Text

Rev 8 to SAR for HN-100 Series 3 Radwaste Shipping Cask
	ML20151R158
Person / Time
Site:	07109151
Issue date:	01/07/1986
From:	; WESTINGHOUSE HITTMAN NUCLEAR, INC.
To:	;
Shared Package
ML20151Q988	List: ML20151Q984; ML20151R158; ML20265F816; ML20265F820; ML20265F824; ML20265F829;
References
	26310, STD-R-02-001, STD-R-2-1, NUDOCS 8602050418
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{{#Wiki_filter:7/- 9/51 Document Numb:r: Rav: Rev Data: WESTINGHOUSE STD-R-02-001 8 1-7-86 HITTMAN NUCLEAR p INCORPORATED

Title:

SAFETY ANALYSIS REPORT FOR THE HN-100 'd kg /c//h,. SERIES 3 RADWASTE SHIPPING CASK O'too/ 9)g // R Rev Date* Director Product QA Engr. Mgr. Mgr. ~ 7*

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1/7/86 'fn A7f) 85-232 - g k 6. WE This Mate iol is Prop letory to the in1 erests of H ttm an and Shol be Preserved in Confidence. O 9602050410 860109 PDR ADOCK 07109151 C PDR l l

- In the righthand margins, the light revision bars indicate rev. 7 and the heavy revision bars indicate rev.

8. (~ ~* On.each revised page, the most recent revision date appears in the upper righthand cornerl FORM-01(B) Page 1 of 148 Jh3/d

9 SAFETY ANALYSIS REPORT FOR THE HN-100 SERIES 3 RADWASTE SHIPPING CASK REVISION 8 Referencing 10 CFR 71 Type "A" Packaging Regulations i STD-R-02-001 Westinghouse Hittman Nuclear Incorporated 9151 Rumsey Road Columbia, MD 21045 O\\.) c Page 2 of 148 l

STD-R-02-001 Page 3 of 148 (:) i i NOTICE This Safety Analysis Report and the associated drawings are the property of Westinghouse Hittman Nuclear Incorporated, Columbia, Maryland. This material is being made available for the purpose of obtaining required certifications by the U.S. Regulatory Commission, enabling utilities and other firms producing radio-active waste to be registered users of equipment and services supplied by Westinghouse Hittman Nuclear Incorporated, and enabling equipment to be manu-factured on behalf of and under contracts with Westinghouse Hittman Nuclear Inco rporated. Parties who may come into possession of this material are cautioned that the information is PROPRIETARY to the interests of Westinghouse Hittman Nuclear Incorporated, is not to be reproduced from this report and the associated drawings, or facsimiles made of these drawings without the express written consent of Westinghouse Hittman Nuclear Incorporated. \\.)

STD-R-02-001 Page 4 of 148 1.0 PURPOSE The purpose of the following document is to provide the information and engineering analysis that demonstrates the performance capability and ~ structural integrity of the HN-100 Series 3 Radwaste Shipping Cask and its compliance with the requirements of 10 CFR 71, Section 71.21 and Appendix A.

2.0 DESCRIPTION

The HN-100 Series 3 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 3 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 diameter by 81-1/2 inches high with internal cavity of 75-1/2 inches inside diameter by 73-3/8 inches high. 2.1 Cask Body l The cask body is a steel-lead-steel annulus in the form of a vertical oriented, right cir:ular cylinder closed on the bottom end. The side walls consist of a 3/8 inch inner steel shell, a 1-7/8 inch thick concentric 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 a gasket seal and the required lid hold-down ratchet binders. Four liftina lugs are welded to the outer steel shell. ^% o (d =

STD-R-02-001 Pg ogl48 O 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 lifting device located at its center to facilitate handling. 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 (affixed 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 secured against movement relative to the cask with an internal struc-tural members such as bottoms and pallets. Basically, the internal's consist of the waste, containers 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. 2.7 Gross Package Weights The respective gross weights of the cask components and its designated radwaste loads are as follows: Cask body 29,030 pounds Closure lid 5,800 pounds f^; \\ Shield plug 370 pounds i ~ -.

STD-R-02-001 Page 6 of 148 JAN 7 tus Total cask (unloaded) 35,200 pounds HN-100 --Large container (s) l and waste 17,800 pounds i HN-100 - 55 gallon size containers (up to 14 drums 15,200 pounds of radioactive waste) HN-100 - 30 gallon size con-tainers (up to 18 drums of 11,000 pounds radioactive waste) 2.8 Radwaste Package Contents 2.8.1 Type and Form of Material The contents of the various internal containers can be process solids in the form of spent ion exchange resins, filter exchange media, evaporator concentrates, and spent filter cartridges. Materials will be either dewatered, 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- {p)T tainers with weights not exceeding 17,800 pounds. 3.0 DESIGN CONDITIONS 3.1 General Standards (Reference 10 CFR 71 Section 72.31) 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 Y compared to the minimum yield of that material. The safety factor is found by dividing the minimum ^ 1

STD-R-02-001 Page 7 cf 148 JAN 7 ggs O m* yield by the calculated stress. A safety factor greater than 1.0 is required for acceptability. (fy = 38,000 for ASTM A516 Grade 70). For the shell, steel having a minimun yield strength of 49,000 psi is specified. For the tie-down lugs, steel having a minimum yiel,d strength of 46,000 psi is specified. 3.1.3.2 Stresses in material due to shearing is analyzed using the " Maximum Energy - Distortion Theory" which states the shearing clastic limit is 1//3=57.7%ofthetensileelasticlimit!. As with 3.1.3.1, a factor of safety greater than 1.0 is required for acceptability. f, = (0.577)(38,000) = 21,926 psi for A516 Grade 70 3.1.3.3 Weld filler material rod is E70 Grade. Analysis is based on American Welding Society Structural Code D1.1-79. For fillet welds, shear stress on effec-tive throat regardless of direction of loading is 30% of specified minimum tensile strength of weld metal. For complete joint penetration groove welds with ten-sion normal to the effective area the allowable st ress () is the same as the base metal. 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. Since all welds have been non-destructively examined, weld efficiency is known to be greater than this. 3.1.4 Lifting Devices 3.1.4.1 Package Weight The package weights used for analysis are as follows: Empty package 35,200 pounds Payload: large con-tainer and waste 17,800 pounds Gross Weight 53,000 pounds 3.1.4.2 Cask Lifting Lugs Material is ASTM A516 Grade 70 with a minimum of 38 KSI Tension Yield Strength and 21,926 PSI Shear Yield Strength (57.7% of 38,000 PSI). 3 \\J Design and Behavior of Steel Structures, Salmon & Johnson, page 47. l

.. _....... ~. _.... STD-R-02-001 Page 8 of 148 JAh 7 ins t)Y Maximum Load 53,000 lbs. and four lugs are used to lift the cask. 53,000 lbs. X 3 g's/4 lugs = 39,750 lbs/ Lug (Vertical) f 2.5 TEAR OW o Sling Angle to Lift 45* Load 39,750 lbs/ Sin 45* 2.5" DIA. Load = 56,215 lbs. Stress = 56,215 lbs/[2 X (2"x2.5")] \\ " THK. 2 o = 5,622 psi I G" O 5,622 PSI << 21,926 F.S. = 3.9 TENSION 3.75"a Sling Angle to Lift 45* Load = 56,215 lbs. ,9' i 3 Stress = 56,215 lbs/(2" x 5.4926") w-i f 2.5 " c = 5,117 psi 3 5,117 PSI << 38,000 psi F.S. = 7.4 4.2426 - 1.25 + 2.5 = 5.4926

STD-R-02-001 Per,e 9 of 148 JAh 7 i385 TEAR 0% Vertical Lift Stress = 39,750 lbs/[2 X (2" x 2.5")] o = 3,975 psi 3,975 PSI << 21,926 F.S. = 5.5 TENSION Vertical Lift i Stress = 39,750 lbs/2" x (6"-2.5") o = 5,679 psi 5,679 PSI << 38,000 psi F.S. = 6.7 3.1.4.3 Cask Lifting Lug Welds All welds 3/4" fillet. Allowable shear stress on effective area of weld is 0.3 times nominal ten-sile strength of weld metal, E 70 rod, or (0.3)(70,000) = 21,000 psi. 2" WELD 775" WELD 6" WELD f-2. 5 Dia. lb l V k 6" 4 E 1 J e .mvesym -se / I G" WELD BOTH SDES L 2" WELD Q _ m ;.......-.. ~ _... ...,. ; 1.....

STD-R-02-001 P ge 10 of 148 I J4N 7 1986 (- U Vert ical Lif t Assume weld to be 85% efficient. Minimum throat of weld - (Sin 45')3/4" = 0.53" Weld strength per inch = 0.85(0.53" x 1")(21,000 psi) = 9460 lbs/in of weld Weld required = 39,750 lbs/9460 lbs/in. of weld = 4.2 in. of weld 4.2 in. << (2"+2"+12"+7.75"+6") F.S. = 7.1 3.1.4.4 Tiedown Lugs for Lifting Cask (inadvertant use) If it is assumed the entire load is carried by the 2 tiedown lug, Section Modulus = (bh )/6; where b=15.5 in., length of tiedown lug; and h = 2", thickness of tiedown lug (See Diagram page 23). Tiedown lug section modulus = 15.5 (2)(2/6) = 3 10.33 in Stress = (39,750 lbs)(3"/10.33 in ) = 8 11.5 KSI where 3" is the distance from the cask surface to the centerline of the hole. 11.5 KSI < 46 KSI F S. = 4.0. ) Weld required on lift lug 4.2 in. 4.2 < 12" F.S. = 2.9 Weld to tiedown lug Shear = 39,750 lbs/(2 x 15.5) = 1282 lbs per inch' l of weld compression or tension due to moment couple 39,750 lbs (3"/15.5 in) of tiedown lug = 7,694 l l lbs in/ inch of lug and moment couple with 2" thick ,~ lug 2"(3847 lbs) = 7,694 lb/in. l Moment of Rotation (39,750)(4.34)/15.5 in = 11,130 lb-in/in. lug For 2" lug, then 5565 lb/in Stress on weld to tiedown lug = 0 = /(1282)2 + (3847)2 + (5565)2 = 6886 6886 lbs per inch of weld < maximum of 9460 lb per inch of weld F.S. = 1.37

STD-R-02-001 Page 11 of 148 JAN 7 1986 3.1.4.5 Lift Lu2s with 45* Sling Angle The forces are 39,750 lbs vertical and 39,750 lbs l horizontal. These forces will be restrained by the 7.75 and 6.0 ioch 3/4 inch welds attaching the lift lugs to the reinforcing plate and the two 6.0 inch 3/4 inch welds attaching the lift lugs to the tiedown lugs. The shear stresses due to the vertical component of the force is: f = 39,750/(7.75+6.0+6.0+6.0)(0.75)(0.707)(0.85) = 39,750/11.6 = 3427 psi l The centroid of the welds is located as follows: y = [7.75(7.75/2)t6(6/2) + 6 (7.75) + 6 x 6] + (7.75 + 6 + 6 + 6) = (30.0 + 18 + 46.5 + 36) + 25.75 = 5.06 in. Total Moment = (2.75 in)(39,750 lb) = 109,313 in-lb Total Moment = Compressive Moment + Tensile Moment Compressive Moment = Tensile Moment 39,75O I b, Total Moment = 2(Tensile Moment) = 2f[(6 x 2.69 + 2.69 x 2.69 x 1/2 + 6 x 0.94 + 0._94 x 0.94 x 1/2)(0.75)(0.707)(0.85)] g ( } - 2f[(16.1 + 3.6 + 5.6 + 0.4)(0.75)(0.707)(0.85)] 2.75" --~ = 2f(25.7)(0.75)(0.707)(0.85) ~~ = 23.29f f = 109,313/23.29 = 4694 psi 7.75" 5.06" @3/4" Combined Stress = J(4694)2 + (3427)2 ^ 6"@3/4" = 5,812 psi o 0.34" F.S. = 9460/5,812 = 1.63 2.69" / Therefore, it can be safely concluded that the ( '6"@3/4n lug will not yield under a load equal to three times the weight of the package. 6"@3/4" -9 w -e e, er e * -s

STD-R-02-001 Page 12 of 148 \\> 3.1.4.6 Lid Lifting Lugs (Secondary and Primary) 1 A. Secondary Lid Lif ting Lug Material is ASTM A516 Grade 70 2" l = = Maximum Load 370 lbs. 6 Carried by one lug 1%" o ip h y TEAR OUT Area = 2[(1-1/2 - 15/16) - 7/32] 3/8 74" DI A. Area = 0.258 in2 3 " THK. /6 Stress = 3 g's x 370/0.258 in2 = 4,310 psi 4,310 psi << 21,926 F.S. = 5.0 j TENSION Area = (2.0 - 7/16) 3/8 = 0.586 in2 Stress = 3 X 370/0.586 = 1900 psi 1900 psi << 38,0C0 psi F.S. = 20 B. Secondary Lid Lif ting Lug Weld r r- ~ 1/2" fillet weld with allowable 21,000 psi Effective size Sin 45" (.5) = 0.353 Area of weld (2 + 2 + 3/8 + 3/8) 0.353 A = 1.68 in2 Stress = 3 X 370/1.43 0 = 660 psi 660 psi << 21,000 psi F.S. = 31.8 Therefore, the secondary lid lug is able to resist a load of three times its weeight without reaching a yield stress. C. Primary Lid Lifting Lugs Material is ASTM A516 Grade 70.

~. ..--..L.:.*L ~. . ----..:.-..-.=. STD-R-02-001 geIgof148 T's I Maximum L ad = Primary Lid 5800 lbs. I" TE Secondary Lid 370 lbs. 6170 lbs. h n A 4 16 2N" l*DIA. m n TEAR OUT (Vertical Lift) 6" = Area = 2 x [(2-3/4 1/2) - 1/2] 1 2 and stress = 6170/1.5 Area = 1.5 in o = 4,113 psi 4,113 psi << 21,926 ps'i F.S. = 5.3 TEAR OUT (45' Sling Angle) 2 2 s Area (short path) = ( J.75 +.75 ) 1 = 1.06 in Load = J(6170)2 X 2 = 8726 lbs. Stress = 1/2 (8726)/1.06 = 4,116 psi. 4,116 psi << 21,926 psi F.S. = 5.3 TENSION (Vertical Lift) 2 Area = (6 - 1) 1 = 5 in Stress = 6170/5 = 1234 psi 1234 psi << 38,000 psi F.S. = 30.8 TENSION (45' Sling Angle) Area (short path) = 42 X (1.25)2 - 1/2 = 1.2678 Stress = 1/2 (8726)/1.2678 = 3441 psi 3441 psi << 38,000 psi F.S. 11.0

STD-R-02-001 Pegg h g l48 ,-0 F. Primary Lid Lifting Lug Weld 1/2" weld at shear of 21,000 psi a) shear stress due to vert = horz component av = oh = 6170/(6+6+1+1)(Sin 45*)(0.5)(0.85) = 1467 psi b) Stress due to moment Total Moment = Compression Moment + Tensile Moment Compression Moment + Tensile Moment Total Moment = 2(Tensile Moment) = 2(2x3x(2/3)xfx0.5x0.707x0.85) = 2.4f f = 9,255/2.4 = 3856 psi Combined stress = 4(3856 + 1467)2 + 14632 = 5521 psi F.S. = 21,000/5521 = 3.8 Therefore, it can be concluded that the lift-ing lugs for the lid are more than adequate to resist a load at three times its weight. 3.1.4.7 Lifting Lug Covers Since the primary and secondary lid lif ting lugs are not capable of resisting the full weight of the package they will be covered during transit. 6 ._,,w

STD-R-02-001 Page 15 of 148 N,' 14 585 (rd 3.1.5 Tie Downs The tie down lug material is ASTM A516 Grade 70 with a minimum yield of 46 KSI and a 26.5 KSI usable shear (57.7% l of 46 KSI). The cask outer shell material has a minimum yield of 49 KSI, which will be specified for fabrication and certified by test of the material. A system of tie downs is provided as part of the package. They will be utilized as follows: 1 X l "Z O s A / i } 1 2, 10g # Worst case accident conditions Sg O '~

STD-R-02-001 Page 16 of 148 JAN 7 Ing - 53.9" u y j 12.0" I' h 69.8" 57.8" = 1 , 2.; ? 81.8" =, A i i h JL w. i< G-d g3 / 23.z " N g ys,_% = g .. n

2. -

VIEW AA r9

STD-R-02-001 Page 17 of 148 Jhd 7ggy; s 3.1.5.1 Cask Center of Gravity item Weight Arm Moment 42.9" = 1,510,080 lb.in. Cask 35,200 lbs. 37.7" = 671,060 lb.in. Liner / waste 17,800 lbs. 53,000 lbs. 2,181,140 Center of Gravity = 2,181,140/53,000 CG = 41.2 inch 3.1.5.2 Tie Down Forces Reference frame with respect to the trailer is shown on the tie down drawing (page 16). l-up - down Y; front - rear X; side - side Z accelerations: Y axis - 2 g's X axis - 10 g's Z axis - 5 g's Tie Down Lengths Long tie downs (low trailer attachment points) length = /63 2 = 101.0 in. 2 + 57.82 + 53.9 Short tie downs (high trailer attachment points) 2 = 99.2 in. length /60 2 + 53.9 2 + 57.8 Tie down tensions Tie down tensions resolved by vector direction Long tie down at tension T L 63 Along Y axis 101.0 L L 53 9 Along X axis 101.0 L L 57 8 Along Z axis 101.0 L L (2) .-_,.__~.en,. ~ - - v-

STD-R-02-001 whgi 7 M86(p e 18 of 148 Short tie down at tension T 3 Along Y axis 9 2 S S Along X axis T = 0.5432 T g g Along Z axis T = 0.5825 T g.2 3 g 10W Force (front-rear) Overturning (front-rear) due to 10W along X axis Overturuing moment (taken about the axis of rotation @ Point A) = 10(53,003 lb)39.2" = 20,776,000 in-lb Each of the two rear (or front) tie downs (cne (~ long and one short) must restrain half the above moment or 10,388,000 in-lb Tension in the long tie down 10,388,000 in-lb = (64")(0.5333 T ) g + (69.8")(0.6233 T ) g Tg = 133,801 lb Tension in the short tie down 10,388,000 in-lb = (64")(0.5432 T ) 3 + (69.8")(0.6047 T ) 3 T = 134,957 lb g i lO .l i .m

STD-R-02-001 P e 19 of 148 7 IN6 i Q V SW Force (side-side) Overturning (side-side) due to SW along 2 axis Overturning moment (taken about the axis of rotation @ Point-A) = 5 (53,000 lb) (39.2") = 10,388,000 in-lb 6 Each of two side tie downs (one long and one short) must restrain half the above moment or 5,194,000 in-lb Teasion in the long tie down 5,194,000 in-lb = (64")(0.5718 T ) l + (12.0")(0.6233 T ) g T = 117,845 in-lb g Tension in the short tie down 5,194,000 in-lb = (64")(0.5825 T ) 3 l + (12.0")(0.6047 T ) 3 T = 116,624 in-lb g 2W Force (up-down) Lifting (up) due to 2W along Y axis Lift = 2 (53,000 lb) - 53,000 = 53,000 lb l Each of two long and two short tie downs will carry a quarter of the load. 13,250 lb = 0.6233 T g Tg = 21,258 lb 13,250 = 0.6047 T3 T = 21,912 lb 3 f 6

_=_ - STD-R-02-001 Page 20 of 148 J44 7 IZ6 1 O t Total Tension Total tension with all forces acting simultaneously T = 133,801 + 117,845 + 21,258 i g I T = 272,904 lb j g i 1 .l T = 134,957 + 116,624 + 21,912 3 I T = 273,493 S i l i i I i i i O i j l 1 I 'l } } l I e t =l r t I 1 lO t t i I i i l l s' 1 -.$ ).....*T Y." 9-1-% [.'r y -'+ i'-'---'v---- 'vw w-vv1 wwe -w-v urv ww y-T w" +-*m-----'we--e--ww-we---www wr e = e m--w-

STD-R-02-001 Page 21 of 148 NN 7 1986 emb 3.1.5.3 Tie-down Lugs i The tie-down lugs are constructed of ASTM A516 Grade' steel having a minimum yield stress of 46 KSI and a maximum ultimate stress of 78 KSI. The following values are used in the design of the tie-down lugs. Tensile Yield = 46,000 PSI Allowable Bearing Stress = 0.9 Tensile Yield Strength Maximum Ultimate Tensile = 78,000 PSI Shear Yield = 0.577 x 46,000 = 26,542 PSI Shear Ultimate = 0.577 x 78,000 PSI = 45,006 PSI Allowable Shear Stress on Welds = 21,000 PSI Bearing Stress on Lug & Pin Maximum Load = 273,493 lbs Diameter of Hole = 2.5 inches Diameter of Pin = 2.25 inches ~ Thi g ess g Lug .0 inches 3 3 4 2 Projected Area of Pin = 3.0 x 2.25 = 6.75 in 1 273 493 Bearing Stress = 6575 = 40,517 psi F.S. = 41,400 + 40,517 = 1.02 Tear Out The shear plane associated with the projected area of the pin is shown as follows: 2 y = /1.252 1.125 = 0.545 o y2 = 1.25 - 0.545 = 0.705" ...2.._., .. - _ -..... - -.. ~.. - _ - . _.. _., _ _ _, _. ~,. _

STD-R-02-001 Pege 22 of 148 JAh 7 1986 l

1. 75 "

l 7 273,493 lb ~ 2.25" Dia. Pin 2.5" Dia. Hole 3.0" thick 273 493 Shear Stress = 2x3.0x1.75 ,047 psi F.S. = 26,542 + 26,047 = 1.02 Tencion i 273,493 lb. Stress = 273,493/3.0"(8.25"-2.5") ~ = 15,855 psi i [~ N 8.25" ( F.S. = 46,000 + 15,855 = 2.90 Reinforcing Plate Weld d 273,493 lb 2.25" Dia. Pin ~ \\e- \\ 2.5" Dia. Hole 6" 7" Thickness main lug 2 inches Thickness reinforcing plates 2 @ 1/2 inch Total thickness 3.0 inches Load on " reinforcing plate = 1/6 x 273,493 = 45,582 psi Length of weld = 6 + 7 + 7 = 20" t Area of shcar = 20 x 1/2 x Sin 45* x 0.85 = 6.01 in Shear = 45,582 + 6.01 = 7,584 psi F.S. = 21,000 + 7,584 = 2.77

STD-R-02-001 Pege 23 of 148 JAll 7 Igg 3.1.5.4 Tie-down Lug Welds a) Pure shear Area of shear = 2(15.5)(Sin 45*)(1)(0.85) T 2" + (6+6+2)(Sin 45*)(0.75)(0.85) = 25 in2 Stress = 273,493 /25 = 10,940 psi b b) Moment forces on weld s' Maximum Moment = 273,493 x 2.75 = 752,106 in-lbs l l 15.5" ~ 6" 6 3/4" ,,,,,,,,,,,,,,,",ns,,,,nn,,,Y 6 e 3/4" in 2"e 34" 11 I I n s"in ", ""F g '"'"""'""8". 3"5 " " ' "'"""f"" " " """' ( ' """ Reference = 15.5" 9 1" tine L 10.95 e , 2x1x3/4+2x15.5x8.75+6x0.75x8.33+6x0.75x10.95+16.5x2x3/4 0.75 x 2 + 15.5 x 2 + 6 x 0.75 + 6 x 0.75 + 2 x 0.75 384.4 = - 8.94 in. j e-*g 43 f / l

2. 0 bni/,isus,sii ii n _ r, s

sins"nnn nii, niinn nin,' 1 l 1 yurunu< usuuuruiruauniouu urn u a < < uisur. s = 7.94 X 6.56 M = compressive moment + tension moment compressive moment = tension moment M = 2(tension moment) M = 2f[(2.01)(6)(0.75)(0.707)(0.85) + (6.56)(2)(1/2)(6.56)(2/3) (0.707)(0.85) + 2(6.56)(0.707)(0.85)(0.75)] M = 57.2f = 752,106 f = 13,149 psi Combined stress f = /10,9402 + 13,1492= 17,105 psi F.S. = 21,000 + 17,105 = 1.23 o

STD-R-02-001 Page 24 of 148 JAN 7 1906 (3 V 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 dis-tance between the line of action of the tiedown force and the attachment plate. ML r i F v z __Mc Offset 4.25 in J 9 Fx F = 273,493 x cos 39* 30' = 211,034 lb F = 273,493 x sin 39* 30' = 173,963 lb i M = Circumferential moment = (211,034 lb)(4.25 in) = 896,895 in-lb M = Longitudinal moment = (173,963 lb)(4.25 in) = 739,343 in-lb Reference for method of calculation: Welding Research Council, Bulletin No 107 (WRC 107), Stress in Cylindrical Pressure Vessels from Struc-tural Attachments. l T = r/t = radius to thickness ratio = 40.9/0.875 = 46.7 l C = 1/2 the circumferential width of the loaded plate = 3 (33*/360*)(2n 40.9)\\ = 11.78 in C = 1/2 the longitudinal width of the loaded plate = 235 in/2 = 11.75 in. 2 B = C /r = 11.78/40.9 = 0.288 3 3 B = C /r = 11.75/40.9 = 0.287 2 2 ~ Check that 5 1 T $ 100 o

STD-R-02-001 P:ge 25 of 148 O w.m.nciaive. a,si, cess. i. cri,as,,ces shesi. V, - concentrated shear load in the cir-E \\z IB cumimnuni direction, ii, i 2 - 1 2 V, - cone: ntrated shear load in the lon- + gitudinal direction, Ib (0.3) 1.2 } M, - enternal overturnmg moment in the circumferential direction with re-spect to the shell, in. Ib M, - external overturning moment in the senee i n.m.aciatue. I ngitudinal direction with re- - normal stress in the ith direction on spect to the shell, m. Ib e, the surface of the shell, poi R. - mean radius of cylindrical shell, in. - shear stress on the ith face of the jth 1 - length of cylindrics1 shell,in. cu direction - halflength of rectangular loading in S - stress intensity - twice maximum c, circumferential direction, in. shear stress, psi - halflength of rectangularloading in N, - membrane force per unit length in c, longitudinal direction, in. the ith direction, Ib/in. - wall thickness of cylindrical shell, M, - bending moment per unit length in 7 in. the ith' direction, in. Ib/in. - coordinate in longitudinal direction K. - membrane stress concentration fae.1 of shell tor (pure tension or compression) - coordinate in circumferential direc-bendirig stress concentration factor y tion of shell K. denotes direction. In the case of i - cylindrical coordinate in circum-spherical shells, this will refer to. ferential direction of shell the tangential and radial direc. ti - -'th > et < t-O norTnal to the shell through the s - attachment parameter center of the attachment as si - ci/R. shown in Fig.1. In the case of B. - c,/R. = R.'T; shell parameter cylindrical shells, this will refer y to longitudinal and circumferen-C,, C, = multiplication factors for N. and N, for rectangular surfaces given tist directions with respect to the in Tab!es 7 and 8 axis of the cylinder as shown in - coemeients given in Tables 7 and 8 K,, K, Fig. 2. M., M, - bending moments in shell wall in denotes tensile stress (when asso-the circumferential and longi- + = ciated with,,) tudinal direction with respect to denotes compressive stress (when the shell. asexisted with,,) - membrane forces in shell wsilin the

- , N, modulus 'cf elasticity, psi

.rcumfmndal and longitenal E = concentrated radial load or total ' " ' " " * '. mpect to W sM P = distributed radial load, Ib - normal stress m the circumferential direction with respect to the shell, twi - nonnat strew in tlw I ngitudinal di-2 General Equation rettion with respect to the nhell, In the analysis of stresses in thin shells, one pro-psi ceeds by considering the reintion between internal shear stree on the z face in the

directi<m with respect to the

, membrane forces, internal bending moments and atress concentrationn in accordance with the follow-sgiegg, p ; - shear stiess on the 4 face in the z e,,- direction with retpect to the mg'

h,a 'T
K' 6 Af, N,

shell, psi T' 1 =-

i '!. M, l 'hLS d. f L ,4 -.Q } l.a - g.i< "l 1-. i- -- g ,, g,,, ;;;W;; ~ --~ 4 T-

, 1*

g -A

tt

-y + r ,44.tt4 . pyg.; p7 c, . 7 g 1--- y.'trq-- - - g- - "i -"-vi -.. ei A -- ---o_..ky. 1 m.j,e,e ~~ ' --"--9 f p 'J - co o )

n. germ rt" m-"-*r a

- I,W H $ 9;! '-'-""-7HT I+" Mg 3 M-' 1 h'f "t" t -" 9-n 6 . i r., 1.t-- 3 r g 1 "" f at*Mtf 'O Titil .-dle'" r r'i. c.. - 7 y.; p .,...}l. o4 "T 7 h eN i .j s O i ,i. <ir n a -. a-L. .,;..5 -_.--.-..-.,,.1 N i---- 2 I N ,.h. t"-' '. "... } N. Fty"a'- 8; p! r .--,y,. j 2 Q-..g-- g i i oe - -. _ - _ s.l g . p % ,6 -. r 7 g, t t y.,, r i+ W-!h ,[ t'rd',, ii d p.

%. }lf ; j

' -; m

-; he

'-- h,- -".- . e 4, a l - m, "**' A ; ' T-t-t-s n e, --...fg l g 40 % ~~ ;j;; q ' ;'y .. 42. .;t .. g -}, ~ 4.- r ;; -.,, 8 .[ t_ .. ~,j;. 3; y y j [.. y [' f J L.._ a 2..pg..y. in di ~"_"."-- M.,.44, ~ ~ r,,,p; 1 M- ~ .i.-

(

.,q '. t " I ! ~ ' 'I' - [ti' Ttv ! .4-- x -T + s .j "-"-t."r9 ""H 7 1 i "T** J --'~-* " ,,:i m H 4..n tt' --t*--' E' t-r a 'r r W,h ',n,,,e,,-{f y:n'd j A f; l ,,+tr : s. h tp --" -ri -r"- r M j-- sy -t 't" E egil ...I i &+ o. [ . y.. d--.. i 5-t .g /. p.f. h a- .... ( a.... - ..o .1 y -t-i ,,,y.' .p /.j,.7 _u --.. ,....'.{.---.. ....e...). .a n ...g, F.4 f.1W. 4

7.

..., i.". tl. g -d ,. 7 r ,..i. p_ _;.t. p . -e ..p.. . a.-...:..y cy j. -... ' 2 yr ,.y , q 3,, ;- O 3 {_;. 3.... !"l 4 --4_- .t.

i..

o E .H.. l i Lj q.- --, ..:h:t}.. _../l f

f. p.y y_

..,... ~...

y. -7 q

y. iIgp-p....... y,_.

t lI" b.t S.'o_

e a.:. . - ) {.- .,,,} 7 - _-. _,.9, -,,f.

7. j.,.

7 rrtt -e -1 o E 4 . 3 . ~-- . g: a.f / i "- T - d l" - : lit ,.* T'- "rTr tt~ g -- TE i ':i: -,. -i. .m .;,tt 7i i j f r, i i i C+ 7-r o-i > r r-s ,iiir i jp.g- _ =t r 1,; 1 i 1 4.ib, ^ k ,~gi j q m_.,- -}-"._ . ;b,, g: 4 -44,4-4 w t-ai r 3 4 1 2-i - - ~ 9 Q Q w.t j1 h. i ., _,.{ 3 9 j-

... TJ u g

.2 --}t " ' + - *, it

1 tv i i,j.

~ t- - * * " - ~ k"*,' S 'rJ le '-"--t-"- -i ti i 'ri r r ~~'"-7 d" <v r" "T-- r-"-- 1t-". t" tt-p _ ' h,,\\.y 5 .j_q.,;.g. q. c,g n . gg., "_f.. ,..g - hu..,n hriiy jt 7 pt, g y rt "

, 'fli r' i

1;;4-J Pt-r 7 u p -.,h;"gg Q' a_- c - p - _q 9":.gf m. m . g.,.w;4g 3. s s i r ' r 2 m ,. r t r _g ::; nn m cr-e-, prN Ih--t-tNr %m. e s.g.. OF ' 'rt-t--~ w:m y ~ r %r E@-M r-T 1 s ..16 M e t v i ~' i i i i + i vm , i,, s i n i p tri e 1,1 "- i'I' ' lt } b M N 'I t ish:...~I-9 "'~ i d/ 5FF '"1 I" 'l-d iI k' e n + a m,. m m, i .'W di ' 'a e - t" r , g r- irt?,:ty! 7 6 E i s li" hot p !M \\,JmTb'!1 4 .,r p 3 " a. 1 'h 7 ,t t'-- Wo. I ~~~ i.wie r m ft R f;... 5 ', l n.'m i. n L 1 ) ~ K'i w piar.r pum il 3 tl g g wi, ir ji, .r n i. i ' ; vippi l III g i . = '"p f - 4g 1; i t i - o - _f,, 1 ,,,,,,,1,. ,' fM m' e [ .t , r c.p o n i e,m 1. 1 c vc. P1 i i ' Il! A copov. m niri i 1.. ' leli 1.! i i Pf u n i..' 5 s d i L un i:t A Hn i 'N

eh X' I

m w e.., m, . "y L Ni i le ii! o 1 , I. T.mqut u i i,i. In s ni 5= i t as F i _2_R <rca g it i po i t "iw r, p, ("'[f ^ 'Y 'PI' 1: s e e S 3 . iN*10 "~" I' !n t'fi hi I. i."'!!1ttti,i, 'i, ! g e i i t i ro ivpu p'f'*f i ( i .p my i,.n si, p "i ar Ww A' 4" W hh Anh N m ' N1 IF 5 ~ yg n.. gi i l"- t i < n i v, ni i ip M.;iH I 'i~ per yi e n yr J {""1 mi m oxvusht. '.V > < y, N o > ' p'i'$ '"t ' N.in,. ' t_ l 7y. on i r.pr,n-n T C N ; ri, rrmn ti .m ' n os i '. 'n utfP TYtt-2 . v v m m ivrWH ff ~ ~, - r i ri,- 1.m F ,e 1 i i e n m.n com W. i.r"ornun 1! _g M r Frir PrTi, i o ri.n f. <"g a ygem n T; i 1 1 M vm 'c, -,i.c pp; ~~1 ari+pi in -$' wr.smo,xo' n4HFr' 9 Newt. ic' y 1" '~'ri' <N in w - i p 1 > *'"t1 ff '"t" g f [N, i : >; T u n, 1 g 1 bw pyn is,m ;

gitT,

"( " r ; i nni, m y y c t=,i w

m. t no w-

'"'t 1, p A ri ! py. j etnijit TTT h. t % <w l ""-" rr a i "N c i, t ri. TK ' 's .rh gr i r i. .L - W"'"f QTT' cu l r s '% gr' i t i, r% i ip,'u r : 1 vr _w- - -. % innun 1 - A-in rw. , i na i g[ ""'"~ rr< g i c i n!an iisi 'lili 1 .,n._ n u ++-c m - R?,, ) -- wnm n i i on!nn lii 11 m n- .+!u iil (III ju al nrIII1111 I m i it [ [ . n, i r! r - w .o..I.I.I.m .i 1 6 1 .i .l .l 1 91 l l s

STD-R-02-001 Page 27 cf 148 s...... Ilt e_ I e- = N e f \\ i. (d g a ,-h,,-- --. -e-f - F1--i d 3 l M.! Ih!h. h.! h. lH,,.. 4 !, t-t ; ii i,i, .. r. s,... 1...i. s. x. s. . s..,

s...ts

_-.,s ~. t _._.i., i.._ T.T F'.r.-*.. T..T.._ 1 - n_, 3_. ._..L_; 4_..u... m_4,_,4. _ a { p.;- p ... p!_,_ _o..,-. .4 . y. t.. -. ya_,. 1 1 _; - 4 u 4,.,., _ A p. i - .a _9 p 1 I I I I 1 3 i l o g. i [ .i ( g _-_ y.._ w, a i ] r = L 'r ' r ' =

-r

. r, ;; 1 :i a i A e, Q / ll q hv t. ep f.i g j t.(!- j e . a m 9 y m 4--- .y t y ' =,_ _3.;,._- s__- -p = = a a. . ' k. ,_{I } e. g. L ,[. I 4:33, p:RhN-j' C y I w ' 327 _3_. / --~ :, _g, F : _?

1-

,c ; 3. I~' s a p! s..-.;g, _ J' }_:3- _ -

i.

. * ~ ~~d I

.:.; :..; ;.Ii '.--r;:.:-d ---

.J

a...

-4:- .1,- a_ u __ ..:_:11:t ; .:p::.;_:- 4. - - **

L u :::_:-

.= : n.

_ :. u::

.: :;:.h r-

_rs. _ .. N,.o._ . _I

e. _.

- 4. v.+.- :_.. o_,_. _ ! T, _,:... _ d,.'4.4_-.. --._e. -- 4 +. 9

m. g.

..a. -,_4 3 I I 2 I I i J _t. __ 1-i 1 g,,_,,,[=ij'-2t.:_- I i b + 9.- b' dIf. ]~ ) N- . I N -fd f a8 ? i I 8 t.~ J g__ n r _~_-.. --. 4__; u.. _ __._tt. v : i I g, ~ i

==.f. 13-. 4i 1 i, -1 - i l

. k.

,: t W.5-ir 4 a, N=m 1,j=g~. = ~=: a -.~.-e+-.==rl-".r.==- 2. _. _:- -- - E ? ?= y- _3__c. ,. -.c. _. x_:-.

=r-c.- m.

...2_-= .:..= :- :- :.

==-2 _.. _ = = ~- e - -- ---e := -- --

==c.r.:.- _ _g : _ _ y-.7 3_ :: ( _::: ;.1..,g v _p.,-._,_ a 3 - ~. _A.__..o ..__&.+.. 4. i.l. w_ u. Y"E D,'- "' N..fNh,_h. $N.diA,,~> -u a I h / m. 3 r_ : _ w h'4 .T M.i r m r i 0.064 8_ = z Ap ,-w

mTn* u.-, -.w.

M=i W mw@W p T ,1 __p 7_._a_- T JT-I~::~. .2 (l. t - . - -=m t -4. ,.,b_ 2,,2.m.. 46 .ge

n

-,- =-. =. 4 = 4, m = e _I..,. j _ _... t I i .~.. MM1_ _ N i _ J..,.. 433'._ _., ,2 g ..g g 3 g, ..{._,. J. j. a =.i 9 - ~~~~"

T.._'.# 55 S #i-~ '. _-h !'i-It.Ed - -% I 155-15 ~,3555 E_-4 E-w- ^

_.r.--, . - _.., _._.,. L _ . = _._.n.,. s 1

4.._.f_4 p-- -

- - w-3 u i 1 _g. _ .,._. + 1. .3 4.,. _._+ 2 I.'., f, 2,-.- pHg ,8 ge.q[ y

. g.,

-p jy _,_. ar_ a a - w _2 a e._, =- y a a r_ l_ :; =..<.l r-s i__.=_. _a = -I 1 q -,2. s_. _: -L. ._=' _ _t = ~ s-- - --=a i v- .4 i,= p f c .x .-1 I =, - +-b9-h R2ik M, J- 'L. 2 j :-'7 :... .{ i p fi]..i-F.:.t.:-). J :r . i. I, L,,1 ' j',!' d, m a_ r.- -u 2. = =. m =-

fd: .1 ~ 2 f r m

--- -, H.. --f2 i ::-. -. - _--H r7 'i 8-- 4-. ~ n - I i _ ~ - ---+:._- - lI l I_! _ _. _ '.. *h. h _ _l $. gl N_ 5dhY I L- ... 1._ R m- +_ . 4.. e.. -e.- . t. j._. . e. t-t f '~ m f r N g -i-* J

r - j +"

y _.y 9 0 ses Die O ss 020 c hS clo oss oso 043 osc Q Fig. IA Moment M,/(M./R.Jr) due to an esternal circumferential moment M. on a circular cyhnder .%Iorna on.<hril. ,~--w

STD-R-02-001 Page 28 of 148 d-l l,l l6lI' e.l e e ,j,\\ 4.. l, l4 s L d.L,.. ) s =~- _u.9'. = v =._=-=' -- --TrfT H 1l H t1 *

r -- - - v'=?=' -, + -+ -rm f'

lhei' ik: r--,4r,N' r.b Q 14 ma Hf it! h i W ' S; 2 7--' t ' - -- t l1 1 . ee: ,_ d_ -.$. A* i. i ?. I ! m =,- ..._10.e. Q. _8 i.

a.. -FEE..-...

2- . -i .f a... b!scK: ch: - J ;/ r f.... .-!-. =. - _ - - _. _

.r.

m,G '.-.n.+q..w. . =...-. = =.. _. . = _.. . y-'- - ~ ' - ' ' ' - * = c'.

- ~ - - -

- ' * - * - ' * - - -*-t-*-~ ,,r .'*--t' -N hh d. i i ! ,Ii l l a, g ,,,EENk

"~

7 -l. i, .c. i i i e [g'. ,' l 4 R 6T - 's. e,_ c / il i r = K 11 $% e i'ss { ) t ,, % %. =.n _gN_ Q i ]' } 3. N s _ _.;, *. wie, _1_4g s_. ..';_".I a 5 m c m ? .t. =6, h., - r -. % q-._% ' s m'T. t- 'E E - {.t b r,..u 1 i a..:i i w. . ; o ;.: ; ;_. 3d. -.fg : i..:.: & g

-1.:_-

. _ = '% T. A w~.. _ 1 5 gu--- '" r ^" s:. - - - .$=[~.-];f _ 4 _._ c 1 mf 4; f.2_. r, p;- Ag ~r - -- -F F - r'-- . ; D TT "E - t

W,. -

5f m- -+- ~ 7 ,q..> x_ ii' ,l1i- -,e* ] ..@y yv' ' s 4.- W 'T i i I iii a a: 7 ' i, 1 a_ s + = h t ! s: ! ! p' ~ i z;.g u - a p= W' sFE '- 2M i- ' ~~.3 -' __1 -n pfdW

J' I 1-# '

a_9 ).i E i,' , t r-.- . -- i i.: i '-f$y_. ~i w.-q g = -v ,- } 14 T-E 2E i,, i m ,+.-, .,. g g_ y 1 i p 7 .i-r 6 /=/. =, .. %u &.', l i_ t. 6 a._ .L.. _u.,_ 2 s.. --? L, 6 /" s. 4 a_ .. = _ .r_. =t - .w3 . r e.,- 1.3 .. u se +m

s/

3 3= ~ ..w p.=,_.:=_2.: gEi Lz_ _ s _ja3gec:i-j= 1. r j. i -!f . =:. =.. = -

=1 -w

.=.: :.: x =. : : .:_=:= : r .=.v = =.. ; = :_: _ g p- = = = =. r. -- t - 1 a. 1_ _ =' .-_4._4...,_, 1 a:n .. r....L,- !1 u_ {~ _ y 7 _..q_ _j. J. 9.;-. } __y.Q 'l [. h' { / i.. 1 t e. b [ -~~ - r=. _' __

{

j. 1-T7 . i g-L._,,_ i +__._ y - r__ _u .,. !.t.... t_ 2 = R_M_ f.-r] i m. 4- . + - - g .r 7 ' + 1.. -.=: J n.. .4_ - _:=- E... : :- g .a.. .a.= a.-- , ^ __.4 ^

==e_=e,e =o.a.e.. .e. _r a.. o 4._ 4 e_e... w. .*i .e 4ei. ..a. .. ~ -.. O ,6' y -] - M "Y-g- g a I A } , i ;.i j y a',. i j) _] M.'Z" 2 jN ' I.'. 3,,. f' '**""7"'9P-" I -aI - 8- . ~

b. d 4

.,1 I - 7 y j = .l 1S. 1 ' .i S_,'2 fE~ E '1 j.. _ ' '- - .. l.. t. .y. a_ c .. _1, ~ l l- )% f ,e g_,, g; 8 : 6. .J P a -f !j' .i ' - 7' 1-.-" s j s. g

1:

I. I' ..p. _.e.... ... a..'. -. a.. .,.1--- ~~ ( . T. u./ .l 1! !t t1* 3 o oos oc ois oro o rs 0 30 0 35 o *o 0 45 oSo Fig. 38-Mambrane force N,/(M /R.'tt) due to an enternallongitudinal moment M6 on a cartular cyht' der i Storases w Shsfl.

STD-R-02-001 Pege 29 of 148 = l( ,4

bhps, nd

. s .... l 1,

1

._ i . ( Q. _ } ._l _1,' ,.,. l, alp I i i Q, ! = r+ +r .P a -t-e i -4 yo_ h a,,ad -,. k-- 2 - 5.,3-.~ g ___l 44 a, a .d iedMs?r - 6/-t -+-- --d 3d M5'- I 6 l ! i t I.I-- .jr' 'j j ! aa c ~~ - '2= J ii _ ' K._. -L _ h.._:. t _._. i' e fl._.Lj.J_1 L jt t9 1 J-.L i i N l 8 i t_i i! i* . !- - i ! . _.f=> ":1 - ? _r.t j 1.:.: fs.:_}.12_.:_ .._~

-[. r.l

.L !.,[. -[ t 1- --+- -.lg1'.!'- :.llj.,I ~=.- _.. s w ....,s .. s-. s l _4 f _._ s,.,.. . !....s + a,...1, 1, 1 1,.. g.

4. s.

. 1 1,. T<m. . 4... 4 ..4.I. ,i. .f ,i e = t yg ,+ -4 s r a e_ 6 - i j-l i \\ e _ =; = ?===: _E ;E '. - 4 l..; _i.-_,,.

.s%

L,y aN m -c- =,; 1 .= =- _-.:=_ a-l -

== ' 4 4-- -l-

=,.

=- .'.-9

== = =-

.=

-=

_-._ _.. _. E _ _i 1 =- .-s ' '~. n!4 l iiIi l' E - i-M'-E=:i-F-iM-. E 2 =5.E+ 93.== _E _=E-._ - {. -4..- -E r- _ :.- __; r; s .,.% :. q!.," -E:._r_;_

r-

= .* {..;,1. r..1- -i :5 :-

n-

... n.

,.:: : r_.x .. ~ ! sl r b.. ;

t.; -t_: -

--'. :=. :_'~- r ' ri-- m ? _. _.. _. 4 ,s.._. j...l.... w._ .+. 4 .l ..Q 1 m, .t l. }.. 4 s. --~ 1 Ct ~ _.__4__ _.1._ ~. i m - -_f 1- . i i a L u -== = ~' .a 1 -~ -r -h %g N,c&w p + 1 D ~ ' ~ ~ 1, ! ', t

= -

a .~5 .= _:

,S~.-

,.14,. 3' s. "n,. f b-w. );}ss- >=-h... M,!, ,.J m

3. ',=h
=

..m. ~ ~s At -n n, a 7 'a, i,! =

=

',, s

(' E; =EE-d,h1.1. k, \\.,E ! i % !i., it i-. " * %g[f-'

-.==

r,, g-- 1 ,s hw,..*:.. ; - :_ _ 2 ~ gL '. D....;g :. .. i s: p e ..y JNck2 3 [:.. A ;,,. y g

s... L.,-..... _. _._. _. _ _. -

..s Nm_%m A....a w .N..., e u. - s.-._ +- h --+'J.~' --:-7:: - ~

- --i

- +,' O C12. H *W. 2 k14'M 3.,N -_( l als.y. -+ c,y ,: ~ x -.= y..2 IT ~-~1 H3 1 1 _'

_ '-+

k s- +- w .? ~ _ _'_..:a._ 41 .i s I E p. ,,f.._ l. .w.. g v rn _ u L t e ~a v+ w2. ,t ,_ m n n .= un = s _ m_ 1 & ,#~ N e si z '_, = - sem ' fa -N 'g 9. 4~_...,7. t, = ~l.*. +_ u.- m .~.n__ X s s ,~ l l 1 7 N 2 q 'l W.- -.' - ~ L_ N ..g..,_ = -.14w.~.-x. c% _. , =o m+=m -a=s = m ~_m.a. m s e = _r-t, pr- . = =cmm r.; a.s .y: ? N. W ag. h %: L: ~: "

-a; w::m -3.-

= 1_ p-. p, - }w s s Em

e. > --
- _gx

, _ _ _=__=_._: - + - = t ri a !.D M!?fE f,:i: A-- -T:

Mir!s ii

,,e = e =_,,-.. r.r - m. - m.... ... __t_..._ .= v._., 1 _.t,_.1..s.,.,...~.. _. t,+_ _.. 4 ._ _u_ 2 s .Q-- - -[!C,[] - D% 4 -] -l p ., =.._- . -p. - 1 wpx-w s e- -- i a y + 4

c:__=n _.

- _+ h p:=4 _. _ -j s. :m: _r- - 3 --{_ 3 ' i'_ .t- = a -=- w -~7..-.~ = 2 r g= J--i 4- .O e m.g =,._ O. -H -- m A+ -t r: c e.----- - =:_._

i,- E3 i -- r-=

.-,i--

-c- _----+- -{

.2 Y

-s

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a.,= g- -.-- y,--

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4 s.. t _g .e O aes eso 4 85 eto 0 PS 0 to o se oto 04S 080 Fig.15 Moment M,/(Ms./R.A) due to en esternet longitudu.at moment M. on a circurer cytender J plane of symmetry) Stresses in Skils

STD-R-02-001 Pcg2 30 of 148 m a I I I <1 iI I 1 ] v 1 y l J* '

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w . ii w. 2 i- , i, i, .4

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.-fNy

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f" T tt--

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=

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n. u -u.

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STD-R-02-001 P.:ge 31 cf 148 n--. !!! !.!). l4j. 11'**I

l. d..11_

1llt t a ' 1 h-n_ i . i. . t. t-i a 2 a =.r T i ql"l iii;i it i>~ l_L' -

i. w !

i.~ /% L T o* .t. . q.4 j 1. 4..,. , a. =.. j 4 . 3.. :.. ; ; i s._/ t t : = = __.4 4

. + -3

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.

}.. - =;-=23.y y-{.. a.. ~ - +. - s =

: _;. N

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i et I

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-- x E EE--

d..:M. 7 2_L r a : - r_N.: - t g b .,.1

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_2 -" :-.~ 3 . 2. e L.~ I ! I ! i I I iiii i ! I I i I I ! I I I 1 ! I IT 4g 0.,3 ( ^ I i i 3 * ! I i ! i ' - 1I' ' iI I! I - I i _L. - _;. _ u_ _ ;_L.:_ _La. .L__ _ __-- - - 1 ._1 : n= . {q w- -_i?- - - - --2_ - _ _~ E__fT] u.4 iM% %-s O.'%V1. - ~.y u,,.-- -s, 3. m; -.,- r,. cy;. g-w ag, -L t - gm4-N-+ -X< < - ^..,-.'+=~-r-- -.=5 r %, Cs 0.026,-~m ~ %I_.%. .g-. S. ,,. m%, =,m.=..+. *.. w _-- 59..i ir, . T.. N=:_ . ~ " - -. - - - =.-t -r

w. -

~. u.:... :_:. =- -. 59..,.. i ii,,...... ~,.5--.'!."O..w...,. -:--.=._=-~ m..__ .__.=.D. -= .-_~.I. 200 g. ."= ; . " _,,, _.-.-..m _ 7a g m., c: ....,..w.... e_-g-3 '- h. 4. l. '. s ",~'l I, T. r- [. I I 11 - 1 i, i, 2, i

a. -.

i 1 'V ' y i i, . _ i f 9. _._w_.- } 6-i -7 3 ,.m. I q l . t *i 0 .. _~-1 :- 5 _ = s_., :..__:., 3.j _,.y....... - ..j. .3 . -.. ~ s= - li'* s1 i = ,1 I' 43 5 . t *1 l- ] ll t 1-.. 1 1. . l. 3 . _p t 8, ~ t-j.] .j - (- g-yy)j

j, o 2s c 30 O SS oeo oes 0 S0 o

0 06 om os c ro ) Fig. 2A-Moment M./(M./R.s)'due to an enternet circumf erential moment M. on a circutar cyfinder .%raar a en.%rlb

STD-R-02-GE1 Paga 32 ef 148 L ,,3 ---r-,.r. --,i.7,,,,- .., n.1. r.._..r_* i *L --' v v. , 2. .6, .i.. .J -._,.e. 1 1 _ __.u, 1 g a ,f 1. ..,.; i p y { {.4 }... l... 8 .._ a pr.-.7 pr i,.i.. rK ...d O. .. - a -. -- n ..t.'.-. i i .t-E _ +g 1 i,.., m._ ..j. g

p.

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.. r. l.c.

n....._ _

.. +..... m %; m.-l_._ .; ht,. ,, !.v'.I t '! Wh y::.: :__l_ ..y n._ ,p w.%.___ ._m n__- fy, :,, a f' - i . u n._,.h.x n. -r i;;.,- r.. ',. - li ,ii.--- ,_1 .u.y - . [i../ / - ' y; {- .i 4 .6 r r: rr mr .i

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)I

,. ] l _- [._ .p Q_ _..ju r -f'..... ... s,. -... -_._j....,.....,3 p _. 3. e'*.*f_-J_j i. / f s-i>i i 1 i 1 1 I 1 I ii . I ',Y . ;a T T_-'., Ii1 i 'I I i 3 'T' Ii f* I 1~ A. g _L.,_ e e _ fI/ / I. I e i r /3 i it -l = - iif + u- --R I J f./ -/.M f. .4 i 1-i y.. _. _, w i;'; ijQ. j __;,j_.j I. _ __ m a [ 72. .. ;.,} jQ i __f. -_.'..;,_ p. / w_ '. y..M_ _i.,[

_r u= -

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=.: .. -

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- y

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i T.

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1 1* ' g. 1 i t-i l*I: . e- .e ... p .I....~ .. + .1,... o aos e f. 8_..-- o so o ss c ro oss o so 0 35 o*C o *S 0 50 g,3 O Fig. 45--Meint>rar*e force N./(M,/R.9) due to an esternallongitudinal moment Mt on a cercular cytinder hirencs an.Wils

STD-R-02-001 Page 33 of 148 l-JTT.T.f 'M y il'Ti{ l g.p i 4 j., ,A .h.:. p, = = -.4 .s e *l } 4' -.l. f e .l ... g --..e 64 -44 p, j w-_ e g (-.s a-.... L j l 3. ; g i, i .g. - llgl,. ] l, i;.y g r__... l i .-..i,.. g .i. . q, q,. g.,.. i - 6 .n - n 9 !.l. .? gm. m2 ) .i e i. Ah = i* t

. ~..

II ! ! 3iI I 2 I. i~k.3 ]. i

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d-O3 1 54Ei{ L-@ "uM'- W9-v -W r rO = =.25 1 33". = r .a 4 1-1 I. 1 i I 1 i 1 v. al E 46 i f[ '_ Qg 1 =_ --]_---' . 3.--.2 i ~ 1 1 N j ]9 ~ h= jjy,hq:gh' f 'j jj [t ~ 3 2;f Q 4.j_ 1 _;l: L _ ; f i b. .-.?..*7 -=: -Etp: ' 1 - =- EOk --- d.. h: '"- m. =~. - * ' 1: ?'s- - : - -- '. * = ' - T--

=

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==-4:: -E ~_~, ^ i =.::,.- n =.r

* =:.=_.~-

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.---- ;.- n:--- :::~. :r-- : ::: :

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o ri

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- (h,D__

M -* = . 6. _,'"4" ~*F L g2 4 -.me-E a Ms e '%=t 1'3pa.,** % 2% r=. s. pa3 t i

=

2 q L a = WO@'<Th _ 2 5 3% Q's )@MM W 357- ' -E E !' m 2 " +,'i d 5 9 [\\p N f L P 'i .%_TT r..%_ _ - -S @r.Me w: .:l w N_ w. i i i : n e e\\ '- g-2 _- a.1 - - ' D a. %--- -M ? .- A r _b__h" .c -s ag== %g' -.u--- a-- L== 7ts'- M : **, m = 1 = = C.= N t 'a - u". We c E i 3 =: = %y.t~.::. s,.. i. T. 5

9. x;

a.mi Q4_.t. _.u_u_=_ _ .%x-m< = EE - - fi - 'I -.1 a -- : - g - 7 m - x =_1 -... __ _~A-. a..:.. m__ ~~ 0.01u s m m,-- r <mm I s - J..a s s..._,,..~ _ _. ___ w ,'.tA ..s,, s. . %.. w m_..g. .....N i t 1 p pw ,, " 3, j-p_ x :T 1 1 M 71 4 3, ,2 _,'l pq4 34 ' q{ j ~ '*i1 @' t m 6 " % LE.5 ' W ? ',J JM _,2 g r S4

=

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5b _ L-M.'E E -EE 2=

3-= = w e,,.7=.- _L,, -M_%.%_. ye. T_ ~.w_~T-"' T F_a_ I-

43. ]=p~_q#

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32 _ -Qm t u r'.

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v. ", _.%'.,'. ^. E.,'Q c

@l ,5 1-,- C,Ti W. . a 3. mu. = s =

== r M r.--T - A. v. sr s=P = - =. ~. rq.-. m -e s ,.s. .w., = = ' f.:Exi'^ =7

.,m.u L

- -u s 2:E _ 42 . m. i 2 = a : a. : p. _ _.. .s

N

= .r

=.-..: = = - =

:-

-+ s s 2 1_Wd_FM _- _ _1.2 r . g:%_.. g,y 3:9,.. g.-._._.. g s.-. ..-.'1* %_ _ i I .* g

I

. __%__..p ...,.w 2 1 s.--d '_..I -,_ j[ f ', ; l* 4 1 i l )

4 kM3k -M

-A m ,.., u .,v s L._ - a c _ bw-e. b-n+. - g *' . t. J u A ~ _T'i'.~1 iW.- -. :.4 -35. ~.. -i = M F e_.- , l-2 _ 5_ M EiL rt. e. R d "*2 ~ g- - T = g e =r 1 -,~;-t= m - p.

L L) ~

-,1J+ y a 3 ..s .fG .d '.1, a;

3 2

m - ~. _ d J. q.. _3=y.c?.-i a_ 5. f ). N-4 ' 4 g j Tj gyg-@;_.: - - .a y 3 :r.J.;.-

a T= =.. n_.. t

_ ~Q-TJ qO = .. :/

m.

e . !..:.:.:7. -.;...... r :;.-_. r

n_ M -: ..~.;-*.;

._.9-h4 .a. . m_ - v.-a 1 1 LL.._ , 7- ._- 1 di O SS 950 Olt O iO 0)$ 0 30 O SS 040 0 45 0 90 Fig 28 Moment M.AW/RJ) Jue to en enternettongitudmei moment M on a circu'ar cyhneer (Stress o piene of symmetry) Nuramr. h, SARI /,

STD-R-02-001 Tcbb 5-Camput:tlin Sh ct f2r Loc:1 Str::ss:s in cylindrirl Shells P:gn 34 of 148 <P N ' M 7 ING v m= l e.h,% 3. c..............s 3%* bq.,,n.dt..d.. 0 _ i6. % 7 vz c.di.u.... e= v6

u. Ay3 in. Ib.

.a. i 6. 1 - o,*g o....., r OUND R ML *EI33 2c) oL' (0.875) 1..: O, L.ag. M .at. - ATTACHMENT . - O - i n.16.

Aug, yg, T.e.l.a M.w.ne, m.. ? W 16.

v. sh L..d. TT g' rg Sh e L..d. V6 ; *, 5..... c.a..a...... d....;, y74 g--pq3 g lg j

2. @..e,y

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6) 6.ad.a. i..d. n n,

s a v....I eks 6.... aa. oT E: E ae., s 's f...... l....a An..h..a

dios,

e., @ 9...

n: ....,4. a... h s., a.. a...... a gggy g v....I div., ST R E 5SE 5 -.61..d es.pp.s.o. ch.:. h..a. e e... s... g a. sh..a Fse. n d eve... C..p we. 6. twt...l v. s.I Fl"s I., .tr....ad.ne.e e.s It

a, AL B.

BL C. CL 09 DL " (,%.). 6 - 'lc - = (?) % = 2 l.*' J2 + + + a + MM)ff))j$ -Hi - 3381 +33ei +33sl .."f..,a "(..,"f.s). 7s1 =. =^ i ..,!.e.4 "(..,. s)..s1. " EMMM 3M F372 +37r4 -372 '^ ".<,..v ... y, = -5922 - 9 22 4922 +5922 Mfj$y'y ( "! ) .f...oA >= ggene mme n::.;;.,.4 -L::a.:.= -ssa +582 sse see '"M.'t.!:.:.Pi : ---- Il78'l 60 li79'i 60 4r367 34545 4Go7 3yy;y '!c". ,,";. =~ ~ (n) 6 = ( "f) 6 = +

f *e. "/

+ + = . "/...,6 -(.., :.y). 4 = l$$$M]$$$M oce6 -2 +%86 +m a MMM M -i5'to7 +tSM el%R -lM7 . 7..s.oh "(..,L ) ..'71 = =^ .J..,,a " (.v.:.u).. % T -3S7 -m +2 m +27plll@@>Jj@ig a -B793 +878 +833 - 8 79.3MMM 0?i .21..p.g3 a (.u..:n)..'"v1 = I '. "x ,"U.' O *'~"" lI580 6006 ll580 6006 VH93 63El 2V'03 632l l + +- r t +- r V r.s = 1,,"b "x:::'::,'~ ,a. -: PM-MMM r.a = Q.7 +SIl8 + 5IIS -Site -SiiB i' "':!.D'L T. o - 4c'r MMM -4230 see e v Au g 6.i..n, r. SilB SliB siis slie, 423o 4230 4230 4230 .e 6,....s....r. } COMBINED T SS NT TY - S 4 { )

1) When T / 0, S = largest absolute maanitude of either 2

2 or /(ox - 04)2 + 47 g

4) 2 + 47 S = 1/2 [c + 0 I /(o 0

x 4 x Yo, Yb

2) When T = 0, S = largest absolute magnitude of either S=0x, 04 or (O

-0) x 4 Y

...~...-.. ~.~. -.... -. STD-R-02-001 Pege 35 of 148 w

m-l C

w The highest stress on the outer shell is 42.3 KSI. The steel used in the outer shell will be specified to have a minimum yield strength of 49 KSI and will be certified by t. sting. 3.1.5.6 Failure Under Excessive Load The tie-down lugs are designed to fail first under excessive load and preclude damage to the package. Based on ultimate strength of the shell material, the force required to cause extensive deformation to the shell would be: 504,313 lbs F = 273,493 x = (where 78,000 si is the ultimate strength of the shell material. The tie-down lugs are designed to fail by tear out when excessive forces are applied. The force required to cause tear out is as follows: 4 F = 45,006 x 2 x 3.0 x 1.75 = 472,563 lbs. Compared to the force required to damage the shell, the factor of safety will be: F.S. = 504,313/472,563 = 1.07 = + .= f 3 5 0.

%..-,r,.-.w+.,,

w mm-.-_.4m . - -.,,e w_, y-y%-., r.m*r=-- e .e-ww-+-ww-w' + w e = wt+ e+ ' ---T--w-e---

STD-R-02-001 Page 36 of 148 JAN 7 gag 3.2 Normal Conditions of Transport (Appendix A-10 CFR 71) 3.2.1 Haat Since the package is constructed of steel and lead, tempera-tures 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 Each binder has an ultimate strength of 135,000 lbs. Therefore, it can be concluded that the reduced pressure will produce no detrimental effects.

6.2.4 Vibration All components are designed for a transportation environ-ment.

No loss of integrity will be experienced. 3.2.5 WaterSprgy jg 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 = 53,000 lb x 12 in. 5 in.lb. Maximum energy = 6.36 x 10

T STD-R-02-001 Page 37 of 148 Energy will be absorbed by crushing of corner. \\ N N g! \\s t '. ", - Q\\ ' *:, : by *. * '., I s*..a *.... s 1 s. l ?. - h \\ BOTTOM l CORNER d g j N/ +. The volume of the crushed ungula, assuming the worst case of a 45* impact angle is calculated by the following equation: v,=,qs1...sy+..cos.3 (> LJ R h s -s \\ A. sge 29 } .t- $? ~,

STD-R-02-001 Page 38 of 148 JAN 7 gg O The maximum amount of steel crushed will be: b = R(1 - Cos () = 1.66 d 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. I\\ S %e N y ;,.,q, c-r 3 .f 1.17 l I ~ ~ / x" i 1.66 j / y The deceleration force exerted on the cask is calculated O' as the product of contact surface area and the yield strength of the steel (38,000 psi). Au = "* - (xy + ab sin~1 *) 2 where for 45* angle = 0: i R = 40.5 in a = R/cos 45' = 57.275 in 7 b = R = 40.5 in 4 h = 1.66 in C = R-b = 40.5-1.66 = 38.84 in 2 - 38.842 = 11.47 in 2 2 440.5 y = JR -C = x = C/cos 45* = 54.928 in f5 9 )( 0.5) Au = n( 2 - (54.928)(11.47)+(57.275)(40.5) sin 5 25 Au = 36.32 in2 F = (36.32)(38,000 psi) = 1,380,160 lb. g force = 1,380,160/53,000 = 26.0 g's -_,_n--.--- ,,,,,m.,,__,,_..,,,-,,m7m ,.w-

m---q,-,-.

e m ,g,--.s

STD-R-02-001 Paje 39 of 148 JAR 73 O 3.2.6.2 Effects of Bottom Corner Drop on Balance of Cask The 26.0 g deceleration will be transmitted to l 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 Deceleration 4 Forces Contents 47

g*%

s + Upper d '/ o Bottom Plate Inner Shell lower Lead Shield Bottom qr Plate Outer Shell Corner Impact M\\Ni kh\\\\\\ Reaction Force Bottom Corner Drop

STD-R-02-001 Pcge 40 of 148 .!A% yg b Summary of cask component weights as used in the following drop analyses Primary Lid 5,800 lb Shield Plug 370 lb Outer Body Shell* 7,224 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 17,800 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: Load from primary lid and shield plug will be distributed to the 1 inner and outer shells in accordance with the shell cross sec-tional areas. The inner shell will receive loadings at its connection to the 2 upper bottom plate consisting of: Load from lid and shield plug p Load from self weight of inner shell d 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. The outer shell will receive loadings at its connection to the 3 3 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. The upper bottom plate will receive loadings consisting of: 4 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.

STD-R-02-001 Page 41 of 148 JAN 7 y Cask Analysis 1 Load from Primary Lid and Shield Plug Deceleration Forces Shear Area of Weld \\ / \\ 9,#,:. A %,, +, [ Deceleration (,) e 4 Forces Det ail "A" / 1/2" b e. 3f pacti Forces Loading = (5800 + 370)26.0 = 160,420 lb Lateral force = 160,420 (sin 45*) = 113,434 Axial force = 160,420 (cos 45*) = 113,434 2 2 2 Inner shell area = n/4(76.25 75.5 ) = 89.388 in 2 2 2 Outer shell area = n/4(81.75 80 ) = 222.317 in Z Total Area = 311.705 in C) Inner area = 89.388/311.705 = 28% Ng Outer area = 222.317/311.705 = 72%

STD-R-02-001 Pcge 42 of 148 JAM 7y O Force on inner shell = (113,434)(0.28) = 31,762 lb lateral and axial Force on outer shell = (113,434)(0.72) = 81,672 lb lateral and axial 2-Stresses Developed in Inner Shell and Attachment Welds Deceleration Forces Inner Container Up Bottom Plah 3/8" N wel Detail "B" (} Reaction on Inner Shell V 1/2" welds Stress in weld around perimeter of inner shell at cask lid 31,762 lb/n (75.5)(3/8)(0.707)(0.85) = 594 psi Total stress = 8 ('s94) = 840 psi F.S. = 21,000/840 = 25.0 Stress in weld connecting inner shell to upper bottom plate Total force = \\ self weight of inner shell + \\ lid and shield plug ( of weight acting on \\ of shell) + waste Total force = (1900/2)(26.0)(sin 45')+(31,762/2)+(17,800)(26.0)(sin 45*) = 360,596 lb. O a

I STD-R=02-001 j Pcg2 43 of 148 ) MN 7 las / \\ U Lateral Weld Stress = 360,596/n(75.5/2)(2)(3/8)(sin 45*)(0.85) = 6745 psi (lateral) Axial weld tress is caused caly hy lid Icad and shc11 celf eight. 31,762 + 35,200 = 66,693 lb Axial weld stress = 66,693 lb/n(75.5)(3/8)(sin 45*)(0.85)(2) = 624 psi Total stress = 467452 + 6242 = 6774 psi 2 2 Axial shell stress = 66,693/(76.25 - 75.5 )(n/4) = 746 psi which is less than weld stress. Shear shell stress = lateral force / area [(1900)(26.0)(sin 45*)+(31,752)+(17,800)(26.0)(sin 45*)] 8814 psi 2 75.5)2 (n/4)(1/2) (76.25 F.S. = 21,926/8814 = 2.49 1 3-Stresses Developed in Outer Shell & Attachment Welds \\_ Deceleration Force f s Inner Shell Upper Bottom Plate Outer Shell Lower Bottom Plate h / / Crush Depth h Weld Area in Shear Stress in weld around perimeter of outer shell at cask lid 81,672 lb/n(80.875)(0.5)(sin 45*)(0.85) = 1070 psi both axial and lateral Total stress = /2 (1070) = 1513 psi (D t/ F.S. = 21,000/1513 = 13.9 I

STD-R-02-001 Page 44 of 148 JAN 7 mg Stress in weld connecting outer shell to lower bottom plate Lateral force = 4 load of outer shell + lead shield + lid and shield plug (the supported by \\ outer shell) (7224/2)(26.0)(sin 45')+(81,672/2)+(14,397/2)(26.0)(sin 45*) l = = 239,585 Lateral stress = 239,585/n(80/2)(0.5)(sia 45*)(0.85) = 6344 psi j Axial Load = (7224)(26.0)(sin 45')+81,672 = 214,484 lb. l Axial stress = 214,484 lb/n(80)(0.5)(sin 45')(0.85) = 2840 psi Total stress in weld = J(6344)2 + (2840)2 = 6951 psi F.S. = 21,000/6951 = 3.02 80 )(n/4) 2 2 Axial stress in outer shell = 214,484/(81.75 = 965 psi < 38,000 psi yield n(,) Lateral shear stress in outer shell t = (239,585)/(81.752 2 80 )(n/4)(1/2) = 2155 psi < 21,926 psi yield 4-Stress in Weld Joining Upper to Lower Bottom Plates 'k - h = Deceleration Forces e Inner Container Upper & Contents Bottom Platt / 3/8" Detail "D" d Reaction on Inner Shell 1/2" welds

STD-R-02-001 Page 45 of 148 JA4 y 135 ~ ) Load on Weld = Upper bottom plate + Inner shell + Shear from lid on inner shell + Waste + \\ lead j Load = [2645 + 1900 + 17,800 + (14,397/2)] (26.0)(sin 45') + 31,762 574,913 lb = e Stress due to lateral load 574,913 lb/(76)(n)(sin 458)(0.5)(0.85) o = g g = 8012 psi. o Since all axial loads are transferred in bearing, the maximum ) weld stress will be equal to 8012 psi. This is within acceptable j' limits. l i b i s i 1 ) I !I 4. w-,-mwae. ,w,-m,,,-u.s.e.,, caw,. ,w.. meg.... . -,,,#-c~. mem-+ y,,,p. .wm - owe - w o-y we me-m --

N v W TN-T

"**-'v-'t'-'

STD-R-02-001 Page 46 of 148 O 3.2.6.3 One Foot Drop on Top Corners* A drop on the upper corners of the cask would decelerate the cask and would result in axial and transverse deceleration forces between the cover and the balance of the cask. / Contents Deceleration O Cask Transverse Deceleration [nner Shell Outer Shell Cask Axial Deceleration Primary Cover 1/2" 1/2" s i F Impact Point

STD-R-02-001 Page 47 cf 148 JH 't in. a The top cover is stepped and the inner plate has a nominal clearance of one eighth inch. Upon impact, this plate would immediately contact the inner shell. The transverse decelcration force ir.ust be reciated by the bearing stress between the inner cover plate and. the cask inner shell and by the weld between the two cover plates. The magnitude of the transverse decelera-tion force will depend upon the orientation of the cask and the corresponding deceleration forces. As shown later, the maximum deceleration force will occur when the cask is dropped on a long flat edge of the primary cover. The maximum deceleration for this case is 17.88 l g's. The weight of the cask less the upper cover plate is 53,000 - 3260 = 49,740 lbs. The transverse decelera-l l tion force acting on the weld between the two plates is 49,740 x 17.88 V2 = 628,866 lbs. The weld is a 1/2 l inch weld, 75-1/4 inches in diameter. The stress in the weld = 8,853 psi '

75.25(g)(0 5)(0.707)

F.S. = 2, = 2.37 The weight of the cask less the cover and shield plug i is 53,000 - 5800 - 370 = 46,830 lbs. The transverse l I force between the inner cover plate and the inner shell of the cask will be: F = 46,830 (17.88)+[2' = 592,075 lbs A 40' are length on the inside diameter of the inner shell plate times the thickness of the lower primary cover plate is assumed as the bearing area between the two surfaces. Area = D ( 40*rr) t = 75.25 (40*t1') (2) = 52.5 in. 2 360 360 The bearing stress between the two plates will be: F = 592,075 52.5 = 11,278 psi Allowable Bearing Stress = 0.9 x Tensile Yield Strength F.S. = 11,278,000) = 3.91 f (

STD-R-02-001 Pcge 48 of 148 [N %.) 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 considered to be worst case. 37.4" 34.4" J r h impact An impact in this orientation will cause minimum bend-ing of the cover and will result in high impact loads on the cover. The majority 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 illus-trated below. Crushing O F k_) .Following impact the edge Fa of the corner will begin crushing until inelastic E Fg '4 rotation around the bend in point occurs. ( Neutral The point at which this Axis will occur is calculated as follows. 1.50" Width at bend = 37.4 in Thickness = 2 in. M = (38,000 psi)(1 in)(37.4 in)(1 in) = 1,421,200 in-lb 38,db0 psi - m ) o

e

[ 1" 38,000 psi ' e 5 1 s_, Y

STD-R-02-001 Page 49 of 148 N 7m O F, = Force required to initiate bending l F = M/X = 1, 1,200 = 947,467 lbs e 4.., 947,467/53,000 = 17.88 g's.(axial and lateral) ll F=F,(5)=1,339,920 1,339,920/53,000 = 25.28 g's (total) 2 Area crushed steel F + 38,000 = 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 = (0.512)(1 025)(34.4) = 9.036 in3 Energy absorbed in cruching 9.036(38,000) = 343,364 in-lb Bending O When the force due to crushing reaches the above value noted, V the cover will bend inelastically. The bending will occur around the impact limiter ring and with the shell of the cask. Fa F M f J 1.5 inch The balance of the energy will be absorbed by bending of the lid (Total Energy) - [ Energy Absorbed in Crushing] [(53,000 lb)(12 in)] - [343,364 in-lb) 292,636 in-lb = Energy absorbed in bend. e O wits a iai rerce or 947.467 is reeeirea te ca e de ai=8. this amount of energy will be absorbed by an axial displace-ment of: 292,636 in-lb/947,467 = 0.309 in

---_A....._. STD-R-02-001 Page 50 of 148 NM 7 Ing The g force developed during the bending process is cal-l culated using a kinematic approach. Velocity at start of bending is: %W - I(2)(292,636)(386.4) '2 KE g _ (53,000) = 65.3 in/sec As calculated before, the inelastic bending deformation is 0.309 in. The time it takes the cask to move this distance, based on average velocity is: AX/V = (0.309 in)/[(65.3)(0.5)in/sec) = 0.00946 sec avg g force = (AV/At)/(386.4 in/sec) I = (65.3/.00946)/(386.4) = 17.86 g's (both axial and lateral direction) The above shows that maximum g force is 17.88 g's in crush-ing and bending in both axial and lateral directions. The force of impact on the corner is 947,467 lb. (axial component) Contents 947,467 lbs 40.2" o l LA h o o 2R 02R 1.5"] 17,800 x 17.88 lbs pR e 23" -34.4" 23" m The loads on the ratchet binders will be proportional to their distance from the pivot point of the cover on the cask.

I STD-R-02-001 Pag _e 51 of 148 m 7 mg Forces tending to open the lid consist of weights from waste, lid, and shield plug. (17,800 + 5800 + 370)(17.88) = 428,584 lb l Summing the moments about point 'A' (947,467)(1.5) + (428,584)(40.2) = 18,650,277 in-Ib 18,650,277 in-lb = 2R(23)(23/80.4) i + 2R(57.4)(57.4/80.4) l + 2R(80.4) 1 18,650,277 in-lb = 255.92 R R = 72,875 lb (in farthest binder from impact) R = 72,875 (57.4/80.4) J = 52,028 lb (in middle binder) R = 72,875(23/80.4) = 20,847 lb (in binder closest to impact) l I !O i h s+ m-w m w mr d t-vw-w wm w,-m1 ww mi-g sw- ,,=y-r,----,w+q e- -m-wr,ww wywg7---w wwy-y,mmwgm-, - - -w wm+ e+'-e rw------~w~-s--sg g--m-p "w==m --~---w< = - - - v-- w'

STD-R-02-001 Page 52 of 148 j The 1/4 inch thick seal ring made of AISI 1020 steel, located on the outer periphery of the top of the cask wall will experience some pres-sure 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 947,467 lb. s .4 ~- _g te a o The yielding surface area reacting against this force is proportional to the angle 0 and the radius. (39.62 ) (35,000 psi) = pressure (psi) ~ 3 (Pressure) n(39.8M ) in / degree = force / degree 9 360 w As seen from Table 3-1, the entire force is distributed over a 88* arc of the ring. This is approximately the angle between two long 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 maximu:n deformation is 0.30 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 88* distributed load, the effective area of these two welds is: Area = nl(79.5)(0.5)+(76.25)(0.375)]2 = 105 360 f = 947,467/105 = 9026 psi The actual stress values will be lower since the upper ring will cause the load on the weld to be distributed over a larger area. The spacer ring has a width of 0.5 inches compared to a combined width of 1.25 I inches for the inner and outer shell. Accordingly, the stresses in the shells will be 40 percent of the stresses in the spacer ring. ,o

STD-R-02-001 Page 53 of 148 pd TABLE 3-1 Pressure Exerted on 1/4" Seal Ring Due to Top Drop on Large Flat Angle Pressure Force / Degree _IF Press /E Strain 1 34,780 12,026 12.026 0.001160 0.000290 2 34,769 12,022 24,048 0.001159 0.000289 3 34,748 12,015 36,063 0,001158 0.000289 4 34,715 12,004 48,067 0.001157 0.003289 5 34,673 11,990 60,057 0.001155 0.000288 6 34,620 11,971 72,028 0.001154 0.000288 7 34,557 11,950 83,978 0.001152 0.000288 8 34,483 11,924 95,902 0.001150 0.000287 9 34,398 11,895 107,797 0.001146 0.000286 10 34,303 11,862 119,659 0.001143 0.000285 11 34,198 11,825 131,484 0.001140 0.000285 12 34,082 11,785 143,269 0.001136 0.000284 13 33,956 11,742 155,011 0.001131 0.000283 14 33,820 11,695 166,706 0.001127 0.000282 15 33,673 11,644 178,350 0.001122 0.000280 16 33,515 11,590 189,940 0.001117 0.000279 17 33,348 11,531 201,471 0.001111 0.000277 18 33,170 11,470 212,941 0.001105 0.000276 19 32,983 11,405 224,346 0.001100 0.000275 20 32,785 11,337 235,683 0.001092 0.000273 21 32,577 11,265 246,948 0.001086 0.000271 22 32,360 11,190 258,138 0.001078 0.000269 23 32,133 11,111 269,249 0.001071 0.000267 24 31,895 11,029 280,278 0.001063 0.000265 25 31,649 10,944 291.222 0.001055 0.000263 26 31,392 10,855 302,077 0.001046 0.000261 27 31.126 10,763 312,840 0.001037 0.000259 28 30,850 10,668 323,508 0.001028 0.000257 29 30,565 10,569 334,077 0.001018 0.000254 30 30,271 10,467 344,544 0.001009 0.000252 31 29,968 10,363 354,907 0.000999 0.000249 32 29,655 10,255 365,162 0.000988 0.000247 33 29,333 10,143 375,305 0.000977 0.000244 34 29,003 10,029 385,334 0.000966 0.000241 35 28,663 9,911 395,245 0.000955 0.000238 l 36 28,315 9,791 405,036 0.000944 0.000236 37 27,958 9,668 414,704 0.000932 0.000233 38 27,593 9,541 424,245 0.000919 0.000229 39 27,220 9,412 433,657 0.000907 0.000226 40 26,837 9,280 442,937 0.000894 0.000223 41 26,447 9,145 452,082 0.000881 0.000220 42 26,049 9,007 461,089 0.000868 0.000217 43 25,643 8,867 469,959 0.000854 0.000213 44 25,229 8,724 478,680 0.000841 0.000210, t Entire force 2(478,680) = 957,360 > 947,467 lb. 0 will be distributed over an area of % 2(44) = 88

STD-R-02-001 geqSWf148 (O m Lid Ratchet Binder Assembly Based on the 72,875 lb developed in the far ratchet binder during a top corner l l 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 135,000 pounds. The binders will generally fail in the threaded portion of the shank. The shank is fabricated from Grade C1040 or equivalent cold worked steel 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\\ inches. The strength of the shank is calculated as follows: 2 Yield Strength = 70,000 x 1.5 x n + 4 = 123,700 pounds 2 Ultimate Strength = 85,000 x 1.5 x n + 4 = 150,207 pounds Based on yield strength the factor of safety will be: 123,700 + 72,875 = 1.70 Ratchet Binder Pin - V 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 72,875 72,875 lb lb 2 1-1/8" 2 y l l Resultant Force = 72,875 lb I I s j s s s j + o, = (72,875/2)/(1.125)2(n/4) = 36,657 psi F.S. = (74,000)(0.577)/(36,657) = 1.16 h v e m --,,m

STD-R-02-001 Page 55 of 148 JAN 71906 Lid Ratchet Binder - 11pper Lug -t 4" b" (--- it \\ I 3/4" h A 1.90' 2" g m-48-DIA _a c. /t h" h" lh" y ---g 1" p-- e 9-3/d Tear Out - 3-3/4" Shear - o, = 72,875.lb/(1.5)(2 - 0.0625 - 0.27)(2) = 14,568 psi 9/16" r FS = (38,000)(0.577)/(14,568) = 1.51 _f 1/16", n Bearing - ~ = 72,875 lb/(1.125)(1.5) = 43,185 psi 2"i l5/8" h R i _ L /d 2/1.125 1; > f 1 ~ * = 2.88 g jj I 43,185/70,000 gjg Tension - = 72,875/(1.9 + 1.375 - 1.25)(1.5) UT = 23,992 psi T i FS = 38,000/23,992 = 1.58 I I

...u. - m ..4, STD-R-02-001 P ge 56 of 148 JAN ? mg Weld \\" double groove weld, complete joint penetration with tension normal to i effective area. Allowable stresses same as base metal. " full groove and h" fillet (both ofdes) b" full groove

/

Y' f \\ " full 1-15/16" groove J 9/14 A-3/8" J e U 72,875 It Tension. Neutral Axis =72,875/[(2)()(1\\)+(2)(\\)([2)(2-3/4)](0.85) oT O = 15,909 psi T Moment - = (72,875)(0.5625) = 2a [( )(1.5)(1.9375)+(2)( )(1.4375)2 o (2/3)(/2)(\\)](0.85) o,= 9,934 psi = 25,843 psi tot m T FS = 38,000/25,843 = 1.47 I i O

STD-R-02-001 Pege 57 of 148 dAR 1m ,s.,L) Lid Ratchet Binder - Lower Lug Tear Out b" = 1-3/8" = Shear - g_ o, = 72,875/(1.5)(1.75-0.0625-0.27)(2) n = 17,137 psi 1-3/4" s FS = 21,926/17,137 = 1.28 Bearing - ik" a . DIA (* (' } B = 43,185 psi (1.75)/(1.125) FS = L /d, (43,185)/(70,000) e = 2.52 o /fu B 1 " Thk 9" Tension = 72,875/(3.5-1.25)(1.5) = 21,593 psi l 6 %- 6b" FS = 38,000/21,593 = 1.76 Weld Shear - o, = 72,875 lb/(9+1.5)(\\)(2)(.85)(2)(.707) y ggn o, = 5775 psi = - zu Moment - (72,875)(2.625) = 1' 2a [( )(1.5)(4.5)+(2)( )(4.5)2(;f3)(g)) 72,875 ll" (0.85)([2) = 7859 psi f, _ _. Neutral m Axis 2-5/3 t- "T *4 = 9,753 psi ~ m s 4g., FS = 21,000/9,753 = 2.15 ~s am

STD-R-02-001 Page 58 of 148 .I/v# 7g 9 Iid Ratchet Binder - Lower Lug (Optional Design) ac 3/16" \\ _ ~ 2a - n 3/16" [ f 9 y n 1-11/1( lh" \\ \\\\ [ 1-11/16" 84N 1/8"( [ NM gg., I-3/Y DIA 1/8" [ Ik" c y ik" N %1 N/e/ 9 \\ \\ 9.. N/ \\ N N N N pP N \\ N4 -.-._, k" % / u Section C-C Wel'dHoldbg("thickplatestolug Assume each circumferential weld must support the load - o = (72,875/2)/(8.5+2.5+0.5+2.25+6.75+0.25)(3/16)(sin 45*)(0.85) o = 15,582 psi FS = 21,000/15,58%.s 1.35 (Optional Desig,n) l Tear Out - l Shear of the sleeve - o = 72,875/(1.6875)(1.25)(2) = 17,274 psi l \\ FS = 21,926/17,274 = r.27 l

STD-R-02-001 Page 59 of 148-JAN yg h l U 7 Shear of the pin - o = 72,875/[(1)(1.6875-0.27-0.0625)+(0.5)(1.6875-0.27-0.25-0.0625)](2) o = 19,102 psi 1" Thk FS = 21,926/19,102 = 1.15 \\ bearing of pin - o = 72,875/(1.125)(1.25) = 51,822 psi 72,875 lb h Yi \\ FS = L /d = 1.6875/1.125 _ e 51,822/70,000 '// 7 c/fu Tension - (Optional Design) g5/8", [ o = 72,875/(3.5 - 1.5)(1.25) = 29,150 psi Neutral l FS = 38,000/29,150 = 1.30 3,3, Weld (Optional Design) Shear '/ o, = 72,875/(9+1)(2)(\\)(/2)(0.85) b" e, = 6062 psi Moment (72,875)(2.625) = m [(1)(fi)( )(4.5)+(2)(4.5)2(2/3) 20 (\\)([2)()}(0.85) o,= 8,841 psi [

4

+ = 10,719 psi T s m FS = 21,000/10,719 = 1.96 l l l 6L) l' l

..:. :a ~ STD-R-02-001 Page 60 of 148 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" \\ d 3.. 5" Impact The energy absorption sequence will be the same as that previously shown for the drop on the long flat edge and will consist of initial crushing, bending, and crushing. Because the cover overhangs 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 inch) Fa F Thickness = 2 in. F Fb '9 e 5" = M = (38,000 psi)(1 in)(1 in)(25.14 in) = 955,320 in-lb Fa = M/x = 955,320 in-lb/5 in. = 191,064 lb F = 191,064 (J2) = 270,205 lb Area of crushed steel = F/38,000 psi = 270,205 lb/38,000 psi 2 = 7.1 in 0

STD-R-02-001 Pe$e61of148 8 7 lug The area of the trapezoid is described by (1.4d + 3.44d ) where d/ 4 = depth of crush. 2 7.1 = 1.4d + 3.44d2 or d = 1.25 in therefore, (d/8) = 0.882 inch Width of crush = 1 + 4.87d = 7.08 in 2 3 Volume of crushed steel = d /2 + 1.925d + 1 = 4.18 in 3 Energy absorbed in crushing = (4.18in )(38,000 psi) = 158,840 in-lb l f" / Deceleration Force During Initial Crushing i d' ~ Energy absorbed 158,840 in-lb i I 5" Energy remaining = 477,160 ( I, \\c 2 Force = (7.1 in )(38,000 psi) / 3-3/4" 3/8"_/ = 270,205/53,000 = 5.1 g's l l n v 120 L 3" l-1/8 Bending of Cover After the initial crushing of the corner and the buildup of force noted above the corner of the cover will bend inelas-tically 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 absorption and deceleration forces will be as follows: 50 Fa j l.3,, Bending of lid (15* until lug hits ( outer shell of cask) Energy remaining = 477,160 in-lb l g Energy absorbed in bending (1.3" travel) E = (Fa)(d) = (191,064 lb)(1.3") O = 248,383 in-lb g Energy remaining = 228,777 in-lb I C n-

STD-R-02-001 Page 62 of 148 JAk 7 ag f.V Failure of the Lug After coming in contact with the shell, the lug will fail due to ten-sile 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 tension W I c mPression d h" weld all around I o/ 4 N 4.5 F / / pL) m [(1.5)( )(1.875)+(\\)(J2)(1.875)2(2/3)(\\)(2)]0.85 M = 2a M = 5.2(21,000 psi) = 109,368 in-lb The compressive strength of the shell of the cask will be equal to or greater than the tensile yield strength of 49,000 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 vill shear the weld is attained.

STD-R-02-001 Page 63 cf 148 UU 7% - 0.375" d y2 I F 72 4 L d F=49,000(1.5)(yg)f=36,750y2 lb Y2 = 4.5 h in 3 M = (F) (y2) (F)(4.5 y2 ) in-lbs = 3 165,375 y2 -12,250y}=109,368in-lbs = 12,250 y} - 365,375,y} + g9,368,= 0 Y1 _ 165,375 - (165,3752 - 4(12,250)(109,368) 2(12,250) y2 = 0.70 inches The depth of shell deformation or, d, will be as follows: d y1 0.70 0.375 4.5-yi 4.5 - 0.70 d = 0.07 inches Deflections or deformations of this magnitude in the shell will not affect the integrity of the cask. r (n) l l

STD-R-02-001 Page 64 of 148 JM 7 156 /b V Secondary Bending During and 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 636,000 in-lbs Less Initial Crushing 158,840 in-lbs Less Initial Bending 248,383 in-lbs Remaining Energy 228,777 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. O d 9. a d 2 M = F f = 955,320 in-lbs f = J25 - d a a 2 F, = 955,320//25 - d 6 -A

STD-R-02-001 Page 65 of 148 JAN 7 INS O~ The energy absorbed ~is computed as follows: displacement 1 Fa Fa(avg) d E IE 1.3 4.83 199,034 1.5 4.77 200,290 199,662 0.45 89,450 89,450 2.0 4.58 208,468 204,380 0.50 102,190 191,640 2.18 4.5 212,306 210,387 0.18 37,870 229,500 The secondary bending is capable of absorbing the remaining energy. The additional displacement of the lid during secondary bending is: 2.18 - 1.3 = 0.88 in. Deceleration Forces - It was calculated that the initial crushing caused a deceleration force of 5.1 g's. Calculate the deceleration forces of secondary bending since this is the shortest distance travelled in any of the phases discussed. 2(KE)(g) = 4[2(228,777)(386.4) = 57.76 in/sec v =} y V 53,000 v = 28.88 in/sec at = AX/v = 0.88/28.88 = 0.0305 sec 2 a = AV/At = 57.76/0.0305 = 1893.8 in/sec 2 2 a = (1893.8 in/sec )/(386.4 in/sec ) = 4.9 g's This indicates that the maximum deceleration force during a 12 inch drop on a short flat corner on the lid is 5.1 g's. l This does not exceed the g forces calculated in the drop on a long flat. Q

...~...:.... ......... ~.,.. STD-R-02-001 Page 66 of 148 1"J ilO 3.2.6.6 Side Drop The cask is dropped on its side. The energy is assumed to be absorbed entirely on the lid edge. Total energy = 636,000 in-lb 636,000 in-lb = 1/2 mv" Initial velocity, v = 96.3 in/sec Let d = depth of crush Voluoe of steel required to absorb energy 636,000 in-lb/38,000 psi = 16.7 in* W = d Tan 67.5 67 hU gq q./ s 3 Volume of steel described by [2(1/7 ow)(2 in)] + [(1 in)(2 in)] Fi.aal depth = 4.82 d* + 2d = 16.7 in* d = 1.67 in As shown on Table 3-2, the highest g force exerted on the lid is 12.0 g. A\\.)

STD-R-02-001 Pege 67 cf 148 JAN 7 ggg Table 3-2 Energy Energy Incremental d Vol. Absorbed Remaining Velocity Time Deceleration (in) (in ) (in-lbs) (in-lb) (in/sec) (sec) (in/sec ) (g's) 2 3 (based on avg. vel.) 0 0 0 636,000 96.3 0.1875 0.54 20,689 615,311 94.7 0.00106 806 2.1 0.375 1.43 54,256 581,744 '92.1 0.00200 1310 3.4 0.5 2.21 83,790 552,210 89.7 0.00137 1752 4.5 0.75 4.21 160,027 475,973 83.3 0.00290 2207 5.7 1.0 6.82 259,160 376,840 74.1 0.00320 2875 7.5 1.25 10.03 381,187 254,813 61.0 0.00370 3540 9.1 1.5 13.85 526,110 109,890 40.0 0.00500 4200 10.9 1.67 16.7 636,000 0 0 0.00850 4705 12.0 () m-- --,--,-w _~.- - - - - -,, +.- ---,,,-.----,----,,g--- --e

STD-R-02-001 Page 68 of 148 4 JAN 7 IM6 mU The weight of the cask less the upper cover plate is 53,000 - 3260 = 49,740 lbs. The deceleration force acting on the weld between the two cover plates will be: F = 49,740(12.0 g's) = 596,880 lbs. The weight of the cask less the cover and shield plug is 53,000 - 5800 - 370 = 46,830 lbs. The deceleration force between the cover plate and the inner shell of the cask will be: F = 46,830 (12.0 g's) = 561,960 lbs. The loads calculated for the weld and the inner shell due to a one foot drop on the top corners in section 3.2.6.3 are higher than the respective loads calculated here. Therefore, the cask will safely survive a side drop as well. 3.2.7 Penetration Impact from a 13 pound rod will have no effect on the package. 3.2.8 Compression This requirement is not applicable since the package exceeds 10,000 pounds. CONCLUSION From the above analysis,.it can be concluded that the HN-100 Series 3 cask is in full compliance with the requirements set forth in 10 CF.R 71 for Type "A" Packaging. O

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i STD-R-02-001 Page 69 of 148 JAN 7 56 4.0 THERMAL EVALUATION The HN-100 Series 3 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 lir.it the amount of these materials that can be transported as fol-lows: Specific ( Total ( } Gamma Isotope Energy Activity Activity Mev pCi/ml Ci Cobalt 60 1.33 5.0 23.2 Cesium 137 0.66 140.0 650 (I) Based on cement solidified 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 cask, the heat generated by the waste will be as follows: ,pd Heat Total ~ Generation Activity Total Heat (Watts / Curie) (Curies) (Watts) (Btu /Hr) Cobalt 0.0154 23.2 0.35 1.19 Cesium 0.0048 650 3.14 10.7 The weight of waste per container will be about 15,700 pounds. Based on a specific heat of 0.156 Btu per degree F., 2449 Btu's or over 9 days with cesium would be required to heat the waste one degree Fahrenheit. Accordingly, the amount of heat generated by the waste is insignificant. /f V

= _.... -we~.. STD-R-02-001 Page 70 of 148 ,50V 141985 9 5.0 CONTAINMENT The shipping cask is a vessel which encapsulates the radioactive material and provides primary containment and isolation of the radioactive material 1 from the atmosphere while being transported. The cask is an upright 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 annulst 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 8 equally spaced 3/4 inch studs on a 20-7/8 inch diameter circle. 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 2 2 2 2 - Ri ) = n(40 - 39.25 ) = 186.73 in Area = n(Ro Gasket is equivalent to 3/8 inch thick by 3/4 inch wide Durometer 40. Based on past experience from the manufacturer, a torque of 175 to 200 ft-lbs 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) + 6,170 lb lid weight = F = 34,170 lb. 1 1 Equivalentpressureongasket=f=34,170lb/186.73in 2 2= 183 lb/in As shown on Appendix D-1, the compression of the gasket due to tighten-ing of the ratchet binder to this minimum is 20% of the gasket thicknes 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 120 ft-lb. O

STD-R-02-001 Page 71 of 148 1i1 The gasket dimensions are 22.75 in. OD, 20.25 in. ID, and 3/8 in. thick. The gasket is equivalen; to Durometer 50. 2 z 2 2 10.125 ) = 84.43 in 2 - Ri ) = n(11.3/5 Area = n(Ro Force downward on the lid is the sum of the weight of the lid plus the force of the studs (P). P= = (120 ft-lb)(12 in/ft)/(0.15)(0.75 in) P = (12,800 lb/ stud)(8 studs) ='102,400 lb W = 370 lb Total force = 102,400 + 370 = 102,770 lb Pressureongasket=f= 102,770 lb/84.43 in2= 1217.2 psi As shown on Appendix D-2, the compression of the shield plug gauket is 33% of initial thickness or 1/8 inch. l 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 of 7.5 psig as required by 10 CFR 71, Appendix A. As described in Section 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 3.2) is 4,113 pounds. The resultant strain on the steel ratchet binder (1-3/4" diameter) is: 8 f P/AE = (4.113)/(1.77)(30 x 10 ) = 0.000093 in./in. l l P = 4,113 lb l 2 A = Area of 1\\" minor diameter = 1.77 in 8 E = Youngs Modulus = 30 x 10 and for a 24 inch long binder, total strain is: (24 in.)(0.000077 in/in.) = 0.0019 in i / This is less than 2% of the initial compression of the gasket. =

STD-R-02-001 Page 72 of 148 ( 5.' 1 4 1985 0 1 5.4 Gasket Compression Test j A compression test to check resiliency was done on Items 4 and 5 on Dwg. C001-5-9138, the. primary lid and secondary shield plug gaskets. The samples were each 1.n by 3/8 inch thick made of Durometer 40 2 neoptene and Durometer 50 neoprene recpectively. Each sample was put in a compression device and compressed. The final results indicated 2 Durometer 40 that it required abcut 4,500 lb to compress the 1 in sample to a thickness of 1/8 inch. After removing the sample from the test stand, the sample returned to its original thickness oi 3/8 inch. 2 Similarly, the 1 in 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 v s removed. The test compressed each gasket material 66 percent of its original height and each survived. The spacer rings have been increased to 1/4 inch which limits gasket compression to 33 percent of the gasket thickness, thereby further reducing possible damage to the gasket material. 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. A cask having an octagonal cover secured by ratchet binders was drop-ped on the extended corner by Nuclear Packaging, -Inc. The identifica- /~} tion number of the package which was dropped is 71-9130. The package, Model No. 50-256, had a weight of 19,160 lbs and was loaded with a liner containing sand with a weight of 4200 lbs for a gross weight of 23,360 lbs. The package was dropped on an essentb1ly unyielding surface from a height of 46 inches. The package was pressure tested before and after the drop test and no leakage was detected. The deformation of the corner subjected to the drop test is shown below: 0" 1" "!" 3" 4" 5" 6" 6.75" I n o o o p Corner j j Deformation / / \\r i / k rNa/hh88 ddd d dd O ~... _ _ _ _

STD-R-02-001 Pn,ge 73 of 148 .m 7 gas /A The energy absorbed in dropping a 23,360 lb package from 46 inches is 8 in-lbs. The energy to be absorbed in a one foot drop of an 1.07 x 10 HN-100 Series 3 cask is 12 x 53,000 = 6.36 x los in-lbs or less than 60 percent of the unit tested. The cover's are the same thickness and the overhang of the corners are approximately the same. Accordingly, the HN-100 Series 3 should experience 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. J i l o ( '. ..-..~____-_,.,__,.._m_._,_,

STD-R-02-001 Page 74 of 148 [U ^T 6.0 OPERATING PROCEDURES Customers that use the KN-100 Series 3 cask are supplied a copy of the Hittman Rad Services Manual. This manual describes the services that will be supplied and contains a section of requisite operating procedures. These procedures describe the inspections of the cask and trailer, handling precautions, loading / unloading procedures, and the surveys and forms that need to be completed prior to the cask leaving the customer's site. Section 6.1 and 6.2 outline the requirements set forth in the operating procedures for the KN-100 Series 3 cask. Upon receipt of the package, the following routine determinations specified in 10CFR71.87 shall be performed: The packaging is proper for the contents to be shipped o The packaging and tie-downs are in unimpaired physical condition o Any external contamination shall be within limits o 6.1 Procedures for Loading the Package 6.1.1 Loosen and remove the ratchet binders, which secure the /s primary lid. bg 6.1.2 Using suitable rigging attacher to the lid lifting lugs, lift and remove the primary cask lid. Care should be taken to ensure that the lid and cask sealing surfaces are not damaged during removal and placement of the primary cask lid. The underside of the lid may be contaminated, exercise appropriate precautions. 6.1.3 Inspect the interior of the cask to ensure that no water, loose articles, nor significant defects are present. 6.1.4 Place the disposable liner, steel drums, high integrity container, or other waste container into the cask. Use shoring to secure all but close fitting contents inside the cask. l 6.1.5 Inspect the primary lid gasket for defects or deterioration which would affect sealing. A seal must be made for ship-ment. 6.1.6 Using suitable rigging attached to the lid lifting lugs, lift the primary lid and place it on the cask. Use the alignment pins or alignment marks for the proper alignment and ensure that the primary lid gasket is not damaged, s'h (_) l

l STD-R-02-001 Page,75 of 148 t.. 2 - L., (3 \\/ 6.1.7 Install and hand-tighten the ratchet binders to achieve a uniformly compressed gasket. Then torque each of the ratchet binders to 175 to 200 ft-lbs to effect the desired compres-sion on the primary lid gasket. 6.1.8 If loading waste through the shield plug (steel liner and high integrity container (HIC) only): Loosen and remove the shield plug holddown nuts, a. b. Lift and remove the shield plug using the shield plug lifting lug. Care should be taken to ensure that the plug and cask sealing surfaces are not damaged. The underside of the plug may be contaminated, exercise appropriate precautions. Load waste into liner or HIC through shield plug opening. c. d. Install the liner or HIC closure components. Inspect the shield plug gasket for defects or deteriora-e. tion which would affect sealing. A seal must be made for shipment. f. Using the shield plug lifting lug, lift the shield plug C,e,) and place it in the primary lid opening. Use the s alignment pins or alignment marks for proper alignment and ensure that the shield plug gaske,t is not damaged. g. Install and tighten the shield plug holddown nuts to 120-135 ft-lbs. 6.1.9 Install tamper proof seals. 6.1.10 Perform a radiation surs ey on the cask and trailer system to ensure that the radiation levels and contamination levels are below specified limits. 6.1.11 Complete the necessary shipping papers, certifications, and pre-release checklist. Placard vehicle and label cask as necessary. 6.2 Procedure for Unloading the Package 6.2.1 Survey the cask and trailer in accordance with applicable site requirements. 6.2.2 Remove the cask primary lid as in Sections 6.1.1 & 6.1.2. 6.2.3 Exercising caution due to possible high radiation dose rate, connect slings of the disposable container or pallet to a { }'

~ STD-R-02-001 Page 76 of 148 (\\t suitable lifting device and remove the disposable container from the cask, s Note: Care should be taken to avoid damage to lid and cask gaskets. j 6.2.4 Install the cask primary lid as in Sections 6.1.6 and 6.1.7. 6.2.5 Survey the cask and trailer for release in accordance with applicable site requirements. 6.3 Preparation of an Empty Package for Transport 6.3.1 Inspect the interior of the cask to ensure that no water or loose articles are present. 6.3.2 Complete the closure and survey of the cask in accordance with Sections 6.1.6 - 6.1.7 and 6.1.9 - 6.1.11. r 4 l I i i

STD-E-02-001 Page 77 of 148 h3V 141985 0 7.0 ACCEPTANCE TESTS & MAINTENANCE PROGRAM 7.1 Acceptance Test Fabrication of the HN-100 Series casks meets the requirements of' Subpart D of 10CFR71. Fabrication is implemented and documented under a Quality Assurance program in accordance with the applicable require-ments of 10CFR71, Subpart H. 7.1.1 Visual Inspection The package will be inspected visually for any adverse condition in materials or fabrication using applicable codes, standards, and drawings. Materials are specified under the ASTM and ASME codes. Welder qualifications and weld procedures are in accordance with ASME, Section IX. Non-destructive testing of welds includes visual, liquid penetrant, and magnetic particle as described in the ASTM code, prior to painting. 7.1.2 Structural and Pressure Tests After fabrication is complete, the cask assembly is sub-jected to a pneumatic pressure test of 8 psig (-0 psig, ) + 0.5 psig). The package is visually inspected after the [(~/ pressure test and must remain in an unchanged condition. 7.1.3 Leak Tests A leak test will be performed using a test fixture (with calibrated pressure gauge and pre-set relief valve) mounted into the shield plug cavity on the primary lid. Air is introduced at a maximum rate of 1/2 psig/ min until the pressure reaches 8 psig (-0 psig, + 0.5 psig). All joints on the test fixture, primary lid and shield plug gaskets are bubble tested. The package is pressurized for 30 minutes and should not decrease in pressure below 7.5 psig 3 8 STD. cm /sec. The for a test sensitivity of at least 10 system will be depressurized at a rate of approximately 2 psig/ min and the test fixture removed and shield plug reinstalled. l 7.1.4 Component Tests 7.1.4.1 Gaskets Prior to painting, seating surfaces are to have a 63 microinch finish. Leak testing (see Section 7.1.3) of the cask will be the final acceptance for gasket design. (u ;/ - x l i-

STD-R-02-001 Pcg2 78 of 148 n.: L..a 7.1.5 Tests for Shielding Integrity Upon completion of the lead shielding pour, a gamma scan is done of the cask wall to determine lead thickness, and an existence of any voids or impurities in the poured lead. The gamma scan procedure contains acceptance criteria for verification that lead thickness is not less than 1-3/4 inches. 7.1.6 Thermal Acceptance Tests No thermal acceptance testing will be performed on the HN-100 Series cask. Refer to the Thermal Evaluation, Sec-tion 4.0, of this report. 7.2 Maintenance Program Maintenance and repair of the KN-100 Series 3 cask is controlled by the Quality Assurance Program. The casks and trailers annually undergo three (3) routine technical inspections spaced within each four months. These inspections are proceduralized in cask maintenance and repair procedures. 7.2.1 Structural and Pressure Tests j_, Inspections will be performed to check the cask for contami-( ) nation, structural damage to interior or exterior, gaskets, studs, signs and placards, shielding, ratchet binders, and tie downs. 7.2.2 Leak Tests A leak test, according to Section 7.1.3, will be performed at least once within the twelve (12) months prior to cny l use. 7.2.3 Subsystem Maintenance The KN-100 Series 3 cask contains no subsystem assemblies. 7.2.4 Valves, Rupture Discs, and Gaskets on Containment Vessel Gaskets which show any visual defects (cracking, gouging, tearing, etc.) will be replaced in accordance with the packaging drawings and cask maintenance and repair pro-cedures. The gaskets will be replaced if inspection shows any defects or every twelve (12) months, whichever occurs first. l O

STD-R-02-001 Page 79 of 148 7.2.5 Shielding Shielding tests will be performed if damage has required repairs affecting shield integrity. Any additional shield testing shall be in accordance with the original criteria specified in Section 7. 3. 51B I f 4 i 2-vt-v-n7e wy-le=<www g, ewe =a------- or----w---_wy,y -w%p ,.w- -7,m.- ,wegm.,y ,.-r-,---- ---.-----mv--g-- y-, w-w

STD-R-02-001 Page 80 of 148 NEV 141%5 O APPENDIX A l O O

STD-R-02-001 V 143NS 'O APPENDIX A ADDENDUM TO SAFETY ANALYSIS REPORT FOR SHIELDING INSERT FOR THE HN-100 SERIES 3 RADWASTE SHIPPING CASK 4 Referencing 10 CFR 71 Type "A" Packaging Regulations t l Westinghouse Hittman Nuclear Incorporated Columbia, Maryland 21045 O

i STD-R-02-001 P:ge 82 of 148 1.0 PURPOSE AND BACKGROUND The purpose of the following document is to provide the informatici and engineering analysis that demonstrates the performance capability and structural integrity of the HN-100 Series 3 Radwaste Shipping Cask and its compliance with the requirements of 10 CFR 71, Section 71.21 and Appendix A with the use of a 1" thick steel shielding insert.

2.0 DESCRIPTION

The HN-100 Series 3 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-100/150 One large disposable container HN-100 RADLOK One large high integrity container HN-100/14 Fourteen 55 gallon drums HN-100/10 Ten 55 gallon drums [ The HN-100 Series 3 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. A 1 inch thick removable steel cylinder with a bottom is placed in the cask body. A 1 inch thick steel plate is bolted to the lid and one to the shield plug. The cask's principal outside dimensions are 81-3/4 inches outside diameter and 81-1/2 inches high. The internal cavity with the insert are 73-1/4 inches inner diameter and 71-5/8 inches inner height. 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 the 1 inch steel insert, a 3/8 inch inner steel shell, a 1-3/4 inch thick concentric lead cylinder, and a 7/8 inch thick outer steel shell. The bottom is five inches thick (two 2 inch thick steel plates welded together and 1 inch removable steel insert bottom) and the 2 inch plates are welded integrally to both the inter-nal 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 f"N and a stainless steel cavity sleeve are optionally provided. /U w-r-. +-, vi-y t

STD-R-02-001 Psge 83 of 148 2.2 Cask Lid The cask lid is five inches thick (two 2 inch thick steel plates welded together and 1 inch removable steel plate) 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. The one inch steel plate is bolted to the underside of the primary lid with eight S/8 inch bolts. 2.3 Shield Plug The shield plug is six inches thick (two 2 inch thick steel plates and one 1 inch thick steel plate welded together and 1 inch removable steel plate) 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 lifting device located at its center-to focilitate bandling. The one inch steel plate is bolted to the underside of the shield plug with four 5/8 inch bolts. 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 beat rate limits are observed, and the material secured against movement relative to the cask. Basically, the inter-nals consist of the waste, contained if process waste is being trans-ported, 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: ) "O GW "* w* D O

STD-R-02-001 h$0* fh30 Shield Insert Component Weight Per Weight with Component Shield Insert Cask body 29,030 lbs 5,995 lbs 35,025lbsl Shield Insert 7,245 lbs Closure lid 5,800 lbs 1,195 lbs 6,995 lbs Shield plug 370 lbs 55 lbs 425 lbs Total cask (unloaded) 42,445 lbs HN-100 - Large con-tainer(s) and waste 10,555 lbs HN-100 - 55 gallon size containers (up to 14 10,555 lbs drums of radioactive waste) 2.8 Radwaste Package Contents 2.8.1 Type and Form of Material The contents of the various internal containers can be process solids in the form of spent ion exchange resins, filter exchange media, evaporator concentrates, and spent filter cartridges. Materials will be either dewatered, 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-l tainers with weights not exceeding 10,555 pounds. Radio-I active materials may include source and transuranic ma-terials in Type A quantities or greater than Type A quan-tities of low specific activity materials. The contents may also include exempt quantities of fissile material as defined in 10 CFR 71.9. 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 h As noted, the primary lid is secured by means of eight high strength ratchet binders. The secondary lid is

STD-R-02-001 5tf"f5d8 affixed with eight 3/4 inch diameter bolts. Therefore, the gl package is equipped with a positive closure system that will (d 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 ASTM A516 Gr 70 with f of 38,000 psi 2 For the shell, steel having a maxilum yield strength of 49,000 psi is specified. For the tie-down lugs, steel having a minimum yield strength of 46,000 psi is specified. 3.1.3.2 Stresses in material due to shearing is analyzed using the " Maximum Energy - Distortion Theory" which statestheshearingelasticlimitis1//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 Struc-tural Code D1.1-79. For fillet welds, shear stress sj on effective throat regardless of direction of load-ing is 30% of specified minimum tensile strength of weld metal. For complete joint penetration groove welds with tension normal to the effective area the allowable stress is the same as the base metal. Fillet weld allowable stress = (70,000 psi) (0.3) = 21,000 psi In order to be more conservative, a weld efficiency of 85% is also added. 3.1.4 Lifting Devices 3.1.4.1 Package Weight The package weights used for analysis are as follows: Empty package 42,445 lbs Payload: large container and waste 10,555 lbs l Gross Weight 53,000 lbs l Design and Behavior of Steel Structures, Salmon & Johnson, page 47. ^ I 2 ASTM Standards, Parts 4 and 5. .k p y ..*7 -p..,, -. - _.

STD-R-02-001 Page 86 cf 148 JM 7 Igg 3.1.4.2 Cask Lifting Lugs Material is ASTM A516 Grade 70 with a minimum of 38 KSI Tension Yield Strength and 21,926 psi Shear Yield Strength (57.7% of 38,000 PSI). Maximam Load 53,000 lbs. and four lugs are used to lift the l cask. 53,000 lbs x 3 g's/4 lugs = 39,750 lbs/ Lug (Vertical) TEAR OUT i Sling Angle to Lift 45' Load 39,750 lbs/ Sin 45* g, E Load = 56,215 lbs. l.I 2.5" DI A. Stress = 56,215 lbs/[2 x (2" x 2 5")] .... :.g /m N c = 5,622 psi N "THK. I I (.) I i 2 3 .I. i s 6" 2 f i i i 5,622 PSI << 21,926 F.S. = 3.9 /N O

STD-R-02-001 Pege 87 of 148 JAK 7 EEE O j i l 1 i 3.75"a I el S* - g,nj TENSION l 3 t' e i i n8 e to Lift 45' 1 i 8 2.5 3 i Load = 56,215 lbs. I i Stress = 56,215 lbs/(2" x 5.4926") o = 5,117 psi 5,117 PSI << 38,000 psi F.S. = 7.5 4.2426 - 1.25 + 2.5 = 5.4926 TEAR OUT Vertical Lift Stress = 39,750 lbs/[2 x (2" x 2.5")]I o = 3,975 psi 3,975 PSI << 21,926 F.S. = 5.6 TENSION Vertical Lift l Stress = 39,750 lbs/2" x (6" - 2.5") o = 5,679 psi-5,679 PSI << 38,000 psi F.S. = 6.7 3.1.4.3 Cask Lifting Lug Welds All welds 3/4" fillet. Allowable shear stress on effective area of weld is 0.3 times nominal tensile strength of weld metal, E 70 rod, or (0.3) (70,000) = 21,000 psi. v w y em

c

STD-R-02-001 Page 88 of 148 JAN 73 2' WF.LD j i i i g I 1 ~ .I 7.75 WELD - 6" WELD l g2.5,, Dia. 6t s I. ;, {,: l k

i. j 6"

( h i i, A l mwevyrsssw2*m / l i I~ 6" WELD BOTH SIDE 5 i L 2" WELD Vertical Lift Assume weld to be 85% efficient. Minimum throat of weld = (Sin 45*) 3/4" = 0.53" Weld strength per inch = 0.85 (0.53" x 1") (21,000 psi) = 9460 lbs/in. of weld Weld required = 39,750 lbs/9460 lbs/in. of weld = 4.2 in. of weld 4.2 in. << (2" + 2" + 12" + 7.75" + 6") F.S. = 7.1 Lift Lugs with 45' Sling Angle The forces are 39,750 lbs vertical and 39,750 lbs horizontal. These forces will be restrained by the 7.75 and 6.0 inch 3/4 inch welds attaching the lift lugs to the reinforcing plate and the two 6.0 inch 3/4 inch welds attaching the lift lugs to the tiedown lugs. The shear stresses due to the vertical component of the force is: f = 39,750/(7.75 + 6.0 + 6.0 + 6.0) (0.75) (0.707) (0.85) = 39,750/11.6 = 3427 psi 6

STD-R-02-001 Pege 89 of 148 JAN 7gg The centroid of the welds is located as follows: \\J y = [7.75(7.75/2) + 6(6/2) + 6 (7.75) + 6 x 6) + (7.75 + 6 + 6 + 6) (30.0 + 18 + 46.5 + 36) + 25.75 = = 5.06 in, b* Total Moment = (2.75 in) (39,750 lb) = 109,313 in-lb h Total Moment = Compressive Moment + Tensile Moment I d Compressive Moment = Tensile Mom'ent 2.75" Total Moment = 2(Tensile Moment) = 2f[(6 x 2.69 + 2.69 x 2.69 x 1/2 + 6 x 0.94 p 7.75" I + 0.94 x 0.94 x 1/2)(0.75)(0.707)(0.85)] 5.06" ' ( +

+

(* - 6"@3/4" '4 = 2f (25.7)(0.75)(0.707)(0.85) O. 34" = 23.29 f 2.69" \\ 6"@3/4" f = 109,313/23.29 = 4,694 psi Combined Stress = 4(4,694)2 + (3,427)2 6"@3/4 = 5,812 psi F.S. = 9460/5,812 = 1.63 Therefore, it can be safely concluded that the lug will not yield under a load equal to three times the weight of the package. 3.1.4.4 Tiedown Lugs for Lifting Cask (inadvertant use) If it is assumed the entire load is carried by 2 the tiedown lug, Section Modulus = (bh )/6; where b = 15.5 in., length of tiedown lug; and h = 2", thickness of tiedown lug, (See Diagram page 98). l Tiedown lug section modulus = 15.5 (2)(2/6) = 10.33 in Stress = (39,750 lbs)(3"/10.33 in ) l l 3 8 = 11.5 KSI where 3" is the distance from the cask surface to the centerline of the hole. 11.5 KSI < 46 KSI F.S. = 4.0 n) t Weld required on lift lug 4.2 in 4.2 < 12" F.S. = 2.9 - ~..,...

_.~.......~. _ _.._....__. _ STD-R-02-001 Pegn 90 of 148 JAN 7 Igg O Weld to tiedown lug Shear = 39,750 lbs/(2 x 15.5) = 1282 lbs per l l inch of weld compression or tension due to moment couple 39,750 lbs (3"/15.5 in) of tiedown lug = 7694 lbs in/ inch of lug and moment couple with 2" thick lug 2" (3847 lbs) = 7694 lb/in Moment of Rotation (39,750)(4.34)/15.5 in = 11,130 4b-in/ in. lug For 2" lug, then 5,565 lb/in Stress on weld to tiedown lug = 0 = J(1282)2 + (3847)2 + (5565)2 = 6885 6885 lbs per inch of weld < maximum of 9460 lb per inch of weld F.S. = 1.37 3.1.4.5 Lid Lifting Lugs (Secondary and Primary) Secondary Lid Lifting Lug V Material is ASTM A516 Grade 70 Maximum Load 425 lbs. a Carried by one lug 8%" o is a I i TEAR OUT Area = 2[(1-1/2 - 15/16) - 7/32] (3/8) l Area = 0.258 in.2 h" DIA* Stress = 3 g's X 425/0.258 in.2 = 4,941 psi 3 " THK. 941 Psi << 21,926 F.S. = 4.4 4 TENSION Area = (2.0 - 7/16) 3/8 = 0.586 in.2 Stress = 3 X 425/0.586 = 2,175 psi 2,175 psi << 38,000 psi F.S. = 17.4 Secondary Lid Lifting Lug Weld l 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) X 0.353 = 1.68 A = 1.68 in.2 Stress = 3 X 425/1.68 o = 759 psi j 759 psi << 21,000 psi F.S. = 27.6 I

STD-R-02-001 Page 91 of 148 h 14125 Therefore, the secondary lid lug is able to resist h a load of three times its weight without reaching a yield stress. C. Primary Lid Lifting Lugs Material A516 Grade 70 1" THK. Maximum Load = Primary Lid 6,995 lbs. I Secondary Lid 425 lbs. 7,420 lbs. I h h 2 N" l'OIA. h.* o y 6" = = TEAR OUT (Vertical Lift) b'd Area = 2 x [(2-3/4) - (1-1/2) - 1/2] (1) Area = 1.5 in.2 and stress = 7420/1.5 a = 4,947 psi 4,947 psi << 21,926 F.S. = 4.4 TEAR OUT (45' sling angle) .75 ) (1) = 1.06 in.2 2 Area (short path) = ( 4 752+ Load = 7,420 5 = 10,493 lbs. Stress = (1/2) (10,493)/1.06 = 4,950 psi 4,950 psi << 21,926 psi F.S. = 4.4 TENSION (Vertical lift) Area = (6 - 1) (1) = 5 in.2 Stress = 7,420/5 = 1,484 psi 1,484 psi << 38,000 psi F.S. = 25.6 v '? = J

i 4 ) STD-R-02-001 Page 92 of 148 NOV 14 s:5 TENSION (45' Sling Angle) (( ) Area (short path) = J2x(1.25)2 - 1/2 = 1.2678 in2 Stress = (1/2) (10,493)/1.2678 = 4,138 psi 4,138 psi << 38,000 psi F.S. 9.2 Primary Lid Lifting Lug Weld 1/2" weld at shear of 21,000 psi a) shear stress due to vert = horz component. av = oh = 7,420/(6+6+1+1)(Sin 45*)(0.5)(0.85) = 1,764 psi b) Stress due to moment Total Moment = Compression Moment + Tensile Moment Compression Moment + Tensile Moment Total Moment = 2(Tensile Moment) = 2(2x3x(2/3)xox0.5x0.707x0.85) F) (,, = 2.40 0 = 11,130/2.4 = 4,638 psi CombinedStress=J~(4,638+1,764)2+ 1,7642 = 6,641 psi F.S. = 21,000/6,641 = 3.2 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 l Since the primary and secondary lid lifting lugs are not capable of resisting the full weight of i the package they will be covered during transit. 3.1.5 Tie Downs The tie down lug material is ASTM A516 Grade 70 with a minimum yield of 46 KSI and a 26.5 KSI usable shear (57.7% of 46 KSI). The cask outer shell steel has a minimum yield of 49 KSI, which will be specified for fabrication and certified by test of the material.

STD-R-02-001 Page 93 of 148 JAN 7m - 53.9" ~, A 12.0" o t 69.8" = = 57.8" dh 3 1 81 '. 8" A = I 16 h h 60" 'n OC.C. G3, 3 ' Yo 7" N psem 6 i 2i VIEW AA g .---.y .__-.._____-,..7 _-..m.., y

1 STD-R-02-001 Pcge 94 of 148 JRM 7g A system of tie downs is provided as part of the p package. They will be utilized as in View A-A. 3.1.5.1 Cask Center of Gravity Item Weight Arm Moment 1,803,913 in-lb Cask 42,445 lbs 42.5" = 367,314 in-lb Liner 10,555 lbs 34.8" = and Waste 53,000 lbs 2,171,227 in-lb Center of Gravity = 2,171,227/53,000 f CG - 41.0 in. 3.1.5.2 Tie Down Forces Reference frame with respect to the trailer is shown on the tie down drawing (Page 93) up - down Y; front - rear X; side - side Z accelerations: Y axis - 2 g's X axis - 10 g's /S Z axis - 5 g's V Tie Down Lengths Long tie downs (high trailer attachment points) length = /632 + 57.82 + 53.9 = 101.0 inches Short tie downs (low trailer attachment points) length = 4602 + 57.82 + 53.9.2 = 99.2 inches Tie Down Tensions Tie down tensions resolved by vector direction Long tie down at tension T Along Y axis {0L = 0.6233 TL Along X axis T = 0.5333 T 10 0 g g

STD-R-02-001 Psge 95 of 148 )id. 7Jg Along Z axis Tg = 0.5718 Tg 10 Short tie down at tension T3 Along Y axis T = 0.6047 T g 3 Along X axis ' T = 0.5432 T 3 g Along Z axis T = 0.5825 T 3 3 10W Force (front-rear) Overturning (front-rear) due to 10W along X axis Overturning moment (taken about the axis of rota-tion @ Point A) = l 10(53,000 lb) 39.0" = 20,670,000 in-lb Each of the two rear (or front) tie down. (one long and one short) must restrain half the above moment or 10,335,000 in-lb Tension in the long tie down 10,335,000 in-lb = (64")(0.5333 T ) g + (69.8")(0.6233 T ) L T = 133,119 lb. g 4 Tension in the short tie down ' t 10,335,000 in-lb = (64")(0.5432 T ) l g + (69.8")(0.6047 T ) g T = 134,268 lb 3 SW Force (side-side) Overturning (side-side) due to SW along Z axis Overturning moment (taken about the axis of rotation @ Point A) = 1' 5(53,000 lb)(39.0") = 10,335,000 in-lb O ++o- .-e,p-.- c- +-*-# 4 . - -. ~. p

STD-R-02-001 Page 96 of 148 i JAN 7lg Each of two side tie downs (one long and one short) must restrain half the above moment or 5,167,500 in-lb Tension in the long tie down 5,167,500 in-lb = (64")(0.5718 T ) g + (12.0")(0.6233 T ) l g T = 117,244 lb l l g Tension in the short tie down 5,167,500 in-lb = (64")(0.5825 T ) l g + (12.0")(0.6047 T ) g T = 116,029 lb 3 2W Force (up-down) I.ifting (up) due to 2W along Y axis Lift = 2 (53,000 lb) - 53,000 = 53,000 lb { Each of the two long and two short tie downs will carry a quarter of the load. 13,250 lb = 0.6233 Tg T = 21,258 lb g 13,250 = 0.6047 Tg T = 21,912 lb g Total Tension Total tension with all forces acting simul-taneously l T = 133,119 + 117,244 < 21,258 g T = 271,621 lb T = 134,268 + 116,029 + 21,912 g T = 272,209 lb 3 6 -i

STD-R-02-001 Pega 97 of 148 JAN 7 gg r3 3.1.5.3 Tie Down Lugs V The tie down lugs are constructed of ASTM A516 GR 70 steel, having a minimum yield of 46 KSI and an ultimate tensile of 78 KSI. The following values are used in the design of the tie down lugs. Tensile Yield = 46,000 PSI Bearing Yield = (46,000)(0.9) = 41,400 PSI Tensile Ultimate = 78,000 PSI Shear Yield = (46,000)(0.577) = 26,542 PSI Shear Ultimate = (78,000)(0.577) = 45,006 PSI Allowable Shear Stress for Welds = 21,000 PSI Hole Diameter = 2.5 in, Pin Diameter = 2.25 in Tear Out - a = 272,209 lb/(3.0)(2)(1.75) 25,925 psi = i' [ F.S. = 26,542/25,925 = l.02 ,u IMS 9 Bearing a _ o = 272,209 lb/(3.0)(2.25) = 40,327 psi '{ /.75 i 83~ F.S. = 41,400/40,327 = 1.02 2.5 h 1r Tension / r, { o = 272,209/(3.0)(8.25-2.5) 8.25" = 15,780 psi F.S. = 46,000/15,780 = 2.92 l Reinforcing Plate Weld Thickness main lug 2 inches Thickness reinforcing plates 2 @ 1/2 inch d 272,209 lb Total thickness 3.0 inches Load on reinforcing plate = 2.25" Dia. Pin 1/6 x 272,209 = 45,368 psi 7 '\\ Length of weld = 6 + 7 + 7 = 20" % 2.5" Dia. Hole Area of shear = 20 x 1/2 x Sin 45' t x 0.85 = 6.01 in (9 6" Stress = 45,368 + 6.01 = 7,549 psi 7" F.S. = 21,000 + 7,549 = 2.78 ~'

STD-R-02-001 Page 98 of 148 JAN 7g 3.1.5.4 Tie Down Lug Welds fV a) Pure Shear [272,209 lb]/[2(15.5)(Sin 45*)(1)(0.85) + (6+6+2)(Sin 45*)(0.75)(0.85)] ] Fl 2" = 10,913 psi / 8 i b) Moment forces on weld T M = 272,209 x 2.75 = 748,575 in-lbs. l s 15.5" 6" @ 3/4" 2" @ 3/4" ~> 6" @ 3/4" pg 1" l Win su nni i nn, g . in>>>>,,rinssinis in ni si,, ~ l 'I W 6 I i Reference _,, "'""""'""j"3 5""'"/""/'f"/""'"""'('"."5"@1" ~1 15 Line ~ ~' 10.95 y 2 x1x3/ 4+2 x15. 5 x 8. 75+6 x0. '/5 x 8. 33+6 x0. 75 x10. 95+ 16. 5 x2x 3/ 4 0.75x2+15.5x2 + 6 x 0.75 + 6 x 0.75 + 2 x 0.75 = 384' 8.94 in. = 3 w$ J f / 4 9 sn uin n ni n n,n nin n n,,,$

2. 0 % sis,issnini sisi r, a

f l v/uinu u u u niuuiussuruu u u n uo uu u u urui, 7.94 6.56 = M = compressive moment + tension moment compressive moment = tension moment M = 2(tension moment) M = 2f[(2.01)(6)(0.75)(0.707)(0.85) + (6.56)(2)(1/2)(6.56)(2/3) (0.707)(0.85) + 2(6.56)(0.707)(0.85)(0.75)] M = 57.2f = 748,575 f = 13,087 psi Combined Stress f = /(10,913)2 + (13,087)2 = 17,040 psi F.S. = 21,000/17,040 = 1.23 (v

STD-R-02-001 Pcge 99 of 148 JAN 7gg 3.1.5.5 Analysis of Tiedown Loads on Cask Shell The tieoown 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. l HL Offset 4.25 in. p = - (, z __> Mc Fx F = 272,209 X cos 39-1/2* = 210,043 lb F = 272,209 X sin 39-1/2' = 173,146 lb M = Circumferential moment = (210,043 lb)(4.25 in) p

= 892,683 in-lb M = Longitudinal moment = (173,146 lb)(4.25 in) =

g 735,870 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 = (33*/360') (2n 40.9) \\ = 11.78 in. C = 1/2 the longitudinal width of the loaded 2 plate = 23.5 in/2 = 11.75 in. B = C /r = 11.78/40.9 = 0.288 3 3 B = C /r = 11.75/40.9 = 0.287 2 2 Check that 5 < y <,100 0

STD-R-02-001 Page 100 cf 148 JAli 7lg The highest stress on the outer shell is 42.1 KSI. { ( The steel used on the outer shell will be speci-f) fied to have a minimum yield strength of 49 KSI and will be certified by testing. 3.1.5.6 Failure Under Excessive Load The tiedown lugs are designed to fail under excessive load and preclude da' mage to the package. Based on ultimate strength of the shell material, the force required to initiate damage to the shell would be: F = (272,209 ) ( 8,000)r 504,330 lb (42,100) The lug is designed such that it will fail before the 504,330 lb limit is reached. l[ The mode of failure will be tear out of the tie-down lug. The force required to cause tear out is: F = (45,006)(2)(3.0)(1.75) = 472,563 lb. Compared to the force required to damage the shell, the factor of safe.ty will be: O(' F.S. = 504,330/472,563 = 1.07 r 6

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i CL D. ot 3C or' M4 " [ Nd ~ ~ ~~" ~ ~ ~ P 4C P/ \\P/.. /. U (.@\\, ) 6P e IC er .o + +. = + + ac.i r = ..N..g,=s "-(..,.Y s)"..$, =. N MN E -3M -3M6 +M6 +33Cs ."",. 6 '"- (..n )..y,. = llll)ffffi$$lV//f -3 w e7ze sme -377c v,3]lglglllg .<?..., n -(.<,":.y)..";, = -Se -ssw + sew.-se iI *i. ".#.s.oh,. "'(m..n)*..[,.= -5931 +583'i +58M -ss2.AQM M M ~ g'yl,',*'.'.*.""'4 1172 8 (c0 1172 8 (oo 4 fit 3 34383 4(It3 343e.3 '" " ~~" l ,,";. =' x- (,,, ) f, 'r 4c l 2e.1 .. s i) - = + + + + l = .r a c. r .. /...o n - (..,":.y ).."h, = ffj]$$$gd F//J

243 -30s eo43 +'ws a

.."'..u.& " (.. A )..'"',. - ll@ll$2l Bffd]$$ -is.miss + ism ->ss => .<,;..e,.c -( v.:.u).. ), = -2.774 -27N +27x +2774MMM t3 f <=

.:;..,.s

. u,).2. - -sm am +sm -e7szgena gglewe ^"/',"' '"l "; "- " tis 2r, 5s7a risz.6 5978 zts7a 6232. 24373 6297_ 1... ,o..= r. o

2...,1

+ +- e e + r e +- ma agg . v.n -.

r.. s

.sm +swn -5 m a Is"7.Tt T'O " g %. hhhh -42/0 J/2)O f#2IO +4'M 0 = =.2 : - sa gay so3y sm wo wo wo we c ""'""dTMSS4"i[Mj" - SRXsRRRRMtM32 s I

1) When T y' 0, S = largest absolute maanitude of either 2

x 4 x 4 r(/(cx - 04)2 + 47 S = 1/2 [c + 0 !(U -0) + 41 g Yd Yb

2) When T = 0, S = largest absolute magnitude of either S=0x, 04 or (O 0) 4 x

STD-R-02-001 Page 102 of 148 j JM 7gg 1 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 Since the package is constructed of steel and lead, tem-peratures down to -40*F will have no effect on the package. 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 Each binder has an ultimate strength of 135,000 lbs. Therefore, it can be concluded that the reduced pressure 3 will produce no detrimental effects. 7 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. l 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 = 53,000 lb x 12 in. 5 Maximum energy = 6.36 x 10 in-lb 5 3 Volume of steel = 6.36 x 10 /38,000 = 16.74 in

I STD-R-02-001 p:ge 103 of 148 JAN 7 tgg Energy will be absorbed by crushing of corner. O s N N N ' r. _ l N

t.

4 'A ,5,'e~ [, N g 9 N

./

/ N N..', 4

y

/ \\g\\ s, '/ / N / \\ / BOTTOM Ng j CORNER Ng / N/ f. The volume of the crushed ungula, assuming the worst case of a 45* impact angle is calculated by the following equation: I" 3 V, = R Sin 4 - - 4Cos4 3 4 = 16.4* when V, = 16.74 in R R h h-s A C 2p

STD-R-02-001 Pcge 104 of 148 JAN 7g The maximum amount of steel crushed will be: 0 b = R(1 - Cos 4) = 1.66 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. cf v.#e q AJ 'c*,/<% .~._:f-? ?~ t =~- 1.19" ~~~~ ~2 1.69" o y n R *e g ~@' ~ g *Tlie deceleration force exerted on the cask is l

s.. h t '...J calculated as the product of contact surface area v

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 b = 1.66 in C = R-h = 40.5-1.66 = 38.84 in y = JR -C2 = 440.52 - 38.842 = 11.47 in 2 x = C/cos 45' = 54.928 in Au = "( }( ) ~I ( } (54.928)(11.47)+(57.275)(40.5) sin 25 Au = 36.32 in2 F = (36.32)(38,000 psi) = 1,380,160 lb. g force = 1,380,160/53,000 = 26.0 g's

STD-R-02-001 Pcge 105 of 148 73 3.2.6.2 Effects of Bottom Corner Drop on Balance of Cask ll The 26.0 g deceleration 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 Deceleration Contents / s +% s 0' Upper 4 Bottom ~ Inner Shell Plate I.ead Shield lower Bottom p Plate Outer Shell Corncr Impact \\M lbh Dh Reaction Force Rottom Corner Drop ) N -e,,<*

STD-R-02-001. Pcge 106 of 148 JAN 7 ggg /- Summary of cask component weights as used in the following drop analyses L.) l Primary Lid 6,995 lb Shield Plug 425 lb i Outer Body Shell* 7,224 lb I Inner Body Shell 1,900 lb Upper Bottom Plate 2,645 lb Lowe'r Bottom Plate 2,864 lb Lead Shield 14,397 lb Waste Contents ** 16,550 lb ll

This includes the weight of lid ratchet binders, tiedown lugs, etc.
- This includes the weight of the cask body portion of the insert plus contents.

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 plat will activate the entire perimeter weldment to the lower bottom plate. S-Weight of primary lid insert plate and secondary lid insert plate are included in weights of their respective lids. Therefore, the attachment j( bolts are analyzed in the corner drop. The cask body shield insert is (')', considered part of the waste and therefore is considered free to move with the waste. These assumptions are conservative for this analysis.

STD-R-02-001 Pege 107 of 148 JAh 7lm Cask Analysis 1 - Load from Primary Lid and Shield Plug Decelerati Forces Shear Area of Weld \\ / \\ 'O e b, 07 / %.;s o- .ocs k Deceleration Forces P Detail "A*' / 1/2" k jg" l y /Renct i ForccM Loading = (6,995 + 425)26.0 = 192,920 lb Lateral force = 192,920 (sin 45*) = 136,415 lb. Axial force = 192,920 (cos 45') = 136,415 lb. 2 2 2 Inner shell area = (n/4)(76.25 - 75.5 ) = 89.388 in 2 - 80 ) = 222.317 in 2 2 outer shell area (n/4)(81.75 2 Total area = 311.705 in Inner area = 89.388/311.705 = 28% Outer area 222.317/311.705 = 72%

...;..- ;u... ) STD-R-02-001 Pn e 108 of 148 i JAE 7 EE6 (q Force on inner shel.1 = (136,415)(0.28) = 38,196 lb lateral and axial s Force on outer shell = (136,415)(0.72) = 98,219 lb lateral and axial 2-Stresses Developed in Inner Shell and Attachment Welds 4 ? Deceleration Forces Inner container Upper & Content s Bottom Plaiz 3/8" N wel Detail "B" Reaction on Inner Shell O \\ 1/2" welds Stress in weld around perimeter of inner shell at cask lid (38,196 lb)/n(75.5)(3/8)(0.707)(0.85) = 715 psi Total stress = 42I(715) = 1,011 psi F.S. = 21,000/1,011 = 20.8 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)(26.0)(sin 45')+(38,196/2)+(16,550)(26;0)(sin 45') = 340,832 lb Lateral Weld stress = (340,832)/n(75.5/2)(2)(3/8)(sin 45')(0.85) = 6,375 psi (lateral) (,)

STD-R-02-001 Pag 2 109 of 1/:8 JAN 7lm Axial weld stress is caused only by lid load and shell self weight. (_) 38,196 + 34,931 = 73,127 Axial weld stress = 73,127 lb/n(75.5)(3/8)(sin 45*)(0.85)(2) = 684 psi Total Stress = 46,3752 + 6842 = 6,412 psi 2 2 Axial shell stress = 73,127/(76.25 - 75.5 )(n/4) = 818 psi Shear shell stress = lateral force / area [(1900)(26.0)(sin 45*)+(38,196)+(16,550)(26.0)(sin 45*)] = 8'444 psi-(76.252 - 75.52)(n/4)(1/2) i F.S. = 21,926/8,444 = 2.6 i 3-Stresses Developed in Outer Shell & Attachment Welds \\- Deceleration Force / s Inner Shel1 Upper tt m Plate \\ Outer Shell Lower Bottom Plate h / / Crush Dept h a Weld Area in Shear Detail "C" Shear of Outer Shell Weld Stress in weld around perimeter of outer shell at cask lid (98,219 lb)/n(80.875)(0.5)(sin 45*)(0.85) = 1,286 psi both axial and lateral Total stress = 8 (1,286) = 1,819 psi F.S. = 21,000/1,819 = 11.5 / v

STD-R-02-001 Page 110 of 148 AA 7lg 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) (7,224)(26.0)(sin 45*)+(98,219/2)+(14,397/2)(26.0)(sin 45*) = 314,264 lb. = Lateral stress = (314,264)/n(80/2)(0.5)(sin 45*)(0.85) = 8,322 psi Axial Load = (7,224)(26.0)(sin 45*) + 98,219 = 231,031 lb Axial stress = 231,031 lb/n(80)(0.5)(sin 45*)(0.85) = 3,059 psi (3,059)2 = 8,866 psi Total stress in weld = ](8,322)2 + F.S. = 21,000/8,866 = 2.4 2. go2)(n/4) Axial stress in outer shell = 231,031/(81.75 1,039 psi < 36,000 psi yield = Lateral shear stress in outer shell 2 2 80 )(n/4)(1/2) (314,264)/(81.75 = 2,827 psi < 20,772 psi yield = 6

l STD-R-02-001 Pone 111 of 148 JAR 7 196 4-Stress in Weld Joining upper to Lower Bottom Plates Deceleration Forces Inner Container Upper Bottom Platu A 4"Ln t s 3/8" Detail "D" Reaction on Inner Shell / Load on weld = Upper bottom plate + Inner shell + Shear from lid on inner shell + Waste + 1/2 lead Load = [2,645 + 1,900 + 16,550 + (14,397/2)](26.0)(sin 45') + 38,196 = 558,366 lb Stress due to lateral load OL = 558,366 lb/(76)(n)(sin 45')(0.5)(0.85) oL = 7,782 psi Since all axial loads are transferred in bearing, the maximum weld stress will be equal to 7,782 psi. This is within acceptable limits. l e i o ---.-nn. ,,e._.,,,,-,__.,---.,.,.. -.,,. _.,,.. _ _, _ _,,, _ _ _., _ _, - _., _ _,

STD-R-02-001 Page 112 of 148 EDV 14155 () 5-Bolts Securing Lid Insert Plates to Lids Primary Lid Weight of Primary Lid Shield Plate = 1,200 lb Use eight 5/8" bolts (fy = 105,000 psi) Initial torque is 100 ft-lb-(max) force due to torque = T/hd (100)(12)/(0.15)(0.5364) = 14,914 lb. = (1,200)(26.2)(Sin 45') force due to deceleration = 22,231 lb. = 1 force on each bolt.=-(22,231/8) + 14,914 17,693 lb. = stress = 17,693 lb/0.226 int = 78,288 psi () F.S. = 105,000/78,288 = 1.34 Secondary Lid Weight of Secondary Lid Shield Plate = 60 lbs. Use four 5/8" bolts (fy = 105,000 psi) Initial torque is 100 ft-lb = 14,914 lb. force due to deceleration = (60)(26.2)(Sin 45') = 1,112 lb. force on each bolt = (1112/4) + 14,914 = 15,192 lb. stress = 15,192 lb/0.226 inz = 67,221 psi F.S. = 105,000/67,221 = 1.56 t C) I s J -.,.---..-.--..,c-.. ..,re,, ,.w.4 .,,a, ,-.-.__,,_.,.L--,--,...----.,..

STD-R-02-001 Pcge 113 of 148 3.2.6.3 One Foot Drop on Top Corners 7 \\' i A drop on the upper corners of the cask would decelerate the cask and would result in axial and transverse deceleration forces between the cover and the balance of the cask. 1 l 1 Contents Deceleration Cask Transverse Deceleration ' Inner Shell ^ Outer Shell rank Avtal Deccleration Primary Cover 3/8" 1/2" n i Impact Point o

=

=-

STD-R-02-001 Pcg2 114 cf 148 JAN 7In 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 deceleration force must be resisted by the bearing stress between the inner cover plate and the cask inner shell and by l the weld between the two cover plates. The magnitude of the transverse deceleration force will depend upon the orientation of the cask and the corresponding deceleration forces. As shown later, the maximum deceleration force will occur when the cask is dropped on a long flat edge of the primary cover. The maximum deceleration for this case is 17.88 g's. The weight of the cask less the, upper cover plate is 53,000 - 3260 = 49,740 lbs. The transverse deceleration force acting on the weld between the two plates is 49,740 x 17.88 (2 = 628,866 lbs. ll The weld is a 1/2 inch weld, 75-1/4 inches in diameter. The stress in the weld: = 8,853 psi '

75.25(1T 5 0.85)(0.707)

F. S. = = 2.37 5 The weight of the cask less the cover and shield plug is 53,000 - 6,995 - 425 = 45,580 lbs. The transverse force between the inner cover plate and the inner shell of the cask will be: F = (45,580)(17.88) (E = 576,271 lbs. A 40' are length on the inside diameter of the inner shell plate times the thickness of the lower pri-mary cover plate is assumed as the bearing area between the two surfaces. Area = D ( 0 ) t = 75.25 ( ) (2) = 52.5 The bearing stress between the two plates will be: f = 576,271 52.5 = 10,977 psi Allowable Bearing Stress = 0.9 x Tensile Yield Strength F.S. = N 2000 = 4.02 10,977

STD-R-02-001 Page 115 of 148 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 considered to be worst case. 37.4" 1 34.4" N w 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 majority 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 7 F Following impact the edge of a the corner will begin crushing 4 ~7l until inelastic rotation around y to the bend point occurs. 1 Neut ral The point at which this will ,,) AxlS occur is calculated as follows: Width at bend = 37.4 in t._ 1.50" x Thickness = 2 in M = (38,000 psi)(1 in)(37.4 in)(1 in) = 1,421,200 in-lb. 38,h0 psi - ] l f 1" 1R,000 psi ' c J. ( o

.u-.~.. ..-....-w..- STD-R-02-001 Pone 116 of 148 l JAN 7lg F, = Force required to initiate bending g F, = 11/X = 1, ,200 = 947,467 lb 947,467/53,000 = 17.88 g's (axial and lateral) F=F,([i)=1,339,920 1,339,920/53,000 = 25.28 g's (total) 2 Area crushed steel F + 38,000 = 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 teel = (0.512)(1.025)(34.4) 2 = 9.036 in3 l Energy absorbed in crushing 9.036(38,000) = 343,364 in-lb Bending l 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 and with the shell of the cask l Ta p l 1.5 inch i The balance of the energy will be absorbed by bending of the lid l [ Total Energy) - [ Energy Absorbed in Crushing] [(53,000 lb)(12 in)) - [343,364 in-lb) 292,636 in-lb = Energy Absorbed in bend l With an axial force of 947,467 lb required to cause bending, this amount of energy will be ab-sorbed by an axial displacement of l 292,636 in-lb/947,467 = 0.309 in. l

... ~.... - - -,.~. STD-R-02-001 Page 117 of 148 JAM 7m The g force developed during the bending process l is calculated using a kinematic approach. Velocity at start of bending is ,/2 keg _(2)(292,636)(386.4) V W V (53,000) = 65.3 in/sec As calculated before, the inelastic bending de-formation is 0.309 in. The time it takes the cask to move this distance, based on average velocity is AX/V

(0.309 in)/[(65.3)(0.5)in/sec)

8 0.00946 see g force = (AV/At)/386.4 in/sec) = (65.3/0.00946)/(386.4) = 17.86 g's (both axial and lateral direction) The above shows the maximum g force is 17.88 g's in crushing and bending in both axial and lateral directions. The force of impact on the corner is 947,467 lb. (axial component) Contents 40.2" 947,467 lbs = 1 A d i o U2R 1.5"j y2R 16.550 X 17.88 lbs - 23" 34.4" 21'i.- The loads on the ratchet binders will be propor-tional to their distance from the pivot point of the cover on the cask.

1 STD-R-02-001 Page 118 of 148 JAN 7 lgg Forces tending to open the lid consist of weights from waste, lid, and shield plug. l (16,550 + 6,995 + 425)(17.88) = 428,584 Summing the moments about point 'A' (947,467)(1.5) + (428,584)(40.2) = 18,650,277 in-lb 18,650,277 in-lb = .2R(23)(23/80.4) + 2R(57.4)(57.4/80.4) + 2R(80.4) 18,650,277 in-lb = 255.92R R = 72,875 lb (in farthest binder from impact) R = 72,875 (57.4/80.4) = 52,029 lb (in middle binder) R = 72,875 (23/80.4) = 20,847 lb (in binder closest to impact) 6

STD-R-02-001 Pcar 119 of 148 l The 1/4 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 947,467 lb. f O b.815' - L 4-y ze I G ZS

g The yielding surface area reacting against this force is proportional to the angle O and the radius.

( ~! } (35,000 psi)1 = pressure (psi) 2 (Pressure) n(39.875 ) in / degree = f ree/ degree 360 As seen from Table 3-3, the entire force is distributed over a 88* arc of the ring. This is less than the angle between three of the large flats. 6 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 0.30 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 88* distributed load, the effective area of these two welds is: t O [( .5)(0.5 H(76.25)(0.375)]2 = 105 in Area = n f = 947,467/105 = 9,026 psi The actual stress values will be lower since the upper ring will cause the load on the weld to be distributed over a larger area. 1 Report of the Iron F. Steel Technical Cossnittee " Estimated Properties and Machinability of Hot Rolled and Cold Drawn Carbon Steel Bars", SAE J414, Feb. 1968.

STD-R-02-001 Page 120 of 148 Table 3-3 () Pressure Exerted on 1/4" Seal Ring Due to Top Drop on Large Flat Angle Pressure Force / Degree IF Press /E Strain 1 34,780 12,026 12,026 0.001160 0.000290 2 34,769 12,022 24,048 0.001159 0.000289 3 34,748 12,015 36,063 0.001158 0.000289 4 34,715 12,004 48,067 0.001157 0.000289 5 34,673 11,990 60,057 0.001155 0.000288 6 34,620 11,971 72,028 0.001154 0.000288 7 34,557 11,950 83,978 0.001152 0.000288 8 34,483 11,924 95,902 0.001150 0.000287 9 34,398 11,895 107,797 0.001146 0.000286 10 34,303 11,862 119,659 0.001143 0.000285 11 34,196 11,825 131,484 0.001140 0.000285 12 34,082 11,785 143,269 0.001136 0.000284 13 33,956 11,742 155,011 0.001131 0.000283 14 33,20 11,695 166,706 0.001127 0.000282 15 33,673 11,644 178,350 0.001122 0.000280 16 33,515 11,590 189,940 0.001117 0.000279 17 33,348 11,531 201,471 0.001111 0.000277 18 33,170 11,470 212,941 0.001105 0.000276 19 32,983 11,405 224,346 0.001100 0.000275 20 32,785 11,337 235,683 0.001092 0.000273 O 21 32,577 11,265 246,948 0.001086 0.000271 22 32,360 11,190 258,138 0.001078 0.000269 23 32,133 11,111 269,249 0.001071 0.000267 24 31,895 11,029 280,278 0.001063 0.000265 25 31,649 10,944 291,222 0.001055 0.000263 26 31,392 10,855 302,077 0.001046 0.000261 27 31,126 10,763 312,840 0.001037 0.000259 28 30,850 10,668 323,508 0.001028 0.000257 29 30,565 10,569 334,077 0.001018 0.000254 30 30,271 10,467 344,544 0.001009 0.000252 31 29,968 10,363 354,907 0.000999 0.000249 32 29,655 10,255 365,162 0.000988 0.000247 33 29,333 10,143 375,305 0.000977 0.000244 34 29,003 10,029 385,334 0.000966 0.000241 35 28,663 9,911 395,245 0.000955 0.000238 36 28,315 9,791 405,036 0.000944 0.000236 37 27,958 9,668 414,704 0.000932 0.000233 38 27,593 9,541 424,245 0.000919 0.000229 39 27,220 9,412 433,657 0.000907 0.000226 40 26,837 9,280 442,937 0.000894 0.000223 41 26,447 9,145 452,082 0.000881 0.000220 42 26,049 9,007 461,089 0.000868 0.000217 43 25,643 8,867 469,959 0.000854 0.000213 44 25,229 8,724 478,680 0.000841 0.000210 Entire force 2(478,680) = 957,360 > 947,467 lb. will be distributed over an area 0 of ~ 2(44) = 88*

STD-R-02-001 Page 121 of 148 JA8 73 I.id Ratchet Binder Assembly Based on the 72,875 lb developed in the far ratchet binder during a ll 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 135,000 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: 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 + 72,875 = 1.70 0 6

STD-R-02-001 Pa e 122 of 148 JN 7 m 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 U E 1h 2 7_2 875 lb 1 2 1-1/8" a 't Resultant Force = 72,875 lb l j j j s s j j + l 0, = (72,875/2)/(1.125)2(n/4) = 36,657 psi F.S. = (74,000)(0.577)/(36,657) = 1.16 Lid Ratchet Binder - Upper Lug O e o 11 u \\ h" l[ h 3/4" h p 1.90' g" g ~ CW 4f DIA Hr / h ',, Hg ,_ is - i-Ib1/d 3-3/4" ( 6

STD-R-02-001 Pege 123 of 148 3R 7 m Tear Out - Shear - o, = 72,875 lb/(1.5)(2 - 0.0625 - 0.27)(2) = 14,568 psi FS = (38,000)(0.577)/(14,568) = 1.51 Bearing - OR = 72,875 lb/(1.125)(1.5) = 43,185 psi 1 73, Le/d _ 2/1.125 = 2.88 43,185/70,000 I o/ou 9/16.. i r Tension - 1/16" OT = 72,875/(1.9 + 1.375 - 1.25)(1.5) YI i ~ d 27a OT = 23,992 psi 2" { 5/8" h il U ^ FS = 38,000/23,992 = 1.58 7j fj Weld O v 1/2" double groove weld, 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) W Y " iull groove f " full groove'"[ (, 1-15/16". n r 'p 9/16" m,i-1/1!". r 8 , u 72,875 lb Neut ra 1 Axis O

STD-R-02-001 Page 124 of 148 JAN 7 ggg i Tension (:i o = 72,875/[(2)(1/2)(1-1/2) + (2)(1/2)(J7)(2-3/4)] (0.A5) g o = 15,909 psi g Moment o = (72,875)(0.5625) = 20m [(1/2)(1.5)(1.9375) + (2)(1/2)(1.4375)2 (2/3)(/7)(1/2)](0.85) 0,= 9,934 psi i Total o =a +o = 15,909 + 9,934 = 25,843 Tot t a l 4 j F.S. = 38,000/25,843 = 1.47 j d 4 1 r b i i 4 i i l i I I r I I I l t l __...,._-._--._..___...._..._z-.___.________---__

STD-R-02-001 Page 125 of 148 JAN 7pg i Lid Ratchet Binder - Lower Lug v Tear Out 1-3/8" \\" Shear - 4' o, = 72,875/(1.5)(1.75-0.0625-0.27)(2) n o, = 17,137 psi is o FS = 21,926/17,137 = 1.28 Bearing - (* }(* n ,DIA B = 43,185 psi I'!d = (1. 5)/(1.125) FS = o /fu (43,185)/(70,000) g 1 " Thk = 2.52 Tension N" 72,875/(3.5-1.25)(1.5) = 21,593 psi FS = 38,000/21,593 = 1.76 sein Shear - o,=72,875lb/(9+1.5)(1/2)([i)(.85)(2) I [l o u Ik" o, = 5,575 psi 4,, Moment - (72,875)(2.625) = l ] 72,875 lj { 20,[ ( 1/ 2 ) (1. 5 ) (4. 5 ) + ( 2 ) ( 1/ 2 ) (4. 5 ) 2 (2/3)(1/2)](0.85)([2) o = 7,859 psi Neutral m Axis a = 4o2 +o = 9,753 psi z f 49, FS = 21,000/9,753 = 2.15 ~ ] O

STD-R-02-001 Pcge 126 of 148 JAN 7g Lid Ratchet Binder - Lower Lug (Optional Design) 3/16" 4_ 1" w 3/16'I n A \\ I'5 " '\\ 1-ll/1f \\ DIA. /h 1-11/16" l u I T-3/M_ Ik" y I/8 DIA 1%,, u C y f [ j 4xxx>oce M *"4 m N N 9" \\ k \\ \\ \\ \\ \\ \\ y \\ b NQ,... [ v 4" c Section C-C Weld holding 1/4" thick plates to lug Assume each circumferential weld must support 1/2 the load - o = (72,875/2)/(8.5+2.5+0.5+2.25+6.75+0.25)(3/16)(sin 45 )(0.85) o = 15,582 psi FS = 21,000/15,582 = 1.35 Tear Out - (Optional design) Shear of the sleeve - o = 72,875/(1.6875)(1.25)(2) = 17,274 psi I-FS = 21,926/17,274 = 1.27 73 V

STD-R-02-001 g e 127 of 148 7lg Shear of the pin - o = 72,875/[(1)(1.6875-0.27-0.0625)+(0.5)(1.6875-0.27-0.25-0.0625)](2) o = 19,102 psi FS = 21,926/19,102 = 1.15 bearing of pin - o = 72,875/(1.125)(1.25) = $1,822 psi FS = Le/d, 1.6875/1.125 = 2.03 o/fu 51,822/70,000 Tension - (Optional Design) f a = 72,875/(3.5 - 1.5)(1.25) = 28,150 psi 1" Thk i FS = 38,000/28,150 = 1.30 Weld (Optional Design) 72,875 lb ['V k Shear h ( il/ o,=72,875/(9+1)(2)(1/2)([2)(0.85) 0, = 6,062 psi 2 - 5 / R-M ~ Moment Ne I (72,875)(2.625) = 20,[(1)(d)(1/2)(4.5)+(2)(4.5)2(2/3) p (1/2)(d)(1/2)](0.85) -s o,= 8,841 psi

["s * "a

= 10,719 psi

4...

OT FS = 21,000/10,719 = 1.96 O

STD-R-02-001 Page 128 of 148 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 [ 1" 5" = 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. C rushing Width at bend = 25.14 in. (Depth of 5.0 inches) I4' I Thickness = 2 in. u, Fb v Ib P 5" M = (38,000 psi)(1 in)(1 in)(25.14 in) = 955,320 in-lb Fa = M/x = 955,320 in-lb/ 5 in. = 191,064 lb. F=191,064(d)=270,205lb Area of crushed steel = F/38,000 psi = 270,205 lb/38,000 psi s = 7.11 in

STD-R-02-001 Pge gof 148 2 The area of the trapezoid is described in (1.4d + 3.44d ) where d/ d = depth of crush. %) 2 or d = 1.25 in. 7.11 = 1.4d + 3.44d therefore, (d/ 8 ) = 0.882 in. Width of crush = 1 + 4.87d = 7.08 in. 2 8 Volume of crushed steel = d /2 + 1.925d + 1 = 4.18 in 3 Energy absorbed in crushing = (4.18 in )(38,000 psi) = 158,840 in-lb Deceleration Force During Initial Crushing h"! Energy absorbed 158,840 in-lb d% +-- Energy remaining = 477,160 I s. b (_ Force = (7.11 in )(38,000 psi) 2 / 3-3/4" = 270,250/53,000 = 5.1 g's l 3'3 " ) 120 3" 1-1/8" Bending of Cover After the initial crushing of the corner and the build-up of force noted above the corner of the cover will 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 deceleration forces will be as follows: Bending of lid (15' until lug hits outer shell of cask) 5 Fa l 1.3 " Energy remaining = 477,160 in-lb o Energy absorbed in bending B' Fb (1.3" travel) E = (Fa)(d) = (191,064 lb) (1.3") = 248,383 in-lb l E _ g,,em.,ni.g = m, m in.lb. 6 l

STD-R-02-001 p gg g of 148 Failure of the Lua O 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" / 7 a - centroid s / tension f compremsfon % d.5" / % g" o,3,,,,,,, O C / l / o D 1 i M = 20, [(1.5)(1/2)(1.875)+(1/2)([2')(1.875)2(2/3)(1/2)(2)]0.85 M = 5.2(21,000 psi) = 109,368 in-lb The compressive strength of the shell of the cask will be equal to or greater than the tensile yield strength of 49,000 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 the weld is attained. 6

- ' :.2 %

. T...a.*

.-.n STD-R-02-001 Page 131 of 148 - 0.375" 3 C ya l '[ I F-P ya b d F=49,000(1.5)(ys)f=36,750y lb y2 = 4.5 - in. M = (F)(y2) = (F)(4.5 - ) in-lbs = 165,375y -12,250y}=109,368in-lbs 12,250y} - 165,375yt + 109,368 = 0 _ 165,375 - J165,3752 - 4(12,250)(109,368) l Y1 ~ 2 (12,250) l yi = 0.70 inches The depth of shell deformation or, d, will be as follows: 0.375,4.5-yg _ 0.70 d y1 4.5-0.70 d = 0.07 inches Deflections or deformations of this magnitude in the shell will not af fect the integrity of the cask. O I ____.m- ...,_.__-,.-.,,---___,_.,r,,,_ --.m..-

STD-R-02-001 Page 132 of 148 JAN 7W r gtef ing of 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 636,000 in-lbs Less Initial Crushing 158,840 in-lbs Less Initial Bending 248,383 in-lbs Remaining Energy 228,777 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. ! 6 5" d F, g I f 1 = 425 - da M = F 1 = 955,320 in-lbs I a a F,=955,320/[25-d 2 l t l

STD-R-02-001 Pese 133 of 148 JAN TE The energy absorbed is cos'puted as follows: displacement k Fa Fa(avg) d E IE 1.3 4.83 199,034 1.5 4.77 200,290 199,662 0.45 89,450 89,450 2.0 4.58 208,468 204,380 0.5 102,190 191,640 2.18 4.5 212,306 210,387 0.18 37,870 229,500 The secondary bending is capable of absorbing the remaining energy. The additional displacement of the lid during secondary bending is 2.18 - 1.3 = 0.88 in Deceleration Forces - It was calculated that the initiel crushing caused a deceleration force of 5.1 s's. Calculate the deceleration forces of secondary bending since this is the shortest distance travelled in any of the phases discussed, y,/2(KE)(g),/2(228,777)(386.4)=57.76in/sec W 53,000 v,y = 28.88 in/sec At = AX/v,yg = 0.88/28.88 = 0.0305 see 2 a = AV/At = 57.76/0.0305 = (1,893.8 in/sec ) 2 2 e a = (1893.8 in/sec )/(386.4 in/sec ) = 4,9 g, This indicates that the maximum deceleration force du. ring a 12 inch drop on a short flat corner on the lid is 5.1 g's. i This does not exceed the a forces calculated in the drop on a long flat.

"1 w-

.-m.

STD-R-02-001 Page 134 of 148 & I hb5 3.2.6.6 Side Drop t~ The cask is dropped on its side. The energy is assumed to be absorbed entirely on the lid edge. Total energy = 636,000 in-lb 636,000 in-lb = 1/2 av' Initial Velocity, v = 96.3 in/see Let d = depth of crush Volume of steel required to absorb energy 3 636,000 in-lb/38,000 psi = 17.7 in W = d Tan 67.5 67 40 2" thick ( 1a Volume of steel described by [2(1/2 dw)(2 in)] + [(1 in)(d)(2 in)] = 4.82 d* + 2d Final depth = 4.82 d' + 2d = 16.7 inE d = 1.67 in As shown on Table 3-2, the highest g force exerted on the lid is 12.0 g's. i

STD-R-02-001 Page 135 of 148 Jfd 7 ING TABLE 3-2 Energy Energy Incremental d Vol. Absorbed Remaining Velocity Time Deceleration I 3 (in) (in ) (in-lbs) (in-lb) (in/sec) (sec) (in/sec ) (g's) (based on avg. velocity) 0 0 0 636,000 96.3 0.1875 0.54 20,689 615,311 94.7 0.00196 860 2.1 0.375 1.43 54,256 581,744 92.1 0.00200 1,310 3.4 0.5 2.21 83,790 552,210 89.7 0.00137 1,752 4.5 0.75 4.21 160,027 475,973 83.3 0.00290 2,207 5.7 1.0 6.82 259,160 376,840 74.1 0.00320 2,875 7.5 1.25 10.03 381,187 254,813 61.0 0.00370 3,540 9.1 1.5 13.85 526,110 109,890 40.0 0.00500 4,200 10.9 1.67 16.7 636,000 0 0 0.00850 4,70.9 12.0 0 1 h q.) g e ,g = = y v., Y '"9W e---- ~_p-m - e-ey-e. m ,__,,y, ,,p,_ y,__,

STD-R-02-001 Pcge 136 of 148 JAN 7 bot r-K O The weight of the cask less the upper cover plate is 53,000 - 3260 = 49,740 lbs. The deceleration force acting on the weld between the two cover plates will be: F = 49,740 (12.0 g's) = 596,880 lbs. The weight of the cask less the cover end shield plug is 53,000 - 6995 - 425 = 45,580 lbs. The deceleration force between the cover plate and the inner shell of the cask will be: F = 45,580 (12.0 g's) = 546,960 lbs The loads calculated for the weld and inner shell due to a one foot drop on the top corners in section 3.2.6.3 are higher than the respective loads calcu-lated here. Therefore, the cask will safely survive a side drop as well. 3.2.7 Penetration Impact from a 13 pound rod will have no effect on the pack-age. /~*/ \\ k-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 3 Cask l is in full compliance with the requirements set forth in 10 CFR 71 for l Type "A" Packaging. l l l l I l Y (. - l ~ - =

STD-R-02-001 P:gn 137 of 148 JAN 7 ggg 4.0 THERMAL EVALUATION The HN-100 Series 3 casks with shield insert 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 (I) Total ( } Gamma Isotope Energy Activity Activity Nev pCi/ml Ci Cobalt 60 1.33 20.8 96.6 Cesium 137 0.66 583.0 2,708 (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 traneported in the HN-100 Series 3 cask, 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 96.6 1.48 5.0 Cesium 0.0048 2,708 13.0 44.2 The weight of waste per container will be about 9,895 pounds. Based on a specific heat of 0.156 BTU per degree F.,1,544 BTU's or over a day with cesium would be required to heat the waste one degree Fahrenheit. Ac-cordingly, the amount of heat generated by the waste is insignificant. 9 / U

STD-R-02-001 Page 138 of 148 liDV 14 M5 5.0 CONTAINMENT 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 upright 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 and 1" thick steel insert. 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 8 equally spaced 3/4 inch studs on a 20-7/8 inch diameter circle. There is a drain plug in the base of each HN-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 2 2 2 2 - 39.25 ) = 186.73 in Area = n(Ro - Ri ) = g(40 Gasket is equivalent to 3/8 inch thick by 3/4 inch wide Durometer 40. Based on past experience f rom the manufacturer, a torque of 175 to 200 ft-Ibs 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) + 7,420 lb lid weight = F = 35,420 lb Equivalentpressureofgasket=f=35,420lb/186.7in 2 2= 190 lb/in As shown on Appendix D-1, the compression of the gasket due to tighten-ing of the ratchet binder to this minimum is 20% of the gasket thick-l ness, or about 3/32 inch. 5.2 Shield Plug Gasket Similarily, the compression for the shield plug is calculated. A Based on the stud torquing procedure for the shield plug, the minimum U torque value is 120 ft-lb. l

STD-R-02-001 Page 139 gf 148 g 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 t Area = n(Ro2 - Riz) = n(11.3752-10.125 ) = 84.43 in Force downward on lid is the sum of the weight of the lid plus the force of the studs (P). P=gf=(120ft-lb)(12in/ft)/(0.15)(0.75in) P = (12,800 lb/ stud)(8 studs) = 102,400 lb W = 425 lb Total force = 102,400 + 425 = 102,825 lb Pressureongasket=f= 102,825 lb/84.43 in2= 1217.9 psi As shown on Appendix D-2, the compression of the shield plug gasket is 33% of the initial thickness or 1/8 inch. l 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 /w will withstand an internal pressure of 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 3.2) is 4,113 pounds. The resultant strain on the steel ratchet binder (1-3/4" diameter) is: P/AE = (7,613)/(1.76)(30 X 10 ) = 0.000144 in./in. 6 P = 4,113 + 3,500 = 7,613 lb 2 A = Area of 1-1/2" minor diameter = 1.76 in 8 E = Youngs Modulus = 30 X 10 and for a 24 inch long binder, total strain is: (24 in.)(0.000144 in./in.) = 0.0035 in This is less than.4% of the initial compression of the gasket. ,e ~ .+ e 9 .,.v, -y,,-_.w.,.,__,,y ~ - -,

STD-R-02-001 Peg? 140 of 148 LdV 141985 x 5.4 Gasket Compression Test f 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. by 3/8 inch. thick made of Durometer 40 t The samples were each 1 in 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 test stand, the sample returned to its original thickness of 3/8 inch. Similarly, the 1 in Durometer 50 sample was compressed to 1/8 inch 2 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 original height and each survived. The spacer rings have been increased to 1/4 inch which limits gasket compression to 33 percent of the gasket thickness, thereby further reducing possible damage to the gasket material. 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. 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 weight of o 23,360 lbs. The package was dropped on an essentially unyielding surface from a height of 46 inches. The package was pressure tested before and after the drop test and no leakage was detected. The deformation of the corner subjected to the drop test is shown below: n., 1" 7" 3" 4" 5" 6" 6.75" e-l lj .6 4, i ' Corner j Deformation / / / k @99@@ e-* O O O O O O I ~

-=,.e.

STD-R-02-001 Pgeg4gef148 The energy absorbed in dropping a 23,360 lb package irom 46 inches is 6 in-lbs. The energy to be absorbed in a one foot drop of an f] 1.07 x 10 5 in-lbs or less than HN-100 Series 3A cask is 12 x 53,000 = 6.36 x 10 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. O Q

STD-R-02-001

8%D)Y4 M 8 6.0 OPERATING PROCEDURES (Removed for Rev. 7)

O 7.0 ACCEPTANCE TESTS & MAINTENANCE PROGRAM (Removed for Rev. 7) G .= .--,-----.a

STD-R-02-001 Page 143 of 148 APPENDIX B Appendix Removed for Rev. 6 6 P

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_ _,, _., _m.,

STD-R-02-001 Page 144 of 148 I O APPEhTIX C Appendix Removed for Rev. 6 O 1-l l l ( l l l O L

STD-R-02-001 Page 145 of 148 b APPENDIX D Gasket Compression Nomograph l l b i l I - - - - - - - - - ~ -..

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O lIII APPENDIX Dl

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STD-R-02-001 Page 147 of 148 h0V 14 l!!! L lll ! a .. l'i s -k.. {.i. n : .e i t, u 8 il lw l 3 1:8 s =u ! lt:lg{t{!I ,\\ y 8.s!! h l!!!glja!.i,an=1 !.!gg ii .a \\ U 's\\ a s# eQ. "I!! a l II!i. l '5 M- g h 6 d s an'- ttis 1 <. t.I s i s ; 1 -

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5 ) STD-R-02-001 Page 148 of 148 Asi 14 IM O APPENDIX E Appendix Removed for Rev. 7 0 l lC L -}}

ML20151R158

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2.0 DESCRIPTION

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2.0 DESCRIPTION

(0.309 in)/[(65.3)(0.5)in/sec)

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