ML20203J385
| ML20203J385 | |
| Person / Time | |
|---|---|
| Site: | Crystal River  |
| Issue date: | 12/09/1997 |
| From: | FLORIDA POWER CORP. |
| To: | |
| Shared Package | |
| ML20203J289 | List:
|
| References | |
| REF-GTECI-A-46, REF-GTECI-SC, TASK-A-46, TASK-OR S97-0542, S97-0542-R00, S97-542, S97-542-R, NUDOCS 9712190198 | |
| Download: ML20203J385 (74) | |
Text
-
l ATTACHMENT D L
FLORIDA POWER CORPORATION CRYSTAL RIVER UNIT 3 DOCKET NUMBER 50-302/ LICENSE NUMBER DPR-72 i
CALCULATION S97-0542, REV. 0 "CR3 Structural Seismic Margin Evaluation (Study)"
J
??A2qg;;gg;;g;go, PDR p
m
' Florida INTEROFFICE CORRESPONDENCE Power a
.-C.M L.RM i
Nuclear Engineering NT6D 1503 v
0"'Ci MAC T elephone
SUBJECT:
Crystal River Unit 3 Ouality Records iransmittal Analysis / Calculation l
TO: Records Managemant NR2A The following analysis / calculation package is submitted as the QA Record copy:
DOCNo D.4 DOCUMENT IDtNf 6FiCAfiON ffJMet41 Rv g ygt t M.bl TOT AL PAGil TRANtM'11tD >
S97 0542 0
TITLt CH3 Structural Selsmlo Margin Evaluation (Study) j i
a WD$ IIDENTIFY 6 4 TwoMDS ROM LAT[4 Aphtf y ALI Seismic, IPEEE, i
DNRtF IntriistNCil C84 FILES + LIST PRIMARf riLI flRafe i
V4ND (VENDOR NAMt)
VINDOR DOCUMENT NVM9tR IDKRIFi stiPtRstCt0 DOCUMtNTS (DARE F)
Duke 2ngineering & Services CDT.1 l SWT+1 l VBTR 10 CHT 1 l SFHE 1 A l
l MUriE 2A l SFHE 1B l
MUHE 2B ll VBTR 1 A l
ll l
COMMtNTS QusAGE RESYNICTIONS. PROPRiti ARY, (TC 1 This calculation is being issued for reference only as an independent study to address plant " extent-of condition.* This calculation is not a design basis calculatiori for any of the above tag numbers.
NOTE:
Use Tog number only for valid tag numbers (i.e., RCV 8, SWV 34, DCH-99), otherwise; use Part number
- field (i.e., CSC14599, AC1459). If more space is required, write See Attachment" and list on separate
- sheet, "FOR RECORDS MANAGEMENT USE ONLY
s Ouality Record Transmittal received and information entered into SEEK.
Entered by:
Date (Return copy of Quality Docment Transmittal to NOE Support Specialist.)
p j o gt/ % DATt vthn AfioN INmN tR Dave sua viss t N.
tt Am r Otse Nmmets c o.,
n 0;te flff/U l&A1
/Jivf 3y ec: Nuclear Projects (if MAR /CGWR!PEERE Calcuistion Review form Part lit actione required Yes @ No Return to Service Related)
Yes @ No (if Yes send copy of the form to Nuclear Regulatory ssurance and a Supervisor Config. Mgt. Info.
copy of the Calculation to the Responsible Organizationis) identified in Mgr., Nucl. Operations Eng. (Originall w/ attach Part til oie the Calculation Rev ow f ormj Rev 9817
(njN CALCULATION REVIEW
_b...___..
cAc =v m Page 1 of 2 I cALCaATION IJO #14V 897 0542. Revision 0 PART I -
DESIGN ASSUMPTION /lNPUT REVIEW: APPLICABLE ] Yes X No The following organtiations have reviewed and concur with the design assumptions and inputs identified for this calculation:
Nuclear Plant Technical Support System Engr w=i,.ro.s.
Nuclear Plant Operations 57.i 4...
e,,5, se...+.i.
s,n.
. +...
PART ll -
RE3ULTS REVIEW: APPLICABLE Yes @ No T.% following organizations have reviewed and concur with the results of this calculation and understand the actions which the organizations must take to imploment the results.
Nuc'ent Plant Technical Su,
.1 System Engr 6*a* * ** '
Nuclear Plant Operations 6,..i
. o.i.
Nuclear Plant Maintenance wn.i
.c.i.
Nuclear Licensed Operator Training s,n.i
.o..
i Manager, Site Nuclear Services 5,n.i..e.i.
Sr. Radiation Protection Engineer son.w..>o.i.
OTHERS:
s.n.i.. o.i.
s,n.t S.i.
% e$?
l _.
,m(d)$
CALCUL'ATION REVIEW Page 2 of 2 CALCutA1IL.4 No /MtV.
897 0542, Revision 0 PART lli CONFIGURATION CONTROL: APPLICABLE Yes y No The following is a list of Plant procedures / lesson plans /other doeurnents and Nuclear Engineering calculations which require updating based on calculation results review:
Document Date Reouired Retoonstbfe Oronniretion Upon completion, forward a copy to the Manager, Nuclear Regulatory Assurance Group for tracking of actions if any item: are identified in Part l11. If calculations are listed, a copy shall be sent to the original file and the calculation log updatid to reflect this Impact.
PART IV - NUCLEAR ENGINEERING DOCUMENTATION REVIEW The responsible Design Engineer must thoroughly review the below listed documents to assess if the calculation requires revalon to these documents if "Yes," the change authorizations must be listed below and issued concurrendy with the calculation.
'V"*
Enhanend Design Basis Document ]Yes@No"C" Vmdor Qualification Package Yes@
No "C*
FSAR Yes x No "'"*
Topical Design Basis Doc.
Yes@
No "C"
improved Tech. Specifiestion
] Yes @ No "*"*
E/SOPM QYesx No Other Documents reviewed:
improved Tech. Spec. Bases
] Yes g t tcwe Config. Mgmt. Inf o, flystem Yes@No y,,
Analysis Basis Document
[ Yes @ No"C" Yes Design Basis Document
] Yes @ No"C*
Yes]
Appenda R Fire Study
] Yes @ No"C*
Yes Firs Harardous Analysis
] Yes @ No"#*
YesQ NFPA Code Conformance Document Q Yes @ No"C" Yes]
-.m.......
PART V PLANT REVIEWSIAPPROVALS FOR INSTRUMENT SETPOINT CHANGE PRC/DNPO approval is required if a setpuint is to be physically changed in the plant through the NEP 213
- process, PRC Review Required
] Yes X No PRC Chantman
/Date DNPO Review Required O Yes
@ No ot hrON
'iNI % C' fl Ot &4uN (Niv,J.(m. 69uNilo NAMt M
Rev. & 97
~
7 (ir:f[ "".'"2')N DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 tw e rou Page i
of i
DOCUU(NT CIWTif tCATCN NQ htVCCN 897 0542 0
SECTION I PURPOSE:
The purpose of this calculation is to place into the CR3 files a copy of the attached evaluation.
This evaluation was done as part of the structural seismic margins assessment effort initiated by FPC management. This effort was part of an " extent of condition" review done of CR3's structural c:mponents and mechanical equipment.
SECTION 11 RESULTS/CONCLUSIONSj Tho attached raview shows all components reviewed have a High Confidence Low Probability Fcilure (HCLPF) value + hat exceeds the CR3 SSE value of 0.1g. Therefore, acceptable.
SECTION lil DESIGN INPUTS:
See pages 62 and 63 of attached review.
SECTION IV ASSUMPTIONS:
Th:re are no assumptions that requiro confirmation.
REG.I,10N V
REFERENCES:
S a pagea 62 and 63 of attached review.
SECTION VI DETAILED CALCULATIONS:
See attached.
SECTIDN Vil ATTACHMENTS:
C:lculations (sixty three pages) prepared by Duke Engineering and Services is follows.
' 8" % %%
METt Lde of ht RtsP: Acesee tagww 4
Oh ANALYSIS / CALCULATION Crystal Riger Unit 3 Page 1 of 63 aws m n j
$17'OSY2. /A< 0 VnOJtC t FPC CRYSTAL RIVER UNIT 3. IPEEE EVALUATION This calculation perfonns a reduced scope seismic evaluation for the IPEEE program. The HCLPF values for the selected components are shown in Section 6.0 of this calculation.
L i
I DESIGN ENGINEER DATE VERJFICATION ENGINEER DATE SUPERVISOR. NUCLEAR ENGINEERING DATE k lllll
/f?
A.VAYsa a.eb fPWtf A. KARArt hk
... ~.... -,
Sg ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet _2_, of13, FPC CRYS'I AL RIVER UNIT 3. IPEEE EVALUATION f77-Offug Table of Contents Page-
- 1. PURPOSE 4
23 ASSUMPTIONS 4
33 METHODOLOGY 4
- 4. ANALYSIS 5
4.1 Condensate Storage Tank (CDT 1) 5 4.1.1 Weight of the Tank's ficad, Wh (kips) 17 4.1.2 Anchor Bolt Reduction Factors 18 4.2 Chilled Water Espansion Tank (CIIT.1) 21 4.2.1 Input Data 21 4.2.2 Anchorage Demand 27 4.2.3 Anchorage Capacity 28 4.3 Evaluation of the RCP Seal Return Cooler MUllE.2A & 2B 31 4.3.1 Input Data 31 4.3.2 Calculation 31 4.4 Nuclear Service Closed Cycle Surge Tank (SWT.1) 36 4.4.1 Design input for SWT l:
36 4.4.2 Anchorage Evaluation of SWT 1:
38 4.4.3 Anchor Bolt Capacity 39 4.4.4 Structural Members Capacity 41 4.5 Spent Fuelllent Exchanger SFilE 1 A & B 42 4.5.1 Design input for SFilE 1 42 4.5.2 Anchorage Evaluation of SFiiE l A & IB:
44 4.5.3 Anchor Bolt Capacity 47 4.5.4 Structural Members Capacity 47 4.6 Regulating Transformer A & il(VBTR.l A & IB 48 4.6.1 input Data 48 4.6.2 Anchorage Capacity 49 4.7 Evaluation of Raceway Supports 51 4.7.1 Sample 1 52 4.7.2 Sample 2 56 MN Nh
@g ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet 3 of.51 FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION J77-Of f'2.
- s. CONCLUSION
'61
- 6. REFERENCES 62 b OM
$E U1
8 Vm ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet 4 of,13.
FPC CRYSTAL RIVER UNIT 3. IPEEE EVALUATION Spy Offy2
- 1. Purpose The purpose of this calculation is to perform a reduced scope seismic evaluation ~ for the IPEEE program. The Safe Shutdown Earthquake (SSE) level is considered as the Review Level Earthquake (RLE). The RLE for CR3 is 0.lg. A High Confidersee Low Probability Failure (HCLPF) value is computed for those cornponents that are selected for the reduced scope.
The following are the components evaluated in this cdculation:
l'.- Condensate Storage tank (CDT 1)
- 2. Chilled Water Expansion Taak (CHT-1) 3.- RCP Seal return Cooler (MUHE 2A & 2B) 4.- Nuclear Service Closed Cycle Surge Tuik (SWT-1) 5.- Spent Fuel Heat Exchanger (SFHE I A & IB)
- 6. Regulating Transforrner A & B (VBTR 1 A & IB) 7.- Raceway Support (Two samples)
- 2. Assumptions There are no assbmptions that require conformation in this calculation.
- 3. Methodology The allowables are based on 1.7 times the AISC nominal allowable as recommended by the GIP-2 Damping values specified by the GIP 2 are used.
Hand calculation is used for all the calculation.
The evaluation of large flat bottom tank is based on EPRI NP 6041 (App. H of Ref.1).
The expansion anchors allowables are based on Appendix 0 of Ref. (1).
m
,, w s
eRl$i ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet 5 of 53.
FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION 777- # 7&
4, Analysis 4.1 Condensate Storaga Tank (CDT-1)
Vertical Flat Bottom Tank (ifCLPF); Appendix 11 of NP 6041: Ref. (1)
CONDENSATE STORAGE TANK Component I.D.:
CDT 1 Florida Power Corp. Crystal River Unit 3 1
Input Tank Mean Radius (ft), R =
16.00 Tank Thickness at base (in), ts =
O.25 Tcnk Bottom Thickness (in), tb =
0.25 Tank fleight (ft),11 =
35.00 Fluid licight (ft),lif =
32.50 See note 2.
Dome lleight (ft), Dh =
1.33 Dome Radius (ft), Dr =
N/A Number of Sections =
4
,See note 1.
, First Section Thickness (in), lieight (ft) 0.25 7
Second Section Thickness (in), lieight (ft) 0.25 7
Third Section Thickness (in),11eight (ft) 0.25 7
Founh Section Thickness (in), lieight (ft) 0.25 14 Note 1: If the number of sections is different than 4, then adjust the thicknesses and heights of the third and founh sections accordingly.
%a im m
i Sg ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet 6 of J 1 FPC CRYSTAL RIVER UNfr 3. IPEEE EVALUATION f ?7 Ofyt Note 2: Fluid lleight, lif = 32.5 ft is based on FPC Doc. No. S 94-0011 (Ref. 3).
flowever, this information requires verification.
Poisson's Ratio, y =
0.3 Modulus of Elasticity for Tank, A285 C (ksi), Es =
29,000 Tank Yield Stress (ksi,, Sye =
30 Density of The Fluid,(Ib/ft^3), Fd =
62.40 Density of The Tank Material (Ib/ft^3), Td =
490 i
Anchor Bolt effective length (in), lib =
47.25 AB length from tank base to top of the chair (in), he =
9 Anchor Bolt Size (in), d =
1.25 No of Anchor Bolt z.
24 Modulus of Elasticity for Bolts, A307 (ksi), Eb =
29,000 Ancho; Bolt Tensile Stress Capacity (ksi), Ult /3 =
20 Horizontal & Vertical PGA @ 5% damp. =
0.20 Av = 0.13 See note 3 Weight of Tank's Head (kips), Wh =
7.90 and C.G. (ft),
Xh*=
35.67 Weight of Tank's Shell(kips). Ws =
35.92 and C.G. (ft),
Xs =
17.50 Weight of Tank's Bottom (kips), Wb =
8.21 and C.G. (ft),
Xb=
0 Weight of Water (kips), Ww =
1,631.01 and C.G. (ft).
Xw=
16.25 Tank weight Head +shell,(kips), Wte =.
41.48 Eq. (H 26) k OE U $.*l
S"g ANALYSIS / CALCULATION CONTINUATION SHEET Crvstol River Unit 3 Sheel 7 of,63.
FPC CRYSTAL RIVER UNIT 3. IPEEE EVALUATION ffJ.pyy2 Conservatively locate C.G. of the roof at 1/2 ofits height, which would increase the moment arm.
The dome weight is conservatively computed =
7.9 kips Note 3: Since the computed impulsive mode frequency (fi = 8.34 Hz with an expected accuracy range of about 7.5 liz to 9.2 liz)is higher than the frequency at the peak of the Ground Response Spectra (f = 1.3 liz, Ground Response Spectra (f = 1.3 Hz, Ref. 4), the PGA is conservatively used in lieu of consideration of soll/ structure interaction effects (Ref. 2).
i HorizontalImpulsive Mode Response Fluid's lleight/ rank's Radius, Hf/R =
2.03125 Tank's average wall thickness,(in) tav =
0.25 Tank's effective uniform wall thickness, (in) te.=
0.25 Horizontal Impulsive Frequency Coefficient Cwi. =
0.103 Table 7.4 cf Ref. (5) of App.
H of Ref. (1)
Horizontal Impulsive Mode Natural Frequency (Hz), fi Equation (H-2) fi=(Cw;/2(3.14)Hf)SQRT(Es/Td)SQRT(0.127Td/Fd) =
8.344335 Horizontalimpulsive Spectral Acceleration,(g) Sai* =
0.20 From Ground R.S.
Conservatively use PGA (see note 3)
For ll/R > 1.5 Use Eqs (H 5) &(H-6)
Impulsive Effective Weight, Wi and its effective Height, Xi =
Eqs. (H 5)& (H 6)
Wi(kips)= Ww(1-0.436(R/Hf)) =
1,280.92 m.
um
9g ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet 8 of.53.
FPC CRYSTAL RIVER U?llT 3. IPEEE EVALUATION f?/.Offy2 Xi (ft)= Hf(0.5-0.I88(R/Hf)) =
13.24 Impulsive mode base shear, VI, and moment, Mi =
Eqs. (H 3) & (H 4)
VI (kips) = Sai(Wh+Ws+Wi) =
264.95 Mi (kips ft) = Sal (WhXh+WsXs+WiXi) =
3.574 Impulsive mode hydrodynamic pressures at depth >0.15Hf, Pi = Eq. (H 8)
Pi (psi) m WiXiSai/(l.36RHf^2) =
1.02 i
llorizontal Convective (Sloshing) Mode Response Horizontal Convective Mode Natural Frequency (Hz), fc Equation (H 10) fc=SQRT('. 5ft/sec^2/R*tanh(1.835(Hf/R))) =
0.306009 Horizontal Convective Spectral Acceleration (g), Sac =
0.16 Ground R.S. @ 0.5% damping Convective Effective Weight, We and its effective Height, Xe =
Eqs. (H 13) & (H 14)
We (kips) = 0.46Ww(R/Hf)tanh(l.835(Hf/R)) =
364.93 Xc (ft) = H(1-(cosh (1,835(H/R)) 1)/1.835(H/R) sinh (l.835(H/R)))
Xc (ft) =
24.19 Convective mode base shear. Ve, and moment, Mc =
Eqs. (H 11) & (H 12)
Vc (kips), SacWe =
59.03 Me (kips ft) = Sac (WcXc) =
1,427.94 MN M NI
~
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal River Unit 3 Sheet 4 of_63.
FPC CRYSTAL RIVER UNIT 3,IPEEE EVALUATION f'l/-OffL Convective mode hydrodynamic pressure, Pc =
Eq. (li 16)
Pc (psi) =(0.267WwSa:)(cosh (l.835(H y)/R))/(RHeosh(1.835(H/R)) = 0 at the base of the tank y is the distance from top of the water to the base.
Fluid slosh height, hs (ft) =0.837R(Sac /g) =
2.14 Eq. (H 17)
Vertical Fluid Mode Response Tank's effective wall thickness / Tank's Radius, t/R =
0.001302 Vertical Fluid Mode Response Co:fnuent Cwv, =
0.091'from (5) of Appendix H Vertical Fluid Mode Natural Frequency (liz), fv Equation (H-19) fv=(Cwv/2(3.14)HOSQRT(Es/Td)SQRT(0.127Td/Fd) =
7.37218 Vertical Fluid Mode Spectral Acceleration (g), Say"=
0.13 Ground R.S.@ 5% damping
" Conservatively use venical PGA Vertical fluid mode hydrodynamic pressure at base, Pv =
Pv (ps!) = 0.8FdHfSavCos(3,14*(Hf y)/2HO =
1.50 Eq.(H 20)
Overturning Moinent and Base Shear
.n
@g ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet _LO.of 61 FPC CRYSTAL RIVER UNIT 3. IPEEE EVALUATION f 7 7-of y4 Combined liorizontal base shear, Vsh, and base moment, Msh =
Vsh (kips) =sqrt( Vi^2+Vc^2)=
271.44 Msh (kips. ft) =sqn( Mi^2+Mc^2)=
3,849.13 Pressure on the Dase of the Tank liydrostatic Pressure (ksi), Pst =
0.01408 = 62.4*lif/144000 llorizontal Seismic Pressure (ksi), Psh =
0.00102
~~
Vertical Seismic pressure (ksi), Psv =
0.00150 Max. Compressive Pressure (ksi). Pe+ =
0.0157 l Pst+Psh+0.4Psv (11 22)
=
\\1in.. Compressive Pressure (ksi), Pc- =
0.01451 Pst+Psh-0.4Psv (H 22)
=
Min. Tension Pressure (ksi), Pt =
0.01246 =
Pst Psh 0.4Psv (H 23)
Min. Average Pressure (ksi), Pa =
0.01348 =
Pst 0.4Psv (11 24)
Pressure Capacity (ksi), Pea = Sye*ts/R =
0.039063 Based on lioop stresses Compare Pressure Capacity, Pca, Ys. Max. Pressure, Pe+
Pea /Pe+ =
2.49 The tank is adequate for hydrostatic and hydrodynamic fluid pressure.
Compressive Buckling Capacity of the Tank Shell El:phant Foot Buckling Stress.(ksi) Spe:
Equation (H 27)
Spe = (0.6Es'ts/R) [I - (Pc+'R/Sye'ts)^2] [1 - 1/(1.12 + S l^1.5)][(S l+(Sye/36ksi))/(S t +1)]
oe
Og ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet _LI.of.fi3.
FPC CRYSTAL RIVER UNIT 3. IPEEE EVALUATION f9/-#f'/1 Where S1 = R/ts/400 Si =
1.92 Spc=
13.17 Dianiond Shape Juckling Stress,(ksi), Spd; Equation (H 30)
Spd =(0.6gm + Dgm)Es/(R/ts)
Where phi =( 1/16)sqrt(R/ts) gm = 1 0.73(l-exp^ phi) and, Dgm is an increase factor for internal pressure obtained from Fig. 2.9 of EPRI Ref.[2]
For (Pc /E) [R/ts]^2 =
0.295063 Input Dgm=
0.16 Spd=
15.08 Allowable Compressive Stress (ksi), Cb =
11.85 Bolt IIold Down Capacity (From Section 4.1.2 of this calculation)
R: duction due to concrete foundation Re =
0.45 Reduction due to top plate, Rip =
1.00 Reduction due to vertical Stiffener, Rvs a 1.00 Reduction due to tank shell stresses, Rshl=
0.57 leduction due to weld capacity to tank shell, Rwid =
1.00 o u,
Og ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet.12. of,$3, FPC CRYSTAL RIVER UNIT 3, IPEFE EVALUATION f 7 7.Off/2 Anchor Bolt Nominal Area (in^2), Ab = PI'd^2/4 =
1.23 Anchor Bolt Nominal Capacity (kips). Tnom.=
41.72
= 1.7 *20* Ab Eq.(H 31)
Anchor Belt Allowable Tensile Capacity (kips) Tbc =
Tbc =Tnom*Rc'(Min.(Rtp,Rvs,Rshl,Rwld)) =
10.70 Fluid liold Down Forces Max. Bolt Elongation (in.), Bem = 1% of lib =
0.4725 e
Allowable Bolt Elongation (in), Be = Tbc* lib /(Eb'Ab) =
0.014 Max. Allowable uplift (in), de0 = Min.(Bem,Be) =
0.014 K (kip in) = Es'ts^3/12(l v^2) =
41.49496 Eq. (1132) k = sqn[(R/ts)sqn(3(1 v^2)))
35.62224 Eq. (ll 32)
Ks = 2Kk/R, (kip-in) =
15.39733 Eq. (1132)
Tank Base (in^3), Ib = tb^3/12(l.v^2) =
0.001431 Eq.(l{ 33)
Mf/P (in^2) = R*ts[1 R/lik}/sqn(12(1 v^2)) =
14.32472 Eq.(H 32)
F (in) = [l+ Ks*IJ2Es*1b]
de (in) = (Pt /Eslb)*[L^4/24 (1/F)(Kst^5/72Esib + Mf*L^2/6P))
Eq. (11-34)
Te (in) = Pt *[U2+(1/F)(Kst^2/12Esib + Mf/PL)]
Eq. (1135)
.w
1 9g ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet.13 of _63.
FPC CRYSTAL RIVER UNIT 3. IPEEE EVALUATION 77 7-07y4 Trial Length for zero uplift Liu(In.)=-
6.4 i L (in) 4.41 5.41 6.41 7.41 8.4 i F(in) 1.818198 2.003731 2.189263 2.3747962.560328 de (in)
-0.00435 0.00333 1.82E 05 0.0065860.017352 For zero uplift, Lzu =
6.41 Fluid lioldown at zero uplift (Kip /in), Tc0 =
0.059872 t
l Trial Length for Uplift of 0.6tb = dee=
0.15 Le (In.) =
12.935 L (in) 10.935 11.935 12.935 13.935 14.935 F (in) 3.028798 3.214331 3.399863 3.585396 3.770928 de (in) 0.070689 0.105483 0.150118 0.206113 0.275086 For 0.6tb uplift, Le =
12.935 Fluid liolddown at 0.6tb Uplift (Kip /in), Tee = 0.103583 Fluid Holddown due to one inch uplift (Kip /in^2), Te1 Te1 = (Tee-Te0)/dec =
0.291409 Total fluid holdown is approximated by: Tee n Te0 + Tel*dee (H 38)
Applicability of the Small Displacement Theoryt
=w-
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal River Unit 3 Sheet 14 of.f0_
FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION
[97-Of'/4 (11R) = 0.07
< 0.15: The small displacement theory solution is applicable.
(de0/tb) =
0.06
< 0.6; The small displ icement theory solution is applicable.
Input Trial angle Beta until C'm equal to compressive capacity of the shell Cm
~*
Angle Deta (rad) =
2.49 See note 4.
N:te 4: The trial angle Beta is fine tuned here for better convergence of C'm and Cm.
Output Fluid lloidown Inci ment (kip /in), dTe =
0.004141
Te l'Hb 180 degree angle
3.141593 rad.
Bolts Stif.(kips), Kb = 10.70228 Eq. (H-43)
Trial beta 2.39 2.44 2.49 2.54 2.59 C1 2.015124 2.155996 2.318712 2.508712 2.733411 C2 0.938149 0.883205 0.826772 0.768938 0.709794 C3 2.828276 3.067048 3.340661 3.657695 4.029833 C4 1.669279 1.662869 1.655714 1.647959 1.639757 Maximum Compressive Shortening (in), dc =
0.0016 Eq. (1140)
Maximum Compressive Capacity (kip /in), Cm = 1.306 Based on Eq. (H 39)
Compressive Capacity of the Shell(kip /in), Cm =2.963 Based on Eqs. (H 28 & H 29)
Cn) based on the minimum obtained, Cm =
1.306 u.n
Og ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet L.of1L FPC CRYSTAL RIVER UNIT 3. IPEEE EVALUATION 79 7-Of77 Anchor Bolts Tension Cos.
alpha (i)
Tb1 0
10.70 10.70 10.70 10.70 10.70 171.2 2Tb2 0.261799 20.98 20.99 21.00 21.00 21.01 324.5 2Tb3 0.523599 19.75 19.78 19.81 19.83 19.86 274.5 2Tb4 0.785398 17.78 17.85 17.91 17.97 18.02 202.7 2Tb5 1.047198 15.22 15.34 15.44 15.54 15.62 123.5 2Tb6 1.308997 12.24 12.41 12.57 12.71 12.84 52.0 ITb7 1.570796 9.04 9.27 9.48 9.67 9.85 0.0 2Tb8 1.832596 5.34 6.13 6.39 6.64 6.85
-26.5 2Tb9 2.094395 2.85 3.20 3.52 3.81 4.07
-28.2 2Tbl0 2.356194 0.29 0.69 1.05 1.38 1.67
-11.9 2Tbil 2.617994 0.00 0.00 0.00 0.00 0.00 0.0 2Tbl2 2.879793 0.00 0.00 0.00 0.00 0.00 0.0 1Tbi3 3.141593 0.00 0.00 0.00 0.00 0.00 0.0 Sum Tbi(kips) =
114.69 116.36 117.87 119.25 120.48 C'm, per Equation (H-46); should be equal to the allowable compression capacity of the shell.
C'm (kip /in) =
1.120 1.214 1.322 1.447 1.593 hisc, per Equation (H-47)
Sum TbixRcos(alphai)(kip-ft) =
1,081.95 hisc = 12*C'm*C2*R^2 + Sum [Tbi'R*cos(alphai)) + 24*Te0*R^2' sin (beta) + 12*dTe*C4*R^2 m so w> w
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal River Unit 3 Sheet 1.6.of.31 FPC CRYSTAL RIVER UNIT 3. IPEEE EVALUATION f77 OfV4 Overturning htoment Capacity (kip ft), hisc =
4,683.12 Eq. (H-47)
Tank Overturning bloment (kip-ft), hish =
3,849.13 Overturning Moment Capacity > Overturning Moment (Demand) 0.K.
Base Shear Capacity Shear Coefficient Friction, COF =
0.7 Vsc (kips)=COF((Wte+Pa(3.14R^2))+st'.mTbi) =
1,204.55 Eq. (H 51) i Tank Dase Shear (kips), Vsh=
271.44 Base Shear Capacity > Hase Shear (Demand) 0.K.
Fluid slosh height, hs (ft) =
2.14 Freeboard = 2.5 ft > slosh height = 2.14 ft 0.K
==
Conclusion:==
The tank meets the requirements of NP 6041 HCLPF = (Msc/Msh)*(ZPA SSE) 0.12
=
DJO O
Og ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet jl.of.31 FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION J77-Ofy2 4.1.1 Weight of the Tank's IIead, Wh (kips) 1 Rafter ciips Number of rafter clips, n =
16.00 Area, A (in^2) =(6.25*6) - 1/2(2.5)^2 =
34.38 Thickness, t (in) =
0.25 Wl(kips) = n' A*t*(490)/(1000* 12^3) =
0.04 2
L21/2x21/2xl/4 Length, L(It) =
100.00 Weight per ft, w (lbs/ft) =
4.10 W2 (kips) = w'L/1000 =
0.41 3
Weight of the cone shape roof Plate thickness, tp (in) =
O.19 Radius of the cone, R (in) =
192.00 lielght of the cone, He (in) = 1/12*R 16.00 W3 = 3.14'R*sqn(R^2 + He^2)*tp*490/(1000* 12^3) =
6.18 4
Weight of the C5x6.7 Length, L(ft) =
160.00
- Weight per ft, w (lbs/ft) =
6.70 W4 (kips) = w*1J1000 =
1.07 5
Misc. weight W5 (kips) =
0.20 Total weight of the tank's head Wh (kips) = W1+W2+W3+W4+W5 7,90 how 6M 9tc 4.'t
@g ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet lfLof fil FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION f 97-Of 74 4.1.2 Anchor Bolt Reduction Factors
.l.i.2.1 Reduction due to concrete foundation. Re (Ref.5. App. C)
Reduction due to anchor bolt embedment (1.25" dia. cast in place J bolts with 90-degree hooks)
Actual embedment L (in) =
38.00 Minimum embedment Lmin (in) (wit't corree configuration) =
68.13 Reduction factor:lem = U62.5d =
0.49 see note 4
~~
Nste 4: The evaluation is modified since no credit is taken for the 90 deg. hook.
Reduction due to concrete strength Reduction factor Rconc = sqn(3000/3500) =
0.93 Reduction due to concrete foundation Rc = Rem
- Rconc 0.45 Pu = Rc*Tnom 18.79
.l.i.2.2 Reduction due to top plate. Rtn (Ref S. Section 7.2. Sten R)
Top plate maximum bending stresses, sigma sigma (ksi) = [(0.375g-0.22d)Pu]/(f*c^2) 6.81 g (in)=
3.00 d (in)=
1.25 e (in) =
1.00 f (in) = b-e-(d+1/16)/2 =
2.34 b (in) =
6.00 e (in) =
3.00 Fy (ksi) =
30.00 Rtp=Fy/ sigma 4.40
> l.0 Reduction due to top plate, Rip 1.00 s.
nw
Og ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet _L.of.51 FPC CRYSTAL RIVER UNIT 3. IPEEE EVALUATION g 7,pfyy i
4.1.2.3 Reduction due to vertical stiffener plate. Res (Ref. S. Sectinn 7.2. Sten 10)
Critical cross section width of the vertical stiffener plate, k =
4.80 k = 6* sin (theta) = 6* sin (53.13) = 4.8 in theta = tan(8/6) = 53.13 deg Thickness of chair venical stiffness plate,j (in) =
0.50 k/j 9.60 95/sqrt(Fy/1000) = 95/sqn(36000/1000) 15.83 k/j=9.6 < 95/sqn(Fy/1000) = 15.83 0.K.
h (in) =
9.00 e (in) =
1.00 0.04(h-c) =
0.32 j = 0.5 in > 0.04(h c) = 0.32 in O.K.
J = 0.5 in > 0.5 in (marginal) 0.K.
Pu (ksi) =
18.79 Pu/(2*k*j) 3.91 Pu/(2*k*j) = 3.91 ksi < 21.0 ksi O.K.
Reduction due to vertical stiffener plate, Rys 1.00
.t.l.2.4 Reduction due to tank shellstress. Rshi(Ref S. Section 7.2. Strp 9)
Width of the chair top plate parallel to shell, a (in) =
5.00 Eccentricity of anchor bolt, e (in) =
3.00 Thickness of base plate of tank, tb (in) =
0.25 Thickness of tank shell, ts (in) =
0.25 Height of the chair, he (in) =
9.00 Tank Radius (in), R =
192.00 Z= 1/[0,177'a'tb'(tb/ tsp 2/sqn(R*ts)+1.0]
0.97 1
F2 421
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal River Unit 3 Sheet.20.of 53.
FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION f 9 7 Of'/4 A o Pu'e/ts^2 901.92 D = 1.43*a*he^2/(R*ts) 12.06563 C a (4 *a*he^2)^0.333 11.71571 D a 0.031/sqn(R*ts) 0.004474 sigma = A*[1.32*Z/(11+C)+D) 52.5481 Sye (ksi) =
30 Rshi = sigma /Syc =
0.570906 Reduction due to tank shell stress, Rshi 0.57 i
.t.l.2.5 Reduction due to chair.tu. tank shell weld, Rwid (Ref. 5 Section 7.2. Step Ii)
Weld size between chair and tank wall, tw (in) 0.25 Ns 'd load, Ww (kips /in) =
1.00 Ww= Pu * (sqn [ l /(a+2he)^2)+[e/(a
- he +0.667
- he^2)^2)
Weld capacity, We (kips /in) =
5.41 We=(30.6*tw)/sqn(2)
Ww = 1.29 kips /in < We = 5.41 kips /in O.K Reduction due to chair to tank shell weld, Rwid 1.00 M *M M 1
i Sg ANALYSIS / CALCULATION CONTINUATION SHEET Crystal Filver Unit 3 Sheet.1L of _S3.
FPC CRYSTAL RIVER UNIT 3. IPEEE EVAL.UATION
.99 7-Ofy2 4
4.2 Chilled Water Expansion Tank (CHT 1)
The CilT 1 Expansion Tank is a horizontal tank suspended from the ceiling (Elevation 183' 0") of the Control Building. The weakest link is judged to be the anchorage of the tank support structure to the ceiling which is evaluated below.
The structural members are judged to be adequate by the SRT.
4,2.1 Input Data Tank dry weight WI = 133 lbs (Ref. 29)
Tank capacity =
60 gals (Ref. 29)
Water Weight Ww = (62,4 lbs/ft')(60 gals)(0.134 ft / gals) = 502 lbs 3
Support frame is made of 2x2xl/4 angles (Ref.10) w = 3.19 lbs/ft A = 0.938 in:
I = 0.348 in d
he, 6 e ers g;1
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal River Unit 3 Sheet.22. of.53.
FPC CRYSTAL RIVER UNIT 3,IPEEE EVALUATION
.N 7 Ofp'2 Calculation i
36*
C.C.
..............q.............
-=
1 12 48*
12 16' O
.=
Elevation looking EAST Elevation looking NORTli Totallength of the L2x2xl/4:
2 2
2 3
L = 2[2*36 + n*16/2 + 2*(16 + 28 )it: + 2*(48 + 28 )ir2)] = $45.5 in = 45.5 ft Weight of the support frame Ws = (3,19 lbs/ft)(45.5 ft) = 145 lbs Miscellaneous (attached small bore piping)
Wm = 50 lbs (conservative)
Total weight W = Wt + Ww + Ws + Wm = 133 + 502 + 145 + 50 = 830 lbs ow
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal River Unit 3 Sheet 21of,31 FPC CRYSTAL RIVER UNIT 3 IPEEE EVALUATION f ? /-Of yl.
Freauenev in the loncitudinal (N S) direction.
Compute stiffness in the N S direction:
K = 2k 12El(I + cos0)
= (12)(29E6)(0.348)(1+cos60*)
(L )'(3L, +4L,(I+cos0)) (8)'[(3)(28)+4(8)(1+cos60*)) 2.15x10'lbs/in k=
=
~~
K = 2(2.15x10') = 4.3x10' Ibs/in 0 = tan(48/28) = 60*
L=48" i-Checker's note: The computed suffnesses were checked using an L =28" alternate method and it was i
0 determined that they are satisfactory.
L:=8" Frequency in the N S direction EN 5j(43x10*)(386.4) i fgg 1
/
830 om
Sg ANALYSIS / CALCULATION CONTINUATION SHEET Crystal Fllver Unit 3 S:,oet.21 of.S3.
FPC CRYSTAL RIVER UNIT 3, IPEEE F, VALUATION f 9 7-bryZ 4
Freauenev in the transverse (E W) dirt _qtion.
Compute stiffness in the E W direction:
K = 2k 12El(I + cos0)
(12)(29E6)(0.348)(I + cos30*)
'~
- (3L, + 4L (1 + cos6),
(8)',(3)(28) + 4(8)(1 + cos30'))
2.46x 10'Ibs / in (L )
3 4
4 k = 2(2.46x10 ) = 4.9x10 lbs/in 0 = tan(16/28) = 30' L=16*
i.
L,= 2 8 "
O L, = 8 "
Fregaency in the E.W direction 2xY IV 2xI(4.9x10 )(386.4) 4 1 IKg
/
830 sn uw
Sg ANALYSIS / CALCULATION -
CONTINUATION SHEET Crystal River Unit 3 Sheet.25_ of J1 FPC CRYSTA). RIVER UNIT 3. IPEEE EVALUATION f77-Off!-
Freauency in the vertical direction a
Compute stiffness in the vertical direction:
K = 4 AE (4)(0.938)(29E6) = 3x10. lbs / m.
=
L 36 f = g gg = g(3x10')(386.4) 1 f
1 u 188 H:
Rigid Summary of the frequencies calculation:
Longitudinal N-S direction:
f = 22.5 Hz Frequency range:
0.8*22.5 = 18 Hz < f = 22.5 Hz < l.2*22.5 = 27 Hz Transverse E-W direction:
f = 24 Hz Frequency range:
0.8*24 = 19.2 Hz < f = 24 Hz < l.2*24 = 28.8 Hz Vertical direction:
f = 188 Hz
/. Rigid i
D The seismic demand is obtained from the Ref. 6 (design horizontal in-structure response spectrum for SSE) for the Control Building at elevation 186.83' for 4% damping.
Longitudinal N-S direction:
as.s = 0.4ag n su n u,
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal River Unit 3 Sheet.26.of.S3.
FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION f 7 7-Off L Transverse E-W direction:
as.w = 0.40g Vertical direction:
av = 2/3*ZPA = 2/3*0.2g = 0.13g Loads annlied at C.G. of the tank:
FN s = 830'0.40 = 332 lbs Ft.w = 830*0.40 = 332 lbs Fv = W(1 + av) = 830(1 + 0.13) = 830 + 108 = 938 lbs i
Loads at C.G. of the basentate:
Due to Fu.s = 332 lbs Tension:
F1 = (1/2)(332*36)/48 = 125 lbs Shear:
S1 = 33?/4 = 83 lbs Due to Ps.w = 332 lbs Tension:
F2 = (1/2)(332*36)/16 = 374 lbs Shear:
S2 = 332/4 = 83 lbs Due to Fv = 830 + 108 = 938 lbs Tension:
F3 = (1/4)(830 + 108) = 208 + 27 = 235 lbs Total tensile load:
2 2
F = 20S + (125 + 374: + 27 )i/: = 603 lbs Total shear load:
2 2
S = (83 + 83 )2 = 117 lbs Sh b N.
Y&
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal River Unit 3 Sheet _22.of.H3.
FPC CRYSTAL RIVER UNIT 3,IPEFE EVALUATION
[72-#ff 2-Basenlate 9"x9"x1/2"(tvn).
By engineering judgment the connection between the angle and the baseplate is adequate due to small loads.
C.L. plate, angle and bolts inn-l 0
i HD--
l t
6" l
,,,, C.L. plate, angle and bolts l
l l
0
-i>--
, in..
l iin-6" iic" 4.2.2 Anchorage Demand Resultant loads:
a
.n a
SFlodd ANALYSIS / CALCULATION Powe CONTINUATION SHEET Crystal River Unit 3 Sheet.28.of13.
FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION.
J7 7-Off 2.
Pullout / bolt P = F/4 = 603/4 = 151 lbs Shear /balt V = S/4 = 117/4 = 30 lbs 4.2.3 Anchorage Capacity Each vertical member uf the expansion tank support structure is anchored to the 3' concrete ceiling (Ref. 7) through 4-1/2" Q anchor bolts. Conservatively assume WEJ If wedge anchors.
Nominal Capacity for 1/2"6 WEJ lT Anchor Bolts with 2" minimum embedment (per GIP, Ref. 5).
P m = 2290 !bs V m = 2380 lbs Concrete Streneth Check f', =3,000 psi RF = 3000/4000 = 0.75 p
RF, = 2000/10000 + 0.65 = 0.95 Check for Anchor type RT = RT = 0.5 p
Check for Tiehtness RI = RI = 0.75 p
Check for Embedment 1.ength RL = RL = 1.0 p
Check for Spacine Spacing S = 6" > Sm = 5" EL nm
l
)
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal River Unit 3 Sheet.29_ of13.
FPC CRYSTAL RIVER UNIT 3. IPEEE EVALUATION f 77 OffL RS o RS, e 1.0 p
Check for Edce Distance The edge distance E > Esitu =5" RE = RE. = 1.0 p
Psi = (0.75)(0.5)(0.75)(2290) = 644 lbs Vei = (0.95)(0.5)(0.75)(2380) = 848 lbs Shear Tension Interaction:
P/Pm + V/Va = 151/644 + 30/848 = 0.27 < l.0 O.K.
Factor of safety = 1/0.27 = 3.7 > 1.5 for reduced inspection.
Nete: since the shear tension I.R = 0.27 << l.0, the possible misalignment of the angles and the baseplate will have no adverse effect on results.
Basentate stresses; M = 2* Pa*b = 2*644*3 = 38M in lbs 2
S =' 9*0.5 /6 = 0.375 in' f = M/S = 38M/0.375 = 10304 psi < 0.9Fy = (0.9)(36000) = 32400 psi O.K.
The capacity of the 1/2"O WEJ IT Anchor Bolts envelopes the seismic demand. Therefore, the anchorage of the Chilled Water Expansion Tank is adequate.
m nw
Sg ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Un!t 3 Sheet 10. of _63.
FPC CRYSTAL RIVER UNIT 3. IPEEE EVALUATION ff7-Off2 Check weld between the expansion tank and L2x2xI/4 Loads at each support:
F1 = F.s / 2 = 332 / 2 = 166 lbs N
F2 = Fe.w / 2 = 332 / 2 = 166 lbs F3 = Fv / 2 = 938 / 2 = 4691bs Total weld load:
i 2
2 2
Fw = (166 + 166 + 469 )ir2 = 524 lbs Weld allowable load (conservatively assume 1/8" fillet weld, total length of 1/2")
FAu.= (l.7)(.707)(l8000 psi)(0.125 in)(0.5 in) = 1352 lbs > 524 lbs O.K.
The remain of the support components including welds are adeqisste due to small loads.
HCLPF = (Safety Factor )(7.PA SSE) = (3.7/1.5)(0.lg) = 0.25g
.m
(
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal Rivet Unit 3 Sheet.)_L of.62.
FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION 57 7-Off'l-4.3 Evaluation of the RCP Seal Return Cooler MUHE-2A & 2B The RCP Seal Return Cooler MUHE-2B located at the Auxihary Building, Elevation 119'. By the SRT judgment the open section cantilever beam (W8x31) supporting the subject cooler need to be evaluated. The subject cantilever beam is welded to the embedded plate.The SRTjudged that the anchorage of the MUHE-2A is adequate. Also, the remainder support components are judged to be acceptable.
4.3.1 Input Data Weight =
1950 lbs (Ref.14)
~~
Tank O.D.
10.75 in (Ref.14)
Beam length =
32 in (walkdown info)
Assume that center gravity of the full tank is located at the middle of the tank:
e = 8" + 8"/2 = 12".
Properties of the V8X31 beam (Ref.10):
2 A = 9.13 in ; r = 2.02 in (min):
b = d = 8 in; tr = 0.435 in;
- t. = 0.285 m lx = 110 in ; Sx = 27.5 in' d
ly = 37.1 in'; Sy = 9.27 in' Torsional constant K = J = 0.54 in' Warping constant Cw = 530 in*
Modulus of Elasticity E = 29E6 lbs/in:
2 Shear modulus of regidity G = 11.2E6 lbs/in 4.3.2 Calculation Weight of the tack:
~ su
,w
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal River Unit 3 Sheet U. of _fi3.
FPC CRYSTAL RIVER UNIT 3,IPEEE EVALUATION 87-Off 4 Use W = 2000 lbs The seismic demand is obtained from the Ref. 3 (design horizontal m structure response spectrum for OBE) for the Auxiliary Building at elevation 119' for 4% damping. Conservatively use peak acceleration :,lue. Use factor of 2 to obtain SSE value.
Longitudinal direction:
a = (2)(0.353g) = 0.71g n
Transverse direction:
a = (2)(0.353g) = 0.71g n
Vertical direction:
av = 2/3*0.71g = 0.47g Loads applied at C.G. of the tank:
Fu = Fu.s = 2000'0.71 = 1420 lbs Fu = Fe.w = 2000'0.71 = 1420 lbs Fv = W(l + av) = 2000(l + 0.47) = 2940 lbs Loads at each support:
F1 = Fu = 1420 lbs F2 = Fa / 2 = 1420 / 2 = 710 lbs F3 = Fv / 2 = 2940 / 2 = 1470 lbs Loads at C.G. of the W8x31 supporting beam:
3 1
2!
F1 = 1420 lbs F2 = 710 lbs MN M M)
ANALYSIS / CALCULATION-g.
CONTINUATION SHEET
-~~~
Crystal River Unit 3 Sheet.33 of 63.
FPC CRYSTAL RIVER UNIT 3 IPEEE EVALUATION f77-Of72 F3 = 1470 lbs M1 = F2*e = 710* 12 = 8520 in lbs M2 = Fl *e = 1420* 12 = 17040 in lbs Bending moments at the end of the cantilever beam (L = 32 in)
Mi = 8520 +F3*L = 8520 + 1470*32 = 55560 in lbs M3 = F1*L = 17040*32 = 45440 in-lbs i
Axial stress:
2 fa = F2/A = 710 lbs / 9.13 in = 78 psi Bending about major axis:
Ibi = M1/ Sx = 55560 in-lbs / 27.5 in' = 2020 psi Bending about minor axis:
lb = M3 / Sy = 45440 in lbs / 9.27 in' = 4902 psi 3
Calculate shear stress due to torsional regidity of the cross section (t ), shear stress due to warping (t ),
3 i
2 and bending stress due to warping (c )(Ref. I1, Tables 21 & 22).
a Parameter (Ref. I1. Table 22) p = (K*G / Cw'E)2 = (0.54* 11.2E6 / 530*29E6) = 0.02 in
ANALYSIS / CALCULATION-g CONTINUATION SHEET Crystal River Unit 3 Sheet 34 of _n FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION ff 7-Off2 Compute angle of rotation 0 and the succesive derivatives O',0" and O'" (R.ef. I1 Table 22, case Ib)'
Case Ib: one end of the beam is free to twist and warp, other end fixed (no twist or warp).
4 tt x = 0:
Omax = To( L - Tanh(pt)) / CwEp3 = 17040*( 0.02*32-Tanh(0.64)) / (530*29E6*0.02 ) = 0.0104 rad 3
where To = M2 = 17040 in lbs - applied torsion moment L = 32 in - cantilever length
~~
L = 0.02*32 = 0.64 at x = 0:
10' maxi = To(1 - 1/ Cosh ( L)) / CwE 2 = 17040(1-1/ Cosh (0.64)) / (530*29E6*0.02 ) = 0.0004 2
at x = L:
4 0" max = To(Tanh(pt)) / CwE = 17040*(Tanh(0.64)) / (530*29E6*0.02) = 3.14x10 et x = L:
4 0"* max = To / CwE = 17040 / (530*29E6) = 1,11x10 Maximum bending stress due to warping, c.
c = h*b*E*0"/4 = 8*8*29E6*3.14x10 / 4 = 14570 psi Maximum shear stress due to warping, T 2 2
2 4
- T = h*b *E*0"*/16 = 8*8 *29E6*1.11x10 /16 = 1030 psi s
Maximum shear stress due to torsional regidity of the cross section, t i a..
,* m
ANALYSIS / CALCULATION g
CONTINUATION SHEET
" ~ - -
Crystal River Unit 3 -
Sheet.31of 32 FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION ff 7.Of y /-
T = tr.G'0' = 0.435*11.2E6*0.0004 = 1948 psi i
Total normal stress:
f = fa + ib + fb3 + o = 78 + 2020 + 4902 + 14570 = 21570 psi < 0.9 Fy = 32400 psi O.K.
i 2
Shear stress due to F1 = 1420 lbs; A1 = 2*b*t = 2*8'O.435 = 6.96 in (average shear) t i
f hi = F1/A1 = 1420 / 6.96 = 204 psi 2
Shear stress due to F3 = 1470 lbs; A3 = h*t.= 8*0.285 = 2.28 in f,u = F3/A3 = 1470 / 2.28 = 645 psi Total normal stress:
f.h = f ni + f u + t + T = 204 + 645 + 1948 + 1030 = 3827 psi < 0.4Fy = 14400 psi O.K.
i i
i2 Check 1/2" thick angle.
M = (1420)(8 - 10.75/2) = 3728 in-lbs 2
S = bt /6 = (8)(0.5)2/6 = 0.33 in' f = M/S = 3728 / 0.33 = 11297 psi < 32400 psi O.K.
HCLPF = (Min. Safety Factor )(ZPA SSE) = (32400/21570)(0.lg) = 0.15g mm aw
m ANALYSIS / CALCULATION Powe CONTINUATION SHEET Crystal River Un!t 3 Sheet.M.of _S1 FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION f77.Oryt-4.4 Nuclear Service Closed Cycle Surge Tank (SWT-1)
This section evaluates the seismic adequacy of the elevated tank anchorage. Only the anchorage (i.e. anchor bolt capacity is checked).
The support legs, the base plate and the tank attachment to the vertical legs are judged to be adequate by the SRT.
4A.1 Design Input for SWT-1:
The following were obtained from field walkdown:
Tank capacity = 10,000 Gallons Location: Auxiliary Building El 95'-0" Two (2) Anchor Bolts for each leg (7/8" $ )
Ring all around on top of the legs.
Pedestal Height = 7" Anchor Bolt embedment = 11.0"(Ref.19 & 24)
Shell Thickness = S/8" The following are from Ref. (20)
Tank Diameter = 132" = 11.0' Tank Length = 192" = 16.0' Baseplate Size = 14x14xl" Bottom of tank = 2.5' of the floor Ring at 7'-10" from the floor Normal water level = 11.5' From Ref. (20)
Dome height = 2'-7.5" = 2.63' Structural legs W12X53 Weld between column and tank are judged adequate Ring and Attachmentjudged adequate.
9CE 521
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal River Unit 3 Sheet 37 of.13.
FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION f77.Ofyt 4.4.1.1 Geometrn
~ Anchorage details is shown in Ref. (18).
Tank & Pedestals l
l' e
l 9'
l 18" Typ.
I.
5' 10.75"(Typ.)
I
(
l 9" Typ.
...p......,,........g - g....... Q. Tant & Pedestals
. g.
g.
yLl l
5' 10.75" (Typ.)
. 2".
(TypJ G of Pedestal e
?
I Nuclear Service Closed Cycle Surce Tank Foundation (Dwg SC-422-042) h
@g ANALYSIS / CALCULATION CONTINUATION SHEET Crve+al River Unit 3 Sheet.31. of _fia.
FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION fp 7 Ofy4, 4.4.2 Anchorage Evaluation of SWT 1:
Volume of steel based on glir. der geometry = (2x66 + x132
- 192)0.625 = 66.869 in' 2
Weight of tank shell(using steel density of 490 lbs/ft') = 66869*490/(12
- 1000) = 18.96 kips 3
Weight of tank empty including some miscellaneous attachment = 1.l(18.96) = 20.9 kips Weight of fluid (using water density of 62.4 lb/ft') = x5.5
- 11.5*62.4/(1000) = 68.2 kips 2
Total Weight, W = 20.9 + 68.2 = 89.1 kips Use 90.0 kips Center gravity of the tank empty is at = 192/2 + 2.5* 12 = 126" from the ground.
Location of the C.G. for the total mass = (20.9* 126 + 68.2*(2.5* 12 + 5.75*12))/89.1 = 105.33" Use 113.0"
~~
Since, the tank is mounted on legs, the natural frequency of the tank is dominated by the stiffness of the legs and the total mass lumped at the center gravity. The sloshing effects for thi:; tank is accounted for by increasing the location of the C.G. by 10% (i.e. 0.1*5.75*12 = 6.9"; C.G. @ 105.3 + 6.9 = 112.2") (Cons.).
The horizontal natural frequency is based on the cantilever length,1 = 113".
2 W12X53 A = 15.6 in I,x = 425 in' lyy = 95.8 in' S = 70.6 in' S = 19.2 in' y
1.732 J29000
- 2(425 + 95.8)386.4 Using Case 3a of Table 36 of Ref. (11), fi = 2x IWl'
= 2.6 Hz
~-
2x V 90 *113' The vertical natural frequency =f' = 1N# =1 29000 *4 *15.6* 386.4 = 41.8 Hz 2x V WI 2xl 90*113 Since, the floor elevation is the top of the mat and the Auxilliary building is founded on the rock, the ground input motion is considered applicable to the floor at elevation 95 -0". Conservatively the peak acceleration of
.the ground spectra from Ref. (3) is used for the horizontal direction. The ground peak acceleration is used with 4% damping value.
Horizontal SSE ground peak acceleration = 2*0.1Ig = 0.22 g Vertical SSE acceleration at (2/3 of horizontal ZPA) = (2/3)*2*0.05g = 0.067g i
%.=
,= w
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal River Unit 3 Sheet 39 af,53.
FPC CRYSTAL RIVER UNIT 3 IPEEE EVALUATION f f 7-Ofy2 4.4.2.1 Seismic Forces:
Resultant horizontal force = (0.22*90*2 n) = 28.0 kips i
Resultant moment = 28.0* 113 = 3164.0 in-kips Vertical force = 0.067 *90 = 6.0 kips Shear / leg = 28.0 kips /4 = 7.0 kips Resisting moment arm for two legs in sension, = (70.75*2'") = 100.0" Tension per leg (assume two legs in tension)
= 3164/2(100) + 6/4 - 90/4
= 15.82 +1.5 - 22.5 = -5.18 kips (compression)
For three legs in tension the distribution is as shown :
oc TI 70 5" 70 75" Load Distribution Equilibrium : 141.5*T + 2(T/2)*70.75 = 3164 in-kips Tension per leg, T = 3164/(141.5 + 70.75) +6/4 - 90/4 = 14.91 + 1.5 - 22.5 = - 6.1 kips (Compression)
The anchor bolts are in compression )
4.4.3 Anchor Bolt Capacity P.. = 20.44 kips V.m = 10.22 kips Check Embedment L = 11"
> L.,. = 8.75" RL = RL, = 1.0 p
Check Spacine Distance S = 7.0" E,n = 11.0" w
4 SM mm
i eg ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet 40 oll3.
FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION fp 7-Of7L The concrete pullout capacity is controlled by the pedestal cross section area, since the shear cone area is not s
anticipated to extend into the floor slab.
pV, = 4
- 0.6543000
- 18
- 18 = 46.1 kips / pedestal 1000 Pullout capacity per bolt = 46.1/2*2 = 11,5 kips (i.e. 2 bolts / pedestal and S.F. of 2 for ductile failure)
RS = 11.5/22.4 = 0.51 p
RS = l.0 for S > 2D Check Edee Distance E = 9.0 - 3.5 = 5.5" <
E. = 7.75" The reduction for the pullout due to the edge distance is not computed here since it was accounted for in the spacing check. Hence, RE = 1.0 p
RE* = 0.0131 g-
- 5.5 ' '
= 0.0131
= 0.j2
.D.
0.875 It should be noted that for any loading direction, the edge distance violation appl" only to four bolts and the other bolts are bearing on the concrete pedestal and wottid not have any reduction. Therefore, only six anchor bolts are considered active for the total shear and no shear reduction is performed.
Concrete Strength Check RF, = RF, = I 3500 13500 = 0.93
=
The pullout capacity is computed above based on 3000 psi compressive strength. Hence, the concrete strength reduction factor is not applied for pullout.
Pall = 0.51*22.44 = 11.4 kips Vail = 0.93* 10.22 = 9.5 kips Shear / bolt, V = 28.0/6 = 4.67 kips Tension / bolt, P = 0 kips Shear-Tension Interaction V/Vm = 4.67/9.5 = 0.49 l.0 0.K.
,, m
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal River Unit 3 Sheet 41 of 31 FPC CRYSTAL R!VER UNIT 3, IPEEE EVALUATION ff;7-OfV2.
4.4,4 Structural Members Capacity The base plate is adequate since there is no tension load on the anchor bolts.
The bas plate isjudged adequate for the compressive loads Check Column Bendine Moment = 3164 in kips (Conservative since it is based on tank C.G. instead oflegs cantilever length)
Considering two legs bending about the strong axis and two legs about the weak axis, the load distribution is as follows:
i Stiffness ratio of strong to weak axis = 425/95.8 = 4.44 Bending about strong axis, M. = (3164/2)(4.44/(1+4.44)) = 1291 in. kips Strong axis stresses = M /S = 1291ri0.6 = 18.3 ksi F = 36 ksi O.K.
y Weak axis stress =M /S = (3164/2 - 1291)/18.7 = 291/19.2 = 15.2 ksi <
F = 36 ksi O.K.
y y y
The allowable stresses is 1.7 times AISC allowables. Conservativdy, the AISC nominal allowables for strong axis and weak axis bending are limited to 0.6F. Hence,1.7*0.6F = 1.02F. The bending stresses are within y
y y
the yield stresses.
Maximum axial load, Fa = 14.91 + 1.5 + 22.5 = 38.91 kips Axial stresses = Fa/A = 38.91/15.6 = 2.5 ksi Negligible.
The actual cantilever length is considered to the C.G. of the weld between the column and the tank.
= 12((2.5+2.63) + (l/2)[7.833 - (2.5+2.63)]} = 12*6.5 = 78" Hence, the bending stresses are reduced to reflect the smaller cantilever length (i.e. 78/l13 = 0,69)
Total stresses = 0.69(18.3 + 15.2) + 2.5 = 25.6 Ksi
, 36.0 ksi O.K.
/. The structural members are adequate.
HCLPF = (Min. Safety Faetor )(ZPA SSE) = (36.0/25.6)(0.1g) = 0.14g s su aw
9m ANALYSIS / CALCULATION Powe CONTINUATION SHEET Crystal River Unit 3 Sheet 42.of.f0.
FPC CRYSTAL RIVER UNIT 3. IPEEE EVALUATION f f 7.Of y 4 4.5 Spent Fuel Heat Exchanger SFHE 1 A & B This section evaluates the seismic adequaq of the Spent Fuel Heat Exchanger anchorage (SFHE 1 A & B).
The SFHE 1 A & B are vertical heat exchangers supported approximately 1/3 down from the top. The support
~
is at elevation 143'-0" in the Auxiliary building. The heat exchangers are anchored to the structural members.
4.5.1 Design Input for SFIIE-1 The following were obtained from Ret. (24):
Tank length = 19'-7.31" Use 20.0' Tank. O.D. = 22.0" Location: Auxiliary Building El.143'-0" One (1) Anchor Bolts for each lug (l" $ )
Anchor bolt material = A307 (Conservative assumption)
Horizontal structural rnembers : W8X28 (Assumption based on walkdown)
Anchor Bolt between structural steel (5/8" $ A307; Material type is conservative)
Brackets between beams are assumed adequate including the embedded plates.
Structural members span = 10.0' (Walkdown info) t i
NM
Ono,$6 ANALYSIS / CALCULATION Powe CONTINUATION SHEET Crystal River Unit 3 Sheet 43 of.fi3.
I'PC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION fp7-Ofyt
.t.5.1.1 Geometry:
Anchorage details are shown in Ref. (24).
Tank & Lugs l
4 A.
l l
~*
l, l '-5.5" (
)
l i
)
...............[......s.......
...... Q.. Tank & Lugs
...ap.
i j
i l
l 1-
" (Typ.)
l
~
Sxnt Fuel Heat Exchancer Foundation h
% 6%
wm elt
A 9no,$4 ANALYSIS / CALCULATION Powe CONTINUATION SHEET Crystal River Unit 3 Sheet 44 of _fi3.
FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION-g 7 0774.
4.5.2 Anchorage Evaluation of SFIIE 1A & IB:
Heat Exchancer Weicht:
Weight of tank = 6.5 kips From Ref. (24)
The support location for the heat ex: hanger is located at: 6.79' + 2.72' = 9.51' from top as shown in setting plan drawing Ref. (24). Therefore, the loads at C.G. are approximately at six inches (6") off from the support location.
h = (20'/2 - 9.5 )12 = 6" The lower nozzle is at: (15.44' - 6.79')* 12 = 104" from the support location.
The upper nozzle is at: (6.79' + 1.25')* 12 = 96.5" from the suppon location.
The nozzle loads are con.puted at the heat exchanger centerline. The nozzle bending moment and torsional moinent due to the eccentricity from the nozzle to the tank centerline are neglected in this analysis.
Nozzle Shear Loads at SFHE-1 A (Lower Nozzle) From Ref. (26)
Pipe size is 10" Sch. STD. weight = 82.9 lbs/ft (including insulation)
Unsupported pipe length:
E-W direction = { l.58 + 0.34 + 6.9 + 2.99 + 1.5 + 4.88 + 2(0.33 + 0.11) + 1.96 + 2.50}/2 = 11.77*
N-S direction = (l.58 + 0.34 + 6.9 + 2.99 + 1.5 + 4.88 + 2(0.33 + 0.11) + 1.96 + 2.50 + 6.0 + 1.55 + 14.5)/2
= 22.8' Vertical = { l.58 + 0.34 + 6.9 + 2.99}/2 = 5.9' Nozzle Shear Loads at SFHE-1B (Lower Nozzl.e_) From Ref. (26)
Unsupported pipe length:
- E-W direction = { l.58 + 0.34 + 6.9 + 3.25 + 1.89 + 1.29 + 4.88 + 2(0.33 + 0.11) + 1.96 + 2.50}/2 = 12.74' N-S direction = { l.58 + 0.34 + 6.9 + 3.25 + 1.89 + 1.29 + 4.88 + 2(0.33 + 0.11) + 1.96 + 2.50 + 6.0
+ 1.55 + 14.5 }/2 = 23.76' Vertical = 11.58 + 0.34 + 6.9 + 3.25 + 1.89 }/2 = 6.98*
N l
Og ANALYSIS / CALCULATION CONTINUATION SHEET Crvetal River Lnit 3 Sheet 45 of.33_
FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION f 7 /.pyy2.
Mozzle Shear Loads at SFHE-1 A (Upper Nozzle) From Ref. (25)
Pipe size is 10" Sch. STD. weight = 82.9 lbs/ft (including insulation)
Unsupported pipe length:
E W direction = { l.58 + 3.0 + 22.1 + 2.0 + 6.0 + 10.9 + 1.75 + 3.5 + 7.0 }/2 = 28.9' N S direction = (1.58 + 3.0 + 12.0 J/2 = 8.3' Venical = { l 58 + 3.0 + 22.1 + 2.0 + 6.0 }/2 = 17.34' Nozzle Shear Loads at SFHE-1B (Upper Nozzle) From Ref. (25)
Unsupported pipe length:
E W direction = (l.58 + 3.0 + 22.1 + 2.5 + 3.65 + 8.2 + 8.3 + 3.1 + 7.0 }/2 = 29.8' N-S direction = { 1,58 + 3.0 + 11.0 }/2 = 8.0' Vertical = (l.58 + 3.0 + 22.1 + 2.5 }/2 = 14.6' It is conciuded that SFHE-1B has the high.'st nozzle shear loads. The three orthogonal directions are summed using the SRSS method. Also, the nozzle loads from different noz des are added to the heat exchanger loads using the SRSS methodology since the maximums are not likely to occur at the same time.
The cooler and attached piping are considered flexible. The attached piping is supported off the floor at elev.
I19'-0" and the heat exchanger supported at an intermediate level between the elev. I19'-0" and elev.143'-0".
Hence, the floor peak acceleration at elevation 119'-0" of the Auxilliary building for 4% damping value is amplified by 1.5 factor to account for multi-mode amplification. The seismic input is obtained from Ref. (3).
Horizontal SSE ground peak acceleration = 1.5*2'0.35g = 1.05 g Vertical SSE acceleration at (2/3 of horizontal) = (2/3)* l.05g = 0.7g Resultant horizontal acceleration = 1.05 (2)as = 1.48g 8* 4 96 9XI 42'
u ANALYSIS / CALCULATION Powe CONTINUATION SHEET Crystal River Unit 3 Sheet 46 of fi3.
FPC CRYSTAL RIVER UNIT 3. IPEEE EVALUATION f7/-Of72
.I.5 2.1 Seismic Forces:
Lower Nozzle Horizontal nozzle seismic loads = 1.05*0.083 (12.74 + 23.76 ) 5 = 2.35 kips 2
2 Vertical nozzle dead load = 0.083*6.98 = 0.58 kips Vertical nozzle seismic loads = 0.7*0.58 = 0.41 kips 1!np_gr Noz71e Horizontal nozzle seismic loads = 1.05*0.083 (29.P + 8.0 )oa = 2.68 kips 2
2 Vertical nozzle dead loads = 0.083*14.6 = 1.21 kips Vertical nozzle seismic loads = 0.7'l.21 = 0.85 kips Heat Exchancer Self1.oads Dead load = 6.5 kips Horizontal force = 1.48*6.5 = 9.6 kips Vertical force = 0.7*6.5 = 4.5 kips ultant Loads Resultant dead load = 0.58 + 1.21 + 6.5 = 8.3 kips 2
2 2
Resultant saw = (9.6 + 2.35 + 2.68 )os = 10.2 kips 2
2 2
Resultant vertical seismic = (0.41 + 0.85 + 4.5 )oa = 4.6 kips Resultant bending mome..t = ((9.6*6)2 + (2.35* 104)2 + (2.68*96.5)2) 5
= 360.5 in-kips Anchor bolt le ylm Shear / bolt = 10.2 kips /4 = 2.6 kips Resisting moment arm for two legs in tension, = (17.5*2"2) = 24.75" Tension / bolt = ([360.5/2(24.75)]2 + (4.6/4)2)"5 - 8.3/4 = 7.4 - 2.1 = 5.3 kips Compression / bolt = 7.4 + 2.1 = 9.5 kip.
..n 4
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal River Unit 3 Sheet 47 of $3_
FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION fy7 OryL 4.5.3 Anchor Bolt Capacity Pm = 1.7* 15.7 (AISC Ref.10) = 26.7 kips 5.3 kips
..O.K.
V,. = 26.7/2 = 13.4 kips 2.6 kips
..O.K.
It sh'ould be noted that the anchor bolts are connected to the structural members. Hence, the allowables are the same as the nominal.
Shear-Tension Linear InteractioD (Conservatively is used)
V/Vai + P/Pai = 2.6/13.4 4 5.3/26.7 = 0.39 <
l.0 0.K.
The anchor bolts between variable structural members are judged adequate by inspection. Also, the connection to the embedded plate is judged to be adequate by the SRT.
4.5.4 Structural Members Capacity Check W8X28 A = 8.25 in:
S = 24.3 in ;
S = 6.63 in' ;
3 y
The heat exchangers are conservatively considered to be located at (1/3 ) of the span from the wall. The load, P is based on the maximum compressive load per bolt comput:d above. Since, this evaluation is performed for IPEEE program, the allowable stresse. are 1.7 time normal allowables. The axial stresses are negligible.
The axial stresses are negligible.
Moment about strong axis = Pab/L = 2*9.5*3.33*6.67* 12/10 = 506.4 in-kips Moment about weak axis = Pab/L = 2*2.6*3.33*6.67* 12/10 = 138.6 in-kips The existing structural members W8X28 are compact sections (i.e. L = 6.67' < Lc = 6.9' Ref.10)
Strong-axis bending stresses = 506.4/24.3 = 20.84 ksi l.7*0.66*36 = 40.4 ksi Weak-axis stresses = 138.6/6.63 = 20.91 ksi l.7*0.75*36 = 45.9 ksi Biaxial Bending :
20.84/40.4 + 20.91/45.9 = 0.97 l.0 O.K.
.. The structural members are adequate.
MCLPF = (Min. Safety Factor )(ZPA SSE) = (1/0.97)(0.1g) = 0.103g
- w
- X1 m
norw ANALYSIS / CALCULATION Powe CONTINUATION SHEET Crystal River Unit 3 Sheet 48 of.31 FPC CRYSTAL RIVER UNIT 3,IPEEE EVALUATION f 97-Of yL 4.6 Regulating Transformer A & B (VBTR-1 A & 1B The Constant Voltage Transformers VBTR -1 A & IB are equipment class 4 as described in the GIP-2 (Ref. 5) located in the Control Building Elevadon 124'. The subject transformers are GE Model 9T91Y724415 KVA floor mounted transformers anchored to the floor with (4) 1/2"O expansion anchor bolts (2 anchor bolts on each side).
4.6.1 Input Data i
Dimensions:
24" wide x 25" deep x 39.5"high Mounted channels:
4" high (assume C4x5.4) based on walkdown data.
Transf~mer weight:
925 lbs Assume that center gravity is located at the geometrical center of the transformer :
y = 39.5"/2 = 19,75" use y = 20".
Calculation Weight cf the transformer ir 850 lbs (Ref. 23). Conservatively use 925 lbs.
W = 925 lbs The seismic deraand is obtained from the Ref. 4 (design horizontal in-stmeture response spectrum for SSE) for the Control Building at elevation 124' at 4% damping. In accordance with GIP-2 (Ref. 5) the transformer is considered flexible with 5% damping. Conservatively use peak acceleration value at 4% damping.
Horizontal direction:
a = 0.4g n
Vertical direction:
av = 2/3*0.4g = 0.27g (conservative since transformer is rigid in vertical direction)
. Seismic loads applied at C.G. of the transformeq Pn = W'a = 925*0.4 = 370 lbs n
Fv = W*av = 925*0.27 = 250 lbs Bending moments at the base:
a nw
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal River Unit 3 Sheet 49 o* f3.
FPC CRYSTAL RIVER UNIT 3 IPEEE EVALUATION f77-Ofy2 M1 = M2 = Fn'y = 370*20,7400 in-lbs Conservatively use the distances between anchor bolts of 18" and 15" respectively ( per Ref. 23, the actual holt spacing is 19" in both directions).
Anchorage Demand Resultant loads:
Pullout / bolt F = [(7400 / (2* 15))2 + (7400 / (2* 18))2 + (250 / 4)2)i/: - 925 / 4= 96 lbs
~
e 2
2 Shear / bolt V = (1/4)(370 + 370 ) = 523/4 = 131 lbs 4.6.2 Anchorage Capacity The transformer is anchored to the concrete floor with (4) 1/2"O expansion anchor bolts (Hilti or Red Head based on walkdown information and SRTjudgment).
Nominal Capacity for 1/2" 4 WEJ IT Anchor Bolts with 2" minimum embedment,(per GIP, Ref. 5).
Pnom = 2290 lbs V. = 2380 lbs Concrete Streneth Check f', =3,000 psi RF, = 3000/4000 x 0.75 RF = 3000/10000 + 0.65 = 0.95 Check for Anchor type RT, = RT. = 1.0 Check for Tinhtness su ex w
9norw ANALYSISICALCULATION Powe CONTINUATION SHEET Crystal River Unit 3 Sheet _5E.of.fi3.
FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION fp 7 -Of yg Ri = RI = 0.75 p
Check for Embedment Leneth RL = RL. = 1.0 p
Check for Spacip_g Spacing S = 15" > S. = 5" RS, = RS. = 1.0 i
Check for Edce Distance The edge distance E > Eum =5" RE = RE = 1.0 p
Pei = (0.75)(0.75)(2290) = 1288 lbs Vei = (0.95)(0 75)(2380) = 1696 lbs Shear-Tension Interaction:
P/Pai + V/Vsi = 96/1288 + 131/1696 = 0.15 < l.0 O.K.
Factor of safety = 1/0.15 = 6.7 > 1.5 for reduced inspection.
The capacity of the 1/2"O Expansion Anchor Bolts envelopes the seismic demand.Therefore, the anchorage of the transformers VBTR -1 A &lB is adequate.
IICLPF = (Min. Safety Factor )(ZPA SSE) = (6.7/1.5)(0.lg) = 0.45g
%e 6%
yn v.
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal River Unit 3 Sheet.it of.33.
FPC CRYSTAL RIVER UNIT 3,IPEEE EVALUATION f9 7. pyy4, 4.7 Evatustion of Raceway Supports Limited Analytical Review (LAR) will be performed for the typical cable tray support shown below using guidelines given in the Ref. 5 & le. Two typical raceway supports will be evaluated. The weights of the cable tray used in this calculation are from Ref. 5, section 8.3.9.
Desien Input 2
tray with 4 inches ( 100%) of cable fill.................................................................. 25 lbs/ft 24" tray with 4 inches (100%) of cable f111.................................... 2 x 25 lbs/ft = 50 lbs/ft Top tray has all around cover 60" length (assume sheet metal 1/16" thick):
2 (60)(0.0625)(490)/12 = 12.8 lbs/ft na uw
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal River Unit 3 Sheet.5.2.of fi3.
FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION f 7 7-Of y 2.
1 4.7.1 Sample 1 d) i l
~
1e - w l
l
~*
gg.
==
1 I
16'
~
I 1
9 16' I
I 36*
Elev. Ikg. North
l Sg ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet _S. of _fil FPC CRYSTAL RIVER UNIT 3. IPEEE EVALUATION f 7 7-Of'/2 Per review of the support configuration the checks to te performed are D.L. vertical and fatigue evaluation of the rod hanger, Determine weights:
Tributary span for trays is 6 ft (1) 24" :np tray witn cover (50 lbs/ft + 12.8 lbs/ft)(6 ft) = 377 lbs (3) 2r' trays (3)(50 lbs/ft)(6 ft) = 900 lbs Miscellaneous assume 30 lbs i
W = 377 A 900 + 30 = 1307 lbs.
Support weight:
T Assume cach rod carries 50% of totalload.
Load per rod: W1 = 1307/2 = 654 lbs Worst case: Additio. d load from the attached tray running Tast (max. span = 7.5 ft).
Additional load from the attached tay running East (max. spen = 7.5 ft).
W = '0.5)(501bs/ft)(7.5 ft) = 188 lbs Assume 50% of the load applied to one rmt.
W2 = 0.5*188 lbs = 94 lbs Total Dead Load on support W or = l'"~ + 188 = 1495 lbs.
Use W or = 1500 lbs i
1 Totalload on the rod:
Wg = Wl + W2 =654 + 94 = 748 lbs.
Use Wa a 750 lbs Check 1:
Dead Load Desien Check Allowables Loads in accordance with Reference 13 5/8" steel threaded rod F u.= 1810 lbs A
Beam clamp (C-clamp, Grinnell Fig. 61)
FAu.= 850 lbs m
wa
eM ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet _14 of.fi3.
FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION fpy - Of yE.
Steel threaded rod:
F = 750 lbs < 1810 lbs O.K J
Beam clarap Fig. 61 F = 750 lbs < 850 lbs O.K Trapeze cross members (P1000 unistrut, use L = 36 in).
~~
Load w = 377 lbs < 900 lbs (Ref. 27)
Therefore suppon meets (cad load design check.
Check 2:
3 X D.L. Vertical Capacity Check:
Check the overhead connections for 3.0 Dead Load 3 X D.L. = 3(800) = 2400 lbs Beam clamp Fig. 61 F = 2400 lbs > (2.0)(850 lbs) = 1700 lbs Note that 2.0 increase is per Ref. 5, section 8.3.8 for manufacturer's published design values.
The support fails vertical capacity check.
Check 3:
Lateral Load Check:
i Since the support is located at Turbine Building, at elevation 145' which is less than about 40 feet above plant grade (119'), dead load plus t transverse acceleration of 2.5 times the floor spectra ZPA (Ref.12).
Floor ZPA = 1.5* 1.25*0.1g = 0,19g Where 0.lg is a ground ZPA (Ref. 4) llorizontal acceleration an = 2.5*0.19g = n.475 g llorizontal(transverse) load om
l Sg ANALYSiGICALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet.51of.51 FPC CRYSTAL RIVER UNIT 3. IPEEE EVALUATION ff 7-OfV&
Fn = (W or)(a ) = (1500)(0.475) = 713 lbs T
n Assume that 2/3 of total transverse (E W) load will be resisted by the support frame action and 1/3 of the load will taken by the E W cable tray Compute an additional load on the beam clamp due to horizontal load of 713 lbs AF = (713 lbs)(24" + 16"*3/2) / 36" = 951 lbs Totalload on the beam clamp F = Wa + AF = 750 + 951 = 1701 lbs = 1700 lbs O.K.
Check 4:
Ductility Check:
This suppon may be considered as inherently ductile. Lateral loading leads to bending of the hanger rods. Therefore fatigue evaluation should be performed.
Support location:
Turbine Building, Elev. I19'-O" (ceiling)
Floor ZPA = 1.5* 1.25'0.lg = 0.19g < 0.50g (Ref. 5)
Where 0 lg is a ground ZPA (Ref. 4)
In accordance with Ref. 5, Fig. 8.13 the maximum acceptable weight for 5/8" threaded rod with the length greater than 17" for 0.50g is 3700 lbs > total dead weight Wror = 1500 lbs O.K Check 5:
Limit State Check I
flastic hinge moment capacity of threaded rod can be determined based on Ref.12.
Using root section, apparent yield of 90 ksi, and shape factor of 1.7:
Sroot = n(0.514)'/32 = 0.0133 in' Mp = (90000 psi)(1.7)( 0.0133 in') = 2035 in lbs P*36 = W or*18 + 14Mp T
P = (1500* 18 + 14*2035) / 36 = 1541 lbs < Allouble = 1700 lbs O.K.
==
Conclusion:==
support is seismically rugged.
)
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal River Unit 3 Sheet.S. cf _fi3.
FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION ff7-Ofy/
4.7.2 Sample 2 Typical support conDguration is shown on the FPC CR3 drawing E 214121 Rev.10 (Ref. 28).
68" L6x4x3/8 t
3
..d q s
l' L4x4x5/16 4-l M
26" 9
%. & M FE 821
Oso,$d ANALYSIS / CALCULATION powe CONTINUATION SHEET
" - - ~
Crystal River Unit 3 Sheet.17.ofIL FI'C CRYSTAL RIVER UNIT 3. IPEEE EVALUATION f8)7-OfyL Fer review of the support configuration the checks to be performed are D.L., vertical and duhtility evaluation.
Determine weights:
Tributary span for trays is 10 ft (3) 24" trays (3)(50 lbs/ft)(10 ft) = 1500 lbs Top tray has a Oreproof insulation which may be conservatively assumed to have the same unit weight by itself as the cable in the tray it covers =
(50 lbs/ft)(10 ft) = 500 lbs Weight of the support structure:
use 350 lbs Total weight: W = 1500 + $00 + 350 = 2350 lbs.
T Check 1:
Dead Load Desien Check Trapeze cross members (L4x4x5/16. L = 30 in, S = 1.29 in' ).
Top tray uniform load w = 2*500 lbs / 30 in = 33.3 lbs/in Section modulus S = 1.29 in' (Ref.10)
M = (w)(L') / 8 = (33.3)(30)2 / 8 = 3746 ih lbs Bending stresses 3
f3 = M/S = 3746 in lbs /1.29 in = 2904 psi Increase stresses by factor vf 1.25 to accaunt for bending about principal axes:
fb = 2904* 1.25 = 3630 psi < Fu = 21600 psi O.K.
Check 3/16 fillet weld between trapeze cross and vertical post Load at each end of the cross member F = 2*500 /2 = 500 lb3 Weld area A = 2x 3 = 6 (Ref. 28) f = F/A = 500/6 = 83 lbs/in Fall = (.707)(21000)(3/16) = 2784 lbs/in > 83 lbs/in O.K.
Check 3/16 fillet weld between vertical post (L4x4x5/16) and L6x4x3/8 attached to the ceiling.
Load per each vertical member F = 2350 /2 = 1175 lbs m
om
~
Sg ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet.53 ofJ1 FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION ff7 Ofyg Weld area A = 2x 3 + 4 = 10 in (Ref. 28) f = F/A = 1175/10 = 118 lbs/in Fall = (.707)(21000)(3/16) = 2784 lbs/in > 118 lbs/in O.K.
Check 3/4" Phillips Red Head Self Drilling Concrete Anchors (Ref. 28)
Tension on bolt including load eccentricity P = 4"x F / 2" = 4xi l75/2 = 2350 lbs d'
^
/
e-
+
3/16 \\
l
/
F 3/16 3/16
[
-%l Nominal Capacity for 3/4"$ Phillips Red Head Anchor Bolts with 3.25" minim 'm embedment ( GIP, Ref. 5, with safety factor of 3).
P.. = 4690 lbs V.. = 5480 lbs In accordance with Appendix 0 of EPRI NP-6041-SL (Ref.1), the safety factor of 2.4 may be used for tension and 2.0 for shear. Therefore:
P.. = (4690)(3/2.4) = 5863 lbs V.. = (5480)(3/2) = 8220 lbs l
a *=
l
,x m
9"g ANALYSIS / CALCULATION CONTINUATION SHEET Crystal niver Unit 3 Sheet,19. of,$1 FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION f f 7-p34,.
Concrete Strencth Check I' =3,000 psi RF = 3000/4000 = 0.75 p
RF = 3000/10000 + 0.65 = 0.95 4
Deck for Anchor type RT, = RT. = 1.0 Check for Tichtness
/
Ri = R1, = 1.0 p
Check for Embehent Leneth RL =RL,=1.0 p
Deck for Spacine Spacing > S,nin = 7.5" RS, = RS, = 1.0 Check for Edce Distance The edge distance E > EhuN =7.5" RE = RE. = 1.0 p
Pm = (0.75)(5863) = 4398 lbs Ve = (0.95)(8220) = 7809 lbs Interaction Ratip; P/Pm = 2350/4395 = 0.53 <
l.0 O.K.
Therefore support meets dead load design check.
w.*
om
@g ANALYSIS / CALCULATION i
CONTINUATION SHEET Crystal River Unit 3 Sheet.fif)_ of.51 FPC CRYSTAL RIVER UNIT 3 IPEEE EVALUATION fp7 07PL Geck 2:
3 X D L Venical Canacity Check:
Load pei cach veni memer F = 3 x D.L. = 3
- 5) = 3525 lbs Check 3/16 fillet weld between vertical member (L4x4x5/16) and L6x4x3/8 attached to the ceiling.
Weld area A = 2x 3 = 10 in f = F/A = 3525/10 = 353 lbs/in Fall = ( 707)(21000)(3/16) = 2784 lbs/in > 353 lbs/in O.K.
i Check the overhead connections for 3.0 Dead Load 3 X D.L. = 3(1175) = 3525 lbs Allowable tension for 3/4" $ Phillips Red Head Anchor Bolts: Psi = 4395 lbs P/P.n = 3525/4398 = 0.8 < l.0 0.K.
The support meets venical capacity check.
Check 3:
Ductility Chec_!g Due to support anchorage details (eccentric loading on the supporting angle) the SRT judged that the raceway suppon is ductile.
Also, the combined weld throat thickness
= (0.707)(3/16")(3 sides) = 0.398" is larger than the angle thickness of 0.313" as shown in Fig. 4 2 of Ref. (12).
==
Conclusion:==
support is seismically rugged.
Ass 6 M 9E 421
~
ANALYSIS / CALCULATION g
CONTINUATION SHEET Crystal River Unit 3 Sheet.6.1.of.fl FPC CRYSTAL RIVER UNIT 3. IPEEE EVALUATION f f / Ofy2,
- 5. Conclusion The components selected for the IPEEE program are scismically adequate and have the following ifCLPF values.
Section No.
Description liCLPF 4.1 Condensate Storage tank (CDT-1) 0.12g 4.2.
Chilled Water Expansion Tank (CliT-1) 0.25g
~
4.3 RCP Seal return Cooler (MUliE 2A & 2B) 0.15g 4.4.
Nuclear Service Closed Cycle Surge Tank 0.14g
/
(SWT-1) 4.5 Spent Fuel licat Exchanger (SFllE 1 A & IB) 0.103g 4.6 Regulating Transformer A & B (VBTR 1 A 0.45
& IB) 4.7 Raceway Support LAR A Limited Analytical Review (LAR) was performed for the raceway supports.
t h
NN k..
9g ANALYSIS / CALCULATION CONTINUATION SHEET Crvstal River Unit 3 Sheet.fi2 of13.
FPC CRYSTAL RIVER UNIT 3, IPEEE EVALUATION ff 7-OM/,
62 References 1.
EPRI NP-6041 SL; A Methodology for Assessment of Nuclear Power Plant Seismic Margin (Revision 1).
2.
EPRI NP 5228 SL; Seismic Verification of Nuclear Plant Equipment Anchorage (Revision 1). Volume 1:
Development of Anchorage Guidelines.
3.
FPC-CR3 Document No.: S 94-0011: Rev. 0; Seismic Verification of Tanks. SQUG Methodology, 4.
FPC CR3 Seismic Evaluation Report for USI A-46, Rev. 0 5.
Generic Implernentation Procedure (GIP) for Seismic Verification of Nuclear Plant Equipment, Rev. 2, Corrected 2/14/92 i
6.
FPC CR3, Seismic Qualification Data, Section 5.0, Rev. 8 August 1995.
7.
FPC CR3, Dwg. SC-405-045, Rev. 2; Roof Slab at Elev.186'-10" Plans, Sections, Details.
8.
FPC-CR3; Dwg.:D 12-4, Rev. 3; Condensate Storage Tank.
9.
ANSI /ACI 349 90 Code Requirements for Nuclear Safety Related Concrete Structures.
- 10. AISC Manual of Steel Construction, Allowable Stress Design Ninth Edition.
- 11. Roark's Formulas for Stress & Strain, Sixth Edition.
- 12. EPRI NP-7151-D, Project SQ01 1 Final Report, March 1991, Cable Tray and Conduit System Seismic Evaluation Guidelines.
- 13. ITT Grinnell Catalog PH83.
- 14. FPC CR3 Dwg.: 50A01 A10192, Rev.4 (B ASCO Inc.); MUHE-2B Layout.
- 15. Edward G. Nawy; Reinforced Concrete A Fundamental Approach, Third Edition, Prentice Hall 1995
- 16. ASME Boiler & Pressure Vessel Code: Section III Appendices; 1993
- 17. E, N. Gaylord and C. N. Gaylord; Structural Engineering Handbook; Third Edition
- 18. FPC Dwg.: SC-122-042; Rev.10; Aux. Bldg. Equipment Foundations (El. 95'-0"), Anchorage details of S W T 1.
- 19. FPC Dwg.: SC-422-026; Floor Layout showing SWT-1 Location.
- 20. FTC Dwg.: 5 315 D1 Rev. 3: Closed Cycle Nuclear Service Surge Tank (SWT-1) Layout and Details.
- 22. FPC-CR3 File No.: 324 HS, VBTR 1 A,lB Layout.
- 23. FPC CR3 Calc. No.: S 96-0013 Rev. 0; Qualification of Tanks per U.S.I. A-46.
As. bla FM W
Og ANALYSIS / CALCULATION CONTINUATION SHEET Crystal River Unit 3 Sheet.fd.of.53.
FPC CRYSTAL RIVER UNIT 3 IPEEE EVALUATION f 7 7-Of yg,
- 24. FPC CR3 Spent Fuel Heat Exchanger SFHEl A & 1D Setting Plan; BADCOCK & WILCOX Dwg.
No.:620-0007: Rev. S.
- 25. FPC-CR3 Dwg. No.: PI 305 822 Rev. 3: Piping Isometrics From Pump 3A to Cooler 3A and Pump 3B to Cooler 3B.
- 26. FPC-CR3 Dwg. No.: PI 305 948. Rev. O, Piping Isometrics.
- 27. Unistrut General Engineering Cctalog No.12
- 28. FPC CR3 Dwg. E 214 121, Rev.10. Electrical Cable Trays Hanger Details.
.e
/
eac w
A'ITACHMENT E FLORIDA POWER CORPORATION CRYSTAL RIVER UNIT 3 DOCKET NUMBER 50-302/ LICENSE NUMBER DPR-72 Screening Evaluation Work Sheets (SEWS) for RCPM-3A and RCPM-3B i
l STATUS:
Y S.CEEI,NING L' VALUATION WORK SHEET (SEWS)
- Equip. ID No.
- RCM-J B Equip. Class: 21 Instruments and control Panels and cabin 113 i
Equip:ent
Description:
REACTOR COOLANT PUNP POWER MONITORING CABINET B Loc 0 tion: Bldg. CONTROL Floor El. Jos Room, Row / Col 303/I Manufccturer, Model, etc (op:ional) Rashester Instrument systems Inc.
CAVEATS - BOUNDING SPECTRUM (Identify with an asterisk (*) those caveats which-aro cet by intent without meeting the spn:ific wording of the caveat rule and explain the reason for this conclusion in the COMMENTS section below) 1.
Equipment is included in earthquake experience equi, ment class?
Y 2.
Steel frame and sheet metal structurally adequate.'
Y 3.
Adjacent cabinets or panels which are close enough to impact, or N/A COctions of multi-bay cabinets or panels, are bolted together if they contain relays?
4.
Drawers and equipment on slides restrained from falling out?
N/A 1
5.
Attached lines have adequate flexibility?
Y 5.
Rolays properly installed and supported?
Y 7.
Anchorage adequate (see checklist below for details)?
Y 8.
Have you looked for and found no other adverse concerns?
Y Is tho intent of all the caveats met for Bounding Spectrum?
Y
\\NCHORAGE
- 1. _B sed on walkdown inspection and judgement, is the installation and factors effecting anchorage capacity adequate?
Y t
2.
Have you looked for and found no other adverse concerns?
Y Arc anchorage requirements met?
Y INTERACTION EFFECTS k/[M
?
luated by:
A a Date:
Ivelunted by:
/
Dato:
/
534 Rev.: 1
/
t w
ws-g
-e w-to
-g~t-w-t vr= s w-
~e-nre-
>vr m-m--i
~- -
- ~
,s STATUS:
Y SCREENING EVALUATION WORK SHEET (SEWS)
,Eauip. ID No.: RCPM-3B Equip. Class: 11 Instruments and control Pancis 3Ad Cabinets r
1.
Soft targets free from impact by nearby equipment or structures?
Y' 2.
If equipment contains relays, equipment free from all impact by Y
ncarby equipment or structures?
3.
Attached lines have adequate flexibility?
Y l
4.
Overhead equipment or distribution systems are not likely to Y
collapse?
- 5. ' Hove you looked for and found no other adverse concerns?
Y 10 cquipment free from interaction effects?
Y I
IS EQUIPMENT SEISMICALLY ADEQUATE?
Y CQMMENTS:
C51 "U"
Access not permitted by operations.
Walkdown deferred to Cutage.
1 RESOLUTION:
Walkdown with access performed on 3/12/96.
Equipment cvaluation revised to seismically adequate.
PHOTOS 1 Rol1# 42 Photo # 4 Roll #
Photo #
Roll #
Photo #
Rol1#
Photo #
Roll #
Photo #
Roll #
Photo #
I lunted by:.
)
M
, Date:
I[
fh-
/
Date:
/Y I Evaluated by:
' ' /
534 Rev.: 1 w-
-v-r y
y
t STATUS Y
SCREENING EVALUATION WORK SHEET (SEWSt coulp. ID No.: RC.tK-3A Equip. Class: 21 I.DA.truments and Control Panels and Cabine1A i
Equip;ent
Description:
REACTOR COOLANT PUNP POWER MONITORING CABINET A Locotion Bldg. CONTROL _._ Floor El. 108 Room, Row / Col 303/I M nufseturer, Model, etc (optional) Rgghtster Instrumant Systems Inc.
~ CAVEATS - BOUNDING SPECTRUM (Identify with an asterisk (*) those caveats which aro not by intent without meeting the specific wording of the caveat rule and explcin the reason for this conclusion in the COMMENTS section below) 1.
Equipment is included in earthquake experience equipment class?
Y 2.
Steel frame and sheet metal structurally adequate?
Y
' 3.-
Adjacent cabinets or panels which are close enough to impact, or N/A osctions of multi-bay cabinets o panels, arn bolted together if they contain relays?
4.
Drawers and equipment on slides restrained from falling out?
N/A 5
Attached lines have adequate flexibility?
Y 6.
Rslays properly installed and supported?
Y 7.
Anchorage adequate (see checklist below for details)?
Y 8.
Have you looked for and found no other adverse concerns?
Y In the intent of all the caveats met for Bounding Spectrum?
Y SECHORAGE 1.
Based on walkdown inspection and judgement, is the installation and factors 4
affecting anchorage capacity adequate?
Y 2.
Have you looked for and found no other adverse concerns?
Y Ara anchorage requirements met?
Y INTERACTION EFFECTS i
lunted byt it Date:
Nbb Evolusted byt
/W Date:
q L533L
~Rev.
1-f/
j l
.~.
-__m.,-:-_-#
m
.a m_:
a a
a, d
%6.
..,m.
-.2.a
_.A m
m w.
_a_
2.
n._
STATUS:
Y SCREENING EVALUATION WORK SifEET (SEWS)
Eauip. ID Ho.: RJPM-3A Equip. Class: 11 Instruments and control Pantig and cabinets
" 1.
Soft targets free from impact by nearby equipment or structures?
Y 2.
If equipment contains relays, equipment free from all impact by Y
[
naaroy equipment or structures?
3.
Attached lines have adnquate flexibility?
Y 4.
Overhead equipment or distribution sy' stems are not likely to Y
collapse?
5.
H:ve you looked for and found no other adverse concerns?
Y Io cquip snt free from interaction effects?
Y IS EQUIPMENT SEISMICALLY ADEQUATE?
Y COMMENTS:
B81:
"Un Access not permitted by operations.
Walkdown deferred to cutage.
RESOLUTION:
Walkdown with access performed on 3/13/96.
Equipment cvaluation revised to seismically adequate.
I2s clearance to wall approximately 1/4".
Conduit on top of cabinet will restrain the cabinet from hitting the wall.
B87:
Only 3 of 4 anchor bolts could be seen.
The fourth is covered by cable.
Judged acceptable.
PilOTOS :
Roll # 42 Photo # 4 Roll #
Photo #
Rol1#
Photo #
Roll #
Photo #
Roll #
Photo #
Roll #
Photo #
/bh.
1 lusted by:
Date:
4, 5, 'V/f[
Evolunted by:
Date:
533 Rev.: 1 g/
ATTACHMENT F 4
FLORIDA POWER CORPORATION CRYSTAL RTVPR C NIT 3 DOCKET NUMBER 50-302/LichhE NUMBER DPR.-72 DESIGN BASIS TANK CM..CULATIONS FOR:
The Pressurizer Quench Tank / Tag Number WDT-5)
Reactor Coolant Bleed Tanks (Tag Numbers WDT-3A, 3B, 3C)
Dedicated Emergency Feedwater Tank (Tag Number EFT-2)
.