ML19305E067
| ML19305E067 | |
| Person / Time | |
|---|---|
| Site: | 07106574 |
| Issue date: | 03/20/1980 |
| From: | Mallory HITTMAN NUCLEAR & DEVELOPMENT CORP. (SUBS. OF HITTMAN |
| To: | Macdonald C NRC OFFICE OF NUCLEAR MATERIAL SAFETY & SAFEGUARDS (NMSS) |
| References | |
| 15815, NUDOCS 8004220274 | |
| Download: ML19305E067 (19) | |
Text
l
}g
/l fC 7/~ dI7I u ea Development Corpor:ti:n Refer to:
9190 Red Branch Road Columbia. Maryland 21045 IIN-E167 301/730-7800 March 20, 1980 Mr. Charles MacDonald Transportation Branch U.S. Nuc1 car Regulatory Commission 9217 Maryland National Bank Bldg.
Washington, D.C. 20555 Subj ect:
Submittal of Requested Information Extension of Expiration Date
Reference:
U.S. NRC letter RCTC:R110 71-6574 Dated December 21, 1979 Certificate of Compliance 6574, Rev. 4 -
I Model No. IIN-200 Shipping Package
Dear Mr. MacDonald:
By enclosure to the referenced letter, information was requested relative to the " Safety Analysis Report for the llN-200 Radwaste Shipping Cask".
Attachment A contains the requested information. The further detail con-tained in Attachments B, C, and D would be used to revise the current Safety Analysis Report.
It is requested that the expiration date on the Certificate of Compliance be extended pending your review of the requested information.
Attached is our check in the amount of $150 to cover the fee for processing this amendment.
It is suggested that the expiration date be extended to September 30, 1980 to allow time for your review and the subsequent re-vision and approval of the Safety Analysis Report.
It is requested that the amendment extending the expiration date of the current certificate of compliance be issued prior to March 31, 1980.
If you have any questions or desire further information please contact me.
Very truly yours, iar es W.
, ory Vice Pres dent, Engineering 1h Attachments A, B, C, 6 D Check # 07907 8 004 e 20 g76 ggff L
8
- O e
P P
- ATTACINENT A 9
s b
.4 w
~
. Radioactive hbterials Package
' Certificate of Compliance No. 6574
!!NDC Model No. IIN-200 Submittal of Requested Information The following information is submi+ted in response to the U.S. Nuclear Regulatory Commission letter FCTC:.d10, 71-6574 dated December 21, 1979.
This information has been prepared in the form of revised sheets to the Safety Analysis Report, Revision 2, dated December 3, 1979.
1.
The g-loads under top-corner impact conditions have been reevaluated to consider the peak conditions.
Attachment B contains replacement pages for pages 72, 73 and 74 a.
of the current Safety Analysis Report.
b.
The analysis of the crushing of the foam under corner impact has i
been revised to climinate the previous assumption that the energy dissipated by internal and external crushing of the foam is equally divided.
l c.
The revised analysis is based on externai crushing occurring until the maximum external force is attained (maximum external deformation of the foam).
This would also be the maximum force of the cask on the internal foam.
This maximum force is used to calculate the maximum g-load 59.63 G, (Cask Inertia Loading, F.1(c)), which is used later to calculate the stress in the bolts,
( F.1(f) ).
d.
Using the revised calculations of internal and external crushing, the amount of foam crushed an,d the remaining uncrushed foam are calculated, (F.1(c)).
2.
The allowable bolt stress has been recalculated based on the maximum g-loads. The maximum stress on the bolts is the minimum yield strength and the maximum bolt elongation is considerably less than the gasket draw.
Attachment C contains replacement pages for pages 74, 75 and 76 a.
of the current Safety Analysis Report.
b.
The analytical model has been corrected to show a singic bolt at position "L 12".
The peak g-load of 59.63 g's has been substituted for the 30 g's c.
j.
previously used.
i d.
The calculated stress of 75,060 psi in the outermost bolt is a l
factor of 1.40 less than yield of bolts of type based on proof l
tests of a number of bolts.
The clongation of the outer most bolt is about a factor of 40 e.
less than the compression of the gasket, f.
This analysis assumes the cover is totally rigid.
If yield of the cover were considered the maximum stress and resulting elon-gation would be less.
3.
The bending of the cask lid assembly under top end impact has been reevaluated using a different model. The outer plate with its rein-forcing is more rigid than the other members and is considered to carry the full load due to the decelleration of the contents and cover including the lead shiciding. The foam at the center is not considered to be a rigid support and only contributes a reaction force which reduces the end moment. The model is not dependent upon shear between
- he two plates.
Attachment D contains six pages which will replace pages 92, 93 a.
and 94, (through paragraph (c)), of the current Safety Analysis Report. When revised, the balance of the pages will be renumbered.
b.
Section F.4(a) has been amended to include a calculation of the "g load" associated with the bending of the cask lid assembly.
c.
The model considers the physical properties of a symetrical segment of the outer reinforced plate or one-thirty sixth of the cover.
The moments of inertia are calculated for various sections, d.
The maximum moment at the load point on the segment is calculated based on limiting the end reaction from the internal foam to the product of the crushed area and the crushing resistance.
c.
The end moment is also calculated using the limited value of the interior reaction with the foam.
f.
The stresses are calculated based on the respective section modulus and are shown to be nominal.
g.
Calculations of the stresses in the welds, plates and Icad show them to be acceptable.
p
f e
ATTACINENT B s
(b). Corner Impact (External crushing)
The "G" loading is dependent upon the maximum crushed area of the foam and the volume of the foam which has been crushed by the corner impact. The interrelation of the crushed area and the "Ungula" volume is illustrated below.
Cyl. C l
/['/,c/y, l
/ Base
'/.
o Att l
11 r
/Of
,' Base Ungula I
I C
y I
volum t
Crushed Ungula R=b
/
/
R Cos 0 " "
)h Gy=
R - (x cos0)2 2
J also e = x cos 01=R-tan 01
" f - (xy + ab sin -
A
=
1
- C u
2cos0 - (xy + cose sin A
=
,R F
/
R-C
,y(R - [3 h-Obtained by trial and G
V 11 2
=
2
-R C sin R.
error calculation of energy absorption as a function of From A and V above, when II = 17.77 in.,
, deflection.
Verification u
u of total crushed volume is shown in F.l(c) and F.1(d)
V = 6264 in with clearance on crushing 3
u of foam.
2 G
A = 1218 in u 3-20-80
s (c)
Corner Impact (Internal Crushing)
Deformed foam is illustrated below by shaded area.
b Limited deformation due o b to relatively large area.
x g e
-r %,
A, 0.7071 o
Vs GW
't 2.4 Y
j
/
Plane of Au o 13" f
Vu 14.96 12.56 h
V 17.7 d
2 F = 2350 #/in x Au in2, gg y R
A = Deformed internal area = 13.0 x dia.
2 A
= 13 x 66.25 2
A
= 861.25 in y
6 GW = UA9 2350 x 861.25
,307# maximum force on cask
=
0.707
- 0.7071 also GW = UA = 2350 x 1218 = 2,862,,300#
maximum force on cask u
(d) Total Crushed Volume KE = DW = 30 x 12 x 48,000 = 17,280,000 in#
KE = U(Vu + V ) = 17,280,000 s
Vu+V 17,280,000
=
s
= 7353 in 2350 Vs = 7353 - Vu = 7353 - 6264 = 1089 in i
= 1.26 21 3-20-80
(c)
Cask Inertia Loading
59.63 G maximum possible "g" load on cask G
O (f) Cask Lid Loading (Bolt Stress)
Impact at the upper corner of the cask will result in the cask contents pushing against the cask lid.
The contents of the cask, it should be noted are positioned to limit actual movement to one (1) inch or less. The loading on the cask lid is realized in the bolts. The bolt stress is therefore equivalent to the inertia load of the contents and the inertia force of the lid itself. The following maximum weights of these constituents have been J
conservatively estimated.
Content wt. = 10,675 Cask lid wt. = 6000#
Total wt.
w = 16,675 I
] W cos$
U
\\ F10
\\
24 N F6 bolts
//////
\\
L2
.4 L6 L8 10,12 Fig. V-16 Impact loading on cask lid closure bolts The maximum loaded bolt is that one furthest from the point of impact. The force acting on this bolt is designated at:
F
= Max. bolt force 12 3-20-80
e t
ATTACilh!ENT C o
e 4
(c)
Cask Inertia Loading G=
= 59.63 G maximum possibic "g" load on cask 0
(f) Cask Lid Loading (Bolt Stress)
Impact at the upper corner of the cask will result in the cask contents pushing against the cask lid. The contents of the cask, it should be noted are positioned to limit actual movement to one (1) inch or less. The loading on the cask lid is realized in the bolts.
The bolt stress is therefore equivalent to the inertia load of the contents and the inertia force of the lid itself. The following maximum weights of these constituents have been conservatively estimated.
Content wt. = 10,675 Cask lid wt. = 6000#
Total wt.
w = 16,675 F12 T N cos$
\\ F10 g
8 N F6
/
bolts
\\F4
/
//////
\\
v L2
.4 f.6 b
10 12 Fig. V-16 Impact loading on cask lid closure bolts The maximum loaded bolt is that one furthest from the point of impact. The force acting on this bolt is designated at:
F
= Max. Bolt force 12 3-20-80
Taking the summation of moments about point "o".
Bolt load 0 P
= Maximum 12 Bolt load 0 L = L x P (Based on deflection assuming lid rigid) 2R 12 i
g!
Bolt moment = L x xP 12 6
- 7. Bolt moments 2 + L *+ L *.... L $)
2(Li
=
2 3 -
i P12 + 2RP12 2R 2
L*+L*....L 2
- 2). R (Li +
6 RGW cos$
=
2 3
ii p
R R-12 i
2 2
l 6P R GW cos &
- 17.000 R' 12
_ R GW cos &
=
(L'+L'+La'....Li l') + R" i
2 1
0 45 P
59.63 6 675 x 0.7071
=
12 41,358#
=
2 6 f
,060 O n t2*
5
=
1 B
where:
l 2
2 L = R(1-sin 75 ) = R x 0.034 L
= R x 0.001
[
i i
2 2
i L2 = R(1-sin 60 ) = R x 0.134 L
= R x 0.018 2
L3 = R(1-sin 45 ) = R x 0.293 L3 = R x 0.086 2
2 z = R x 0.250 2
L% = R(1-sin 30 ) = R x 0.500 Lu 2
2 L5 = R(1-sin 15 ) = R x 0.741 Ls = R x 0.549 2
2 t
j Ls = R(1-sin 0 )
= R x 1.000 Ls = R x 1.000 2
L7 = R(1+ sin 15 ) = R x 1.259 L7 = R x 1.585 2
2 Le = R(1+ sin 30 )'= R x 1.500 Le = R x 2.250 2
2 L9 = R(1+ sin 45 ) = R x 1.707 L
= R x 2.914 9
i L o= R(1+ sin 60 ) = R x 1.866 L o, g2x 3.482 2
i i
2, g2x 3.865 Lit = R(1+ sin 75 ) = R x 1.966 Lit (L$+ Lj+ L.... L j) 2 2R x 16.00
=
2 R
, g2xf 3-20-80 (L[+ Lj+ L 2,,,,L2 3) + R2 = R2x 17,000 l
75
Factor of Safety (A320 Grd L7 or Grd L43)
Y(
l P
= 1.40 on yield (1.67 on Ult.)
s
=
5 t12 P2 41,358 x t
P Bolt clongation e=
12 7g- = 0.7854 x 29 x 10*
e = 0.0018 1 = 0.0018 x 1.75
= 0.0032 inches Note that gasket draw is approximately 0.125", therefore will remain intact.
(g)
Bolt Spacing The center-to-center bolt spacing is:
_Dg (3.14)(61.25 in) = 8.02 in 3
actual N
24 bolts The minimum suggested bolt spacing is m = gasket factor = 1.0 t = flange thickness = 1.5 in a = major diameter of bolt = 1.0 in 1)
Cask designers guide S.
min 6t 6 x 1.5 S.
+ 2a
+ 2 = 8 in.
=
=
min m+0.5 1.0+0.5 llence, the bolt spacing satisfies the minimum suggested spacing and is acceptable. 3-20-80 t
6 e
J ATTACllMENT D C
r l
J
t.
The compressive strength of the foam is equal to 2350 psi-therefore:
i y + V )2350
[
K.E. = (V 2
K.E.
n(R
- I13 )X + nR1 '.
350
=
2 K.E. = 2350 n X(R a -R "1 )
j 3
i X
= K.E./(2350)(n)(R
- Il
+R
)
2 3
1 2
2 2
X
= 17,280,000/(2350)( n)( 32.125
- 26.55
+ 11.5 )
j l
X
= - 5. 00 in
- 4 Based on the foam compaction graph at 70% compaction, the_
's
. foam compressive strength begins to drastically inscrease.
t i
5.09/8.00 0.G4 <.7
=
4 4
f Margin of safety = ( 0.70/0.G4 - 1)' = 0.109 j
Free Fall Distance 30 x 12 70'.73 g,
=
Crush Distance 5.09 i
\\
O i
i h
I i
e I
i f
-92.
3-20-80 w.
p p-
+.e---
(b)
Ilending of Cask Lid Assembly Since the outer plate is much more rigid than the inner plate and/or the lead shieli.ing, this analysis considers that the outer plate carries the full load due to an.celeration of the flat drop lid.
The p? ate is analyzed as a section of the lid having an end moment at the outer edge, simple support from the foam at the outer and inner edges, with the uniform load concentrated at the centroid of the section as illustrated below.
P = Contents + Load 36
?
P o
eat i
l
[
w w w w w wsu
(}of cask
_y
_ g, Mo h
a R.
I foam 0
foam
)
(
i t
O gb '
g i
15.05 = t
/
11 N
r l
's Y
Y Ib l '2 2'3
_ 3'2 __
r la 1 in.
i i 3-20-80 l
L l
R. is developed from crushing of the inner foam as follows:
1 2
UA; " 2350 x 0.7854 x 11.5 i
36 36 R. = 6780#
1 0.73 (10675 + 6000)
R -
- 6780
- 6780 o
36 36
= 32762 - 6780 = 25982#
GW Note P = g = 32,762#
Geometrical Properties ( a in radians)
Y
=R 1-1 1+
la 3a R
2-t j
R
.J sin 5
.05 1
Y
= 26.55 1
3 x 0.0873 1
26.55 2-15.05 Y
= 6.583 in la 26.55j
= 15.05 - 6.583 = 8.47 in Y 1b "
~
1a Chord I.ength at Centroid Chord = 2(ylb + 11.5) tan 5 = 3.49 in.
Properties of outer plate 0 P 1.
At Centroid Point of bl max.
h B
j
- 3 1/2 1.19 o
x 2 3/4._
- 1 1.06 4
4 A
I cif 1o n
1/L
' 'I 9 3-20-80 r
01
bd
~ ~,
Item Area YA y A 12 o
A 3.49 1.25 3.92 0.29 B
3.50 9.63 4.96 3.,57 4
6.99 10.88 8.88
+
3.86 = 12.74 in L
10.88
= 1.56 in 6.99 S = g = 12.74 I
. 3 4.33 in
=
g4 Properties of Outer Plate @ M
- i Flange Width = 2 x 26.55 sin 5 = 4.63 inches
?
lI
~
i
- 3 1/2 h
- 1. 2 2 --
2 3/4 -
- 1. 03 -;
1 1
I-w t-i.,
1/2 4.63 L
Item Area YA Y 'A bd'3/12
?
o A
4.63 2.32 4.91 0.39 B
3.50 9.63 5.21 3.57 4
7.83 11.95 10.12 +
3.96 14.08 in
=I
=
X 11.95 14.08
. 3 I
1.53 S
4.74 in =
=
7.83 2.97 L
8.47 6.58 Load 6 Movement Diagram 57,427 # in.
P
[ "o i
p n
U 11 11 1
o 15.05 i
T 113,535 # in.
i From E M
=0 M = 6.58P - 15.05 11.
l o
0 1
= (6.58 x 32,762) - (15.50 x 6780) i 113,535 # in.
M
=M
=
max o
3-20-80 -
Verification 13y Calculation bl P
bl = 8.47 R. = 8.47 x 6780 p
i bl = 57427 < 113,535 = big Stress in outer plate 0 P f
= 13.263 #/in b
4 33 Stress in outer plate 9 Fig
= 23,953 #/in f
b
.N
,000 F.S. = 23,953 2.30
=
Weld Verification @ bi g
4.63 in.
N
.4 o
'/
3, j
l i
f 1.5 x f A = bl 6
f 16,347 #/in t"
5A "1
=
x 4.63 0
F.S. =
= 3.67 Weld 0 intersection of 1" plates Q = 1.03 x 4.63 = 4. 77
= $ = 25,982
[-
,g,77 s
I 14.08 G
F = 8800 #
3
=1, 5 #/in Weld stress f s
2x 071 x.5 F. S. = -@12,4 45
,000 4.82
= 3 20_go
Approximate Stress Levels in Plates of the Lid It can be shown that for simple beams, loaded in the same manner, that the stress is proportional to the product of the deflection,nodulus j
of elasticity and depth of the beam in question.
As simple beams with a concentrated load from Pab I
M
=f
=
max f
b -C Pab (a+2b) $3a(a+2b) 1_
6 A
=
max
/
27 EI 6 all beams loaded in the same manner i
I I
A f
max bC
- EI fh ExC i
A
= Kf 6
max b Exc Amax
.i Note K is equal for all b cams in the lid and A,,
is identical for l
all beams in the lid. Therefore:
6 fha s Eo x Co 29 x 10 2.97 f
Ei x.Ci 29 x 10 0.75 bi fh i
6 f
b 97
%_0.25 fbo bi i.
6 fo f
_ Eo xC 29 x 10 x 2.97 b
1 n
bLD "LD xC 2 x 10 x 1.25 gp
- I
.03 f bLD 2 x 2.(
b bo-l 1
5 f
i I
3-20-80 97
/;5rlil!i?
_