ML20132B493

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Forwards Rev 6 to SAR in Response to Request for Addl Info Re Application for Certificate of Compliance for AP-300 Shipping Container Per R Odegaarden Telcon.Text Expanded to Show That Container Design Satisfies Required G Loadings
ML20132B493
Person / Time
Site: 07109192
Issue date: 08/12/1985
From: Jerome Murphy
ANEFCO, INC.
To: Macdonald C
NRC OFFICE OF NUCLEAR MATERIAL SAFETY & SAFEGUARDS (NMSS)
References
25646, NUDOCS 8509260207
Download: ML20132B493 (35)
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Text

{{#Wiki_filter:. 7/- f/9> A v NEFCO -~ ANEFCO INC. Ef"!!- oe877 203 43i.32se August 12, 1985 U.S. Nuclear Regulatory Commission Transportation Certification Branch d Division of Fuel Cycle and Material Supply w g, Washington, D.C. 20555 Attention: Charles E. MacDonald, Mail Stop 396 SS gggg AUG141985>

Subject:

Model AP-300 Type A Package

g. a.Nucttu sttutAitu Docket No. 71-9192 conuesu le tms

'b ust ht-ca h N

Dear Mr. MacDonald:

Attached are revisions to the SAR for our AP-300 shipping container which has been submitted to you with our application for a certificate of Compliance. The revisions have been made to satisfy some questions which were posed by your Mr. Richard Odegarden in my telephone conversation with him about two weeks ago. The responses are related to the-structural aspects of the report and have been marked as revision 6 in the text for_ easy identification. We have expanded the text further.to show in detail that the structure of the-shipping container is adequate, with a margin of safety, to satisfy the required simultaneous "g" loadings at the center of gravity of the cask during transport. We have also provided a toroidal ring et the top of the container and a square tube ring at the base of the container. The calculations show that these will reduce the "g" loading by absorbing the energy of a one -foot drop.and assure the continued integrity of the container during a presumed one' foot drop. We hope that these revisions clarify the design of the cask and meet with' your approval, so that we may be granted a Certificate of Compliance at your earliest convenience. Very truly yours, p g ANEFCO, Inc. O l g ( John D. Murp y 2 -?. President AUG 151985 > wu, 'a t_,h. h \\A) ~ 2 nu o. Ad6[a JDM:sm ~N N, 8509260207 850812 DR ADOCK 07109192 h7 encls. C PDR !b

REVISION 6 - 8/7/85 ) INSERT PAGES REMOVE PAGES 2.4-9 2.4-9 2.4-10 2.4-11 Drawings 133]l 2.4-12 Drawings 133-1 2.4-11 2.4-lla 2.4-13 134-1 138-1 2.4-12 2.4-14 138-1 2.4-13 2.4-14 2.6-13 2.6-13b 2.6-13 2.6-14 2.6-13b 2.6-15 2.6-14 2.6-16 2.6-15 2.6-17a 2.6-16 2.6-17b 2.6-17 2.6-18 2.6-17a 2.6-20 2.6-17b 2.6-17c 4.2-3 2.6-17d 2.6.18 2.6-20 2.6-21 4.2-3 l

i ~ i 2.4.4 TIE-DOWN DEVICE In order to satisfy the requirements of 173.412 (d) the< cask tie-down blocks were designed to meet Title 10 of the Code of Federal Regulations 71.31 (d) which stipulates that the tie-down structure be capab3e of sustaining at the center of gravity of the cask a 'g' loading component of: s 2 g's = Vertical 10 g's Forward horizontal = 5 g's Sidevard horizontal = 2.4.4.1 TIE-DOWN FORCES y -27L "[ N g 73' I 70* ._ 7 ()* " l -( k / 14.7 5" t a-41.81*--+ 47.7* 96* .- a ~ 1 1 a46.67 72 \\ - 42.3* 42.3'N FIGURE 2 4-3 TIE-DOWN CONFIGURATION 47.66 in.- (See Section 2.2) 4 Cask center of gravity '= = Cask radius ilA.8125 in. /= 9 6 i n. Overall, Height ~= Ibs. 4 Weight' 66,720. c., 2.4-8 Revision 4-4/15/85

For a laskd cask weight of 66,720 lbs, it is considered that the following forces will act simultaneously at the center of the cask in the following directions: Vertical - V 2g will act in the upward direction and lg will act in the downward direction for a net lg force or 66,720 lbs. upward. Forward Horizontal Hp 10g will act in the_ forward horizontal direction or 667,200~lbs. Sideward Horizontal - HS Sg will act in the sideward horizonta 3 direction or 333,600.Jbs. Load in Tie-Down Rods Vertical As shown in Figure 2.4-3, the vertical component of.the g force in one rod is = F cos 47.7'. y Where F = the tension force in a tie-r6d due to the vertical y 6 g' force Therefore, 4F cos 47.7* =.66,720 lbs. y Thus, the tension in a tie-rod due to the vertical g force is: F = 24,784 lbs. y + i i 4 v 2.4-9 Revision 6-8/7/85

~ u ,f"' / s N / F 10 ru - 3l v'i ~ \\ s-F, g +-48.sl + 27.414 p 2 F 51r 47.7"c 3f g (ob]10 b0) C.G v 2 F,w, 47.7 A p e % 47" 77- [ T i O v u-y fF3mp7 Ficure 2.4-3a Forward Horizontal Tie-Down Forward Horizontal' Only the two rear rods are effective in resisting the 10g forward horizontal force, as shown in Figure 2.4-3a. c Let FFH = tension force in a tie-rod for. forward horizontal g force l IMoments at Pt. O at bottom f (assume this is pt of rotation)- '66,720(10)(46.67) = 66,720 (41.81) + 2F sin 47.7 cos 35 (72) FH +2F s 47.7(41.81+27.42) FH 2.4-10 Revision 6-8/7/85

31,138,224 = 2,789,563.2 + 87.'245FH+ FH 180.43F = 28,348,660.8 FH Thus, the tensioniin a tie rod due to forward horizontal g force is F = 157,ll7.Ibs. FH Sideward Horizontal Only two rods are effective in resisting the 5g sideward force as shown in Figure 2.4-3b. rA ri Sg a ~ + ;. j p ss (s ~ (, wh +---- 4 I. s -, %- h4iy ZF A 47 7'ce35 sw 6 bl77_o(d) C'O ~If 73' 2 F.,a tod77 o . 3 MWi" 'I @p2D O y F %re.ex View A-A F Ficure 2.4-3b - Sideward Horizontal Tie-Down 2,4-11~ Revision 6-8/ 7 /85 s

.~. 1 ension.in a tie-rod for sideward-horizontal 9 force Let F = SH Moments at Pt. 0 at bottom 66,720(5)(46.67) = 66,720(41.81) +2F sin.47.7 sin 35.0 D2) SH i +2F cos 4 7. 7 (41. 8.1-2 7. 4 2) SH 15,569,112 = 2,789,563.2 + 61.1 FSH + 19.4 FSH 80.5 FSH " Thus, the tension in a tie-rod due to sideward horizontal g force is F = 158/752 lbs. ~ SH Summary of Tie-Rod Forces The maximum tensile force in a rod kom the simultaneous' application of the 3 g' forces is F =Fy+FFH + SH 1 = 24,784 + 157,117 +'158,752 F F7 = 340,653 = 340.7 kip 2.4.4.5 Tie-Down Pads for Cask Assembly 2.4.4.5.1 Loading l The tie-down pads must be capable of: sustaining the total force of l the maximum force previously calculated, viz. 340.7 kip. .The tie-j g down pads are designed to resist this maximum force. Each pad is designed for 340.7 kip. 2.4.4.5.2 Tie-Down Pad Design' ksi Use steel of Fy - 38 min. weld with low hydrogen electrode. Steel to be noted in Group II Table 4.2 of AWS Dl.1-80 Structural . Welding Code. Check hole for bearing and shear. (assume 2" min. for Pin) 4 = 42.6 ksi (actual bearing stress) F = p 1 2,4-lla Revision 6-8/7/85

Allowable AISC (8th Edition specification para.1.5.1.5 Bearing) (1.5.1.1 Tension) F

1.5 Fy = 1.5 x 38

57 kai (1.5.1.1 Shear) p ' Safety Factor to allowable AISC bearing: SF = 57/42.6 = 1.34 Check te.ar out of hold down device along lines 1-2 and 3-4, Section A-A (page 2.4-13) Av = 2 x 3 x 4 = 24 sq. in. The shear' capacity of structural steel is 2/3 of the tensile capacity. Therefore, the tear out capacity is: 0 2/3 x 38 x 24 = 608 kip; SF = " 1*7S 0.7 At the highest loaded lug the tension force = 340.7 k. Therefore, horizontal component = 340.7 cos 42.3' = 252.0 k and the vertical component = 340.7 sin 42.3* =~229.3 k h Bending on plane 1-2 M = 6" x 340.7 = 2044.2 kip-in' l Assume weld pattern is 22" x 4" pattern. 2 Area of weld material = 2 (22+4) (.75 x.707) =-27.6 in l_ (.53) (22) 3 + 4 ( 53) (11) 2 = 1453.6 in 4 I=2 12 4 C = 11" .6 3 S = I/C = 132.15 in = 4 Tension stress on weld = F3, = 2 5 = 15.47 ksi 4 Shearstressonwelp=FSW 27 j = 12.3 ksi Combined stress = f (15.47)2 + (12.3)2 = 19.76 ksi comb Allowable tensile stress

21 ksi S.F. on tension

.063 = 9.76 2.'4 Revision 6 - 8/7/85

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i l -Tie-Down Pads-As shown in the drawing, four tie-down pads are provided to attach the shipping cask to.the truck bed. A 7" square, 2" high plate is 4 welded to the cask body using a 3/4" fillet weld all around. A 0 plate,-4" thick, 22" long and 10%" high is welded at right angles to the base plate and the cask body.using a 3/4" fillet weld all i around. A 2 " diameter hole is provided in the latter plate and 3 used to enable tie-down. Both plates will be fabricated from a steel having_ a minimum, Fy = 38 ksi and the lug plate will be b welded using a E70 ksi low hydrogen electrode. The cask will be installed on the truck bed so that the center line of the 110* l angle between adjacent pads is parallel with the direction of -travel. The transverse line is parallel to the center line of the 70' angle between adjacent pads, The lif ting / tie down pads and the welds that attach.them to the t cask shell are therefore adequate with a safety factor in excess of 1.06 to resist the assumed simultaneous maximum g loading during transportation. i 1 3 i i i i 4 4 ? I l l i I ( 1 l Re ision 6 - 8/7/85 v 2.4-14

2.6.5 Vibration The approximate natural frequency of the loaded cask is based on the concentric steel shells. Using Roark

  • page 576, Case Ib, considering the cask a uniform beam with both ends simply supported, and a uniform load w per unit length including the cask weight.

f = (9.87/2 tf) (EIg/wl )0.5 4 n Where: E= Modulus of Elasticity I= Area of Moment of Inertia L= Cask height = 96 in. i l= Distance between supports = 83.5 in, w= 65,710.80 (lb) 684.49 lb/in = 96 (in) I 6 E = 29 x 10 1b/in R= Outside Radius of Cask Shell = = 41.8125 in. R = Inside Radius of Cask Shell = R - 1.25 = 40.5625 in. g 4 6 3 12 3h- (R -Ry) = 29 x 10 x 274 x 10 = 7.946 x10 EI = E x 32.2 x 12/684.49 x 83.5 )b 4 1 fn = (9.87/2gT) (7.946x 10 fn = 477 HZ This natural frequency is satisfactory for truck transport, since it is well above the low frequency range of truck suspension systems (1-20 HZ)

  • R. J. Roark & W.C. Young, " Formulas for Stress & Strain" Fifth edition, McGraw Hill, 1975 i

2.6-12 Revision 1 7/31/84 l

2.6.6 WATER SPRAY A heavy water spray on the package will not harm the package because it is constructed of ASTM A516, Grade 70 steel. In addition, no water will leak into the primary containment because of the bolted closure and seals. Therefore, the only possible effect of the water spray would be to lower the cask te=perature. 2.6.7 FREE DEDT In designing a cask for transport of radioactive material, the regulations require that a free drop of the cask through a predetercined distance (in -feet) ento a flat, essentially unyielding surface cust be investfrated. A segmented toroidal ring will be attached to the ANEFC0 AP-300 cask to absorb the energy which is generated by a 1 foot drop of the cask, as shown in Figure 2.6-1. As shown in ORNL-NSIC-68 in Section 2.8.2, this ring will protect the cask closure not only in an end drop, but will operate properly regardless of the anE e at which the cask impacts on a l horizontal surface. The damage evaluated here is that due to defortation at the impact plane and indirect damage due to deceleration. The toroidal ring will.be a 3" O.D. x \\" thick steel tube made from b 1020 steel hot rolled and electric resistance welded. Using equation 2.17 on page 67 in ORNL-NSIC-68, the energy absorbing character-istics in a crushing impact are correlated by: 2 E = Syt L [ js + 0.4 ts R R Where: E = Energy absorbed (in-lb) Sy = Yield strength of steel (psi) t - Tube thickness (in) SI L = Tube length (in) R = Mean radius (in) A = Deformation of tube (in) and as shown in Figure 2.6-2. The energy which cust be absorbed by the tube for a 1 foot drop E - 66,720 lb x 12 in = 800,640 in-lb For the 3" tube described above l l Sy - 38,000 psi t = 0.25 in L = 83.625 - 262.7 in R = 3 + 2. 5 1.375 g 4 Substituting in the above equation: m 38,000(0.25)2(262.7) ~ 0.4 2 = - 6 800,640 1.373 1.373 ~ /i = 1.28 Since the 1.D. of the tube is 2.5 inchesthe tube can accommodate the deformation caused by a 1 foot drop of the cask. l l 2.6-13 kevision 6 8/7/85

ToKoD,AL GING Et4EF61 Ab50E6C f. ~ R \\tIG W EL,I) 4., ...__-__..g_, 6[ l \\ 'l bi'G U)Eup j Eo:i R su c. l CEf, 00TER i sect FIGURE 2.6-1 TOROIDAL RING ENERGY ABSORBER \\ t x s.,,, s s s,,,,., 3 ,x~ FIGURE 2.6-2 DEFINITIONS OF TERMS USED IN EO. IN SEC. 2.6.7 2.6-13a Revision 5 - 6/10/85

k i i - The tube will be welded to the cask outside shell with two \\" fillet . welds using E70 kai' low hydrogen electrode, which has an allowable tensile stress of 21 ksi, as shown in Figure 2.6-1. I The force due to the deceleration of the cask can be determined using equation 2.4 on page 36 of ORNL-NSIC-68 (Cask Designer's Guide) F - 2h (W) 3 Where: W = weight of the leaded cask I j Ng = the mean no. of g's the. cask subject upon impact 9.375 b. Ng = 1.., = Ng can be calculated by dividing the drop height by the stopping distance in accordance with the statement in section 2.7 of the Cask Designer's j Guide. 1.251 x 10 kip Therefore, F=' (66,720)1bs = This force must be. resisted by the welds with which the tube is attached to the cask shell. The total area of the 3/4" weld that will resist the } force can be calculated as: i i A = 2 Y(D)(b) I Where: D = 0.D. of cask b = effective throat of weld 4 2 A.= 21Y(83.625)(0.354) = 186 in 4 i i The tensile stress on the veld area is: [ 4 3 l 5 f. 1.262 x 10 kip 6.73 ksi = i 186 }' i i p 3.12 [ 4 Safety Factor = = 6.73 Since allowable stress for the weld is 21 ksi, the tube welds will resist the force of the drop impact with a safety factor of 3.12. t r t l A square tube ring, fabricated from ASTM A500-CRB steel, 2 inches on a side and 0.154" thick, will be installed on the bottom of the cask to absorb t b the energy during a bottom drop of the AP-300 cask through a height of 12". This arrangement is shown in Fig. 2.6-2a. r I i l i l i I 2.6-13b Revision 6 - 8/7/85' \\ l _m ..m..,-_

. - -. _. _. ~. _ - - - - -..... - _ _. M .4 1 t f p,. m e.g use asiv h sco i. h L / Ga s stTt t i i } 2.b j Figure 2.6-2a AP-300 Cask _ eny i i Bottom Desien j The energy of the end drop is: j l-E= W = 66,720 lbs x 12 in. = 800,640 in-lb l ) 7 The volume of ASIM A-500 Gr5 steel to absorb this energy is: i f .y, Ef, 800,640 in-Ib 17.41 in3. = j 46.000 lb/in The surface of the' wall of.the tube ring which will be impacted by a bottor-end dron is: i 2 j A= 7T Dt = 7T 82.375 (0.154) = 39.9 in i { The depth of the ring vall deformed to absorb the-drop energy 15: 3 y 17.t.1 in 6 in d = 7 = 39.9 inz = s 2 4 The g loading on the cask lid under the impact conditions can be calculated by dividing the drop height by the stopping distance, as shown in Section 2.7 l of the Cask Designer's Guide (OR'1-NSIC-68). [ I 12 .5 =

6 E ".436 The g loading on the. lid will therefore bc b

FS = 2Wg = (2) 9.46 kip x 27.5 = 520.3 kip l The force will be distributed on the 2.5" lip (Part No. B-5) where the lid is bolted to-the cask, whose area is: i 2 f(81.125 - 76 ) 632.5 in A= = s Subtracting the area of the 36 bolts, the net area is: 2 2 16.6 in .A = 36 x (0.7656 ) = 3 632.5 - 16.6 = 615.9 in Revision 6 - 8/7/85 l 2.6-14 t t

~ i The stress on the plate will be: f=6 0.84 ksi = 9 The SA-240 plate, with a yield strength of 30 ksi, will not bend out of j shape under a stress of 5.11 kai with a safety factor of 35.7. = 35.7 SF = 84 2 'At impact during the bottom drop the force on the impact ring can be t expressed as = M = (2)66,720g 515.63g lb/in 'N = f 1D 82.375 1 Where g = 27.5 as shown above Ng = 14.18 kip /in l The stress in :he wall of the ring will be f 14.18 kip /in 92.1 ksi f= = = t .154 in The stress at which the wall of the tube will buckle or crush can bc l i calculated * = 0.6 ( f er where'g" = knockdown factor which is a function of R/t b for R/t 267.45

3' = 0.4

= = E cr = 0 6_(0.4) (2 5 ) + 25,665 psi Therefore 2 ( ( I , %re NL E!u } {. [ ese ASTV A 5bo i j { k / G8 t $TTib i i le na tw 3 F l i s' .- C.lty' --+ J k l Es4l.lfir' ~' u r.u ,u. a.,- s

  • Buckling of Bars, Plates and Shells, Brush & Almroth, McGraw 11111, 1975 1

2.6-15 Revision 6 - 8/7/85 -.m ,y- _m,_. ,.--,-__,v.,_, ,-..__,,.....,r .,-4%,..

~ _ l $6 77bM IMFK 7~ f

o. g q ; 2 H

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3 f / / / f l/ / // A f A T; 2 , ^ s o,, t l.75 i D m a.n ut u u it u 1141um c i For bottom, flat impact (g=27.5), the loading on outside bottom plate l is due to the weight of the contents, and the weight of the bottom plates h and lead. 7/(81.125)2(2.0625)(710)3 - 4380 lbs. Weight of lead = 4 (12) 81.125)2-(1.75) (.283) Weight of steel = = 2560 lbs. contents = 20,000 lbs 26,940 lbs I 2

5.21 lbs/in q

1 125)g 2 at impact q3 = 27.5 (5.21) = 143.28 lb/in 4 For uniformily loaded circular plate, with fixed edges reference to Roark & Young, page 363. The edge bending moment is~ t M a " qa2 r g t 2.6-16 Revision 6 - 8/7/85 I i

(81.125) 2 I Therefore M = 29,468 in 1bs/in / Calculate effective bending moment of botton l plate .5 Assume the inside and outside act together 2.0 61 '. 0.5(1.0)(3.563) + 1.25(1.0)(.625) i. ' 7 0.5(1.0) + 1.25(1.0) 2 ~

[.

= 1.46" l 1 ~.5(2.099)2 + 1.25(.835)2 = 3.074 in /in 4 t = h 12(1) = 3.33" )

22,551.55 psi Stress top fiber

= 3 074) b Assume that the weld yield strength for dynamic impact is 25% greater than for static loading } Thus FS.= = 1.16 2 and the bottom plate welds are adequate with a factor of safety of 1.16. 4 { The plate bending stress at plane A-A for 1" strip M = 840 (2.5)2 =2,625 in 1bs/in f=h = 3.938 psi = s Edge rotation = h = " 1+ ) 114.7 lb/in 9~ = O 8 7 =0 (2 0 )(1.3) 8 There will be no releases because of compression maintained on the seal. 2.6.8 CORNER DROP As indicated in Section 2.6.7, the segmented toroidal ring, whose design is evaluated above, will protect the cask closure regardless of the angic at which the cask impacts on a horizontal surface (Cask Desi ner's Guide, F ORNL NSIC-68, page 66). 2.6-17 Revision 6 - 8/7/85

[ t From a physical standpoint, the maximum direct damage in a corner drop would occur with the cask so oriented that the line passing between the center of gravity and the point of impact coincides with the direction of the fall. The geometrical representation of the corner drop is shown in Figure 2.6-3. The ideaIiration of deformation, the external damage after impact, is indicated as Z. S' \\ / V / .e l / i if/HN Hf HfM///ista 2 9 FIGURE 2.6-3 CORNER DROP DEFORMATION CEOMETRY Using Figure 2.6-3, the angle A (angle of impact) can be seen to be: 4 Tan A = R/H Where: R= Cask Radius 41.8125" [ H= Center of Gravity with relation to the top (lid) end = 49.33" [ Therefore, the angle of impact (a) is: 41.8125/49.33" .8476 Tan A = = 40.28* A = Impact on the top corner causes an axial force component and a lateral force component. The axial force acts on the cask. lid in the same way as analyzed previously. The lateral force component on the contents bears against the sides of the cask before any shear load is transmitted to the. bolts connecting the lid to the cask, g The behavior response for a corner drop should result in a greater deformation of either the toroidal ring for a top drop or the square ring for the bottom drop. It is speculated that.the crushing under the point of impact will be extensive with less deformation occurring away from the point of impact. This greater deformation would result in "g" factors less than those for a flat drop. 2.6-17a Revision 6 - 8/7/85

TOP CORNER DROP 40.28* from vertical Drop angle A = Bolt Tension 9.375 (same as for flat drop. This is believed Assume g = to be conservative) = 2Ng(W + W ) c s 40.28' F 1 c y F - 2(9.375)(29,462) cos 40.28* - 421,432 lbs. y Fp.(from internal pressure) = 52,500 'Therefore Fb" 6 F, = - = 38,726 psi Bolt Shear' -Even'though the design is such that bolt shear should not occur, we will assume that the bolts do take the. shear induced by the in-plane' component h of"g" forces on the lid. This is very conservative. F = 2(9.375)9,462 sin 40.28' = 114,701 Ibs ~ s 01 3186 lbs/ bolt F, = = fs" g_ = 7211 Psi For bolts in combined tension and shear

Reference:

Guide to Design Criteria for-Bolted and Riveted Joints, Fisher and Struik, John Wiley and Sons, 1974. 2 + y = 1.0 (0 62)2 Where x = ratio of the shear stress on the shear plane to the tensile ~

strength, y=

' ratio of the tensile stress to the tensile strength Therefore x='1 000 = 0.058 38,726 Y" = 0.310 125,000 ( 2.6 17b Revision 6 - 8/7/85 i L^ [. '

Therefore- + (.310)2 A1,o 0.009 + 0.096 h-1.0 h' O.105 62-1.0 .Therefore, Bolt is 0.K. If 50% of the tensile strength is used (62,500 psi) 0.034 + 0.384 1 1.0 - 0;418 C 1.0 Bolt is still O.K. Therefore, the:11d remains in place. 2.t.E.1 Eolting A sign i The bolts screwed into the helicoil inserts are designed to withstand the expected decelerating forces resulting fro: the'iepect resulting fro: a one f oot drop at_the impact velocity. [ Tnt heliceil it inserted inte the bolt. ring which.,is fabricated cf SA-240 Type 304 with a u.inicus yield of 30 kip /in'. The root diateter of the 1.5 inch long helicoil is 0.79 inches. Therefore, at. yield, the tensile strength of the helicoil in the retal ring is: 2 F = 30,000 lb/in x 1Y(1.5)(.79)' - 112 kip Tr.e t e r.s ile strength of t hcliccil assembly basedion the ranufacturer's data, shown in figure 2.6. , is 125 kip. The bolt which will be used is a 3/41 th SA-320 Grade 1.7A bolt with a 3 cinicut tensile strength of 125 kip /in. Therefore, it will be capabic of resisting a force T = 125,000 lb/in x f (.75) = 55.2 kip Theref ore the bolt will yield before either the helicoil, or the ring material when an excessive force is applied to the assembly. This is the basis of selection of the bolt - specification. 2.6-17c Revision 6 - 8/7/85 _a

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2. 6. 8. 3. 2 BOLTS IN TENSION Solving for the force in tension on the bolt due to

'g' component using ORNL-NSIC-6E P.37 the Forrule (2.7) the r.inimun bolt area for tensile is: A. = FW 4 FG 4 TE r,.. g, g Where: FW = Tension on the bolts due to the decelerating drop force FG = Tension on the bolts due to the gasket using Fsg or Foc, whichever is greater htere: Tsg = Tension due to gasket seating Foc = Tension due to Inaintenance of a tight seal on the gasket Fp = Tension due to internal pressure at reduced atmospheric pressure Sa = Ultimate strength of the bolts Using ORNL-NSIC-68 P.36 Formula (2.4), the tension on the bolts due to the drop (FW) is found to be: l FW = W (2Ng) htere: W = cask lid and contents weight = 29,462 lbs. cl Ng = "g" loading due to the drop - 9.375 Therefore: FW = 29.462 { 2 (9.375))! 0 FW = 552.413 lbs. Using OPSL-NSIC-68 P.36 Forv.ula (2.5), the tension load due to gasket seating (Fsg) is found to be: Fsg = b%dy htere: d = Mean diameter of the gasket, in = 77.25 y = Minimum yield design seating streng'th (ASMr Section VIII, Table VA-49.1 (1974) for self scaling type gaskets (Neoprene), Ib/in2=0 b = Effective gasket seating width, in = 1.875 Since the y value is "0" the Tsg value is negligible and can be considered "0". 2.6 Revision 6 8/7/85

t Using ORNL-NSIC-68 P.36 Formula (2.6) to determine the ~ tensile load on the bolt 2 created to maintain a tight seal on a f 3 at gasket,(roc) is found to be: To: = bder Where:

r. = gasket factor (AS.ul Section VIII Tah3e VA-49.1)=0 b = effectivc seating vidth in = 3.675 d = nean dia.ctcr of the gasket in = 77.25 (See Section 3.5.2) p = differential pressure, psi = 13.2 Since the n. value is
  • 0', the Toe value is neg3icib3e

~ and can be considered "O*. Using ORNL-NSIC-68 P.35 (Fortrula 2. 3) the force due to the internal pressure as a result of the case given in Section 1.5.2 yields: f Tp = Trp (d)2 4 Where: p = dif ferential pressure across the gasket psf - 12.2 d = mean diameter of the gasket, in = 77.25 Fp = 'b'(11.2) (7 7. 2 5) 2 4 ( l Fp

  • 52,500 lbt.

Substituting the derived values into the Torru3a (2.7) for the n.inimun. bolt area it is found to bc: A.*: = FW 4 F_p Sa Where S,a = U3 tir. ate tensi3c strength of the bo3 tr ( A320-L7a f ror. Table 1. 3.1) = 125 K pr i. 2.6-19 Revision 5 - 6/10/85 I l

r T Therefore:

  1. "(552,500 + 52,500) lbs 125,000 lb/inz 6

AM = 5.76 sq. in. Evaluations for the 3/4" bolts yicids NB = AB k*here: B = number of bolts required N Amin = minimum bolt area due to tension = r t Area of the bolts =.34 AB Therefore: 5 - 16.9 NB, L Ng = 17 bolts Since the cask uses 36 - 3/4" bolts, the lid will remain in place during the drop. Check load / bolt: 52,500 + 552,500 = 16,806 lbs yb 6, = 49,429 psi Tensile stress = 9, Trom ORNI.-NSIC-68. Stress should not execed the yield strength or,50% r of the ultimate strength F = 62,000 psi TU/2 ( FS = 1.26 = 9 Thus, 36 bolts is adequate with a 1.26 safety factor. Check local bending of 2" thick flange of lid near a bolt. 6(1 Of>) bending stress in f. = 25,209 psi f = = = 1*l9 (F.S.) yield = 2 0 .6-20 Revision 6 - 8/7/85

e 4 L Check elongation of bolt due to tensile load PL 16,806 (2.75) 0.005 in A, AE 0.34(28.5 x 10" = Rotation at edge of plate (Roark & Young, Table 24, case 10) )p52,500 M2, , 133,4 p,g 3 ga q, 9 a 8D(1+ ) 3 28.5 x 10 (2)3 7 6 Et = 2.09 x 10 p. b 12(1-8) 12(1.34) 133.4(39)3. ,a 8(2.09 x 10')(1.3) 0.0364 radians a (2.1') NOTE: This rotation assumes no restraint from bearing of lid flange on the outer shell extension. This ef fect would reduce the plate edge rotation. 2.8.8.1.3 Bolts in Shear The plug portion of the lid has a radini cicarance less than that of the g lid bolts clearance holes, preventing contact of the lid with the closure bolts during the hypothetical drop conditions which would put a shear load on the closure bolts. t k 2.6-21 Revision 6 - 8/7/85

2.6.8.1.3 Conclusions Since all the damage vill occur in the toroidal segeented ring and the bolts are sufficient to retain the lid, 8 no release of radioactive contents will occur due to direct or indirect damage in the (1) one foot corner drop. 2.6.9 Compression The AP-300 cask weighs in excess of 5000 Kg. Therefore, no compression load was considered. 2.6.10 Penetration The regulations in 49 CFR 173.465 (c) stipulate that the cask must be able to withstand the impact of a 1.25 inch diameter bar with hemispherical end weighing 13.2 lbs. being dropped from a height of 3.3 feet on the mor.t vulnerable region of the cask. The most vulnerable region of the cask is the 1.25 inch thick outer stec1 shell. If the ASTM-A516 Grade 70 steci plate is assumed perfectly rigid, the kinetic energy of the falling bar must be absorbed by the shear deformation of the plate. This is conservative because any bending deformation of the plate will also absorb energy and reduce the tendency for shear failure. The energy required to cause shear can be expressed as: Es = KPD t S3 Where: K = ductility factor =.60 S = Ultimate strength in shear 27,000 psi 3 D = Bar diameter = 1.25 in. t = Plate thickness = 1.25 in. Thus the energy the outer shell can absorb is: Es" 99,400 in lbs. The kinetic energy of the falling bar is found to be Eb = Wh Eb = 13.2 (3.3 x 12) Eb = 523 in-lbs. Thus, the most vulnerabic part of the cask will not be penetrated by the falling bar. 2.6-22 Revision 6/3@/ne m

4.2.2 Pressurization of Containment Vessel The contents of the containment vessel will be only solids, with no possibility of gas release. The only conditions for pressure formation above. ambient atomspheric pressure would be exposure of the cask to 100*F temperatures in the shade. Assuming a maximum internal temperature of 180'F the maximum absolute pressure within the containment v'essel, assuming that it is loaded at 70*F - 530*R, is: 2 4 14.71b/in x = 17.7 psia 3 The structural analysis of the cask in Section 2 demonstrates the capacity of the AP-300 cask to withstand an internal pressure of 3 psi gauge. 4.2.3 Coolant Contamination There will be no coolant used in the AP-300 cask. 4.2.4 Coolant Loss There will be no coolant used in the AP-300 cask. 4.3 Containment Reauirements for Hypothetical Accident Conditions The hypothetical accident conditions need not be considered for the AP-300 cask, when it is used to transport greater than A i quantity LSA waste packages. 1 I 4.2-2

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