ML22063B161

SAR Update

ML22063B161
Person / Time
Site:	07103036
Issue date:	02/04/2022
From:	Office of Nuclear Material Safety and Safeguards
To:
B. White NMSS/DFM
Shared Package
ML22063B158	List: ML22063B159 ML22063B160 ML22063B161
References
Download: ML22063B161 (307)

Text

-A Structural analysis

()

()-A. Structural analysis A.1 Structural design A.1.1 General description A Type B(U) packaging consists of an inner shell, an outer shell, and a fuel basket as shown in ()-Fig.C.1.

The inner shell consists of a shell containing a fuel basket and a lid.

Fuel basket No.1 is the rectangular type as shown in ()-Fig.C.8. A basket for rectangular elements or wrapped KUCA fuels can contain up to ten elements.

Fuel basket No.2 is also the rectangular type as shown in ()-Fig.C.9. The basket can contain up to 1 spectrum converter.

After being placed in the fuel basket, the fuel elements, wrapped KUCA fuels and spectrum converter are fixed by a spacer made of silicone rubber.

Inner shell combined with its lid forms containment boundary as shown in

()-Fig.C.3 and also works as a pressure vessel against inner pressure. Inner lid attached to inner shell by inner lid bolts keep containment of its joint using double O-ring system.

Outer shell with its lid forms containment boundary as shown in ()-Fig.C.4.

Heat insulator and shock absorber are filled between inner shell and outer shell.

Outer lid attached to outer shell with outer lid bolts keep containment of its joint using Gasket.

Inner lid would never be opened by any possible contingency since it is covered by outer lid during transport. Outer lid bolt has a lock and a seal so that they would show evidence that it has not been opened.

This packaging is lifted and tied down with 4 eye-plates shown in ()-Fig.C.5.

The package is tie_down to tie down device shown in ()-Fig.C.2 with eye-plates during transport.

()1

A.1.2 Design standards The design standards for the packaging are based on the Public Notification and Section -Subsec, NB of ASME. Analytical standard are determined for each set of test conditions.

(1) Analytical standards

()-Table A.1 shows the various test conditions for the design standards corresponding to the items being analyzed. The analytical standards will be determined on the basis of the mechanical properties of the materials shown in section ()-A.3 and the temperatures shown in section ()-B.

The design standard value in which distortion level has no influence on the packaging's containment under accident test conditions is used for the inner lid clamping bolts, which are essential for the containment boundary.

Yield stress is used as the analytical standard for the hoisting and clamping device in accordance with the Public Notification". Penetration resistance is chosen as the analytical standard for the collision during the penetration test.

Welding efficiency is 1.0 for welding parts inspected by radiation method and 0.45 for other welding parts.

Symbols of the design standard value in Tables are as follows; Sm  ; Design stress intensity value Sy  ; Yield point of the designh Su  ; Design tensile strength Sa  ; Alternative peak stress N  ; Number of cycles Na  ; Allowable number of cycles DF  ; Accumulative usage factor (= N/Na)

()2

(2) Combinations of design load Combinations of design load are determined on conditions (structure temperature, material, safety factor, etc.) of each components shown in

()-Table.A.2 and ()-Table.A.3.

(3) Margin for safety Margin of safety (MS) is obtained as follows.

The designs standard Analytical tan dard value value Margin for safety (MS)= -1 Analytical value According to the design standards described above, ()-Table A.4 (1/24)

(24/24) shows conditions of structural analysis, analytical item and method, etc.

()3

()-Table A.1 Design standard for structural analysis Pm; General primary membrane stress Q ; Secondary stress PL; Local primary membrane stress F ; Peak stress Pb; Primary bending stress DF; Accumulative usage factor Component Primary+secon Primary+secon Primary stress -dary+peak Condition Item -dary Stress Position stress to be evaluated Pm(PL) PL+Pb PL+Pb+Q PL+Pb+Q+F Lifting device Eye plate <Sy <Sy Routine Tie-down device Eye plate <Sy <Sy transport Pressure Package Withstanding the effect of changing ambient pressure.

Vibration Package Withstanding the effect of vibration during transport.

Inner shell <Sm <1.5Sm <3Sm Fatigue Thermal test Inner lid evaluation

<2/3Sy <Sy <Sy Inner lid bolt (DF<1)

Water spray test Package Withstanding the water spray test.

Normal Inner shell

<Sm <1.5Sm <3Sm conditions Fuel basket Free drop test of Inner lid (1.2m height) transport Inner lid bolt <2/3Sy <Sy <Sy Fuel element/plate Inner shell <Sm <1.5Sm <3Sm Stacking test Inner lid <2/3Sy <Sy <Sy Penetrating test Outer shell No penetration Inner shell 2

< Su <Su Fuel basket 3 Drop test Inner lid (9m height)

Inner lid bolt <2/3Sy <Sy Fuel element/plate Outer shell No penetration Accident Drop test 2

conditions (1m height Inner shell < Su <Su 3

of penetration) <2/3Sy Inner lid <Sy transport 2 Inner shell < Su <Su 3

Thermal test Inner lid

<2/3Sy <Sy Inner lid bolt 2

Water immersion Inner shell < Su <Su 3

(15m depth) <2/3Sy Inner lid <Sy Note: The same criteria for stress evaluation are used for both Type B(U) packages and fissile packages.

fuel plate includes spectrum converter

()4

()-Table A.2 Design load, combination of load (1/2)

Component Load Require-Condition Item Position Internal External Thermal ment Mass* Other to be evaluated pressure pressure expansion Lifting device Eye plate Routine Tie-down device Eye plate transport Pressure Package Vibration Package Inner shell Thermal test Inner lid Inner lid bolt Water spray test Package Normal Inner shell conditions Fuel basket of Free Drop test Inner lid transport (1.2m height)

Inner lid bolt Fuel element/plate B(U) Penetrating test Outer shell package Stacking test Inner shell Inner shell Fuel basket Drop test Inner lid (9m height)

Inner lid bolt Fuel element/plate Accident Drop test Outer shell conditions (1m height Inner shell of penetration) Inner lid transport Inner shell Thermal test Inner lid Inner lid bolt Water immersion Inner shell (15m depth) Inner lid

Analyzed under combination of load.  : Analyzed under single load.

• : Mass does not mean weight simply but means mass (force) considering impact force such as given (mass) x (acceleration).

Note: fuel plate includes spectrum converter

()5

()-Table A.2 Design load, combination of load (2/2)

Component Load Require-Condition Item Position Internal External Thermal ment Mass* Other to be evaluated pressure pressure expansion Water spray test Package Inner shell Normal Fuel basket Free drop test conditions Inner lid (1.2m height) of Inner lid bolt transport Fuel element/plate Stacking test Inner shell Penetrating test Outer shell Inner shell Fuel basket Fissile Drop test Inner lid packages (9m height)

Inner lid bolt Fuel element/plate Accident Drop test Outer shell conditions (1m height Inner shell of penetration) Inner lid transport Inner shell Thermal test Inner lid Inner lid bolt Water immersion Inner shell (0.9m depth) Inner lid

Analyzed under combination of load.  : Analyzed under single load.

• : Mass does not mean weight simply but means mass (force) considering impact force such as given (mass) x (acceleration).

Note: fuel plate includes spectrum converter

()6

()-Table A.3 Load conditions (1/2)

Component Load Require

-ment Condition Item Position Internal External Thermal Mass* expan-s Other to be evaluated pressure pressure ion x3 times Lifting device Eye plate =6.99x103N x2[g]

(up, down, front, Tie-down device Eye plate back)

Routine x1[g]

(Left, right) transport Initial clamping Pressure Package 9.81x10-2MPa 60kPa force 5.89x104N Vibration Package Inner shell 9.81x10-2MPa Inner lid 9.81x10-2MPa Thermal test Initial clamping Inner lid bolt 9.81x10-2MPa 75()

force 5.89x104N Water spray test Package Normal Inner shell xAcceleration 9.81x10-2MPa conditions Fuel basket =254.1[g]

of Inner lid (for horizontal drop) 9.81x10-2MPa transport =250.6[g] Initial Free drop test (for vertical drop) clamping (1.2m height) Inner lid bolt 9.81x10-2MPa 75()

=90.8[g] force (for corner drop) 5.89x104N B(U) Fuel Package element/plate Stacking test Inner shell x5 times+Self weight 9.81x10-2MPa Penetrating test Outer shell 6kg Bar drop Inner shell xAcceleration 9.81x10-2MPa Fuel basket =367.0[g]

Inner lid (for horizontal drop) 9.81x10-2MPa

=388.4[g] Initial Drop test (for vertical drop) clamping (9m height) Inner lid bolt 9.81x10-2MPa

=310.9[g] force (for corner drop) 5.89x104N Fuel element/plate Self weight Accident Outer shell x1m drop on mild conditions Drop test steel bar of (1m height Inner shell xAcceleration 9.81x10-2MPa transport penetration)

=72.1g(Horizontal)

Inner lid 9.81x10-2MPa

=147.1[g](Vertical)

Inner shell 9.81x10-2MPa Inner lid 9.81x10-2MPa Thermal test Initial clamping Inner lid bolt 9.81x10-2MPa force 5.89x104N Water immersion Inner shell 147kPa (15m depth) Inner lid 147kPa

• : Mass does not mean weight simply but means mass (force) considering impact force such as given (mass) x (acceleration).

()7

()-Table A.3 Load conditions (2/2)

Component Load Require Thermal Condition Item Internal External

-ment Position Mass* expan-s Other to be evaluated pressure pressure ion Water spray test Package Water spray Inner shell xAcceleration 9.81x10-2MPa Fuel basket =254.1[g]

Inner lid (for horizontal drop) 9.81x10-2MPa Normal =250.6[g] Initial Free drop test conditions (for vertical drop) clamping (1.2m height) Inner lid bolt =90.8[g] 9.81x10-2MPa 75()

of force transport (for corner drop) 5.89x104N Fuel element/plate Stacking test Inner shell x5 times+Self weight 9.81x10-2MPa Penetrating test Outer shell 6kg bar drop Inner shell 9.81x10-2MPa Fuel basket xAcceleration Inner lid =379.0[g] 9.81x10-2MPa (for horizontal drop) Initial Drop test

=446.2[g] clamping Fissile (9m height) Inner lid bolt 9.81x10-2MPa (for vertical drop) force package

=332.3[g] 5.89x104N Fuel (for corner drop) element/plate Outer shell Self weight Accident 1m drop on mild Drop test conditions steel bar (1m height of Inner shell xAcceleration 9.81x10-2MPa penetration) transport =72.1g(Horizontal) -2 Inner lid 9.81x10 MPa

=147.0[g](Vertical)

Inner shell 9.81x10-2MPa Inner lid 9.81x10-2MPa Initial Thermal test clamping Inner lid bolt 9.81x10-2MPa force 5.89x104N Water immersion Inner shell 9 kPa (0.9m depth) Inner lid 9 kPa

• : Mass does not mean weight simply but means mass (force) considering impact force such as given (mass) x (acceleration).

()8

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (1/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element B(U) package Routine 1. Chemical and galvanic transport reaction Activation difference of (1) Chemical reaction Corrosion electric no chemical reaction Nil (2) Galvanic reaction Corrosion position no galvanic reaction Nil

2. Strength at low temperature No (1) Body SUS304 -40 Material 1 Degradation Allowable lowest temperature brittle (2) Bolt SUS630 -40 Material 1 Degradation Allowable lowest temperature fracture (3) O-ring Silicon- -40 Material 1 Degradation Allowable lowest temperature -40 rubber

3. Containment system 9

(1) Inner lid ()-Fig.C.3 SUS630 75 Opening due to Possibility of Nil contingency contingency

4. Lifting device M:Bending moment t:Plate thickness b:Width of eye plate (1) Eye plate ()-Fig.A.9 SUS304 75 Mass of package 3 Bending stress 6M Sy b = 2 tb F

3 Shear stress = 0.6Sy A

Combined stress = b 2 + 4 2 S

5. Tie-down device 6M (1) Eye plate ()-Fig.A.11 SUS304 75 Mass of package 2 Bending stress b = 2 Sy tb F

()-Fig.A.12 2 Shear stress = 0.6Sy A

Combined stress = b 2 + 4 2 S

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (2/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element B(U) package Routine 6. Pressure transport Note 1:

PDm (1) Frame of SUS304 75 1 Combined stress = Design Inner shell 2t standard of P Dm Formula for each stress z =

4t thin cylinder component is P determined r =

2 Note 1 using Sm.

Reduction of Note 2:

P a2 (2) Inner bottom plate SUS304 75 ambient 1 Combined stress = +/-0.225 Analysis h 2 Formula pressure 60kPa standard of P a2 for r = +/-0.75 each stress fixed disc h2 component is z = P determined 10 Formula using Sy.

P a2 for (3) Inner shell lid SUS630 75 1 Combined stress = r = 1.24 simply Note 3:

h2 Initial margin supported z = P disc of tightening Note 2 is about 1.1mm.

F (4) Inner shell SUS630 75 Initial bolt 1 Tensile stress t =

Ar load lid bolt F Internal 1 Tensile stress t =

pressure n Ar P a4 SUS630 75 Internal 1 Displacement = Formula (5) Displacement of inner 64 O-ring part of inner pressure for 2

shell lid x 1 r 2 displace- Note 3 a ment of O-ring 5+ r 2 x part 1+ a 2

7. Vibration No (1) Package ()-Fig.A.14 SUS304 75 Vibration 1 Resonance 1 k resonance fu =

2 m (2) Fuel basket fu :characteristic frequency

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (3/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element B(U) package Normal 1. Thermal condition test 1.1 Thermal expansion conditions approx. Thermal (1) Gap between basket ()-Fig.A.15 SUS304 62/63 expansion 1 Compression Presense of gap between inner Free and inner shell shell and basket.

1.2 Stress Calculation Note 1:

(1) Frame of Inner shell ()-Fig.A.16 SUS304 75 Internal 1 Combined Stress Formula for thin cylinder Design

()-Fig.A.17 pressure standard of Note 1 each stress (2) Inner bottom plate ()-Fig.A.18 SUS304 75 Internal 1 Combined Stress Formula for fixed disc component is pressure determined using Sm.

(3) Inner shell lid ()-Fig.A.19 SUS630 75 Internal 1 Combined Stress Formula for simply supported 11 pressure disc Note 2:

Analysis F

(4) Inner shell lid bolt ()-Fig.A.21 SUS630 75 Initial bolt 1 Tensile stress t = standard of Ai Note 2 load each stress Internal Tensile stress F component is t =

pressure n Ai determined Thermal Tensile stress Negrigible using Sy.

expansion Note 3:

Initial margin of tightening (5) Displacement of ()-Fig.A.20 SUS630 75 Internal 1 Displacement Formula for displacement of is about 1.1mm.

Note 3 O-ring part of inner pressure O-ring part lid

2. Water spray test Water spray 1 Absorption Absorption Nil Water-repellent Water-repellent Good

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (4/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element B(U) package Normal 3. Free drop test 3.1 Horizontal drop Note 1:

conditions (1) Deformation of shock ()-Fig.A.35 Horizontal drop 1 Deformation OH Effect of absorber ()-Fig.A.36 from 1.2m height O: Minimum thickness before deformation drop will be judged H: Deformation Note 1 in thermal

Thickness after drop test.

M b = Note 2:

(2) Frame of Inner shell ()-Fig.A.37 SUS304 75 ditto 1 Bending stress Z Analytical F standard of (3) Inner bottom plate ()-Fig.A.38 SUS304 75 ditto 1 Shear stress =

A Note 2 each stress F component is (4) Upper part of inner ()-Fig.A.39 SUS630 75 ditto 1 Shear stress = determined A

shell (Inner lid) using Sm.

12 ML (5) Inner shell lid ()-Fig.A.40 SUS630 75 ditto 1 Bending stress b = max 1 Note 3:

bolt M b = Analysis (6) Fuel basket ()-Fig.A.41 SUS304 75 ditto 1 Bending stress Z standard of M each stress b =

(7) Fuel element/plate ()-Fig.A.42 AG3NE 75 ditto 1 Bending stress Z Note 3 component is 44 determined W

1 Compression c = using Sy.

A6061P a (h 2 h 1 )

(Spectrum Converter) stress (T6) y = cr (1 + sec e L cr )

1 Buckling stress r 2K E Note 4 Note 4:

yYield stress Analysis crbuckling stress standard is Sy.

E modulus of direct elasticity Kradius-of-gyration of area Note 5:

Llength Analysis r section modulus/cross standard of section each stress eeccentricity component is (8) Fuel element hold down M determined b = Note 5 part ()-Fig.A.45 A6061P 75 ditto 1 Bending stress Z using Sy.

(T6)

Note: Bolt stress due to internal pressure and initial bolt load is obtained from the design condition and formula described in1.2 Stress calculation.

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (5/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element B(U) package Normal 3.2 Vertical drop ()-Fig.A.46 test (Bottom side) Note 1:

conditions Effect of (1) Deformation of shock ()-Fig.A.47 Vertical drop 1 Deformation OV deformation absorber (Bottom side) O: Minimum thickness before will be judged from 1.2m height drop Note 1 in thermal V: Deformation test.

Thickness after drop Note 2:

Analytical standard of F each stress (2) Frame of Inner shell ()-Fig.A.48 SUS304 75 ditto 1 Compression c = component is A

stress Note 2 determined using Sm.

13 (3) Inner bottom plate ()-Fig.A.49 SUS304 75 ditto 1 Combined stress Formula for fixed disc Note 3:

Analysis standard of each stress (4) Inner shell lid ()-Fig.A.50 SUS630 75 ditto 1 Combined stress Formula for simply supported component is disc Note 3 determined using Sy.

(5) Inner shell lid bolt SUS630 75 ditto 1 F

=

(6) Fuel Element/plate ()-Fig.A.51 AG3NE 75 ditto 1 Shear stress 2 (h 2 h 1 ) b 53 Wo t =

(Spectrum Converter) A6061P 1 Tensile stress A (T6)

W 1 Compression c = Note 3 A

stress W

(7) Fuel element hold ()-Fig.A.54 A6061P ditto Compression c =

75 1 A down part (T6) stress

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (6/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element B(U) package Normal 3.3 Vertical drop ()-Fig.A.55 test (Lid side) Note 1:

condition (1) Deformation of shock ()-Fig.A.56 Vertical drop 1 Deformation OV Effect of absorber (Lid side) from O: Minimum thickness before deformation 1.2m height drop Note 1 will be judged V: Deformation in thermal

Thickness after drop test.

F Note 2:

(2) Frame of Inner shell ()-Fig.A.57 SUS304 75 ditto 1 Compression c = Analytical A

stress Note 2 standard of each stress (3) Inner bottom plate ()-Fig.A.58 SUS304 75 ditto 1 Combined stress Formula for fixed disc component is determined (4) Inner shell lid ()-Fig.A.59 SUS630 75 ditto 1 Combined stress Formula for simply supported using Sm.

14 disc Note 3 R Note 3:

(5) Inner shell lid bolt SUS630 75 ditto 1 Tensile stress t =

n Ai Analysis standard of F each stress (6) Fuel Element/plate ()-Fig.A.60 AG3NE 75 ditto 1 Shear stress = component is 2 (h 2 h 1 ) b 62 determined 1 Tensile stress Wo using Sy.

t =

(Spectrum Converter) A6061P A Note 3 (T6) 1 Compression W stress c =

A (7) Fuel element hold ()-Fig.A.63 A6061P 75 ditto 1 Compression W down part (T6) stress c =

A 3.4 Corner drop ()-Fig.A.63 Corner drop from Analyzed for each item of para.5.15.3 above from 1.2m drop horizontal and vertical component of impact (1) Inner shell lid bolt ()-Fig.A.64 SUS630 75 (Lid side) 1 Bending stress max N V W L V VMAX V =

2 2A r Note 3 N H W L H HMAX H =

2 2 A r 3.5 Inclined drop ()-Fig.A.65 Inclined drop Analyzed for each item of para.5.15.3 above from 68 from 1.2m height horizontal and vertical component of impact

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (7/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element B(U) package Normal 4. Stacking test test F + mg Note 1:

(1) Frame of Inner shell ()-Fig.A.72 SUS304 75 Mass of package x5+Self weight Bending stress Z = Note 1 condition A Design standard of each stress component is determined using Su.

(2) Inner shell lid ()-Fig.A.71 SUS630 75 Mass of package x5+Self weight Combined stress Formula for simply supported Note 2 disc Note 2:

Analysis standard of each stress component is determined 15 using Sy.

5. Penetration test (1) Outer shell ()-Fig.A.72, SUS304 75 Impact of mild x1 Absorbed energy 1 No E2 = t Cr d t 2 73 steel bar 2 penetra-(Cr: Shear strength)=0.6Su tion

6. Free drop on each Not applicable corner or each rim

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (8/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element B(U) package Accident 1. Drop test test 1.1 Vertical drop Note 1:

conditions (Bottom side) Effect of (1) Deformation of shock ()-Fig.A.75 Vertical drop 1 Deformation OV deformation absorber (Bottom side) O: Minimum thickness before Note 1 will be judged from 9m height drop in thermal V: Deformation test.

Thickness after drop (2) Frame of Inner shell SUS304 75 ditto 1 Compression W Note 2:

c =

stress A Analytical Note 3 standard of (3) Inner bottom plate SUS304 75 ditto 1 Combined stress Formula for fixed disc each stress component is determined (4) Inner shell lid SUS630 75 ditto 1 Combined stress Formula for simply supported using Su.

16 disc Note 2 Note 3:

(5) Inner shell lid bolt SUS630 75 ditto 1 Analysis standard of each stress F component is

=

(6) Fuel element/plate AG3NE 75 ditto 1 Shear stress 2 (h 2 h 1 ) b determined using Sy.

(Spectrum Converter) A6061P Wo (T6) 1 Tensile stress t =

A 1 Compression W Note 2 c =

stress A W

(7) Fuel element hold down A6061P 75 ditto 1 Compression c =

A part (T6) stress

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (9/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element B(U) package Accident 1.2 Vertical drop test (Lid side) Note 1:

conditions Effect of (1) Deformation of shock ()-Fig.A.76 Vertical drop 1 Deformation OV deformation absorber (Lid side) from O: Minimum thickness before will be judged 9m height drop Note 1 in thermal V: Deformation test.

Thickness after drop Note 2:

Analytical F standard of (2) Frame of Inner shell SUS304 75 ditto 1 Compression c = each stress A

stress Note 2 component is determined (3) Inner bottom plate SUS304 75 ditto 1 Combined stress Formula for fixed disc using Su.

17 Note 3:

Analysis (4) Inner shell lid SUS630 75 ditto 1 Combined stress Formula for simply supported standard of disc each stress F component is (5) Inner shell lid bolt SUS630 75 ditto 1 Tensile stress t =

n Ai determined using Sy.

F (6) Fuel element/plate AG3NE 75 ditto 1 Shear stress =

2 (h 2 h 1 ) b Note 3 (Spectrum Converter) A6061P (T6) Wo 1 Tensile stress t =

A Compression W 1 c =

stress A W

(7) Fuel element hold down 75 ditto 1 Compression c =

A6061P A part stress (T6)

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (10/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element B(U) package Accident 1.3 Horizontal drop test OH Note 1:

conditions (1) Deformation of shock ()-Fig.A.77 Horizontal drop 1 Deformation O: Minimum thickness before Effect of absorber from 9m height drop Note 1 deformation

Thickness after drop will be judged H: Deformation in thermal test.

M (2) Frame 0f Inner shell SUS304 75 ditto 1 Bending stress b = Note 2:

Z Note 3 Analytical standard of (3) Inner bottom plate SUS304 75 ditto 1 Combined stress Formula for fixed disc each stress component is F determined (4) Upper part of inner SUS630 75 ditto 1 Shear stress =

using Su.

18 shell (Inner lid) A (5) Inner shell lid bolt SUS630 75 ditto Bending stress M 1 b = max Note 3:

I Analytical M standard of (6) Fuel basket SUS304 75 ditto 1 Bending stress b = Note 2 each stress Z

component is M determined (7) Fuel element/plate AG3NE 75 ditto 1 Bending stress b =

(Spectrum Converter) A6061P Z using Sy.

(T6) 1 Compression W stress c =

a (h 2 h 1 )

1 Buckling stress y = cr (1 + sec e L cr )

r 2K E Note 4:

yYield stress Analysis crbuckling stress standard is E modulus of direct cr.

elasticity Note 4 Kradius-of-gyration of area Llength r section modulus/cross section eeccentricity (8) Fuel element hold down A6061P 75 ditto 1 Bending stress part M Note 2 (T6) b =

Z

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (11/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element B(U) package Accident 1.4 Corner drop ()-Fig.A.78 Corner drop 1 Analyzed for each item of para.8.18.3 above from test from 9m height horizontal and vertical component of impact Note 1:

conditions Analytical (1) Inner lid bolt SUS630 75 Corner drop 1 Bending stress max standard of from 9m height N V W L V VMAX each stress V =

(Lid side) 2 2A r component is Note 1 determined N H W L H HMAX H = using Sy.

2 A r 2

19 1.5 Inclined drop ()-Fig.A.79 Inclined drop 1 Analyzed for each item of para.8.18.3 above from

()-Fig.A.82 from 9m height horizontal and vertical component of impact

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (12/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element B(U) package Accident 2. Drop test test 2.1 Penetration ()-Fig.A.83 conditions (1) Outer lid ()-Fig.A.84 SUS304 75 Drop onto a mild 1 Penetration No bar from 1m energy Penetra-height tion (2) Outer bottom plate SUS304 75 ditto 1 Penetration energy (3) Frame of Outer shell SUS304 75 ditto 1 Penetration energy

3. Thermal test 3.1 Thermal expansion Note 1:

20 Analytical (1) Gap between inner SUS304 500/ Thermal 1 Compression Presense of gap between inner free standard of shell and fuel basket 225 expansion shell and basket each stress component is 3.2 Stress by pressure determined using Su.

(1) frame of Inner shell SUS304 500 Internal pressure 1 Combined Stress Formula for thin cylinder Note 2:

(2) Inner bottom plate SUS304 500 Internal pressure 1 Combined Stress Formula for fixed disc Note 1 Analytical standard of (3) Inner shell lid SUS630 225 Internal pressure 1 Combined Stress Formula for simple support disc each stress component is (4) Inner shell lid bolt SUS630 225 Initial torque 1 Tensile stress F determined t =

Ai using Sy.

Note 2 F

225 Internal pressure 1 Tensile stress t =

n Ai Note 3:

Initial margin Formula for displacement of of tightening (5) Displacement of SUS630 225 Internal pressure 1 Displacement Note 3 is about 1.1mm.

O-ring part of inner O-ring part lid

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (13/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element B(U) package Accident 4. Water immersion test ()-Fig.A.85 test 4.1 Water immersion Note 1:

conditions (15m depth) Analytical standard of (1) Frame of Inner shell ()-Fig.A.88 SUS304 External pressure 1 Combined Stress Formula for thin cylinder each stress Note 1 component is (2) Inner bottom plate ()-Fig.A.89 SUS304 External pressure 1 Combined Stress Formula for fixed disc determined using Su.

(3) Inner shell lid ()-Fig.A.90 SUS630 External pressure 1 Combined Stress Formula for simply supported Note 2 disc Note 2:

4Bt Analytical (4) Buckling of inner ()-Fig.A.86 SUS304 External pressure 1 Buckling stress Pe = Note 1 standard of 2 Do shell each stress B : Buckling factor component is DO: Outer diameter of inner determined 21 shell using Su.

Note 3:

(5) Displacement of ()-Fig.A.91 SUS630 External pressure 1 Displacement Formula for displacement of Note 3 Initial margin O-ring part of inner O-ring part of tightening lid is about 1.1mm.

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (14/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element Fissile Normal 1. Water spray test Water spray 1 Absorption Absorption Nil package test Water-repellent Water-repellent Good conditions 22

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (15/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element Fissile Normal 2. Free drop package test 2.1 Horizontal drop Note 1:

conditions OH Effect of (1) Deformation of shock ()-Fig.A.95 Horizontal drop 1 Deformation O: Minimum thickness before deformation absorber from 1.2m height drop will be judged H: Deformation Note 1 in thermal

Thickness after drop test.

M Note 2:

(2) Frame of Inner shell SUS304 75 ditto 1 Bending stress b =

Z Analytical standard of (3) Inner shell bottom =

F Note 2 SUS304 75 ditto 1 Shear stress each stress plate A component is (4) Upper part of inner F determined SUS630 75 ditto 1 Shear stress =

shell (Inner lid) A using Sm.

23 ML (5) Inner shell lid bolt SUS630 75 ditto 1 Bending stress b = max Note 3:

1 Analysis M

(6) Fuel basket SUS304 75 ditto 1 Bending stress b = standard of Z

each stress M Note 3 component is (7) Fuel element/plate AG3NE 75 ditto 1 Bending stress b =

Z determined (Spectrum Converter) A6061P 1 Compression W using Sy.

c =

(T6) stress a (h 2 h 1 )

1 Buckling stress cr Note 4:

y = cr (1 + sec )

e L r 2K E Note 4 Analysis yYield stress standard is crbuckling stress cr.

E modulus of direct elasticity Note 5:

Kradius-of-gyration of area Analysis Llength standard of r section modulus/cross each stress section component is eeccentricity determined using Sy.

(8) Fuel element hold A6061P b =

M Note 5 75 ditto 1 Bending stress down part (T6) Z

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (16/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element Fissile Normal 2.2 Vertical drop package test 2.2.1 Vertical drop Note 1:

conditions (Bottom side) Effect of deformation (1) Deformation of shock ()-Fig.A.96 Vertical drop 1 Deformation OV will be judged absorber (Bottom side) O: Minimum thickness before in thermal from 1.2m height drop Note 1 test.

V: Deformation

Thickness after drop Note 2:

Analytical standard of each stress F

(2) Frame of Inner shell SUS304 75 ditto 1 Compression c = component is A

stress determined Note 2 using Sm.

24 (3) Inner shell bottom SUS304 75 ditto 1 Combined stress Formula for fixed disc Note 3:

plate Analysis standard of each stress (4) Inner shell lid SUS630 75 ditto 1 Combined stress Formula for simply supported component is disc Note 3 determined using Sy.

(5) Inner shell lid bolt SUS630 75 ditto 1 F

(6) Fuel element/plate AG3NE 75 ditto 1 Shear stress =

2 (h 2 h 1 ) b (Spectrum Converter) A6061P Wo (T6) 1 Tensile stress t =

A Note 3 1 Compression W c =

stress A (7) Fuel element hold A6061P 75 ditto 1 Compression W c =

down part (T6) stress A

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (17/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element Fissile Normal 2.2.2 Vertical drop package test (Lid side) Note 1:

condition Effect of (1) Deformation of shock Vertical drop 1 Deformation OV deformation absorber (Lid side) from O: Minimum thickness before will be judged 1.2m height drop Note 1 in thermal V: Deformation test.

Thickness after drop Note 2:

F Analytical (2) Frame of Inner shell SUS304 75 ditto 1 Compression c = standard of A

stress Note 2 each stress component is (3) Inner shell bottom SUS304 75 ditto 1 Combined stress Formula for fixed disc determined plate using Sm.

25 (4) Inner shell lid SUS630 75 ditto 1 Combined stress Formula for simply supported Note 3:

disc Analysis Note 3 R standard of (5) Inner shell lid bolt SUS630 75 ditto 1 Tensile stress t =

n Ar each stress component is F determined (6) Fuel Element/plate AG3NE 75 ditto 1 Shear stress = using Sy.

2 (h 2 h 1 ) b (Spectrum Converter) A6061P 1 Tensile stress t =

Wo (T6) A 1 Compression W Note 3 c =

stress A W

(7) Fuel element hold A6061P 75 ditto 1 Compression c =

A down part (T6) stress 2.3 Corner drop Corner drop from Analyzed for each item of para.3.13.3 above from 1.2m drop horizontal and vertical component of impact (1) Inner lid bolt SUS630 75 Corner drop from 1 Bending stress max 1.2m drop N V W L V VMAX V =

(Lid side) 2 2A r Note 3 N W L H HMAX H = H 2 2 A r

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (18/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element Fissile Normal 3. Stacking test package test F + mg Note 1:

(1) Frame of Inner shell SUS304 75 Mass of package x5+Self weight Bending stress Z = Note 1 condition A Analysis standard of each stress component is determined using Su.

(2) Inner shell lid SUS630 75 Mass of package x5+Self weight Combined stress Formula for simply supported Note 2 disc Note 2:

Analysis standard of each stress component is determined 26 using Sy.

4. Penetration test (1) Outer shell SUS304 75 Impact on mild 1 Absorbed energy 1 No E2 = t Cr d t 2 steel bar 2 penetra-(Cr: Shear strength)=0.6Su tion

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (19/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element Fissile Accident 1. Drop test package test 1.1 Vertical drop Note 1:

conditions 1.1.1 Vertical drop Effect of (Bottom side) deformation will be judged (1) Deformation of shock Vertical drop 1 Deformation OV in thermal absorber (Bottom side) O: Minimum thickness before test.

from 9m height drop Note 1 V: Deformation Note 2:

Thickness after drop Analytical standard of each stress W component is (2) Frame of Inner shell SUS304 75 ditto 1 Compression c = determined A

stress Note 3 using Su.

27 (3) Inner shell bottom SUS304 75 ditto 1 Combined stress Formula for fixed disc Note 3:

plate Analysis standard of each stress (4) Inner shell lid SUS630 75 ditto 1 Combined stress Formula for simply supported component is disc Note 2 determined using Sy.

(5) Inner shell lid bolt SUS630 75 ditto 1 F

(6) Fuel element/plate AG3NE 75 ditto 1 Shear stress =

2 (h 2 h 1 ) b (Spectrum Converter) A6061P (T6) Wo 1 Tensile stress t =

A 1 Compression W c = Note 2 stress A (7) Fuel element hold down A6061P 75 ditto 1 Compression W part (T6) c =

stress A

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (20/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element Fissile Accident 1.1.2 Vertical drop package test (Lid side) Note 1:

conditions (1) Deformation of shock Vertical drop 1 Deformation OV Effect of absorber (Lid side) from O: Minimum thickness before deformation 9m height drop will be judged V: Deformation Note 1 in thermal

Thickness after drop test.

(2) Frame of Inner shell SUS304 75 ditto 1 Compression F stress c = Note 2:

A Note 2 Analytical (3) Inner shell bottom SUS304 75 ditto 1 Combined stress Formula for fixed disc standard of plate each stress (4) Inner shell lid SUS630 75 ditto 1 Combined stress Formula for simply supported component is disc determined F using Su.

28 (5) Inner shell lid bolt SUS630 75 ditto 1 Tensile stress t =

n Ar Note 3:

F (6) Fuel element/plate AG3NE 75 ditto 1 Shear stress = Analysis 2 (h 2 h 1 ) b standard of (Spectrum Converter) A6061P Wo Note 3 each stress 1 Tensile stress t =

(T6) A component is determined Compression W 1 c = using Sy.

stress A (7) Fuel element hold down A6061P Compression W 75 ditto 1 c =

part (T6) stress A

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (21/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element Fissile Accident 1.2 Horizontal drop package test Note 1:

conditions (1) Deformation of shock Horizontal drop 1 Deformation OH Effect of absorber from 9m height O: Minimum thickness before deformation drop Note 1 will be judged

Thickness after drop in thermal H: Deformation test.

M (2) Frame of Inner shell SUS304 75 ditto 1 Bending stress b = Note 2:

Z Note 2 Analytical (3) Inner shell bottom SUS304 75 ditto 1 Combined stress Formula for fixed disc standard of plate each stress component is F

(4) Upper part of inner SUS630 75 ditto 1 Shear stress = determined shell (Inner lid) A using Su.

29 M

(5) Inner shell lid bolt SUS630 75 ditto 1 Bending stress b = max Note 3:

I Analytical M

(6) Fuel basket SUS304 75 ditto 1 Bending stress b = standard of Z each stress Note 3 component is M

(7) Fuel element/plate AG3NE 75 ditto 1 Bending stress b = determined Z using Sy.

(Spectrum Converter) A6061P 1 Compression W (T6) c =

stress a (h 2 h 1 )

1 Buckling stress cr ) Note 4 y = cr (1 + sec e L Note 4:

r 2K E Analytical yYield stress standard is crbuckling stress cr.

E modulus of direct elasticity Kradius-of-gyration of area Llength r section modulus/cross section eeccentricity Note 3 A6061P M (8) Fuel element hold down 75 ditto 1 Bending stress b =

part (T6) Z

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (22/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element Fissile Accident 1.3 Corner drop Corner drop 1 Analyzed for each item of para.1.1 and 1.2 above package test from 9m height from horizontal and vertical component of impact Note 1:

conditions Analytical standard of each stress component is determined using Sy.

(1) Inner shell lid bolt SUS630 75 Corner drop 1 Bending stress max from 9m height N W L V VMAX V = V (Lid side) 2 2Ai Note 1 N H W L H HMAX H =

2 2 Ai 30

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (23/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Temp. Applied formula or element Standard figure Type Loading factor Element Fissile Accident 2. Drop test package teset 2.1 Penetration conditions (1) Outer shell lid SUS304 75 Drop onto a mild 1 Penetration No bar from 1m energy Penetra-height tion (2) Outer shell bottom SUS304 75 ditto 1 Penetration No plate energy Penetra-tion (3) Frame of Outer shell SUS304 75 ditto 1 Penetration No energy Penetra-tion 31

3. Thermal test 3.1 Thermal expansion Note 1:

Analytical (1) Gap between inner SUS304 500/ Thermal 1 Compression Presence of gap between shell free standard of shell and fuel basket 225 expansion and basket each stress component is 3.2 Stress by pressure determined using Su.

(1) Frame of Inner shell SUS304 500 Internal pressure 1 Combined Stress Formula for thin cylinder Note 2:

(2) Inner shell bottom SUS304 500 Internal pressure 1 Combined Stress Formula for fixed disc Note 1 Analytical plate standard of (3) Inner shell lid SUS630 225 Internal pressure 1 Combined Stress Formula for simply supported each stress disc component is determined (4) Inner shell lid bolt SUS630 225 Initial torque 1 Tensile stress F using Sy.

t =

Ai Note 2 F Note 3:

225 Internal pressure 1 Tensile stress t =

n Ai Initial margin of tightening Formula for displacement of is about 1.1mm.

(5) Displacement of SUS630 225 Internal pressure 1 Displacement Note 3 O-ring part of inner O-ring part lid

Symbols;

Principal stress t  : Torsional stress b : Bending stress F  : Load

()-Table A.4 Design conditions, analytical methods of structural analysis (24/24) c : Compressive stress P  : Pressure

Shear stress A  : Cross section Design condition Analytical methods Requirement Condition Item Reference Design load Remark Material Applied formula or element Standard figure Type Loading factor Element Fissile Accident 4. Water immersion test package test 4.1 Water immersion Note 1:

conditions (0.9m depth) Analytical standard of (1) Frame of Inner shell SUS304 External pressure 1 Combined stress Formula for thin cylinder each stress Note 1 component is (2) Inner shell bottom SUS304 External pressure 1 Combined stress Formula for fixed disc determined plate using Su.

(3) Inner shell lid SUS630 External pressure 1 Combined stress Formula for simply supported Note 2 disc Note 2:

4Bt Analytical (4) Buckling of inner SUS304 External pressure 1 Buckling stress Pe = Note 1 standard of 2 Do shell each stress B : Buckling factor component is DO: Outer diameter of inner determined 32 shell using Su.

Note 3:

(5) Displacement of SUS630 External pressure 1 Displacement Formula for displacement of Note 3 Initial margin O-ring part of inner O-ring part of tightening lid is about 1.1mm

A.2 Weight and center of gravity As indicated in ()-Table-C.3, the package weighs 950 kg in maximum. Its center of gravity is shown in ()-Fig.A.1.

27.6° 1800 22.8° 997 840

()-Fig.A.1 Position of center of gravity A.3 Mechanical properties of materials

()-Table.A.5 is a list of the mechanical properties of the materials used in the analysis.

()-Table.A.6 shows the mechanical properties of the materials to be used as analytic references.

In addition, the value based on the current appropriate source is indicated in (). Even in a case where values based on these current, appropriate sources are used for mechanical property of major members etc. of this shipping cask, it is confirmed that the impact on the analysis result will be minimal, and there will be no problem for safety.

() 33

Mechanical properties of stainless steel and aluminum alloy versus temperature is indicated in ()-Fig.A.2, ()-Fig.A.3, ()-Fig.A.4, and

()-Fig.A.5.

()-Fig.A.6 and ()-Fig.A.7 show a design fatigue curve for the analysis.

A stress-strain curve of balsa used as a shock absorber is indicated in

()-Fig.A.8. The figures are quoted from references shown later.

() 34

()-Table.A.5 Mechanical properties of materials Modulus of Linear Design Design yield Design stress longitudinal expansion tensile strength Sy intensity Poisson's Stress-strai Material Code Main application parts elasticity E factor strength Su Sm ratio n diagram 2 2 2 2

[N/mm ] [l/] [N/mm ] [N/mm ] [N/mm ]

Main body of inner shell

[2]

Main body of outer shell ()-Fig.A.2 ()-Fig.A.2 ()-Fig.A.2 ()-Fig.A.2 ()-Fig.A.2 Stainless steel SUS304 0.3 and outer lid (4/5) (5/5) (1/5) (2/5) (3/5)

(austenitic)

Fuel basket

[2]

35 Inner lid Stainless steel SUS630 ()-Fig.A.3 ()-Fig.A.3 ()-Fig.A.3 ()-Fig.A.3 ()-Fig.A.4 Inner lid clamping bolt 0.3 precipitation H1150 (3/4) (4/4) (1/4) (2/4) (1/1)

Outer lid clamping bolt hardened type

[14]

Fuel element (A) ()-Fig.A.5 Aluminum alloy AG3NE 0.3 Fuel plate (1/1)

[4]

Balsa Shock absorber ()-Fig.A.8 Stainless steel: see Literature [2] Numbers shown in brackets ( )

Aluminum alloy : see Literature [14] indicate the number of the sheets Balsa  : see Literature [4] for the Figure No.

()-Table.A.6 Mechanical properties of materials to be used as design standards

- Normal conditions

- Normal test conditions Accident test conditions

- Accident test conditions (excluding thermal (only for thermal tests)

Evaluated position Material tests)

T Sm Sy Su E T Sm Sy Su E 183 1.92 16.71 1 Inner shell main body SUS304 75 137 466 500 387 (180) (1.91) (15.9) 311 688 847 1.99 9.38 2 Inner lid SUS630 75 225 612 (310) (687) (846) (1.92) (11.3) 183 1.92 16.71 3 Fuel basket SUS304 75 137 466 (180) (1.91) (15.9) 183 1.92 16.71 36 4 Outer shell main body SUS304 75 137 466 (180) (1.91) (15.9) 183 1.92 16.71 5 Outer lid SUS304 75 137 466 (180) (1.91) (15.9) 688 1.99 9.38 6 Inner lid clamping bolt SUS630 75 229 847 225 612 (687) (1.92) (11.3) 688 1.99 9.38 7 Outer lid clamping bolt SUS630 75 229 847 (687) (1.92) (11.3)

Fuel element (A) 63.8 8 AG3NE 75 167 0.697 25.7 Fuel plate 63.7 JRR-4B type 9 Fuel element (B) 75 63.8 88.3 fuel plate Fuel element hold down part 10 A6061P(T6) 75 245 295 Spectrum Converter T: Temperature [] Sm: Design stress intensity [N/mm2] Sy: Design yield point [N/mm2] Su: Design tensile strength [N/mm2]

E: Modulus of longitudinal elasticity [x105N/mm2]  : Linear expansion factor [x10-6-1]

( ): Code for Nuclear Power generation Facilities: Rules on of Materials Nuclear Power Plants (2012 edition) of the Japan Society of mechanical Engineers

37

()-Fig.A.2 Variations in mechanical properties of SUS304 according to changes in temperature (1/5)

38 Design yielding strength

()-Fig.A.2 Variations in mechanical properties of SUS304 according to changes in temperature (2/5)

39 Design yielding strength

()-Fig.A.2 Variations in mechanical properties of SUS304 according to changes in temperature (3/5)

Elastic coefficient long 40

()-Fig.A.2 Variations in mechanical properties of SUS304 according to changes in temperature (4/5)

Liner expansion coefficient 41

()-Fig.A.2 Variations in mechanical properties of SUS304 according to changes in temperature (5/5)

42

()-Fig.A.3 Variations in mechanical properties of SUS630 according to changes in temperature (bolt material) (1/4)

43

()-Fig.A.3 Variations in mechanical properties of SUS630 according to changes in temperature (bolt material) (2/4)

Elastic coefficient long 44

()-Fig.A.3 Variations in mechanical properties of SUS630 according to changes in temperature (bolt material) (3/4)

Liner expansion coefficient 45

()-Fig.A.3 Variations in mechanical properties of SUS630 according to changes in temperature (bolt material) (4/4)

46

()-Fig.A.4 Variations in mechanical properties of SUS630 according to changes in temperature (1/1)

Design yielding strength 47

()-Fig.A.5 Variations in mechanical properties of AG3NE according to changes in temperature (1/1)

Sa Repetition peak stress strength Na The permission repetition number of times

()-Fig.A.6 Design fatigue curve (austenitic type stainless steel and high nickel alloy)[2]

Sa Repetition peak stress strength Na The permission repetition number of times

()-Fig.A.7 Design fatigue curve (high tensile strength bolt)[2]

48

()-Fig.A.8 Stress-strain curve of shock absorber[4]

49

A.4 Requirements of the package A.4.1 Chemical and electrical reactions

()-Table A.7 is a list of the different materials that come in contact with each other in this package. The materials used in this package, being chemically stable in air, will not trigger any chemical or electrical reaction when coming in contact with one another.

()-Table A.7 List of different materials contacted Positions Materials Inner shell Shock absorber Stainless steel Timber Outer shell Inner shell Heat insulator Stainless steel Hardened Outer shell polyurethane Inner shell main body O-ring Stainless steel Silicone rubber Inner lid Fuel basket Spacer Stainless steel Silicone rubber Inner lid Protective sheets Spacer Polyethylene Silicone rubber Protective sheets Fuel basket Polyethylene Stainless steel Protective sheets Peripheral Polyethylene Polyurethane foam shock absorber Protective sheets Fuel element Polyethylene Aluminum alloy Peripheral shock Fuel element Polyurethane foam Aluminum alloy absorber Cushion rubber Lower part of Silicone rubber Stainless steel the fuel basket Inner shell Gasket Stainless steel Ethylene propylene Outer shell rubber Fitting bracket Fusible plug Stainless steel Solder 50

A.4.2 Low temperature strength This package is a BU type package, as is indicated in ()-B. This section will demonstrate the reliability of the packaging in ambient conditions of -

40.

The minimum temperatures of each part of the package and the materials involved are shown in ()-Table A.8.

()-Table A.8 Minimum temperatures of parts of package Minimum Brittleness transition Citation, Evaluated position Material temperature temp./min. service literatures and

() temperature () references Aluminum 1 Content Aluminum alloy -40 No brittle fracture Hand Book20 Austenitic 2 Inner shell -40 No brittle fracture JIS B 8270 stainless steel Stainless Steel Austenitic 3 Outer shell -40 No brittle fracture Manual16 stainless steel Precipitation Stainless steel 4 Inner lid hardened stainless -40 Below -40 Heat steel Treatment18 Austenitic 5 Outer lid -40 No brittle fracture JIS B 8270 stainless steel Stainless Steel Austenitic 6 Fuel basket -40 No brittle fracture Manual16 stainless steel Precipitation Inner lid clamping 7 hardened stainless -40 Below -40 bolt Stainless steel steel Heat Precipitation Outer lid clamping Treatment18 8 hardened stainless -40 Below -40 bolt steel Summary of technology for 9 Inner lid O-ring Silicone rubber -40 Below -40 hybrid materials21 10 Shock absorber Balsa -40 Below -40 Appendices A.10.4 Hardened Internal data of 11 Heat insulator -40 Below -40 polyurethane foam manufacturers22 The austenitic stainless steels of the inner and outer shells, as shown in

()-Fig.A.103 and the precipitation hardened stainless steels of the inner lid and bolts as shown in ()-Fig.A.104 can maintain adequate value of strength 51

endurable to impulse at the temperature -40, and also the Aluminum alloy used for fuel elements is free from any brittle fracture at the temperature -40, as show in ()-Table.A.8.

The tolerable temperature for the silicone rubber used for the O-ring is lower then -40. The O-ring preserves full sealing performance at -40.

The Balsa wood used for the shock absorber, as shown in ()-Fig.A.100, can maintain the function as the shock absorber sufficiently at the temperature

-40, since the material properties are free of any significant error at each temperatures, at room temperature, -20 and -40.

Therefore, at -40, this package is completely functional.

A.4.3 Sealing device After the fuel elements are stored in the main body of the inner shell, the inner lid is clamped with bolts and then secured with the outer lid. Thus, the inner lid cannot be opened inadvertently. Similarly, the outer lid cannot be easily opened as it is locked and sealed after being fixed to the main body of the outer shell.

If opened, it will easily be detected.

52

A.4.4 Hoisting accessory The hoisting accessory described in this section is a hoisting eye-plate fixed to the side of the main body of the outer shell. For design standard of the stress generated at the hoisting accessory, the yield stress Sy at the temperature of 75 is employed with safety margin, in consideration of 65, the maximum temperature at the point of eye-plate on the outer surface of the packaging on normal transportation, obtained by (II) -B Thermal Analysis.

()-Fig.A.9 shows an analytical model of an eye-plate of the hoisting accessory for the main body.

29 15 45 90 32 Eye-plate 10 50

()-Fig.A.9 Analytical model for eye-plate The gross weight of a package lifted (mo) on a hoisting eye-plate of the main body is 950 kg at the maximum, as indicated in ()-Table C.3.

A maximum load F(N) applied on one of four eye-plates when lifting a package is given by the following equation, with the load factor of 3.

1 3 F= (3xgxmo)= x9.81x950=6.99x103 [N]

n 4 53

where g: gravitational acceleration; g=9.81 [m/sec2]

Therefore, when the upward vertical load , F=6.99x103 [N] as shown in

()-Fig.A.9 works on the eye-plate, stress on each cross section is analysed as follows.

(1) Section A-A The shearing stress [N/mm2] generated in the shaded portion (section A-A) of the eye-plate shown in ()-Fig.A.9 is given by the following equation.

F F

A th where

Shearing stress [N/mm2]

F: Maximum load, F=6.99x103 [N]

t: Plate thickness, t=15 [mm]

h: Height, h=29 [mm]

Therefore, 6.99 x 10 3

= =16.1 [N/mm2]

29 x 15 So it is less than the design standard value allowable correspond to shearing stress on the eye-plate material (SUS 304) (0.6sy=108 N/mm2). And the margin of safety (MS) is 0.6Sy 110 MS= -1= -1=5.70 16.1 (2) Section B-B The bending stress b [N/mm2] generated at the fixing point of the eye-plate as indicated in ()-Fig.A.9 is given by the following equation.

M F 1 b = = 2 Z tb / 6 where M: Bending moment [N/mm2]

54

Z: Section modulus [mm3]

l: Moment arm, l=50 [mm]

b: Width of eye-plate, b =90 [mm]

t: Plate thickness, t=15 [mm]

Therefore, 6.99 x 10 3 x 50 b = =17.3 [N/mm2]

15 x 90 2 / 6 and it is less than the design yield strength (Sy=180N/mm2) of the eye-plate material (SUS304).

The margin of safety (MS) turns out Sy 183 MS= -1= -1=9.40 b 17.3 And the shearing stress generated in the section B-B is given by the following equation.

F F 6.99 x 10 3

= = = =5.18 [N/mm2]

A txb 15 x 90 It is therefore less than the design standard value allowable correspond to shearing stress on the eye-plate material (SUS304).

The margin of safety (MS) is 0.6Sy 110 MS= -1= -1=19.8 5.18 The composite stress [N/mm2] of the above-mentioned bending stress b and shearing stress is given by the following equation.

= b 2 + 4 2 = 17.3 2 + 4 x 5.18 2 =20.2 [N/mm2]

It is less than the yield point of the design of the eye-plate material (SUS304).

The margin of safety (MS) is Sy 183 MS= -1= -1=7.91 20.2 55

(3) Welded part on the section B-B 15 90 50

()-Fig.A.10 Analytical model of welded part on eye-plate.

The bending stress b [N/mm2] generated on the welded fixing part of the eye-plate shown in ()-Fig.A.10 is given by the following equation M F 1 b = =

Z Z where Z: Section modulus of the welded part [mm3]

1 Z= 2ab2 6

a: Weld-throat thickness, a=7 [mm]

b: Width of a plate, b=90 [mm]

Therefore, b will be 6.99 x 10 3 x 50 b = =18.5 [N/mm2]

1 x 2 x 7 x 90 2 6

This is less than the design standard value on the welded part (0.45Sy=81.0N/mm2).

The margin of safety (MS) is 0.45Sy 0.45 x 183 MS= -1= -1=3.37 b 18.5 56

The shearing stress generated on the welded part of the section B-B is given by the following equation.

F F 6.99 x 10 3

= = = =5.55 [N/mm2]

A 2a b 2 x 7 x 90 This is less than the design standard value allowable correspond to shearing strength on the welded part (0.45xO.6xSy=48.6 N/mm2).

The margin of safety (MS) is 0.45 x 0.6 x Sy 0.45 x 0.6 x 183 MS= -1= -1=7.75 5.55 The composite stress [N/mm2] of the bending stress mentioned above b and the shearing stress is given by the following equation

= b 2 + 4 2 = 18.5 2 + 4 x 5.55 2 =21.6 [N/mm2]

It is less than the design standard value on the welded part (0.45Sy=81.0N/mm2).

The margin of safety (MS) is, 0.45Sy 0.45 x 183 MS= -1= -1=2.75 21.6 The results of the analysis mentioned above is outlined in ()-Table A.9.

As indicated in ()-Table A.9, the margin of safety (MS) in every analysis is positive and the eye-plate is sound during hoisting.

If the expiration year of use is 40 years, the number of transport is 3 times per year and the number of times of handling per transport is 10 times, the number of times of lifting is 12,000 times. The maximum stress in the evaluations (1) to (3) above is 21.6 N/mm2, and the repeated stress is 10.8 N/mm2. This is lower than 4.0 x 102 N/mm2 which is repeated peak pressure strength at 12000 times shown in () -Fig.A.6, and the allowable number of repetitions is larger than the number of repetitions during the planned use period.

57

A.4.5 Tightening device This packaging is transported after being tightened by a device, shown in

()-Fig.A.11.

The packaging and the tightening device are secured with an eye-plate and a turnbuckle.

()-Fig.A.11 Acceleration during transportation 58

The acceleration which occurs during transportation is 2G from front to rear, 1G from left to right, 1G towards the top and 3G towards the bottom, as indicated in ()-Fig.A.11.

After taking the combined force of these factors into consideration, the tensile strength applied to the turnbuckle due to the overturning moment around the supporting points and as indicated in ()-Fig.A.11 is as follows:

2 HG + R TA= xmoxg [N]

2H T sin sec + 2 cos {R (1 + cos ) + E cos }

3 H G cos + R TB= xmoxg [N]

H T sin + (2R + E ) cos where TA : Tensile force of the turnbuckle taking as the supporting point.

TB : Tensile force of the turnbuckle taking as the supporting point.

HG : Gravity height, HG= 997 [mm]

HT : Height to the center of the eye-plate, HT= 1320 [mm]

R: Outer radius of the packaging, R= 420 [mm]

E: Length where the eye-plate is fixed: E= 50 [mm]

Angle of the turnbuckle, = 15°

Direction angle of the eye-plate, = 45° mo: Weight of the package, mo= 950 [kg]

g: Gravitational acceleration, g = 9.81 [m/s2]

The following equations are given, 2 x 997 + 420 TA=

2 x 1320 x sin 15° sec 45° + 2 cos 15° {420(1 + cos 45°) + 50 x cos 45°}

x950x9.81=9.30x103 [N]

3 x 997 x cos 45° + 420 TB= x950x9.81=1.97x104 [N]

1320 x sin 15° + (2 x 420 + 50) cos 15° Therefore, the tensile force is greater when point is taken as the supporting point T=TB=1.97x104 [N]

Thus, the stress analysis is conducted at this load level.

59

The following equations demonstrate the horizontal and the vertical components of force (F and V) when the eye-plate of the packaging receives the maximum tensile force T from the tie-down turnbuckle during transport.

T= 1.97x104 [N]

F= Tsin=1.97x104xsin15°=5.10x103 [N]

V= Tcos=1.97x104xcos15°=1.90x104 [N]

The analytical model for this case is displayed in ()-Fig.A.12 15 45 90 32 29 Eye-plate 10 15° 50

()-Fig.A.12 Analytical model for eye-plate 60

The following is an analysis of the stress generated in each cross section when the directional load of the turnbuckle T =1.97x104 [N] is applied to the eye-plate as indicated in ()-Fig.A.12.

(1) A-A cross section The following equation demonstrates the shearing stress (N/mm2) generated in the shaded portion (A-A cross section) of the eye-plate shown in ()-Fig.A.12.

T T

A th where

shearing stress [N/mm2]

T: maximum load, T=l.97xl04 [N]

t: board thickness, t =15 [mm]

h: height, h=29 [mm]

Therefore 1.97 x 10 4

= =45.3 [N/mm2]

29 x 15 It is less than the design standard value allowable correspond to shearing strength (0.6Sy =l08N/mm2 of the eye-plate material (SUS 304).

The margin of safety MS is 0.6Sy 110 MS= -1= -1=1.38 45.3 (2) B-B cross section The following equation demonstrates the bending stress b(N/mm2) generated in the fixed part (B-B cross section) of the eye-plate shown in

()-Fig.A.12.

M V 1 b = =

Z tb 2 / 6 where M: bending moment [Nmm]

61

z: section modulus [mm3]

V: vertical component force, V=1.9OxlO4 [N]

t: eye-plate board thickness, t=15 [mm]

l: moment arm, l =50 [mm]

b: eye-plate width, b=90 [mm]

Therefore, 1.90 x 10 4 x 50 b = =46.9 [N/mm2]

15 x 90 2 / 6 is obtained, and it is less than Yield point of the design (Sy=180N/mm2) of the eye-plate material (SUS 304).

The margin of safety MS is Sy 183 MS= -1= -1=2.83 b 46.9 The shearing stress generated in the B-B cross section is given by the following equation:

V V 1.90 x 10 4

= = = =14.1 [N/mm2]

A txb 15 x 90 It is less than the design standard value allowable correspond to shearing stress (0.6Sy=108N/mm2) of the eye-plate material (SUS 304).

The margin of safety (MS) is 0.6Sy 110 MS= -1= -1=6.65 14.1 The composite stress (N/mm2) of the bending stress b (N/mm2) mentioned above and the shearing stress is given by the following equation

= b 2 + 4 2 = 46.9 2 + 4 x 14.12 =54.7 [N/mm2]

It is less than Yield point of the design (Sy =180N/mm2) of the eye-plate material (SUS 304).

The margin of safety (MS) is Sy 183 MS= -1= -1=2.29 54.7 62

(3) Welded part of B-B cross section 15 90 50

()-Fig.A.13 Analytical model for welded part of eye-plate The following equation demonstrates the bending stress b(N/mm2) generated in the welded part of the fixed part of the eye-plate shown in

()-Fig.A.13.

M V 1 b = =

Z Z where Z: Section modulus of the welded part, 1

Z= 2ab2 [mm3]

6 a: Throat depth, a = 7 [mm]

b: Board width, b = 90 [mm]

Therefore, b is 1.90 x 10 4 x 50 b = =50.3 [N/mm2]

1 x 2 x 7 x 90 2 6

This is less than the design standard value (0.45Sy= 81.0N/mm2) of the welded part.

63

The margin of safety (MS) is 0.45Sy 0.45 x 183 MS= -1= -1=0.61 b 50.3 The shearing stress generated at the welded part of the B-B cross section is given by the following equation V V 1.90 x 10 4

= = = =15.1 [N/mm2]

A 2a b 2 x 7 x 90 This is less than the design standard value allowable correspond to shearing stress (0.45x0.6xSy=48.6N/mm2) of the welded part.

The margin of safety (MS) is 0.45 x 0.6 x Sy 0.45 x 0.6 x 183 MS= -1= -1=2.21 15.1 The composite stress (N/mm2) of the bending stressb and the shearing stress is given by the following equation

= b 2 + 4 2 = 50.3 2 + 4 x 15.12 =58.7 [N/mm2]

This is less than the design standard value (0.45Sy=81.0N/mm2) of the welded part.

The margin of safety (MS) is 0.45Sy 0.45 x 183 MS= -1= -1=0.37 58.7 A summary of the results of the above-mentioned analyses is given in

()-Table A.9.

As shown in ()-Table A.9, the margin of safety (MS) of the results of the analyses being positive in each case, the eye-plate is sound when tied down.

64

()-Table.A.9 Summary of analyses under routine transport The design Analysis Conditions Analysis item Type of load Design standard standard value result Margin of safety MS N/mm2 N/mm2 Hoisting accessory 1.Eye-plate during hoisting Weight of the packagex3 A-A cross section (1)Shearing stress 0.6Sy 108 16.1 5.70 B-B (1)Bending stress Sy 180 17.3 9.40 cross section (2)Shearing stress 0.6Sy 108 5.18 19.8 (3)Composite stress Sy 180 20.2 7.91 B-B (1)Bending stress 0.45Sy 81.0 18.5 3.37 65 cross section (2)Shearing stress 0.27Sy 48.6 5.55 7.75 (welded part) (3)Composite stress 0.45Sy 81.0 21.6 2.75 Routine transport Tightening device Acceleration 2.Eye-plate in tie-down position Left-right:1G Front-rear:2G A-A Top :1G Cross section (1)Shearing stress Bottom :3G 0.6Sy 108 45.3 1.38 B-B (1)Bending stress Sy 180 46.9 2.83 cross section (2)Shearing stress 0.6Sy 108 14.1 6.65 (3)Composite stress Sy 180 54.7 2.29 B-B (1)Bending stress 0.45Sy 81.0 50.3 0.61 cross section (2)Shearing stress 0.27Sy 48.6 15.1 2.21 (welded part) (3)Composite stress 0.45Sy 81.0 58.7 0.37

A.4.6 Pressure We shall analyze the soundness and sealing performance of the packaging in the case where external pressure would decrease to 60 kPa. The analysis is performed assuming that minimum temperature if the package components is -40 and the maximum temperature is 65 .

When external pressure decreases to 60 kPa, the pressure in the inner shell is P2 = P0 - Pa = 0.1013 - 0.060 = 0.0413 [MPa]

where P0 : Inner shell initial internal pressure (atmospheric pressure),P0= 0.1013 [MPa]

Pa  : External pressure after pressure decrease, Pa= 0.060 [MPa]

For purposes of stress evaluation, in A.5.1.3 Stress Calculation, the internal pressure utilized in the packaging is 9.81xlO-2MPa. In this section, we will analyze the internal pressure, utilizing the total of differential pressure P = P1 + P2 = 0.0981 + 0.0413 0.140 [MPa]

The stress evaluation parts and the analysis method are the same as in section A.5.1.3 and the results of the stress evaluation are shown in ()-Table A.10.

()66

Stress units

-Table A.10 Stresses evaluation under changed pressure ;N/mm2 Stress Stress Primary+secondary Stress due Primary stress Fatigue Stress at due to stress to internal Position initial thermal Pm(PL) Sm MS PL+Pb 1.5Sm MS PL+Pb 3Sm MS PL+Pb Sa Na DF MS pressure to be evaluated clamping expansion +Q +Q+F

-0.070 Frame of Inner min 1 3.29 3.36 137 39.7 3.36 411 121 3.36 1.68 500 5x10-4 2x10 shell 10 1.65 Inner Surface 4.53 1.36 Bottom plate -0.140 min 2 of the inner 0.140 137 977 4.67 205 42.8 4.67 411 87.0 4.67 2.34 500 5x10-4 2x10 10 Outer Surface shell -4.53 67

-1.36 0

Inner Surface

-4.66

-4.66 Inner shell -0.140 2/3Sy Sy Sy min 3 0.140 3270 4.66 146 4.66 146 4.66 2.33 500 5x10-4 2x10 lid 458 687 687 10 Outer Surface 4.66 4.66 0

Inner shell Sy/1.5 Sy 4 lid 174 4.59 180 1.55 180 2.82 720* 360 500 4000 0.125 7.0 clamping bolt 459 688 Displacement of the (1) Displacement 1.72x10-2mm (3) Residual margin of tightening of O-ring 5

inner lid O-ring (2) Initial clamping value of the O-ring 1.1mm t1.082mm PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; t; Ability of bolt stress; NaPermissible number of repetition; DFCumulative fatigue coefficient; SmDesign stress intensity value; SyYield point of the design; MSMargin of safety;

• Stress concentration factor = 4; r; Diameter direction stress; o;Periphery direction stress; t; Axial stress;

A.4.7 Vibration This package is secured with a turnbuckle on a tightening device, as indicated in ()-Fig.C.2. The turnbuckle is safely secured in order to avoid loosening due to vibration from the transport vehicle. Hence, we shall assume that no vibrations will be caused by this. Below, we shall calculate the natural frequency of the package itself, which will be compared to the vibration caused by the vehicle or ship of transport, and demonstrate that this will not cause the package to resonate during transport.

(1) Vibrations of the packaging

()-Fig.A.14 shows an analytical model for the vibration of the packaging.

4 4

3 3

2 1 1 2

()-Fig.A.14 Vibration analytical model of packaging (68

As is indicated in ()-Fig.A.14, by assuming that the packaging is a mass system supported by four types of parallel springs, the natural

[8]

frequency at that time can be given by the following equation  :

K 0= x 10 3 [rad/sec]

m Therefore, 0 1 K x 10 3 f0= = [Hz]

2 2 m where 0 : Natural angular frequency of the packaging [rad/sec]

f0  : Natural frequency of the packaging [Hz]

m  : Package mass, m=950 [kg]

K  : Parallel spring constant[kg/mm]

K= 4 = ki = K1 + K2 + K3 + K4 i =1 A i E i A1E1 A 2 E 2 A 3 E 3 A 4 E 4

= 4

= + + +

i =1 1i 11 12 13 14 A1 : Cross section of the reinforcement, A1 =2.83x103 [mm2]

A2 : Cross section of the balsa, A2 =4.76xl05 [mm2]

A3 : Cross section of the outer shell board, A3 =7.92x103 [mm2]

A4 : Cross section of the turnbuckle, A4 =2.83xl03 [mm2]

E1 : Modulus of longitudinal elasticity of the reinforcement; E1 =1.92xl05 [N/mm2]

E2 : Modulus of longitudinal elasticity of the balsa, E2 =98.1 [N/mm2]

E3 : Modulus of longitudinal elasticity of the outer shell board, E3 =1.92xl05 [N/mm2]

E4 : Modulus of longitudinal elasticity of the turnbuckle, E4 =2.04xl05 [N/mm2]

l1 : Length of the reinforcement, l1 =194 [mm]

l2 : Length of the balsa, l2 =341 [mm]

(69

l3 : Length of the outer shell board, l3 =1320 [mm]

l4 : Length of turnbuckle, l4 =470 [mm]

Therefore, k1 : Spring constant of the reinforcement, k1 =2.79x106 [N/mm]

k2 : Spring constant of the balsa, k2 =0.137x106 [N/mm]

k3 : Spring constant of the inner shell board, k3 =1.15x106 [N/mm]

k4 : Spring constant of the turnbuckle, k4 =1.23x106 [N/mm]

K=(2.79+0.14+1.15+1.23)x106 =5.31x106 Therefore, the natural frequency is 1 5.32 x 10 6 x 10 3 f0= =377 [Hz]

2 950 This natural frequency of 377 Hz is outside the vibration range of 0 to 50 Hz which will present in the vehicle or ship during transport. Therefore, there is no possibility of coincidental vibration. In addition, the frequency expected during transportation is about 0 to 50 [Hz], which is different from the natural frequency, so that the input excitation force is not amplified. Therefore, since the acceleration generated by the main body during transportation is sufficiently included in the free fall under general test conditions, the package will not be cracked or damaged.

(2) Fuel basket The fuel basket is supported by a spacer in the inner shell, and will not receive directly any external vibration.

The fuel element and so on is also protected at top and bottom by a silicone foam spacer, and will not receive any vibrations.

(3) Evaluation The natural frequency of this packaging is higher than the vibration generated by the transport vehicle, and so, coincidental resonance will not occur.

Therefore, the inner lid clamping bolt and other clamping devices will (70

not loosen during transport, and sealing performance will be fully preserved.

In addition, the fuel basket and the fuel element are supported by rubber inside the inner shell, and soundness will be fully preserved despite the vibrations during transport.

(71

A.5 Normal test conditions This package is a BU type package. Therefore, the normal test conditions defined on the regulation are as follows.

(1) Water spray test The following tests shall be performed after test (1).

(2) Free drop test (3) Stacking test (4) Penetration test The following test shall be performed after tests (1) to (4).

(5) One week period placed in an environment of -40 to 38.

The following section will analyze the effect to the package caused by the tests mentioned above. The results of this analysis shall demonstrate that the design standards for normal test conditions are satisfied.

A.5.1 Thermal test A.5.1.1 Outline of temperature and pressure This section is a summary of the pressure and temperature used for design analysis under normal test conditions.

(1) Design temperature As determined in ()-B.4.2 Maximum Temperature, the package temperature may rise to a maximum of 65 (minimum -40) . Therefore, the design temperature under normal test conditions shall be conservatively determined to be 75, adopting a margin of safety, as indicated in ()-Table A.11, for both the inner and outer shells.

()-Table A.11 Design temperature under normal test conditions Part Design temperature()

1 Fuel element 75 2 Fuel basket 75 3 Inner shell main body 75 4 Inner lid 75 5 Outer shell 75 (72

(2) Design pressure As determined in ()-B.4.4 Maximum Internal Pressure, the internal pressure of the inner shell may increase up to 0.016 MPa in gauge pressure.

Even when the temperature changes from -40 to 65, the pressure inside the container is 0.046MPa [gauge]. Therefore, the design pressure in normal test conditions shall be conservatively determined as 0.0981 MPa, adopting a margin of safety, as indicated in ()-Table A.12.

()-Table A.12 Design pressure under normal test conditions Portion Design pressure 1 In the inner shell 9.81x10-2MPaG (73

A.5.1.2 Thermal expansion This section will assess the stress generated when differential thermal expansion causes the inner shell and fuel basket to come into contact. The analytical model is shown in ()-Fig.A.15 Rubber packing Inner shell Fuel basket Rubber packing

()-Fig.A.15 Analytical model of thermal expansion The increase in temperature in the fuel basket and the inner shell is 75, as indicated in ()-B Thermal Analysis. There is no temperature difference, where thermal expansion does not occur, since the two parts are made of the same material (SUS 304).

There is also practically no temperature difference between the outer and inner shells. The inner shell will not be influenced by thermal expansion of the outer shell.

(74

Therefore, no stress will be generated by thermal expansion in the fuel basket and inner shell.

A.5.1.3 Stress calculation Stress calculation shall be conducted in this section.

Temperature gradient, loads from the outside and pressure may generate stress in each part of the package.

The ratio of the inner shell's inner radius to the board thickness is higher than 10 and can be considered as a thin cylinder. Therefore, temperature differences will little occur inside the board thickness of the shell. Also, although the inner lid and the bottom plate of the inner shell are thicker than the other parts, temperature differences will have little possibility of occurring since these parts are protected by heat insulators and shock absorbers, as in the outer lid.

The same applies to the fuel basket, where the board thickness is 3 or 3.4mm.

This thinness will make it improbable for temperature differences to occur.

Therefore, since the thermal stress due to temperature differences in the plate thickness of the parts of the packaging is minimal, this stress is not calculated in this section.

Next, we shall analyze the stress generated in each part by internal pressure, keeping in mind the fact that the internal pressure of the inner shell is the pressure used in the package.

We shall also analyze the inner lid clamping bolt, which is a crucial part in the sealing boundary, after taking into consideration the initial clamping strength and thermal expansion.

(75

(1) Stress evaluation positions The stress evaluation position of the inner shell under normal test conditions is shown in ()-Fig.A.16. In this section, the main stress shall be determined, the different types of stress being shown in ()-Table A.13.

A stress evaluation will be conducted in section A.5.1.4.

Code Evaluation position Frame of inner shell Bottom plate of inner shell Inner lid Inner lid O-ring displacement Inner lid clamping bolt

()-Fig.A.16 Stress evaluation position under normal test conditions 76

Inner shell In the center of the inner shell, pressure inside the inner shell shall be utilized as internal pressure.

The analytical model of the stress generated in the center of the inner shell which subjected to internal pressure is shown in ()-Fig.A.17. The stress

(,z,r) generated in the center of the shell is given as a thin cylinder

[7]

by the following equations  :

()-Fig.A.17 Stress analysis model of inner shell center portion PD m

=

2t PD m z=

4t P

r=-

2 where

Circumferential stress [N/mm2]

z : Axial stress [N/mm2]

r : Radial stress [N/mm2]

P: Design pressure inside the inner shell, P =9.81x10-2 [MPagauge]

Dm: Frame of inner shell mean diameter, Dm =D + t =460 +1O=47O [mm]

t: Frame of inner shell board thickness, t=10.0 [mm]

D: Frame of inner shell bore, D =460 [mm]

Thus, the stresses are 9.81 x 10 2 x 470

= =2.31 [N/mm2]

2 x 10 9.81 x 10 2 x 470 z= =1.15 [N/mm2]

4 x 10 r=-0.0491 [N/mm2]

77

Bottom plate of the inner shell

()-Fig.A.18 shows an analytical model for the stress on the bottom plate of the inner shell when receiving internal pressure.

The stress generated in the fixed part of the peripherally supported disc is, P a2 Inside P

=+/-0.225 h2 Outside 2 P a2 r=+/-0.75 h2

()-Fig.A.18 Stress analysis model of inner z=-P (Inner surface) shell bottom plate where

Circumferential stress [N/mm2]

z : Axial stress [N/mm2]

r : Radial stress [N/mm2]

P: Design pressure inside the inner shell, P =9.8lx10-2 [MPagauge]

a: Radius of inner shell bottom plate, a =230 [mm]

h: Wall thickness of inner shell bottom plate, h =35 [mm]

Therefore, the stresses are 9.81 x 10 2 x 230 2

=+/-0.225 = +/-0.953 [N/mm2]

35 2 9.81 x 10 2 x 230 2 r=+/-0.75 = +/-3.18 [N/mm2]

35 2 z=-0.098 (Inner Surface) [N/mm2]

The double signs of the stress values correspond to the inner and outer surface respectively.

78

Inner lid

()-Fig.A.19 shows an analytical model of the stress on the inner lid when receiving internal pressure.

The stress (, r, z ) generated in the peripherally simply supported disc is maximum at the center P a2

=r= 1.24 h2 z=-P (Inner surface) where

Circumferential stress [N/mm2]

r : Radial stress [N/mm2]

z : Axial stress [N/mm2]

P: Design pressure inside the inner shell, P =9.81x10-2 [MPagauge]

a: Radius of inner shell bottom plate, a =285 [mm]

h: Wall thickness of inner shell bottom plate, h =55 [mm]

Therefore, the following values are obtained, 9.81 x 10 2 x 285 2

=r= 1.24 = 3.27 [N/mm2]

55 2 z=-0.098 (inner surface) [N/mm2]

The double sign indicates the inside for the top, the outside for the bottom.

Bolt circle Inside Outside 2a

()-Fig.A.19 Stress analysis model of inner lid center portion 79

Inner lid O-ring displacement An analytical model of the inner lid O-ring displacement is shown in

()-Fig.A.20.

An displacement (mm) of the simply supported disc shown in ()-Fig.A.20 can

[7]

be determined by the following equations  :

P a4 r2 5 + r2

= 1 2 64D a 1 + a 2 where P: Design pressure in the inner shell, P =9.81x10-2 [MPagauge]

Poisson's ratio, =0.3 a: Radius of the support points circle of the inner lid, a =285 [mm]

r: Distance from the center to the evaluation point, ri : radius of inner O-ring groove, ri =237.5 [mm]

D: Inner lid bending stiffness, E h3 D = [Nmm]

12(1 2 )

E: Modulus of longitudinal elasticity E =1.99x105 [N/mm2]

h: Minimum plate thickness of the inner lid, h=36.7 [mm]

Therefore, the displacementi of the groove portion of the inner O-ring is 9.81 x 10 2 x 285 4 x 12 x (1 0.3 2 ) 237 2 i = x 1 64 x 1.99 x 10 5 x 36.7 3 285 2 5 + 0.3 237 2 x = 1.17 x 10 2 2

1 + 0.3 285 [mm]

i is sufficiently smaller than the initial clamping value 1.1mm Groove for (Difference of O-ring groove inner O-ring depth and O-ring diameter)

Bolt circle p

()-Fig.A.20 Analytical model of inner lid O-ring displacement 80

Inner lid clamping bolt The stress generated by initial clamping stress, internal pressure and thermal expansion shall be analyzed regarding the inner lid clamping bolt (hereinafter referred to as bolt).

(a) Initial clamping stress The analytical model figure of the stress generated by the initial clamping force in the bolt is shown in ()-Fig.A.21.

The tensile stress t generated in the Lid bolt as shown in ()-Fig.A.21 Upper end is given by the equation F

t =

Ai ()-Fig.A.21 Stress analysis model of bolt of inner lid (initial clamping stress)

Where F : Initial clamping force of the bolt, T 2.825 x 10 5 F= = =5.89x104 [N]

kd 0.2 x 24 T : Initial clamping torque, T =2.825x1O5 [Nmm]

k : Torque coefficient, k=0.2 d : Nominal diameter of the bolt, d=24 [mm]

Ai: Cross section of the trough radius of the bolt (M24),

Ai = di2 = x20.7522 =338.2 [mm2]

4 4 di: Minimum diameter of the bolt, di =20.752 [mm]

Therefore, the following va1ue is obtained 5.89 x 10 4 t = =174 [N/mm2]

338.2 81

(b) Stress due to internal pressure The analytical model of the stress generated by the internal pressure in the bolt is shown in ()-Fig A.22.

The tensile stresst generated in the bolt as shown in ()-Fig.A.22 is given by the following equation ri P 2

t =

n Ar where ri : Radius of the surface receiving pressure, ri =237.5 [mm]

P  : Design pressure in the inner shell, P =9.81x10-2 [MPa[gauge))

Ar: Cross section of the minimum diameter of the bolt M24, Ar=338.2 [mm2]

n  : Number of bolts, n =16 Therefore, the tensile stress is, x 237 2 x 9.81 x 10 2 t = =3.20 [N/mm2]

16 x 338.2 Bolt circle Inner radius of the Bolt inside O-ring grove 16-M24 P

2

()-Fig.A.22 Stress analysis model of bolt of inner lid (stress due to internal pressure) 82

(c) Stress due to thermal expansion The analytical model of the stress generated by thermal expansion in the bolt is shown in ()-Fig.A.23.

The temperature of the bolt and of the inner lid is 75, in accordance with

()-B Thermal Analysis, and there is no temperature difference. The material also is the same, the SUS63O, Stress due to thermal expansion is negligible.

M24

()-Fig.A.23 Stress analysis model of bolt of inner lid (stress due to thermal expansion)

A.5.1.4 Comparison of allowable stress The results of stress evaluation related to each of the analyses conducted in section ()-A.5.1.3 are summarized in ()-Table A.13.

As is shown in this table, the margin of safety against the design standard value allocated to each case, whether they are simple or multiple loads, is positive. Even when the ambient temperature changes from -40 to 38, there is no effect on thermal expansion and thermal stress.

Therefore, under normal test conditions (thermal test), the soundness of the package can be maintained. In addition, in the case where the number of usage of the package is set at 500*, the margin of safety in regard to allowable cycles is, as shown in ()-Table A.13, positive. Therefore, the soundness of the packaging will not be lost through repeated loads.

Times of use N = 3/yearx40 yearsxtolerance ratio 500 times 83

Stress units

-Table A.13 Stress evaluation under normal test conditions (thermal test) ;N/mm2 Stress Stress Stress Primary+secondary Primary stress Fatigue Stress at due to due to stress Position initial internal thermal Pm(PL) Sm MS PL+Pb 1.5Sm MS PL+Pb 3Sm MS PL+Pb Sa Na DF MS to be evaluated clamping pressure expansion +Q +Q+F

-0.0491 min 1 Frame of Inner shell 2.31 2.36 137 57.0 2.36 411 173 2.36 1.18 500 5x10-4 2x10 10 1.15 Inner Surface 3.18 0.953 Bottom plate -0.098 min 2 of the inner 0.098 137 1396 3.28 205 61.5 3.28 411 124 3.28 1.64 500 5x10-4 2x10 10 Outer Surface shell -3.18 84

-0.953 0

Inner Surface

-3.27

-3.27

-0.098 2/3Sy Sy Sy min 3 Inner shell lid 0.098 4672 3.27 209 3.27 209 3.27 1.64 500 5x10-4 2x10 458 687 687 10 Outer Surface 3.27 3.27 0

Inner lid 2/3Sy Sy 4 174 3.22 177 1.58 177 2.89 708 354 500 4000 0.125 7.0 clamping bolt 458 687 Displacement of the (1) Displacement 1.16x10-2mm (3) Compression dipth of O-ring 5 inner shell lid (2) Initial clamping value of O-ring 1.1mm t1.088mm O-ring PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; t; Ability of bolt stress NaPermissible number of repetition; DFCumulative fatigue coefficient; SmDesign stress intensity value; SyYield point of deesign; MSMargin of safety;

• Stress concentration factor = 4 rDiameter direction stress oPeriphery direction stress

A.5.2 Water spray The outside surface of this packaging is made of stainless steel, and there is no water absorption. Therefore, there is no possibility of degradation of the material due to the spraying of water.

In addition, the outer lid is secured to the main body with a outer lid clamping bolt using a washer. This is waterproof and presents no risks of water entering inside the packaging.

A.5.3 Free drop The weight of this package is maximum 950 kg. Since it is below 5000 kg, the free drop height under normal test conditions is determined by regulation standards as 1.2 m.

The free drop posture is analyzed for the following four cases:

1) Horizontal drop

2) Vertical drop (lid side and bottom side)

3) Corner drop (lid side and bottom side)

4) Inclined drop (lid side and bottom side)

The purposes of this analysis are,

1) To demonstrate that the sealing performance of the inner shell is preserved by demonstrating that the deformation wrought by a free drop do not extend to the inner shell which is the sealing boundary.

2) The inner shell will not be damaged by the shock caused by the free drop, and will preserve full leak tightness.

3) There is no damage of the contained material.

(1) Analysis method The following are the analysis conditions for the stress generated in the contained material, the fuel basket, the main body of the packaging and for the deformation of the transport packaging in the case where the package would 85

be subjected to a free drop test of 1.2m.

(a) Deformation

1) The drop energy of the package will be completely absorbed by the shock absorber in the case where the shock surface is a rigid body. Therefore, the deformation of the outer shell will be the deformation of the shock absorber.

This is conservative assumption ignoring absorption by the steel plate or the heat insulator.

2) The deformation and acceleration caused by the shock absorber shall be calculated on the basis of the shock absorbing function analysis program CASH- indicated in A.10.1.

(b) Stress

1) The drop energy of the package shall be absorbed by the deformation of the steel plate utilized in the shock absorber, the main body of the outer shell and the outer lid.

2) The acceleration utilized in the stress analysis (hereafter referred to as design acceleration) shall be 1.2 times the calculation value (acceleration generated in the shock absorber) of "CASH- (this value was determined through comparison with test results as indicated in section A.10.1) plus the acceleration of the steel plate.

This is a safety evaluation since the shock strength present in the package will be combined to the acceleration of the shock absorber and the acceleration of the steel plate.

Design acceleration = calculation results of CASH- x 1.2

+ acceleration due to steel plate.

3) Generated acceleration of the steel plate will be determined using simplified calculations.

86

(2) Drop energy The weight of the package utilized in the analysis is 960 kg as indicated in A.2 Weight and Center of Gravity. The drop energy is Ea = Ev = mgh where Ea: Energy absorption of the shock absorber [J]

Ev: Drop energy of the package [J]

m: Package mass, m=950 [kg]

h: Drop height, h=1.20 [m]

g: Gravitational acceleration, g =9.81 [m/s2]

Therefore, the following value is obtained Ea = Ev = 960x9.81x1.2 =1.12x104 [J]

=1.12x107 [Nmm]

(3) Performance of the shock absorbers obtained by means of the CASH- analysis program The results of the deformation in the shock absorber and of the acceleration through the shock absorbers performance analysis program CASH- are shown in

()-Table A.14.

The acceleration which is 1.2 times the results of the CASH- program utilized in the analysis is also shown in the above table.

87

()-Table A.14 Deformation and acceleration of shock absorber under normal test conditions Acceleration (xg)

Deformation Drop posture Calculation (mm) x 1.2 value Horizontal 20.9 89.3 107.1 Lid side 24.1 58.8 70.5 Vertical Bottom side 18.2 78.9 94.6 Lid side 27.6° 58.6 16.3 19.6 Corner Bottom side 22.8° 50.3 17.3 20.8 5° 21.5 14.1 16.9 15° 41.5 13.0 15.6 30° 60.8 17.1 20.5 Lid side 45° 65.8 21.7 26.0 60° 59.3 25.7 30.8 75° 46.9 34.5 41.4 85° 27.4 36.5 43.8 Inclined 5° 22.2 5.96 7.15 15° 40.1 16.8 20.2 30° 56.2 19.5 23.4 Bottom side 45° 60.4 22.5 27.0 60° 61.4 24.9 29.9 75° 44.4 29.0 34.8 85° 25.0 30.2 36.2

• This is the angle of the center line of the package to the drop direction.

(same below) where g: Gravitational acceleration, g = 9.81 [m/s2]

88

(4) Increase in acceleration caused by steel plate (i) Horizontal drop We will obtain the increase in acceleration caused by the steel plate during a horizontal drop.

The position of evaluation is shown in ()-Fig.A.24.

Code Position of evaluation Outside cylinder steel plate Outer lid flange Stiffening ring Outer shell panel Partition Eye-plate Eye-plate fixation plate Flange of the main body of the outer shell Eye-plate fixation leg

()-Fig.A.24 Acceleration evaluation position of steel plate for horizontal drop 89

Outside cylinder steel plate An analytical model of the outside cylinder steel plate as an annulus ring on which the whole weight of the package rests uniformly is shown in ()-Fig.A.25.

()-Fig.A.25 Acceleration analysis model of outer shell plate for horizontal drop As is indicated in ()-Fig.A.25, the bending moment of the annulus ring on which the uniform load w rests, can be given by the following equation.

1 M= wR2 cos + sin + cos sin 2 + cos + ( ) sin 2

In the above equation, M is maximum at =, and the following is obtained, 3

M= wR2 + sin 2 cos + (2 ) sin 2

When the stress generated by the bending moment becomes equal to the deformation stress s, the maximum resistance force F may be generated.

3 wR 2 + sin 2 cos + (2 p) sin M 2 s = =

Zp Zp Therefore, the uniform load w at this time is given by the following 90

Z W=

3 R 2 + in 2 co + (2 ) in 2

Therefore, the maximum resistance force is the following 2 Z F= 2wR=

3 R + in 2 co + (2 ) in 2

where M: Bending moment of the annulus ring [Nmm]

w: Uniform load [N/mm]

F: Maximum resistance force [N]

R: Radius of the annulus ring, R =420 [mm]

s: Deformation stress (at ordinary temperatures),s =520 [N/mm2]

Arbitrary angle based on OC [rad]

Radius of the deformed part, R 1 420 20.9

= cos 1 = cos = 18.15° = 0.317 [rad]

R 420

Deformation, =20.9 [mm]

Zp: Plasticity section modulus, 1 1500 x 3 2 Zp= bh 2 = = 3375 [mm3]

4 4 b: Annulus ring width, b =1500 [mm]

h: Annulus ring thickness, h=3 [mm]

Therefore, the maximum resistance force is, 2 x x 520 x 3375 F=

3 420 x + sin 2 18.15° cos 18.15° + (2 x 0.317 ) sin 18.15° 2

= 3.57x104 [N]

91

The equation of the increase in acceleration NH1 caused by the outside cylindrical steel plate is, F 3.57 x 10 4 NH1= = =37.6=3.83g [m/s2]

m 950 where m: Weight of the package, m=950 [kg]

92

Flange of the outer lid The analytical model is shown in ()-Fig.A.25, as with section A.5.3 (4)(i)

. But, since the deformation has not reached the annulus ring, the in the moment equation is given as 0.

The cross section of the flange of the outer lid is given in ()-Fig.A.26.

5 106 367

()-Fig.A.26 Cross section of outer shell lid flange

[10]

The maximum resistance force is given by the following 4 4x F= sZp= x520x1.40x104 =8.31x104 [N]

3R 3 x 367 where F: Maximum resistance force [N]

R: Radius of the annulus ring, R =367 [mm]

s: Deformation stress (at ordinary temperatures), s =520 [N/mm2]

Zp: Section modulus of plasticity, 1 2 5 x 106 2 Zp = bh = =1.40x104 [mm3]

4 4 b: Annulus ring width, b =5 [mm]

h: Annulus ring thickness, h =106 [mm]

Therefore, the increase in acceleration NH2 caused by the flange of the outer 93

lid is, F 8.31 x 10 4 NH2= = =87.5=8.92g [m/s2]

m 950 94

Stiffening ring The analytical model is shown in ()-Fig.A.25, as with section A.5.3(4)(i)

[10]

. The maximum resistance force is given by the following equation 2 Z F=

3 R + in 2 co + (2 ) in 2

where F: Maximum resistance force [N]

R: Radius of the annulus ring, R =406 [mm]

s: Deformation stress (ordinary temperature), s =520 [N/mm2]

Deformation amount, =6.9 [mm]

Half angle of the deformed part, R 1 406 6.9

= cos 1 = cos = 10.58° = 0.185 [rad]

R 406 Zp: section modulus of plasticity [mm3],

h Zp= {(b-h)2+h2 }

4 3

= {(40-3)2+32 }

4

=1.03x103 [mm3]

b: Ring width, b=40 [mm]

h: Ring thickness, h=3 [mm]

Therefore, F is 2 x x 520 x 1.03 x 10 3 F=

3 406 x + sin 2 10.58° cos 10.58° + (2 x 0.187 ) x sin 10.58° 2

=8.30x103 [N]

Therefore, the increase in acceleration NH3 due to the stiffening ring is F 8.30 x 10 3 NH3= = =8.74=0.891g [m/s2]

m 950 95

Panel of the outer lid The analytical model is shown in ()-Fig.A.27.

()-Fig.A.27 Acceleration analysis model of outer shell head plate for horizontal drop As indicated in ()-Fig.A.27, bending moment is generated by the reaction force of the drop in the outer lid panel at the curved point of the head. When the stress produced by this bending moment becomes equal to the deformation stresss, the maximum resistance force F, assuming that it is generated, is given by the equation ss ss C h 2 F= Zp =

r r 4 where F: Maximum resistance force [N]

s: Deformation stress (room temperature), s =520 [N/mm2]

Zp: Section modulus of plasticity, C h2 Zp= [mm3]

4 C: Shock absorber deformation width, C =262 [mm]

h: Panel thickness, h=3 [mm]

r: Radius of the corner, r=150 [mm]

96

Therefore, the following equation is given.

520 262 x 3 2 F= x =2.04x103 [N]

150 4 Two panels are provided in the packaging, and the increase in acceleration NH4 caused by the outer lid panel is 2 F 2 x 2.04 x 10 3 NH4= = =4.29=0.437g [m/s2]

m 950 97

Partition The analytical model is shown in ()-Fig.A.25, as with section A.5.3(4)(i)

.But, since the deformation has not reached the annulus ring, in the moment equation is given as 0.

The cross section of the partition is given in ()-Fig.A.28.

3 h=97 368.5

()-Fig.A.28 Cross section of partition plate The maximum resistance force is given by the following equation.

4 4x F= zZp= x520x7.06x103 3R 3 x 368.5

=4.17x104 [N]

where F: Maximum resistance force [N]

R: Radius of the annulus ring, R =368.5 [mm]

z : Deformation stress (at ordinary temperatures),z =520 [N/mm2]

Zp: Plasticity section modulus, 1 3 x 97 2 Zp= b h2 = =7.06x103 [mm3]

4 4 b: Annulus ring width, b=3 [mm]

98

h: Annulus ring thickness, h =97 [mm]

Therefore, the increase in acceleration NH5 caused by the partition is obtained by the following.

F 4.17 x 10 4 NH5= = =43.9=4.48g [m/s2]

m 950 99

Eye-plate The analytical model is shown in ()-Fig.A.29.

90 32 15 X-X section

()-Fig.A.29 Deformation analysis model of eye plate As is indicated in ()-Fig.A.29, when the eye-plate is hit by a direct force, maximum compression stress is generated at the cross section X-X. When this stress is equal to the deformation stress s, maximum resistance force F is generated, shown by the following equation F=sA =s(b-d) t where F: Maximum resistance force [N]

s: Deformation stress (at room temperatures),s =520 [N/mm2]

A: Evaluated cross sectional area [mm2]

b: Eye-plate width, b=90 [mm]

t: Eve-plate board thickness, t=15 [mm]

d: Eye-plate hole radius, d=32 [mm]

Therefore, F=520x(90-32)x15=4.52x105 [N]

The increase in acceleration NH6 due to the eye-plate is obtained by the following equation.

F 4.52 x 10 5 NH6= = = 476 = 48.5g [m/s2]

m 950 100

Eye-plate fixation plate The analytical model is shown in ()-Fig.A.30.

(Unit: mm)

()-Fig.A.30 Analytical model of eye-plate fixing-plate As is indicated in ()-Fig.A.30, the fixed bridge beams which receive the concentrated load in their center generate maximum bending moments on both extremities.

When this stress is equal to the deformation stress s, maximum resistance

[7]

force F is generated, shown by the following equation 8

F= sZp 1

where F: Maximum resistance force [N]

s: Deformation stress (at ordinary temperatures),s =520 [N/mm2]

Zp: Plasticity section modulus, 1

Zp= b h2 4

b: Eye-plate width, b=110 [mm]

h: Eye-plate board thickness, h=10 [mm]

l: Distance between fixed points, l=130 [mm]

Therefore, 8 1 F= x520x x110x102 =8.80x104 [N]

130 4 The increase in acceleration NH7 due to the eye-plate fixation plate is obtained by the following equation 101

F 8.80 x 10 4 NH7= = =92.6 [m/s2] =9.44g m 950 102

Flange of the outer shell The analytical model is shown in ()-Fig.A.31.

(Unit: mm)

()-Fig.A.31 Analytical model of flange of outer shell As indicated in ()-Fig.A.31, the fixed beam, having a long thin rectangular cross section, suffers side buckling when receiving the concentrated load on its center. If this buckling load is equal to the maximum resistance force F, it is given by the following equation[19]

16.93 B y C F=

2 where F: Maximum resistance force [N]

Distance between supported points, =130 [mm]

By : Bending rigidity on Y axis, 1 1 By = Ebh 3 = x1.95x105x113x53=2.30x108 [Nmm2]

12 12 E: Modulus of longitudinal elasticity (at ordinary temperatures);

E=1.95x105 [N/mm2]

h: Flange board thickness, h =5 [mm]

b: Flange point width, b =113 [mm]

C: Twisting rigidity, 103

bh 3 h C= 1 0.630 G 3 b 113 x 5 3 5 1 0.630 x x 7.51 x 10 4

=

3 113

=3.44x108 [N/mm2]

G: Modulus of transverse elasticity (at ordinary temperatures);

G=7.51x104 [N/mm2]

Therefore, F is 16.93 2.30 x 10 8 x 3.44 x 10 8 F= 2

=2.82x105 [N]

130 The increase in acceleration NH8 caused by the flange in the main body of the outer shell is F 2.82 x 10 5 NH8= = =297 [m/s2] = 30.3g [m/s2]

m 950 104

Eye-plate fixation lug The analytical model is shown in ()-Fig.A.32.

(Unit: mm)

()-Fig.A.32 Analytical model of eye-plate fixing lug As indicated in ()-Fig.A.32, when the compression stress at the X-X cross section is equal to the deformation stress s, maximum resistance force F is generated and given by F=sA =s2h(b1 + b2) where F: Maximum resistance force [N]

s: Deformation stress (at room temperatures);s =520 [N/mm2]

A: Evaluated cross sectional area [mm2]

b1 : Plate width, b1 =50 [mm]

b2 : Plate width, b2 =40 [mm]

h: Plate thickness, h=4 [mm]

Therefore, F=520x2x4x(50+40)=3.74x105 [N]

The increase in acceleration NH9 due the eye-plate fixation leg is, F 3.74 x 10 5 NH9= = = 394 = 40.2g [m/s2]

m 950 Based on the results mentioned so far, the equation for the total increase in 105

acceleration caused by the steel plate during the horizontal drop is NH = NH1 + NH2 + NH3 + NH4 + NH5 + NH6 + NH7 + NH8 + NH9

= (3.83+8.92+0.891+0.437+4.48+48.5+9.44+30.3+40.2)xg

= 147.0g [m/s2]

106

(ii) Vertical drop We shall obtain the increase in acceleration caused by the steel plate during a vertical drop. An analytical model is given in ()-Fig.A.33.

420 115 3

Corner of r150 outer cylinder Reinforcing plate 15.5° 15.5° 3 Deformation 166.5

()-Fig.A.33 Acceleration analysis model of steel plate for vertical drop As indicated in ()-Fig.A.33, the resistance force is the addition of the strength F1 which compresses the outside cylinder corner and the strength F2 which compresses the conical reinforcement plate. The deformation of the steel plate is equal to the deformation of the shock absorber indicated in ()-Table A.14. The resistance forces F1 and F2, which arise when the stress is equal to the deformation stress, can be obtained by the following equations.[17]

F1=2hrsin2s F2=2h(R2+tan)coss where F1 : Outside cylinder corner resistance force [N]

F2 : Conical reinforcement plate resistance force [N]

107

h : Board thickness, h =3 [mm]

r : Radius of the outside cylinder corner, r=150 [mm]

Angle for deformation ,

=cos-1 1 r

deformation, Lid side vertical drop : 1 = 24.l [mm]

Bottom side vertical drop : 2 = 18.2 [mm]

24.1 1 = cos-1 1 =32.9° 150 18.2 2 = cos-1 1 =28.5° 150 R2 : Radius of the upper part of cone, Lid side vertical drop : R2 = 115 [mm]

Bottom side vertical drop : R2 = 113 [mm]

Conical angle, =15.5° s: Flow stress (at room temperatures), s =Su=52O [N/mm2]

Therefore, F1 and F2 in a lid side vertical drop are as follows, F1 = 2x3x150xsin2 32.9°x520=4.34x105 [N]

F2 = 2x3x(115+24.1xtan15.5°) xcos15.5°x520

=11.49x105 [N]

and in a bottom side vertical drop, F1 = 2x3x150xsin2 28.5°x520=3.35x105 [N]

F2 = 2x3x(113+18.2xtan15.5°) xcos15.5°x520

=11.15x105 [N]

Hence, the acceleration generated by these can be determined by the following equation, F F1 + F2 NV = =

m m In a lid side vertical drop, 4.34 + 11.49 NV = x105 =1.67x103 [m/s2]=170.2g [m/s2]

950 In a bottom side vertical drop, 3.35 + 11.15 NV = x105 =1.53x103 [m/s2]=156.0g [m/s2]

950 where g: Gravitational acceleration, g=9.81 [m/s2]

108

(iii) Corner drop We shall determine the increase in acceleration caused by the steel plate during a corner drop.

The analytical model is shown in ()-Fig.A.34.

Fall angle Volume of deformation

()-Fig.A.34 Acceleration analysis model of steel plate for corner drop.

109

As indicated in ()-Fig.A.34, the maximum resistance force caused by the outer steel plate during a corner drop is given by the following[15]

x tanxsinBcosB 3 3 R R i F o x R o x sin where F : Maximum resistance force [N]

Ro: Cylindrical steel plate outer radius, R 420 [mm]

Ri: Cylindrical steel plate inner radius, R 417 [mm]

h : Cylindrical steel plate board thickness, h 3 [mm]

Drop angle Lid side corner drop: 27.6°O.482 [rad]

Bottom side corner drop: 22.8°0.398 [rad]

Deformation Lid side corner drop 58.6 [mm]

Bottom side corner drop 50.3 [mm]

B : Angle cos-1 R o sin Lid side corner drop, 58.6 cos-1 45.7°0.797 [rad]

420 x sin27.6° Bottom side corner drop, 50.3 cos-1 46.3°0.808 [rad]

420 x sin22.8° s: Deformation stress (at room temperatures),s=520 [N/mm2]

Therefore, F is in the lid side corner drop, 420 3417 3 x tan27.6° x0.797sin45.7° x cos45.7° F x520 420 x sin27.6° 6.55x105 [N]

and in the bottom side corner drop, 420 3417 3 x tan22.8° x0.808sin46.3° x cos46.3° F x520 420 x sin22.8° 6.53x105 [N]

Therefore, the acceleration generated by these is given by the following equation.

F N

m In the lid side corner drop, 6.55 x 10 5 N 689 [m/s2]70.3[m/s2]

950 and in the bottom side corner drop, 6.53 x 10 5 N 687 [m/s2]70.0[m/s2]

950 110

(5) Design acceleration As with the corner drop, we shall determine the acceleration during an inclined drop. This is shown in ()-Table A.15. In addition, we shall calculate the design acceleration utilized in the drop stress analysis which will be summarized in the same table.

Design acceleration = Calculation results of CASH-x1.2Acceleration due to steel plate

()-Table A.15 Design acceleration under normal test conditions Acceleration Design CASH-Drop posture due to steel acceleration x1.2 plate (xg) (xg)

Horizontal 107.1 147.0 254.1 Lid side 70.5 170.2 240.7 Vertical Bottom side 94.6 156.0 250.6 Lid side 27.6° 19.6 70.3 89.9 Corner Bottom side 22.8° 20.8 70.0 90.8 5° 16.9 161.7 178.6 15° 15.6 90.7 106.3 30° 20.5 67.9 88.4 Lid side 45° 26.0 56.3 82.3 60° 30.8 50.7 81.5 75° 41.4 58.9 100.3 85° 43.8 75.6 119.4 Inclined 5° 7.15 169.2 176.4 15° 20.2 86.4 106.6 30° 23.4 60.5 83.9 Bottom side 45° 27.0 49.6 76.6 60° 29.9 53.3 83.2 75° 34.8 54.3 89.1 85° 36.2 65.9 101.8 where g: Gravitational acceleration, g = 9.81 [m/s2]

111

(6) Stress analysis of 1.2m horizontal drop The stress analysis of the 1.2 m horizontal drop are conducted separately with the main body, the fuel basket and the fuel element. In addition, as for the stress analysis in each of these sections, the only principal stress will be determined, the evaluation of the stress intensity and the stress classification shall be conducted in section A.5.3(6)(d).

(a) Main body of the packaging The stress evaluation positions of the main body of the packaging during the 1.2 m horizontal drop are determined as shown in ()-Fig.A.35 from a sealing performance preservation.

Symbol Evaluation position Shock absorber (deformation quantity)

Inner shell Bottom plate of the inner shell Top part of the inner shell (Inner lid)

Inner lid clamping bolt

()-Fig.A.35 Stress evaluation position for 1.2m horizontal drop (main body of inner shell) 112

Deformation of the shock absorber We shall determine that even if the shock absorber is deformed by the 1.2 m horizontal drop, this deformation will not reach the inner shell nor to the inner lid.

The analytical model is shown in ()-Fig.A.36.

Shock absorber

()-Fig.A.36 Analytical model of interference to inner shell due to shock absorber deformation for 1.2 m horizontal drop As is indicated in ()-Fig.A.36, the remaining thickness (mm) of the shock absorber after the 1.2 m horizontal drop can be given by the following equation O

where OMinimum thickness of the shock absorber before the test, O104 [mm]

HDeformation of the shock absorber,H 20.9 [mm]

Therefore, the remaining thickness is 104 20.983.1 [mm]

This determines that the deformation caused by the 1.2 m horizontal drop will concern the shock absorber only, and will not reach the main body of the inner shell nor to the inner 1id.

113

Inner shell

()-Fig.A.37 shows an analytical model of the stress on the inner shell for the 1.2 horizontal drop.

()-Fig.A.37 Stress analysis model of inner shell for 1.2m horizontal drop As is indicated in ()-Fig.A.37, the inner shell is supported at both ends, the beam is assumed to support the uniform load, the bending stress b is at its maximum in the center of the supporting points and can be given by the following equation M

Z where M: Bending moment, F 1 1 M= = mNl [Nmm]

8 8 F: Impact load, F mN [N]

m: Load between the supporting points of the package, m700 [kg]

N: Design acceleration, N =254.1g [m/s2]

l: Length between the supporting points, l 1359 [mm]

1 M x700x254.1x9.81x13592.96x108 [Nmm]

8 Z: section modulus, d 2 d 4 4 Z= x [mm3]

32 d2 114

d2: Outside diameter of the inner shell, d2 480 [mm]

d1: Inside diameter of the inner shell, d1 460 [mm]

480 4 460 4 Z= x = 1.70 x 10 6 [mm3]

32 480 Therefore, the bending stress is given by the following equation.

2.96 x 10 8 174 [N/mm2]

1.70 x 10 6 Bottom plate of the inner shell An analytical model of the stress on the bottom plate of the inner shell for the 1.2 horizontal drop is shown in ()-Fig.A.38.

R1 R2

()-Fig.A.38 Stress analysis model of inner shell bottom plate for 1.2m horizontal drop As is indicated in ()-Fig.A.38, the A-A cross section of the inner shell's bottom plate receives the drop force of the fuel basket for horizontal drop. The stress generated at this time is, F

A where F: Impact force, 1

F mBmFxN [kg]

2 115

mB : Mass of fuel basket, mB 138 [kg]

mF : Mass of contents, mF 92 [kg]

N: Design acceleration, N254.1 g [m/s2]

1 F 13892x254.1 x9.812.87x105 [N]

2 A: Cross sectional area of the inner shell's bottom plate (shaded portion in ()-Fig.A.38)

A R R tan 2 2 R1: Outside radius of inner shell's bottom plate outside the protruding section, R1l3O [mm]

R2: Inside radius of inner shell's bottom plate inside the protruding section, R2105 [mm]

R

Angle, cos-1 2 36.1°0.631 [rad]

R1 A 130x 0.631 105x tan 36.1° 2 2 6.60x103 [mm2]

Therefore, the shearing stress is, 2.87 x 10 43.5 [N/mm]

6.60 x 10 Upper part of the inner shell An analytical model of the stress on the upper part of the inner shell for the 1.2 horizontal drop is shown in ()-Fig.A.39.

N.W 0.5 1.0

()-Fig.A.39 Stress analysis model of inner shell upper part 116

of 1.2m horizontal drop As indicated in ()-Fig.A.39, the inner lid slides to the drop direction and comes in contact with the upper part of the inner shell at point .

Shearing stress is generated in the inner lid, F

A where, F: Impact strength, FNm [N]

m: Weight of the inner lid, m120 [kg]

N: Design acceleration, N254.1g [m/s2]

F 254.1x9.81x120 2.99x105 [N]

A: Cross sectional area of the inner shell's upper part (shaded portion in ()-Fig.A.39),

AR12 R tan 2 2 R1: Outside radius of the inner shell flange, R1307 [mm]

R2: Inside radius of the inner shell, R2230 [mm]

Angle, R

cos 41.5°0.724 [rad]

R A 307 2 x 0.724 230 2 x tan 41.5 2 2 4.35x104 [mm]

Therefore, the shearing stress is, 2.99 x 10 5

= 6.87 [N/mm]

4.35 x 10 4 117

Inner lid clamping bolt An analytical model of the stress on the inner lid clamping bolt for the 1.2 m horizontal drop is shown in ()-Fig.A.40.

()-Fig.A.40 Stress analysis model for inner lid clamping bolt for 1.2m horizontal drop As indicated in ()-Fig.A.40, the momentum of the inner lid acts on the clamping bolts of the inner lid for the 1.2m horizontal drop.

Bending stress b [N/mm2] is thus generated in the clamping bolt, and this is given by the following equation ML max NmLL max LAr 2

I where M: Angular momentum M = NmL [Nmm]

N: Design acceleration, N254.1g [m/s2]

m: Weight of the inner lid, m120 [kg]

L: Moment arm, L 18.0 [mm]

L: Distance from each bolt to the overturning point [mm]

L1 42.5 L5 437.8 L2 121.2 L6 516.5 L3 223.9 L7 559.O L4 335.1 118

Lmax: Distance from the overturning point to the farthest bolt, LmaxL 559 Ar: Cross section of the groove of the inner lid clamping bolt (M24),

Ar dr x20.752 338.2[mm2]

4 4 Therefore, the stress is, 254.1 x 9.81 x 120 x 18.0 x 559 2 x (42.5 121.2 223.9 2 335.12 437.8 2 516.5 2 559 2 ) x 338.2 2 2 4.68 [N/mm]

(b) Fuel basket In this section, we shall analyze the stress generated in the fuel basket at the 1.2m horizontal drop. The fuel basket is the rectangular type. We shall determine the section modulus for this type.

The stress shall be evaluated according to the axial strength of the pipe.

(1) Section modulus of square fuel basket We shall determine the section modulus of the square fuel basket.

The analytical model is shown in ()-Fig.A.41.

()-Fig.A.41 Analytical model of section modulus of rectangular fuel basket 119

(i) Section modulus regarding X-X axis As indicated in ()-Fig.A.41, the section modulus regarding the X-X axis is given by the following equation, 2

10 6Ay1 Z

ey where Zx : Section modulus regarding the X-X axis [mm3]

Io : Second moment of area of a single square pipe, 1 1 I h24h14 1004944 12 12 1.83x106 [mm4]

h1 : Outside dimension of the square pipe, h1100 [mm]

h2 : Inside dimension of the square pipe, h294 [mm]

A : Cross sectional area of the square pipe, Ah12 h22 1002 942 1.16x103 [mm2]

y1 : Distance to the center of the square pipe, y1lOO [mm]

ey : Distance to the top surface of the fuel basket, ey150 [mm]

Therefore, the section modulus is 10 x 1.83 x 10 6 6 x 1.16 x 10 3 x 100 2 Z 5.86x105 [mm3]

150 (ii) Section modulus regarding Y-Y axis As indicated in ()-Fig.A.41, the section modulus regarding the Y-Y axis is given by the following equation 2 2 2 102A 1 22 3 Zy e

where Zy : Section modulus regarding the Y-Y axis [mm3]

Io : Secondary moment of the cross section of a single square pipe, I1.83x106 [mm4]

A: Cross sectional area of the square pipe, A1.16x103 [mm2]

X1 : Distance to the center of the pipe, x1 50 [mm]

X2 : Distance to the center of the pipe, x2 100 [mm]

120

X3 : Distance to the center or the pipe, x3 150 [mm]

ex : Distance to the top part of the fuel basket, ex200 [mm]

Therefore, the following equation is obtained.

10 x 1.83 x 10 6 2 x 1.16 x 10 3 x50 2 2 x 100 2 x 150 2 Zy 200 6.14x105 [mm3]

(iii) Section modulus regarding U-U axis As indicated in ()-Fig.A.41, the section modulus regarding the U-U axis is given by the following equation.

2 2 2 2 10 I 2A v1 v 2 v 3 v 4 Z

ev where Zu : Section modulus regarding the U-U axis [mm3]

Io : Second moment of area for a single square pipe, I 1.83x106 [mm4]

A : Cross sectional area of the square pipe, Al.16xl03 [mm2]

V1 : Distance to the center of the pipe, v1 25 2 35.4 [mm]

V2 : Distance to the center of the pipe, v2 50 2 70.7 [mm]

V3 : Distance to the center of the pipe, v3 75 2 106 [mm]

V4 : Distance to the center of the pipe, v4 100 2 141 [mm]

ev : Distance to the top of the fuel basket, ev 150 2 212 [mm]

Therefore, the sectional modulus is 10 x 1.83 x 10 6 2 x 1.16 x 10 3 x35.4 2 70.7 2 106 2 1412 Zu 212 4.95x105 [mm3]

Of the values mentioned above, the smallest shall be adopted ZminZxZyZu4.95x105 [mm3]

(2) Axial strength of square fuel basket The analytical model is the same as in ()-Fig.A.41.

The bending stress generated in the fuel basket reaches its maximum in the center and is given by the following equation.

121

2 M f p NL b

Z 8Z where Bending stress [N/mm2]

M: Maximum bending moment [Nmm]

f p NL2 M

8 Wf : Uniform weight due to the fuel element (This uniform load should be of the maximum weight per unit length among all square fuel elements (JRR-3 Standard type))

m 92 W 0.08 [kg/mm]

1150 mf : Weight of the fuel element, mf92 [kg]

l: Length of the fuel element, l1150 [mm]

Wp : Uniform weight due to the individual weight of the fuel basket, mp 138 Wp 0.115 [kg/mm]

L 1200 mp : Weight of the fuel basket, mp138 [kg]

L: Length of the supporting point, L1200 [mm]

N: Acceleration, N254.1 g [m/s2]

Z: Section modulus of the fuel basket, Z4.95x105 [mm3]

Therefore, the bending stress is, 0.080.115 x 254.1 x 9.81 x 1200 2 177 [N/mm2]

8 x 4.95 x 10 5 122

(c) Fuel elements and fuel plate (Non-irradiated fresh fuels for research reactor)

(c)-1. Fuel elements In this paragraph, an analysis of stress is performed on fuel elements for the 1.2 m horizontal drop. As indicated in (I)-D, specifications of the rectangular fuel elements.

(1) Evaluation of the fuel elements for a drop case Fuel elements are evaluated for two cases of horizontal drop as shown in

()-Fig.A.42.

Horizontal drop to the direction perpendicular to the fuel plate Rectangular fuel elements Horizontal drop to the direction parallel to the fuel plate Rectangular fuel elements

()-Fig.A.42 Evaluation of fuel elements for 1.2 m horizontal drop 123

(2) Fuel elements (i) Fuel plate As shown in ()-D with regard to the rectangular fuel element, there are 11 types of fresh fuel elements including 3 KUR fuel elements, and there are 9 types of lowly irradiated fuel elements. In this section, horizontal drop to the direction perpendicular to fuel plate and to the direction parallel to the fuel plate are treated separately. Furthermore, it is assumed that uranium aluminum alloy has the same strength as the covering material.

Horizontal drop to the direction perpendicular to fuel plate In this section, the analysis method for JRR-3 standard type is shown and the analysis result for the other 19 types, using the same analysis method, is shown in ()-Table A.16.

The analytical model is shown in ()-Fig.A.43.

1.52 0.76 66.6

()-Fig.A.43 Analytical model of rectangular fuel elements for 1.2 m horizontal drop perpendicular to fuel plate.

As indicated in ()-Fig.A.43, a beam with both ends fixed and receiving uniform load due to dead load will receive maximum bending moment at its fixed end. The bending stress is, M

Z where M: Bending moment per unit [Nmm/mm]

wl 2 M

12 w: Uniform load [N/mm2]

m 0.279 w N x254.1x9.811.36x10-2 ba 66.6 x 770 m: Weight of the fuel plate, m0.279 [kg]

124

N: Design acceleration, N254.1 g [m/s2]

a: Length of the fuel plate, a770 [mm]

l: Distance between fixed points, l66.6 [mm]

Z: Cross sectional area per unit width, 3 3 3 3 1 h 2 h 1 1 1.27 0.51 Z x 0.251 [mm3/mm]

6 h2 6 1.27 h2 : Fuel plate thickness, h2 1.27 [mm]

h1 . Fuel plate core thickness, h1 0.51 [mm]

Therefore, the bending stress is, 1.36 x 10 -2 x 66.6 2 20.0 [N/mm2]

12 x 0.251 Horizontal drop to the direction parallel to the fuel plate As shown in ()-D with regard to the fuel elements, the KUR fuel elements consists of curved fuel plates, so the inertial force of the fuel plates and side plates may cause both compressive and buckling stress. Therefore, buckling stress analysis is performed for KUR fuel elements.

For the KUR special fuel element, the two middle plates of 3.1mm thickness placed in parallel to the fuel plates receive the inertial force of the side plates. Therefore, for the fuel plates of the KUR special element, stress analysis of the middle plates will be perfomed, followed by stress analysis of the fuel plates.

For the fuel elements of other reactor types, the analysis method for JRR-3 standard type is shown in the following and the analysis result for the other 16 types, using the same analysis method, is shown in ()-Table A.16.

The analytical model is shown in ()-Fig.A.44.

()x

()-Fig.A.44 Analytical model of rectangular fuel element for 1.2m horizontal drop parallel to fuel plate 125

As indicated in ()-Fig.A.44, the rectangular plate which receives its dead load and the partial weight of the side plate generates compressive stress .

m Fm xN

[N/mm2]

A ah 2h 1 where N: Design acceleration, N254.1g [m/s2]

mF: Weight of the fuel plate, m 0.279 [kg]

ms: Partial weight of the side plate, ms 0.038 [kg]

a: Length of the fuel plate, a770 [mm]

h2: Fuel plate thickness, h2 1.27 [mm]

h1: Fuel plate core thickness, h10.51 [mm]

Therefore, the compressive stress is, 0.2790.038 x 254.1 x 9.81 1.35 [N/mm2]

770 x1.270.51 Next, the analysis for the KUR fuel elements are summarized as following.

Firstly, KUR standard and half-loaded fuel elements are analyzed. Here, the analysis method for KUR standard type is shown, and the half-loaded fuel elements are analyzed using the same method.

For a curved beam subject to compressive axial load, the maximum bending moment occurs at the concave surface of the beam. Here, the buckling stress is analyzed based on the following formula by Southwell [23], where the combined compressive stress and yield stress are correlated; e L cr y = cr 1 + sec ---------------------

r 2k E where P

cr:bucking stress of the beamaveraged axial compressive stress), [N/mm2]

A P compressive load of the beam [N]

A cross section of the beam [mm2]

y yield stress [N/mm2]

E modulus of direct elasticity [N/mm2]

beam length [mm]

e eccentricity of the beam [mm]

I k radius-of-gyration of area [mm]

A I geometrical moment of inertia [mm4]

Z modulus of section [mm3]

126

Z r [mm]

A The corresponding values for KUR standard elements are as follows; modulus of direct elasticity at 75E = 6.97x104 [N/mm2]

yield stress at 75y = 63.7 [N/mm2]

cross section per fuel plate unit width A = ( 1.52 0.5 )x1= 1.02 [mm2 / mm]

geometrical moment of inertia per fuel plate unit width I = 0.282 [mm4 / mm]

modulus of section per fuel plate unit width Z = 0.371 [mm3 / mm]

eccentricitye = 4 [mm]

widthL = 66 [mm]

Then we obtain; I

k =

A 0.282

=

1.02

= 0.526 [mm]

Z r =

A 0.371

=

1.02

= 0.364 [mm].

By substituting the above values to RHS of Eq., cr could be obtained as cr = 4.67 [N/mm2].

The compressive stress is obtained using the following formula as in the case of JRR-3 standard fuel element; W (mf+ms)N c = = [N/mm2]

A a(h2-h1) where N Design acceleration, N = 254.1g [m/s2]

mfWeight of the fuel plate, mf = 0.235 [kg]

Ms msPartial weight of the side plate,ms =

n Msweight of side plate 0.650 [kg]

nnumber of fuel plates 18 [plates]

127

0.650 ms = = 0.036 [kg]

18 a Length of the fuel plate a = 625 [mm]

h2Fuel plate thickness h2 = 1.52 [mm]

h1Fuel plate core thickness h1 = 0.50 [mm]

Therefore, compressive stress is (0.235+0.036)x254.1x9.81 c =

625x(1.52 - 0.50)

= 1.07 [N/mm2].

This gives c = 1.07 [N/mm2] cr = 4.67 [N/mm2],

which shows that the integrity of fuel plates of KUR standard fuel elements are maintained under the horizontal drop condition to the direction perpendicular to the fuel plate.

Next, the KUR special elements will be analyzed as follows.

The KUR special fuel elements have two middle plates (thickness 3.18mm) placed in parallel to the fuel plates. In the analysis, we first assume that the inertial force of the side plates are totally received by these two middle plates and verify the integrity of the middle plates.

Then, the integrity of the fuel plates will be analyzed.

The compressive stress of the middle plates c are given as follows; W (m+ms)N c = = [N/mm2]

A at where N Design acceleration, N = 254.1g [m/s2]

mmweight of middle plate, mm= at [kg]

a length of middle plate, a = 721 [mm]

tthickness of middle plate, t= 3.18 [mm]

distance between fixed edge, = 66 [mm]

density, = 2.7x10-6 [kg/mm3]

mm= 721x3.18x66x2.7x10-6

= 0.409 [kg]

Ms msweight of side plate section, ms =

n Msweight of side plate, 0.650 [kg]

nnumber of middle plates, 2 [plates]

0.650 ms = = 0.325 [kg]

2 128

Therefore, the compressive stress is (0.409+0.325)x254.1x9.81 c =

721x3.18

= 0.80 [N/mm2],

which is less than the design yield strength of the material at 75 (63.7N/mm2). Thus the integrity of the middle plates are maintained.

The integrity of the fuel plates are analyzed as follows. As the fuel plates of the KUR special type elements are identical to those of the KUR standard fuel elements,the buckling stress cr is identical to that of KUR standard fuel element, i.e.

cr = 4.67 [N/mm2].

The compressive stress c is given as follows; W mfN c = = [N/mm2]

A a(h2-h1) where, N Design acceleration, N = 254.1g [m/s2]

mfWeight of the fuel plate, mf = 0.235 [kg]

a Length of the fuel plate a = 625 [mm]

h2Fuel plate thickness h2 = 1.52 [mm]

h1Fuel plate core thickness h1 = 0.50 [mm].

Therefore, 0.235x254.1x9.81 c =

625x(1.52 - 0.5 )

= 0.92 [N/mm2],

This gives c = 0.92 [N/mm2] cr = 4.67 [N/mm2],

which shows that the integrity of fuel plates of KUR special fuel elements are maintained under the horizontal drop condition to the direction perpendicular to the fuel plate.

129

(ii) Fuel element hold down part The lowly irradiated fuel element, as shown in section I-D, is cut at the portion of the lower adapter and the upper holder in order to reduce the weight. Therefore, since the total length becomes short, a hold-down part is provided to adjust the length. In this section, the stress analysis method and the stress generated at the hold-down. part are shown, the result is described in ()-Table A.16, and the stress analysis model is described in ()-Fig.A.45.

Hold down part Fuel element

()-Fig.A.45 Analytical model of holder As shown ()-Fig.A.45, the hold-down part is considered to be a beam supported at the both end, subjected to the uniform load of its own weight, the maximum bending moment occurs at the center of the beam, and the stress is given as follows.

M Z

where M: Bending moment per unit length [Nmm]

wl 2 M

8 w: Uniform load [N/mm2]

w m z x N l

1.4 x254.1x9.8117.1 204 1

M = x17.1x2042=8.90x104 [N mm]

8 mz: Mass of the hold down part, mz1.4 [kg]

N : Design acceleration, N=254.1g [m/s2]

l : Length of hold down part, l=204 [mm]

z : Modulus of elasticity 4

h h 4

Z= o i 32 ho

=9.242x103 [mm3]

ho: Outside diameter of hold down part; ho60 [mm]

hi: Inside diameter of hold down part; hi52 [mm]

Therefore, 8.90 x 104 b = = 9.63 [N/mm2]

9.242 x 103 130

(c)-2. Fuel plate (for Critical Assembly fuel (KUCA fuel))

In this paragraph, an analysis of stress is performed on KUCA fuel for the 1.2 m horizontal drop. As indicated in (I)-D, specifications of the rectangular fuel plate.

The analytical model is shown in ()-Fig.A.46.

x a

()-Fig.A.46 Analytical model of the fuel plate for 1.2m horizontal drop parallel to fuel plate As indicated in ()-Fig.A.46, the fuel plate which receives its dead load generates compressive stress .

m Fm xN

[N/mm2]

A ah 2h 1 where mF: Weight of the fuel plate [kg]

a: Length of the fuel plate [mm]

h2: Fuel plate thickness [mm]

h1: Fuel plate core thickness [mm]

h2-h1:Cladding thickness [mm]

N: Design acceleration [m/s2]

131

In the case of horizontal drop, the design acceleration is applied to coupon fuel in the direction shown in ()-Fig.A.47 Aluminum Aluminum Case Cover

()-Fig.A.47 Analytical model of the coupon fuel for 1.2m horizontal drop In this case, the total thickness of cladding is 0.8 mm, which is the total of the bottom plate of aluminum case (thickness 0.4 mm) and the aluminum cover (thickness 0.4mm).

mF: Weight of the fuel plate mf=0.036 [kg]

l: Length of the fuel plate l=50.8 [mm]

h2-h1:Cladding thickness h2-h1= 0.8 [mm]

N: Design acceleration N=254.1g [m/s2]

Therefore, the compressive stress c are given as follows.

(0.036x254.1xg)/(50.8x0.8) 2.21 N/mm2 Regarding flat fuel plate, two cases are conceivable: one is horizontal drop in the plane direction of the fuel plate and the other horizontal drop in the 132

direction parallel to the fuel plate.

In the case of horizontal drop in the plane direction of the fuel plate

(()-Fig.A.48), the compressive stress c is obtained by the fuel plate width (62 mm) , fuel core width (56 mm) and as follows.

mF: Weight of the fuel plate mf=0.23 [kg]

l: Length of the fuel plate l=600 [mm]

h2: Fuel plate thickness h2 =62 [mm]

h1: Fuel plate core thickness h1= 56 [mm]

N: Design acceleration N=254.1g [m/s2]

Therefore, the compressive stress c are given as follows.

(0.23x254.1xg)/(600x(62-56))

0.16 N/mm2

()-Fig.A.48 Analytical model of the flat fuel plate for 1.2m horizontal drop in the plane direction of the fuel plate In the case of horizontal drop in the direction parallel to the fuel plate

(()-Fig.A.49),

mF: Weight of the fuel plate mf=0.23 [kg]

l: Length of the fuel plate l=600 [mm]

133

h2: Fuel plate thickness h2 =1.5 [mm]

h1: Fuel plate core thickness h1= 0.5 [mm]

N: Design acceleration N=254.1g [m/s2]

Therefore, the compressive stress c are given as follows.

(0.23x254.1xg)/(600x(1.5-0.5))

0.96 N/mm2

()-Fig.A.49 Analytical model of the flat fuel plate for 1.2m horizontal drop in the direction parallel to the fuel plate 134

(c)-3. Fuel plate (for Spectrum Converter)

In this paragraph, an analysis of stress is performed on Spectrum converter for the 1.2 m horizontal drop. As indicated in (I)-D, the specifications of it is the disk-shaped fuel plate.

The analytical model is shown in ()-Fig.A.50.

b d

R2 R R1

()-Fig.A.50 Analytical model of the spectrum converter for 1.2m horizontal drop parallel to disk face As indicated in ()-Fig.A.46, the torus which receives its dead load receives the maximum bending moment at the bottom surface, and the bending stressb generated is given by the following equation.

6 b

2 where M: Bending moment M = 1.5w 2 Nmm wEqually distributed load wFxxg2R mass of spectrum converter (excluding top and bottom aluminum plate 135

F2.293 kg R Average radius of the inner and outer diameters of the spectrum converter casemm 1Spectrum converter outer diametermm 2Spectrum converter inner diametermm 1-2/22)/2(310-254/2254)/2 141mm Design acceleration 254.12 M = 1.5 x 2.293 x 254.1 x 9.81 x 1412 192.4 kNmm thickness of the spectrum converter 10.62 mm Width of the outer ring aluminum part of the spectrum converter 1 - 2/2310-254/228 mm Therefore, b 6x192.4x103/10.62x282 138.6 mm2 136

(d) Comparison of the allowable stress A summary of the stress evaluation results obtained for each analysis in section

()-A.5.3(6) is given in ()-Table A.16.

As demonstrated in this table, the margin of safety in regard to the design standard value is positive for individual or multiple loads.

Therefore, the soundness of this package is maintained under test conditions of the 1.2m horizontal drop.

137

Stress units

-Table A.16 Stress evaluation for 1.2 m horizontal drop (1/6) ;N/mm2 Stress Stress Stress Primary+secondary Impact Primary stress Fatigue at due to due to stress Stress initial internal thermal Position Pm(PL) Sm MS PL+Pb 1.5Sm MS PL+Pb 3Sm MS PL+Pb Sa N Na DF MS stress +Q +Q+F to be evaluated clamping pressure expansion

-0.0491 1 Frame of Inner shell 2.31 2.36 137 57.0 175 205 0.171 1.15 174 3.18 Bottom plate of 0.953 2 0.098 137 1396 87.1 205 1.35 the inner shell -0.098 138 43.5

-0.0491 Upper part of 2.31 3 the inner shell 2.36 137 57.0 13.9 205 13.7 1.15 (Inner lid) 6.87 174 3.20 Inner shell lid 2/3Sy Sy Sy 4 4.68 177 1.58 182 2.77 182 2.77 clamping bolt 458 687 687 5 Square fuel basket 177 177 205 0.158 PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFCumulative fatigue coefficient; SmDesign stress intensity value; SyYield point of the design; MS; Margin of safety r ; Diameter direction stress o ; Periphery direction stress 2 ; Axial stress b ; Bending stress  ; Shear stress t ; Ability of bolt stress

Stress units

-Table A.16 Stress evaluation for 1.2 m horizontal drop (2/6) ;N/mm2 Stress Stress Stress Primary+secondary Stress Impact Primary stress Fatigue at due to due to stress Position initial internal thermal stress Pm(PL) 2/3Sy MS PL+Pb Sy MS PL+Pb Sy MS PL+Pb Sa N Na DF MS to be evaluated clamping pressure expansion +Q +Q+F JRR-3 standard Surface element 20.0 20.0 63.8 2.19 direction 1 (Uranium silicon aluminum Axial 1.35 1.35 63.8 46.2 dispersion alloy) direction JRR-3 follower Surface element 13.2 13.2 63.8 3.83 direction 2 (Uranium silicon aluminum Axial 1.13 1.13 63.8 55.4 dispersion alloy) direction Surface 16.0 16.0 63.8 2.98 JRR-4 B type direction 3

element Axial 1.17 1.17 63.8 53.5 direction Surface 16.4 16.4 63.8 2.89 JRR-4 L type direction 139 4

element Axial 1.62 1.62 63.8 38.3 direction JRR-4 Surface 22.0 22.0 63.8 1.90 (Uranium silicon direction 5

aluminum Axial dispersion alloy) direction 1.50 1.50 63.8 41.5 Surface JMTR 20.3 20.3 63.8 2.14 direction 6 standard Axial element 1.37 1.37 63.8 45.5 direction Surface JMTR 13.9 13.9 63.8 3.58 direction 7 follower Axial element 1.18 1.18 63.8 53.0 direction PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFCumulative fatigue coefficient; SmDesign stress intensity value; SyYield point of design; MSMargin of safety b ; Bending stress c ; Compression stress

Stress units

-Table A.16 Stress evaluation for 1.2 m horizontal drop36 ;N/mm2 Stress Stress Stress Primary+secondary Stress Impact Primary stress Fatigue at due to due to stress Position initial internal thermal stress Pm(PL) 2/3Sy MS PL+Pb Sy MS PL+Pb Sy MS PL+Pb Sa N Na DF MS to be evaluated clamping pressure expansion +Q +Q+F KUR standard Surface element 13.9 13.9 63.7 3.58 direction 1 (Uranium silicon aluminum Axial 1.06*1 1.06 4.67 3.40 dispersion alloy) direction KUR Special Surface element 13.9 13.9 63.7 3.58 direction 2 (Uranium silicon aluminum Axial 1.06*1 1.06 4.67 3.40 dispersion alloy) direction KUR half-loaded Surface 13.9 13.9 63.7 3.58 element direction 3 (Uranium silicon aluminum Axial 0.92*1 0.92 4.67 4.07 dispersion alloy direction 140 PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFCumulative fatigue coefficient; SmDesign stress intensity value; SyYield point of design; MSMargin of safety b ; Bending stress c ; Compression stress

• 1axial compression stress

Stress units

-Table A.16 Stress evaluation for 1.2 m horizontal drop (4/6) ;N/mm2 Stress Stress Stress Primary+secondary Stress Impact Primary stress Fatigue at due to due to stress Position initial internal thermal stress Pm(PL) 2/3Sy MS PL+Pb Sy MS PL+Pb Sy MS PL+Pb Sa N Na DF MS to be evaluated clamping pressure expansion +Q +Q+F JMTRC Surface 15.5 15.5 63.8 3.11 Standard fuel direction 1 element Axial (A,B,C type) 1.10 1.10 63.8 57.0 direction JMTRC Surface Standard fuel 15.4 15.4 63.8 3.14 direction 2 element (2.2pin,fix type) Axial (B,C type) 1.09 1.09 63.8 57.5 direction JMTRC Surface 23.2 23.2 63.8 1.75 Special fuel direction 3 element Axial (Special A type) 1.10 1.10 63.8 57.0 direction JMTRC Surface 15.9 15.9 63.8 3.01 Special fuel direction 141 4 element Axial (Special B type) 1.37 1.37 63.8 45.5 direction JMTRC Surface Special fuel 23.1 23.1 63.8 1.76 direction 5 element (Special C, Axial 1.65 1.65 63.8 37.6 Special D type) direction Surface JMTRC 9.92 9.92 63.8 5.43 direction 6 fuel follower (HF type) Axial 0.89 0.89 63.8 70.6 direction JMTRC Surface 15.4 15.4 63.8 3.14 Standard fuel direction 7 element Axial (MA,MB,MC type) 1.09 1.09 63.8 57.5 direction JMTRC Surface Special fuel 23.0 23.0 63.8 1.77 direction 8 element Axial (Special MB, 1.08 1.08 63.8 58.0 Special MC type) direction Surface JMTRC 10.0 10.0 63.8 5.38 direction 9 fuel follower Axial (MF type) 0.91 0.91 63.8 69.1 direction PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFCumulative fatigue coefficient; SyYield point of design; MSMargin of safety b ; Bending stress c ; Compression stress

Stress units

-Table A.16 Stress evaluation for 1.2 m horizontal drop (5/6) ;N/mm2 Stress Stress Stress Primary+secondary Stress Impact Primary stress Fatigue at due to due to stress Position initial internal thermal stress Pm(PL) 2/3Sy MS PL+Pb Sy MS PL+Pb Sy MS PL+Pb Sa N Na DF MS to be evaluated clamping pressure expansion +Q +Q+F JMTRC Special fuel element hold 1 down part 9.63 9.63 245 24.4 (Special A type)

JMTRC Special fuel element hold 2 down part c 15.6 15.6 245 14.7 (Special B type)

JMTRC Special fuel element hold 3 down part c 9.63 9.63 245 24.4 (Special C,Special D type)

JMTRC 142 Special fuel element hold 4 down part c 9.63 9.63 245 24.4 (Special MB,Special MC type)

PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFCumulative fatigue coefficient; SyYield point of design; MSMargin of safety b ; Bending stress c ; Compression stress

Stress units

-Table A.16 Stress evaluation for 1.2 m horizontal drop (6/6) ;N/mm2 Stress Stress Stress Primary+secondary Stress Impact Primary stress Fatigue at due to due to stress Position initial internal thermal stress Pm(PL) 2/3Sy MS PL+Pb Sy MS PL+Pb Sy MS PL+Pb Sa N Na DF MS to be evaluated clamping pressure expansion +Q +Q+F 1 KUCA coupon c 2.21 2.21 63.7 28.8 Surface 0.16 0.16 63.7 398 direction 2 KUCA flat plate Axial c 0.96 0.96 63.7 66.4 direction 3 Spectrum Converter b 138.6 138.6 245 0.77 PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFCumulative fatigue coefficient; 143 SyYield point of design; MSMargin of safety b ; Bending stress c ; Compression stress

(7) Analysis of stress for 1.2 m bottom side vertical drop Analysis of stress for 1.2 m bottom side vertical drop is separately performed for the main body of the packaging and fuel elements. Stress analysis in each portion should be performed in order to determine the principal stress. Classification of stress and evaluation of stress intensity are conducted in A.5.3 (7)(c).

(a) Main body of the packaging The positions for stress evaluation of the main body of the packaging for 1.2 m bottom side vertical drop are given as in ()-Fig.A.50 from the standpoint of maintaining sealing performance.

Symbol Evaluation position Shock absorber (for quantity of deformation)

Barrel of the inner shell Bottom plate of the inner shell Inner lid Inner lid clamping bolt

()-Fig.A.50 Stress evaluation position for 1.2m lower side vertical drop (main body of packaging) 144

Deformation of the shock absorber Even if the shock absorber is deformed for 1.2 m bottom side vertical drop, the deformation should not reach the bottom of the inner shell.

An analytical model is shown in ()-Fig.A.51.

()-Fig.A.51 Analytical model of interference to inner shell due to shock absorber deformation for 1.2 m lower side vertical drop As is indicated in ()-Fig.A.51, residual quantity (mm) of the shock absorber for 1.2 m bottom side vertical drop is, 0 v where 0 Minimum thickness of the shock absorber before deformation, 0 194 [mm]

v Deformation of the shock absorber, v 18.2 [mm]

Therefore, the deformation is 19418.2175.8 [mm]

Thus, for 1.2 m bottom side vertical drop, deformation suffered by the shock absorber does not reach the bottom of the inner shell.

145

Frame of Inner shell

()-Fig.A.52 shows an analytical model for the stress on the frame of inner shell for 1.2 m bottom side vertical drop.

()-Fig.A.52 Stress analysis model of inner shell for 1.2m lower side vertical drop As shown in ()-Fig.A.52, compression force, due to dead weight and the weight of the peripheral part of the inner lid, acts on the inner shell.

The stress resulting from the compression is, F

A where F: Compressive force acting on the side of inner shell, Fm 1 m 2 m 3 m 4 m 5 N [N]

m 1 : Weight of the inner shell (side and flange), m1200 [kg]

m2 : Weight of the inner lid m2120 [kg]

m3 : Weight of the fuel basket m3138 [kg]

m4 : Weight of the content m4 92 [kg]

m5 : Weight of the outer lid m5120 [kg]

N  : Design acceleration, N250.6g[m/s2]

F(20012013892120)x250.6x9.811.65x106 [N]

A: Cross sectional area of the inner shell 146

A d 2 2d 1 2 [mm2]

4 d 2 : Outer diameter of the inner shell, d2=48O [mm]

d1 : Inner diameter of the inner shell, d 1 =460 [mm]

A 480246021.48x104 [mm2]

4 Therefore,c due to compression is, 1.65 x 10 6 c 111 [N/mm2]

1.48 x 10 4 Bottom plate of the inner shell

()-Fig.A.53 shows the stress analysis model of the inner shell's bottom plate for 1.2 m bottom side vertical drop.

Inner surface Outer surface

()-Fig.A.53 Stress analysis model of inner shell bottom plate for 1.2m lower side vertical drop As indicated in ()-Fig.A.53, the weight of the fuel baskets contents and the dead weight of the inner shell's bottom plate act uniformly on the bottom of the inner shell. The stress, generated on the disc which receives uniform load, reaches its maximum at the fixing point of the circumferentially fixed disc.

147

The stress is w a 2

+/-0.225 h2 w a 2 r +/-0.75 h2 z winner surface where

Circumferential stress [N/mm2]

r: Radial stress [N/mm2]

Z: Axial stress [N/mm2]

a: Inner radius of inner shell's bottom plate, a230 [mm]

h: Thickness of the inner shell's bottom plate, h 35 [mm]

w: Uniform load, m 3 4 7 N w=

a 2 m3 :Weight of the fuel basket, m3 138 [kg]

m4 :Weight of the content m4 92 [kg]

m7 :Weight of the inner shell's bottom plate, m7 55 [kg]

N: Design acceleration, N250.6 g [m/s2]

1389255 x 250.6 x 9.81 w 4.22 [N/mm2]

x 230 2 Therefore, 4.22 x 230 2

+/-0.225x 2

+/-41.0 [N/mm2]

35 4.22 x 230 2

+/-0.75x 2

+/-137 [N/mm2]

35 4.22inner surface [N/mm2]

For the double signs of the stress values, the upper sign () corresponds to the inner surface and the lower sign () to the outer surface 148

Inner lid

()-Fig.A.54 shows the stress analytical model of the inner lid for 1.2 m bottom side vertical drop.

Outer surface Inner surface 2

()-Fig.A.54 Stress analysis model of inner lid for 1.2m lower side vertical drop As indicated in ()-Fig.A.54, the dead weight acts uniformly on the inner lid.

The stress, generated in the disc which receives uniform load, reaches its maximum in the center of the disc.

The stress is, wa 2 1.24 h2 wouter surface where

Radial stress [N/mm2]

Circumferential stress [N/mm2]

Axial stress [N/mm2]

a: Radius of the circle of the inner lid supporting points, a285 [mm]

h: Thickness of the inner lid, h55 [mm]

w: Uniform load resulting from dead weight of the lid, whN7.93x10-6x55x250.6x9.811.07 [N/mm2]

149

N: Design acceleration, N 250.6 g [m/s2]

Density of the inner lid, 7.93x10-6 [kg/mm3]

Hence, 1.07 x 285 2 1.24 = 35.6 [N/mm2]

55 2 1.07outer surface [N/mm2]

For the double signs of the stress values, the upper sign () corresponds to the outer surface and the lower sign () to the inner surface.

Inner lid clamping bolt In a bottom side vertical drop, no load is received by the inner lid clamping bolt.

Therefore, no stress is generated.

150

(b) Fuel elements and fuel plate (b)-1. Fuel elements (Non-irradiated fresh fuels for research reactor)

(1) Fuel plate This section analyzes the stress generated on the rectangular fuel element for 1.2 m bottom side vertical drop.

() In case of calking both ends of fuel plate With regard to the rectangular fuel element, there are 11 types of fresh fuel elements including the follower type, and there are 9 types of lowly irradiated fuel elements. In this section the analysis method is described for the JRR-3 standard, the same analysis was conducted for the other 11 types, and the result is shown in the ()-Table A.17.

However, analysis is performed on the assumption that uranium-aluminum alloy has the same strength as the covering material.

()-Fig.A.55 shows an analytical model.

()-Fig.A.55 Stress analysis model of rectangular fuel element for 1.2m lower side vertical drop.

As indicated in ()-Fig.A.55, the fuel plate is caulked and fixed at both extremities. Its sustaining force is, F H 2b where F H : Strength to sustain the fuel plate [N]

f: Sustaining force per unit length; f26.5 [N/mm]

151

b: Length of the fuel plate; b770 [mm]

Therefore, F H 26.5x2x7704.08x104 [N]

Thus, the force for dropping of the fuel plate is FmN where F: Force for dropping [N]

m: Weight of the fuel plate, m0.279 [kg]

N: Design acceleration, N250.6 g [m/s2]

Therefore, F0.279x250.6x9.81686 [N]

Thus, the fuel plate does not slip down since the force to sustain the fuel plate exceeds the force for dropping of the plate.

As shown above, when a force for dropping due to the dead weight of the fuel plate which is fixed on its extremities acts on it, the shearing stress occurs. This stress is, F

2 h 2 h 1b where

Shearing stress [N/mm2]

F: Force for dropping of the fuel plate, F686 [N]

h2: Thickness of the fuel plate, h21.27 [mm]

h1: Thickness of fuel plate core, h10.51 [mm]

b: Length of the fuel plate, b770 [mm]

Thus, the shearing stress is 686 0.586 [N/mm2]

2 x1.270.51 x 770 152

() In case of fuel plate fixed by pin The stress of the pin fixing of the fuel plate of the lowly irradiated fuel element, generated at 1.2m vertical drop, is analyzed. There are 6 types of lowly irradiated fuel elements including follower types, in this section, the stress analysis method for the pin fixing type fuel element is described and its result is shown in the ()-Table A.17.

The uranium aluminum alloy is treated to have the same strength as the clad material in the analysis.

The analytical model is shown in ()-Fig.A.56.

Pin

()-Fig.A.56 Analytical model of 1.2m lower portion vertical drop of lowly irradiated fuel element As shown in ()-Fig.A.56, the fuel plate is fixed with pin at the side plate.

This retaining force is given as follows.

F H = a x A [N]

Where, FH  : Force for retaining fuel plate [N]

a : Allowable shear stress of pin = 36.8 [N/mm2]

A  : Sectional area of pin [mm2]

= xd2xn 4

d  : Pin diameter = 2.2 [mm]

n  : No. of pin = 62 [-]

F H = a x xd2xn 4

153

= 36.8x x2.22x62=8.67x103 [N]

4 The force acting on the fuel plate due to the acceleration is given as

follows, F = mN Where, F : Force acts on fuel plate when dropping [N]

m : Weight of the fuel plate = 0.217 [kg]

N : Design acceleration = 250.6 [m/s2]

Therefore the following value is obtained.

F = 0.217x250.6x9.81 = 533 [N]

From the above, the retaining force of the pin is larger than the dropping force of the fuel plate by the acceleration. The fuel plate does not slide from fixing.

When the fuel plate, fixed with pin at the both ends, is freely dropped, the tensile force occurs at the pin portion of the fuel plate and is given as follows, Wo t =

A Where, WxN Wo =

n/2 0.217 x 250.6 x 9.81 Wo = = 17.2 62 / 2 t : Stress of fuel plate pin [N/mm2]

Wo : Load acting on fuel pin portion [kg]

n : No. of pin, n = 62 N : Design acceleration, N = 250.6g [m/sec2]

A : Effective sectional area of pin [mm2]

A = ((L 1 - L 2 )/2 - d)xt 1 L 1 : Width of fuel plate, L 1 =70.6 [mm]

L2 : Width of fuel plate core, L2 =61.8 [mm]

154

t1 : Thickness of fuel plate, t =1.27 [mm]

d  : Pin diameter, d =2.35 [mm]

A = ((70.8 - 60.4)/2 - 2.35)x1.27 = 2.60 [mm2]

Therefor the following value of stress is obtained.

17.2 t = = 6.61 [N/mm2]

2.60

() In case of fuel plate not fixed by the side plate The stress of the lowly irradiated fuel element, when dropped vertically from 1.2m height, is analyzed. There are five types of lowly irradiated fuel elements including follower types, in this section. The fuel element not fixed by the side plate is analyzed and the result is shown in ()-Table A.17.

The uranium aluminum alloy is treated to have the same strength as the clad material in this analysis. The analytical model is shown in ()-Fig.A.57.

x

()-Fig.A.57 Analytical model of 1.2m lower portion vertical drop of lowly irradiated fuel element As shown in ()-Fig.A.57, the compressive stress is generated in the rectangular plate subjected to its own weight of the fuel element, and is given as follows, W mF x N c = =

A l (h 2 h1)

Where, m F : Fuel plate mass, m F = 0.223 [kg]

155

l  : Width of fuel plate, l =66.6 [mm]

h2 : Thickness of fuel plate, h2 =1.27 [mm]

h1 : Thickness of fuel plate core, h1 =0.51 [mm]

N  : Design acceleration, N =250.6g [m/s2]

Therefore, the following value of stress is obtained.

0.223 x 250.6g c = = 10.8 [N/mm2]

66.6 x (1.27 0.51)

(2) Fuel element hold-down part As shown in ()-D section, the lowly irradiated fuel elements are cut at the lower adapter portion and the upper holder portion in order to reduce the weight, therefore the total length becomes short, A hold-down part is provided to adjust the length. In this section, the stress analysis method for the stress occurs in the hold down part is shown and the result is summarized in ()-Table A.17.

The analysis model is shown in ()-Fig.A.58.

Fuel element Hold down part

()-Fig.A.58 Analytical model of hold down part As shown in ()-Fig.A.58, the hold-down part is subjected to the own weight and the fuel element weight, and the compressive stress c is generated as follows, W (m z + mf ) x N c = =

A 1 (h o 2 hi 2 )

4 Where, 156

m z : Mass of hold down part, mz = 1.3x2 [kg]

mf : Mass of hold element, mf = 2.0 [mm]

N  : Design acceleration, N =250.6g [m/s2]

ho : Outside diameter of hold down part, ho =60 [mm]

hi : Inside diameter of hold down part, h i =52 [mm]

Therefore, following is obtained.

(2.6 + 2.0) x 250.6 g c = = 16.1 [N/mm2]

1 x (60 2 52 2 )

4 157

(b)-2. Fuel plate for the critical assembly fuel (KUCA fuel)

This section analyzes the stress generated in the fuel plate for the critical assembly at the time of 1.2 m vertical drop. The analysis model is the same as the lowly irradiated fuel elements shown in ()-Fig.A.57 (when the fuel plate and side plate are not fixed).

For the coupon fuel in vertical drop, the design acceleration is applied to the perpendicular direction to the plane of the fuel as shown in ()-Fig.A.59.

hl hg

()-Fig.A.59 Analytical model of 1.2m vertical drop: coupon fuel In the case, the total thickness of cladding is 6 mm, which is the twice width of aluminum case.

m F : Weight of the fuel plate m f =0.036 [kg]

l: Length of the fuel plate l=50.8 [mm]

h 2 -h 1 :Cladding thickness h 2 -h 1 = 6 [mm]

N: Design acceleration N=250.6g [m/s2]

Therefore, the compressive stress c are given as follows.

(0.036x250.6xg)/(50.8x6) 0.30 N/mm2 158

For the flat fuel in vertical drop, the design acceleration is applied to the parallel direction to the plane of the fuel plate as shown in ()-Fig.A.60.

hg hl

()-Fig.A.60 Analytical model of 1.2m vertical drop: flat fuel In the case, m F : Weight of the fuel plate m f =0.23 [kg]

l: Length of the fuel plate l=62 [mm]

h 2 : Fuel plate thickness h 2 =1.5 [mm]

h 1 : Fuel plate core thickness h 1 = 0.5 [mm]

N: Design acceleration N=250.6g [m/s2]

Therefore, the compressive stress c are given as follows.

(0.23x250.6xg)/(62x(1.5-0.5))

9.12 N/mm2 159

(b)-3. Fuel plate for Spectrum converter This section analyzes the stress generated in the spectrum converter at the time of 1.2 m vertical drop. For the spectrum converter in vertical drop, the design acceleration is applied to the perpendicular direction to the plane of the fuel as shown in ()-Fig.A.62.

R2 R1

()-Fig.A.62 Analytical model of 1.2m vertical drop: Spectrum converter In the case, the spectrum converter receives its own weight, and the compressive stress c generated at that time is given by the following equation.

= = x (1 2 2 2 )

4 where mass of spectrum converter 2.49 kg 1 Spectrum converter outer diameter R 1 310 mm 2 Spectrum converter inner diameter R 2 254 mm Design acceleration 250.6 2 Therefore, the compressive stress is the following values.

2.49x250.6x9.81/(/4x31022542) 0.24 mm2 160

(c) Comparison of allowable stress Results of the stress evaluation in each analysis item in ()-5.3 (7) are shown together in ()-Table A.17.

As shown in this table, the margin of safety in regard to analysis reference is positive even if each or combined load is applied.

Therefore, the integrity of this package is maintained under the condition of the 1.2 m bottom side vertical drop test.

161

Stress units

-Table A.17 Stress evaluation for 1.2 m bottom side vertical drop (1/6) ;N/mm2 Stress Stress Stress Primary+secondary Impact Primary stress Fatigue Stress at due to due to stress Position initial internal thermal stress Pm(PL) Sm MS PL+Pb 1.5Sm MS PL+Pb 3Sm MS PL+Pb Sa N Na DF MS to be evaluated clamping pressure expansion +Q +Q+F

-0.0491 1 Frame of Inner shell 2.31 112 137 0.223 1.15 -111 Inner Surface 3.18 137 0.953 41.0 4.32 137 30.7 144.5 205 0.418 Bottom plate of -0.098 -4.22 2

Outer Surface the inner shell -3.18 -137 162

-0.953 -41.0 0 137 140 205 0.464 0 0 Inner Surface

-3.27 35.6 2/3Sy Sy

-3.27 35.6 0.098 4672 32.4 20.2 458 687 Upper part of

-0.098 0 3 the inner shell Outer Surface 3.27 -35.6 (Inner lid) 2/3Sy Sy 3.27 -35.6 1.07 427 31.3 20.9 458 687

-1.07 174 3.20 Inner shell lid 2/3Sy 4 177 1.58 clamping bolt 458 PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFCumulative fatigue coefficient; SmDesign stress intensity value; SyYeild point of the design; MSMargin of safety r Diameter direction stress o Periphery direction stress 2 Axial stress b Bending stress Shear stress t Ability of bolt stress

Stress units

-Table A.17 Stress evaluation for 1.2 m bottom side vertical drop (2/6) ;N/mm2 Stress Stress Stress Primary+secondary Impact Primary stress Fatigue Stress at due to due to stress Position initial internal thermal stress Pm(PL) 2/3Sy MS PL+Pb Sy MS PL+Pb Sy MS PL+Pb Sa N Na DF MS to be evaluated clamping pressure expansion +Q +Q+F JRR-3 standard 1 (Uranium silicon aluminum 0.586 0.586 63.8 107 dispersion alloy)

JRR-3 follower element 2 (Uranium silicon aluminum 0.479 0.479 63.8 132 dispersion alloy) 3 JRR-4 B type element 0.439 0.439 63.8 144 163 4 JRR-4 L type element 0.693 0.693 63.8 91.0 JRR-4 5 (Uranium silicon aluminum 0.603 0.603 63.8 104 dispersion alloy)

JMTR 6 standard element 0.595 0.595 63.8 106 JMTR 7 follower element 0.498 0.498 63.8 127 PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFCumulative fatigue coefficient; SyYield point of design; MSMargin of safety Shear stress

Stress units

N/mm2

-Table A.17 Stress evaluation for 1.2 m bottom side vertical drop (3/6)

Stress Stress Stress Primary+secondary Impact Primary stress Fatigue Stress at due to due to stress Position initial internal thermal stress Pm(PL) 2/3Sy MS PL+Pb Sy MS PL+Pb Sy MS PL+Pb Sa N Na DF MS to be evaluated clamping pressure expansion +Q +Q+F KUR standard 1 (Uranium silicon aluminum 0.454 0.454 63.7 139 dispersion alloy)

KUR Special element 2 (Uranium silicon aluminum 0.454 0.454 63.7 139 dispersion alloy)

KUR half-loaded element 3 (Uranium silicon aluminum 0.454 0.454 63.7 139 dispersion alloy) 164 PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFCumulative fatigue coefficient; SyYield point of design; MSMargin of safety Shear stress

-Table A.17 Stress evaluation for 1.2 m bottom side vertical drop (4/6)

Stress Stress Stress Primary+secondary Impact Primary stress Fatigue Stress at due to due to stress Position initial internal thermal stress Pm(PL) 2/3Sy MS PL+Pb Sy MS PL+Pb Sy MS PL+Pb Sa N Na DF MS to be evaluated clamping pressure expansion +Q +Q+F JMTRC 1 Standard fuel element 0.45 0.45 63.8 140 (A,B,C type)

JMTRC Standard fuel element 2 (2.2pin,fix type) t 6.61 6.61 63.8 8.65 (B,C type)

JMTRC 3 Special fuel element c 10.8 10.8 63.8 4.90 (Special A type)

JMTRC 4 Special fuel element c 0.38 0.38 63.8 166 (Special B type) 165 JMTRC 5 Special fuel element c 11.0 11.0 63.8 4.80 (Special C,Special D type)

JMTRC 6 fuel follower 0.36 0.36 63.8 176 (HF type)

JMTRC 7 Standard fuel element 0.45 0.45 63.8 140 (MA,MB,MC type)

JMTRC Special fuel element 8 (Special MB,Special MC c 10.7 10.7 63.8 4.96 type)

JMTRC 9 fuel follower 0.36 0.36 63.8 176 (MF type)

PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFCumulative fatigue coefficient; SyYield point of the design; MSMargin of safety c Compression stress Shear stress t Stress of the part of fuel plate pin

-Table A.17 Stress evaluation for 1.2 m bottom side vertical drop (5/6)

Stress Stress Stress Primary+secondary Impact Primary stress Fatigue Stress at due to due to stress Position initial internal thermal stress Pm(PL) 2/3Sy MS PL+Pb Sy MS PL+Pb Sy MS PL+Pb Sa N Na DF MS to be evaluated clamping pressure expansion +Q +Q+F JMTRC Special fuel element hold 1 down part c 16.1 16.1 245 14.2 (Special B type)

PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFFatigue accumulation coefficient; SyYield point of the design; MSMargin of safety c Compression stress 166

-Table A.17 Stress evaluation for 1.2 m bottom side vertical drop (6/6)

Stress Stress Stress Primary+secondary Impact Primary stress Fatigue Stress at due to due to stress Position initial internal thermal stress Pm(PL) 2/3Sy MS PL+Pb Sy MS PL+Pb Sy MS PL+Pb Sa N Na DF MS to be evaluated clamping pressure expansion +Q +Q+F 1 KUCA Coupon type b 0.30 0.30 63.7 212 2 KUCA Flat type b 9.12 9.12 63.7 7.0 3 Spectrum converter c 0.24 0.24 245 1020 PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFFatigue accumulation coefficient; SyYield point of the design; MSMargin of safety b Bending stress c Compression stress 167

(8) Stress analysis for 1.2m lid side vertical drop This section analyzes the stress for 1.2 m lid side vertical drop separately for the main body of the packaging and fuel basket. Stress analysis in each item is performed for the purpose of determining the principal stress.

Classification of stress and evaluation of stress intensity are conducted in A.5.3(8)(c).

(a) Main body of the packaging Stress evaluation positions of the main body of the packaging for 1.2 m lid side vertical drop are determined as shown in ()-Fig.A.61 from the viewpoint of maintaining the containment.

Symbol Evaluation position Shock absorber (Quantity of deformation)

Barrel of an inner shell Bottom plate of an inner shell Inner lid Inner lid clamping bolt

()-Fig.A.61 Stress evaluation position for 1.2m lid side vertical drop (main body of a packaging) 168

Deformation of the shock absorber Even if the shock absorber is deformed for 1.2 m lid side vertical drop, the deformation should not reach the bottom of the inner shell.

An analytical model is shown in ()-Fig.A.62.

()-Fig.A.62 Analytical model of interference to inner shell due to shock absorber deformation for 1.2m lid side vertical drop As is indicated in ()-Fig.A.62, the remaining quantity (mm) of the shock absorber in the 1.2 m lid side vertical drop is, 0

where 0 : Minimum thickness of shock absorber before deformation, 0186 [mm]

v: Deformation of the shock absorber, 24.1 [mm]

Hence, the remaining thickness is, 18624.1161.9 [mm]

Therefore, for 1.2 m lid side vertical drop, it is only the shock absorber that suffers deformation and the deformation does not attain the inner lid.

169

Frame of Inner shell

()-Fig.A.63 is a stress analysis model of the frame of inner shell for 1.2 m lid side vertical drop.

()-Fig.A.63 Stress analysis model of inner shell for 1.2m lid side vertical drop As shown in ()-Fig.A.63, compression, due to dead weight and the weight of the peripheral part of the inner lid, acts on the frame of inner shell.

The stress generated from the compression is given by the following equation.

F A

where F: Compression force acting on the inner shell Fm1m3m6N [N]

m1 : Weight of the inner shell (side and flange), m1210 [kg]

m3 : Weight of the fuel basket, m3138 [kg]

m6 : Weight of the outer shell, m6225 [kg]

N : Acceleration, N240.7g [m/s2]

F(210138225)x240.7x9.811.35x1O6 [N]

A: Cross sectional area of the frame of inner shell, A d 2 2d 1 2 [mm2]

4 170

d2  : Outer diameter of the inner shell, d2 480 [mm]

d1  : Inner diameter of the inner shell, d 1 460 [mm]

480246021.48x104 [mm2]

4 Therefore, 1.35 x 10 6 91.2 [N/mm2]

1.48 x 10 4 Bottom plate of the inner shell

()-Fig.A.64 shows the stress analysis model of the inner shell's bottom plate for 1.2 m lid side vertical drop.

Outer surface Inner surface

()-Fig.A.64 Stress analysis model of inner shell bottom plate for 1.2m lid side vertical drop As indicated in ()-Fig.A.64, the dead weight of both the bottom of the outer shell and the bottom plate of the inner shell act uniformly on the bottom of the inner shell. Stress generated in the disc which receives the uniform load reaches its maximum at the fixed end.

The stress is wa 2

+/-0.225 h2 171

wa 2 r +/-0.75 h2 wOuter surface where

Circumferential stress [N/mm2]

Radial stress [N/mm2]

Axial stress [N/mm2]

a: Inner radius of the inner shell's bottom plate, a 230 [mm]

h: Plate thickness of the inner shell's bottom plate, h35 [mm]

w: Uniform load, m 8 N w 7 a 2 m 7 : Weight of the inner shell's bottom plate, m7 55 [kg]

m 8 : Weight of the bottom of the outer shell, m8 120 [kg]

N: Acceleration, N240.7g [m/s2]

55120 x 240.7 x 9.81 w 2.49 [N/mm2]

x 230 2 Therefore, 2.49 x 230 2

+/-0.225x +/-24.2 [N/mm2]

35 2 2.49 x 230 2

+/-0.75x +/-80.6 [N/mm2]

35 2 2.49outer surface For the double signs of the stress values, the plus sign corresponds to the outer surface and the minus sign to the inner surface.

172

Inner lid

()-Fig.A.65 shows a stress analysis model of the inner lid for 1.2 m lid side vertical drop.

()-Fig.A.65 Stress analysis model of inner lid for 1.2m lid side vertical drop 173

As indicated in ()-Fig.A.65, the weight of the contents and the fuel basket act uniformly on the center of the inner lid, and the dead weight of the inner lid also acts uniformly, while the latter is being supported by the circular reaction of the shock absorber and the inner lid clamping bolt.

The stress, generated on the circumferentially supported disc under these loads, reaches its maximum in the disc center. It can be given by superposing the analysis results of each of the models (1),(2),(3) and (4) shown in ()-Fig.A.65.

Contents and fuel basket As shown in ()-Fig.A.65-(1), the stress, generated within the concentric circle of the circumferentially supported disc under uniform load, reaches its maximum in the disc center. It is given by the following equation[]

2 b 3P 1 b 4 (1 ) ln a

4 (1 )

8 h2 b a P 1 inner surface where

Radial stress [N/mm2]

Circumferential stress [N/mm2]

Axial stress [N/mm2]

a: Radius of supporting points circle on inner lid, a285 [mm]

b: Radius of load, b230 [mm]

h: Plate thickness of the inner lid, h55 [mm]

P1 Uniform load resulting from content and fuel basket, m m P 1 3 2 4 N [N/mm2]

b m 3 : Weight of the fuel basket, m3138 [kg]

m4 : Weight of the bottom of the outer shell, m492 [kg]

N: Acceleration, N240.7g [m/s2]

13892 P1 x 240.7 x 9.813.27 [N/mm2]

x 230 2

Poisson's ratio, 0.3 Therefore, 3 x 3.27 x 230 2 285 230 2 4(1 0.3 ln 4 (10.3 8 x 55 2 230 285 2 174

99.9 [N/mm2]

3.27 inner surface For the double sign of the stress value, the upper sign () corresponds to the inner surface and the lower sign () to the outer surface.

The supporting point reaction R 1 in this case is, R 1 m 3 m 4 N 13892x240.7x9.815.43x105 [N]

Dead weight of inner lid As indicated in ()-Fig.A.65-(2), the stress, generated on the circumferentially supported disc under the uniform load resulting from the disc's dead weight, reaches its maximum at the disc center. It is given by the following equation P2a 2 1.24 h2 P 2 inner surface where Radial stress [N/mm2]

Circumferential stress [N/mm2]

Axial stress [N/mm2]

a: Radius of supporting points circle on inner lid, a285 [mm]

h: Plate thickness of the inner lid, h55 [mm]

N: Acceleration, N240.7g [m/s2]

Density of the inner lid, 7.93x10-6 [kg/mm3]

P2 : Uniform load resulting from the lid's dead weight, P2hN 7.93xl0-6x55x240.7x9.811.03 [N/mm2]

Hence, the stress on the lid is 1.03 x 285 2 1.24 2 34.3 [N/mm2]

55 1.03inner surface The upper sign () of the stress value corresponds to the inner surface and the lower sign () to the outer surface.

175

The supporting points' reaction force R 2 in this case is as follows.

R 2 P 2 a2 1.03xx28522.63x105 [N]

Deduction of the shock absorber's reaction As shown in ()-Fig.A.65-(3), the stress, generated within the concentric circle of the circumferentially supported disc under uniform load, reaches its maximum at the disc center. It is given by the following equation[7]

3P 2 a c2 3

4 1 ln 4 (1 8h 2 c a2 P 3 inner surface where

Radial stress [N/mm2]

Circumferential stress [N/mm2]

Axial stress [N/mm2]

a: Radius of supporting points circle on inner lid, a 285 [mm]

c: Radius of load; ccOtan11524.1xtan15.5°122 [mm]

cO: Upper radius of the circular cone, cO 115 [mm]

Circular cone angle, 15.5°

Deformation thickness in the shock absorber, 24.1 [mm]

h: Plate thickness of the inner lid, h 55 [mm]

Poisson's ratio, 0.3 P3 : Compressive stress on the shock absorber, P 3 0.932 [N/mm2]

Therefore, 3 x 0.932 x 122 2 285 122 2 4(1 0.3)ln 4 (10.3 8 x 55 2 122 285 2 14.2 [N/mm2]

0.932inner surface[N/mm2]

For the double sign of the stress value, the upper sign () corresponds to the inner surface and the lower sign () to the outer surface.

The supporting points' reaction in this case is R 3 P 3 c20.932xx12224.36x104 [N]

176

Reaction of the shock absorber As shown in ()-Fig.A.65-(4), the stress, generated in the circumferentially supported disc under uniform load of the shock absorber's reaction, reaches its maximum at the disc center, and it is given by the following equation.

P4a 2 r 1.24 h2 P 4 outer surface where

Radial stress [N/mm2]

Circumferential stress [N/mm2]

Axial stress [N/mm2]

a: Radius of supporting points circle on inner lid, a285 [mm]

h: Plate thickness of the inner lid, h55 [mm]

P4 : Compressive stress on the shock absorber, P 4 0.932 [N/mm2]

Hence, the stress on the lid is 0.932 x 285 2 1.24 = 31.0 [N/mm2]

55 2 0.932outer surface [N/mm2]

For the double sign of the stress value, the upper sign () corresponds to the inner surface and the lower sign () to the outer surface.

The supporting points' reaction force R 4 in this case is, R 4 P 4 a2-0.932xx2852-2.38x105 [N]

On the basis of the results mentioned above, the superposed reaction is, 99.9 34.3 14.2 +/- 31.0 = 117 [N/mm2]

3.271.030.9325.23inner surface[N/mm2]

The upper signs of these terms correspond to the inner surface and the lower signs to the outer surface.

The combined reaction of the supporting points is, R5.432.630.442.38x1056.12x105 [N]

177

Inner lid clamping bolt As indicated in A.5.3 (8)(a)(D), the dead weight of the contents, the fuel basket and the inner lid act on the inner lid. On the other hand, the inner lid is supported by the reaction of the shock absorber, the reaction of the conical reinforcing plate and the inner lid clamping bolt.

The supporting point reaction R works on the inner lid clamping bolts.

Therefore, the tensile stress arising in these bolts is, R

nA where

Tensile stress [N/mm2]

R: Supporting points reaction, R 6.12x105 [N]

n: Number of inner lid clamping bolts, n 16 Ar: Root thread area of the clamping bolt M 24, 2

A r d x 20.752 2 338.2 [mm2]

4 4 d r : Minimum diameter of the clamping bolt, d r 20.752 [mm]

Therefore, 6.12 x 10 5 113 [N/mm2]

16 x 338.2 178

(b) Fuel elements, fuel plate (b)-1. Fuel element

() Fuel plate In this section, the stresses of the rectangular fuel elements are analyzed for 1.2 m lid side vertical drop.

(i) In case of calking both ends of fuel plate With regard to the rectangular fuel element, there are 7 types of fresh fuel elements including the follower type, and there are 9 types of lowly irradiated fuel elements. In this section, the analysis method is described for the JRR-3 standard, the same analysis was conducted for the other 8 types and the result is shown in the ()-Table A.18.

However, the analysis is performed on the assumption that uranium-aluminum alloy has the same strength as the covering material.

()-Fig.A.66 shows an analytical model.

()-Fig.A.66 Stress analysis model of rectangular fuel element for 1.2m lid side vertical drop As indicated in ()-Fig.A.66, the fuel plate is caulked and fixed at both ends and its retaining strength is, F H f2 b where 179

F H : Strength to sustain the fuel plate [N]

f: Sustaining strength per unit length, f26.5 [N/mm]

b: Length of the fuel plate, b770 [mm]

Therefore, F H 26.5x2x7704.08x104 [N]

On the other hand, the force for dropping of the fuel plate is FmN where F: Dropping force of the fuel plate [N]

m: Weight of the fuel plate, m0.279 [kg]

N: Design acceleration, N240.7g [m/s2]

Therefore, F 0.279x240.7x9.81659 [N]

Thus, the fuel plate does not slip down since the strength to sustain the fuel plate exceeds the dropping force.

As shown above, when the fuel plate which is fixed at its both ends suffers a dropping force due to its own dead weight, a shearing stress arises.

F 2 h 2h 1b where

Shearing stress [N/mm2]

F: Dropping force of the fuel plate, F659 [N]

h2: Thickness of the fuel plate, h2l.27 [mm]

h1: Thickness of the core of the fuel plate, h10.51 [mm]

b: Length of the fuel plate, b770 [mm]

Therefore, the shearing stress is, 659 0.563 [N/mm2]

2 x1.270.51 x 770

() In case of fuel plate fixed by pin 180

The stress of the pin-fixing the fuel plate of the lowly irradiated fuel element, generated at 1.2m vertical drop, is analyzed.

There are 6 types of lowly irradiated fuel elements including follower types, in this section, the stress analysis method for the pin fixing type fuel element is shown and it's result is shown in ()-Table A.18.

The uranium aluminum alloy is treated to have the same strength as the clad material in the analysis.

The analytical model is shown in ()-Fig.A.67.

Pin

()-Fig.A.67 Analytical model of 1.2m upper portion vertical drop of lowly irradiated fuel element As shown in ()-Fig.A.67, the fuel plate is fixed with pin at the side plate.

This retaining force is given as follows.

F H = a x A [N]

Where, FH  : Force for retaining fuel plate [N]

a : Allowable shear stress of pin = 36.8 [N/mm2]

A  : Sectional area of pin [mm2]

= xd2xn 4

d  : Pin diameter = 2.2 [mm]

n  : No. of pin = 62 181

Therefore, the following value is obtained.

F H = a x xd2xn 4

= 36.8x x2.22x62=8.67x103 [N]

4 The force acting on the fuel plate due to the acceleration is given as follows, F = mN Where, F : Force acts on fuel plate when dropping [N]

m : Weight of the fuel plate = 0.217 [kg]

N : Design acceleration = 240.7g [m/s2]

Therefore the following value is obtained.

F = 0.217x240.7x9.81 = 512 [N]

From the above, the retaining force of the pin is larger than the dropping force of the fuel plate by the acceleration, the fuel plate does not slide from fixing.

When the fuel plate, fixed with pin at the both ends, is freely dropped, the tensile force occurs at the pin portion of the fuel plate and is given as follows, Wo t =

A Where, WxN Wo =

n/2 0.217 x 240.7 x 9.81 Wo = = 16.5 [N]

62 / 2 t : Stress of fuel plate pin [N/mm2]

Wo  : Load acting on fuel pin portion [kg]

n  : No. of pin, n = 62 N  : Design acceleration, N = 240.7g [m/sec2]

A  : Effective sectional area of pin [mm2]

A = ((L1 - L2)/2 - d)xt1 182

L1 : Width of fuel plate, L1 =70.6 [mm]

L2 : Width of fuel plate core, L2 =61.8 [mm]

t1 : Thickness of fuel plate, t =1.27 [mm]

d  : Pin diameter, d =2.35 [mm]

A = ((70.6 - 61.8)/2 - 2.35)x1.27 = 2.60 [mm2]

Therefore, the following value of stress is obtained.

16.5 t = = 6.35 [N/mm2]

2.60

() In case of fuel plate not fixed by the side plate The stress of the lowly irradiated fuel element, when dropped vertically from 1.2m height, is analyzed. Among five types of lowly irradiated fuel element including follower types, in this section, the fuel element not fixed by the side plate is analyzed and the result is shown in ()-Table A.18.

The uranium aluminum alloy is treated to have the same strength as the clad material in this analysis. The analytical model is shown in ()-Fig.A.68.

x

()-Fig.A.68 Analytical model for 1.2m upper portion vertical drop of lowly irradiated fuel element 183

As shown in ()-Fig.A.68, the compressive stress is generated in the rectangular plate subjected to the fuel element own weight, and is shown as follows, W xN c = = mF A l (h 2 h1)

Where, m F : Fuel plate mass, mF = 0.223 [kg]

l  : Width of fuel plate, l =66.6 [mm]

h2 : Thickness of fuel plate, h2 =1.27 [mm]

h1 : Thickness of fuel plate core, h1 =0.51 [mm]

N  : Design acceleration, N =240.7g [m/s2]

Therefore, the following value of stress is obtained.

0.223 x 240.7g c = = 10.4 [N/mm2]

66.6 x (1.27 0.51)

() Fuel element hold down part As shown in ()-D section, the lowly irradiated fuel elements are cut at the lower adapter portion and the upper holder portion in order to reduce the weight, Therefore the total length becomes short. A hold-down part is provided to adjust the length. In this section, the stress analysis method for the stress occurs in the hold-down part is shown and the result is summarized in ()-Table A.18.

The analysis model is shown in ()-Fig.A.69.

Fuel element Hold down part

()-Fig.A.69 Analytical model of hold down part 184

As shown in ()-Fig.A.69, the hold down part is subjected to the own weight and the fuel element weight, and the following compressive stress is generated.

W ( m z + mf ) x N c = =

A (h o 2 h i 2 )

4 Where, m z : Weight of the hold down part = 1.4 [kg]

mf : Weight of the fuel element = 6.6 [kg]

N : Design acceleration = 240.7g [m/sec2]

ho : Outer diameter of the hold down part = 60 [mm]

hi : Inner diameter of the hold down part = 52 [mm]

Therefore, (1.4 + 6.6) x 240.7g c = = 26.8 [N/mm2]

(60 2 52 2 )

4 185

(b)-2. Fuel plate for the critical assembly fuel (KUCA fuel)

This section analyzes the stress generated in the fuel plate for the critical assembly at the time of 1.2 m vertical drop. The analysis model is the same as the lowly irradiated fuel elements shown in ()-Fig.A.68 (when the fuel plate and side plate are not fixed).

For the coupon fuel in vertical drop, the design acceleration is applied to the perpendicular direction to the plane of the fuel as shown in ()-Fig.A.59 and the total thickness of cladding is 6 mm, which is the twice width of aluminum case.

m F : Weight of the fuel plate m f =0.036 [kg]

l: Length of the fuel plate l=50.8 [mm]

h 2 -h 1 :Cladding thickness h 2 -h 1 = 6 [mm]

N: Design acceleration N=240.7g [m/s2]

Therefore, the compressive stress c are given as follows.

(0.036x240.7xg)/(50.8x6) 0.28 N/mm2 For the flat fuel in vertical drop, the design acceleration is applied to the parallel direction to the plane of the fuel plate as shown in ()-Fig.A.60.

In the case, m F : Weight of the fuel plate m f =0.23 [kg]

l: Length of the fuel plate l=62 [mm]

h 2 : Fuel plate thickness h 2 =1.5 [mm]

h 1 : Fuel plate core thickness h 1 = 0.5 [mm]

N: Design acceleration N=240.7g [m/s2]

186

Therefore, the compressive stress c are given as follows.

(0.23x240.7xg)/(62x(1.5-0.5))

8.76 N/mm2 (b)-3. Fuel plate for the critical assembly fuel (KUCA fuel)

This section analyzes the stress generated in the spectrum converter at the time of 1.2 m vertical drop. The analysis model is the same as the model shown in ()-Fig.A.62.

mass of spectrum converter 2.49 kg 1 Spectrum converter outer diameter R 1 310 mm 2 Spectrum converter inner diameter R 2 254 mm Design acceleration 240.7 2 Therefore, the compressive stress is the following values.

2.49x240.7x9.81/(/4x31022542) 0.23 mm2 187

(c) Comparison of the allowable stresses The results of the stress evaluation concerning each analyzed item in()-5.3(8) are shown together in ()-Table A.18.

As shown in this table, the margin of safety in regard to the analysis reference is positive for individual and combined loads.

Therefore, the integrity of this package is maintained under the 1.2 m lid side vertical drop test conditions.

188

Stress units

-Table A.18 Stress evaluation for 1.2 m lid side vertical drop (1/6) ;N/mm2 Stress Stress Stress Primary+secondary Impact Primary stress Fatigue Stress at due to due to stress Position initial internal thermal stress Pm(PL) Sm MS PL+Pb 1.5Sm MS PL+Pb 3Sm MS PL+Pb Sa N Na DF MS to be evaluated clamping pressure expansion +Q +Q+F

-0.0491 1 Frame of Inner shell 2.31 92.4 137 0.482 1.15 -91.2 Inner Surface 3.18 -80.6 0.953 -24.2 0.098 137 1396 77.3 205 1.65 Bottom plate of -0.098 0 2

Outer Surface inner shell -3.18 80.6 189

-0.953 24.2 2.49 137 54.0 79.9 205 1.56 0 -2.49 Inner Surface

-3.27 -117 2/3Sy Sy

-3.27 -117 5.33 84.9 115 4.97 458 687

-0.098 -5.23 3 Inner shell lid Outer Surface 3.27 117 2/3Sy Sy 3.27 117 0.932 490 121 4.67 458 687

-0.932 174 3.20 113 Clamping bolt of 2/3Sy 4 290 0.579 inner shell lid 458 PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFCumulative fatigue coefficient; SmDesign stress intensity value; SyYield point of the design; MSMargin of safety t ; Ability of bolt stress r ; Diameter direction stress o ;Periphery direction stress 2 ; Axial stress b ;Bending stress  ; Shear stress

Stress units

-Table A.18 Stress evaluation for 1.2 m lid side vertical drop (2/6) ;N/mm2 Stress Stress Stress Primary+secondary Impact Primary stress Fatigue Stress at due to due to stress Position initial internal thermal stress Pm(PL) 2/3Sy MS PL+Pb Sy MS PL+Pb Sy MS PL+Pb Sa N Na DF MS to be evaluated clamping pressure expansion +Q +Q+F JRR-3 standard element 1 (Uranium silicon aluminum 0.563 0.563 63.8 112 dispersion alloy)

JRR-3 follower element 2 (Uranium silicon aluminum 0.460 0.460 63.8 137 dispersion alloy) 3 JRR-4 B type element 0.422 0.422 63.8 150 190 4 JRR-4 L type element 0.666 0.666 63.8 94.7 JRR-4 5 (Uranium silicon aluminum 0.579 0.579 63.8 109 dispersion alloy)

JMTR 6 standard element 0.572 0.572 63.8 110 JMTR 7 follower element 0.479 0.479 63.8 132 PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFCumulative fatigue coefficient; SyYield point of the design; MSMargin of safety  ; Shear stress

Stress units

N/mm2

-Table A.18 Stress evaluation for 1.2 m lid side vertical drop (3/6)

Stress Stress Stress Primary+secondary Impact Primary stress Fatigue Stress at due to due to stress Position initial internal thermal stress Pm(PL) 2/3Sy MS PL+Pb Sy MS PL+Pb Sy MS PL+Pb Sa N Na DF MS to be evaluated clamping pressure expansion +Q +Q+F KUR standard 1 (Uranium silicon aluminum 0.436 0.436 63.7 145 dispersion alloy)

KUR Special element 2 (Uranium silicon aluminum 0.436 0.436 63.7 145 dispersion alloy)

KUR half-loaded element 3 (Uranium silicon aluminum 0.436 0.436 63.7 145 dispersion alloy)

PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; 191 SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFCumulative fatigue coefficient; SyYield point of the design; MSMargin of safety  ; Shear stress

-Table A.18 Stress evaluation for 1.2 m lid side vertical drop (4/6)

Stress Stress Stress Primary+secondary Impact Primary stress Fatigue Stress at due to due to stress Position initial internal thermal stress Pm(PL) 2/3Sy MS PL+Pb Sy MS PL+Pb Sy MS PL+Pb Sa N Na DF MS to be evaluated clamping pressure expansion +Q +Q+F JMTRC 1 Standard fuel element 0.44 0.44 63.8 144 (A,B,C type)

JMTRC Standard fuel element 2 (2.2pin,fix type) t 6.35 6.35 63.8 9.04 (B,C type)

JMTRC 3 Special fuel element c 10.4 10.4 63.8 5.13 (Special A type)

JMTRC 4 Special fuel element c 0.37 0.37 63.8 171 (Special B type) 192 JMTRC 5 Special fuel element c 10.4 10.4 63.8 5.13 (Special C,Special D type)

JMTRC 6 fuel follower 0.34 0.34 63.8 186 (HF type)

JMTRC 7 Standard fuel element 0.43 0.43 63.8 147 (MA,MB,MC type)

JMTRC Special fuel element 8 (Special MB,Special MC c 10.3 10.3 63.8 5.19 type)

JMTRC 9 fuel follower 0.35 0.35 63.8 181 (MF type)

PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFCumulative fatigue coefficient; SyYield point of the design; MSMargin of safety  ; Shear stress c ; Compression stress t ; Stress of the part of fuel pin

-Table A.18 Stress evaluation for 1.2 m lid side vertical drop (5/6)

Stress Stress Stress Primary+secondary Impact Primary stress Fatigue Stress at due to due to stress Position initial internal thermal stress Pm(PL) 2/3Sy MS PL+Pb Sy MS PL+Pb Sy MS PL+Pb Sa N Na DF MS to be evaluated clamping pressure expansion +Q +Q+F JMTRC Special fuel element hold 1 down part c 26.8 26.8 245 8.14 (Special A type)

JMTRC Special fuel element hold 2 down part c 15.4 15.4 245 14.9 (Special B type)

JMTRC Special fuel element hold 3 down part c 27.9 27.9 245 7.78 (Special C,Special D type)

JMTRC 193 Special fuel element hold 4 down part c 27.9 27.9 245 7.78 (Special MB,Special MC type)

PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFCumulative fatigue coefficient; SyYield point of the design; MSMargin of safety c ; Compression stress

-Table A.18 Stress evaluation for 1.2 m lid side vertical drop (6/6)

Stress Stress Stress Primary+secondary Impact Primary stress Fatigue Stress at due to due to stress Position initial internal thermal stress Pm(PL) 2/3Sy MS PL+Pb Sy MS PL+Pb Sy MS PL+Pb Sa N Na DF MS to be evaluated clamping pressure expansion +Q +Q+F 1 KUCA coupon type b 0.28 0.28 63.7 227 2 KUCA Flat type b 8.76 8.76 63.7 7.3 3 Spectrum converter c 0.23 0.23 245 1065 PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; 194 SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFCumulative fatigue coefficient; SyYield point of the design; MSMargin of safety b ; Bending stress c ; Compression stress

(9) Corner drop A corner drop is a special case of inclining drops The package is made to drop with its corner directed downwards, as shown in ()-Fig.A.70, where the line drawn from the center of gravity to the point which first touches the ground meets at a right angle to the solid plane.

(a) Deformation of shock absorber

()-Fig.A.70 shows the relationship between the deformation of the shock absorber and its remaining thickness.

This figure shows that deformation only occurs in parts of the shock absorber and does not reach the inner shell.

()-Fig.A.70 Analytical model of interference to inner shell due to shock absorber deformation for 1.2 m corner drop 195

(b) Stresses on the packaging and content

()-Table A.19 shows the horizontal and vertical components of the design acceleration for the corner drops (see ()-Table A.15).

()-Table A.19 Design acceleration for corner drops (xg)

Drop type Total acceleration Vertical component Horizontal component for specimen (N) (N V = Ncos) (N H = Nsin)

Lid side 89.9 79.7 41.7 Corner Bottom side 90.8 83.7 35.2

()-Table A.19 shows that each accelerating component is smaller than the acceleration recorded in the vertical and horizontal drop. Hence, stress is not analyzed here.

The analyses of the inner lid clamping bolts of different kinds other than those shown in section A.5.3(6) to (8) are described in the following paragraphs.

196

(c) Stress on the inner lid clamping bolts for corner drop During the drop of the bottom side corner, the acceleration of the vertical component is far greater than that of the horizontal component. For this reason, the stress due to momentum on the bolts of the lid can be neglected.

During the drop of the upper corner, stress occurs on the bolts due to momentum of the inner lid. The stress is analyzed in this section.

()-Fig.A.71 shows an analytical model of the stress.

G: Center of gravity of the inner lid H: Horizontal direction pivoting point V: Vertical direction pivoting point

()-Fig.A.71 Analytical model of stress on inner lid clamping bolts for lid side corner drop 197

Bending stress occurs on the inner lid clamping bolts due to the momentum of the inner lid when the package is made to fall with its lid side corner facing downwards (see ()-Fig.A.71).

The maximum bending stress that occurs on the bolt (8) and (8') during this drop is obtained as follows:

max N mLl 8 2 x11 12 13 14 15 16 17 18 Ar 2 2 2 2 2 2 2 2 N HmL Hl'8 2 x1' 2 1' 3 1' 4 1' 5 1' 6 1' 7 1'8 Ar 2 2 2 2 2 2 2 where max : Maximum bending stress on bolts 8 and 8' [N/mm2]

Stress due to vertical acceleration component [N/mm2]

Stress due to horizontal acceleration component [N/mm2]

N  : Vertical acceleration component Nv = 79.7g [m/s2]

N  : Horizontal acceleration component NH = 41.7g [m/s2]

m  : Load applied on the inner lid m = 350 [kg]

L : Vertical momentum arm Lv = 310 [mm]

L : Horizontal momentum arm LH = 18.6 [mm]

l : Distance from pivoting point V to a bolt [mm]

l': Distance from pivoting point H to a bolt [mm]

l 1 = 30.5 1 2 = 73.0 l' 2 = 42.5 1 3 = 151.7 l' 3 = 121.2 1 4 = 254.4 l' 4 = 223.9 1 5 = 365.6 l' 5 = 335.1 1 6 = 468.3 l' 6 = 437.8 1 7 = 547.0 l' 7 = 516.5 1 8 = 589.5 l' 8 = 559.0 Ar: Area of core section of bolt(M24);

Ar = dr 2 = x 20.752 2 = 338.2 [mm2]

4 4 198

Hence, the stresses are obtained as follows:

=

79.7 x 9.81 x 350 x 310 x 589.5 2 x (30.5 73.0 151.7 2 254.4 2 365.6 2 468.3 2 547.0 2 589.5 2 2 2 x 338.2

= 67.6 N/mm2 41.7 x 9.81 x 350 x 18.6 x 559.0

=

2 x42.5 121.2 223.9 2335.12437.8 2516.5 2559 2 2 2 x 338.2

= 2.32 [N/mm2]

()-Table A.20 shows an evaluation of the stresses on the inner lid clamping bolts for lid side corner drop.

199

Stress units

-Table A.20 Stress evaluation for 1.2 m lid side corner drop

N/mm2 Stress Stress Stress Impact Primary+secondary Stress Primary stress Fatigue at due to due to stress stress initial internal thermal Position Horizontal Vertical Pm(PL) Sm MS PL+Pb 1.5Sm MS PL+Pb 3Sm MS PL+Pb Sa N Na DF MS to be evaluated clamping pressure expansion Component Component +Q +Q+F 174 3.20 Inner lid 2/3Sy Sy 1 2.32 67.6 177 1.58 247 1.78 clamping bolt 458 687 PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFCumulative fatigue coefficient; SmDesign stress intensity value; SyYield point of the design; c ; Compression stress MSMargin of safety t ; Ability of bolt stress b ; Bending stress,  ; Shear stress 200

(10) Bottom side inclined drop (a) Deformation in shock absorber

()-Fig.A.72 shows the relationship between the angle of dropping and the deformation for various types of bottom side inclined drop.

Angle of Minimum thickness Deformation of Remaining thickness dropping of shock absorber shock absorber of shock absorber before deformation 5° 211.9 22.2 189.7 15° 236.8 40.1 196.7 30° 260.2 56.2 204.0 45° 265.7 60.4 205.3 60° 252.9 61.4 191.5 75° 222.6 44.4 178.2 85° 193.7 25.0 168.7

()-Fig.A.72 Analytical model of interference with inner shell due to shock absorber deformation for 1.2 m lower side inclined drop

()-Fig.A.72 shows that at the drop, deformation only occurs in parts of the shock absorber and does not reach the inner shell.

201

(b) Stresses on packaging and content

()-Table A.21 shows the horizontal and vertical components of the design acceleration at the bottom side inclined drop (()-Table A.15).

()-Fig.A.73 shows the relationships between the angle of dropping and the acceleration.

()-Table A.21 Relationship between drop angle and acceleration Angle at Acceleration (G) dropping Acceleration Vertical component Horizontal component (degrees) (N) (Ncos) (Nsin) 5 176.4 175.7 15.4 15 106.6 103.0 27.6 30 83.9 72.7 42.0 45 76.6 54.2 54.2 60 83.2 41.6 72.1 75 89.1 23.1 86.1 85 101.8 8.9 101.4

()-Fig.A.73 Relationship between acceleration and drop angle for 1.2 m lower side inclined drop

()-Table A.21 shows that each acceleration component is smaller than the acceleration recorded in the horizontal and vertical drop.

Hence, stress is not analyzed here.

202

(11) Lid side inclined drop (a) Deformation of the shock absorber

()-Fig.A.74 shows the relationship between the dropping angle and the deformation.

Angle at Minimum thickness Deformation in Remaining thickness dropping of shock absorber shock absorber of shock absorber

() before deformation 5° 201.1 21.5 179.6 15° 210.5 41.5 169.0 30° 212.2 60.8 151.4 45° 199.1 65.8 133.3 60° 171.9 59.3 112.6 75° 132.7 46.9 85.8 85° 101.1 27.4 73.7

()-Fig.A.74 Analytical model of interference with inner shell due to shock absorber deformation for 1.2 m upper side inclined drop

()-Fig.A.74 shows that deformation only occurs in parts of the shock absorber and does not reach the inner shell.

203

(b) Stresses on the packaging and content

()-Table A.22 shows the vertical and horizontal components of the design acceleration for lid side inclined drop (see ()-Table A.15).

()-Fig.A.75 shows the relationships between the angle of drop and the acceleration.

()-Table A.22 Relationship between drop angle and acceleration Angle at Acceleration (G) dropping Acceleration Vertical component Horizontal component (degrees) (N) (Ncos) (Nsin) 5 178.6 177.9 15.6 15 106.3 102.7 27.5 30 88.4 76.6 44.2 45 82.3 58.2 58.2 60 81.5 40.8 70.6 75 100.3 26.0 96.6 85 119.4 10.4 118.9

()-Fig.A.75 Relationship between acceleration and drop angle for 1.2 m upper side inclined drop

()-Table A.22 shows that each acceleration component is smaller than the acceleration recorded in the horizontal and vertical drop. Hence, stress is not analyzed here.

204

A.5.4 Stacking test We will analyze here the stresses that may occur on the package when a compressive load on technical standards is applied on it.

In the analysis of the stresses, the principal stress is obtained. The stress classifications and stress intensity evaluations are shown in section A.5.4(3).

(1) Compressive load The specified load to be applied to the specimen under the test condition is ;

the greater of the two, the compressive load W 1 five times as high as the weight of the package, or the load W 2 obtained by multiplying the projected area A of the package by the pressure of 0.013 [MPa] (any which is larger).

For the package in question, these loads are respectively W1 = 5m o g [N]

W2 = 0.013A [N]

Where, m o : Weight of the package, mo = 950 [kg]

A: Projected area of the package, A = D2 = x8402 =5.54x105 [mm2]

4 4 D: Outer diameter of the package, D = 840 [mm]

g: Gravitational acceleration, g = 9.81 [m/s2]

Thus, W 1 = 5x950x9.81 = 4.66x104 [N]

W2 = 0.013x5.54x105 = 7.20x103 [N]

and W 1 > W 2 .

The compressive load F is defined.

F = W 1 =4.66x104 [N]

(2) Analysis of the stresses We will analyze stresses that may be generated when a compressive load is applied for a period of twenty-four hours to different parts of the packaging.

()-Fig.A.76 shows the positions where stresses under the compressive load are to be evaluated.

205

Inner shell lid part The inner shell trunk

()-Fig.A.76 Stress evaluation position for compressive load (A) Inner lid

()-Fig.A.77 shows an analytical model of the inner lid.

()-Fig.A.77 Analytical model of inner lid under compressive load 206

()-Fig.A.77 shows that both its own weight and the compressive load act uniformly on the inner lid which, supported on its circumference, suffers the maximum stress at the center. The stress results are as follows.

w a2 r = = 1.24 h2 z = - w (outer surface) where r : Radial stress [N/mm2]

Circumferential stress [N/mm2]

z  : Axial stress [N/mm2]

a: Diameter of inner lid supporting points, a = 285 [mm]

h: Thickness of inner lid, h = 55 [mm]

w: Uniform load, (m 2 + m 5 ) g + F w = [N/mm2]

a 2 m 2 : Weight of inner lid, m2 = 120 [kg]

m5: Weight of outer lid, m5 = 120 [kg]

F : Compressive load, F = 4.66 x 104 [N/mm2]

(120 + 120) x 9.81 + 4.66 x 10 4 W = =0.192 [N/mm2]

x 285 2 Thus, the stress to be obtained is, 0.192 x 285 2 r = = 1.24 x 2

= 6.39 [N/mm2]

55 z = - 0.192 (outer surface) [N/mm2]

The upper and lower parts of the double sign correspond to the outer and inner surfaces respectively.

207

(B) Inner shell

()-Fig.A.78 shows an analytical model of the inner shell.

()-Fig.A.78 Analytical model of inner shell under compressive load

()-Fig.A.78 shows that both the weight of the inner shell and compressive load act on the inner shell. The stress z which is generated by this compressive force is, F+ mg z =

A where z : Compressive load [N/mm2]

F  : Compressive force [N] F = 4.66x104 [N]

m  : Weight of the inner shell m = m 1 +m 2 +m 3 +m 4 +m 5 +m 6 m 1 : Weight of inner shell (barrel and flanges), m 1 = 200 [kg]

m 2 : Weight of inner lid, m 2 = 120 [kg]

m 3 : Weight of fuel basket, m 3 = 138 [kg]

m 4 : Weight of content, m 4 = 92 [kg]

208

m 5 : Weight of outer shell, m 5 = 120 [kg]

m 6 : Weight of outer lid, m 6 = 120 [kg]

m = 200 + 120 + 138 + 92 + 120 + 120 = 790 [kg]

g : Gravitational acceleration, g = 9.81 [m/s2]

A : Cross section of inner shell, A = (d 2 2 - d 1 2) 4 d 2 : Outer diameter of inner shell, d2 = 480 [mm]

d1 : Inner diameter of inner shell, d 1 = 460 [mm]

A = (4802 - 4602) = 1.48x104 [mm2]

4 Thus, the stress to be obtained is, 4.66 x 10 4 + 895 x 9.81 z = = 3.74 [N/mm2]

1.48 x 10 4 (3) Comparison of allowable stress The results of the stress evaluation from the analyzed items defined in section A.5.4 are put together in ()-Table A.23.

This table shows that in relation to the reference values, a positive margin of safety can be achieved when single or superposed loads are generated.

Therefore, the soundness of the package can be maintained under normal test conditions (compression).

209

Stress units

-Table A.23 Stress evaluation for stacking test ;N/mm2 Stress Stress Stress Primary+secondary Impact Primary stress Fatigue Stress at due to due to stress initial internal thermal Position stress Pm(PL) Sm MS PL+Pb 1.5Sm MS PL+Pb 3Sm MS PL+Pb Sa N Na DF MS to be evaluated clamping pressure expansion +Q +Q+F Inner Surface

-3.27 6.39 2/3Sy Sy

-3.27 6.39 0.098 4672 3.22 212 458 687

-0.098 0 1 Inner shell lid Outer Surface 3.27 -6.39 2/3Sy Sy 210 3.27 -6.39 0.192 458 2384 2.93 687 233 0 -0.192 0.0491 2 Frame of Inner shell 2.31 4.83 137 27.3 1.15 -3.74 PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; Q Secondary stress; F Peak stress; SaRepeated peak stress; N Number of uses; NaPermissible number of repetition; DFCumulative fatigue coefficient; SmDesign stress intensity value; SyYield point of the design; MSMargin of safety r ; Diameter direction stress o ;Periphery direction stress 2 ;Axial stress

A.5.5 Penetration The penetration test is carried out to demonstrate that a bar of 32 mm in diameter and 6kg in weight dropped vertically from a height of 1 meter with its hemispherical end downwards does not penetrate the weakest part of the package.

In the analyses, the contributions from the shock absorber and heat insulator under the outer shell is neglected on the assumption that the entire energy will be consumed in the deformation of the outer steel plate of the outer shell. Thus, the evaluation will ensure the maximum in safety.

The inner shell and lid that form the main structure of the containment system is covered with an outer shell and lid. The thickness of the outer steel sheet is 3 mm. ()-Fig.A.79 shows an analytical model of the package.

()-Fig.A.79 Penetration model We will describe below the case where the testing bar drops and reaches the object in such an orientation that the outer steel sheet is penetrated with the greatest of ease (see ()-Fig.A.79).

The potential energy E 1 [Nmm] of the bar before the drop is obtained as follows.

E 1 = mgh where m: Weight of the bar, m = 6 [kg]

h: Drop height, h = 1000 [mm]

g: Gravitational acceleration, g = 9.81 [m/s2].

211

Thus, E 1 = 6x9.81x1000 = 5.89x104 [Nmm]

The energy E 2 which is necessary to permit the bar to penetrate the 3 mm steel sheet is obtained as follows.

t E 2 = o cr d(t-y)dy Where, cr : Shearing strength of the outer steel sheet, cr = 0.6xSu = 0.6x466 = 280 [N/mm2]

Su: Design tensile strength, Su =466 [N/mm2]

d: Diameter of the bar, d = 32 [mm]

t: Thickness of the outer steel sheet, t = 3 [mm].

When the equation is integrated with the above values, 1

E 2 = cr dx xt2 2

1

= 280xx32x x32 = 1.27x105 [Nmm]

2 E 1 = 5.89x104 [Nmm] E 2 = 1.27x105 [Nmm]

Therefore, the dropping bar does not penetrate the outer steel sheet.

()-Fig.A.80 shows an analytical model for this test.

y cr dy

()-Fig.A.80 Shearing model This concludes that the dropping bar does not adversely affect the containment system or the soundness of the package.

212

A.5.6 Corner or edge drop These requirements should be applied for the wooden or fiber plate made rectangular parallels piped shapes weighting less than 50kg and the cylindrical objects made of fiber plate weight less than 100kg. This packages, weighting 950kg, will be excluded from those requirements.

A.5.7 Summary of results and evaluation An outline of the test results for the package under normal test conditions is given below.

(1) 1.2 m drop As shown in section A.5.3, deformations in the shock absorber in different cases of 1.2 m drop come within the range from 18.2 mm (vertical drop) to 58.6 mm (corner drop). Hence, deformation in each orientation does not affect the inner shell.

The impact accelerations occurring come within the range from 89.9G to 254.1G. Stresses occurring are lower than the analytical reference values. Hence, the package retains its soundness and containment.

(2) Other analyses The tests concerning the pressures at drop, vibration, water spraying, stacking test, and the analyses for penetration, prove that the inner shell constituting the containment barrier retains its sound containment and leaktightness.

(3) Comparison with the allowable stresses The analyses conducted in consideration of the composite effect of different loads described in section A.l.2-(2) show that the package conforms with all the items of the design reference in section A.1.2-(1).

The package retains its sound containment and leaktightness.

213

A.6 Accident test conditions This package is classified as B(U) type, and has the following test conditions set out in the relevant technical standards.

(1) Drop test After the drop test I, package must be exposed to the following conditions.

(2) Drop test (3) Thermal test (4) Water immersion test After these tests (1) to (4) the package must be exposed to the following test conditions.

This section analyzes the effects that the preceding test conditions have on the package and shows how test results satisfy the design standards for the accident test conditions.

A.6.1 Mechanical test - Drop test I (9 m drop) or mechanical test-Drop test (dynamic pressure pickles)

This section describes the effects at 9 m drop that has on the package and covers the following four types of drop, which shows this package can maintain its soundness at 9m drops.

1) Vertical drop (lid side, bottom side)

2) Horizontal drop

3) Corner drop (lid side, bottom side)

4) Inclined drop (lid side, bottom side)

(a) Analysis model Analysis illustrates the stresses generated in these drop tests.

The energy generated by a 9 m drop is absorbed by the deformation of the shock absorber installed at the top and bottom sections of the outer shell.

This section evaluates the shock force applied to the package and analyzes its effects.

(b) Prototype test The details are given in the accompanying document.

214

(c) Model test Not applicable.

This analysis is intended to ensure ;

1) The deformation in outer shell, caused by the 9 m drop, is not transmitted to the sealed inner shell, thus precluding breaking of the containment

2) The impact of the 9 m drop does not damage the inner shell and break the seal.

3) No damage to package content.

(1) Analysis methods The following characteristics of deformation and stress generated in the packaging, fuel baskets and content are analyzed when the 9 m drop tests performed on the package.

(a) Deformation

1) It is assumed that impact is with a rigid surface and the drop energy of the package is absorbed only by the shock absorber. This means the volume of outer shell deformation is equivalent to the extent of shock absorber deformation. It ignores absorption by the metal plating and heat insulator, and leads to the higher deformation values, safety evaluation.

2) The acceleration and volume of deformation caused by the shock absorbing material are calculated using the CASH- absorption performance program described in section A.10.1.

(b) Stress

1) The drop energy of the package is absorbed by the deformation of the shock absorber and the metal plating that constitutes the outer shell body and outer lid.

2) The design acceleration used for analyzing stress is a summation of the acceleration of the metal plating and the CASH- value 215

(acceleration generated in the external shock absorber) multiplied by 1.2 (factor established through comparisons with test results shown in section A.10.1.)

As this acceleration combines the acceleration factors of both the shock absorbing material and metal plating, it is used for safety evaluations of impact generated in the package.

Design accelerations = CASH- resultxl.2 + metal plating acceleration

3) The acceleration generated in the metal plating is obtained by simple calculations.

(2) Drop force As indicated in section A.2 Weight and Center of Gravity, the weight of the package used for analysis is 950kg and drop force is calculated using the following equation:

Ua = Uv = mgh where Ua: Energy absorbed by shock absorber [J]

Uv: Drop energy of the package [J]

m: Mass of transportation packaging, m = 950 [kg]

h: Height of drop, h = 9 [m]

g: Gravitational acceleration g = 9.81 [m/s2]

And drop energy is Ua = Uv = 950x9.81x9 = 8.39x104 [J]

= 8.39x107 [Nmm]

(3) Results of CASH- shock absorber analysis program

()-Table A.24 shows the results of CASH- program calculations of the values for acceleration and deformation generated in the shock absorbing material.

The table also lists the acceleration of the CASH- values multiplied by 1.2, which are used in stress analysis.

216

(4) Design acceleration

()-Table A.25 lists the CASH- calculation code values multiplied by 1.2, shown in ()-Table A.24, and the acceleration factors for identical metal plating described in section A.5.3(4) and calculated using identical procedures.

The design acceleration factors, used for drop stress analysis, are calculated according to the following equation and are also listed in the table.

Design acceleration = CASH- resultx1.2 + metal plating acceleration.

217

()-Table A.24 Deformation and acceleration of shock absorber under accident test conditions Volume of Acceleration [xg]

Drop posture Deformation Calculated

[mm] x1.2 value Horizontal 81.6 162.6 195.1 Lid side 126.7 110.4 132.5 Vertical Bottom side 106.3 102.4 122.9 Lid side 27.6° 128.6 61.8 74.2 Corner Bottom side 22.8° 111.3 65.1 78.1 5° 35.7 34.6 41.5 15° 85.2 50.2 60.2 30° 133.9 62.6 75.1 Lid 45° 145.2 76.5 91.8 side 60° 129.6 81.3 97.6 75° 98.4 81.5 97.8 85° 49.5 61.0 73.2 Inclined 5° 36.0 21.4 25.7 15° 84.6 60.3 72.4 30° 127.1 65.2 78.2 Bottom 45° 135.6 74.8 89.8 side 60° 133.5 76.5 91.8 75° 93.7 67.7 81.2 85° 44.8 45.9 55.1 where g: Gravitational acceleration, g=9.81 [m/s2]

218

()-Table A.25 Design acceleration under accident test conditions Acceleration Design CASH-due to steel acceleration Drop posture plate x1.2

[xg] [xg]

Horizontal 195.1 171.9 367.0 Lid side 132.5 277.3 409.8 Vertical Bottom side 122.9 265.5 388.4 Lid side 27.6° 74.2 225.6 299.8 Corner Bottom side 22.8° 78.1 232.8 310.9 5° 41.5 342.8 384.3 15° 60.2 296.1 356.3 30° 75.1 214.8 289.9 Lid 45° 91.8 176.2 268.0 side 60° 97.6 158.7 256.3 75° 97.8 175.5 273.3 85° 73.2 181.1 254.3 Inclined 5° 25.7 349.2 374.9 15° 72.4 291.7 364.1 30° 78.2 194.9 273.1 Bottom 45° 89.8 160.0 249.8 side 60° 91.8 165.6 257.4 75° 81.2 163.3 244.5 85° 55.1 156.1 211.2 where g: Gravitational acceleration, g=9.81 [m/s2]

219

A.6.1.1 Vertical drop (1) Bottom side vertical drop Shock absorber deformation is 106.3 mm as shown in ()-Table A.24 and acceleration is 388.4g as shown in ()-Table A.25 when package is dropped 9 m vertically onto its bottom.

(a) Deformation in shock absorber This shows that deformation in the shock absorber caused by at 9 m vertical drop onto its bottom is not transmitted to the bottom of the inner shell. ()-Fig.A.81 shows an analytical model.

()-Fig.A.81 Analytical model of interference to inner shell due to shock absorber deformation for 9 m lower side vertical drop As shown in ()-Fig.A.81, the remaning mm of shock absorber after the package is dropped vertically 9 m onto its bottom is calculated using the following equation.

= o - v where o : Minimum thickness of shock absorber prior to deformation, o = 194 [mm]

v : Deformation of shock absorber, v = 106.3 [mm]

The remained thickness is,

= 194 - 106.3 = 87.7 [mm]

The deformation produced by dropping the package vertically 9 m onto its bottom is limited to the shock absorber and is not transmitted to the bottom of the inner shell.

(b) Stress generated in various parts of package The analysis procedures and evaluation positions described in section ()

A.5.3 (7) are used and the analysis and evaluation results are both listed in ()-Table A.26.

220

Stress units

-Table A.26 Stress evaluation for 9 m lower side vertical drop (1/6) ;N/mm2 Stress Stress Impact Primary stress Stress at due to Position initial internal stress Pm(PL) 2/3 Su MS PLPb Su MS to be evaluated clamping pressure

-0.0491 1 Frame of Inner shell 2.31 173 310 0.791 1.15 -172 Outer Surface Inner Surface Outer Surface Inner Surface 3.18 211 0.953 63.4 6.63 310 45.7 221 466 1.10 Bottom plate of -0.098 -6.53 2

inner shell -3.18 -211 221

-0.953 -63.4 0 310 214 466 1.17 0 0

-3.27 55.3 2/3 Sy Sy

-3.27 55.3 0.098 458 4672 52.1 687 12.1

-0.098 0 3 Inner shell lid 3.27 -55.3 2/3 Sy Sy 3.27 -55.3 1.66 458 274 50.4 687 12.6 0 -1.66 174 3.20 Inner shell lid 2/3 Sy 4 177 458 1.58 clamping bolt PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; SyYield point of the design; SuDesign tensile strength; MSMargin of safety r ; Diameter direction stress o ; Periphery direction stress 2 ;Axial stress b ; Bending stress ; Shear stress t ; Ability of bolt stress

Stress units

-Table A.26 Stress evaluation for 9 m lower side vertical drop (2/6) ;N/mm2 Stress Stress Impact Primary stress Stress at due to Position initial internal stress Pm(PL) 2/3 Sy MS PLPb Sy MS to be evaluated clamping pressure JRR-3 standard element 1 (Uranium silicon aluminum 0.908 0.908 63.8 69.2 dispersion alloy)

JRR-3 follower element 2 (Uranium silicon aluminum 0.742 0.742 63.8 84.9 dispersion alloy) 3 JRR-4 B type element 0.680 0.680 63.8 92.8 222 4 JRR-4 L type element 1.074 1.074 63.8 58.4 JRR-4 5 (Uranium silicon aluminum 0.935 0.935 63.8 67.2 dispersion alloy)

JMTR 6 standard element 0.922 0.922 63.8 68.1 JMTR 7 follower 0.772 0.772 63.8 81.6 PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; SyYield point of the design; MSMargin of safety  ; Shear stress

Stress units

-Table A.26 Stress evaluation for 9 m lower side vertical drop (3/6) ;N/mm2 Stress Stress Impact Primary stress Stress at due to Position initial internal stress Pm(PL) 2/3 Sy MS PLPb Sy MS to be evaluated clamping pressure KUR standard 1 (Uranium silicon aluminum 0.703 0.703 63.7 89.6 dispersion alloy)

KUR Special element 2 (Uranium silicon aluminum 0.703 0.703 63.7 89.6 dispersion alloy)

KUR half-loaded element 3 (Uranium silicon aluminum 0.703 0.703 63.7 89.6 dispersion alloy)

PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; 223 SyYield point of the design; MSMargin of safety  ; Shear stress

Stress units

-Table A.26 Stress evaluation for 9 m lower side vertical drop (4/6) ;N/mm2 Stress Stress Impact Primary stress Stress at due to Position initial internal stress Pm(PL) 2/3 Sy MS PLPb Sy MS to be evaluated clamping pressure JMTRC 1 Standard fuel element 0.71 0.71 63.8 88.8 (A,B,C type)

JMTRC Standard fuel element 2 (2.2pin,fix type) t 10.2 10.2 63.8 5.25 (B,C type)

JMTRC 3 Special fuel element c 16.8 16.8 63.8 2.79 (Special A type)

JMTRC 4 Special fuel element c 0.59 0.59 63.8 107 (Special B type)

JMTRC Special fuel element 224 5 c 17.1 17.1 63.8 2.73 (Special C,Special D type)

JMTRC 6 fuel follower 0.56 0.56 63.8 112 (HF type)

JMTRC 7 Standard fuel element 0.70 0.70 63.8 90.1 (MA,MB,MC type)

JMTRC Special fuel element 8 (Special MB,Special MC c 16.6 16.6 63.8 2.84 type)

JMTRC 9 fuel follower 0.56 0.56 63.8 112 (MF type)

PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; SyYield point of the design; MSMargin of safety t ; Ability of bolt stress c ; Compression stress  ; Shear stress

Stress units

-Table A.26 Stress evaluation for 9 m lower side vertical drop (5/6) ;N/mm2 Stress Stress Impact Primary stress Stress at due to Position initial internal stress Pm(PL) 2/3 Sy MS PLPb Sy MS to be evaluated clamping pressure JMTRC Special fuel element hold 1 down part c 24.9 24.9 245 8.83 (Special B type)

PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; SyYield point of the design; MSMargin of safety 2 ;Axial stress 225

Stress units

-Table A.26 Stress evaluation for 9 m lower side vertical drop (6/6) ;N/mm2 Stress Stress Impact Primary stress Stress at due to Position initial internal stress Pm(PL) 2/3 Sy MS PLPb Sy MS to be evaluated clamping pressure 1 KUCA coupon type 0.46 0.46 63.7 138 2 KUCA flat type 14.2 14.2 63.7 4.5 3 Spectrum converter c 0.38 0.38 245 644 PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; 226 SyYield point of the design; MSMargin of safety ; Shear stress

(2) Lid side vertical drop Shock absorber deformation is 126.7 mm as shown in ()-Table A.24 and acceleration is 409.8g as shown in ()-Table A.25 when the package is dropped 9 m vertically onto its top.

(a) Deformation in shock absorber This shows that deformation in the shock absorber caused by a 9 m vertical drop onto its top is not transmitted to the top area of the inner shell. ()-Fig.A.82 shows an analytical model.

()-Fig.A.82 Analytical model of interference to inner shell due to shock absorber deformation for 9 m upper side vertical drop As shown in ()-Fig.A.82, the remaining mm of shock absorber after the package is dropped vertically 9 m onto its top is calculated using the following equation.

= o - v where o : Minimum thickness of shock absorber prior to deformation, o = 186 [mm]

v : Deformation of shock absorber, v = 126.7 [mm]

The remained thickness is,

= 186 - 126.7 = 59.3 [mm]

This shows that deformation in the shock absorber caused by a 9 m vertical drop onto its top is not transmitted to the inner shell lid.

(b) Stress generated in various parts of package The analysis procedures and evaluation positions described in section

()A.5.3(8) are used and the analysis and evaluation results are both listed in ()-Table A.27.

227

Stress units

-Table A.27 Stress evaluation for 9 m upper side vertical drop (1/6) ;N/mm2 Stress Stress Impact Primary stress Stress at due to Position initial internal stress Pm(PL) 2/3 Su MS PLPb Su MS to be evaluated clamping pressure

-0.0491 1 Frame of Inner shell 2.31 156 310 0.987 1.15 -155 Outer Surface Inner Surface Outer Surface Inner Surface 3.18 -137 0.953 -41.1 0.098 310 3162 134 466 2.47 Bottom plate of -0.098 0 2

inner shell -3.18 137

-0.953 41.1 4.23 310 72.2 138 466 2.37 228 0 -4.23

-3.27 -193 2/3 Sy Sy

-3.27 -193 10.1 458 44.3 186 687 2.69

-0.098 -10.0 3 Inner shell lid 3.27 193 Sy 3.27 193 196 687 2.50 0 0 174 3.20 150 Inner shell lid 2/3 Sy 4 327 458 0.400 clamping bolt PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; SyYield point of the design; SuDesign tensile strength; MSMargin of safety r ; Diameter direction stress o ; Periphery direction stress 2 ; Axial stress b ; Bending stress t ; Stress of the part fuel plater pin

Stress units

-Table A.27 Stress evaluation for 9 m upper side vertical drop (2/6)

N/mm2 Stress Stress Impact Primary stress Stress at due to Position initial internal stress Pm(PL) 2/3 Sy MS PLPb Sy MS to be evaluated clamping pressure JRR-3 standard element 1 (Uranium silicon aluminum 0.958 0.958 63.8 65.5 dispersion alloy)

JRR-3 follower element 2 (Uranium silicon aluminum 0.783 0.783 63.8 80.4 dispersion alloy) 3 JRR-4 B type element 0.718 0.718 63.8 87.9 4 JRR-4 L type element 1.133 1.133 63.8 55.3 229 JRR-4 5 (Uranium silicon aluminum 0.987 0.987 63.8 63.6 dispersion alloy)

JMTR 6 standard element 0.973 0.973 63.8 64.5 JMTR 7 follower 0.815 0.815 63.8 77.2 PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; SyYield point of the design; MSMargin of safety  ; Shear stress

Stress units

-Table A.27 Stress evaluation for 9 m upper side vertical drop (3/6)

N/mm2 Stress Stress Impact Primary stress Stress at due to Position initial internal stress Pm(PL) 2/3 Sy MS PLPb Sy MS to be evaluated clamping pressure KUR standard 1 (Uranium silicon aluminum 0.741 0.741 63.7 84.9 dispersion alloy)

KUR Special element 2 (Uranium silicon aluminum 0.741 0.741 63.7 84.9 dispersion alloy)

KUR half-loaded element 3 (Uranium silicon aluminum 0.741 0.741 63.7 84.9 dispersion alloy)

PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; SyYield point of the design; MSMargin of safety  ; Shear stress 230

Stress units

-Table A.27 Stress evaluation for 9 m upper side vertical drop (4/6) ;N/mm2 Stress Stress Impact Primary stress Stress at due to Position initial internal stress Pm(PL) 2/3 Sy MS PLPb Sy MS to be evaluated clamping pressure JMTRC 1 Standard fuel element 0.74 0.74 63.8 85.2 (A,B,C type)

JMTRC Standard fuel element 2 (2.2pin,fix type) t 10.8 10.8 63.8 4.90 (B,C type)

JMTRC 3 Special fuel element c 17.7 17.7 63.8 2.60 (Special A type)

JMTRC 4 Special fuel element c 0.63 0.63 63.8 100 (Special B type)

JMTRC 5 Special fuel element c 18.1 18.1 63.8 2.52 (Special C,Special D type) 231 JMTRC 6 fuel follower 0.59 0.59 63.8 107 (HF type)

JMTRC 7 Standard fuel element 0.73 0.73 63.8 86.3 (MA,MB,MC type)

JMTRC Special fuel element 8 (Special MB,Special MC c 17.6 17.6 63.8 2.62 type)

JMTRC 9 fuel follower 0.59 0.59 63.8 107 (MF type)

PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; SyYield point of the design; MSMargin of safety t ; Stress of the part fuel plater pin  ; Shear stress c ; Compression stress

Stress units

-Table A.27 Stress evaluation for 9 m upper side vertical drop (5/6) ;N/mm2 Stress Stress Impact Primary stress Stress at due to Position initial internal stress Pm(PL) 2/3 Sy MS PLPb Sy MS to be evaluated clamping pressure JMTRC Special fuel element hold 1 down part c 45.7 45.7 245 4.36 (Special A type)

JMTRC Special fuel element hold 2 down part c 26.3 26.3 245 8.31 (Special B type)

JMTRC Special fuel element hold 3 down part c 47.4 47.4 245 4.16 (Special C,Special D type)

JMTRC Special fuel element hold down part 232 4 c 47.4 47.4 245 4.16 (Special MB,Special MC type)

PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; SyYield point of the design; MSMargin of safety t ; Compression stress

Stress units

-Table A.27 Stress evaluation for 9 m upper side vertical drop (6/6) ;N/mm2 Stress Stress Impact Primary stress Stress at due to Position initial internal stress Pm(PL) 2/3 Sy MS PLPb Sy MS to be evaluated clamping pressure 1 KUCA coupon type 0.50 0.50 63.7 127 2 KUCA flat type 15.0 15.0 63.7 4.2 3 Spectrum converter c 0.40 0.40 245 612 PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; SyYield point of the design; MSMargin of safety  ; Shear stress 233

A.6.1.2 Horizontal drop The deformation of 81.6 mm as shown in ()-Table A.24 and acceleration of 367.0g as shown in ()-Table A.25 are generated in the shock absorber when horizontal drop is 9 m.

(1) Deformation in shock absorber This shows that the deformation generated in the shock absorber by a 9 m horizontal drop is not transmitted to the inner shell. ()-Fig.A.83 shows an analytical model.

Shock absorber

()-Fig.A.83 Analytical model of interference to inner shell due to shock absorber deformation for 9 m horizontal drop As shown in ()-Fig.A.83, the remaining mm of shock absorber after a 9 m horizontal drop is calculated by the following equation,

= o - H where o : Minimum thickness of shock absorber prior to deformation, o = 104 [mm]

H : Deformation of shock absorber, H = 81.6 [mm]

The remaining thickness is,

= 104 - 81.6 = 22.4 [mm]

The deformation produced by a 9 m horizontal drop is limited to the shock absorber and is not transmitted to the inner shell.

(2) Stress generated in package and content The analysis procedures and evaluation positions described in section ()

A.5.3(6) are used and both the analysis and evaluation results are listed in ()-Table A.28.

234

Stress units

-Table A.28 Stress evaluation for 9 m horizontal drop (1/6) ;N/mm2 Stress Stress Impact Primary stress Stress at due to Position initial internal stress Pm(PL) 2/3Su MS PLPb Su MS to be evaluated clamping pressure

-0.0491 1 Frame of Inner shell 2.31 2.36 310 130 253 466 0.841 1.15 252 3.18 Bottom area of 0.953 2 0.098 310 3162 125 466 2.72 inner shell -0.098 62.6

-0.0491 Top area of 2.31 2/3 Sy Sy 235 3 2.36 120 50.6 20.0 180 8.00 inner shell 1.15 9.93 174 3.20 Inner shell lid 2/3 Sy Sy 4 6.78 177 458 1.58 184 687 2.73 clamping bolt Rectangular 5 255 255 466 0.827 fuel basket PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; SyYield point of the design; SuDesign tensile strength; MSMargin of safety r ; Diameter direction stress o ; Periphery direction stress 2 ; Axial stress b ; Bending stress t ; Stress of the part fuel plater pin  ; Shear stress

Stress units

-Table A.28 Stress evaluation for 9 m horizontal drop (2/6) ;N/mm2 Stress Stress Impact Primary stress Stress at due to Position initial internal stress Pm(PL) 2/3 Sy MS PLPb Sy MS to be evaluated clamping pressure JRR-3 standard element Surface 28.8 28.8 63.8 1.21 (Uranium silicon directio 1 aluminum Axial 1.94 1.94 63.8 31.8 dispersion alloy) directio JRR-3 follower element Surface 19.0 19.0 63.8 2.35 (Uranium silicon directio 2 aluminum Axial 1.63 1.63 63.8 38.1 dispersion alloy) directio Surface 23.1 23.1 63.8 1.76 directio 3 JRR-4 B type element Axial 1.68 1.68 63.8 36.9 directio Surface 24.6 24.6 63.8 1.59 directio 4 JRR-4 L type element Axial 2.34 2.34 63.8 26.2 directio 236 JRR-4 Surface 31.8 31.8 63.8 1.00 (Uranium silicon directio 5 aluminum Axial 2.17 2.17 63.8 28.4 dispersion alloy) directio Surface 29.6 29.6 63.8 1.15 JMTR directio 6 standard element Axial 1.99 1.99 63.8 31.0 directio Surface 20.1 20.1 63.8 2.17 JMTR directio 7 follower Axial 1.71 1.71 63.8 36.3 directio PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; SyYield point of the design; MSMargin of safety b ; Bending stress c ; Compression stress

Stress units

-Table A.28 Stress evaluation for 9 m horizontal drop (3/6) ;N/mm2 Stress Stress Impact Primary stress Stress at due to Position initial internal stress Pm(PL) 2/3 Sy MS PLPb Sy MS to be evaluated clamping pressure KUR standard Surface 20.1 20.1 63.7 2.16 (Uranium silicon directio 1 Axial aluminum 1.54 *1 1.54 4.67 2.03 dispersion alloy) directio KUR half-loaded Surface 20.1 20.1 63.7 2.16 element directio 2

(Uranium silicon Axial 1.54 *1 1.54 4.67 2.03 aluminum directio KUR Special element Surface 20.1 20.1 63.7 2.16 (Uranium silicon directio 3 Axial aluminum 1.33 *1 1.33 4.67 2.51 dispersion alloy) directio PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; SyYield point of the design; SuDesign tensile strength; MSMargin of safety b ; Bending stress c ; Compression stress

• 1axial compression stress 237

Stress units

-Table A.28 Stress evaluation for 9 m horizontal drop (4/6) ;N/mm2 Stress Stress Impact Primary stress Stress at due to Position initial internal stress Pm(PL) 2/3 Sy MS PLPb Sy MS to be evaluated clamping pressure Surface JMTRC 22.4 22.4 63.8 1.84 directio 1 Standard fuel element Axial (A,B,C type) 1.59 1.59 63.8 39.1 directio JMTRC Surface 22.2 22.2 63.8 1.87 Standard fuel element directio 2 (2.2pin,fix type) Axial (B,C type) 1.58 1.58 63.8 39.3 directio Surface JMTRC 33.4 33.4 63.8 0.91 directio 3 Special fuel element Axial (Special A type) 1.59 1.59 63.8 39.1 directio Surface JMTRC 22.9 22.9 63.8 1.78 directio 4 Special fuel element Axial (Special B type) 1.98 1.98 63.8 31.2 directio 238 JMTRC Surface 33.4 33.4 63.8 0.91 Special fuel element directio 5 (Special C, Axial Special D type) 2.39 2.39 63.8 25.6 directio Surface JMTRC 14.3 14.3 63.8 3.46 directio 6 fuel follower Axial (HF type) 1.29 1.29 63.8 48.4 directio Surface JMTRC 22.3 22.3 63.8 1.86 directio 7 Standard fuel element Axial (MA,MB,MC type) 1.57 1.57 63.8 39.6 directio JMTRC Surface 33.2 33.2 63.8 0.92 Special fuel element directio 8 (Special MB, Axial Special MC type) 1.56 1.56 63.8 39.8 directio Surface JMTRC 14.4 14.4 63.8 3.43 directio 9 fuel follower Axial (MF type) 1.31 1.31 63.8 47.7 directio PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; SyYield point of the design; MSMargin of safety b ; Bending stress c ; Compression stress

Stress units

-Table A.28 Stress evaluation for 9 m horizontal drop (5/6) ;N/mm2 Stress Stress Impact Primary stress Stress at due to Position initial internal stress Pm(PL) 2/3 Sy MS PLPb Sy MS to be evaluated clamping pressure JMTRC Special fuel element hold down 1 part 13.9 13.9 245 16.6 (Special A type)

JMTRC Special fuel element hold down 2 part 22.5 22.5 245 9.88 (Special B type)

JMTRC Special fuel element hold down 3 part 13.9 13.9 245 16.6 (Special C,Special D type)

JMTRC Special fuel element hold down 4 part 1.39 1.39 245 16.6 (Special MB, Special MC type) 239 PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; SyYield point of the design; MSMargin of safety b ; Bending stress

Stress units

-Table A.28 Stress evaluation for 9 m horizontal drop (6/6) ;N/mm2 Stress Stress Impact Primary stress Stress at due to Position initial internal stress Pm(PL) 2/3 Sy MS PLPb Sy MS to be evaluated clamping pressure 1 KUCA coupon type c 3.19 3.19 63.7 20.0 Surface direction 0.24 0.24 63.7 265 2 KUCA flat type Axial direction c 1.39 1.39 63.7 45.8 3 Spectrum converter 200.2 200.2 245 0.23 240 PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; SyYield point of the design; MSMargin of safety b ; Bending stress c ; Compression stress

A.6.1.3 Corner drop (1) Deformation of shock absorber

()-Fig.A.84 shows the deformation and the remaining thickness of the shock absorber. Deformation affects only the external shock absorber and is not transmitted to the inner shell.

Lid side corner drop Bottom side corner drop

()-Fig.A.84 Analytical model of interference to inner shell due to shock absorber deformation for 9m corner drop 241

(2) Stresses of packaging and content

()-Table A.29 shows the design acceleration factors for the corner drop, listed in ()-Table A.25, separated into vertical and horizontal elements.

()-Table A.29 Design acceleration for corner drop (xg)

Vertical acceleration Horizontal acceleration Drop type Acceleration (N)

(N V = Ncos) (N H = Nsin)

Corner Lid side 299.8 265.7 138.9 Bottom side 310.9 286.6 120.5 As ()-Table A.29 shows, acceleration components for all directions are smaller than those for vertical and horizontal drop. For this reason stress analysis is omitted.

The procedures described in section () A.5.3 (9) are used for the inner lid clamping bolts and the analysis and evaluation results are both listed in ()-Table A.30.

242

Stress units

-Table A.30 Stress evaluation for 9 m upper corner drop ;N/mm2 Stress Stress Impact Primary stress Stress at due to stress Position initial internal Horizontal Vertical Pm(PL) 2/3 Sy MS PLPb Sy MS to be evaluated clamping pressure Component Component 174 3.20 Inner lid 1 clamping bolt 7.73 226 177 459 1.58 411 687 0.671 PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; SyYield point of the design; MSMargin of safety t ; Ability of bolt stress b ; Bending stress  ; Shear stress 243

A.6.1.4 Inclined drop (1) Bottom side inclined drop (a) Deformation of shock absorber

()-Fig.A.85 shows the relationship between the angle at dropping and the deformation.

Residual volume Volume of deformation Angle at Minimum thickness Deformation of Remaining thickness dropping of shock absorber shock absorber of shock absorber before deformation (mm) (mm)

(mm) 5° 211.9 36.0 175.9 15° 236.8 84.6 152.2 30° 260.2 127.1 133.1 45° 265.7 135.6 130.1 60° 252.9 133.5 119.4 75° 222.6 93.7 128.9 85° 193.7 44.8 148.9

()-Fig.A.85 Analytical model of interference to inner shell due to shock absorber deformation for 9m lower side inclined drop 244

()-Fig.A.85 shows that in the drop, deformation occurs only in parts of the shock absorber and does not reach the inner shell.

(b) Stresses of packaging and content

()-Table A.31 shows the horizontal and vertical components of the design acceleration for the bottom side corner drop ()-Table A.25.

()-Fig.A.86 shows the relationships between the angle of drop and the acceleration.

()-Table A.31 Relationship between drop angle and acceleration Angle at Acceleration (G) dropping Acceleration (N) Vertical component Horizontal component (Ncos) (Nsin) 5 374.9 373.5 32.7 15 364.1 351.7 94.2 30 273.1 236.5 136.6 45 249.8 176.6 176.6 60 257.4 128.7 222.9 75 244.5 63.3 236.2 85 211.2 18.4 210.4

()-Fig.A.86 Relationship between acceleration and drop angle for 9m lower side inclined drop 245

()-Table A.31 shows that each accelerating component is sma11er than the acceleration recorded for the vertical and horizontal drop. Hence, stress is not analyzed here.

(2) Lid side inclined drop (a) Deformation of the shock absorber

()-Fig.A.87 shows the relationship between the angle at dropping and the deformation.

Residual volume Volume of deformation Angle at Minimum thickness Deformation in Remaining thickness dropping of shock absorber shock absorber of shock absorber before deformation (mm) (mm)

(mm) 5° 201.1 35.7 165.4 15° 210.5 85.2 125.3 30° 212.2 133.9 78.3 45° 199.1 145.2 53.9 60° 171.9 129.6 42.3 75° 132.7 98.4 34.3 85° 101.1 49.5 51.6

()-Fig.A.87 Analytical model of interference to inner shell due to shock absorber deformation for 9 m upper side inclined drop 246

()-Fig.A.87 shows that deformation only occurs in parts of the shock absorber and does not reach the inner shell.

(b) Stresses on the packaging and content

()-Table A.32 shows the horizontal and vertical components of the design acceleration when dropped on the lid side corner ()-Table A.25.

()-Fig.A.88 shows the relationships between the angle of dropping and the acceleration.

()-Table A.32 Relationship between drop angle and acceleration for drop test Angle at Acceleration (G) dropping Acceleration (N) Vertical component Horizontal component (Ncos) (Nsin) 5 384.3 382.8 33.5 15 356.3 344.2 92.2 30 289.9 251.1 145.0 45 268.0 189.5 189.5 60 256.3 128.2 222.0 75 273.3 70.7 264.0 85 254.3 22.2 253.3

()-Fig.A.88 Relationship between acceleration and drop angle for 9 m upper side inclined drop

()-Table A.32 shows that each acceleration component is smaller than the acceleration recorded for the horizontal and vertical drop. Hence, stress is not analyzed here 247

A.6.1.5 Summary of the results We will describe here what deformations occur on the package observed in the mechanical test (drop I). The analysis will evaluate the possibility of the inner shell being damaged.

()-Table A.33 shows the deformations in various drop tests.

()-Table A.33 Relationship between drop angle and acceleration for drop test Analyzed part of Minimum Deformation Remaining Design Item shock absorber thickness in shock thickness acceleration of shock absorber of absorber shock before absorber deformation Drop type (mm) (mm) (mm) xg(m/s2)

Lid side end 186 126.7 59.3 409.8 Vertical drop Bottom side end 194 106.3 87.7 388.4 Cylindrical Horizontal drop 104 81.6 22.4 367.0 part Lid side end 218.9 128.6 90.3 299.8 Corner drop Bottom side end 254.1 111.3 142.8 310.9 Lid side end 201.1 35.7 165.4 384.3 5° Bottom side end 211.9 36.0 175.9 374.9 Lid side end 210.5 85.2 125.3 356.3 15° Bottom side end 236.8 84.6 152.2 364.1 Lid side end 212.2 133.9 78.3 289.9 30° Bottom side end 260.2 127.1 133.1 273.1 Inclined Lid side end 199.1 145.2 53.9 268.0 45° drop Bottom side end 265.7 135.6 130.1 249.8 Lid side end 171.9 129.6 42.3 256.3 60° Bottom side end 252.9 133.5 119.4 257.4 Lid side end 132.7 98.4 34.3 273.3 75° Bottom side end 222.6 93.7 128.9 244.5 Lid side end 101.1 49.5 51.6 254.3 85° Bottom side end 193.7 44.8 148.9 211.2 248

()-Table A.33 shows that deformation occurs only in parts of the shock absorber and does not reach the inner shell in a bottom side corner drop.

()-Tables A.26, A.27, A.28 and A.30 show that stress occurring on the packaging and content for each drop does not exceed the standard value, and therefore does not cause any damage to them.

Thus, the package do not affect the containments and shielding performance of the packaging.

249

A.6.2 Mechanical test - - - Drop test (1 m drop)

In this section we will analyze the package on the assumption that drop test is carried out after drop test I.

We will examine here how the package is affected when it is dropped from the height of one meter onto a mild steel cylinder with a diameter of 150 mm.

()-Fig.A.89 shows the package to be examined in this section for three different drops:

(a) Vertical lid side drop (direct hit on to the outer lid)

(b) Vertical bottom side drop (direct hit on to the outer shell bottom plate)

(c) Horizontal drop (direct hit on to the outer shell).

Package Package Center of Center of gravity gravity Mild steel bar (a) Vertical lid side drop (b) Vertical bottom side drop Center of Package gravity (c) Horizontal drop (direct hit on to the outer shell)

()-Fig.A.89 Analytical model for drop test 250

(1) Penetration We will demonstrate in this section that the evaluated portions shown in

()-Fig.A.89 are not penetrated.

In the analyses, the contributions from the shock absorber and heat insulator under the outer shell is neglected on the assumption that the entire energy will be consumed in the transformation of the outer steel plate of the outer shell. Thus, the evaluation will ensure the maximum of safety.

(a) Direct hit of outer lid onto test cylinder (vertical drop) with the lid side end directed downwards.

()-Fig.A.82(a) shows the case of a direct hit of the outer lid onto the mild steel test cylinder, the dropping energy Uo of the package is obtained by the equation.

Uo = mgH where m: Weight of the package, m = 950 [kg]

H: Height from which the package is dropped, H = 1000 [mm]

g: Gravitational acceleration, g = 9.81 [m/s2]

Thus, Uo is, Uo = 950x9.81x1000 = 9.32x106 [Nmm]

The deformation (U) is obtained on the assumption that the dropping energy Uo is equal to the deforming energy U.

U = s V where s : Stress on the panel, s = 466 [N/mm2]

V: Volume of panel deformed, V = {(d + t)t} [mm3]

d: Diameter of mild steel cylinder, d = 150 [mm]

t: Thickness of the panel, t = 6 [mm]

Deformation [mm]

On the assumption that Uo is equal to U, 9.32x106 = 466x{(150 + 6)6}

251

Hence,

= 6.8 [mm]

When the deformation of 126.7 mm caused in drop test I is added to the above value, we obtain 133.5 mm. As the minimum thickness before deformation of the heat insulator is 186 mm, its remaining thickness after deformation is 52.5 mm. Therefore, deformation does not reach the inner shell.

The strength of the outer lid panel is evaluated on the assumption that the deformational strain is smaller than the specified elongation of the material. Drop test will not cause any penetration in the panel.

()-Fig.A.90 shows an analytical model of the panel.

()-Fig.A.90 Analytical model for penetration strength under conditions of drop test As ()-Fig.A.90 shows, the elongation (1) of the outer lid panel under the conditions of drop test is obtained by the equation 1 = 1'- 1 where 1': Length of the panel after deformation, 1'= 2 x + d [mm]

2 1: Length of the panel before deformation, 1 = 2 x + d [mm]

Deformation, = 6.8 [mm]

d: Diameter of the mild steel bar, d = 150 [mm]

Therefore, 252

1 = 2 x + d - (2 + d) = 1.14 2

The strain in the case of such an elongation is 1 1.14d 1.14 x 6.8

= = = = 0.047 1 2d + d 2 x 6.8 + 150 The strain in head plate is 4.7 (%). Because the outer lid head plate of type SUS 304 has a specified elongation of more than 40 % before penetration, no real penetration can occur.

(b) Direct hit of the outer shell bottom plate onto the mild steel bar (Vertical drop, bottom side down)

As shown in ()-Fig.A.89 (b), the deformation produced when the bottom plate of the outer shell directly hits the mild steel bar, is 6.8mm because the thickness and materials of the bottom plate and head plate are the same as those described in the preceding section.

When the above value is added to the deformation value of 106.3 mm obtained in drop test I, 113.1 mm is obtained. As the minimum thickness of the heat insulator before deformation is 194 mm, its remaining thickness after deformation is 80.9 mm. Therefore, deformation does not reach the inner shell.

The strain is 4.7 %, the same as that described in the preceding section, and likewise, the elongation before penetration is 40 %. Therefore, no penetration occurs in the outer shell bottom plate.

(c) Direct hit of the outer shell on to mild steel bar (horizontal drop)

As ()-Fig.A.89(c) shows, the deformation which occurs when the outer shell directly hits the mild steel bar is obtained by the equation Uo = s {(d + t)t }

Where Uo : Dropping energy, Uo = 9.32x106 [N/mm]

d: Diameter of the mild steel bar, d = 150 [mm]

t: Thickness of shell plate, t = 3 [mm]

s: Deforming stress on the shell, s = 466 [N/mm2]

Hence, 9.32x106 = 466x{(150 + 3)x3 }x

= 13.9 [mm]

253

When the above value is added to the value of deformation 81.6 mm obtained in drop test I, 95.5 mm is obtained. As the remaining thickness of the heat insulator before deformation is 177 mm, its remaining thickness after deformation is 81.5 mm. Therefore, deformation does not reach the inner shell.

As in the preceding cases, the strain is obtained by the following equation.

1 1.14d

1 2d + d where 1: Elongation [mm]

1: Length before deformation [mm]

Deformation, = 13.9 [mm]

d: Diameter of the mild steel bar, d = 150 [mm]

Hence, 1.14 x 13.9

= =0.089 2 x 13.9 + 150 The strain in the shell sheet is 8.9 %. Because the outer lid head plate of type SUS304 has an elongation of more than 40 % before penetration, no real penetration can occur.

(2) Study of the packaging The packages acceleration which occurs at the 1m drop will be obtained in this section.

(a) Lid side vertical drop The acceleration, N, of the package which occurs when the outer lid directly hits the mild steel bar (see ()-Fig.A.89(a)) is obtained by using the analytical model (see ()-Fig.A.90) and the following equation, F

N = [m/s2]

m where F: Reaction force in the deformation of the panel, F =s(d + t)t [N]

254

s: Deforming stress in the panel, s = 466 [N/mm2]

d: Diameter of the mild steel bar, d = 150 [mm]

t: Thickness of the panel, t = 6 [mm]

m: Weight of the package, m = 950 [kg]

Therefore, N is, 466 x x (150 + 6) x 6 N = = 1442 = 147.0g [m/s2]

950 (b) Bottom side vertical drop The acceleration, N, of the package which occurs when the outer lids bottom plate directly hits the mild steel bar (see ()-Fig.A.89(b)) is 147.0 g because the thickness and material of the head plate is the same as those described in the preceding section.

(c) Horizontal drop The acceleration, N, of the package which occurs when the outer shell directly hits the mild steel bar (see ()-Fig.A.89(c))is obtained by using the analytical model (see ()-Fig.A.90) and the following equation, F

N = [m/s2]

m where F: Reaction force in the deformation of the panel, F =s(d + t)t [N]

s: Deforming stress in the panel, s = 466 [N/mm2]

d: Diameter of the mild steel bar, d = 150 [mm]

t: Thickness of the panel, t = 3 [mm]

m: Weight of the package, m = 950 [kg]

Hence, N is 466 x x (150 + 3) x 3 N = = 707 = 72.1g [m/s2]

950 This result of the analysis is smaller than the design acceleration obtained in drop test I (()-Table A.33 shows horizontal: 367.0g, vertical/lid side end: 409.8g; vertical/bottom side end: 388.4g). For this reason, stresses are not analyzed in this section.

255

A.6.2.1 Summary of the results

()-Table A.34 shows the results of the analyses and evaluation of drop test

/mechanical test.

()-Table A.34 Evaluation of penetration for drop test (1) Deformation Minimum insulator Deformation Evaluated Deformation Remaining thickness in drop test position in drop test thickness before deformation (mm) (mm)

(mm) (mm) 1 Outer shell lid 186 126.7 6.8 52.5 Outer shell 2 194 106.3 6.8 80.9 bottom plate Frame of outer 3 177 81.6 13.9 81.5 shell (2) Deformed strain Reference Evaluated Reference in Margin of value Result position analysis safety in analysis 1 Outer shell lid Rupture strain 40 % 4.7 % 7.51 Outer shell 2 Rupture strain 40 % 4.7 % 7.51 bottom plate Frame of outer 3 Rupture strain 40 % 8.9 % 3.49 shell (3) Acceleration Reference Evaluated Reference in Margin of value Result position analysis safety in analysis Acceleration 1 Outer shell lid 409.8g 147.0g 1.79 in drop test I Outer shell Acceleration 2 388.4g 147.0g 1.64 bottom plate in drop test I Frame of outer Acceleration 3 367.0g 72.1g 4.09 shell in drop test I

()- of the packaging and contents are not damaged because the acceleration Table A.34 shows that the deformed strain of different parts observed in drop test is smaller than the reference elongation of SUS304. Therefore, no penetration occurs and the damage in this case does not reach the inner shell.

The acceleration occurring at drop test is lower than that which occurs at drop test I.

Thus, dropping conditions that may cause maximum damage to the package do not affect the containment and shielding performance of the packaging.

256

The main body on is lower than that at drop test I.

A.6.3 Thermal test A.6.3.1 Summary of temperatures and pressure In this section, we will describe the outline of the temperatures and pressures to be used in the designing and analysis of the behavior of the package under accident test conditions.

(1) Design temperatures The evaluation of ()-B.5.3 revealed that the temperature rises up to 209.9 in the fuel basket, 483.2 in the inner shell and 187.8 in the inner lid.

Therefore, the design temperature under accident conditions is evaluated in the manner that contributes to ensuring the maximum safety as shown in ()-Table A.35.

()-Table A.35 Design temperatures used for accident test condition Position Temperature ()

1 Fuel basket 225 2 Inner shell 500 3 Inner lid 225 (2) Design pressure As was evaluated in the section ()-B.5.4, the pressure in the inner shell can rise up to 0.065 MPa (measured at the gauge). Hence, the design pressure in the package under accident test conditions is evaluated to achieve maximum safety on the assumption that a pressure difference of 0.0981 MPaG occurs (see ()-Table A.36).

()-Table A.36 Design pressure of package under accident condition Position Design pressure 1 Inner shell inside 9.81 x 10-2 MPa A.6.3.2 Thermal expansion Stress due to the difference of thermal expansion between the inner surface of the inner shell and the outer surface of the fuel basket will be described here.

The temperature of fuel basket and the inner shell may rise to 225 and 500 respectively (see ()-Table A.35).

However, stress is generated by difference of thermal expansion because the fuel basket is not fixed to the inner shell.

257

A.6.3.3 Comparison of allowable stresses (1) Stress calculation Stress generated on different parts of the package due to the design pressure will be analyzed for the same parts as those described in section A.5.1.3, using the same method.

In this analysis, the temperatures shown in ()-Table A.35 will be used on the parts of the package.

(2) Displacement of the O-rings of inner lid Displacement that can be generated at the O-rings due to the design pressure will be analyzed for the same parts as those described in section A.5.1.3(1) ,

using the same method.

(3) Stress analysis and evaluation

()-Table A.37 shows the results of the stress analyses.

These results demonstrate that the integrity of the package can be maintained under accident test conditions (thermal test).

258

Stress units

-Table A.37 Stress analysis and evaluation under accident test conditions (thermal test) ;N/mm2 Stress Stress Stress Stress at due to due to Primary stress Position initial internal thermal to be evaluated clamping pressure expansion Pm(PL) 2/3Su MS PLPb Su MS

-0.0491 1 Frame of Inner shell 2.31 2.36 258 108 1.15 Outer Surface Inner Surface Outer Surface Inner Surface 3.18 0.953 0.098 258 2631 3.28 387 116 Bottom plate of -0.098 2

inner shell -3.18

-0.953 0 258 3.18 387 120 259 0

-3.27 2/3 Sy Sy

-3.27 0.098 408 4162 3.17 612 192

-0.098 3 Inner shell lid 3.27 2/3 Sy Sy 3.27 0 408 3.27 612 186 0

Inner shell lid 2/3 Sy 4 174 3.22 177 408 1.30 clamping bolt Interior : 1) Displacement = 1.29x10-2 mm Displacement of the 5 2) Initial clamping value of the O-ring = 1.1 mm inner lid O-ring 3) Remaining height of O-rings l = - 1.087 mm PmGeneral primary membrane stress; PLLocal primary membrane stress; PbPrimary bending stress; SyYield point of the design; SuDesign tensile strength; MSMargin of safety r ; Diameter direction stress o ;Periphery direction stress 2 ;Axial stress t ; Ability of bolt stress

A.6.4 Water immersion In this section we will demonstrate that when immersed 15 m under water, the package can sufficiently endure the external pressure of 147 kPa.

We supposed here that the inner shell is subjected to this pressure. ()-Fig A.91 shows the parts evaluated for stress.

Since the radioactivity of this package will not exceed 105 times A 2 , then water immersion test is not required.

Symbol Evaluated position Frame of inner shell Bottom plate of inner shell Inner shell lid Displacement of O-rings on inner shell lid

()-Fig A.91 Stress evaluation position of inner shell for 15 m immersion test 260

Frame of inner shell The frame of inner shell suffering external pressure is evaluated for its buckling and for the stress that may occur at its center.

(a) Buckling

()-Fig.A.92 Analytical model shows the permissible buckling pressure for the frame of inner shell under external pressure.

()-Fig.A.92 Analytical model of allowable buckling pressure for frame of inner shell The allowable buckling pressure Pe (()-Fig.A.92) for the frame of inner shell

[1]

is obtained by the following equation The formula and figure for finding the respective allowable bucking stress Pe are applied also to the current, appropriate source.

4B t Pe =

2 Do where P e : Allowable buckling pressure [MPa]

Do : Outer diameter of inner shell, Do = 480 [mm]

t: Wall thickness of frame of inner shell, t = 10 [mm]

B: Factor obtained from ()-Fig.A.93, B = 650 1: Length of the inner shell, 1 = 1324 [mm]

Hence, 4 x 650 x 10 x 9.81 Pe = = 1.77 [MPa]

3 x 480 x 100 Therefore, the margin of safety MS for the external pressure P = 0.147 MPa which the frame of inner shell suffers is, 261

Pe 1.77 MS = 1 111.0 P 0.147 Hence, the inner shell does not buckle under external pressure.

Stainless Steel (SUS304)

(Remarks)

1. The intermediate value shall be obtained by proportional calculation.

2. The way of application of this figure shall be given in the following, In case of the cylinder shape subjected to a pressure on the outer surface (1) Take a value, 1/Do, on the axis of ordinates.

(2) Calculate the value, Do/t, assuming the thickness, t, of the plate to be used.

(3) Draw a horizontal line from the point responding to 1/Do and obtain the crossing point of the horizontal line with the curve responding to Do/t.

(4) Draw a vertical line through the crossing point obtained in (c), and obtain the crossing point of the vertical line with the curve corresponding to the operating temperature.

(5) Draw a horizontal line from the crossing point obtain in (d), obtaining B which is the crossing point of the said horizontal line with the axis of ordinates.

()-Fig.A.93 Curve representing backling behavior factor of inner shell under external pressure 262

(b) Center of inner shell

()-Fig.A.94 shows an analytical model for the stresses occurring at the center of the inner shell under external pressure. The stress that may occur at the center of the inner shell is supposed to be a thin cylindrical wall and is obtained by the following equation.

()-Fig.A.94 Stress analysis model of center of inner shell P Dm

= -

2t P Dm z = -

4t P

r = -

2 where

Circumferential stress [N/mm2]

z  : Axial stress [N/mm2]

r  : Radial stress [N/mm2]

P: External pressure, P = 0.147 [MPa]

Dm: Average diameter of frame of inner shell, Dm = D + t = 460 + 10 = 470 [mm]

t: Wall thickness of frame of inner shell, t = 10.0 [mm]

D: Inner diameter of frame of inner shell, D = 460 [mm]

Hence, the following values are obtained.

0.147 x 470

= - = -3.45 [N/mm2]

2 x 10 0.147 x 470 z = - = -1.73 [N/mm2]

4 x 10 263

r = - 0.0735 [N/mm2]

Bottom plate of inner shell

()-Fig.A.95 Analytical model shows the stresses on the bottom plate of the inner shell under external pressure.

Assuming that the bottom plate of the inner shell is a disk fixed on its circumference, the stress on this fixed part is, P a2

= +/-0.225 h2 P a2 r = +/-0.75 h2 z = - P (outer surface)

()-Fig.A.95 Stress analysis model of bottom plate of inner shell where

Circumferential stress [N/mm2]

r  : Radial stress [N/mm2]

z  : Axial stress [N/mm2]

P: External pressure, P = 0.147 [MPa]

a: Diameter of the bottom plate of inner shell, a = 230 [mm]

h: Wall thickness of the bottom plate of inner shell, h = 35 [mm]

Hence, 0.147 x 230 2

= +/-0.225 = +/-1.428 [N/mm2]

35 2 0.147 x 230 2 r = +/-0.75 = +/-4.76 [N/mm2]

35 2 z = - 0.147 (outer surface) [N/mm2]

For the double sign of the stress value, the upper sign (-) corresponds to the inner surface and the lower sign (+) to the outer surface respectively.

264

Inner lid

()-Fig.A.96 Analytical model shows the stresses that may occur on the inner lid under external pressure.

The stress [N/mm2] that may occur on the disk supported on its circumference is at a maximum in the center (see ()-Fig.A.96) and is obtained as follows.

P a2

= r = 1.24 h2 z = - P (outer surface)

()-Fig.A.96 Stress analysis model of center of inner lid where

Circumferential stress [N/mm2]

r  : Radial stress [N/mm2]

z  : Axial stress [N/mm2]

P: External pressure, P = 0.147 [MPa]

a: Diameter of the bottom plate of inner shell, a = 285 [mm]

h: Wall thickness of the bottom plate of inner shell, h = 55 [mm]

Hence, 0.147 x 285 2

= r = 1.24 = 4.89 [N/mm2]

55 2 z = - 0.147 (outer surface) [N/mm2]

For the double sign of the stress value, the upper sign (-) corresponds to the inner surface and lower sign (+) to the outer surface respectively.

265

Displacement of the O-rings of inner lid

()-Fig.A.97 Analytical model shows the displacement of the O-rings on the inner lid under external pressure.

Outside O-ring groove Inside O-ring groove

()-Fig.A.97 Displacement analysis model of O-rings of inner lid under external pressure The outer O-ring is at a distance of 1 from the supporting point of the disk suffering the uniform load. Its displacement is obtained as follows:

P a a3

= 1 = x1 [mm]

8D (1 + )

where

Displacement of the outer O-ring [mm]

Angle of deflection at supporting point [rad];

P a a3

=

8D (1 + )

P: External pressure, P = 0.147 [MPa]

Factor of safety, = (R/a)2 a : Distance from the center of inner lid to the supporting point, a = 230 [mm]

R : Radius of the inner lid, R = 310 [mm]

D : Bending stiffness, E t3 D = [Nmm]

12(1 2 )

E : Longitudinal elastic modulus of the inner lid, E = 1.92 x 105 [N/mm2]

t : Minimum wall thickness of the inner lid, t = 36.7 [mm]

Poisson's ratio, = 0.3 1 : Distance from the supporting point to the outer O-ring, 1 =30.1 [mm]

Hence, the displacement of the outer O-ring is 266

0.147 x (310 / 230) 2 x 230 3 x 12 x (1 0.3 2 )

= x30.1 8 x 1.99 x 10 5 x 36.7 3 x (1 + 0.3)

= 0.0104 [mm]

This value is far smaller than the initial clamping value of the O-ring (

= 1.1 mm). For this reason, the packaging cannot be adversely affected when exposed to external pressure.

267

()-Table A.38 shows the test results of items to .

()-Table A.38 Stresses evaluated for 15 m water immersion test Stress Primary stress Stress Position Pm (PL) 2/3 Su Ms Pl+Pb Su Ms r -0.0735 Center of 3.38 310 90.7 inner shell -3.45 z -1.73 Outer Surface Inner Surface Outer Surface Inner Surface r -4.76

-1.428 Bottom plate z 0 of 0.147 310 2107 4.91 466 93.9 inner r 4.76 shell 1.428 z -0.147 r 4.89 4.89 Inner z 0 2/3 Sy Sy 0.147 3114 4.89 139 lid 458 687 r -4.89

-4.89 z -0.147

-External pressure P = 0.147 MPa Buckling of

-Allowable external pressure Pe = 1.77 MPa the inner shell

-Margin of safety MS = 11.0 Displacement of

-Displacement of outer O-ring = 0.0104 mm O-rings on inner

-Initial clamping value of O-rings = 1.1mm lid Note. Stress and stress intensity units: N/mm2 These figures show that the package can maintain the integrity for its containment.

268

A.6.5 Summary of result and evaluation The tests under accident conditions were examined by analytical methods. The results of the mechanical test (drop test ) revealed that only the outer shell suffered deformation.

The results of the mechanical test (drop test ) revealed that only the outer shell suffered local deformation.

In addition, the stress that occurs on each part of the inner shell does not exceed the allowable value, so the containment interface, suffering no damage, is not adversely affected.

In the thermal test, the stress that occurs on each part of the inner shell does not exceed the allowable value, so the containment interface, suffering no damage, is not adversely affected.

In the water immersion test, the inner shell can endure an external pressure of 147 KPa and maintain its soundness. Further, the fuel elements will never get fractured in the strength test, and the stress generated is not more than the allowable value.

The results of the evaluation of the outer shell, inner shell and content will be used for (B) Thermal analysis, (C) Containment analysis, (D) Shielding analysis, and (E) Criticality analysis.

In the (B) Thermal analysis, (C) Containment analysis, (D) Shielding analysis, and (E) Criticality analysis, the results of the (A) Structural analysis were taken into consideration as follows.

(1) Thermal analysis Those parts of the packaging which are essential to the thermal analysis are represented by the inner shell and inner lid.

The inner lid is covered with the outer lid.

In the structural analysis, the deformation of the lid side shock absorber is 126.7 mm at the vertical drop and 81.6 mm at the horizontal drop, while the thickness before deformation of the material is respectively 186 mm and 104 mm. So the deformation does not occur in the inner shell.

No penetration occurs in the outer shell at drop test .

The outer lid does not come off, sufficiently maintaining its functions as 269

a heat insulator.

We therefore suppose that in the thermal analysis, the inner shell is not damaged, and that the remaining thickness of the heat insulator and the shock absorber are determined to ensure the maximum in safety.

(2) Containment analysis In the structural analysis, both the containment system of the packaging and the fuel elements suffer no damage and maintain their integrity.

In the containment analysis, the results are used to evaluate the leakage of radioactive material.

(3) Shielding analysis In the shielding analysis, damage of either the outer shell, inner shell or fuel elements will influence the results.

In the structural analysis, the thickness of the lid side and bottom side shock absorber is 186 mm in the axial direction and 104 mm in the radial direction. Thus, deformation does not reach the inner shell and the packaging maintains its integrity.

In drop test , the outer shell is locally deformed, but the inner shell is not deformed.

Thus, in the shielding analysis we supposed that the inner shell would not be deformed, and, in order to ensure the maximum in safety, that the package has no outer shell, no heat insulator, and no shock absorbers.

(4) Criticality analysis As in the case of the shielding analysis, we supposed here that the inner shell would not be deformed, and, in order to ensure the maximum in safety, that the package has no outer shell, no heat insulator, and no shock absorbers.

270

A.7 Reinforced immersion test The maximum quantity of radioactivity of these transported articles is less than 100,000 times of the A2 level, which is not considered relevant.

A.8 Radioactive content The fuel element, the radioactive content in the package consists of laminated fuel plates supported by the side plates on its ends (see ()-Fig.D.1). The fuel is located between aluminum alloy plates.

The specifications of the fuel element are shown in ()-Table D.

Structural analyses of the fuel elements are carried out under normal and accident test conditions on the assumption that they will suffer the same impact acceleration as that in the transport packaging. Therefore, the stress generated in any of the fuel elements is not more than the allowable stress under general and specific testing conditions, so that the fuel element are free from getting fractured.

271

A.9 Fissile package This package, under the category of the fissile package in the Regulations, is used at an ambient temperature of more than -40. It is very unlikely that the package, as described in A.4.2, will be damaged or cracked at operating temperatures between

-40 and 38.

Therefore, here is analyzed the damage of the package under the following test conditions, which is assumed for criticality analysis in (II)-E Criticality Analysis.

A.9.1 Normal test conditions In consideration of (II) E Criticality Analysis, damage of the package is analyzed on the results of A.5 and A.9.2 as show in ()-Fig.A.98.

1) water 2) 1.2m free 3) stacking 4) steel bar spraying drop penetration Steel bar 1.0 Package Package Package Package 1.2

()-Fig.A.98 Normal test conditions A 9.1.1 Continuous test (1) Water spray The same as A 5.2, there is no damage to the package.

(2) 1.2m free drop(1.2m drop)

The same as the normal test conditions for the B(U) type package, there is no damage to the inner cell of criticality system as described in A 5.3 A 9.1.2 Stacking test The same as A 5.4 there is no damage to the inner cell of criticality model.

272

A9.1.3 Penetration test The same as A 5.5, there is no damage to the inner cell of criticality model.

With the results above, the damages of the package are summarized as shown in (II)-Table A.39. This package, as shown in (II)-Table A40, meets the requirements for the fissile package under the normal tests conditions stipulated by the regulation and the notification.

(II)-Table A 39 Damages of the fissile package under the normal test conditions Test conditions Damage to the package Note Water spray No damage _______________

1.2m drop Deformation of outer shell, Outer shell, shock absorber and shock absorber and heat heat insulator are neglected in insulator criticality analysis. Eye-plate has possibility to be deformed, but it is neglected in criticality analysis.

Acceleration, stress at each part of the package, etc. do not exceed the value of 9m drop test respectively.

Stacking No damage _______________

6kg penetration No damage _______________

(II)-Table A40 Compliance with requirements for fissile package under normal test conditions Requirements for fissile package Evaluation The structure should not be made a dent The outer shell, shock absorber and heat which contains a cube of 10cm. insulator are deformed, but the deformation of inner shell, constituting criticality system, is not deformed with a dent which contains a cube of 10cm.

The package shall preserve the minimum The external dimensions of the inner overall outside dimensions of the shell, which is a system subject to package to at least 10cm. criticality assessment, are 48 cm in outer diameter and 140 cm in length, and each side of the circumscribed rectangular solid is 10 cm or more.

273

A.9.2 Special test conditions for fissionable transported articles The accident test conditions for the fissile packages are given as the testing procedures shown in ()-Fig.A.99, as A and B, i.e.,

The damage incurred under normal test conditions and composite effect caused by the different tests including 9 m drop, 1 m penetration, fire test (800 for 30 minutes) and 0.9 m immersion.

The damage incurred under normal test conditions and 15 m immersion test.

Among the above given A and B, the safety evaluation is to be executed under the condition A, in which the composite effect is taken into account considering 9 m drop test which is presumed having significant effect on the critical system and the fire test where the shock absorber burns out and adjacent packages come to be placed closer to each other.

[a]

1) Normal test 2) Drop test 3) Drop test 4) Fire test 5) Water conditions immersion test (A.9.1)

Water 800x30min. 0.9 Package Package Package Package

[b]

1) Normal test conditions 2) Immersion test (A.9.1)

Water 15 Package

()-Fig.A.99 Accident test condition

()274

Here is employed as normal test conditions a continuous test accompanying damage, as shown in (II)-Table A 39.

In consideration of criticality analysis in () E, damage affected package is evaluated as follows.

1. Continuous test of normal test conditions Damage of the package under the mentioned test conditions is as shown in

()-Table 39.

2. drop test(9m)

(1)Dropping attitude and the order of the drop test Dropping attitude and the order of the drop test are given in ()-Fig.A.100.

In case the dropping directions of 1.2m drop and 9m drop test are the same, deformation of the shock absorber will be considered the greatest, and thus here is considered that case.

()275

Deformation area (Horizontal)

(Vertical) 1.2m drop (Corner)

(Horizontal)

(Corner)

(Vertical) 9m drop

()-Fig.A.100 Drop attitude and test order

()276

(2)Deformations and design accelerations Deformations and design accelerations of the fissile package produced in the drop test I (1.2 drop test and the consecutive 9m drop test) for fissile package are analyzed by the method described in section A.5.3.

()-Table A.41 shows the results of the analyses.

()-Table A.41 Deformations and design accelerations of shock absorber under accident test conditions (combined evaluation)

Acceleration and Acceleration (g) Rate of Deformation acceleration to Drop Deformation design CASH- Steel plate Design height (mm) acceleration due x1.2 acceleration acceleration Drop to drop test attitude (9m drop only)

Horizontal 206.2 172.8 379.0 88.8 1.033 Upper 161.8 284.4 446.2 136.6 1.089 Portion Vertical Lower 131.4 276.2 407.6 117.6 1.049 9m portion Upper 85.6 238.2 323.8 133.9 1.080 Portion Corner Lower 87.1 245.2 332.3 115.7 1.069 Portion

• 1: 9 m drop is evaluated by considering the deformation by 1.2 m drop.

(3)Evaluation of damages of the package Design acceleration of the drop test for the fissile package, as shown in (II)-Table A.41, increases by 9% at the maximum in comparison with that of the drop test 1 for the B(U) fissile package. Among the structural evaluation results of drop test 1 of the fissile package, the part of the smallest safety margin is the spectrum converter on the horizontal drop, as shown in (II)-Table A. 28. The safety margin is 0.23 or 23%.

In structural evaluation of the package, the increasing rate of acceleration is the same as that of the generated stress. Even when the design acceleration and the generated stress increases by 9%, the smallest safety margin is 0.19, which shows that the structural integrity of the packaging and its contents is maintained.

()277

3. 1m penetration test In the drop test of A 9.2.1 and A 9.2.2 above, the outer shell, shock absorber and heat insulator are deformed, but these are not related to evaluation of 1m penetration test, as shown in A 6.2. Therefore, the damage of the package on the present test will be the same as the results in A 6.2(See the summary A 6.2).

4. Thermal test In the thermal test, deformation of outer shell, shock absorber and heat insulator is taken into account, but effect of their deformation is considered negligible. Thus, damage evaluation of the package under this test will be the same as A 6.3.3(3).

5. Immersion test(0.9m)

As proved by 15m immersion test, the package damage in 0.9m immersion test will not expand.

6. Summary of the package damage Summary of damage to the package under special test conditions is described here.

(II)-Table A. 42 Damage of the fissile package under special test conditions Conditions Damage of the package Notes drop(9m) Deformation of outer shell, shock Outer shell, shock absorber and absorber and heat insulator heat insulator are neglected in criticality analysis.

Penetration(1m) Deformation of outer shell, shock Outer shell, shock absorber and absorber and heat insulator heat insulator are neglected in criticality analysis.

Thermal Partly damaged by a fire In criticality analysis, heat test(fire) Rise in temperature for each part insulator is neglected and water density is set at 1.0g/cm3 Immersion(0.9m) No damage In criticality analysis, assessed for the package filled with water

()278

A.10 Appendix A.10.1 Analysis program for absorbing performance of shock Absorber : CASH- ***** ()-A-280 A.10.2 Validity of the free drop analyses of JRF-90Y-950K package ****************** ()-A-286 A.1O.3 Displacement of inner lid O-rings ******************************************* ()-A-287 A.10.4 Stress/strain characteristics of the shock absorber at low temperatures ***** ()-A-282 A.10.5 Stress/strain characteristics of hard polyurethane foam ********************* ()-A-293 A.10.6 Low temperatures strength of SUS 304 **************************************** ()-A-294 A.lO.7 Low temperature impact value of SUS 304 ************************************* ()-A-295 A.10.8 Low temperature impact value of SUS 630/H1150 ******************************* ()-A-296 A.10.9 Method for calculating the torque of inner lid clamping bolt **************** ()-A-297 A.10.10 Mechanical characteristics of JRR-4B fuel plate ***************************** ()-A-303 A.10.11 Literature ****************************************************************** ()-A-305

()279

A.10.1 Analysis program for the absorbing performance of shock absorber :

CASH-(1) General CASH- is a calculation code which is used to analyze the shock absorber by an uniaxial displacement method (U.D.M) when the package equipped with shock absorber on its top and bottom is dropped.

The deformation, the energy absorbed, and the impact force (acceleration and g value) occurring in the package when dropped with various postures (vertical, horizontal, and inclined).

As shown in ()-Fig.A.101, this code can be applied to shock absorbers consisting of areas (called material areas) of different mechanical characteristics (stress/strain relationships).

A, B, and C represent material areas.

()-Fig.A.101 Analytical model of shock absorber (2) Analysis theory The CASH- code is a program for analyzing the impact performance of the packages shock absorber in various inclined drop tests (inclination = 0 degrees: vertical drop, inclination = 90 degrees: horizontal drop) in a uniaxial displacement method (U.D.M.) which is based on the following two basic principles.

a) Energy absorbing characteristics are analyzed by a U.D.M.;

()280

b) Uniaxial bars with inclined orientation is replaced with an equivalent couple of uniaxial bars of horizontal and vertical orientation.

The analysis theory of the CASH- code based on these principles is described below.

a) Uniaxial displacement method (U.D.M.)

This is a theory which assumes that each area subject to deformation absorbs the deforming energy in a uniform and uniaxtial manner. Areas subject to deformation such as shock absorber are replaced with a number of uniaxial bars. The energy absorbing characteristics of the entire shock absorber is evaluated on the basis of the energy absorbing characteristics of the uniaxial bars.

We will consider here a case where a mass which weighs W and has an energy E O hits the structure shown in ()-Fig.A.102.

l0 : Initial length l : Final Leigh l : Displacement A : Cross section Uniform deformation Before deformation After deformation

()-Fig.A.102 Analytical model by uniaxial displacement method

()281

The compressive stress/strain relationship of the structure is supposed to appear as shown in ()-Fig.A.103

()-Fig.A.103 Compressive stress/strain relationship of material The deformation l of the structure and the acceleration a which occurs in the mass are obtained as follows.

The strain that is generated when a structure suffers a l deformation is,

= l / l 0 (A.10-1)

The stress is,

= f() = f(l / l 0 ) (A.10-2)

Hence, the force F that occurs when the structure suffers a l deformation is, F = A = Axf(l / l 0 ) (A.10-3)

The energy E that is absorbed by the structure when it suffers a l deformation is,

/ 0 E = o Fdl= l 0 o A()d (A.10-4)

When the energy E 0 that the structure has to absorb is given, the final deformation l* is determined using formula A.10-4,

• / 0 E0 = l0 o A()d (A.10-5)

When l* is substituted in formula(A.10-3), we obtain as follows, F* = Af(l* / l0) (A.10-6)

Therefore, the acceleration a* is, a* = F* / W (A.10-7)

()282

b) Uniaxial bar with inclined orientation We will describe in this section how to handle the uniaxial bars inclined orientation based on a uniaxial displacement method.

Calling the inclined drop angle, we suppose that the following equation is valid among the stresses with inclined direction, vertical z and horizontal direction x for the same strain ,

() =z() cosm+ x() sinm (A.10-8)

Where m is the constant for inclination of the material.

In this case, there is approximately the following relationship between E

, Ez, and Ex, E = Ez cosm-2+ Ex sinm-2 (A.10-9) also, approximately the relationship between F, Fz, and Fx, F = Fz cosm-1+ Fx sinm-1 (A.10-10) where E and F are respectively the energy and force generated when the uniaxial bars oriented to the inclination suffer, while the energy and force generated in Ez and Fz when the uniaxial bars are vertically oriented suffer, and the energy and force generated in Ex and Fx when the uniaxial bars are horizontally oriented suffer (see the following charts).

Uninaxial Uninaxial bar bar Uninaxial bar F Fz Fx Uniaxial bars Uniaxial bars Uniaxial bars oriented to the inclination vertically oriented horizontally oriented

()283

(3) Demonstration of CASH- code To demonstrate the validity of the CASH-, drop tests carried out for four kinds of casks were analyzed. The comparison of the analytical and experimental values are shown in ()-Table A.43.

()-Table A.43 shows that, a) The deformation of the shock absorber was found to be greater in the analytical values based on the CASH- code than in the experimental values, thus ensuring the maximum safety.

b) The design value of the acceleration based on the CASH- was found to be equal to, or greater than, the experimental value, thus ensuring valid results.

The weight of the package is 95O kg which remains within the weight range of the four different casks.

()-Fig.A.104 shows that the shock absorber used in the package is in the same proportion as those of other packagings and cause no problems in applying the analysis code.

These results permit us to suppose that evaluation of the shock absorber performance based on the CASH- code will lead to justifiable results.

However, in the designing of the shock absorber, the following points are taken into account, i) A design acceleration + 20 % of the value based on the CASH- code is adopted as the acceleration that can occur.

ii) Calculated values are adopted as the deformation of the shock absorber because the CASH- code leads to higher values.

()284

()-Table A.43 Comparisons of analytical values by CASH- and experimental values Type of cask TYPE 1 TYPE 2 TYPE 3 TYPE 4 Weight (kg) 62,000 43,500 710 9,600 Outer dimensions (mm) 6,080x2,400 6,220x1,800 3,960x566 3,290x1,080 Posture at droping Verti- Hori-z Verti- Hori-z Corner Verti- Hori-z Verti- Hori-z Corner cal ontal cal ontal cal ontal cal ontal Analytical value 78.2 90.8 95.4 112.4 115.4 131.4 274.9 167 128 73.3 Acceleration (g)

Design value 93.8 109.0 115 135 138.4 158 330 201 152 88.0 (g)

Experimental value 70 67 114 117 73 135 320 200 150 51.5 (g)

Deformation Analytical value 172 190.3 187 156 383 189.4 50.0 63.4 120 310.1 (mm)

Experimental value 131 117 88 73 155 68.5 16.3 50.0 73.7 22.4 (mm)

• The design values which are equal to the values of the analytical value multiplied by a factor of 1.2 are used in the designing, taking possible variations of test results into account.

The main body shock absorber of shell L

L D

D TYPE 1 TYPE 2 TYPE 3 TYPE 4 Package L 1 /L 2 0.42 0.53 0.56 0.68 0.43 D 1 /D 2 0.52 0.67 0.56 0.57 0.57 Shock Balsa Plywood Plywood Balsa Balsa absorber +Redwood

()-Fig A.104 Proportion of shock absorbers

()285

A.10.2 Validity of the free drop analyses of the JRF-9OY-950K package

()-Table A.44 compares the results of a drop tests and the analytical results obtained from a prototype packaging.

Generally, the analytical results were obtained so as to ensure the maximum in safety.

()-Table A.44 Comparison of analytical and experimental results Ratio of Analytical Item Test results analyses/ Remarks results test Acceleration Drop test 367.0 366 1.003 (G) Drop test 80.4 18.3 4.393 Deformation Drop test 94.0 46 2.043 (mm)

()286

A.10.3 Displacement of inner lid O-rings

()-Fig.A.105 shows an analytical model showing the displacement of the O-rings in the 1.2 m lid side vertical drop of the package.

()-Fig.A.105 Analytical model of inner lid for 1.2 m lid side vertical drop

()287

()-Fig.A.105 shows that the uniform load consisting of the weight of the content and that of the fuel basket act at the center of the inner lid, and that the uniform load due to the dead weight of the inner lid acts on the lid.

On the other hand, the inner lid is supported by inner lid clamping bolt and the shock absorber which has a circular reaction force.

Displacement of the O-rings fixed on the inner lid which suffer these loads can be calculated by combining the results of the analyses using the ,, and models (see ()-Fig.A.99).

Contents + fuel basket The displacement that can occur in the disk suffering a uniform load on its concentric circle(see ()-Fig.A.105) is, P1 b 4 r4 4a 2 (1 )b 2 r 2 r 2 a 1 = 4 2 2 2 + 1 1n +

16 D 4b 2(1 + )a 2 b b b 4(3 + )a 2 (7 + 3)b 2 4(1 + )b 2 where 1 : Displacement of the inner O-ring [mm]

Poisson's ratio, =0.3 a: Radius of the supported points of the inner lid, a = 285 [mm]

b: Radius of the loads, b = 230 [mm]

r: Radius of the inner O-ring groove, r = 237.5 [mm]

m3 : Weight of the fuel basket, m3 = 138 [kg]

m4 : Weight of the content, m4 = 92 [kg]

N: Acceleration, N = 240.7g [m/s2]

h: Minimum wall thickness of the inner lid, h = 36.7 [mm]

E: Longitudinal modulus of elasticity, E = 1.99x1O5 [N/mm2]

P1 : uniform load of the content/fuel basket, (m 3 + m 4 ) (138 + 92)

P1 = N= x240.7x9.81 = 3.27 [N/mm2]

b 2 x 230 2 D: Bending stiffness of the inner lid, E h3 1.99 x 10 5 x 36.7 3 D = = = 9.01x108 [Nmm]

12(1 2 ) 12(1 0.3 2 )

Hence the displacement 1 due to the content + fuel basket is,

()288

3.27 x 230 4 237.54 4 x 2852 (1 0.3) x 230 2 237.52 1 = x 16 x 9.01 x 108 4 x 230 4 2 x (1 + 0.3)2852 230 2 237.52 285 4(3 + 0.3)2852 (7 + 3 x 0.3)230 2 230 2 2 + 1 n +

230 4(1 + 0.3)230 2

= 0.341 [mm]

Weight of the inner lid The displacement 2 (mm) that can occur in the disk suffering a uniform load (see ()-Fig.A.105 ) is, P2 a 4 r 2 5 + r 2 2 = 1 2 64D a 2 1 + a where 2 : Displacement of the inner O-ring [mm]

Poisson's ratio, = 0.3 a: Radius of the supported points of the inner lid, a = 285 [mm]

r: Radius of the inner O-ring groove, r = 237.5 [mm]

h: Wall thickness of the inner O-ring groove, h = 55 [mm]

N: Acceleration, N = 240.7g [m/s2]

Density of the inner lid, = 7.93x10-6 [kg/mm3]

D: Bending rigidity of the inner lid, D = 9.01x108 [Nmm]

P2 : Uniform load due to the dead weight of the inner lid, P2 =hN = 7.93x10-6x55x240.7x9.81 = 1.03 [N/mm2]

Hence the displacement 2 due to the weight of the inner lid is, 1.03 x 2854 237.52 5 + 0.3 237.52 2 =

64 x 9.01 x 108 1 2852 1 + 0.3 2852 = 0.122 [mm]

Reaction force of the shock absorber to be subtracted The displacement 3 (mm) that can occur in the disk suffering a uniform load on its concentric circle (see ()-Fig.A.105 ) is, P3 C 4 r 4 4a 2 (1 )C 2 r 2 r 2 a 3 = 4 2 2 2 + 1 1n +

16 D 4C 2(1 + ) 2 C C C 4(3 + )a 2 (7 + 3)C 2 4(1 + )C 2 where 3 : Displacement of the inner O-ring [mm]

Poisson's ratio, = 0.3 a: Radius of the supported points of the inner lid, a = 285 [mm]

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C: Radius of the load, C = C0 +tan = 115 + 24.1tan15.5°= 122 [mm]

C0: Upper radius of the circular cone, C0 = 115 [mm]

Circular cone angle, = 15.5° [degrees]

Deformation of the shock absorber, = 24.1 [mm]

D: Bending rigidity of the inner lid, D = 9.01x108 [Nmm]

P3 : Compressive stress on the shock absorber, P3 = 0.932 [N/mm2]

r: Radius of the inner O-ring groove, r = 237.5 [mm]

Hence, 3 is, 0.932 x 122 4 237.54 4 x 2852 (1 0.3) x 122 2 237.52 3 = x 16 x 9.01x 108 4 x 122 4 2 x (1 + 0.3)2852 122 2 237.52 285 4(3 + 0.3)2852 (7 + 3 x 0.3)122 2 2 + 1 n +

122 2

122 4(1 + 0.3)122 2

= 0.0370 [mm]

Reaction force of the shock absorber The displacement 4 (mm) that can occur in the disk which suffers a uniform load on its concentric circle (see ()-Fig.A.105 ) is, P4 a 4 r 2 5 + r 2 4 = 1 2 64D a 2 1 + a where, 4 : Displacement in the inner O-ring [mm]

Poisson's ratio, = 0.3 a: Radius of the supported points of the inner lid, a = 285 [mm]

r: Radius of the inner O-ring groove, r = 237.5 [mm]

D: Bending rigidity of the inner lid, D = 9.01x108 [Nmm]

P 4 : Compressive stress on the shock absorber, P 4 = 0.932 [N/mm2]

Hence the displacement 4 due to the reaction force of the shock absorber is, 0.932 x 2854 237.52 5 + 0.3 237.52 2 =

64 x 9.01x 108 1 2852 2

= 0.110 [mm]

1 + 0.3 285 Thus, the total displacement is,

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= 1 + 2 + 3 - 4 = 0.341 + 0.122 + 0.0370 - 0.110 = 0.390 [mm]

Incidentally, as for the 9 m lid side vertical drop test replacing the values of acceleration, (409.8g), compressive stress of shock absorber, (2.66N/mm2) and displacement, (127mm), with the corresponding value for 1.2 m lid side vertical drop test, the same analysis is conducted and the results of evaluation are given in ()-Table A.45.

()-Table A.45 Analysis results of displacement of inner O-rings of inner lid Name of Total

• Analysis condition Displacement Remaining height displacement displacement Normal condition (internal pressure) 0 0.0116 1 0.341 1 1.2 m lid side 0.402 0.698 2 0.122 vertical drop 3 0.0370

- 4 -0.110 Normal condition (internal pressure) 0 0.0116 1 0.581 2 9 m lid side 0.636 0.464 2 0.207 vertical drop 3 0.151

- 4 -0.315

• Note: Residual tightening interference = Initial clamping value (1.1 mm) - Total displacement As shown in ()-Table A.45, remaining height of the inner O-ring in each of the cases of 1.2 m and 9 m lid side vertical drop tests is always positive so that it can be granted that the containment of packages will be duly maintained.

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A.10.4 Stress/strain characteristics of the shock absorber at low temperatures

()-Fig.A.106 shows the stress/strain characteristics of the shock absorber at low temperatures.

(1) Direction perpendicular to the wood grain of the shock absorber (2) Direction parallel to the wood grain of the shock absorber

()-Fig.A.106 Stress/strain characteristics curves for shock absorber at low temperatures

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A.10.5 Stress/strain characteristics of hard polyurethane foam

()-Fig.A.107 shows the stress/strain characteristics of the hard polyurethane foam.

()-Fig.A.107 Stress/strain curves for hard polyurethane foam

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A.10.6 Low temperature strength of SUS 304

()-Fig.A.108 shows the mechanical characteristics of the material SUS 304 at low temperatures.

[16]

()-Fig.A.108 Low temperature strength of SUS 304

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A.10.7 Low temperature impact values of SUS 304

()-Fig.A.109 shows the low temperature impact values of the material SUS 304.

[16]

()-Fig.A.109 Low temperature impact value of SUS 304

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A.10.8 Low temperature impact Value of SUS 630/H1150

()-Fig.A.110 shows the low temperature impact values of the material.

[18]

()-Fig.A.110 Low temperature impact value of SUS 630H1150

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A.10.9 Method for calculating torque of inner lid clamping bolts In this section, we will analyze the initial clamping force of the inner lid clamping bolt (called the bolt below).

B42 a=32 d27 k=40 s=10 L=30 M24 d24

()-Fig.A.111 Analytical model for initial clamping force of inner lid clamping bolts The minimum required clamping force for the bolt shown in ()-Fig.A.111 is, F min = F C + F G + F H where F min : Minimum force required for tightening the bolt [N]

FC : Loss of compressive force in the inner lid when external force is applied

[N]

FG : Clamping force assured by the O-rings [N]

FH : Decrease of clamping force due to differential thermal expansion [N]

These three values will be analyzed below.

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(1) F C , the loss of compressive force in the inner lid when external force is applied is F C = (1-)W a = (1-)(W 1 + W 2 )/n where W a : Axial external force, Wa = (W1 + W2)/n

= (0.309 + 8.13)x105/16 = 5.27x104 [N]

W1 : Load due to internal pressure, 2

W 1 = P G 1 = 0.175x x4742 = 3.09x104 [N]

4 4 where P: Maximum internal pressure, P = 0.175 [MPa]

G1 : Inner O-ring diameter, G1 = 474 [mm]

W2 : Load occurring at 9 m lid side vertical drop, W2 = 8.13x105 [N]

n: Number of bolts, n = 16

Internal force factor of the bolt, Ft Kt 1.50 x 10 6

= = = = 0.178 [-]

Wa K t + K C (1.50 + 6.92) x 10 6 where K t : Tension spring constant of the bolt, l l + l K t = E b / a + s = 1.45x106 [N/mm]

b A A s where l a : Length of the bolt cylinder, la = 32 [mm]

ls : Length of the thin bolt cylinder, ls = 10 [mm]

Ab : Cross section of the bolt cylinder, 2 2 Ab = d = 24 = 452 [mm2]

4 4 A s : Effective cross section, 2

As = d2 = 22.0512 = 382 [mm2]

4 4

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Where d 2 : Core diameter of the bolt, d2 = 22.051 [mm]

Eb : Longitudinal modulus of elasticity of the bolt, Eb = 1.99x105 [N/mm2]

l: Length equivalent to the elastic displacement in the fitting parts of the nut, l = 0.57d = 13.7 [mm]

KC : Compression spring constant of the inner lid, EC KC = [d m 2-d 1 2] = 6.68x106 [N/mm2]

lK 4 Where l K : Tightening length, lK = 40 [mm]

d1 : Diameter of bolt hole, d1 = 27 [mm]

B: Diameter of the contact surface of the bolt head, B = 42 [mm]

dm : Diameter of equivalent cylinder, lK 40 dm = B + = 42 + = 50 [mm]

5 5 EC : Longitudinal modulus of elasticity of the inner lid, Ec = 1.92x105 [N/mm2]

Hence, F C = (1-) W a = (1 - 0.178)x5.27x104 = 4.33x104 [N]

The tensile force F t in the bolt due to external force is, F t = W a = 0.178x5.27x104 = 9.38x103 [N]

(2) Clamping force for the O-rings The clamping force F G for the O-rings is, F G = (G 1 + G 2 )xq / n where G 1 : Diameter of the inner O-ring, G 1 = 474 [mm]

G 2 : Diameter of the outer O-ring, G 2 = 514 [mm]

q: Linear load of the O-rings, q = 14.3 [N/mm]

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Hence, F G = x(474 + 514)x14.3 / 16 = 2.77x103 [N]

(3) Decrease of clamping force F H due to differential thermal extension F H is 0 because the material of the inner lid is the same as that used for the bolts.

Thus, the minimum required clamping force is, F min = F C + F G + F H = (4.33 + 0.277 + 0)x104 = 4.61x104 [N]

(4) The initial clamping force for the bolt The initial clamping force F 0 of the bolt is a little more than the minimum required force.

F 0 = 5.89x104 [N] = 6.0x103 [kgf]

(5) Initial torque for the bolt The initial torque for the bolt is, T = kdF 0 = 0.2x24x5.89x104 = 2.83x105 [Nmm]

= 28.8 [kgfm]

where k is the torque coefficient (k = 0.2).

(6) Bolt clamping triangle The above analysis results are shown in the bolt clamping triangle (see ()-Fig.A.112).

The following is the symbols used in ()-Fig.A.112.

F 0 : Initial clamping force of bolt, F 0 = 5.89x104 [N]

F min : Minimum required force for clamping the bolt, F min = 4.61x104 [N]

W a : Axial external force, W a = 5.27x104 [N]

F t : Increment of the bolts tensile force when external force is applied, F t = 0.94x104 [N]

F C : Loss in the lids compressive force when external force is applied, F C = 4.33x104 [N]

F C ' : Residual compressive force in the inner lid, F C ' = 1.56x104 [N]

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F H : Decrease of clamping force due to differential thermal extension, F H = 0 [N]

F b : Bolt tensile force, F B = 6.83x104 [N]

F G : O-rings clamping force, F G = 0.28x104 [N]

()-Fig.A.112 shows that the residual compressive force F C ' on the inner lid is higher than the O-ring's clamping force F G .

Therefore, the containment of the O-rings can be maintained by the initial clamping force F 0 .

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Load

[x10]

7 Fo Ft 6

5 Fmin 4 Wa Fc Kt Kc Fb 3

2 FG 1

Fc 0 0.01 0.02 0.03 0.04 0.05 0.06 Growth [mm]

()-Fig.A.112 Triangle diagram for inner lid clamping bolt Explanation of ()-Fig.A.112 (1) This illustration shows that even if axial external force W a acts from the initial clamp force of bolt F 0 , the residual compressive force in the inner lid F C would be larger than O-ring clamping force F G .

(2) On the axial part of the bolts a tensile force F o will be imposed by the initial clamping, and on the body to be clamped, (that is the lid part), a compressive force F o will be generated, two forces being in balance with each other at point

, the status of which is shown in the illustration.

(3) When axial external force W a , acts on any of the bolts in axial direction, the status of the bolt and lid will be moved to point and point .

Point will be removed from Point by means of elongation being generated, by a tensile force F t acting on the bolt axial part, and point will be removed from point to point by means of clamping length being extended as much as according to the compressive force, F C , being lost from

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the body to be clamped, (that is the lid part).

(4) That is to say, on the bolt a tensile force Ft is added, from the body to be clamped, (the lid part) a compressive force F C being removed, and the clamping length will be extended as much as where the compressive force remaining on the body to be clamped (The lid part).

A.10.10 Mechanical characteristics of JRR-4B fuel plate In order to define the analysis criteria by which the plastic deformation will never be generated in the analysis of fuel plate, the proof stress is taken as the analysis standard value.

The proof stress of the JRR-4B fuel plate which is the material for the fuel element (B) shall be specified as given below, (1) The mechanical property of AG3NE, which is the material of fuel element (A) is shown in IAEA Guide Book, Vol.2 (referential document [14]), in which it is specified that the design yielding point (Sy) is not less than 63.8 N/mm2 at the evaluating temperature 75.

(2) JRR-4B fuel has been subjected to a tensile strength test on the basis of a tensile strength test piece which is manufactured from a sheet of fuel plate sampled from each roll badge, the criteria of the test being 88.3 (N/mm2) in the tensile strength; that is <9kgf/mm2>.

This test is considered as one of the subjects of precommissioning test of the nuclear reactor facility.

(3) Besides the above, the results of measurements on 20 pieces of samples in the tensile strength test cited in the proceeding articles (2) are as follows.

Results of measurements Minimum Maximum Average Proof stress 97.1 135 114 (0.2%)(N/mm2)

Tensile strength 108 143 122 (N/mm2)

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(4) H12 materials of JIS 1100P and A1200P, which are the raw materials of JRR-4B type fuel plate cladding material, are deemed to have the strength more than those figures shown in the table given below.

JIS A 1100P H12 A 1200P H12 Proof stress (N/mm2) 73.6 Tensile strength (N/mm2) 93.2128 (5) Looking from the above, it can be deemed as the safety side estimation that the proof stress for the fuel plate having the tensile strength of 88.3N/mm2 as previously cited in the article (2) as the proof stress of the mechanical property of JRR-4B type fuel may be adopted as 63.8N/mm2 (yield point of the design) which is equal proof stress mentioned in proceeding article (1).

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A.10.11 Literature

[1] ASME Sec. Subsec. NB (1974).

[2] Technical Standards for Atomic Energy Installation for Power Generation Including Standards for Structure, ministerial Notice No.501, 1980.

[3] Commentary on Standards for the Structure of Boilers and Pressure Vessels, Japan Boiler Association, 1980.

[4] In-house data of Mitsubishi Heavy Industries. Ltd.

[5] Roark, J.R., Formulas for Stress and Strain (4th edition), Mc'Graw-Hill International Book Company, 1965.

[6] Timoshenko, S.P., Theory of Plate and Shell (I); Japanese translation version by Hasegawa, T.

[7] Manual for Mechanical Engineering, 6th revised edition, Japan Society of Mechanical Engineering, 1977.

[8] Den Hartog, J.P., Mechanical Vibrations, Mc'Graw-Hill Book Co.

[9] Mizuhara, A. et al., Handbook for Structural Calculation, Sangyo Tosho Publishing, 1965.

[10] Formulas Used in Structural Dynamics, compiled by Japan Society of Civil Engineering.

[11] Sekiyu, T. et al., Handbook for Flat Structure Strength, Asakura Shorten.

[12] Handbook for Elastic Stability, Long Column Research Committee, Corona.

[13] Report on Development and Arrangement of Structural Analysis of Transport Packaging for Used Nuclear Fuel , Japan

[14] IAEA Guide Book Vol.2: Research Reactor Core Conversion Safety Analysis and Licensing Issues Fuels.

[15] On the Prediction of Deformation and Deceleration of a Composite Cylindrical Body for the Corner Drop Case, CONF-710801 (Vol.2), 1971, pp.733-776.

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[16] Hasegawa, M., Manual for Stainless Steel, Nikkan Kogyo Shinbun.

[17] Data Book for Strength Designing, compiled by Editorial Committee for data book for strength designing.

[18] Fujita, T., Thermal Processing for Stainless Steel, Nikkan Kogyo Shinbun.

[19] Timoshenko, S.P., Buckling Theory; Japanese translation version by Naka, I. et al., Corona.

[20] Aluminum Hand Book (4th edition), Light Metal Society, (1990)

[21] Summary of Technology for Hybrid Materials, Industrial Technology Center, 1990

[22] In-house data of Nichias Co.,Ltd

[23] Code for Nuclear Power Generation Facilities: Rules on Materials Nuclear Power Plants (2012 edition) of The Japan Society of Mechanical Engineers

[24] Code for Nuclear Power Generation Facilities: Rules on Design and Construction for Nuclear Power Plants (2012 edition) of The Japan Society of Mechanical Engineers

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