ML20133E704

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Forwards Rev 7 to Page 2.4-14f to Correct Formula Originally Submitted on 850823.Correct Number Results in Greater Factor of Safety
ML20133E704
Person / Time
Site: 07109192
Issue date: 08/27/1985
From: Jerome Murphy
ANEFCO, INC.
To: Macdonald C
NRC OFFICE OF NUCLEAR MATERIAL SAFETY & SAFEGUARDS (NMSS)
References
25702, NUDOCS 8510090505
Download: ML20133E704 (4)


Text

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LNEFCQ ANEFCOINC. l %?som oe877 203.233 33s8 August 27, 1985 RETURN TO 396-SS U.S. Nuclear Regulatory Commission

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Transportation Certification Branch Division of Fuel Cycle and Material Supply Washington, D.C.

20555 Attention:

Charles E. MacDonald Mail Stop 396 SS 3

35

Subject:

Model AP-300 Type A Package g

Docket No. 71-9192 3

5

Dear Mr. MacDonald:

i Disregard page 2.4-14f which was sent to you on August 23 and replace the enclosed page 2.4-14f. revision 7.

Inadvertently an incorrect hof.X was inserted in the formula.

The correct number results in a greater safety factor.

Thank you for your cooperation and time.

Very truly yours, ANEFCO Inc.

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S ohn D. Mu'rphy DocKam President 6

USMRC

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The shell material can resist the stresses due to the maximum tie-down circumferential moment with a factor of safety of 1.23.

Stresses Resulting From the Longitudinal' Moment, M g

The circumferential stresses [/ at Point B are obtained as follows:

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

Nj Mj 0.015 From Graphs:

= 3.0

=

M /R,p L M i

g h*

(

L 1183.5

(.0) (41.1875)2(1.25)(.29)

=

M(/R,2y R,2Ty T

4 l

N4 j

p p = 5.77 ksi l

4 1

6(1183.5)

=

=(0.015) 2 (41.1875)(.29)(1.25)'

P /R,p R

pT T

g m

6M4=

5.71 ksi 2

T

= 5.77 + 5.71

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

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

N M

1*4

.025

=

From Graphs:

y jp g

p 2

F /R k

1383.5 x

x

" 1*4 y=

P /R,2p R,2pT (41.1875)2(.29)(1.25) g 1

N[

= 2.69 ksi i

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

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L

- '~

6(1183.5

= 0.25 (41.1875)(.29)(1.25)2

=

T

'L!m E R,E T 6M

= 9.51 ksi 2

T

{

= 2.69 + 9.51 = 12.2 ksi The shear stress resulting from the circumferential load at Point B is:

V

= 4.82 ksi y/

" 4C T (4

2 0)(1.25) y The combined stress intensity at Point B is calculated:

_vz

+4[x4

+

(Q 4)

S = b (6j + 6x

- v,

~P S = h (11. 48 + 12. 2

+

(11.48 - 12.20)

+ 4(4.82)2_

S = 16.67 ksi

=3*

= 2.28 (FS)Y 16.67 The shell material can resist the stresses due to the maximum tie-down longitudinal moment with a factor of safety of 2.28.

Therefore, the governing stress in the shell resulting from tie-down loads is the combined stress intensity at Point A due to the circtunferential moment.

The stress calculations assume that there is no support from the inner shell or the lea 6 between t2e shells when the maximum tie-down loading is applied.

The factors of safety calculated are therefore conservative.

I i

i 2.4-14f Revision 7 - 8/22/85

.