ML20148M999
| ML20148M999 | |
| Person / Time | |
|---|---|
| Site: | 07109092 |
| Issue date: | 11/06/1978 |
| From: | HITTMAN NUCLEAR & DEVELOPMENT CORP. (SUBS. OF HITTMAN |
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| Shared Package | |
| ML20148M997 | List: |
| References | |
| NUDOCS 7811220260 | |
| Download: ML20148M999 (24) | |
Text
'
O PitO. LECT COVElt SilEICT Document Title
" ANALYSIS OF ONE FOOT DROP ACCIDENT" Project Document Number ADDENDUM NO. 1 Rev. 1 for SAFETY ANALYSIS REPORT For The HN-300 SERIES RADWASTE SHIPPING CASK liittman Nuclear & Development Corporation 0100 lied Hranch fload Columbia, Maryland 21045 7811220 M o it ef: Std. Doc.
N/A
- Itev, O
g5 4*-"+.w,,, ja
4
>b 11EVISION 1 OG 11 EV.
D NI'E ENGINEERING Q. A.
PROJ. MGil.
ECN #
{\\fkal ],
- }f 0
6l5/78 jf,
[ } Q f
1 10/Gl78 k), p,,,,.' & f Q, hf' ;
), k),,qt X),LJ.78-15o j
J r
9 4
l l
Prepared by. John Jennings Checked by George Trigilio
TABLE OF CONTENTS Page I.
Structural Analysis 1
i i
A - One Foot Drop Onto Side of Cask 1
i l
B - One Foot Drop Onto Closed End 5
1 l
C - One Foot Drop Onto Long Edge 7
1 l
D - One Foot Drop Onto Door Face 10 i
i(1)
E - One Foot Drop Onto Bottom Edge of Door 12 l
F - One Foot Drop Onto Side Edge of Door 15 I
i i
G - Weld Londing Due to Bolt Tension / Shear Laminated 10 i
l Plate Construction
'3 I
i i
II, Containment Analysis t
17 l
'III.
Shielding Analysis 6
17 APPENDIX A - Figure 1 i
REFERENCES i
i ii
I.
S t.r uc t u al Analysis A.
One Foot. Drop Onto Side of Cask 1.
Impact Energy of Loaded Cask E = Wh W = 43,000 lbs 4 3,000 x 12
=
h = 12 in.
516,000 in - Ib
=
2.
Steel flequired to Absorb Impact V7
=E
&cr = 50,000 lbs/in er 516,000 V
= 10,000
(#"I'
)
= 10.32 in 3.
Stopping Distance The Kinetic energy is absorbed by the deformation of 48 each studs, embedded in the 2" thick top and bottom plates.
_l1 n
_O F1 F1 F1 nn
~
A I'd J
s y
i
\\4S h" 47" i
a 4
4 i
i l
1 of 17 i
I I
n" I
g g"
~
r
(~,',._--
-j.q qy... g y,,-
n g., N ' N-r,
., y \\
f',
, /,./ '
f s.-
( --
'\\
,/IN N \\
/
/
/
Q
/ f\\
' r l/,,, ',/. !.
/,
O
/ / /Nx
,/../.
x
,/
.<r y.N/
./
. /
cu-L
' NN
's s
i 7
' -/'
- i x s l
O
,/j' /. ', // ',/ ',-
N
/
/
, 'N i j / - - -
- --e - y t - (-- -* - %-
--Y-a
- ../
N O
l
's
's v
,'x(-
NNN 1 --
. N b
i' N (N N N,\\ \\
'g l x
p A--
e y \\ a m._.s_
n s.c.s..s._s '..a _ u.
N N 4 r. - -
9
[/ ss's
- .r
',3
'..N Typical Stud Installation m
i, s
- 1.. _ v... __.
s
.PT_.
,,4 w..
\\\\
~
_ _r..
lc' l
(M__.
4 n
l r.-)..
{/ h v
t=A
_ 10.32
~ 48 x 3.1416 x (0.5)"
= 0.274 in 4.
Applied g Load 2h E ~_ T (ref. 1)
~_ 2 x 12 0.274 i
I
= 87.6 5.
Force Applied to 24" Thick Side Plate Weight of 2.1" Thick Plate 3,802
=
Weight of Cask Contents 8,000
=
11',802 l
F = Wg l
l
- 11,802 x 87.6
= 1,033.855 lbs l
i I
{
(
l 2 of 17
6.
Strength of 23" Thick Side Plate Welds Neglecting support by the 1" thick plate beneath:
e = joint efficiency = 0.85 a = 0.5 in.
2
/ = 2(144+40) = 368 in.
S
= 0.6 x 36,000 = 21,600 lbs/in y
F
= e (Vf/2 ) afS y
y
= 0.85 x 0.707 x 0.5 x 308 x 21,000
= 2,388,416 lbs
,388,410 Margin of Safety =
_y 1,033,855 1.31
=
7.
Force Applied to Weld Jointing 13" Thick to 2" Thick Top and Bottom Plates.
Weight of Cask & Contents
= 4 3,000 lbs Weight of 24" Thick Plates 10,072 lbs
=
32,928 lbs F = Wg
= 32,928 x 87.6
= 2,884,492 lbs l
8.
Strength of Welds Joining 13" Thich to 2" Thick Top and Hottom Plates F
= 2 x 2 x 0.85 x 0.75 x 146 x 21,600 y
= 8,041,680 lbs 8,041,080 Margin of Safety
=
- 1 2.884,492 1.79 9.
Maximum Stress In 23" Thick Side Plate Uniformly distributed load, one edge free, remaining edges fixed.
See Reference No.
2, Table 26, Case 10a-3 o f 17
s
,'?/ h';W.////./ '
/ ' ':D b
a/b = 40.5/144
= 0.2812
[ 40.5 4
1,033,855-
= 177.3 lbs/in 2
k = 144 x 40.5 f,e.-,.Yb.-.
J B
= 0.0270 t = 2.25 i
~BlI D Max 0-
=
t2
, -0.027G x 177.3 x (144)2 (2.25)2
= -20,044 This is a compressive stress on the bottom of the plate at the center line of the fixed end ("a" side).
The support i
I contributed by the 1" thick plate below has been ignored.
1 1
l Maximum Shear Theory of Failure Combined Shear j
and Axial Stress O'
3G,000 >
(0{+40")
o o
y s
i i
Where O't = 20,044
- /in Ref. A-9 I
I o
and O'
= 21,600
= 9,351 #/in" Ref. A-5
!(1) 8 1 + 1T31 j
O'y = 36,000
> [20.0442 + (4 x 9.3512)) i
=
x 1000 1
l
>= (401.76
+ 349.7'6)3 x 1000 I
i 30,000 27,414 /. OK t
i i
i Margin of Safety a
i j
36,000 1
-1
= 0*313 27,414 i
i l
i l
l l
1 4 of 17
o
' B.
One Foot Drop Onto Closed End 1.
Impact Energy of Loaded Cask E = 510,000 in-lbs (see A-1 above) 2.
Steel Deformation Required to Absorb Impact i
V = 10.32 in3 (see A-2, above) 3.
Stopping Distance b\\ F "
i
-- y-s O
O O
O O
i Total No. Studs l
l(1)
=g O
O
= 20 s
-y 1
O o
o O
O y
~.-..
f w,v'f r'/,
s
,/
\\
= f.
_ l-
/n'... _..__. A '
t i
i
- 10.32 t
~ e I
F.' '
l 2G x 3.1416 x (0.5)"y
..._ __y_
q a_
l(1) i
= 0.505 in (gpT7_%_
i c.
1 I
..--.~W..
4.
Applied g Load Kl g"
l
-9 g=
i
_ 2 x 12 i
t 0.505 l(1)
= 47.5 5 of 17
s 5.
Force Applied to 2:l" Thich End Plate Weight of Plate 1,355 lbs
=
8,000 lbs j
Weight of Cask Contents
=
Total 9,355 lbs
=
I F = 9,355 x 47.5 (1) l i
444,363 lbs 1
=
6.
Strength of 24" Thick End Plate Welds Neglecting the support contributed by the 1" thick plate beneath:
a = 0.5 in
/= 2(52.5 + 40.5) = 186 in F
= 0.85 x 0.707 x 0.5 x 186 x 21,600 y
= 1.207,188 lbs l
Margin of Safety
- 1,207,188
- 1 444,363 k1) 1.72
=
.. ~ - -
7.
Maximum Stress in 24" Thick End Plate l
We neglect the support contributed by the 1" thick plate, l
beneath.
The plate is uniformly loaded and fixed at all edges.
See reference No.
2, Table 26, Case Sa, 52.5" a/b 52.5
,1,3g 40.5 405 b
l i
B1
= 0.4095 i
t 2.25
=
,(1)
O.
B2
= 0.1944 q= 210 0 lbs/in2
% = 0.0207 6 of 17
~
2
.- B,9 b Max 0-t2
_ -0.4095 x 210.0 x (40.5)
(2.25)2 o
27,8G2 lbs/in'
=
i i
I(1) 3G,000 Margin of Safety = 27,862
-1 8
- 0.0292 8.
Maximum Shear Theory of Failure Combined Shear and Axial St' cess O'
= 36,000 d
(O2 + 40
)
5 Y
t s
i l
Where O'
= 27,862 Ref. B-7 I
and O'
= 21,G00 l
2.72 s
= 7,941 Ref. B-6 d
(27,862)2 = 4(7.941) I $ x 1000 O'
= 36,000 l
i(1) y i
g
.j (776.29 + 252.24) x 1000
=
l l
36,000 > 32,071 OK l
Margin of Safety =
36,000 -1
= 0.123 l
32,071 i.
I I
C.
ONE FOOT DROP ONTO LONG EDGE 1.
Stopping Distance Tan &
- 47 k '/ : *x 62.5
\\
\\
N Yj
' N,
= 0.7520 j
V
's s
$3I N \\,,,
s 0- = 30. 9 4 s
Ng
,N N
'N
'N
/ = 148 in
'1 3
V
= 10.32 in (see A-2, above) 7 of 17
s
/
- 53. 0$4 g
$ !o.9 4'
/
/
u 36.94 h +h (t(t AreaofTriangle=t(t(t cot 53.06 cot 2
2
= 1.32994 t
+ 0.75191 t 8
2
= 2.08185 t 2
2 Volume of Prism
= 2.08185 t 2
o 308/11 t~
= 154.06
=
Z V = 10.32 = 154.06 t2 0.258 in t =
2.
Applied g Load E"
2 x 12 0.258 92.89
=
3.
Component g Load To Side Plate gs = g cos<>
92.89 x 0. 79926
=
74.24
=
4.
Force Applied To Side Plate F = WgS W= 11,802 lbs 11,802 x 74.24 (see A-5, above)
=
876,233
=
2,388,416 lbs Fy
=
(see A-G above) 2,388,416 Margin of Safety
=
~
876.233 1.73
=
8 o f 17
t 5.
Maximum Stress In Side Plate 2
3-
= 115. 90 lbs/in 52.5 x 144 See-A-9 above for remaining parareters y,
y _ -0.048 x 115.90 x (144)2 (2.25)2 2
=- 22,787 lbs/in 3G,000 Margin of Safety
=
-1 22,787
= 0 58 6.
Component g Load to Bottom Plate g3 = g S I N 13-(see C-2, above)
= 92.89 x 0.60098
= 55.83
)
7.
Force Applied To Bottom Plate i
Weight of lt" thick plate
= 2,678 Weight of Cask Contents
= 8,000 10, G78 F = Wgb 10,678 x 55.83
=
596,100 lbs
=
8.
Strength of Bottom Plate Welds Fy = 2,388,416 (see A-G above)
,388,416 Margin of Safety _
_ y 596,100
= 3.01 9 of 17
a 9.
' Maximum Stress In Bottom Plate Reference No. 1 Table 26 Case 10,a (see A-9, above)
-596,100 o
q=-
= 78.85 lbs/in" 144 x.52.5 14 tt a/b = 0.364G S2.5 a
B1 = 0.048 t=
2.00 b
o
-B1 4 b" Max O'=
2 t
o
_ - 0.048 x 78.85 x (144)"
(2.00) 2
=-19,620 lb/in 30,000 Margin of Safety
_y
=
27,695
= 0.83 D.
ONE FOOT DROP ONTO DOOR FACE 1.
Stopping Distance Volume of Steel rgquired to absorb impace = 10.32 in (see A-2 above).
See Figure 1 Assume that wolds securing the hinges to the cash body shear without absorbing energy.
The hinge " hooks" will remain
,ff,/,
,,,, 7 captured under the cask, because the impact force passed between the centroid of the " hook" and the cask body.
The depth of deformation of each " hook", dividing energy absorption equally between the four, is 0.445 inch, i
2.
Applied g Load E"
, 2 x 12 0.445 10 of 17
53.9
=
3.
Force Applied to Doors Assume the weight of the cask contents is equally divided between the two doors.
Weight of Door
= 1350 Weight of Cask Contents
= 4000 5,350 F= Wg
= 5,350 x 53.9
= 288, 365 lbs 4.
Closure Bolt Load b
Assume the door is a rigid 7g p4 3
body.
Also assume the bolts are strained in proporation to r3 E2 their distance from the hinge Fo Fi t
and to the number of bolts in t
I the reaction plane.
F7 would bez l
the highest bolt loading.
Moment applied to door = 288,365 x 15.5 in.
= 4,469,658 lb-in Summation of Forces:
- "7 Fo+
= 288,365 28.75 Summat. ion of Moments:
b nF7
= 4,409,658 28.75 l
1 11 of 17 i
X Force (inches) n fxn = 213.5 o
F1 1,75 2
F 6.75 2
2 F3 10.75 2
2
{x n =
4402.875 F4 15.25 2
F 19.75 2
5 F6 24,25 2
_ p.
F 28.75 2
7 F7 _ 28.75 x 4,469,678
= 29,186 lbs 4402'.875 h!ax 0t
= 36,945 lbs/in
=
=
0.790 Minimum Tensile Strength of Bolt material is 105,000 lbs/in (Ref. 3) 05,000 Margin of Safety
_y
=
36,944
= 1.84 14.5 x 29,272 Fo = 288,365 -
28.75 69,970 lbs
=
E.
ONE FOOT DROP ONTO BOTTOM EDGE OF DOOR I
I 1.
Stopping Distance
.0 Tan e
=
151.6875
,s p
'\\
%s 6'
= 0.30984 l
s s.q G = 17.220
$ gy \\x
\\
'N
-g \\)
)
N
} = Gl.5 in N
\\
i N/
/
12 of 17
I e
- 72. B- __ 4 f.22 -
17 s
/k t
t7 (t cot 17.22)
,/
Area =
g(t cot 72.78) +
/
t 9
= 1.76821 t~
..___._.__._..___.___._.V 2
Volume = 61.5 x 1.76821 t l
2 3
= 108.75 t in 10.32
= 108.75 t (see A-2 above) t
= 0.308 in.
2.
Applied g Load g=b t
, 2 x 12 0.308
= 77.9 3.
Normal Force Applied to Door Assume the weight of the cask contents is evenly divided between the two doors.
Weight applied to each door is 5,350 lbs.
(see D-3 above) t Wg cos 9-F
=
5,350 x 77.9 x 0.95518
=
= 398, 086 lbs 4.
Roar Force Applied to Door Wg SIN 6 F
=
y 5,350 x 77.9 x 0.29604
=
123,379 lbs
=
5.
Bolt Reactions Making the same assumptions as in D-4 above:
13 of 17
' Summation of Forces:
" 7
= 398.084 F
o + 8. 7 5 7
398.,086 Fo+
=
28.75 F
= 398,084 -7.4609 F7 o
Summation of Moments:
j[x nF7
= 398,086 x 15.5 28.75 398,086 x 15.5 x 28.75 y
4402.875
= 40,291 lbs 1
Max O
=
t
_ 40,291 0,790 o
51,002 lbs/in"
=
O'y =fv (see E-4 above)
, 123,379 1 :c 0.790 o
= 11,155 lbs/in" Combined Stress = ( O'tE+4 T 2)6 y
= ((( 51,002)2 +
(11,155)2f1 2
52,P08 lbs/in Margin of Safety
- 1 (ref. 3) 52,208 1.011
=
14 of 37
i F.
ONE FOOT DROP ONTO SIDE EDGE OF DOOR l
1.
Stopping Distance D
61'5 g
,g'4>.
Tan 6. =
'J 15' 6875 8
N
'N gs 14
=
\\
')
\\gs
&=,
u
\\
'/
See D-1 abu. and Figure 1 7
,r,c 7,
, 7 y,,r y--- y The cask will impact on two of the door hinges.
The welds securing the hinges to the cask body are assumed to fail without absorbing any energy.
The two hinge " hooks" impacted will be captured beneath the cask.
The depth of deformation of each " hook", dividing the energy equally between them, is 0.714 inch.
2.
Applied g Load
.g = Sb t
_ 2 x 12 0.714 33.6
=
Since this g loading is much less than any other calculated above, the cask will withstand this drop without failure.
l
)
l l
l 15 of 17
G.
WELD LOADING DUE TO BOLT TENSION /SIIEAR AND LAMINATED PLATE CONSTRUCTION i
1.
Weld Loading due to Tension i
i i
1
- k i
eL9so l
I i
I I
l Ak l
\\E 87._,950 l
=
l e,2.,9 s o #
l 1-1/8 diameter A-325 Bolt - Minimum Tensile Strength l
I i
= 82,950 lbs.
i i
l Separating Force = 82,950 sin 30 cos 30
= 35,918 lbs.
l' Weld Length Between Bolts = (4.50 - 1.25) = 3.25" 1
I i
Assuming 1/8 Machine Allowance on 1/2" groove weld.
l l
F 3.25 (.500
.125) 36,000 = 43,875
=
1 l
Margin of Safety = 43,875 - 1 = 0.222 l
35,D18 l
2.
-Weld Loading due to Shear 1
I Ref. E-4 I
I l
Shear Force Applied to Door = 123,379 i
Shear Force Per Bolt = 123,379-8,813 lbs.
14 I
i Weld Strength per Bolt = 43,875 lbs.
i l
l Margin of Safety = 43,875 - 1 = 3.98 i
8,813 16 of 17
II.
CONTAINiiENT ANALYSIS j'
Positive containment is provided by the cask internal disposable j
containers which have leak tight seals and closures.
t l(1)
III.
SIIIELDING ANALYSIS All of the deformations to shielding calculated above are small.
No deformation extends into the effective shield thickness used in the initial dose calculations.
17 of 17
.- - - - nn~..~.
. ~... - -. - - - ~. ~. -.. - - - ~.. - ~.. - ~ -
n.-----.~.~
. - -. - - - - -.... -.,. -. - - ~,. - -. - - -.._.. - -... - ~
n re -
9 I
n k
d I
I i
t l
E a
i i
I '
e i
l 1-t i
1 i
I l,
f APPENDIX A
l a
l l
i
(
i l
i i
l 3
1 P
I r
4 i
)
l I
l l
i I
I i
I i
t.
a 2
1 A
i i
(
i a
l,-.-
x
--x---.,
..-,_.na.
..,_,e-.w--w-
-..,.,-m,nn.,
,m-,~,---nm.,---<
FIGURE 1 Reference Drawing C-001-4-011 l
g(('N'4tg//D77K'" " k 'N,\\
///
!?
"~
+
\\
Nw\\
-./
f'
/N
'N N
\\
,N y\\
'p '
\\
,s
'd
\\
x 6 y'*
\\
/
N )
\\
\\
b-
%O__.
12341_'_.g y
e/
h, et Y
=c
$L
(
Cx Of
., gA
\\
k i
b>
N U
\\sm-
\\
v y
-/
n.0w Sheet 1 of 2
Figure 1 -' Sheet 2
+ c 9
N Area of Shaded Segment f
2
=R (9 - 1/2 SIN 2 9) e s /
Where:
COS 9 = R-t =1-Hinge is 3 inches thick.
2 Volume = LR (9 - 1/2 SIN 29)
= 18.75(9 - 1/2 SIN 29)
For "D",
V = 10.32 4
0.1376 = (9 - 1/2 SIN 29) t (9 - 1/2 SIN 29) 0.200 0.0422 0.400 0.1177 0.440 0.1354 0.445 0.1377 For "F"
V = 10.32 2
0.2752 = (9 - 1/2 SIN 29) t (9 - 1/2 SIN 2G) 0.720 0.2785 0.715 0.278C 0.714 0.2751 l
Sheet 2 of 2
M l
References 1.
Cask Designers Guide 2.
Formulas For Stress & Strain, Fifth Edison Raymond J.
Roark and Warren C.
Young.
3.
ASME Boiler and Pressure Vessel Code, Article SA-325
)
m.
O f
I n.=mme g
h m
l 1
beams m
humus
{
O. n -ouo
/
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