ML20042C252
| ML20042C252 | |
| Person / Time | |
|---|---|
| Site: | 07109074 |
| Issue date: | 02/02/1982 |
| From: | ANEFCO, INC. |
| To: | |
| Shared Package | |
| ML20042C250 | List: |
| References | |
| 20406, NUDOCS 8203310111 | |
| Download: ML20042C252 (16) | |
Text
.
?
/*
CONSOLIDATED REPORT ANEFCO, INC.
AP-100 CASK USA /9074/A Dated February 2, 1982 4
4
.j i
3 4
i i
m 6m@A yh3 fl y$h g
,M J f=23 j= y
{_h o
=.
un
.aw. a e
.. ~ _
General Information By application dated October 26, 1972, ANEFCO, Incorporated requested approval for shipment of low. specific activity radio-active material in the Model AP-100 packaging.
Based on the statements and representations contained in the application, as supplemented and compiled oith this consolidated report the thermal and structural requirements of 10 CFR part 71 are analyzed to demonstrate adequacy for normal conditions of transport.
The Model AP-100 was licensed by the NRC under docket number 71-9074.
The package identification number as approved is USA /9074/A.
PACKAGING DESCRIPTION A steel encased, lead shielding cask for low specific activity non-fissile radioactive material.
The cask is a right circular cylinder 92-1/4 inches high by 64 inches in diameter.
The cask consists of two 3/8 inch thick cylindrical steel shells surrounding a 1-1/4 inch thick lead shield.
A 64 inch diameter by 3/8 inch thick steel basc plate is welded to the cylindrical steel shells using full penetration welds.
A 60 inch diameter by 3/8 inch thick steel plate is separated from the base plate by a 1 1/4 inch thick circular lead plate, and welded to the inner steel shell using full penetration welds.
'The cask cover is a welded steel construction which encases a 1/2 inch thick lead plate.
The 64 inch diameter by 1 1/2 inch thick cover has a 60 inch diameter by 1 inch thick lip which fits into the cavity on the closure.
All welds on the cover assembly are full penetration welds.
The. cover is secured to the cask body using full penetration welds.
The cask closure is sealed by (2) ethylene propylene gaskets.
Four symmetrically spaced lugs are welded to the cask body for lifting and tie-down.
Three removable, threaded eyes are attached to the' cover for lifting.
The package empty gross weight is approximately 17,700 pounds.
The package gross weight is approximately 28,000 pounds.
CONTENTS The packaging is for the shipment of radioactive material, as a solid or solidified material which meets the requirements of low specific activity radioactive material.
Solidified waste shall be enclosed in containers.
The maximum content weight shall not exceed 10,000 pounds.
,k i
mm mm_
d CONTAINMENT The structural and thermal evaluation of the AP-100 packaging shows that adequate containment is provided for normal conditions of transport.
Since the package is used for shipment of low specific activity radioactive material, it need not satisfy the hypothetical accident conditions when the' package is transported on a motor vehicle, railroad car, aircraft, inland water craft, or hold or deck of a scagoing vessel assigned for sole use of the licensee.
The radioactive waste material must b.e. solid or solidified.
Solidified' waste shall be enclosed in containers within the cask during shipment.
In accordance with 10 CFR 71 Appendix A " Normal Conditions of Transport" each of the following conditions of transport were separately analyzed to determine its effect on the AP-100 package.
1.
Heat:
Direct sunlight at ambient temperatures of 130 F in still air.
2.
Cold:
An ambient temperature of 40 F in still air and shade.
3.
Pressure:
Atmosphere pressure of 1.5 times standard atmosphere pressure.
4.
Vibration:
Normal to transport mode.
5.
Water Spray:
30 minutes spray.
6.
Free Drop:
Two foot as in excess of 20,000 pounds.
7.
Penetration: 1 1/4 inch diameter, 13 pounds. 40 inches 1.
1.JSolar Heat Load According to L.B.
Shappert (ORNL-NSIC-68) the solar heat load is 442 BTU /hr-sq.ft.
This value must be multiplied by the local mass transmittance of the earth's atmosphere and a value of relative total radiation factor obtained from figure 5.4 (see l
Appendix) from ORNL-NSIC-68.
The areas for heat absorbtion are:
A.=?OH (Area of Vertical Cylinder) 7 3.14 (65 (95) l
=
1 19,400 sq. inches.
=
A,= 28 sq. feet Ag=.785D (Area of flat top)
Ff jt::::
r=::s 3 ic::l3 2=4 4
1
..,;ht MM*, QlMJ r
'/ e * "ii
_]
u
.785 (65)2
=
3320 sq. inches
=
f 1.92 sq. feet A
=
The solar heat load at 90 degrees latitude is:
442 (.7) 0.3 qf
=
93 BTU /hr - sq. ft.
~ =
~~
Q = q,,A 93 (28)
=
f f
2600 BTU /hr.
=
442 (.7)
.4 qp
=
124 BTU /hr-sq. ft.
=
Q4 =qp Ap= 124 (1.9) 236 BTU /hr.
=
Total heat load is:
Qg 2600 + 236
=
2836 BTU /hr.
=
For an unfinned cask hA (Ts-Ta)
Q
=
g g
where:
(= total heat rejected (BTU /hr) h = heat transfer coeff,for radiation and convection t
A = Area of cask (sq. ft.)
T=
Surface temperature ( F) s T = Air temperature (130 F) h
=h,
+h g
e
. r c
where:
hr coeff.for radiative heat transfer
=
hc coeff.for convective-heat transfer
=
(Ts-Ta)1/3 (McAdam - Heat Transfer h
=C C
McGraw Hill, N.Y.,
1954) where C= 0.19.for vertical surfaces
-= 0.22 for horizontal. surfaces maq
/L m&1-
.'a=
1 l*L
&A
- Et ML2VA#4M4ffd'sI3 This equation caa be used for laminar flow conditions with little loss in accuracy.
Weighting the coeff,by the areas vertical and flat:
.19 (28) +.22 (1.92) h; 29.92
=
__5.32 +.423 29.92
=
5.74 29.92
=
0.192
=
Since we can neglect decay heat the total heat rejection load is that due to solar heating so that using figure 5.5 from ORNL-NSIC-68 (see Appendix) for C = 0.19, E=
.8 and T, 130 F
=
we can solve for T by trial and error.
g Ts fgr _
hr he ht Qt (2836) 180 50 1.4
.7 2.1 3140 170 40 1.27
.65 1.92 2290 175 45 1.38
.68 2.06 2780
'176 46 1.38
.68 2.06 2830 Thus there is no effect from 130 F ambient temperature in still air.
2.1 Ambient temperature of-40 Fin still air; the vertical and horizontal surfaces for laminar heat flow (C = 0.19H C = 0.22).
The temperature lower scale would likewise not be effected by the strength of materials curve so attached.
3.1 Factor.5 x Atmosphere Pressure ASTMA-193-55T Grad B-7 Minimum mechanical properties for either hot or cold rolled steel unless 2.5 inches.
125,000 psi Tensile strength 105,000 psi Yield pt 269/321 Brindle hardness 14.64 lbs/in.
at sea level atmosphere pressure
=
/.
=m q.
w=a 6 1
5
,-r ma rN T h.<i'h } } U E M A h Ei-iv
- d- }
~
'This is less-than 1% of material specification used in the 100 series work.
An analysis of the one-half atmosphere shows that the containment. system can sustain this pressure condition S = dx14._ 7x32
=627 psi (S (25,000)
.372 insignificant ~p4fc#ct Pressure of
.5 times _ standard atmosphere has an on the cask material.
4.1
~ Vibration:
the vibrational wave ef fect can'be calculated for a body with an applied force; harmonic or an inertia-force method.
Trailer Weight = 8,000~lbs' Cask Weight 42,000 lbs.
-K=
800 3200
=
1/4 f= 1 K
1 3200 x (12 x 32.2)
=
jyp 8/W 2]T 42,000 l
f= 0.517 Oscillations per second I
Thus there is no vibration deformation as the g load factor is many times lower.
5.1 Water Spray The design against.the thirty (30) minute water _ spray is the step design built into the-cJosure and machine fit flatness.
There are no other surfaces to be effected by the water spray.
6.1 Free Drop The analysis of a free drop follows that of L.B.
Shappert in ORNL-NSIC-68 (pertinent pages have been made an appendix of-this report).
R2 Te d's (0 - 1/2 sin 2 0)
E (steel)
=
R L @e-(9 - 1/2 sin 2 _0)
E (lead)
=
t where:
E= energy absorbed in lecd or steel ends (inch-lbs) l R' = Radius of Cask-(inches)
L =_ Cylinder length between end-plates (inches)
- lEi P4MM 'e w
z:z
/
'Gx dE:Z25'h d
i ii43%' SUN?'th&M ? 3 Te thickness of end plates (inches)
=
fs - dynamic flow stress of steel (inch-lb/ cubic inch) 6'e = dynamic flow stress of lead (inci-lb/ cubic inch) 0 deformation angle (radians)
=
E
= WH Where:
~
W = weight of cask (1bs)
H = height of drop (inches)
R 32.5 inches
=
T 10 inches
=
L 95 - Te = 95 - 10 = 85 inches
=
6's 60,000 inch-lba/ cubic inch
=
Ge 10,000 inch-lbs/ cubic inch
=
W 46,874 lbs
=
H
= 360 inches Energy to be absorbed is:
WH = 46,874 x 360 16,900,000 inch-lbs.
=
E (steel)
(32.5)
(10) (60,000) (0 - 1/2 sin 2 0)
=
2 E (lead)
(32.5)
(85) (10,000) (0 - 1/2 sin 2 0)
=
E (steel)
(10550) (60000) (0 - 1/2 sin 2 0)
=
E (lead)
(89750) (10000) (0
- 1/2 sin 2 0)
=
E (steel) = 635,000,000 (0 - 1/2 sin 2 0)
E (lead)
= 897,500,000 (0 - 1/2 sin 2 0)
-Since all the drop energy is absorbed by the deformations:
E=E (steel)+ E (lead)
E=
1,532,500,000 (0 - 1/2 sin 2 0)
,k
- n_nn,
_7_
E (0 - 1/2 sin 24) 1,532,500.000 l
=
16,900,000 1,532,500,000
=
16.9 0-1/2 sin 2 0 1,532.5
=
.1/2 sin'2 0
...011 l
0-
=
solving for 0 by trial and error:
l 0
20 sin 20 1/2 sia 20 0 (radians) 9-1/2 sin 20 20
-40
.6428
.3214
.3491
.0277 10 20
.3420
.1710
.1745
.0035 1
15 30
.5000
.2500
.2618
.0018 14 30' 29
.4848
.2424
.2530
.0106 14 45' 29 30'
.4924
.2462
.2574
.0112 the angle 0 is therefore 14 45' sin 0 =.2546 cos 0=
.9670' s
h R
l
\\
'N/
t if
\\
/
d A
the deformation d is therefore:
l d=R (1 - cos 9) j
= 32.5 (1 -.9670)
U59
32.5 (.033) i d
1.07 inches I
l i
rY
-=~=-~
.i H
the velocity Vo at the moment of impact is:
Vo= V2 gh Since the velocity at impact is zero the cask is subject to a negative acceleration which reduces Vo to zero within the stopping distance d Vo= -V.2 Gd where:
G = Ng N = number of g's causing deceleration g= gravitational constant' h = height of free fall (inches) d = stopping distanc (inches) 2 gh = 2 Gd 2 gh = -2 (N) gd therefore:
2 gh N = -2 gd h
N=
-d
'360 N= -1.07 N=
336 g's PENETRATION OUTER SHELL PUNCTURE In ORNL-NSIC-68 " Cask Designers Guide" by L.
B.
Sheppert, the minimum out'er shell thickness required to wit h s tand the
~ test is given by the equation:
penetration A NBE= C O
__m __.,
0.71 t = (W/ S )
t = Thickness in inches W= Cask weight in pounds with full load Ultimate tensile strength of the outer shell S
=
in psi.
W= 27,700 pounds 60,000 psi for A516 steel (lower limit)
S
=
0.71 t =(27,700) 60,000
= 0.58 t
t = 0.75 inches 3
The cask outer shell thickness is;p inch and therefore, is sufficient.
r LIFTING DEVICES Cask lifting devices, which are a structural part of the package, provide a lifting system capable of supporting three times the weight of the loaded package without generating stress in any material of the package in excess of its yieldz strength.
When the weight (W) of-the loaded cask is taken as 27,700 pounds, 3W is 83,100 pounds.
Lifting devices are fabricated from two (2) inch' plate carbon steel with yield strength of 36,000 psi.
The bearing stress between the lifting pin and the hole for one lug is:
i C= 1.5 (27,700/4) 2,600 psi
=
2x2 The design safety factor is 36,000/2,600 = 14.
In any event, failure of the lifting device under excessive load will not impair the cask-integrity or shielding.
LID LIFTING SYSTEM r
l The lid liftin~g device consists of three 'l 1/4 inch removable threaded' eyes located symmetrically on the cover.
The direct tensile stress in each eye is:
T=3 13200) 26000 psi
=
3'
(.785) (1.25)2 l.
uNE=CO
.8 The shear stress has a maximum value where F=F V
H The effect stress integrity is:
(2600f 1/2 = 5800 psi S= 2 26002
+
The stress, in the lifting devices, is considerably less than the yield strength of the materials involved and do, comply with the requirements of 10CFR71.
TIE-DOWN The cask anchor points to the transport vehicle or tie-down devices are the same four-lifting lugs that are structural parts of the package.
There is no yielding with LOG longitudinal, 2-G vertical, and 5-G transverse forces.
The maximum resultant force acting on each lifting lug is:
[102+52 + 2 ] 1/2 W/4=2.84W=79,000 lbs 2
F
=
Use of the resultant tie-down force of 2.84W acting on each lug is more conservative than the regulatory design load of 3W/2 for lifting devices.
The bending stresses out the base of the lugs are:
Sb=ML,= 1.5x2.84x27,700 Z
29.5 T
Where Z= 29.5 in. 4 section modulus Sb =4000 psi.
The shear stress at the base of the lug determined by:
The weld fastening the lug to the shell skin can support 78,700 pounds.
Full rigidy of the tie-down system results in equal re-actions from the maximum resultant force since each tie-down lug will have both a tension and compression tie-down.
The failure of the tie-down device will not reduce the effectiveness of the cask, since the cask has'an inner shell wrapped with a lead shield capable of maintaining the radiation and containment integrity.
L muz esa
/
2mm 4mm
/*
' ' was e
The cask anchor points to the transport vehicle are the four lifting lugs.
The resultant. force on each of the four lugs is obtained by a superposing the root-mean square value of the three directional loadings on each of the four lifting lug points, the resultant force acting on each lug is given by:
2 +5+2)
W/4 = 2.84W= 79,000 lbs 2
F=[,10 Use of the resultant tie-down force of 2.84W acting on each lug is more conservative,than he regulatorv design load of 3W/2 for lifting devices.
~
The bending stresses out the base of the lugs are:
S - ML = 1.5 x2.84 x 27,000 bZ 29.5 Where Z= 29.5 in.
section modulus T
SB = 4000 psi.
the shear stress at the base of the lug is determined by
{i F
= 2.84 x 27,700 = 3000 psi
=
T 24 A
T The above stresses combine to give an effective stress of:
S= 2 [-(4000/2)
+ (3300) 3I 7700 psi 44 S (30,000)
=
y The shear stresses on the body material are h abm 2.84 x 27,700
= 8750 psi 49C 25,000 psi.
3/8 x 24 The stresses are less than the stress strength of the material and the velds involved, and therefore satisfy the regulations.
The shear stress for the loaded resultant force per lifting lug is less than the shear stress cf the outer shell.
sbm = 78,700 16,000 p,si
<( 25,000 psi
=
.(3/8) 4.5 2+4.5+2
/
/t
- =cx:s n =s1&
's:me,a p
- mi.#W2nWWWmGM
.1
'l DROP TEST (2 FOOT)
An analysis of the impact of the two foot drop of the cask (without an overhang) demonstrates that integrity of the shielding and containment properties of the package are not significantly impaired.
This analysis indicates the capability of the cask to perform adequately without damage to the shielding and containment.
The integrity of the cask is established by.this analysis for a drop even in the most. damaging drop orientation.
The deceleration is:
a=u d
Where d is the crush depth and F.= ma = w a and F = wh g
d C. G.
+
C.G.
p,-
6.
R,
- f.. ;'
-,- - k ' '
-s
\\
g g
9
- v. - W N-m
\\
L__
a.
\\p s
- m-R=
a'+
b a=
d b = R cos B sin A tan A=
32
.711 A= 35
=
?
The deformed section volume is:
3 V. = R tan A (sin B - sin B
-Bcos B) 3 O
l M
?M
.a La w
- D 2 0c: 9 9 X*D W 5^ % :L% % b A
-l i
The volume required to absorb the impact energy from a two foot drop, assuming a crush stress of 60,000 psi, is:
V = wh
=__27,700 x 2x 12 = 11 in.
l 6'
60,000 then:
K=
(sin B - sin B B cos B)
V
=
=
3 RJ tan A
~
K=
11
=. 0 0 0'4 -
3 32 x.711 By trial and error calculation:
o B = 18 The crush depth, by this deformation analysis, is d=R (1 - cos B) sin A 32 (1 -.9511)
(.5736)
.9 inches
=
=
The effective deceleratio.n is a=
~60,000 ( 11/.9) g or 21g
_ 27,700 Since the force on the cask is the crush stress time the cross sectional area being crushed d = wh
[
So that the deceleration is:
FV~
A Mfi g = IjI_
g w
Lw_
The analysis for the crush depth from a corner drop produces no reduction of cask effectiveness.
It is obvious that the cask thickness can withstand a two foot corner drop effectively from this analysic without rupture and impairment of the shielding and containment properties.
Some local permanent deformation of the corner drop would.
result, but the corner welds are of such design and quality that rupture would not occur.
Containment would be maintained and the deformation would not effect the shielding and containment properties.
-asas ra:s d'4::iB Y,gszig JMG E
s q f y w, 4: q 3 4 ~-KEx x T'ik w R
.=
ITEM 2 The evaluation of the lid corner drop should consider realistic absorption of energy (see item 1).
In addition, the lid and lid attachment-loads resulting from the corner impact inertial loads of the lid and contents should be considered.
The lid and lid attachment ' ads resulting from the corner impact. inertia loads of the lid and contente is considered as follows: -
A lid flush with the cask outer diameter will have a realistic absorption energy characteristic.
The crush deformation depth for a two-foot' drop on the
[
lid corner is
.9 inches as-shown in the bottom drop analysis in response.
This local deformation will result in minor damage to the 1 1/2 inch outer steel plate of the co.ver lid at the point of impact.
The integrity of the shielding.and containment will be maintained.
t Justify the assumption ti.a t V/t is the maximum crush area
[
for the deceleration calculation presented on page 6 (December 16, 1976 letter).
The calculated ahear stresses for the side drop should consider.a reduced sh' ear area.
Alsofjustify the load re-sistance offered by.the lid bolt in maintaining closure for the side drop condition.
In considering the differences in flexibility be-i tween the bolts and weldments as'well as both clearances, the i
following was performed.
I The crush deformation analysis is shown in Response 1.
A side drop will produce forces on the closure lid due to the accelera-tion of the lid and the force on the shield closure shield.
Assuming i
that the lid can' move sideways, it will strike up against the closure ring.- The bolts.will not be subjected to loading since the clearance between the lid and the closure ring (0.045 to.065 in.) is less that the clearance between the bolts and the lid holes (0.075 in.).
The lid. assembly weighs 2600 lbs.jat a 21 g side-load, the total
~ bearing load is 55,000 lbs..Taken over the closure ring depth of 1.
in.,;the load is equal to 55,000 lbs/in.
l l
From case'2c, page 517. Roark, the formula for a cylinder'in a-cylindrical. socket is:
1/2 (f'c =.591 (PE/K )
D i
c I
L
- dNEEPCO, l
=------m r
where DD
=(60.035) (59.970)
K 55,400
=
=
D 1 2 (60.035
- 59.970)
DpD2 IE =.591 (55,0'00 x 29 x 10 )
~
55,400
( = 3200 psi, which is acceptable The lid lifting eyes used to support three times the loaded packaging is calculated to consider the induced bending loads on the lifting eyes and bolts, and the plane stresses on the lid.
The Vertical Force on each eye (considering three times the 2,600 lb weight of the cask lid) is:
F
=2600 F
(3/3)
(2600) 2.600lbs
=
=
q' ~
V U.634 V
A
= 4100 psi b
the shear stress through the root area has a maximum value of rf" AFa B
where F
=F based on the lifting spider geometry.
y e"
J =6 4100 psi.
The effective stress intensity is 2
[~ (T/2)
+T
- J 1/2 Si
=
[(1540)2 (4100)((1/2
+
2
=
20406
='8760 psi <[S 105,000 psi
=
Y The lifting eyes can only be installed in the lid closure subsequent to the removal of the lid closure bolts.
The closure assembly. lifting, when engaged, will lift only the lid since the. lid will no longer be secured'to the' cask body.
r=
h.
lj Mum l:x=s%
- i fu
==
m::9 wmrEw m2Waalw M sJ