RTL-001-CALC-ST-0203, Rev. 6 - 'RT-100 Cask Bolting ... · 3.12 ASME B1.13M-2005, "Metric Screw...

CALC. NO. RTL-001-CALC-ST-0203

..J E N E R C O N CALCULATION COVER SHEET REV. 6Excellence-Every projecL Eveiy day.

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Title: RT-100 Cask Bolting Evaluation Client: Robatel Technologies, LLC

Project: RTL-001

Item Cover Sheet Items Yes No

1 Does this calculation contain any open assumptions that require confirmation? LII(If YES, Identify the assumptions)

2 Does this calculation serve as an "Alternate Calculation"? (If YES, Identify the 0design verified calculation.)Design Verified Calculation No.

3 Does this calculation Supersede an existing Calculation? (If YES, identify the LIsuperseded calculation.)Superseded Calculation No.

Scope of Revision:

Revision 6 is being made to update revision of reference drawings. Reference 3.14 deleted because it is a repeat ofReference of 3.19, and the Reference 3.14 version is not used in the calculation. Several editorial changes were alsomade in the calculation. Updated calculations due to typographical errors.

Revision Impact on Results:

Calculations updated due to typo corrections.

Study Calculation EL Final Calculation [

Safety-Related M Non-Safety Related LI

(Print Name and Sign)

Originator: David Hartmangruber Date: 01/03/2014

Design Verifier: Curt Lindner - ' •- Date: 01/0312014

Approver: Nand Lambha Date: 0110312014

CALC. NO. RTL-001-CALC-ST-0203i~~j~ ENERCON ~ CALCULATION _ _ _ _ _ _ _ _

EcellenCe-Evety pojet. Evey dqy REVISION STATUS SHEET REV. 6

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CALCULATION REVISION STATUS

REVISION DATE DESCRIPTION

0 8-24-2012 Initial Issue

1 10-8-2012 Revised in its entirety to match SAR Review Plan Format

2 11-28-2012 Added Sections 7.1.4, 7.1.4.1, 7.1.4.2, Refs. 3.20 and 3.21, andAppendix 1 Fig. 2 to address bolt plastic deformation allowed byNUREG/CR-6007 and seal integrity; corrected typo in modulus ofelasticity units throughout; minor editorial changes.

3 8-28-2013 Added Sections 7.3 to add calculation for port seal cover toaddress RAI 4.6. Corrected page sequence by renumbering page57A.

4 9-04-2013 Incorporated customer comments.

5 9-12-2013 Updated drawing reference and deleted unused Ref. 20.

6 01-03-2014 Updated drawing references. Corrected errors in equations andYoung's modulus values.

PAGE REVISION STATUS

PAGE NO. REVISION PAGE NO. REVISION

All 6

APPENDIX REVISION STATUS

CALC. NO. RTL-001-CALC-ST-0203EJEN ER C ON CALCULATION _ _ _ _ _ _ _

DESIGN VERIFICATION REV. 6PLAN AND SUMMARY SHEET PAGE NO. 3 of 65

Calculation Design Verification Plan:Calculation to be reviewed for correctness of inputs, design criteria, analytical methods, acceptance criteria and numericalaccuracy.

Stated objectives and conclusions shall be confirmed to be reasonable and valid.

Any assumptions shall be clearly documented and confirmed to be appropriate and verified based on sound engineeringprinciples and practices.

(Print Name and Sign for Approval - mark "N/A " if not required)

Approver: Nand Lambha Date: 01-03-2014

Calculation Design Verification Summary:

Calculation has been designated as Safety Related as noted on the cover sheet.

Calculation has been verified to be mathematically correct and performed in accordance with appropriate design inputs,assumptions, analytical methods, design criteria and acceptance criteria.

The conclusions developed in the calculation are reasonable, valid and consistent with the purpose and scope.

Assumptions are appropriate and correct.

Based On The Above Summary, The Calculation Is Determined To Be Acceptable.

.4nae and Sign)

Design Verifier: Curt Lindner Date: 01-03-2014

Others: Date:

E2J ENERCONExcellence-Every projeci. Every doy

CALCULATIONDESIGN VERIFICATION

CHECKLIST

CALC. NO. RTL-001-CALC-ST-0203

REV. 6

PAGE NO. 4 of 65

CHECKLIST ITEMS Yes No N/A

Design Inputs - Were the design inputs correctly selected, referenced(latest revision), consistent with the design basis, and incorporated in the Xcalculation?

Assumptions - Were the assumptions reasonable and adequately Xdescribed, justified and/or verified, and documented?

Quality Assurance - Were the appropriate QA classification andrequirements assigned to the calculation?

Codes, Standards, and Regulatory Requirements - Were the applicablecodes, standards, and regulatory requirements, including issue and Xaddenda, properly identified and their requirements satisfied?

Construction and Operating Experience - Have applicable constructionand operating experience been considered?

Interfaces - Have the design-interface requirements been satisfied, Xincluding interactions with other calculations?

Methods - Was the calculation methodology appropriate and properly Xapplied to satisfy the calculation objective?

Design Outputs - Was the conclusion of the calculation clearly stated, didit correspond directly with the objectives, and are the results reasonable Xcompared to the inputs?

Radiation Exposure - Has the calculation properly considered radiationexposure to the public and plant personnel?

Acceptance Criteria - Are the acceptance criteria incorporated in thecalculation sufficient to allow verification that the design requirements have Xbeen satisfactorily accomplished?

Computer Software- Is a computer program or software used, and if so, Xare the requirements of CSP 3.02 met?

COMMENTS

i (Prin ame and Sign)

Design Verifier: Curt Lindner Date: 01-03-2014

Others: Date:

CALC. NO. RTL- 001-CALC-ST-0203

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TABLE OF CONTENTSCalculation Cover Sheet ................................................................................................................................ 1Calculation Revision Status Sheet ................................................................................................................. 2Calculation Design Verification Plan and Summary Sheet ....................................................................... 3Calculation Design Verification Checklist ................................................................................................. 4

1 Purpose and Scope ...................................................................................................... 6

2 Sum m ary of Results and Conclusions ............................................................................ 6

3 References ........................................................................................................................ 7

4 Assum ptions ..................................................................................................................... 8

5 Design Inputs .................................................................................................................... 9

6 M ethodology .................................................................................................................. 10

7 Calculations ..................................................................................................................... 107.1 Closure Bolt Evaluation ................................................................................................... 11

7.1.1 M ethodology ......................................................................................................................... 117 .1 .2 Lo a d s ..................................................................................................................................... 1 17.1.3 Load Com binations ................................................................................................................ 357.1.4 Seal Integrity ......................................................................................................................... 59

7.2 Vibration .............................................................................................................................. 607.2.1 Vibration Evaluation of the RT-100 Cask Primary Lid Bolts ............................................. 607.2.2 Vibration Evaluation of the RT-100 Cask Secondary Lid Bolts .......................................... 61

7.3 Port Cover Seal and Bolt Evaluation ................................................................................. 637.3.1 Port Cover Seal ...................................................................................................................... 637.3.2 Bolt Evaluation ...................................................................................................................... 64

8 Appendix ........................................................................................................................... 18.1 Elastomer O-Ring Data ................................................................................................. 3 Pages8.2 Cask Design Input Email ............................................................................................... 2 Pages


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1 PURPOSE AND SCOPE

Robatel Technologies is designing the RT-100 transport cask, containing radioactive waste in theform of dewatered resins and filters, to meet the requirements of 10CFR Part 71 (Ref. 3.1). TheNRC requirements in Reference 3.1 state that any package must be evaluated for the normalconditions of transport and for the hypothetical accident conditions. These evaluations for the caskbody, cask lid, impact limiters, etc. have been carried out in References 3.3, 3.4, 3.15, 3.16 and3.17. The NRC requirements in Section 71.43(c) of Reference 3.1 state that a package mustinclude a containment system that can be securely closed through the use of a positive fasteningdevice that cannot be opened unintentionally or by a pressure that may arise within the package.The RT-100 package is designed with two sets of closure bolts: 18 M36 hex head bolts at thesecondary lid and 32 M48 hex heads at the primary lid. These two sets of bolts are credited withmaintaining positive closure of the package under all accident conditions. The purpose of thiscalculation is to structurally qualify the fully-loaded RT-100 cask for the loadings associated with thenormal conditions of transport and the hypothetical accident conditions, as determined inReferences 3.3, 3.4, 3.15, 3.16 and 3.17.

2 SUMMARY OF RESULTS AND CONCLUSIONS

All structural bolts have a factor of safety of greater than 1.0 under the most adverse effects fromthe normal conditions of transport and the hypothetical accident conditions. The minimum factor ofsafety is 1.16 for the M36 bolts at the secondary lid. The minimum factor of safety is 1.01 for the IM48 bolts at the primary lid. The results of the analysis show that the RT-100 cask can withstandthe normal conditions of transport and the hypothetical accident conditions while maintainingpositive closure of the containment vessel.

Therefore, the closure bolts of the RT-100 transport cask are adequate for their design function.


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3 REFERENCES

3.1 Nuclear Regulatory Commission, 10 CFR Part 71, "Packaging and Transportation ofRadioactive Material"

3.2 Drawings:a. ROBATEL Transport Package RT100 - General Assy, RT100 PE 1001-1 Hb. ROBATEL Transport Package RT100 - General Assy, RT100 PE 1001-2 H

3.3 ENERCON Calculation RTL-001-CALC-ST-0201 Rev. 5, "RT-100 Lifting Structural Evaluation"3.4 ENERCON Calculation RTL-001-CALC-ST-0202 Rev. 4, "RT-100 Tie-Down Structural

Evaluation"3.5 NUREG/CR-6007, "Stress Analysis of Closure Bolts for Shipping Casks", 19923.6 ASME B&PV Code, Section II, 20073.7 ASME B&PV Code, Section III, 20073.8 Nuclear Regulatory Commission Regulatory Guide 7.6 Revision 1, "Design Criteria for the

Structural Analysis of Shipping Cask Containment Vessels"3.9 EPRI "Good Bolting Practices", Volume 1, Large Bolt Manual, 19873.10 Joseph Edward Shigley & Larry D. Mitchell, "Mechanical Engineering Design", 4th Edition3.11 Erik Oberg, et. al., "Machinery's Handbook", 2 6th Edition3.12 ASME B1.13M-2005, "Metric Screw Threads: M Profile"3.13 NUREG/CR-0128, "Shock and Vibration Environments for a Large Shipping Container During

Truck Transport (Part II)", 19783.14 [Deleted]3.15 ENERCON Calculation RTL-001-CALC-ST-0403 Rev. 4, "RT-100 Pin Puncture Evaluation"3.16 ENERCON Calculation RTL-001-CALC-TH-0102 Rev. 6, "RT-100 Cask Maximum Normal

Operating Pressure Calculation"3.17 ENERCON Calculation RTL-001-CALC-ST-0401 Rev. 6, "RT100 Cask Impact Limiter Drop

Evaluation"3.18 Roark's Formulas for Stress and Strain, Sixth Edition, McGraw-Hill, New York, 1989.3.19 AISC Guide to Design Criteria for Bolted and Riveted Joints, AMERICAN INSTITUTE OF

STEEL CONSTRUCTION, 2nd Ed. 2001.3.20 [Deleted]3.21 Parker O-Ring Handbook, 5 0 th Anniversary Ed., ORD 5700, Copyright 2007, Parker Hannifin

Corporation, Cleveland, OH3.22 Erik Oberg, et. al., "Machinery's Handbook", 2 5 th Edition


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4 ASSUMPTIONS

4.1 The weight of the cask for the analytical evaluation of the hypothetical accident is considered asthe total weight of the cask and the maximum payload. The damage sustained by the cask andthe impact limiters during the free drop evaluations does not result in any significant reduction inload, so no reduction is considered. This assumption is acceptable without further evaluation.

4.2 It is assumed that the bolts will experience high-cycle fatigue range (> 108 cycles). Therefore,an endurance limit that is appropriate for high cycle fatigue is used to determine the allowablefatigue stresses (Ref. 3.19). Stresses due to vibration are sufficiently low that frequentreplacement is not required. Bolts are inspection prior to each use and defective bolts arereplaced accordingly. This assumption is acceptable without further evaluation.

4.3 It is possible for the center of gravity of the payload to shift ±10% of the interior dimensions ofthe cask containment (Ref. 3.5). This shift has no effect on the bolt loading conditions exceptfor the end drop, corner drop and pin puncture load cases. The bolt load equations used inaccordance with Reference 3.5 consider the payload weight as a pressure load distributedacross the cask lid interior surface in order to maximize the combination of bolt tension andprying loads (see Section 4.6, Ref. 3.5). This load application, along with the assumptionsmade in the derivation of the bolt loads, is very conservative (see Section 2.2, Ref. 3.5). Inaddition, if the cask-cavity radius is much larger than the cask wall thickness, the location of theapplied load does not have a significant impact on the resulting average and prying loads (seeAppendix Ill, Ref. 3.5). This shift of the center of gravity will therefore have no significant impacton the results of the analysis and the methodology of NUREG/CR-6007 (Ref. 3.5) is still validand acceptable for use. This assumption is acceptable without further evaluation.

There are no unverified assumptions in this calculation. Other design assumptions used, if any, willbe noted and referenced as needed in the body of the calculation.


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5 DESIGN INPUTS

5.1 The total weight of the fully assembled cask is 34,696kg (Ref. 3.2).5.2 The maximum payload weight is 15,000 Ibs, or 6,803.9 kg (see Appendix 2). Conservatively, a

value of 7,060kg will be used for the evaluation.5.3 The weight of the upper impact limiter is 2,541kg (Ref. 3.2).5.4 The weight of the primary lid is 3,648kg (Ref. 3.2).5.5 The weight of the secondary lid is 857kg (Ref. 3.2).5.6 The material properties used for the cask shell and the lid bolts shall be as given in Table 1,

unless noted otherwise.5.7 All allowable stresses for the cask shell and the lid bolts shall be as given in Table 2, unless

noted otherwise.5.8 A value of 9.81 m/s2 will be used for the gravitational acceleration.5.9 A value of 0.31 will be used for the Poisson's Ratio of all stainless steel in accordance with

Table NF-2 of Reference 3.6.5.10 Nut factors (k) for bolt preloads will be in accordance with Table G of Reference 3.9. For the

non-lubricated condition, the 0.30 mean value of as-received stainless steel fasteners will beused. For the lubricated condition, the 0.15 mean value of Fel-Pro N 5000 (paste) will beused.

Table 5.1 - Material Properties

Strength (MPa) Young's CoefficientMaterial Temp. Yield Ultimate Membrane Modulus of Thermal(IC) Expansion(C) (SY) (S.) Allowable (GPa) (10E6 mam)

(Sin)

-30 172 483 115 198 -20 172 483 115 195 15.365 157 463 115 192 15.8TypeS304L 100 146 452 115 189 16.2150 132 421 115 186 16.6

200 121 406 110 183 17.0250 114 398 103 179 17.4

-30 896 1030 343 = 1/3Su 199 -20 896 1030 343 202 11.5

ASME SA-354 65 855 1030 343 199 11.8Grade BD 100 816 1030 343 197 12.1

150 792 1030 343 194 12.4200 768 1030 343 191 12.7250 737 1030 343 188 13.0

Notes:1. Material properties are taken from ASME B&PV Code, Section II, Part D (Ref. 3.6) by

interpolation.


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Table 5.2 - Allowable Stresses

Material

Design Criteria ASME SA-240 ASME SA-354Type 304L Grade BD

MPa MPa

Yield Stress, Sy 146 816Tensile Strength, Su 452 1030Design Stress Intensity, Sm 115 299

Pm 115 299Normal

Conditions Pm + Pb 173 449Pm + Pb + Q 345 897

Hypothetical Pm 276 718

Accident Pm + Pb 414 1030Conditions Total Stress 904 2060

Notes:1. Allowable stresses are taken from ASME B&PV Code, Section II, Part D (Ref. 3.6), by

interpolation (see Table 1).2. Allowable stresses are established per Regulatory Guide 7.6 (Ref. 3.8).3. Allowable stresses are established per Regulatory Guide 7.6 (Ref. 3.8), Regulatory Position 4

and ASME, Section III, Division 3 (Ref. 3.7), WB-3200 criteria. The limit on this stress componentis 3S,.

4. Regulatory Guide 7.6 does not provide criteria. ASME Section III, Appendix F (Ref. 3.7) hasbeen used to establish these criteria.

5. Regulatory Guide 7.6 (Ref. 3.8), Regulatory Position 7 and ASME Section III, Division 3, (Ref.3.7) WB-3221.9 criteria for limiting these stresses to 2Sa at ten (10) cycles results in higherallowable stresses than 2S,. The limits for peak stresses are therefore conservatively set at 2S,.

6 METHODOLOGY

The RT-100 transport cask will be a safety-related structure in accordance with 10 CFR Part 71(Ref. 3.1). The cask consists of a stainless steel containment structure with a lead shielding panelbetween the inner and outer cask walls, a pair of ductile steel and foam impact limiters and two setsof stainless steel closure bolts at the primary and secondary closure lids (Ref. 3.2). The boltloadings under the various normal and accident conditions are determined, where applicable, inaccordance with the recommendation of NUREG/CR-6007 (Ref. 3.5). The loads determined inaccordance with this methodology are very conservative. In addition, extremely conservativedynamic amplification factors will be used for determination of the bolt loads under the accident dropload cases. Therefore, factors of safety only slightly above 1.0 are considered to have ampleconservatism and are therefore acceptable. For additional information on the conservatism used inthe various calculations, see the specific evaluations in the following section.

7 CALCULATIONS

NOTE: In many cases, calculations are made using exact values, not the rounded numbers shown.Therefore, in certain situations, the numbers displayed may not be capable of providing the finalsolution. Using exact numbers, however, provides the most accurate solution possible.


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7.1 Closure Bolt EvaluationThe RT-100 package is designed with two sets of closure bolts: 18 M36 hex head bolts at the secondarylid and 32 M48 hex head bolts at the primary lid. These two sets of bolts are credited with maintainingpositive closure of the package under all accident conditions. The purpose of this calculation is tostructurally qualify these bolts for the loadings associated with the normal conditions of transport and thehypothetical accident conditions.

7.1.1 MethodologyBolt loadings under the various normal and accident conditions are determined in accordance with therecommendations of NUREG/CR-6007. Stresses resulting from these loads are compared with thedesign criteria per NUREG/CR-6007 (Ref. 3.5).

7.1.2 LoadsThe following loads are evaluated in this section:

* Internal pressure loads* Temperature loads* Bolt preload* Impact loads* Puncture loads* External pressure loads" Gasket seating load

These loads are combined per NUREG/CR-6007 in Section 7.1.3.

7.1.2.1 Internal Pressure LoadsPer NUREG/CR-6007 (Table 4.3), the forces and moments generated under the internal pressure loadare a tensile load, Fap, a shear load, Fsp, a fixed edge closure force, Ffp, and a fixed edge closuremoment, Mfp. These are evaluated as follows for the primary and secondary lid bolts.

7.1.2.1.1 Internal Pressure Loads for Primary Lid Closure BoltsThe tensile force per bolt due to internal pressure, Fap, is:

F cxD2 x(Ph -Po)F - g

4xNhwhere,

DIg = Outer Seal Diameter= 1835 mm (Ref. 3.2)

Nb = Number of Bolts= 32 (Ref. 3.2)

Pli = Internal Pressure= 35 psi = 241.3 kN/m 2 use 250 kN/m 2 (Ref. 3.16)

Plo = External Pressure= 0 kN/m 2 (conservative)

Thus,

F 7 x 1..8352 x (250 -O). = 4x32

= 20.7 kN/bolt

The shear force per bolt due to internal pressure, Fsp, is:


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F. TrxE, xt, x(P,- P) x Dlb2xNb xEc xtC x (1-Nu,)

where,El = Primary Lid Material Elastic Modulus,

(SA 240 TYPE 304/304L per Table 2)= 195 GPa at 20' C (Table 1)

DIb = Primary Lid Bolt Circle Diameter= 1920 mm (Ref. 3.2)

Nul = Primary Lid Material Poisson's Ratio,(SA 240 TYPE 304/304L per Table 2)

= 0.31 (Table 1)Ec = Cask Material Elastic Modulus,

(SA 240 TYPE 304/304L per Table 2)= 195 GPa at 200 C (Table 1)

tl = Primary Lid Thickness= 210 mm (Ref. 3.2)

tc = Cask Wall Thickness= 65 mm (neglecting lead) (Ref. 3.2)

The remaining terms are as previously defined. However, this expression for shear force does not applyto the RT-100 cask design because the maximum gap between the lid and cask body (just 4 mm =1741- 1737 per Ref. 3.2, Detail 1) is less than the minimum gap between the bolt clearance holes and boltshank ( 5.5 mm = 52.5- 47 per Ref. 3.2 and Machinery's Handbook). Thus, the RT-100 primary lid boltswill not be subjected to any shear loads. Therefore,

FP = 0.0 kN/bolt.

The fixed edge closure force, Ff, and moment, Mf,, are:DR, x (P, - P,) 1.920x (250 - 0)

FP 4 - 4 = 120.0 kN/m

and,(Pli - P1o)X D 2 (250 - 0) x 1.920 2

M fp = - lb _ 32 28.8 kN-m/m.32 32

7.1.2.1.2 Internal Pressure Load for Secondary Lid Closure BoltsThe secondary lid closure bolt forces and moments are determined using the same methodology andequations as shown for the primary lid bolts (Section 7.1.2.1.1) except that the secondary lid features areused.

The tensile force per bolt due to internal pressure, Fas, is:

F00 x DIx X(Pi - P1 )

4xNb

where,Dig Outer Seal Diameter

= 850 mm (Ref 3.2)Nb = Number of Bolts

= 18 (Ref. 3.2)


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P1 = Internal Pressure= 35 psi = 241.3 kN/m 2 use 250 kN/m 2 (Ref. 3.16)

P1. = External Pressure= 0 kN/m 2 (conservative)

Thus,

Fr x 0.8502 x (250- 0)FaS •

4x18

= 7.9 kN/bolt

The maximum gap between the lid and cask body (just 4 mm =748 - 744 per Ref. 3.2, Detail 2) is lessthan the minimum gap between the bolt clearance holes and bolt shank (5.5 mm = 40.5- 35 per Ref. 3.2and Machinery's Handbook). Thus, as with the primary lid (Section 7.1.2.1.1), the shear force per bolt dueto internal pressure, F,,, is:

Fss= 0.0 kN/bolt.

The fixed edge closure force, F1s, and moment, Mfs, are:

Ftý = Dlb X(Pi -Po)4

and,M =(Pli - P,°)× D2b

32

where,Dib = Secondary Lid Bolt Diameter

= 926 mm (Ref. 3.2)

All other terms are previously defined. Thus,

Dib x (pi - PJ) 0.926 x (250-0)=5. kN/m4 4

and,(Pli - P,0 )x D 2 (250 - 0)x 0.926 2

M f. = b - 6.7 kN-m/m.32 32

7.1.2.2 Temperature Loads

Temperature differentials and/or differences in the thermal-expansion coefficients of the joint componentsinduce bolt loads. These forces are evaluated Per NUREG/CR-6007 (Table 4.4).

7.1.2.2.1 Temperature Loads for Primary Lid Closure BoltsThe tensile force per bolt due to temperature, Fatp, is:

F = 0.25xTrxxEx(caxT, -c xTb)

where,Db = Nominal Bolt diameter

= 48 mm (Ref 3.2)


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Eb = Bolt Material Elastic Modulus, (SA 354 Grade BD)= 202 GPa at 20' C

Ck = Primary Lid Material Coefficient of Thermal Expansion= 16.6x10-6 m/m/°C

C = Bolt Material Coefficient of Thermal Expansion= 11.5 x106 m/m/°C

T = Maximum Primary Lid Temperatureunder NCT conditions

= 71'C (conservatively use 100 'C) (Table 3.1-2)Tb = Minimum Bolt Temperature

under NCT conditions-20 0 C (10 CFR 71.71)

Thus,

Fa =0.25xitx0.0482 x202x106 x(16.6x10' x100-11.5x10' x-20)

= 690.9 kN/bolt

Shear force per bolt due to temperature, Fstp, is considered zero because the clamped components (theprimary lid and the cask forged ring) have essentially the same temperature.

7.1.2.2.2 Temperature Loads for Secondary Lid Closure BoltsThe secondary lid closure bolt forces determined using the same methodology and equations as shownfor the primary lid bolts (Section 7.1.2.2.1) with the secondary lid geometry, material properties andtemperatures. Since the primary and secondary lids are constructed of the same materials andexperience essentially the same temperatures, only the bolt diameter in the previous equation must bechanged. The secondary bolt diameter is 36 mm. Thus, the tensile force per unit bolt due to temperature,Fstl, is:

F,=0.25x 7T x 0.0362 x 202x10' x (16.6 10' x100 -11.5x 10" x -20)

= 388.6 kN/bolt

As with the primary lid, the shear force per bolt due to temperature, Fsts, is considered zero because theclamped components (the primary and secondary lids) have essentially the same temperature.

7.1.2.3 Bolt PreloadsTightening torques for the primary and secondary lid bolts are, respectively, 850 N-m +/-10 % and 350 +/-10 % per Ref 3.2. The method of analysis is described in NUREG/CR-6007, Table 4.1.

7.1.2.3.1 Bolt Preload for Primary Lid Closure BoltsThe primary lid bolt preload, Fpl, is determined as follows (NUREG/CR-6007, Table 4.1):

F = TK, ---where,

Db Nominal Bolt diameter= 48 mm (Ref 3.2)

K Nut Factor for empirical relation betweenapplied torque and the achieved preload0.15 (lubricated) minimum (Ref. 3.9)0.30 (dry) maximum


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T = Applied Torque= 850 N-m +/-10% (Ref 3.2)

To determine the maximum preload, Fplmax, for the primary lid bolts, the minimum nut factor, K, of 0.15(lubricated) and maximum tightening torque of 940 N-m (conservatively bounds the 850 N-m+10%maximum):

F Tmax 940 l kNK- LxK xDb 0.15x0.048 1000N

= 130.6 kN

The residual torsion moment, Mrn, is.

M 0l =0.5XTma =0.5x940= 470 N-m

The residual tensile bolt force, F. 1, is

Fa,= Fplmax = 130.6 kN

7.1.2.3.2 Bolt Preload for Secondary Lid Closure BoltsThe maximum secondary lid bolt preload, Fpsmax, is detennined hi a manner similar to the primary bolt lids (Section7.1.2.3.1). Thus,

Fpsina. - TmLXb

K lL x D b

where,Db = Nominal Bolt diameter

= 36 mm (Ref 3.2)T Applied Torque

= 350 N-m +/-10% (Ref 3.2)

Other terms are as previously defined. Using a nut factor, K, of 0.15 (lubricated) and a tightening torqueof 390 N-m (conservatively bounding the 350 N-m+10% maximum) yields the maximum preload, Fpsmax,for the secondary lid bolts:

TmaF x 390Fps.,.. --

KLXbD 0. 15x0.036= 72.2 kN

The residual torsion moment, Ms, is:

Mrs =0. 5XTm, = 0.5x 390= 195.0 N-m

The residual tensile bolt force, Fars, is:

Fars = Fpsmax = 72.2 kN


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7.1.2.4 Impact LoadsMaximum tension and shear loads in the closure bolts due to the regulatory impact drops are evaluated inaccordance with NUREG/CR-6007. Using the NUREG terminology, the primary lid bolts are evaluated asclosure bolts for an unprotected lid and the secondary bolts are evaluated as components of a protectedlid. This means the primary bolt evaluation includes the impact or inertial forces of the entire cask and thesecondary lid bolts are evaluated only for the forces due to the inertia of the secondary lid.

7.1.2.4.1 Dynamic Load FactorsDrop impact loadings are generally considered triangular or half sine loadings and NUREG/CR-3966presents dynamic load factor (DLF) charts for either pulse shape. For this analysis, we compare resultsand utilize the loading with the higher DLF.

Dynamic load factors for triangular and half sine loadings are shown in Figures 2.3 and Figure 2.15(NUREG/CR-3966). This information is presented as graphs where the DLF is the ordinate and theabscissa is td/T. This latter quantity, td/T, is the ratio of the impact duration, td, and the natural period ofthe impacting object, T.

For bolt closure analyses, the period of the lids, T, is considered. T is determined from the lid lowestmode frequency.

7.1.2.4.1.1 Dynamic Load Factors for Primary Lid Closure BoltsTo determine the primary lid frequency, the primary lid and secondary lid are considered a single simply-supported flat circular plate. Thus, (Roark, Table 36, Case 1la):

Resonant Frequency of Primary Lid (with secondary lid attached):f = 4.99 .Dg:

2n V wr'where,

D Lid Flexural Rigidity

- E 1 1

12( 1- Nu 2

g = conversion factor= 1000 kg-mm/s 2-N

w = weight per unit arear = lid bolt radius

= DIb/2= 960 mm

Per 3.2, the primary lid weighs 3648 kg and the secondary lid weighs 857 kg. Thus,

(3648+857)w 0r0-5960 2

0.001556 kg/mm2

D may be determined from previously defined values:


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D 195x103 . 210'

12(1-0.3 12)

= 1.665x10 11 N-mm

The frequency of the primary lid (with attached secondary lid) is:

= 0.7942. 1.665 x 10' .1000" 0.001556.960'

= 282 Hz

The period of the primary lid is equal to 1/fl, or T = 1/282 = 0.00354 s. Impact durations for the NCT andHAC impacts range from 0.012 s to 0.045 s (ST-0401). Thus, the smallest value of the ratio td/T is 3.389and the largest is 12.71. With these values, the maximum DLF was determined from Figures 2.3 and2.15 (NUREG/CR-6007) to be less than 1.2. Thus, it is concluded that the DLF for the primary lid boltsmay be conservatively bounded by a value of 1.2 for both NCT and HAC drops.

7.1.2.4.1.2 Dynamic Load Factors for Secondary Lid Closure BoltsTo determine the secondary lid frequency, the secondary lid is considered a simply-supported flat circularplate. Thus, (Roark, Table 36, Case 1la):

Resonant Frequency of Secondary Lid:S = 0.7942 Dg

Nwr 4

where,D = Lid Flexural Rigidity

-E 1 t•

120- Nul2

gc = conversion factor= 1000 kg-mm/s2 -N

w = weight per unit arear = lid bolt radius

= DIb/2= 463 mm

Per 3.2, the secondary lid weighs 857 kg. Furthermore,

El Secondary Lid Material Elastic Modulus,(SA 240 TYPE 304/304L)

= 195 GPa at 20 0 CDib Secondary Lid Bolt Circle Diameter

= 926 mm (Ref. 3.2)N.= Secondary Lid Material Poisson's Ratio,

(SA 240 TYPE 304/304L per Table 2)0.31

= Secondary Lid Thickness= 110 mm (stainless steel only) (Ref. 3.2)

Thus,

857w7.4632


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= 0.001272 kg/mm2

D may be determined from previously defined values:

D = 195x103. 110'12(I-0.3 12)

= 2.395 x10 10 N-mm

The frequency of the secondary lid is:

f 0.7942. 2.395 x10'0 1000V0.001272o 1463

= 508 Hz

The period of the secondary lid is equal to 1/fl, or T = 1/508 = 0.00197 s. Impact durations for the NCTand HAC impacts range from 0.012 s to 0.045 s (ST-0401). Thus, the value of td/T is 4.6 or more. Withthis value, the maximum DLF can be determined from Figures 2.3 and 2.15 (NUREG/CR-6007) to beapproaching unity. However, for consistency with the primary lid bolt analyses, the DLF for the secondarylid bolts will be conservatively set to 1.15 for both NCT and HAC drops.

7.1.2.4.2 End Drop Loads

7.1.2.4.2.1 Primary Lid BoltsImpact loads in the primary lid bolts due to an end drop are determined using the formulas for evaluatingbolt forces/moments generated by impact load applied to an unprotected closure in NUREG/CR-6007Table 4.5. An acceleration of 125 g is used in this analysis (which bounds the 123 g maximum reported inRef. 3.17).

The non-prying tensile bolt force per primary lid bolt, Ftp, is:

- 1.34 xsin(xi)x DLF xai x(WL + WJ)xgNb

where,x = End Drop Impact Angle

= 900

DLF = 1.15 (Section 7.1.2.4.1)a, = Maximum Impact Acceleration

= 123 g (use 125 g) (3.17)WL = Closure Lid Weight

= 3648 kg (use 3650 kg) (3.2)we= Cask Payload Weight

= 6804 kg (use 7000 kg) (3.2)Nb = Number of Bolts

= 32 (3.2)Thus,

1.34 x sin(90.0)x 1.15 x 125 x(3650 + 7000) x9.81 1 kNFtp x

32 1000 N= 628.9 kN/bolt

As discussed in Section 7.1.2.1.1, the RT-100 primary lid bolts will not be subjected to any shear loads.Thus,


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FP = 0.0 kN/bolt.

The fixed edge closure lid force, Ff, is:

F 1.34 xsin(x,)x DLFxa ×(WL +W:)xgrx Dib

where,

Dib = Primary Lid Bolt Diameter= 1920 mm (Ref. 3.2)

The remaining terms are as previously defined. Thus,

Fr 1.34xsin(90.0)xl.15x125x(3650+7000)x9.81 lkNK.= X -- _

7x 1.920 1000N

3336.4 kN/m

The fixed edge closure lid moment, Mf, is:

1.34x sin(xi)x DLFx ai x(WL + ±W)xg7Ex8

All other terms are as previously defined. Thus,

1.34xsin(90.0)xl.15x125x(3650+ 7000)x9.81 lkN71x8 1000N

= 800.7 kN-m/m

The additional tensile bolt force per bolt, Ftp, caused by the prying action of the primarylid is (NUREG/CR-6007 Table 2.1):

Nb C1 + C2

where,P = Bolt Preload per unit Length of Bolt Circle

Fpimax Dib

Cl =

C2 =

Non-prying Tensile Bolt ForceMAX(Ff, P)Force Constant1.0Second Force Constant


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8 (D.D)2 E +(D,-D,)xE,× xt,.3 x (D ,o D ,, 2 1- N ., D lb

(N, x D' x Eb

Do. Closure Lid Diameter at Outer Edge= 2016 mm (3.2)

Dj= Closure Lid Diameter at Inner Edge= 1730 mm (3.2)

t = Closure Lid Flange Thickness= 120 mm (3.2)

Elf = Primary Lid Flange Material Elastic Modulus,(SA 240 TYPE 304/304L)195 GPa at 20°C

Lb Bolt length between the top and bottom surfaces of theclosure lid at the bolt circle67 mm (3.2)

Other terms are as previously defined. Thus,

8 x (D.I-bDJ -E, x t.

- 2 x xEb

C2 =

= 3.47

P = 130.6x 32x 1.920

= 692.9 kN/mand

2 x 800.7 -1 X (3336.4 - 3336.4)- 3.47 x (3336.4 - 692.9)7r x .920 2.016 -1.92032 ) 1+3.47

316.2 kN/bolt

The total tension force, Fa, is

Fa Ft + Ftp= 628.9 + 316.2= 945.1 kN/bolt

The shear force, F8, is 0.

The maximum bending moment generated by the applied loads, Mbb, is:


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Mbb = lb-D+ Kbi I MfNb Kb+K1)

where,

Kb =(-x Eb x4)

(32 )x202A10 6) x0.048')

.0.067) 1.920 64

4,167 kN

K1 = E, x t

3 [(1- N,) + (1 -N) 2 D )oo D× b

195x 10' x 0.210'

3x[(I -0.31:)+(1-0.31)2 x 1.920)j 1.920, (2.01'6) . 192

= 234,719 kNThus,

Mbb x~l Kb ')xMfMbb = Nb ) Kb+K1)

rnx1.920)( 4.1,67 x800.732 ) 4,167 +234,719

= 2.6 kN-m

7.1.2.4.2.2 Secondary Lid BoltsImpact loads in the secondary lid bolts due to an end drop are determined similarly as for the primary lidbolts in Section 7.1.2.4.2.1.

The non-prying tensile bolt force per secondary lid bolt, Ftf, is:

. 1.34xsin(x,)xDLFxa, x(WL +Wj)xgIs -

Nb


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= 900DLF = 1.15 (Section 7.1.2.4.2.1)ai = Maximum Impact Acceleration

= 123 g (use 125 g) (3.17)WL = Closure Lid Weight

= 857 kg (use 860 kg) (3.2)Wcs= Payload Weight borne by Secondary LidNb) Number of Bolts

= 18 (3.2)

Since the payload weight is assumed to be evenly distributed across both the primary and secondarylids, the weight borne by the secondary lid can be obtained by multiplying the payload weight by theratio of areas, i.e.,

A0 785 x7000

- 3.192)= 1674 kg

Thus,1.34 x sin(90.0) x 1.15x 125 x (860 +1674) x 9.81 1 kN

18 1000 N

= 266.0 kN/bolt

As discussed in Section 7.1.2.1.1.2, the RT-100 secondary lid bolts will not be subjected to any shearloads. Thus,

F, =0.0 kN/bolt

The fixed edge closure lid force, Ff, is:1.34 x sin(xi)x DLF x ai x (WL + W)x g

' 7x Dlb

where,Dib = Secondary Lid Bolt Diameter

= 926 mm (Ref. 3.2)

The remaining terms are as previously defined. Thus,1.34x sin(90.0) x 1.15 x125x(860 +1674)x 9.81 1kN

,x 0.926 1000 N= 1646.1 kN/m


S1.34x sin(x,)xDLFxa, x(WL + W. )xg7zx8

where all terms are as previously defined. Thus,


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M 1.34xsin(90.0)xl.15x125x(860+1674)x9.81 lkN11x8 1000 N

= 190.5 kN-m/m

The additional tensile bolt force per bolt, Fin, caused by the prying action of the secondary

lid is (NUREG/CR-6007 Table 2.1):

2xM o l-CX(B-F)-C2x(B-P)1

Nb C1+C2


Fplmax x Nb

7rx Db

B = Non-prying Tensile Bolt Force= MAX(Ff, P)

Cl = Force Constant= 1.0

C2 = Second Force Constant

E, +t• (D, -D,)xE, xt']3<(D,o - D,b 1 - Nu, Dib

x( Lb

,Nb x D 2b

D=. Closure Lid Diameter at Outer Edge1000 mm (3.2)

D,= Closure Lid Diameter at Inner Edge= 745 mm (3.2)

t = Closure Lid Flange Thickness= 80 mm (3.2)

Elf = Secondary Lid Flange Material Elastic Modulus,(SA 240 TYPE 304/304L)195 GPa at 20 * C

Lb Bolt length between the top and bottom surfaces of theclosure lid at the bolt circle43 mm (3.2)


3×(D+o.D. -D xx1-N xDt"

C2 = x(Do Db [ - N , DE

(NxD xEb bLb b b


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- 3x(1.000- 0. 9 2 6 )2

-195x10 6 x0.11 (1.000-.745)x195xlO6xO.0803

1 -0.31 0.926

X( 0.043 .x18x0.036 2x202x1 06)

= 1.79

P = 72.2 x 18= x 0.926

= 446.7 kN/m

and

7E x 0 .926

2x0190.5 16x(1646.1-1646.1)-1.79x(1646.1-446.7)j1.00-0.926 " i"

= 173.5 kN/bolt


Fa = Ft + Ft,= 266.0 + 173.5

439.5 kN/bolt

The shear force, Fs, is 0.

The maximum bending moment generated by the applied loads, Mbb, is (NUREG/CR-6007, Table2.2):

Mbb = "nx> Kb"++Kb xMf

where,

Kb X ( b

Lb CDlb) "i(18 lx(202xlO')./x¢.036 4

0 .043 k 0.926 ) •, 64

= 2,396.5 kN


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K1 =3

E2 x (t• b)

3x (I-N 'j,+ (11 N,,,) x xD,,

195 x106 x0.1103

3 (1-0.3 12) +(1- 2 0.31)) x 0.926(1 1.000 ,j

Thus,= 71,203 kN

Mbb - "l'bNbKb xMf

Kb + K1)

_ itx0.926 2.40 ) x 190.518 ) 2.40 + 71.2)

= 1.0 kN-m

7.1.2.4.3 Corner Drop EvaluationsThe closure bolt evaluations for the corner drop impact are conducted very similarly to the end dropanalyses in Section 7.1.4.1. The cask body acceleration is changed and the impact angle, xi, is set equalto 52.5' (corresponding to a 37.50 angle between cask axis and vertical line). Additionally, an accelerationof 120 g is used in this analysis (which bounds the 116 g maximum reported in Ref. 3.17). Results aresummarized in Table 7.1.2.4.3-1.

Table 7.1.2.4.3-1 Closure Bolt Loads for 9.0 m Corner-Drop

Non-Prying Prying Bending Shear Force,BOLTILOCATION Tensile Force, Ft Force, Ftp Moment, Mbb Fs (kNlbolt)

(kN/bolt) (kN/m) (kN-mlbolt)

M48x170 Bolts/Primary Lid 479.0 265.0 2.0 0.0

M36x120 Bolts/Secondary Lid 211.4 147.4 0.8 0.0

7.1.2.4.4 Side Drop EvaluationsAs shown in Sections 7.1.2.1.1 and 7.1.2.1.2, the gap between the cask body and the primary andsecond lids are smaller than the gap between the bolts and the bolt clearance holes. Therefore, no shearload are imparted to the bolts from the cask body. Since the side impact drop will primarily generate shearloads with respect to the bolts, the primary and secondary closure lid bolts will not see any significantloading from the side impact drop and are acceptable with respect to the end and corner impact drop.

7.1.2.5 Puncture Loads

7.1.2.5.1 End PuncturePuncture loads in the primary and secondary closure lid bolts due to a puncture are determined using theformulas for evaluating bolt forces/moments in NUREG/CR-6007 Table 4.7.


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7.1.2.5.1.1 Primary Lid BoltsThe non-prying tensile bolt force per primary lid bolt, Fto, is:

Ftp = sin(xi)x P.Nb

where,x, = End Drop Impact Angle

= 900PunNb

MIN(Pun1 , Pun2)Number of Bolts32

The term, Pun, is the maximum impact force that can be generated by the puncture pin during a normalimpact. It is the smaller of:

Punl1 0.75 x 7ExD 2b X y1

0.6 x nt x Dph X ti x S,11where,

Dpb = Puncture bar diameter= 150 mm (10 CFR 71.73(c)(3))= Closure Lid Thickness

110mm (Ref 3.2)(the secondary lid thickness neglecting the lead)

Syl Yield Strength of Closure Lid Material(SA 240 304L)172 MPa at 20 ' C

Sul Ultimate Strength of Closure Lid Material(SA 240 304L)483 MPa at 20 ' C

Thus,Pun1 0.75 x7r x 0-150 2 x 172000

= 9,118.5 kNPun2 0.6xirx0.150x0.110x483000

= 15,022 kNPun= MIN(9,118.5, 15,022)

9,118.5 kN

andFtp sin(90)x9118.5

32= 285.0 kN/bolt.

As shown in Sections 7.1.2.2.1.1 and 7.1.2.2.1.1, the design of the primary and secondary lids preventsshear loads being applied to the bolts. Thus,

Fsp = 0.


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It is noted that the equation given for Fsp in NUREG/CR-6007 also shows Fp = 0.

The fixed edge closure lid force, Ff, is:

Ff = sin(xj)x P.irt x Dlb

where,Dib = Primary Lid Bolt Circle Diameter

= 1920 mm (Ref. 3.2)


Ff = sin(90) x9118.5

Ffx 1.920

= 1511.7 kN/m


=f sin(x )x Pn4xn

Thus,

Mf sin (90) x 9118.54xit

= 725.6 kN-m/m

The additional tensile bolt force per bolt, Ftp, caused by the prying action of the primarylid is (NUREG/CR-6007 Table 2.1):

2__ x -ClX (B -Ff)-C2 x(B -P)]F,, x D,0x D, -D.F o - -N , C 1+ C 2


692.9 kN/m (as shown in Section 7.1.2.4.2.1)B Non-prying Tensile Bolt Force

= MAX(Ff, P)Cl Force Constant

= 1.0C2 = Second Force Constant

S 8 ), [E, xt' +.(DI.o- Dj),,)x ttilf..3 x (D,.-- ,-ix/ -- +Dix Lb Dbbj) --uIDl b

X•N xD D2 x E"

= 3.47 (as shown in Section 7.1.2.4.2.1)Dio = Closure Lid Diameter at Outer Edge

2016 mm (3.2)


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Di= Closure Lid Diameter at Inner Edge= 1730 mm (3.2)

t= Closure Lid Flange Thickness= 120 mm (3.2)

Elf = Primary Lid Flange Material Elastic Modulus,(SA 240 TYPE 304L)

= 195 GPa at 20 0CLb = Bolt length between the top and bottom surfaces of the

closure lid at the bolt circle67 mm (3.2)


Ftp=

2x725.6 - II 11.7- 6;r_192 2.016 - 1.920

32 + 3.47

= 517.3 kN/bolt


Fa = Ft + Ftp= 285.0 + 517.3= 802.3 kN/bolt



Mbb -( rtxDb+ Kb ')xMfNb Kb+K1)

where,

Kb tXý+.EbJX[.DiJ

= 4,167.8 kN (as shown in Section 7.1.2.4.2.1)

K1 = E xt•

3 -19 N)+ (Ia- Nsuho x S 7...42 D .b

=234,719 kN (as shown in Section 7.1.2.4.2.1)


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Thus,

Mbb =bx Kb x Mf

Irx1. 9 20). 4.167 725.6~32) 4.167 +234.7

= 2.4 kN-m

7.1.2.5.1.2 Secondary Lid BoltsThe non-prying tensile bolt force per secondary lid bolt, Fs,, is:

Fts = sin(xj)x P.Nb


= 900Pun = MIN(Punl , Pun2)Nb = Number of Bolts

= 18

Pun was evaluated in Section 7.1.2.5.1.1:Pun = 9,118.5 kN

The primary and secondary lids act together under the pin puncture load. Therefore, the secondary lidwill only see a portion of the impact load from the pin, so Pun will be reduced by the ratio of the secondarylid volume to the total lid volume.

V = Secondary Lid Volume

2 Xt= --xD~hxt1

4

Vt = Total Lid Volume

= -x D 2 xt4 P

where,D~bp = Closure Lid Bolt Diameter at Primary Lid Bolts

= 1920 mm (3.2)t• = Closure Lid Thickness at Primary Lid Bolts

= 210 mm (3.2)t~a = Average Lid Thickness

ti + tip

2

Thus,110+210

ta =

2= 160 mm


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V = -- x 0.9262x0.1104

= 0.067 m3

71Vt = -x1.92W ×x0.160

4= 0.463 m 3

Pun = r X 1;Vt

0.067Pun = 9118.5×X0

0.463= 1325.6 kN

and

sin (90) x 1325.618

73.6 kN/bolt.

As shown in Sections 7.1.2.2.1.1 and 7.1.2.2.1.1, the design of the primary and secondary lids prevents

shear loads being applied to the bolts. Thus,

Fss = 0.

It is noted that the equation given for F,, in NUREG/CR-6007 also shows F,, = 0.

The fixed edge closure lid force, Ff, is:Ff sin(xi )x Pun

7r x Dlbwhere,

Dib Secondary Lid Bolt Circle Diameter926 mm (Ref. 3.2)


Ff = sin(90) x 1325.6×T x 0.926

455.7 kN/m


= sin(xi ) x Pun4xir

Thus,

Mf : sin(90) x 1325.64x iT

= 105.5 kN-m/m


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The additional tensile bolt force per bolt, Ft, caused by the prying action of the secondary lid is (Ref. 3.5Table 2.1):

2xMf[ -CIx(B- F,)C2x(B-P)]

Nb CI+C2

where,P Bolt Preload per unit Length of Bolt Circle

= 446.7 kN/m (as shown in Section 7.1.2.4.2.2)B = Non-prying Tensile Bolt Force

= MAX(Ff, P)Cl = Force Constant

= 1.0C2 = Second Force Constant

8 x[, l"(I. i f X t•]"

3 x(DIO-Dlb) IL-N. Dlb

Nb xD 2xE

= 1.79 (as shown in Section 7.1.2.4.2.2)Dio = Closure Lid Diameter at Outer Edge

1000 mm (3.2)Di= Closure Lid Diameter at Inner Edge

= 745 mm (3.2)ti = Closure Lid Flange Thickness

= 80 mm (3.2)Elf = Secondary Lid Flange Material Elastic Modulus,

(SA 240 TYPE 304L)= 195 GPa at 20 ' C

Lb = Bolt length between the top and bottom surfaces of theclosure lid at the bolt circle43 mm (3.2)


Fts =

2 x 105.5 1 7x7 179x -

(nt x 0 - 6" (455.7 -0455.7)0- (455.7 -446.7)

= 164 kN/bolt


Fa = Ft +Ft.


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= 73.6 + 164= 237.6 kN/bolt



Mbb - Olb K 1 Mfirx D _____

Nb wer

where,

Kb = × 5-7

= 2,396.5 kN (as sh own in Section 7.1.2.4.2.2)E, x t'K1 =

3 x 2

71,03 kN U +71,203 kN (as shown in Section 7.1.2.4.2.2) I

Thus,

Mbb = 7txDlbjx(Kb >Mf

(n x 0.926)kN=0.6 kN-m

S 2.40 x105.5.2.40 + 71.2)

7.1.2.5.2 Side PunctureIn Section 7.1.2.1.1, the gap between the cask body and the primary lid is shown to be smaller than thegap between the M48 bolts and the bolt clearance holes. Therefore, no shear load will be imparted to thebolts from the cask body. Further, there are no other loads resulting from side puncture at the bolts. Thus,no significant loads are imparted to the primary and secondary closure lid bolts during a side punctureevent.

7.1.2.6 External Pressure

Loads in the primary and secondary closure lid bolts due to external pressure are evaluated using theformulas for evaluating bolt forces/moments in Ref. 3.5 Table 4.3.

7.1.2,6.1 Primary Lid BoltsThe pressure outside the cask, Pl0 , in the case of immersion is assumed to be 350 kPa (3.16). Thepressure inside the cask, P1 , is taken to be 0 kPa.

The axial force per bolt due to external pressure is:

Fa7ExD. X(Pu -P )

4xNb


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where,Dig = Outside Seal Diameter


= 32 (Ref 3.2)

Thus,Fx 1.835 x (0-350)Fa =

4x32

= -28.9 kN/bolt

Since this force is negative (inward acting), the actual resulting bolt force is Fa = 0 because theapplied load is supported by the cask wall and not by the bolts.

The fixed edge closure lid force is:

Ff - Dlb x(Pli -P 1 o)

4where,

Db= Bolt Circle Diameter= 1920 mm (Ref 3.2)

Thus,1.92 x(0-350)EF =

4

= -168.0 kN/m

The fixed edge closure lid moment is:

Mf (Pli - Pl) x Dlb32

(0- 350)x 1.92 2

32

-40.32 kN-m/m

The shear bolt force per bolt is:Fs 7cr x E, X t, x (Pli - Plo)x D lb

2xNEb xtE( x(t, x-(l- N)

All terms are as previously defined.

Thus,

7r x 195000x 0.2 1Ox (0- 350) x 1.922F-s - 2 x 32 x195000x0.065x (1 -0.3 1)


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= -296.5 kN/bolt

The maximum gap between the lid and cask body is less than the minimum gap between the boltclearance holes and bolt shank (see Section 7.1.2.1.1). Thus, the RT-100 primary lid bolts will not besubjected to any shear loads. Therefore,

k =0.0 kN/bolt.

7.1.2.6.2 Secondary Lid BoltsThe pressure outside the cask, P10, in the case of immersion is assumed to be 350 kPa (3.16). Thepressure inside the cask, P,, is conservatively taken to be 0 kPa.

The axial force per bolt due to external pressure is:

Fa = xDjg X(Ph -i14xNb

where.Dt0 = Outside Seal Diameter


18 (Ref 3.2)Thus,

7x0.8502 x(0-350)Fa = 4x18

= -11.0 kN/bolt

Since this force is negative (inward acting), the actual resulting bolt force is Fa = 0 because the load is

supported by the cask wall and not by the bolts.

The fixed edge closure lid force is:

f D~b x(Pli-P o)4

where,Dib = Bolt Circle Diameter

= 926 mm (Ref 3.2)

Thus,0.926x (0- 350)

Ff =

4= -81.0 kN/m

The fixed edge closure lid moment is:

- (Pli- P)x D~b32

= (0- 350)x 0.9262

32-9.4 kN-m/m

The shear bolt force per bolt is


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7rxEl xt] x(P1 i -PJxDt'Fs

2 xNb xE,, xt, x (I-NJ1

All terms are as previously defined. Thus,

7r x 195000x 0.11 lOx (0 -350) x0.926 2FS ,

2 x 18 x 195000x 0.065x (I -0.3 1)

= -64.2 kN/bolt

The maximum gap between the lid and cask body is less than the minimum gap between the boltclearance holes and bolt shank (see Section 7.1.2.1.2). Thus, the RT-100 secondary lid bolts will not besubjected to any shear loads. Therefore,

, =0.0 kN/bolt.

7.1.2.7 Gasket Seating LoadA small closure force is required to maintain a positive seal between the cask lid and the cask body.However, this closure force is much less than the minimum preloads provided for the closure bolts at theprimary and secondary lids. Therefore, the gasket seating load is negligible, and F, = 0.

7.1.3 Load CombinationsThe loadings in Section 7.1.2 are combined to form load cases for the closure bolt analysis per Ref. 3.5.The corresponding bolt stresses are obtained and compared to the criteria defined in Ref. 3.5. Asummary of the loads on the bolts for the primary and secondary lids under the normal conditions oftransport and the hypothetical accident conditions is presented in Tables 7.1.3-1 and 7.1.3-2,respectively.

Table 7.1.3-1: Primary Lid Bolt Load Summar

Non-Prying Torsional Prying Force PryingLoad Case Applied Load Tensile Force Ft Moment Mt Ff Moment Mf

(kN/bolt) (kN-m/bolt) (kN/m) (kN-mlm)

Minimum 52.8 0.2 0.0 0.0Preload Residual Torque Maximum 130.6 0.5 0.0 0.0

Gasket Seating Load 0.0 0.0 0.0 0.0

IntemalPressr 250 kN/m 2 (35 psi) 20.7 0.0 120.0 28.8Pressure

Thermal 100°C 690.9 0.0 0.0 0.0

Puncture Drop on 15 cm diameter pin 285.0 2.4 1511.7 725.6ExternalPressur 350 kPa pressure 0.0 0.0 -168.0 -40.3PressureFree Drop Drop from 9 m height 628.9 2.6 3336.4 800.7


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Table 7.1.3-2: Sccondarv Lid Bolt Load Summary

Non-Prying Torsional Prying Force PryingLoad Case Applied Load Tensile Force Ft Moment Mt Ff Moment Mf

(kN/bolt) (kN-mlbolt) (kN/m) (kN-m/m)

Minimum 29.6 0.1 0.0 0.0P Maximum 72.2 0.2 0.0 0.0

Gasket Seating Load 0.0 0.0 0.0 0.0

Internal 250 kN/m 2 (35 psi) pressure 7.9 0.0 57.9 6.7Pressure

Thermal 100 0C 388.6 0.0 0.0 0.0

Puncture Drop on 15 cm diameter pin 237.6 0.6 455.7 105.5 I

External 350 kPa pressure 0.0 0.0 -81.0 -9.4 IPressure

Free Drop Drop from 9 m height 266.0 1.0 1646.1 190.5 I

7. 1.3.1 Primary Lid Closure Bolt Evaluation under Normal Conditions of TransportThe maximum tension, shear and bolt bearing loads in the primary lid bolts due to the combined NCTloads will be evaluated in accordance with Ref. 3.5, with due consideration given to the prying effects onthe fixed lid. Since the prying forces act inward, normal to the cask lid, an additional prying force isgenerated (Ref. 3.5). For the NCT condition, the controlling load case is the summation of the bolt Ipreload, the internal pressure load and the thermal expansion load. The maximum bolt tension load (Ft),shear load (Fs) and torsional moment (Mt) for the primary lid bolts are (see Table 7.1.3-1):

F, = , + F., + F.•,

Fs = FJ + FS,

Mt = p, +M +, +M

where,Fp -

Fap --

Fatp =

Fp =

Mpt =

Mat

Maximum Bolt Preload130.6 kN/bolt (Table 7.1.3-1)Bolt Internal Pressure Load20.7 kN/bolt (Table 7.1.3-1)Bolt Temperature Load690.9 kN/bolt (Table 7.1.3-1)Bolt Internal Pressure Shear Load0.0 kN/ bolt (Table 7.1.3-1)Bolt Temperature Shear Load0.0 kN/bolt (Table 7.1.3-1)Maximum Bolt Preload Torsional Moment0.5 kN-m/bolt (Table 7.1.3-1)Bolt Internal Pressure Torsional Moment0.0 kN-m/bolt (Table 7.1.3-1)


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W Bolt Temperature Torsional Moment0.0 kN-m/bolt (Table 7.1.3-1)

Thus,Ft = Fp +Fap + Fý,

= 130.6+20.7+690.9= 842.1 kN/bolt

Fs = Fý, +F,

= 0.0+0.0= 0.0 kN/bolt

M = Mp, + M.a + Mt

= 0.5+0.0+0.0= 0.5 kN-m/bolt

Conservatively, the fixed-edge closure lid prying is taken from the external pressure load case. Thisaccident load case bounds all normal conditions and provides a conservative result. The additionaltensile bolt force per bolt, Ftp, caused by the prying action of the primary lid is (Ref. 3.5 Table 2.1):

[ 2xM -CIx(B-Ff)-C2x(B-P)

F N 1EDb ) C +C2

where,Ff Fixed Edge Closure Force

= -168.0 kN/mMf = Fixed Edge Closure Moment

= -40.32 kN-m/m

(Table 7.1.3-1)

(Table 7.1.3-1)

Other terms are as previously defined.

Thus,

2 x (-40.32) -1 x (692.9 - (-168.0))- 3.47 x

aF 1921+ 3.47

= -18.4 kN/bolt

Since this bolt load is less than the load generated by the minimum bolt preload (130.6 kN > -18.3 kN), the pryingforce generated by the external pressure is not critical with respect to bolt stress and will not result in the loss oflid closure seal.

The total tension force, Fa, is:


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Fa = F, + F

= 842.1 + (-18.4)823.7 kN/bolt


Mbb 7E xl" K


Mbb (.•txg ,0t Kb X MfNb Kb+K1)

(7rx 1920 x 4.17 x (-40.32.)32.-. 4.17+ 234.7

= -0.13 kN-m

The average bolt stresses from the combined NCT loads are determined in accordance with Table 5.1 ofRef. 3.5. The bolt stress diameter, D, is:

D = D, - 0. 9 3 8 2 x pwhere,

p = Bolt Pitch= 5.0 mm (Ref. 3.11)

Thus,D = D, - 0.9382 x p

= 0.048 - 0.9382 x 0.0050= 0.043 m

The average tensile stress, Sba, average shear stress, Sbs, maximum bending stress, Sbb, andmaximum torsional stress, Sbt, are:

1.2732 x FaSba = D

1.2732 x823.7 1iMPa= JX

0.0432 1000 ki/ ,in-)

559.1 MPa

Sbs - 1.2732 x Fs[02


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1.2732 x 0.0 1 MPax

0.043 x 1000 kN/

= 0.0 MPa

Sbb 10.186xMbbD'b b

10.186 x (-0.13) IAIMPa

0.048' X00kNI

-12.2 MPa

Sbt 5.093 x M,

5.093 x 0.5 1 MPa0.048 3 100 k3m2

= 21.6 MPa

The allowable stresses for the bolts are:

ta= Allowable Tensile Stress= 0.7xS.=sa Allowable Shear Stress= 0.42 x S.

Gba = Allowable Bending Stress= 1.5xSr

where,Su= Primary Bolt Ultimate Stress

(SA 354 Grade BD)1030 MPa at 20 'C

Smn = Primary Bolt Membrane Stress(SA 354 Grade BD)

= 343 MPa at 20'CThus,

Gta - 0.7 x Su= 0.7x1030= 721 MPa

0sa = 0.42 x S.

= 0.42 x 1030= 432.6 MPa

0ba = 1.5xS.

= 1.5x343= 514.5 MPa


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The maximum interaction ratio for the combined shear and tension loads is therefore:

1. R. Sb C _rY+Cs0 i(Ft. ) (Ts

The minimum factor of safety is:

____

= 0.601

1I.R.

1.P

0.601= 1.66 > 1.0

+ ( 0 )

FS

The maximum interaction ratio for the bending load is therefore:

1.R. = (Sbb)2

(- 12.2 )2

514.5)= 0.00056


1FS = 1

1.R.,

I

0.00056

1774.7 > 1.0

The maximum stress intensity in the primary lid bolts under the combined loads, Sbi, is:

Sbi = J(Sba + Sbb)2 + 4 x (Sbs + Sbt) 2

= = V(559.1 + (-12.2))2 +4x(O+ 21.6)2

= 548.6 MPa

The primary lid closure bolts utilize a custom washer for the bolts with an outer diameter, do,, of 130 mmand a hole diameter, doh, of 52 mm. Therefore, the bearing stress under the bolt head, Sbrg, is:


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FSbrg

=

A, b9

Where,Abrg Bolt Bearing Area

0 2 2.= - x °d -d °,)

= 4x(o.13ao-o.o052)

= 0.0111 m2

Thus,823.7 IMPa

Sbrg- - x0.0ll 1 0 0 0 kN/2

/M 2

73.9 MPa

The allowable normal condition bearing stress on the lid is taken to be the yield stress of the lid materialat 250 *C. The maximum interaction ratio for the bearing load is therefore:

1.R. - SbrgSyl

where,Sy = Primary Lid Material Yield Stress,

(SA 240 TYPE 304L)= 114 MPa at 25 0 ' C

Thus,

1. R. - SbrgSyl

73.9

114= 0.65


1FS

I.R

0.65= 1.54 > 1.0

Because the cask material is weaker than the bolting material, failure will occur at the root of the caskmaterial threads. The minimum required thread engagement length to prevent cask material failure isdetermined in accordance with Ref. 3.11. Since the constants in the equation assume customary units,


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the metric units used for the cask design will be converted for determination of the required engagementlength. Thus, the minimum engagement length, Le, for the cask is:

L Sub x 2 x Ab

Sul )< 7r X n x D0 min X[ I + 0. .5 77 35x (Dsmin - En~max)

where,Sub Primary Bolt External Thread Tensile Strength

(SA 354 Grade BD)149,389 psi (1,030 MPa) at 20' C (used 150,000 psi)

Sul Cask Internal Thread Tensile Strength(SA 240 TYPE 304L)

= 65,558 psi (452 MPa) at 200 C (used 70,000 psi)Ab Stress Area of Primary Bolt External Threads

= 2.28 in2 (1,470 mm2) (Ref. 3.11)p Bolt Pitch

= 0.197 in (5.0 mm) (Ref. 3.11)n " Number of Threads per Inch

1

pDs,min = Minimum Major Bolt Diameter

1.866 in (47.399 mm) (Ref. 3.12)En.max = Maximum Pitch Diameter of Internal Thread

= 1.705 in (43.297 mm) (Ref. 3.12)Lep = Provided Engagement Length

= 72.0 mm (3.2)

Thus,I

n = -

p1

0.197= 5.08 threads/in

and

Le = x2x A,

S, x7r x n xD.., [rnX1 + 0. 57735 x(D ý,ru -E._)4

150000x 2 x 2.28

70000x7tx5.08x1.866xL5 -1 0.57735x(1.866 1.705)]

25.4mmx

in

=43.5 mm <Lep =72.0Omm


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Therefore, the primary closure lid bolts are acceptable for the normal conditions of transport.


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7.1.3.2 Secondary Lid Closure Bolt Evaluation under Normal Conditions of TransportThe maximum tension, shear and bolt bearing loads in the secondary lid bolts due to the combined NCTloads will be evaluated in accordance with Ref. 3.5, with due consideration given to the prying effects onthe fixed lid. Since the prying forces act inward, normal to the cask lid, an additional prying force isgenerated (Ref. 3.5). For the NCT condition, the controlling load case is the summation of the boltpreload, the internal pressure load and the thermal expansion load. The maximum bolt tension load (Ft),shear load (Fs) and torsional moment (Mt) for the primary lid bolts are (see Table 7.1.3-2):

Ft = F + F.ý + F,,

F, F=Mt = M p,+NMal +MVs,

where,Fp = Maximum Bolt Preload

= 72.2 kN/bolt (Table 7.1.3-2)Fas = Bolt Internal Pressure Load

= 7.9 kN/bolt (Table 7.1.3-2)Fats= Bolt Temperature Load

= 388.6 kN/bolt (Table 7.1.3-2)Fss = Bolt Internal Pressure Shear Load

= 0.0 kN/bolt (Table 7.1.3-2)Fass = Bolt Temperature Shear Load

0.0 kN/bolt (Table 7.1.3-2)Mpt = Maximum Bolt Preload Torsional Moment

= 0.2 kN-m/bolt (Table 7.1.3-2)Mat = Bolt Internal Pressure Torsional Moment

= 0.0 kN-m/bolt (Table 7.1.3-2)M = Bolt Temperature Torsional Moment

= 0.0 kN-m/bolt (Table 7.1.3-2)Thus,

Ft = Fý + F~p+ F'=,

= 72.2 + 7.9 + 388.6

= 468.7 kN/bolt

Fs = F p + F.,

= 0.0+0.0= 0.0 kN/bolt

Mt = Mp, + M., + Ml,

= 0.2+0.0+0.0

= 0.2 kN-m/bolt

Conservatively, the fixed-edge closure lid prying is taken from the external pressure load case. Thisaccident load case bounds all normal conditions and provides a conservative result. The additionaltensile bolt force per bolt, Ftp, caused by the prying action of the secondary lid is (Ref. 3.5 Table 2.1):


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FP a!x D~bj XDli -- Dlb

C1 + C2

where,Ff = Fixed Edge Closure Force

= -81.0 kN/mMf = Fixed Edge Closure Moment

-9.4 kN-m/m

(Table 7.1.3-2)

(Table 7.1.3-2)


Thus,

F-p (r x 0.926) x18

2 x (-9.4) - I x (446.7 - (-81.0))- 1.79 x (446.7 - 446.70.754 - 0.926

1 + 1.79

= -24.5 kN/bolt

Since this bolt load is less than the load generated by the minimum bolt preload (69.5 kN > -24.5 kN), the pryingforce generated by the external pressure is not critical with respect to bolt stress and will not result in the loss oflid closure seal.


Fa = F, + Fv

= =468.7 + (-24.5)

= 444.1 kN/bolt


Mbb Kb X blx MfN× Nb Kb+K1


Mbb = r 2T)x(l Kb )~N Nb Kb+K1,

=(rxO.926)' 2.40 x(94- r.x (-9.4)18 )J 2.40 +71.2)


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= -0.05 kN-m

The average bolt stresses from the combined NCT loads are determined in accordance with Table 5.1 ofRef. 3.5. The bolt stress diameter, D, is:

D Db - 0.9382 x p

where,p = Bolt Pitch

= 4.0 mm (Ref. 3.11)Thus,

D = Db - 0.9382 x p

= 0.036 - 0.9382x 0.0040= 0.032 m

The average tensile stress, Sba, average shear stress, Sbs, maximum bending stress, Sbb, andmaximum torsional stress, Sbt, are:

1.2732 x FaSba 2

1.2732 x 444.1 IMPax0.032?2 1000 kv// ,

/m-543.8 MPa

Sbs 1.2732 x FsD 2

1.2732 xO.0 1 MPa

0.032 2 1000 kN/'/m 2

0.0 MPa

Sbb - 10.I186OxMbbSEbb

10.186 x (-0.05) x IMPa

0.036' 1000 kN///M -

-10.8 MPa

Sbt = 5.093 xMSbt

5.093 x 0.2 1 MPa

0.0363 1000 kN/2 m2

21.3 MPa


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The allowable stresses for the bolts are as previously defined. The maximum interaction ratio for thecombined shear and tension loads is therefore:

1.R. = ISba +lSbs

721 \ (3.)

= 0.569

The minimum factor of safety is:1

FS =I.R

= 1.8 > 1.0


1. R. = Sbb )

Y ba )t , 1 0 .8 ) 2

= I-0-](514.5)

= 0.00044


FS =I.R.

1

0.00044

= 2279.9 > 1.0

The maximum stress intensity in the primary lid bolts under the combined loads, Sbi, is:

Sbi = J(Sba + Sbb)2 + 4 x (Sbs + Sbt)2

= = V(543.8 + (-10.8))2 + 4x(0+21.3) 2

= 534.7 MPa


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The primary lid closure bolts utilize a custom washer for the bolts with an outer diameter, dow, of 90 mmand a hole diameter, dOh, of 40 mm. Therefore, the bearing stress under the bolt head, Sbrg, is:

Sbrg =F,

A gWhere,

Abrg Bolt Bearing Area

= 1 2~~ 092 ~.4

- x(dow-doll)

4

= x.00 -0.044= 0.0051 M2

Thus,444.1 IMPaSbrg = .0.0051 1000k/

87.0 MPa


1. R. Sbrg

S)l


Thus,

1. R. - SbrgS•l

87.0

114.4

0.76


FS1

I.R

0.76


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= 1.31 > 1.0

Because the cask material is weaker than the bolting material, failure will occur at the root of the caskmaterial threads. The minimum required thread engagement length to prevent cask material failure isdetermined in accordance with Ref. 3.11. Since the constants in the equation assume customary units,the metric units used for the cask design will be converted for determination of the required engagementlength. Thus, the minimum engagement length, Le, for the cask is:

Le S.b x 2 xAb

S. X7 Xn ,.n +l 0. 5773 5x (D,,,i,- Eý,'5u1 7

tfI~~ mi L2xn

where,Ab = Stress Area of Primary Bolt External Threads

= 1.27 in2 (817 mm 2) (Ref. 3.11)p Bolt Pitch

= 0.157 in (4.0 mm) (Ref. 3.11)n = Number of Threads per Inch

P

Ds~min = Minimum Major Bolt Diameter1.396 in (35.465 mm) (Ref. 3.12)

Enmax = Maximum Pitch Diameter of Internal Thread= 1.270 in (32.270 mm) (Ref. 3.12)

Lep = Provided Engagement Length= 54.0 mm (3.2)

All other terms are as previously defined. Thus,

n =

p

0.157= 6.35 threads/in

and

Suhx2xAhLe S•x~xnxDum1 "X ±•O'57735x(Dsmin-Eni''a\)1'L×

150000 x 2 x 1.27

= 70000 xinx6.35 x 1.396 x 2x6.35 + 0.57735 x (1.396-1.270)

25.4 mmx

in= 32.7 mm < L•p = 54.0 mm

Therefore, the secondary closure lid bolts are acceptable for the normal conditions of transport.


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7. 1.3.3 Primary Lid Closure Bolt Evaluation under Hypothetical Accident Conditions

The maximum tension, shear and bolt bearing loads in the primary lid bolts due to the combined HACloads will be evaluated in accordance with Ref. 3.5, with due consideration given to the prying effects onthe fixed lid. Since the prying forces act inward, normal to the cask lid, an additional prying force isgenerated (Ref. 3.5). For the HAC condition, the controlling load case is the summation of the boltpreload, the internal pressure load and the end drop load. Since the internal pressure load will actcounter to the drop load, the internal pressure load will be considered as negative for determination of themaximum bolt tension load. The maximum bolt tension load (Ft), shear load (F,) and torsional moment(MI) for the primary lid bolts are (see Table 7.1.3-2):

F, ~F., + F~d

F, ±Fd

lV P, .M tM dt

where,Fp

Fap

Fad =

Fsp=

Fsd =

Mpt =

Mat =

Mdt =

Maximum Bolt Preload130.6 kN/bolt (Table 7.1.3-1)Bolt Internal Pressure Load20.7 kN/bolt (Table 7.1.3-1)Bolt End Drop Load628.9 kN/bolt (Table 7.1.3-1)Bolt Internal Pressure Shear Load0.0 kN/ bolt (Table 7.1.3-1)Bolt End Drop Shear Load0.0 kN/bolt (Table 7.1.3-1)Maximum Bolt Preload Torsional Moment0.5 kN-m/bolt (Table 7.1.3-1)Bolt Internal Pressure Torsional Moment0.0 kN-m/bolt (Table 7.1.3-1)Bolt End Drop Torsional Moment0.0 kN-m/bolt (Table 7.1.3-1)

Thus,Ft = - F.+F.

130.6-20.7+628.9= 738.8 kN/bolt

Fs, = F + F',

= 0.0+0.0= 0.0 kN/bolt

Mt = M P1 + M., + M di

= 0.5+0.0+0.0= 0.5 kN-m/bolt

Conservatively, the fixed-edge closure lid prying is taken from the end drop load case. The additionaltensile bolt force per bolt, Ft,, caused by the prying action of the primary lid is (Ref. 3.5 Table 2.1):


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CI+C2

where,Ff = Fixed Edge Closure Force

= 3336.4 kN/mM= Fixed Edge Closure Moment

= 800.7 kN-m/m

(Table 7.1.3-1)

(Table 7.1.3-1)


Thus,

F,P R X1.920

2.016 .7 -x (3336.4- 3336.4) -3.47< (3336.4-692.9)1x2.016 -1.920

1+ 3.47

316.2 kN/bolt

This bolt load is greater than the load generated by the minimum bolt preload. However, the drop load is aninward load, which will press the closure lid against the sealing gasket. The prying force generated by the dropload will therefore not result in the loss of lid closure seal. The outward load of the internal pressure has alreadybeen evaluated in Section 7.1.3.1 and found to be acceptable. All other accident loads are acceptable bycomparison.

The total tension force, F,, is:

Fa = , +F,,

= 738.8+316.2= 1055.1 kN/bolt


b lxDb x( Kb )xMfMbb . Nb Kb+K1)


Mbb =(-TxD, )br Kb 'xMfN b Kb+•+K1 )

n ×x1.920 4.17 x800732 4.172.6-34.7)

=2.6 kN-m


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The average bolt stresses from the combined HAC loads are determined in accordance with Table 5.1 ofRef. 3.5. The bolt stress diameter is as defined previously. The average tensile stress, Sba, averageshear stress, Sbs, maximum bending stress, Sbb, and maximum torsional stress, Sbt, are:

1.2732 x FaSba = ____

1.2732x 1055.1 1 MPa

0.0432 1000 kN 2'

716.2 MPa

1.2732xFsSbs D2

1.2732 xO.0 1MPa0.043 2 1000 kN/// Ym2

0.0 MPa

Sbb 10.186xM bbD'bb

10.186 x 2.63 1MPa0.048 x 1000 2

242.5 MPa

Sbt 5.093 x M,Sbt

5.093 x 0.5 1 MPa-X

0.048 1000 kN/

21.6 MPa


1. R. = b Sbs )

(t, ,. 0s, a<

716.2- ( 0.0

(721) +,432.6)

= 0.987


FSI.R.

0.987


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= 1.01 > 1.0


I.R . = (S b b j2

2425 )2

= 0.222


FS1

I.R.1

0.22= 4.50 > 1.0

The primary lid closure bolts utilize a custom washer for the bolts with an outer diameter, d,,, of 130 mmand a hole diameter, doh, of 52 mm. Therefore, the bearing stress under the bolt head, Sbrg, is:

Sbrg =Fa

Abrg


Thus,1055.1 1 MPab 0.0111 1000 kN/m2

94.6 MPa


1.R. = SbrgSYl


I.R. SbrgSYI

94.6

114.4= 0.83



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1FS =

I.R.

0.83= 1.21 > 1.0


S2xnflAll terms are as previously defined.

Thus,

Le = S, x 2 xAbLeF 1

S., X7rXnxD s~mn Xl 1+0. 57735 x (Dmn-En.

150000 x 2 x 2.28

70000xnx5.08x1.866X 2x ±+0.57735x(1.866-1.705)]

25.4 mmx

in43.5 mm < Lep = 72.0 mm

Therefore, the primary closure lid bolts are acceptable for the hypothetical accident conditions.

7.1.3.4 Secondary Lid Closure Bolt Evaluation under Hypothetical Accident ConditionsThe maximum tension, shear and bolt bearing loads in the secondary lid bolts due to the combined HACloads will be evaluated in accordance with Ref. 3.5, with due consideration given to the prying effects onthe fixed lid. Since the prying forces act inward, normal to the cask lid, an additional prying force isgenerated (Ref. 3.5). For the HAC condition, the controlling load case is the summation of the boltpreload, the internal pressure load and the end drop load. Since the internal pressure load will actcounter to the drop load, the internal pressure load will be considered as negative for determination of themaximum bolt tension load. The maximum bolt tension load (Ft), shear load (F,) and torsional moment(Mt) for the primary lid bolts are (see Table 7.1.3-2):

Ft = Fp - F s + F.d

Fs= ss + FsdM = M P, + M , + MId,

where,Fp = Maximum Bolt Preload


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Fas

Fad =

F.sd -

Fsd

Mpt=

Mat =

Mdi =

72.2 kN/boltBolt Internal Pressure Load7.9 kN/boltBolt End Drop Load266.0 kN/boltBolt Internal Pressure Shear Load0.0 kN/ boltBolt End Drop Shear Load0.0 kN/bolt

(Table 7.1.3-2)

(Table 7.1.3-2)

(Table 7.1.3-2)

(Table 7.1.3-2)

(Table 7.1.3-2)Maximum Bolt Preload Torsional Moment0.2 kN-m/bolt (Table 7.1.3-2)Bolt Internal Pressure Torsional Moment0.0 kN-m/bolt (Table 7.1.3-2)Bolt End Drop Torsional Moment0.0 kN-m/bolt (Table 7.1.3-2)

Thus,= F..F ++Fx

= 72.2 - 7.9 + 266.0= 330.4 kN/bolt

Fs = F' + F,

= 0.0+0.0= 0.0 kN/bolt

Mt = Mp, + Mat + Md,

= 0.2+0.0+0.0= 0.2 kN-m/bolt

Conservatively, the fixed-edge closure lid prying is taken from the end drop load case. The additionaltensile bolt force per bolt, Ftp, caused by the prying action of the primary lid is (Ref. 3.5 Table 2.1):

2 xMf

ltv x D, X1D1 i - C IX(B- F,) -C2 x(B -P)jNb C1+C2

where,F= Fixed Edge Closure Force

= 1646.1 kN/mM= Fixed Edge Closure Moment

= 190.5 kN-m/m

(Table 7.1.3-2)

(Table 7.1.3-2)


F x 0_.926

18 )

2x190.5 -1 x(1646.1-1646.1)-1.79x(1646.1-446.7'r 1.000-0.926× 1+1-.-.9

= 173.5 kN/bolt


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This bolt load is greater than the load generated by the minimum bolt preload (130.6 kN < 178.0 kN). However,the drop load is an inward load, which will press the closure lid against the sealing gasket. The prying forcegenerated by the drop load will therefore not result in the loss of lid closure seal. The outward load of the internalpressure has already been evaluated in Section 7.1.3.2 and found to be acceptable. All other accident loads areacceptable by comparison.


Fa = F +Fv

= 330.4 + 173.5503.8 kN/bolt


Mb (7XDb KbMbb Nb Kb+KI)


Mbb =ix---b J K--KIb XMf

(Trx0.926) ( 2.40 x-I lx x190.518 2.40+71.2)

= 1.0 kN-m

The average bolt stresses from the combined HAC loads are determined in accordance with Table 5.1 ofRef. 3.5. The bolt stress diameter is as defined previously. The average tensile stress, Sba, averageshear stress, Sbs, maximum bending stress, Sbb, and maximum torsional stress, Sbt, are:

Sba 1.2732 x Fa[)2

= 616.8 MPa

Sbs = 1.2732 x FsD 2

0.0 MPa

Sbb - 10.186xMbbD 3

= 218.9 MPa

Sbt 5.093 x M,D 3= 21.3 MPa



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1.R. = (Sba) 2 +(Sbs j2

( ",. CF-,:

= 0.732

The minimum factor of safety is:1

FS =I.R.

1

0.732= 1.37 > 1.0


I.R. = Sbb

= 0.113


1FS =

I.R.= 8.86 > 1.0

The primary lid closure bolts utilize a custom washer for the bolts with an outer diameter, do,, of 90 mmand a hole diameter, doh, of 40 mm. Therefore, the bearing stress under the bolt head, Sbrg, is:

FSbrg =

Abr•


Sbrg = 98.7 MPa


1. R. - SbrgS•1


Thus,

1. R. - SbrgSIA

98.7

114.4


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= 0.86


1FS

1I.R.

1

0.86= 1.16>1.0


Sub x2xAbLe


Le =

S~xtxnD~,, K2xn 073(

Su x2xA b

S. T tn xD~,,_. x[ + 0.57735 x(D

150000x 2x 1.27

70000x t x 6.35x 1.396xL 1 + 0.57735x (1.396-1.270)]12 x6.35

25.4 mmx in

=32.7 mm < Lep=54.0 mm

Therefore, the secondary closure lid bolts are acceptable for the hypothetical accident conditions.


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7.1.4 Seal IntegrityThe maximum stress analyses in the previous sections are based on criteria for the accident conditionsintended to prevent failures by excessive plastic deformation or by the rupture of the bolt. Using the yieldstress as the stress limit for average tensile bolt stress implies that a small amount (0.02%) of plasticdeformation is permitted (Ref. 3.5). The following calculations show that the o-rings will continue toprovide positive sealing of the closure lids even with this small plastic deformation.

7.1.4.1 Primary Lid SealsThe 0.02% bolt plastic deformation permitted in Ref. 3.5 is distributed over the 67 mm bolt shankdimension shown in Detail 1 of Ref. 3.2. This may result in a separation between the primary lid and caskflange mating surfaces of 0.0134 mm (= 67 mm x 0.0002). However, the primary lid seals are 12 +/-0.3mm diameter EPDM rubber and the grooves for these seals are 9.4 +/- 0.15 mm deep (see Ref 3.2).Thus, the seal is minimally compressed 2.15 mm (= (12 - 0.3) - (9.4 + 0.15)). Considering that EPDM 0-rings have a compression set of up to 45% (Appendix 1, Figure 2) at 150 'C, the minimum compressionin the seal is 1.18 mm (= 2.15 - 0.45*2.15). Since the minimum seal compression greatly exceeds theseparation due to possible plastic deformation, the primary lid/cask flange containment boundary willremain sealed following an HAC drop event.

7.1.4.2 Secondary Lid Seals

The 0.02% bolt plastic deformation permitted in Ref. 3.5 is distributed over the 43 mm bolt shankdimension shown in Detail 2 of Ref. 3.2. This may result in a separation between the secondary andprimary lid mating surfaces of 0.0086 mm (= 43 mm x 0.0002). However, the secondary seals are 12 +/-0.3 mm diameter EPDM rubber and the grooves for these seals are 9.4 +/- 0.15 mm deep (see Ref 3.2).Thus, the seal is minimally compressed 2.15 mm (= (12 - 0.3) - (9.4 + 0.15)). Considering that EPDM 0-rings have a compression set of up to 45% (Appendix 1, Figure 2) at 150 0 C, the minimum compressionin the seal is 1.18 mm (= 2.15 - 0.45*2.15). Since the minimum seal compression greatly exceeds theseparation due to possible plastic deformation, the primary to secondary lid containment boundary willremain sealed following an HAC drop event.


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7.2 Vibration10 CFR 71.71 (c)(5) requires that "vibration normally incident to transport" be evaluated. The RT-100package consists of thick section materials that will be unaffected by vibration normally incident totransport, such as over the road vibrations.

7.2.1 Vibration Evaluation of the RT-100 Cask Primary Lid BoltsRT-100 package may be subjected to a cycle range typically associated with high-cycle fatigue (> 108

cycles). The endurance limit of the material for the high cycle fatigue can be approximated by using a60% reduction, r,, of the ultimate tensile strength (Ref. 3.19) with an additional 10% reduction, rg, for theconnection surface (Ref. 3.11). Thus the endurance limit for the material is:

Sa = (1-r). ×(1-r) S. b

where,Sub = Bolt Ultimate Stress

= 1030 MPa (ASTM A354 Grade B)

Sa = (1-O.60)x(1-0.1O)x1030= 370.8 MPa

NUREG 0128 gives the following RMS vibration load factors for the road travel:

f = Vertical Vibration Load Factor= 0.52

fL = Longitudinal Vibration Load Factor= 0.27

ft Transverse Vibration Load Factor0.19

The RT-100 package is transported in the vertical orientation. The cask lid will be subjected to vibrationin the vertical direction. A notch factor, fN, of 3.0 will be used, conservatively (Ref. 3.19). The vibrationstress in the bolts is:

=v Fhxf,Ab

where,Fb = Bolt Force due to Vibration

Nb

Ab = Bolt Stress Area1470 mm 2 (Ref. 3.11)

WLp = Cask Lid Weight= 3648 kg, use 3650 kg (Ref. 3.2)



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0.52 x 3650 x 9.81 1 kNFb = X

32 1000 N= 0.58 kN

0.58 x 3.0 1 MPaSv

0.001470 10 2

= 1.19 MPa << Sa = 370.8 MPa

Since the stress in the bolts is well below the endurance limit of the material, the primary lid bolts will notbe subjected to transportation-related fatigue damage during their service life.

The maximum shock loading coefficient for the three orthogonal directions is specified as 2.9 (Ref. 3.13).The cask primary lid will be subjected to shock loading during transport. The primary lid closure boltshave been shown to withstand a 125 g impact load, which is much larger than the 2.9W shock loadingduring transport. The primary lid closure bolts are therefore acceptable for shock loading by comparison.

7.2.2 Vibration Evaluation of the RT-100 Cask Secondary Lid BoltsPer Section 7.2.1, the components of the package are into high-cycle fatigue range (> 108 cycles). Theendurance limit of the material for the high cycle fatigue for the secondary lid bolts is the same as for theprimary lid bolts. The cask lid will be subjected to vibration in the vertical direction. A notch factor, fN, of3.0 will be used, conservatively (Ref. 3.19). The vibration stress in the bolts is:

Sv = Fb, f.Ab

where,Fb Bolt Force due to Vibration

fx WLpX g

Nb

Ab = Bolt Stress Area= 817 mm 2 (Ref. 3.11)

WLS = Cask Lid Weight= 857 kg (Ref. 3.2)


All other quantities are as defined in Section 7.2.1.

0.52x857x9.81 1 kNFb = 1 -818 1000 N

= 0.24 kN0.24 x 3.0 1 MPa0.000817 1000 kN,/ 2

m0.89 MPa << Sa = 370.8 MPa

Since the stress in the bolts is well below the endurance limit of the material, the secondary lid bolts willnot be subjected to transportation-related fatigue damage during their service life.


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The maximum shock loading coefficient for the three orthogonal directions is specified as 2.9 (Ref. 3.13).The cask primary lid will be subjected to shock loading during transport. The secondary lid closure boltshave been show to withstand a 125g impact load (Section 7.1.2.4.2.2), which is much larger than the2.9W shock loading during transport. The secondary lid closure bolts are therefore acceptable for shockloading by comparison.

The RT-100 satisfies the requirements for normal vibration incident to transport as required by 10 CFR71.71 (c)(5) [Ref. 3.1].

II


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7.3 Port Cover Seal and Bolt Evaluation

The RT-100 cask port cover utilizes a double polymer (EPDM) 0-ring configuration face seal to protectthe leak test port. For this evaluation the diameter of the outer O-ring is considered to maximize theseating force. The port cover is seal with six DIN912 M10 x 30-A4-70 bolts.

7.3.1 Port Cover Seal

7.3.1.1 0-ring Sealing ForceThe 0-ring requires a minimum 3.7 N/mm sealing force (Appendix 8.1, Trelleborg). The force required toseat the polymer 0-ring seal is (Described in Appendix 8.1, "Compression Forces"):

= ,xC -3.7x(Tx136.6)-1,587.8 NWhere,

Yf = Sealing force = 3.7 N/mmC 0-ring circumference = 136.6 mm

7.3.1.2 Port Cover PreloadThe preload force available to maintain a tight seal that accounts for reduction in preload during HAC is:

PL = F - Pc = 45,662 N

Where,Fc = Available closure force

Pmin x Nb= 48,600 N

Pmin = Minimum preload per boltTmin / k / d

= 8,100 NTmin = Minimum torque (-10%) [Ref. 3.2(b)]

= 24,300 N-mmk = Nut factor - non-lubricated condition (Ref 3.9, Table G)

= 0.3d = Nominal bolt diameter

= 10 mmNb = Number of bolts

= 6PHAC = Loss of preload during HAC (Ref. 3.5)

= 0.0002 x E x AT= 2,938 N

E = Modulus of elasticity of 304L= 1.89 x 10 11 Pa @ 100-C

A = Tensile area of the bolt (Ref. 3.22, p. 1416)

0.7854!d 0.97431)2

n)= 77.6386 mm 2

n = Number of threads per inch= 16.93


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7.3.1.3 Factor of Safety to Maintain a Tight SealComparing the available preload force to the load required to maintain a tight seal, the factor of safety is:

FS = P_ _ 45,662

F 1587.8

= 28.8

7.3.2 Bolt Evaluation

7.3.2.1 Thread EngagementFor the port cover, the mating intemal and external threads are manufactured of materials of equal tensilestrengths. To prevent stripping of the external threads, the minimum engagement length, Le, is [Ref. 3.22,p.1415, Equation (1)]:

Le 2x A,

rrxK n~m 1+0.57735 xn x(E,5 - nK.ý)

2.46 mmWhere,

Knmax = Maximum minor diameter of internal thread8.676 mm (Ref. 3.22, p. 1694)

Esmin = Minimum pitch diameter of internal thread8.862 mm (Ref. 3.22, p. 1696)

The available thread length based on the drawings is 15.5 mm (Ref. 3.2b). Since 15.5 mm > 2.46 mm,there are sufficient threads to prevent stripping of the bolts.

7.3.2.2 Thread Shear Evaluation

The load necessary to shear the external threads due to the tensile force is:

Ps = 0.6 x As x Sy85,789 N

From [Ref. 3.22, p 1416, Equation (5)][1

= 979.3 mm2

Le = 15.5 mmSY= 1.46 x 108 Pa, SS 304L @ 100*C

The tensile force generated in the bolt is [Ref. 3.22, p 1412, Equation (22)]:= T

PB =dT+ (d+b)V

(2n 2cos A 4 )= 16,026 N

Where,T = Torque (+10%) [Ref. 3.2(b)]

= 29,700 N-mmd2 = Min major diameter [Ref. 3.22, p. 1687 and Table 2]

= d +EI


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d = Bolt basic major diameter= 10 mm

b = Bold head dimension= 16 mm

El = Fundamental deviation (Ref. 3.22, Table 5, p. 1685)= 0.032

a• = Half thread angle= 300

p = Coefficient of friction (Ref. 3.22, p. 1413)= 0.15

Comparing the load required to shear the external threads with the tensile force generated in the bolt, thefactor of safety is:

FS = 85,78916,791

= 5.4

7.3.2.3 Load to Break BoltThe load necessary to break the bolt is:

P = S x At35,069 N

Where,Su,= 4.52 x 108 Pa @ 100'C

Since the load required to break the bolt is greater than the applied force (35,069 N > 16,791 N), the boltswill not fail.


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PAGE NO. 1 of 3

8 APPENDIX

8.1 Elastomer O-Ring Data


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PAGE NO. 2 of 3

Joints toriques

Rydroa~lque. dymique. staticiue

30

2 0:15E6 1 0

c a" t"'1 d... .. ..... " ' "

ci•'am(•tre do tore d&, mm

07

Figure 16 Plage admissible de compression initiale enfonction du dlamere de tore, montage radialstatique ou axial

Efforts de compression

Les efforts de ddformatinn variant avec r'ecrasemen't initialet [a duretW Shore. La figure 17 montre I'effort decompression spictfique en N par cm de la circonfdrencedu joint on fonction du diamrtre de tore.

Les efforts de compression indiquts peuvent itre utilis•spour estimer la force totale 4 appliquer pour le montagestatique des joints torlques.

Compression forces

The defornmation forces vary depending on the extent ofthe initial squeeze and the Shore hardness. Figure 16shows the specific compression force per cm of the sealcircumference as a function of The cross section.

The compression forces shown can be used to estimatethe total force to be applied for static installation ofO-Rings.

Figure 17 Efforts de romprossion sur la circonfdrence dujoint torique salon le matdriau

B.2.3 Allongement- compressionDens le cas d'une 4tanchdit6 radiate, le joint torique mont6dans une gorge interne - "dtanch~it6 extfrieure" - doitkre ddr6 sur tout le diamatre de la gorge. lallongementmaximal au montage est do 6% pour los joints toriques dediam6tre intdricur > 50 mm et de 8% pour los jointstoriques de diam6tre Intdrieur c 50 mm.

Avec des gorges extemoes - "tanch6it6 intidreure" -le lointtorique est comprimO de prdf6rence sur sa p~riphdrie. Lacompression p6rlph~rique maximale au montage est de3%.

Le d~passement de ces veteurs se traduira par uneaugmentation ou une diminution importante du diamktrede tore du joint torique, ce qul risque de r6duire [a dureede vie du joint.

On peut calculer la r~ductlon du dlam~tre de tore (d.) AI'aide de Ia formula suivante

Rackldion_- d10. *(cl,,eAavec dtlinil = diam~tre Intdrieur minimal du Joint tarique

dzmn = diam~tre de tore minimal du joint toriquedjras = diam6tre maximal du logement

mals on peut dire qu'il eat approximativement dgal b lamorid de I'allongement. Un allongement de 11% corres-pond 6 une rdduction d'environ 0,5% du diamftro de toreIdQ.

Dernibre information disponible ur www.tss.trelleborq.cormfcdition Avfil 2008 TRELLEBORG

WA ýMe U1s 39


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PAGE NO. 3 of 3

can~xu~n ~g1031

e-AK-2

vW0 -4

C4"~ Ctanaki

S11 I!1 I

rMa~qwoft"

F~igur Z-1-3: Cacr~weviion St4 vis Pukljww Famifjý

Figure 1. Compression set for EPDM 0-ring (Ref. 3.21)


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PAGE NO. 1 of 2

8.2 Cask Design Input Email


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PAGE NO. 2 of 2

Johnathon McFarland

From: Curt Lindner [[email protected]]Sent: Friday, July 06,2012 4:04 PMTo: John Staples; Johnathon McFarlandSubject: RT-100 - Maximum Payload and Gross Weights

For the purposes of preparing the lifting, tie-down and bolting analyses, consider a maximum payload weight as 15000lbs (6820 kg) and a maximum total weight of 41,000 kg (90,200 Ibs). These values bound the sum of the maximum grossweight in the procurement agreement and the maximum cask weight of 34054 kg (74,920 Ibs) per Robatel drawirg RT-100 PE 1001-1, Rev. C.

Curt Lindner

Lead Engineer, Advanced Analysis

I1ý E N E R C 0 N FEDERAL SERVICES, INC.

clindnerý,enerconcornoffice: 770-497-8818 x237cell: 678-362-7110

RTL-001-CALC-ST-0203, Rev. 6 - 'RT-100 Cask Bolting ... · 3.12 ASME B1.13M-2005, "Metric Screw...

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