ML20028B935
| ML20028B935 | |
| Person / Time | |
|---|---|
| Site: | 07109151 |
| Issue date: | 11/23/1982 |
| From: | HITTMAN CORP. |
| To: | |
| Shared Package | |
| ML20028B934 | List: |
| References | |
| STD-R-02-004, STD-R-2-4, NUDOCS 8212130061 | |
| Download: ML20028B935 (90) | |
Text
{{#Wiki_filter:-... i i ADDENDUM TO SAFETY ANALYSIS REPORT FOR THE HN-100 SERIES 3A RADWASTE SHIPPING CASK Docket Number 71-9151 l i Referencing 10 CFR 71 Type "A" Packaging Regulations STD-R-02-004 l l t i EFT 3 Is. i CLp-I ILi L wq g-8212130061 821123 PDR ADOCK 07109131 PDR /
l ADDENDUM TO SAFETY ANALYSIS REPORT FOR THE HN-100 SERIES 3A RADWASTE SHIPPING CASK Docket Number 71-9151 Referencing 10 CFR 71 Type "A" Packaging Regulationr i l STD-R-02-004 i l i Hittman gu g gr,& gvg}gggeng3g gporation i WID00/iW
1 Document Number: Rev: Rev Date: HITTMAN NUCLEAR & STD-R-02-004 1 11-19-82 DEVELOPMENT CORPORATION
Title:
ADDENDUM 'ID SAFETY ANALYSIS REPORT FOR THE HN-100 SERIES 3A RADWASTE SHIPPING CASK Supervisor Prepared Director Trans-QA R3v. Rev Date by Engineering portation Manager Dr ting h.f J tt' ( 323 0 7-13-82 k, [f// 4 .p 8 -260 1 11-19-82 HNDC-Ol(A) Page of RID &9N
PROPRIETARY DATA NOTICE This Safety Analysis Report and the associated drawings are the property of Hittman Nuclear & Development Corporation, Columbia, Maryland. This material is being made available for the purpose of obtaining required certifications by the U. S. Nuclear Regulatory Commission, enabling utilities and other firms producing radioactive waste to be registered users of equipment and services supplied by Hittman Nuclear & Development Corporation, and enabling equipment to be manufactured on behalf of and under contracts with Hittman Nuclear & Development Corporation. Parties who may come into possession of this material are cautioned that the information is PROPRIETARY to the interests of Hittman Nuclear & Development Corporation, is not to be reproduced from this report and the associated drawings; or facsimiles made of these drawings without the express written consent of the Hittman Nuclear & Development Corporation. libc0PN
PROPRIETARY DATA 1.0 PURPOSE AND BACKGROUND The purpose of this suppliment is to provide the information and engineer-ing analyses to demonstrate that the HNDC HN-100 Series 2 radwaste shipping casks (Certificate of Compliance 71-9079) can be upgraded to meet the per-formance capabilities and structural integrity of the HN-100 Series 3 Rad-waste Shipping Cask (Docket 71-9151). The HN-100 Series 2 cask when modi-fied will be designated as HN-100 Series 3A casks and will be operated under the same Certificate of Compliance as the HN-100 Series 3 casks. The HN-100 Series 3A casks (formerly HN-100 Series 2) were fabricated in 1976 and 1979. The designation for the individual units are as follows: Unit Designation Year of Fabrication KN-100-6 1976 HN-100-7 1976 HN-100-8 1979 HN-100-9 1979 HN-100-10 1979 KN-100-11 1979 The casks are constructed primarily of A36 carbon steel.
2.0 DESCRIPTION
The HN-100 Series 3A Shipping Cask is a top-loading, shielded container de-signed specifically for the safe transport of Type "A" quantities and greater than Type "A" LSA radioactive waste materials between nuclear facilities and waste disposal sites, The radioactive materials can be packaged in a variety of different type disposable containers. Typical configurations for the internals and their model designations are as fol-lows: Model Number Cask Internals HN-100/170 One large disposable container HN-100/2-80 Two large stackable liners HN-100/18 Eighteen 30 gallon drums HN-100/14 Fourteen 55 gallon drums HN-100/8 Eight 55 gallon drums The HN-100 Series 3A Shipping Cask is a primary containment vessel for radioactive materials. It consists of a cask body, cask lid, and a shield plug being basically a top-opening right circular cylinder which is on its vertical axis. Its principal dimensions are 81-3/4 inches outside by 81-1/2 inches high with internal cavity of 75-1/2 inches inside diameter by 73-3/8 inches high. obOhb
PROPRIETARY DATA 2.1 Cask Body The cask body is a steel-lead-steel annulus in the form of a vertical oriented, right circular cylinder closed on the bottom end. The side walls consist of a 3/8 inch inner steel shell, a 1-3/4 inch thick con-centric lead cylinder, and a 7/8 inch thick outer steel shell. The bottom is four inches thick (two 2 inch thick steel plates welded together) and is welded integrally to both the internal and external steel body cylinders. The steel shells are further connected by welding to a concentric top flange designed to receive a gasket type seal. Positive cask closure is provided by the gasket seal and the required lid hold-down ratchet binders. Four lifting lugs are welded to the outer steel shell. A plugged drain in the base and a stainless steel cavity sleeve are optionally provided. 2.2 Cask Lid The cask lid is four inches thick (two 2 inch thick steel plates welded together) which is stepped to mate with the upper flange of the cask body and its closure seal. Three steel lug lifting devices are welded to the cask lid for handling. The cask lid also contains a " shield plug" at its center. 2.3 Shield Plug The shield plug is five inches thick (two 2 inch thick steel plates and one 1 inch thick steel plate welded together) fabricated in a design similar to the cask lid. It has a gasket seal and uses eight hold-down bolts to provide positive cask closure. The shield plug also has a lif ting device located at its center to facilitate han-dling. 2.4 Cask closure The shipping cask has two closure systems: (1) the cask lid is closed with eight high-strength ratchet binders and a gasket seal, (2) the shield plug is closed with eight 3/4 inch bolts and the same seal system used for the cask lid but smaller. 2.5 Cask Tiedown System The shipping cask tiedown system consists of two sets of crossed tiedown cables (totaling 4). Four shear blocks or a shear ring (af-fixed to the vehicle load bed) firmly position and safely hold the cask during transport. 2.6 Cask Internals The internals of the HN-100 shipping cask can be any one of an exten-sive variety of configurations. Some examples are given in terms of weight in 2.7. Other arrangements are possible, providing the gross weight and the decay heat rate limits are observed, and the material.
PROPRIETARY DATA secured against movement relative to the cask with an internal struc-tural members such as bottoms and pallets. Basically, the internals consist of the waste, contained if process waste is being transported, and the structures used to fix the waste relative to the cask. The container may be constructed of high integrity plastics, steel or other metals. Shoring is used with small secondary container to prevent movement during normal conditions of transport. Shoring is not required for containers and pallets designed to fit the cavity. 2.7 Gross Package Weights The respective gross weights of the cask components and its designated radwaste loads are as follows: Cask body, including 28,797 pounds reinforcing plates Closure lid 5,537 pounds Shield plug 366 pounds Total cask tealoaded) 34,700 pounds HN-100 - Large container (s) and waste 14,000 pounds
- HN-100 - 55 gallon size containers (up to 14 14,000 pounds drums of radioactive waste)
HN-100 - 30 gallon size con-tainers (up to 18 drums of 8,100 pounds radioactive waste) 2.8 Radwaste Package Contents 2.8.1 Type and Form of Mt :erial The contents of the various internal containers can be process solids in the form of spent ion exchange resins, j filter exchange media, evaporator concentrates, and spent filter cartridges. Materials will be either dewatered, i solid, or solidified. 2.8.2 Maximum Quantity of Material Per Package Type A materials and greater than Type A quantities of low specific activity radioactive materials in secondary con-tainers with weights not exceeding 14,000 pounds. Radio-active materials may include source and transuranic ma-terials in Type A quantitics or greater than Type A quan-tities of low specific activity materials. The contents may also include exempt quantities of fissile material as de-fined in 10 CFR 71.9. l
- Although gross weight is limited to 48,700 lb, 53,005 lb is used in structural analysis.
PROPRIETARY DATA I 3.0 DESIGN CONDITIONS 3.1 General Standards (Reference 10 CFR 71 Section 71.31) 3.1.1 Chemical Corrosion The cask is constructed from heavy structural steel plates. All exterior surfaces are primed and painted with high quality epoxy. There will be no galvanic, chemical, or other reaction among the packaging components. 3.1.2 Positive Closure System As noted, the primary lid is secured by means of eight high strength ratchet binders. The secondary lid is affixed with eight 3/4 inch diameter bolts. Therefore, the package is equipped with a positive closure system that will prevent inadvertant opening. 3.1.3 Design Criteria on which Structural Analysis is Based 3.1.3.1 Stresses in material due to pure tension are com-pared to the minimum yield of that material. The safety factor is found by dividing the minimum yield by the calculated stress. A safety factor greater than 1.0 is required for acceptability. Material used is A36 with f of 36,000 psi, A516 Gr 70 with f of38,000 psi {andA311 Grade 1137 with f of8$,000 psi.2 Y 3.1.3.2 Stresses in material due to shearing is analyzed using the " Maximum Energy - Distortion Theory" which states the shearing elastic limit is 1//3 = 57.7% of the tensile elastic limit.1 As with 3.1.3.1, a factor of safety greater than 1.0 is required for acceptability. 3.1.3.3 Weld filler material rod is E70 Grade or better. Analysis is based on American Welding Society Structural Code D1.1-79. For fillet welds, shear stress on effective throat regardless of direction of loading is 30% of specified minimum tensile strength of weld metal. For complete joint pene-tration groove welds with tension normal to the effective area the allowable stress is the same as the base metal. I Design and Behavior of Steel Structures, Salmon & Johnson, page 47. 2ASTM Standards, Parts 4 and 5. PROPRIETARY DATA Fillet weld allowable stress = (70,000 psi)(0.3) 21,000 psi = In order to be more conservative, a weld ef-ficiency of 85% is also added. 3.1.4 Lifting Devices 3.1.4.1 Package Wei g The package weights used for analysis are as follows: Empty package 34,700 lbs Payload: large 1 container and waste 18,305 lbs Gross Weight 53,005 lbs 3.1.4.2 Cask Lifting Lugc Material A-516 Grade 70 with a minimum of 38 KSI Tension Yield Strength. 21,926 psi shear yield strength (57.7% of 38,000 psi). There are two types of lifts that the cask may undergo. The cask will be designed for a 3 g lif t for either case. The first is a vertical lift using a 2 point lift beam. The second case is a 4 point lift at a 45* sling angle. At no time will the cask be lifted by a 2 point /45 sling angle. l 2 point / vertical lift 1' (53,005 lb)(3 g)/(2 lugs) = 79,510 lb. 2,, thick 4 point /45* sling angle (53,005 lb)(3g)/(4 lugs)(sin 45 ) = 56,220 lb l 3\\" 4 For the lug, the two point vertical lift is worst case. l 2h" Dia. l Tear-out ' 3-1/4". _ 2 -3 /4 ". l ~ o = (79,510 lb)/(2)(2)(3.25-1.25) 1 W o = 9,938 psi e S.F. = 21,926/9,938 = 2.2 4 1The weight of contents will be limited to 14,000 lbs. even though 18,305 lbs. has been used in the analysis. llucMW
PROPRIETARY DATA Bearing a = 79,510 lb/(2)(2.5) = 15,902 psi S.F. = (38,000)(.9)/15,902 = 2.15 Tension (79,510 lb)/(2)(6-2.5) = 11,358 psi F.S. = 38,000/11,358 = 3.35 2" thick 7 3.1.4.3 Cask Lifting Lug Welds (Worst Case) 79,510 lb Pure shear - (3/4" groove weld) Total weld length = 28 inches t { o = (79,510 lb)/(28)(.75)(.85) 2-3/4" Sy" o = 4,454 psi 3/4" groove %{ Moment - weld l- - --' Moment = (79,510 lb)(2.75 in) = 218,652 in-lb i y Centroid of weld = (10)(11) + (7)(12.5) + (9)(4.5) + (6)(1 n 10 + 7 + 9 + 6 l lv a. y = 8 inches g 8" 218,652 in-lb = 2a [(1/2)2(1/2)(2)(2/3)+(5-1/2)2(1/2) (2/3)(2)+(2)(8)] (.85)(.75) { 5" o = 4,7?0 psi / Total a = ](4,454)2 + (4,720)Z = 6,490 psi h F.S. = 21,000/6,490 = 3.2 r r 3.1.4.4 Tie Down Lugs for Lifting Cask (Inadvertant Use) If it is assumed the entire load is carried by the 2 tiedown lug, the section modulus (S) = (b h )/6; where b=2 inch, thickness of tiedown lug; and h=6 3 inch, width of tiedown lug. S=12 in. Material is A311, Gr 1137 with fy = 80,000 psi. The maximum distance from the applied load to the edge of the backing plate and tiedown lug weld is 4 inches. Mc _ M _ (79,510)(4 in) - 26,503 psi I S 12 F.S. = 85,000/26,503 = 3.2 - -
PROPRIETARY DATA Weld - (see 3.1.5.4) Shear - 79,510 lb/[(21.5)(/2)(.85)(.75)+(13)(.75) (0.85)(sin 45)] = 3,150 psi i = [(.707)(6.5)(2)(8+3.25)+(7.75)(2)(14.5+3.375) ([2)+(6)(J2)(22.25)+(5.5)(1)(14.5)+(8.5)(1) (11.9)] + [(6.5)(2)(.707)+(7.75)(2)(J2)+(6) (J2)+5.5+8.5] i = 865/53.6 = 16.13 in. I = (6)([2')(.75)(6.15)2+(2)(7.75)([2)(.75)(2.275)2+ (5.5)(1)(1.6)2(.75)+(.75)(8.5)(1)(4.2)2+(6.5) (sin 45)(2)(.75)(4.8)2 = 607 in 4 o = b = (79,510)(16.1)(Cos 37 )(6.15)/607 = 10,358 psi I o Total = V3,150z + 10,3582 = 10,826 psi 3.1.4.5 Lid Lifting Lugs (Secondary and Primary) A. Secondary Lid Lifting Lug Material A36 oy = 36,000 psi u o shear = 20,785 psi Maximum Load 370 lbs. q, Carried by one lug u n TEAR OUT Area = 2[(1-1/2 - 15/16) - 7/32] (3/8) %," DI A. Area = 0.258 in.2 3 " THK. Stress = 3 g's X 370/0.258 in.2 = 4,310 psi 4 4,310 psi << 20,785 F.S. = 4.8 TENSION Area = (2.0 - 7/16) 3/8 = 0.586 in.2 Stress = 3 X 370/0.586 = 1,900 psi 1900 psi << 36,000 psi F.S. = 18.9 B. Secondary Lid Lifting Lug Weld 1/2" fillet weld with allowable 21,000 psi Effective size Sin 45 (.5) = 0.353 in Area of weld (2 + 2 + 3/8 + 3/8) = 0.353 A = 1.68 in.2 Stress = 3 X 370/1.43 o = 660 psi 660 psi << 21,000 psi F.S. = 31.8 llb0dh
.t PROPRIETARY DNTA Therefor.t, the secondary lid lug is able to resist a load of three times its weight without reaching a yield stress. C. Primary Lid Lifting Lugs Material A36. 1" THK. Maximum Load = Primary Lid 5,537 lbs. Secondary Lid 336 lbs. 5,903 lbs. h A -r n 2k" l"DIA. n a 4 6" z TEAR OUT (Vertical Lift) Area = 2 x [(2-3/4) 1/2) - 1/2] (1) Area = 1.5 in.2 and stress = 5903/1.5 o = 3,940 psi 3,940 psi << 20,785 F.S. = 5.2 TEAR OUT (45* sling angle) Area (short path) = ( 4 75Z +.75Z) (1) = 1.06 in.2 Load = 5,903 [i = 8,350 lbs. Stress = (1/2) (8350)/1.06 = 3,940 psi 3,940 psi << 20,785 psi F.S. = 5.2 TENSION (Vertical lift) Area = (6 - 1) (1) = 5 in.2 Stress = 5903/5 = 1,181 psi 1,181 psi << 36,000 psi F.S. = 30.5 PROPRIETARY DATA TENSION (45* Sling Angle) Area (short path) = (J2) (1.25)2 - 1/2 = 1.2678 in2 Stress = (1/2) (8350)/1.2678 = 3,295 psi 3,295 psi << 36,000 psi F.S. 10.9 F. Primary Lid Lifting Lug Weld 1/2" weld at shear of 21,000 psi a) shear stress due to vert = horz component. oy = ch = 5903/(6+6+1+1)(Sin 45')(0.5)(0.85) = 1403 psi b) Stress due to moment Total Moment = Compression Moment + Tensile Moment Compression Moment + Tensile Moment Total Moment = 2(Tensile Moment) = 2(2x3x2/3xfx0.5x0.707x0.85) o = 3.6c o = 8,855/3.6 = 2,460 psi Combined Stress = J (2460 + 1403)2 + 1403* = 4110 psi F.S. = 21,000/4110 = 5.1 Therefore, it can be concluded that the lifting lugs for the lid are more than adequate to resist a load at three times its weight. 3.1.4.6 Lifting Lug Covers Since the primary and secondary lid lifting lugs are not capable of resisting the full weight of the package they will be covered during transit. 3.1.5 Tie Downs Lug material is A311 Crade 1137 with a minimum yield of 85 KSI and a 49 KSI usable shear (57.7%). The cask shell has a minimum yield of 36 KSI. BluBOfDiff
PROPRIETARY DATA' A E 1.12 14.03 26.84 91.38 I_ 64.54 7.430 ~ i m h.9 33.44 a D 19.0 ,r Z 18.12 33.44 l 7.43 65.66 = 14.03 26.84 A A n Jh 70.12 I n 66.87 Y n 40.1 Z 3$8 5.5 I u a VIEW A-A -
PROPRIETARY DATA A system of tie downs are provided as part of the package. They will be utilized as in View A-A. 3.1.5.1 Cask Center of Gravity Item Weight Arm Moment Cask 34,700 lbs
- 41.36" 1,435,192 in-lb
= Liner 1,325 lbs 35.70" = 47,302 in-lb Waste 18,305 lbs 35.20" 644,336 in-lb = 53,005 lbs 2,126,830 in-lb Center of Gravity = 2,126,830/53,005 CG - 40.1 in. 3.1.5.2 Tie Down Forces, Reference frame with respect to the trailer is shown on the tie down drawing (Page 10) up - down Y; front - rear X; side - size Z accelerations: Y axis - 2 g's X axis - 10 g's Z axis - 5 g's Tie Down Lengths Long tie downs (high trailer attachment points) length = 465.62 + 52.42 + 64.62 = 106.0 inches Short tie downs (low trailer attachment points) length = V64.52 + 51.52 + 63.52 = 104.1 inches Tie Down Tensions Tie down tensions resolved by vector direction Long tie down at tension Tg 64'6 Along Y axis Tg = 0.6096 Tg 106.0 65.6 Along X axis Tg = 0.6188 Tg 106.0 aIbONhhAf[
PROPRIETARY DATA 52*4 Along Z axis T = 0.4943 T g g 106.0 Short tie down at tension Tg 63.5 Along Y axis T = 0.6100 T g 3 104.1 64*5 Along X axis T = 0.6196 T g 3 104.1 51.5 Along Z axis T = 0.4947 T 3 3 104.1 l 10W Force (front-rear) Overturning (front-rear) due to 10W along X axis Overturning moment = 10(53,005 lb) 40.1" = 21,255,000 in-lb Each of the two rear (or front) tie downs (one long and one short) must restrain half the above moment or 10,627,500 in-lb Tension in the long tie down l 10,627,500 in-lb = (70.12)(0.6188 T ) + (67.7)(0.6094 T ) l Tg = 125,551 l'o. Tension in the short tie down i 10,627,500 in-lb = (66.87)(0.6196 T )g + (67.7)(0.6100 T ) S T = 128,460 lb g SW Force (side-side) Overturning (side-side) due to SW along Z axis Overturning moment = I 5(53,005 lb)(40.1") = 10,627,500 in-lb l l l l l l 1 l
PROPRIETARY DATA Each of tws side tie downs (one long and one short) must restrain half the above moment or 5,313,750 in-lb Tension in the long tie down 5,313,750 in-lb = (70.12)(0.4943 T )g + (7.43)(0.6094 T ) g T = 135,595 lb Tension in the short tie down 5,313,750 in-lb = (66.87)(.4947 T )g + (7.43)(0.6100 T ) 3 T
- I I' I
S 2W Force (up-down) Lifting (up) due to 2W along Y axis Lift = 2 (53,005 lb) - 53,005 = 53,005 lb Each of two long and two short tie downs carry the load or quarter the load per tie down. 13,251 lb = 0.6094 T L Tg = 21,745 lb 13,251 = 0.6100 T3 T = 21,725 lb g Total Tension Total tension with all forces acting simul-taniously T = 125,551 + 135,595 + 21,745 g T = 282,891 lb g T = 128,460 + 141,275 + 21,725 S T = 291,460 lb g ilboGWR
PROPRIETARY DATA 3.1.5.3 Tie Down Lugs The tie down lugs are constructed of A311 GR 1137 steel, having a minimum yield of 85 KSI and an ultimate tensile of 90 KSI. The following values are used in the design of the tie down lugs. Tensile Yield = 85,000 psi Bearing Yield = (85,000)(0.9) = 76,500 psi Tensile Ultimate = 90,000 psi Shear Yield = (85,000)(0.577) = 49,075 psi Shear Ultimate = (90,000)(0.577) = 51,962 psi Allowable Shear Stress for Welds = 21,000 psi Hole Diameter = 2.25 in, pin diameter = 2 in Tear Out - o = 291,460 lb/(2)(2)(1.985) [ f 1 h 36,710 psi i = f q/o.5 F.S. = 49,075/36,710 = 1.3 b(A-C 5 h l Bearing jg t9 g l l 9), / b E N o = 291,460 lb/(2)(2) = 72,865 f 7 i F.S. = 76,500/72,865 = 1.05 Tension o = 291,460/(2)(6-2.25) = 41,637 psi
- -82.
b" ( i F.S. = 85,000/38,851 = 2.15 = ,jg [ 3.1.5.4 Tie Down Lug Welds / 0 l a) Pure Shear [291,460 lb)/[(21.5)(4)(.85)(.75)+(13)(.75) (.85)(sin 45)] = 11,545 psi l :
PROPRIETARY DATA b) Homent forces (lug is 2 inches thick, therefore moment arm is 1 inch.) Moment = (291,460 lb)(1 inch) = 291,460 in-lb 61" Min. X, (2)(6 )(11) + 2 (7.75)(3.875) 5 6" 2(6-1/2 + 7-3/4) y 3/4" g.M X = 7.125 ~ 7" f 3/4" , 7,, 291,460 = 2a [(6)(7)+(7)2(2)(2/3)(1/2)] (.75)(.85)(J2) o = 2,165 psi Combined stress = 4 (11,545)d + (2,165)Z = 11,746 psi S.F. = 21,000/11,746 = 1.78 3.1.5.5 Analysis of Tiedown Loads on Cask Shell The tiedown loads are transmitted into the cask shell as external moments. These moments are the product of the tiedown forces and the offset distance between the line of action of the tiedown force and the attachment plate. 7 l Offset 2.0 in. ML F z __N U C e Fx ll000aWs f
PROPRIETARY DATA r, = 291,460 X Cos 37-1/2' = 231,230 lb F = 291,460 X sin 37-1/2' = 177,430 lb x M = Circumferential moment = (231,230 lb)(2.0 in) c = 462,460 in-lh M = Longitudinal moment = (177,430 lb)(2.0 in) = g 354,860 in-lb Reference for method of calculation: Welding Re-search Council, Bulletin No. 107 (WRC 107), " Stress in Cylindrical Pressure Vessels from Structural Attachments." y = r/t = radius to thickness ratio = 40.9/0.875 = 46.7 C = 1/2 the circumferential width of the loaded 3 plate = 18/2 = 9 in C = 1/2 the longitudinal width of the loaded 2 plate = 24 in/2 = 12 in. B = C /r = 9/40.9 = 0.220 3 3 B = C /r = 12/40.9 = 0.293 2 2 Check that 5 $ y $ 100 PROPRIETARY DATA Ntemenclature Applicable to Cylindrecal Shells V, concentrated shear load in the cir- [ B \\2 [B \\2 cumferential direction, Ib 1 2 - 1 V,. concentrated shear load in the lon- + = (0.3) (1.2)- gitudinal direction, Ib Af, external overturning nmment m the circumferential direction with re-spect to the shell, in. Ib General Nomewlature Af., external overturning moment in the = long tudinal disection with re-e, - normal stress in the ith direction on spect t the shell, m. Ib the surface of the shell, psi R. mean radius of cylindncal shell, in. ru - shear stress on the ith face of thejth length of cylindr, cal shell,,. i i m = direction halflength of rectangular loading in S = stress intensity - twice maximum e, shear stress, psi circumferential direction, in. halflength of rectangular loading in N. - membrane force per unit length in c, the ith direction, Ib/in. longitudinal direction, in. wall thickness of cylindrical shell, Af = bending moment per unit length in T the ith' direction, in. Ib/in. in. coordinate in longitudinal direction K. - membrane stress concentration fac-x tor (pure tension or compression) of shell coordinate in circumferential dirce-K. - bending stress concentration factor y i - denotes direction. In the case of tion of shell cylindrical coordinate in circum-spherical shells, this will refer to 4 the tangential and radial direc. ferential direction of shell I R., tions with respect to an axis a e normal to the shell through the p attachment parameter center of the attachment as si ci/R. shown in Fig.1. In the cane of s, c, / R. = cylindrical shells, this will refer y R.'T; shell parameter multiplication factors for N, and to longitudinal and circumferen-C,, C, = tial directions with respect to the N, for rectangular surfaces given axis of the cylinder as shown in in Tables 7 and 8 coefficients given in Tables 7 and 8 Fig. 2. K,, K, bending moments in shell wall in + = denotes tensile stress (when asso-Af,, Af, ciated with,,) the circumferential and longi- - denotes compressive stress (when tudinal direction with respect to associated with,,) the shell
- '.nbrane forces in shell wall in the E
- modulus 'of elasticity, psi N** N' P - concentrated radial load or total C'""".##"'N*I "" I""" ' * "'" " I dimtion widi wsput to th sheH distributed radial load, Ib normal st ress,m the circumferential direction with respect to lhe shell, psi """"at stre+ in the longit udinal di-2 General Equation rei tion with respect to the shell, in the analysis of stresses in thin shells, one pro-psi cceds by considering the relation between internal shear stress on the x fare in the + , membrane forces, internal bending nunnents und siirection with respect to the stress concentrations in accordance with the follow-si,yn, psi 8ng: shear stiew on the 4 face in the 1 r., ,, - K. T
- K.G f.
~ direction with re<pect to the T shell, psi MMQ. 'SW
PROPRIETARY DATA u a e a E s EL hj.- .= 23-1 -$~ 1 -- ' d '1-- .1'- r t_ 3 :ik: g 4 N= - -tr 4Mr% t -Mi-4 -42 9e-C =r d=F
=
+=v=::i c =. -+ = = g-4 g g -.m. f .c._ m ---.-+- - . r_ =- I Y . -j y i_ = = { -',: y-1 {- 7_. L:
- ==
..+w t P 21 -.t.
- .= :trr: t :
_:-r.-2. ':_T:t _k r ?- 45-2:
- 1: -'.,.: X
.: =: sc.z xt d:2 -- L-"1 D A[wl. _.-- ;__._ -2= =_pta-_ rt-~t-et-; c:- = =--' g"[' ~'=:== _ : - + =... ; e+- -=' - ' ~ ~ ' -' - t- =t 1=...~T.Z.. _. _. ~ '- - - ~ 7J:~***~**' ~*71'1**-O:.'T _".'T. --.-_. - ~ - - - ~. ~ ~ + ~4_ - - + - - + - - + - E.-- ++ ; +. 'l : i j- % _.-k d h +- U _ i .. I -+-- -+- I I I = x _FMMip ' s = a FL1-E _T'= r m y y [T_% f,p w Ff =
- p i-'-
7:_. =.+_:.- =l
- =
i 'E W [.
- #1 f } ^ )) & :,
-%.:.:= ' m M s = - += 14.Mt 4; 4 - 4 =M = - 4 2= C = = H= 4
- 4=E+M *-@
N W JtEf a x w m - w 5 gr~- ' ". - ~' M 's d., '.% .. =----j - 4 -5 F if hm 16'- =3 -- $ 1 - -. a ~ g _p g .k._,'.4-M4_M ~ - ~ T ^ s p ._= hem i-M% =~ . = g; 23: n z, .n = /', % 23 :i"-ad[iM d'.22 5;. 2; - '=i 1-MN T- =$b C=2dM-l:MItTNI Ma[2 '= df-[ ? = r a :: _;==r.v =t==.: w=u:-=a _: M- =
- ==. M+ ~=~1-.
- u-
_ m -. -=m-c.m~_.r.. ::w-m z.-' /.9_=L- +s =a= = u -_4-_ x x;q._ T.a*.
- =
= w-pp-/C r t-ti;f M 24 i_i i /C._ u.-. . _. -. _..M -= --. CSw 1-W ;--(h4 I.*"7 %*TDb%- --.V I swf-.r., 1-~ u -M m '% -'N . _1l a .#"t--+ t -:-9**' 7% .O l E! !!/ i/-. E ' I I ! A- ~ v4'*% I EI1 T I I J l I './' !Y_.<- 'h i_ : ** L I I 1 I s E E I: I 'I I /I I . I ~ l I 'r-'>eAe -==i,, e'% 1 ~ I 3 eI IJ ! J' f f f r ~y'{ "he a= .F 1 wpu l 2* Mi g l ,f 3 1'-~ ~ h n,,,,- .J ,- [
- _ 5 b
-T
- '~
C EN= 9 /5U5E E E 5N 5 ~~: i= ~ ' =-' ~~~T' = T-N-F a a j ) Nh.bM'E'If-_2 k kW E rI-ME ~ I ~ ~ - ' ~-EN -i'b-t ~-'! m u . p E 4 -- ?P z EE r y 7 = a r i 3 + z= a 1--- n - "=- _. 2 s 1 -2 gj ~1i j ' ' - -E-=.3F = U E-=:1
- 5-1-f
= =-- ..= .r y-.p_.. y g---- _4 -.-=EE l 55E -- -TE = =_ i= EEE~ M-MEf #7 -fEit=_= i =s%=45EE+- E i i, ; M :+ u: r" ? gl l f YW 1 Y t I II I 't i ! J . E ' . I. ~# I El f I II q my 3g f 7. I. I a 1 i I I 1 r 5 y rp r (' 33%' ' '
- 1 I
F -9 ' '~ i_ g ' J r -- 7_ }g. di" a-e 1F.i_ if k 1 I 4 I e ~ ~ J = 5 4 = = cf %=- = = 4' i: = = p i f V
- 9 Ji.;6-Mf-
' - - E:m a= ? EJ W. ! -'-w # =Mt m <-I-4 - -i= E25H fF j 19 s' -. i = =W "= = e r i g, Mr5-$% 5
==M'~~i =NC =i N '='-N 'M I M-sM=, 'r M554-55 E5=j 5bME=[NMi:3~ e , s-f r -.-wn IT g;,- .i i, t 7 -r 1
- g
- a q_
m. =bt-G e
- +-I H
--i - 4:
- i-1
= =:: e
- - E i f
. j'= f4- --'--+- 7 _y='+- 1 =..
- 4+
E - 4 5:=_==4 1 2 ,_ gpj j Li9Q -j%-g E-gry. +E-tp q- -_fi .= = + _ isi~ss 3-N=h= Wi- =-h 5-I--i Yi-N= ;h - $ri- _.i ii@d- .Eh [5i~~ NE 2-t~N
- E_..T6.
- i. _-. E_. ; = iT-i.5__cE.__2.~_...2 : n j= g-._.. 3;34_ =g
- ,a=-l__t g_
O- -C1-2T
- sfu f
' 5 _,_. E ~. ~M i..;_.-.-::- =--.-; p-- vt-t : H-U - 1
- .:a_n" :=7 :ct::
._w-K I EE l =3,l l,f,.,f. " ~ ._r_,._. t I j A.I t' i rI i f, ~ ..( 4,_ l.1.. F .I _l ' i~ m"- y E _ 1 - --+; u--
1
- h 21
+ r =, -L = m s i -i-j p -jT -_s ,J pF_ _g = g-- ^ y- +,Er-- _=.- =y . c% 2-: im m= 4g =, p
=
==
- -m & #i:
y .ih +g. = = :=i=4= -W'r += _-i-=-fW.Mn=N-=h
- M
-D Fri =iif fiMer= -b ,E b ~ r 9: = =- g.1;M..e s = 3 j p g g 4- _ y-Q J 'f _ y 4 j i ~ rA idi 4 n= - 04:- i&e --f = Y- - _f. =Q]^^-= 1 .y l_i_ i.--. -ly--i - i=fi } E- ~ = -f ;: y-4 = - l. 'Ei 5 =- 5 . +.: Wf =li E:E t-EF M& 2 E i4: _ ri @ i~i 4W = =_=F 4-!i s? ilik -5*~ I=M-2 - -- r-T=5 - = =
==f;'- =s__E=2_a._=r.- n. 2._- a ' Cr._L 2=t== = _. ~ -. + = -.= ^, - t 2
===;
. =s is>. = _.E_m_ - - - e= =._= = n = E..a_E_E._.=_ + = F , f - -_.. _ ~ ~
==-m -r-_._-- \\ p i_._i '.'-*-*-* " F- ". .4 l 1 --r 7-9 I I i I.. i I i I l I ! I A ! 1 A A 3 I I i 1 i, , I I i I i 0' 0 0' 0 0 45 0'0 0 35 0 0 06 050 OIS O' 0 025 5 4 3 2 Fig. 3A-Membrane force N,/(M./R.,9) due to an external circumferential moment M on a circular cylinder stresses in Shells PROPRIETARY DATA e a. L. -I I I ! I I I 1 I - F a 1-I I i i y_
- p.
.I .j-1 2 -i i 4.
- ( -
_i -j. .j. s-t t .j
- - 1
- t-..
q j., .(_ i 8 ._._..u_, ,I' ..r_ i.__ I. .e = J .._ f 4 - {.4-L =-- = . - l. = , a .-i. .i2
- 1..
t. .x.. Lt e G .-i_=_-_..:=_.._. =. i u. y _: _ t... *; .2 _.,_;;,, j .3 ;g,; _,,3 a..__. ,..-{g; ;._, 4_.. ,j_j,g g. .'_30_._-.1. _;. _.._J._.;O._. __ C. . _ L. ;~..l.. l. ' J. '.. _1..: ; _i.
- _.* -. _ P_* ::.'.* +.C.
_. _1_- *f._1.. ._2+J._'.; $ T-C .....e.. _ _... _. _ -. _. _. _ t_ a _ -4, T _&_au ' r_... J. I_L IA r 'Y ~ i . - +. _. _ i+.%._. + I -I -s-+_. i_ i n l t 8-y. p g' .:12- . z. -q. rg-Ig :.
- j..
- 4 -
-t-2 c ,1 - 1. - .1.. -!.:]:t..} Li.
- .=Jp1 T
- .=
4-A .I Y._ t. k}_n. - A. _r..
- 3..-
2. . t.. r .=. -t 4 ^ ..t_.- i.'.:i.. ..M_..*-*-... = -d'IN j ~ _!rl-f; }- =1 ?Is:-N. N.=__ E-I= ~il-h.I'-~k=h--...... I.. - d 2' ~3 - -. Mf' E ...'...M E:. .. :. '.g=g:.%. _. ^,.. :2_ +. :. ' -~-_:.2.-_._. ..T.-. =-.u.'..:~:_~.."._;._.=._'.._::t_..
- .1--*__
- :^_ _._ n.._- _: t :. ;_::
g ..._.-.+-. _.... _. ~. ~ _.--... ..w . - ~ - _ ~ '--.m_+... .D.-
- ._._ w
-.-a g f I g .I I I 1. a -i_
- l A
^1- = - 1 -a =
;
_ y. =- _ s- . -j _ 2.
- y=1;p
_g -g+ -t= 3-__. =
- -g.
==;,==. = =
=. =
- py
=-
== = M =iE'th=13-Qsh + TNT =+- % = =*= r:52=f #4 7f = ~ 3 -1 1-iti-i-Ib'! @ 474r:r=d -= = = 2 t .-2_ g ~ ~ _t - [ r =- c=: q -E
===t2 - + .g- += = d_- .r 4g 7 q _p. +-p =. a _.+-5 = q_ _ L_:c___.= _..-i. _i_:.. _-2__-d_.-' J:2_f.ij h._n.'t--- .i:_ _..: t: :: E-2 --r. t _=,95-2. - 9 '.ic.f. _. -2 n2_-- _.I'_ d._:_-_-- _ -_ _L. 3:_ H_.i.i_ 1 -4 ._:=.., =..x_- _-1_21: 2 =ma n -! _n m
- = rs
- n=
n +- 21=-2_: u2 22--1r qu
=
-+ -m - _ _ c -.-+- 4_.__ L 4 I y . I. __4 l I ' I I I ._ c. i _ o,w i _1 ____c-- m ---om % g' y 1 1 i .~ w,. r r .m e r__ - i :D' _a __. w.' Tw_-%4.,, L - -rm k,r. g y
- d-'--
M.- M i ]- e .m w._J .u. 7 "' N + b. _ _1 %g g_
- g q 3. _g g_
g . =:=:= =: =:-m m ,,,o.-- Y'- :_sWq~ = m- ~_ xrm-- . = . z _ t.:. _.=r= 2 9 = = * - T- " ' ' * * ~ 7-7- y ---%- g'"- g -. , ",.'.'.'_'-r_'*'"; =- f.: =- T- "W
== .s =
== _ L==- 2 = 4 .I -:_7- +-.= q: r } -}.f '}, --. _Q i1-h 3.- 5 [ j . _ = Tf 7 "-' rM#_g - ~ '= 2-i
- =
g .p .g_- -i-hg _2 : u z.t. 2r_ -.z -: y 3. 2:
- g gs
.2.
- .2 2t--
--:::: I _..22' _I r
- 2372-.'
2 3... = ' t-f. g a. 1 I~:_}; ;. 2.- ._.1 -4: 1 z 4:t _.. : 22.x
- rt
==2:2:_ 1= =
== m -+:. n: :::=:-cnn n:::2 = 2 _.
- :::-rn _ _:;
..=
2
- u. 2
.mn::t y: _ =.,. _ . =.~r. t -
- n : et. a- _.~-v r=:-
-r_ u ._i_._.._ _..a; _.i _... - +. - + ~ - ~ * - + E . _
- g
,.-.., :_.____-+-.-n n - + -. _.. _*~+-'. + -.. -. +: : ; _ + - _ ~., - + _. _.. ..w 4 1 -'-W~*._-~~.L 1 i # ' i -'" ~ +- 3 u - i ^ t t-t + 4 s_ 4 -.. 3 - l e%.e, y m g 1 - j--- 6 -I~ I N ~*- 'I' ~ I- ~ 1 + q_ =4. :_.. -- r.iz.z{__t. p. =..-E..
- E. s.= t.: E.
- j:t
= ..2 2..-- _f-3: .Ej.. r.2.. 4#.. 4.. =. = g 3 _~ g _b .7-q- i-j - 1_.. .u .-~- - wm. u.. m,T,. = r.
- n. -t
- m. -
=1
- +.H-
-fi-! rd-4 tid-I M cl:' -4F - ~rS P 1 i -. _.l'.4: - E. : _1 r ; -f: .u. r._:_ __:...- 4..,__3...-r, _,e -:-21 a. _.:.-t. _:.. i. rt. _. u. -;
- .L. _:. -
- d. d.. : p....
o r -::r..._ ~.----. 2.:...t..g..d. r a_. _-...t=_-- _2_ g r +~ _.; :_. _ _- _.n--.. ~n r:_:.. r.*-+; n p n._- g r._._ + : n_+. _n.__.. _... W._I._r. _.n_n. _. _ _. ~ _. - _ =n m
- _*=:.c
++ ._._._._.. __._._ p _- g ._._~:__._.. .. - ~ _ .__ m_._._ 4_.,...,, - + _ _ _m_ r ___ -4_. _.r, _ r i'
- i.
+_ _ _7 % 4_,_ i i. t -+ N -+ 1-t-t-t-+ W-*---+r-- -+-t-+-+-4-- +-t-1 + 0 oce 0 io ois 050 o h5 030 055 040 045 0' 0 5 Fig. lA-Moment M,/(M,/R 4) due to an external circumferential rnoment M, on a circular cyhnder Stresses in Shells O ! b b,, -
PROPRIETARY DATA - a i iJ i I i -I a___ .I I i r 1 y__ e s 4 2 L s 8 -- ~ i 5 - ~rn d 5 '1 M 4-n __ l i = 4 mj i' s 3 4 _2-_ f , :r:. '- 1 4 ' -i
- - 4
.=%-4--i L = = -..r 't= t @ Ef 2-i d-' 5? (MSI* m,,!-' 34b ~'
- -~~-
= d 2 4-E jd - - E M 1h=~. [~.':-$- I-}_ --f.'-jf5-ic' g siMs i -F49% =iE5 ink- .=i-ist ~ii--il/- -4ti- :-I:Mu== -- + N=1--
- ] isici-4536-2_-P
= . y w._ = %_-
- - = =::....:.
= = = - - -.=._._-z.- 1,- m. _ _- _ ;_ _=. -= = - - - ,_:==. ,== r m.m,m x y Ii II. 2 v2 mu m x x. ' ^ 1 I ! II .. ' _ - wT%._-s
- w.
'+ ^ r j II: 1. I ' ' t4E u ty Lg A I I I I I-. ' t.k '%~ i g-1 l' '7 w' A. s. m.~ qm.'l m&I -- -a rd f mm'.,C g J ~ Tihn' q._'" l-" l .I 'l ]- I.' ] =]:. I ? C aVT g F-m,U ~ um,_ sc s 'l. -1. -1 V
- i_-
Q s a y 1 / i ; 7
- ? =4 !-
%g - N.?='t,C s -P m = = v m= -i: '. 4- =d-:. ~ c 25 f '.. 'J':- )5---.I5'd 8 ON 5r!2i ^~ % m,,j~'Ydjg" fNi'%e,-[(- i i f =-"- --Nd ! - j-i. jcf j.Diriz L g -J.- } W =' i-@ z= +zmi,4,,,,,rF%AM-7 C,s @ = w -i-t-t1 - r :; :L-A:Nr M = r =--r _- p c - C g i 1 O + -.O 7%-
- =,
w % ~N-D p, ' 4 ii! n,- 3. arsn 2 ,,,,J 'N .'4 O %. N.?i11 . m -r,- 1 3 . n. -A_=' % df F"% d_%, _%,. y' =h'. __'T %'bi M,4-a=-i-. p,6 ; -+ W .mi T - p-M= L fM = = .r#_. = =.-3a gjJ t-g . fth_- 5, E. /.-- I'-~;'g= ;;M'""" L 1; -.4= 7"' mas 17 %13-i. d-~:- M Q_ _m b w g_. ' s. y ,. y,_ '. T 't '-*-r--I-,- g-g-p=j-n-y~*'-**'~-~.'_z-q~-=-;- g. Ff Wl ., _,. '~-- *anzmg x.,_ -v L - ~ %T' 1.~ I '] Q ,,,,4_ - H f. -.m ~ 4 - m--- i I I i I. ;
- Al'_f ar'- w
.s - = :~s- %m.v_-h - -- o Q R R A Er, e r E 1, x 1 Q a ^ 4 3 p I F !f II -/ ' ' ',,7 8 I - = = =, '/ i hWE '_,t ,'N_ t'h gm II f a "- 2F - ~ lI ~ IJuF _. 1 '. -~ .:L p 1? N Y'"" E" a ~ " 2 LF. 3 #1i 5 d_aHb -=-'. = W =4.J-
- 1--
=j - ,a+=&
- E I'~I7 f.
= +=.#Ms i =- =._F ""'=+ = E= = i - E =r 22:: =;, s = '= _4 s 't : =- ~~ f b b' lN
- - $-b b9~~5
- 95 I T-' MTE MN ~T'~ i MF f 5 T a 5'-4h5 3E-h -= 7:=_*.Ef.__. _i ;J ea*-E+ +="=. # = +4;F-.-M-i "- --i-]=f;f 3 c '""=- a r = a m a r m __. - m_2 - 1 432 _m. a un. m A J 2 F-7!' = a d= .I #d" e _-J k+ If - -n];
- 7. i; F_
'.Y_-4.. _;*
- ) ' I n ( 1;
-s e __: U'_p ~ 7 E + F.1=a r_ = =- + +4 z'i#" - - Ei -24 i=_ ---r i= - ? H :-D ~ F1s 4. c2 -:'- . J -...=,En+ e,- 'l-- N5 f s5h5i 'N 2
- - -" h-- E$E
$ E ~'i-
- (IN
~~ N -3 ) 2 M O %1E %,-.-W@. " - - d 3 l = a-e_ _.==r- =g = =_ =1=_ _==. r; ;. .m-._ 7:._:,.=== n.: : --___.
== u;_.,=_ ,.-.:=1-= +--
== .r- - = - - - - , --j - F-m - _ -. - +. - - -3 --2:-.=;:ran.22--1 :_:m._: _.*r_,.f I : -. _ _ _. _. = -t..:._:__ _ -.:,----- AE E AAv E 1 l m - r I J I I XI N I I. 1 g l J K .I :I'
- I. I I
l E i I 'r I f 1 I e_ r i r i i aN g g l7 : --'i --"I.~ j_2 ' II I-E1. j 1_ p = r -- - - a- - g 1 1 z_ { y 'E r p -- e jgz} q 5 g p.; tg p . _i ; 1 .'j]' ~g J ^' y T J F R 1 +? + r i tihi- - 0 +s9s- =: -i~4 -== ? -4 4 -.= +&- =4+.i - t t_
- -' s=-
.= i+ w':-+.2 if h4] .c_
- 1. ~. ~ 'O
- 7. 1'.-
~. M.5 E7 --3M WWib =.=L =MI ~ -i=1=du t: =1 - iy-= 51f 5;= =-; e i 3'-
- v.. - -
_=n. _.3 _ ;1= ^ ' I= _=__ n i= t== = ~ '===E
- ~~==_ =
== = s = EE i - T= 7
==== :- ~:-~*~' _--=.u==.i s - i -- r - ~ h-" 4;it IJag .w + - as !- lli ~ L'*" r -I.- -f5-. I -4 ~ ra_
- 7-m
+3 !- r ~ = i't* i ~ i%- V- + + - '.=m-V- P r - 5f:.=-4 4 -~ f -i- ~= = --i= r_ - 5fi '. = 5r v.-+ < = v - + + - ui j _. p _ z_
- - w;!
g j b e -- 2
- -'~
1=-M -F;I gi=_y.E: ge
- =_:24:.=2. L
- 2 -i_a :
.72._q. p.4___ ; 1l jr[
- 5=r's M-1.Mi-N--' i-F f@
2:
- i~"
T :- 2 E i*-?-j si - -i,ld:.- V = };' ~Q55=Q =gi.5c.: 4; J_.W._ h.-i-Q -l _. _q. ifi-ii=i-.q_E; ~;.-f ~ ~~ ~i _ ::5. 5 =~=.~: - =~ m .H_ -_.~ili =. - 4 I' F' 1 L I : I i 1 i a '" F' T - F E Elli ~~ .3_~ .3-I:I l-I~~ .-l_.._' I. l~ d ' 3- .3. } _3 .II i ci~'i 'f* Y I---' { f -' _~ _ -Q-i iM m' -* b i !P i4 -f TO s-A 7 -.=51 sL E i -i._i= _. - }.' y4 'i ~ f -"4 F-- .+ ---E:z: ~ -TT .. 2 - =-
=
2 _i u.. ca p: - E."-- T ~t' u. _^Y +' *- - 7. = Nl -~[715'? s 3 ---:.1.25' 'h a
- - D
_W-g r--- 1 i s , " -"-==-4 -1 d-i ii 'l - i. = > a u ---. =-=- # =. L .=t & k . 7 1.=_ 5 '. .- f. q- _., :.M 4 y- ,ti p: . ; i -. i_ q_:_c hfikEI.If LI-id'! : _=i 4. E i4. I 27E - i-@s -
- r. - -14;r-
-._..;::a;=2 : _. ;;; -- : 2.- - _;_ ;.; z _ __. un. M :. c - - ' -:-----~ ~ au :. .._ _. n. --'-~; r_._-*._1.;.- M_r:.*_.._ r:-
2.
- :n a.-. =n - ~.. f _._ y .4. i_... f. .. 1r :- .. _. _. %_. j_ T. ;.- :._ __.a_r i _.m i .~ ._4 -,.__.4_, . ~.. _. _ , 7.,_. q -._ ... _; % _9.% 0 0 05 OiO OIS O20 O '25 0 30 0 35 0 40 O'45 OSO Fig. 38-Membrane fore N./(M,jR,,,9) due to an extemallongitudinal moment M ort a circular cylinder L Stresses in Shells..
PROPRIETARY DATA I I I. a . h' !^ L= V t R__ g y {- '-j_ h:-~ h --2 )- -'i -~ -ItT l' ~l - '~ '~ hibi-- ~ 'N i g_. k-- -~-d-- ' { -----r-IIs'+ =- r +,--+ 4 _'+ ? ~ 2, . p - ((c{J -j - Y ~.)_. d2 [.h _-'t_' n'5 _.] ' ~ x__ '._". 1-J TI 3: - t rq : T-di - '1 h.i - - I: dtd-Yi '.' [ 1 d 2 15 - i-I ~. _ c t-3 3--'
- t. - -
. 6_ j--t".. {. ; j i_ , i. T. .. i. !_f ${ ..i 3 4 d._ 4..'t { ! !
- i1.
.3.. _ :41 - i .j. O. f. _. _= I_: _. - 7 ,__ r-+gt: n 44.c.-:_. ._u _;. t. ::.:.. u.. 1:. :.u ..:. i : n..tt .. : : t. _2 _ ; _ _u t t ; i_ 1 - ^ ^ +-+- : -:- tm. _Lt__t mm.:r= = ; += 1:q =tx -:_r: L t _n+ =t J .y _. _ +. _. _. _. _.w--+_++-+.. +-+-.-
- : :+
4. -+ 4 +H -+ .: -4 +H_.+_e,.4... . +. .p 4.- -*-+-+-+-ep+,._-+-.' +-+--+-*4- ' ^ y l 41~ 4.-_-.e.. 4..g-- ^+_4._.._H .. i., .-.4_+ ._-+- t-4 -" l l L4-j { -o_L._+_r--. + i > +-
- * ~ -
- g
-i I I ~5' . _h = a 3-4 3 -3. p: - -j H: j - -3 g;ji-47
- j ;- rj i;
_j. j r t: 3 zi 'i- --tijiEri:::r;.-sjif g:qt-f i 2 @4 c r M =M a 4-j- e =i .+ c =_ = - *a = 2 = = q:'2_ h -.j - -f - i _t- -'? pi-:: 2 ?- -- ;;y- - - -. - --- ~- +l; N
- Fr -l-h
~ :: 2C t. - W t.- .a -i. _irj= -.i4 4-1 2 I51.? - V 9-Mi --.E N r .2
- ._ *. _,, _u._...'t-a_
!*_+'". _.___-.$~*...". ~~.',-*'-~,t-~, g _.e_+_ ; -.__+.. ~..~_4..t. _ _"I--+t _T '. T !- 14 1 -% _4.-. '--" Z$ 3J: _1--T'.-.._7..+-_~. -'7 4._ g_-.. p. -- _._-. % _.- . _.m_._._.-_._... _._.3-.- N-+=__.___._;_+- ~-+- -_e. i T ! I 1 i 1 g, I e" -R 'l ,3_ A = T = y f_--l2- = _ '
=='- = =}- -.q--- a ~ I T l' ~ ~ -~ 3 - E '! = ' * ~ 2 1 --h-.b 4=J 4? ME""9W2"ew-wm:-6t
t
==- a 4 =1- -i= - 02s=-- r#
i=
.g g" 9"to~ i 'l.' ~~ h ]Ibq %=g,,'%' A i 5- = 1 'D'~%?f* hmmm _YI numE., '%'. --] -i a:- _ - -4 32 l E = 1k'i [$%.'%) h %m' 1 %w eu_l' Yh is iW "h ?A . 4.~iG W% -i = s.T %f"*' '%
- m:
iE 'i '-9c 1 "-E33 J MGN4.Mi4-G-@jl . = [.--DMm - U: ,f""t is.-u = 2-:E = 5 -EF =*
- .' 3-2
- 4 l
bb
- .A_ gr:&:_.5 W_'_
_.m%w-::_yE..% p%- 35 3 [I-
- '-'=
5 k, ~3 r[-~t 5 +- 5 p g =i ww.- - +- ._ _=_ w;. m _ _ = _y -. _ M _a=== = -: =. ^ m.,, m s+~,.-. tu
- *^-
C -.m 44-me-h.-. x. ~ _. N ** :,i~: ~ - +,T ^ N N_ ~., Y, _p- - s.Ns m'. sN O2014 1
- ~ ' ' - * -.
N
- ' '> e w
'L ' ! s_ : i - 1 i ' 1- 't Ni 3
- m
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- ~*
i r w,
- w i.
am s-. L - I . 'a. 3 .1. I E lrel' c -~~ ~ T y x: h. 'b ~ ~ -i 3 l 2 . ml- -:^- N. -.am 3 ' dQ. i h 'll a" -_.i 6 =-"%],N . %hi - "y['h = = ~
- ~-
i-* 2% J-y .E- 0 E
- 5
-
E 4N ' 5 - 7 =-~ = gg h
- ]3 Nj-
}.; 't h ( pp%, r'--' - [- 2 j 7 y = E = = 2 + g g'- 4. ~ ~ - # =fe %.42 T :=M#-N 7= s s=h tN E=i =F +Me =:s 3 M 14 a- = E = = g_ - l 2 r= = - %J _.'i? E% i.._ yg -'qL.ic 21_ - ~ ~ %# = g y hI E 5-i 5 -~'b j%E." INN 2. 2% M,! [m,, W N_. O.. f I %F. yy. _-[--.- m - I-~ 5 2 53 ~'--- - I m
- ,. -F
= f# M4 'E % 6 = ?f1EtE= i -~*i-9' -~ 'i = = 1 g g. _ -pm- -=.=.. = g r====r g"_ >;g;_._3. g gg_ ___S_ 2.f _c_ _.7 2_g_._ 3-2.. rag - *: N,k-h;,. g_ = e+]S e 3-.-e-%x,_._._ m.._ 3-.
- ^
a _,_.i__. 2::T g__ 4 mw-s e-- .--mg -x i. %.--* m,+,..- .-+ b % m-
- --+-+-_
- +-
+%.--+-y -+-- 7 7 y N I u ~ E. 7 P { g a ? l ^= h'*- d' ?'% ^' = y a j= '. 7 ~ ~---- Q-_f ( ,.-~I_ ~ y__ i ch
- j i 7 _a p 2
. - ~ g '~ - _.g - 7 7 _-{.:. -.",g r; .. _ :3 2 g y 3 - Ii ' - I;2{ 5 2-fU!ZliT . - ii.' N= 42' 2 f h~ 2'N b' T ~ ! BIEN Nk'-' ~ Ji" ' _1 i'
- -'2 I
^ 5 b=*i2 --.i-= =T T =3 _: v 7 q ~ '~ ~_ : ~ - 3 = } . +. _..s c___ a M ~=N M1. '2 I-I N 5 [ i - MD - [' ! - -t ' 5
- i-
=' ; =' ~ 9 :r-r I = ' ~ - f = T _4 I: : 2. 3._-p...i.- 3.:_i. :3 :
- 6 z:i_:: r1
__.+_._.4.-*c. -& - i-_m =K_E_.l.L. E.iE._ };1-+=.P_..-p_.u,_i. j-
- r. r* =:i_..ir? 1
- 4- -44
+ u-1 =t:rt.
- m. _ t-
+--.--._p-._ .. -. _ +
- +.
4
- +_
b. 4 L._. I i. 3 I 1. 9 ,1 I i 7 I.. i ~ T I I ' I, i. I I I I i i I i ' I I 0' 5 0 50 O Ob6 OIO CIS 020 0 325 0 30 0 35 040 4 Fig. IB-Moment M,/(M /R,,J) due to an externallongitudinal moment M on a circular cyhnder JStress on the longitudinal t t plane of symmetry) Stresses in Snells T. p d VJ J C' g 'r-11 l] O
PROPRIETARY DATA e - -l 'l8(' 'ffTI~~ iI 2 7 f _7 - Jl- .'f '~- -) 1 - T ] t ' Y'.." _ {-~ ?I ';i. 7)L' 'et a '. Mi i=== - + - E M-' m' =i-M - 'lii_ 2 _L.. Jl. 4 - ki i T
- a 4 1--
i- -ri-y "= 5. " " =- i EI=Mij gjf E"- Il i-MiT4.L'4 - 9 5 5-E--F ~ % - id4-5-L --!_1~=---- =E-E - -5d:5 -?+'E iE-' s z N rN D ^^ 5.-DI i 55 - 15!_(-5M g _.~.n.. _ ~ _-'_'. -E. _, 5 5 N mM2 ~- - ~ ~ -S _-_' 5
- _ - _2'#l54 n.
-~M d- ---_ -d c. 'L :: V - -. ' " - -. Ii'_'4E=J~LJ* -E-r-ALT-s T- '.I. 4 'I-'__ 'L----5 '4 5-l- 'M._1 ^ ' _, - - e!r -ie' = m ~~f'i Ni '. -5.' -r { 3 -^l--' _~ i E- .1 'TW .F 't3' a_:.- ;
- - = :=- a =.
a_.. ..=s:
== __~- =r-,- ' i-'. ~J ~- J = , b [ '"h g_ =- ~~;_L:: e-inity
= = =;= 2 y, 1=1
k-v '---: ; g. - - -gg - : , = _.==:- = =j ~ ~ f --4% -.__ _:-E- % IT 9 n y B% . l - s'~ C '~
L
' "Q '~ !I r w ,h .mC 'Ia II md" I I 1 I -'-8h .__ ! [ 'k7I_ m - MnM1'"u l El iHm r-i= g m sur suwwmu @, mnumre u-u-= mm b g m MlE D I ! Nm._ sg y a em i - - n -n e- -- -m _. - -m m= uum =% y - - = - e. m_ e_ g I Y -' -b~-h. _. _..._%_--^- 1 2 ~? ---~ ._..t =- g y=- g u.. -. = _ -_._ _g 5_ 5 -.. 8_ .:f _t=.= 3;;_- ~r ; 3 1 r. .. g -- P e 7 s i i .-E -mR.5 :,.-g--um- -. - ---g_-g g g g.- __= = _ ?--_.- q - j--' . l^ 1:
- =~~'_
m q q e _ g._. _ --_g = =. 4.Ms-I na I g u i""lE a e s i i u m _ mg P p u A u = -- c l g. g g p ..A e -i m = g gu ie i -n - w a .a l e i n = ~ r'; e e J 55 -i-'% e d# c i' = i ='.' J-t E ~' =_ _-- b-j.=f:$ h-d/E-Nyl"" U ~~ '^-~ ~ -"~ ' ' ' '.._ Z. ! -- r =_ _; , _ _, z.
- ~
~~ gif-/ T,_iE m-#-C ~ ~ ~ ~ ~ U -'MA__ U =5 "E ~_ . a.=r-T-- -w : e.= = l X' N. F F I r J-a w 1 1 - 'I . /! I I 1. I i asuas- __[ m -~ YI :,I .J.
- J 1
.I i adF I 'f 1 ] l l r v. e"- ~- -Mb ~ iJ' .' i E'=:_%E 4F-
==i t ~ ~ uv m -tM .' E-g 4 2i = -Er - b_-b' 'dE
- 2 i'i-- SP-
-__.- A ~ ~ ~ - - =- l ^. I- ~ 'i_ I ,=- '-'[ w ~ui ~ 'h [ - _' mu-i g ~ y =MIb
- Sb5 N5 s d1i.,.si 'E -N-'.
._' 1 5. f--- -i E ~- S A r.- 1-F k,= N[ '-Y N b Nb . 2si C--_--.$'M_ -M LmE #I-? r A_ U 3E@.555 -- j_ __ r =~ cf.- -- ,y.,_ _ -.. --n e=- r- -.--2.. . w-n -= -JW fM-4i_ .UE i'in h. 'i.ar r-EL'r iLaC. A_ t 0 im = a nr1 :rs - r mw w-M-m .- - u n-- wn+= ...2.- 2 ~- ar ri2 = i m -= g, -n = _: ;3c- = = {gff '-f f. 4 -fQE id -Q:].=f-,. pQ3) 7 F 92.- .= . f_ ~ ;,j h.f -, m 9 6:_ =.2 1Bl /+ /- k W -b : - W E -?. 2Fr psi'[1_- j t-g _- { 4-- ~1m - f -2;-E; 1 I i-; = &l 12/=-1;/ H - &^ f 4 24M:4=fi i -L -- -*H== -*= -2=^^ = ~ ~' = -- D ^ ~ = $$$N?~? W = ;3K W 5_ ^'* %M.T T ~ ( .w+. r .f...
- fff-f.=_W- / g.
f -~- 2_ A r-' D _ g.- g '/: /W ' f il i, ; l 1 - --+ +.: 4 2 1 l rv 1 1 i / d"
- Ti 1.4
(: * /t 4: ^f E T 4- 'n > 1 H - tC ~ -I =I I _i _ V--- Cr' -t .-A r r ers a n i n' % ,w_ - m-mse +x w e
- u ri-m n rw
- = n I U i 2 ->/
+ Y3m =a n-!=;= - iw =w e + -==- + 1 ~= tw r w-- _.
== J e 1 s =2 s ' It+ F fn-f= m-+;--+. k 1-i=t L =^ m-+ =,v+ 3 -W-I= aM i-t2M- = = Ff f /: Lf 2=..wi-. - +. ._ ; = y _+ u- -n I:. m -t-i_t. _:4 =i-
====)- - Eg=g=_
== = M 1 TT I i 1" r c i, =# I i f! mr .T 1 : 11t W W- -i we =
- ler.
r i t 4 h <+ c + m e = = +
- 4
== ~ ]-].f -fr a -4:- ^' pf t n p s,y- ...L Q:j _. jy s.". =- ^ 5 _ Q J._-=. f; Q Ij./: /. - i..
- u. n+
i-.= a +2=y w w i +p =
- p.,(
- -1
+.% _ :..- = - - .. =. = - - - Hg _=g ^ g;..- = q d~[$ 7: W
- 55 5#-1.
^ '-@ 5 @2'E5%'"-A-KP =
- , II 2 l[f.%1._....
- 2 =. = -
- _.:.=._.
~- - =~ 2 .:--- = = --..:---
===:-. -:.
- ,,f,j_.
.{=-- ~~ 1_. = th II I $ ~ II I ~
- 2
-N 'I i I -~I l
- 1 l-
"~ L' g_ If r N -4 + 1~ i - 4 2-- -i i.' ' Q -]l y l [j: 1 G a. 4.- .L=&-QJ --n + a nwg
- = q+
- g =
-F= Ilu==V--.,, SE M mi- =: L =- L: ?M=M# & L = ,_ - hf5::';5^ E Z U _+ l- --.d'q = -l
- ~ '--fg^- -QQ#: ;&' :_ --: :.
a ti 2 J.r 'Tr_ ? L+ n-r- [f ' _i r l:r N: f,- Q I --Y ' =i'-.--Q Q s i -..z - s -n t- -k i=1+- a I +=4= >i== ?i: =i=~k: = r q,- g. N p_; l .:12 --- m :. :rr -...r. -- n _s : :.-r; : : L _ ~ , y .._ 9. _ - ..,.- -.'r ; . m -* :... _1- ,1 qr - _2. 1 1 o' 5 OIo CIS o20 0'5 0 30 0 35 ohc oAS o 0 o'o 2 S Fig. 4A-Membrane force NJ(%/R.M) due to an extemal circi.mferential moment M, on a circubt cyhnder Stresses in Shells PROPRIETARY DATA F "E J--'r [ T [ ] 'l pr 7-g g y g [ a. i 6 f i 3 4 y a u _4.. 7 2 s r =
- :.=
-1.. .1 - -~
- 3..z 5
y_ -3
- ,1
-- {. .-j I- .' -..-f .h: n . 7- -.t.-<= 1 1 . II 1 ir-i a .-1-I g_ s ..g. j _ r ra {c . - f. 7 -L o.) 4 1 = - L;. .q '
- - j;_.;
.j:g....-+- 5 =b Q g ^ g__ . 5I$
- H.
- _f4"
---'-t-4' 'E-J- 4 ? I i 5- -- =d T~d N- ~ ^ L - EN-II-2d f IS 4:- g _}_ J. -j1_ p_ l j.I.;-,. . Qi. ..L_. p _ :. r 4.-- "it - g-[j{ -+ ~ ~ 3 -- _._,!.7 7'-F. :* i L :- . :r* T. :-;r. 7 ~ - -
- -
- T
- C.
~*-*-*-* :l - n
- 2. :: r
.n-*J'_-*-*._T* 1 '.. *.7 U.-*.!_.**_I* 'J..*.
- _L.J T_'. l. :. -* 7
_ ". _ _ ~. - 4_..
- o. 4. -+. 4+
_e._+.+_w .,m.i, r _a-..-+_4,4 i t 44 t 94 4. Y I i f g +y y.1 -, 1 g
- q -
- j g
.y- .g .I
- r y-s-r- _ -
1-T- !i1 1- :J .}. 14.i 6 -1 : -{_
- _{ - { ^<
q - )._ _- 1 y_ ~ 1 '5 - - i.- Q -t m .. -j _:~J -._-i.- j i
- 4
- j :. g.
- -g fi. -+;.+
y i
== g __ -= = a' = _j .:j 3 ; =p ;d t._. _;= } _. f g-- i = g :: - ry-;_Q . 'i._, =.h :
- _-E. }
- - _f-
^ 5=- _~-i' 2i~ 5-5 -3 _ - 7-- g_ = 7 4-i ::1 bi E
== -L:-- - ' ' ':. E =- = =. :' : +:
- z..:
=%-_:'=_- =s=.
- ^
= ~..: '--1 3 3 .._: _... - -'_i. z: =-
- d -..
n
- 3--
w - +_-; r5 --3.5 .%.i :n...=W'. = = =_ . = - '='--'=
- - 2.
- -
' -==-- 2 ^^ v-2 a .r Z. _. . _i:- zdi' L-z ~ 1-gi: = i 7 2 _' _ _:--~.IM.- A. r 2 .. p { .;;-.y.'i.e -i 5 4 2 -- _p_
- g_. g:, L: p yy
- - -- -gg3. _ ; _;.L..:..g _ r.:.- -.,.-3_--._ ._arr- * - + I i I I I 1 l l 1 [ ] 1 3-a t--r _ I-i .Z
l
11 - I '" _1. .I f _7-I __1 : '_f---- ?T - _I- 'r i r e a = 7'lill I -f. OF -I = V-' d '- -j 1-i - %_ _!-J -- - --f-Irh ~- -i 4-.'41. ii .- p _J W _ i2 Q-,. __ q' gli-45 -.f iy-._g70 i-i i _ = = ij 1:7._ . _r-.n 2 g_ .~_- [ (- -m~' 5 hy-~._..- l t -- . 3Q ' If i j -ia.@p- . Q'-Er_= IEI_ $5~ T' -9455I";-.- It NUM z$~ t' b- $- ?--Nr -I2N Y~ ~ 8~ - k-iN'i I b.s a_ a ' p. -l ? I 3.' l
- f. L
~. '1 r! 1 %b -T_ -- 4 + L ic
- --- 14.
-i]
- a-4_{
- q:
4
- _L;.= 4_
^ g =.- .=
'
Q..- p. 3 ^^ y;i._ : _S.-k y=.- j3 i__: -Q__t_Q g __1 1-i 5 ..si . --Qi-: -' i: t2.. E. .. f : .--p.-j.. _j - -f-A _Q-.ip p'..*g; {FQEi:- 5NrN52'I 5i-':202-!.- -f- -' 'kl _.1 '- 4.:72-2 'z -ii f4: E i- = -t .F.E'EMd-- ~M:-5 ~~ G Ii- - + -,2:.. u-r.r:; .c 27_ n r-..- .. : _;;: s; g . :--*: * - t ._k _. :::::1,: 5 '* -
- Ct::--*7'"-*-'*- ----.**:**-
4- .. 1..* ; ~. -. ; *- 2: r ~_. *:' F '**J*..*'t. J___*' ~-*- - .~__f.-. .e ..+_.- _. . s.w _ ..-4_+-. 9.,_ w.. r 1 r a' ~ = s- -1 -f-F = ' Y: " I~ z4. = 1- -F ~~ 4r 4- -i = F l= 1- -1 = y_ z 7 = +p --.:7 .p 2: =p
- _1_
= q. -_y. = ,=2 ,..._u.L =.-ss.;.-1::==.3-m x+ = w = 7_1 -- y- .==. . gy. = = S-CkW5YN h' -i. ' 'rE 5- "N= C:.N*- NM EN s- < aq.-D((~ -= y ~ M!SN N. :_ ' T" W +L1 ' T '- g_ - c a' LU' = U g_p-L ~e ! + + t k L w-1 T*=4,, y g Cat-Q-, c3- _L=T- '- F 4 7-*, i==.-% m. =-y g Q -r- ? j. L,2,s su m_ ,.j - g_= 3sg 12 7p ;n-pe: - W-d-N. J, - Nf*=' 4J CM i - ~N
- i. -E3=i
=+ d - - ~ m ~ ...r 0.026 -w:. ~ = ~ = -~,.._.: _,.hd- - ~ - ~.,,, q x ~ ~ ~ - -r =r -a-a = t N .w - - c , c 3; . v,._p s --1--~__~__- m ~ w e w ---e.&Q-ar_:_:. :._ = =-::.+_r =:- _::r -. q a.~. --= g ~- - + ~ .._ h.%_ _d_: r -M_...;. x.. w_ . Wt.9 _,,.; /yy. + t , g y g --g .c E .. - -., ~..n_n _.. :.a... cc ._+ s g ["g 1.I ~ ~t t I D5
- 1-
'I T 'Et 7 r= i -r-i f I .2 e r u i i ++; w m .c T--VT 6 r g__ _ j._ i ,-kJ_ i~- =., 'E-~ i .r- ..a g =-i )y.- 13 N:* -f-l - i d. Fi
- =3 &
y. ..L. 7 +i 2
- --l,,
a Ji _:53= = hi _i: & i:.+: { ; j'. 2 --j.. --gq--y g_:. - y _4 7- ( 4 ;;.y 1 'I ' s ,1. r r, c y. 4 1 ! I IT3 F7 a-i > t i 4 s t 2. .s ..: _% :1 7 [' '5 M.4 dis t 5 ^' - T55E 4 ._ _ _ _. _. _...~. r ~._2.__.t..1 - :.... -L~-... _ n *_ - . c..: _- -...5%.'._. .-- @. _..N 1 J- -9 C: 2 n._..,_._ __.~._/_-_.=. _. n. _. . ~ _ _ _ _ 010 O tS 020 02S 030 035 040 045 O SO n 0 0 05 ~ Fig. 2A-Moment M./(MJRJ)*due to an external circumferential moment M, on a circular cylinder Stresses in Shells 0!bObh
E PROPRIETARY DATA .i 1 r, , i. y v l 3 + 4 t 3 Ii p_ __ - - -, -.g i
- r. +
-y-13 r +-M-. , - y-t - r pta. L . wit a 8.==a.M
.
g
- a, 2
i j i. .j.._ --r af
- 4.
- .
p.. [ -y r . -Rfi = - - t 1 1 1 -f 4 I m i * ' i- -i- = h] 4 , s v 4: y{ Q5 L U"WT"-- ].. g ..-E $N cF rb-i=# ._ {Tjj-i.... iy ji{.i%'.-jygj.' .y- _:t_ : _2 f - -. _y H,.-i jQ-hpig h. 2T.jl _._-._.y.;- -_- y - -. a-.2 = _w 7-. 7_y-- . = - - 22
- m___.- =
.ct = .____-.-. u.__ r :-- - ::_. t --. -:-. : 1:=::2._-1 =- z.:-- g ~W. -1 2 w,_. - c; _ h_-. : 2-* T --- - - ..m.--- 4 . m. k 'hv; " ~ * ' i i I I lg a [2 .T1(I " I I 1 ir i I i I. I.. 1 [ I i g I I ] I I T T ] J g* ';ff : t' ..iw 7 '. - - [& '~h , J,a,b '%. { - {- ' 1 _l [ ].] . L' = 1; vt I. si
- c. { -_
g_
- -l. 2f: #'T-
._- A% 4. 9 i
4
= 1
- =+1
=:: r-y_ s_5
- -W44^= Mw%,?&
- H -- ! '-V v= --= q=z+.Q=
= -' r=.3 ' +'- .5 g_ 1'Mi~ridi-N =N N [NNiif 2' 18-I--@ -~~M=.:~-- ~I-M-$fIrEC' 'h : ~ 'J I s' hel..~l I l6 'Y:1\\ -- L { T [- - ~ 2M 4 ~ --id--- ~ ~ N5s 0' =_ -~- "N'+E INN \\ -I I 'i I -~ 5 I T ~ ~ 4- ' IM/?- r 2 -+ an_'TQ 4 :o %D-N1- . = Fh = ~
- -~ =
- 4li5 t2 - + = = = y[tf -f@ W t-iw-r-+= -m %g-t--p,q4. - wj: %+t 4 -* g g qm =
- p WMWPM -t=+il,h
- id -D'%L%;=Th@ t#EibMW.tE =15 += =, + = z Ect4# E w -J-/2.gy;ggE._yyi =_ = -3== 2- ~~- ^ 2
- - r --+
%%pgp=j:3_4 LLF E --~ g l,D m. _ pg.E, E ss 1 z r Q ". ~ w, m 4 ~- _ m,~_ y,1I-E YI [
- ^
[_ .h . AX-ja IIf f I s".F I l _ %hi"C = i '%O ' ' I W 1%, if f I I I . i F [ g 83IL ~ 1 i f I
- d -m. %M%N_
i g "'-] a 3 I -- 'i-Je"" 1_] ;_ T_.%U -"hu dT4ingm ^ = = a ' I J.'I I -LIi 7 i ' I I :- 1 '- f-i/ W-r 2D' ? -V ~ =~-.fi._. -% %a: =--- p. -b N' h'Y ^W ~Ys Th * ' ~.f ~ 1 -'; V % ' N. =k ; ~ ~ --Y- '=-- =_ ^ $j 2 5 ~' ~ - 5 m =-n s I:iw-fet= W .. ri= m+ &+ = + 4 P-N1== wi+ 9 = ~
- Whs/ M=7=t= =Mt h =W r-e -4WriE--E= e===5E =-- p
=====- - - ~ 2,_:; a wns i r ws we Ld J,J 4f -7L - - -2 ..E f :* f&I E e ;= n +3 Y' - ~ .= r f F = 7 i-f.9 = M~iL . i : Ji .-. fm _---s F
- -.- =
= + 5 5 h. [? h 'k=, h0 &r -') --'=*iS / ^ $= ' s__ ='-'-~ =?=%- - ~ - ~~--~=- ~' # _L = -- ~_ 7 ^* E fWh-=/ tr-iVi - =c+ 9C wM-91;;. *- " ~Ei- +4=i=r-i -m hf 1 - '#t-5 = = h h db -![-I ~ 2NMI N.D_ h 1 E 5 g_- ; p. . c., a, -1 f ,2~ -~- / -d' 1i :. :. 1" g--- 4] ~ ~ ~ - - 1;.,
- z 1.
'I I' / r I Am I 1 1 1 r I T f t.. T F '/ 1 I .r i s i t t i a '_" l' ~ f IIi;
- il
f.' 1 ..n#' r i. 'f: ? 'O 1-4 r-a-e 3 _. It f =f h I- " 4 - 1. -t UF a i- ;; 1. -i l-- mi = = = i y_ $ $ Wh ~Y (- ]
$
.. ?~ /"' i EES ~7Y +V $ t'^~^ - ^ ^= ^-15W-' ~^ = -: = -~ =-== =-'- + l t i ~^ 5 iT 'ili f- ^ ^ g_ fN~ M - ~ ~[-bh [F I-.9" 'f5 MN -I-~ 5I6-Ei NY5' =- -Y- ^$.I T $k51-~'i E ~., i 2' ~- ~ T g_ -- f 2f!I-[ ~[4 Nh'=I=!NM -~ I ~-N5@ IN _h 53-' '=$ YM__ ~T[-i lig j E11 i I_# j S 'TT ~
- sT n vw a-r u n
f . ^... =- q__ l I.. / F ;' i. -Y f 1 '-f <f t. I xe .t i T %. +_ 1 3e 4-?: - -W ~ i. i . +~? ---M= ' fli-f-4/5* M/ @ K= t Lr E -4 '.- s f.k4-i y_. Nffi[}f[ O[ ; H+ f ' r.8c.i 4^ }5 ': r~ .i 'r ij ai?1
- i
_qg ' -j. 'T5p ..}4x._i j qM:_= H-= j i p /iff -f:- -- rf i 2+.+:.; .g4 c -Mi. - = g_g51 -i '.pyi.41 =p, -E=+ pg-;2 p. p ~_ =.q. (:-...- - _ -. _...__=.._.:_-_. : :_ : =.:._:..- A-- - - - ~ - - - - _.. _ ~... _ -. _. yy. _ _. _. g f -f h I (f/f--/__
- ..c
- ;--.:_..
+n--
- -, _:. = --t-----4 2
= --= --- r y,,; a. ,' h' i f. 1 i-1 i 4 a oi r t r 1 e __ f1^ 4 f + + p __ ll l. =. 6-1:1 :=s E 5 per: L m r
- ; R
.s== - v=y m=,m w g= I-t w==- = r1 a a ff 2 1 1 c i e M.- r ,+ _l- =- = 4 j h
- 4
- - (EEU.-~u_
^ _ f ' ~ - ~ * - [--- hi[:.. 'EN~' .2 i g_ j ~.* :* Z M--'**- n _.,s. ~'--'- B s:.._ ._.- T. :
- 7. z..
...a_ .._ _ :- ' :-~=_ -- ~~- _ - - - 3 5 g T 9 w w o oos o so o 15 0.20 021 0 30 o 35 040 045 0 50 v Fig. 48-Membrane force N./(MdRM) due to an externallongitudinal moment Mr. on a cirMar cyhnder Stresses in Shells PROPRIETARY DATA a _u_ ~I. I rI I ' .1 I I i i iI I T I I-IsI- 'I ! 3 7 iiI i -I a . I -- t 6 3 ' ' i t ' (, !.]- ?, r-I e p: y -+ gh =- m.- g2_ a._ s _.j .i 3- ) { g i-n x =: 2.r-1 N.{ " i 6 - d. ,.. qgj {7 --=ic -{.'; '.i _ qt : J_- I-I i .1 .I 1 6 I1 -=_._g t ( 3 M i ~f t_ t -i !=+ t. = 2 4 r I-I. : = = = - 1 i : =-.1 cf. - i- - j_ f'E_j. -j~g-i- , * = _.i ~- 3-;--i= '.i:i .ry y.'. -igf.j. -- .I-- { -_3c [ - .[=- j fj f
-
g . U=iri--
- t ij _ s 3._{ p 11=!i_:t = =_3d=_ci; _2.0 _._._r
= E 3=.=i. +I= .--Ei si=.1= + I rE- .. _. ^ . c r. : :.... :_ -. : r. =..t: -.-_ _1 -.. _x-s-. - _.-,t..;_._.__._-. ..n:. - _. _._b..._..rr - - - - ~.. _. =... - - - ...,.:_=
- -.=.== uu w
7:.__ . m. :--. : :- - :cr ;---- ._._._a. PM4 M a"4u -@g. .W ._ I I 1-1 T-E g _l _I -Li. 1: 't-E -- 1: -4 -'=- P
:
e _ 2 r 1 f:: ?i = i== - }i =' y_i5ff i&E5 = = i-b = -
- i= F 1
=i-E l- + --2=- E. ^ 5:' =iW -k - i t =5 it ti = = =-i 1 -= = Mt =-
- M
= - E ? =-4 i=~
- =. -
=4-M=M !! EEE E ~' t m _ l ---#:*=T t=+H-:=l=i: - -i-iEl'IM: Pi i ~ = - E E. I i. I-J.i. -1 1- - m - - 'c 4 5 i =h 2 E + ' 79-I } .I - I 'l - -f'- - h I-~ -t i -i.:
- e
.= = - dr - i-r= = = .--+d i '. -r i s = -- i= 21 if = = =J E :- E e i
- {a E --I i =
s 4= +1L --- +H'W - =i ~+ 1- .^ t= T =
- =%
=- = IX.= & b-== - n ^1==- = = = M=;= P
- E#Rs Mi-5 T =E4:Mih ni r.b.
?s?---- =_=_---- C,-'_t .. -. + -.. =- =33E@?a-EsiMi s = ...~ -.r--t ::m --.__ I f g_ 1- :!u-r_. ! aq _ i nt_
- j_3n 7. _t 1 m m
m m. = = j_ -l- -b i' .iii i L 4 --*- 1. 'Id= skT" h
- 3=
m. "" ' '='. i 2 = s.J-d- ' = :-i-M i W +t + = =f- -f= W 4 - a c. J 5~7 -~96 %u --um.ii= "h'-?m d =nE7 "' % ?% _si ifT k-i' RP2%.-*.I9_i
EEi%45k=F '"m6=?%, = W!?5fiM:_5 d "i=5 7 % n..r t1*3 :*-+= =P Lr M-~s
--N MF='% i:=4:4J i Jik ='A = a ~ E=h-kM R,.M
- f ::E N-" k
- 1--il @M*- =l--
. x m m x-. ur- %.a4
- -w:n___ w.-
2mr r- -i c = : s + mn 4 mi 1 = -.a. + =,emi i w -m i
== u.: r a a n n.- n a c_2-- - F -.' *fe 1 ;- t -a w .-e --M-s-=.\\Til 1 ' f61 -?6 n' =+ M ~-3 2 = gjj.g[+.- i s( ?E*N=: 3E-3542 Vlit 1=i %-- M tr- C -f'-- ' r'h L.
- r. 'N =t "h rl -#
== y ..m=isE -h 5
- .L=i=__
aW1hhNL EisiT-W #Elisii-== 1 *%., -Wn:~ f-D.*4.F 1s Z_~ l_ i hij = y. _ _ _ y- _c:. = -x OD2l. m y -, w -
- -: y
..i
- s '
. N,m ~Neg _, _ m w. 'm n t .. z y 6 w _ -" [,f m - s:. s :,-- 3
- w
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.A A I I I .Y Y s'- A .A-s. m1, w i 5m w . j- -. i. i .E'.
- f. L -
4.~i i u= _u_ 1 y t-m g 1= = m-- e t -P h . 13 = l-~~-E i ] =: i.J. J?k-i h.i = 3 4 __ LP:.-i d2 - ='" g_g % =4 4'.= "-4 =r*'P '-J "' - -M F* =-
- =P-'J== *
=--
- 'i ' +=
~= ~- 5i 1 E! E ' ---= =- [yEl '=. y y ri'+4 '_I' = % '-J= W- %cI i--i- -i "-M 5 i's WiW b =c-i -i's M
=
E~=-=
- =
= 0: 1 ~g g N :==22=r --= N...-.:,. % g=.79 - a _ r2- : _2.- .3 .a ...jg3_.gg
- _a_ m.._2nLV a..._,;=.g _qa
..=2-g- _l! m vrc i i r u. _r-r.. a s m w_s.- B 3 i -r- -= r = 1 u.
- r 1 2 er
=
- m
-er-w: r %._+-i-v E" +- d=m } isTe t 1 = s = a~ 4 =ic:i->:"W.== = - ~. =L 22_=# -=k y t i :#?= - _=&4 i .- 2? D: 4-+=-W = = = YE EU= e 1_ -- = -E MM - 21=" . =--i- =Ei =E t_f -iN=Mf +1 - G@=MT E=~=i ' = ~ 1= i ?i -i_~% =
1 r
= ~~~ Z ~~~'-_T_t-.J _ iM =M -E C E5;-J1id:M-# #-D4:=Ri@r-0W=2MF 3 =s=f=' .__-m_
- -_..p_
%.. _ _ _ _ _s___-- -- -- ;.... _. = _ _ - __ - =--_. - ' = - - - - -- % __g = _ . _ =x w.
- q. _ _m c_ _.
.._.m. v L____
- m.. _ _ _
a.+. _ _ _. - ._.y. K e.-- c r r' na 1: Ni. -i ; is2. _:a 1 - I' - 12~ !~'I- '"V 1 Y J._i 1 -I I T" r = =c s. + ei r 2 + 4 e, i + n. r ~.- ! E = ( 4 .M;~- 1 - --~ j ~-i-=' ~:j - i. d-D s 9 + ++g
- 1 7
- ..=p j%
a-pg. = - Og igi x
== = = u: _i-; :-
-=*
j===2= %.7.y 4:=2 E Mi fft-~ 15: Y 2-I q-
== -y m. =+. g w_ a _.y L' { ' '.= i : {.E -[;;i : -ir, t - {=;5Q= q2.,.zi3- :-.:Qi - -i-
- .= :: :-
p m--- -u.. : _r: r _ ' ' ' " ^ -__:--_=._. _.......= _ _.. _ - -
- _ _.. _.. _ = _ _ _. =._...
u _ = _ _.... =. = _. ~... _..
- __._.l _ __
w__ _ _ . _ + _. -. _+. - 3 0 00S 010 O!S 0 20 025 0 30 03S 040 0 45 0 50 Fig. 28-Moment M,/(MUR,,J) due to an externallongitudinal moment Ms.on a circular cylinder (Stress on longitudinal plaq of symmetry) Stresses in Shells r PROPRIETARY DATA - Table 5-Computation Sheet for Local Stresses in Cylindrical Shells
- /[P L
.V d>
- 1. A,,t..d L.
3. c mes e. .c... e M _ L/ [" a.d..i i.. #. P-T T Du* F ~'
- c.....
~ Ib. ROUHD [g, Dl r L.a... a.16. p :O Au' ATTACHMENT T.........ae.
- a. Ib.
a fgu sh L d.
- v..
b. g 's sk L..d, v<- 6. e g8-yq s..... c.a..a....... d... : / T i
- 2. c mer
.). ib.. l..d. za f T N6 b) 6.ad..e i. d. Kb Rm L Cu C v....i ihs k.... L i R.z M"" in. NOT E: E. wee.li b.....l.. 6. An.ch..as.edi.., '. a .a. o ne, den...ieh... .....a.... v.u.I d.... WMCE Sh F. n d... cm,w..b i.e...i....g ST R E 5 5 E S - f I..d e..pp...t e thet .h.=a. e.. ... ea e.ke== i ,3 e. ........d.a.......l.. c. o o< "~ c ,,. = - (;.) 6 = i:." r. - (?) 5 = + + + + .. :..w -(..L)..;,=. VAB7/AVA RRt -ww5 +msas ~ .h -(..:o).:";,.= lEl VAg VJfA 75fA -ateyunenyt . "*...as -(~..y).."'y,= -ve -yus +554 +35cBBB pfffiggl BRg a ';:;.".aes ~k:.;)..;',.= -sui +3ut +3a.t -3u.I Rlll 888 RRl l$G ',"'2.~.""'+'"~~~~'" 7204 llb 7Zo4 lI6 2fTSa 2@e2fSSe 24ke j + '!c" . ;. = - ( "d.) 6 - a (?) 5 - ',7-lc. p + + + = ..l. egg "- (.., :.o).. ', = fffA @ffjil fff4 fffjll -53 4 -53k+5344 +S.s4 "a ~ .."'..e a, " ( k )..,. = lffff Vfff Ffff(d lflff4 -V42 +9541 6 542 -%R i
== .O..o in "(.23). 3, = -k,3 -/43.ty,3 +/4m ///M ////M ////A Ffjil a ';::.:l es, " k.":n) .'"o",. = -5491 -t5n +54y -5h lf!ffj}l V///4 ffffl VffA f ' " '. ', ' ' ~ " ". ' ' ' ' " " * " " 1I69 3 813 7lt,9 3ats 14888 41 % I4882 41 % \\ l ' i;.':.'". e s., r. + - i O t +- r t. +. r e r.s=h,, +1340734 -1340 -73#V/jfA ///j fffff V///) U W""
- "' ~ N L r M M V5fhlflA 4225 '422S *4 225 +422F l
'," 's"'.~ "".'t.' '-~ 734d 7340 7340 732 4225 4 225 421,5 4225' COMBINED STRESS INTENSITY - E \\b l4,1 4 5.1 4.7 N5.1 Niz.s 3 z.s .3 (Units in KSQ 4. Y 9. 14, 95N 26 A 21 26. 21.
- 1) When C pf 0, S = larcest absolute maanitude of either 2 [c + 04 _I do
-. 0 )2 + 4T or[(Ox-04)2+4Tj. S= 4 x x _V. b o i
- 2) When T = 0, S = 1argest absolute magnitude of either S=0x, 04 or (O
-0) x 4 N,/(M /R.9) so determined by (C ) from Table 4.s Calculation of Stresses t t 8 (see para. 4.3). 4.3.1 STRESSES RESULTING FROM RADIA! LOAD. 4.2.2.5.2: When considering bending moment P. (M,): 0 - Kz. fig'~ where Ks, is given in Table 4.3.1.1 Circumferential Stnsses (a.): 8. Step 1. Using the applicable values of 4 and y Stresses in Shells -.
PROPRIETARY DATA At the highest stress location on the uuter shell, the stress is 26.1 KSI; for ASTM A36, the steel used on the outer shell, generic minimum yield is 36 KSI. The outer shell will not be overstressed due to the forces calculated in 3.1.5.2. 3.1.5.6 Failure Under Excessive Load The tiedown lugs are designed to fail under exces-sive load and preclude damage to the package. Based on ultimate strength of the shell material, the force required to initiate damage to the shell would be: F = (291,460) (58,000)r 647.690 lb (26,100) The lug is designed such that it will fail before the 647,690 lb limit is reached. The mode of failure will be tear out of the tie-down lug. The force required to cause tear out is: F = (51,962)(2)(2)(1.985) = 412,578 lb. Compared to the force required to damage the shell, the margin of safety will be: MS = 647,690/412,578 = 1.57 l 1 l l i ha RM
PROPRIETARY DATA 3.2 Normal Conditions of Transport (Appendix A-10 CFR 71) 3.2.1 Heat Since the package is constructed of steel and lead, tem-peratures of 130*F will have no effect on the package. 3.2.2 Cold Same as 3.2.1, above. 3.2.3 Reduced Pressure An 0.5 atmosphere pressure will produce an equivalent inter-nal pressure of 7.35 psi. This pressure acting over the lid will produce a load of: F = (75.5)2 (n) (7.35)/4 = 32,906 lbs Since there are eight binders, the load per binder will be: P = 32,906/8 = 4,113 lbs/ binder l Each binder has an ultimate strength of 125,000 lbs. Therefore, it can be concluded that the reduced pressure will produce no detrimental effects. 3.2.4 Vibration All components are designed for a transportation environ-ment. No loss of integrity will be experienced. 3.2.5 Water Spray Not applicable. 3.2.6 Free Drop Since the package weighs in excess of 30,000 lbs., it must be able to withstand a one foot drop on any surface, without loss of contents. 3.2.6.1 One Foot Drop on Bottom Corner Energy to be absorbed = 53,005 lb X 12 in. Maximum energy = 6.36 X 10s in-lb i 5 Volume of steel = 6.36 x 10 /36,000 = 17.66 in8 1 l
PROPRIETARY DATA Energy will be absorbed by crushing of corner. N\\ I NN f \\ P. ~ N
- b <.
N y\\ ...,(4y e_ -} [ ' \\' ', :',j. /, j \\ %..- 4 o r N ' o:7 / \\ / BOTTOM 's\\ CORNER \\ / A NN f. The volume of the crushed ungula, assuming the worst case of a 45* impact angle is calculated by the following equation: V =R3 Sin 4 - - $ Cost s 3 3 $ = 16.6* when V, = 17.66 in P \\a \\ + 1 s \\ A. 29 } 1
PROPRIETARY DATA i The maximum amount of steel crushed will be: b = R(1 - Cos $) = 1.69 l The effect on the cask body due to the corner impact event is shown below. Even though the weld will be crushed locally, there will be no loss of the cask's integrity.
- e
- f%
6 nj #7 ? >, \\ t - = - = - 1.19" -52r2_E '- 22 - i 1.69" o 'e *#o l The decelleration force exerted on the cask is calculated as the product of contact surface area and the yield strength of the steel (36,000 psi) Au = ** - (xy + ab sin ~
- )
2 where for 45* angle = 0: R = 40.5 in a = R/cos 45* = 57.275 in b = R = 40.5 in h = 1.69 in C = R-h = 40.5-1.69 = 38.81 in y = JR -C 2 2 440.5Z - 38.8* = 11.58 in = x = C/cos 45* = 54.90 in -I ( ) Au = n(57.275)(40.5) - (54.9)(11.58)+(57.275)(40.5) sin (57.275) (2) Au = 34.60 in2 F = (34.6)(36,000 psi) = 1,245,611 lb. g force = 1,245,611/53,005 = 23.5 g's PROPRIETARY DATA 3.2.6.2 Effects of Bottom Corner Drop on Balance of Cask The 23.5 g decelleration will be transmitted to the outer portions of the cask. This force will be composed of two components, one force will act laterally with respect to the bottom plate. The other component will act axially with respect to the plate. / Cask Cover Inner Container .Decelleration Fo rces Contents ~*e% s Q e, s Upper e, D ^ o Bottom Plate Inner Shell Lead Shield Bottom y Plate Outer Shell Corner Impact \\\\\\N hkhD\\ Reaction Fo rce Bottom Corner Drop - - - -
PROPRIETARY DATA Summary of cask component weights as used in the following drop analyse s Primary Lid 5,563 lb Shield Plug 366 lb Outer Body Shell* 6,965 lb Inner Body Shell 1,900 lb Upper Bottom Plate 2,645 lb Lower Bottom Plate 2,864 lb Lead Shield 14,397 lb Waste Contents 18,305 lb
- This includes the weight of lid ratchet binders, tiedown lugs, etc.
The following design criteria and assumptions are the basis for the bottom corner drop analysis. The following load distributions are considered: 1-Load from primary lid and shield plug will be distributed to the inner and outer shells in accordance with the shell cross sectional areas. 2-The inner shell will receive loadings at its connection to the upper bottom plate consisting of: Load from lid and shield plug Load from self weight of inner shell Load from waste considered to act on one-half of the shell perimeter nearest corner of impact. Load from one-half lead shield considered to act on the half of inner shell perimeter not receiving waste loading. All other loads on the inner shell will be considered to act uniformly around shell perimeter. 3-The outer shell will recieve loadings at its connection to the lower bottom plate consisting of: Load from lid and shield plug Load from self weight of outer shell Load from one-half of the lead shield considered to act on that half of the shell perimeter nearest the corner of impact. 4-The upper bottom plate will receive loadings consisting of: Loads transferred through the inner shell weld Load from self weight of the upper bottom plate. Due to the rigidity of the upper bottom plate, all loadings on this plate will activate the entire perimeter weldment to the lower bottom plate. 4 PROPRIETARY DATA Cask Analyis 1 - Load from Primary Lid and Shield Plug ./ Dece11eratto Forces Shear Area of Weld \\ / / s '4 o n O b y 04, % V Decelleration 4 Forces Detail "A" / 1/2" j l l 3/4" i Reacti Fo rces Loading = (5,563 + 366)23.5 = 139,331 lb Lateral force = 139,331 (sin 45*) = 98,522 lb. Axial force = 139,331 (cos 45') = 98,522 lb. 2 z 2 Inner shell area = (n/4)(76.25 - 75.5 ) = 89.388 in 2 80 ) = 222.317 in Outer shell area (n/4)(81.752 - 2 Total area = 311.704 in2 Inner area = 89.388/311.704 = 28% Outer area 222.317/311.704 = 72%.
PROPRIETARY DATA Force on inner shell = (98,522)(0.28) = 27,586 lb lateral and axial Force on outer shell = (98,522)(0.72) = 70,936 lb lateral and axial 2-Stresses Developed in Inner Shell and Attachment Welds Decelleration Forces Inner Container ,n en s Bott m Platu 1/R" N wel Detail "B" Reaction on Inner Shell 1/2" welds Stress in weld around perimeter of inner shell at cask lid (27,586 lb)/n(75.5)(3/8)(0.707)(0.85) = 516 psi Total stress = /I'(516) = 730 psi Safety Factor = 21,000/730 = 28.7 Stress in weld connecting inner shell to upper bottom plate Total force = 1/2 self weight of inner shell + 1/2 lid and shield plug (1/2 of weight acting on 1/2 of shell) + waste Total force = (1900/2)(23.5)(sin 45 )+(27,586/2)+(18,305)(23.5)(sin 45 ) = 333,753 lb Lateral Weld stress = (333,753)/n(75.5/2)(2)(3/8)(sin 45 )(0.85) = 6,243 psi (lateral)
PROPRIETARY DATA Axial weld stress is caused only by lid load and shell self weight. 27,586 + 31,572 = 59,158 lb Axial weld stress = 59,15E lb/n(75.5)(3/8)(sin 45 )(0.85)(2) = 553 psi Total Stress = d6,243 + 5532 = 6,267 psi 2 2 2 Axial shell stress = 59,158/(76.25 - 75.5 )(n/4) = 662 psi which is less than weld stress. Shear shell stress = lateral force / area [(1900)(23.5)(sin 45*)+(27,586)+(18,305)(23.5)(sin 45*) = 8,129 psi 2 - 75.5 )(n/4)(1/2) 2 (76.25 Safety Factor = 20,772/8,129 = 2.5 3-Stresses Developed in Outer Shell & Attachment Welds \\_ Decelleration Force / j Inner Shell Upper Bottom Plate Outer Shell Lower Bottom Plate / / Crush Depth a Weld Area in Shear Detail "C" Shear of Oater Shell Weld Stress in weld around perimeter of outer shell at cask lid (70,936 lb)/n(80.875)(0.5)(sin 45 )(0.85) = 929 psi both axial and lateral Total stress = [2"(929) = 1,314 psi Safety Factor = 21,000/1,314 = 16.0 L
PROPRIETARY DATA Stress in weld connecting outer shell to lower bottom plate Lateral force = 1/2 load of outer shell + 1/2 lead shield + 1/2 lid and shield plug (the 1/2 supported by 1/2 outer shell) (6965/2)(23.5)(sin 45*)+(70,936/2)+(14,397/2)(23.5)(sin 45 ) = 212,954 lb. = Lateral stress = (212,954)/n(80/2)(0.5)(sin 45 )(0.85) = 5,639 psi Axial Load = (6,965)(23.5)(sin 45 ) + 70,936 = 186,673 lb Axial stress = 186,673 lb/n(80)(0.5)(sin 45 )(0.85) = 2,471 psi Total stress in weld = 4(5,639)Z + (2,471)2 = 6,156 psi Safety Factor = 21,000/6,156 = 3.4 2 - 80 )(gj4) 2 s Axial stress in outer shell = 186,673/(81.75 = 840 psi < 36,000 psi yield Lateral shear stress in outer shell 2 2 (212,954)/(81.75 - 80 )(n/4)(1/2) = 1,915 psi < 20,772 psi yield = PROPRIETARY DATA 4-Stress in Weld Joining Upper to Lower Bottom Plates necelleration Forces Inner container Upper & Contents Bottom Platu 1/8" Detail "D" ..6, we i, Reaction on Inner Shell Load on weld = Upper bottom plate 1/2" we litt. + Inner shell + Shear from lid on inner shell + Waste + 1/2 lead Load = [2,645 + 1,900 + 18,305 + (14,397/2)](23.5)(sin 45*) + 27,586 = 526,902 lb Stress due to lateral load OL = 526,902 lb/(76)(n)(sin 45*)(0.5)(0.85) OL = 7,343 psi Since all axial loads are transferred in bearing, the maximum weld stress will be equal to 7,343 psi. This is within acceptable limits..
PROPRIETARY DATI 3.2.6.3 One Foot Drop on Top Corners A drop on the upper corners of the cask would decellerate the cask and would result in axial and transverse decelleration forces between tne cover and the balance of the cask. Contents Decelleration Cask Transverse Decellerati inner Shell m Outer Shell Cank Axial Decelle ra t ion Primary Cover 3/8" % 1/2" d Impact Point PROPRIETARY DATA The top cover is stepped and the inner plate has a nominal clearance of one-eighth inch. Upon im-pact, this plate would immediately contact the inner shell. The transverse decelleration force must be resisted by the bearing stress between the inner cover plate and the cask wall and by the weld between the two cover plates. The magnitude of the transverse decelleration force will be equal to the axial force which will depend upon the orientation of the cask and the corresponding decelleration forces. As shown later, the maximum decelleration force will occur when the cask is dropped on the long flat edge. The maximum decel-1eration for this case is 16.3 g's. The weight of the cask less the upper cover plate is 53,005 - 3,260 = 49,745 lbs. The transverse decelleration force acting on the weld between the two plates is 49,745 x 16.93 = 842,183 lbs. The weld is a 1/2 inch weld, 75-1/4 inches in diameter. The stress in the weld: 842,183 75.25(n)(0.5)(0.85)(0.707) = 11,856 psi Safety Factor = 21'000 = 1.77 11,856 The weight of the cask less the cover and shield plug is 53,005 - 5,563 - 366 = 47,076 lbs. The transverse force between the inner cover plate and the inner shell of the cask will be: F = 47,076(16.93) = 796,997 lbs. The bearing area between the two surfaces will equal the diameter of the inner plate times the thickness of the plate. Area = 75.25(2) = 150.5 in2 The bearing stress between the two plates will be: f = 796,997 + 150.5 = 5,296 psi llbONfhE[
PROPRIETARY DATA 3.2.6.4 One Foot Drop on Top Corner of the Long Flat Edge In a top drop on a corner, one of the extreme con-ditions would be the impact of the cask along the top edge on one of the long flat sides of the cover. Angle drop of 45* is con-idered to be worst case. 37.4" 1 34.4"
- s h impact An impact in this orientation will cause minimum bending of the cover and will result in high im-pact loads on the cover. The major;ty of the energy will have to be absorbed by crushing of the steel. The bending and crushing of the cover will occur in steps as illustrated below.
Crushing p Following impact the edge of F the corner will begin crushing a until inelastic rotation around 1 the bend point occurs. pg 7 I,, I The point at which this will occur is calculated as follows: Neut rcl ^ ~' I " Width at bead = 37.4 in Thickness = 2 in i - 1.50" M = (36,000 psi)(1 in)(37.4 in)(1 in) = 1,346,400 in-lb. l 38,0p0 psi j l'h 38,000 psi PROPRIETARY DATA F = M/X = = 897,600 lb 1.5 897,600/53,005 = 16.93 g's (axial and lateral) F, = Force required to initiate bending F = F, (fl) = 1,269,400 1,269,400/53,005 = 23.95 g's (total) 2 Area crushed steel F + 36,000 = 35.26 in Width of crushed steel 35.26 + 34.4 = 1.025 in Depth of crushed steel 1.025/2 = 0.512 in Volume crushed steel = (. )(.025)(34.4) 2
9.036 ins Energy absorbed in crushing 9.036(36,000)
325,296 in-lb Bcading When the force due to crushing reaches the above value noted, the cover will bend inelastically. The bending will occur around the impact limiter ring Fa and with the shell of the cask F r 1.5 inch The balance of the energy will be absorbed by bending of the lid [ Total Energy) - [ Energy Absorbed in Crushing] [(53,005 lb)(12 in)] - [325,296 in-lb] 310,864 in-lb = Energy Absorbed in bend With an axial force of 897,600 lb required to cause bending, this amount of energy will be ab-sorbed by an axial displacement of 310,864 in-lb/897,600 = 0.346 in. [%WR
PROPRIETARY DATA The g forces developed during the bending process is calculated using a kinematic approach. Velocity at start of bending is 2 KE g _ (2)(310,864)(386.4) s W ~( (53,005) = 67.3 in/sec As calculated before, the inelastic bending de-formation is 0.346 in. The time it takes the cask to move this distance, based on average velocity is AX/V,y = (0.346 in)/[(67.3)(0.5)in/sec) = 0.0103 sec g force = (AV/at)/386.4 in/sec) = (67.3/0.0103)/(386.4) = 16.93 g's (both axial and lateral direction) The above shows the maximum g force is 16.93 g's in crushing and bending in both axial and lateral directions. The force of impact on the corner is 89/,600 lb. (axial component) Contents 897,600 lbs 40.2" o , A n < i o 2R U 2R
- 1. 5" /'
2R 18,305 x 16.93 lbs 3 -+-- 2 3 " 34.4" 23" The loads on the ratchet binders will be propor-tional to their distance from the pivot point of the cover on the cask. -
PROPRIETARY DATA Forces tending to open the lid consists of weights from waste, lid, and shield plug. (18,305 + 5,563 + 366)(16.93) = 410,281 Summing the moments about point 'A' (897,600)(1.5) + (410,281)(40.2) = 17,839,721 in-lb 17,839,721 in-lb = 2R(23)(23/80.4) + 2R(57.4)(57.4/80.4) + 2R(80.4) 17,839,721 in-lb = 255.92R R = 69,708 lb (in farthest binder from impact) R = 69,708 (57.4/80.4) = 49,766 lb (in middle binder) R = 69,708 (23/80.4) = 19,941 lb (in binder closes': to impact) l t l ! hlb0dthkI[
PROPRIETARY DATA l The 1/8 inch thick seal ring, made of AISI 1008 steel, located on the outer periphery of the top of the cask wall will experience some pressure resulting from a top corner drop. This worst case appears in a top corner drop on a large flat. The force exerted is equal to that which is required to bend the lid, or 897,600 lb. f /; -V _l ze 3 f-2S
- I The yielding surface area reacting against this force is proportional to the angle 0 and the radius.
( 25,000 psi)1 = pressure (psi) 2 2 (Pressure) n(39.875 39.375 ) in / degree = force / degree 2 As seen from Table 3-3, the entire force is distributed over a 132* arc of the ring. This is less than the angle between three of the large flats. 8 By dividing the incremental pressure by Young's Modulus (E=30 x 10 ), the ratio of the strain may be calculated, and by multiplying by the ring thickness, an actual deformation may be predicted. As seen in Table 3-3, the maximum deformation is 0.10 mils. This causes no great deformation or damage to the spacer ring. The force will be transmitted to the shell by the double one-half inch weld to the outer shell and the double three-eighth inch weld on the inner shell. Based on a 132* distributed load, the effective area of these two welds is: 2 Area = n[(79.5)(0.5)+(76.25)(0.375)]2 = 157 in 0 f = 897,600/157 = 5,700 psi The actual stress values will be lower since the upper ring will cause the load on the weld to be distributed ovee a larger area. 1Report of the Iron & Steel Technical Committee " Estimated Properties and Machinability of Hot Rolled and Cold Drawn Carbon Steel Bars", SAE J414, Feb. 1968. -
PROPRIETARY DATA TABLE 3-3. PRESSURE EXERTED ON 1/8" SEAL RING DUE TO TOP DROP Angle Pressure Force / Degree JF Press /E Strain (Degrees) (psi) (psi / Degree) (11,) (Percent) (in) 1 24,842 8,590 8,590 0.000830 0.000103 2 24,835 8,587 17,178 0.000828 0.000103 3 24,819 8,582 25,760 0.000827 0.000103 4 24,797 8,575 34,335 0.000826 0.000103 5 24,766 8,564 42,900 0.000825 0.000103 6 24,728 8,550 51,450 0.000824 0.000103 7 24,683 8,535 59,985 0.000823 0.000103 8 24,630 8,517 68,502 0.000821 0.000102 9 24,570 8,496 77,000 0.000819 0.000102 10 24,502 8,473 85,473 0.000816 0.000102 11 24,427 8,447 93,920 0.000814 0.000101 12 24,344 8,418 102,338 0.000811 0.000101 13 24,254 8,387 110,725 0.000808 0.000101 14 24,156 8,353 119,080 0.000805 0.000100 15 24,051 8,317 127,397 0.300802 0.000100 16 23,940 8,278 135,675 0.000798 0.000099 17 23,820 8,237 143,912 0.000794 0.000099 18 23,693 8,193 152,105 0.000789 0.000098 19 23,559 8,146 160,252 0.000785 0.000098 20 23,418 8,098 168,350 0.000780 0.000097 21 23,276 8,049 176,400 0.000780 0.000097 22 23,115 7,993 184,393 0.000770 0.000096 23 22,952 7,937 192,330 0.000765 0.000095 24 22,782 7,878 200,208 0.000759 0.000095 25 22,606 7,817 208,025 0.000753 0.000094 26 22,423 7,753 215,778 0.000747 0.000093 27 22,233 7,688 223,466 0.000741 0.000092 l 28 22,036 7,620 231,086 0.000734 0.000091 29 21,832 7,550 238,635 0.000727 0.000090 t 30 21,622 7,477 246,111 0.000720 0.000090 31 21,405 7,402 253,512 0.000713 0.000089 32 21,182 7,325 260,837 0.000706 0.000088 33 20,952 7,245 268,082 0.000698 0.000087 34 20,716 7,163 275,245 0.000690 0.000086 35 20,474 7,079 282,325 0.000682 0.000085 36 20,225 6,993 289,318 0.000674 0.000084 37 19,970 6,905 296,223 0.000665 0.000083 38 19,709 6,815 303,038 0.000657 0.000082 39 19,442 6,723 309,761 0.000648 0.000081 40 19,169 6,628 316,390 0.000639 0.000080 41 18,890 6,532 322,921 0.000630 0.000078 42 16,606 6,433 329,354 0.000620 0.000077 43 18,316 6,333 335,687 0.000610 0.000076 (Continued) ! h0bb
PROPRIETARY DATA TABLE 3-3. (Continued) Angle Pressure Force / Degree IF Press /E Strain (Degrees) (psi) (psi / Degree) (1b) (Percent) (in) 44 18,020 6,231 341,918 0.000600 0.000075 45 17,719 6,127 348,045 0.000590 0.000074 46 17,412 6,020 354,066 0.000580 0.000072 47 17,100 5,913 359,980 0.000570 0.000071 48 16,783 5,803 365,/83 0.000559 0.000070 49 16,461 5,692 371,475 0.000548 0.000068 50 16,134 5,579 377,054 0.000538 0.000067 51 15,802 5,464 382,518 0.000526 0.000066 52 15,465 5,347 387,865 0.000515 0.000064 53 15,123 5,229 393,094 0.000504 0.000063 54 14,777 5,109 398,204 0.000492 0.000061 55 14,426 4,988 403,192 0.000480 0.000060 56 14,071 4,865 408,058 0.000470 0.000058 57 13,711 4,741 412,800 0.000457 0.000057 58 13,348 4,615 417,415 0.000445 0.000055 59 12,980 4,488 421,903 0.000432 0.000054 60 12,608 4,360 426,263 0.000420 0.000052 61 12,233 4,230 430,493 0.000408 0.000051 62 11,854 4,099 434,592 0.000395 0.000049 63 11,471 3,966 438,559 0.000382 0.000047 64 11,085 3,833 442,392 0.000370 0.000046 65 10,695 3,698 446,090 0.000356 0.000045 66 10,302 3,562 449,652 0.000343 0.000043 The spacer ring has a minimum width of 0.5 inches compared to a combined width of 1.25 inches for the inner and outer shell. Accordingly, the stresses in the shells will be 40 percent of the stresses in the spacer ring. Lid Ratchet Binder Assembly Based on the 72,100 lb developed in the far ratchet binder during a top corner drop, the ratchet binder, the ratchet binder pin, and lug assemblies are analyzed as follows: Ratchet Binders The ratchet binder will have a shank diameter of 1-3/4 inches and rated generically for an ultimate failure load of 125,000 pounds. The binders will generally fail in the threaded portion of the shank. The shank is fabricated from Grade C-1040 cold worked steel or equivalent having a generic yield strength of 70,000 psi and an ultimate strength of 85,000 psi. The minimum root diameter of the thread portion of the shank is 1-1/2 inches. The strength of the shank is calculated as follows: PROPRIETARY DATA 2 Yield Strength = 70,000 x 1.5 x n + 4 = 123,700 lbs 2 Ultimate Strength = 85,000 x 1.5 x n + 4 = 150,207 lbs Based on yield strength the factor of safety will be: 123,700 + 72,100 = 1.71 Ratchet Binder Pin Pin is 1-1/8 inch diameter bolt made of SAE Grade 5 or equivalent having a yield strength of 74,000 psi. Based on double shearing of' the bolt during loading, A 72,100 lb lb 2 2 1-1/8-( Resultant Force = 76,105 lb f j j j 4-o, = (69,708/2)/(1.125)2(n/4) = 35,063 psi Safety Factor = (74,000)(0.577)/(35,063) - 1.21 Lid Ratchet Bindet - Upper L.tg - y s, 3/4" ! \\ l[ h 4 1.90' g 2 (4- ~ 4s DIA JL %/ -1 s is, 1-D-3/d ( 3-3/4 ) 81000FNi
PROPRIETARY DATA Tear Out - Shear - o, = 69,708 lb/(1.3)(2 - 0.0625 - 0.27)(2) = 13,934 psi SF = (38,000)(0.577)/(13,934) = 1.57 Bearing - 9/16 oR = 69,708 lb/(1.125)(1.5)= 41,308 psi r 37, Le/d _ 2/1.125 = 3.0 1/16" o/fu 41,308/70,000 2., 'a .27 Tension - 5/8 4 oT = 69,708/(1.9 + 1.375 - 1.25)(1.5) // ,y SF = 38,000/22,950 = 1.6 Weld 1/2" double groove wc1d, complete joint penetration with tension normal to effective area. Allowable stresses same as base metal. 1/2" full groove and 1/2" fillet (both sides M 'd " full groove f x 1 1.83 n ,p 455;
- 1-1/8"7 e
i " 76,105 lb Neutre 1 Axis ( ( ( (~ ~ = 1.83 Neutral Axis 1 + (2)(3k) ({2 ) PROPRIETARY DATA Tension - oT = 69,708/[(1/2)(1-1/2)+(2)(1/2)(v'E)(3.25)](0.85) OT = 15,340 psi Moment - (69,708)(0.455) = 2a [(1/2)(1.5)(1.83)+(2)(1/2)(1.33)2 (I/3)((f)(1/2)](0.85) o,= 8,'456 psi tot * "m * "T = 23,796 psi F.S. = 36,000/23,796 = 1.51 BlbadNT
I PROPRIETARY DATA Lid Ratchet Binder - Lower Lug g_3 fan gn Tear out h --+ Shear - gn [ \\l o, = 69,708/(1.5)(1.75-0.0625-0.27)(2) 1-3/4 o = 16,393 psi SF = 21,926/16,393 = 1.33 m 1\\" Bearing - l o, = s,708/(1.12 m.n "^ = 41.308 psi SF = _le/d _ (1.75)/(1.125) 0 /fu T41,308)/ (70,000) 3 1h" Thk 9" = 2.63 i Tension 6h" 69,708/(3.5-1.25)(1.5) = 20,654 psi SF = 38,000/20,654 = 1.84 Weld Shear - o = 69,708 lb/(9+1.5)(1/2)([i)(.85)(2) s Ik" "s = 5,522 psi 4" = Moment - -p (69,708)(2.625) = 76,105 li i ' k b" 2e [(1/2)(1.5)(4.5)+(2)(1/2)(4.5)2 n (273)(1/2)](0.85)(4) 0,= 7,517 psi l I Neutral T = Voy + y = 9,327 psi Axis o 2-1 h SF = 21,000/9,327 = 2.25 4h" PROPRIETARY DATA Lid Ratchet Binder - Lower Lug (Optional Design) aC 3/16" % 1" w 3/16"N n a 9 \\ 1-11/1( lh" \\ '\\. M DIA. \\ l [ 1-11/16" u )h" 1 Ik" y I-3/Y DIA I/8 ik,, u C y q,, ik" y AMXX N %1L 3/16" N 9" \\ \\ \\ \\ \\ \\ \\ N N \\ NkMs.. / 4" ? Section C-C i Weld holding 1/4" thick plates to lug Assume each cit cumferential weld must support 1/2 the load - o = (69,708/2)/(8.5+2.5+0.5+2.25+6.75+0.25)(3/16)(sin 45 )(0.85) o = 14,905 psi SF = 21,000/14,905 = 1.4 Tear Out - (Optional design) Shear of the sleeve - o = 69,708/(1.6875)(1.25)(2) = 16,523 psi SF = 21,926/16,523 = 1.32 'Y
PROPRIETARY DATA Shear of the pin - o = 69,708/[(1)(1.6875-0.27-0.0625)+(0.5)(1.6875-0.27-0.25-0.0625)](2) a = 18,272 psi SF = 21,926/18,272 = 1.19 bearing of pin - o = 69,708/(1.125)(1.25) = 49,570 psi SF _ Le/d _ 1.6875/1.125 - 2.11 c/fu 49,570/70,000 Tension - (Optional Design) o = 69,708/(3.5 - 1.5)(1.25) = 27,883 psi SF = 38,000/27,883 = 1.32 Weld (Optional Design) j., g Shear \\ = 69,708/(9+1)(2)(1/2)(Jf)(0.85) as o, = 5,800 psi 3 g 7 '1" Moment (69,708)(2.625) = 2-5/8" 20,[ ( 1 ) ([i) ( 1/ 2 ) (4. 5 ) + (2) (4. 5 ) 2 ( 2/ 3 ) ~ (1/2)(/2)(1/2)](0.85) Neutral o,= 8,456 psi ^*I" l T
- 4"sz+
= 10,255 psi ~ O m SF = 21,000/10,255 = 2.04 g s. g 5%\\4 PROPRIETARY DATA 3.2.6.5 One Foot Drop on Top Corner at One Inch Flat In a top drop on a one inch corner, the other extreme condition would be the impact of the cask on one of the one inch corners of the cover over one of the tie down lugs. 25.14" N 2 1., 3 Impa ct The energy absorption sequence will be the same as that previously shown for the drop on the long flat edge and will consist of the intial crushing, bending, and crushing. Because the cover over-hangs the cask to a greater extent, the cover will act more as an energy absorber. Crushing Width at bend = 25.14 in. (Depth of 5.0 inches) Fa F Thickness = 2 in. Fb "y 5" M = (36,000 psi)(1 in)(1 in)(25.14 in) = 905,040 in-lb Fa = M/x = 905,040 in-lb/ 5 in. = 181,008 lb. F = 181,008(8) = 255,984 lb Area of crushed steel = F/36,000 psi = 255,984 lb/36,000 psi = 7.11 in2 libad@T
~ PROPRIETARY DATA 2 The area of the trapazoid is described in (1.4d + 3.44d ) where d/ 4 = depth of crush. 7.11 = 1.4d + 3.44d2 or d = 1.25 in. therefore, (d/4) = 0.882 in. Width of crush = 1 + 4.87d = 7.08 in. 2 Volume of crushed steel = d /2 + 1.925d + 1 = 4.18 in8 3 Energy absorbed in crushing = (4.18 in )(36,000 psi) 150,480 in-lb = Decelleration Force During Initial Crushing } [ Energy absorbed 150,480 in-lb y d ' Energy remaining = 485,580 5" > (, Force = (7.11 in )(36,000 psi) 2 - ( 3-3/4" = 255,960/53,005 = 4.83 g's / 3/8"_/ 32" 3 120 3" 1-1./8-Bending of Cover l After the initial crushing of the corner and the build-up of force noted above the corner of the cover will l bend inelastically until the lug under the corner contacts the shell of the cask. The amount of axial displacement will be 1.05 inches and the energy absorp-tion and decelleration forces will be as follows: 50 Fa Bending of lid (15* until lug i 1,3 hits outer shell of cask) y Energy remaining = 485,580 in-lb Energy absorbed in bending (1.3" travel) l E = (Fa)(d) = (181,008 lb) (1.3") 1 l - 235,310 in-lb Energy remaining = 250,270 in-lb. -5 4 -- l
PROPRIETARY DATA Failure of the Lug After coming in contact with the shell, the lug will fail due to tensile shear in the weld to the cask cover. The moment which will cause failure of the weld is calculated as follows: 3.75" centroid ~~ i t ens ion I compression % a.5 [ O F / / 8 M = 2a, [(1.5)(1/2)(1.875)+(1/2)([i)(0.875)2(2/3)(1/2)(2)]0.85 M = 5.2(21,000 psi) = 109,368 in-lb The compressive streagth of the shell of the cask will be equal to or greater than the tensile yield strength of 36,000 psi. The lug is 1.5 inch wide and will come in contact with the cask about 4.5 inches from the spacer ring. The lug will locally deform the shell until the moment that will shear t.he weld is attained. IslD00sM
PROPRIETARY DATA- - 0.375" h y2 I y n yi L d F=36,000(1.5)yt)f=27,000yt lb y2 = 4.5 - } in. M = (F)(y2) = (F)(4.5 - ) in-lbs = 121,500y1 - 9,000 y} = 109,368 in-lbs 9,000y{ 121,500yt + 109,368 : O 2 _ 121,500 - 4121,500 - 4(9,000)(109,368) Yi ~ 2 (9,000) l yi = 0.97 inches l The depth of shell deformation or, d, will be as follows: d y,_ 0.97 0.375 4.5-yt 4.5-0.97 d = 0.10 inches Deflections or deformations of this magnitude in the shell will not affect the integrity of the cask. l
PROPRIETARY DATA Secondary Bending During the following the shearing of the tiedown lug weld, the corner of the cask cover will continue to bend and absorb energy. Neglecting the energy that would be absorbed in the shearing of the tiedown lug from the cover, the amount of energy to be absorbed in secondary bending will be: Initial Kinetic Energy 636,060 in-lbs Less Initial Crushing 150,480 in-lbs Less. initial Bending 235,310 in-lbs Remaining Energy 250,270 in-lbs In secondary bending, the bending of the corner of the cover will reduce the moment arm for the axial force and the force required to cause bending will increase. 5" d E Fa 2 = 425 - da M, = F,2 = 905,040 in-lbs F, = 905,040/425 - dZ alb0hlkk [
PROPRIETARY DATA The energj absorbed is computed as follows: displacement E Fa Fa(avg) d E IE 1.3 4.83 187,454 1.5 4.77 189,748 188,600 .2 37,720 37,720 2.0 4.58 197,495 193,622 .5 96,811 134,531 2.5 4.33 209,010 203,252 .5 101,626 236,157 2.6 4.27 211,911 210,460 .1 21,046 257,203 (103%) The secondary bending is capable of absorbing the remaining energy. The additional displacement of the lid during secondary bending is 2.6 - 1.3 = 1.3 in Decelleration Forces - It was calculated that the initial crushing caused a decelleration force of 5.0 g's. Calculate the decelleration forces of secondary bending since this is the shortest distance travelled in any of the phases discussed. ( )(E) = ( 0,270)(386.4) = 60.41 in/sec v= W 53,005 v = 30.2 in/sec av8 at = AX/v = 1.3/30.2 = 0.043 see 2 2 a = AV/At = 60.41/0.043 = (1,403 in/sec )/(386.4 in/sec ) = 3.63 g. This indicates that the maximum decelleration force during a 12 inch drop on a short flat corner on the lid is 4.83 g's. This does not exceed the g forces calculated in the drop on a long flat. _ _ _ _ _ _ _ _ _ _ _ _ _
PROPRIETARY DATA 3.2.6.6 Side Drop on Ratchet Binder The cask is dropped on its side. The energy is assumed to be absorbed entirely on the lid edge. This is the worst case with respect to the ratchet binders in that their load consists of the waste and the lid weight. Total energy = 636,060 in-lb 2 Initial velocity 636,060 in-lb = 1/2 mv v = 96.3 in/sec Let d = depth of crush Volume of steel required to absorb energy 3 636,060 in-lb/36,000 psi = 17.66 in W = d Tan 67.5 o d oc % 2" thick m 1. Volume of steel described by [2(1/2 dw)(2 in)] + [(1 in)(d)(2 in)] = 4.82 d2 + 2d 2 Final depth = 4.82 d + 2d = 17.66 ina d = 1.72 in As shown on Table 3-2, the highest g force exerted on the lid is 11.83 g, say 12 g's. libnWI
n l l 1 l TABLE 3-2 Energy Energy Incremental d Vol. Absorbed Remaining Velocity Time Decelleration l (in) (in ) (in-lbs) (in-lb) (in/sec) (sec) (in/sec ) (g's) Z 3 (based on avg. velocity) 0 0 0 636,060 96.3 0.1875 0.54 19,600 616,460 94.8 0.00196 765 1.98 l 0.375 1.43 51,400 584,660 92.33 0.0020 1,235 3.2 f 0.5 2.21 79,380 556,680 90.09 0.00137 1,635 4.23 I 0.75 4.21 151,605 484,455 84.04 0.0029 2,086 5.4 1.0 6.82 245,520 390,540 75.46 0.0031 2,767 7.16 1.25 10.03 361,125 274,935 63.31 0.0036 3,375 8.73 1.5 13.85 498,420 137,640 44.8 0.0046 4,023 10.4 1.72 17.66 636,060 0 0 0.0098 4,571 11.83 = W 3> lc W 4:3 lkm t
PROPRIETARY DATA If we conservatively assume that the total payload and lid weight to be solely reacted by the binders then each must carry the following: p _ (18,305 lb payload + 5,903 lb lid) 12 g's 7 binders P = 41,500 lbs/ binder The ratchet binders are designed to take a load greater than those calculated for a top corner drop on a long flat (69,708 lb). Therefore, the binders will withstand a load of 41,500 lb from a side drop. 3.2.7 Penetration Impact from a 13 pound rod will have no effect on the pack-age. 3.2.8 Compression This requirement is not applicable since the package exceeds 10,000 pounds. CONCLUSION From the analysis, it can be concluded that the HN-100 Series 3A Cask is in full compliance with the requirements set forth in 10 CFR 71 for Type "A" Packaging. 8lU00iTN
PROPRIETARY DATA 4.0 THERMAL EVALUATION The HN-100 Series 3A casks will be used to transport waste primarily from nuclear electric generating plants. The principal radionuclides to be transported will be Cobalt 60 and Cesium 137. The shielding on the cask will limit the amount of these materials that can be transported as follows. Specific ( Total ( ) Gamma Isotope Energy Activity Activity Mev pCi/ml Ci Cobalt 60 1.33 5.0 23.2 Cesium 137 0.66 140.n 650 (1) Based on cement solidification waste and 10 mR at six feet from cask. (2) Based on 164 cubic feet of solidified material. With the maximum amount of these materials that can be transported in the HN-100 Series 3 cash. the heat generated by the waste will be as follows: Heat Total Generation Activity Total Heat (Watts / Curie) (Curies) (Watts) (BTU /HR) Cobalt 0.0154 23.~ 0.35 1.19 Cesium 0.0048 650 3.14 10.7 The weight of waste per container will be about l',500 pounds. Based on a specific heat of 0.156 BTU per degree F., 2,730 BTU's or over 10 days with cesium would be required to heat the waste one degree Fahrenheit. Ac-cordingly, the amount of heat generated by the vaste is insignificant. _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ____- _______________
l c PROPRIETARY DATA 5.0 CONTAINMENT 4 The shipping cask is a vessel which encapsulates the radioactive material and provides primary containment and isolation of the radioactive material from the atmosphere while being transported. The cask is an upcight circular cylinder composed of layers of structural steel with lead for radiation shielding, between the steel sheets. The lamina are of 3/8 inch inner steel, 1-7/8 inch of lead shield and a 7/8 inch outer steel shell. The heavy steel flange connecting the annular steel shells at the top provides a seat for a Neoprene gasket seal used to provide positive atmospheric isolation when the lid is closed by tightening the eight (8) ratchet binders which are equally spaced at 45* intervals on the outer circumference of the cask. The shield plug is located in the center of the cask lid, has a Neoprene gasket and is bolted to the outer portion of the lid with 0 equally spaced 3/4 inch studs on a 20-7/8 inch diameter circle. There is a drain plug in the base of each IIN-100 Series 3A cask consisting of either 1/2 in. or 1-1/4 in. pipe plug, depending on the unit. 5.1 Primary Lid Gasket Determine the amount of compression of the primary lid gasket due to tightening of the ratchet binders. Gasket 0.D. = 80 inches I.D. = 78.5 inches 2 - 39.25 ) = 186.7.1 in2 2 2 - Ri ) = n(40 2 Area = n(Ro Gasket is equivalent to 3/8 inch thick by 3/4 dach wide Durometer 40. Based on past experience from the manufacturer, a 100 pound force exerted on the handle of the ratchet binder will develop about 3,500 pounds of tension in the binder. Therefore, force downward on lid compressing the gasket (8 binders)(3,500 lb/ binder) + 6.000 lb lid weight = F = 34,000 lb Equivalent pressure of gasket = - = 34,000 lb/186.7 in2= 182.1 lb/in2 As shown on Appendix D-1, the comprecsion of the gasket due to tighten-ing of the ratchet binder is 20% of the gasket thickness, or about 3/32 inch. 5.2 Shield Plug Gasket Similarily, the compression for the shield plug is calculated. Based on the stud torquing procedure for the shield plug, the minimum torque value is 80 ft-lb. llbudtT9??
PROPRIETARY DATA The gasket dimensions are 22.75 in. OD, 20.25 in. ID, and 3/8 in. thick. The gasket is equivalent to a Durometer 50. 2 2 2 2 Area = n(Ro2 - Ri ) = n(11.375 - 10.125 ) = 84.43 in Force downward on lid is the sum of the weight of the lid plus the force of the studs (P). P= = (80 ft/lb)(12 in/ft)/(0.15)(0.75 in) P = (8,533 lb/ stud)(8 studs) = 68,266 lb W = 400 lb Total force = 68,266 + 400 = 68,666 lb Pressureongasket=f=68,666lb/84.43ta2 = 813.3 psi As shown on Appendix D-2, the compression of the shield plug gasket is 25% of the initial thickness or 1/8 inch. 5.3 Seal with Internal Pressurization The inner steel shell is designed to act as a pressure vessel when the cask lid is in place and tightened. As shown in Section 3.2, the cask will withstand an internal pressure c f 7.5 psig as required by 10 CFR 71, Appendix A. As described in Seciton 1.0, the nature of the waste being transported is such that phase change or gas generation which may over pressurize the cask, will not occur. The stepped flange surface at the end of the cask body has been designed to minimize effects of columnated radiation streaming and problems associated with gasket damage during impact. If the cask is pressurized to 7.5 psig, the resultant force on each ratchet binder (as calculated in Section 3._) is 4,113 pounds. The resultant strain on the steel ratchet binder (1-3/4" diameter) is: 6 P/AE = (4,113)/(1.76)(30 X 10 ) = 0.000078 in./in. P = 4,113 lb 2 A = Area of 1-1/2" minor diameter = 1.76 in 6 E = Youngs Modulus = 30 X 10 and for a 24 inch long binder, total strain is: (24 in.)(0.000078 in./in.) = 0.0018 in This is less than 2% of the initial compression of the gasket. - - - - - - - - - - - - - - - - l
I l PROPRIETARY DATA The full compression of the gasket from a free drop on the lid is prevented tv a 1/8 inch thick and 1/2 inch wide steel ring running the circumference of the lid. If the lid is compressed more than 1/4 inch, the steel will absorb the energy of the fall. 5.4 Gasket Compression Test A compression test to check resiliency was done on Items 4 and 5 on Drawing STD-02-020, the primary lid and secondary shield plug gaskets. The samples were each 1 inn by 3/8 inch thick made of Durometer 40 neoprene and Durometer 50 neoprene respectively. Each sample was put in a compression device and compressed. The final results indicated 2 Durometer 40 that it required about 4,500 lb to compress the 1 in sample to a thickness of 1/8 inch. After removing the sample from the Similarly,the1iglereturnedtoitsoriginalthicknessof3/8 inch. test stand, the sa n Durometer 50 sample was compressed to 1/8 inch thick, and it required 10,000 lb. It also returned to its original thickness when the load was removed. The test compressed each gasket material 66 percent of its origirial height and each survived. Tim spacer rings have a thickness to 1/8 inch which limits the compression of the gasket which is the value demonstrated in the test. 5.5 Warping of Covers The possible distortion of the cover and possible leakage due to dis-tortion has been addressed on a cask of nearly identical design fea-tures. A cask having an octagonal cover secured by ratchet binders was dropped on the extended corner by Nuclear Packaging, Inc. The identification number of the package which was dropped is 71-9130. The package, Model No. 50-256, had a weight of 17,160 lbs and was loaded with a liner containing sand with a weight of 4,200 lbs for a gross veight of.'3,360 lbs. The package was dropped on an essentially unyielding surface from a height of 46 inches. The package was pres-sure tested before and after the drop test and no leakage was de-tected. The deformation of the torner subjected to the drop test is shown below: 0" 1" 2" 3" 4" 5" 6" 6.75" I n i o Corner j Defermation f f / / 9 @E9$$ -o o o o o o h00bkEI{
PROPRIETARY DATA The energy absorbed in dropping a 23,360 lb package from 46 inches is 1.07 x 100 in-lbs. The energy to be absorbed in a one foot drop of an 5 HN-100 Series 3A cask is 12 x 53,005 = 6.36 x 10 in-lbs or less than 60 percent of the unit tested. The covers are the same thickness and the overhang of the corners are approximately the same. Accordingly, the HN-100 Series 3A should ex-perience less deformation. All of the deformation occured outside of the impact limiter and no deformation of the cover was found in the areas which could affect the seal. ___-
L PROPRIETARY DATA 6.0 OPERATING PROCEDURES Customers that use the HN-100 casks are supplied a copy of the Rad Services Manual. This manual describes the services that will be supplied and con-tains a section on operating procedures. Included in this manual are the weight limitations, and type and quantity of licensed material limitations. The operating procedures describe the inspection of the trailer and cask upon arrival at the site, the opening, loading, closing procedures, and the forms that need to be filled out prior to the cask leaving the customer's site. An example is shown in Appendix A of this document. This is all in accordance with Subpart D to 10 CFR 71. Inspections performed under the operating procedure are done by the custo-mer prior to loading the cask, by the driver prior to leaving the site, at scheduled stops during transit, and after arriving at the cosignee's site. Inspection includes that cask has not been rigificantly damaged, closure of the package and any sealing gaskets are present and free of any defects, checking of the maximum loose and fixed contamination levels on the cask, and that the cask has been loaded and closed in accordance with written procedures. This is all in accordance with Section 71.54 of Title 10. Radioactive Shipment Record describing the shipment and giving the informa-tion required by Section 71.62 of the Title 10 are required to be filled out in triplicate. One copy is telecopied to HITTHAN prior to shipment leaving the site and a copy is mailed to HITTMAN as soon as possible after the shipment leaves the site. The other two copies accompany the shipmcat to the cosignee. An example is contained in Appendix B of STD-R-02-001. , libadfDiff
PROPRIETARY DATA 7.0 ACCEPTANCE TESTS & MAINTENANCE PROGRAM The HN-100 Series 3A casks have been in service from four to six years (as HN-100 Series 2 cask) have been subject to the requirements of Subpart B of 10 CFR 71. The materials used in the iriginal construction and in the modification of these casks are specified under ASTM and ASME codes. Welder qualifications and weld procedures are in accordance with ASME or equivalent codes. The modifications to these casks will be performed using the same pro-cedures and standards used in the construction of the RNDC HN-100 Series 3 ci,s ks. Specifically: Welding qualifications and weld procedures vill be in accordance with o ASME Code, Section IX. o The tie-down lugs, lift lugs and ratchet binder lugs will be removed from the cask body and lid. Reinforcing plates with lift lugs and tie-down lugs will be installed o on the cask. Ratchet binder lugs will be installed on the cask body and lid and the o present ratchet binders will be replaced with larger units. Upon completion of the modifications, a gamma scan will be conducted o of the cask wall in those areas which may have been affected by weld-ing. Acceptance criteria fo/ the gamma scan will be based on maintaining a o nominal lead thickness of 1-3/4 inches with no individual reading greater than a factor of three greater than the average readings for the overall cask. o After modifications are complete, the cask assembly will be subjected to a pneumatic pressure test of 8 psig using the test fixture and procedures contained in Appendix E.1 While under pressure, seating surfaces and gaskets will be checked for leakage. If leakage is fcund and cannot be eliminated, the cask and/or covers will be modified or replaced to attain a satisfactory seal repeatedly. The alignment of the ratchet binders and lugs will be checked to assure o that the binders opecate smoothly and without excessive friction. The gaskets will be placed on the cask body and shield plug opening, o the lid and shield plug will be installed in accordance with the operat-ing procedures and checked to assure that the seals have been compressed uniformly. _ _ _ _ _ _ _ _
c PROPRIETARY DATA The modifications of the HN-100 Series 3A cask will be implemented and documented under aa approved Quality Assurance program in accordance with the renuirements of 10 CFR 71, Appendix E.1 If any cask is modified in the future, undergoes extensive repair or is in-volved in an accident which might conceivably damage the sealing surfaces or closure devices, the cask assembly will be subjected to a pneumatic pressure test of 8 psig. Cask maintenance and repair is controlled by the Quality Assurance Program. The casks and trailers undergo a routine technical inspection at least once every four months. These inspections involve checking cask for contamina-tion, damage to interior or exterior, gaskets, studs, signs and placards, shielding and tiedowns. These inspections are covered by Cask Maintenance and Repair procedures. An example is shown in Appendix C.t 1STD-R-02-001 I llbudfdiff
i d APPENDIX A Cask Handling Procedure leinc-o-oot-2/3/3A ilDAVR
=.... Pag 2 1 of 14 4 PROJECT COVER SHEET Document Title Cask Handling Procedure Project Document Number HNDC-0-001-2/3/3A, Rev. 5 for HN-100 Series 2, Series 3 and 3A Shielded Transport Casks Hittman Nuclear & Development Corporation i 9190 Red Branch Road { Columbia, Maryland 21045 l l l l Ref: Std. Doc. N/A Rev. --- l 006M D
Spec. No. Procedure No. IINDC-0-001-2 /3 Page 2 of 14 r IlEVISION LOG llEV. DATE , ENGINEEITING Q. A. PItOJ. MGIt. ECN # [ 2 1/17/80 ,,, [ ,LhkC A4 %.. wr, <-s-8 c ' -" " 4 9-10-82 $ b L,b J-/' 82-167 --. w, 5 11-22-82 82-261 ~ \\L / v e e
t ~ k Procedure HNDC-0-001-2/3/3A Page 3 of 14 1. PURPOSE The purpose of this procedure is to provide. instructions for loading / unload-ing the HN-100 Series 2, Series 3 and 3A radioactive waste shipping casks. II. RESPONSIBILITY It is the responsibility of the user of 'a United States Nuclear Regulatory Commission (USNRC) certified package (cask) to assure the following: 1. He has the Certificate of Compliance for the cask and all referenced documents. 2. He is a registered user of the certified cask. 3. Under his Quality Assurance Program, the cask is inspected to verify its compliance with the terms and conditions of the Certificate of Compliance. 4. The cask is loaded and closed in accordance with an apprcpriate written procedure. 5. The cask is loaded in accordance with the Certificate of Compliance. 6. The shipment meets all the Department of Transportation, U.S. Nuclear Regulatory Commission, Burial Site Disposal Criteria and Burial Site License requirements. NOTE: If there is a problem meeting any of the above requirements, immediately notify the regional HNDC Operations Office. II
I. PROCEDURE
1.0 When ordering the cask, assure the following: 1.1 Waste to be shipped in the cask is either Low Specific Activity 49 CFR 173.389(c) or Type A quantities of Normal or Special Form 49 CFR 173.389(d), 49 CFR 173.389(g) and 49 CFR 173.389 (1). 1.2 Burial Site Disposal Criteria and/or Licenses and current copies of 10 CFR and 49 CFR are in your possession. 1.3 Waste is packaged or will be packaged in an acceptable manner in accordance with the-Department of Transportation (49 CFR), U.S. Nuclear Regulatory Commission (10 CFR), and the applicable burial site require-ments (Burial Site Disposal Criteria and/or Licenses). 1.4 Certificate of Compliance USA /9079/A (Series 2) and USA /9151/A (Series 3 and 3A) and all referenced documents are in your possession and your. site is a registered user of the cask. IID00tPTff
Procedure HNDC-0-001-2/3/3A Page 4 of 14 1.5 Your site has an approved U.S. Nuclear Regulatory Commission Quality Assurance Program in accordance with 10 CFR 71.51. NOTE: If there is a problem assuring any of the above, immediately notify the regional HNDC Operations Office. 2.0 Receipt Inspection 2.1 Survey the empty cack and the vehicle to determine the maximum loose and fixed contamination levels. 2 Loose contamination levels should be gess than 2,200 DPM/ 100 cm Beta-Gamma and less than 120 DPM/100 cm Alpha. Fixed contamination levels should be less than 0.5 mrem /hr. NOTE: Fixed contamination greater than 0.5 mrem /hr but less than 50 mrem /hr require the cask to have a Yellow II label. Under such conditions the empty cask must be a Radioactive Shipment and be accompanied by properly completed Radioactive Shipment Records. NOTE: If cask is received with contamination levels in excess of those above, immediately notify the regional HNDC Operations Office. 2.2 Inspect Tiedowns 2.2.1 Inspect tiedown lugs and shackles on cask and trailer for cracks and wear which would affect their strength. 2.2.2 Inspect tiedown cables to assure they are not loose or damaged (crimped, frayed, etc.). 2.2.3 Inspect tiedown ratchets / turnbuckles to assure they are in proper working condition. NOTE: If there is a problem with any of the items inspected, immediately notify the regional HNDC Operations Office. 2.3 Inspect Cask 2.3.1 Inspect ratchet birders that hold lid to body of cask to assure they are in p oper vorking condition. 2.3.2 Inspect retaining pins and locking pins for cracks and wear which would affect their strength. 2.3.3 Check to assure that cask lid (primary lid and shield plug) lifting lug covers are with the cask. 2.3.4 Check to assure drain line plug (HN-100 Series 2 and 3A only) has been properly installed. 4
~ Procedure KNDC-0-001-2/3/3A Page 5 of 14 2.3.5 ~ Remove cask lid in accordance with step 4.1. 2.3.6 ' Inspect primary lid gasket for cracks or tears which would affect proper sealing. NOTE: Cask must be properly sealed prior to shipment. 2.3.7 Inspect interior of caek for standing water. NOTE: Water must be removed prior to shipment. 2.3.3 Inspect interior of cask for obstructions to loading. 2.3.9 Inspect interior of cask for defects which might affect the cask integrity or shielding afforded by cask. 2.3.10 Inspect the shield plug holddown nuts to assure they are all present and not damaged. 2.3.11 Unless. it can be verified through other means, verify that the shield plug gasket has no cracks or tears which would affect proper sealing as follows: 2.3.11.1 Remove the shield plug from the primary cask lid in accordance with steps 4.2.3.6, 4.2.3.7 and - 4.2.3.8. ,t, - 2.3.11.2-Inspect the shield plug holddown studs for damage. 2.3.11.3 Inspect the shield plug gasket for cracks or tears which would affect proper sealing. NOTE: Cask must be properly sealed prior to shipment. ~ I 2.3.12 If loading drums, install shield plug (if removed) onto primary lid in accordance with steps 4.2.3.11, 4.2.3.12 and 4.2.3.13 and proceed to step 4.2.1 or 4.2.2. 2.3.13 If loading preloaded liners, install shield plug (if' removed) onto primary lid in accordance with steps 4.2.3.11, 4.2.3.12 and 4.2,3.13 and proceed to step 4.2.4. 2.3.14 If loading waste into liner inside cask, proceed to step 4.2.3 (omitting steps 4.2.3.6, 4.2.3.7 and 4.2.3.8 if shield plug was removed). NOTE: If there is a problem with any of the items inspected above, immediately notify the regional HNDC Operations Office. 3.0 Removal of Cask from trailer llD00fD5f?
Procedure EKDC-0-001-2/3/3A Page 6 of 14 NOTE: If it is necessary to remove cask from trailer proceed as follows: 3.1 If cask is equipped with rain-cover, and it has not been removed, remove the rain-cover from the cask. 3.2 - Loosen ratchet binders as necessary to remove pins from shackles at cask end of tiedown system. 3.3 Remove pins from shackles. 3.4 Loosen cask shear blocks as necessary. 3.5 Using the four (4) cask lift lugs and suitable rigging (minimum sling length of 58 in.), lift cask off trailer and place cask in proper position for loading. Cask weight - HN100 Series 2 - 33,800 lbs. HN100 Series 3 - 35,290 lbs. HN100 Series 3A - 34,700 lbs. NOTE: Do not use cask lid lifting lugs to lift cask. 4.0 Loading Cask 4.1 Remove the primary full diameter cask lid as follows: 4.1.1 If cask is equipped with a raincover, and it has not been removed, remove the raincover from the cask. 4.1.2 Disconnect the cask lid from the cask as follows: 4.1.2.1 Release the ratchet-binder handle from its storage position. l 4.1.2.2 Engage the flip block to the cprocket wheel in the-direction necessary to loosen the ratchet binder (see Sketch 1). 4.1.2.3 Loosen the ratchet binder by pulling the handle in the appropriate direction. 4.1.2.4 Remove the ball-lock pin by depressing the top of the pin and pulling the pin through the hole in the threadless bolt (see Sketch 2). 4.1.2.5 Remove the threadless bolt by pulling the bolt through the holes in the upper ratchet binder connector and lid closure lug (see Sketch 2).
Procedure HNDC-0-001-2/3/3A Page 7 of 14 E cAsg yD RATU4ET S PROCKET Bl4 pER i WHEEL LID 4 -y,g ANDLE ( Gg lla gsuee ^ t C Pig 7 THREADu!S g HANDLE FLIP Bot.T STORAGE Bt.ocK V WPER. b RATCHET POSITtog blNOGR N = SKETCH 1 SKETCH 2 4.1.3 Remove the three (3) cask lid lifting lug covers. 4.1.4 Using the three (3) lifting lugs on the cask lid to accom-modate suitable rigging and exercising caution in the handling of the cask lid due to possible contamination of the underside of the lid, remove the cask lid. 4.2 Loading can be accomplished by one of the following methods. 4.2.1 In cask loading of seven (7) drum pallets: NOTE: Review Pre-release Checklist (Attachment 1) or similar site document and shipping papers to assure that inspections required on the checklist or site document are performed during the cask loading process as necessary and that the information required on the shipping papers is determined as necessary. l l 4.2.1.1 Using the slings provided and exercising caution i in the handling of the pallet due to possible l contamination, remove the top pallet from the l cask. Pallet Weight - 750 lbs. 4.2.1.2 Exercising caution to avoid placing drums on the pallet lift slings, load seven (7) drums on the pallet in the cask. (See Sketch 3 for dr place-m u ment on pallet.) 1lh0 W M
Procedure KNDC-0-001-2/3/3A Page 8 of 14 Maximum Drum Weight - 800 lbs. RIBS RlBBED PM.LET ~ D RtAMS NOTE 1: Newer pallets are ribless, but should be loaded with the same drum pattern. NOTE 2: For maximum shielding, load higher dose rate drums in the center position and the positions toward the front and rear of the trailer. 4.2.1.3 Place the top pallet into the cask. Pallet Weight - 750 lbs. 4.2.1.4 Exercising caution to avoid placing drums on the pallet lift slings, load seven (7) drums on the pallet in the cask (see Sketch 3). l 4.2.1.5 Proceed to Step 4.2.5. 4.2.2 Loading the Seven (7) Drum Pallets Outside the Cask NOTE: Review Pre-release Checklist (Attachment 1) or similar site l document and the shipping papers to assure that inspections required on the checklist or site document are performed during the cask loading process as necessary and that infor-mation required on the shipping papers is determined ' as l necessa ry. 1 4.2.2.1 Using slings provided and exercising caution in the handling of the pallet due to possible con-tamination, remove both the pallets from the cask. Pallet Weight - 750 lbs. l 4.2.2.2 Load seven (7) drums onto each pallet (see Sketch 3). Maximum Drum Weight - 800 lbs. 4.2.2.3 Lift one of the loaded pallets and place it inside the cask. For maximum shielding, assure proper orientation of pallet (see Note 2 of Sketch 3).
Procedure HNDC-0-001-2/3/3A Page 9 of 14 Maximum Loaded Pallet Weight - 6400 lbs. 4.2.2.4 Lift the other loaded pallet and place it inside the cask on the top of the first pallet. For maximum shielding, assure proper orientation of pallet (see Note 2 of. Sketch 3), Maximum Loaded Pallet Weight - 6400 lbs. -4.2.2.5 Assure easy access to the pallet lifting slings for removal of pallet at burial site. 4.2.2.6 Proceed to Step 4.2.5. 4.2.3 In Cask Loading of Liner NOTE: Review Pre-release Checklist (Attachment 1) or similar site document and shipping papers to assure that inspections required on the checklist or site do.ument are performed during 'the cask loading process as necessary and that the information required on the shipping papers is determined as necessary. 4.2.3.1 If necessary remove cask from trailer in accordance with Steps 3.1 through 3.4. 4.2.3.2 Using the slings provided, place liner in the cask. . Empty Liner Weight - Plain .1540 lbs. Mixer - 2050 lbs. Underdrain - 2140 lbs. Mixer /Underdrain - 2650 lbs. 4.2.3.3 Install shims / shoring between liner and cask as necessary to secure in position. 4.2.3.4 Using the three (3) lifting lugs on the cask lid to accommodate suitable rigging, place cask lid on cask-using alignment pins to assure proper position-ing. Lid Weight - 5,600 lbs. 4.2.3.5 Secure the cask lid to the cask as follows: 4.2.3.5.1 Install the threadless bolt through the upper ratchet binder connector and the lid closure lug (see Sketch 2). BibadfD19f
Precedure HNDC-0-001-2/3/3A Page 10 of 14 4.2.3.5.2 Install the ball lock pin by press-ing down on the top of the pin and inserting the pin through the hole in the threadless bolt. 4.2.3.5.3 Tighten the ratchet binder by. engaging the flip block to the sprocket wheel. 4.2.3.5.4 Tighten ratchet binders so lid is snug against gasket without com-pressing it. The sequence of tightening the binders consists of tightening a binder, then the binder 180* opposite, then the binders 90* to the original, etc. 4.2.3.5.5. Once all binders are snug, hand tighten all binders to compress the gasket. Use the same tightening sequence outlined in 4.2.3.5.4. 4.2.3.$.6 Visually inspect the primary lid to assure uniformity of closure. 4.2.3.5.7 Disengage the flip block and rotate and secure the handle to its storage position (see Sketch 1). 4.2.3 5.8 Install the three (3) cask lid lifting covers. 4.2.3.6 Remove the 8-3/4" shield plug holddown nuts. 4.2.3.7 Remove the shield plug lifting lug cover. 4.2.3.8 Exercising caution due to possible contamination of the underside of the shield plus, remove the shield plug. Shield Plug Weight - 400 lbs. 4.2.3.9 Load the waste into the liner through the shield plug opening. 4.2.3.10 Instell the liner lid, plugs or caps onto the liner. 4.2.3.11 Place the shield plug on the cask using the shield plug guide pins for proper positioning.
i Procedure HNDC-0-001-2/3/3A Page 11 of 14 NOTE: Care should.be taken to avoid damage to the gasket. Shield Plug Weight - 400 lbs. 4.2.3.12 Secure the shield plug by installing and tightening the 8-3/4" shield plug holddown nuts in accordance with Torquing Procedure HNDC-0-1002/3/3A. 4.2.3.13 Install the shield plug lifting lug' cover. 4.2.3.14 If cask is equipped with raincover and the cask was not removed from trailer, install raincover. 4.2.3.15 Proceed to step 5.0 of this procedure if cask was removed from trailer. Otherwise, proceed to step 6.0. 4.2.4 Loading preloaded liner. NOTE: Review Pre-release Checklist :(Attachment 1) or similar site document and -the shipping papers to assure that inspections required on the checklist or site document.are performed. during the cask loading process as necessary and that infor-mation required on the shipping papers is. determined as necessary. 4.2.4.1 If necessary, remove cask from trailer in accord-ance with step 3.1 through 3.4. 4.2.4.2 Assure lid, plugs or caps are installed on liner. 4.2.4.3 Using the lifting slings provided, place liner into the cask. Fr.11 Liner Weight - 17,715 lbs. maximum 4.2.4.4 Install shims / shoring between liner and cask as necessary to secure in position. 4.2.5 Install the primary full diameter cask lid as follows: 4.2.5.1 Using the'three (3) lifting lugs on the cask lid to accommodate suitable rigging, lift the cask lid and place it on the cask using the alignment pins for proper positioning. Lid Weight - 5,600 lbs. 4.2.5.2 Secure the cask lid in accordance with step 4.2.3.5. hb0dfhSbI
Procedure HNDC-0-001-2/3/3A Page 12 of 14 4.2.5.3 Install tamper proof seals. 4.2.5.4 If cask is equipped with raincover, install rain-Cover. 5.0 If cask was removed from trailer, proceed as follows: 5.1 Using the four (4) cask lift lugs and suitable rigging lift cask and place cask in proper position on trailer. See Sketch 4 for proper orientation. Series 2 - Loaded Cask Weight - 49,000 lbs. Series 3 - Loaded Cask Weight - 53,005 lbs. Series 3A - Loaded Cask Weight - 48,700 lbs. cA5LES " *W Yo cY ~ Touch uSE. MATCH MARKS ON SIDE os cask" $. Alp DEClC OG TKAILEK IF PRESEnlT SKETCH 4 5.2 Install shackles through the end of the tiedown cables and attach to cask tiedown lugs by screwing pin through shackle and hole in lug. 5.3 Tighten the cask shear blocks to secure the cask in position. 5.4 Tightcn ratchet binders / turnbuckles as necessary to secure cask on trailer. 5.5 If cask is equipped with raincover, install raincover. 6.0 Prepare cask and vehicle for shipment as follows: 6.1 Perform radiation surveys of cask and vehicle and complete the necessary shipping papers, certification, and Pre-release Checklist'(Attachment 1) or site equivalent. 6.2 Placard vehicle and label cask as necessary.
Procedure HNDC-0-001-2/3/3A Page 13 of 14 7.0. Unloading Cask 7.1 Survey the cask and trailer in accordance with applicable site require-ments. 7.2 Remove cask lid in accordance with step 4.1 of this procedure. 7.3 Exercising caution due to possibly high dose rate,' connect slings from liner or pallet to a suitable lifting device. Series 2 - Maximum Liner Weight - 14,000 lbs. Series 2 - Maximum Pallet Weight - 6,400 lbs. Series 3 - Maximum Liner Weight - 17,715 lbs. Series 3 - Maximum Pallet Weight - 6,400 lbs. Series 3A - Maximum Liner Weight - 14,000 lbs. Series 3A - Maximum Pallet Weight - 6,400 lbs. 7.4 Exercising caution due to' possible high dose rate, lift liner or pallet clear of the cask and place in disposal area. NOTE: Care should be taken to avoid damage to lid gasket. 7.5 Repeat steps 7.3 and 7.4 for second pallet. 7.6 Install cask lid in accordance with step 4.2.5 of this procedure. 7.7 Survey the cask and trailer for release in accordance with applicable site requirements. I I i l l l i llD00/DTfl
a Precedura HNDC-0-001-2/3/3A Page 14 of 14 PRE-RELEASE CHECKLIST Date Shipment No. Transport Co. Time of Arrival at Site Time of Departure from Site Initial 1. Inner Container (s) Sealed 2. Inner Container (s) Secured in Place 3. All gaskets and gasket sealing surfaces inspected and free of defects 4. Cask Lid and/or Shield Plug Closure Bolts Torqued / Ratchet Binders i Tightened 5. Tamper proof Seal Inspected 6. Drain Line Plug Installed
- 7. ~ Lift Lugs Covers Installed 8.
Cask Tiedowns Inspected l 9. Cask Properly Labeled 10. Vehicle Properly Placarded l 11. Surveys Completed and Recorded 12. Shipping Papers Properly Filled i Out and Signed i l Signature l l Title Date
P b 'J a_ c/ UNITED STATES g% NUCLEAR REGULATORY COMMISSION ~+ o WASHtNCTON, D.C. 20555 5
- j
%,...../ FCTC:RH0 JAN 311983 71-9151 MEMORANDUM FOR: Steve Scott, Chief Document Management Branch, TIDC THROUGH: Lea G. Walker, Section Leader Licensing Assistance Section, FCMS, NMSS FROM: Charles E. MacDonald, Chief Transportation Certification Branch, FCMS, NMSS
SUBJECT:
PROPRIETY DATA - PUBLIC DOCUMENT ROOM The following application and/or drawings marked "Proprietory" should be placed in the NRC Public Docun.ent Room: Hittman Nuclear and Development Corporation application dated November 23, 1982 and all associated safety analysis reports and drawings pertaining to m the Model No. HN-100 Series 3A package design. Authorization: Hittman Nuclear and Development Corporation letter dated November 23, 1982. Charles E. MacDonald, Chief l Transportation Certification Branch Division of Fuel Cycle and Material Safety, NMSS
Enclosure:
Hittman Nuclear and .~' Development Corp. application dtd 11/23/82 l 1 l}}