Text

Rev 9 to SAR for HN-100 Series 3 Radwaste Shipping Cask
	ML20215A682
Person / Time
Site:	07109151
Issue date:	05/27/1987
From:	; WESTINGHOUSE HITTMAN NUCLEAR, INC.
To:	;
Shared Package
ML20215A674	List: ML20215A673; ML20215A682; ML20266H441; ML20266H445; ML20266H451; ML20266H455; ML20266H459;
References
	STD-R-02-001, STD-R-02-001-R09, STD-R-2-1, STD-R-2-1-R9, NUDOCS 8706170031
	Download: ML20215A682 (149)
	v • d • e

{{#Wiki_filter:- _ _ _ _ _ _ _ _ _ _ - L ) 1 Document Numb:r: Rev: Rev Dete: l WESTINGHOUSE STD-R-02-001 9 5-27-87 ) HITTMAN NUCLEAR INCORPORATED M *8 SAFETY ANALYSIS REPORT FOR THE HN-100 l SERIES 3 RAP"ASTE SHIPPING CASK s Director Product QA

Rey, Rev Date Engr.

Mgr. Mgr. ~ g EC 7** 11/12/85 h s ECN-8** 1/7/86 $n p7 f) 85-232 l g y y 6. U"*E h~~ " - b ECN 9-5/27/87 kA UJ'/#f 87-072 i l ( I ) f

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

and the heavy revision bars indicate rev. 8. I '* On.each revised page, the most recent revision date appears in the upper righthand cornert 1 FORM 01(B) p.g. 1 or 148 8706170031 870526 PDR ADOCK 07109151 C PDR b __m____z___________

SAFETY' ANALYSIS REPORT FOR THE-HN-100 SERIES 3 RADWASTE SHIPPING CASK ~ REVISION 9 Referencing 10 CFR 71 Type "A" Packaging Regulations STD-R-02-001 l Westinghouse Hittman. Nuclear Incorporated 1256 N. Church Street 1 Moorestown, NJ 08075 i l i Page 2 of 148 -)

STD-R-02-001 Page 3 of-148 MAY 2 7 N NOTICE This' Safety Analysis Report and the associated' drawings are the property ^ of Westinghouse Hittman Nuclear Incorporated, Moorestown,-New Jersey. This l material.is being made available for the purpose of' obtaining required -certifications by the U.S. Regulatory Commission, enabling utilities.and other firms producing radioactive waste to be registered users of equipment and i services supplied by Westinghouse Hittaan Nuclear Incorporated, and enabling equipment to be manufactured on-behalf of and under contracts with Westinghouse -Hittman Nuclear Incorporated. Partics who may come into possession of this material are cautioned that the information is PROPRIETARY to the-interests of Westinghouse.Hittinan Nuclear Incorporated, is not to be reproduced from this. report and the' associated drawings, er facsimiles made of these drawings without ~the express written-consent of Westinghouse Hittman Nuclear Incorporated. .) i 1

STD-R-02-001 Pass 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 q 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. ]J

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 j configurations for.the internals and their model designations are as fol-( lows: I Model Number Cask Internals HN-100/170 One large disposable container l 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 j plus 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 disseter by 73-3/8 inches high. 2.1 Ce,'k Body The cask body is a steel-lead-steel annulus in the form of a vertical oriented, right circular cylinder closed on the bottom end. The side walls consist of a 3/8 inch inner steel shell, a 1-7/8 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 lifting luss are welded to the outer steel shell.

m. STD-R-02-001 i _Paar 5 of.J48 JM 7g 2.2 Cask Lid The cask 114 is four inches thick (two 2 inch thick steel plates welded toget.ber) which is stepped to mate with tb6 upper flange of the cask body and ite closure seal. Three steel lug liftias devices are welded to the cask lid.for handling. The cask lid elso contains a " shield plug" at its center. 2.3 Shield Pluz-The shield plug is five ' inches thick.(two 1 inch thick steel plates i and one 1 inch thick steel plate weided together) fabricated it, a design similar to the cask lid. It has a gasket seal and uses eight. hold-down bolts to previde positive cask closure. The shield plug also han a lifting device located at its center to facilitate handling. 2.4 _ Cask Closure The shipping rask has two closure systems: (1) the cask lid is closed with eight high-strength retchet bindere; and a gasket seal, (2) the shield plug is closed with eight 3/4 inch bolts and the same seel .ystem used for th6 cask lid but smaller, 2.5 Cask Tiedown System The shipping cask tiedown system consiwts 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 examp k s are given in terms of Other arran' ements are possible, prcviding the gross weig'nt in 2.7. g weight and the dacay heat rate limits are observed, and the materlal secured against movement relative to the cask with an internal struc-tural members such as bottoms and pallets. Basically, tu internal's consist of the vaste containers if process waste is being transported, e 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 Packmae Weishts The respect 13e grons weights of the cask components and its designated radwaste loads are a6 follows: Cask body 29,030 pounds i Closure lid 5,800 pcunds Shield plug 370 pounds ]

' ~ " ~ ,.1 i, 4' c g

i 1

STD-R-02-001 Pass 6 of 148-JAD TW a. ( u Total: cask (unloaded) 35,200 pounds 10i-200 Large. container (s) 17,800 pounds-and wasta IDI-100 - 55 gallon size ^ % containers (up to 14 Gruss 15,100 pounds of radioactive waste) 1D1-100 - 30 gallon size con-tainers'(up to 18 drums of 11,000 pounds radioactive waste) 2.8 Radwaste Packame 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. I 1 2.8.2 Maximum Quastity of Material Per Packste Type A materials and greater than Type A quantities of low specific activity radioactive materials in secondary con-tainsrs with weights r_ot exceeding 17,800 pounds. (_ i 3.' O DESIGN CONDITIONS 3.1 General Standards d (Reference 10 CFR 71 Section 71.31) 1 3.1.1 Chemical Corrosion l cask is constructed from heavy structural steel plates. ' That All exterior surfaces are primed and painted with high There will be no asivanic, chemical, or l quality epoxy. l other reaction among the packaging components. 1 3.1.2 Positiv2 Closure System As noted, the primary lid is secured by means of eight high f f. strength ratchet binders. The secondary lid is affired with eight 3/4 inch diameter bolts. Therefore, the package is equipped with a positive closure system that will prever,t inadvertant opening. l Desian CriterCa on which istructural Analysis is Based '3,1.3 3.1.3.1 Stresses in material due to pure tension are ( ccmpared to the minimum yield of that material. Tue safety.facter is found by dividing the minimum

( q STD-R-02-001, k i Pag 2 7 or 148 i Mt27 W y yield by the calculated stress. A safety factor.- greater than.1.0 is required for acceptability.. (fy = 38,000 psi for ASTM A516 Grade 70).. For the shell _ l 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 ~1s specified. 3.1.3.2 Stresses in material due to shearing is analyzed using the " Maximum Energy - Distortion Theory" which states. i the shearing elastic: limit is.1//i = 57.7% of the tensile elastic'limitl. As with 3.1.3.1, a factor of safety greater than 1.0 is required for acceptability. f,y = (0.577)(38,000) = 21' 926 psi'for A516 Grade-70 3.1.3.3-Wald filler material rod is E70 Grade. Analysis.is base'd on American Welding Society Structural-Code D1.1-79. For fillet welds,. shear stress on effective throat regardless.of direction of loading is.30% of' specified minimum tensile strength of weld metal.- For-complete joint 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. Since all welds have been nondestructively 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 wasta 17,000 pounds Gross Weight 53,000 pounds 3.1.4.2 Cask Lifting Lugs Material is ASTM A516 Grade 70 with a minimum of 38 kai Tens 11e' Yield Strength and 21,926 psi Shear Yield l Strength (57.7% of 38,000 pai). I Design and Behavior of Steel Structures, Salmon & Johnson, page 47 )-

STD-R-02-001 -Pege 8 of 148 Mir271981 . Maximum Load 53,000 lbs. and four lugs ~are used to lift the cask. 1 53,000 lbs. X 3 g's/4 lugs = 39,750 lbs/ Lug (Vertical) E*0 ' TEAR OUT Sling Angle to Life 45' 2.5' DIA. Load 39,750 lbs/ Sin 45' Load = 56,215 lbs. N "THK. 2 Stress - 56,215 lbs/(2 X (2" x 2.5")) I o = 5,622 psi .I 1 1 6" ~ l 5.622 psi << 21.926 psi F.S. - 3.9 TENSION 3'15"" Sling Angle to Life 45' I Load = 56,215 lbs. Y l, 3 s-Stress = 56,215 lbs/(2" x 5.4926") ......I 2.5" 3 o = 5,117 psi 5.117 psi << 38,000 psi F.S. = 7.4 4.2426 - 1.25 + 2.5 = 5.4926 in l l. L i I

i STD-R-02-001 Pago 9 of 148 MAY 2 7 IM TEAR OUT Vertical Lift-Stress = 39,750 lbs/[2 X (2" x 2.5")] o = 3,975 psi 3,975 psi << 21,926 psi F S. = 5.5 l TENSION Vertical Lift Stress = 39,750 lbs/2" x (6" - 2.5") o = 5,679 psi 5,679 psi << 38,000 psi F.S. = 6.7 l 3.1.4.3 Cask Lifting Lug Welds All welds 3/4" fillet. Allowable shear stress on j effective area of weld is 0.3 times nominal tensile i strength of weld metal, E 70 rod, or (0.3)(70,000) = j 21,000 psi. j i 2" WELD { 1 1 J 775" WELD [6" WELD g2.5,,Dia. lt

/

6" i i j mmsem==uw 6 / G" WELD BOTH SIDES b 2" WELD ~

e -) STD-R-02-001 Fego 10 of 148' MM 2 71987 Vertical Lift Assume weld to 85% efficient. Minimum throat of weld = (Sin' 45') 3/4 = 0.53" I^ Weld strength per. inch = 0.85(0.53" x 1")(21,000 psi) = 9460 lbs/in'of weld Wald 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. = M 3.1.4.4 Tiedown Luas for Liftina Cask (inadvertent use) If it is assumed the entire load is carried by the tiedown lug, Section Modulus = (bha)/6; where b=15.5", I. length of tiedown lug; and h = 2", thickness of tiedown lug (See Diagram page 23). s Tiedown lug section modulus = 15.5 (2) (2/6) = 10.33 in 8 Stress - (39,750 lbs)-(3"/10.33'in ) = 11.5 kai .j where 3" is the distance from the' cask surface to'the centerline of the hole. 11.5 kai < 46 kai F.S. = 4.0 Weld required on lift lug 4.2 in.y 4.2" < 12" l F.S. = 2.9 Weld to tiedown lug-Shear = 39,750 lbs/(2 x 15.1) = 1282 lbs per. inch of weld compression or tension due to moment i couple 39,750 lbs (3"/15.5") of tiedown lug = 7,694 lbs l in/ inch of lug and moment couple with 2 thick lug 2"(3847 lbs) = 7,694 lb/in. 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 = a = v/(1281)3 + (3847)a + (5565)" = 6886 lbs per inch l-6886 lbs per inch of weld < maximum of 9460 lb per inch of weld F.S. = 1.37 I l

STD-R-02-001-Pago 11 of 148 HA127 INT 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 g j 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 3/4 inch long welds attaching the lift lugs'to the l i tiedown lugs. The shear stresses due to the vertical component of the force ist f = 39,750/(7.75+6.0+6.0+6.0)(0.75)(0.707)(0.85) = 39,750/11.6 = 3,427 psi l-The centroid of the welds is located as follows: 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. 4 Total Moment - (2.75 in)(39,750 lb) = 109,313 in-lb Total Moment = Compressive Moment + Tensile Moment Compressive Moment = Tensile Moment 1 Total Moment = 2(Tensile Moment) l = 2f(6 x 2.69 + 2.69 x 2.69 x 1/2 + 6 x 0.94 + $9,7.TO I b, 0.94 x 0.94 x 1/2)(0.75)(0.707)(0.85)) l y = 2f((16.1 + 3.6 + 5.6 + 0.4)(0.75)(0.707)(0.85)] l l j' = 2.f(25.7)(0.75)(0. 707)(0.85) U 2.75" = 23.29f L5.06" f = 109,313/23.29 - 4,694 psi 75" Combined Stress = /(4694)a + (3427)2 ^ 6"f3/4" = 5,812 psi p ( - ~ ~ ~ l0 F.S. = 9460/5,812 = 1.63 l f .)4" 2.69" Therefore, it can be safely concluded that the lug o ~ \\ 6"f3/4" will n e yield under a load equal to three times the weight of the package. 6"f3/4' l

i i STD-R-02-001 Pags 14 of 148 MAY 2 7 IK F. Primary Lid Lifting Lug Weld 1/2" weld.at shear of 21,000 psi a) shear stress due to vert. = horz. component l -j cv = ch = 6170/(6+6+1+1)(Sin 45')(0.5)(0.85) = 1467 psi i b) ' Stress due to moment Total Moment.= Compression Moment + Tensile Moment Compression Moment + Tensile Moment-Total Moment = 2(Tensile Mcment) = 2(2x3x(2/3)xfx0.5x0.707x0.85) i = 2.4f f = 9,255/2.4 = 3856 psi Combined. stress = /(3856 + 1467)2.+.1467 3- = 5521 psi F.S. - 21,000./5521 =.3.8 Therefore, it can be concluded that the lifting lugs for the lid are more than l adequate to resist a load at three times its g weight. 3.1.4.7 Lifting Lug Covers Since the primary and secor.dary lid lifting lugs are I not capable of resisting the full weight of the package they will be covered during transit. i l l

GTD-R-02-001 Pegs 12 of.148 HAY 2 7 5 l; 3.1,4.6 Lid Lifting Lugs (Secondary and Primary) A. Secondary Ltd Lifting Lug'- Material is ASTM A516 Grade 70 2* l = = Maximum Load 370 lbs. n Carried by one lug - l q-o o o- ~M TEAR OUT Area = 2((1-1/2 - 15/16) - 7/32] 3/8 %" THK. Area = 0.258 In: Stress = 3 g's x 370/0.258 ina = 4,310 psi-4,310 psi << 21,926 F.S. = 5.0 TENSION Area = (2.0 - 7/16) 3/8 = 0.586 in8 l Stress = 3 X 370/0.586 = 1,900 psi l 1,900 psi << 38,000 psi F.S. = 20 h i B. Secondary Lid Lifting Lug Weld 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 Stress = 3 X 370/1.43 i s A = 1.68 inz o = 660 psi 660 psi << 21,000 psi F.S. = 31.8 t Therefore, the secondary lid lug is able to resist a load of three times its weight without reaching a yield stress. C. Primary Lid Lifting Lugs Material is ASTM A516 Grade 70. i 1 I .h

STD-R-02-001 Pags 13 of 148 ) MAY 27 F Maximum Load = Primary Lid 5800 lbs. I I" THK.3 ' secondary tid-370 lbs. \\ 6170 lbs. ') ~ n 22k" q. l'DIA. n n 6*' { i TEAR OUT (Vertical Lift) Area = 2 x [(2-3/4 1/2) - 1/2] 1 Area = 1/5 ina and stress = 6170/1.5 o = 4,113 psi 4,113 psi << 21,926 psi F.S. = 5.3 j TEAR OUT (45* Sling Angle) T .75 ) 1 = 1.06 in: Area (short path) = (/.758 + Ioad = /(6170)2 X 2 = 8726 lbs. ,ttress = 1/2 (8726)/1.06 = 4,116 psi 4,116 psi << 21,926 psi F.S. - 5.3 TEESION (Vertical Lift) s Area = (6 - 1) 1 = 5 in Stress = 6170/5 = 1,234 psi 1,234 psi << 38,000 psi F.S. 30.8 TENSION (45* Sling Angle) . Area (short path) = /2 X (1.25)2 - 1/2 = 1.2678 Stress - 1/2 (8726)/1.2678 = 3,441 poi 3,441 psi << 38,000 psi F.S. = 11.0 I

STD-R-02-001 Pear 15'of.148 NY t 4 9 3.1.5 Tie Downs The tie down lug material is ASTM A516 Grade 70 with a j minis m 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. t A system of tie downs is provided as part of the package. They will be utilized as follows: iq i I k i i I } "2 'd i I A s i \\ l { Y E9 ) 0 ) log = 1 { Worst case accident conditions Sg

._a_..._. STD-R-02-001 j P 32 16 cf 148 1 M ff L M l 53.9" z- = A 12.0" .a 'i 69.8" I 57.8" = z ' 'i < v.lj io a'! 1(4-c 2 u.e-p y =

sc.

's tr '. L 1 'I n IL x. o .3 l<< / 3,.,,, e s 1' iP $5 Meat RtMG) ., c o e-VIEW AA C 4

x + I V ,ySTD-R-02-001 e \\ c t.p,gy 17 og 14g, X } DAN 7m' s m. g I g _[ g l 't i-3.1.5 1. ' CgCenter of Gravity . (f K .s s ~ "1 tee ' We g, Are temeg IN

CI%

35,200 lbs., - 42.9" = 1,510.080 lb.in. 37.7{' = ,;gl.050lb.in. fi Linar/wasta 17.800 lbs.9 2;332,140 53,000 lbs,. t !y' , ( 4[ s., Centerf of i Gravity % 2,181,140/33,000 I \\]h^ \\ /J s w s y n. x, s h CG = 41.2 ihch N ' N', b i tp xs 4r 7 w <q E W.. i, 'Md.5.2 Tie Down Fo:.*ces ', s __m c ( /h j 2 y 1. 1 t' Reference fr ' tidhletespecttothIt trabc. f s nhown f\\ 9 {g,[s Mr p on the tie dois. de M (page 16). 3 \\, .')1 ms C< e . g'~ .i. s

q. /

p.',$t A i i p., . N1 a Je f "d i }s, t Op - { J L e ]'.3 9 - down Y; front rearsX; side - side Z d it j j dr g y

p..

-% d j q , pf ' iaccelerations: Y axis ' 2 3's j a t 4 I axis - 4 3's -\\L g q> '.;.3 t g 'h ~ Z n is ' S s's,

\\.

s\\ l' r!i t' M,t g t . u.g y s M. d t c 4 j j' \\,( eJ'y%1 ,.[- h ,4 / ' M ,/ ** Tie Down Lengths j-. 1 ( [ J. chmentgoints) lg,, Long tie downs (low trailer t h. '.[ Cr " : }_ / t p. *h f [<, 1engthm/538 7.88+LN398 ~i l s 101.0 in. l i ys .w. .t

.i Short tie d was$ (high' trailer attachment points)

[', lengthy 502 + 57.88 + 53.98 = 99.2 in. g j \\, Q. .J I s .1 N vA 4 / fie down tensions .)%. - (V

o r

'/ y + er 7,, Tie down tens de res edbyvectordiyection [y d. '\\ Q" 3 S Long tie down et tension'TL g i i T t x \\c-i \\h 4 l. ,63 4 %i Along Y axis { 101.0r T, = 0.6233 T 1 L d q i 4 t c u i L Along X axis 53*9 TL = 0;3333 7L i l-101.0 i i 4, kl V' c' e 1 h i W 4., 0 .4 57.8 d \\ i

- f' 101.0 L = 0.5718 f j 1

b "I $I W. %s (f 'l. j ( / O.'; \\q = $' O..,Q g,y) 3%. W. -{ t a m .f g\\ n P,,' 'N k( 1s, (, i E. ( (- -ih}e l5 \\ s3 M. i's, r, [v \\ y' y L 8...Y.s ,h. g, V r >\\ -) t

n.

+- .( . >..a@c.. sg m :.r. 3 s.- :-=- -.:=. - - :. -..; :

~ j STD-R-02-001 Pan 2 18 cf 148 JAN 7.358. f 1 i Short tie down at tension T ~ g Along 7 axis TS = 0.6047 Tg 9 2 Along I exis 53;92 g = 0.5432 Tg T Along Z axis 57l8 Tg = 0.5825 Tg g 10W Force (front-rear). l Overturning (front-rear) due to 10W along X axis j Overturning moment (taken about the axis of rotation e Point A) = j 1 10(53,000 lb)39.2" = 20,776,000 in-lb 'Esch of the two rear (or front) tie downs (one 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 ) f g + (69.8")(0.6233 T ) g g = 133,801'lb T Tension in the short tie down 10,388,000 in-lb = (64")(0.5432 T ) g + (69.8")(0.6047 T ) g T3 = 134,957 lb I l t 4 0

L STD-R-02-001' l. Peg 2 19.of 148-MAY 2 7 IN7 I SW Force'(side-side)' -j 'l Overturning (side-side) due to SW along Z axis i Overturning moment (taken about the axis of rotation @ Point A) =' '5 (53,000 lb) (39.2") = 10,388,000 in-lb Each of two side tie downs (one long and one-short) must. restrain half the above moment or 5,194,000 in-lb Tension in the long tie down 5,194,000 in-lb = (64")(0.5718 T ) t + (12.0")(0.6233 T ) t l T = 117,845 lb t Tension in the short tie down 5,194,000 in-lb = (64")(0.5825 T ) g + (12.0")(0.6047 T ) g -T = 116,624 lb l g 2W Force (up-down) i Lifting (up) due to 2W along Y axis Lift = 2 (53,000 lb) - 53,000 = 53,000 lb i Each of two long and two short tie downs will carry a quarter of the load. 13,250 lb = 0.6233 Tt T = 21,258 lb t 13,250 = 0.6047 T g T = 21,912 lb g

STD-R-02-001 Peg 2 20 of 148-MAY 2 7 IN7 Total Tension j Total tension with all forces acting simultaneously. - T = 133,801 +.117,845 + 21,258 t T = 272,904 lb t T =L134,957 + 116,624 + 21,912 g l-T = 273,493 lb g t 1 l l l l l _f

STD-R-02-001 Page 21 of 148 M 27 MF 3.1.5.3-Tie-down Luas .The tie-down lugs are constructed of ASTM A516 Grade steel having a minimum yield stress of 46 kai and a ~ maximum ultimate stress of 78 kai. The following values are used in the design of the tie-down lugs. l Tensile Yield = 46,000 psi Allowable Bearing Stress = 0.9 Tensile Yield Strength Maximum Ultimate Tensile = 78,000 poi 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 l Diameter of Hole = 2.5 inches Diameter of Pin = 2.25 inches Thickness of Lug = 3.0 inches j a Projected Area of Pin = 3.0 x 2.25 = 6.75 in j Bearing Stress = = 40,517 psi F.S. = 41,000 + 40,517 - 1.02 I Tear Out l The shear planc associated with the projected area of the pin is shown as follow: i 2 71 = /1.25 2 1.125 = 0.545" l 7 = 1.25 - 0.545 = 0.705" 2 i 1

STD-R-02-001 Pasa 22 of 148 { 37g l {

1. 75 "

y 273,493 lb 2.25" Dia. Pin \\ 2.5" Dia. Hole = 3.0" thick Shear Stress = 2x3 0 = 26,047 psi F.S. = 26,542 + 26,047 = 1.02 Tension 273,493 lb. Stress = 273,493/3.0"(8.25"-2.5") = 15,855 psi 8.25" F.S. = 46,000 + 15,855 = 2.90 i Reinforcing Plate Weld d 273,403 lb 2.25" Dia. Pin 'i ec \\,, \\ 2.5" Dia. Mole 6p ,.. J r Thickness main lug 2 inches Thickness reinforcing pistes 2 9 1/2 inch Total thickness 3.0 inches Load on in" reinforcing plate = 1/6 x 273,493 = 45,582 psi Length of weld = 6 + 7 + 7 = 20" 8 Area of shear = 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 m

I 1 g STD-R-02-001 P2 sit 23 ef 148 JAN 7M i (: ~ 3.1.5.4 Tie-down Lua Welds a) Pure shear Area of shear = 2(15.5)(Sin 45')(1)(0.85) 2" + (6+6+2)(Sin 45*)(0.75)(0.85) = 25 ine / Stress = 273,493 /25 = 10,940 psi l 6 b) Moment forces on weld s' Maximum Moment = 273,493 x 2.75 = 752,106 in-lbs i m :_ a une:_ 1-l1 1 F Reference "'""""'"g"3'5'""'""'""-'"""'""""(""."""t1" { m 5 Line 1 10.95 ,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. e 43 y 2.01 onquin,nsn n nnnn ninnu,nnnnion, h I' in,~,,n, n,n n,n,n-n,n,nn,,n,n,~,,,n,,,,,n, (. 7.94 M 6.56 M = compressive moment + tension moment I compressive moment = tension mossent l 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)) I M = 57.2f = 752,106 l f = 13,149 psi Combined stress f = /10,940: + 13,1492 37,3o5 p,i I F.S. = 21,000 + 17,105 = 1.23 1 -.... -. - _..... _.. =. -. =.,.

I STD-R-02-001 Pass 24 of 148 M.I N 3.1.5.5 Analysis of Tiedown ! N ds on Cask Shell The tiedown loads are transmitted into the cask snell 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. 1 I NL F =b g > Mc Offset 4.25 in e Fx F,= 273,493 x cos 39' 30' = 211,034 lb F, = 273,493 x sin 39' 30' = 173,963 lb t M,'s 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 g Reference for method of calculation: Welding Research Council, Bulletin No 107 (WRC 107), Stress in Cylindrical Pressure Vessels from Struc-tural Attachments. T = r/t = radius to thickness ratio = 40.9/0.875 = 46.7 3 = 1/2 the circumferential width of the loaded plate = C (33*/360')(2n 40.9)% = 11.78 in C = 1/2 the longitudinal width of the loaded plate = 23\\ in/2 = 11.75 in. 2 B3 = C /r = 11.78/40.9 = 0.288 3 B2 = C /r = 11.75/40.9 = 0.287 2 Check that 5 i T 1 100 o o.

STD-R-02-001 Page 25'et 148 .l ] t Nomenstetwee s e senseis cr ndekes smen. i . (; e n u concentrated ahear load in the cir. - J Y, = cumfmntial direction, Ib- ~ (.Bhn [B concentrated shear load in the lon. 3 2 a <g Va = situdinal direction,16 I - (0.3) *(12) external overturning moment in the ' N, = circumferential direction with re-spect to the shell,in. Ib - / external overturning moment in the

4 seneeet mome.siet.e.

longitudinal direction with re-j = 1 normal stress in the ith direction on spect to the shell, in. Ib a, = mean redme of cylindrical shell, in. the surface of the she!!, psi R. = length of cylindnest shell, in. shear stress on h ith face of thefth f = = ru direction half length of rectangular loading in = stress intensity - twice maximum e, circumfmntial direction, in. 8 = shear stress, psi half length of rectangular loading in = membrane force per unit length in e, longitudinal direction, in. N, = the ith direction, Ib/in.~ wall thickness of cylindrical shell, tending moment per unit length in T = M, in. = the ith' direction, in. Ib/in. coordinate in longitudinal direction / = membrane stress concentration fae. of shell { K. = tor (pure tension or compression) coordinate in circumferential direc-J = bending stress concentration factor y tion of shell K. = denotas direction. In the case of cylindrical cuordinate in circum-3 = = spherical shells, this will refer to e fmntial direction of shell the tangential and radial direc. l R. i = (. tions with' respect to an axis attachment parameter a normal to the shell through the s = ci/R. contar ' of the attachment as si = c /A. shown in Fig.1. In the case of A, = = R. 'T; shell persmuter = multiplication factors for N. and cylindrical shells, this w'ill refer v to longitudinal and circumferen. C,, C, N, for metangular surfaces given tial directions with respect to the in Table 7 and 8 axis of the cylinder as shown in = coefficients given in Tables 7 and 8 K,, K, Fig. 2. M., M, = bending moments in shell. wall in the circumferential and longi- + = denotas tensile strees (when asso. ciated with e,) tudinal direction with respect to = denotas compressive strees (when the shell. "" I*" I" 0'" **b" '"' / associated with e,)'

- - N' i

E = modulusIf elasticity, pel '*'"I""."A respect to the WH q F = concentrated radial load or total dimtion wi distributed radial load, Ib = normal stree in the circumfmntial direction with respect to the l inhell,3mi f " '*""8l stre n in the lengitudinal di-rection with ecogaet to the shell, h e 2 General Equation poi la the analysis of stresses in thin shells, one pro-coeds by considering the reistion between internal , Acar stetw on the z face in the

directiven with rempoet to the

, membrane forces, internal bending moments and nheH psi strema concentrutions in accordance with the follo** shear stiess on the e face in the a

8(

~r.~, direction with redpect to the - (" e,

K. N,
N* SM.

sheH poi (; T p' A s t' mo eneene. mee + euw ene eween so v. soes e e n

.. +.... ~~ T I. J STD-R-02-001 Page 26 of 148 9__. p iii il i .p .L . i e j d . 1 m 2: -. "E r ,. l.,.T

a.

. I.I L E l9 9 9'~ 3 \\ c- .M.,. l C'...P...

1.. l..

t t r ..i a. i V.. H. ~,, 1, .4 wg a .t-- u. = x.11=. : a. :.+.:.s... y::.. :n ~ w nr ~?: - gy t w,. .,4. ,. n ,4,,,,,4,,,,,4,,,,, ) r i 1 '} c,,n e u ~ eme - m., s., a w = f m y 1 - s .i x a > y, m . c u.. n --a...,e nk ;J a a% q.:, 1 r yA e a v s _- a ' 'D'

, '- t r. ;

'-'- -msrm f z;.y -9 s pu s U K.- 3 f' g 3 i 5 = Enss t r; : a ,- in rye w_.; .,,y, g gs ,7;,g,- y. hig,, , m.w 7 3,. r < 't .r=- o, 1 ns 2 ,. s 4 h ,.r 4 E. i.- 2 a _.q JJ 'd ~ a . r. ,. r, 3 DI ;1 yaw e .,.,y ..1 . i,,, y n, 4 g,., r -s M %M.w% 'Iv.Ng r m a +,..J m + .m 4.9

. %.3, s

I Ij

. m

= a%. n

r.

E q.. 4. u.,,. p _< 'a.m + vn.,..? a.,m,,..u,g_ =,.~ r 5.: m , EH.1.5p 5. ? (_1 3 e-i.:.- 7-u n<-.;. g:=.._-.4-1.6 m m = .,,g. _.,s.l m 5.: _~-- %;r 1- ,m..u , %,1__.;.=__. %_, F. 31..r a _1 ,m, c um_ z m m i e-w_ , c m ._, u n t,a,r a u -x

_v
.-. y

' w c~' 4 a., ~y a?.~ .%.,,.n f. r. r m.-.)NJ 3P'I a M j . a: n p ..b I.; ad$I - 4 ./ . 2f.' ~ 2

-D

= -} g.,f $f q .~

g @ W V I

(A! $ $ d h MS 'i"l7N". u, 3,, s l [ i._ [.^ i J. n.? I VE n

4.v H..D.

.Pt . d . 3 3d b [.: O 3 e u EM 7'E=W-M v.=J --jry 7 u r T,e,,,,,,qj ]i,,Lp

3

,p,4 g q.g + 5::..ii:.. j y 1 - . 13+ ..:.: h! - 7: :.

.a_

-: r m a:. i,jgp; ;...;/,. ~ ru p.s.:a..:.v 9.-=a --h 4 - r- .,- -s.. ... ~ ~ *. =.. ".. *... '. ". *.

. *. =. =,.. -

- -t * -t- '--1*==- H,_,t : a r-u,1 1, ,s. ,.1, f.., 4 i 1 s. 1 y.m.w f., - m o, 1 34 I i

1 i

ii 1 3 g 4, g,4,' 8# 'E e, .~ j3, , ii ..' t.Ad j e 'd 3 4 "a"U-E'. pd d'T.],50M ;1 "..WJi i i:, ,,e i.i.i - a. s .w.g i i ; i 'r 2 0 4 v.m = g.---1.,- n 4.t 4i i j j u.L. F. j' g q.1,6.q. ..f.-[- 7 2 ,i i),e- .m z w . :. "a q:. ) i 'i v 5 l u 2 p-I e t 1 ij l ) { ijnt s =

g it,,

L =I N. kMl I r![ d.r'

s.

' -k E$ +4-i !1 :,. t t. t [ j . t. a. l *. ',/. 9 - ..:. up.

3. 4 -

t .g3 .2 .t V, f.s t) .t

a- :. t ;.: :

.. t :..*:.".:.,;:=.; :-i . '.+ t 1.: a

1 :4 -

a.3.. : '

::.?:

. !^;* 't *.'.t:- '.. : J 7-

g

. J:P.jM- * + - - --' ~ - ' ' "r'..,.. ;;.;ta-y H,,,y >, 4 .....g.. g .$a.6... eg. G g 2,. ....4..... J.,,, W. ...W m a(f.F s ,J j g, I /j 3 '4 2.pt -' a. ..aL . it sad,r 3* H as .ArW 'u a: - IaA.P 2 .1; q. y I;< ~ }. ..t 1g 7' i' t I 'k 1- !_ E; ; y . i OH,a j 3 ? ;i,F,! L,.M. l!,~ if, I . i q.

y.

4 .i} L

.m

-1" .l. M !-.. I'.;.,I' O w.. -~4.- 4_ i

t...

L t<h .48mi J.iq u--. 4-J J d

..+.,.

? .. i:224.-1Mi ..I. J

:.s = :.

'1 JH d . m r -s- ]<.(. C.-, ~ 3.

3. 5~A" -EF-

.jQj'j i i~ j i5:,= s n

r.., g F..g.. '.ig:5.F. 2 EI

.;.y=.,i g 3,.=.i...

s.

.y p7.

C

. x:

i a d.L "l .r a ",". .m 4 W' i

13i...... d* *.!r l.

.n- - 1

-::=

I

g. 3

..m .4 + 9 i - a a1 7 '" - '*I i "f" I i . i tes ese se age oss eso ese etc ees ese e Fig. 3A Wembrane force N /(M./R.W) due to en estemel cucumfere@el moment s $kteers in.% Lis i

STD-R-02-001. Pags 27 of 148 4 l t I s. t e_ a = i 9J. .i !i 3 g .i.. !I I i .e I i i m i .-.i.. 1 --1.- . 4. ..e m ..41 . u i..,. .s. .4. t 3.J 4.,.,__. qt ..{ .4. g ..,.y .m-... .u... m i -a 1 ,q- .e 1 i e, .- -aj.- i W d m 1,i r a

a. a.

- i. ;,.. g, a, i ir 6-.- r l 4. 1'. r n j l',. 1,. g J a v . a a r a nL'-- A-

w

,.r t g g ?:~ r L.

4.

.P 6 3 4. g. t 9=== L I'? ^l 6' 4," / 4*!. 14 3 ,: La.: ,.a, { g. -a g.. f-, i ar.- ? = !jl..t 4

l. g.

.s-.: a

.v :

i "1 ...s. c... y 4..... ... I.. 3. s.. ...3.. . 4.. ,g, o ........... 3 ;g, .5.... m i .g.., a. ,...4.. ..l.,... .w .4... ..J. i a 1 ~ ; i s i. at" . N4 f. A o r 4 en, j a a .w a a % I kE.- ..I F1

4.

L 3 8 -."Y I-m C '-

.': .a--

... V n ,a g. ..3 y, r. -m l I 3l., 9 .:.r.. f

s. h..,

f u

3..

g ..=. =. -

= =..= m=

. p.1.{.a J. = ..{..4,j, J

z. -u...:

2 .: -ea .., =.:=.=... .,g.. m.. g 2. ,;;=_ .; = ) 1 ....g _.g ...g,y, - _..,,.... ,.= g,g.,, p... g.. i 1 4-a l ".7.,4 3.f7 6 l. - _. )l.4, 4."!".4 6 ( ;. ".I.". ....'I* a a4 M. 4... g ~.. .y u -m m

'm 4i v

m 1 0 h, I h ' Nw""",* D"i.-

1...'. I M'. N I

q '2-'

T 9 l' i cm i l } 0*064 e-a.. w.. u.=- =

+ +

m m,.l. m i,,- ,T w .. u..+......._... .. -...q--. _-}.. h 3 + -) m n m .; t s E 1 .a E,.-. .J. <i.a,ai,=q. s. r 2- .l-v. " j a I l a..,.. MR. .j - + l t i 1. .) li. i.4 i <m - --, _ .3 ..:1.... s..r.. -.. r: .. :s.. 4 "3 .r. r. +. ...,,m o

a

., 9. . "C,. .,./3.. . T". 1 .+. g.3 4.. v.4 5

4. a.d:

L 4 .3 .r ,y r -a ug .n e gmg t i s. , + a.. s . i.r. - n &

2. -7, a

m a a- , c = i R s. a r? 5 t 1.W M5 r 'a.d e vi w J a 1 lt '1 =.-L C a e_ - .,.. 3 ,-rl. F., a )

9.d '.I,.(E y 1

a 1 E = -. k = . +.. = = 4$. I i q' 1 5 .- m.v..,c 9-i t 1.'1. . a . =. l 1 .==".:,.~,- ? 5 ;Ei nEr . 3 -F. 4," W i s.'t y4-.. i h .u 3-a d $. ib.. i s br b. F t. 4.1 1 r a

4..J e

p 1t s. .}..... .~..4..,.,. !..... 1.1... j... g.. 14........

1..1.

n. 9

, t r-f ; t-rt1 g-4 'd -'9-*- p. _1

r-j

.m. e nas ois es eso ens eso one soo oes eso j Mg.1A Mement M,/(M./oJ) shee to en esteenal ewsumletential moment M. en e awcuter cyteneer .%r. ors on.. Gell. \\

.., 2 STD-R-02-001 has 28 of 148 .I i .I.l . ={.4-t

s..

a jg - 11 .t._. 8 i

.6 l--

r. L,, ,.. n.:..,.., 1 u 3,..... , w ...I J. s , mg n. i 33 i..-..f. .i s. !t t'1 1 i.* 1 IE,. FC h )%..

i. !.,

. t ... a.E.- E., . i .s a.o. 1 .i. !.. ~ ~ i..:: ::7..f...h..!:.M..:.i.:.

x I.Q._....

i p.: e ~ e'" *% .1 1 a. .=. h ,,.e.g.. 4 woi O I s. .m< e. em 1C I I i. I I I I I W 3 1 I .sW ) r-5 41.m 1u L I' ei_. I ii I ! I ' f g %I' sP-i i i i c r u w. / - 2 .q. R*Kg j!,' i 'j ;! i -@n%_.... Q j ]- i kW 'L w 'MMP** "d L m ~ ,l E. h Y ?,n.QMFWo = a 3.4 i , - '.m .,c m. # ' M.

1. ".',". i. i u

m. Y.

. h. F: J a# y A 8 8 N ,w%_%. a m-r.- m** V,,'* '. ow ., i :g k n. 1..., i -/ %.".%.:.5,...4.< .:_4.: ;..:. y

1 w.

.a 4 m. .g-y-,,,; g,. , q.... .._.r p ~' " * ' E..

- ""~i*

' ~ ~ ~ ' ' "'~~ "'E "" r i i1 1 a T 5"1 i2' - =- 'ii i i i o . 'p. . ;as 4 1 i, i $ k[ I 1 I .i .I i. f gM a%/ 4'-

J i88*~ e 3-I a -. '

' & 3 a 'l a W a > a!- ~.1,.'d.3 '. ~n.H. :. f. < 2 : T ~Fi<r '~. o.- -' *r * !&'.k-!. ' 8 C-ab,.#. l . d,,- ,:2

=

.m i. i 1 ii 1 ( . 2 *w 3 a. t-2 2 p A,- c i i ,i. i i ,ii W. .<t 1 ,Li j k + ,.,,,4,,, r 4* . g. a Mgg: J In + - 8 i ' 't 6 4 ( - 4 P M.. 1mm. r ' 4. - .. ".. 6-., .a 13 5 h i i +1t .i .L c t. .Ed-T l-4 ,H- : .ra,d nr.u. j

i r

n ..'.2 h.. O.. 3 A.. : :.:!.". .9 :!.T_.F.-1. 0 . I..i.:.' ..i.. : i.. i M.Gi:7 ~ j

r. :_n

... ::: 2_ e' >.+q .~g .,g p. p q - .. f, y

; f.

I' I t ia 1 I i I .f R '~a Y_ l. 41 J d ~.a j i

I 3

S* e i.u ..,.4.. e '.y f p ,. t. . 4.,.. - f* s a s t r r i4~ S l 1-8 ' .i i 1 l- _ a 1.4. .. _.s. . i. s ., J-;..,- P-

r.,H q

t-

n.

4. p. 8' .j (- .t. ' lt+ .',,:. y.:'- M.:l..).i. j l[ ".3.*.N

g

~ % M. n..[. i.

i :

' ~ "C...6. -= .. 4.. g I 4.'. !.a i . i i i, 6 ,ii' 7. Is i n .i , i 1 fIJ J w.L ei. I. 4 i 3 a p - ]. ". . ']M." r E i i i = 'r j ,j frj l a t i s =.,..q. n t

2d

? G%.' m: j ,.....i ..1.. g, e d.m g7 . yjg sa eq I! L ..e $,,d,'l 14 1 e i r

':: d.

-) : i.-.a '.6 ".. .4 .4 .J '. '. .. $'.[., I 1' ..-.e.. C, a. 1 1."*I. h.I... a g9., 4 9 Ob 9'S O'S 9 80 0 39 0 80 0 99 0 'ee c et 030 Fig. W Memetene leste N,/(M /R.9) fue to en estemallongilweenal moment M en a sistweet splendet 6 9,rasen in.Wila

l i STD-R-02-001 Pag 29 of 148 y. ?A.< 1 a w _m, .h j l 1 a - ir,, ad: .l a p. 4___ im ~ . y u -.. _ff-$-O s f, Ig g jf f ]M EJ $E d n._ _ - I g.1 ,,,m g -s Q. lNI

'sJ

~ 1 l I b.. r .nr a h i i M I.t -$ M M I'i. ] I. ld !}!,; ![jij i l.! !) i !d = a

- ~~ ' " '

-5~= 5 ~ :-N '. l'.,.....l... ,. '.t" w ..1 q e{. .I.. , } a }. i. e. i. A E I . 4 3 a w, a* , e i s 3 ~~ ~ y up~ I e W ~... -.,,,: g,,,._ g d r,. y .j 1 a ' -:.s; a; n 7i.,..g,a-' J-g i '. 3,y rg .{. Mg.

i,r f.

. 9 +-. . p-g; M 4 w s.

-..e..

-.1 - .',:.m-y- j- -.1r -=,-.. = -:,.-) ; r .r .3_ _ :.a:

.,:..a. h,:~

s. I. l _,.-.r.] ; R. s .:5, L. j.,

... a. n: :-..l

s.

_= s. .My.- ~~ e e -e e. ., s, 4 ,,.!.... "1,. _._".T"- ..o... ...)... i 4 FI T 1 a El; l C". i e M E ~ .4 a.n v..t s.3 a g-y n,.*w-Q - d'," EF H

- sau,

3 / d c W"%.s"W.r?,, $$ Y

~'~'=

'm ~ ,,, N% ~ t % '. g I'%. 4 4 e 'g p '. Q.i r i%[mCk %d. Nd %. L i ~[ [e.,,,, J' 4.) + g, 1 iT *"'M b i.r.n a - ,.5 d %'!"%%.i@- l'- @~ l. "N 7 j j.i 7 N 8 C u-i!A '" aim ..' "f..!8, .!r C-Y. E ~ = I t :4=b t ng %; ' il f . '.. :.: : re! !i. \\.f %g. N,1)%

d. -d

- w!. y h. %e. h...A: 8 .g A 3 e 3--- w--- . A. a ..'=eA., .y-, M..3,.'.. g.. ; ?,==. g. s C-3 _. +%... -'7.,4. s.L, - ^ -- M. - -.. * % - c _ ^.- W..- 1 . +.. . ~... ._g .%^ .U. m, 0.0L2,,,.-~ , m a -e,...< ,.m -~ _ %.n d. v. a . s c. n,.. es i . m = w e en y 1 4 e. ..y..p.W,,a I g p 'N s r i.' st. '.y 4 sJ; %.i .3 i S3 ~ "- =d i -;.ym6

== 3 A N -y{ i 3 3 pem

q...._ ' '. 9.

m. ' n. .j.N, 5 m g i .m a ? QP. * * " 8' . a. wd, ,....a,y s E N!. y gg pqy e _ 3.- "- u p ;p !. . q' FN$i i% a ! i.} }.. - 'N. i:g_. r.t i. = ..r.(*a u.{. y .c...,.r, = c .:i.p A:i.;S [sJT q H.4. f i} EE.- E+Jii m ri ~ a . E -..'. --- - -. _-E i-9.-. e----_- .. ~ ~.,. - _e,. ... -...:. E.g. u.......w, .., 4...., w 1-

1. 4 e'? Q.,.

1 -.=.l...f.. . 4.l 1 1 ,,.. w_ 1 J;; ~ yq' -44%gg .".a. ggf q 5 2 g,; i L. .i -nim Y U "".~.]'" m

s z.

,2 s f E1 3 d A2 jWcC- .s 1 S %s

.- r.^

~ a u '2, m.. i ? a r% -, 7-g; t <vn. a F. u I z g t

...4J i,s

' ', = -., <. R U ~ T! ! 5 4.e f;=. .=.j F 3 ~ t . p - f - P.(- '-~ i- ., ;J 4..IL e, an - 1 W 2"" e I t */ l- 's,- r .p-. ,--, ].g )_. ,a e+ $ij r .":,J g ; ; y - 1 3. r s . i.c .].l. b g 6 . ig a. i9 ..t. ,,j,.

,,, y,7.,g

.g E .. c=.:.d.y e l a.,. ~ .1 .: : :- u.:-. '..,. B,t,4 ~.... n .. 4,. .]. e ass eJe es eao eas ese eas see een ese Ms.15-.Mement M.MMdRJ) due to en estemelloag'tudmet moment IA en a sucesor cytend piene of symmetryn Sktsats in Skrils 1 ~ - ' ~ - - ~ ~'-~-

STD @ @& 801 P:st 30 of M8-I k . I I I I ! I I I I 3 I I . I. Es. i 1 ' ' We.. a I-I I I I gy sd 'l I 3 IY & P .'._ p' h<

r. u.

i . i... 6 < 3. . y.y w . c. -4 r 7 @il%- -t Tist' fu.a.sk; e.. he s.-4, WM 48'- Eins' ? r.W ' f i'- a.:.i s v-Q -e = ' =- " .' ; --l l j q..y,. g 1 } ,n _.:a ; 1 _. 7"'. ay J i sw 9.g . m 6:.i.y ;.T 1 4 .; & c [ w i 3 g L-w - i n i;W' .t.,., t 4. f e 6h 3 Las .e a' 19e W, i . r.$: '3mp l-. 9 -s -+4y sr. s .t. 4. :. r er:. .r::. r.u..,m. 1 r n c.. sr55s..:: r. :-- , f. 4.. .::l:,.1.; r..... ~ ~,...,. f ,ame c-j ~ ~ y ' .u. .u., s. ~ a .z f 4. J gpN mumpu mmmmm.m... mggggg ---m==-asmu na e" L -===== - gg 7 7 s sp mm . q --_ p

s, _

E. wh 2 .. -==--- -... . ngg= m%- = d y,r i a. = .amm 1 W. no - 4.e. p _..n. mu 5 .w -w w%- v 5.~ m_. 1 e a ~ i.sn Jt+

4r -Hd i

1 =4 g. r D " ^*- 4.3 ...s:.s w.%,d n - o w - r .."D a y y wra :,- ws -e .t,.r; , r . s . m c s a .= H es.2 7 7 F::RQ r. HP3 1 1 % LR W 'umE

. = C 3 "

= 3 . - v, a, _- EM:,,,.m.,.W -=- .:. m _.. - =

.r....tr. -~

1 2 - =u. = ..pt:: :' . :.b..l.1

' *.. h,$.

.../! < T V. kb M..e".T.T.'T.. l 2..* **" E p' _e. ......c ..o , W.We.. .e-I. .p. ys id,. 3pd p. ..-j 4 i(, 9 . y. . gum 4, ,r 3 jri i;- i g g m,ggr [ j

F r 9

GP8" III"~ .A n f,y }". ,,nner~. , ""~ QJ f.g r ad / ! l i. i m M" Ie ss J f 11. / i /' 3 e: %W 1* i .,o****~ e '.i a- '4 - R4.4 fW@ f:'.'.h'% g l u-1 I I a.' 1M' _ l l E. .J' i 6 I r.<. p yr j k k [ [ 1. p 7 t- -k ~ ~ ~ f E ii l!I 4 [ f..[ /- - (- --- M -i ' r : i-

1. - T' 1

i ' C. / I-Li/ r .u.t ,~ .~ . -) an

O... :.

r .*. g-. :.,. .~@ y. ~

J.* t'.:. ":..:l *. l.: *~.. * * "..* :. :'..

a, .,i / -'7- ,,,r....... i G li I ' 4,.- ....pc,,.. y, q I f f g, [ v 14 r w mm. 5.m3 ri o = i : m< i;,i o r r s.,a P' fmy-rs.: r y ( d I 9 fm f,. 5 i, e -u f /. / / 8 a Id'3 "8 J'r

* * *M '

' ' i 4~ - *d-S ' i bJr a' [v 'D

'**i 1

Y E E F 1 n = P d 4 I I l .J ~s- .' r F

,. ' r

~ f J J E 9' s' ii

< f i

w R1 1 I-e ' l. r .. i p

, 'ai.. 1 M

'I/' / m y f g i,f p ja i c sukC. V pg i ir.l M.. J 't

P.

2.t.:- a t,. sz r..s.

.:, w rn..:: : : q rx r v I g - "3 ..=.:. a. ..'..=:. *s= I R.a .W p-T.T.. * * , i. :: ::=..t 2:.: - ::: - . =.. - -.

.i,

=...

=. F . _... -.... g,... 1 a i . I i1. ?'> EII-I LJ ' g gy_ l1

'.dk- '.d

.',e pp s i E I 1. I I i ..n aus d% l v ' &. '.4 M d 1 g f. M w g s s-Tt M l' a 3.n . M ) g [ J s=. g

l. 4 mr g-:

gg i w um u ': m- __ yze 'g 1. i n a r,1 e ? .f.M?

4

)

M.*g

j 1

i t 1 ....;.s, ..,.... s...r. :.:.:.:.: ~ .. 1- ..,......~i.. 4 4.. ..... i. 1 C. n e o'en sie eis ese ein e se ist eis ses s'ee Ms. 4A-Membrane force M./(M,/R.ty) fue to en esternal sistumfetential moment M, en a siscular sylinder .Whostais W E . t ,..,._... _....,.. m...s.. . ~.._. e,.. ~ _.......,.. _ -. .. ~_.... -.

-- STD-R-02-0014 Pag 31 of 143-J. ...l !...! i...,e.ri l.l,p i..,i..ii.i_._,, y,....., r..._ r-, ..e 9 t ,..,..g e s 1 h i !I 1 i !I a I i i a. !.* ! '.i.= T".o.T.~ 1 r i j i at si l' a l y '~ ~T ^ 6 .s. e . 4 i . -.m g g.

. p,. e, 4

.-.i a j . l.t ;e ..a.gg.e 4 e. 3.4 e. i B m.. i e e +. J i I 4.= luh I}!i. - l

, Ii

= .== = 1 9 ,i;I "s '* ' j=' '1 f.' f. a r . t. f. =

1

, t ! !. . l. '. !. } * ,. 8..:. I. ...4...

1 s...

e.... e s.s.. e.a.4.4 s. e e.e. e e.. Y.. ..e. o.. e.5..,.e., e. e..0e. p, e e.- e... e.,$. 6,"d..e..... 2. e. 5...e....,..e. e.., ..i, 4

W 3.. e...I..

e* ..6 .m.. .. I d... .m. g==. s g ...I...e. m. 6 e o.. 4- .y. a E-E 4 ? I I 1 1 i I. t ' I 1 i 1 i 1 I r i i J t 6.. a "J tt a. i 4 8' ]r i in1 l j .4 % 1 ii. j

I a.

.I y .t g , g p .( i' ..af i 'l .4 9 7 6%..i r'. i .=.0 i. '. n a i g~.~_ t t ..+ I. f ' .I i i. _i i y., .. m s a 1 i1 th f 4i i' ! 6 _, # p. i J. l ? > t. t".. :

i. s.. :

,.t-

t... 't...

t i i ! li!% ..am..... ~ ../ l i 8 l a.-r..- .a i J.

. J..

4.. . :..'"*. *.; i _: ..."l* /.. n

4..*."; * *... _...

.. L '. 3 ~4. *: "L I... o 2.".1.'.4.C.7..i.. "..*."O. 1".1.r., ... !".M2 4.s .g-. --..... J.4.{ .q. {.. ( .. e. y.. .,q 4... i i i 1 1 1 1 1 1 1 1 a toaE'. sG-i i e I t l I 1 a r 8" ya . i l 1 . 4 4, f". i 9 Ie R. ';? . f. f a ? 8 l e I 8 I a i: & -d i8 4 I L' 1Ii {414 S I8 f W t"8*- e 4 .ti I i.t i L 21@- .1 3 4 I i l i i I i i I e i I. I. i, , r t.r. s i i .t 6 + 1 y _ 1 - j g, 3,l . i : J 1 =6 m. . i 1 .-..-s. s'!, . + t is e ),,a i t i ie i ~- - s' , t 1 e.) i I! r t 1. I'i !i

8. i. M *i4J#.ld-u!!-

ju t. ' h-if...m.. 'f I ! i I - N..; t ' i '. E i.

f.. *

,} F' s. .a., i :. 3 g.4 -..}'. --n .,..'"2"."... L.. i p I i i1 I I 1 1 I e e I J '1 I I I .#7 1 am g-R 1 2 I I i

I

. I I i 1 4 1 i I A n 15 C gjg w .:.., 2,_ -.a. _ i 4 i 1 . L. i i __..- _ _ o _ u. _ ~ _., ~ ~ ~ - -.... oi+ ,,j,h,. i l mA, i Y%?4%:54:_4.u 4_ _,,& "f+t w.%.te--c.~ t-em.Mw.. -t,.H. m- "'@ eM-2-Y ' ,,'2 . 7 NMSO -8" @-M Mh'N . _m.. ep ~ m ~ ~ ~ a-. De026. =.. e. ) . =.... y --x., w e-.. - -~-. .,.. _~-, .m.n.

.

,r. .. =.... ..::.:. r.s _uw _ - sw.. s i r.....e.e: :.. ~ ~. n. e:= } "a..._ i en .~...... .X

4......

. i . 1 ,1 1 .o ,,1 I,. 3 i, 1 84 6 1 m ,a s i ii, 1 r. i a w e i i .W eiii. u ,-- 4,5.Em r. s.- 4ta i i 1 .... ~ d I' t (. ~7 .r.e-*- ..Q.4 3 .7 L u. r m.? a e_ ~ h ._i .s.. p.. it3e _ _*: 1 i - I. 7 i_ I I 8 .J,. 1... 't. e. e.. "1 1 - 1til' '1.*

t-t*

o'e r

in sie ein obe she ese oss sie ees s

e e Fig. RA Mement Me/(M./RJf,ews to en esteens: siisumletential moment 64, en a sistwise cyha eQrranra ire.%ribi 9

STD.R-02-001 Page 32 cf 144 ...,1...,..,., s_...i .....,.y.,...,. 1 ......s ...i 7.,, p 7, L7 ..W.L,, ,.L;, a._~. a. 1 i 4..u.:. :,. g 3.3..j 4, J..l , ~... = 4 b. .+ ? \\ .L .a. ,,,.1

47..y

{. .44 ,. a ; i -.- aw '; %. -,--n:- 'a-1 m.,.'.r.9lT.9N.l } 7,.r tz.- :w

l

.1 r L'N'1 - '

+ I ,. -l..p -~ -] q4 i. i i. , {.l 4.. . j a. ,l .t .i.,+ if! i! l!! !i f. I 1 a .; i.,

.:;[.:

r.: -.L..._-

s. .r t-1 . a. 'l'... ;. t.

?:

. -. +. .*:.!3..

1. 4 : : ;

1 1.,...... 1 8.. ..,. l...1.1e,...

g..,1

{ .. I... .....r. p.t.889. ..l...,,.,.g I..l. % -.. }.- ,. = %. - . g..... 4 g m m ] f I 1 3 8 I,! j 'I i 4 d T l t i w' a e ax w i ,i> - i i = ] y 2 w... r n .t i i l .u..

n. g.1 6 2 pf,.r

, = e b~ntT- ;,.y. q 4-a> m r m.

f.. rum..

y c 1 i em.: -17 a. = ] u r T 9 -M I i L -W4'.I i 7 i!t !t

i.r-E;st i

ii rrt l.. i J /4/" W ii 3.se ' 'i i !!2 : "*%.h.:,, N. i:* .. i",. F. H!:i.ds 1.6 ' r", ', -.. "..,,",~,e_..... i N' ~ u..

m

..r. p 1.... h.t.i. 9"iig: [.... t.3. .-L - t.

Tr_;,,, I t'.I,e: .;.,i 3 i::F7- ;:

, e._. . a.. F.u _n..~ s._ . _.. t.,... 2. .t. q (u m : i i ~ 1 a._ l' I. f.*.If a i..i.... i.. =. .,i. t4.T--..I, i . ;[.g.F. ,l . [.1. L.. f..

/ !. I in].3 I"

..T' I I 346

3. *! '

,4 ...i

s a

..l. 4l$ .ma*"""~~ T',[. ,.J. [ ..l. .j gg ,i l ,j 'L i# {i.

i. ' :'

If _N[% i iiifI...L{I ' - r.i..I.8 _' J l 'l4 l 5 r f f f T/ 'ly,.:. .1 15 p { s. g_ 41. t .!.R

. :4

r...-

o, f../../:..../ J, g: : 2 =__

w..

....j.. .. T . f..... ..\\ e+ - g . Il ...i.. j ~ r a 11 m i .i. w i.i i i ii n ii ir 1 b*- i s rr' s .I e. m i ii. 4,,,J,4,,,,,,'I ii i /\\ t p II f */.a e s .. n . t <, pt 'd. .-4 y PL f I:.J /i. L t..4 i.., e i.. i a ,.= f +- 'j' pi j i

t. ) 9, i e t

ru.'_ f

i.

,.u.. _., i m..i.. u.. =. ..j

y.f g.:._:.,; g

, i. e,

g., ;

.l .e ai+ i! ( t t f. f -=

. i..

i. e...' H.a-

ii:.i.

i. ! :. i.'

/._+.r .. t. L.

I iiT NN

_ _.. :-.i_!._!!: [ /((.)! h-3NI.E -

s i.:

i.

i. 14. .? p ... + ...,..u- ...g.....q.;.t.. -..---_ a p If m$ f us 1*l. f = i 1 j k .h a a f t. I t =l..'* ) __a g... s Lf _ _. ,4..'J .--4....-- ... i _

d. --- 4..--~=

4..J.. L l I f ....a.

i i, !.

.i ..n 7,i. u. [ . r. {...-- .i s.... j-1 l8: . -..I... .......1,.... C y se es e se est o ne e ns e s o oes ese e eos y w y w Fig. eS_Memerene force M./(M /R.9) swe to en esternallongitwemet moment M en a swtweer crimsef 6 j. i i sonners in skeun A

STD-A-02-003 Paar 33 of 148

u
F-

. GT-TT0 Tn ri i my _i -.t ..g.p _:..g.; p... ;a..:.. - - n a. a___ . u i +i j i. . 44 i. .ll g, .. s.. .I ..g -. 4 o g. I f.: .ter i._p, e 8 .f-F -.-.. - 6 .E g,i.4 p4,;. .j.b*..( 31 -v - >,--i r t ,i in . e 3

1lt i

6J h 1 Ex

i a

.lu I y.m'-lu' swer:. ia. : = e

.r.r.;.s.e

.t._ i

i. i. ; ! i lt.+;

5 + i r .l.l [j j i l. ,_ _E_ E E

*!: i. :.t./.i. ', U..l.i h?.:.i:1.

....i. M::::::5?I::

a..

:.) :

~ "_ 77 .-. y.,. i ;,1. 1 .6. .r. A I I I j: I y 3 j g s-. y_ I I m a m , r. J

3.,

a 1 iq i q ._o g . _c m i = i i .- irs v . e.. m-- +. 1 e-r.- w.r;: v., :

=
s gr, r

..6r f, lgis; e ' O't ='E M; =- : 3 - if es !==

w. mo

. EE "i;s

: e i m ac.

s h 4 4

r. a

= a 5 0 '.i m - v ; n w r2 t.m s j j i J d, i r as a s x .m .t ....r m.. . p ,. m _,. -y:.13,.,m r. 3 1 1 L. : .d m i,r. -nie V.2*1%.!

s' M,. Vi M. *4 ! ~in G D. -V.W

'. ~, -u's.+..' t: & M..f."~: ' - =.,. ~= e +~- .i. + i. "

*
N,. P L'% = -.-. : * ? :

a ..:=:._. ..,2? fi!'E 8 3

== ... u... ;;. _.: : =..=.._:~__-- .. =_ ~e... ; =::.:- + "..; r.u. j ~. ~ - = J _ y; a_-.. i=.;. F g?Ig t I t, I a i .,,i ."M i i. i . sni ;; q y m ilase =, 11 li m

9N - e..! __i !-i

~ * *.. s W*I

i. um.4 t. 6a.

.Il am n r i a 'm F hJ ~Qal ser gir{ i j ~ j ' 3*a. EEghgg,M q imir' W

m. 2 3" n. s _

.T,'T= jur.4j# a H.- a - =._ m.m ..a.. i i 1 h m.i. . ~ m . m. _. n. Tj i w. u ,. =. - ,,i .i, i =, -, s. # SP.eur "'W _ i i,,. '.,r-g g E. a, m,, * *

l*'

t* ,g Af LD 2 %.*W -7>.~ 1.- Dir. e .m E (($ kk ! $bhMN 8[! @c.[ ~ ' Y"' m.

d I,
- C" i:IdJMP "t s.Y II a

E1 L e n;6 m 1 -q <' %-.. h-z.. '.. 3

3. a. * =-

. :.a sa ;. j g. %g % L .{ T.,,,, s 3. .n,, ~ s.-. :.m. '= .. -.. r 4:p: ~ ~- n . - as.g,n._....-

ti* g. m.:;

=

= :.* u e u

. =.. -., - -- %e 0.01c 2.-+ +. 5 .- a g yv ., h ag. .- -. nq L: ' i. - =- ~., %<...s. .4.. 4 1 i m,. 4. N -- -; g._ &s Y j t 1, 1 D_ f .,',J gJg,,' } 1.1.,. ,'L, " G.L ,I M t,_ .-W"-h*E.fb 9 ~ . ~. M .. m.. ' L ",. 8, a = m.4 .R,"; -'~ i a.. ' - - j i'ir b Li-ha 3, 6mi. (-a's s. i3.% J,L 98-a E = E33] i ~uh

_r., g

v. =.

.3.,.y w..; _:E. e- '-Wr. a g _ =- 1 5 s i a ^ ' ~ r m a m la N' ! 3 7i A, i f'tb.is 1 -T *' ' s m - JISCR EW4E.fiil ? t' -d i i< r*+.u M ai.36?

- o s. i A.*4,.' *hm ' d,".w.e.. wc

~ m~i i J.. Jer.A..w.wn > rv. ER9I Z-m.x m aa -,e %e N.y c.e - w. i. .t w. rc. l e_- - u-w

* - Ei.

z. _ _m t ur = ---... _ =. :.:m... n. _.. n .:ws..... =_ =... -.. _... ) +. ~ -2 OUkiSC _3T s

i '!

$. t + , i.

4.,,g..

.A_..A..... o,, eg . =. - e; .e,. e... g~ ammi 5-. ;i.62 . I i i i,, 1, i r i -u m ' - 4 i, r- ..t f 4 y 1 ,2 r ci ' -w s.w e. ..v. m.. C ' e.. g g 'a .M's r m1 w Ir-. -u.is -v 4 7.,7. .r y m )

C's a

m j y

"' 'lV5.n. '

.t*1 ; i s '. ( 3 g 9 ll. y, y.I ' '.l' H M.f ?& -si (( s .I j - "j 2 d _. t.;- si.1. - *.a t.. .r u i ; ..i 4 3, H l 'q j 2 j k -M.. H. D a f 8 ' - 9 5 %DM1'"T-1 'd = _ _ _. '. ... 4 4.,e m k r5i I -- t l 284F.: f rj j q q g i r-l 7__

=i

-7 j

~' N..E ."~.2#- * ~ i 5 ? 1 2.. * ...l . :. ' i..*. a 1 _.,-, p: ~ ~*"9 h,.; **/l:"L 4..* *'. ". ". 3 +-4.: .. 0_ f M.*..*. i g. r ) W ...~ ~ ~. - - L.- -n-- 1 C' ~ ~ - ,a-4 eas sie als ~ e.he ehe eso ese eso oes ees 1 e, 1 1, i Fig. Is moment u.Auga,J) eve to en este,nenensitudinet moment u, en e eieswie, e,s. nee,Istres piene of symmet,y) i $leraars in %rlia I

STD-R-02-001 Table 5-Computation She:t for Local Strass:s la cylindric:I Sh:lle pag, 34 gg 34g ]

;f<. &= .. ( l,,). 6 - 5A y-a (?) " - + + .."/.36 -(. J.u)..L = F/Ai/A Pf/A R$R1.90e6.m+m oca f[Ml[//jdEE -1 9/07 +tSW *t5E.R -19/07 . 7..u.o"u, " (. 2 4 ). "'u = ..e & -(.v".a)..% ' -2% -m a a VffffV/4lV/AW/(ff ~

i

.L,.s -G: o) ..'"s. - -Bn3 +sm + ens -s,ssl$glVffAtlll>Tf4 11586 6 006, f(5B0 6006 1@3 6321 ze3 6321 '*J.','O7 $ ~~"" P.E.'"".' '

va. < r.o - T."@

+ +- t-e +- r +- /J[glMMk Di..U',' ' . d'.i. +SI8 +5IIs -5tle -SitB r.4 trn.:- ~ -s pyAgyAVApsyfpeso -a o e y'd..7.f".*~~'" SilB 5tlB siis 5119, 4230 4230 4230 4230 ~' """ dIESS4"y$5d" ~ SMRSXDPtMMM29&M i s (;' 1)Whent#[0,s=laraestabsolutemaanitudeofeitherx - U$)2 + 47 ~ x 4 x 2}or(/(o S = 1/2 c+c I /(c ~U) + 47 Yo, Yb

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

-0) x 4 STD-R-02 001 Page a5 of 148 JXR - 7 W ( The highest stress on the outer shell is 42.3 KSI. 1 The steel used in the outer shell will be specified i-to have a minimum yield strength of 49 KSI and j will be certified by testing. l l 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: F=273,493xy = 504,313 lbs j (where 78,000 psi is the ultimate strength of the shell material). The tie-down lugs are designed to fail by teer out when excessive forces are applied. The force i required to cause tear out is as follows: F = 45,006 x 2 x 3.0 x 1.75 = 472,563 lbs. (" Compared to the force required to damage the i l shell, the factor of safety will be: F.S. = 504,313/472,563 = 1.07 i 1 t l 1 1 1

e, = ~ ;. ~::.- _ :

=~. _.:... ;,...,;~..............a ;..:.

I STD-R-02-001 P:32 36 of 148 Ju 7 as ,( 3.2 Normal Conditions of Transport (Appendix A-10 CFR 71) 3.2.1 Heat 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-nel pressure of 7.35 psi. This pressure acting over the lid will produce a load of: F = (75.5)? (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 1 Each binder has an ultimate strength of 135,000 lbs. y C Therefore, it can be concluded that the reduced pressure will produce no detrimental effects. 3.2.4 Vibration All components are designed for a transportation environ-ment. No loss of integrity will be experienced. Water Sprh, 4 3.2.5 Not applicable. 3.2.6 Free Drop Since the package weighs in excess of 30,000 lbs., it must q be able to withstand a one foot drop on any surface, without g loss of contents. 3.2.6.1 One Foot Drop on Botton Corner Energy to be absorbed = 53,000 lb x 12 in. 5 in.lb. Maximum energy = 6.36 x 10 C STD-R 02-001 Pass 37 of 148 Energy will be absorbed by crushing of corner. N N f N l N . P. Q, .h.., *. ;. ** ,s........- s\\(Nr.,t '?.!' 4 BOTTOM s N d CORNER g N/j f. I The volume of the crushed ungula, assuming the t worst case of a 45' impact angle is calculated by the following equation: . ( 8 Sin $ Sin $ - $Cos$ V =R8 3 s J R \\ l y l i \\ \\ A 1 29 } W O p I . ;p... ..v.. - s re =:.. 4 x,,. _y,.y* v: : -;y,-w e.e = -. * * ' ; ;> r.* ., ~. .?'* STD-R-02-001 J Pas 2 38 of 148 ) JAI 7 g j L l ) E 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 j shown below. Even though the weld will be crushed locally, j i there will be no loss of the cask's integrity. Ih

c g

Y4fc.\\ I,4 \\s,.z.g:.-e i

1. r-w

/ Si. e ( The deceleration force exerted on the cask is calculated J as the product of contact surface area and the yield strength j ( of the steel (38,000 psi). ] Au = "# - (xy + ab sin ~1 *) 2 where for 45' angle = 6: i R = 40.5 in

y a = R/cos 45' = 57.275 in

] I b = R = 40.5 in h = 1.66 in C = R-h = 40.5-1.66 = 38.84 in y = M = 440.5 2 - 38.842= 11.47 in x = C/cos 45' = 54.928 in ~1 Au = n(57. )(40.5) - (54.928)(11.47)+(57.275)(40.5) sin 7,' 5 Au = 36.32 int F = (36.32)(38,000 psi) = 1,380,160 lb. ] 3 force = 1,380,160/53,000 = 26.0 g's i ... _....... _....... ~. a STD-R-02-001 1 Paan 39 cf 148 JAlf f g k-3.2.6.2 Effects of Bottos Corner Drop on Balance of Cask The 26.0 g deceleration will be transmitted to l l-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 j other component will act axially with respect to the plate. Cask Cover Inner Container i Deceleration Torces Contents k e 4 4 d, '4 9 Upper go p Bottom Inner Shell Plate 6 toad Shield tower Botton qr Plate Outer Shell Corner Impact kkN d bYD\\ l Reaction Force l C' Botton Corner Drop

r--

- -- '~ ~' )

STD-R-02-001 Paga 40 of 148 JAN 7g C Summary of cask component weights as used in the following drop analyses Frimary Lid 5,800 lb -Shield Plug 370 lb Outer Body Shell* 7,224 lb ' Inner Body Shell 1,900 lb Upper Botton Plate 2,645 lb Lower Botton Plate 2,864 lb Lead Shield 14,397 lb Waste Contents 17,800 lb. l

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 plus will be distributed to the ] 1 inner'and outer shells in accordance with the shell cross sec-tional~ areas. 2 The inner shell will receive loadings at its connection to the upper bottom plate consisting of: Load from lid and shield plus 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. A11'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 lower bottom plate consisting of: Load from lid and shield plug. Load from self weight of outer shell. i 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 j this plate will activate the entire perimeter veldsent to the j lower bottom plate. )

STD-R-02-001 Paga 41 of 148 M 71g Q Cask Analysis Load from Primary Lid and Shield Plua 1 Deceleration Forces Shear Area of Weld \\ / Deceleration ( Forces Detail "A" / 1/2" b3/4" Raset 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 . J 4 \\ f 2 8 2 75.5 ) = 89.388 in Inner shell area = n/4(76.25 8 - 80 ) = 222.31i in 8 8 Outer shell area = n/4(81.75 3 Total Area = 311.705 in Inner area = 89.388/311.705 = 28% 4 .C i Outer area = 222.317/311.705 = 72% j

STD-R-02-001 P:ga 42 of 148 JAf yg C Force on inner shell = (113,434)(0.28) = 31,762 lb interal and axial Force on outer shell = (113,434)(0.72) = 81,672 lb Interal and axial 2-Stresses Developed in Inner Shell and Attachment Welds Deceleration Forces Inner Container i Bottom ai. 3/8" J wel N 4 Detail "E" Reaction on Inner Shell 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 (594) = 840 psi l 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 (4 of weight acting on 4 of shell) .J + waste y e,- Total force = (1900/2)(26.0)(sin 45')+(31,762/2)+(17,800)(26.0)(sin 45') )!.Y' = 360,596 lb. 1 ) r (* c ' e N:

- :;- :. :u.,,-, n. 2 :::- :.

- A ~ - - - - -

STD-R-02-001 Page 43 of 148 N iM I-I 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 lb Axial weld stress = 66,693 lb/n(75.5)(3/8)(sin 45')(0.85)(2) = 624 psi Total stress = 467452 + 624a = 6774 psi 2 8 - 75.5 )(n/4) = 746 psi which is less Axial shell stress = 66,693/(76.25 than weld stress. Shear shell stress = lateral facce/ area f(1900)(26.0)(sin 45')+(31,752)+(17,800)(26.0)(sin 45')] = 8814 psi I (76.25z. 75.5)3 (n/4)(1/2) l F.S. = 21,926/8814 = 2.49 ) 3-Stresses Developed in Outer Shell & Attachment Welds \\_ Deceleration Force t / I Inner Shell Upper Botton Plat" Outer Shell Lower Bottom Plate / / Crush Depth h ~ Wald 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, t Total stress = JT (1070) = 1513 psi F.S. = 21,000/1513 = 13.9 _____=**._=se j , e e, e e ,e

+

STD-R-02-001 P:32 44 ef 148 JM 7 36 k Stress in weld connecting outer shell to lower bottom plate Lateral force = load of outer shell + lead shield + 4 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 l = 239,585 l Lateral stress = 239,585/n(80/2)(0.5)(sin 45')(0.85) = 6344 psi Axial Load = (7224)(26.0)(sin 45')+81,672 = 214,484 lb. Axial stress = 214,484 lb/n(80)(0.5)(sin 45')(0.85) = 2840 psi Total stress in weld = /(6344)2 + (2840)2 = 6951 psi F.S. = 21,000/6951 = 3.02 1 2 8 - 80 )(n/4) Axial stress in outer shell = 214,484/(81.75 = 965 psi < 38,000 psi yield ( Lateral shear stress in outer shell 2 80 )(n/4)(1/2) 2 = (239,585)/(81.75 ~ = 2155 psi < 21,926 psi yield i 4-Stress in Weld Joining Upper to Lower Bottom Plates 1 k : : h 'd Deceleration Forces Inner Container I l upper I Contents Botton Plat. / 3/8" wel Nr Detail "D" l Reaction on Inner Shell 1/2" welds -^ .mm_ m nn a ser

STD-R-02-001 Paga 45 of 148 Load on Reid = Upper bottom plate + Inner shell + Shear from lid on inner shell + Waste + i lead Load = [2645 + 1900 + 17,800 + (14,397/2)] (26.0)(sin 45') + 31,762 I = 574,913 lb Stress due to lateral load o = 574,913 lb/(76)(tr)(sin 45")(0.5)(0.85) g o = 8,012 psi g Since all axial loads are transferred in bearing, the maximum weld stress will be equal to 8,012 psi. This is within acceptable limits. l t i I

I STD-R-02-001 Page 46 ef 148 I' i' 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 I transverse deceleration forces between the cover and the balance of the cask. l 1 i Contents Deceleration Cask Transverse Deceleration [nner Shell Outer Shell / / Cask Axial ~ Deceleration Pri.ary _ Cover 1/2" 1/2" n (

1. pact,oint
.;,.g,-: a. - ~

.w m .) STD-R-02-001 Page 47 of 148 JAN 7h 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 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 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 ( s's. i 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 U's 628,866 lbs. The weld is a 1/2 l ll inch weld, 75-1/4 inches in diameter. The stress in the weld 628.866 '

75.25(v)(0.5)(0.85)(0.707) = 8,853 psi 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 l

l force between the inner cover plate and the inner shell of the cask will be: F = 46,830 (17.88)+8 = 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. )t=75.25(f60)(2)=52.5in' 0 Area = D ( 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. = 000), L9l_

STD-R-02-001 Paga 48 of 148 I b 3.2.6.4 One Foot Drop on Top Corner of the Lona Flat Edae 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" ) 34.4a J i h impact An impact in this orientation will cause minisua 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-t trated below. Crushina C.- F Following impact the edge Fa of the corner will begin g Fg crushing until inelastic 7 rotation around the bend gn point occurs. 9, 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 3 38,db0 psi m 8 o m 38,000 psi [ l'k - ry i a---- 5 )

~ m.. STD-R-02-001 Page 49 ef 148 N 75 l F,= Force required to initiate bending l F, = M/X = I' j200 = 947,467 lbs 947,467/53,000 = 17.88 g's (axial and lateral) ll F=F,(d)=1,339,920 1,339,920/53,000 = 25.28 s's (total) s 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 in8 Energy absorbed in crushing 9.036(38,000) = 343,364 in-Ib Bending 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. Fa F 3 1.5 inch j The balance of the energy will be absorbed by bending of the lid { [ Total Energy) - [ Energy Absorbed in Crushing] [(53,000lb)(12in)]-[343,364in-lb) 292,636 in-lb = Energy absorbed in bend. ( With an axial force of 947,467 lb required to cause bending, this amount of energy will be absorbed by an axial displace-ment of: 292,636 in-lb/947,467 = 0.309 in f I

STD-R-02-001 I Paga 50 ef 148 N 7g C The g force developed during the bending process is cal- [ culated using a kinematic approach. i Velocity at start of bending is: )2KEa,h(2)(292,636)(386.4) W (53,000) = 65.3 in/sec As calculated before, the inelastic bending deformation j is 0.309 in. The time it takes the cask to move this distance, based j on average velocity is: i i AX/V,yg = (0.309 in)/[(65.3)(0.5)in/sec] = 0.00946 see g force = (AV/At)/(386.4 in/sec) = (65.3/.00946)/(386.4) = 17.86 g's (both axial and lateral direction) 1 The above shows that maximum g force is 17.88 g's in crush-( ing and bending in both axial and lateral directions. l i The force of impact on the corner is 947,467 lb. (axial component) I Contents 40.2" 947,467 lbe 2 4 l A }2R n 4 6 U2R l R 1.5"j 17,800 x 17.88 lbs .-- 2 3 " 34.4" 2 3"--. The loads on the ratchet binders will be proportional to their distance from the pivot point of the cover (.. on the cask. .D =m =n

STD-R-02-001 P e 51 cf 148 1M Forces tending to open the lid consist of weights from~ waste, lid, and shield plug. J (17,800 + 5800 + 370)(17.88) = 428,584 lb Summing the moments about point 'A' I (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 = 255.92 R R = 72,875 lb (in farthest binder from impact) { } R = 72,875 (57.4/80.4) = 52,028 lb (in middle binder) R = 72,875(23/80.4) = 20,847 lb (in binder closest to impact) ( C L=====:: - = = =.==== =.= -.,

== i STD-R-02-001 Fase 52 of 148 { l ( 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 Israe flat. The force exerted is equal to that which is required to bend the lid, or 947,467 lb. ey f* ~ \\ -. 4, v 20 e g k The yielding surface area reacting against this force is reportional to the angle 8 and the radius. (39.625 s 6-4*) (35,000 psi) = pressure (psi) (Pressure) n(39.8752 - 39.375 ) 8 g,g/ degree = force / degree. -( As seen from Table 3-1, the entire force is distributed over a 88' are 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 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: Area = n[(79.5)(0.5)+(76.25)(0.375)]2 = 105 360 f = 947,467/105 = 9026 psi h The actual stress values will be lower since the upper ring will cause the load on the veld 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 shells will be 40 percent of the stresses in the spacer ring. .h

STD-R-02-001 Pagt 53 of 148 TABLE 3-1 Fressure Exerted on 1/4" Seal Rina Due to Top Drop on Larne Flat M Pressure Force / Degree IF Press /E Strain 1 34,780 12.026 12.026 0.001160 0.000290' l 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 1 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 1 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 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, C Entire force 2(478,680) = 957,360 > 947.467 lb. } will be distributed over an area of 4 2(44) = 88' 4 m ..n a _ <WMr> _n er.\\ Q g ....Q" n-Cee* ,q e W4 g,, p, y

9900 g, dD *

'_g a wp e 9

STD-R-02-001 g2 5Wf 148 3 Lid Ratchet Binder Assembly Based on the 72,875 lb developed in the far ratchet binder during a top corner ll 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 8 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 - 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 i r l 1 / j s s s s ) Resultant Force = 72,875 lb i o, = (72,875/2)/(1.125)2(n/4) = 36,657 psi F.S. = (74,000)(0.577)/(36,657) = 1.16 ~ j

STD-R-02-C01 Pa82 55 ef 148 { JAN 7legg l ( Lid Ratchet Binder - Upper Lug ---4 h" h" t~--- 1r u hG A 3f 4.. / \\ 4" " JF n p 1.90 2" J ( a-ts-DIA ' fr / h v -iit # ,_1s y 1 D-3/N Tear Out - 3,374 c Shear - o, = 72,875,1b/(1.5)(2 - 0.0625 - 0.27)(2) = 14,568 psi 9/16" FS = (38,000)(0.577)/(14,568) = 1.51 IW ~~ Bearina - ~ 2"' IS/" h R = 72,875 lb/(1.125)(1.5) = 43,185 psi O 2/1.125 FS = L /d = 43,185/70,000 ,4 ./ { = 2.88 I~ e

gjf, Tension -

T = 72,875/(1.9 + 1.375 - 1.25)(1.51 l O f = 23,992 psi o FS = 38,000/23,992 = 1.58 _ ~ _ _ _ _ _ _. _. _ _ _,.... _ _,,........

STD-R-02-001 Pag 2 56 cf 148 i' OhR 1m I L [-5f j f Weld 4" double groove weld, complete joint penetration with tension normal to effective area. Allowable stresses same as base metal. %" full groove and h" fillet (both opdes) h" full groove f / f i h" full / 1-15/16" groove P 9/% d-3/8" .P y 72,875 lb Neutral Tension. Axis 1 T=72,875/[(2)()(14)+(2)()(4)(2-3/4)](0.85) 0 O = 15,909 psi l l T Homent - = (72,875)(0.5625) = 20 [(4)(1.5)(1.9375)+(2)(4)(1.4375): o (2/3)(JI)(4)](0.85) o,= 9,934 psi a,g = o,+ of = 25,843 psi g FS = 38,000/25,843 = 1.47 L ~ l l\\C A

l STD-R-02-001 Pas 2 57 cf 148 M 73 k Lid Ratchet Binder - Lower Lug Tear Out 1-3/8" h" - Shear - (_ c. o, = 72,875/(1.5)(1.75-0.0625-0.27)(2) s

II'I37 P'i 1-3/4" s

FS = 21,926/17,137 = 1.28 3 earing - ( gga 6 .DIA o = 72,875/(1.125)(1.5) = 43,185 psi (1.75)/(1.125) yS = L /d, (43,185)/(70,000) e = 2.52 3 g j g, B 1h" Thk gn Tension = 72,875/(3.5-1.25)(1.5) = 21,593 psi l { 6h" l FS = 38,000/21,593 = 1.76 i i i Weld Shear - o, = 72,875 lb/(9+1.5)(h)(2)(.85)(2)(.707) gga e, = 5775 psi 4a Moment - (72,875)(2.625) = j 20,((4)(1.5)(4.5)+(2)(%)(4.5)2(2/3)(4)) 72,8751] l n (0.85)(4') i "a = 7859 psi Neutral _f Axis of* # z * "s = 9,753 psi 2-d m (. 4" FS = 21,000/9,753 = 2.15 N

STD-R-02-001 Past 58 of 148 M 7g ( Lid Ratchet, Binder - Lower Lug (Optional Design) 3/16" g _ ac ga 1 n 3/16"/ n 9 y 1-11/11 lh" DIA. l l \\ 1-11/16" l h \\ Y ik" y T-3/Y' DIA 1/8" /~ l Ik" l u. T C ik"~ jL 9" \\ \\ \\ \\ \\ \\ C \\ \\ \\ i _s k- - ir ? Section C-C We1d Hold k k" hick plates to lus ~ Assume each circumferential weld must support 4 the load - o = (72,875/2)/(8.5+2.5+0.5+2.25+6.75+0.25)(3/16)(sin 45*)(0.85) j i i o = 15,582 psi FS = 21,000f,15,53%.a M 6 4 TearOut-(Opti7 ting 0e gn) Shear of the sleeve - o = 72,875/(1.6875)(1.25)(2) = 17,274 psi ( FS = 21,926/17,274 = t.27 i .= =. ~ :.. :.=. =.

STD-R-02-001 P32 59 of 148 NN Thg J Shear of the pin - l 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 I" M FS = 21,926/19,102 = 1.15 bearing of pin - c = 72,875/(1.125)(1.25) = 51,822 psi 72,875 lb L \\ 73, L,/d, 51,822/70,0001.6875/1.125 = 2.03 eff 7

gjf, j

Tension - (Optional Design) l 5/8" ( a = 72,875/(3.5 - 1.5)(1.25) = 29,150 psi Neutral l FS = 38,000/29,150 = 1.30 Axis ~ o Weld (Optional Design) Shear a, = 72,875/(9+1)(2)(\\)(/2)(0.85) 4" o, = 6062 psi Moment (72,875)(2.625) = 20,[(1)(fi)()(4.5)+(2)(4.5)2(2/3) ()(4)(\\)](0.85) o,= 8,841 psi T

4"s * #m = 10,719 psi l

{ FS = 21,000/10,719 = 1.96 1 l 4 0 1 i

u-. STD-R-02-001 Pasa 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. 25.14d \\ [ 1" _ 5" o 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. 1 Crushina Width at bend = 25.14 in.. ( (depth of 5.0 inch) p, p Thickness = 2 in. F Fb 7 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 (/l) = 270,205 lb Area of crushed steel = F/38,000 psi = 270,205 lb/38,000 psi 8 = 7.1 in C

STD-R-02-001 Pa 61 of 148 7 26 The area of the trapezoid is described by (1.4d + 3.44d ) where d//i = depth of crush. 2 8 or d = 1.25 in 7.1 = 1.4d + 3.44d therefore,(d//5)=0.882 inch Width of crush = 1 + 4.87d = 7.08 in 8 8 Volume of crushed steel = d /2 + 1.925d + 1 = 4.18 in s Energy absorbed in crushing = (4.181n )(38,000 psi) = 158,840 in-lb l h " / Deceleration Force During Initial Crushing Energy absorbed 158,840 in-lb 5" Energy remaining = 477,160 . h 8 Force = (7.1 in )(38,000 psi) ) 3-3/4" 3/8"J = 270,205/53,000 = 5.1 s's l w ( l 120 3" Bendian of cover 1-1/8 After the initial crushing of the corner and the buildup of force noted above the corner of the cover wil: 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: 1.3,, Bending of lid (15' until lug hits '8 Fa outer shell of cask) o Energy remaining = 477,160 in-lb g 73 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 ]' r - :r :-:... :..:: - :- -. q- -

l STD-R-02-001 I Pas 2 62 of 148 j JAN 1 g ( Failure of the Lua After coming in contact with the shell, the lug will fail due to ten-sile sheer in the weld to the cask cover. The soment which will cause failure of the weld is calculated as follows: I l 3.75" cent'roid tension compression g 4 4" weld all around % 4.5 / O f P I / ) o. l (. l r M = 20, [(1.5)(4)(1.875)+( )(JI)(1.875)2(2/3)(4)(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. i -... ~ -

STD-R-02-001 Pegs 63 of 148: M 1m [.? s. ~ 0.375"' L i i j 7I I F U Yi 4 IL i d ~ -l 1 l F=49,000(1.5)(yg)f=36,750yaIb y2 = 4.5 - gi g, j 3 i j M = (F) (y2) = (F)(4.5 - y2 ) in-lbs 3 ) =165,375y2-12,250y{=109,368in-lbs 12,254 y{ , 65,375 y{ + g9,368 = 0 Y1

165,375 - (165,3758 - 4(12,250)(109,368) 2(12,250) ys = 0.70 inches I

The depth of shell deformation or, d, will be as follows: d y, 0.70 0.375 ,4.5-yg, 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. I

i STD-R-02-001 Past 64 of 148 JA4 7 gg 'b. 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 e' < bed in secondary bending will be: Initial Kinetic Energy 636,000 in-lbs I,ess 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 are for the axial force and the force required to cause bending will increase. C l + 3.. p / b N

t. '

F, d A = J25 - d M,= F,1 = 955,320 in-lbs 8 8 F,= 955,320//25 - d l --- ~~ - ..n...om..

STD-R-02-001 Paga 65 cf 148 N '1m e r The energy absorbed is computed as follows: w-t - displacement A Fa Fa(ava) d E ZE 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 l 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: i 2.18 - 1.3 = 0.88 in. Deceleration Forces - It was calculated that the initial crushing caused a deceleration force of 5.1 3's. Calculate the deceleration forces of secondary bending since this is the shortest distance travelled in any of the phases discussed. 2(KE)(a), 2(228 7 386.4) = 57,76 in/see v,y = 28.88 in/sec At = AX/v,q = 0.88/28.88 = 0.0305 see 8 a = AV/At = 57.76/0.0305 = 1893.8 in/sec 8 8 a = (1893.8 in/sec )/(386.4 in/sec ) = 4.9 s's This indicates that the maximum deceleration force during a 12 inch drop on a short flat corner on the lid is 5.1 3's. l l, This does not exceed the 3 forces calculated in the drop on a long flat. I [ I (-

STD-R-02-001 Page 66 ef 148 Jf.h i 1%3

f. 'b 3.2.6.6 Side Drop The cash 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 a depth of crush Volume of steel required to absorb energy 636,000 in-lb/38,000 psi = 16.7 in* 4 W = d Tan 67.5' 67 hD 2" thick 1" I Volume of steel described by [2(1/2 dw)(2 in)] + [(1 in)(2 in)] Final 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.

1 STD-R-02-001 Paga 67 of 148 j N TM C Table 3-2 Inergy Energy Incremental d Vol. Absorbed Remaining Velocity Time Deceleration l (in) (in ) (in-lbs) (in-lb) (in/sec) (sec) (in/sec ) (g's) 3 3 (based on l avg. vel.) j 4. -; 0 0 0 636,000 96.3 0.1875 0.54 20,689 615,311 94.7 0.00196 806 2.1 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 l 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 C. l l 1 1 j C l -_n_n_w_m--=_-=_-___::______~:.-_---_-_____'L-__:-___::____l___-_-__-___-__-__--__-___

STD-R-02-001 Pag 2 68 ef 148 JAN 7 1986 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: i F = 49,740(12.0 s'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: 1 F = 46,830 (12.0 s's) = 561,960 lbs. The loads calculated for the veld and the inner shell due to a one foot drop on the top corners in r,ection 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 1 is in full compliance with the requirements set forth in 10 CFR 71 for Type j "A" Packaging, j l i

....-.. -.._~,._- ; _..;-

}

STD-R-02-001 Page 69 of 148 JAN 7 136 C 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 limit the amount of these materials that can be transported as fol-lows: II) Total (2) Gamma Specific Isotope Energy Activity Activity Mev pCi/ml Ci Cobalt 60 1.33 5.0 23.2 Cesium 137 0.66 140.0 650 ) II) 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: 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.

e) I STD-R-02-001 Peg 2 70 of 148 ) NAY 27 IN7 i 5.0 _ CONTAINMENT The shipping cask is-a vessel which' encapsulates the radioactive material 3 and provides primary containment and.fsolation of the radioactive material l from the atmosphere while being transported. ThecaskisanSprightcircularcylindercomposedoflayersofstructural 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 i inch outer steel shell.- The heavy steel flange connecting the annular steel shells at the top provides_a seat for a Neoprene gasket seal used to provide positive atmospheric isolation when the lid is closed by tightening the eight (8) ratchet binders which are equally spaced at 45' intervals on the outer circumference of the cask. The shield plug is located in the center of.the cask lid, has a Neoprene gasket and is bolted to thq outer portion of the lid with 8 equally spaced 3/4 inch studs on a 20-7/iFinch diameter circle. 5.1 Frimary Lid Gasket c t . Determine the amount of compression of the primary lid gasket due 'to tightening of the ratchet binders. Gasket 0.D. = 77.75 inches 1.D. =.76.25 inches 1 i, Area = Tr(Ro2 - Rin) = 1Y(38.8752-2 a 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 about 3,500 pounds of tension in the binder. l TherefoSe, force downward on lid compressing the gasket (8 binders)(3,500 lb/ binder) + 6,170 lb lid weight = F = 34,170 lb. 8 z 34,170 lb/181.43 in - 188 lb/in Fquivalent pressure on gasket = = m 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 ,b gasket thickness, or about 3/32 inch. ) 5.2 Shield Plug 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. ') i )

i STD-R-02-001 Pcge 71 of 148 MAY 2 71H7 .The gasket dimensions are 22.75 in. OD, 20.25 in. ID, and 3/8 in. thick. The gasket is Durometer 50. l Area = 1Y(Roz - Ris),79.(11.3758 2 s - 10.125 ) = 84.43 in Force downward on the lid is the sum of the weight of the lid plus the force of the studs (P). P= = (120 f t-lb)(12 in/f t)/(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 2 Pressure on gasket = = 102,770 lb/84.43 in = 1217.2 psi As shown on Appendix D-2, the compression of the shield plug gasket is 33% of initial thickness or 1/8 inch. 5.3 Seal with Internal Pressurization The inner steel shell is designed to act as a pressure vessel when the cask lid is in place and tightened. As shown in Section 3.2, the cask will withstand an internal pressure 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 no 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: 6 P/AE = (4.113)/(1.77)(30 x 10 ) = 0.000093 in./in. P = 4,113 lb A = Area of 1 " Minor diameter = 1.77 ins E = Youngs Modulus = 30 x 106 p,i l and for a 24 inch long binder, total strain is: (24 in.)(0.000077 in/in.) = 0.0019 in This is less than 2% of the initial compression of the gasket. -I

STD-R-02-001 Pega 72 of 148 W5114 Y \\ l 5.4 Casket compression Test A compression test to check resiliency was done on Items 4 and 5 on Dws. C001-5-9138, the primary lid and secondary shield plug gaskets. The samples were each 1 in8 by 3/8 inch thick made of Durometer 40 neoprene and Durometer 50 neoprene respectively. Each sample was put in a compression device and compressed. The final results indicated 8 Durometer 40 that it required about 4,500 lb to compress the 1 in sample to a thickness of 1/8 inch. After removing the sample from the Similarly,the1iglereturnedtoitsoriginalthicknessof3/3 inch. test stand, the sa Durometer 50 sample was compressed to 1/8 inch n 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 Warpina 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 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: nn in.!" 3" 4" 5" 4" 6.75" I a, o i, 9 V Corner j j Deformation / i ( e / hbhha8 l 4se4 es

~., _. STD-R-02-001 1 Pa 2 73 ef 148 1 7 as [. I The eners absorbed in dropping a 23,360 lb package from 46 inches is t 1.07 x 10 in-lbs. The energy to be absorbed in a one foot drop of an 5 in-lbs or less than 101-200 Series 3 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 IDi-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. l i e e 6

x STD-R-02-001 Pass 74 of 148 6.0 OPERATING PROCEDURES Customers that use the HN-100 Series 3 cask are supplied a copy of the Mittman 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 HN-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 Packsne 6.1.1 Loosen and remove the ratchet binders, which secure the 4 primary lid. i 6.1.2 Using suitable rigging attached 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, f 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 j shoring to secure all but close fitting contents inside the / cask. 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, f 1 lift the primary lid and place it on the cask. Use the I alignment pins or alignment marks for the proper alignment and ensure that the primary lid gasket is not damaged.

STD-R-02-001 P s.t,75.(f 148 s.. .( 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): a. Loosen and remove the shield plug holddown nuts. 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 plus 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. and place it in the primary lid opening. Use the alignment pins or alignment marks for proper alignment and ensure that the shield plug gasket is not damaged. 3 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 survey on the cask and trailer system to l ensure that the radiation levels and contamination levels are below specified limits. f l 6.1.11 Complete the necessary shipping papers, certifications, and pre-release checklist. Placard vehicle and label cask as necessary. 6.2 Procedure for Uniondina the Package i Survey the cask and trailer in accordance with applicable 6.2.1 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 ratte, ( connect slings of the disposable container or pallet to a l l l

STD-R-02 001 Pas 2 76 of 148 suitable lifting device and remove the disposable container from the cask. Note: Care should be taken to avoid damage to lid and cask gaskets. 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 j J loose articles are present. i 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. l (.

r STD-R-02-001 Pass 77 cf 148 h0V 14 M85 (1-7.0 ACCEPTANCE TESTS & MAINTENANCE PROGRAM 7.1 Acceptance Test' Fabrication of the NN-100 Series casks meets the requirements of Subpart'D of 10CFR71. Tabrication 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 ASTH' code, prior to painting. i 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, i (. + 0.5 psig). The package is visually inspected after the ( pressure test and must remain in an unchanged condition. 1 7.1.3 Leak Tests A leak test will be performed using a test fixture.(with calibrated pressure gauge and pre-set relief valve) reounted into the shield plus cavity on the primary lid. Air is introduced'at a maximum rate of 1/2 psig/ min until the pressure reaches 8 psig (-0 pois, + 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 i a for a test sensitivity of at least 10 a STD. co /sec. The system will be depressurized at a rate of approximately-2 psig/ min and the test fixture removed and shield plus reinstalled. 7.1.4 Component Tests 7.1.4.1 Gaskets 1 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. (.- . _ ~.- .g.._.,,..

e STD-R-02-001 Pag 3 78 of 148-MAY 2 7 HE 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 veids or impurities in the poured lead. The gamma scan procedare contains acceptance criteria for verification that lead thick'.ess is not less than 1-3/4 inches. 7.1,6 Thermal Acceptance Tests No thermal acceptance testing will be performed on the EN-100 Series 3 cask. Refer to the Thermal Evaluation, Section 4.0, of l this report. 7.2 . Maintenance Program Maintenance and repair of the HN-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 Inspections will be performed to check the cask for contamination, 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 any use. 7.2.3 Subsystem Maintenance The HN-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, gauging, tearing, etc.) will be replaced in accordance with the packaging drawings and cask maintenance and repair procedures. The gaskets will be replaced if inspection shows any defects or every twelve (12) months, whichever occurs first.

I STD-R-02-001 P:ss 79 of 148 7.2.5 Shieldina Shielding tests will be performed if damage has required repairs affecting shield integrity. Any additional shield j testing shall be in accordance with the origins 1 criteria specified in Section 7.1.5. 518 i i \\ ~ \\ i

STD-R-02-001 Peg 2 80 of 148 NBV 14 LM5 I J ( APPENDIX A l l 9 1 -~~~~~. - =

STD-R-02-001 l$y 14 5 ( APPENDIX A ADDENDUM TO SAFETY ANALYSIS REPORT FOR SHIELDING INSERT FOR THE HN-100 SERIES 3 RADWASTE SHIPPING CASX Referencing ( 10 CFR 71 Type "A" Packaging Regulations I t Westinghouse Hittman Nuclear Incorporated-Columbia, Maryland 21045 i 9

gTD-R-02-001 Pcg2 82 of 148 1.0 PURPOSE AND BACKGROUND e 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 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 rle safe transpcrt of Type "A" quantities and greater than Type "A" i.St. 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 botton) 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 and a stainless steel cavity sleeve are optionally provided. ..s......._..._...., I

STD-R-02-001 Page 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 14fting 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 5/8 inch bolts. 2.3 Shield Plus 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 i gasket seal and uses eight hold-down bolts to provide positive cask closure. The shield plus also has a lifting device located at its center to facilitate handling. The one inch steel plate.is bolted to the underside of the shield plus with four 5/8 inch bolts. 2.4 Cask closure l The shipping cask has two closure systees: (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 c.ssk lid but smaller. s 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 RN-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. 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 Packste Weights The respective gross weights of the cask components and its designated redwaste loads are as follows: 1 .......,....-.u,..u..-7.- .m.

STD-R-02-001 5Af' fist' 2'8 Shield Insert Component . Weight with Weight Per Component Shield Insert I . Cask body 29,030 lbs 5,995 lbs 35,025 lbs shield Insert 7,245 lbs Closure lid-5,800 lbs 1,195 lbs 6,995 lbs 4C Shield plug 370 lbs 55 lbs 425 lbs Total cask (unloaded)- 42,445 lbs HN-100 - Large con-toiner(s) and weste; 10,555 lbs HN-100 - 55 gallon size containers (up to 14 10,555 lbs drums of radioactive waste) 2.8 Radwaste Packste Contents j 2.8.1 Type and Form of Material 1 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 Packase 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. l 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 j quality epoxy. There will be no galvanic, chemical, or i other reaction among the packaging components. 3.1.2 Positive Closure System i As noted, the primary lid is secured by means of eight high strength ratchet binders. The secondary lid is j I -_m.._.__.._.

STD-R-02-001 Psg2 85 of 148 MAY 2 7 IN7 affixed with eight 3/4 inch diameter bolts. Therefore, the package is equipped'with a positive closure system that will prevent inadvertent opening. 3.1.3 Design Criteria on which Structural Analysis is Based 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. Material used is ASTM A516 Gr 70 with f of 38,000 psi. For the shell, steel having a minlmum 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 states the shearing elastic limit is 1//3 = 57.7% of the tensile elastic limit.1 As with 3.1.3.1, a factor of safety greater than 1.0 is required for acceptability. 3.1.3.3 Weld filler material rod is E70 Grade or better. Analysis is based on American Welding Society Structural Code D1.1-79. For fillet welds, shear stress on effective throat regardless of direction of loading is 30% of specified minimum tensile strength of weld metal. For complete joint penetration groove i 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 Gross Weight 53,000 lbs ' Design and Behavior of Steel Structures, Salmon & Johnson, page 47. 2ASTM Standards, Parts 4 and 5.

l' STD-R-02-001 g Paga 86 of 148 i MAY 2 7157 3.1.4.2 ' cask Lifting Lugs Material is ASTM A516 Grade 70 with a minimum of 38 kai Tensile Yield Strength and 21,926 psi Shest Yield Strength (57.7% of 38,000 psi). 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) i f 2.5 " TEAR OUT fi g Sling Angle to Lift 45'

*, t '

i Load 39,750 lbs/ Sin 45' l l.I 2.5" DtA. Load = 56,215 lbs. .. I.) ~%, i ij! '*%-Z'THK Stress = 56,215 lbs/[2 x (2" x 2.5")] !i' o = 5,622 psi .I l G" l i f 5,622 psi << 21, 926 psi F.S. = 3.9 i

STD-R-02-001 Pag 2 87 of 148 E 27 M81 l i l 8 -l l r: l TENSION +! gY, j, Sling Angle to Life 45* g ,3, h-t ii Load 56,215'1bs. d- -,--l 2.5" Stress = 56,215 lbs/(2" x 5.4926") .3' 8 l o = 5,117. psi 5,117 psi << 38,000 psi F.S. - 7.5 4.2426 - 1.25 + 2.5 = 5.4926 in. TEAR OUT Vertical Lift Stress = 39,750 lbs/[2 x (2" x 2.5")] o = 3,975 psi 3,975 psi << 21,926 psi F.S. = 5.6 TENSION Vertical Lift Stress = 39,750 lbs/2" x (6" - 2.5") o = 5,679 psi 5,679 psi << 38,000 pai F.S. = 6.7 l 3.1.4.3 Cask Lif ting 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, E70 rod, or (0.3) (70,000) = 21,000 psi

STD-R-02-001 i Psga 88 cf 148 JAN 7 136 2" WE.LD I ~ k 'i;i; j i !.l 7.75" WELD 6" WELD l E2.5" Dia. l l:l I j/

1. !

4 i.j 6" i ( W a i! .m#mmmmma / l 1 i 1 i G" WELD BOTH SDE5 t i b 2" WELD .i r i 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-LiftLuaswith45'SlinaAnsie 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 1 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 c .. - - -. u

.a

r=.-.

., 2..

STD-R-02-001 Pege 89 of 148 MAY 2 7 IN1 The centroid of the welds is located as follows: y = (7.75(7.75/2) + 6(6/2) + 6(7.75) + 6 x 6] .t'(7.75 + 6 + 6 + 6) (30 0 + 18 + 46.5 + 36) + 25.75 = 5.06 in. = 39/750' Ib. Total Moment = (2.75 in) (39,750 lb) = 109,313 in-lb ~~~ j Total Moment = Compressive Moment + Tensile Moment 2.75" Compressive Moment = Tensile Moment Total Moment = 2(Tensile Moment) - 2f((6 x 2.69 + 2.69 x 2.69 x 1/2 + 6 x 0.94 75,, 5.06" + 0.94 x 0.94 x 1/2)(0.75)(0.707)(0.85)] 6"63/4" = 2f ((16.1+3.6+5.6+0.4)(0.75)(0.707)(0.85)) 0.34" = 2f (25.7)(O.75)(O.707)(O.85) 2.69" = 23.29 f \\ 6"e3/4" y f = 109,313/23.29 = 4,694 psi m ned tress = M,6 W + (3,427F 6"f3/4" = 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 Tie Down Lugs for Lifting Cask (inadvertent use) If it is assumed the entire load is carried by the tie down lug, Section Modulus = (bha)/6; where b = 15.5", length of tie down lug; and h = 2", thickness of tie down lug, (See Diagram page 98). Tie down lug section modulus = 15.5 (2)(2/6) = 10.33 8 8 in ; Stress = (39,750 lbs)(3"/10.33 in ) = 11.5 KSI l where 3" is the distance from the cask surface to the l centerline of the hole. 11.5 KSI < 46 KSI F.S. = 4.0 Weld required on lift lug 4.2 in., 4.2" < 12" l F.S. = 2.9 l \\

1 i STD-R-02-001 Pega 90 of 148 MAY 27 E7 Wald to tie down lug Shear = 39,750 lbs/(2 x 15.5) = 1282 lbs per inch of weld compression or tension due to moment couple 39,750 lbs (3"/15.5 in) of tie down 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 '1b-in/ in. lug For 2" lug,-then 5,565 lb/in Stress on weld to tie: down lug = a = /(1282)2 + (3847): + (5565): - 6885 lbs per inch. l 6885 lbs per inch of weld < maximum of 9460 lb per inch of weld F.S. = 1.37 1 3.1.4.5 Lid Lifting Lugs (Secondary and Primary) Secondary Lid Lifting Lug 2" Material is ASTM A516 Grade 70 r Maximum Load 425 lbs. 6 Carried by one lug i o Q" iga )n y o M M* TEAR OUT Area = 2((1-1/2 = 15/16) - 7/32] (3/8) Area = 0.258 in.2 j %" THK. Stress = 3 g's x 425/0.258 in.2 = 4,941 psi 4,941 psi << 21,926 psi F.S. - 4.4 TENSION Area = (2.0 - 7/16) 3/8 = 0.586 in." Stress = 3 x 425/0.586 = 2,175 psi 2,175 psi << 38,000 psi F.S. = 17.4 Secondary Lid Lifting Lug Weld 1/2" fillet veld 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 759 psi << 21,000 psi F.S. = 27.6 1 ]

STD-R-02-001 Pags 91 of'148 MAY 2 71m Therefore, the. secondary lid lug is able to resist a load of three times its weight without reaching a yield stress. C. Primary Lid Lifting Lugs Material.A516 Grade 70 Maximum Load = Primary Lid 6,995 lbs. 1,THK.]. Secondary Lid -425 lbs. \\ 7,420 lbs. 2( l'DIA. o o a 6" TEAR OUT (Vertical Lift) Area = 2 x ((2-3/4) - (1-1/2) - 1/2) (1) Area = 1.5 in.8 and stress = 7420/1.5 i o = 4,947 psi 4,947 psi << 21,926 psi F.S. = 4.4 TEAR OUT (45' sling angle) 2 Area (short path) = (/.75 +.75 ) (1) = 1.06 in.2 Load = 7,420 /2 - 10,493 lbs. Stress = (1/2) (10,493)/1.06 = 4,950 psi 4,950 psi << 21,926 psi F.S. = 4.4 TENSION Area - (6 - 1) (1) = 5 in.2 Stress = 7,420/5 = 1,484 psi 1,484 psi << 38,000 psi F.S. - 25.6 1

STD-R-02-001; Pege 92 of-148 MAY 2 7157 TENSION (45' Sling' Angle) Area (short path) = /2x(1.25)2 - 1/2 = 1.2678 in Stre'esf= (1/2) (10,493)/1.2678 = 4,138 psi 4,138 psi u.38,000 psi F.S. ='9.2 Primary Ltd Lifting Lug Weld i' 1/2" weld,at shear of 21,000 psi a) shear stress due to-vert = horiz component. oh = 'ch = 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(Tens 11e Moment) = 2(2x3x(2/3)xox0.5x0.707x0.85) = 2.4a-o = 11,130/2.4 = 4,638 psi. Combined Stress = / (4,638 + 1,764)2 + 1,764 2 = 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 Since the primary and secondary lid lifting lugs are ) not capable of resisting the full weight of the package i they will be covered during transit. g 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 f will be specified for fabrication and certified by test of'the 3 material. 1 /

f ~ STD-R-02-001 Pass 93 of 14g JAN 7g V 1 u r 53.9" 1' A f 12.0" l L 69.8" = 57.8" l Ri 1 4 A 81.8" 7 i L L is 60" gy W.Q. n g. 8 L .,9,O" U 8 y 9,/6HEAlfam) F h 81 VIEW AA [ ~- ~

f-STD-R-02-001 Past 94 cf 148 JAN 1m A system of tie downs is provided as part of the (.' package. They will be utilized as in View A-A. 3.1.5.1 Cask Center of Gravity Item Weiaht Are Moment 1,803,913 in-lb Cask 42,445 lbs

42.5"

= 367,314 in-lb j Liner 10,555 lbs 34.8" = 53,000 lbs 2,171,227 in-lb l and Waste 1 Center of Gravity = 2,171,227/53,000 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) i up - down Y; front - rear X;_ side - side Z accelerations: Y axis - 2 g's X axis - 10 3's ( Z axis - 5 s's Tie Down Lenaths Long tie downs (high trailer attachment points) length = 4632 + 57.82 + 53.92 = 101.0 inches i Short tie downs (low trailer attachment points) length = 460 +57782 + 53.9.2 = 99.2 inches Tie Down Tensions Tie down tensions resolved by vector direction Long tie down at tension Tg iOg=0.6233Tg T Along Y axis 5 Along X axis Tg=0.5}33Tg 30 0

c

STD-R-02-001 Paga 95 of 148 JM. 7Jg Along Z axis Tg = 0.5718 Tg 10 Short tie down at tension T I g Along T axis Tg = 0.6047 Tg Along X axis 3*' TS = 0.5432 Tg 92 Along Z axis Tg = 0.5825 T3 10W Force (front-rear) Overturning (front-rear) due to 10W along X axis i Overturning moment (taken about the axis of rota-1 tion e Point A) = 10(53,000 lb) 39.0" = 20,670,000 in-lb l .I Each of the two rear (or front) tie downs (one long and one short) must restrain half the above i ) moment or i 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 ) g Tg = 133,119 lb. 4 Tension in the short tie down 10,335,000 in-lb = (64")(0.5432 T ) l g + (69.8")(0.6047 T ) g Tg = 134,268 lb SW Force (side-side) Overturning (side-side) due to 5W along Z axis Overturning moment (taken about the axis of rotation e Point A) = 5(53,000 lb)(39.0") = 10,335,000 in-lb w- ,pesmee m4e. en e ' mm < 4 ,e e

STD-R-02-001 Paga 96'ef 148 JAN 7W Each of two side tie downs (one long and one I short) must restrain half the above soment or l 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 Tg = 117,244 lb l -l Tension in the short tie down ) 5,167,500 in-lb c (64")(0.5825 T ) l 3 + (12.0")(0.6047 T ) S T3 = 116,029 lb 2W Force (up-down) i Lifting (up) due to 2W along Y axis Lift = 2 (53,000 lb) - 53,000 = 53,000 lb l Each of the two long and two short tie downs will carry a quarter of the load. 13,250 lb = 0.6233 Tg Tg = 21,258 lb 13,250 = 0.6047 T3 TS = 21,912 lb Total Tension Total tension with all forces acting sinul-taneously Tg = 133,119 + 117,244 + 21,258 Tg = 271,621 lb Tg = 134,268 + 116,029 + 21,912 T, = 272,209 2 j ..... _............ - - - - ~

STD-R-02-001 Paga 97 of 148 JAN 1 mg 3.1.5.3 Tie Down Luas { 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 Tenrile Ultimate = 78,000 PSI Shear d eid = (46,000)(0.577) = 26,542 PSI Shear Ultimate = (78,000)(0.577) s 45,006 PSI ~ A11ow6ble Shear Strews 1or 1.' elds r 21,000 FSI ' %cle Diameter n 2.5 in, Pin Diameter = 2.25 in Te,ar Out - i a = 272,209 lb/(3.0)(2)(1.75) 25,925 psi = (, F.S. = 26,542/25,925 = 1.02 I 145 Bearina I n - o = 272,209 lb/(3.0)(2.25) = 40,327 psi' (T. dht' y' 2.5 i /.7J F.S. = 41,400/40,327 = _1.0_2 h di Tension de r, o = 272,209/(3.0)(8.25 2.S) 8.25" { z 15,730 psi s* F.S; e 46,000/15,780 = 2,.JJ l Reinforcina Plate Weld Thickness ma'in lug. 2 inches [ Thickness reinfotcing plates 2 9 1/2 inch d ,;.72.,209 lb Total thickness 3.0 inches Load on reinforcing plate = ,2.25" m. tin 1/6 x 272,209 si 45,368 psi Length of weXd = 6 + 7 + 7 = 20" % 2.5" Dia. Role Area of shear r. 20 x 1/2 x Sin 45' M x 0.85 = 6 d3 in8 { 6,, Stress b 45,368 + 6.01 = 7,549 psi 7,, f' F.S. = 21,000 + 7,549 = 2.78 w N nu

STD-R-02-001 Pa83 98 of 148' JAN 7g s: 3.1.5.6-Tie Down Lua Welds a)- Pure.8 hear i 1272,209 lb]/12(15.5)(Sin 45*)(1)(0.85) + (6+6+2)(Sin 45')(0.75)(0.85)).

-jl 2,,

= 10,913 psi \\ q / b) . Moment forces on weld / -M = 272,209 s 2.75 m 748,575 in-lbs. q s 15.5" 6" 9 3/4" 6".G'3/4" 2" 6 3/4" In ,,,,,,,,,,,,,,,,,,,,,,,in,,, h, e ,in ninin,,sn. V g t _a "'""'""""Y"*" s. ~""""""m ". 35'" " ' ' " """r \\15.5" 9 1" i Reference 8 10.95_ y i 2 x1x 3/ 4+2x 15. 5 x8. 75+6 x0. 75 x8. 33+6 x0. 75 x10.95+16. 5 x2x3/ j 0.75x2+15.5x2 + 6 x 0.75 + 6 x 0.75 + 2 x 0.75 i 4 = '384.4TT~ = 8.94 in.:. at ', s nnnn,,,,, <, e in,n,nsu,n,,,>n s,n,ns,,, 2.0\\ tsi f r surunsun,nunua<n,uu,ur un an, e n n,,,,,,,n 7.94-0 6.56 q 1 ~l M = compressive moment + tension moment j compressive moment = tension moment M = 2(tension moment) M = 2fl(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)8 + (13,087)8 = 17,040 psi F.S. = 21.000/17,040 = 1.23 2.. :,.-. =,2 :.. . ; : =..:;. _:,,, :.:- =., ::.~.:

.,,_,__,~.,

y

STD-R-02-001 Pasa 99 of 148 Jtd 1g 3.1.5.5 Analysis of Tiedown Loads on Cask Shell (' The tiedown loads are transmitted into the cask shell as external moments. These moments.are the product of the tiedown forces and the offset distance between the line of action of the tiedown force and the attachment plate. l ML offset 4.25 in. .r F v 2 > Me 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) l { c = 892,683 in-lb g = Longitudinal moment = (173,146 lb)(4.25 in) = M 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 g = p/2 the circumferential width of the loaded 1 C late = (33*/360') (2n 40.9) 4 = 11.78 in. i 1/2 the longitudinal width of the loaded C 2 = P ate = 23.5 in/2 = 11.75 in. l B = C /r = 11.78/40.9 = 0.288 3 3 B2 = C /r = 11.75/40.9 = 0.287 2 Check that 5 1 y i 100 C a:= : u a. :. x..: :...

STD-R-02-001 Page 100 cf 148 JAN 7W The highest stress on the outer shell is 42.1 KSI. { The steel used on the outer shell will be speci-fied to have a minimus 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 damage 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 ) (78,000)- 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: l (45,006)(2)(3.0)(1.75) = 472,563 lb. F = Compared to the force required to damage the shell, the factor of safety will be: F.S. = 504,330/472,563 = 1.07 ( 94es ,g ..-.e..oe

- - - 'e**

STD-R-02-001 Table 5-C:mputstlan Shcot fsr Loc.I Str:ss:s la cylindrint Shsils Page 101 of 148 l Mg, '88 73 "8 V,Mc4 e 3. c -.i. e........ t A,,es.d L..d.- i O ..,, %.7 L/ ] (I Ps i a.ds.: i d, = e vs

s. gr T--

. s6. c. m e.e, so

-.ROUN D to.srsi jt.- :o,gg g*

0 mi. ia, eb. Leae.me ae., TTACHME.NT Tw.iaa m...ae, 4i = la. Ib. ,s, si L d. v. is. ~- ~ 1 - 1 s ' g' 3 Si e L d, VL

16.

sere.. co.. eu.es.. d se,< s (' s rT

r..

.i. ....i j l at s ,eevs1. > 6.... i.. a a,< s C,, ab ...,s..... in. NOT En Ene e,.fi f.e.. f... sa Ap..h..no neds.., .. a E in. ...ah.ip....; W ..I da.., n.a s,. Esses as..e... .i..h. .h............;,...h... and... p.e. b f.e..s v...I P,s ......d..... s. .. (,,,. ). ud e 3, or'. 4 ("4 ) eP 4 + e 1C o,r < 7 + + " *..,,. -(....)..L =. ]lRB V//A BRE HEl[ -s.s -s5 +wsess .,.6 -(..,;*a). n,. = M F/fA F/ffd Fff/4 -m +3ws eme -s77e 'a . " *., a -(.,.:.,i) L - -SM -5Bw +saw +se ]lRljgll lff/4 7/fff @?i.,..,..A a L.".*y ).'"/,, = -Sei +sas't +5se _ssu R$gyfff pfff ffj i ,. 4.',~: '?A -- it728 Go n7za c.o atn3 syss3 atns 393e3 1

f. & =-

.. (,,,). 6,., = ( "f) 5 - + +- + 2r,2 ..!.,6 -(.., y)..% = lll$lilR l$Ri lRil.56 .%s +9c43 +.m.s <a .%.,.oh a( a ).7,. - R$ll $$ll llR$ BBR -isn5 usns +. sus -> sus na ..., a

-(. L e)..% -

-trM -2m am +2 m ////f 'f//fglllQ ? 'l = FffA lglC-i::t. l,.g. a L;:y).'"/,, - -su.+sm +s75z -wsz llBB 05 % 5s7B H5% @ 78 2.s7a 6292. z'I378 6292. 'f 0"', '".".' 0' "'"""" I + +- t-t. + r V +- 2C'"*.t*. ,a. < r.+ = r,%- ME F//[Jg ' 072';" r.a = 4,,. +509'l +5094 -5t94 son

7C't

T** =& fff)l V)fffll Y{jf)jf gYffff -42Io -'I210 +42io +4Do} "'2*""".'.5." ~~'" 5094 5014 505y 5094 4ZIO 4210 4Lto 42tb (,b fi ek 8 4 I ~ /

1) When T pf 0, S = laraest absolute maanitude of either x

2}or/(ex - 04)2 + 47 ) 2 I S = 1/2 [0x+0g I /(e ~U) + 47 g Ya Yb

2) When T = 0, S = largest absolute magnitude of either j

l 5 = 0, 04 or (0, - 0 ) x 4 "--. '; :.n : ::: ' ::

= =: :,.

STD-R-02-001 Pag 2 102 of 143 JAN 7g 3.2 Normal Conditions of Transport (Appendix A-10 CFR 71) t 3.2.1 Heat Since the package is constructed of steel and lead, ten-peratures of 130*F will have no effect on the package. '3.2.2 Cold Since the package is constructed of steel and leaJ, tes-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-nel 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 p-(. will produce no detrimental effects. 3.2.4 Vibration All components are designed for a transportation environ-ment. No loss of integrity will be experienced. 3.2.5 Water Spray Not applicable. 3.2.6 Free Drop Since the package weighs in excess of 30,000 lbs., it must be able to withstand a one foot drop on any surface, without loss of contents. ] 3.2.6.1 One Foot Drop on Botton Corner Energy to be absorbed = 53,000 lb x 12 in. i 5 in-lb ) Maximum energy = 6.36 x 10 2 8 Volume of steel = 6.36 x 10s/38,000 = 16.74 in .:.".72.* *l.C.' ". ' *. -

. ~.7.1.. :.. *........ _. '..'l

l i STD-R-02-001 Paga 103 cf 148 JAk 7y { Energy will be absorbed by crushing of corner. %g\\ l %s t. P. ,- e., g Nl*.',/,f,k',.,.,[, ~ N %s% f.. '., '. d :',f ../ / f a*'r / f N ~~7 BOTTOM N ,4 CORMER s\\s/ / Y+ i i " The volume of the crushed ungula, assuming the i ( worst case of a 45' impact angle is calculated by the following equation: 1 Sin 4 II" V =R8 - $Cos$ 3 8 $ = 16.4* when V, = 16.74 in I t R l J l s N . k-- C n} .=.-==_, ..-.=... _ =..

STD-R-02-001 Pogo 104 of 148 MAY27IN7 The maximum amount of steel crushed will be: ( b = R(1 - Cos $) = 1,66 inches The effect on the cask body due to the corner impact event is shown below. Even though the weld will be crushed loc 11y, there will be no loss of the cask's integrity. i f l g ,,;'t, ?*. i ~ 1 19" 4.,, 1 6'" The decerlation force exerted on the cask is calculated as the product of contact surface area and the yield strength of the steel (36,000 psi) Au = frab (xy + ab sin ~l 1) where for 45* angle = 0: R = 40.5 in a = R/cos 45' = 57.275 in b = R = 40.5 in h = 1,66 in C = R-h = 40.5-1.66 = 38.84 in y = /R -C2, /40.5 2 2 38.842 11.47 in = x = C/cos 45' = 54.928 in 1(5 8 Au = Sr(57.275 (40.5) [ (54. 928) ( 11. 47) +(57. 275 ) (40. 5) sin Au = 36.32 in 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 Peg 2 105 ef 148 78 3.2.6.2 Effects of Botton 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 aris11y with. respect to the plate. Cask Cover Inner Container Deceleration Forces p. k, Contents l>p*.

e,

+ + Upper go ^e, Inner Shell Bottom Plate Lead Shield Iower Bottom y [ Outer Shell Flate / I Corner Impact MD)I kh\\D Reaction Force Bottom Corner Drop

..i..-a.~~-- STD-R-02-001 Pass:106 of 143 JMt3-h Summary of cask component weights as used in the following drop analyses Primary Ltd 6,995 lb Shield Flug 425 lb' i Outer Body Shell* 7,224 lb I Inner Body Shell 1,900 lb Upper Botton Plate 2,645 lb Lower Botton Plate 2,364 lb Lead Shield 14,397 lb i 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.

1 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 plus will be distributed.to the inner and outer shells in accordance with the she11' cross sectional areas. 2-The inner shell will receive loadings at its connection to the upper bottom i 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 f= A 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 6 elf 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 weldsent to the lower bottom plate. 5-Weight of primary lid insert plate and secondary lid insert plate are included in weights of their respective lids. Therefore, the attachment 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 weste. These assumptions are conservative for this analysis. l

STD-R-02-001 Paga 107 ef 148 JAN 7g Cask Analysis 1 - Load from Primary Lid and Shield Plus or ~. Decelerati Forces Shear Area of Weld i N 'O e 4,

- v%

%,,.S z Deceleration Forces l Detail "A" 1/2" / b 3/4" fitcart t Forcea 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. 8 8 i 8 - 75.5 ) = 89.388 in Inner shell area = (n/4)(76.25 8 2. go ),222.317 in s Outer shell area (n/4)(81.75 8 Total area = 311.705 in Inner area = 89.388/311.705 = 28% h outer area 222.317/311.705 = 72% i -v - ...oma e, e e --*me=* mo e e e -e a eagn. = m e e o

STD-R-02-001 g2 Igof148 Force on inner shell s (136,415)(0.28) = 38,196 lb Interal and axial j ( Force on outer shell = (136,415)(0.72) = 98,219 lb interal and axial 2-Stresses Developed in Inner Shell and Attachment Welds l Deceleration Forces Inner Container Upper & Contents Bottoin PlaW I 3/8" wel N Detail "B" Reaction on Inner Shell 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 = [2 (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)(si,n 45')(0.85) = 6,375 psi (lateral) C 1 c

STD-R-02-001 P:82 109 et 148 JAN 7g l Axis 1 weld stress is caused only by lid load and shell self wei8ht. 38,196 + 34,931 = 73,127 i Axial weld stress = 73,127 lb/n(75.5)(3/8)(sin 45')(0.85)(2) = 684 psi Total Stress = 46,375z + 6842 = 6,412 psi 2 75.5 )(n/4) = 818 psi 2 Axial shell stress = 73,127/(76.25 Shear shell stress = lateral force / area [(1900)(26.0)(sin 45')+(38,196)+(16.550)(26.0)(sin 45*)] _ 8,444 psi (76,253 - 75.53)(n/4)(1/2) F.S. = 21,926/8,444 = 2.6 (- 3-Stresses Developed in outer Shell & Attachment Welds \\- Deceleration Force f [ Inner Shell Upper Bottom Plat" Outer Shell Lower Bottom Plate / / Crush Depth Weld Area in Shear i Detail "C" Shear of Outer Shell Weld l Stress in weld around perimeter of outer shell at cask lid 4 (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

STD-R-02-001 Pant 110 of 148 JKk 7 g ( Stress in weld connectin8 outer shell to lower bottom plate Lateral force = 1/2 load of outer shell + 1/2 lead shield j + 1/2 lid and shield plug (the 1/2 supported by 1/2 { outer shell) j (7,224)(26.0)(sin 45')+(98,219/2)+(14,397/2)(26.0)(sin 45') = f 314,264 lb. = ( Lateral stress = (314,264)/n(80/2)(0.5)(' sin 45')(0.85) = 8,322 psi i l 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 Total stress in weld = 4(8,322)2 + (3,o39): = 8,866 psi F.S. = 21,000/8,866 = 2.4 \\ s 2. go )(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 = J l l 1 e C N h

STD-R-02-001 f 148 Pag 1}1 4-Stress in Weld Joinina Upper to Lower Botton Plates Deceleration Forces Inner Container Upper A Contenta Bottom Flat. III" Detail "D"

d Reaction on Inner Shell Load on weld = Upper bottom plate

+ Inner shell (- + Shear from lid on inner shell + Waste + 1/2 lead 1 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 i 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 1 i 1 C

.,.. = =

STD-R-02-001 Pag 2 112 of 148 EV14 ES 5-Bolts Securina 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 (nex) force due to torque = T/kd = (100)(12)/(0.15)(0.5364) = 14,914 lb. force due to deceleration = (1,200)(26.2)(Sin 45') = 22,231 lb. force on each bolt = (22,231/8) + 14,914 = 17,693 lb. stress = 17,693 lb/0.226 inz = 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 in8 = 67,221 psi i F.S. = 105,000/67,221 = 1.56 i ( ) l a. ....-... =. .c . _.::.g 2-

STD-R-02-001 Page 113 af 148 3.2.6.3-One Foot Drop on Top Corners .C l A drop on the upper corners of the cask would j decelerate the cask and would result in axial and j transverse deceleration forces between the cover i and the balance of the cask. I i 1 l Contents Deceleration Cask be$e$ft!ation' Inner Shell Outer Shell Cank Axini Deceleration Primary Cover 3/8" % 1/2" J L Impact Point L l................-.-~...---.-..

STD-R-02-001 Peg 2 114 of 148 MAY 2 7 M f " { The top cover-is stepped and the inner plate has a-l nominal clearance of one-eighth. inch. Upon impact, 4 jl this plate would immediately contract the inner shell. The'transvev;se decerleration 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 q cover plates. The magnitude of the transverse j decerleration force will depend upon the orientation of 1 the' cask and the corresponding decerleration forces.- 3 l. As shown:later, the maximum deceleration force will L occur when the cask is dropped on a long flat edge of- ) h the primary cover. The maximum decerleration for this .l case is 17.88 g's, j J 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. The. weld is a 1/2 inch. weld, 75-1/4 inches in diameter. The stress in the weld: 628.866 " 75.25(4r)(0.5)(0.85)(0.707) = 8,853 psi-F.S. = 8 = 2.37 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)+/2 = 576,271 lbs. A 40* arc 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. l Area = D ( ) t = 75.25 ( j'(2)=52.5in8 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 l

r.S. = 9(- ) = 4.02 10,977 i j l

7 'STD-R-02-001 Pass 115 cf 148 3.2.6.4 One Foot Drop on Top Corner of the I,ona Flat Edne 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 s u h impact An impact in this orientation will cause minimum bending of the cover and will result in high in-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. Crushina 7 F Following impact the edge of a the corner will begin crushing 4 3 7A until inelastic rotation around ~ gn the bend point occurs. 1 Nctit rn) The point at which this will , ) Axis occur is calculated as follows: Width at bend = 37.4 in D* Thickness = 2 in M = (38,000 psi)(1 in)(37.4 in)(1 in) = 1,421,200 in-lb. 0 38, O ps1 - ~~~"""5 1" 38.000 psi, e I. ( ,~ ..- = :.: ; -...:....::.:.:.... ~ :.... m,

STD-R-02-001 Pag 2 116 of.148 MAY Z 7 INT F,= Force required to_ initiate bending F, = M/X = = 947,467 lb 947,567/53,000 - 17.88 g's (axial and lateral) F = F, (/2) = 1,339,920 lb 1,339,920/53,000 = 25.28' g's (total) 8 ' Area crushed steel F + 38,000 = 35.26 in Width of crushed steel 35.26

34.4 = 1.025 in i

Depth of crushed steel 1.025/2 = 0.512 in Volume crushed steel - (0.512)(1.025)(34.4) 2

9.036 in8 Energy absorbed in crushing 9.036(38,000)

'343,364 in-lb Bending 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 y, p 4 I 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 i 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-lb/947,467 = 0.309 in.

STD-R-02-001 Page 117 of 148 JM 7 5 The 3 force developed during the bending process l' .( is calculated using a kineastic approach. Velocity at start of bending is [2)(292,636)(386.4) , /2 KE a,V VW (53,000) s.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 5 AX/V8 = (0.309 in)/[(65.3)(0.5)in/sec) =- 0.00946 sec g force = (AV/At)/386.4 in/sec) = (65.3/0.00946)/(386.4) = 17.86-g's (both axial and lateral i direction) The above shows the maximum s' force is 17.88 s's in crushing and bending in both axial and lateral. (. directions. l The force of. impact on the corner is 947,467 lb. (axial component)- . Contents 40.2" 947,467 lbs = q A U p U2R y2R 1.5"j 16.550 X 17.88 lbs =-- 2 3 " 34.4" 21' 6 l The loads on the ratchet binders'will be propor-tional to their distance from the pivot point of the cover on the cask. l...'_.. ......w.,a.. .......+...............a._y.-............

i STD-R-02-001 Pass 118 of 148 JAN 7 gg Forces tending to open the lid consist of weights (. ' from waste, lid, and shield plu8 (16,553 +'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) ^ i R = 72,875 (23/80.4) = 20,847 lb (in binder closest to impact) ( l i m, e.c o- .ww., .......a

Os@-R-@8-@@l Page 119 of 148- [, The'1/4. inch thick' seal ring, made of AISI 1008 steels, located'on the outer periphery of the top of the cask well will esperience 'some j pressure resulting from a top corner drop. This worst case. appears in L l a top corner drop on a large flat. The force exerted is equal to that l l~ which is required to bend the lid, or 947,467 lb. l-l 4 V I ze I ~ The yielding surface area reacting against this force is proportional l to the angle 8 and the radius. (39.62 0 / ) (35,000 psi)1 = pressure (psi) 5 (Pressure) n(39.8 W ) in / degree = force / degree I 360 As seen from Table 3-3, the entire force'is distributed ove'r a 88* cd are of the ring. This is less:than the angle.between three of the large flats. s 'By dividing the incremental pressure by Young's Modulus (E=30 x lo ), 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: 8 Area = n[(79.5)(0.5)+(76.25)(0.375)]2 = 105 in 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. Report of the Iron & Steel Technical Cosunittee " Estimated Properties and 1 Machinability of Hot Rolled and Cold Drawn Carbon Steel Bars", SAE J414, Feb. 1968. _j \\.........

1 STD-R-02-001 Pass.120 of 148. Table 3-3 Pressure Ezerted on 1/4" 8eal Rina Due to Top' Drop on Lerne Flat A dn. . Pressure Force /Dearee. E Press /E Strain ~* 1 34,780 12,026 12,026 0.001160 0.000290-l '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 'O.000288 6 34,620 11,971 72,028 0.001154 'O.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-Il 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 33,20 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 j 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' O.000273 l 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 'O.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 A of ~ 2(44) = 88* I - -. ~. ~....

s. :.

, w _.. m.

STD-R-02-001' Pass 121 of 148 JAX Y m. Lid 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 8 x n + 4 = 150,207 lbs Ultimate Strength = 85,000 x 1.5 Based on yield strength the factor of safety will be: 123,700 + 72,875 = 1,.7,0 t {.. j s---.:..,..._.a.,

STD-R-02-001 Pag 2 122 of 148 MAY 2 7 IN7 Ratchet Binder Pin Pin'is 1-1/8 inch diameter bolt made of SAE Grade 5 or equivalent having a j yield strength of 74,000 pai. Based on double shearing of the bolt during . loading, 72,875 72,875 lb lb 1-1/8" Resultant Force = 72,875 lb } s i s s s s } a, = (72,875/2)/(1.125)2(41'/4) = 36.657 psi F.S. = (74,000)(0.577)/(36,657) = 1.16 l -Lid Ratchet Binder - Upper Lug ---9 Y Y' I 3/4" ! \\ A g 90' 24" 1 ~~~~~ D (gi,.. .s.. nu 9 / v -H 3s us", i.. D-3/d 3-3 /4"

STD-R-02-001 Pass 123 cf 148 JAN 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 I 73 = Le/d,2/1.12543,185/70,000 = 2.88 l gj, r Tension - f 1/16" ' l~ cT = 72,875/(1.9 + 1.375 - 1.25)(1.5) 2"i L,5/8" h cT = 23,992 psi 4 FS = 38,000/23,992 = 1,58 7j fj Weld ( 1/2" double groove weld, complete joint penetration with tension normal to effective area. Allowable stresses same ,j as base metal. l 1/2" full groove and j 1/2" fillet (both sides) H--- 3 9 %" full groove f h" full groove'"[ I \\ 7, 1-15/16", n W ,p 9/16" 4 va" ; 1 d 72,875 lb !!eutra 1 Axis C

STD-R-02-001-Pcge 124 of 148 MAY27Igri ' Tension o = 72,875/[(2)(1/2)(1-1/2) + (2)(1/2)([i)(2-3/4)] (0.85) g o = 15,909 psi g Moment = (72,875)(0.5625) = 2am (f)1/2)(1.5)(1.9375) + (2)(1/2)(1.4375): a" (2/3)((2 (1/2))(0.85) e,=_9,934 psi Total Tot " "t * #m = 15,909 + 9,934 = 25,843 psi l a F.S. = 38,000/25,843 - 1.47 -l l 1 ) )

l STD-R-02-001 i Pag 2 125 cf 148 ) JAN 7m i Ltd Ratchet Binder - Lower Lug ( ] Tear Out h" 1-3/8" Shear - gi o* = 72,875/(1.5)(1.75-0.0635-0.27)(2) 4" [ I n o, = 17,137 psi ] FS = 21,926/17,137 = 1.28 Bearina - g,, B = 72,875/(1.125)(1.5) n .DIA O = 43,185 psi le/d, (1.75)/(1.125) i 73, O /fu (43,185)/(70,000) B 1\\" Thk = 2.52 9,, Tension 6h" 72,875/(3.5-1.25)(1.5) = 21,593 psi ( FS = 38,000/21,593 = 1.76 Weld Shear - _ /l o,=72,875lb/(9+1.5)(1/2)(d)(.85)(2) o o II" o, = 5,575 psi Homent - (72,875)(2.625) = l 72,875 li 20,[(1/2)(1.5)(4.5)+(2)(1/2)(4.5)2 y (2/3)(1/2)](0.85)([2') l o = 7,859 psi m Neutral, Axis e a goa+oz = 9,753 ysi 2-1 ~ FS = 21,000/9,753 = 2.15

STD-R-02-001 Paga 126 of 148 JAN gg Lid Ratchet Binder - Lower Lug (Optional Design) l I 3/16"g, s 1" -4 aC ~ 3/16'I n , '5- \\p 1-11/1f I'8 ""' \\ l DIA. \\ [ h 1-11/16" l _Jr [ %N 1/8'k [ " 9 P tica T-3/Y I/8 DIA 14,, C gg, y f [ i - s ir R/ k 9" k I N \\ k \\N \\ \\ \\ ( N h %/ 4" Section C-C Weld holding 1/4" thick pistes to lug Assume each circumferential weld must vupport 1/2 the load - o = (72,375/2)/(8.5+2.5+0.5+2.25+6.75+0.25)(3/16)(sin 45')(0.85) l i l, 0 = 15,582 psi q l l 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 [ FS = 21,926/17,274 = 1.27 j I ~ a < .e s.

STD-R-02-001 gs127cf148 Tm Shear of the pin - a = 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) = 51,822 psi 1.6875/1.125 FS =.Le/d,51,822/70,000 = 2.03 o/fu Tension ~- (Optional Design) o = 72,875/(3.5 - 1.5)(1.25) = 28,150 psi 1" Tlik F3 = 35,000/28,150 = 1.30 Weld (Optional Design) 72,875 lb Shear n o,=72,875/(9+1)(2)(1/2)(8)(0.85) e, = 6,062 psi 2 '>/ A Homent (72,875)(2.625) = "*[j ', ' 20,((1)(8)(1/2)(4.5)+(2)(4.5)2(2/3) (1/2)(d)(1/2)](0.85) I l o,_a B,841 psi of* #s = 10,719 psi q.. m j FS = 21,000/10,719 = 1.96 I C i .s...--

- ---- * --. ~... c,: : 21,

~ ~.' . *... =..:. l * ~ ~.. -..... -. -. ~. -.'... :. '... ~ 2- - -

STD-R-02-001 1 iPage 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 i on one of the one-inch cornets of the cover over -one of the tie-down lugs. 25.14" 3. s Impact J The energy absorption sequence will be the same as that previously shown for the drop on the long j flat edge and will consist of the intial crushing, i bending, and crushing. Because the cover over-hangs the cask to a greater extent, the cover will act more as an energy absorber. j Crushing Width at bend = 25.14 in. (Depth of 5.0 inches) Fa F Thickness = 2 in. o Fb' 1 ~ l 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([2)=270,205lb Area of crushed steel = F/38,000 psi = 270,205 lb/38,000 psi = 7.11 in8 C.e w 9 "*M-4.M***a*. at er pseuw +-me m.4w&,,, 9,'y?g gg 4,

'STD-R-02-001 Pgn gof 148 l 8 The area of the trapezoid is described in (1.4d + 3.44d )-- r where d/]T = depth of crush. (L.. l 8 or d = 1.25 in. 7.11 = 1.4d + 3.44d therefore,(d//l)=0.882in. Width of crush = 1 + 4.87d = 7.08 in. 8 s Volume of crushed steel = d /2 + 1.925d + 1 = 4.18 in Energy ab' sorbed in crushing = (4.18 in )(3g,oon p,g) s = 158,840 in-1b Deceleration Force Durina Initial Crushina k' ' Energy absorbed 158,840 in-lb I d3

Energy remaining = 477,160 l

5" 8 Force = (7.11 in )(38,000 psi) b, = 270,'250/53,000 = 5.1 s's l / 3 3/4" 3/8".) 3's" ( 120 o 3" ' I~ Bendian 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: I Bending of lid (15' until lug hits outer shell of cask) f 1.3 " Energy remaining = 477,160 in-lb l) o 4 p, h: Energy absorbed in bending a y (1.3" travel) E = (Fa)(d) = (191,064 lb) (1.3")

248,383 in-lb Energy remaining = 228,777 in-lb.

j ~. _.. _.

, -. ~ -.~_--n.~.~.- STD-R-02-001 fjgg g of 148 Failure of the Luz {- 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" al M - centroid I tension y U I' compresafon S 4'.5" [

Ja w,3,,,,,,,,

O J / / c M=20,[(1.5)(1/2)(1.875)+(1/2)([2')(1.875)2(2/3)(1/2)(2)]0.85 l 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 soment that will shear the weld is attained.

STD-R-02-001 P:ss 131 af 148 - 0.375" ~ d l l ya j ~~~I I F -- P ys L e .d F=49,000(1.5)(ys)f=36,750.y lb l ya = 4.5 - % in. M = (F)(ys) =(F)(4.5-y)in-lbs =165,375ys-12,250y{=109,368in-lbs -) 12,250y{ - 165,375ys + 109,368 = 0 165,375 - J165,3758 - 4(12,250)(109,368) I 71

2 (12,250) l l

l y2 = 0.70 inches The depth of shell deformation or, d, will be as follows: d 0.70 0.375,4.5-ygvi, 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. C _~~.. v=.: = :.. =.= -~;.==.n._ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

STD-R-02-001 Pasa 132 of 148 JAN 7W Secondary Bending (. During the following 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 ic-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 are for the axial force and the force required to cause bending will increase. C 3.. h L a d J l M, = F,1 = 955,320 in-lbs 1 = J25 - d2 l ~ F,=955,320/[25-d 8 I .w.1 _.

STD-R-02-001 1 Page 133 ef 148 I JAN 'i m The energy absorbed is computed as follows: r displacement i Fa Fa(ava) d E-E { 1 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 1 l The secondary bending is capable of absorbing the remaining energy. The additional displacement of the lid during secondary bending is j i 2.18 - 1.3 = 0.88 in Deceleration Forces - It was calculated that the initial crushing caused a deceleration force of f 5.1 s's. Calculate the deceleration forces of secondary bending since this is the shortest distance travelled in any of the phases discussed. ,,[2(KE)(g),[M223,777)(386.4)=57.76in/see W 53,000 s f v,y = 28.88 in/sec at = AX/v,yg = 0.88/28.88 = 0.0305 see a = AV/At = 57.76/0.0305 = (1,893.8 in/sec ) j s t 8 a = (1893.8 in/sec )/(386.4 in/sec ) = 4.9 s's This indicates that the maximum deceleration force during a 12 inch drop on a short flat corner on the lid ie 5.1 s's. l l This does not exceed the g forces calculated in the drop on a long flat. ( (

STD-R-02-001 Pasa 134 of 148'. A 7 INS 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. l Total energy = 636,000 in-lb j 636,000 in-lb = 1/2 av"' Initial Velocity, y = 96.3 in/sec Let d = depth of crush Volume of steel required to absorb energy I 636,000 in-lb/38,000 psi = 17.7 in l W = d Tan 67.5' 67 %.0 2" thick 1" I I Volume of steel described by j [2(1/2'dw)(2 in)] + [(1 in)(d)(2 in)] J = 4.82 d" + 2d Final depth = 4.82 d' + 2d = 16.7 inI j j d = 1.67 in As shown on Table 3-2, the highest 3 force exerted on j the lid is 12.0 s's. I t l l l N ~~ . :......::.... - = =.: :.. =-..=.:= [ - -......

STD-R-02-001 Pa8a 135 of 148 JM 7E -{ TABLE 3-2 Energy -Ener8y Incremental d .Vol. Absorbed Ressining Velocity Time Deceleration (in) (M (in-lbs) (in-lb) (in/sec) (sec) (in/sec ) (g's) 3 (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.S 2.21' 83,790 552,210 89.7 0.00137 1,752 4.5 l 4'5,973 83.3 0.00290 2,207 5.7 7 0.75 4.21 160,027 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 4,705 12.0 i j l i l 4 C .,.=:.-=.--.=:c..= .=.-. :. e.. + -.. -

i STD-R-02-001 Pagt 136 of 148 JAN 719h .h - 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 s's) = 596,880 lbs. i The weight of the cask less the cover and shield plus is 53,000 - 6995 - 425 = 45,580 lbs. The i deceleration force between the cover plate and the inner shell of the cask will be: F = 45,580 (12.0 3'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 fros' a 13 pound rod'will have no effect on the pack-age. (. 3.2.8 Compression This requirement is not applicable since the package exceeds 10,000 pounds. CONCLUSION From the analysis, it can be concluded that the HN-100 Series 3 Cask l is in full compliance with the requirements set forth in 10 CFR 71 for Type "A" Packaging. Y

STD-R-02-001 P:ga 137 of 148 JAN 73 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. II) Total (2) Gamma Specific Isotope Energy Activity Activity Mev pCi/ml C1 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. l With the maximum amount of these materials that can be transported in the l HN-100 Series 3 cask, the heat generated by the waste will be as follows: Heat Total f-l ( Generation Activity Total Heat (Watts / Curie) (Curies) (Watts) (BTU /HR) Cobalt 0.0154 96.6 1.48 5.0 1 I Cesium 0.0048 2,708 13.0 44.2 l f i 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. 1 f l -

m m :.. 3.

..g;..

g. p

STD-R-02-001 Pag 2 138 of 148 MAY 2 7 IN7 4 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. i 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. j The shield plug is located in the center of the cask lid, has a Neoprene l 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. = 77.75 inches I.D. = 76.25 inches 2 - R1 ) = W(38.8758-2 2 a Area = TT(Ro 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 j 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 = 35,340 { l Equivalent ressure of gasket = 1 = 35,420 lb/181.43 ins A = 195 lb/in i 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 Plug 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.

n 3

STD-R-02-001:

Paga'139 of 148 "~ .MAY 2 7 liff The~ gasket dimensions are 22.75 in. 0D, 2025~1n. ID, and 3/8.in. ' thick. :The gasket is Durometer 50. l s - R1 )14Y (11,3752 _.10.125 ) = 84.43 in8 8 8 1 Area = 1Y (Ro Forcedownwardonlidisthesumof.the' weight.ofthelid'hlus. .the force of.the studs'(P). ~P= = (120' f t-lb)(12 in/f t)/(0.15)(0.75 in) P.= (12,800.~1b/ stud)(8 studs) = 102,400 lb-q l W = 425 lb J Total force = 102,400 +L425 ='102',825 lb Pressure on gasket.=---= 102,825 lb/84.43 ina .1217.9. psi-As shownLon Appendix D-2,.the compression of the. shield. plug' gasket.is 33% of the initial thickness or 1/8, inch. 5.3 Seal with Internal Pressurization. f 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. l required by-10.CFR 71. Appendix'A.- As described in Section l'.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 j been: designed to minimize effects of columnated radiation streaming and problems ~ associate 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: 6 P/AE.= (7,613)/(1.76)(30 x 10 ) = 0.000144 in./in. P = 4,113 + 3,500 = 7,613 lb 8 A = Area of 1-1/2" minor diameter = 1.76 in E = Youngs Modulus = 30 x 10 p.i l 6 and for a 24 inch long binder, total strain is: 1 (24 in.)(0.000144 in./in.) = 0.0035 in This is less than 4% of the initial compression of the gasket. 1

STD-R-02-001 .{ Pass 140 of 148 l ) U114 C5 5.4 Gasket compression Test-A compression test to check resiliency was done on Items 4 and 5 on Drawing STD-02-020,.the primary. lid and secondary shield plug asskets. The samples were each 1 in8 by 3/8 inch thick made of Durometer 40 Each sample was put neoprene and Durometer 50 neoprene respectively.The final results-indicated-3 in a compression device and compressed. 8 ) that it required about 4,500 lb to compress the 1 in.Durometer 40 sample to a thickness of 1/8 inch. After removing the. sample from the l f 3/8 inch. Siellarly, the 1 g le returned to its original thickness oDuromet test stand, the s ) in It also returned to its original thick, and it required 10,000 lb. thickness when the load was removed..The test compressed each gasket esterial 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 J reducing possible damage to the gasket material. 5.5 Warpina of covers l .The possible distortion of the cover and possible leakage due to dis-j A tortion has been sddressed on a cask of nearly identical design. cask having an octagonal cover secured by ratchet binders was dropped The identification on the extended corner by Nuclear Packaging, Inc. 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 Li 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: i n.. 1" p" 3" 4" 5" 4" 6.75" - l g 9 o Corner j l Deformation f j j 3 / e l / / 9_. $. 9 9 3! 8 p 1

-STD-R-02-001 Pago 141 of 148

48M 2 7 Ilgf -

The energy absorbed in' dropping a 23,360 lb package from 46' inches is 1.07 x 106 in-lbs. The energy to be absorbed in a one foot 5 in-lbs l drop of an HN-100 Series 3 cask is 12 x 53,000 = 6.36 x 10 or less than 60 percent of the unit tested. The covers are the same thickness and the overhang of the corners are approximately the same. Accordingly, the HN-100 Series 3 l should experience less deformation. All of the deformation occurred outside of the impact limiter and no deformation of the cover was found in areas which could affect the seal. l

STD-R-02-001

810}Y4N'8 6.0 OPERATING PROCEDURES (Removed for Rev. 7)

..(' 1 i I l 2 (. 1 7.0 ACCEPTANCE TESTS & MAINTENANCE PROGRAM (Removed for Rev. 7) 1 ..:....+.. .......a...-,,......,: - <. - :... r a

STD-R-02-001 Pasa 143 cf 148 C APPENDIX B Appendix Removed for Rev. 6 e a j i I ,...x..... .u .-:..a . :r.

STD-R-02-001 Paga 144 of 148 APPENDIX C Appendix Removed for Rev. 6 l r 4 i-l l ) I t 9 ) 1

STD-R-02-001 Pasa 145 ef 148 oifu me APPENDIX D Gasket Compression Nomograph I i I } l 4 i ,r. lt i 1 I i j 1 J j i l I I 1 OOP@ Oe OO @^ M 4 gw ee

W WW O W98 M apg W p g4,ggp g,

STD-R-02-001 Paan 146 of.148 MAY 2 7137 - t ll 1f"Il!lt!'- h~'11]!!!ld(!b I w!lljjjl h8 o il r . r.! t L,,L,,L M J,. Y qx ,4 ,,,p; rr \\, /.. m* I i., 1 E I ii? I iI i .,an Sh / 8 ,l,,83 58??848 8.' 8 I em L "4 mm' s-c, x .t j f*Iey \\, g39 = \\,+ .\\ l k-$:.i.... s .x sit 7 4n 88 $1 E., p SE aI g i: ea 1 \\s I y g .54 ll 1

i. E; le 333

E81 l

0 y'. .h$gf8 b l ho alas.ils i l L; I i 11 APPENDl1 D-l- ..,.,....,s. mm -w - s e--- - m = n-

> ~ ~ * ~ - ' ~ ~ ~ ~ ~ ' ' ' ' ~ ^ * -

s w-a-ve uva q Page 147 of 148 fl0V 14 Bt5 i,- (5-iiii 111111 t # l i \\. il.l. il.1 si gal11 i. M- ,11.'iIlllj;01 p % ;_% t 1,- s ll!! I> " il,ii i 1 ,s = iI Ig g 3 e gg,i i a og lg _, g i i i 7 i 7 7 7..py; g ; gg ;_.g.. p- \\ i l $>? ? .v 3 ~ v v,w e v. s = \\ (. .);. \\ j ,3,.

A m n = =. p y m

+ , vi,. + + s7,, j \\ 8;e 3- { \\ s... o lIF 5.i ,be \\, 3/ =,' \\ t '\\, 2, x 'lil l k lIF 3 Il l \\. 3 g.- I; i. + y :: J a.s.i, 5 $a'm. I, I >( t \\ 2 3 ~ r-mpc,io l' i g 11g d &n 3 g I n1 1 1.+ 4 .1 1 f ArrEAD11,0 2 il [-

,.-.=.,,:=---=~-=.=--.------:-:----~~-----

~

STD-R-02-001 Paga 148 ef 148 M/14125 APPENDIX E Appendix Removed for Rev. 7 1 I i ( i i!i I l t } O ( C ,.-..........:.-s.. ,,e

o. c. LOVERSIZE DOCUMENT PAGE PULLED g: SEE APERTURE CARDS n.

NUMBER OF OVERSIZE PAGESiFILMED ON APERTURE CARDS-I APERTURE CARD /HARD COPY AVAILABLE FROM RECORD SERVICES: BRANCH,TIDC' FTS'492 -8989 i}}

ML20215A682

Text

2.0 DESCRIPTION

.

2.0 DESCRIPTION

.,.. = =

9.036 in8 Energy absorbed in crushing 9.036(38,000)

Navigation menu

Search