ML20080H889
| ML20080H889 | |
| Person / Time | |
|---|---|
| Site: | Hatch  |
| Issue date: | 09/26/1994 |
| From: | Cimorelli S, Rodabaugh J, Sridhar B GENERAL ELECTRIC CO. |
| To: | |
| Shared Package | |
| ML20080H886 | List: |
| References | |
| DRF-B11-00604, DRF-B11-604, GENE-771-39-079, GENE-771-39-0794-R01, GENE-771-39-79, GENE-771-39-794-R1, NUDOCS 9502240066 | |
| Download: ML20080H889 (54) | |
Text
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ATTACHMENT 1 GENE-771-39-0794, REVISION 1 HATCH UNIT 1 SHROUD REPAIR HARDWARE STRESS ANALYSIS t
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The following pages labeled Class II are NOT PROPRIETARY.
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G D E.**l.39 o*94 ca g ff DRFBil unnut Sarouc Mecaanica Repair Program Hatch Unit 1 Shroud Repair Hardware Stress Analysis September,1994
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Prepared by:
Or V. A Av M2(,/94 B. N. Sridhar, Senior Engineer Reactor and Plant Design ngineering Verilled by:
' M4/94 S. S. Cimorelli, Engineer Reactor and Plant Design Engineering Approved by:
~7T)./l.. b h I 9 !24 I,/
J. F. Rodabaujh, ProjectNanager '
Shroud Repair Projects #
Hakh Und 1 Shroud Repair Hardware Stress Analysis Page1
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Revdwoe I Clan 11 DRF 81100604 ABSTMCT This document provides the results of the stress analysis of Hatch Unit I Shroud -
Repair Hardware during seismic, LOCA, and other loading. The objective of the analysis is to demonstrate the structuralintegrity of the shroud and repair hardware under critical loading conditions.
t-The results show that the shroud and repair hardware meet the requirements of the Design Specification 25 A5572, Rev. 2.
l Hatch Umt i Shroud Repair Hardware Stress Analysis Page 2
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. GE.t E.?T139-QT94 Revuwn I Clas !!
DRF 81I.00604 Executive Summary This report provides the results of the design a'id stress analysis of Hatch Unit I Shroud and
- Repair Hardware during seismic and other lor. ding. The shroud repair hardware is to be installed based on assumed cracking of the horizontal shroud welds. This design has the following features:
It consists of a lower spring and an upper spring which support the shroud under seismic, LOCA and other loading. It limits the displacements and stresses of the shroud to optimal values which are within the requirements of the design specification 25A5572, Reference 8.1. The hardware is acceptable for the shroud with and without cracks in the horizontal welds.
. A 3 D linear static Finite Element Analysis (FEA) of the overall shroud and spring assembly (a 180 or one-half model), a 2-D/3 D linear /non-linear static FEA of the lower spring, upper spring, and upper support bracket and hand calculations of the other components (except for the tie rod) have been performed. The FEA was done using FEA software - COSMOS /M,1.70 version, Reference 8.2, which has been validated for this application using test cases. The tie rod mode frequency analysis was done using ANSYS, Reference 8.6. All the FEA results have been independently verified by using handbook f
analysis methodsc As a result of these analyses, the designs of the repair hardware components have been optimized in order to meet stiffness and stress intensity requirements.
A design specification, Reference 8.1 has been developed to establish the load combinations and criteria for allowable stresses for structural analysis. Several different load cases were analyzed.. The most severe case for the shroud is the faulted or 1/2 SME plus steam line LOCA plus dead weight. For the other components, the critical cases vary and-are discussed in the report.
Based on FEA and hand calculation results, it is concluded that all the repair hardware components and the shroud meet the requirements of the design specification.
The forces applied to the ASME Code Section III reactor pressure vessel are analyzed and shown to be acceptable in Reference 8.5.
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Hatch L' rut i Shroud Repair Hardware Stress Analysis Page 3
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Revason i Class 11 DRFBII.00604 (t
y PROPRIETARYINFORMATION NOTICE This document contains proprietary information of General Electric Company and isfurnished to Southern Nuclear Corporation in confidence solelyfor the purpose or purposes stated in the transmittalletter. No other use, direct or indirect of the document or the information it contains is authori:ed. The recipient shall not publish or otherwise disclose it or the information to others without written consent of General Electric, and shall return the document at the request of GeneralElectric.
IMPORTANTNOTICE REGARDING CONTENTS OF THIS REPORT t
The only undertakings of Generel Electric Company respecting information in this document are contained in the contract between Southern Nuclear Corporation (SNC) and General E:ectric Company, and nothing contained in this document shall be construed as changing the contract. The use of this information by anyone other than SNC, orfor anypurpose other than thatfor which it is intended, is not authori:ed; and with respect to any unauthori:ed use, General Electric Company makes no representation or warranty, and assumes no liability as to the completeness, accuracy, or usefulness of the information containedin this document.
Hatch Urut I Shroud Repair Hardwan Stress Analysss Page4
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GE \\ E.??!.39-0 ?94 s-Rrchiun i Class 11 DRF Bil.00604 Table of Contents Section Description Page 1.0 Introduction 8
2.0 Shroud Repair Hardware Design Description 8
3.0 Material Properties 11 4.0 Loading Conditions 12 Upset Operating Basis earthquake (OBE) +
Normal Pressure + Dead Weight -
1 Thermal Normal and Upset Emergency Design Basis earthquake (DBE) + Normal Pressure +
Dead Weight Emergency LOCA + Dead Weight Emergency 1/2 SME + Normal Pressure + Dead Weight, Appendix A-2 1
Faulted DBE + LOCA + Dead Weight Faulted 1/2 SME + htSL LOCA + Dead Weight, Appendix A 2 i
5.0 Analysis Criteria 12 6.0 Overall Shroud Model 12 6.1 Model Description 12 6.2 Boundary Conditions 13 6.3 Applied Loads 13 6.4 FEA Model Results, Shroud 20 7.0 Component FEA Models and Hand Calculations 25 7.1 Lower Spring FEA 25 7.2 Upper Spring FEA 29 7.3 Upper Support Bracket FEA 32 7.4 Tic Rod Analysis 35 7.5 Thermal Stress Analysis (Hand Calculations) 40 7.6 Summary of Hand Calculations of Other Components 43 7.7 Fatigue Evaluation 44 7.8 Gusset Pin Bearing Stress 44 8.0 References 45 Hateh L'mt I Shroud Repair Hardware Stress Analysis Page s
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'8-GE3f.??l.J9 4?94 Rouson i Class ll DRf B11-00604 List of Figures Figure Description Page 1.0 Pictorial View of Hatch Unit 1 Shroud Repair Hardware 9
2 Shroud Weld Joint Designation 10 3
Shroud FEA Model(Element Plot) 14 4
Gap Elements in Shroud Model to Depict Cracks 15 5
Cosine Loading at Core Plate 17 6
Cosine Loading at Top Guide & Shroud Head 18 7
Shroud Stress Intensity Plot (Faulted, DBE + LOCA) 22 8
Lower Spring Element Plot 26 9
Stress Intensity Plot (Lower Spring) 27 10 Element Plot (Upper Spring) 30 11 Stress Intensity Plot (Upper Spring) 31 12 Element Plot (Upper Support Bracket) 33 4 13 Stress Intensity Plot (Upper Support Bracket) 34' 14 Tie Rod FEA Model(ANSYS) 36 15 Tie Rod FIV Model 39 A-1.0 Top Guide Ring Rotations 48 Hateh Umt i Shroud Repair Hardwart Stress Analysa Page 6
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List of Tables Table Description Page 1.0 Tie Rod Mode Frequency List 37 V
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DRF8tl.00604 -
1.0 INTRODUCTION
Cracks have been found during both visual and ultrasonic examination (UT) in the shroud weld joints in several Boiling Water Reactors (BWR's). As a result, for Hatch Unit 1 Shroud, SNC is taking a pre-emptive measure by implementing a corrective action using the design modification developed by GENE without performing inspections. The report deals with an analysis of the GENE design modification.
2.0 SHROUD REPAIR HARDWARE DESIGN DESCRIPTION The repair hardware and shroud are shown in Figure 1.
Four sets of radially acting stabilizers 90 degrees apart are used to maintain the alignment of the core shroud to the reactor pressure vessel (RPV) during seismic loading. The set of stabilizers replace the structural functions of the shroud horizontal welds that are postulated to contain cracks. Each stabilizer assembly consists of a tie rod, an upper spring, a lower g
spring, an upper bracket, and other minor pans. The tie rod provides the vertical load i
1 carrying ability from the upper bracket to the RPV gusset attachment as well as suppon for the springs. The upper spring provides radial load carrying ability from the shroud, at the top guide elevation, to the RPV. - The lower spring provides radialload carrying ability from the shroud, at the core support plate elevation, to the RPV. The upper bracket provides an _
attachment feature to the top of the shroud as well as restraint of the upper shroud weld.' A' middle support is also provided for the tie rod so that its natural frequency wil_1 be higher than '
that of forcing frequency due to flow induced vibrations. Wedges between the core suppon plate and shroud are also provided at each stabilizer location to prevent relative motion of the core plate and the shroud.
There are 9 horizontal welds in the Hatch Unit I shroud, (Figure 2). These welds are called Hi-H8, with HI being the uppermost and H8 being the attachment of the shroud support cylinder to the shroud support plate. Each cylindrical section of the shroud is prevented from unacceptable motion by the stabilizers. The motion of the sections above H1, between H1 and H2, and between H2 and H3 are restrained by the upper bracket. The upper bracket contacts the shroud and is radially supported by the upper spring which contacts the RP_V.
There is a feature on the upper bracket which prevents unacceptable motion of the section.
between H3 and H4. The section between H4 and H5 is prevented from unacceptable motions by the middle suppon for the tie rod. The lower spring contacts the shroud such that it prevents unacceptable motion of the sections between H5 and H6A, as well as H6A and E
H6B and H7. The section between H7 and H8 is prevented from unacceptable motion by the gussets.
Hatch Umt i Shroud Repair Hudwwe Stress Analyus Page8
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Reu.co l CLus11 DRF BII.00604 3.0 MATERIAL PROPERTIES 3.1 The following material properties for the shroud,304SST,730E854 are obtained from ASME Code, Reference 8.3.
At 550 F 16900 psi S.,
=
- 25. 8 x 10' PSI' E
Modulus of Elasticity
=
=
4 a = 9.46 x 10 in/in/*F Coefficient of Expansion
=
3,2 The tie rod,316SST, ll2D6312, has the following properties which are obtained from References 8.1 (design specification) and 8.3.
At 525*F Ooerating Temocrature) i.t Sm = 22,200 PSI E = 25.8 x 10' in/in 4
a = 9.45 x 10 in/in/ F t
3.3 Lower spring, (Ni Cr Fe - X750),112D6314, and upper spring,112D6315, and upper spring bracket, ll2D6316 have the following properties which are obtained from References 8.1 and 8.3.
At 525 F (OperatingTemperature)
Sm *' 47,500 psi E = 28.4 x 10' psi 4
a = 7.50 x 10 in/in/*F i
1 Page11 Hakh Unit ! Shroud Repair Hardware Sitess.kulyus
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' Clan 11 DRTBIl.00604 4.0. LOADING CONDITIONS The applied loads which the repair hardware and the adjunct shroud must be capable of withstanding are defined in Reference 3.1. These loads are defined for the following conditions:
Condition
' Descriotion Upset Operating basis earthquake (OBE) + normal pressure + dead weight Thermal Normal and Upset Emergency Design basis earthquake (DBE) + normal pressure + dead weight Emergency LOCA + dead weight Emergency 1/2 ShE + Normal Pressure + Dead Weight Faulted DBE + LOCA + dead weight Faulted 1/2 ShE + MSL LOCA + Dead Weight The seismic loads for this analysis were obtained from Reference 8.1. The thermal conditions for the thermal stress analysis are also defined in Reference 8.1.
5.0 ANALYSIS CRITERIA t
The structural analysis criteria are defined in Reference 8.1, Basically, the primaiy and secondary stresses (Pm, Pm + Pb and Pm + Pb + Q) are compared with the allowable and the stress margins are calculated for the various loading conditions 6.0 OVERALL SYSTEM MODEL 6.1 Model Description A detailed 3 D FEA model, Figure 3 of Hatch I shroud was developed for stress analysis purposes to fully evaluate all of the loading conditions specified in Reference 8.1. The model consists of a 180 degree shroud segment composed of 3-D brick, gap,3-D beam, and spring elements. A 180 degree segment was required in order to evaluate non-symmetric seismic loads specified in Reference 8.1 The model was created using the Geostar Module of COSMOS /M FEA software,1.70 version. Dimensions of the shroud were obtained from 730E854 and those of the repair hardware were obtained from 112D6309 and component drawings.
The top, middle, and bottom sections of the shroud are modeled as 3-D brick (solid) elements The core support ring and the top guide ring were also modeled as 3-D brick elements. The
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core plate was modeled as 3 D beam elements connected like the spokes of a wheel.
At any time during a seismic event, only 1 out of a set of 4 springs is active. These 2 springs (upper and lower) are modeled as spring elements and are connected to the brick elements on the sh' oud by means of 3-D Beam elements with low moments ofinertia and high areas. This r
Hatch L' rut i Shroud Repair Hardware Stress Analyss Page12
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. Cracks in the shroud 'are simulated using gap elements which are shown ni Figure 4. The nodes at the ends of the gap elements are physically separated by 0.001 inch..Thus, a crack is simulated 360 degrees on the inner edge and the only connectivity between the two sections of the shroud is at the outer edge. This method of modeling the cracks in the shroud permits relative displacement between the nodes at the ends of the gap elements in the horizontal plane.
It also permits a ' slight rolling of the rings adjacent to the gap elements..This modeling method is also conservative for the shroud stresses, since it allows some moment transfer, unlike a hinged joint in a cracked shroud.
As shown in Figure 4, only two sets of gap elements are used: one set at the core support ring and another set at the top guide ring. This scheme assumes that welds H2 and H6A have cracked 360 degrees through the wall thickness, except at the outer edges. The entire shroud analysis in this report is based on this assumption and is the worst case for the shroud. The reason for this conclusion is that if there are cracks at any other locations, the shroud sections would move radially and would not impose any stresses on the shroud.
The load from the fuel passes to the top guide and core plate and is then transferred through H2 and H6A weld joints. This is the main reason why the springs are located at these two
't elevations only.
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6.2 BOUNDARY CONDITIONS Symmetry boundary conditions were applied to all nodes on the x-y plane, i.e. UZ = RX = RY
0 (UZ = Z displacement, RX, RY are rotations). The anchor points for the spring elements which represent the reactor pressure vessel wall are fixed in all directions, i e. UX = UY =UZ
RX =RY = RZ = 0. The nodes on the bottom surface of the shroud are fixed in the vertical direction only, i e. UY = 0.
6.3 APPLIED LOADS One load case was analyzed in the computer model, i.e. DBE per Reference 8.1. The other load cases are scaled by the ratio of the lower spring forces. Using iterative techniques, the loads were adjusted such that the spring forces from the COSMOS /M model were close to the following spring forces which are obtained from Reference 8.1, for DBE Case:
32,096 lbs.
Upper spring force
=
83,257 lbs.
Lower spring force
=
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9 GAP ELEMENTS TOP GUIDE RING
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Hatch Uma i Shroud Repair Hardware stress Anaipis
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Clan 11 DRF 8110060_4 The other case when the springs are at 45 degrees is not as severe for the shroud and therefore is not analyzed. It is, however, considered in the lower spring analysis.
A cosine distribution was assumed for the seismic loads Fig. 5 & 6 and applied along the core support location. Because of the way the core support ring is connected to the shroud, the load is maximum at 0 degrees and 0 at 90 degrees location. The same loads are applied for the nodes from 90 degrees to 180 degrees location.
Because of the way the top guide ring is connected to the shroud, the load is maximum at 90 degrees location and drops to zero at 0 degree location. This is true for shroud head loads also. For the top guide and shroud head, the loads are applied only for the quadrant from 0 to 90 degrees.
Tie rod tensile load for various cases are calculated as follows:
From Reference 81, the net uplift pressures are:
Core Plate Shroud Head APeo osi APsh osi LOCA 28 30 Normal 23.8 8.4
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The tie rods are in tension for the case of a 360 degree through wall shroud crack FOR DBE + LOCA 2
Area of the core plate, in Acp
=
2 2
(H/4)(166 -137 x (H/4)(10.875 ), 761E462 H
=
2 8917 in
=
z Area of Shroud Head, in Ash
=
2 H/4 (183.5 ),730E854
=
26,446 in
=
Net upward force = F tocr APep Acp + APsh Ash - Wsh
=
Weight of Shroud (730E854) + weight of shroud head (732E109) + weight of Wsh
=
top guide (730E852) + weight of core plate (732E109) 76,500 + 102,200 + 11,300 + 17,500
=
207,500 lbs, neglecting weight of peripheral fuel
=
28 x 8,917 + 30 x 26,446 - 207,300 Ftoc4
=
835,556 lbs
=
or per rod, F oc4 = 835,556/4 t
209,000 lbs
=
Hatch Unst i Shroud Repair Hardware Stress Analpis Page 16
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GE.\\E.??I.3s.0704
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'T 60*
/e9-Pt 30'
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90*
I 180*
180*
TOP GUIDE TOP OF SHROUD HEAD Figurt 6. Cosine Londing at the Top Guide & Shroud Head Py n Ham Una 1 SW Repaar Hudware Stras Analyus l
GCAT.**l.19 0*94 Revssnon i Class ll DRF Bil.00604 From Reference 8.1, for DBE, Tie rod mcment = M = 19.87 E6 in-Ibs Assume that the moment produces Forces F in the tie rods at the mean radius R between the gusset pin and the shroud outside diameter.
Reactor pressure vessel (RPV) 0.D. = 218 inches. Shroud head 0.D. = 189.5 inches.
(l89.5/2 + (218/2-4))/2 = 99.9 inches R
=
M/2R = 19.87E6 x 106/(2x99.9) = 99,450 lbs F
=
Vertical Seismic = i (2/3)(Ground Horizontal Seismic)
(2/3)(0.15) =
.lg
=
0.1 x Wsh = 0.1 x 207,300
=
20,730 lbs
=
Per rod FVDBE = 20,730/4 = 5,200 lbs 99,450 + 5,200 = 104,650
=
For DBE + LOCA, maximum tensile load Ft 209,000 + 99,450 + 5,200
=
313,650 lbs
=
FOR DBE + NORMAL t
F normat = 23.8 x 8,917 + 8.4 x 26,446 - 207,300 227,071 lbs
=
or per rod F normal = 227,071/4 = 56,768 lbs 99,450 + 5,200 + 56,768 = 161,500 lbs Ft
=
FOR OBE + NORMAL 13.97 E6 in Ibs M
=
M/2R = 13.97 x 10'/(2x99.9) = 66,920 lbs f
F
=
FVoss (2/3)x0.08 x 207,300 = 11,056 lbs/4 = 2,764 or 2,800 lbs
=
66,920 + 2,800 + 56,768 = 126,500 lbs FT
=
1/2 SME + NORMAL + W 34.3 E6 in Ibs M
=
34.3 E6/(3x99.9) = 114,500 lbs, The factor 3 is used per Reference 8.9 F
=
F 1/2 SME =
114,500 + 56,768 + 5,200 = 176,500 lbs 1/2 SME + MSL LOCA + W 22.7 E6 in Ibs M
=
22.7 E6/(2x99.9) = 113,614 lbs F
=
113,614 + 209,000 + 5,200 = 327,900 lbs Hat.;h Unit i Shroud Repair Hardware Stress Anal >1ts Page 19
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GENE.??l-300*90 Revewn i Gass II DRF BI100604 VERTICAL LOADS LN SHROUD MODEL DBE + LOCA 1/2 x 313,650 =156,825 lbs At 0 degrees
=
F.ocA = 209,000 lbs 90 degrees
=
t (F.ocA - F - FVDBE)/2 180 degrees
=
t (209.000 - 99,450'- 5200)/2
=
52,175 lbs
=
Each of these loads is distributed over 4 nodes at the top of the shroud flange.
Since the objective of this analysis is only to evaluate the effect of the repair hardware, only the above loads are included in the model. Therefore, the pressure loads need not be considered in evaluating the shroud in this analysis. The vertical pressure force is included in the tic rod loading. The hoop stresses due to internal pressure are included in the SI values.
6.4 FEA MODEL RESULTS, SHROUD 6.4.1 Faulted Case (DBE + LOCA)
The nodal stress intensity (SI) (Pm + Pb) plot from COSMOS /M is shown in Figure 7. The maximum SI = 27,400 psi at 0 degree location on the core support ring. This is a peak value of SI and is calculated for a lower spring force of 92,480 lbs. versus required value of 83,25710sf Ratio = 83,257/92,480 = 0.9 Hoop stress = pr/t = (29x177. 5/2)/1.5 =1657 psi SI Max = 21,828 psi, after linearizing the nodal stresses across the thickness Total SI Max = 0.9 x 21,828 + 1,657 = 21,302 psi 3Sm.
Allowable SI
=
3 x 16,900 = 50,700 psi
=
or, SI Max < SI allow Hence the shroud stresses are acceptable.
The displacement of the shroud in X - direction = UX = 0.9x0.62 = 0.56 inch at the lower spring location This is < 1,49 inches, UX allow, Reference 8.1, Hence O.K.
The < ping forces are :
22,876 lb.
Upper Spring
=
92,480 lb.
Lower Spring
=
Since this case is used only for shroud stresses, these spring forces are within an acceptable range for all practical purposes.
Page 20 Hatch (l rut i Skoud Repaar Hardware Stress Analysis
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GENE ??!.30-n~94 Revawa i Class 11 DRFBIl.00604
- It can be seen that the calculated upper spring force of 22,886 lbs from COSMOS model is less -
- than.32,096 lbs (required) at the top guide ring. This is acceptable since the stresses in the top guide location are much smaller than those at the core plate ring. Thus, the stresses in the core.
- plate ring are governing.
1
- Pm = 9,163 x 0.9 = 8,247 psi for the shroud, after linearizing the stresses.
-Sm = 16,900 psi
.SI allow = 2 x 16,900 = 33,800 psi.
. Pm < 2 Sm, Hence OK
- Also, at the final iteration, the following horizontal shroud loads were used:
In Figure 6, P = 2,642 lbs Thus, at the core plate ring, total load applied for the full model = 2x2x(P+Pcos60* + Pcos 30")
= 2 (2(2,642 + 2,288 + 1,321)) = 25,000 lbs.
In Figure 6, top guide, P1 = 9,547 lbs
- or Load for full model = 2(P1 + Pl Cos 60* + P1 cos 30*)
= 2(9,547 + 8,268 + 4,774)
= 45,178 lbs t
9 In Figure 5, shroud head, P2 = 9,547 lbs or Load at shroud head = 45,178 lbs
= 115,356 lbs This is equal to Sum of spring forces,
= (22,876 + 92,480) lbs Hatch L' rut 1 Shroud Repaar Hardware Stress Analyeu Page21
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DETAIL "A"-
i l
i l
i Figure 7. Shroud Stress Intensity Plot (Faulted, DBE + LOCA)
.i l
Hatch Unit i Shroud Repair Hardware $wu kulpes Pap n 1
le I GE.VE. "I.3 9 4 *94 Revasos I C%s 11 -
+
DRF 811006M 6.4.2 Upset Case (OBE + Normal Pressure)
I Since the stresses at the core plate ring are governing, the stresses can be scaled down for this case.
39,437/92,480 =0.43 Ratio
=
9,163 x 0.43 = 3.940 psi < Sm or 16,900 psi, Hence OK Pm
=
0.43 x 21,828 = 9,386 psi from previous section.
Pm + Pb
=
pr/t = (23.8 x 177.5)/(2x1.5) = 1,408 psi Hoop stress
=
9,386 + 1,408 = 10,794 psi < l.5 Sm or 25,350 psi, Hence OK Total Pm + Pb
=
Pm + Pb + Q = 10,794.+ 9,126 (thermal stress, from Appendix A-1)
SI MAX
=
19,920 psi < 3 Sm or 50,700 psi, Hence OK
=
The same comment as in the previous section is valid for the top guide ring stresses.
< 0.75 UX allow, Reference 8.1 UX = 0.27
=
1.
e r
l
'I l
Hatch Umt i Shroud Repair Hardware Stress AnaI>iis Page 23 i
y GE.VE.??l.19 0794 Revaion I Class 11 DRFB1100604 6.4.3 Emergency (DBE and Normal Pressure)
This Emergency use is a DBE with normal operating pressure differences. The same FEA results for the shroud can be used as for the faulted event of DBE + LOCA SI Max = 21.828 psi, from Section 6.4.1 Maximum Pm + Pb
=
2.25 Sm (Reference 8.1)
SI Allow
=
38,000 psi
=
St Max < SI Allow 8.247 psi and less than 1.5 Sm.
Pm
=
Hence the shroud meets the Emergency requirements.
UX = 0.56 inches, UX allow = 1.49 inches Hence UX < UX allow OK 6.4.4 Emergency (LOCA Only)
This Emergency case is a LOCA without any seismic.
LOCA Load for tie rod = Fooca = 209,000 lbs., from Section 6.3.
This load acts on the two plates of the upper suppon bracket and is then transferred to the f;
shroud flange.
2.00 x 2 x 5.25 from 730E854 and 112D6316 Bearing Area
=
z 21 in
=
Bearing Stress SB =
209,000/21 = 9,952 psi S Allow = SY = 17,800 psi, Reference 8.3 SB < S Allow Hence ok.
Using the method of appendix A-1 for conservatism.
SI Max = (Pm +Pb) = (9,126 x 209,000)/57,571 = 33,130 psi for F = 209,000 x 4=
836,000 lbs.
SI allow = 2.25 Sm = 38,000 psi i
Hence SI Max <SI allow. Hence ok.
Pm is very small and <l.5 Sm.
UX is very small for LOCA and < l.12 inches Hence OK l
Thus, the design is acceptable.
Hatch Umt i Shroud Repair Hardware Stress Analysts Page 24
GE.\\ E **I.39 4 *94 e
e Reawni Class 11 DRFBil.0060t 7.0 COMPONENT FEA MODELS AND HAND CALCULATIONS 7.1 Lower spring,(ll2D6314) FEA The element plot is shown in Figure 7 The lower spring is modeled as shell 4T (thick shell) elements with varying thicknesses Three different load cases were analyzed.
7.1.1 Combined Radial and Axial Loads (faulted case, DBE + LOCA)
As axialload of 313,650 lb is applied at thejunction between the tie rod and the spring. At the same time, a radialload of 83,257 lb. (Ref. S 1)is applied at the point of contact between the shroud and the spring.
For boundary conditions, the node at the block (near the tie rod) is fixed in Y direction, i.e.
UY = 0. Also, for one node at the pin at the lower end, UX = 0 and for another node at the pin', UY = UZ = RX = RY = RZ = 0. The nodal stress intensity (Pm + Pb) plot is shown in Figure 8.
100,843 psi, after linearizing the stresses.
SI Max
=
Sm for spring material =
47,500 psi at 525'F, Reference 8.1 1
Allowable SI =
3 x Sm, faulted conditions 142,500 psi.
=
Hence, SI Max < Si allow.
Pm = 36,553 psi, from COSMOS Model, cfter linearizing the stresses SI allow = 2.0 Sm = 95,000 psi Hence, Pm <2.0 Sm Hence, the lower spring meets the design specification requirements.
The X - displacement at the node where the axial load is applied = UX = 0.3448 inches from COSMOS /M model.
Load = 313,650 lb.
Axial stiffness = 313,650/0.3448 = 909,724 lblin This is used in Section 7.5.1 to calculate the overall axial stiffness of the tie rod assembly.
7.1.2 Radisi Load + Normal Axial Load (Emergency, DBE + Normal)
A radial load of 83,257 lb. and a normal axial load of 161,500 lbs (Section 6.3) are applied at the shroud contact point.
Pm + Pb
= 90,821 psi, after linearizing the stresses from COSMOS model Allowable SI
= 2 25 Sm, emergency case SI
= 2.25 x 47,500 106,875 psi
=
Pass 23 Hatch Umt 15hroud Repait Hardware Stress.4ul>us
I t
- 2
.v G Lvt-m a u m -
Reklon 1 l
Class H DRFB11-H6N i
i I
F i
Y i
t it'i NM M)
!t Y
n s
x Q
)--(p/s.g g ;j ii
)-", IIIT s
F q
y
)_-
a f*
[
h 1
N
~
M s,,
j i 5 a V /~
i s:b r
- t i
.ss
.[
e 4
Figure 8. Lower Spring E'emect l
Hatch Utut i Slutud Rgair !!ardware Stress Analysis Page 26 i
l
-]
GLVE-1714MTN Revkion 1 Class!!
DRFBil.006N i
h Lit STRESS Lc=1 Intens 1.47E+S5 1.32E+S5 1.1GE+S5 1.83E+S5 8.86E+S4 7.4GE+S4 5.94E+S4 4.48E+S4 3.62E+S4 GE+S4 E+83 1
l Figure 9.
Stress Intensity Plot (lower spring).
l l
llatch Unit i Shroud Repair Flardware Str.sa Azul,wis Page 27 l
i
~
~
( g-GD E. * *).3 9-u *9o ReuwnI Chusil DRF RIl-onnot Hence, SI Max < Si allow, thus OK.
Pm = 32,077 psi from COSMOS Model, after stress linearization S allow = 1.5 Sm = 71,250 psi Pm < S allow Thus ok The Y displacement at the load = UY = 0.56 inches.
Or radial stiffness = (83,257/0.56) = 148,526 lb /in.
This stiffness is close to th: required value of 150,000 lbs/in, and hence is acceptable.
UY allow = 1.12 :nches Reference 81 Hence UY < UY allow 7.1.3 Lower spring at 45 degree location In this case, the bottom spring will slide and this case is very similar to Section 7.1.2, thus the spring stiffness would be the same as if the force acts on spring axis. Hence this case is acceptable.
7.1.4. Emergency (LOCA only)
Axial load on spring = 209,000 lbs. Use 211,000 lbs for conservatism.
I Pm = 45,504 psi from COSMOS Model SI allow = 1.5 Sm = 71,250 psi Pm < SI allow Hence, ok Pm + Pb = 77,500 from COSMOS Model Si allow = 2.25 Sm = 106,875 psi Pm + Pb < Si allow Hence OK 7.1.5 Upset (OBE + Normal Pressure) 4 Axial Load
= 126,500 lbs, Section 6 3 Radial Load
= 39,437 lbs, Reference 8.1 Pm + Pb = SI max = 67,500 psi from COSMOS model Si allow = 1.5 Sm = 71,250 psi > SI max. Hence OK.
Pm = 35,500 psi from COSMOS model Sm = 47,500 psi > Pm. Hence OK.
Hence, the design is acceptable for all the load cases.
Hat ;h Unn i Stroud Repaa Hardware Sevis Analysa Page 28
f
]
CE.VE.??/.19 0?9d Revi.non i Class il DRF BIl.00604 7.2 UPPER SPRING FEA The upper spring (one half model)is modeled as SHELL 4T elements. Symmetry boundary conditions were applied on the plane of symmetry. A unit ! cad of 32,210 lb. is applied at the point which represents the reactor pressure vessel wall.
The element plot is shown in Figure 10 and stress intensity plot is shown in Figure !!.
7.2.1 Emergency & Faulted Cases Load = 32,096 lbs, Reference 8.1, for faulted SI Max (Pm + Pb) = (94,982 x 32096)/32,210 psi = 94,700 psi, after linearizing the stresses.
SI allow = 2.25 Sm = 106,875 psi, using Emergency allowables for conservatism SI Max < SI allow Pm = (39,693 x 32,096)/32,210 = 39,500 psi SI allow = 1.5 Sm = 71,250 psi Pm < SI allow, hence, ok Hence, the design is acceptable.
Under the load. UX = 0.636 inches i
~
Radial stiffness = 32,210/0.636 = 50,645/2 = 25,322 lblin.
Stiffness for both halves of spring in series = 50,645/2 = 25,322 lblin.
Radial Shroud stiffness = 136,406 lblin from a separate model of the top section of the shroud.
Stiffness of the shroud and upper spring (25,322 x 136,406) / (25,322 + 136,406)
=
21,357 lblin
=
This is very close to the required stiffness of 20,000 lblin.
UX for top guide =32,210/21,357 = 1.51, for faulted case.
Hence the design meets the requirements of the design specification, Reference 8.1.
7.2.2 Upset Case (OBE + Normal Pressure)
For OBE case, load = 16,113 lbs., reference 8.1 Pm + Pb = SI Max = (16,113/32,210) x 94,982
= 46,600 psi SI allow = 1.5 Sm = 71,250 psi SI < SI allow Hence OK.
Pm = (16,113/32,210) x 39,693 = 19,900 psi SI allow = 1.0 Sm = 47,500 psi Pm < SI allow, hence ok Page 29 Hatch llnd i Shroud Repair Hardware stress Analysis
p I
- '((
,.p e
- s' car m.n.om Reinken 2 CIsse #
DRFBH.ONN t
- -- __-.g c- :::
c.: :::
c: :::
iE: :EE e-- --t.
c::::
i
- ::s c:. :
- 2: Es
- :e c: :
-i:
)[:
c:-:
- 2 c:
c:
' Y
- m. X Figure 10. Element Plot (Upper Spring).
~
Hatdi Unit 1 Sivoud Repair Harduare Stress Analpis Page 30 l
l pg.
,m___
,..gg 7
e
+
GLVE-711-3 M 7N Revulon 1 Class 11 DRFBf100604 Intene 2.34E+6 2.11E+6 I
-- -- 1.97 E +6 1.64E+6 1.4GE+6 1.17E+6 DETAIL "A"-
7.S2E+G DETAIL V i, 4.60E+6 A
2.34E+6 1.12s06 X
3 Figure 11. Stress Intensity Plot (Upper Spring).
itatch Unit i Shroud Repair Hardware Stros Analysis Page 31
GE.\\E ??l.19-0794 Revuwn i Cim Il DRF R1100604 7.3 ' Upper Support Bracket FEA 7.3.1 Faulted (LOCA + DBE)
The element plot is shown in Figure 12. The model consists of shell 4T elements. A downward unit load of 326,500/2 = 163,250 lb. is applied on the bottom plate and is distributed over 3 nodes Because of geometry and loading symmetry, only one half of the upper support bracket is modeled. The calculated load for this load case for this load case = 313,650 lbs, Section 6.3 For boundary conditions, symmetry boundary conditions were applied on the plane of symmetry. For one node at the top where the bracket is reacted upon by the shroud flange, UX = UY = RX = RY = 0. For one node on the vertical leg near the inside diameter of the shroud flange, UX = 0. Also, UX = 0 for one node where the support bracket contacts the shroud on the outside diameter.
The stress intensity (Pm + Pb) plot is shown in Figure 13.
SI Max = (90,832 x 313,650)/326,500 = 87,257 psi, after linearizing the stresses.
Sm
= 47,500 psi SI Allow = 3 Sm(Faulted)
= 142,500 psi SI Max < SI Allow Pm = 35,436 psi SI allow = 2.0 Sm = 95,000 psi or Pm < SI allow, hence ok.
Thus, the design is acceptable.
0.0883 inches under the load
- Also, UY
=
Axial Stiffness = (326,500)/ 0.08829 = 3.692E6 lblin.
7.3.2 Emergency 2 (LOCA) 209,000 lb.
Ft.oca
=
(209,000/326,500) x 90,832 = 58,144 psi St Max
=
2.25 Sm = 106,875 psi SI Allow
=
23,600 psi Pm <l.5 Sm or 71,250 psi Pm
=
Hence, the design is acceptable.
7.3.3 Upset (OBE + Normal Pressure)
Load =
126,500 lbs, Section 6 3 SI
= 35,200 psi:
SI allow = 1.5 Sm = 71,250 psi SI
< SI allow, hence ok Pm = 14,300 psi < l.0 Sm ei 47,500 psi 7.3.4 Emergency 1 (DBE + Normal Pressure)
SI max = 44,900 psi, SI allow = 2.25 Sm = 106,875 psi > SI max Hence O.K.
Pm = 18,300 psi < l.5 Sm or 71,250 psi Hence O.K.
Hatch 'Jmt i Shroud Repair Hardware Stras Arulys s Page 32
fG p
e' GLV&771-3M7N -
a m i-Clan 11 DRFB11-686et p(
-A 2: ;:W
..A
, ::pj C ;::
I t
x 5g- :
-4 Figure 12. Element Plot (Upper Support Bracket)
Hatch Unit 1 Shroud Repair Hardware Stress Analyss Page 33
GENE.7713D 0194 Reision !
Class 11 DRFB1100604 1ns..
1.83E+S'5 1
S.27E+S4 9.24E+S4 7.21E+S4 S.18E+S4 5.15E+S4 4.12E+S4
... 3.SSE+S4 2.86E+S4 W.83E+S4 1
S.SGG12S Figure 13. Stress Intensity Plot (Upper Support Bracket) l 1
o Hatch Unit 1 Shroud Repair Hardware Stress Analyss Page 34 l
I
GD E ~~l-Jo.n*tt Reruwe i Class ll DRF BII.onnat 7.4 Tie Rod Analysis 7.4.1 Tie Rod Hand Calculations Faulted (DBE + LOCA)
The maximum tensile axial load on tie rod = 313,650 lb. from Section 6.3 2
Thread Relief Area = (FI/4) x 3.33 112D6312 2
8.709 in
=
Tensile Stress
= Pm = (313,650)/8.709
~
36,014 psi
=
2 Sm faulted case Si Allow
=
2 x 22.200, from Section 3.0
=
45,600 PSI
=
Pm < SI Allow. Hence, OK.
7.4.2 For Emergency (LOCA only) 209,000 lb., Section 6.3 Axial Load
=
209,000/8.709 Pm
=
24,000 psi
=
1.5 Sm S Allow e
1.5 x 22,200
=
34,200
=
Pm < S Allow. Hence, O.K.
e 7.4.3 For Upset Case (OBE + Normal Pressure) 126,500 lbs., Section 6.3 Axial Load
=
126,500/8.709 Pm
=
14,600 psi
=
Sm SI Allow
=
22,800 psi
=
Pm < Si allow. Hence, O.K.
7.4.4 Emergency (DBE + Normal Pressure)
Axial Load
= 161,420 lbs, Section 6.3 Pm
= 18,600 psi < l.5 Sm or 34,200 psi Hence OK.
7.4.2 Tie Rod FEA and Natural Frequency.
The tie rod FEA was done using FEA software ANSYS. It was modeled as 2-D beam elements (StifD) and the element plot is shown in Figure 14. Included in the model are the upper support bracket and lower spring both modeled as 2.D beam elements. A load of 57,571 lb. (Tie Rod Preload) was applied at the Upper support bracket.
For boundary conditions, for the node at the clesis pin, UX = UY = UZ = ROTX = 0 For the nodes at the tie rod to lower spring junction, mid support and at the top of the upper support bracket, UY = UZ = ROT X = 0 (Rollers)
A mode frequency analysis was performed using ANSYS. The frequencies for the first 6
)
modes are shown in Table 3. The hydrodynamic mass was included in the model.
Hakh Lan i Shroud Repar Hardware $ trees Analp.
Page M
.fi
~
GE \\ E. * ~l.p.,o ao Revuwoe I Cl.us s ll DRF Bil.00snt a
F=575 7 i Las
[
10 UY=UZ=ROTX=0 9
o 8
o 7
o
[
6 UY=UZ=ROTX=0 1
5 o
l 4
o E
'3 UY=UZ=ROTX=0 2
o g1 UX=UY=UZ=ROTX=ROTY=0 7-Y
'Z Figure 14. Tie Rod FEA Model(ANSYS).
~
Hateh Unit I Shroud Repair Hardware Sitess Analpis Page 36
)
~
cr.sr.cel.;e.,c,4 Revum I Class il DRF 8II.00604 Table 1. Tie Rod Mode Frequency List.
Mode Freauency (Cycles / Time) 1 17.15 2
41,81.
3 82.73 4
97.91 5
100.87 6
135.28 The natural frequency = 17.15 Herz. This is compared with the forcing frequency due to flow induced vibrations (FIV).
e e
I t
Hatch Umt i Shroud Repair Hardware Stress Analysis Pay 37
GDEJ*l.)9 0 *94 Revuwe i Class 11 DNF BII 00604 7.4.3 EITect of FIV on Tie Rod Assembly.
The tie rods are 3.5 inches diameter bars that are 172.65 inches long They are threaded on
' both ends, have a tension preload due to difTerential thermal expansion of 57571 lbs. and there is a mid-span support.
The tie rods have the same diameter as the guide rods and have approximately one fourth the unsupported length compared to the guide rods which have not experienced FIV.-
The calculated lowest natural frequency of the tie rod is 17 Hz.
The potential excitation forces would come from the shroud, has a lowest natural frequency well below that of the tie rod and the fluid flow.
The tie rods are installed at 45,135,225, and 315 degree locations. The flow in these regions is primarily parallel to the tie rods At the limiting elevation, the bulk flow velocity is approximately 6.3 feet per second. Even if this was a cross flow velocity, the lowest tie rod frequency is well above the classic excitation f
.22 v/d
=
f-
.22 x 6.3 12/3.5 = 5 Hz
=
The only remaining concern is the cross flow adjacent to thejet pump inlet.. The tie rod will i be approximately 5.5 inches from the nearest approach of thejet pump inlet. Figure 17 predicted the cross flow velocity near the jet pump inlet at the tie rod.
The resulting cross flow is approximately 3 feet per second, which yields a frequency of approximately 2 Hz. Note that this cross flow velocity only exists for approximately 1 foot of length. This frequency is less than 17 hertz, hence OK.
Therefore, the design is acceptable.
Hatch Umt i Shroud Repair Hardware stress Analvus Page 38 i
GUE.W1.j s.4794 RI Chas //
i DRFBl100644 I
=
CROSS 45-FLOW k
~
=
n.
10.5
=
INLET TIE MIXER ROD W
Figure 15. Tie Rod FIV Model.
Hatch Umt i Shroud Repair Hardware Svens Analyus Page 39
i.
l
.,g' g '
GE.%E HI.39 4*94 Revaion I Class 11 DRF 81100604 7.5 ThermalStress Analysis l
7.5.1 Overall Axial Stiffness Of Tie Rod Assembly Axial stiffness of a upper support bracket = kb = 3.692 E5 lb./in, from Section 7.3.
Axial Stiffness of Lower Spring = ks = 9.097 E5 lb./in, from Section 7.1.1.
For Tie Rod 2
(fl/4) x 3.5,112D6312 A
=
2 9.62 in
=
25.8 E6 psi, from Section 3.2 E
=
177.65 - 5.00, i 12D6312 L
=
172.65 inches i
=
Stiffness = kr = (AE)/L = (9 62 x 25.8E6)/172.65
= 1.44E6 lb /in k
= Axial stiffness of Tie Rod assembly (1/k) = (1/ks) + (1/kr) + (1/kb)
Substituting and simplifying
?
(1/k) = 1.0993E-6 + 0.6969E-6 + 2.7086E-7 or k = 483.790 lblin This stiffness is very close to the required value of 500,000 lb./in, Reference 8.4 and hence is acceptable.
7.5.2 Normal Conditions This represents the case when the shroud is at 534*F and the tie rod assembly is at 522*F.
- Since the coefficient of thermal expansion for Inconel X-750, the material for the upper support bracket and the lower spring is less than that of the shroud material (304SS), the shroud grows more than the tie rod assembly. This produces differential thermal expansion and a tensile load on the tie rod assembly.
Unner Summort Bracket Exonnsion:
48.9 - 4.50, i12D6317 and 112D6318 L1
=
44.4 in.
=
7.50 E - 6 in/in/F cx
=
(522 - 70) = 452 *F AT
=
Hence. AL1 =
44.4 x 7.50 E 6 x 452 0.151 inches
=
Hatch Urut i Shroud Repait Hardware Sven Analms Page 40 l
e G D il.* I.10.n'et Revuwn i Cl.Ln il I)RF BII.nnnat Tie Rod Exnansion-172.65 inches, from Section 7.5.1.
L2
=
9.45 E-6 in/in/'F i
=
a 452'F AT
=
AL2 =
172.65 x 9.45 E-6 x 452 -
0.737 inches
=
Lower Sprina Exnansion:
65.50 inches, i 12D6314 L3
=
7.50 E-6 in/in /'F
=
a 464 F AT
=
AL3 =
65 50 x 7.5 E-6 x 452 0 222 inches
=
For Total Tie Rod Assembiv:
LTA = 44.4 + 172.65 + 65.50
= 282.55 inches ALTA
= 0.151 + 0.737 + 0.222
. = 1. I 1 inches 1.
Shroud and Inconel 600 Ennansions:
Shroud Length Ls = 267.44 inches from 730E854 Length ofInconel 600 piece = 282.55 - 267.44 L1= 15.11 inches 7.85 E - 6 in/inf'F, Reference 8.3 a 600
=
9.46E-6 in/in/ F, Section 3.1 a Shroud
=
ATs
= 534 - 70 = 464*F ALs
= 267.44 x 9.46E - 6 x 464 1.1739 inches
=
all = 15.11 x 7.85E - 6 x 464
= 0.055 inches Total ALSA for shroud assembly = 1.229 inches Net differential expansion = 1.229 - 1.11 = 0. I19 inches Stiffness of Tie Rod assembly:
= k = 489,570 lblin., from Section 7.5.1
- c. Force in Tie Rod = 483,790 x 0.119
= 57,571 lb.
2 2
Tie Rod area at the thread relief = (FU4) x 3.33 = 8.709 in 57,571/8.709 = 6,610 psi Tensile Stress in Tie Rod
=
HA:h (*mt i Shroud Repaar Huduwe Strees Analpis Page 41
's e
GDE.??) 1s.079a Resume !
Class 11 DRF BII-00604 57,571/8.709 = 6,610 psi Tensile Stress in Tie Rod
=
6610 psi or, Pm
=
22,200 psi, from Section 3.2 Pm < Sm, Hence, O.K.
Sm
=
7.5.3 Upset Conditions This is the case when the shroud temperature is at 430*F and the tie rod assembly is at 300.*F. Using the method of Section 7.5.2.
Unner Sunnort Bracket Exnnnsion:
48.9 - 4.50, i12D6317 and 112D6318 LI
=
44 4 in.
=
7.20E - 6 in/in. /*F
=
a (300 - 70) = 230 F AT
=
Hence, ALI =
44.4 x 7.20 E-6 x 230 0.0735 inches
=
Tie Rod Exonnsion:
L2
=
172.65 inches, fiom Section 7.5.1 9.23 E-6 in/in/*F a
=
230*F AT
=
AL2 =
172.65 x 9.23 E 6 x 230 0.3665 inches
=
Lower Snrine Exnnnsion-65.50 inches,112D6314 L3
=
7.20 E-6 in/in/*F a
=
467*F AT
=
AL3 =
65.50 X 7.2 E-6 X 230 0.1085 inches
=
For Total Tie Rod Assembiv:
LTA =
44.4 + 172.65 + 65.50 282.55 inches
=
0.0735 + 0.3665 + 0.1085 ALTA
=
0.5485 inches
=
Shrotid and Inconel 600 Exnansions:
267.44 inches from 730E859 Shroud Length Ls
=
282.55 - 267.44 Length ofInconel 600 piece
=
15.11 inches LI
=
7.73 E-6 in/in/*F, Reference 8.3 a 600
=
Page 42 Hatch Umt i Shroud Repair Hardware Stress Analysis l
c o e. i.r:.u os Revuws !
Clus ll DRF B1100604 a Shroud =
9 23 E 6 in/in/*F, Section 3.1 ATs 400 - 70 = 330 F
=
ALS 267.44 x 9 23 E-6 x 330
=
ALS 0.8146 inches
=
- 15. Il x 7.73 E-6 x 330
=
0 0385 inches
=
Total a LSA for shroud Assembly = 0.853 I inches Net differential expansion.
0.8531 - 0.5485 = 0.3046 inches
=
Stiffness Of Tie Rod Assembly k
483,790 lblin.
=
=
/. Force in Tie Rod 483,790 x 0.3046 =
147,362 lb.
=
Tensile Stress in Tie Rod =Pm=
147,362 / 8.709 = 16,921 psi.
Sm = 22,200 psi, Section 3.2 Pm < Sm, Hence, O K.
Hence, the tie rod meets the design specification requirements for both normal and upset conditions.
7.6 Summary of Hand Calculations of Other Component These calculations are filed in DRF B l 1-00604.
7.6.1 Mid Support Bracket,112D6330 t
The load on this bracket = 25,000 lb., Reference 8.1, Emergency Case The maximum bearing stress = 10,246 psi Sy
= 17,800 psi, Reference 8.3, for 316SST S allow
= 1.5 Sy, faulted case
= 1.5 x 17,800 = 29,025 psi Pm + Pb < S allow, Hence, the design is acceptable.
7.6.2 Core Plate Spacer / Wedge,112D6333,112D6334.
Load on Spacer = 83,257 lbs, Reference 8.1 Bearing Stress Sb
=
7,177 psi
=
Sy 17,800 psi
=
S Allow 2.0 Sy = 35,600 psi, Faulted Case
=
Pm + Pb < S allow.
Hence, the design is acceptable.
7.6.3 Shroud Head Modification (Cut Out)
Pm + Pb = SI = 26,439 psi during lifting.
Sallow = 1.5 Sm = 30,000 psi at 70 degrees F Hence, Pm < S allow Hence, the shroud head meets the design specification requirements.
Hateh Unit I shroud Repair Hardware Stress Analysis Page 43
GE.\\f. * *l.19-0 *94 Reuwn!
Clan 11 DRFB1100404 7.7 Fatigue Evaluation For the shroud, since the number of applied load cycles for faulted and other load cases is very small, no formal fatigue analysis is required.
The only critical case is for the tie rod thermal upset case.
From Section 7.5.3, S a = (Pm/2) x k = 17123/2 = 8562 x 4 = 34,248 psi From Figure I 9. I of Reference 8.3, N allow = 2 x 10' cycles.
N actual = 10 cycles, Reference 8.1 or N actual < N allow, Hence, O.K.
For the shroud, the same conclusion holds true. Thus, fatigue requirements are satisfied.
7.8 Gusset Pin Hole Bearing Stress and Pin Shear Stress Maximum Load = Tie Rod Load = P For faulted (DBE + LOCA)
P = 313,650 lb.
2 Bearing area
= 2 x 3 = 6in 1
Bearing Stress = Sb = 52,275 psi S allow
= 2 x 1.5 x Sy = 2 x 1.5 x 28,580 = 85,740 psi, Reference 8.3, alloy 600 Sb < S Allow The other cases are not as severe.
The gusset pin is in double shear.
Shear area
= (2x3.14)/4) x 9 = 14.1372 sq.in.
Shear Stress
= Tau = 313,650/14.1372 = 22,200 psi S allow
= 2 x 0.5Sm = 2 x 0.5x47,500, (X-750, same as lower spring material)
= 47,500 psi Tau < Sallow., Hence, o.k.
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Hatch Unit I shroud Repair Hardware Stress Analyus Page 44
cese.nl.so.0 se Rouwni Clan ll DRTBII.00604 8.0 References 8.1 25A5572, Revision 2, Shroud Repair Hardware Design Specification.
8.2 COSMOS /M, Finite Element Structural Analysis Computer Code, Structural Research and Analysis Corporation, Low Angeles, California 8.3 ASME, Boiler & Pressure Vessel Code (B&PV),Section III, Appendices,1989 Edition.
8.4 GENE-771-48-0894, Revision 1, Seismic Design Report For Hatch Unit 1, Nuclear Power Plant With Shroud Repair, GENE, 8.5 225A5594, Revision 1, Reactor Pressure Vessel Code Stress Report, GENE 8.6 ANSYS, Finite Element Structural Analysis Code, Swanson Analysis Systems, Houston, Texas.
8.8 Formulas for stress and strain, RJ. Roark & W.C. Young, fourth & fifth Editions, McGraw Hill Book Co_, New York, NY 8.9 GENE-771-43-0894 Rev. O, Shroud Repair,1/2 SME Time History Seismic Analysis, GENE.
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Hatch Unit i Shroud Repair Hardware Stress Analpas Page 45
GD E. "I.3 9 0 *9 ReuwnI Class ll DRF 8tl.00604 Appendix A-1
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DRFBII.00604 Shroud Stresses due to difTerential thermal expansion Normal Case Bearing area = 21 sq. in from Section 6.4.4 Tie rod load = 57,571.lbs, Section 7.5 2 Bearing stress = Sbrg = 58,258/21 = 2741 psi S allow = Sy = 17,800 psi, Reference 8 3 l
Sbrg < Sy Hence ok The stress in the top guide ring, Figure A-1.0 with failed H2 & H3 welds is obtained from Reference 8.8, page 334 Sb = (MRc)/I M = 6F/2f1R c = 6 inches R= 94.75 inches,730E854 F = 57,571 x 4 = 230,284 lbs.
If the ring is disconnected by failure of welds H2 & H3, ring rotation will occur and the load 2
will be reduced. From Reference 8.8, theta = (MR )/EI and Sb = (6Fx 6)/(2fl t b) 2 1.
F = 230,284, R = 94.75, t = 3, b = 7.5,. E = 25.8E6 theta = 0.0723 radians = 4.14*
UY = (7.5/2) tan 4.14* = 0.272 inches Thus, the ring will rotate with a stiffness of k= F/(2 UY) or k = 230,284/(2 x 0.272) = 423,726 lbs /in.
S. for shroud = (Sb x k)/(k + k tie rod)
Sb
= (6 x 230,284 x 6)/(2 x fl x 32 x7.5)
= 19,546 psi k tie rod = 489,570 lbslin S = (483,790/(483,790 + 489,570)) x 19,546
= 9126 psi This is used in Section 6.4.2 Thermal Unset Using the method of the previous section, Sbs = 23,088 psi for F = 147,362 x 4 = 589,448 lbs.
or Pm + Pb + Q = 23,088 psi e
S allow = 3Sm = 3 x 16,900= 50,700 psi Pm + Pb + Q < S allow, hence ok.
Page 47 Hatch Unit i Shroud Repair Hardware Stress Analysis
GE.\\ E. **1<3 %e194 Restan i Cims il DRFBil.00604 h
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GENE.") je.o=0.s Rouw:e t Class ll DRFBil.own Appendix A 2 1/2 SME LOADING t..
Hatch Unit i Shroud Repair Harduare Stress Anahsis Page 49 b
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GEAE.??!.39-0794 Revuunei DRT 811 to A-2.1 SHROUD.
1/2 SME + P + W From shroud model used in Section 6 4, the lower spring force = 92,480 lbs. From Reference 8.1, required Lower Spring Force = 78,000 lbs Ratio
= 78,000/92,480 = 0.84 Pm
= 9,163 x 0.84 = 7,730 psi <l.SSm or 25,350 Pm + Pb
= 21,828 x 0.84 + 1,657 = 20,000 psi < 2.25 Sm or 38,000 psi 1/2 SME + MSL LOCA + W Using the method discussed in the previous paragraph, Ratio
= 189/92.4 = 2.04 Pm
= 18,700 psi < 2Sm or 33,800 psi Pm + Pb
= 46,200 psi < 3Sm or 50,700 psi A 2.2 UPPER BRACKET Scaling the stresses by the ratio of axial forces the following results are obtained:
t 1/2 SME + P + W Axial Load = 257,800 lbs Ratio = 257,800/326,500
= 0.79 Pm
= 29,200 psi < l.SSm or 71,250 psi, Hence OK Pm + Pb
= 71,800 psi < 2.25 Sm or 106,875 psi, Hence OK 1/2 SME + MSL LOCA + W Axial Load = 313,650 lbs Pm
= 33,500 psi < 2Sm or 95,000 psi Pm + Pb
= 94,000 psi < 3Sm or 142,500 psi A-2.3 TIE ROD 1/2 SME + P + W Pm
= 114,500/8.709 Pm
= 13,147 psi < l.SSm or 34,200 psi, Hence OK Pm + Pb
= 13,147 psi < 2.25 Sm or 51,300 psi, Hence OK 1/2 SME + MSL LOCA + W Pm
= 327,900/1i.709 Pm
= 37,650 psi < 2Sm or 45,600 psi Pm + Pb
= 37,650 psi < 3Sm or 68,400 psi Hauh Umi i Shroud Repair Hardware Stress Analysis Page 50
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GENE.??I.39 0*9,s Revuwn I Class 11 DRF81100604 A-2.4 UPPER SPRING To analyze the 1/2 SME cases, the design of the deflection limiter (stop) was considered. The two halves of the spring close at a displacement of 2.0 inches. Through linear analysis,it was found that Pm + Pb was greater than the yield strength (Sy) when the spring halves closed.
Thus, a non linear plasticity analysis was performed using COSMOS. The von-mises isotropic hardening option for material non-linearity was chosen for the shell 4T elements.
Also, the bilinear plasticity method was used. For this option, the following material properties were used:
Sy
= 92,300 psi from Reference 8 3 Tangent modulus is obtained as follows:
Su
= 142,300 psi Elongation at yield = 0.2%
Hence, slope of the stress strain curve beyond yield = (142,300 - 92,300)/(0.3 - 0.002)
= 1.6879E5 Both loading and unloading was modeled by specifying the load in dummy time steps as follows:
Time Time Load Step h
0 0
0 1
10 Full value 2
20 0
By trial and error the following results were obtained:
Time Step Time Force Pm + Pb Displacement Ibs psi inches (both halves) 10 1
52,700 111,000 2.54 9
0.9 47,430 109,000 2.04 8
0.8 42,160 108,000 1.74 Hence, it was concluded that when the spring closes at the stop or at 2.00 inch displacement, Pm + Pb = 109,000 psi (peak value) and Load = 47,000 lbs.
The peak stress occurs at the top corner. After linearization, the following results are obtained:
Pm 45,300 psi < l.SSm or 71,250 psi for 1/2 SME + P + W
=
Pm + Pb 106,300 psi < 2.25Sm or 106,875 psi
=
Hatch Unit i Shroud Repair Harduare Streu Anal >m Page$1
m-cese.vil-io o os RenswnI Class ll DRF BIl.04644 Thus for both 1/2 SME cases, Load = 49.700 lbs > 47,000 lbs and hence the spring will bottom out.
1/2 SME + MSL LOCA + W The faulted allowables are higher and Pm and Pm + Pb are the same as above. Also, from COSMOS model, at time step 20, corresponding to zero load (unloading), displacement =
0 22 inches, under the load. This is a permanent displacement and is < l.4 inches allowable for 1/2 SME + P + W & <186, allowable for 1/2 SME -MSL LOCA + W.
Hence, the upper spring meets the requirements for both 1/2 SME cases.
A-2.5 LOWER SPRING The lower spring also has a stop in the Y Section. This is considered in the (1/2 SME + MSL LOCA + W) case.
1/2 SME + NORMAL PRESSURE Using the method of Section 7.1, radialload
= 78,000 lbs I
axialload
= 176,500 Pm
= 30,800 psi after linearizing the stresses
< l.5 Sm or 71,250 psi Hence OK Pm + Pb
= 87,000 psi < 2.25Sm or 106,875 psi, Uy at the load
= 0.51 inch < l.12 Uy allow Hence OK There is no permanent displacement for this load case 1/2 SME + MSL LOCA + W For this case, axial load = 327,900 lbs 189,000 lbs Radialload
=
A linear analysis shows that the lower spring goes plastic under this load, but its deflection is limited by the stop. A plasticity analysis similar to that of the upper spring was performed.
The following results were obtained by trial and error.
Time Step Force Pm + Pb Displacement Ibs psi inches 10 138,000 99,200 1.31 9
124,200 95,300 0.925 8
110,400 88,800 0.772 Page 52 Hat.;h Crut i Shroud Repair Hardware Stress Analysis 1
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- The stop limits the displacement to 0 95 inches. For this displacement, by approximate
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interpolation, -
Pm + Pb = 96,000 psi. < 3Sm or 142,500 psi.
39,600 psi < 2.0 Sm or 95,000 psi Hence OK Pm
=
Also, during unloading when the force is reduced to zero, Uy = 0.4 inches under the load.
This is a permanent displacement < 0 66 Uy allowable Hence OK Thus, the lower spring is l
acceptable for both 1/2 SME load cases.
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