load Case: Fig. 8 Vessel Critical Load Factor and Buckling Mode Shape with Stiffening Rings Fig. 6 Vacuum Ring Stress Contour under Load Case: BUCKLING ANALYSIS WITH FEA The worst condition of buckling is when the vessel is fully filled with test water and under external load of 15 psi. The critical buckling load factors, named Buckling Load Multipliers in FEA application, of the vessel and saddle are shown in Fig.7 and Fig. 8, respectively. The critical load factor of the saddle is , which means that when load is increased to times the applied load, buckling will happen in the red area in Fig. 7. Fig. 8 shows the buckling modal shape of the vessel and WELD DESIGN OF VACUUM RINGS The weld size design of stiffening rinds is based on the load case of weight, vacuum (per ASME VIII-1 UG30), which is the load case of filled product under external pressure load of 15 psi. In terms of the load case, the reaction force contour of a selected ring is as shown in Fig. 9. From the result, it can be seen that the reaction force distribution is almost uniform and symmetrical to the center of ring. This further means that the reaction force is mainly generated by external pressure of 15 psi. The summation of whole ring edge reaction forces is 450,318 lbs, and the ring is continually welded at both sides. Therefore, we have: f = /( ) = 512 lbs/in Minimum weld size w = f/s a = in Where, S a is able stress 11,200 w lbs/in [3] Since the vessel is normally operated under internal pressure load, the weld size design of internal stiffening rings should consider an extreme internal pressure case of 175 psi that occurs during a hydro-test. 4 Copyright 2005 by ASME

5 prescribed structure and boundary conditions than the analytical model (e.g., Zick s method), which has led to more accurate results. Furthermore, the rings play important roles in the vessel object; especially it was found that absence of the ring would cause the buckling of the vessel object. With FEA we were able to more accurately define the weld size for the ring. ACKNOWLEDGMENTS The authors are grateful to the writing assistance from professor, Chris Zhang at the University of Saskatchewan, John Martin at Lockheed Martin, Inc., and the advice from MEG Worley Limited in design and calculation. Fig. 9 Reaction Force under Weight Vacuum REFERENCES [1] L. P. Zick, 1971, Stresses in Large Horizontal Cylindrical Pressure Vessels on Two Saddle Supports, Welding Research Supplement, p [2] Constantine C. Spyrakos & John Raftoyiannis, 1997, Linear and Nonlinear Finite element Analysis in Engineering Practice, Algor Inc. Publishing Division, Pittsburgh, PA, p [3] The James F. Lincoln Arc Welding Foundation, 1986, Design of Welded Structures, the James F. Lincoln Arc Welding Foundation, Cleveland, OHIO, p Fig. 10 Reaction Force under 175 psi Internal Pressure The summation of reaction force of whole ring edge is 5225,648 lbs under the maximum internal pressure load of 175 psi. The reaction force contour shown as in Fig. 10 indicates that shell rounds up under internal pressure and stiffening rings constrains the rounding up by the welds. In addition, almost all of the reaction forces are generated by internal pressure. Since the stiffening rings are continually welded both sides, we have: f = /( ) = 5941 lbs/in Therefore, minimum weld size w = f/s a = 0.36 in Where: S a is weld able stress. S a = S m = ,000=16,500 w lb/in (per ASME- VIII-1 UW18(d)) It is noted that the selected ring here has the maximum summation of reaction force among the stiffening rings. CONCLUSION In this paper, we employed an FEA application (Algor) and a compress calculation application (Compress CodeWare) to verify the design of a large horizontal storage vessel supported by two saddles. The FEA model is more refined in the 5 Copyright 2005 by ASME

6 APPENDIX: Weight, Operating Longitudinal stress between saddles (right saddle loading and geometry govern) = +- 3 K 1 Q (L/12) / (p R 2 t) = ,691 (2,068/12) / (p ) = 4, psi S p = P R/(2 t) = /( ) = 9, psi Maximum tensile stress t = + S p = 14, psi Maximum compressive stress (shut down) c = = 4, psi Longitudinal stress at the right saddle L e = 2 (Left head depth)/3 + L + 2 (Right head depth)/3 = /3 + 2, /3 = 2, in Bending moment at the right saddle: M q = w [2 H A r /3 + A 2 r /2 - (R 2 - H 2 )/4] = [ / /2 - ( )/4] = 36,919,228 lb-in = +- M q K 1 '/ (p R 2 t) = 36,919,228 1/ (p ) = 3, psi S p = P R/(2 t) = /(2 0.75) = 8, psi Maximum tensile stress t = + S p = 11, psi Maximum compressive stress (shut down) c = = 3, psi Tangential shear stress in the shell Q shear = Q - w (a + 2 H/3) = 432, ( /3) = 257,744.8 lb S 3 = K 2.2 Q shear /[R (t + t p )] = ,744.8/[ ( )] = 1, psi Tangential shear stress in the shell at the wear plate edge (right saddle) S 3 = K 2.2 Q shear /(R t) = ,744.8/( ) = 2, psi Ring compression in shell over right saddle S 5 = K 5 Q/{(t + t p ) [t s (R o t c ) 1/2 ]} = ,691.0/{( ) [ ( ) 1/2 ]} = 11, psi Saddle splitting load (right) Area resisting splitting force = Web area + wear plate area A e = H eff t s + t p W p = = in 2 S 6 = K 8 Q / A e = ,691.0 / = 4, psi Stiffener ring calculations, stress in shell (right) Ring description: 1.25x12 Flat Bar Ring location: inside shell Ring material: SA-36 Ring separation: 16 in Ring moment of inertia: I r = 360 in 4 Ring cross sectional area: A = 30 in 2 Combined shell/ring inertia: I = in 4 NA distance to shell extreme fiber: C 1 = in NA distance to ring extreme fiber: C 2 = in Hoop stress in shell: S c = 6, psi S 7 = -K 7 Q/A - K 6 Q R C 1 /I = ,691.0/ , / = -6, psi Stiffener ring calculations, stress in ring (right) S 7 = -K 7 Q/A + K 6 Q R C 2 /I = ,691.0/ , / = 6, psi Load on stiffener due to vacuum Q e = (P e D o L s )/2 = ( )/2 = 137,962.5 lb Total load on stiffener Q = 432, ,962.5 = 570,653.5 lb S 7 = -K 7 Q/A - K 6 Q R C 1 /I + S c = ,653.5/ , / ,400 = -10, psi Stiffener ring calculations, stress in ring (Weight, Vacuum, right) S 7 = -K 7 Q/A + K 6 Q R C 2 /I = ,653.5/ , / = 9, psi Shear stress in anchor bolting, one end slotted 6 Copyright 2005 by ASME

7 Maximum seismic or wind base shear = 0 lb Thermal expansion base shear = W m = 441, = 198,624.1 lb Corroded root area for a 2.25 inch series 8 threaded bolt = in 2 (4 per saddle ) Bolt shear stress = 198,624.1/( ) = psi Web plate buckling check Allowable compressive stress S c is the lesser of 20,000 or 33, psi: (20,000) S c = K i p 2 E/[12 ( ) (d i /t w ) 2 ] = 1.28 p 2 29,000,000/[12 ( ) (23.75/0.75) 2 ] = 33, psi Allowable compressive load on the saddle b e = d i t s /[d i t s + 2 t w (b - 1)] = /[ (22-1)] = F b = n (A s + 2 b e t w ) S c = 7 ( ) 20,000 = 2,307,106 lb Saddle loading of 687,040 is <= F b ; satisfactory. Primary bending + axial stress in the saddle due to end loads (assumes one saddle slotted) S b = V (H s - x o ) y / I + Q / A = 0 ( ) 11 / 4, / = 0 psi The primary bending + axial stress in the saddle <= 26,000 psi; satisfactory. Secondary bending + axial stress in the saddle due to end loads (includes thermal expansion, assumes one saddle slotted) S b = V (H s - X o ) y / I + Q / A = 198,624.1 ( ) 11 / 4, ,691 / = 14, psi The secondary bending + axial stress in the saddle < 2 S y = 76,000 psi; satisfactory. Saddle base plate thickness check where a = 23.75, b = in t b = [b 1 q b 2 /(1.5 S a )] 1/2 = [ /(1.5 20,000)] 1/2 = in The base plate thickness of 1.5 in is adequate. 7 Copyright 2005 by ASME