ML20148M588
| ML20148M588 | |
| Person / Time | |
|---|---|
| Site: | 07109080 |
| Issue date: | 11/08/1978 |
| From: | Jennings J, Trigilio G HITTMAN NUCLEAR & DEVELOPMENT CORP. (SUBS. OF HITTMAN |
| To: | |
| Shared Package | |
| ML20148M574 | List: |
| References | |
| NUDOCS 7811220073 | |
| Download: ML20148M588 (22) | |
Text
{{#Wiki_filter:_ F Q 1l % E e RECEWED PROJECT COVER StIEET k n i,978
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.y.n Document Title \\ u s *m -.~ 'Ng. e # w e.*** N t s Y " ANALYSIS OF ONE FOOT DROP ACCIDENT Project Document Number ADDENDUM NO. 1 Rev. 1 for i SAFETY ANALYSIS REPORT For The HN-400 SERIES RADWASTE SIIIPPING CASK IIittman. Nuclear & Development Corporation 0100 Red Branch Road Columbia, Maryland 21045 k ' 's. i s k j m. .e d R e f: Std. Doc. N/A Rev. __ l 11202 l 13 i 78112200 $
Rl: VISION LOG R EV. DATE ENGINEERING Q. A. PROJ. MGR. ECN # 0 5/23/78 lj u Q f f bhf f.,, (j), g ,./,[ (dh ,c, b, d, 1 11/8/78 z sg r.' 7s_,go Calculations prepared by John Jennings Checked Bf George Triailio 1 i
TABLE OF CONTENTS Page I. STRUCTUAL ANALYSIS 1 A. One Foot Drop Onto Side of Cask 1 J l B. Ono Foot Drop Onto Closed End 5 l l l C. One Foot Drop Onto Long End 8 i l D. Side Drop, Drop Reaction 9 s l E. One Foot Drop Onto Bottom Edge of Door 11 l(1) l F. One Foot Drop Onto Short Edge of Door 14 ) l l G. One Foot Drop Onto Door Hinge. 16 l l II. CONTAINMENT ANALYSIS 18 l 18 III. SHIELDING ANALYSIS l i 19 l REFERENCES I ii w-
I. STRUCTURAL ANALYSIS A. One Foot Drop Onto Side of Cask 1. Imoact enercy of loaded cask E = Wh W = 43,000 lbs. = 43,000 x 12 h = 12 in. = 516,000 in -lb 2. Steel Deformation reauired to absorb imnact. V 6cr = E 2 V = 516,000 6cr = 50,000 lb/in 50,000 (ref. 1) 10.32 in3 = 3. Stooninc Distance The kinetic energy is absorbed by the deformation , of 48 each steel studs, embedded in the side walls of the cask as shown below: l4g ~ n nnnn nnn n n n J T i 00 / u u u u uu u u u u u l l(1) - 18@ 7 0.C. - = l Total number of studs, each side = 38 s 1 of 19 ~
// _x g N N ( a [ TYPICAL r' i / /, // STUD u INSTALLATION a 1 l s Vk 3 e L ~ --- l 't-Y ^ a 10.32 38 x 3.1416 x ( 0. 5 ) 2 l = 0.346 in g y a 4. Applied g load 2h 8,' t (ref. 1) ,2 x 12 l 0.346 (I) _=-69.4 i l 2 of 19 - I w w e-
.j 5. Force applied to side plate on the impacted side Weight of side plate 4,116 lbs Weight of cask contents 12,000 lbs = Total 16,116 lbs l l l F=Wg l(1) 16,116 x 69.4 l = i l l 1.118.450 lbs_ = 6. Strength of side plate welds al = 0.125 in P. = length of weld bead = 2(149+39) in a1 = 0.500 in e = joint efficiency = 0.85 Sv = 0.6 Sy = 21,600 Fy= e(U /2)(al + a2) SV = 0.85 x 0.707 x (0.125 + 0.5) x 376 x 21,600 = 3,050,422 lbs l Safety Margin. 3,050,422, 3 X1) 1,118,450 l
- 1_71 7.
Maximum stress in side plate All edges fixed, uniformly distributed load, / Reference 3 - Table X, Case 41 / a = 149 in t = 2.5 in b = 39 in (y j b/a = = 0.262 118xj =192 5 lbs/in* w= Max S = Sb = 0; 1 623' T l = 0.5 x 192.5 x (39)2 {(3) (2.5) G 1 + 0.623(0.262)Cf t = 23,423 lbs/in l Safety Margin = 36,000 ,) l 23,423 l = 0. 537 3 nf M l
1 8. Maximum Shear Theory of Failure l 5 = 36,000 >(S2 + 4S2)12 t s y = i 2 Where S = 23,423 lbs/in t l l and S = 21,600 2 s = 7912 lbs/in 1 + 1.73 l = 36,000 2i(23,423)2 + 4(7,912)2 12 x 1000 (1) and Sy ??(548.64 + 250.40)
- x 1000 j
2 > 28,267 lbs/in /. 0K l l Margin of Safety = 36,000 7g 737 - 1 = 0.274 i l e i l i d ) 1, d 1 i 4 l 4 l 4 of 19
B. One Foot Drop Onto Closed End 1. Impact Energy of Loaded Cask E = 516,000 in -lb (see A-1 above) 2. Steel Deformation Required to Abosrb Impact V = 10.32 inz (see A-2 above) 3. Stopping Distance g ~b 8a. = -O.-. /2) ir (f N) u e N Total No. Studs T = 18 e U C O O O 6 @ l 2., 'O.C. ~ / 4 7 / { t= l l h N l , 10.32 18 x 3.1416 x (0.5625)2 =_ l 0.577 in y '(1 ) l i ~ l '/9 m l Typical Stud Installation 5 of 19
4. Applied g Load 2h g= t ,2x 12 l(I) 0.577 = 41.6 5. Force Applied to Closed End Plate 1 Weight of End Plate 2,141 lbs = 12,000 lbs Wei,ght of Cask Contents = Total 14,141 lbs F = Wg l(I) = 14,141 x 41. 6 I = 588,266 lbs 6. Strength of End Plate Welds ai = 0.125 in 1= 2(77.5 + 39) = 233 in a.2 = 0.500 in e= 0.85 Sv= 21,600 lb/in 2 Fv = e(Vi/2)(ai +a )tSv z =,0.85 x 0.707 x (0.125 + 0.500) x 233 x 21,600 = 1,890,288 lbs (1) Safety Margin = 1,89588,'266 ~1 = 2.21 7. Maximum Stress in End Plate (see A-7, above) a = 77.5 in t 2.5 in = b = 39.0 in w=F 88,26 2 194.6 lb/in j 7.5 = l M = b/a = 0.503 2 Max S = Sb = 0.5 wb t 2(1 = 0.623w') (39f 0.5 x 194.6 x ()) l ( 2. 5 )2 Il + 0.623(0.503)'I i 2 = 23,446 lb/in 6 of 19
l 8. Maximum Shear Theory of Failure l S = 36,000 d(52 + 4Ss)' y t 2 l Where S = 23,446 lbs/in Ref. B-7 t and S = 21,600 = 6729 lbs/in Ref. B-6 s 1 + 2.21
- (1)
N (23,446)2 + 4(6,729) 1
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x 1000 S = 36,000 y l 2 (549.7 + 181.1 ) x 1000 l 3'6,000 > 27,034 lbs/in2 i Margin of Safety = 36,000 - 1 = 0.332 27,034 t 7 of 19 l. 1
C. One Foot Drop Onto Long Edge 1. Stopping Distance (A 43 \\ Tane = 84 CG 95^ 0.51190 = 'IF /, e= 27.10 f / / / / / / / / / //
- 2. = 149 in
) / 10.32 i n) V = u /['27.1 (see A-2, above) o 62.9 h t t Area of Triangle -,2-(t cot 62.9) + (t cot 27.1 )] = 2 2 0.256 t + 0.977 t2 = 2 2 1.233 t in = 1.233t3 Volume of Prism = 2 183.72 t in3 = 3 V= 10.32 = 183.62 t 0.237 in t = 2. Applied g Load 9" t ,2x 12 0.237 101.2 = 8 of.J 9
4 ,wpe ope ++pmm =r*oM nw,: n= as hwe <v b wem-se e mam-u l 3. Component'g Load to Si_de Plate i 9 s = g - c o s o. 101.2 x 0.89021 = = 90.1 4. Force Applied to Side Plate F = Wgs W = 16,116 lbs (see A-5, above) 16,116 x 90.1 = 1.452,052 lbs = j c 5. Safety Margin of Side Plate Welds ] Fv = 3,050,422 lbs (See A-6, above) 3,050 2 2, ), ] Safety Margin = 1.1 = l 4 6. Maximum Stress In Side Plate W = 3h x j 2 249.8 lbs/in = i For all otl t-r parameters, see A-7 above. 0.5 wb' Itax S = St i. l (1+0.623~') t2 4 _ 0.5x249.Sx(39)' ~ (2.5)'l1+0.623(0.262)') 2 = 30,389 lbs/in Safety Margin = 36,000 - ) 30,389 0.18 = D. Side Drop, Door Reactions 1. Weight of Door W = weight of plate and weight of hinges 2509 + 337 = 2846 lbs = 9 of 19
2. Applied g Load Note - The g load for a long edge drop exceeds that for a side drop. Therefore we will use the g value. for a long edge drop. g = 101.2 (see C-2 above) 3. Force Applied to Door F= Wg = 2846 x 101.2 = 288,015 lbs 4. Strength of Shear Bar Weld I 1 4 d
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% x l', FLAT BAR n I 78 l dgdn[d. :h..3..@O3 n1 -w h -- Y Q z 10 of 19
Fv = 0.85 0.707 x (0.75+0.5) x 35.0 x 0.6 x 36,000 = 56Ls 3 lbs 567,898 Margin of Safety = 788,015 - I = 0.97 E. One Foot Drop Onto Bottom Edge of Door 1. Stopping Distance A 43 Tan + = T5T.5 \\S\\.$ CG 0.28382 = O i + = 15. 8 50 'I F 2=84in ei 10.32 in' V = (see A-2, above) Area of Triangle =f t7(tcot74.15)]+]c' (t cot 15. 8 5 )'j 0.14196 t2+ 1.76110t2 74.15 15.85 = ) [ 2 1.903t in2 = q 1.903 t2 Volume of Prism = 159.86 t in 2 = i I V= 10.32 = 159.86 t2 t = 0.254 in 2. Applied g Loading 2h 9=7 =2x 12 0.254 94.5 = 11 of 19
3. Normal Force Applied to Door F, Wg = 14,846 x 94.5 = 1,402,947 lbs = 4. Shear Force Applied to Door Fv = Wg SIN 4w 43,000 x 94.5 x 0.27312 = Fv = 1,109,822 lbs 5. Strength of Shear Bar Weld Fv = 0.85 x 0.707 x (0.75 + 0.5) x (78 x 2)x 0.6 x 36,000 2,531,201 lbs = Margin of Safety = 2,531,201 ~) 1,109,822 = 1.18 6. Bolt Tensile Load J H G 1 F E U DA CA -7 Fo a / Assume the plate is a rigid body. Also assume the bolts are strained in proportion to their distance from the hinge and to the number of bolts in the reaction plane. The 7 end bolts see the largest reaction F 3E"i F = x> Normal Force applied t1 door = 1,109,822 lbs (see E-3, above) 12 of 19
Moment applied to door = 1,109,822 x 41.875 = _4_6. 473,795 lb-in 2} Reaction q. Plane (inches) I A 0.875 7 8 9.0 2 C 14.75 2 I x; n; = 1235.625 D 20.5 2
- 7. x; *n j = 82,4 6 4. 4 6 E
34.6875 2 F 47.6875 2 G 61.875 2 H 67.625 2 l I 73.375 2 J 81.5 7 Summation of Forces: [F = 1,109,822-F - [ * "E F, =0 o F, +8 (*fn;) = 1,109,822 Summation of fio.ments: t1 = 4 6,4 2 3,7 9 5 -k5 4.- ( xl n;)
0 1011.8340 F,
46,473,795 F,= 45,930 lbs F= 1,109,822 - 696,348 lbs = 413,474 lbs From Reference 4, Article SA - 325: 1 - 8 UNC Minimum tensile strength is 72,700 lbs. .70 Margin of Safety = ~ l 3 0.58 = 13 of 19
F. One Foot Drop Onto Short Edge of Door i 1. Stopping Distance \\ TAN = 84/151.5 A.- = 0.55858 = 29.18 t9 = 43 in. .'e / / [g V= 10.32 in3(see A-2, above) Area of Triangle = f(tcot 60.82) f(tcot 29.18) + 1.1746 t1 = 1.1746 x 43 x t = 10.32 i n' 1 Volume of Prism = 60.82 29.18 t = 0.452 n / l i 2. Applied g Loading g / I 9"Y a \\ ,2x 12 0.452 53.1 = 3. Normal Force Applied to Door F = Wg 14,846 x 53.1 = 788,333 lbs = 4. $ hear Force Applied to Door Fv = Wg SIN G 14 of 19
= 43,000 x 53.1 x 0.48755 = 1,113,324 lbs 5. Strength of Shear Bar Weld Fv = 0.85 x 0.707 x (0.75 + 0.5) x 35 x 2 x 0.6 x 36,000 1,135,796 lbs = 1,135,796 Margin of Safety = _ ) 1,113,234 = 0.02 -E 6. t1oment Applied to Door M=Wgf 14,846 x 53.1 x 21.5 = 16,948,936 lb-ini = i 7. ' Bolt Tensile Load Reaction Xi Plane (inches) ni A 1.0 7 B 3.5 2 C 9.5 2 xn = 645 D 15.5 2 {x'n= 22,607.5 E 21.5 2 F 27.5 2 G 33.5 2 H 39.5 2 I 42.0 7 Summation of Forces: E +4j[(x;n;)=788,333 Fo F = 708,333 - 15.3571 Fr o Summation of Moments [M = 16,948,936-{1~ ( * "i ) 0 = 2 15 of 19
F = 31,488 lbs y Margin of Safety = 7 -I 3 8 = 1.30 Fo = 304,768 lbs G. One Foot Drop Onto Door Hinge 1. Stopping Distance Assume that the top of the hinge member is solid rather than slotted. This is conservative, since it decreases t and hence increases the g load.
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.x-LD Q li l e' I 14 _ 3'// = Area of Triangle.= f(tcot6.12)- f(4 cot 72.90) + 6.12 72.9 0 4.98765 t l ( p = 7 t i Volume of Prism a 10.32 = 4.98765 x 6 x 2.5 x t* 0.371 in t = 2. Applied g Loading 9" 2x 12 , 0.371 64.6 g = 16 nr lo l
3. Force Applied to Door F = Wg 14,846 x 64.6 = 959,052 lb = 4. Moment Applied to Door M = F1 959,052 x 49.875 = s 47,832,718 lb-in = 5. Bolt Tensile Load j The analysis is identica.1 to that done in E-6, above. Rea' tion Xi Plane (inches) ni A 8.875 7 8 17.0 2 C 22.75 2 0 28.5 2 ~ 2 39.6875 2 F 55.6875 2 G 69.875 2 H 75.625 2 I 81.375 2 J 89.5 7 s [q ni 1469.625 = ,lx{ni 103,660.2188 = Summation of Forces: Fo = 959,052 - 16.4204 F3 Summation of Moments 1158.2147 F; = 47,832,718 _3 = 41,293 lbs 17 of 19
Margin of Safety = - 1 = 0.76 Fo = 380,922 lbs II. CONTAINMENT ANALYSIS l Positive containment is provided by the cask internal disposable l80 containers which have leak tight seals and closures. 1 i III. SHIELDING ANALYSIS All of the deformations to shielding calculated above are small. No de formation extends into the effective shield thickness used in the initial dose calculations. See sketch below. Dotted line indicates boundary of the N effective shield \\ thickness. \\x \\ x 1 N._ ( l l DEF O RM ATioN -de l I ] 1 1 1 1 18 of 19
REFEREflCES 1. ORNL - NSIC-68, " Cask Designer Guide", L. B. Shappert, Oak Ridge National Laboratory 2. " Manual of Steel Construction", Sixth Edition, American Institute of Steel Construction. 3. " Formulas for Stress And Strain", Third Edition, Raymond J. Roark, McGraw-Hill Book Co, Inc. 4. "ASME Boiler and Pressure Vessel Code". 4 i 19 of 19
i OCU man Pi' W humult ainages A O. un-, 5 NO. OF PAGES REASON: l 0 PAGE ILLEGIBLE: O HARD COPV FILED AT: PDR OTHER 1 1 O BETTER COP / REQUESTED ON GE 100 LARGE TO FILM: C HARD COPV FILED AT: PDR CF OTHER ON APERTURE CARD NO. 7/ / d O #d 5,7 +b - y A oy -__----_-___--_--_-----------_-------_a
%g UNITED STATES 3% 7' t NUCLEAR REGULATORY COMMISSION fI 1 0, WASHINGTON, D. C. 205SS o,.> DEC 0 61978 Mr. Dixon Hoyle, Director Office of Program Implementation Bureau of Oceans and International Environmental and Scientific Affairs U.S. Department of State Washington, D.C. 20520
Dear Mr. Hoyle:
Enclosed please find an application from B.K. Sweeney Manufacturing Company for a license to export byproduct material to Condition 2 countries. Before taking action on this license application, we would appreciate your views, in accordance with established procedures and from the overall perspective of the Executive Branch, as to whether the issuance of the requested license would be inimical to the interests of the United States, including the common defense and security. Sincerely, C j d t oerald G. Spl-inger,' Assi a rector Export / Import and International Sa fegua rds Office of International Programs
Enclosure:
Appl. dtd 11/29/78 (XB001000) cc w/ enclosure: Ar. Holsey G. Handyside, DOE Mr. Richard L. Williamson, ACDA/NP/NX Ms. Sheila Buckley, D0D Mr. Duane Sewell, DOE Mr. Rauer H tieyer, DOC Mr. Robin DeLaBarre, 005 Vig 781213oc)52
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