ML20133G360

Forwards Revs to SAR for AP-300 Shipping Container Submitted W/Application for Certificate of Compliance.Revs Include Calculations Showing That Shell of Cask Can Withstand Stresses Imposed by G Loads

ML20133G360
Person / Time
Site:	07109192
Issue date:	08/23/1985
From:	Jerome Murphy ANEFCO, INC.
To:	Macdonald C NRC OFFICE OF NUCLEAR MATERIAL SAFETY & SAFEGUARDS (NMSS)
References
25683, NUDOCS 8510150395
Download: ML20133G360 (33)

Text

7/- TMa

' NEFCO ANEFCO INC. R O/s'mo 0 7

203 4a m asa TURN TO 396-SS August 23, 1985 P9L U.S. Nuclear Regulatroy Commission g

Transportation Certification Branch Division of Fuel Cycle and Material Supply 6

Washington Dv C.

20555 D

W Attention:

Charles E. MacDonald, Mail Stop 396 SS 2-.E.r

Subject:

Model AP-300 Type A Package C

Docket No. 71-9192

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 and by your Mr. Henry Lee in his conversation with our Mr. Zwickler.last week.

The responses are related to the structural aspects of the report and have been marked as revision 7 in the text for easy identification.

We have expanded the text further to show in detail that the struc-ture 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.

The revisions include calculations which show that the shell of the cask can withstand the stresses imposed by the required simultaneous "g"

loads during transportation.

Calculations are also provided which show that the bottom plate of the cask can safely withstand the "g" loadings imposed following a presumed one foot drop of the cask.

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, 4

AN FC INC.

occmg g

[l Guff lMf1

~ '

Mw:

M AUG 2 61985 "

ohn D. Murphy

-2 97

~ ? 7 y).

President

• c' # 9

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$ N L) *n JDM/Ja b,

i' O!!!O150395 050023 h((]

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{DR ADOCK 071 2

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DATE OF D0C.

((!N' DATE RCVD.

j FCU7 PD FCAF LPDR m

I&E REF.

WUR SAFEGUARDS _

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DESCRIPTION:

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REVISIN 7 - 8/22/85 INSERT PAGES REMOVE PAGES 2.4-3 2.4-3 2.4-4 2.4-4 2.4-5 2.4-5 2.4-10 2.4-10 2.4-11 2.4-11 2.4-lla 2.4-lla 2.4-12 2.4-12 2.4-13 2,4-13 2.4-14 2.4-14 2.4-14a 2.4-15 2.4-14b 2.4-14c 2.6-13b 2.4-14d 2.6-14 2.4-14e 2.6-15 2.4-14f 2.6-16 2.4-15 2.6-17 2.6-13b 2.6-14 2.6-15 2.6-16 2.6-17 Drawings 133-1 134-1 138-1

Tha maxirum stress and strain the ganket ccn withstand without fcilure is in excess of 5000 psi. Therefore, a safety factor in excess of 9 is available when the bolts are torqued to 115 ft.lb. The frequency of leak j

test and gasket replacement schedule is based on the following.

The fabricator data indicates that under 5000 psi the gasket material has

)

a compressibility of 15 - 35% and a minimum recovery of 40%.

Assuming a maximum compressibility of 35% at 500 psi and. a linear relation-ship between stress and compressibility, the compressibility of the gasket material at 552.6 psi can be calculated.

I 552.6 compressibility = 35% x

" 3'07I i

5000 3

If 40% of compressibility is recovered, then only 60% of the compressi-

)

bility is lost for each compression.

1 1

l compressibility = 0.6(3.87%) = 2.32% per use 1

l The gasket will be replaced after six (6) sequential uses and a leak test will be performed whenever a gasket replacement takes place or at a l

minimum of one leak test per year.

2.4.3 Lifting Devices i

2.4.3.1 Lifting Devices for Cask Assembly The cask lifting device consists of a hoisting beam, four wire rope I

slings and a total of 8 shackles. The schematic outline is shown in I

figure 2.4-1.

2.4.3.1.1 Loading The empty cask weight is 46,717 lbs.

It is assumed that the cask 5'

is loaded with its payload of 20,000 lbs, in evaluating the design of the lifting / tie down pad eye.

4 Total lifting weight is therefore:

j Wt - 46,717 lbs. + 20,000 lbs. = 66,717 lbs.

i According to Title 10 of the Code of Federal Regulations, Part 71.31

)

(c), the lifting system should be capable of lifting three times the j

4 expected load.

D = 3(Wt)

Design load = 3(66,717) = 200,151 lbs.

f 6

Each pair of lifting / tie down pad eyes are designed to take the 1

entire load, therefore, each car is designed to take h the load or 100,075 lbs.

i i,

l 2,4-2 Fevision 5 - 6/10/85 i

2.4.3.1.2 Lifting Pad Design The lifting / tie down pad is constructed of A516 which has a yield Fy = 38 kip and will be welded to the outer steel shell with a low hydrogen electrode.

(See Figure 2.4-2)

Check the hole in the lifting pad for bearing and shear. (assuming 2.75" minimum for pin)

= 9.1 ksi (actual bearing stress)

F

=

p 2.75"x4 Safety factor to yicid:

38 SF =

4.18

=

9.1 Check tear out of lif ting device A = 2 x 3 x 4 = 24 sq. in.

y The shear capacity of structural steel is 2/3 of the tensile

' capacity. Therefore, the tear out capacity is:

5 2/3 x 38 x 24 = 608 kip SF =

= 6.07 0.1 Check bending of pad Sm-fIh -Tir

= 22(8

, T (1. 5 - )

3

= 232.0 in 4

F = Y, 3(66.72) = 100.1 kip 3g 3

Bending moment = 100.1 kip x 4 " = 450.5 kip-in

'i" Bending stress = N 1.94 ksi

=

232.0 in Safety factor SF = 38

= 19. 6 1.94 Check welds bending along plane 1-2 (see Figure 2.4-2)

M = 4.5" x 100.2 = 450.3 kip-in Sm for 1" fillet veld A=

422 + 4}(1 x 0.707) v.36.8 in

(.707)(22)3 +4(.707)(11)2 3

Sm = 2

,3 11 2.4-3 Revision 7 - 8/22/85

rv

\\

^

x d

/ )

F= M

~

N I4 l

<2s 2r i

/

?

/

A A

\\

i r__

SIDE ELEVATION s

2C s'

~

V 3*D 4ER

\\

6, t

.,I

/

l I4'Y k

y

\\

k oO d) i F

b0kb bb gg T \\Otl A-A

%p 1FTINGffIE-DOWN PAD AND LUG Figure 2.4-2 u

'2.4-4 Revicion 7 - 8/22/85

i I

l9 3.56 ksi Fweld

=

=

tension Fweld 100.1 2.74 ksi 4

=

stress 36.5 Combined stress =

(2.56 + 2.74 )b = 3.75 ksi Allowable tensile stress for E70 ksi low hydrogen electrode is 21 ksi.

21

.6 l

I SF to tensile stress for lifting

=

=

3.75 l

The strength of the lifting / tie down pads and the welds with which they are attached to the cask shells are therefore i

adequate to lift the filled cask. It should be remembered that the evaluation is based on using only two of the four I

pads.

In practice, the lifting device will be attached to all four pads.

1 i

1 I

1 j

l 1,

l t

I 2.4-5 Revision 7 - 8/22/85 3

i

g'

,/

j i

N e

I W

\\

cH l

d' 15*

~

+f F, g

. ~.._.

(

+-41.11

+ 27.4[-W 3 > 2F sic 477'enV I

y i

(ob37.0 b5)

C..G g

2F,v, 0.7, 1

4 i

60" 77 2

6

,0 v

1r v

fFmqr Figure 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.

Let F

• "8 FH (Moments at Pt. O at bottom (assume this is pt of rotation)

Q 66,720(10) (46.67 ) =

2F sin 47.7 cos 35 (72)

FH

+2F c s 4 7. 7 ( 41. 81+ 2 7. 4 2 )

FH j

2.4-10 Revision 7 - 8/22/85

31,138,224 =

87.245FH+'

FH 180.43F FH Thns, the tensioni_in a tie rod due to forward horizontal g force is F

= 172,578 lbs.

FH Sidevard Horizontal only two rods are effective in resisting the 5g sideward force as shown in Figure 2.4-3b.

rA 55g i

1r

+ c.

sn F

6

.i

~G

\\

s._

l y

4 !. 7 {--w

+ lvii 4, o

x 2.F S 47.7% 35 5H 002.o(d >~*0 6

9f 0

-4 72' 2 F,a tos4.7.7 7

Mv7' y

v y

A k%cT j

VIEW A-A F

Figure 2.4-3b - Sideward Horizontal Tie-Down 2.4-11 Revision 7 - 8/22/85

l Let F

= tension in a tie-rod for sidcward horizontal g force SH Moments at Pt. O at bottom i

66,720(5) (46.67 ) = 2F sn sin 35.0 02)

SH i

+2F cos 47.7 Hl.8.1-27.42) gg i

15,569,112 =

61.1 F

+ 19.4 F i

gg SH l

y l

80.5 F 15,569,112 SH Thus, the tension in a tie-rod due to sideward herizontal 9 force j

is F

= 19 3,4 05 lbs.

SH j

Summary of Tie-Rod Forces The maximum tensile force in a rod tom the simultaneous application i

of the 3 g forces is 1

1 -

F

=.Fy+FFH +

SH y

F

= 24,784 + 172&S78 + 193,405 y

1=

390,767

= 390.8 kip F

2.4.4.5 Tie-Down Pads for Cask Assembly 2.4.4.5.1 Loading The tie-down pads must be capable of sustaining the total force of the maximum force previously calculated, viz. 390.8 kip.

The tie-

=p down pads are designed to resist this maximum force.

Each pad is designed for 390.8 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 2.75" nin. for Pin) 9

~T = 35. 5 ksi (actual bearing stress)

F

=

g p

2.75 2.4-lla Revision 7 - 8/22/85 e

C Safety Factor '

SF = 38/35.5 = 1,07 Check tear out of hold down device along lines 1-2 and 3-4, Section A-A (page 2.4-13) 4 Av = 2 x 3 x 4 = 24sq. in.

The shear capacity of structural steel is 2/3 of the tensile capacity.

Therefore, the tear out capacity is:

08 2/3 x 38 x 24 = 608 kip; SF = 390.8 1.56 At the highest loaded lug the tension force = 390.8 kip Therefore, horizontal component = 390.8 cos 42.3* =

289 kip and the vertical component = 390.8 sin 42.3* =

263 kip Bending on plane 1-2 Li -in k

M = 4g" x 39 0. 8 = 1758.6 P

Assume weld pattern is 22" x 4" pattern.

Area of weld material

= 2 (22+4) ( l x.707) = 36.8 in

(.707) (22)3 + 4 (.707,)(11) 2

= 1938.1 in#

I=2 12 C = 11" 3

176.19 in S = I/C =

=

4 Tension stress on weld = F,3, =

f76.'9 5

= 10.0 ksi 390 8 0.6 ksi

=

Shearstressonwelp=Fgg, =

36.8

( 10.0 ) 2 + (10.6)2 14.6 ksi

=

Combined stress = fcomb Allowable tensile stress

= 21 ksi S.F. on tension = 21- - -

=

1.438 14.6 2.4-12 Revision 7 - 8/22/85

t.

i Fj p

T-A

'N k

/ )

K i

42,3*

24' r

/

?

e r

/

N e

N I

.1 '._

_ SIDE _ ELEVATIOh!

6 lc 2.8

)

/

3*D 4fR \\

h

\\

\\

l h

F 10 4I h

k V

O J

1"

\\

k q @,,)

855 ? '

i.us sec nog g_g pgp TIE-DOWN PAD AND LUG 2.4-13 EcVision 7 - 8/22/05

Tie-Dcwn Pads As shown in the drawing, four tie-down pads are provided to attach.the shipping cask to the truck bed.

Each pad, a plate 24" square and 1" thick, is formed with the curvature of the vessel in the vertical direction.

The lug, a plate 4" thick, 22" long and 8" high, is welded at right angles to the pad with a 1" fillet weld all around.

The pad is welded to the cask body using a 1" fillet weld all around.

A 3" diameter hole provided in the lug is used to enable. tie-down.

Both plates will be fabricated from a steel having a minimum, Fy qT

= 38 ksi and the lug plate will be welded using a E70 ksi low hydrogen electrode.

The cask will be installed on the truck bed so that the center line of che'110* 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 lifting / tie-down pads and the welds that attach them to the cask'shell are therefore adequate with a safety factor in excess of 1.43 to resist the assumed simultaneous maximum g loading during transportation.

2.4.4.5.5 Excessive Loading of the Tie-Down j.

. y

/

e g a.e

~, W:

K s.,e

. y F= 3Sc. E legW.I.

Tie-Down Fitting (Stresses in Shell)

The stresses induced into the outer shell of the vessel by the forces, calculated in Section-2.4.4.1 due to loadings of 29, 10g and 59 in the vertical, forward horizontal and sideward horizontal adirectionsrespectively, are very localized in the vicinity of the fitting.

It is conservatively assumed that only the outer shell will react the maximum tie-down fitting load of 390.8 kip.

The analysis approach is taken from Welding Research Council Bulletin 2.4-14 Revision 7 - 8/22/85 l

No. 107* which is based on the classic analytical studies of P.P. Biglaard,

c. w Sw = W d M en ( sess (

a an y ( v e w As ste ;< x A A I h' 4 %

• 2' I,

Pa.d

)

['e a

/

e N

j

& %. M ust s.,'

4 ne ck.e <

M j

Ver4 fs tw$

}

2-I s

e I

n. e a s v s.u sf g u v n.nw f

af er -

.I i

n

(

'4Y 1

& 'P u\\

v 6

(AW w/

n eTe

=o co m y The bending moment and shears in the shell circumferential(c) and longitudial (L) directions due to the maximum resultant force of 390.8 kip will be assumed to react into the shell by the pad attachments.

ir 1

JC[ '

Y'

/

Vg m

W J.

t Pr E t

Vu U

Nic.

l

• " Local Stresses in Spherical and Cylindrical Shells due to j

Dcternal Loadings", by Wichman, Hopper, and Mershon, NRC l

Bulletin No. 187, August 1965.

"evisien 7 8/22/P5

(

2.4-14a

The shear forces are:

i 289.05 kip l

V

= 390.8 sin 47.7

=

263.01 kip V

= 390.8 cos 47.7

=

3 f

The bending moments are:

i l

F( =

4.5V

= 1183.5 in-kip 3

i 4.5V

= 1300.7 in-kip M

=

c c

i i

The shell geometric parameter used is:

[ = R,

, 41.1875 32.95

=

1.25 T

The pad geometric parameter used is:

2Cy = 24.0 therefore, C

= 12.0 1

2C

= 24.0 2

therefore, C

= 12.0 2

(

s The combined shell/ pad geometric parameters used are:

i.

C l

1 12.0

= 0.29 i

E l "R

41.1875 m

i E2 41.18h5

= 0.29 4

m The constants for the circumferential moment (M ) f r this design c

are:

Ys p=

fp. p2

= 0.29 For membrane forces

=

y For bending moments E

= 0.32 p=K El

=

2 c

(K =

1.~ 1 )

The constants for the longitudial m8 ment (M ) for this design g

are:

2 'l3

('pS

0.29 p

i For membrane forces

=

y2 2

" 0'29 For bending moments E=

g p p2

=

2.4-14b.

nevision 7 8/22/85

_ _ _ _.. _ _ _ _ -.. _, ~. _. -. _ _. _ _ _ _ - _ _ _

e Stresses resulting from the circumferential moment,l'c J

The circumferential stresses (64 )at Point A are obtained as follows :

Step 1 -

Read from graph in reference the value for i

NL i

(M /R p)

C m

i i

Step 2 -

Read from graph in reference the value for 4

M_4

= 0.07 (M

mb c

Step 3 -

Circumferential membrane stress:

V 3,

4 c

1300.7

=(1.3)

T 2

( 41.18 75 f(. 29 ) (1. 25 )

(M /Rm B)

R BT c

4 2.75 ksi

=

Step 4 -

Circumferential bending stress:

M 6M l

6M 4 g

c (6)'1300.7

= (0.07) 41.18 7 5 (1. 25 )'l. 3 2 )

2 M!M RTy T

c r

m 6Mf

=

26.53 ksi i

2 T

i l

Step 5 -

Combine the circumferentisl membrane and bending stress 2

t b+

(I h

k

=

T 2

j T

I' 6

= 2.75 + 26.53

= 29.28 ksi 4

l

?

2.4-14c Revision 7 - 8/22/85 l

(

i

=_

. - ~ -

l 4

i The loncitudinal stresses (fy) at Point A are obtained as follows:

The same steps are repeated using different graphs from the i

reference.

i x

x l

u,, 2, i

M iR,p

= 0.03 t

c c

m i

N N

M 1300.7 x

c (3 3hu 1875,2 (.29) (1.25)

,x x

l

,,,,, 2,

,,2,,

r

)

I N

i

[

6.98 ksi l

=

6 F.

M 6M x

x c

6(1300.7)

(0.03) n

=

T M /R,E pT A (41.1875)(1.25)2(.32) 2 c

m j

6M l

11.37 ksi

=

2 T

N 6M I

18.35 ksi

[

gx=

T 2

= 6.98 + 11.37

=

T I

The shear stress resulting from the longitudinal load at L

l Point A is b

VL 263.01 4.38 kai

=

4C T

" 4 (12) (1. 25) 2

-The combined stress intensity at Point A is calculated:

F t/

S=

($ +6x+

(6p - b )

+ 4Zax

~

_(

[

j S = b (29.28 + 18.35 +

29. 28 - 18.35) 2 + 4 (4238) 2

)

e d

I j

S = 30.82 ksi Material yield stress

= 38 ksi

[

1 8

1.23 (FS)y

= 30.82

=

2.4-14d Pcvision 7 - 8/22/85

__.,..m....-._,.__,,,

._ =.

~

I The shell material can resist the stresses due to the maximum I

tie-down circumferential moment with a factor of safety of 1.23.

3 l

Stresses Resulting From'the Longitudinal Moment, M g The circumferential stresses 6~/ at Point B are obtained as follows:

The above steps are repeated using the appropriate graphs from j

the reference.

Ng Mj From Graphs:

= 3.0

= 0.015 L M M/R,p L

N.

N M

4 L

1183.5 3.0) ( 41.18 75 )A(1. 25 ) (. 29 )

=

M /R,2p R,2Ty T

L

]

i, N4 p

g- = 5.77 ksi 6M 4

6(1183.5) 2

=(0*015) 2 (41.1875)(.29)(1.25)'

M /R,p R

pT T

g 4=

5.71 ksi 2

T I

= 5.77 + 5.71

= 11.48 ksi The longitudinal stresses 6 x at Point B are obtained as follows:

The above steps are repeated using the appropriate graphs from the reference.

i N

M x

y P /R, f Ym E i

From Graphs:

"L 1183.5 b,

x

,y,4 2

M /R,2p p pT (41.1875) (. 29 ) (1. 25) i 3

N[

= 2.69 ksi i

2.4;14e Revision 7 - 8/22/85

t I

6M M

6M l

x_

x L

6(1183.5 0.25 M /R,y p

T (41.1875)(.29)(1.25)2 2

3 T

m 4

6M

= 9.51 ksi 2

T t

f

{

= 2.69 + 9.51 = 12.2 ksi j

i The shear stress resulting from the circumferential load at i

Point B is:

a V

= 4.82 ksi

/

" 4C T (4

2 0)(1.25) 1 i

The combined stress intensity at Point B is calculated:

1

)

(% d)

+ 4[g. '/s )

lf

( C 'y + 6 x S=

+

-v 2

)

(11.48 + 18.35 +

(11.48 - 18.35)2 + 4(4.82)2 j

l S

=

1 S = 20.83 ksi j

(FS)y

=

1. 1*0 20.83 The shell material can resist the stresses due to the maximum i

tie-down longitudinal moment with a factor of safety of 1.82.

i l

j Therefore, the governing stress in the shell resulting from tie-i' down loads is the combined stress intensity at Point A due to the j'

circumferential moment.

f The stress calculations assume that there is no support from the inner shell or the lead between he shells when the maximum tie-

[

down loading is applied.

The factors of safety calculated are j

l therefore conservative.

p i

L r

i 1

I o

'l 2.4-14f Revision 7 - 8/22/85 s

i e

i

e 1

2.4.4.7 CHOCKING RING r.

78 Cask ;j-

),2 e

s 10W h#

E

,.............9,,--

[

f i

Trailer Side Frame l

8' MC 8.5 j

h:a FIGURE 2.4-10 FORCE DI AGRAM CHOCKING RING I

i l

2.4-15 Revision 7 - 8/22/85 l

t 1

i The chocking ring welded to the steel deck plate is designed l

to prevent sliding of the cask due to the forces imposed during l

the conditions of transport stated in Title 10 of the Code of l

Federal Regulations 71.31 (d).

i Making the conservative assumptions that the friction force between the cask and support is neglible, the maximum force that would be transferred at the base (using the analogy of a simple beam on two supports) is:

4 Max. Reaction =

102+52 373.0 kip x 66.72 kip

=

The chocking (restraining) ring consists of a 7/8 thick ring welded to the base by two inch fillet welds, one on each face of the ring.

The shear restraint of the steel. ring is:

13.3 kip /in Fv = 0.4 x 38 x 7/8 x 1

=

allow.

}

The welding electrode has a tensile strength of 80 ksi.

0.3 x 80 x 2 x

.5 16.97 kip /in

=

Fv

=

E allcw.

The shear of the steel ring governs.

Width of ring, required to restrain base of cask is:

33 28.05 in 4

=

3,3 The minimum diameter of the ring = 0.D. of the cask = 83.625" Therefore, the cask is safely restrained by the ring, assuming friction is neglible.

,83.625"-

Cask

~

W = 66.72 kip Ecaction

(

m s

(tic down)

I 10W

. SW/

Reaction y

s (Ring) 2.4-16 Revision 4 - 4/15/65

Toro.pbu kur4G Et4 EPA'l 4650R6E C

~

RING WELy h,

60' Rig wtup ]

~

BOLT Rit'6

/

s C,ASYs 00TER 5 @'t-FIGURE 2.6-1 TOROIDAL RING ENERGY ABSORBER X

te, Jr:

\\.

s

*4e, FIGU.1E 2.6-2 DEFINITIONS OF TERMS USED IN EO. IN SEC. 2.6.7 2.6-13a Revision 5 - 6/10/85

l The tube will be welded to the cask outside shell with two h" fillet welds using E70 ksi lov hydrogen electrode, which has an allowable tensile stress of 21 ksi, as shown in Tigure 2.6-1.

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) t i

F = 2Ng(W)

Where:

W = weight of the leaded cask l

Eg = the mean no. of g's the cask subject upon impact

{

9.375

(

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 l

Guide.

3

/

Therefore, T=

b (66,720)1bs 1.251 x 10 kip

=

.26

(

3 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:

A = 2 Y(D)(b)

Where:

D = 0.D. of cask b = effective throat of weld 2

A = 2 W(83.625)(0.354) = 186 in The tensile stress on the weld area is:

3 b

f = 1.262 x 10 kip 6.73 ksi

=

i 2'

4 Safety Tactor = 6.73 i

Since allowabic stress for the weld is 21 ksi, the tube welds will resist the force of the drop icpact with a safety factor of 3.12.

A square tube ring, fabricated from ASTM A500-GRB stec1, 2h inches on a side and 0.0F3" thick, will be installed on the bottom of the cask to absorb l

the energy Juring a bottom drop of the AP-300 cask through a height i

of 12".

This arrangement is shown in Tig. 2.6-2a.

2.6-13b Revision 7 - 6/22/85

^

,q y rs N e Rs*3 l

• se AS T M A-500 L

G R 3 3TAEL l

i Figure 2.6-2A AP-300 Cask Bottor. Desien 2.5' n o.osi

=

s

=

l a

v

.y i

The force on the impact ring, the stress in the wall of the rina and the stress at which the wall of the tube will buckle or crush t

can be expressed as follows*:

C

=0.6[E

F=I Ng=

cr Nj== Force on impact ring Where:

g-Gravity ' Force 3

R

= Radius of the ring t

= thickness of the tube f

= stress in the tube of the wall (T = stress at which wall tube will buckle er 6

= knockdown factor which is a function of R/t E

= Yourg 's modulus y

8 for f =

I

= 496.2

[=03 0

I I

= 10,339 psi Therefore F

=

cr Under stress conditions the stress in the wall of the tube will be:

(66,720) 1

= 3106 psi 2 Trt 41.1875)

.083 The wall will not buckle under static load.

However, the wall will not be abic to support more than:

10,339

= 3.33 g 3106 Assuming that the 12" drop will cause the tube wall to collapse, then the stopping distance will be the height of the tube minus twice its thickness.

2.334 d = 2.5 = 2(.083)

=

• Buckling of Bars and Shells, Brush & Almroth, McGraw 11111, 1975 2.6-14 Revision 7-8/22/85

Th2 g loadint cn the cask lid undar th? 1;ip:ct conditions ccn be calculoted by dividing th2 drop height by the etcppint distence, at shown in 52ction 2.7.

l, cf th2 Ccsk Designer's Guid2 (OF5L-NSIC-66).

12 h

E 5.14 2.334 The g loading on the lid will therefore be b

Fg = 24 = (2) 9.46 kip x 5.14 = 97.2 kip The force vill be distributed on the 2.5" lip (Part No. B-5) where the lid is i

bolted to the cask, whose area is:

2 2

2 f(61.125 632.5 in

- 76 )

A=

=

Subtracting the area of the 36 bolts, the net area is:

2 2

A = 36 x (0.7656 ) = 16.6 in 3

632.5 - 16.6 = 615.9 in The stress on the lid plate will be:

0.16 ksi 61S 9 Y

The SA-240 plate, with a yield strength of 30 ksi, will not bend out of shape under a stress of 0.16ksi with a safety f actor of 187.5.

= 187.5 SF

=

0

(

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(.,.. u, c c, sse os w s sec i

L

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G5 t situ le nwlese 3

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