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Rev 0 to Sar,Cns14-170 Series III Type a Radwaste Shipping Cask
	ML20045F621
Person / Time
Site:	07109249
Issue date:	06/30/1993
From:	; CHEM-NUCLEAR SYSTEMS, INC.
To:	;
Shared Package
ML20045F620	List: ML20045F619; ML20045F621;
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{{#Wiki_filter:- O SAFETY ANALYSIS REPORT CNS 14-170 SERIES 111 TYPE A RADWASTE SHIPPING CASK REVISION 0 g JUNE 1993 O CHEM-NUCLEAR SYSTEMS, INC. 140 Stoneridge Drive Columbia, SC 29210 $07 Soo![ $$$$$49 C PDR L,._

_~ -_ CNS 14-170 Series lll SAR Docket No. 71-9249 Revision O June 1993 TABLE OF CONTENTS i 1.0 PURPOSE 1-1 t L 2-1

2.0 DESCRIPTION

2.1 Cask Body.... 2-1 2.2 Cask Lid....... 2-1 2.3 Shield Plua 2-1 2.4 Cask Closure................ 2-1 i 2.5 Cask Tiedown System 2-1 2.6 Cask Internals...................... 2-2 2.7 Gross Packaae Weiahts 2-2 2.8 Radwaste Packaae Contents............................ 2-2 1 2.8.1 Type and Form of Material........................ 2-2 2.8.2 Maximum Quantity of Material Per Package. 2-2 e ) i 3.0 DESIGN CONDITIONS............. 3-1 i 3.1 G e ne ral St a nd a rd s..................................... 3-1 e f 3.1.1 Chemical Corrosion.............................. 3-1 i 3-1 3.1.2 Positive Closure System............................ 3.1.3 Design Criteria on which Structural Analysis is Based........ 3-1 t 3.1.4 Lif ting Devices................................ 3-2 3.1.5 - Tie Downs........... 3-9 3.2 Normal Conditions of Transoort (Accendix A-10 CFR 71) 3-19 3.2.1 Heat........................... 3-19 3-19 i

3. 2. 2 C ol d..............

3-19 l 3.2.3 Reduced Pressure.... 3.2.4 Vibration. 3-19 l 3.2.5 Water Spray 3-10 3.2.6 Free Drop 3-19 i i 3.2.7 Penetration...... 3-48 t 3.2.8 Compression.......... 3-48 e 4.0 THERMAL EVALUATION....... 4-1 ) I I e i

. ~. - N i CNS 14-170 Series ill SAR Docket No. 71-9249 a Revision 0 June 1993 { i 4 TABLE OF CONTENTS (CONTINUED) { L i 5-1 5.0 CONTAINMENT.... 1 5-1 5.1 Primary Lid Gasket.. 5-1 5.2 Shield Plug Gasket. 5-2 5.3 Seal With Internal Pressuri7ation 5-2 i ] 5.4 Gasket Comoression Test 5.5 Warcina of Covers. 5-3 i 6.0 OPERATING PROCEDURES 6-1 l 6.1 Loadina Procedure For 55-Gallon Drums 6-1 1 6.2 Unicadina Procedure 6-2 6.3 Processino Liners inside Cask 6-3 i 7.0 ACCEPTANCE TEST AND MAINTENANCE PROGRAM 7-1 { 7-1 7.1 initial Acceotance Test. l 7-1 7.2 Maintenance Proaram. t 7-2 '7.2.1 Painted Surfaces 7-2 7.2.2 Structural Members and Welds 7.2.3 Gaskets and Test For Seal integrity (Leak Test) 7-3 7-3 t 7.2.4 Fasteners '............ 7-3 1 7.2.5 Ratchet Binders.... t 8-1 8.0 OUALITY ASSURANCE... l f I ) h i i i i

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision O June 1993 PURPOSJi O 1.0 i The purpose of the following document is to provide the information and engineering analysis that demonstrates the performance capability and structural integrity of the CNS 14-170 Series ill Radwaste Shipping Cask and its compliance with the requirements of 10 i CFR 71, Section 71.21 and Appendix A. t 3 9 i r O i i I t i k 1-1 I i

=. i CNS 14-170 Series ill SAR Docket No. 71-9249 Revision O June 1993 l l

2.0 DESCRIPTION

The CNS 14-170 Series ill Shipping Cask is a top-loading, shielded container designed specifically for the safe transport of Type "A" quantities and greater than Type "A" 1.SA radioactive waste materials between nuclear facilities and waste disposal sites. The radioactive materials can be packaged in a variety of different type disposable containers. The CNS 14-170 Series ill Shipping Cask is a primary containment vessel for radioactive l materials. It consists of a cask body, cask lid, and a shield plug being basically a top-- i opening right circular cylinder which is on its vertical axis. Its principal dimensions are 81-l 1/2 inches outside diameter by 81-1/2 inches high with an internal cavity of 75-1/4 ' inches inside diameter by 73-3/8 inches high. i i 2.1 Cask Body The cask body is a steel-lead-steel annulus in the form of a vertical oriented, right circular cylinder closed on the bottom end. The side walls consist of a 3/8-inch inner steel shell, a 17/8-inch thick concentric lead cylinder, and a 7/8-inch thick outer steel shell. The. i bottom is four inches thick (two 2-inch thick steel plates welded together) and is welded integrally to both the internal and extemal 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 lifting lugs are welded to the outer steel shell. 2.2 Cask tid 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 i lug lifting devices are welded to the cask lid for handling. The cask lid also contains a 9 " shield plug" at its center. 2.3 Shield Plug i 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 l and uses eight hold-down bolts to provide positive cask closure. The shield plug also has l a lifting device located at its center to facilitate handling. { 2.4 Cask Closure I The shipping cask has two closure systems: (1) the cask tid is closed with eit)ht I 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 use ' for the cask lid but smaller. i 2.5 Cask Tiedown System The shipping cask tiedown system consists of two sets of crossed tiedown cables (totaling [ O 4). Four shear blocks or a shear ring (affixed to the vehicle inad bed) firmly position and safcly hold the cask during transport. I 2-1 P

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision O June 1993 l 2.6 .Qask Internals O l The internals of the CNS 14-170 Series 111 shipping cask can be any one of an extensive vahty of configurations. Some examples are given in terms of weight in 2.7. Other 3 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 l structural members such as bottoms and pallets. Basically, the internal's consist of the i 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. l 2.7 Gross Packaae Weiahts The respective gross weights of the cask components and its designated radwaste loads are as follows: Cask body 29,030 pounds l Closure lid 5,800 pounds Shield plug 37Qoounds Total cask (unloaded) 35,200 pounds Large container (s) and waste 17,800 pounds l Total cask (loaded) 53,000 pounds 2.8 Radwaste Packaoe Contents i 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 containers with weights not exceeding 17,800 pounds. f l h O 9 2-2 = =

i CNS 14-170 Series til SAR Docket No. 71-9249 Revision O June 1993 j 3.0 DESIGN CONDITIONS j 3.1 ' General Standards i (Reference 10 CFR 71 Section 71.31) l l 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 i 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 inadvertent opening. P 3.1.3 Design Criteria on which Structural Analysis is Based l 1 3.1.3.1 Stresses in material due to pure tension are compared 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. (fy = 38,000 psi fcr ASTM A516 Grade 70). For the shell, O steel having a minimum 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. t 3.1.3.2 Stresses in material due to shearing is analyzed using the " Maximum Energy I - Distortion Theory" which states the shearing elastic limit is 1/v 3 = 57.7% of the tensile elastic li'mit'. 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 A5'i6 Grade 70) 3.1.3.3 The weld filler material rod is E70 Grade. Analysis is based on American l Welding Society Structural Code D1.1-79. For fillet welds, shear stress on effective throat regardless of direction of loading is 30% of specified minimum tensile strength of weld metal. For complete joint 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 f in order to be more conservative, a weld efficiency of 85% is also added. l Since all welds have been nondestructively examined, weld efficiency is known to be greater than this. l ) O 'Desian and Behavior of Steel Structures, Salmon & Johnson, page 4-7. 3-1 i

CNS 14-170 Series til SAR Docket No. 71-9249 Revision O June 1993 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 container and waste 17.800 pounds Gross Weight 53,000 pounds 3.1.4.2 Cask Lifting Lugs 1 The cask lifting lug materialis ASTM A516 Grade 70 with a minimum of 38 ksiTensile Yield Strength and 21,926 psi shear yield strength (57.7% of 38,000 psi). The maximum load is 53,000 lbs, and four lugs are used to lift the cask. (53,000 lbs)(3 g's/4 lugs) = 39,750 lbs/ lug (Vertical) TEAR OUT N Sling Angle to Lift 45 "3 (. 'N Load = 39,750 lbs/ sin (45 ) ' 23*' Decrneter Load = 56,215 lbs , Stress = 56,215 lbs/l(2)(2 in)(2.5 in)] s 2 inches Thic$ cr = 5,622 psi i l 5,622 psi < < 21,926 psi F.S. = 3a e ingn l 1 i 4 ? O 3-2

CNS 14-170 Series til SAR Docket No. 71-9249 Revision O June 1993 i

TENSION O-Sling Angle to Lift 45 x Load = 56,215 lbs. / '\\ y N 'f Stress = 56,215 lbs/[(2 in)(5.4926 in)) a = 5,117 psi t 3' 5,117 psi < < 38,000 psi F.S. = 2 4 t 4.2426 - 1.25 + 2.5 = 5.4926 in i TEAR OUT Vertical Lif t l Stress = 39,750 lbs/(2(2 in)(2.5 in)] a = 3,975 psi 3,975 psi < < 21,926 psi F.S. = 5J TENSION Vertical Lif t i Stress = 39,750 lbs/I(2 in)(6 in - 2.5 in)] a = 5,679 psi 5,679 psi < < 38,000 psi F.S. = 6.7 i t 3.1.4.3 Cask Lifting Lug Welds 1 All welds are 3/4-inch fillet. The allowable shear stress on the effective area of the weld is 0.3 times the nominal tensile strength of the weld metal, E 70 rod, or (0.3)(70.000) = 21,000 psi. i 3-3 i

1 CNS 14-170 Series til SAR Docket No. 71-9249 Revision O June 1993 7 Y sea \\,a \\ / A_ 's A_ 7.>s te,4 \\ Tx\\ es ec y-22 D,ameter \\ \\ 1/ N 3 x-r una ' - 6_ s O f4 N 6" werd Bett sides -T M L Vertical Lift Assume weld to be 85% efficient. Minimum throat of weld = sin (45 )(3/4 in) = 0.53 inch Weld strength per inch = 0.85(0.53 in)(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 in + 2 in + 12 in + 7.75 in + 6 in) F.S. = 2d 3.1.4.4 Tiedown Lugs for Lifting Cask (inadvertent use) If it is assumed the entire load is carried by the tiedown lug, the section modulus equals 2 (bh )/6; where b = 15.5 inches (length of tiedown lug); and h = 2 inches (thickness of tiedown lug). (See Diagram on page 3-15). 15.5 (22 6) = 10.33 in8 Tiedown lug section modulus = / Stress = (39,750 lbs)(3 in/10.33 in ) = 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_._Q Weld required on lift lug is 4.2 inches: 4.2" < 12" F.S. = 2_a 3-4

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision 0 June 1993 l Weld to Tiedown Luo l Shear = 39,750 lbs/(2)(15.5 in) = 1282 lbs/ inch of weld compression or tension due to moment couple 39,750 lbs(3 in/15.5 in) tiedown lug = 7,694 lbs-in/ inch of lug and moment couple with 2-inch thick lug ( 2 inch (3847 lbs) = 7,694 lb/in i 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 l o = s/12812 + 38472 55652 + = 6886 lbs per inch 6886 lbs per inch of weld < maximum of 9640 lb per inch of weld F.S. = 1.37 3.1.4.5 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 long, 3/4-inch welds attaching the lift lugs to the reinforcing plate and the two 6.0-inch long, 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 in + 6.0 in + 6.0 in + 6.0 in)(0.75)(0.707)(0.85) = 39,750/11.6 = 3,427 psi The centroid of the welds is located as follows: y = (7.75(7.75/2) + 6(6/2) + 6(7.75) + (6)(6)] + (7.75 + 6 + 6 + 6) in = (30.0 + 18 + 46.5 + 36) + 25.75 in = 5.06 in Total Moment = (2.75 in)(39,750 lb) = 109,313 in-Ib O 3-5

I CNS 14-170 Series ill SAR Docket No. 71-9249 Revision 0 June 1993 ) I p Total Moment = Compressive Moment + Tensile Moment ] U Compressive Moment = Tensile Moment j Total Moment = 2(Tensile Moment) = 2f[(6)(2.69) + (2.69)(2.69)(1/2) + (6)(0.94) + "j" " (0.94)(0.94)( 1/2)(0.75)(0.707)(0.85 )] / i = 2f[(16.1 + 3.6 + 5.6 + 0.4) (0.75)(0.707)(0.85)] Ur 1- = 2f(25.7)(0.75)(0.707)(0.85) y,4;.q i u,c C'I@ r--- = 23.29f n{ (, S /\\ f = 109,313/23.29 = 4,694 psi (!*I.7* Combined Stress = / 46942 + 34272 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. 3.1.4.6 Lid Lif ting Lugs (Secondary and Primary) A. Secondarv Lid Liftina Lu_g Materialis ASTM A516 Grade 70 f-~~~~ ~ 2 " - i l l l Maximum load carried by one lug is 370 lbs. CI;.- y_, i i\\ \\ t: n/W b 1 1 i L ___. J I_._: _ _I \\ w% w uc m TEAR OUT Area = 2[(1-1/2 - 15/16) - 7/32][3/81 f 3-6

CNS 14-170 Series lll SAR Docket No. 71-9249 Revision 0 June 1993 2 Area = 0.258 in ] Stress = (3 g's)(370/0.258 in') = 4,310 psi 4,310 psi < < 21,926 F.S. = LQ i TENSION a Area = (2.0 - 7/16)(3/8) = 0.586 in Stress = (3)(370)/0.586 = 1,900 psi 1,900 psi < < 38,000 psi F.S. = 2Q B. Secondarv Lid Liftina Lua Weld 1/2" fillet weld with allowable stress of 21,000 psi Effective size = sin (45 )(0.5) = 0.353 in Area of weld = (2 + 2 + 3/8 + 3/8)(0.353) = 1.68 in: Stress = (3)(3701/1.43 o = 660 psi 660 psi < < 21.000 psi F.S. =.313 Therefore, the secondary lid lug is able to resist a load of three times its weight without reaching a yield stress. t C. Prirnary Lid Liftina Luos Materialis ASTM A516 Grade 70. Maximum Load = Primary Lid 5800 lbs [ -p+-- i Secondary Lid 370 lbs 1 1 2# 6170 lbs u. i i \\ s : s L/ I _J __._.J___.i l \\ t, mo ' e ce 4 TEAR OUT (Vertical Lift) 2 Area = (2)(2.75 - 1.5 - 0.5)(1) = 1.5 in 3-7 t' I

. _ _ _ _ _ _ _~ CNS 14-170 Series ill SAR Docket No. 71-9249 Revision 0 June 1993 3 Stress = 6,170/1.5 O a = 4,113 psi 4,113 psi < < 21,926 psi F.S. - M TEAR OUT (45 Sling Angle) { i Area (short path) = ( / 0.752 + 0.752 )( 1 ) = 1.06 inch 2 j 8,726 lbs Load = / (61702)( 2 ) = Stress = (0.5)(8726)/1.06 = 4,116 psi 4,116 psi < < 21,926 psi F.S. = M TENSION (Vertical Lift) 2 Area = (6 - 1) 1 = 5 in Stress = 6170/5 = 1,234 psi i 1,234 psi < < 38,000 psi F.S. = 30.8 [ TENSION (45 Sling Angle) Area (short path) = / ( 2 )( 1.252 ) - 0.5 = 1.2678 inch 2 Stress = (0.5)(8726)/1.2678 = 3,441 psi i 3,441 psi < < 38,000 psi F.S. = 11.0 F. Primarv Lid Liftina Lua Weld 1/2" weld at shear of 21,000 psi (a) Shear stress due to vertical = horizontal component I i oy = oh = 6170/(6 + 6 + 1 + 1)[ sin (45 )l(0.5)(0.85) = 1467 psi 3-8 i j

=- CNS 14-170 Series lil SAR Docket No. 71-9249 Revision 0 June 1993 J (b) Stress due to moment i Total Moment = Compression Moment + Tensile Moment l Compression Moment = Tensile Moment j Total Moment = 2 (Tensile Moment) = 2 (2)(3)(0.67)(f)(0.5)(0.707)(0.85) j = 2.4f f = 9,255/2.4 = 3,856 psi l Combined Stress = / 38562 + 14672 + 14672 i cr = 5521 psi [ F.S. = 21,000/5521 = 3.8 Therefore, it can be concluded that the lifting logs 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 lifting lugs are not capable of resisting the full weight. r of the package, they will be covered during transit. 3.1.5 Tie Downs l The tie down lug materialis 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 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: l s s O 3-9 f

CNS 14-170 Series ill SAR. Docket No. 71-9249 Revision O June 1993 I o E /\\o y 3 + e 4 \\. .p . 2, w,tk a ~., 1 O l hb.V m s. x ./[ 'x., / A p / ,/ \\ L i_, ./ N., .a j [ I T I - 17]'3d<'1 lT- ] l 57"- ]'~ssj .l =m=.s,o .Ii VIEW A A O 3-10

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision O June 1993 3.1.5.1 Cask Center of Gravity item Weicht Arm Blower Cask 35,300 lbs X 42.9" 1,510,080 lbs-in Liner / Waste 17.700 lbs X 37.7" 671.060 lbs-in 'l 53,000 lbs 2,181,140 lbs-in Center of Gravity - 2,181,140/53,000 CG = 41.2 inches 3.1.5.2 Tie Down Forces Reference frame with respect to the trailer is shown on the tie down drawing. i up-down Y; front-rear X; side -side Z Accelerations: Y axis - 2 g's O X axis - 10 g's Z axis - 5 g's Tie Down Lenaths Long tie downs (Iow trailer attachment points) length = /632 + 57.82 + 53.92 = 101.0 inches Short Tie Downs (high trailer attachment points) length = /602 + 57.82 + 53.92 = 99.2 inches Tie Down Tensions e Tie down tensions resolved by vector direction f Long tie down at tension Tt Along Y axis: (63.0/101.0)(T ) = 0.623 - T, t Along X axis: (53.9/101.0)(T ) = 0.5333 T t t 3-11 f

CNS 14-170 Series til SAR Docket No. 71-9249 ) Revision 0 June 1993 Along Z axis: (57.4/101.0)(T ) = 0.5718 T t t Short tie down at tension T s Along Y axis (60/99.2)(T ) = 0.6047 T s s Along X axis (53.9/99.2)(T ) = 0.5432 T. 3 ' t Along Z axis (57.8/99.2)(Ts) = 0.5825 Ts 10W Force (Front-Rear) Overturning (front-rear) due to 10W along X axis Overturning moment (taken about the axis of rotation @ Point A) l = 10 (53,000 lb) 39.2 in = 20,776,000 in-lb Each of the two rear (or front) tie downs (one long and one short) must restrain half the above moment of 10,388,000 in-Ib. Tension in the long tie down 10,388,000 in-Ib = (64 in)(0.5333 T ) + (69.8 in)(0.6233 T ) f t t i 133,801 lb T = t Tension in the short tie down 10,388,000 in-Ib = (64 in)(0.5432 T ) + (69.8 in)(0.6047 T ) s 3 Ts = 134,957 lb 5W Force (Side-Side) Overturning (side-side) due to 5W along Z axis Overturning moment (taken about the axis of rotation @ Point A) = 5 (53,000 lb)(39.2 in) = 10,388,000 in-Ib Each of two side tie downs (one long and one short) must restrain half the above moment, 5,194,000 in-lb. Tension in the long tie down 5,194,000 in-Ib = (64 in)(0.5718 T ) + (12.0") (0.6233 T ) O t t T 117,845 lb = t 3-12 m -u

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision 0 June 1993 Tension in the short tie down i 5,194,000 in-lb = (64 in)(0.5825 T ) + (12.0 in)(0.6047 Ts) s Ts = 116,624 lb 2W Force (Uo-Down) Lifting (up) due to 2W along Y axis i Lif t = 2 (53,000 lb) - 53,000 lb = 53,000 lb l Each of two long and two short tie downs will carry a quarter of the load,13,250 lb. 13,250 lb = 0.6233 To i T = 21,258 lb i 13,250 lb = 0.6047 T. T. = 21,912 lb i Total Tension r Total tension with all forces acting simultaneously: T = 133,801 + 117,845 + 21,258 t = 272,904 lb [ i Ts = 134,957 + 116,624 + 21,912 = 273,493 lb I 3.1.5.3 Tie Down Lugs 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 l in the design of the tie down lugs: j Tensile Yield - 46,000 psi Allowable Bearing Stress - 0.9 Tensile Yield Strength Maximum Ultimate Tensile - 78,000 psi Shear Yield = (0.577)(46,000) = 26,542 psi P Shear Ultimate = (0.577)(78,000 psi) = 45,006 psi i 3-13 +

CNS 14-170 Series ill SAR Docket No. 71-9249 . Revision 0 June 1993 Allowable Shear Stress on Welds - 21,000 psi Bearina Stress on Luo and Pin Maximum Load - 273,493 lbs Diameter of Hole - 2.5 inches Diameter of Pin - 2.25 inches Thickness of Lug - 3.0 inches Projected Area of Pin = (3.0)(2.25) = 6.75 in' Bearing Stress = (273,493/6.75) = 40,517 psi F.S. = 41,000/40,517 = 1.Q2 Tear Out The shear plane associated with the projected area of the pin is shown as follows: i y, = / 1.252 - 1.1252 = 0.545 inch y, = 1.25 - 0.545 = 0.705 inch i f. / 273,493 Id ,/ ^ V /$/ $g[, 5 I b N, [/ 2.29 Dio, F en ',x f, '- 2 5" Dic Hcde (y s 3 o~ in:ci Shear Stress = 273,493/(2)(2.0)(1.75) = 26,047 psi F.S. = 26,542/26,047 = 1.02 3-14 I

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision 0 June 1993 l ) 1 - O Tension Stress = 273,493/(3.0)(8.25 - 2.5) y,,,, = 15,855 psi /}) '"( F.S. = 46,000 /15,855 = 21Q Reinforcina Plate Weld _ ,_ / } / Thickness main lug 2 inches Thickness reinforcing plates 2 @ 1/2 inch j K, 33.,,, Total thickness 3.0 inches p'J' : s t - Load on 1/2" reinf. plate = (1/6)(273,493) 'g,L vr n.c. a = 45,582 psi Length of weld = 6 + 7 + 7 = 20 inches 2 Area of shear = (20)(0.5) sin (45 )(0.85) = 6.01 in Shear = 45,582 / 6.01 = 7,584 psi F.S. = 21,000 / 7,584 = 2.77 3.1.5.4 Tie Down Lug Welds y,1r[2 (a) Pure shear f',,f; .;v' p / Area of shear = 2 (15.5) sin (45 )(1)(0.85) + s 's/ /, / (6 + 6 + 2) sin (45 )(0.75)(0.85) = 25 in' f/ e ..s/gf 1s t ghy / Stress = 273,493 / 25 = 10,940 psi t 7 (b) Moment forces on weld Maximum Moment = (273,493) (2.75) = 752,106 in-lbs O 4 3-15

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision 0 June 1993 i O 3[4" Weld j 6 Long 3[4" Weld 2 Long q ga .f '11111111111111111111111!!!1111 411191 ! !! 1111111111111!! f l 1111ll l l l b illfilfl ililliliill filli f filillit lillbilflifilli!!Ill tilfllf fillisill 8.35" i 10.95" = t 1 ? X = 2(1)(0.75) + (?"15.5)(8.75) + (6)(0.75)(8.33) 4 (6)(0.75)(10.95) + (16.5)(2)(0.75) (0.75)(2) + (15.5)(2) + (6)(0.75) + (6)(0.75) + (2)(0.75) l = 8.94 in .-- 2.01 " ',////////////////////////////// If/////////////////////////////// ///////////////////////////////////! [/////////////////////////////////// 7.94"

6. 56

= = = i M = compressive moment + tension moment compressive moment = tension moment M = 2(tension moment) M= 2f 1(2.01)(6)(0.75)(0.707)(0.85) + (6.56)(2)(0.5)(6.56)(0.67)(0.707)(0.85) l + 2(6.56)(0.707)(0.85)(0.75)) M = 57.2f = 752,106 in-Ib j 3-16 i )

CNS.14-170 Series ill SAR Docket No. 71-9249 Revision O June 1993 I 'e f = 13,149 psi i Combined stress i f = / 10,9402 + 13,1492 = 17,105 psi F.S. = 21,000 + 17,105 = 1.2S i 3.1.5.5 Analysis of Tie Down Loads on Cask Shell The tie down loads are transmitted into the cask shell as external moments. These moments are the product of the tie down forces and the offset distance between the line of action of the tie down force and the attachment plate. The offset distance between the line of action of the C 3 tie-down force and the attachment plate is 4.25 inches. F, = 273,493 cos (39) 30' = 211,034 lb O p rz - 9 F, = 273,493 sin (39 ) 30' = 173,963 lb j C v c M, = Circumferential moment i (211,034 lb)(4.25 in) = 896,895 in-lb t = rx Longitudinal moment '~ M = t (173,963 lb)(4.25 in) = 739,343 in-Ib l = Reference for method of calculation: Welding Research Council, Bulletin No.107 (WRC 107), Stress in Cylindrical Pressure Vessels from Structural Attachments. T = r/t = radius to thickness ratio = 40.9/0.875 = 46.7 l ? C, = 1/2 the circumferential width of the loaded plate = (33 /360 )(2r7)(40.9)(0.5) = 11.78 in C, = 1/2 the longitudinal width of the loaded plate = 23.5/2 = 11.75 inches B, = C,/r = 11.78/40.9 = 0.288 I 3-17 l

~ CNS 14-170 Series ll1 SAR ' Docket No. 71-9249 .I Revision 0 June 1953 B, = C,/r = 11.75/40.9 = 0.287 O Check that 5 s T s 100 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 testing. 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: c F = (273,493) (78,000/42,300) = 504,313 lbs (Where 78,000 psi is the ultimate strength of the shell material.) The tie down lugs are designed to fail by tear out when excessive forces are app!ied. The force required to cause tear out is as follows: F = (45,006)(2)(3.0)(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 p 5 1 4 O 3-18 i

CNS 14-170 Series til SAR Docket No. 71-9249 Revision 0 June 1993 3.2 Normal Conditions of Transoort (Accendix A-10 CFR 71) p) - \\. 3.2.1 Heat Since the package is constructed of steel and lead, temperatures 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 A 1/2-atmosphere pressure will produce an equivalent internal pressure of 7.35 psi. This pressure acting over the lid will produce a load of: F = (75.5)2 (tr)(7.35)/4 = 32,906 lbs i Since there are eight binders, the load per binder will be: F = 32,906/8 = 4,113 lbs/ binder Each binder has an ultimate strength of 135,000 lbs. O Therefore,it can be concluded that the reduced pressure will produce no detrimental effects. 3.2.4 Vibration All components are designed for a transportation environment. No loss of integrity will be experienced. i 3.2.5 Water Spray Not applicable. 3.2.6 Free Drop Since the package weighs in excess of 30,000 lbs. it must be able to withstand a one foot drop on any surface, without loss of contents. 3.2.6.1 One Foot Drop on Bottom Corner Energy to be absorbed = 53,000 lb x 12 in. 5 Maximum energy = 6.36 x 10 in. Ib. .p Energy will be absorbed by crushing of comer. ( l 3-19 a +

i l CNS 14-170 Series ill SAR Docket No. 71-9249 Revision O June 1993 ,es i P J N c N j i 4 () T'

E-;r'

~ Ef i=w w A ' 4 Groce 7C# s ~. The volume of the crushed ungula, assuming the worst case of a 45 impact angle, is calculated by the following equation: V, = R 3 { sin 0 sin 3 p - @ cos o } 3 i ss ,/ _._f.-.__ g.j 's / ../ ,R ,.)._. t \\ Y g.. 'y \\,,. 's~ \\-4 \\u t % d L:~-_/ The maximum amount of steel crushed will be: b = R(1 - Cos () = 1.66 The effect on the cask body due to the corner impact event is shown below. Even though 's _ the weld will be crushed locally, there will be no loss of the cask's integrity. s 3-20

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision 0 June 1993 f3 t) N [s / N x N 2k / s N z 's 2 [k s's / s 7 }- ]} 7 / N s '~T C N r' j \\ _/ l / ', - / 7 i 1[,, /p KGY K /N N -/ /Nx 1.65? N 1.6t> ,/y N 9e 3,/ Darnage Area The deceleration force exerted on the cask is calculated as the product of contact surface area, and the yield strength of the steel (38,000 psi). Au = pab - (xy + ab sin x) 2 a 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 = / R2 -C2 = / 40.52 - 38.842 ' 1.47 in = x = C / cos 45 = 54.928 in Au = n(57.275)(40.5)/2 - (54.928)(11.47) + (57.275)(40.5) sin ~'(54.928/57.275) 3-21

CNS 14-170 Series til SAR Docket No. 71-9249 Revision 0 June 1993 p As, = 36.32 in' \\ F = (36.32)(38,000 psi) = 1,380,160 lb. g Force = 1,380.160/53,000 = 26.0 g's 3.2.6.2 Effects of Bottom Corner Drop on Balance of Cask 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. yask Cover inner Containem7' \\N / DecWwatesR.catet _._A..._. O / \\ E inner Shell Lead Shield a-nu J ' x Outer Shell Plate Comer hpeC1 s\\\\\\\\\\\\\\'c\\\\\\\\\\\\\\N' ..c__. 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. 3-22 'l l ~

t CNS 14-170 Series til SAR Docket No. 71-9249 Revision O June 1993 4 The following design criteria and assumptions are the basis for the bottom comer 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 ccnsisting of: 9 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. j Load from one-half lead shield considered to act on the half of inner shell perimeter not receiving waste loading. { Al' other loads on the inner shell will be considered to act uniformly around shell perimeter. [ i 3) The outer shell will receive loadings at its connection to the lower bottom f plate consisting of: Load from lid and shield plug Lead from self weight of outer shell [ t d from one-half of the shield considered to act on that half e shui' peri' eter nearest the corner of impact. ei m 4) The upper botton, plate will receive loadings consisting of-l Loads transferred through the inner shell weld l t 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. l ~ i i 3-23 i s

CNS 14-170 Series til SAR Docket No. 71-9249 Revision O June 1993 (~g Cask Analysis NI 1) (pad Frorn Primarv Lid and Shield Plua i Decelerotion Forces 'N / Sheor-Arco of wefd Compressive Stress / N in inner and Outer SheUs + 'N \\ \\ N. \\N Deccleroticrt / Forces \\ ,f,. x)<^ g x o x Reaction Forces Loading = (5800 + 370)26.0 = 160,420 lbs Lateral force = 160,420 (sin 45 ) = 113,434 lbs Axial force = 160,420 (cos 45 ) = 113,434 lbs 2 2 Inner shell area = n/4(76.252 - 75.5 ) = 89.388 in 2 Outer shell area = n/4 B1.75 - 802) = 222.317 in2 i Total Area = 89.388 + 22.317 = 311.705 in' inner area = 89.388 / 311.705 = 28% Outer area = 222.317 / 311.705 = 72% Force on inner shell =_ (113,434)(0.28) = 31,762 lbs lateral and axial Force on outer shcIl = (113,434)(0.72) = 81,672 lbs lateral and axial O 3-24

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision 0 June 1993 2) Stresses Develooed in Inner Shell and Attachment Welds O l Y f g \\ inner N Container Upper Bott \\ Picte v / \\ )/ 3/8" we!ds (typicot) K / 1/2" Welds (typical) Deceleration rorces Reaction on Inner Shell O 4 Stress in weld around perimeter of inner shell at cask lid: 31,762 lb/ir (75.5)(3/8)(0.707)(0.85) = 594 psi Total stress = / 2 594 = 840 psi t t F.S. = 21.000/840 = 25.0 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 = (2900/2)(26.0) sin 45 +(31,762/2)+ (17,800)(26.0) sin 45 = 360,596 lb O 3-25 1

i CNS 14-170 Series ill SAR Docket No. 71-9249 Revision O June 1993 Lateral weld stress = 360,596/n(75.5/2)(2)(3/8)(sin 45 )(0.85) = 6745 psi (lateral) Axial weld stress is caused only by lid load and shell self weight: 31,762 + 35,200 = 66,693 lbs Axial weld stress = 66.693/in(75.5)(3/8) sin 45 (0.85)(2)] = 624 psi Total stress = v/67452 + 6242 = 6774 psi 2 2 Axial shell stress = 66,693/(76.25 - 75.5 )(n/4) = 746 psi (less than the weld suess) Shear shell stress = lateral force / area (1900)(26.0)(sin 45 )+(31.752)+07.800)(26.0)(sin 45 ) = 8814 psi 2 2 (76.25 - 75.51 (n/4)(1/2) F.S. = 21,926 / 8814 = 2.49 3) Stresses Devetooed in Outer Shell & Attachment Welds O m er b' O t i Om s. CONIC n( .- nner Sh ei' U p Of f p Picte \\p.e M-p-twa 7 LC ner t-Ott om 3 j T / /k Picte 'i, j, \\ \\ ,g O s * <.'r SW

q. '

i cec epr o t.on Depth Fer c e ~<m r ~-J s Wem Anc m %r i [ Stress in weld around perimeter of outer shell at cask lid 81,672 lb/In(80.875)(0.5) sin 45 (0.85)] = 1070 psi (axial and lateral) ) 3-26 [

l CNS 14-170 Series til SAR Docket No. 71-9249 1 Revision 0 June 1993 l i Total stress = / 2 (1070) = 1513 psi F.S. = 21,000/1513 = m Stress in weld connecting outer shell to lower bottom plate: i Lateral force = 1/2 load of outer shell + 1/2 load shield + 1/2 lid and shield plug (the 1/2 supported by 1/2 outer shell)- l = (7224/2)(26.0) sin 45 + (81,672/2) + (14,397/2)(26.0) sin 45 l = 239,585 lbs j Lateral stress = 239,585/In(80/2)(0.5) sin 45 (0.85)) = 6344 psi Axial load = (7224)(26.0) sin 45 + 81,672 = 214,484 lbs Axial stress = 214,484 lb/in(80)(0.5) sin 45 (0.85)] = 2840 psi Total stress in weld = / 63442 + 28402 = 6951 psi j F.S. = 21,000/6951 = 3.02 Axial stress in octer shell = 214,484/(81.752 - 80 )(yj4) [ 2 = 965 psi < 38,000 psi vield Lateral shear stress in outer shell = (239,585)/(81.752 - 802)(n/4)(1/2) = 2155 psi < 21,926 psi yield i l l t I [ l l O s 3-27 J. f

l CNS 14-170 Series til SAR Docket No. 71-9249 Revision 0 June 1993 4) Stress in Weld Joinina Uoner to Lower Bottom Plates (fg N inner N Container Upper Botto Picte ') 3/8' Weids (typical) / 1/2" Weids (typicol) s ete!cration Forces ( Reaction on inner Shell 'A Load on weld = Upper bottorn plate + Inner shell + Shear from lid on inner shell + Waste + 1/2 lead Load = (2645 + 1900 + 17,800 + (14,397/2)] (26.0) sin 45 + 31,762 = 574,913 lb Stress due to lateralload: at = 574,913 lb/l(76)(rr) sin 45 (0.5)(0.85)] a = 8,012 psi t Since all axialloads are transferred in bearing, the maximum weld stress will be equal to 8,012 psi. This is within acceptable limits. f s 3-28

CNS 14-170 Series ill SAR Docket No. 71-9249 Rev;sion O June 1993 l 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 / g NCosk Transverse Deceleration N A f Cask Axio!

f f

l/ Deceleration 4 Primary Cover \\ 1/2" N Impact Point. The top cover is stepped and the inner plate has a nominal clearance of 1/8-inch. Upon impact, 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 the weld between the two cover plates. The magnitude of the l 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. '%J The weight of the cask less the upper cover plate is 53,000 - 3250 = 49,740 lbs. The 3-29 j

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision 0 June 1993 ) O transverse deceleration force acting on the weld betwe - * *'-e two plates is: l (49,740)(17.88)Vf = 628,866 lbs The weld is a 1/2-inch weld,751/4 inches in diameter The stress in the weld is: t i. f= 628.866 = 8,853 psi 75.25(n)(0.5)(0.85)(0.707) F.S. = 21.000 = 2.37 8,853 The weight of the cask less the cover and shield plug is 53,000 - 5800 - 370 = 46,830 lbs. The transverse force between the inner cover plate and the inner shell of the cask will be: p = 46,830(17.88) = 592,075 lbs /2 A 40 arc length on the inside d ameter of the inner shell plate times the thickness of the lower primary cover plate is assus.:ed as the bearing area between the two surfaces. Area = D (i'O n) t = 75.25 (40 n) (2) = 52.5 in' l 360 360 The bearing stress between the two plates will be: F = 592,075 / 52.5 = 11,278 psi Allowable Bearing Stress = (0.9) Tensile Yield Strength F.S. = 0.9(49,000) /11,278 = 3.91 j 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 conditions would be the impact of the cask along the top edge on one of the long flat sides of the cover. An angle drop of 45 is considered to be worst case. l 1 O 3-30

CNS 14-170 Series ill SAR Docket No. 71-9249 l Revision O June 1993 l i f~s ?\\,_) 37.4" 34.4" N / Impact An impact in this orientation will cause minimum bending 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 illustrated below: Crushina Following impact, the edge of the corner will begin crushing until inelastic rotation around the bend point occurs. The point at which this will occur is calculated as r follows: ,r Width at bend = 37.4 in _. 9_v. _. 3'- r, rL Thickness = 2 in ~-- M = (38,000 psi)(1 in)(37.4 in)(1 in) = 1,421,200 in-lb L' v-38,000 psi j) j t ( h. 'y" k 38,000 psi ~ ) F, = Force required to initiate bending F, = M/X = 1,421,200 = 947,467 lbs J 3-31

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision 0 June 1993 r] 947,467/53,000 = 17.88 g's (axial and lateral) V F = F, / 2 = 1,339,920 lbs 1,339,920 / 53,000 = 25.28 g's (total) 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 i Volume crushed steel = (0.512)(1.025)(34.4) = 9.036 in8 2 Energy absorbed in crushing = 9.03G (38,000) = 343,364 in-Ib Bendina 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. 0 F - m -. - __i i I -m _a The balance of the energy will be absorbed by bending of the lid. - m, Total Energy - Energy Absorbed in Crushing = Energy Absorbed in Bend j (53,000 lbs)(12 in) - 343,364 in-lbs = 292,636 in-lbs 3-32

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision O June 1993 O With an axial force of 947,467 lb required to cause bending, this amount of energy will be absorbed by an axial displacement of: 292,636 in-lbs / 947,467 lbs = 0.309 in The g force developed during the bending process is calculated using a kinematic approach. The velocity at start of bending is: 2 KE g (2)(292.636)(386.4) y, W 53,000 v = 65.3 in/sec i 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)/l(65.3)(0.5)in/sec) = 0.00946 sec g force = (AV/At)/(386.4 in/sec) = (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 crushing and bending in both axial and lateral directions. The force of impact on the corner is 947,467 lbs (axial component), contents

4 0 2' --

947,467 pm 0 lA ,_j I 1w w ia mem. ms m } i .- y.. - p y ya.c.-. 2,.~.. J O 3-33 , _. _.. ~

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision 0 June 1993 O The loads on the ratchet binders will be proportional to their distance from the pivot point of the cover on the cask. Forces tending to open the lid consist of weights from waste, lid, and shield plug. (17,800 + 5800 + 370)(17.88) = 428,584 lbs Summing the moments about point 'A' = (947,467)(1.5) + (428,584)(40.2) = 18,650,277 in-Ibs 18,650,277 in-Ibs = 2R(23)(23/80.4) + 2R(57.4)(57.4/80.4) + 2R(80.4) 18,650,277 in-Ibs = 255.92 R R = 72,875 lbs 'in farthest binder from impact) R = 72,875 (57.4/80.4) = 52,028 lbs (in middle binder) R = 72,875(23/80.4) = 20,847 lbs (in binder closest to impact) The 1/4-inch thick seal ring made of AISI 1020 steel, located on the outer periphery of the i 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 lbs. O Of \\ 39.E 75('f/ ./ \\ [g/, \\ s { k! ,-a ~ ~ ~'<:gl x,3p: ~~ } x ' ~ '~~~. q; l 39.6 N \\ x ix,)

Vklx, i

The yielding surf ace area reacting against this force is proportional to the angle O and the radius. -3 l 3-34

CNS 14-170 Series lil SAR Docket No. 71-9249 Revision 0 June 1993 i (39.625)(cos 0-1/2 ) (35,000 psi) = pressure (psi) O' 39.875 2 2 2 (Pressure) tr(39.875 - 39.375 ) in / degree = force / degree 360 i 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. By dividing the incremental pressure by Young's Modulus (E x 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 mits. 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 an 88* distributed i load, the effective area of these two welds is: Area = _8_a, n{(79.5)(0.5)+(76.25)(0.375)}2 = 105 in' 360 f = 947,467 /105 = 9023 psi The actual stress values will be lower since the upper ring will cause the load on the weld O 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 inches for the inner and outer shell. Accordingly, the stresses in the shell will be 40 percent of the stresses in the spacer ring. i ? -[ O 0 3-35

\\ CNS 14-170 Series ill SAR Docket No. 71-9249 l Revision O June 1993 Table 3 Pressure Exerted on 1/4" Seal Rina due to Too Droo on Larae Flat ANGLE PRESSURE ' FORCE / IE PRESSURE STRAIN DEGREE i 1 34,780 12,026 12,026 0.001160 0.000290 2 34,769 12,022 24,048 0.001159 0.000289 i 3 34,748 12,015 36,063 0.001158 0.000289 4 34,715 12,004 48,067 0.001157 0.000289 i { 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 f 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 L t 16 33,515 11,530 189,940 0.001117 0.000279 f 17 33,348 11.531'- 201,471 0.001111 0.000277 18 33,170 11.470 212,941 0.001105 0.000276 l 19 32,983 11,405 224,346 0.001100 0.000275 20 32,785 11,337 235,683 0.001092 0.000273 P 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 l 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 j 29 30,565 10,569 334,077 0.001018 0.000254 30 30,271 10,467 344,544 0.001009 0.000252 3-36

CNS 14-170 Series til SAR Docket NO, 71-9249 Revision O June 1993 O Table 3 Pressure Exerted on 1/4" Se81 Rina due to Too Drop on Larae Flat (cont.) ANGLE - PRESSURE FO3CE/ IE PRESSUR STRAIN DEGREE E 31 29,968 10,363 354,907 0.000999 0.000249 32 29,655 10,255 365,162 0.000988 0.000247 i 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 t 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 l 44 25,229 8,724 478,680 0.000841 0.000210 Entire force 2(478,680) = 957,360 > 947,467 lbs will be distributed over an area Of - 2(44) = 88 W E i O 3-37 i

CNS 14-170 Series 111 SAR Docket No. 71-9249 Revision O June 1993 Lid Ratchet Binder Assembly Based on the 72,875 lb developed in the far ratchet binder during a top corner drop, the ratchet binder, the ratchet binder pin, and lug assemblies are analyzed as follows: Ratchet Binders The ratchet binder will have a shank diameter of 13/4 inches and rated generically for an ultimate failure load of 135,000 pounds. The binders will generally failin 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 1/2 inches. The strength of the shank is calculated as follows: 2 Yield Strength = 70,000 (1.5 ) rr + 4 = 123,700 lbs Ultimate Strength = 85,000 (1.5 ) rr + 4 = 150.207 lbs I 2 Based on yield strength, the factor of safety will be: F.S. = 123,700 / 72,875 = 1.70 I Ratchet Binder Pin O The pin is a 1 1/8 inch diameter bolt, made of SAE Grade 5 or equivalent, having a yield 'i strength of 74,000 psi based on double shearing of the brlt during loading. t i 72.675 72.875 i Ib lb 2. 1 1/8" 2 I I I r Resultant Force = 72875 lb d o as[ i a = (72,875/2)/(1.125)2(rr/4) = 36,657 psi F.S. = (74,000)(0.577) / (36,657) = 1.16 i O 3-38

CNS 14-170 Series ill SAR Docket No. 71-9249 i Revision O June 1993 Lid Ratchet Binder - Uoner 1.ua ( - 0.5" p0.5" 0.75" f 2.5"

1. 9"

=j 4.5" ~ ~ ~ ~ ~ ~ ' \\ \\ s /1. 25" 0.5" 1 -- 1 " -! MO.5" - - 1. 3 75" - f 3.75" = Tear Out Shear: a sus-t o, = 72,875 lbs / (1.5)(2 - 0.0625 - 0.27)(2) ( -m. A c ', 0.06:5~ i rg!M-4. r, 'l I I = 14,568 psi .1 "-+ i' ,,8 o.ri 1 ( W( R 2 F.S. = (38,000)(0.577) / (14,568) = 1.51 l l l / o.625* Bearina: 2" ) ( = 's" on = 72,875 lb / (1.125)(1.5) = 43,185 psi j I.. $ _., J' ? F.S. = Lid = 2/1.125 = 2.68 a/fu 43,185/70,000 T 3-39

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision 0 June 1993 Tension o = 72,875 / (1.9 + 1.375 - 1.25)(1.5) r l o = 23,992 psi r f F.S. = 38,000 / 23,992 = 1.5S Welcf l 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) / 7 1/2" Full Groove 1/2" f tt Groove ) u 1.9375" p 0.5625" I - 1.375" Neutrol Axis 72.875 lbs I Tension f o = 72,875 / ((2)(0.5)(1.5) + (2)(0.5)Vf)(2.75)) (0.85) r o = 15,909 psi r Moment (72,875)(0.5625) = 20,1(0.5)(1.5)(1.9375) + (2)(0.5)(1.4375)' (0.67)VE)(0.5)](0.85) o = 9,934 psi m o,,, = o + o = 25,843 psi m r F.S. = 38,000 / 25,843 = 1.47 P 3-40 3 Vy V

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision 0 June 1993 Lid Ratchet Binder - Lower Lua (p/' Tear Out , g3,. m3 g_3 IC S' ~ Shear 1 L u s-o.s-o, = 72,875 / (1.5)(1.75 - 0.0625 - 0.27)(2) ]__ o, = 17.137 psi \\ ,-vr w F.S. = 21,926 /17,137 = 1.28 9' s.3-Bearina o, = 72,875 / (1.125)(1.5) x- \\ 1 n 43,185 psi .J us7 - = (1.75)/(1.125)

2.52 F.S.

LJd e = o fu (43,185)/(70,000) i e Tens:on o, = 72,875 / (3.5-1.25)(1.5) = 21,593 psi F.S. = 38,000 / 21,593 = 1.76 W eld Shear a, = 72,875 / (9 + 1.5)(0.5)(2)(0.85)(2)(0.707) o, = 5775 psi Moment (72,875)(2.625) = 20,,J(0.5)(1.5)(4.5) + (2)(0.5)(4.5)2(0.67)(0.5)) (0.85)Vf) 7859 psi o = m OT = 0m + o = 9,753 psi f 2 i F.S. = 21,000 / 9,753 = 2.15 3-41 l

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision O June 1993 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 tha cover over one of the tiedown lugs. '= 25.14" 1" 5" t 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. Crushina Width at bend = 25.14 in (depth of 5.0 inch) Thickness = 2 in M = (38,000 psi)(1 in)(1 in)(25.14 in) = 955,320 in-Ibs Fa = M/x = 955,320 in-lbs / 5 in = 191,064 lbs F = 191,064 Vf) = 270,205 lb i 2 Area of crushed steel = F / 38,000 psi = 270,205 lbs / 38,000 psi = 7.1 in The area of the trapezoid is described by (1.4d + 3.44d2) where dVf = depth of crush. 2 7.1 = 1.4d + 3.44d or, d = 1.25 in 1 3-42

CNS 14-170 Series til SAR Docket No. 71-9249 Revision O June 1993 ,o Therefore: dVf = 0.882 inch Width of crush = 1 + 4.87d = 7.08 in Volume of crushed steel = d2/2 + 1.925d + 1 = 4.18 in Energy absorbed in crushing = (4.18 in')(38,000 psi) = 158,840 in-lb F Deceleration Force Durina initial Crushina / d )) Energy absorbed 158,840 in-Ibs g 1 Energy remaining = 477,160 in-lbs 'r--- 3 75' l Force = (7.1 in )(38,000 psi) ~ 2 = 270,205/53,000 = 5.1 g's _,. y I" \\ _t __ r -tur l Bendina of Cover Af ter the initial crushing of the corner and the buildup 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 absorption, and deceleration forces will be as follows: \\ 0 Bending of lid (15 until lug hits outer shell of p==cd 4 y / cask). 1._ pe ^ ^ 4, % f ']} ~.~: : - f Energy remaining = 477,160 in-lbs hl f) Energy absorbed in bending (1.3" travel): s.- E = (Fa)(d) = (191,064 lbs) (1.3") l = 248,383 in-lbs ' ' ~ Energy remaining = 228,777 in-Ibs l %) 3-43

CNS 14-170 Series 111 SAR Docket No. 71-9249 Rovision O June 1993 N Failure of the t.ua 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' - Cen tr oid Tension Compression l '1/2" Weld oil Around ~45" ) j V / l<l M=2am [(1.5)(0.5)(1.875) + (0.5)Vf)(1.875)2(0.67)(0.5)(2)] O.85 M = 5.2 (21,000 psi) = 109,368 in-Ibs 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 inches wide and will contact with the .i cask about 4.5 inches from the spacer ring. The lug willlocally deform the shell until the moment that will shear the weld is attained. l F = 49,000 (1.5)(yi) 0.5 = 36,750 yi Ibs y2 = 4.5 - yi/3 in t M = (F) (y2) = (F)(4.5 - yi/3) in-lbs t 2 = 165,375 yi - 12,250 y 3 = 109,368 in-lbs i 2 2 12,250 y 3 - 165.375 y i + 109,368 = 0 ' _ 165,375 - / 165,3752 - 4(12,250)(109.368) ~ ~ 2(12,250) yi = 0.70 in i 3-44 =

CNS 14-170 Series !!! SAR Docket No. 71-9249 Revision 0 June'1993 0 x w --0.375" N. y2 r L f F t L / %.~d --_ m-d The depth of shell deformation or, d, will be as follows: d = __y,__ = 0.70 0.375 4.5-yi 4.5-0.70 d = 0.07 in Deflections or deformations of this magnitude in the shell will not effect the integrity of the cask. Secondary Bendina 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-Ibs 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 fcrce and the force required to cause bending willincrease. 3-45

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision 0 June 1593 ,r3 h .A \\ ,/ \\ f - / 25 - d2 s s M, = F f = 955,320 in Ibs j .-~/\\ ,e \\ p _ 955,320 1 ,'\\ / 25 - d2 \\/ Y F \\ a O)

\\/

The energy absorbed is computed as fvflows: 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 displacernent 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 traveled in hJ any of the phases discussed. 3-46

CNS 14-170 Scrins 111 SAR Docket No. 71-9249 Revision 0 - June 1993 l i g) 2 KE g 2 (228,777) 3864 57.76 in/sec y, = W 53,000 v., = 28.88 in/sec i at = AX/v, = 0.88 / 28.88 = 0.0305 sec a = AV/At = 57.76/0.0305 = 1893.8 in/sec2 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. This does not exceed the g forces calculated in the drop on a long flat. 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-lbs 2 636,000 in-lbs = 1/2 mv ~ Initial velocity, v = 96.3 in/sec Let d = depth of crush Volume of steel required to absorb energy: 636,000 in-Ibs/38,000 psi = 16.7 in: v.=c. 10, m \\, ', / '\\ cud a j-I N N - V../ i Lrd The volume of steel can be described by: L V, = 2 (0.5 dw)(2 in) + (1 in)(2 in) Final depth = 4.82 d2 + 2d = 16.7 in3 d = 1.67 in I O As shown on Table 3-2, the highest g force exerted on the lid is 12.0 g's. 3-47

CNS 14-170 Series ill SAR Docket No. 71-9249 - Revision O June 1993 TABLE 3-2 Incremental-e Energy Energy Time (sec). d-Vol. Absorbed Remaining Velocity (based on Deceleration 2 2 (in) (in ) - (in-Ibs) (in-lb) (in/sec) avg. vel.) (in/sec ) (g's) 0 0 0 636,000 96.3 O.1875 0.54 20,689 615,311 94.7 0.00190 BOS 2.1 0.375 1.43 54,256 581,744 92.1 0.00200 13iO 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 I 1.5 13.85 526,110 109,890 40.0 0.00E00 4200 10.9 1.67 16.7 636,000 0 0 0.00850 4705 12.0 The weight of the cask less the upper cover plate ic 53,000 -3260 = 49,740 lbs. The deceleration force acting on the weld between the two cover plates will bo: 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 10 ads 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 A From the above analysis, it can be concluded that the CNS 14-170 Series 111 cask is in full () compliance with the requirements set forth in 10 CFR 71 for Type "A" Packaging. '1 3-48 j )

CNS 14-170 Series lli SAR Docket No. 71-9249 Revision 0 June 1993 4.0 THERMAL. EVAL.UATION O The CNS 14-170 Series ill 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 willlimit the amount of these materials that can be transported as follows: Gamma Specific" Totalt2i Energy Activity Activity isotone Mev uCi/mi Ci Cobalt 60 1.33 5.0 23.2 Cesium 137 0.66 140.0 650 'DBased 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 CNS 14-170 Series ill 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 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. A;cordingly, the amount of heat generated by the waste is insignificant. O 4-1

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision O June 1993 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 ir.ner steel,1-7/8 inch of lead shield and a 7/8 inch outer steel shell. The heavy steel flange connecting the annular steel shells at the top provides a seat for a neoprene gasket seal used to provide positive atmospheric isolation when the Ed 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 O.D. = 77.75 inches I.D. = 76.25 inches 2 2 2 2 Area = ir(Ro - Ri2) = rr(38.875 - 38.125 ) = 181.43 in Gasket is a 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 Obout 3,500 pounds of tension in the binder. Therefore, the force downward on lid compressing the gasket is: (8 binders)(3,500 lb/ binder) + 6,170 lb lid weight = F = 34,170 lb. Equivalent pressure on gasket = E = 34,170 lb/181.43 in2= 188 lb/in' A As shown on Appendix D-1, the compression of the gasket due to tightening of the ratchet binder to this minimum is 20% of the gasket thickness, or about 3/32 inch. 5.2 Shield Pluo Gasket Similarly, 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. The gasket dimensions are 22.75 in. O.D., 20.25 in. l.D., and 3/8 in, thick. The gasket is Durometer 50. 2 2 2 Area = rr(Ro - Ri2) = r7(11.375 - 10.1252) = 84.43 in 5-1 i..il

. L CNS 14-170 Scrias ill SAR - Docket No. 71-9249 Revision 0 Juns 1993 i Force downward on the lid is the sum of the weight of the lid, plus the force of the studs i O (P). j P = _L = (120 ft-Ib)(12 in/ft)/(0.15)(0.75 in) KD P = (12,800 lb/ stud)(8 studs) = 102,400 lb l W = 370 lb Total Force = 102,400 + 370 = 102,770 lb 1217.2 psi 2 Pressure on gasket = E = 102,770 lb/84.43 in = A l As shown on Appendix D-2, the compression of the shield plug gasket is 33% of initial thickness or 1/8 inch. l 5.3 Seal With internal Pressurization 1 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 flanged surface at the end of the O 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 = (4.113)/(1.77)(30 x 105) = 0.000093 in./in. P = 4,113 lb A = Area of 1-1/2" minor diameter = 1.77 in' I E = Youngs Modulus = 30 x 105 psi and for a 24 inch long binder, total strain is: (24 in.)(0.000077 in/in.) = 0.0019 in. q This is less than 2% of the initial compression of the gasket. i } i 5.4 Gasket Comoression Te_s_1 A compression test to check resiliency was done on items 4 and 5 on CNSI Drawing C 110-D-0016, Revision C, the primary lid and secondary shield plug gaskets. The t 5-2

CNS 14-170 Series til SAR Docket No. 71-9249 Rcvision O June 1993 r samples were cach 1 in: by 3/8 inch thick made of Durometer 40 neoprene and Durometer O 50 neoprene, respectively. Each sample was put in a compression device and compressed. The final results indicated that it required about 4,500 It: to compress the 1 2 in Durometer 40 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 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 Waroina of Covers The possible distortion of the cover and possible leakage due to distortion has been addressed on a cask of nearly identical design. A cask having an octagonal cover 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 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 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: O 0" 1" 2" 3" 4" 5" 6"6.75" l l 't I I f U f f f A I I I I I ( l \\ O M O O h C O eq W o DJ O O O O O O O O O l 5-3

CNS 14-170 S rias ll! SAR Dock *t No. 71-9249 Revision O June 1993 t' ' The energy absorbed in dropping a 23,360 lb package from 46 inches is 1.07 x 10* in-Ibs. ( The energy to be absorbed in a one foot drop of an CNS 14-170 Series ill cask is 12 x 53,000 = 6.36 x 10 in-lbs or less than 60 percent of the unit tested. 5 The covers are the same thickness and the overhang of the corners are approximately the same. Accordingly, the CNS 14-170 Series lit should experience less deformation. All of the deformation occurred outside of the impact limiter and no deformation of the cover was found in the areas which could affect the seal. O b r f 1e 5-4 i

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision O June 1993 ./ 6.0 OPERATING PROCEDURES This section describes the general procedures for preparation, loading, and unloading of the CNS 14-170 Series til Cask. Detailed procedures developed, reviewed, and approved in accordance with the requirements of the CNSI Quality Assurance Program are issued to authorized users. 6.1 Loadino Procedure For 55-Gallon Drums (or for installation of pre-filled liners) 6.1.1 Remove cask rain cover, if installed. 6.1.2 Loosen and disconnect ratchet binders from cask lid. 6.1.3 Using a lifting sling attached to the three symmetrically located primary lid lif ting lugs, lift the primary lid. 6.1.4 Lay down lid on a suitable protected surface, treating lid underside as potentially contaminated. 6.1.5 Using crane and suitable rigging, remove pallet (s) and any shoring materials from cask cavity. 6.1.6 Visually inspect cask caGty to verify integrity. Clean and inspect gasket sealing surfaces. w NOTE: REMOVE ANY MATERIAL OR MOISTURE FROM CASK CAVITY UNDER THE SUPERVISION OF QUALIFIED HEALTH PHYSICS PERSONNEL USING THE NECESSARY H.P. MONITORING AND RADIOLOGICAL l HEALTH SAFETY PRECAUTIONS AND SAFEGUARDS. 6.1.7 Load each pallet with a maximum of seven (7) 55-gallon drums. Shielding effectiveness may be optimized by placing drums with highest surface dose rate near the center to the pallet. 6.1.8 Attach crane to the lifting ring on the pallet slings and carefully lower into cask cavity, use caution to prevent damage to gasket sealing surfaces in inner walls of cask cavity. 6.1.9 Place shoring where appropriate between drums and cask cavity walls to prevent movement during transport. 6.1.10 Repeat the loading procedure for the next layer of palletized drums. NOTE: FOR PRE-FILLED LINERS, USE CRANE TO LOWER g LINER INTO CASK CAVITY. USE CAUTION TO l. PREVENT DAMAGE TO GASKET SEALING SURFACE OR INNER WALLS OF CASK CAVITY. 6-1

CNS 14-170 Series 111 SAR Docket No. 71-9249 R0 vision O June 1993 (' 6.1.11 Lift the primary lid onto cask and position using the alignment pins ( and the alignment marks painted on cask body and lid. 6.1.12 Replace the lid ratchet binders and tighten to 175-200 ft.-lbs. torque. Return the handles to their storage position. 6.1.13 Install anti-tamper seal wires on cask lid or ratchet binders. 6.1.14 Perform cask survey and verify that the following requirements are satisfied: A. Cask external radiation levels do not exceed 200 mR/ hour on contact,10 mR/ hour at 2 meters, and 2 mR/ hour in the tractor cab in accordance with 10 CFR 71.47 and 49 CFR 173.441. B. Cask external removable contamination is ALARA, and does 2 2 not exceed 22 dpm/cm beta-gamma and 2.2 dpm/cm alpha in accordance with 10 CFR 71.87 and 49 CFR 173.443. C. That trailer placarding and cask labeling meet DOT i specifications 4r CFR Part 172. l 6.1.15 Install cask rain cover (if required) and secure to cask. O 6.2 Unioadina Procedurg NOTE: UPON RECEIPT OF CASK, PERFORM SURVEY FOR DIRECT RADIATION AND REMOVABLE CONTAMINATION USING APPROVED PROCEDURES TO ASSURE COMPLIANCE WITH APPLICABLE i REQUIREMENTS OF 10 CFR 20.205. 6.2.1 Remove cask rain cover, if installed. 6.2.2 Loosen and disconnect ratchet binders from cask lid. i 6.2.3 Using a lifting sling attached to the three symmetrically located primary lid lifting lugs. lift the primary lid. j 6.2.4 Inspect and verify integrity of lid gasket. f 6.2.5 Lay down lid on a suitable protected surface, treating lid underside as potentially contaminated. 6.2.6 After survey by Health Physics personnel, attach crane to liner or drum pallet (s) and remove from cask cavity. 6.2.7 Visually inspect cask cavity to verify integrity. Clean and inspect O gasket sealing surfaces. I 6-2

I ~ CNS 14-170 Series ill SAR Docket No. 71-9249-1 ' Revision 0 Junn 1993'. NOTE: O. . REMOVE ANY MATERIALS OR MOISTURE FROM. i CASK CAVITY UNDER THE SUPERVISION OF QUALIFIED HEALTH PHYSICS PERSONNEL USING THE NECESSARY H.P. MONITORING AND RADIOLOGICAL HEALTH SAFETY PRECAUTIONS A.ND SAFEGUARDS. i i 6.2.8 Install new liner or drum pallet (s) in cask. l l NOTE: IF LINER OR DRUMS CONTAIN MATERIALS THAT 'i COULD PRODUCE HYDROGEN GAS AS DESCRIBED IN j THE CERTIFICATE OF COMPLIANCE, ASSURE PROPER -- VENTING OR OTHER PROTECTIVE MEASURES ARE l TAKEN. 6.2.9 Lift the primary lid onto cask and position using alignment pins and the alignment marks pointed in the cask body and lid. ] j 6.2.10 Replace the lid ratchet binders and tighten to 175-200 f t.-lbs. torque. .l Return the handles to their storage position. t 6.2.11 if CNSI seal is broken on secondary lid, remove lid and check integrity of gasket. Reinstalllid and torque nuts to 120 - 10ft.-Ibs., using a. star pattern. O i 6.2.12 . Install anti-tamper seals on primary and secondary lids. .l 6.2.13 Prior to departure from site, ensure that exterior radiation levels (fixed

f and removable) are below site release and applicable Federallimits.

i f 6.2.14 Ensure proper placarding of cask and trailer. 6.3 Processino Liners inside Cask i (For empty liners pre-installed in cask cavity.) 6.3.1 Remove cask rain cover, if installed. [ i 6.3.2 Loosen and remove the eight (8) nuts that secure the secondary lid. l [ 6.3.3 Attach a lifting sling to the lifting lug on the secondary lid, and lift off l lid. j 6.3.4 Inspect and verify integrity of gasket. ~ 6.3.5 Lay down lid on a suitable protected surface, treating lid underside as f potentially contaminated. 1 i 6.3.6 Inspect and clean gasket sealing surfaces. 6-3 1 L

CNS 14-170 Series ill SAR Dock:t No. 71-9249 Revision 0 June ~993 f 6.3.7 Proceed with filling the liner following appropriate personnel ^ .) precautions and operational procedures. NOTE: PRIOR TO CLOSING THE LINER LID AND CASK, ASSURE TEN DAY VENT PERIOD CAN BE MET PRIOR TO SHIPPING THE CASK. (SEE HYDROGEN GENERATION CONDITIONS OF CERTIFICATE OF COMPLIANCE.) 6.3.8 Lift the secondary lid onto"the primary lid ard position using indicated alignment marks and alignment pins. 6.3.9 Replace the eight (8) hex nuts and torque to 120 1 10 ft.-Ibs., using a star pattern. 6.3.10 Ensure primary lid ratchet binders are tightened to 175-200 ft.-lbs. torque and their handles are returned to the storage position. 6.3.11 Install anti-tamper seal on primary and secondary lids. 6.3.12 Perform cask survey and verify that the following requirements are satisfied. A. Cask external radiation levels do not exceed 200 mR/ hour on contact,10 mR/ hour at 2 meters, and 2 mR/ hour in the tractor L cab in accordance with 10 CFR 71.47 and 49 CFR 173.441. B. Cask external removable contamination is ALARA, and does 2 2 not exceed 22 dpm/cm beta-gamma and 2.2 dpm/cm alpha in accordance with 10 CFR 71.87 and 49 CFR 173.443. C. That trailer placarding and cask labeling meet DOT specifications 49 CFR Part 172. 6.3.13 Install cask rain cover (if required) and secure to cask. 6.3.14 Check the cavity drain line to determine that the drain ple' % p aperly installed using a pipe thread sealant. I(_ S-4 U

l CNS 14-170 Series lll SAR Docket No. 71-9249 ) Revision O June 1993 7.0 ACCEPTANCE TEST AND MAINTENANCE PROGRAM q j V 7.1 Initial Acceptance Test During and after fabrication of the cask, various tests and inspections are required prior to the first use of the cask. The tests and inspections required to be performed are enumerated below: 7.1.1 Visual inspection and dimensional verification of the entire cask and its accessories shall be performed to verify compliance with the requirements of appropriate drawing, specifications, applicable codes and other pertinent data indicating qualitative and quantitative acceptable criteria. All visual inspection shall be carried out either directly or remotely as described hereafter. The following shall be inspected: surface condition, dimension, finishes, shape, locations, details, size of holes, cleanlines, etc. 7.1.2 Lifting Lugs Load Test shall also be performed. The Load Test of all lifting lugs shall be performed in accordance with a procedure approved by Chem-Nuclear Systems, Inc. Each lug shall be load tested to one and a half times the calculated load capacity of the lug and held ten minutes as a minimum. After the load test, all welding on the lugs will be examined by magnetic particle testing using the procedure approved by Chem-Nuclear Systems, Inc. and performed O. by qualified personnel. Magnetic Particle Test shall meet the requirements of ASME Code Section lil, Division i Subsection NF, Article NF-5000 and Section V, Article 7. Test Reports shall be documented and included to Quality Assurance Records of the cask. 7.1 3 Test for Shielding integrity shall be performed after the lead pouring operations. A Gamma Scan shall be applied to verify lead thickness, shielding capacity and to determine existence of any voids or impurities in the poured lead. The Gamma Scan procedure shall specify an acceptance criteria for verification that the lead thickness is not less than 1-3/4 inches. The Gamma Scan must show no grater than ten percent loss of shielding at any point based on a four inch grid spacing. The Gamma Scan shall be performed by qualified personnel in accordance with a procedure approved by Chem-Nuclear Systems, Inc. Results of the Gamma Scan shall be documented and included in the Quality Assurance Records of the cask. Test for Sealintegrity shall also be performed prior to first use of the cask. A Leak Test shall be applied to the cask to assure leak tightness of the seals. The cask shall be pneumatically pressurized to 8 psi and while under pressure, ceals are soap bubble checked for leakage acceptance criteria - no visible bubbles. Test Report shall be documented and included in the Quality Assurance Records of the cask. 7.2 Maintenance Program The Chem-Nuclear Systems, Inc. maintenance requirements for the CNS 14-170 Series ill 7-1

CNS 14-170 Series ill SAR Docket No. 71-9249 Revision 0 June 1953 Cask are articulated in this section. These requirements reflect the specific operating p(). conditions, limitations and regulatory requirements. 7.2.1 Painted Surfaces A. Painted surfaces shall be cleaned by steam or pressurized hot water using standard commercial equipment, chemical solutions, and procedures. There are no special precautions required in this cleaning operation. B. Chipped or scratched surfaces shall be retouched or repainted using Chem-Nuclear Systems, Inc. approved paints. i' C. Alignment stripes shall be repainted when they are chipped, peeled off, faded, or not legible. Only localized surface preparation (sanding and cleaning) is required prior to repainting of alignment stripes. 7.2.2 Structural Members and Welds A. All structural members and welds shall have been checked prior to initial use of the cask. Inspections of structural members and welds are not required during routine use unless the cask has been involved in an accident or has been lifted improperly or in an overload condition. In those cases, inspection must be performed as follows: N o Drop or Accident - All accessible structural members, welds, ratchet binders, shall be visually inspected. In addition, all accessible welds must be magnetic particle tested. The Magnetic Particle Test shall be performed by qualified personnel using a Chem-Nuclear Systems, Inc. approved procedure. Magnetic Particle ^ Test shall meet the requirements of ASME Code, Section lit, Division 1 Subsection NF, Article NF-5000 and Section V, Article 7. The Gamma Scan must be repeated and evaluated to the initial acceptance criteria. Test reports shall be documented and included in the Quality Assurance Records of the cask. o Imoroner or Overload Lifj - All welds on the primary or secondary lid which were used during the time of improper or overload lift shall be load tested and magnetic particle tested. Load and Magnetic Particle Test shall be performed in the same manner as delineated in the above paragraphs. B. Whenever the cask requires total repainting and is sandblasted, all structural members and welds shall be visually inspected for any indications. Suspect members and welds shall be magnetic particle tested. Magnetic Particle Test shall be performed in the same manner y as delineated in the above paragraphs. 7-2

CNS 14-170 Serics ill SAR Docket No. 71-9249 Revision O June 1993 g C. Weld repairs if any shall be performed by qualified personnel using i Chem-Nuclear Systems, Inc. approved procedures. Welding shall meet the requirements of ASME Code Section IX and/or AWS D1.1. 7.2.3 Gaskets and Test For Seal Integrity (Leak Test) k A. All Gaskets shall be inspected for proper installation prior to each cask loading. B. All Gaskets shall be replaced once a year as a minimum regardless of condition. C. Gaskets which cannot be sealed or are damaged must be replaced or repaired. Damages may include cuts, nicks, chips, indentations, or any other defects apparent to the naked eye which would affect sealing functions. i D. Any painted surfaces in contact with the gaskets shall be maintained 'j in good condition. Any paint surface defect shall be properly retouched or repaired per Paragraph 4.2.1.B. E. Tests for sealintegrity shall be performed after the annual replacement of all gaskets as a minimum. A Leak Test shall be performed in the same manner as delineated in above Paragraph 4.1.4. 7.2.4 Fasteners f A. All fasteners shall be inspected for damage after each use. Fasteners i shall be replaced if the' following conditions exist: o Deformed or stripped threads. o Cracked or deformed hexes on bolt heads or nuts. o Elongated or scored grip length area on bolts or studs. o Severe rusting or corrosion pitting. B. Fasteners shall be inspected for clean lines and presence of lubricant in the threads prior to use. Any fastener found dirty shall be cleaned and relubricated. 7.2.5 Ratchet Binders A. The ratchet binders are designed for long rugged use with minirnal maintenance. All ratchet binders shall be inspected for operation and i general condition prior to use. B. Lubrication is required very infrequently and can be achieved by 7-3

[ CNS 14-170 Series ill SAR Dockst No. 71-9249 Revision O June 1993 disassembly of the binder and lubricating the threads. A good (p^) indication for the need to lubricate the ratchet binder will be dry threads on the joining bolt or hard operation. f 'i\\ i i O l 7-4 l

CNS 14-170 Series lll SAR Docket No. 71-9249 Revision O June 1993 8.0 QUALITY ASSURANCE pd 8.1 On December 21,1978, Chem-Nuclear Systems, Inc. filed with the NRC a description of its Quality Assurance Program in accordance with 10. CFR 71.51(a). The Chem-Nuclear Systems, Inc. Quality Assurance Program was initially approved by the NRC on October 26,1979 and Approval Certificate No. 0231 was issued. The Chem-Nuclear Systems, Inc. Quality Assurance Program was subsequently re-approved on September 6,1983, January 23, 1985, and January 17,1990. Under this certificate, the continued use of the program is authorized. 8.2 The Licensed CNS 14-170 Series ill Casks owned by Chem-Nuclear Systems, Inc. which are certified under the provisions of 10 CFR 71 and built after January 1979 are designed, fabricated, assembled, tested, modified, maintained and repaired in accordance with a Nuclear Regulatory Commission approved Quality Assurance Program (Docket #71-0231). l 0 i 1 8-1

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