Text

Rev 1 to S-96-0013, Qualification of Tanks Per USI A-46
	ML20203J370
Person / Time
Site:	Crystal River
Issue date:	12/04/1997
From:	Mary Thompson; FLORIDA POWER CORP.
To:	;
Shared Package
ML20203J289	List: 3F1297-24, Provides Supplemental Info to Support Util Response to NRC Request for Addl Info,Dtd 970128.Clarification to Util RAI Responses,Discussed During NRC Audit USI A-46 & Calculations Encl also.W/10 Oversize Drawings; ML20203J303; ML20203J315; ML20203J370; ML20203J385; ML20203J400; ML20203J417; ML20266D724; ML20266D729; ML20266D734; ML20266D740; ML20266D744; ML20266D748; ML20266D752; ML20266D758;
References
	REF-GTECI-A-46, REF-GTECI-SC, TASK-A-46, TASK-OR S-96-0013, S-96-0013-R01, S-96-13, S-96-13-R1, NUDOCS 9712190195
	Download: ML20203J370 (151)
	v • d • e

{{#Wiki_filter:__ i. ~ INTEROFFICE CORRESPONDENCE u e r A<-u,~ { Nuclear Engineering NOTD 240 1634 OH4c e MAC Telephc.nc I ( SututCT: Crystal River Unit 3 l Quality Document Transmittal Analysis / Calculation t i i 1 10; Records Management NR2A L The followmg analysis / calculation package is submitted as the OA Record copy: i DOCND itt'C DDcVMtNT #Dff(fif eCat'ON MJMat41 Alv $vsflM l! t ot At P4Gts f atANSuf11lD i S 96 0013 1 See Attached Ob 3 l titti j Oualification of Tanks per U.S.I. A 40 l l I e wns iiot Ntre r a t ewcaos toa tat tn etta tvau SOUG, Tank, Seismic i Damt8 tatitNINCtl DA Pillt List PaiMaar fitt tensta i S 91 0003 1 I l I % fND evt4 Don NAMil l vlNDom boCUMtvt NUMatt iDAht's l DVPinstDED DOCUMENTS IDaht*1 FPC lN/A I N/A l SWHE 1 A l WDT 3A l C.DHE 4 A f SWHE 1B l WDT 38 l CDHE-40 SWHE 1C l WDT 3C l SWHE 10 fWDTS l PART NO. i i l couwtms iusAct atsf a.crioNt eno.a,f,Anv, tic, l 1, i i NOTE: Uso Tag number only for vahd tag numbers (i.e.. RC%S, SWV 34. DCH 99), otherwar.c: use Part number i field (i.s.. CSC14599. AC1459). If more space is required, write *See Attachment" and list on separate sheet. Dt tPGN ING'Nt ta oat t ytMsC Ato% (NO)httm DAtt eG Daft ,!MAik'04t# Pud IM,. ) if /jM /t.07 97 j k. cc: Nuclear Protects (if MAR /COWR/PEERE Calculation Review form Part lli actions required Oves @ No Return to Service Related! O voi e no ' (* Supervisor Config. Mgt, tr.to. itt ves. send cony oi ihe sorrn to Nucica, Reguiatory Assurance and a Mgr.. Nuct. Operations Eng. (Original) w/ attach copy of the Caiculation to the Responsible Organisation(s) identified in Part lit on the Calculation Review totrn ) i~ 3 4712190195 971216 [ ADOCK0500g2 n.. PDR P

9.......... ANALYSIS / CALCULATION

SUMMARY

. c suo eau D850PtINt CONTROL 60 MWi& ION ttyIL 00cuMINT 10tNTinCAfl0N NUMatR S 96 0013 1 ttf t L CL At$:FeC At 0N (CM(Of Q*4t 1 @ Safety Related Qualification of Tanks per U.S.I. A 46 O Non $sfety nelated uany cownettatvsvete N/A VtN004 00CINt4t Nvvein N/A APPROVAL PRINTED SIGNATURES NAME Design Engineer /7/- M s Mark Thornpson D*'* Illih7 Verihcation Engineer g[ yg/4 g_h Bryan K. Henshaw Date 91 c. m., Supervisor Q ( D. L. Jopling ~ 4 cM / o:ie 11 if LM$ ngy!bep f RenInnna all ringna in attnehment M, Alan revinnd Pagen 7 and 1 ,,julachrpeta P_ Q. A _ A R wntn added in ententatinn. euarv6s 6vuvaar The ouroote of this revision is to annivre selected tanks and condenst(s and ensure the seismic,ggpacity is .grantnr than thn enltmic demand Thn a:nnnndary,nutnnen of thief alenlatinn it in mnkn n enrrnntinn in ihm _wgight nf PAWiEd A,.1 B

10. and.1D-t a wt16 $vuu*a,

.JMapks and condensers analvred in this calculation are seisimicaffv austified for tiesian basis loadina at CR 3. Thn inerensadaginht inr SWHF 1 A 1B_.10. and 1D han nnt nffnetnet nriginal cnnetutinna nf R98-0013 Ros. 0 whi.ah detCIDMnd that the Nent Frchangnrn wntn naimmiently nonlifinifnr CR.3 deeign hatin innding. e T-

~ CALCULATION REVIEW c w ev m Page 1 of 2 meaatoev7msv S 96 0013, Rev.1 PARTI. DESIGN ASSUMPTl0N1NPUT REVIEW: APPLICABLE O Yes @ No The following or0anizations have reviewed and concur with the design assumptions and inputs identified for this calculation: Nuclear Plant Technical Support System Engr s,-w. um Nuclear Plant Operations e,,,, s,*w.om s,w.um i.e.w.om ^ MRT 11 RESULTS REVIEW: APPLICABLE O Yes @ No The following organitations have reviewei and concur with the results of this calculation and understand the actions which the organl ations rnust take to implement the results, Nuclear Plant Technical Support System Engr s, w.cn Nuclear Plant Operations s,n.w.om Nuclear Plant Maintenance sv.w.vm y yg Nuclear Licensed Operator Training *'*~ O ves O N/A Manager, Site Nuclect Services O Yes C NM Sr. Radiation Protection Engineer O Yes O N/A OTHERS: 54,.w. vm $9n.w. D.i. h s11

noride ,.. m... e..r. CALCULATION REVIEW Page 2 of 2 c atCaat:08eho Atv S 96 0013, Rey,1 PART lil. CONFIGURATION CONTROL: APPLICABLE O Yes @ No ) The following is a list of Plant procedures / lesson plans /other documents and Nuclear Engineering calculations which require updating based on calculation results review: Document pate Agqvirad Responsible Ornanimig M d4 Upon completion, forward a c9py to the Manager, Nuclear Regulatory Assurance Group for tracking of actions if cny items are Identified in Part lil, if calculations are listed, a copy shall be tent to the original file and'the c"Iculation log updated to reflect this impact. PART IV - NUCL2AR ENGINEERING DOCUMENTATION REVIEW The responsible Design Engineer must thoroughl) review the below liste(8 documents to assess if the calculation requires revision to these documents. 'l "Yes," the change authorizations must be listed below and issued concurrently with the calculation, inhanced Design Basis Document O ves @ No "C'l Vendor Quahfication Package fv0i FSAR O ves @ No "'"* Topical Det6gn Basis Doc. O ves SNo FCai improved Tech. $pecification O s es @ No "*"* EISOPM O ves SNo PCa! Improved Tech. Spec. Bases C vos @ No "'"*'8 Dther Documents reviewed; Config. Mgmt. Info. System O ves @ No scic*a O ves O No Anatysis Basis Document O ves @ No "C O ves O No tea.AssOn Dog Melm=cgn De: ton Dosis Document O vos @ No "c O ves O No comespec ape==ui i Appendia R Fere Studv O ves @ No "c O ves O No ecw=os poc nosae=cs. Firs Hetardous Analysis O vei S No "c O ves O No icwww wc mei==ce, NrPA Code Conformare Docement O vos @ No "c* O ves O No ic==a poc met==cti PART V PLANT REVIEWS / APPROVALS FOR INSTRUMENT SETPolNT CHANGE PRC/DNPO approvalis required if a setpoint is to be physically changed in the plant through the NEP 213 i process. PRC Review Required O Yes @ No l DNPO Review Required O Yes @ No l DNPO /Date mmwiewiwan m ev.ua.e.ono wt I hef fIS i "Jr"M Whh MW N* *#Ng*th,0 ] MArcx TveuproA/ l een s 91 e

( )g CALCULATION VERIFICATION REPORT / Crystal Ftiver Unit 3 L/ cavinne anu Page 1 of I w ros. w un S 96-0013, Revision 1 PROACIMN[ '" Ouahfication of Tanks per U.S.I. A-46 YES NO WIA

1. M O

O Are inputs inuuding codes, standardF, regulatory requirements, procedures, data, and Engineering methodology correctly selected and apphed? 2, 3 O O Have assumptions been identified? Are they reasonab!e and justified? (See NEP 101, V.c., for discussion on assumptions and justificationd

3. E O

O Are referer.ves properly identified, correct, mnd complete? (See NEP 101, V.c, for discussion on referencesh

4. S O

O Have applicable construction and operating experiences been considered?

5. K 0

0 Was an appropriate Design Analysis / Calculation method used?

6. O O

E fn cases where computu softwain was used. has tae program been verified or reverified in accordance with NEP 135 for safety related design applications andtor are inputs, formulas, and outputs associated with spreatisheets accurate?

7. S O

O is the output reasonable comrared to inputs?

8. O O

B Has technical desion information provioed via letter, REA, IOC or telecon by other disciplines or programs been verified by that discipline orbro'gra'm'? 98 O O Has technical design information p*ovided via letter or telecon from an external Engineering Organization or vendor been confirmed and accepted by FPC7

10. O S

O Do the calculation results indicato a non conforming condition exists? If 'Yes,* immediately notify the rtst ansible Supervisor.

11. O 2

O Do the results require a changc to other Engineering documents? If "Yes," have these documents been identified for revision on the Calculation Review Form? I hive performed a verification on the subject calculatioa package and find the results acceptable, viancatoi sboa rm w spa. e a tenan oats new 3"J

9, Florida INTEROFFICE CORRESPONDENCE Nuclear Engineenno Desion (NED) NA1E 3581 Oftce MAC T.. phone

SUBJECT:

Crystal River Unit 3 Ouahty Document Transmittal. Analysis /Calctilation, Page 1 of 2 TO. Recorris Management. NR2A The following anaysis/ calculation package it submitted as the OA Record copy: (OCNO (rPC DOCLfMENT 4DE Nip Catch nth 0t h) Riv SY5f t Mt3) fotAL PAGES dANSMitTED S-06-0013 0 See Attached 4) flTLt Qualification of Tanks per U.S.I. A 46 ' ' Gs cotNirneywonoi ron utta attaicvAu SQUG, Tank DAMtf lR(f(Mt NCf 8 oA FILE 8 087 PRNARY flLE FiRST) SP-83-033 vt NV qvt.NuvH hAML) VLNDUM DUi VMLhl NUMULH 4 DANT.fi DVVLN6LDLD MUMLhl4 tbAktf J Programmatic Solutions, Inc None n/a N

*511 mmmmmmamanduim! U TAG l

See Attached l l l l n \\1 \\1 l l l u_, ~,. ~~ m.,-, _o... o c, This calculation provides seismic qualification of tanks per the "SQUG" guidelines. This calculation was done by Programmatic Solutions, Inc. under Contract N00931AA. NOTE: Use Tag nuniber only for valid tag numbers (i e, RCV-8, SVN 34, DCH 99). otherwise; use Part number fiela (i e., CSC14599. AC1459).)f rnore space is required, write "See Attachment" and Qt op sepafate sheet. M t Datt ptw n -- DAtk e 8h-tNdb,r /Iki vtweAtou t wGo ttrb.d'i, vet $s.cnhiTl(lthMk12kl% ~ cc: MAR Ofr,ce (if MAR Related) 0 Yes E No Plant Document Updates Required 0 Yes 5 No (if Yes send copy of th'> Mgr. Nuct Confg Mgt Calculaton Review form to Nuclear Lk:ensing and a copy of the Calculaton to Mgr Nuct Eng. Desgn the ResponsNe Orgavaton(s)Identfed in Puft lli on the Cakulaton Revew form) 4 (Orginal) w/ attach A/E N&dA**//// INtT#dt,/(E Yes 0 No [/M (4M) (if yes, Transmit w/ attach) l Mew 499 RtT; Lde of Pla1 RESP: Nucie.# (mgewemg

9' Florida INTEROFFICE CORRESPONDENCE Nuclear Encineering Design (NED) NA1E 3581 e Oke MAC Tekphnne suaxet: Crystal River Unit 3 Quahty Document Transmittal. Analysis / Calculation, Sheet 2 of 2 To Records Management - NR2A DOCNO (FPC DOCUMENT IDENTIFICATION NUMBER): S 96-0013, REV.O Systems: Tag Nunibers: CA CAT 5A CH CAT 5B DC CHT1 DH. DCT 1 A DL DCT1B EG DHHE 1A lA DHHE 1B MS DLHE 1A MU DLeiE 1B SF DLHE 2A SW DLHE 2B WD EGT 1 A EGT 1B EGT 2A '~ EGT 28 1ADR 1 lAT 1 A lAT 1B MSV411 AR1 MSV 411 AR2 MSV 411 AR3 MSV-412 AR1 MSV 412-AR2 MSV 412 AR3 MSV 413 AR1 MSV 413 AR2 MSV-413 AR3 MSV 414-AR1 MSV 414 AR2 MSV-414-AR3 MUHE 2A MUHE 2B MUT1 SFDM 1 SWtlE 1 A SWHE 10 SWHE 1C SWHE1D SWT1 WDT 1 A WDT1B WDT 1C no s*s ut; t,. e eu use: w, ce; =,

sorida DESEN ANALYSIS / CALCULATION

[rygel Crystal River Unit 3 Page f of ) Wlut eutuv caro 4 acs St v se,% S96 0013 0 1, PURPOSE The purpose of this calculation is to evaluate the anchorage adequacy under seismic loading of those 1: ems of equipment identified as Class 21 " Tanks and IIcat Exchangers" according to the SQUG GIP (Reference 1). The. methodology of Section 7 of the PSP for Seismic Verincation of Nuclear Power Plant Equipment (Reference 2) was used where applicable (i.e., For Oat bottomed vertical tanks or for horizontal tanks / heat exchangers supponed on saddles), For equipment items not meeting the intent of the PSP Section 7 methodology (for example, small vertical tanks supr',rted en legs), an evaluation of the anchorage is performed using extremely conservative values to assure anc!.orage adequacy. 2. DESIGN INPUTS Design input values were obtained from u,e vendor equipment drawings, foundation drawings and anchorage drawings referenced in the individual evaluations (Attachments A thiough Appendix 0). Acceleration values used to define the seismic demand for each specific item evaluated were obtained from Section 5.0 of the E/SQPM (Refercnce 3). 3. ASSUMPTIONS In any instance where required information was not available (such as dimensions, anchor bolt type, etc.), appropriate conservative assumptions were made and are documented in the individual evaluations. Foi example, if required dimensions were not available, field measurements may have been obtained and these data were referenced when used; or if the anchor bolt type was unknown, a conservative anchor bolt dimension, type, and GIP reduction factor would be documented and used in the evaluation. 4. REFERENCES (1) Generic Implementation Procedure (GIP) for Seismic Verification of n : lear Power Plant Equipment, Revision 2, SQUG, Febmary 1992. (2) Florida Power Corporation Plant Specific Procedure for Seismic Verification of Nuclear Power Plant Equipment, Revision 0, a m.v.- - - w y

lorida DESIGN ANALYSIS / CALCULATION

$9uM Crys %I River Unit 3 Page.h of ) we,.i u wu.u,, u a vi..u,, S96ON3 0 (3) Florida Power Corporation "Envirotunental and Seismic Qualification Program hianual (E/SQPhi)", Rev. 8. Section 5.0, Seismic Qualification Data. 5.0 TANK AND IIEAT EXCllANGER cal.CULATIONS Individual anchorage adequacy evaluations of the following Class 21 items of equipment were performed and are documented in Attachments A through 0, respectively, of this calculation: Boric Acid Storage Tanks.......... CAT 5A, CAT 5B Chilled Water Expansion Tank............................... CllT 1 Decay lleat Closed Cycle Surge Tank.................. DCT-1 A, DCT-1B Decay IIcat Removal IIcat Exchangers........... .... DilllE-1 A, DilllE 1B Emergency Diesel Generator Lube Oil Cool @LilE 1 A, DLilE-1B, DLIIE-A, DLilE 2B Emergency Diesel Generator Air Receivers... EGT-1 A, EGT-1B, EGT-2A, EGT 2B instrument Air Dryer ...................................IADR1 Instmment Air Receivers........................... I AT-1 A, IAT-1B hiain Steam Valve Air Reservoirs........ htSV-411 AR1 Through htSV-414-AR3 RCP Seal Return Coolers... .................... h1UllE 2A, hiUllE 2B hi a ke Up Ta nk.................................. h1UT-1 Spent Fuel Coolant Demineralizer.............. SFDht 1 Nuclear Service CCC lleat Exchangers. SWilE.l A, SWilE 1B, SWilE-lC, SWilE-lD Nuclear Service Closed Cycle Surge Tank..... SWT-1 Waste Gas Decay Tanks.................... WDT 1 A, WDT-1B, WDT-lC

5.0 CONCLUSION

S It was determined that the anchorage for all of the equipment presented in the individual anchorage evaluations included in Attachments A thorough O are adequate. For the equipment where the methodology of Section 7 of the PSP for Seismic Verification of Nuclear Power Plant Equipment (Reference 2) was applied, all requirements of this procedure were met. l For equipment for which simplified analysis methods were used to evaluate the anchorage it was found in all cases that significant margins remained between calculated seismic demand and anchorage capacity even though various conservative assumptions j were used. For example, seisnue accelerations were overestimated because 2% damping { curves were used when 4% damping should be applied (twcause 4% curves were not available), i c., ,, v.w ~ n n. n e m

lorid, DESIGN ANALYSIS / CALCULATION ME Cystal River Unit 3 Page k of3 m,.mm w,-

S96 0013 A (3) Florida Power Corporation "Enviromnental and Seismic Qualification Program blanual (E/SQPht)", Rev. 8, Section 5.0, Seismic Quanfication Data. 5.0 TANK AND llEAT EXCilANGER CALCULATIONS Individual anchorage adequacy evaluations of the following Class 21 items of(quipment were performed and are documented in Attachments A throu.1d, respectively, of this l calculation: Boric Acid Storage Tanks. CAT 5A, CAT 5B Chilled Water Expansion Tank............................... CllT 1 Decay Heat Closed Cycle Surge Tank.................. DCT-1 A, DCT 1B Decay Heat Removal lleat Exchangers................ DHHE-1 A, DHHE-1B Emergency Diesel Generator Lube Oil Cool &LilE-1 A, DLHE 1B, DLilE A, DI.HE 28 Emergency Diesel Generator Air Receiv:rs... EGT 1 A, EGT-1B, EGT-2A, EGT 2B i instrument Air Dryer ............. I ADR 1 Instrument Air Receivers.. I AT-1 A, IAT-1B hiain Steam Valve Air Reservoirs........ htSW411 AR1 Through htSV-414 AR3 RCP Seal Return Coolers........................ hiUllE-2 A, hiUHE-2B h i a k e Up Ta nk....................................... hi UT-1 Spent Fuel Coolant Demineralizer........................... SFDht1 Nuclear Service CCC Heat Exchangers. SWHE 1A, SWHE-1B, SWHE 1C, SWHE-1D Nucleir Service Closed Cycle Surge Tank....,.....,............ SWT1 Waste Gas Decay Tanks................... WDT 1 A, WDT-1B, WDT lC R CAc. TO R. C o ot. ANT

W L Esb "rANXS.. i......,,. W DT *3 A, WDT.*55, WDT *5 C N EA c. tor s eel ANT' DRAW TANM......,,.. W D T. G MANE.oNetNsen.

,.........,,,.,,c.ogc..qA,cayg.4g

5.0 CONCLUSION

S It was determined that the ancharage for all of the equipment presented in the individual anchorage evaluations included in Attachments A thorough O are adequate, For the equipment where the methodology of Section 7 of the PSP for Seismic Verification of Nuclear Power Plant Equipment (Reference 2) was applied, all requirements of this procedure were met. For equipment for which simplified analysis methods were used to evaluate the anchorage it was found in all cases that significant margins remained between calculated y seismic demand and anchorage capacity even though various conservative assumptions were used. For example, seismic accelerations were overestimated because 2% damping W curves were used when 4% damping should be applied (because 4% curves were not available). >M R& ], 683 W Vg5 RL 4.'. .A Gb & fp49'

@ u:aorida l DESlEN ANALYSISICALCULATION ' owe r. Cintal River Unit 3 em Page 3 of 3 kww a,.w m. a os,a S96 0013 7. ATTACllMENTS A. Boric Acid Storage Tank Six Pages 11. Chilled Water Expansion Tank... Five Pages C. Decay lleat Closed Cycle Surge Tank Six Pages D. Decay l{ eat Removal Heat.i changers................. Six Pages t E. Ernergency Diesmi Generator Lube Oil Coolers.......... Six Pages F, Emergency Diesel Generator Air Receivers Six Pages G. Instrument Air Dryer.................. Six Pages 11. Instrument Air P.eceivers Six Pages 1. Main Steam Valve Air Reservoirs.................. Five Pages J. RCP Seal Return Coolers........................ Six Pages K. M a ke Up Ta nk.............. Six Pages L. Spent Fuel Coelant Demineralizer................... Si Pages t M. Nuclear Service CCC licat Exchangers................ Six Pages N. Nuclear Service Closed Cycle Surge Tank.............. Six Pages O. Waste Gas Decay Tanks......................... Six Pages P. R EACTbM cootAVT-15LEED tanks 22 FAGES G. RCA cWR C o o LA NT' D R AIN 'rA NK.......... 10 PAGES R. MknN coNucNsens. ,,~g pggg$ S. sIXC CE P7' fro AA 'civm cNG/Ngottg4 g gyctggy, pgwgg c.gpgggggg g pqg, ML l. kd6 Y F ON R$ %F, N%K.59# k' $46#h

9 $aMSI

lorida DESIGN ANALYSISICALCULATION Crystal River Unit 3 I of b Page

_, m., m....e ~_ S96-0013 0 Attachment "A" Boric Acid Storage Tank Six Pages Total MLI.L4408Fei N&br NA tw kN w<=N _t

Cclerlati:n: S96-0013 rev, 0 FPC - Crystal River Unit 3 Seismic Verification of Tanks nev By cate I chk d ay i cai. Calculation For: l Horizontal Tank l _i i l Equip. ID: CAT-SA Equipment

Description:

Building: AUXlLIARY Elevation: 119 BORIC ' ACID STORAGE TANK A Rm Row / Col: 302 / O Also Apolicable for: I_ CAT-5B l Tank Drawing: M-6063 Rev. 2 Anch. Drw.: SC-422-010 and SC-422-043 Vendor: Babcock & Wilcox, Buffalo Tank Div. Model: Step 1: (1) Input Data See Figure 7-13 of Florida Power Plant Specific Procedure for ' Seismic Verificaticr: of Nuclear Plant Equipment", Rev. 1, 9/12/94 - At>plicable? Tank: Diameter (ft) D 9.00 OK Length (ft) L 17.08 OK Thickness of tank shell(in) 1 0.27 min. thick. cale. Weight of tank plus fluid (Ibf) W tt 73000.00 Weight density (Ibf/ft ^ 3) Gm 61.16 OK Height of c,g. above anchorage (ft) H eg 5.28 OK Saddle: Spacing (ft) S 9.92 OK Height of sadd!e plate from bottom of h 12.00 the tank to tm base plate (in) Shear modulus (psi) G 1.12E+ 07 Eastic modulus (psi) E 2.9.0E + 07 Number of Saddles Ns 2,00 OK Base Plate: Thickr.spirwc ? late under saddle (in) tb 0.75 Min. yield :.trenptn (psi) fy 30000.00 Thickness of leg of weld tw 0.25 Assumed Eccentricity from anchor bolt CL to es 2.70 Assumed the vertical saddle plate Bolts. . Number of locations, each saddle NL 2.00 OK l Number et anchor bolts per location Ne 2.00 OK i Diameter :sf anchorbolt (in) d 1.00 L Distance between extreme anchor D' 8.50 OK bolts in base plate of saddle (ft) Loading: SSE Floor reponse spectra at 4% damping CAT-5A Page 1 of 5

Cciculation S96-0013 rev. O FPC - Crystal River Unit 3 Seismic Verification of Tanks nev' ey oat. chk d ay I cate Pds 1/1 11 % _ o '7)HX ol9/9f Calculation For: 1 l Horizontal Tank l t I i i Step 2: (2) Anchor Bolt Allowables (From GIP Section 4.4 and Apperdix C) Allowables for 1.0" Cast-in-Place Bolts Pnom = 26.69 ksi Voom = 13.35 ksi RLp = 1.00 embedment red. factor RLs ' 1.00 RSp = 1.00 spacing red. factor RSs = 1.00 rep =' 1.00 edge distance red. factor REs = 1.00 RFp = 0.93 for fc=3000 psi concrete RFt: 0.93 = RCp = 1.00 cracked concrete red. fact. RCs = 1.00 Pu' = Pnom (RLp)(RSp)(rep)(RFp)(RCp) = Pnom (0.93) = 24.71 Kip Vu' = Vnom (RLs)(RSs)(REs)(RFs)(RCs) = Vaom (0.93) = 12.36 Kip Step 3: (3) Base Plate Bendina Strenath Reduction Factor (RB) RB = Bending strength reduction factor = (fy) (tb ^ 2) 0.23 (3) (Pu') Step 4: (4) Base Plate Weld Strenath Reduction Factor (RW) RW = Weld strength red. fact. = (tw) (es) (30600) (2.83) 2.36 Pu' E op 9 g3) 6.r.chor Tension and Shear Allowable Pu = (Pu') (smaller of RB, RW) = 5.63 Kip Vu = Shear allowable anchor load = (Vu') = 12.36 Kip Step 6: (6) Calculated Ratios Alp = (Pu') / (Vu') = 0.46 Wb = (Wtf) / [ (NS) * (NL) * (NB) ] = 9125.00 lbs Vu / Wb = 1.35 Heg / D' = 0.62 Heg / S = 0.53 F1 = SQRT [ (NS ^ 2) + 1 ) 2.24 = F2 = SQRT [ (NL ^ 2) * (Hcg / D') ^ 2 + (.667 ^ 2) + ((Heg/S) ^ 2) * ( (NS ^ 2) / (NS-1) ^ 2 ) ) = 1.77 CAT-5A Pace 2 of 5

C:lculatiouf S96-0013 rev. O FPC - Crystal River Unit 3 Seismic Verification of Tanks sev sy ' cate chk'e sy cate i 0 @ft aMMI \\lnt.3 L Calculation For: I l Hor;zontal Tank L I I Step 7: (7) Determ!ne Acceleration Capecity of Tank Anchorage Llow = [ (Vu) / (Wb) ) * [(1) /(F1)) 0.61 g = Lup [ (Vu) / (Wb) + (0.7) / (Alp) ) 0.58 g = = [ (0.7) / (Alp) J * (F2) + (F1) Lamb = Smaller of Llow or Lup 0.58 g = Step 8: . (8) is Tank / Heat Exchanaer Riaid or Flexible in Transverse or Vertical? Sc = From Figure 7-14 for Tanks = 20.00 ft ( Use D = 9.000 ft ) ( Use t = 0.269 in ) is Tank / Heat Exchanger Rigid or Flexib'e ? Rigid if Sc >or= S, Flexible if Se < S From Step 1, S = 9.917 ft Tank in Transverse or Vertical Direction is Rigid Step g: (9) is Tank / Heat Exchanaer Riaid or Flexible in Lonoitudinal Direction? Flong. = [ (1) / (2PI) )

SORT [ (ks)*(g) / (Wtf) )

where ks = 1 (h ^ 3) + (h) (3

E
lyy)

(As

G)

( Use lyy = 183.31 in ^ 4 ) ( Use As = 23.91 in ^ 2 ) therefore, ks = 6.52E+ 06 Fiong. = 29.5685 Tank in Longitudinal Direction (see Note below): Rigid (Rigid if Flong >or= 33, Flexible if Flong < 33) Note: The preceeding evaluation of ks is for unbraced saddles. The saddles for the horizontal tank in question are braced by two cross members connecting the top and bottom extremes of the saddles on each side of the tank..This . cross bracing supplies significant stiffening to the bending' resistance of the saddles. The calculated longitudinal frequency (29.6 Hz) underestimates L the actual frequency and the tank will be assumed as rigid (Flong > 33). It is also noted that the maximum acceleration in the range above 20 Hz is much less than the spectral peak ( < 0.15 g vs. 0.71 g). CAT-sA Pace 3 of s .i-

C:lculation: S96-0013 rev. O FPC - Crystal River Unit 3 Seismic Verification of Tanks nev: By i _cate chk d ay cate 0 11M $M/ar Peis t/12/otL . Calculation For: l Horizontal Tank l Step 10: (10) _ Compare Seismic Demand to Capacity Acceleration From Steps 8 and 9, if tank / heat exchanger is: rigid - Use Zero Period Acceleration (ZPA) of 4% damped floor response spectrum flexible - Use Peak Spectral Acceleration (SPA) of 4% damped floor response spectrum The seismic loading is the 4% damped SSE spectra for the Auxiliary Building at 119' which is determined in FPC calculation S-94-0011, " Seismic Verification of Tanks - SQUG Methodology", Rev. O,1/19/94. From calculation S-94-0011 pages 27 and 28: OBE FRS Peak (4% damping) = 0.353 g OBE FRS ZPA (4% damping) = 0.050 g Since tank is rigid, use 4% SSE ZPA (SSE ZPA = 2 times OBE ZPA), therefore, Horizontal 4% SSE ZPA = 0.10 g Vertical 4% SSE ZPA (2/3 Horiz.) = 0.07 g Anchorage is Adequate if: (1) Lamb > ZPA (for rigid tanks / heat exchangers) or (2) Lamb > SPA (for flexible tanks / heat exchangers) 0.10 g ZPA (use ZPA as explained above) = 0.58 g Anchorage Capacity, Lamb, from Step 7 = Check if Capacity (Lamb) > Demand (ZPA) ? OK Step 11: (11) C_onfirm Stresses in the Saddle are Acceptable The saddle and stiffners are only about 5" deep (between the Saddle pad and top of the plinth), in addition the saddles are well braced laterally (two cross members connecting top and bottom of each saddle on each side of tank). Bending of the braced saddle is adequate by inspection. For shear the anchorage has been. determined as adequate and the arnount of shear area in the stiffened saddles is much greater than the area of the anchor bolts (4 1" anchor bolts per plinth). The shear capacity of the saddles is also adequate by inspection. CONCLUSION The Horizontal Tanks under evaluation: l CAT-5A l l CAT-5B l are acceptable in accordance with Section 7 of the FPC PSP for " Seismic Verification - of Nuclear Plant Equipment". CAT-sA Page 4 of s

Ccicul: tion: S96-0013 rev,0 FPC - Crystal River Unit 3 Seismic Verification of Tanks nev By oft. I c8k d By oate o ' 7227 EtGU Pds i \\ lt.t/% _ Calculation For: { Horizontal Tank l I ~i[ORIDA POTVER C6D. aDRATION 0 03 s 4ts 043 0 sc 4ts ns ~~ it nvainvio, no m. CRY 3 fat RIVnt PLANI eassat AssocIATIT, x. ~ spot eso 3 anim tw lNQNtitt AND CON 5Jtf ANf 3 'i $wvetwol Archer Bolt lis' TilXooN/ BM u ht. 5t40 ct. its

0 ' I eT.' i f,' l*t!'s, {' H, mwou s soowN oN owo no sMos on nc s'kn STANDARD TYPE 3 SPECIAL TYPES TYPE l

I 1 4 I _._n~ 4 - mo m g g I: :.,g g _7 u 1 i=d .. H 7":d = h _1 8 v m r a j[ NM i! y hx7 Ln. (Y d j au W Fis, L TW ye g y i l $8[ se etArt 3 i pire m,rne piAm Ltact> twetAes

=

a O L A B C 18 T Ti so ne P G = s oia L tasta 4 f20 /4 /* It' if 5" It &~ /r' 4' 14 2' o*4 ~ /! ' t' it 3~ S' 4* 14 /t " ly't t Ell /6 Si 6 a .y g,y..q. 4 9 7 ,9. -. -- 3 4 ~ t its J IV 2:6'

Y 4*

f' fi e 2li 1:s\\- e i G 114 4 /4~ 2*t* 2*C .y V e Sinn FL6. f" 8 !!{ fitV a i Ets 4 l'i 3*S* lis,' s* s' It sesu ns, fi is tY aspf i II 4' YJ* t 'e' T 5* 44 etAn !L G. 'kl 94 l't

119V 6 116 2i l

1 try 24 I %* s'.I

2*%~

4* e #* is sun it.s, 4C is 3" /: sin 4 pro o t'l sn-l' 4~ + s y r' ki o r sen a S !!3 4 4~ Itt' l'/ t* t' 8 SIAu m. 1* 4 !* O' RT 4 l E30 4 35 l*- l ' Y t* i t l l l e,,,, a, ACMS:- Au Marten to se ts.w. 4.sc sim. W seost8issLt14 STInttettf ssof t0 CAT-SA Pace 5 of 5

c8MSA lorida DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 Page l of f LKX tmst N1 fut hid K.A fiO8v hO Nt vi$loh S96-0013 0

1 Attachment "B" Chilled Water Expansion Tank Five Pages Total m na i. v. o, v.m nur. m.., a v~.,

Cciculati:n SM-0013. rev. O FPC -- Crystal River Unit 3 Seismic Verification of Tanks nov sy cate i chk d ay -i cate 0_GHA IWs/fr td5 i V2V_gL Calculation For: [ Tank suspended on 4 Legs] Equip. ID: CHT-1 Equipment

Description:

Building: CONTROL Elevation:- 181 CHILLED WATER EXPANSION TANK Rm Row / Col: 302 / G Also Applicable for: I i H,rizontal Tank Suspended on 4 Legs Drawing: 55-144-1-0 (SD 400-3) Anch. Drw.: SC-405-045 and SC-423-037 Vendor: Tacoinc. Model: ASME Expansion Tank SD 400-3 Methodoloay Tank CHT-1 is a small tank suspended from the ceiling on four legs with cross bracing in both directions. Each leg is attached to the reinforced concrete ceiling with 4 approx.1/2" diameter anchor bolts. Field inspection verified that the tank is welded to the saddles which prevents longittIdinal motion and (1) prevents the tank from falling and (2) prevents pipe break. The SOUG methodology given in Section 7 of the Florida Power PSP " Seismic Verification of Nuclear Plant Equipment", Rev. 1,9/12/94 is not applicable. The following simplified calculation uses conservative assumptions for the anchor bolt size and type, overestimates appUcable seismic accelerations, and uses conservative values for uncertain dimensions to determine the adequacy of the tank anchorage. Dimensions Dimensions are obtained from the referenced drawing when possible and from conservative measurements obtained during walkdowns (as noted). Tank: Outside Diameter (in) D 16.00 from drawing OverallLength (in) L 72.00 from drawing Weight of the tank (galvanized) (Ib) Wt ' 133 lb from drawing Tank Capacity (gal) C 60 from drawing Weight of water (Ib/ gal) Ww 8.34 Distance base plate to tank c.g. (in) h 48.00 field estimate Anchorage: (Assume worst case = unknown 3/8" expansion anchors) D,ameter Anchor Bolt _ (in) bd 0.375 Assumed Number anchor bolt total Nb 16.00 Number bolt per leg N ieg 4.00 Bolt Embedment (approx.) (in) Lb 3.75 to x oiam. Concrete strength (psi) I' c 3000.00 Base Plate: Thickness base plate each leg (in) t bp 0.50 field estimate Side Dimensions (in) Ibp 8.00 field estimate Base plate spacing (narrow) (in) s 20.00 field estimate CHT-1 Pace 1 of 4 a

- C:lculation: S96 0013 rev. O FPC - Crystal River Unit 3 Seismic Verification of Tanks nev! By j oat. - cnk d ay i cate Tl GBA' /0/fMr T& -i/u/_cic_ Calculation For: lTarik suspended on 4 Legsj l l i 4 Calculation (1) Weight The tank welght consists of the empty tank and the contents: W tank = Wt + W contents = Wtank = ( Wt ) + ( C ) x ( W w ) = 633 lb Use W tank = 650 lb (2)C.G. The tank C. G. was estimated during the walkdown to be less than 4' from the anchorage base-plate. C.G. - 48.00 in

(3) Loading To determine the Seismic Demand should use Control Complex Elev.181' SSE floor reponse spectra at 4% damping. The spectra for tne Control Complex at 193' are obtained from Figure 19A in the FPC ' Environmental and Seismic Qualification Program Manual', (E/SOPM), Rev. 8, Section 5.0, Seismic Qualification Data.

SSE Spectrum Peak (3% damping) = 1.35g ZPA for 3% = 0.25g SSE Spectrum Peak (5% damping) = 1.10g ZPA for 5% = 0.25g Tank is cross braced in both horizontal directions and is probably rigid; however, conservatively use peak floor response spectrum applies and further, conservatively use 3% SSE values as 4% SSE values: Horizontal 4% SSE Peak = 1.35 g Vertical 4% SSE Peak (2/3 Horiz.) = 0.90 g (4) Overturning Worst case will be for horizontal earthquake acting in the narrow tank leg direction (see figura). Vertical seismic force act in a downward direction assisting pullout. At-tA o o Section A A O D = 16' l L = 72' l CHT-1 Pace 2 of-4

Ccic lati:n: S%0013 rev. 0 FPC--- Crystal River Unit 3 Seismic Verification of Tanks-nov sy oate chk d ey o ie Pds-1/2.24% o /0/thf i Calculation For: l l [ Tank suspended on 4 Legl I- ~ I i Overturning (Continued) (a)- Conservatively determine the pullout force per leg as the sum of the pullout per leg due to vertical seismic loads plus the two pullout loads due to horizontal seismic loads acting parallel and perpendicu!ar to the tank axis, i.e., P1 + P2 + P3, where Pullout / leg =

1. Pullout due to vertical earthquake per leg:

P1 = ( W) * ( 1.0 + SSE vert) / 4 = 309lb 2 Pullout due to worst horiz. earthquake per leg: P2 = ( W) * ( SSE hor) * ( h ) / ( D

2 ) =

1316 lb

3. Pullout due to other horiz earthquake per leg:

P3 = ( W ) * ( SSE hor) * ( h ) / ( Arm

2 ) =

439 lb (estimate Arm = 48'from field measurement) Therefore, Pullout / leg = P = P1 + P2 + P3 = 2064 lb (b) Determine anchor bolt pullout forces, Pu: Each tank leg has 4 anchor bolts, maximum pullout per anchor bolts: Pu = (P)/4 = 516 lb (c) Determine anchor bolt shear forces, Vu: 878 lb Total shear = ( W ) ( SSE Horiz. ) = Vu = ( Total shear ) / ( 16 bolts ) = 55 lb (5) Anchor Bolt Allowables (Assume = unknown 3l3" expansion anchors) Conservatively assume that the anchor bolts are 3/8" expansion anchors of - unknown type to minimize allowables (bolts are probably 1/2" diameter cast-in-place bolts). Allowables for 3/8" Expansion Anchor Bolts (From GlP Table C.2.1) Pnom = 1.46 ksi Vnom = 1.42 ksi RTp = 0.60 type red. factor RTs = 0.60 RLp = 1.00 embedment red. factor RLs = 1.00 i RSp = 1.00 spacing red, factor RSs = 1.00 rep = 1.00 edge distance red. factor REs = 1.00 'RFp = 0.93 for fc=3000 psi concrete RFs = 0.93 RCp = 1.00 cracked concrete red, fact. RCs = 1.00 Pu' = Pnom (RTp) (RLp) (RSp) (rep) (RFp) (RCp) = 0.81 Kip Vu' = Vnom (RTs) (RLs) (RSs) (REs) (RFs) (RCs) = 0.79 Kip CHT-1 Page 3 of 4

Cciculatiom S96-0013 rev. O FPC - Crystal River Unit 3 ~ By oat. i chk d ay oat. Seismic Verification of Tanks nev Calculation For:

l Tank suspended on 4 Legsj i

I I (6) Evaluate Anchorage Allowable Maximum? Maximum anchor bolt pulloat 811 lb 516lb OK Maximum anchor bolt shear 789 lb 55 lb' OK Interaction: The linear interaction formula for expansion bolts is taken from Section C.2.11 of the GIP: Pu + Vu 1 P all V all 0.71 OK 0.64 + 0.07 = CONCLUSION This extremely conservative analysis demonstrates that the tank under evaluation would be adequately anchored even if the worst case assumption of 3/8" expansion anchors of unknown type and manufacture is imposed. The tank under evaluation: I CHT-1 l 1 I is acceptable. CHT-1 Page 4 of 4

ce98<l glorida DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 Page l of _ b _,...<.e S96-0013 0 i 9 4 Attachment "C" Decay Heat Closed Cycle Surge Tank Six Pages Total m na v. a r.,

w. w-nn,m

Cciculati:n S96-0013 rev. O FPC -- Crystal River Unit 3 . Seism'o Verification of Tanks E~ev By Qate l Chk'd By j Date 0 6 /# fi/ I Pd.s !l?.hs/_ss_ Calculation For: I l l l l Vertical Tank on 4 Legs l j l t i Equip. ID: DCT-1A Equipment

Description:

Building: AUXILIARY DECAY HEAT CLOSED CYCLE SURGE Elevation: 095 TANK Rm Row / Col: 306/S Also Applicable for: lDCT-1B l Vertical Tank on 4 Wide Flange Legs Drawing: 5-315-D2 Rev. 3 (M001859) Anch.Drw.: SC-422-042 and SC-423-026 Vendor: Plant City Steel Co. Model: . Vert: cal Dished Head 5000 Gal Tank Methodology Tank DCT-1 A is not a flat bottomed vertical tank so the SOUG methodology given in Section 7 of the Florida Power PSP " Seismic Verification of Nuclear Plant Equipment", Rev. 1,9/12/94, is not applicable. DCT-1 A is supported by

a wide flange sections (8WF31) spaced at 90% around the perimeter. Since the tank anchorage will be the critical element, this calculation will focus on the anchorage.

Dimensions Dimensions are obtained from the referenced drawings Tank: Outside Diameter (in) D 90.00 OverallHeight (in) H 195.00 Thickness of tank shell (in) ts 0.250 Thickness of tank head (top / bottom) (in) th 0.375 Weight density steel (Ibf/in ^ 3) W st 0.2840 Weight density fluid (Ibf/in ^ 3) W fi 0.0361 Height of shell portion (in) hs 157.00 Height of heads (top & bottom) (in) hh 19.00 Nominal Height of water (in) hw 159.00 Anchorage: Cast-in-Place Bolts (see SC-423-026) Diameter Anchor Bolt (in) bd 0.875 Number anchor bolt total Nb 8.00 Number bolt per leg Nseg 2.00 Bolt Embedment (approx.) (in) Lb 10.00 i Bolt Spacing (center to center) (in). Sb 5.00 Bolt Edge Distance (in) Eb 6.50 Concrete strength (psi) f'c 3000.00 ) Base Plate: Thickness base plate each leg (in) t bp 1.00 Side Dimensions (in) I bp 12.00 Paae 1 of 5 ~ -..

C:lculation S96-0013 rev. O FPC - Crystal River Unit 3 Seismic Verification of Tanks nev! By I cate. chk e sy o.t. 0 TE -i /d@H Mb I2/W95 ~ ' ' Calculation For: l ly Vertical Tank on 4 Legs,_j C_ alculation. (1) Weight The tank weight consists of the shell portion and the tpp and bottom heads. The tank is consarvatively asrumed to be cylindrical with top and bottom circular disks: W tank = W shell + 2 (W head) + W contents = W sheli = (pi) (D) (hs) (ts) (Wst) = 3152 lb W head = 2 (pi) (th) [(D) (hh) + (D/2) ^ 2] (Wst) = 2499 lb W contents = (pl) (D/2) ^ 2 (hw) (W fi) = ' 36516 lb W tank = 42167 lb (2)C.G. The tank is located 2'-6" above the anchorage. The C. G. is calculated from the anchorage base-plate. Tank cg = (W Steel) (H/2) + (W water) (hw/2) 81.91 in (W tank) C.G. = ( Tank cg ) + ( 2'-6") = 111.91 in (3) Loading To determine the Seismic Demand should use Auxiliary Building Elev. 95' SSE floor reponse spectra at 4% damping. The spectra for the Auxiliary Building at 95' are identical to the Ground Response spectra. (

Reference:

' Environmental and Seismic Qualification Program Manual", (E/SOPM), Rev. 8, Section 5.0 Seismic Qualification Data, Figure 22]. OBE Spectrum Peak (2% damping) = 0.135g ZPA for 2% = 0.05g OBE Spectrum Peak (5% damping) = 0.100g ZPA for 5% = 0.05g Conservatively for 4% SSE use 2 times the 2% OBE; therefore, Horizontal 4% SSE Peak = 0.27 g Vertical 4% SSE Peak (2/3 Horiz.) = 0.18 g. Assume tank is flexible, use Spectral Peak as acceleration (4) Overturning Worst case will be for horizontal earthquake at 45 degrees to tank legs. Therefore d'etermine overturning for horizontal along 45 deg to legs and vertical earthquake acting upward (assisting overturning). Let F1 and F2 each represent vertical force in two legs (see Figure); i.e., F1 is the upward force resisting overturning and f2 is the force assisting overturning: 1 Pace 2 of 5 1

Cciculati:m S96-0013 rev. O FPC - Cryst:1 Riv3r Unit 3 Seismic Verification of Tanks a.v sy. pai. chk d By oat. O ft)/f/ Calculation For: ~ /D/ffif Pk l2/l3fqs 1 ~ l Vertical Tank on 4 Legs l Overturning (Continued) f L 8v d 195" Cg e >X m 9 y \\ k j @i F I Y l SEcrtw A A J -d j p$ A A F, q (a) Moment arm = Arm = [ ( D / 2) + (I bp / 2 ) ) / sqrt( 2 ) = 36.06 in (b) Sum Forces vertical: F1 + F2 = (W tank) (1.0 - SSE vert) = F1 + F2 = 42167 lb * (0.82) = 34577 lb (c) Sum Moments about Z: F1 (Arm) = F2 (Arm) + (W tank) (SSE hor) (cg) = F1 - F2 = (W tank) (SSE hor) (cg) / (Arm) = 35331 lb (d) Solve equations (c) and (b) for F1: F1 + ( F1 - 35331 ) 34577 lb = F1 = 34954 lb F2 = -377 lb (e) Determine anchor bolt pullout forces: Each force (F1 and F2) represent two of the tank legs and each leg has two 7/8" diameter anchor bolts. The maximum and minimum forces are: Max. anchorage vertical forco (F1/4) = 8738 lb Min, anchorage vertical force (F2/4) = -94 lb Since negative anchorage forces represent bolt pullout, only the minimum force needs to be considered for this tank. Pu = 94 lb (f) Determine anchor bolt shear forces: Total shear = ( W tank) ( SSE Heriz. ) 11385 lb = Bolt shear = ( Total shear ) / ( 8 bolts ) = Vu = 1423 lb Pace 3 of S

Cciculati:m S96-0013 rev. O FPC - Crystal River Unit 3 Seismic verification of Tanks nov sy _ pa.i. chk d ey oate Calculation For: ~ l Vertical Tank on 4 Legs l = (5) Anchor Bolt Allowables (From GlP Section 4.4 and Appendix C) Allowables for 7/B' Cast-in-Place Bolts (From SC-423-026) Pnom = 20,44 ksi Vnom = 10.22 ksi RLp = 1.00 embedment red. factor RLs = 1.00 .RSp = 0.80 spacing red. factor RSs = 1.00 rep = 0.94 edge distance red. factor REs = 0.72 RFp = 0.93 for fc=3000 psi cor. crete RFs = 0.93 RCp = 1.00 cracked concrete red. fact. RCs = 1.00 Pu' = Pnom (RLp) (RSp) (rep) (RFp) (RCp) = 14.31 Kip Vu' = Vnom (RLs) (RSs) (REs) (RFs) (RCs) = 6.84 Kip (6) Evaluate Anchorage Allowable Maximum? Maximum anchor bolt pullout 14308 lb 94 lb OK Maximum anchor bolt shear 6840 lb 1423 lb OK Interaction: The interaction curves for cast-in-place bolts are taken from Section C.3.7 and Figure C.3-2 ohhe GIP. Since the GIP anchorage critoria for cast-in-place bolts and headed studs ensure that failure does not occur in concrete, the interaction formulation for steel failure is recommended: for 0.0 < (V/Va) < 0.3, (P/Pa) < 1 for 0.3 < (V/Va) < 1.0, 0.7 x (P/Pa) + (V/Va) < 1 thereforo, Since (V/Va) 0.21 = (P/Pa) 0.01 OK = CONCLUSION The tanks under evaluation: L DCT-1A I DCT-1B J are acceptable. Page 4 of 5

Cciculati:n: S96-0013 rev. O FPC - Crystal River Unit 3 Seismic Verification of Tanks Revi By Date Chk'd By i Date TN/5' '@}YIf Pd$ ! lVt&~_ Calculation For: l Vertical Tank on 4 Legs l ~~ I I SC-42b " 026 FLORIDA POWER CORPORATION 4203 0 0 S 423 c24 o ft /210/4 - St.PltitSSUto,f10tioA woes omose use esaw o me,

c ma*wi=o ao CRY $fAL RIVER PLANT onstat AsRRTItal. ING.

UNif NO. 3 555.000 Kw (NGINtit,' AND CONSUtfANTS t t ADING. Pt NN A. Structurel-Anchor Bolt lisi a s i... x n. A 1'LIARf _3'CJ U.2.iSJ. E u$UEVT /*Jur'C. JYt lc}% lXsa yk.%( '* ' * '" ' " "" # 8 MtilMAL AS SHOWN ON DWC.NO((.///*,%f//((42fOff __ STANDARD TYPES SPECML TYPES tot i 2 3 e 5 I,o """l A,,, a F-P,, d 4 -O Dh ,p -D -L D{ g Os o s 1.,, Y b5 Y .D su.?l 1 ::- m , - < d~,. e one t e st.t k instaos ",0,0,' g j pin stun so etan BroO e E i ol' LtacTH O L A B C H T Ti 50 NEx P G = 4 O!

)

14' 29' J' 23 6' %' fl6' 20 16' l'6$ ' 4 02 32 4' i J' 25' 32 4' // * !s ' 32 2' l' 0 ' f Jo p (,

gg Ay' g 2lfj!;&*

4 C3 tt 2 c' i j* 4

4 66

/~ j'. 6 fy' 63 j' $) ' jj' Sg 2' o f.jy r t- l !!)' 8 J' 4' !)' 8 19 ' llfj ' 4 DS 6 1 2 4 36 b* I !' 2' /2 4' Q' Ps ' l2 2' tich' 4 07 63 b* l S' l 2' 68 4' ! 0$' 4' ' Ps' 88 2' g 4 08 24 I4' 2' ! 24' 24 S' }j ' t 24 24' :;if' V' S 09 BJ l' 2'C' 2fj' 30 5' It' &* 80 4 D/C 32 h' l' 2' ih' 32 3' Q' 44' 32 /* OW ~ 4 D // /2 It ' /* 5' /%' /2 3' Q' ft ' /2 lh' ! Cfj' Page 5 of 5 l

c$aDid

lorida DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 Page l cf b

_,m,~,_ Attachment "D" Decay Heat Removal Heat Exchangers Six Pages Total l. W3 Mk I. ( te of V4N Mt 47, het krvpswed'ry

Calculati as S96-0013 rev. 0 FPC - Crystal River Unit 3 Seistnic Verification of Tanks n.v sy . Dat. Chk'd By Dat. i 0 7 89 lb/3thr Rts \\/ W % _ Calculation For: ~ l Horizontal Heat Exchanger l Equip. ID: DHHE-1A Equipment

Description:

Building: AUXlLIARY DECAY HEAT REMOVAL HEAT Elevation: 075 EXCHANGER A Rm Row / Col: _304/P-O__ Also Applicable for: FDHHE-18 l Heat Exchanger Drawing: 68-G-198 2-1 Rev 5 Anch. Drw.: SC-422-003, SC-422-004 and SC-423-021 Vendor: Yuba Heat Transfer Co. Model: BEU-37-237 Step 1: (1) Input Data See Figure 7-13 of Florida Power Plant Specific Procedure for " Seismic Verification of Nuclear Plant Equipment", Rev. 1, 9/12/94 _ Appliicable? Tank: Diameter (ft) D ' 3.12 OK Length (ft) L ' 28.83 OK Thickness of tank shell(in) t 0.24 Weight of tank plus fluid (Ibf) W tt 30800.03 Weight density (Ibf/ft ^ 3) G am 139.49 OK Height of c.g. above anchorage (ft) H co 2.32 OK Saddle: Spacing (ft) S ' 14.67 OK Height of saddle plate from bottom of h 9.14 the tank to the base plate (in) Shear modulus (psi) G 1.12E: Elastic modulus (psi) E 2.90E + Number of Saddles N. 2.L OK Base Plate: Thickness base plate under saddle (in) tb 0.63 Min. yield strength (psi) fy 30000.00 Thickness of leg of weld tw 0.25 A. g Eccentricity from anchor bolt CL to e: 3.00 the vertical saddle plate Bolts: Number of locations, each saddle NL 2.00 OK Number of anchor bolts per location NB 1.00 OK Diameter of anchorbolt (in) d 1.00 Distance between extreme anchor D'

2.25 OK bolts in base plate of saddle (ft)

Loading: SSE Floor reponse spectra at 4% damping DHHE-1A Pa ce _1 of s

C:le:lation: S96-0013 rev, O FPC - Crystal Hiver Unit 3 Seismic Verification of Tanks nv sy oni chkd By oat. O M/ /C/36M f PA s lin/S_ _ Calculation For: l Horizontal Heat Exchangar l Stop 2: (2) Anchor Bolt Allowe5les (From GIP Section 4.4 end Appendix C Table C.3-1) Allowables for 1.0" Cast-In-Place Bolts (Mark D49 type 2 from SC-423-021) (also from SC-422-041) Actual Embedment = 19.00 Allowable = 10.00 in Actual Spacing = 27.00 Allowable = 12.63 in Actual Edge Distance = 9.00 Allowable = 8.75 in Pnom = 26.69 ksi Vnom = 13.35 ksi RLp = 1.00 embedment red, factor RLs = 1.00 RSp = 1.00 spacing red, factor RSs = 1.00 rep = 1.00 edge distance red, factor REs = 1.00 RFp = 0.93 for fc=3000 psi concrete RFs = 0.93 RCp = 1.00 cracked concrete red, fact. RCs = 1.00 Pu' = Pnom (RLp)(RSp)(rep)(RFp)(RCp) = 24.71 Kip Vu' = Vnom (RLs)(RSs)(REs)(RFs)(RCs) = 12.36 Kip Step 3: (3) B_ase Plate Bendino Strenath Reduction Factor (RB) RB = Bending strength toduction factor = {fyUtb ^ 2) ' O.16 (3) (Pu') Step 4: (4) _ Base Plate Weld Strenoth Reduction Factor (RW) RB = Weld st.angth red. fact.= (tw) (es) (30600) (2.83)

2.63 Pu' Step 5:

(5) Anchor Tension and Shear Allowable Pu = (Pu') (smaller of RB, RW) = 3.91 Kip Vu = Shear allowable archor load = (Vu') =

12.36 Kip 3 Step 6:

.; Calculated Ratlos Alp = (Pu') / (Vu') = 0.32 Wb = (Wtf) / [ (NS) * (NL) * (NB) ] = 7700.00 lbs Vu / Wb = 1.61 Mcg / D' = 1.03 Heg / S = 0.16 F1 = SORT [ (NS ^ 2) + 1 ) ' 2.24 = F2 = SQRT [ (NL ^ 2) * (Heg / D') ^ 2 + (.667 ^ 2) + ((Hcg /S) ^ 2) * ( (NS ^ 2) / (NS-1) ^ 2) ] = 2.19 DHHE -1 A Paae 2 of S

C:lculati:m S96-0013 rev. O FPC - Crystal River Unit 3 Seismic Verification of Tanks sivT sy o.t. enk d ey o.t. 0 '71/# /o/,30/ff 7e 1/2.2LE CrJculation For: fiiorizontal Heat Exchanger J Step 7: (7) Determine Acceleration Capacity of Tank Anchorace Llow [ (Vu) / (Wb) ) * [ (1) /(F1)) ' O.72 g = = Lup jyu) / (Wb) + (0.7) / (Alp))

0.54 g

= = [ (0.7) / (Alp) ) * (F2) + (F1)

0.54 g Lamb =

Smaller of Llow or Lup = Step 8: (8) is Tank / Heat Fxchanaer Riald or Flexible in Transverse or Vertical? Sc = From Figure 7-15 for Heat Exch. = ' 12.00 ft ( Use D = 3.123 ft ) ( Use t = 0.237 in) ~ is Tank / Heat Exchanger Rigid or Flexible ? Rigid if Sc >or= $, Flexible if Se < S From Step 1, S = 14.667 ft Tank / Heat Exchanger in Transverse or Vertical = Flexible 1 Step 9: (9) is Tank / Heat Exchancer Riald or Flexible in Lonoitudinal Direction? Flong. = [ (1) / (2PI) )

SQRT[ (ks)*(g) / (Wtf) )

where ks = 1 (h ^ 3) + (h) (3

E
lyy)

(As

G)

( Use lyy = 14.15 in ^ 4 ) ( Use As = 30.25 in ^ 2 ) therefore, ks = 1.55E+ 0C Flong. = 22.1603 Tank / Heat Exchanger in Longitudinal Direction = Flexible Step 10: (10) Compare Seismic Demand to Capacity Acceleration From Steps 8 and 9, if tank / heat exchanger is: rigid - Use Zero Period Acceleration (ZPA) of 4% damped floor response spectrum flexible - Use Peak Spectral Acceleration (SPA) of 4% damped floor response spectrum DHHE-1A Pace 3 of 5

Ccic:lation: 596-0013 rev. O FPC - Crystal River Unit 3 C 31smic Verification of Taiiks Rev [4y, _ Cat _e Chk'd By Dato 0

6) tut

/Wsokf Pdt \\h,IqL. Cale,alation For: [Iiorizontal Heat Exchanger j Step 10 (Continued): The seismic loading is the 4% damped SSE spectra for the Auxiliary Building at 75' which are identical to the ground response spectra. [

Reference:

" Environmental and Seismic Qualification Program Manual", (E/80PM), Rev. 8, Section 5.0 Seismic Qualification Data, Figure 22), From E/SOPM Section 5.0; l

- OBE Spectrum Peak (2% damping) = 0.135g ZPA for 2% = 0.05g OBE Spectrum Peak (5% damping) = 0.100g ZPA for 5% = 0.05g Conservatively, for 4% SSE take 2 times 2% OBE Peak =

therefow, Horizontal 4% SSE Peak =

0.27 g Vertical 4% SSE Peak (2/3 Horiz.) = 0.18 g Anchorage is Adequate if: (1) Lamb > ZPA (for rigid tanks / heat exchangers) or (2) Lamb > SPA - (for flexible tanks / heat exchangers) SPA (use peak as specified above) 0.27 g = Anchorage Capacity, Lamb, from Step 7 0.54 g = Check if Capacity (Lamb) > Demand (SPA) ? OK Step 11: (11) _ Confirm Stresses in the Saddle are Acceptable The saddle and stiffners are only about 6" deep (between the Saddle pad and top of the plinth). Bendir.; of the stiffened saddle is adequate by inspection. For shear th9 anchorage has been determined as adequate and the amount of shear area in the stiffened saddles is much greater than the area of the anchor bolts (2 1" anchor bolts per plinth) and is therefore also adequate. C__ONCLUSION The Heat Exchangers under evaluation: l DHHE-1A I I DHHE-1B l are acceptable in accordance with Section 7 of the FPC PSP for " Seismic Verification ,f c;uclear Plant Equipment" DHHE-1A Page 4 of s

Calcul: tion: S96-0013 rev. 0 FFPC - C stal River Unit 3 Seismic rification of Tanks Rvi sy Dat. chk'd By Date Calculation For: [_ Horizontal Heat ExchangerJ l l FLORIDA POWER CORPORATION cm S An-ott o SC 4n on it. Ptteessues. nossa w so m-% CRY 3fAL RfVIE PLANT eausi anocwn, suc. mett een 3 set,000 tw ENG8 Melts AND CONSUtf ANTl Stevetwola Arachor Soll list DECW *E.hT P.7 EQ'J\\*MERT F MD'S. G lr,5t l v*s Pfah[ *.f.Q wew as sews on owc eSt. 4tt-003@cA " ' " = " =a = "" STANDA10 TYPES SPECIAL TYPES TYPI 1 3 3 4 l n. n. ag .o - -o g 4 t r-a > -+< t u l DIAM LimCTM TWataM WI Ptait e a PE t E ttyt eAM eged l DIA LINSD O L A 8 C N Y T, n men P G e t 4* 14 40 4* i [ 40 1" l'-o* 1 0 46 40 t' i 1 D4B l' 140' ti 5" t' 1" b 1" 14' I l i 1 i I

UTFSI-Akt BO\\T-MATEA\\M

~TO BE-A 6 T M AYo emmisasau m ooss eum i SDIOWWE NOf t9 i DHHE-1A Pace 5 of 5

l l

:rida DESIGN ANALYSIS / CALCULATION

'.mDid Crystal River Unit 3 e Prge of b ".ocweut otniv caio =o ut vis.o. S96 0013 0 f Attachment "E" Emergency Diesel Generator Lube Oil Coolers Six Pages Total bM RL [. L.fe C4 V4M 5E 4 37. F%4 eer &r W AN

Ccicul:ti:n: 596-0013 rev. 0' FPC - Crystcl Riv:r Unit 3 ' Seismic Verification of Tanks Revi ey

Dat, i

chk'd By Date 0 Q l0/W4( Adt ! I2/lWW Calculation For: I DLHE-1A l j I Horizontal Heat exchan_ger] Equip. ID: DIHE-1A Equiprnent

Description:

Building: DIESEL EMERGENCY DIESEL GENERATOR Elevation: l 119 LUBE OIL COOLER 1 A Rm Row / Col:l 301/ Q Also Applicable for: lDLHE-1B, DLHE-2A ard DLHE-2B 'l Horizontal Heat Exchanger Drawing: 5- 047-19-142- 061 Anch. Drw.: SC-421-176 Vendor: Colt Industries Model: 19142 "CPK" Stacking Methodology: The DLHE heat exchangers are stacked one on top of the other. In the following calculation it is assumed that the length of the " saddle" member extends from the bottom of the upper heat exchanger to the top of the base plate at the foundation _, The weight is also increased by the ratio of the moments about the base to adjust for the l presence of the bottom heat exchanger; that is, W' = [ (h1 + h2) / h2 ) x W, where h1 is the height to the cg of the lower heat exchanger, h2 is the height of the cg of the upper heat exchanger, and W is the weight of one heat exchanger This is very conservative since the heat exchangers are actually " connected" by four sections along the length (2 pipe sections and two stiffened saddles about 10" high) and will be much stiffer than the assumed configuration. The anchorage configuration beneath the two caddle base plates are different (see drawing SC-421-176 for details). One configuration uses four 1" diameter by 1' long, Phillips Wedge anchors WS-100120 with 7" minimum embedment while the other uses four 3/4" by 1'-3.5"long Maxi-bolts MB-750 with 9.25" minimum embedment. This calculation presents the Phillips configuration since the saddle stiffness is lower and the anchorage capacity reduction factors are greater; however, both configurations are OK. Step 1: (1) [nput Data See Figure 7-13 of Florida Power Plant Specific Procedure for " Seismic Verification of Nuclear Plant Equipment", Rev. 1, 9/12/94 (Notes are at the bottom of page 4.) Appliicable? Tank: Diameter (ft) D 1.67 OK Length (ft) L 14.78 OK Thickness of tank shell (in) t 0.38 3/8 in Weight of tank plus fluid (Ibf) W tt 8000.00 See Note 1 Weight density (Ibf/ft ^ 3) G am 186.84 See Note 1 Height of c.g. above anchorage (ft) H eg 3.58 Saddle: Spacing ft) S N 7.21 OK 1 Height of (saddle plate from bottom ofh 33.00 the tank to the base plate (in) Shear modulus (psi) G 1.12E + 07 Elastic modulus (psi) E 2.90E + 07 Number of Saddles Ns 2.00 OK DLHE-1 A Paae 1 of 5 1

Cciculation: S%0013 rev. O FPC - Crystcl River Unit 3 Seismic Verification of Tanks nev _s ,oate I chk d ey I cate o G / $y ~ ' Calculation For: F DIRE-1A l ~ f}p]g( l QS ty/n[gy_ j [ Horizontal Heat Exchanger l ~ i i i Base Plate: Thickness base plate under saddle (in) tb 0.75 Min, yield strength (psi) fy 30000.00 Thickness of leg of weld tw 0.38 Eccentricity from anchor bolt CL to e 3.00 the vertical saddle plate Bolts: Nurnber of locations, each saddle NL 2.00 OK Number of anchor bolts per location NB 2.00 OK Diameter of anchor bolt (in) d 1.00 Distance between extreme anchor D' 1,42 OK bolts in base plate of saddle (ft) Loading: SSE Floor reponse spectra at 4% damping Step 2: (2) Anchor Bolt Allowables (From GlP Section 4.4 and Appendix C Sble C.2-1)- Allowables for 1.0" Phillips Wedge WS-100120 Expansion Anchors ;see SC-421-176) Actual Embedment = 7.00 Min. Allow = 4.50 in Actual Spacing = 9.00 Min. Allow = 10.00 in Actual Edge Distance = 12.00 Min. Allow = 10.00 in Pnom n 6.95 ksi Vnom = 9.53 ksi RTp = 1.00 manuf. type red. factor RTs = 1.00 RLp = 1.00 embedment red. factor RLs = 1.00 RSp = 0.90 spacing red, factor RSs = 1.00 rep = 1.00 edge distance red. factor REs = 1.00 RFp = 0.87 for fc=3000 psi concrete RFs = 0.95 RCp = 1.00 cracked concrete red, fact. RCs = 1.00 RRp = 1.00 essential relays red, factor RRs = 1.00 Pu' = Pnom (RTp)(RLp)(RSp)(rep)(RFp)(RCp)(RRp) = 5.42 Kip Vu' = Vnom (RTs)(RLs)(RSs)(REs)(RFs)(RCs)(RRs) = 9.05 Kip Step 3: (3) Base Plate Bending Strenath Reduction Factor (RB.) RB = Bending strength reduction factor = (fy) (tb ^ 2) 1.04 (3) (Pu') Step 4: (4) Base Plate Weld Strenath Reduction Factor (RW) RB = Weld strength red fact.= {twl(es) (30600) (2.83) 17.97 Pu' DLHE-1A Pane 2 of 5

Calculati:m S96-0013 rev. 0 FPC - Cryst:1 Riv r Unit 3 Sel.=,1c Verification'of Tanks nov! ey_ _ont ci.k d By cate olG11l Calculation For: l DL!-lE .1 A l l ~ INWG < PA5 t2ASL95-l Horizontal Heat Exchanger l Step 5: (5) Anchor Tension and Ghear Allowable Pu = (Pu') (smaller of RB, RW) = 5.63 Kip Vu = Shear allowable anchor load = (Vu') = 9.05 Kip Step 6: (6)' Calculated Ratios Alp = (Pu') / IVu') = 0.62 Vu / Wb(W=tf) / L (NS) * (NL) * (NB) ) = Wb = 1000.00 lbs 9.05 Heg / D' = 2.53 Heg / S = 0.50 F1 = SORT ' NS ^ 2) + 1 F2 = SORT ' ((NL ^ 2) */ D') ^ 2 + (.667 ^ 2) 2.24 = + ((Hcg /S) ^ 2) * (

2) / (NS-1) ^ 2 ) ] =

5.20 Step 7: (7) Determine Acceleration Capacity of Tank Anchorage Llow [(Vu) / (Wb) ) * [(1) /(F1) ) 4.05 g = = Lup J[Lvu))/ (Wb) + (0.7)])_(A,j u 1.26 g = = (0.7 / (Alp) ] * (F2 + 1j Lamb = Smaller of Llow or Lup 1.26 g = Step 8: (8) is Tank / Heat Exchanaer Riaid or Flexible in Transverse or Vertical? Sc = From Figure 7-15 for Heat Exch. = 11.00 ft (Use D = 0,375 in )) 1.667 ft ( Use t = Is Tank / Heat Exchanger Rigid or Flexible ? Rigid if Sc >or= S, Flexible if Sc < S From Step 1, S = 7.208 ft Tank / Heat Exchanger in Transverse or Vertical = Rigid Step 9: (9) is Tank / Heat Exchanaer Ricid or Flexible in Lonaitudinal Direction? Flong. = [ (1) / (2PI)]

SORT [ (ks)*(g) / (Wtf) ]

where ks = 1 (h ^ 3) + (h) (3

E
lyy)

(As

G)

( Use lyy = 243.24 in ^ 4 ) ( Use As = 20.50 in ^ 2 ) therefore, ks = 5.43E+ 05 Fiong. = 25.7684 Tank / Heat Exchanger in Longitudinal Direction = Flexible DLHE-1A Page 3 of s

Cakulati:nt S96-0013 rev. 0 FPC - Cryst:1 Riv r Unit 3 Seismic Verification of Tanks nov sy, opte. chkday . cate O GJf to&M Pds i2/JWBs- ' ~' Calculation For: l DLHE-1A l I Horizontal Heat' Exchanger l Step 10: (10) Compare Seismic Demand to Capacity Acceleration-From Steps 8 and 9, if tank / heat exchanger is: rigid - Use Zero Penod Acceleration (ZPA) of 4% damped floor response spectrum flexible - Use Peak Spectral Acceleration (SPA) of 4% damped floor response spectrum The seismic loading is the 4% damped SSE spectra for the Diesel Generator Building at 119' elevation. According to sketch A on page 5 -20 of the " Environmental and Seismic Qualifiution Program Manual", (E/SOPM), Rev. O, Section 5.0 Seismic Qualification Data, t,'.is elevation is represented by Figure 22. From E/SOPM Section 5.0 Figure 22; OBE Spectrum Peak (2% damping) = 0.135g ZPA for 2% = 0.05g OBE Spectrum Peak (5% damping) = 0.100g ZPA for 5% = 0.05g Conservatively, for 4% SSE take 2 times 2% OBE Peak = therefore, Horizontal 4% SSE Peak = 0.27 g Vertical 4% SSE Peak (2/3 Horiz.) = 0.18 g Anchorage is Adequate if: 1) Lamb > ZPA (for rigid tanks / heat exchangers) or (2) Lamb > SPA (for flexible tanks / heat exchangers) ( 0.27 g SPA (use peak as specified above) = 1.26 g Anchorage Capacity, Lamb, from Step 7 = Check if Capacity (Lamb) > Demand (SPA) ? OK Step 11: (11) Confirm Stresses in the Saddle are Acceptable The saddle and stiffners are only about 3.5" deep (between the Saddle pad and top of the plinth) Bending of the stiffened saddle is adequate by inspection. For shear the anchorage has been determined as adequate and the amount of shear area in the stiffened saddles is much greater than the area of the anchor bolts (4 1" anchor bolts per plinth) and is therefore also adequate. CONCLUSION The Heat Exchangers under evaluation: I DLHE-1A l l DLHE-2A l l DLHE-1B l l DLHE-2B-l ~ are acceptable in accordance with Section 7 of the FPC PSP for " Seismic Verification of Nuclear Plant Equipment". NOTES (1) As noted under

Methodology
the subject heat exchangers are stacked one on top of the other.

The total

wet
weight from the vendor drawing is 12oso lbs, and the ' equivalent
weight was calculated using 1/2 this weight as the weight of each heat exchanger lW'= W * (h1 + h2) / h2].

The 'eqlvalent* weight density falls outside the applicable range in Tab'e 7-6 of the PSP (see step 1 reference), but the weig'l 1ensity for a single heat exchanger meets these requirements. DLHE-1A Pace 4 of 5 i

Ccic:lations S96-0013 rev. O t FPC - Crystal River Unit 3 Seismic Verification of Tanks Rev By _Date Chk'd By Date / Calculation For: f' T)IFTE-1 A,j ~ /#3"/W Pdi ' v463_95~ O 71/// l Horizontal Heat Exchanger l se I 'e I t l,ll T I Il II 4_ o j,c - Section A-A ~ lM l h I te ~~ A o o-Aj g "r r o o. 4 / \\6 Section B B B j t 5 lilllillifililllllilli i 1 DLHE-1 A Paae 5 of 5

l @ [:clorida DESIGN ANALYSIS / CALCULATION UU Crystal River Unit 3 Page __b of k. w..., u,,, n,. S96 0013 0 i I Attachment "F" Emergency Diesel Generator Air Receivers Six Pages Total .s~ Et I. L d4 0' P e 1 HL tr. h44e L fg.teerr$

C lerlation: 596-0013 rev. O FPC - Crystcl Riv:r Unit 3 Seismic Ver."! cation of Tanks nev ey I p.9 i enkday i cate O EM/ i /NM/f( l Pdi JMV95 Calculation For: I~ Vertical Tank on Skirt j i l i Equip. ID: EGT-1A Equi ment

Description:

p Building: DIESEL EMERGENCY DIESEL GENERATOR A Elevation: 119 AIR REOElVER 1 A Rm Row / Col: 301/ N Also Applicable _for: fliGT-18, EGTT2A and EGT-28 l Vertical Tank (Air Receiver) on Skirt Drawing: PT-8027-X (4203-86-034-0) Anch. Drw.: SC-421 - 171, SC-421 - 172, SC.-42:3-044 Vendor: Morrison Brothers Co. Model: 30 x 103 Air Receiver hiethodology EGT -1 A is a vertically oriented air receiver for which the SOUG methodology given in Section 7 of the Florida Power PSP ' Seismic Verification of Nuclear Plant Equipment', Rev. 1,9/12/94, is not applicablo. EGT-1 A is welded to a 15-1/2' high cylindrical skirt that is anchored to a reinforced concrete plinth by four 3/4* diameter cast-in-place bolts spaced at 90 degrees around the perimeter. Since the tank anchorage is the critical element, this calculation will focus on the anchorage. Dimension _s Dimensions are obtained from the referenced drawings Tank: Outside Diameter (in) D 30.00 Overall Height (in) H 103.00 Thickness of tank shell(in) t 0.437 Thickness of tank head (top / bottom) (in) th 1375 Weight density steel (Ibf/in ^ 3) W st 0.2840 Weight density contents (Ibf/in ^ 3) Wn 0.0001 air Height of shell portion (in) h: 85.00 Height of heads (top & bottom) (in) hh 9.00 Nominal Height of contents (in) hw 0.00 not applicable Anchorage: Cast-in-Place Bolts, type B-13 (see SC-423-044) Diameter Anchor Bolt (in) bd 0.75 Number anchor bolt total Nb 4.00 Number bolt per leg N ico 1.00 Bolt Embedment (in. ) (type B-13 has Lb 16.00 minimum from an embedment > 16') SC-423-044 Bolt Spacing (center to center) (in) Sb 21.00 Bolt Edge Distance (in) Eb 10.00 minimum Concrete strength (psi) f' c 3000.00 Base Plate: Thickness angle (4 welded to skirt) (in) t bp 0.25 Estimated Angle Dimensions (square) (in) I bp 3.00 Estimated EGT-1A Pace 1 of s +-

Calculatlans S96-0013 r;v. O FPC - Cryst:1 Fliv:r Unit 3 Seismic Verification of Tanks nev e pate chk'd ay i cat. O iBl $/3/91 Pds Mlad5-Calculation For: l l Vertical Tank on Skirt I I Calculatio3 (1) Weight The tank weight consists of the shell portion and the top and bottom heads. The tank is conservatively assumed to be cylindrical with top and bottom circular disks: W tank = W shell + 2 (W head) + W contents - W shell = (pl) (D) (hs) (ts) (Wst) = 994 lb W head = 2 (pl) (th) [(D) (hh) + (D/2) ^ 2) (Wst) = 331 lb W contents = (pi) (D/2) ^ 2 (hw) (W fl) = 0 lb W skirt (stand) (from drawing) = 182 lb W tank = 1507lb (2)C.G. The bottom of the tank is 6.0 in above the anchorage. The C.G. is calculated from the anchorage base-plate as: Tank cg = (W Steel) (H/2) + (W contents)(hw/2) 51.50 in (W tank) C.G. = ( Tank cg ) + ( dist. to bottom ) = 57.50 in (3) Loading To determine the Seismic Demand use Diasel Generator Building spectra for elevation 119' (SSE 4% damping). Tne spectra for the Diesel Generator Building at 119' are identical to the Ground Response spectra. [

Reference:

' Environmental and Seismic Qualification Program Manual", (E/SOPM), Rev. 8, Section 5.0 Seismic Qualification Data, Figure 22]. OBE Spectral Peak (2% damping) = 0.135g ZPA for 2% = 0.05g OBE Spectral Peak (5% damping) = 0.100g ZPA for,5% = 0.05g Conservatively use 2*(2% OBE Peak) as 4% SSE Peak = therefore, Horizontal 4% SSE Peak = 0.27 g Vertical 4% SSE Peak (2/3 Horiz.) = 0.18 g (4) Overturning Worst case will be foi horizontal earthquake at 45 degrees to tank legs. -Therefore determine overturning for horizontal along 45 deg to legs and vertical earthquake acting upward (assisting overturning). Let F1 and F2 each represent vertical force in two legs (see Figure on next page); i.e., F1 is the upward force resisting overturning and F2 is the force assisting overturning: EGT-1A Paae 2 of s

Cciculatiom SE0013 rev. O FPC - Crystal Riv:r Unit 3 Seismic Verification of Tanks n.v! e T o i. chk d ay i o i. O G/} D/2lG( Pds O2&Lqs_ Calculation For: l Vertical Tank on Skirt l Overturning (Continued) '- -- D - - 7 S, f N.,,,' ce *

s, g

,/. w I Arm I w J P T-Z Hou,t L_ n n r, i r, (a) Moment arm = Arm = [ ( D / 2 ) + (I bp / 2 ) ) / sqrt( 2 ) = 11.67 in (b) Sum Forces vertical: F1 + F2 = (W tank) (1.0 - SSE vert) = F1 + F2 = 1507 * (1-SSEv) = 1236 lb (c) Sum Moments about Z: F1 (Arm) = F2 (Arm) + (W tank) (SSE hor) (eg) = F1 - F2 = (W tank) (SSE hor) (eg) / (Arm) = 2006 lb (d) Solve equations (c) and (b) for F1: F1 + ( F1 - 2006 ) 1236 lb = F1 = 1621 ib F2 = -385 lb (e) Determine anchor bolt pullout forces: Each force (F1 and F2) represent two of the tank legs and each leg has one 3/4" diameter anchor bolts. The maximum and minimum forces are: Max anchorage vertical force (F1/2) = 811 lb Min, anchorage vertical force (F2/2) = -192lb Since negative anchorage forces represent bolt pullout, only the minimum force needs to be considered for this tank. Pu = 192 lb (f) Determine anchor bolt shear forces: Total shear = ( W tank ) ( SSF Horiz. ) = 407 lb Bolt shear = ( Total shear) / ( 4 bolts ) = Vu = 102 lb EGT-1A Pace 3 of 5 I

Ocdc StL,-coe tw. o FPC - Crystal River Unit 3 nev! a . cate enk d ay o.i. Seismic Verification of Tanks 3.jMf// m/VMi 9ds Gams. Calculation For: [ Vertical Tank on 4 Legs l t i 1 i Overturning (Continued) D i lF D S y H C9 *

8:

q as' W f mk A" Z c A ( _ y l u Ys Y1 Section A. A (a) Moment arm = Arm = 15.29 in [ ( D / 2 ) + (I bp / 2 ) ) / sqrt( 2 ) = (b) Sum Forces vertical: F1 + F2 = (W tank) (1.0 - SSE vert) = F1 + F2 = 3540 * (1 -SSEv) = 1874lb (c) Sum Moments about Z: F1 (Arm) = F2 (Arm) + (W tank) (SSE hor) (cg) = F1 - F2 = (W tank) (SSE hor) (cg) / (Arm) = 10647 lb (d) Solve equations (c) and (b) for F1: 1874 lb F1 + ( F1 - 10647 ) = F1 = 6261 lb F2 = -4387 lb (e) Determine anchor bolt pullcut forces: Each force (F1 and F2) represent two of the tank legs and each leg has two 1' diameter anchor bolts. The maximum and minimum forces are: Max, anchorage vertical force (F1/4) = 1565 lb Min, anchorage vertical force (F2/4) = -1097 lb Since negative anchorage forces represent bolt pullout, only the minimum force needs to be considered for this tank. Pu = 1097 lb (f) Determine anchor bolt shear forces: 2499lb Total shear = (W tank) ( SSE Horiz. ) = Bolt shear = (Total shear ) / ( 8 bolts ) = Vu = 312 lb SFDM-1 Pace 3 of 5 ,v- .r-

bt SM, co e, rtv.O FPC - Cryst:1 Riv r Unit 3 Seismic Verification of Tanks n_ev sy paie chk'd By Date i o @D1 .123/19f 96s Lth3/As_ Calculation For: I Vertical Tank on'4 Legs j ] I (5) Anchor Bolt Allowables (From GIP Section 4.4 and Appendix C) Allowables for 1' Cast-in-Place Bolts (Type D-53 from SC-423-027) Pnom = 26.69 ksi Vnom = 13.35 ksi RLp = 1.00 embedment red. factor RLs = 1.00 RSp = 0.79 spacing red. factor RSs = 1.00 rep = 0.88 edge distance red factor REs = 0.47 RFp = 0.93 for fc=3000 psi concrete RFs = 0.93 RCp = 1.00 cracked concrete red fact. RCs = 1.00 Pu' = Pnom (RLp)(RSp)(rep)(RFp)(RCp) = 17.17 Kip Vu' = Vnom (RLs)(RSs)(REs)(RFs)(RCs) = 5.83 Kip (6) Evaluate Anchorage Au wable Maximum? o Maximum anchor bolt pullout 17172 lb 1097 lb OK Maximum anchor bolt shear 5829 lb 312 lb OK Interaction: The interaction curves for cast-in-place Lalts are taken from Section C.3.7 and Figure C.3-2 of the GlP Since the GIP anchorage criteria for cast-in-place bolts and headed studs ensure that failure does not occur in concrete, the interaction formulation for steel failure is recommended: for 0.0 < (VNa) < 0.3, (P/Pa) < 1 for 0.3 < (VNa) < 1.0, 0.7 x (P/Pa) + (VNa) < 1 there' ore, since (VNa) 0.05 = 0.06 < 1 OK (P/Pa) = CONCLUS!ON The tank (s) under evaluation: [, SFDM-1 J l I is/are acceptable. t SFDM-1 Page 4 of s

(Cdc5%-co3to.O FPC - Crystal River Unit 3 Seismic Verification of Tanks ned oy I cai. i enkday 1 cat. ~Illi i/P/]l'$f I %$ T'2ll5L% 0 1 I Calculation For: I Vertical Tank on 4 Legs l l l i l FLORIDA POWER CORPO.tATION 0 03 t 4.!..~ 0

; E.'

.1 it eitmins esoiea 4.

CtVlf At i1VII PL ANi uiiffilTiiMTITil *c uni, => s attooo. tanc+tlet a=> contats*ett staorwc tt w a _Structwel. # ackee lett t,e ..... ~., l')8iJ1E M W _ h _ll 'll C' f.:5 ~ H) dMr./*( l..' jif 1 ' ' c "d martest at twowu ow owc. e.o * *.jgi.l; STANDARn TYPES SPECIAL. TYPES 3l ) a tvet i 1 8! 4 O [ A,, .,L.,., .s F .s - -p v f....

h

~ a

0,,c

~ 5' ' " ' ' h""" o.a. tt.6,,

t. :aen

,,",0, i 1 g sau cia Lt.sta to als P G O L A 8 C H T Ti 4 0% 11 I" I' It' 1Y 11 5' 'i '4' 11 Y l'55' 4 0 $t; 10 lW lh* 6 F Si '4' 6 lY l' N 4 c.% 11 'g 8 iT l'i il 4' h' '4' 11 T l' W I o a: MTS!- Att. MATT 9I At. TO EE A51M A% STEEL m see:vntase etueevne motso SFDM-1 Page 5 of 5

I @ [:c83U lorida DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 Page f of _ h st os,u tu. vin si w,is t.i.o wo S96-0013 0 Attachment "G" Instrument Air Dryer Six Pages Total R$ 5, lid @f F#d ME &F.1%A4% L@**$

Cale lation: 596-0013 rev. 0 FPC - Cryst:1 Rivcr Unit 3 Selsrnic Verification of Tanks nev ey . of te chk d ay i cate t 0 48// /BIE/6f Wi !I?,d3/A5 Calculation For: I Vertical Tanks on Skid j _I i l i Equip. ID: lADR-1 Equipment

Description:

Building: TURBINE. Elevation: 095 INSTRUMENT AIR DRYER 1 Rm Row / Col: 302 / A Also licable for: Vertical Tanks (Air Dryer) on Skid Drawing: 4203-75-326-0 Anch. Drw.: SC-405-011, SC-405-012, SC-408-102 and SC-423-033 Vendor: Lectrodryu Division, McGraw-Edison Co. Model: Model 6581 Methodology lADR-1 has two vertical dryer tanks on a common skid for which the SOUG methodology given in Section 7 of Florida Power PSP ' Seismic Verification of Nuclear Plant Equipment', Revision 1,9/12/M, is not applicable, lADR-1 includes the two dryer tanks and associated piping mounted on two 56.5'long angles (longitudinally), which are in turn welded to two 27'long angles in the shallow or transverse directiun. This skid (at the 27' angles) is mounted on a reinforced concrete plinth by four 3/4' diameter cast-in-place bolts spaced at 22'(narrow dimension) and 54.25', Since the tank anchorage is the critical element, a conservative calculation assuming the c.g. of the air dryer to be 3/4 of the height dryer tanks is used to check the anchorage adequacy. Dimensions Dimensions are obtained from the referenced drawings Tank: Onerall Length (in) L 56.50 OverallHeight (in) H 100.00 Estimated Weight of the Instrument Dryer (Ib) W 4500 from drawing Weight density contents (Ibf/in ^ 3) W ti 0.0001 air Height of c. g. (3/4 H) (in) h eg 75.00 Anchorage: Cast-in-Place Bolts, type AB-213 (see SC-423-033) Diameter Anchor Bolt (in) bd 0.75 Number anchor bolt total Nb 4.00 Number bolt per corner Nieg 1.00 Bolt Embedment (in,) (type AB-213 has Lb 13.00 minimum from an embedment > 13") SC-423-033 Bolt Spacing (minimum) (in) Sb 22.00 SC-408 - 102 Bolt Edge Distance (in) Eb 4.88 See NOTE 1 Concrete strength (psi) f' c 3000.00 Allowables: GlP Table C,3-1 Cast-in-place Bolt Allowables Pullout Capacity (kip) P nom 15.03 Shear Capacity (kip) V nom 7.51 lADR-1 Paae 1 of s

Ccle"latiom SM-0013 rev. O a FPC - Cryst:1 Riv r Unit 3 Seismic Verification of Tanks nev' By , oei. I chk d By o ie O G1M ~IDIGii Pdt \\2LaL95- ~ ~ Calculation For: ~ I Vertical Tanks on Skid l Allowables: GIP Table C.3-1 Cast-in-place Bolt Allowables (Continued) Minimum Embedment (in) L min 7.50 Minimum Spa:ing (in) S min 9.50 Minimum Edge Distance (in) E min 6.63 Calculation (1) Weight The Instrument Dryer weight is taken from the vendor drawing as: W dryer = 4500 lb Conservatively use for totalweight W = 5000 lb (2)C.G. The c.g. of the dryers'is conservatively taken to be at 3/4 of the height above the concrete p,linth. C.G. = ( 3/4 ) x Height 75.00 in = (3) Loading To determine the Seismic Demand use 4% SSE Turbine Building spectra for elevation 95', The spectra for the Turbine Building at 95' are identical to the Ground Response spectra. (

Reference:

Environmental and Seismic Qualification Program Manual", (E/SOPM), Rev 8, Section 5.0 Seismic Qualification Data, Figure 22]. Although the Instrument Air Dryer appears to relatively rigid (freq. > 33 Hz.), conservatively assume the system is flexible and use the spectral peak as the seismic demand.

OBE Spectral Peak (2% damping) = 0.1359 ZPA for 2% = 0.05g OBE Spectral Peak (5% damping) = 0.100g ZPA for 5% = 0.05g Conservatively use two times the 2% OBE Peak as 4% SSE Peak: therefore, Horizontal 4% SSE Peak = 0.27 g Vertical 4% SSE Peak (2/3 Horiz.) = 0.18 g (4) Overturning Worst case overturning will be for horizontal earthquake normal to the long axis of the base (i. e., earthquake acting transverse to the dryers). Therefore determine overturning for horizontal earthquake acting N-S (parallel to the short axis) combined with vertical earthquake acting upward (assisting the overturning). Let F1 and F2 each represent vertical force in two anchor points (see Figure on next page); i.e., F1 is the upward force resisting overturning and F2 is the force ' assisting' overturning: lADR-1 Pace 2 of S

Cciculation: S96-0013 rev. 0 , FPC - Cryst:1 Riv r Unit 3 Seismic Verification of Tanks nevt ey o.t. cnk d ay o t. O TA 8Mf( Pdi 12 /13 49 5-Calculation For: { Vertical Tanks on Skid I Overturning (Continued) s, C9 ""* 8: W H H, l ....I l I I l L r, r, o o bA 1s854 x x - J i dgx x a o 0 11.00 in (from drawing) (a) Moment arm = Arm = (b) Sum Forces vertical: F1 + F2 = (W dryer) (1.0 - SSE vert) = F1 + F2 = 5000 * (1 -SSEv) = 3690 lb (c) Sum Moments about Z: F1 (Arm) = F2 (Arm) + (W dryer) (SSE hor) (cg) = F1 - F2 = (W dryer) (SSE hor) (cg) / (Arm) = 9205 lb (d) Solve equations (c)(and (b) for F1: F1 + F1 - 9205 ) 3690lb = F1 = 6447 lb F2 = -2757 lb (e) Determine wichor bolt pullout forces: Each force (F1 and F2) represents two of the dryer anchorages and each anchorage has one 3/4" diameter anchor bolt. The maximum and minimum forces per anchor point are: Max. anchorage vertical force (F1/2) = 3224 lb Min. anchorage vertical force (F2/2) = -1379 lb Since negative anchorage forces represent bolt pullout, only the minimum force needs to be considered for this tank. Pu = 1379lb (f) Determine anchor bolt shear forces: Total shear = { W tank ) ( SSE Horiz. ) 1350 lb = Bolt shear = ( Total shear ) / ( 4 bolts ) = Vu = 338lb IADR-1 Page 3 of s i

Cciculati:n: SU-0013 rev, U FPC - Crystti Riv r Unit 3 Seismic Verification of Tanks nev sy I .oate I chk'd By Date Calculation For: ~ l Vertical Tanks on Skid l i I (5) Anchor Bolt Allowables (From GIP Section 4.4 and Appendix C) Allowables for 3/4' Cast-in-Place Bolts (AB-213 Type 2 from SC-423-033) Pnom = 15.03 ksi Vnom = 7.51 ksi (Table c.3-1) RLp = 1.00 embedment red. factor RLs = 1.00 Lmin = 7.s0' RSp = 1.00 spacing red. factor RSs = 1.00 Smin = 9.50' rep = 0.90 edge distance red. factor REs = 0.55 See NOTE 1 RFp = 0.93 for fc=3000 psi concrete RFs = 0.93 RCp = 1.00 cracked concrete red. fact. RCs = 1.00 Pu' = Pnom (RLp)(RSp) (rep)(RFp)(RCp) = 12.57 Kip g Vu' = Vnom (RLs)(RSs)(REs)(RFs)(RCs) = 3.85 Kip (6) Evaluate Anchorage Allowable Maximum? Maximum anchor bolt pullout 12570 lb 1379 lb DK Maximum anchor bolt shear 3848 lb 338 lb OK Interaction: The interaction curves for cast-in-place bolts are taken from Section C.3.7 and Figure C 3-2 of the GIP. Since the GIP anchorage } criteria for cast-in-place bolts and headed studs ensure that failure does not occur in concrete, the interaction formulation for steel failure is recommended: for 0.0 < (V/Va) < 0.3, (P/Pa) < 1 for 0.3 < (V/Va) < 1.0, 0.7 x (P/Pa) + (V/Va) < 1 i therefore, since (V/Va) 0.09 = 0.11 < 1 OK (P/Pa) = CONCLUSION The Instrument Air Dryer under evaluation: l IADR-1 I I l is acceptable. MoltLS (1) The calculated edge distance reduction factor is extremely conservative. The top of the reinforced concrete plinth is approximately s' above the concrete floor (see drawing SC-408-102) and the embedment length (> 13') exceeds the minimum embedment (6.62s') by more than 6*. Therefore there is only edge distance consideration over part of the anchor bolt length. The true edge distance reduction factors for this a whorage are closer to 1.0; however, values based on the conservative interpretation of edge distance (4-7/8' to plinth edge) are used in this calculation, t IADR-1 Page 4 of s ~y

Calculati::n: S96-0013 rev. O l I FPC - Crystti Rivar Unit 3 Seismic Verification of Tanks nev ey I _ Date i Chk'd By Date t ~' Calculation For: ~ l Vertical Tanks on Skid l l ~~ l i FLORIDA POWER CORPORATION no) 5 423.C33 o Sc 413.o33 n ettiesevos. etossa -..c... CtY11AL RIYtt PL ANT MHilTilRTIlls. iac. ueni ama a us ooo sw ING'NiI88 AND CONsottAuf t ,,,7 $ltveturel

Anchot Bell L}'

10R.81NE ROOM EQU1PMENT FOUNDS. wt*Q FylJl'f,v7i'.".,;if,,fg f '" " *" * * * "' o e' "'s asArtnat 44 snow ou twe e.o g.40610 W405-102 STANDARD TYPES SPECIAl. TYPES ttPt i I 3 4 ~ r, D,, ~~ ~ a a _s s u ~ b b / h a I-L tr,- > "E'] Q,',' so Ptatt a, j pipesterve ciam tensta instics i utt ..o a so ac t P G U i oia tensta D L A 4 C H T Ti 1 4 101 4 la l' 6' T 4 y go 4 g gig. 1 13 ?01 44 I" l'. 5' 1(a 44 ga g,

4 7.

g1gg. I 22e. 31 3' l'.G" 14" st 4' q-M 1* SF' I 4M B / l' 4' 14' 8 4' 4" 8 2' Ito!' t stos it i* t. i" ti" it g-e it 1 isoy 1 88 M 11 l' It 9' Z" 11 4' q" 11 t' It09' gioh 1 4 I01 44 l' !! 4* l' 44 4* y 44 2 I 4* 2' lG 3' i' 16 Iga stog. 1 4100 tG f 1 4 205 42 ) itta y iga 41 47 ltG* I Q* a 3 (= 3 g, I!O-1 4 110 8 t t{' 4 g= g. 4 i t Io* tI 3' 1 A8til 4 is' 1" 10 4' 4" ic, 7" Itd f 1 2112 16 l' t 1 4113 4 l'.4' t' 4 4" h 4 2.* It0T 1 4M 4 F lt r,' t. 4 g. g. 4 d" fOf i NOTES:.ALL MAM. RIAL To E A.S.T.M. A% STEEL es secues unteis etweama mot o IADR-1 Page 5 of 5

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DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 Page_ ofJ m._,,,,~.,_ $96 0013 0 Attachment "H" Instrument Air Receivers Six Pages Total kT) RL I, L Ag gr VW4 Rt tr. NA4e L Pg W4y ~-

% -co S rw. O FPC - Crystal Riv:r Unit 3 Seismic Verification of Tanks n.v o i pai. chk d ay I o.t. '7h//g J/hN/ Pdi j I?/13/AL 0 Calculation For: 1 ~ I Vertical Tank on Skirt l I i i C_alculation (1) Weight The tank weight consiste of the shell portion and the top and bottom heads. The tank is conservatively assumed to be cylindrical with top and bottom circular disks: W tank = W shell + 2 (W head) + W contents = W shell = (pl) (D) (hs) (ts) (Wst) = 915 lb W head = 2 (pl) (th) [(D) (hh) + (D/2) ^ 2] (Wst) = 453lb W contents - (pi) (D/2) ^ 2 (hw) (W fl) = 0 lb W skirt (stand) (estimated) = 150 lb W tank = 1517 lb Conservatively use for W tank = 1600 lb (2)C.G. The bottom of the tank is 6.0 in above the anchorage. The C.G. is calculated from the anchorage base-plate as: Tank cg = (W Steel) (H/2) + (W contents){hw/2) 64.50 in w (W tank) 70.50 in C.G. = ( Tank cg ) + ( dist. to bottom ) = (3) Loading To determine the Seismic Demand use 4% SSE Turbine Building spectra for elevation 95'. The spectra for the Turbine Building at 95' are identical to the Ground Response spectra. [

Reference:

' Environmental and Seismic Qualification Program Manual", (E/SOPM), Rev. 8, Section 5.0 Seismic Qualification Data, Figure 22].

OBE Spectral Peak (2% damping) = 0.135g ZPA for 2% = 0.05g OBE Spectral Peak (5% damping) = 0.100g ZPA for 5% = 0.05g Conservatively use 2 x (2% OBE Peak) as 4% SSE Peak; therefore, Horizontal 4% SSE Peak = 0.27 g Vertical 4% SSE Peak (2/3 Horiz.) = 0.18 g i i (4) Overturning Worst case will be for horizontal earthquake at 45 degrees to tank legs. Therefore determine overturning for horizontal along 45 deg to legs and vertical earthquake acting upward (assisting overturning). Let F1 and F2 each represent vertical force in two legs (see Figure on next page)I l.e., F1 le te upward force resisting overturning and F2 is the force i assisting overturning: lAT-1A Paae ? of s

6QG-cn s a n FPC - Crystal River Unit 3 Seismic Verification of Tanks nev: eg 1 o.ie enx d ay i cai. AL 8 .1/d/78 Calculation For: J-- PdA ! 12/13/A5_ [ Vert: cal Tank on Skirt j L i Equip. ID: ~iX~T-1A Equipment

Description:

Building: TURBINE Elevation: 095 INSTRUMENT AIR RECElVER A ~~ Rm Row / Col: 303 / A Also Ap_plicable for: [~IAT _10 l Vertical Tank (Air Receiver) on Skirt Drawing: FN-83-27 (83-124-0) and 4203-83-047-A Anch. Drw.: SC-405-011, SC-405-012, SC-408-101 and SC-423-033 Vendor: American Welding & Tank Co. Model: 650 Gal. Vertical Air Receiver (200 psl ASME Design) MethodoloJy lAT-1 A is a vertical air receiver for which the SQUG methodology given in Section 7 of the Florida Power PSP ' Seismic Verification of Nuclear Plant Equipment', Rev. 1,9/12/94, is not applicable. lAT-1 A is welded to a 34' diameter cylindrical skirt that is anchored to a reinforced concrete plinth by four angle tabs welded to the skirt at 90 degrees around the perimeter with one 5/8' diameter cast-in-place bolt per angle. Since the tank anchorage is the critical element, this calculation focuses on adequacy of the anchorage. Dimensions Dimensions are obtained from the referenced drawings Tank: Outside Diameter (in) D 41.00 Overall Height (in) H 129.00 Thickness of tank shell (in) t. 0.250 40.s' inside dia. Thickness of tank head (top / bottom) (in) th 0.250 40.s' inside dia. Weight density steel (Ibf/in ^ 3 W st 0.2840 Weight density contents (Ibf/in ^ 3) W ti 0.0001 air Height of shell portion (in) ha 100.00 approx. Weif t of Wt (top 4-botto,,,, v.,, ,,. ~ ogem. Nominal Height of contents (in) hw 0.00 not applicable Anchorage: Cast-in-Place Bolts, typo AB-208 (see SC-423-033) Diameter Anchor Bolt (in) bd 0.63 Number anchor bolt total Nb 4.00 Number bolt per leg Nieg 1.00 Bolt Embedment (in. ) (type AB-208 has Lb 13.00 minimum from an embedment > 13") .cn-423-033 Bolt Spacingt(ggef(igghcentlig6@013 rev @ b 31.48 sc-408-101 Bolt Edge Distance in) Eb 4.00 Sc-404-101 Concrete strength (psi) fc 3000.00 Base Plate: Thickness angle (4 welded to skirt) (in) t bp 0.25 Estimated Angle Dimensions (square) (in) Ibp 6.00 so-408-101 1AT-1A Paae 1 of s v.

h(t3L tfRt.tct1 ru.C) FPC - Cryst ~l Alvar Unit 3 Seismic Verification of Tanks nov sy pai. enk d ay o.i. 4 0 GG? /PB/H PA1 l 1742hL Calculation For: I l_ Vertical Tank on Skirt l l Ov0rturning (Continued) { ( D 8 lr cg

n + s, y
E

-} \\ ^"" w / z H.u,g L. n IL t r, i r, (a) Moment arm = Arm = ( 37' diam / 2 )

sin (45) =

15.74 in (b) Sum Forces vertical: F1 + F2 = (W tank) (1.0 - SSE vert) = F1 + F2 = 1600 * (1 -SSEv) = 1244 lb (c) Sum Moments about 2: F1 (Arm) = F2 (Arm) + (W tank) (SSE hor) (eg) = F1 - F2 = (W tank) (SSE hor) (eg) / (Arm) = 1935 lb (d) Solve equations (c) and (b) for F1: F1 + ( F1 - 1935 ) 1244 lb = F1 = 1589lb F2 = -345 lb (e) Determine anchor bolt pullout forces: Each force (F1 and F2) represent two of the tank legs and each leg has one S/8' diameter anchor bolt. The maximum and minimum forces are: Max, anchorage vertica! force (F1/2) = 795 lb Min, anchcratje vertical force (F2/2) = -173 lb Since negative anchorage iorces represent bolt pullout, only the minimum force needs to be considered for this tank. Pu = 173 lb (f) Determine anchor bolt shear forces: Total shear = ( W tank ) ( SSE Horiz. ) = 410 lb Bolt shear = (Total shear) /(4 bolts ) = Vu = 102 lb IAT-1A Pace 3 of s

bc 6C0 -cotB ttu.O FPC - Cryst-l Riv r Unit 3 l /((9.i.,, ._Rdf {igt3h13. ~ Seismic Verification of Tanks RR 87 chk d ay oat. i jff/J 0_ Gef Calculation For: i [_ Vertical Tank on Skirt __] I I (5) Anchor Bolt Allowables (From GIP Section 4.4 and Appendix C) Allowables for 5/8' Cast-In-Place Bolts (AB-208 Type 2 from SC-423-033) Pnom = 10.44 ksi Vnom = 5.22 ksi (Table c.s-1) RLp = 1.00 embedment red, factor RLs = 1.00 Lmin = e.2s' RSp = 1.00 spacing red, factor RSs = 1.00 smin =7.87s' rep = 0.90 edge distance red. factor REs = 0.54 Emin = 5.5' RFp = 0.93 for fc=3000 psi concrete RFs = 0.93 RCp = 1.00 cracked concrete red, fact. RCs = 1.00 Pu' = Pnom (RLp)(RSp)(rep)(RFp)(RCp) = 8.68 Kip Vu' = Vnom (RLs)(RSs)(REs)(RFs)(RCs) = 2.59 Kip (6) Evaluate Anchorage Allowable Maximum? Maximum anchor bolt pullout 8683lb 173 lb OK Maximum anchor bolt shear 2593lb 102lb OK Interaction: The interaction curves for cast-in-place bolts are taken from Section C.3.7 and Figure C.3-2 of the GIP, Since the GIP anchorage criteria for cast-in-place bolts and headed studs ensure that failure does not occur in concrete, the interaction formulation for steel failure is recommended: for 0.0 < (VNa) < 0.3, (P/Pa) < 1 for 0.3 < (V/Va) < 1.0, 0.7 x (P/Pa) + (VNa) < 1 therefore, since (VNa) 0.04 = 0.02 < 1 OK (P/Pa) = CONCLUSION The tanks under evaluation: I lAT-1A l l lAI-1B l are acceptable. IAT-1A Page 4 of - s

bdt Stu -cct 'S rw.C FPC - Cryst:1 River Unit 3 Seismic Verification of Tanks nej e cai. chk d ay I o t. ~b ~ JWMGf FB5 l?/13/AS: Calculation For: l Vertical Tank on Skirt ] i I FLORIDA POWER CORPORATION I420) 5 413 633 o 5c413 o33 ft.ettitsaveo.eiossa . c... w. CRYliAt klVIt Pt AH1 ~Uiilil'ATIS&TlllDuc w sea 3 66.10cc ew INCe'Nf f el AND CO*d5VLI ant 1 ltulwola Ardet Self Lis' % J.7 TURBINE Gi'OOM (QUtPMENT FOUNDS.

tLQ., Q" R L_,.

i;;g;d N- <~ < - - ~. a,n As wo ~ oN ~ ~ax es m mot STANOMO TYPES SPECIAL TYPES ftPt I i 8 e p. 6 .m,,p. ~, -p g Jt t v ~ O l _3! h h i P l..., .s AT e. Q,',' { Part 3Lityt cias i.f mCin thetADS 60 PLaft a t mast Otto A O L A 4 C H T T 60 Mts P G 2 cia Leasta 1 A&tol 4 I" l' B" 1( 4 5* (" 4 L" l'oV 1 ABNI 44 1" l' 5' t(' 44 5' (* 44 t' 110 1' t Aar101 31 L' l' G" t(" 31 4' {* 31 l' l'+ 0* 2 2-M 8 n' l' 4' It' 8 4' l* 8 t' It0!' ? @t0S 11 I" Tl" t (" 11 5" t,' It 2" It0V 1 WM 11 j l' 9' Z 11 4' f it t' It0f 1 A&tol 44 h* I!4' t' 44 4' l' 44 2' I 0L' Q t SM IG f it " t' IG 3' q' 10 ll" stol' 4 i em 41 r ist" r i.t' 41 41 ItG' I li' 8 3' h' 8 l-Itog" 1 etto 8 E A6tIl 4 I 3' i I [' 4 3' 4 I l' iko (* t J t AM.ll ifs l' l' 5' 1" IG 4' ll" ICs 2" M L 4 '4" l'.04' 1 ett3 4 I 4' t' 4 4" l 1 A824 4 i" l' re' t' 4 3" l' 4 l{' tIo ( l m ,,,,,4 AUTFSt-ALL MAff.RtAL TO N. A.S.T.M. A% STEEL m usoes mLtis 97Weenes uptle IAT-1A Page 5 of 5

i 9 i$93SI

lorida DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 j

Page l of 1 tu.we, u =is,cti.o., o aa v.aw S96-0013 0 Attachment "1" Main Steam Valve Air Reservoirs Five Pages Total R R I. L ?g gf V g 1 Rg ) F, F$f $@ g*s

(bJc S%ccia tw.o FPC - Crystal River Unit 3 Seismic Verification of Tanks ne/ By I cate enk d By cat. F7/2 T97//Ef j pas i htL9L_ l Calculation For: I i l Air Reservoir l i i l Equip.10: MSV-411 - AR1 Equipment

Description:

Building: INTERMEDIATE Elevation: 119 ME,V-411 AIR RESERVOIR 1 Rm Row / Col: 309 /4 Also_A plicable for: J MSV-411 - ARl, MSV-412-ARl, MSV-413-ARl, MSV-414-ARI, i = 1 to 3 Air Reservoir Data Drawing (s): l-308-335 Anch. Drawing: 1. INTRODUCTION Because structural drawings and details for these Air Reservoirs were not available, a bounding calculation using very conservative assumptions is required. The following calculation demonstrates that the capacity of the anchorage for the Main Steam Valve Air Reservoirs greatly exceeds postulated seismic loading. The calculation covers the following Air Reservoirs: Main Steam Valve Associated Air Reservoirs MSV - 411 MSV - 411 - AR1 MSV - 411 - AR2 MSV - 411 - AR3 MSV - 412 MSV - 412 - AR1 MSV - 412 - AR2 MSV - 412 - AR3 MSV - 413 MSV - 413 - AR1 MSV - 413 - AR2 MSV - 413 - AR3 MSV - 414 MSV - 414 - AR1 MSV - 414 - AR2 MSV - 414 - AR3 II. METHODOLOGY All of the above Air Reservoirs are well mounted on walls near the associated valves. Since the only significant seismic effects are the weight of the air reservoirs themselves, the air reservoirs are conservatively assumed to be 7 feet long by 3 feet in diameter with 0.375 inch wall thickness, To demonstrate that the anchorage capacity greatly exceeds selcmic demand, it was further asaumed that the anchorage configuration consists of four 3/8 inch diameter expansion anchors in a square pattern spaced 30 inches apart. This anchor bolt size and type underestimates the actual anchorage capacity and the anchorage spacing is less than the actual spacing (minimizing the resisting moment arm). A square pattern also allows a single calculation to be valid for wall mounted air reservoirs with the long axis (cylinder centerline) oriented in either the vertical or the horizontal direction (see the attached figuic fur tne assumed geometry). M SV-411 - AR 1 Paco 1 of 4

(b30 Ml. CO6 ttv. O FPC - Cryst:1 Rivcr Unit 3 Seismic Verification of Tanks nev! sp o i. I _ chk d By j cat. ~bMIA . 3 M M Pe u nh.1.ts-Calculation For: I I l [ Air Reservoir l _1 I i i i i i Expansion anchors were assumed (rather than Cast-in-place bolts) to minimize anchorage capacity. The shear and pullout capacities of these anchors were obtained from the SQUG GlP (Reference 1) Table C.2-1 ar i were further reduced by a generic reduction factor of 0.6 per Table C.2-2 of the GIP to account for " Unknown" concrete fasteners. A maximum acceleration (demand) of 0.3g is assumed to act simultaneously in both horizontal direct'ons and will be combined with a vertical acceleration of 2/3 the horizontal (or 0.2g) by absolute summation. Since the Air Reservoirs are rigid (that is, frequency > 30 Hz. since loaded only by self-weight), the actual seismic demand for these air reservoirs would be the Zero Period Acceleration (ZPA) of the 4% damped SSE spectra at the 119' elevation of the Intermediate Building which is about 0.1g; therefore, the assumed acceleration values used in the calculation are extremely conservative. All of the assumptions described above and used in the following anchorage calculation conservatively bound the actual configurations, weights, seismic peak accelerations and load combination methods and will therefore overestimate the seismic demand. Ill. CALCULATION (1) Air Reservoir parameters: h = reservoir height 84.00 in = r = reservior radius 18.00 in = 0.375 in t = reservior thickness = ws a weight density (steel) 0.284 lb / in ^ 3 = W = totalweight = [ (h) (2 pl r) (t) + (2 pl r ^ 2) (t) ) (ws) 1229 lb = use W = 1250 lb (2) Anchorage Configuration: n = number of anchor bolts 4 = s = anchor spacing (horizontal or vertical) 30.0 in = d = diameter of anchor bolt 0.375 in = Pa = pullout capacity (GIP) 1460 lb x 0.6 876 lb = Va = shear capacity (GIP) 1420 lb x 0.6 852 lb = (3) Calculated Seismic Demand: ah = horizontalacceleration 0.3 g = av = vertical acceleration 0.2 s = V = Shear (vertical) per bolt: (1 + av ) * ( W) / n 3 R Ib = P = Pullout (horizontal) = P1 + P2 + P3, where P1 = ( r ) * ( 1 + av ) * ( W ) / [( s ) * ( n/2 )) 113 lb = P2 = ( r ) * ( ah) * ( W ) / [( s ) * ( n/2 )) 450 lb = P3 = ((sh) * (W) ) / 4 94 lb = P = P1 + P2 + P3 656 lb, = M SV-411 - AR 1 Pace 2 of 4

0.lc SQL,-cog s.v.o FPC - Cryst:1 Riv:r Unit 3 Seismic Verification of Tanks nevl ey oat. I chk d ay oat. 0 }], fgl[~ Pdy VNZ%_ Calculation For: l Air Reservoir l I lil. CALCULATIOS (Continued) (4) Capacity Check: Shear: V < Va 375 < 852 OK Pullout: P < Pa 656 < 876 OK 4 Interaction (bilinear): for 0.3 < V / Va < 1.0 then 0.7 * ( P / Pa ) + V / Va < 1.0 0.96 < 1.0 OK IV. CONCLUSION A bounding calculation for the Air Reservoir anchorage has been performed using very conservative assumptions: (1) Assumed air reservoir weight and dimensions exceed actual (2) Assumed seismic demand exceeds actual seismic demand (3) Assumed 3 simultaneous seismic loads combined by absolute sum (4) Assumed anchor bolt spacing (anchorage moment arm) is less than actual (5) Assumed enchor bolt type and size underestimate actual capacity Based upon these very conservative assumptions, the counding calculation demonstrates that the seismic capacity of the subject Air Reservoir anchorage exceeds the expected seismic demand. REFERENCES (1) Generic implementation Procedure (GIP)'For Seismic Verification of Nuclear Plant Equipment", Revision 2, February 1992. f M SV-411 - AR1 Pace 3 of 4

a,__,, -= blC 81LeCol3 to O FPC - Cryst:1 River Unit 3 Seismic Verification of Tanks nv sy o.i. enk d ay i o t. I.lii/.hi~ 1All W/AL... Pdc i Cf ation For: [~ Air Reservoir l m., l l E l SV -*.= top w I 4 _. ? A i - ;+ \\ g _.a.............a. = 9 M SV-411 - aft 1 P ae 4 of 4 a

9 <$8MU

lorida DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 Page - I of I a -, -,m..

S96-0013 0 I P Attachment "J" RCP Seal Return Coolers Six Pages Total b b F&[ (d$ $4 f.$'I RLMF NM Oh ( "40' #j

h(dC6CL-co G nV,0 FPC - Crystal River Unit 3 Seismic Verification of Tanks By oat. i c2vd ey l . Dai. n.] 7pi O r nhr/9f Vdj j 1/2,8_ Calculation For: [ Horizontal Heat Exchanger j Equip. lD: MUHE-28' Equipment

Description:

Building: AUXILIARY Elevation: 119 RCP SEAL RETURN COOLER A Rm Row / Col: 302 / L-M Also Applicable for: [ MUHE-2A J Seal Cooler Data Drawing: Anch.Drw.: SC-421-110, SC-422-043, and SC-423-032 Vendor: Introduction Vendor drawings for the Seal Coolers wero not available so field measurements were obtained during a walkdown on 10/24/95. The Seal Coolers are smaller than typical heat exchangers and do not meet all of the conditions for heat exchanger calcula4ons according to Table 7.6 of the Florida Power Plant Specific Procedure for " Seismic Verification of Nuclear Plant Equipment", Rev. 1,9/12/94 (for example, the 10" outer diameter is below the applicable 1 ft to 14 ft range); therefore a conservative calculation of the anchorage is performed. Since the Seal Cooler is effectively a rugged 10" diameter pipe section that is rigid in the longitudinal direction, the calculation focuses on anchorage capacity to resist demand due to overturning from hctrizontal seismic loads acting transverse to the seal cooler combined with vertical seismic loads acting upward (to assist in overturning). Use of field measurements and conevative assumptions in ta calculations to confirm anchorage adequacy are acceptable as long as the results show significant margins above allowable values. (1) Input Data ( Notes appear at the bottom of page 2 ) Tank: Diameter (in) D 10.00 Length (ft) L 17.83 Thickness of tank shell (in) t 0.125 See Note 1 Weight of Seal Cooler (Ib) W tf 2000 see step 4 Weight density (Ibf/ft ^ 3) G am 180 See step 4 Height of c.g. above anchorage (in) H eg 7.500 Saddle: Spacing (ft) S 14.917 Height of saddle plate from bottom of h 2.500 the tank to the base plate (in) Shear modulus (psi) G 1.12E + 07 Elastic modulus (psi) E 2.90E+ 07 Number of Saddles Ns 2.00 MUHE-28 Page 1 of s

hE.Sm<.ct% av. C FPC - Cryst:1 RivCr Unit 3 Seismic Verification of Tanks siv' e osie Chk'd By I Date 0R sinki _PdS lliHcs_ ~ I Calculation For: I Horizontal Heat Exchanger l Base Plate: Thickness base plate under saddle (in) tb 0.375 Depth of the base plate (in) hb 7.00 Width of the base plate (in) bb 5.00 Anchor bolt spacing (in) sb 6.00 Bolts: Number of locations, each saddle N t. 2.00 Number of anchor bolts per location Ns 1.00 Diameter of anchor bolt (in) d 0.75 Loading: SSE Floor reponse spectra at 4% damping (2) Anchor Bolt Allowables (From GIP Section 4.4 and Appendix 0 Table C.3-1) Allowabies for,75' Cast-In-Place Bolts (Mark D60 type 4 from SC-423-032) (edge distances from SC -422-043 arc minimum in transverse direction) Actual Embedment = 7,50 Allowable = 7,50 in See Note 2 Actual Spacing = 6.00 Allowable = 9.50 in Actual Edge Distance = 4.00 Allowable = 6.63 in Pnom = 15.03 ksi Vnom = 7.51 ksi RLp = 1.00 embedment red, factor RLs = 1.00 RSp = 0.86 spacing red. factor RSs = 1.00 rep = 0.84 edge distance red, factor REs = 0.37 RFp = 0.93 for fc=3000 psi concrete RFs = 0.93 RCp = 1.00 cracked concrete red. fact. RCs = 1.00 Pu' = Pnom (RLp)(RSp)(rep)(RFp)(RCp) = 10.05 Kip Vu' = Vnom (RLs)(RSs)(REs)(RFs)(RCs) = 2.59 Kip (3) Seismic Demand The seismic loading is the 4% damped SSE spectra for the Auxilia.ry Building at 119' which is determined in FPC calculation S-94-0011, "Sei9mic Veilfication of Tanks - SOUG Methodology', Rev. O,1/19/94. From calculation S-94-0011 pages 27 and 28: OBE FRS Peak (4% damping) = 0.353 g OBE FRS ZPA (4% damping) = 0.050 g Assume Seal Cooler is flexible in transverse direction, therefore take 4% SSE SPA as demand (SSE SPA = 2 times OBE SPA), or: Sh = Horizontal 4% SSE SPA = 0.706 g Sv = Vertical 4% SSE SPA (2/3 Horiz.) = 0.471 g NOTES (1) Minimum Thickness Calculation: T'LP } * {_Do ;P.F 0.112, use 0.125' + t-add = Sm+y~ (with P = 200 psi, Sm = 20000 psi, y =( 0.4 and additional thickness, t-add, = 1/16') (2) is at least 18', however, the minimum from the GlP Table C.3-1 is assumed for simplicity.The oc MUHE-?B Pace P of 5

Qiu s%.cos o.c FPC - Crystal Rivor Unit 3 Seismic Verification of Tanks nev e 1 o.i. chk d ay , oate Calculation For: [ Horizontal Heat Exchangerj 1 (4) Determine Total Welchi The weight of the seal cooler is determined by assuming that the weight density of the heaviest heat exchanger from Table 7.6 of the Florida Power Plant, Specific Procedure for 'Selsmic Verification of Nuclear Plant Equipment", Rev. 1,9/12/94 is applicable. This weight density is then multiplied by the seal cooler volume: W = ( G am) * ( Volume ) = ( 180 lb/ft ^ 3 ) * ( Volume ) = (G am) * [(D/ 2) ^ 2 * (n) * (L)) = 1750.78 lb Conservatlively take W = 2000.00 lb (5) C_etculate Overturnino due to Transverse Load _in_g Worst case overturning will be for horizontal earthquake acting transverse; i.e., resisted by the anchor bolt with 6 in center to center spacing. Let F1 and F2 each represent vertical force in anchor bolts at each s.de of the saddle base plate (see Figure on next page). Thus, F1 is the upward force resisting overturning and F2 is the force assisting overturning: (a) Sum forces vertical: F1 + F2 = ( W) * ( 1.0 - Sv ) = 1058.67 lb (b) Sum moment about base plate edge (see Figure on page 4): ( F2

a2 ) + ( F1
a1 ) + ( Sh
Heg
W) = ( W) * ( 1 - Sv ) * ( hb/2 )

where a1 = ( hb - sb ) / 2 0.50 in = a2 = ( sb) + (bb - sb ) / 2 6.50 in = (c) Solve equation (a) for F1, substitute into (b) and solve for F2: F1 = ( W ) * ( 1.0 - Sv ) - ( F2 ) = ( WV ) - ( F2 ) F2 = { ( W ) * [ ( 1 - Sv ) * ( hb/2 ) - ( Sh

Heg ) ) -

[ ( W) * ( 1.0 - SV ) * ( a1 ) ) ) / ( a2 - a1) therefore, F2 = -1235.67 lb F1 = 2294.33 lb (d) Determine anchor bolt pullout forces: Each force (F1 and F2) represent two af the neat exchanger ariehor bolts. The maximum and minimum anchor bolt forces are therefore: P max = 1147.16 lb P min = -617.83 lb Since negative anchorage forces represent bolt pullout, only the minimum force needs to be considered. That is, P 617.83 lb = (e) Determine anchor bolt shear forces: Totalshear = ( W) * ( SSE Horiz. ) 1412.00 lb = Bolt shear = ( Total shear ) / ( 4 bolts ) = V 353.00 lb = M UHE-2ti Pace 3 of s

bC Mtt-ColB rev.O FPO - Crystal River Unit 3 Seismic Verification of Tanks nev sy I oaie i chk'd By i cat. O!'Dfi //hr/6 Hli ! l[_22/& I Calculation For: fHorizontal Heat Exchanger] I I i i I (6) Figu.re_1: Calculation oeometry NN j 17'10* j 8, 10' ..,h o-o (( s- __H \\\\ M . 3, Ii Section B S F, F 3 (7) Evaluate Anchorage Allowable Maximum? Maximum anchor boli pullout 10050.09 lb 617.83 lb OK Maximum anchor bolt shear 2590.81 lb 353.00 lb OK Interaction: The interaction curves for cast-in-place bolts are taken from Section C.3.7 and Figure C.3-2 of the GlP. Since the GIP anchorage criteria for cast-in-place bolts and headed studs ensure that failure does not occur in concrete, the interaction formulation for steel failure is recommended: for 0.0 < (V/Va) < 0.3, (P/Pa) < 1 for 0.3 < (V/Va) < 1.0, 0.7 x (P/Pa) + (V/Va) < 1 0.136 therefore, since (V/Va) = 0.061 < 1 OK (P/Pa) = (8) Confirm Stresses in the Saddle are Acceptable The saddle and stiffners are ony about 2.5" deep (between the Saddle pad and top of the plinth). Bending of the stiffened saddle is adequate by inspection. For shear the anchorage has been determined as adequate and the amount of shear area in the stiffened saddles is much greater than the area of the anchor bolts (2 3/4" anchor bolts per plinth) and is therefore also adequate. CONCLUSION The Seal Coolers under evaluation:. I MUHE 2A l MUHE-28 I are acceptable since anchorage capacity greatly exceeds demand. MUHE-28 Page 4 of 5 l_

l IC S itt C O G av. o FPC - Crysici Rivar Unit 3 Solsmic Verificetion of Tanks nev sy o t. enk d ey i cat. O M !!/2 sM/ %t V12./%_ Calculation For: I l Horizontal He t Exchanger l l I FLORIDA PCWER CORPOR ATION 4m P.r.1;!i 'c t:.:U..:'. st ntlassuso stotina -=.r.. ps. .x.m. m. CRY $1AL RIVII PLANI eaasti 7sociates.,NC Wurt e 3 411000 re INC;Nttts aNo CONAttaNis etasseo 8!N'ea si.ve...i.i. n.. son t.- .~ e.> tium E'.CG touip stia FL Et i.o. o' ,x*tir,'tl.LV.c; * ' t.;f manoa as www on two No 5C 4M*043 STANDARD TYPES $PECIAL ^I?S) itet i 1 3 4 E: A E 74 4 n. - n. ~ f s" PtrHi ?trij :: o w; p s r .0 W.e. hnietseeve I

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on t.t=ctu to Aos soptatt a ecco i D L A B C H T Ti so nas P G i cia i tach 4 0% 16 '5' l' to' l'4' 16 4* Si 'l t16 1* l'5'i i t' 11 5 % i V 11 l'f $5V 4 M9 11 'O l' 9' t 4 D<.0 4 3 *' tto' r 4 4'

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'4 ' 4 l' 1 5Y 4 4 0 61 IG l l'lO' l'i IG 'f '4 ' 16 r l'5'i -~ i 6 l i 4 -...t_.

0TES*-k.L VATEQtAL to 'E A SIM. A% STEEL

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9

lorida DESIGN ANALYSIS / CALCULATION 9E'TI Crystal River Unit 3

~ eaa',I or 6 _,-~..<.,_e S96 0013 o 1 Attachment "K" Make-Up Tank Six Pages Total 4 R5 E. L *It Cl V WI HL bY, kupC*@@ L % W& > ~s h

(llt Skr CCG nV O FPC - Crystcl Rivar Unit 3 Seismic Verification of Tanks nevl ey i _ Date chk'd 8 y i Date o ) 7M ! @lNHf PdS WDLes_ Calculation For: i I l Vertical Tank on 4 Legs l 1 i Equip. ID: MUT-1 Equipment

Description:

Building: AUXlLlARY Elevation: 119 MAKE-UP TANK Rm Row / Col: 302 / L Also Applicable for: I Vertical Tank on 4 Legs: Drawing: M-6057 (Babcock & Wilcox) & SK-7848-1 Anch. Drw.: SC-421 -110, SC-412-111, SC-422-043, SC-423-027 Vendor: Babcock & Wdcox Model: ASME Section ill Dished Head Methodolony Tank MUT-1 is not a flat bottomed vertical tank so the SOUG methodology given in Section 7 of the Florida Power PSP " Seismic Verification of Nuclear Plant Equipment", Rev. 1,9/12/94, is not applicable. MUT-1 is supported by 4 legs (6" x 6" x 1/2" angle 5'-11" long welded to an 8" x 8" x 1" base plate) spaced at 90% around the perimeter. Each leg is anchored to a reinforced concrete plinth (12" x 12" x 5" high) by one 1" diameter cast-in-place bolt. Since anchorage is the critical element, calculation focuses on anchorage. Dimensions Dimensions a.e obtained from the referenced drawings Tank: Outside Diameter (in) D 96.00 OverallHeight (in) H 161.00 Thickness of tank shell (in) ts 0.438 See Note 1 Thickness of tank head (top / bottom) (in) th 0,438 See Note 1 Weight density steel (Ibf/in ^ 3) W st 0.2840 Weight density fluid (Ibf/in ^ 3) W ti 0.0361 Height of shell portion (in) hs 123.00 Height of heads (top & bottom) (in) hh 19.00 Nominal Height of water (in) hw 123.00 Assume height of Shell Anchorage: Cast-in-Place Bolts, type D-53 (see SC-423-027) Diameter Anchor Bolt (in) bd 1.000 Number anchor bolt total Nb 4.00 Number bolt per leg Nieg 1.00 Bolt Embedment (in. ) (type D-53 has Lb 10.00 to x out. Diam. an embedment > 10 x 0.D.) Bolt Spacing (center to center) (in) Sb 68.00 Bolt Edge Distance (in) Eb > 8.75 See Note 2 Concrete strength (psi) f' c 3000.00 Base Plate: Thickness base plate each leg (in) t bp 1.00 Side Dimensions (square,in) I bp 8.00 Page 1 of 5

b l c S G Lt - o o G tt v. O FPC - Crystcl Riv r Unit 3 Seismic Verification of Tanks nev! B l . osie cnk d By I cat. 0 G 1$ff/'/1{ i 9dj )'t/Ji95 Calculation For: l l Vertical Tank on 4 Legs l l I' t i I Calculation (1) Weight The tank weight consists of the shell portion, the top and bottom heads, and the leg attachments (4 leg angles, a steel attachment ring, cross bracing, etc.). The tank is conservatively assumed to be cylir.drical with top ano bottom. circular disks. The calculated weight uses an estimated thickness (see Note 1) to further overestimate weight to account for legs and miscellaneous steel. W tank = W shell + 2 (W head) + W contents = W shell = (pi) (D) (hs) (ts) (Wst) = 4609 lb W head = 2 (pi) (th) ((D) (hh) + (D/2) ^ 2] (Wst) = 3223 lb W contents = (pi) (D/2) ^ 2 (hw) (W ft) = 32140 lb Calculated W tank = 39972 lb Use as W tank = 40000 lb (2)C.G. The bottom of the tank is 16.5 in above the anchorage. The_C.G. is calculated from the anchorage base-plate as: Tank cg = (W Steel) (H/2) + (W water) (hw/2) 65.22 in (W tank) C.G. = ( Tank cg ) + ( dist to bottom ) 81.72 in = (3) Loading To detarmine the Seismic Demand use Auxiliary Building Elevation 119' SSE fk or reponse spectra at 496 damping. These spectra are determined in FPC calculation S-94-0011, " Seismic Verification of Tanks - SOUG Methodology", Rev. O,1/19/94. From S-94-0011 pages 27 and 28: OBE FRS Peak (4S6 damping) = 0.353 g OBE FRS ZPA (4% damping) = 0.05 g Conservatively assume tank is flexible and use peak spectral acceleration. For 4% SSE FRS peak use 2 times OBE Peak, therefore: Horizontal 4% SSE Peak = 0.71 g Vertical 4% SSE Peak (2/3 Horiz.) = 0 47 g (4) Overturning Worst case will be for horizontal earthquake at 45 degrees to tank legs. Therefore determine overturning for horizontal along 45 deg to legs and vertical earthquake acting upward (assisting overturning). Let F1 and F2 each represent vertical force in two legs (see Figure on next page); i.e., F1 is th:'. award fcree resisting overturning and F2 is the force assisting overturning: Pace 2 of 5

bGC fab -CrJB ttV.C; FPC - Crystcl Riv0r Unit 3 Seismic Verification of Tanks nov sy_ , oat. I cnx d By oni. Pdf 11/n/qs 0 GN2 /b&M/ ~ ~ Calculation For: l Vertical'. ink on 4 Legs l 1 l I[- Overtuming (Continued) s f, H cg *

S, X

h W a f} A o z o A 1 o F F-i Section A A (a) Moment arm = Arm = from drawing SC-422-043 = 2'- 10" = 34.00 in (b) Sum Forces vertical: F1 + F2 = (W tank) (1.0 - SSE vert) = F1 + F2 = 40000 * (1 -SSEv) - 21173 lb (c) Sum Moments about Z: F1 (Arm) = F2 (Arm) + (W tank) (SSE hor) (cg) = F1 - F2 = (W tank) (SSE hor) (cg) / (Arm) = 67878 lb (d) Solve equations (c) and (b) for F1: F1 + ( F1 - 67878 ) 21173 lb = F1 = 44526 lb F2 = -23352 lb (c) Determine anchor bolt pt:llout forces: Each force (F1 and F2) represents two of the tank legs and each leg has one 1" diameter anchor bolt. The maximum and minimum forces are: Max. anchorage vertical force (F1/2) = 22263 lb Min. anchorage vertical force (F2/2) = -11676 lb Since negative anchcrage forces represent bolt pullout, only the minimum force needs to be considered for this tank. Pu = 11676 lb (f) Determine anchor bolt shear forces: Total shear = ( W tank ) ( SSE Horiz. ) 28240 lb = Bolt shear = { Total shear ) / ( 4 bolts ) = Vu = 70SO lb Paco 3 of 5

h.lC 6 %.i-c_ct %x v. O FPC - Crystol River Unit 3 ~ Seismic Verification of Tanks nev: _s ,Date _ ! Chk d By oate 0 l Tl$y _ $8ll$f 26S W5W~AL Calculation For: I l Vertical Tank on 4 Legs l l l (S) Anchor Bolt Allowables (From GIP Section 4.4 and Appendix C) Allowables for l' Cast-in-Place Bolts (Type D-53 from SC-423-027) Pnom = 26.69 ksi Vnom = 13.35 ksi RLp = 1.00 embedment red. factor RLs = 1.00 RSp = 1.00 spacing red. factor RSs = 1.00 rep = 1.00 edge distance red, factor REs = 1.00 (Gee Note 2) RFp = 0.93 for fc=3000 psi concrete RFs = 0.93 RCp = 1.00 cracked concrete red. fact. RCs = 1.00 Pu' = Pnom (RLp)(RSp)(rep)(RFp)(RCp) = 24.71 Kip (See Note 2) Vu' = Vnom (RLs)(RSS)(REs)(RFs)(RCs) = 12.36 Kip (See Note 2) (6) Evaluate Anchorage Allowable Maximum? Maximum anchor bolt pullout 24710 lb 11676 lb OK Maximum anchor bolt shear 12360 lb 7060 lb OK Interaction: The interaction curves for cast-in-place bolts are taken from Section C 3.7 and Figure C.3-2 of the GIP. Since the GIP anchorage criteria for cast-in-place bolts and headed studs ensure that failure does not occur in concrete, the interaction formulation for steel failure is recommended: for 0.0 < (VNa) < 0.3, (P/Pa) <1 c 0.3 < (VNa) < 1.0, 0.7 x (P/Pa) + (VNa) < 1 therefore, since (VNa) 0.57 n 0.90 OK 0.7 x (P/Pa) + (VNa) = CONCLUSION The tank (s) under evaluation: l MUT -I I is/are acceptable. NOTES: (1) The specified minimum thickness given on the drawing (FA A057) is not legible. A value of 7/16* is assumed for the shell and head thicknesses. A minimum thickness calculation using the ASME formula: tm = [P DJ / (2(Sm + yP)] + a where P = internal design pressure (psi) O = outside diameter (in) Sm = allowable stress intensity (psi) y = 0.4 a = additionalthickness (corrosion, etc.) (in) yields a thickness of about 0.365'(with P = 100 and a = 1/8'). Since these values are only used to determine the weight of the tank,7/16'is a conservative estimation. (2) The reinforced plinth is 5' high (see drawing SC-422-043). The minimum embedmont for the D-53 type 4 cast-in-place bolt (see SC-423-027) is (L - T - G - bt - 1' = 23' - 2.S* .5* -1* - 1' = 18*). Therefore, even if the plinth is ignored, the remaining embedmont exceeds the Emin (18' - 5* = 13' > 10'), and the edge reduction factor can be taken as 1.0 (i.e., ignore the plinth edge). Paco 4 of S

fJC SClV -CC4 3 ru/. O a 1 FPC - Crystal River Unit 3 Seismic Verification of Tanks aef sy I-,Date i Chk'd By - 1 Date t ! ' Gd/ /D/3/K PdL 17AMM l Calculation For: ~ ~ l -Vertical Tank on 4 Legs - l 4 ] FLORIDA POWER CORPORA *llON - nu lifb:'J o

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4.m o CRY $ fat Elvit Pt ANT oittest associatts. inc Whet see 3 651000 tw INO'Nttfl AND CON $uitaNtl j i at A0iNG Ptpem A E stewe,oi. we, to: L., s o.. ~.., r)#1. C < t'. t G F $ EL 'P-Q'(IC5iH) &\\ I r.'*c. L* Vi') ,..' anArtedt al see0wN ON DwG 6 0 (C*.j*(l.lli l STANDARO TYPES SPEC /AL TYPES l 2 4 ties i i o I ., l F"' A,, ',:,P 4 -o . _,g ,qi _,., u g r 1 C" "# bi h 3 fr i % y.1.Mii ? l c.,,, Duu LlNGtN IMS{a$$ $$ P(&Il 0, Pel$(llY$ g I g mani .O L A 8 C H T Ti i D6A LIM 6h se ass P G = 4 0% 11 l* ' l' it' W 11 5' Y '4' l'1 1* l' %* 1 l 4 0% S TP M' i\\* 6 Y Y '4* 6 l\\" MV I* 't 11 T MY i 4 t.% 11 W8 Ur h* 11 4' t \\ f I I. .i l i emensens'as MNS!- Alt. MATE 9IAL TO BE A.51M. A% STEEL ei noe:mais 0148tButetWO100 Page 5 nr 5 l t..~-s -w-, rw --ee+---M ++w-,w-m=-- m a--+w,-n-rw er-r--w 31 - - -, -, - - - - - - - * + - -, ---i

9 c:lorida DESIGN ANALYSIS / CALCULATION b?M,U Crystal River Unit 3 Page l of 5 _..g _, _, ~, _ L L Attachment "L" Spent Fuel Coolant Demineralizer 5-Six Pages Total 4 6 4 d ML I. LJw @ VWE R&&F. M 4e L V'M W j

h Salt-Coi3 o O FPC - Crystrl Rivsr Unit 3 Seismic Verification of Tanks sv sy, , oate i chk'd By i Date 0 'D/M _jM 991f l Pdt i IVaaL95_ Calculation For: 1 l Vertical Tank on 4 Legs l l l l i i i Equip. ID: SFDM-1 Equipment

Description:

Building: AUXILIARY Elevation: 119 SPENT FUEL COOLANT DEMINERALIZER E Rm Row / Col: 302/J Also Applicable for: I \\ Verto Tank on 4 Legs Drawing: 40-45 00106 (FPC M001160) Anch. Drw.: SC-421-110, SC-412-111, SC-422-043, SC-423-027 Vendor: Babcock & Wilcox Model: Methodoloay Tank SFDM-1 is not a flat bottomed vertical tank so the SOUG methodology given in Section 7 of the Florida Power PSP " Seismic Verification of Nuclear Plant Equipment", Rev. 1, 9/12/94, is not applicable. SFDM-1 is supported by 4 legs (3" Sch 40 pipe welded to a 7-1/4" diameter x 1/4" circular base plate) spaced at 90% around the perimeter. Since the tank anchorage is the critical element, this calculation will focus on the anchorage. Dimensions Dimensions are obtained from the referenced drawings Tank: Outside Diameter (in) D 36.00 Overall Height (in) H 195.00 Thickness of tank shell(in) ts 0.313 s/16 drawing Thickness of tank head (top / bottom) (in) th 0.313 5/16' assumed Weight density steel (Ibf/in ^ 3) W st 0.2840 Weight density fluid (Ibf/in ^ 3) W: 0.0361 Height of shell portion (in) h3 48.00 Height of heads (top & bottom) (in) hh 7.88 Nominal Height of water (in) hw 74.00 Anchorage: Cast-in-Place Bolts, type D-53 (see SC-423-027) Diameter Anchor Bolt-(in) bd 1.000 Number anchor bolt total Nb 8.00 Number bolt per leg N ieg 2.00 Bolt Embedment (in. )- (type D-53 has Lb 10.00 10 x out. Diam. an embedment > 10 x O.D.) Bolt Spacing (center to center) (in) Sb 6.00 Bolt Edge Distance (in) Eb 6.00 Concretc strength (psi) f' c 3000.00 Base Plate: Thickness base plate each leg -(in) t bp 0.25 Side Dimensions (cire, diam., in) Ibp 7.25 SFDM Paan 1 of 5

OllC Nls -COG to.O FPC - Crystcl Riv;r Unit 3 - Seismic. Verification 'of Tanks aevi ey _ oate. I cnk d By cate "U @ /E IDl.3//71 i 961 \\2A V95_ + Calculation For: i l Vertical Tank on 4 Legs I. I Calculation (1) Weight The tank weight consists of the shell portion and the top and bottom heads. The tank is conservatively assumed to be cylindrical with top and bottom circular disks: W tank = W shell- + 2 (W head) + W contents = W shell = (pi) (D) (hs) (ts) (Wst) = 482lb W head = 2 (pi) (th) [(D) (hh) + (D/2) ^ 2] (Wst) = 339lb W contents = (pi) (D/2) ^ 2 (hw) (W fl) = 2719 lb-W tank = 3540 lb (2)C.G. The bottom of the tank is 14.125 in above the anchorage. The C.G. is calculated from the anchorage base-plate as: Tank cg = (W Steel) (H/2) + (W water) (hw/2) 51.02 in,, (W tank) C.G. = ( Tank cg ) + ( dist, to bottom ) = 65.15 in g (3) Loading To determine the Seismic Demand use Auxiliary Building Elevation 119 SSE floor reponse spectra at 4% damping. These spectra are determined in FPC calculation S-94-0011, " Seismic Verification of Tanks - SQUG Methodology", Rev. O,1/19/94. From S-94-0011 pages 27 and 28: OBE FRS Peak (4% damping) = 0.353 g OBE FRS ZPA (4% damping) = 0.05 g Conservatively assume tank is flexible and use peak spectral acceleration. For 4% SSE FRS peak use 2 times OBE Peak, therefore: Horizontal 4% SSE Peak = 0.71 g Vertical 4% SSE Peak (2/3 Horiz.) = 0.47 g (4) Overturning Worst case will be for horizontal earthquake at 45 degrees to tank legs. Therefore determine overturning for horizontal along 45 deg to legs and vertical earthquake acting upward (assisting overturning). Let F1 and F2 each represent vertical force in two legs (see Figure on next page); i.e., F1 is the upward force resisting overturning and F2 is the force assisting overtuming: SFDM - 1 Pace 2 of 5

9 c:m98.*d brida DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 9 Page of b ~ ~, _,.., _ S96-0013 1 Y q Attachment "M" Nuclear Service CCC Heat Exchangers Six Pages Total e% .e m .......,-...... -.. m m,

--1%chmenF V-cse. 5%-0018 2tv. -1 g FPC - Crystal Rivsr Unit 3 - Seismic Verification of Tanks av er oat. cneo ev oat. 0 Calculation For: 1 -WM nhz/9 afNN n n 97 Horizontal Heat Exchanger Equip, ID: SWHE 1A Ec uipment

Description:

Building: AUXILIARY N JCLEAR SERVICE-CLOSED CYCLE Elevation: 095 COOLING HEAT EXCHANGER 3A Row / Col: Sea Water Also Applicable for: ISWHE 1C; for SWHE 1B and 1D sue S97-0002 l Heat Exchanger - Drawing: 3-69-06-30275-D1 Rev. 4 and 3-69-06-30225-B1 (M1307) Anch.Drw.: SC-422-028, SC-422-041 and SC-423-026 Vendor: Struther Wells Corp. Model: Type 37-41NX32-6H-EXCH Step 1: (1) Input Data See Figure 7-13 of Florida Power Plant Specific Procedure for " Seismic Verification of Nuclear Plant Equipment", Rev. 1,9/12/94 (Notes are at the bottom of page 4.) Appliicable? Tank: Diameter (ft) D 3.16 oK Length (ft) L 41.00 oK Thickness of tank shell(in) t 0.44 7 tis in Weight of tank plus fluid (Ibf) W if 70000.0 Rev.1 change Weight density (Ibf/ft^3) Gu 218.21 o Height of c.g. above anchorage (ft) Hw 2.08 Saddle: Spacing (ft) S 22.00 see Note i Height of saddle plate from bottom of h 6.06 the tank to the base plate (in) Shear modulus (psi) G 1.12E+07 Elastic modulus (psi) E 2.90E+07 Number of Saddles Ns 2.00 oK Base Plate: Thickness base plate under saddle (in) tb 1.00 i Min, yield strength (psi) fy 30000.0 Thickness ofleg of weld - tw 0.50 i ucentricity from anchor bolt CL to es 5.00 the vertical saddle plate Bolts: Number of locations, each saddle N i. 2.00 oK Number of anchor bolts per location Ne 1.00 oK Diameter of anchor bolt (in) d 1.00 l a Distance between extreme anchor D' 1.83 oK bolts in base plate of saddle (ft) Loading: SSE Floor reponse spectra at 4% damping SWHE-1 A Page 1 of 5 r----.

ATT9MME!JT V s.9to noIS Rev I tv1 Solk U V FPC - Crystal Riv;r Unit 3_ Seismic Verification of Tanks new er

oat, enk d ey cate 0-Calculation For:

1 & #( N/u/07 w ru-ww Horizontal Heat Exchanger Step 2: (2) Anchor Bolt Allowables (From GlP Section 4.4 and Appendix C Table C.3-1) Allowables for 1.0" Cast-In-Place Bolts (Mark D4 type 4 from SC-423-026) (also from SC-422-041) Actual Embedment = 13.00 Min Allow. 10.00 in Actual Spacing = 22.00 Min Allow. 12.63 in Actual Edge Distance = 9.00 Min Allow. 8.75 in Pnom = 26.69 ksi Vnom 13.35 ksi RLp = 1.00 embedment red, factor RLs = 1.00 RSp = 1.0L spacing red, factor RSs = 1.00 rep = 1.00 edge distance red. factor REs = 1.00 RFp = 0.93 for fc=3000 psi concrete RFs= 0.93 RCp = 1.00 cracked concrete red. fact. RCs = 1.00 Pu' = Pnom (RLp)(RSp)(rep)(RFp)(RCp) = 24.71 Kip Vu' = Vnom (RLs)(RSs)(REs)(RFs)(RCs) = 12.36 Kip - Step 3: (3). Base Plate Bendino Strenoth Reduction Factor (RB) RB = Bending strength reduction factor = (fv) (tb^2) 0.40 (3) (Pu') Step 4: (4) Base Plate Weld Strennth Reduction Factor (RW) RB = Weld strength red, fact.= (tw) (es) (30600) (2.83) 8.76 Pu' Step 5: (5) Anchor Tension and Shear Allowable Pu = (Pu')(smaller of RB, RW) = 10.00 Kip Vu = Shear allowable anchorload = (Vu') = 12.36 Kip Step 6: (6) Calculated Ratios Alp = (Pu') / (Vu') = 0.81 Wb = (Wtf) / [(NS) * (NL) * (NB)] = 17500.0 lbs Vu / Wb = 0.71 Hcg / D' = 1.14 Hcg / S = 0.09 F1 = SQRT [(NS^2) + 1 ] = 2.24 F2 = SQRT [ (NL^2) * (Hcg / D')^2 + (.667^2) + ((Hcg /S)^2) * ( (NS^2) / (NS-1)^2 ) } = 2.38 SWHE-1A Page 2 of 5

AHetchmentM + F$PC - Crystal River Unit 3 S.9(o CD 3, Rex).l-- Pepe40Ro-Seismic Verification of Tanks a.v. sv o.t.- cak o ey o.t. 0 Calculation For: 1- //MMM ////rh> seg - er/vs/9 r Horizontal Heat Exchanger- ~ Step 7: (7) Determine Acceleration Capacity of Tank Anchoraae Llow = -[(Vu) / (Wb)] * [(1) /(F1)] = 0.32 g-Lup [ (Vu) / (Wb) + (0.7) / (Alp)] 0.37 g = = [ (0.7) / (P, ) * (F2) + (F1) _ Lamb = ' Smaller-of Llow or Lup = 0.32 g Step 8: (8) is Tank / Heat Exchancer Riold c. Flexible ln Transverse or Vertical? Sc = From Figure 7-15 for Heat Exch. = 13.00 ft . ( Use D = 3.156 ft ) ( Use t = 0.438 in ) is Tank / Heat Exchanger Rigid or Fjexible ? - Rigid if Sc >or= S, Flexible if Sc < S From Step 1, S = 22.000 ft Tank / Heat Exchanger in Transverse or Vertical Flexible Step 9: -(9) is Tank / Heat Exchanaer Riaid or Flexible in Lonaitudinal Direction? Flong. = [(1)/(2PI)]

SQRT[(ks)*(g)/(Wtf)]

where ks = 1 (h^3) + (h) (3 *. E

lyy)

(As

G)

( Use lyy = 907.08 in^4 ) ( Use As 29.25 in^2 ) therefore,_ Flong. = 80.8208342 ks =- 4.67E+07 Tank / Heat Exchangerin Longitudinal Direction Rigid Step 10:: (10)- Comoare Seismic' Demand to Capacity Acceleration From Steps 8 and 9,if tank / heat exchangeris: rigid - - Use Zero Period Acceleration (ZPA) of 4% damped floor response spectrum flexible - Use Peak Spectral Acceleration (SPA) of 4% dampec floor response spectrum SWHE.1A Page 3 of 5

fidehMenW FPC - Crystal Rivor Unit 3 S 4 (.o 0 0 13. 2 9 0.l Pace 6db 1 Seismic Verification of Tanks. Rev sy-oate chxo ey . o t. o _ Calculation For:. 1 yh;l:V n/n/C7 TkW /Wo3/v i Horizonta' Heat Exchanger 7//t - ~ Step 10 (Continued): The seismin loading is the 4% damped SSE spectra for the Auxiliary Building at 75' which are identical to the ground response spectra. [

Reference:

" Environmental and Seismic Qualification Program Manual", (E/SOPM), Rev. 8, Section 5.0 Seismic Qualification Data, Figure 22]. From E/SOPM Section 5.0; - OBE Spectrum Peak (2% damping) = 0,135g ZPA for 2% = 0.05g OBE Spectrum Peak (5% damping) = 0.100g ZPA for 5% = 0.05g Conservatively, for 4% SSE take 2 times 2% OBE Peak = therefore, Horizontal 4% SSE Peak = 0.27 g Vertical 4% SSE Peak (2/3 Horiz.) = 0.18 g 7 Anchorage is Adequate if: (1) Lamb > ZPA (for rigid tanks / heat exchangers) or (2) Lamb > SPA (for flexible tanks / heat exchangers) SPA (use peak as specified above) = 0.27 g Anchorage Capacity, Lamb, from Step 7 = 0.32 g Check if Capacity (Lamb) > Demand (SPA) ? OK Step 11: (11) Confirm Stresses in the Saddle are Acceptable The saddle and stiffners are only about 6" deep (between the Saddle pad and top of the plinth), Bending of the stiffened saddle is adequate by inspection, For shear the anchorage has been determined as adequate and the amount of si. ear area in the stiffened saddles is much greater than the area of the anchor bolts (2 1" anchor bolts per plinth) and is therefore also adequate. CONCLUSION The Heat Exchangers under evaluation: l SWHE-1A I ~~ l SWHE1C 1 are acceptable in accordance with Section 7 of the FPC PSP for " Seismic Verification of Nuclear Plant Equipment". For the following Heat Exchangers mounted on steel frames see S97-0002: l SWHE-1B l l SWHE-1D _J NOTES - (1) < The spacing between saddles (22') slightly exceeds the SQUG exclusion rule d 70' max. spacing. - However, the heat exchanger is rigid longitudinally, the results vary only slightly versus the same heat exchanger with 20' saddle spacing, the capacity >> demand, and SWHE.1 A is located at the base elevation in a low seismic stoa. It is concluded that the intent of the caveat is met. SWHE.1A Page 4 of 5 ,--r-

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.. - ~ gyg y FPC - Crystal River Unit 3 S % 0013, Ged.1 PcM e to 4 b~ .Seistnic Verification of Tanks sovi ey I cat. e c hk'o e y " i cat. - t/2 0 i 6& I v///d / i Pdr !i /do-Q Calculation For: / !W I p/gfo I pfg yg l Horizontal Heat Exchanger l i I I I I i FLORIDA POWER-CORPORClON 4m JJ !. J.: llj ! o /J. 4lj J4: if etisISwec, stesica ' 3, r. r,, ~~ ClT17At ttVlt PL ANT oios : a s ses.ala s. o.; vu.' M 3 Lil000 re lw*( t # * * ** CcNIVL' a % ?$ s t asia 4C et%=a l' esc %#s!*4*dce 804 La u,,...,,

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l 9 ce9M,"<d glorida DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 Page l of b w_,_,~,_ S964013 o 1 Attachment "N" Nuclear Service Closed Cycle Surge Tank Six Pages Total l l l l RL 1, L#9 6f VdM M& DP. Pee 44e tr96ggwog

s (Cdc ML,-cot 's rev O L FPC - Crystcl Rivar Unit 3 Seismic Verification of Tanks a.v ly , Da3 Chk'd By Date. ~ 0 M)f) II/2/G> \\7/RfAC Calculation For: l-Vertical Tank on 4 Legs..... ] l ~ l i Equip. ID: SWT-1 Equipment

Description:

Building: AUXILIARY NUCLEAR SERVICE CLOSED CYCLE Elevation: 095 SURGE TANK Rm Row / Col: 307/S Also Applicable for: l I Vertical Tank on 4 Wide Flange Legs Drawing: 5-315-D1 Rev.3 Anch. Drw.: SC-422-042 and SC-423-026 Vendor: Plant City Steel Co. Model: Methodoloay Tank SWT-1 is not a flat bottomed vertical tank so the SOUG methodology given in Section 7 of the Florida Power PSP " Seismic Verification of Nuclear Plant Equipment", Rev. 1,9/12/94, is not applicable. SWT-1 is supported by 4 wide flange sections (12WF53) spaced at 90% around the perimeter. Since the tank anchorage will be the critical element, this calculation will focus on the anchorage. Dimensions Dimensions are obtained from the referenced drawings Tank: Outside Diameter (in) D 132.00 Overall Height (in) H 192.00 Thickness of tank shell (in) ts 0.625 Thickness of tank head (top / bottom) (in) tn 0.875 Weight density steel (Ibf/in ^ 3) W st 0.2840 Weight density fluid (Ibf/in ^ 3) W fi 0.0361 Height of shell portion (in) ha 129.00 Height of heads (too & bottom) (in) hh 31.50 Nominal Height of water (in) hw 138.00 Anchorage: Cast-in-Place Bolts (see SC-423-026) Diameter Anchor Bolt (in) bd 0.875 Number anchor bolt total Nb 8.00 Number bolt per leg N ieg 2.00 Bolt Embedment (approx.) (in) Lb 12.00 Bolt Spacing (center to center) (in) Sb 7.00 Bolt Edge Distance (in) Eb 7.00 Concrete strength (psi) f' e 3000.00 Base Plate: Thickness base plate each leg (in) t bp 1.00 Side Dimensions (in) I bp 14.00 Paae 1 of s

$ L, - oce rev. O FPC - Crystcl Riv r Ur.it 3 Seismic Verification of Tanks nev sy I . cate I chk d ay cate Calcu'3 tion For: l Vertical Tank on 4 Legs l Calculation (1) Weight The tank weight consists of the shell portion and the top and bottom heads. The tank is conservatively assumed to be cylindrical with top and bottom circular disks: W tank = W shell + 2 (W head) + W contents = W shell = (pi) (D) (hs) (ts) (Wst) = 9495 lb W head = 2 (pi) (th) [(D) (hh) + (D/2) ^ 2] (Wst) = 13294 lb W contents = (pi) (D/2) ^ 2 (hw) (W fl) = 68175 lb W tank = 90964 lb (2)C.G. The tank is located 2'-6' above the anchorage. The C. G. is calculated from the anchorage base-plate. Tank cg = (W Steel) (H/2) + (W water) (hw/2) 75.76 in (W tank) 105.76 in C.G. = ( Tank cg ) + ( 2'-6") = (3) Loading To determine the Seismic Demand should use Auxiliary Building Elev. 95' SSE floor reponse spectra at 4% damping, The spectra for the Auxiliary Building at 95' are identical to the Ground Response spectra. [

Reference:

" Environmental and Seismic Qualification Program Manual", (E/SOPM), Rev. 8, Section 5.0 Seismic Qualification Data, Figure 22). OBE Spectral Peak (2% damping) = 0.135g ZPA for 2% = 0.05g OBE Spectral Peak (5% damping) = 0.100g ZPA for 5% = 0.05g Conservatively use 2*(2% OBE Peak) as 4% SSE = therefore, Horizontal 4% SSE Peak = 0.27 g Vertical 4% SSE Peak (2/3 Horiz.) = 0.18 g Assume tank is flexible, use Spectral Peak as acceleration (4) Overturning Worst case will be for horizontal earthquake at 45 degrees to tank legs. Therefore determine overt".iing for horizontal along 45 deg to legs and vertical earthquake ac'..ag upward (assisting overturning). Let F1 and F2 each represent vertical force in two legs (see Figure); i.e., F1 is the up vard force resisting overturning and f2 is the force assisting overturning: Pace 2 of s

Ccd c. SQL, -tes rev.C FPC - Cryst:1 Riv r Unit 3 Seismic Verification of Tanks nev ey I . D_ ate, l Chk'd By Date O WM W1Ml' HH M./B/35_ Calculation For: ~ l Vertical Tank on 4 Legs l Overturning (Continued) t r M V a 192" cs, & x i >X m V f 'b y \\ z v M i SECTW h*h J .d .p$ -- p e u. = A A F, q (a) Moment arm = Arm = [ ( D / 2) + (I bp / 2 ) ] ^ 2 / sqrt( 2 ) S1.62 in = (b) Sum Forces vertical: F1 + F2 = (W tank) (1.0 - SSE ved) = F1 + F2 = 90964 lb * (0.82) = 74590 lb (c) Sum Moments about Z: F1 (Arm) = F2 (Arm) + (W tank) (SSE hor) (cg) = F1 - F2 = (W tank) (SSE hor) (cg) / (Arm) = 50323 lb (d) Solve equations (c) and (b) for F1: F1 + ( F1 - 50323 ) 74590 lb = F1 = 62456 lb F2 = 12134 lb (e) Determine anchor bolt pullout forces: Each force (F1 and F2) represent two of the tank legs and each leg has two 7/8" diameter anchor bolts. The maximum and minimum bolt forces are: Max. anchor bolt axial force (F1/4) = 15614 lb Min. anchor bolt axial force (F2/4) = 3033 lb Since all anchor forces are positive, bolt pullout (negative force) does not occur for this tank. Therefore, pullout is zero, Pu = 0 lb (f) Determine anchor bolt shear forces: 'iotal shear = ( W tank ) ( SSE Horiz. ) = 24560 lb Bolt shear = ( Total shear ) / ( 8 bolts ) = Vu = 3070 lb Pace 3 of 5

Ocdc. SCtb-cc>thrwo FPC - Crystal Rivor Unit 3 Seismic Verification of Tanks Eev sy. cat _ ; chk d ay cate. O Mf/ Il/2/O Pd$ \\%3/A5-Calculation For: I Vertical Tank on 4 Legs I i i (5) Anchor Bolt Allowables (From GIP Section 4.4 and Appendix C) Allowables for 7/8" Cast-In-Place Bolts (From SC-422-026) Pnom = 20.40 ksi Vnom = 10.20 ksi RLp = 1.00 embedment red. factor RLs = 1.00 RSp = 1.00 spacing red. factor RSs = 1.00 rep =- 1.00 edge distance red. factor REs = 0.84 RFp = 0.93 for fc=3000 psi concrete RFs = 0.93 RCp = 1.00 cracked concrete red. fact. RCs = 1.00 Pu' = Pnom (RLp)(RSp)(rep)(RFp)(RCp) = 18.89 Kip Vu' = Vnom (RLs)(RSs)(REs)(RFs)(RCs) = 7.92 Kip (6) Evaluate Anchorage Allowable Maximum? Maximum anchor bolt pullout 18887 lb 0 lb OK Maximum anchor bolt shear 7917 lb 3070 lb OK Interaction: ( P / Pa) + (V/ Va) < 1 1 0.39 OK The interaction curves for cast-in-place bolts are taken from Section C.3.7 and Figure C.3-2 of the GlP. Since the GIP enchorage criteria for cast-in-place bolts and headed studs ensure that failure does not occur in concrete, the interaction formulation for steel failure is recommended: for 0.0 < (VNa) < 0.3, (P/Pa) <1 for 0.3 < (VNa) < 1.0, 0.7 x (P/Pa) + (VNa) < 1 therefore, since (VNa) 0.39 = 0.7 x (P/Pa) + (VNa) 0.39 < 1 OK = CONCLUSION The Tank under evaluation: l SWT-1 l is acceptable. Page 4 of 5 I

CGAt SRt -ccL% c.O FPC - Cryst:1 Riv0r Unit 3 i Scismic Vorification of Tcnks nev! By i . Data i Chk'd By i D::13 Calculation For: F Vertical Tank on 4 Legs l I SC-42b - 02b h FLORIDA POWER CORPORATION d203 03 S 423 024 o ft #23 0/4 St. Pt18056U00, fLOllDA woes onese s as esaw.mo w, esc esawn.o o CRYSTAL tlVER - PLANT ostost! Assoc 1Afss, mr UN11 NO 3 855 000 Ing (NGIN((t,' AND CONSULTANTS - Seriectural-Aacher Boll list

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@co8M*d gi:rida DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 Page_ l of 6 oocwe ~i a =in oi.e. no n #3,% S96-0013 0 Attachment "O" Waste Gas Decay Tanks Six Pages Total f hva Mt l. LJe of V4M R&&P. hene ENNW

hC Sb-CCG rev.O FPC - Crystd Riv:;r Unit 3 Seismic Verification of Tanks Revi By _ o t. i chk d er I cat. OM NOffff 1 Pd5 WD/AS-Calculation For: I l Vertical Tank ori 4 Legs l l Equip. ID: WDT-1 A Equipment

Description:

Building: AUXILIARY Elevatiu ' 095 WASTE GAS DECAY TANK A Rm Row / Col: 302 / O Also Applicable for: I WDT-18, WDT-1C l Vertical Tank on 4 Wide Flange Legs Drawing: M-6122 Rev. 4 and SK-8079-1 Anch. Drw.: SC-422-028, SC-422-041 and SC-423-026 Vendor: Buffalo Tank Div., Bethlehem Steel Corp. Model: PLS 135-1/2" x 455" x.655" M_ethodoloay ~ Tank WDT-1 A is a vertical tank supported on 4 WF legs with a 135" high shell and ASME elliptical heads (top and bottom). The SOUG methodology given in Section 7 of the Florida Power PSP ' Seismic Verification of Nuclear Plant Equipment", Rev. 1,9/12/94, is for flat bottomed vertical tanks and-is therefore not applicable. WDT-1 A is supported by 4 wide flange sections (10 WF 45) 7'-5.25"long spaced at 90% around the perimeter. Each leg is welded to a 10" x 12"x 3/4" base plate with 2 holes for 1 -1/8" diameter anchor bolts. Since the tank anchorage will be the critical element, this calculation will focus o:1 the anchorage. Dimensions Dimensions are obtained from the referenced drawings Tank: Outside Diameter (in) D 145.31 OverallHeight (in) H 210.00 Thickness of tank shell (in) t. 0.655 Thickness of tank head (top /bottcm) (in) th 0.655 Weight density stect (Ibf/in ^ 3) W st 0.2840 Weight density contents (Ibf/in ^ 3) W ti 0.0001 waste gas? Height of shell portion (in) ha 135.50 Height of heads (top & bottom) (in) hh 37.25 Nominal Height of contents (in) hw 210.00 assume full neignt Anchorage: Cast-iri-Place Bolts (see SC-423-026) Diameter Anchor Bolt (in) bd 1.125 Number anchor bolt total Nb 8.00 Number bolt per leg N ieg 2.00 Bolt Embedment (in) (type D8 has Lb 11.25 10 x outside diam. an embedment > 19") Bolt Spacing (center to center) (in) Sb 5.50 from sc-422-041 Bolt Edge Distance (in) Eb 6.25 from sc-422-041 Concrete strength (psi) f'c 3000.00 Base Plate: Thickness base plate each leg (in) t bp 0.75 - Side Dimensions (in) Ibp 10.00 110' x 12') WDT-1A Paae 1 of 5 4

00]G S%c-cc3 c.O FPC - Cryst:1 Riv r Unit 3 SeismicVerification of Tanks-neu ey 1: _ oai. l Chk'd By _ Date 0 ef//// - I t/1/,/6/ Pd( l@_q3_ Calculation For: l Vertical Tahk on 4 Legs l-I I Calculation (1) Weight The tank weight consists of the shell portion and the top and bottom heads. The tank is conservatively assumed to be cylindrical with top and bottom circular disks: W tank = W shell- + 2 (W head) + W contents + W legs = W shell = (pi) (D) (hs) (ts) (Wst) = 11507 lb W head = 2 (pi) (th) [(D) (hh) + (D/2) ^ 2] (Wst)_ = 12496 lb W contents = (pl) (D/2) ^ 2 (hw) (W ft) = 348 lb W legs = (4) [(45 lbs per WF) + (W base pl.) ] = 416 lb W tank = 24767 lb (calculated) use W tank = 25180 lb (from drawing) (2)C.G. The tank is located 1'-8.75" above the anchorage. The C. G. is caTculated from the botton of the anchorage basJ-plate. Tankcg = (W Steel) (H/2) + (W contents) (hw/2J 105.00 in (W tank) C.G. = ( Tank cg ) + ( 1'-S.75") = 125.75 in . (3) Loading To determine the Seismic Demand should use Auxiliary Building Elev. 95' SSE floor reponse spectra at 4% damping. The spectra for the Auxiliary Building at 95' are identical to the Ground Response spectra. [

Reference:

' Environmental and Seismic Qualification P ogram Manual", (E/SOPM), Rev. 8, Section 5.0 Seismic Qualification Data, Figure 22]. 7 OBE Spectral Peak (2% damping) = 0.135g - ZPA for 2% = 0.05g OBE Spectral Peak (5% damping) = 0.100g ZPA for 5% = 0.05g Conservatively '.se 2*(2% OBE Peak) as 4% SSE; therefore, Horizontal 4% SSE Peak = 0.27 g Vertical 4% SSE Peak (2/3 Horiz.) = 0.10 g Assume tank is flexible, use Spectral Peak as acceleration . (4) Overtuming Worst case will be for horizontal earthquake at 45 degrees to tank legs. L Therefore determine overturning for horizontal along 45 deg to legs and vertical earthquake acting upward (assisting overturning). Let F1 and F2 each represent vertical force in two legs (see Figure); i.e., l F1 is the upward force resisting overturning and f2 is the force assisting overturning: WDT-1A Pace 2 of s V u u L l

hCdC Ecilt -cog av. O FPC -- Cryst:1 Rivar Unit 3 Seismic Verification of. Tanks n.vi By o t. _ cnk d By oat. M" @ Jo/EN} Yd' 1 1263/AS-O Calculation For: l Vertical Tank on 4 Legs l Overturning (Continued) l D j l D S y f H cg *

S,

=x } ,..-, j,. ] w.. ^ " Z A o F F a i Section A-A (a) Moment arm = Arm = [ ( D / 2 ) + (I bp / 2 ) ) ^ 2 / sqrt( 2 ) 54.91 in = (b) Sum Forces vertical: F1 + F2 = (W tank) (1.0 - SSE vert) = F1 + F2 = 25180 lb * (0.82) = 20648 lb (c) Sum Moments about Z: F1 (Arm) = F2 (Arm) + (W tank) (SSE hor) (cg) = F1 - F2 = (W tank) (SSE hor) (eg) / (Arm) = 15569 lb (d) Solve equations (c) and (b) for F1: F1 + ( F1 - 15569 ) 20648 lb = F1 = 18109 lb F2 = 2539 lb (e) Determine anchor bolt pullout forces: Each force (F1 and F2) represent two of the tank legs and each leg has two 1 -1/8" diameter anchor bolts. The maximum and minimum bolt forces ar~ Max. anchor bolt axial force (F1/4) = 4527 lb Min. anchor bolt axial force (F2/4) = 635 lb Since all anchor forces are positive, bolt pullout (negative force) does not occur for this tank. Therefore, pullout is zero, Pu = 0 lb (f) Determine anchor bolt shear forces: Total shear = ( W tank ) ( SSE Horiz. ) S799 lb = Bolt shear = ( Total shear ) / ( 8 bolts ) = Vu = 850 lb WnT-1A Paae 3 of 5 4

boJc SQL. -cot 3, ev.O FPC - Cryst:1 River Unit 3 Seismic Verification of Tanks Revi By . caie enkdBy i cate Calculation For: '7XM EM/91 Td 5 LL7/al.B5_ 0 l l Vertical Tank on 4 Legs l l i i (5) Anchor Bolt Allowables (From GIP Section 4.4 and Appendix C) Allowables for 1-1/8" Cast-in-Place Bolts (D8 type 4 from SC-423-026) Pnom = 33.80 '<s; Vnom = 16.90 ksi RLp = 1.00 embedment red. factor RLs = 1.00 RSp = 0.75 spacing red, factor RSs = 1.00 rep = 0.85 edge distance red. factor REs = 0.40 RFp = 0.93 for fc=3000 psi concrete RFs = 0.93 RCp = 1.00 cracked concrete red. fact. RCs = 1.00 Pu' = Pnom (RLp)(RSp)(rep)(RFp)(RCp) = 20.13 Kip Vu' = Vnom (RLs)(RSs)(REs)(RFs)(RCs) = 6.33 Kip (6) Evaluate Anchorage A!!owable Maximum? Maximum anchor bolt pullout 20130 lb 0 lb UK Maximum anchor bolt shear 6326 lb 850 lb OK Interaction: The interaction curves for cast-in-place bolts are taken from Section C.3.7 and Figure C.3-2 cf the GIP. Since the GIP anchorage criteria for cast-in-place bolts and headed studs ensure that failure does not occur in concrete, the interaction formulation for steel failure is recommended: for 0.0 < (VNa) < 0.3, (P/Pa) < 1 for 0.3 < (VNa) < 1.0, 0.7 x (P/Pa) + (VNa) < 1 therefore, since (VNa) 0.13 = (P/Pa) 0.00 < 1 OK = C_ONCLUSIO_N The vertical tanks under evaluation. l WDT-1 A l l WDT-13 l l WDT-1C l are acceptable. WDT-1A Page 4 of s

r,MC.$la -CC6 tt31.O FPC - Crystcl Riv r Unit 3 ~ Seismic Verification of Tanks nev' e o.i. i enk d ey. cat. 0 [iGl 10/%/][ W < h\\W$.L. 7 Calculation For: l Vertical Tank on 4 Legs j J _[ l i i i FLORIDA POWER CORFOR ATION mb i.' 1 5 423 0/4 de Sc f/J v h it et$esinso enoties .t. mee i... ..m... Cl*$1At livil PL ant ~ ~$iGTfrJECUilfrig u,,,, o i sin ooo.w eso ~ tis suo rowswa~t. tiaoewo Ptuwe Siece ewel.AeoeBeh(+ ~ E. J J - 3.,*J *s.'.J.' f.,*."I dy_* ! tC. & M fG $W.'). waras as s own on ovo uol.'.dll. p5l.'il.:v's STENDARQ.LYPES SPfC/AL TYPf$ / vest i L F ~ A,, 8i A., .s. ,. o s,, .. ;.-pj s I .o -aL.pg g. f ! L2 _..z.,,, - > + - a > ~%% e oa L e i,r t a tresset ",',,'[ so r.att e,

.et ativt I

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li' l]

lh' lI0h' 4 Oil 4 l 'I ' 2.C' 24' 4 5' 4' 12 ' 4 if' (* 6 ' 1' 8 l'* l'0k' 4 01) 6 /* o' 3' 2h' 8 5' Q* 4 i-i I I.IS*-Al L T MATERIML TO BC A S.TM. A 36 emenue.s aeg m acnes vaais atuee ms metto WDT-1 A W.ce 5 of 5 _,.J_-.__

C:lculati:n: S96-0013 rev. 0 FPC - Cryst:1 Riv:r Unit 3 Seismic Verification of Tanks nev! By oi cnk d ay o i. O G7]ll d(1YGf Pdt \\?/J3/AL Calculation For: l Vertical Tank on Skirt ] o 7 (5) Anchor Bolt Allowables (From GlP Section 4.4 and Appendix C) Allowables for 3/4' Cast-in-Place Bolts (B-13 Type 2 from SC-423-044) Pnom = 15.03 ksi Vnom = 7.51 ksi RLp = 1.00 embedmont red. factor RLs = 1.00 RSp = 1.00 spacing red. factor RSs = 1.00 rep = 1.00 edge distance red. factor REs = 1.00 RFp = 0.9S for fc=3000 psi concrete RFs = 0.93 RCp - 1.00 cracked concrete red. fact. RCs >= 1.00 Pu' = Pnom (RLp)(RSp)(rep)(RFp)(RCp) = 13.92 K'p Vu' = Vnom (RLs)(RSs)(REs)(RFs)(RCs) = 6.95 Kip (6) Evaluate Anchorage Allowable Maxirnum? Maximum anchor bolt pullout 13915 lb 192 lb DK Maximum anchor bolt shear 6953 lb 102 lb OK Interaction: The interaction curves for cast-In-place bolts are taken from Section C.3.7 and Figure C.3-2 of the GIP. Since the GIP anchorage criteria for cast-in-place bolts and headed studs ensure that failure does not occur in concrete, the interaction formulation for steel failure is recommended: for 0.0 < (VNa) < 0.3, (P/Pa) < 1 for 0.3 < (VNa) < 1.0, 0.7 x (P/Pa) + (VNa) < 1 therefore, since (V/Va) 0.01 = (P/Pa) 0.01 < 1 OK = CONCLUSION The tanks under evaluation: 1 EGT-1A l l EGT-2A l I EGT-1B l l EGT-2B l are acceptable. EGT-1A Page 4 of s )

C:lculati:n! 596 0013 rev, O FPC - Crystal Riv r Unit 3 i cat. i enk day ~i \\?h' 3L95. Seismic Verification of Tanks sev' oat. Ofhsb-JCfRN'f Hii~ } Calculation For: [ Vertical Tank on Skirt I FLORIDA POWER CORPORATION d20 N / J C'! 'ti e // ti .c. u. .a. .s.. it teini p uto. etosion Caf8tAl flVit PLAlll ciL646r 41 E T Hig. enc ING Ntitt AND Cow &VtteNil 811000 gw Wes.1 ts0 $ ....['" D,, 9,wswat. Anckee lett in J. ( ' . f} lt *8 l.*" . i!!. sisir t' s i-S autism as s,.o-w ou owo a.o fit _ w // STANDARD TYPES SPECIAL TYPES ~ 8 8 8 8 1 Fl. & t 5' S 7.; w *p< L' s j sn g l -,p'.snlir n . m.o. x Eq(.. ; .o 1.o ; u wr 0: 11 I 5 h j5 't-d .d { #11N al[ ldh yf y 3 s 5 l f r_. i ",0,,*' so etatt e, y, tiee aeeve . Dias Lt me.t. tuetaos I A t s O L A B C H T Tg se at* P G i

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Ltasta et 0 I

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~ I E.:J J i 't ]Q' 3" th' ? s 4 b *lf 'Vi l' 9' fp' 26 J' ij' '4 ' i 8.'s !b k's Cit' th 4 A.11 214 ' l'l ' 2' 0' Zis T If 't ' C4 I l'Il ' 74' h'i l b ' l' O' W l'f is 74' ' NOTES:- KL MA TICIC ~; of 451M 'A'3(' 3'5E5 UV' m,g, .wltwt AS Sl@d CN WJ5: SC'421*.'7/ N'/:t*:12. en ucnts vuttu 't ""*# "'" ' St.42/.17J / Jc.42t+ /74 EGT-1A Page 5 of 5

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(LttMltmwtt 'N SM DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 sw PZ eA 2 vut=, munncano =vuote arm v mvvase =vvetana $-%-001b I NA PE A C.TD R COO L AWT B L E E D. TA4E5 wbT-3 A, w b 7 - 5 S, w b r - S c. C H E.c t S E ls.m ic ( A PA cit y or 'bLEE D T/W ES. vslu b PS P SEtrivy ~7 t^1E THODo to 6 4, (, NOTE ; state T#E TAf4(1 A 4 C 140 r. F L AT Bo rToevi T A u vs, A w d ra c y. si r o g TA n ic.s tM.T.s, no r A LL rn e src/s s y THE PkP A RE APPLIL ASL E ) P A c u m e.r c e ?>nr.ui c3vs. i tad 4 7-l g,ptf ), L 6 a0 E.typuo< tic At.. T'An. .YES .idwG.lsoosur.e). Aus of.sym%crcy u vcA.ric% .Tcs (ow6.#8 iso.sur.,)~ p t Ar Borrow . Wo .(.twosecosHrd. Auteouo to.coau2c rc: fab, ME.s. I Dwb. 5C 411-Atl) TA4y. tAATEAi4L SA.140 Tf 304t ..,., N E: 5 ' ( bw 6 16l30.alTU. A9 ncA sotTs EvcuM. 5fo.'Co M ts.._..( b w 0. st-42 t o+ d A,po A.BotTs c Asr lu PL ALE N ES, (bvyG X-412 04lp 5c CPf24) regio i w A t t,t e r. si m u ^ A N ES. ( Eb4D. G//i, PG ?.t ) e s T A u a ^ oiv3l R e 5 fo ns ( & NE5 CD wb 1 g:3 o 387Q. TA n Htibw1.H't.so 10 80 (t. yEs (bw(3 #6 8.10 s '# r s), gaa avio Hpwr:H S 80 ro Fo 4t 3 E3 (. ow 0. le l30 3H r.i)

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P. : 12.0 "

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A77R6HMtIJ T"P S"h DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 gg _ p4. g 3 7 mv.,i o,..,,,c.,,o, ~ v.. .n,. o~ .vo.m.a S-b col 3 I M/A siate THc hstcc o raues A 4.c Not FLAr BortoMc.0 TAuv.s J EPRI R c POR.T EPRI uP-g 2.e LL m i yotvinc 4 ar eomu c aos I D C T E R MIUi4 (,7 Au EQvgvALcur cyLipoticAL redir., IT Sv6 G C t. r3 FINDlW6 Au EQvsv A LE ur.T/4 (. VyTH THE SAME CA 0lvJlVOLVMg ( P6 2*39 ) Feom essa rat e u.'- 5. r tie eceso ru ts ne v e.: Cerr % Pl9) tat 4 %, VOLVM E; liOSO..St ) 3 .Liquib voLvM6,: lOl50.Vt EOQt V AL Cu l-hcl (,Hr D F A Th 4 K MJ tT d.A 4ADiv5 O l' /O ' e' E': votust /w.

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R T Tr9 0t4 M C f) T f DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 gg pg,, gg avoi=1 we w. cat,o vo.. aim,w ac uoin, avo.ta,ia S- % - 0 013 1 4/A ST E. P

7. '

DE TE4Mt u t,Tl4 6 FOLLo W8M6 ,P 4710,.5 H/R.: 388/12.0): 3. Z 3 E /(2. I O.34.4fl10." ( O. 00 E.9 3 E, = (0.tsslog 4- $ 0. 8 fl05) t(Olbl15 4 lo s f( 0.344 4105') g 4.21 boy : 0. 7 9 5 '.' l ;:. t.s.& b, a h 2. bO: 25 %E5 k e O.2726 ". 12o.,_:.O.OOk$ .he 4 /R : 0 7.725 Trdf4 : Tr(1,z57/4 .l.2.3,/ 2ccou sccroh Aaci AuL906,Ect. A b: ~ t' : M A6 E t, 204l.15 BoEG ~' 2 TT A E, _ } U ito ,26.3E4 4 s

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I A7 TAC 4r1.cn7 9 DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 Sheet P(o of22 4Cuad[N f IDtNilFICA faON NuwSta Rt viS40N REl/M ARS# NUMDC49ILE S-% 0o13 l M/A. STEP 3: . Fb/iD Sicatisf AL MoDet Ff sQv cuq DE7Edwuc f. 4 .F A.c m TABLE 7.-2. UM? A/ R.3 3,21 Lecj p_ : o,.0 b.1. F.' = o. 9 p h.:F[ .l.2co/R OM _ lM0//to) ~ F,

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ATTAQ4 Ment P Sh DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 sheet P7 of 2 2. 4CVWINT IDENilFCAItON NUW8th Rivt&!ON MillW A%$P NUW$[RFILE 5-9(.g-co l3 / N/A i STEP 5: D E T E 2 / M M E. B ASE .5 HEM LOAD (O) F6u F16 7-5 I w8Eu kg/tk : 0.005 H

1.23 Q.' :.. O. 7 5 3.

Q: _ Q' W t Sqc .o. 753 d te79 %o + 0.M

. 71 X l.00 0 S.TE.P to,'

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o. 7n

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o. 69 (a.

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477EfdM6AIT P DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 sheet P8 of 22. me..,,n.n,mo. u. an.. .n-- vo.-u S- %- OOl3 I N/A STE F 6; .cou rluvED m : 6,33 % o *3a s. (o.i + + o.s314 o. i A )%.(o.ut 4 c.m3 3 0.14 ) ' + L .( o.esao.471. - c.nt + o s93)i.19 P ].iI t. = 633360 48B h o.oW57+(.o..of 7 )' +[. oci 47fI4N) !2-l,% E.7.tv4 L.., a STE P l.' D E.TE E MI M E. SOLT~ TE NSI.L E. LOAD.CAFottry, C R c r,16 i e. sf e riog 4,( AP F, t.,) i e Fo R..l'l4"$ t.45 T I N PL $t E_._SOLT), (CEC TA BL E C.3,d pr., C,3d,_. _ l l ..;.a -.-4-... Pv Llour c A P. : 4). 7 2. K/.6o lt 41 p2./,83o# 3 %,S8 Esi 1 Ii s " <. de' (ep/,89a;.suca u.9. pa.; me.4 10.86'

13. 4 s. 4y' ss au cy, 20.eup.Is mw wsw =

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0 857.

5500 1 ~ P uu. ou r e.v : 4),3t+ o.as? : 35.75 r/ tog sacAa. coe : zo.sa. o.as 7 -

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_4 i l f J 6 04 D00 SF1

QTiinHt-1ELtT P SM DESIGN ANALYSIS /CALOULATION Crystal River Unit 3 sue P4 Mll MCUMINI 4DENIlFiCaflON kVWetR mt vibtON REtiMANSP NUWSt% FILE S'Ho-0013 l N/A STEP 8 C H Ec t'. TO, P PL A.TF PCA DwG 18130 87 3 Tyirvor hAye m A VC d DC 4 L. t (L n 6 PL A L 4'r DfL.0v7Cl C06 E o f Ty E TC P PL A TE, \\ T TH E F.t rore, 70 ocrfamc. Tac owe st<.gre, .t4 T H E TO P f Lq T E, E E FE 4. TO ^ fotmqt4 r0 A g rrL41, . $ LTIL A s y 5," c o. By 8evhoyo 7 coa /t.it. t wM.Ac q e, yoyac, . fA GL E 2.(, 0.^5 E. 8 b (pj s95}. _ 11IIIlit11 i / .- g ..i Oiv)Is.llh._.9.g, 7p 3 \\ l 21YL. '1T rb. s a E 3. h, '/ ' ~ o % = 3

417&o_. (l to.3)ln [zia M + o. cta 3

3( 75) ' L ' 9.(. ,..j \\ !_ 'i j / s i / '.. %: 295i8 g ( S ocoopp' .... \\ 4-f ><io uisa. b/t 4 ' /. " / TO P.,PLi\\ TE lb Qll. l d_. t \\NIud E/L Ose% crc 4. Fo tt l%"6 80 LT, : 3" '.. ch :. I 'I " t

t. :. 3/4."

4/b : 4. O 4, :. o.* ok 7 . l'O. Iks.. 4' . w. t K ' Q, 3 C3 pl % 30 Vsl ,.3.y-. _a .._.,.,..). t i ..i 900 SF1

ATTFNM6MT P SM DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 Sheet 2_ICL. of 2 2. XWAlWINT ID(NilFtCAHON NUWS(R RE Vl&#0N R(l'WA%$P NUWBENFtLE S -% -oot 3 I A/A ST E.9 9 ', C H E C t'.

s. K i cT' srn.Esscc Av ts A A ArrAc4~ mr HOTC t bu E. TO TH E co Ge *Goot Arw *4 o C THE TO P P L A T E,. THC

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l. 3 2 2 o.03 e'

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h. C % c. HEIGHT-8'/4" f..

R. = t'om& A L 00% o f S4LT = 12 o Yt. b r,.* BOTTOM.pt. ATE TditWC.53. 0 3/4" s - y, : 1.0 0 40 . O.177+11 e o 75*. 0,75 ) 2., f, o . [l20.5 + o.37.d'/L O' W 4t hn-+ 2.125

l. 3 2.

O A D o.o31 Gzo.s 4 oa,5) 9 (4415 s.(sN) 0:o.5=.375 I, h (19 +[8.257

. (o,175)I .' 6b 39 P ( hc 3 0 0 0 0 p s e' S XIR.T. $ h ELL. AT-C. A A c, /\\ rTA t,g g ey r,13, o k. ;, i i -s g ,.3 f i D9O 900 871

ArrMArvur P DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 sheet _Pil at _12 >OCUWf 8st 80(NfiF8CAflON NUMBER RtvissON REUWA4%P NVWBEMILE 5- % - 0 o13 l 4/A STE P 10.' c.14 E c t. v c a.r e t. 57mrc ac/2 pcn res uorE' TMil u A to 4 c/2VA T'V E C H Ei l'- O'Nt f. 78 C. V EdLit L4 L STirfE.uctl At,4C H 4 v e~ A Cl u h Il. 67-C ON r(p g, t Ov? C/2 .E 06 E. o 3 L : vccr, Pt.cE wcrs 2 3 /.t" , l) Y-- 95 T ' V.U3. ft.MC T@ l.. N8 " T - h/foov]. t-t ' O Haitt T6 E : 3A" 'I 10 L I1.b4 . '. 0? . 2.) . T > Q.04 ( h - c ) 1 J > 0,5 " J: 0,MI h O. T. O.M (6.25.75) . Lowescc. 5sNvE.TNE AE.1.5 A-R.375 > q.3 9 ~rmes o f c q u o. 4.t r,s c.. t ~ 0VT4eAE _ EC6C,0f THE STIfftuqh T.M.N4bG.72.\\TMCll.T#$ 'S ACC E.PTA 6L E .3) Pu q 2.ioco est.

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ATYACMHellTP DESIGN ANALYSIS / CALCULATION Crystal Hiver Unit 3 sheet Pil of Zf socuwtut ios=micatc= auwata armon atvuaw auwst=ne 5-W ools i 4/A S T E P lll C 8 N e. T o s Kle.T watt WELb MOTE .iH t$ Is, A VELY.CousEa V A TiyE C HW-As THE CNAIL l$. A. CC M 7tlNfv!, CtMW A C O J t4 b T~t/ E S l't/2 T, W w ~ Pu I ) *.+ C h 4+1h / Q rs t 0.fa*7 h k. I 407 g [ l \\,1 2*l15 g..,j IS i,(8 25) /,,,{,,lqy,3,gg + c bM(8.2.5)' 1 i ~l15b15/t9 ~ Nw (, %Op h w.6 1154 ( 3oGcohd3/25[I 1254 b/in (G761. l.6//o ~ W E L B. L ETW E E 9 C H A i /l 5 s El.cr W A u., ij, qK G STEPS

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6 q A h I E 6 46 Dec 0F1

fiTYhtut1ENT f Oh DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 smt Pl3 M 22 )OCW(ht IDENilf,CATaON NUMS(R A(Vl$ EON PlvWA%$P NUWSI% flit S-%-ool3 1 4 /A STEP 15'. bcreawue dismous.saave svt.eui% sru_.c C. A AA9T.y FATo A foe sg.i0-r pd *. { O. &5 f. dy Es T F / ta' ~ 1-o.in1 Q e?.) =.Il-o.mOsW,) ? o ee . I: .a. k*.__,..P-lf$.!IL..1]l .l2.0 o,33) !.h s12 - le L. ......t m._,.. . 6l

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0 lot 11 tit ? Shr DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 sheet Pl4 of it 'rQCVWE.NT lOINT4flCAf TON NUW9(81 R(yt$10N AL& WAR $P NUWBE4 8 8Lt 5-% - Col 3 I 4/A C H E c.E. T'A M .5 L i c7 STP_ESS E.s C099LCR . to t A pn.cz,y t ur(1E4), To 8E P//* S t all' A Y.@r s (ILO.S Y (ILO*l2Sf 9

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t W O4 W_D T 9 M DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 g g y,, p avoi.a emi,wo., ~vo.o .im ~ . ~m. vo.wu 5,- 9 (0- D OI 3 / N/A STEP r7//s; NsTE AD OF CHEt(Mb Foc. OVEAr@u90 CA#Acri1, D ETE 4 4E. n 9sio n su Bott.s put to ven.rret LoAba ovra vguiss ocu,- i ( cEr A rrAt u $ foc octrtr~uiwb Bocr sre!.ca.s cs,) TotAt yeaTot toncP: 0. 743 E4 las (, cec Pc, R&) rorAt <:Ncc rucuius momc ar M l,53 E 7// 2 5 l WEG H Ib Grreo es9) ro r% f er so cis : (.2o ) 14'p rocts 4.44 4 0 A m 4 T c 4. p - 20,'- $ '/ 4 " #w G 16@sa r 3 teoriou P-r), 6CAdlub. PLAff W orH,h,.:Q " (D W6,18f 30 54f,3 SE tr c-C), YG. Boer A g= o.890;,,t 9.ccr peg, R ATio. o f moovtus oF 'E LN.TaTy _ Fod. 5 7Ect. To T//A r,ol: co atec 7E, _ is n io. ( ger fe 64-1,4 or.ortex s ) e q .p6 6-J-/h) e./D. i, 16 /h, , ( c.y A r t 6 co nwe v r E. t J u. s h%4 M/ P : 1. FS EG.o. 70.cb. c: E/b : 171b /20,4.4 : o.08 4 to *. 8.9 0 ft(re 1o,44) 6, .14 As fr o = . ot3 m t! l<.. ( n t h f i. _._ G+flo-r).0L3 t 10 .Gll1% ~ E /L,, .617[*013) z. ? 2.7 t ,t 900 SF1

ATTACHnenTp 9"""** DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 sheet.PN, or Jz

CCuw(NT IDENTIFICAflON NVWS(M p(VI&oON ALVW A%SP hJW8t% FILE 5 - % - 0 011 I

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i Engineering instructi:n No. 2 WMCI r ec. cryst.i n'.e. unn s 'hb l # 't h PAM jg) seism. venraaenof Tanke oc ssaist.oSt OF

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,,,,j d,j d i CALCULAT10N ,4ces W u,ennntyyn ononatoa SJ.Sanan Wo. DATF 01/18/84 Using the Power Method described in Calculation DC-5520127.0SE. the 4.0% damped ground spectrum is derrved as follows: pl := 3 02:: 5 03 := 4 ANALYSIS / CALCULATION

DI n = 0.563 n: =

In [2 DOC ID # S-% 00l3 ATT # P REV l SHEET _P/7 OF tt i A4Gnal; * ' A3nnal; f 8'"I

A5nnal; 0.15 u

F g\\ N u

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N g \\ Acceleration g J 2,3 0.05 /f /$ 0.1 1 10 100 IJ Frequency (H2) Figure - OBE Ground Response Spectra for 3.0%,4.0%, and 5.0% equipment damping. Hothental Direction. GAH46 2-92 THIS IS... PERI.'ANENT RECORD DO NOT DESTROY

Engine: ring instruction Ns. 2 6vWI

* '".m.ose rpc. Crysul Riv.e unit 3

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s..,n..v.a... an. n.n6.

oc.ssao REV. ( g ] d M cat. Cut.ATION PAGFS # ucnnnt uto ' nainwAvon s.A seenen. wo. ] _ /10/94 01 DATF Usting of the 4.0% Damped Horizontal OBE Ground Spectrum J. 1 35 Frequency (Hz) Acceleration (g) ANALYSIS / CALCULATION DOC iD # S~%-MG ATT # P~ F, AMnal REV 1 SHEET P!3 OF 7L ]0.4 0.02 0.036 E 0.053 EN 0.071 I 0.089 D 0.108 id 0.109 1.6 0.107 ~ U U"TLI 2 0.101 T.2 M I4i 0.097 3 0.095 D 0.094 3 0.092 T' O.091 h 0.09 n m IN 0.087 ~ 4 0.086 IJ 0.085i 4.4 0.084' G 0.083 I.N 0.082 T 0.081 D

6. irs-I..

0.079 5.6 0.078 D 0'UTE T 0.U n 6J 0.07J 6.4 0.On U 0.072 U 0.071 T 0.07 G 4446 2 92 THIS IS A PERMANf_'NT RECORD DO NOT DESTROY

4 TABLE 11-5 Disposal Systen Component Data 3 CAFACITY. FT (each tank) CESIGN MATERIAL Tamp. Press YttifED CtstGN Sil5MIC I!!2 MO. MAME rret Tota l tiquid F pste Body Lintag TO CDDE ctstGs CUMMER15 907-3A R.C. Bleed Tads V/5 7 050 10.150 250 25 $5 Mone V.H. ASME Ill-C Class 1 Mawlm,= crerating te v/ press is WOT-36 WDf-3C 15CF/e3 psig. Contates ceaely one primary syste= voi WDT-4 Misc. Vaste Storage H/5 3.150 2.750 250 25 Rene V.M. AsME til-C Class 1 Martswr or* rat tag teep / press is Tank 150F/e3 ps69 WOT-5 Reactor Coolant

/L 831 561 300 100 55 None V.M.

ASM tit-C Class I Ruptere disk provides Drain Tank overpressure relief. Internal plate coils previde coollag. UDT-6 Spent Resin Storage r/t 920 860 150 15 $5 Mone $g ASME III-C Class I theinal resia capacity 800 f t I fank or two year's retent los as des 6,n tasis. WDi-TA Concentrated Vaste V/t 920 728 200 15 SS sone V.M. ASME Ill-C Class i Moeinal can year's retention of b01-18 Storage Tank s evaporator concentrate. WDT-8A Concentrated Boric v/t 920 728 200 15 51 None V.M. ASME III-C Class I nominal one year's storag* rer WDf-88 Acid f anb s Table 11-3. It em f. t. WUT-9 Mestrailier Ta4 V/t' 530 470 150 15 C5 Rubber Atm. ASME III-C Class Closed Vent System Legend: V/S - vertical skirt 11/5 - horizontal saddle V/L - vertical legs ANALYSIS / CALCULATION V/II - vent licader DOC ID # SSfo@'3 ATT # b av i SHEET Pl9 OF Zr 11-48 (Rev. 19) I f e

FPC - Crystal River Unit 3 1 Seismic Verification of Tanks nev' ey 1 osi. i enk e ey oate ~ 0 l 7/T f '?$/$ Yid QA( .' \\h5 Calculation For: I' I [ Vertical Tank on 4 Legs 1 I t i ~ t l l + _S_C - 42b - 02fo FLORlDA POWER CORPORATION 403 03 5 JJJ c14 o f.' /21 v/4 1 St. Pttlesnuto, sto ic A .... ;w, u. .m u... C2Y$7AL RIY!2 PL ANT Gaut a t AIR <IAtss. mc. UNif NO $ gj! (<4 rw INCINt!!.* AND CcN!Vtt ANil g( 9$' $#ttf f 9tol-Aachot Bolf Uit p.,,, .J. ! 'UJ P i lJ C i SJ ! : W OE L i f. T A'C. )?f W/1l A t19pyt '".. ~ u esv.at as snowH on owo.Nofl.411.1s!!$&J Nl

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l " ' " '" ** n '

,_ STANDARD TYPES SPECIAL TYPES fut t 1 3 4 5 ~ ~ T"" 6} r 1 i '".. 3 t , 6,, s T'.S si _ s r u L[ E y~ , o., p u w,, WM, )Q ~ o' l I.1m _1 # O] 3 k i tLi 3 l3[ d,.r> -. J'_jm , _.e,,, s;,7 4

i, -

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/' .'J J,' 69 S' fj' ))' pg 2' o M{' 4 0$ 6 '1 ' l ' ' l 19' 8 j' Jy fj' 8 l}f //?)' 1 1 06 /! b* l 2' /2 $' lj ' Pj ' /2 2' 1;Ch' 4 O t! 4' / * ! l 2' $3 j'.!y )) ' gg 2' I!O$' 2 4 DS 2J ; 12' 2 *,1 * !. 2f' 24 S' $; s Q' 2j 24' t' 6ll* f 0.9 6J l' z' O' 2)y 30 s* Igo.y* 99 \\ A' l' th' J2 j' J)

Q* jj t'

O'9" 4 CIC E: b 4 2//,I .h' l'* E ' IIY l2 J' Q' fj ' /2 llj' ! C')' l ANALYSIS / CALCULATION ooc 10. 5.9o-coq m n P REV l SHEET.P)Q.0F 21 Pano 5 of 5

Table 12.2 SECTION lli, DIYislON I - APPENI' ICES 1983 Ldition TABLE I.2.2 o, YlELD STRENGTH VALUES S, FOR AUSTEN! TIC STEELS, HIGH N!CKEL ALLOYS, I AND COPPER-NICKEL ALLOYS 5pecified Type Product M4 Yseld Spec. Nominal or Form

Strength, No.

Composititen Grade Class (Nett (1)) ksi High Ahoy 5teels Type 304 5tainless Steels sa) SA 182 F304L Forg. ~ W85 SA 21) TP)o4L Smts. Tube 5 A.240 304 L P!*te },,,- SA 249 T P304 L Wid. Tube 5A.312 TP)D4L Wid. & Smts. Pipe SA.358 304L 1 Wid. Pipe SA 40) WP304 L Fitting - 25 5A 40) WP)D4 LW Wid. Fitting 5A 479 304L Bar SA 688 T P)04L Wid. Tube SA 81) TP304L Ydd. Tube k SA 814 TP)D4L Wid. Tube ga) 5A.1P2 F304LN Forging t u. Q = W85 5A.21) TP)v4LN Smls. Tube 5A 240 304LN Plate g G 5A 249 TP304LN Wid. Tube O 'L SA-312 TP304LN Wid. & Smis. Pipe h 5A.336 F)04LN F orging <b 5A.)$8 304 LN Wid. Pipe M ul 5A.376 TP304LN Smis. Pipe h 5A 40) WP)D4LN Fitting a 5A-40) WP)D4LNW Wid. Fitti.ig eq-3 5A 479 304LN Bar O o-SA 688 TP304L N Wid. Tube hh SA 81) TP304LN Wid. Tube y) % SA 814 TP)04LN Wid. Tube g SA.)$1 CF) Casting 4 SA 351 CF8 Casting 2 h@ SA 451 CPF) Cast Pipe < o gg 5 A-4 51 CPF8 Cast Pipe SA 182 F304 & F30e" Forg 5A 21) TP)D4 & T ( H Smis. Tube - 30 5A 240 304 & 304h Plate SA-240 305 Plate 5A 249 TP304 & TP304H Vad. Tube SA 312 TP)o4 & TP)D4H Smls. & W18. Pipe SA.))6 F304 & F)C4H Forg $4 358 304 & 304H 1. Wid. P.pe SA 376 TP)D4 & TP)S4H Smis, Pipe SA 40) WP304 & WP304H Fitting SA-40) WP304W & WP)o4HW Wid. Fitting 5 A-430 FP304 & FP)D4H Forg. Pipe SA 451 CPF8 Cast Pipe S A-4 $2 T P)D4 H Cast Pipe S A-479 302 Bar SA 479 304 & 304H Bar SA 479 ER308 Bar SA 688 TP304 Wid. Tube SA-81) TP304 4 i# 304H Wid. Tube 8 SA-814 TP304 & TP304H Wid. Tube 56

__. _ - - ~ ~. - - -. ~. - -. - _ --_-.- -.

i l 1983 Eeden APPENDIX l Table 16.0 l TABLE l-6 0 i M00VLt OF ELASTICITY f 0F MATERIALS FOR CIVEN TEMPERATURES t Modulus of Elasticity ( = Value Given x 106 pst for Temp Y of t f Matenal -325 -200 -100 70 200 300 400 500 600 700- 800 Ferrout Materials Car. 4 Steels with 31.4 30.8 30.2 29.5 28 8 28.3 27.7 27.3 26.7 25 5 24.2 v C s 0,30% s i Larbon steels with 31.2 30.6 30.0 29.3 28.6 28.1 27.S '27.1 26.5 25.3 24.0 i' C > 0.30% l Carbows'ybdenum steels 31.1 30.$ 29.9 29.2 28.5 28 0 27.4 27.0 26.4 25,3 23.9 Nickel steels 29.6 29.1 28.S 27.8 27.1 26.7 26.1 25.7 !$.2 24.6 23.0 i Chrome-fnolybdenum steels %-2 Cr 31.6 31.0 30 4 29.7 29 0 28 5 27.9 27.8 26.9 26.3 25.S t 2%-3 i,/ 32.6 32.0 31.4 30.6 29.8 29 4 28.8 28.3 27.7 27.1 26.3 3-9 Cr 32.9 32.3 31.7 30.9 30.1 29.7 29.0 28.6 !$.0 27.3 26.1 Straight thromium steels 31.2 30.7 30.1 29.2 28.S 27.9 27.3 26.7 26.1 2$.6 24.7 i Austerstic, precip4ation 30.3 29.7 29.1 28.3 27.6 27.0 26.8 25.8 2SJ 24 8 24.1 hardened, and other = hi9h alloy steels i Nonferrous Materlats High Nickel Alloys

~'

N02200 (200) N02201 (201) 32.1 31.S 30.9 30.0 29.J 28.8 28.5 28.1 27.8 U.3 26.7 ~~ N04400 (400) N04405 (405) 27.8 27.3 26.8 26.0 2$.4 25.0 24.7 24.3 24.1 23.7 23.1 N07750 (750) 33.2 32.6 31.9 31.0 30.2 29.8 29.S 29.0 28.7 28.2 27.6 [ NOM 18 (718) 31.0 30.S 29.9 29.0 28.3 27.8 27.6 27.1 26.8 264 25.8 N06002 (X) 30.S 29.9 29.4 28.S 27.8 27.4 27.1 26.6 26.4 25 9 25.4 N06600 (600) 33.2 32.6 31.9 31.0 30.2 29.9 29.3 29.0 28.7 28.2 27.6 N06625 (625) 32.1 31.5 30.9 30.0 29.3 28.8 28.5 28.1 27.8 27.3 26.7 N08020 (20Cb-3) 30.0 29,4 28 8 28.0 27.3 26.9 26 6 26 2 25.9 25.5 24.9 ~" N08800 (800) N08810 (800H) 30.5 29.9 29 4 28.5 27.8 27.4 27.1 26 6 26.4 25,9 25.4 N08025 (825) 30.0 29.4 28.8 28.0 27.3 26.9 26.6 26.2 25.9 25.8 24.9 M 0001 (8) 33 3 32.7 32.0 31.1 30.3 29.9 29.5 29.1 28.8 28.3 27.7 885 N1045 (8 7) 33.6

3) 0 32.3 31.4

'30.6 30.1 29.8 29.) 29.0 28 6 27.9 585 N10276 (C-276) 31.9 31.7 30,7 29.8 ~ 29.1 28.t 28.3 27.9 27.6 27.1 26.5 Aluminum and Aluminum Alloys A03560 056) ~ - A95085 (5083) A95086 ($086) 11.4 11.1 10.8 10,3 g.8 g.S 9.0 8.1 A95456 (5456) A'NALYSIS/ CALCULATION ooc to a S-%ooG ATT # P REV I SHEET 22 OF 22 -w. ~,..y-e e -.,-,,--e i--%,e-,-, ,,-,-,,,m_._3_. vw....,.----. ,,...c, . ~.. ,s ---.=e v 3

S M.......... DESIGN ANALYSIS / CALCULATION Crystal River Unit 3 Sheet.h_L of M />CUWINT ID&NilF8CATON NUWS( A M(Yl&lON Pit'W A%$P NVWh(%Fitt S -%-oo IS I a/A B s A.T T. AC. H. M.E t.4T. . m. ..-. - ~... I s I .,i r.1 a ._...m._ m.. _.-. R E A.CTO.R. CCOLAWT. DRp! TAN K. .r. a W DT-5. 2. i I I I l l l i i .i._ .... # r... .. 7._.. l 1 ..,-.9 4 8 I l ) _m... I u o -.n = i i l 1 .e k f 6 g g. 3 1 4

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h DES!GN ANALYSIS / CALCULATION M+ochment Q Crystal River Unit 3 % g g _jo 'CCuMF.a i iD(NTIFICAtlO4 NUMBEM PCVISION R[l/WAM/SP NUW8(R/ FILE s 5-%-ooIS I H/A i i i 4 _ p L.. _.;.., _. _ ] i i 4 I I ' $ _,. _[,d.Li,scb. p 4 /o k.p v = 4 J;' i i i i i_. i i_ p _._7._.___ u. _ i -, y.,._a _y_dre.vis vs.__ f Tf GL /t sa Le.r. ~.716.+ 3 7 T;ot gL.L.cp i gassis!MW Lu i f*. i $ U-(. l.$ N h', N O T_ j k,_ d QN( $ kk,. L -,_.s. i .. _. [.! _ l_ ! _i u d i u w; _ t _ L. - . _, _. q. b w a, n wo n .miec".._.--. n-I f I b__L_j_._d _b._.f._. _l__ L _ !__.)% dx-d e h n % a_ M i zy L j _ L1_ nLJ I fe_JillI ).ddrdrd_lcw.7)L mala!.ph_._a._.111I q.44..J1U._p_q,rpi_J%&Midisea. replak).pegru W...o i

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h.......... DESIGN ANALYSIS / CALCULATION %chment 9 Crystal River Unit 3 Sheet 06 of _10 >OCUWINT IDENilFICAllON NUMBEA MtVISION REVWAR/SP NUW8ER/ FILE 5 - % -oo 13 I td/A _ _ _f._. _j __.._ {_j _. j _J__L.J_' ..i. -_ w,,. x w knuv.m aa e c ' _a = Jgui _iDhv.l_ktdiuh V . ; lc+t..).sfyi 4djscLcL._h.e.Lt_1 es.. _, _ ;.. '. _.. 3a ao o + d.c4tsYiP ' _ ; A.... = 2k ozo A.spo/c.. i s. _ !. [ o 37.5 t,4. ..C'. 22.tl.25 j _..ii_q.._ : _;_ _. _ _.

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O h.......... DESIGN ANALYO ' .CULATION Auochme+ Q Crystal a 3 Sheet M of_ 10 EUWENT IOENTIFICATIOM 84WMeta nrytscq 8s19 ;45u #tLE $-Qs-OOl3 l N }? ~.. ' c'd_ i k_ _.I-_ _.)kan_IM'.14 iszusu ksh 2 I I a i u nk o-i f - j

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TMcchmmt M DESIGN ANALYSIS / CALCULATION Q Crystal River Unit 3 Sheet _Q 7 of3 0 accourut m=rvicatioN wwweta navist x atuwA m a wvust uitt 5-90-ool.3 i U/A .B O TS _ Du t. t' O V E #.TitA L. LOAD A tib DCTE AMLM E @At TENJ;tod ON L OV E R.Tv # 'Jiu k M0m E W (CEF,ArTALH $,) TOTAL yEntgAL toA,b P: I,325011 (RE F P6 p t) m : /5/;O,900 #6-16//:.,; 130075' S4/l,e (Cff PG Q 2.) TO T6L OVE /2.rs 20 :49 iworeE4't 'UNSE(2 oF 80l75 W~8 Rnm ein+cTEc D: 9'- o'/t " .Br. Ac.mb etarc.wiory ( c 1.4 4 i l'l4.'4 BOLT At: 9 89 0 m.I. (coor $4@. j _ _. j_.., i ....,L_...6.- .-..-.4...- i e: M/P :.130015 /.,32.5 0

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TABLE 11-5 Disposal Sv5 tem Co ponent Data 3 CAPACITY. FT i (each tank) DESIGft MATERIAL t Temp. Press VENTED DE51GN SEI5MIC IT(M NO. NAME TYPE Total Liquid F pstg Body Lining TO CODE DESIGit COMMENTS r l WDT+3A R.C. Bleed Tanks V/5 11.050 10.150 250 25 55 None V. H. ASPC III-C. Class I. Maximum operating tev/ press is WDT-38 WOT-3C 150F/e3 psig. Contains nearly one pr!mery system volume. WOT-4 Hisc. Vaste Storage H/5 3.150 2.750 250 25 55 None V.H. ASME Ill-C Class I Marleum operating temp / press Is' Tan 8t 150F/e3 psig. WDT-5 Reactor Coolant V/L 831 561 - 300 100 55 None V.H. A5ME Ill-C Class ! Rupture disk provides. g - Drain Tank overpressure relief. Internal l plate coils previde coolle.g. WDI-6 Spent Resin Storage V/L 920 8G0 150 15 55 None Sump ASME Ill-C Class i Nominal

esin capacity 800' f t Tank or two year's retent ion as.

design basis, t WDI-7A Concentrated Vaste. V/L 920 228 200 15 55 None V.M. ASME III-C Class ] WDT-78 Storage Tanks Moreinal one year's retention of. evaporator concentrate. WOT-BA Concentrated Boric V/L 220 728 200 15 55 None V.H. ASME Ill-C Class I Nominal one year's storage per. [ WDT-88 Acid Tanks Table 11 3. Item I.I. i WDT-9 Neutrallier Tank V/L* 530 470 150 15 C5 Rubber Atm. ASME Ill-C Class Closed Vent i System legend: V/S - vertical skirt 11/5 - horizontal saddle V/L - vertical legs V/il - vent header I ANALYSIS / CALCULATION ooc to a s-Wcos ATT s G l 11-48 REV l SHEETG_0_ OF /0 (Rev. 19) I I f

i 4

.. ~

@l A 285/A 285M TABLE 2 Tensile RequiG,ments ~ Grade A Grace B Graos C kSa (MPa) ksi (MPa) ksa (MPal fensde strengm 45-65 1310-450) 50-70 (345-485] 5s-75 (380-515} d y,o strengm. rne 24 [165) 27 l18s) 30 ' (205) gwgaten m 8 m. or [200 mm). mm 58 27 25 23 gengaton m 2 m. or (50 mm), me t 30 29 2r ~d Determeed Dy either the 0 2 % onset method or the 0 $ % entenson-unceread inetted 8 See Spec 4aten A 20/A 20M SUPPLESIENTARY REQUIRESIENTS n . tion shall Supplementary requirements shall not apply unless specified in the order. A list of standardi:.ed supplementary requirements for use at the option of the purchaser are utrements rocedures, neluded in Specification A 20/A 20M. Those which are considered suitable for use with this Jality and specification are hsted below by title.

53. Simu!.ited Post Weld Heat Treatment of Mechanical Test Coupons, te i

S4. Additional Tension Test, and S14. Bend Test. I specifica-able when ADDITIONAL SUPPLEMENTARY REQUIREMENTS t! to meet ; the lis;ed Also listed below are additional optional supplementary requirements suitable for this md to the specification: 1. (n conmet I 557. Copper Bearing S58. Restricted Copper the 557.1 The copper content, by heat analysis shall be 558.1 The maximum incidental ecpper content by heat 0.20-0.35 % and by product analysis 0.18-0.37 % analysis shall not exceed 0.25 E L g g;, The Amerncan Society for Testong and Matenats takes no smasst on respectmg the vak#ty at any patent ryhts asserted e conne w@ any item mentuoned M ttus standard, Users of tho9 standard O s espres,tly advised that detertrunaten of the valo&ty of any or strCss patertt r9 hts, and the rtsk of mirangement of such rQhts, are entwery their owet responsblety. T!us standard ss subrect to revrsion at any teve by the responsible techancal cornmottee and must be reveened enety l at not revesed, e*ther reapproved or withr*rawn You comtnants are ant'ted eether for revissort of thss 41&ndard or to* ad6ton r 'nts as to. and srsoukt be addressed to A$rM treadquarters tour comments wua receove careful considerettort at a sneetmg of the resportsb techmcat commottee, wruch r% may attend. Il you feet that your comments have not receoved a tarr heareg you snould en v<ews known to the ASTM Corrmttaa on Standards, t9 t6 'lacs St.. P?ulacetprua. PA 19I03 i r nsion test 199C A w At. B oo t'. of ASTtA STAu TAR.es S Ef.TIo 4 i VOLWE O'* 0 k iin Table orace c ANALYSIS / CALCULATION DOC ID # S-%<n3 ATT # O 0 90 O ss REV l SHEET H OF 10 0.035 1 _0040 1 187 . J

l i j FLORIDA POWER CORPORATION 4 03 s 423 014 I St 423 024 St. PWtE30ues. PseeSA woesemme em enasses o. sec neue me~ I CmfAL RfWR PLANT = us. inst osa a ens,ees sw eMomeets AMo cc>egintaNes swwhssi-AmW loll be ,,,,,,m REACTDR BLh SLAB EL.95t0* m t'l Ar,ilv'l7w de)E '/' h o'

a m=
- " ^5 8ao*N ON **c' NoE 421008 4009 T j& &.*"A STANNRO TYPES S M'CIAL TkosS tTP:

1 1 3

4. '

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m:x ~ WW %&rn n;t ca - g 'd-V .Qt C,ULATION OF SEISMIC CAPACITY OF 4 "*'" 017 133 442 7 c w m i. m a c. w orusr a t ,,, _ g g n,

The purpose of this calculation is to show the scismic adequacy of the turbine condensers at the Crystal River plant. The condensers have not been analyzed for seismic loads. Numerous condensers have L survhtd major canhquakes significantly larger than the Cry stal River earthquake design basis event.
This concluston is based on and documented by the work performed by Stevenson & Associates in post.

carthquake (Loma 'ricta 1989) reconnaissance at Meer Island as well as similar findings /conchisions s documented by the BWR Owners Group in support of their evaluations for alternative main steam - isolation valve (MSIV) leakage pathway. The MSIV leakage pathway study has necessitated seismic evaluation of turbine condensers. The BWR plants that have performed these evaluations have used experience data to verify the adequacy of the condenser itself and performed simple anchorage evaluations to demonstrate the anchorage adequacy of the condenser, The same approach is beinS used for the r evaluation of the Crystal River condensers. 4 Summtv Since the condenser cannot overturn due to design basis seismic forces for normal condenser water Icvels, the analysts considered only the shear resistance for this design analysis. The evahiation shows the shcar capacity of the anchorage for the condensers of 371 kips in the governing N.S direction c.-ceeds the demand seismic shear load of 256 kips. References

1. BWR Owners Group MSIV Leakage Pathway Study and Methodology,1995

2. S&A Meer Island Post Loma Pricta Study of Turbine Condensers,1996

3. Foster. Wheeler Corporation Crystal River Condenser Drawing 93 817-3 101

4. TPC Calculation S.910003, Rev. O, 3/22/91 Evaluation i

Per reference 3, the total operating bad for condenser with a normal w ater level of 8'-4" is 2,324 kfps. The horizontal reaction loads at the shear key (" anchor T') and the support at the southeast corner (for condenser 38) due to pressure drop are 604 kips maximum These loads are taken by the so-called anchor T and in bearing against the concrete wall against w hich the condenser bears on its castern boundary, As such, they are net resolved into the existing anchor bolts The thermal expansion loads are self relieved by the slotted base plate cenfigurations of the western two supports. As such, thermal loads are also not - - taken by the existing anchorage.

The existing anchorage was originally designed to be a tension-only anchorage to resist " uplift forces

that could occur at start-up under certain assumptions. As such, this design condition is clearly not concunent with operation of the plant and does not need to be considered concurrently with a seismic 1 design basis carthquake. Checking the overturning moment (OTM) and using the peak of the 4% ground spectrum factored by for multi modal effectst c OTM = 2,324k x 0.22g x 1.5 x 43'/2 = 16,500 ft k The restoring moment (RM) reduced to include the clTect of venica carthquake f2/3 of honzontal peak) is' RM = 12,324k - 0,67(0.22g)(2,324k)] x 15' - 29,700 ft L t i ANALYSIS / CALCULATION g

)OC ID # S *k-ooa ATT # R l,ev-

/ SHEET R2 OF1_ t

~- - -.. .~

Doc y 04.+ S7 - 08: 36c.

. S tovoris on 8. Ascociatos (6171933-4428-P.2 i L The two western supports (for Condenser 3B example) are slotted to relieve thermal expansion, bun y - are slotted at approximately a 45' angle to the N-S and E W directions and since rotation is preven::a by the configuration the supports are actm for both directions of earthquake The two eastern supports (for condenser 3B examp!c) are slotted in the N 5 direction, so they can only resist an E.W earthquake. Thus, all'of the N S carthquake loads are taken by the western supports and the anchor T; therefore, only two supports (thus,12 anchors) are active.. All of the scismic lateral force is taken by the Iwo supports and the anchor T. Ilse the peak of the response ? spectrum sin:e the fundamental frequency is not known. %= 2,324k x 0.22g = 311 kips The anchor T can 'ake shear in toth directions The shear area e rhe T in any ork nal direction r utiliziAg both legs of the T is 42 in x 2 in - 84 in2. The shear capacit3 of the shear LL for A36 steel using a 0 4Fy for fauhed loads is: bw = 84 x 0.4 x 36ksi = 1210 kips Checking the weld capacity b=r a = 82" x 3/4" x 0.707 x 0.3 x 70 ksi x 1.7 = 1550 kips > 1210 kips so 1210 kips governs. The allowable for the anchor T must be reduced by the 204 kips needed for operating loads kving a capacity of approximately 600 kips for the anchor T for the N.S carthquake. Since the reactiot, from the N S earthquake is one half of 511 Lips, the anchor T can sustam its component of reaction from the N S earthquake. The six added 1.25" diameter bolts reside in oversized 1.75" wide slots, so they must be considered inactive. Checking the original 6 anchor bolts (6 l.75" diameter bolts)in the western supports: The total shear force resistance of the six anchors is based on the fact that the shear plane is not through the threaded portion of the studs is: 3 2 6(n d / 4) 0.17Fu x 1.7 - J(n[1.75 ]/4) 0 22 (58) x 1.7 = 313 tips Ec shear reaction on the Iwo western supports is one-half of 511 kips, or 256 kips, which is less 'han 313 - kips, so OK. Therefore, the anchorage of the Crystalliner condenscrs is adequate. PERFORMED BY; WA!!IliR DJORDJEVIC Stevenson & Associates - 11/21/97 l ANALYSIS / CALCULATIONm..- j DOCIDtl_S.m.caf3 m y lREV / c._ SHEET R3 op 3-4 ~-c. -~

l j\\ ,lIljl!!l 1 s tcu GR- ,f E nod E T gsC I ~ T NA i i S M s sA R M G e yV O E J I E. Dl E t aH IN C E; t ER n G S t SE la LN OSE 0 L R rA d NsA OTI EN e8 N IEG t I J u n EoRI Ct I e9 EO S da YLlSEI EN G s1 CSE t I BV TVS Et s AC

cn I vAI TE a7 DY n

y DCiDM EF 1 n VEU N I ua L 1 D E . r a EF L M%' r R O LDALLOV e5 NR l t n O ANCAYC T1 f o S gT SYR AJ VT OE GINN O4" i . r CF V NT U f I F lee i ET SHESO lvm EN dsle OI CECER i b SO PC RE SE T neb SO U x C UTSVN ot e aDa S R GD TOEI E sgC NSCE N N np T M Ke U A UGN T S lanr C R EER NN I T THA o R S T P i irdf E E R D e us M A EA t l t A E acr L Mno CU p i N p u S i yiI I j ll !i if I l !! i! i sykCEC bOC ~OZ D 8OU*?hd2g 2' P i4 t a< mIm$ 6-Q 3 ~ ,o,- .* ~

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v-3 M ANALYSIS! CALCULATION . il DOC ID # 6% COO ATT # S REV I SHEET 3 2 OF 17 A NEW METNGO FOR CALCt'LATING AfEHOR-80LT STFtS5ES % N By ifarold 5. Davis' Q ^$ 0 h J g/ lE# INTR 000CTION O O $0 f The determination of stresses in a ctrtular e ay of anchor bolts may' appear 8 YI g 0 id l$ to te an easy, routine task. However, just the opposite is true because basic equations are contersome and their solution time-consuming. In this paper the if 0 ,/g8 I basic equations are siglified by espressing then in terms of trigonometric 3 f / 'UC Hf functions called 8-factors. By using these factors and the nicerical proced- / (' ures described belou, stresses can be obtained with a hand toget.r in a g c (' 3 g! fraction of the time previm.31y required. Or, the re9 sired number and site hd of anchor bolts can be determined directly if external loads and sesign stresses are knoun. The sig11fied equations can also be used to calculate longitudinal stresses in reinforced concrete towers and chimneys, aM at the [JP!ca 31 lure Mechanisms base of steel storage vessels. 'a A review of aecher-bolt theory is presented first. This is followed by a

i

.a Fi g u re 6 ' Consulting Engineer. Pra8ects section; tRK suelear Industries. Inc.; 8 i Richland,ifA. ^ 6-1-17 g 6-2-1 's .4

i) description of the steplified equations and 8-factors. Their application in ( N l solving anc%or-bolt problems is then illustrated with several nurerical w.k D. 1 1 czampics. The basic equatlens are presented in the Appendia. I t i 2 AMCHOR BOLT THfMY l Compr4SSIOR s In the typical anchor-bolt problem, vertical tefision forces are talen 8f88 solely by the anchor bolts located on one side of the neutral ants while COSg 9 .cospressive forces are provided t,y the bearing surface located on the cther i side of the neutral asis. In general, it is assumed that a) plane sections T l rec.als plar.e af ter bending and b) stresses vary linearly with distance from s / (a) Crosa j the narral asis As shown in Fig. 1. the anchor bolts are replaced by an / Section / ,8l equivalent steel area, or rino, having a thickness ti and a diameter of D. g. dA / dh Equivalerit / f The compression side of the ring has an coutvalent steel thickness of tz-tension stea l e Area The neutral ants is located at a distance of ED from the conipression side D-Cos 4 2 a is cf the ring section; or by the angle e. Aere o

ces., (1 - 2k). Yte y

. distance between the resultant of the tension forces (T) and the resultant M cf the compressive forces (C), acting on the equivalent ring section. is [N' equal to jD. The distance from the central asis to the resultant of the --* k D + compressive forces is defined as z. The eccentricity of loading (e) equals P the tending ewwent (M) at the section divided by the antal load (P). Addit-1 10,. and (b) Loade T L ional information on anchor-bolt theory is presented in the references S stressee b s C listed in Appendia 1. b As noted above, the equations and procedures introduced in this paper may \\\\ \\\\ \\ / -

h

} also be used to crunputa longitudinal stresses in cylindrical str6 tures such T'P g s 3 g ts reinforced-concrete towers, silos and chimneys, or at the base of st?el Z D -- - \\ storage tanks. In order to use the equations, geometrical parameters for e these structures esist correspond to the conditions of fig.1. FIG. 1. - Anchor-Bolt Theory g 6-2-3 j. 6-2-2 ANALYSIS / CALCULATION l DOC ID # S-Hvo 13 ATT # 5 t' REV I SHEETS 3 OF O

SASIC f @tATIOM A.%D 8-FACTOP5 G The basic equations are presented in the appendia. Scee of these equa-tions 1.are teen published previously II; whereas. cthers are unirsue. Only the simplif ted equations, required for natin9 ancher-bolt calculations, tre discussed below. TASTE 1(a).-falues of S. j and A/jD then k Is 8etween 0.0 and 0.5 The followin9 equation is of prime interest $1mce it can tre used to locate k 8 J r/jD k 8 j 2/j0 the neutral aufs or to deterwitne required thictnesses: 0.75000 0.66667 0.25 5.5R72 0.77E66 0.57587 0 = \\ 0.01 1154.3782 0.75245 0.66184 0.26 5.14731 0.77922 0.57271 0.02 400.0562 0.75452 0.65736 0.27 4.75135 0 J1974 0.56958 8= 8 II I I j (g) 0.03 213.4778 0.75638 0.65309 0.28 4.39372 0.78024 0.56645 l tt \\ e + r ) i \\ }ii ~ If/ t 0.04 135.9344 0.75808 0.62896 0.29 4.06970 0.78072 0.56334 0.05 95.3555 0.75967 0.64414 0.30 3.77524 0.78117 0.56025 The two upper s1 ns must be used when P is upward; whereas, the 1, 0.06 71.1112 0.76115 0.64102 0.31 3.50691 0.78 M9 0.M6 s 9 1 sl9ns apply when P is dowrmeard. Factor B is defined as: 0.07 55.3166 0.76254 0.63719 0.32 3.26176 0.78199 0.55409 0.08 44.3783 0.76386 0.63343 0.33 3.03725 0.78237 0.55103 i 0.09 36.4506 0.76510 0.62973 0.34 2.83115 0.78272 0.54798 0.10 30.4968 0.76628 0.62609 0.35 2.64164 0.78305 0.54494 8 = +1 (pg 0.11 25.9034 0.76741 c.62250 0.36 2.46696 0.78i36 0.54i,0 tan a - a 0.12 22.2732 0.76843 0.61896 0.37 2.30568 0.78364 0.53888 0.13 19.3502 0.76950 0.61546 0.38 2.15650 0.78391 0.53586 0.14 16.9586 0.77047 0.61200 0.39 2.01831 0.78415 0.53285 Values of j and r/jD are also comples functions of a and are defined by 0.15 14.9747 0.77140 0.60858 0.40 1.80009 0.78437 0.52984 equations given in the appendia. Numerical values of B. J. and r/jo are 0.16 13.3094 0.77228 0.60519 0.41 1.77094 0.78456 0.52684 0.17 11.8968 0.77313 0.60!83 0.42 1.66009 0.78474 0.52385 tabulated in Table 1(a) for values ch 1 between 0 and 0.5, arr! in Table I(b) 0.18 10.6877 0.77394 0.59850 0.43 1.55681 0.7848) 0.52086 for values of k between 0.5 and 1.0. 0.19 9.6442 0.77471 0.59520 0.44 1.46047 0.78503 0.51787 i 0.20 8.7372 0.77545 0.59192 0.45 1.37050 0.78514 0.51489

8. J and r/jD are explicit functions of k.

Therefore, the thickness ratio 0.21 7.9436 0.77615 0.58861 0.46 1.28640 0.78523 0.51191 l (t,/ts) can be cornputed with equation 1. for ariy given value of k and selec-0.22 7.2452 0.77682 0.58544 0.47 1.20770 0.78531 0.50813 0.23 6.6273 0.??747 0.58723 0.48 1.13397 0.78536 0.50595 tid values of eccentricity ratio (e/D). Tables 2 and 3 IIst values of thich-0.24 6.087? 0.77808 0.57904 0.49 1.0648f 0.78539 0.50298 mess ratto cbtained in this manner for k between 0 and 1.0 and a nueber of 0.25 5.5872 0.77866 0.57587 0.50 1.00000 =/4 0.50000 eccentricity ratios between 0.25 and infinity. Thic& ness ratios listed in Table 2 result when the axial load produces coepression o*: the free-t>ody sectiont whereas, values IIsted in Table 3 result when the antal 1 cad is in I tenston. The general relationshfp between these factors is shown in flg. 2. l (. J j ANALYSIS / CALCULATION DOC ID # $-QG-oo t 3 ATT # 5 REV I SHEET 5 4 OF 17

r TA6tt 2(a)-Values of Thic6 nets Ratio (tit,) nthen Antal Lead is in Coetression TA8tE 1(b).-Values of 8. j and f/JD When k is 8etween 0.5 and 1.0 e/O - 0.35 0.40 0.45 0.50 0.55 0.60 0.70 0.40 0.90 k k 8 j r/jD t 8 J 2/ja 0.00 0.M 7M7 N 754 440 3M 272 6.50 1.00000 n/4 0.8;0000 0.75 0.17898 0.77866 0.42413

0. M 1186 363 225 137 101 86.1 0.51 0.93909 0.78539 0.49702 0.76 0.16453 0.77808 0.42096 0.15 393 158 103 65.1 49.9 41.8 0.52 0.88185 0.78536 0.49405 0.77 0.15089 0.77747 0.41777 0.20 174 83.2 56.8 36.9 28.6 24.1 0.53 0.82802 0.78531 0.49107 0.78 0.13802 0.77682 0.41456 0.25 2737 89.9 48.4 34.3 22.9 18.0 15.2 0.54 0.77736 0.7P523 0.48809 0.79 0.12589 0.77615 0.41133 0.30 24; 51.8 30.4 22.2 15.2 12.1 10.3 0.55 0.72966 0.78514 0.48511 0.80 0.11445 0.N545 0.40808 0.35 31.5 30.9 19.4 14.6 10.2 8.18 7.01 0.56 0.68471 0.78503 0.48213 0.81 0.103689 9.77471 0.40480 0.40 44.9 19.5 12.9 9.93 7.10 5.75 4.95 0.57 C.64234 0.78489 0.47914 0.82 0.093566 0.77394 0.40155 0.45 24.9 12.6 8.75 6.87 5.01 4.09 3.54 0.58 0.60238 0.78474 0.47615 0.83 0.084056 0.U313 0.39817 0.50 109 14.7 8.32 5.99 4.79 3.56 2.93 2.55 0.59 0.56467 0.78456 0.47316 0.84 0.075135 0.77228 0.39481 0.55 30.7 9.01 5.54 4.12 3.34 2.52 2.10 1.93 0.60 0.52908 0.78437 0.47016 0.85 0.066779 0.77140 0.39142 0.60 13.8 5.64 3.69 2.82 2.32 1.78 1.49 1.31 0.61 0.49546 0.78415 0.46715 0.86 0.058967 0.77047 0.38800 0.65 7.17 3.54 2.44 1.91 1.60 1.24 1.05 0.924 0.62 0.46371 0.78391 0.46414 0.87 0.051679 0.76950 0.38454 0,.70 32.5 3.93 2.21 1.59 1.27 1.07 0.845 0.718 0.637 0.63 0.43371 0.78364 0.46112 0.88 0.044897 0.76848 0.38104 0.75 7.24 2.18 1.34 1.00 0.813 0.696 0.556 0.476 0.424 0.64 0.40536 0.78336 0.45810 0.89 0.038605 0.76741 0.37750 0.80 2.76 1.18 0.780 0.598 0.495 0.428 0.346 0.2*8

0. "%7 0.65 0.37855 0.78305 0.45506 0.90 0.032788 0.76628 0.37391 0.63 1.14 0.592 0.415 0.327 0.274 0.240 0.126 0.170 0.;53 1

0.66 0.35321 0.78272 0.45702 0.91 0.027434 0.76510 0.37027 0.90 0.429 0.254 0.186 0.150 0.128 0.113 0.094 0.082 0.ca 0.1J 0.32925 0.78237 0.44897 0.92 0.022534 0.76386 0.36657 0.95 0.110 0.012 0.055 0.045 0.039 0.034 0.029 0.025 0.9.'3 0.68 0.30658 0.78:99 0.44591 C.93 0.018078 0.76254 0.362E1 1.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.69 0.28'515 0.78159 0.44284 0.94 0.014062 0.76115 0.35298 0.70 0.26488 0.78117 0.43975 0.95 0.010487 0.75967 0.35506 0.71 0.24572 0.70072 0.43666 0.95 0.007356 0.75808 0.35104

0. n 0.2n60 0.78024 0.43355 0.97 0.0046s4 0.75528 0.3469 ANALYSIS / CALCULATION 0.73 0.21047 0.77974 0.43042 0.99 0.002500 0.75452 0.34264 0.74 0.19428 0.77922 0.42729 0.99 0.000366 0.75245 0.33816 E O # 5-9r006 ATT #

5 0.75 0.17896 0.77866 0.42413 1.D0 0.000 0.75000 0.33333 REV I SHEET S 6OF C I 6-2-7 626 I I

m en N N N m ao se w Nw C. . N... M. *. *. ?. B. R. 3. R. R. R.. S. E. E. B. E. E. 8 a mm te m N a= - Q COO COO C00 00 we e en O - N.= m w w .c a - Om L9 m M m. . N. N. E. m. O m. e. e.. v. m. N. .. O. se e in e. Aww N a3 - 8. 8 . O. O. 0 0 .I N en m N *= c O OOO OOO OOO OO t.= io .= A w N en N . e. N. M. $. k. M. ge. 8b. N.. . 8 E. 0 0 8. 8 O er o. in n ,. O O OOO OOO OOO OO .= ma g O N -.O. mNO eee - en d en e w= O w m=w mNm eOw O+w O. O Oo. O. o .o to. a e 5 em. sa. se. m. w. e. w. w. N. N. .==. 0, O. O. O. O. w e C em wN-COO C00 OJO COO DO k O =. g E e,. em m N., e-m -N m-F en C wNm ed N en W.a N en m N ** OOO d" en. t N. G3 @ O. N. N. en. M. N. e= C. O. O. O. O. O. O. O. { s o% O e. .s=. 0 en N M N== 0CO O C4 c. OOO OOO. C2 0

=

s ~ 2.: ~ w 00 Ce ~No - en e.. p ( ,o T. S. $. A NO. O. O. d'. 8. *8. > 8 9 N. w.

8. P.o.= 8.

m. w N...O.

m N es N N sa l t,- g O W W O N 48 m*=*= O0O OOO OOO O O 63 OO ) 7W .A MNN e* C.t N a= gr er en N s g) e en p N e= w N *= Q a $P 4 w. g M. m et. N. N. w v g a=.

==.. O. O. O. O. .. t 8 8 O. W< O r g.e ^ g O N-C C00 000 000 0 0W m a T C en O N m. Em @.e ' O. Er.n te. a. w. N. a=. 0 O. O. e . 4 9 e a a e 4# o town mO O =" M s C. Q w J a.d C er m -OO O Cs O OO y a N J< . ~.o.R i8 2 -m N m m. .m O,. o JN ddd d < Q CC O en O. W O nn OeC en o @ O ha O nn Q na omO O. O. O. w. em. N. N.

m. 9. w.

w ed'. to. @ te. N. N. w. e. @ sm. O. og 4 000 000 000 000 000 000 00-o e.s in u - O.3 m.= e e... to w em N e th nom %w m. I e w en. O. em. en. es. id. e3 et O. N. en. M. N. . O. O. O. O. an O

n e3 in mN-

--O OOO OOO !.'l O O +m

=

00 N e to MwN. en a= en w mee e s2 ch e= O e3 cm N > m-O O N. w. O. . ea. O.

P. Q. N.

to. w. N.

.O.

C. O. O. O. ' m m. tes e wNf4 - - sD -OOO OOO OOO O .O. m M so N e m as es N w Pee O N en ** w O MN M-O I O r=. en. N. eh. w O. N. no.. e) to. w. m. . n=. O. O. O. O. i g C - to Ps O to wmN e= ** O OOO OOO OOO 4 l sn sa e M N*m -Og Nm O O e On O ar e0 e er to. en. w. cr. en to ms w-c 8 .w as. N. m. er. ta3. er. m. N. o.= 0

0. C. O.

w e C g =g we e-N wmN --O COO 000 000 Nm -- g en es 6C m em O N N mi. N en .V. E G

Ft.

to. s=. thP%. 0%. m3 o. w. O. N. te. m'.N.==. Q. O. O. O. e. m go m v Es so te c to @ Ch w=0 . M. O 6 ene N= m in N m so en m N N - a= QOO O C3 O OOO JE ww N-N

== gg R.e WNO e w e' Ne e m de N eOw

mO to - *** m3 as O nn - O tog

.O.# 0%.e N. m. m. er. N. O. N. ma. so. w.N. e=. O. O. O. w 5 SD. ,,a =en= en w e to w m N**- OOO OOO OOO s'= e. Ne ~ w

='. 8%

ON O. nn w e J3 er N e3 NNm et m m O m=N eO

=* E N W O.

M. Er.t w P' M. es, Ps. 94 et. N O. P. e as go m-- N== ===. O O.

n. N.

. O. O. g N. Nm O N== en m OOO OOO haa 0'* e= e. e" (> O N w em %N M att M M e en Ern et 60 40 er w so N O 8 N. m. en. e=. N. er. c.' 0%e %O.** e3. en. f9. N. e=. 000 er .N m ..w mN- -QC C00 000 gN gN-t t O SwO ehm Oe m mms + 51 +n O. m..o N8e R w.n 8 O -~N v. Y b b bYY b$b bYY bbY j en =.~, ... ~ J

it - slay geJin*N lo vollesol O'& 4*0 8'O l'0 g*O S*0 t'0 C'0 t*o &'O o \\\\ \\\\\\\\ \\ \\ \\ \\ ) } _ \\ \\ \\ N 's .+ d. \\ \\ \\\\\\\\ \\ \\ \\ \\ "8 E \\ \\\\ \\\\ \\ \\ N N \\ \\ - O N N N N,o-o

! E a

o =\\ .* I f N 'N s.e 3 N N 3

g s

s 4 ?m_ \\ \\ ~ \\ =L? 00li Y x Rst_ \\\\ \\-,'" 5 3 ~ N t\\ \\ \\\\4 V \\ \\ \\\\- "4 \\ \\ x.od x /x x x \\\\

';(

i E b'L s l st 0 \\ \\ \\ C -0 i.* o \\ \\ \\ \\ u". o og 3g'0 - 3 1 \\ \\ \\ \\\\ \\\\\\ c 0*09 e s '. *. *IE 555 E8 $k pg.. m~- --o coo ooo coo

E.E $5$ $

. k a. i e gg goe mm- -oo o c ci coo oco 8 e~~ .8: 499 :.45 E@s 2.E% 538 '~ q g gg paw mu- -oo coo ooo ooo O M o f$

E E E.55 $.

kk k.8 o =p g gm -., ~~- -oo com oco ooo 2 = q

n O y 1s

re: e88 5 e< b H

~- . - ~. E =*.

E :. :

o 3 e"-

~

" N -- . c o. "

a.

m o m - g gg omm ~~- ooo com ooo ooo y D 0 m o = a 3 O O 5 E.55 EE$ . coo oco ooo 8 s ". ": O c. g. em ~~- ooo s ~ T* U) 7* - m X$ 30$ SAS~ M~SA ~ #$0he~ 588 .e [G l o8 ~ w ad a4 eJd add add did dad J ~

~

5 0 < o >C 2 w ~.0 82* 23: : ~8 on 28 R 88 .k i 8 e '. " ~ - a. o o g gw NwN --Q ooo ooo coo coe 1 I O .ns 8e ooo oGo oCo coo oCo coo o O== \\ l f* R

t* m.:

T i f i When the axial load is Jero and e/D is equal to infinity.1/t is equal 13 with bolts located cri the other side. During each cycle the neutral ants 2 g This pure eccent relationship is defined by the darter curve in Fig. ?. noves back and forth across the section. The anchor-bolt ecuations descrit 4 B. [ Cur =*:; Mated above this B curve correspend to downward loads, which pro. above cs' be used to locat- 'he neutral amis at any particular point in the eisce compression on the section; whereas. curves located below this curve cycle if the eccentricity ratio (c/D) is equal to er greater thac 1/4 When corresperti to epward, or tensile loads. When drawn to a larger scale, a chart t = 1.0 ce rere, the lietting value of J = 0.75 and the corresponding value cf this type can be used f a locate the neutral arts if the thickness and cf e/j0 is 1/3. (See Table 1.) eccentricity ratios are known. On the other hand, k can te obtained from When e/0 is equal to or less than I/4 the neutral an N is located at the data presented in Tables 2 and 3. or by using the r.umerical procedure diametral amis of the ring section. in this case stresses can be deterefned described later in the paper. The latter method is the most precise. from the b. sic equation: Once k is known stresses may be computed using the following 8-factor: l 5 _P_ ! Mc (6) = A 1 } g, (tana -a) + e l sec e + 1 The section modulus (1/c) for a cylindrical ring is equal to = (D/2) t or e '8 ADft. Substitutleg 12 latter factor in the above equat ten, it becomes: where e = ces. (1-k) in radians. The tensile stress (5 ) in the most-distant b encnor bolt and the manirun compressive stress ($ ) on the opposite side of e g p g, gyg the neutral amis are given by the following equations: A D This sigle equation ces be used to compute peak stresses =Aen the section t D 8 is subjected entirely to compressive stresses or Aen it is subjected I i entirely to tensile stresses. If the asf al load is downward, the area (A) to be used in the above fons:Ta is equal to eDts; whereas. A is equal to y

C05
S 5,

= ($) g eDt when the asial load is upward. j, c,, ILLt!$TilATIVE ANCHOR-BOLT PROBttM These equivalent steel stresses occur on the circle defined by D. The use

Process Equipment Design." by Brownell and Young (Ref. 3). Includes an gf the above equations for computing anchor-bolt stresses and for determin-encellent chapter on the theor; and design of anchor bolts for vertical ing the required number ard size of bolts is described below.

vessels. A typical anchor-bolt problem is solved on paga 189. using the LIZITING VAttES OF ECCENTRICITY FATIO basic equations. Design parameters from this illustrative problem will be When a cylirdrical structure oscillates in a vertical plane about its used as input in the numerical examples discussed beles. The results foundation, anchor bolts on one side are subjected to tension, alternately g ANALYSIS /CAlbbbTION DOC ID # S~%~OOG ATT # REV I SHEET S 8 OF fl

~ I' cbtafned herein are compared with these presented in Ref. 3. In order to

2) locate the neutral a:15.

demonstrate the accuracy and merits of the new method. Deteestne k corresponding to values of e/0 and tr/t, obtained above. g. -L-In.the ancher-bolt probles of Ref. 3. 24 steel anchor bolts. 21/2 in in By inspection of Fig. 2. t = 0.321. (or see page 27.) '8 diameter, are located uniformly around a 11.0 ft bolt circle. The weight of

3) Determine values of coefficients and compute facter_. 8,.

the cylindrical tower, which is 10.0 ft in disseter, is 600.000 lb. The Determine j and 2/jD fross Table 1. corresponding to the above k. tower is subjected to an overturning moment of 8 x 10' f t Ib. about its base ey interpolation: j = 0.752 and r/j0

0.554 when k = 0.321 caused by wind. The bearing plate is 12 in, wide and rests on a con: rete Therefore: e/jo = 1.550.

foundation. The coepressive strength (f *) of the concrete is 3000 psi. Coapste 8, for e = cos~'(1-2k)

1.2047 rad, with equation 3.

The ratto (n) of the sedulus of elasticity for steel to that of concrete is 1.4035 + = = 1.1942. (taa -e )

g,

]l equal to 10. The allowable tensile stress (5,) is M.M psi at the bolt seca e 2M3

I

.a g I threads; whereas. the allowable compressive stress (f,,) is 1200 pst in the s j

4) Concute tensile stress (5 ) in sest-distant bolt estnq equation 4 concrete.

b n ,} COMPUTATION of A1CHOR-80LT 5iPESSE5 e r b ~ The new method for calculating anchor-bolt stresses consists of the follow-S 8 t, D 8, 0.2151 (11 z 12)(1.190) ing steps. The anchor-bolt array and loads described in the precec'ing para-graph. are used in the saeple calculations.

5) Cescute maniews empression stress (5,) en' opposite side of bol.t etrcle vsine m ation 5.

1) Comeste e/D and 1/ts.

2 f1-ces. 1 - OJ580 . g79s,,g s S =S T 0.6 x 10' T "~l l 'd'~ (I*C85* II'552 l(1 + 0.3580) - 8 x 10 e 13.333 ft e = M 13.333 = I~212 c bl e = = = 1/4 ard

6) Cospute manicum stress in concrete.

g"n 24 (3,72) 8 tg 0.2153 in. = b = t = - I

6.474 5

l T 3(11 a 12) t, f,

2 = 8.299 830 pst (at D = 11.0 f t)

= n in = 4 0 ;, bid =84.5p*,.0.5)=.3,s. i f c. (n-ut,. m 5t. t ,.3,3,,o. an, 3 t, = ( 4 (at 0 = 12.0 f t) n 10 8 - note: 1 in. = 25.4 m; I ft = 0.305 ma i Ib = 0.453 kg; I pst = 6.89 kPa a l 6-2-M j ANALYSIS / CALCULATION I DOC ID # 5-90-0013 ATT # ~ S REV l SHEET S 9 OF 1 ~1

I I i 8l The results ottained above are almost identical tc trese ca?culated ty are equal to 1.212 and 6.4N. respectively. Reference to rig. 2. er Table 2. Srtwnell and Young in 2ef 3: however, the above ressits are more precise indicates that k is probably between 0.33 and 0.32. Values of tr/t. corres-i 't, amt were obtained with less effort. (These results are suvmsrized late. in ponding to these two t-values, are ccurated with equation 1 as follows: Table 6 as Case 1.) As defined in Fig. f. tg represents the etsivalent thickness of a ring r/jD f-8 fr/t, Pgving a diameter of D and the sane area as the anchcr bolts. On page 181 c;f Sef. 3. Brownell and Young used the root area of the bolt threads to 0.33 0.78.237 1.54914 0.55tn3 0.99811 3.03725 6.0R0 coepute t,. In order to be censistent with the conditions of the reference 0.32 3.78199 1.54989

0. 5540g o.995M 3.26176 6.537 probles. this procedure is used in the above calculation. It is based on t

the assumtion that the free-body section cuts thr. ugh the bolt threads. If Valuas of J r/jD and B. In the above chart, are selected from Table 1. A the free-body section cuts through the unthreaded length of the telts. then more accesrate value of k is then determned by interpolating between the the gross bolt area should be used to crnpute t. In this casc the result-resulting thickness ratios. For exag le: i ing value of S is the average tensile stress in the gross section of the b

0. 32 + (0.33-0.31 (6.537-6.474) =

0. 321 k

= bolt. The canimum stress will occur at the threads, arvi is equal to S Ib 7'f' 0303 b times the gross bolt area divided by the net area at the threads. Depending e value of k. cibtained in this r.anner, is usually accurate enough for most upon the job specifications, the net area may be defined as the root area or anchor-bolt calculations. as the tensile stress area. Values of tr/t, and k from the above table are riotted in rig. 2 for in-LOCAfl% THE fif0fRAL Atl5 formation. The dashed line between points define a localfred design curve In order to coropute ancher-belt sttesses accurately, it is necessary to for e/d = 1.212. Such a curve could have been used to obtain k = 0.321. first locate the neutral amis. The easiest way to do this is to select L from a chart similar to flg. 2. However the eccentricity curves rey be so When 6 is very small, or very larga. it may be necessary to repeat the far apart that interpolation is dif ficult and accuracy poor. This is espec-above calculations in order to locate the neutral amis accurately. As an f ally the case when e/D is very ses11 On the other hand, the numerscal example, assure that it is necessary to determine 6 when the eccentricity procedure described below may be used to determine k. It is based upon the ratio is 0.34 instead of 1.212. Reference to Fig. 2. or Tahle 2. Indicates fact that any e/D curve is essentially a straight line between consecutive that k is probably between 0.75 and 0.80, when e/D = 0.34 and tr/t, = 6.474. k-values. if the difference between these k-values is small. Calculations, similar to those described above, are sumarized in theupper tilth reference to the anchor-telt problem discussed above, e/D and tr/t s tabular section of Fig. 3. The objective of these initial calculations is 6-2-1E 'l " ANALYSIS / CALCULATION ooc to # S-DOOG ATT # 3 REV I SHEET S 10 OF 17 r--

I =

t t-e

+ 1 / s a + s the first appressmation of k M779) is obtained try Innearly saterpolatteep \\ JIT O F/ tan e-. ~ between thickness ratios of 6.108 and 7.869. An inspection of e/D curves fJ jh" b 8 (r d) k .I 8 /t, T plotted 19 flg. 2 tr:dtcates that the true value of n will be slightly 2 smaller than 0.778. but larger than 0.777 (These values provide a at of 0.60 0.77545 0.438 4 0.40808 0.03038 0.11485 3.882 0.001, although some ether at-value could be used to the nest Iteratiosi, if 0.73 0.77682 0.43768 0.41456 0.0231? 0.13802 5 108 desired.) Coefficients j and r/jD. corresponding to these two k-values, are 0.77 0.77747 0.43732 0.41777 0.01955 0.15Fe9 7.869 1 obtained by linear Interpolation of watoes recorded prevfously for k = 0.78 s. o,og 00065 .00321 is recorded on the sidole itne of the upper tabu'sr section, as well as the 2.16036 0.778 0.77695 0.a3761 0.41520 0.02241 0.14054 6.412 2 algebraic difference (-0.00321) for respective values of r#jD. These differ-2.15796 0.777 0.77702 0.43757 0.41552 0.02205 0.14130 6.573 1 ence values are used to cbtain valves of j and r/jD for the neat fleration. e e. a ewal 0.778 J = 0.77H1 + M MmM= 0.776E 2.15945 0.77762 0.77697 0.43759 0.41532 0.02227 0.18102 6.4733 Or Pa an ein va of act r y interpo a i n. i is bHter peference Data + (6 -k,)(T,-T)/[T.T ) to compute them with equation 2. Values of thtctness ratto, corresponding k,*6: 2 i to k-values of 0.778 and 0.777, are then cenputed with equation 1. They k, k, - k g T -T Ti -T k i eqJa1 6.41224 end 6.57258 respectlwely. tinearly interpolating between 0.77 .01 1.395 1.761 0.7779 these two value of thickness ratto indicates that k is equal to 0. 77762. when 0.777 001 0.0% 0.1 61 0.77762 1 /tt 6.474 and e/D = 0.3's. This k-value is used to conpute a final value = ry 2 y $3 C.ad for e and the corresponding value of t,/te. This computed value is equal to e a ,,J 777E? the prescribed thickness ratle (6.474); thus, the solution for k is verlfted. k The final valt;e of k obtained using the fornut of flg. 315 quite enact. As noted earlier, ors iteration is all that is usually required for most FIG. 3. - Format For 1.ocating Neutral Axis 6-2-16 ~ A.NALYSIS/ CALCULATION 00c to # S- % oor3 ATT # REV l SHEETS fl OF 17

u. =..__...

V (j anchor-bolt prcelees. No.ever, te iterations and several significant figures This equation is exact i. hee $ a and 5,, occ;r on the cirtla defined by D. If b _q ar2 vsed in the computations presented in flg. 3. because e/D is small. the cenpression forces do not lie enactly on the bolt circle. an average [ Greater precision can be obtained with a third iteration, but is probably of diameter may be used in the calculations. ne red dr N d aach W can M MmM ddy b Academic interest only. However, when performing scre than tuo iterations, v lues of j and r/jo should be corputed with the basic equations given in the the following fonula: appendiz, tising these enact values of j and z/jD. a third iteration produced b (cah!C2) ~ the following results. starting with tr/t, = 6.474 and e/D = 0.34 O.777624 j 0.776960 3 0.141012 = = 1/js and c2 = r/jDB. However, ca and Ca equation is exact because c a a D are then 6.4740 can t,e obtained from Table 4. These design valves of c, and ca 'l 2.15947 rad r = 0.415326 t

= E N T inverted in the above equation to determine N and A. The adeguacy of a 8 b ,3 The required precision will depend upon requisites of the particular particular anchor-bolt array should then be verified using the analysis anchor-bolt problers. The procedure described above, and incorporated in the procedure described in the first part of this paper, j i firmit of fig. 3 will slapitfy and standarsfire the calculations, andproduce A general calculation procedere is illustrated by the following example. 's Design requirenents are assumed to be the same as for the illustrative the desired precistop with a sinisum of ef fort. probles considered earifer, escept that the rester of bolts is not known DESIGi APPLICAT!ONS initially. Also. it is assumed that the free-Sady section does not cut Calcula:lons 111ustrated above are typical of analyses required to locate througta the bolt threads. In this case. S equals 15.155 pst. which g the neutral amis and to determine stresses when the ancher-bolt layout and caternal leads are 6nown. In the typt' cal design problem, the external loads and allowable stresses are given and it is necessary to determine the size ca # and nueber of anchor-bolts, and perhaps the bearino width. For balanced e design conditions. the masimum tensile stress (S ) in the bolts and the b a .u a a s as ect % n

l maxismar compressive stress (5 ) in the foundation are equal to their respec-g at M.)

% exact apulaattons cMd be used ca tive design limits. Sba and 5,,. In this special case. t.t neutral amis is for Sba **I Sca; however such precisfon is usually not required at this located by the fc11owing equation: e stage of the design caleviations. a I

cos*

ba - S The design calculations may be performed in the following manner: ca rad (8) abal ( ba ca 4-2 21 6 2 20 ANALYSIS / CALCULATION ~' DOC ID # S-9/o COL 3 ATT f; 3 REV l SHEET 812. OF ll

o

1) locate neotral eats operosisately, based wron allewable design stresses.

} $ a = Alttwable tenstle TASt[ 4.-Values of Design CNfficients f3be, 3 jrad h tes ce = L .gj g y,,, ( be

ca )

S, = A11ewable coseressive g stress t 8 J r/jD ci=1/jB c =r/jf*a 0.00 s/2 0.;500 0.6667 0.84 5 0.4244 For teie design stresses 9 een in Asstne 2-1/2 in. f Bolts t 0 05 1.5019 0.1597 0.6449 0.8753 0.4288 A C.10 1.4436 0.7663 0.6261 0.9040 0.4337 the reference prc81ee: l set e A = 4.909 te ' O.15 1.3362 0.7714 0.5086 0.9352 0.*190 = Cos i a 9ross 0.20 1.3304 0.7755 0.5919 0.9693 0.4449

bal

= 20,000 m h = 15,155 pst 0.25 1.2755 0.7787 0.5759 1.0068 0.4515 ,j 5 = 1.36 rae f. 0.30 1.2210 0.7812 0.5603 1.04B4 0.4589 4 0.35 1.1666 0.1831 0.5449 1.0946 0.4671 (D = 11.0 ft) k = O-cose)/2 = 0.395 a 0.40 1.1119 0.7844 0.5298 1.1466 0.4765 i 0.45 1.0564 0.7851 0.5149 1.2057 0.4874 9 4.50 1.0000 e/4 0.5000 1.2732 0.5000

2) Determine desion paraneters.

.s 0.55 0.9471 0.7851 0.4851 1.3520 0.5149 { "M. 8 s 10' = 13.333 e = 13.333 = 1.212 d 0.60 0.M21 4 a.E44 0.4702 1.4453 0.5330 1 0.65 0.8201 i 8831 0.4551 1.5571 0.5549 P 0.6 z 10' T 11.0 , ifs 0.70 0.7547 0.7812 0.4398 1.6961 0.5827 0.75 0.634 0.7787 0.424I

1. 0 50 0.6192 0.80 0.6091 0.7755 0.4081 2.1170 0.6700 0.85 0.5245 0.7714 0.3914 2.4716 0.7462

3) Determine required nuM>er of 2-1/2 In. 9 bolts using equation 9.

0.90 0.4260 0.7663 0.3739 3.0633 0.8777 0.95 0.2996 0.7597. 0.3551 4.3936 1.1852 =P . MOO. m O M m 1. H 2 -0. 4 W t c,) = l' O.3333 Sg, (c3 T' ~+ MA, 1.00 0.00 0.7500 = = 15.155 = 112.8 in. i 112.8 = 22.98

23 18

f 4.909 j s However, consider esf ag 22 teolts, because 12-in. bearing tdth is 9teater 3 than req;lred for tsalanced design.

4) Verify adequacy of 22 ancher t>cits if 21/2 in._ in__ diameter.

e The bearing plate. er concrete ring wall. is 12 in. wide. It has an average diameter of 11.0 ft a.*5 an outside diaeter of 12.0 ft. The a 3 7 73 r-2 22 i, ANALYSIS / CALCULATION b DOC ID # S-9/o-OOG ATT # REV I SHEET _S 13. OF 17

2 eccentricity rette eqi.als 1.212. Using the gress bolt area, ta = 22 x 8 is cavuted with equation 2; ehereas. values of j and t/j0 are taken from I (4.909)/132s = 0.2f,Ot in. The compression thickness = t = (12 + 9t l/10 Table 1. (Dr. en approntmete value of tytti could be selected directly from i f ~ 1.4344 In. and 1 /t = 5.508. The neutral axis, corresposiding to these Fig. 2 or Table 2. correspondtag to t = 0.4 and e/D = 1.2.) ! N equivalent 2 parameters, is located usin3 the fermat of Fig. 3.

Thus, thickness of the bolts = to

g/=D = 112.8/132= = 0.272 in.

Thus, ta 1.25323 red, re<ssired for balanced 4*esign is equal to 0.272 (3.87), or 1.053 in. Yhts k 0.343S7 and a = = 1.1733. equivalent steel thickriess conforos to a concrete bearing width of: 2.759 and 8 9 = = Ancher-bolt stresses are computed using equations 4 and 5-t = st, - (n-1)t = 10(1.053) - 9(0.272) = 8.:08 in. g 8808 psi (D = 11.0 f t) Ordinarily. It is not feastble to provide an anchor. belt array having the 14.8 % pst and 5, S = = b f exact values of g and t required by balanced design cond'tfons. However. g f 14.898

19.f46 psi (9 = 11.0 ft) these parameters may be used as a tyside in selectf ag nominal values fot, belt 5

t 3.715 1 $1re, nurter of bolts and bearing wid*h. For eravle, requirements of the 884 psi (D = 12.0 f t) 7808

3. 783 + 0. 5 f,

= 3 10 \\ L783 / bearing width. On the other hand, the soecified bearing width (12 in.) is significantly greater than 8.08 in.. so that 22 holts may be adequate.

5) Cosipare manizuss working stresses with allowable design values validity of a particular choice should be vertfled by locating the neutral 19,686 ps1 S

20.000 psi = = t aufs and then computing stresses. The results for 22 bolts and a 12-in. f, 684 pst l.200 psi bearing width are sumartred in Table 6 as Case 2. Thus. 22 anchor bolts. 21/2 in. in diameter, are satisfactory. TABtt 6.-Stress Samary ' C0>9fMTS (W DESIGN PROCEDtrPE Bolts Unit Stresses - pst The required bolt area (g), defined by equation 9, will cause a tensile Case (2-1/2 in. ts/t k 3 diameter) b t c f stress in the most-distant anchor bolt equal to the manisarn allowable stress, only if the correspondfwj value of 1 /t is satisfied. For eraeple, when Pef. 3 24 6.474 0.32 17.450 17.450 965 2 l k = 0.395 and e/D = 1.'.12. the correspendtng thickness ratio can be obtained 1 24 6.474 0.321 17.552 17.552 9 alt from equation 1 as follows: 2 '22 5.508 0.3419 14.898 19,686 884 [ j 3 24 5.124 0.3542 13.852 18.264 855 p. 1.9481 1+ " 3 87-a j t ( l.54541 - 0.53135 flote: 1 in. = 25.4 syn 1 pst = 6.89 kPa andg=1.212. I I 6-2-25 6-2-24 ANALYSIS / CALCULATION i Doc to n 5-Wool 3 ATT # S REV I SHEE' S I4 OF I7

A MMIX IWERf m ^ On the cther hand. the designer may prefer to ese 24 bolts In order to 1. Taylor. F.W., Thengson. 5.E., and 5mm15kl. E.. Cenerete. Plain and _j Provide an even mmeer of belts with a nostinal 15* angular spacing. For -f 8'I""'d* I'I I

4th ed.. John uitey S Sons. Inc.. New fort.1925.

24 bolts, t,/ts = 5.124. using the gross area of the belts and a 12-in. 2. %1Headon for tk My and hmetles of WW knente bearingutdth[ The results for this anchor-belt array are sumartred in Noneys.* A staMard W-54. hai of W Mcan heu Wm. Table 6 as Case 3. The ef fects on the location of the neutral aufs and epen working stresses, caused by vsfrM3 ?8 troits instead of 22. are indicated by cesparfeg respective parameters arwf stmss values for Case 3 with those for

I """'"* L* * *"

'""I* Wiley 1 Sons. fac.. New Yort. 1959. pp. 183-19tt. Case 2. '9"" * *

The ef ft<:t of using the gross-bolt ares to cospate ti and to locate the f'

neutral amis, instead of the root area, is shown by cospai e g stress values 5. Cartner. A.I.. "fi<nnegrams for the Sole *fon of Anchor Bolt Preblees.* ' fa Table 6 for Case 3 and Case 1. respectively. The mastause tevistle stress 08.254 pst) obtained for Case 3 is greater than that (17.552 pst) obtair:=4 for Case 1. Therefore. it is more accurate and conservative to use the gmss-APP [MCII II A1til5 5 bolt area to locate the neutral amis rather than the root area. celess the The following raties and factors are dirensionle'ss ind say be used directly free-body section actually passes through the bolt threads. with either U.S. Carst:nnary or 51 valts: GNCLU5 forts Eccentricity ratto = e/D; a = E,/E ; and g The simpitfled equations. B-factors and nisserical data presented in this Thickness ratio = t /t t Factors:

8. Bs. J. t. e/jD and r/jD.

2 paper provide a rapid, accurate ar.d convenient Mthod for performing ancher-However, the U.S. Customary units are utilized in the paper in order to confom bolt calculations. If any two of the anchor-bolt parameters (e/D. ta/ts and k) =lth those used in tft-illustrative problem of Ref. 3. - For this reason, the are known. the other one can be determined using the equations and computation following cceversfon factors may be of interest: cids described above. Anchor-bolt stresses, produced by the external leads 1 in. = 25.4 m; 1 lb = 0.453 tg; and actfag on a given belt circle. can be computed accurately once k is known. On Ift = 0.305 m: 1 pst = 6.89 kPa. the 9ther hand. if the allowable stresses and external loads are given, the required nup6er of anchor bolts can be obtained directif by using equations 8 and 9. which required little effort to solve. I 8 I

C?

6-2-26 ANALYSIS / CALCULATION DOC ID # 596-coI3 ATT a S 1 REV / SHEET S 16 OF M

7 AMihDIZ !Il-to7ATICff APPDtDII IV.-Ceriyation of fquatIwas The followfog symbols are used in this paper: Anchor-bolt paraseters are defined in f fg.1. Let 5 equal stress ta. A e. cress-sectional area of one anchor bolt; ring at angle 4. $ lace $ vartes linearly with distance from mestral ents: 8-factor; 8 = I S $c (A-1) b width of bearing ring; = b = cos (s-)}

cos a 1
cose I - cose resultant of coepression forces' C e c e a coefficient e'

cos{.,), ces.\\['Dt j d, (A.r) 2 S T 2 5 dA = = b oe diseeter of telt circles. g g K i + ces

sodulus of elasticity; E

= S Dti (8 ) (A-3) S Dt (tan. - el + w oe eccentricity; ..T = = 3 b b sec e + 1 i compressive stress in concrete; / f a g a factor used to empress distance between T and C; j = Stallarly: a factor used to locate neutral axis; t = 1 \\ ds (A-4) C 2 5 dA =2 S ' " * ~ *** D = 2 Q= bending sesent; b i l ] g y 1

tos e number of anchor bolts Q

= a* ratio i.'seds11 of elasticity (steel to concrete); i .~. C s ot, t an = -. ) = axlat lead; see. + 3 _s,ot, s, (A.5j = h P = equivalent steel stress; = resultant of tensile forces; l#ren P Y 0 T C and t:8 = t 8, = = = 3 equivalent ring thickness; t e distance between centroidal arts and C. . t, e 81 = (tan e - s) + e (A-6) 8 2 = = 'T B2 (tan = - e) The following subscripts are used in the paper: compression bal = balanced design o e allowable c = ' ' ~ (* 0) = tension v bolt t = b = (or) D$ t:8

P - DSb '2 872 0

(A.7) = b E-2-2C e ANALYSIS / CALCULATION DOC ID # S-9(o-0013 ATT# 6 REV f SHEET Slb OF G

. ] s, P-(4-s) tr r - t Bi a f EPPECT OF PLATE FLEXI8ILITY ON DESICM 0)- M-Pr-TjD " = 0, (r M = OF EXPANSION AMCHOPED RASE PLATES By Surendra 5t. Goel,I M. ASCE, Tara'P. shatua,I' (DS t:8 ) JD. = 0 (or) P.'(e - 2).- b 3 g- ) (TD e r (A-9) = P e-r]_ P -~ 5 D - ~ ID INTm710N T2 Dt B i l Expansion anchored plate'~ assemblies, which are com-fewate equations (A.8) and (A 9) and solve for 8: nonly used to attach pipe supports to hardened concrete, are designed with the assumption that the base plate is rigid. 8I t 1 There has been a concern that the r891d plate assweption ear {A.10) g'. ] ]' y result in an underestimation of forces in some anchors. The ( e 2 . underestimation is due to the prying action beteneen the S-s \\ N - Y' ~ plate.and the supporting concrete and due to an unequal dis-tribution of loads among the anchors because of.the anchor If P is epuard. In tension, the above equation becomes: 8 7' configuration with respect to the goint of application of the f I load. The intent of this paper is to investigate the effect" I . 4 1 of these f actors on the final anchor loads to be used in the y e. r .de.1,n. Y N The prying action results in additional force on Basic equations for touputing j and are presented below: (i) anchors because due to plate flexibility, the plate pushes against the concrete in the vie"-ity of the anchor and, to - satisfy local equilibrium, increases the anchor force. The 1.5 cose sina + (= - elcos,= s.e + 3-7- + prying action force in the anchor is a function of the rela-- j 2 Slaa * (=-elc85* tive stiffness of the base plate with respect to that of the 1.5 cose sine + acos's anchors. For a given base plate, the stiffer.the anchors, t .T (A.12) the greater the prying action,'and vice-versa. a { The original concern with the prying action was due sin e - a cose to an erroneous use of high anchor bolt stiffness based'on-s - - - 1.5 cosa sine + ecos e D cos a + (A.13) anchor material. 'The stiffness obtained from tests on espan-z = T sine - acose sion anchors installed in concrete is much lower. tfse of ~ this test stiffness gives a realistic anchor force which-F t 1. sargent & Lundy, Engineers,. Chicago, Illinois F.2 30 g Goel/Whatua/Longlais ANALYSIS / CALCULATION j DOC ID # S-Woof 3 gr, 3 W - l _ SHEET S 17of I] .}}

ML20203J370

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19.f46 psi (9 = 11.0 ft) these parameters may be used as a tyside in selectf ag nominal values fot, belt 5

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SUMMARY

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5.0 CONCLUSION

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19.f46 psi (9 = 11.0 ft) these parameters may be used as a tyside in selectf ag nominal values fot, belt 5

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