ML19242B689
| ML19242B689 | |
| Person / Time | |
|---|---|
| Site: | 07109097 |
| Issue date: | 06/21/1979 |
| From: | Hansen L NUCLEAR PACKAGING, INC. |
| To: | Macdonald C NRC OFFICE OF NUCLEAR MATERIAL SAFETY & SAFEGUARDS (NMSS) |
| References | |
| NUDOCS 7908090294 | |
| Download: ML19242B689 (8) | |
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NUCLEAR
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315 SO 28TH STREF". TACOM A, W ASHINGTC'. 38405 120t. 572 7775 da G43
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7 June 21,197 9 k.}
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Mr. Charles E. MacDonald, Chief \\ Transportation Branch % 2 d," \\ Division of Fuel Cycle and Material Safety 9, S U.S. Nuclear Regulatory Commission /s -j Washington, DC 20555 ~
REFERENCE:
Docke* Number 71-909.2.
SUBJECT:
Certificate of Compliance for CNSI Cask 14-195L
Dear Mr. MacDonald:
In response to ycur letter of December ll, 1978, we are providing detailed answers to the noted questions. Should additional information or clarifi. cation be required, please do not hesitate calling. Sincerely yours, NUCLEAR ACKAGIN INC. A M4v Larry
- ansen LJH/dmm Enclosure 7908000:27f' gov
'.l 13303
o Chen-Nuclear Systens, Inc. Model No. CNSI-14-195L Docket No. 71-9097 Question No. 1 The analysis presented in the July 31, 1978 transmittal assumes a Rigid Body Cask Method. For this approach a lid response acceleration was found to be: R = 109.9 g's (Ref. page 7j, July 31, 1978) A second and totally different approach can be taken to calcu-late the response amplitude of the lid. Both methods calculate an estimate of foundation acceleration U for the lid. 20 U l A A u n _ sa.s _ ; _ sa.s, Given the foundation acceleration, U, the acceleration response amplitude, R, of the lid is found as: R = 20 sin (nt/T) where: T = period of lid, sec = 1/f = 1/70.49 T = duration of foundation acceleration pulse, u 1 2h see derivation in July 31, 1978 transmittal =- -, u g C,J'j I 7 J.. c_3 s.
2 The first approach to calculate d, assumed a rigid body cask. The second method shown below assumes an elastic cask. Elastic Cask Method As a moving elastic body (rod) impacts a rigid boundary, an elastic stress wave is developed. This stress wave propagates from the impacted end to the opposite end (where the lid is located) where it is reflected back down the rod. The lid foundation accele-ration, u, will be computed from the magnitude of this stress wave. The magnitude of the stress wave is: where: E = modulus of elasticity ]2gh, impact velocity v= = d2 ( 3 8 6. 4 ) (12_) - >5.3 in/sec c= sonic velocity of material The corresponding force is: F = cA where: A = Area For the cask sidewalls: l p 2 E lbs-sec c Area Force 2 Component Material psi in' (psi) (in ) (lbs) 6 6 Outer Shell Steel 28.5x10 .294/386.4 13938 191.44 2.668x10 6 Inner Shell Steel 13938 30.23 0.421x10 6 Shield Lead 2x10 .410/386.4 4436 402.66 1.736x10 6 TOTAL: 4.875x10 Eo7 ry J/J ssJ
) 3 6 The elastic stress wave force, 4.875x10 lbs, is applied to the foundation of the lid. This is equivalent to a foundation acceleration of:
- 1 u = h = 4100 g's t
Where: N = weight of lid s (38.5') ( 1. 2 5 ) (. 2 8 4 ) + ( 1. 6 2 5 ) (. 410 )j =- 4 1188.89 lbs. = 6 F= 4.875x10 lbs. The associated response acceleration of the lid is found as follows. The pulse duration is: 1 2h 1 2 (1) -6 = 60.79x10 sec. T =- = p ; g 4100 32.2 The lid response acceleration is: R = 20 sin (nT/T) 2(4100) sin y(60.79x10-6)(70.49 110.1 g's = = This is very close to the prediction for the rigid body case. The reason can be seen by examining the above two equations. 2h ' 1 r R = 2psin _p T g_ sinx=x-{(thefirst two terms of the sine expansion) Let: n 2h. aT - 55.19 g Then: x = [_2 C (# ) 1 (^)3 6 s ~ 1 a 3-2 = _a6 110.4 280 g.32 = d ~ e u >) J.N Ji.
4 It can be seen that the maximum response acceleration of the lid cannot exceed 110.4 g's. As the foundation acceleration decreases, so does the response acceleration. To explore this, let us calculate the foundation acceleration that reduces the response acceleration by 5%: 80}8.32 (. 0 5) (110. 4 ) = u 28018.32 lb = 71.24 g,s _(. 05) (110. 4) j This is well below the foundation acceleration value predicted by the rigid body method. This comparison demonstrates why both the rigid body and elastic body methods give nearly identical results. Question No. 2 Both the primary and secondary lid of the cask have been structuraly revised. Please note changes shown in Dwg. 1-295-100, Rev. H. Should the cask impact onto the primary lid end of the cask, loads would be reacted across a number of contact points. From the drawing it can be seen that a heavy 7/3 inch thick steel plate has been added to the secondary lid to bring its surface up to and flush with the outer cask rim. Therefore, the secondary lid will react inertial and payload impact forces in direct compression. Loads across the primary lid will be reacted at the outer rim by the 12 spacer block (i.e., Item 15). In addition, eight F ri 7 r JJJ Jv3
5 e heavy gauge square tubes have been added to the lid to form radial spokes. These tubes are also flush with the top surface of the cask and will provide uniform support. Therefore, inpact forces will be reacted in direct compression across the primary lid and spacers, thus resulting in little or no bending stresses. Question No. 3 As noted above, the secondary lid has been modified to include an additional 7/8 inch thick plate to the outside surface. This plate does not completely cover the central portion of the lid and is, therefore, conservatively neglected from the following analysis. Loads across the secondary lid could result from inertial forces of the lid and payload. This load for the total primary lid was calculated to 400337 lbs. or 86 psi. Therefore: =e-- 2 6
== n tvji W = 86 PSI a A , - = 3/o r From Roark 4th Edition, Case 2, the maximum stress is given as: S= (U/4in) ( 6/t2) m+ (m+1) log a/r- (m-1) r /4a' 2 CG7 7 r J?J J00
6 Where: W= (86 psi) (262)n/4 = 45,600 lb. m= 3 r= 13 a= 18 S= (45660/4n 3) (6/t ) 3+4 log 18/13 -2(13)2/4(18)2] = 1211.6(6/t ) ( 4. 04) 4896(6/t ) = Since: S = Mc/I = M(6/t ) S = 4896 c/I Where: if / //////- ,s / (o ) t g I. fo 2 5 2 i (2') aaaaaa b .17.5 A Y AY Ig 1 .75 2.125 1.593 .0351 2 .125 .062 .008 .875 1.601 7 = 1.601/ 875 = 1.83 in. 4 I + IAD .491 in I = T o c = 7 = 1.83 S= (4896)(1.83)/(.491) S= 18247 psi E i; 7 " ~; 7 J l.) ' J ,1 f = ~, s.
y Margin of Safety (Secondary Lid Bending) Ty/S~1 M.S. =F (36000 psi)/(18247 psi)-1 = = +.97 Stresses calculated on page 6 of 8 of previous submittal provide stresses for the inner edge of the primary lid. Secondary lid bolt loads are: 2 (86 psi) (26 ) (n/4)/18 bolts P = b 2536 lbs/ bolt = By inspection, this is well below the capcity of a standard 3/4" diameter bolt. Bolt pads are secured to the primary lid by a 3/8 in. weld. f = P !A C y b w Where: 2 (1. 5) (n) (. 375) (sin 45 ) = 1.25 in A = y c = 50% weld efficiency 2 ~ f (2536 lbs)/(1.25 in ) (. 50) = y = 4058 psi Margin of Safety (Boss Attachment) M.S. =F /f -1 = 36000 psi /4058 -1 = +7.8 Therefore, it can be concluded that stresses in the secondary lid and inner edge of the primary lid are well within the acceptance criteria. CO7 "EO JiJ JsU 13303 ..x :. .}}